Step-by-step solution
Steps 67
Estimated time 108 – 162 min
Hardest technique X-Wing
Clues 23
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R3C3 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C4 = 6.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R4C5 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C5 = 7.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R5C7 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R6C5 = 9.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R6C7 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C5 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C2 = 2.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R9C1 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C8 = 6.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R1C6 = 2.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R2C1 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R2C2 = 6.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R2C3 = 5.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R3C9 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C3 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C5 = 2.
(1) Spot R3C2 and R3C5 on this row. (2) Spot R1C2 and R3C2 on this column.
(1) These two cells can only hold 1 and 4: they reserve those digits. (2) These two cells can only hold 1 and 4: they reserve those digits.
(1) 1 and 4 are therefore removed from the other cells of the row. (2) 1 and 4 are therefore removed from the other cells of the column.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R4C1 = 1.
Look at where 7 can go in this box.
All its spots lie on a single row (R7C1 and R7C2).
7 is therefore removed from that row outside the box.
Find 1 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 1.
R8C5 is seen by both free ends: 1 is therefore eliminated there.
Spot R8C5 and R9C6 in this box.
These two cells can only hold 3 and 4: they reserve those digits.
3 and 4 are therefore removed from the other cells of the box.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 5.
Spot R8C4 and R9C4 on this column.
These two cells can only hold 1 and 8: they reserve those digits.
1 and 8 are therefore removed from the other cells of the column.
(1) Look at where 5 can go on this row. (2) Look at where 5 can go on this column.
(1) All its spots lie within a single box (R4C7, R4C8 and R4C9). (2) All its spots lie within a single box (R4C9 and R5C9).
(1) 5 is therefore removed from the rest of the box. (2) 5 is therefore removed from the rest of the box.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R4C9 = 5.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C9 = 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.
R9C3 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.
R2C7 is seen by both free ends: 3 is therefore eliminated there.
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.
R1C4 is seen by both free ends: 3 is therefore eliminated there.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 4.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 1.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 1.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 3.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 8.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 4.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 4.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 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 R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R3C7 = 7.
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 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C2 = 3.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C7 = 9.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C8 = 7.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C9 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C1 = 7.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C3 = 4.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C8 = 3.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R7C1 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C2 = 7.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R7C3 = 1.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C7 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C9 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C3 = 3.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C4 = 8.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C7 = 5.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R8C8 = 1.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C3 = 8.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C4 = 1.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C7 = 4.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 3.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 9.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 5.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 3.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 5.