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 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 7 could go.
Across the whole column, 7 has only one possible spot left.
So R2C5 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R2C7 = 5.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R2C9 = 3.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R4C1 = 3.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R4C4 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R5C2 = 1.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R1C7 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R2C1 = 1.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R3C7 = 9.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R4C9 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C9 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C7 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C8 = 9.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R3C2 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C1 = 5.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R9C1 = 7.
(1) Find where 3 and 4 can go on this row. (2) Find where 3 and 4 can go in this box.
(1) These two digits only fit in R8C4 and R8C5 on the row. (2) These two digits only fit in R8C4 and R8C5 in the box.
(1) The other candidates in R8C4 and R8C5 are therefore eliminated. (2) The other candidates in R8C4 and R8C5 are therefore eliminated.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R7C4 = 1.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R8C2 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C9 = 1.
Find where 6 and 9 can go in this box.
These two digits only fit in R7C5 and R7C6 in the box.
The other candidates in R7C5 and R7C6 are therefore eliminated.
Spot R7C5 and R7C6 on this row.
These two cells can only hold 6 and 9: they reserve those digits.
6 and 9 are therefore removed from the other cells of the row.
(1) Look at where 6 can go in this box. (2) Look at where 8 can go in this box.
(1) All its spots lie on a single row (R6C5 and R6C6). (2) All its spots lie on a single row (R5C7 and R5C8).
(1) 6 is therefore removed from that row outside the box. (2) 8 is therefore removed from that row outside the box.
Look at where 6 can go on this column.
All its spots lie within a single box (R1C4 and R2C4).
6 is therefore removed from the rest of the box.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 2.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R5C4 = 5.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C4 = 2.
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.
Find 2 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 2.
R1C2 and R8C1 is seen by both free ends: 2 is therefore eliminated there.
Look at where 2 can go in this box.
All its spots lie on a single column (R1C1 and R3C1).
2 is therefore removed from that column outside the box.
Find 2 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 2.
R4C7 is seen by both free ends: 2 is therefore eliminated there.
Look at cell R4C7.
Every other digit already appears in its row, column or box.
Only one option remains: R4C7 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R1C2 = 4.
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 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C1 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C3 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C5 = 4.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 6.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 9.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 3.
Look at cell R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 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 R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 2.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 6.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 9.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 3.
Find 8 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 8.
R2C2 and R8C1 is seen by both free ends: 8 is therefore eliminated there.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 6.
Look at cell R2C4.
Every other digit already appears in its row, column or box.
Only one option remains: R2C4 = 8.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 2.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 7.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 7.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 8.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 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 R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 2.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 8.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 6.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 8.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 6.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 2.
Look at cell R5C8.
Every other digit already appears in its row, column or box.
Only one option remains: R5C8 = 8.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 2.
Look at cell R3C1.
Every other digit already appears in its row, column or box.
Only one option remains: R3C1 = 2.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 8.