Assassin 42

Para · Post by **Para** » Wed Mar 21, 2007 11:09 am

One more elimination

56. 45 on C9: R34C9 - R8C9 = 9
56a. R9C8 = 1 -->> R34C9 = {46} -->> R56C9 = {15}: 22(4) can't be {1489/1579/1678}
56b. R9C8: no 1

Para

rcbroughton · Post by **rcbroughton** » Wed Mar 21, 2007 7:48 pm

Continuing on from the good work around n6 . . .

somebody check step my step 58. Can I use UR in this instance? not one of my stronger techniques but it looks right.

[edited to recognise the fact that 57 duplicated one of Para's moves and also to add a cleanup on step 59]

[edit - missed Para's step 52 that already did the work of 57.]
57. 45 on n6. innies=30
57a. {35679} not possible because it clashes with 9(2)
57b. {25689} ditto
57c. {45678} ditto
57d. {24789} - if r6c7 was 4, r5c6 would need to be 7 which doesn't fit the combos for 18(3)n56
57e. {34689} - if r6c6 was 4, r5c6 would need to be 3 which doesn't fit the combos for 18(3)
57f. {15789} - no 4
57g. conclusion -> no 4 in r6c7

58. unique rectangle at r7c34 r9c34 with {79} in 2 row, 2 columns, 2 nonets and 2 cages
58a. can't have 7 or 9 at r9c3 otherwise puzzle has non-unique solution

59. Can't have 1 at r1c9 as it eliminates all places for 1 in n6
59a. r1c9=1 -> no 1 at r56c9
59b. r1c9=1 -> no 1 at r1c45 -> r3c6=1 -> no 1 at r4c78
no other places in n6 for 1
59c. clean-up - no 7 at r2c9

Richard

Andrew · Post by **Andrew** » Wed Mar 21, 2007 8:03 pm

Just to let you know that I've added to step 55, eliminating more combinations from the innies for N5. I've made the changes in my previous message so that anyone starting from the beginning should see the steps in the correct order.

BTW, just for information, Para's step 56a also works as
56a. R9C8 = 1 -->> R34C9 = {46} -->> R56C9 = {15} -->> no valid combinations for R12C9

Thought it worth mentioning in case the interactions in C9 are helpful later.

Para · Post by **Para** » Wed Mar 21, 2007 8:07 pm

Hi richard.

Yes that UR is correct. I spotted it before but i tend to avoid UR's in Killers.

Para

Andrew · Post by **Andrew** » Wed Mar 21, 2007 9:33 pm

rcbroughton wrote:[edited to recognise the fact that 57 duplicated one of Para's moves and also to add a cleanup on step 59]

Step 57 was still useful because it eliminated half of the combinations for the innies in N5.

BTW {35679} could also have been eliminated from the n6 innies because they must contain 8.

I suppose it's a matter of personal preference whether one uses a UR step. It's something I've never used myself. I don't doubt that Ruud has checked that this puzzle only has one solution. Therefore I have made that move in my diagram to be consistent with the thread. However it would be nice if subsequent moves can be used to show that one can logicially reach the solution without using a UR move.

I think the combinations for the cages and hidden cages that we have worked on so far are

13(4) in N14 13{27/45}
24(5) in N12 {12678/14568/23478}
26(4) in N36 9{278/368/458/467}
20(4) in N254 27{38/56}
18(3) in N56 {279/369/378/567}
14(4) in N7 24{17/35}
23(4) in N78 69{17/35}

R4C123 37{1/2/4}
R4C789 {158/159/189/248/249/468/469/489}

5 innies in N5 {12569/12578/13568/23468}
5 innies in N6 89{157/247/346}

I haven't included 13(3) in N2 and 22(4) in N9 which have only had candidates eliminated from individual cells. I think all possible combinations for these cages are still valid.

Can someone confirm that the above combinations are correct and are the only valid ones for these cages and hidden cages?

sudokuEd · Post by **sudokuEd** » Thu Mar 22, 2007 11:36 am

OK Andrew. They all look good. And you've opened up a little teeny crack. My final step isn't as productive as first thought: found a mistake. So, don't get a headache over them: it makes you really concentrate.

Marks pic [edited out] didn't include 7 and 9 in r9c3: some disquiet about this - must not be desperate enough yet.

First, one extra part of step 52c. R6C6 = 6 -->> R4C789 = {189/468}
60. The {189} is blocked.
60a. since r6c6 = 6 -> r5c2 = 6 -> 6 for n6 must be in r4c789 = {468} only
60b. R6C6 = 9 -->> R4C789 = {489}
60c R6C6 = 7 -->> R4C789 = {469}
60d. R6C6 = 6 -->> R4C789 = {468}
60e. R6C6 = 3 -->> R4C789 = {159/249}
60f. R6C6 = 2 -->> R4C789 = {158/248}

61. Now combining this with 5 innies n5 = {12569/12578/13568/23468}
61b. R6C6 = 9 -->> R4C789 = {489} and rest of innies n5 = {1256}
61c R6C6 = 7 -->> R4C789 = {469} and rest of innies n5 = {1258}
61d. R6C6 = 6 -->> R4C789 = {468} and rest of innies n5 = {1259} ({1358/2348} both blocked by 8 in r4c789)
61e. R6C6 = 3 -->> R4C789 = {159} and rest of innies n5 = {2468}
61f. R6C6 = 3 -->> R4C789 = {249} and rest of innies n5 = {2468}
61g. R6C6 = 2 -->> R4C789 = {158} and rest of innies n5 = all blocked by 5 and 8 in r4c789
61h. R6C6 = 2 -->> R4C789 = {248} and rest of innies n5 = {1569} ({1578/3468} blocked by 8 in r4c789

62. However,61h is also blocked
62a. 2 in r6c6 and other 4 innies = {1569}
62b. -> 1 in r6c4 -> r4c123 = 11 = [1]{37} (step 27b)
62c. -> rest of 20(4) = {28} only (not 5,6 or 9!):
62d. -> r6c6 !=2

63. In summary
63a. no 2 r6c6
63b. r4c789 = {489/469/468/159/249}
63c. 5 innies n5 = {12569/12578/23468}
63d. 2 locked n5

64. 22(4)n = {1579/3469/3478} ({1489/1678/3568/4567} blocked by 5 innies)

65. No 2 in r9c8
65a. "45"c9 -> r9c8 + 9 = r34c9
65b. r9c8 = 2 -> r34c9 = 11 = {38} -> r239c8 = {69}[2} -> 9(2)n6 Clash
..............................{47} -> r239 = {69}[2} -> 9(2)n6 Clash
65b.-> no 2 r9c8

66. Building more on steps 43a, 41 and 47
66a. r2346c4 = [3{28}4] and r1c2 = 2 -> 13(3)n2 = [148]/{157}
66b. r2346c4 = [4382] and r1c2 = 2 -> 13(3)n2 = {157}
66c. r2346c4 = [4382] and r1c2 = 7 -> 13(3)n2 = {56[2]56}
66d. r2346c4 = [4{56}2] and r1c2 = 2 -> 13(3)n2 = {139/157}
66e. r2346c4 = [4{56}2] and r1c2 = 7 -> 13(3)n2 = {139/38[2]38} ({256}blocked by r3c4)
66f. r2346c4 = [6{28}1] and r1c2 = 2 -> 13(3)n2 = [319/418/517]
66g. r2346c4 = [6{28}1] and r1c2 = 4 -> 13(3)n2 = [319/517] ({238} blocked by r3c4)

66h. In summary 13(3)n2 = {139/148/157/238/256} ({247/346} eliminated)

Andrew · Post by **Andrew** » Thu Mar 22, 2007 3:41 pm

First I wasn't saying that we shouldn't use the UR, just that it is a matter for philosophical debate. If others want to go ahead with it, I'll work on that basis. However if we want to keep the 7,9 in R9C3, as Ed has said he's done in his latest diagram, that's also good; we will know that we can always use the UR later if we need it.

We are going out for the day so have only had a quick glance at Ed's moves. I think he missed one extension.

67. R4C789 = {489/469/468/159/249} (step 63b) [4/5], killer pair 4/5 in R4C789 and R56C9 for N6, no 5 in 18(3) cage in N56 = {279/369/378}

rcbroughton · Post by **rcbroughton** » Thu Mar 22, 2007 6:15 pm

Another couple building from Ed's position.

68. no 2 at r4c7 or r4c8 as it removes all placements for 2 in r6
68a. r4c78=2 -> no 2 in n6r6
68b. r4c78=2 -> no 2 in n6r5 -> r5c3=2 -> r6c34<>2
68c. no other 2's in r6 - so can't have a 2 at r4c7 or r4c8

Para · Post by **Para** » Thu Mar 22, 2007 7:21 pm

OK a bit more. We'll get there.

69. 45 on C9: R9C8 + 9 = R34C9
69a.R9C8 = 3 -->> R34C9 = {39}: {48} clashes with 22(4) in N9; 8 then locked in R789C9, now R9C3 = {75} clashes with 22(4) in N9
69b.R9C8 = 4 -->> R34C9 = {49}: {58/67} blocked because R23C8 would be {49} and then 2 4's in C8.
69c.R9C8 = 5 -->> R34C9 = {59}: {68} blocked: R34C9 = {68} -->> R23C8 = {39} -->> R56C8 = {27} -->> R56C9 = {15} -->> no options left for 8(2)N1.
69d.R9C8 = 6 -->> R34C9 = {69}: [78] blocked: R34C9 = [78] -->> R23C8 = {29}: no options left for 9(2) in N6
69e.R9C8 = 7 -->> R34C9 = {79}
69f.R9C8 = 8 -->> R34C9 = {89}
69g. Conclusion 9 locked in r34C9 for C9 and 26(4) in R2C8.

70. 45 on C9: R9C8 + 9 = R34C9
70a. R9C8 = 3 -->> R34C9 = {39} -->> R23C8 = {68} -->> R56C8 = {27} -->> R56C9 = {15}: No options left for 8(2) in N1
70b. R9C8 = 4 -->> R34C9 = {49} -->> R56C9 = {15} -->> R12C9 = {26} -->> R789C9 = {378}
70c. R9C8 = 5 -->> R34C9 = {59} -->> R23C8 = {48} -->> R56C9 = {24} -->> R789 = {368} -->> R12C9 = {17}
70d. R9C8 = 6 -->> R34C9 = {69} -->> R56C9 = {24}(no {26} in R12C9, only place left for 2) -->> R56C8 = {36} -->> R789C9 = {358/178}
70e.R9C8 = 7 -->> R34C9 = [79] -->> R23C8 = {28/46} -->> R12C9 = {35} -->> R56C9 = {24} -->> R56C8 = {36} -->> R23C8 = {28} -->> R789C9 = {168}
70f. R9C8 = 8 -->> R34C9 = {89} -->> R789C9 = {167/347/356}

71. Conclusions:
71a. R3C9 + R9C8: no 3
71b. 22(4) in R7C9: no {4567}
71c. 8 locked in 22(4) in R7C9 for N9

Para

rcbroughton · Post by **rcbroughton** » Thu Mar 22, 2007 8:29 pm

Another move or two.

Also noticed in Ed's mark pic I think there should be a 15 removed from r1c2. (At least they're not there on the version I was working from previously)

I'm also assuming 7,9 are gone from r9c3 - no sense ignoring a valid technique that gets rid of a couple of candidates, is there?

[edit - added an extra bit following step 73]

71. 45 on c9 - outies r239c8 total 17(3)={278}/{368}/{458} - must use 8 -locked for c8 ({467} blocked by 9(2)n6 no other combos)

72. Can't have an 8 at r8c7 or r9c7 as it eliminates all placements for 8 in c8
72a. r89c7=8 -> r9c8<>8
72b. r89c7=8 -> 456c7<>8 -> r4c9=8 -> r23c8<>8
72c. no places left with 8 in c8

73. 45 on n9 innies total 23
73a. can only have {12569} or {23459} - no 7 so remove 7 from 20(3)n89
73b. {13469} - blocked by 22(4)n9
73c. {14567} - blocked by 22(4)n9
73d. {23567} - blocked by 22(4)n9
73e. {12479} - blocked because r7c78 would need to be {41} -> rest={279} which would require r9c6 being 2 which breaks the 20(4)n89

74. 45 on c89. innies r1478c8=19(4)={1459}/{1369}/{1279} - {1279} requires 7 at r1c8 - so no 2 at r1c8
74a. other combos {1567}/{2359}/{2647}/{3754} blocked by 9(2)n6

75. adding to step 73: can't have a 3 in r8c8
75a. consider only combo with 3 is {23459} - r7c78={23}/{25}/{34}/{35}
1) 9(2)n6 restricts {23} in r8c78
2) 17(3) from step 71 and 9(2) restrict {24}/{25}/{34}/{35} in r8c78
75b. {23}/{34}/{35} -> no 3 in r8c8
75c. {25} -> {349} in remaining innies. with 2 in r7c8 can only have 9 in r8c8, with 5 in r7c8 can only have 4 or 9 in r8c8

76. {259} now locked in these innies in n9 - not possible in 22(4)

Richard

Para · Post by **Para** » Thu Mar 22, 2007 9:14 pm

Hi

I think that we are waiting to see if it is really necessary to use that UR as we are not really working down there to find eliminations momentarily.
They are still in my grid, but i have them in the back of my mind for when we get back there.

Para

rcbroughton · Post by **rcbroughton** » Thu Mar 22, 2007 9:41 pm

True - we'll see when we get there.

Just spotted another couple of steps stemming from the work I just did in n9.

77. 45 on c9 innies total 31.
77a. 6(2)n6 blocks {45679}
77b. 8(2)n3 blocks {35689}
77c. {25789} not possible because there's no 2 left
77c. only combos {16789}/{34789} - no combo with 5, so no 5 in r34c9

78. (cleanup) 45 on n3 - outies= 13 - no 8 in r2c6

79. 45 on n36 - outies r2346c6=18(4)={1269}/{1359}/{1467}/{2349}/{2358}/{2367}/{2457}/{3456}
79a. 45 on c6789 innies r158c6=20(3) - {569} blocked by the 18(4)
79b. only valid combos left in r158c6=20(3) are {389}/{578}/{479} - no 6

Para · Post by **Para** » Thu Mar 22, 2007 9:52 pm

Hi

quickie

80. 8(2) in R1C9: no [71]
80a. 2 and 5 locked in cages 8(2) + 6(2) in C9.
80b. 6(2) = {15} -->> 8(2) = {26} (needs to use 2)
80c. 6(2) = {24} -->> 8(2) = {35} (needs to use 5)

81. 8 in C89 locked in cages 26(4) R2C8 and 22(4) in R7C9 so both need 8.
81a. 26(4): no {4679}

Para

Para · Post by **Para** » Thu Mar 22, 2007 10:01 pm

Ok two more quick simple ones.

82. 24(5) in R1C7 can't have combo of {23/36} because of 8(2) in R1C9.
82a. 24(5): no {13569/23469/23478/23568}

83. 45 on N3: 2 outies = 13; R2C6: no 8.

Para

Para · Post by **Para** » Thu Mar 22, 2007 10:11 pm

OK a few more.

84. 26(4) in R2C8: no {3689}: clashes with 8(2) in R1C9.
84a. 26(4): no 3 or 6
84b. Clean up: R9C8: no 6 (step 70), R2C6: no 7

85. 24(5) in R1C7 requires one of 3,6 in N3: no {12489/12579}
85a. (oops, should have seen that before) 1 locked in 24(5) in N3: no {24567}

Para