FPTP family theory, REDLOG shadowing

[Craig Carey]

At 06:33 15.12.99 , Markus Schulze wrote:
>Dear Craig,
>
>you wrote (13 Dec 1999):
>> Mr Schulze, are satisfied with my comments on geometry?
>
>Unfortunately, there are still many questions waiting to
>be answered.
>
>******
>
>Craig Carey wrote (10 Dec 1999):
...
>The reason why so many paradoxa stay undiscovered (e.g.: The Dodgson method
>has been proposed in 1873; but the first person who discovered that the
>Dodgson method violates monotonicity was Fishburn in 1977.) is the fact
>that many paradoxa occur only when the number of candidates is large
>(e.g.: The Dodgson method violates monotonicity only if there are at least
>five candidates.). [Another example: Ossipoff's subcycle rules violate
>Pareto only if there are at least six candidates.]
>
>It seems to me that a problem of your geometrical interpretation of election
>methods is the fact that -if you want to define an election method implicitely
>by describing its simplex- you have a different simplex for every possible
>number of candidates so that even if you have proven the absence of a given
>paradox for up to 102 candidates (or whatever your favorite number of
>candidates was) you still don't know whether this paradox occurs for 103
>candidates unless you have also defined and investigated its simplex for 103
>candidates. Do you agree?

(If the boundaries were placed in a Simplex, it would be explicit rather
 than implicit.) The 102 vs 103 argument is flawed in implying people
 can't construct inductive proof with 100 formula in a sequence.
"Do you agree?" No. There did not seem to be a paradox there.

These simplexes can be just created into existence by the results of
 algebra. For example: if a method satisfies SPC, then the 4 candidate
 method can be simplified down from 40 vertices (39 dimensions) to a 6
 vertex simplex in 5 dimensions. Surely there is a better term than
 "SPC". What about 'STV-truncation-resistance'?.

>******
...
>I don't remember that ever somebody said "that pairwise comparing
>of two candidates had some mathematical importance." Those who support
>Condorcet methods do this because their supported Condorcet method meets
>monotonicity and is very difficult to manipulate by running additional
>candidates. It never happened that anybody said that he prefers a
>Condorcet method simply because it is a pairwise comparison method.

Surely they hold a belief that if A beats B in a pairwise comparing
 then A ought beat B in the election.

>Do you want to say that pairwise methods are eo ipso worse than
>non-pairwise methods? If the answer is "Yes!": Does your criticism
>of Condorcet methods also include those methods that on the one side
>meet the Condorcet criterion but that on the other side do not depend
>only on the matrix of pairwise comparisons (e.g. the Dodgson method)?

It seems to me obvious that if A beats B in a pairwise comparing, there
 is no good reason why A should win before (or if) B does.
 

>******
>A weaker version of the participation criterion has been
>introduced by Fishburn and Brams [Peter C. Fishburn, Steven J. Brams,
>"Paradoxes of Preferential Voting," Mathematics Magazine, vol. 56,
>p. 207-214, 1983].

Those two have a page on a method called the "Approval Vote":
  http://bcn.boulder.co.us/government/approvalvote/altvote.html
The page says "simple". It is a bit more than simple.

>You wrote (10 Dec 1999):
>> I do not know if the AV is passed or failed by the participation
>> axiom. It will be passed in the 3 candidates case. An interesting
...

>It has already been proven by Fishburn and Brams that Alternative
>Voting violates the participation criterion even in the 3 candidate
>case.

There rule isn't know to me perhaps.
My formula allowed the last preference to be considered.
Not that allowing full preference lists instead of lists with
 at least the last preference missing, will lead to FPTP
 in the 3 candidate 1 winner case. That is not obvious from
 you folowing example:

>Example:
>
>  7 voters vote A > B > C.
>  6 voters vote B > A > C.
>  8 voters vote C > B > A.
>  
>  If Alternative Voting is used, then candidate A is elected.
>  But if three of the eight CBA voters didn't go to the polls,
>  then candidate B would be elected.
>
Adjust the example and have the problem in IFPP rather than
 STV. It is going to be the case that IFPP nc=3 will be robust
 against problems/objections (but that can bechecked):

7 A
6 BA
7 CBA
AV: A
Q=6.66, IFPP: A

7 A
6 BA
4 CBA
AV: B
Q=5.66, IFPP: B

(4:CB+A)--(7:CBA+)  Fail
(4:CB+)--(7:CB {A+) Pass

So it turns out that the rule would seem to be wrong, rather than
 STV/IFPP. The boundaries can be found as follows...

Plot the boundaries in the triangle (a 2D simplex) and label the
 corners:

   A

BA   CBA

B wins (a:A, b:BA, c:CBA) if and only if it wins this:
  c:CB
  b:B
  a:A
C has its FPTP shaped region and B region expands into the A region
 without the slopes exceeding those of the BA-CB and A-CB edges.
 
Similarly, C wins (a:A, b:BA, c:CBA) if and only if it wins this:
  b:BA
  a:A
  c:C
So now the B & C wins region of the original problem is known.

The divide between BA and CBA is  a single line segment and the
 other two divides are each 2 lines segments.
Etc.

Hence if it is said that FPTP is not good enough, then the
 participation axiom as Markus and I stated it, cannot consider
 what happens to the n-th preference when there are n candidates.
 So Brams et all may have defined it wrong or wanted to fail the
 'axiom'.

I now consider the participation axiom to be uninteresting...,
 t is likely to start doing something that (P1) doesn't, only when the
 systems are large enough that the correct winners are neither obvious
 nor known.




Craig Carey