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sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Tue Jun 12, 2007 12:26 pm    Post subject: Texas Jigsaw Killer 18

This is a real toughie. See what you mean Mike and Para. First attempt couldn't get anywhere, so had to put the creative hat on.

Seems like I've been able to make some progress with some interesting techniques. Can someone check through to make sure its all valid? [EDIT: Rewritten from step 22 + new marks pic. Thanks Mike for finding the mistake. EDIT2: problem with new step 27 and other typos fixed: thanks Para and Mike] Add some more too if you like . I probably won't be able to look at it again till the weekend.

Seems like we need to add Killer LoL to our jargon to account for the effect of cage combinations on LoL. I also use 'overlap' to describe the congruence between combinations. Might need to find a different name so it is not confused with the normal 'overlap' technique. Names and the need for them is not my forte.....

Cheers
Ed

Texas Jigsaw Killer 18
1. 23(3)r3c6 = {689}: all locked for n3

2. "45" r123: r4c6 - 7 = r3c9
2a. r3c9 = {12}
2b. r4c6 = {89}
2c. 6 in n3 only in r3: 6 locked for r3

3a. no 1 r2c1. Here's how.
3b. 1 in r2c1 -> r3c12 = {89}
3c. But this clashes with r3c67.

4. Killer LoL (KLoL) chain r123: no 2 r2c1. Here's how.
4a. LoLr123: 4 innies r2c1 + r3c129 = 4 outies r4c456 + r5c5
4b. only one 2 is possible in outies -> only one 2 is possible in innies.
4c. 2 in r2c1 -> r3c12 = {79} -> r4c6 = 9 -> r3c9 = 2 (step 2)
4d. but this means two 2's in r123 innies. Not possible.

5.KLoL r123: no {567} combo in 18(3)r4c6. Here's how.
5a.LoLr123 same as step 4a. Outies must have 8/9 (r4c6) -> innies must have 8/9
5b. ->18(3)r2c1 {567} combo. blocked

6. KLoL r123: no 2 r3c12. Here's how.
6a. 2 in r3c12 -> rest of 18(3) = {79}(no 8)
6b. 2 in r3c12 -> r3c9 = 1 -> r4c6 = 8 (step 2)
6c. But we know from LoLr123 that there can be no 8 in outies when 2 in r3c12.

7. 22(3)r5c3 = 9{58/67}
7a. 9 locked for n6(r5c3)

8. "45" c12: r4c3 + 4 = r6c2
8a. r4c3 = 1..5

9. "45" r89: 3 innies r8c136 = 8 = h8(3)r8
9a. = 1{25/34}
9b. 1 locked for r8

10. 1 in n9(r8c2) only in r9: 1 locked for r9

11. 21(3)r8c8 = {489/579/678} = [4/7..]

12. KLoL r89: no {457} combo in 16(3)r8c7. Here's how.
12a. LoL r89: 6 innies r8c1367 + r9c78 = 6 outies r56c67 + r7c78
12b. and both = 24 ie h24(6)r89innies = 24(6)r89outies
12c. 6 outies must have 1,2 & 3 for n8(r5c6)
12d. -> 6 innies must have 1,2 & 3
12e. note: 6 innies cannot have any repeats since the 6 outies are all in the same nonet.
12f. -> when h8(3)r8 = {125}, 16(3)r8c7 must be 3....
12g. -> when h8(3)r8 = {134}, 16(3) must be 2...
12h. -> {457} blocked from 16(3)

13. KLoL r89: no {358} combo. in 16(3)r8c7
13a. since the 6 outies in LoL r89(step 12) are all in the same nonet, there can be no repeats in the 6 innies
13b. since h8(3)r8 = {125/134} = [3/5..]
13c. -> {358} combo blocked from 16(3)
13d. -> the valid combinations in the h24(6)r89innies are
i. {125/349} = {123459}
ii. {125/367} = {123567}
iii. {134/259} = {123459}
iv. {134/268} = {123468}
13e. = 123{459/468/567} = [4/7..] not both.

14. KLoL r89: {147} combo blocked from both 12(3) cages in n8(r5c6): clash with 21(3)r8c8 = 4/7..] step 11 (thanks Mike for this much easier way).

15. 16(3)r8c7 = {259/268/349/367}
15a. ->{456} blocked from 15(3)r7c5
15b. ->{239/356} blocked from 14(3)r5c4

16. "45" r789: r6c7 + 3 = r7c9
16a. min r7c9 = 4, max r6c7 = 6

17. "45" c1234: r6c5 - 2 = r4c4
17a. max r4c4 = 7, min. r6c5 = 3

18. "45" c789: r135c7 = 19 = h19(3)c7
18a. no 1

19. LoLc789: 4 innies r189c7 + r9c8 = 4 outies r3456c6
19a. no 1 in innies -> no 1 in r56c6 [edit typo]

20. 1 in n8 only in 12(3)r6c7 = {129/138/156}(no 4,7)

21. 12(3)r5c6 = {237/246/345}(no 8,9)

22. KLoL Over-lap c789: no {259} combo. in 16(3)r8c7. Here's how.
22a. Lol c789: 4 innies r189c7 + r9c8 = 4 outies r3456c6
22b. the 12(3)r5c6 has 2/4 cells in the outies.
22c. the 16(3)r8c7 has 3/4 cells in the innies
22d. -> the valid combination in the 12(3) must overlap with 2 of the candidatates from the 16(3) combination
(IF that combo has at most 1 of 6/8/9 corresponding to r34c6 in outies: this is what I missed first WT)
22e. since 12(3) combinations are {237/246/345} and since {259} combo in 16(3)r8c8 has ONLY 1 of 6/8/9 AND does not have 2 candidates overlapping with 12(3)
22f. ->{259} combo blocked

23. 16(3)r8c8 = {268/349/367}(no 5)

24. KLoL + (double)Overlap c789: r56c6 no 5 [edit out invalid part]. Here's how.
24a...d (same reasoning as steps 22a..d)
24e. the 12(3) combinations are {237/246/345}
24f. the {349/367} combo's from 16(3) must have a double-overlap candidates with the 12(3) -> r56c6 = {34/37}
24g. for the {268} combo from 16(3), r3456c3 = {68}2{3/4/7} -> r56c6 = {23/24/27}(no 5)
24h. finally, the {268} combo from 16(3) could also double-overlap with 12(3) -> r56c6 = {26}
24h. In summary: r56c6 = {23/24/26/27/34/37}(no 5)

25. LoLc789: no 5 in outies -> no 5 in r1c7

26. "45" c6789: 3 innies r789c6 = 13 = h13(3)c6 [edit:typo]
26a. = {139/148/157/256} others blocked by r56c6 (step 24h.)

27. {148} combo blocked from h13(3)c7. Here's how. [edit: true but flawed reason: see Mikes preamble to step 37]
27a. 15(3)r7c5 = {159/168/249/258/267/357} ({348} blocked by 16(3)r8c7)

[edit: following steps are still valid]
28. h13(3)c6 = {139/157/256}(no 4,8)
28a. r78c6 + r9c6 = [913]/{15}[7]/[751]/{25}[6]/[625]
28b. ({13}[9]/[931]/{17}[5]/{56}[2] blocked by combinations unavailable in 15(3))
28c. no 3 r78c6, no 2 or 9 in r9c6 [edit:typos]

29. 15(3)r7c5 = [591]/9{15}/[375]/8{25}/[762] = {159/258/267/357} = [5/7..]
29a. r7c5 = {35789}

30. 14(3)r5c4: {257} blocked by 15(3)

 Code: .-----------------------.-----------------------.-----------.-----------------------.-----------------------. | 123456789   123456789 | 12345678    12345678  | 123456789 | 123456789   2346789   | 123457      123457    | :-----------.           :-----------.           |           |           .-----------'-----------.           | | 3456789   | 123456789 | 123456789 | 12345678  | 123456789 | 123456789 | 123457      123457    | 123457    | |           '-----------:           '-----------:           :-----------'-----------.           :-----------: | 1345789     1345789   | 12345789    12345789  | 12345789  | 689         689       | 123457    | 12        | :-----------------------'-----------.-----------'-----------:           .-----------'-----------:           | | 123456789   123456789   12345     | 1234567     12345678  | 89        | 123456789   123456789 | 123456789 | :-----------------------.-----------+-----------.           :-----------'-----------.           |           | | 123456789   123456789 | 56789     | 123456789 | 12345678  | 23467       234567    | 123456789 | 123456789 | |           .-----------'           |           '-----------:           .-----------+-----------'-----------: | 123456789 | 56789       56789     | 123456789   3456789   | 23467     | 12356     | 123456789   123456789 | :-----------'-----------.-----------'-----------.-----------'-----------:           '-----------.           | | 12345678    12345678  | 12345678    12345678  | 35789       125679    | 1235689     1235689   | 456789    | |           .-----------:           .-----------+-----------.           :-----------.-----------'-----------: | 12345     | 23456789  | 12345     | 456789    | 23456789  | 125       | 2346789   | 456789      456789    | :-----------'           :-----------'           |           '-----------:           '-----------.           | | 123456789   123456789 | 456789      456789    | 123456789   13567     | 2346789     2346789   | 456789    | '-----------------------'-----------------------'-----------------------'-----------------------'-----------'

Last edited by sudokuEd on Thu Jul 05, 2007 4:24 am; edited 1 time in total
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sun Jul 01, 2007 4:38 pm    Post subject: Hi Ed haven't picked up all the techniques for Jigsaw Killers yet. I might need to review the tutorials. Picking up from you marks pic, 31. 45 Rule on R789. r7c9 = r6c7+3 - eliminates 7 from r7c9 32. 45 on r123. outies total 19. 32a when r4c6=8 r4c9<>8 = {29}/[38]/{47}/{56} 32b when r4c6=9 r4c9<>9 = [19]/{37}/{46} (can't be {28} because of 12(3) cage) 32c. r4c9 can't be 8 and r5c9 can't be 1 33. 45 on c789 - innes total 19 33a. r1c7 can't be 2 34. 45 on r1234 outies r5c58 minus innies r34c9 equals 5 34a no combos with 1 allowed : out [71] - in {12} - 2 1's in nonet out [81] - in {13} - 2 1's in nonet 35. 45 on c6789 - outies r789c5 total 14 35a. no combo with 8 at r9c5 36. 45 on c9 - outies r68c8 minus innies r12c9 equals 3 - innies limited to total of 11,10,9,8,7 or 5. 36a. 11 - 8 - no 1 in r6c8 36b. 7 - outies = [19] - but innies can only be {34} - {34} and 1 blocked by 12(3) r3459 36c. 5 - outies = [17]/[26]/[35] - but [17] blocked because r1c8 needs to be 7 to fit 12(3) 36d. -> no 1 at r6c8 Rgds Richard
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Wed Jul 04, 2007 10:09 pm    Post subject:

Hi guys,

 sudokuEd wrote: 27b. Each combo in the h13(3)c7 must have 2 cells from the 15(3). No available combination in 15(3) has 2 of {148} -> {148} not possible

How about {168} - that has 2 of {148}?

Fortunately, h13(3)c6 = {148} implies 15(3)r7c5 = {168} and vice versa.
Due to observation in step 37a below, h13(3)c6 = {148} and 15(3)r7c5 = {168} together block all possible combinations for 16(3)r8c7
Therefore, we can indeed deduce that h13(3)c6 <> {148} and 15(3)r7c5 <> {168}.
The walkthrough is therefore not invalidated.

Here are some further steps...

Texas Jigsaw Killer 18 Walkthrough (continued)

37. Because 16(3)r8c7 maps to 3 of r3456c6 (LoL c789), of which r9c6 is a peer...
37a. ...r9c6 cannot duplicate any digit in 16(3)r8c7
37b. -> {367} combo for 16(3)r8c7 blocked by h13(3)c7 (see step 28)
37c. -> no 7 in 16(3)r8c7

38. 16(3)r8c8 = {268/349}
38a. -> blocks {248} combo from 14(3)r5c4
38b. -> 14(3)r5c4 = {149/158/167/347} (no 2)

39. LoL c6789: outies r56c4+r67c5 = innies r129c6+r1c7
39a. no 2 in outies -> no 2 in innies r12c6

40. LoL r89: outies r56c67+r7c78 = innies r8c1367+r9c78
40a. no 7 in innies -> no 7 in outies (r5c67+r6c6)
40b. -> 12(3)r5c6 = {246/345} = {(5/6)..}
40c. -> 4 locked for n8

41. {156} combo for 12(3)r6c7 blocked by 12(3)r5c6 (step 40b)
41a. -> 12(3)r6c7 = {129/138} (no 5,6)

42. LoL c789 (see step 19)
42a. no 7 in outies -> no 7 in innie r1c7

43. 7 no longer available for h19(3)c7 (step 18)
43a. -> no 3 in r15c7

44. LoL r789: outies r5c34+r6c2345 = innies r7c789+r8c89+r9r9
44a. no 2 in outies -> no 2 in innies r7c78
44b. -> no 8,9 in r7c9

45. LoL r789: outies r5c467+r6c4567 = innies r7c12349+r8c13
45a. no 9 in innies -> no 9 in outies
45b. -> 14(3)r5c4 = {158/167/347} (no 9)
45c. -> no 7 in r4c4 (step 17)

46. {357} combo for 15(3)r7c5 blocked by 14(3)r5c4 (step 45b)
46a. -> 15(3)r7c5 = {159/258/267} (no 3)

47. 15(3)r7c5 must contain one of {12} in r78c6
47a. -> no 1 in r9c6, otherwise h13(3)c6 cage total can't be reached (see step 26)

48. LoL c1234: outies r4589c5+r9c6 = innies r1c3+r1256c4
48a. no 9 in innies -> no 9 in outies r89c5

49. h19(3)c7 (step 18) = {289/469/568} (3,7 unavailable)
49a. -> if 16(3) = {268}, 6 must go in r9c8
49b. -> no 6 in r89c7, and no 2,8 in r9c8

50. n9(r8c2) confined to r89
50a. -> remaining cells in r89 (r8c136789+r9c789) must contain the digits 1..9 (no duplicates)
50b. -> r8c9 <> r9c8
50c. -> r89c89+r7c8 (innies, LoL c89) contain 5 different digits
50d. -> r34c67+r2c7 (outies, LoL c89) contain (same) 5 diffent digits
50e. -> no 6,8,9 in r4c7 (since these digits are already taken in 23(3)r3c6
50f. -> no 1 in r4c8
50g. no 2 in innies (step 50c)
50h. -> no 2 in outies r24c7

51. 6 in c7 now only in h19(3)c7 (step 18)
51a. -> h19(3)c7 = {469/568}
51b. -> no 2 in r5c7

52. LoL c89: outies r56c6+r4567c7 = innies r1239c8+r12c9
52a. no 8 in innies -> no 8 in outie r7c7
52b. 8 in 12(3)r6c7 now only in r7c8
52c. -> no 3 in r7c8

53. no 6 in r9c5 (due to {124} unavailable in r9c6 and 1 unavailable in r8c5)

54. no 8 in r12c6 (no permutations for 16(3)r1c6 w/ 8 in r12c6)
54a. h13(3)c7 (step 28) blocks {16} combo in r12c6
54b. -> no 6 in r12c6 (no remaining permutations available for 16(3)r1c6)

55. 8 in c6 locked in n3(r1c8)
55a. -> no 8 in r3c7

marks pic after step 55a:

 Code: .-----------------------.-----------------------.-----------.-----------------------.-----------------------. | 123456789   123456789 | 12345678    12345678  | 123456789 | 134579      4689      | 123457      123457    | :-----------.           :-----------.           |           |           .-----------'-----------.           | | 3456789   | 123456789 | 123456789 | 12345678  | 123456789 | 134579    | 13457       123457    | 123457    | |           '-----------:           '-----------:           :-----------'-----------.           :-----------: | 1345789     1345789   | 12345789    12345789  | 12345789  | 689         69        | 123457    | 12        | :-----------------------'-----------.-----------'-----------:           .-----------'-----------:           | | 123456789   123456789   12345     | 123456      12345678  | 89        | 13457       23456789  | 12345679  | :-----------------------.-----------+-----------.           :-----------'-----------.           |           | | 123456789   123456789 | 56789     | 1345678   | 12345678  | 2346        456       | 23456789  | 23456789  | |           .-----------'           |           '-----------:           .-----------+-----------'-----------: | 123456789 | 56789       56789     | 1345678     345678    | 2346      | 123       | 23456789    123456789 | :-----------'-----------.-----------'-----------.-----------'-----------:           '-----------.           | | 12345678    12345678  | 12345678    12345678  | 5789        125679    | 139         189       | 456       | |           .-----------:           .-----------+-----------.           :-----------.-----------'-----------: | 12345     | 23456789  | 12345     | 456789    | 2345678   | 125       | 23489     | 56789       56789     | :-----------'           :-----------'           |           '-----------:           '-----------.           | | 123456789   123456789 | 456789      456789    | 123457      3567      | 23489       3469      | 56789     | '-----------------------'-----------------------'-----------------------'-----------------------'-----------'

_________________
Cheers,
Mike
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Thu Jul 05, 2007 12:26 am    Post subject: I have no vision today. Can only see these 2 steps. Maybe better tomorrow evening. 56. 15(3) at R7C5 only has 6 in R7C6 -> R7C6: no 7 57. Hidden13(3) at R78C6 + R9C6 = [913]/{15}[7]/{25}[6]/[625] : {3/5...}/{3/6/7}/{1/6/9} 57a. 16(3) at R1C6: {358} blocked, R12C6 would be {35}, blocked by hidden 13(3) 57b. 16(3) at R1C6: {367} blocked, R12C6 would be {37}, R1C7 = 6 -> R3C6 = 6(hidden single), blocked by hidden 13(3) 57c. 16(3) at R1C6: {169} blocked, R12C6 = {16}, R1C7 = 9 -> R34C6 = {89} or R12C6 = {19}, R1C7 = 6 -> R3C6 = 6, either way would force {169} in C6, but blocked by hidden 13(3) 57d. Conclusion: 16(3) at R1C6 = {17}[8]/{57}[4]/{349}: no 6 Para
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Thu Jul 05, 2007 9:07 am    Post subject:

Thanks Para.

Here's the next small batch of moves (still got that sinking feeling about this puzzle though ). Hopefully, one of us will have a "Eureka!" moment:

58. 11(3)r1c3 = {128/137/146/236/245}
58a. {137} and {146} both blocked by 16(3)r1c6 (step 57d)
58b. -> 11(3)r1c3 = {128/236/245}
58c. -> no 7, 2 locked for n2

59. Nishio: if r6c7 = 3, then
59a. 3 in n7(r5c4) forced into r9c8 (cannot go in r5c4 due to LoL r789)
59b. -> 3 in n3(r1c8) forced into c9
59c. but this would leave nowhere to place the 3 in n5(r3c9)
59d. -> no 3 in r6c7
59e. -> no 6 in r7c9 (step 16)

60. 18(3)r4c7 = {189/279/369/459/378/468/567}
60a. {459} blocked by r7c9
60b. -> only combo with 4 is {468}
60c. {68} only in r45c8 -> no 4 in r45c8

61. h15(3)r7 (at r7c789 = innies r789) = [384]/{19}[5]
61a. -> {519} blocked from 15(3)r4c8
(would require 9 in both r6c8 and r7c7, leaving nowhere to place 9 in c9)
61b. -> no 1 in r6c9

62. 15(3)r4c8 = {249/258/267/348/357/456}
62a. {258} blocked because of r789 i/o difference "pincer" (r7c9 = 5 -> r6c7 = 2)
62b. only other combo with 8 is {348}
62c. but when [348}, 8 must go in r6c9, otherwise clash w/ h15(3)r7 (step 61)
62d. Conclusion: no 8 in r6c8

63. 12(3)r7c3 = {138/147/237/246/345} ({156} blocked by 22(3)r5c3)
63a. either r7c9 = 5, or r7c789 = [384] -> {345} combo blocked for 12(3)r7c3
63b. Either way, no 5 in r7c34

New marks pic after step 63b:

 Code: .-----------------------.-----------------------.-----------.-----------------------.-----------------------. | 123456789   123456789 | 1234568     1234568   | 13456789  | 134579      489       | 123457      123457    | :-----------.           :-----------.           |           |           .-----------'-----------.           | | 3456789   | 123456789 | 123456789 | 1234568   | 13456789  | 134579    | 13457       123457    | 123457    | |           '-----------:           '-----------:           :-----------'-----------.           :-----------: | 1345789     1345789   | 12345789    12345789  | 1345789   | 689         69        | 123457    | 12        | :-----------------------'-----------.-----------'-----------:           .-----------'-----------:           | | 123456789   123456789   12345     | 123456      12345678  | 89        | 13457       2356789   | 12345679  | :-----------------------.-----------+-----------.           :-----------'-----------.           |           | | 123456789   123456789 | 56789     | 1345678   | 12345678  | 2346        456       | 2356789   | 23456789  | |           .-----------'           |           '-----------:           .-----------+-----------'-----------: | 123456789 | 56789       56789     | 1345678     345678    | 2346      | 12        | 2345679     23456789  | :-----------'-----------.-----------'-----------.-----------'-----------:           '-----------.           | | 12345678    12345678  | 1234678     1234678   | 5789        12569     | 139         189       | 45        | |           .-----------:           .-----------+-----------.           :-----------.-----------'-----------: | 12345     | 23456789  | 12345     | 456789    | 2345678   | 125       | 23489     | 56789       56789     | :-----------'           :-----------'           |           '-----------:           '-----------.           | | 1234578     123456789 | 456789      456789    | 123457      3567      | 23489       3469      | 56789     | '-----------------------'-----------------------'-----------------------'-----------------------'-----------'

_________________
Cheers,
Mike
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Thu Jul 05, 2007 12:06 pm    Post subject:

TJK18 - Never-Ending Story

Found some more, although it doesn't make the puzzle any less impossible!

I can't even see how it can be solved using hypotheticals...

The reason for this pessimism is that it's possible to fill up over two-thirds of the grid with a near solution (consisting of almost all the wrong digits), including every cell outside of r12345c12345. Therefore, any contadiction of this "near-solution" must involve pulling in at least one cell from this region. Unfortunately, there's very little we have to go on there. The 2 locked in 11(3)r1c3 doesn't help (both solution and near-solution have 2 in the outies if doing LoL on c1234 or c12345), and the innie/outie difference on c12 (step 8) is too weak to have a significant short range effect in the absence of a massive hypothetical.

64. 21(3)r8c8 cannot contain both of {89}
64a. -> r456c9 must contain at least one of {89}
64b. -> 18(3)r4c7 cannot contain both of {89}
64c. -> no 1 in r4c7

65. 1 in n5(r3c9) now locked in 12(3)r3c9 -> not elsewhere in c9
65a. 12(3)r3c9 = {129/138/147/156}
65b. 8 only in r5c9 -> no 3 in r5c9

For reference, here's the near-solution I was talking about above:

 Code: .-------------.-------------.------.-------------.-------------. | 2689   29   | 16     128  | 1689 | 5      4    | 3      7    | :------.      :------.      |      |      .------'------.      | | 4689 | 349  | 36   | 8    | 3689 | 7    | 5      1    | 2    | |      '------:      '------:      :------'------.      :------: | 28     235  | 37     258  | 38   | 6      9    | 4    | 1    | :-------------'------.------'------:      .------'------:      | | 469    459    1    | 15     16   | 8    | 7      2    | 3    | :-------------.------+------.      :------'------.      |      | | 2      23   | 5    | 7    | 13   | 4      6    | 9    | 8    | |      .------'      |      '------:      .------+------'------: | 7    | 8      9    | 3      4    | 2    | 1    | 5      6    | :------'------.------'------.------'------:      '------.      | | 1      7    | 2      6    | 5      9    | 3      8    | 4    | |      .------:      .------+------.      :------.------'------: | 3    | 6    | 4    | 9    | 2    | 1    | 8    | 7      5    | :------'      :------'      |      '------:      '------.      | | 5      1    | 8      4    | 7      3    | 2      6    | 9    | '-------------'-------------'-------------'-------------'------'

Hopefully, someone can prove me wrong. Otherwise I'll be considering having "TJK18" engraved on my tombstone!

R.I.P.
_________________
Cheers,
Mike

Last edited by mhparker on Thu Jul 05, 2007 3:10 pm; edited 1 time in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Thu Jul 05, 2007 2:38 pm    Post subject: Hi Just a little bit more. Only 2 digits eliminated in three steps. 66. 18(3) at R1C5 = {189/369/459/567}: {378/468} blocked by 16(3) at R1C6 67. Hidden13(3) at R78C6 + R9C6 = [913]/{15}[7]/[256]/[625]: [526] blocked by h15(3) at R7C789(needs one of {58}): R78C6 = [52] -> R7C5 = 8 68. LOL R89: R8C6789 + R9C789 = R5C3 + R6C23 + R7C1234: all different digits outies are in same house 68a. 21(3) at R8C8 = {579} -> 16(3) at R8C7 = {268} - > R7C7 = 1 -> R7C56 = {59} 68b. 21(3) at R8C8 = {678} -> h 15(3) in R7C789 = {19}[5] 68c. Conclusion: 5 in R7 locked for R7C569: R7C12: no 5 Para
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

 Posted: Thu Jul 05, 2007 3:18 pm    Post subject: Another tiny increment: 65b. sub-step added (elimination of 3 in r5c9) - see corrected walkthrough above ... 69. r7c123456 = h30(6)r7 = {125679/234678} 69a. {59} only in r7c56 69b. -> no 1 in r7c6_________________Cheers, Mike
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Thu Jul 05, 2007 4:21 pm    Post subject: With this pace i think we are gonna go on till step 200. But here's a bit more again. 70. 12(3) at R8C5 = {138/147/156/237/246/345} 70a. {345} blocked, this is how: 70aa. R9C6 = 3 -> R89C5 = {45} -> R7C5 = 5: 2 5's in C5 70ab. R9C6 = 5 -> R89C5 = {34} -> R7C5 = 7 -> 18(3) at R1C5 = {189} -> 16(3) at R1C6 = {457} -> R12C6 = {57}: 2 5's in C6 70b. 12(3) at R8C5 = {138/147/156/237/246}: only place for 1 is R9C5 -> R9C5: no 5 70c. 12(3) at R8C5 = {138/147/156/237/246}: {1/2...} 71. 12(3) at R8C2: {129} blocked by 12(3) at R8C5 -> no 9 in 12(3) at R8C2 71a. 9 in N9 locked in 21(3) at R8C4 = {489/579}: no 6; {4/5...}/{4/7...} 71b. 12(3) at R8C2 = {138/156/237/246}: {147/345} blocked by 21(3) at R8C4 71c. 12(3) at R8C5 = {138/156/237/246}: {147} blocked by 21(3) at R8C4 Para
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Thu Jul 05, 2007 6:44 pm    Post subject: I share Mike's pessimism. This one is a stinker. 72. 45 rule on c9 - innies total 33 72a. 2 in r6c9 means r7c9=4 - r1289c9=27 = {37}{89} or {57}{69} 72b. {37}{69} - blocked because can't make cage sum in 21(3)r8c8 72c. {57}{69} - blocked because can't make cage sum in 12(3)r1c8 73. 15(3)r6c8 - only combo with 9 is {249} - no 9 at r6c8 74. 45 rule on r1234 r5c58 = r34c9 + 5 74a. Innies total 10,8,7,6,5,4,3 - outies total 15,13,12,11,10,9,8 74b. 10 - 15 - requires 2 9's in nonet r3c9 74c. 8 - 13 - can't have 13=[67] in outies 74d. 7 - 12 - can't have {66} in outies 74e. 6 - 11 - can't have 6 in outies because need 6 in r5c9 74f. 5 - 10 - no 6 at r5c5 (missing 4 at r5c8) 74g. 4 - 9 - outies can't be [63] as it makes two 3's in nonet r3c9 74h. 3 - 8 - outies can't be [62] as it makes two 2's in nonet r39 74i. (phew) no 6 at r5c5 Lots of work for not very much
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Thu Jul 05, 2007 7:27 pm    Post subject: Just a little bit.... 75. 15(3) in R6C8 = {249/348/357/456}(needs 4 or 5 in R7C9): {4/7...} 75a. 12(3) in R3C9 = {129/138/156}({3/6/9}: {147} blocked by 15(3) -->> no 4,7 75b. 18(3) in R4C7 = {279/378/468/567}: {369} blocked by 12(3) 76. LOL R789: R5C467 + R6C4567 = R7C12349 + R8C13: All cells in innies in same house except R7C9, which contains only {45} so only {45} can appear twice in the outies. 76a. 12(3) at R5C6 = {246/345} = {3/6..} 76b. 14(3) at R5C4 = {158/167/347}: {356} blocked by LOL R789 + 12(3) at R5C6 77. R5C4: no 5, this is how. 77a. R5C4 = 5 -> R6C45 = {18} -> R6C7 = 2 -> 12(3) at R5C6 = {345} except no room left for 5 in 12(3) Para
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Thu Jul 05, 2007 9:53 pm    Post subject:

TJK18: The next action-packed episode

More candidates than usual gone this time. Fun, isn't it? .

78. 18(3)r2c1 = {189/369/378/459/468}
78a. {459} blocked. Here's how:
78b. LoL r123: r2c1+r3c129 = r4c456+r5c5
78c. outies include 11(3) cage
78d. -> 3 of the digits in innies must add up to 11
78e. -> if r3c9 = 1, 2 of 18(3)r2c1 must sum to 10 and...
78f. ...if r3c9 = 2, 2 of 18(3) must sum to 9
78g. now, if 18(3) = {459}, then 14(3)r4c1 = {167} (only non-conflicting combo)
78h. -> r3c9 = 1 (hidden single c9)
78i. but this contradicts assertion in step 78e
78j. -> 18(3)r2c1 <> {459}
78k. -> 18(3)r2c1 = {189/369/378/468} (no 5) = {(3/8)..}
78l. -> {238} combo blocked from 13(3)r5c1

79. LoL r123 (see step 78b): no 5 in innies -> no 5 in outies
79a. -> no 5 in 11(3)r4c4, no 7 in r6c5 (step 17)
79b. -> 11(3)r4c4 = {128/137/146/236} = {(1/6)..}
79c. -> {169} combo blocked for 16(3)r2c3

80. no 4 in r2c1 (requires {68} in r3c12 -> conflict w/ r3c67)

81. r3c679 (innies r123) = h16(3)r3 = {169/268} = {(1/8)..}
81a. -> r3c12 <> {18}
81b. -> no 9 in r2c1

82. LoL r12: r1c89+r2c1789 = r3c345+r4c45+r5c5
82a. no 9 in innies -> no 9 in outies r3c345

83. 9 in c4 locked in n9
83a. -> no 9 in r9c3

84. 18(3)r1c5 = {189/369/459/567} (step 66)
84a. {69} only in r12c5
84b. -> no 3 in r12c5

85. no 2 in r2c3 (would require one of {69} in r3c34 - unavailable)

86. CPE: r7c4 sees all 7's in n7(r5c4)
86a. -> no 7 in r7c4

87. no 1 in r7c3 (would require 8 in r7c4 -> clash w/ r7c789, which needs 1 of {18} (step 61))

88. 14(3)r5c4 = {158/167/347} (step 45b)
88a. {17} only in r56c4
88b. -> no 6 in r56c4

89. {345} combo blocked from r456c4 (innies c1234) = h(12)c4. Here's how:
89a. 5 only within 14(3)r5c4 -> 14(3) = {158}
89b. -> can't get 2 of {345} in r56c4
89c. -> h(12)c4 = {138/147/156/237} ({246} blocked because {26} only in r4c4)
89d. -> h(12)c4 = {(1/2)..}
89e. -> r12c4 <> {12}
89f. -> no 8 in r1c3

New marks pic after step 89f:

 Code: .-----------------------.-----------------------.-----------.-----------------------.-----------------------. | 123456789   123456789 | 123456      1234568   | 1456789   | 134579      489       | 123457      23457     | :-----------.           :-----------.           |           |           .-----------'-----------.           | | 3678      | 123456789 | 13456789  | 1234568   | 1456789   | 134579    | 13457       123457    | 23457     | |           '-----------:           '-----------:           :-----------'-----------.           :-----------: | 134789      134789    | 1234578     1234578   | 134578    | 689         69        | 123457    | 12        | :-----------------------'-----------.-----------'-----------:           .-----------'-----------:           | | 123456789   123456789   12345     | 12346       1234678   | 89        | 3457        2356789   | 123569    | :-----------------------.-----------+-----------.           :-----------'-----------.           |           | | 123456789   123456789 | 56789     | 13478     | 123478    | 2346        456       | 2356789   | 25689     | |           .-----------'           |           '-----------:           .-----------+-----------'-----------: | 123456789 | 56789       56789     | 134578      34568     | 2346      | 12        | 234567      3456789   | :-----------'-----------.-----------'-----------.-----------'-----------:           '-----------.           | | 1234678     1234678   | 234678      123468    | 5789        2569      | 139         189       | 45        | |           .-----------:           .-----------+-----------.           :-----------.-----------'-----------: | 12345     | 2345678   | 12345     | 45789     | 2345678   | 125       | 23489     | 56789       56789     | :-----------'           :-----------'           |           '-----------:           '-----------.           | | 12345678    12345678  | 4578        45789     | 12347       3567      | 23489       3469      | 56789     | '-----------------------'-----------------------'-----------------------'-----------------------'-----------'

_________________
Cheers,
Mike
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Fri Jul 06, 2007 5:37 am    Post subject: Morning all, here's last nights moves from when internet went down here. That's 4 digits in 6 moves. 90. 18(3) at R2C1 = {189/369/378/468} = {1/6/7..} 90a. 14(3) at R4C1: {167} blocked by 18(3) 91. 14(3) at R4C1: {356} blocked; Here's how: 91a. 14(3) = {356}: R4C45 and R4C78 need {47} but can't contain both 91aa. 11(3) at R4C7 = {137}({146 blocked cause no 6 available) -> R5C5 = 3; R4C4 = 7; R4C3 = 1-> R6C5 = 3(step 17): 2 3's in C5 91b. 14(3) = {149/158/239/248/257/347}: no 6 92. 14(3) at R4C1: {347} blocked; here's how 92a. 14(3) = {347} -->> R2C7 = 7(hidden single) -->> R23C8 = {12} -->> naked pair {12} in R3C89(nowhere else in R3), which leaves no options for 18(3) at R2C1 92b. 14(3) = {149/158/239/248/257}: {1/2..} 93. R5C5: no 8 93a. R5C5 = 8 -> R4C45 = {12}: clashes with 14(3) cage at R4C1 94. outies R1234: R5C589 = 17: R8C59 = 15/13/12/11/10/9/8, so R8C8: no 3 95. 14(3) at R4C1: {158} blocked; this is how a.14(3) = {158} -> R4C6 = 9 -> R3C9 = 2("45" R123): no place for 1 in C9 b. 14(3) = {149/239/248/257} = {2/4..} c. 13(3) at R5C1: {247} blocked by 14(3) So at a pace of eliminating an average of 1 digit per move how many moves do we have to make before we are done? Para
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Fri Jul 06, 2007 7:05 am    Post subject:

First: a very big thankyou to Mike for getting around my mistake way back forever ago. A really neat move it was too. Para noticed the same mistake. Thanks to you both.

96. no 6 r6c5. Here's how.
96a. "45" c1234: 3 outies r456c5 = 13 = h13(3)r4c5
96b. = {148}/{17}[5]/{238}/[625]/[634]/[643] ({{27}[4]/{34}[6} blocked by combo's in 11(3))
96c. -> no 6 r6c5
96d.-> no 4 r4c4 (I/O c1234)

97. 14(3)r5c4 = {158/347}

98. {258} combo. blocked from 15(3)r7c5. Forces 3 into both 14(3) and 16(3) in this nonet.
98a. 15(3)r7c5 = {159/267}(no 8)
98b. -> no 6 r9c6 (I/O c6789)
98c. 15(3)r7c5:6 only in r7c6 -> no 2 r7c6

99. 12(3)r8c5 = {138/156/237}(no 4)({246} blocked by r9c6)
99a. 1 only in r9c5 -> no 5 r8c5

100. from step 67. Hidden13(3) at R78C6 + R9C6 = [913]/[517]/[625]
100a. -> r7c5 + r9c6 =
i.[5][3]
ii.[9][7]
iii.[7][5]
100b. -> r789c5 =
i.[581]/[5]{27}
ii.[9]{23}
iii.[761]
100c. = [2/5/6..]

101. from step 96b. h(13)r456c5 = {148}/{17}[5]/{238}/[634]/[643] ([625] blocked by r789c5)(step 100c.)

102. {17}[5] blocked from r456c5. Here's how.
102a. r456c5 = {17}[5] forces 9 into both r789c5 (step 100b.) and 18(3)r1c5
102a. h(13)r456c5 = {148}/{238}/[634]/[643] (no 5 or 7 r456c5)
102b. -> no 3 r4c4 (I/O c1234)

103. 14(3)r5c4 = {158/347}.
103a. {15} only in r56c4 -> no 8 r56c4

104. 12(3)r8c5 = {138/156/237} = [1/3..]
104a. -> 12(3)r8c2 = {156/237/246}(no 8) ({138} blocked by other 12(3))

105. 11(3)r4c4 = {128/146/236} = [6/8..]
105a. -> {268} blocked from 16(3)r2c3
105b. -> {468} blocked from 18(3)r1c1

106.from step 89c. h(12)r456c4 = [1]{47}/[615]/[2]{37}
106a. no 1 r6c4

107. weak links on 1s in r7 & 12(3)r6c7
107a. -> no 1 r8c1 (1 in r6c7 -> 1 in r7 in n6r4c3 -> no 1 in 8c1: 1 in r7c1 -> no 1 in r8c1))

108. 11(3)r7c1 {128} blocked: r7c12 = {18} -> 12(3)r6c7 = {138}:but 2 8s r7
108a. = {137/146/236/245}(no 8) = [3/4..]

109. 12(3)r7c3 = {138/147/237/246}(no 5)({345} clashes with 11(3) step 108a)
i. {138}
ii. [714] only. 7 in r7c3 -> 15(3) r7c5 = {159} with 1 in r8c6 -> no 1 in r8c3
iii. [7]{23}
iv. {26}[4]. 6 in r7 -> 15(3)r7c5 = {59}[1] -> r7c9 = 4 -> no 4 possible in r7c34
109a. -> no 4 r7c34

 Code: .-------------------------------.-------------------------------.-------------------------------. | 123456789 123456789 123456    | 1234568   1456789   134579    | 489       123457    23457     | | 3678      123456789 13456789  | 1234568   1456789   134579    | 13457     123457    23457     | | 134789    134789    1234578   | 1234578   134578    689       | 69        123457    12        | :-------------------------------+-------------------------------+-------------------------------: | 12345789  12345789  12345     | 126       123468    89        | 3457      2356789   123569    | | 123456789 123456789 56789     | 1347      1234      2346      | 456       256789    25689     | | 123456789 56789     56789     | 3457      348       2346      | 12        234567    3456789   | :-------------------------------+-------------------------------+-------------------------------: | 123467    123467    23678     | 12368     579       569       | 139       189       45        | | 2345      234567    1234      | 45789     23678     125       | 23489     56789     56789     | | 1234567   1234567   4578      | 45789     1237      357       | 23489     3469      56789     | '-------------------------------.-------------------------------.-------------------------------'
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

 Posted: Fri Jul 06, 2007 10:20 am    Post subject: Light at the end of the tunnel Thanks Ed, that made a big difference. A couple of overlooked moves first... 110. no 5 in r8c6 (from permutations for h13(3) listed in step 100) 111. 2 in r7 locked in n6(r5c3) 111a. -> no 2 in r8c13 Now to get down to business... 112. LoL r123 (see step 78b): no 7 in outies -> no 7 in innies r2c1+r3c12 112a. -> 18(3)r2c1 = {189/369/468} = {(4/9)..} 112b. 6 only in r2c1 -> no 3 in r2c1 113. 14(3)r4c1 = {149/239/248/257} (step 95b) 113a. {149} blocked by 18(3)r2c1 (step 112a) 113b. -> 14(3) = {239/248/257} (no 1) 113c. 2 locked for r4 and n4 114. h12(3)r456c4 (step 106) = {147/156} (no 3) 114a. 1 locked for c4 115. CPE: r5c5 sees all 1's in c4 115a. -> no 1 in r5c5 116. LoL r1234: r2c1+r3c12+r4c123 = r5c589+r6c89+r7c9 116a. no 1 in outies -> no 1 in innies r3c12 117. 18(3)r2c1 = {369/468} 117a. 6 locked, only in r2c1 117b. -> r2c1 = 6 117 moves to make a placement! (never would have thought first to go would be a cell in the top-left of the grid). Time for a handover (marks pic to follow)._________________Cheers, Mike
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