Step-by-step solution
Steps 69
Estimated time 112 – 168 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 2 could go.
Across the whole row, 2 has only one possible spot left.
So R1C3 = 2.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R1C7 = 1.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R3C5 = 3.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R5C1 = 6.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R5C6 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C8 = 3.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C9 = 2.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R7C3 = 3.
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 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C9 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C4 = 3.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C8 = 7.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R9C2 = 2.
(1) Spot R3C2 and R3C3 on this row. (2) Spot R3C2 and R3C3 in this box.
(1) These two cells can only hold 1 and 5: they reserve those digits. (2) These two cells can only hold 1 and 5: they reserve those digits.
(1) 1 and 5 are therefore removed from the other cells of the row. (2) 1 and 5 are therefore removed from the other cells of the box.
Look at cell R3C1.
Every other digit already appears in its row, column or box.
Only one option remains: R3C1 = 7.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R1C6 = 5.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R2C4 = 1.
Spot R1C1 and R9C1 on this column.
These two cells can only hold 8 and 9: they reserve those digits.
8 and 9 are therefore removed from the other cells of the column.
(1) Find where 1 and 8 can go on this row. (2) Find where 2 and 7 can go on this row. (3) Find where 1 and 8 can go in this box. (4) Find where 2 and 7 can go in this box.
(1) These two digits only fit in R4C2 and R4C3 on the row. (2) These two digits only fit in R7C4 and R7C5 on the row. (3) These two digits only fit in R4C2 and R4C3 in the box. (4) These two digits only fit in R7C4 and R7C5 in the box.
(1) The other candidates in R4C2 and R4C3 are therefore eliminated. (2) The other candidates in R7C4 and R7C5 are therefore eliminated. (3) The other candidates in R4C2 and R4C3 are therefore eliminated. (4) The other candidates in R7C4 and R7C5 are therefore eliminated.
(1) Look at where 8 can go in this box. (2) Look at where 5 can go in this box. (3) Look at where 4 can go in this box.
(1) All its spots lie on a single column (R1C5 and R2C5). (2) All its spots lie on a single row (R5C2 and R5C3). (3) All its spots lie on a single column (R7C2 and R8C2).
(1) 8 is therefore removed from that column outside the box. (2) 5 is therefore removed from that row outside the box. (3) 4 is therefore removed from that column outside the box.
Spot R5C7 and R9C7 on this column.
These two cells can only hold 4 and 9: they reserve those digits.
4 and 9 are therefore removed from the other cells of the column.
Spot R7C2 and R7C8 on this row.
These two cells can only hold 4 and 9: they reserve those digits.
4 and 9 are therefore removed from the other cells of the row.
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.
R1C5 and R4C6 is seen by both free ends: 6 is therefore eliminated there.
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.
R1C8 is seen by both free ends: 9 is therefore eliminated there.
Spot R1C8 and R3C9 in this box.
These two cells can only hold 4 and 6: they reserve those digits.
4 and 6 are therefore removed from the other cells of 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.
R5C2 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 5.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 1.
Look at cell R3C3.
Every other digit already appears in its row, column or box.
Only one option remains: R3C3 = 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 R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R1C1 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C4 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C5 = 8.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C8 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C9 = 7.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R2C3 = 8.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R2C5 = 7.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C6 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R3C9 = 4.
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 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 1 could go.
Across the whole box, 1 has only one possible spot left.
So R7C6 = 1.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C7 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R8C1 = 1.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R8C6 = 8.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C7 = 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 R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 2.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 6.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 2.
Find 4 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 4.
R4C8 is seen by both free ends: 4 is therefore eliminated there.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C7 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R7C2 = 9.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R7C8 = 4.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C2 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C5 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C6 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C7 = 9.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 9.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 5.
Look at cell R4C9.
Every other digit already appears in its row, column or box.
Only one option remains: R4C9 = 6.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 9.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 4.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 6.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 9.
Look at cell R2C8.
Every other digit already appears in its row, column or box.
Only one option remains: R2C8 = 9.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 5.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 4.