Sudoku Solution No. 47

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 24
Given digit (clue) Solved digit Placement (✓ RlCc=d) Elimination (✗ RlCc≠d)
1 Hidden Single ✓ R2C6=9
Scan the empty cells in this row where 9 could go. Across the whole row, 9 has only one possible spot left. So R2C6 = 9.
2 Hidden Single ✓ R4C5=4
Scan the empty cells in this column where 4 could go. Across the whole column, 4 has only one possible spot left. So R4C5 = 4.
3 Hidden Single ✓ R5C3=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R5C3 = 4.
4 Hidden Single ✓ R5C7=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R5C7 = 2.
5 Hidden Single ✓ R6C5=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R6C5 = 2.
6 Hidden Single ✓ R7C8=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R7C8 = 2.
7 Hidden Single ✓ R8C6=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R8C6 = 7.
8 Hidden Single ✓ R9C1=4
Scan the empty cells in this box where 4 could go. Across the whole box, 4 has only one possible spot left. So R9C1 = 4.
9 Hidden Single ✓ R9C6=2
Scan the empty cells in this box where 2 could go. Across the whole box, 2 has only one possible spot left. So R9C6 = 2.
10 Hidden Single ✓ R9C7=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R9C7 = 8.
11 Hidden Single ✓ R1C5=8
Scan the empty cells in this column where 8 could go. Across the whole column, 8 has only one possible spot left. So R1C5 = 8.
12 Hidden Single ✓ R4C2=2
Scan the empty cells in this row where 2 could go. Across the whole row, 2 has only one possible spot left. So R4C2 = 2.
13 Hidden Pair ✗ 4 eliminations
Find where 3 and 9 can go on this column. These two digits only fit in R4C1 and R6C1 on the column. The other candidates in R4C1 and R6C1 are therefore eliminated.
14 Naked Pair ✗ 4 eliminations
Spot R4C1 and R6C1 in this box. These two cells can only hold 3 and 9: they reserve those digits. 3 and 9 are therefore removed from the other cells of the box.
15 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.
16 Hidden Single ✓ R8C7=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R8C7 = 9.
17 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.
18 Hidden Single ✓ R9C2=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R9C2 = 9.
19 Hidden Single ✓ R1C7=4
Scan the empty cells in this row where 4 could go. Across the whole row, 4 has only one possible spot left. So R1C7 = 4.
20 Pointing ✗ 3 eliminations
(1) Look at where 7 can go in this box. (2) Look at where 5 can go in this box. (1) All its spots lie on a single column (R1C8 and R2C8). (2) All its spots lie on a single column (R7C9 and R8C9). (1) 7 is therefore removed from that column outside the box. (2) 5 is therefore removed from that column outside the box.
21 Hidden Single ✓ R4C6=5
Scan the empty cells in this row where 5 could go. Across the whole row, 5 has only one possible spot left. So R4C6 = 5.
22 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. R3C6 is seen by both free ends: 3 is therefore eliminated there.
23 Hidden Single ✓ R2C5=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R2C5 = 3.
24 Hidden Single ✓ R3C9=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R3C9 = 3.
25 Hidden Single ✓ R9C4=3
Scan the empty cells in this box where 3 could go. Across the whole box, 3 has only one possible spot left. So R9C4 = 3.
26 Naked Single ✓ R8C5=6
Look at cell R8C5. Every other digit already appears in its row, column or box. Only one option remains: R8C5 = 6.
27 Naked Single ✓ R7C4=1
Look at cell R7C4. Every other digit already appears in its row, column or box. Only one option remains: R7C4 = 1.
28 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.
29 Hidden Single ✓ R4C4=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R4C4 = 9.
30 Hidden Single ✓ R6C1=9
Scan the empty cells in this box where 9 could go. Across the whole box, 9 has only one possible spot left. So R6C1 = 9.
31 Hidden Single ✓ R6C4=6
Scan the empty cells in this box where 6 could go. Across the whole box, 6 has only one possible spot left. So R6C4 = 6.
32 Hidden Single ✓ R8C3=3
Scan the empty cells in this row where 3 could go. Across the whole row, 3 has only one possible spot left. So R8C3 = 3.
33 Box/Line ✗ 2 eliminations
Look at where 6 can go on this column. All its spots lie within a single box (R2C7 and R3C7). 6 is therefore removed from the rest of the box.
34 Skyscraper ✗ 2 eliminations
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. R6C2 and R9C3 is seen by both free ends: 1 is therefore eliminated there.
35 Hidden Single ✓ R8C2=1
Scan the empty cells in this column where 1 could go. Across the whole column, 1 has only one possible spot left. So R8C2 = 1.
36 Naked Single ✓ R8C9=5
Look at cell R8C9. Every other digit already appears in its row, column or box. Only one option remains: R8C9 = 5.
37 Two-String Kite ✗ 1 elimination
Find 6 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 6. R9C3 is seen by both free ends: 6 is therefore eliminated there.
38 Naked Single ✓ R9C3=7
Look at cell R9C3. Every other digit already appears in its row, column or box. Only one option remains: R9C3 = 7.
39 Hidden Single ✓ R1C1=7
Scan the empty cells in this column where 7 could go. Across the whole column, 7 has only one possible spot left. So R1C1 = 7.
40 Hidden Single ✓ R2C8=7
Scan the empty cells in this box where 7 could go. Across the whole box, 7 has only one possible spot left. So R2C8 = 7.
41 Hidden Single ✓ R6C2=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R6C2 = 7.
42 Hidden Single ✓ R7C9=7
Scan the empty cells in this row where 7 could go. Across the whole row, 7 has only one possible spot left. So R7C9 = 7.
43 Naked Single ✓ R1C8=1
Look at cell R1C8. Every other digit already appears in its row, column or box. Only one option remains: R1C8 = 1.
44 Naked Single ✓ R9C8=6
Look at cell R9C8. Every other digit already appears in its row, column or box. Only one option remains: R9C8 = 6.
45 Naked Single ✓ R9C9=1
Look at cell R9C9. Every other digit already appears in its row, column or box. Only one option remains: R9C9 = 1.
46 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.
47 Naked Single ✓ R3C6=1
Look at cell R3C6. Every other digit already appears in its row, column or box. Only one option remains: R3C6 = 1.
48 Naked Single ✓ R4C9=6
Look at cell R4C9. Every other digit already appears in its row, column or box. Only one option remains: R4C9 = 6.
49 Naked Single ✓ R5C8=3
Look at cell R5C8. Every other digit already appears in its row, column or box. Only one option remains: R5C8 = 3.
50 Naked Single ✓ R6C8=5
Look at cell R6C8. Every other digit already appears in its row, column or box. Only one option remains: R6C8 = 5.
51 Naked Single ✓ R4C3=1
Look at cell R4C3. Every other digit already appears in its row, column or box. Only one option remains: R4C3 = 1.
52 Naked Single ✓ R5C6=8
Look at cell R5C6. Every other digit already appears in its row, column or box. Only one option remains: R5C6 = 8.
53 Naked Single ✓ R6C3=8
Look at cell R6C3. Every other digit already appears in its row, column or box. Only one option remains: R6C3 = 8.
54 Naked Single ✓ R6C6=3
Look at cell R6C6. Every other digit already appears in its row, column or box. Only one option remains: R6C6 = 3.
55 Naked Single ✓ R6C7=1
Look at cell R6C7. Every other digit already appears in its row, column or box. Only one option remains: R6C7 = 1.
56 Naked Single ✓ R5C2=6
Look at cell R5C2. Every other digit already appears in its row, column or box. Only one option remains: R5C2 = 6.
57 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.
58 Naked Single ✓ R2C7=6
Look at cell R2C7. Every other digit already appears in its row, column or box. Only one option remains: R2C7 = 6.
59 Naked Single ✓ R3C3=6
Look at cell R3C3. Every other digit already appears in its row, column or box. Only one option remains: R3C3 = 6.
60 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.
61 Naked Single ✓ R7C2=8
Look at cell R7C2. Every other digit already appears in its row, column or box. Only one option remains: R7C2 = 8.
62 Naked Single ✓ R7C3=5
Look at cell R7C3. Every other digit already appears in its row, column or box. Only one option remains: R7C3 = 5.
63 Naked Single ✓ R3C1=8
Look at cell R3C1. Every other digit already appears in its row, column or box. Only one option remains: R3C1 = 8.
64 Naked Single ✓ R7C1=6
Look at cell R7C1. Every other digit already appears in its row, column or box. Only one option remains: R7C1 = 6.