Step-by-step solution
Steps 64
Estimated time 104 – 156 min
Hardest technique X-Wing
Clues 23
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R1C4 = 1.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C7 = 2.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R1C8 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R2C5 = 4.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R2C6 = 2.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R2C7 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C3 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C3 = 1.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C9 = 6.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R6C3 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C9 = 4.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R7C2 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R7C8 = 4.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R7C9 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C4 = 9.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C8 = 9.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R3C5 = 9.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R4C1 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C5 = 6.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R4C6 = 4.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R4C7 = 1.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C8 = 8.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R5C1 = 4.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C8 = 2.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C6 = 1.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R6C8 = 5.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R7C3 = 3.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R7C4 = 2.
Spot R9C3 and R9C9 on this row.
These two cells can only hold 2 and 7: they reserve those digits.
2 and 7 are therefore removed from the other cells of the row.
Look at where 5 can go in this box.
All its spots lie on a single column (R8C6 and R9C6).
5 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.
R3C4 and R5C5 is seen by both free ends: 7 is therefore eliminated there.
Find 7 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 7.
R8C6 is seen by both free ends: 7 is therefore eliminated there.
Look at where 7 can go in this box.
All its spots lie on a single row (R7C5 and R7C6).
7 is therefore removed from that row outside the box.
Find 7 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 7.
R2C3 is seen by both free ends: 7 is therefore eliminated there.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C5 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R3C6 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C2 = 3.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R5C2 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C7 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C1 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C4 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C5 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C6 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C1 = 8.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 5.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 7.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 3.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 5.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 5.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 7.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 7.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 3.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 5.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 3.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 6.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 5.
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 7.
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 5.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 6.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 2.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 2.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 3.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 7.