Sudoku Solution No. 52

Hard Score 72 / 100
Digits appear one by one.

Step-by-step solution

Steps 65 Estimated time 104 – 156 min Hardest technique X-Wing Clues 23
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Naked Single ✓ R8C8=4
Look at cell R8C8. Every other digit already appears in its row, column or box. Only one option remains: R8C8 = 4.
2 Hidden Single ✓ R1C5=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C5 = 1.
3 Hidden Single ✓ R1C7=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R1C7 = 4.
4 Hidden Single ✓ R3C1=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R3C1 = 3.
5 Hidden Single ✓ R3C5=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R3C5 = 4.
6 Hidden Single ✓ R3C9=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R3C9 = 6.
7 Hidden Single ✓ R4C4=3
Scan the empty cells in this column where 3 could go. Across the whole column, 3 has only one possible spot left. So R4C4 = 3.
8 Hidden Single ✓ R4C7=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R4C7 = 8.
9 Hidden Single ✓ R4C8=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R4C8 = 6.
10 Hidden Single ✓ R5C2=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R5C2 = 4.
11 Hidden Single ✓ R5C7=2
Scan the empty cells in this column where 2 could go. Across the whole column, 2 has only one possible spot left. So R5C7 = 2.
12 Hidden Single ✓ R5C8=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R5C8 = 3.
13 Hidden Single ✓ R6C2=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R6C2 = 3.
14 Hidden Single ✓ R6C9=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R6C9 = 1.
15 Hidden Single ✓ R7C3=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R7C3 = 1.
16 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.
17 Hidden Single ✓ R8C6=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R8C6 = 1.
18 Hidden Single ✓ R8C9=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R8C9 = 8.
19 Hidden Single ✓ R9C3=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R9C3 = 4.
20 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.
21 Hidden Single ✓ R2C7=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R2C7 = 1.
22 Hidden Single ✓ R4C1=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R4C1 = 1.
23 Hidden Single ✓ R5C6=6
Scan the empty cells in this column where 6 could go. Across the whole column, 6 has only one possible spot left. So R5C6 = 6.
24 Pointing ✗ 3 eliminations
Look at where 5 can go in this box. All its spots lie on a single column (R4C6 and R6C6). 5 is therefore removed from that column outside the box.
25 Skyscraper ✗ 1 elimination
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. R1C4 is seen by both free ends: 2 is therefore eliminated there.
26 Pointing ✗ 1 elimination
Look at where 2 can go in this box. All its spots lie on a single column (R1C6 and R2C6). 2 is therefore removed from that column outside the box.
27 Hidden Single ✓ R6C5=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R6C5 = 2.
28 Two-String Kite ✗ 1 elimination
Find 5 on one row and one column, each with two candidate positions, sharing a box. The two strands meet in the shared box. One of the two free ends must hold 5. R6C1 is seen by both free ends: 5 is therefore eliminated there.
29 Naked Single ✓ R6C1=8
Look at cell R6C1. Every other digit already appears in its row, column or box. Only one option remains: R6C1 = 8.
30 Hidden Single ✓ R5C5=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R5C5 = 8.
31 Hidden Single ✓ R7C4=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R7C4 = 8.
32 Pointing ✗ 1 elimination
Look at where 7 can go in this box. All its spots lie on a single row (R4C5 and R4C6). 7 is therefore removed from that row outside the box.
33 Skyscraper ✗ 1 elimination
Find 9 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 9. R4C5 is seen by both free ends: 9 is therefore eliminated there.
34 Naked Single ✓ R4C5=7
Look at cell R4C5. Every other digit already appears in its row, column or box. Only one option remains: R4C5 = 7.
35 Hidden Single ✓ R8C4=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R8C4 = 7.
36 Naked Single ✓ R8C2=2
Look at cell R8C2. Every other digit already appears in its row, column or box. Only one option remains: R8C2 = 2.
37 Hidden Single ✓ R1C1=2
Scan the empty cells in this column where 2 could go. Across the whole column, 2 has only one possible spot left. So R1C1 = 2.
38 Hidden Single ✓ 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 R1C2 = 6.
39 Hidden Single ✓ R1C6=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R1C6 = 8.
40 Hidden Single ✓ R2C6=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.
41 Hidden Single ✓ R3C3=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R3C3 = 8.
42 Hidden Single ✓ R3C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R3C6 = 7.
43 Hidden Single ✓ R9C4=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R9C4 = 2.
44 Skyscraper ✗ 1 elimination
Find 9 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 9. R5C3 is seen by both free ends: 9 is therefore eliminated there.
45 Hidden Single ✓ R1C3=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C3 = 5.
46 Hidden Single ✓ R1C4=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R1C4 = 9.
47 Hidden Single ✓ R2C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R2C9 = 5.
48 Hidden Single ✓ R3C4=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R3C4 = 5.
49 Hidden Single ✓ R3C7=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R3C7 = 9.
50 Hidden Single ✓ R4C2=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C2 = 9.
51 Hidden Single ✓ R4C6=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R4C6 = 5.
52 Hidden Single ✓ R5C1=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R5C1 = 5.
53 Hidden Single ✓ R5C3=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R5C3 = 7.
54 Hidden Single ✓ R5C9=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R5C9 = 9.
55 Hidden Single ✓ R6C6=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R6C6 = 9.
56 Hidden Single ✓ R6C8=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R6C8 = 5.
57 Hidden Single ✓ R7C1=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R7C1 = 7.
58 Hidden Single ✓ R7C2=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R7C2 = 5.
59 Hidden Single ✓ 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 R7C8 = 9.
60 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.
61 Hidden Single ✓ R9C7=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R9C7 = 5.
62 Naked Single ✓ R2C2=7
Look at cell R2C2. Every other digit already appears in its row, column or box. Only one option remains: R2C2 = 7.
63 Naked Single ✓ R2C3=9
Look at cell R2C3. Every other digit already appears in its row, column or box. Only one option remains: R2C3 = 9.
64 Naked Single ✓ R7C5=6
Look at cell R7C5. Every other digit already appears in its row, column or box. Only one option remains: R7C5 = 6.
65 Naked Single ✓ R9C1=6
Look at cell R9C1. Every other digit already appears in its row, column or box. Only one option remains: R9C1 = 6.