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

Ernie Prabhakar drernie at mac.com
Fri Jan 9 00:58:02 PST 2004

```Hi again,

Sorry, I shouldn't try to write at midnight - I realized I got step 5
backwards. Of course, I may be making an even worse mistake now that's
1 AM....

Rather than LeastSchwartzBeat, I think I actually want
LeastSchwartzBeatBy (LSBB).   That is, the candidate which loses to
other Schwartz set members by the smallest amount, will be the first
one to have all the beats against him or her dropped, and is thus the
winner.    If a candidate has more than one beat against them, then use
the largest (since that's the last dropped).

However, I can't (at this hour) figure out if this method is equivalent
to SSD or RP (MAM), or even how exactly the two differ.   Help!

Thanks,
- Ernie P.

On Jan 9, 2004, at 12:04 AM, Ernest Prabhakar wrote:

> 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.  If there's a tie in absolute votes, compare relative votes.
>  If 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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