[EM] Multiwinner participation rule. Geometric descriptions

Craig Carey research at ijs.co.nz
Sun Dec 12 06:58:10 PST 1999


At 09:57 12.12.99 , Markus Schulze wrote:
...
>Markus Schulze wrote (11 Dec 1999):
>> At 03:03 20.10.99 , Craig Carey wrote:
>> > At 22:57 19.10.99 , Markus Schulze wrote: ...
>> > > Do you question that some election methods sometimes punish voters
>> > > for going to the polls and voting sincerely? Or do you question that
>> > > a voter rather wants to have no influence on the election result than
>> > > to worsen the election result?
...
>Suppose that candidate A is elected if voter V1 doesn't go to the polls.
>Suppose that voter V1 strictly prefers candidate A to candidate B.
>Suppose that candidate B is elected if voter V1 goes to the polls.
>Then voter V1 is punished.
>
>Can you give a different definition of "punish"?
>
>Markus Schulze

Regarding the word "punish", my answer would be "no" provided the topic is
 is about papers being (partially or fully) discarded or added, and the
 number of winners equals one.



I'll simplify that "multiwinner participation axiom" definition I put
 in a previous message.

I had written this:
-------------------------------------------------------------------------
Definition: (Q2), "multiwinner participation (axiom)" rule. (11-Dec-99):

(For All V)(For All p)(For All Q)[
   [(For All q)[(q in Q) => (p = trunc(q,length(p)))]] =>
      (For All t) [(0<=t)(#t=#Q)(#(p.W(V))=#(p.W(V+t*Q))) .=>
         (Satisf(W(V),p) <= Satisf(W(V+t*Q),p))]]
-------------------------------------------------------------------------

If 'SPC' holds then Q can be removed.
SPC says that a candidate's win-lose state is unaffected by subsequent
 preferences.
For example, p=(ABC), Q=((ABCD),(ABCE)), t=(2,5)
   V+t*Q=             V+(1*t)p=
   2 ABCD               7 ABC
   5 ABCE                S
    S
In these 2 examples, if A (or B, or C) wins one then it wins the other.
 Similarly for B and C but not D
 and E.
Hence p.W(V+t*Q) = p.W(V+(1*t)p), in this case:
  {A,B,C}.W(V+t*Q) = {A,B,C}.W(V+7p)

Quite probably those persons that would want to impose a participation
 axiom would also impose SPC too. If SPC is not imposed then the rule
 has been made weaker.

Also the rule can be altered and made more restrictive, by altering
 #(p.W(V))=#(p.W(V+tp))  into  #(p.W(V))<=#(p.W(V+tp)).  That would
 would require that the satisfaction value increases when the number
 of winners increases.

-------------------------------------------------------------------------
Definition: (Q2), "multiwinner participation axiom" rule. (13-Dec-99):

(For All V)(For All p)(For All t) [
   (0<=t)(#(p.W(V))<=#(p.W(V+tp))) .=>
      (Satisf(W(V),p) <= Satisf(W(V+tp),p))]]
-------------------------------------------------------------------------

The Satisf function can be removed.

(Satisf(W,p) <= Satisf(X,p)) = (For All j=1..length(p)).[ G(W,X,p,j) ]

G(W,X,p,j) = ((W.trunc(p,j-1) = X.trunc(p,j-1)) => (W.{p[j]} <= X.{p[j])})

("<=" is subset, so: (W.{x} <= X.{x}) = ((x in W) => (x in X))

The rule as defined may not be beyond being improved.



Here are some example alterations. (Papers are not remarked.)

Write a n:A+B+CDEF to mean that the paper (ABCDEF) has weight n and that
  {A,B} = ({A,B,C,D,E,F}.Winners).

Let Z(p) = [#(p.W(V)), #(p.W(V+tp)); Satisf(W(V),p), Satisf(W(V+tp),p)]
The rule fails the particular alteration if this is false:
   (z1<=z2)=>(z3<=z4).

(0:AB)--(9:AB+): Z(AB)=(0,1; 0,1/4), Allowed
(0:AB+)--(9:AB): Z(AB)=(1,0; _,_), Allowed
         Allow the adding of AB to causes B to start losing

(0:ABC+DE+FG)--(9:ABC+DE+FG+): Z(ABCDEFG)=(2,3;5/32,5/32+1/128), Allowed
(0:ABC+DE+FG+)--(9:ABC+DE+FG): Z(ABCDEFG)=(3,2; _, _), Allowed
(0:A+BCD)--(9:AB+C+D+):
  Z(AB)  =(1,1; 1/2, 1/4), Not allowed
  Z(ABCD)=(1,3; 1/2, 1/4+1/8+1/16), Not allowed

I do consider this rule to be worth imposing. It doesn't actually
 stop voters being punished when going to voting booths, because
 Mr Schulze's participation axiom allows  (0:AB+)--(9:AB).

Mr Schulze, are satisfied with my comments on geometry?.
I haven't heard much about Condorcet in quite a while (in this
 mailing list).



-------------------------------------------------------------

Mr Catchpole wrote "Regularity is cool". I believe that that indicates
 Mr Catchpole ignored or did not get (or read) my proof that his
 "Regularity" idea is a useless and valueless and impossible to satisfy.
 I shall repeat the comments I made:


At 20:48 09.12.99 , Craig Carey wrote:
...
>1. GEOMETRIC ARGUMENT Allowing REJECTION Of The 'Regularity' Rule:
...
>(2) near (a,b,c)=(1/3,1/3,1/3) (the triangle centre), the {B},{C} boundary
>    is b=1/3.  No problem near the centre of the triangle.
>
>An unfixable problem is that the two lines, b=1/2 and b=1/3, have to be
> joined, and that can't be done without the rule being violated. So it ...
...

The problem is joining the parallel lines. The rule doesn't allow them
 to be joined.

The periphery of the B-wins region (for a system with papers re (AB), (B),
 (C)) should be closed. The rule doesn't allow that.

No less alarming (not the same as "cool") is how the rule rejects FPTP for
 being insufficiently lacking in proportionality. If the papers were such
 that FPTP would have a tie between B and C, then this regularity rule
 would require the divide between B and C be (Zb+Zac<Zc+Zab) (if Zac was
 small enough and the papers were {Zab:AB, Zb:B, Zc:C} and Zac=0).


                                 + + + + +
                            E-mail reader programs

I got 6 e-mail messages from Mr Catchpole (via the mailing list program)
 and I argue for a case that less would have been better.

Here's the messages:

Message 1: ---------------------------------------------------------------
At 13:25 12.12.99 , David Catchpole wrote:
>On Wed, 8 Dec 1999, Craig Carey wrote:
>
>> >The addition (removal) of a candidate does not, for any other candidate, 
>> >increase (decrease) the probability of that other candidate winning.
>> >
>> 
>> That is clear enough. It is not something people will agree is desirable
...
>No, here we're talking about non-deterministic voting systems where some
>form of "lottery" is involved. For instance, an example of a

By message 4, probability theory seemed to be redundant.

Message 2: ---------------------------------------------------------------
Subject: Re: FPTP family theory, REDLOG shadowing
At 13:31 12.12.99 , David Catchpole wrote:
>PS. I've decided to call it the Saari Octahedron from now on, seeing as
>though it really isn't cubic or even rhomboid.
>

Message 3: ---------------------------------------------------------------
Subject: RE: The family of "regular" probabilistic (stochastic) electoral systems
At 13:48 12.12.99 , David Catchpole wrote:
>On Wed, 8 Dec 1999 DEMOREP1 at aol.com wrote: ...
>> D-
>> A simple example-
>> 
>> 2 A>B>C
>> 1 C>B>A
>> 
>
>well, without C: p(A)=2/3, p(B)=1/3
>without B: p(A)=2/3, p(C)=1/3
>without A: p(B)=2/3, p(C)=1/3
...

Six unexplained assertions. They need to be explained.

Message 4: ---------------------------------------------------------------
Subject: RE: The family of "regular" probabilistic (stochastic)  electoralsystems
At 14:08 12.12.99 , David Catchpole wrote:
>Ja, vat ist das problem? Obviously, A's removal will _increase_  B's
>probability of winning to unity. Regularity is cool with this- it would
...
>> 
>> 1 A
>> 1 B
>>...

If the probability of a a single candidate winning a 'one winner one
 candidate' election is exactly one ("unity"), then isn't the probability
 of a winner winning any election exactly "unity"?. I presume every
 subscriber has this figured out.

Message 5: ---------------------------------------------------------------
Subject: RE: The family of "regular" probabilistic (stochastic)   electoralsystems
At 14:12 12.12.99 , David Catchpole wrote:
...
>
>Well, it's not a cube per se, but rather an octahedron formed by the
>intersection of 
>
>1<=x+y+z<=2
>
>and
>
>0<=x,y,z<=1

What are x and y and z ? (they were not defined).

Message 6: ---------------------------------------------------------------
Subject: Re: FPTP family theory, REDLOG shadowing
At 14:17 12.12.99 , David Catchpole wrote:
>Saari also uses a triangle, but the triangle doesn't go far enough in an
>analysis of what's a possible configuration of votes and what's not.

The words "possible configuration of votes" suggests a simplex (and nothing
 else). But Mr Catchpole makes a blunder, uses the word octahedron.
 Surely Mr Saari would prefer student Catchpole tried harder.

>That's where the Saari octahedron kicks in in his analysis.
>


Craig Carey, 13 December 1999




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