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 Post subject: Assassin 443
PostPosted: Sun Oct 01, 2023 6:54 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
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Assassin 443

Got so beguiled by the many interesting interactions possible that took a long time to see the "easy" way to crack this. Really enjoyed it. JSudoku has a hard time but SS gives it 1.40 [edited to correct numbering on image]
triple click code:
3x3::k:8448:8448:8448:3841:3841:8194:8194:5635:5635:8448:3844:8448:8448:8194:8194:5635:9733:9733:3844:3844:4614:8194:8194:5635:5635:9733:9733:3079:3079:4614:8194:3336:3336:9733:9733:9733:3079:6665:4614:4614:2826:2826:2059:5388:5388:6665:3853:3853:3853:6670:2831:2059:5388:2320:4625:6665:3602:3602:6670:2831:2831:2323:2320:4625:6665:3602:3602:6670:3092:6933:2323:6933:4625:6665:6665:6670:6670:3092:6933:6933:6933:
solution:
+-------+-------+-------+
| 6 5 9 | 7 8 2 | 4 3 1 |
| 2 8 7 | 4 1 3 | 5 6 9 |
| 4 3 1 | 9 5 6 | 7 2 8 |
+-------+-------+-------+
| 3 2 6 | 8 4 9 | 1 7 5 |
| 7 1 8 | 3 6 5 | 2 9 4 |
| 5 9 4 | 2 7 1 | 6 8 3 |
+-------+-------+-------+
| 8 4 2 | 1 9 7 | 3 5 6 |
| 1 7 5 | 6 3 8 | 9 4 2 |
| 9 6 3 | 5 2 4 | 8 1 7 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 443
PostPosted: Thu Oct 05, 2023 1:29 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for this Assassin! Loved it, my walkthrough kept flowing!

Here's how I solved Assassin 443:
Prelims

a) R1C45 = {69/78}
b) R4C56 = {49/58/67}, no 1,2,3
c) R5C56 = {29/38/47/56}, no 1
d) R56C7 = {17/26/35}, no 4,8,9
e) R67C9 = {18/27/36/45}, no 9
f) R78C8 = {18/27/36/45}, no 9
g) R89C6 = {39/48/57}, no 1,2,6
h) 21(3) cage at R5C8 = {489/579/678}, no 1,2,3
i) 11(3) cage at R6C6 = {128/137/146/236/245}, no 9
j) 14(4) cage at R7C3 = {1238/1247/1256/1346/2345}, no 9

1a. 45 rule on C123 5 outies R25678C3 = 16 = {12346}, locked for C4, clean-up: no 9 in R1C5
1b. 45 rule on N1 1 outie R2C4 = 1 innie R3C3 + 3 -> R2C4 = {46}, R3C3 = {13}
1c. 45 rule on R6789 1 innie R6C8 = 2 outies R5C27 + 5 -> R6C8 = {89}, R5C28 = {12/13}, 1 locked for R5, clean-up: no 1,2,3 in R6C7
[Ed and wellbeback both extended this with what is effectively the short forcing chain
21(3) cage at R5C8 = {489/579/678}
Consider combinations for R5C27 = {12/13}
R5C27 = {12} => R6C7 = {67}, R6C8 = 8 => 21(3) cage = {489} (cannot be {678} which clashes with R6C7)
or R5C27 = {13} => R6C7 = {57}, R6C8 = 9 => 21(3) cage = {489} (cannot be {579} which clashes with R6C7)
-> 21(3) cage = {489}, locked for N6
It didn’t matter that I didn’t spot that, I got that result in step 4g without trying; I don’t usually look for early forcing chains.]

1d. 1 in N5 only in R6C456, locked for R6, clean-up: no 8 in R7C9
1e. 45 rule on R1234 4(1+3) innies R3C3 + R4C123 = 12
1f. R3C3 = {13} -> R4C123 = 9,11, no 9
1g. 45 rule on R1234 3 outies R5C134 = 18 = {279/369/378/459/468/567}
1h. 2 on {279} must be in R5C4 -> no 2 in R5C13
[Note. 18(4) cage at R3C3 and R5C134 = 18 share R5C34 -> R34C3 = R5C1 may be useful later.]

2a. 45 rule on N1236 2(1+1) outies R4C4 + R7C9 = 1 innie R3C3 + 13
2b. R3C3 = {13} -> R4C4 + R7C9 = 14,16 = [77/86/95/97] -> R4C4 = {789}, R7C9 = {567}, R6C9 = {234}
2c. 45 rule on N9 1 outie R6C9 = 1 innie R7C7 = {234}
2d. 32(7) cage at R1C6 = {1234589/1234679/1235678}
2e. R1C45 = {78} (cannot be [96] which clashes with 32(7) cage), locked for R1 and N2
2f. 32(7) cage must contain at least one of 7,8, only in R4C4 -> 32(7) cage = {1234589/1234679}, R4C4 = {78}
2g. Naked pair {78} in R14C4, locked for C4
2h. R4C4 + R7C9 = [77/86], R7C9 = {67} -> R6C9 = {23}, clean-up: no 4 in R7C7 (step 1j)
2i. R4C4 + R7C9 = [77/86] = 14 -> R3C3 = 1, R2C4 = 4 (step 1b)
2j. Hidden killer triple 1,2,3 in R4C789, R5C7 and R6C9 for N6, R5C7 = {123}, R6C9 = {23} -> R4C789 must contain one of 1,2,3
2k. Hidden killer triple 1,2,3 in R4C123 and R4C789 for R4, R4C789 contains one of 1,2,3 -> R4C123 must contain two of 1,2,3
2l. Killer triple 1,2,3 in R4C123 and R5C2 for N4
2m. R3C3 = 1 -> R4C123 (step 1e) = 11 = {128/137/236} (cannot be {146/235} which only contain one of 1,2,3), no 4,5
2n. 12(3) cage at R4C1, R4C123 = 11 -> R5C1 = R4C3 + 1, no 2 in R4C3, no 5,6 in R5C1
2o. R4C123 = {128/137/236}
2p. 8 of {128} must be in R4C3 -> no 8 in R4C12
2q. 4 in N1 only in R3C12, locked for R3
2r. 15(3) cage at R2C2 contains 4 = {249/348/456}, no 7
2s. 7 in N1 only in R2C13, locked for R2
2t. 32(7) cage = {1234589/1234679} -> R1C7 = 4
2u. 32(7) cage = {1234589/1234679}, 2,3,9 locked for N2

3a. 45 rule on N5 5 innies R45C4 + R6C456 = 21 contains 1 for N5 = {12378/12567/13458/13467} (cannot be {12369/12459} because R4C4 only contains 7,8, cannot be {12468} which clashes with R4C56), no 9
3b. 9 in N5 only in R4C56 = {49} or R5C56 = {29} -> R5C56 = {29/38/56} (cannot be {47}, locking-out cages), no 4,7

4a. R4C4 + R7C9 (step 2h) = [77/86], R6C9 = R7C7 (step 2c)
4b. Consider permutations for R4C4 + R7C9
R4C4 + R7C9 = [77] => R1C45 = [87], R6C9 = 2, R7C7 = 2 => R67C6 = 9 = {18/36/45} => 7 in C6 only in R89C6 = {57}, 5 locked for C6 => R67C6 = {18/36}
or R4C4 + R7C9 = [86], R1C45 = [78], R6C9 = 3, R7C7 = 3 => R67C6 = 8 = {17/26} => 8 in C6 only in R89C6 = {48}
-> R89C6 = {48/57}, no 3,9, no 4,5 in R67C6
4c. Consider placements for R3C6 = {56}
R3C6 = 5 => R89C6 = {48}, locked for C6
or R3C6 = 6, R67C6 = {17/18}, killer pair 7,8 in R67C7 and R89C6, locked for C6
-> no 8 in R45C6, clean-up: no 5 in R4C5, no 3 in R5C5
4d. Consider again placements for R4C4 + R7C9
R4C4 + R7C9 = [77] => R1C45 = [87], R6C9 = 2, R7C7 = 2 => R67C6 = 9 = {18/36} => 7 in C6 only in R89C6 = {57}, 5 locked for C6
or R4C4 + R7C9 = [86]
-> R4C56 = {49/67}, no 5,8
4e. R4C123 (step 2o) = {128/137/236}
4f. R4C56 = {49} (cannot be {67} because R4C456 = 8{67} clashes with R4C123), locked for R4 and N5, clean-up: no 2 in R5C56
4g. 4,9 in N6 only in 21(3) cage at R5C8 = {489}, 4 locked for R5, 8 locked for N6
4h. 38(7) cage at R2C8 = {1256789}, no 3, 8,9 locked for N3
4i. 22(5) cage at R1C8 = {13567}, no 2 -> R3C7 = 7, clean-up: no 1 in R5C7
4j. 7 in N6 only in R4C89, locked for R4 -> R4C4 = 8, R1C45 = [78], R4C123 = {236}, locked for N4, 2,6 locked for R4, clean-up: no 3 in R5C6
4k. R6C7 = 6 (hidden single in N6) -> R5C7 = 2, R6C9 = 3 -> R7C9 = 6, R7C7 = 3
4l. 22(5) cage = {13567} -> R1C8 = 3, R3C6 = 6, R5C56 = [65],
4m. R7C7 = 3 -> R67C7 = 8 = {17}, locked for C7

5a. Naked pair {48} in R89C6, 4 locked for C6 and N8 -> R4C56 = [49], R12C6 = [23]
5b. R34C3 + R5C4 = [163] = 10 -> R5C3 = 8 (cage sum)
5c. Naked triple {569} locked for N1, 5 locked for R1
5d. R1C9 = 1 -> R24C7 = [51]
5e. R3C45 = {59}, 9 locked for R3 and N2 -> R2C5 = 1
5f. Naked pair {28} in R3C89, locked for R3 and N3 -> R2C89 = [69], R5C89 = [94], R6C8 = 8, R3C89 = [28], R5C1 = 7, clean-up: no 1,7 in R78C8
5g. R3C12 = {34} = 7 -> R2C2 = 8 (cage sum), R2C13 = [27], R4C12 = [32], R3C12 = [43]
5h. Naked pair {45} in R78C8, locked for N9, 5 locked for C8
5i. 18(3) cage at R7C1 = {189} (only remaining combination), 1,9 locked for N7, 9 locked for C1
5j. R6C23 = {49} (hidden pair in R6) -> R6C4 = 2 (cage sum)
5k. R6C1 = 5 -> R789C2 = {467}, 4 locked for C2 and N7
5l. R78C4 = [16] = 7 -> R78C3 = 7 = {25}

and the rest is naked singles.


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 Post subject: Re: Assassin 443
PostPosted: Tue Oct 10, 2023 6:35 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Great you enjoyed this one so much Andrew! Enjoyed your solution! I missed your really helpful 2e but managed to get the same two next placements as you. I went a different way after that.

Alternate middle to A443
After Andrew's step 3b:
.-------------------------------.-------------------------------.-------------------------------.
| 23569 23569 23569 | 78 78 123569 | 4 123569 123569 |
| 2356789 235689 2356789 | 4 123569 123569 | 1235689 1235689 1235689 |
| 2345689 2345689 1 | 59 23569 56 | 2356789 2356789 2356789 |
:-------------------------------+-------------------------------+-------------------------------:
| 12367 12367 3678 | 78 456789 456789 | 12356789 123456789 123456789 |
| 4789 123 456789 | 236 235689 235689 | 123 456789 456789 |
| 456789 456789 456789 | 1236 123456789 12345678 | 567 89 23 |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 2345678 | 1236 123456789 12345678 | 23 12345678 67 |
| 123456789 123456789 2345678 | 1236 123456789 345789 | 12356789 12345678 123456789 |
| 123456789 123456789 23456789 | 59 123456789 345789 | 12356789 123456789 123456789 |
'-------------------------------.-------------------------------.-------------------------------'


End of Andrew's step 3b above. Paste candidates into A443 in SudokuSolver

4. "45" on r6789: 4 outies r5c2789 = 16 and must have 1 and one of 2,3
4a. but {1267/1357} blocked by combo crossover clash with r6c7 (CCC)
4b. and {1258} blocked by 21(3)r5c8 can't be {58}[8]
4c. = {1249/1348}(no 5,6,7)
4d. must have 4 -> 21(3)n6 = {489} only: all locked n6, 4 locked for r5

5. 4 and 9 in r4 only in 13(2)r4c5 = {49}: both locked for n5

6. h18(3)r5c134 must have 7 for r5 = {279/378/567}
6a. 7 locked for n4
6b. but {567} as [756] blocked by must also have 6 in r4c3 (cage sum)
6c. = {279/378}(no 5,6)

7. "45" on r1234: 1 remaining innie r4c3 + 1 = 1 outie r5c1 = [67/89]

8. 12(3)n4 = {129/237}(no 6)
8a. 2 locked for r4, n4

9. 5 in r5 only in 11(2)n5 = {56}: both locked for n5

10. naked pair {56} in r35c6: both locked for c6

11. 5 in r4 only in r4c789: locked for n6 and 38(7)r2c8
11a. no 3 in r5c7

12. 11(3)r6c6 = {128/137}
12a. -> r67c6 = {18/17}(no 2,3,4)
12b. 1 locked for c6

13. 12(2)n8 = {39/48}(no 7) = 4 or 9
13a. -> killer pair 4,9 with r4c6: 9 locked for c6

14. naked pair {23} in r12c6: both locked for n2, 3 for c6

15. 12(2)n8 = {48}

on from there
Cheers
Ed


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 Post subject: Re: Assassin 443
PostPosted: Wed Oct 11, 2023 3:05 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
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Thanks Ed! I too missed the first easy step (Andrew Step 1a!) until very late. So here is my WT without using it...
(Started by using your first step in the alternate middle :))

Assassin 443 WT:
1. IOD r6789 -> r6c8 = r5c27 + 5
-> Either r5c27 = {12} and r6c8 = 8, in which case r6c7 from (67) -> 21(3)n6 = [{49}8]
Or r5c27 = {13} and r6c8 = 9, in which case r6c7 from (57) -> 21(3)n6 = [{48}9]
I.e., 21(3)n6 = {489} with 4 in r5c89

2. IOD n9 -> r6c9 = r7c7
-> r7c7 not from (489}
38(7)n36 must have (89) which can only go in n3 in r23c89
-> (HP (89) in c7) r89c7 = {89}
Also -> 4 in c7 only in r123c7
-> 4 not in 38(7)n36
-> 38(7)n36 = {1256789}

3. 32(7)r1c6 is missing two values = +13(2)
Since r1c4 sees all cells in 32(7), that must be one of the missing values.
Since r1c45 = +15(2) -> value in r1c5 must be in 32(7), which can only be in r4c4
-> r4c4 = r1c5 and be from (6789)

4. Innies r1234 = r3c3 + r4c123 = +12(4)
3 in r4 only in r4c123
-> No (89) in r4c123
-> (89) in r4 in r4c567
-> 11(2)n5 = {56}
-> r4c567 = [8{49}]
-> 15(2)n2 = [78]
[Moderator’s note. Therefore 32(7)r1c6 = {1234589}. This makes step 5 easier to follow.]

5. IOD n36 -> r3c6 = r1c7 + r6c9 - 1
Since r67c9 cannot be [18] -> r6c9 is 2 or more -> r3c6 > r1c7
-> Whatever value is in r1c7 can only go in n2 in r2c4
-> value in r3c6 not in 32(7)
-> The two values not in 32(7) = [r1c4,r3c6] = [76]
-> 6 in n3 in r2c89
Also -> 22(5)r1c8 = {367(15|24)) with 6 in r3c6

6. Value in r1c7 less than r3c6. I.e., is max 5
The same value is in r2c4
IOD n1 -> r2c4 = r3c3 + 3
-> [r2c4,r1c3] from [41] or [52]

7. 7 in n3 in the 22(5)
-> 7 in 38(7) must be in n6 in r4c89
-> 8(2)n6 not {17}
-> r5c2 = 1 (Step 1)

8. Remaining IOD n7 -> r6c1 + 2 = r7c3 + r8c3
Since the value in r6c1 must be the same as one of the values r7c3 or r8c3 -> the other one in r78c3 must be 2.
I.e., 2 in r78c3
-> r3c3 = 1
-> r2c4 = r1c7 = 4
-> 22(5)r1c8 = {13567}
-> r23c89 = {2689}
-> r4c789 = {157}
-> 8(2)n6 = [26] and 21(3)n6 = [{49}8]
-> 9(2)c9 = [36] and r7c7 = 3
-> (HS 3 in n5) r5c4 = 3
-> 18(4)r3c3 = [1683]
-> 12(3)n4 = [{23}7]
etc.


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