At 06:34 24.02.00 , Craig Carey wrote:
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>Voting against candidates using negative votes.
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>
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>At 23:05 23.02.00 , Donald E Davison wrote:
...
To some extent, any election method can be made to handle negative
votes, just by preprocessing the data using P2. That would keep negative
numbers out of the method. However for methods that are not so well
designed (...), the sets of winners may be different.
Suppose the ballot counts are as follows (with the candidates being
A,B,C,D,E):
A a0
AB ab0
ABC abc0
ABCD abcd
ABCE abce
ABD abd0
ABDC abdc
ABDE abde
ABE abe0
ABEC abec
ABED abed
AC ac0
ACB acb0
... ...
Let g1=min(a0,b0,c0,d0,e0)
Let g2=min(ab0,ac0,ad0,ae0, ba0,bc0,bd0,be0, ca0,cb0,cd0,ce0, ,da0,... ,...)
Let g3=min(abc0,abd0,abe0, acb0,acd0,ace0, adb0,adc0,ade0, aeb0,aec0,aed0,
bac0,bad0,bae0, bca0,bcd0,bce0, bda0,bdc0,bde0,... ,... edc0)
Let g4=min(abcd,abce,abde, acbd,acbe,acde, adbc,adbe,adce, aebc,aebd,aecd,
bacd,bace,bade, bcad,bcae,bcde, bdae,bdce,bdce,... ,... edba)
Then for each of {g1,g2,g3,g4}, a constant is added to very paper from which
they were derived, just to make none of the ballot papers have a negative
count. The example would be generalised.
An example. This election
A. 2
AB -5
B. 6
C. 7
can be changed into this:
A. 2
AB 0
AC 5
B. 6
BA 5
BC 5
C. 7
CA 5
CB 5
That could be changed into this (if allowed):
A. 2
AC 5
B. 16
C. 17
-------------------------------
Correction
>A check of STV against P2 with votes allowed to be negative.
...
> A. a0-2t+u, a = a0-x
> AB -x+t
> AC t
> B. b+u
> C. c+u
>
>"Quota" = Q = (a+b+c+3u)/K. "Sum" = S = (a+b+c+3u).
>K=3 in "Droop-STV", and K=2 in "Hare-STV".
...
>Case (b+u<Q)(c+u<Q)(Q<a+u):
...
> YBC = 0 < c - b + x(a+b+c+3u)/a
> = 0 < a.c - a.b + x(a+b+c+3u) (provided 0<a)
...
> I much suspect that YBC (which decides whether the winners are
> {A,B} or {A,C}) is not dependent on u, despite it containing u.
I guess that "is independent" statement ought be constrained by saying
that all ballot counts need to be positive.
Mr. G. A. Craig Carey
