Interview Question for Senior Software Engineer at Amazon.com:

The simple solution is to implement RandBin() using Rand5().

int RandBin()
{

}

Sorry this UI posts without warning

int RandBin()
{
   int rand5Res = Rand5();

   return rand5Res < 2 ? 0 : rand5Res < 4 ? 1 : RandBin();
}

And then use RandBin() to implement Rand7;

int Rand7()
{
   int rand = RandBin() << 2 | RandBin() << 1 | RandBin();

   return rand < 7 ? rand : Rand7();
}

int rand7()
{
  while(1)
  {
    int n = ((rand5()%2)*4 + (rand5()%2)*2 + (rand5()%2)*1);
    if(n == 0)
      continue;
     return n;
  }
}

The rand5()%2 will generate 0 and 1 with equal probability and we need 3 bits since we are going from 000 upto 111. So we call this function thrice for each bit position.