Fwah Ruud: you're torturing me.
That is a fantastic puzzle. Had to use a chain move to unlock it. But perhaps someone can find a better way. Found some really interesting ways to use LoL - hope they are valid. Couldn't find a way to use Jean-Christophe's really great use of LoL and cage sums from TJK28 though. Perhaps that's why I found this puzzle so hard.
Won't put the walk-through into tiny text this time. Please let me know of any shortcuts missed, or improvements/corrections.
[Thanks to Glyn for some improvements/corrections and naming ] [some more typos fixed: thanks Mike][And more improvements and corrections from Andrew. Thanks!]Enjoy! Ed
Texas Jigsaw Killer 291. "45" r12: 3 outies r3c456 = 7 = {124}
1a. {124} locked for r3 and not elsewhere in 17(5) cage
2. "45" r12: 2 innies r2c46 = 10 = {37}
2a. {37} locked for r2
2b. Common Peer Elimination (CPE) -> no 3 or 7 r1c6
3. 13(2)r1c5 = {49/58}/[76](no 6 r1c5)
4. 24(3)r3c8 = {789}
4a. -> no 789 in r12c9 or r4c78
5. 14(4)r1c8: no 9
6. 8(3)r5c7 = 1{25/34}
6a. 1 locked for r5
6b. CPE ->no 1 r6c7
7. 15(3)r3c7. All combinations have 7/8/9 which are only in r3c7. The exception is {456}
7a. -> r3c7 = 5..9
8. 7 and 9 now only in r3 in n4(r1c9)
8a. 7 & 9 locked for r3
8b. 17(3)r3c1 - no 1,2,5,7 r4c1
8c. 17(3) = {359/368/458}
8d. deleted
8e. 18(3)r3c3: {279} combo blocked by r3c3 -> no 2 r4c23
9. 3 in r3 only in c123
9a. CPE: no 3 r1c1 or r4c2
10. 6(2)r8c5 = {15/24}
10a. -> 1 in c5 locked in 6(2) or r3c5 (6(2) = {24} -> r3c5 = 1)
11. "45" c123: r19c4 = 17 = {89}: locked for c4
12. 14(3)r1c3 must have 8/9
12a. = {149/158/239/248}
12b. cannot have both 8 and 9 -> r12c3 = {14/15/23/24} = 1..5
13. 16(3)r8c3 must have 8/9
13a. {367/457} blocked
13b. cannot have both 8 & 9 -> no 8,9 r89c3
14. "45" c123: 4 innies: r1289c3 = 13 = h13(4)c3
14a. must have 1: 1 locked for c3
14b. -> {259} combo blocked from 16(3)r8c3: (
because 13(4) cannot contain both of 2&5)
14c. -> 16(3)r8c3 = {169/178/268/349/358}
14d. -> r12-89c3 = {14-26/14-35/15-34/23-17/24-16}
15. deleted
16. "45" r1234: 3 innies = 9 = h9(3)r4
16a. = {126/135/234}(no 789) = [2/5..]
16b. ->{258} combo blocked from 15(3)r3c7
17. 15(3)r3c7: all combo's without 1 have 2/3/4 in r4
17a. -> 1 in r4 in h9(3) or r4c78
18. 1 in n5(r3c1) only in
r67c1 or r6c2
18a. CPE: no 1 r7c2
19. LoL r6789: 3 innies
r67c1+r6c2 = 3 outies r4c9+r5c89
19a.innies must have 1 (step 18) -> outies must have 1
19b. 1 in
r5c8919c. 1 locked for r5 (no 1 r5c7) & 1 locked for n6(r4c9)
19d. deleted, faulty
[Thanks Andrew]20. 9(3)r6c7 = {126/135/234} -> r7c7 = 1..4
21. 16(3)r5c1 = {259/268/349/367} ({358/457} blocked by 8(3)r5c7)
21a.
16(3) = [2/3..], 8(3)r5c7 = [2/3]: Killer pair 2/3: 2 & 3 locked for r5
[Andrew noticed that “45” on r5 3 innies r5c456 = 21 is a simpler way to do the above and I use that "45" in the next step. Thanks]21b. 2 in n5(r
3c1) is either in the innies of LoLr6789 (step 19) -> must be in r5c89 for outies OR 2 in n5 is in 16(3)r5c1.
21c. Either way, no 2 r5c7. Is there a name for this move?
( LoLR - "Law of Left(&Right)-overs")21d. -> no 5 r5c89 ({125} combo must have 5 in r5c7)
21e. -> no 5 in innies of LoL r6789 (step 19) in r6c12+r7c1
(rest of this step deleted)21f
. no 6 in the outies -> no 6 in innies-> no {156} combo in 12(3)r7c122. "45" r5: r5c456 = 21 = h21(3)r5
22a. = {489/579/678}
23. "45" r6789: 3 innies r6c456 = 15 = h15(3)r6
23a. = {159/168/249/267/348/357} ({258/456} blocked by h9(3)r4 in 45(9) see step 16a.)
24. "45" r89: 2 innies r8c46 = 8 = h8(2):
24a. = {17/26/35}(no 4,8,9)
24b.-> 2 locked for n9(r7c5) in h8(2) or 6(2) ({15} in 6(2) -> {26} in h8(2))
25. "45" r89: 3 outies r7c456 = 20 = h20(3)
25a. = {389/479/569/578}(no 1,2,
no 3 in r7c56)
25b. 28(5)r7c4: all combinations must work for both the h8(2) and h30(3)
25c. {14689/24589/34678} all blocked
26. "45" c89: 1 outie r5c7 + 5 = 2 innies r46c8
26a. min r5c7 = 3 -> min r46c8 = 8 -> no 1 r4c8
27. "45" c89: 5 outies = 19
27a. -> must have 1:1 locked for c7
28. "45" c789: 2 outies r19c6 = 9.
28a. no 9, no 6 r9c6
29. "45" n4(r1c9): r1c8 + 16 = 3 innies
29a. Max. 3 innies = {589} = 22
29b. max, r1c8 = 6
29c. CPE: no 8 r1c6 since it sees all 8's in n3(r1c7)
29d. no 1 r9c6 (h9(2)c7)
30. 17(3)r1c
7 must have 8/9 for n3(r1c6): r5c6 only other place with 8/9
30a. 17(3) = {179/269/278/359/368/458} = [8/9..] not both
30b. only other place for 8 or 9 is in r5c6 = {89}
31. [18] blocked from h9(2)c6.
31a. 1 in r1c6 -> r23c7 = [79] -> r5c6 = 8: clash with r9c6
31b. no 1 r1c6, no 8 r9c6
32. LoL c1234: 4 outies r1c56+r25c5 = 4 innies r8c4+r9c234
32a. no 1,3 in 4 outies -> no 1 or 3 in 4 innies
32b. 16(3)r8c3 = [1/2/3] only in r8c3 = {123}
33. LoL c6789: 3 outies r346c5 = 3 innies r189c6
33a. no 8 or 9 in innies -> no 8 or 9 in r6c5
Now - a contradiction chain. Is there another way?
34. no 2 r1c6 because of LoL c6789 (step 33)
34a. 2 in r1c6 -> r9c6 = 7 (h9(2)c6) -> {27} in innies
34b. -> {27} in outies r346c5 -> 6(2)r8c5 = {15} -> 13(2)r1c5 = {49}
34c. but {249} in n2 forces 1 into both r3c4 and {158} in 14(3)r1c3
34d. no 2 r1c6
34e. no 7 r9c6 (h9(2)c6)
34e. no 2 r8c4 (LoL c1234 step 32)
34f. oops should have done this earlier too: h8(2)r8 =
[53/62/71]35. 17(3)r1c6 must have 4,5,6
35a. = {269/359/368/458}
36. LoL c1234: Outies r125c5+r1c6 must have 4/5. Here's how.
36a. since 13(2)r1c5 = [4/5] or {67}
36b. if {67} -> r1c6 = 4/5
36c. -> LoLc1234: innies r8c4 + r9c234 must have 4/5
36d. -> Killer pair 4/5 with 6(2)r8c5: 4 and 5 locked for n9(r7c5)
37. r9c6 = 3 finally!
37a. r89c7 = {69/78} = [8/9..]
37b. no 5 r8c4 (h8(2)r8)
38. r2c46 = [37]
39. r1c6 = 6 (h9(2)c6)
39a. r12c7 =
{29}/[38]=[8/9..]
40. Killer Pair 8/9 in r1289c7:8 and 9 locked for c7
41. 9 in r3 only in 24(3)r3c7 -> no 9 r4c9
41a. -> no 9 in r7c12 + r7c1 (innies LoL r6789)
42. 13(2)r1c5 = {49/58}(no 7) = [4/5..]
43. Killer Pair 4/5 in 13(2) & 6(2)c5: 4 and 5 locked c5
44. r1c2 = 7 (hsingle n2)
45. 15(3)r3c7 = {267/357/456}(no 1)
46. r7c7 = 1 (hsingle c7)
46a. r6c78 = {26/35}(no 4)
47. 1 in r4 in h9(3) = 1{26/35}(no 4)
47a. 1 locked for 45(9)
48. h15(3)r6 = {249/267/348/357} ({258/456}blocked by r6c78)
49. h13(4)c3: r12-89c3
49a. from step 14d ={14-35/14-26/15-34/24-16}(no 7 r9c3)
49b. = 14{35/26}
49c. 4 locked c3
49b. CPE for 7's n9(r7c5): no 7 r7c4
50. Killer pair 1/2 in c5 in r3c5 and 6(2)
50a. 2 locked c5
50b.
[edit:this step added]. LoLc6789: no 7 in 3 innies -> no 7 r6c5. (or 3 and 6 placed in innies -> outies r347c5 must have {36} for c5)
50c.{36} pair r46c5: 3 and 6 locked c5 and 45(9)
51. 28(5)r7c4 = 67{159/249/258}
51a.
deleted52. r4c5 = 6 (hsingle n3) How long has that been there?
52a. r6c5 = 3
52b. h9(3)r4: r4c46 = {12}:
52c. 1 and 2 locked r4 & 45(9)
53. naked pair {12}:locked for n3
53a. r3c6 = 4
53b. r12c7 = [38]
the rest is cage sums, hidden/naked singles.