[EM] MinMax (Pairwise Opposition) A-wins equation

[Kevin Venzke]

Craig,

 --- Craig Carey <research at ijs.co.nz> a écrit : 
> >We are short on rules satisfying Mono-raise-random...
> 
> You meant to say "short on methods".

"Rules" means the same thing.

> (1) Your own MMPO method was passed by Mono-Raise-Random when 3 candidates
> · unless I was sent a wrong A-wins equation.

It makes no difference, since I'm thinking about the general case, for which
we don't have any methods other than FPP. Are you saying you would like the
Schulze method if it satisfied MRRandom with three candidates?

> (2) Could MMPO-whatever provide some trick to make STV variants be monotonic ?.
> · No indication of that.

By "STV variants" do you mean methods that satisfy Truncation Resistance?
If so, it shouldn't be possible without sacrificing Majority and Clone-Winner.

> · You don't seem to desire to hae a 1/3 quota in a 3 candidate 1 winner
> · election, so it could be time for you to start answering questions about
> · Truncation Resistance or Later No Harm.

Why, what questions?

> >> · · / ·|/ · A-loses
> >> · ·/· ·Q
> >>
> >> Between P and Q there is an infinitesimally boundary between 2 of Mr Schulze's
> >> cases. It must be checked before it can be concldued that the slopes are
> >> within the bounds.
> >
> >Diagrams are created by following the rules of the method. If you prove
> >e.g. that raising a candidate can't make him lose, according to the rules
> >of the method, then you also prove that this can't occur on the diagram
> >which is supposed to be based on the method.
> 
> It is cute to have found a way to found a numerical argument about the surface
> of a polytope that happens to run correctly when many more than 3 candidates.

A limit on the number of candidates is a weakness of the diagrams, not of
proofs in general. My argument that MMPO satisfies Mono-raise makes no
assumptions about the number of candidates.

You do agree that the diagrams are created based on the methods, right?

> >> A better looking fairer Condorcet method than the Schulze method, might be
> >> Mr Venkze's MMPO (1-winner) Condorcet variant, since seeming to not fail
> >> Monotonicity-2 when 3 candidates.
> >
> >Hmm, that's interesting. I wonder what circumstances you checked?
> 
> Here is the simplified MMPO 3 candidate A-wins equation that I sent to you.
> Why not simply look at it ?:

That's not what I meant. I mean, did you for instance check all instances of
replacing ballot BA with AC, to find that it never made A lose? If so, I'll
have to try to create a failure scenario by hand...

> AW6 :=
> · · (b + cb < a + ba and b + cb < a + ca and c + bc < a + ba and c + bc < a + ca)
>  or (b + cb < a + ca and c + bc < a + ca and c + bc < b + ab and cb < ab)
>  or (b + cb < a + ba and c + bc < a + ba and b + cb < c + ac and bc < ac)
>  or (c + bc < b + ab and cb < ab and b + cb < c + ac and bc < ac) ;
> ------------------------------------------------------------------------------------
> 
> It is of no interest to the more intelligent STV community since one candidate
> can win when holding 0% of the vote (and none of the votes are negative).

I don't think it's intelligence so much as a fetish for first preferences.

> Here is an example showing that changing (AB) into (A) causes A to change
> from a winner into a loser.
> 
> ---------------------
> (A) · · · 0 · · · 2 · · · <- increased by 2
> (AB)· · · 2 · · · 0 · · · <- decreased by 2
> (AC)· · · 1 · · · 1
> (B) · · · 3 · · · 3
> (C) · · · 4 · · · 4
> ---------------------
> · · · ·A wins ·A loses

Do you consider this unfair to someone?

> >Suppose A wins and his score is from opposition from C. Suppose C is close,
> >and gets max opposition from B. Then changing the ballot BA to AC might
> >have no effect but to decrease C's score (which is good for C).
> >
> >So I think it can't be right, that 3-candidate MMPO satisfies Mono-2.
> 
> I don't see wrong term in AW6 and I don't follow your claim.

This is the equation I originally sent you:

A-wins :=
((b+cb)>=(c+bc) && (a+ca)>=(c+ac) && (a+ba)>=(b+ab) && (b+cb)<=(a+ca) && (b+cb)<=(a+ba))
||
((b+cb)>=(c+bc) && (a+ca)>=(c+ac) && (b+ab)>=(a+ba) && (b+cb)<=(a+ca) && (b+cb)<=(b+ab))
||
((b+cb)>=(c+bc) && (c+ac)>=(a+ca) && (a+ba)>=(b+ab) && (b+cb)<=(c+ac) && (b+cb)<=(a+ba))
||
((b+cb)>=(c+bc) && (c+ac)>=(a+ca) && (b+ab)>=(a+ba) && (b+cb)<=(c+ac) && (b+cb)<=(b+ab))
||
((c+bc)>=(b+cb) && (a+ca)>=(c+ac) && (a+ba)>=(b+ab) && (c+bc)<=(a+ca) && (c+bc)<=(a+ba))
||
((c+bc)>=(b+cb) && (a+ca)>=(c+ac) && (b+ab)>=(a+ba) && (c+bc)<=(a+ca) && (c+bc)<=(b+ab))
||
((c+bc)>=(b+cb) && (c+ac)>=(a+ca) && (a+ba)>=(b+ab) && (c+bc)<=(c+ac) && (c+bc)<=(a+ba))
||
((c+bc)>=(b+cb) && (c+ac)>=(a+ca) && (b+ab)>=(a+ba) && (c+bc)<=(c+ac) && (c+bc)<=(b+ab));

So my scenario is: Let's say A wins in one of the two ||'d conditions where
(c+bc >= b+cb) and (b+ab >= a+ba). Those are the 6th and 8th conditions. So
suppose you change ba into ac. Clearly, to me, that could cause the failure
of the fifth "subconditions" within them (e.g. c+bc<=a+ca).

Kevin Venzke



	

	
		
Découvrez le nouveau Yahoo! Mail : 250 Mo d'espace de stockage pour vos mails ! 
Créez votre Yahoo! Mail sur http://fr.mail.yahoo.com/