[EM] Exact spatial model probabilities?

[Forest Simmons]

Very interesting!

El sáb., 22 de ene. de 2022 1:45 a. m., Richard Lung <
voting at ukscientists.com> escribió:

>
>
> Roger Penrose did a proof of super-luminal connections, using "magic
> dodecahedra" popularised in his book, Shadows of the Mind. Previously, as
> in the Bell theorem, the proof had only been statistical. Penrose offered a
> deterministic proof.
>
> I offered my own explanation of super-luminal connections, in my free
> e-book, Statistical Relativity Elections. Essentially, I argued it was a
> geometric mean of super-luminal connections and sub-luminal connections.
> This comes out of the special relativity situation, which allows, in
> theory, for tachyons, as well as tardyons. (Quantum theory does not allow
> for individual tachyons.)
>
> I call myself a naive physicist (an amateur, I scarcely pretend that much).
>
>
> Regards,
> Richard Lung.
>
>
>
>
> On 22 Jan 2022, at 5:53 am, Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
> These are fantastic questions.
>
> The spherical geometry makes sense for Universal Domain methods because,
> as Saari pointed out in his Geometry of Voting, the natural distance
> between two permutations is the number of transpositions it takes to
> convert one into the other... the Kemeny distance.
>
> There are 12 permutations of three candidates ... which arrange themselves
> naturally on the face of a clock.  Any "Gray Code" where successive
> permutations differ by only one transposition will match the twelve hours
> with the twelve permutations:
>
> For example, ABC, ACB, CAB, CBA, BCA, BAC, and back to  ABC. Each
> permutation will be diametrically opposite to its reverse order.
>
> For permutations of four candidates you get 24 points on the surface of a
> sphere.
>
> Each permutation is adjacent to three others. For example, ABCD is next to
> BACD, ACBD, and ABDC. When you arrange all 24 of these permutations
> symmetrically on the surface of a sphere, so that the distance between any
> two adjacent permutations is the same, you will find that reversed
> permutations are polar opposites.
>
> It's harder to visualize the arrangement of the 120 permutations of five
> symbols, but topologically they  reside symmetrically on the surface of a
> 4-dimensional ball. The surface of a four dimensional compact manifold is
> in general a three dimensional manifold, in this case a 3-sphere, just as
> the surface of a 3-dimensional ball is an ordinary 2-dimensional sphere ...
> 2D manifolds have area, while 3D manifolds have volume. A circle, the
> boundary of a disc, is a "1-sphere" because it has length, namely its
> circumference.
>
> General Relativity (without black holes or worm holes) models the Big Bang
> as a four dimensional space-time manifold whose temporal cross sections are
> 3-spheres ... precisely the kind of spheres on which the 120 distinct
> complete rankings of five candidates must be arranged if you require
> adjacent rankings to be the same distance apart.
>
> So the space of permutations of n symbols consists of n factorial points
> distributed evenly on the boundary of an (n-1) dimensional ball. A
> Euclidean ball of radius R consists of all vectors v having L_2 norms ||v||
> less than or equal to R.The set where equality holds is its boundary. If
> the vector space has a basis of n vectors, the ball of radius R will be n
> dimensional, and its boundary will be (n-1) dimensional.
>
> These high dimensional vector spaces are ubiquitous in science, especially
> in quantum mechanics and other applications of partial differential
> equations ... the general mathematical setting being "Functional Analysis."
>
> So Saari had the right idea topologically, but not metrically ...the same
> topological space can have many different metrics that yield the same
> topology, and the Kemeny metric is not quite right; by making all adjacent
> permutations the exact same distance apart, the Kemeny metric introduces
> clone dependence into methods based on it ... most notably the Kemeny-Young
> method, and Borda as related to Saari's geometry of voting.
>
> In my de-cloned Kemeny-Young posts, I have shown how to modify the Kemeny
> metric to make it respect the natural clustering of clone sets, while
> preserving the natural Kemeny generated topology.
>
> Topology is all about local stretching and shrinking without ripping apart
> or smashing together either the natural proximities or the large scale
> connections.
>
> Now we start to see the difficulty of trying to work within Universal
> Domain rules: UD requires ballots that are permutations of the candidates,
> so if we identify voters with their ballots, voters are distributed on
> spheres. But issue spaces are products of line segments, like rectangles,
> blocks, etc.
>
> If there are two issues, then issue space is a rectangle. If there are
> four candidates, then ballot space is a two dimensional sphere, so
> according to UD rules, that only allow voter influence through their ballot
> rankings, there is a topological mismatch of voter/ballot space and issue
> space.
>
> For example, the Borsuk-Ulam theorem shows that any continuous map from an
> n-sphere into a rectangle (or any other lower dimensional product of
> intervals) must have a pair of polar opposite points that get mapped to the
> same point; i.e. this requires that two polar opposite rankings correspond
> to the exact same point in issue space.
>
> That's one reason I take Universal Domain rules with a grain of salt, and
> feel free to supplement rankings with ratings, approval cutoffs, etc.
>
> I know that's not any help with your main question, but it may give some
> insight into Durand's spherical distribution suggestion.
>
>
> I'll try to give a practical suggestion next time!
>
>
> El vie., 21 de ene. de 2022 7:02 a. m., Kristofer Munsterhjelm <
> km_elmet at t-online.de> escribió:
>
>> My minimum-manipulability optimization system can find exact
>> probabilities that a voter will vote according to a particular ranking,
>> under impartial culture.
>>
>> It's easy to approximately find such probabilities for spatial models as
>> well by just creating an instance and randomly placing voters on it to
>> determine what they vote, and hence in the limit of number of trials
>> going to infinity, get the probability.
>>
>> But I was wondering, is it possible to do so exactly? Let's say that in
>> a spatial model, first c candidates are placed uniformly at random in an
>> unit d-cube. Then voters are also placed uniformly at random in this
>> cube and they prefer candidates closer to them to candidates further away.
>>
>> Now I would suppose that the probability that a voter votes A first is
>> equal to the volume of the Voronoi cell that belongs to A. (And the
>> probability that a voter will vote A>B>C is the volume of the
>> intersection of the closest-neighbor Voronoi cell for A with the
>> second-closest-neighbor for B and the third-closest-neighbor for C.)
>>
>> Well, that eliminates one source of randomness -- assuming I can exactly
>> calculate the vertices of these regions. But there's still the second
>> source in the randomness of the candidates. Do you know of any calculus
>> tricks to get the probabilities over every possible candidate position,
>> or is this simply too hairy?
>>
>> One benefit of the spatial model over impartial culture is that it's
>> much more clear how one may assign utilities to each ranking, so the
>> optimization system could also be used to investigate tradeoffs between
>> VSE and manipulability. (Durand suggested sampling on a sphere for
>> associating utilities with impartial culture, but I haven't read the
>> suggestion in detail. https://arxiv.org/pdf/1511.01303.pdf)
>>
>> -km
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>>
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