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 Post subject: Assassin 391
PostPosted: Sat Feb 01, 2020 8:09 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 837
Location: Sydney, Australia
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Killer x. 1-9 cannot repeat on either diagonal.
NOTE: disjoint 17(3)r1c6 + r2c46

Assassin 391
SudokuSolver messes this up but JSudoku goes well. Took me a good while to find the key step but pretty obvious and easy once you see it. Felt a lot easier than the last monster! Certainly looks easy in an optimised WT. Mine is cracked at step 12. Hope its still enough of a challenge for you guys.
code: triple click:
3x3:d:k:6144:3329:3329:3329:3329:4354:2307:3076:4357:6144:6144:6144:4354:5126:4354:2307:3076:4357:3079:3079:10504:5126:5126:5126:5641:2314:4357:3079:10504:10504:10504:10504:5641:5641:2314:2314:7179:4883:4883:10504:10504:5641:5641:9485:9485:7179:4883:4883:5903:5903:5903:9485:9485:9232:7179:7179:8977:8977:5903:9485:9485:9232:9232:7179:8977:8977:2066:2062:3852:9232:9232:9232:7179:8977:8977:2066:2062:3852:3852:9232:9232:
solution:
Code:
+-------+-------+-------+
| 3 4 1 | 2 6 5 | 7 8 9 |
| 7 8 6 | 3 1 9 | 2 4 5 |
| 9 2 5 | 4 7 8 | 6 1 3 |
+-------+-------+-------+
| 1 7 8 | 9 4 3 | 5 2 6 |
| 4 5 3 | 6 2 7 | 1 9 8 |
| 6 9 2 | 5 8 1 | 3 7 4 |
+-------+-------+-------+
| 2 3 7 | 8 9 6 | 4 5 1 |
| 5 1 9 | 7 3 4 | 8 6 2 |
| 8 6 4 | 1 5 2 | 9 3 7 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 391
PostPosted: Sun Feb 09, 2020 8:26 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 204
Location: California, out of London
Another fine puzzle - thanks Ed. Certainly easier than the previous one as you say, but still provides opportunities for some cool moves! Love the big cages :)
Assassin 391 WT:
1. 36(8)r6c9 has no 9
-> 9 in n9 in r79c7
-> 9 in n6 in 37(6)
-> r9c7 = 9
-> r6c9 = r7c7

2. IOD n8 -> r7c456 = {689} with r7c6 from (68)
-> The two 8(2)s in n8 are {17} and {35}
-> r89c6 = {24}

3! Innies n2 = r1c45 = +8(2)
This cannot be the same as either of the two 8(2)s in n8
-> r1c45 = {26}
-> 13(4)r1 = [{14}{26}]

4. Remaining outies c789 = r457c6 = +16(3)
Since r7c6 from (68) and 2 already in c6 -> 6 not in r45c6
IOD c6789 -> r2c4 + 6 = r3c6 + r7c6
-> (Since r2c4 and r3c6 see each other) r7c6 cannot be 6
-> HS 6 in c6 -> r7c6 = 6
-> r45c6 = +10(2) = {19} or {37}

5! 1 not in 37(8)
-> 1 in n6 in r4c789 or r5c7
-> r4c6 cannot be 1

41(7)r3c3 = {2456789}
-> Trying r4c6 = 9 puts r3c3 = 9 which leaves no solution for 12(3)r3c1
-> r45c6 = {37}

Also trying r4c6 = 7 puts r3c3 = 7 which also leaves no solution for 12(3)r3c1
-> r45c6 = [37]

6. 3 in n6 only in 37(6) or r6c9
If the latter -> r7c7 = 3
-> 37(6) = {346789}
-> r4c789 + r5c7 = {1256}

7. 1 in n5 only in r6c456
-> 23(4)r6c4 = {1589} with r7c5 from (89)
-> 41(8) = {57(9|8)} in n14 and {246(8|9} in n5

8. Remaining outies r1234 = r5c457 = +9(3)
This can only be [{26}1]
-> r4c789 = {256}
-> r3c3 = 5
-> r4c1 = 1
-> r34c7 = +11(2) can only be [65]
-> 9(3)r3c8 = [1{26}]

9. Also 9(2)n3 = [72]
-> 12(3)n1 = [{29}1]
Also 5 in c6 only in r12c6
-> 1 in n2 only in r2c5
-> 8(2)r8c5 = {35} and 8(2)r8c4 = {17}
-> 20(4)n2 = [1478]

10. Also HS 3 in n2 -> r2c4 = 3 and r12c6 = {59}
-> 24(4)n1 = [3{678}]
-> r3c9 = 3
-> 12(2)n3 = [84]
Also r6c4 = 5 and r6c6 = 1
-> 17(3)n3 = [953]
Also (HS 9 in D\) -> r4c4 = 9
-> r67c5 = [89]
-> 41(7) = [5{78}9462]
-> 19(4)n4 = [5{239}]
-> r56c1 = [46]
etc.


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 Post subject: Re: Assassin 391
PostPosted: Mon Feb 17, 2020 6:49 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 837
Location: Sydney, Australia
Really pleased you like it! I generally try and get at least a couple of big cages in....just for you!

I used the identical pathway to wellbeback (but written long before!) so perhaps its a very narrow solution. Miss one key step and it could get very, very hard. The SudokuSolver score of 2.00 might feel more realistic in that case.

a391 WT:
Assassin 391

Preliminaries courtesy of SudokuSolver
Cage 8(2) n8 - cells do not use 489
Cage 8(2) n8 - cells do not use 489
Cage 12(2) n3 - cells do not use 126
Cage 9(2) n3 - cells do not use 9
Cage 9(3) n36 - cells do not use 789
Cage 13(4) n12 - cells do not use 89
Cage 37(6) n689 - cells do not use 1
Cage 41(7) n145 - cells ={2456789}
Cage 36(8) n69 - cells ={12345678}

1. r7c7 sees all 9s in n6 -> no 9 (Common Peer Elimination CPE)
1a. r9c7= 9 (hsingle n9)

2. "45" on n8: 3 innies r7c456 = 23 = {689} only: all locked for r7 and n8
2a. no 2 in two 8(2) cages n8

3. naked quad 1,3,5,7 in r89c45: all locked for n8

4. naked pair 2,4 in r89c6: locked for c6

5. 9 in n6 only in 37(6) -> no 9 in r7c6

key step. Took quite a while to see it even though it is the type of step I like to look for.
6. "45" on n2: 2 innies r1c45 = 8
6a. since it is a h8(2), and since both 8(2) cages in n8 partially see it, they must all have different combos
6b. -> h8(2)r1c45 = {26} only: both locked for n2 and r1
6c. -> r1c23 = 5 (cage sum) = {14} only: both locked for r1 and n1

really glad the disjoint cage gets a look-in
7. "45" on c6789: 1 outie r2c4 + 6 = 2 innies r36c6.
7a. since one innie and 1 outie are in the same nonet and can't be equal -> no 6 in r6c6 (IOU)

8. "45" on c789: 3 outies r457c6 = 16 and must have 6 or 8 for r7c6: but can't have both since no 2
8a. -> r7c6 = 6 (hsingle c6)
8b. -> r45c6 = 10 = {19/37}(no 5,8)
8c. -> r345c7 = 12

9. "45" on n9: 1 outie r6c9 = 1 innie r7c7 (no 1,6,8 in r6c9)

10. r4c6 sees all 1 in n6 -> no 1 in r4c6
10a. no 9 in r4c6 (h10(2))

11. "45" on n12: 3 innies r3c123 = 16 = {259/268/358/367}
11a. but {25}[9] and {36}[7] blocked by 5 and 3 in r4c1 respectively
11b. -> no 7,9 in r3c3

Cracked

12. 41(7) must have 7 & 9. r4c6 sees all of those -> no 7,9 in r4c6
12a. r4c6 = 3 (placed for d/), r5c6 = 7 (h10(2))
12b. 7 in 41(7) only in n4: locked for n4

13. "45" on r56789: 3 remaining innies r5c457 = 9 = {26}[1] only
13a. 2 & 6 locked for 41(7), r5 and n5
13b. 4 & 9 in 41(7) only in r4: locked for r4

14. "45" on n12: 1 innie r3c3 - 4 = 1 outie r4c1 = [51] only, 5 placed for d\

15. r34c7 = 11 (cage sum) = [65] only, 6 placed for d/, r5c45 = [62], 2 placed for both D, r1c45 = [26]

16. 9(2)n3 = [72]

17. 1 innie n3: r3c8 = 1

18. r3c12 = 11 (cage sum) = {29/38}(no 7)

19. killer pair 8,9 in r3c126: both locked for r3

20. "45" on r12: 1 innie r2c5 + 2 = 1 outie r3c9
20a. = [13] only

21. 12(2)n3 = [84], 4 placed for d/

22. r3c45 = {47}, 7 locked for n2
22a. -> r3c6 = 8 (cage sum)

23. naked pairs {59} in r12c69: locked for r12
23a. r1c1 = 3 (placed for d\), r2c4 = 3, r7c7 = 4, placed for d\, r6c9 = 4 (iodn9=0)

24. "45" on c1: 2 outies r37c2 + 2 = 1 innie r2c1
24a. -> r3c2 = 2, r3c1 = 9
24b. -> r2c1 - 4 = r7c2 = [73] only

25. 8(2)r8c5 = {35}: 5 locked for c5 and n8

26. naked pair {89} in r67c5: both locked for c5 and 23(4) cage
26a. -> r6c46 = [51]: 5 placed for their D

27. r6c8 = 7 (hsingle n6)

Lots of naked now
Cheers
Ed


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 Post subject: Re: Assassin 391
PostPosted: Tue Feb 25, 2020 2:45 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1703
Location: Lethbridge, Alberta, Canada
Thanks Ed for this latest Assassin! It could be considered to be a "Human Solvable".

It took me even longer to find the key step; therefore my solving path was rather different because I did more work before I found it.

Here is my walkthrough for Assassin 391:
Prelims

a) R12C7 = {18/27/36/45}, no 9
b) R12C8 = {39/48/57}, no 1,2,6
c) R89C4 = {17/26/35}, no 4,8,9
d) R89C5 = {17/26/35}, no 4,8,9
e) 9(3) cage at R3C8 = {126/135/234}, no 7,8,9
f) 13(4) cage at R1C2 = {1237/1246/1345}, no 8,9
g) 37(6) cage at R5C8 = {256789/346789}, no 1
h) 41(7) cage at R3C3 = {2456789}, no 1,3
i) 36(8) cage at R6C9 = {12345678}, no 9

1a. 13(4) cage at R1C2 = {1237/1246/1345}, 1 locked for R1, clean-up: no 8 in R2C7
1b. 45 rule on N2 2 outies R1C23 = 5 = {14/23}
1c. R1C23 = 5 -> R1C45 = 8 = {17/26/35}, no 4
1d. 45 rule on N3 2 innies R3C78 = 7 = {16/25/34}, no 7,8,9
1e. 45 rule on N12 1 innie R3C3 = 1 outie R4C1 + 4, R3C3 = {56789}, R4C1 = {12345}
1f. 2,4 in 41(7) cage at R3C3 only in R4C2345 + R5C45, CPE no 2,4 in R4C6

2a. 9 in N9 only in R79C7, locked for C7
2b. 9 in N6 only in R5C89 + R6C8, locked for 37(6) cage at R5C8
2c. R9C7 = 9 (hidden single in C7) -> R89C6 = 6 = {15/24}
2d. 45 rule on N8 3 remaining innies R7C456 = 23 = {689}, locked for R7, 6 locked for N8, clean-up: no 2 in R78C45
2e. 9 in R7 only in R7C45, CPE no 9 in R6C4
2f. R78C6 = {24} (hidden pair in N8), locked for C6
2g. 45 rule on N9 1 outie R6C9 = 1 remaining innie R7C7 -> R6C9 = {23457}
2h. 37(6) cage at R5C8 = {256789/346789}, CPE no 7 in R45C7
2i. 1 in N6 only in R4C789 + R5C7, CPE no 1 in R4C6
2j. 7,9 in R4 only in R4C23456, CPE no 7,9 in R5C45

3. 45 rule on C789 3 remaining outies R457C6 = 16 = {169/178/358/367}
3a. R7C6 = {68} -> no 6,8 in R45C6
3b. R45C6 = {35/37}/[71/91], no 9 in R5C6
3c. Consider placement for 8 in N6
8 in R45C7 => R7C6 = 8 (because 37(6) cage at R5C8 contains 8) => R45C6 = 8 = {35}/[71] => R345C7 = 14 = {248} (cannot be {158} which clashes with R45C6)
or 8 in R5C89 + R6C78 => R7C6 = 6, R45C6 = 10 = {37}/[91], R345C7 = 12 = {156/246/345}
-> R345C7 = {156/246/248/345}
3d. 45 rule on R1234 4 outies R5C4567 = 16 = {1258/1267/1348/1456/2356} (cannot be {1357} because R5C45 only contain one odd number, cannot be {2347} = {24}[73] which clashes with R45C6)

4. 45 rule on C6789 2 innies R36C6 = 1 outie R2C4 + 6, IOU no 6 in R6C6

[Only just spotted what seems to be a key step; it proves to be the key step. With hindsight it could have been used immediately after step 2d.]
5. R89C4 and R89C5 are both {17/35}, preventing R1C45 being {17/35} -> R1C45 = {26}, locked for R1 and N2, clean-up: no 3 in R1C23 (step 1b), no 3,7 in R2C7
5a. Naked pair {14} in R1C23, locked for N1, 4 locked for R1, clean-up: no 5 in R2C7, no 8 in R2C8
5b. R7C6 = 6 (hidden single in C6) -> R45C6 (step 3) = 10 = {37}/[91], no 5
5c. 37(6) cage at R5C8 = {256789/346789}, 8 locked for N6
5d. Naked pair {89} in R7C45, CPE no 8 in R6C4
5e. Hidden killer pair 1,3 in R45C6 and 23(4) cage at R6C4 for N5, R45C6 contains one of 1,3 -> 23(4) cage must contain one of 1,3 = {1589/3479/3569/3578} (cannot be {1679} which clashes with R45C6, other combinations don’t contain 1 or 3), no 2
5f. 2 in N5 only in R45C45, locked for 41(7) cage in R3C3
5g. 12(3) cage at R3C1 = {129/138/237/246/345} (cannot be {147} because 1,4 only in R4C1, cannot be {156} which clashes with R3C3 + R4C1 = [51], step 1e)
5h. 4 of {345} must be in R4C1 -> no 5 in R4C1, clean-up: no 9 in R3C3 (step 1e)
5i. 41(7) cage at R3C3 = {2456789}, 9 locked for R4, clean-up: no 1 in R5C6 (step 5b)
5j. Naked pair {37} in R45C6, locked for C6, 7 locked for N5, 3 locked for 22(5) cage at R3C7, clean-up: no 4 in R3C8 (step 1d)
5k. R345C7 (step 3c) = {156/246}, 6 locked for C6, clean-up: no 3 in R1C7
5l. R12C7 = [72/81] (cannot be [54] which clashes with R345C7), no 4,5
5m. Killer pair 1,2 in R2C7 and R345C7, locked for C7, clean-up: no 2 in R6C9 (step 2g)
5n. 23(4) cage at R6C4 = {1589}, 1,5 locked for R6, 5 locked for N5, clean-up: no 5 in R7C7 (step 2g)
5o. Killer pair 1,5 in R6C4 and R89C4, locked for C4
5p. 41(7) cage = {2456789}, CPE no 5,7 in R56C3
5q. 4,6 in N5 only in R45C45, locked for 41(7) cage, clean-up: no 2 in R4C1 (step 1e)
5r. 17(3) disjoint cage at R1C6 = {179/359/458}
5s. 3,4,7 of {179/359/458} only in R2C4 -> R2C4 = {347}
5t. 45 rule on N1 3 remaining innies R3C123 = 16 = {259/268/358} (cannot be {367} because 12(3) cage at R3C1 cannot contain both of 3,6), no 7, clean-up: no 3 in R4C1 (step 1e)
5u. 41(7) cage = {2456789}, 7 locked for R4 and N4 -> R4C6 = 3, placed for D/, R5C6 = 7, clean-up: no 9 in R1C8

6a. 1,6 in N6 only in R4C789 + R5C7
6b. 45 rule on N69 4 remaining innies R4C789 + R5C7 = 14 = {1256}, 2,5 locked for N6
6c. R5C6 = 7 -> R5C4567 = {1267} (step 3d) -> R5C7 = 1, R5C45 = {26}, locked for R5 and N5, R2C7 = 2 -> R1C7 = 7, clean-up: no 5 in R12C8, no 5 in R3C7, no 5,6 in R3C8 (both step 1d), no 7 in R6C9 (step 2g)
6d. R45C6 = [37], R5C7 = 1 -> R34C7 = 11 = [65], 6 placed for D/, R3C8 = 1 (step 1d), R5C5 = 2, placed for both diagonals
6e. Naked quad {4789} in R4C2345, 4 locked for R4, 8 locked for 41(7) cage at R3C3 -> R4C1 = 1, R3C3 = 5, placed for D\
6f. 2 in R3 only in R3C12, R4C1 = 1 -> R3C12 = 11 = {29}, 9 locked for R3 and N1 -> R3C6 = 8
6g. 7 in R3 only in R3C45, locked for N2
6h. 17(3) disjoint cage at R1C6 (step 5r) = {359} (only remaining combination) -> R2C4 = 3, R12C6 = {59}, locked for N2, 9 locked for C6
6i. R6C6 = 1, placed for D\, R6C4 = 5, placed for D/
6j. Naked pair {17} in R89C4, locked for N8, 7 locked for C4 -> R3C45 = [47]
6k. R2C9 = 5 (hidden single in N3), R12C6 = [59], R2C8 = 4, placed for D/, R1C8 = 8, R1C9 = 9, placed for D/, R3C9 = 3, R56C9 = [84], R67C7 = [34], R56C8 = [97], R8C7 = 8
6l. R1C1 = 3, placed for D\, R8C8 = 6, R9C9 = 7, both placed for D\, R2C2 = 8, placed for D\, R89C4 = [71], R9C1 = 8, R8C2 = 1, R7C3 = 7, R1C2 = 4, R8C9 = 2, R89C6 = [42]
6m. R9C23 = [64] (hidden pair in R9)

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

Rating Comment:
I'll rate my walkthrough for A391 at Easy 1.5. Even if I'd found the key step sooner I'd still have given the same rating since the key step is a short implied forcing chain.


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