[EM] A distance based method

[fsimmons at pcc.edu]

Here's something close to what I had in mind originally:

Let p be the probability distribution for a monotone, clone free, lottery like random favorite.

Define the distance between rating ballots b1 and b2 as

d(b1, b2) = [sum (over the alternatives X) of the product p(X)*|b1(X)-b2(X)|^n]^(1/n),

where we can choose n greater than or equal to one for various possibilities.

Now for each alternative A, construct a representative ballot b as the average of all of the ratings ballots 
that rate A at the top.

Then the (voter perceived) distance between two candidates is the distance between their representative 
ballots.

The use of p to give weights to the "components" of the ballot vectors is what de-clones the metric.

----- Original Message -----
From: Kristofer Munsterhjelm 
Date: Monday, July 11, 2011 10:46 am
Subject: Re: [EM] A distance based method
To: fsimmons at pcc.edu
Cc: election-methods at lists.electorama.com

> fsimmons at pcc.edu wrote:
> > First find a clone consistent way of defining distance between 
> candidates.
> This could be an interesting algorithm problem in itself. It is 
> possible 
> to triangulate points in space (assuming Euclidean distances) if 
> you 
> have the exact distances; but what if you have only the rank 
> order of 
> distances? You get a (usually) underconstrained problem of the sort:
> 
> Given v voters, n candidates, and a dimension integer k > 1,
> find v + n coordinates in k-dimensional space so that the 
> Euclidean 
> distance from V's coordinate to C's coordinate for some voter V 
> and 
> candidate C is less than the Euclidean distance from V's 
> coordinate to 
> D's coordinate, for the same voter V and another candidate D, 
> iff V 
> ranks C ahead of D. Break ties by assigning coordinates so that 
> the sum 
> of the Euclidean distances to the candidates from the origin is 
> minimized.If it is not possible to make such an assignment, make 
> one that 
> contradicts as few candidate rankings as possible.
> 
> I have no idea how you would actually do this, though, and it 
> would be 
> prone to overfitting. It might not be cloneproof, either, since 
> differences in clone rankings could eliminate some rotations 
> that would 
> otherwise be picked as the best choice by the tiebreaker. It 
> would 
> definitely not pass IIA, as the addition of "superbad" 
> candidates could 
> serve as anchors.
> 
> > Then while two or more candidates remain
> > of the two with the greatest distance from each other
> > eliminate the one with the greatest pairwise defeat
> > EndWhile.
> 
> Or, if you have the voters' coordinates too, you could use 
> histograms, 
> kernel density estimation, or some other estimation to try to 
> reconstruct opinion space, and then pick a subset of the 
> candidates that 
> "reproduces" that opinion space as best as is possible. E.g. if 
> each 
> candidate (and voter) is a Gaussian in opinion space and you 
> want p 
> seats, find the p Gaussians where the difference between the 
> space given 
> by the sum of the p Gaussians (of the prospective council) and 
> of the v 
> Gaussians (of every voter), normalized, is minimized.
> 
> Bandwidth selection would be a pain, as would finding the 
> "right" number 
> of dimensions.
> 
>