Sudoku Solution No. 1

Hard Score 61.3 / 100
Digits appear one by one.

Step-by-step solution

Steps 64 Estimated time 65 – 98 min Hardest technique Box/Line Clues 26
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Hidden Single ✓ R1C3=4
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.
2 Hidden Single ✓ R1C6=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R1C6 = 1.
3 Hidden Single ✓ R1C7=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R1C7 = 6.
4 Hidden Single ✓ R2C4=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R2C4 = 9.
5 Hidden Single ✓ R3C9=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R3C9 = 4.
6 Hidden Single ✓ R4C8=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R4C8 = 2.
7 Hidden Single ✓ R5C4=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R5C4 = 1.
8 Hidden Single ✓ R5C7=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R5C7 = 4.
9 Hidden Single ✓ R6C1=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R6C1 = 6.
10 Hidden Single ✓ R6C2=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R6C2 = 9.
11 Hidden Single ✓ R6C4=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R6C4 = 4.
12 Hidden Single ✓ R7C1=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R7C1 = 4.
13 Hidden Single ✓ R8C8=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R8C8 = 4.
14 Hidden Single ✓ R8C9=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R8C9 = 6.
15 Hidden Single ✓ R9C2=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R9C2 = 1.
16 Pointing ✗ 1 elimination
Look at where 2 can go in this box. All its spots lie on a single row (R3C4 and R3C5). 2 is therefore removed from that row outside the box.
17 Box/Line ✗ 3 eliminations
Look at where 8 can go on this column. All its spots lie within a single box (R1C8 and R2C8). 8 is therefore removed from the rest of the box.
18 Naked Triple ✗ 6 eliminations
Spot R7C3, R7C6 and R7C8 on this row. These three cells share only 3, 5 and 7. 3, 5 and 7 are therefore removed from the rest of the row.
19 Skyscraper ✗ 2 eliminations
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. R2C3 and R8C2 is seen by both free ends: 3 is therefore eliminated there.
20 Box/Line ✗ 1 elimination
Look at where 3 can go on this column. All its spots lie within a single box (R1C2 and R2C2). 3 is therefore removed from the rest of the box.
21 Skyscraper ✗ 2 eliminations
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. R6C3 and R4C4 is seen by both free ends: 3 is therefore eliminated there.
22 Hidden Single ✓ R4C1=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R4C1 = 3.
23 Hidden Single ✓ R6C5=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R6C5 = 3.
24 Hidden Single ✓ R7C3=3
Scan the empty cells in this column where 3 could go. Across the whole column, 3 has only one possible spot left. So R7C3 = 3.
25 Hidden Single ✓ R7C6=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.
26 Hidden Single ✓ R8C2=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R8C2 = 7.
27 Naked Single ✓ R7C8=5
Look at cell R7C8. Every other digit already appears in its row, column or box. Only one option remains: R7C8 = 5.
28 Naked Single ✓ R9C1=5
Look at cell R9C1. Every other digit already appears in its row, column or box. Only one option remains: R9C1 = 5.
29 Hidden Single ✓ R4C4=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R4C4 = 7.
30 Hidden Single ✓ R1C5=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R1C5 = 7.
31 Naked Pair ✗ 1 elimination
Spot R1C8 and R2C8 in this box. These two cells can only hold 3 and 8: they reserve those digits. 3 and 8 are therefore removed from the other cells of the box.
32 Skyscraper ✗ 3 eliminations
Find 8 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 8. R9C4, R9C5 and R8C7 is seen by both free ends: 8 is therefore eliminated there.
33 Naked Single ✓ R8C7=3
Look at cell R8C7. Every other digit already appears in its row, column or box. Only one option remains: R8C7 = 3.
34 Hidden Single ✓ R3C4=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R3C4 = 2.
35 Hidden Single ✓ R9C4=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R9C4 = 3.
36 Skyscraper ✗ 1 elimination
Find 8 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 8. R3C3 is seen by both free ends: 8 is therefore eliminated there.
37 Hidden Single ✓ R1C4=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C4 = 5.
38 Hidden Single ✓ R3C5=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R3C5 = 8.
39 Hidden Single ✓ R4C6=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R4C6 = 8.
40 Hidden Single ✓ R4C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R4C9 = 5.
41 Hidden Single ✓ R5C5=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R5C5 = 5.
42 Hidden Single ✓ R6C3=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R6C3 = 5.
43 Hidden Single ✓ R6C7=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R6C7 = 8.
44 Hidden Single ✓ R7C7=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R7C7 = 1.
45 Hidden Single ✓ R7C9=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R7C9 = 9.
46 Hidden Single ✓ R8C4=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R8C4 = 8.
47 Hidden Single ✓ R8C6=5
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.
48 Hidden Single ✓ R9C5=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R9C5 = 9.
49 Hidden Single ✓ R9C7=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R9C7 = 2.
50 Hidden Single ✓ R9C9=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R9C9 = 8.
51 Naked Single ✓ R1C2=3
Look at cell R1C2. Every other digit already appears in its row, column or box. Only one option remains: R1C2 = 3.
52 Naked Single ✓ R1C8=8
Look at cell R1C8. Every other digit already appears in its row, column or box. Only one option remains: R1C8 = 8.
53 Naked Single ✓ R2C8=3
Look at cell R2C8. Every other digit already appears in its row, column or box. Only one option remains: R2C8 = 3.
54 Naked Single ✓ R2C9=1
Look at cell R2C9. Every other digit already appears in its row, column or box. Only one option remains: R2C9 = 1.
55 Naked Single ✓ R3C3=7
Look at cell R3C3. Every other digit already appears in its row, column or box. Only one option remains: R3C3 = 7.
56 Naked Single ✓ R3C7=5
Look at cell R3C7. Every other digit already appears in its row, column or box. Only one option remains: R3C7 = 5.
57 Naked Single ✓ R5C2=2
Look at cell R5C2. Every other digit already appears in its row, column or box. Only one option remains: R5C2 = 2.
58 Naked Single ✓ R5C3=8
Look at cell R5C3. Every other digit already appears in its row, column or box. Only one option remains: R5C3 = 8.
59 Naked Single ✓ R7C5=2
Look at cell R7C5. Every other digit already appears in its row, column or box. Only one option remains: R7C5 = 2.
60 Naked Single ✓ R2C1=8
Look at cell R2C1. Every other digit already appears in its row, column or box. Only one option remains: R2C1 = 8.
61 Naked Single ✓ R2C2=5
Look at cell R2C2. Every other digit already appears in its row, column or box. Only one option remains: R2C2 = 5.
62 Naked Single ✓ R2C3=2
Look at cell R2C3. Every other digit already appears in its row, column or box. Only one option remains: R2C3 = 2.
63 Naked Single ✓ R2C7=7
Look at cell R2C7. Every other digit already appears in its row, column or box. Only one option remains: R2C7 = 7.
64 Naked Single ✓ R5C1=7
Look at cell R5C1. Every other digit already appears in its row, column or box. Only one option remains: R5C1 = 7.