Step-by-step solution
Steps 66
Estimated time 68 – 102 min
Hardest technique Box/Line
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 9.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R4C6 = 6.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R5C8 = 7.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R6C2 = 6.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C6 = 4.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R8C3 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R9C2 = 8.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 8.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 9.
Scan the empty cells in this row where 8 could go.
Across the whole row, 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 R4C8 = 8.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R4C9 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R5C3 = 8.
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.
Look at cell R4C3.
Every other digit already appears in its row, column or box.
Only one option remains: R4C3 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R6C8 = 5.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C9 = 5.
(1) Spot R5C5 and R5C6 on this row. (2) Spot R6C7 and R8C7 on this column.
(1) These two cells can only hold 1 and 5: they reserve those digits. (2) These two cells can only hold 2 and 3: they reserve those digits.
(1) 1 and 5 are therefore removed from the other cells of the row. (2) 2 and 3 are therefore removed from the other cells of the column.
Find where 4 and 9 can go on this row.
These two digits only fit in R7C3 and R7C8 on the row.
The other candidates in R7C3 and R7C8 are therefore eliminated.
Spot R7C8 and R9C7 in this box.
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 box.
(1) Look at where 9 can go in this box. (2) Look at where 2 can go in this box.
(1) All its spots lie on a single column (R7C3 and R9C3). (2) All its spots lie on a single row (R9C5 and R9C6).
(1) 9 is therefore removed from that column outside the box. (2) 2 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.
R1C3 and R9C1 is seen by both free ends: 4 is therefore eliminated there.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R1C7 = 7.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C8 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R2C1 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R2C4 = 7.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R3C1 = 7.
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 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 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C7 = 4.
Look at cell R3C7.
Every other digit already appears in its row, column or box.
Only one option remains: R3C7 = 9.
Look at cell R7C3.
Every other digit already appears in its row, column or box.
Only one option remains: R7C3 = 4.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R9C3 = 9.
Look at where 2 can go in this box.
All its spots lie on a single column (R1C3 and R3C3).
2 is therefore removed from that column outside the box.
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.
R3C8 and R7C9 is seen by both free ends: 6 is therefore eliminated there.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R1C9 = 6.
Scan the empty cells in this row where 6 could go.
Across the whole row, 6 has only one possible spot left.
So R3C4 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C4 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C5 = 6.
Scan the empty cells in this column where 6 could go.
Across the whole column, 6 has only one possible spot left.
So R9C8 = 6.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 1.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C3 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R1C5 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C6 = 3.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R2C5 = 9.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R2C8 = 1.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R2C9 = 2.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R3C3 = 1.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R3C6 = 2.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R3C8 = 3.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C1 = 2.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C6 = 5.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C9 = 3.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C1 = 1.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R6C3 = 3.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R6C7 = 2.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C2 = 1.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C7 = 3.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C8 = 2.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R9C1 = 3.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C5 = 2.
Look at cell R1C2.
Every other digit already appears in its row, column or box.
Only one option remains: R1C2 = 9.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 3.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 5.
Look at cell R5C5.
Every other digit already appears in its row, column or box.
Only one option remains: R5C5 = 1.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 1.