[EM] Clock Methods

[Forest Simmons]

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 <gervase.lam at group.force9.co.uk>
> 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.
>