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[Marty R.]

I'm unable to re-create the grid, but I hope this is enough to make my point. Below are columns 4-5-6 of row 9 in a puzzle I had. I believe the other numbers in the cells are not relevant to my question.

2+ 2 2-

I did coloring on the 2s; it was fully legitimate, with all the rows/columns/boxes in the chain containing just two 2s, except where it ended up in this row with three 2s. So my question is, is it legitimate to eliminate the center "2" based on the + and - on either side of it?

In a way, I want to say "yes" because the chain was valid, but in another way, I want to say "no" because if I change the left "2" to a -, then the right "2" is not forced to be +.

Of course, my real interest is not just this particular example, but the broader principle of the legitimacy (or lack thereof) of the candidate to be eliminated, and the + and - all residing in the same group.

[Steve R]

I am afraid I do not quite understand your question.

My first thought was that you had identified a chain of cells with the property that either:
(a) every cell marked + contains a 2 and no cell marked – contains a 2; or
(b) every cell marked - contains a 2 and no cell marked + contains a 2.

If this had been the case, one or other of the cells in row 9 which are marked + and - must contain 2 so 2 can be excluded from any other cell in the row.

However, if the polarity of the left-hand left cell in row 9 is changed from + to – in a chain of this type, the right hand cell must change polarity too. It is simply a matter of going back round the chain changing + to – and – to +.

With luck someone else on the forum will see more clearly and provide a more helpful response.

Steve

[TKiel]

If I'm following the discussion correctly, that result is a fairly common colouring exclusion and it is certainly valid. Changing the polarity on the leftmost cell does not allow one to directly change it in the rightmost (since they are not conjugate), but if the chain that connects the two cells is a valid conjugate chain, changing the polarity of one element must change the polarity of all and the exclusion is still valid.

[Marty R.]

Thanks. I think we're saying the exclusion of the center "2" is valid. The original coloring chain was valid; if the leftmost "2" were changed from a + to a -, then every other "2" in the original chain would change to the opposite sign. But, of course, what had me wondering was that if the leftmost "2" had been changed, it would only force the rightmost "2" if I backtracked on the original chain, not if I tried a new chain.

I hope this is making some sense.

[keith]

Marty,

From your original post:

[Marty R.]

Yes Keith, the chain was a loop that did not involve that row or box except as a starting and ending point. My confusion was not knowing if the oppositeness principle applied only to the original chain or also applied to any other like numbers in the group that were outside the chain.