Date: Tue, 28 Dec 2004 13:40:07 -0800 From: Ted Stern <[EMAIL PROTECTED]> Subject: [EM] Re: Sprucing up MMPO and other methods
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I have a question about the first stage, eliminating covered candidates:
On 21 Dec 2004 at 16:09 PST, Forest Simmons wrote:1. Eliminate covered candidates until each remaining candidate has a short (length one or two) beat path to each of the other remaining candidates.
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In step one form a matrix M whose (i,j) element is one if candidate i beats candidate j pairwise (as well as in the case if i=j), but is zero otherwise. Then form the matrix A=M^2.
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This implies that M equals
0 1 1 0 0 0 1 0 0 0 0 1 1 1 0 0
Actually M has 1's down the main diagonal because of the i=j proviso in my definition of M.
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I think it is possible to show that every 4-candidate cycle can be reduced to 3 candidates this way. Can you verify this?
Yes, as I mentioned before, if you put arrows on the edge of a tetrahedron so that no vertex is a source or a sink, then all such arrangements of arrows are isomorphic. So once you've shown that one of these reduces to a three cycle, then you've shown that all of them do.
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