<div dir="auto">Now I want to point out how the HMC ergodic idea makes it simple, efficient, and extremely precise to find the volume of each preference order region:<div dir="auto"><br></div><div dir="auto">Initialize an algebraic variable 'BallotProfile' with the value zero.<br><div dir="auto"><br></div><div dir="auto">Then as you evolve numerically the wandering test point ODE,</div><div dir="auto"><br></div><div dir="auto">(d/dt)^2 r= -grad V(r),</div><div dir="auto"><br></div><div dir="auto">at each successive point r of the numerical solution, calculate all of the distances from r to the respective candidate positions, and sort them to find the ranked preference ballot for that position in the form of a string, such as, "A>B=C>D". Before resuming the ODE evolution, add that string into the variable 'BallotProfile', treating it as a variable name, i.e.an algebraic monomial.</div><div dir="auto"><br></div><div dir="auto">After a few billion steps, according to the ergodic theorem that applies to this kind of energy conserving dynamical system, the 'BallotProfile' collected over time along the trajectory of the wandering point r, will also represent the spatial distribution of preference ballots.</div><div dir="auto"><br></div><div dir="auto">Practical suggestions ... convert the second order vector differential equations into a system of two first order vector ODE's ...</div><div dir="auto"><br></div><div dir="auto">dr/dt=v, </div><div dir="auto">dv/dt=-grad V(r)</div><div dir="auto"><br></div><div dir="auto">Solve numerically using RK4 or even simple Euler's method ... after all we're just using the path to sample the space. Try several initial conditions ... nothing too far outside the support of the voter distribution.</div></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mié., 26 de ene. de 2022 11:30 a. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com">forest.simmons21@gmail.com</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto">I should probably give more of an explanation for the choice of the atomic electo-potential, i.e. the scalar potential function V(r) for one voter located at R:<div dir="auto"><div dir="auto"><br></div><div dir="auto">V(r)=||r-R||</div><div dir="auto"><br></div><div dir="auto">The force=-grad V(r) represents the influence of the voter pushing or pulling in one direction or another. The true winner should be at a point where the resultant force of the entire electorate is balanced at as close to zero as possible.</div><div dir="auto"><br></div><div dir="auto">So what is the force of one voter under this choice of potential V(r)?</div><div dir="auto"><br></div><div dir="auto">It is -grad V(r) anchored at the voter position R. </div><div dir="auto"><br></div><div dir="auto">-grad|r-R|||=-(r-R)/||r-R||, which is a unit vector pointing from r to R.</div><div dir="auto"><br></div><div dir="auto">In other, words each voter exerts a pull of one dyne towards herself in the tug of war to find the winning position.</div><div dir="auto"><br></div><div dir="auto">One voter one vote translates to one voter one dyne. Each voter is allotted </div><div dir="auto"> the same amount of pull ... but different</div><div dir="auto">voters pull in different directions. Sincere voters pull towards their own location in in voter space. </div><div dir="auto"><br></div><div dir="auto">If there exists a unique equilibrium point where all of the pulls balance, that point is the geometric median of the voter distribution.</div><div dir="auto"><br></div><div dir="auto">It is easy to see that in any distribution of voters along a straight line, this balance occurs where r is at the ordinary median of the didtribution,with half of the voters pulling with same strength in each direction.</div><div dir="auto"><br></div><div dir="auto">That's the heuristic for the choice of "electo-potential" ... one person, one dyne of pull towards their own position.</div></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mié., 26 de ene. de 2022 10:23 a. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" target="_blank" rel="noreferrer">forest.simmons21@gmail.com</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div>Just one correction ****<br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mié., 26 de ene. de 2022 10:02 a. m., Forest Simmons <<a href="mailto:forest.simmons21@gmail.com" rel="noreferrer noreferrer" target="_blank">forest.simmons21@gmail.com</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">El mié., 26 de ene. de 2022 3:12 a. m., Kristofer Munsterhjelm <<a href="mailto:km_elmet@t-online.de" rel="noreferrer noreferrer noreferrer" target="_blank">km_elmet@t-online.de</a>> escribió:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 26.01.2022 04:23, Forest Simmons wrote:<br>
<br>
> The dirac delta is the convolution identity distribution ... convolving<br>
> it with another distribution leaves it unchanged with cliffs and sharp<br>
> corners intact. <br>
> <br>
> But if you convolve with a smooth approximation of Dirac, like a<br>
> gaussian with tiny variance, you get an infinitely differentiable<br>
> approximation of the "horrible" function.<br>
<br>
Right. I once wrote a fully differentiable genetic algorithm (I was<br>
intending to use it for hyperparameter tuning). There were two problems.<br>
First, local optima. Second, suppose that you have a plain statement like:<br>
<br>
if (x>y) {<br>
return z;<br>
} else {<br>
return f(z);<br>
}<br>
<br>
When smoothing, this turns into something like<br>
<br>
return z * sig(x-y, k) + f(z) * sig(y-x, k)<br>
<br>
where sig(x, k) is a sigmoid function that returns 1 if x>>0, 0 if x<<0,<br>
and k is a steepness parameter controlling sig'(x) around 0.<br>
<br>
If k is too high, then gradient descent fails because there's no<br>
noticeable slope down to the optimum; it's flat almost everywhere and<br>
then goes to zero exactly at the optimum (vanishing gradient problem).<br>
<br>
So we need some smoothness, which means that both branch values come<br>
into play. But this makes the function slow to evaluate as the program<br>
must "go both ways" on every conditional.<br></blockquote></div></div><div dir="auto"><br></div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"></blockquote></div></div><div dir="auto">Here's a better way to get a smooth potential function V(x,y,z,) in the present context:</div><div dir="auto">First the potential V(x,y,z)for all voters concentrated at one point (X,Y,Z):</div><div dir="auto"><br></div><div dir="auto">V(r)=||r-R||, where r=(x,y,z) and R=(X,Y,Z).</div><div dir="auto"><br></div><div dir="auto">To smooth, replace this with</div><div dir="auto"><br></div><div dir="auto">V(r)=(epsilon+||r-R||^2)^(1/2)</div><div dir="auto"><br></div><div dir="auto">This is more realistic anyway because you cannot truly have all voters concentrated at one point. The epsilon smear factor would be a function of the variance of the voters clustered around the point R.</div><div dir="auto"><br></div><div dir="auto">That variance could be given exactly for a uniform distribution on a ball of radius delta centered at R, for example.</div><div dir="auto"><br></div><div dir="auto">Then for multiple factions...</div><div dir="auto"><br></div><div dir="auto">V(r)=Sum(over R in Omega) of</div><div dir="auto">f(R)*sqrt(epsilon(R)+||R-r||^.5),</div><div dir="auto"><br></div><div dir="auto">where Omega is the set of faction centers, and f(R) is the fraction of voters clustered near R, i.e. in a delta(R) [related to epsilon(R)] neighborhood of R.</div><div dir="auto"><br></div><div dir="auto">If we have a smooth (i.e. pre-smeared) pdf f(R), we can write the "electo-potential" V(r) as an integral</div><div dir="auto">V(r)=Integral(over R in Omega)of</div><div dir="auto"> ||r-R||f(r)dxdydz </div><div dir="auto"><br></div><div dir="auto">The hamiltonian for the wandering voter particle is the total energy, kinetic + potential, E=T+V, where T =.5 ||dR/dt||^2.</div><div dir="auto"><br></div><div dir="auto">The system (in vector form) of ODE's for the motion of the particle is</div><div dir="auto"><br></div><div dir="auto">(d/dr)^2=-grad V(r)</div></div></blockquote></div></div><div dir="auto">*** Should be</div><div dir="auto"><span style="font-family:sans-serif">(d/dt)^2=-grad V(r)</span><br></div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div dir="auto"><div dir="auto"><br></div><div dir="auto">I need to check the HMC link that Daniel gave me to see what notation they are using.</div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
> Electrical engineers have a vast library of standard test patterns to<br>
> use as input signals for use in designing and testing circuits.<br>
> <br>
> We need a similar library of test distributions for use in designing and<br>
> testing election methods.<br>
> <br>
> Election methods could even be profiled by their responses to these test<br>
> patterns.<br>
<br>
This, I do agree with. There have been a lot of voting method proposals<br>
lately, and we need some way to easily determine:<br>
<br>
- what is its VSE (under what models)<br>
- what is its voter-strategy susceptibility<br>
- what is its candidate-strategy susceptibility (cloning)<br>
- what criteria does it definitely fail<br>
<br>
and it would be nice to also know what criteria it definitely passes,<br>
though that requires formal verification, which is much harder than just<br>
testing a bunch of cases.<br>
<br>
(I remember using REDLOG to come up with BTV once, but I don't remember<br>
the details. Perhaps I should look into the current state of the art for<br>
provers, like Z3... so many things to do and so little time in which to<br>
do them!)<br>
<br>
-km<br>
</blockquote></div></div></div>
</blockquote></div></div></div>
</blockquote></div>
</blockquote></div>