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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sun Dec 12, 2010 9:41 pm Post subject: |
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| peterj wrote: | In the BUG+1 scenario regardless of whether there is one or more extra candidates you know that you can eliminate the two BUG candidates. If there is only one extra candidate that will of course solve the cell. If there is more than one it creates a pair/triple etc. which might allow further eliminations such as a locked set etc.
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Your evaluation is too complicated. Consider my 4-candidate BUG+1 from above. For S and T to be part of a BUG, they must exist as bilocations in [r5], [c5], and [b5]. So, if you perform r5c6<>ST, then there are going to be three Hidden Singles exposed for each of S and T. I doubt if any BUG+1 grid can survive that.
| Code: | +-----------------------------------------------+
| . . . | . . . | . . . |
| . . . | . SA . | . . . |
| . . . | . . . | . . . |
|---------------+---------------+---------------|
| . . . | TE . . | . . . |
| SB . . | . ST+YZ . | . . TF |
| . . . | SC . . | . . . |
|---------------+---------------+---------------|
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . TD . | . . . |
+-----------------------------------------------+
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| Quote: | In a BUG+n (n>1) scenario, you know that the logic is going to include some sort of Kraken/"pincer" type logic - because you cannot make any immediate eliminations of the BUG candidates (as you don't know which cell or both to use).
So to me the cell-based (as opposed to candidate-based) numbering offers more information to the reader - just! |
I believe the most likely option in this scenario is to treat the additional candidates as streams in a forcing chain looking for a common elimination. In this case, knowing the number of cells is irrelevant to knowing the number of streams/candidates!
Regards, Danny
Bottom Line: I've never see a solution that says, "I have two poly-valued cells and can thus make this deduction". What I have seen is a solution where it's said, "there is a BUG except for these candidates, and they (the candidates) let me make such-'n-such deduction". |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Sun Dec 12, 2010 10:05 pm Post subject: |
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| daj95376 wrote: | | I doubt if any BUG+1 grid can survive that. |
Probably right!
| daj95376 wrote: | | In this case, knowing the number of cells is irrelevant to knowing the number of streams/candidates! |
Also right!
So the most informative and least consistent approach might be... 
* BUG+1 for any BUG with one cell regardless of number of extras since we can eliminate the BUG candidates and almost always solve it.
* BUG+n for more than one cell where n is number of extra candidates as this defines the number of streams/Kraken-like options to consider.
Nah  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun Dec 12, 2010 10:26 pm Post subject: |
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| peterj wrote: | So the most informative and least consistent approach might be...
* BUG+1 for any BUG with one cell regardless of number of extras since we can eliminate the BUG candidates and almost always solve it.
* BUG+n for more than one cell where n is number of extra candidates as this defines the number of streams/Kraken-like options to consider.
Nah |
I can see how you might interpret my comments as such. That's what I get for using BUG+1 for conformity in my first comment ... when it's obvious from my second comment that I don't support conformity in the BUG+n labelling cited by ronk.
Regards, Danny |
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