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Feasibility Results

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I’ve been able to run through the full combinations for two-, three-, and four-candidate ballots, with specific winners for the plurality and/or Borda Count methods.  The last one in particular is a huge mess of data, but some interesting patterns emerge.

The first thing of note is that a problem is never infeasible solely due to a plurality winner, except in the two-candidate cases.  That means that under the most commonly used voting method, if there are more than two candidates, the winner may have very little to with the actual preferences of voters.  As we saw in the AC>AD>BC>BD>CD” href=”https://eswald.wordpress.com/2011/07/19/abacadbcbdcd/”>very first case, a candidate can win even when it would lose to each of the others in a one-on-one situation.

Borda, on the other hand, isn’t nearly so susceptible to such messiness.  There are still over eleven thousand four-candidate preference permutations that can be manipulated to have more than one winner, but that’s less than half of them.  Fewer than two thousand admit more than two winners, and fewer than a hundred can swing all four ways.  In addition, the potential Borda winners in each permutation frequently participate in cycles on the preference graphs, meaning that they’re each in the Smith set.

Not always, however.  In particular, the Condorcet criterion counter-example listed on Wikipedia was reproduced exactly, but with different names, in the AB>CA=CB,A row.  For my next trick, I would like to figure out when exactly a non-Smith candidate can win a Borda election, and exactly how bad those results feel.

Sometime I’ll also compare the pairwise methods (Ranked Pairs, Beatpath, River, Minimax, and Kemeny-Young) to each other, but at least part of that work has already been done.  Meanwhile, I’m spending a ridiculously large amount of processing power on the five-candidate ballots, just to ensure that my four-candidate results remain valid.


Written by eswald

23 Aug 2011 at 10:07 pm

Posted in Mathematics

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