Step-by-step solution
Steps 63
Estimated time 65 – 98 min
Hardest technique Box/Line
Clues 26
Given digit (clue)
Solved digit
Placement (✓ RlCc=d)
Elimination (✗ RlCc≠d)
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R1C6 = 9.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R1C8 = 5.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R1C9 = 4.
Scan the empty cells in this row where 7 could go.
Across the whole row, 7 has only one possible spot left.
So R2C9 = 7.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R3C1 = 5.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R3C2 = 8.
Scan the empty cells in this row where 9 could go.
Across the whole row, 9 has only one possible spot left.
So R3C3 = 9.
Scan the empty cells in this column where 5 could go.
Across the whole column, 5 has only one possible spot left.
So R4C5 = 5.
Scan the empty cells in this column where 9 could go.
Across the whole column, 9 has only one possible spot left.
So R4C7 = 9.
Scan the empty cells in this box where 4 could go.
Across the whole box, 4 has only one possible spot left.
So R5C8 = 4.
Scan the empty cells in this row where 5 could go.
Across the whole row, 5 has only one possible spot left.
So R6C9 = 5.
Scan the empty cells in this column where 1 could go.
Across the whole column, 1 has only one possible spot left.
So R8C1 = 1.
Scan the empty cells in this box where 5 could go.
Across the whole box, 5 has only one possible spot left.
So R8C2 = 5.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R9C5 = 4.
Scan the empty cells in this row where 4 could go.
Across the whole row, 4 has only one possible spot left.
So R2C2 = 4.
Scan the empty cells in this column where 4 could go.
Across the whole column, 4 has only one possible spot left.
So R4C3 = 4.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R2C4 = 3.
Find where 1 and 9 can go on this row.
These two digits only fit in R9C4 and R9C9 on the row.
The other candidates in R9C4 and R9C9 are therefore eliminated.
(1) Look at where 7 can go in this box. (2) Look at where 8 can go in this box. (3) Look at where 8 can go in this box.
(1) All its spots lie on a single column (R5C7 and R6C7). (2) All its spots lie on a single column (R4C9 and R5C9). (3) All its spots lie on a single column (R7C3 and R8C3).
(1) 7 is therefore removed from that column outside the box. (2) 8 is therefore removed from that column outside the box. (3) 8 is therefore removed from that column outside the box.
(1) Look at where 2 can go on this row. (2) Look at where 6 can go on this column.
(1) All its spots lie within a single box (R9C2 and R9C3). (2) All its spots lie within a single box (R7C8 and R9C8).
(1) 2 is therefore removed from the rest of the box. (2) 6 is therefore removed from the rest of the box.
Spot R8C3 and R8C8 on this row.
These two cells can only hold 7 and 8: they reserve those digits.
7 and 8 are therefore removed from the other cells of the row.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R7C5 = 7.
Find 1 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 1.
R3C9 and R7C7 is seen by both free ends: 1 is therefore eliminated there.
Look at cell R3C9.
Every other digit already appears in its row, column or box.
Only one option remains: R3C9 = 6.
Look at cell R7C7.
Every other digit already appears in its row, column or box.
Only one option remains: R7C7 = 3.
Look at cell R1C7.
Every other digit already appears in its row, column or box.
Only one option remains: R1C7 = 1.
Look at cell R3C5.
Every other digit already appears in its row, column or box.
Only one option remains: R3C5 = 1.
Scan the empty cells in this box where 3 could go.
Across the whole box, 3 has only one possible spot left.
So R1C2 = 3.
Scan the empty cells in this row where 3 could go.
Across the whole row, 3 has only one possible spot left.
So R6C3 = 3.
Look at where 7 can go in this box.
All its spots lie on a single column (R4C2 and R5C2).
7 is therefore removed from that column outside the box.
Spot R5C2, R5C5 and R5C7 on this row.
These three cells share only 2, 6 and 7.
2, 6 and 7 are therefore removed from the rest of the row.
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.
R4C1 is seen by both free ends: 2 is therefore eliminated there.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R1C3 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R1C4 = 2.
Scan the empty cells in this column where 2 could go.
Across the whole column, 2 has only one possible spot left.
So R2C1 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R2C5 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R4C2 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R5C2 = 7.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R5C5 = 2.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R6C7 = 7.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R7C3 = 8.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R7C6 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R7C8 = 6.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R7C9 = 1.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R8C3 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R8C4 = 9.
Scan the empty cells in this box where 8 could go.
Across the whole box, 8 has only one possible spot left.
So R8C8 = 8.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R8C9 = 2.
Scan the empty cells in this box where 6 could go.
Across the whole box, 6 has only one possible spot left.
So R9C2 = 6.
Scan the empty cells in this box where 2 could go.
Across the whole box, 2 has only one possible spot left.
So R9C3 = 2.
Scan the empty cells in this box where 1 could go.
Across the whole box, 1 has only one possible spot left.
So R9C4 = 1.
Scan the empty cells in this box where 7 could go.
Across the whole box, 7 has only one possible spot left.
So R9C8 = 7.
Scan the empty cells in this box where 9 could go.
Across the whole box, 9 has only one possible spot left.
So R9C9 = 9.
Look at cell R4C6.
Every other digit already appears in its row, column or box.
Only one option remains: R4C6 = 3.
Look at cell R4C9.
Every other digit already appears in its row, column or box.
Only one option remains: R4C9 = 8.
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 = 1.
Look at cell R5C7.
Every other digit already appears in its row, column or box.
Only one option remains: R5C7 = 6.
Look at cell R5C9.
Every other digit already appears in its row, column or box.
Only one option remains: R5C9 = 3.
Look at cell R6C4.
Every other digit already appears in its row, column or box.
Only one option remains: R6C4 = 6.
Look at cell R4C1.
Every other digit already appears in its row, column or box.
Only one option remains: R4C1 = 6.
Look at cell R4C4.
Every other digit already appears in its row, column or box.
Only one option remains: R4C4 = 7.
Look at cell R6C1.
Every other digit already appears in its row, column or box.
Only one option remains: R6C1 = 8.