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 Post subject: Assassin 416
PostPosted: Sun May 01, 2022 8:19 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 940
Location: Sydney, Australia
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Diagonals so 1-9 cannot repeat on either of them.

Assassin 416
Held me up on the first attempt but then realised it is an easier one. Reluctant to make zero killers these days since they potentially close up solution paths, unless they really prolong the puzzle. Not sure this one does, unlike a415....Anyway, SS gives it 1.50 and JSudoku uses 3 "advanced" steps.

triple click code:
3x3:d:k:0000:0000:0000:0000:1795:1796:5125:5125:7174:0000:0000:0000:0000:1795:1796:5125:7174:3080:0000:0000:0000:0000:2313:7174:7174:3080:3080:2314:4875:0000:8204:2313:8204:7174:1805:1805:2314:4875:1294:1294:8204:3087:3087:7440:7440:4875:4875:6161:8204:2834:8204:5651:1812:7440:3605:3605:6161:6161:2834:5651:5651:1812:7440:3605:6161:5378:4615:1537:1537:4608:5651:7440:6161:5378:5378:4615:4615:1537:4608:4608:5651:
solution:
+-------+-------+-------+
| 3 9 1 | 7 2 4 | 5 6 8 |
| 8 7 4 | 6 5 3 | 9 1 2 |
| 6 2 5 | 1 8 9 | 4 3 7 |
+-------+-------+-------+
| 5 8 9 | 2 1 7 | 6 4 3 |
| 4 1 2 | 3 6 5 | 7 8 9 |
| 7 3 6 | 9 4 8 | 2 5 1 |
+-------+-------+-------+
| 9 4 3 | 8 7 6 | 1 2 5 |
| 1 5 7 | 4 3 2 | 8 9 6 |
| 2 6 8 | 5 9 1 | 3 7 4 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 416
PostPosted: Thu May 05, 2022 4:43 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1812
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. This zero killer is fine, it presumably took away some easy steps but still leaves plenty of useful 45s.

Here's my walkthrough for Assassin 416:
Prelims

a) R12C5 = {16/25/34}, no 7,8,9
b) R12C6 = {16/25/34}, no 7,8,9
c) R34C5 = {18/27/36/45}, no 9
d) R45C1 = {18/27/36/45}, no 9
e) R4C89 = {16/25/34}, no 7,8,9
f) R5C34 = {14/23}
g) R5C67 = {39/48/57}, no 1,2,6
h) R67C5 = {29/38/47/56}, no 1
i) R67C8 = {16/25/34}, no 7,8,9
j) 20(3) cage at R1C7 = {389/479/569/578}, no 1,2
k) 21(5) cage at R8C3 = {489/579/678}, no 1,2,3
l) 6(3) cage at R8C6 = {123}
m) 32(5) cage at R4C4 = {26789/35789/45689}, no 1

1a. Naked triple {123} in 6(3) cage at R8C6, locked for N8, clean-up: no 8,9 in R6C5
1b. 32(5) cage at R4C4 = {26789/35789/45689}, 8,9 locked for N5, clean-up: no 1 in R3C5, no 3,4 in R5C7
1c. Killer triple 1,2,3 in R12C6 and R89C6, locked for C6, clean-up: no 9 in R5C7
1d. 45 rule on N8 3 innies R7C456 = 21 = {489/579/678}
1e. 45 rule on N4 3 innies R456C3 = 17 = {179/269/278/359/368/458/467}
1f. R5C3 = {1234} -> no 1,2,3,4 in R46C3
1g. 45 rule on N3 2 outies R3C6 + R4C7 = 15 = {69/78}
1h. R3C6 + R4C7 = 15 -> R1C9 + R2C8 + R3C7 = 13
1i. 45 rule on N7 2 outies R6C3 + R7C4 = 14 = {59/68}
1j. R6C3 + R7C4 = 14 -> R7C3 + R8C2 + R9C1 = 10 = {127/136/145/235}, no 8,9
1k. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 22 = {589/679}, 9 locked for N5 and D/
1l. R4C6 + R5C5 + R6C4 = 22 -> R4C4 + R6C6 = 10 = [28/37/46/64]
1m. 45 rule on C5 3 innies R589C5 = 18 = {189/239/369/378} (cannot be {459/468/567} because R8C5 only contains 1,2,3), no 4,5

2a. 45 rule on R456789 3 innies R4C357 = 16 = {169/178/259/268/349/358/367/457}
2b. 1,2,3,4 only in R4C5 -> R4C5 = {1234}, clean-up: no 2,3,4 in R3C5
2c. 45 rule on N5 4 innies R46C5 + R5C46 = 13 = {1237/1246/1345}
2d. 7 of {1237} must be in R5C6 -> no 7 in R6C4, clean-up: no 4 in R7C5
2e. Consider placement for 5 in C5
R12C5 = {25}, 2 locked for C5, R46C5 cannot be [13] because R34C5 = [81] clashes with R67C5 = [38] => R46C5 + R5C46 cannot be {1237}
or R34C5 = [54] => R46C5 + R5C46 = {1246/1345}
or R67C5 = {56} => R46C5 + R5C46 = {1246/1345}
-> R46C5 + R5C46 = {1246/1345}, no 7, 4 locked for N5, clean-up: no 6 in R4C4 + R6C6 (step 1l), no 5 in R5C7
2f. 45 rule on C789 3 outies R357C6 = 20 = {479/569} (cannot be {578} which clashes with R6C6), no 8, 9 locked for C6, clean-up: no 7 in R4C7 (step 1g)
2g. R5C6 = {45} -> no 4,5 in R7C6
2h. 8 in C6 only in R46C6, locked for N5
2i. Consider combinations for R357C6 = {479/569}
R357C6 = {479}, locked for C6 => R6C6 = 8, R4C6 + R5C5 + R6C4 (step 1k) = {679} => R4C6 = 6
or R357C6 = {569} => R5C6 = 5
-> R46C5 + R5C46 = {1345}, 3,5 locked for N5, clean-up: no 7 in R3C5, no 3 in R5C3, no 5,9 in R7C5
[Cracked. Fairly straightforward from here.]
2j. R4C6 + R5C5 + R6C4 = {679}, 6,7 locked for D/, 7 locked for N5
2k. R4C4 = 2, R6C6 = 8, both placed for D\, clean-up: no 5 in R4C89, no 7 in R5C1, no 6 in R7C4 (step 1i)
2l. Naked triple {679} in R347C6, 6 locked for C6, clean-up: no 1 in R12C6
2m. 1 in C6 only in R89C6, locked for N8
2n. 8 on D/ only in R1C9 + R2C8 + R3C7, locked for N3 and 28(5) cage at R1C9, clean-up: no 7 in R3C6 (step 1g)
2o. Naked pair {69} in R3C6 + R4C7, CPE no 6 in R4C6 -> R4C6 = 7, clean-up: no 2 in R5C1
2p. R37C6 = {69} -> R5C6 = 5, R5C7 = 7, clean-up: no 2 in R12C6, no 4 in R4C1, no 6 in R7C5
2q. 20(3) cage at R1C7 = {479/569}, no 3, 9 locked for N3
2r. 7 of {479} must be in R1C8 -> no 4 in R1C8
2s. 10(3) cage at R7C3 = {145/235}, 5 locked for N7 and D/
2t. Naked pair {34} in R12C6, locked for N2, 3 locked for C6
2u. R8C5 = 3 (hidden single in N8), R6C5 = 4 -> R7C6 = 7, R4C5 = 1 -> R3C5 = 8, R5C4 = 3 -> R5C3 = 2, clean-up: no 6 in R12C5, no 6 in R4C1, no 6 in R4C89, no 8 in R5C1, no 3 in R7C8
2v. Naked pair {25} in R12C5, 5 locked for N2
2w. R456C3 (step 1e) = 17, R5C3 = 2 -> R46C3 = 15 = {69}, locked for C3 and N4, clean-up: no 3 in R4C1, no 9 in R7C4 (step 1i)
2x. Naked pair {34} in R4C89, locked for R4 and N6, clean-up: no 4 in R7C8
2y. Naked pair {69} in R7C6 + R8C5, locked for N8
2z. 21(3) cage at R8C3 = {489/678} -> R9C2 = {69}, R89C3 = {478}, 8 locked for C3 and N8
2aa. Naked pair {69} in R9C25, locked for R9
2ab. Naked pair {58} in R4C12, locked for N4
2ac. Naked pair {14} in R5C12, 1 locked for R5 and N4
2ad. Naked pair {69} in R6C34, 6 locked for R6, clean-up: no 1 in R7C8
2ae. 8 in N6 only in R5C89, locked for 29(5) cage at R5C8
2af. 8 in N9 only in 18(3) cage at R8C7 = {189/378/468}, no 2,5
2ag. 6,9 of {189/468} must be in R8C7 -> no 1,4 in R8C7
2ah. 7 of {378} must be in R9C8 -> no 3 in R9C8

3a. R7C4 = 8 (hidden single in R7) -> R6C3 = 6 (step 1i), R4C3 = 9, R6C4 = 9 -> R5C5 = 6, placed for D\, R4C7 = 6 -> R3C6 = 9, R7C6 = 6, R9C5 = 9, clean-up: no 1 in R6C8
3b. R9C2 = 6 -> R89C3 = 15 = {78}, 7 locked for C3 and N7
3c. Naked pair {89} in R5C89, 9 locked for 29(5) cage at R5C8
3d. R5C9 = 9 (hidden single in C9) -> R5C8 = 8
3e. R8C9 = 6 (hidden single in R8), R5C89 = 17 -> R67C9 = 6 = [15/24/51]
3f. R7C8 = 2 (hidden single in N9) -> R6C8 = 5 -> R67C9 = [15/24]
3g. Naked pair {12} in R6C79, CPE no 1 in R9C9
3h. R8C7 = {89} -> R9C78 = 9,10 = {18}/[37], no 4
3i. Killer pair 7,8 in R9C3 and R9C78, locked for R9
3j. Naked triple {345} in R479C9, locked for C9
3k. R1C9 = 8 (hidden single in C9)
3l. R2C1 = 8 (hidden single in N1), R4C1 = 5 -> R5C1 = 4, R5C2 = 1
3m. 7 in C9 only in R23C9, locked for N3
3n. R23C9 = {17/27} -> R3C8 = {34}
3o. 20(3) cage at R1C7 = {569} (only remaining combination) -> R12C7 = {59}, locked for C7
3p. R8C7 = 8 -> R9C78 = 10 = [37]

4a. Naked pair {34} in R34C8, locked for R8 -> R2C8 = 1, R3C7 = 4 (cage sum), both placed for D/
4b. R67C7 = [21], R8C8 = 9, R9C9 = 4 (cage sum), 1,4,9 placed for D\

and the rest is naked singles.


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 Post subject: Re: Assassin 416
PostPosted: Sun May 08, 2022 5:34 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 247
Location: California, out of London
My path is quite different from Andrew's I think. Again it took me a while to find the way in. Here's how I did it. None of the steps were particularly advanced ...
Assassin 416 WT:
1. Innies n3 (= D/n3) = +13(3)
Innies n7 (= D/n7) = +10(3)
-> D/ in n5 = +22(3) = {589} or {679}
-> r4c4,r6c6 = +10(2)

2. 6(3)n8 = {123}
Two of (123) in the two 7(2)s in n2
-> Those same two of (123) in n5/c4 in r45c4
-> [r4c4,r6c6] either [28] or [37]

3! Given (789) in n5 all in the 32(5) and 3 locked in c6 in r1289c6
-> 12(2)r5c6 either [48] or [57]
Outies c789 = r357c6 = +20(3)
-> Either r5c6 = 4 and r357c6 = <749>, or r5c6 = 5 and D/n5 = {679}
In neither case can r6c6 = 7
-> 32(5)n5 = [2{679}8]

4. -> 8 in D/ in n3 = {148} or {238}
-> 5 in D/ in n7 = {235} or {145}
-> Outies n7 = r6c3,r7c4 = +14(2) = [68]
-> (HS 8 in c5) 9(2)r3c5 = [81]
-> 1 in n8 in r89c6
-> The 7(2)s in n2 are {25} and {34}
-> (HS 3 in c4) 5(2)r5c3 = [23]
-> (Innies n4) r4c3 = 9

5. Also (HP 45 in c4) 18(3)n8 = [{45}9]
-> 9 in n5 only in r6c4
-> 9 in n2 only in r3c6
-> (Outies n3) r4c7 = 6
-> r5c5 = 6 and r4c6 = 7
-> 11(2)r6c5 = [47]
-> 12(2)r5c6 = [57]
Also 7(2)n6 = {34}
-> (89) in n6 only in r5c89
Also (NS) r7c6 = 6
Also 7(2)r1c5 = {25}
-> 7(2)r1c6 = {34}
-> 6(3)n8 = [3{12}]
Also r6c789 = {125}
-> r6c12 = {37}
Also r4c12 = {58} and r5c12 = {14}

6. Since r7c456 = [876] -> At least two of (678) in 21(3)n7
-> 21(3)n7 = <768> (I.e., r9c2 = 6)
-> 6 in n9 only in r8c9
-> r67c9 = +6(2) = {15} or [24]
-> 8 in n9 in 18(3)n9
-> 2 in n9 only in r7c8
-> r6c8 = 5 and r6c79 = {12}
-> 1 not in r9c9

7. 8 in D/ in r1c9 or r2c8
Since (67) already in c7 -> 20(3)n3 = <569> or <479> (I.e., 9 in n3 in r12c7)
-> 9 in n9 only in r8c8
-> 22(5)r6c7 can only be [26194]
-> r67c9 = [15]
-> 18(3)n9 = [837]
-> r1c8 = 6 and r12c7 = {59}

Straightforward from here


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 Post subject: Re: Assassin 416
PostPosted: Wed May 11, 2022 6:18 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 940
Location: Sydney, Australia
This ended up quite a decent puzzle. My key steps are quite different.

First though, really enjoyed wellbeback's step 2. Had to think hard about it. Here's the way it made sense to me.
step 2 unpacked:
a. Killer triple 123 in r1289c6: all locked for c6
b. 6(3)n8 from {123} -> whichever of 1,2,3 is in r12c6 must be in r8c5
c. -> that same digit must be in n5 only in r45c4

d. also, the 1,2 or 3 in r12c5 can only be in n5 in r45c4
e. -> r45c4 from {123}
My start to a416:
Preliminaries from SudokuSolver
Cage 5(2) n45 - cells only uses 1234
Cage 7(2) n2 - cells do not use 789
Cage 7(2) n2 - cells do not use 789
Cage 7(2) n6 - cells do not use 789
Cage 7(2) n69 - cells do not use 789
Cage 12(2) n56 - cells do not use 126
Cage 9(2) n25 - cells do not use 9
Cage 9(2) n4 - cells do not use 9
Cage 11(2) n58 - cells do not use 1
Cage 6(3) n8 - cells ={123}
Cage 21(3) n7 - cells do not use 123
Cage 20(3) n3 - cells do not use 12
Cage 32(5) n5 - cells do not use 1

This is a highly optimised solution so any clean-up needed is stated.
1. "45" on n7: 2 outies r6c3 + r7c4 = 14 = {59/68} (h14(2)n48)
1a. -> r7c3 + r8c2 + r9c1 = 10 (h10(3)n7)
1b. = {127/136/145/235}(no 8,9)

2. "45" on n3: 2 outies r3c6 + r4c7 = 15 = {69/78} (h15(2)n26)
2a. -> r1c9 + r2c8 + r3c7 = 13 (h13(3)n3)

3. "45" on d/ with h13(3)n3 and h10(3)n7 -> 3 remaining innies r4c6 + r5c5 + r6c4 = 22 + {589/679}
3a. 9 locked for n5, d/, and 32(5)
3b. -> r4c4 + r6c6 = 10 (h10(2)n5), no 5

4. 6(3)n8 = {123}: all locked for n8

5. "45" on c56789: 4 outies r4689c4 = 20
5a. min. r689c4 = {456} = 15 -> max. r4c4 = 3 (ie, no 4 or 5 since they are already taken in the min. = 15)
5b. -> h10(2)n5 = [28/37]
5c. and 32(5)n5 = {26/35}{789}: 7,8 locked for n5
5d. no 4 in r7c5

6. "45" on c789: 3 outies r357c6 = 20
6a. max. r5c6 = 5 -> min. r37c6 = 15
6b. -> no 4,5 in r7c6

7. 4 in n8 only in 18(3) = {459/468}(no 7)

8. "45" on n4: 3 innies r456c3 = 17
8a. min. any two cells (17 - 9) = 8 -> can't have two of {1234}
8b. -> r4c3 from (5..9)

9. "45" on r4..9: 3 innies r4c357 = 16
9a. min. r4c37 = {56} = 11 -> max. r4c5 = 4

key steps
10. 7 in n8 only in r7c5 -> rc5 = 4; or 7 in n8 in r7c6 -> r6c6 = 8
10a. note: r6c56 = 4 or 8 (no eliminations yet)

11. alternating hidden killer pair 6,7 in n5. h22(3) has both 6,7; or they are both must be in r6c56 which is contrary to step 10a.
11a. -> h22(3) = {679}: 6,7 locked for n5: all locked for d/
11b. -> r6c6 = 8, placed for d\
11c. -> r4c4 = 2 (h10(2)n5), placed for d\

12. h13(3)n3 = {148/238}(no 5)
12a. 8 locked for d/, n3 and 28(5)
12b. -> h15(2)n26 = {69}
12c. -> no 6,9 in r4c6
12d. -> r4c6 = 7

13. 5 on d/ only in n7: locked for n7 and 24(5)
13a -> h14(2)n48 = [68]
13b. -> r5c5 + r6c4 = [69], 6 placed for d\

14. r7c5 = 7 (hsingle n7)
14a. -> r6c5 = 4

15. "45" on n8: 1 innie r7c6 = 6
15a. -> r3c6 = 9
15b. -> r4c7 = 6 (h15(2)n26)

16. 9(2)r3c5 = [81] (only permutation)
16a. -> r5c3467 = [2357]

17. 7(2)r1c5 = {25}: both locked for n2 and c5
17a. -> r89c5 = [39]
17b. and 7(2)r1c6 = {34}: 4 locked for n2

continue from here with Wellbeback's step 6
Cheers
Ed


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