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Congratulations to Afmob on finding the "easy" way to do this puzzle (A144V2) - which is the same way as SS does. No wonder I couldn't find it! Am very glad I ditched this one as the V1 as I dislike that sort of combo analysis and would rate it much higher than Afmob and SS have! But as I said, really well done to Afmob for finding it.

BTW - by altering the code to block Afmob's way to this,
3x3::k:2817:5890:5890:7683:7683:3076:8197:2054:2054:2817:5890:5890:7683:7683:3076:8197:8197:8197:6919:4360:7945:7945:7683:3076:4618:4618:8197:6919:4360:4360:7945:7945:7945:5899:4618:8197:6919:6919:6919:780:6413:7945:5899:5899:5899:6926:5135:6919:780:6413:6413:4880:4880:5899:6926:5135:5135:5393:5906:6413:6413:4880:5899:6926:6926:6926:5393:5906:5906:5906:4627:2068:2325:2325:6926:5393:5906:4627:4627:4627:2068:
..it now gets a score of 2.26. However, it appears that SudokuSolver is unable to do one key move (see my WT below step 20). If I give SS that move when it's first available, it does 18 fewer steps, so the score could come down below 2. Still very tough and very close to my original estimation of 2.0.

Short WT for A144 v2
Prelims
i. 11(2)n1: no 1
ii. 26(4)n1: no 1
iii. 8(2)n3: no 4,8,9
iv. 23(6)n6: no 9
v. 3(2)n5 = {12}
vi. 20(3)n4: no 1,2
vii. 19(3)n6: no 1
viii. 21(3)n8: no 1,2,3,
ix. 8(2)n9: no 4,8,9
x. 9(2)n7: no 9

1. 3(2)n5 = {12}: both locked for n5 & c4

2. "45" n1247: 2 innies r3c34 = 4 = [13]
2a. split cage 27(4)n5 = {4689/5679}
2b. 6 & 9 locked for n5

3. 3 remaining innies n5 = 15 and must have 3 for n5
3a. = 3{48/57}
3b. 3 locked for 25(5)n5

4. "45" r1234: 1 outie r5c6 - 2 = 3 innies r3c1 + r4c17
4a. min. 3 innies = [2][12] = 5 -> min r5c6 = 7
4b. max. 3 innies = 7 -> max. r4c17 = 5 (no 5..9)
4c. max. 3 innies = 7 and min. r4c17 = 3 -> max. r3c1 = 4

5. "45" n47: 2 outies r3c12 = 10 = [28/46]

6. 6 in n5 only in r4: locked for r4

7. 23(6)n6 = 1234{58/67}
7a. r4c9 "sees" all of the 23(6) -> no 1,2,3,4 in r4c9 (CPE)

8. "45" n3: 2 outies r4c89 = 13 (no 1,2,3)
8a. = [49]{58}(no 7)

Now time for a new move, IOE (see image below) [edit: Changed wording for clarity, Thanks Afmob and Andrew!]
When the Outies "see" all of a candidate for a house except for the Innies, the 3rd cell must Equal the in/out difference (IOD).
i. For example, "45" n3: 2 innies r3c78 - 5 = 1 outie (r4c9)
ii. the 1 outie (r4c9) sees all of 4, 8 & 9 for n3 (they can't be in the 8(2)n3), except for the innies r3c78.
iii. so, when 8 is in r4c9, 8 must be in the innies, r3c78. Since 1 innie = 1 outie -> the 3rd cell (in this case, one of r3c78) must equal the IOD for n3 = 5
iv. same thing for 9 since it sees all 9s in n3 except r3c78. So, 9 in r4c9 -> r3c78 = {59}(IOE)
v. of course, this doesn't work for the 5 in r4c9 because it doesn't see all 5s in n3 apart from the innies. One 5 can hide from r4c9 in r1c8.

9. So, "45" n3: 1 outie + 5 = r3c78
9a. r4c9 = (589). If not 5, it must be 8 or 9 -> 5 locked in r3c78 (IOE).
9b. -> 5 locked in innies of n3 OR r4c9
9c. -> no 5 in common peers in r4c8, r1c7, r2c789, r3c9 & r1c9.



NOTE: IOE is not enough on its own to make eliminations in this case. Only if r4c9 = (89) would we have been able to lock 5 in r3c78 for n3 & r3 using IOE only.
9e. no 8 in r4c9 (h13(2))

10. 18(3)n3 must have 4/8 = {459/468}(no 2,7)
10a. {68}[4] blocked by r3c2 -> no 8 r3c78
10b. note: 4 locked in 18(3). Important for later.

11. 1 in n6 only in 23(6) -> no 1 in r7c9

12. "45" n36: 2 outies r7c89 = 10 (no 5)
12a. = {28/37/46}(No 9)

This move took a long time to spot.
13. 19(3)n6, {289} blocked since the 8 & 9 sees both 8 & 9 in h13(2)n6 = [49/85]
13a. 19(3) = {379/469/568}(no 2) ({478} blocked by r4c8)
13b. = [8/9..]

14. 2 in n6 only in 23(6) -> no 2 in r7c9
14a. h10(2)r7c89 = {37/46}(no 8)

15. Killer pair {89} in h13(2)n6 & r6c78 (step 13b): 8 locked for n6

16. 23(6)n6 = {123467}(no 5)

17. split cage 27(4)n5 = {4689/5679}
17a. {4689} must have 8 in r5c6 because r4c8 = (48) NOTE: r4c8 = r5c6
17b. {5679} must have 9 in r5c6 because r4c9 = (59) NOTE: r4c9 = r5c6
17c. r5c6 = (89)
17d. r4c456 = {469/567}(no 8)

Another one that took a long time to get since it meant seeing the cell cloning in steps 17a & b
18. since r5c6 = one of r4c89 (steps 17a & b) = (89) -> implied killer pair on (89) with r6c78 = [8/9..](step 13b) -> no 8 in r6c56
18a. 8 in n5 only in r5: locked for r5

19. h15(3)n5 = {348/357}
19a. 8 in {348} must be in r5c5 -> no 4 in r5c5

Now time for the final puzzle breaker.
20. no 4 in r3c1 because of 4s in r5 & 18(3)n3: Grouped Turbot Fish. Like this.
20a. 4 in r5 is in n4 or n6.
20b. If in n4 it must be in the 27(6)n1 -> no 4 in r3c1
20c. If in n6 -> 4 locked in 18(3)n3 must be in r3 -> no 4 in r3c1.
20d. r3c1 = 2

Puzzle is cracked. Very satisfying solution.

Cheers
Ed

Another puzzle from my Unfinished folder "bites the dust"; I finished this one about a week ago but only later found time to go through the posted WTs.

Thanks Ed for a challenging V2 and also for your interesting post about IOE; I think I now understand that technique although I haven't yet been able to apply it to any puzzles.

I seem to have found another way for the key breakthrough, possibly not as technically simple as the way used by Afmob and SS but I feel more elegant, to use udosuk's word.

I took a long time to find the breakthrough but when it came I enjoyed it and felt that it was worth the effort.


Here is my walkthrough for A144 V2.

Prelims

a) R12C1 = {29/38/47/56}, no 1
b) R1C89 = {17/26/35}, no 4,8,9
c) R56C4 = {12}
d) R89C9 = {17/26/35}, no 4,8,9
e) R9C12 = {18/27/36/45}, no 9
f) 20(3) cage at R6C2 = {389/479/569/578}, no 1,2
g) 19(3) cage at R6C7 = {289/379/469/478/568}, no 1
h) 21(3) cage at R7C4 = {489/579/678}, no 1,2,3
i) 26(4) cage at R2C1 = {2789/3689/4589/4679/5678}, no 1
j) 23(6) cage at R4C7 = {123458/123467}, no 9, CPE no 1,2,3,4 in R4C9

Steps resulting from Prelims
1a. Naked pair {12} in R56C4, locked for C4 and N5
1b. 23(6) cage at R4C7 = {123458/123467}, CPE no 1,2,3,4 in R4C9

2. 9 in C9 only in R234C9, locked for 32(6) cage at R1C7, no 9 in R1C7 + R2C78

3. 45 rule on N3 2 outies R4C89 = 13 = [49]/{58/67}, no 1,2,3,9 in R4C8

4. 45 rule on N7 2 outies R6C12 = 11 = [29]/{38/47/56}, no 1,9 in R6C1

5. 45 rule on N36 2 outies R7C89 = 10 = {28/37/46}/[91], no 5

6. 45 rule on N47 2 outies R3C12 = 10 = {19/28/37/46}, no 5

7. 45 rule on N1247 2 innies R3C34 = 4 = [13], clean-up: no 7,9 in R3C12 (step 6)
7a. R3C34 = 4 -> R4C456 + R5C6 = 27 = {4689/5679}, 6,9 locked for N5

8. 3 in N5 only in R5C5 + R6C56, locked for 25(5) cage at R5C5, no 3 in R7C67
8a. R5C5 + R6C56 = {348/357}
8b. 45 rule on N5 2 remaining outies R7C67 = 10 = {19/28/46}, no 5,7

9. 45 rule on N2 1 remaining innie R1C4 = 1 outie R2C3 + 3, no 4 in R1C4, no 7,8,9 in R2C3

10. 1 in N6 only in R4C7 + R5C789 + R6C9, locked for 23(6) cage, no 1 in R7C9, clean-up: no 9 in R7C8 (step 5)

11. 17(3) cage at R3C2 = {269/278/368/458/467} (cannot be {179/359} because R3C2 only contains 2,4,6,8), no 1

12. 18(3) cage at R3C7 = {279/459/468/567}
12a. 8 of {468} must be in R4C8 (R3C78 cannot be {48/68} which clash with R3C12), no 8 in R3C78

13. 45 rule on C789 2 innies R78C7 = 1 outie R9C6 + 9
13a. Max R78C7 = 17 -> max R9C6 = 8

14. R123C6 = {129/147/156/246}, no 8

[This was how far I managed to get when this puzzle first appeared. When I came back to it, I still took some time to find my next step

9 in C9 only in R234C9, R4C89 = 13 (step 3) -> 18(3) cage at R3C7 (step 12) can only contain 9 if it also contains 4 in R4C8 (otherwise cannot place 9 in C9) -> 18(3) cage at R3C7= {459/468/567}, no 2
I’ve changed this to a comment because after the next step and then some more routine ones, the 18(3) cage is directly reduced to two combinations.

Next I studied the 27(6) cage at R3C1 and the 23(6) cage at R4C7 because R3C1 “sees” all of N4 except for R4C23 + R6C2, similarly for R7C9 and N6.
However this didn’t lead to anything. Then I spotted the next step which seems to be the key breakthrough.]

15. R1C4 = R2C3 + 3 (step 9)
15a. R2C3 cannot be 2, 4 or 6 because then 2,4,6,8 in R12C1, R2C3 and R3C12 form killer quad for N1, also if R2C3 is even then R1C4 is odd -> all numbers in 26(4) cage at R1C2 would be odd but this isn’t possible because {3579} only totals 24
15b. -> R2C3 = {35}, R1C4 = {68}
[Maybe step 15a seems to be a contradiction move but I prefer to consider it as creative analysis of even and odd numbers using an innie-outie difference for this purpose.

While checking my walkthrough I noticed a more direct way to write this step.
15. Hidden killer quad 2,4,6,8 in R12C1, R12C23 and R3C12 for N1, R12C1 contains one of 2,4,6,8, R3C12 contains two of 2,4,6,8 -> R12C23 must contain one of 2,4,6,8
15a. 26(4) cage at R1C2 must contain two even numbers (because {3579} only totals 24), R1C23 + R2C2 can only contain one even number -> R1C4 = {68}, R2C3 = {35} (step 9)]

16. 21(3) cage at R7C4 = {489/579} (cannot be {678} which clashes with R1C4), no 6, 9 locked for C4 and N8, clean-up: no 1 in R7C7 (step 8b)

17. 27(5) at R1C5 = {12789/14589/14679/23589/23679/24579/34569/34578} (cannot be {13689/15678/24678} which clash with R1C4
17a. R123C6 (step 14) = {129/147/156} (cannot be {246} which clashes with 27(5) cage), 1 locked for C6 and N2, clean-up: no 9 in R7C7 (step 8b)

18. R5C5 + R6C56 (step 8a) = {357} (only remaining combination, cannot be {348} which clashes with R7C67), 5,7 locked for N5

19. 21(3) cage at R7C4 (step 16) = {579} (only remaining combination, cannot be {489} which clashes with R14C4, ALS block), 5,7 locked for C4 and N8

20. R1C4 = R2C3 + 3 (step 9) -> R1C4 + R2C3 = [63/85]
20a. 27(5) at R1C5 (step 17) = {23589/24579/34578} (cannot be {23679/34569} which clash with R1C4 + R2C3), no 6, CPE no 5 in R2C6

21. R4C89 (step 3) = {58/67} (cannot be [49] which clashes with R4C456, ALS block), no 4,9
21a. Killer pair 6,8 in R4C456 and R4C89, locked for R4
21b. Hidden killer pair 6,8 in R4C456 and R4C89 for R4, R4C89 contains one of 6,8 -> R4C56 only contains one of 6,8 -> R5C6 = {68}
21c. 4,9 in N5 only in R4C456, locked for R4

22. 9 in C9 only in R23C9, locked for N3
22a. 18(3) cage at R3C7 (step 12) = {468/567}, no 2, CPE no 6 in R12C8, clean-up: no 2 in R1C9

23. 17(3) cage at R3C2 (step 11) = {278} (only remaining combination, cannot be {368/458/467} because 4,6,8 only in R3C2) -> R3C2 = 8, R4C23 = {27}, locked for R4 and N4, R3C1 = 2 (step 6), clean-up: no 3,9 in R12C1, no 6 in R4C89 (step 3), no 3,4 in R6C1 (step 4), no 4,9 in R6C2 (step 4), no 1 in R9C1, no 7 in R9C2

[It’s no longer relevant but earlier I’d written the note
1 in N4 only in 27(6) cage at R3C1
R4C23 must contain one odd and one even number, R6C12 must contain one odd and one even number, R3C1 only has even numbers -> 27(6) cage must contain three odd and three even numbers
27(6) cage = {123489/123678/124569/124578/134568} (cannot be {123579} which has four odd numbers).]

24. Naked pair {58} in R4C89, locked for R4 and N6, CPE no 5,8 in R2C8
24a. R5C6 = 8 (hidden single in N5), clean-up: no 2 in R7C7 (step 8b)
24b. 8 in C4 only in R12C4, locked for N2

25. 18(3) cage at R3C7 (step 22a) = {468/567}, 6 locked for R3 and N3, clean-up: no 2 in R1C8

26. Killer pair 3,5 in R6C12 and R6C56, locked for R6
26a. 7 in R5C5 + R6C56 must be in R6C56 (R6C56 cannot be {35} which clashes with R6C12), 7 locked for R6 and N5
26b. 7 in N6 only in R5C789, locked for 23(6) cage at R4C7, no 7 in R7C9, clean-up: no 3 in R7C8 (step 5)

27. 9 in N6 only in R6C78, locked for R6
27a. 19(3) cage at R6C7 = {289/469}, no 7, clean-up: no 3 in R7C9 (step 5)
27b. 8 of {289} must be in R7C8 -> no 2 in R7C8, clean-up: no 8 in R7C9 (step 5)

28. Naked quad {2468} in R7C6789, locked for R7

29. 20(3) cage at R6C2 = {569} (only remaining combination) -> R6C2 = 6, R7C23 = {59}, locked for R7 and N7 -> R7C4 = 7, R6C1 = 5 (step 4), clean-up: no 6 in R12C1, no 3,4 in R9C1, no 4 in R9C2
[At first I spotted that R7C23 are both odd -> R6C2 must be even -> R6C2 = 6 ...]

30. Naked pair {47} in R12C1, locked for C1 and N1, clean-up: no 2 in R9C2

31. R1C3 = 6 (hidden single in N1), R1C4 = 8, R2C4 = 4, R4C4 = 6, R2C3 = 5 (step 9)

32. Naked triple {279} in R123C5, locked for C5 and N2 -> R3C6 = 5, R12C6 = [16], R4C56 = [49], R6C56 = [37], R5C5 = 5, R7C5 = 1, clean-up: no 7 in R1C89, no 4 in R7C7 (step 8b)

33. Naked pair {68} in R89C5, locked for 23(5) cage at R7C5, no 6,8 in R8C7
33a. R789C5 = 1{86} = 15 -> R8C67 = 8 = [35], clean-up: no 3 in R9C9

34. Naked pair {35} in R1C89, locked for R1 and N3 -> R12C2 = [93], R9C2 = 1, R9C1 = 8, R78C1 = [36], R89C5 = [86], R45C1 = [19], R5C23 = [43], R6C3 = 8, R4C7 = 3

36. R89C7 = {17} (only remaining combination) -> [17]

37. Naked pair {24} in R9C36, locked for R9 -> R9C78 = [93]

38. Naked pair {24} in R6C79, locked for R6 and N6 -> R5C9 = 6

39. Naked pair {24} in R67C9, locked for C9 -> R3C9 = 9

and the rest is naked singles.