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Ed
 Post subject: Assassin 205
PostPosted: Thu Feb 03, 2011 10:02 pm 
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Found a really neat way to make big inroads near the beginning. Unfortunately, the end is a bit tedious but hopefully I just missed something.

Assassin 205
note: it has a remote 6(2) cage at r5c46
Image
code:select, copy and paste into solver:
3x3::k:5121:2562:2562:2562:2563:11524:11524:11524:11524:5121:3845:3845:9990:2563:2563:2563:11524:11524:5121:5121:9990:9990:3079:3079:4616:11524:11524:2313:2313:9990:9990:5898:3079:4616:4616:11524:3851:3851:9990:1548:5898:1548:2061:2830:2830:4111:7184:7184:7184:5898:5898:2061:3857:3857:4111:4111:7184:7184:7184:5650:2061:5139:5139:2324:4117:5142:5142:5142:5650:5650:5650:5139:2324:4117:4117:4117:5142:4631:4631:4631:5139:
solution: 
+-------+-------+-------+
| 2 4 1 | 5 3 6 | 8 9 7 |
| 7 9 6 | 8 4 1 | 2 3 5 |
| 8 3 5 | 9 2 7 | 6 4 1 |
+-------+-------+-------+
| 1 8 4 | 6 9 3 | 5 7 2 |
| 9 6 7 | 4 5 2 | 1 8 3 |
| 5 2 3 | 7 1 8 | 4 6 9 |
+-------+-------+-------+
| 4 7 9 | 1 6 5 | 3 2 8 |
| 6 5 2 | 3 8 9 | 7 1 4 |
| 3 1 8 | 2 7 4 | 9 5 6 |
+-------+-------+-------+
Cheers
Ed
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Frank
 Post subject: Re: Assassin 205
PostPosted: Fri Feb 04, 2011 5:44 pm 
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Tough one Ed. I liked the fact it was asymmetric.

My 2 key moves were:

:
to show that the {234} in r2c7 copied over to r4c9, and the 11/2 cage in N6 was {38}


Thanx for the puzzle :)
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Andrew
 Post subject: Re: Assassin 205
PostPosted: Mon Feb 07, 2011 12:11 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
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Thanks Ed for a fun puzzle.

I found my first important step fairly quickly but it took me some time to find my other key moves;
with hindsight:
steps 13, 14 and 17 could have been done a lot earlier.

Rating Comment:
I'll rate my walkthrough for A205 at 1.5. I used a couple of short forcing chains, the first of which used killer pairs (which is why I didn't say Easy 1.5), law of leftovers and a hidden killer triple. Edit. I've now shown that the first forcing chain is better written as a hidden killer quad.

Here is my walkthrough for A205:
Prelims

a) R2C23 = {69/78}
b) R4C12 = {18/27/36/45}, no 9
c) R5C12 = {69/78}
d) R5C46 = {15/24}
e) R5C89 = {29/38/47/56}, no 1
f) R6C89 = {69/78}
g) R89C1 = {18/27/36/45}, no 9
h) 10(3) cage at R1C2 = {127/136/145/235}, no 8,9
i) 8(3) cage at R5C7 = {125/134}
j) 10(4) cage at R1C5 = {1234}
k) 39(6) cage at R2C4 = {456789}, no 1,2,3
l) and of course 45(9) cage at R1C6 = {123456789}

Steps resulting from Prelims
1a. 8(3) cage at R5C7 = {125/134}, 1 locked for C7
1b. 10(4) cage at R1C5 = {1234}, 1 locked for N2, CPE no 4 in R2C4
1c. 1 in N3 only in 45(9) cage -> no 1 in R4C9

[There must be interactions between R2C23 and 39(6) cage at R2C4 but I can’t yet see how to use them.]

2. R2C7 = {234} -> R2C7 + 8(3) cage at R5C7 = 10,11,12 = {1234/1235/1245}, 2 locked for C7

3. 45 rule on R1234 1 innie R4C5 = 1 outie R5C3 + 2 -> R4C5 = {6789}, R5C3 = {4567}

4. R5C3 = {4567} -> either R5C3 = {45}, killer pair 4,5 in R5C3 and R5C46, locked for R5 or R5C3 = {67}, killer pair 6,7 in R5C12 and R5C3, locked for R5 -> R5C89 = {29/38}, no 4,5,6,7
[I later realised that this step is better written as
Hidden killer quad 4,5,6,7 in R5C12, R5C3, R5C46 and R5C89 for R5, R5C12 contains one of 6,7, R5C3 contains one of 4,5,6,7, R5C46 contains one of 4,5 -> R5C89 cannot contain more than one of 4,5,6,7 -> R5C89 = {29/38}, no 4,5,6,7]

5. Killer pair 8,9 in R5C12 and R5C89, locked for R5
5a. Killer pair 8,9 in R5C89 and R6C89, locked for N6

6. 18(3) cage at R3C7 = {279/369/378/459/468/567} (cannot be {189} because 8,9 only in R3C7), no 1
6a. 8,9 of {369/378/459/468} must be in R3C7 -> no 3,4 in R3C7
6b. 5 of {459/567} must be in R4C78 (R4C78 cannot be {67} which clashes with R6C89) -> no 5 in R3C7

7. 1 in N6 only in R56C7, locked for 8(3) cage at R5C7 -> no 1 in R7C7

8. 12(3) cage at R3C5 = {129/138/147/156/237/246/345}
8a. 1 of {129/138/147/156} must be in R4C6, 5,6,7 of {237/246/345} must be in R3C56 (R3C56 cannot be {23/24/34} which clash with 10(4) cage at R1C5, ALS block) -> no 5,6,7,8,9 in R4C6

9. 45 rule on N5 3 innies R4C46 + R6C4 = 16 = {169/178/268/349/358/367} (cannot be {259/457} which clash with R5C46)
9a. 1,2 of {169/179/268} must be in R4C6 -> no 1,2 in R6C4

10. Law of Leftovers for N3, R1C6 + R4C9 must contain exactly the same pair of candidates as R23C7, no 5 in R23C7 -> no 5 in R1C6 + R4C9

11. 45 rule on N1 1 outie R1C4 = 1 innie R3C3, no 2,3 in R1C4, no 8,9 in R3C3

12. 10(3) cage at R1C2 = {127/136/145/235}
12a. 6,7 of {127/136} must be in R1C4 -> no 6,7 in R1C23

13. Hidden killer triple 1,2,3 in R4C12, R4C6 and R4C789 for R4, R4C789 can only contain one of 2,3 (cannot contain both of 2,3 which would clash with R5C89) -> R4C12 and R4C6 must each contain one of 1,2,3 -> R4C6 = {123}, R4C12 = {18/27/36}, no 4,5
13a. Killer pair 2,3 in R4C789 and R5C89, locked for N6

14. 8(3) cage at R5C7 = {125/134}
14a. 2,3 only in R7C7 -> R7C7 = {23}

15. 45 rule on N6 2 outies R37C7 = 1 innie R4C9 + 7
15a. Max R37C7 = 12 -> no 6,7 in R4C9

16. Law of Leftovers for N3, R1C6 + R4C9 must contain exactly the same pair of candidates as R23C7
16a. R2C7 = R4C9 = {234} -> R1C6 = R3C7 = {6789}

17. Consider placements for 3 in N6
3 in R4C789 => R4C12 = {18/27} => R5C12 = {69} (cannot be {78} which clashes with R4C12
3 in R5C89 = {38}, locked for R5 => R5C12 = {69}
17a. -> R5C12 = {69}, locked for R5 and N4, clean-up: no 3 in R4C12, no 8 in R4C5 (step 3), no 2 in R5C89

18. Naked pair {38} in R5C89, locked for R5 and N6, clean-up: no 3 in R2C7 (step 16), no 7 in R6C89
18a. Naked pair {69} in R6C89, locked for R6 and N6

19. 7 in N6 only in R4C78, locked for R4 and 18(3) cage at R3C7, no 7 in R3C7, clean-up: no 7 in R1C6 (step 16), no 2 in R4C12, no 5 in R5C3 (step 3)
19a. Naked pair {18} in R4C12, locked for R4 and N4
19b. R89C1 = {27/36/45} (cannot be {18} which clashes with R4C1), no 1,8

20. R4C6 = 3 (hidden single in R4), R3C56 = 9 = {27/45}, no 6,8,9
20a. Killer pair 2,4 in 10(4) cage at R1C5 and R3C56, locked for N2, clean-up: no 4 in R3C3 (step 11)
20b. 3 in N2 only in R12C5, locked for C5

21. R4C45 = {69} (hidden pair in R4)

22. Killer pair 2,4 in R2C7 and 8(3) cage at R5C7, locked for C7

23. 18(3) cage at R3C7 (step 6) must contain 7 in R4C78 = {279/567}, no 4,8, clean-up: no 8 in R1C6 (step 16)

24. 4 in 39(6) cage at R2C4 only in R45C3, locked for C3 and N4
24a. 8,9 in 39(6) cage at R2C4 only in R234C4, locked for C4
24b. 2 in N6 only in R6C123, locked for R6

25. 10(3) cage at R1C2 = {127/136/145/235}
25a. R1C4 = {567} -> no 5 in R1C23

26. 2 in R4 only in R4C89, CPE no 2 in R123C8
26a. 2 in 45(9) cage at R1C6 only in R1234C9, locked for C9

27. 8 in N5 only in 23(4) cage = {1589/2489/2678} (cannot be {4568} which clashes with R5C46)
27a. 2 of {2489/2678} must be in R5C5 -> no 4,7 in R5C5

28. R5C3 = 7 (hidden single in R5), R4C3 = 4 (hidden single in N4), R4C9 = 2, R2C7 = 2 (hidden single in N3), R7C7 = 3, clean-up: no 8 in R2C2

29. Naked pair {57} in R4C78, locked for N6, R3C7 = 6 (step 23), R1C6 = 6 (step 16), R3C3 = 5, R1C4 = 5 (hidden single in N2), R4C4 = 6 (hidden single in 39(6) cage at R2C4), R4C5 = 9, clean-up: no 4 in R3C56 (step 20), no 1 in R5C6
29a. Killer pair 8,9 in R2C23 and R2C4, locked for R2

30. Naked pair {27} in R3C56, locked for R3
30a. Naked triple {235} in R6C123, locked for R6

31. 23(4) cage in N5 = {1589/2489}, no 7
31a. 2,5 only in R5C5 -> R5C5 = {25}
31b. R6C4 = 7 (hidden single in N5)

32. 28(6) cage at R6C2 must contain 7 = {123679/124579/134578} (cannot be {124678} because R6C23 must contain two of 2,3,5), 1 locked for R7
32a. R6C23 must contain two of 2,3,5 -> no 2,5 in R7C345
32b. 9 of {123679} must be in R7C3 -> no 6 in R7C3

33. 20(4) cage in N1 = {1469/1478/2369/2378} (cannot be {1289/1379/2468/3467} which clash with R2C23
33a. 2 of {2369/2378} must be in R1C1 -> no 3 in R1C1
33b. 6,7 of {2369/2378} must be in R2C1 -> no 3 in R2C1
33c. 9 of {1469/2369} must be in R3C2 (R123C1 cannot contain both of 6,9 which would clash with R5C1) -> no 9 in R13C1

34. 9 in C6 only in R789C6
34a. 45 rule on N9 3 outies R789C6 = 18 = {189/459} (cannot be {279} which clashes with R3C6), no 2,7
34b. Killer pair 1,4 in R2C6 and R789C6, locked for C6 -> R6C6 = 8, clean-up: no 1 in R89C6

35. Naked triple {459} in R789C6, locked for C6 and N8 -> R2C6 = 1, R5C6 = 2, R5C4 = 4, R56C5 = [51], R3C56 = [27], R7C4 = 1

36. 28(6) cage at R6C2 (step 32) = {123679} (only remaining combination) -> R7C3 = 9, R7C5 = 6, R6C23 = {23}, locked for R6 -> R6C1 = 5, clean-up: no 6 in R2C2, no 4 in R89C1

37. R6C1 = 5 -> R7C12 = 11 = {47} (only remaining combination), locked for R7 and N7, R7C6 = 5, R7C9 = 8, R7C8 = 2, R5C89 = [83], clean-up: no 2 in R89C1
37a. R7C89 = [28] = 10 -> R89C9 = 10 = {19/46}, no 5,7
37b. Killer pair 6,9 in R5C9 and R89C9, locked for C9
37c. Killer pair 1,4 in R3C9 and R89C9, locked for C9 -> R1C9 = 7, R2C9 = 5
37d. R1C7 = 8 (hidden single in N3)

38. Naked pair {36} in R89C1, locked for C1 and N7 -> R5C12 = [96]

39. 9 in N3 only in R13C8, locked for C8 -> R6C89 = [69], clean-up: no 1 in R89C9 (step 37a)
39a. Naked pair {46} in R89C9, locked for C9 and N9 -> R3C9 = 1

40. R89C5 = {78} = 15 -> R8C34 = 5 = [23], R6C23 = [23], R1C3 = 1, R1C2 = 4 (step 25)

and the rest is naked singles.
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Ed
 Post subject: Re: Assassin 205
PostPosted: Thu Feb 10, 2011 9:41 am 
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Frank wrote:
Tough one Ed. I liked the fact it was asymmetric.
Thanks. Nice to know some of us are asymmetrics - it's not everyone's taste. The next couple I'm working on are also. Been aiming for simpler cage structures and using asymmetry to give that extra bit of a twist.

I used a variation on Andrew's "Killer pairs" move in his step 4 to make big inroads elsewhere (my step 5). Andrew's is a blocking move whereas mine is a locking move....perhaps Alternating Killer pairs (or AK blocks in Andrew's)???? If anyone can remember seeing it used before please let me know. It may be a good one for the techniques forum.

Thank you to Andrew for his walk-through, very easy to read. Appears that Frank used a similiar approach. Andrew's step 34 makes a huge difference to the ending of the puzzle compared to how I found it.

A205 Alt start
10 steps:
(Optimised walk-through so some obvious eliminations are not done. However, I try and do clean-up as I go. Please tell me about any mistakes or things that could be clearer)

Prelims courtesy of SudokuSolver
i. Cage 6(2) n5 - cells only use 1,2,4,5
ii. Cage 15(2) n1 - cells only use 6,7,8,9
iii. Cage 15(2) n4 - cells only use 6,7,8,9
iv. Cage 15(2) n6 - cells only use 6,7,8,9
v. Cage 9(2) n7 - cells do not use 9
vi. Cage 9(2) n4 - cells do not use 9
vii. Cage 11(2) n6 - cells do not use 1
viii. Cage 8(3) n69 - cells do not use 6,7,8,9
ix. Cage 10(3) n12 - cells do not use 8,9
x. Cage 10(4) n23 - cells ={1234}
xi. Cage 39(6) n1245 - cells ={456789}

1. "45" on r1234: 1 innie r4c5 - 2 = 1 outie r5c3 = [64/75/86/97] = [6/7..](last bit important for later)
1a. r4c5 = (6..9), r5c3 = (4..7)

2. "45" on r5: 3 innies r5c357 = 13
2a. {139/238} in that h13(3) don't work with r5c3 from (4..7)
2b. {148/256} blocked by 6(2) at r5c46 = [1/4, 2/5..]
2c. h13(3) = {157/247/346}(no 8,9)

3. 15(2)n4 can only have one of 8/9 -> Hidden Killer pair 8,9 in r5
3a. -> 11(2)n6 must have 8/9 = {29/38}(no 4,5,6,7)

4. 15(2)n6 can only have one of 6/7 -> Hidden killer pair 6,7 in n6
4a. -> r4c789 must have 6/7 for n6 (no eliminations yet)

5. from step 1, either r4c5 is from (67) -> Killer pair 6,7 in r4 with r4c789 (step 4a)
5a. or r5c3 is from (67) -> Killer pair 6,7 in n4 with 15(2)
5b. -> no 6,7 in r4c123 (Alternating Killer pair..???)
5c. 9(2)n4 = {18/45} (no 2,3)

6. r4c789 can't have more than one of 2 or 3 since 11(2)n6 must have one
6a. -> hidden killer pair 2,3 in r4 -> r4c6 = (23)
6b. and r4c789 must have one of 2/3 for r4
6c.-> Killer pair 2,3 with 11(2)n6: both locked for n6

7. 8(3)r5c7 = {125/134}: must have 2/3 -> r7c7 = (23)
7a. must have 1 -> 1 locked for n6 and c7

8. 1 in r4 only in 9(2)n4 = {18} only: both locked for n4 and 8 for r4
8a. 15(2)n4 = {69} only: both locked for n4 & r5
8b. 11(2)n6 = {38} only: both locked for n6 & 3 for r5
8c. 15(2)n6 = {69} only: both locked for n6 & r6

9. r4c6 = 3 (hsingle r4)

10. "45" on n6: 2 outies r37c7 - 7 = 1 innie r4c9
10a. max. 2 outies = 12 -> max. r4c9 = 5
10b. 7 in n6 only in r4c78: locked for r4 and for 18(3) at r3c7
10c. 18(3) can't have {378} since no 3,8 in r4c78
10d. 18(3) = {279/567}(no 3,4,8)
10e. must have 6/9 -> r3c7 = (69)

Keep on from there. If your finding it hard going at the end, see Andrew's powerful step 34. Missed that.
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 205
PostPosted: Fri Feb 11, 2011 6:17 am 
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Thanks for the kind comments, Ed, but your clever step 5 is a lot more powerful than my step 4. You were helped a lot by your step 2; I don't think I seriously looked at that "45". As a result you got one of your key results in your step 8; I didn't reach the same naked pair until my step 19a.

Thanks for commenting about my step 34. Yes, it was very useful.
Maybe you missed the fact that:
the outies from N9 at that stage had to contain 9.

Ed wrote:
Nice to know some of us are asymmetrics - it's not everyone's taste.
I'm also happy with asymmetrics. I never assume that because I've found a "45" there must be a corresponding one on the opposite side of the grid.

I think if there's a feature of cage patterns that I'm not particularly keen about it's disjoint/remote cages. The one in A205 wasn't a problem, since it was completely enclosed in both R5 and N5; for this puzzle it didn't affect "45s" involving multiple columns. More generally disjoint/remote cages do tend to reduce the number of available "45s", often significantly.
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