Step-by-step solution
Steps 67
Estimated time 108 – 162 min
Hardest technique X-Wing
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R2C8 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R3C3 = 1.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R4C1 = 1.
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 box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C3 = 9.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R4C9 = 6.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R5C4 = 9.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C1 = 9.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R7C2 = 6.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C3 = 5.
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 R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 2.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R3C2 = 3.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C5 = 1.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R6C5 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C8 = 1.
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 3 could go.
Across the whole row, 3 has only one possible spot left.
So R9C3 = 3.
Look at cell R8C2.
Every other digit already appears in its row, column or box.
Only one option remains: R8C2 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R7C8 = 2.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C4 = 1.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C5 = 3.
Spot R3C4 and R7C4 on this column.
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 column.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 3.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C4 = 6.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R1C6 = 3.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R1C8 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R2C4 = 5.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R2C9 = 3.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R3C9 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R4C7 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C7 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C8 = 3.
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 row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R2C3 = 6.
Find 4 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 4.
R2C1 is seen by both free ends: 4 is therefore eliminated there.
Look at where 4 can go in this box.
All its spots lie on a single row (R1C1 and R1C3).
4 is therefore removed from that row 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.
R9C1 and R8C7 is seen by both free ends: 7 is therefore eliminated there.
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.
R3C8 is seen by both free ends: 8 is therefore eliminated there.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C9 = 8.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 7.
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.
R2C5 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.
R3C6 is seen by both free ends: 7 is therefore eliminated there.
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.
R1C5 is seen by both free ends: 7 is therefore eliminated there.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 2.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 4.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 7.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 8.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 4.
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 = 8.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 4.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 4.
Look at cell R9C8.
Every other digit already appears in its row, column or box.
Only one option remains: R9C8 = 8.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 2.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 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 R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 7.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 8.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 7.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 7.
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 4.
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 = 4.
Look at cell R8C1.
Every other digit already appears in its row, column or box.
Only one option remains: R8C1 = 8.