# Tiebreakers, Subcycle Rules

Norman Petry npetry at sk.sympatico.ca
Fri Aug 7 12:57:55 PDT 1998

```Mike,

On 5AU98 (Condorcet Versions--some equivalent?), you wrote:

>All 4 of those procedures can't deal with a subcycle example
>in which one member of the subcycle is winner with respect to
>all non-subcycle alternatives, but not within the subcycle.
>
>A subcycle  rule or tiebreaker needed.

And then, on 6AU98 (Tiebreakers, Subcycle Rules), you wrote:

>Here's my indecisiveness example:
>
>{A,B,C} is a subcycle, but not a clone set.
>
>A>B 2
>B>C 3
>C>A 1
>
>{A,B,C}>D 7
>
>E>A 8
>E>B 5
>E>C 4

And finally, on 6AU98 (Re: Oops, it isn't Condorcet(EM)), you wrote:

>In my indecisiveness example, LOP picks B. Sequential Dropping
>picks E. My version of Petry pics E. There's no Schulze winner,
>and I don't know if there is a set with un-countered beat-paths
>to all the rest.
>
>It seems fairer to pick B than E in that example. It seems
>important, to me, to pick something that's a winner in a
>1st order cycle among the Smith set. I'd like to define
>a criterion about that. (I just have). But that criterion
>is incompatible with Pareto, as is LOP.

I've had difficulty verifying your result, as you neglected to specify D vs.
E in your example.  To test Schulze for indecisiveness, I assumed:

D>E: 2

as in your subsequent clone-set example.  When I applied Schulze to the
problem, it decisively chooses E as the winner.  Here are the beat-path
comparisons I got:

A>>D: 7:2
B>>C: 3:2
B>>D: 7:2
C>>D: 7:2
E>>A: 8:2
E>>B: 5:2
E>>C: 4:2
E>>D: 5:2

Looking at it again, I assumed (based on your description of the problem)
that you must have accidentally reversed B vs E, since none of {A,B,C} _are_
winners with respect to all of {D,E}.  Since B is not a winner of the
subcycle, I changed your example to:

B>E: 5

to try to create the problem you described.  Having done this, I applied
Schulze's method and obtained the following:

A>>D: 7:2
B>>A: 5:2
B>>C: 4:2
B>>D: 7:2
B>>E: 5:2
C>>D: 7:2
E>>A: 8:2
E>>C: 4:2
E>>D: 7:2

The Schulze method is decisive in this case also, choosing B.  So far, I've
yet to see an example where Schulze's method is less decisive than any of
the other methods being discussed.

Could you please re-post the example and/or your results, that show Schulze
to be indecisive in these cases?

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: August 6, 1998 1:38 PM
Subject: Tiebreakers, Subcycle Rules

>
>I hadn't remembered Markus's tie-solution for Schulze. But
>that reassures me that the situation I described hasn't been
>ignored. But what if the Schwartz set is the whole set
>of candidates, and you can't separate out a set that has
>un-countered beat-paths to the rest? I don't know if my
>indecisiveness example is such a case. I haven't gotten anywhere
>with it yet.
>
>Here's my indecisiveness example:
>
>{A,B,C} is a subcycle, but not a clone set.
>
>A>B 2
>B>C 3
>C>A 1
>
>{A,B,C}>D 7
>
>E>A 8
>E>B 5
>E>C 4
>
>As I said, I don't yet know if Markus's tiebreaker
>solves this, or if this set of candidates can be divided
>into 1 subset that has un-countered beat-paths to all the
>others.
>
>But surely that can happen, can't it? If so, then, in that
>instance, a further tiebreaker or subcycle rule is needed.
>
>***
>
>Trying those 4 methods on Markus's recent 4-candidate example,
>I got a single winner with Schulze, but got a 2-candidate tie
>with the other 3. So either Tideman isn't equivalent to Schulze,
>or I don't yet know Tideman's own rule. David, would you post
>the entire wording of Tideman's rule?
>
>But though Petry's method does seem the same as the brief
>wording of it that I suggested, it isn't the same as successively
>dropping the smallest defeats till something is unbeaten. That
>would be plain Condorcet(EM).
>
>Try them on this:
>
>{A,B,C} is a clone set subcycle.
>
>A>B 20
>B>C 30
>C>A 40
>
>E>{A,B,C} 1
>{A,B,C}>D 3
>D>E 2
>
>***
>
>Must quit before the keyboard decides to inexplicably quit.
>
>Mike
>
>
>

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