Bristol variations

Advanced methods and approaches for solving Sudoku puzzles

Bristol variations

Postby ravel » Thu Mar 09, 2006 8:26 pm

When i tried to find alternatives to the shame of the giant Bristol sudoku, it turned out that there are 23 unique sudokus, that you can get by changing 4 digits (which keeps the nice pattern). According to susser this is the hardest one (a swordfish, 2 6-link and a 7-link forcing chain). Are there more elegant solutions ?
Code: Select all
------------
4.6|.2.|9.8
..8|...|5..
...|...|...
------------
6..|285|..9
...|9.3|...
8..|761|..4
------------
...|...|...
..5|...|3..
2.1|.9.|6.7
------------
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Postby Havard » Thu Mar 09, 2006 9:19 pm

well... there is a nice ALS-xz in R2345C2 and R357C3 that solves the puzzle by elimination a lot of 4's...:)

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Postby Wolfgang » Fri Mar 10, 2006 9:55 am

Havard, it seems, that i am still ALS-blind. Is it here, where you are ?
Code: Select all
4      35     6   |  35     2    7    |  9    1     8         
1379   1279   8   |  146    13   469  |  5    2467  236       
13579  1279   279 |  14568  135  4689 |  47   2467  236       
-------------------------------------------------------
6      147    347 |  2      8    5    |  17   37    9         
157    127    27  |  9      4    3    |  178  5678  156       
8      59     39  |  7      6    1    |  2    35    4         
-------------------------------------------------------
379    46789  479 |  14568  1357 2468 |  148  24589 125       
79     46789  5   |  1468   17   2468 |  3    2489  12         
2      348    1   |  3458   9    48   |  6    458   7   
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Postby Carcul » Fri Mar 10, 2006 11:27 am

Hi Wolfgang.

You have a naked triple in box 2 that eliminates the "1s" from r23c4 and the "1s" from r78c5. Then, r8c5=7 and r8c1=9.

Regards, Carcul
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Postby Wolfgang » Fri Mar 10, 2006 11:45 am

Thanks Carcul, does it solve the puzzle for you ?
Bob Hansons solver showed me the marked cells, which allow the elimination of 17 in r4c2 (what solves the puzzle). But i dont know how to argue it as an ALS.
Code: Select all
4      35     6   |  35     2    7    |  9    1     8         
1379   1279*  8   |  146    13   469  |  5    2467  236       
13579  1279*  279 |  14568  135  4689 |  47   2467  236       
-------------------------------------------------------
6      -147   347 |  2      8    5    |  17   37    9         
157^   127*   27^ |  9      4    3    |  178  5678  156       
8      59*    39  |  7      6    1    |  2    35    4         
-------------------------------------------------------
379    46789  479 |  14568  1357 2468 |  148  24589 125       
79     46789  5   |  1468   17   2468 |  3    2489  12         
2      348    1   |  3458   9    48   |  6    458   7   

[Edit:]
Is this a valid ALS reasoning ?
A and B have a common cell (r5c2):
A=1257 in r5c123
B=12579 in r2356
1 and 7 in r4c2 both lock 5 in A (r5c1) and in B (r5c2)
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Postby tarek » Fri Mar 10, 2006 12:41 pm

hmmmm,

if you combine the extra cells of r4c7 & r4c8........

Then it can be achieved using the ALS-xy rule........

Set 1 (59) in r6c2
Set 2 (1279) in r5c2,r2c2,r3c2
Set 3 (1357) in r6c8,r4c8,r4c7

Restricted candidates would be 9 & 5
common non restricted candidates would be 1 & 7 achieving that
elimination.......


On the other hand, as there are many simpler moves that don't achive solution, I was wondering if that solver actually looks for "The Best next move", skipping those non productive essential simpler moves

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Postby Carcul » Fri Mar 10, 2006 1:04 pm

Wolfgang wrote:Thanks Carcul, does it solve the puzzle for you ?


No, but these moves then allow us to apply the ALS xz rule pointed out by Havard:

A=r2345c2, {1,2,4,7,9}
B=r357c3, {2,4,7,9}

x=9, z=4 => r789c2<>4 which solve the puzzle.

Regards, Carcul
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Postby ronk » Fri Mar 10, 2006 1:04 pm

Wolfgang wrote:Bob Hansons solver showed me the marked cells, which allow the elimination of 17 in r4c2 (what solves the puzzle). But i dont know how to argue it as an ALS
..............
A and B have a common cell (r5c2):
A=1257 in r5c123
B=12579 in r2356 (sic)

Sets A and B cannot have a common cell, and Bob Hanson's posts (and I suspect solver too) rarely has an error ... so I think something got lost in transcription. But here's an illustration for the ALS Havard identified above.
Code: Select all
 4     35    6   | 35   2   7    | 9   1    8
 137  A1279  8   | 46   13  469  | 5   2467 236
 1357 A1279 B279 | 468  135 4689 | 47  2467 236
-----------------+---------------+--------------
 6    A147  -347 | 2    8   5    | 17  37   9
 157  A127  B27  | 9    4   3    | 178 678  156
 8     59    39  | 7    6   1    | 2   35   4
-----------------+---------------+--------------
 37   -4678 B47  | 1468 35  2468 | 148 9    125
 9    -468   5   | 1468 7   2468 | 3   248  12
 2    -348   1   | 35   9   48   | 6   458  7

Set A = {r2c45c2} = {12479}
Set B = {r357c3} = {2479}
    x = 9
    z = 4
    eliminating all 4s that can "see" both 4s in r4c3 and r7c4

Ron
Last edited by ronk on Fri Mar 10, 2006 9:34 am, edited 1 time in total.
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Postby Wolfgang » Fri Mar 10, 2006 1:08 pm

tarek wrote:On the other hand, as there are many simpler moves that don't achive solution, I was wondering if that solver actually looks for "The Best next move", skipping those non productive essential simpler moves

No, it did not, it was me, who skipped the other moves. Also it came 2 times with the same sets for eliminating 1 and 7 and had an extra move between them.
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Postby tarek » Fri Mar 10, 2006 2:15 pm

Wolfgang wrote:No, it did not, it was me, who skipped the other moves. Also it came 2 times with the same sets for eliminating 1 and 7 and had an extra move between them.


so we are left with this to explain the kill (Which is fantastic IMO).....
Code: Select all
4      35     6   |  35     2    7    |  9    1     8         
1379   1279^  8   |  146    13   469  |  5    2467  236       
13579  1279^  279 |  14568  135  4689 |  47   2467  236       
-------------------------------------------------------
6      -147   347 |  2      8    5    |  17%  37%   9         
157    127^   27  |  9      4    3    |  178  5678  156       
8      59*    39  |  7      6    1    |  2    35%   4         
-------------------------------------------------------
379    46789  479 |  14568  1357 2468 |  148  24589 125       
79     46789  5   |  1468   17   2468 |  3    2489  12         
2      348    1   |  3458   9    48   |  6    458   7   

Eliminating 17 from r4c2(ALS-XY  A=59 in r6c2(*)   B=1279 in r5c2,r2c2,r3c2(^)   C=1357 in r6c8,r4c8,r4c7(%)   x=9 y=5 z=1&7)


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Postby Wolfgang » Fri Mar 10, 2006 2:47 pm

Thanks, Carcul and Ron, for pointing me to see Havards ALS.
ronk wrote:Sets A and B cannot have a common cell...

But i still think, my reasoning above is correct, isnt it ? So is it no ALS because of the common cell, but the argumentation is the same ?
PS: As i see it, it must only hold that x is not in the common cell.
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Postby ronk » Fri Mar 10, 2006 3:30 pm

Wolfgang wrote:But i still think, my reasoning above is correct, isnt it ? So is it no ALS because of the common cell, but the argumentation is the same ?
............
A and B have a common cell (r5c2):
A=1257 in r5c123
B=12579 in r2356
1 and 7 in r4c2 both lock 5 in A (r5c1) and in B (r5c2)

Well, maybe set overlap (cell sharing) is actually valid for the ALS xz-rule ... under certain circumstances.

For your defined sets, we know either set A or set B cannot contain the 5. When A doesn't contain a 5, A is locked to 127. When B doesn't contain a 5, B is locked to 1279. So any 1s (or alternately, the 2s and 7s) candidates that "see" all 1s (or 2s and 7s) of both sets may be excluded. That sure sounds like the ALS xz-rule to me, where x=5 and z is alternately 1, 2, and 7. And it would hold even if the 5s were weakly linked.

Maybe the sets can share cells as long as the shared cell doesn't contain 'x'. If true, that would certainly open up additional exclusions from the ALS xz-rule.

In any case, your reasoning for this particular case is definitely valid IMO, despite the common cell.

Ron
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Postby Wolfgang » Fri Mar 10, 2006 3:39 pm

Wolfgang wrote:PS: As i see it, it must only hold that x is not in the common cell.

ronk wrote:Maybe the sets can share cells as long as the shared cell doesn't contain 'x'.

We had the same idea:)
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Postby aeb » Fri Mar 10, 2006 4:05 pm

Wolfgang wrote:
Code: Select all
4      35     6   |  35     2    7    |  9    1     8         
1379   1279*  8   |  146    13   469  |  5    2467  236       
13579  1279*  279 |  14568  135  4689 |  47   2467  236       
-------------------------------------------------------
6      -147   347 |  2      8    5    |  17   37    9         
157^   127*   27^ |  9      4    3    |  178  5678  156       
8      59*    39  |  7      6    1    |  2    35    4         
-------------------------------------------------------
379    46789  479 |  14568  1357 2468 |  148  24589 125       
79     46789  5   |  1468   17   2468 |  3    2489  12         
2      348    1   |  3458   9    48   |  6    458   7   

Is this a valid ALS reasoning ?
A and B have a common cell (r5c2):
A=1257 in r5c123
B=12579 in r2356
1 and 7 in r4c2 both lock 5 in A (r5c1) and in B (r5c2)

The conclusion is certainly correct. Beautiful! This case is more subtle than what we have seen earlier.
Do the standard rewrite where no A and B are distinguished. The set is r2356c2,r5c13 of size 6. Digits 12579 with max multiplicities 22121 for a total of 8, so each choice that eliminates three digits can be removed. Now each of the three possibilities for (5,2) already eliminates one digit, and a choice (4,2)1 or (4,2)7 eliminates two further digits, impossible.
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Postby ronk » Fri Mar 10, 2006 4:08 pm

Wolfgang wrote:PS: As i see it, it must only hold that x is not in the common cell.
Sorry, I stopped reading too soon, Ron
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