# Generating Uniform Random Numbers on [0,1]

Chris Lomont, March 2017

Common advice for generating uniform random numbers in [0,1] in many languages looks like this [1,2,3]:

This however does not generate many possible floating-point numbers in $[0,1]$, leaving gaps – there are many floating-point numbers that should occur but cannot due to this method. To understand why, you need to understand how floating-point numbers are (usually) stored on computers.

Throughout we’ll assume the random number generators produce uniformly random integers in a known range.

## IEEE 754 floating-point format

The most common way to store floating-point numbers is the IEEE 754 format, which has 1 sign bit, $e$ exponent bits, and $m$ mantissa bits. I’ll write $S$ for the sign bit, $E$ for the exponent bit pattern and $M$ for the mantissa bit pattern. For a 64 bit double used in C, C#, JavaScript, and many similar languages, it has this layout:

sign exponent mantissa
1 bit 11 bits 52 bits

When $E\neq 0$ and $E\neq 2047$, this represents the floating point value $f=(-1)^S \times 2^{E-bias}\times 1.M$.

When $E=0$ and $M=0$, then $f=0$.

These two cases are called normal numbers, and most numbers you’ll use fall in this category.

When $E=0$ and $M\neq 0$, then the value $f$ is $f=(-1)^S \times 2^{1-bias}\times 0.M$.

This case is called a subnormal number. It lets you use smaller numbers than naively using the above format for all bit patterns. Note the implicit 0 in the mantissa and the 1 in the exponent, instead of 0. These ensure that all values behave nicely.

When $E=2047$ and $M=0$, the bit patterns represents signed $\infty$.

When $E=2047$ and $M\neq0$, the bit pattern represents NaNs (Not a Number), used for error conditions.

The bias for a 64-bit IEEE 754 floating-point number is 1023, allowing positive and negative exponents. The $1.M$ symbols mean there is an implied 1 bit, then a decimal point, then $M$ is the 52 bits following the decimal. Thus if $M=1010…000_2$, $1.M$ is $1+1\times\frac{1}{2}+0\times\frac{1}{4}+1\times\frac{1}{8}+0…0=1.625$. When a number is stored in this format the hardware determines position of the leading 1 bit, from that computes the exponent bits $E$, and if $E>0$, removes the implicit leading bit, and stores the next 52 bits for $M$. If $E>2046$, then one of the infinities is stored. If $E=0$, then subnormal numbers are stored.

For the rest of this article we’ll use the non-negative floating-point values, i.e., $S=0$.

## The problem

So, now that you understand the format, why is the original code bad? The first problem is rand() in many languages does not generate enough values; in many C environments RAND_MAX was (is?) 32767, meaning there are only 32767 possible output values. This is fine for a few uses, but if you really need a large number of floating-point values with more variety for modern simulations, you need more possible output values.

How many IEEE 754 floating-point values are there in $[0,1]$? For each exponent in $E=0,E=1,…,E=bias-1$ there are $2^{52}$ distinct floating-point values, and the final value $f=1$ is the exponent $E=bias$ with $M=0$. This is $1+1023\times 2^{52}$ which is almost $2^{62}$. So there is no way to generate every possible floating-point number in [0,1] with the appropriate odds using less than a 62 random bits.

So the next attempt might be something like

The problem with this (and all such attempts), is this assumes the possible floating-point values are equally spaced, which they are not. For example, when $E=1$, the lowest bit of $M$ is a change of $2^{-1074}$, whereas when $E=1000$, the lowest bit of $M$ is a change of $2^{-75}$.

Thus all such attempts will fail at hitting all possible floating-point values in $[0,1]$, lowering the quality of the random number generation for demanding simulations.

## The solution

To generate all possible floating-point bit patterns with the appropriate probabilities, consider generating arbitrarily long random bit sequences $0.b_1b_2b_3…$, and rounding to the nearest representable floating-point numbers. This terminates since the smallest possible non-zero positive IEEE 754 64-bit number is $E=0$ and $M=1$, which is $2^{1-1023-51}=2^{-1074}$.

To do rounding, note that the probability of a trailing infinite sequence of 0s or 1s is each zero. (Note this is not the same as saying that it cannot happen; it is saying the probability of the infinite sequence happening is zero, which are two distinct things. As long as your simulation uses a finite number of random values, which seems pretty likely, then this is mathematically correct.).

To generate numbers in $[0,1)$, round down; to generate in $(0,1]$, round up, and for our case $[0,1]$, we’ll round up if the first bit past $M$ is 1, else we’ll round down. This will generate all possible floating-point numbers, with the correct probabilities.

In pseudo code (note this does not work in JavaScript since there is no 64 bit integer type) this is (we assume rand64 returns a uniform random 64 bit integer, over the whole range. Using other sources can be made to fit by concatenating bits as needed.):

Finally, specific language tricks/support allow shortcuts for some of these steps.

## More problems

Do the same analysis for the interval $[a,b]$, $a\leq b$ arbitrary. It’s messy but doable from the techniques explained above.