Dear Mike,
you wrote (2 Apr 2001):
> Blake Cretney wrote (1 Apr 2001):
> > Of course Condorcet didn't really propose Ranked-Pairs
> > (defeat-support). That method was invented recently by Mike,
> > based on the method Tideman invented around 1985.
>
> Invented recently by Mike? Steve Eppley defined it before I did.
> But thanks for trying to give me credit for that better version.
Actually, it was me who introduced this method to this mailing
list in 1997. Steve Eppley didn't promote Ranked Pairs
(defeat-support) before 2000.
I wrote (4 Oct 1997):
> Suppose, there are not two (or more) pairwise comparisons,
> where the winner of the one pairwise comparison gets exactly
> as many votes as the winner of the other pairwise comparison.
> And suppose, that there are no pairwise ties. Then, Tideman
> can be explained very simply:
>
> Step1:
>
> In the pairwise-comparisons matrix, look for the highest
> number in a pairwise comparison, that hasn't been locked
> or skipped yet! If this pairwise comparison would create a
> tie, skip it! If this pairwise comparison would not create
> a tie, lock it!
>
> Step2:
>
> Repeat Step1, until every pairwise comparison is either
> locked or skipped! Then: The winner is that candidate, who
> is not "dominated" (i.e. who wins every locked pairwise
> comparison).
>
> Example:
>
> 46 voters vote A.
> 10 voters vote BAC.
> 10 voters vote BCA.
> 34 voters vote CBA.
>
> The pairwise-comparison matrix looks as follows:
>
> A:B=46:54
> A:C=56:44
> B:C=20:34
>
> The highest number in the pairwise-comparisons matrix is 56.
> Thus: A > C is locked.
>
> Now, the highest number of a pairwise comparison, that hasn't
> been locked or skipped yet, is 54. Thus: B > A is locked.
>
> Now, the highest number of a pairwise comparison, that hasn't
> been locked or skipped yet, is 34. But C > B would create a
> tie (A > C > B > A). Thus: C > B is skipped.
>
> Thus: We have A > C and B > A. B is the only candidate, who
> is not dominated.
************
You wrote (2 Apr 2001):
> Blake Cretney wrote (1 Apr 2001):
> > I argued that since no one had come up with an EXAMPLE where
> > Condorcet had considered incomplete rankings, he hadn't.
>
> So much for Blake's claim that Condorcet didn't consider incomplete
> rankings. We covered this some time ago, but Blake repeated his
> refuted statement again.
I guess that Blake rather wants a concrete example from Condorcet
with incomplete rankings. Although Condorcet mentions the
possibility of incomplete rankings in his paper "Sur les Elections,"
it isn't clear how he actually handles these incomplete rankings.
Markus Schulze
