Step-by-step solution
Steps 64
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 column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R1C1 = 5.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R1C7 = 2.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R1C8 = 9.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R3C3 = 2.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R4C2 = 3.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R4C4 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R6C1 = 9.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C7 = 3.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R7C2 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C3 = 9.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C6 = 8.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C9 = 5.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R8C3 = 3.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C4 = 9.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C6 = 2.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R9C1 = 1.
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 9 could go.
Across the whole row, 9 has only one possible spot left.
So R2C6 = 9.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R4C5 = 5.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R5C5 = 9.
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 7 could go.
Across the whole row, 7 has only one possible spot left.
So R8C1 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R9C6 = 5.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R4C8 = 7.
Spot R5C4 and R7C4 on this column.
These two cells can only hold 4 and 6: they reserve those digits.
4 and 6 are therefore removed from the other cells of the column.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 7.
Find where 1 and 3 can go on this row.
These two digits only fit in R3C4 and R3C9 on the row.
The other candidates in R3C4 and R3C9 are therefore eliminated.
Look at where 7 can go on this row.
All its spots lie within a single box (R3C5 and R3C6).
7 is therefore removed from the rest of the box.
Find 4 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 4.
R5C2 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 (R4C1 and R4C3).
4 is therefore removed from that row outside the box.
Find 6 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 6.
R5C2 is seen by both free ends: 6 is therefore eliminated there.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C5 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C9 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C9 = 8.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R3C1 = 8.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R4C7 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C8 = 8.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 6.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 1.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 3.
Look at cell R2C3.
Every other digit already appears in its row, column or box.
Only one option remains: R2C3 = 6.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 1.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 4.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 4.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 1.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 3.
Look at cell R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 6.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 7.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 6.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 4.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 6.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 6.
Look at cell R9C5.
Every other digit already appears in its row, column or box.
Only one option remains: R9C5 = 4.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 7.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 4.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 4.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 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 R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 6.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 7.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 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 R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 6.