[EM] Condorcet vs Approval

[Forest Simmons]

Thanks to Martin Harper and Craig Layton for valuable critiques.

In particular, Martin is right. The voters should be able to make
distinctions among their unapproved candidates, too.

Here's a more ideal version of a compromise between Condorcet and
Approval, which could be considered a dyadic refinement of Approval.
A voted ballot would look like this:

ABC > DE >> F > GHIJ >>> KLMNOP > QRS >> TUVW > XYZ

Or expressed differently

111 ABC
110 DE
101 F
100 GHIJ
011 KLMNOP
010 QRS
001 TUVW
000 XYZ

Candidates are compared head-to-head as in Condorcet.  Candidates in the
same group get scores of zero against each other. When candidates are in
distinct groups, the more approved candidate (the one further to the left
in the first representation or with the higher binary label in the second
representation) gets a score corresponding to how big the biggest
inequality symbol is that separates the two candidates being compared. In
the second representation, the score would be the place of the most
significant digit in which the two binary labels differ.

For example, between C and F, the biggest inequality is a double
inequality >> , so C beats F by two points.  In the binary labels for C
and F the most significant digit in which they differ is the second digit,
so again, C beats F by two points.

In this example, any letter before K in the alphabet beats any letter
after J by three points because to get from one group to the other you
have to cross the triple inequality >>> and (from the other point of view) 
all such pairs differ in their most significant digit which is the third
binary place (from the right). (Significance increases from right to
left.) 


The >>> marks the fundamental boundary between approved and unapproved.
the >> marks the boundary between a refined approval within those two
groups.  Similarly, the > gives further refinement.

For strategy, figure the >>> location first, as you would in normal
approval.  Then figure each >> as you would in normal approval if the only
candidates running were the ones in the group to be divided by the >> .
Similarly, put in each > by pretending nothing exists outside the group
being divided.

That's it except for deciding how to break Condorcet cycles. I'm still not
sure of the best method, but one possibility is to first ignore the single
inequalities, and if that doesn't do it, then ignore the double
inequalities.

I have some other ideas for avoiding cycles which I will explain later.

Obviously, this method can be extended to many levels of refinement. For
practical purposes, I think three would be maximum. However, theoretical
purposes abound, so we shouldn't neglect it for cumbersomeness alone.

Forest