Step-by-step solution
Steps 66
Estimated time 108 – 162 min
Hardest technique X-Wing
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 3.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R1C2 = 2.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R1C4 = 3.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R3C2 = 8.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R3C7 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C9 = 6.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R6C9 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C8 = 3.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R8C1 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C2 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C6 = 3.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R3C1 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C3 = 7.
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.
(1) Spot R5C5 and R6C5 on this column. (2) Spot R5C5 and R6C5 in this box.
(1) These two cells can only hold 2 and 4: they reserve those digits. (2) These two cells can only hold 2 and 4: they reserve those digits.
(1) 2 and 4 are therefore removed from the other cells of the column. (2) 2 and 4 are therefore removed from the other cells of the box.
Find where 2 and 5 can go in this box.
These two digits only fit in R7C7 and R8C8 in the box.
The other candidates in R7C7 and R8C8 are therefore eliminated.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R1C8 = 4.
(1) Look at where 7 can go in this box. (2) Look at where 1 can go in this box.
(1) All its spots lie on a single row (R2C4 and R2C5). (2) All its spots lie on a single column (R4C8 and R5C8).
(1) 7 is therefore removed from that row outside the box. (2) 1 is therefore removed from that column 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.
R9C4 and R8C9 is seen by both free ends: 8 is therefore eliminated there.
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.
R2C1 is seen by both free ends: 5 is therefore eliminated there.
Look at where 5 can go in this box.
All its spots lie on a single row (R1C1 and R1C3).
5 is therefore removed from that row outside the box.
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 2.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R1C1 = 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 box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R2C5 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R2C6 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R3C6 = 9.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R3C8 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C4 = 8.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C6 = 5.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R6C2 = 5.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R7C4 = 2.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R7C6 = 4.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R8C3 = 5.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C5 = 9.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C6 = 8.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 1.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 7.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 4.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 1.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 6.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R4C1 = 6.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R9C3 = 4.
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.
R4C2 and R6C7 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 1.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 9.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 4.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 2.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 8.
Look at cell R5C8.
Every other digit already appears in its row, column or box.
Only one option remains: R5C8 = 1.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 9.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 4.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 2.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 1.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 8.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 9.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 7.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 9.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 8.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 9.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 7.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 1.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 9.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 1.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 7.