Step-by-step solution
Steps 69
Estimated time 112 – 168 min
Hardest technique X-Wing
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R1C1 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R2C4 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C1 = 8.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C4 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C7 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C4 = 2.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R6C3 = 2.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R6C9 = 8.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R7C6 = 8.
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 row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R9C2 = 4.
(1) Find where 5 and 6 can go on this column. (2) Find where 2 and 4 can go in this box.
(1) These two digits only fit in R1C4 and R6C4 on the column. (2) These two digits only fit in R7C5 and R8C6 in the box.
(1) The other candidates in R1C4 and R6C4 are therefore eliminated. (2) The other candidates in R7C5 and R8C6 are therefore eliminated.
(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 (R4C8 and R6C8). (2) All its spots lie on a single row (R7C2 and R7C3).
(1) 3 is therefore removed from that column outside the box. (2) 7 is therefore removed from that row outside the box.
Look at where 3 can go on this row.
All its spots lie within a single box (R1C7 and R1C9).
3 is therefore removed from the rest of the box.
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.
R1C6 and R7C5 is seen by both free ends: 2 is therefore eliminated there.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 4.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 2.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R2C3 = 3.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R2C5 = 2.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R3C5 = 3.
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 9 could go.
Across the whole box, 9 has only one possible spot left.
So R5C2 = 9.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C1 = 3.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R6C6 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C1 = 9.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R8C9 = 4.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C8 = 2.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R2C9 = 9.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R3C3 = 9.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R7C7 = 2.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C8 = 9.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C9 = 5.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R1C9 = 3.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R7C4 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C7 = 3.
Spot R4C5 and R6C5 in this box.
These two cells can only hold 1 and 7: they reserve those digits.
1 and 7 are therefore removed from the other cells of the box.
Look at where 7 can go on this row.
All its spots lie within a single box (R5C7, R5C8 and R5C9).
7 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.
R5C7 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 column (R5C8 and R6C8).
5 is therefore removed from that column outside the box.
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 and R5C1 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 = 6.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 5.
Look at cell R3C1.
Every other digit already appears in its row, column or box.
Only one option remains: R3C1 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R5C8 = 5.
Find 6 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 6.
R1C4 and R4C6 is seen by both free ends: 6 is therefore eliminated there.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 5.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 5.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 6.
Look at cell R6C8.
Every other digit already appears in its row, column or box.
Only one option remains: R6C8 = 1.
Look at cell R9C8.
Every other digit already appears in its row, column or box.
Only one option remains: R9C8 = 7.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 6.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 7.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 7.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 1.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 1.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 7.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 7.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 1.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 1.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 6.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 5.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 7.
Look at cell R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 6.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 7.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 6.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 1.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 6.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 1.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 7.