# Perfect Method Criteria

Steve Eppley seppley at alumni.caltech.edu
Thu Aug 8 14:24:24 PDT 1996

```Mike O wrote:
-snip-
>As I've said though, some refinements, in the form of
>anti-order-reversal options, which I've hinted at on this list,
>without stating a definite definition--some such refinements may
>well be able to make Condorcet's method near-perfect enough so that
>we could say that it virtually meets those 2 perfect method
>criteria. But I can't say that for sure, and I haven't even
>completely defined those refinements. I believe that's the next
>place to look, if anyone wants to find still better 1-balloting sw
>methods.
-snip-

Ok, here's an attempt which I haven't put much thought into.  It's
a variation of Condorcet which automatically elevates one's lesser
of evil somewhat in one's ballot if the greater evil is winning.
Kind of a last-ditch attempt to "approve" the compromise as much as
one's true favorite, when it's necessary based on the actual votes
cast.

It's an iterative Condorcet where after each round, the nonequal
rankings of candidates preferred more than the round-winner are
converted to equal rankings.  The iterations proceed until there are
no more ballots that can be converted this way.

To illustrate, suppose there's a 3-candidate election:  L vs M vs R.
Suppose that after the initial round, the winner is L.  Then at the
start of the next round all the {R>M>L} and {M>R>L} ballots are
converted to {R=M>L} and a new round-winner is calculated.

If someone objects that this ballot manipulation might not be
appreciated by every voter, this could be dealt with by including
in every ballot a checkbox where the voter could grant or deny
permission.

Here's an example where plain Condorcet fails the strict LOE
principle:
24: A>B>C
22: A>C>B
10: B>A>C
10: B>C>A
34: C=B>A
This is an election which B wins by plain Condorcet, and according
to my formulation of the strict LOE principle, changing ballots from
B>C or C=B to C>B must result in either B or C winning.  But in this
example, changing the 34 C=B votes to C>B will result in A winning by
plain Condorcet.

The iterative process I mentioned will not let A keep the win, though:
24: A>B>C
22: A>C>B
10: B>A>C
10: B>C>A
34: C>B>A
A wins round 1:
A<B 46<54
C<A 44<56
B<C 44<56
The ballots convert to:
24: A>B>C
22: A>C>B
10: B>A>C
10: B=C>A       <---  these changed
34: C=B>A       <---  these changed
B wins round 2 (since C no longer beats B):
A<B 46<54
C<A 44<56
C<B 22<34
The ballots convert to:
24: A>B>C
22: A=C>B       <---  these changed
10: B>A>C
10: B=C>A
34: C=B>A
B wins round 3:
A<B 46<54
A<C 34<44
C<B 22<34
No more ballots can be converted.

That was just an example in which the iterations "succeeded" in
electing B.  I haven't tried yet to come up with a "bad" example
showing how this method might violate some important criteria or
standards.  Maybe it even violates LOE in other scenarios.
Unfortunately, I'm not very good at developing instructive examples.

* *

It might also be interesting to see how this iterative technique
would affect other methods, like Instant-Runoff-1:
46: A>B>C
20: B
34: C>B>A
A wins round 1 (since B is eliminated, then C).
The ballots convert to:
46: A>B>C
20: B
34: C=B>A   <--- these changed
B wins round 2 (since C is eliminated, then A).
No more ballots can convert, so B is the winner.

* *

Another iterative technique which might work would be similar to
the above, except only convert nonequal rankings to equal when the
nonequal rankings are part of the lesser choice's largest pair-loss
in the round.  In other words, if A wins the round, convert C>B>A to
C=B>A if B's largest pair-loss is to C.

If it turns out that some iterative LOE technique works well,
perhaps the Smith filter will no longer make a difference and can