[EM] IFPP for more than three candidates?
Markus Schulze
markus.schulze at alumni.tu-berlin.de
Tue Sep 18 09:45:56 PDT 2001
Dear Craig,
you wrote (18 Sep 2001):
> One possibility is to eliminate candidate d only if (2d<a+b)(2d<a+c)(2d<b+c).
> 'Adding' the 3: (2d<a+b)(2d<a+c)(2d<b+c) => (6d<2(a+b+c)) = (d<1/4). Thus:
> (1/4<d) => not[(2d<a+b)(2d<a+c)(2d<b+c)]
> A candidate is not eliminated in stage 1 when it has more an 'eliminate quota'
> full of votes.
>
> -----------------------
>
> Apply that to the 1st example, Example 1:
>
> 10 AB : Total = 48; 48/4 = 12; 48/3 = 16
> 15 B : Candidates under the "1/4" quota are: {A,D}
> 23 C
> 0 D
>
> Only need to consider the candidates under the '1/4' quota:
> (2d<a+b) = (0<25)
> (2d<a+c) = (0<33)
> (2d<b+c) = (0<38) : All 3 are True, so eliminate D in stage 1
>
> (2a<b+c) = (20<38) = True
> (2a<b+d) = (20<15) = False
> (2a<c+d) = (20<23) = True : Not all 3 are True, so A is not eliminated.
>
> Eliminating D gives:
> 10 AB : Quota = 48 / 3 = 16
> 15 B : A and B are under the quota so candidate C wins.
> 23 C
>
> Example 2 proceeds just like before, so C wins that. Since the winner does not
> change, P1 (monotonicity) is held.
Situation 1:
7:A>B>C>D
3:D>A>B>C
15:B>A>C>D
23:C>A>B>D
The quota is 12. D is eliminated since (2*d=6 < 22=a+b), (2*d=6 < 30=a+c)
and (2*d=6 < 38=b+c). A is eliminated since (2*a=14 < 38=b+c),
(2*a=14 < 18=b+d) and (2*a=14 < 26=c+d). Then B beats C.
Situation 2:
7:A>B>C>D
3:D>A>B>C
15:B>A>C>D
20:C>A>B>D
3:A>C>B>D
The quota is 12. D is eliminated since (2*d=6 < 25=a+b), (2*d=6 < 30=a+c)
and (2*d=6 < 35=b+c). A is not eliminated since (2*a=20 > 18=b+d).
In the second round, the quota is 16 so that A and B are eliminated and C
wins.
IFPP still violates monotonicity.
Markus Schulze
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