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 Post subject: Assassin 388
PostPosted: Fri Nov 15, 2019 7:56 am 
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Assassin 388

This is a hard-at-the-beginning 1.70 (SSscore) puzzle. Quite like it since I found a really nice step. I tried making it into a zero killer but the SS score went berserk though I could still solve it with the same steps. So, have left all the cages in to increase the chance of multiple solution paths. Not an x-killer!
JSudoku has no more trouble with either puzzle so perhaps it used my solving path.
triple click code:
3x3::k:2304:2304:6657:6657:6657:2050:7683:7683:3332:4869:2304:6657:3078:6657:2050:7683:7683:3332:4869:4869:5895:3078:3078:9480:9480:9480:3332:2825:2825:5895:9482:6667:6667:6667:9480:3332:2825:5895:5895:9482:6667:4364:9480:9480:9480:3085:3086:3086:9482:6667:4364:4364:4364:4367:3085:3085:9482:9482:2576:5137:4367:4367:4367:4626:9482:9482:3859:2576:5137:5137:1812:4373:4626:4626:3859:3859:3859:2326:2326:1812:4373:
solution:
Code:
+-------+-------+-------+
| 4 3 9 | 8 5 2 | 6 7 1 |
| 5 2 1 | 7 3 6 | 8 9 4 |
| 8 6 7 | 1 4 9 | 3 5 2 |
+-------+-------+-------+
| 7 1 5 | 9 8 3 | 2 4 6 |
| 3 9 2 | 5 6 4 | 1 8 7 |
| 6 4 8 | 2 7 1 | 9 3 5 |
+-------+-------+-------+
| 1 5 4 | 6 9 8 | 7 2 3 |
| 2 8 3 | 4 1 7 | 5 6 9 |
| 9 7 6 | 3 2 5 | 4 1 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 388
PostPosted: Fri Nov 15, 2019 1:42 pm 
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I just went through this on JSudoku and I've not seen so many small fishes before 6 and all on the same number

Was that how you solved it Ed?


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 Post subject: Re: Assassin 388
PostPosted: Fri Nov 15, 2019 9:05 pm 
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Nope. My way is very different. Haven't stepped through a puzzle with JS for a long time. Don't find it easy. It found one really nice step early on even though I can't understand how it did it. I had to work backwards from its result to be able to understand what it must have done.

It is unusual to have so many simple fish in JS but for A385 it used 10 complex x-y chains, with 16 chains (including fish) altogether!

Sorry I don't get to try your puzzles HATMAN. These Assassins take all my available solving time.

Cheers
Ed


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 Post subject: Re: Assassin 388
PostPosted: Sat Nov 16, 2019 9:06 am 
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Please try the Triankles Ed as they are very interesting to create and I'd like to solve one created by someone else.

Maurice


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 Post subject: Re: Assassin 388
PostPosted: Thu Nov 21, 2019 6:37 am 
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Thanks Ed for another excellent Assassin. I wonder whether my key step was the same as your really nice step; possibly not.

Thanks Ed for pointing out an error part way through my step 9. I've now reworked from there, including finding a nice forcing chain.
Here is my walkthrough for Assassin 388:
Prelims

a) R12C6 = {17/26/35}, no 4,8,9
b) R6C23 = {39/48/57}, no 1,2,6
c) R78C5 = {19/28/37/46}, no 5
d) R89C8 = {16/25/34}, no 7,8,9
e) R89C9 = {89}
f) R9C67 = {18/27/36/45}, no 9
g) 9(3) cage at R1C1 = {126/135/234}, no 7,8,9
h) 19(3) cage at R2C1 = {289/379/469/478/568}, no 1
i) 11(3) cage at R4C1 = {128/137/146/236/245}, no 9
j) 20(3) cage at R7C6 = {389/479/569/578}, no 1,2
k) 30(4) cage at R1C7 = {6789}
l) 13(4) cage at R1C9 = {1237/1246/1345}, no 8,9

Steps resulting from Prelims
1a. Naked pair {89} in R89C9, locked for C9 and N9, clean-up: no 1 in R9C6
1b. Naked quad {6789} in 30(4) cage at R1C7, locked for N3
1c. 13(4) cage at R1C9 = {1237/1246/1345}, 1 locked for C9
1d. 6,7 of {1237/1246} must be in R4C9 -> no 2 in R4C9
[Ed pointed out that I missed something else here; I found it at step 10.]

2. 45 rule on N12 2 innies R3C36 = 16 = {79}, locked for R3
2a. 45 rule on N4 1 outie R3C3 = 1 innie R6C1 + 1 -> R3C3 + R6C1 = [76/98]
2b. 19(3) cage at R2C1 = {289/469/478/568} (cannot be {379} because 7,9 only in R2C1), no 3
2c. 7,9 of {289/469/478} must be in R2C1 -> no 2,4 in R2C1
2d. 5 of {568} must be in R23C1 (R23C1 cannot be {68} which clashes with R6C1) -> no 5 in R3C2
2e. 4 of {469/478} must be in R3C1 (cannot be [964] which clashes with R3C3 + R6C1 = [76], cannot be [784] which clashes with R3C3 + R6C1 = [98]) -> no 4 in R3C2
2f. 12(3) cage at R6C1 = {138/156/246} (cannot be {129/147/237/345} because R6C1 only contains 6,8), no 7,9
2g. R6C1 = {68} -> R7C12 = {13/15/24}, no 6,8
2h. 37(7) cage at R3C6 must contain 8, locked for N6

3a. 45 rule on C6789 2 innies R4C67 = 5 = {14/23}
3b. 11(3) cage at R4C1 = {128/137/146/236/245} cannot be {12}8/{13}7/{24}5 (which clash with R4C67) -> no 5,7,8 in R5C1
3c. 45 rule on C6789 3 outies R456C5 = 21 = {489/579/678}, no 1,2,3
3d. Combined cage R456C5 + R78C5 = {489}{37}/{579}{28}/{579}{46}/{678}{19}, 7,9 locked for C5

4. 45 rule on C12 3 innies R568C2 = 21 = {489/579/678}, no 1,2,3, clean-up: no 9 in R6C3

5. 20(3) cage at R7C6 = {389/479/569/578}
5a. 3 of {389} must be in R8C7 -> no 3 in R78C6
5b. 4 of {479} must be in R78C6 (R78C6 cannot be {79} which clashes with R3C6) -> no 4 in R8C7

6. 45 rule on C1234 4 outies R1239C5 = 14 = {1238/1256/1346/2345}
6a. 8 of {1238} must be in R123C5 (R123C5 cannot be {123} which clashes with R12C6) -> no 8 in R9C5

7. 45 rule on R89 2 outies R7C56 = 2 innies R8C23 + 6
7a. Min R8C23 = 6 (cannot be [41] which clashes with R7C12) -> min R7C56 = 12, no 1,2 in R7C5, clean-up: no 8,9 in R8C5

8. R568C2 (step 4) = {489/579/678}
8a. Consider placements for R3C2 = {268}
R3C2 = 2 => R23C1 = [98], R6C1 = 6, R3C3 = 7 => 23(4) cage at R3C3 = {2579/3479} (cannot be {3578} = 7{358} which clashes with R6C23), no 6,7,8 in R5C2 => R568C2 = {489/579}
or R3C2 = 6 => R568C2 = {489/579}
or R3C2 = 8 => R568C2 = {579}
-> R568C2 = {489/579}, no 6, 9 locked for C2
[First key step.]

9. 45 rule on R6789 7 (6+1) outies R4C4567 + R5C456 = 37
9a. R4C67 (step 3a) = 5 -> R4C45 + R5C456 = 32 = {26789/35789/45689}, no 1, 8,9 locked for N5
[Second key step.]
9b. R456C5 (step 3c) = {489/579/678}
9c. 4 of {489} must be in R6C5 -> no 4 in R45C5
9d. 8 on R6 only in R6C123, locked for N4
9e. 11(3) cage at R4C1 (step 3b) = {137/146/236/245}
9f. Hidden killer pair 1,2 in 11(3) cage and 23(4) cage at R3C3 for N4, 11(3) cage contains one of 1,2 -> 23(4) cage must contain one of 1,2 = {1679/2579}, no 3,4
[Alternatively 23(4) = {1679/2579} (cannot be {3479/3569} which clashes with 11(3) cage).
First time through I made a careless mistake here by either overlooking a digit in N4 or carelessly deleting it from my worksheet, so reworked from here.]
9g. 11(3) cage = {137/146/245} (cannot be {236} which clashes with 23(4) cage)
9h. Hidden killer pair 3,4 in 11(3) cage and R6C23 for N4, 11(3) contains one of 3,4 -> R6C23 must contain one of 3,4 = {48}/[93], no 5,7
9i. Consider combinations for R6C23
R6C23 = {48}, 4 locked for N4 => 11(3) cage = {137}
or R6C23 = [93] => R3C3 = 9, 19(3) cage at R2C1 = {568}, 9(3) cage at R1C1 = {234} => R12C3 = {17}, locked for C3 => 23(4) cage = {2579} => 11(3) cage = {146}
-> 11(3) cage = {137/146}, no 2,5, 1 locked for N4
9j. 23(4) cage = {2579}, no 6, 2 locked for C3
9k. R4C12 = {14/16/17/37/46}, R4C67 (step 3a) = {14/23} -> combined cage R4C1267 must contain 3, locked for R4

[Just spotted.]
10. R12C9 + R3C789 = {12345}, CPE no 2,3,4,5 in R5C9

11. 37(7) cage at R4C4 must contain 9
11a. Consider placements for 9 in N2
9 in R12C4, locked for C4 => 9 in 37(7) cage in R7C3 + R8C23, locked for N7 => R2C1 = 9 (hidden single in C1) => R3C3 = 7
or 9 in R3C6 => R3C3 = 7
-> R3C3 = 7, R3C6 = 9
[Final key step. With hindsight this step was available after step 6, but I only saw it while doing the rework.
Back to my original steps, slightly modified where appropriate.]
11b. 23(4) cage at R3C3 (step 9j) = {2579}, 9 locked for N4, clean-up: no 3 in R6C3
11c. Naked pair {48} in R6C23, locked for N4, 4 locked for R6 -> R6C1 = 6
11d. Naked triple {137} in 11(3) cage, 7 locked for R4
11e. R456C5 (step 3c) = {579/678}, 7 locked for C5 and N5, clean-up: no 3 in R78C5
11f. 20(3) cage at R7C6 = {578} (only remaining combination), 8 locked for C6 and N8, clean-up: no 2 in R8C5, no 1 in R9C7
11g. 20(3) cage at R7C6 = {578}, CPE no 5,7 in R8C4
11h. R1239C5 (step 6) = {1238/2345} (cannot be {1256} which clashes with R456C5, cannot be {1346} which clashes with R78C5), no 6
11i. 9 in N6 only in R6C78 -> 17(4) cage at R5C6 = {1259/1349}, no 6,7, 1 locked for R6
11j. 4 of {1349} must be in R5C6 -> no 3 in R5C6
11k. R4C45 + R5C456 (step 9a) = {26789/45689} (cannot be {35789} which clashes with R6C5), no 3
11l. R6C1 = 6 -> R7C12 = 6 = {15/24}, no 3
11m 45 rule on N1 2 innies R12C3 = 10 = {19/46}, no 3,5,8
11n. 3 in N1 only in 9(3) cage at R1C1 = {135/234}, no 6
11o. Killer pair 1,4 in 9(3) cage and R12C3, locked for N1
11p. R568C2 (step 8a) = {489} (only remaining combination, cannot be {579} because R6C2 only contains 4,8) -> R5C2 = 9, R68C2 = {48}, locked for C2
11q. Naked pair {25} in R45C3, 5 locked for C3
11r. 8 in N1 only in R23C1, locked for C1
11s. 19(3) cage at R2C1 (step 2b) = {289/568}
11t. R3C2 = {26} -> no 2 in R3C1
11u. 13(4) cage at R1C9 = {1246/1345}, 4 locked for C9
11v. 17(4) cage at R6C9 = {1367/1457/2357} (cannot be {2456} = 2{456}/5{246} which clashes with R89C8)
11w. 17(4) cage = {1367/2357} (cannot be {1457} which clashes with R7C12), no 4

12. 45 rule on N9 2 innies R89C7 = 1 outie R6C9 + 4
12a. Min R89C9 = 7 -> min R6C9 = 3
12b. R6C9 = {357} and R8C7 = {57} are both odd -> R9C7 must be even = {246} -> R9C6 = {357}
12c. 4 in C6 only in R45C6, locked for N5

13. 17(4) cage at R5C6 (step 11i) = {1259/1349}
13a. Consider combinations for 17(4) cage
17(4) cage = {1259}, CPE no 2,5 in R6C4 => R6C4 = 3
or 17(4) cage = {1349}, 3 locked for R6
-> R6C9 = {57}
13b. Naked pair {57} in R6C59, 5 locked for R6
13c. 17(4) cage at R6C9 (step 11w) = {1367/2357}, 3 locked for R7 and N9, clean-up: no 4 in R89C8
13d. R9C7 = 4 (hidden single in N9) -> R9C6 = 5, clean-up: no 3 in R12C6, no 1 in R4C6 (step 3a)
13e. Naked pair {78} in R78C6, 7 locked for C6, N8 and 20(3) cage at R7C6 -> R8C7 = 5
13f. R89C8 = {16} (only remaining combination), locked for C8 and N9
13g. Naked triple {237} in R7C789, 2,7 locked for R7, 7 locked for 17(4) -> R6C9 = 5, R6C5 = 7, R78C6 = [87]
13h. 17(4) cage at R5C6 = {1349} (only remaining combination) -> R5C6 = 4, R6C4 = 2 (hidden single in R6), R4C6 = 3 -> R4C7 = 2 (step 3a), R6C6 = 1, R45C3 = [52]
13i. Naked pair {17} in R4C12, 1 locked for R4 and N4 -> R5C1 = 3
13j. R5C7 = 1 (hidden single in R5) -> R3C7 = 3, R6C78 = [93], R7C789 = [723]
13k. Naked triple {124} in R123C9, 4 locked for C9 and N3 -> R3C8 = 5, R4C9 = 6, R5C89 = [87], R4C8 = 4, R3C1 = 8
13l. Naked pair {26} in R12C6, locked for N2
13m. Naked pair {14} in R3C45, locked for R3 and N2, R2C4 = 7 (cage sum), R12C8 = [79], R2C1 = 5, R3C29 = [62]
13n. 9(3) cage (step 11n) = {234} (only remaining combination) -> R1C1 = 4, R12C2 = {23}, locked for C2, R12C3 = [91], R7C12 = [15], R9C2 = 7
13o. R9C5 = 2 (hidden single in C5) -> R89C1 = [29], R89C9 = [98]
13p. 15(4) cage at R8C4 = {2346} (only remaining combination) -> R8C4 = 4, R78C5 = [91]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough at 1.5; I used some forcing chains, I think the 'round the houses' step 11a needs a full 1.5.


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 Post subject: Re: Assassin 388
PostPosted: Tue Nov 26, 2019 8:12 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
HATMAN wrote:
Please try the Triankles Ed
I would have to just post 1 Assassin per month in that case. You are prolific Will have a think about it.

Andrew found the really nice chain that I referred to in JSudoku. Much better than my way.

Here's how JSudoku expresses that chain of Andrew's, his step 11. Also, check about Andrew's step 12b. Love it! I'm determined to find one of those one day.
JSudoku:
Grouped Turbot Fish on 9 with 5 links (9): R457C4=R78C3, R8C2-R89C1=R2C1-R3C3=R3C6 -> R1C4 <> 9
My WT:
Preliminaries
Cage 17(2) n9 - cells ={89}
Cage 7(2) n9 - cells do not use 789
Cage 8(2) n2 - cells do not use 489
Cage 12(2) n4 - cells do not use 126
Cage 9(2) n89 - cells do not use 9
Cage 10(2) n8 - cells do not use 5
Cage 9(3) n1 - cells do not use 789
Cage 20(3) n89 - cells do not use 12
Cage 11(3) n4 - cells do not use 9
Cage 19(3) n1 - cells do not use 1
Cage 30(4) n3 - cells ={6789}
Cage 13(4) n36 - cells do not use 89

1. naked quad 6,7,8,9 in n3: all locked for n3

2. r5c9 sees all 1,2,3,4,5 in n3 -> not in r5c9

3. 17(2)n9 = {89} both locked for c9 and n9

4. "45" on n12, 2 innies r3c36 = 16 = {79} only: both locked for r3

Important step for later. Liked finding it.
5. "45" on n4, 1 outie r3c3 - 1 = 1 innie r6c1 = [76/98]
5a. -> r3c6 + r6c1 = [96/78]
5b. -> r5c9 + r6c1 = [66/76/68]
5c. -> 6 locked: no 6 r6c789 nor r5c123

6. 37(7)r3c6 must have 8 which is only in n6: locked for n6

7. "45" on c6789: 2 innies r4c67 = 5 = {14/23}
7a. -> r456c5 = 21 = {489/579/678}(no 1,2,3)

8. "45" on r6789: remembering h5(2)r4c67 -> 5 remaining outies r45c45+r5c6 = 32 = {26789/35789/45689}(no 1)
8a. 8 & 9 locked for n5

9. 8 in r6 only in n4: locked for n4

10. 11(3)n4 = {137/146/236/245}
10a. {13} blocked from r4c12 by h5(2)r4c67 needs 1 or 3
10b. -> no 7 in r5c1

11. hidden killer pair 1,2 in n4 since 11(3) can't have both
11a. -> 23(4)r3c3 must have 1 or 2 = {1679/2579}(no 3,4)

12. hidden killer pair 3,4 in n4 since 11(3) can only have one of them (step 10.)
12a. 12(2)n4 must have 3 or 4 = {39/48}(no 5,7)

a key step
13. 7 in n4 is only in 23(4)r3c3 or r4c12 -> r3c3 + r4c9 cannot be [77]
13a. 13(4)r1c9 = {1237/1246/1345}: 1 locked for c9
13b. but {123}[7] -> {45}[6] in r3c78 (hidden pair n3) + r5c9 -> 37(7) = {2345689}(no 7) -> r3c6 = 9 -> r3c3 = 7: but leaves no 7 for n4
13c. -> 13(4)r1c9 = {1246/1345}(no 7)
13d. must have 4: locked for c9

And another one, why no 6 in r6c9 is helpful (step 5c)
14. "45" on c9: 3 innies r567c9 = 15
14a. must have 7 for c9 = {267/357}
14b. -> r67c9 = {27}/[26]/{35}
14c. -> 17(4)r6c9: {1367/1457} blocked
14d. = {2357/2456}(no 1)
14e. must have both 2 & 5 -> {25} blocked from 7(2)n9
14f. = {16/34}(no 2,5) = 4 or 6
14g. -> {2456} blocked from 17(4)r6c9 since both 4 & 6 are in n9
14h. = {2357} only (no 4,6)
14i. -> r67c9 = {27}/{35}

15. 12(3)r6c1 must have 6 or 8 for r6c1
15a. but [8]{13} blocked by 3 in 17(4)r6c9 forced into r6c9 but [83] in r6c19 clashes with 12(2)n4
15b. -> r6c1 = 6
15c. r7c12 = 6 = {15/24}(no 3,7,8,9) = 2 or 5
15d. -> 2 or 5 in 17(4)r6c9 must be in r6c9
15e. -> 3 and 7 only in r7c789: locked for r7 and n9 and 17(4)r6c9
15f. killer pair 2,5 in r7c12789: both locked for r7

16. r3c3 = 7 (1 outie n4), r3c6 = 9

17. 20(3)r7c6 = {578} only = [875]
17a. r9c67 = [54/36]

18. 7(2)n9 = {16} only: both locked for c8, 6 for n9
18a. r9c67 = [54]

19. 8 in n4 only in 12(2) = {48}: 4 locked for r6 and n4

20. naked triple {237} in r7c789: 2 locked for r7 and 17(4)r6c9
20a. r6c9 = 5, r6c5 = 7
20b. r7c12 = 6 = {15} only: 1 locked for r7 and n7

21. 9 in n6 only in 17(4)r5c6 = {1349} only = [4]{139}
21a. 1 & 3 locked for r6 -> r6c4 = 2

22. r4c67 = [32] (h5(2)), r6c6 = 1

23. 13(4)r1c9 = {1246} only -> r4c9 = 6; 1,2,4 locked for n3
23a. r3c78 = [35], r5c789 = [187]

24. 37(7)r4c4 = {2345689} only: 5 locked for n5 and c4, 3 locked in r8c23 for r8 and n9

25. r45c5 = 14 (cage sum) = [86] only

26. r45c4 = {59}: 9 locked for c4 and 37(7) cage
26a. r7c34 = {46}: 4 locked for r7 and both for 37(7) cage

27. "45" on n1: 2 remaining innies r12c3 = 10
27a. {46} blocked by r7c3 = (46)
27b. = {19/28}(no 3,4,5,6)

Pretty easy now
Cheers
Ed


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