>robert bristow-johnson <r...@audioimagination.com>
(please ignore my horrible editing in webmail.)
>hi,
>this is my first post to this list. i subscribed to it a while ago after some
>internet searching on
>issues regarding multi-candidate elections which i did after our recent
>mayoral race in Burlington
>Vermont (dunno if you heard about it or not).
Welcome!
>supposedly the Schulze beatpath method is supposed to be the most optimal for
>theoretical
>reasons (i understand what the goal of the Schulze method is, but not all of
>the mathematical
>steps), but it is also important to have a deterministic and monotonic measure
>of voter support
>that is understandable to the less scholarly.
Schulze Beatpath is an excellent method, and I'd be happy if it were adopted,
though optimality
depends on what criteria you prefer. For example, I also like Kemeny-Young,
though having the number
of calculations go up as N! (N being the number of candidates) tends to be a
problem. :)
>> Step 1: For each ranked ballot, create a matrix for each pairwise vote,
>> based on the distance and direction between each candidate. For example,
>> on the ballot A>B>C, you would get:
>>
>> -2 -1 0 1 2
>> AB 0 0 0 1 0
>> BA 0 1 0 0 0
>> AC 0 0 0 0 1
>> CA 1 0 0 0 0
>> BC 0 0 0 1 0
>> CB 0 1 0 0 0
>like Borda, i wonder what eternal principle determines the weightings you've
>assigned. all we
>know is this voter likes A better than C and B better than C. but we do not
>know how much more
>this voter prefers B over C. maybe this voter thinks that both B and C are two
>pieces of crap, but
>flipped a coin as to which was worse (which is the main reason i see no
>reason, whether it's
>Condorcet or IRV, that ties shouldn't be allowed, even ties for 1st choice).
>or maybe the voter
>only thinks that C is a piece of crap and flipped a coin between A and B.
>there is no way to know
>(without range voting which i also don't like).
Actually, the method described allows ties, I just didn't want to make my
initial ballot even more
complicated (grin). Here are the thirteen possible ballots with three
candidates, including ties (but
excluding range type ballots):
1: A>B>C
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 0 0 0 1
CA 1 0 0 0 0
BC 0 0 0 1 0
CB 0 1 0 0 0
1: B>C>A
-2 -1 0 1 2
AB 1 0 0 0 0
BA 0 0 0 0 1
AC 0 1 0 0 0
CA 0 0 0 1 0
BC 0 0 0 1 0
CB 0 1 0 0 0
1: C>A>B
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 1 0 0 0
CA 0 0 0 1 0
BC 1 0 0 0 0
CB 0 0 0 0 1
1: A>C>B
-2 -1 0 1 2
AB 0 0 0 0 1
BA 1 0 0 0 0
AC 0 0 0 1 0
CA 0 1 0 0 0
BC 0 1 0 0 0
CB 0 0 0 1 0
1: C>B>A
-2 -1 0 1 2
AB 0 1 0 0 0
BA 0 0 0 1 0
AC 1 0 0 0 0
CA 0 0 0 0 1
BC 0 1 0 0 0
CB 0 0 0 1 0
1: B>A>C
-2 -1 0 1 2
AB 0 1 0 0 0
BA 0 0 0 1 0
AC 0 0 0 1 0
CA 0 1 0 0 0
BC 0 0 0 0 1
CB 1 0 0 0 0
1: A>B=C
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 0 0 1 0
CA 0 1 0 0 0
BC 0 0 1 0 0
CB 0 0 1 0 0
1: B>C=A
-2 -1 0 1 2
AB 0 1 0 0 0
BA 0 0 0 1 0
AC 0 0 1 0 0
CA 0 0 1 0 0
BC 0 0 0 1 0
CB 0 1 0 0 0
1: C>A=B
-2 -1 0 1 2
AB 0 0 1 0 0
BA 0 0 1 0 0
AC 0 1 0 0 0
CA 0 0 0 1 0
BC 0 1 0 0 0
CB 0 0 0 1 0
1: A=B>C
-2 -1 0 1 2
AB 0 0 1 0 0
BA 0 0 1 0 0
AC 0 0 0 1 0
CA 0 1 0 0 0
BC 0 0 0 1 0
CB 0 1 0 0 0
1: B=C>A
-2 -1 0 1 2
AB 0 1 0 0 0
BA 0 0 0 1 0
AC 0 1 0 0 0
CA 0 0 0 1 0
BC 0 0 1 0 0
CB 0 0 1 0 0
1: C=A>B
-2 -1 0 1 2
AB 0 0 0 1 0
BA 0 1 0 0 0
AC 0 0 1 0 0
CA 0 0 1 0 0
BC 0 1 0 0 0
CB 0 0 0 1 0
A=B=C
-2 -1 0 1 2
AB 0 0 1 0 0
BA 0 0 1 0 0
AC 0 0 1 0 0
CA 0 0 1 0 0
BC 0 0 1 0 0
CB 0 0 1 0 0
(I hope I did that right.)
Theoretically, you could have a range-type ballot as well. For the range 0-4,
you might have:
A=4
B=3
C=0
For this ballot, you would have a matrix like
-4 -3 -2 -1 0 1 2 3 4
AB 0 0 0 0 0 1 0 0 0
BA 0 0 0 1 0 0 0 0 0
AC 0 0 0 0 0 0 0 0 1
CA 1 0 0 0 0 0 0 0 0
BC 0 0 0 0 0 0 0 1 0
CB 0 1 0 0 0 0 0 0 0
Now, I have not attempted to figure out what adding range-like ballots might
look like. The
results may be incredibly bizarre. Which actually sounds rather fun, come to
think of it.
I'll close for now, since this is a very long post that will probably make
everyone's eyes glaze over.
(Sorry!) I might add something when I'm not at work. :)
Michael Rouse
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