Step-by-step solution
Steps 67
Estimated time 108 – 162 min
Hardest technique X-Wing
Clues 23
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R1C2 = 1.
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 8 could go.
Across the whole row, 8 has only one possible spot left.
So R4C2 = 8.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R4C6 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C3 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C7 = 4.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R5C8 = 1.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R6C2 = 5.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R6C7 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C1 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C2 = 9.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R9C3 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C7 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C6 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R8C1 = 4.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R8C7 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R9C6 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R5C4 = 9.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R5C5 = 8.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R8C4 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R2C8 = 4.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R2C9 = 9.
Spot R4C8 and R6C8 on this column.
These two cells can only hold 3 and 7: they reserve those digits.
3 and 7 are therefore removed from the other cells of the column.
Look at where 7 can go in this box.
All its spots lie on a single column (R5C6 and R6C6).
7 is therefore removed from that column outside the box.
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.
R1C9 and R7C7 is seen by both free ends: 7 is therefore eliminated there.
Find 2 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 2.
R8C3 is seen by both free ends: 2 is therefore eliminated there.
Look at where 2 can go in this box.
All its spots lie on a single column (R7C2 and R9C2).
2 is therefore removed from that column outside the box.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C1 = 2.
Find where 2 and 6 can go on this row.
These two digits only fit in R3C6 and R3C8 on the row.
The other candidates in R3C6 and R3C8 are therefore eliminated.
Find 2 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 2.
R1C9 and R7C8 is seen by both free ends: 2 is therefore eliminated there.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 3.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 7.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 5.
Look at cell R7C8.
Every other digit already appears in its row, column or box.
Only one option remains: R7C8 = 8.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 8.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 3.
Find 3 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 3.
R9C4 is seen by both free ends: 3 is therefore eliminated there.
Find 3 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 3.
R9C2 is seen by both free ends: 3 is therefore eliminated there.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C8 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R2C4 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C6 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R3C8 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C1 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C8 = 3.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R5C2 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C6 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R6C4 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C6 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C8 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C4 = 5.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C5 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C3 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C9 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C2 = 7.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R9C5 = 3.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 9.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 2.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 5.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 2.
Look at cell R3C1.
Every other digit already appears in its row, column or box.
Only one option remains: R3C1 = 3.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 9.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 2.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 4.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 2.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 4.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 2.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 7.