[EM] RE : clone immunity definitional problems
Kevin Venzke
stepjak at yahoo.fr
Sat Jan 27 12:23:01 PST 2007
Warren,
I suggest you use whatever definition makes it possible to prove
something interesting.
--- Warren Smith <wds at math.temple.edu> a écrit :
> Problem1: Say in a vote A=B and then we clone B. Can we get the new
> vote B1>A>B2?
I believe the answer should be no, since A is voted "between" the clones
as under Mike's definition.
> Problem2: Say in a vote A=B and we clone A=B1=B2. That can convert an
> election with 50-50
> win probability into 33-33-33. But in a probability-conserving clone
> definition,
> Prob(A wins) = 50 should be unaltered by cloning B.
By Woodall's definition the win probability of candidates outside the
new clone set is never allowed to *increase*. Only when cloning a
candidate with some win probability may the win odds of other candidates
decrease.
> Perhaps the answer is to say that A and B were ALREADY cloned... but if
> so you have to make the definition reflect that.
I don't think this is promising. The definition of "to clone" would
have to specify that other candidates can't qualify as clones of the
new candidates.
> Markus Schulze's pdf paper he cited gives a hairy definition of
> "independence of clones" in section 4.5 page 53.
>
> Trying to decode it, he defines "cloning B" to mean (in the below, A,B,
> and C are distinct):
> * iff A>B in in an old vote in the old election, then A>(all B clones)
> in the new one.
> * iff B>A in in an old vote in the old election, then (all B clones)>A
> in the new one.
> * iff C>A in in an old vote in the old election, then C>A in the new
> one.
> * Schulze does not say anything explicitly about equalities but some
> facts can be deduced
> because he used "iff" rather than "if." (By the way, it might be
> better merely to use "if"
> because "iff" may lead to insurmountable problems...). Here are
> deduced facts:
> * iff A=B in in an old vote in the old election, then A=(all B clones)
> in the new one.
> That is a very strong demand by Schulze, and one I feel should be
> avoided if we can avoid
> it.
Hmm, does it really demand this? I think it only demands that if A=B
and you clone B, A cannot be ranked strictly above or strictly below
the entire set of clones.
That is A=B could become A=B1>B2 or B1>A=B2 or maybe even B1>A>B2, but
not A>B1>B2 or B1>B2>A.
Kevin Venzke
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