[EM] Inferring a method from an MMC axiom

[Craig Carey]

At 03\03\10 14:13 +0100 Monday, Markus Schulze wrote:
 >Dear Craig,
 >
 >you wrote (10 March 2003):
 >> To Mr Schulze: did you find out why Mr Dummett had such a
 >> weak rule ?. That rule had a Floor(x) function on it, reducing
...
 >
 >Let's say that V > 0 is the number of voters, S > 0 is the number
 >of seats, and C > S is the number of candidates.
 >
 >A "solid coalition" is a set of candidates with a set of voters
 >such that every voter in this set of voters strictly prefers every
 >candidate of this set of candidates to every candidate outside this
 >set of candidates.
 >

The word "voters" should be changed into papers. Doing that discards
any intended unwritten constraint that the count of the papers be
non-negative. I.e. it is proper to fail the rule if mishandling
negative quantities.

If a 'solid coalition' has k candidates in it, then each paper in that
'solid coalition' must name all of those k candidates.

Argument for that:
(1) suppose there are no candidates outside of the 'solid coalition'.
Then it is naming all the candidates.
(2) In the alternative case. there is a candidate outside of the
'solid coalition'. Call that candidate "y".
Suppose a paper does not name a candidate the coalition. Call that
candidate "x".
There are 2 cases:
(2.1) Both the preferences x and y are off the edge and not showing on
the paper.
(2.2) The paper does not name x, but it does name y. A candidate
outside of the coalition is appearing in the coalition's papers.

For sub-case (2.1), a guess is made that it fails the test of 'strictly
prefers'
For sub-case (2.2), the test is failed since a paper would not strictly
prefer a candidate off the edge over named on the paper.
Infinitesimally-size cases of a candidate off being no less preferred
by the paper than one on, could arise. That seems OK.


That is quite different from MMC. Here the MMC "set of candidates" is
{A, C, D}:

--------------------------------------------------------------------------
At 03\03\04 11:20 +1300 Tuesday, Craig Carey wrote:
 >At 2003\03\03 04:59 +0000 Monday, MIKE OSSIPOFF wrote:
 >...
 > >Blake's MMC:
 > >
 > >A majority of voters are agreed
 > >that they prefer any one of a set of candidates over any candidate
 > >outside that set, but disagreee over who should win from inside the
 > >set. It seems to follow from majoritarian principles that someone from
 > >inside the set should win. Further, this principle can't lead to a
 > >contradiction.
...
 >     AD  25
 >     B.  27
 >     C.  24
 >     D.  24
 >
 >The 1 winner IFPP method finds that candidate B is the winner.
--------------------------------------------------------------------------

MMC seems to be too strong to be desirable. The solid coalitions rule
could be too reluctant to impose a constraint to be acceptable. I.e.



 >Let's say that a given solid coalition consists of C0 candidates
 >and V0 > 0 voters. Then "Dummett's proportionality criterion for
 >solid coalitions" says that at least min{ceil((S+1)*V0/V)-1; C0}
 >candidates of this solid coalition must be elected.
 >

I consider it correct to delete the "S > 0" constraint. That reduces
the possibility/certainty of the rule being failed.

Here is an example:

AB  32
BA  33
C_  35

3W0 IFPP Winners = {}
3W1 IFPP Winners = {C}  . [since Max(32,33) < (32+33+25)/3 = 100/3]
3W2 IFPP Winners = {B, C} [derived at the bottom of the message]
3W3 IFPP Winners = {A, B, C}

Solid coalition = {A, B}
V0 = 65 = total for {A, B}
V = 100 = total
C0 = 2 = #{A, B}

Rule says: number of winners in the {A, B} set is greater than or
equal to, MNW, MNW = Min{Ceil((S+1)*V0/V)-1; C0}

Case S = 0 : MNW = Min{Ceil(1*0.65)-1; 2} = min{0; 2} = 0
Case S = 1 : MNW = Min{Ceil(2*0.65)-1; 2} = min{Ceil(0.3); 2} = 1
Case S >= 2: MNW = Min{Ceil(3*0.65)-1; 2} = min{Ceil(1.95); 2} = 2

Hence the extended version of the supplied rule of Dummett implies:

3W1 case: not ({} = {A, B}.Winners)
3W2 case: {A, B} = Winners

Therefore if rejects IFPP. Hence I was wrong to say that the
rule was "weak".


 >The reason why Dummett's criterion is so weak is that it is a
 >conditio sine qua non: When a given multi-winner election method
 >doesn't meet Dummett's proportionality criterion for solid
 >coalitions then it isn't even a proportial election method due
 >to Dummett's theory.
 >


Notes on the topic:
(1) The rule has very little to say about the entire non-negative
solution space since requiring so many preferences. I suppose the
rule is not good enough. Maybe something better could be thought
up. However my P2 considers few papers too.

(2) By the argument I provided before, MMC can be replaced with
the new rule. It can be made an axiom that gives way to P2,
right number of winners, and truncation resistance and that
requires that monotonicity give way to itself.

  . . .a(AB)
  . . . /\
  . . ./ .\
  . . /. . \s
  . ./ . . .\
  . /. . . . \
  .+----r-----+
  b(BA). . . .c(C)

Case S = 1 and (1/2) < (a+b) < 1 and 1=a+b+c :
    MNW = Min{Ceil(2*(a+b))-1; 2} = 1
So: 1 winner and c < (a+b+c)/2 implies C loses.

Monotonicity has an M bulging towards C with 2 expansions of the
C loses region. At the centre lines are parallel to the (AB)-(B)
and (BA)-(C) faces (i.e. edges).

The rule of Dummett places the C-wins-loses divide on the
line 2c=a+b.

Going back the to the "10% .. 25% checkbox" comment, what would
Mr Schulze want when the rule is weakened using a single real
parameter ?:
(a) shift the line c=1/2 over to c=1/3; or:
(b) break the line c=/12 and shift its two pieces over to the
  two lines (1/3 = b), (1/3 = a).

Also the  2 candidate case can be considered.

Also you reject P2 (though this list can't comment on it despite
its simplicity). E.g. I was saying that if the rule disagrees
with P2 (no change on adding k((A)+(B)+(C))+m(2(A)-(AB)-(AC))+...)
then it gives way to it.

I guess you are content to have the 1/3 quota (i.e. opt for
option (a): shift the c line.

Condorcet was demonstrably very bad when one candidate was way
behind, and with this new idea, there seems to be a hope that
the results can be politically unaccepable and far worse than
IFPP when there is a close race between popular candidates.




__________________________________________

The two winner solution of this
|  AB  32
|  BA  33
|  C_  35
can be obtained using the 'duality' rule
|  AB  -32
|  BA  -33
|  C_  -35
Total = -100. Quota = -33.3333..
Only C is under the quota so the winner of the negated election is
found to be A by use of the Alternative Vote.