如何在开放区间（x，y）获得下一个随机翻倍

Question

I know the basic algorithm for a random number in a half-closed interval is:

Random rand = new Random();
double increment = min + (max - min) * rand.nextDouble();

This will give you a random number on the interval [min, max) because nextDouble includes 0 in the range of results ( [0.0,1.0) ) that it returns. Is there a good way to exclude the minimum value and instead provide a random number on (min, max) ?

Answer 1

In theory , calling Math.nextUp(double d) should do it.

double minUp = Math.nextUp(min);
double increment = minUp + (max - minUp) * rand.nextDouble();

In reality, rounding after multiplication may still cause min to be returned, so a retry loop would be better. Given the rarity of an exact min value, performance won't suffer.

double increment;
do {
    increment = min + (max - min) * rand.nextDouble();
} while (increment <= min || increment >= max);

Just for heck of it, I also added a max check.

Answer 2

Rather than asking for the minimum exclusively, you could ask for it inclusively -- but have the minimum be the next value of double using Math::nextUp :

min = Math.nextUp(min);

Doubles are discrete, so this is analogous to in integer-land, rephrasing (0, 10) as [1, 10) .