Step-by-step solution
Steps 66
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 row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R1C5 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C6 = 8.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R2C5 = 7.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R2C7 = 2.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C9 = 8.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R6C2 = 8.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R7C2 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R9C6 = 2.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 5.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 1.
Look at cell R8C2.
Every other digit already appears in its row, column or box.
Only one option remains: R8C2 = 6.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 2.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R1C9 = 1.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C5 = 8.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C5 = 2.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R7C4 = 1.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R7C5 = 6.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C7 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C4 = 8.
Scan the empty cells in this row where 6 could go.
Across the whole row, 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 R9C8 = 4.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R4C1 = 1.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R4C3 = 2.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C7 = 4.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C3 = 4.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C7 = 1.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R6C5 = 1.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C3 = 1.
Look at where 3 can go in this box.
All its spots lie on a single row (R6C7 and R6C9).
3 is therefore removed from that row outside the box.
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.
R7C8 is seen by both free ends: 3 is therefore eliminated there.
Look at where 3 can go on this column.
All its spots lie within a single box (R1C8 and R2C8).
3 is therefore removed from the rest of the box.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R3C6 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C4 = 3.
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.
R7C6 and R8C9 is seen by both free ends: 7 is therefore eliminated there.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R7C8 = 7.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R8C4 = 7.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R4C9 = 7.
Find 5 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 5.
R6C9 is seen by both free ends: 5 is therefore eliminated there.
Look at where 5 can go on this column.
All its spots lie within a single box (R8C9 and R9C9).
5 is therefore removed from the rest of the box.
Find where 5 appears on two rows: altogether, only two columns are involved (R5C1, R5C6, R7C1 and R7C6).
On each covered column, 5 must lie within one of these two rows.
5 is therefore removed from those two columns outside the two X-Wing rows.
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 9.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 5.
Find 9 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 9.
R8C5 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 5.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 9.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 5.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 9.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 3.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 3.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 9.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 5.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 9.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 3.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 9.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 5.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 7.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 9.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 3.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 5.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 6.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 9.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 7.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 5.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 6.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 5.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 3.