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Marty R. · Posted: Sun Jul 29, 2007 3:04 pm Post subject: BUG+2 pattern

The grid below contains a BUG+2 pattern, i.e., all bivalue cells except for two trivalue. I have seen before, someone posting that puzzles with this pattern can be solved by the logic that one or the other BUG candidates (both are 6 in this case) must be present. In this case the 6 in r4c3 led to an invalid solution, so the puzzle was solved with a 6 in r9c8.

Was that just a coincidence? Is there any reason that both of the candidates can't be present? If so, then it seems to me that BUG+2 is not a valid solving technique.

(Note: I am aware that the 16 rectangle easily solves the puzzle).

Steve R · Posted: Sun Jul 29, 2007 5:17 pm Post subject:

Marty,

I don’t really understand your concern here. All these bivalue grave patterns work in the same way. For example, the BUG+2 is no different from a UR with two roof cells each containing 6 as the single non-bg candidate. In either case at least one of the two cells concerned must contain 6 if the puzzle is to have a unique solution. The rest of the grid is used to determine whether both cells contain 6 or whether it is just one.

This is exactly how you applied the principle, isn’t it?

Steve

Marty R. · Posted: Sun Jul 29, 2007 7:17 pm Post subject:

Steve,

I don't know if I worded my question well. What I am trying to determine is if the two trivalue cells are an "either/or" situation, in the sense that a bivalue cell must be one or the other, or two 9s in a box must be one or the other. In this particular grid, is there any reason that both of the trivalue cells couldn't be = 6?

What I am trying to determine is what approach to take when I'm left with this BUG+2 pattern.

Steve R · Posted: Sun Jul 29, 2007 7:54 pm Post subject:

I also feel I’m missing the point somewhere.

All the BUG+2 says is that precisely one of the following three statements is true:
(a) r4c3 contains 6;
(b) r9c8 contains 6;
(c) both r4c3 and r9c8 contain 6.

Perhaps you’re looking for a quick fix. If so, I know the feeling. I did try playing with conjugates here but the best I could manage was a nice loop which eliminates 6 from r9c9. This may sound cleverer than your approach but in principle it is no better.

I seem to recall somebody once using a BUG+7 but generally speaking even BUG+3 can be very hard going. It is only when the multivalent cells form a useful pattern that the conclusions are obvious. Otherwise all you can do is fiddle about with locked sets, almost locked sets and conjugates elsewhere until something turns up

Of course the setters of most normal puzzles are programmed to keep things relatively simple.

Steve

Marty R. · Posted: Sun Jul 29, 2007 9:23 pm Post subject: