[EM] maybe a new variant of Condorcet

[peter barath]

When some weeks ago I started to read electo pages,
I was both fascinated an overwhelmed by the information.
One day at home thinking I realized I didn't know whether
in "clone-proof" and other phrases "clone" means a
candidate which relates (wins/ties/loses) to every
other candidate as its "clone-companion", or (which
would be a narrower definition) a candidate which
is placed next to its companion on every vote.

The next day at my workplace by continuing the surf
I realized that the latter case is true, but it was
too late. I created a (maybe) new variant for
single-winner elections, which I name quasi-cloning.

(Let me note that for single winner, I very quickly
became pro-Condorcet.)

I call a subset of candidates a quasi-clone set, if:

1. they don't make up the whole set of candidates
2. for every candidate out of the set they are in
the same winning relation with (all beat / all tie /
all lose)

(You can ask why to make the subsets at all, but I think
this Rubicon is already crossed with the Smith-set,
which is a special kind of quasi-clone sets.)

When evaluating a vote, we make up the most coarse
quasi-clone distribution of the whole candidate set.
Then compute the winner quasi-clone set. Then consider
the winner set as a whole candidate set and if it
contains more than one candidate, do the procedure
again for it. And so on, while it is necessary.

To run the contest between the quasi-clone sets, I
propose the Schwarz method. The number of members
in a quasi-clone set, of course, can vary from set
to set, so to measure winning stengths, we may
use some average of margins, winning votes, or
proportions or whatever we want.

It's a question whether the most coarse (the one
with the least number of members) distribution is
unique. I found that in most cases it is.

Let's take a case where candidate x is a member
of both A and B (different) quasi-clone sets. If one
of the two is a strict part of the other, only
the bigger one "has the right" to take part in
the most coarse distribution. If they both have
at least one member outside the other, their
union have to be related uniformly to outsiders,
so, it "has the right".

But what if the union of A and B constitutes the
whole candidate set? In that case: let's call
Ap the part of A outside B; let's call Bp the
part of B outside A and let's call M the common part.

Let a be a candidate in Ap. If x (which is in M)
beats a, then (beacuse M is part of B) anyone in B
must beat a, therefore anyone in Bp must beat anyone
in A. In this case the most coarse distribution is not
unique, but in any case the winner subset will be Bp
sooner or later.

The case is similar with a beating x. If a ties x,
it makes unsolvable ties anyway.

(If all this is hard to follow, draw it on paper.)

Peter Barath

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