Sudoku Solution No. 11

Hard Score 71 / 100
Digits appear one by one.

Step-by-step solution

Steps 65 Estimated time 65 – 98 min Hardest technique Box/Line Clues 23
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Naked Single ✓ R9C3=2
Look at cell R9C3. Every other digit already appears in its row, column or box. Only one option remains: R9C3 = 2.
2 Naked Single ✓ R9C1=4
Look at cell R9C1. Every other digit already appears in its row, column or box. Only one option remains: R9C1 = 4.
3 Hidden Single ✓ R1C5=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R1C5 = 4.
4 Hidden Single ✓ R1C7=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R1C7 = 8.
5 Hidden Single ✓ R2C4=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R2C4 = 8.
6 Hidden Single ✓ R4C1=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C1 = 9.
7 Hidden Single ✓ R4C2=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R4C2 = 4.
8 Hidden Single ✓ R4C8=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R4C8 = 7.
9 Hidden Single ✓ R6C4=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R6C4 = 4.
10 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.
11 Hidden Single ✓ R8C5=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R8C5 = 8.
12 Hidden Single ✓ R9C8=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R9C8 = 6.
13 Hidden Single ✓ R9C9=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R9C9 = 7.
14 Naked Single ✓ R2C5=1
Look at cell R2C5. Every other digit already appears in its row, column or box. Only one option remains: R2C5 = 1.
15 Naked Single ✓ R9C4=9
Look at cell R9C4. Every other digit already appears in its row, column or box. Only one option remains: R9C4 = 9.
16 Hidden Single ✓ R1C1=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R1C1 = 7.
17 Hidden Single ✓ R3C1=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R3C1 = 8.
18 Hidden Single ✓ R3C5=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R3C5 = 7.
19 Hidden Single ✓ R4C3=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R4C3 = 8.
20 Hidden Single ✓ R5C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R5C6 = 7.
21 Hidden Single ✓ R7C4=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R7C4 = 6.
22 Hidden Single ✓ R7C6=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R7C6 = 4.
23 Hidden Single ✓ R8C6=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R8C6 = 1.
24 Hidden Single ✓ R8C9=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R8C9 = 4.
25 Naked Single ✓ R1C4=2
Look at cell R1C4. Every other digit already appears in its row, column or box. Only one option remains: R1C4 = 2.
26 Naked Single ✓ R2C6=9
Look at cell R2C6. Every other digit already appears in its row, column or box. Only one option remains: R2C6 = 9.
27 Naked Single ✓ R1C6=6
Look at cell R1C6. Every other digit already appears in its row, column or box. Only one option remains: R1C6 = 6.
28 Naked Single ✓ R4C6=2
Look at cell R4C6. Every other digit already appears in its row, column or box. Only one option remains: R4C6 = 2.
29 Hidden Single ✓ R3C3=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R3C3 = 6.
30 Hidden Single ✓ R4C5=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R4C5 = 6.
31 Hidden Single ✓ R5C1=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R5C1 = 6.
32 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.
33 Hidden Single ✓ R8C7=2
Scan the empty cells in this column where 2 could go. Across the whole column, 2 has only one possible spot left. So R8C7 = 2.
34 Hidden Single ✓ R2C1=2
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.
35 Hidden Single ✓ R5C2=2
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.
36 Naked Pair ✗ 4 eliminations
Spot R2C8 and R8C8 on this column. These two cells can only hold 3 and 5: they reserve those digits. 3 and 5 are therefore removed from the other cells of the column.
37 Hidden Pair ✗ 1 elimination
Find where 3 and 5 can go in this box. These two digits only fit in R1C9 and R2C8 in the box. The other candidates in R1C9 and R2C8 are therefore eliminated.
38 Pointing ✗ 1 elimination
Look at where 5 can go in this box. All its spots lie on a single row (R6C1 and R6C3). 5 is therefore removed from that row outside the box.
39 Skyscraper ✗ 1 elimination
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. R7C2 is seen by both free ends: 3 is therefore eliminated there.
40 Pointing ✗ 1 elimination
Look at where 3 can go in this box. All its spots lie on a single column (R7C1 and R8C1). 3 is therefore removed from that column outside the box.
41 Pointing ✗ 1 elimination
Look at where 3 can go in this box. All its spots lie on a single column (R5C3 and R6C3). 3 is therefore removed from that column outside the box.
42 Skyscraper ✗ 1 elimination
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. R7C2 is seen by both free ends: 5 is therefore eliminated there.
43 Naked Single ✓ R7C2=1
Look at cell R7C2. Every other digit already appears in its row, column or box. Only one option remains: R7C2 = 1.
44 Naked Single ✓ R3C2=9
Look at cell R3C2. Every other digit already appears in its row, column or box. Only one option remains: R3C2 = 9.
45 Hidden Single ✓ R1C3=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C3 = 1.
46 Hidden Single ✓ R1C8=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R1C8 = 9.
47 Hidden Single ✓ R1C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C9 = 5.
48 Hidden Single ✓ R2C2=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R2C2 = 5.
49 Hidden Single ✓ R2C8=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R2C8 = 3.
50 Hidden Single ✓ R3C8=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R3C8 = 1.
51 Hidden Single ✓ R3C9=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R3C9 = 2.
52 Hidden Single ✓ R4C4=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R4C4 = 3.
53 Hidden Single ✓ R4C7=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R4C7 = 5.
54 Hidden Single ✓ R4C9=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R4C9 = 1.
55 Hidden Single ✓ R5C3=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.
56 Hidden Single ✓ R5C4=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R5C4 = 1.
57 Hidden Single ✓ R6C1=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R6C1 = 1.
58 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.
59 Hidden Single ✓ R6C8=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R6C8 = 2.
60 Hidden Single ✓ R6C9=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R6C9 = 3.
61 Hidden Single ✓ 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 R7C7 = 3.
62 Hidden Single ✓ R8C8=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R8C8 = 5.
63 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.
64 Naked Single ✓ R7C1=5
Look at cell R7C1. Every other digit already appears in its row, column or box. Only one option remains: R7C1 = 5.
65 Naked Single ✓ R8C1=3
Look at cell R8C1. Every other digit already appears in its row, column or box. Only one option remains: R8C1 = 3.