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Sashimi UR ?

 
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Sat Jul 26, 2008 4:27 pm    Post subject: Sashimi UR ? Reply with quote

In a recent puzzle posted by ravel (view the original thread here) I made a mistake (instead of removing 5 from r4c4 and r4c6, I only removed it from r4c4) and came to this position, which is - if not impossible, then very hard to solve:
Code:

+--------------------------+--------------------------+--------------------------+
| 1       68      2        | 68      3       7        | 4       9       5        |
| 7       3       469      | 459     469     456      | 8       1       2        |
| 5       489     489      | 2       489     1        | 7       3       6        |
+--------------------------+--------------------------+--------------------------+
| 49      15      7        | 149     469     456      | 2       8       3        |
| 8       269     69       | 39      7       23       | 5       4       1        |
| 24      15      3        | 145     248     458      | 6       7       9        |
+--------------------------+--------------------------+--------------------------+
| 6       24      45       | 7       1       9        | 3       25      8        |
| 29      89      589      | 34      246     346      | 1       56      7        |
| 3       7       1        | 68      5       268      | 9       26      4        |
+--------------------------+--------------------------+--------------------------+



play online

Looking at boxes 4 and 5, we might see that there is an "almost UR": If r4c4 had a "5" in it, we could remove 5 from r4c4 and r6c4.

Would it be correct, then, in those cases where there is no rectangle but only 3 cells like this
Code:


. * * |
. . . |
------+
. . . |
. ? * |

to simply add the missing candidate to the fourth cell and see if a UR elimination can be made ... ????

Maybe this is just a ridiculous idea but I'd be very interested to hear your opinions on it, especially why it should not work ...
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wapati



Joined: 10 Jun 2008
Posts: 472
Location: Brampton, Ontario, Canada.

PostPosted: Sat Jul 26, 2008 8:24 pm    Post subject: Reply with quote

AFAIK, if a given does not prevent you from adding the candidate then UR patterns are valid. I do do this, in fact did it today in a puzzle I was solving. A more obvious "proof" is the advice people give that you should look at every possibility before you do UR eliminations as there may be overlapping cells. Putting them back in for the second is the same as "saving up" the eliminations until you have gotten the ones from the other UR(s).
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Sat Jul 26, 2008 8:53 pm    Post subject: Reply with quote

thx, wapati, my thinking exactly.

Still, there is that nagging doubt: in a sense, all candidates but the correct ones are "prevented by a given".

But if there were a real danger, one could never use any uniqueness argument before the puzzle is solved... Experience tells us that it does not matter when we use uniqueness arguments - there has never been a case where a UR elimination found at the very beginning of a solution turned out wrong in the end.

Even when I sometimes fail to remove a candidate from a cell (when it is a given in the same row, for example) that has never led to wrong solutions, only to a slowdown in the solving process.

Which could be formulated as
"nataraj's conjecture": it is always possible (and will not lead to contradictions) to add arbitrary candidates to any cell.

In other words,
"cell a is X or Y" is never wrong, provided we have established "cell a is X".
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Sat Jul 26, 2008 10:17 pm    Post subject: Reply with quote

"Adding a candidate back in" is an acceptable way to go about it if you like. However, it might help to keep in mind that it is the non-unique solution that is the deadly pattern, and not the configuration of candidates that contain such a possible solution.

In this case, if r6c4 were <5>, then we would have <5>s in r4c2 and r6c4 and <1>s in r4c4 and r6c2 (due to the various links present). This result is not unique because the <1>s and <5>s can be xx without any impact on the rest of the grid. So, r6c4 cannot be <5>.
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Sat Jul 26, 2008 10:39 pm    Post subject: Reply with quote

Asellus,
great to have confirmation for the "add-in" conjecture. Thanks! And at the same time you proved it was unnecessary Very Happy

The elimination of 5 in r6c4 is actually what I was after...
I was trying to do it by re-introducing the "missing" 5, but you showed how the elimination can be done without the 5 in r4c4. That's good to know.

So instead of trying to complete the UR with missing candidates in some corners, I'll go try and make the eliminations without those "ghosts"...
Still thinking about this: maybe it is easier for the brain to recognize a pattern with the added candidate than to do a (perfectly valid) elimination in a situation that is so obviously no UR ...

--------

Guess archeologists would have an advantage here: they are supposed to be looking at ruins and to imagine how the building looked when it was still in use Laughing
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