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 Post subject: Assassin 347
PostPosted: Thu Mar 01, 2018 1:37 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1039
Location: Sydney, Australia
Assassin 347
Attachment:
a347.JPG
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This is an X puzzle so 1-9 cannot repeat on the two diagonals.

An easy start but then quite resistant. I used one difficult move but then found a really nice follow-up step that left me very satisfied. SudokuSolver scores it at 1.40, JSudoku has to use some chains. Good luck!

code::
3x3:d:k:4608:4865:4865:4865:2818:3587:3587:3844:3844:4608:4608:4357:4865:2818:3587:1542:3844:4615:6920:4105:4608:4357:1290:1542:3844:2827:4615:6920:3852:4105:5901:1290:5901:2827:3854:4615:6920:3852:3852:3852:5901:4367:4367:3854:3854:6920:3852:1040:5901:8465:5901:3346:4371:4371:6920:1040:5652:3093:8465:1558:5655:3346:4371:5652:5652:3093:8465:8465:8465:1558:5655:4371:5652:7448:7448:7448:7448:7448:7448:5655:5655:
solution:
Code:
+-------+-------+-------+
| 9 6 2 | 7 5 8 | 4 3 1 |
| 1 3 8 | 4 6 2 | 5 9 7 |
| 4 7 5 | 9 3 1 | 2 8 6 |
+-------+-------+-------+
| 8 4 9 | 1 2 7 | 3 6 5 |
| 6 5 1 | 3 4 9 | 8 7 2 |
| 7 2 3 | 5 8 6 | 9 1 4 |
+-------+-------+-------+
| 2 1 6 | 8 9 5 | 7 4 3 |
| 5 8 4 | 6 7 3 | 1 2 9 |
| 3 9 7 | 2 1 4 | 6 5 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 347
PostPosted: Mon Mar 05, 2018 9:23 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 Ed for your latest Assassin.

Yes, after the easy start, it was quite resistant for a bit.
After nibbling for a bit I found:
combined cages and half-cages, then a combined innies-outies which cracked the puzzle. I also used hidden killer pairs quite often.

Here is my walkthrough for Assassin 347:
Prelims

a) R12C5 = {29/38/47/56}, no 1
b) 17(2) cage at R2C3 = {89}
c) 6(2) cage at R2C7 = {15/24}
d) 16(2) cage at R3C2 = {79}
e) R34C5 = {14/23}
f) 11(2) cage at R3C8 = {29/38/47/56}, no 1
g) R5C67 = {89}
h) 4(2) cage at R6C3 = {13}
i) 13(2) cage at R6C7 = {49/58/67}, no 1,2,3
j) 12(2) cage at R7C4 = {39/48/57}, no 1,2,6
k) 6(2) cage at R7C6 = {15/24}
l) 15(5) cage at R4C2 = {12345}
m) 33(5) cage at R6C5 = {36789/45789}, no 1,2

Steps resulting from Prelims
1a. Naked pair {89} in R5C67, locked for R5
1b. Naked pair {89} in 17(2) cage at R2C3, CPE no 8,9 in R2C456 + R3C123, clean-up: no 2,3 in R1C5
1c. R3C2 = 7 -> R4C3 = 9, R2C3 = 8 -> R3C4 = 9; clean-up: no 2 in R2C5, no 2 in R3C8, no 2,4 no R4C7, no 3,4 in R7C4, no 3 in R8C3
1d. Naked pair {13} in 4(2) cage at R6C3, CPE no 1,3 in R456C2 + R79C4
1e. Naked triple {245} in R456C2, locked for C2, N4 and 15(5) cage at R4C2
1f. Naked pair {13} in R5C34, locked for R5
1g. Naked pair {13} in R56C3, locked for C3 and N4
1h. Naked triple {678} in R456C1, locked for C1
1i. R456C1 = {678} = 21 -> R37C1 = 6 = {15/24}
1j. 33(5) cage at R6C5 = {36789/45789}, CPE no 7,8,9 in R9C5
1k. 8 in R3 only in R3C789, locked for N3

2. 45 rule on N5 4 innies R46C5 + R5C46 = 22
2a. Max R4C5 + R5C4 = 7 -> min R5C6 + R6C5 = 15, no 3,4,5 in R6C5
2b. Max R5C6 + R6C5 = 17 -> min R4C5 + R5C4 = 5, max R5C4 = 3 -> min R4C5 = 2, clean-up: no 4 in R3C5
2c. R4C5 + R5C4 cannot total 6 -> R4C5 + R5C4 = 5,7 = [23/41/43] -> R4C5 = {24}, clean-up: no 2 in R2C5

3. 15(3) cage at R4C8 = {258/267/357/456} (cannot be {168/348} because 1,3,8 only in R4C8), no 1
3a. 8 of {258} must be in R4C8, 2 of {267} must be in R5C89 (R5C89 cannot be {67} which clashes with R5C1) -> no 2 in R4C8
3b. 1 in N6 only in R4C9 + R6C89, CPE no 1 in R78C9

4. 45 rule on R12 3 outies R3C367 = 1 remaining innie R2C9 + 1
4a. Min R3C367 = 7 (cannot be {123} which clashes with R3C5) -> min R2C9 = 6
4b. Max R3C367 = 10 -> no 8 in R3C7
4c. R2C9 = {679} -> R3C367 = 7,8,10 = {124/125/145/235} (cannot be {134/136} which clash with R3C5), no 6 in R3C37
4d. Killer pair 1,3 in R3C367 and R3C5, locked for R3, clean-up: no 8 in R4C7, no 5 in R7C1 (step 1i)
4e. R3C89 = {68} (hidden pair in R3), 6 locked for N3, clean-up: no 6,7 in R4C7
4f. R2C9 = {79}, R3C9 = {68} -˃ 18(3) cage at R2C9 = {189/378/567} (cannot be {369} = [963] which clashes with 11(2) cage at R3C8) -˃ R4C9 = {135}
4g. R2C9 = {79} -> R3C367 = 8,10 = {125/145/235}, 5 locked for R3, clean-up: no 1 in R7C9 (step 1i)
4h. Naked pair {24} in R37C1, locked for C1
4i. Combined cage 18(3) at R2C9 + 11(2) at R3C8 = [765]+[83]/[783]+[65]/[981]+[65] -> 5 in R4C79, locked for R4 and N6, clean-up: no 8 in R7C8
4j. Naked pair {24} in R4C25, locked for R4
4k. 15(3) cage at R4C8 (step 3) = {267} (only remaining combination), locked for N6, 2 also locked for R5, clean-up: no 6,7 in R7C8
4l. 4 in N6 only in R6C789, locked for R6
4m. 4 in N5 only in R45C5, locked for C5, clean-up: no 7 in R12C5
4n. Hidden killer pair 1,3 in R4C46 and R4C79 for R4, R4C79 contains 5 and one of 1,3 -> R4C46 must contain one of 1,3
4o. Killer pair 1,3 in R4C46 and R5C4, locked for N5
4p. Hidden killer pair 2,5 in R6C2 and R6C46 for R6, R6C2 = {25} -> R6C46 must contain one of 2,5
4q. Hidden killer pair 2,5 in R45C5 and R6C46 for N5, R6C46 contains one of 2,5 -> R45C5 must contain one of 2,5
4r. R45C5 contains 4 and one of 2,5, R4C5 = {24} -> R5C5 = {45}
4s. 4 in R45C5 -> combined half cage R34C5 + R4C5 = [145/324]
4t. Killer pair 3,5 in R12C5 and R34C5 + R4C5, locked for C5, also 3 in R23C5, locked for N2
4u. 7,9 in C5 only in R678C5, locked for 33(5) cage at R6C5, no 7 in R8C4, no 7,9 in R8C6

5. 45 rule on N1 3 remaining innies R1C23 + R3C1 = 12 = {246/345} (cannot be {129} because 1,9 only in R1C2, cannot be {156} because 1,5,6 only in R1C23), no 1,9, 4 locked for N1
5a. R1C2 = {36} -> no 6 in R1C3
5b. 6 in N1 only in R12C2, locked for C2
5c. Killer pair 2,5 in R1C23 + R3C1 and R3C3, locked for N1
5d. 5 in C1 only in R89C1, locked for N7, clean-up: no 7 in R7C4
5e. 22(4) cage at R7C3 contains 5 = {1579/3568} (cannot be {2569/2578} because 2,6,7 only in R7C3, cannot be {4567} which clashes with R8C3), no 2,4
5e. 8 of {3568} must be in R8C2 -> no 3 in R8C2
5f. Hidden killer pair 8,9 in 22(4) cage and R9C2 for N7, 22(4) cage contains one of 8,9 -> R9C2 = {89}

6. 45 rule on N3 2 innies R12C7 = 2 outies R4C79 + 1
6a. R4C79 = {15/35} = 6,8 -> R12C7 = 7,9, no 9 in R1C7
6b. 18(3) cage at R2C9 (step 4f) = {189/378/567}, R4C79 = [35/51/53] -> combined innies+outies R12C7 + R4C79 = {45}[35]/[72][35]/[34][51] (cannot be [72][53] which clashes with 18(3) cage = [783] -> R1C7 = {3457}, R2C7 = {245}, R4C9 = [15], 3 in R14C7, locked for C7, clean-up: no 5 in R3C6
[Cracked. The rest is straightforward.]
6c. R3C5 = 3 (hidden single in R3) -> R4C5 = 2, clean-up: no 8 in R1C5
6d. Naked pair {56} in R12C5, locked for C5 and N2 -> R5C5 = 4, placed for both diagonals, R9C5 = 1, clean-up: no 5 in R8C7
6e. 33(5) cage at R6C5 = {36789/45789}, R678C5 = {789} -> R8C46 = {36/45}, no 8
6f. R46C5 + R5C46 = 22 (step 2), R4C5 = 2 -> R5C46 + R6C5 = 20 -> R5C4 = 3, R5C6 + R6C5 = {89}, locked for N5, R56C3 = [13] -> R7C2 = 1
6g. R4C79 = [35] (hidden pair in R4) -> R3C8 = 8, R3C9 = 6, R2C9 = 7 (cage sum), R5C9 = 2
6h. R4C79 = [35] = 8 -> R12C7 = 9 = {45}, locked for C7 and N3, clean-up: no 4 in R3C6, no 2 in R7C6, no 9 in R7C8
6i. Naked pair {12} in R3C67, locked for R3 -> R3C1 = 4, R3C3 = 5, placed for D\
6j. R79C7 = {67} (hidden pair in C7), locked for N9
6k. Naked pair {67} in R6C6 + R7C7, locked for D\ -> R4C4 = 1, placed for D\
6l. Naked pair {39} in R1C1 + R2C2, locked for N1 and D\
6m. R8C8 + R9C9 = [28] = 10 -> R7C7 + R9C8 = 12 = [75], 7 placed for D\, R7C8 = 4 -> R6C7 = 9, R7C6 + R8C7 = [51], R7C4 = 8 -> R8C3 = 4
6n. R3C6 = 1 (hidden single in R3) -> R2C7 = 5

and the rest is naked singles, without using the diagonals.
Thanks Ed for pointing out that I'd omitted one combination in step 4f. I've corrected that step with resulting changes to steps 4i and 6b.

Rating Comment:
I'll rate my walkthrough for A347 at 1.25, using the techniques mentioned above.


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 Post subject: Re: Assassin 347
PostPosted: Thu Mar 15, 2018 6:55 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
Thanks Ed. I found this incredibly difficult! Much harder than the score suggests.
Congratulations to Andrew for his solution. I failed to spot what he did in lines 4a through 4g so my solution path was a lot more tortuous.
I haven't got a walkthrough - but here is a summary of the first few steps.

Assassin 347 Initial Steps:
0. Prove 17(2)r2c3 = [89], 16(2)r3c2 = [79].
1. Prove n4. r456c1 = {678}, r456c2 = {245}, r4c3 = 9. r45c4 = {13}
2. Prove r4c5 from (24)
3. Prove 9 in n1 not in r1c2.
4. Prove 9 not in r2c9 (Innies r12)
5. Prove 1 in r4 only in n5 (r4c46) -> r5c34 = [13], 4(2)r6c3 = [31]
6. Prove 5 in n1 not in r123c1
7. Prove 5 in n7 not in r7c1
8. Prove 22(4)n7 = [6583] or [6385], r9c2 = 9
9. Prove 15(4)n3 = {1239} (9 in D/ only in n3)
10. Prove (47) in D/ in n5 -> r4c5 = 2.
etc.


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 Post subject: Re: Assassin 347
PostPosted: Sat Mar 17, 2018 4:38 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1039
Location: Sydney, Australia
This ended up a very nice Assassin and not that difficult! Loved Andrew's cascades of hidden killer subsets. We found the same key eliminations but saw them very differently. My one difficult step, Andrew saw as 3 'simple' steps (his 4c,d,e) so well done there. My optimised solution then can immediately do the breakthrough step (my step 10)- but like Andrew, first solve took a long time to find. The benefit of having both types of WTs.

Wellbeback would have cracked the puzzle by his step 4 if he had found the same early steps as me and Andrew.
Assassin 347 start & middle
20 steps:
Thanks to Andrew for help to make this WT clearer
Preliminaries courtesy of SudokuSolver
Cage 4(2) n47 - cells ={13}
Cage 16(2) n14 - cells ={79}
Cage 17(2) n12 - cells ={89}
Cage 17(2) n56 - cells ={89}
Cage 5(2) n25 - cells only uses 1234
Cage 6(2) n23 - cells only uses 1245
Cage 6(2) n89 - cells only uses 1245
Cage 12(2) n78 - cells do not use 126
Cage 13(2) n69 - cells do not use 123
Cage 11(2) n2 - cells do not use 1
Cage 11(2) n36 - cells do not use 1
Cage 15(5) n45 - cells ={12345}
Cage 33(5) n58 - cells do not use 12

No Routine clean-up done unless specified
1. 16(2)r3c2 = {79} -> no 7,9 in r123c3
1a. r2c3 = 8, r3c4 = 9, r3c2 = 7, r4c3 = 9

2. 4(2)r6c3 = {13} -> no 1,3 in r456c2 nor r789c3

3. naked triple {245} in r456c2: locked for c2, n4 and 15(5) -> no 2,4,5 in r5c4

4. naked pair {13} in r5c34: both locked for r5
4a. naked pair {13} in r56c3: both locked for c3 and n4

5. naked triple {678} in r456c1: all locked for c1
5a. sum to 21 -> r37c1 = 6 (cage sum) = {15/24}(no 3,9)

6. 23(5)n5: must have a 1 or 3 since {24567} = 24 -> killer pair 1,3 with r5c4: both locked for n5
6a. r3c5 (13)

7. "45" on r12: 3 remaining outies r3c367 - 1 = 1 innie r2c9
7a. min. 3 outies = {124} = 7 because {123} blocked by r3c5 -> min. r2c9 = 6
7b. max r2c9 = 9 -> max r3c367 = 10 -> no 8 in r3c7

8. 8 in r3 only in r3c89: locked for n3

The difficult step
9. "45" on r123: 5 remaining innies r3c1589 + r2c9 = 28 and must have 8 for r3 -> 4 remaining innies = 20 (and can't have 8!) = [23]{6}+[9]/[41]{6}+[9]/[43]{6}+[7]
9a. r3c1 = (24), r3c89 = {68}, r2c9 = (79)
9b. 6 locked for r3 and n3
9c. r4c7 = (35)

Really like this one.
10. r123c7 can't have more than one of 3,5 because of r4c7 -> the 15(4)n3 must have at least one of 3,5 in r1c89 + r2c8 for n3 (hidden killer pair)
10a. = {1239/1257/1347} = [7/9..]
10b. can't have more than one of 3,5 -> no 3,5 in r3c7
10c. 15(4) must have 1: locked for n3
10d. no 5 in r3c6
(note: 10b. also means 3/5 must be in r12c7 -> killer pair with r4c7: but not needed for this optimised solution)

11. killer pair 7,9 in 15(4)n3 and r2c9: locked for n3

12. naked triple {124} in r3c167: all locked for r3
12a. r3c3 = 5: placed for D\; r34c5 = [32]

13. h6(2)r37c1 = {24} only: both locked for c1

14. 18(4)n1 = {1359} only: 1,3,9 all locked for n1
14a. r1c2 = 6

15. 6 and 7 in c7 only in r679c7 -> no 6,7 in r7c8 since it sees all those cells (CPE)
15a. no 6,7 in r6c7
15b. r79c7 = {67} only (hidden pair in c7): both locked for n9

16. 2 only D\ only in n9: 2 locked for n9
16a. 22(4)n9 must have 2, and 6/7 for r7c7 = {2479/2569/2578}(no 1,3)

17. 23(5)n5: can't have both 1,3 because of r5c4 = {14567} only: all locked for n5
17a. r5c4 = 3, r56c3 = [13], r7c2 = 1

18. 8 on D\ only in n9: 8 locked for n9
18a. r8c8 + r9c9 = {28} only = 10 -> r7c7 + r9c8 = 12 = [75]; 7 placed for D\
18b. r9c7 = 6

19. 13(2)r6c7 = {49} only: -> no 4,9 in r456c8 nor r8c7
19a. r8c7 = 1, r7c6 = 5

20. 12(2)r7c4 = [84] only permutation
much easier now. Don't forget the diagonals!
Cheers
Ed


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