EXERCISE 6.2: generating a random Double from a random Int

[Adam Mackler]

Exercise 6.2 on page 83:

Write a function to generate a Double between 0 and 1 , not including 1.

The given hint and answer indicate generating a non-negative Int and then dividing it by the one more than the maximum possible Int value.

Maybe I'm missing something, but it seems to me that since a Double is represented using 64 bits and an Int 32, that using the amount of randomness in an Int to generate a random Double will mean that the results will be unevenly distributed, perhaps with possible values of the Double never returned.  Here's what I came up with:

  def double(rng: RNG): (Double, RNG) = {

    val (int1, rng2) = rng.nextInt

    val (int2, rng3) = rng2.nextInt

    val long = (int1.toLong << 32) | int2            // this is a random Long

    val nnLong = if (long < 0) -(long + 1) else long // non-negative random Long

    val rLong = -(nnLong.toDouble / Long.MinValue)   // random double

    (rLong, rng3)

  }

I generate two random integers, convert one to a Long and bit-shift it 32 bits to the left and then OR it with the other Int.  This should give me a random Long.  Then I make it non-negative, divide it by the minimum Long value, and negate that.  Intuitively this seems to me that it would give a more even distribution since there are extra bits of randomness from the second Int.  Is my thinking correct?  Is this worse or better than the answer given in the book?

[Ben Hutchison]

On Tue, Apr 10, 2018 at 8:24 AM, Adam Mackler <adamm...@gmail.com> wrote:

Maybe I'm missing something, but it seems to me that since a Double is represented using 64 bits and an Int 32, that using the amount of randomness in an Int to generate a random Double will mean that the results will be unevenly distributed, perhaps with possible values of the Double never returned.  

I agree with your analysis. Generating Doubles from a single Int is no a good real world strategy.  I guess it was just posed as an exercise.

-Ben

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