[EM] Re: Counting Time

Paul Kislanko kislanko at airmail.net
Thu Jan 20 11:49:26 PST 2005


Forest Simmons wrote (Thursday, January 20, 2005)

> Here's a quick way to find the Condorcet Winner if there is one:
> 
> Use Rob LeGrand's ballot by ballot approval idea, but instead 
> of ballot by 
> ballot, use voter by voter.
> 
> For fairness, either randomize the order of polling the 
> voters or else 
> poll them twice, once from left to right, and once from right 
> to left, so 
> that each voter gets to vote twice.

 This is an example of how terminology can confuse me.

In what way is "voter by voter" different from "ballot by ballot"? Until I
read this I would have considered those phrases synonymous.

But, Mike Ossippoff was kind enough to provide me the other day a very
efficient way to find the CW if there is one, and more generally how to find
the Smith Set which will have one member if there's a CW.

1. For each ballot, for each pair of candidates {X, Y} count defeat.X =
defeat.X if the voter ranked Y better than X, defeat.Y = defeat.Y +1 if the
voter ranked X better than Y. If equal rankings are allowed, leave both
defeat.X and defeat.Y at their current values if on this ballot the voter
specified X=Y. 

2. Sort the candidates by defeat.candidate in ascending order. The candidate
with the lowest number of these pairwise-by-voter defeats are in the Smith
Set, and if no alternative pairwise-defeats (according to the
pairwise-matrix SUM over all ballots) then the that identifies the CW.

If there's not a CW you can continue to construct the Smith Set by the
simple rule "all X with defeat.X equal to or higher than the most recent
addition to the Smith Set and have a pairwise-win over the most recent
addition to the Smith Set are also in the Smith Set", and iterate until no
remaining candidate has a pairwise win over any of the previously identified
candidates.

This is very fast, as it can almost be done as a byproduct of constructing
the pairwise matrix. The key is you have to "do over ballots" to count
pairwise defeats. You cannot get that information from the pairwise matrix.





More information about the Election-Methods mailing list