# [EM] Saari and Cyclic Ambiguities

Alex Small asmall at physics.ucsb.edu
Thu Mar 28 17:07:20 PST 2002

```I'll begin this message concerning Saari's insights by saying that, for the
record, I am AGAINST the Borda Count for real-world elections (although I
understand it has engineering and sports applications).  That said, let me
share a thought:

Saari pointed out that cyclic ambiguities come from a "Condorcet profile"
or "symmetric profile".  If the electorate consists of 3 groups

35 A>B>C
33 B>C>A
32 C>A>B

we can "decompose" the electorate into

32 A>B>C  +  3 A>B>C
32 B>C>A  +  1 B>C>A
32 C>A>B

The total electorate has no Condorcet winner.  The "residual" does.

It's my understanding that, in general, if there is no Condorcet winner,
subtracting out the symmetric profiles will leave a "residual" profile that
DOES have a Condorcet winner.

Caveat:  If a Condorcet winner does exist, subtracting out the symmetric
profile might yield a different Condorcet winner.

So, this leads me to propose the following possible election method:

1)  If a Condorcet winner exists then simply elect him.  End of story.
2)  If not, eliminate from all ranking orders any candidate NOT in the
Smith set.  So, if A, B, and C are in the Smith set, a person whose
preference order is A>D>C>B is now considered to have the preference order
A>C>B.
3)  Having narrowed down the list, subtract out the symmetric profiles.
4)  There will now be a "Condorcet winner."  Elect him.

A few weeks ago I discussed Saari's insight, and DEMOREP suggested that
discarding ballots like that is of dubious merit.  Here's my defense:

Suppose that there's a conference on the mathematics of voting, and being
the election geniuses that we are we all give invited talks.  Ten of us
decide to go out to dinner together.  After we've delivered our talks, nine
people are waiting for me to show up (I'm running late, because I had to
call my fiance).

Joe suggests that those present pick a restaurant.  The results are:

3 Thai>Pizza>Steak
3 Pizza>Steak>Thai
3 Steak>Thai>Pizza

We have an exactly symmetric profile.  Demorep suggests approval voting,
but at that minute I walk through the door.  I maintain that _if_ we only
go by preference orders, then my vote should decide all.  Not because I'm
special, but because if you're only one vote away from an exact tie, that
single vote must be the deciding factor for any reasonable method.

Since my preference order is Thai>Pizza>Steak I maintain that Thai should
be the winner.  With the method I proposed above Thai would win.  With SSD
the weakest defeat is Thai vs. Steak (margin of 2).  All other defeats have
a margin of 4.  SSD would hence select Thai.  Not knowing much about other
methods I can't say which others would pick Thai.

So, it isn't such a stretch to conclude that cyclic ambiguities should be
resolved by whatever ballots are in excess of a tie.

Of course, Demorep's idea of using approval to pick the winner (presumably
from among the members of the inner-most unbeaten set in the general case)
is far easier.  I merely throw this out for examination.  I haven't thought
out what flaws it might have.

Alex

```