The Great Solo Challenge

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1				7			6
2			5						7
3		8			2	4
4				5				9
5	1	3				9
6		4					3	6
7	7			2					1
8	3				5		7
9				3	7		4

Why this level?

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 1 intertwined hypotheses, examines 3 states and closes 1 branches by contradiction. The Expert level asks to keep a chain of possibilities in addition to the interactions. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-001

View the solution to SOLO-001 →

SOLO-002 · Classic Sudoku

EXTREME

The Hidden Crossroads

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1		4				3			9
2				9					6
3								5
4	2		9			4
5		1						7
6			7	8	2				4
7	3				8		2	1
8					3			8
9		5				1

Why this level?

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 6 intertwined hypotheses, examines 29 states and closes 14 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be made by the two sides. be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-002

View the solution to SOLO-002 →

SOLO-003 · Classic Sudoku

EXPERT

The Inverted Mesh

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1					7		9	1
2	7				3
3				4	9
4			9						8
5			3			4		5
6		8		5		2		7
7	2	1				7			5
8				8			2
9			6				7

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-003

View the solution to SOLO-003 →

SOLO-004 · Classic Sudoku

EXTREME

Branches of Silence

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	7		2		8			4
2	3			1		2
3				6	5
4
5	8	5			7				2
6							9	6
7			8	4			5	3
8						5	6
9		3				1

Why this level?

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 8 intertwined hypotheses, examines 81 states and closes 40 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be made by a single candidate, single places, bare pairs/triples and interactions between units. be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-004

View the solution to SOLO-004 →

SOLO-005 · Diagonal Sudoku

EXPERT

The Two Crossings

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.
The two large diagonals each contain numbers 1 to 9 without repetition. This constraint adds to the blocks.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	2	3	4	5	7	8	9
1				5
2	9	3			7
3	4	5					8
4	7		4		6
5			1
6							3
7			5	9		4
8				4
9		7				3

Why this level?

After unique candidates, unique places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 2 intertwined hypotheses, examines 5 states and closes 2 branches by contradiction. The Expert level asks to keep a chain of possibilities in addition to the interactions. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-005

View the solution to SOLO-005 →

SOLO-006 · Diagonal Sudoku

EXTREME

The Stubborn Mirror

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.
The two large diagonals each contain numbers 1 to 9 without repetition. This constraint adds to the blocks.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	3	4	5	6	7	8	9
1					2		1	3
2	4		6			8		9
3				4
4
5					6
6	7	1	4
7	5			6	9	7
8
9			1	2	5

Why this level?

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 9 intertwined hypotheses, examines 37 states and closes 18 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be made to the extent that it is possible to be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-006

View the solution to SOLO-006 →

SOLO-007 · Irregular Sudoku

EXPERT

The Living Border

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
The nine regions delimited by a thick line also contain figures 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1				9	1				8
2			3			4			5
3		2			3
4			5						9
5							2	8
6	4		6
7
8	9			7		8
9						6			4

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-007

View the solution to SOLO-007 →

SOLO-008 · Irregular Sudoku

EXTREME

The Nine Territories

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
The nine regions delimited by a thick line also contain figures 1 to 9 without repetition.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1				7
2	9						6	3
3		3	1
4			3			4	8		7
5
6	8				2
7	2			5
8			5		3	8
9					1	9	4

Why this level?

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 8 intertwined hypotheses, examines 27 states and closes 13 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be made to the extent that it is possible to determine the number of empty boxes. be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-008

View the solution to SOLO-008 →

SOLO-009 · Killer Sudoku

EXPERT

Captive Sums

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.
Each dotted cage carries a total. The numbers of the cage have this sum and are all different; this prohibition also applies between two cells of the cage that do not share a line, column or block. The few pre-filled figures are part of the problem.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	208			13			12		22
2		18	24		11
3					16				17
4	124		16		21	3		14
5	15						13
6		10			18
7	15		8	18		12	17		14
8	19				9
9	3					15

Thick stroke: region. Dotted inside the cells: cage and its total. A tint on a diagonal reminds the diagonal constraint, if it applies.

Exact list of cages (coordinates and totals)

Cage 1: L1C1, L1C2, L1C3, L2C1; total 20.
Cage 2: L1C4, L1C5, L1C6; total 13.
Cage 3: L1C7, L1C8; total 12.
Cage 4: L1C9, L2C8, L2C9, L3C8; total 22.
Cage 5: L2C2, L3C1, L3C2, L3C3; total 18.
Cage 6: L2C3, L2C4, L3C4; total 24.
Cage 7: L2C5, L2C6, L2C7; total 11.
Cage 8: L3C5, L3C6, L3C7, L4C7; total 16.
Cage 9: L3C9, L4C9, L5C9; total 17.
Cage 10: L4C1, L4C2, L5C2; total 12.
Cage 11: L4C3, L4C4, L5C3, L5C4; total 16.
Cage 12: L4C5, L5C5, L5C6; total 21.
Cage 13: L4C6; total 3.
Cage 14: L4C8, L5C8, L6C8, L6C9; total 14.
Cage 15: L5C1, L6C1; total 15.
Cage 16: L5C7, L6C7; total 13.
Cage 17: L6C2, L6C3, L6C4; total 10.
Cage 18: L6C5, L6C6, L7C5; total 18.
Cage 19: L7C1, L7C2; total 15.
Cage 20: L7C3, L8C3; total 8.
Cage 21: L7C4, L8C4, L9C4, L9C5; total 18.
Cage 22: L7C6, L8C6; total 12.
Cage 23: L7C7, L7C8, L8C7, L8C8; total 17.
Cage 24: L7C9, L8C9, L9C9; total 14.
Cage 25: L8C1, L8C2, L9C2, L9C3; total 19.
Cage 26: L8C5; total 9.
Cage 27: L9C1; total 3.
Cage 28: L9C6, L9C7, L9C8; total 15.

Why this level?

After unique candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive proof reaches 3 intertwined hypotheses, examines 9 states and closes 4 branches by contradiction. The Expert level asks to keep a chain of possibilities in addition to the interactions. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-009

View the solution to SOLO-009 →

SOLO-010 · Killer Sudoku

EXTREME

Unyielding Totals

Complete the grid completely. Only one solution respects all constraints.

Materials: Pencil and gum; no commercial material.

Rules you need

Fill in the 81 boxes with numbers from 1 to 9. Each row and column contains each of the nine digits exactly once.
The coordinates L1C1 to L9C9 are read from the top left corner. No calculation is made on the coordinates numbers.
Each of the nine blocks of 3 × 3 also contains numbers 1 to 9 without repetition.
Each dotted cage carries a total. The numbers of the cage have this sum and are all different; this prohibition also applies between two cells of the cage that do not share a line, column or block. The few pre-filled figures are part of the problem.

The puzzle

The empty boxes are to be completed. The given numbers are fixed. The answer and proof are on another page.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	9		14				13	22
2	16	18		27		19
3					18			20
4		17							8
5	13		10	18		21
6								19
7	19	19	8		27				156
8					12		11
9		5					2		7

Thick stroke: region. Dotted inside the cells: cage and its total. A tint on a diagonal reminds the diagonal constraint, if it applies.

Exact list of cages (coordinates and totals)

Cage 1: L1C1, L1C2; total 9.
Cage 2: L1C3, L1C4, L1C5, L1C6; total 14.
Cage 3: L1C7, L2C7, L2C8; total 13.
Cage 4: L1C8, L1C9, L2C9, L3C9; total 22.
Cage 5: L2C1, L3C1, L4C1; total 16.
Cage 6: L2C2, L2C3, L3C2, L3C3; total 18.
Cage 7: L2C4, L2C5, L3C4, L4C4; total 27.
Cage 8: L2C6, L3C6, L3C7, L4C6; total 19.
Cage 9: L3C5, L4C5, L5C5; total 18.
Cage 10: L3C8, L4C7, L4C8, L5C8; total 20.
Cage 11: L4C2, L4C3, L5C2; total 17.
Cage 12: L4C9, L5C9; total 8.
Cage 13: L5C1, L6C1, L6C2; total 13.
Cage 14: L5C3, L6C3; total 10.
Cage 15: L5C4, L6C4, L6C5, L6C6; total 18.
Cage 16: L5C6, L5C7, L6C7, L7C7; total 21.
Cage 17: L6C8, L6C9, L7C8, L8C8; total 19.
Cage 18: L7C1, L8C1, L9C1; total 19.
Cage 19: L7C2, L8C2, L8C3, L9C3; total 19.
Cage 20: L7C3, L7C4, L8C4; total 8.
Cage 21: L7C5, L7C6, L8C6, L9C6; total 27.
Cage 22: L7C9, L8C9; total 15.
Cage 23: L8C5, L9C4, L9C5; total 12.
Cage 24: L8C7, L9C7, L9C8; total 11.
Cage 25: L9C2; total 5.
Cage 26: L9C9; total 7.

Why this level?

After unique candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 9 intertwined hypotheses, examines 127 states and closes 63 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path Maybe it's shorter.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-010

View the solution to SOLO-010 →

SOLO-011 · Advanced Mastermind

EXPERT

The Eight Inks

Find the only ordered code of 5 digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each position contains a number from 0 to 7. Repeats are allowed. The code has exactly 5 positions; an initial zero is allowed.
"Good place" includes the positions whose number and place are correct. "Unplaced" counts the other numbers present, respecting the multiplicity.
To count the wrong placed, first matches the well placed numbers, then each remaining occurrence at the most once. The indices do not give the location of the incorrectly placed numbers.
All indices are simultaneously true. There is no hidden information about colors, parity or missing numbers.

The puzzle

The table reads from top to bottom. The proposals are tests and are not already confirmed number positions.

Write down the possible multiplicities of each number, then look for arrangements compatible with all scores.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Test	Guess	Exact matches	Misplaced matches
2	1	02553	0	2
3	2	62532	1	0
4	3	05062	0	2
5	4	73375	1	2

Why this level?

The reference method leaves 5 multi set(s) after solving multiplicity constraints; then it is necessary to cross their arrangements with all the position constraints. Repeats prevent the numbers from being treated as a simple set. The reference method combines occurrence counting and configuration removal.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-011

View the solution to SOLO-011 →

SOLO-012 · Advanced Mastermind

EXPERT

The Silent Lock

Find the only ordered code of 5 digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each position contains a number from 0 to 7. Repeats are allowed. The code has exactly 5 positions; an initial zero is allowed.
"Good place" includes the positions whose number and place are correct. "Unplaced" counts the other numbers present, respecting the multiplicity.
To count the wrong placed, first matches the well placed numbers, then each remaining occurrence at the most once. The indices do not give the location of the incorrectly placed numbers.
All indices are simultaneously true. There is no hidden information about colors, parity or missing numbers.

The puzzle

The table reads from top to bottom. The proposals are tests and are not already confirmed number positions.

Write down the possible multiplicities of each number, then look for arrangements compatible with all scores.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Test	Guess	Exact matches	Misplaced matches
2	1	22543	1	3
3	2	41050	1	0
4	3	70302	1	2
5	4	37433	1	1

Why this level?

The reference method leaves 1 multi set(s) after solving multiplicity constraints; then it is necessary to cross their arrangements with all position constraints. Repeats prevent treating the numbers as a simple set. The reference method combines occurrence counting and configuration removal.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-012

View the solution to SOLO-012 →

SOLO-013 · Advanced Mastermind

EXTREME

Six Slots Without an Echo

Find the only orderly code of 6 digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each position contains a number from 0 to 7. Repeats are allowed. The code has exactly 6 positions; an initial zero is allowed.
"Good place" includes the positions whose number and place are correct. "Unplaced" counts the other numbers present, respecting the multiplicity.
To count the wrong placed, first matches the well placed numbers, then each remaining occurrence at the most once. The indices do not give the location of the incorrectly placed numbers.
All indices are simultaneously true. There is no hidden information about colors, parity or missing numbers.

The puzzle

The table reads from top to bottom. The proposals are tests and are not already confirmed number positions.

Write down the possible multiplicities of each number, then look for arrangements compatible with all scores.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Test	Guess	Exact matches	Misplaced matches
2	1	345125	1	1
3	2	073421	0	4
4	3	573237	2	1
5	4	531562	2	0

Why this level?

The reference method leaves 16 multi set(s) after solving multiplicity constraints; then it is necessary to cross their arrangements with all position constraints. The resolution combines equations of multiplicity caps, permutations with repetition and interlocking position constraints. The corrected provides the complete cutting of this exploration. The Extreme level remains an appreciation of this manual effort, without calibration on players.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-013

View the solution to SOLO-013 →

SOLO-014 · Advanced Mastermind

EXTREME

The Chamber of Duplicates

Find the only orderly code of 6 digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each position contains a number from 0 to 7. Repeats are allowed. The code has exactly 6 positions; an initial zero is allowed.
"Good place" includes the positions whose number and place are correct. "Unplaced" counts the other numbers present, respecting the multiplicity.
To count the wrong placed, first matches the well placed numbers, then each remaining occurrence at the most once. The indices do not give the location of the incorrectly placed numbers.
All indices are simultaneously true. There is no hidden information about colors, parity or missing numbers.

The puzzle

The table reads from top to bottom. The proposals are tests and are not already confirmed number positions.

Write down the possible multiplicities of each number, then look for arrangements compatible with all scores.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Test	Guess	Exact matches	Misplaced matches
2	1	474563	1	2
3	2	013237	1	3
4	3	163003	2	3
5	4	635050	0	3
6	5	042546	1	2

Why this level?

The reference method leaves 1 multiset(s) after solving multiplicity constraints; then it is necessary to cross their arrangements with all position constraints. The resolution combines equations of multiplicity ceilings, permutations with repetition and intertwined position constraints. The corrected provides the complete cutting of this exploration. The Extreme level remains an appreciation of this manual effort, without calibration on players.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-014

View the solution to SOLO-014 →

SOLO-015 · Cryptarithm

EXPERT

The Twin Carries

Determine the value of each of the ten letters, then write the two additions in digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each letter A to J represents a number from 0 to 9. Two different letters always represent different numbers.
The ten digits are therefore used exactly once in the substitution; they may reappear several times in the numbers by repetition of the letters.
No number starts with zero. Both additions use exactly the same substitution. The notation is ordinary decimal.

The puzzle

The two additions are simultaneously true. The letter names do not bear any value index.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Addition	First term	Second term	Result
2	1	DDFB	JHEG	AHIBD
3	2	EFIF	JEEC	ADBFE

Why this level?

Ten letters and two coupled additions require a combination of retention, injectivity and transport of constraints between equations. A first correct equation leaves several assignments: resolution requires a table of deductions and a systematic exploration, not a simple operation to complete.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-015

View the solution to SOLO-015 →

SOLO-016 · Cryptarithm

EXPERT

The Second Addition

Determine the value of each of the ten letters, then write the two additions in digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each letter A to J represents a number from 0 to 9. Two different letters always represent different numbers.
The ten digits are therefore used exactly once in the substitution; they may reappear several times in the numbers by repetition of the letters.
No number starts with zero. Both additions use exactly the same substitution. The notation is ordinary decimal.

The puzzle

The two additions are simultaneously true. The letter names do not bear any value index.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Addition	First term	Second term	Result
2	1	DBBG	GGJE	CGDF
3	2	GCIF	FHGC	DIIAG

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-016

View the solution to SOLO-016 →

SOLO-017 · Cryptarithm

EXTREME

Ten Letters, Two Locks

Determine the value of each of the ten letters, then write the two additions in digits.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each letter A to J represents a number from 0 to 9. Two different letters always represent different numbers.
The ten digits are therefore used exactly once in the substitution; they may reappear several times in the numbers by repetition of the letters.
No number starts with zero. Both additions use exactly the same substitution. The notation is ordinary decimal.

The puzzle

The two additions are simultaneously true. The letter names do not bear any value index.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4
1	Addition	First term	Second term	Result
2	1	JIIB	FHBF	IDIF
3	2	CGJG	EAII	IDHE

Why this level?

Ten letters and two coupled additions require a combination of restraints, injectivity and transport of stresses between equations. The large number of surviving substitutions to the first equation makes manual exploration particularly long; the restraint controls must be intertwined in both additions.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-017

View the solution to SOLO-017 →

SOLO-018 · Futoshiki

EXPERT

The Seven Staircases

Complete square 7 × 7 with figures from 1 to 7.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each row and column contains exactly once every 1-7 digit. There are no 3 × 3 regions and no diagonal constraints.
The printed values are fixed. All inequalities in the list are strict and must be respected simultaneously.
LrCc is the line r and column c, counted from the top and left. A box marked " ·" is to be filled in.

The puzzle

All useful inequalities are given below; the neighbourhoods not mentioned do not bear any additional constraints.

L5C1 < L4C1

L6C6 < L5C6

L3C6 < L3C7

L2C6 < L2C7

L6C2 < L7C2

L4C1 < L3C1

L7C7 < L7C6

L5C7 < L6C7

L5C2 < L4C2

L1C4 < L1C5

L1C4 < L2C4

L6C1 < L7C1

L6C4 < L7C4

L6C6 < L6C5

L4C2 < L3C2

L4C5 < L3C5

L2C7 < L3C7

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7
1	·	·	·	2	·	5	·
2	6	·	·	·	4	·	·
3	·	·	3	·	5	·	·
4	·	·	·	6	·	·	·
5	·	·	·	·	·	·	·
6	·	·	·	·	·	·	7
7	·	·	5	·	·	6	3

Why this level?

The crossover of single lines, single columns and inequities is not closed by propagation of the only forced numbers in the reference solver. The Expert level is based on the combination of single terminals and locations, followed by controlled exploration; it is not only a result of the size of the square.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-018

View the solution to SOLO-018 →

SOLO-019 · Futoshiki

EXTREME

The Eight Passages

Complete square 8 × 8 with figures from 1 to 8.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each row and column contains exactly once every 1 to 8 digit. There are no 3 × 3 regions and no diagonal constraints.
The printed values are fixed. All inequalities in the list are strict and must be respected simultaneously.
LrCc is the line r and column c, counted from the top and left. A box marked " ·" is to be filled in.

The puzzle

All useful inequalities are given below; the neighbourhoods not mentioned do not bear any additional constraints.

L4C1 < L4C2

L5C6 < L5C7

L8C2 < L7C2

L1C5 < L1C4

L6C2 < L6C3

L1C2 < L2C2

L5C1 < L4C1

L4C5 < L3C5

L2C3 < L2C2

L4C3 < L4C2

L6C1 < L6C2

L3C6 < L3C7

L7C3 < L6C3

L6C4 < L5C4

L8C8 < L8C7

L5C2 < L4C2

L4C2 < L3C2

L1C2 < L1C3

L6C1 < L5C1

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1	·	5	·	·	2	7	·	·
2	·	·	2	3	·	·	·	·
3	6	·	·	·	·	·	·	·
4	·	·	·	·	·	·	·	·
5	·	·	4	·	·	·	8	·
6	·	·	·	·	·	8	·	1
7	·	·	·	5	8	2	·	·
8	·	·	·	6	3	·	·	·

Why this level?

The crossover of single lines, single columns and inequities is not closed by propagation of the only forced numbers in the reference solver. Entwined hypotheses and non-local contradictions are necessary in its method; this motivates the Extreme level, without pretending to exclude a shorter human trick.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-019

View the solution to SOLO-019 →

SOLO-020 · Futoshiki

EXTREME

Columns Under Tension

Complete square 8 × 8 with figures from 1 to 8.

Materials: Pencil, gum and gridded paper; no commercial game required.

Rules you need

Each row and column contains exactly once every 1 to 8 digit. There are no 3 × 3 regions and no diagonal constraints.
The printed values are fixed. All inequalities in the list are strict and must be respected simultaneously.
LrCc is the line r and column c, counted from the top and left. A box marked " ·" is to be filled in.

The puzzle

All useful inequalities are given below; the neighbourhoods not mentioned do not bear any additional constraints.

L4C5 < L4C4

L8C6 < L7C6

L1C4 < L1C3

L2C3 < L1C3

L2C2 < L2C3

L3C6 < L3C7

L1C7 < L2C7

L6C6 < L6C7

L6C6 < L5C6

L8C1 < L8C2

L8C3 < L7C3

L7C4 < L7C5

L3C1 < L2C1

L6C2 < L5C2

L6C4 < L6C5

L5C6 < L4C6

L6C2 < L7C2

L7C1 < L8C1

L7C5 < L7C6

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1	·	·	·	·	·	2	·	·
2	·	·	·	5	·	·	·	·
3	·	·	·	4	·	·	5	·
4	·	5	·	·	·	·	8	4
5	·	4	3	6	2	·	·	·
6	·	·	·	·	6	3	·	·
7	·	·	·	·	·	·	·	6
8	·	·	·	·	1	·	·	5

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-020

View the solution to SOLO-020 →

SOLO-021 · Nonogram

EXPERT

Crossing Intervals

Black out the boxes of this grid 12 × 12 to satisfy all series.

Materials: Pencil and gum; grid below.

Rules you need

A series shows the lengths of the consecutive black blocks in the order, from left to right for a line and from top to bottom for a column.
Two adjacent blocks are separated by at least one empty box. Non-blacked boxes must remain empty. "0" indicates a completely empty line.
The LrCc coordinates refer to line r, column c, counted from the upper left corner.

The puzzle

The series and possible pre-filled boxes are the only data of the problem. ■ = already black box, · = already proven empty box; the white boxes without symbol remain unknown. No symmetry or recognizable image should be assumed.

Coordinates: lines from top to bottom, columns from left to right.
L / C	C1133	C21212	C312111	C4223	C533	C61312	C71113	C8121	C911111	C101111	C11131	C1212111
L11 · 5 · 1
L21 · 2 · 1 · 2
L31 · 1 · 1
L43 · 1 · 1 · 3
L53 · 2 · 1 · 2
L61 · 1 · 1
L73 · 1 · 1 · 1
L81 · 3
L91 · 5
L101 · 2 · 1 · 1 · 1
L112 · 1 · 1 · 2
L122 · 1 · 2 · 1

Why this level?

The simple intersection blocks; the crosses require an analysis of nested cases (depth 2, 6 branches impossible in the solver). Editorial level based on this need for coordination, not on the only format of the grid.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-021

View the solution to SOLO-021 →

SOLO-022 · Nonogram

EXPERT

The Suspended Weave

Black out the boxes of this grid 12 × 12 to satisfy all series.

Materials: Pencil and gum; grid below.

Rules you need

A series shows the lengths of the consecutive black blocks in the order, from left to right for a line and from top to bottom for a column.
Two adjacent blocks are separated by at least one empty box. Non-blacked boxes must remain empty. "0" indicates a completely empty line.
The LrCc coordinates refer to line r, column c, counted from the upper left corner.

The puzzle

Coordinates: lines from top to bottom, columns from left to right.
L / C	C1152	C2172	C3242	C41131	C51111	C6211	C71213	C822	C922	C1021111	C1113	C1242
L12 · 2 · 1 · 1
L21 · 1 · 1 · 1
L34 · 1 · 1
L42 · 3 · 3
L52 · 2 · 2
L63 · 1 · 2
L74 · 1 · 1
L84 · 1 · 1
L94 · 1 · 2
L101 · 2
L115 · 2 · 1
L123 · 1 · 1

Why this level?

The simple intersection blocks; the crosses require an analysis of nested cases (depth 5, 8 branches impossible in the solver). Editorial level based on this need for coordination, not on the only format of the grid.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-022

View the solution to SOLO-022 →

SOLO-023 · Nonogram

EXTREME

The Fragmented Grid

Black out the boxes of this 16 × 16 grid to satisfy all series.

Materials: Pencil and gum; grid below.

Rules you need

A series shows the lengths of the consecutive black blocks in the order, from left to right for a line and from top to bottom for a column.
Two adjacent blocks are separated by at least one empty box. Non-blacked boxes must remain empty. "0" indicates a completely empty line.
The LrCc coordinates refer to line r, column c, counted from the upper left corner.

The puzzle

Coordinates: lines from top to bottom, columns from left to right.
L / C	C211111111	C321111	C1011131
L11 · 1 · 2
L21 · 2 · 2 · 1 · 1 · 1
L32 · 1 · 2
L41 · 1 · 2 · 2 · 2			·
L52 · 1 · 1 · 1 · 2
L61 · 1 · 3 · 1
L73 · 2 · 3 · 3
L82 · 2 · 1
L91 · 3 · 1 · 4
L101 · 1 · 3 · 1 · 2	■
L111 · 1 · 1 · 2
L121 · 1 · 3 · 2
L131 · 2 · 2 · 1 · 3		·
L141 · 1 · 1 · 1 · 1 · 2
L151 · 2 · 2 · 1 · 1
L163 · 1 · 3 · 3

Why this level?

The simple intersection blocks; the crosses require an analysis of nested cases (depth 7, 34 branches impossible in the solver). Editorial level based on this need for coordination, not on the only format of the grid.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-023

View the solution to SOLO-023 →

SOLO-024 · Nonogram

EXTREME

The Hidden Border

Black out the boxes of this 16 × 16 grid to satisfy all series.

Materials: Pencil and gum; grid below.

Rules you need

A series shows the lengths of the consecutive black blocks in the order, from left to right for a line and from top to bottom for a column.
Two adjacent blocks are separated by at least one empty box. Non-blacked boxes must remain empty. "0" indicates a completely empty line.
The LrCc coordinates refer to line r, column c, counted from the upper left corner.

The puzzle

Coordinates: lines from top to bottom, columns from left to right.
L / C	C94311	C1141111	C122111
L11 · 3 · 3 · 1 · 1
L29 · 1
L32 · 1 · 1 · 1 · 3
L42 · 1 · 1 · 1 · 3 · 2
L52 · 2 · 1 · 1 · 2
L65 · 3 · 2		■
L72 · 3 · 2 · 2 · 1
L81 · 2 · 5 · 2			·
L91 · 2 · 1 · 1 · 1
L101 · 1 · 1 · 2 · 3
L112 · 2 · 1	·
L121 · 1 · 1 · 1 · 2
L131 · 2 · 1 · 2 · 1
L141 · 1 · 1 · 1 · 3
L151 · 3
L161 · 3 · 1

Why this level?

The simple intersection blocks; the crosses require an intertwined case analysis (depth 10, 16 impossible branches in the solver). Editorial level based on this need for coordination, not on the only format of the grid.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-024

View the solution to SOLO-024 →

SOLO-025 · Hamiltonian path

EXPERT

The Silent Checkpoints

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

Materials: Pencil, gum and grill; no connection required.

Rules you need

The successive numbers occupy two boxes sharing one side; no diagonal displacement.
Each box receives exactly a number of 1 to 36. Print marks remain in their place.
The course starts in 1 and ends in 36; it does not have to go back to the start.

The puzzle

The white boxes are all usable. The boxes without clues are not obstacles.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	5	6
1	1
2				8
3
4			28
5	15	18		33	36
6					35

Why this level?

Coordinate the intervals between markers, the separations of regions and the constraints of parity. The solver must explore 613 prefixes; the level reflects this combined exploration, without automatically absorbing time of calculation and human difficulty.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-025

View the solution to SOLO-025 →

SOLO-026 · Hamiltonian path

EXPERT

The One-Way Crossing

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

Materials: Pencil, gum and grill; no connection required.

Rules you need

The successive numbers occupy two boxes sharing one side; no diagonal displacement.
Each box receives exactly a number of 1 to 36. Print marks remain in their place.
The course starts in 1 and ends in 36; it does not have to go back to the start.

The puzzle

The white boxes are all usable. The boxes without clues are not obstacles.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	5	6
1				36
2	28
3	21		1
4
5		12		4
6		9	6

Why this level?

Coordinate the intervals between markers, the separations of regions and the constraints of parity. The solver must explore 423 prefixes; the level reflects this combined exploration, without automatically absorbing time of calculation and human difficulty.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-026

View the solution to SOLO-026 →

SOLO-027 · Hamiltonian path

EXTREME

The Maze Without Walls

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

Materials: Pencil, gum and grill; no connection required.

Rules you need

The successive numbers occupy two boxes sharing one side; no diagonal displacement.
Each box receives exactly a number of 1 to 36. Print marks remain in their place.
The course starts in 1 and ends in 36; it does not have to go back to the start.

The puzzle

The white boxes are all usable. The boxes without clues are not obstacles.

Coordinates: lines from top to bottom, columns from left to right.
L / C	2	3	5
1
2	23
3
4	1
5	36		17
6	33	32

Why this level?

Coordinate the intervals between markers, the separations of regions and the constraints of parity. The solver must explore 5682 prefixes; the level reflects this combined exploration, without automatically absorbing time of calculation and human difficulty.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-027

View the solution to SOLO-027 →

SOLO-028 · Pentomino tiling

EXPERT

The Eight Shapes

Pave the rectangle 5 × 8 completely with the 8 pentominos given, without overlap.

Materials: Pencil and gum; eventually reproduce the 8 silhouettes on gridded paper to cut them.

Rules you need

Each room contains exactly five boxes bound by their sides. Use each of the 8 letters exactly once.
Rotations and turns are allowed. A piece cannot be deformed or cut.
Any box containing a letter must be covered by the part of this letter. The other boxes of the rectangle must also be covered.
In the silhouettes below, "/" separates rows, the dots are outside the room.

The puzzle

Silhouettes exact, to be read row by row:

I : I / I / I / I / I

L : L· / L· / L· / L· / LL

P : PP / PP / P·

T : TTT /T· /T·

V.V.V.V.V.

W : W·· / WW· / ·WW

X : X· / XXX / X·

Z : ZZ· / ·Z· / ·ZZ

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1			I
2
3
4
5

Parts to be placed

Exhibit I

Exhibit L

Exhibit P

Exhibit T

Exhibit V

Exhibit W

Exhibit X

Exhibit Z

Why this level?

Combining orientations, constraints of indices and impossibility of coverage; 980 sub-pavements are examined with choice of the most constrained box. The level is editorial and does not rely on an external standard.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-028

View the solution to SOLO-028 →

SOLO-029 · Pentomino tiling

EXTREME

The Locked Mosaic

Completely paste the rectangle 5 × 10 with the given 10 pentominos, without overlap.

Materials: Pencil and gum; eventually reproduce the 10 silhouettes on gridded paper to cut them.

Rules you need

Each room contains exactly five boxes bound by their sides. Use each of the 10 letters exactly once.
Rotations and turns are allowed. A piece cannot be deformed or cut.
Any box containing a letter must be covered by the part of this letter. The other boxes of the rectangle must also be covered.
In the silhouettes below, "/" separates rows, the dots are outside the room.

The puzzle

Silhouettes exact, to be read row by row:

F : ·FF / FF· / ·F·

L : L· / L· / L· / L· / LL

N : N· / NN / ·N / ·N

T : TTT /T· /T·

U : U·U / UUU

V.V.V.V.V.

W : W·· / WW· / ·WW

X : X· / XXX / X·

Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y: Y:

Z : ZZ· / ·Z· / ·ZZ

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10
1
2
3
4
5							F

Parts to be placed

Exhibit F

Exhibit L

Exhibit N

Exhibit T

Exhibit U

Exhibit V

Exhibit W

Exhibit X

Exhibit Y

Exhibit Z

Why this level?

Combining orientations, constraints of indices and impossibility of coverage; 6105 sub-pavements are examined with choice of the most constrained box. The level is editorial and does not rely on an external standard.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-029

View the solution to SOLO-029 →

SOLO-030 · Pentomino tiling

EXTREME

The Stubborn Rectangle

Completely paste the rectangle 5 × 10 with the given 10 pentominos, without overlap.

Materials: Pencil and gum; eventually reproduce the 10 silhouettes on gridded paper to cut them.

Rules you need

Each room contains exactly five boxes bound by their sides. Use each of the 10 letters exactly once.
Rotations and turns are allowed. A piece cannot be deformed or cut.
Any box containing a letter must be covered by the part of this letter. The other boxes of the rectangle must also be covered.
In the silhouettes below, "/" separates rows, the dots are outside the room.

The puzzle

Silhouettes exact, to be read row by row:

F : ·FF / FF· / ·F·

I : I / I / I / I / I

L : L· / L· / L· / L· / LL

N : N· / NN / ·N / ·N

P : PP / PP / P·

T : TTT /T· /T·

V.V.V.V.V.

W : W·· / WW· / ·WW

X : X· / XXX / X·

Z : ZZ· / ·Z· / ·ZZ

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10
1
2
3
4							T
5

Parts to be placed

Exhibit F

Exhibit I

Exhibit L

Exhibit N

Exhibit P

Exhibit T

Exhibit V

Exhibit W

Exhibit X

Exhibit Z

Why this level?

Combining orientations, constraints of indices and impossibility of coverage; 13455 sub-pavements are examined with choice of the most constrained box. The level is editorial and does not rely on an external standard.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-030

View the solution to SOLO-030 →

SOLO-031 · Othello / Reversi

EXPERT

The Final Tempo

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

Trait : Black. There are 8 empty boxes left.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	B	N	N	N	·	N	B	N
2	B	B	N	N	·	B	B	N
3	B	N	N	N	B	N	B	N
4	N	N	N	N	B	B	B	N
5	N	B	B	N	B	B	B	N
6	·	B	B	B	N	N	B	N
7	·	·	B	B	B	B	B	N
8	·	B	·	B	·	B	B	N

Why this level?

Several first moves are legal; an estimate by the immediate gain does not replace the calculation until the end. The resolution requires a minimum tree on several floors, with verification of the turns and mobility. Editorial evaluation based on the required techniques; not calibrated by a human panel.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-031

View the solution to SOLO-031 →

SOLO-032 · Othello / Reversi

EXPERT

The Deceptive Corner

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

Trait : Black. There are 8 empty boxes left.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	B	B	B	B	B	B	·	N
2	B	B	N	B	B	B	N	B
3	·	B	B	B	B	N	B	·
4	N	N	N	N	N	B	B	B
5	N	N	N	B	B	N	B	B
6	N	N	B	N	B	B	B	B
7	N	N	B	B	·	·	B	B
8	N	N	B	N	·	·	·	B

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-032

View the solution to SOLO-032 →

SOLO-033 · Othello / Reversi

EXPERT

Mobility in Reserve

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

White to move. There are 9 empty squares.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	N	N	N	N	N	N	B	·
2	·	N	N	N	N	N	N	N
3	·	N	B	B	N	B	N	N
4	B	N	B	B	B	N	B	·
5	B	N	N	N	B	N	N	B
6	B	N	B	N	B	N	B	B
7	B	N	B	B	N	N	·	B
8	·	N	·	·	B	N	·	B

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-033

View the solution to SOLO-033 →

SOLO-034 · Othello / Reversi

EXTREME

The Defense Tree

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

Trait : Black. There are 10 empty boxes left.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	B	N	N	·	B	·	·	N
2	B	N	N	B	·	·	·	N
3	·	B	B	N	N	N	N	N
4	N	B	B	N	B	N	B	N
5	B	N	N	B	N	N	B	N
6	·	N	B	B	B	N	B	N
7	·	B	N	N	N	N	N	·
8	B	B	N	N	N	N	N	N

Why this level?

End of part requiring a ramified defense shaft on ten or eleven boxes, with possible beating orders, mobility and passes. It is necessary to combine retrograde calculation, control of adverse responses and exact comparison of alternatives; no local corner or immediate majority rule is sufficient. Editorial evaluation based on the required techniques; not calibrated by a human panel.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-034

View the solution to SOLO-034 →

SOLO-035 · Othello / Reversi

EXTREME

Paths to a Majority

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

Trait: White. There are 11 empty boxes left.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	·	·	B	B	B	B	B	B
2	·	B	B	B	N	N	B	·
3	B	B	B	N	N	N	·	·
4	B	B	N	B	B	N	N	B
5	B	N	B	B	B	N	·	N
6	B	N	N	B	N	N	N	B
7	B	N	N	N	N	N	B	·
8	B	N	·	B	B	B	·	·

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-035

View the solution to SOLO-035 →

SOLO-036 · Othello / Reversi

EXTREME

The Last Word

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

Materials: Pencil, gum; tray 8 × 8 and optional reversible pions.

Rules you need

Tray 8 × 8; columns A to H from left to right, lines 1 to 8 from top to bottom. N = Black, B = White, · = empty box.
A shot sets a pawn on an empty box and must frame at least one opposing pawn between this pawn and a friendly pawn, in a straight horizontal, vertical or diagonal line. All frames created are turned.
Without a legal blow, you have to pass. If no camp can play, the game stops. Score: number of his pawns less number of opposing pawns; empty boxes do not count. The goal is to maximize this differential, not just to win.

The puzzle

Trait: White. There are 11 empty boxes left.

For the evidence, also give the best defense after your first blow, then examine the other legal blows. No second person is necessary: analyze both sides on paper.

Coordinates: lines from top to bottom, columns from left to right.
L / C	A	B	C	D	E	F	G	H
1	·	N	·	N	N	N	·	·
2	·	N	B	B	B	B	B	·
3	B	N	B	B	B	B	B	N
4	B	N	B	N	B	N	B	·
5	B	N	B	B	B	N	B	·
6	B	N	B	B	B	N	B	N
7	B	N	N	N	N	N	N	N
8	·	·	B	B	B	·	B	N

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-036

View the solution to SOLO-036 →

SOLO-037 · Nim — pair subtraction

EXPERT

Two Workshops

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

Materials: Pencil and paper; optional chips, all pile numbers are given.

Rules you need

Original variant "pair subtraction". Each component is a pair of heaps A and B. The pairs remain independent.
At each turn, choose exactly ONE pair, then remove 1, 2 or 3 chips from A alone, OR B alone, OR exactly the same number (1, 2 or 3) of both piles. No pile can become negative. Never remove two different pairs in the same round.
The two sides alternate. The player without a legal blow loses: taking the last chips makes it win. There is neither pass nor rule misery. The tokens of the same pile are indistinguishable; the pairs and piles A/B are labeled.

The puzzle

You play first. To analyze alone, calculate the winning and losing positions of both sides.

Gives a complete list of the winners and a general proof of answer: only one cooperative suite is not enough.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3
1	Pair	Pile A	Pile B
2	1	8	11
3	2	10	7

Why this level?

Combine mex recurrence and XOR sum of non-standard games; demonstrate a valid strategy for all responses. Sizes do not treat like those of the ordinary Nim. Editorial evaluation based on the required techniques; not calibrated by a human panel.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-037

View the solution to SOLO-037 →

SOLO-038 · Nim — pair subtraction

EXPERT

The Broken Balance

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

Materials: Pencil and paper; optional chips, all pile numbers are given.

Rules you need

Original variant "pair subtraction". Each component is a pair of heaps A and B. The pairs remain independent.
At each turn, choose exactly ONE pair, then remove 1, 2 or 3 chips from A alone, OR B alone, OR exactly the same number (1, 2 or 3) of both piles. No pile can become negative. Never remove two different pairs in the same round.
The two sides alternate. The player without a legal blow loses: taking the last chips makes it win. There is neither pass nor rule misery. The tokens of the same pile are indistinguishable; the pairs and piles A/B are labeled.

The puzzle

You play first. To analyze alone, calculate the winning and losing positions of both sides.

Gives a complete list of the winners and a general proof of answer: only one cooperative suite is not enough.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3
1	Pair	Pile A	Pile B
2	1	12	9
3	2	7	13

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-038

View the solution to SOLO-038 →

SOLO-039 · Nim — pair subtraction

EXTREME

Three Linked Fronts

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

Materials: Pencil and paper; optional chips, all pile numbers are given.

Rules you need

Original variant "pair subtraction". Each component is a pair of heaps A and B. The pairs remain independent.
At each turn, choose exactly ONE pair, then remove 1, 2 or 3 chips from A alone, OR B alone, OR exactly the same number (1, 2 or 3) of both piles. No pile can become negative. Never remove two different pairs in the same round.
The two sides alternate. The player without a legal blow loses: taking the last chips makes it win. There is neither pass nor rule misery. The tokens of the same pile are indistinguishable; the pairs and piles A/B are labeled.

The puzzle

You play first. To analyze alone, calculate the winning and losing positions of both sides.

Gives a complete list of the winners and a general proof of answer: only one cooperative suite is not enough.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3
1	Pair	Pile A	Pile B
2	1	17	19
3	2	14	18
4	3	20	15

Why this level?

Build a Grundy table with two variables, assemble three components by XOR, list all the winning hits and prove the universal answer. The difficulty is on several combined tools, not on a long calculation of isolated chips. Editorial evaluation based on the required techniques; not calibrated by a human panel.

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-039

View the solution to SOLO-039 →

SOLO-040 · Nim — pair subtraction

EXTREME

The Great Pile Board

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

Materials: Pencil and paper; optional chips, all pile numbers are given.

Rules you need

Original variant "pair subtraction". Each component is a pair of heaps A and B. The pairs remain independent.
At each turn, choose exactly ONE pair, then remove 1, 2 or 3 chips from A alone, OR B alone, OR exactly the same number (1, 2 or 3) of both piles. No pile can become negative. Never remove two different pairs in the same round.
The two sides alternate. The player without a legal blow loses: taking the last chips makes it win. There is neither pass nor rule misery. The tokens of the same pile are indistinguishable; the pairs and piles A/B are labeled.

The puzzle

You play first. To analyze alone, calculate the winning and losing positions of both sides.

Gives a complete list of the winners and a general proof of answer: only one cooperative suite is not enough.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3
1	Pair	Pile A	Pile B
2	1	21	16
3	2	18	23
4	3	22	19

Why this level?

Your workspace

Write down your assumptions separately from deductions you have proved.

Notes for SOLO-040

View the solution to SOLO-040 →

Solutions to the Great Solo Challenge

The 40 solutions begin here.

SOLO-001 · Classic Sudoku

EXPERT

The Ninth Lock

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

423715689 / 615893247 / 987624513 / 876532194 / 132469875 / 549187362 / 758246931 / 364951728 / 291378456

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	4	2	3	7	1	5	6	8	9
2	6	1	5	8	9	3	2	4	7
3	9	8	7	6	2	4	5	1	3
4	8	7	6	5	3	2	1	9	4
5	1	3	2	4	6	9	8	7	5
6	5	4	9	1	8	7	3	6	2
7	7	5	8	2	4	6	9	3	1
8	3	6	4	9	5	1	7	2	8
9	2	9	1	3	7	8	4	5	6

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 3 states; the 1 contradictory branches are closed. Finally check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L3C3: only place for 7 in unit 3 → 7
L5C8: only place for 7 in unit 5 → 7
L6C6: only place for 7 in unit 6 → 7
L4C2: only place for 7 in unit 4 → 7
L7C8: only place for 3 in unit 7 → 3
L3C9: only place for 3 in unit 3 → 3
L1C3: only place for 3 in unit 12 → 3
L4C6: only place for 2 in unit 15 → 2
L4C5: only place for 3 in unit 4 → 3
L2C6: only place for 3 in unit 2 → 3
L4C7: only place for 1 in unit 4 → 1
L4C9: only place for 4 in unit 4 → 4
L1C6: only place for 5 in unit 15 → 5
L6C1: only place for 5 in unit 22 → 5
L6C3: only place for 9 in unit 6 → 9
L6C9: only place for 2 in unit 6 → 2
L5C3: only place for 2 in unit 5 → 2
L2C7: only place for 2 in unit 16 → 2
Nude subset in Unit 5: L5C7, L5C9 reserve [5, 8]; remove these numbers from L5C4, L5C5.
Nude subset in unit 26: L7C6, L8C6, L9C6 reserve [1, 6, 8]; remove these numbers from L7C5, L8C4.
Units 12 and 25 overlap. In unit 12, 1 must lie in the overlap. Remove it from L8C2, L9C2.

L1C9 = 8

Assumption: L1C9 = 8

L5C9: only candidate → 5
L5C7: only candidate → 8
L3C7: only place for 9 in unit 21 → 9
L3C1: only candidate → 6
L3C4: only candidate → 1
L1C5: only candidate → 9
L3C8: only candidate → 5
L4C1: only candidate → 8
L4C3: only candidate → 6
L6C4: only candidate → 8
L2C4: only candidate → 6
L2C5: only candidate → 8
L5C4: only candidate → 4
L5C5: only candidate → 6
L6C5: only candidate → 1
L7C5: only candidate → 4
L7C3: only candidate → 8
L7C6: only candidate → 6
L7C7: only candidate → 5
L7C2: only candidate → 9
L2C2: only candidate → 1
L1C2: only candidate → 2
L1C1: only candidate → 4
L1C8: only candidate → 1
L2C1: only candidate → 9
L2C8: only candidate → 4
L8C2: only candidate → 6
L8C4: only candidate → 9

contradiction: a box no longer has no candidate

L1C9 = 9

Assumption: L1C9 = 9

L3C7: only candidate → 5
L3C8: only candidate → 1
L5C7: only candidate → 8
L5C9: only candidate → 5
L7C7: only candidate → 9
L7C2: only place for 5 in unit 7 → 5
L9C8: only place for 5 in unit 9 → 5
L2C5: only place for 9 in unit 14 → 9
L3C4: only candidate → 6
L3C1: only candidate → 9
L5C4: only candidate → 4
L5C5: only candidate → 6
L8C3: only place for 4 in unit 8 → 4
L7C5: only place for 4 in unit 7 → 4
L9C2: only place for 9 in unit 9 → 9
L8C4: only place for 9 in unit 8 → 9
L9C1: only place for 2 in unit 9 → 2
L1C1: only candidate → 4
L1C8: only candidate → 8
L1C5: only candidate → 1
L1C2: only candidate → 2
L2C1: only candidate → 6
L2C2: only candidate → 1
L2C4: only candidate → 8
L2C8: only candidate → 4
L4C1: only candidate → 8
L4C3: only candidate → 6
L6C4: only candidate → 1
L6C5: only candidate → 8
L7C3: only candidate → 8
L7C6: only candidate → 6
L8C2: only candidate → 6
L8C8: only candidate → 2
L8C9: only candidate → 8
L8C6: only candidate → 1
L9C3: only candidate → 1
L9C6: only candidate → 8
L9C9: only candidate → 6

valid complete solution

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 3,
  "branches": 1,
  "contradictions": 1,
  "depth": 1,
  "pairs": 2,
  "locked": 1
}

← Return to puzzle SOLO-001

SOLO-002 · Classic Sudoku

EXTREME

The Hidden Crossroads

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

745163829 / 123958746 / 986247351 / 269714538 / 418395672 / 537826194 / 374689215 / 691532487 / 852471963

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	7	4	5	1	6	3	8	2	9
2	1	2	3	9	5	8	7	4	6
3	9	8	6	2	4	7	3	5	1
4	2	6	9	7	1	4	5	3	8
5	4	1	8	3	9	5	6	7	2
6	5	3	7	8	2	6	1	9	4
7	3	7	4	6	8	9	2	1	5
8	6	9	1	5	3	2	4	8	7
9	8	5	2	4	7	1	9	6	3

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 29 states; the 14 contradictory branches are closed. Finally check each unit and each starting index.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L1C8: only candidate → 2
L5C9: only place for 2 in unit 5 → 2
L6C7: only place for 1 in unit 6 → 1
L3C9: only place for 1 in unit 18 → 1
L4C9: only place for 8 in unit 18 → 8
L9C9: only place for 3 in unit 18 → 3
Nude subset in Unit 4: L4C2, L4C8 reserve [3, 6]; remove these numbers from L4C4, L4C5, L4C7.
L4C7: only candidate → 5
L5C4: only place for 3 in unit 13 → 3
Nude subset in Unit 5: L5C5, L5C6, L5C7 reserve [5, 6, 9]; remove these numbers from L5C1, L5C3.
L6C1: only place for 5 in unit 22 → 5
Nude subset in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2, L7C2, L8C2.
Nude subset in Unit 27: L7C9, L8C9 reserve [5, 7]; remove these numbers from L8C7, L9C7.
Nude subset in unit 16: L5C7, L8C7, L9C7 reserve [4, 6, 9]; remove these numbers from L2C7, L3C7.
L2C8: only place for 4 in unit 21 → 4
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C1, L1C3, L2C1, L2C3, L3C1, L3C3.
L1C7: only place for 8 in unit 1 → 8

L2C1 = 1

Assumption: L2C1 = 1

L8C3: only place for 1 in unit 8 → 1
Nude subset in Unit 4: L4C2, L4C8 reserve [3, 6]; remove these numbers from L4C4, L4C5.
Nude subset in Unit 5: L5C5, L5C6, L5C7 reserve [5, 6, 9]; remove these numbers from L5C1, L5C3.
Nude subset in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2, L7C2, L8C2.
Nude subset in unit 16: L2C7, L3C7 reserve [3, 7]; remove these numbers from L8C7, L9C7.
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L2C3, L3C1, L3C3.

L1C1 = 6

Assumption: L1C1 = 6

L1C3: only candidate → 5
Nude subset in Unit 4: L4C2, L4C8 reserve [3, 6]; remove these numbers from L4C4, L4C5.
Nude subset in Unit 5: L5C5, L5C6, L5C7 reserve [5, 6, 9]; remove these numbers from L5C3.
Nude subset in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2, L7C2, L8C2.
Nude subset in unit 13: L1C4, L4C4 reserve [1, 7]; remove these numbers from L3C4, L7C4, L8C4, L9C4.
Nude sub-assembly in unit 14: L1C5, L4C5 reserve [1, 7]; remove these numbers from L2C5, L3C5, L9C5.
L2C5: only candidate → 5
L5C6: only place for 5 in unit 5 → 5
Nude subset in unit 16: L2C7, L3C7 reserve [3, 7]; remove these numbers from L8C7, L9C7.
L9C1: only place for 7 in unit 9 → 7
L7C2: only candidate → 9
L8C1: only candidate → 4
L5C1: only candidate → 8
L3C1: only candidate → 9
L5C3: only candidate → 4
L7C3: only candidate → 6
L7C6: only candidate → 7
L7C9: only candidate → 5
L7C4: only candidate → 4
L8C2: only candidate → 2
L8C9: only candidate → 7
L9C3: only candidate → 8
L3C5: only place for 4 in unit 3 → 4
L8C4: only place for 5 in unit 8 → 5
L9C4: only place for 2 in unit 9 → 2
L3C4: only candidate → 6
L9C7: only place for 4 in unit 9 → 4

L1C4 = 1

Assumption: L1C4 = 1

L1C5: only candidate → 7
L4C4: only candidate → 7
L4C5: only place for 1 in unit 4 → 1
Nude subassembly in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2.

L2C2 = 7

Assumption: L2C2 = 7

L2C7: only candidate → 3
L2C3: only candidate → 2
L2C6: only candidate → 8
L3C3: only candidate → 3
L3C2: only candidate → 8
L3C6: only candidate → 2
L3C7: only candidate → 7

L4C2 = 3

Assumption: L4C2 = 3

L4C8: only candidate → 6
L5C7: only candidate → 9
L5C5: only candidate → 6
L6C2: only candidate → 6
L6C6: only candidate → 9
L6C8: only candidate → 3
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L4C2 = 6

Assumption: L4C2 = 6

L4C8: only candidate → 3
L6C2: only candidate → 3

L5C5 = 6

Assumption: L5C5 = 6

L5C7: only candidate → 9
L6C6: only candidate → 9
L6C8: only candidate → 6
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L5C5 = 9

Assumption: L5C5 = 9

L5C7: only candidate → 6
L6C6: only candidate → 6
L6C8: only candidate → 9
L8C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 8

Assumption: L2C2 = 8

L2C6: only candidate → 2
L2C3: only candidate → 3
L2C7: only candidate → 7
L3C2: only candidate → 7
L3C3: only candidate → 2
L3C6: only candidate → 8
L3C7: only candidate → 3

L4C2 = 3

Assumption: L4C2 = 3

L4C8: only candidate → 6
L5C7: only candidate → 9
L5C5: only candidate → 6
L6C2: only candidate → 6
L6C6: only candidate → 9
L6C8: only candidate → 3
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L4C2 = 6

Assumption: L4C2 = 6

L4C8: only candidate → 3
L6C2: only candidate → 3

L5C5 = 6

Assumption: L5C5 = 6

L5C7: only candidate → 9
L6C6: only candidate → 9
L6C8: only candidate → 6
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L5C5 = 9

Assumption: L5C5 = 9

L5C7: only candidate → 6
L6C6: only candidate → 6
L6C8: only candidate → 9
L8C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L1C4 = 7

Assumption: L1C4 = 7

L1C5: only candidate → 1
L4C4: only candidate → 1
L4C5: only place for 7 in unit 4 → 7
Nude subassembly in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2.

L2C2 = 7

Assumption: L2C2 = 7

L2C7: only candidate → 3
L2C3: only candidate → 2
L2C6: only candidate → 8
L3C3: only candidate → 3
L3C2: only candidate → 8
L3C6: only candidate → 2
L3C7: only candidate → 7

L4C2 = 3

Assumption: L4C2 = 3

L4C8: only candidate → 6
L5C7: only candidate → 9
L5C5: only candidate → 6
L6C2: only candidate → 6
L6C6: only candidate → 9
L6C8: only candidate → 3
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L4C2 = 6

Assumption: L4C2 = 6

L4C8: only candidate → 3
L6C2: only candidate → 3

L5C5 = 6

Assumption: L5C5 = 6

L5C7: only candidate → 9
L6C6: only candidate → 9
L6C8: only candidate → 6
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L5C5 = 9

Assumption: L5C5 = 9

L5C7: only candidate → 6
L6C6: only candidate → 6
L6C8: only candidate → 9
L8C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 8

Assumption: L2C2 = 8

L2C6: only candidate → 2
L2C3: only candidate → 3
L2C7: only candidate → 7
L3C2: only candidate → 7
L3C3: only candidate → 2
L3C6: only candidate → 8
L3C7: only candidate → 3

L4C2 = 3

Assumption: L4C2 = 3

L4C8: only candidate → 6
L5C7: only candidate → 9
L5C5: only candidate → 6
L6C2: only candidate → 6
L6C6: only candidate → 9
L6C8: only candidate → 3
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L4C2 = 6

Assumption: L4C2 = 6

L4C8: only candidate → 3
L6C2: only candidate → 3

L5C5 = 6

Assumption: L5C5 = 6

L5C7: only candidate → 9
L6C6: only candidate → 9
L6C8: only candidate → 6
L8C6: only candidate → 6

contradiction: a box no longer has no candidate

L5C5 = 9

Assumption: L5C5 = 9

L5C7: only candidate → 6
L6C6: only candidate → 6
L6C8: only candidate → 9
L8C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L1C1 = 7

Assumption: L1C1 = 7

Nude subset in Unit 4: L4C2, L4C8 reserve [3, 6]; remove these numbers from L4C4, L4C5.
Nude subset in Unit 5: L5C5, L5C6, L5C7 reserve [5, 6, 9]; remove these numbers from L5C1, L5C3.
Nude subset in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L2C2, L3C2, L7C2, L8C2.
Nude subset in unit 16: L2C7, L3C7 reserve [3, 7]; remove these numbers from L8C7, L9C7.
Units 9 and 26 overlap. In unit 9, 7 must lie in the overlap. Remove it from L7C4, L7C6, L8C4, L8C6.
Nude subset in unit 15: L5C6, L6C6, L7C6 reserve [5, 6, 9]; remove these numbers from L2C6, L3C6, L8C6.
L8C6: only candidate → 2
L9C3: only place for 2 in unit 9 → 2
L2C2: only place for 2 in unit 2 → 2
L3C4: only place for 2 in unit 3 → 2
L3C5: only place for 4 in unit 3 → 4
L9C1: only place for 8 in unit 9 → 8
L5C1: only candidate → 4
L5C3: only candidate → 8
L2C6: only place for 8 in unit 2 → 8
L3C6: only candidate → 7
L2C5: only candidate → 5
L2C3: only candidate → 3
L2C7: only candidate → 7
L3C3: only candidate → 6
L1C3: only candidate → 5
L3C1: only candidate → 9
L3C2: only candidate → 8
L3C7: only candidate → 3
L7C3: only candidate → 4
L8C1: only candidate → 6
L5C6: only place for 5 in unit 5 → 5
Units 7 and 26 overlap. In unit 7, 6 must lie in the overlap. Remove it from L9C4, L9C5.

L1C4 = 1

Assumption: L1C4 = 1

L1C5: only candidate → 6
L5C5: only candidate → 9
L5C7: only candidate → 6
L4C8: only candidate → 3
L4C2: only candidate → 6
L4C4: only candidate → 7
L4C5: only candidate → 1
L6C2: only candidate → 3
L6C6: only candidate → 6
L6C8: only candidate → 9
L7C6: only candidate → 9
L7C2: only candidate → 7
L7C9: only candidate → 5
L7C4: only candidate → 6
L8C2: only candidate → 9
L8C7: only candidate → 4
L8C4: only candidate → 5
L8C9: only candidate → 7
L9C4: only candidate → 4
L9C5: only candidate → 7
L9C7: only candidate → 9
L9C8: only candidate → 6

valid complete solution

L1C4 = 6

Assumption: L1C4 = 6

L1C5: only candidate → 1
L4C4: only place for 1 in unit 4 → 1
L4C5: only place for 7 in unit 4 → 7
L7C6: only place for 6 in unit 7 → 6
L6C6: only candidate → 9
L5C5: only candidate → 6
L5C7: only candidate → 9
L8C7: only candidate → 4
L9C5: only candidate → 9
L9C7: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C1 = 7

Assumption: L2C1 = 7

L2C7: only candidate → 3
L3C7: only candidate → 7
L3C3: only place for 3 in unit 12 → 3
Nude subset in Unit 4: L4C2, L4C8 reserve [3, 6]; remove these numbers from L4C4, L4C5.
Nude subset in Unit 5: L5C5, L5C6, L5C7 reserve [5, 6, 9]; remove these numbers from L5C1, L5C3.
Nude subset in Unit 11: L4C2, L6C2 reserve [3, 6]; remove these numbers from L3C2, L7C2, L8C2.
Units 9 and 26 overlap. In unit 9, 7 must lie in the overlap. Remove it from L7C4, L7C6, L8C4, L8C6.

contradiction: the number 7 can no longer be included in unit 15

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 29,
  "branches": 14,
  "contradictions": 14,
  "depth": 6,
  "pairs": 25,
  "locked": 5
}

← Return to puzzle SOLO-002

SOLO-003 · Classic Sudoku

EXPERT

The Inverted Mesh

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

345278916 / 791635824 / 862491537 / 529713468 / 673984152 / 184562379 / 218347695 / 937856241 / 456129783

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	3	4	5	2	7	8	9	1	6
2	7	9	1	6	3	5	8	2	4
3	8	6	2	4	9	1	5	3	7
4	5	2	9	7	1	3	4	6	8
5	6	7	3	9	8	4	1	5	2
6	1	8	4	5	6	2	3	7	9
7	2	1	8	3	4	7	6	9	5
8	9	3	7	8	5	6	2	4	1
9	4	5	6	1	2	9	7	8	3

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 3 states; the 1 contradictory branches are closed. Finally check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L2C2: only place for 9 in unit 2 → 9
L3C9: only place for 7 in unit 3 → 7
L5C5: only place for 8 in unit 5 → 8
L6C9: only place for 9 in unit 6 → 9
L5C4: only place for 9 in unit 5 → 9
L5C2: only place for 7 in unit 5 → 7
L4C4: only place for 7 in unit 4 → 7
L5C9: only place for 2 in unit 5 → 2
L4C2: only place for 2 in unit 4 → 2
L4C1: only place for 5 in unit 4 → 5
L6C7: only place for 3 in unit 6 → 3
L4C6: only place for 3 in unit 4 → 3
L7C8: only place for 9 in unit 7 → 9
L7C4: only place for 3 in unit 7 → 3
L8C3: only place for 7 in unit 8 → 7
L9C5: only place for 2 in unit 14 → 2
L9C4: only candidate → 1
L8C9: only place for 1 in unit 8 → 1
L8C5: only place for 5 in unit 14 → 5
L9C6: only candidate → 9
L8C6: only candidate → 6
L7C5: only candidate → 4
L7C3: only candidate → 8
L7C7: only candidate → 6
L5C7: only candidate → 1
L4C7: only candidate → 4
L4C8: only candidate → 6
L4C5: only candidate → 1
L5C1: only candidate → 6
L6C5: only candidate → 6
L3C2: only place for 6 in unit 3 → 6
L8C1: only place for 9 in unit 8 → 9
L9C8: only place for 8 in unit 9 → 8
L9C2: only place for 5 in unit 9 → 5
Nude subassembly in Unit 2: L2C4, L2C8, L2C9 reserve [2, 4, 6]; remove these figures from L2C3.
Units 2 and 21 overlap. In unit 2, 4 must lie in the overlap. Remove it from L1C9.

L1C2 = 3

Assumption: L1C2 = 3

L8C2: only candidate → 4
L8C8: only candidate → 3
L3C8: only candidate → 2
L2C8: only candidate → 4
L1C9: only candidate → 6

contradiction: a box no longer has no candidate

L1C2 = 4

Assumption: L1C2 = 4

L8C2: only candidate → 3
L8C8: only candidate → 4
L2C8: only candidate → 2
L2C4: only candidate → 6
L1C4: only candidate → 2
L1C3: only candidate → 5
L1C6: only candidate → 8
L1C1: only candidate → 3
L1C9: only candidate → 6
L2C3: only candidate → 1
L2C6: only candidate → 5
L2C7: only candidate → 8
L2C9: only candidate → 4
L3C1: only candidate → 8
L3C3: only candidate → 2
L3C6: only candidate → 1
L3C7: only candidate → 5
L3C8: only candidate → 3
L6C3: only candidate → 4
L6C1: only candidate → 1
L9C1: only candidate → 4
L9C9: only candidate → 3

valid complete solution

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 3,
  "branches": 1,
  "contradictions": 1,
  "depth": 1,
  "pairs": 1,
  "locked": 1
}

← Return to puzzle SOLO-003

SOLO-004 · Classic Sudoku

EXTREME

Branches of Silence

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

762983145 / 395142876 / 481657329 / 146298753 / 859376412 / 273514968 / 618429537 / 924735681 / 537861294

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	7	6	2	9	8	3	1	4	5
2	3	9	5	1	4	2	8	7	6
3	4	8	1	6	5	7	3	2	9
4	1	4	6	2	9	8	7	5	3
5	8	5	9	3	7	6	4	1	2
6	2	7	3	5	1	4	9	6	8
7	6	1	8	4	2	9	5	3	7
8	9	2	4	7	3	5	6	8	1
9	5	3	7	8	6	1	2	9	4

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 81 states; the 40 contradictory branches are closed. Finally check each unit and each starting index.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L5C8: only candidate → 1
L1C9: only place for 5 in unit 1 → 5
L1C2: only place for 6 in unit 1 → 6
L1C7: only place for 1 in unit 1 → 1
L2C3: only place for 5 in unit 2 → 5
L2C9: only place for 6 in unit 2 → 6
L6C4: only place for 5 in unit 6 → 5
L4C8: only place for 5 in unit 4 → 5
L9C1: only place for 5 in unit 9 → 5
L3C6: only place for 7 in unit 20 → 7
L2C5: only place for 4 in unit 20 → 4
Nude subset in unit 13: L1C4, L5C4 reserve [3, 9]; remove these numbers from L4C4, L8C4, L9C4.
L8C5: only place for 3 in unit 8 → 3
Nude subset in unit 26: L7C5, L7C6, L9C5 reserve [2, 6, 9]; remove these numbers from L8C4, L9C4.
L4C4: only place for 2 in unit 13 → 2
L6C5: only candidate → 1

L1C4 = 3

Assumption: L1C4 = 3

L1C6: only candidate → 9
L5C4: only candidate → 9
L4C5: only candidate → 6
L7C6: only candidate → 6
L5C3: only place for 6 in unit 5 → 6

contradiction: the number 6 can no longer appear in unit 9

L1C4 = 9

Assumption: L1C4 = 9

L1C6: only candidate → 3
L5C4: only candidate → 3
L5C7: only candidate → 4
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.

L2C2 = 8

Assumption: L2C2 = 8

L2C7: only candidate → 7
L2C8: only candidate → 9
Nude subset in Unit 9: L9C4, L9C7, L9C8 reserve [2, 7, 8]; remove these numbers from L9C3, L9C5, L9C9.
L7C5: only place for 2 in unit 14 → 2
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Nude subassembly in Unit 18: L3C9, L4C9, L6C9 reserve [3, 7, 8]; remove these numbers from L7C9, L8C9.
L7C2: only place for 7 in unit 7 → 7
Nude subassembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these numbers from L6C3, L6C6.
L6C6: only candidate → 8
L4C6: only candidate → 4
Nude subassembly in Unit 4: L4C1, L4C2, L4C5 reserve [1, 6, 9]; remove these numbers from L4C3.
Units 9 and 27 overlap. In unit 9, 2 must lie in the overlap. Remove it from L8C8.

L3C8 = 2

Assumption: L3C8 = 2

L9C7: only place for 2 in unit 9 → 2
Nude subassembly in Unit 4: L4C1, L4C2, L4C5 reserve [1, 6, 9]; remove these numbers from L4C3.
Nude sub-assembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these figures from L6C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L3C7 = 3

Assumption: L3C7 = 3

L3C9: only candidate → 8
L4C7: only candidate → 8
Nude subassembly in Unit 4: L4C1, L4C2, L4C5 reserve [1, 6, 9]; remove these numbers from L4C3.
Nude sub-assembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these figures from L6C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L4C2 = 1

Assumption: L4C2 = 1

Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Nude sub-assembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these figures from L6C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L3C2 = 4

Assumption: L3C2 = 4

L6C2: only candidate → 2
L6C1: only candidate → 4
L8C2: only candidate → 9
L8C1: only place for 2 in unit 8 → 2
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 9
L5C3: only candidate → 6
L4C1: only candidate → 9
L4C5: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

L3C1 = 9

Assumption: L3C1 = 9

L3C3: only candidate → 1
L4C1: only candidate → 6
L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 1
L7C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C2 = 9

Assumption: L3C2 = 9

Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Nude sub-assembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these figures from L6C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 4
L7C9: only place for 1 in unit 7 → 1
L8C3: only place for 1 in unit 8 → 1
L9C9: only place for 4 in unit 9 → 4
L8C9: only place for 9 in unit 18 → 9
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 4

Assumption: L3C1 = 4

L3C3: only candidate → 1
L6C1: only candidate → 2
L6C2: only candidate → 4
L8C2: only candidate → 2
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Nude subset in Unit 8: L8C4, L8C8 reserve [7, 8]; remove these numbers from L8C9.
Nude sub-assembly in Unit 9: L9C4, L9C8 reserve [7, 8]; remove these numbers from L9C9.

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C6: only candidate → 9
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 9

Assumption: L4C2 = 9

L4C5: only candidate → 6
L4C1: only candidate → 1
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 9
L3C1: only candidate → 4
L3C2: only candidate → 1
L3C3: only candidate → 9
L6C1: only candidate → 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 8

Assumption: L3C7 = 8

L3C9: only candidate → 3
L4C7: only candidate → 3
L6C9: only candidate → 7
L4C9: only candidate → 8
L4C3: only place for 7 in unit 4 → 7
L6C3: only place for 3 in unit 6 → 3

L4C2 = 1

Assumption: L4C2 = 1

L3C2 = 4

Assumption: L3C2 = 4

L6C2: only candidate → 2
L6C1: only candidate → 4
L8C2: only candidate → 9
L8C1: only place for 2 in unit 8 → 2

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 9
L5C3: only candidate → 6
L4C1: only candidate → 9
L4C5: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

L3C1 = 9

Assumption: L3C1 = 9

L3C3: only candidate → 1
L4C1: only candidate → 6
L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 1
L7C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C2 = 9

Assumption: L3C2 = 9

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 4
L7C9: only place for 1 in unit 7 → 1
L8C3: only place for 1 in unit 8 → 1
L9C9: only place for 4 in unit 9 → 4
L8C9: only candidate → 9

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 4

Assumption: L3C1 = 4

L3C3: only candidate → 1
L6C1: only candidate → 2
L6C2: only candidate → 4
L8C2: only candidate → 2

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C6: only candidate → 9
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 9

Assumption: L4C2 = 9

L4C5: only candidate → 6
L4C1: only candidate → 1
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 9
L3C1: only candidate → 4
L3C2: only candidate → 1
L3C3: only candidate → 9
L6C1: only candidate → 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C8 = 8

Assumption: L3C8 = 8

L3C9: only candidate → 3
L3C7: only candidate → 2
L6C9: only candidate → 7
L4C9: only candidate → 8
L4C7: only candidate → 3
L9C7: only candidate → 8
L9C4: only candidate → 7
L8C4: only candidate → 8
L9C8: only candidate → 2
L8C8: only candidate → 7
L4C3: only place for 7 in unit 4 → 7
L6C3: only place for 3 in unit 6 → 3

L4C2 = 1

Assumption: L4C2 = 1

L3C2 = 4

Assumption: L3C2 = 4

L6C2: only candidate → 2
L6C1: only candidate → 4
L8C2: only candidate → 9
L8C1: only place for 2 in unit 8 → 2

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 9
L5C3: only candidate → 6
L4C1: only candidate → 9
L4C5: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

L3C1 = 9

Assumption: L3C1 = 9

L3C3: only candidate → 1
L4C1: only candidate → 6
L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 1
L7C6: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C2 = 9

Assumption: L3C2 = 9

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 4
L7C9: only place for 1 in unit 7 → 1
L8C3: only place for 1 in unit 8 → 1
L9C9: only place for 4 in unit 9 → 4
L8C9: only candidate → 9

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 4

Assumption: L3C1 = 4

L3C3: only candidate → 1
L6C1: only candidate → 2
L6C2: only candidate → 4
L8C2: only candidate → 2

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C6: only candidate → 9
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 9

Assumption: L4C2 = 9

L4C5: only candidate → 6
L4C1: only candidate → 1
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 9
L3C1: only candidate → 4
L3C2: only candidate → 1
L3C3: only candidate → 9
L6C1: only candidate → 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 9

Assumption: L2C2 = 9

L3C2: only place for 8 in unit 11 → 8
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.

L2C7 = 7

Assumption: L2C7 = 7

L2C8: only candidate → 8
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Units 17 and 27 overlap. In unit 17, 7 must lie in the overlap. Remove it from L7C9, L8C9, L9C9.
L7C2: only place for 7 in unit 7 → 7
Nude subassembly in Unit 6: L6C1, L6C2 reserve [2, 4]; remove these numbers from L6C3, L6C6.
L6C6: only candidate → 8
L4C6: only candidate → 4
L4C2: only candidate → 1
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 4
L7C9: only place for 1 in unit 7 → 1
L8C3: only place for 1 in unit 8 → 1
L9C9: only place for 4 in unit 9 → 4
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Units 17 and 27 overlap. In unit 17, 7 must lie in the overlap. Remove it from L8C9.

L3C7 = 2

Assumption: L3C7 = 2

L3C8: only candidate → 9
L3C9: only candidate → 3
L6C9: only candidate → 7
L4C9: only candidate → 8
L4C7: only candidate → 3
L6C3: only candidate → 3
L8C9: only candidate → 9
L9C7: only candidate → 8
L9C4: only candidate → 7
L8C4: only candidate → 8
L9C8: only candidate → 2
L8C8: only candidate → 7
L4C3: only place for 7 in unit 4 → 7
L7C5: only place for 2 in unit 14 → 2

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 3

Assumption: L3C7 = 3

L3C9: only candidate → 9
L3C8: only candidate → 2
L4C7: only candidate → 8
L9C7: only candidate → 2
L9C4: only place for 8 in unit 9 → 8
L8C4: only candidate → 7
L8C8: only candidate → 9
L8C9: only candidate → 8
L9C8: only candidate → 7
L7C5: only place for 2 in unit 14 → 2
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 4

Assumption: L3C1 = 4

L3C3: only candidate → 1
L6C1: only candidate → 2
L6C2: only candidate → 4
L8C2: only candidate → 2
L7C5: only place for 2 in unit 7 → 2
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Units 17 and 27 overlap. In unit 17, 7 must lie in the overlap. Remove it from L8C9, L9C9.

L3C7 = 2

Assumption: L3C7 = 2

L3C8: only candidate → 9
L3C9: only candidate → 3
L6C9: only candidate → 7
L4C9: only candidate → 8
L4C7: only candidate → 3
L6C3: only candidate → 3
L8C8: only candidate → 7
L8C4: only candidate → 8
L9C4: only candidate → 7
L9C7: only candidate → 8
L9C8: only candidate → 2
L4C3: only place for 7 in unit 4 → 7

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C6: only candidate → 9
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 3

Assumption: L3C7 = 3

L3C9: only candidate → 9
L3C8: only candidate → 2
L4C7: only candidate → 8
L7C9: only candidate → 1
L9C7: only candidate → 2
L8C1: only place for 1 in unit 8 → 1
Nude subassembly in Unit 4: L4C1, L4C5 reserve [6, 9]; remove these numbers from L4C3.
Nude subset in unit 18: L4C9, L6C9 reserve [3, 7]; remove these numbers from L8C9, L9C9.

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C7 = 8

Assumption: L2C7 = 8

L2C8: only candidate → 7
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Units 17 and 27 overlap. In unit 17, 8 must lie in the overlap. Remove it from L8C9, L9C9.

L3C1 = 1

Assumption: L3C1 = 1

L3C3: only candidate → 4
L9C9: only place for 4 in unit 9 → 4
Nude subset in Unit 7: L7C1, L7C5, L7C6 reserve [2, 6, 9]; remove these numbers from L7C2, L7C9.
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Units 17 and 27 overlap. In unit 17, 8 must lie in the overlap. Remove it from L8C9.

L3C7 = 2

Assumption: L3C7 = 2

L3C8: only candidate → 9
L3C9: only candidate → 3
L9C7: only candidate → 7
L4C7: only candidate → 3
L9C4: only candidate → 8
L8C4: only candidate → 7
L9C8: only candidate → 2
L8C8: only candidate → 8
L6C3: only place for 3 in unit 6 → 3
L7C2: only place for 7 in unit 7 → 7
L6C9: only place for 7 in unit 6 → 7
L4C9: only candidate → 8
L4C3: only place for 7 in unit 4 → 7
L4C2: only place for 1 in unit 4 → 1
L6C6: only place for 8 in unit 6 → 8
L7C9: only place for 1 in unit 7 → 1
L8C9: only candidate → 9
L8C3: only candidate → 1
L7C5: only place for 2 in unit 14 → 2
L4C6: only place for 4 in unit 15 → 4

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 3

Assumption: L3C7 = 3

L3C9: only candidate → 9
L3C8: only candidate → 2
L4C7: only candidate → 7
L9C7: only candidate → 2
L7C5: only place for 2 in unit 14 → 2
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Nude sub-assembly in Unit 18: L4C9, L6C9 reserve [3, 8]; remove these numbers from L8C9.

L4C2 = 1

Assumption: L4C2 = 1

L7C2: only candidate → 7
L7C9: only candidate → 1
L6C3: only place for 7 in unit 6 → 7
L6C9: only place for 3 in unit 6 → 3
L4C9: only candidate → 8
L8C9: only candidate → 7
L8C4: only candidate → 8
L8C8: only candidate → 9
L8C3: only candidate → 1
L9C4: only candidate → 7
L9C8: only candidate → 8
L4C3: only place for 3 in unit 4 → 3
L6C6: only place for 8 in unit 6 → 8
L4C6: only place for 4 in unit 15 → 4

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 4

Assumption: L4C2 = 4

L6C1: only candidate → 2
L6C2: only candidate → 7
L6C3: only candidate → 3
L6C9: only candidate → 8
L4C9: only candidate → 3
L6C6: only candidate → 4
L7C2: only candidate → 1
L7C9: only candidate → 7
L8C2: only candidate → 2
L8C9: only candidate → 1
L4C6: only place for 8 in unit 4 → 8
L4C3: only place for 1 in unit 4 → 1
L8C1: only place for 4 in unit 8 → 4

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 4

Assumption: L3C1 = 4

L3C3: only candidate → 1
L6C1: only candidate → 2
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Units 17 and 27 overlap. In unit 17, 8 must lie in the overlap. Remove it from L8C9, L9C9.

L3C7 = 2

Assumption: L3C7 = 2

L3C8: only candidate → 9
L3C9: only candidate → 3
L9C7: only candidate → 7
L4C7: only candidate → 3
L9C4: only candidate → 8
L8C4: only candidate → 7
L9C8: only candidate → 2
L8C8: only candidate → 8
L6C3: only place for 3 in unit 6 → 3
L7C2: only place for 7 in unit 7 → 7
L6C2: only candidate → 4
L4C2: only candidate → 1
L6C6: only candidate → 8
L6C9: only candidate → 7
L4C9: only candidate → 8
L8C2: only candidate → 2
L4C6: only place for 4 in unit 4 → 4
L4C3: only place for 7 in unit 4 → 7
L7C5: only place for 2 in unit 7 → 2

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C6: only candidate → 9
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C6: only candidate → 6
L7C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 3

Assumption: L3C7 = 3

L3C9: only candidate → 9
L3C8: only candidate → 2
L4C7: only candidate → 7
L9C7: only candidate → 2
L8C2: only place for 2 in unit 8 → 2
L7C5: only place for 2 in unit 7 → 2
Nude subset in Unit 7: L7C2, L7C9 reserve [1, 7]; remove these numbers from L7C1.
Nude subset in unit 15: L5C6, L7C6 reserve [6, 9]; remove these figures from L4C6.
Nude subset in Unit 18: L4C9, L6C9 reserve [3, 8]; remove these numbers from L8C9, L9C9.
Units 11 and 22 overlap. In unit 11, 4 must lie in the overlap. Remove it from L4C3, L6C3.

L4C2 = 1

Assumption: L4C2 = 1

L7C2: only candidate → 7
L6C2: only candidate → 4
L6C6: only candidate → 8
L6C9: only candidate → 3
L4C9: only candidate → 8
L6C3: only candidate → 7
L7C9: only candidate → 1
L4C3: only place for 3 in unit 4 → 3
L4C6: only place for 4 in unit 4 → 4
L8C1: only place for 1 in unit 8 → 1

L4C1 = 6

Assumption: L4C1 = 6

L4C5: only candidate → 9
L5C3: only candidate → 9
L5C6: only candidate → 6
L7C1: only candidate → 9

contradiction: a box no longer has no candidate

L4C1 = 9

Assumption: L4C1 = 9

L4C5: only candidate → 6
L5C3: only candidate → 6
L5C6: only candidate → 9
L7C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 4

Assumption: L4C2 = 4

L6C2: only candidate → 7
L6C3: only candidate → 3
L6C9: only candidate → 8
L4C9: only candidate → 3
L6C6: only candidate → 4
L7C2: only candidate → 1
L7C9: only candidate → 7
L8C1: only candidate → 9
L7C1: only candidate → 6
L4C1: only candidate → 1
L7C6: only candidate → 9
L5C6: only candidate → 6
L4C5: only candidate → 9
L4C3: only candidate → 6
L4C6: only candidate → 8
L5C3: only candidate → 9
L8C8: only candidate → 8
L8C4: only candidate → 7
L8C3: only candidate → 4
L8C9: only candidate → 1
L9C3: only candidate → 7
L9C4: only candidate → 8
L9C5: only candidate → 6
L9C8: only candidate → 9
L9C9: only candidate → 4

valid complete solution

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 81,
  "branches": 40,
  "contradictions": 40,
  "depth": 8,
  "pairs": 49,
  "locked": 8
}

← Return to puzzle SOLO-004

SOLO-005 · Diagonal Sudoku

EXPERT

The Two Crossings

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

716854392 / 893612754 / 245973168 / 371489625 / 624135879 / 589726413 / 138597246 / 962341587 / 457268931

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	7	1	6	8	5	4	3	9	2
2	8	9	3	6	1	2	7	5	4
3	2	4	5	9	7	3	1	6	8
4	3	7	1	4	8	9	6	2	5
5	6	2	4	1	3	5	8	7	9
6	5	8	9	7	2	6	4	1	3
7	1	3	8	5	9	7	2	4	6
8	9	6	2	3	4	1	5	8	7
9	4	5	7	2	6	8	9	3	1

The reasoning

Note candidates in each box by crossing lines, columns and regions, diagonals.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly one complete grid. It visits 5 states and rules out the 2 contradictory branches. Check each unit and starting clue.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L9C1: only place for 4 in unit 9 → 4
L9C7: only place for 9 in unit 9 → 9
L9C2: only place for 5 in unit 9 → 5
L5C5: only place for 3 in unit 28 → 3
L4C1: only place for 3 in unit 4 → 3
L7C2: only place for 3 in unit 11 → 3
L1C7: only place for 3 in unit 16 → 3
L1C6: only place for 4 in unit 1 → 4
L2C9: only place for 4 in unit 2 → 4
L2C8: only place for 5 in unit 2 → 5
L4C9: only place for 5 in unit 4 → 5
L6C1: only place for 5 in unit 6 → 5
L5C6: only place for 5 in unit 5 → 5
L8C7: only place for 5 in unit 8 → 5
L6C4: only place for 7 in unit 29 → 7
L1C1: only place for 7 in unit 1 → 7
L3C5: only place for 7 in unit 14 → 7
L5C8: only place for 7 in unit 17 → 7
L4C6: only place for 9 in unit 23 → 9
L5C9: only place for 9 in unit 18 → 9
L6C3: only place for 9 in unit 6 → 9
L6C7: only place for 4 in unit 6 → 4
L5C3: only place for 4 in unit 5 → 4
L8C1: only place for 9 in unit 8 → 9
Units 5 and 22 overlap. In unit 5, 6 must lie in the overlap. Remove it from L6C2.
Units 9 and 26 overlap. In unit 9, 8 must lie in the overlap. Remove it from L7C6, L8C4, L8C6.
Units 24 and 17 overlap. In unit 24, 1 must lie in the overlap. Remove it from L1C8, L3C8, L8C8.
Units 21 and 29 overlap. In unit 21, 1 must lie in the overlap. Remove it from L7C3, L8C2.
Units 27 and 28 overlap. In unit 27, 8 must lie in the overlap. Remove it from L6C6.
Units 23 and 14 overlap. In unit 23, 8 must lie in the overlap. Remove it from L2C5, L9C5.
Units 28 and 27 overlap. In unit 28, 1 must lie in the overlap. Remove it from L7C9, L8C9.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L7C1, L8C3.

L3C7 = 1

Assumption: L3C7 = 1

Units 1 and 19 overlap. In unit 1, 1 must lie in the overlap. Remove it from L2C1.
L7C1: only place for 1 in unit 10 → 1
Units 5 and 22 overlap. In unit 5, 6 must lie in the overlap. Remove it from L6C2.
Units 9 and 26 overlap. In unit 9, 8 must lie in the overlap. Remove it from L7C6, L8C4, L8C6.
Units 18 and 27 overlap. In unit 18, 1 must lie in the overlap. Remove it from L8C8.
L9C9: only place for 1 in unit 28 → 1
L8C6: only place for 1 in unit 8 → 1
L2C5: only place for 1 in unit 2 → 1
L8C4: only place for 3 in unit 8 → 3
L3C6: only place for 3 in unit 3 → 3
L8C9: only place for 7 in unit 8 → 7
L7C6: only place for 7 in unit 7 → 7
Units 27 and 28 overlap. In unit 27, 8 must lie in the overlap. Remove it from L6C6.
Units 23 and 14 overlap. In unit 23, 8 must lie in the overlap. Remove it from L9C5.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3.

L1C9 = 2

Assumption: L1C9 = 2

L7C9: only candidate → 6
L7C3: only candidate → 8
L7C7: only candidate → 2
L5C7: only candidate → 8
L8C2: only candidate → 6
L5C2: only candidate → 2
L4C3: only candidate → 1
L1C3: only candidate → 6
L1C8: only candidate → 9
L1C4: only candidate → 8
L1C2: only candidate → 1
L3C1: only candidate → 2
L2C1: only candidate → 8
L3C8: only candidate → 6
L3C4: only candidate → 9
L4C8: only candidate → 2
L4C5: only candidate → 8
L5C1: only candidate → 6
L6C2: only candidate → 8
L6C6: only candidate → 6
L2C6: only candidate → 2
L2C4: only candidate → 6
L6C5: only candidate → 2
L6C8: only candidate → 1
L8C3: only candidate → 2
L8C8: only candidate → 8
L9C4: only candidate → 2
L9C5: only candidate → 6
L9C6: only candidate → 8

valid complete solution

L1C9 = 6

Assumption: L1C9 = 6

L7C9: only candidate → 2
L7C3: only candidate → 8

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C7 = 2

Assumption: L3C7 = 2

L5C7: only candidate → 8
L7C7: only candidate → 1
L8C8: only place for 8 in unit 17 → 8
L1C9: only place for 1 in unit 18 → 1
L8C2: only candidate → 6
L5C2: only candidate → 2
L1C2: only candidate → 8
L5C1: only candidate → 6
L3C1: only candidate → 1
L2C1: only candidate → 2
L1C3: only candidate → 6
L1C8: only candidate → 9
L1C4: only candidate → 2
L3C8: only candidate → 6
L3C6: only candidate → 3
L3C4: only candidate → 9
L6C2: only candidate → 1
L4C3: only candidate → 8

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 5,
  "branches": 2,
  "contradictions": 2,
  "depth": 2,
  "pairs": 0,
  "locked": 15
}

← Return to puzzle SOLO-005

SOLO-006 · Diagonal Sudoku

EXTREME

The Stubborn Mirror

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

679852413 / 432671859 / 815943672 / 356297148 / 284516937 / 791438265 / 548369721 / 123784596 / 967125384

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	6	7	9	8	5	2	4	1	3
2	4	3	2	6	7	1	8	5	9
3	8	1	5	9	4	3	6	7	2
4	3	5	6	2	9	7	1	4	8
5	2	8	4	5	1	6	9	3	7
6	7	9	1	4	3	8	2	6	5
7	5	4	8	3	6	9	7	2	1
8	1	2	3	7	8	4	5	9	6
9	9	6	7	1	2	5	3	8	4

The reasoning

Note candidates in each box by crossing lines, columns and regions, diagonals.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 37 states; the 18 contradictory branches are closed. Finally, check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L1C7: only place for 4 in unit 1 → 4
L8C6: only place for 4 in unit 15 → 4
L9C9: only place for 4 in unit 28 → 4
L7C2: only place for 4 in unit 7 → 4
L7C9: only place for 1 in unit 7 → 1
Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C2, L3C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.

L6C6 = 3

Assumption: L6C6 = 3

Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C2, L3C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.

L2C6 = 1

Assumption: L2C6 = 1

L5C5: only place for 1 in unit 28 → 1
L4C7: only place for 1 in unit 4 → 1
L8C1: only place for 1 in unit 8 → 1
L3C2: only place for 1 in unit 3 → 1
Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.
Units 29 and 21 overlap. In unit 29, 5 must lie in the overlap. Remove it from L3C8, L3C9.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L2C2 = 2

Assumption: L2C2 = 2

L5C4: only place for 2 in unit 13 → 2
L4C1: only place for 2 in unit 10 → 2
Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C3.
Units 5 and 24 overlap. In unit 5, 7 must lie in the overlap. Remove it from L4C8, L4C9.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.
Units 29 and 21 overlap. In unit 29, 5 must lie in the overlap. Remove it from L3C8, L3C9.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L2C8 = 5

Assumption: L2C8 = 5

Nude subset in Unit 3: L3C7, L3C8, L3C9 reserve [2, 6, 7]; remove these numbers from L3C1, L3C3, L3C4, L3C6.
L3C6: only candidate → 8
L4C6: only candidate → 7
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C3.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L2C3 = 3

Assumption: L2C3 = 3

L2C5: only candidate → 7
L3C4: only place for 3 in unit 3 → 3
L7C4: only candidate → 8
L7C3: only candidate → 2
L3C7: only candidate → 6
L3C1: only candidate → 9
L3C3: only candidate → 5
L4C4: only candidate → 9
L1C4: only candidate → 5
L1C5: only candidate → 9
L7C8: only candidate → 3
L8C4: only candidate → 7
L8C5: only candidate → 3
L9C1: only candidate → 8
L1C1: only candidate → 6
L5C1: only candidate → 3
L8C2: only candidate → 9
L8C3: only candidate → 6
L8C8: only candidate → 8
L9C3: only candidate → 7
L1C3: only candidate → 8
L1C2: only candidate → 7
L4C3: only candidate → 4
L4C8: only candidate → 6
L5C3: only candidate → 9
L5C7: only candidate → 5
L4C9: only candidate → 8
L4C2: only candidate → 5

contradiction: a box no longer has no candidate

L2C3 = 7

Assumption: L2C3 = 7

L2C5: only candidate → 3
L3C1: only place for 3 in unit 3 → 3
L9C2: only place for 7 in unit 9 → 7
Nude sub-assembly in Unit 3: L3C7, L3C8, L3C9 reserve [2, 6, 7]; remove these numbers from L3C3, L3C4.
Units 11 and 22 overlap. In unit 11, 3 must lie in the overlap. Remove it from L4C3, L5C3.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3, L9C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C3.

L3C3 = 5

Assumption: L3C3 = 5

L3C4: only place for 9 in unit 3 → 9
L4C4: only candidate → 8
L7C4: only candidate → 3
L8C4: only candidate → 7
L1C4: only candidate → 5
L1C5: only candidate → 7
L8C5: only candidate → 8
Nude subset in Unit 8: L8C2, L8C8 reserve [6, 9]; remove these numbers from L8C3, L8C7, L8C9.
Units 11 and 22 overlap. In unit 11, 3 must lie in the overlap. Remove it from L4C3, L5C3.
Units 18 and 24 overlap. In unit 18, 8 must lie in the overlap. Remove it from L5C8, L6C8.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L9C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L9C3.

L1C1 = 6

Assumption: L1C1 = 6

L8C8: only candidate → 9
L8C2: only candidate → 6
L3C7: only candidate → 2
L7C3: only candidate → 8
L1C3: only candidate → 9
L1C2: only candidate → 8
L7C8: only candidate → 2
L8C9: only candidate → 5
L4C9: only candidate → 6
L3C9: only candidate → 7
L3C8: only candidate → 6
L5C9: only candidate → 8

contradiction: a box no longer has no candidate

L1C1 = 9

Assumption: L1C1 = 9

L5C1: only candidate → 8
L8C8: only candidate → 6
L8C2: only candidate → 9
L9C1: only candidate → 6
L3C7: only candidate → 2
L3C8: only candidate → 7
L3C9: only candidate → 6
L4C9: only candidate → 5
L4C5: only candidate → 9
L5C9: only candidate → 7
L6C5: only candidate → 5
L6C2: only candidate → 6
L1C2: only candidate → 8
L1C3: only candidate → 6
L4C2: only candidate → 3
L4C3: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C3 = 9

Assumption: L3C3 = 9

L3C4: only place for 5 in unit 3 → 5
L4C4: only candidate → 8
L1C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C8 = 7

Assumption: L2C8 = 7

L4C6: only candidate → 8
L3C6: only candidate → 7
L7C3: only candidate → 2
L5C9: only place for 7 in unit 5 → 7
L4C5: only place for 7 in unit 4 → 7
L8C4: only place for 7 in unit 13 → 7
L3C7: only place for 5 in unit 29 → 5
Nude subset in Unit 3: L3C8, L3C9 reserve [2, 6]; remove these numbers from L3C1, L3C3.
Nude subset in Unit 25: L8C2, L9C1 reserve [6, 9]; remove these numbers from L8C3, L9C2, L9C3.
Nude subassembly in Unit 8: L8C3, L8C5 reserve [3, 8]; remove these numbers from L8C7, L8C8, L8C9.
L6C9: only place for 8 in unit 18 → 8
Units 28 and 19 overlap. In unit 28, 8 must lie in the overlap. Remove it from L1C2, L1C3, L3C1.

L2C3 = 3

Assumption: L2C3 = 3

L2C5: only candidate → 5
L6C5: only candidate → 9
L1C5: only candidate → 8
L1C4: only candidate → 9
L1C1: only candidate → 6
L3C4: only candidate → 3
L4C4: only candidate → 5
L4C9: only candidate → 6
L3C9: only candidate → 2
L3C8: only candidate → 6
L6C7: only candidate → 2
L6C8: only candidate → 5
L6C2: only candidate → 6
L7C4: only candidate → 8
L7C8: only candidate → 3
L8C2: only candidate → 9

contradiction: a box no longer has no candidate

L2C3 = 5

Assumption: L2C3 = 5

L2C5: only candidate → 3
L8C5: only candidate → 8
L7C4: only candidate → 3
L7C8: only candidate → 8
L3C1: only place for 3 in unit 3 → 3
Nude subassembly in Unit 3: L3C8, L3C9 reserve [2, 6]; remove these figures from L3C3.
Nude subset in Unit 9: L9C1, L9C7, L9C8 reserve [3, 6, 9]; remove these numbers from L9C2, L9C3.
L8C3: only place for 3 in unit 25 → 3
Units 28 and 19 overlap. In unit 28, 8 must lie in the overlap. Remove it from L1C2, L1C3.

L1C5 = 5

Assumption: L1C5 = 5

L6C5: only candidate → 9
L4C4: only candidate → 5
L4C9: only candidate → 6
L3C9: only candidate → 2
L3C8: only candidate → 6
L6C7: only candidate → 2
L6C8: only candidate → 5
L6C2: only candidate → 6
L8C2: only candidate → 9

contradiction: a box no longer has no candidate

L1C5 = 9

Assumption: L1C5 = 9

L3C4: only candidate → 8
L1C4: only candidate → 5
L4C4: only candidate → 9
L3C3: only candidate → 6
L1C1: only candidate → 8
L1C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 5

Assumption: L2C2 = 5

L3C7: only place for 5 in unit 29 → 5
L8C9: only place for 5 in unit 8 → 5
Nude sub-assembly in unit 29: L2C8, L4C6, L7C3 reserve [2, 7, 8]; remove these numbers from L8C2, L9C1.
Nude subset in Unit 25: L8C2, L9C1 reserve [6, 9]; remove these numbers from L8C3, L9C2, L9C3.
Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C3.
Units 11 and 22 overlap. In unit 11, 2 must lie in the overlap. Remove it from L4C1, L4C3, L5C1, L5C3.
L3C1: only place for 2 in unit 10 → 2
L2C8: only place for 2 in unit 2 → 2
L7C3: only candidate → 8
L3C3: only candidate → 9
L4C6: only candidate → 7
L3C6: only candidate → 8
L7C4: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C6 = 7

Assumption: L2C6 = 7

L8C2: only place for 7 in unit 29 → 7
L1C3: only place for 7 in unit 1 → 7
L8C1: only place for 1 in unit 8 → 1

contradiction: the number 7 can no longer be included in unit 9

all the possibilities of this bifurcation have been examined

L6C6 = 8

Assumption: L6C6 = 8

Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C2, L3C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.
Units 15 and 20 overlap. In unit 15, 3 must lie in the overlap. Remove it from L2C5, L3C4.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C1, L8C3, L9C2, L9C3.

L1C1 = 6

Assumption: L1C1 = 6

Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C2, L8C3.
Units 15 and 20 overlap. In unit 15, 3 must lie in the overlap. Remove it from L2C5, L3C4.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C1, L8C3, L9C2, L9C3.

L4C6 = 1

Assumption: L4C6 = 1

L5C7: only place for 1 in unit 5 → 1
L8C1: only place for 1 in unit 8 → 1
L3C2: only place for 1 in unit 3 → 1
L2C5: only place for 1 in unit 2 → 1

contradiction: the number 1 can no longer appear in unit 28

L4C6 = 7

Assumption: L4C6 = 7

Nude sub-assembly in unit 20: L2C6, L3C6 reserve [1, 3]; remove these numbers from L2C5, L3C4.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C3.
Units 21 and 3 overlap. In unit 21, 7 must lie in the overlap. Remove it from L3C2, L3C4.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C1, L8C3, L9C2, L9C3.

L2C5 = 5

Assumption: L2C5 = 5

L2C8: only candidate → 2
L7C3: only candidate → 8
L7C4: only candidate → 3

contradiction: a box no longer has no candidate

L2C5 = 7

Assumption: L2C5 = 7

L8C4: only place for 7 in unit 13 → 7
Nude sub-assembly in unit 20: L2C6, L3C6 reserve [1, 3]; remove these figures from L3C4.
Units 1 and 19 overlap. In unit 1, 7 must lie in the overlap. Remove it from L3C2.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C1, L8C3, L9C2, L9C3.

L2C6 = 1

Assumption: L2C6 = 1

L3C6: only candidate → 3
L5C5: only place for 1 in unit 28 → 1
L4C7: only place for 1 in unit 4 → 1
L8C1: only place for 1 in unit 8 → 1
L3C2: only place for 1 in unit 3 → 1
Units 10 and 22 overlap. In unit 10, 3 must lie in the overlap. Remove it from L4C2, L4C3, L5C2, L5C3, L6C2.
Units 29 and 21 overlap. In unit 29, 5 must lie in the overlap. Remove it from L3C8, L3C9.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L2C8 = 2

Assumption: L2C8 = 2

L7C3: only candidate → 8
L7C4: only candidate → 3

contradiction: a box no longer has no candidate

L2C8 = 5

Assumption: L2C8 = 5

Nude subassembly in Unit 3: L3C7, L3C8, L3C9 reserve [2, 6, 7]; remove these numbers from L3C1, L3C3.
Nude subset in Unit 10: L3C1, L9C1 reserve [8, 9]; remove these numbers from L4C1, L5C1.
Nude sub-assembly in unit 22: L4C1, L5C1 reserve [2, 3]; remove these numbers from L4C2, L4C3, L5C2, L5C3, L6C2.
Units 6 and 24 overlap. In unit 6, 2 must lie in the overlap. Remove it from L4C8, L4C9, L5C7, L5C8, L5C9.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L2C2 = 2

Assumption: L2C2 = 2

L2C3: only candidate → 3
L5C4: only place for 2 in unit 13 → 2
L4C1: only place for 2 in unit 10 → 2
L5C1: only place for 3 in unit 10 → 3
L9C2: only place for 3 in unit 11 → 3
L9C3: only place for 7 in unit 9 → 7
L1C2: only place for 7 in unit 1 → 7
Units 9 and 27 overlap. In unit 9, 6 must lie in the overlap. Remove it from L8C7, L8C9.
Units 11 and 22 overlap. In unit 11, 5 must lie in the overlap. Remove it from L4C3, L5C3.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3.
Units 29 and 25 overlap. In unit 29, 9 must lie in the overlap. Remove it from L8C3.

L3C1 = 8

Assumption: L3C1 = 8

L9C1: only candidate → 9
L9C7: only candidate → 6
L3C7: only candidate → 2
L7C3: only candidate → 8
L7C4: only candidate → 3
L7C8: only candidate → 2
L8C2: only candidate → 6
L8C3: only candidate → 2
L8C5: only candidate → 8
L8C9: only candidate → 5
L9C8: only candidate → 8
L1C4: only place for 8 in unit 1 → 8
L6C9: only place for 2 in unit 6 → 2
L6C8: only place for 6 in unit 6 → 6
L3C8: only candidate → 7
L3C9: only candidate → 6
L4C9: only candidate → 8
L5C9: only candidate → 7
L4C3: only place for 6 in unit 4 → 6
L4C8: only place for 4 in unit 4 → 4
L5C8: only candidate → 9
L5C7: only candidate → 5
L5C2: only candidate → 8
L5C3: only candidate → 4
L6C7: only candidate → 3
L8C7: only candidate → 9
L8C8: only candidate → 3
L4C5: only place for 3 in unit 4 → 3

L1C3 = 5

Assumption: L1C3 = 5

L1C5: only candidate → 9
L3C3: only candidate → 9
L3C4: only candidate → 5

contradiction: a box no longer has no candidate

L1C3 = 9

Assumption: L1C3 = 9

L1C5: only candidate → 5
L3C3: only candidate → 5
L3C4: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C1 = 9

Assumption: L3C1 = 9

L3C3: only candidate → 5
L1C3: only candidate → 8
L3C4: only candidate → 8
L7C3: only candidate → 2
L3C7: only candidate → 6
L7C4: only candidate → 3
L4C4: only candidate → 9
L1C4: only candidate → 5
L1C5: only candidate → 9
L7C8: only candidate → 8
L8C5: only candidate → 8
L8C2: only candidate → 9
L8C3: only candidate → 6
L4C3: only candidate → 4
L5C3: only candidate → 9
L5C7: only candidate → 5
L5C2: only candidate → 8
L5C9: only candidate → 7
L3C9: only candidate → 2
L3C8: only candidate → 7
L5C8: only candidate → 4
L6C9: only candidate → 6
L4C8: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 3

Assumption: L2C2 = 3

L2C3: only candidate → 2
L7C3: only candidate → 8
L7C4: only candidate → 3
L7C8: only candidate → 2
L8C5: only candidate → 8
L8C8: only candidate → 9
L3C3: only candidate → 5
L4C4: only candidate → 2
L9C1: only candidate → 9
L3C1: only candidate → 8
L3C4: only candidate → 9
L1C5: only candidate → 5
L1C4: only candidate → 8
L4C1: only candidate → 3
L4C5: only candidate → 9
L5C1: only candidate → 2
L5C4: only candidate → 5
L6C5: only candidate → 3
L6C8: only candidate → 6
L3C8: only candidate → 7
L5C9: only place for 7 in unit 5 → 7
L8C2: only place for 2 in unit 8 → 2
L3C7: only candidate → 6
L3C9: only candidate → 2
L6C9: only candidate → 5
L4C9: only candidate → 8
L4C8: only candidate → 4
L4C3: only candidate → 6
L4C2: only candidate → 5
L5C8: only candidate → 3
L5C7: only candidate → 9
L5C2: only candidate → 8
L5C3: only candidate → 4
L6C2: only candidate → 9
L1C2: only candidate → 7
L1C3: only candidate → 9
L6C7: only candidate → 2
L8C3: only candidate → 3
L8C7: only candidate → 5
L8C9: only candidate → 6
L9C2: only candidate → 6
L9C3: only candidate → 7
L9C7: only candidate → 3
L9C8: only candidate → 8

valid complete solution

all the possibilities of this bifurcation have been examined

L2C6 = 3

Assumption: L2C6 = 3

L3C6: only candidate → 1
L2C2: only place for 1 in unit 2 → 1
L5C7: only place for 1 in unit 5 → 1
L4C5: only place for 1 in unit 4 → 1
L8C1: only place for 1 in unit 8 → 1

contradiction: the number 1 can no longer appear in unit 29

all the possibilities of this bifurcation have been examined

L1C1 = 9

Assumption: L1C1 = 9

L3C4: only place for 9 in unit 3 → 9
L8C2: only place for 9 in unit 29 → 9
L8C1: only place for 1 in unit 8 → 1
Units 1 and 19 overlap. In unit 1, 6 must lie in the overlap. Remove it from L3C1, L3C2, L3C3.
L8C8: only place for 6 in unit 28 → 6
Units 3 and 19 overlap. In unit 3, 8 must lie in the overlap. Remove it from L1C2, L1C3.
Units 9 and 25 overlap. In unit 9, 7 must lie in the overlap. Remove it from L8C3.
Units 15 and 20 overlap. In unit 15, 3 must lie in the overlap. Remove it from L2C5.
Units 25 and 12 overlap. In unit 25, 2 must lie in the overlap. Remove it from L2C3, L3C3, L4C3, L5C3.
Units 29 and 23 overlap. In unit 29, 1 must lie in the overlap. Remove it from L4C5.
Units 29 and 25 overlap. In unit 29, 8 must lie in the overlap. Remove it from L8C3, L9C2, L9C3.

L3C3 = 3

Assumption: L3C3 = 3

L2C6: only place for 3 in unit 15 → 3
L9C2: only place for 3 in unit 25 → 3
L9C7: only candidate → 9
L9C8: only candidate → 8
L9C1: only candidate → 6
L9C3: only candidate → 7
L3C9: only place for 6 in unit 21 → 6
L7C3: only place for 8 in unit 29 → 8
L7C4: only candidate → 3
L7C8: only candidate → 2
L8C3: only candidate → 2
L2C3: only candidate → 5
L1C3: only candidate → 6
L2C8: only candidate → 7
L2C5: only candidate → 1
L2C2: only candidate → 2
L3C1: only candidate → 8
L1C2: only candidate → 7
L3C2: only candidate → 1
L3C6: only candidate → 7
L3C8: only candidate → 5
L3C7: only candidate → 2
L4C4: only candidate → 5

contradiction: a box no longer has no candidate

L3C3 = 5

Assumption: L3C3 = 5

L5C5: only candidate → 1
L4C6: only candidate → 7
L2C6: only place for 1 in unit 2 → 1
L3C6: only candidate → 3
L3C2: only place for 1 in unit 3 → 1
L3C1: only place for 8 in unit 3 → 8
L9C1: only candidate → 6
L3C7: only candidate → 2
L2C8: only candidate → 5
L2C5: only candidate → 7
L3C8: only candidate → 7
L3C9: only candidate → 6
L7C3: only candidate → 8
L7C4: only candidate → 3
L4C4: only candidate → 2
L2C2: only candidate → 3
L2C3: only candidate → 2
L4C1: only candidate → 3
L5C1: only candidate → 2
L5C4: only candidate → 5
L1C4: only candidate → 8
L1C5: only candidate → 5
L4C5: only candidate → 9
L5C2: only candidate → 8
L5C9: only candidate → 7
L6C5: only candidate → 3
L7C8: only candidate → 2
L6C8: only candidate → 9
L5C7: only candidate → 3
L5C8: only candidate → 4
L4C8: only candidate → 8
L4C9: only candidate → 5
L4C2: only candidate → 6
L1C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 37,
  "branches": 18,
  "contradictions": 18,
  "depth": 9,
  "pairs": 15,
  "locked": 58
}

← Return to puzzle SOLO-006

SOLO-007 · Irregular Sudoku

EXPERT

The Living Border

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

564912378 / 783694125 / 128537946 / 215873469 / 379465281 / 496281537 / 637149852 / 942758613 / 851326794

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	5	6	4	9	1	2	3	7	8
2	7	8	3	6	9	4	1	2	5
3	1	2	8	5	3	7	9	4	6
4	2	1	5	8	7	3	4	6	9
5	3	7	9	4	6	5	2	8	1
6	4	9	6	2	8	1	5	3	7
7	6	3	7	1	4	9	8	5	2
8	9	4	2	7	5	8	6	1	3
9	8	5	1	3	2	6	7	9	4

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 3 states; the 1 contradictory branches are closed. Finally check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L1C6: only place for 2 in unit 1 → 2
L2C8: only place for 2 in unit 2 → 2
L7C7: only place for 8 in unit 16 → 8
L3C3: only place for 8 in unit 12 → 8
L2C4: only candidate → 6
L3C4: only candidate → 5
Units 1 and 21 overlap. In unit 1, 3 must lie in the overlap. Remove it from L4C7.
Units 5 and 23 overlap. In unit 5, 4 must lie in the overlap. Remove it from L4C5.
Units 5 and 23 overlap. In unit 5, 5 must lie in the overlap. Remove it from L6C5, L6C6.
Units 6 and 23 overlap. In unit 6, 2 must lie in the overlap. Remove it from L4C5.
Units 4 and 22 overlap. In unit 4, 2 must lie in the overlap. Remove it from L7C1.
Units 8 and 27 overlap. In unit 8, 6 must lie in the overlap. Remove it from L7C8, L7C9.
L7C1: only place for 6 in unit 7 → 6
L1C2: only place for 6 in unit 11 → 6
L1C1: only place for 5 in unit 1 → 5
L1C3: only place for 4 in unit 19 → 4
Nude subset in Unit 21: L1C7, L1C8 reserve [3, 7]; remove these numbers from L2C7, L3C7, L3C9, L4C7.
Units 11 and 25 overlap. In unit 11, 4 must lie in the overlap. Remove it from L7C4.
L5C4: only place for 4 in unit 13 → 4
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L6C4.
L4C4: only place for 8 in unit 13 → 8
L4C1: only place for 2 in unit 4 → 2
L6C5: only place for 8 in unit 6 → 8
L6C4: only place for 2 in unit 6 → 2
Units 9 and 26 overlap. In unit 9, 2 must lie in the overlap. Remove it from L7C5, L8C5.
L9C5: only place for 2 in unit 14 → 2
Units 17 and 27 overlap. In unit 17, 9 must lie in the overlap. Remove it from L9C7.
Units 23 and 15 overlap. In unit 23, 3 must lie in the overlap. Remove it from L7C6.
L9C4: only place for 3 in unit 26 → 3
L7C4: only candidate → 1
L8C3: only candidate → 2
L7C3: only candidate → 7
L9C1: only candidate → 8
L9C2: only candidate → 5
L2C2: only place for 8 in unit 2 → 8
L7C9: only place for 2 in unit 7 → 2
L5C1: only place for 3 in unit 10 → 3
L9C3: only place for 1 in unit 26 → 1
L5C3: only candidate → 9
L9C7: only candidate → 7
L1C7: only candidate → 3
L1C8: only candidate → 7
L9C8: only candidate → 9
L6C2: only place for 9 in unit 6 → 9

L2C1 = 1

Assumption: L2C1 = 1

L2C7: only candidate → 9
L2C5: only candidate → 7
L3C1: only candidate → 7
L3C6: only candidate → 9
L4C5: only candidate → 6
L5C5: only candidate → 5
L7C6: only candidate → 5
L7C8: only candidate → 3
L7C2: only candidate → 4
L7C5: only candidate → 9
L8C2: only candidate → 3
L8C5: only candidate → 4
L4C6: only place for 3 in unit 4 → 3
L4C2: only place for 7 in unit 4 → 7
L5C2: only candidate → 1
L5C6: only candidate → 7
L5C9: only candidate → 6
L3C9: only candidate → 1

contradiction: a box no longer has no candidate

L2C1 = 7

Assumption: L2C1 = 7

L2C5: only candidate → 9
L2C7: only candidate → 1
L3C1: only candidate → 1
L3C6: only candidate → 7
L3C9: only candidate → 6
L3C8: only candidate → 4
L3C7: only candidate → 9
L4C7: only candidate → 4
L6C7: only candidate → 5
L8C7: only candidate → 6
L5C5: only place for 6 in unit 5 → 6
L4C5: only candidate → 7
L4C2: only candidate → 1
L4C6: only candidate → 3
L4C8: only candidate → 6
L5C2: only candidate → 7
L5C9: only candidate → 1
L5C6: only candidate → 5
L6C6: only candidate → 1
L6C8: only candidate → 3
L6C9: only candidate → 7
L7C6: only candidate → 9
L7C8: only candidate → 5
L7C5: only candidate → 4
L7C2: only candidate → 3
L8C2: only candidate → 4
L8C5: only candidate → 5
L8C8: only candidate → 1
L8C9: only candidate → 3

valid complete solution

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 3,
  "branches": 1,
  "contradictions": 1,
  "depth": 1,
  "pairs": 1,
  "locked": 11
}

← Return to puzzle SOLO-007

SOLO-008 · Irregular Sudoku

EXTREME

The Nine Territories

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

486753129 / 924875631 / 531962748 / 613294857 / 759381264 / 847126593 / 298547316 / 165438972 / 372619485

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	4	8	6	7	5	3	1	2	9
2	9	2	4	8	7	5	6	3	1
3	5	3	1	9	6	2	7	4	8
4	6	1	3	2	9	4	8	5	7
5	7	5	9	3	8	1	2	6	4
6	8	4	7	1	2	6	5	9	3
7	2	9	8	5	4	7	3	1	6
8	1	6	5	4	3	8	9	7	2
9	3	7	2	6	1	9	4	8	5

The reasoning

Note the candidates in each box by crossing the line, column and region.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 27 states; the 13 contradictory branches are closed. Finally check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L9C1: only place for 3 in unit 10 → 3
L7C5: only place for 4 in unit 26 → 4
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 23 and 5 overlap. In unit 23, 8 must lie in the overlap. Remove it from L5C8, L5C9.
L3C9: only place for 8 in unit 24 → 8
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L4C1 = 5

Assumption: L4C1 = 5

Units 3 and 21 overlap. In unit 3, 5 must lie in the overlap. Remove it from L1C9, L2C6, L2C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L2C6 = 1

Assumption: L2C6 = 1

Units 3 and 21 overlap. In unit 3, 5 must lie in the overlap. Remove it from L1C9, L2C9.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 20 and 1 overlap. In unit 20, 1 must lie in the overlap. Remove it from L1C1, L1C2.
L8C1: only place for 1 in unit 10 → 1
L5C1: only place for 7 in unit 10 → 7
L4C2: only place for 1 in unit 11 → 1
Nude subset in unit 19: L1C1, L3C1 reserve [4, 6]; remove these numbers from L1C2, L1C3, L2C2.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 1 must lie in the overlap. Remove it from L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L1C1 = 4

Assumption: L1C1 = 4

L3C1: only candidate → 6
L1C9: only place for 6 in unit 1 → 6
Units 1 and 20 overlap. In unit 1, 9 must lie in the overlap. Remove it from L3C4.
Units 3 and 21 overlap. In unit 3, 5 must lie in the overlap. Remove it from L2C9.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
L9C9: only candidate → 5
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 1 must lie in the overlap. Remove it from L6C7.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 14 and 23 overlap. In unit 14, 6 must lie in the overlap. Remove it from L5C4, L6C3, L6C4, L6C6.
Nude subset in Unit 6: L6C3, L6C6, L6C7 reserve [5, 7, 9]; remove these numbers from L6C2, L6C4, L6C8, L6C9.
Nude subset in Unit 23: L6C3, L6C6, L6C7 reserve [5, 7, 9]; remove these numbers from L4C5, L5C4, L5C5.
L4C5: only candidate → 6
L5C5: only candidate → 8
L2C4: only place for 8 in unit 13 → 8
Units 27 and 17 overlap. In unit 27, 6 must lie in the overlap. Remove it from L5C8, L6C8.
L6C2: only place for 6 in unit 6 → 6
L5C6: only place for 6 in unit 5 → 6
L7C6: only candidate → 7
L6C6: only candidate → 5
L3C6: only candidate → 2
L1C6: only candidate → 3
L2C9: only candidate → 4
L3C4: only candidate → 4
L6C8: only place for 4 in unit 6 → 4
L8C2: only place for 4 in unit 8 → 4
L5C3: only place for 4 in unit 5 → 4
L9C2: only place for 7 in unit 11 → 7
L9C8: only place for 8 in unit 9 → 8

L1C3 = 2

Assumption: L1C3 = 2

L2C2: only candidate → 5
L1C2: only candidate → 8
L2C3: only candidate → 7

contradiction: a box no longer has no candidate

L1C3 = 8

Assumption: L1C3 = 8

L7C2: only place for 8 in unit 7 → 8
L5C2: only place for 9 in unit 11 → 9
L4C4: only candidate → 2
L4C8: only candidate → 9
L9C4: only candidate → 6
L8C4: only candidate → 9
L7C3: only candidate → 6
L7C8: only candidate → 1
L8C9: only candidate → 2
L8C7: only candidate → 7
L8C8: only candidate → 6
L9C3: only candidate → 2
L2C3: only candidate → 7
L2C5: only candidate → 5
L1C5: only candidate → 9
L1C8: only candidate → 2
L1C2: only candidate → 5
L1C7: only candidate → 1
L2C2: only candidate → 2
L3C5: only candidate → 7
L3C8: only candidate → 5

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L1C1 = 6

Assumption: L1C1 = 6

L3C1: only candidate → 4
Nude subset in Unit 13: L3C4, L4C4, L9C4 reserve [2, 6, 9]; remove these numbers from L2C4, L5C4, L6C4, L8C4.
L8C4: only candidate → 4
L2C4: only candidate → 8
L5C5: only place for 8 in unit 5 → 8
Nude sub-assembly in Unit 23: L5C4, L6C4 reserve [1, 3]; remove these numbers from L6C6, L6C7.
Units 3 and 21 overlap. In unit 3, 5 must lie in the overlap. Remove it from L1C9, L2C9.
Units 5 and 24 overlap. In unit 5, 5 must lie in the overlap. Remove it from L6C8, L6C9.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 22 overlap. In unit 11, 4 must lie in the overlap. Remove it from L5C3.
L2C3: only place for 4 in unit 12 → 4
L2C9: only candidate → 2
L2C2: only candidate → 5
L2C5: only candidate → 7
L1C9: only place for 4 in unit 1 → 4
L3C4: only place for 2 in unit 3 → 2
L9C4: only candidate → 6
L4C4: only candidate → 9
L4C5: only candidate → 6
L4C8: only candidate → 2
L7C6: only candidate → 7
L6C6: only candidate → 5
L1C6: only candidate → 3
L3C6: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C6 = 2

Assumption: L2C6 = 2

L3C4: only place for 2 in unit 3 → 2
L9C4: only candidate → 6
L4C4: only candidate → 9
L4C5: only candidate → 6
L7C6: only candidate → 7
L9C3: only candidate → 2
L9C9: only candidate → 5
L1C2: only place for 2 in unit 1 → 2
L4C2: only candidate → 1
L4C8: only candidate → 2

contradiction: the number 2 can no longer be included in unit 5

all the possibilities of this bifurcation have been examined

L4C1 = 6

Assumption: L4C1 = 6

Units 3 and 21 overlap. In unit 3, 6 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L3C1 = 4

Assumption: L3C1 = 4

Nude subset in Unit 13: L3C4, L4C4 reserve [2, 9]; remove these numbers from L2C4, L5C4, L6C4, L8C4, L9C4.
L9C4: only candidate → 6
L7C6: only candidate → 7
L9C3: only candidate → 2
L9C9: only candidate → 5
Units 3 and 21 overlap. In unit 3, 5 must lie in the overlap. Remove it from L2C6.
Units 3 and 21 overlap. In unit 3, 6 must lie in the overlap. Remove it from L1C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 19 and 11 overlap. In unit 19, 2 must lie in the overlap. Remove it from L5C2.
L4C4: only place for 2 in unit 22 → 2
L3C4: only candidate → 9
L1C9: only place for 9 in unit 1 → 9
L1C8: only place for 4 in unit 1 → 4
L2C9: only place for 4 in unit 2 → 4
L6C2: only place for 4 in unit 6 → 4

contradiction: the number 4 can no longer be included in unit 5

L3C1 = 5

Assumption: L3C1 = 5

Units 3 and 21 overlap. In unit 3, 6 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L1C1 = 1

Assumption: L1C1 = 1

L4C2: only candidate → 2
L4C4: only candidate → 9
L4C5: only candidate → 5
L4C8: only candidate → 1
L5C4: only place for 1 in unit 5 → 1
L5C5: only place for 8 in unit 5 → 8
L1C5: only candidate → 9
L2C5: only candidate → 7
L3C5: only candidate → 6

contradiction: the number 1 can no longer appear in unit 6

L1C1 = 4

Assumption: L1C1 = 4

L5C1: only candidate → 7
L8C1: only candidate → 1
Units 3 and 21 overlap. In unit 3, 6 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L4C2 = 1

Assumption: L4C2 = 1

Units 3 and 21 overlap. In unit 3, 6 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 27 and 17 overlap. In unit 27, 8 must lie in the overlap. Remove it from L1C8.

L2C2 = 2

Assumption: L2C2 = 2

Nude subset in unit 1: L1C2, L1C3 reserve [6, 8]; remove these numbers from L1C5, L1C8, L1C9.
Nude subset in unit 14: L1C5, L4C5 reserve [5, 9]; remove these numbers from L2C5, L3C5, L5C5.
Units 2 and 21 overlap. In unit 2, 5 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.

L1C2 = 6

Assumption: L1C2 = 6

L1C3: only candidate → 8
Nude subset in unit 14: L1C5, L4C5 reserve [5, 9]; remove these numbers from L2C5, L3C5, L5C5.
Units 2 and 21 overlap. In unit 2, 5 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.

L1C5 = 5

Assumption: L1C5 = 5

L4C5: only candidate → 9
L4C4: only candidate → 2
L4C8: only candidate → 5
L9C4: only candidate → 6
L7C6: only candidate → 7
L9C3: only candidate → 2
L9C9: only candidate → 5
L2C6: only place for 5 in unit 2 → 5
L5C2: only place for 5 in unit 5 → 5
L6C7: only place for 5 in unit 6 → 5
L6C3: only place for 7 in unit 6 → 7
L2C3: only candidate → 4
L2C9: only candidate → 1
L2C4: only candidate → 8
L2C5: only candidate → 7
L3C4: only candidate → 9
L3C5: only candidate → 6
L3C6: only candidate → 2
L1C9: only candidate → 9
L3C7: only candidate → 7
L3C8: only candidate → 4
L5C3: only candidate → 9
L5C5: only candidate → 8
L6C2: only candidate → 4
L7C3: only candidate → 6
L7C9: only candidate → 3
L6C9: only candidate → 6
L8C4: only candidate → 4
L8C9: only candidate → 2
L5C9: only candidate → 4
L8C7: only candidate → 9
L7C7: only candidate → 1
L7C8: only candidate → 8
L7C2: only candidate → 9
L8C2: only candidate → 7
L8C8: only candidate → 6
L9C2: only candidate → 8
L9C8: only candidate → 7

contradiction: the number 6 can no longer appear in unit 5

L1C5 = 9

Assumption: L1C5 = 9

L4C5: only candidate → 5
Units 2 and 21 overlap. In unit 2, 5 must lie in the overlap. Remove it from L1C9.
Units 3 and 21 overlap. In unit 3, 7 must lie in the overlap. Remove it from L2C6.
Units 8 and 27 overlap. In unit 8, 2 must lie in the overlap. Remove it from L9C8, L9C9.
Units 11 and 25 overlap. In unit 11, 7 must lie in the overlap. Remove it from L7C3.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.

L1C9 = 1

Assumption: L1C9 = 1

L2C4: only place for 1 in unit 2 → 1
L2C5: only place for 8 in unit 2 → 8
L5C5: only candidate → 6
L3C5: only candidate → 7
L2C6: only candidate → 5
L2C9: only candidate → 4
L2C3: only candidate → 7
L6C3: only candidate → 9
L6C4: only candidate → 3
L5C4: only candidate → 8
L7C3: only candidate → 6
L7C6: only candidate → 7
L6C6: only candidate → 1
L6C7: only candidate → 7
L9C3: only candidate → 2
L5C3: only candidate → 4
L6C2: only candidate → 5
L5C2: only candidate → 9
L4C4: only candidate → 2
L3C4: only candidate → 4
L4C8: only candidate → 9
L6C9: only candidate → 6
L6C8: only candidate → 4
L7C2: only candidate → 8
L7C8: only candidate → 1
L8C4: only candidate → 9
L8C7: only candidate → 2

contradiction: a box no longer has no candidate

L1C9 = 2

Assumption: L1C9 = 2

L3C4: only place for 2 in unit 3 → 2
L4C4: only candidate → 9
L4C8: only candidate → 2
L9C4: only candidate → 6
L7C6: only candidate → 7
L3C6: only candidate → 6
L3C5: only candidate → 7
L2C5: only candidate → 8
L3C7: only candidate → 9
L3C8: only candidate → 4
L5C5: only candidate → 6
L8C4: only candidate → 4
L2C4: only candidate → 1
L1C8: only candidate → 5
L1C6: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L1C2 = 8

Assumption: L1C2 = 8

L1C3: only candidate → 6
L9C8: only place for 8 in unit 9 → 8
L7C3: only place for 8 in unit 7 → 8
L9C9: only place for 5 in unit 9 → 5
L2C6: only place for 5 in unit 21 → 5
Nude subset in unit 14: L1C5, L4C5 reserve [5, 9]; remove these numbers from L3C5, L5C5.
Units 11 and 25 overlap. In unit 11, 6 must lie in the overlap. Remove it from L8C4.
Nude subset in Unit 13: L3C4, L4C4, L8C4 reserve [2, 4, 9]; remove these numbers from L2C4, L5C4, L6C4, L9C4.
L9C4: only candidate → 6
L7C6: only candidate → 7
L9C2: only candidate → 7
L9C3: only candidate → 2
L4C4: only place for 2 in unit 22 → 2
Units 3 and 21 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C9.
Units 13 and 23 overlap. In unit 13, 3 must lie in the overlap. Remove it from L6C6, L6C7.
Units 21 and 18 overlap. In unit 21, 1 must lie in the overlap. Remove it from L5C9, L6C9, L7C9.

L1C5 = 5

Assumption: L1C5 = 5

L4C5: only candidate → 9
L4C8: only candidate → 5
L6C3: only candidate → 7
L2C3: only candidate → 4
L2C9: only candidate → 1
L2C4: only candidate → 8
L2C5: only candidate → 7
L3C4: only candidate → 9
L3C5: only candidate → 6
L3C6: only candidate → 2
L1C9: only candidate → 9
L3C7: only candidate → 7
L3C8: only candidate → 4
L5C3: only candidate → 9
L5C5: only candidate → 8
L8C4: only candidate → 4
L5C2: only place for 5 in unit 5 → 5
L6C2: only candidate → 4
L5C9: only place for 4 in unit 5 → 4
L6C7: only place for 5 in unit 6 → 5
L6C8: only place for 9 in unit 6 → 9
L8C8: only place for 7 in unit 8 → 7
L8C9: only place for 2 in unit 18 → 2
L8C7: only candidate → 9
L8C2: only candidate → 6
L7C2: only candidate → 9
L6C6: only place for 6 in unit 23 → 6
L6C9: only candidate → 3
L5C6: only candidate → 1
L1C6: only candidate → 3
L5C4: only candidate → 3
L5C7: only candidate → 2
L1C7: only candidate → 1
L1C8: only candidate → 2
L5C8: only candidate → 6
L6C4: only candidate → 1
L7C7: only candidate → 3
L7C8: only candidate → 1
L7C9: only candidate → 6

valid complete solution

L1C5 = 9

Assumption: L1C5 = 9

L3C4: only candidate → 4
L2C3: only candidate → 7
L2C5: only candidate → 8
L2C4: only candidate → 1
L2C9: only candidate → 4
L4C5: only candidate → 5
L4C8: only candidate → 9
L5C5: only candidate → 6
L3C5: only candidate → 7
L6C3: only candidate → 9
L5C3: only candidate → 4
L6C2: only candidate → 5
L5C2: only candidate → 9
L6C4: only candidate → 3
L5C4: only candidate → 8
L6C6: only candidate → 1
L6C7: only candidate → 7
L6C9: only candidate → 6
L6C8: only candidate → 4
L7C2: only candidate → 6
L7C8: only candidate → 1
L8C2: only candidate → 4
L8C4: only candidate → 9
L8C7: only candidate → 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C2 = 8

Assumption: L2C2 = 8

L9C8: only place for 8 in unit 9 → 8
L1C5: only place for 8 in unit 1 → 8
L5C4: only place for 8 in unit 5 → 8
L7C3: only place for 8 in unit 7 → 8
L9C9: only place for 5 in unit 9 → 5
L6C4: only place for 3 in unit 13 → 3
L2C4: only place for 1 in unit 13 → 1
L1C9: only place for 1 in unit 1 → 1
L2C6: only place for 5 in unit 21 → 5
L2C5: only candidate → 7
Nude subassembly in Unit 1: L1C2, L1C3 reserve [2, 6]; remove these numbers from L1C6, L1C7, L1C8.
L1C6: only candidate → 3
Nude subassembly in unit 20: L1C7, L1C8 reserve [5, 9]; remove these numbers from L3C4.
Units 5 and 24 overlap. In unit 5, 1 must lie in the overlap. Remove it from L6C8.
Units 14 and 23 overlap. In unit 14, 5 must lie in the overlap. Remove it from L6C7.

L1C2 = 2

Assumption: L1C2 = 2

L1C3: only candidate → 6
Nude subassembly in unit 20: L1C7, L1C8 reserve [5, 9]; remove these numbers from L3C4.
Units 5 and 24 overlap. In unit 5, 1 must lie in the overlap. Remove it from L6C8.
Units 11 and 25 overlap. In unit 11, 6 must lie in the overlap. Remove it from L8C4.
L9C4: only place for 6 in unit 13 → 6
L7C6: only candidate → 7
L9C2: only candidate → 7
L9C3: only candidate → 2
L2C3: only candidate → 4
L2C9: only candidate → 2
L3C4: only candidate → 2
L3C6: only candidate → 6
L3C5: only candidate → 9
L3C7: only candidate → 7
L3C8: only candidate → 4
L4C4: only candidate → 9

contradiction: a box no longer has no candidate

L1C2 = 6

Assumption: L1C2 = 6

L1C3: only candidate → 2
L2C3: only candidate → 4
L2C9: only candidate → 2
L5C3: only candidate → 9
L4C4: only candidate → 2
L3C4: only candidate → 9
L1C7: only candidate → 5

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C2 = 2

Assumption: L4C2 = 2

L4C4: only candidate → 9
L4C5: only candidate → 5
L4C8: only candidate → 1
L5C4: only place for 1 in unit 5 → 1
L5C5: only place for 8 in unit 5 → 8
L1C5: only candidate → 9
L2C5: only candidate → 7
L3C5: only candidate → 6

contradiction: the number 1 can no longer appear in unit 6

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 27,
  "branches": 13,
  "contradictions": 13,
  "depth": 8,
  "pairs": 14,
  "locked": 78
}

After single candidates, single places, bare pairs/triples and interactions between units, the solver still reaches bifurcations. The exhaustive evidence reaches 8 intertwined hypotheses, examines 27 states and closes 13 branches by contradiction. The Extreme level combines eliminations of candidates and deep branched analysis; this is not a ranking based on the number of empty boxes. Editorial evaluation according to the protocol of the collection, without calibration on a panel of players; another logical path can be made to the extent that it is possible to determine the number of empty boxes. be shorter.

← Return to puzzle SOLO-008

SOLO-009 · Killer Sudoku

EXPERT

Captive Sums

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

861472395 / 537916428 / 294835176 / 415283769 / 678149532 / 923567841 / 782654913 / 146398257 / 359721684

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	208	6	1	134	7	2	123	9	225
2	5	183	247	9	111	6	4	2	8
3	2	9	4	8	163	5	1	7	176
4	124	1	165	2	218	33	7	146	9
5	156	7	8	1	4	9	135	3	2
6	9	102	3	5	186	7	8	4	1
7	157	8	82	186	5	124	179	1	143
8	191	4	6	3	99	8	2	5	7
9	33	5	9	7	2	151	6	8	4

Thick stroke: region. Dotted inside the cells: cage and its total. A tint on a diagonal reminds the diagonal constraint, if it applies.

Exact list of cages (coordinates and totals)

Cage 1: L1C1, L1C2, L1C3, L2C1; total 20.
Cage 2: L1C4, L1C5, L1C6; total 13.
Cage 3: L1C7, L1C8; total 12.
Cage 4: L1C9, L2C8, L2C9, L3C8; total 22.
Cage 5: L2C2, L3C1, L3C2, L3C3; total 18.
Cage 6: L2C3, L2C4, L3C4; total 24.
Cage 7: L2C5, L2C6, L2C7; total 11.
Cage 8: L3C5, L3C6, L3C7, L4C7; total 16.
Cage 9: L3C9, L4C9, L5C9; total 17.
Cage 10: L4C1, L4C2, L5C2; total 12.
Cage 11: L4C3, L4C4, L5C3, L5C4; total 16.
Cage 12: L4C5, L5C5, L5C6; total 21.
Cage 13: L4C6; total 3.
Cage 14: L4C8, L5C8, L6C8, L6C9; total 14.
Cage 15: L5C1, L6C1; total 15.
Cage 16: L5C7, L6C7; total 13.
Cage 17: L6C2, L6C3, L6C4; total 10.
Cage 18: L6C5, L6C6, L7C5; total 18.
Cage 19: L7C1, L7C2; total 15.
Cage 20: L7C3, L8C3; total 8.
Cage 21: L7C4, L8C4, L9C4, L9C5; total 18.
Cage 22: L7C6, L8C6; total 12.
Cage 23: L7C7, L7C8, L8C7, L8C8; total 17.
Cage 24: L7C9, L8C9, L9C9; total 14.
Cage 25: L8C1, L8C2, L9C2, L9C3; total 19.
Cage 26: L8C5; total 9.
Cage 27: L9C1; total 3.
Cage 28: L9C6, L9C7, L9C8; total 15.

The reasoning

Note the candidates in each box by crossing the line, column and region and amounts of cages.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 9 states; the 4 contradictory branches are closed. Finally check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L4C6: only candidate → 3
L8C5: only candidate → 9
L9C1: only candidate → 3
Nude sub-assembly in Unit 10: L5C1, L6C1 reserve [6, 9]; remove these numbers from L2C1, L3C1, L7C1, L8C1.
L7C1: only candidate → 7
L7C2: only candidate → 8
Nude subset in Unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3, L4C3, L5C3, L6C3.
Nude sub-assembly in unit 22: L5C1, L6C1 reserve [6, 9]; remove these numbers from L4C2, L4C3, L5C2, L5C3, L6C2.
L6C2: only place for 2 in unit 22 → 2
Units 4 and 24 overlap. In unit 4, 9 must lie in the overlap. Remove it from L5C7, L5C9, L6C7.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 13 and 20 overlap. In unit 13, 9 must lie in the overlap. Remove it from L1C6, L3C6.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.

L2C3 = 7

Assumption: L2C3 = 7

Nude subset in Unit 10: L5C1, L6C1 reserve [6, 9]; remove these numbers from L2C1, L3C1.
Nude subset in Unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3, L4C3, L5C3.
Nude subset in Unit 13: L2C4, L3C4 reserve [8, 9]; remove these numbers from L1C4, L4C4, L5C4, L8C4, L9C4.
Nude subset in unit 20: L2C4, L3C4 reserve [8, 9]; remove these numbers from L1C6, L2C5, L2C6, L3C5, L3C6.
Nude subassembly in unit 22: L5C1, L6C1 reserve [6, 9]; remove these numbers from L4C3, L5C3.
Units 4 and 24 overlap. In unit 4, 9 must lie in the overlap. Remove it from L5C7, L5C9, L6C7.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L5C6, L6C6.

L2C4 = 8

Assumption: L2C4 = 8

L3C4: only candidate → 9
Nude subset in Unit 10: L5C1, L6C1 reserve [6, 9]; remove these numbers from L2C1, L3C1.
Nude subset in Unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3, L4C3, L5C3.
Nude subassembly in unit 22: L5C1, L6C1 reserve [6, 9]; remove these numbers from L4C3, L5C3.
Units 4 and 24 overlap. In unit 4, 9 must lie in the overlap. Remove it from L5C7, L5C9, L6C7.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L5C6, L6C6.

L5C1 = 6

Assumption: L5C1 = 6

L6C1: only candidate → 9
L5C6: only place for 9 in unit 15 → 9
Nude sub-assembly in unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 7 and 27 overlap. In unit 7, 1 must lie in the overlap. Remove it from L8C7, L8C8, L8C9, L9C7, L9C8, L9C9.
Units 8 and 25 overlap. In unit 8, 1 must lie in the overlap. Remove it from L9C2, L9C3.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L6C6.
Units 26 and 9 overlap. In unit 26, 2 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
L8C6: only place for 8 in unit 15 → 8
L7C6: only candidate → 4
L9C5: only candidate → 2
L9C6: only place for 1 in unit 9 → 1
L7C5: only place for 5 in unit 26 → 5
L4C5: only candidate → 8
L5C5: only candidate → 4
L5C3: only place for 8 in unit 12 → 8
L1C4: only place for 4 in unit 13 → 4
L3C6: only place for 5 in unit 15 → 5
L2C2: only candidate → 9
L1C2: only candidate → 6
L1C5: only candidate → 7
L1C6: only candidate → 2
L2C6: only candidate → 6
L1C9: only candidate → 9

contradiction: a box no longer has no candidate

L5C1 = 9

Assumption: L5C1 = 9

L6C1: only candidate → 6
L6C6: only place for 9 in unit 15 → 9
Nude sub-assembly in unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3.
Nude sub-assembly in Unit 23: L4C5, L5C5, L5C6 reserve [6, 7, 8]; remove these numbers from L4C4, L5C4, L6C4, L6C5.
L6C3: only candidate → 3
L6C4: only candidate → 5
L6C5: only candidate → 4
L4C8: only candidate → 2
L4C4: only candidate → 1
L4C2: only candidate → 7
L5C2: only candidate → 1
L5C4: only candidate → 2
L7C5: only candidate → 5
L7C6: only candidate → 4
L8C6: only candidate → 8
L9C5: only candidate → 2
L9C3: only place for 9 in unit 12 → 9
L8C1: only candidate → 1
L1C4: only place for 4 in unit 13 → 4
L1C3: only candidate → 1
L3C3: only candidate → 4
L4C9: only place for 9 in unit 4 → 9
L5C8: only place for 4 in unit 5 → 4
L5C9: only place for 3 in unit 5 → 3
L3C9: only candidate → 5
L2C2: only candidate → 9
L1C2: only candidate → 6
L1C5: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C4 = 9

Assumption: L2C4 = 9

L3C4: only candidate → 8
Nude subset in Unit 10: L5C1, L6C1 reserve [6, 9]; remove these numbers from L2C1, L3C1.
Nude subset in Unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3, L4C3, L5C3.
Nude subassembly in unit 22: L5C1, L6C1 reserve [6, 9]; remove these numbers from L4C3, L5C3.
Units 4 and 24 overlap. In unit 4, 9 must lie in the overlap. Remove it from L5C7, L5C9, L6C7.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L5C6, L6C6.

L5C1 = 6

Assumption: L5C1 = 6

L6C1: only candidate → 9
L5C6: only place for 9 in unit 15 → 9
Nude sub-assembly in unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3.
Units 7 and 27 overlap. In unit 7, 9 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
Units 23 and 13 overlap. In unit 23, 1 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 7 and 27 overlap. In unit 7, 1 must lie in the overlap. Remove it from L8C7, L8C8, L8C9, L9C7, L9C8, L9C9.
Units 8 and 25 overlap. In unit 8, 1 must lie in the overlap. Remove it from L9C2, L9C3.
Units 23 and 13 overlap. In unit 23, 2 must lie in the overlap. Remove it from L1C4, L7C4, L8C4, L9C4.
Units 14 and 23 overlap. In unit 14, 8 must lie in the overlap. Remove it from L6C6.
Units 26 and 9 overlap. In unit 26, 2 must lie in the overlap. Remove it from L9C7, L9C8, L9C9.
L8C6: only place for 8 in unit 15 → 8
L7C6: only candidate → 4
L9C5: only candidate → 2
L9C6: only place for 1 in unit 9 → 1
L7C5: only place for 5 in unit 26 → 5
L4C5: only candidate → 8
L5C5: only candidate → 4
L5C3: only place for 8 in unit 12 → 8
L1C4: only place for 4 in unit 13 → 4
L3C6: only place for 5 in unit 15 → 5
L4C9: only place for 9 in unit 4 → 9
L6C6: only place for 7 in unit 15 → 7
L6C3: only candidate → 3
L6C4: only candidate → 5
L4C3: only candidate → 5
L6C5: only candidate → 6
L5C7: only candidate → 5
L6C7: only candidate → 8
L9C7: only candidate → 6
L2C6: only candidate → 6
L1C5: only candidate → 7
L1C6: only candidate → 2
L9C4: only candidate → 7
L9C8: only candidate → 8
L8C9: only candidate → 7
L3C9: only candidate → 6
L2C9: only candidate → 8
L5C9: only candidate → 2
L4C4: only candidate → 2
L4C8: only candidate → 6
L5C4: only candidate → 1
L4C2: only candidate → 1
L1C3: only candidate → 1
L3C1: only candidate → 2
L1C2: only candidate → 6
L2C1: only candidate → 5
L4C7: only candidate → 7
L5C2: only candidate → 7
L5C8: only candidate → 3
L1C7: only candidate → 3
L1C8: only candidate → 9
L1C9: only candidate → 5
L2C8: only candidate → 2
L2C5: only candidate → 1
L2C7: only candidate → 4
L2C2: only candidate → 3
L3C5: only candidate → 3
L3C7: only candidate → 1
L3C8: only candidate → 7
L7C7: only candidate → 9
L7C8: only candidate → 1
L6C8: only candidate → 4
L6C9: only candidate → 1
L7C9: only candidate → 3
L7C4: only candidate → 6
L7C3: only candidate → 2
L8C1: only candidate → 1
L8C3: only candidate → 6
L8C4: only candidate → 3
L8C7: only candidate → 2
L8C8: only candidate → 5
L8C2: only candidate → 4
L3C2: only candidate → 9
L3C3: only candidate → 4
L9C2: only candidate → 5
L9C3: only candidate → 9
L9C9: only candidate → 4

valid complete solution

L5C1 = 9

Assumption: L5C1 = 9

L6C1: only candidate → 6
L6C6: only place for 9 in unit 15 → 9
Nude sub-assembly in unit 12: L7C3, L8C3 reserve [2, 6]; remove these numbers from L1C3, L3C3.
Nude sub-assembly in Unit 23: L4C5, L5C5, L5C6 reserve [6, 7, 8]; remove these numbers from L4C4, L5C4, L6C4, L6C5.
L6C3: only candidate → 3
L6C4: only candidate → 5
L6C5: only candidate → 4
L4C8: only candidate → 2
L4C4: only candidate → 1
L4C2: only candidate → 7
L5C2: only candidate → 1
L5C4: only candidate → 2
L7C5: only candidate → 5
L7C6: only candidate → 4
L8C6: only candidate → 8
L9C5: only candidate → 2
L1C4: only place for 4 in unit 13 → 4
L3C6: only place for 5 in unit 15 → 5
L4C9: only place for 9 in unit 4 → 9
L3C8: only place for 6 in unit 3 → 6
L1C3: only place for 1 in unit 1 → 1
L3C1: only candidate → 2
L1C2: only candidate → 6
L1C5: only candidate → 7
L1C6: only candidate → 2
L2C1: only candidate → 5
L1C9: only candidate → 5
L3C5: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L2C3 = 9

Assumption: L2C3 = 9

L9C2: only candidate → 9
L1C4: only place for 9 in unit 13 → 9
L1C5: only candidate → 3
L1C6: only candidate → 1
Nude subassembly in Unit 1: L1C7, L1C8 reserve [5, 7]; remove these numbers from L1C2, L1C3, L1C9.
Nude subset in Unit 10: L5C1, L6C1 reserve [6, 9]; remove these numbers from L2C1, L3C1.
L2C1: only candidate → 2
L2C7: only candidate → 1
L3C7: only candidate → 3
L1C9: only place for 2 in unit 1 → 2
L2C2: only place for 3 in unit 2 → 3
L3C1: only candidate → 5
L8C1: only candidate → 1

contradiction: the number 1 can no longer appear in unit 3

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 9,
  "branches": 4,
  "contradictions": 4,
  "depth": 3,
  "pairs": 22,
  "locked": 34
}

← Return to puzzle SOLO-009

SOLO-010 · Killer Sudoku

EXTREME

Unyielding Totals

Complete the grid completely. Only one solution respects all constraints.

This page contains the solution.

The answer

637412598 / 981375624 / 245896731 / 528961473 / 174238965 / 396754182 / 712589346 / 463127859 / 859643217

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9
1	96	3	147	4	1	2	135	229	8
2	169	188	1	273	7	195	6	2	4
3	2	4	5	8	189	6	7	203	1
4	5	172	8	9	6	1	4	7	83
5	131	7	104	182	3	218	9	6	5
6	3	9	6	7	5	4	1	198	2
7	197	191	82	5	278	9	3	4	156
8	4	6	3	1	122	7	118	5	9
9	8	55	9	6	4	3	2	1	77

Thick stroke: region. Dotted inside the cells: cage and its total. A tint on a diagonal reminds the diagonal constraint, if it applies.

Exact list of cages (coordinates and totals)

Cage 1: L1C1, L1C2; total 9.
Cage 2: L1C3, L1C4, L1C5, L1C6; total 14.
Cage 3: L1C7, L2C7, L2C8; total 13.
Cage 4: L1C8, L1C9, L2C9, L3C9; total 22.
Cage 5: L2C1, L3C1, L4C1; total 16.
Cage 6: L2C2, L2C3, L3C2, L3C3; total 18.
Cage 7: L2C4, L2C5, L3C4, L4C4; total 27.
Cage 8: L2C6, L3C6, L3C7, L4C6; total 19.
Cage 9: L3C5, L4C5, L5C5; total 18.
Cage 10: L3C8, L4C7, L4C8, L5C8; total 20.
Cage 11: L4C2, L4C3, L5C2; total 17.
Cage 12: L4C9, L5C9; total 8.
Cage 13: L5C1, L6C1, L6C2; total 13.
Cage 14: L5C3, L6C3; total 10.
Cage 15: L5C4, L6C4, L6C5, L6C6; total 18.
Cage 16: L5C6, L5C7, L6C7, L7C7; total 21.
Cage 17: L6C8, L6C9, L7C8, L8C8; total 19.
Cage 18: L7C1, L8C1, L9C1; total 19.
Cage 19: L7C2, L8C2, L8C3, L9C3; total 19.
Cage 20: L7C3, L7C4, L8C4; total 8.
Cage 21: L7C5, L7C6, L8C6, L9C6; total 27.
Cage 22: L7C9, L8C9; total 15.
Cage 23: L8C5, L9C4, L9C5; total 12.
Cage 24: L8C7, L9C7, L9C8; total 11.
Cage 25: L9C2; total 5.
Cage 26: L9C9; total 7.

The reasoning

Note the candidates in each box by crossing the line, column and region and amounts of cages.
In the certificate below, units 1 to 9 are lines, 10 to 18 columns, 19 to 27 regions. Units 28 and 29, if present, are diagonals.
A nude subset of two or three cells reserves as many numbers: these figures are excluded from the other cells of the unit. An interaction limits one digit in the intersection of two units and excludes from the rest of the second.
Each hypothesis opens a separate branch. Never keep in the neighbouring branch a value found under an abandoned hypothesis. The contradictions and all the figures deduced appear in the tree of evidence.
The search finds exactly a complete grid. It runs through 127 states; the 63 contradictory branches are closed. Finally, check each unit and each starting point.

Common mistakes and false leads

Consolidate a temporary hypothesis with a demonstrated value.
Use a subset of three boxes with four digits in the union.
Forget a variant constraint or keep data from a refuted branch.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

L8C9: only candidate → 9
L9C2: only candidate → 5
L9C9: only candidate → 7
Nude subset in unit 18: L4C9, L5C9 reserve [3, 5]; remove these numbers from L1C9, L2C9, L3C9, L6C9.
L1C8: only candidate → 9
L6C9: only place for 2 in unit 18 → 2
Nude subset in Unit 21: L1C9, L2C9, L3C9 reserve [1, 4, 8]; remove these numbers from L1C7, L2C7, L2C8, L3C7, L3C8.
L2C8: only candidate → 2
Nude subset in unit 16: L1C7, L2C7 reserve [5, 6]; remove these numbers from L3C7, L4C7, L5C7, L6C7, L7C7, L8C7.
Nude subset in Unit 21: L1C7, L2C7 reserve [5, 6]; remove these numbers from L3C8.
Nude subassembly in Unit 3: L3C7, L3C8 reserve [3, 7]; remove these numbers from L3C1, L3C2, L3C3, L3C4, L3C5, L3C6.
Nude subset in Unit 24: L4C9, L5C9 reserve [3, 5]; remove these numbers from L4C7, L4C8, L5C7, L5C8, L6C7, L6C8.
Nude subset in Unit 27: L8C7, L9C8 reserve [1, 8]; remove these numbers from L7C7, L7C8, L8C8.
L6C8: only candidate → 8
L3C8: only candidate → 3
L3C7: only candidate → 7
L8C7: only candidate → 8
L9C8: only candidate → 1
L4C7: only candidate → 4
L5C6: only candidate → 8
L7C7: only candidate → 3

L1C7 = 5

Assumption: L1C7 = 5

L2C7: only candidate → 6

L4C8 = 6

Assumption: L4C8 = 6

L5C8: only candidate → 7

L4C9 = 3

Assumption: L4C9 = 3

L5C9: only candidate → 5

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
L3C6: only candidate → 6
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L3C4 = 8

Assumption: L3C4 = 8

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L3C9 = 1

Assumption: L3C9 = 1

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L1C9 = 4

Assumption: L1C9 = 4

L1C3: only candidate → 8
L2C9: only candidate → 8
L9C3: only candidate → 9
L7C1: only candidate → 7
L7C6: only candidate → 9
L8C1: only candidate → 4
L8C6: only candidate → 7
L9C1: only candidate → 8
L9C6: only candidate → 3
L1C2: only place for 7 in unit 1 → 7
L1C1: only candidate → 2

contradiction: a box no longer has no candidate

L1C9 = 8

Assumption: L1C9 = 8

L2C9: only candidate → 4
L6C1: only place for 1 in unit 10 → 1
L5C1: only candidate → 9
L6C2: only candidate → 3
L7C1: only candidate → 7
L2C1: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C9 = 4

Assumption: L3C9 = 4

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1.
Nude subset in unit 14: L5C5, L9C5 reserve [4, 6]; remove these numbers from L1C5, L6C5.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 1 must lie in the overlap. Remove it from L1C1, L1C3, L2C3.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Nude sub-assembly in unit 1: L1C1, L1C2 reserve [3, 6]; remove these numbers from L1C3, L1C4, L1C5.
L1C9: only place for 8 in unit 1 → 8
L2C9: only candidate → 1
Nude subset in unit 19: L1C1, L1C2 reserve [3, 6]; remove these numbers from L2C1, L2C2, L2C3.
L3C2: only candidate → 2
Nude sub-assembly in unit 12: L3C3, L4C3 reserve [1, 5]; remove these numbers from L6C3.
L9C3: only place for 9 in unit 12 → 9
L7C1: only candidate → 7
L4C1: only candidate → 2
L2C6: only candidate → 5
L2C1: only candidate → 9
L3C1: only candidate → 5
L3C3: only candidate → 1
L3C5: only candidate → 9
L4C3: only candidate → 5

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C4 = 9

Assumption: L3C4 = 9

L2C4: only candidate → 8
L2C5: only candidate → 3
L4C4: only candidate → 7
L2C1: only place for 9 in unit 2 → 9
L7C1: only candidate → 7
L7C6: only candidate → 9
L8C1: only candidate → 4
L8C6: only candidate → 7
L9C1: only candidate → 8
L9C3: only candidate → 9
L4C3: only candidate → 5
L2C2: only candidate → 7
L1C5: only candidate → 7
L2C3: only candidate → 1
L1C3: only candidate → 4
L1C4: only candidate → 2
L1C6: only candidate → 1
L1C9: only candidate → 8
L2C6: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
L3C6: only candidate → 6
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L3C4 = 8

Assumption: L3C4 = 8

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L3C9 = 1

Assumption: L3C9 = 1

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L1C9 = 4

Assumption: L1C9 = 4

L1C3: only candidate → 8
L2C9: only candidate → 8
L7C2: only candidate → 9
L3C2: only candidate → 2
L4C2: only candidate → 8
L4C3: only candidate → 5
L3C5: only candidate → 5
L2C6: only candidate → 4
L2C2: only candidate → 3

contradiction: a box no longer has no candidate

L1C9 = 8

Assumption: L1C9 = 8

L2C9: only candidate → 4
L6C1: only place for 1 in unit 10 → 1
L5C1: only candidate → 9
L6C2: only candidate → 3
L7C1: only candidate → 4
L2C1: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C9 = 4

Assumption: L3C9 = 4

Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3.
Nude subset in unit 14: L5C5, L9C5 reserve [4, 6]; remove these numbers from L1C5, L6C5.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Units 3 and 19 overlap. In unit 3, 1 must lie in the overlap. Remove it from L1C1, L1C2, L1C3, L2C2, L2C3.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Nude sub-assembly in unit 1: L1C1, L1C2 reserve [3, 6]; remove these numbers from L1C3, L1C4, L1C5.
L1C9: only place for 8 in unit 1 → 8
L2C9: only candidate → 1
Nude subset in unit 19: L1C1, L1C2 reserve [3, 6]; remove these numbers from L2C1, L2C2, L2C3.
Nude subset in unit 19: L1C3, L2C2, L2C3 reserve [4, 7, 8]; remove these numbers from L2C1.
L6C1: only place for 1 in unit 10 → 1
L5C1: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L3C4 = 9

Assumption: L3C4 = 9

L2C4: only candidate → 8
L2C5: only candidate → 3
L4C4: only candidate → 7
L2C1: only place for 9 in unit 2 → 9
L7C1: only candidate → 4
L8C1: only candidate → 7
L7C2: only candidate → 9
L7C6: only candidate → 7
L8C6: only candidate → 3
L9C1: only candidate → 8
L9C3: only candidate → 3
L9C6: only candidate → 9
L4C6: only place for 1 in unit 4 → 1
L2C6: only candidate → 5
L3C5: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L4C6: only place for 1 in unit 15 → 1
L5C4: only place for 2 in unit 23 → 2
Nude subset in Unit 4: L4C3, L4C4, L4C5 reserve [5, 7, 9]; remove these numbers from L4C1, L4C2.
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.
L2C6: only candidate → 5
L3C6: only candidate → 6
L3C5: only candidate → 9
L3C4: only candidate → 8
L4C4: only candidate → 9
L4C2: only candidate → 8
L2C1: only candidate → 9
L3C1: only candidate → 5
L2C2: only candidate → 7

contradiction: a box no longer has no candidate

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L4C6: only place for 1 in unit 15 → 1
L5C4: only place for 2 in unit 23 → 2
Nude subset in Unit 4: L4C3, L4C4, L4C5 reserve [5, 7, 9]; remove these numbers from L4C1, L4C2.
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.
L2C6: only candidate → 5
L3C6: only candidate → 6
L3C5: only candidate → 9
L3C4: only candidate → 8
L4C4: only candidate → 9
L4C2: only candidate → 8
L2C1: only candidate → 9
L3C1: only candidate → 5
L2C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C9 = 5

Assumption: L4C9 = 5

L5C9: only candidate → 3
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
L3C6: only candidate → 6
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.
L3C2: only place for 2 in unit 3 → 2
L4C6: only place for 2 in unit 4 → 2
L2C6: only candidate → 4
L1C4: only candidate → 2
L1C6: only candidate → 1
L6C6: only candidate → 5
L1C5: only place for 7 in unit 1 → 7
L1C3: only candidate → 4
L1C9: only candidate → 8
L2C9: only candidate → 1
L3C9: only candidate → 4
L4C4: only candidate → 7
L2C1: only candidate → 7

contradiction: a box no longer has no candidate

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
L3C6: only candidate → 6
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.
L3C2: only place for 2 in unit 3 → 2
L4C6: only place for 2 in unit 4 → 2
L2C6: only candidate → 4
L1C4: only candidate → 2
L1C6: only candidate → 1
L6C6: only candidate → 5
L1C5: only place for 7 in unit 1 → 7
L1C3: only candidate → 4
L1C9: only candidate → 8
L2C9: only candidate → 1
L3C9: only candidate → 4
L4C4: only candidate → 7
L2C1: only candidate → 7
L5C2: only candidate → 6
L1C1: only candidate → 6

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1
L6C1: only place for 5 in unit 22 → 5
Units 6 and 23 overlap. In unit 6, 3 must lie in the overlap. Remove it from L4C4, L4C5, L4C6.
Nude subassembly in Unit 4: L4C4, L4C5 reserve [7, 9]; remove these numbers from L4C1, L4C2, L4C3, L4C6.
L5C2: only candidate → 6
L5C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C8 = 7

Assumption: L4C8 = 7

L5C8: only candidate → 6

L4C9 = 3

Assumption: L4C9 = 3

L5C9: only candidate → 5

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.

L4C4 = 6

Assumption: L4C4 = 6

L9C4: only candidate → 4
L9C5: only candidate → 6
L1C5: only place for 3 in unit 14 → 3
L1C9: only place for 4 in unit 1 → 4

contradiction: the number 6 can no longer appear in unit 1

L4C4 = 9

Assumption: L4C4 = 9

L3C5: only place for 9 in unit 14 → 9
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.

L3C4 = 6

Assumption: L3C4 = 6

L9C4: only candidate → 4
L5C4: only candidate → 2
L9C5: only candidate → 6
L4C5: only candidate → 5
L2C4: only candidate → 8
L2C5: only candidate → 4

contradiction: a box no longer has no candidate

L3C4 = 8

Assumption: L3C4 = 8

Nude subassembly in Unit 2: L2C4, L2C5 reserve [3, 7]; remove these numbers from L2C1, L2C2, L2C3, L2C6.
Nude subset in unit 2: L2C1, L2C2 reserve [8, 9]; remove these numbers from L2C9.
L1C9: only candidate → 8
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L6C6.
Nude sub-assembly in unit 20: L2C4, L2C5 reserve [3, 7]; remove these numbers from L1C4, L1C5.
L3C2: only place for 4 in unit 19 → 4
L2C2: only candidate → 8
L2C1: only candidate → 9
L2C9: only candidate → 4
L3C9: only candidate → 1
L2C3: only candidate → 1
L2C6: only candidate → 5
L3C3: only candidate → 5
L3C6: only candidate → 6
L1C3: only candidate → 7
L3C1: only candidate → 2
L4C1: only candidate → 5
L4C5: only candidate → 6
L4C2: only candidate → 2
L4C3: only candidate → 8
L4C6: only candidate → 1
L1C5: only candidate → 1
L5C2: only candidate → 7
L5C3: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.

L4C4 = 6

Assumption: L4C4 = 6

L9C4: only candidate → 4
L9C5: only candidate → 6
L1C5: only place for 3 in unit 14 → 3
L1C9: only place for 4 in unit 1 → 4

contradiction: the number 6 can no longer appear in unit 1

L4C4 = 9

Assumption: L4C4 = 9

L3C5: only place for 9 in unit 14 → 9
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.

L3C4 = 6

Assumption: L3C4 = 6

L9C4: only candidate → 4
L5C4: only candidate → 2
L9C5: only candidate → 6
L4C5: only candidate → 5
L2C4: only candidate → 8
L2C5: only candidate → 4

contradiction: a box no longer has no candidate

L3C4 = 8

Assumption: L3C4 = 8

Nude subassembly in Unit 2: L2C4, L2C5 reserve [3, 7]; remove these numbers from L2C1, L2C2, L2C3, L2C6.
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 20: L2C4, L2C5 reserve [3, 7]; remove these numbers from L1C4, L1C5, L1C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these figures from L6C6.
Units 14 and 23 overlap. In unit 14, 5 must lie in the overlap. Remove it from L4C6, L6C6.

L1C3 = 3

Assumption: L1C3 = 3

L2C3: only place for 7 in unit 12 → 7

contradiction: a box no longer has no candidate

L1C3 = 7

Assumption: L1C3 = 7

L3C6: only place for 6 in unit 20 → 6
L4C6: only place for 1 in unit 4 → 1
L1C5: only candidate → 1
L2C6: only candidate → 5
L1C9: only place for 8 in unit 1 → 8
L1C6: only place for 2 in unit 15 → 2
L1C4: only candidate → 4
L9C4: only candidate → 6
L5C4: only candidate → 2
L9C5: only candidate → 4
L4C5: only candidate → 6
L5C5: only candidate → 3
L2C4: only candidate → 3
L2C1: only candidate → 9
L2C5: only candidate → 7
L3C3: only candidate → 5
L3C1: only candidate → 2
L4C1: only candidate → 5
L4C2: only candidate → 2
L4C3: only candidate → 8
L2C2: only candidate → 8
L5C1: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L4C6: only place for 1 in unit 15 → 1
L5C4: only place for 2 in unit 23 → 2
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.
L2C6: only candidate → 5
L3C6: only candidate → 6
L3C5: only candidate → 9
L3C4: only candidate → 8
L4C4: only candidate → 9
L6C4: only candidate → 7
L2C4: only candidate → 3
L1C3: only candidate → 7
L1C4: only candidate → 4
L1C6: only candidate → 2
L1C9: only candidate → 8
L2C2: only candidate → 8
L2C1: only candidate → 9
L2C3: only candidate → 1
L2C5: only candidate → 7
L2C9: only candidate → 4
L3C2: only candidate → 4
L3C3: only candidate → 5
L3C1: only candidate → 2
L3C9: only candidate → 1
L4C1: only candidate → 5
L4C3: only candidate → 8
L4C5: only candidate → 6
L4C2: only candidate → 2
L5C2: only candidate → 7
L5C3: only candidate → 4
L5C1: only candidate → 1
L5C5: only candidate → 3
L6C1: only candidate → 3
L1C1: only candidate → 6
L1C2: only candidate → 3
L6C2: only candidate → 9
L6C3: only candidate → 6
L6C5: only candidate → 5
L6C6: only candidate → 4
L7C1: only candidate → 7
L7C6: only candidate → 9
L8C1: only candidate → 4
L8C2: only candidate → 6
L8C3: only candidate → 3
L8C6: only candidate → 7
L9C1: only candidate → 8
L9C3: only candidate → 9
L9C4: only candidate → 6
L9C5: only candidate → 4
L9C6: only candidate → 3

valid complete solution

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L4C6: only place for 1 in unit 15 → 1
L5C4: only place for 2 in unit 23 → 2
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.
L2C6: only candidate → 5
L3C6: only candidate → 6
L3C5: only candidate → 9
L3C4: only candidate → 8
L4C4: only candidate → 9
L6C4: only candidate → 7
L2C4: only candidate → 3
L1C3: only candidate → 7
L1C4: only candidate → 4
L1C6: only candidate → 2
L1C9: only candidate → 8
L2C5: only candidate → 7
L3C3: only candidate → 5
L6C5: only candidate → 5
L4C5: only candidate → 6
L2C1: only candidate → 9
L3C1: only candidate → 2
L4C1: only candidate → 5
L4C2: only candidate → 2
L4C3: only candidate → 8
L2C2: only candidate → 8
L5C2: only candidate → 7
L5C5: only candidate → 3
L6C6: only candidate → 4
L7C1: only candidate → 4
L5C1: only candidate → 1
L5C3: only candidate → 4
L2C3: only candidate → 1
L2C9: only candidate → 4
L3C2: only candidate → 4
L3C9: only candidate → 1
L6C1: only candidate → 3
L1C1: only candidate → 6
L1C2: only candidate → 3
L6C2: only candidate → 9

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C9 = 5

Assumption: L4C9 = 5

L5C9: only candidate → 3
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L7C1 = 7

Assumption: L7C1 = 7

L7C6: only candidate → 9
L8C6: only place for 7 in unit 8 → 7
L9C6: only candidate → 3
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.
L4C3: only place for 8 in unit 4 → 8
L4C2: only candidate → 2
L5C2: only candidate → 7
L9C3: only candidate → 9
L5C3: only candidate → 4
L5C1: only candidate → 9

contradiction: a box no longer has no candidate

L7C1 = 9

Assumption: L7C1 = 9

L7C6: only candidate → 7
L8C6: only candidate → 3
L9C3: only candidate → 8
L9C6: only candidate → 9
L4C1: only place for 8 in unit 4 → 8
L8C1: only place for 7 in unit 8 → 7
L2C1: only candidate → 3

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C1, L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.

L7C6 = 7

Assumption: L7C6 = 7

L8C6: only candidate → 3
L9C6: only candidate → 9
L8C1: only place for 7 in unit 8 → 7
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C1, L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L2C1, L3C1.
L4C3: only place for 8 in unit 4 → 8
L4C2: only candidate → 2
L5C2: only candidate → 7
L7C2: only candidate → 9
L7C1: only candidate → 4
L5C1: only candidate → 9

contradiction: a box no longer has no candidate

L7C6 = 9

Assumption: L7C6 = 9

L9C3: only candidate → 8
L9C1: only candidate → 9
L4C1: only place for 8 in unit 4 → 8
L8C6: only place for 7 in unit 8 → 7
L9C6: only candidate → 3
Units 4 and 23 overlap. In unit 4, 3 must lie in the overlap. Remove it from L6C4, L6C5.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L3C1.
L2C1: only candidate → 7
L3C1: only candidate → 1
L7C1: only candidate → 4
L5C1: only candidate → 2
L5C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1
L6C1: only place for 5 in unit 22 → 5
L5C3: only place for 4 in unit 22 → 4
L6C3: only candidate → 6
L5C1: only candidate → 1
L5C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L1C7 = 6

Assumption: L1C7 = 6

L2C7: only candidate → 5

L4C8 = 6

Assumption: L4C8 = 6

L5C8: only candidate → 7

L4C9 = 3

Assumption: L4C9 = 3

L5C9: only candidate → 5

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L1C1 = 1

Assumption: L1C1 = 1

L1C2: only candidate → 8
L1C9: only candidate → 4

contradiction: a box no longer has no candidate

L1C1 = 5

Assumption: L1C1 = 5

L1C2: only candidate → 4

contradiction: the number 7 can no longer appear in unit 1

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.

L4C4 = 7

Assumption: L4C4 = 7

L5C1: only place for 2 in unit 5 → 2
L4C3: only candidate → 5
L4C5: only candidate → 9
L2C5: only candidate → 3
L3C5: only candidate → 5
L4C2: only candidate → 8
L4C1: only candidate → 1
L4C6: only candidate → 2
L1C4: only candidate → 2
L5C2: only candidate → 4

contradiction: a box no longer has no candidate

L4C4 = 9

Assumption: L4C4 = 9

L3C5: only place for 9 in unit 14 → 9
L4C5: only candidate → 5
L5C5: only candidate → 4
L5C4: only candidate → 2
L1C6: only candidate → 2
L2C6: only candidate → 6
L3C4: only candidate → 8
L3C6: only candidate → 5
L4C6: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L1C1: only place for 5 in unit 1 → 5
L1C2: only candidate → 4
L1C9: only candidate → 8

contradiction: a box no longer has no candidate

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L1C1: only place for 5 in unit 1 → 5
L1C2: only candidate → 4
L1C9: only candidate → 8

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C9 = 5

Assumption: L4C9 = 5

L5C9: only candidate → 3
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.
L1C1: only candidate → 1
L1C2: only candidate → 8
L1C9: only candidate → 4

contradiction: a box no longer has no candidate

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 3 and 19 overlap. In unit 3, 2 must lie in the overlap. Remove it from L1C1, L1C2.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.
Nude subset in unit 1: L1C1, L1C2 reserve [1, 8]; remove these numbers from L1C3, L1C4, L1C5, L1C6, L1C9.
L1C9: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1
L6C1: only place for 5 in unit 22 → 5
Units 6 and 23 overlap. In unit 6, 3 must lie in the overlap. Remove it from L4C4, L4C5, L4C6.
Nude subassembly in Unit 4: L4C4, L4C5 reserve [7, 9]; remove these numbers from L4C1, L4C2, L4C3, L4C6.
L5C2: only candidate → 6
L5C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C8 = 7

Assumption: L4C8 = 7

L5C8: only candidate → 6

L4C9 = 3

Assumption: L4C9 = 3

L5C9: only candidate → 5

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.

L4C4 = 6

Assumption: L4C4 = 6

L9C4: only candidate → 4
L2C5: only candidate → 4
L3C5: only candidate → 6

contradiction: a box no longer has no candidate

L4C4 = 9

Assumption: L4C4 = 9

L3C5: only place for 9 in unit 14 → 9
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.

L4C5 = 5

Assumption: L4C5 = 5

L5C5: only candidate → 4
L5C4: only candidate → 2
L1C6: only candidate → 2
L3C6: only candidate → 5
L9C4: only candidate → 4
L3C4: only candidate → 8
L1C5: only candidate → 1
L1C1: only candidate → 5
L1C2: only candidate → 4
L1C3: only candidate → 8

contradiction: a box no longer has no candidate

L4C5 = 6

Assumption: L4C5 = 6

L5C4: only candidate → 2
L1C6: only candidate → 2
L5C5: only candidate → 3
L9C4: only candidate → 6
L2C4: only candidate → 3
L2C5: only candidate → 7
L3C4: only candidate → 8
L1C4: only candidate → 4
L1C1: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.

L4C4 = 6

Assumption: L4C4 = 6

L9C4: only candidate → 4
L2C5: only candidate → 4
L3C5: only candidate → 6

contradiction: a box no longer has no candidate

L4C4 = 9

Assumption: L4C4 = 9

L3C5: only place for 9 in unit 14 → 9
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L6C6.

L4C5 = 5

Assumption: L4C5 = 5

L5C5: only candidate → 4
L5C4: only candidate → 2
L1C6: only candidate → 2
L3C6: only candidate → 5
L9C4: only candidate → 4
L3C4: only candidate → 8
L1C5: only candidate → 1
L1C1: only candidate → 5
L1C2: only candidate → 4
L1C3: only candidate → 8

contradiction: a box no longer has no candidate

L4C5 = 6

Assumption: L4C5 = 6

L5C4: only candidate → 2
L1C6: only candidate → 2
L5C5: only candidate → 3
L9C4: only candidate → 6
L2C4: only candidate → 3
L2C5: only candidate → 7
L3C4: only candidate → 8
L1C4: only candidate → 4
L6C4: only candidate → 7
L9C5: only candidate → 4
L6C5: only candidate → 5
L1C3: only candidate → 7
L1C5: only candidate → 1

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L1C1: only place for 5 in unit 1 → 5
L1C2: only candidate → 4
L1C9: only candidate → 8

contradiction: a box no longer has no candidate

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
L1C5: only place for 1 in unit 14 → 1
L1C1: only place for 5 in unit 1 → 5
L1C2: only candidate → 4
L1C9: only candidate → 8

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L4C9 = 5

Assumption: L4C9 = 5

L5C9: only candidate → 3
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L5C7 = 1

Assumption: L5C7 = 1

L6C7: only candidate → 9
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L7C8 = 4

Assumption: L7C8 = 4

L8C8: only candidate → 5
L7C4: only candidate → 5
L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L8C4: only candidate → 1
L7C2: only candidate → 1
Nude sub-assembly in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L7C1 = 7

Assumption: L7C1 = 7

L7C6: only candidate → 9
L8C6: only place for 7 in unit 8 → 7
L9C6: only candidate → 3
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L1C1 = 1

Assumption: L1C1 = 1

L1C2: only candidate → 8
L1C9: only candidate → 4

contradiction: a box no longer has no candidate

L1C1 = 2

Assumption: L1C1 = 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C1 = 9

Assumption: L7C1 = 9

L7C6: only candidate → 7
L8C6: only candidate → 3
L9C3: only candidate → 8
L9C6: only candidate → 9
L4C1: only place for 8 in unit 4 → 8
L8C1: only place for 7 in unit 8 → 7
L9C1: only candidate → 3
L2C1: only candidate → 6
L3C1: only candidate → 2
L2C3: only candidate → 1
L1C1: only candidate → 5

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C8 = 5

Assumption: L7C8 = 5

L8C8: only candidate → 4
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L8C4 = 3

Assumption: L8C4 = 3

L8C5: only candidate → 2
L7C2: only place for 2 in unit 7 → 2
L8C6: only place for 5 in unit 8 → 5
L9C6: only candidate → 6

contradiction: a box no longer has no candidate

L8C4 = 5

Assumption: L8C4 = 5

L7C5: only candidate → 8
L8C5: only candidate → 2
L7C3: only candidate → 2
L7C4: only candidate → 1
Nude subset in Unit 9: L9C4, L9C5 reserve [4, 6]; remove these numbers from L9C1, L9C3, L9C6.
Nude subset in unit 26: L9C4, L9C5 reserve [4, 6]; remove these numbers from L7C6, L8C6.
Nude subset in unit 15: L7C6, L8C6, L9C6 reserve [3, 7, 9]; remove these numbers from L1C6, L2C6, L3C6, L4C6, L6C6.
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C1, L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L7C6 = 7

Assumption: L7C6 = 7

L8C6: only candidate → 3
L9C6: only candidate → 9
L8C1: only place for 7 in unit 8 → 7
Units 11 and 19 overlap. In unit 11, 8 must lie in the overlap. Remove it from L1C1, L1C3, L2C1, L2C3, L3C1, L3C3.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L1C1 = 1

Assumption: L1C1 = 1

L1C2: only candidate → 8
L1C9: only candidate → 4

contradiction: a box no longer has no candidate

L1C1 = 2

Assumption: L1C1 = 2

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L7C6 = 9

Assumption: L7C6 = 9

L9C3: only candidate → 8
L9C1: only candidate → 9
L4C1: only place for 8 in unit 4 → 8
L8C6: only place for 7 in unit 8 → 7
L9C6: only candidate → 3
Units 4 and 23 overlap. In unit 4, 3 must lie in the overlap. Remove it from L6C4, L6C5.
Units 12 and 19 overlap. In unit 12, 5 must lie in the overlap. Remove it from L1C1, L3C1.

L2C1 = 6

Assumption: L2C1 = 6

L3C1: only candidate → 2
L3C3: only candidate → 5
L1C1: only candidate → 1

contradiction: a box no longer has no candidate

L2C1 = 7

Assumption: L2C1 = 7

L3C1: only candidate → 1
L1C1: only candidate → 5
L1C2: only candidate → 4

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

L5C7 = 9

Assumption: L5C7 = 9

L6C7: only candidate → 1
L6C1: only place for 5 in unit 22 → 5
L5C3: only place for 4 in unit 22 → 4
L6C3: only candidate → 6
L5C1: only candidate → 1
L5C2: only candidate → 7

contradiction: a box no longer has no candidate

all the possibilities of this bifurcation have been examined

How we verified this answer

Exact enumeration, shutdown after two solutions, terminal verification of all units and cages.

{
  "nodes": 127,
  "branches": 63,
  "contradictions": 63,
  "depth": 9,
  "pairs": 102,
  "locked": 55
}

← Return to puzzle SOLO-010

SOLO-011 · Advanced Mastermind

EXPERT

The Eight Inks

Find the only ordered code of 5 digits.

This page contains the solution.

The answer

Code: 70730.

The reasoning

Separates two constraints for each line: the number of equal positions per position, and then the total of common occurrences. This total is well placed + wrong placed; for each number, the minimum of the two multiplicities is taken.
To cut down the work, ignore the positions for now. Let n₀,…,n₇ be the counts of the eight digits. Their sum is 5.
Proposal 02553 requires min(n0, 1) + min(n2, 1) + min(n3, 1) + min(n5, 2) = 2.
Proposal 62532 requires min(n2, 2) + min(n3, 1) + min(n5, 1) + min(n6, 1) = 1.
Proposal 05062 requires min(n0, 2) + min(n2, 1) + min(n5, 1) + min(n6, 1) = 2.
Proposal 73375 requires min(n3, 2) + min(n5, 1) + min(n7, 2) = 3.
These limitations of occurrences alone leave 5 multi-unit(s).Here they are, written in ascending order: 00337; 00377; 01577; 04577; 05777.
Multiassembly 00337 : after successive application of the numbers of property placed, there remains 12 → 5 → 2 → 0 arrangement(s).
Multiassembly 00377 : after successive application of the numbers of property placed, there remains 15 → 3 → 1 → 1 arrangement(s).
Multi ensemble 01577 : after successive application of the numbers of property placed, there is left 30 → 0 → 0 → 0 arrangement(s).
Multi ensemble 04577 : after successive application of the numbers of property placed, there remains 30 → 0 → 0 → 0 arrangement(s).
Multiassembly 05777 : after successive application of the numbers of property placed, there remains 10 → 0 → 0 → 0 arrangement(s).
So we can make two separate tables, occurrences and then places. Only one line in the second table keeps an arrangement; that's the code announced. This separation avoids blindly browsing all the ordered words.
Starts with the 32.768 words ordered length 5 out of 0 to 7 and do not remove the initial zeros or the words with repetition.
Index 1, proposal 02553: retains exactly the configurations with 0 position equal(s) and 2 common occurrence(s). There are 7.051 candidates left.
Index 2, proposal 62532: keeps exactly the configurations with 1 position equal(s) and 1 common occurrence(s). There are 319 candidates left.
Index 3, proposal 05062: retains exactly the configurations with 0 position equal(s) and 2 common occurrence(s). There are 16 candidates left.
Index 4, proposal 73375: retains exactly the configurations with 1 position equal(s) and 3 common occurrence(s). There is 1 candidate remaining.
In index 1, candidate 00000 would give 1 well placed and 0 badly placed, instead of 0 and 2: it is eliminated.
In Index 2, candidate 10005 would give 0 well placed and 1 ill placed, instead of 1 and 0: it is eliminated.
In index 3, candidate 10002 would give 2 well placed and 1 ill placed, instead of 0 and 2: it is eliminated.
This filtering can be done on paper in branches: first fixes the possible occurrences, then affects their places and cuts a branch as soon as a line exceeds its number of places or can no longer reach its common total. You must not choose a candidate only because it satisfies the last line.
Final check against 02553: 0 well placed, 2 wrong placed.
Final check against 62532: 1 well placed, 0 wrong placed.
Final check against 05062: 0 well placed, 2 wrong placed.
Final check against 73375: 1 well placed, 2 badly placed.
The exhaustive check takes all possible words and has exactly one solution. The numbers are therefore determined at all places; it is not just a compatible code found at random.

Common mistakes and false leads

Consolidate wrongly placed and figure absent.
Count twice an occurrence of the code when the proposal repeats a number.
Prohibit repetitions or the initial zero without the rules requiring it.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "multisets_and_position_counts": [
    {
      "bag": "00337",
      "remaining_after_position_constraints": [
        12,
        5,
        2,
        0
      ]
    },
    {
      "bag": "00377",
      "remaining_after_position_constraints": [
        15,
        3,
        1,
        1
      ]
    },
    {
      "bag": "01577",
      "remaining_after_position_constraints": [
        30,
        0,
        0,
        0
      ]
    },
    {
      "bag": "04577",
      "remaining_after_position_constraints": [
        30,
        0,
        0,
        0
      ]
    },
    {
      "bag": "05777",
      "remaining_after_position_constraints": [
        10,
        0,
        0,
        0
      ]
    }
  ]
}

How we verified this answer

Complete listing of all codes; calculation independent of each score; deletion of redundant indices.

{
  "candidates": 32768,
  "solutions": 1,
  "clues": 4,
  "survivors_after_clues": [
    7051,
    319,
    16,
    1
  ]
}

← Return to puzzle SOLO-011

SOLO-012 · Advanced Mastermind

EXPERT

The Silent Lock

Find the only ordered code of 5 digits.

This page contains the solution.

The answer

Code: 32752.

The reasoning

Separates two constraints for each line: the number of equal positions per position, and then the total of common occurrences. This total is well placed + wrong placed; for each number, the minimum of the two multiplicities is taken.
To cut down the work, ignore the positions for now. Let n₀,…,n₇ be the counts of the eight digits. Their sum is 5.
Proposal 22543 requires min(n2, 2) + min(n3, 1) + min(n4, 1) + min(n5, 1) = 4.
Proposal 41050 requires min(n0, 2) + min(n1, 1) + min(n4, 1) + min(n5, 1) = 1.
Proposal 70302 requires min(n0, 2) + min(n2, 1) + min(n3, 1) + min(n7, 1) = 3.
Proposal 37433 requires min(n3, 3) + min(n4, 1) + min(n7, 1) = 2.
These limitations of occurrences alone leave 1 multi set(s).Here they are, written in ascending order: 22357.
Multi ensemble 22357 : after successive application of the numbers of property placed, there remains 26 → 6 → 2 → 1 arrangement(s).
So we can make two separate tables, occurrences and then places. Only one line in the second table keeps an arrangement; that's the code announced. This separation avoids blindly browsing all the ordered words.
Starts with the 32.768 words ordered length 5 out of 0 to 7 and do not remove the initial zeros or the words with repetition.
Index 1, proposal 22543: keeps exactly the configurations with 1 position equal(s) and 4 common occurrence(s). There are 652 candidates left.
Index 2, proposal 41050: retains exactly the configurations with 1 position equal(s) and 1 common occurrence(s). There are 39 candidates left.
Index 3, proposal 70302: keeps exactly the configurations with 1 position equal(s) and 3 common occurrence(s). There are 4 candidates left.
Index 4, proposal 37433: retains exactly the configurations with 1 position equal(s) and 2 common occurrence(s). There are 1 candidate(s).
In Index 1, candidate 00000 would give 0 well placed and 0 badly placed, instead of 1 and 3: it is eliminated.
In Index 2, candidate 02234 would give 0 well placed and 2 badly placed, instead of 1 and 0: it is eliminated.
In index 3, candidate 23252 would give 1 well placed and 1 ill placed, instead of 1 and 2: it is eliminated.
This filtering can be done on paper in branches: first fixes the possible occurrences, then affects their places and cuts a branch as soon as a line exceeds its number of places or can no longer reach its common total. You must not choose a candidate only because it satisfies the last line.
Final check against 22543: 1 well placed, 3 badly placed.
Final check against 41050: 1 well placed, 0 wrong placed.
Final check against 70302 : 1 well placed, 2 wrong placed.
Final check against 37433 : 1 well placed, 1 wrong placed.
The exhaustive check takes all possible words and has exactly one solution. The numbers are therefore determined at all places; it is not just a compatible code found at random.

Common mistakes and false leads

Consolidate wrongly placed and figure absent.
Count twice an occurrence of the code when the proposal repeats a number.
Prohibit repetitions or the initial zero without the rules requiring it.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "multisets_and_position_counts": [
    {
      "bag": "22357",
      "remaining_after_position_constraints": [
        26,
        6,
        2,
        1
      ]
    }
  ]
}

How we verified this answer

Complete listing of all codes; calculation independent of each score; deletion of redundant indices.

{
  "candidates": 32768,
  "solutions": 1,
  "clues": 4,
  "survivors_after_clues": [
    652,
    39,
    4,
    1
  ]
}

← Return to puzzle SOLO-012

SOLO-013 · Advanced Mastermind

EXTREME

Six Slots Without an Echo

Find the only orderly code of 6 digits.

This page contains the solution.

The answer

Code: 331037.

The reasoning

Separates two constraints for each line: the number of equal positions per position, and then the total of common occurrences. This total is well placed + wrong placed; for each number, the minimum of the two multiplicities is taken.
To cut down the work, ignore the positions for now. Let n₀,…,n₇ be the counts of the eight digits. Their sum is 6.
Proposal 345125 requires min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n4, 1) + min(n5, 2) = 2.
Proposal 073421 requires min(n0, 1) + min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n4, 1) + min(n7, 1) = 4.
Proposal 573237 requires min(n2, 1) + min(n3, 2) + min(n5, 1) + min(n7, 2) = 3.
Proposal 531562 requires min(n1, 1) + min(n2, 1) + min(n3, 1) + min(n5, 2) + min(n6, 1) = 2.
These limitations of occurrences alone leave 16 multi-group(s).Here they are, written in ascending order: 000237; 001277; 001337; 001377; 002237; 011277; 011337; 011377; 012277; 012777; 013337; 013777; 022237; 024677; 033467; 034677.
Multi ensemble 000237: after successive application of the numbers of property placed, there remains 32 → 13 → 1 → 0 arrangement(s).
Multi ensemble 001277 : after successive application of the numbers of property placed, there remains 48 → 10 → 0 → 0 arrangement(s).
Multi ensemble 001337 : after successive application of the numbers of property placed, there remains 66 → 28 → 2 → 0 arrangement(s).
Multi ensemble 001377 : after successive application of the numbers of property placed, there remains 48 → 19 → 1 → 0 arrangement(s).
Multi ensemble 002237 : after successive application of the numbers of property placed, there remains 66 → 15 → 2 → 0 arrangement(s).
Multi ensemble 011277 : after successive application of the numbers of property placed, there remains 66 → 21 → 0 → 0 arrangement(s).
Multi ensemble 011337 : after successive application of the numbers of property placed, there remains 72 → 24 → 2 → 0 arrangement(s).
Multi ensemble 011377 : after successive application of the numbers of property placed, there remains 66 → 23 → 3 → 0 arrangement(s).
Multi ensemble 012277 : after successive application of the numbers of property placed, there remains 66 → 10 → 0 → 0 arrangement(s).
Multi ensemble 012777 : after successive application of the numbers of property placed, there is left 32 → 6 → 0 → 0 arrangement(s).
Multi ensemble 013337 : after successive application of the numbers of property placed, there is left 56 → 21 → 3 → 1 arrangement(s).
Multi ensemble 013777 : after successive application of the numbers of property placed, there is left 32 → 11 → 1 → 0 arrangement(s).
Multi ensemble 022237 : after successive application of the numbers of property placed, there is left 56 → 7 → 1 → 0 arrangement(s).
Multi ensemble 024677 : after successive application of the numbers of property placed, there remains 96 → 39 → 4 → 0 arrangement(s).
Multi ensemble 033467 : after successive application of the numbers of property placed, there remains 132 → 58 → 3 → 0 arrangement(s).
Multi ensemble 034677 : after successive application of the numbers of property placed, there remains 96 → 45 → 4 → 0 arrangement(s).
So we can make two separate tables, occurrences and then places. Only one line in the second table keeps an arrangement; that's the code announced. This separation avoids blindly browsing all the ordered words.
Starts with the 262.144 ordered words of length 6 out of 0 to 7. The initial zeros or words with repetition are not removed.
Index 1, proposal 345125: retains exactly the configurations with 1 position equal(s) and 2 common occurrence(s). There are 26,034 candidates left.
Index 2, proposal 073421: retains exactly the configurations with 0 position level(s) and 4 common occurrence(s). There are 2,490 candidates left.
Index 3, proposal 573237: retains exactly the configurations with 2 positions and 3 common occurrence(s). There are 65 candidates left.
Index 4, proposal 531562: retains exactly the configurations with 2 positions and 2 common occurrence(s). There are 1 candidate(s) left.
In index 1, the candidate 000000 would give 0 well placed and 0 badly placed, instead of 1 and 1: it is eliminated.
In index 2, candidate 000015 would give 1 well placed and 1 ill placed, instead of 0 and 4: it is eliminated.
In index 3, candidate 100137 would give 2 well placed and 0 badly placed, instead of 2 and 1: it is eliminated.
This filtering can be done on paper in branches: first fixes the possible occurrences, then affects their places and cuts a branch as soon as a line exceeds its number of places or can no longer reach its common total. You must not choose a candidate only because it satisfies the last line.
Final check against 345125: 1 well placed, 1 wrong placed.
Final check against 073421 : 0 well placed, 4 wrong placed.
Final check against 573237: 2 well placed, 1 wrong placed.
Final check against 531562: 2 well placed, 0 wrong placed.
The exhaustive check takes all possible words and has exactly one solution. The numbers are therefore determined at all places; it is not just a compatible code found at random.

Common mistakes and false leads

Consolidate wrongly placed and figure absent.
Count twice an occurrence of the code when the proposal repeats a number.
Prohibit repetitions or the initial zero without the rules requiring it.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "multisets_and_position_counts": [
    {
      "bag": "000237",
      "remaining_after_position_constraints": [
        32,
        13,
        1,
        0
      ]
    },
    {
      "bag": "001277",
      "remaining_after_position_constraints": [
        48,
        10,
        0,
        0
      ]
    },
    {
      "bag": "001337",
      "remaining_after_position_constraints": [
        66,
        28,
        2,
        0
      ]
    },
    {
      "bag": "001377",
      "remaining_after_position_constraints": [
        48,
        19,
        1,
        0
      ]
    },
    {
      "bag": "002237",
      "remaining_after_position_constraints": [
        66,
        15,
        2,
        0
      ]
    },
    {
      "bag": "011277",
      "remaining_after_position_constraints": [
        66,
        21,
        0,
        0
      ]
    },
    {
      "bag": "011337",
      "remaining_after_position_constraints": [
        72,
        24,
        2,
        0
      ]
    },
    {
      "bag": "011377",
      "remaining_after_position_constraints": [
        66,
        23,
        3,
        0
      ]
    },
    {
      "bag": "012277",
      "remaining_after_position_constraints": [
        66,
        10,
        0,
        0
      ]
    },
    {
      "bag": "012777",
      "remaining_after_position_constraints": [
        32,
        6,
        0,
        0
      ]
    },
    {
      "bag": "013337",
      "remaining_after_position_constraints": [
        56,
        21,
        3,
        1
      ]
    },
    {
      "bag": "013777",
      "remaining_after_position_constraints": [
        32,
        11,
        1,
        0
      ]
    },
    {
      "bag": "022237",
      "remaining_after_position_constraints": [
        56,
        7,
        1,
        0
      ]
    },
    {
      "bag": "024677",
      "remaining_after_position_constraints": [
        96,
        39,
        4,
        0
      ]
    },
    {
      "bag": "033467",
      "remaining_after_position_constraints": [
        132,
        58,
        3,
        0
      ]
    },
    {
      "bag": "034677",
      "remaining_after_position_constraints": [
        96,
        45,
        4,
        0
      ]
    }
  ]
}

How we verified this answer

Complete listing of all codes; calculation independent of each score; deletion of redundant indices.

{
  "candidates": 262144,
  "solutions": 1,
  "clues": 4,
  "survivors_after_clues": [
    26034,
    2490,
    65,
    1
  ]
}

← Return to puzzle SOLO-013

SOLO-014 · Advanced Mastermind

EXTREME

The Chamber of Duplicates

Find the only orderly code of 6 digits.

This page contains the solution.

The answer

Code: 140633.

The reasoning

Separates two constraints for each line: the number of equal positions per position, and then the total of common occurrences. This total is well placed + wrong placed; for each number, the minimum of the two multiplicities is taken.
To cut down the work, ignore the positions for now. Let n₀,…,n₇ be the counts of the eight digits. Their sum is 6.
Proposal 474563 requires min(n3, 1) + min(n4, 2) + min(n5, 1) + min(n6, 1) + min(n7, 1) = 3.
Proposal 013237 requires min(n0, 1) + min(n1, 1) + min(n2, 1) + min(n3, 2) + min(n7, 1) = 4.
Proposal 163003 requires min(n0, 2) + min(n1, 1) + min(n3, 2) + min(n6, 1) = 5.
Proposal 635050 requires min(n0, 2) + min(n3, 1) + min(n5, 2) + min(n6, 1) = 3.
Proposal 042546 requires min(n0, 1) + min(n2, 1) + min(n4, 2) + min(n5, 1) + min(n6, 1) = 3.
These limitations of occurrences alone leave 1 multiset(s).Here they are, written in ascending order: 013346.
Multi ensemble 013346 : after successive application of the numbers of property placed, there remains 144 → 64 → 19 → 5 → 1 arrangement(s).
So we can make two separate tables, occurrences and then places. Only one line in the second table keeps an arrangement; that's the code announced. This separation avoids blindly browsing all the ordered words.
Starts with the 262.144 ordered words of length 6 out of 0 to 7. The initial zeros or words with repetition are not removed.
Index 1, proposal 474563: retains exactly the configurations with 1 position equal(s) and 3 common occurrence(s). There are 42,344 candidates left.
Index 2, proposal 013237: retains exactly the configurations with 1 position equal(s) and 4 common occurrence(s). There are 3,996 candidates left.
Index 3, proposal 163003: retains exactly the configurations with 2 positions and 5 common occurrence(s). 58 candidates remain.
Index 4, proposal 635050: retains exactly the configurations with 0 position equal(s) and 3 common occurrence(s). There are 5 candidates left.
Index 5, proposal 042546: retains exactly the configurations with 1 position equal(s) and 3 common occurrence(s). There are 1 candidate(s) left.
In index 1, the candidate 000000 would give 0 well placed and 0 badly placed, instead of 1 and 2: it is eliminated.
In index 2, candidate 000364 would give 1 well placed and 1 ill placed, instead of 1 and 3: it is eliminated.
In Index 3, candidate 001473 would give 1 well placed and 3 badly placed, instead of 2 and 3: it is eliminated.
This filtering can be done on paper in branches: first fixes the possible occurrences, then affects their places and cuts a branch as soon as a line exceeds its number of places or can no longer reach its common total. You must not choose a candidate only because it satisfies the last line.
Final check against 474563: 1 well placed, 2 badly placed.
Final check against 013237: 1 well placed, 3 badly placed.
Final check against 163003: 2 well placed, 3 badly placed.
Final check against 635050: 0 well placed, 3 badly placed.
Final check against 042546: 1 well placed, 2 badly placed.
The exhaustive check takes all possible words and has exactly one solution. The numbers are therefore determined at all places; it is not just a compatible code found at random.

Common mistakes and false leads

Consolidate wrongly placed and figure absent.
Count twice an occurrence of the code when the proposal repeats a number.
Prohibit repetitions or the initial zero without the rules requiring it.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "multisets_and_position_counts": [
    {
      "bag": "013346",
      "remaining_after_position_constraints": [
        144,
        64,
        19,
        5,
        1
      ]
    }
  ]
}

How we verified this answer

Complete listing of all codes; calculation independent of each score; deletion of redundant indices.

{
  "candidates": 262144,
  "solutions": 1,
  "clues": 5,
  "survivors_after_clues": [
    42344,
    3996,
    58,
    5,
    1
  ]
}

← Return to puzzle SOLO-014

SOLO-015 · Cryptarithm

EXPERT

The Twin Carries

Determine the value of each of the ten letters, then write the two additions in digits.

This page contains the solution.

The answer

A = 1, B = 3, C = 9, D = 5, E = 6, F = 7, G = 2, H = 4, I = 0, J = 8

5573 + 8462 = 14035

6707 + 8669 = 15376

The reasoning

Each column carries a hold of 0 or 1. For an addition of two numbers to four digits, a possible fifth position of the result is a final hold, so is 1.
Write a column equation, from right to left: first-term digit + second-time digit + incoming hold = result digit + 10 × outgoing hold. The same letter must keep its value in both additions.
Addition 1, column 1: B + G + r1_0 = D + 10 × r1_1.
Addition 1, column 2: F + E + r1_1 = B + 10 × r1_2.
Addition 1, column 3: D + H + r1_2 = I + 10 × r1_3.
Addition 1, column 4: D + J + r1_3 = H + 10 × r1_4.
Addition 2, column 1: F + C + r2_0 = E + 10 × r2_1.
Addition 2, column 2: I + E + r2_1 = F + 10 × r2_2.
Addition 2, column 3: F + E + r2_2 = B + 10 × r2_3.
Addition 2, column 4: E + J + r2_3 = D + 10 × r2_4.
In these equations, r1_0 and r2_0 are zero. Each other retention is 0 or 1; the same letter always refers to the same number.
Explore possible pairs of digits in the column of units, calculate the restraint and bar any branch giving the same number to two different letters. Repeat for tens, hundreds and thousands; a letter already affected no longer receives any new choice.
The first addition admits 16 partial assignments respecting these constraints. After assigning letters still absent to free numbers, 16 substitutions are checked against the second addition.
Crossing the two additions is essential: the entire substitution must be transported, not separately resolve each line and then juxtapose two incompatible responses.
Addition 1: DDFB + JHEG = AHIBD, i.e. 5573 + 8462 = 14035.
Column 1 from the right: 3 + 2 + carry 0 = 5. Write 5 and carry 0.
Column 2 from the right: 7 + 6 + carry 0 = 13. Write 3 and carry 1.
Column 3 from the right: 5 + 4 + carry 1 = 10. Write 0 and carry 1.
Column 4 from the right: 5 + 8 + carry 1 = 14. Write 4 and carry 1.
Addition 2: EFIF + JEEC = ADBFE, i.e. 6707 + 8669 = 15376.
Column 1 from the right: 7 + 9 + carry 0 = 16. Write 6 and carry 1.
Column 2 from the right: 0 + 6 + carry 1 = 7. Write 7 and carry 0.
Column 3 from the right: 7 + 6 + carry 0 = 13. Write 3 and carry 1.
Column 4 from the right: 6 + 8 + carry 1 = 15. Write 5 and carry 1.
False track of the first addition: A=1, B=2, D=7, E=9, F=3, G=5, H=6, I=4, J=8. It does not extend into any substitution that satisfies the second addition; keeping only an exact first sum would therefore be insufficient.
The solver lists all the deductions and all the injection assignments compatible with the first addition, then tests the second without approximation. Only one complete substitution remains.

Common mistakes and false leads

Attribute a number to a letter forgetting that it is already taken.
Allow an initial zero.
Forget to postpone a deduction or change the value of a letter between the two additions.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

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      "extensions_satisfying_second": 0
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      "extensions_satisfying_second": 0
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}

How we verified this answer

Complete search by columns, binary deductions, rotation of unallocated numbers and arithmetic control of both equals.

{
  "first_equation_solutions": 16,
  "complete_substitutions_checked": 16,
  "column_nodes": 1059,
  "solutions": 1
}

← Return to puzzle SOLO-015

SOLO-016 · Cryptarithm

EXPERT

The Second Addition

Determine the value of each of the ten letters, then write the two additions in digits.

This page contains the solution.

The answer

A = 0, B = 9, C = 7, D = 1, E = 3, F = 8, G = 5, H = 6, I = 4, J = 2

1995 + 5523 = 7518

5748 + 8657 = 14405

The reasoning

Each column carries a hold of 0 or 1. For an addition of two numbers to four digits, a possible fifth position of the result is a final hold, so is 1.
Write a column equation, from right to left: first-term digit + second-time digit + incoming hold = result digit + 10 × outgoing hold. The same letter must keep its value in both additions.
Addition 1, column 1: G + E + r1_0 = F + 10 × r1_1.
Addition 1, column 2: B + J + r1_1 = D + 10 × r1_2.
Addition 1, column 3: B + G + r1_2 = G + 10 × r1_3.
Addition 1, column 4: D + G + r1_3 = C + 10 × r1_4.
Addition 2, column 1: F + C + r2_0 = G + 10 × r2_1.
Addition 2, column 2: I + G + r2_1 = A + 10 × r2_2.
Addition 2, column 3: C + H + r2_2 = I + 10 × r2_3.
Addition 2, column 4: G + F + r2_3 = I + 10 × r2_4.
In these equations, r1_0 and r2_0 are zero. Each other retention is 0 or 1; the same letter always refers to the same number.
Explore possible pairs of digits in the column of units, calculate the restraint and bar any branch giving the same number to two different letters. Repeat for tens, hundreds and thousands; a letter already affected no longer receives any new choice.
The first addition admits 39 partial assignments respecting these constraints. After assigning letters still absent to free numbers, 234 substitutions are checked against the second addition.
Crossing the two additions is essential: the entire substitution must be transported, not separately resolve each line and then juxtapose two incompatible responses.
Addition 1: DBBG + GGJE = CGDF, i.e. 1995 + 5523 = 7518.
Column 1 from the right: 5 + 3 + carry 0 = 8. Write 8 and carry 0.
Column 2 from the right: 9 + 2 + carry 0 = 11. Write 1 and carry 1.
Column 3 from the right: 9 + 5 + carry 1 = 15. Write 5 and carry 1.
Column 4 from the right: 1 + 5 + carry 1 = 7. Write 7 and carry 0.
Addition 2: GCIF + FHGC = DIIAG, 5748 + 8657 = 14405.
Column 1 from the right: 8 + 7 + carry 0 = 15. Write 5 and carry 1.
Column 2 from the right: 4 + 5 + carry 1 = 10. Write 0 and carry 1.
Column 3 from the right: 7 + 6 + carry 1 = 14. Write 4 and carry 1.
Column 4 from the right: 5 + 8 + carry 1 = 14. Write 4 and carry 1.
False track of the first addition: B=9, C=6, D=4, E=2, F=3, G=1, J=5. It does not extend into any substitution that satisfies the second addition; keeping only an exact first sum would therefore be insufficient.
The solver lists all the deductions and all the injection assignments compatible with the first addition, then tests the second without approximation. Only one complete substitution remains.

Common mistakes and false leads

Attribute a number to a letter forgetting that it is already taken.
Allow an initial zero.
Forget to postpone a deduction or change the value of a letter between the two additions.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

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How we verified this answer

Complete search by columns, binary deductions, rotation of unallocated numbers and arithmetic control of both equals.

{
  "first_equation_solutions": 39,
  "complete_substitutions_checked": 234,
  "column_nodes": 1544,
  "solutions": 1
}

← Return to puzzle SOLO-016

SOLO-017 · Cryptarithm

EXTREME

Ten Letters, Two Locks

Determine the value of each of the ten letters, then write the two additions in digits.

This page contains the solution.

The answer

A = 7, B = 0, C = 1, D = 5, E = 4, F = 3, G = 8, H = 9, I = 6, J = 2

2660 + 3903 = 6563

1828 + 4766 = 6594

The reasoning

Each column carries a hold of 0 or 1. For an addition of two numbers to four digits, a possible fifth position of the result is a final hold, so is 1.
Write a column equation, from right to left: first-term digit + second-time digit + incoming hold = result digit + 10 × outgoing hold. The same letter must keep its value in both additions.
Addition 1, column 1: B + F + r1_0 = F + 10 × r1_1.
Addition 1, column 2: I + B + r1_1 = I + 10 × r1_2.
Addition 1, column 3: I + H + r1_2 = D + 10 × r1_3.
Addition 1, column 4: J + F + r1_3 = I + 10 × r1_4.
Addition 2, column 1: G + I + r2_0 = E + 10 × r2_1.
Addition 2, column 2: J + I + r2_1 = H + 10 × r2_2.
Addition 2, column 3: G + A + r2_2 = D + 10 × r2_3.
Addition 2, column 4: C + E + r2_3 = I + 10 × r2_4.
In these equations, r1_0 and r2_0 are zero. Each other retention is 0 or 1; the same letter always refers to the same number.
Explore possible pairs of digits in the column of units, calculate the restraint and bar any branch giving the same number to two different letters. Repeat for tens, hundreds and thousands; a letter already affected no longer receives any new choice.
The first addition admits 102 partial assignments respecting these constraints. After assigning letters still absent to free numbers, 2448 substitutions are controlled against the second addition.
Crossing the two additions is essential: the entire substitution must be transported, not separately resolve each line and then juxtapose two incompatible responses.
Addition 1: JIIB + FHBF = IDIF, i.e. 2660 + 3903 = 6563.
Column 1 from the right: 0 + 3 + carry 0 = 3. Write 3 and carry 0.
Column 2 from the right: 6 + 0 + carry 0 = 6. Write 6 and carry 0.
Column 3 from the right: 6 + 9 + carry 0 = 15. Write 5 and carry 1.
Column 4 from the right: 2 + 3 + carry 1 = 6. Write 6 and carry 0.
Addition 2: CGJG + EAII = IDHE, or 1828 + 4766 = 6594.
Column 1 from the right: 8 + 6 + carry 0 = 14. Write 4 and carry 1.
Column 2 from the right: 2 + 6 + carry 1 = 9. Write 9 and carry 0.
Column 3 from the right: 8 + 7 + carry 0 = 15. Write 5 and carry 1.
Column 4 from the right: 1 + 4 + carry 1 = 6. Write 6 and carry 0.
False track of the first addition: B=0, D=7, F=1, H=4, I=3, J=2. It does not extend into any substitution that satisfies the second addition; keeping only an exact first sum would therefore be insufficient.
The solver lists all the deductions and all the injection assignments compatible with the first addition, then tests the second without approximation. Only one complete substitution remains.

Common mistakes and false leads

Attribute a number to a letter forgetting that it is already taken.
Allow an initial zero.
Forget to postpone a deduction or change the value of a letter between the two additions.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

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        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 8,
        "H": 1,
        "D": 9,
        "J": 3
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 8,
        "H": 3,
        "D": 1,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 8,
        "H": 6,
        "D": 4,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 8,
        "H": 9,
        "D": 7,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 9,
        "H": 2,
        "D": 1,
        "J": 3
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 9,
        "H": 7,
        "D": 6,
        "J": 3
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 5,
        "I": 9,
        "H": 8,
        "D": 7,
        "J": 3
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 7,
        "H": 2,
        "D": 9,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 8,
        "H": 1,
        "D": 9,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 8,
        "H": 4,
        "D": 2,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 8,
        "H": 5,
        "D": 3,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 8,
        "H": 7,
        "D": 5,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 8,
        "H": 9,
        "D": 7,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 9,
        "H": 4,
        "D": 3,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 9,
        "H": 5,
        "D": 4,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 6,
        "I": 9,
        "H": 8,
        "D": 7,
        "J": 2
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 7,
        "I": 9,
        "H": 3,
        "D": 2,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 7,
        "I": 9,
        "H": 4,
        "D": 3,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 7,
        "I": 9,
        "H": 5,
        "D": 4,
        "J": 1
      },
      "extensions_satisfying_second": 0
    },
    {
      "first_addition_mapping": {
        "B": 0,
        "F": 7,
        "I": 9,
        "H": 6,
        "D": 5,
        "J": 1
      },
      "extensions_satisfying_second": 0
    }
  ]
}

How we verified this answer

Complete search by columns, binary deductions, rotation of unallocated numbers and arithmetic control of both equals.

{
  "first_equation_solutions": 102,
  "complete_substitutions_checked": 2448,
  "column_nodes": 682,
  "solutions": 1
}

← Return to puzzle SOLO-017

SOLO-018 · Futoshiki

EXPERT

The Seven Staircases

Complete square 7 × 7 with figures from 1 to 7.

This page contains the solution.

The answer

Grid, line by line:

3 6 4 2 7 5 1

6 2 1 7 4 3 5

7 4 3 1 5 2 6

5 3 7 6 2 1 4

4 1 6 5 3 7 2

1 5 2 3 6 4 7

2 7 5 4 1 6 3

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7
1	3	6	4	2	7	5	1
2	6	2	1	7	4	3	5
3	7	4	3	1	5	2	6
4	5	3	7	6	2	1	4
5	4	1	6	5	3	7	2
6	1	5	2	3	6	4	7
7	2	7	5	4	1	6	3

The reasoning

First enter the candidates from 1 to 7 in each empty box. Remove any numbers already fixed in its line or column. The rows and columns each require a complete permutation.
Owns inequalities in both directions. If A < B, A can no longer take a value greater than or equal to the greatest candidate of B, and B can no longer take a value less than or equal to the smallest candidate of A.
After each reduction, check the unique locations of each digit in each row and column. A growing chain of k boxes consumes at least k separate values; its ends also give bounds.
The reference method combines these reductions and then controlled assumptions. On this problem, the exact search visits 13 knots, encounters 6 contradictions and reaches a depth of 3 nested choices. These numbers describe this solver and not a guaranteed human duration.
A hypothesis is treated as a temporary branch: notes its box and its number, spreads all the constraints, then returns to the last choice if a box loses all its candidates or if a line can no longer contain a number. Do not erase a hypothesis without noting the contradiction.
Here are the first assignments of the path that leads to the solution. "Assumption" means a choice to be confirmed in exploration, not a proven local deduction:
L3C6 = 2 — single location.
L7C5 = 1 — single location.
L3C4 = 1 — single location.
L2C6 = 3 — hypothesis.
L2C7 = 5 — inequality.
L3C7 = 6 — inequality.
L2C4 = 7 — row/column.
L7C4 = 4 — row/column.
L6C4 = 3 — inequality.
L5C4 = 5 — single location.
L1C7 = 1 — hypothesis.
L2C2 = 2 — hypothesis.
L2C3 = 1 — row/column.
L7C2 = 7 — row/column.
L4C6 = 1 — single location.
L4C3 = 7 — single location.
L7C1 = 2 — row/column.
L3C2 = 4 — row/column.
L6C6 = 4 — row/column.
L5C6 = 7 — single location.
L6C1 = 1 — inequality.
L6C5 = 6 — inequality.
L4C2 = 3 — inequality.
L3C1 = 7 — row/column.
L4C5 = 2 — row/column.
L4C1 = 5 — single location.
L6C2 = 5 — row/column.
L6C3 = 2 — row/column.
L1C2 = 6 — row/column.
L5C2 = 1 — row/column.
Example of a rebutted branch: supposes L2C6 = 1, then L1C5 = 3. The propagation results in: The number 5 can no longer be placed in the group L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7 The certificate of exploration gives all branches, with indices of boxes counted from zero.
Example of a rebutted branch: supposes L2C6 = 1, then L1C5 = 7, then L2C3 = 2. The propagation results in: Two identical fixed digits in boxes L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7 The certificate of exploration gives all branches, with indices of boxes counted from zero.
Example of a rebutted branch: supposes L2C6 = 1, then L1C5 = 7, then L2C3 = 7. The propagation results in: L-inequality L6C4 < L7C4 no longer has compatible torque values. The exploration certificate gives all branches, with indices of cells counted from zero.
The final grid allows line-by-line and column-by-column control: each one contains once all the numbers. Then check the inequalities one by one:
L5C1 < L4C1 : 4 < 5.
L6C6 < L5C6 : 4 < 7.
L3C6 < L3C7 : 2 < 6.
L2C6 < L2C7 : 3 < 5.
L6C2 < L7C2 : 5 < 7.
L4C1 < L3C1 : 5 < 7.
L7C7 < L7C6 : 3 < 6.
L5C7 < L6C7 : 2 < 7.
L5C2 < L4C2 : 1 < 3.
L1C4 < L1C5 : 2 < 7.
L1C4 < L2C4 : 2 < 7.
L6C1 < L7C1 : 1 < 2.
L6C4 < L7C4 : 3 < 4.
L6C6 < L6C5 : 4 < 6.
L4C2 < L3C2 : 3 < 4.
L4C5 < L3C5 : 2 < 5.
L2C7 < L3C7 : 5 < 6.
The exact counting continued after finding this grid to look for a second solution. All other branches fail: the solution is unique.

Common mistakes and false leads

Add the Sudoku regions: they are not part of Futoshiki.
Forget that inequality forces its two ends.
Take a search hypothesis for a figure deducted unconditionally.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "indexing": "Case i = ligne i//n+1, colonne i%n+1. Les indices bruts commencent à zéro.",
  "events": [
    {
      "event": "branch",
      "assumptions": [],
      "cell": 12,
      "values": [
        1,
        3
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 12,
          "value": 1
        }
      ],
      "cell": 4,
      "values": [
        3,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 1
        },
        {
          "cell": 4,
          "value": 3
        }
      ],
      "reason": "Le chiffre 5 ne peut plus se placer dans le groupe L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 12,
          "value": 1
        },
        {
          "cell": 4,
          "value": 7
        }
      ],
      "cell": 9,
      "values": [
        2,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 1
        },
        {
          "cell": 4,
          "value": 7
        },
        {
          "cell": 9,
          "value": 2
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 1
        },
        {
          "cell": 4,
          "value": 7
        },
        {
          "cell": 9,
          "value": 7
        }
      ],
      "reason": "L’inégalité L6C4 < L7C4 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        }
      ],
      "cell": 6,
      "values": [
        1,
        4
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 1
        }
      ],
      "cell": 8,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 8,
          "value": 1
        }
      ],
      "reason": "La case L6C3 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "solution",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 8,
          "value": 2
        }
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 4
        }
      ],
      "cell": 8,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 8,
          "value": 1
        }
      ],
      "reason": "La case L4C7 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 12,
          "value": 3
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 8,
          "value": 2
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L1C1, L1C2, L1C3, L1C4, L1C5, L1C6, L1C7"
    }
  ]
}

How we verified this answer

Comprehensive research with different domains, constraints, unique locations and inequalities; independent control of the final grid.

{
  "nodes": 13,
  "branches": 6,
  "contradictions": 6,
  "max_depth": 3,
  "forced": 136,
  "solutions": 1,
  "size": 7,
  "givens": 11,
  "inequalities": 17
}

← Return to puzzle SOLO-018

SOLO-019 · Futoshiki

EXTREME

The Eight Passages

Complete square 8 × 8 with figures from 1 to 8.

This page contains the solution.

The answer

Grid, line by line:

1 5 6 8 2 7 4 3

8 6 2 3 4 5 1 7

6 8 3 2 7 1 5 4

4 7 5 1 6 3 2 8

3 2 4 7 1 6 8 5

2 3 7 4 5 8 6 1

7 4 1 5 8 2 3 6

5 1 8 6 3 4 7 2

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1	1	5	6	8	2	7	4	3
2	8	6	2	3	4	5	1	7
3	6	8	3	2	7	1	5	4
4	4	7	5	1	6	3	2	8
5	3	2	4	7	1	6	8	5
6	2	3	7	4	5	8	6	1
7	7	4	1	5	8	2	3	6
8	5	1	8	6	3	4	7	2

The reasoning

First enter the candidates from 1 to 8 in each empty box. Remove any numbers already fixed in its line or column. The rows and columns each require a complete permutation.
Owns inequalities in both directions. If A < B, A can no longer take a value greater than or equal to the greatest candidate of B, and B can no longer take a value less than or equal to the smallest candidate of A.
After each reduction, check the unique locations of each digit in each row and column. A growing chain of k boxes consumes at least k separate values; its ends also give bounds.
The reference method combines these reductions and then controlled assumptions. On this problem, the exact search visits 43 knots, encounters 21 contradictions and reaches a depth of 9 nested choices. These numbers describe this solver and not a guaranteed human duration.
A hypothesis is treated as a temporary branch: notes its box and its number, spreads all the constraints, then returns to the last choice if a box loses all its candidates or if a line can no longer contain a number. Do not erase a hypothesis without noting the contradiction.
Here are the first assignments of the path that leads to the solution. "Assumption" means a choice to be confirmed in exploration, not a proven local deduction:
L5C4 = 7 — inequality.
L5C1 = 3 — inequality.
L6C1 = 2 — inequality.
L6C4 = 4 — row/column.
L1C4 = 8 — row/column.
L1C3 = 6 — row/column.
L4C8 = 8 — single location.
L1C1 = 1 — hypothesis.
L1C7 = 4 — hypothesis.
L1C8 = 3 — row/column.
L3C2 = 8 — hypothesis.
L8C3 = 8 — single location.
L2C1 = 8 — single location.
L2C2 = 6 — hypothesis.
L4C2 = 7 — row/column.
L6C2 = 3 — row/column.
L7C2 = 4 — row/column.
L7C1 = 7 — row/column.
L7C8 = 6 — row/column.
L8C7 = 7 — single location.
L2C7 = 1 — hypothesis.
L7C7 = 3 — row/column.
L7C3 = 1 — row/column.
L2C6 = 5 — hypothesis.
L2C5 = 4 — hypothesis.
L2C8 = 7 — row/column.
L3C4 = 2 — hypothesis.
L3C7 = 5 — row/column.
L3C6 = 1 — only location.
L3C3 = 3 — single location.
Example of a rebutted branch: supposes L1C1 = 1, then L1C7 = 3, then L3C2 = 7. The propagation results in: The box L3C4 no longer has any candidate after excluding the numbers set in its line/column. The exploration certificate gives all branches, with indices of boxes counted from zero.
Example of a rebutted branch: supposes L1C1 = 1, then L1C7 = 3, then L3C2 = 8, then L2C2 = 6, then L2C8 = 5. The propagation results in: The box L3C5 no longer has any candidate after excluding the numbers set in its line/column. The certificate of exploration gives all branches, with indices of cells counted from scratch.
Example of a rebutted branch: supposes L1C1 = 1, then L1C7 = 3, then L3C2 = 8, then L2C2 = 6, then L2C8 = 7, then L3C4 = 1. The propagation results in: The number 2 can no longer be placed in group L1C7, L2C7, L3C7, L4C7, L5C7, L6C7, L7C7, L8C7 The certificate of exploration gives all branches, with indices of cells counted from scratch.
The final grid allows line-by-line and column-by-column control: each one contains once all the numbers. Then check the inequalities one by one:
L4C1 < L4C2 : 4 < 7.
L5C6 < L5C7 : 6 < 8.
L8C2 < L7C2 : 1 < 4.
L1C5 < L1C4 : 2 < 8.
L6C2 < L6C3 : 3 < 7.
L1C2 < L2C2 : 5 < 6.
L5C1 < L4C1 : 3 < 4.
L4C5 < L3C5 : 6 < 7.
L2C3 < L2C2 : 2 < 6.
L4C3 < L4C2 : 5 < 7.
L6C1 < L6C2 : 2 < 3.
L3C6 < L3C7 : 1 < 5.
L7C3 < L6C3 : 1 < 7.
L6C4 < L5C4 : 4 < 7.
L8C8 < L8C7 : 2 < 7.
L5C2 < L4C2 : 2 < 7.
L4C2 < L3C2 : 7 < 8.
L1C2 < L1C3 : 5 < 6.
L6C1 < L5C1 : 2 < 3.
The exact counting continued after finding this grid to look for a second solution. All other branches fail: the solution is unique.

Common mistakes and false leads

Add the Sudoku regions: they are not part of Futoshiki.
Forget that inequality forces its two ends.
Take a search hypothesis for a figure deducted unconditionally.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "indexing": "Case i = ligne i//n+1, colonne i%n+1. Les indices bruts commencent à zéro.",
  "events": [
    {
      "event": "branch",
      "assumptions": [],
      "cell": 0,
      "values": [
        1,
        4
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        }
      ],
      "cell": 6,
      "values": [
        3,
        4
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        }
      ],
      "cell": 17,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        }
      ],
      "cell": 9,
      "values": [
        6,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        }
      ],
      "cell": 15,
      "values": [
        5,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 5
        }
      ],
      "reason": "La case L3C5 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        }
      ],
      "cell": 19,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 1
        }
      ],
      "reason": "Le chiffre 2 ne peut plus se placer dans le groupe L1C7, L2C7, L3C7, L4C7, L5C7, L6C7, L7C7, L8C7"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        }
      ],
      "cell": 20,
      "values": [
        5,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 5
        }
      ],
      "reason": "La case L7C7 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 7
        }
      ],
      "cell": 22,
      "values": [
        4,
        5
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 22,
          "value": 4
        }
      ],
      "cell": 14,
      "values": [
        1,
        5
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 22,
          "value": 4
        },
        {
          "cell": 14,
          "value": 1
        }
      ],
      "reason": "Le chiffre 3 ne peut plus se placer dans le groupe L1C6, L2C6, L3C6, L4C6, L5C6, L6C6, L7C6, L8C6"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 22,
          "value": 4
        },
        {
          "cell": 14,
          "value": 5
        }
      ],
      "reason": "La case L3C6 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 15,
          "value": 7
        },
        {
          "cell": 19,
          "value": 2
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 22,
          "value": 5
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 3
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        }
      ],
      "cell": 17,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        }
      ],
      "cell": 9,
      "values": [
        6,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        }
      ],
      "cell": 14,
      "values": [
        1,
        5
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        }
      ],
      "cell": 13,
      "values": [
        4,
        5
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 4
        }
      ],
      "cell": 12,
      "values": [
        5,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 12,
          "value": 5
        }
      ],
      "cell": 19,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 12,
          "value": 5
        },
        {
          "cell": 19,
          "value": 1
        }
      ],
      "reason": "Le chiffre 2 ne peut plus se placer dans le groupe L1C7, L2C7, L3C7, L4C7, L5C7, L6C7, L7C7, L8C7"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 12,
          "value": 5
        },
        {
          "cell": 19,
          "value": 2
        }
      ],
      "reason": "La case L4C1 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 12,
          "value": 7
        }
      ],
      "reason": "La case L3C3 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        }
      ],
      "cell": 12,
      "values": [
        4,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 4
        }
      ],
      "cell": 19,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 19,
          "value": 1
        }
      ],
      "reason": "Le chiffre 2 ne peut plus se placer dans le groupe L1C7, L2C7, L3C7, L4C7, L5C7, L6C7, L7C7, L8C7"
    },
    {
      "event": "solution",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 19,
          "value": 2
        }
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 7
        }
      ],
      "cell": 18,
      "values": [
        3,
        5
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 18,
          "value": 3
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 1
        },
        {
          "cell": 13,
          "value": 5
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 18,
          "value": 5
        }
      ],
      "reason": "L’inégalité L3C6 < L3C7 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 14,
          "value": 5
        }
      ],
      "reason": "L’inégalité L4C5 < L3C5 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 1
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        }
      ],
      "cell": 17,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        }
      ],
      "cell": 9,
      "values": [
        6,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        }
      ],
      "cell": 19,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 19,
          "value": 1
        }
      ],
      "reason": "Le chiffre 2 ne peut plus se placer dans le groupe L1C7, L2C7, L3C7, L4C7, L5C7, L6C7, L7C7, L8C7"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 6
        },
        {
          "cell": 19,
          "value": 2
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 17,
          "value": 8
        },
        {
          "cell": 9,
          "value": 7
        }
      ],
      "reason": "La case L3C4 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    }
  ]
}

How we verified this answer

Comprehensive research with different domains, constraints, unique locations and inequalities; independent control of the final grid.

{
  "nodes": 43,
  "branches": 21,
  "contradictions": 21,
  "max_depth": 9,
  "forced": 285,
  "solutions": 1,
  "size": 8,
  "givens": 15,
  "inequalities": 19
}

← Return to puzzle SOLO-019

SOLO-020 · Futoshiki

EXTREME

Columns Under Tension

Complete square 8 × 8 with figures from 1 to 8.

This page contains the solution.

The answer

Grid, line by line:

4 8 6 3 5 2 1 7

7 1 2 5 8 4 6 3

6 3 8 4 7 1 5 2

2 5 1 7 3 6 8 4

8 4 3 6 2 5 7 1

5 2 7 1 6 3 4 8

1 7 5 2 4 8 3 6

3 6 4 8 1 7 2 5

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1	4	8	6	3	5	2	1	7
2	7	1	2	5	8	4	6	3
3	6	3	8	4	7	1	5	2
4	2	5	1	7	3	6	8	4
5	8	4	3	6	2	5	7	1
6	5	2	7	1	6	3	4	8
7	1	7	5	2	4	8	3	6
8	3	6	4	8	1	7	2	5

The reasoning

First enter the candidates from 1 to 8 in each empty box. Remove any numbers already fixed in its line or column. The rows and columns each require a complete permutation.
Owns inequalities in both directions. If A < B, A can no longer take a value greater than or equal to the greatest candidate of B, and B can no longer take a value less than or equal to the smallest candidate of A.
After each reduction, check the unique locations of each digit in each row and column. A growing chain of k boxes consumes at least k separate values; its ends also give bounds.
The reference method combines these reductions and then controlled assumptions. On this problem, the exact search visits 55 knots, encounters 27 contradictions and reaches a depth of 9 nested choices. These numbers describe this solver and not a guaranteed human duration.
A hypothesis is treated as a temporary branch: notes its box and its number, spreads all the constraints, then returns to the last choice if a box loses all its candidates or if a line can no longer contain a number. Do not erase a hypothesis without noting the contradiction.
Here are the first assignments of the path that leads to the solution. "Assumption" means a choice to be confirmed in exploration, not a proven local deduction:
L4C5 = 3 — inequality.
L4C4 = 7 — inequality.
L3C6 = 1 — inequality.
L5C6 = 5 — inequality.
L4C6 = 6 — row/column.
L8C4 = 8 — single location.
L1C4 = 3 — hypothesis.
L3C5 = 7 — hypothesis.
L2C5 = 8 — hypothesis.
L7C6 = 8 — single location.
L1C5 = 5 — hypothesis.
L7C5 = 4 — row/column.
L2C6 = 4 — hypothesis.
L8C6 = 7 — row/column.
L7C3 = 5 — single location.
L6C1 = 5 — single location.
L4C1 = 2 — hypothesis.
L4C3 = 1 — row/column.
L8C2 = 6 — inequality.
L2C1 = 7 — hypothesis.
L2C7 = 6 — single location.
L2C3 = 2 — row/column.
L8C3 = 4 — row/column.
L2C2 = 1 — inequality.
L2C8 = 3 — row/column.
L8C1 = 3 — row/column.
L8C7 = 2 — single location.
L3C1 = 6 — row/column.
L7C1 = 1 — row/column.
L1C1 = 4 — single location.
Example of a rebutted branch: supposes L1C4 = 1. The propagation results in: Two identical fixed digits in boxes L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8 The certificate of exploration gives all branches, with indices of squares counted from zero.
Example of a rebutted branch: supposes L1C4 = 3, then L3C5 = 7, then L2C5 = 4, then L1C8 = 1. The propagation results in: L-inequality L8C3 < L7C3 no longer has compatible torque values. The exploration certificate gives all branches, with indices of squares counted from zero.
Example of a rebutted branch: supposes L1C4 = 3, then L3C5 = 7, then L2C5 = 4, then L1C8 = 7. Propagation results in: The number 4 can no longer be placed in the group L7C1, L7C2, L7C3, L7C4, L7C5, L7C6, L7C7, L7C8 The certificate of exploration gives all branches, with indices of boxes counted from zero.
The final grid allows line-by-line and column-by-column control: each one contains once all the numbers. Then check the inequalities one by one:
L4C5 < L4C4 : 3 < 7.
L8C6 < L7C6 : 7 < 8.
L1C4 < L1C3 : 3 < 6.
L2C3 < L1C3 : 2 < 6.
L2C2 < L2C3 : 1 < 2.
L3C6 < L3C7 : 1 < 5.
L1C7 < L2C7 : 1 < 6.
L6C6 < L6C7 : 3 < 4.
L6C6 < L5C6 : 3 < 5.
L8C1 < L8C2 : 3 < 6.
L8C3 < L7C3 : 4 < 5.
L7C4 < L7C5 : 2 < 4.
L3C1 < L2C1 : 6 < 7.
L6C2 < L5C2 : 2 < 4.
L6C4 < L6C5 : 1 < 6.
L5C6 < L4C6 : 5 < 6.
L6C2 < L7C2 : 2 < 7.
L7C1 < L8C1 : 1 < 3.
L7C5 < L7C6 : 4 < 8.
The exact counting continued after finding this grid to look for a second solution. All other branches fail: the solution is unique.

Common mistakes and false leads

Add the Sudoku regions: they are not part of Futoshiki.
Forget that inequality forces its two ends.
Take a search hypothesis for a figure deducted unconditionally.

Show the full search proof

Each branch is independent. The tree records the deductions and contradictions found with the method used here.

{
  "indexing": "Case i = ligne i//n+1, colonne i%n+1. Les indices bruts commencent à zéro.",
  "events": [
    {
      "event": "branch",
      "assumptions": [],
      "cell": 3,
      "values": [
        1,
        3
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 1
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        }
      ],
      "cell": 20,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        }
      ],
      "cell": 12,
      "values": [
        4,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 4
        }
      ],
      "cell": 7,
      "values": [
        1,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 7,
          "value": 1
        }
      ],
      "reason": "L’inégalité L8C3 < L7C3 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        }
      ],
      "reason": "Le chiffre 4 ne peut plus se placer dans le groupe L7C1, L7C2, L7C3, L7C4, L7C5, L7C6, L7C7, L7C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        }
      ],
      "cell": 4,
      "values": [
        4,
        5
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        }
      ],
      "cell": 6,
      "values": [
        1,
        6
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        }
      ],
      "cell": 7,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        }
      ],
      "cell": 1,
      "values": [
        6,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 6
        }
      ],
      "cell": 0,
      "values": [
        5,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 6
        },
        {
          "cell": 0,
          "value": 5
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C2, L2C2, L3C2, L4C2, L5C2, L6C2, L7C2, L8C2"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 6
        },
        {
          "cell": 0,
          "value": 8
        }
      ],
      "reason": "Le chiffre 7 ne peut plus se placer dans le groupe L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7, L6C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        }
      ],
      "cell": 0,
      "values": [
        5,
        6
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 5
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 6
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C2, L2C2, L3C2, L4C2, L5C2, L6C2, L7C2, L8C2"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 8
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 6,
          "value": 6
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C3, L2C3, L3C3, L4C3, L5C3, L6C3, L7C3, L8C3"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        }
      ],
      "cell": 13,
      "values": [
        4,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 4
        }
      ],
      "cell": 24,
      "values": [
        1,
        2
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 24,
          "value": 1
        }
      ],
      "reason": "Le chiffre 8 ne peut plus se placer dans le groupe L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7, L6C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 24,
          "value": 2
        }
      ],
      "cell": 8,
      "values": [
        6,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 24,
          "value": 2
        },
        {
          "cell": 8,
          "value": 6
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L8C1, L8C2, L8C3, L8C4, L8C5, L8C6, L8C7, L8C8"
    },
    {
      "event": "solution",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 4
        },
        {
          "cell": 24,
          "value": 2
        },
        {
          "cell": 8,
          "value": 7
        }
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 7
        }
      ],
      "cell": 6,
      "values": [
        1,
        4
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 7
        },
        {
          "cell": 6,
          "value": 1
        }
      ],
      "reason": "La case L2C3 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 7
        },
        {
          "cell": 12,
          "value": 8
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 13,
          "value": 7
        },
        {
          "cell": 6,
          "value": 4
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        }
      ],
      "cell": 12,
      "values": [
        4,
        7
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 4
        }
      ],
      "cell": 4,
      "values": [
        5,
        7
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 4,
          "value": 5
        }
      ],
      "reason": "La case L3C3 ne possède plus aucun candidat après exclusion des chiffres fixés dans sa ligne/colonne."
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 4,
          "value": 7
        }
      ],
      "cell": 7,
      "values": [
        1,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 4,
          "value": 7
        },
        {
          "cell": 7,
          "value": 1
        }
      ],
      "reason": "L’inégalité L8C3 < L7C3 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 4
        },
        {
          "cell": 4,
          "value": 7
        },
        {
          "cell": 7,
          "value": 8
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L7C1, L7C2, L7C3, L7C4, L7C5, L7C6, L7C7, L7C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        }
      ],
      "cell": 4,
      "values": [
        4,
        5
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        }
      ],
      "cell": 7,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        }
      ],
      "cell": 1,
      "values": [
        6,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 6
        }
      ],
      "reason": "Le chiffre 7 ne peut plus se placer dans le groupe L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        }
      ],
      "cell": 0,
      "values": [
        5,
        6
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 5
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 6
        }
      ],
      "reason": "Le chiffre 8 ne peut plus se placer dans le groupe L1C3, L2C3, L3C3, L4C3, L5C3, L6C3, L7C3, L8C3"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 4
        },
        {
          "cell": 7,
          "value": 8
        }
      ],
      "reason": "Le chiffre 6 ne peut plus se placer dans le groupe L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        }
      ],
      "cell": 6,
      "values": [
        1,
        4
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        }
      ],
      "cell": 7,
      "values": [
        7,
        8
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        }
      ],
      "cell": 1,
      "values": [
        6,
        8
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 6
        }
      ],
      "reason": "Le chiffre 7 ne peut plus se placer dans le groupe L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        }
      ],
      "cell": 0,
      "values": [
        4,
        6
      ]
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 4
        }
      ],
      "cell": 10,
      "values": [
        2,
        4
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 10,
          "value": 2
        }
      ],
      "reason": "Le chiffre 5 ne peut plus se placer dans le groupe L6C1, L6C2, L6C3, L6C4, L6C5, L6C6, L6C7, L6C8"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 4
        },
        {
          "cell": 10,
          "value": 4
        }
      ],
      "reason": "Le chiffre 1 ne peut plus se placer dans le groupe L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 7
        },
        {
          "cell": 1,
          "value": 8
        },
        {
          "cell": 0,
          "value": 6
        }
      ],
      "reason": "L’inégalité L8C3 < L7C3 ne possède plus de couple de valeurs compatible."
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 1
        },
        {
          "cell": 7,
          "value": 8
        }
      ],
      "reason": "Le chiffre 7 ne peut plus se placer dans le groupe L3C1, L3C2, L3C3, L3C4, L3C5, L3C6, L3C7, L3C8"
    },
    {
      "event": "branch",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 4
        }
      ],
      "cell": 10,
      "values": [
        2,
        4
      ]
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 10,
          "value": 2
        }
      ],
      "reason": "Le chiffre 5 ne peut plus se placer dans le groupe L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    },
    {
      "event": "contradiction",
      "assumptions": [
        {
          "cell": 3,
          "value": 3
        },
        {
          "cell": 20,
          "value": 8
        },
        {
          "cell": 12,
          "value": 7
        },
        {
          "cell": 4,
          "value": 5
        },
        {
          "cell": 6,
          "value": 4
        },
        {
          "cell": 10,
          "value": 4
        }
      ],
      "reason": "Deux chiffres fixés identiques dans les cases L1C1, L2C1, L3C1, L4C1, L5C1, L6C1, L7C1, L8C1"
    }
  ]
}

How we verified this answer

Comprehensive research with different domains, constraints, unique locations and inequalities; independent control of the final grid.

{
  "nodes": 55,
  "branches": 27,
  "contradictions": 27,
  "max_depth": 9,
  "forced": 493,
  "solutions": 1,
  "size": 8,
  "givens": 16,
  "inequalities": 19
}

← Return to puzzle SOLO-020

SOLO-021 · Nonogram

EXPERT

Crossing Intervals

Black out the boxes of this grid 12 × 12 to satisfy all series.

This page contains the solution.

The answer

The grid below is the only solution; ■ = black, · = empty.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10	11	12
1	■	·	■	■	■	■	■	·	·	■	·	·
2	·	■	·	■	■	·	·	·	■	·	■	■
3	■	·	·	·	■	·	■	·	·	·	·	·
4	■	■	■	·	·	■	·	■	·	■	■	■
5	■	■	■	·	·	■	■	·	■	·	■	■
6	·	·	·	■	·	■	·	·	·	·	■	·
7	·	■	■	■	·	·	·	■	·	■	·	■
8	·	·	·	·	■	·	■	■	■	·	·	·
9	■	·	■	■	■	■	■	·	·	·	·	·
10	■	·	·	■	■	·	■	·	■	·	·	■
11	■	■	·	■	·	■	·	·	·	■	■	·
12	·	■	■	·	·	■	·	■	■	·	·	■

The reasoning

Start by listing the investments in each series. Keep only those compatible with the already fixed crosses: any box common to all investments can be fixed.
This intersecting closure is not enough here. Case analyses should be conducted on a line or column, each choice should be passed on to the perpendicular lines and a choice should be rejected as soon as a series can no longer be placed.
The exhaustive control visited 9 states, encountered 6 contradictions and reached 2 levels of assumptions. These measures describe this solver; they do not represent a universal ranking of players.
L1: 20 compatible placements; L1C6 black, L1C7 black.
L4: 5 compatible placements; Black L4C2, Black L4C3, Black L4C10, Black L4C11.
L5: 5 compatible placements; L5C2 black, L5C3 black, L5C6 black, L5C11 black.
C2: 12 compatible placements; L3C2 empty, L6C2 empty.
C3: 8 compatible placements; L3C3 empty, L6C3 empty.
C6: 10 compatible placements. L2C6 is empty.
C7: 20 compatible placements. L2C7 is empty.
C10: 40 compatible placements; L3C10 empty, L5C10 empty.
L2: 9 compatible placements; L2C4 black, L2C11 black.
L5: 4 compatible placements. L5C12 is filled.
C11: 5 compatible placements; L1C11 empty, L3C11 empty, L6C11 black, L7C11 empty.
L6: 16 compatible investments; L6C10 empty, L6C12 empty.
C12: 8 compatible placements; L3C12 empty, black L4C12.
L4: 4 compatible placements. L4C9 is empty.
Case analysis on L4 : placement 3/4 = ■^^··^^^^^^.
C4: 12 compatible placements. L10C4 is filled.
Case analysis on L5 : placement 3/4 = ■^^^^^^^^^^^.
C4: 6 compatible placements; L7C4 black, L11C4 black.
C7: 7 compatible placements. L6C7 is empty.
C8: 17 compatible placements. L2C8 is empty.
C9: 22 compatible placements. L6C9 is empty.
L2: 3 compatible placements; L2C9 black, L2C10 empty, L2C12 black.
L7: 11 compatible placements. L7C1 is empty.
C1: 9 compatible placements. L10C1 is filled.
C9: 4 compatible placements; L1C9 empty, L3C9 empty.
C12: 4 compatible placements. L1C12 is empty.
L1: 3 compatible placements; L1C4 black, L1C5 black, L1C10 black.
L10: 14 compatible placements. L10C2 is empty.
C2: 6 compatible placements; black L11C2, black L12C2.
C4: 3 compatible placements. L3C4 is empty.
C5: 5 compatible placements; L2C5 black, L3C5 black.
L2: 2 compatible placements. L2C3 is empty.
L3: 2 compatible placements; L3C1 black, L3C6 empty.
L11: 10 compatible placements; L11C1 black, L11C3 empty, L11C5 empty.
C1: 2 compatible placements; black L1C1, empty L2C1, empty L6C1, empty L8C1.
C3: 3 compatible placements; L1C3 black, L12C3 black.
C5: 3 compatible placements; L8C5 black, L12C5 empty.
C6: 3 compatible placements; black L6C6, empty L7C6, black L11C6.
L1: 1 compatible placements; L1C2 empty, L1C8 empty.
L2: 1 compatible placements. L2C2 is filled.
L6: 2 compatible placements. L6C5 is empty.
L7: 6 compatible placements; L7C3 black, L7C12 black.
L11: 4 compatible placements. L11C7 is empty.
L12: 4 compatible placements; L12C1 empty, L12C4 empty, L12C9 black.
C1: 1 compatible placements. L9C1 is filled.
C3: 2 compatible placements. L8C3 is empty.
C4: 1 compatible placements; L6C4 black, L8C4 empty, L9C4 black.
C5: 2 compatible placements. L9C5 is filled.
C7: 2 compatible placements; L3C7 black, L8C7 black, L9C7 black, L12C7 empty.
C8: 10 compatible placements. L3C8 is empty.
C9: 3 compatible placements. L11C9 is empty.
C12: 3 compatible placements. L8C12 is empty.
L6: 1 compatible placements. L6C8 is empty.
L8: 2 compatible placements; L8C10 empty, L8C11 empty.
L9: 2 compatible placements; L9C2 empty, L9C6 black, L9C9 empty, L9C10 empty, L9C11 empty, L9C12 empty.
L11: 2 compatible placements; L11C8 empty, L11C11 black.
L12: 3 compatible placements. L12C6 is filled.
C6: 1 compatible placements; L8C6 empty, L10C6 empty.
C9: 2 compatible placements. L10C9 is filled.
C11: 1 compatible placements; L10C11 empty, L12C11 empty.
C12: 1 compatible placements; L10C12 black, L11C12 empty, L12C12 black.
L8: 1 compatible placements; L8C2 empty, L8C8 black, L8C9 black.
L10: 2 compatible placements; L10C7 black, L10C8 empty, L10C10 empty.
L11: 1 compatible placements. L11C10 is filled.
C2: 1 compatible placements. L7C2 is filled.
C7: 1 compatible placements. L7C7 is empty.
C8: 2 compatible placements. L12C8 is filled.
C9: 1 compatible placements. L7C9 is empty.
C10: 1 compatible placements; L7C10 black, L12C10 empty.
L7: 1 compatible placements; L7C5 empty, L7C8 black.
C5: 1 compatible placements. L10C5 is filled.
C8: 1 compatible placements. L9C8 is empty.
L9: 1 compatible placements. L9C3 is filled.
L10: 1 compatible placements. L10C3 is empty.
Examples of contradictions encountered:
C7 has no legal placement left at depth 1.
L2 has no legal placement left at depth 1.
C5 has no legal placement left at depth 2.
C5 has no legal placement left at depth 2.
C3 has no legal placement left at depth 2.
L12 has no legal placement left at depth 1.

Common mistakes and false leads

Do not confuse an unknown box with an empty proven box.
A hypothesis must be cancelled with all its consequences if a series becomes impossible.
The form suggested by the first black boxes is not evidence.

How we verified this answer

Enumeration of placements by line, propagation by intersections and exhaustive search for up to two solutions; new verification of response series.

{
  "nodes": 9,
  "contradictions": 6,
  "depth": 2,
  "forced_cells": 366,
  "branches": 2
}

← Return to puzzle SOLO-021

SOLO-022 · Nonogram

EXPERT

The Suspended Weave

Black out the boxes of this grid 12 × 12 to satisfy all series.

This page contains the solution.

The answer

The grid below is the only solution; ■ = black, · = empty.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10	11	12
1	■	■	·	·	·	■	■	·	■	·	·	■
2	·	·	■	·	·	■	·	·	■	·	·	■
3	·	■	■	■	■	·	·	·	·	■	·	■
4	■	■	·	·	·	■	■	■	·	■	■	■
5	■	■	·	■	■	·	■	■	·	·	·	·
6	■	■	■	·	·	■	·	·	■	■	·	·
7	■	■	■	■	·	·	■	·	■	·	·	·
8	■	■	■	■	·	·	·	·	·	■	·	■
9	·	■	■	■	■	·	■	·	·	·	■	■
10	·	·	·	·	·	·	■	·	·	■	■	·
11	■	■	■	■	■	·	■	■	·	·	■	·
12	■	■	■	·	·	·	·	■	·	■	·	·

The reasoning

Start by listing the investments in each series. Keep only those compatible with the already fixed crosses: any box common to all investments can be fixed.
This intersecting closure is not enough here. Case analyses should be conducted on a line or column, each choice should be passed on to the perpendicular lines and a choice should be rejected as soon as a series can no longer be placed.
The exhaustive control visited 14 states, encountered 8 contradictions and reached 5 levels of assumptions. These measures describe this solver; they do not represent a universal ranking of players.
L4: 10 compatible placements; L4C6 black, L4C10 black.
L9: 20 compatible placements. L9C4 is filled.
L11: 10 compatible placements; Black L11C3, Black L11C4, Black L11C5.
C1: 10 compatible placements; black L5C1, black L6C1, black L7C1.
C2: 1 compatible placements; black L1C2, empty L2C2, black L3C2, black L4C2, black L5C2, black L6C2, black L7C2, black L8C2, black L9C2, black L10C2, empty L11C2, black L12C2.
C3: 9 compatible placements; L6C3 black, L7C3 black.
C4: 6 compatible placements; L6C4 empty, L7C4 black, L8C4 black, L10C4 empty, L12C4 empty.
C5: 35 compatible placements; L10C5 empty, L12C5 empty.
C7: 15 compatible placements. L10C7 is filled.
C10: 11 compatible placements. L5C10 is empty.
L3: 25 compatible placements; L3C3 black, L3C4 black.
L5: 8 compatible placements. L5C3 is empty.
L7: 15 compatible placements. L7C5 is empty.
L8: 25 compatible placements. L8C3 is filled.
L9: 16 compatible placements. L9C3 is filled.
L12: 15 compatible placements; black L12C1, black L12C3.
C1: 6 compatible placements; L10C1 empty, L11C1 black.
C3: 2 compatible placements; L1C3 empty, L10C3 empty.
C4: 2 compatible placements; L2C4 empty, L4C4 empty.
L1: 20 compatible placements. L1C1 is filled.
L4: 6 compatible placements; L4C7 black, L4C11 black.
L10: 3 compatible placements; L10C6 empty, L10C8 empty.
L11: 6 compatible placements. L11C6 is empty.
C1: 3 compatible placements. L2C1 is empty.
Case analysis on C3 : placement 1/2 = * * * * * * * * * * *
L4: 3 compatible placements. L4C1 is filled.
C1: 2 compatible placements. L9C1 is empty.
L9: 6 compatible placements; L9C5 black, L9C6 empty.
C5: 10 compatible placements. L8C5 is empty.
L8: 15 compatible placements. L8C1 is filled.
C1: 1 compatible placements. L3C1 is empty.
L3: 10 compatible placements; L3C5 black, L3C6 empty.
C5: 3 compatible placements; L2C5 empty, L4C5 empty.
L4: 1 compatible placements; L4C8 black, L4C9 empty, L4C12 black.
Case analysis on C4 : placement 2/2 = * * * * * * * * * * *
L5: 3 compatible placements; L5C5 black, L5C6 empty.
C5: 1 compatible investments; L1C5 empty, L6C5 empty.
C6: 4 compatible placements; L1C6 black, L2C6 black.
L1: 3 compatible placements; L1C7 black, L1C8 empty.
L2: 6 compatible placements. L2C7 is empty.
C8: 7 compatible placements. L2C8 is empty.
Case analysis on L1: placement 2/3 = ■·····················································································································································································································································································································
C9: 7 compatible placements; L2C9 black, L3C9 empty.
C10: 1 compatible placements; L2C10 empty, L3C10 black, L6C10 black, L7C10 empty, L8C10 black, L9C10 empty, L10C10 black, L11C10 empty, L12C10 black.
C11: 6 compatible placements. L3C11 is empty.
C12: 6 compatible placements; L2C12 black, L3C12 black, L5C12 empty.
L2: 1 compatible placements. L2C11 is empty.
L3: 1 compatible placements; L3C7 empty, L3C8 empty.
L5: 2 compatible placements; L5C8 black, L5C11 empty.
L6: 5 compatible placements. L6C12 is empty.
L8: 4 compatible placements; L8C9 empty, L8C11 empty.
L9: 3 compatible placements; L9C11 black, L9C12 black.
L10: 2 compatible placements. L10C12 is empty.
L11: 4 compatible placements. L11C8 is filled.
L12: 4 compatible placements; L12C9 empty, L12C11 empty.
C7: 3 compatible placements; L5C7 black, L6C7 empty, L11C7 black.
C8: 1 compatible placements; L6C8 empty, L7C8 empty, L8C8 empty, L9C8 empty, L12C8 black.
C11: 1 compatible placements; L6C11 empty, L7C11 empty, L10C11 black, L11C11 black.
C12: 1 compatible placements; L7C12 empty, L8C12 black, L11C12 empty, L12C12 empty.
L5: 1 compatible placements. L5C9 is empty.
L6: 1 compatible placements; L6C6 black, L6C9 black.
L7: 2 compatible placements. L7C9 is filled.
L8: 1 compatible placements; L8C6 empty, L8C7 empty.
L10: 1 compatible placements. L10C9 is empty.
L11: 1 compatible placements. L11C9 is empty.
L12: 1 compatible placements; L12C6 empty, L12C7 empty.
C6: 1 compatible placements. L7C6 is empty.
C7: 1 compatible placements; L7C7 black, L9C7 black.
C9: 1 compatible placements. L9C9 is empty.
Examples of contradictions encountered:
C10 has no legal placement left at depth 2.
L9 has no legal placement left at depth 3.
C10 has no legal placement left at depth 4.
C12 has no legal placement left at depth 5.
L7 has no legal placement left at depth 5.
C7 has no legal placement left at depth 5.
L7 has no legal placement left at depth 4.
L1 has no legal placement left at depth 1.

Common mistakes and false leads

Do not confuse an unknown box with an empty proven box.
A hypothesis must be cancelled with all its consequences if a series becomes impossible.
The form suggested by the first black boxes is not evidence.

How we verified this answer

Enumeration of placements by line, propagation by intersections and exhaustive search for up to two solutions; new verification of response series.

{
  "nodes": 14,
  "contradictions": 8,
  "depth": 5,
  "forced_cells": 375,
  "branches": 5
}

← Return to puzzle SOLO-022

SOLO-023 · Nonogram

EXTREME

The Fragmented Grid

Black out the boxes of this 16 × 16 grid to satisfy all series.

This page contains the solution.

The answer

The grid below is the only solution; ■ = black, · = empty.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
1	·	■	·	·	·	·	■	·	·	·	·	■	■	·	·	·
2	■	·	■	■	·	■	■	·	·	·	■	·	■	·	■	·
3	·	■	■	·	·	·	·	·	·	·	·	■	·	■	■	·
4	■	·	·	·	·	■	·	■	■	·	·	■	■	·	■	■
5	·	■	■	·	■	·	·	·	·	■	·	■	·	·	■	■
6	·	·	·	·	■	·	·	·	■	·	■	■	■	·	·	■
7	■	■	■	·	·	■	■	·	·	■	■	■	·	■	■	■
8	·	·	·	■	■	·	·	·	·	·	■	■	·	·	■	·
9	·	·	■	·	■	■	■	·	■	·	■	■	■	■	·	·
10	·	■	·	·	·	■	·	·	■	■	■	·	■	·	■	■
11	·	·	·	·	■	·	·	■	·	·	■	·	■	■	·	·
12	·	■	·	·	■	·	·	■	■	■	·	·	·	■	■	·
13	■	·	·	·	·	■	■	·	■	■	·	■	·	■	■	■
14	·	■	·	·	■	·	■	·	·	■	·	■	·	■	■	·
15	■	·	·	·	·	·	·	■	■	·	■	■	·	■	·	■
16	■	■	■	·	■	·	·	·	■	■	■	·	■	■	■	·

The reasoning

Start by listing the investments in each series. Keep only those compatible with the already fixed crosses: any box common to all investments can be fixed.
This intersecting closure is not enough here. Case analyses should be conducted on a line or column, each choice should be passed on to the perpendicular lines and a choice should be rejected as soon as a series can no longer be placed.
The exhaustive control visited 52 states, encountered 34 contradictions and reached 7 levels of assumptions. These measures describe this solver; they do not represent a universal ranking of players.
L7: 15 compatible placements; L7C3 black, L7C10 black, L7C14 black.
L10: 35 compatible placements; L10C1 empty, L10C3 empty.
L13: 31 compatible placements. L13C14 is filled.
C2: 5 compatible placements; L9C2 empty, L11C2 empty, L12C2 black, L13C2 empty, L14C2 black, L15C2 empty, L16C2 black.
C3: 74 compatible placements. L8C3 is empty.
C10: 62 compatible placements. L6C10 is empty.
C11: 56 compatible placements. L8C11 is filled.
C12: 20 compatible placements; L6C12 black, L7C12 black, L8C12 black, L9C12 black.
C14: 23 compatible placements; L6C14 empty, L8C14 empty, L11C14 black, L12C14 black, L14C14 black.
C15: 6 compatible placements; L2C15 black, L3C15 black, L4C15 black, L7C15 black, L12C15 black, L13C15 black.
L2: 21 compatible placements; L2C14 empty, L2C16 empty.
L7: 6 compatible placements; L7C9 empty, L7C11 black, L7C13 empty, L7C16 black.
L8: 12 compatible placements; L8C10 empty, L8C13 empty.
L9: 25 compatible placements. L9C16 is empty.
L11: 155 compatible placements. L11C16 is empty.
L12: 15 compatible placements; L12C1 empty, L12C3 empty, L12C13 empty, L12C16 empty.
L14: 20 compatible placements; L14C1 empty, L14C3 empty, L14C16 empty.
L16: 30 compatible placements. L16C3 is filled.
C1: 113 compatible placements; L15C1 black, L16C1 black.
C3: 10 compatible placements; L6C3 empty, L15C3 empty.
C13: 4 compatible placements; L1C13 black, L2C13 black, L3C13 empty, L4C13 black, L5C13 empty, L6C13 black, L9C13 black, L10C13 black, L11C13 black.
C16: 4 compatible placements; L1C16 empty, L3C16 empty, L5C16 black, L6C16 black, L10C16 black, L13C16 black.
L1: 81 compatible placements. L1C15 is empty.
L2: 15 compatible placements. L2C12 is empty.
L3: 45 compatible placements. L3C14 is filled.
L4: 20 compatible placements; L4C11 empty, L4C12 black, L4C14 empty, L4C16 black.
L5: 70 compatible placements; L5C14 empty, L5C15 black.
L6: 22 compatible placements; L6C11 black, L6C15 empty.
L10: 10 compatible placements; L10C12 empty, L10C14 empty, L10C15 black.
L11: 63 compatible placements; L11C12 empty, L11C15 empty.
L13: 11 compatible placements. L13C13 is empty.
L16: 20 compatible placements. L16C4 is empty.
C11: 21 compatible placements; L9C11 black, L10C11 black.
C12: 3 compatible placements; L1C12 black, L3C12 black, L5C12 black, L14C12 black.
C14: 2 compatible placements; L15C14 black, L16C14 black.
C15: 1 compatible placements; L8C15 black, L9C15 empty, L14C15 black, L15C15 empty, L16C15 black.
C16: 2 compatible placements. L8C16 is empty.
L1: 36 compatible investments; L1C11 empty, L1C14 empty.
L3: 9 compatible placements. L3C11 is empty.
L5: 35 compatible placements. L5C11 is empty.
L9: 10 compatible placements. L9C5 is filled.
L10: 4 compatible placements; L10C8 empty, L10C9 black, L10C10 black.
L14: 10 compatible placements; L14C11 empty, L14C13 empty.
L15: 25 compatible placements. L15C13 is empty.
C9: 17 compatible placements. L8C9 is empty.
C11: 1 compatible placements; L2C11 black, L11C11 black, L12C11 empty, L13C11 empty, L15C11 black, L16C11 black.
C13: 1 compatible placements. L16C13 is filled.
C14: 1 compatible placements. L9C14 is filled.
L2: 10 compatible placements. L2C10 is empty.
L9: 7 compatible placements. L9C10 is empty.
L11: 22 compatible placements. L11C10 is empty.
L12: 6 compatible placements; L12C8 black, L12C12 empty.
L16: 3 compatible placements; L16C8 empty, L16C9 black, L16C10 black, L16C12 empty, L16C16 empty.
C9: 12 compatible placements; L14C9 empty, L15C9 black.
C10: 3 compatible placements; L12C10 black, L13C10 black, L14C10 black, L15C10 empty.
C12: 1 compatible placements; L13C12 black, L15C12 black.
C16: 1 compatible placements. L15C16 is filled.
L12: 3 compatible placements; L12C7 empty, L12C9 black.
L13: 4 compatible placements; L13C8 empty, L13C9 black.
L15: 1 compatible placements; L15C4 empty, L15C5 empty, L15C6 empty, L15C7 empty, L15C8 black.
C7: 71 compatible placements. L16C7 is empty.
C8: 9 compatible placements; L11C8 black, L14C8 empty.
C9: 10 compatible placements; L9C9 black, L11C9 empty.
L9: 4 compatible placements. L9C8 is empty.
L11: 5 compatible placements. L11C7 is empty.
Case analysis on L16: placement 1/2 = ■^^^^··^^^^^^^··.
Case analysis on L12: placement 2/3 = * * * * * * * * * *
C5: 14 compatible placements. L13C5 is empty.
L13: 2 compatible placements; L13C6 black, L13C7 black.
C6: 35 compatible placements. L14C6 is empty.
C7: 35 compatible placements. L14C7 is filled.
Case analysis on L13: placement 1/2 =·······················································································································································································································································································································
Case analysis on L14: placement 2/2 = ■·····················································································································································································································································································································
C5: 5 compatible placements; L7C5 empty, L8C5 black, L10C5 empty, L11C5 black.
L7: 4 compatible placements; L7C2 black, L7C7 black.
L8: 2 compatible placements; L8C1 empty, L8C2 empty, L8C7 empty, L8C8 empty.
L11: 1 compatible placements; L11C1 empty, L11C3 empty, L11C4 empty, L11C6 empty.
C2: 1 compatible placements; black L1C2, empty L2C2, black L3C2, empty L4C2, black L5C2, empty L6C2.
C3: 3 compatible placements; L2C3 black, L9C3 black.
C7: 11 compatible placements. L6C7 is empty.
L1: 7 compatible placements; L1C1 empty, L1C3 empty.
L3: 2 compatible placements; L3C4 empty, L3C5 empty, L3C6 empty, L3C7 empty, L3C8 empty, L3C9 empty, L3C10 empty.
C3: 1 compatible placements; L3C3 black, L4C3 empty, L5C3 black.
C6: 8 compatible placements. L9C6 is filled.
L3: 1 compatible placements. L3C1 is empty.
L5: 10 compatible placements; L5C1 empty, L5C4 empty.
L9: 1 compatible placements; L9C1 empty, L9C4 empty, L9C7 black.
C1: 2 compatible placements; L2C1 black, L4C1 black.
C7: 2 compatible placements. L10C7 is empty.
L2: 3 compatible placements; L2C4 black, L2C5 empty.
C4: 5 compatible placements. L1C4 is empty.
C5: 2 compatible placements; L1C5 empty, L5C5 black.
L5: 4 compatible placements. L5C6 is empty.
C6: 6 compatible placements. L4C6 is filled.
L4: 2 compatible placements. L4C5 is empty.
C5: 1 compatible placements. L6C5 is filled.
L6: 3 compatible placements; L6C4 empty, L6C6 empty.
C6: 2 compatible placements; black L7C6, empty L8C6, black L10C6.
L7: 2 compatible placements. L7C8 is empty.
L8: 1 compatible placements. L8C4 is filled.
L10: 1 compatible placements. L10C4 is empty.
C4: 1 compatible placements; L4C4 empty, L7C4 empty.
L4: 1 compatible placements; L4C7 empty, L4C8 black, L4C9 black.
L7: 1 compatible placements. L7C1 is filled.
C1: 1 compatible placements. L6C1 is empty.
C7: 1 compatible placements; L1C7 black, L2C7 black, L5C7 empty.
C8: 1 compatible placements; L1C8 empty, L2C8 empty, L5C8 empty, L6C8 empty.
C9: 3 compatible placements. L5C9 is empty.
L1: 1 compatible placements; L1C6 empty, L1C9 empty, L1C10 empty.
L2: 1 compatible placements; L2C6 black, L2C9 empty.
L5: 1 compatible placements. L5C10 is filled.
L6: 1 compatible placements. L6C9 is filled.
Examples of contradictions encountered:
L7 has no legal placement left at depth 3.
C5 has no legal placement left at depth 4.
L2 has no legal placement left at depth 6.
C3 has no legal placement left at depth 6.
C5 has no legal placement left at depth 6.
L10 has no legal placement left at depth 7.
C5 has no legal placement left at depth 7.
C5 has no legal placement left at depth 7.
C1 has no legal placement left at depth 5.
L2 has no legal placement left at depth 6.
C3 has no legal placement left at depth 6.
C7 has no legal placement left at depth 6.
L10 has no legal placement left at depth 7.
C7 has no legal placement left at depth 7.
C8 has no legal placement left at depth 7.
C7 has no legal placement left at depth 4.
C3 has no legal placement left at depth 6.
C3 has no legal placement left at depth 6.
C5 has no legal placement left at depth 6.
L10 has no legal placement left at depth 7.

Common mistakes and false leads

Do not confuse an unknown box with an empty proven box.
A hypothesis must be cancelled with all its consequences if a series becomes impossible.
The form suggested by the first black boxes is not evidence.

How we verified this answer

Enumeration of placements by line, propagation by intersections and exhaustive search for up to two solutions; new verification of response series.

{
  "nodes": 52,
  "contradictions": 34,
  "depth": 7,
  "forced_cells": 1687,
  "branches": 17
}

← Return to puzzle SOLO-023

SOLO-024 · Nonogram

EXTREME

The Hidden Border

Black out the boxes of this 16 × 16 grid to satisfy all series.

This page contains the solution.

The answer

The grid below is the only solution; ■ = black, · = empty.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
1	■	·	■	■	■	·	■	■	■	·	■	·	·	·	·	■
2	·	·	■	■	■	■	■	■	■	■	■	·	·	·	■	·
3	·	■	■	·	·	·	■	·	■	·	■	·	■	■	■	·
4	·	■	■	·	■	·	■	·	■	·	■	■	■	·	■	■
5	■	■	·	·	■	■	·	■	·	·	·	■	·	·	■	■
6	·	■	■	■	■	■	·	·	■	■	■	·	■	■	·	·
7	■	■	·	■	■	■	·	·	■	■	·	·	■	■	·	■
8	■	·	·	■	■	·	■	■	■	■	■	·	·	·	■	■
9	■	·	■	■	·	·	·	■	·	■	·	■	·	·	·	·
10	■	·	■	·	·	■	·	·	■	■	·	·	·	■	■	■
11	·	■	■	·	·	·	·	·	·	■	■	·	■	·	·	·
12	·	·	·	·	·	■	·	■	·	■	·	·	■	·	■	■
13	·	■	·	■	■	·	■	·	■	■	·	·	■	·	·	·
14	·	■	·	■	·	■	·	·	·	■	·	■	■	■	·	·
15	·	·	·	·	·	·	·	■	·	·	·	·	■	■	■	·
16	·	·	·	·	·	■	·	·	·	■	■	■	·	·	·	■

The reasoning

Start by listing the investments in each series. Keep only those compatible with the already fixed crosses: any box common to all investments can be fixed.
This intersecting closure is not enough here. Case analyses should be conducted on a line or column, each choice should be passed on to the perpendicular lines and a choice should be rejected as soon as a series can no longer be placed.
The comprehensive control visited 29 states, encountered 16 contradictions and reached 10 levels of assumptions. These measures describe this solver; they do not represent a universal ranking of players.
L2: 21 compatible placements; L2C6 black, L2C7 black, L2C8 black, L2C9 black.
L4: 7 compatible placements; L4C2 black, L4C11 black, L4C12 black, L4C15 black.
L6: 23 compatible placements. L6C5 is filled.
L7: 21 compatible placements. L7C6 is filled.
L8: 12 compatible placements; L8C4 black, L8C7 black, L8C8 black, L8C9 black, L8C10 black.
C2: 74 compatible placements. L5C2 is filled.
C6: 6 compatible placements; L1C6 empty, L3C6 empty, L4C6 empty, L6C6 black.
C7: 15 compatible placements; L3C7 black, L4C7 black, L7C7 empty, L9C7 empty.
C9: 35 compatible placements; L3C9 black, L4C9 black.
C10: 20 compatible placements; L6C10 black, L7C10 black, L9C10 black, L10C10 black, L11C10 black.
L1: 20 compatible placements; L1C1 black, L1C2 empty, L1C3 black, L1C4 black, L1C5 black.
L3: 24 compatible placements; L3C8 empty, L3C10 empty.
L4: 3 compatible placements; L4C8 empty, L4C10 empty, L4C13 black, L4C14 empty, L4C16 black.
L6: 12 compatible placements; L6C1 empty, L6C4 black.
L7: 4 compatible placements; L7C1 black, L7C2 black, L7C3 empty, L7C4 black, L7C5 black, L7C8 empty, L7C13 black.
L9: 78 compatible placements. L9C11 is empty.
L11: 39 compatible placements. L11C12 is empty.
C1: 8 compatible placements. L2C1 is empty.
C2: 25 compatible placements; L2C2 empty, L6C2 black.
C3: 22 compatible placements; L2C3 black, L3C3 black, L4C3 black, L5C3 empty.
C4: 11 compatible placements; black L2C4, empty L3C4, empty L4C4.
C5: 18 compatible placements; L2C5 black, L3C5 empty, L8C5 black.
C8: 24 compatible placements. L1C8 is filled.
C10: 6 compatible placements; L12C10 black, L13C10 black.
L1: 16 compatible placements. L1C9 is filled.
L2: 4 compatible placements; L2C10 black, L2C11 black, L2C12 empty.
L3: 3 compatible placements; L3C1 empty, L3C2 black, L3C14 black, L3C15 black.
L4: 1 compatible placements; L4C1 empty, L4C5 black.
L5: 70 compatible placements. L5C1 is filled.
L6: 5 compatible placements; L6C3 black, L6C7 empty, L6C8 empty.
L8: 6 compatible placements; L8C3 empty, L8C6 empty, L8C11 black.
C1: 1 compatible placements; L8C1 black, L9C1 black, L10C1 black, L11C1 empty, L12C1 empty, L13C1 empty, L14C1 empty, L15C1 empty, L16C1 empty.
C2: 15 compatible placements. L8C2 is empty.
C5: 7 compatible placements; L5C5 black, L9C5 empty.
C6: 5 compatible placements. L5C6 is filled.
C9: 23 compatible placements. L5C9 is empty.
C10: 3 compatible placements. L1C10 is empty.
C11: 15 compatible placements; L1C11 black, L3C11 black, L5C11 empty, L7C11 empty.
C12: 74 compatible placements. L1C12 is empty.
C14: 56 compatible investments; L1C14 empty, L2C14 empty.
L1: 3 compatible placements. L1C7 is filled.
L3: 2 compatible placements. L3C12 is empty.
L5: 10 compatible placements; L5C4 empty, L5C7 empty.
L7: 3 compatible placements. L7C9 is filled.
L9: 18 compatible placements. L9C2 is empty.
L10: 39 compatible placements. L10C2 is empty.
L12: 86 compatible placements. L12C9 is empty.
C4: 5 compatible placements; L9C4 black, L10C4 empty.
C12: 29 compatible placements; L5C12 black, L6C12 empty.
L6: 3 compatible placements. L6C9 is filled.
L9: 17 compatible placements; L9C3 black, L9C9 empty.
C3: 1 compatible placements; L10C3 black, L11C3 black, L12C3 empty, L13C3 empty, L14C3 empty, L15C3 empty, L16C3 empty.
Case analysis on L3 : placement 1/2 = * * * * * * * * * * *
C13: 7 compatible placements; L1C13 empty, L2C13 empty, L5C13 empty, L12C13 black, L13C13 black.
L5: 4 compatible placements. L5C15 is filled.
L12: 25 compatible placements. L12C11 is empty.
C15: 50 compatible placements. L1C15 is empty.
L1: 1 compatible placements. L1C16 is filled.
C16: 15 compatible placements; L2C16 empty, L5C16 black, L6C16 empty.
L2: 1 compatible placements. L2C15 is filled.
L5: 2 compatible placements. L5C14 is empty.
L6: 2 compatible placements. L6C14 is filled.
C14: 15 compatible placements; L7C14 black, L8C14 empty.
C15: 35 compatible placements. L6C15 is empty.
L6: 1 compatible placements. L6C13 is filled.
L7: 1 compatible placements; L7C12 empty, L7C15 empty, L7C16 black.
L8: 1 compatible placements; L8C13 empty, L8C15 black, L8C16 black.
C15: 10 compatible placements. L9C15 is empty.
C16: 10 compatible placements. L9C16 is empty.
Case analysis on L5: placement 1/2 = ■······················································································································································································································································································································
C8: 10 compatible placements; L9C8 black, L10C8 empty.
C10: 1 compatible placements; L14C10 black, L15C10 empty, L16C10 black.
L10: 15 compatible placements. L10C14 is filled.
L14: 33 compatible placements. L14C9 is empty.
C14: 4 compatible placements; L9C14 empty, L11C14 empty.
Case analysis on L9: placement 2/3 = '·················.
C6: 1 compatible placement; L10C6 black, L11C6 empty, L12C6 black, L13C6 empty, L14C6 black, L15C6 empty, L16C6 black.
C12: 6 compatible placements. L10C12 is empty.
C13: 3 compatible placements. L14C13 is filled.
L10: 4 compatible placements; L10C5 empty, L10C7 empty, L10C15 black.
L11: 8 compatible placements. L11C5 is empty.
L12: 9 compatible placements; L12C5 empty, L12C7 empty.
L14: 6 compatible placements; L14C5 empty, L14C7 empty, L14C11 empty, L14C14 black, L14C16 empty.
C14: 2 compatible placements; L12C14 empty, L16C14 empty.
C15: 6 compatible placements. L11C15 is empty.
Case analysis on C14: placement 2/2 =······················································································································································································································································································································
Case analysis on C5 : placement 1/3 = ■^^^^^^^^·············································································································································································································································································································
Case analysis on C13: placement 2/3 = * * * * * * * * * * *
L10: 2 compatible placements. L10C16 is filled.
L11: 2 compatible placements; L11C7 empty, L11C8 empty, L11C11 black, L11C16 empty.
L15: 11 compatible placements. L15C16 is empty.
C11: 3 compatible placements. L10C11 is empty.
C16: 2 compatible placements. L16C16 is filled.
L10: 1 compatible placements. L10C9 is filled.
L16: 3 compatible placements; L16C2 empty, L16C4 empty, L16C7 empty, L16C15 empty.
C2: 3 compatible placements. L14C2 is filled.
Case analysis on L11: placement 1/2 = '···················
C2: 2 compatible placements. L12C2 is empty.
Case analysis on C2 : placement 1/2 = * * * * * * * * * * * *
L13: 5 compatible placements. L13C4 is filled.
C4: 2 compatible placements. L15C4 is empty.
Case analysis on C4: placement 2/2 = ■·····················································································································································································································································································································
L12: 1 compatible placements; L12C8 black, L12C12 empty, L12C15 black, L12C16 black.
L14: 2 compatible placements. L14C8 is empty.
C8: 2 compatible placements. L13C8 is empty.
C15: 2 compatible placements. L13C15 is empty.
C16: 1 compatible placements. L13C16 is empty.
L13: 2 compatible placements; L13C7 black, L13C12 empty.
C7: 1 compatible placements. L15C7 is empty.
C12: 1 compatible placements; L14C12 black, L15C12 empty, L16C12 black.
L14: 1 compatible placements. L14C15 is empty.
L15: 3 compatible placements. L15C15 is filled.
L16: 1 compatible placements; L16C8 empty, L16C9 empty, L16C11 black.
C8: 1 compatible placements. L15C8 is filled.
C11: 1 compatible placements; L13C11 empty, L15C11 empty.
L13: 1 compatible placements. L13C9 is filled.
L15: 1 compatible placements. L15C9 is empty.
Examples of contradictions encountered:
C15 has no legal placement left at depth 4.
C16 has no legal placement left at depth 5.
C15 has no legal placement left at depth 5.
C12 has no legal placement left at depth 5.
C15 has no legal placement left at depth 4.
C16 has no legal placement left at depth 6.
L13 has no legal placement left at depth 10.
C12 has no legal placement left at depth 10.
C4 has no legal placement left at depth 8.
C8 has no legal placement left at depth 7.
C16 has no legal placement left at depth 6.
C4 has no legal placement left at depth 5.
L14 has no legal placement left at depth 5.
C14 has no legal placement left at depth 3.
L10 has no legal placement left at depth 2.
C15 has no legal placement left at depth 1.

Common mistakes and false leads

Do not confuse an unknown box with an empty proven box.
A hypothesis must be cancelled with all its consequences if a series becomes impossible.
The form suggested by the first black boxes is not evidence.

How we verified this answer

Enumeration of placements by line, propagation by intersections and exhaustive search for up to two solutions; new verification of response series.

{
  "nodes": 29,
  "contradictions": 16,
  "depth": 10,
  "forced_cells": 658,
  "branches": 12
}

← Return to puzzle SOLO-024

SOLO-025 · Hamiltonian path

EXPERT

The Silent Checkpoints

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

This page contains the solution.

The answer

The full numbering below determines the only compatible route.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6
1	1	2	3	4	5	6
2	12	11	10	9	8	7
3	13	20	21	22	23	24
4	14	19	28	27	26	25
5	15	18	29	32	33	36
6	16	17	30	31	34	35

The reasoning

From marker 1 to marker 8, the portion retained is: 1 (L1C1) → 2 (L1C2) → 3 (L1C3) → 4 (L1C4) → 5 (L1C5) → 6 (L1C6) → 7 (L2C6) → 8 (L2C5). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 8 to marker 15, the portion retained is: 8 (L2C5) → 9 (L2C4) → 10 (L2C3) → 11 (L2C2) → 12 (L2C1) → 13 (L3C1) → 14 (L4C1) → 15 (L5C1). Check each orthogonal displacement, then mark the boxes already consumed before continuing.
From marker 15 to marker 18, the portion retained is: 15 (L5C1) → 16 (L6C1) → 17 (L6C2) → 18 (L5C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 18 to marker 28, the portion retained is: 18 (L5C2) → 19 (L4C2) → 20 (L3C2) → 21 (L3C3) → 22 (L3C4) → 23 (L3C5) → 24 (L3C6) → 25 (L4C6) → 26 (L4C5) → 27 (L4C4) → 28 (L4C3). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 28 to marker 33, the portion retained is: 28 (L4C3) → 29 (L5C3) → 30 (L6C3) → 31 (L6C4) → 32 (L5C4) → 33 (L5C5). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 33 to marker 35, the portion retained is: 33 (L5C5) → 34 (L6C5) → 35 (L6C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 35 to marker 36, the portion retained is: 35 (L6C6) → 36 (L5C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
The exploration of all compatible continuations gives 613 states. It rejects 46 separations of the remaining cells and 221 violations of distance or parity. Only one complete visit remains.
To reproduce this proof on paper, starting from 1, try each free neighbour in the order high, low, left, right, and keep a tree. Immediately reject a path that reaches a marker in the wrong rank, isolate an unvisited area or cannot reach the next marker in as many steps as required. The last complete path is that displayed; the validator provided performs exactly this exploration.

Common mistakes and false leads

Local continuity is not enough: a detour can cut the remaining boxes in half.
Two boxes of the same color on a checker require an even number of movements; check this parity between markers.
Reaching a landmark too early is as forbidden as missing it.

How we verified this answer

DFS comprehensive simple paths, with necessary controls of connection, benchmarks, distance and parity; comparison of all solutions up to two.

{
  "nodes": 613,
  "dead_ends": 6,
  "connectivity_rejections": 46,
  "distance_rejections": 221
}

← Return to puzzle SOLO-025

SOLO-026 · Hamiltonian path

EXPERT

The One-Way Crossing

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

This page contains the solution.

The answer

The full numbering below determines the only compatible route.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6
1	29	30	31	32	33	36
2	28	27	26	25	34	35
3	21	22	23	24	1	2
4	20	19	18	17	16	3
5	11	12	13	14	15	4
6	10	9	8	7	6	5

The reasoning

From marker 1 to marker 4, the portion retained is: 1 (L3C5) → 2 (L3C6) → 3 (L4C6) → 4 (L5C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 4 to marker 6, the portion retained is: 4 (L5C6) → 5 (L6C6) → 6 (L6C5). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 6 to marker 9, the portion retained is: 6 (L6C5) → 7 (L6C4) → 8 (L6C3) → 9 (L6C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 9 to marker 12, the portion retained is: 9 (L6C2) → 10 (L6C1) → 11 (L5C1) → 12 (L5C2). Check each orthogonal displacement, then mark the boxes already consumed before continuing.
From marker 12 to marker 21, the portion retained is: 12 (L5C2) → 13 (L5C3) → 14 (L5C4) → 15 (L5C5) → 16 (L4C5) → 17 (L4C4) → 18 (L4C3) → 19 (L4C2) → 20 (L4C1) → 21 (L3C1). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 21 to marker 28, the portion retained is: 21 (L3C1) → 22 (L3C2) → 23 (L3C3) → 24 (L3C4) → 25 (L2C4) → 26 (L2C3) → 27 (L2C2) → 28 (L2C1). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 28 to marker 36, the portion retained is: 28 (L2C1) → 29 (L1C1) → 30 (L1C2) → 31 (L1C3) → 32 (L1C4) → 33 (L1C5) → 34 (L2C5) → 35 (L2C6) → 36 (L1C6). Check each orthogonal shift, then mark the boxes already consumed before continuing.
The exploration of all compatible continuations gives 423 states. It rejects 93 separations of the remaining cells and 83 violations of distance or parity. Only one complete visit remains.
To reproduce this proof on paper, starting from 1, try each free neighbour in the order high, low, left, right, and keep a tree. Immediately reject a path that reaches a marker in the wrong rank, isolate an unvisited area or cannot reach the next marker in as many steps as required. The last complete path is that displayed; the validator provided performs exactly this exploration.

Common mistakes and false leads

Local continuity is not enough: a detour can cut the remaining boxes in half.
Two boxes of the same color on a checker require an even number of movements; check this parity between markers.
Reaching a landmark too early is as forbidden as missing it.

How we verified this answer

DFS comprehensive simple paths, with necessary controls of connection, benchmarks, distance and parity; comparison of all solutions up to two.

{
  "nodes": 423,
  "dead_ends": 17,
  "connectivity_rejections": 93,
  "distance_rejections": 83
}

← Return to puzzle SOLO-026

SOLO-027 · Hamiltonian path

EXTREME

The Maze Without Walls

Number all boxes from 1 to 36 to form a single orthogonal path without repetition.

This page contains the solution.

The answer

The full numbering below determines the only compatible route.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6
1	5	6	7	8	9	10
2	4	23	22	21	20	11
3	3	24	25	26	19	12
4	2	1	28	27	18	13
5	35	36	29	30	17	14
6	34	33	32	31	16	15

The reasoning

From marker 1 to marker 17, the portion retained is: 1 (L4C2) → 2 (L4C1) → 3 (L3C1) → 4 (L2C1) → 5 (L1C1) → 6 (L1C2) → 7 (L1C3) → 8 (L1C4) → 9 (L1C5) → 10 (L1C6) → 11 (L2C6) → 12 (L3C6) → 13 (L4C6) → 14 (L5C6) → 15 (L6C6) → 16 (L6C5) → 17 (L5C5). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 17 to marker 23, the portion retained is: 17 (L5C5) → 18 (L4C5) → 19 (L3C5) → 20 (L2C5) → 21 (L2C4) → 22 (L2C3) → 23 (L2C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 23 to marker 32, the portion retained is: 23 (L2C2) → 24 (L3C2) → 25 (L3C3) → 26 (L3C4) → 27 (L4C4) → 28 (L4C3) → 29 (L5C3) → 30 (L5C4) → 31 (L6C4) → 32 (L6C3). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 32 to marker 33, the portion retained is: 32 (L6C3) → 33 (L6C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
From marker 33 to marker 36, the portion retained is: 33 (L6C2) → 34 (L6C1) → 35 (L5C1) → 36 (L5C2). Check each orthogonal shift, then mark the boxes already consumed before continuing.
The exploration of all compatible continuations gives 5682 states. It rejects 1042 separations of the remaining cells and 856 violations of distance or parity. Only one complete visit remains.
To reproduce this proof on paper, starting from 1, try each free neighbour in the order high, low, left, right, and keep a tree. Immediately reject a path that reaches a marker in the wrong rank, isolate an unvisited area or cannot reach the next marker in as many steps as required. The last complete path is that displayed; the validator provided performs exactly this exploration.

Common mistakes and false leads

Local continuity is not enough: a detour can cut the remaining boxes in half.
Two boxes of the same color on a checker require an even number of movements; check this parity between markers.
Reaching a landmark too early is as forbidden as missing it.

How we verified this answer

DFS comprehensive simple paths, with necessary controls of connection, benchmarks, distance and parity; comparison of all solutions up to two.

{
  "nodes": 5682,
  "dead_ends": 102,
  "connectivity_rejections": 1042,
  "distance_rejections": 856
}

← Return to puzzle SOLO-027

SOLO-028 · Pentomino tiling

EXPERT

The Eight Shapes

Pave the rectangle 5 × 8 completely with the 8 pentominos given, without overlap.

This page contains the solution.

The answer

The paving below is unique for the indicated pieces and clues. A letter identifies an entire pentomino.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8
1	I	I	I	I	I	X	L	L
2	V	V	V	W	X	X	X	L
3	V	Z	Z	W	W	X	T	L
4	V	Z	P	P	W	W	T	L
5	Z	Z	P	P	P	T	T	T

The reasoning

The silhouettes are defined by the squares bearing their letter. The dots are not part of the room. Build possible orientations: four rotations, then their mirror, by removing the duplicates.
Start with the boxes bearing an index. For each orientation of the piece concerned, check that it covers all the indices of its letter and no indication of another letter. Then choose an empty box with the least possible investments.
Place I on L1C1, L1C2, L1C3, L1C4, L1C5. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place L on L1C7, L1C8, L2C8, L3C8, L4C8. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place T on L3C7, L4C7, L5C6, L5C7, L5C8. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place X on L1C6, L2C5, L2C6, L2C7, L3C6. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place W on L2C4, L3C4, L3C5, L4C5, L4C6. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place P on L4C3, L4C4, L5C3, L5C4, L5C5. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place Z on L3C2, L3C3, L4C2, L5C1, L5C2. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place V on L2C1, L2C2, L2C3, L3C1, L4C1. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
The proof of uniqueness explores 980 subpaves and eliminates 634 dead ends where a free box no longer accepts any remaining parts. Only one complete cover remains.
To independently check the answer, count five boxes per letter, compare each group to its close rotation/turn silhouette, and then check the clues. To prove uniqueness on paper, take the placement tree by the most constrained box and bar each branch that leaves an isolated box or impossible to cover. The validation script performs this exhaustive exploration; the displayed groups give the winning path.

Common mistakes and false leads

A mirror is allowed, but not a partial displacement of only one box.
Two groups of the same range are not necessarily of the same form.
A cavity of five cells may remain unfilled if its shape does not correspond to any remaining room.

How we verified this answer

Accurate coverage of boxes and coin identities, all free orientations; counting up to two solutions and independent verification of groups of five boxes.

{
  "nodes": 980,
  "dead_ends": 634,
  "max_depth": 8
}

← Return to puzzle SOLO-028

SOLO-029 · Pentomino tiling

EXTREME

The Locked Mosaic

Completely paste the rectangle 5 × 10 with the given 10 pentominos, without overlap.

This page contains the solution.

The answer

The paving below is unique for the indicated pieces and clues. A letter identifies an entire pentomino.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10
1	L	L	X	U	U	W	Y	Y	Y	Y
2	L	X	X	X	U	W	W	Y	Z	Z
3	L	T	X	U	U	F	W	W	Z	V
4	L	T	N	N	F	F	F	Z	Z	V
5	T	T	T	N	N	N	F	V	V	V

The reasoning

The silhouettes are defined by the squares bearing their letter. The dots are not part of the room. Build possible orientations: four rotations, then their mirror, by removing the duplicates.
Start with the boxes bearing an index. For each orientation of the piece concerned, check that it covers all the indices of its letter and no indication of another letter. Then choose an empty box with the least possible investments.
Place F on L3C6, L4C5, L4C6, L4C7, L5C7. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any room already laid.
Place N on L4C3, L4C4, L5C4, L5C5, L5C6. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place T on L3C2, L4C2, L5C1, L5C2, L5C3. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place L on L1C1, L1C2, L2C1, L3C1, L4C1. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place X on L1C3, L2C2, L2C3, L2C4, L3C3. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any room already laid.
Place U on L1C4, L1C5, L2C5, L3C4, L3C5. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place W on L1C6, L2C6, L2C7, L3C7, L3C8. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place Y on L1C7, L1C8, L1C9, L1C10, L2C8. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place V on L3C10, L4C10, L5C8, L5C9, L5C10. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place Z on L2C9, L2C10, L3C9, L4C8, L4C9. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any room already laid.
The proof of uniqueness explores 6105 subpaves and eliminates 4131 dead ends where a free box no longer accepts any remaining parts. Only one complete cover remains.
To independently check the answer, count five boxes per letter, compare each group to its close rotation/turn silhouette, and then check the clues. To prove uniqueness on paper, take the placement tree by the most constrained box and bar each branch that leaves an isolated box or impossible to cover. The validation script performs this exhaustive exploration; the displayed groups give the winning path.

Common mistakes and false leads

A mirror is allowed, but not a partial displacement of only one box.
Two groups of the same range are not necessarily of the same form.
A cavity of five cells may remain unfilled if its shape does not correspond to any remaining room.

How we verified this answer

Accurate coverage of boxes and coin identities, all free orientations; counting up to two solutions and independent verification of groups of five boxes.

{
  "nodes": 6105,
  "dead_ends": 4131,
  "max_depth": 10
}

← Return to puzzle SOLO-029

SOLO-030 · Pentomino tiling

EXTREME

The Stubborn Rectangle

Completely paste the rectangle 5 × 10 with the given 10 pentominos, without overlap.

This page contains the solution.

The answer

The paving below is unique for the indicated pieces and clues. A letter identifies an entire pentomino.

Coordinates: lines from top to bottom, columns from left to right.
L / C	1	2	3	4	5	6	7	8	9	10
1	F	F	Z	Z	P	P	V	V	V	I
2	L	F	F	Z	P	P	P	N	V	I
3	L	F	X	Z	Z	W	T	N	V	I
4	L	X	X	X	W	W	T	N	N	I
5	L	L	X	W	W	T	T	T	N	I

The reasoning

The silhouettes are defined by the squares bearing their letter. The dots are not part of the room. Build possible orientations: four rotations, then their mirror, by removing the duplicates.
Start with the boxes bearing an index. For each orientation of the piece concerned, check that it covers all the indices of its letter and no indication of another letter. Then choose an empty box with the least possible investments.
Place T on L3C7, L4C7, L5C6, L5C7, L5C8. Verify that these five boxes form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place I on L1C10, L2C10, L3C10, L4C10, L5C10. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place N on L2C8, L3C8, L4C8, L4C9, L5C9. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place V on L1C7, L1C8, L1C9, L2C9, L3C9. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place P on L1C5, L1C6, L2C5, L2C6, L2C7. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place F on L1C1, L1C2, L2C2, L2C3, L3C2. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place L on L2C1, L3C1, L4C1, L5C1, L5C2. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
Place Z on L1C3, L1C4, L2C4, L3C4, L3C5. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place W on L3C6, L4C5, L4C6, L5C4, L5C5. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any rooms already laid.
Place X on L3C3, L4C2, L4C3, L4C4, L5C3. Verify that these five cells form a rotation or mirror of its silhouette and do not cover any parts already laid.
The proof of uniqueness explores 13455 subpaves and eliminates 8821 dead ends where a free box no longer accepts any remaining parts. Only one complete cover remains.
To independently check the answer, count five boxes per letter, compare each group to its close rotation/turn silhouette, and then check the clues. To prove uniqueness on paper, take the placement tree by the most constrained box and bar each branch that leaves an isolated box or impossible to cover. The validation script performs this exhaustive exploration; the displayed groups give the winning path.

Common mistakes and false leads

A mirror is allowed, but not a partial displacement of only one box.
Two groups of the same range are not necessarily of the same form.
A cavity of five cells may remain unfilled if its shape does not correspond to any remaining room.

How we verified this answer

Accurate coverage of boxes and coin identities, all free orientations; counting up to two solutions and independent verification of groups of five boxes.

{
  "nodes": 13455,
  "dead_ends": 8821,
  "max_depth": 10
}

← Return to puzzle SOLO-030

SOLO-031 · Othello / Reversi

EXPERT

The Final Tempo

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

Black must play E8: differential guaranteed +6.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defence: E1: +2; E2: -4; A6: +2; A7: -4; B7: -8; C8: -10; E8: +6. The maximum is unique: E8.
After E8, all the opposing legal answers must be dealt with. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp that was to play at the start:
Adverse response E1: N E2, B C8, N A7, B B7, N A8, B A6. End: 37 black, 27 white; initial differential +10.
Adverse response E2: N E1, B B7, N A8, B C8, N A7, B A6. End: 40 black, 24 white; initial differential +16.
Adverse response A6: N C8, B B7, N A7, B A8, N E2, B E1. End: 35 black, 29 white; initial differential +6.
Adverse response B7: N A6, B C8, N E2, B A7, N A8, B E1. End: 35 black, 29 white; initial differential +6.
Adverse response C8: N E1, B E2, N A7, B B7, N A8, B A6. End: 37 black, 27 white; initial differential +10.
The opponent retains the answer giving the smallest of these differentials: 6. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Complete main line: N E8, B A6, N C8, B B7, N A7, B A8, N E2, B E1. Final check: 35 blacks and 29 whites.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 8,
  "legal_first_moves": 7,
  "first_move_scores": {
    "E1": 2,
    "E2": -4,
    "A6": 2,
    "A7": -4,
    "B7": -8,
    "C8": -10,
    "E8": 6
  },
  "states": 12145,
  "history": [
    "E6",
    "F6",
    "D3",
    "D6",
    "F5",
    "C3",
    "E7",
    "F8",
    "D7",
    "F4",
    "G4",
    "G6",
    "G5",
    "H4",
    "B2",
    "C4",
    "G7",
    "D8",
    "G3",
    "F2",
    "B4",
    "B3",
    "A2",
    "G8",
    "H8",
    "C5",
    "H6",
    "A1",
    "C7",
    "B8",
    "C2",
    "H5",
    "H3",
    "C1",
    "F3",
    "H2",
    "B1",
    "F7",
    "C6",
    "A3",
    "F1",
    "H7",
    "H1",
    "D2",
    "G2",
    "E3",
    "D1",
    "B6",
    "A5",
    "B5",
    "A4",
    "G1"
  ],
  "player": 1,
  "best_score": 6,
  "principal_variation": [
    "N E8",
    "B A6",
    "N C8",
    "B B7",
    "N A7",
    "B A8",
    "N E2",
    "B E1"
  ]
}

← Return to puzzle SOLO-031

SOLO-032 · Othello / Reversi

EXPERT

The Deceptive Corner

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

Black must play G1: differential guaranteed +24.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defense: G1: +24; A3: +22; H3: +14; E7: +6; F7: +0; E8: +6; G8: -4. The maximum is unique: G1.
After G1, all opposing legal responses must be addressed. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp that was to play at the start:
Adverse response H3: N A3, B E8, N F8, B passes, N E7, B F7, N G8. End: 44 black, 20 white; initial differential +24.
Adverse response E8: N H3, B passes, N A3, B passes, N F8, B passes, N E7, B F7, N G8. Finish: 46 black, 18 white; initial differential +28.
The opponent holds the answer giving the smallest of these differentials: 24. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Full main line: N G1, B H3, N A3, B E8, N F8, B pass, N E7, B F7, N G8. Final check: 44 black and 20 white. The blanks are not assigned to anyone.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 8,
  "legal_first_moves": 7,
  "first_move_scores": {
    "G1": 24,
    "A3": 22,
    "H3": 14,
    "E7": 6,
    "F7": 0,
    "E8": 6,
    "G8": -4
  },
  "states": 5711,
  "history": [
    "D3",
    "E3",
    "F4",
    "C5",
    "D2",
    "G3",
    "C4",
    "E2",
    "G4",
    "B5",
    "C2",
    "C1",
    "G2",
    "H2",
    "E1",
    "F3",
    "D6",
    "D1",
    "G5",
    "F1",
    "A6",
    "C6",
    "F6",
    "B3",
    "B1",
    "C7",
    "B6",
    "H4",
    "A4",
    "C3",
    "B2",
    "F2",
    "B8",
    "F5",
    "E6",
    "H5",
    "G6",
    "D7",
    "B7",
    "A1",
    "D8",
    "A7",
    "A8",
    "H6",
    "H1",
    "A2",
    "H7",
    "H8",
    "A5",
    "C8",
    "B4",
    "G7"
  ],
  "player": 1,
  "best_score": 24,
  "principal_variation": [
    "N G1",
    "B H3",
    "N A3",
    "B E8",
    "N F8",
    "B passe",
    "N E7",
    "B F7",
    "N G8"
  ]
}

← Return to puzzle SOLO-032

SOLO-033 · Othello / Reversi

EXPERT

Mobility in Reserve

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

White must play H1: differential guaranteed +16.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defence: H1: +16; A2: -20; A3: -16; G7: +0; A8: -12; C8: -22; G8: +2. The maximum is unique: H1.
After H1, all the opposing legal answers must be processed. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp which was to play at the start:
Adverse response H4: B A8, N G7, B G8, N D8, B A3, N A2, B C8. End: 24 black, 40 white; initial differential +16.
Opposite response G7: B H4, N D8, B A8, N pass, B C8, N pass, B G8, N pass, B A3, N A2. Finish: 22 black, 42 white; initial differential +20.
Adverse response C8: B H4, N D8, B A8, N G7, B G8, N pass, B A3, N pass, B A2. Finish: 15 black, 49 white; initial differential +34.
Adverse response D8: B H4, N G7, B A8, N pass, B C8, N pass, B G8, N pass, B A3, N A2. End: 22 black, 42 white; initial differential +20.
The opponent retains the answer giving the smallest of these differentials: 16. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Complete main line: B H1, N H4, B A8, N G7, B G8, N D8, B A3, N A2, B C8. Final check: 24 black and 40 white. The blank boxes are not assigned to anyone.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 9,
  "legal_first_moves": 7,
  "first_move_scores": {
    "H1": 16,
    "A2": -20,
    "A3": -16,
    "G7": 0,
    "A8": -12,
    "C8": -22,
    "G8": 2
  },
  "states": 16489,
  "history": [
    "D3",
    "E3",
    "F3",
    "C3",
    "C5",
    "B5",
    "D6",
    "C7",
    "F5",
    "F4",
    "G5",
    "F2",
    "D2",
    "D1",
    "B6",
    "A7",
    "B8",
    "G6",
    "B3",
    "E6",
    "F7",
    "H5",
    "B4",
    "G3",
    "C1",
    "B1",
    "B7",
    "E2",
    "F1",
    "C6",
    "C4",
    "C2",
    "D7",
    "A5",
    "F6",
    "G1",
    "H3",
    "G2",
    "E1",
    "G4",
    "A1",
    "A6",
    "H6",
    "E7",
    "H2",
    "A4",
    "H7",
    "E8",
    "B2",
    "H8",
    "F8"
  ],
  "player": -1,
  "best_score": 16,
  "principal_variation": [
    "B H1",
    "N H4",
    "B A8",
    "N G7",
    "B G8",
    "N D8",
    "B A3",
    "N A2",
    "B C8"
  ]
}

← Return to puzzle SOLO-033

SOLO-034 · Othello / Reversi

EXTREME

The Defense Tree

Black plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

Black must play H7: differential guaranteed +12.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defense: D1: -10; E2: -14; A3: -14; A6: +4; A7: +4; H7: +12. The maximum is unique: H7.
After H7, all the opposing legal answers must be dealt with. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp which was to play at the start:
Adverse response D1: N A3, B E2, N A7, B A6, N G2, B G1, N F1, B passes, N F2. Finish: 45 black, 19 white; initial differential +26.
Adverse response E2: N D1, B A3, N F2, B F1, N A6, B A7, N G1, B G2. Finish: 38 black, 26 white; initial differential +12.
Opposite response G2: N A3, B D1, N E2, B F2, N G1, B A6, N F1, B passes, N A7. Finish: 45 black, 19 white; initial differential +26.
Adverse response A3: N D1, B E2, N A6, B G2, N G1, B A7, N F2, B passes, N F1. Finish: 40 black, 24 white; initial differential +16.
Adverse response A6: N A3, B F2, N D1, B E2, N A7, B G2, N G1, B F1. Finish: 44 black, 20 white; initial differential +24.
The adversary retains the answer giving the smallest of these differentials: 12. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Complete main line: N H7, B E2, N D1, B A3, N F2, B F1, N A6, B A7, N G1, B G2. Final check: 38 blacks and 26 whites.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 10,
  "legal_first_moves": 6,
  "first_move_scores": {
    "D1": -10,
    "E2": -14,
    "A3": -14,
    "A6": 4,
    "A7": 4,
    "H7": 12
  },
  "states": 67481,
  "history": [
    "F5",
    "F6",
    "E6",
    "D6",
    "C4",
    "C3",
    "F7",
    "G5",
    "D7",
    "G7",
    "H5",
    "H4",
    "B2",
    "F4",
    "H3",
    "D3",
    "G3",
    "C6",
    "D2",
    "C2",
    "H6",
    "B5",
    "H8",
    "G6",
    "C7",
    "A1",
    "A4",
    "D8",
    "B3",
    "F3",
    "C1",
    "H2",
    "C8",
    "C5",
    "E8",
    "E7",
    "B1",
    "A5",
    "E3",
    "G4",
    "B6",
    "B8",
    "B7",
    "G8",
    "H1",
    "A8",
    "B4",
    "A2",
    "F8",
    "E1"
  ],
  "player": 1,
  "best_score": 12,
  "principal_variation": [
    "N H7",
    "B E2",
    "N D1",
    "B A3",
    "N F2",
    "B F1",
    "N A6",
    "B A7",
    "N G1",
    "B G2"
  ]
}

← Return to puzzle SOLO-034

SOLO-035 · Othello / Reversi

EXTREME

Paths to a Majority

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

White must play H7: differential guaranteed +16.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defense: G3: +12; G5: +8; H7: +16; C8: +10; G8: +12. The maximum is unique: H7.
After H7, all the opposing legal answers must be dealt with. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp which was to play at the start:
Adverse response A1: B G5, N B1, B A2, N G3, B C8, N H8, B G8, N H3, B H2. Finish: 19 black, 45 white; initial differential +26.
Adverse response B1: B A2, N A1, B G5, N G3, B C8, N H8, B G8, N H3, B H2. End: 20 black, 44 white; initial differential +24.
Adverse response A2: B B1, N A1, B G5, N G3, B C8, N H8, B G8, N H3, B H2. Finish: 19 black, 45 white; initial differential +26.
Adverse response H2: B H3, N G5, B G3, N H8, B C8, N A1, B G8, N B1, B A2. End: 20 black, 44 white; initial differential +24.
Adverse response H3: B H2, N H8, B C8, N G5, B G8, N A1, B G3, N B1, B A2. Finish: 15 black, 49 white; initial differential +34.
Opposite response G5: B C8, N H2, B G3, N H3, B G8, N A1, B B1, N A2, B passes, N H8. End: 24 black, 40 white; initial differential +16.
Adverse response H8: B G5, N G3, B C8, N H3, B H2, N A1, B G8, N B1, B A2. End: 20 black, 44 white; initial differential +24.
The opponent retains the answer giving the smallest of these differentials: 16. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Complete main line: B H7, N G5, B C8, N H2, B G3, N H3, B G8, N A1, B B1, N A2, B passes, N H8. Final check: 24 blacks and 40 whites.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 11,
  "legal_first_moves": 5,
  "first_move_scores": {
    "G3": 12,
    "G5": 8,
    "H7": 16,
    "C8": 10,
    "G8": 12
  },
  "states": 440058,
  "history": [
    "D3",
    "C5",
    "B6",
    "C3",
    "E3",
    "B5",
    "F6",
    "G7",
    "C6",
    "D2",
    "B4",
    "F4",
    "F3",
    "A5",
    "D1",
    "E1",
    "E6",
    "C7",
    "B7",
    "C1",
    "G6",
    "D7",
    "A7",
    "D6",
    "C4",
    "H6",
    "A6",
    "F2",
    "E8",
    "D8",
    "C2",
    "F8",
    "E7",
    "A8",
    "B8",
    "F5",
    "B3",
    "B2",
    "G4",
    "E2",
    "G1",
    "H4",
    "A4",
    "G2",
    "F1",
    "A3",
    "F7",
    "H1",
    "H5"
  ],
  "player": -1,
  "best_score": 16,
  "principal_variation": [
    "B H7",
    "N G5",
    "B C8",
    "N H2",
    "B G3",
    "N H3",
    "B G8",
    "N A1",
    "B B1",
    "N A2",
    "B passe",
    "N H8"
  ]
}

← Return to puzzle SOLO-035

SOLO-036 · Othello / Reversi

EXTREME

The Last Word

White plays. Find his only best first shot and the final differential that he can guarantee against any optimal defense.

This page contains the solution.

The answer

White must play A8: guaranteed differential +6.

The reasoning

Start by establishing the legal blows by looking, in each of the eight directions, for an unempty adverse suite ended with a pawn of your color. A blow in a corner is not automatically the best.
Each sheet is evaluated by the number of pawns of the camp at the beginning minus those of the other camp. At a node where this camp plays, one keeps the maximum; at an opposing node, the minimum. A pass changes the line without consuming any box; two camps without legal blow complete the part.
The exact values of the first shots, after the best defense: A1: -6; C1: +4; A2: -30; A8: +6; B8: -2; F8: -6. The maximum is unique: A8.
After A8, you have to deal with all the opposing legal answers. Here is an optimal continuation of the two camps (N = Black, B = White); the score is always brought back to the camp that was to play at the start:
Adverse response C1: B B8, N F8, B A1, N H1, B A2, N G1, B H2, N H4, B H5. End: 29 black, 35 white; initial differential +6.
Opposite response G1: B B8, N C1, B A1, N A2, B H1, N H2, B H4, N F8, B H5. End: 22 black, 42 white; initial differential +20.
Adverse response H2: B B8, N F8, B A2, N H4, B H5, N pass, B A1, N pass, B C1, N pass, B G1, N pass, B H1. End: 16 black, 48 white; initial differential +32.
Adverse response H4: B B8, N F8, B H5, N G1, B A1, N A2, B C1, N pass, B H1, N H2. End: 24 black, 40 white; initial differential +16.
Adverse response H5: B B8, N H4, B A1, N F8, B A2, N G1, B C1, N pass, B H1, N H2. End: 27 black, 37 white; initial differential +10.
Adverse response B8: B F8, N H1, B C1, N A1, B A2, N G1, B H2, N H5, B H4. End: 28 black, 36 white; initial differential +8.
Adverse response F8: B B8, N H5, B H4, N C1, B A1, N G1, B H1, N H2, B A2. Finish: 26 black, 38 white; initial differential +12.
The opponent retains the answer giving the smallest of these differentials: 6. The other answers therefore do not refute the first blow; they obtain an equal or better result for you.
Complete main line: B A8, N C1, B B8, N F8, B A1, N H1, B A2, N G1, B H2, N H4, B H5. Final check: 29 blacks and 35 whites. The blanks are not assigned to anyone.
Technique: Minimal retrograde calculation, mobility and pass management, then comparison of filling orders. Note for each branch available blows rather than assuming an uninterrupted alternation. The table of values is an exhaustive search-assisted proof; continuations alone would not prove that the omitted defences are inferior.

Common mistakes and false leads

Count only the pawns returned by the next shot.
Forget a diagonal or a mandatory pass.
Consolidate a main line with evidence against all defences.

How we verified this answer

Legal part generation from the standard initial position; exhaustive minima stored to the end, without truncated depth or leaf heuristic. All the values of the first strokes are checked.

{
  "empty": 11,
  "legal_first_moves": 6,
  "first_move_scores": {
    "A1": -6,
    "C1": 4,
    "A2": -30,
    "A8": 6,
    "B8": -2,
    "F8": -6
  },
  "states": 688850,
  "history": [
    "C4",
    "C5",
    "C6",
    "E3",
    "F5",
    "B6",
    "A6",
    "D6",
    "E6",
    "E7",
    "F7",
    "C3",
    "D3",
    "G5",
    "D7",
    "C7",
    "F6",
    "G8",
    "F2",
    "E2",
    "D8",
    "C8",
    "D2",
    "E8",
    "B3",
    "B2",
    "B4",
    "A5",
    "F3",
    "E1",
    "B1",
    "G6",
    "A4",
    "A3",
    "H7",
    "F4",
    "D1",
    "G7",
    "G4",
    "A7",
    "H8",
    "C2",
    "H6",
    "G3",
    "F1",
    "B5",
    "H3",
    "G2",
    "B7"
  ],
  "player": -1,
  "best_score": 6,
  "principal_variation": [
    "B A8",
    "N C1",
    "B B8",
    "N F8",
    "B A1",
    "N H1",
    "B A2",
    "N G1",
    "B H2",
    "N H4",
    "B H5"
  ]
}

← Return to puzzle SOLO-036

SOLO-037 · Nim — pair subtraction

EXPERT

Two Workshops

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

This page contains the solution.

The answer

Winning shots: 1: A−1 and B−1 → (7, 10); 1: A−3 → (5, 11); 2: B−3 → (10, 4).

The reasoning

Each pair is a finished impartial game: the total number of chips decreases strictly at each stroke. At the position (0, 0), no movement exists; its number of Grundy is 0.
For a pair (a, b), build all the numbers that can be reached by removing 1, 2 or 3 A alone, B alone, or both simultaneously. The number of Grundy is the mex: the smallest non-negative integer absent from this set. Calculates by increasing amounts a+b, as all successors have a smaller sum.
Start of the table, useful for checking the convention: a\b | 0 1 2 3 4 5 6 7 ; 0 | 0 1 2 3 0 1 2 3 ; 1 | 1 2 0 4 1 2 0 4 ; 2 | 2 0 1 5 3 0 1 2 ; 3 | 3 4 5 6 2 7 4 5 ; 4 | 0 1 3 2 0 1 5 3 ; 5 | 1 2 0 7 1 2 0 4 ; 6 | 2 0 1 4 5 0 1 6 ; 7 | 3 4 2 5 3 4 6 5.
The values of the given pairs are: g(8,11)=2, g(10,7)=4. Their XOR is 6. It is not null: the position is winning, provided you join a null XOR.
For each move in pair k, replace its old value g with the new value g′: new XOR = old XOR ⊕ g ⊕ g′. The winning moves are exactly: 1: A−1 and B−1 → (7, 10); 1: A−3 → (5, 11); 2: B−3 → (10, 4).
Control of all other hits: 1: A−1 → (7, 11) : XOR 1 ; 1: B−1 → (8, 10) : XOR 1 ; 1: A−2 → (6, 11) : XOR 3 ; 1: B−2 → (8, 9) : XOR 5 ; 1: A−2 and B−2 → (6, 9) : XOR 4 ; 1: B−3 → (8, 8) : XOR 4 ; 1: A−3 and B−3 → (5, 8) : XOR 5 ; 2: A−1 → (9, 7) : XOR 5 ; 2: B−1 → (10, 6) : XOR 3 ; 2: A−1 and B−1 → (9, 6) : XOR 2 ; 2: A−2 → (8, 7) : XOR 1 ; 2: B−2 → (10, 5) : XOR 2 ; 2: A−2 and B−2 → (8, 5) : XOR 3 ; 2: A−3 → (7, 7) : XOR 7 ; 2: A−3 and B−3 → (7, 4) : XOR 1. winner exists for the opponent.
Why this works: when the XOR of the components is zero, a move changes one Grundy number to a different value, making the XOR nonzero. When the XOR is nonzero, choose a component with the highest set bit of the XOR. Its target g′ = g ⊕ XOR is less than g. By the definition of mex, every value below g is reachable. This lets you restore the XOR to zero after each opposing move.
The game ends because the total number of tokens decreases. Keeping the XOR at zero after each of your moves ensures you are never the player unable to move: (0,0) for every pair gives XOR zero. You therefore take the final token. This reasoning covers every opponent response.
For the verification of high values, apply exactly the recurrence, up to a maximum of 11 of this exercise. A symmetry g(a,b)=g(b,a) is permitted. Does not assume periodicity from the first eight lines: the evidence here uses the complete table.

Common mistakes and false leads

Apply the XOR of the heap sizes: it is the Grundy numbers of the PAIRS that combine.
Forget simultaneous withdrawals, or allow two different quantities when withdrawing simultaneously.
Transpose a poverty rule to this normal rule.

How we verified this answer

Accurate dynamic programming of Grundy numbers, control of the mex and all the first moves. The sum property is demonstrated in the fix.

{
  "pairs": [
    [
      8,
      11
    ],
    [
      10,
      7
    ]
  ],
  "grundy_values": [
    2,
    4
  ],
  "xor": 6,
  "winning_moves": [
    "1: A−1 et B−1 → (7, 10)",
    "1: A−3 → (5, 11)",
    "2: B−3 → (10, 4)"
  ],
  "legal_first_moves": 18,
  "max_component": 11,
  "table": [
    [
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3
    ],
    [
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      1,
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      1,
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      7,
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      2,
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      3,
      4,
      6,
      5,
      3,
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      4,
      5
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      2
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      1,
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      4,
      7,
      5,
      2,
      3,
      4,
      5
    ]
  ]
}

← Return to puzzle SOLO-037

SOLO-038 · Nim — pair subtraction

EXPERT

The Broken Balance

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

This page contains the solution.

The answer

Winning shots: 2: A−3 → (4, 13).

The reasoning

Each pair is a finished impartial game: the total number of chips decreases strictly at each stroke. At the position (0, 0), no movement exists; its number of Grundy is 0.
For a pair (a, b), build all the numbers that can be reached by removing 1, 2 or 3 A alone, B alone, or both simultaneously. The number of Grundy is the mex: the smallest non-negative integer absent from this set. Calculates by increasing amounts a+b, as all successors have a smaller sum.
Start of the table, useful for checking the convention: a\b | 0 1 2 3 4 5 6 7 ; 0 | 0 1 2 3 0 1 2 3 ; 1 | 1 2 0 4 1 2 0 4 ; 2 | 2 0 1 5 3 0 1 2 ; 3 | 3 4 5 6 2 7 4 5 ; 4 | 0 1 3 2 0 1 5 3 ; 5 | 1 2 0 7 1 2 0 4 ; 6 | 2 0 1 4 5 0 1 6 ; 7 | 3 4 2 5 3 4 6 5.
The values of the given pairs are: g(12,9)=1, g(7,13)=6. Their XOR is 7. It is not null: the position is winning, provided you join a null XOR.
For each move in pair k, replace its old value g with the new value g′: new XOR = old XOR ⊕ g ⊕ g′. The only winning move is 2: A−3 → (4, 13).
Check of all other moves: 1: A−1 → (11, 9) : XOR 5 ; 1: B−1 → (12, 8) : XOR 6 ; 1: A−1 and B−1 → (11, 8) : XOR 4 ; 1: A−2 → (10, 9) : XOR 6 ; 1: B−2 → (12, 7) : XOR 4 ; 1: A−2 and B−2 → (10, 7) : XOR 2 ; 1: A−3 → (9, 9) : XOR 4 ; 1: B−3 → (12, 6) : XOR 5 ; 1: A−3 and B−3 → (9, 6) : XOR 6 ; 2: A−1 → (6, 13) : XOR 1 ; 2: B−1 → (7, 12) : XOR 3 ; 2: A−1 and B−1 → (6, 12) : XOR 2 ; 2: A−2 → (5, 13) : XOR 3 ; 2: B−2 → (7, 11) : XOR 4 ; 2: A−2 and B−2 → (5, 11) : XOR 5 ; 2: B−3 → (7, 10) : XOR 5 ; 2: A−3 and B−3 → (4, 10) : XOR 3. They each leave a non-zero sum: a winning answer exists for the opponent.
Why this works: when the XOR of the components is zero, a move changes one Grundy number to a different value, making the XOR nonzero. When the XOR is nonzero, choose a component with the highest set bit of the XOR. Its target g′ = g ⊕ XOR is less than g. By the definition of mex, every value below g is reachable. This lets you restore the XOR to zero after each opposing move.
The game ends because the total number of tokens decreases. Keeping the XOR at zero after each of your moves ensures you are never the player unable to move: (0,0) for every pair gives XOR zero. You therefore take the final token. This reasoning covers every opponent response.
For the verification of high values, apply exactly the recurrence, up to a maximum of 13 of this exercise. A symmetry g(a,b)=g(b,a) is allowed. Does not assume periodicity from the first eight lines: the evidence here uses the complete table.

Common mistakes and false leads

Apply the XOR of the heap sizes: it is the Grundy numbers of the PAIRS that combine.
Forget simultaneous withdrawals, or allow two different quantities when withdrawing simultaneously.
Transpose a poverty rule to this normal rule.

How we verified this answer

Accurate dynamic programming of Grundy numbers, control of the mex and all the first moves. The sum property is demonstrated in the fix.

{
  "pairs": [
    [
      12,
      9
    ],
    [
      7,
      13
    ]
  ],
  "grundy_values": [
    1,
    6
  ],
  "xor": 7,
  "winning_moves": [
    "2: A−3 → (4, 13)"
  ],
  "legal_first_moves": 18,
  "max_component": 13,
  "table": [
    [
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
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      1
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      0,
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      2,
      0,
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      2,
      0,
      7,
      1,
      2
    ]
  ]
}

← Return to puzzle SOLO-038

SOLO-039 · Nim — pair subtraction

EXTREME

Three Linked Fronts

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

This page contains the solution.

The answer

Winning shots: 1: A−3 → (14, 19); 2: A−3 → (11, 18).

The reasoning

Each pair is a finished impartial game: the total number of chips decreases strictly at each stroke. At the position (0, 0), no movement exists; its number of Grundy is 0.
For a pair (a, b), build all the numbers that can be reached by removing 1, 2 or 3 A alone, B alone, or both simultaneously. The number of Grundy is the mex: the smallest non-negative integer absent from this set. Calculates by increasing amounts a+b, as all successors have a smaller sum.
Start of the table, useful for checking the convention: a\b | 0 1 2 3 4 5 6 7 ; 0 | 0 1 2 3 0 1 2 3 ; 1 | 1 2 0 4 1 2 0 4 ; 2 | 2 0 1 5 3 0 1 2 ; 3 | 3 4 5 6 2 7 4 5 ; 4 | 0 1 3 2 0 1 5 3 ; 5 | 1 2 0 7 1 2 0 4 ; 6 | 2 0 1 4 5 0 1 6 ; 7 | 3 4 2 5 3 4 6 5.
The values of the given pairs are: g(17,19)=5, g(14,18)=1, g(20,15)=2. Their XOR is 6. It is non-zero: the position is winning, provided you join a null XOR.
For each move in pair k, replace its old value g with the new value g′: new XOR = old XOR ⊕ g ⊕ g′. The winning moves are exactly: 1: A−3 → (14, 19); 2: A−3 → (11, 18).
Control of all other hits: 1: A−1 → (16, 19) : XOR 1 ; 1: B−1 → (17, 18) : XOR 3 ; 1: A−1 and B−1 → (16, 18) : XOR 7 ; 1: A−2 → (15, 19) : XOR 11 ; 1: B−2 → (17, 17) : XOR 1 ; 1: A−2 and B−2 → (15, 17) : XOR 4 ; 1: B−3 → (17, 16) : XOR 2 ; 1: A−3 and B−3 → (14, 16) : XOR 1 ; 2: A−1 → (13, 18) : XOR 7 ; 2: B−1 → (14, 17) : XOR 7 ; 2: A−1 and B−1 → (13, 17) : XOR 5 ; 2: A−2 → (12, 18) : XOR 4 ; 2: B−2 → (14, 16) : XOR 5 ; 2: A−2 and B−2 → (12, 16) : XOR 7 ; 2: B−3 → (14, 15) : XOR 1 ; 2: A−3 and B−3 → (11, 15) XOR 2 ; 3: A−1 → (19, 15) : XOR 12 ; 3: B−1 → (20, 14) : XOR 1 ; 3: A−1 and B−1 → (19, 14) : XOR 7 ; 3: A−2 → (18, 15) : XOR 2 ; 3: B−2 → (20, 13) : XOR 5 ; 3: A−2 and B−2 → (18, 13) : XOR 4 ; 3: A−3 → (17, 15) : XOR 3 ; 3: B−3 → (20, 12) : XOR 4 ; 3: A−3 and B−3 → (17, 12) : XOR 5. They each leave a non-zero sum: a winning answer exists for the adversary.
Why this works: when the XOR of the components is zero, a move changes one Grundy number to a different value, making the XOR nonzero. When the XOR is nonzero, choose a component with the highest set bit of the XOR. Its target g′ = g ⊕ XOR is less than g. By the definition of mex, every value below g is reachable. This lets you restore the XOR to zero after each opposing move.
The game ends because the total number of tokens decreases. Keeping the XOR at zero after each of your moves ensures you are never the player unable to move: (0,0) for every pair gives XOR zero. You therefore take the final token. This reasoning covers every opponent response.
For the verification of high values, apply exactly the recurrence, up to a maximum of 20 of this exercise. A symmetry g(a,b)=g(b,a) is allowed. Does not assume periodicity from the first eight lines: the evidence here uses the complete table.

Common mistakes and false leads

Apply the XOR of the heap sizes: it is the Grundy numbers of the PAIRS that combine.
Forget simultaneous withdrawals, or allow two different quantities when withdrawing simultaneously.
Transpose a poverty rule to this normal rule.

How we verified this answer

Accurate dynamic programming of Grundy numbers, control of the mex and all the first moves. The sum property is demonstrated in the fix.

{
  "pairs": [
    [
      17,
      19
    ],
    [
      14,
      18
    ],
    [
      20,
      15
    ]
  ],
  "grundy_values": [
    5,
    1,
    2
  ],
  "xor": 6,
  "winning_moves": [
    "1: A−3 → (14, 19)",
    "2: A−3 → (11, 18)"
  ],
  "legal_first_moves": 27,
  "max_component": 20,
  "table": [
    [
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
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      1,
      2,
      3,
      0,
      1,
      2,
      3,
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    ],
    [
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    ],
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      0
    ],
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      2,
      0,
      7,
      1,
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      0,
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      0,
      4,
      1
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      4,
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      3,
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      0
    ]
  ]
}

← Return to puzzle SOLO-039

SOLO-040 · Nim — pair subtraction

EXTREME

The Great Pile Board

Find all the first shots that guarantee victory, and demonstrate strategy against any adverse response.

This page contains the solution.

The answer

Winning shots: 1: B−1 → (21, 15); 2: A−2 → (16, 23); 2: A−3 and B−3 → (15, 20).

The reasoning

Each pair is a finished impartial game: the total number of chips decreases strictly at each stroke. At the position (0, 0), no movement exists; its number of Grundy is 0.
For a pair (a, b), build all the numbers that can be reached by removing 1, 2 or 3 A alone, B alone, or both simultaneously. The number of Grundy is the mex: the smallest non-negative integer absent from this set. Calculates by increasing amounts a+b, as all successors have a smaller sum.
Start of the table, useful for checking the convention: a\b | 0 1 2 3 4 5 6 7 ; 0 | 0 1 2 3 0 1 2 3 ; 1 | 1 2 0 4 1 2 0 4 ; 2 | 2 0 1 5 3 0 1 2 ; 3 | 3 4 5 6 2 7 4 5 ; 4 | 0 1 3 2 0 1 5 3 ; 5 | 1 2 0 7 1 2 0 4 ; 6 | 2 0 1 4 5 0 1 6 ; 7 | 3 4 2 5 3 4 6 5.
The values of the given pairs are: g(21,16)=1, g(18,23)=4, g(22,19)=3. Their XOR is 6. It is non-zero: the position is winning, provided you join a null XOR.
For each move in pair k, replace its old value g with the new value g′: new XOR = old XOR ⊕ g ⊕ g′. The winning moves are exactly: 1: B−1 → (21, 15); 2: A−2 → (16, 23); 2: A−3 and B−3 → (15, 20).
Check of all other moves: 1: A−1 → (20, 16) : XOR 7 ; 1: A−1 and B−1 → (20, 15) : XOR 5 ; 1: A−2 → (19, 16) : XOR 5 ; 1: B−2 → (21, 14) : XOR 7 ; 1: A−2 and B−2 → (19, 14) : XOR 4 ; 1: A−3 → (18, 16) : XOR 3 ; 1: B−3 → (21, 13) : XOR 5 ; 1: A−3 and B−3 → (18, 13) : XOR 7 ; 2: A−1 → (17, 23) : XOR 1 ; 2: B−1 → (18, 22) : XOR 3 ; 2: A−1 and B−1 → (17, 22) : XOR 2 ; 2: B−2 → (18, 21) : XOR 2 ; 2: A−2 and B−2 → (16, 21) : XOR 3 ; 2: A−3 → (15, 23) : XOR 7 ; 2: B−3 → (18, 20) : XOR 4 ; 3: A−1 → (21, 19) : XOR 2 ; 3: B−1 → (22, 18) : XOR 4 ; 3: A−1 and B−1 → (21, 18) : XOR 5 ; 3: A−2 → (20, 19) : XOR 7 ; 3: B−2 → (22, 17) : XOR 5 ; 3: A−2 and B−2 → (20, 17) : XOR 4 ; 3: A−3 → (19, 19) : XOR 1 ; 3: B−3 → (22, 16) : XOR 3 ; 3: A−3 and B−3 → (19, 16) : XOR 7. They each leave a non-zero sum: a winning answer exists for the opponent.
Why this works: when the XOR of the components is zero, a move changes one Grundy number to a different value, making the XOR nonzero. When the XOR is nonzero, choose a component with the highest set bit of the XOR. Its target g′ = g ⊕ XOR is less than g. By the definition of mex, every value below g is reachable. This lets you restore the XOR to zero after each opposing move.
The game ends because the total number of tokens decreases. Keeping the XOR at zero after each of your moves ensures you are never the player unable to move: (0,0) for every pair gives XOR zero. You therefore take the final token. This reasoning covers every opponent response.
For the verification of high values, apply exactly the recurrence, up to a maximum of 23 of this exercise. A symmetry g(a,b)=g(b,a) is allowed. Does not assume periodicity from the first eight lines: the evidence here uses the complete table.

Common mistakes and false leads

Apply the XOR of the heap sizes: it is the Grundy numbers of the PAIRS that combine.
Forget simultaneous withdrawals, or allow two different quantities when withdrawing simultaneously.
Transpose a poverty rule to this normal rule.

How we verified this answer

Accurate dynamic programming of Grundy numbers, control of the mex and all the first moves. The sum property is demonstrated in the fix.

{
  "pairs": [
    [
      21,
      16
    ],
    [
      18,
      23
    ],
    [
      22,
      19
    ]
  ],
  "grundy_values": [
    1,
    4,
    3
  ],
  "xor": 6,
  "winning_moves": [
    "1: B−1 → (21, 15)",
    "2: A−2 → (16, 23)",
    "2: A−3 et B−3 → (15, 20)"
  ],
  "legal_first_moves": 27,
  "max_component": 23,
  "table": [
    [
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3
    ],
    [
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4
    ],
    [
      2,
      0,
      1,
      5,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2
    ],
    [
      3,
      4,
      5,
      6,
      2,
      7,
      4,
      5,
      6,
      7,
      4,
      5,
      6,
      7,
      4,
      5,
      6,
      7,
      4,
      5,
      6,
      7,
      4,
      5
    ],
    [
      0,
      1,
      3,
      2,
      0,
      1,
      5,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3
    ],
    [
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4
    ],
    [
      2,
      0,
      1,
      4,
      5,
      0,
      1,
      6,
      2,
      0,
      1,
      7,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2
    ],
    [
      3,
      4,
      2,
      5,
      3,
      4,
      6,
      5,
      3,
      7,
      4,
      5,
      2,
      6,
      4,
      5,
      7,
      6,
      4,
      5,
      7,
      6,
      4,
      5
    ],
    [
      0,
      1,
      3,
      6,
      0,
      1,
      2,
      3,
      0,
      1,
      5,
      2,
      0,
      1,
      5,
      3,
      0,
      1,
      2,
      3,
      0,
      1,
      2,
      3
    ],
    [
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      3,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4
    ],
    [
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      4,
      5,
      0,
      1,
      4,
      6,
      0,
      1,
      7,
      2,
      0,
      1,
      6,
      3,
      0,
      1,
      2
    ],
    [
      3,
      4,
      2,
      5,
      3,
      4,
      7,
      5,
      2,
      3,
      4,
      5,
      2,
      7,
      3,
      5,
      4,
      6,
      7,
      5,
      2,
      6,
      4,
      5
    ],
    [
      0,
      1,
      3,
      6,
      0,
      1,
      3,
      2,
      0,
      1,
      6,
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      3,
      2,
      0,
      1,
      5,
      3
    ],
    [
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      6,
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      3,
      1,
      2,
      0,
      4,
      1,
      2,
      0,
      4
    ],
    [
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      4,
      5,
      0,
      1,
      3,
      4,
      0,
      1,
      6,
      2,
      0,
      1,
      3,
      5,
      0,
      1,
      6
    ],
    [
      3,
      4,
      2,
      5,
      3,
      4,
      2,
      5,
      3,
      4,
      7,
      5,
      2,
      3,
      6,
      4,
      5,
      7,
      6,
      8,
      2,
      7,
      3,
      5
    ],
    [
      0,
      1,
      3,
      6,
      0,
      1,
      3,
      7,
      0,
      1,
      2,
      4,
      0,
      1,
      2,
      5,
      0,
      1,
      4,
      2,
      0,
      1,
      6,
      2
    ],
    [
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      6,
      1,
      2,
      0,
      6,
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      5,
      1,
      2,
      0,
      3
    ],
    [
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      7,
      3,
      0,
      1,
      6,
      4,
      0,
      1,
      3,
      6,
      0,
      1,
      4
    ],
    [
      3,
      4,
      2,
      5,
      3,
      4,
      2,
      5,
      3,
      4,
      6,
      5,
      2,
      4,
      3,
      8,
      2,
      5,
      3,
      4,
      2,
      7,
      3,
      5
    ],
    [
      0,
      1,
      3,
      6,
      0,
      1,
      3,
      7,
      0,
      1,
      3,
      2,
      0,
      1,
      5,
      2,
      0,
      1,
      6,
      2,
      0,
      1,
      4,
      2
    ],
    [
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      6,
      1,
      2,
      0,
      6,
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      7,
      1,
      2,
      0,
      3
    ],
    [
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      4,
      2,
      0,
      1,
      4,
      5,
      0,
      1,
      3,
      6,
      0,
      1,
      3,
      4,
      0,
      1,
      6
    ],
    [
      3,
      4,
      2,
      5,
      3,
      4,
      2,
      5,
      3,
      4,
      2,
      5,
      3,
      4,
      6,
      5,
      2,
      3,
      4,
      5,
      2,
      3,
      6,
      4
    ]
  ]
}

← Return to puzzle SOLO-040

L / C	1	2	3	4	5	6	7	8	9
1	208			13			12		22
2		18	24		11
3					16				17
4	124		16		21	3		14
5	15						13
6		10			18
7	15		8	18		12	17		14
8	19				9
9	3					15

L / C	1	2	3	4	5	6	7
1	·	·	·	2	·	5	·
2	6	·	·	·	4	·	·
3	·	·	3	·	5	·	·
4	·	·	·	6	·	·	·
5	·	·	·	·	·	·	·
6	·	·	·	·	·	·	7
7	·	·	5	·	·	6	3

L / C	1	2	3	4	5	6	7	8
1	·	5	·	·	2	7	·	·
2	·	·	2	3	·	·	·	·
3	6	·	·	·	·	·	·	·
4	·	·	·	·	·	·	·	·
5	·	·	4	·	·	·	8	·
6	·	·	·	·	·	8	·	1
7	·	·	·	5	8	2	·	·
8	·	·	·	6	3	·	·	·

L / C	1	2	3	4	5	6	7	8
1	·	·	·	·	·	2	·	·
2	·	·	·	5	·	·	·	·
3	·	·	·	4	·	·	5	·
4	·	5	·	·	·	·	8	4
5	·	4	3	6	2	·	·	·
6	·	·	·	·	6	3	·	·
7	·	·	·	·	·	·	·	6
8	·	·	·	·	1	·	·	5

L / C	A	B	C	D	E	F	G	H
1	B	N	N	N	·	N	B	N
2	B	B	N	N	·	B	B	N
3	B	N	N	N	B	N	B	N
4	N	N	N	N	B	B	B	N
5	N	B	B	N	B	B	B	N
6	·	B	B	B	N	N	B	N
7	·	·	B	B	B	B	B	N
8	·	B	·	B	·	B	B	N

L / C	A	B	C	D	E	F	G	H
1	B	B	B	B	B	B	·	N
2	B	B	N	B	B	B	N	B
3	·	B	B	B	B	N	B	·
4	N	N	N	N	N	B	B	B
5	N	N	N	B	B	N	B	B
6	N	N	B	N	B	B	B	B
7	N	N	B	B	·	·	B	B
8	N	N	B	N	·	·	·	B

L / C	A	B	C	D	E	F	G	H
1	N	N	N	N	N	N	B	·
2	·	N	N	N	N	N	N	N
3	·	N	B	B	N	B	N	N
4	B	N	B	B	B	N	B	·
5	B	N	N	N	B	N	N	B
6	B	N	B	N	B	N	B	B
7	B	N	B	B	N	N	·	B
8	·	N	·	·	B	N	·	B

L / C	A	B	C	D	E	F	G	H
1	B	N	N	·	B	·	·	N
2	B	N	N	B	·	·	·	N
3	·	B	B	N	N	N	N	N
4	N	B	B	N	B	N	B	N
5	B	N	N	B	N	N	B	N
6	·	N	B	B	B	N	B	N
7	·	B	N	N	N	N	N	·
8	B	B	N	N	N	N	N	N

L / C	A	B	C	D	E	F	G	H
1	·	·	B	B	B	B	B	B
2	·	B	B	B	N	N	B	·
3	B	B	B	N	N	N	·	·
4	B	B	N	B	B	N	N	B
5	B	N	B	B	B	N	·	N
6	B	N	N	B	N	N	N	B
7	B	N	N	N	N	N	B	·
8	B	N	·	B	B	B	·	·

L / C	A	B	C	D	E	F	G	H
1	·	N	·	N	N	N	·	·
2	·	N	B	B	B	B	B	·
3	B	N	B	B	B	B	B	N
4	B	N	B	N	B	N	B	·
5	B	N	B	B	B	N	B	·
6	B	N	B	B	B	N	B	N
7	B	N	N	N	N	N	N	N
8	·	·	B	B	B	·	B	N

L / C	1	2	3	4	5	6	7	8	9
1	208			13			12		22
2		18	24		11
3					16				17
4	124		16		21	3		14
5	15						13
6		10			18
7	15		8	18		12	17		14
8	19				9
9	3					15

L / C	1	2	3	4	5	6	7
1	·	·	·	2	·	5	·
2	6	·	·	·	4	·	·
3	·	·	3	·	5	·	·
4	·	·	·	6	·	·	·
5	·	·	·	·	·	·	·
6	·	·	·	·	·	·	7
7	·	·	5	·	·	6	3

L / C	1	2	3	4	5	6	7	8
1	·	5	·	·	2	7	·	·
2	·	·	2	3	·	·	·	·
3	6	·	·	·	·	·	·	·
4	·	·	·	·	·	·	·	·
5	·	·	4	·	·	·	8	·
6	·	·	·	·	·	8	·	1
7	·	·	·	5	8	2	·	·
8	·	·	·	6	3	·	·	·

L / C	1	2	3	4	5	6	7	8
1	·	·	·	·	·	2	·	·
2	·	·	·	5	·	·	·	·
3	·	·	·	4	·	·	5	·
4	·	5	·	·	·	·	8	4
5	·	4	3	6	2	·	·	·
6	·	·	·	·	6	3	·	·
7	·	·	·	·	·	·	·	6
8	·	·	·	·	1	·	·	5

L / C	A	B	C	D	E	F	G	H
1	B	N	N	N	·	N	B	N
2	B	B	N	N	·	B	B	N
3	B	N	N	N	B	N	B	N
4	N	N	N	N	B	B	B	N
5	N	B	B	N	B	B	B	N
6	·	B	B	B	N	N	B	N
7	·	·	B	B	B	B	B	N
8	·	B	·	B	·	B	B	N

L / C	A	B	C	D	E	F	G	H
1	B	B	B	B	B	B	·	N
2	B	B	N	B	B	B	N	B
3	·	B	B	B	B	N	B	·
4	N	N	N	N	N	B	B	B
5	N	N	N	B	B	N	B	B
6	N	N	B	N	B	B	B	B
7	N	N	B	B	·	·	B	B
8	N	N	B	N	·	·	·	B

L / C	A	B	C	D	E	F	G	H
1	N	N	N	N	N	N	B	·
2	·	N	N	N	N	N	N	N
3	·	N	B	B	N	B	N	N
4	B	N	B	B	B	N	B	·
5	B	N	N	N	B	N	N	B
6	B	N	B	N	B	N	B	B
7	B	N	B	B	N	N	·	B
8	·	N	·	·	B	N	·	B

L / C	A	B	C	D	E	F	G	H
1	B	N	N	·	B	·	·	N
2	B	N	N	B	·	·	·	N
3	·	B	B	N	N	N	N	N
4	N	B	B	N	B	N	B	N
5	B	N	N	B	N	N	B	N
6	·	N	B	B	B	N	B	N
7	·	B	N	N	N	N	N	·
8	B	B	N	N	N	N	N	N

L / C	A	B	C	D	E	F	G	H
1	·	·	B	B	B	B	B	B
2	·	B	B	B	N	N	B	·
3	B	B	B	N	N	N	·	·
4	B	B	N	B	B	N	N	B
5	B	N	B	B	B	N	·	N
6	B	N	N	B	N	N	N	B
7	B	N	N	N	N	N	B	·
8	B	N	·	B	B	B	·	·

L / C	A	B	C	D	E	F	G	H
1	·	N	·	N	N	N	·	·
2	·	N	B	B	B	B	B	·
3	B	N	B	B	B	B	B	N
4	B	N	B	N	B	N	B	·
5	B	N	B	B	B	N	B	·
6	B	N	B	B	B	N	B	N
7	B	N	N	N	N	N	N	N
8	·	·	B	B	B	·	B	N

L / C	1	2	3	4	5	6	7	8	9
1	208			13			12		22
2		18	24		11
3					16				17
4	124		16		21	3		14
5	15						13
6		10			18
7	15		8	18		12	17		14
8	19				9
9	3					15

L / C	1	2	3	4	5	6	7
1	·	·	·	2	·	5	·
2	6	·	·	·	4	·	·
3	·	·	3	·	5	·	·
4	·	·	·	6	·	·	·
5	·	·	·	·	·	·	·
6	·	·	·	·	·	·	7
7	·	·	5	·	·	6	3

L / C	1	2	3	4	5	6	7	8
1	·	5	·	·	2	7	·	·
2	·	·	2	3	·	·	·	·
3	6	·	·	·	·	·	·	·
4	·	·	·	·	·	·	·	·
5	·	·	4	·	·	·	8	·
6	·	·	·	·	·	8	·	1
7	·	·	·	5	8	2	·	·
8	·	·	·	6	3	·	·	·

L / C	1	2	3	4	5	6	7	8
1	·	·	·	·	·	2	·	·
2	·	·	·	5	·	·	·	·
3	·	·	·	4	·	·	5	·
4	·	5	·	·	·	·	8	4
5	·	4	3	6	2	·	·	·
6	·	·	·	·	6	3	·	·
7	·	·	·	·	·	·	·	6
8	·	·	·	·	1	·	·	5

L / C	A	B	C	D	E	F	G	H
1	B	N	N	N	·	N	B	N
2	B	B	N	N	·	B	B	N
3	B	N	N	N	B	N	B	N
4	N	N	N	N	B	B	B	N
5	N	B	B	N	B	B	B	N
6	·	B	B	B	N	N	B	N
7	·	·	B	B	B	B	B	N
8	·	B	·	B	·	B	B	N

L / C	A	B	C	D	E	F	G	H
1	B	B	B	B	B	B	·	N
2	B	B	N	B	B	B	N	B
3	·	B	B	B	B	N	B	·
4	N	N	N	N	N	B	B	B
5	N	N	N	B	B	N	B	B
6	N	N	B	N	B	B	B	B
7	N	N	B	B	·	·	B	B
8	N	N	B	N	·	·	·	B

L / C	A	B	C	D	E	F	G	H
1	N	N	N	N	N	N	B	·
2	·	N	N	N	N	N	N	N
3	·	N	B	B	N	B	N	N
4	B	N	B	B	B	N	B	·
5	B	N	N	N	B	N	N	B
6	B	N	B	N	B	N	B	B
7	B	N	B	B	N	N	·	B
8	·	N	·	·	B	N	·	B

L / C	A	B	C	D	E	F	G	H
1	B	N	N	·	B	·	·	N
2	B	N	N	B	·	·	·	N
3	·	B	B	N	N	N	N	N
4	N	B	B	N	B	N	B	N
5	B	N	N	B	N	N	B	N
6	·	N	B	B	B	N	B	N
7	·	B	N	N	N	N	N	·
8	B	B	N	N	N	N	N	N

L / C	A	B	C	D	E	F	G	H
1	·	·	B	B	B	B	B	B
2	·	B	B	B	N	N	B	·
3	B	B	B	N	N	N	·	·
4	B	B	N	B	B	N	N	B
5	B	N	B	B	B	N	·	N
6	B	N	N	B	N	N	N	B
7	B	N	N	N	N	N	B	·
8	B	N	·	B	B	B	·	·

L / C	A	B	C	D	E	F	G	H
1	·	N	·	N	N	N	·	·
2	·	N	B	B	B	B	B	·
3	B	N	B	B	B	B	B	N
4	B	N	B	N	B	N	B	·
5	B	N	B	B	B	N	B	·
6	B	N	B	B	B	N	B	N
7	B	N	N	N	N	N	N	N
8	·	·	B	B	B	·	B	N