Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
660 views
in Technique[技术] by (71.8m points)

random - How many double numbers are there between 0.0 and 1.0?

This is something that's been on my mind for years, but I never took the time to ask before.

Many (pseudo) random number generators generate a random number between 0.0 and 1.0. Mathematically there are infinite numbers in this range, but double is a floating point number, and therefore has a finite precision.

So the questions are:

  1. Just how many double numbers are there between 0.0 and 1.0?
  2. Are there just as many numbers between 1 and 2? Between 100 and 101? Between 10^100 and 10^100+1?

Note: if it makes a difference, I'm interested in Java's definition of double in particular.

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Java doubles are in IEEE-754 format, therefore they have a 52-bit fraction; between any two adjacent powers of two (inclusive of one and exclusive of the next one), there will therefore be 2 to the 52th power different doubles (i.e., 4503599627370496 of them). For example, that's the number of distinct doubles between 0.5 included and 1.0 excluded, and exactly that many also lie between 1.0 included and 2.0 excluded, and so forth.

Counting the doubles between 0.0 and 1.0 is harder than doing so between powers of two, because there are many powers of two included in that range, and, also, one gets into the thorny issues of denormalized numbers. 10 of the 11 bits of the exponents cover the range in question, so, including denormalized numbers (and I think a few kinds of NaN) you'd have 1024 times the doubles as lay between powers of two -- no more than 2**62 in total anyway. Excluding denormalized &c, I believe the count would be 1023 times 2**52.

For an arbitrary range like "100 to 100.1" it's even harder because the upper bound cannot be exactly represented as a double (not being an exact multiple of any power of two). As a handy approximation, since the progression between powers of two is linear, you could say that said range is 0.1 / 64th of the span between the surrounding powers of two (64 and 128), so you'd expect about

(0.1 / 64) * 2**52

distinct doubles -- which comes to 7036874417766.4004... give or take one or two;-).


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...