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 Post subject: Assassin 333 X 15s
PostPosted: Sat Mar 19, 2016 12:13 am 
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Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Assassin 333 X 15s

Managed to do it with one cage total.

SS gives it 1.55, JS uses a lot of fishes.


Image

JS Code:
3x3:d:k:3841:3841:3841:3855:3855:3849:3849:3844:3843:3842:3842:3842:3855:3856:3856:3849:3844:3843:3850:3850:23:3861:3857:3856:24:3844:3843:3850:3861:3861:3861:3857:3858:3858:3866:25:3860:3862:3862:3853:3858:28:29:3866:3866:3860:30:3862:3853:31:3867:3867:3866:3852:3847:3848:3854:3854:3854:3854:3854:3852:3852:3847:3848:3851:3859:3859:32:3845:3845:3845:3847:3848:3851:3851:3859:33:3846:3846:3846:

Solution:
348917526
672584391
591263748
136492875
829751634
754836912
983145267
215679483
467328159


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 Post subject: Re: Assassin 333 X 15s
PostPosted: Sun Mar 27, 2016 12:51 am 
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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
Approaching HS with this one. I found it much easier than 1.55. Thanks HATMAN!

Assassin 333X 15s WT:
0. Each row in n1,n9 and each column in n3,n7 = +15(3)
-> Each of those rows and columns must have exactly one of the numbers (789)

1. 15(5)r7 = {12345}
-> r7c1289 = {6789}
Innies n9 -> r7c789 = +15(3)
-> Can only be [1{68}] or [2{67}]
-> r6c9 = r7c7 from (12)

2. Innies - Outies n7 -> r7c3 = r9c4 (from 12345)
-> 15(3)n8 can only have one number from (12345). (Same as in r7c7)
-> 15(3)n8 is {168} or {267} and must be the same three numbers as in r7c789.
-> 9 in n8 in r89c6

3. r7c12 = {79} or {89}
-> Whichever of (78) is in r7c89 can only go in n7 in r89c3
Innies n7 -> r789c3 = +15(3) and must be different from r7c789 = +15(3)
-> 6 cannot go in n7 in r89c3
-> 6 in n7 in r89c12
Since 6 cannot go in same +15(3) as 9 in n7
-> Either 9 in r7c1 or 6 in r89c1
-> 15(2)r5c1 = {78}
-> r7c1 = 9, r7c2 from (78)
Also -> 15(2)r6c6 = [69]
-> 15(2)r5c4 = {78}
-> 15(2)r3c5 = [69]
-> r8c4 = 6
-> r9c2 = 6

4. 6 in D/ only in r12
-> 6 in n6 in r45c7
-> Since r7c9 from (12) -> 15(4)n6 = {1347}
-> r6c9 = 2
-> r7c789 = [2{67}]
-> 15(3)r7c2 = [816]
-> 15(3)r7c1 = [9{24}]
-> Whichever 15(3) in n9 contains 9 must be {159}
-> 15(3)r9c7 = {159}
-> 15(3)r8c7 = {348}
-> r89c1 = [24]
-> r89c5 = [72]
Also -> r89c6 = [98]
-> r789c3 = [357] and r9c4 = 3

5. HS 7 in 15(4)n6 -> r4c8 = 7
-> r7c89 = [67]
-> HS 6 in D/ -> r1c9 = 6
-> HS 9 in D/ -> r2c8 = 9
15(3)r9c7 = {159} -> 15(3)r1c8 = [294] or [492]
-> r5c9 = 4 and r56c8 = {13}
Also -> 15(3)r1c9 = [6{18}]
-> r123c7 = {357}
-> 15(3)r9c7 - [159]
-> r4c9 = 5
-> r4c7 = 8
-> r5c7 = 6
Also 15(3)r8c7 = [483]

6. On D/ r4c6,r5c5 = +7(2) can only be [25]
-> NS 7 in r3c7
-> r12c7 = {35} and r1c6 = 7
-> 15(3)r2c1 = [672]
-> 15(3)r1c1 = [348]
-> r3c123 = [591]
etc.


Happy Easter to all.


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 Post subject: Re: Assassin 333 X 15s
PostPosted: Thu Mar 31, 2016 3:34 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
Thanks HATMAN for your latest Assassin! Maybe my walkthrough seems even more HS, for my key step 8 and the possibly unnecessary step 5.

There was some similarity with wellbeback's walkthrough, in which I particularly liked the middle of step 3.

Here is my walkthrough for Assassin 333 X 15s:
Prelims

a) R34C5 = {69/78}
b) R56C1 = {69/78}
c) R56C4 = {69/78}
d) R6C67 = {69/78}
e) 15(5) cage at R7C3 = {12345}

Steps resulting from Prelims
1. Naked quad {6789} in R6C1467, locked for R6
1a. Naked quad {6789} in R4C5 + R56C4 + R6C6, locked for N5
1b. 15(5) cage at R7C3 = {12345}, locked for R7
1c. Max R4C6 + R5C5 = 9 -> min R4C7 = 6

2. 15(3) cage at R6C9 = {168/267}, 6 locked for R7 and N9, R6C9 = {12}
2a. 45 rule on N9 1 outie R6C9 = 1 innie R7C7, R6C9 = {12} -> R7C7 = {12}
2b. 15(4) cage at R4C8 = {1347/1356/2346} (cannot be {1239/1248/1257} which clash with R6C9), no 8,9, 3 locked for N6
2c. Killer pair 1,2 in 15(4) cage and R6C9, locked for N6
2d. 9 in R7 only in R7C12, locked for N7
2e. 15(3) cage at R7C1 can only contain one of 7,8,9 -> no 7,8 in R89C1
2f. 15(3) cage at R7C2 can only contain one of 7,8,9 -> no 7,8 in R89C2
2g. 45 rule on N7 1 innie R7C3 = 1 outie R9C4 -> R9C4 = {12345}
2h. Min R567C1 = 22 = {679/689/789}, 9 locked for C1
2i. 15(3) cage at R7C1 = {159/249/258/348/357} (cannot be {168/267} which clashes with R56C1, cannot be {456} because R7C1 only contains 7,8,9), no 6 in R89C1

3. 15(3) cage at R8C4 = {168/267} (cannot be {159/249/258/348/357/456} which clash with R7C456 + R9C4), 6 locked for N8
3a. 1,2 of 15(3) cage must be in R89C5 (R89C5 cannot be {67/68} which clash with R34C5) -> no 1,2 in R8C4
3b. 9 in N8 only in R89C6, locked for C6, clean-up: no 6 in R6C7
3c. Hidden killer pair 7,8 in 15(3) cage and R89C6 for N8 -> R89C6 = {79/89}
3d. 45 rule on N3 1 outie R1C6 = 1 innie R3C7, no 9 in R1C6 -> no 9 in R3C7
3e. R56C4 = {69/78}, R8C4 = {678} -> R568C4 = {69}7/{69}8/{78}6, 6 locked for C4

4. 45 rule on C89 1 innie R4C9 = 2 outies R89C7, max R89C7 = 9, no 9 in R89C7

5. R6C9 = R7C7 (step 2a) -> whichever of 1,2 is in these cells must also be in 15(3) cage at R1C8 = {159/168/249/258/267}, no 3

6. Naked quad {6789} in R4C5 + R56C4 + R6C6, R56C4 = {69/78} -> R4C5 + R6C6 = [78/87/96], no 6 in R4C5, clean-up: no 9 in R3C5
6a. Consider combinations for R56C4 = {69/78}
R56C4 = {69}, locked for C4 => 6 in N8 only in R89C5, locked for C5 => 6 in N2 only in R123C6
or R56C4 = {78} => R4C5 + R6C6 = [96] => R3C5 = 6
-> no 6 in R12C5

7. 45 rule on N23 3 innies R3C457 = 15, min R3C5 = 6 -> max R3C47 = 9, no 9 in R3C4

8. Where there are three 15(3)s, including hidden 15(3)s, in the same row or nonet, the sets of combinations must be {159/267/348} or {168/249/357}
8a. 15(3) cage at R1C1 ‘sees’ all of R1C789 -> 15(3) cages in N3, including the hidden 15(3) cage, must have the opposite set of combinations than the 15(3) cages in N1
8b. Similarly 15(3) cage at R1C9 ‘sees’ all of R789C9 -> 15(3) cages in N3 must have the opposite set of combinations than the 15(3) cages in N3 -> 15(3) cages in N1 and N9 must have the same set of combinations and 15(3) cages in N3 and N7 must have the same set of combinations

9. R6C9 = R7C7 (step 2a), 15(3) cage at R6C9 (step 2) = {168/267} -> R7C789 = {168/267}
9a. Consider combinations for R7C789
R7C789 = {168}, locked for R7 => R7C12 = {79}, set of combinations in N7 = {159/267/348} (step 8b) => 15(3) cage at R7C1 = {159} (cannot be {267} which clashes with R56C1), 15(3) cage at R7C2 = {267}, R789C3 = {348}
or R7C789 = {267}, locked for R7 => R7C12 = {89}, set of combinations in N7 = {168/249/357} => 15(3) cage at R7C1 = {249} (cannot be {168} which clashes with R56C1), 15(3) cage at R7C2 = {168}, R789C3 = {357}
-> 15(3) cage at R7C1 = {159/249}, 15(3) cage at R7C2 = {168/267}, R789C3 = {348/357}
9b. 15(3) cage at R7C1 = {159/249} -> R7C1 = 9, R89C1 = {15/24}, clean-up: no 6 in R56C1
9c. R789C3 = {348/357}, 3 locked for C3 and N7 -> 15(3) cage at R8C3 = {348/357}
9d. 15(3) cage at R7C2 = {168/267}, 6 locked for C2
9e. Naked pair {78} in R56C1, locked for C1 and N4
9f. R6C67 = [69] (cannot be {78} which clashes with R6C1), 6 placed for D\, clean-up: no 9 in R5C4
9g. Naked pair {78} in R56C4, locked for C4 and N5 -> R4C5 = 9, R3C5 = 6, R8C4 = 6, R9C2 = 6 (hidden single in N7)
9h. Naked pair {78} in R5C14, locked for R5
[Cracked, the rest is fairly straightforward; clean-ups omitted.]

10. R1C9 + R2C8 = {69} (hidden pair on D/), locked for N3
10a. R3C7 + R6C4 = {78} (hidden pair on D/)
10b. R3C457 = 15 = [168/267]
10c. 9 in C4 only in 15(3) cage at R1C4 = {159/249}
10d. Killer pair 1,2 in 15(3) cage and R3C4, locked for N2
10e. R1C6 = R3C7 (step 3d) = {78}
10f. Naked triple {789} in R189C6, locked for C6
10g. R1C6 + R2C5 = {78} (hidden pair in N2)

11. 15(3) cage at R1C6 = {258/348/357} -> R12C7 = {25/34/35}
11a. R1C9 = {69} -> 15(3) cage at R1C9 = {159/168/249/267} (cannot be {357/456} which clash with R12C7), no 3
11b. Killer pair 1,2 in 15(3) cage and R6C9, locked for C9
11c. 3 in N3 only in R12C7 = {34/35}, locked for C7
11d. 1,2 in C7 only in R789C7, locked for N9

12. 9 in N4 only in 15(3) cage at R5C2 = {159/249}
12a. 6 in N4 only in R4C13, locked for R4
12b. Naked pair {78} in R34C7, locked for C7
12c. R4C7 = {78} -> 15(3) cage at R4C6 = {258/348/357} -> R4C6 + R5C5 = {25/34/35}
12d. 1 on D/ only in R8C2 + R9C1, locked for N7
12e. R5C7 = 6 (hidden single in C7)

13. 45 rule on N1 1 outie R4C1 = 1 innie R3C3 = {1245}
13a. R4C3 = 6 (hidden single in N4)
13b. 15(4) cage at R3C4 = {1356/2346}, 3 locked for R4
13c. R3C4 = {12} -> no 1,2 in R4C24
13d. 3 in N4 only in R46C2, locked for C2

14. 45 rule on N6 3 remaining innies R4C79 + R6C9 = 15 = [852] (only possible combination)
14a. R3C7 = 7, placed for D/ -> R56C4 = [78], R56C1 = [87]
14b. 15(3) cage at R1C9 (step 11a) = {168} (only remaining combination) -> R1C9 = 6, R23C9 = {18}, locked for C9 and N3, R2C8 = 9, R7C9 = 7, R7C8 = 6 (cage sum), R7C2 = 8, R8C2 = 1 (cage sum)
14c. R7C1 = 9 -> R89C1 = {24}, locked for C1 and N7
14d. R4C1 = 1 -> R3C12 = [59], R1C1 = 3, placed for D\, R4C4 = 4, placed for D\
14e. R4C6 = 2, placed for D/, R5C5 = 5, placed for both diagonals
14f. R1C1 = 3 -> R1C23 = 12 = [48], R5C2 = 2
14g. R2C1 = 6 -> R2C23 = 9 = [72]

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

Rating Comment:
I'll rate my walkthrough for A333 at 1.5 for my step 8, plus I used a short forcing chain.


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