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 Post subject: Assassin 371
PostPosted: Sat Mar 02, 2019 7:17 am 
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Posts: 1040
Location: Sydney, Australia
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Assassin 371

I found this one fairly easy. A couple of advanced steps popped out quickly but perhaps got lucky. Never know till you guys try. SudokuSolver and JSudoku both have to work (1.75 and 1 longish chain)
code:
3x3::k:6144:6144:6144:4865:3074:3331:3331:5145:4357:6144:6144:4865:4865:3074:3846:5145:5145:4357:3847:2568:1033:1033:3846:3846:5145:5898:4357:3847:2568:1547:1547:3084:3084:5145:5898:5898:3847:4891:2573:2573:3342:3342:3342:5898:4111:2844:2844:4891:4891:4891:2832:2832:4111:4111:2065:2065:4370:4370:3866:3866:3859:3859:3859:1812:4373:3606:4370:5399:3866:2072:2072:3332:1812:4373:3606:3606:5399:5399:5399:3332:3332:
solution:
Code:
+-------+-------+-------+
| 5 1 9 | 2 8 6 | 7 4 3 |
| 7 2 8 | 9 4 3 | 1 5 6 |
| 4 6 3 | 1 7 5 | 2 9 8 |
+-------+-------+-------+
| 2 4 1 | 5 3 9 | 8 6 7 |
| 9 7 6 | 4 2 8 | 3 1 5 |
| 8 3 5 | 6 1 7 | 4 2 9 |
+-------+-------+-------+
| 3 5 2 | 7 9 4 | 6 8 1 |
| 1 9 7 | 8 6 2 | 5 3 4 |
| 6 8 4 | 3 5 1 | 9 7 2 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 371
PostPosted: Tue Mar 05, 2019 4:53 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed! It was a little tricky but not 1.75 worth I think.

Here's how I did it.

Step 3 rewritten and other clarifications added in red. Corrections too! Thanks Andrew, Ed.
Assassin 371 WT:
(Note: I use the notation <xyz> to mean [xyz] or [zyx])

1. Outies n78 -> r9c7 = 9
-> 17(2)n7 = [98]
-> 9 in n8 in r7c456

2. 4(2)r3 = {13}
Outies n2 = r23c3 + r1c7 = +18(3)

Trying 4(2) = [13]
puts r2c3 = 9, r1c7 = 8, -> r1c6 = 5
and (HS 1 in n2) r2c6 = 1
which leaves no solution for n2.
-> 4(2)r3 = [31]

3. Outies c1234 -> r6c5 = 1
-> 1 in c6 in n8 in r789c6
Innies r89 -> r8c46 = +10(2)
Since 9 already in r8 -> H10(2)r8c46 cannot be [91] -> 1 not in r8c6
Also r8c6 cannot be any of (159) -> 15(3)n8 cannot be {159}
Outies r89 = r7c3456 = +22(4) (including a 9 in r7c456).
-> 15(3)n8 cannot be [618] since that puts r7c4 = 9 and another 6 in r7 in r7c3.
Also -> 15(3)n8 cannot be [816] since that puts r7c4 = 9 and 4 in both r7c3 and r8c4.
-> 1 not in r7c6

-> HS 1 in c6 -> r9c6 = 1

4! Remaining innies n1 = r2c3 + r3c1 + r3c2 = +18(3)
Remaining outies n2 = r1c7 + r2c3 = +15(2)
-> Whatever is in r1c7 cannot be the same as r2c3
Also cannot be the same as r3c3 (= 3)
Also cannot be the same as either r3c1 or r3c2 since that leaves the other of r3c12 = 3.
-> Whatever goes in r1c7 goes in n1 in r2c12
-> it goes in n2 in r3c56
Since that number + r1c6 = +13(2) -> 2 cannot go in 15(3)n2
-> 2 in n2 in r12c4
-> 19(3)r1c4 = {289}
-> r2c3 from (89) and r1c7 from (76)
-> 13(2)r1 = {67}
Also since r1c7 + r2c3 = +15 and since whatever is in r1c7 is in r3c56 in the 15(3) -> whatever is in r2c3 can only go in n2 in r1c5
-> 12(2)n2 from [84] or [93]

5! 12(2)n2 from [84] or [93]
-> H+11(2)r89c5 cannot be {38}
-> H+11(2)r89c5 from {47} or {56}
2 in n8 in 15(3)n8
-> 3 in n8 in r789c4
Outies n7 -> r789c4 = +18(3) = {378} or {369}
-> (45) in c4 only in r456c4
-> 12(2)n5 = {39}
-> 12(2)n2 = [84]
-> r2c3 = 8, r12c4 = {29}
Also -> 15(3)n2 = [3{57}]
-> 13(2)r1 = [67]

6. Also r789c4 = {378}
-> r89c5 = {56}
-> 15(3)n8 = {249}
r8c46 = +10(2) can only be [82]
-> 15(3)n8 = [942]
-> 12(2)n5 = [39]

7. Also r456c4 = {456}
-> 10(2)r5c3 = {46}
Also HS 2 in n5 -> r5c5 = 2
-> 13(3)r5c5 = [283]
-> 11(2)r6c6 = [74]
Outies r6789 = r5c29 = +12(2) = {57}
But no solution for 19(4)r5c2 with r5c2 = 5
-> r5c29 = [75]

8. 10(2)r3c2 = {46}
-> r4c2,r5c3 in n4 = {46}
-> 19(4)r5c2 = [7561]
-> 10(2)r5c3 = [64]
-> 6(2)r4c3 = [15]
Also
-> 10(2)r3c2 = [64]
-> Remaining Innie n1 -> r3c1 = 4

9. HS 9 in c3 -> r1c3 = 9
-> r789c3 = {247}
-> 8(2)n7 = {35}
-> 7(2)n7 = [16]
Also 17(3)r7c3 = [278]
-> 14(3)r8c3 = [{47}3]

10. 16(3)r5c9 can only be [5{29}]
-> 11(2)n4 = [83]
-> 15(3)r3c1 = [429]
etc.


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 Post subject: Re: Assassin 371
PostPosted: Fri Mar 08, 2019 12:21 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for continuing to create Assassins! :D

As wellbeback said, this one wasn't as hard as the SS score.

Thanks Ed for pointing out a couple of eliminations that I'd missed, and correcting a typo.
Here is my walkthrough for Assassin 371:
Prelims

a) R12C5 = {39/48/57}, no 1,2,6
b) R1C67 = {49/58/67}, no 1,2,3
c) R34C2 = {19/28/37/46}, no 5
d) R3C34 = {13}
e) R4C34 = {15/24}
f) R4C56 = {39/48/57}, no 1,2,6
g) R5C34 = {19/28/37/46}, no 5
h) R6C12 = {29/38/47/56}, no 1
i) R6C67 = {29/38/47/56}, no 1
j) R7C12 = {17/26/35}, no 4,8,9
k) R89C1 = {16/25/34}, no 7,8,9
l) R89C2 = {89}
m) R8C78 = {17/26/35}, no 4,8,9
n) 19(3) cage at R1C3 = {289/379/469/478/568}, no 1

Steps resulting from Prelims and Initial Placements
1a. 45 rule on N9 1 innie R9C7 = 9 -> R89C2 = [98], clean-up: no 4 in R1C6, no 1,2 in R34C2, no 2,3 in R6C1, no 2 in R6C6
1b. 45 rule on C1234 1 outie R6C5 = 1, clean-up: no 5 in R4C3, no 9 in R5C3
1c. Naked pair {13} in R3C34, locked for R3, clean-up: no 7 in R4C2

2. 45 rule on R6 2 outies R5C29 = 12 = [39/48]/{57}, no 1,2,6, no 3,4 in R5C9
2a. 45 rule on R56 2 innies R5C18 = 10 = {19/28/37/46}, no 5
2b. 45 rule on R89 2 innies R8C46 = 10 = {28/37/46}, no 1,5

3a. 45 rule on N7 1 outie R9C4 = 1 innie R7C3 + 1, no 7 in R7C3, no 1 in R9C4
3b. Max R7C3 = 6 -> min R78C4 = 11, no 1,2 in R7C4
3c. R3C4 = 1 (hidden single in C4) -> R3C3 = 3, clean-up: no 7 in R5C4, no 4 in R9C4
3d. 45 rule on C12 1 outie R1C3 = 1 innie R5C2 + 2, R5C2 = {3457} -> R1C3 = {5679}
3e. 45 rule on N2 1 outie R2C3 = 1 remaining innie R1C6 + 2 -> R1C6 = {567}, R2C3 = {789}, clean-up: no 4,5 in R1C7
3f. 45 rule on N36 3 innies R156C7 = 14, min R1C7 = 6 -> max R67C7 = 8, no 7,8 in R5C7, no 8 in R6C7, clean-up: no 3 in R6C6
[Ed pointed out that I’d overlooked no 6 in R5C7; I hadn’t looked at the min-max options carefully enough.]
3g. 45 rule on N1 3 remaining innies R2C3 + R3C12 = 18 = {279/459/468/567}
3h. 2 of {279} must be in R3C1, 7,8,9 of {459/468/567} must be in R2C3 -> no 7,8,9 in R3C1

4. R1C3 = R5C2 + 2 (step 3d), R2C3 = R1C6 + 2 (step 3e), R2C3 + R3C12 (step 3g) = {279/459/468/567}
4a. Consider placements for R1C3 = {5679}
R1C3 = 5 => R5C2 = 3 => R46C2 = {46} => R2C3 + R3C12 = {468} (cannot be {279} because 7,9 only in R2C3), no 9 in R2C3
or R1C3 = {67} => R1C67 = [58] (cannot be {67} which clashes with R1C3) => R2C3 = 7
or R1C3 = 9
-> no 9 in R2C3, clean-up: no 7 in R1C6, no 6 in R1C7
[Re-worked after I realised that I’d found an improvement on my original forcing chain starting from R2C3.]
4b. R2C3 + R3C12 = {468/567}, no 2, 6 locked for R3 and N1, clean-up: no 4 in R5C2, no 8 in R5C9 (step 2)
4c. R2C3 = {78} -> no 7 in R3C2, clean-up: no 3 in R4C2
4d. Naked pair {46} in R34C2, locked for C2, clean-up: no 5,7 in R6C1, no 2 in R7C1
4e. 45 rule on N2 2 outies R1C7 + R2C3 = 15 = naked pair {78}, CPE no 7,8 in R1C123 + R2C789, clean-up: no 5 in R5C2, no 7 in R5C9 (step 2)
[I overlooked no 7,8 in R1C4 from the same CPE. Thanks Ed. That would have simplified step 4i.]
4f. 7,8 in N1 only in R2C123, locked for R2, clean-up: no 4,5 in R1C5
4g. R5C34 = [19]/{28/46} (cannot be [73] which clashes with R5C2), no 3,7
4h. R5C18 (step 2a) = {19/28/46} (cannot be {37} which clashes with R5C2), no 3,7
4i. 19(3) cage at R1C4 = {289/379/478} (cannot be {469} because R2C3 only contains 7,8, cannot be {568} which clashes with R1C6), no 5,6
4j. 4 of {478} must be in R2C4 -> no 4 in R1C4
4k. 6 in N2 only in R12C6, locked for C6, clean-up: no 5 in R6C7, no 4 in R8C4 (step 2b)
4l. R156C7 = 14 (step 3f), min R1C7 = 7 -> max R56C7 = 7, no 6 in R5C7, no 7 in R6C7, clean-up: no 4 in R6C6

5. 15(3) cage at R3C1 = {159/168/249/258/267/348} (cannot be {357} = 5{37} which clashes with R5C2, cannot be {456} which clashes with R89C1)
5a. R3C1 = {456} -> no 4,5,6 in R45C1, clean-up: no 4,6 in R5C8 (step 2a)
5b. 5 in N4 only in R6C23, locked for R6, clean-up: no 6 in R6C7
5c. Max R16C7 = 12 -> min R5C7 = 2

[Clean-ups more limited from here.]
6. 1 in R5 only in R5C18 (step 2a) = {19} or R5C34 = [19] (locking cages) -> 9 in R5C148, locked for R5-> R5C9 = 5, R5C2 = 7 (step 2), R1C3 = 9 (step 3d), clean-up: no 3 in R2C5, no 4 in R6C1, no 1 in R7C1
[Cracked.]
6a. R5C2 = 7, R6C5 = 1 -> R6C34 = 11 = [29/56/83], no 4,6 in R6C3, no 2,4,8 in R6C4
6b. R5C9 = 5 -> R6C89 = 11 = {29/38/47}, no 6
6c. 4 in R6 only in R6C789, locked for N6
6d. R156C7 = 14 (step 3f) = [734/824] -> R6C7 = 4, R6C6 = 7, clean-up: no 5 in R4C56
6e. 3 in R5 only in 13(3) cage at R5C5 = {238/346}
6f. R4C56 = {39} (cannot be {48} which clashes with 13(3) cage), locked for R4 and N5 -> R6C4 = 6, R6C3 = 5 (cage sum), clean-up: no 1,4 in R5C3
6g. 13(3) cage = {238} (only remaining combination) -> R5C56 = {28}, locked for R4 and N5 -> R5C34 = [64], R5C7 = 3, R1C7 = 7, R1C6 = 6, R2C3 = 8 (step 4e), clean-up: no 8 in R6C89
6h. Naked pair {29} in R6C89, locked for R6 and N6 -> R6C12 = [83]
6i. R4C4 = 5 -> R4C3 = 1, R45C1 = [29] -> R3C1 = 4 (cage sum)
6j. R89C1 = {16} (only remaining combination), locked for C1 and N7 -> R12C1 = [57], R7C1 = 3 -> R7C2 = 5
6k. R9C4 = R7C3 + 1 (step 3a) -> R7C3 = 2, R9C4 = 3, R12C4 = [29]
[Alternatively 17(3) cage at R7C3 = {278} (only remaining combination) …]
6l. R8C46 = 10 (step 2b) = [82], R7C4 = 7
6m. R12C5 = [84], R23C6 = [35]
6n. 15(3) cage at R7C7 = {168} (only remaining combination), locked for R7 and N9, R8C7 = 5 -> R8C8 = 3
6o. R5C8 = 1 -> R3C8 + R4C89 = 22 = {679}, R3C8 = 9, R4C89 = {67}, locked for N6
[A long time ago I’d spotted 45 rule on N3 2 innies R1C7 + R3C8 = 1 outie R4C7 + 8, no 8 in R3C8 (IOU) but didn’t use it as it didn’t lead to anything at the time.]
6p. R4C7 = 8 -> R3C7 = 2, R3C9 = 8 -> R12C9 = 9 = [36]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A371 at Easy 1.5. I used a short forcing chain and locking cages.


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 Post subject: Re: Assassin 371
PostPosted: Sun Mar 10, 2019 7:28 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Really enjoy making these, in big part to see the variety of approaches we come up with. Mine is closest to Andrew's this time but still quite different because of my step 5b. Really enjoyed wellbeback's step 4 though had to really think about parts of it. A buzz when they click!

I'll stick to two puzzles a month to allow others to jump in with puzzles. .... :)
a371 start:
Preliminaries courtesy of SudokuSolver
Cage 4(2) n12 - cells ={13}
Cage 17(2) n7 - cells ={89}
Cage 6(2) n45 - cells only uses 1245
Cage 7(2) n7 - cells do not use 789
Cage 8(2) n7 - cells do not use 489
Cage 8(2) n9 - cells do not use 489
Cage 12(2) n2 - cells do not use 126
Cage 12(2) n5 - cells do not use 126
Cage 13(2) n23 - cells do not use 123
Cage 10(2) n45 - cells do not use 5
Cage 10(2) n14 - cells do not use 5
Cage 11(2) n56 - cells do not use 1
Cage 11(2) n4 - cells do not use 1
Cage 19(3) n12 - cells do not use 1

Note: no clean-up done unless stated
1. "45" on n9: 1 innie r9c7 = 9
1a. r89c2 = [98]
1b. no 1,2 in r34c2

2. "45" on c1234: 1 outie r6c5 = 1

3. "45" on n7
3a. 1 outie r9c4 - 1 = 1 innie r7c3
3b. no 1 in r9c4, no 7 in r7c3
3c. max. r7c3 = 6 -> min. r78c4 = 11 (no 1)

4. hidden single 1 in c4 -> r3c4 = 1, r3c3 = 3

5. "45" on n2: 2 outies r1c7 + r2c3 = 15 = [69]/{78}
5a. 13(2)r1c6 = [58]{67}
5b. -> 7 locked in one of r1c67 + r2c3 -> no 7 in r1c1234

6. "45" on r6789: 2 outies r5c29 = 12 = [39/48]/{57}(no 1,2,6; no 3,4 in r5c9

7. "45" on c12: 1 outie r1c3 - 2 = 1 innie r5c2
7a. = [53/64/97](r1c3 = (569), r5c2 = (347)
7b. no 7 in r5c9 (h12(2))

8. "45" on n1: 3 remaining innies r2c3 + r3c12 = 18
8a. but {279} as [9]+[27] only, blocked by 6 in r1c7 (outiesn2=15) -> 5 in r1c3 (naked single) -> 3 in r5c2 (iodc12=-2) which clashes with [73] in r34c2 (two 3s)
8b. but {459} as [9]+[54] blocked by [64] in r1c3 + r5c2 (two 4s)
8c. = {468/567}(no 2,9)
8d. cannot have both 7,8 -> no 7,8 in r3c12
8e. must have 6: locked for n1 and r3
8f. no 4 in r5c2 (iodc12=-2), no 8 in r5c9 (h12(2)), no 6 in r1c7 (h15(2)), no 7 in r1c6

9. outies n2 = 15 = {78} only: no 8 in r1c14, no 7,8 in r2c789

10. 10(2)r3c2 = {46} only: both locked for c2

11. 7,8 in n1 only in r2: 7,8 both locked for r2

12. 19(3)r1c4 must have 7,8 for r2c3
12a. {56}[8] blocked by r1c6 = (56)
12b. {478} blocked by 7,8 only in r2c3
12c. = {289/379}(no 4,5,6)
12d. must have 9: locked for c4 and n2
12e. r12c5 = [75/84]

13. naked pair {78} in r1c57: both locked for r1
13a. but [78] blocked by [55] in r1c6 + r2c5
13b. = [87], r2c5 = 4, r1c6 = 6, r2c3 = 8 (outiesn2=15)

14. 15(3)n2 = [3]{57} only: 5 locked for r3
14a. {29} in r12c4: 2 locked for c4

15. {37} blocked from 10(2)r5c3 by r5c2 = (37); no 3,7 in r5c34, no 1,9 in r5c3

16. "45" on c12: 3 remaining outies r1c3 + r6c34 = 20
16a. max. r16c3 = 16 -> min. r6c4 = 4

17. "45" on n7: 3 outies r789c4 = 18 and must have 3 for c4 = {378} only: all locked for n8, 7,8 for c4

18. "45" on n7: 1 innie r7c3 + 1 = 1 outie r9c4 = [23/67]

19. "45" on n7: 3 innies r789c3 = 13. Must have 2,6 for r7c3
19a. = {247/256}(no 1)
19b. must have 2: locked for c3 and n7

20. 6(2)r4c3 = [15] only

21. 12(2)n5 = [39] only

22. "45" on n3: 1 remaining innie r3c8 - 1 = 1 outie r4c7 = [98] only

23. r5c34 = {46}: both locked for r5

24. r5c7 <> 5 since r5c56 <> 8

25. "45" on r56: 2 innies r5c18 = 10 (no 5)

26. hidden single 5 in r5 -> r5c9 = 5, r5c2 = 7 (h12(2), r1c3 = 9 (iodc12=-2)

cracked. Sorry, lazy to give anymore.
Cheers
Ed


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