Solving Hardest Killer Sudoku
Killer Sudoku Solving Using The I-O-45 Strategy
Solved the killer sudoku puzzle from conceptispuzzles.com's hardest 10 logic puzzle list. The last one I did was the regular sudoku, involved some guessing. I assumed it was a pure logic puzzle and spent many hours over a few months between doing other puzzles.
Finally solved it and then discovered that you had to make a guess at least two places when I checked to see how others solved it. This was when I wrote about it and wanted to show how others solved it. So, must have made those guesses not realizing it at the time.
The killer Sudoku
The second puzzle is a killer sudoku and is solved by logic. No guessing. I used a solver after solving it because I couldn't remember all the steps and didn't want to take a long time to resolve it. The solver ran out of strategies in a few steps and I had to do the rest manually anyway using the I-O-45 strategy that the solver didn't use.
The I-O-45 Strategy
The strategy I named I-O-45. Innie-outtie-45. A sudoku unit, when adding all nine of the numbers = 45. When there is one innie and one outtie and the sum of the cages that originate in the unit = 45 they will both be the same number. When the sum is over 45the outtie will be higher than the innie by whatever the sum is over 45. When the sum s less then 45 then the outtie is smaller by the difference between 45 and the sum. The innie or outtie can be more than 1 cell. You just add them together and treat them as one cell. I don't think I would have solved this puzzle as easy if I had not used this strategy.
The puzzle
Original
The Start
This is how the puzzle starts. Below is all the further the solver at, sudokuwiki.org, got. I wrote 18 steps. The major steps and some of the minor ones the solver did.
Did not list every deletion. Most eliminations are simple sudoku strategies. The red numbers are the steps
Steps done by the solver
After steps done by solver
Solver steps
1. 4 cage in B2(box2) can only be 1 or 3
2. cages in this block sum to 43 leaving 45-43=2
3. only one 2 allowed in acage
4. 2 outtie from all cages being 92 C(column)s 8 and 9
5. are only 2s in B2 other 2s C6 gone
6. cage of 3 cells = 19 can not contain a 1
7. cage of 3 cells = 16 can not contain a 6,8 or 9
8. 4,5,7 triple removes from cells
9. 1-5 pr can remove 1 and 5 in many cells
10. 2 cell 7 cage no 7,8or 9,
11. cage of size 3 stlls 12 can not be 8 or 9
12. 2 cell 10 cage can not have a 5
13. 2 at 4-4 unique in Column and Box
14. cage of 2 cells 14 can only be 5,6,8 or 9
15. 2 is only # that works in 5-8
16. 6 cage can only have 1,2,4 or 5
17. 5 is only # that works, same with 1-3 pr
18. 2 only # after remove 6 and 9 that works in the cage of 16
Doing the manual solving part
The pic below is the rest of the steps listed in green from 19 to 71. The cage value has been added and the yellow row and column numbers added to make solving and following easier. The small red numbers are the step that lead to that number's elimination.
All done
The solution
Steps 19 through 71
19. When I took over and used two outties at B1 and B2. This resulted in
the only # that can be at 7-3 being a 4. The two outties = 13. so a
9 goes in 1-4.
20. My I-O-45 strategy in B3 = 45. both innie and outtie are the same. 6 is the only # that fits. leaving a 4 as the only # that can go in 4-8, makes new 12 cage at B3
21. 8 at 3-5 only # left after 6 gone, 6 at 4-5
22. 4 only # that can go in 6-6 removes 4 at 6-4,
23. 3 at 7-6
24. 4-6 pr B8 C4
25. 1 7-5 26.
26. 3 6-5
27. cages in B9 = 60 - 45 = 15. in outties only 7 or 8 will fit. 9s out,
28. 9s needed in B8 R8, other 9s out in R8
29. 7-8 pr C4
30. 5 in 5-5
31. 4-7 pr B2 =5 in 1-7 =9 in 8-5
32. 1 ni 18, 5 in 2-8
33.1-9 pr C6
34. 9 in 8-5
35. 5-9 pr C3
36. 4-3 pr C1
37. 3-9 pr C7
38. 7 in 3-9
39. 4-8 pr C9
40. pseudo 9 cage B9 can only be: 2-6 pr and 1 in 9-9
41. 5-9 pr C9, = 3 in 5-9
42. 8 not in 17 cage in B5 = 1-9 pr and 7 in 4-7
43. 8 in 7-7, 4 in 8-7, 5 in 9-7, 7 in 9-6
44. 8 in 6-8,
45. 9 in 9-8,7 in 7-8, 3 in 8-8
46. 6 in 8-4, 4 in 9-4,
47. 3-6-8 B7 R9
48. 2 in 9-9, 6 in 7-9
49. 8- in 8-6 50. 2-9 pr B7
51. 1 in 8-3 is only # that works with cage
52. 7 in 6-4, = 8 in 5-4
53. only 8 left in B4
54. 4 in 5-2, 6 in 5-3
55. 1 in 5-7, 7 in 4-7
56. 5-9 pr C4, = 1 in 4-6, 9 in 5-6
57. 7 in 5-1
58. 5 in 8-1, 7 in 8-2
59. 6-8 pr
60. 2 in 2-2, 5 in 3-2
61. 6 in 1-2, 8 in 1-1, 1 in 2-1
62. 3 in 9-2, 8 in 9-3, 6 in 9-1
63. 9 in 7-2, 2 in 7-1
64. 9 in 6-1, 1 in 6-2
65. 5 in 4-3, 9 in 3-3
66. 3 in 2-4, 1 in 3-4
67. 9 in 2-7, 3 in 3-7
68. 7 in 2-3, 3 in 1-3
69. 4 in 2-5, 7 in 1-5
70. 8 in 2-9, 4 in 1-9
71. 9 in 4-9, 5 in 6-9 END
If you see any errors: Please let me know
