Step-by-step solution
Steps 68
Estimated time 112 – 168 min
Hardest technique X-Wing
Clues 24
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 2.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R1C6 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C7 = 8.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R2C1 = 2.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R5C3 = 8.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C6 = 2.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R8C5 = 5.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R9C1 = 8.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 3.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 9.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 5.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R4C1 = 5.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C9 = 9.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R5C9 = 5.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C1 = 1.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C8 = 5.
(1) Spot R7C6 and R8C6 on this column. (2) Spot R7C9 and R8C8 in this box.
(1) These two cells can only hold 1 and 4: they reserve those digits. (2) These two cells can only hold 1 and 3: they reserve those digits.
(1) 1 and 4 are therefore removed from the other cells of the column. (2) 1 and 3 are therefore removed from the other cells of the box.
Find where 2 and 3 can go on this column.
These two digits only fit in R7C2 and R8C2 on the column.
The other candidates in R7C2 and R8C2 are therefore eliminated.
Spot R7C2 and R8C2 in this box.
These two cells can only hold 2 and 3: they reserve those digits.
2 and 3 are therefore removed from the other cells of the box.
Look at where 4 can go in this box.
All its spots lie on a single column (R4C7, R5C7 and R6C7).
4 is therefore removed from that column outside the box.
(1) Look at where 6 can go on this column. (2) Look at where 3 can go on this column.
(1) All its spots lie within a single box (R4C2 and R6C2). (2) All its spots lie within a single box (R4C7 and R6C7).
(1) 6 is therefore removed from the rest of the box. (2) 3 is therefore removed from the rest of the box.
Find 1 on two rows, each with exactly two candidate positions.
A shared column connects one end of each pair. Whichever way the logic falls, one of the two free ends must hold 1.
R2C4 and R4C5 is seen by both free ends: 1 is therefore eliminated there.
Find 9 on one row and one column, each with two candidate positions, sharing a box.
The two strands meet in the shared box. One of the two free ends must hold 9.
R3C2 is seen by both free ends: 9 is therefore eliminated there.
Look at where 9 can go on this column.
All its spots lie within a single box (R5C2 and R6C2).
9 is therefore removed from the rest of the box.
Find where 6 and 9 can go on this row.
These two digits only fit in R6C2 and R6C6 on the row.
The other candidates in R6C2 and R6C6 are therefore eliminated.
Find 7 on two rows, each with exactly two candidate positions.
A shared column connects one end of each pair. Whichever way the logic falls, one of the two free ends must hold 7.
R1C1, R1C3 and R3C7 is seen by both free ends: 7 is therefore eliminated there.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 1.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 4.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 9.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 1.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 6.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 9.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R1C4 = 9.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R1C8 = 7.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R2C5 = 1.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R3C1 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R4C8 = 1.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C1 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C2 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C3 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R7C6 = 4.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R7C9 = 1.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C1 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C3 = 9.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R8C6 = 1.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C7 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C8 = 3.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 6.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 2.
Look at cell R8C2.
Every other digit already appears in its row, column or box.
Only one option remains: R8C2 = 2.
Find 4 on two rows, each with exactly two candidate positions.
A shared column connects one end of each pair. Whichever way the logic falls, one of the two free ends must hold 4.
R1C3, R2C3 and R4C2 is seen by both free ends: 4 is therefore eliminated there.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 3.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 4.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 7.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 6.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 3.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 4.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 8.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 7.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 7.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 8.
Look at cell R4C7.
Every other digit already appears in its row, column or box.
Only one option remains: R4C7 = 3.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 4.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 3.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 7.
Look at cell R2C4.
Every other digit already appears in its row, column or box.
Only one option remains: R2C4 = 4.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 6.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 4.