Step-by-step solution
Steps 65
Estimated time 104 – 156 min
Hardest technique X-Wing
Clues 25
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C2 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R2C5 = 6.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R3C2 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R4C1 = 1.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C5 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R4C6 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C9 = 9.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R5C3 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R6C4 = 6.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R6C6 = 1.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R7C3 = 1.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R9C7 = 9.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R8C1 = 3.
(1) Spot R2C3 and R2C7 on this row. (2) Spot R4C2 and R6C2 on this column.
(1) These two cells can only hold 4 and 5: they reserve those digits. (2) These two cells can only hold 3 and 8: they reserve those digits.
(1) 4 and 5 are therefore removed from the other cells of the row. (2) 3 and 8 are therefore removed from the other cells of the column.
(1) Find where 5 and 8 can go in this box. (2) Find where 4 and 6 can go in this box.
(1) These two digits only fit in R8C3 and R9C1 in the box. (2) These two digits only fit in R8C6 and R9C6 in the box.
(1) The other candidates in R8C3 and R9C1 are therefore eliminated. (2) The other candidates in R8C6 and R9C6 are therefore eliminated.
(1) Spot R9C1 and R9C5 on this row. (2) Spot R8C6 and R9C6 on this column.
(1) These two cells can only hold 5 and 8: they reserve those digits. (2) These two cells can only hold 4 and 6: they reserve those digits.
(1) 5 and 8 are therefore removed from the other cells of the row. (2) 4 and 6 are therefore removed from the other cells of the column.
(1) Look at where 3 can go in this box. (2) Look at where 3 can go in this box. (3) Look at where 7 can go in this box.
(1) All its spots lie on a single row (R1C5 and R1C6). (2) All its spots lie on a single column (R4C8 and R6C8). (3) All its spots lie on a single row (R7C4 and R7C6).
(1) 3 is therefore removed from that row outside the box. (2) 3 is therefore removed from that column outside the box. (3) 7 is therefore removed from that row outside the box.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 7.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 9.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 9.
Look at cell R2C4.
Every other digit already appears in its row, column or box.
Only one option remains: R2C4 = 2.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 3.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R3C9 = 2.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R8C5 = 9.
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.
Spot R3C6 and R7C6 on this column.
These two cells can only hold 5 and 7: they reserve those digits.
5 and 7 are therefore removed from the other cells of the column.
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.
R1C9 is seen by both free ends: 5 is therefore eliminated there.
Look at where 5 can go on this column.
All its spots lie within a single box (R7C9 and R8C9).
5 is therefore removed from the rest of the box.
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.
R3C1 is seen by both free ends: 5 is therefore eliminated there.
Find where 5 appears on two columns: altogether, only two rows are involved (R1C1, R9C1, R1C5 and R9C5).
On each covered row, 5 must lie within one of these two columns.
5 is therefore removed from those two rows outside the two X-Wing columns.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C8 = 3.
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 column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R6C8 = 5.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 8.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: 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 R1C1 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R1C4 = 9.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R1C5 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C6 = 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 box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R1C9 = 1.
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 4 could go.
Across the whole box, 4 has only one possible spot left.
So R2C7 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R3C1 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R3C4 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R3C6 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R3C7 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R5C4 = 8.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C5 = 3.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R5C6 = 9.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C4 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C6 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C9 = 8.
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 box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C3 = 8.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C8 = 6.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C9 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R9C1 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C5 = 8.
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.
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.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 4.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 4.
Look at cell R9C2.
Every other digit already appears in its row, column or box.
Only one option remains: R9C2 = 4.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 6.