[EM] Factorising methods (Was: IFPP for more than three candidates?

[Craig Carey]

At 01.09.17 01:03 -0700 Monday, Rob LeGrand wrote:
 >Well, it's a good thing Craig doesn't define his IFPP method for more than
 >three candidates!

First sentence botched. A definition can be implicit. It seems to be rather
definitely defined. implicitly, now that I have a definition of P4 at
http://www.ijs.co.nz/one-man-one-vote.htm  (the definition is worded with a
scratchiness that could hinder others' comprehending of it).

 >
 >Suppose there's an election among A, B and C:
 >
 >10:A>B>C
 >15:B>A>C
 >23:C>A>B
 >
 >C wins in IFPP after A and B are eliminated in the first round (quota 16).  But
 >say universally despised candidate D runs:
 >
 >10:A>B>C>D
 >15:B>A>C>D
 >23:C>A>B>D
 >
 >If IFPP is generalized for four candidates, the quota becomes 12; A and D are

It is a generalising of the Alternative Vote. Proving that 1/n quota without
solving IFPP explicitly presumably can be done.

...
 >in the first round if they all had the same first-place totals.  Looks like the
 >quota fix to IRV only "works" for three candidates, making it all but useless.
 >
Just how useless?. It could be that the 'Q' method is overall improvement on the
Alternative Vote method.

 >On his page http://www.ijs.co.nz/irv-wrong-winners.htm Craig calls this IFPP
 >generalization the 'Q' method.
 >

The Q method certainly can be criticised a lot. At the moment I don't see how
to improve it.

Here is a quick line of reasoning that ends in a dead end. I could say that
it just a matter of casting shadows, starting with votes being about equal
and the region small and centred around the simplex-centre. However there are
problems with corners of rectangular faces casting shadows with 3 or surfaces.
Also the attempt can't be completed without a definition of all the rules. A
problem ought occur if the "power<=1" rule isn't defined, and when I got that
defined, then the shadow [of the infinitesimally applying 'P4' rule] was
umbral and casting 'reducing in size' shadows which could be reverted.
The rejecting of consideration of pairs can be retained but unlike with
monotonicity with its concern for a single candidate, keeping the powers of
factions (of papers) constrained can involve a consideration of more than
one candidate.

The presumption of having k stages (or <=k stages) in an election with k
candidates is so unsound that it might even favour with the CVD crabs
scuttling to produce a spirit of reform in others.

My method below is the use of guessing, and the aim is to find the H terms.

Ma = (a<1/4)  :  Mb = (b<1/4)  :  Mc = (c<1/4)  :  Md = (d<1/4)

Aw = (1/4<a)[(b<1/4) or Hba][(c<1/4) or Hca][(d<1/4) or Hda]

Assume solution is of this form, which is in the style of 3 candidate
1 winner IFPP:
    Aw = -Ma[Mb or Hba][Mc or Hca][Md or Hda]  : Aw = (A wins), etc.
The terms Hba, etc., are unknown.

3 candidate IFPP:
    Aw' = (1/3<a')[(b'<1/3)or(b'+cb'<a'+ca')][(c'<1/3)or(c'+bc'<a'+ca')]

Aw = -Ma[Mb or Hba][Mc or Hca][Md or Hda]
Bw = -Mb[Mc or Hcb][Md or Hdb][Ma or Hab]
Cw = -Mc[Md or Hdc][Ma or Hac][Mb or Hbc]
Dw = -Md[Ma or Had][Mb or Hbd][Mc or Hcd]

CASE d<c<1/4<b<a : True = Md = Mc, False = Mb = Ma

    Aw = Hba
    Bw = Hab
    Cw = Dw = False

   Maybe replace Hba with Hba = (0 < Tba), etc.

CASE d<1/4<c<b<a : True = Md, False = Mc = Mb = Ma

    Aw = Hba.Hca
    Bw = Hcb.Hab
    Cw = Hac.Hbc
    Dw = False

   That does not suggest rejecting Hba=(0<Tba) idea

CASE d<c<b<1/4<a : True = Md = Mc = Mb, False = Ma

    Aw = True
    Bw = Cw = Dw = False

But what to do now is not obvious: the 3 candidate Aw IFPP formula
is not in a form that appears to factorize into Hba.Hca,
  = (0<Tba)(0<Tca).

That does not seem to lead anywhere obvious. Defining the T values
looks like something that will not happen

So that's a weak attempt to justify the absence of an improvement to
the Q method. A fix to Q might involve regarding the 1st stage as
being 4-fold. There is the option of truncating, in addition to the
known method of eliminating and transferring papers.

It would be nice to hear from the CVD so called "I.R.V." experts
provide the mathematical arguments that led them to see the obviously
obvious Q method as being inferior. Fixing Q might take years of
research, but getting intelligence from the CVD could take decades,
if ever achievable. I fear they might one day say that they want to
know what IRV is, but so far no sign of such a threat against experts.