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 Post subject: Assassin 418
PostPosted: Wed Jun 01, 2022 9:33 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
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Assassin 418
16 years since the Assassin series was started by Ruud! I found this new puzzle quite pleasant with a couple of hard-to-see steps. SudokuSolver gives it 1.75 but JSudoku found it okay with 3 'advanced' steps which is why I tried it.
triple click code:
3x3::k:4608:4608:4608:6401:6401:6401:6401:1538:1538:3843:4608:2564:3589:4102:4102:5383:5383:1538:3843:2564:2824:2824:3589:4102:1801:5383:3082:2564:6923:6923:6923:6923:3589:1801:8716:3082:1559:2564:5133:5902:3589:1559:8716:8716:8716:6927:5133:5133:5902:5902:4368:8716:4625:8716:5650:6927:5133:5902:4368:4115:2580:4625:4625:5650:5650:6927:4368:4115:4115:2580:2580:4625:5650:6927:2069:2069:4368:5654:5654:5654:5654:

solution:
+-------+-------+-------+
| 4 6 5 | 8 1 7 | 9 2 3 |
| 7 3 2 | 5 4 9 | 8 6 1 |
| 8 1 9 | 2 6 3 | 5 7 4 |
+-------+-------+-------+
| 3 5 7 | 6 9 1 | 2 4 8 |
| 1 4 8 | 3 2 5 | 7 9 6 |
| 9 2 6 | 4 7 8 | 3 1 5 |
+-------+-------+-------+
| 5 7 4 | 9 3 6 | 1 8 2 |
| 6 9 3 | 1 8 2 | 4 5 7 |
| 2 8 1 | 7 5 4 | 6 3 9 |
+-------+-------+-------+


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 Post subject: Re: Assassin 418
PostPosted: Sat Jun 04, 2022 4:13 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Wow! 16 years! I started solving Assassins later that month and posted my first walkthrough at about Assassin 12.

Thanks Ed for this latest Assassin, a fun one. I found it much more enjoyable than the way I solved the previous one.

Here's how I solved Assassin 418:
Prelims

a) R23C1 = {69/78}
b) R3C34 = {29/38/47/56}, no 1
c) R34C7 = {16/25/34}, no 7,8,9
d) R34C9 = {39/48/57}, no 1,2,6
e) R5C16 = {15/24}
f) R9C34 = {17/26/35}, no 4,8,9
g) 6(3) cage at R1C8 = {123}
h) 21(3) cage at R2C7 = {489/579/678}, no 1,2,3
i) 10(3) cage at R7C7 = {127/136/145/235}, no 8,9
j) 10(4) cage at R2C3 = {1234}
k) 14(4) cage at R2C4 = {1238/1247/1256/1346/2345}, no 9
l) 27(4) cage at R4C2 = {3789/4689/5679}, no 1,2
m) 27(4) cage at R6C1 = {3789/4689/5679}, no 1,2

1a. 45 rule on R1 2 outies R2C29 = 4 = {13}, locked for R2
1b. 6(3) cage at R1C8 = {123}, locked for N3, 2 locked for R1, clean-up: no 4,5,6 in R4C7, no 9 in R4C9
1c. 10(4) cage at R2C3 = {1234}, CPE no 3 in R4C2
1d. 27(4) cage at R4C2 = {3789/4689/5679}, 9 locked for R4
1e. 27(4) cage at R6C1 = {3789/4689/5679}, CPE no 9 in R789C1

2a. 2 in N1 only in R2C3 + R3C23
2b. 45 rule on N1 3 innies R2C3 + R3C23 = 12 = {129/237/246} -> R3C3 = {679}, R3C4 = {245}
2c. 5 in N1 only in R1C123, locked for R1
2d. 2 in N1 only in R2C3 + R3C2, locked for 10(4) cage, no 2 in R4C1 + R5C2
2e. R35C2 must contain one of 1,3 (because no 1,3 in R2C3) but not both because R2C2 = {13}) -> R4C1 = {13}
2f. Killer pair 1,3 in R2C2 and R35C2, locked for C2
2g. 10(4) cage = {1234}, CPE no 4 in R1C2
2h. 45 rule on N9 2(1+1) outies R6C8 + R9C6 = 5 = {14/23}
2i. 45 rule on C789 1 innie R1C7 = 1 outie R9C6 + 5 -> R1C7 = {6789}
2j. 45 rule on N6 3 innies R4C79 + R6C8 = 11 = {128/137/245} -> R4C9 = {578}, R3C9 = {457}
2k. Consider combinations for R4C79 + R6C8
R4C79 + R6C8 = {128} => R3C9 = 4, R3C7 = {56} => R3C34 = [92] (cannot be [65] which clashes with R3C7
or R4C79 + R6C8 = {137} => R3C9 = 5 => R3C34 = [74/92]
or R4C79 + R6C8 = {245} => R3C7 = 5 => R3C34 = [74/92]
-> R3C34 = [74/92], no 5,6
2l. R2C3 + R3C23 = {129/237}, no 4
2m. 10(4) cage = {1234} -> R2C3 = 2, R5C2 = 4, clean-up: no 2 in R6C16, no 6 in R9C4
2n. Naked pair {13} in R23C2, locked for N1
2o. Killer pair 7,9 in R23C1 and R3C3, locked for N1
2p. 18(4) cage at R1C1 = {1458/3456}, 4 locked for R1
2q. Naked pair {15} in R5C16, locked for R5
2r. 7,9 in R1 only in 25(4) cage at R1C4 = {1789/3679}
2s. Min R2C4 = 4 -> max R35C5 + R4C6 = 10, no 8 in R35C5 + R4C6
2t. 45 rule on R1234 1 innie R4C8 = 1 remaining outie R5C5 + 2 -> R4C8 = {458}, R5C5 = {236}

3a. R4C79 + R6C8 (step 2j) = {128/137} (cannot be {245} because R4C79 + R6C8 + R4C8 = {245}8 clashes with 27(4) cage at R4C2), no 4,5, 1 locked for N6, clean-up: no 7 in R3C9, no 1 in R9C6 (step 2h), no 6 in R1C7 (step 2i)
3b. 27(4) cage at R4C2 = {4689/5679} (cannot be {3789} which clashes with R4C9), no 3
3c. R4C167 = {123} (hidden triple in R4)
3d. 45 rule on N4578 4(1+2+1) remaining innies R4C1 + R4C6 + R5C5 + R9C6 = 10
3e. Min R4C1 + R49C6 = 6 -> no 6 in R5C5, clean-up: no 8 in R4C8 (step 2t)
[Or, more simply, killer pair 7,8 in 27(4) cage and R4C9, locked for R4. I’d seen that 45 earlier, so wanted to use it, and it helps with the continuation]
3f. Max R4C1 + R4C6 + R5C5 = 8 but only as [323] -> no 2 in R9C6, clean-up: no 7 in R1C7 (step 2i), no 3 in R6C8 (step 2h)
3g. 21(3) cage at R2C7 = {579/678} (cannot be {489} which clashes with R1C7), no 4
3h. 4 in N3 only in R3C79, locked for R3 -> R3C4 = 2, R3C3 = 9, clean-up: no 6 in R23C1, no 6 in R9C3
3i. Naked pair {78} in R23C1, locked for C1, 8 locked for N1
3j. Naked triple {456} in R1C123, 6 locked for R1
3k. R1C123 = {456} -> R2C2 = 3 (cage sum), R2C9 = 1, R3C2 = 1 -> R4C1 = 3, clean-up: no 4 in R3C7
3l. R3C9 = 4 (hidden single in N3) -> R4C9 = 8
3m. 27(4) cage at R4C2 = {5679}, 5 locked for R4
3n. 25(4) cage at R1C4 = {1789} (only remaining combination), no 3, 7 locked for N2

4a. R4C8 = 4 -> R5C5 = 2 (step 2i), R4C6 = 1, R6C16 = [15], R4C7 = 2 -> R3C7 = 5, R6C8 = 1 -> R9C6 = 4 (step 2h, or 45 rule on N4578), R1C7 = 9 (step 2i)
4b. Naked triple {678} in R2C178, 6,8 locked for R2, 6 locked for N3
4c. R2C6 = 9 -> R2C5 + R3C6 = 7 = [43], R2C4 + R3C5 = [56], clean-up: no 3 in R9C3
4d. 2 in C6 only in R78C6 -> 16(3) cage at R7C6 = {268} (cannot be {259} because 5,9 only in R8C5) -> R78C6 = {26}, 6 locked for C6 and N8, R8C5 = 8
4e. 17(4) cage at R6C6 = {1358} (only remaining combination) -> naked triple {135} in R79C5 + R8C4, locked for N8, R9C4 = 7 -> R9C3 = 1
4f. R7C4 = 9, R6C4 = 4 (hidden single in N5) -> R5C4 + R6C5 = 10 = [37]
4g. Naked triple {679} in R5C789, 6,7 locked for R5, 6,9 locked for N6 -> R6C79 = [35]
4h. R56C3 = [86] -> R6C2 + R7C3 = 6 = [24]
4i. Naked triple {256} in R789C1, 5,6 locked for C1 and N7
4j. 27(4) cage at R6C1 = [9738]
4k. R9C67 = [46] -> R9C89 = 12 = {39}, locked for N9, 3 locked for R9

and the rest is naked singles.


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 Post subject: Re: Assassin 418
PostPosted: Fri Jun 10, 2022 8:24 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Really nice pure solution by Andrew. I used a similar path to get started but then a different middle. From the end of his step 2,
end step 2 candidates- paste into a418 in SudokuSolver:
.-------------------------------.-------------------------------.-------------------------------.
| 4568 568 4568 | 136789 136789 136789 | 6789 123 123 |
| 6789 13 2 | 45678 456789 456789 | 456789 456789 13 |
| 6789 13 79 | 24 1234567 123456789 | 456 456789 457 |
:-------------------------------+-------------------------------+-------------------------------:
| 13 56789 356789 | 3456789 3456789 1234567 | 123 458 578 |
| 15 4 36789 | 236789 236 15 | 236789 236789 236789 |
| 356789 256789 1356789 | 123456789 123456789 123456789 | 123456789 1234 123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 12345678 56789 13456789 | 123456789 123456789 123456789 | 1234567 123456789 123456789 |
| 12345678 256789 3456789 | 123456789 123456789 123456789 | 1234567 1234567 123456789 |
| 12345678 56789 13567 | 123567 123456789 1234 | 123456789 123456789 123456789 |
'-------------------------------.-------------------------------.-------------------------------'


3. "45" on n1: 2 innies r3c23 = 10 = [37/19]
3a. -> combined cages 4(2)r3c2,r4c1 + 11(2)r3c3 = [3741/1923]
3b. but [3741] puts 7(2)r3c7 = [52] which leaves nothing for r3c9 = (457)
3b. -> = [1923] only

cracked
Cheers
Ed


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 Post subject: Re: Assassin 418
PostPosted: Tue Jun 14, 2022 7:23 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed! Agree with Andrew's comments. I started a different way...
Assassin 418 WT:
1. 6(3)n3 = {123}
Outies r1 = r2c29 = +4(2) = {13}
-> 2 in r1c89

2. 10(4)n14 = {1234}
-> Whichever of (13) is in r2c2 also goes in r4c1
Innies n1 = r2c3 + r3c23 = +12(3)
-> r3c3 is Min 5
-> 2 in n1 in 10(4) in r2c3 or r3c2

3! IOD n7 -> r6c1 = r79c3 + 4
-> r6c1 is Min 7 and that value in n7 must go in r8c2
-> r6789c1 = +22(4)
-> r145c1 = +8(3)
Since r14c1 is min {13} -> r5c1 is Max 4
-> 6(2)r5 cannot be [51]

4! 4 in 10(4)n14 either in n1 which puts Innies n1 = [{24}6]
or 4 in 10(4)n14 in r5c2 which puts 6(2)r5 = [15] and Innies n1 = [{12}9]
I.e., r3c3 from (69)

-> 15(2)n1 = {78}
-> r6c1 = r8c2 = 9
-> Innies n7 = r79c3 = +5(2)
-> (78) in n7 only in 27(4)
-> 27(4)r6c1 = [9{378}]
-> r79c3 = [41]
-> r789c1 = {256}
Also r9c4 = 7

5! In n4 (r4c1, r5c1, r5c2) all from (134)
-> Remaining Innies n4 = r4c23 = +12(2) = {57}
-> 27(4)r4 = [{57}{69}]
Also 20(4)n4 = [{268}4]

6. Innies r9 = r9c125 = +15(3) with no (17)
Given r9c1 from (256) and r9c2 from (38)
-> r9c125 only [285] or [582]
-> 22(4)r9 = {3469}

7. Outies n9 = [r6c8,r9c6] = +5(2) = [14] or [23]
IOD c789 -> r1c7 = r9c6 + 5
-> r1c7 from (89)
-> (Given 7 already on r4) Innies n3 either [954] or [864]
-> Innies n6 either [281] or [182]

8. r13c7 either [95] which puts HS 9 in n1 in r3c3
or r13c7 is [86] which puts NS 9 in r3c3
Either way 11(2)r3 = [92]
-> 10(4)n14 = [2134]
-> r6c2 = 2
-> Innies n6 = [281]
etc.


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