Rand(7) from Rand(5)

Question

simple_coder · Accepted Answer

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.

Anonymous · Answer

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();
}

Anonymous · Answer

Above answer is not correct because 0 is not returned.  We need equal probability for all 0, 1,2,3,4,5,6.  The modified would be:

int rand7()
{
    while(1)
   {
       int n = ((rand5()%2)*4 + (rand5()%2)*2 + (rand5()%2)*1;
       if(n==0)
           continue;
       return n-1;
   }
}

David · Answer

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

ravi · Answer

what about calling rand(5) 7 times adding it then % 7 
(  rand5() + rand5() + rand 5() + rand5() + rand5 () + rand5() + rand5()   )  % 7

-suhane

Anonymous · Answer

int r = rand5() *5 + rand5()
while (r>=21){
 r =  rand5() *5 + rand5();
}
return r%7;

Anonymous · Answer

Java:
Random r = new Random();
r.nextInt(5)+r.nextInt(5)%3

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