[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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