Wolfram Mathematica

2:49 AM

@rasher I am not saying that your answer doesn't demonstrate good practice. But generating a random number is not "doing nothing"--it changes the RNG state. Hoewever, precisely because of your insistence on re-seeding every time, BlockRandom becomes essentially pointless in the code and your answer fails to show clearly why it didn't work in the OP's example.

Michael Hale

5:41 AM

Ok, I added random movement and collision detection with the lakes to the current community game. Suggest what you want added next or add it and post a new version.

landscape =
  Map[If[# < .47, 2, 1] &,
   GaussianFilter[RandomReal[1, {100, 100}], 5], {2}];
people = RandomChoice[Position[landscape, 1], 200];
pos = RandomChoice[Position[landscape, 1]];

validQ[p : {x_, y_}] :=
  If[Clip[p, {1, Length@landscape}] == p && landscape[[x, y]] == 1,
   True, False];
move[p_, o_] := If[validQ[p + o], p + o, p];
randomMove[p_] :=
  If[RandomInteger[3] == 0,
   RandomChoice[move[p, #] & /@ {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}],
   p];
plotLandscape[pos_] :=
 ArrayPlot[ReplacePart[ReplacePart[landscape, people -> 3], pos -> 4],

Michael Hale

7:11 AM

Oops. You don't need the Refresh.

Dynamic[people = randomMove /@ people; Dynamic@plotLandscape[pos],
 TrackedSymbols -> {}, UpdateInterval -> .1]

Jacob Akkerboom

11:06 AM

@MichaelHale, maybe we can add food sources and hunger. The goal would be to outsmart the randomly moving black dots to get to the food first.

Here is some nice lisp code, from the book land of lisp. There is no player there, but there are randomly moving animals that evolve as time goes by: http://landoflisp.com/evolution.lisp. A poor demonstration: http://www.youtube.com/watch?v=snsVmSjbKqQ. I believe an M is an animal and a * is a food source. The funky part is that basically two types of animals evolve, a horse and an elephant. Elephants try to stay near the jungle, where more food spa…

Michael Hale

12:44 PM

@JacobAkkerboom I like it. Here's the part of the book describing the experiment: books.google.com/…

When I used Lisp in my AI class I hadn't been exposed to Mathematica yet. A lot of code to generate a weighted random integer, heh.

This:

(defun turn (animal)
  (let ((x (random (apply #'+ (animal-genes animal)))))
    (labels ((angle (genes x)
               (let ((xnu (- x (car genes))))
                 (if (< xnu 0)
                     0
                     (1+ (angle (cdr genes) xnu))))))
        (setf (animal-dir animal)
              (mod (+ (animal-dir animal) (angle (animal-genes animal) x)) 8)))))

To this:

turn[animal_] :=
 setTurn[animal, RandomChoice[genes@animal -> Range@8]]

I'm about to sleep, and then I have a visitor coming over during the afternoon, but I will have time tomorrow evening and night.

Maybe Pacman type controls (you keep moving until you change direction) would be better because Mathematica doesn't work well with holding down arrow keys and there is no reason with those rules to ever stand still.

So the score would be how long you could survive, while the others multiply and eventually food isn't produced fast enough so it gets harder and harder until you die?

Jacob Akkerboom

1:22 PM

@MichaelHale ah, I like your response! First of all I was just trying to figure out how to work with the categorical distribution (http://en.wikipedia.org/wiki/Categorical_distribution) in Mathematica. That is a bit of a coincidence. Unfortunately RandomChoice does not help me, because I would like to use functions like Probability and Mean on a distribution. Maybe I will ask a question about it.

The pacman type controls are a good idea I think. It is unfortunate that Mathematica does not work well with holding down arrow keys, I did not know this.

kirma

3:35 PM

What is the distribution of angles of in equilateral convex polygons?

Searke

@kirma I have the feeling the question suffers from the same problems as buffon's needle problem. It's important to state the problem formally so that there aren't multiple answers depending on how you choose to interpret the answer

kirma

@Searke I guess so... :)

Searke

Sorry. Multiple answers depending on how you choose to interpret the question.

In a sense, we need to know what it means to generate a random equilateral convex polygon. There are different ways of generating a random equilateral convex polygon which might favor some angles more than others. So interpreting the question is really asking, what is a "fair" way of generating random equilateral convex polygons.

This kind of nuance doesn't happen with most probability problems because there's a natural sense of what the fair (even) distribution of outcomes is.

kirma

I have been pondering about that for days. This is not professional interest, so I'm not awfully concerned of it all, but makes an interesting problem to think of.

Searke

For more complicated objects like graphs though, it becomes a problem. My experience has taught me to hold off on answering such a question till I can figure out what the intended application is.

It is an interesting problem. Even given a formal definition of it, I'm not sure how I would solve it.

kirma

3:47 PM

I'm absolutely certain I can't do it. :)

Related to this:

5

Q: Generating a (random) 2-dimensional cellular network

I want to generate a network of random cells like the following picture: The general idea is to generate some random points (x,y) as coordinate of a vertex and then randomly choose some of these points to connect them to achieve (almost) a random cellular network since there two constraints: ...

Searke

You could post the question to math.stackexchange's chat, but they'd likely say what I said and then throw you into the math dungeon to be eaten by the math wolfs.

kirma

That's the reason why I don't dare to trek there. :)

Searke

I would start by asking myself what I thought the problem meant and what it's solution was for a four sided figure like a rhombus or square.

belisarius

"the volume is changing with respect to the user login height"

It deserves a thousand stars

kirma

My horribly informal intuition says that distribution for a triangle is single value, for rhombus it's uniform distribution, and beyond that point, it's something similar to normal distribution.

belisarius

3:51 PM

source: mathematica.stackexchange.com/q/43243/193

Searke

There are likely multiple reasonable answers that could be given

kirma

@Searke I have no idea where this intuition stems from.

Searke

buffon's needle problem has mutliple answers depending on what you mean by "randomly placed"

so too your question has multiple answers depending on "randomly selected"

Here. This is the classical example

Bertrand paradox (probability)

The Bertrand paradox is a problem within the classical interpretation of probability theory. Joseph Bertrand introduced it in his work Calcul des probabilités (1889) as an example to show that probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined. Bertrand's formulation of the problem The Bertrand paradox goes as follows: Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle? Bertrand gave thr...

kirma

I think my logic is something like "if I assume all angles are uniformly distributed, but constrained by geometric constraint of equilateralness and convexness, observe real distribution of angles in such polygons, replace initial distribution by these, and repeat this forever, do distributions converge to a specific distribution?" Or something.

Searke

But the distribution of angles will be dependent on one another in some way. I'm not as confident that your specification is unambiguous.

kirma

3:59 PM

Of course, I sort of define the distribution in this case through the method I'm trying to hunt it, which is wrong.

Searke

An adequate description would be an algorithm which produced such a random equilateral convex polygon

kirma

At least I know the distribution for triangles. :)

Searke

And of course if you had such an algorithm, you could then always monte-carlo the problem and try to find a solution.

But I think you'll find that the problem mostly lies in trying to fully define the problem.

kirma

Yep.

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