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 Post subject: Assassin 332 X Twelves
PostPosted: Thu Mar 10, 2016 12:53 pm 
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Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Assassin 332 X Twelves

I tried to do it all with 12-cages, but could not get a unique solution.

SS gives it 1.5 but JS fails to solve - so perhaps harder?


Image

JS Code:
3x3:d:k:3086:3086:11:13:19:20:21:22:23:3087:3087:24:25:26:27:28:29:30:31:3088:3088:32:33:34:3084:35:36:3089:3088:3088:37:38:39:3084:3084:40:3089:41:42:43:44:45:3082:3084:46:47:48:49:50:51:52:3082:3081:3081:3602:3079:3079:3077:3077:3076:3076:3076:3076:53:3602:3080:3080:54:3078:3075:3074:3073:3602:55:3080:3080:56:3078:3075:3074:3073:

Solution:
576312948
398746152
214958376
425879613
831564729
769231584
957483261
642197835
183625497


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PostPosted: Mon Mar 14, 2016 5:01 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
Thanks again HATMAN. A fun puzzle. Not too difficult.

Assassin 332 X Twelves WT:
1. Naked sextet (!) r89c789 = {345789}
-> r7c789 = {126}
-> r7c6 = 3
-> 12(2)s in n8 are {48} and {57}
-> whichever of {48} or {57} is in r89c6 is also in r7c23
-> NS r7c1 = 9
-> 9 in n8 in r89c5

2. Since one of (45) is in r89c6 and also in r7c23 -> 12(4)r8c3 cannot be {1245}
-> 12(4)r8c3 = {1236} with 3 in r89c3
-> 14(3)n7 = [9{14}]
-> 12(2)r7c2 = {57}
-> 12(2)r7c4 = {48}
-> 12(2)r8c6 = {57}
-> 8 in n7 in r8c1 or r9c2

3! Since 4 and 8 are on different rows in n9 -> 4 and 8 cannot both be in the same row in n7
-> either r89c1 is [84] or r89c2 is [48]
-> 12(2)s in n1 cannot be {48}
-> 12(2)s in n1 are {57} and [39]
-> 12(2)r4c1 = {48}
-> r89c2 = [48]
-> r9c1 = 1
-> r8c4 = 1

4. Whichever of (57) is in r12c2 must go in n7 in r7c3
-> it must also go in c1/n4 in r6c1
-> NP r38c1 = {26}

5. Given:
a) One of (26) is in r89c3 and also in r3c1, and
b) 3 in n1 in r12c1 and 3 in n7 in r89c3 -> 3 in c2/n4 in r456c2
-> 12(4)r3c2 cannot be {1236}
(3 would have to go in r4c2 which leaves no place for the (2 or 6) that's in r3c1 and r89c3)
-> 12(4)r3c2 = {1245}
-> r3c3 = 4
Also -> 5 in r4c23
-> 5 in n1 in r12c1
-> 7 in n1 in r12c2
-> r7c23 = [57]
-> r4c3 = 5
-> r34c2 = {12}
Also -> r6c1 = 7
Also -> r56c2 = {36}

6. Innies - outies n6 -> r45c9 = r3c7 + 9
r3c7 from (2356) (1,4 already in D/)
r3c7 cannot be 5 (would put r45c9 = +14(2) which leaves no place for 7 in n6)
-> r3c7 from (236)
a) 2 in r3c7 puts r45c9 = {29}
b) 3 in r3c7 puts either 12(2)r8c8 = {39} or 12(2)r8c9 = {39}
puts 12(2)r6c8 not = {39}
puts r45c9 = {39}
c) 6 in r3c7 puts r45c9 = {69}

-> r45c9 = {(2|3|6)9}
-> HS 7 in n6 -> 12(2)r5c7 = [75]
-> 12(2)r6c8 = {48}
-> 12(2)r8c7 = [84]
-> 12(2)r8c8 = {39}
-> 12(2)r8c9 = {57}

7. Naked Quad in D\ in r1289 = {3579}
-> 8 in D\ in r4c4 or r5c5
Given that and also given:
a) r45c1 = {48}, and
b) r7c45 = {48}, and
c) r6c89 = {48}, and
d) 4 already in both diagonals
-> 4 and 8 in n5 are in not in the same row or column.
-> Only places for them in n5 are r4c4 = 8 and r5c6 = 4
-> r45c1 = [48], r7c45 = [48]
-> 4 in n2 in r23c5 and 8 in n2 in r123c6

8. -> HS 7 in n5 -> r4c5 = 7
Also 8 in D/ in r1c9 or r2c8
-> HS 8 in n2/r3 -> r3c6 = 8
Also 9 in n8 in r89c5
Also 9 in n3 in r12c7
-> HS 9 in n2/r3 -> r3c4 = 9
-> HS 9 in D/ -> r4c6 = 9
-> HS 9 in n6 -> r5c9 = 9
-> HS 9 in n4/c3 -> r6c3 = 9
Also HS 5 in n5/r5 -> r5c4 = 5 (cannot be in r5c5 since NQ in D\ r1289 = {3579})
-> NS 5 in D/ -> r2c8 = 5
-> r1c12 = [57]
-> r2c12 = [39]
-> r89c8 = [39]
-> r89c9 = [57]
-> r89c6 = [75]
Also r9c3 = 3

9. Finally
HS 8 in D/ -> r1c9 = 8
-> r2c3 = 8
Also HS 5 in n2/r3 -> r3c5 = 5
Also 3 in n5 in r6c45
-> HS 3 in n4/c2 -> r5c2 = 3
-> r6c2 = 6
-> r5c5 = 6
-> r9c45 = [62]
-> r8c3 = 2
-> r38c1 = [26]
-> r34c2 = [12]
etc.

Rating:
1.25. The (first) key move was my Step 3.


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PostPosted: Thu Mar 24, 2016 11:04 pm 
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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 another interesting puzzle.
Loved the interactions:
even though it took me a very long time to spot the most important one. An ideal puzzle for wellbeback who spotted that a lot quicker than I did.
It felt like a Human Solvable; maybe it would have been if the SS score had been higher. It is, of course, also acceptable as an Assassin.

Here is my walkthrough for Assassin 332 X Twelves:
Prelims

a) R1C12 = {39/47/57}, no 1,2,6
b) R2C12 = {39/47/57}, no 1,2,6
c) R45C1 = {39/47/57}, no 1,2,6
d) R56C7 = {39/47/57}, no 1,2,6
e) R6C89 = {39/47/57}, no 1,2,6
f) R7C23 = {39/47/57}, no 1,2,6
g) R7C45 = {39/47/57}, no 1,2,6
h) R89C6 = {39/47/57}, no 1,2,6
i) R89C7 = {39/47/57}, no 1,2,6
j) R89C7 = {39/47/57}, no 1,2,6
k) R89C7 = {39/47/57}, no 1,2,6
l) 12(4) cage at R3C2 = {1236/1245}
m) 12(4) cage at R3C7 = {1236/1245}
n) 12(4) cage at R7C6 = {1236/1245}
o) 12(4) cage at R8C3 = {1236/1245}

1. 45 rule on N9 1 outies R7C6 = 3, clean-up: no 9 in R7C2345, no 9 in R89C6
1a. Naked quad {4578} in R7C45 + R89C6, locked for N8
1b. R7C1 = 9 (hidden single in R7) -> R8C2 + R9C1 = 5 = {14/23}, clean-up: no 3 in R12C2, no 3 in R45C1
1c. 9 in N8 only in R89C5, locked for C5
1d. 12(4) cage at R8C3 = {1236} (cannot be {1245} which clashes with R7C23), 3 locked for C3 and N7, clean-up: no 2 in R8C2 + R9C1
1e. Naked pair {14} in R8C2 + R9C1, locked for N7 and D/, clean-up: no 8 in R7C23
1f. Naked pair {57} in R7C23, locked for R7 and N7
1g. Naked pair {48} in R7C45, locked for R7 and N8
1h. Naked pair {57} in R89C6, locked for C6
1i. 12(4) cage at R8C3 = {1236}, 1 locked for C4 and N8

2. 12(4) cage at R3C7 = {1236/1245}, 1 locked for N6
2a. 5 of {1245} must be in R3C7 (cannot be 2{145} which clashes with R56C7 + R6C89) -> no 5 in R4C78 + R5C8

3. R89C8 and R89C9 must be different from R6C89 -> whichever combination is in R6C89 must be in R89C7
3a. 12(4) cage at R3C7 = {1236/1245}
3b. Consider combinations for R6C89 = {39/48/57}
R6C89 = {39}, locked for N6 => R56C7 = {48/57} => 12(4) cage = {1236}
or R6C89 = {48}, locked for N6 => 12(4) cage = {1236}
or R6C89 = {57} => R89C7 = {57}, locked for C7 => 12(4) cage = {1236}
-> 12(4) cage = {1236}
3c. R56C7 = {48/57} (cannot be {39} which clashes with 12(4) cage)
3d. R6C89 = {48/57} (cannot be {39} because R6C89 {39} + R89C7 = {39} clash with 12(4) cage)
3e. R6C89 = {48/57} -> R89C7 = {48/57}
3f. Naked quad {4578} in R56C7 + R6C89, locked for N6
3g. Naked quad {4578} in R5689C7, locked for C7
3h. 9 in N6 only in R45C9, locked for C9, clean-up: no 3 in R89C9
3i. R89C8 = {39} (hidden pair in N9), locked for C8
3j. 12(4) cage = {1236}, 3 locked for C7
3k. Naked triple {126} in R457C8, locked for C8
3l. 9 on D/ only in R4C6 + R6C4, locked for N5

[It seems likely that 5,7 in N9 must be in R89C9, since R89C6 and R89C7 both {57} would appear to be a Killer UR. I’ll leave that for now and only use it if needed. I prefer puzzles which can be solved uniquely, rather than relying on uniqueness to solve them.]

4. In the same way as for step 3, R1C12, R2C12 and R45C1, which are all 12(2) cages and ‘see’ at least one cell of each other, must have different combinations -> one of R1C12 and R2C12 must be {39}, locked for N1
4a. Either R1C1 = 3 or R2C2 = 9 -> killer pair 3,9 in R1C1 + R2C2 and R8C8, locked for D\
4b. 3 in N1 only in R12C1, locked for C1
4c. 9 in N1 only in R12C2, locked for C2
4d. 9 in R3 only in R3C46, locked for N2

5. Taking step 4 a bit further, consider combinations for R45C1 = {48/57}
R45C1 = {48} => R1C12 and R2C12 = {39/57} => killer pair 5,7 in R12C2 and R7C2, locked for C2
or R45C1 = {57}, locked for N4
-> no 5,7 in R456C2
5a. R45C1 = {48} => R1C12 and R2C12 = {39/57}, 5,7 locked for N1 => hidden killer pair 5,7 in R12C1 and R6C1
or R45C1 = {57}, locked for N4
-> 5,7 in R12456C1, locked for C1

5. R1C12 + R2C12 + R45C1 must have different combinations (step 4) -> one of them must be {48}
5a. Grouped X-Wing for 4 in R1C12 + R2C12 + R45C1 and R8C2 + R9C1, no other 4 in C12
5b. Grouped X-Wing for 8 in R1C12 + R2C12 + R45C1 and R8C1 + R9C2, no other 8 in C12

6. Consider combinations for R45C1 = {48/57}
R45C1 = {48} => R1C12 and R2C12 = {39/57} => hidden killer pair 5,7 in R12C1 and R6C1 => R6C1 = {57} => R6C89 = {48} => R89C9 = {57}, killer pair 5,7 in R1C1 + R2C2 and R9C9, locked for D\ => R4C35 = [57] (hidden pair in R4)
or R45C1 = {57} => R45C2 + R6C12 = {1236} => R4C3 = 4
-> R4C3 = {45}
6a. R4C3 = {45} -> 12(4) cage at R3C2 = {1245}, 4 locked for C3
6b. Killer pair {45} in R45C1 and R4C3, locked for N4
6c. Caged X-Wing for 5 in 12(4) cage and R7C23, no other 5 in C23, clean-up: no 7 in R12C1
6d. 7 in C1 only in R456C1, locked for N4
6e. R45C1 = {48} => R4C3 = 5
or R45C1 = {57}, locked for C1, no 7 in R12C2 => 7 in N1 only in R12C3, locked for C3 => R7C3 = 5
-> no 5 in R3C3
6f. R45C1 = {48} => R4C3 = 5, R34C2 = {12} => R56C2 = {36}, locked for N4
or R45C1 = {57} => R45C2 + R6C12 = {1236}, locked for N4
-> no 6 in R56C3

7. Consider combinations for R89C9 = {48/57}
R89C9 = {48} => R6C89 = {57} => 7 in N4 only in R45C1 = {57} => R12C12 = {39/48} => killer pair 4,8 in R1C1 + R2C2 and R9C9, locked for D\
or R89C9 = {57} => R6C89 = {48}, locked for R6
-> no 4,8 in R6C6

[I should have spotted this earlier …; with hindsight it could have been done after step 1.]
8. R8C12 cannot be [84] which clashes with {48} in R89C7 or R89C9, R9C12 cannot be [48] which clashes with {48} in R89C7 or R89C9, R89C1 cannot be [84] which clashes with {48} in R1C12, R2C12 or R45C1 -> R89C2 = [48], R9C1 = 1, clean-up: no 4,8 in R12C1
8a. R45C1 = {48} (hidden pair in C1), locked for N4, R4C3 = 5, R7C3 = 7, placed for D/
8b. Naked pair {12} in R34C2, locked for C2 and 12(4) cage at R3C2 -> R3C3 = 4, placed for D\, clean-up: no 8 in R8C9
[So Killer UR wasn’t necessary ;-)]
8c. Naked pair {57} in R89C9, locked for C9 and N9 -> R89C7 = [84], clean-up: no 5,7 in R6C8
8d. Killer pair 5,7 in R1C1 + R2C2 and R9C9, locked for D\
8e. 8 on D\ only in R4C4 + R5C5, locked for N5
8f. R6C1 = 7 (hidden single in C1) -> R56C7 = [75]
8g. Naked pair {48} in R6C89, locked for R6

9. R2C8 = 5 (hidden single on D/), R2C1 = 3 -> R2C2 = 9, placed for D\, R1C1 = 5, placed for D\, R1C2 = 7
9a. R3C8 = 7 (hidden single in C8)
9b. R4C5 = 7 (hidden single in R4)
9c. R2C4 = 7 (hidden single in C4)
9d. R5C4 = 5 (hidden single in N5)
9e. R5C6 = 4 (hidden single in N5), R45C1 = [48]
9f. R4C4 = 8 (hidden single in N5) -> R7C45 = [48]
9g. R1C9 = 8 (hidden single of D/) -> R1C8 = 4
9h. R23C5 = [45] (hidden pair in C5)

10. R3C46 = [98] (hidden pair in R3)
10a. R4C6 = 9 (hidden single in N5)
10b. R5C9 = 9 (hidden single in N6)
10c. R5C2 = 3 (hidden single in R5), R6C2 = 6
10d. R5C5 = 6 (hidden single in N5), placed for both diagonals
10e. Naked pair {29} in R89C5, locked for C5 and N8 -> R89C4 = [16]
10f. Naked pair {23} in R89C3, locked for C3 and N7
10g. R5C3 = 1, R5C8 = 2, R3C7 = 3, placed for D/, R6C4 = 2, R6C6 = 1, placed for D\

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

Rating Comment:
I'll rate my walkthrough for A332X at least Hard 1.5.
I'm not sure why the SS score is only 1.5. However I noticed from the SS scoring summary that it used UR shortcuts.


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