Step-by-step solution
Steps 69
Estimated time 112 – 168 min
Hardest technique X-Wing
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 6.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 1.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R1C3 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R3C1 = 9.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R3C5 = 6.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C1 = 5.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R6C9 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C6 = 6.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R9C5 = 5.
Look at cell R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R2C2 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R2C5 = 9.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R2C6 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R3C2 = 4.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R3C9 = 2.
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 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C2 = 8.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R9C3 = 2.
Spot R5C6 and R6C6 in this box.
These two cells can only hold 1 and 9: they reserve those digits.
1 and 9 are therefore removed from the other cells of the box.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C4 = 8.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R7C5 = 1.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C4 = 2.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C5 = 4.
Find where 2 and 9 can go in this box.
These two digits only fit in R5C8 and R6C8 in the box.
The other candidates in R5C8 and R6C8 are therefore eliminated.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R6C7 = 6.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R9C8 = 6.
Spot R5C8 and R6C8 on this column.
These two cells can only hold 2 and 9: they reserve those digits.
2 and 9 are therefore removed from the other cells of the column.
Look at where 4 can go in this box.
All its spots lie on a single row (R4C7 and R4C8).
4 is therefore removed from that row outside the box.
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.
R2C3 and 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 = 4.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 8.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 9.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 3.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 8.
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 7.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 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 R7C8.
Every other digit already appears in its row, column or box.
Only one option remains: R7C8 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R4C7 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R6C3 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R5C6 = 9.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C8 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R6C5 = 2.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R6C6 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R6C8 = 9.
Spot R3C8 and R4C8 on this column.
These two cells can only hold 1 and 3: they reserve those digits.
1 and 3 are therefore removed from the other cells of the column.
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.
R1C1 and R2C4 is seen by both free ends: 7 is therefore eliminated there.
Look at where 7 can go in this box.
All its spots lie on a single row (R2C1 and R2C3).
7 is therefore removed from that row outside the box.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R3C7 = 7.
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.
R2C4 is seen by both free ends: 3 is therefore eliminated there.
Find 3 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 3.
R2C3 is seen by both free ends: 3 is therefore eliminated there.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C1 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C4 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C5 = 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 3 could go.
Across the whole box, 3 has only one possible spot left.
So R2C1 = 3.
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 box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C8 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R3C8 = 3.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R5C5 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C7 = 3.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R6C1 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C4 = 3.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 1.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 1.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 1.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 1.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 3.