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Analyzing the Pairwise Matrix

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In the course of researching various voting methods, I came across a note that there are 720 unique orderings for pairwise defeats among four candidates, accounting for symmetry.  I have also noticed that Wikipedia’s convenient summary table of criteria met by various methods contains several unsupported claims, including at least one which has been questioned on the discussion page of another article.

I, for some crazy reason, have become interested enough to attempt to analyze as many situations as I conveniently can.  This may possibly provide objective research for the issues at hand, but is in large part just an itch to scratch.

For this kind of research, it is important to have a convenient notation.  To enumerate the possible ballots, I will call my candidates Adam, Bill, Cody, Dave, and Erin; more frequently, however, I will simply use the first initial.  To specify a pairwise ordering, CD>BD>DA>AC>AB>BC, for example, shows that Cody was ranked higher than Dave more frequently than Bill was ranked higher than Dave, and so on.

I will probably at some point consider equal rankings, as well, or at least equal placement within the rankings.  As an example, CD>BD>AC=DA>AB>BC=0 shows that Adam beat Cody exactly as often as Dave beat Adam, and that Bill and Cody are completely tied.

Enumerating all of the possibilities is relatively easy, but trimming out the symmetrical positions will be much easier than trying to analyze each one separately.  It seems to me that an easy way to eliminate symmetric alternatives is to ensure that each candidate’s first appearance in the ordering string is in alphabetical order; sets of equally dominant pairs should also appear in alphabetical order, and only alphabetical pairs should appear in the fully-tied set.  This gives Adam a distinct advantage, but I don’t think the others will mind too much.

That said, I’m not entirely certain where the 720 figure came from, other than as the number of permutations of the six pairs.  However, each pair can be reversed, so the number of actual possibilities is multiplied by a power of 2.  After symmetry reduction, my script finds 1920 possibilities to analyze, from AB>AC>AD>BC>BD>CD to AB>CD>DB>CB>DA>CA.  Adding a fifth candidate raises that total to almost 31 million.  Adding the equal rankings and tied pairs adds yet another order of magnitude; 26,452 four-candidate cases.  Even the three-candidate cases start looking interesting with 34 possibilities.

I could be analyzing a single ballot each day for the next several years, but to be honest, I would get bored long before then.  Instead, I expect to find several broad groups, based on how differently the various voting methods react to them.  That may involve calculating things like the Smith set, finding minimal voter counts and rankings that would result in such a matrix, and determining how reasonable it could be for each candidate to win.

Whether any of this will end up answering my original question is up for grabs, but I think it could.


Written by eswald

17 Jun 2011 at 4:17 pm

Posted in Politics

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