[EM] Tideman & Schwartz set

MIKE OSSIPOFF nkklrp at hotmail.com
Wed Mar 13 20:11:40 PST 2002

I'd said:

My example in which Tideman chooses outside the initial
Schwartz set doesn't have any equal defeats.

Markus replied:

Could you please post a concrete example where Tideman chooses
decisively outside the Schwartz set?

I reply:

Sure. AB10, BC8, CD7, DB9 The remaining pairs are pair-tied. In
other words, D & C are pair-tied with A.

The Schwartz set is {A}. RP returns a tie between A & D. If, as is
considered by most to be most democratic, ties are solved in some
random way such as Random Ballot, D could win.

I'd said:

>What you're saying is that it's possible to write a class of examples
>in which Tideman won't choose outside of the initial Schwartz set.
>That doesn't contradict my statement that Tideman can choose outside
>the initial Schwartz set.

Markus replied:

Actually no. What I say is that when you choose the TBRC in such
a manner that all Schwartz winners are ranked ahead of all other
candidates then Tideman never chooses decisively or randomly
outside the Schwartz set.

I reply:

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.

So, if aren't psychic, and don't have an oracle to ask, and haven't
rigged the ballot box, then actually yes, it's only in special
fortuitous examples that the Schwartz set will be ranked over everyone
else in the TBRC.

Markus continued:

You wrote (12 Mar 2002):
>What's the TBRC? Transitive B_____ Ranking of Candidates? The
>ordering of candidates that's consistent with the defeats kept by
>the Tideman procedure?

The TBRC (= Tiebreaking Ranking of Candidates) says to whose favour
indecisive situations are resolved.

I reply:

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

Most likely, the TBRC is typically the ranking voted by the chairman
of the committee.

I call that way  of solving ties Chairman Ballot, to name it in
a way related to Random Ballot. It's a good way for e-mail or postal
ballots in which it would be difficult to verify a randomizing
process for all the voters, who can't all be present at a coin-flipping
or number-drawing, etc.

For the polls that we did last summer, I specified a different
tie solution, which could be called pseudorandom ballot--an
elaborate way of choosing based on presumably unpredictable
aspects of the poll result.

Mike Ossipoff

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