Step-by-step solution
Steps 62
Estimated time 63 – 95 min
Hardest technique Box/Line
Clues 23
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Look at cell R9C1.
Every other digit already appears in its row, column or box.
Only one option remains: R9C1 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R4C1 = 6.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R4C8 = 3.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R5C3 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C9 = 6.
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 column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R8C5 = 2.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R8C8 = 1.
Look at cell R7C1.
Every other digit already appears in its row, column or box.
Only one option remains: R7C1 = 5.
Look at cell R8C1.
Every other digit already appears in its row, column or box.
Only one option remains: R8C1 = 8.
Scan the empty cells in this column where 3 could go.
Across the whole column, 3 has only one possible spot left.
So R1C1 = 3.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R3C5 = 6.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R3C6 = 3.
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 column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R6C3 = 9.
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 row where 2 could go.
Across the whole row, 2 has only one possible spot left.
So R7C7 = 2.
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 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C2 = 9.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R8C4 = 3.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C6 = 5.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R9C2 = 4.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R9C9 = 3.
Scan the empty cells in this row where 1 could go.
Across the whole row, 1 has only one possible spot left.
So R1C9 = 1.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R3C3 = 2.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R6C4 = 5.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R1C5 = 5.
Spot R1C3 and R2C3 in this box.
These two cells can only hold 4 and 8: they reserve those digits.
4 and 8 are therefore removed from the other cells of the box.
Find where 2 and 8 can go on this row.
These two digits only fit in R6C2 and R6C9 on the row.
The other candidates in R6C2 and R6C9 are therefore eliminated.
Look at where 4 can go on this row.
All its spots lie within a single box (R3C8 and R3C9).
4 is therefore removed from the rest of the box.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R8C7 = 4.
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.
Find 7 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 7.
R3C2 is seen by both free ends: 7 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 R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 4.
Look at cell R3C8.
Every other digit already appears in its row, column or box.
Only one option remains: R3C8 = 7.
Look at cell R6C8.
Every other digit already appears in its row, column or box.
Only one option remains: R6C8 = 4.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 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 R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 7.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C3 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R1C4 = 8.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R1C6 = 7.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R2C1 = 1.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R2C2 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R2C3 = 8.
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 4 could go.
Across the whole box, 4 has only one possible spot left.
So R2C6 = 4.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R4C2 = 8.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R4C4 = 4.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R4C5 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R4C9 = 9.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R5C2 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R5C4 = 9.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R5C5 = 8.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C9 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C1 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R6C2 = 2.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R6C6 = 1.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R6C9 = 8.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C5 = 1.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C6 = 9.