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 Post subject: Assassin 430
PostPosted: Wed Mar 15, 2023 7:06 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Attachment:
a430.png
a430.png [ 57.64 KiB | Viewed 2868 times ]
note the 15(2)r6c3+r7c1

Assassin 430
Andrew encouraged me to still post this sooner rather than wait till start of April. Very happy to!

I found this really resistant to getting the final crack but could always keep making progress. Really liked it. Unfortunately, I couldn't make use of a feature of the cage shape but hope someone can! It gets a score of 1.90 and JSudoku has a harder time than usual. I wouldn't have tried this one if an earlier version of the puzzle hadn't been a bit easy.
triple click code:
3x3::k:1792:0000:0000:6163:6163:6163:6404:8453:8453:1792:0000:2054:2054:2054:6404:6404:6404:8453:0000:0000:0000:0000:6663:6404:6663:6404:8453:5640:5640:1801:0000:6663:6663:6663:5642:8453:6411:5640:1801:5132:5132:5642:5642:5642:8453:6411:6411:3853:6926:5132:5642:2063:2063:4880:3853:6411:6411:6926:6926:1809:7698:7698:4880:5123:5123:6926:6926:6926:1809:7698:4880:4880:5123:5123:5123:1794:1794:7698:7698:2561:2561:
solution:
+-------+-------+-------+
| 2 1 6 | 9 7 8 | 4 5 3 |
| 5 9 3 | 1 4 6 | 2 7 8 |
| 8 4 7 | 3 2 5 | 9 1 6 |
+-------+-------+-------+
| 9 7 2 | 5 6 1 | 8 3 4 |
| 4 6 5 | 8 3 2 | 1 9 7 |
| 1 3 8 | 4 9 7 | 6 2 5 |
+-------+-------+-------+
| 7 8 9 | 2 5 4 | 3 6 1 |
| 6 2 1 | 7 8 3 | 5 4 9 |
| 3 5 4 | 6 1 9 | 7 8 2 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 430
PostPosted: Mon Mar 20, 2023 4:05 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 an enjoyable puzzle. I took a while to find the start of my step 3 but then continued to find some fun steps; definitely more fun steps than usual.

Here's how I solved Assassin 430:
Prelims

a) R12C1 = {16/25/34}, no 7,8,9
b) R45C3 = {16/25/34}, no 7,8,9
c) Disjoint 15(2) cage at R6C3 = {69/78}
d) R6C78 = {17/26/35}, no 4,8,9
e) R78C6 = {16/25/34}, no 7,8,9
f) R9C56 = {16/25/34}, no 7,8,9
g) R9C89 = {19/28/37/46}, no 5
h) 24(3) cage at R1C4 = {789}
i) 8(3) cage at R2C3 = {125/134}
j) 22(3) cage at R4C1 = {589/679}
k) 20(3) cage at R5C4 = {389/479/569/578}, no 1,2

1a. Naked triple {789} in 24(3) cage at R1C4, locked for R1 and N2
1b. 22(3) cage at R4C1 = {589/679}, 9 locked for N4
1c. 45 rule on N4 3 outies R7C123 = 24 = {789}, locked for R7 and N7
1d. 25(5) cage at R5C1 cannot contain more than two of 7,8,9 -> no 7,8 in R5C1 + R6C12
1e. Hidden killer pair 7,8 in 22(3) cage and R6C3 for N4, 22(3) cage contains one of 7,8 -> R6C3 = {78}, clean-up: no 9 in R7C1
1f. R347C1 = {789} (hidden triple in C1)
1g. Naked pair {78} in disjoint 15(2) cage at R6C3, CPE no 7,8 in R4C1 + R7C3 -> R4C1 = 9, R7C3 = 9
1h. R36C3 = {78} (hidden pair in C3)
1i. Naked pair {78} in R3C13, locked for C3 and N1
1j. 45 rule on N47 1 innie R8C3 = 1, clean-up: no 6 in R45C3, no 6 in R7C6
1k. 8(3) cage at R2C3 = {125/134}, 1 locked for R2 and N2, clean-up: no 6 in R1C1
1l. 45 rule on N9 2(1+1) outies R6C9 + R9C6 = 14 = {59/68} (cannot be [77] because R6C9 + R9C6 ‘see’ all the cells in N9)
1m. 7 in R9 only in R9C789, locked for N9
1n. 45 rule on N8 1 innie R9C6 = 1 remaining outie R6C4 + 5 -> R6C4 = {34}, R9C6 = {89}, clean-up: no 8,9 in R6C9
1o. 9 in R6 only in R6C56, locked for N5
1p. 1,2 in N5 only in R4C456 + R5C56, CPE no 1,2 in R4C8
1q. 20(3) cage at R5C4 = {389/479/569/578}
1r. 9 of {389/479/569} only in R6C5 -> no 3,4,6 in R6C5
1s. 45 rule on N78 2 outies R6C49 = 9 = [36/45]
1t. R6C78 = {17/26} (cannot be {35{ which clashes with R6C49), no 3,5
1u. Hidden killer triple 7,8,9 in R9C6, R9C7 and R9C89 for R9, R9C6 contains one of 8,9, R9C7 and R9C89 can each only contain one of 7,8,9 -> R9C7 = {789}, R9C89 = {19/28/37}, no 4,6
1v. R9C6, R9C7 and R9C89 each contain one of 7,8,9, CPE no ,8,9 in R8C7
1w. R8C45 + R9C6 = {789} (hidden triple in N8)
1x. 20(3) cage = {389/479/569/578} must have one of 7,8,9 in C5 (cannot have two which would clash with R18C5
1y. Killer triple 7,8,9 in R1C5, 20(3) cage and R8C5, locked for C5

2a. 1 in N8 only in R7C6 + R9C45 -> combined cage R78C6 + R9C45 = {16}{25}/{16/34}, 6 locked for N8
2b. 7 in N8 only in 27(6) cage at R6C4 = {123579/124578}
2c R6C4 = {34} -> no 3,4 in R7C45
2d. Naked pair {25} in R7C45, locked for R7 and N8
2e. 6 in R7 only in R7C789, locked for N9
2f. 30(5) cage at R7C7 = {35679/45678} (cannot be {15789/24789} because 7,8,9 only in R9C67, cannot be {25689} because 2,5 only in R8C7, cannot be {34689} which clashes with R9C89), no 1,2, 5,6 locked for N9
2g. 30(5) cage at R7C7 = {35679/45678} -> R89C7 = [57], clean-up: no 1 in R6C8, no 3 in R9C89
2h. 45 rule on C9 3 outies R189C8 = 17 = {269/359/368/458}, no 1, clean-up: no 9 in R9C9
2i. 5,6 only in R1C8 -> R1C8 = {56}
2j. 1 in N9 only in R79C9, locked for C9
2k. Hidden killer pair 5,6 in R12345C9 and R6C9 for C9, R6C9 = {56} -> R12345C9 must contain one of 5,6
2l. 33(6) cage at R1C8 contains both of 5,6 = {235689/245679/345678}
2m. 19(4) cage at R6C9 = {1369/1459/2368/2458} (cannot be {1468/2359} which clash with R9C89)
2n. Consider combinations for R9C89 = {28}[91]
R9C89 = {28}, locked for N9, 8 locked for R9 => R9C6 = 9 => 30(5) cage = {35679}, 3 locked for N9 => 19(4) cage = {1459}
or R9C89 = [91], R9C6 = 8 => 30(5) cage = {45678}, 4 locked for N9 => 19(4) cage = {2368}
-> 19(4) cage = {1459/2368}
2o. 1 of {1459} must be in R7C9 -> no 4 in R7C9
2p. 3 of {2368} must be in R7C9 -> no 3 in R8C89
2q. 3 in N9 only in R7C789, locked for R7, clean-up: no 4 in R8C6
2r. 5,6 in C9 only in R123456C9 -> R123456C9 forms hidden 33(6) cage
2s. 45 rule on C9 3 remaining innies R789C6 = 12 must contain 1 for C9 = {129/138} = [192/381], R8C9 = {89}, R9C9 = {12}, clean-up: no 2 in R9C8
2t. Naked pair {89} in R8C9 + R9C8, locked for N9

[It took a while to find the next step.]
3a. Consider placement for 6 in N1
6 in R1C23 => R1C8 = 5 => R6C9 = 5 (hidden single in C9)
or R2C1 = 6 => R1C1 = 1 => R6C2 = 1 (hidden single in N4) => R6C78 = {26}, 6 locked for R6 => R6C9 = 5
or 6 in R23C2, R45C2 = 13 = {58}, 8 locked for C2 => R7C12 = [87], R6C3 = 7 => R6C78 = {26}, 6 locked for R6 => R6C9 = 5
-> R6C9 = 5, R6C4 = 4 (step 1s), R7C9 = 1, R8C89 = [49] (all step 2n), R9C89 = [82], R9C6 = 9, 6 in C9 only in R12345C9, locked for 33(6) cage -> R1C8 = 5, R7C6 = 4 (hidden single in R7) -> R8C6 = 3, naked pair {16} in R9C45, 6 locked for R9, clean-up: no 2 in R2C1
3b. R1C3 = 6 (hidden single in C3), clean-up: no 1 in R1C1
3c. 1 in N4 only in 25(5) cage at R5C1 = {12679/13489} (cannot be {13579} which clashes with R45C3, cannot be {12589} = 5{12}[89] which clashes with R12C1 + R9C1, killer ALS block), no 5
3d. Whichever value is in R7C2 must also be in R6C3 because of 15(2) disjoint cage at R6C3 -> R5C1 + R6C123 + R7C3 form hidden 25(5) cage with the same combination as 25(5) cage at R5C1
3e. Hidden 25(5) cage = {13489} (cannot be {12679} = {126}[79] which clashes with R6C78) -> R5C1 = 4, R6C12 = {13}, locked for R6, 3 locked for N4, R6C3 = 8 -> R7C12 = [78], clean-up: no 3 in R12C1, no 7 in R6C8
3f. Naked pair {25} in R45C3, locked for C3, 5 locked for N4
3g. R12C1 = [25] -> R9C1 = 3, R9C3 = 4, R2C3 = 3 -> R2C45 = 5 = [14], R2C2 = 9, R9C45 = [61]
3h. Naked pair {26} in R6C78, locked for R6 and N6
3i. R4C8 ‘sees’ all remaining cells in N5 except for 20(3) cage at R5C4, R4C8 = {37} -> 20(3) cage must contain one of 3,7 = {389/578}, no 6
3j. Naked triple {789} in R168C5, locked for C5 -> R5C4 = 8, R8C45 = [78], R1C456 = [978], R56C5 = [39]

4a. R6C6 = 7 -> R4C8 = 3, R7C78 = [36], R6C78 = [62]
4b. R4C8 = 3, R6C4 = 7 -> R5C678 = 12 = 2{19}, R4C456 = [561]
4c. R23C6 = [65], R2C8 = 7 -> R12C7 + R3C8 = 7 = [421]

and the rest is naked singles.


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 Post subject: Re: Assassin 430
PostPosted: Sat Mar 25, 2023 8:17 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Amazed that Andrew found that start of step 3. Really nice start to the puzzle too. Have pinched that up to 2m. And particularly enjoyed 3e in the fun steps.

Here's how I did the crucial middle.
a430 mid:
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 123456    123456    23456     | 789       789       789       | 12346     56        23456     |
| 123456    1234569   2345      | 12345     12345     123456    | 1234689   123456789 23456789  |
| 78        1234569   78        | 123456    123456    123456    | 123469    1234569   234569    |
:-------------------------------+-------------------------------+-------------------------------:
| 9         5678      2345      | 12345678  123456    12345678  | 123468    345678    2345678   |
| 123456    5678      2345      | 345678    345678    12345678  | 1234689   123456789 23456789  |
| 123456    123456    78        | 34        5789      123456789 | 126       1267      56        |
:-------------------------------+-------------------------------+-------------------------------:
| 78        78        9         | 25        25        134       | 346       346       134       |
| 2346      2346      1         | 789       789       346       | 5         23489     23489     |
| 23456     23456     23456     | 1346      1346      89        | 7         289       128       |
'-------------------------------.-------------------------------.-------------------------------'

Paste above code into A430 in SudokuSolver

Andrew's step 2m to above then
3. 2 outies n9 = 14 and r9c6 must repeat in n9 in 19(4)r6c9 -> a h14(2) in that cage -> the other two cells = 5 = [1]{4}/[3]{2}
3a. ie 19(4) = {1459/2368} with r7c9 = (13)
3b. note: 3 in r7c9 must have 6 in r6c9

4. "45" on r6789: 2 innies r6c56 - 12 = 1 outie r5c1
4a. -> min. r6c56 = 13 (no 1,2,3)
4b. max. r6c56 = 17 -> max. r5c1 = 5

Took a very long time to find this.
5. Consider 6 in n8
i. 6 in 7(2)r9c4 -> 6 in c3 only in r1c3
ii. or 6 in 7(2)r78c6 = [16] -> [63] in r67c9 (step 3b), -> 8(2)r6c7 = {17}: 1 locked for r6 -> 1 in n4 only in r5c1
5a. -> {16} blocked from 7(2)n1 = {25/34}

6. 25(5)r5c1 must have 9 and one of 7,8 for r7c2 and 1 for n4
6a. = {12589/12679/13489/13579}
6b. and 3 in r6 only in h9(2)r6c49 = [36] or in 25(5)
6c. -> 6 in 25(5) must also have 3 (Locking out cages) but this is not possible -> no 6 in r6c12

7. 6 in n4 only in r45c2 = {67} only: both locked for c2, 7 for n4

8. r1c3 = 6 (hsingle n1)

Much easier now but still have to think about it.
Cheers
Ed


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