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 Post subject: Assassin 336 X
PostPosted: Thu Jun 09, 2016 5:26 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Assassin 336 X

A serendipity one, SS gives it 1.5 but JS uses 24 fishes - a lot of work.

Image

JS Code:
3x3:d:k:4609:4609:4098:4098:2819:2820:2820:5125:5125:4609:4609:4614:4098:2819:2820:3847:3847:5125:4360:4614:4614:1545:1545:4874:3847:3847:5387:4360:4360:2316:8461:8461:4874:4874:5387:5387:3342:3342:2316:8461:8461:8461:2063:1296:1296:2833:2833:3346:3346:8461:8461:2063:3603:3603:2833:6932:6932:3346:3605:3605:4630:4630:3603:2327:6932:6932:4120:3097:3354:4630:4635:4635:2327:2327:4120:4120:3097:3354:3354:4635:4635:

Solution:
784361259
213958476
596247138
932514867
857629341
641873592
175486923
469132785
328795614


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 Post subject: Re: Assassin 336 X
PostPosted: Fri Jun 10, 2016 6:11 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Assassin 336 X E

I tried for a slightly easier one. JS does it without fishes (still a lot of work), but SS gives it 1.7 - strange.

Image

JS Code:
3x3:d:k:4609:4609:4098:4098:2819:2820:2820:5125:5125:4609:4609:4614:4098:2819:2820:3847:3847:5125:4360:4614:4614:1545:1545:4874:3847:3847:5387:4360:4360:2316:8461:8461:4874:4874:5387:5387:3342:3342:2316:8461:8461:8461:2063:1296:1296:2833:2833:3346:3346:8461:8461:2063:3603:3603:2833:4124:2845:3346:3605:3605:4630:4630:3603:2327:2845:4124:4120:3097:3354:4630:4635:4635:2327:2327:4120:4120:3097:3354:3354:4635:4635:


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 Post subject: Re: Assassin 336 X
PostPosted: Tue Jun 21, 2016 4:45 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
I wrote this a week ago but somehow forgot to post it. Thanks for the puzzle HATMAN!

Hidden Text:
1. Where does 9 go in c23?
Innies c12 -> r378c2 = +22(3) = {679} or {589}
-> 9 in c2 in r378c2

Innies n1 -> r1c3,r3c1 = +9(2) -> No 9 in r1c3
27(4)n7 contains a 9 -> No 9 in r9c3
Innies n5 -> r4c6,r6c4 = +12(2) -> min r6c4 = 3
Trying 9 in r6c3 puts r67c4 = [31] puts r4c6 = 9 leaves no place for 9 in r5
-> 9 in c3 in r2378c3
-> 18(3)r2c3 contains a 9

2. Innies r89 -> r8c237 = +22(3) = {679} or {589}
Innies r789 -> r7c149 = +8(3) = {125} or {134}
If the former puts 14(2)n8 = {68}
If the latter puts 2 in r7 in r7c78 puts 18(3)r7c7 = {279} -> 9 locked in r78 in 27(4) and 18(3) puts 14(2)n8 = {68}
Either way -> 14(2)n8 = {68}

3. +22(3)r8c237 cannot contain both 6 and 8
-> 27(4)n7 cannot contain both 6 and 8
-> 27(4)n7 from {3789} or {5679}
If the former puts 3 in r7c3 -> 9(3)n7 = {126} -> (innies r789) +8(3)r7c149 = [5{12}] -> r9c3 = 4
If the latter puts 8 in r9c3 -> 1 in r7c1

-> Innies n7 r7c1,r9c3 either [54] or [18]

But if the former - where does 5 go in c3?

a) 5 cannot go in r789c3
b) Since 4 in r9c3 -> 5 cannot go in r45c3
c) Remaining innies c123 are r16c3 = +9(2) -> 5 cannot go in r16c3
d) Since 4 cannot go anywhere in 18(3)r2c3 -> neither can 5

-> Innies n7 r7cr9c3 = [18]

4. Essentially cracked now. E.g.,
9(3)n7 = {234}
27(4)n7 = {5679} with 6 in r8c23
-> 5 in r7c23 (Since H22(3) cannot contain both (56))
-> r7c49 = {34}
-> 2 in r7 in r7c78
-> 18(3)n9 = {279}
Also -> r9c7 from (56)
-> HS 9 in r9 -> 12(2)n7 = [39]
-> r7c49 = [43]
-> r9c7 = 6
-> r89c6 = {25}
-> r89c4 = [17]
-> Since Innies r1234 = r4c345 = +8(3) = {125} or {134} -> r4c5 = 1
-> HS 1 in n4 -> r6c3 = 1
-> r6c4 = 8
-> r4c6 = 4
Also (Innies c123) -> r1c3 = 4
-> r3c1 = 5
etc.


Last edited by wellbeback on Thu Sep 29, 2016 6:06 pm, edited 1 time in total.

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 Post subject: Re: Assassin 336 X
PostPosted: Sat Jun 25, 2016 5:25 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Well spotted - an excellent breakthrough sequence - I felt there was something in that area but could not see it.


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 Post subject: Re: Assassin 336 X
PostPosted: Mon Aug 29, 2016 2:36 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
I tried the easier version first, finishing it three weeks ago. At that time I wasn't going to post this walkthrough until I'd tried the full Assassin, but things have been so busy for me in the past three weeks including watching a couple of sports of the Rio Olympics that I haven't had time to look at any puzzles on this site.

So here's how I did the easier version; as usual in the order I saw things - I know it could be optimised somewhat.

The SS score for this puzzle is incomprehensible. This puzzle MUST be easier that the full version, since the 4-cell cage in N7 has been split into two 2-cell cages, one of which has only one combination.

Here is my walkthrough for Assassin 336 X E:
Prelims

a) R12C5 = {29/38/47/56}, no 1
b) R3C45 = {15/24}
c) R45C3 = {18/27/36/45}, no 9
d) R5C12 = {49/58/67}, no 1,2,3
e) R56C7 = {17/26/35}, no 4,8,9
f) R5C89 = {14/23}
g) 16(2) cage at R8C2 = {79}
h) 11(2) cage at R8C3 = {29/38/47/56}, no 1
i) R7C56 = {59/68}
j) R89C5 = {39/48/57}, no 1,2,6
k) 11(3) cage at R1C6 = {128/137/146/236/245}, no 9
l) 20(3) cage at R1C8 = {389/479/569/578}, no 1,2
m) 19(3) cage at R3C6 = {289/379/469/478/568}, no 1
n) 21(3) cage at R3C9 = {489/579/678}, no 1,2,3
o) 11(3) cage at R6C1 = {128/137/146/236/245}, no 9
p) 9(3) cage at R8C1 = {126/135/234}, no 7,8,9

1. Naked pair {79} in 16(2) cage at R8C2, locked for N7, clean-up: no 2,4 in 11(2) cage at R8C3
1a. 45 rule on N7 2 innies R7C1 + R9C3 = 9 = {18/45} (cannot be {36} which clashes with 11(2) cage)
1b. Killer pair 5,8 in R7C1 + R9C3 and 11(2) cage, locked for N7

2a. 45 rule on N1 2 innies R1C3 + R3C1 = 9 = {18/27/36/45}, no 9
2b. 45 rule on N3 2 innies R1C7 + R3C9 = 10 = [19/28/37/46/64]
2c. 45 rule on N9 2 innies R7C9 + R9C7 = 9 = {18/27/36/45}, no 9

3a. 45 rule on R123 3 innies R3C169 = 20 = {389/479/569/578}, no 1,2, clean-up: no 7,8 in R1C3 (step 2a)
3b. 45 rule on R1234 3 innies R4C345 = 8 = {125/134}, 1 locked for R4, clean-up: no 1,2,3 in R5C3
3c. 45 rule on R789 3 innies R7C149 = 8 = {125/134}, 1 locked for R7, clean-up: no 1 in R9C3 (step 1a), no 1,2,3 in R9C7 (step 2c)

4. 45 rule on C12 3 innies R378C2 = 22 = {589/679}, 9 locked for C2, clean-up: no 4 in R5C1, no 8 in R7C3
4a. 6 of {679} must be in R8C2 -> no 6 in R3C2
4b. Killer triple 3,5,6 in R7C149, R7C3 and R7C56, locked for R7

5. 45 rule on N5 2 innies R4C6 + R6C4 = 12 = {39/48/57}, no 1,2,6

6. 18(3) cage at R7C7 = {279/459/468} (cannot be {189} which clashes with R7C56, cannot be {369/567} because 3,5,6 only in R8C7, cannot be {378} which clashes with R7C2 + R7C56, killer ALS block), no 1,3
6a. 2 of {279} must be in R7C78 (R7C78 cannot be {79} which clashes with R7C2) -> no 2 in R8C7
6b. 5,6 of {459/468} must be in R8C7 -> no 4,8 in R8C7

[I ought to have seen this 45 earlier …; at least it avoids the messy combined cage for R7C5678 which I’d been looking at]
7. 45 rule on R89 3 innies R8C237 = 22 = {589/679}, 9 locked for R8, clean-up: no 3 in R9C5
7a. R8C237 = 6{79}/[895] -> R8C2 = {68}, no 6 in R8C7, clean-up: no 6 in R7C3
7b. 6 in R7 only in R7C56 = {68}, locked for R7 and N8, clean-up: no 4 in R89C5
7c. 18(3) cage at R7C7 (step 6) = {279/459}, 9 locked for N9
7d. R378C2 (step 4) = {589/679}, R8C2 = {68} -> no 8 in R3C2

8. R3C169 (step 3a) = {389/569/578} (cannot be {479} which clashes with R3C2 + R3C45, killer ALS block), no 4, clean-up: no 5 in R1C3 (step 2a), no 6 in R1C7 (step 2b)
8a. Combined half cage R3C139 + R3C2 must contain 9, locked for R3

[Taking steps 4 and 7 further … cracks this puzzle]
9. R378C2 (step 4) = {679}, R8C237 (step 7) = {679} (they cannot be [598] and [895] which clash with 16(2) cage at R7C3) -> R8C2 = 6, R7C3 = 5, both placed for D/, R37C2 = {79}, locked for C2, R8C37 = {79}, locked for R8, clean-up: no 4 in R45C3, no 7 in R4C6 + R6C4 (step 5), no 6,7 in R5C1, no 2 in R7C149 (step 3c), no 4 in R7C1 + R9C3 (step 1a), no 5 in R9C5, no 4,7 in R9C7 (step 2c)
9a. R7C9 + R9C7 = [18], clean-up: no 1 in R4C3
9b. Naked pair {34} in R7C49, locked for R7
9c. Naked triple {279} in 18(3) cage at R7C7, locked for N9
9d. R6C3 = 1 (hidden single in N4) -> R67C4 = 12 = [84/93], clean-up: no 8,9 in R4C6 (step 5), no 7 in R5C7
9e. R9C3 = 8 -> R89C4 = 8 = [17] (cannot be {35} which clashes with R8C5), R9C5 = 9 -> R8C5 = 3, R7C4 = 4 -> R6C4 = 8 -> R4C6 = 4 (step 5), both placed for D/
9f. Naked pair {25} in R89C6, locked for C6 and 13(3) cage at R8C6 -> R9C7 = 6, R7C9 = 3
9g. R7C1 = 1 -> R6C12 = 10 = [64] (cannot be [73] which clashes with R45C3), R5C3 = 7 -> R4C3 = 2, R8C3 = 9, R7C2 = 7, R3C2 = 9, R8C7 = 7
9h. R3C2 = 9 -> R23C3 = 9 = {36}, locked for N1 -> R1C3 = 4, R3C1 = 5 (step 2a)
9i. R3C1 = 5 -> R4C12 = 12 = [93], 9(3) cage at R8C1 = [432], 3 placed for D/, R3C4 = 2 -> R3C5 = 4, R3C7 = 1, placed for D/, R5C5 = 2, placed for both diagonals
9j. R2C2 + R3C3 + R9C9 = [164] (hidden triple on D\)

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough for A336 X E at 1.25. Maybe that's a bit on the high side; it's based on using killer ALS blocks and on the interaction in step 9 which cracks the puzzle.


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 Post subject: Re: Assassin 336 X
PostPosted: Thu Sep 22, 2016 6:03 pm 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Quite a long time since I did the easier version; I've now done the Assassin level one.

Nice solving by wellbeback! More direct that mine; I used a lot of the same steps.

Here is my walkthrough for Assassin 336X:
Prelims

a) R12C5 = {29/38/47/56}, no 1
b) R3C45 = {15/24}
c) R45C3 = {18/27/36/45}, no 9
d) R5C12 = {49/58/67}, no 1,2,3
e) R56C7 = {17/26/35}, no 4,8,9
f) R5C89 = {14/23}
g) R7C56 = {59/68}
h) R89C5 = {39/48/57}, no 1,2,6
i) 11(3) cage at R1C6 = {128/137/146/236/245}, no 9
j) 20(3) cage at R1C8 = {389/479/569/578}, no 1,2
k) 19(3) cage at R3C6 = {289/379/469/478/568}, no 1
l) 21(3) cage at R3C9 = {489/579/678}, no 1,2,3
m) 11(3) cage at R6C1 = {128/137/146/236/245}, no 9
n) 9(3) cage at R8C1 = {126/135/234}, no 7,8,9
o) 27(4) cage at R7C2 = {3789/4689/5679}, no 1,2

1. 27(4) cage at R7C2 = {3789/4689/5679}, 9 locked for N7
1a. 45 rule on N1 2 innies R1C3 + R3C1 = 9 = {18/27/36/45}, no 9
1b. 45 rule on N3 2 innies R1C7 + R3C9 = 10 = [19/28/37/46/64]
1c. 45 rule on N7 2 innies R7C1 + R9C3 = 9 {18/27/45} (cannot be {36} which clashes with 27(4) cage), no 3,6
1d. 45 rule on N9 2 innies R7C9 + R9C7 = 9 = {18/27/36/45}, no 9

2. 45 rule on N5 2 innies R4C6 + R6C4 = 12 = {39/48/57}, no 1,2,6

3a. 45 rule on R123 3 innies R3C169 = 20 = {389/479/569/578}, no 1,2, clean-up: no 7,8 in R1C3 (step 1a)
3b. 45 rule on R1234 3 innies R4C345 = 8 = {125/134}, 1 locked for R4, clean-up: no 1,2,3 in R5C3
3c. 45 rule on R789 3 innies R7C149 = 8 = {125/134}, 1 locked for R7, clean-up: no 1,2 in R9C3 (step 1c), no 1,2,3 in R9C7 (step 1d)
3d. Min R9C7 = 4 -> max R89C6 = 9, no 9 in R89C6

4a. 45 rule on C12 3 innies R378C2 = 22 = {589/679}, 9 locked for C2, clean-up: no 4 in R5C1
4b. 45 rule on R89 3 innies R8C237 = 22 = {589/679}, 9 locked for R8, clean-up: no 3 in R9C5
4c. Consider combinations for 27(4) cage at R7C2 = {3789/4689/5679}
27(4) cage = {3789} = [7398/8397] (R78C2 and R8C23 cannot be {78} because 22(3) hidden cages only contain one of 7,8)
or 27(4) cage = {4689} = [6498/8496] (R78C2 and R8C23 cannot be {68} because 22(3) hidden cages only contain one of 6,8)
or 9 of 27(4) cage = {5679} must be in R7C2 + R8C23 (R7C2 + R8C23 cannot be {567} because 22(3) hidden cages cannot contain 5 and one of 6,7)
-> no 8,9 in R7C3, no 8 in R8C2
4d. Consider combinations for R7C149 (step 3c) = {125/134}
R7C149 = {125}, 5 locked for R7 => R7C56 = {68}, locked for R7 => 27(4) cage = [7398]/{5679}
or R7C149 = {134}, 3,4 locked for R7 => 27(4) cage = {5679}
-> 27(4) cage = {3789/5679}, no 4, no 8 in R7C2, 7 locked for N7, clean-up: no 2 in R7C1 (step 1c)
4e. 9(3) cage at R8C1 = {126/234} (cannot be {135} which clashes with 27(4) cage), no 5
4f. 8 in N7 only in R89C3, locked for C3, clean-up: no 1 in R4C3
4g. 8 in N7 only in R89C3, CPE no 8 in R8C4
4h. 1 in R4 only in R4C45, locked for N5
4i. 1 in R5 only in R5C789, locked for N6, clean-up: no 7 in R5C7
4j. 8 of R378C2 = {589} must be in R3C2 -> no 5 in R3C2

5. R7C149 (step 3c) = {125/134}
5a. Hidden killer pair 2,4 in R7C149 and 18(3) cage at R7C7 for R7, R7C149 contains one of 2,4 -> 18(3) cage must contain one of 2,4 = {279/459/468}, no 3
5b. 18(3) cage = {279/459} (cannot be {468} which clashes with R7C149 + R7C56, killer ALS block), no 6,8, 9 locked for N9
5c. 5 of {459} must be in R8C7 (R7C78 cannot be {45} which clashes with R7C149) -> no 5 in R7C78
5d. 8 in R7 only in R7C56 = {68}, locked for R7 and N8, clean-up: no 4 in R89C5

6. 45 rule on C123 3 innies R169C3 = 13 = {148/157/238/247/256/346} (cannot be {139} because R9C3 only contains 4,5,8), no 9
6a. 5 of {157/256} must be in R9C3 -> no 5 in R16C3, clean-up: no 4 in R3C1 (step 1a)
6b. 9 in N4 only in R45C1, locked for C1

7. 27(4) cage at R7C2 (step 4d) = {3789/5679}
7a. Consider combinations for R45C3
R45C3 = [27]/{45} => 2 or 4 locked for N4 => 11(3) cage at R6C1 cannot be {24}5
or R45C3 = [36] => 27(4) cage = {5679}, 5 locked for N7
-> no 5 in R7C1, clean-up: no 4 in R9C3 (step 1c)

8. R7C149 (step 3c) = {125/134}, R7C1 + R9C3 (step 1c) = [18/45], 27(4) cage at R7C2 (step 4d) = {3789/5679}
8a. Consider combinations for 18(3) cage at R7C7 = {279/459}
18(3) cage = {279}, 2 locked for R7 => R7C149 = {134}, locked for R7 => 27(4) cage = {5679}
or 18(3) cage = {459}, 4 locked for R7 => R7C1 = 1, R9C3 = 8 => 27(4) cage = {5679}
-> 27(4) cage = {5679}, locked for N7 -> R9C3 = 8, R7C1 = 1
8b. R6C3 = 1 (hidden single in R6), R9C3 = 8 -> R1C3 = 4 (step 6), R3C1 = 5 (step 1a)
8c. 6 in N7 only in R8C23, locked for R8 -> R8C237 (step 4b) = {679}, 7 locked for R8
8d. 27(4) cage = {5679}, 5 locked for R7 => R7C149 = 1{34}, 4 locked for R7
8e. Naked triple {279} in 18(3) cage at R7C7, locked for N9
8f. R9C3 = 8 -> R89C4 = 8 = [17] (cannot be {35} which clashes with R89C5)
8g. R9C5 = 9 (hidden single in R9) -> R8C5 = 3, R7C4 = 4, R6C4 = 8 (cage sum), R4C6 = 4 (step 2), both placed for D/, R7C9 = 3 -> R9C7 = 6 (step 1d), R3C4 = 2 -> R3C5 = 4
8h. R12C5 = {56} (only remaining combination), locked for C5 and N2 -> R7C56 = [86]
8i. Naked pair {39} in R12C4, locked for C4 and N2 -> R4C4 = 5, placed for D\, R5C4 = 6
8j. Naked pair {27} in R56C5, locked for N5
8k. 1 in N2 only in R12C6, locked for 11(3) cage at R1C6 -> R1C7 = {23}
8l. Clean-ups: no 6,9 in R3C9 (step 1b), no 3 in R4C3, no 7 in R5C1, no 7,8 in R5C2, no 5 in R5C3, no 2 in R56C7, no 2 in R5C8

9. R3C6 = {78}, R4C6 = 4 -> R4C7 = {78} (cage sum)
9a. Naked pair {78} in R3C69, locked for R3
9b. R378C2 (step 4a) = {679} (only remaining combination), locked for C2
9c. R45C3 = [27], R7C3 = 5, placed for D/, R5C5 = 2, placed for both diagonals, R6C5 = 7, R9C1 = 3, placed for D/, clean-up: no 1 in R5C7, no 3 in R5C8
9d. Naked pair {14} in R5C89, locked for R5 and N6, R5C2 = 5 -> R5C1 = 8, R4C2 = 3, R6C12 = [64], R4C1 = 9, R56C7 = [35], R1C7 = 2 -> R3C9 = 8 (step 1b)
9e. Naked pair {79} in R78C7, locked for C7 and N9
9f. R23C7 = [41] = 5 -> R23C8 = 10 = [73], 7 placed for D/
9g. R1C1 = 7, placed for D\, R7C7 = 9, placed for D\
9h. R8C8 = 8 (hidden single in C8), placed for D\ -> R2C2 = 1, placed for D\

and the rest is naked singles, without using the diagonals.

Rating Comment:
I'll rate my walkthrough at 1.5; I used several fairly short forcing chains.


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