Step-by-step solution
Steps 63
Estimated time 65 – 98 min
Hardest technique Box/Line
Clues 24
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
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 R2C4 = 7.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R2C7 = 1.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R3C6 = 2.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R4C4 = 5.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C9 = 6.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R5C8 = 5.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C9 = 1.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R6C9 = 2.
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 column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R7C8 = 2.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C2 = 2.
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.
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.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R2C3 = 5.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R3C7 = 5.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R6C3 = 1.
Look at where 4 can go in this box.
All its spots lie on a single column (R5C7 and R6C7).
4 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.
R1C1 and R7C3 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 R9C1 = 6.
Spot R1C1 and R1C8 on this row.
These two cells can only hold 3 and 9: they reserve those digits.
3 and 9 are therefore removed from the other cells of the row.
Spot R8C4, R8C6 and R9C4 in this box.
These three cells share only 3, 4 and 9.
3, 4 and 9 are therefore removed from the rest of the box.
Find 9 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 9.
R4C8 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R4C8.
Every other digit already appears in its row, column or box.
Only one option remains: R4C8 = 3.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 9.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 3.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 1.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 6.
Look at cell R1C1.
Every other digit already appears in its row, column or box.
Only one option remains: R1C1 = 3.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 8.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 4.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 9.
Look at cell R6C7.
Every other digit already appears in its row, column or box.
Only one option remains: R6C7 = 8.
Look at cell R4C2.
Every other digit already appears in its row, column or box.
Only one option remains: R4C2 = 7.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 8.
Look at cell R4C7.
Every other digit already appears in its row, column or box.
Only one option remains: R4C7 = 9.
Look at cell R5C3.
Every other digit already appears in its row, column or box.
Only one option remains: R5C3 = 3.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 7.
Look at cell R7C6.
Every other digit already appears in its row, column or box.
Only one option remains: R7C6 = 1.
Look at cell R7C4.
Every other digit already appears in its row, column or box.
Only one option remains: R7C4 = 8.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R1C4 = 1.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R7C3 = 7.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R7C7 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C2 = 3.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R9C7 = 7.
Find 9 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 9.
R8C3 is seen by both free ends: 9 is therefore eliminated there.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 4.
Look at cell R1C3.
Every other digit already appears in its row, column or box.
Only one option remains: R1C3 = 6.
Look at cell R1C6.
Every other digit already appears in its row, column or box.
Only one option remains: R1C6 = 4.
Look at cell R2C5.
Every other digit already appears in its row, column or box.
Only one option remains: R2C5 = 3.
Look at cell R2C6.
Every other digit already appears in its row, column or box.
Only one option remains: R2C6 = 9.
Look at cell R3C3.
Every other digit already appears in its row, column or box.
Only one option remains: R3C3 = 9.
Look at cell R3C4.
Every other digit already appears in its row, column or box.
Only one option remains: R3C4 = 6.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 4.
Look at cell R7C2.
Every other digit already appears in its row, column or box.
Only one option remains: R7C2 = 9.
Look at cell R7C9.
Every other digit already appears in its row, column or box.
Only one option remains: R7C9 = 4.
Look at cell R8C6.
Every other digit already appears in its row, column or box.
Only one option remains: R8C6 = 3.
Look at cell R9C9.
Every other digit already appears in its row, column or box.
Only one option remains: R9C9 = 9.
Look at cell R2C2.
Every other digit already appears in its row, column or box.
Only one option remains: R2C2 = 4.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 3.
Look at cell R6C6.
Every other digit already appears in its row, column or box.
Only one option remains: R6C6 = 6.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 9.
Look at cell R9C4.
Every other digit already appears in its row, column or box.
Only one option remains: R9C4 = 4.