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Ed
 Post subject: Assassin 445
PostPosted: Wed Nov 01, 2023 8:38 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
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Note: cells with no cage can have repeats as per standard sudoku rules.

Assassin 445
Long time since I made a "zero" killer. Jsudoku has a terrible time with this puzzle but SudokuSolver gives it 1.55. I found it a pretty quick one but with enough interesting steps to make it worthy.

triple click code:
3x3::k:3584:3584:3584:6145:3842:3843:2820:4613:4613:5382:3335:3335:6145:3842:3843:2820:9736:4613:5382:3335:6145:6145:3842:3843:9736:9736:9736:5382:3337:6145:2314:2314:9736:9736:4875:4875:3337:3337:1292:1292:5133:9736:9736:4875:0000:5135:5135:5135:5135:5133:5133:5648:0000:0000:2322:4115:7956:7956:7956:7956:5648:0000:0000:2322:4115:5649:5649:3086:7956:5648:0000:3605:2322:5649:4115:3086:3086:7956:5648:0000:3605:
solution: 
+-------+-------+-------+
| 8 2 4 | 1 6 3 | 5 9 7 |
| 5 3 1 | 9 7 8 | 6 4 2 |
| 7 9 6 | 5 2 4 | 1 3 8 |
+-------+-------+-------+
| 9 5 3 | 8 1 6 | 2 7 4 |
| 1 7 2 | 3 4 5 | 9 8 6 |
| 6 4 8 | 2 9 7 | 3 5 1 |
+-------+-------+-------+
| 4 1 5 | 6 8 9 | 7 2 3 |
| 2 8 9 | 7 3 1 | 4 6 5 |
| 3 6 7 | 4 5 2 | 8 1 9 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 445
PostPosted: Sat Nov 04, 2023 4:10 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for a challenging Assassin. My solving path had an easy start, an easy finish but a pretty hard middle. I hope someone else managed to find a better way to solve the middle.

Which both Ed and wellbeback did. I've added a note about what I missed; if I'd found it, my solving path could have been like theirs, rather than harder and quite different.

Here's how I solved Assassin 445:
Prelims

a) R12C7 = {29/38/47/56}, no 1
b) R4C45 = {18/28/36/45}, no 9
c) R5C34 = {14/23}
d) R89C9 = {59/68}
e) 21(3) cage at R2C1 = {489/579/678}, no 1,2,3
f) 19(3) cage at R4C8 = {289/379/469/478/568}, no 1
g) 20(3) cage at R5C5 = {389/479/569/578}, no 1,2
h) 9(3) cage at R7C1 = {126/135/234}, no 7,8,9
i) 22(3) cage at R8C3 = {589/679}
j) 38(8) cage at R2C8 = {12345689}, no 7

1a. 45 rule on C1234567 3 outies R2C8 + R3C89 = 15, 45 rule on N3 4 innies R2C8 + R3C789 = 16 -> R3C7 = 1
1b. 45 rule on C7 2 remaining innies R45C7 = 11 = {29/38/56}, no 4
1c. 45 rule on C123456 2 innies R45C6 = 11 = {29/38/56}, no 4
1d. 38(8) cage at R2C8 = {12345689}, 4 locked for N3, clean-up: no 7 in R12C7
1e. 45 rule on N12 two outies R4C13 = 12 = {48/57}/[93], no 1,2,6, no 9 in R4C3
1f. 45 rule on N2 two outies R34C3 = 9 = [27/45/54/63], clean-up: no 4 in R4C1
1g. 45 rule on N124 two outies R56C4 = 5 = {14/23}
1h. Whatever is in R5C3 must be in R6C4 and in N6 in R4C789, no 1 in R4C789 -> no 1 in R5C3, no 1 in R6C4, clean-up: no 4 in R5C4
1i. 45 rule on R12345 2 innies R5C59 = 10 = {37/46}/[82/91], no 5, no 8,9 in R5C9
1j. 1 in N5 only in R4C45 = {18} or in R56C4 = [14] -> no 4 in R4C45 (locking-out cages), clean-up: no 5 in R4C45
[I ought also to have looked at 1 in N5 only in R4C45 = {18} or R5C34 -> R4C13 = [57/75/93] (cannot be {48}), no 4,8. Then I could have looked at combinations for the 15(3) cages in N2 including R123C4 = 15.]
1k. 22(3) cage at R8C3 = {589/679}, CPE no 9 in R8C2
1l. 45 rule on N7 1 outie R8C4 = 1 innie R7C3 + 2, no 1,2,8,9 in R7C3

2a. R4C13 = 12 (step 1f) = [57/75/84/93], R5C59 = 10 (step 1i)
2b. R5C34 = 5, R56C4 = 5 (step 1g) -> R5C3 + R6C123 = 20(4) cage at R6C1
2c. 20(4) cage = R5C3 + R6C123 = {1289/1379/1469/1478/2369/2468/2567} (cannot be {2378/2459/3458/3467} which clash with R4C13, cannot be {1568} because R6C4 only contains 2,3,4)
2d. Consider combinations for R56C4 = [14]/{23}
R56C4 = [14] => R4C2 = 1 (hidden single in R4) => 20(4) cage = R5C3 + R6C123 = {2369/2468/2567}
or R56C4 = {23}, 1 then only in R4C45 = {18}, locked for R4 => 20(4) cage = R5C3 + R6C123 = {1289/2369/2468/2567} (cannot be {1379} which clashes with R4C13 = [57/75/93] cannot be {1469/1478} because R6C4 then only contains 2,3)
-> 20(4) cage = R5C3 + R6C123 = {1289/2369/2468/2567}, 2 locked for R6 and N4
[Alternatively the eliminations of {1469/1478} can be made in a similar way to step 1h, cannot contain both of 1,4.]
2e. Consider placement for 1 in N5
1 in N5 only in R4C45 = {18}, 8 locked for R4
or in R56C4 = [14] => 20(4) cage = {268}4, 8 locked for N4)
-> R4C13 = [57/75/93], no 4,8, clean-up: no 5 in R3C3 (step 1f)
2f. Consider combinations for R45C7 (step 1b) = {29/38/56}
R45C7 = {29} blocks R5C3459 (step 1i) = {23}[91] => 1 in N6 only in R6C89, locked for R6
or R45C7 = {38}, locked for N6 and 3 in R6 in 20(4) cage = {2369}
or R45C7 = {38}, locked for N6 and 3 in R6 in 20(3) cage at R5C5, locked for N5 => R56C4 = [14] => 20(4) cage = {2468}
or R45C7 = {56} => R45C6 = {29/38} => R56C4 = [14] (cannot be {23} which then clashes with R56C4) => 20(4) cage = {2468}
-> 20(4) cage = R5C3 + R6C123 = {2369/2468/2567}, no 1, 6 locked for R6 and N4
2g. 1 in R6 only in R6C89, locked for N6, clean-up: no 9 in R5C5 (step 1i)
2h. Consider placement for 4 in N5
4 in 20(3) cage = {479}, 9 locked for R6
or R6C4 = 4 => 20(4) cage = {268}4
-> 20(4) cage = R5C3 + R6C123 = {2468/2567}, no 3,9, clean-up: no 2 in R5C4, no 3 in R5C3
2i. Consider permutations for R56C4 = [14/32]
R56C4 = [14] => 20(4) cage = {268}4
or R56C4 = [32] => R45C7 = {56}, 6 locked for N5, R5C3 = 2, R5C59 = [46] => 20(3) cage = 4{79}, 7 locked for R6
-> 20(4) cage = R5C3 + R6C123 = {2468}, 4,8 locked for R6 and N4
2j. 20(3) cage at R5C5 = {389/479/569/578}
2k. 4,6,8 only in R5C5 -> R5C5 = {468}, clean-up: no 3,7 in R5C9
2l. Killer pair 2,4 in R5C3 and R5C59, locked for R5, clean-up: no 9 in R4C6, no 9 in R4C7
2m. 4 in R4 only in R4C89, locked for N6, clean-up: no 6 in R5C5
2n. 19(3) cage at R4C8 = {469/478}, no 2,3,5
2o. 6 in N5 only in R4C45 = {36} or R45C6 (step 1c) = {56} -> no 3 in R45C6 (locking-out cages), clean-up: no 8 in R45C6
2p. 38(8) cage at R2C8 = {12345689}, CPE no 8 in R12C7, clean-up: no 3 in R12C7
2q. 6 in N5 only in R4C45 + R45C6, CPE no 6 in R4C7, clean-up: no 5 in R5C7
2r. 4 in C7 only in R789C7, locked for N9
2s. Consider combinations for 19(3) cage = {469/478}
19(3) cage = {469}, 6,9 locked for N6, R5C9 = 2 => R5C5 = 8, R45C7 = [83], R5C4 = 1, R4C23 = [13] (hidden pair in N4) => R4C1 = 9
or 19(3) cage = {478} => 9 in R4 only in R4C12
-> 9 in R4C12, locked for R4 and N4
[The last hard step, maybe the hardest step]
2t. 13(3) cage at R4C2 = {157} (cannot be {139} = 9{13} which clashes with R5C4), locked for N4 -> R4C13 = [93], R3C3 = 6, clean-up: no 6 in R4C45, no 8 in R5C7, no 5,8 in R8C4 (step 1l)
2u. 6 in N5 only in R45C6 = {56}, locked for C6 and 38(7) cage at R2C8, 5 locked for N5
2v. 9 in N5 only in R6C56, locked for R6
2w. Killer pair 2,8 in R4C45 and R4C7, locked for R4
2x. 19(3) cage = {469/478} -> R5C8 = {89}

[Fairly straightforward from here]
3a. 15(3) cage at R1C6 = {249/348}, no 1,7, 4 locked for C6 and N2
3b. 1 in C6 only in R789C6, locked for N8
3c. 12(3) cage at R8C5 = {237/246/345}, no 8,9
3d. 24(5) cage at R1C4 = {13569/23568}, no 7, 5 locked for C4 and N2
3e. Killer pair 8,9 in 24(5) cage and 15(3) cage at R1C6, locked for N2
3f. 7 in N2 only in 15(3) cage at R1C5 = {267} (only remaining combination), locked for C5, 2 locked for N2, clean-up: no 8 in 24(5) cage, no 9 in 15(3) cage at R1C6, no 2,7 in R4C45
3g. Naked triple {159} in R123C4, 1,9 locked for C4 -> R4C45 = [81], R5C3456789 = [2346986], R4C67 = [62], R6C456 = [297], clean-up: no 7 in R7C3 (step 1l)
3h. Naked pair {56} in R12C7, locked for C7 and N3 -> R6C7 = 3
3i. Naked triple {478} in R789C7, 7,8 locked for N9
3j. Naked pair {59} in R89C9, locked for C9 and N9 -> R6C89 = [51]
3h. Naked pair {35} in R89C5, locked for N8, R9C4 = 4 (cage sum)
3i. R7C5 = 8, R789C6 = {129} = 12 -> R7C34 = 11 = [56] (cannot be [47] which clashes with R7C7)
3j. R8C4 = 7 -> R8C3 + R9C2 = 15 = [96], R89C9 = [59], R89C5 = [35]
3k. 9(3) cage at R7C1 = {234} (only remaining combination), locked for C1 and N7
3l. R4C1 = 9 -> R23C1 = 12 = {57}, locked for N1, 7 locked for C1 -> R5C12 = [17], R78C2 = [18], R6C2 = 4
3m. R1C1 = 8 -> R1C23 = 6 = [24]
3n. R1C48 = [19] (hidden pair in R1) -> R12C9 = 9 = [72]

and the rest is naked singles.
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wellbeback
 Post subject: Re: Assassin 445
PostPosted: Tue Nov 07, 2023 7:31 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks again Ed! You always provide puzzles with interesting combinations!
My middle bit was quite different from Andrew's. Here's how I did it...
Assassin 445 WT:
1. Innies n2 = r123c4 = +15(3)
-> r34c3 = +9(2)
-> Remaining Outies n1 = r4c13 = +12(2)
-> Remaining Outies n4 = r56c4 = +5(2)
-> Remaining Innies n5 = r45c6 = +11(2)
-> Remaining Outies n3 = r45c7 = +11(2)
-> Remaining Innie c7 = r3c7 = 1
-> 1 in n6 in uncaged area in n6
Also 1 in n9 in uncaged area in n9

2. 38(8) has no 7
-> 4 in 38(8) in n3
-> 7 in n3 in 18(3)n3
-> Two values in r45c7 are in n3 in 18(3)n3
-> Two values in r45c6 are in n3 in 11(2)n3

3. Since r45c4 = +5(2) -> r5c3 = r6c4
-> Whatever that value is goes in n6 in r4c789
-> That value not 1
-> 5(2)r5 = [41] and r6c4 = 4 - or - 5(2)r5 = {23} and r45c4 = {32}
Combined cage H+11(2)r56c7 + 19(3)n6 = +30(5)
That cannot have two values = +5
-> (Since value from r5c3 is in r4c789) -> Value from r5c4 is in n6 in r6c789
-> Value from r5c4 is in n4 in r4c123

4. 1 in n5 only in r5c4 (-> 5(2)r5 = [41]) or in 9(2)n5 = {18}
Either way r4c13 = +12(2) cannot be {48}
-> r4c13 = [93] or {57}
-> r34c3 from [63], [27], or [45]

5! Each column in n2 is a +15(3)
Possible arrangements are {159}, {267}, {348} - or - {168}, {249}, {357}
Consider H+15(3)r123c4
It cannot be {357} since r4c3 from (357)
It cannot be {348} since one of (34) in r45c4
It cannot be {249} since one of (24) in r45c4
It cannot be {267} since that puts 5(2)r5 = [41] leaving no solution for r34c4 = +9(2)
-> H+15(3)r123c4 from {1(59|68)}
-> r45c4 = {23}
-> (HS 1 in n5) 9(2)n5 = [81]
-> H+15(3)r123c4 = {159}
-> r789c4 = {467}

6. Remaining 15(3)s in n2 are {267} and {348}
-> (Since one of (47) in r123c6) H+11(2)r45c6 = {56}
-> 15(3)r1c6 = {348} and 15(3)r1c5 = {267}
-> 20(3)n5 = [{49}7]
-> r789c6 = {129}
-> 12(3)n8 = <345>
-> r7c5 = 8
-> r78c4 = {67}
Also 11(2)n3 = {56}

7. Trying 5(2)r5 = [32] puts r4c13 = [57] and r4c2 = 2 (Step 3 last line) which leaves no solution for 13(3)n4
-> 5(2)r5 = [23]
-> r6c4 = 2
Also r34c3 = [63]
-> r4c1 = 9

8. Innies r12345 = r5c59 = +10(2) = [19] or [46]
-> 7 in n6 in 19(3)
-> (HS 2 in r4) r45c7 = [29]
-> r5c59 = [46]
-> r6c5 = 9 and r45c6 = [65]
Also 19(3)n6 = [{47}8]
-> 13(3)n4 = [5{17}]
-> r6c123 = {468} and r6c789 = [3{15}]
-> r789c7 = {478}
-> 14(2)n9 = {59}
-> 18(3)n3 = [9{27}]
-> r4c89 = [74]
-> r3c9 = 8 and r23c8 = {34}
Also -> 13(3)n1 = {139}
-> 15(3)r1c6 = [384]
-> 14(3)n1 = {248}
etc.
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Ed
 Post subject: Re: Assassin 445
PostPosted: Fri Nov 10, 2023 8:33 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Thanks for you kind observations wellbeback! We are so lucky to have two high quality software solvers so I don't have to put in too much effort to find interesting cage structures.

I was very close to wellbeback's solving path but didn't need his step 7. Found a different way (esp steps 21,22).
A445:
Preliminaries
Cage 5(2) n45 - cells only uses 1234
Cage 14(2) n9 - cells only uses 5689
Cage 9(2) n5 - cells do not use 9
Cage 11(2) n3 - cells do not use 1
Cage 22(3) n78 - cells do not use 1234
Cage 9(3) n7 - cells do not use 789
Cage 21(3) n14 - cells do not use 123
Cage 20(3) n5 - cells do not use 12
Cage 19(3) n6 - cells do not use 1
Cage 38(8) n356 - cells ={12345689}

Andrew did his start a more direct way

1. "45" on c7: 3 innies r345c7 = 12

2. "45" on whole grid -> 7 zero cells = 24
2a. "45" on n69 (remembering h12(3)c7) -> 1 outie r3c7 = 1
2b. -> r45c7 = 11 = {29/38/56}(no 4)

3. "45" on c123456: 2 innies r45c6 = 11 (no 4)

4. "45" on n124: 1 innie r5c3 = 1 outie r6c4
4a. but can't both be 1 since there is no 1 available for r4c789 (Empty Rectangle)
4b. -> are both (234)
4c. no 4 in r5c4
4d. also r56c4 = 5 = [14]/{23}

5. 1 in n5 only in 9(2) = {18} or in r5c4 -> r5c34 = [41]:
5a. note = 4 or 8

6. "45" on n12: 2 outies r4c13 = 12
6a. but {48} leaves no 1 for n5 (bit more direct than Andrew's 2e)
6b. = [93]/{57}

7. "45" on n2: 2 outies r34c3 = 9 = [27/45/63]

8. 3 innies n2, r123c4 = 15
8a. {267} blocked by [44] in r35c3 (step 4.)
8b. {249/348} blocked by r56c4 (step 4d)
8c. {357} blocked by r4c3 = (357)
8d. = {159/168/258/456}(no 3,7)

9. 3 in n2 must be in one of the other 15(3) = {348/357}
9a. -> {258/456} blocked from r123c4
9b. = {159/168}(no 2,4)
9c. 1 locked for n2 and c4

The original version of this puzzle was now completely cracked, hence, the zero killer version.
10. r56c4 = {23}: both locked for c4 and n5
10a. no 4 in r5c3

11. r4c5 = 1 (hsingle n5) -> r4c4 = 8
11a. -> r123c4 = {159} only (step 9b): 5 and 9 locked for n2, c4, and no 5 in r4c3 -> no 4 in r3c3
11b. no 7 in r4c1 (h12(2))

12. h11(2)r45c6 = {56} only: both locked for c6, n5 and 38(8)

13. h11(2)r45c7 sees all n3 apart from 18(2) so must repeat there along with 7 (no 5,6)

14. 11(2)n3 = {56} (hpair): both locked for c7

15. 5(2)r5c3 = {23}: both locked for r5

16. 22(3)r8c3 must have 6 or 7 for r8c4 = {679} only: 9 locked for n7, no 6,7 in r8c12

17. 15(3)r1c6 = {348} only: all locked for c6 and n2

18. 15(3)r1c5 = {267}: all locked c5

19. 20(3)n5 = {49}[7]: 4,9 locked for c5

20. 12(3)n8 = {345} only -> r9c4 = 4
20a. 3,5 locked for n8
20b. r7c5 = 8

21. killer pair 2,3 in r345c3: both locked for c3, 3 for n4

22. 13(3)n4: [2]{56} blocked by r5c6 = (56)
22a. [2]{47} blocked by r45c3 = {237}
22b. = {148/157}(no 2,6,9)
22c. 1 locked for n4 and r6

23. "45" on r12345: 2 innies r5c59 = 10 = [46] only

On from there
Cheers
Ed
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