Sudoku Solution No. 68

Hard Score 72 / 100
Digits appear one by one.

Step-by-step solution

Steps 64 Estimated time 104 – 156 min Hardest technique X-Wing Clues 23
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Hidden Single ✓ R4C1=7
Scan the empty cells in this column where 7 could go. Across the whole column, 7 has only one possible spot left. So R4C1 = 7.
2 Hidden Single ✓ R4C2=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R4C2 = 4.
3 Hidden Single ✓ R4C3=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C3 = 9.
4 Hidden Single ✓ R5C9=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R5C9 = 6.
5 Hidden Single ✓ R6C4=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R6C4 = 4.
6 Hidden Single ✓ R8C6=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R8C6 = 2.
7 Hidden Single ✓ R8C8=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R8C8 = 9.
8 Hidden Single ✓ R9C8=7
Scan the empty cells in this column where 7 could go. Across the whole column, 7 has only one possible spot left. So R9C8 = 7.
9 Hidden Single ✓ R9C9=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R9C9 = 4.
10 Naked Single ✓ R8C4=1
Look at cell R8C4. Every other digit already appears in its row, column or box. Only one option remains: R8C4 = 1.
11 Hidden Single ✓ R4C5=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R4C5 = 5.
12 Hidden Single ✓ R8C7=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R8C7 = 6.
13 Naked Single ✓ R5C4=8
Look at cell R5C4. Every other digit already appears in its row, column or box. Only one option remains: R5C4 = 8.
14 Naked Single ✓ R2C4=5
Look at cell R2C4. Every other digit already appears in its row, column or box. Only one option remains: R2C4 = 5.
15 Hidden Single ✓ R1C9=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R1C9 = 8.
16 Hidden Single ✓ R2C6=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R2C6 = 8.
17 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.
18 Hidden Single ✓ R4C9=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R4C9 = 2.
19 Hidden Single ✓ R6C2=8
Scan the empty cells in this box where 8 could go. Across the whole box, 8 has only one possible spot left. So R6C2 = 8.
20 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.
21 Hidden Single ✓ R1C5=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R1C5 = 7.
22 Hidden Single ✓ R3C1=8
Scan the empty cells in this row where 8 could go. Across the whole row, 8 has only one possible spot left. So R3C1 = 8.
23 Hidden Single ✓ R3C2=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R3C2 = 9.
24 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.
25 Hidden Single ✓ R6C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R6C6 = 7.
26 Hidden Single ✓ R7C1=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R7C1 = 9.
27 Hidden Single ✓ R7C4=7
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.
28 Hidden Single ✓ R7C6=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R7C6 = 6.
29 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.
30 Naked Single ✓ R9C1=1
Look at cell R9C1. Every other digit already appears in its row, column or box. Only one option remains: R9C1 = 1.
31 Naked Single ✓ R9C2=6
Look at cell R9C2. Every other digit already appears in its row, column or box. Only one option remains: R9C2 = 6.
32 Hidden Single ✓ R2C5=6
Scan the empty cells in this row where 6 could go. Across the whole row, 6 has only one possible spot left. So R2C5 = 6.
33 Pointing ✗ 1 elimination
Look at where 5 can go in this box. All its spots lie on a single column (R1C7 and R3C7). 5 is therefore removed from that column outside the box.
34 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. R5C1 is seen by both free ends: 3 is therefore eliminated there.
35 Two-String Kite ✗ 1 elimination
Find 1 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 1. R3C8 is seen by both free ends: 1 is therefore eliminated there.
36 X-Wing ✗ 3 eliminations
Find where 1 appears on two columns: altogether, only two rows are involved (R1C6, R4C6, R1C8 and R4C8). On each covered row, 1 must lie within one of these two columns. 1 is therefore removed from those two rows outside the two X-Wing columns.
37 Skyscraper ✗ 1 elimination
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. R3C3 is seen by both free ends: 1 is therefore eliminated there.
38 Hidden Single ✓ R2C2=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R2C2 = 1.
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. R7C7 is seen by both free ends: 3 is therefore eliminated there.
40 Naked Single ✓ R7C7=1
Look at cell R7C7. Every other digit already appears in its row, column or box. Only one option remains: R7C7 = 1.
41 Hidden Single ✓ R1C1=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R1C1 = 4.
42 Hidden Single ✓ R1C3=6
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.
43 Hidden Single ✓ R1C6=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R1C6 = 3.
44 Hidden Single ✓ R1C7=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R1C7 = 5.
45 Hidden Single ✓ R1C8=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R1C8 = 1.
46 Hidden Single ✓ R2C1=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R2C1 = 3.
47 Hidden Single ✓ R2C7=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R2C7 = 4.
48 Hidden Single ✓ R3C5=1
Scan the empty cells in this row where 1 could go. Across the whole row, 1 has only one possible spot left. So R3C5 = 1.
49 Hidden Single ✓ R3C7=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R3C7 = 3.
50 Hidden Single ✓ R3C8=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R3C8 = 6.
51 Hidden Single ✓ R4C6=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R4C6 = 1.
52 Hidden Single ✓ 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 R4C8 = 3.
53 Hidden Single ✓ R5C3=1
Scan the empty cells in this box where 1 could go. Across the whole box, 1 has only one possible spot left. So R5C3 = 1.
54 Hidden Single ✓ R5C5=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R5C5 = 3.
55 Hidden Single ✓ R6C3=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R6C3 = 3.
56 Hidden Single ✓ R6C9=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R6C9 = 1.
57 Hidden Single ✓ R7C2=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R7C2 = 3.
58 Hidden Single ✓ R7C9=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R7C9 = 5.
59 Hidden Single ✓ R8C1=5
Scan the empty cells in this box where 5 could go. Across the whole box, 5 has only one possible spot left. So R8C1 = 5.
60 Hidden Single ✓ R8C9=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R8C9 = 3.
61 Naked Single ✓ R1C2=2
Look at cell R1C2. Every other digit already appears in its row, column or box. Only one option remains: R1C2 = 2.
62 Naked Single ✓ R3C3=5
Look at cell R3C3. Every other digit already appears in its row, column or box. Only one option remains: R3C3 = 5.
63 Naked Single ✓ R5C1=2
Look at cell R5C1. Every other digit already appears in its row, column or box. Only one option remains: R5C1 = 2.
64 Naked Single ✓ R5C2=5
Look at cell R5C2. Every other digit already appears in its row, column or box. Only one option remains: R5C2 = 5.