Given normally distributed integers with a mean of 0 and a standard deviation $\sigma$ around 1000, how do I compress those numbers (almost) perfectly? Given the entropy of the Gaussian distribution, it should be possible to store any value $x$ using $$\frac{1}{2} \mathrm{log}_2(2\pi\sigma^2)+\frac{x^2}{2\sigma^2}\rm{log}_2\rm{e} \;\text{ bits.}$$
Arithmetic coding should provide perfect compression. In principle it's not too hard. I can calculate the interval boundaries from the cumulative distribution function of the Gaussian. In practice, I hit considerable difficulties because when using floating point operations I cannot achieve perfect reproduction of the results, and I have no idea how to do this without FP operations. Perfect reproduction is necessary because the uncompressing code must come up with exactly the same interval boundaries as the compressing code.

So my question is: How do I compute the interval boundaries? Or is there any other way to achieve (near) perfect compression of such data?

Strictly speaking, the normal distribution is defined only for continuous variables. So what I mean here are integers x with a probability distribution function $$P(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\rm e^{-\frac{x^2}{2\sigma^2}}.$$
This distribution does not sum up exactly to 1. However, for $\sigma>100$ the difference from 1 is less than $10^{-1000}$ and hence it's more precise than any practical calculation would be.


2 Answers 2


In what follows, I'm going to assume that $x \ge 0$, and then try to put that back later.

First, we'll simplify the notation a little. The length of the code for $x$ should be:

$$L_m(x) = a + b \left(\frac{x}{\sigma}\right)^2 + \log_2 \sigma$$


$$a = \frac{\log_2 2\pi}{2} \approx 1.325$$ $$b = \frac{1}{2 \ln 2} \approx 0.721$$

This suggests that you should divide $x$ by $\sigma$, encode the remainder in $\log_2 \sigma$ bits, then encode the quotient $q$ in $a + bq^2$ bits (say, using arithmetic coding). Instead of storing $20002$ codes using Yuval's method, you now only need $20$ or so.

In a sense, what you're doing here is converting the normal distribution to a standard distribution. Moreover, if you move $\sigma$ to the nearest power of 2, the division can be done using bit shifts and masks. $\sigma=1024$ is close enough to $1000$, so that part of the operation is extremely cheap.

However, there is a catch. To arithmetic code $L_0(10)$, you need about 60 bits. On a 64-bit machine, this is pushing the practical limit of an efficient arithmetic coder. So you should probably consider partitioning the space for more efficient encoding.

One possibility is to use binary arithmetic coding, with one binary symbol per $\sigma$. The probability that $x$ is within $\sigma$ of the mean is $0.6826895$ (or so). The probability that $x$ is within $2\sigma$ given that it is not within $\sigma$ is $0.85660651$ (or so), and so on. I calculate the code length for the $8\sigma$ symbol to be around 27.3 bits, and trying to calculate the length for $9\sigma$ breaks into numeric instability using the simple calculation that I did.

No matter what you try, representing numbers that are more than about $8\sigma$ from the mean is going to be difficult to do well, but given that those numbers are extremely rare, perhaps you should just fall back to a universal code (e.g. one of the Elias codes) for that case.

So in summary, here's the representation that I'd try.

  1. Divide $\left| x \right|$ by $\sigma$. Call the quotient $q$ and the remainder $r$.
  2. Store the sign of $x$ as a single bit.
  3. Arithmetic encode $q$ using a sequence of binary variables:
    1. Use a single binary symbol with $p_0 = 0.6826894921370859$. Store $0$ if $q=0$, or $1$ if $q>0$.
    2. If you stored $1$, use another binary symbol with $p_0 = 0.8566065013011934$. Store $0$ if $q=1$, or $1$ if $q>1$.
    3. ...and so on. Stop when you get to $q \ge 8$).
  4. If $q<8$, binary encode $r$ in $\log_2 \sigma$ bits.
  5. If $q\ge 8$, use a universal binary code of choice (Elias gamma is a popular choice) to store $x - 8\sigma$.

Now having said all this, a common approach used in many compression standards, which works fairly well when $\sigma$ is small, is to use exp-Golomb coding. It's not theoretically perfect, but seems to work well in practice.

  • $\begingroup$ Thanks, that seems quite practical. Is there any advantage in picking multiples of $sigma$ to split that interval? I tend to prefer splitting it into pieces with p_0=0.5 each, would that work as well? $\endgroup$ Mar 4, 2014 at 8:31
  • $\begingroup$ By al means try it! The only reason why I used $\sigma$ is the conspicuous $\log \sigma$ term; storing a remainder modulo $\sigma$ is exactly that size. That's often a strong hint in designing close-to-optimal compression schemes. One other thing to keep in mind is that you shouldn't be afraid of overhead if it's asymptotically well-behaved. Overhead is a fact of life. $\endgroup$
    – Pseudonym
    Mar 4, 2014 at 11:00
  • $\begingroup$ How do you 60 bits in your example. I get something close to 73. $\endgroup$
    – Isaac Gerg
    Feb 5, 2021 at 16:59
  • $\begingroup$ Also, what does the subscript m in $L_m$ mean? $\endgroup$
    – Isaac Gerg
    Feb 5, 2021 at 17:06
  • $\begingroup$ @IsaacGerg It's been 6 years since I wrote this, and I'm not completely sure. However... good catch. I think that should have been $L_0$ everywhere, to denote zero-order compression. And I think that should have read $L_0(10\sigma)$. Does that check out correctly? $\endgroup$
    – Pseudonym
    Feb 7, 2021 at 22:54

Let $P(x)$ be the quantity mentioned in your question, define a sequence $a_0,\ldots$ by $$ 0,1,-1,2,-2,3,-3,4,-4,\ldots, $$ and let $Q(n) = P(0) + \cdots + P(n-1)$. You're interested in computing the values $Q(n)$ in a consistent way across different computers. Your actual implementation is going to store these values only up to a certain accuracy. You can compute the exact numbers to that accuracy using software floating-point arithmetic. You can then store the results in some table. The table won't be too long, since the probability that $|x| > 10\sigma$ is smaller than $2^{-75}$, and so it is enough to store $20002$ values.

By the way, if you want to store the numbers as floating-point reals, you are constrained by the size of the mantissa: a long double, for example, has a mantissa of length $63$ bits on Intel CPUs, and so you cannot handle events with probably on the order of $2^{-64}$ or smaller (which is probably not a problem). If you are adamant about handling such events, you will need to store these values as fixed-point reals, which is perhaps what your arithmetic coding code expects anyway.


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