[EM] Cloneproof SSD
MIKE OSSIPOFF
nkklrp at hotmail.com
Sun Jan 21 21:01:38 PST 2001
Thanks for the further information about the Schwartz set method,
Cloneproof SSD, as regards simultaneous dropping & dropping in all
orders. I'm interested in anything you can tell me about how that
method does. I have a few comments, below:
>Example:
>
> 30 voters vote A > B > C.
> 30 voters vote B > C > A.
> 30 voters vote C > A > B.
>
> Suppose, that candidate C is substituted by a large number
> of clones and that the strength of a pairwise defeat
> between two clones is always below 60:30.
>
>If pairwise defeats were dropped randomly then the probability
>that one of the clones was elected would increase with the
>number of clones. The same is true when the strength of the total
>violation was measured in a lexicographical manner (as Steve
>Eppley suggested for his MTM method).
When what method is used? With Cloneproof SSD? What if the defeats
are all dropped simultaneously in Cloneproof SSD? The defeats among
the C clone set are the equally weakest defeats in the current Schwartz
set, and so we drop all of those defeats simulaneously. Then we
simultaneously drop all the remaining defeats. So the C clone set
and A & B are all pairtied now. When we randomly choose a tiebreaking
ballot, the members of the C clone set, no matter how many of them
there are, are no more likely to be at the top of that ballot than
A & B are. A, B, and the C clone set are equally likely to be at the
top of that tiebreaking ballot, as they should be. Adding those C
clones hasn't increased the probability of the winner being a C clone.
>
>******
>
>Suppose that there are n pairwise defeats of equal strength.
>
>Suppose that Tideman's Ranked Pairs method is used. The reason why
>you have to check all n! ways to consider one defeat after the other
>and cannot simply consider all n pairwise defeats simultaneously is
>simply the fact that it could happen (1) that these pairwise defeats
>lock in a cycle when you lock all them simultaneously and (2) that
>the Tideman method doesn't terminate when you skip all these pairwise
>defeats simultaneously.
I no longer advocate Tideman, but as I define it, Tideman terminates
as soon as there are no more cycles. Doesn't the standard wording
terminate as soon as every defeat has been either skipped or locked?
But I emphasize that I no longer advocate Tideman, because Cloneproof
SSD seems so much better for small committees, and because even ordinary
SSD does better in public elections (where its clone-criterion
violation has negligible likelihood).
>
>However, when the Schwartz set heuristic is used then it is the
>same whether all n! ways to drop one pairwise defeat after the
>other is checked or whether all n pairwise defeats are dropped
>simultaneously.
Thanks for that information. That was one of the things that I was
wanting to find out.
>
>******
>
>The winner of the Schwartz set heuristic and the winner of the
>beat path heuristic can differ only when there is at least one
>pairwise tie.
>
>Example:
>
> A:B=50:50
> A:C=35:25
> B:C=40:60
>
>The Schwartz set heuristic would choose candidate A because
>candidate A is the unique Schwartz winner. The Schwartz set
>heuristic treats all pairwise ties in the same manner.
>
>The beat path heuristic would choose candidate C because beat
>paths must be able to contain pairwise ties. Otherwise it would
>be possible that there is neither a beat path from candidate X
>to candidate Y nor from candidate Y to candidate X.
But A has beatpaths to C & B, and neither has a beatpath to A.
Doesn't a beatpath have to consist only of pairwise defeats?
Taken literally, a beatpath is a path of beats, not ties.
Mike Ossipoff
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