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Ed
 Post subject: Assassin 449
PostPosted: Mon Jan 01, 2024 3:34 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
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a449.png
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An x puzzle, so no repeats on either diagonal.

Assassin 449
Happy New Year!!

A repeat of some of the big cages in a442 but that is where the similarities end. Hopefully I'll be over my infatuation with those shapes now! This gets 1.60 but a warning: JSudoku has to struggle and struggle which suggests a very narrow solving path to get started. I was stuck for some time but wasn't too complicated once I saw the way. Actually quite neat.
triple click code:
3x3:d:k:3840:3585:3585:3585:5122:4099:4099:6404:6404:2821:3840:3078:2823:5122:10760:4099:6404:6404:2821:3078:3078:2823:5122:10760:6404:2825:2825:2826:2826:5131:3078:10760:10760:10760:10760:2825:2826:5131:5131:7948:10760:10760:2829:2829:5398:2575:2575:7948:7948:7948:4368:5398:5398:5398:2575:7442:7948:7948:2323:2323:4368:5398:3092:7442:7442:7442:5141:2577:6926:6926:4368:3092:7442:7442:5141:5141:2577:2577:6926:6926:4368:
solution: 
+-------+-------+-------+
| 9 7 5 | 2 8 6 | 1 3 4 |
| 8 6 2 | 4 3 1 | 9 5 7 |
| 3 4 1 | 7 9 5 | 6 2 8 |
+-------+-------+-------+
| 2 3 7 | 5 6 8 | 4 9 1 |
| 6 9 4 | 1 7 2 | 3 8 5 |
| 5 1 8 | 9 4 3 | 2 7 6 |
+-------+-------+-------+
| 4 5 3 | 6 2 7 | 8 1 9 |
| 7 2 6 | 8 1 9 | 5 4 3 |
| 1 8 9 | 3 5 4 | 7 6 2 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 449
PostPosted: Thu Jan 04, 2024 10:37 pm 
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Quite neat indeed! Thanks Ed and Happy New Year to all.
Here's how I did it...
Assassin 449 WT:
1. Outies c1234 -> r6c5 = 4

2. Outies r6789 = r5c49 = +6(2) = {15} or [24]
-> 3 in n5 only in r4c4, r6c4, or r6c6
12(4)r2c3 from {1245} or {1236}
6 cannot be in 12(4) in n1 since that puts 15(2)n1 = {78} which leaves no place for both (45) in n1.
-> r4c4 cannot be 3
-> 3 in n5 in r6c46

3! Outies n1 -> r1c4 + r4c4 = +7(2)
Since r4c4 not 3 -> whatever value is in r4c4 must be in 42(8) in r23c6
Innies n2 = +14(4)
-> Remaining Innies n2 = r1c6 + other of r23c6 = +7(2)
Outies n3 = r1c6 + r4c9 = +7(2)
-> Whatever value is in r4c9 goes in n2 in r23c6. I.e., is in the 42(8)
-> r4c9 not 3
-> (HS 3 in n6) 11(2)n6 = {38}

4. From Step 2 we know the values in r4c49 are the same as in r23c6 and are both maximum 6.
-> 8 in 42(8) only in r4c56
-> (HS 8 in n4/r6) r6c3 = 8
-> (Remaining Innies n4) 10(3)r6c1 = [{15}4]
-> 11(3)n4 = {236} with 3 in r4c12
-> 20(3)n4 = {479}

5. Since Outies r6789 = r5c49 = +6(2) -> (15) in r5 either both in r5c49 or both in r5c56
But the latter case leaves no place for both (15) in r4 in n6
-> r5c49 = {15}
-> 4 in r4/n6 in r4c789
-> 4 in n4 in r5c23

6. Remaining Innies n7 = r79c3 = +12(2) = {57} or {39}
-> r5c2 from (79)
-> r5c3 = 4
-> 4 in n1 in r13c2

7. Innies D\ = [r3c3,r4c4,r5c5] = +13(3)
Max r3c3 + r4c4 = +9
-> Min r5c5 = 4
But since 4 already in n5 and 5 already in r5 -> Min r5c5 = 6
-> Max r4c4 = 5
-> 6 not in 12(4)r2c3
-> 12(4)r2c3 = {1245} with 4 in r3c2

8. 21(5)r5c9 cannot contain both (79)
-> One of (79) in r6 in r6c46.
-> r6c456 = <347> or <349>
-> r6c789 = {269} or {267}
(34) on D\ only in 17(4)D\ (With 4 in r8c8 or r9c9)
IOD n9 -> r6c6 + r8c6 = r7c8 + 11
-> r7c8 not the same as r6c6
-> r7c8 not 3 (Leaves no place for 3 in 17(4)D\)
-> 21(5)r5c9 = {12567} with r5c9,r7c8 = {15} and r6c789 = {267} (Only solution)
-> r6c456 = <349>

9. Whether r5c9 = 5 or r7c8 is 5 -> 12(2)n9 not {57}
-> 12(2)n9 = {39}
-> (Since 27(4) must contain a 9) r8c6 = 9
-> r6c46 = [93]
Also 12(2)n9 = [93]
Also -> (IOD n9) r7c8 = 1
-> r5c49 = [15]
-> r4c789 = {149}
-> (Outies n3) r1c6,r4c9 = [61]
-> 20(3)n2 = {389}
Also -> 9(2)n8 = {27} (Only solution)
-> r7c34 = [36]
Also 17(4)D\ = [3{248}]
-> 15(2)D\ = {69}
-> H+13(3)D\ = [157]
Etc.
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Andrew
 Post subject: Re: Assassin 449
PostPosted: Sun Jan 07, 2024 12:22 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. A tough one to find a way in; I spent too long looking to extend steps I was working on, although that's not obvious from my WT, before I found the way to crack it.

Happy New Year all, particularly to regular posters Ed and wellbeback!

Here's how I solved Assassin 449:
Prelims

a) 15(2) cage at R1C1 = {69/78}
b) R23C1 = {29/38/47/56}, no 1
c) R23C4 = {29/38/47/56}, no 1
d) R5C78 = {29/38/47/56}, no 1
e) R7C56 = {18/27/36/45}, no 9
f) R78C9 = {39/48/57}, no 1,2,6
g) 20(3) cage at R1C5 = {389/479/569/578}, no 1,2
h) 11(3) cage at R3C8 = {128/137/146/236/245}, no 9
i) 11(3) cage at R4C1 = {128/137/146/236/245}, no 9
j) 20(3) cage at R4C3 = {389/479/569/578}, no 1,2
k) 10(3) cage at R6C1 = {127/136/145/235}, no 8,9
l) 20(3) cage at R8C4 = {389/479/569/578}, no 1,2
m) 10(3) cage at R8C5 = {127/136/145/235}, no 8,9
n) 12(4) cage at R2C3 = {1236/1245}, no 7,8,9
o) 27(4) cage at R8C6 = {3789/4689/5679}, no 1,2
p) 42(8) cage at R2C6 = {12456789}, no 3

1a. 45 rule on C1234 1 outie R6C5 = 4, clean-up: no 5 in R7C6
1b. 45 rule on R6789 2 outies R5C49 = 6 = [15/24/51]
1c. 45 rule on N1 2 outies R14C4 = 7 = {16/25}/[43], no 3,7,8,9 in R1C4
1d. 45 rule on N3 2 outies R1C6 + R4C9 = 7 = {16/25/34}, no 7,8,9
1e. 45 rule on N4 1 innie R6C3 = 1 outie R7C1 + 4, no 1,2,3 in R6C3, no 6,7 in R7C1
1f. 45 rule on N2 4 innies R1C46 + R23C6 = 14 contains 1 = {1238/1247/1256/1346}, no 9
1g. 45 rule on N8 4 innies R789C4 + R8C6 = 26 must contain 9 for N8 = {2789/3689/4589/4679}, no 1
1h. 45 rule on N9 2 outies R68C6 = 1 innie R7C8 + 11
1i. Min R68C6 = 12, no 1,2 in R6C6
1j. Max R68C6 = 17 -> max R7C8 = 6
1k. 45 rule on D\ 1 innie R5C5 = 2 outies R2C3 + R3C2 + 1
1l. Min R2C3 + R3C2 = 3 -> no 1,2 in R5C5
1m. 45 rule on N47 3 innies R679C3 = 20 = {389/479/569/578}, no 1,2
1n. Hidden killer pair for 1,2 in R3C3 + R4C4 and 17(4) cage at R6C6, R3C3 + R4C4 must contain at least one of 1,2 (1,2 cannot both be in R2C3 + R3C2 because min R5C5 = 5), 17(4) cage must contain at least one of 1,2 -> R3C3 + R4C4 and 17(4) cage must each contain one of 1,2
1o. 17(4) cage = {1349/1358/1457/2348/2357/2456} (cannot be {1367} which clashes with 15(2) cage at R1C1)
1p. Hidden killer pair for 17(4) cage and R7C8 in N9, 17(4) contains one of 1,2 in N9 -> R7C8 = {12}

2a. 45 rule on D\ 3 innies R3C3 + R4C4 + R5C5 = 13 = {139/148/157/238/247/256} (cannot be {346} = {34}6 because 12(4) cage at R2C3 cannot contain both of 3,4)
2b. Consider combinations for 15(2) cage at R1C1 = {69/78}
15(2) cage = {69}, locked for D\
or 15(2) cage = {78}, locked for N1 => R23C1 = {29/56} => R2C3 + R3C2 cannot be {26} which clashes with R23C1 => R3C3 + R4C4 + R5C5 cannot be {13}9
-> no 9 in R5C5
2c. Consider again combinations for 15(2) cage at R1C1 = {69/78}
15(2) cage = {69}, locked for N1 and D\ => 12(4) cage = {1245} => R3C3 + R4C4 cannot be {23}
or 15(2) cage = {78}, locked for D\
-> R3C3 + R4C4 + R5C5 = {148/157/247/256} (cannot be {238}), no 3 in R3C3 + R4C4, clean-up: no 4 in R1C4 (step 1c)
2d. 3 on D\ only in 17(4) cage at R6C6 (step 1o) = {1349/1358/2348/2357}, no 6
2e. R23C4 = {29/38/47} (cannot be {56} which clashes with R14C4), no 5,6
2f. 3 in N5 only in R6C46, locked for R6
2g. 21(5) cage at R5C9 = {12459/12468/12567}, CPE no 1,2 in R45C8, clean-up: no 9 in R5C7
2h. R5C78 = [29]/{38/56} (cannot be {47} which clashes with 21(5) cage), no 4,7
2i. Combined half cage R14C4 + R5C4 = {16}2/{16}5/{25}1, 1 locked for C4
2j. Consider placement for R7C8
R7C8 = 1 => 17(4) cage = {2348/2357}
or R7C8 = 2 => R68C6 = 13 (step 1h) => 17(4) cage = {1349/1358} with 9 of {1349} in R6C6
-> no 9 in R7C7 + R8C8 + R9C9

3a. R14C4 (step 1c) = 7 = {16/25}, R3C3 + R4C4 + R5C5 (step 2c) = {148/157/247/256}
3b. Consider combinations for 20(3) cage at R1C5 = {389/569/578}
20(3) cage = {389/569}, 9 locked for N2
or 20(3) cage = {578}, locked for C5 => R5C5 = 6, R4C4 = {25} => R14C4 = [25]
-> R23C4 = {38/47}, no 2,9
3c. 9 in N2 only in 20(3) cage = {389/569}, no 7, 9 locked for C5
3d. R1C46 + R23C6 (step 1f) = {1238/1247/1256} (cannot be {1346} which clashes with R23C4)
3e. 21(5) cage at R5C9 (step 2g) = {12459/12468/12567}
3f. Consider placement for 7 in N6
7 in R4C78 => no 7 in R23C6 => R1C46 + R23C6 = {1238/1256} => 4 in 42(8) cage at R2C6 only in R4C78 = {47}, locked for N6
or 7 in 21(5) cage = {12567}
-> no 4 in R5C9, clean-up: no 2 in R5C4 (step 1b)
[Looks like a key step.]
3g. 21(5) cage = {12567} (only remaining combination), 5,6,7 locked for N7, 6,7 locked for R6, clean-up: no 1,2 in R1C6 (step 1d)
3h. Naked pair {15} in R5C46, locked for R5
3i. Killer pair 1,5 in R14C4 and R5C4, locked for C4
3j. 7 in N5 only in R45C56, locked for 42(8) cage, no 7 in R23C6
3k. 7 in N2 only in R23C4 = {47}, locked for C4, 4 locked for N2, clean-up: no 3 in R4C9 (step 1d)
3l. 3 in N6 only in R5C78 = {38}, locked for R5, 8 locked for N6
3m. 9 in N6 only in R4C78, locked for R4 and 42(8) cage
3n. Killer pair 6,7 in 15(2) cage at R1C1 and R5C5, locked for D\, clean-up: no 1 in R1C4
3o. 12(4) cage at R2C3 = {1245} (cannot be {1236} because R3C3 + R4C4 only can only contain one of 1,2, step 1n), 4 locked for N1, clean-up: no 7 in R23C1
3p. R6C46 = {39} (hidden pair in N6), 9 locked for R6 -> R6C3 = 8 (hidden single in R6)
3q. 8 in N5 only in R4C56, locked for 42(8) cage, no 8 in R23C6
3r. R1C46 + R23C6 = {1256} (only remaining combination), 5,6 locked for N2, clean-up: no 4 in R4C9 (step 1d)
3s. R4C78 = {49} (hidden pair in N6), 4 locked for R4
3t. Naked triple {389} in 20(3) cage, 3,8 locked for C5, clean-up: no 1,6 in R7C6
3u. R4C6 = 8 (hidden single in N5), placed for D/, clean-up: no 1 in R7C5
3v. 8 in C4 only in 20(3) cage at R8C4 = {389} (cannot be {578} because 5,7 only in R9C3) -> R89C4 = {389}, R9C3 = {39}, CPE no 3 in R9C6

4a. R6C3 = 8 -> R7C1 = 4 (step 1e), R6C12 = 6 = {15}, locked for R6 and N4, clean-up: no 5 in R7C5, no 8 in R8C9
4b. Naked triple {267} in R4C789, 2 locked for N6 and 21(5) cage at R5C9 -> R45C9 = [15], R7C8 = 1, R5C4 = 1, R1C6 = 6 (step 1d), clean-up: no 9 in R2C2, no 7 in R78C9
4c. 12(4) cage at R2C3 (step 3o) = {1245}, 1 locked for N1
4d. 45 rule on N7 2 remaining innies R79C3 = {39}, locked for C3 and N7
4e. Naked pair {39} in R6C4 + R7C3, locked for D/
4f. R6C4 + R7C3 = {39}, R5C4 = 1, R6C35 = [84] -> R7C4 = 6 (cage sum), clean-up: no 3 in R7C6
4g. 4 on D/ only in R1C9 + R2C8 + R3C7, locked for N3
4h. 17(4) cage at R6C6 (step 2d) = {2348} (only remaining combination) -> R6C6 = 3, R7C7 + R8C8 + R9C9 = {248}, locked for D\, 4,8 locked for N9, clean-up: no 7 in 15(2) cage at R1C1
4i. R6C4 = 9, R79C3 = [39], R78C9 = [93], R89C3 = [83]
4j. R5C5 = 7 (hidden single on D\), placed for D/
4k. R5C1236 = [6942], R4C3 = 7
4l. R4C4 = 5, R1C34 = [52] -> R1C2 = 7 (cage sum)
4m. R23C1 = {38} (hidden pair in N1), 3 locked for C1
4n. R1C6 = 6, R1C7 = 1 (hidden single in R1) -> R2C7 = 9 (cage sum)
4o. Naked pair {38} in R15C8, locked for C8
4p. Naked pair {38} in R2C15, locked for R2
4q. R1C8 = 3 (hidden single in N3)
4r. R3C9 = 8 (hidden single in N3), R4C9 = 1 -> R3C8 = 2 (cage sum)
4s. R8C2 + R9C1 = {12} (hidden pair on D/), locked for N7

and the rest is naked singles, without using the diagonals.
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Ed
 Post subject: Re: Assassin 449
PostPosted: Thu Jan 11, 2024 8:42 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Thanks for your NY wishes! Love having you guys do my puzzles. Such variety. I touched base with wellbeback's solution a few times but for different reasons. Really enjoyed Andrew's WT to see a more conventional way, and just how many areas there are to explore.
a449 WT:
Preliminaries from SudokuSolver
Cage 15(2) n1 - cells only uses 6789
Cage 12(2) n9 - cells do not use 126
Cage 9(2) n8 - cells do not use 9
Cage 11(2) n1 - cells do not use 1
Cage 11(2) n2 - cells do not use 1
Cage 11(2) n6 - cells do not use 1
Cage 10(3) n8 - cells do not use 89
Cage 10(3) n47 - cells do not use 89
Cage 20(3) n78 - cells do not use 12
Cage 20(3) n4 - cells do not use 12
Cage 20(3) n2 - cells do not use 12
Cage 11(3) n4 - cells do not use 9
Cage 11(3) n36 - cells do not use 9
Cage 12(4) n15 - cells do not use 789
Cage 27(4) n89 - cells do not use 12
Cage 42(8) n256 - cells ={12456789}

No clean-up done unless stated.
1. "45" on c1234: 1 outie r6c5 = 4

2. "45" on r6789: 2 outies r5c49 = 6 = {15}/[24]

3. "45" on n4: 3 innies r6c123 = 14 = {158/167/239/257/356} = 1 or 3 or 7

4. 11(3)n4: {137} blocked by h14(4)n4
4a. = {128/146/236/245}(no 7)

5. "45" on n3: 2 outies r1c6 + r4c9 = 7 (no 7,8,9)

Took the longest time to see this
6. if 7 in r4 is in r4c3 -> r5c23 = 13 = {49/58} -> killer pair 4,5 with h6(2)r5 (step 2) -> no 7 in 11(2)n6 -> 7 in r5 only in r5c56
6a. or 7 in r4 is in 42(8)
6b. -> 7 locked for 42(8) -> no 7 in r23c6

7. "45" on n1: 2 outies r14c4 = 7 (no 7,8,9, no 3 in r1c4)

8. 42(8)r2c6 must have a 4
8a. -> r1c4 + r4c9 <> [44]
8b. -> r1c4 + r1c6 <> [43] (from outiesn3=7)
8c. -> both 3 and 7 are forced into r123c5 when r14c4 = [43] since 11(2) can't be 7
8d. but then can't make a 20(3)r1c5
8e. -> no 4 r1c4
8f. -> no 3 r4c4

9. r4c4 is from (1256)
9a. 42(8)r2c6 must have each of 1,2,5,6
9b. -> r4c4 repeats in r23c6
9c. "45" on r123: 2 outies r4c49 = 2 innies r23c6
9d. -> r4c9 must also repeat in r23c6
9e. -> no 8,9 in r23c6
9f. and no 3 in r4c9

10. 3 in n5 only in r6: locked for r6

11. 3 in n6 only in 11(2) = {38}: both locked for r5, 8 for n4

12. 8 must be in 42(8) and only in r4c56: locked for r4 and n5

13. r6c3 = 8 (hsingle n4)
13a. -> r6c12 = 6 = {15} only (h14(3) cage sum): both locked for r6 and n4
13b. r7c1 = 4 (cage sum)

14. hidden killer pair 7,9 in n2
14a. 20(3) has at most one of 7,9 since no 4
14b. -> 11(2) must have one
14c. = {29/47}(no 3,5,6,8)

15. 8 in n2 only in 20(3) = {389/578}(no 6)
15a. 8 locked for c5
15b. -> r4c6 = 8 (hsingle n5), locked for d/

16. 8 in c4/n8 on in 20(3)r8c4 = {389/578}(no 4,6)

17. 4 in c4 only in 11(2)r2c4 = {47}: both locked for n2, 7 for c4

18. 20(3)n2 = {389} only: 3,9 locked for c5, 3 for n2
18a. no 4 in r4c9 (outiesn3=7)

19. "45" on n7: 2 remaining innies r79c3 = 12 = {39/57}(no 1,2,6)

20. 21(5)r5c9 must have three of {2679} for r6c789
20a. = {12369/12567}(no 4,8)
20b. can't have four of {2679} -> no 2,6,7,9 in r7c8
20c. must have 2,6 both locked for r6 and n6

21. h6(2)r5 = {15} only: both locked for r5

22. 4 on d/ only in n3: locked for n3

23. 11(3)r3c8 must have 1 or 5 for r4c9 = {128/137}(no 5)
23a. -> r45c9 = [15]
23b. -> 21(5)r5c9 = [5]{267}[1]: 7 locked for r6, n6
23c. and r1c6 = 6 (outiesn3=7)

24. naked pair {49} in r4c78: both locked for r4 and 42(8) cage

25. hpair {49} for n4 in r5c23 -> r4c3 = 7 (cage sum)

26. h12(2)r79c3 = {39} only: both locked for c3 and n7

27. naked pair {39} in r6c4 + r7c3: both locked for d/ and 31(5)
27a. r5c4 = 1 -> r7c4 = 6 (cage sum)

28. 9(2)n8 = {27} only: both locked for r7 and n8

on from there
Cheers
Ed
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