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Random number generator in arbitrary probability distribution fashion

  • Difficulty Level : Hard
  • Last Updated : 16 Jun, 2021

Given n numbers, each with some frequency of occurrence. Return a random number with probability proportional to its frequency of occurrence.

Example: 

Let following be the given numbers.
  arr[] = {10, 30, 20, 40}  

Let following be the frequencies of given numbers.
  freq[] = {1, 6, 2, 1}  

The output should be
  10 with probability 1/10
  30 with probability 6/10
  20 with probability 2/10
  40 with probability 1/10

It is quite clear that the simple random number generator won’t work here as it doesn’t keep track of the frequency of occurrence. 

We need to somehow transform the problem into a problem whose solution is known to us.

One simple method is to take an auxiliary array (say aux[]) and duplicate the numbers according to their frequency of occurrence. Generate a random number(say r) between 0 to Sum-1(including both), where Sum represents summation of frequency array (freq[] in above example). Return the random number aux[r] (Implementation of this method is left as an exercise to the readers).



The limitation of the above method discussed above is huge memory consumption when frequency of occurrence is high. If the input is 997, 8761 and 1, this method is clearly not efficient.

How can we reduce the memory consumption? Following is detailed algorithm that uses O(n) extra space where n is number of elements in input arrays.
1. Take an auxiliary array (say prefix[]) of size n. 
2. Populate it with prefix sum, such that prefix[i] represents sum of numbers from 0 to i. 
3. Generate a random number(say r) between 1 to Sum(including both), where Sum represents summation of input frequency array. 
4. Find index of Ceil of random number generated in step #3 in the prefix array. Let the index be indexc
5. Return the random number arr[indexc], where arr[] contains the input n numbers.
  Before we go to the implementation part, let us have quick look at the algorithm with an example: 
      arr[]: {10, 20, 30} 
      freq[]: {2, 3, 1} 
      Prefix[]: {2, 5, 6} 
  Since last entry in prefix is 6, all possible values of r are [1, 2, 3, 4, 5, 6] 
         1: Ceil is 2. Random number generated is 10. 
         2: Ceil is 2. Random number generated is 10. 
         3: Ceil is 5. Random number generated is 20. 
         4: Ceil is 5. Random number generated is 20. 
         5: Ceil is 5. Random number generated is 20. 
         6. Ceil is 6. Random number generated is 30. 
  In the above example 
      10 is generated with probability 2/6. 
      20 is generated with probability 3/6. 
      30 is generated with probability 1/6.

How does this work? 
Any number input[i] is generated as many times as its frequency of occurrence because there exists count of integers in range(prefix[i – 1], prefix[i]] is input[i]. Like in the above example 3 is generated thrice, as there exists 3 integers 3, 4 and 5 whose ceil is 5. 

C++




// C++ program to generate random numbers
// according to given frequency distribution
#include <bits/stdc++.h>
using namespace std;
 
// Utility function to find ceiling of r in arr[l..h]
int findCeil(int arr[], int r, int l, int h)
{
    int mid;
    while (l < h)
    {
        mid = l + ((h - l) >> 1); // Same as mid = (l+h)/2
        (r > arr[mid]) ? (l = mid + 1) : (h = mid);
    }
    return (arr[l] >= r) ? l : -1;
}
 
// The main function that returns a random number
// from arr[] according to distribution array
// defined by freq[]. n is size of arrays.
int myRand(int arr[], int freq[], int n)
{
    // Create and fill prefix array
    int prefix[n], i;
    prefix[0] = freq[0];
    for (i = 1; i < n; ++i)
        prefix[i] = prefix[i - 1] + freq[i];
 
    // prefix[n-1] is sum of all frequencies.
    // Generate a random number with
    // value from 1 to this sum
    int r = (rand() % prefix[n - 1]) + 1;
 
    // Find index of ceiling of r in prefix array
    int indexc = findCeil(prefix, r, 0, n - 1);
    return arr[indexc];
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4};
    int freq[] = {10, 5, 20, 100};
    int i, n = sizeof(arr) / sizeof(arr[0]);
 
    // Use a different seed value for every run.
    srand(time(NULL));
 
    // Let us generate 10 random numbers accroding to
    // given distribution
    for (i = 0; i < 5; i++)
    cout << myRand(arr, freq, n) << endl;
 
    return 0;
}
 
// This is code is contributed by rathbhupendra

C




//C program to generate random numbers according to given frequency distribution
#include <stdio.h>
#include <stdlib.h>
 
// Utility function to find ceiling of r in arr[l..h]
int findCeil(int arr[], int r, int l, int h)
{
    int mid;
    while (l < h)
    {
         mid = l + ((h - l) >> 1);  // Same as mid = (l+h)/2
        (r > arr[mid]) ? (l = mid + 1) : (h = mid);
    }
    return (arr[l] >= r) ? l : -1;
}
 
// The main function that returns a random number from arr[] according to
// distribution array defined by freq[]. n is size of arrays.
int myRand(int arr[], int freq[], int n)
{
    // Create and fill prefix array
    int prefix[n], i;
    prefix[0] = freq[0];
    for (i = 1; i < n; ++i)
        prefix[i] = prefix[i - 1] + freq[i];
 
    // prefix[n-1] is sum of all frequencies. Generate a random number
    // with value from 1 to this sum
    int r = (rand() % prefix[n - 1]) + 1;
 
    // Find index of ceiling of r in prefix array
    int indexc = findCeil(prefix, r, 0, n - 1);
    return arr[indexc];
}
 
// Driver program to test above functions
int main()
{
    int arr[]  = {1, 2, 3, 4};
    int freq[] = {10, 5, 20, 100};
    int i, n = sizeof(arr) / sizeof(arr[0]);
 
    // Use a different seed value for every run.
    srand(time(NULL));
 
    // Let us generate 10 random numbers accroding to
    // given distribution
    for (i = 0; i < 5; i++)
      printf("%d\n", myRand(arr, freq, n));
 
    return 0;
}

Java




// Java program to generate random numbers
// according to given frequency distribution
class GFG
{
 
// Utility function to find ceiling of r in arr[l..h]
static int findCeil(int arr[], int r, int l, int h)
{
    int mid;
    while (l < h)
    {
        mid = l + ((h - l) >> 1); // Same as mid = (l+h)/2
        if(r > arr[mid])
            l = mid + 1;
        else
            h = mid;
    }
    return (arr[l] >= r) ? l : -1;
}
 
// The main function that returns a random number
// from arr[] according to distribution array
// defined by freq[]. n is size of arrays.
static int myRand(int arr[], int freq[], int n)
{
    // Create and fill prefix array
    int prefix[] = new int[n], i;
    prefix[0] = freq[0];
    for (i = 1; i < n; ++i)
        prefix[i] = prefix[i - 1] + freq[i];
 
    // prefix[n-1] is sum of all frequencies.
    // Generate a random number with
    // value from 1 to this sum
    int r = ((int)(Math.random()*(323567)) % prefix[n - 1]) + 1;
 
    // Find index of ceiling of r in prefix array
    int indexc = findCeil(prefix, r, 0, n - 1);
    return arr[indexc];
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = {1, 2, 3, 4};
    int freq[] = {10, 5, 20, 100};
    int i, n = arr.length;
 
    // Let us generate 10 random numbers accroding to
    // given distribution
    for (i = 0; i < 5; i++)
    System.out.println( myRand(arr, freq, n) );
}
}
 
// This code is contributed by Arnab Kundu

Python3




# Python3 program to generate random numbers
# according to given frequency distribution
import random
 
# Utility function to find ceiling of r in arr[l..h]
def findCeil(arr, r, l, h) :
 
    while (l < h) :   
        mid = l + ((h - l) >> 1); # Same as mid = (l+h)/2
        if r > arr[mid] :
            l = mid + 1
        else :
            h = mid
     
    if arr[l] >= r :
        return l
    else :
        return -1
 
# The main function that returns a random number
# from arr[] according to distribution array
# defined by freq[]. n is size of arrays.
def myRand(arr, freq, n) :
 
    # Create and fill prefix array
    prefix = [0] * n
    prefix[0] = freq[0]
    for i in range(n) :
        prefix[i] = prefix[i - 1] + freq[i]
 
    # prefix[n-1] is sum of all frequencies.
    # Generate a random number with
    # value from 1 to this sum
    r = random.randint(0, prefix[n - 1]) + 1
 
    # Find index of ceiling of r in prefix array
    indexc = findCeil(prefix, r, 0, n - 1)
    return arr[indexc]
 
# Driver code
arr = [1, 2, 3, 4]
freq = [10, 5, 20, 100]
n = len(arr)
 
# Let us generate 10 random numbers accroding to
# given distribution
for i in range(5) :
    print(myRand(arr, freq, n))
 
    # This code is contributed by divyesh072019

C#




// C# program to generate random numbers
// according to given frequency distribution
using System;
 
class GFG{
     
// Utility function to find ceiling
// of r in arr[l..h] 
static int findCeil(int[] arr, int r,
                    int l, int h) 
    int mid;
    while (l < h) 
    
         
        // Same as mid = (l+h)/2 
        mid = l + ((h - l) >> 1);
         
        if (r > arr[mid]) 
            l = mid + 1;
        else
            h = mid; 
    
    return (arr[l] >= r) ? l : -1; 
}
 
// The main function that returns a random number
// from arr[] according to distribution array 
// defined by freq[]. n is size of arrays. 
static int myRand(int[] arr, int[] freq, int n) 
     
    // Create and fill prefix array 
    int[] prefix = new int[n];
    int i; 
    prefix[0] = freq[0]; 
     
    for(i = 1; i < n; ++i) 
        prefix[i] = prefix[i - 1] + freq[i]; 
   
    // prefix[n-1] is sum of all frequencies.
    // Generate a random number with 
    // value from 1 to this sum
    Random rand = new Random();
    int r = ((int)(rand.Next() * (323567)) %
                      prefix[n - 1]) + 1; 
   
    // Find index of ceiling of r in prefix array 
    int indexc = findCeil(prefix, r, 0, n - 1); 
    return arr[indexc]; 
}
 
// Driver Code
static void Main()
{
    int[] arr = { 1, 2, 3, 4 }; 
    int[] freq = { 10, 5, 20, 100 }; 
    int i, n = arr.Length; 
     
    // Let us generate 10 random numbers
    // accroding to given distribution 
    for(i = 0; i < 5; i++) 
        Console.WriteLine(myRand(arr, freq, n)); 
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
 
// JavaScript program to generate random numbers
// according to given frequency distribution
 
    // Utility function to find ceiling of r in arr[l..h]
    function findCeil(arr, r, l, h)
    {
        let mid;
        while (l < h)
        {
            mid = l + ((h - l) >> 1); // Same as mid = (l+h)/2
            (r > arr[mid]) ? (l = mid + 1) : (h = mid);
        }
        return (arr[l] >= r) ? l : -1;
    }
 
    // The main function that returns a random number
    // from arr[] according to distribution array
    // defined by freq[]. n is size of arrays.
    function myRand(arr, freq,  n) {
        // Create and fill prefix array
        let prefix= [];
        let i;
        prefix[0] = freq[0];
        for (i = 1; i < n; ++i)
            prefix[i] = prefix[i - 1] + freq[i];
 
        // prefix[n-1] is sum of all frequencies.
        // Generate a random number with
        // value from 1 to this sum
        let r = Math.floor((Math.random()* prefix[n - 1])) + 1;
 
        // Find index of ceiling of r in prefix array
        let indexc = findCeil(prefix, r, 0, n - 1);
        return arr[indexc];
    }
 
    // Driver code
    let arr = [1, 2, 3, 4];
    let freq = [10, 5, 20, 100];
    let i;
    let n = arr.length;
 
    // Use a different seed value for every run.
 
    // Let us generate 10 random numbers accroding to
    // given distribution
    for (i = 0; i < 5; i++)
      document.write(myRand(arr, freq, n));
       
      // This code is contributed by rohitsingh07052.
</script>

Output: May be different for different runs 

4
3
4
4
4

This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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