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 C++ implementation of above idea –

// C++ Program to print 0, 1 or 2 with equal // probability #include <iostream> using namespace std; // Random Function to that returns 0 or 1 with // equal probability 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. 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 functions int main() { // Intialize random number generator srand(time(NULL)); for(int i = 0; i < 100; i++) cout << rand3(); return 0; }

Output:

21110111011120021110020202101120220220222111001001 21202021102100010200121121210122011022111020011012

**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.

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