[EM] Strategy-proof vs Monotone, IFPP, Mono-add-top

Kevin Venzke stepjak at yahoo.fr
Sun Jan 23 11:08:04 PST 2022


Hi Kristofer,

> I'd say that IFPP's improvement for three candidates (and moreso fpA-fpC
> or Smith,IFPP) is that it has the kind of strategy resistance that's
> only (out of methods commonly discussed here) shared by IRV and
> Smith-IRV hybrids respectively; and it does so while being monotone. So
> it's evidence that you can have a monotone method that's robust to
> strategy, and since every three-candidate minimally strategic method
> seems to lie pretty close to it, it would presumably be a good building
> block for a fully general such method.

Sure. Now, for whatever this is worth:

I was looking back and to my surprise, I actually once asked Craig why IFPP was
better than FPP. And specifically why should A lose this election:

49 A
40 B>C
11 C>B

He replied:
1. Since all the axioms are relative, he can't make any judgment about an
isolated scenario.
2. Since IFPP elects B, electing A must violate an axiom, but he "lack[s] an
argument saying which."
3. "Proportionality" is an objective. This is probably the best answer, because
he seemed hostile to the (rather related) concept of clone independence.

Le samedi 22 janvier 2022, 16:46:38 UTC−6, Kristofer Munsterhjelm <km_elmet at t-online.de> a écrit :
> Craig, of course, was not trying to find a Condorcet method and IIRC he
> considered the Condorcet criterion to be inexpressible in his framework
> (although I think that if I had been around then and had known what I
> know now, I could've explained it in terms he'd have understood).

It was the definition of Schulze that he couldn't adapt, which led him to
criticize Markus for inadequately defining it and writing proofs about it.

Off-list I learned that he needed an equation for each candidate. So I sent him
definitions for 3-candidate MinMax(WV), MMPO, and DSC. However, he was reluctant
to accept from me that MinMax(WV) could stand in for Schulze(WV) with three
candidates.

Craig lays out his final take in this interesting 2005 post:
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-January/079981.html

Craig Carey wrote:
> The main idea here is that a novel way to get answers out of believers in
> Condorcet variants, is to ask the wrong people. It seems that they can
> undo the problem of methods being fully undefined by using guessing or
> soem other technique.

> Also, Mr Venkze probably IGNORED the whole PDF article of Mr Schulze since
> his "A-wins" expression had only 5 lines. Here is my argument:
> The 3 "for loops" of the Schulze thing would lead to an expression that
> would be nearer 90 lines long.
> That polytope expression should simplify greatly, since it has to simplify
> into the "ha2" expression (unless ha2 is wrong).

> Anyway, to conclude, the Schulze (1-winner) method is rejected since it fails
> Dr D. Woodall's Mono-Raise-Random. Ditto with Mr Heitzig's method etc.
> 
> The Alternative Vote is failed too. I guess the Alternative is better than
> the Schulze method.

> A better looking fairer Condorcet method than the Schulze method, might be
> [the] MMPO (1-winner) Condorcet variant, since seeming to not fail
> Monotonicity-2 when 3 candidates.

That last paragraph is odd because MMPO certainly doesn't seem to satisfy that
standard. But if MMPO can get a good review (despite Craig being aware that it
was indecisive) I regret that he didn't give the DSC equations a try. Maybe he
would have liked it.

On Mono-add-top:

>> To me, it's too much to hope, that there is some technicality that
>> will let us do this.
>
> Right. Perhaps I don't have an intuitive sense of just how restrictive
> it is, and so I think it's possible where it's not :-)
> 
> But if it's not possible, then there should be a Moulin-style proof of
> it. Perhaps there's a relation between the kind of differential
> constraint observations I've been doing and such proofs. A good first
> step/example of this would be to convert my proof ideas for showing
> unmanipulable majority incompatible with Condorcet, into a Moulin-style
> exhaustive proof. I'd have to develop my theory a lot more before I
> could do that, though!

For sure, a Moulin-style proof must be possible. But it will have a lot of steps
and scenarios. I could imagine that you start with a four-candidate Smith set
and show that each one cannot be the winner.

If one can prove that they *are* compatible, then that would be the event of the
decade for me. That would be a shock. And this doesn't take a proof, hopefully,
just a method definition. I'd accept results from simulations.

If they are compatible, studying this method might reveal a trick we could use
to squeeze other strategy guarantees out of Smith methods that I would currently
judge to be impossible. (Not sure what those could be, though.)

Kevin


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