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.
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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 <[EMAIL PROTECTED] >> > 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.
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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).
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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:
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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;
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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.
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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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