Opensource Implementation of the Alias Method

By : Chii
Source: Stackoverflow.com
Question!

I am doing a project at the moment, and in the interest of code reuse, I went looking for a library that can perform some probabilistic accept/reject of an item:

i.e., there are three people (a, b c), and each of them have a probability P{i} of getting an item, where p{a} denotes the probability of a. These probabilities are calculated at run time, and cannot be hardcoded.

What I wanted to do is to generate one random number (for an item), and calculate who gets that item based on their probability of getting it. The alias method (http://books.google.com/books?pg=PA133&dq=alias+method+walker&ei=D4ORR8ncFYuWtgOslpVE&sig=TjEThBUa4odbGJmjyF4daF1AKF4&id=ERSSDBDcYOIC&output=html) outlined here explained how, but I wanted to see if there is a ready made implementation so I wouldn't have to write it up.

By : Chii


Answers

Here is a Ruby implementation: https://github.com/cantino/walker_method

By : Andrew


i just tested out the method above - its not perfect, but i guess for my purposes, it ought to be enough. (code in groovy, pasted into a unit test...)

    void test() {
        for (int i = 0; i < 10; i++) {
            once()
        }
    }
    private def once() {
        def double[] probs = [1 / 11, 2 / 11, 3 / 11, 1 / 11, 2 / 11, 2 / 11]
        def int[] whoCounts = new int[probs.length]
        def Random r = new Random()
        def int who
        int TIMES = 1000000
        for (int i = 0; i < TIMES; i++) {
            who = selectPerson(probs, r.nextDouble())
            whoCounts[who]++
        }
        for (int j = 0; j < probs.length; j++) {
            System.out.printf(" %10f ", (probs[j] - (whoCounts[j] / TIMES)))
        }
        println ""
    }
    public int selectPerson(double[] probabilies, double r) {
        double t = r
        double p = 0.0f;
        for (int i = 0; i < probabilies.length; i++) {
            p += probabilies[i];
            if (t < p) {
                return i;
            }
        }
        return probabilies.length - 1;
    }

outputs: the difference betweenn the probability, and the actual count/total 
obtained over ten 1,000,000 runs:
  -0.000009    0.000027    0.000149   -0.000125    0.000371   -0.000414 
  -0.000212   -0.000346   -0.000396    0.000013    0.000808    0.000132 
   0.000326    0.000231   -0.000113    0.000040   -0.000071   -0.000414 
   0.000236    0.000390   -0.000733   -0.000368    0.000086    0.000388 
  -0.000202   -0.000473   -0.000250    0.000101   -0.000140    0.000963 
   0.000076    0.000487   -0.000106   -0.000044    0.000095   -0.000509 
   0.000295    0.000117   -0.000545   -0.000112   -0.000062    0.000306 
  -0.000584    0.000651    0.000191    0.000280   -0.000358   -0.000181 
  -0.000334   -0.000043    0.000484   -0.000156    0.000420   -0.000372
By : Chii


Would something like this do? Put all p{i}'s in the array, function will return an index to the person who gets the item. Executes in O(n).

public int selectPerson(float[] probabilies, Random r) {
	float t = r.nextFloat();
	float p = 0.0f;

	for (int i = 0; i < probabilies.length; i++) {
		p += probabilies[i];
		if (t < p) {
			return i;
		}
	}

	// We should not end up here if probabilities are normalized properly (sum up to one)
	return probabilies.length - 1;		
}

EDIT: I haven't really tested this. My point was that the function you described is not very complicated (if I understood what you meant correctly, that is), and you shouldn't need to download a library to solve this.

By : finalman


This video can help you solving your question :)
By: admin