Tiebreakers, Subcycle Rules

[Norman Petry]

Mike,

I'm still getting decisive results for Schulze.  Here's my analysis:

A>>D: 7:6 (A>D vs. D>E>A)
B>>A: 6:5 (B>D>E>A vs. A>D>E>B)
B>>D: 7:5 (B>D vs. D>E>B)
B>>E: 6:5 (B>D>E vs. E>B)
C>>A: 6:4 (C>D>E>A vs. A>D>E>C)
C>>B: 5:4 (C>D>E>B vs. B>D>E>C)
C>>D: 7:4 (C>D vs. D>E>C)
C>>E: 6:4 (C>D>E vs. E>C)
E>>A: 8:6 (E>A vs. A>D>E)
E>>D: 7:6 (E>A>D vs. D>E)

Therefore, the unique Schulze winner  for this example is C.  Your results
seem quite different (you record beat-path wins D>>A, A>>B, A>>C).

Am I applying Schulze's method incorrectly?


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 7, 1998 2:35 PM
Subject: Re: Tiebreakers, Subcycle Rules


>
>
>Sorry--I did neglect to say the magnitude of D>E. It's 6, as
>I was doing the example.
>
>In case I miswrote it before, let me write the example in total:
>
>{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
>
>D>E 6
>
>***
>
>B & C have Schulze wins over D & E, so D & E can't be Schulze
>winners.
>
>D & E have Schulze wins over A, so A can't be a Schulze wioner.
>
>A has Schulze wins over B & C, so they can't be Schulze winner.
>
>So there is no Schulze winner.
>
>***
>
>My fault, having not specified the magnitude of D>E. Though
>my recent statements about which methods do what in
>that example seem correct (but I've missed errors lots of times),
>I was mistaken previously when I said that none of those 4
>methods can deal with that example. It's a Schulze indecisiveness
>example. It also isn't terribly favorable for Sequential Dropping,
>which declines to choose from the subcycle for the same reason
>that Schulze doesn't choose. But, considering the relative rarity
>& unpredictability of subcycle problems, that's forgivable. And,
>in regards to Schulze's result there, it isn't that it's a problem
>to sometimes use a tie-breaker--it's just that the need to have
>one adds complication & words to the definition. There may well,
>for all I know, be reasons why Schulze, with the right tie-breaker,
>is better than Sequential Dropping. If that were determined for
>sure, then maybe Schulze would be a contender for proposing in
>meetings of an initiative committee, and it could be determined
>how people would react. But explaining any but the simplest
>rules to a political office-holder, or to the person-on-the-street,
>is _daunting_.
>
>In fact, when considering that task, that makes plain Condorcet(EM)
>look pretty good. For the properties it affords, that method
>is as simple as it can get.
>
>Mike
>
>