Implement rand3() using rand2()

• Difficulty Level : Medium
• Last Updated : 02 Feb, 2021

Given a function rand2() that returns 0 or 1 with equal probability, implement rand3() using rand2() that returns 0, 1 or 2 with equal probability. Minimize the number of calls to rand2() method. Also, use of any other library function and floating point arithmetic are not allowed.

The idea is to use expression 2 * rand2() + rand2(). It returns 0, 1, 2, 3 with equal probability. To make it return 0, 1, 2 with equal probability, we eliminate the undesired event 3.
Below is the implementation of above idea –

C++

 // C++ Program to print 0, 1 or 2 with equal// probability#include using namespace std; // Random Function to that returns 0 or 1 with// equal probabilityint rand2(){    // rand() function will generate odd or even    // number with equal probability. If rand()    // generates odd number, the function will    // return 1 else it will return 0.    return rand() & 1;}  // Random Function to that returns 0, 1 or 2 with// equal probability 1 with 75%int rand3(){    // returns 0, 1, 2 or 3 with 25% probability    int r = 2 * rand2() + rand2();         if (r < 3)        return r;         return rand3();} // Driver code to test above functionsint main(){    // Initialize random number generator    srand(time(NULL));      for(int i = 0; i < 100; i++)        cout << rand3();      return 0;}

Java

 // Java Program to print 0, 1 or 2 with equal // probabilityimport java.util.Random;class GFG{   // Random Function to that returns 0 or 1 with  // equal probability  static int rand2()  {     // rand() function will generate odd or even    // number with equal probability. If rand()    // generates odd number, the function will    // return 1 else it will return 0.    Random rand = new Random();     return (rand.nextInt() & 1);  }   // Random Function to that returns 0, 1 or 2 with   // equal probability 1 with 75%  static int rand3()  {     // returns 0, 1, 2 or 3 with 25% probability    int r = 2 * rand2() + rand2();     if (r < 3)      return r;    return rand3();  }   // Driver code  public static void main(String[] args) {    for(int i = 0; i < 100; i++)      System.out.print(rand3());  }}// This code is contributed by divyesh072019.

Python3

 # Python3 Program to print 0, 1 or 2 with equal# Probability import random# Random Function to that returns 0 or 1 with# equal probability def rand2():     # randint(0,100) function will generate odd or even    # number [1,100] with equal probability. If rand()    # generates odd number, the function will    # return 1 else it will return 0    tmp=random.randint(1,100)    return tmp%2     # Random Function to that returns 0, 1 or 2 with# equal probability 1 with 75%def rand3():         # returns 0, 1, 2 or 3 with 25% probability    r = 2 * rand2() + rand2()    if r<3:        return r    return rand3()     # Driver code to test above functionsif __name__=='__main__':    for i in range(100):        print(rand3(),end="")         #This code is contributed by sahilshelangia

C#

 // C# Program to print 0, 1 or 2 with equal // probabilityusing System;class GFG{   // Random Function to that returns 0 or 1 with  // equal probability  static int rand2()  {     // rand() function will generate odd or even    // number with equal probability. If rand()    // generates odd number, the function will    // return 1 else it will return 0.    Random rand = new Random();    return (rand.Next() & 1);  }   // Random Function to that returns 0, 1 or 2 with   // equal probability 1 with 75%  static int rand3()  {     // returns 0, 1, 2 or 3 with 25% probability    int r = 2 * rand2() + rand2();     if (r < 3)      return r;    return rand3();  }   // Driver code  static void Main()  {    for(int i = 0; i < 100; i++)      Console.Write(rand3());  }} // This code is contributed by divyeshrabadiya07.

PHP



Output :

2111011101112002111002020210112022022022211100100121202021102100010200121121210122011022111020

Another Solution –
If x = rand2() and y = rand2(), x + y will return 0 and 2 with 25% probability and 1 with 50% probability. To make probability of 1 equal to that of 0 and 2 i.e. 25%, we eliminate one undesired event that’s resulting in x + y = 1 i.e. either (x = 1, y = 0) or (x = 0, y = 1).

int rand3()
{
int x, y;

do {
x = rand2();
y = rand2();
} while (x == 0 && y == 1);

return x + y;
}

Please note above solutions will produce different results every time we run them.