[EM] Tideman & Schwartz set

Markus Schulze markus.schulze at alumni.tu-berlin.de
Thu Mar 14 01:39:17 PST 2002

Dear Mike,

you wrote (13 Mar 2002):
> Do you advocate choosing the TBRC after the count results are in?
> If so, that sounds rather like a cheat. What you're then saying
> is that you avoid the election of someone outside the Schwartz set
> by making sure that you declare them loser when there's a tie :-)
> If that's what you meant, then I must admit that you were not talking
> about a special class of examples in which Tideman won't choose outside
> the Schwartz set. Instead, you were talking about preventing the
> victory of someone outside the Schwartz set by making sure that
> you declare him loser if he's in a tie. That's of course what
> you're doing if you write the TBRC intentionally so that all Schwartz
> set members are ranked over everyone else.
> But suppose you advocate choosing the TBRC before the election.
> Now, unless you've got a pre-rigged ballot box ready, how do you
> choose the TBRC in such a manner that all the Schwartz winners are
> ranked over all the other candidates? One would hope that before
> the election, you don't know who will be in the Schwartz set.
> The TBRC is written before the election, isn't it. In that case,
> as I said: Actually yes, your condition for your guarantee that
> no one outside the Schwartz set can win is actually a special
> fortuitous situation in which the Schwartz set just happens to
> be a set of candidates who are all ranked over everyone in the

I don't advocate that the TBRC should be chosen before the election
or after the count. And I don't know why I should do so. I suggest
that the TBRC should be chosen after the election and before the
count. Blake Cretney writes (http://www.condorcet.org/rp/details.html):

> To handle ties, you'll want a complete ranking of the candidates to
> act as a tie-breaker. In the simplest case, you could just randomly
> order the candidates. For example, by a draw, or using a random
> number generator.  Or, you could designate one person to provide
> the ranking, preferably as soon as the nominations are in.
> Once you have the tiebreaking ranking of the candidates, it allows
> you to designate a precedence between victories at the same margin.
> Then, you can use the normal procedure or definition.
> Let's say you have two different victories, like A vs. B and C vs. D
> this involves the candidates A, B, C, and D. One of these candidates
> is ranked highest on the tiebreaker. Whichever victory involves that
> candidate is considered higher after the tiebreaker. The same candidate
> can appear in both victories, as long as it isn't highest. For example,
> given the tiebreaker A>B>C>D>E, C>B would beat D>E, since B is ranked
> highest on the tiebreaker.
> But it might be, that you have the same candidate repeated, and this
> candidate is the highest. Not surprisingly, you look at the candidates
> the victories don't share, and see which is higher of those. That
> victory is considered higher after the tiebreaker. For example, with
> the tiebreaker A>B>C>D>E, C>D would beat E>C, since D is higher than E
> on the tiebreaker.
> Next, you consider pairwise ties. Score those as a victory for the
> higher ranked over the lower ranked candidate. So, with tiebreaker
> A>B>C>D>E, B=E would cause B to be ranked over E.

I suggest that --when you consider the Schwartz criterion important--
then you should calculate the Schwartz set and choose the TBRC in
such a manner (1) that all Schwartz winners are ranked ahead of all
other candidates in the TBRC and (2) that the other rankings are
filled by random ballots so that independence from clones is met.

Markus Schulze

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