[EM] Fwd: Fwd: Two MMV definiions (brief, and ordered-procedure)

Michael Ossipoff email9648742 at gmail.com
Tue Dec 10 13:46:35 PST 2013


I want to clarify that this posting, forwarded now as a new posting,
is an answer to your posting in which you spoke of nonmonotonicity
with my 3-stage MMV version.

---------- Forwarded message ----------
From: Michael Ossipoff <email9648742 at gmail.com>
Date: Tue, Dec 10, 2013 at 1:53 PM
Subject: Re: Fwd: Two MMV definiions (brief, and ordered-procedure)
To: Anders Kaseorg <andersk at mit.edu>,
"election-methods at electorama.com" <election-methods at electorama.com>

There are no stage 1 discards

Stage 2 discards AB and BC, because they contradict eachother (equal)
and CA (stronger).

There are no stage 3 discards.

B and C have no kept defeats, andso  they both win.

So C is helped, by being in a 2-way, instead of a tie among all the
alternatives, by the strengthening of the CA defeat.

Michael Ossipoff

On Mon, Dec 9, 2013 at 8:29 PM, Anders Kaseorg <andersk at mit.edu> wrote:
> On 12/09/2013 08:15 AM, Michael Ossipoff wrote:
>> Keep every defeat that doesn't contradict a set of kept stronger defeats.
>> Then, among the kept defeats, un-keep each defeat that contradicts a
>> set consisting of defeats equal to it, and of kept defeats stronger
>> than it is.
> So this is different from my fix: you’re dividing the process into two
> stages such that defeats in an equal-strength cycle can still be used to
> reject strictly weaker contradicting defeats in stage 1, before being
> discarded themselves in stage 2.  That makes sense.  It does seem to avoid
> the monotonicity failure cases discussed before.
> I’ll abandon my earlier fix in favor of this one.
>> Maybe there should be a 3rd stage that discards any set of equal
>> defeats that contradict only eachother.
> No, that would introduce a new monotonicity failure in the case of multiple
> overlapping cycles.  Suppose there are equally strong defeats
> A > B, B > C, C > A, B > D, D > A
> Then the extra stage rejects the equal-strength cycles A > B > C > A and A >
> B > D > A.  However, if support for C > A were increased, then the extra
> stage would no longer be able to reject the first cycle, which means B > C
> and C > A would be affirmed, harming C.
> Anders

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