# Implement rand3() using rand2()

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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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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