[EM] Symmetric cyclic AB+BC+CA rule rejects STV

[Craig Carey]

At 20:42 08.03.00, Norman Petry wrote:
 >List members interested in reading some nonsense about the merits of Borda's
 >method might be interested in:
 >
 >http://www.colorado.edu/education/DMP/voting_b.html
 >
 >In this paper, Saari argues against Condorcet's method by providing the
 >following "bad example":
 >
 >ABC 29
 >BAC 28

-----------------------------------------------------
At 20:48 08.03.00, Norman Petry wrote:
 >That should have been:
 >
 >ABC 20
-----------------------------------------------------

 >
 >so far, so good.  He then introduces what he considers a group of "confused
 >and irrational" voters with the following (cyclic) preferences:
 >
 >ABC 9
 >BCA 9
 >CAB 9
 >
 >When these voters are added to the election the results are:


:... Saari seems to
 >think that a good method should discount the effect of the addition of a
 >contrived group of ballots which happen to form a symmetric voting cycle.
 >Presumably, such sets of ballots represent a kind of "noise" input which the
:counting system should cancel, if possible.  ...
...


The idea might be a lot sounder if [Mr Saari] had instead expected
  invariance of the winners in 3 candidate election had 6, not 3, papers been
  added. I.e. these papers: ABC, BCA, CAB, ACB, CBA, BAC.
(That would be just ruled implied by rule P2).


Both STV and IFPP are failed by this test when it is applied to only those
  ratios of ballot counts where the 2 methods find the exact same winners.

Proof: A 1 winner 3 candidate example:

AB 3  | 3
BC 0  | 0
C  2  | 2    IFPP & STV Winner = A

Add (2:AB + 2:BC + 2:CA):

AB 5  | 5
BC 2  | 2
C  2  | 4    IFPP & STV Winner = C,  (IFPP Quota=3.666...)
CA 2

The test would require A = C.

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Here is a little REDLOG REDUCE program that I used to find the above
  example:

load_package rl;  % Load REDLOG into REDUCE (running in PSL Lisp).
rlset ofsf; % mode = logic of inequalities over real numbers
n:=9;
for v:=0:5 do for a:=0:n do for b:=0:n do for c:=0:n do begin
  wa := (b+c<2a)and((b<a+u)or(2b<a+c))and((c+u<a)or(2c<a+b));
  p:=sub({u=0},wa);  q:=sub({u=v},wa);  % substitute u in wa with numbers
  q1:=(a+b+c)/3;  q2:=q1+v;
  r:=(p and not q);  % the winner A must change
  if ('true = rlsimpl r) and (a+b+c+v)<=9
     and ( (a>q1 and b>q1 or b>q1 and c>q1 or c>q1 and a>q1) %IFPP=STV
        or (a>q2 and b>q2 or b>q2 and c>q2 or c>q2 and a>q2) )
     and (not (a=b+c or b=c+a or c=a+b))  % exclude STV internal draws
     and (not (a=b+c+v or b=c+a+v or c=a+b+v))
  then write "{(a+b+c+v),{a,b,c,v}}= ",{(a+b+c+v),{a,b,c,v}}
end;