Step-by-step solution
Steps 65
Estimated time 104 – 156 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 1 could go.
Across the whole column, 1 has only one possible spot left.
So R1C1 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R3C7 = 1.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R3C9 = 5.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R5C4 = 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 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 5 could go.
Across the whole box, 5 has only one possible spot left.
So R7C2 = 5.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R7C5 = 4.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R8C3 = 3.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C4 = 1.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C8 = 3.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R1C9 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R2C3 = 4.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R2C7 = 3.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R2C9 = 9.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 8.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R6C7 = 9.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R7C7 = 8.
(1) Find where 3 and 8 can go on this column. (2) Find where 1 and 5 can go on this column. (3) Find where 1 and 5 can go in this box.
(1) These two digits only fit in R1C5 and R4C5 on the column. (2) These two digits only fit in R5C8 and R6C8 on the column. (3) These two digits only fit in R5C8 and R6C8 in the box.
(1) The other candidates in R1C5 and R4C5 are therefore eliminated. (2) The other candidates in R5C8 and R6C8 are therefore eliminated. (3) The other candidates in R5C8 and R6C8 are therefore eliminated.
(1) Look at where 8 can go in this box. (2) Look at where 6 can go in this box.
(1) All its spots lie on a single row (R4C4 and R4C5). (2) All its spots lie on a single column (R4C7 and R5C7).
(1) 8 is therefore removed from that row outside the box. (2) 6 is therefore removed from that column outside the box.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R5C3 = 8.
Look at where 6 can go on this row.
All its spots lie within a single box (R6C5 and R6C6).
6 is therefore removed from the rest of 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.
R9C6 is seen by both free ends: 2 is therefore eliminated there.
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.
R4C6 is seen by both free ends: 2 is therefore eliminated there.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 3.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 8.
Look at cell R1C5.
Every other digit already appears in its row, column or box.
Only one option remains: R1C5 = 3.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R1C2 = 8.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R2C4 = 8.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C1 = 9.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C1 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R6C8 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C1 = 8.
Look at cell R5C8.
Every other digit already appears in its row, column or box.
Only one option remains: R5C8 = 1.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 6.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 1.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 2.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 5.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C7 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R4C9 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R7C8 = 2.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R9C5 = 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.
Look at where 2 can go in this box.
All its spots lie on a single row (R3C4 and R3C5).
2 is therefore removed from that row outside the box.
Find 7 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 7.
R3C2 and R3C3 is seen by both free ends: 7 is therefore eliminated there.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C6 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C8 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R2C1 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R2C2 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R2C8 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R3C2 = 9.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C3 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R3C4 = 2.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R3C5 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C3 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C7 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C2 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C7 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C3 = 9.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C6 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C1 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C5 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C2 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C4 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C6 = 9.