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)
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C1 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R1C4 = 5.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R2C4 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R2C9 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R3C1 = 5.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R3C6 = 1.
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 row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R6C7 = 2.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C9 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C6 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C1 = 3.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 9.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 7.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 6.
Look at cell R3C3.
Every other digit already appears in its row, column or box.
Only one option remains: R3C3 = 3.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R1C5 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C6 = 7.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R5C4 = 3.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R9C8 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R1C2 = 9.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R1C3 = 6.
Look at cell R8C2.
Every other digit already appears in its row, column or box.
Only one option remains: R8C2 = 1.
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 4.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 8.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 4.
(1) Spot R7C3 and R8C3 on this column. (2) Spot R4C1 and R5C1 in this box.
(1) These two cells can only hold 7 and 9: they reserve those digits. (2) These two cells can only hold 6 and 9: they reserve those digits.
(1) 7 and 9 are therefore removed from the other cells of the column. (2) 6 and 9 are therefore removed from the other cells of 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.
R7C4 is seen by both free ends: 1 is therefore eliminated there.
Spot R7C3 and R7C4 on this row.
These two cells can only hold 7 and 9: they reserve those digits.
7 and 9 are therefore removed from the other cells of the row.
Spot R6C4 and R9C4 on this column.
These two cells can only hold 1 and 4: they reserve those digits.
1 and 4 are therefore removed from the other cells of the column.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C2 = 4.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R4C7 = 1.
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 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C8 = 5.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C9 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C3 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C4 = 4.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C5 = 1.
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 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C7 = 5.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R7C8 = 1.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C4 = 1.
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.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 8.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 4.
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.
R9C5 is seen by both free ends: 7 is therefore eliminated there.
Look at cell R9C5.
Every other digit already appears in its row, column or box.
Only one option remains: R9C5 = 9.
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 R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 7.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 8.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 7.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 7.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 8.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 9.
Look at cell R4C9.
Every other digit already appears in its row, column or box.
Only one option remains: R4C9 = 7.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 6.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 9.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 7.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 9.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 6.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 8.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 9.