Random double

28 Jan 2015.

I need to generate a random floating point number.

I know I don't want to do something like rand() % 100 / 100.

A double is just 64-bits, so generate a 64-bit string that looks like

0x3FFn nnnn nnnn nnnn

Where the ns are random hex digits.

As a double, this will be a random number in the range $[1, 2)$.

Floating point numbers generated this way have the same distribution as the underlying generator used.

Why not zero to one?

The range $[1, 2)$ uses a single exponent, whereas the range $[0, 1)$ spans 1074 different exponents.

Because of the way the floating point encoding works, if I use uniformly distributed random numbers to generate all the bits of the double (the exponent as well as the significand) up to 1.0, I will end up with the same number of samples in the range $[2^{-1074}, 2^{-1023})$ as I will in the range $[0.5, 1.0)$.

The benefit of using the range $[1, 2)$ is that I can use a fixed exponent and generate just the significand, and the distribution stays the same.

Floating point encoding

IEEE 754-1985 double-precision floating point numbers are encoded as:

Sign bit.
11 bits of exponent.
52 bits of significand (sometimes called fraction or mantissa).

Laid out like this:

6  6       5 5         4         3         2         1         0
3210987654321098765432109876543210987654321098765432109876543210
+EEEEEEEEEEEssssssssssssssssssssssssssssssssssssssssssssssssssss
|          |                                                   |
sign bit   exponent                                  significand

The exponent is encoded with an offset:

A zero value means the exponent is $2^{-1022}$ and the number is a "denormal."
A nonzero value, $E$, means the exponent is $2^{E-1023}$ and the significand is "normalized."

How to interpret the significand:

For example, the bit string 1011100000000000000000000000000000000000000000000000.
This corresponds to the binary number $1.10111_2$.
The leading 1. is implicit in the encoding:
- The exponent is chosen so that the first significant bit is 1.
- Since the first bit is known, we can elide it.
- If it were a denormal number, the significand would be $0.10111_2$.
$1.10111_2$ means $1 + \frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32}$.
Or, in decimal: $1.718750_{10}$.

Floating point 0.0 is all zeroes:

+EEEEEEEEEEEssssssssssssssssssssssssssssssssssssssssssssssssssss
0000000000000000000000000000000000000000000000000000000000000000

The next number after zero (i.e. nexttoward(0., 1.)) is:

+EEEEEEEEEEEssssssssssssssssssssssssssssssssssssssssssssssssssss
0000000000000000000000000000000000000000000000000000000000000001
                     1         2         3         4         5
bit index:  1234567890123456789012345678901234567890123456789012

Here the exponent is $2^{-1022}$, the number is denormal, and the significand has the least significant bit set, so the number is:

$$ 2^{-1022} \times 0.0000000000000000000000000000000000000000000000000000000000000001_2\\ = 2^{-1022} \times 2^{-52}\\ = 2^{-1074} $$

The floating point representation of 1.0 is:

+EEEEEEEEEEEssssssssssssssssssssssssssssssssssssssssssssssssssss
0011111111110000000000000000000000000000000000000000000000000000

The exponent is encoded as $01111111111_2 = 1023_{10}$ which corresponds to $2^0$.

The significand is $1.0_2$ because a normalized number has the implicit bit and the encoded significand is all zeroes after it.