Positive and Negative preferences (was Re: [EM] MC -1, 0, 1 --- STV hybrid

Craig Carey research at ijs.co.nz
Thu Mar 6 00:53:01 PST 2003



Is that something like this?:

(+1 for A 1st; +1 for B 2nd; -1 for C 3rd; ...)

That could be written as (+A, +B, -C).

The leading +s can be missed:

If C loses and no other information is given, then that
can be written as (A B- -C)

A generalized montonicity rule would imply:

   Prohibited:  (A C B+) <---> (A B- C)

(It says: shifting B to the left doesn't cause it start losing)

For the negatively orientated preference this could be done:

. . . .Allowed:  (A C +B+) <---> (A C -B-)
Hence: Allowed:  (A C B+ ) <---> (A -B- C)

For the power<=1 rule that enforces One Man One Vote:

The set of Winners intersected with {A,B,C} when this paper is
added:

    1(A -B C)

can be reproduced when these papers are added:

    w.(A C) + x.(A) + y.(C) - z.(B)

for some w+x+y+z=1, 0<=(w,x,y,z).

This could be defined x.(A B C D) = -x.(-A -B -C -D).

So the negation of a paper p is paper p with the same weight but
with the preferences negated.

The IFPP Duality principle (which produces the same method when
switching winners with losers whilst also negating weights) has
an interpretation.

The mix of the power<=1 (no-corruptness) power rule and duality,
could lead to unconvincing alteration-examples.

It is the strange problems with power that may arise that could
be of concern. I have not considered the problem.

When all the preferences are positive and there are only 3
candidates, power has nothing to impose after monotonicity has
been imposed. It is not obvious at the outset if examples
showing a failure to constrain power would show up if there are
only 3 candidates. The 3 candidate case can't be that difficult.

For P2, the smudging of the weight over the next preference
could have the added next preferences be either all +1 preferences
or else all -1 type preferences.

-------------






At 03\03\06 09:45 +1030 Thursday, Chris Benham wrote:
 >This is my first posting here. I have just composed  a very expressive
 >(elect-one person) election method , and not as a joke.
 >Voters number  candidates in order of preference, and also mark
 >candidates as Approved  and up to
 >number of candidates minus 2  as  Disapproved (or  Unacceptable).
 >1. A candidate who is ranked number 1 on more than half  the ballots wins.
 >2. If no result, if there is one and only one candidate who marked as
 >Approved by the majority, then that candidate wins. If the set of
 >candidates Approved by the majority has more than one member, then all
 >non-members are eliminated and their preferences are transferred. If  a
 >candidate now has a majority  then that candidate wins. If  not, then
 >the numbered rankings are abandoned, and the winner is the candidate
 >with the highest  Approvals minus  Disapprovals tally.
 >3. If  there were no candidates who were Approved by the majority, then
 >eliminate all candidates marked as Disapproved by the majority and
 >transfer their preferences  and proceed as before.
 >If  a candidate has a majority they win. If no result, then abandon the
 >numbered rankings  and elect the candidate with the highest  Approvals
 >minus Disapprovals score.
 >A vote with as little as just the number 1 or a single Approval or
 >Disapproval will be valid. The restriction on the number of
 > Disapprovals is mainly to make it impossible for  all the candidates to
 >be Disapproved  by the majority.
 >I  think  that there should be few strategy problems. I  am  interested
 >in any comments, but I don't promise to (immediately) understand them.
 >               Chris Benham
 >

That is not all that clear and an example could have got more to take
a meaning from the text.


Craig Carey




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