Step-by-step solution
Steps 63
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 box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C2 = 3.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R1C5 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R2C3 = 5.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R2C9 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R4C1 = 2.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R4C4 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C7 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C1 = 3.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C8 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R6C2 = 5.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R6C9 = 2.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R7C5 = 5.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R8C6 = 2.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C9 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C5 = 3.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C6 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C4 = 8.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R2C4 = 1.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R3C1 = 1.
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 8 could go.
Across the whole box, 8 has only one possible spot left.
So R3C8 = 8.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R4C6 = 1.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 7.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 6.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 9.
Look at where 9 can go in this box.
All its spots lie on a single column (R4C2 and R5C2).
9 is therefore removed from that column outside 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 R8C4 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C8 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C4 = 6.
Look at cell R9C8.
Every other digit already appears in its row, column or box.
Only one option remains: R9C8 = 7.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 4.
Find 8 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 8.
R9C1 is seen by both free ends: 8 is therefore eliminated there.
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 9.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 6.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R6C6 = 6.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R7C9 = 9.
Spot R4C3 and R6C1 in this box.
These two cells can only hold 7 and 8: they reserve those digits.
7 and 8 are therefore removed from the other cells of the box.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 9.
Find where 7 appears on two rows: altogether, only two columns are involved (R3C3, R3C5, R4C3 and R4C5).
On each covered column, 7 must lie within one of these two rows.
7 is therefore removed from those two columns outside the two X-Wing rows.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R8C2 = 7.
Find 8 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 8.
R4C3 is seen by both free ends: 8 is therefore eliminated there.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 7.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 8.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 9.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 7.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 8.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 7.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 7.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 6.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 9.
Look at cell R3C3.
Every other digit already appears in its row, column or box.
Only one option remains: R3C3 = 6.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 7.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 4.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 8.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 4.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 1.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 8.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 8.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 6.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 1.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 4.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 1.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 8.