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)
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R1C3 = 5.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R1C6 = 4.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R1C7 = 3.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C8 = 8.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R3C5 = 1.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R3C7 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R5C1 = 4.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R5C7 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C7 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C9 = 4.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R8C8 = 5.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C9 = 7.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R9C4 = 4.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C8 = 1.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 9.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 6.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 2.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 7.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R2C8 = 7.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R6C8 = 2.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R8C4 = 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.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R4C2 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C4 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C3 = 7.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R7C1 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C5 = 7.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C3 = 1.
Look at where 8 can go in this box.
All its spots lie on a single row (R5C5 and R5C6).
8 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.
R4C1 is seen by both free ends: 6 is therefore eliminated there.
Find 9 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 9.
R4C1 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 8.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R7C3 = 8.
Look at where 9 can go on this column.
All its spots lie within a single box (R4C3 and R5C3).
9 is therefore removed from the rest of the box.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 1.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 3.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 5.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 9.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 6.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 8.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 3.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 8.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 1.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 2.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 5.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R3C1 = 2.
Find 6 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 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 = 2.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 9.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 2.
Look at cell R8C1.
Every other digit already appears in its row, column or box.
Only one option remains: R8C1 = 6.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 9.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 2.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 6.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 9.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 6.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 6.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 9.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 6.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 9.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 9.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 6.