[EM] P4 sketched: a constraining of powers of papers

[Craig Carey]

At 22:33 04.03.00 , Bart Ingles wrote:
>
>Craig Carey wrote:
>
>> The weights of the Approval Voting method may as well be
>>  multiplicatively normalised so as to give each voter equal
>>  power. However that seemed to be opposed.
>
>As I understand your definition of power, this would turn approval
>voting into cumulative voting, or equivalently FPP in a single candidate
>election.
>
>
:> Regarding "power", a person should be able to, as exactly as
>>  possible, oppose the vote of a neighbour that lives over the
>>  fence or any judge across the street.
>
>Can you show an example in approval voting where a single voter is
>unable to cancel the vote of any other single voter?
>

The paragraph does no make the statement I intended. Something like
 this could be ruled out:
    If the prime minister across the road cast an Approval Vote using
     a single sub-vote, then other voters should not be able to both
     neutralise that and other votes at the with a single vote.

No votes should have more influence than a FPTP/SNTV vote (a vote with
 no 2nd preferences).


I sketch an outline of some different ways to define this idea. This
 is a definition of a rule named P4.

A P4 rule is probably one that I would not regard as optional: in
 any deriving of IFPP I would need it. I have guessed I would need it
 just to derive a solution to the 2 winner 4 candidate formula.

P4 reject the Approval Voting method without rejecting the possibility
 that the rule itself would be rejected for rying to make voters have
 equal influence.

Has someone defined a rule constraining powers of preferential voting
 papers already?. I suppose there is a chance that this P4 won't
 be in complete agreement with what results from applying
 "Min(Sum[max-Vi],Sum[Vi-min])" to Borda (normalizing weights).


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

Here is a check that shows that P1 does not imply this P4 rule.

  A P4 rule would prevent the following:

   - In the collection of papers, neither A nor B win.
   - If the (1:AB) is added, then both A and B then win. [In the Approval
      Voting method, this is the same as adding (1:A + 1:B).]
   - Candidate A loses when the vote added is (x:A + y:B), x<0.9; 0<=y.
   - Candidate B loses when the vote added is (y:A + x:B), x<0.9; 0<=y.
   - There are 80 candidates.
   - The number of winners is 1 or more the same in all cases.

P1 does not prevent that. Apply P1 in all the cases where it may say
 something:

[E1]: A loses (1:AB) implies A loses x:A+y:B,  all 0<=x+y<=1, 0<=x,y
[E2]: B loses (1:AB) implies B loses x:A+y:AB, all 0<=x+y<=1, 0<=x,y
[E3]: A loses (y:A) implies A loses 1:AB,  for all 1<=y
[E4]: B loses (y:B) implies B loses 1:AB,  for all 1<=y

[E1]: A wins x:A+y:B  implies A wins (1:AB), all 0<=x+y<=1, 0<=x,y
[E2]: B wins x:A+y:AB implies B wins (1:AB), all 0<=x+y<=1, 0<=x,y
[E3']: A wins 1:AB implies A wins (y:A), for all 1<=y
[E4']: B wins 1:AB implies B wins (y:B), for all 1<=y

Equations E1,E2,E1',E2' are true and do not rule out the example, and
 equations E3,E4,E3',E4' are irrelevant to the example and do not
 prohibit it.

My P2 would be introducing at least another 75 other papers. And P3
 says the number of winners must be correct.

So a new rule could be defined.

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

(From the viewpoint of imagined STV-style transfer values, the Approval
 Voting method has transfer values that are larger than 1, and it has
 negative vote wastage).



This P4 rule could be worded like this:

 Definition of P4-1:
   'A vote p can always be replaced with votes that do not have second
   preferences and that have a sum of weights that is not greater than
   the weight of vote p, and that cause the same winners, of those
   named in preference list p, to be elected.

 Restated:
  (All t, p)(All U)(Exists X) [ (All g)(g=sign(t)).(All c)(0<=g.X(c)) .
     (g.(Sum(c))X(c)<=g.t).(p.W(U + t:p) = p.W(U+X(A):(A)+X(B):(B)+...)]

   * "X(A):(A)" is a vote for candidate A, and the vote has no 2nd
      preference and it has weight equal to X(A). #X = the number of
      candidates
   * g is 1 or -1.

 Example:
   There exists some x,y,z, (x+y+z<=4)(0<=x,y,z), that allows
     {A,B,C}.Winners(U + 4:ABC) = {A,B,C}.Winners(U + x:A + y:B + z:C)

The definition P4-1 doesn't seem to apply to enough papers at once.


Second attempted definition: P4-2: Stripping off the last preference

 Example:
   For all U, x, y, (0<=x,y)(x+y<=1)
    {A,B,C,D,E}.Winners(U+1:ABCDE) = {A,B,C,D,E}.Winners(U+y:ABCD+x:E)

  If that is applied again wrt. candidate D, this results:

   For all U, x, y, (0<=x,y,z)(x+y+z<=1)
    {A,B,C,D,E}.Winners(U+1:ABCDE)
    = {A,B,C,D,E}.Winners(U+z:ABCD+x:E+y:D)

  After enough applications, a single paper can be decomposed into
   1st preference papers.

  More than one paper can have trailing preferences removed. For
   example: 

   For all U, x, y, (0<=x,y)(x+y<=1)                        # strip E
      {A,B,C,D,E}.Winners(U + 3:ABDE + 5:ACED)
    = {A,B,C,D,E}.Winners(U + 3y:ABD + 3x:E + 5:ACED)

   For all U, x, y, (0<=x,y,z)(x+y+z<=1)(0<=f,g)(f+g<=1)    # strip D
      {A,B,C,D,E}.Winners(U + 3y:ABD + 3x:E + 5:ACED)
    = {A,B,C,D,E}.Winners(U + 3y:ABD + 3x:E + 5g:ACE + 5f:D)

  A 2nd stripping of E might occur. Why that set would be chosen, given
   that ACED doesn't contain B, for example, is hardly obvious. This is
   not adequately checked out.


Third attempted definition: P4-3: Grouping papers by leading
 preferences and spreading their votes over single preferences.

 Example (P4-3):
   If a given list equals (ABC), then ({A,B,C}.Winners) is unchanged
    when all/some papers having the same leading preferences as
    that list (i.e. (ABC), (ABCD),...), have those 1st 3 preferences
    deleted.
   For example, there always exists x,y,z such that:
    (  {A,B,C}.Winners(U + 1:ABCDE)
     = {A,B,C}.Winners(U + x:A + y:B + z:C)) & (0<=x,y,z)(x+y+z<=1).

  The above can followed by a step where the list is (BAC), and that
   would decompose/spread the paper 1:BACDE over single preference
   papers. So the matching of leading preferences might as well be
   by sets.

  The papers perhaps need not be decomposed down to 1st preference
   papers. Once that is done, some constraints for preferences lists
   naming exactly 2 candidates are lost.

  [This resembles that minor/unimportant Q1 I mentioned, where papers
   matching a given list, could not be spread in any way that
   increased a binary satisfaction value calculated rom that list and
   the winners.]


It is easy to guess that one of P4-1, P4-2, P4-3 is a dud. In my
 opinion, the next step is to check them against 3 candidate IFPP.