# [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'
>
> -----------------------
>
> 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

```