Date: Mon, 14 Feb 2005 01:46:38 +0000 From: Gervase Lam <[EMAIL PROTECTED]>
<snip>
A . A>C>B . .[4] A>B>C
Not(B) . . Not(C)
C>A>B [2]. . B>A>C
C . . B
C>B>A . .[3] B>C>A .
Not(A)
[C>A>B was written at the 7 o'clock position. I have now corrected it to C>B>A.]
I don't think the centre of mass is on B.
The total of all the votes is 2 + 3 + 4 = 9.
Draw a vertical and horizontal lines so that they are perpendicular to each other and cross each other at the centre of the circle. With respect to these lines, the co-ordinates of C>A>B is obviously (-1, 0).
The A>B>C and B>C>A voters are respectively 60 degrees above and below the horizontal line. Therefore, their x co-ordinates are respectively cos +60 and cos -60 (i.e. 1/2 and 1/2). Also, their y co-ordinates are respectively sin +60 and sin -60 (i.e. sqrt(3)/2 and -sqrt(3)/2).
The x co-ordinates of each of the rankings is weighted by the number of votes cast for each ranking. This is then used to work out the weighted mean x co-ordinate of the clock or spinning-top.
((2 * -1) + (4 * 1/2) + (3 * 1/2)) / 9 = 3/18
The weighted mean y co-ordinate of the spinning-top is worked out in the same way except that the y co-ordinates of the rankings are used instead of the x co-ordinates.
((2 * 0) + (4 * sqrt(3)/2) + (3 * -sqrt(3)/2)) / 9 = sqrt(3)/18
Therefore, the co-ordinates of the centre of mass is (3/18, sqrt(3)/18)).
Now to work out the angle of the line from the centre of the circle to the centre of mass:
Tan [Angle] = (sqrt(3)/18) / (3/18) = sqrt(3)/3 = 1/sqrt(3).
Therefore, the angle of the line is 30 degrees above the horizontal line. This means the line points at Not(C).
This is not nice in this case because the voters are equally split between candidates A and B. There is nothing between them.
Here's another way to arrive at this same result:
The respective Borda counts of A, B, and C are 10, 10, and 7.
Think of A, B, and C as being vectors, and find the resultant of
10A+10B+7C
Which can also be expressed as 3(A+B)+7(A+B+C) which equals 3(A+B) since A+B+C=0 by symmetry.
Of course A+B is the same as not(C), so the resultant vector points in the direction of not(C), as you also found.
If we interpret not(C) as a toss up between A and B, then this result makes sense.
The problem with Borda (hence this center of gravity approach) is that it is clone dependent.
But sprucing up removes the clone dependence and reduces public elections to the three candidate case, so this method is still worthy of consideration.
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