[EM] RE : clone immunity definitional problems

[Kevin Venzke]

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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