Re: [EM] Condorcet Flavored PR Methods

[Adam Tarr]

Forest, I finally got around to reading this series of posts. It's very interesting stuff and you've obviously made a lot of progress on this. A few comments:

- I'd imagine you're aware of this, but this approach passes the "sanity check" of reducing to a regular pairwise matrix when the size of the "circles" is only one candidate. if A>B on a ballot, then kA = .5, kB = 0, jA = 0, jB = .5, kA+jB=1, kB+jA=0, so 1/1=1 votes for A, and 1/0 -> 0 votes for B. So, only one method need be defined for single and multi-winner, just as for IRV->STV or Approval->PAV.

- The one thing that bugged me about this approach was the decision to use the medians of the symmetric differences as the "exemplar" of each set. (By exemplar, I mean the two candidates that produce the "M" and "m" candidates.) This seemed sort of arbitrary, even if it also seems reasonable. You note that the "sanity check" of matching PAV when all of A is preferred to all of B, but this would also be true if you just picked the best candidate in A-B and B-A as the exemplars. Obviously, you also get the same results in your second example where only one candidate differs in each set.

But on the other hand, you do get different results in your A1>B1>B2>B3>B4>A2>A3>A4 example; choosing the exemplar as the best candidate in A-B and B-A makes you give 1/4 more votes to the A set, rather than 1/2 + 1/3 + 1/4 more votes to the B set. Seems like your approach is a clear winner here.

Was this choice of examplar the only one that gave you the reverse-symmetry property? It seems to be that way but I'm not sure. If that's the case then I'm sold on the merits of this choice.

- There may be one small numerical issue to tackle with the medians, though. You say, "Let m1 and m2 be the respective medians of the sets (A-B) and (B-A)." Later you say, "When M or m falls right on a member of A or B (rather than in the space between two candidates), then such a member adds 1/2 to the count..." But what about this case:

A1>B1>B2>B3>A2>B4>A3>A4

mB is between B2 and B3. But mA is between A2 and A3... which falls on B4.

A1>B1>B2>M>B3>A2>B4=m>A3>A4

kA = 1
kB = 2
jA = 2
jB = 1/2

kB + jA = 4
kA + jB = 1.5

Essentially, since one median falls between candidates, and one falls exactly on a candidate, all the 1/2's don't add up in a tidy manner. So the number of votes to cast for the A slate is the sum of 1/n for n ranging from zero to 1.5, which doesn't really make sense. Now, this can be worked around by using logarithms, and we know that this should be a number around 1.3 or so. Is this the right way to approach this, or am I missing something?

-Adam


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