Tideman Problem

[Norman Petry]

Mike,

Thanks for your replies concerning Tideman.  The problem you identified with
his method in your example certainly does indicate that his procedure is
irrational to some extent.  You wrote:

>Also, I forgot to tell what I don't like about Tideman's
>reason for electing B instead of C in that example:
>
>C loses to B for no other reason than because A & D lose to B.
>
>Because the cycle in which B wins is the cycle with the biggest
>defeats, the cycle of the biggest losers, that means that
>B wins for sure, since its only defeat, the one in that
>big-losers' cycle has been skipped.

I agree.  However, for this to be a serious problem with the method, we
would need to know how likely such a result would be in practice and/or how
difficult it would be for insincere voters to manipulate outcomes by
exploiting this weakness.  A better test of Tideman would be to see if it
violates any important criteria (for example, could the defect you
identified lead to non-compliance with GMC?).

Your bad example just gave me an idea though -- what would happen if we
turned Tideman on it's (his?) head to create a "Reverse-Tideman"?  By this I
mean: basically, Tideman's method is to begin with the strongest
propositions and ignore contradictions.  Your example shows that this might
result in an important, strong proposition being ignored.  What would happen
if we instead started with the weakest propositions, and reversed our
earlier judgements as often as necessary as we worked our way up to the
strongest?  That might break your bad example, and we might not need to
resort to the "Improved Tideman" which needs to backtrack over the
previously skipped propositions to be (potentially) as good as Schulze.

It's just a thought... I haven't even tried one example with such a method,
though, so it's probably complete nonsense!

I'll look into it.


Norm Petry


-----Original Message-----
From: Mike Ositoff <ntk at netcom.com>
To: election-methods-list at eskimo.com <election-methods-list at eskimo.com>
Cc: ntk at netcom.com <ntk at netcom.com>
Date: July 31, 1998 11:44 PM
Subject: Tideman Problem


>
>
>Norm asked how Tideman is inferior to Smith//Condorcet(EM), and
>what worse violation by Tideman's method I was referring to.
>
>Answering the 2nd question answers the first.
>
>***
>
>I'll give 2 examples, but first let me state the problem briefly:
>
>According to Tideman's rules, as I interpreted them, the least
>beaten alternative in a cycle of heavy defeats gets its defeat
>skipped by Tideman. The defeats in that cycle are locked-in or
>skipped before the others are gotten to, because all the defeats
>in that cycle are greater than the other defeats.
>
>So then we start locking in the other defeats. It happens that
>there's another cycle containing the alternative whose defeat
>was skipped. This other cycle has extremely light defeats. But
>they all get locked-in, because none of them lock-in a cycle,
>because of that first skipped defeat, which is common to both
>cycles.
>
>I'll give an example now. Bruce Anderson proved that, for any
>set of pairwise vote totals, there's a set of rankings that is
>consistent with those pairwise vote totals, and which will
>yield those pairwise vote totals. That means that, for methods
>based on pairwise vote totals, there's no need for an example
>to actually specify all the rankings, just the pairwise vote totals.
>
>And the only important part of the pairwise vote totals, when we're
>going by votes-against, is the votes-against total in a defeat.
>I'll specify those. For the opposite pairwise vote total in each
>pairing, you can of course fill in any number between zero &
>1 less than the votes-against figure. For emphasis, I've sometimes
>used votes-against totals of just 1. Of course for that to be
>a defeat, it must be that if A beats B with a votes-against total
>of 1, then zero people have ranked B over A.
>
>The example:
>
>A beats D beats B beats A.
>
>D beats B beats C beats D.
>
>D beats B.
>
>A beats C.
>
>The numbers written after the defeats are the votes-against totals:
>
>A beats D 12
>D beats B 10
>B beats A 11
>
>B beats C 1
>C beats D 2
>
>C beats A 5 (this defeat isn't really relevant to the problem)
>
>***
>
>First we lock-in the biggest defeat, A>D 12. Then B>A 11.
>then we skip D>B 10, because that would lock-in a cycle.
>
>Then we lock-in C>A 5, then C>D 2, then B>C 1. That doesn't lock
>in a cycle, because D>B 10 has been skipped.
>
>***
>
>C, the least beaten alternative, which would win in Schulze's
>method or Condorcet(EM) or Smith//Condorcet(EM) loses, and B,
>with 10 votes against it in a defeat, wins.
>
>***
>
>I first noticed that problem in an example with a subcycle,
>but, as the above example shows, it doesn't require a subcycle.
>
>***
>
>Sure, you could add a rule that a skipped defeat will be
>"unskipped" before other defeats in a cycle involving that
>skipped defeat, other than the cycle that resulted in the
>skipping, are locked-in.
>
>Good luck. That makes an already-messy procedure messier still,
>and complicated.
>
>***
>
>More about the intuitive justification for Schulze:
>
>If you feel that A is better than B, and that B is better than
>C, then usually that means that A is better than C, and usually
>that's so.
>
>If public preferences are cyclic, then there are mutually-
>contradictory sequences of preferences.
>
>If, as indicated by pairwise vote-counts, the public prefers
>A to B, B to C, and C to A, then they're saying they like A
>better than B, but they're also saying that B is better than C
>and that C is better than A.
>
>If someone says that A>B is so conclusive that B>C>A doesn't
>count, remind them that everything is beaten, and so that
>assumption gets us nowhere.
>
>So if the defeat A>B is weaker than B>C and also weaker than
>C>A, then, though the public is contradicting itself, the
>implied preference for B>A is based on more nearly unanimous
>vote totals than is A>B.
>
>As convincing as the A>B defeat might seem, for comparing A
>& B, every alternative is a loser if we go by that, in a
>cycle. So, for comparing 2 alternatives, it makes sense to
>compare those mutually-contradictory sequences of preferences.
>
>So Schulze's method seems intuitive & natural for comparing
>pairs of alternatives.
>
>***
>
>Mike Ossipoff
>
>