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Ed
 Post subject: Assassin 435
PostPosted: Thu Jun 01, 2023 9:31 am 
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Posts: 1044
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An X-puzzle so 1-9 cannot repeat on either diagonal

Assassin 435
17 years of Assassins! Our own wonderful archive of all those original puzzles and many more here

For this new Assassin, thought I'd try more small cages so took a puzzle generated by Cross+A. It got a 1.40 but was a little easy for an A so joined two cages. Its now 1.60 according to SudokuSolver but not hard for JSudoku. Some key steps were quite hidden to me. Nothing too tricky so get a nice aha when they reveal.

triple click code:
3x3:d:k:6912:6912:6679:6679:6679:3331:3331:4100:4100:6912:6912:6679:4613:6679:3331:3331:4100:2822:3079:3079:4104:4613:4613:6153:2570:2570:2822:3079:2571:4104:4104:4104:6153:6153:6153:4620:3085:2571:2571:8462:1551:4368:4368:4620:4620:3085:4370:8462:8462:1551:4368:5393:5393:4620:2323:4370:2836:8462:8462:8470:5393:5393:3074:2323:4370:2836:8470:8470:8470:4353:3074:3074:2323:4117:4117:4117:8470:8470:4353:4353:4353:
solution: 
+-------+-------+-------+
| 6 5 8 | 2 9 4 | 3 1 7 |
| 7 9 4 | 6 3 1 | 5 8 2 |
| 2 1 3 | 5 7 8 | 4 6 9 |
+-------+-------+-------+
| 9 2 1 | 8 4 3 | 6 7 5 |
| 4 3 5 | 7 1 6 | 9 2 8 |
| 8 7 6 | 9 5 2 | 1 4 3 |
+-------+-------+-------+
| 1 4 2 | 3 8 5 | 7 9 6 |
| 3 6 9 | 4 2 7 | 8 5 1 |
| 5 8 7 | 1 6 9 | 2 3 4 |
+-------+-------+-------+

Cheers
Ed
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Andrew
 Post subject: Re: Assassin 435
PostPosted: Mon Jun 05, 2023 4:25 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. I had one "aha" moment; maybe step 2a was also one but step 3b was the one where I felt it.

On checking my walkthrough one more time, I've simplified the end of step 1 and made some other detail changes.

Thanks Ed for some corrections plus some of my own after going through my WT again.
Here's how I solved Assassin 435:
Prelims

a) R23C9 = {29/38/47/56}, no 1
c) R3C78 = {19/28/37/46}, no 5
d) R56C1 = {39/48/57}, no 1,2,6
e) R56C5 = {15/24}
f) R78C3 = {29/38/47/56}, no 1
g) 10(3) cage at R4C2 = {127/136/145/235}, no 8,9
h) 9(3) cage at R7C1 = {126/135/234}, no 7,8,9
i) 27(4) cage at R1C1 = {3789/4689/5679}, no 1,2
j) 13(4) cage at R1C6 = {1237/1246/1345}, no 8,9
k) 33(5) cage at R5C4 = {36789/45789}, no 1,2

1a. 27(4) cage at R1C1 = {3789/4689/5679}, 9 locked for N1
1b. 45 rule on N3 2 outies R12C6 = 5 = {14/23}
1c. 45 rule on N3 2 innies R12C7 = 8 = {17/26/35}, no 4
1d. 45 rule on R12 2 innies R2C49 = 8 = [17]/{26/35}, no 4,8,9, no 7 in R2C4, clean-up: no 2,3,7 in R4C9
1e. 45 rule on R1234 2 innies R4C29 = 7 = {16/25/34}, no 7,8,9
1f. 45 rule on N7 2(1+1) outies R6C2 + R9C4 = 8 = {17/26/35}/[44], no 8,9
1g. 45 rule on N9 2 outies R6C78 = 5 = {14/23}
1h. 45 rule on N9 2 innies R7C78 = 16 = {79}, locked for R7 and N9, clean-up: no 2,4 in R8C3
1i. 33(5) cage at R5C4 = {36789/45789}, CPE no 7 in R6C6
1j. 45 rule on N69 3 innies R4C78 + R5C7 = 22 = {589/679}, 9 locked for N6
1k. 9 in C9 only in R13C9, locked for N3, clean-up: no 1 in R3C78
[I originally moved on to step 2, having spotted that important group of outies, but later came back here to work further in this area as part of step 1.]
1l. Combined cage R12C7 + R3C78 = {17}{28}/{17}{46}/{35}{28}/{35}{46} (cannot be {26}{37} which clashes with R23C9) -> R12C7 = {17/35}, no 2,6, R3C78 = {28/46}, no 3,7
1m. 16(3) cage at R1C8 = {169/178/259/349/358} (cannot be {268} which clashes with R3C78, cannot be {367/457} which clash with R12C7)
1n. Hidden killer pair 8,9 in 16(3) cage and R3C789 for N3, 16(3) cage contains one of 8,9 -> R3C789 must contain one of 8,9
1o. 18(3) cage at R2C4 = {279/369/378/459/567} (cannot be {189} = 1{89} which clashes with R3C789, cannot be {468} = 6{48} which clashes with R3C78), no 1, clean-up: no 7 in R2C9, no 4 in R3C9
1p. 2 of {279} must be in R2C4 -> no 2 in R3C45
1q. Combined cage R23C9 + R3C78 = [29]/{46}/[38]{46}/{56}{28}, 6 locked for N3
1r. 45 rule on R12 3 outies R3C459 = 21 = {489/579} (cannot be {678} which clashes with R3C78), no 3,6, 9 locked for R3, clean-up: no 5 in R2C9, no 3 in R2C4
[Simpler than my original steps for eliminating {369} and particularly {378} from 18(3) cage.]

2a. 45 rule on N78 4(1+3) outies in two different ways R5C4 + R6C234 = 29 must contain 7 in R5C4 + R6C34 for 33(5) cage = 5{789}/7{589/679}/8{579/678}/9{479/578}, no 3 -> R5C4 = {5789}, R6C2 = {57}, R9C4 = {13} (step 1c)
2b. R5C4 + R6C234 = 5{789}/7{589/679}/8{579}/9{578} (cannot be 8{678}/9{479} because R5C4 + R6C34 cannot contain two 8s or two 9s), no 4
2c. Max R9C4 = 3 -> min R9C23 = 13, no 1,2,3 in R9C23
2d. 45 rule on C12 1 innie R9C2 = 1 outie R5C3 + 3, no 7 in R5C3
2e. 45 rule on C6789 3 innies R789C6 = 21 = {489/579/678}, no 1,2,3
2f. 45 rule on C6789 3 outies R8C45 + R9C5 = 12 must contain 2 for N8 = {129/237/246}, no 5,8
2g. 9 of {129} must be in R89C5 (R89C5 cannot be {12} which clashes with R56C5), no 9 in R8C4
2h. 45 rule on N8 3 innies R7C45 + R9C4 = 12 = {138/345} (cannot be {156} because 33(5) cage at R5C4 cannot contain both of 5,6), no 6, 3 locked for N8, clean-up: no 7 in R8C45 + R9C5
2i. R789C6 = {579/678} (cannot be {489} which clashes with R7C45 + R9C4), no 4, 7 locked for C6
2j. R7C45 = {38/45} -> R5C4 + R6C34 = {679/789}, no 5
2k. R5C4 + R6C34 = {679/789}, CPE no 9 in R6C6
2l. R56C1 = {39/48} (cannot be {57} which clashes with R6C2), no 5,7
2m. 9(3) cage at R7C1 = {126/135} (cannot be {234} which clashes with R56C1), no 4, 1 locked for C1 and N7
2n. R78C3 = [29/38/47/83] (cannot be {56} which clashes with 9(3) cage), no 5,6
2o. Min R34C1 = 8 (cannot be {23/25} which clash with 9(3) cage, cannot be {24} which clashes with R56C1 + 9(3) cage, cannot be {34} which clashes with R56C1) -> max R3C2 = 4
2p. 12(3) cage at R3C1 = {129/147/237/246/345} (cannot be {138} which clashes with R56C1, cannot be {156} which clashes with 9(3) cage), no 8
2q. 6 of {246} must be in R3C12 (R3C12 cannot be {24} which clash with R3C78, or [62] but that’s not currently relevant), no 6 in R4C1
2r. 17(3) cage at R6C2 = {278/359/458/467} (cannot be {269/368} because R6C2 only contains 5,7)
2s. R6C2 = {57} -> no 5,7 in R78C2
2t. 9 of {359} must be in R8C2 -> no 3 in R8C2
2u. Consider combinations for 17(3) cage
17(3) cage = {278/359/458} => R78C3 = [29/47] (cannot be {38} which clashes with 17(3) cage
or 17(3) cage = {467}, 6 locked for N7 => 9(3) cage = {135}, 3 locked for N7
-> R78C3 = [29/47], no 3,8
2v. 16(3) cage at R9C2 = {169/178/358/367} (cannot be {349} which clashes with R78C3), no 4, clean-up: no 1 in R5C3 (step 2d)
2w. Consider combinations for 17(3) cage
17(3) cage = {278}, 8 locked for N7 => 16(3) cage = {169/367}
or 17(3) cage = {359/458}, R6C2 = 5
or 17(3) cage = {467}, 6 locked for N7 => 9(3) cage = {135}, 5 locked for N7
-> no 5 in R9C2, clean-up: no 2 in R5C3 (step 2d)
2x. Min R5C3 = 3 -> max R45C2 = 7, no 7 in R5C2
2y. 45 rule on N4 4 innies R4C13 + R6C23 = 23 must contain 7 for N4 = {1679/2579/2678} (cannot be {3479/3578} which clash with R56C1), no 3,4

[Now maybe one of Ed’s “aha” moments.]
3a. 17(3) cage at R6C2 (step 2r) = {278/359/458/467} with {359} = [539]
3b. Consider combinations for R7C45 (step 2j) = {38/45}
R7C45 = {38}, 3 locked for R7 => no 3 in R7C2
or R7C45 = {45}, 4 locked for R7 => R7C3 = 2, R8C3 = 9
-> 17(3) cage = {278/458/467}, no 3,9
3c. 3 in N7 only in 9(3) cage (step 2m) = {135}, 3,5 locked for C1, 5 locked for N7, clean-up: no 9 in R56C1
3d. Naked pair {48} in R56C1, locked for C1 and N4, clean-up: no 3 in R4C9 (step 1e), no 7 in R9C2 (step 2d)
3e. 7 in N7 only in R89C3, locked for C3
3f. 2 in C1 only in R34C1, locked for 12(3) cage at R3C1, no 2 in R3C2
3g. 27(4) cage at R1C1 = {3789/5679} (cannot be {4689} which clashes with 17(3) cage), no 4
3h. 45 rule on C1 3 outies R123C2 = 15
3i. 27(4) cage = {5679} (cannot be {3789} because R123C2 = {38}4 clashes with 17(3) cage), 5,6,7 locked for N1, 5 locked for C2, clean-up: no 2 in R4C9 (step 1e)
[Cracked. The rest is fairly straightforward.]

4a. R6C2 = 7 -> R9C4 = 1 (step 1f)
4b. R34C1 = [29] -> R3C2 = 1 (cage sum), clean-up: no 8 in R3C78, no 6 in R4C9 (step 1e)
4c. R6C3 = 6, naked pair {23} in R45C2, locked for N4, 2 locked for C2 -> R45C3 = [15], clean-up: no 1 in R6C5
4d. R6C2 = 7 -> R78C2 = 10 = {46}, locked for N7, 6 locked for C2
4e. R7C3 = 2, placed for D/, R8C3 = 9, R9C23 = [87], clean-up: no 4 in R6C5
4f. R8C45 + R9C5 (step 2f) = {246} (only remaining combination), 4,6 locked for N8 -> R789C6 (step 2e) = {579} (only remaining combination) = [579]
4g. Naked pair {38} in R7C45, locked for R7, 8 locked for 33(5) cage at R5C4 -> R56C4 = [79], 9 placed for D/
4h. Naked pair {46} in R3C78, locked for R3 and N3, clean-up: no 2 in R2C4 (step 1d), no 5 in R3C9
4i. Naked pair {38} in R3C36, locked for R3 -> R3C9 = 9, R2C9 = 2, 18(3) cage at R2C4 = [657]
[Routine clean-ups omitted from here.]

5a. R5C7 = 9 (hidden single in N6) -> R56C6 = 8 = [62], 2 placed for D\, R7C7 = 7, placed for D\, R12C1 = [67], 6 placed for D\
5b. R12C6 (step 1b) = 5 = {14}, locked for N2, 4 locked for C6, 1 locked for 13(4) cage at R1C6
5c. Naked pair {35} in R12C7, locked for C7 and N3
5d. R4C78 = [67] (hidden pair in R4)
5e. R3C7 = 4, placed for D/, R5C5 = 1, placed for both diagonals, R2C8 = 8, placed for D/, R4C6 = 3, placed for D/, R3C6 = 8, R3C3 = 3, placed for D\
5f. R9C1 =5 -> R8C8 + R9C9 = [54], both placed for D\

and the rest is naked singles, not using the diagonals.
When I found time to look at wellbeback's and Ed's walkthroughs I found that they were both simpler and more direct so didn't need to use my interesting start to step 2.
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wellbeback
 Post subject: Re: Assassin 435
PostPosted: Tue Jun 06, 2023 10:00 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
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This one went very smoothly for me. Thanks Ed! As you wrote, not too tricky.
Assassin 435 WT:
1. Innies n9 = r7c78 = +16(2) = {79}
33(5)r5c4 = {(36|45)789}
-> Only possibilities for Innies n8 = [r7c45,r9c4] = +12(3) are [{38}1] and [{45}3]

2. -> Outies n7 = r6c2,r9c4 = +8(2) = [71] or [53]
-> 12(2)n4 from {39} or {48}
-> 9(3)n7 not {234} - must be {126} or {135}

3! 5 in n7?
-> 11(2)n7 not {56}
Outies c12 = r5c3 + r9c3 + r9c4 = +13(3)
Since Outies n7 = r6c2,r9c4 = +8(2) -> r5c3 + r9c3 = R6c2 + 5
-> r9c3 not 5
17(3)r6c2 cannot contain both (57) -> 5 not in r78c2
Either r6c2 is 5 or r9c4 is 1 which puts r9c2 as min 6
-> r9c2 not 5
-> 5 in n7 in 9(3)n7 = {135}
-> 12(2)n4 = {48}

4. 27(4)n1 must contain one of (58) in r12c2
-> 17(3)c2 not [5{48}]
-> No solution for 17(3)c2 with 5 in r6c2
-> r6c2 = 7
-> r9c4 = 1
-> 33(5)r5c4 = [7{69}{38}]

5. Innies r1234 = r4c29 = +7(2) (No 789)
-> 7 in r4 in r4c78
-> Innies n69 = r4c78,r5c7 = +22(3) = {679}
-> r458c7 = {679} with 6 in r45c7
-> Innies n3 = r12c7 = +8(2) = {35}
-> r12c6 = {14}

6. Innies r12 = r2c49 = +8(2)
(79) in n3 both in c9
Since 1 already in n2 -> r123c9 = [729] and r2c4 = 6
-> 10(2)n3 = [46]
-> r12c8 = {18}
Also -> r6c34 = [69]
Also r3c45 = [57]

7. NS r3c1 = 2
-> 12(3)r3c1 = [219] (Only remaining solution)
-> r12c1 = [67] and r12c2 = {59}
Also (remaining Innie n4) r4c3 = 1
-> 10(3)n4 = [{23}5]
-> 17(3)c2 = [746]
-> 11(2)n7 = [29]
-> 16(3)r9 = [871]

8. Innies c6789 = +21(3) = [579]
Also r5c7 = 9 and r4c78 = [67]
-> r56c6 = [62] and r34c6 = {38}
-> r6c78 = [14]

9. (HS 7 in D\) r7c78 = [79]
-> (HS 9 in D\) r12c2 = [59]

10. 6(2)n5 = [15]
-> r12c8 = [18]
-> r34c6 = [83]
etc.
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Ed
 Post subject: Re: Assassin 435
PostPosted: Fri Jun 09, 2023 6:46 pm 
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Joined: Wed Apr 16, 2008 1:16 am
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Love that IOU in Wellbeback's step 3 (line 4&5)!! Very clever. His step 2 (Andrew's step 2h) was an aha step for me. Took me time to see that. Also really like their nice move in the key area (Andrew's step 2w, Wellbeback's step 3 line 7)

My start after Wellbebacks step 2: 
.-------------------------------.-------------------------------.-------------------------------.
| 3456789   3456789   123456789 | 123456789 123456789 1234567   | 1234567   123456789 123456789 |
| 3456789   3456789   123456789 | 123456789 123456789 1234567   | 1234567   123456789 23456789  |
| 23456789  123456789 123456789 | 123456789 123456789 123456789 | 12346789  12346789  23456789  |
:-------------------------------+-------------------------------+-------------------------------:
| 23456789  1234567   123456789 | 123456789 123456789 123456789 | 123456789 123456789 123456789 |
| 3489      1234567   1234567   | 3456789   1245      123456789 | 123456789 123456789 123456789 |
| 3489      57        3456789   | 3456789   1245      123456789 | 1234568   1234568   123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 12356     234568    234568    | 3458      3458      124568    | 79        79        1234568   |
| 12356     23456789  23456789  | 12456789  12456789  12456789  | 1234568   1234568   1234568   |
| 12356     23456789  23456789  | 13        12456789  12456789  | 1234568   1234568   1234568   |
'-------------------------------.-------------------------------.-------------------------------'

End step 2 above: paste into A435 in SudokuSolver

3. 17(3)r6c2: can't have both 5,7 since {575} = 17
3a. no 5,7 in r78c2

4. 7 in n7 only in 11(2) = {47} or in 16(3)
4a. -> 4 in 16(3) must also have 7 (Locking-out cages)
4b. -> {349} blocked
4c. -> 16(3) which must have 1,3 for r9c4 = {169/178/358/367}(no 2,3,4 in r9c23)

Next aha step
5. "45" on c12: 1 outie r5c3 + 3 = 1 innie r9c2
5a. but [69] blocked by [961] in 16(3)r9c2
5b. = [58/47/36/25]
5c. r5c3 = (2345), r9c2 = (5678)

6. 17(3)r6c2 must have 5 or 7 for r6c2
6a. = {278/359/458/467} = 2 or 5 or 6
6b. (no eliminations yet)

7. "45" on c12: 3 innies r459c2 = 13
7a. but {157} blocked by r6c2 = (57)
7b. {256} blocked by 17(3)r6c2
7c. {247} and {346} as {24}[7] and {34}[6] only, are both blocked by 10(3) can't be {244} nor {343}
7d. = {148/238} only
7e. -> r9c2 = 8
7f. -> r5c3 = 5 (iodc12=+3)

much easier now
Cheers
Ed
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