Step-by-step solution
Steps 64
Estimated time 65 – 98 min
Hardest technique Box/Line
Clues 22
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 7.
Look at cell R8C7.
Every other digit already appears in its row, column or box.
Only one option remains: R8C7 = 2.
Look at cell R2C7.
Every other digit already appears in its row, column or box.
Only one option remains: R2C7 = 9.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R1C3 = 4.
Scan the empty cells in this row where 8 could go.
Across the whole row, 8 has only one possible spot left.
So R1C6 = 8.
Scan the empty cells in this column where 7 could go.
Across the whole column, 7 has only one possible spot left.
So R2C5 = 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 9 could go.
Across the whole box, 9 has only one possible spot left.
So R3C3 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R3C4 = 4.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C8 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C5 = 4.
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 box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R6C8 = 4.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R7C3 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C9 = 7.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C1 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C2 = 7.
Spot R4C9 and R6C9 on this column.
These two cells can only hold 2 and 8: they reserve those digits.
2 and 8 are therefore removed from the other cells of the column.
Find where 1 and 9 can go on this column.
These two digits only fit in R5C6 and R9C6 on the column.
The other candidates in R5C6 and R9C6 are therefore eliminated.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C2 = 6.
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 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C4 = 6.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R5C2 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R5C3 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C6 = 6.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C9 = 6.
Look at cell R1C9.
Every other digit already appears in its row, column or box.
Only one option remains: R1C9 = 5.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 1.
Look at cell R1C8.
Every other digit already appears in its row, column or box.
Only one option remains: R1C8 = 3.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 2.
Scan the empty cells in this row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R2C6 = 2.
Find 3 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 3.
R4C1, R6C4 and R6C5 is seen by both free ends: 3 is therefore eliminated there.
Find 5 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 5.
R4C1, R6C4 and R6C5 is seen by both free ends: 5 is therefore eliminated there.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 8.
Look at cell R4C9.
Every other digit already appears in its row, column or box.
Only one option remains: R4C9 = 2.
Look at cell R6C9.
Every other digit already appears in its row, column or box.
Only one option remains: R6C9 = 8.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 1.
Look at cell R5C4.
Every other digit already appears in its row, column or box.
Only one option remains: R5C4 = 8.
Look at cell R5C6.
Every other digit already appears in its row, column or box.
Only one option remains: R5C6 = 9.
Look at cell R6C5.
Every other digit already appears in its row, column or box.
Only one option remains: R6C5 = 2.
Look at cell R9C6.
Every other digit already appears in its row, column or box.
Only one option remains: R9C6 = 1.
Look at cell R5C1.
Every other digit already appears in its row, column or box.
Only one option remains: R5C1 = 1.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R2C3 = 1.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R6C1 = 9.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R8C8 = 1.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R9C5 = 9.
Find 3 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 3.
R8C4 is seen by both free ends: 3 is therefore eliminated there.
Look at cell R8C4.
Every other digit already appears in its row, column or box.
Only one option remains: R8C4 = 5.
Look at cell R2C4.
Every other digit already appears in its row, column or box.
Only one option remains: R2C4 = 3.
Look at cell R3C6.
Every other digit already appears in its row, column or box.
Only one option remains: R3C6 = 5.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 3.
Look at cell R2C1.
Every other digit already appears in its row, column or box.
Only one option remains: R2C1 = 5.
Look at cell R3C2.
Every other digit already appears in its row, column or box.
Only one option remains: R3C2 = 3.
Look at cell R4C5.
Every other digit already appears in its row, column or box.
Only one option remains: R4C5 = 5.
Look at cell R6C2.
Every other digit already appears in its row, column or box.
Only one option remains: R6C2 = 5.
Look at cell R6C3.
Every other digit already appears in its row, column or box.
Only one option remains: R6C3 = 3.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 3.
Look at cell R7C5.
Every other digit already appears in its row, column or box.
Only one option remains: R7C5 = 8.
Look at cell R7C8.
Every other digit already appears in its row, column or box.
Only one option remains: R7C8 = 5.
Look at cell R8C3.
Every other digit already appears in its row, column or box.
Only one option remains: R8C3 = 8.
Look at cell R8C5.
Every other digit already appears in its row, column or box.
Only one option remains: R8C5 = 3.
Look at cell R9C3.
Every other digit already appears in its row, column or box.
Only one option remains: R9C3 = 5.
Look at cell R9C8.
Every other digit already appears in its row, column or box.
Only one option remains: R9C8 = 8.