[–] [deleted] 0 points 2 points (+2|-0) ago 

[Deleted]

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[–] Nietzsche__ 0 points 1 point (+1|-0) ago 

I agree. The statistical correlation isn't reflected in that formula.

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[–] NotAnOctopus 0 points 1 point (+1|-0) ago 

As already stated your problem is that you are treating dependent variables as independent. Dist.=-Aln(rand(x)), Wt=-Bln(rand(t)) but y is a function of both x and t. You need to come up with an algorithm that factors one given the other. Something like Y(W,D) is given by Wt=-Cln(rand(f(Dist))) filling in the necessary variables and equations for D.

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[–] bikergang_accountant [S] ago 

I'm not quite sure what people are getting at with this independent vs dependent problem with my question.

So I'm calculating distance and weight independently and the Dist*Wt is just an approximation of difficulty. Obviously as an approximation it's pretty lousy and the difficulty would be a manifold of some kind that would take a lot of human factors and could never really be that precise.

But the point is that if we did want a distribution where we could multiply to factors in that distribution to come out the same as -A*ln(rand(0,1)).

Maybe it's the coder in me but I noticed your rand function has an argument. I meant for it to be assumed that was an equal distribution from 0 to 1.

But maybe what you're getting about them not being independent is that we have three numbers we're interested in Dist,Weight,Difficulty, all of which we want to be random but we only have two degrees of freedom.

So your solution is to have the distance effect the actual random number feed into the natural log? Interesting. You wouldn't be able to effect the range because then weight wouldn't be an exponential distribution if there are numbers it can't reach but distance would be so the solution wouldn't be symmetric.

The thought I had at one point is to decide the difficulty at random and then have a random way of divvying up the proportionality of distance to weight randomly as well.

What I also found but didn't share because I didn't want to spoil is that Asqrt(-ln(rand())) solves the stability issue when multiplying it with an independent sibling but it does it too well. It's not the exact same distribution as -Aln(rand()). It's actually more forgiving. Not that that a problem for hiking.

What I suspect if I look into it more is that it may be the same distribution as (ln(rand())+ln(rand())/-2 which is also more forgiving. If so that raises another interesting question. If D(d,n) is sig(i=[1,n],d(rand()))/n, it is easy to find D(d,2), D(d,3) where d=D(ln,2), but what about D(d,1/2). If someone wanted to find D(ln,1/2) that's independently interesting.

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[–] PRODEATH 1 point -1 points (+0|-1) ago 

NERD!!!

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[–] Kael_thas_Sunstrider ago 

Go tell mommy you contributed something today. She might even give you a star!