[EM] Ignore the margins!
Roy
royone at yahoo.com
Wed Aug 15 14:18:38 PDT 2001
Consider this margin-of-victory matrix, where positive values have
been replace with a +:
A B C D X
A - + - - -
B - - - - -
C + + - - -
D + + + - -
X + + + + -
It is still possible to know the complete ranking, just by looking.
The one with the most wins, wins. Then the one with the next-most,
etc.
Now consider this matrix:
A B C D X
A - + - - -
B - - + + -
C + - - + -
D + - - - -
X + + + + -
We still know who wins, but we do not know who comes next. The order
of precedence forms a loop, and when it is broken, one of the
candidates is going to have to be at the bottom. Let's say we choose
A. How many preferences will that disregard? Only those who voted for
A>B. If we chose B, we would be ignoring B>C and B>D, which,
regardless of the number of A>B, B>C, and B>D votes, is a larger
number of preferences discarded than if we just discard A>B.
Clearly, we want candidates who have two victories to be ranked above
those who only have one. In general, the candidates can be grouped by
how many opponents they have defeated, and the new matrices look like
this:
one victory < two victories < four victories
A D B C X
A - - B - +
D + - C - -
Those with two victories outrank those with only one, so we have a
complete ranking, without any concern for margins of victory:
X>B>C>D>A
There are still situations where the margins come into play. If you
have a cycle wherein all candidates have the same number of wins
against each other:
A B C
A - + -
B - - +
C + - -
you still have to choose a bottom. The logical choice is the smallest
margin. If two tie for smallest margin, choose the largest margin to
be the top, instead. The chosen one is ranked below (or above, if you
chose a top) the rest of the matrix, and the matrix can re-drawn to
continue ranking.
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