[EM] Voting by selecting a published ordering
Simmons, Forest
simmonfo at up.edu
Fri May 12 19:38:52 PDT 2006
Paul,
Actually, I did not assume that there was any linear or two dimensional relationship. You can use any measure of closeness that you want, linear, non-linear, ten dimensional, or infinite dimensional (for example the norm of the difference of the quantuum mechanical wave functions of the candidates, if you like).
My only assumption is that there is some measure d of distance that (for each pair of candidates A and B) satisfies d(A,B)=d(B,A) , and that when d(A,B)<d(A,C), consistency requires that candidate A rank candidate B ahead of candidate C .
These distances are numbers, so they can be linearly ordered.
In fact, when there are n candidates, there are C(n,2) of these pairwise distances, which can be ordered in C(n,2)! ways. Each of these orderings is in an equivalence class of size n! obtained by permuting the candidates. So there are at most C(n,2)!/n! distinct geometrically consistent cases.
If n=3, then C(n,2)=3, and C(n,2)!=n!, so there is only one distinct geometrically consistent case.
If n=4, then C(n,2)=6, and C(n,2)!=720, and n!=24, so that leaves 720/24=30 geometrically distinct cases. If you will write out all thirty of these cases (as I have), you can verify for yourself that only one of them fails to always have a Condorcet Winner.
In the process you will see that only one of them (a different one) fails to always have a Condorcet Loser, as well.
So most of the time, in the context of Candidate Published Orderings, Concorcet will yield an unambiguous social ordering of the candidates, with no cycles to resolve.
More importantly (and precisely) it means that if the four candidates accurately judge their relative closeness to each other (and vote sincerely) then there is only one chance in 240 that there will not be a Condorcet Winner.
Why 1/240 instead of 1/30? Because, as I mentioned in my previous message, even in the one case where there is not always a Condorcet Winner, only one eighth of the distributions of the voters among the four factions will result in a cycle.
I would say that's amazing, and extremely relevant to the topic of this thread.
Forest
________________________________
From: Paul Kislanko [mailto:kislanko at airmail.net]
Sent: Wed 5/10/2006 7:35 PM
To: Simmons, Forest
Subject: RE: [EM] Voting by selecting a published ordering
The problem is you just translated a multidemionsal problem into a 2-dimensional one by assuming there was a linear relationship among the candidates, when in fact the social choice problem is that assigning a metric that relates A & B is impossible if there's more than 1 issue and more than 3 voters.
________________________________
From: election-methods-bounces at electorama.com [mailto:election-methods-bounces at electorama.com] On Behalf Of Simmons, Forest
Sent: Wednesday, May 10, 2006 5:19 PM
To: election-methods at electorama.com
Subject: Re: [EM] Voting by selecting a published ordering
I've been exploring certain questions in relation to this topic of Candidate Published Orderings.
First of all, which combinations of candidate orderings make sense geometrically?
This question is not hard to answer in the case of three candidates, A, B, and C.
Suppose that candidates A and C are the furthest apart from each other as measured by some symmetric metric (Euclidean or not).
Then geometric consistency requires that the A and C orderings are ABC and CBA, respectively.
The B ordering could be either BAC or BCA.
Either of these can be obtained from the other by a permutation of the letters.
In summary, there is essentially only one case (up to permutations of the letters) that is geometrically consistent, when there are three candidates:
x:ABC, y:BAC, z:CBA ,
and no matter the sizes of the factions x, y, and z, there will always be a Condorcet Candidate.
There are many more cases to consider when there are four candidates A, B, C, and D.
However, it turns out that (up to permutations of the letters) there are only thirty (30) cases that are geometrically consistent, and 29 (all but one) of these are like the above case in that there will always be a Condorcet Candidate, no matter how the faction sizes w, x, y, and z are distributed.
The geometrically consistent case that does not always have a Condorcet Winner is the following:
w:ACDB, x:BCDA, y:CBDA, z:DBCA.
Even this case has a Condorcet Winner for precisely 87.5 percent of the distributions of faction sizes w, x, y, and z. The exceptions occur when all of the following relations hold simultaneously:
x+y<50%, z<50%, x+z>50%
which describes precisely one eighth of the volume of the tetrahedron given by
x+y+z < 100%
in the first octant of the x, y, z, coordinate system. (Note that w is determined from x, y, and z by w+x+y+z=100% .)
To see that this case is geometrically consistent, locate candidates A, B, C, and D respectively, at the points
(4,2), (0,0), (1,0), and (0,2) in a Cartesian Coordinate Plane.
The important thing about these positions is that
d(A,B)>d(A,D)>d(A,C)>d(C,D)>d(B,D)>d(B,C) .
I'll stop here for now, to give you all a chance to digest these interesting facts.
To me they are rather amazing facts, and potentially useful, IF we can figure out how to incorporate them appropriately into this candidate published ordering setting, or more generally, into a one faction per candidate setting.
More on that later.
Forest
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