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Sorry, I still haven't got the hang of Wiki editing. The edit I made on 7th March 07 was to correct 'r8c5=2' to 'r5c3=2' in the BUG+1 example. - PaulS

Thanks for the corrections, I was very careless in typing in the coordinates. See Help:Editing if you need some help in Wiki editing. --unkx80 09:05, 8 March 2007 (CET)

I'm concerned about this statement: "In a BUG, each cell, row, column and box constraint has 2 candidates left."

This isn't true - in the BUG+1 example, row 7 has three candidates (1, 3 and 6). In fact each row, column and box will have as many different candidate values as it has unsolved cells.

As I understand it, the requirement for a BUG is that:
a. Each unsolved cell has two candidates.
b. In each row, column and box, each candidate value occurs exactly twice.

There seems to be ambiguity between candidate values and occurrences here. I'd propose rewriting this definition as:

"In a BUG, each unsolved cell has two candidates, and in each row, column and box constraint, each candidate value occurs exactly twice." PaulS 09:25, 8 March 2007 (CET)

I think your definition is a lot clearer, so I have replaced the original sentence by Ruud. Thanks. --unkx80 14:56, 8 March 2007 (CET)

Hi again. I see that this topic is marked 'under debate', and I do have a question about it.

A friend has given me a definition of BUG+1 which effectively contradicts this page's statement that 'The term BUG+k is used to denote a BUG grid with k extra candidates.' He says that BUG+k means 'BUG plus k extra [polyvalue] squares', that the polyvalue square in a BUG+1 can contain >3 candidates (so >1 extra candidates), and that the general rule is to eliminate from this square candidates which occur exactly twice in the square's parent row, column and box (which, of course, results in an exposed single when the square is trivalue).

I understand the logic of BUG+1 as described here - I see it as 'if the non-BUG candidate was removed that would produce a BUG, so removing it is impossible in a single-solution puzzle'. However I can't quite see the logical proof of the assertion that all twice-occurring candidates can be removed, even when the square has >3 candidates. It obviously can't be that none of the (multiple) 3-occurrence candidates can be eliminated, so what is it?

So is this alternative description valid, and if so, should this page be changed to reflect it (i.e. to describe the possibility of finding BUG+1 situations where the polyvalue square has 4 or more candidates)? PaulS 11:44, 9 March 2007 (CET)

The reason why this was marked "debate" mainly because we thought that people might not agree to use techniques based on uniqueness. Otherwise, there is no true debate about this technique right now.

As for the description on BUG+k... Ruud, can you help on this? --unkx80 14:16, 9 March 2007 (CET)