SudokuSolver Forum

Continuing working through my backlog, the next variant I tried was A112 V2 where I'd originally got stuck at step 13; when I resumed I edited that step before continuing from that position.

I finished this puzzle about a week ago and glanced at the walkthroughs posted by udosuk, Mike and Afmob but I only went through their walkthrougs today.

I was puzzled by the discussion about a possible weakness in A112 V2 and its focus on the hidden 19(3) cage in R12 which I found very early or, in Mike's case, the three outies for R89. I don't see that as a weakness and I very much doubt that SudokuSolver missed that. I also doubt that any software solver would have found udosuk's step 3; while adding the same thing to both sides of an equation is a known mathematical technique, it's unusual and ingenious for a sudoku solving step. In my view the weakness, if there is one, is that the solving path becomes a lot easier after eliminating 9 from R1C1 or fixing R23C5. I took a very long time to reach that position. Udosuk, Mike and Afmob all post optimised walkthroughs so either they made that elimination a lot earlier than I did or they subsequently optimised their walkthroughs to focus on it. Out of the three walkthroughs posted already, Afmob used the most direct way to make that elimination but all three used difficult steps to achieve it so I don't really think that this puzzle can be considered to have a weakness.

My solving path was very different from the three already posted. I hope that you will find some of my steps interesting.


Here is my walkthrough for A112 V2.

Prelims

a) R56C4 = {16/25/34}, no 7,8,9
b) R56C6 = {18/27/36/45}, no 9
c) R7C34 = {17/26/35}, no 4,8,9
d) R7C67 = {29/38/47/56}, no 1
e) 10(3) cage in N1 = {127/136/145/235}, no 8,9
f) 21(3) cage at R1C3 = {489/579/678}, no 1,2,3
g) 10(3) cage at R1C4 = {127/136/145/235}, no 8,9
h) 9(3) cage at R1C7 = {126/135/234}, no 7,8,9
i) 21(3) cage in N3 = {489/579/678}, no 1,2,3
j) 24(3) cage at R3C3 = {789}
k) 10(3) cage at R3C6 = {127/136/145/235}, no 8,9
l) 20(3) cage at R6C5 = {389/479/569/578}, no 1,2
m) 19(3) cage in N7 = {289/379/469/478/568}, no 1
n) 10(3) cage at R8C3 = {127/136/145/235}, no 8,9
o) 21(3) cage at R8C7 = {489/579/678}, no 1,2,3
p) 13(4) cage at R6C8 = {1237/1246/1345}, no 8,9

1. 45 rule on C5 3 innies R159C5 = 7 = {124}, locked for C5

2. 45 rule on N5 2 innies R46C5 = 16 = {79}, locked for C5 and N5, clean-up: no 2 in R56C6
2a. 9 in N2 locked in R23C4, locked for C4

3. 45 rule on R1289 2 innies R28C5 = 9 = {36}, locked for C5

4. R678C5 = {389/569} (cannot be {578} because 5,8 only in R7C5) -> R6C5 = 9, R4C5 = 7
4a. 7 in 24(3) cage at R3C3 only in R3C34, locked for R3
4b. R23C5 = 11 = [38/65]

5. 45 rule on R12 3 innies R1C19 + R2C5 = 19
5a. R2C5 = {36} -> R1C19 = 13 or 16 = {49/58/67/79}, no 1,2,3
5b. R1C19 = 13 or 16 -> R9C19 = 6 or 9, no 9 in R9C19

6. 10(3) cage at R1C4 = {127/145} (cannot be {136} which clashes with R2C5, cannot be {235} which clashes with R23C5), no 3,6, 1 locked for R1 and N2

7. 9(3) cage at R1C7 = {126/135/234}
7a. 1 of {126/135} must be in R2C7 -> no 5,6 in R2C7

8. 21(3) cage at R1C3 = {489/579/678}
8a. 4,5,6 must be in R12C3 (R12C3 cannot be {78/79/89} which clash with R34C3, ALS block) -> no 4,5,6 in R2C4
8b. Killer triple 7,8,9 in R12C3 and R34C3, locked for C3, clean-up: no 1 in R7C4
8c. 21(3) cage at R1C3 and 24(3) cage at R3C3 must both contain 9 (because there’s no other 9 in C34) -> 21(3) cage = {489/579}, no 6
8d. 7 in C3 only in R123C3, locked for N1, clean-up: no 6 in R1C9 (step 5a)

9. 10(3) cage in N1 = {136/235} (cannot be {145} which clashes with 21(3) cage at R1C3), no 4, 3 locked for N1

10. 9 in N2 only in R23C4
10a. 45 rule on N2 4 innies R23C46 = 24
10b. R23C4 = {79/89} = 16,17 -> R23C6 = 7,8 = {26/34} (cannot be {25} which clashes with 10(3) cage at R1C4, cannot be {35} which clashes with R23C5), no 5
10c. R56C6 = {18/45} (cannot be {36} which clashes with R23C6), no 3,6

11. 13(3) cage in N5 = {238/256/346} (cannot be {148} which clashes with R56C6), no 1
11a. R5C5 = {24} -> no 2,4 in R4C46

12. 10(3) cage at R3C6 = {136/145/235}
12a. 4 of {145} must be in R3C6 -> no 4 in R34C7

13. R23C6 (step 10b) = {26/34} = 7 or 8 -> R1234C7 = 11 or 12 (from sum of cage totals) = {1235/1236/1245}, 1,2 locked for C7, clean-up: no 9 in R7C6
13a. {1235} can only be [21]{35} (because no 1,5 in R23C6 and cannot be {23}{15} because R23C6 cannot both be 4, which can be seen alternatively as R34C7 cannot total 1 more than R12C7)
{1236} can only be {23}{16} (because no 1,5 in R23C6)
{1245} can only be {24}{15} (because 4 only in R12C7)
13b. -> R12C7 = [21]/{23}/{24}, no 5,6, 2 locked for C7 and N3, CPE no 2 in R2C6, clean-up: no 6 in R3C6 (step 10b)
13c. 6 in N2 only in R2C56, locked for R2
13d. 9 in N8 only in R89C6, CPE no 9 in R8C7

14. Hidden killer pair 1,2 in R2C12 and R2C7 for R2, R2C12 contains one of 1,2 -> R2C7 = {12}
14a. 10(3) cage in N1 (step 9) = {136/235}
14b. R2C12 contains 1,2 -> no 2 in R1C2

15. 9(3) cage in R1C7 = {126/234}
15a. 4 of {234} must be in R2C6 (R2C67 cannot be [32] which clashes with R2C12) -> no 4 in R1C7, no 3 in R2C6, clean-up: no 4 in R3C6 (step 10b)
15b. R1234C7 (step 13) = {1235/1236}, 3 locked for C7, clean-up: no 8 in R7C6

16. 3 in R1 only in R1C27, 1 in R2 only in R2C127
16a. 9(3) cage in R1C7 must contain 3 in R1C7 or 1 in R2C7 -> 10(3) cage in N1 must contain 3 in R1C2 or 1 in R2C12 -> 10(3) cage = 3{25}/6{13} (cannot be 5{23}), no 5 in R1C2

17. 45 rule on R67 4 innies R6C3467 = 1 outie R8C5 + 18
17a. R8C5 = {36} -> R6C3467 = 21,24 = {1578/2478/2568/3468/3567/3678/4578}
17b. 7,8 of {1578} must be in R6C67 -> no 1 in R6C6, clean-up: no 8 in R5C6

18. Hidden grouped killer triple 7,8,9 in R1C19, 10(3) cage at R1C4 and 21(3) cages at R1C3 and R1C8 for R12, the two 21(3) cages must each have two of 7,8,9 -> R1C19 + 10(3) cage at R1C4 must contain two of 7,8,9
18a. R1C19 (step 5a) = {49/58}/[67/97] + 10(3) cage at R1C4 (step 6) = {127/145} must contain two of 7,8,9 -> combined cage R1C14569 = {49/58}{127}/[97]{145} (cannot be [67]{145} which only contains one of 7,8,9), no 6 in R1C1, 7 locked for R1

19. 4 in R3 only in R3C1289
19a. 21(3) cages at R1C3 and R1C8 must have different combinations (to avoid clash between R2C34 and R2C89, ALS block)
19b. Only one of the 21(3) cages can contain 7 (because no 7 in R1C38) -> one of the 21(3) cages must be {489}
19c. 4 locked in one of the 21(3) cages and R3C1289 for N13, no 4 in R1C19
19d. R1C19 (step 18a) = {58}/[97], no 9 in R1C9
19e. Killer pair 5,7 in R1C19 and 10(3) cage at R1C4, locked for R1

20. 21(3) cage at R1C3 = {489/579}
20a. 5 of {579} must be in R2C3 -> no 7 in R2C3
20b. R3C3 = 7 (hidden single in C3)
20c. R3C4 + R4C3 = {89}, CPE no 8 in R4C4
20d. 8 in N5 only in R46C6, locked for C6

21. 45 rule on N8 4 remaining innies R79C46 = 22 = {1579/2479/2569/4567} (cannot be {3469} which clashes with R8C5), no 3, clean-up: no 5 in R7C3, no 8 in R7C7
21a. 3 in N8 only in R8C456, locked for R8

22. 12(3) cage in N8 = {129/138/147/156/246} (cannot be {237/345} which clash with R79C46)
22a. 8 of {138} must be in R8C4 -> no 3 in R8C4
22b. 3 in C4 only in R456C4, locked for N5

23. 10(3) cage at R8C3 = {127/136/145/235}
23a. 3 of {136} must be in R9C3 -> no 6 in R9C3

24. 21(3) cage at R1C3 = {489/579}
24a. Cannot be {579}, here’s how
{579} = [957] => R1C19 = {58} clashes with 10(3) cage at R1C4 = {145}
24b. 21(3) cage at R1C3 = {489}, no 5,7

25. Naked pair = {89} in R23C4, locked for C4 and N2 -> R3C5 = 5, R2C5 = 6 (step 4b), R2C6 = 4, R3C6 = 3 (step 10b), R78C5 = [83], clean-up: no 5 in R56C6, no 7 in R7C7
25a. Naked pair {89} in R2C34, locked for R2 and 21(3) cage at R1C3 -> R1C3 = 4

26. Naked pair {57} in R2C89, locked for R2 and N3 -> R1C9 = 8, R1C1 = 5 (step 19d)

27. Naked triple {127} in 10(3) cage at R1C4, locked for R1 -> R1C7 = 3, R2C7 = 2 (step 15), R1C2 = 6, R1C8 = 9
27a. Naked pair {13} in R2C12, locked for N1

28. R56C6 = [18], clean-up: no 6 in R56C4
28a. 6 in N5 only in R4C46, locked for R4

29. R3C6 = 3 -> R34C7 = 7 = [61], clean-up: no 5 in R7C6

30. R3C89 = {14} = 5 -> R45C9 = 15 = [96], R4C3 = 8, R3C4 = 9, R2C34 = [98]

31. R3C12 = {28} = 10 -> R45C1 = 13 = [49]

32. R1C19 = [58] = 13 -> R9C19 = 9 = [27/63/72], no 1,4, no 3 in R9C1

33. 45 rule on N6 2 remaining innies R6C89 = 7 = {25/34}, no 7
33a. R6C89 = 7 -> R7C89 = 6 = {15} (cannot be {24} which clashes with R6C89), locked for R7, N9 and 13(4) cage at R6C8, clean-up: no 2 in R6C89, no 3 in R7C3, no 7 in R7C4, no 6 in R7C6

34. Naked pair {34} in R6C89, locked for R6 and N6, clean-up: no 3,4 in R5C4
34a. Naked pair {25} in R56C4, locked for C4 and N5 -> R4C6 = 6, R4C4 = 3, R5C5 = 4, R7C34 = [26], R7C6 = 7, R7C7 = 4, clean-up: no 7 in R9C9 (step 32)

35. 21(3) cage at R8C7 = {579} (only remaining combination) -> R8C7 = 7, R9C7 = 9, R9C6 = 5

and the rest is naked singles.

_________________
Quis custodiet ipsos custodes?

Normal: [D  Y-m-d,  G:i]     PM->email: [D, d M Y H:i:s]

Software can have bugs, you know! . Furthermore, the hidden cage in question is neither of the ones you listed, but the innies of R1289 at R28C5 = h9(2). The latest SS version still scores this puzzle at 2.48 , because it misses this hidden cage and hence fails to make the first two easy placements. One can see this quite clearly by stepping through the puzzle after scoring it. But the interesting thing is that, if the scoring options are not loaded, SS finds this hidden cage without difficulty!


They don't need to. They just permute the cage, taking into account any blocking constraints. Indeed, by doing so, and taking into account that R3C3+R45C1 cannot contain {89} due to R4C3, SS was able to show (without using udosuk's step 2) that R1C1 not only cannot be a 9, but cannot be a 7, either. Also, if one looks at the solver log (see below), which also lists two other constraints (23(4) cage and R3C5) that were not necessary to get the eliminations, then it's pretty clear that it's doing a full permutational analysis:




Yes, it was a clever way of deterministically formulating it. Most of us lesser mortals would have (with the same IOD cage) probably have reached the same result via induction (i.e., "R1C1 cannot be 9. Here's how: if R1C1 = 9, then ...").


Nor do I. The fact that it's sometimes possible to find a quick way in doesn't IMO necessarily make a puzzle poorer. Indeed, if the "trick" is quite clever, it can be quite motivating for the solver to find it, and arguably be more enjoyable than endlessly shaving off candidates without encountering any really obvious "breakthrough".

_________________
Cheers,
Mike

Thanks Mike, and Afmob by PM, for clarifying which hidden cage was missed; it hadn't been clear from the earlier posts in the thread. It's interesting that SS missed it when using scoring options but found it without scoring options loaded.

I'll have to remember udosuk's algebraic method of adding the same thing to both sides of an equation next time I come across innie-outie differences where one side has multiple cells and the other side only one cell; it seems that it might be useful for that case.

True but it misses it because of a different hidden cage.Getting this the next step is So, by using the Scoring settings and just increasing the "hidden cages" to 5 (from 4) it gives it 1.02! (down from 2.48). Of course, to make this one change to the Scoring routine means other puzzles get messed up (not this dramatically though!) Just did a batch score on the main 115 puzzles that Richard and I used for v3.3.0 and the overall correlation drops about 1.5% but this is not significant. It still means that about 20 out of 100 Assassins will be more than 0.20 score off.

Haven't thought about SS this deeply for ages!

Cheers
Ed