[EM] Simple Election Methods and Geometry

[Kevin Venzke]

Alex,

 --- Alex Small <alex_small2002 at yahoo.com> a écrit : 
> Let's stick to 3 candidates and ignore equal rankings, to keep it simple.  There are 6 possible
> types of voters, and so there are 6 variables to consider:  The fraction of the electorate with
> each possible preference order. These variables live in a 6D space, but because the fractions
> add up to 1 we're dealing with a finite (and 5-dimensional) region of space.  Also, I assume
> that the method treats all candidates equally, so that if every voter swapped the winner (let's
> call him candidate A) with the same other candidate (let's call him B) then B would win.

Thanks to information from Forest, I was able to write a program that plots winners
in a pyramid (a slice of which can be viewed at a time), for three candidates and
4 of the 6 ballot types at one time.

> Conjecture #1:  I conjecture that the methods which minimize the total area of the boundaries
> are those where the boundaries are "straight lines" (formally, 4D hypersurfaces defined by a
> linear equation, so there aren't any kinks or curves).

The only method I've seen which produced curved boundaries in my program is Douglas 
Woodall's QLTD method, a Bucklin variant. The tie-breaker (when two candidates
achieve a majority at the same rank) asks which candidate needs the lowest proportion
of his last-rank votes to hit the 50% mark.

> The notion of drawing "straight lines" in 5D might seem complicated, but the methods that
> correspond to "straight line" boundaries are easy to understand.  They're what Saari would call
> "positional methods".  Candidates get a certain number of points from each voter depending on
> how the voter ranked the candidates.

> In one of my favorite positional methods (mostly
> because of its importance in the strong FBC problem that I continue to pursue on the side), a
> candidate gets 1 point from each voter who ranks him 1st and 2nd, and 0 points from each
> candidate who ranks him 3rd.

In my program, I have implemented this rule as "Approval." (There is no requirement
to vote for a 2nd, correct?)

> Anyway, I'm not saying these are the best methods to use, but simply posing a question as a
> warm-up for something that people on this list might find more practical:
>  
> Question:  Which Condorcet completion method minimizes the total area of the boundaries?

I think it's probably straight Approval (in the above sense), as in: If there is
no CW, elect the Approval winner, even if he's not in the top tier.

I get the impression because, if I take an arbitrary slice of a pyramid, with the
four dimensions chosen arbitrarily and the static dimension set at 25% or 30%, I
get the impression that "Approval" is much more likely to have huge win regions and
shorter total boundary length than other methods.

> My hunch is that it would be some sort of positional method, and that there would be more than 1
> such positional method.  However, it could just as easily turn out that methods which compare
> the magnitude of victory minimize the area of the boundaries and are therefore simpler and less
> prone to manipulation in some sense.

No, I don't think so, since you're inviting "kinks" in the boundaries. You want
a less sensitive method.

Actually, "If no CW, Random Candidate" is probably the best.

Kevin Venzke



	

	
		
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