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 Post subject: Assassin 60 Revisit
PostPosted: Tue Jun 01, 2021 6:43 pm 
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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 60 Revisit.
No less than 5 Wts on the archive. Surely we can't find a new way. This is the 2nd of the V1 Assassins that we've Revisited. It gets a score of 1.60.

Code: Select, Copy & Paste into solver:
3x3::k:1536:1536:5890:5890:5890:5890:5382:3591:3591:1536:5642:5642:3340:3340:3854:5382:5382:3591:5906:5642:2324:2324:3340:3854:3854:4121:4121:5906:2844:2324:3358:3358:3616:3616:3874:4121:5906:2844:2844:3358:4392:4392:3616:3874:3874:5906:4398:4398:3888:4392:2610:2610:4148:4148:4150:4150:4398:3888:3888:2610:5180:5180:4148:5183:4150:3393:5186:5186:2628:5180:4166:4166:5183:5183:3393:3393:5186:2628:2628:4166:4166:
Solution:
+-------+-------+-------+
| 1 3 5 | 6 8 4 | 9 2 7 |
| 2 6 9 | 3 1 7 | 8 4 5 |
| 8 7 4 | 2 9 5 | 3 1 6 |
+-------+-------+-------+
| 6 2 3 | 1 5 8 | 4 7 9 |
| 4 1 8 | 7 6 9 | 2 5 3 |
| 5 9 7 | 4 2 3 | 1 6 8 |
+-------+-------+-------+
| 7 4 1 | 8 3 6 | 5 9 2 |
| 3 5 2 | 9 7 1 | 6 8 4 |
| 9 8 6 | 5 4 2 | 7 3 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 60 Revisit
PostPosted: Tue Jun 08, 2021 8:15 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
Quite tricky to write a clear WT on this one I thought. So many interlocking dependencies. Here's one way...
Corrections thanks to Andrew.
Assassin 60 Revisit WT:
1. 6(3)n1 = {123}
22(3)n1 = {(67|58)9}
-> Innies n1 = {4(58|67)}
Also r3c3 from (456)
-> Both r3c4 and r4c3 from (123)

2. Innies n3 = r4c789 = +10(3)
IOD n3 -> [r3c7,r4c9] from [17], [28], [39]
Innies n7 = r789c3 = +9(3)
Outies n124 = r3c7 + r7c3 = +4(2)
-> r7c3 is max 3
-> IOD n7 -> [r7c3,r9c4] from [15], [26], [37]

3! H+10(3)r3c789 must contain at least one of (15)
-> 9(3)r3c3 cannot be [513]
-> If r3c4 = 1 then r34c3 must be [62]
Similarly if r7c3 = 1 this puts r9c4 = 5 and r89c3 = {26}
-> r3c4 and r7c3 cannot both be 1.

4. IOD n14 -> r1c3 = r3c4 + r7c3 + 2.
Since r3c4 + r7c3 are at least 1+2 = 3 -> r1c3 is min 5.
-> 4 in n1 in r3c13
-> r3c789 cannot be {145}. It must contain two of (123).
-> In r3 one of (123) in r3c4, one in r3c7, and one in r3c89.

5. Trying r3c7,r7c3 = {22} puts 9(3)r3c3 = [531]
and therefore puts (IOD n14) r1c3 = 7.
But Innies n1 cannot contain both (57)
-> [r3c7,r7c3] = {13}

6! H+10(3)r3c789 cannot be [1{27}] (Puts r4c9 = 7)
-> H+10(3)r3c789 = [3{25}] or {136}
But [3{25}] puts both r3c4 and r7c3 = 1 which was disproved in Step 3.
-> H+10(3)r3c789 = [1{36}] or [3{16}]
-> 9(3)r3c3 = [423]
-> H+9(3)r789c3 = [1{26}] and r9c4 = 5
-> H+10(3)r3c789 = [3{16}] and r4c9 = 9

Pretty straightforward from here. E.g.,

7. IOD n14 -> r1c3 = 5
-> r3c1 = 8 and 22(3)n1 = [6{79}]
Also r6c23 = {79}
-> (NS 8 in c3) 11(3)n4 = [{12}8]
-> r12c1 = {12} and r1c2 = 3
Also r456c1 = {456}
Also 3 in n2 only in r2c45 -> 13(3)n2 = [{13}9]
-> 22(3)n1 = [697] and r12c1 = [12]
-> r6c23 = [97]
Also (NS 5 in n2) r23c6 = [75]
-> r1c456 = {468}
Also r1c789 = [9{27}]
-> r2c789 = [{48}5]

8. Innies c6789 = r1c6 + r5c6 = +13(2) can only be [49]
-> HS 8 in c6 -> r4c6 = 8
-> 7 in n6 only in 15(3)
-> 8 in n6 only in r6c89
-> 4 in n6 only in r456c7
-> 21(3)n3 = [984]
-> Remaining Innie c89 -> r7c8 = 9
-> r78c7 = {56}
-> r45c7 = {24}
Also r9c7 = 7
etc.


Last edited by wellbeback on Thu Jun 10, 2021 4:36 pm, edited 1 time in total.

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 Post subject: Re: Assassin 60 Revisit
PostPosted: Wed Jun 09, 2021 10:36 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
Ed commented that there are no fewer than five walkthroughs in the archive for when this Assassin was originally posted by Ruud. Now two more by wellbeback and myself. We both used mostly the same features but saw and analysed them in very different ways. As I've commented previously, I treat these Revisits as new puzzles so don't look back at the archives to see how they were previously solved; however I know that Ed does, so he will know whether we've found new ways to solve this one.

Clarification add for my first key step 3c.
Here is my walkthrough for Assassin 60 Revisited:
Prelims

a) 6(3) cage at R1C1 = {123}
b) 21(3) cage at R1C7 = {489/579/678}, no 1,2,3
c) 22(3) cage at R2C2 = {589/679}
d) 9(3) cage at R3C3 = {126/135/234}, no 7,8,9
e) 11(3) cage at R4C2 = {128/137/146/236/245}, no 9
f) 10(3) cage at R6C6 = {127/136/145/235}, no 8,9
g) 20(3) cage at R7C7 = {389/479/569/578}, no 1,2
h) 20(3) cage at R8C1 = {389/479/569/578}, no 1,2
i) 20(3) cage at R8C4 = {389/479/569/578}, no 1,2
j) 10(3) cage at R8C6 = {127/136/145/235}, no 8,9

1a. Naked triple {123} in 6(3) cage at R1C1, locked for N1
1b. 22(3) cage at R2C2 = {589/679}, 9 locked for N1
1c. 9(3) cage at R3C3 = {126/135/234}
1d. R3C3 = {456} -> no 4,5,6 in R3C4 + R4C3

2a. 45 rule on R89 2 innies R8C27 = 11, no 1,9 in R8C2
2b. 45 rule on C89 2 innies R27C8 = 13 = {49/58/67}, no 3
2c. 45 rule on N3 3 innies R3C789 = 10 = {127/136/145/235}, no 8,9
2d. 45 rule on N3 1 outie R4C9 = 1 innies R3C7 + 6 -> R3C7 = {123}, R4C9 = {789}
2e. 45 rule on N7 3 innies R789C3 = 9 = {126/135/234}, no 7,8,9
2f. Killer triple 1,2,3 in R4C3 and R789C3, locked for C3
2g. Min R5C3 = 4 -> max R45C2 = 7, no 7,8 in R45C2
2h. 45 rule on N7 1 outie R9C4 = 1 innie R7C3 + 4 -> R7C3 = {12345}, R9C4 = {56789}
2i. 45 rule on N124 2(1+1) outies R3C7 + R7C3 = 4 = [13/22/31], clean-up: no 8,9 in R9C4
2j. Max R7C3 = 3 -> min R6C23 = 14, no 1,2,3,4 in R6C2
2k. 15(3) cage at R2C6 = {159/168/249/258/267/348/357} (cannot be {456} because R3C7 only contains 1,2,3)
2l. R3C7 = {123} -> no 1,2,3 in R23C6
2m. 45 rule on R6789 2 innies R6C15 = 7 = {16/25/34}, no 7,8,9
2n. 45 rule on C6789 2 innies R15C6 = 13 = {49/58/67}, no 1,2,3
2o. 45 rule on N9 2 innies R7C9 + R9C7 = 9, no 1,9 in R7C9
2p. 45 rule on R12 4 outies R3C2567 = 24
2q. Max R3C7 = 3 -> min R3C256 = 21, no 1,2,3 in R3C5
2r. Hidden killer triple 1,2,3 in R3C4 and R3C789 for R3, R3C4 = {123} -> R3C789 must contain two of 1,2,3 = {127/136/235}, no 4 in R3C89
2s. 45 rule on C12 2 outies R25C3 = 1 innie R6C3 + 8, IOU no 8 in R2C5

[The first key step.]
3a. 45 rule on N4 2 outies R3C1 + R7C3 = 1 innie R4C3 + 6, IOU no 6 in R3C1
3b. R3C7 + R7C3 = [13/22/31] (step 2i)
3c. Consider combinations for 9(3) cage at R3C3 = {126/135/234}
9(3) cage = {126/234} => 2 placed in R3C4 or R4C3 => R3C7 + R7C3 = [13/31]
or 9(3) cage = {135} => R3C3 = 5 => R3C1 = {48} (R3C13 cannot be [75] which clashes with 22(3) cage at R2C2), R4C3 = {13} => R7C3 must be the reverse {13}
[Note that with R3C1 = {48} and the outie-innie difference of 6, R4C3 and R7C3 have to differ by +2 or -2 in this path]
-> R3C7 + R7C3 = {13}, no 2, clean-up: no 8 in R4C9 (step 2d), no 6 in R9C4 (step 2h)
3d. Naked pair {13} in R3C7 + R7C3, CPE no 3 in R7C7
3e. R3C789 (step 2r) = {136/235} (cannot be {127} = 1{27} because 16(3) cannot be {27}7), no 7, 3 locked for R3 and N4
[Taking things a bit further.]
3f. 45 rule on N5689 2(1+1) innies R4C9 + R9C4 = 14 = [77/95]
3g. Consider placements for R4C9
R4C9 = 7 => R3C89 = 9 = {36}
or R4C9 = 9 => R9C4 = 5 => R89C3 = 8 = {26}, locked for C3 => 9(3) cage at R3C3 = [423/513] => R3C789 = {136}
-> R3C789 = {136}, 1,6 locked for R3 and N3
[Fairly straightforward from here; no further clean-ups.]
3h. R3C4 = 2 -> R34C3 = 7 = [43], R7C3 = 1, R3C7 = 3, R3C89 = {16} = 7 -> R4C9 = 9 (cage sum), R9C4 = 5
3i. Naked pair {26}, locked for N7, 6 locked for C3
3j. R2C2 = 6 (hidden single in N1) -> R2C3 + R3C2 = 16 = {79}, 7 locked for N1
3k. R47C3 = [31] -> R3C1 = 8 -> R1C3 = 5
3l. 11(3) cage at R4C2 = {128} (only remaining combination, cannot be {146/245} because R5C3 only contains 7,8) -> R5C3 = 8, R45C3 = {12}, locked for C3 and N4 -> R1C2 = 3
3m. R7C3 = 1 -> R6C23 = 16 = {79}, locked for R6 and N4
3n. Naked triple {456} in R456C1, 4,5 locked for C1
3o. Naked pair {79} in R36C2, locked for C2

4a. R3C7 = 3 -> R23C6 = 12 = {57} (cannot be {48} because 4,8 only in R2C6), locked for N2 -> R3C5 = 9, R3C2 = 7 -> R23C6 = [75], R2C3 = 9
4b. R3C5 = 9 -> R2C45 = 4 = {13}, 1 locked for R2 -> R2C1 = 2
4c. 21(3) cage at R1C7 = {489} (only remaining combination, cannot be {579} because 7,9 only in R1C7) -> R1C7 = 9, R2C78 = {48}, locked for N3 -> R2C9 = 5
4d. 1 in N8 only in R89C6, locked for C6 and 10(3) cage at R8C6
4e. 10(3) cage at R8C6 = {127/136}, no 4
4f. 7 on {127} must be in R9C7 -> no 2 in R9C7
4g. R9C7 = {67} -> no 6 in R89C6
4h. R7C9 + R9C7 (step 2o) = 9 = [27/36]
4i. R7C9 = {23} -> R6C89 = 13,14 = {58/68}, 8 locked for R6 and N6
4j. R5C6 = 9 (hidden single in C6) -> R56C5 = 8 = {26/35}/[71], no 4, no 1 in R5C4
4k. 10(3) cage at R6C6 = {136} (only remaining combination, cannot be {145} because 1,5 in R6C7, cannot be {235} because R67C6 = {23} clashes with 10(3) cage at R8C6) -> R6C7 = 1, R67C6 = {36}, locked for C6
4l. Naked pair {12} in R89C6, 2 locked for C6, R9C7 = 7 -> R7C9 = 2
4m. R7C9 = 2 -> R6C89 = 14 = {68}, 6 locked for R6 and N6 -> R67C6 = [36], R6C145 = [542], R5C5 = 6 (cage sum)
4n. R4C6 = 8 -> R45C7 = 6 = {24}, locked for N6, 4 locked for C7 -> R2C7 = 8
4o. R78C7 = [56] -> R7C8 = 9 (cage sum)
4p. R6C4 = 4 -> R7C45 = 11 = {38}, locked for R7 and N8

and the rest is naked singles.


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 Post subject: Re: Assassin 60 Revisit
PostPosted: Fri Jun 11, 2021 9:34 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
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8! Another way to solve this puzzle! The key was using one of the newer techniques we've started using since this puzzle came out (step 10). But is quite hidden.

It could even be 9 or 10 so feel free to keep posting. HINT: taking out the 9(3) cage only takes the SSscore up to 1.70 and JSudoku still only uses 4 'complex intersections'.
start to A60 Revisit:
Preliminaries from SudokuSolver
Cage 6(3) n1 - cells ={123}
Cage 22(3) n1 - cells do not use 1234
Cage 9(3) n124 - cells do not use 789
Cage 21(3) n3 - cells do not use 123
Cage 10(3) n568 - cells do not use 89
Cage 10(3) n89 - cells do not use 89
Cage 20(3) n8 - cells do not use 12
Cage 20(3) n7 - cells do not use 12
Cage 20(3) n9 - cells do not use 12
Cage 11(3) n4 - cells do not use 9

1. 6(3)n1 = {123}: all locked for n1

2. "45" on r12: 1 innie r2c6 + 9 = 2 outies r3c25
2a. r2c6 & r3c5 can't be equal -> no 9 in r3c2 (IOU)

3. 22(3)n1 = {589/679}: 9 locked for r2 and n1

4. "45" on n124: 2 outies r3c7 + r7c3 = 4 = {13/22}

5. "45" on n3: 3 innies r3c789 = 10 (no 8,9)

6. 9 in n3 only in r1: 9 locked for r1

7. 23(4)r1c3 = {2678/3578/4568}(no 1)
7a. 8 locked for r1

8. "45" on r1: 1 innie r1c7 - 2 = 2 outies r2c19
8a. min. 2 outies = 3 -> min. r1c7 = 5
8b. max. 2 outies = 7 (no 7,8)

9. 8 in n3 only in 21(3) = (489/678}(no 5)
9a. 8 locked for r2

Key step
10. 4 in r1 only in 23(4) = {4568} or in 14(3)n3
10a. and a 14(3) can't be {455/464}
10b. -> no 5 or 6 in r1c89 (Locking out cages)

11. 5 in r1 only in 23(4)r1c3 = {3578/4568}(no 2)

12. 15(3)r2c6 must have 1,2,3 for r3c7 but can't have more than one of 1,2,3
12a. -> no 1,2,3 in r23c6

13. hidden killer pair 1,2 in n2 since a 13(3) can't have both 1,2
13a. -> r3c4 = (12)

Cracker
14. if 22(3)n1 = {589} -> 5 & 8 in 23(4)r1c3 in n2 -> 15(3)r2c6 = [492]/{67}[2]: ie must have 2 in r3c7
14a. -> r7c3 = 2 (step 4)
14b. -> 9(3)r3c3 = [513]: but this means two 5s in n1
14c. -> 22(3)n1 = {679}: 6 and 7 locked for n1

15. 9(3)r3c3 must have 4 or 5 for r3c3 = {135/234}(no 6)
15a. -> r4c3 = 3

16. "45" on n4: 2 outies r3c1 + r7c3 = 9 = [81] only

Pretty easy now. Lots of "45", cage sums etc
Cheers
Ed


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