# [EM] Fix of 1/4 quota 'Q' AV method (was: IFPP for more than three candidates?

Craig Carey research at ijs.co.nz
Mon Sep 17 16:41:28 PDT 2001

```At 01.09.17 17:41 +0200 Monday, Markus Schulze wrote:
>Dear Rob,
>
>of course, this IFPP generalization also violates the monotonicity
>criterion. When 3 CABD voters are changed to ACBD in your example,
>then the winner is changed to C.
>
>Situation 1:
>
>  10:A>B>C>D
>  15:B>A>C>D
>  23:C>A>B>D
>
>  The quota is 12. A and D are eliminated in the first round, then
>  B beats C.
>
>Situation 2:
>
>  10:A>B>C>D
>  15:B>A>C>D
>  20:C>A>B>D
>   3:A>C>B>D
>
>  D is eliminated in the first round. In the second round, the quota
>  is 16 so that A and B are eliminated and C wins.
>

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

Stage 1:  eliminate {A,D} to give:
25  B
23  C       :   Winners = {B}

That ought be made wrong. Example 1 can be plotted in a tetrahedron and it
is desired to put the already solved 3 candidate solution into the d=0 face
(given that the votes are all non-negative). Maybe two quotas are needed.
(1) The "1/4" quota that marks A and D out as losers, and less easily passed
test saying when A's votes are to be transferred.

----

Reduce the support for C using alteration: 3:(C) -> 3:(AB)

Example 2:
13  AB      :   Total = 48;  48/4 = 12;  48/3 = 16
15  B       :   Candidates under the "1/4" quota are: {D}
20  C
0  D

Stage 1:  eliminate only {D} to give:
13  AB      :   The "1/3" quota = 48/3 = 16
15  B
20  C

Stage 2:  both A and B are under the 48/3 (=12) quota. Eliminate both.
Thus the 'Q' method winners = {C}.

So the method does not comply with P1.

Generalise the method.   Assume 1 = 1+b+c+d

AB   a     :  a = ab
B    b
C    c
D    d

Ex.1:   (d<a<1/4<b<c)(c<a+b)           => (B wins)
Ex.2:   (d<1/4<a<b<c)(a<1/3)(b<1/3)    => (C wins)

Let a=1/4 and change some '<' into '<='. Find the intersection, G, of the
2 regions on the flat a=1/4:

G = (a=1/4).(d<1/4<b<c)(c<a+b).(b<1/3)

The boundary inside G is tilted wrongly since it lies on (1/4=a),
i.e. (a+b+c+d<4a). I.e. [(a=1/4).G] == approx ==> [(C wins) = (b+c+d<3a)]
---------------

Suppose d = 0. then
Aw = (1/3<a)[(b<1/3) or Tcba][(c<1/3) or Tbca]
Bw = (1/3<b)[(c<1/3) or Tacb][(a<1/3) or Tcab]
Cw = (1/3<c)[(a<1/3) or Tbac][(b<1/3) or Tabc]

Aw = (b+c<2a)[(2b<a+c) or Tcba][(2c<a+b) or Tbca]
Bw = (c+a<2b)[(2c<b+a) or Tacb][(2a<b+c) or Tcab]
Cw = (a+b<2c)[(2a<c+b) or Tbac][(2b<c+a) or Tabc]

Tcba = (b+cb<a+ca)  :  Tbca = (c+bc<a+ca)

Aw = (b+c<2a).[(2b<a+c) or (b+cb<a+ca)].[(2c<a+b) or (c+bc<a+ca)]

A   a0      DA   da0
AB  ab      DAB  dab
AC  ac      DAC  dac
B   b0      DB   db0
BA  ba      DBA  dba
BC  bc      DBC  dbc
C   c0      DC   dc0
CA  ca      DCA  dca
CB  cb      DCB  dcb
D    d0

If possible, aim for having D simply 'eliminated' when terms are small.
So aim for this:
Aw = (b+db+c+dc<2(a+da)).
[(2(b+db)<a+da+c+dc) or (b+db+cb+dcb<a+da+ca+dca)].
[(2(c+dc)<a+da+b+db) or (c+dc+bc+dbc<a+da+ba+dba)]

Can't see a nonmonotonicity problem with that expression, which resulted
from 100% eliminating with transferring, the votes of D. In the style of
AV and IFPP, it seems to be safe to simply eliminate a candidate if its
quantity of votes is small enough.

However the boundaries project inwards and not towards the centre of the
simplex and they need to bend and end up at the centre. The bend could
be complex. For a while, the best that would appear would be an
approximate method.
----------------------------------

Guess at a 2nd quota for the 1st stage. The 2nd stage is made to be the
applying of the 1 winner 3 candidate IFPP formula.

Example 1 had these numbers. Want to deem A and D losers and transfer
votes from D but not A (in the 1st stage):

10  AB   a = 10/48 = 0.2083
15  B    b = 15/48 = 0.3125
23  C    c = 23/48 = 0.4792
0  D    d = 0

Aim: find an algorithm that decides to eliminate D and not A.

1/3<a , 1/3<d.  No.

(2a-b-c) = -0.3751
(2a-b-d) =  0.1041
(2a-c-d) = -0.0626

(2b-a-c) = -0.0625
(2b-a-d) =  0.4167
(2b-b-d) =  0.3125

(2c-a-b) =  0.4376
(2c-a-d) =  0.7501
(2c-b-c) =  0.1667

(2d-a-b) = -0.5208
(2d-a-c) = -0.6875
(2d-b-c) = -0.7917

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.

There is still a detail of stage 2 finding the winner to be a candidate deemed
to have lost in stage 1 but that was not eliminated in stage 1. Another detail
is that with that checked or fixed, the method would doubtless getting the
winner wrong sometimes.

G. A. Craig Carey
Auckland

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