Gervase,

Thanks for taking time to explore.  And nice text graphics for the clock!

It turns out that as long as you allow only strict rankings, the center of gravity of the distribution will fall in the the four hour (i.e. 120 degree) sector of the clock face centered on the Borda winner, so that's why I didn't suggest using the center of gravity approach.

And yes, it does generalize naturally to higher dimensional spheres.

But, because of "sprucing up" I am mostly interested in the three candidate case.

My Best,

Forest

From: Gervase Lam <[EMAIL PROTECTED]>
Subject: Re: [EM] Clock Methods (for Three Candidates)

<snip>


A clock method for three candidates is a method that assigns winners (or
winning probabilities) based on the distribution of ballots around the
clock.

As you mentioned later, you can use this to work out the Kemeny-Young Ranking for a 3 candidate election. A while back, I used Kemeny to analyse the following:

4: A>B>C
3: B>C>A
2: C>A>B

I thought I would use the clock method in order to analyse it this time to
see what I would get.

                A
                .
      A>C>B .       .[4] A>B>C

  Not(B) .              . Not(C)

C>A>B [2].                . B>A>C

       C .              . B

      C>A>B .       .[3] B>C>A
                .

              Not(A)

(One thing that I did not realise until I made up this diagram was that
the ballots are equally opposed of each other.)

The Kemeny Ranking is A>B>C.  To a certain extent, this is not surprising.

Imagine the above were a spinning-top, with masses 2, 3 and 4 in the
relevant places on the spinning-top.  My guess is that the position of the
spinning-top's centre of mass would make in tip near Not(C).  This is the
boundary between A and B winning.  It is quite close.

I tried to calculate exactly where the spinning top would tip.  I thought
could get something like a "centre of mass" using polar co-ordinates.  I
couldn't.  I couldn't be bothered to use cartesian co-ordinates to work
out the centre of mass of the spinning-top and therefore to where the
spinning-top would tip....

What if this were a 2-winner election?  I think the problem would be like:
Find the position of two equal masses on the spinning-top that would cause
the spinning-top to tip almost like how the spinning-top in the above
diagram would tip.  The position of the two masses may be at A, B or C.

This could be extended further.  What if the clock were a sphere instead
(i.e. a 3D clock)?  I am not very good at visualising these things, so,
how many candidates can be accommodated this way?

Like the spinning-top, what I imagine is that the centre of mass of the
sphere would make it "tip" at the winning candidate in the 1-winner case.
To find N-winners, find the arrangement of N equal masses on the surface
of the sphere that matches the "tip" the closest.

However, there may be many such arrangements.  So I would only allow
arrangements where the N masses are positioned as far apart as possible.
This makes the N-winners represent a broad "spectrum" of opinion.

Thanks,
Gervase.

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