Step-by-step solution
Steps 66
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 row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R3C4 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C1 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C3 = 1.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R6C7 = 6.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R7C5 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C2 = 7.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R8C3 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R8C9 = 1.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R9C4 = 1.
Find where 2 and 4 can go on this column.
These two digits only fit in R4C7 and R7C7 on the column.
The other candidates in R4C7 and R7C7 are therefore eliminated.
(1) Look at where 2 can go in this box. (2) Look at where 5 can go in this box. (3) Look at where 8 can go in this box.
(1) All its spots lie on a single column (R1C6 and R3C6). (2) All its spots lie on a single row (R3C8 and R3C9). (3) All its spots lie on a single column (R4C9 and R5C9).
(1) 2 is therefore removed from that column outside the box. (2) 5 is therefore removed from that row outside the box. (3) 8 is therefore removed from that column outside the box.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C8 = 8.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R4C7 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R6C4 = 2.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R6C8 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R7C7 = 4.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C2 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C2 = 2.
(1) Look at where 3 can go in this box. (2) Look at where 7 can go in this box.
(1) All its spots lie on a single column (R4C9 and R6C9). (2) All its spots lie on a single column (R4C9, R5C9 and R6C9).
(1) 3 is therefore removed from that column outside the box. (2) 7 is therefore removed from that column outside the box.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C5 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C9 = 3.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R5C9 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C2 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C9 = 7.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R7C1 = 3.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 4.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C5 = 3.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R2C4 = 7.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R9C5 = 4.
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.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 7.
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 column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R4C3 = 7.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R3C6 = 2.
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.
R7C3 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 (R9C1 and R9C3).
5 is therefore removed from that row outside the box.
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.
R2C1 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R1C7 = 1.
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.
R7C3 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R1C1 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R9C9 = 2.
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.
R2C5 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 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 R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 5.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 5.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 9.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 5.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 5.
Look at cell R8C8.
Every other digit already appears in its row, column or box.
Only one option remains: R8C8 = 3.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 5.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 9.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 9.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 9.
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 5.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 8.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 8.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 3.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 9.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 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 R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 8.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 9.