Proof Vermont method isn't mopnotonic (Re: [EM] "More often" (was: IRV and Condorcet operating identically)

[Craig Carey]

I found the mistake in my argument that the Vermont method is
monotonic. That argument appeared in my Politicians and Polytopes
mailing list.

The error occurs in the line "(A wins) = (Hb < Ha)".

At 2003\02\28 11:20 +0000 Thursday & at 02\28 14:25 +0000 Friday,
   I wrote:
-------------------------------------------------------------------
 > (Case 1) Candidate A ranked 3rd or worse.
 > There are 2 candidates ranking above candidate A:
 >
 > Case =. (a<b)(a<c) or (a<b)(a<d) or (a<b)(a<e) or ...
 > · · · · · · · · · · · (a<c)(a<d) or (a<c)(a<e) or ... or ...
 >
 > (A wins) = False
 > -----------------------------
 > (Case 2) Candidate A ranked 1st or 2nd.
 > Case =. (b<a or c<a)(b<a or d<a)(b<a or e<a) and ...
 > · · · · · · · · · · (c<a or d<a)(c<a or e<a) and ...
 > · · · · · · · · · · · · · · · · (d<a or e<a) and ...
 > (A wins) = (Hb < Ha)
 > Ha=(sum of weights of papers naming A, except when B is before A)
-------------------------------------------------------------------

Unfortunately the more complex method is now assumed by me to be
a method that will Not be monotonic.

Sorting cuts into case, and when the cases are reassembled, then
the method very easily  fails the test of monotonicity at the
boundaries between the cases. The Vermont method does a lot less
sorting than the Alternative Vote.

With this being the EM List, we can say that "research never starts".

The SSD method uses sorting so it is easily cutting and joining
while not trying to make the joins pass the test of  monotonicity.

Possibly .. either sorting is removed or the cuts are tilted.

----

At 03\02\28 12:36 +0100 Friday, Markus Schulze wrote:
 >Jan Kok wrote (25 Feb 2003):
 >> "(3)  The instant runoff count committee shall sort and count votes for
 >> candidates.  If, in the first round, no candidate received a majority of
 >> first choices, all candidates shall be eliminated except the two candidates
 >> with the greatest number of first choices.  Ballots which rank eliminated
 >> candidates and which indicate one of the final candidates as an alternate
 >> choice shall be counted as votes for whichever of the final candidates is
 >> ranked higher for that office on each ballot.  In each round, each ballot
 >> is counted as one vote for the highest ranked advancing candidate on that
 >> ballot."
...
 >> ------------------------------------------------------------------
 >> > From:  Markus Schulze <markus.schulze at ...
 >> > Date:  Wed Feb 26, 2003  12:09 pm
 >> > Subject:  Re: [EM] Might IRV adoption be inevitable?
...
 >> > And in so far as there is no known version of proportional
 >> > representation by the [Alternative Vote method] that has been
 >> > proven to meet monotonicity,
 >> ...
 >> ------------------------------------------------------------------
 >>
...
 >Now, candidate A is the winner. This is a clear violation of
 >monotonicity.
 >

Correction: Mr Schulze was right in saying that an AV-like method
that passes the test of monotonicity and that is defined explicitly
for all numbers candidates, and that need not be optimal, is not known.


I tried to improve on the Vermont method and did not seem to succeed.

The Vermont method might be better than the Alternative Vote. It is
not shown and starting off as very unimpressive method since being
identical to the Alternative Vote, it can wrap 33% of the voters into
a monotonicity defect. A huge fraction particularly when the exact
solution is already known for that case.


//////////////////////////



The 3 candidate Alternative Vote has "(A wins)" defined by either of
these:

(1): (A wins) = (b<a)(b<c)(c+bc<a+ba) or (c<a)(c<b)(b+cb<a+ca)

(2): (A wins) = (b<a or c<a) and (c+bc<a+ba or c<b) and (b+cb<a+ca or b<c)

The 2nd form resembles the 3 candidate 1 winner IFFP equation for when
candidate A wins:

(A wins) = (b+c<2a) and (c+bc<a+ba or c<b) and (b+cb<a+ca or b<c)

On sorting that and showing the c<b<a case winners, this is IFPP:

(A wins) = (b+cb<a+ca) or (2b<a+c)
(B wins) = (a+ca<b+cb)(a+c<2b)
(C wins) = False

Below I construct variants of the Vermont method; and make them resemble
the IFPP style.

None of the variants pass the test of monotonicity.

The Alternative Vote worsens at a fairly aggressive pace as candidates
are added (especially when checking for how forcefully the method is
demanding insincerity). The Vermont method has a lot less sorting.
STV has lacks proportionality and the Alternative Vote certainly
does not when 3 candidates (and maybe in general).

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

Assume that the papers are these, and there are 3 candidates.
Ties are mishandled (shapes of different dimensions are union-ed).

Sort so that c<b<a.
___   z0
A__   a0, a=a0+ab+ac
AB_   ab
AC_   ac
B__   b0, b=b0+ba+bc
BA_   ba
BC_   bc
C__   c0, c=c0+ca+cb
CA_   ca
CB_   cb

A variant of the "Vermont" method is created.
In that method, candidate C is eliminated in stage 1, leaving this:
___   z0
A__   a0, a=a0+ab+ac
AB_   ab
A__   ac
B__   b0, b=b0+ba+bc
BA_   ba
B__   bc
___   c0, c=c0+ca+cb
A__   ca
B__   cb

The above is an identical election point/example to this:
___   z0+c0
A__   a0+ac+ca
AB_   ab
B__   b0+bc+cb
BA_   ba

I create a parameterized formula, and then use the rules of
truncation resistance and monotonicity, to remove all of the parameters
except one. This is tried:
  (A wins) =
. . [(b0+bc+cb) + p*ba < (a0+ac+ca) + p*ab]
. .                       OR [2(b0+bc+cb) + q*ba < 2r*(a0+ac+ca) + s*ab]

Assuming truncation resistance implies that shifting votes between
(A) to (AB) won't alter A's win-lose status. That implies that p=1 and
2r = s. That leads to this:

(A wins) = (b+cb < a+ca) or [2(b0+bc+cb) + q*ba < 2r*(a+ca)]

C is already eliminated and there is 1 winner, so (B loses) = (A wins).

(B loses) = (b+cb < a+ca) or [2(b0+bc+cb) + q*ba < 2r*(a+ca)]

Truncation resistance has B's win-lose status be unaffected by all
shifting of votes/weight between the papers (BA) and (B).

Hence q =2. That leads to this:

(A wins) = (b+cb < a+ca) or (b+cb < r*(a+ca))

When r=1, then varian is the original EM List Vermont method.
It is identical to the Alternative Vote.

After re-lettering is done to restore the other 5 cases:

a<b<c ==> (A wins) = False
a<c<b ==> (A wins) = False
b<a<c ==> (A wins) = (c+bc < a+ba) and [r*(c+bc) < a+ba]
b<c<a ==> (A wins) = (c+bc < a+ba) or [c+bc < r*(a+ba)]
c<a<b ==> (A wins) = (b+cb < a+ca) and [r*(b+cb) < a+ca]
c<b<a ==> (A wins) = (b+cb < a+ca) or [b+cb < r*(a+ca)]

When r=1, then
I tested with these values of r:

0/1, 1/8, 1/6, 1/5, 1/4, 1/3, 3/8, 2/5, 1/2, 3/5, 5/8,
2/3, 3/4, 4/5, 5/6, 7/8, 8/7, 6/5, 5/4, 4/3, 3/2, 8/5,
5/3, 2/1, 5/2, 8/3, 3/1, 4/1, 5/1, 6/1, 8/1,

For all those values of r, the (A wins) expression was not
monotonic. They all have 5 or 9 faces (allowing negatives).

Here is the code that I used:
-------------------------------------------------------
procedure win_a(a,b,ba,bc,c,ca,cb); begin
  return
    (b<a)and(a<c) and ( (c+bc < a+ba) and (r*(c+bc) < a+ba) ) or
    (b<c)and(c<a) and ( (c+bc < a+ba) or (c+bc < r*(a+ba)) ) or
    (c<a)and(a<b) and ( (b+cb < a+ca) and (r*(b+cb) < a+ca) ) or
    (c<b)and(b<a) and ( (b+cb < a+ca) or (b+cb < r*(a+ca)) );
end;
procedure wr(rr); begin
     % see if A becomes a loser when changing (B) into (A)
   Pos:=(0<=a)and(0<=b)and(0<=ba)and(0<=bc)and
                 (0<=c)and(0<=ca)and(0<=cb);  % ineffective
   g1 := sub({r=rr}, win_a(a,b,ba,bc,c,ca,cb));
   g2 := sub({r=rr}, win_a(a+t,b-t,ba,bc,c,ca,cb));
   h1 := (0<t) and Pos and g1 and (not g2);
   h1 := (100 = a+b+ba+bc+c+ca+cb) and stri h1;
   tt := rlqea ex ({t,a,b,ba,bc,c,ca,cb}, h1);  % (Exists t,..)
   write "tt = ",tt;
end;
-------------------------------------------------------


The tangent polytopes constraining the normal vectors of
wins-regions are of a quantity that equals the number of
winners. I deleted that since cutting out the wrong argument
saying the Mr Schulze was wrong.

Perhaps Mr Schulze could venture to give some ideas on how to
tilt the divides between the cases?.

E.g. devise linear combinations prior to the sorting or
something.


The CVD has a new plan. Rob Ritchie decided it was too
risky to say that "we too regard that you are right to get no
winner but the best winner that you sought". Instead the
latest brochure is gilded like something from an early
Anold movie and talking about rates and what it means to
the voter. It might mean very little if each candidate for
mayorship offers a similar rate over the next few terms.

That aside, they say that the definition of IRV is a first
principle. The majority principle. IRV is not a church being
built over common interests.

Megaphone prblems: : what about the single day when
y'all vote for the mayor?. Interested in controlling the
selection of the mayor using a fair method ?.

Duh!. the CVD must be thinking of IRV promoting referenda
elections since using a right wing theme to advance it
random-side-of-left model of greatness.


___________________________________________________________________

Corrections:

At 03\02\28 21:05 +1300 Friday, Craig Carey wrote to EM List;
...
 > >Jan Kok said:
 > >> I'm curious if anyone can mathematically justify such statements as
 > >> "Voting method A exhibits property P 'more often' than method B"?
...
 >The method would be perfectly stable and unchanging and the
 >statement to be justified    did presume that.

I'd reword that to:

. . . . . . . . . . . .... did impose a presumption that that was not so.




G. A. Craig Carey


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