# [EM] New (?) Condorcet method: LeastSchwartzBeat

Ernest Prabhakar drernie at mac.com
Fri Jan 9 00:05:05 PST 2004

```Hi all,

I've been working on how best to implement a Condorcet-compatible
algorithm such as Ranked Pairs (or should I say  MAM?) and SSD.   I
think I agree with Mike that the most important critieria -- given that
several algorithms are about equally good -- is simplicity and
transparency, so that it can be explained to ordinary voters (and
politicians!)

I've come up with an algorithm which is similar to MAM and SSD, and
easy to implement (and I hope explain) but I'm not sure if its
identical, or has the same beneficial properties. I call it
LeastSchwartzBeat. Perhaps someone here could help me analyze it.

The basic steps are:

1.   Calculate the pairwise 'beats' between each pair of candidates

2.  Identify a 'least beaten' candidate (L)
- The one with the greatest number of wins
- If several have the same # of wins, pick the one which beats the
others
- If several with the same # of wins form an internally unbeaten cycle,
pick one at random

3.  Collect all candidates which recursively beat L
- e.g., if A & B beat L, C beats A, F & G beat B, and  C, F & G are
beaten by L, then the set = A, B, C, F, G, L
- I believe this is identical to the Schwartz set - is that correct?

4.  If L is the only member, that's the Condorcet winner, and the
algorithm ends.  Otherwise:

5.  For each candidate in the Schwartz set, calculate his/her lowest
beat against other members of the Schwartz set (LeastSchwartzBeat, or
LSB)

6.  The candidate with the highest LSB in absolute votes is the winner.
still tied, compare the Next Least Schwartz Beat, etc.  If all their
beats are equal, declare a tie.

While the terminology might be a little confusing, this would be easy
to demonstrate graphically, and to even calculate manually (given the
Condorcet matrix, and a reasonably small Schwartz set).   I think it
would seem intuitively reasonable to most people,

I have a suspicion this is close to RP (or MAM), in that  you are
effectively locking the first candidate to get all four beats.  The
only area where I think it differs is how it handles 'inner loops'
within the Schwartz set, but I'm not sure how much that matters.

Can anyone help me figure this out?

Thanks,
- Ernie P.
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