[EM] Binary Tree Method

Forest Simmons forest.simmons21 at gmail.com
Wed Sep 13 06:43:17 PDT 2023


This method starts with a list of candidates and organizes them into a
binary tree whose leaves are the listed candidates.

It is convenient to consider each candidate as an embrionic tree with the
candidate itself as both the only leaf and the only node, so that at each
stage we have a list of trees. Then we can proceed (at each stage) by
joining the two trees in the updated list that together form the
"strongest" tree that can be formed by joining two such trees (and there by
updating the list).

So at each stage the number of trees goes down by one as two of them are
joined into one.

Therefore in n-1 stages the candidates are completely organized into a
binary tree.

All we need now is a way to gauge the "strength" of a tree.

There are certainly many ways to do that but it seems to me that a fruitful
way might be to consider the ability of the leaf set of one branch to cover
or nearly cover the leaf set of the other.

So now the question becomes ... how do we gauge the ability of one set of
candidates to cover or nearly cover another?

Suppose the two sets are H and K. The ability of H to cover or nearly cover
K, denoted by cov(h,K), may be defined as follows:

First, for each h in H, let cov(h, K) be defined as max over k in K of
[h>k], where [h>k] represents the number of ballots on which candidate h
outranks candidate k.

Then cov(H,K) is defined by the min over h in H of cov(h, K).

So the very first join in the binary tree construction will be the joining
of the two candidates h and k such that [h>k] is maximal, i.e. the pair
where the defeat of k by h is strongest (when gauged by winning votes).

We should establish a convention to  locate the stronger branch to the left
of the weaker branch at each join.

If we do so, then we can find the winner of this method by starting at the
root node of the completed tree and taking the left branch at each node,
until we reach one of the original leaves ... i.e. one of the candidates in
the original list.

I have to stop here for now, but examples are coming soon!

fws
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