[EM] Improvement on Jobst's Chain climbing method

[Forest Simmons]

A few years back Jobst suggested "chain climbing" as a seamless, Condorcet
compliant way of selecting an alternative from a given ordered list.

For example electing a winner from a list of candidates c1, c2, ... given
in decreasing order of approval.

Chain Climbing initializes a chain of candidates with the last (least
approved in this case) candidate in the list.  Then moving up the list each
successive candidate "climbs the chain" as far it can before being bumped
off by a chain member that defeats it. If it makes it all of the way to the
top, it is added to the top of the chain.

The candidate who ends up at the top of the chain is elected.

Since a beats all candidate will never be defeated, the method is Condorcet
compliant. It also turns out to be clone resistant and monotonic.

Another nice property is that it always selects from the Banks set, a nice
game theoretic subset set of the set of uncovered candidates.

The biggest objection to this method is that when applied to a list a list
where c1 beats c2 beats c3, and c3 beats c1, it elects c2.

Here's my proposed improvement:

Initialize the chain with c1.  Move down the list instead of up.  For each
successive candidate x (as we move down the list) if possible, insert that
candidate into the chain at a point where it is beaten by every candidate
above it and is not defeated by any candidate below it.  If not possible,
discard it.

After going through the entire list (top to bottom) inserting new
candidates where possible into the totally ordered chain, we end up with a
maximal totally ordered chain of candidates (ordered by pairwise defeat)
The candidate at the top fo the completed chain (the one who is not
defeated by any of the others) is elected.

It is easy to show that this method has all of the nice properties of chain
climbing, but retains more of the spirit of the original list..

For example in the A>B>C example above it elects A.

What do you think?

Forest
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