# [EM] Cloneproof SSD

Markus Schulze schulze at sol.physik.tu-berlin.de
Wed Jan 17 10:03:16 PST 2001

```Dear Mike,

you wrote (17 Jan 2001):
> If we have a set of candidates none of whom are beaten by anyone
> outside the set, then those candidates are obviously more qualified
> to win. Condorcet suggested his procedure because the propositions
> (pairwise defeats) cannot all exist together. The ones that cannot
> all exist together are the ones among the current Schwartz set.
> The defeats by Schwartz set candidates against non-Schwartz-set
> candidates are not contradicted by any other defeats. The only
> defeats that are contradicted, for the purpose of choosing a winner,
> are those among the current Schwartz set. The people are saying
> that the Schwartz set candidates are the ones that aren't bested by
> anyone. So it's intuitively natural & obvious to only drop defeats
> from among the current Schwartz set.

I don't consider the Schwartz criterion to be "intuitive" and
"obvious."

The main justification of the Smith criterion is the fact that
an additional candidate can change the Smith set only if he is a
Smith winner. But it is possible that an additional candidate can
change the Schwartz set without being a Schwartz winner.

Example:

A:B=50:50
A:C=58:42
B:C=45:55

When candidate C doesn't run, then the Schwartz set consists
of candidate A and candidate B. When candidate C does run, then
the Schwartz set consists only of candidate A.

******

Both heuristics, the beat path heuristic and the Schwartz set heuristic
for the Schulze method, have advantages and disadvantages. Therefore I
suggest that one should always use both heuristics simultaneously. The
main disadvantage of the Schwartz set heuristic is that one has to
explain and justify too many tiny little details.

Example 1: Most people will simply proceed only until there is an
unbeaten candidate (especially in so far as the runtime of the Schwartz
set heuristic is giantic when the number of candidates is large). They
will not understand why they have to proceed until there are no cycles
any more.

Example 2: Suppose that there are 5 pairwise defeats of equal
strength. Then some people will suggest that all these pairwise
defeats should be dropped simultaneously. Some people will suggest
that all 5!=120 ways to drop one defeat after the other should be
checked to calculate the winners. Some people will suggest that one
should decide randomly which of these defeats should be dropped.
Some people will suggest that the pairwise defeats should be
dropped in such a manner that the strengths of the then following
dropped pairwise defeats are lexicographically minimal.

Example 3: It isn't clear (and as far as I remember correctly you
haven't explained) what has to be done when the winner of the
Schwartz set heuristic is not unique.

All these tiny little details have to be formulated when a given
organization wants to adopt this method. Actually I believe that
it will be significantly more simple to define the beat path
heuristic than the Schwartz set heuristic in an exact and
justifiable manner.

Markus Schulze

```