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Ed
 Post subject: Assassin 211
PostPosted: Thu Apr 28, 2011 11:26 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Technically much easier than the last one. Technically. All common assassin level moves but heaps of them. Takes me nearly 40 optimised steps to crack it. Perhaps there's a shortcut or I missed something. If not, great assassin practice.

Assassin 211

NOTE 1-9 cannot repeat on the diagonals
Image
solution: 
+-------+-------+-------+
| 2 6 9 | 7 3 4 | 8 1 5 |
| 8 1 4 | 2 6 5 | 7 3 9 |
| 3 7 5 | 8 9 1 | 6 2 4 |
+-------+-------+-------+
| 1 4 6 | 3 8 2 | 9 5 7 |
| 9 8 2 | 5 7 6 | 3 4 1 |
| 7 5 3 | 1 4 9 | 2 6 8 |
+-------+-------+-------+
| 5 2 8 | 6 1 7 | 4 9 3 |
| 6 9 7 | 4 5 3 | 1 8 2 |
| 4 3 1 | 9 2 8 | 5 7 6 |
+-------+-------+-------+
Code; copy & paste into solver:
3x3:d:k:4889:4889:5122:5122:3843:3843:3843:3844:3844:4889:5914:5122:5914:4614:4614:4614:2823:3844:4889:5914:5914:5914:4616:4616:2823:1545:1545:2570:5387:5387:5387:4616:2823:3596:3596:8205:2570:5387:1806:1806:6415:4880:2833:8205:8205:3090:2579:2579:6415:4880:4880:2833:2833:8205:3090:2579:6415:3604:3604:3604:3093:8205:8205:3350:6415:4375:3352:3352:3352:3352:3093:3845:3350:3350:4375:4375:3841:3841:3841:3845:3845:
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 211
PostPosted: Sat Apr 30, 2011 6:19 am 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Ed wrote:
Technically much easier than the last one. Technically. All common assassin level moves but heaps of them. Takes me nearly 40 optimised steps to crack it. Perhaps there's a shortcut or I missed something. If not, great assassin practice.
Thanks Ed for a nice Assassin. While I enjoyed my early steps for A210, and definitely your breakthrough step for that puzzle, I enjoyed A211 all the way through.

Thanks Ed for pointing out an error in my step 27a. I've now re-worked from there; I hope it's now correct.

My walkthrough now has 34 non-optimised steps; maybe I've grouped some steps together more than Ed has for his walkthrough. As I've commented in my walkthrough, if I'd spotted step 24 earlier it would have been shorter.

Rating Comment:
I'll rate my walkthrough for A211 at Easy 1.5, after the re-work. This is based on the short forcing chain in step 28.

As I've commented in my walkthrough, step 22 is a fairly easy one which many people would just "talk" their way through as 3 or 4 must be in ... and ... and 8 or 9 must be in ... and ... -> 32(6) cage at R4C9 must contain 3,4,8,9; it only looks complicated when written formally.

Here is my walkthrough for A211:
Prelims

a) R3C89 = {15/24}
b) R45C1 = {19/28/37/46}, no 5
c) R4C78 = {59/68}
d) R5C34 = {16/25/34}, no 7,8,9
e) R67C1 = {39/48/57}, no 1,2,6
f) 12(2) cage in N9 = {39/48/57}, no 1,2,6
g) 20(3) cage at R1C3 = {389/479/569/578}, no 1,2
h) 11(3) cage at R2C8 = {128/137/146/26/245}, no 9
i) 19(3) cage in N5 = {289/379/469/478/568}, no 1
j) 11(3) cage in N6 = {128/137/146/236/245}, no 9
k) 10(3) cage at R6C2 = {127/136/145/235}, no 8,9
l) 13(4) cage at R8C4 = {1237/1246/1345}, no 8,9

1. 13(4) cage at R8C4 = {1237/1246/1345}, 1 locked for R8

2. 45 rule on R89 2 innies R8C28 = 17 = {89}, locked for R8, CPE no 8,9 in R2C28 + R5C5 using diagonals
2a. R8C8 = {89} -> R7C7 = {34}

3. 45 rule on R123 2 outies R4C56 = 10 = {28/37/46}/[91], no 5, no 1 in R4C5

4. 45 rule on C1 2 outies R19C2 = 9 = {18/27/36/45}, no 9

5. 45 rule on D/ 2 innies R1C9 + R9C1 = 9 = {18/27/36/45}, no 9

6. 45 rule on N489 1 innie R9C4 = 9
6a. R9C4 = 9 -> R89C3 = 8 = {26/35}/[71], no 4,8, no 7 in R9C3

7. 9 on D/ only in R7C3 + R8C2, locked for N7, clean-up: no 3 in R6C1

8. 45 rule on N8 2 outies R89C7 = 6 = {15/24}

9. 45 rule on N6 2 outies R7C89 = 12 = {39/48/57}

10. 6 in N9 only in 15(3) cage = {168/267} (cannot be {456} which clashes with R89C7), no 3,4,5
10a. 3 in N9 only in R7C789, locked for R7, clean-up: no 9 in R6C1
10b. 9 in N9 only in R7C89 + R8C8, CPE no 9 in R5C8

11. 45 rule on R1234 2 outies R5C12 = 1 innie R4C9 + 10
11a. Min R5C12 = 11, no 1, clean-up: no 9 in R4C1
11b. Max R5C12 = 17 -> max R4C9 = 7

12. 45 rule on N1 3 outies R123C4 = 17 = {278/368/458/467}, no 1

13. 45 rule on N3 2 innies R12C7 = 1 outie R4C6 + 13
13a. Min R12C7 = 14, no 1,2,3,4
13b. Max R12C7 = 17 -> max R4C6 = 4 -> min R4C5 = 6 (step 3)

14. 45 rule on R12 3 innies R2C248 = 1 outie R3C1 + 3
14a. Min R2C248 = 6 -> min R3C1 = 3

15. 32(6) cage at R4C9 contains at least one of 1,2 which must be in N6
15a. 11(3) cage in N6 = {137/146/236/245} (cannot be {128} which clashes with 32(6) cage at R4C9), no 8

16. 14(3) cage in N8 = {158/167/248/257}
16a. 9 in N7 only in R7C3 + R8C2
16b. 45 rule on N7 4 remaining innies R7C123 + R8C2 = 24 = {1689/2589/2679} (cannot be {4569} = {456}9 which clashes with 14(3) cage in N8), no 4, clean-up: no 8 in R6C1
16c. 7 of {2679} must be in R7C1 -> no 7 in R7C23

17. R89C3 (step 6a) = {35}/[71] (cannot be {26} which clashes with R7C123 + R8C2), no 2,6

18. 45 rule on N47 2 innies R7C3 + R8C2 = 2 remaining outies R45C4 + 9
18a. Min R45C4 = 3 -> min R7C3 + R8C2 = 12, no 1,2 in R7C3
18b. Max R7C3 + R8C2 = 17 -> max R45C4 = 8, no 8 in R4C4

19. R7C123 + R8C2 (step 16b) = {1689/2589/2679} -> R7C2 = {12}
19a. 14(3) cage in N8 = {158/167/257} (cannot be {248} which clashes with R7C123 + R8C2, noting that {1689} can only be [816]9), no 4

20. 4 in R7 only in R7C789, locked for N7, clean-up: no 2 in R89C7 (step 8)
20a. Naked pair {15} in R89C7, locked for C7 and N9, clean-up: no 9 in R4C8, no 7 in R7C89 (step 9)
20b. Killer pair 8,9 in R7C89 and R8C8, locked for N9
20c. 8 in N9 only in R7C89 + R8C8, CPE no 8 in R5C8

21. 11(3) cage in N6 (step 15a) = {137/146/236/245}
21a. 1,5 of {137/245} must be in R6C8 -> no 4,7 in R6C8

22. Hidden killer pair 8,9 in R4C78 and R5C8 + R456C9 for N6, R4C78 contains one of 8,9 -> R5C8 + R456C9 must contain one of 8,9
22a. Hidden killer pair 3,4 in 11(3) cage and R5C8 + R456C9 for N6, 11(3) cage contains one of 3,4 -> R5C8 + R456C9 must contain one of 3,4
22b. R7C89 (step 9) = {39/48}
22c. Taking these steps together 32(6) cage at R4C9 must contain 3,4,8,9 = {134789/234689}, no 5
[This step is actually fairly easy, it just looks more complicated when written formally. Ed should be able to “see” this step in his head. ;-) I probably wouldn’t have used this step if I’d seen step 24 earlier.]

23. 5 in N6 only in R46C8, locked for C8, clean-up: no 1 in R3C9

24. 45 rule on R9 3 outies R8C139 = 15 = {267/456} (cannot be {357} = {35}7 which clashes with R89C3, CCC), no 3, 6 locked for R8, clean-up: no 5 in R9C3 (step 6a)
24a. R8C3 = {57} -> no 5,7 in R8C19
24b. 3 in N7 only in R9C123, locked for R9
24c. 7 in N9 only in R9C89, locked for R9, clean-up: no 2 in R1C2 (step 4), no 2 in R1C9 (step 5)
[With hindsight this 45 could have been used a lot earlier; it was only while checking my walkthrough that I realised that after step 6a it could have been used to eliminate 7 from R8C9, because of a CCC between R8C13 and R89C3. However it’s much more powerful now, after the eliminations in step 17.]

25. 15(3) cage at R9C5 = {168/258/456}
25a. R9C7 = {15} -> no 1,5 in R9C56
25b. Killer pair 2,6 in 15(3) cage at R9C5 and 15(3) cage in N9, locked for R9, clean-up: no 3,7 in R1C2 (step 4), no 3,7 in R1C9 (step 5)

26. 4 in N7 only in 13(3) cage = {148/346}, no 2,5, clean-up: no 4 in R1C2 (step 4), no 4 in R1C9 (step 5)

27. R7C2 = 2 (hidden single in N7)
[Ed pointed out that my original step 27a was flawed so I’ve re-worked from here.]
27a. 10(3) cage at R6C2 = {127/235}, no 4,6

[Maybe the next step is a bit heavier than necessary but it gets me back toward my original solving path quickly].
28. R7C2 = 2 -> R7C123 + R8C2 (step 16b) = {2589/2679}
28a. Consider permutations for R7C123 + R8C2
R7C123 + R8C2 = {2589}, locked for N7 => R89C3 = [71] (step 6a) => 10(3) cage at R6C2 = {235} (only combination, cannot be {127} because 1,7 only in R6C2)
or R7C123 + R8C2 = {2679} = [726]9 => R6C1 = 5
28b. -> 5 must be in R6C123, locked for R6 and N4, clean-up: no 2 in R5C4

29. R4C8 = 5 (hidden single in N6), R4C7 = 9, clean-up: no 1 in R4C6 (step 3)

30. R12C7 = R4C6 + 13 (step 13)
30a. Max R12C7 = 15 -> max R4C6 = 2 -> R4C6 = 2, R4C5 = 8 (step 3), clean-up: no 2,8 in R5C1
30b. R4C6 = 2 -> R12C7 = 15 = {78}, locked for C7 and N3, clean-up: no 1 in R9C1 (step 5)

31. R4C6 = 2 -> 11(3) cage at R2C8 = {236} (only remaining combination), 3,6 locked for N3 and D/
31a. R3C89 = {24} (only remaining combination, cannot be {15} which clashes with R1C9), locked for R3 and N3

[Rather than reverting to the last remaining combination for R7C123 + R8C2, this way looks more interesting.]
32. 7 on D/ only in R5C5 + R6C4, locked for N5
32a. 19(3) cage in N5 = {469} (only remaining combination), locked for N5, clean-up: no 1,3 in R5C3

33. R9C1 = 4 (hidden single on D/), R1C9 = 5 (step 5), placed for D/, R8C1 = 6, R9C2 = 3 (step 26), R1C2 = 6 (step 4), R8C9 = 2, R9C3 = 1, R8C3 = 7 (step 6a), R89C7 = [15]

34. 10(3) cage at R6C2 (step 27a) = {235} (only remaining combination), cannot be {127} because 1,7 only in R6C2) -> R6C23 = [53], R6C1 = 7, R7C1 = 5, R6C4 = 1, R5C5 = 7, placed for D\, R9C9 = 6, placed for D\, R4C4 = 3, placed for D\, R7C7 = 4, R8C8 = 8, both placed for D\, R6C6 = 9, placed for D\

and the rest is naked singles without using the diagonals.
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Ed
 Post subject: Re: Assassin 211
PostPosted: Thu May 05, 2011 9:50 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Thanks for your WT Andrew! I could see everything in my head. I took a long time to find your step 18 when I first did the puzzle. A really great way you did that step 22. :) Thanks to Andrew for some extra observations.
Another way to see Andrew's step 24:
Step 24 can also be seen as IOU.

24. "45" on r9: 2 outies r8c19 - 7 = 1 remaining innie r9c3
24a. -> no 7 in r8c9 (IOU)
Then, using Andrew's start to step 24
24b. "45" on r9: 3 outies r8c139 = 15 = {267/456} (cannot be {357} because r8c9 only contains 2,6), no 3, 6 locked for r8
returns to the rest of Andrew's step 24
Alternate way to see Andrew's step 28:
This makes use of the interesting feature of this puzzle that Andrew found in his step 28.
28. 1 cell of the split 8(2)r6c23 sees all of the split 8(2)r89c3 and both have the same cage total -> they must have different combos -> {1357} locked in those 4 cells -> no 1,3,5,7 in r45c3
(Note, that these elims are not essential to the solution for A211 but are still really neat!)
28a. ->Combined split-cage 16(4)r6c23+r89c3 = {1357} = [7->5..](no eliminations yet)
28b. 7 in n7 in r8c3 in Combined split-cage 16(4)(-> 5 in r6c23 step 28a) or in r7c1 in 12(2)r6c1 (->5 in r6c1)-> 5 locked in r6c123 (Locking cages)
My original solution from Andrew's step 27: 4 steps:
End of Andrew's step 27 here: select marks and use "Paste Into" A211 in SudokuSolver.
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 123456789 1568      3456789   | 345678    123456789 123456789 | 6789      12346789  1568      |
| 123456789 134567    3456789   | 2345678   123456789 123456789 | 6789      123467    123456789 |
| 3456789   13456789  123456789 | 2345678   123456789 123456789 | 234678    124       245       |
:-------------------------------+-------------------------------+-------------------------------:
| 1234678   13456789  123456789 | 1234567   6789      1234      | 689       568       123467    |
| 2346789   3456789   123456    | 123456    1234567   23456789  | 23467     123467    12346789  |
| 457       1357      1357      | 12345678  23456789  23456789  | 23467     12356     12346789  |
:-------------------------------+-------------------------------+-------------------------------:
| 578       2         5689      | 15678     15678     15678     | 34        3489      3489      |
| 46        89        57        | 123457    123457    123457    | 15        89        26        |
| 1348      1348      13        | 9         2468      2468      | 15        267       267       |
'-------------------------------.-------------------------------.-------------------------------'


28. Hidden killer 2,7 in D/ -> 11(3) at r2c8 must have 2/7 (can't have both) and 25(4)r5c5 must have 2/7 (incredibly, it can't have both - couldn't see that in my head but the combo chart says it's true!
Andrew points out: "is incredible...but it's not really that hard to see in one's head. 25(4) = 1+24, 2+23, 3+22 or 4+21 (anything higher repeats numbers) so 2+23 = 2+{689}". )
28a. 11(3) = {128/137/236}(no 4)
28b. no 6 in r4c5 (h10(2)r4c56)

29. "45" on n3: 4 innies r123c7+r2c8 = 24 = {1689/2679/3678}
29a. must have 6 -> 6 locked for n3
29b. no 3 in r9c1 (h9(2)r1c9+r9c1)

30. 13(3)n7 = {148/346} = [4]{18}/[6][43]
30a. 4 locked in r89c1 for n7 & c1
30b. no 8 in r7c1

31. Killer pair 5,7 in r6c123 since r6c23 = {17/35}: both locked for n4 and r6
Back to Andrew's step 29
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