"Reverse-Tideman" Idea Explained

[Norman Petry]

Based on Mike's recent post, I re-examined my example and found that I'd
inconsistently applied the method I described, with the result that my
answer of D appeared unambiguous.  Here's how it should have looked:

A>D: 52          A>(52)D
C>D: 53          A>(52)D; C>(53)D
C>A: 65          C>(65)A>(52)D
A>B: 67          C>(65)A>(52)D; C>(65)A>(67)B
B>C: 68          A>(52)D; A>(67)B>(68)C
D>B: 77          D>(77)B>(68)C; A>(67)B

Result: D>B>C, A>B

This is, of course, exactly what Markus was showing in his example: an
ambiguous result between D,A.

So it looks like I've just succeeded in re-inventing the wheel (and a broken
wheel, at that).  Mike, please stop referring to this as "Petry's method" --
I'd prefer to give Peyton Young the credit!


Norm Petry


-----Original Message-----
From: Norman Petry <npetry at sk.sympatico.ca>
To: Markus Schulze <schulze at sol.physik.tu-berlin.de>
Cc: election-methods-list at eskimo.com <election-methods-list at eskimo.com>
Date: August 4, 1998 10:41 AM
Subject: "Reverse-Tideman" Idea Explained


>Dear Markus,
>
>Thank you for your reply.  You wrote:
>
>>As far as I understand you correctly, you propose
>>that the weakest pairwise defeats should be ignored successively
>>until there are no cycles anymore.
>
>The idea I had was a little different from this.  I'll use your example to
>demonstrate what I had in mind.  First, I'll begin by listing the
>"propositions" suggested by the pairwise comparison matrix, from weakest to
>strongest:
>
>A>D: 52
>C>D: 53
>C>A: 65
>A>B: 67
>B>C: 68
>D>B: 77
>
>Now, I'll try to establish a rank ordering of candidates, by beginning with
>the weakest proposition, and incorporating new information as I work
towards
>the strongest.  Where necessary, I'll reverse my previous judgements, since
>the stronger propositions must outweigh the weaker ones.  I'm not going to
>think about cycles, except to avoid creating them (as in Tideman's method).
>Unlike Tideman's method, I'm going to keep track of the "strength" of the
>propositions, so I can know how to reverse previous opinions.  I'll show
>this as: "A>(52)D" to indicate that the A beats D by 52.
>
>Here's the analysis:
>
>A>D: 52          A>(52)D
>C>D: 53          A>(52)D; C>(53)D
>C>A: 65          C>(65)A>(52)D
>A>B: 67          C>(65)A>(52)D; C>(65)A>(67)B
>B>C: 68          B>(68)C>(65)A>(52)D
>D>B: 77          D>(77)B>(68)C>(65)A
>
>The final ranking is D>B>C>A, so the single winner is D (unambiguously, I
>think).  Note that in steps 3 and 5, I discarded useless information (C>D,
>A>B), since new information allowed me to place C and B in the overall rank
>order.  In step 6, D had to be placed at the beginning of the chain, since
>the evidence for D>B strongly outweighs the evidence for A>D.
>
>I also calculated the single-winner in this example using your beat-path
>method, and produced the same result.
>
>To me, this "Reverse-Tideman" method appears promising, at least in the 2
>cases I've examined (yours and Mike's Tideman bad example).  Is this method
>(as I've described it here) equivalent to yours?
>
>Looking at it, it's almost certain that "Reverse-Tideman" is technically
>inferior to the beat-path method, and will probably produce different
>(worse) results in some cases.  Also, it might be possible to create
>examples where it's impossible to decide how to incorporate new
>propositions.  Yet, it might provide a reasonable approximation in most
>cases.  Unfortunately, the procedure is messier than Tideman's, and because
>we're working from the bottom-up, we couldn't claim that this _is_
>Condorcet's method, so we lose some of the practical advantages that I
>earlier suggested Tideman's method might have over yours.
>
>Without a doubt, your beat-path method has all the elegance,
>comprehensiveness, and consistency that this "Reverse-Tideman" method
lacks.
>Technically, this method offers nothing.  However, the approach itself is
>fairly intuitive, so it might be a useful approximation of the beat-path
>method for hand counts, or for people who have difficulty accepting the
>validity of "beat-paths" in determining winners.
>
>Any comments?
>
>
>Norm Petry
>
>
>-----Original Message-----
>From: Markus Schulze <schulze at sol.physik.tu-berlin.de>
>To: npetry at sk.sympatico.ca <npetry at sk.sympatico.ca>
>Date: August 4, 1998 6:31 AM
>Subject: Re: Tideman Problem
>
>
>>Dear Norman,
>>
>>you wrote (3 Aug 1998):
>>> 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.
>>
>>As far as I understand you correctly, you propose
>>that the weakest pairwise defeats should be ignored successively
>>until there are no cycles anymore.
>>
>>In literature, this has been proposed many times (especially by
>>Peyton Young). Unfortunately, it doesn't work as expected.
>>
>>Example:
>>
>>  19 voters vote A > D > B > C.
>>  17 voters vote B > C > A > D.
>>  16 voters vote C > A > D > B.
>>  16 voters vote D > A > B > C.
>>  16 voters vote C > D > A > B.
>>  10 voters vote D > B > C > A.
>>  06 voters vote B > C > D > A.
>>
>>  A:B=67:33
>>  A:C=35:65
>>  A:D=52:48
>>  B:C=68:32
>>  B:D=23:77
>>  C:D=53:47
>>
>>  There are still cycles (e.g. A > B > C > A,
>>  D > B > C > D, A > D > B > C > A).
>>  A:D=52:48 is the lowest defeat in a cycle,
>>  thus A > D is ignored.
>>
>>  Now, we have:
>>
>>  A:B=67:33
>>  A:C=35:65
>>  B:C=68:32
>>  B:D=23:77
>>  C:D=53:47
>>
>>  There are still cycles (e.g. A > B > C > A,
>>  D > B > C > D).
>>  C:D=53:47 is the lowest defeat in a cycle,
>>  thus C > D is ignored.
>>
>>  Now, we have:
>>
>>  A:B=67:33
>>  A:C=35:65
>>  B:C=68:32
>>  B:D=23:77
>>
>>  There is still a cycle (A > B > C > A).
>>  A:C=35:65 is the lowest defeat in a cycle,
>>  thus A < C is ignored.
>>
>>  Now, we have:
>>
>>  A:B=67:33
>>  B:C=68:32
>>  B:D=23:77
>>
>>  Now, we have no cycles anymore.
>>  But there are two candidates (A and D), who
>>  each wins every not-ignored head-to-head pairing.
>>  Who should win?
>>
>>By the way, as far as I know, the successive ignorance
>>of weakest defeats has already been proposed by
>>Condorcet. In "Essai sur la probabilite des decisions;
>>rendues a la pluralite des voix" (1785), Condorcet wrote:
>>
>> Create an opinion of those n*(n-1)/2 propositions, which
>> win most of the votes. If this opinion is one of the n*(n-1)*...*2
>> possible, then consider as elected that subject, with which this
>> opinion agrees with its preference. If this opinion is one of the
>> (2^(n*(n-1)/2))-n*(n-1)*...*2 impossible opinions, then eliminate
>> of this impossible opinion successively those propositions, that
>> have a smaller plurality, & accept the resulting opinion of the
>> remaining propositions.
>>
>>But then, Condorcet found that this formulation leads to the
>>above mentioned problem. That's why in later writings
>>Condorcet proposed a method identical to Tideman's method.
>>
>>Markus Schulze
>>
>>
>
>