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PsyMar
Hooked

Joined: 17 Nov 2006
Posts: 32
Location: The Triad, North Carolina, US

Posted: Sun Dec 03, 2006 9:15 pm    Post subject: Texas Jigsaw Killer 20 solved!

Turns out there's only a couple hard steps -- after that the whole puzzle falls quite easily.
...Or maybe not. I recall it being easier the time I wasn't writing a walkthrough.

Here's the layout for how I number the nonets:
 Code: 111123333 111123333 441222366 444252666 445252566 445555566 778888899 777888999 777789999

1. Eliminate candidates from cages:
1a. 5/2 in C6 = {14|23}
1b. 7/2s in C1 and R5/N4 and C4 and C5/N2 and R1/N3 = {16|25|34}
1c. 9/2s in C5 and C8 <> 9
1d. 12/2 in R2/N1 = {39|48|57}
1e. 13/2 in C2 = {49|58|67}
1f. 14/2s in R5/N6 and R9 = {59|68}
1g. 15/2s in R1/N1 and R2/N3 = {69|78}
1h. 17/2 in C5/N8 = {89}
1i. 10/3 in R34C34 = {127|136|145|235}
1j. 11/3s in N1 and N3 <> 9
1k. 19/3 in N7 <> 1
1l. 21/3 in R34C67 = {489|579|678}
2. R78C5 = naked pair (89) on C5/N8 -> 9/2 in C5 <> 1
3. 9 of R1 locked in cage 15/2 -> no other 9 in N1, cage 15/2 in R1 = {69} naked pair in N1/R1
4. Outies of R1 = R2C159 = 6 = {123} naked triple in R2 -> 7/2 in C5/N2 = [43|52], 9/2 in C8 = [45|54|63|72|81], 12/2 in R2 = {48|57}
5. LOL -> N13=R12 -> R3C37 = 7 = [25|34] and furthermore R3C3=R2C5 and R3C7 = R1C5
6. Here's one big step.
6a. Combinations for 17/4 in R34C456: {1259|1268|1349|1358|1367|1457|2348|2357|2456}
6b. All cells in cage 17/4 in R34C456 can see either R3C3 or R2C5, these cells are either both 2 or both 3, so 17/4 in R34C456 cannot contain both 2 and 3. Remaining combinations: {1259|1268|1349|1358|1367|1457|2456}
6c. All cells in cage 17/4 in R34C456 can see either R3C7 or R1C5, these cells are either both 4 or both 5, so 17/4 in R34C456 cannot contain both 4 and 5. Remaining combinations: {1259|1268|1349|1358|1367|1457} -> there is a 1 in 17/4 in R34C456
7. 7/2 in R1 <> 1 (combinations)
8. 1s in R12 are locked in the two 11/3 cages -> those cages = {128|137}
9. 9 in R2 locked in cage 15/2 -> 15/2 in R2 = {69}
10. 4 and 5 of N1 locked in R2 -> no 4 or 5 in R2C8
11. Combinations for 9/2 in C8 = {72|81}
12. Innies for N1 = R2C2+R3C3 = 7 -> R2C2 = {45}
13. Outies for R12 = R3C28 = 10 -> R3C2 = {89}
14. LOL -> C12 = N47 -> 6 and 9 not in R489C3+R9C4
15. LOL -> C89 = N69 -> 6 and 9 not in R489C7+R9C6
16. Combinations for 21/3 in R34C67: [498|597} -> R4C6 = 9 -> 15/2 in R2/N3 = [69]
17. 9 in N5 locked in C3 -> 15/2 in R1/N1 = [69]
18. 9/2 in C5 cannot be {45} due to R1C5
19. R2C5 and 9/2 in C5 must contain both 2 and 3 -> elim from rest of C5
20. LOL -> C89 = N69 -> 1 locked in R489C7+R9C6; not in R4C7 -> 1 locked in R89C7+R9C6 -> not in rest of N9
21. R5C89+R67C9 = naked quad elims {5689} from R34C9 (R7C9 cannot be the same as one of R5C89 as if it were, then so would R6C9, thus all 4 are different and this is a valid naked quad)
22. 9 of R3 locked in N4 -> not in R6C2
23. Innies of R6789 = R6C357 = 21 = {579|678} -> no 7 in rest of N5/R6
24. Innies of N36 = R6C89 = 8 = [26|35]
25. 9 of N6 locked in 14/2 in R5 -> R6C3 = 9, 14/2 in R5/N6 = {59} naked pair -> 11 naked singles/last-digit-in-cage moves
26. Recalling step 5, R3C37 = [34] -> R3C9 = 7, R4C7 = 8, R6C7 = 5
27. 7/2 in R1/N3 = {25} naked pair -> 8 NS/LDIC moves
28. 15/3 in C8 = [249] -> 14/2 in R5 = [59] -> R9C8 = 6
29. combinations for 15/4 in R789C67 -> R8C6 = 4
30. 5/2 in C6 = [32] -> R4C6 = 8, 13/2 in C2 = [49]
31. 17/4 in R34C456 = [2681] -> naked singles/LDIC moves solve it.
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