Generate a random binary number with a variable pr

Question

I need a function to generate random integers. (assume Java long type for now, but this will be extended to BigInteger or BitSet later.)

The tricky part is there is a parameter P that specifies the (independent) probability of any bit in the result being 1.

If P = 0.5 then we can just use the standard random number generator. Some other values of P are also easy to implement. Here's an incomplete example:

Random random = new Random();

// ...

long nextLong(float p) {
    if      (p == 0.0f)   return 0L;
    else if (p == 1.0f)   return -1L;
    else if (p == 0.5f)   return random.nextLong();
    else if (p == 0.25f)  return nextLong(0.5f) & nextLong(0.5f);
    else if (p == 0.75f)  return nextLong(0.5f) | nextLong(0.5f);
    else if (p == 0.375f) return nextLong(0.5f) & nextLong(0.75f); // etc
    else {
      // What goes here??
      String message = String.format("P=%f not implemented yet!", p);
      throw new IllegalArgumentException(message);
    }
}

Is there a way to generalise this for any value of P between 0.0 and 1.0?

Answer 1

Here's how I solved it in the end.

1. Generate an integer N between 0..16, following the binomial distribution. This gives the number of '1' bits in the 16-bit partial result.
2. Randomly generate an index into a lookup table that contains 16-bit integers containing the desired number of '1' bits.
3. Repeat 4 times to get four 16-bit integers.
4. Splice these four 16-bit integers together to get a 64-bit integer.

This was partly inspired by Ondra Žižka's answer.

The benefit is that it reduces the number of calls to Random.nextLong() to 8 calls per 64 bits of output. For comparison, rolling for each individual bit would require 64 calls. Bitwise AND/OR uses between 2 and 32 calls depending on the value of P

Of course calculating binomial probabilities is just as expensive, so those go in another lookup table.

It's a lot of code, but it's paying off in terms of performance.

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Update - merged this with the bitwise AND/OR solution. It now uses that method if it guesses it will be more efficient (in terms of calls to Random.next().)