Step-by-step solution
Steps 67
Estimated time 68 – 102 min
Hardest technique Box/Line
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R1C7 = 9.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R3C6 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R3C7 = 8.
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 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C6 = 6.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R6C2 = 9.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C7 = 1.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R6C9 = 6.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R7C2 = 3.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R7C7 = 6.
Scan the empty cells in this column where 8 could go.
Across the whole column, 8 has only one possible spot left.
So R8C1 = 8.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R8C3 = 6.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C9 = 8.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R1C5 = 6.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R2C1 = 9.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R3C1 = 6.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C7 = 7.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R5C1 = 7.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R2C5 = 3.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C6 = 8.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C5 = 8.
(1) Spot R1C4 and R1C6 on this row. (2) Spot R1C4 and R7C4 on this column.
(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 column.
Find where 2 and 9 can go in this box.
These two digits only fit in R4C5 and R4C6 in the box.
The other candidates in R4C5 and R4C6 are therefore eliminated.
Spot R4C5 and R4C6 on this row.
These two cells can only hold 2 and 9: they reserve those digits.
2 and 9 are therefore removed from the other cells of the row.
(1) Look at where 4 can go in this box. (2) Look at where 5 can go in this box.
(1) All its spots lie on a single column (R1C9 and R2C9). (2) All its spots lie on a single row (R6C5 and R6C6).
(1) 4 is therefore removed from that column outside the box. (2) 5 is therefore removed from that row outside the box.
Look at where 5 can go in this box.
All its spots lie on a single row (R5C2 and R5C3).
5 is therefore removed from that row outside the box.
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.
R6C6 and R8C4 is seen by both free ends: 4 is therefore eliminated there.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 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 = 1.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 7.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 2.
Look at cell R9C5.
Every other digit already appears in its row, column or box.
Only one option remains: R9C5 = 9.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 4.
Look at cell R1C4.
Every other digit already appears in its row, column or box.
Only one option remains: R1C4 = 5.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 2.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 9.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 1.
Look at cell R7C8.
Every other digit already appears in its row, column or box.
Only one option remains: R7C8 = 5.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R3C9 = 3.
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 4 could go.
Across the whole box, 4 has only one possible spot left.
So R8C7 = 4.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R8C8 = 9.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R9C2 = 7.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 2.
(1) Find where 1 appears on two rows: altogether, only two columns are involved (R2C2, R2C9, R8C2 and R8C9). (2) Find where 1 appears on two columns: altogether, only two rows are involved (R3C3, R9C3, R3C8 and R9C8).
(1) On each covered column, 1 must lie within one of these two rows. (2) On each covered row, 1 must lie within one of these two columns.
(1) 1 is therefore removed from those two columns outside the two X-Wing rows. (2) 1 is therefore removed from those two rows outside the two X-Wing columns.
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.
R3C2 and R9C3 is seen by both free ends: 2 is therefore eliminated there.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 5.
Look at cell R5C2.
Every other digit already appears in its row, column or box.
Only one option remains: R5C2 = 4.
Look at cell R5C8.
Every other digit already appears in its row, column or box.
Only one option remains: R5C8 = 3.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 3.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 4.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 1.
Look at cell R9C8.
Every other digit already appears in its row, column or box.
Only one option remains: R9C8 = 2.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 1.
Look at cell R2C9.
Every other digit already appears in its row, column or box.
Only one option remains: R2C9 = 4.
Look at cell R3C3.
Every other digit already appears in its row, column or box.
Only one option remains: R3C3 = 2.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 1.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 3.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 4.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 5.
Look at cell R8C2.
Every other digit already appears in its row, column or box.
Only one option remains: R8C2 = 2.
Look at cell R8C9.
Every other digit already appears in its row, column or box.
Only one option remains: R8C9 = 1.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 4.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 2.