# [EM] Seven +/- Two

Forest Simmons fsimmons at pcc.edu
Thu Sep 20 17:34:40 PDT 2001

Roy,

Thanks for the positive feedback and the good idea about graduating only
the current Smith set from one stage to the next, so that the natural
attrition in candidates reduces the computational burden in a potentially
intractable method.

Thankfully, the excellence of your suggestion doesn't depend on the

On ballots of the following two types A > B is counted for the set {A,B}
but not for the set {A,B,C} (in Martin Harper's Universal Approval
reckoning):

C >> A > B
A > B >> C

If A beats B in the size two set reckoning only because of ballots like
these, then even one ballot of the following type could bring about B
beats A in the size three set reckoning:

B >> C > A

If these were the only three ballot types, and each type made up fewer
than half of the total ballots, then there would be no Condorcet Winner in
the size two set reckoning, since A beats B beats C beats A.

In that case the entire set is the Smith set, so all candidates graduate
to the size three set reckoning which picks the Approval winner B.

suggestion in the least.

Besides reducing the computational burden your suggestion may have the
effect of changing the method into one that satisfies Reverse Symmetry.

Forest

P.S.  On a side note, it is interesting to me that most of the methods
that I like best reduce to Approval completed Condorcet (Demorep sans 50%
quota) in three way contests.

I have a lot of respect for Demorep's long standing promotion of ACMA.

Perhaps a theorem is lurking there: A deterministic (no coin tossing)
method that specializes in three way contests only, while satisfying the
Condorcet Criterion and the FBC must be some version of Approval completed
Condorcet.

What qualifying clauses would make this statement into a theorem?

On Thu, 20 Sep 2001, Roy wrote:

> Forest Simmons wrote:
> > of size two. If there is no CW, then do "the count" on subsets of
> > size three. If there is no "beats all" candidate on that basis,
> > then go to subsets of size four. If it doesn't terminate before
> > reaching the whole set count, then it will terminate at that stage
> > because that stage is equivalent to Approval which doesn't suffer
> > from the problem of cycles.
>
> Groovy. The complexity could be somewhat reduced by only advancing
> members of the Smith Set. A win of A>B eliminates B as a possible
> winner in any set containing both A and B, right?
>