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Ed
 Post subject: Assassin 273
PostPosted: Thu Oct 24, 2013 10:09 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
This started out with an even higher SudokuSolver score than the current 1.70. Hopefully now it has a couple of ways to solve it. I used one advanced step. JSudoku thinks its dead easy, very benign, which is why I tried it and indeed, it did come out without too much trouble.

Assassin 273

Image
code, paste into solver:
3x3::k:3584:3584:3584:5908:5908:5908:6146:1027:1027:3076:5381:5381:5908:2837:6146:6146:4103:4103:3076:5381:5638:5638:2837:6146:6146:4103:7944:5381:5381:5638:5903:5903:5903:7177:1546:7944:4107:4107:5638:5638:5903:2316:7177:1546:7944:4107:5901:5134:5134:5134:2316:7177:7944:7944:5901:5901:4886:4886:7177:7177:7177:1552:1552:3089:5901:4886:5394:5394:6675:6675:6675:2817:3089:3089:1815:1815:5394:5394:5394:6675:2817:
solution: 
+-------+-------+-------+
| 5 2 7 | 4 6 8 | 9 3 1 |
| 4 6 1 | 5 9 3 | 7 8 2 |
| 8 3 9 | 7 2 1 | 4 6 5 |
+-------+-------+-------+
| 7 4 2 | 6 8 5 | 3 1 9 |
| 6 9 3 | 1 4 2 | 8 5 7 |
| 1 5 8 | 9 3 7 | 2 4 6 |
+-------+-------+-------+
| 3 7 6 | 8 5 9 | 1 2 4 |
| 9 8 5 | 2 1 4 | 6 7 3 |
| 2 1 4 | 3 7 6 | 5 9 8 |
+-------+-------+-------+
Cheerio
Ed
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wellbeback
 Post subject: Re: Assassin 273
PostPosted: Sat Oct 26, 2013 10:07 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
This is how I did this one. One little contra-chain opened it up for me. I'm sure there's a way around that :)
Hidden Text:
1. 4/2@r1c8 = {13}
Innies c89 -> r89c8 = {79}
-> r8c67 = {28} or {46}

2. At least one of (68) must go in c8 in r23c8
-> 16/3@r2c8 = {268}

3. 14/3@r1c1 = {248} or {257} - No 6.
-> 6 in r1 in n2 in r1c456.

4. 24/5@r1c7 contains 3 of (4579) (in n3) and no 6.
-> can only be (In n2-n3) {12-579} or {13-479}
-> r3c9 from (45)
Also r23c6 = {12} or {13}
Also (Innies n2 = +11) r3c4 from (87)

5. Putting 6/2@r4c8 = {24}...
puts 6/2@r7c8 = {15} and 11/2@r8c9 = {38} and r456c7 = {138}
which leaves no solution for 28/6@r4c7

-> 6/2@r4c8 = {15}
-> 6/2@r7c8 = {24}
-> r8c67 = [28] or [46]

6. r8c6,r789c7 is either 2,{138} or 4,{156}
Both options prevent 24/5@r1c7 from being {12-579}

-> 24/5@r1c7 = {13-479} with the {13} in r23c6 and {479} in r123c7
-> r3c49 = [75]
-> r89c9 = {38}
-> r8c67 = [46]
-> r79c7 = {15}
-> r456c7 = {238}
Also 9/2@r5c6 = {27}

7. 11/2@r2c5 can only be {29}
-> 23/4@r1c4 = {4568}
-> 14/2@r1c1 = {257}
-> r2c4 = 5 and r1c456 = {468}
-> r123c7 = [974]

8. Innies - Outies n5 -> r5c4 is max 2.
Since 2 already in n5 -> r5c4 = 1
-> r6c3 = 8
-> 20/3@r6c3 = [893]
-> r6c7 = 2
-> 9/2@r5c6 = [27]
-> r45c9 = {79} and r6c89 = {46}
Also 6/2@r4c8 = [15]

9. Outies n6789 -> r6c2 + r3c9 = +10
-> r6c2 = 5
-> r6c1 = 1
-> r5c12 = {69}
-> r45c9 = [97]

10. Outies n7 = +16. Since r6c2 = 5 -> r79c4 = +11. Can only be [83].
-> r9c3 = 4
-> HP n4 -> r4c12 = {47}
-> r45c3 = [23]
-> r3c3 = 9
-> 11/2@r2c5 = [92]
-> r78c3 = [65]
-> r12c3 = [71]
-> r23c6 = [31]
-> r23c2 = [63]
-> r3c8 = 6
-> r6c89 = [46]
-> r7c89 = [24]
-> r2c89 = [82]
-> r23c1 = [48]
etc.


CPFC Now 1-0-8 :(
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HATMAN
 Post subject: Re: Assassin 273
PostPosted: Sun Oct 27, 2013 8:46 am 
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
At least you are in the top flight - I'm Leeds and it has been a long 10+ years.
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Andrew
 Post subject: Re: Assassin 273
PostPosted: Mon Oct 28, 2013 2:45 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
And I still support Swindon from a distance; they only made the top flight once but did also win the League Cup once, beating Arsenal in the final.
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Andrew
 Post subject: Re: Assassin 273
PostPosted: Mon Oct 28, 2013 10:55 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Getting back on topic. ;)

Thanks Ed for another interesting Assassin. I particularly enjoyed my early steps.

Here is my walkthrough for Assassin 273:
Prelims

a) R1C89 = {13}
b) R23C1 = {39/48/57}, no 1,2,6
c) R23C5 = {29/38/47/56}, no 1
d) R45C8 = {15/24}
e) R56C6 = {18/27/36/45}, no 9
f) R7C89 = {15/24}
g) R89C9 = {29/38/47/56}, no 1
h) R9C34 = {16/25/34}, no 7,8,9
i) 20(3) cage at R6C3 = {389/479/569/578}, no 1,2
j) 19(3) cage at R7C3 = {289/379/469/478/568}, no 1
k) 26(4) cage at R8C6 = {2789/3689/4589/4679/5678}, no 1

1. Naked pair {13} in R1C89, locked for R1 and N3
1a. 14(3) cage at R1C1 = {248/257}, no 6,9, 2 locked for R1 and N1

2. 6(2) pairs at R4C8 and R7C8 must have different combinations (locking cages) forming naked quad {1245}, CPE no 2,4,5 in R89C8
2a. Caged X-Wing for 1 in R1C89 and 6(2) pairs at R4C8 and R7C8, no other 1 in C89

3. 45 rule on N5 1 outie R6C3 = 1 innie R5C4 + 7 -> R5C4 = {12}, R6C3 = {89}

4. 45 rule on R1 1 innie R1C7 = 1 outie R2C4 + 4, no 4 in R1C7, no 6,7,8,9 in R2C4

5. 45 rule on R89 2 innies R8C23 = 13 = {49/58/67}, no 1,2,3

6. 45 rule on C89 2 innies R89C8 = 16 = {79}, locked for C8, N9 and 26(4) cage at R8C6, no 7,9 in R8C6, clean-up: no 2,4 in R89C9
6a. 45 rule on C89 2 outies R8C67 = 10 = {28/46}, no 3,5

7. 16(3) cage at R2C8 = {259/268/457}
7a. Hidden killer pair 6,8 in 16(3) cage at R6C8 for C8, 16(3) cage contains both or neither of 6,8, R6C8 cannot contain both of 6,8 -> 16(3) cage must contain both of 6,8 = {268}, locked for N1, clean-up: no 2,4 in R2C4 (step 4)
7b. 6 in R1 only in R1C456, locked for N2, clean-up: no 5 in R23C5

8. 6(2) pairs at R4C8 and R7C8 form naked quad {1245}
8a. Caged X-Wing for 2 in 16(3) cage at R2C8 and 6(2) pairs at R4C8 and R7C8, no other 2 in C89

9. 45 rule on C12 2 outies R12C3 = 8 = [26/53/71]

10. 24(5) cage at R1C7 = {12579/13479} (cannot be {12489/13578/23478} because there need to be three of 4,5,7,9 in R123C7), no 8
10a. 1,2,3 only in R23C6 = {12/13}, 1 locked for C6 and N2, clean-up: no 5 in R1C7 (step 4), no 8 in R56C6
10b. 24(5) cage = {12579/13479}, 7,9 locked for C7 and N3

11. 45 rule on N6789 2(1+1) outies R3C9 + R6C2 = 10 = [46/55]
11a. 45 rule on N235 2 innies R3C49 = 12 = [75/84]

12. 45 rule on N1457 3 outies R379C4 = 18
12a. Max R39C4 = 14 -> min R7C4 = 4

13. 6 in R1 only in 23(4) cage at R1C4 = {3569/4568}, no 7
13a. R23C5 = {29/47} (cannot be {38} which clashes with 23(4) cage), no 3,8 in R23C5
13b. R3C49 (step 11a) = [75/84] -> R23C5 = [29/47/92], no 7 in R2C5, no 4 in R3C5
13c. 23(4) cage = {3569/4568} -> R23C4 = [38/57]

14. R379C4 = 18 = {189/279/378/468} (cannot be {369/459} because R3C4 only contains 7,8, cannot be {567} which clashes with R23C4 = [57]), no 5, clean-up: no 2 in R9C3
14a. 9 of {279} must be in R7C4, 7 of {378} must be in R3C4 (cannot be [873] which clashes with R23C4 = [38]), no 7 in R7C4

15. 20(3) cage at R6C3 = {389/479/578} (cannot be {569} which clashes with R6C2), no 6 in R6C45
15a. Combined cage 20(3) + R56C6 = {27/36/45} = {389}{27/45}/{479}{36}/{578}{36}, 3 locked for N5

16. R7C89 = {15/24}, R89C9 = {38/56} -> combined cage R7C89 + R89C9 = {15}{38}/{24}{38}/{24}{56}
16a. 45 rule on N9 2 innies R79C7 = 1 outie R8C6 + 2
16b. R8C6 = {2468} -> R79C7 = 4,6,8,10 but cannot be {28} which clashes with combined cage, no 8 in R79C7

17. R8C23 (step 5) = {49/58/67}, R8C67 (step 6a) = {28/46} -> combined cage R8C2367 = {49}{28}/{58}{46}/{67}{28}, 8 locked for R8, clean-up: no 3 in R9C9

18. 3 in N5 only in 20(3) cage at R6C3 = {389}
or in R56C6 = {36}, locked for C6 => R23C6 = {12}, locked for N2 => R23C5 = [47]
-> no 4,7 in R6C5
18a. 20(3) cage = {389/578} (cannot be {479} because 4,7 only in R6C4), no 4, 8 locked for R6
18b. 7 of {578} must be in R6C4 -> no 5 in R6C4
[I can see a longer forcing chain to eliminated {578}, but I’ll leave that for now.]

19. 7,9 in C9 only in 31(5) cage at R3C9 = {34789/45679} (cannot be {35689} which doesn’t contain 7)
19a. 7,8,9 of {34789} must be in R456C9 -> no 3 in R456C9

20. R3C9 + R6C2 (step 11) = [46/55], 31(5) cage at R3C9 (step 19) = {34789/45679}
20a. 3 in N5 only in 20(3) cage at R6C3 = {389}, locked for R6 => 31(5) cage = {45679}
or in R56C6 = {36}, R3C9 + R6C2 = [46] => R56C6 = [63] => no 3 in R6C8 (while R3C9 + R6C2 = [55] eliminates {34789} which doesn’t contain 5)
-> 31(5) cage = {45679}, no 3,8, 6 locked for N6
20b. Killer pair 4,5 in 31(5) cage and R45C8, locked for N6

21. 3,8 in N6 only in R45C7, locked for C7 and 28(6) cage at R4C7, no 3,8 in R7C56, clean-up: no 2 in R8C6 (step 6a)
21a. R9C9 = 8 (hidden single in N9), R8C9 = 3, R1C89 = [31], clean-up: no 5 in R7C8

22. 3,8 in N6 only in 28(6) cage at R4C7 = {123589} (cannot be {134578/234568} which clash with R7C89), no 4,6,7
22a. R7C567 must be {159} (cannot be {259} which clashes with R7C89), locked for R7, 9 also locked for N8 -> R456C7 must be {238}, locked for C7 and N6, clean-up: no 4 in R45C8, no 8 in R8C6 (step 6a)

23. Naked pair {24} in R7C89, locked for R7 and N9 -> R8C67 = [46], clean-up: no 5 in R56C6, no 3 in R9C3
23a. Naked pair {15} in R79C7, locked for C7
23b. R3C9 = 5 (hidden single in N3), R3C4 = 7 (step 11a), R6C2 = 5 (step 11), clean-up: no 5,7 in R2C1, no 4 in R2C5

24. Naked triple {389} in 20(3) cage at R6C3, locked for R6, 3 also locked for N5 -> R6C7 = 2, clean-up: no 6,7 in R5C6, no 6 in R6C6
24a. R56C6 = [27], R5C4 = 1, R6C3 = 8 (step 3), R45C8 = [15]
24b. Naked pair {39} in R6C45, locked for N5

25. R6C1 = 1 (hidden single in R6), R5C12 = 15 = {69}, locked for R5 and N4

26. Naked pair {13} in R23C6, locked for C6 and N2 -> R2C4 = 5, R1C7 = 9 (step 4)
26a. 14(3) cage at R1C1 = {257} (hidden triple in R1), locked for N1

27. R379C4 = 18 (step 14), R3C4 = 7 -> R79C4 = 11 = [83], R9C3 = 4

28. R45C3 = [23], R3C3 = 9 (cage sum), clean-up: no 5 in R1C3, no 6 in R2C3 (both step 9), no 3 in R23C1
28a. Naked pair {48} in R23C1, locked for C1 and N1 -> R4C1 = 7

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A273 at 1.5; I used two forcing chains, with the second one being harder than as Easy 1.5.
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Ed
 Post subject: Re: Assassin 273
PostPosted: Sun Nov 03, 2013 1:17 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Once again, another puzzle with 3 very different ways to crack it. Unusual for wellbeback to use combinations to crack a puzzle (start of his step 5). Loved Andrew's start!! I used a different way (less fun) to make those eliminations. Once again have used wellbeback's start before we went different ways ...

A273 alt crack
From wellbeback's end of step 4:
Paste candidates into A273 in SudokuSolver or download attached file and open with SudokuSolver: 
.-------------------------------.-------------------------------.-------------------------------.
| 24578     24578     24578     | 456789    456789    456789    | 4579      13        13        |
| 345789    13456789  13456789  | 2345789   2345789   123       | 4579      268       268       |
| 345789    13456789  13456789  | 78        2345789   123       | 4579      268       45        |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 123456789 | 123456789 123456789 23456789  | 123456789 1245      123456789 |
| 123456789 123456789 123456789 | 123456789 123456789 234567    | 123456789 1245      123456789 |
| 123456789 123456789 3456789   | 3456789   3456789   234567    | 123456789 1234568   123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 23456789  | 23456789  123456789 23456789  | 1234568   1245      1245      |
| 123456789 123456789 23456789  | 123456789 123456789 2468      | 2468      79        234568    |
| 123456789 123456789 123456    | 123456    123456789 23456789  | 1234568   79        234568    |
'-------------------------------.-------------------------------.-------------------------------'
End wellbeback step 4 in hide above.

Loved the way Andrew did this next step, see his steps 2 & 8
5. 11(2)n9 = {38/56}(no 2,4) and 6(2)n9 = one of 1 or 2
5a. -> r789c7 must have one of 1 or 2 for n9 (Hidden killer pair: no eliminations yet)
5b. -> r456c7 must have one of 1 or 2 for c7 & 6(2)n6 = 1 or 2 -> 1 and 2 locked for n6 (Hidden killer pair)

6. 7 in c9 only in r456c9: 7 locked for n6
6a. 31(5)r3c9 must have 7 for c9 = {34789/45679} = [note: cannot have both 5 & 8]

The cracker
7. 8 in n6 in c7
7a. or 8 in 31(5) -> 4 in r3c9 (step 6a) -> r23c6 = 3 = {12} (IODn3=+1)
7b. ie, has 8 in r456c7 or 2 in r23c6 -> [28] blocked from h10(2)r8c67 -> = [82]/{46}

8. "45" on n9: 3 innies r789c7 = 12 (remembering 16(2)r89c8)
8a. but {138} blocked since no 1,3,8 in r8c7
8b. {345} blocked by 6(2)n9 = {15/23} = [3/5..]
8c. h12(3) = {156/246}(no 3,8}
8d. must have 6: 6 locked for n9 and c7
8e. 11(2)n9 = {38} only: both locked for c9
8f. r1c89 = [31]
8g. no 5 in r7c8

9. 5 in c8 only in n6: 5 locked for n6

10. 31(5)r3c9 = {45679} only (no 8)

11. 3 & 8 in n6 only in r456c7 in 28(6) = {123589/134578/234568}: 3 & 8 locked for that cage
11a. no 3,8 in r7c56
11b. 28(6) must have 5 which is only in r7: 5 locked for r7
11c. 6(2)n9 = {24} only: both locked for r7 and n9
11d. r8c67 = [46] (h10(2))


continue from there.
Cheerio
Ed


Attachments:
A273 wellbeback WT end step 4 Alt cracker.ssv [91.45 KiB]
Downloaded 758 times
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