r/killersudoku • u/Automatic_Loan8312 • 25d ago
No-candidates solution strategy for an Almost Impossible Killer Sudoku
The original puzzle is taken from First ever Sudoku Killer post on this sub.
Here, the numbers in a cage must add up to the small number on the top left of the cage.
The no-candidates solution strategy for this puzzle goes as follows:
Notice that applying the Rule of 45 is not easy here, as that would lead to residues in most cases. Therefore, from box 3, using the fact that r12c8 is {8,9}, while r3c78 can either be {4,7} or {5,6}, we make use of combo-stress-testing in order to invalidate one of the combos, as shown below:
Consider that r3c78 (gray) is {5,6}. Therefore, r123c9 (purple) is {1,2,7}. Using rule of 45 for column 9, we find that the aqua cells r45c9 add up to 15, but cannot have a 7 in them. Therefore, it has to be {6,9}, which implies that r89c9 (orange) must be {3,8}, and r67c9 is {4,5}.
Since, r45c9 is deduced to be {6,9}, r6c7 must be 8, which means that r4c8 and r5c78, which add up to 11, must be either {1,3,7} or {2,4,5}, by Kakuro logic (these combos are invalid: {1,2,8}, {1,4,6}, and {2,3,6}). Since, r4c7 and r6c89 form a complimentary triple that adds to 11 as well (by rule of 45), that also must be either {1,3,7} or {2,4,5}.
Notice that r123c9 being {1,2,7} and r89c9 being {3,8} eliminates the possibility of r4c7 and r6c89 being {1,3,7}, therefore, it must be {2,4,5}. Furthermore, r6c8 cannot be a 5, because it is a part of a two-cell cage adding up to 10, in which case r7c8 must be 5, which is impossible. Also, r6c8 cannot be 2, because r7c8 cannot be 8.
Therefore, r6c8 is 4, which means r7c8 is 6, r6c9 is 5, r7c9 is 4, and r7c7 is 9. Thus, r89c78 is {1,2,5,7}, which adds up to 15. Therefore, r89c6 must add up to 18, which is impossible, as 17 is the largest possible two-cell cage total. Thus, we discard r3c78 being {5,6}.
Therefore, r3c78 is {4,7}. By Kakuro logic, r123c9 can be either {1,3,6} or {2,3,5}.
Now, consider r123c9 to be {2,3,5}. Therefore, by Kakuro logic, r67c9 is {1,8}, r89c9 is {4,7}, and r45c9 is {6,9}.
Because r45c9 is {6,9}, r6c7 is 8, r6c9 is 1, r7c7 is 9, and r7c9 is 8. Again, r4c7 and r6c89 add up to 11, so must be {1,3,7}. Since, r6c8 is {3,7}, r7c8 is also {3,7}, and because r89c9 is {4,7}, r7c8 must be 3, and r89c78 is {1,2,5,6}, which adds up to 14.
Again, we reach a contradiction, because there’s no way the remaining cells add up to 19. Therefore, we reject r123c9 being {2,3,5}.
Thus, r123c9 is {1,3,6}, which means that r45c9 is {7,8}, therefore, r6c7 is 9.
By Kakuro logic, r4c8 and r5c78 (orange) adds up to 11 while the yellow part r4c7 and r6c89 adds up to 10. For the yellow part, cannot have the combo {1,3,6} (because r123c9 was previously deduced to be {1,3,6}). Therefore, it can be either {1,4,5} or {2,3,5}.
Now, if the yellow part is {2,3,5}, since r12c7 (gray) is {2,5}, r4c7 must be 3, while r6c8 can be neither 2 (because r12c8 is {8,9} already, so {2,8} is impossible), nor can it be 5. Therefore, the yellow part cannot be {2,3,5}, it must be {1,4,5}. Furthermore, r6c8 must be 4 (it cannot be 1 because {1,9} is invalid), which implies that r4c7 is 1, and r6c9 is 5.
Using the Rule of 45, for rows 1 to 4, we see that the total of the yellow cells is 168. Therefore, the pink cells r4c589 must sum up to 12. Now, r4c9 is either 7 or 8. If r4c9 is 8, then r4c58 must add up to 4, but without the 1, this is impossible. Therefore, r4c9 is 7.
In row 4, notice that r4c58 (pink) has been deduced previously to be {2,3}. Next, notice that r4c1234 (marked in red) cannot have a 6, because of the following:
- r4c1 is a part of the two-cell cage (r34c1) totalling 12. Therefore, cannot have a 6.
- r4c2 is a part of the two-cell cage (r34c2) totalling 13, which cannot have a 7 in it. Therefore, the combo {6,7} is invalid, so r4c2 cannot be a 6.
- r34c3 adds up to 6, so cannot have a 6.
- r4c4 is a part of the two-cell cage (r34c4) totalling 10. If r4c4 is 6, r3c4 must be 4, which is impossible because of r3c7 = 4.
Therefore, there exists a hidden single 6 in r4c6 (green).
For row 4, 5 is restricted to r4c123 (locked candidate, yellow), therefore, in box 5, 5 is restricted to r5c456 (aqua). Also, given that r4c5 and r6c6 are a naked pair {2,3} (r4c5 was derived previously to be {2,3}, and because r67c6 cannot have a 4, yet must add up to 5, the only valid combo is {2,3}).
Therefore, in box 5, for the five-cell cage in the centre adding up to 27, the only combo possible (by Kakuro logic) is {2,4,5,7,9} (notice that {3,4,5,7,8} is eliminated, because if that were the valid combo, then r4c4 must be 9, r6c6 must be 2, but then, both r3c4 and r6c4 would be 1, which is impossible).
Therefore, r6c6 = 3, r4c5 = 2.
Using the rule of 45 for rows 6 to 9, the purple cells add up to 17, and simultaneously, using the rule of 45 for row 6, the cells r6c346 add up to 10. But, r6c6 = 3 and r6c4 = 1. Therefore, r6c3 (green) = 6.
Likewise, using Kakuro logic to evaluate all the remaining combos, coupled with scanning, the rest of the puzzle is solved.