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)
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R2C2 = 6.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R3C2 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C7 = 6.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R4C3 = 8.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R6C5 = 6.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R7C2 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C8 = 7.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R7C9 = 1.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R8C9 = 6.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R9C2 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C6 = 6.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 3.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 8.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 2.
Look at cell R5C8.
Every other digit already appears in its row, column or box.
Only one option remains: R5C8 = 5.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 5.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 1.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R1C6 = 2.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R1C9 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R4C4 = 5.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R6C7 = 3.
Spot R1C7 and R2C7 in this box.
These two cells can only hold 4 and 9: they reserve those digits.
4 and 9 are therefore removed from the other cells of the box.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R3C4 = 9.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C5 = 9.
Spot R1C5 and R4C5 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 9 can go in this box.
All its spots lie on a single column (R1C3 and R2C3).
9 is therefore removed from that column outside the box.
Find 1 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 1.
R5C1 is seen by both free ends: 1 is therefore eliminated there.
Look at where 1 can go on this column.
All its spots lie within a single box (R2C1 and R3C1).
1 is therefore removed from the rest of 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.
R2C1 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.
R1C3 is seen by both free ends: 7 is therefore eliminated there.
(1) Find where 7 appears on two rows: altogether, only two columns are involved (R1C1, R1C5, R4C1 and R4C5). (2) Find where 7 appears on two columns: altogether, only two rows are involved (R2C3, R6C3, R2C4 and R6C4).
(1) On each covered column, 7 must lie within one of these two rows. (2) On each covered row, 7 must lie within one of these two columns.
(1) 7 is therefore removed from those two columns outside the two X-Wing rows. (2) 7 is therefore removed from those two rows outside the two X-Wing columns.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 8.
Look at cell R3C1.
Every other digit already appears in its row, column or box.
Only one option remains: R3C1 = 1.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 8.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 7.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 1.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 8.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R5C2 = 1.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R6C6 = 1.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R7C4 = 8.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R8C3 = 1.
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.
R4C2 is seen by both free ends: 3 is therefore eliminated there.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R7C3 = 4.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C6 = 3.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R8C2 = 3.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C4 = 4.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 4.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 7.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 2.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 9.
Look at cell R4C9.
Every other digit already appears in its row, column or box.
Only one option remains: R4C9 = 2.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 3.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 4.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 4.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 9.
Look at cell R2C4.
Every other digit already appears in its row, column or box.
Only one option remains: R2C4 = 7.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 3.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 7.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 2.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 3.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 4.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 9.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 7.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 9.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 4.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 3.