[EM] Clock Methods (for Three Candidates)

[Gervase Lam]

> Date: Sat, 22 Jan 2005 16:37:12 -0800 (PST)
> From: Forest Simmons
> Subject: [EM] Clock Methods (for Three Candidates)

> Take a clock face and put labels A, B, and C at 12:00, 4:00, and 8:00,
> respectively. At 2:00, 6:00, and 10:00 put the labels not(C), not(A),
> and not(B), respectively.
>
> Then on the intervals between the hour marks put the labels
>
>     A>>B>C (between 12:00 and 1:00),
>     A>B>>C (between 1:00 and 2:00),
>     B>A>>C (between 2:00 and 3:00),
>     B>>A>C (between 3:00 and 4:00),
>     B>>C>A (between 4:00 and 5:00), etc.

> 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.