Step-by-step solution
Steps 66
Estimated time 108 – 162 min
Hardest technique X-Wing
Clues 24
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R1C4 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C5 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C3 = 8.
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 column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R2C8 = 4.
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 row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R4C3 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R5C1 = 8.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R5C7 = 2.
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 row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R6C5 = 2.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R7C2 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R7C8 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C4 = 4.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R8C6 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C1 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C4 = 6.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 9.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 1.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R1C2 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C7 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C8 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R7C7 = 1.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R8C7 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C8 = 7.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C2 = 1.
Look at where 1 can go in this box.
All its spots lie on a single row (R6C8 and R6C9).
1 is therefore removed from that row outside the box.
Look at where 9 can go on this row.
All its spots lie within a single box (R3C4 and R3C6).
9 is therefore removed from the rest of the box.
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.
R1C3 and R8C1 is seen by both free ends: 9 is therefore eliminated there.
(1) Spot R4C1 and R8C1 on this column. (2) Spot R1C3 and R2C2 in this box.
(1) These two cells can only hold 3 and 5: they reserve those digits. (2) These two cells can only hold 3 and 5: they reserve those digits.
(1) 3 and 5 are therefore removed from the other cells of the column. (2) 3 and 5 are therefore removed from the other cells of the box.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 6.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 6.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R1C6 = 6.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R5C8 = 6.
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.
R1C9 is seen by both free ends: 3 is therefore eliminated there.
Look at where 3 can go in this box.
All its spots lie on a single column (R1C7 and R2C7).
3 is therefore removed from that column outside the box.
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.
R8C9 is seen by both free ends: 3 is therefore eliminated there.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C8 = 3.
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.
R8C9 is seen by both free ends: 5 is therefore eliminated there.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 9.
Look at cell R9C7.
Every other digit already appears in its row, column or box.
Only one option remains: R9C7 = 5.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 9.
Find 5 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 5.
R2C2 is seen by both free ends: 5 is therefore eliminated there.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 3.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 9.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 5.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 3.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 5.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 3.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 1.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 1.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 5.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 5.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 3.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 1.
Look at cell R6C8.
Every other digit already appears in its row, column or box.
Only one option remains: R6C8 = 1.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 5.
Look at cell R8C1.
Every other digit already appears in its row, column or box.
Only one option remains: R8C1 = 5.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 3.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 9.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 9.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 5.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 3.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 9.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 3.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 1.