Texas Jigsaw Killer 30
Posted: Sat May 26, 2007 11:06 am
What a beautiful puzzle. This one will surely be up in the top 5 of my all time favs. Ruud, you continue to astound me with your puzzle-making skill. Thanks.
No chains this time. . Please let me know of any improvements/corrections. Let us all know of any shortcuts.
Cheers
Ed
Texas Jigsaw Killer 030
1...23...
..4......
...5....6
.........
..7......
.8......9
.........
.........
.........
n2(r1c5)
n3(r1c6)
n4(r2c3)
n5(r3c4)
n6(r3c9)
n7(r5c3)
n8(r6c2)
n9(r6c9)
1. "45" r6789: 3 outies r5c138 = 23 = h23(3)r5
1a. = {689}:locked for r5
2. "45" n5(r3c4): 2 innies r3c4 + r7c6 = 15 = h15(2)n5
2a. = {69/78}
3. 10(2) cages n5(r3c4): no 5
4. 5 in n5 only in 10(3) = 5{14/23}(no 7)
4a. 5 locked for r5
5. "45" r1234: 3 outies r5c279 = 12 = h12(3)r5
5a. = {147/237}
6.19(5)r1c1 must have 1
6a. -> no 1 r2c24
7. 13(4)r2c6 must have 1: no 8,9
7a. -> no 1 r2c5 or r45c7
8. (not essential: but couldn't resist putting it in)
Triangular connection between 1 required in the 19(5) and 12(4) cages -> no 1 r3c1.Here's how.
8a. 1 in r2c3 -> 1 in 13(4) in r3 -> no 1 r3c1
8b. 1 required in 19(5) in n1 -> no 1 r3c1
9. 16(5)r8c7 = {12346}
9a. -> no {12346} in r8c68
10. 28(4)r6c9 = 89{47/56}
Now, time to get serious
11. LoL r789: 5 outies(r5c3 + r6c2349) = 5 innies(r7c1678 + r8c7)
11a. 27(4)r5c3 = 9{378/468/567}(no 1,2)
11b. -> no 1 or 2 in outies
11c. -> no 1 or 2 in innies
12. -> 1 and 2 required in 16(4)r8c7 are no only in n9
12a. 1 & 2 locked n9
13. 27(4)r5c3 must have 9
13a. from LoL r789 (step 11): outies must have at least one 9
13b. -> innies must have at least one 9
13c. -> 9 in innies in r7c678
13d. -> 9 locked for r7
14. but only one 9 is possible in these innies (all in same row)
14a. -> only one 9 is possible in outies (must be in 27(4))
14b. -> no 9 r6c9
15. 28(4)r6c9 must have 9: only in c8
15a.9 locked c8
16. 9 in r5 only in r5c13
16a. CPE on 9's in r5 -> no 9 r234c3
17. 3 in c9 only in r345 or r89
17a. CPE on 3's in c9 -> no 3 r8c7
17b. 3 required in 16(5)r8c7 only in n9
17c. 3 locked n9
18. r8c7 = {46} in n6(r3c9) -> the same digit 4/6 is in r67c9. Here's how.
I'll explain this move by looking at the 4 first.
18a.-> when r8c7 = 4 the 4 for c8 and c9 must come from 2 nonets apart from n6(r3c9)
18b. -> 4 must be in c89 in n3(r1c6) and n9(r6c9) in r67c9 (can't be in r89c9 or r9c8 because same cage as r8c7)
18c. same logic applies when r8c7 = 6
19. -> 4/6 required in 28(4)r6c9 in r67c9
19a. -> no 4/6 in r7c8 (since 28(4) cannot be both 4 and 6)
19b. -> hidden killer pair 4/6 in n9 in r67c9 and 4 cells from 16(5) that are in n9
19c. -> no 4/6 in r9c6
20. "45" n6(r3c9) + n9(r6c9): r3c8 +8 = r89c6
20a. min. r89c6 = {57} = 12 -> min r3c8 = 4
21. 15(4)r3c8: {1239} combo blocked by r3c8
21a. = {1248/1257/1347/1356/2346}(no 9)
22. 9 in c9 only in r12c9 in n3:
22a. 9 locked for n3
22b. 21(3)r1c9 = {189/579}(no 1..4,6)
23. 15(4)r3c8 must have exactly two of 1,2,3 in r345c9 (step 21a)
23a. the only other place for 1,2 or 3 in c9 is in r89c9
23b. r89c9 cannot have more than 1 of 1,2,3 because of r345c9
23c. -> r89c9 must have 4/6
23d. -> {46} Very Hidden Killer Pair (VHKP) in r6789c9 (remembering step 19b).
23e. 4 and 6 locked in r6789c9 for c9 and n9
23f. 4 in 21(3)r1c9 only in r2c8: no 8 r2c8
23g. r9c78 = {123}
24. "45" n23: 2 innies r3c58 = 12 = h12(2)r3
24a. = {48/57}(no 1,2,3,6,9)
25. 15(4)r3c8 now = {1248/1257/1347} = 1{248/257/347}
25a. 1 locked for c9 and n6
26. 1 in n9 now only in r9: 1 locked for r9
27. 4 in 15(4) only in r3c8 -> no 8 r3c8
27a. -> no 4 r3c5 (h12(2)r3)
27b. and no 8 r3c9 (since 8 already in r3c5 in h12(2) for r3)
28. 4 locked in r23c8. Here's how.
28a. 21(3)r1c9 = {579} -> r3c8 = 4
28b. 21(3) = {489} -> 4 in r2c8
28c. 4 locked in r23c8 for c8 and n3
29. 4 in n6(r3c9) only in c7
29a. 4 locked for c7
29b. no 9 r4c6 (since 4 in 16(4)r4c6 only in r4c6
30. "45" n6(r3c9): r3c8 + 7 = 2 innies r7c8 + r8c7
30a. -> 2 innies = 11, 12 or 14
30b. -> no 9 possible r7c8 (means 2 innies = 13/15}
31. r8c8 = 9 (hsingle c8)
32. 9 in n6(r3c9) only in c7
32a. 9 locked for c7
32b. 23(4)r5c8 must have 9 and 6/8
32c. 23(4) = 9{248/356}(no 7)
32d. 23(4) = [6/8] not both -> only in r5c8 in this cage
33. 9 in n2(r1c5) only in c5: locked for c5
33a. no 1 r4c4 or r6c6
34. 9 in r9 on in n8(r6c2)
34a. no 9 r6c2
35. Generalised X-wing on 9's in n23
35a. 9 locked for r12 in c59
36. 9 in n1 only in c1 in r34c1
36a. 9 locked c1
36b. no 9 r3c2 (same cage)
37. r4c2 = 9, r5c3 = 9, r9c4 = 9, r3c1 = 9 (hsingles)
38. 28(4)r6c9 now = 8{47/56}
38a. CPE on 8's in this cage -> no 8 r4c9
39. "45" n6(r3c9): r3c8 + 7 = 2 innies r7c8 + r8c7
39a. max. 2 innies = [76] = 13 ([86] blocked by r5c8)
39b. -> max r3c8 = 6 (no 7)
39c. no 5 in r3c5 (h12(2)r3)
40. Killer pair {45} in 21(3)r1c9 and r3c8.
40a. 4 & 5 locked n3
41. 15(4)r3c8 = {1257/1347} = 17{25/34}
41a. 7 locked for c9 and n6
41b. no 5 r2c8 (7 only in 21(3) in r2c8)
42. 28(4)r6c9 now = split 19(3) = {568}(no 4)
42a. 6 only in r67c9: 6 locked for n9
43. 6 in 16(5)r8c7 only in r8c7
43a. r8c7 = 6
44. r57c8 = [85]
44a. r67c9 = {68}:locked for c9 and n9
Now, the final move, back to where things started.
45. LoL r789: 5 outies(r5c3 + r6c2349) = 5 innies(r7c1678 + r8c7)
45a. 5 innies must have 5 (r7c8) -> outies must have 5
45b. only 5 available is in r6c234 in 27(4)
45c. r6c234 = {567}: locked for r6
All the rest is on the back straight.
No chains this time. . Please let me know of any improvements/corrections. Let us all know of any shortcuts.
Cheers
Ed
Texas Jigsaw Killer 030
1...23...
..4......
...5....6
.........
..7......
.8......9
.........
.........
.........
n2(r1c5)
n3(r1c6)
n4(r2c3)
n5(r3c4)
n6(r3c9)
n7(r5c3)
n8(r6c2)
n9(r6c9)
1. "45" r6789: 3 outies r5c138 = 23 = h23(3)r5
1a. = {689}:locked for r5
2. "45" n5(r3c4): 2 innies r3c4 + r7c6 = 15 = h15(2)n5
2a. = {69/78}
3. 10(2) cages n5(r3c4): no 5
4. 5 in n5 only in 10(3) = 5{14/23}(no 7)
4a. 5 locked for r5
5. "45" r1234: 3 outies r5c279 = 12 = h12(3)r5
5a. = {147/237}
6.19(5)r1c1 must have 1
6a. -> no 1 r2c24
7. 13(4)r2c6 must have 1: no 8,9
7a. -> no 1 r2c5 or r45c7
8. (not essential: but couldn't resist putting it in)
Triangular connection between 1 required in the 19(5) and 12(4) cages -> no 1 r3c1.Here's how.
8a. 1 in r2c3 -> 1 in 13(4) in r3 -> no 1 r3c1
8b. 1 required in 19(5) in n1 -> no 1 r3c1
9. 16(5)r8c7 = {12346}
9a. -> no {12346} in r8c68
10. 28(4)r6c9 = 89{47/56}
Now, time to get serious
11. LoL r789: 5 outies(r5c3 + r6c2349) = 5 innies(r7c1678 + r8c7)
11a. 27(4)r5c3 = 9{378/468/567}(no 1,2)
11b. -> no 1 or 2 in outies
11c. -> no 1 or 2 in innies
12. -> 1 and 2 required in 16(4)r8c7 are no only in n9
12a. 1 & 2 locked n9
13. 27(4)r5c3 must have 9
13a. from LoL r789 (step 11): outies must have at least one 9
13b. -> innies must have at least one 9
13c. -> 9 in innies in r7c678
13d. -> 9 locked for r7
14. but only one 9 is possible in these innies (all in same row)
14a. -> only one 9 is possible in outies (must be in 27(4))
14b. -> no 9 r6c9
15. 28(4)r6c9 must have 9: only in c8
15a.9 locked c8
16. 9 in r5 only in r5c13
16a. CPE on 9's in r5 -> no 9 r234c3
17. 3 in c9 only in r345 or r89
17a. CPE on 3's in c9 -> no 3 r8c7
17b. 3 required in 16(5)r8c7 only in n9
17c. 3 locked n9
18. r8c7 = {46} in n6(r3c9) -> the same digit 4/6 is in r67c9. Here's how.
I'll explain this move by looking at the 4 first.
18a.-> when r8c7 = 4 the 4 for c8 and c9 must come from 2 nonets apart from n6(r3c9)
18b. -> 4 must be in c89 in n3(r1c6) and n9(r6c9) in r67c9 (can't be in r89c9 or r9c8 because same cage as r8c7)
18c. same logic applies when r8c7 = 6
19. -> 4/6 required in 28(4)r6c9 in r67c9
19a. -> no 4/6 in r7c8 (since 28(4) cannot be both 4 and 6)
19b. -> hidden killer pair 4/6 in n9 in r67c9 and 4 cells from 16(5) that are in n9
19c. -> no 4/6 in r9c6
20. "45" n6(r3c9) + n9(r6c9): r3c8 +8 = r89c6
20a. min. r89c6 = {57} = 12 -> min r3c8 = 4
21. 15(4)r3c8: {1239} combo blocked by r3c8
21a. = {1248/1257/1347/1356/2346}(no 9)
22. 9 in c9 only in r12c9 in n3:
22a. 9 locked for n3
22b. 21(3)r1c9 = {189/579}(no 1..4,6)
23. 15(4)r3c8 must have exactly two of 1,2,3 in r345c9 (step 21a)
23a. the only other place for 1,2 or 3 in c9 is in r89c9
23b. r89c9 cannot have more than 1 of 1,2,3 because of r345c9
23c. -> r89c9 must have 4/6
23d. -> {46} Very Hidden Killer Pair (VHKP) in r6789c9 (remembering step 19b).
23e. 4 and 6 locked in r6789c9 for c9 and n9
23f. 4 in 21(3)r1c9 only in r2c8: no 8 r2c8
23g. r9c78 = {123}
24. "45" n23: 2 innies r3c58 = 12 = h12(2)r3
24a. = {48/57}(no 1,2,3,6,9)
25. 15(4)r3c8 now = {1248/1257/1347} = 1{248/257/347}
25a. 1 locked for c9 and n6
26. 1 in n9 now only in r9: 1 locked for r9
27. 4 in 15(4) only in r3c8 -> no 8 r3c8
27a. -> no 4 r3c5 (h12(2)r3)
27b. and no 8 r3c9 (since 8 already in r3c5 in h12(2) for r3)
28. 4 locked in r23c8. Here's how.
28a. 21(3)r1c9 = {579} -> r3c8 = 4
28b. 21(3) = {489} -> 4 in r2c8
28c. 4 locked in r23c8 for c8 and n3
29. 4 in n6(r3c9) only in c7
29a. 4 locked for c7
29b. no 9 r4c6 (since 4 in 16(4)r4c6 only in r4c6
30. "45" n6(r3c9): r3c8 + 7 = 2 innies r7c8 + r8c7
30a. -> 2 innies = 11, 12 or 14
30b. -> no 9 possible r7c8 (means 2 innies = 13/15}
31. r8c8 = 9 (hsingle c8)
32. 9 in n6(r3c9) only in c7
32a. 9 locked for c7
32b. 23(4)r5c8 must have 9 and 6/8
32c. 23(4) = 9{248/356}(no 7)
32d. 23(4) = [6/8] not both -> only in r5c8 in this cage
33. 9 in n2(r1c5) only in c5: locked for c5
33a. no 1 r4c4 or r6c6
34. 9 in r9 on in n8(r6c2)
34a. no 9 r6c2
35. Generalised X-wing on 9's in n23
35a. 9 locked for r12 in c59
36. 9 in n1 only in c1 in r34c1
36a. 9 locked c1
36b. no 9 r3c2 (same cage)
37. r4c2 = 9, r5c3 = 9, r9c4 = 9, r3c1 = 9 (hsingles)
38. 28(4)r6c9 now = 8{47/56}
38a. CPE on 8's in this cage -> no 8 r4c9
39. "45" n6(r3c9): r3c8 + 7 = 2 innies r7c8 + r8c7
39a. max. 2 innies = [76] = 13 ([86] blocked by r5c8)
39b. -> max r3c8 = 6 (no 7)
39c. no 5 in r3c5 (h12(2)r3)
40. Killer pair {45} in 21(3)r1c9 and r3c8.
40a. 4 & 5 locked n3
41. 15(4)r3c8 = {1257/1347} = 17{25/34}
41a. 7 locked for c9 and n6
41b. no 5 r2c8 (7 only in 21(3) in r2c8)
42. 28(4)r6c9 now = split 19(3) = {568}(no 4)
42a. 6 only in r67c9: 6 locked for n9
43. 6 in 16(5)r8c7 only in r8c7
43a. r8c7 = 6
44. r57c8 = [85]
44a. r67c9 = {68}:locked for c9 and n9
Now, the final move, back to where things started.
45. LoL r789: 5 outies(r5c3 + r6c2349) = 5 innies(r7c1678 + r8c7)
45a. 5 innies must have 5 (r7c8) -> outies must have 5
45b. only 5 available is in r6c234 in 27(4)
45c. r6c234 = {567}: locked for r6
All the rest is on the back straight.