[EM] Re: Counting Time

[Paul Kislanko]

Correcting -  I wrote in too much haste:
> 1. For each ballot, for each pair of candidates {X, Y} count 
> defeat.X =
> defeat.X if the voter ranked Y better than X

if it were not obvious, I meant 
  defeat.X = defeat.X +1

Apologies.

> -----Original Message-----
> From: election-methods-electorama.com-bounces at electorama.com 
> [mailto:election-methods-electorama.com-bounces at electorama.com
> ] On Behalf Of Paul Kislanko
> Sent: Thursday, January 20, 2005 1:49 PM
> To: 'Forest Simmons'
> Cc: election-methods-electorama.com at electorama.com
> Subject: RE: [EM] Re: Counting Time
> 
> 
> 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.
> 
> 
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