[EM] Condorcet vs Approval

[Forest Simmons]

One other thing.  In a zero information election, start by expressing your
utilities in binary rounded to three binary digits, this takes you
directly to the second representation of the dyadic refined approval
ballot below, bypassing the strategic <<<, <<, and < boundary
calculations. 

Forest


On Fri, 2 Mar 2001, Forest Simmons wrote:

> 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
> 
>