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 Assassin 70 Goto page Previous  1, 2
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Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Tue Oct 02, 2007 8:27 pm    Post subject: After taking a short break, I solved A70 V3 today. I say it was nearly as difficult as V2 though the moves were less complicated. I guess rating would be 1.25. Walkthrough for Assassin 70 V3: 1. C789 a) 17(2) = {89} locked for C8 + N3 b) 21(3) @ N3 must have 8 or 9, only possible @ R4C7 -> R4C7 = (89) c) 21(3) @ N3 = 7{59/68} <> 4 because of b), 7 locked for R3 + N3 d) 8(3) = 1{25/34} -> 1 locked for C8 e) Innies+Outies C9: 5 = R9C8 - R5C9 -> R9C8 = (67), R5C9 = (12) f) 21(3) @ N9 <> 4 because {489} is impossible since R9C8 = (67) g) 3,4 locked in 14(4) = 34{16/25} <> 7,8 h) 21(3) @ N9: R89C9 <> 6 since R89C9 would be {68} which is blocked by killer pair (68) of 15(2) 2. C789 a) 21(3) = 7{59/68} -> 7 locked for N9 b) 15(2): R6C9 <> 8; 13(2): R6C7 <> 6 c) 8 locked in R456C7 for C7 d) 13(2): R6C7 <> 5 e) Innies+Outies C89: 18 = R345C7 - R4C8 -> R4C8 <> 7 -> R4C8 = (23456) -> R345C7 = 20/21/22/23/24 f) Innies+Outies C89: R5C7 <> 1,2,3 since R345C7 would be smaller than 20 3. R5 a) 22(3) = 9{58/67} -> 9 locked for R5 + N5 b) 11(2) @ N5: R7C4 <> 2 c) 11(3) <> 8 because {128} blocked by R5C9 = (12) d) 12(3) <> {345} because of Killer pair (34) of 11(3) e) Killer pair (12) of 11(3) + R5C9 = (12) -> R5C8 <> 2 f) 12(3) <> 5 -> {156} blocked by Killer pair (56) of 22(3) 4. R89 a) Innies = 10(3) <> 8,9; R8C2 <> 1 since R8C5 + R8C8 <= 8 (3+5) 5. C56 a) Innies C6 = 27(4) <> 1,2 b) 9(2) <> 3,6 since {36} is blocked by Killer pair (36) of 9(3) c) 6(3) = {123} locked for C5 6. C12 a) Innies+Outies C1: -2 = R9C2 - R5C1 -> R5C1 <> 1,2; R9C2 <> 6,7,8,9 b) 18(3) <> 1 because R8C2 has no 1,8,9 7. N3 a) Innies+Outies: -1 = R1C7 + R4C9 - R3C78; R3C78 = 12/13 (see step 1c) -> R1C7 + R4C9 = 11/12 -> R1C6 <> 3,4 since R1C6 + R4C9 would be smaller than 11 8. C789 ! a) Innies = 14(4+1): 1,2,3 locked in R1289C7 = 123{4/5/6} = 10/11/12 -> R4C8 = (234) 9. C789 a) 14(4): R4C9 <> 1,2 since 14(3) has 1 xor 2, so if R4C9 = (12) R12C7 @ 13(3) must have 1 and 2 which is impossible (13(3) = 1 + 2 + ?) -> 13(3) must have 1 xor 2 b) 11(3): R8C6 <> 8 because R89C7 would be {12} so that R12C7 <> 1,2 c) ! 13(2) <> 6,7 because if 13(2) = [76] -> 15(2) = {69} -> no possible combination for 21(3) @ N9 (no 6,9) d) Killer pair (89) in 13(2) + R4C7 = (89) locked for C7 10. N5 a) 8 locked in 22(3) -> 22(3) = {589} locked for R5 + N5 b) 11(2): R6C4 <> 3,6 c) 9(2): R6C6 <> 1,4 11. N6 ! a) 8(3): R6C8 <> 5 because R78C8 would be {12} -> blocked by Killer pair (12) of 11(3) @ N9 b) Hidden Single: R4C9 = 5 12. C789 a) 21(3) @ N9 = {678} locked for N9 b) R9C8 = 6, R7C9 = 9 -> R6C9 = 6 c) 14(4) = {2345} -> 2,3,4 locked for C9 + N3 d) R5C9 = 1 e) 13(3) = {157} -> R1C6 = 7; 1,5 locked for C7 + N3 f) R3C8 = 7, R3C7 = 6, R4C7 = 8, g) Hidden Single: R6C7 = 9 @ N6 -> R7C7 = 4, R5C7 = 7, R5C8 = 4 h) 11(3) = {236} -> R8C6 = 6; 2,3 locked for N9 i) 8(3) = {125} -> R6C8 = 2 -> R4C8 = 3 13. C456 a) 9(3) = 3{15/24} -> Killer pair (45) blocks {45} of 9(2) b) R6C6 = 1, R7C6 = 8, R6C5 = 3 c) 11(2) = [47] -> R6C4 = 4, R7C4 = 7 d) R4C6 = 2 -> R4C4 = 6 -> R4C5 = 7 -> R4C2 = 4 e) 6(3) = {123} -> 1,2 locked for N8 f) 16(3) = {349} locked for R9 + N8 g) R8C4 = 5, R9C7 = 2, R8C7 = 3, R8C8 = 1, R7C8 = 5, R8C5 = 2, R7C5 = 1 h) 17(3) @ N8 = {458} -> R8C3 = 4, R9C3 = 8 14. Rest is clean-up and singles I still hope to see more walkthroughs for V2.Last edited by Afmob on Tue Oct 09, 2007 6:35 am; edited 2 times in total
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Tue Oct 02, 2007 11:57 pm    Post subject: I'm just checking that I'm now using tiny text the key move to solve this assassin is to show that r4c8=2/3/4 Seem OK hopefully no more glitches Gary
sudokuEd
Grandmaster

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

Posted: Wed Oct 03, 2007 5:19 am    Post subject:

OK - this is as far as I could get on the V2. Afmob's step 4e doesn't look correct, so maybe we can put our heads together to try and work this one out.

Don't like tt for tag, so hope no-one minds staying big.

I wouldn't normally go this many steps - but its already typed, so.... BTW - some of the combo clashes I could never have found without Sudoku Solver's help in showing all the mutual exclusions.

Assassin 70V2
i. 3(2)n4 = {12}
ii. 5(2)n3 & n5 = {14/23}
iii. 14(2)n8 = {59/68}
iv. 7(2)n1: no 789
v. 9(2)n4 & n6: no 9
vi. 11(2)n6: no 1
vii.9(3) n7: no 789
viii. 21(3)n8: no 123
ix. 10(3)n1: no 89
x. 20(3)n2, n4 & n6: no 12
xi. 11(3)n8: no 9
xii. 19(3)n1 & n3 &n7:no 1
xiii.14(4)n6: no 9
xiv. 26(4)n4:no 1

1. 3(2)n4 = {12}: both locked for c3

2. min r34c3 in 10(3) = {35} = 8 -> max r3c2 = 2 ({34} blocked by 3 required in r3c2)
2a. 10(3)n1 can only have 1 of 1/2 = {136/145/235}(no 7)
2b. r3c2 = {12}
2c. r34c3 = {36/45/35}

3. "45" n147: r18c4 = h10(2)
3a. = {28/37/46}(no 5,9)

4. "45" n369: r18c6 = h3(2) = {12}: both locked for r6
4a. 15(3)r1c6 must have 1/2 = {159/168/249/258/267}(no 3) = [1/2] but not both
4b. ->no 1,2 in r12c7

5. 14(2)n5 = {59/68} = [5/6..]

6. 14(3)n2 = {347} ({356} clashes with [5/6..] in 14(2)n5)
6a. 3,4,7 all locked for c6
6b. 17(3)n2 can have at most 1 of 3,4,7 because of r23c6 -> {467} blocked

7. "45" c6: r59c6 = h14(2) = {59/68}

8. "45" c4: r59c4 = h10(2)
8a. no 5
8b. min r9c4 = 4 -> max r5c4 = 6

9. 5 in c4 only in 20(3) = 5{69/78}(no 3,4)

10. "45" r89: r8c258 = h15(3)
10a. min r8c28 = {34} = 7 -> max r8c5 = 8

11. "45" c1: r9c2 + 4 = r5c1
11a. max r9c2 = 5
11a. min r5c1 = 5

12. 9(3)n7 = {135/234}(no 6) = [1/2..] ({126} blocked by r7c3)
12a. = 3{..}: 3 locked for n7
12aa. add clean-up -> no 6 r6c1 (from Afmob's next post: thanks)
12b. no 1 in r9c2 since this would force 5 into both r5c1 (step 11) and r89c1 ({135} combo in 9(3))
12c. no 5 r5c1 (step 11)
12d. Killer pair [1/2] in 9(3) and r7c3: both locked for n7
12e. r6c1, no 78

13. 1 in c2 only in n1: 1 locked for n1

14. "45" c9: r5c9 = r9c8
14a. no 9 in r5c9 -> no 9 in r9c8
14b. no 1 in r123c7 -> no 1 in r5c9 or r9c8 (nishio?)

15. "45" c12: r5c3 - 4 = r3c2.
15a. r3c2 = {12} -> r5c3 = {56}
15b. -> {145} combo blocked from 10(3)n1 since it would clash with r5c3
15c. 10(3) = {136/235}(no 4)
15d. r345c3 = {356} naked triple: all locked for c3

16. 19(3)n1 & n7: must have 2 of {4789}
16a. = {289/379/469/478}
16b. h10(2)r18c4 must have 1 of {4789}
16c. -> when 20(3)n2 = {569} -> 5(2) n5 must have the remaining {4789} for c34 = {14}
16d. -> when 20(3)n2 = {578} -> all {4789} taken for c34 -> 5(2)n5 = {23}. These might be important later.

17. "45" c89: r5c7 + 6 = r3c8
17a. max. r5c7 = 3
17b. min r3c8 = 7
[I think it's here I should have seen CPE on 9's in c7 -> no 9 in r3c8. I get rid of it sometime anyway - but you can save a bit of head room by doing it now. I should have bookmarked each step on Sudoku Solver to check this. Learned now!]

18. "45" c89: r345c7 = h13(3)
18a. = {39/48}[1]/{47/56}[2]/{28/46}[3] ({57}[1] blocked by 7 in r3c8 step 17; {38}[2] blocked by 8 in r3c8; {19}[3] blocked by 9 in r3c8))
18b. = {139/148/238/247/256/346}

19. "45" n3: 3 outies = 13
19a. r1c6 = {12} -> r4c79 = 12/11
19b. no 1 r4c9

20. "45" n36: r1c6 + 18 = r6c789
20a. r6c789 = 19/20
20. -> no 1 r6c7; no 8 r7c7

21. 1 in n6 only in 14(4)
21a. {2345} combo blocked

22. "45" r1234: r4c258 = h12(3)

23. 21(3)n8 = {489/579/678} & r9c3 = {4789}
23a. -> 7 locked in these 4 cells for r9
23b. Common Peer Elimination (CPE): r8c4 'sees' all 7's in r9 -> no 7 r8c4 [SS found this one for me - annoyed that I missed it and annoyed that I forgot to turn off "Placement feedback".]
23c. no 3 r1c4
23d. {379} blocked from 19(3)n1

24. Since 21(3)n8 = [4/7,5/8,6/9,7/8,..]
24a. -> 19(3)n7, r8c3 + r8c4 + r9c3 = [469/478/874] all clash with 21(3)
24b. r8c3 + r8c4 + r9c3 = [829/928/739/937/964/748/784]
24c. no 4 r8c3
24d. 9 locked in r9c3456 for r9
24e. 15(3)n9; 7/9 in {249/267} must be in r8c9 & for {258},[2]{58} would clash with 21(3)n8)
24f. -> no 2 in r8c9

25. no 1 in r6c5 because of 1's in c4. Here's how.
25a. 1 in r7c4 -> 1 in 3(2)n4 in r6c3 -> no 1 r6c5
25b. 1 in r56c4 -> no 1 r6c5

26. "45"c789: r1289c7 = h23(4)
26a. r12c7 + r89c7 = {59-18/59-36/68-45/49-28/49-37/58-37/58-46/67-28} ({59-27/68-27} blocked by combo's in 11(3)n8)
26b. h23(4) = {1589/2489/2678/3479/3569/3578/4568} = [2/3/8..]

27. h13(3)r345c7 = {139/148/247/256/346} ({238} blocked by h23(4) step 26b)
27a. from 18a. = h13(3) {39/48}[1]/{47/56}[2]/{46}[3]
27b. no 2 r34c7
27c. no {289} combo in 19(3)n3
27d. no 2 in r5c9 or r9c8 (step 14b)
27e. h13(3) = [3/5/7/8..]

28. 19(4)n3 must have 1/2 for n3
28a. {3457} blocked

29. from step 26b. h23(4) = {1589/2489/2678/3479/3569/4568} ({3578} blocked by h13(3) step 27e)
29a. = r12c7 + r89c7 = {59-18/59-36/68-45/49-28/49-37/58-46/ }

30. when h13(3)r345c7 = {346} = [3]{46} the only combo in h23(4)r1289c7 = {1589} = {59-18}. But this means 2 9s in n3 (i/oc89)
30a. h13(3) = {139/148/247/256} = [2/4/9..]
30b. = h13(3) = {39/48}[1]/{47/56}[2]
30c. r5c7 = {12}
30d. no 9 r3c8
30e. 19(3)n3 = {379/478/568} = [6/7..]

31. 15(3)n2 = {159/168/249/258}(no 7) ({267} blocked by 19(3)n3 step 30e)

32.from step 29 h23(4) = {1589/3479/3569/4568}(no 2) ({2489} blocked by h13(3) step 30a; {2678} blocked by no 7 r12c7 step 29a)
32a. = r12c7 + r89c7 = {59-18/59-36/68-45/49-37/58-46}
32b. no 2 r89c7
32c. h23(4) = [4/5..] -> {45} blocked from 9(2)n6

33. r45c8 must have [1/2] for c8
33a. Killer pair [12] with r5c7: both locked for n6
33b. no 7 r7c7
33c. no 9 r7c9

34. 9 in c8 in 20(3)n6
34a. = 9{38/47/56}

35. 15(3)n9 = {159/168/249/258/267/357/456} ({348} clashes with r89c7 step 32a)

36. "45" n3: 3 outies = 13 -> r4c79 = 11/12
36a. min r4c79 = 3 -> r4c79 = 3/4/5..]

37. h12(3)r4c258 = {129/138/147/156/237/246} ({345} blocked by r4c79 step 36a)

38. "45" n2: r4c46 - 6 = r1c46

39.19(4)n3 {1279} must have 1 & 2 in n3 which clashes with 5(2)
39a. {1378} must have 3 in r4c9 so that it won't clash with 5(2): but {78} clashes with r3c8
39b. {2458} must have 4 in r4c9 so that it won't clash with 5(2): but {258}+{14} in 5(2) clashes with [4/5/8/] needed in r12c7 (step 32a)
39c. = {1369/1459/1468/1567/2359/2368/2467} = [5/6;4/5/6..]

40. 11(2) = {29/38/47} ({56} blocked by 19(4))

41. OK - should have seen this ages ago! Since r5c9 = r9c8: if r9c8 = 5/6 -> r5c9 = 5/6 -> r5c8 <>5/6 as this would clash with r5c3
41a. r5c8: no 5,6
41b. no 7 in r5c9 since no 7 in r9c8

42. 15(3)n9: {456} is the only combo with 4 in r89c9: but {456} in r5c9 + r89c9 (remembering that r5c9 = r9c8) clashes with 19(4)n3
42a. no {456} combo
42b. 15(3) = {159/168/249/258/267/357}
42c. no 4 r89c9

43. Better try the same thing for c1
43a. rmembering r9c2 + 4 = r5c1
43b. r589c1 = [6]{34}/[7]{15/24}/[8]{23}/[9]{13}

44. 23(4)n1 = {1589/1679/2489/2579/2678/3569} ({3479/3578} blocked by r589c1)

45. "45" n1: r4c1 + 15 = 4 innies in n1
45a. r12c3 = 11, 12, 13, 15, 17 (cage sum - r1c4)
45b. r3c23 = 4, 5 or 7 (cage sum - r4c3)
45c. cannot sum to 23 -> no 8 r4c1

46. 26(4)n4 must have 5/6 = {3689/4589/4679/5678}(no 2)

47. From step 36. "45" n3: 3 outies = 13 -> r4c79 = 11/12
& 36a. min r4c79 = 3 -> r4c79 = 3/4/5..]
47a. -> when h12(3)r4c258 + r4c46 = {147}[53/63] leaves no combinations with 3/4/5 in r4c79 that sum to 11/12 ({38/47/56/39/48/57})
47b. -> h12(3)r4c258 = {129/138/156/237/246}
47c. 9 in {129} must be in r4c2 -> no 9 r4c5

48. 19(3)n1 = {289/469/478}
48a. -> r12c3 = {89/78/49/48/47} = [4/8..]
48b. -> combo's with 4 & 8 in 23(4)n1 must have 4 in r4c1 -> r12c3 = [7/9..] not 2 of -> the 23(4) must have 1 of [7/9] for n1
48c. -> {4568} combo blocked from 23(4)n1. Whew.

49. r89c3 = {89/79/78/49/47} = [7/9..]
49a. -> for {479} combo in 20(3), must have 7/9 in r6c2
49b. no other combo with 4 -> no 4 in r6c2

50. "45" c2: r3459c2 = h18(4) = {1269/1278/1368/1467/2349/2358/2457} ({1359/1458} clashes with 7(2)n1);{2367} clashes with 6 in r5c3 when r3c2 = 6(i/oc12)

Marks here. Updated to include cleanup;no 6 in r6c1. Thanks Afmob.
 Code: .-------------------------------.-------------------------------.-------------------------------. | 23456789  123456    4789      | 24678     123456789 12        | 45689     1234      123456789 | | 23456789  123456    4789      | 56789     123456789 347       | 45689     1234      123456789 | | 23456789  12        356       | 56789     123456789 347       | 3456789   78        123456789 | :-------------------------------+-------------------------------+-------------------------------: | 12345679  3456789   356       | 56789     12345678  347       | 3456789   12345678  3456789   | | 6789      3456789   56        | 12346     123456789 5689      | 12        123478    34568     | | 12345     356789    12        | 1234      23456789  5689      | 3678      3456789   34789     | :-------------------------------+-------------------------------+-------------------------------: | 45678     456789    12        | 1234      123456789 5689      | 1236      3456789   23478     | | 12345     456789    789       | 23468     12345678  12        | 1345678   3456789   1356789   | | 12345     2345      4789      | 46789     456789    5689      | 134568    34568     123568    | '-------------------------------.-------------------------------.-------------------------------'

Last edited by sudokuEd on Wed Oct 03, 2007 6:54 am; edited 1 time in total
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Wed Oct 03, 2007 6:41 am    Post subject: You're right, step 4e is wrong. I'll see if I can find a valid walkhrough for V2. BTW, R6C1 <> 6 since R7C1 has no 3 (both in 9(2) @ N7).
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Wed Oct 03, 2007 8:18 am    Post subject: 51. C123: 26(4) <> {5678} because - If 26(4) = {5678} -> R4C3 = 3 -> 9(2) = [17/28/45] -> R6C2 = (49) Now we show that both candidates for R6C2 are impossible: a) R6C2 = 4 -> R78C2 = {79} @ 20(3) -> R89C3 = {48} @ 19(3) -> impossible since R8C4 <> 7 b) R6C2 = 9 -> R78C2 = {47/56} @ 20(3) -> Killer pair (45) in 9(3) + 20(3) @ N9 locked -> 9(2) = [18/27] + R89C3 <> 4 i. If R78C2 = {47} -> R7C1 = 8 -> R8C3 = 9 = R9C3 ii. If R78C2 = {56} -> 7(2) = {34} -> C3 <> 4 Therefore 26(4) <> {5678}. 52. 26(4) = 9{368/458/467} -> 9 locked for N4 PS: I'll update my walkthrough for V2. If my walkthrough for V3 is correct then V2 is more difficult than V3 because of step 51 which is quite complex. Assassin V2 should be cracked now if you apply Killer pairs but that's up to someone else .
goooders
Regular

Joined: 18 Feb 2007
Posts: 16
Location: london

 Posted: Wed Oct 03, 2007 9:42 am    Post subject: v2 for what its worth i did v2 by realising the 20 cage in column 2 couldnt be 956 as it led to a contradiction this meant the only place for a 6 in the bottom left nonet was in the 9 cage after that its fairly straightforward however for me that was a bit too much like trial and error unless anyone disagrees
sudokuEd
Grandmaster

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

Posted: Wed Oct 03, 2007 12:08 pm    Post subject:

 Afmob wrote: 26(4) <> {5678}

Hmm. OK. I've tried very hard to avoid these (looking both ways) sort of contradiction moves. But if there is no nicer way to proceed, how about expressing it this way.

i. r6c2 = {356789}(no 4 see step 49)
ii. 9 in r6c2 -> r78c2 = {47} only. {56} in r78c2 is blocked because it forces 4 in n9 into 9(3) (c12) and 4 into 7(2)(c2) in n1: but, this leaves no 4 for c3.
iii. [9]{47} in 21(3) -> r89c3 = {89} -> r8c4 = 2 (cage sum)
iv. "45" n7: 4 outies = 15
v. 9 in r6c2 -> 2 in r8c4 -> r6c13 = 4 = [31]
vi. In summary, 9 in r6c2 -> 3 in r6c1
vii. when 26(4)n4 = {5678} -> r4c3 = 3 -> r6c2 = 9 (last candidate)
viii. but from V. when r6c2 = 9 -> r6c1 = 3: -> 2 3's in n4
ix. -> {5678} blocked from 26(4).

When I break it down like this I can 'see' it in my head better and makes it more satisfying. Would rather find another way though.

Cheers
Ed
Para
Yokozuna

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

Posted: Wed Oct 03, 2007 12:28 pm    Post subject:

Hi

Surely there is a nicer way. At least i think so.
Let's turn gooders solution into something a bit less T&E.(how to make that elimination in a logical way). Also it looks a bit better than Afmob's breaking move, which is going a bit towards T&E as well.
I did say no walk-throughs(which i didn't keep while solving this V2), but let's just call this a TAG instead

52. 6's in N7: Either R67C1 = [36] or 20(3) at R6C2 = {569} -> R6C2: no 3.
52a. 20(3) = {479/569/578}

53. 45 on C2: 4 innies: R3459C2 = 18. (A quick note, this hidden cage translates into the 26(4) cage at R4C2 through I/O difference of C1 and C12)
53a. R3459C2 = [1]{78}[2]/[1]{69}[2]/[1]{68}[3]/[2]{38}[5]/[2]{58}[3]/[2]{39}[4]/[2]{49}[3](other combinations blocked by other cages in C2)
53b. Translating to 26(4) cage: 26(4) at R4C2(R5C1,R45C2,R5C3) = [6]{78}[5]/[7]{68}[5]/[9]{38}[6]/[7]{58}[6]/[8]{39}[6]/[7]{49}[6] = {3689/4679/5678}= {3|7..},{4|8..},{3|4|5..}: 6 locked for N4; R3459C2: [1]{69}[2] blocked.

54. 45 on N1: 3 outies: R1C4 + R4C13 = 14 = [2][93/75]/[6]{35}/[7][25/43]/[8][15]: [4][73] blocked by 26(4) at R4C2: R1C4: no 4
54a. Clean up: R8C4: no 6

55. 19(3) at R8C3 = {89}[2]/{79}[3]/{78}[4]/{47}[8]: R89C3 = {47/78/79/89} = {7|9..}
55a. 20(3) at R6C2: [4]{79} blocked by R89C3: R6C2: no 4

Actually Ed already did this, i just took it from approximately Ed's position. Too lazy to read his walk-through.

56. 45 on N7: 4 outies: R6C123 + R8C4 = 15
56a. 26(4) at R4C2 = {3|7..},{4|8..},{3/4/5..}; R4C3 = {35}: R6C123 can't have any combination with both {37},{48},{34},{35} or {45}
56b. R6C123 + R8C4 = [182/281][4]/[192/291/381/471][3]/[391/382/472/571][2]: [371][4]/[372][3]/[481][2] blocked by 26(4), [452][4] blocked by 26(4) + R4C3: R6C2: no 5; R8C3: no 8
56c. Clean up: R1C4: no 2

57. 19(3) at R8C3 = {89}[2]/{79}[3]/{78}[4]: R89C3: no 4
57a. 4 in C3 locked for N1
57b. Clean up: R12C2 = {16/25} = {5|6..} : no 3

58. 20(3) at R6C2 = {479/578}:{569} blocked by R12C2: no 6
58a. R7C1 = 6(hidden); R6C1 = 3; R4C3 = 5

And now it is just basics(singles and last cage combos) till the end.

Solution:

 Quote: 964872513 517693842 823514976 145927368 796348125 382165794 651439287 478251639 239786451

greetings

Para

Last edited by Para on Thu Oct 04, 2007 10:07 pm; edited 2 times in total
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Thu Oct 04, 2007 9:43 am    Post subject: assassin v2 Obviously this one turned out to require afearsome amount of combination work and I gave up on it... however,on route,I did find a rather nice contradiction path that you might (or might not!!) want to see. 1.Usual prelims on inies/outies.Also the 14/3 cage N2/5 =3/4/7 ..no other possibilities given other cages in row and innie/outie totals.r6c123+r8c4=15. 2.In c4 9 must be in the 20/3 cage or at r9 3.IF at r4c9 a) 20/3 cage c4 is 5/7/8 b) r4c18 =46 or 64 c) 5/2 cage c4=2/3 -> kp 2 in r67c34 -> 11/2 cage r67c9 <> 2/9 d) Now consider the two 19/3 cages N1/7.From 2. above 4 is either at r8c4 or r1c4. If at r1c4 ->4 at r9c3 (can't be anywhere else in r3)Either way a 4 "looks" at the 21/3 cage N8. 4.So,the 21/3 cage N8 =9 (postulate 3)75 (57 blocked by 14/3 cage c6) 5.With r9c6=5 -> r5c6=9 (these 2 cells=14 from prelim work). 6.With r8c4=a 4/6 r6c2<>9 (otherwise r6c123+r8c4>15..thus 9 is in 26/4 cage N4.Thus,from 5. above r4c2=9 -> for N7 9 at r8c3. 7.Now the 9s placed in N56 and N78 force a 9 into,respectively,r6c8 and r7c8 ( cannot go into 11/2 cage..3c above).Contradiction. 8.Therefore postulate 3 is wrong the 9 in c4 must be in the 20/3 cage which is then 5/6/9. After this still not easy but does solve fairly readily. Wouldn't normally post a hypothetical like this but I was struck by its rather lovely symmetry re the 4/6 @ r18c4,the KP on the 2 eliminating a 9 from the 11/2 cage c9 and then the 4 9s -> a contradiction. The other thing I noted was the two totally different soving pathways in v1 and v3 just determined by the broken cage at r4c258.Killers are fascinating!! Regards Gary [/b]
sudokuEd
Grandmaster

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

Posted: Thu Oct 04, 2007 11:09 am    Post subject: Re: assassin v2

 gary w wrote: the two totally different soving pathways in v1 and v3 just determined by the broken cage at r4c258. Killers are fascinating!!
Why not have a look at Para's A44V1.5? It's got a nicely broken cage. I plan to work on it next week.

 Para about V2 wrote: Surely there is a nicer way
Wow - really clever work Para. Took some time to follow, but it works. BTW - the combo work in 56a & b are not necessary - a simple outies sum on n7 in r6c123 means no 8 possible in r8c4 which then solves it. A bit less combo work. Enjoy Brazil!

 Afmob wrote: solved A70 V3 today...I guess rating would be 1.25.
Really good solution Afmob. 9c is a ripper! Good call on the rating - once you find one of the keys it falls very easily. Found another place to solve it: 4 innies c89. So, maybe even a 1.00 for some.

Very excited about your enormous contribution the last few weeks Afmob. Thanks.

Cheers
Ed
Para
Yokozuna

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

Posted: Thu Oct 04, 2007 12:53 pm    Post subject: Re: assassin v2

sudokuEd wrote:
 Para about V2 wrote: Surely there is a nicer way
Wow - really clever work Para. Took some time to follow, but it works. BTW - the combo work in 56a & b are not necessary - a simple outies sum on n7 in r6c123 means no 8 possible in r8c4 which then solves it. A bit less combo work.

Thanks.
I know it wasn't necessary. But my actual intention was to eliminate the 5 from R6C2 to create a killer triple {789} in N4. But the elimination of the 8 sped things up. Sometimes we just do redundant things.

 Quote: You've hit my conscience hard Para - thanks. As punishment, I won't publish any more not-solved-first puzzles.

Was merely joking about that. But i think you would have gotten there with this V3 eventually. Probably not looking at the right place. I used the opening you mentioned.

Got to pack.

greetings

Para

ps. Gary, there are more of these types of V2's on this forum. Changing just 1 cell can be murderous on a killers complexity. You should check A41V2 for that.
For remote killer cages you could also check out the Black-hole Killer by Ed and the CDK Killer by Nasenbaer.
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Thu Oct 04, 2007 5:01 pm    Post subject: Re: assassin v2

 gary w wrote: Obviously this one turned out to require a fearsome amount of combination work

Or does it? I discovered that this puzzle could be unlocked without an awful lot of combination crunching by looking at the N4 innies. Therefore, here are some alternative steps, starting from Ed's marks pic. Note that the first step in the sequence happens to be the same as Para's step 52.

51. 6 in n7 locked in r7c1+r78c2
51a. -> either 6 in r78c2, or...
51b. ...6 in r7c1 -> 3 in r6c1
51c. Either way: no 3 in r6c2

52. Innies n4: r3c46+r6c123 = 19(5) = {12349/12358/12367/12457}
(Note: {13456} blocked by r5c3)
52a. if {12349}, 9 must go in r6c2
52b. -> no 9 in r4c1
52c. if {12358}, 8 must go in r6c2
52d. if {12457}, 5 must go in r4c3
52e. -> no 5 in r6c2

53. 20(3)n47 = {479/569/578} = {(4/5)..}
53a. {45} of 20(3)n47 now only available within r78c2
53b. -> r78c2 and 9(3)n7 (step 12) form killer pair on {45} within n7
53c. -> no 4,5 elsewhere in n7
53d. cleanup: no 4,5 in r6c1

54. 4 in c3 locked in n1 -> not elsewhere in n1
54a. 4 locked within 19(3)n12 within r12c3
54b. -> no 2,4 in r1c4 (no 2 because otherwise 19(3) cage sum unreachable)
54c. cleanup: no 3 in r12c2; no 6,8 in r8c4 (step 3)
54d. 7(2)n1 = {16/25} = {(5/6)..}

55. 7(2)n1 blocks {569} combo for 20(3)n47
55a. -> 20(3)n47 = {479/578} (no 6)
55b. 7 locked for c2

56. Hidden single in n7 at r7c1 = 6
56a. -> r6c1 = 3

and so on...
_________________
Cheers,
Mike

Last edited by mhparker on Sat Oct 20, 2007 8:10 pm; edited 1 time in total
azpaull
Regular

Joined: 27 Aug 2007
Posts: 14
Location: Arizona

 Posted: Fri Oct 05, 2007 2:04 pm    Post subject: Wow, I couldn't keep track which version you guys were talking about. No matter, I only completed 1.0. Finished it yesterday - when a judge was REALLY late and I had plenty of time to find that C789 key. Thanks, your Honor! I didn't think it was doable without that key, but since it was difficult for (most of) us to find that key, I think we need to give it at least a 1.25. Edit: Make that a 1.5, I just reviewed the rating scale.
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Fri Oct 05, 2007 9:25 pm    Post subject:

 azpaull wrote: Wow, I couldn't keep track which version you guys were talking about.

A good point!

Para has told me that he always starts his walkthroughs by saying which puzzle it was.

I don't always do that. If I post a walkthrough when there is still only one puzzle on the thread, then I don't say which one. If there are multiple puzzles then I do. For example there have been a number of times when I've posted a walkthrough for the original Assassin after variants have been posted in the thread.

It's also good in discussions to say which version of the puzzle.

 azpaull wrote: No matter, I only completed 1.0. Finished it yesterday - when a judge was REALLY late and I had plenty of time to find that C789 key. Thanks, your Honor! I didn't think it was doable without that key, but since it was difficult for (most of) us to find that key, I think we need to give it at least a 1.25. Edit: Make that a 1.5, I just reviewed the rating scale.

It was doable without the key. If you read my second message on page 1 you will see that's what I did. I still rated it as 1.0 to 1.25 that way.
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