Erratum Re: [EM] (P1) and monotonicity for single-winner election systems and Condorcet.

[Craig Carey]

Mr Catpole's theorem is far from being as bad I had written.
I said the rule failed to pass 2 candidate FPTP. However, the
 W(V)<>W(V') is the left of the "implies".

I wrote a REDUCE program to see what 3 candidate methods were 
 passed and failed by the rule Mr Catchpole wrote and which
 is immediately below. The problem was solved using symbolic
 algebra. REDUCE is much slow when doing 19 for loops.
What I found was that the rule failed FPTP, IFPP, and Borda,
 for both 1 & 2 winners. The rule passed Condorcet (1 and 2
 winners) with the regions where Condorcet picks the wrong number
 of winners, being made regions where the rule held (and
 similarly for ties, in all methods). To get that result, I had
 made Pi be Condorcet pairwise comparison. Doing the created
 a rule that failed methods including FPTP, and passed Condorcet.




At 11:46 26.10.99 , Craig Carey wrote:
>
>A method of Mr Catchpole that falls apart over two candidate election
> is studied here.
It didn't. I apologise to Mr Catchpole. It needs 3 candidates before
 the rule to be found to be be of no great value. The rule seems to be
 a way to test if the method Pi-like nature if a pairwise comparison is
 done. There is no need to write "not Pi(c2,c1,V)" instead of "Pi(c1,c2,V)"
 presumably. 

(The following has been edited and A and B have been replaced with c1 and c2,
 and "[" and "]" have been added.)

>
>At 19:46 25.10.99 , David Catchpole wrote:
>>(Further down...)
>>
>>On Sat, 23 Oct 1999, Craig Carey wrote:
>>
>>> >Let Pi(c1,c2:V) represent the truth of whether voter i prefers candidate c1
>>> >over c2 in voting schema V.
>>> >
>>> >Let W(V) represent the set of winners of voting schema V
>>> >
>>> >For all V, all V', W(V)<>W(V') implies [
>>> >
>>> >there exists i, c1, c2 such that c1 is an element of W(V), c1 is not an
>>> >element of W(V'), c2 is an element of W(V'), c2 is not an element of
>>> >W(V), and-
>>> >
>>> >(i) not Pi(c2,c1,V) and Pi(c2,c1,V')
>>> >or
>>> >(ii) Pi(c1,c2,V) and not Pi(c1,c2,V') ]

An example showing that the rule fails FPTP. c1=B, c2=A

V, A:B:C = 33:34:31, FPTP Winner = B
   A   19
   AB  5
   AC  9
   B.  8
   BC  22
   BA  4
   C.  5
   CA  18
   CB  8
condorcet(1, 33,5,9, 34,22,4, 31,18,8) = {c} = Winner
[A<B] = (-9), [B<C] = (1), [C<A] = (-16)

V'=U, A:B:C = 51:47:1, FPTP Winner = A
   A   2
   AB  48
   AC  1
   B.  3
   BC  43
   BA  1
   C.  0.2
   CA  0.6
   CB  0.2

condorcet(1, 51,48,1, 47,43,1, 1,0.6,0.2) = {a} = Winner
[A<B] = (-22/5), [B<C] = (-93), [C<A] = (8)

>
>>
>>> Pi hasn't been defined. It can't assumed that the words "the truth of" mean
>>>  Condercet because introducing that would be a most arbitrary step and
>...
...
>
>>Consider voter i as an element of the set of voters. Pi(A,B,V) is true iff
>>(if and only if) voter i prefers candidate A over candidate B in scenario
>>V. (More further down)
>
>What is Pi?. Unnecessarily adding the words, "the truth of", and then with
...
>My guess is that you would make Pi to be Condorcet pairwise comparison.
>But that has not been stated. 
>
>
>>Nothing's wrong... thorough checks convince me there is nothing to
>>criticise in this definition. ...

Thorough checks using what method?: this rule of yours seems to fail
 far too many methods. Pi not defined (and "i" is not used in the
 formula).

The rule is approximately useless. A rule ought not pass too many or too
 few methods, and impose some useful quality.

It is a very hard to understand way to test if a method is the
 Condorcet method, but that's Pi is a Condorcet pairwise comparison.

>
>V =
>   99 C
>   03 D
>
>V' =
>   99 C
>   01 D
>
>-------------------------------------------------------------------------
> "For all V, all V', W(V)<>W(V') ... there exists i, A, B such that A is

I'd deleted the word "implies"

>  an element of W(V), A is not an element of W(V'), [and] ...
>-------------------------------------------------------------------------


load_package rl;  % load up REDLOG
rlset ofsf;

...
procedure winset(nw); begin
   win:={};
   wa:=rlsimpl wa;  wb:=rlsimpl wb;  wc:=rlsimpl wc;
   if not('false = wa) then win:=append(win,{if 'true = wa then a else {a,wa}});
   if not('false = wb) then win:=append(win,{if 'true = wb then b else {b,wb}});
   if not('false = wc) then win:=append(win,{if 'true = wc then c else {c,wc}});
   if nw = 1 then begin
      wbad:=rlsimpl not((wa and not wb and not wc) or (wb and not wc and not wa)
                     or (wc and not wa and not wb));
   end else if nw = 2 then begin
      wbad:=rlsimpl not((wa and wb and not wc) or (wb and wc and not wa)
                                       or (wc and wa and not wb));
   end else begin
      wbad:=true; wa:=false; wb:=false; wc:=false;
      write "WinSet: Error with nw";
   end;
   write "win = ",win;
   return win;
end;

............


met:=7;
% eliminate the 18 variables using rlqe
rlqe rlall cp(va,vab,vac,vb,vbc,vba,vc,vca,vcb,ua,uab,uac,ub,ubc,uba,uc,uca,ucb);
% if rlqe returns false then rule failed method and then search for
% a numerical example. If true then no example can be found.


G. A. Craig Carey