# [EM] Cloneproof SSD

Markus Schulze schulze at sol.physik.tu-berlin.de
Sat Jan 27 13:05:14 PST 2001

```Dear Mike,

you wrote (21 Jan 2001):
> Markus wrote (21 Jan 2001):
> > 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.

To guarantee that LIIAC is met, it is only necessary that a beat
path consists of wins or ties. It is not necessary that a beat path
consists only of wins.

In the example above:

A:B=50:50 via beat paths.
A:C=35:50 via beat paths.
B:C=35:60 via beat paths.

You wrote (23 Jan 2001):
> Then there's the fact that, in small committee votes, Tideman will
> sometimes embarrass us by choosing outside the Schwartz set. Not
> a serious problem, but definitely an aesthetic gaffe.

I don't consider a violation of the Schwartz criterion to be
"definitely an aesthetic gaffe." To my opinion, it is more
important that as few voters as possible are overruled. In the
example above, candidate A is rejected by 50 voters in his worst
pairwise comparison while candidate C is rejected only by 40
voters in his worst pairwise comparison.

The Schwartz set heuristic unnecessarily chooses winners whose worst
pairwise comparison are strong. This is caused by the fact that
the Schwartz set heuristic doesn't take the "strengths" of pairwise
ties into consideration.

******

Blake wrote (17 Jan 2001):
> Here's my problem with that.  I agree that we have to drop defeats.
> I also agree that we should not drop defeats unnecessarily.  However,
> the process of iterating Schwartz imposes some requirements.  We
> often have to drop defeats just to keep the process going.  So, we
> might have to drop defeats that would be unnecessary to drop if we
> only wanted to impose consistency.
>
> It's a similar problem to IRV.  Obviously some candidates are going
> to be dropped (in the sense that not all candidates can win).
> It's also obvious that if you only look at first place votes among
> non-dropped candidates, that IRV picks the right candidate to drop.
> However, if we don't assume these restrictions, it isn't clear that
> IRV is dropping the right candidates.

I consider iterative heuristics to be less intuitive than non-iterative
heuristics. The problem with iterative heuristics is: Even when we know
that this heuristic works properly in each step, we still don't know
whether the final winner is the right candidate.

I also think that -in so far as there is no general idea behind
the Schwartz set heuristic- too many details have to be explained.
Example: It isn't clear why only circular ties between current Schwartz
winners were contradictory; to my opinion, circular ties are always

******

You wrote (21 Jan 2001):
> 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).

What does Steve Eppley say about the fact that you don't promote the
Tideman method any more?

Markus Schulze

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