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 Post subject: Assassin 355
PostPosted: Wed Aug 01, 2018 8:14 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Attachment:
a355DisjointCage3ZeroCells.JPG
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Disjoint 16(4)r2c23+r3c35: 3 zero cells (no cage and no relationship) at r1c4, r3c4 and r5c6.

Took 5 versions of this puzzle to get it interesting enough to be an Assassin. Glad I persisted. Really like it. No real advanced steps needed. SudokuSolver gets a score of 1.70 with a very long solution. It can't find my key step. JSudoku has no trouble with the puzzle.

Assassin 355
code:
3x3::k:4105:4105:4105:0000:3586:3586:3843:3843:3843:2564:4117:4117:3586:3586:4613:3843:3334:6663:2564:4872:4117:0000:4117:4613:4613:3334:6663:3850:4872:4872:4363:2572:2572:4613:6663:6663:3850:8973:3342:4363:4363:0000:3087:3087:6663:3850:8973:3342:3342:2064:2064:2064:5649:5649:5138:8973:8973:7955:5396:5396:5649:5649:768:5138:5138:8973:7955:7955:5396:6657:6657:768:5138:8973:8973:7955:7955:5396:5396:6657:6657:
solution:
Code:
+-------+-------+-------+
| 2 8 6 | 9 4 7 | 5 1 3 |
| 7 5 4 | 1 2 3 | 6 9 8 |
| 3 9 1 | 8 6 5 | 2 4 7 |
+-------+-------+-------+
| 5 7 3 | 6 1 9 | 8 2 4 |
| 1 6 2 | 4 7 8 | 9 3 5 |
| 9 4 8 | 3 5 2 | 1 7 6 |
+-------+-------+-------+
| 8 1 9 | 7 3 6 | 4 5 2 |
| 6 2 7 | 5 9 4 | 3 8 1 |
| 4 3 5 | 2 8 1 | 7 6 9 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 355
PostPosted: Sun Aug 05, 2018 10:33 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin. There was one difficult bit in my solving path; I hope there's a better way than my step 6e for making those important eliminations.

Here is my walkthrough for Assassin 355:
Prelims

a) R23C1 = {19/28/37/46}, no 5
b) R23C8 = {49/58/67}, no 1,2,3
c) R4C56 = {19/28/37/46}, no 5
d) R5C78 = {39/48/57}, no 1,2,6
e) R78C9 = {12}
f) 19(3) cage at R3C2 = {289/379/469/478/568}, no 1
g) 8(3) cage at R6C5 = {125/134}
h) 14(4) cage at R1C5 = {1238/1247/1256/1346/2345}, no 9
h) 26(4) cage at R8C7 = {2789/3689/4589/4679/5678}, no 1
i) 31(5) cage at R7C4 = {16789/25789/34789/35689/45679}
j) 35(7) cage at R5C2 = {1235789/1245689/1345679/2345678}

Steps resulting from Prelims
1a. Naked pair {12} in R78C9, locked for C9 and N9
1b. 8(3) cage at R6C5 = {125/134}, 1 locked for R6
1c. 31(5) cage at R7C4 contains 9, locked for N8

2. 45 rule on N8 1 outie R9C7 = 7, clean-up: no 5 in R5C8
2a. 45 rule on N9 2 innies R7C78 = 9 = {36/45}, no 8,9
2b. 45 rule on N9 2 outies R6C89 = 13 = {49/58/67}, no 2,3
2c. R14C8 = {12} (hidden pair in C8)
2d. 45 rule on N7 2 outies R56C2 = 10 = [19]/{28/37/46}, no 5, no 9 in R5C2
2e. 35(7) cage at R5C2 contains 5, locked for N7

[This will be the “human solvable” step which SudokuSolver isn’t programmed to find.]
3. 45 rule on the whole grid 3(2+1) innies R13C4 + R5C6 = 25
3a. Max R13C4 = 17 -> min R5C6 = 8
3b. R5C6 = {89} -> R13C4 = 16,17 = {79/89}, 9 locked for C4 and N2
3c. 9 in N8 only in R89C5, locked for C5, clean-up: no 1 in R4C6
3d. 17(3) cage at R4C4 = {278/368/458/467}, no 1

4. 45 rule on N47 2(1+1) outies R3C2 + R6C4 = 12 = {48/57}/[66/93], no 2,3 in R3C2, no 2 in R6C4
4a. 45 rule on N1 1 innie R3C2 = 1 outie R3C5 + 3, R3C2 = {456789} -> R3C5 = {123456}
4b. 45 rule on N36 using R6C89 = 13 (or, if preferred, 45 rule on N3689) 3 innies R346C7 = 11 = {128/146/236/245}, no 9
4c. {128} must have 8 in R4C7 (cannot be [821] which clashes with R4C8), no 8 in R3C7

5. 45 rule on N5 4 innies R5C6 + R6C456 = 18 = {1269/1278/1359/1458/2358} (cannot be {1467/2367/2457/3456} because R5C6 only contains 8,9, cannot be {1368/2349} which clash with R4C56)
5a. R5C6 = {89} -> no 8 in R6C4
5b. 9 in N5 only in R4C56 or R5C6 + R6C456 -> R5C6 + R6C456 = {1269/1359/2358} (cannot be {1278/1458} which clash with R4C56 = [19] (blocking-cages), no 4,7 in R6C456
[Ed told me this is locking-out cages; I’ve never been sure of the difference.]
5c. R6C4567 = [6{12}5]/[3{15}2]/[5{13}4][3{25}1], 5 locked for R6, no 3 in R6C7, clean-up: no 8 in R6C89 (step 2b)
5d. R6C4 = {356} -> R3C2 = {679} (step 4), R3C5 = {346} (step 4a)

6. 45 rule on C1 1 innie R1C1 = 1 outie R8C2, no 5 in R8C2 -> no 5 in R1C1
6a. 5 in C1 only in 15(3) cage at R4C1, locked for N4
6b. 19(3) cage at R3C2 = {289/379/469/478}
6c. 13(3) cage at R5C3 = {139/157/238/256/346} (cannot be {148/247} because R6C4 only contains 3,5,6)
6d. 45 rule on N47 4 innies R4C23 + R56C3 = 20
6e. 19(3) cage cannot be [649/694] with R4C23 totalling 13 because R56C3 = 7 doesn’t contain 5, 13(3) cage cannot be [166] and 13(3) cage = {346} clashes with 19(3) cage -> no 6 in R3C2, clean-up: no 3 in R3C5 (step 4a), no 6 in R6C4 (step 4)
6f. R5C6 + R6C456 (step 5b) = {1359/2358}, 3,5 locked for R6 and N5, clean-up: no 7 in R4C56, no 7 in R5C2 (step 2d)
6g. 13(3) cage = {139/157/238/256/346}
6h. 1 of {139/157} must be in R5C3, 3 of {139/238/346} -> no 3,7,9 in R5C3

7. 1 in R5 only in R5C123, locked for N4, 5 in N4 only in R45C1
7a. 45 rule on R67989 4 outies R4C1 + R5C123 = 14 must contain 1 and 5 = {1256}, locked for N4, clean-up: no 7 in R6C2 (step 2d)
7b. 3 in N4 only in R4C23, locked for R4 -> 19(3) cage at R3C2 = {379}, CPE no 9 in R6C2, clean-up: no 1 in R5C2 (step 2d)
7c. 4 in N4 only in R6C123, locked for R6, clean-up: no 9 in R6C89 (step 2d)
7d. Naked pair {67} in R6C89, locked for R6, N6 and 22(4) cage at R6C8, clean-up: no 5 in R5C7, no 3 in R7C78 (step 2a)
7e. Naked pair {45} in R7C78, locked for R7 and N9
7f. Killer pair 8,9 in R5C6 and R5C78, locked for R5
7g. Naked triple {489} in R6C123, locked for N4
7h. Naked pair {37} in R4C23, locked for R4 and 19(3) cage at R3C2 -> R3C2 = 9, R3C5 = 6 (step 4a), placed for disjoint cell at R2C2, R6C4 = 3 (step 4), clean-up: no 1,4 in R2C1, no 4,7 in R2C8, no 1 in R3C1, no 4 in R4C6
7i. R1C4 = 9 (hidden single in C4)
7j. R3C5 = 6 -> R2C23 + R3C3 = 10 = {127/145/235}, no 8

8. R346C7 (step 4b) = {128} (only remaining combination, cannot be {245} which clashes with R7C7) -> R4C7 = 8, R36C7 = {12}, locked for C7, clean-up: no 2 in R4C56, no 4 in R5C78
8a. Naked pair {39} in R5C78, locked for R5 and N6 -> R5C6 = 8
8b. R13C4 + R5C6 = 25 (step 3), R1C3 = 9, R5C6 = 8 -> R3C4 = 8, clean-up: no 2 in R2C1, no 5 in R2C8
8c. R4C6 = 9 (hidden single in N5) -> R4C5 = 1, R14C8 = [12], R36C7 = [21], clean-up: no 8 in R2C1
8d. R34C7 = [28] = 10 -> R23C6 = 8 = {17/35}, no 4
8e. Naked pair {45} in R45C9, locked for C9
8f. R4C8 = 2, R45C9 = {45} = 9 -> R23C9 = 15 = [87], R6C89 = [76], R19C9 = [39]
8g. R1C89 = [13] = 4 -> R12C7 = 11= {56}, locked for C7 and N3 -> R23C9 = [94], R3C1 = 3 -> R2C1 = 7
8h. R23C6 = [35] (only remaining permutation), R6C56 = [52]
8i. Naked triple {146} in R789C6, locked for C6 and N8
8j. R789C6 = {146} = 11, R9C7 = 7 -> R7C5 = 3 (cage sum)
8k. R3C35 = [16] = 7 -> R2C23 = 9 = {45}, locked for R2 and N1
8l. R45C1 = [51] (hidden pair in N4) -> R6C1 = 9 (cage sum)
8m. R789C1 = {2468} -> 20(4) cage at R7C1 = {2468} (only possible combination), locked for N7, 4 also locked for C1
8n. Naked quad {2468} in R1568C2, locked for C2 -> R2C23 = [54]
8o. R6C34 = [83] -> R5C3 = 2 (cage sum)

and the rest is naked singles.
Thanks Ed for your comment about what my step 5b is, and for correcting a couple of typos. Also for suggesting in your post that the best way to do step 6e is to go directly to step 7, when those key eliminations will come out more easily.

Rating Comment:
I'll rate my WT at 1.5. If I'd spotted Ed's suggestion to omit step 6e and go straight to step 7, then I'd have rated it at Easy 1.5, but I didn't spot that.


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 Post subject: Re: Assassin 355
PostPosted: Thu Aug 09, 2018 8:58 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Andrew wrote:
I hope there's a better way than my step 6e for making those important eliminations
There is! Just don't do step 6e! Drop straight to step 7 and that's the important one. Just a couple of routine clean-ups after step 7a. makes 7b. work and on from there.

I worked in similar areas to Andrew but quite differently.

A355 WT
37 steps:
Preliminaries courtesy of SudokuSolver
Cage 3(2) n9 - cells ={12}
Cage 12(2) n6 - cells do not use 126
Cage 13(2) n3 - cells do not use 123
Cage 10(2) n1 - cells do not use 5
Cage 10(2) n5 - cells do not use 5
Cage 8(3) n56 - cells do not use 6789
Cage 19(3) n14 - cells do not use 1
Cage 14(4) n2 - cells do not use 9
Cage 26(4) n9 - cells do not use 1

No routine clean-up unless stated.
1. "45" on n8: 1 outie r9c7 = 7

Andrew correctly guessed that SS isn't coded to do this step. It is one of my key eliminations since it makes step 11c. work.
2. "45" on whole grid: 3 zero cells r13c4 + r5c6 = 25 = {89}[8]/{79}[9]
2a. must have 9 in r13c4: locked for n2 and c4
2b. r13c4 = {789}, r5c6 = (89)

3. "45" on n7: 2 outies r56c2 = 10 (no 5)

4. 35(7)r5c2 must have 5 which is only in n7: 5 locked for n7

5. "45" on c1: 1 innie r1c1 = 1 outie r8c2
5a. no 5 r1c1

6. 5 in c1 only in 15(3)n4 = {159/258/357/456}
6a. 5 locked for n4

7. "45" on n89: 2 outies r6c89 = 13 = {49/58/67}(no 1,2,3)

8. 8(3)r6c5 = {125/134} = [3/5,4/5..]
8a. must have 1: 1 locked for r6

9. "45" on n5: 1 outie r6c7 + 10 = 2 innies r5c6 + r6c4
9a. -> max. 2 innies = 15
9b. no 8 in r6c4

10. "45" on n47: 2 outies r3c2 + r6c4 = 12 (no 2)
10a. no 3,4 in r3c2

11. r6c89 = 13 (step 7): "45" on r6 -> 4 innies r6c1234 = 24
11a.but {3579/4569/4578} all blocked by 8(3) (step 8)
11b. = {2589/2679/3489/3678}
11c. 5 in {2589} must be in r6c4 -> no 5 in r6c1

12. "45" on r6789: 4 outies r4c1 + r5c123 = 14 and must have 1 & 5 for n4
12a. = {1256} only
12b. 2 & 6 locked for n4
12c. r6c2 = (489) (h10(2)r56c2)

13. from step 11, r6c1234 = 24 = {3489/3678}(no 5)
13a. consider {3678} with r6c34 as (378)[6]
i. [86/76] blocked since they exceed the cage total in 13(3)r5c3
ii. [36] blocked since no 4 in r5c3
13b. -> {3678} blocked from r6c1234
13c. = {3489} only: all locked for r6, 8 & 9 locked for n4
13d. no 5,6,7 in r3c2 (h12(2)r3c2+r6c4)

14. 7 in n4 only in r4: locked for r4
14a. no 3 in 10(2)n5

15. h13(2)r6c89 = {67} only: both locked for n6 and no 6,7 in r7c78
15a. no 5 in 12(2)n6

16. 7 in c6 only in r123c6: 7 locked for n2

17. naked pair {89} in r13c4: 8 locked for n2 and c4

18. r13c4 = 17 -> r5c6 = 8 (3 zero cells = 25)

19. 12(2)r5c78 = {39} only: both locked for r5 and n6

20. 7 in r5 only in 17(3)n5 = {467} only: 4 and 6 locked for n5

21. r6c4 = 3
21a. -> r3c2 = 9 (h12(2)r3c2+r6c4)
21b. -> r4c23 = 10 = {37} only
21c. r13c4 = [98]
21d. no 1 in r5c2 (h10(2)r56c2)

22. "45" on n1: 1 outie r3c5 = 6 (Placed for 16(4)r2c2))

23. 3(2)n9 = {12} only: both locked for c1 and n9

24. hidden pair {12} in c8 -> r14c8 = {12} only

25. {67} blocked from 13(2)n3 by r6c8 = (67) -> r23c8 = [94/85]

26. r6c8 = 7 (hidden single c8), r6c9 = 6

27. 10(2)n5 = {19} only: 1 locked for r4 and n5
27b. -> 1 in 8(3) only in r6c7 -> r6c7 = 1

28. r4c8 = 2, r1c8 = 1

29. 6 in n3 only in 15(4) = {1356} only: 3,5,6 locked for n3, 6 for c7
29a. r23c8 = [94], r3c79 = [27]

30. 26(5)r2c9 = {24578} only
30a. -> r2c9 = 8, r45c9 = {45} only: both locked for c9 and n6
30b. r4c7 = 8

31. r23c1 = [73] (only permutation)

32. r34c7 = 10 -> r23c6 = 8 = [35] only permutation

33. r6c89 = 13 -> r7c78 = 9 = [45] only permutation

34. r9c7 = 7 -> r7c56 + r89c6 = 14 = {1238/1346} only (no 9)
34a. must have 3 -> r7c5 = 3
34b. must have 1: locked for c6 and n8

35. r3c3 = 1
35a. r3c35 = 7 in 16(4)r2c2 -> r2c23 = 9 = {45} only: both locked for r2 and n1

36. r1c1 = (268) -> r8c2 = (268) (IODc1 = 0)

37. naked triple {268} in r158c2: all locked for c2
37a. r56c2 = [64] (h10(2)), r56c3 = [28] (cage sum)

Rest is naked singles. Don't get to say that very often!
Cheers
Ed


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 Post subject: Re: Assassin 355
PostPosted: Sat Aug 11, 2018 7:50 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed. Here's how I did it. Not the most elegant this time :)

Assassin 355 WT:
1. Outies n8 -> r9c7 = 7
-> 7 in n8 in r7c4 or r8c45

2. Innies whole puzzle r13c4 + r5c6 = +25(3)
-> (a) r13c4 = {89}, r5c6 = 8
or (b) r1c34 = {79}, r5c6 = 9
Either way -> 9 in 31(5)n8 in r89c5
-> 9 in n5 in r45c6

-> if (a) 10(2)n5 = [19]
-> if (b) r89c5 = [79] -> r45c6 = [79] -> 10(2)n5 = [37]

3. Small chain
I had a really nice move without the chain which I found on my checking read through relied on a completely false premise. Urgh!

If (b) above is right -> remaining Innies n5 = r6c456 = +9(3) = [6{12}]
But Outies c123 -> r3c5 + r6c4 = +9
So (b) would put 3 both in r3c5 and r4c5
-> (a) is correct.

-> r13c4 = {89}
-> r5c6 = 8
-> 10(2)n5 = [19]

4. Outies n7 = r56c2 = +10(2)
Those values in n7 must go in r789c1
-> 20(4)n7 = 10(2) + 10(2)
-> 5 in n7 in 35(7) in c23
Innies - Outies c1 -> r1c1 = r8c2 (no 5)
-> HS 5 in c1 -> 5 in n4 in r456c1

5. Remaining Innies n5 = r6c456 = +10(3) can only be [3{25}]
-> 17(3)n5 = {467} and r6c7 = 1

6. Outies n47 -> r3c2 = 9
-> r13c4 = [98]
Innies - Outies n1 -> r3c5 = 6
-> 6 in n5 in r45c6
-> 6 in n8 in r789c6
-> 21(5)n8 must have a 1 in r789c6
-> HS 1 in c4 -> r2c4 = 1
-> Remaining Innies n2 -> r23c6 = +8(2) = {35}
-> HS 7 in c6 -> r1c6 = 7
-> r12c5 = {24}

7. Given r1c4 = 9 -> Outies r1 = r2c457 = +9(3)
Can only be r2c457 = [126]
-> r1c5 = 4
-> r1c789 = {135}
-> r1c123 = {268}
-> 10(2)n1 = {37}
-> 16(4)n1 = [{45}16]
-> r23c6 = [35]
-> 10(2)n1 = [73]

8. Also 8(3)n5 = [521]
-> r789c4 = {257} and r89c5 = {89}
Also r7c5 = 3 and r789c6 = {146}
Also 17(3)n5 = [{46}7]

9. 3(2)n9 = {12}
-> Remaining Innies n9 = r7c78 = +9(2) = {45}
-> 22(4)n69 = [{67}{45}]
Also 12(2)n6 = {39}
-> (37) both in r4 in n4 -> 19(3)r3c2 = [9{37}]

10. 13(2)n3 can only be [94]
-> r7c78 = [45]
-> 18(4)r2c6 = [3528]
-> 26(5)n36 = [872{45}]
Also 12(2)n6 = [93]
Also r1c789 = [513]
Also r6c89 = [76]
-> NS r9c9 = 9
-> r89c5 = [98]
-> 26(4)n9 = [3869]

11. r56c3 = +10(2)
Also 56c2 = +10(2)
Since r3c23 = [91]
-> neither of those +10(2)s can be {19}
-> 15(3)n4 = [519]
-> r45c9 = [45]
-> 17(3)n5 = [647]

12. r1789c1 = {2468}
-> 20(4)n7 = {2468}
-> HS 9 in r7/n7 -> r7c3 = 9
-> HS 8 in r7/n7 -> r7c1 = 8

13. 1 in n7 in r79c2
Whichever of {37) is in in r4c3 must also go in n7 in r79c2
-> 5 in n7 in r89c3
-> r2c23 = [54]
-> 13(3)n4 = [283]
-> r56c2 = [64]
-> 16(3)n1 = [286]
-> 20(4)n7 = [8624]
-> 21(5)n8 = [36417]
Also 3(2)n89 = [21]
Also r7c2 = 1
-> HS r8c3 = 7
-> 19(3)n14 = [973]
-> r9c23 = [35]
-> r789c4 = [752]


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