Given a string of values $AAAAAAAABC$, the Shannon Entropy in log base $2$ comes to $0.922$. From what I understand, in base $2$ the Shannon Entropy rounded up is the minimum number of bits in binary to represent a single one of the values.

Taken from the introduction on this wikipedia page:


So, how can three values be represented by one bit? $A$ could be $1$, $B$ could be $0$; but how could you represent $C$?

Thank you in advance.


3 Answers 3


The entropy you've calculated isn't really for the specific string but, rather, for a random source of symbols that generates $A$ with probability $\tfrac{8}{10}$, and $B$ and $C$ with probability $\tfrac1{10}$ each, with no correlation between successive symbols. The calculated entropy for this distribution, $0.922$ means that you can't represent strings generated from this distribution using less than $0.922$ bits per character, on average.

It might be quite hard to develop a code that will achieve this rate.* For example, Huffman coding would allocate codes $0$, $10$ and $11$ to $A$, $B$ and $C$, respectively, for an average of $1.2$ bits per character. That's quite far from the entropy, though still a good deal better than the naive encoding of two bits per character. Any attempt at a better coding will probably exploit the fact that even a run of ten consecutive $A$s is more likely (probability $0.107$) than a single $B$.

* Turns out that it isn't hard to get as close as you want – see the other answers!


Here is a concrete encoding that can represent each symbol in less than 1 bit on average:

First, split the input string into pairs of successive characters (e.g. AAAAAAAABC becomes AA|AA|AA|AA|BC). Then encode AA as 0, AB as 100, AC as 101, BA as 110, CA as 1110, BB as 111100, BC as 111101, CB as 111110, CC as 111111. I've not said what happens if there is an odd number of symbols, but you can just encode the last symbol using some arbitrary encoding, it doesn't really matter when the input is long.

This is a Huffman code for the distribution of independent pairs of symbols, and corresponds to choosing $n = 2$ in Yuval's answer. Larger $n$ would lead to even better codes (approaching the Shannon entropy in the limit, as he mentioned).

The average number of bits per symbol pair for the above encoding is $$\frac{8}{10} \cdot \frac{8}{10} \cdot 1 + 3 \cdot \frac{8}{10} \cdot \frac{1}{10} \cdot 3 + \frac{1}{10} \cdot \frac{8}{10} \cdot 4 + 4 \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot 6 = 1.92$$ i.e. $1.92/2 = 0.96$ bits per symbol, not that far from the Shannon entropy actually for such a simple encoding.


Let $\mathcal{D}$ be the following distribution over $\{A,B,C\}$: if $X \sim \mathcal{D}$ then $\Pr[X=A] = 4/5$ and $\Pr[X=B]=\Pr[X=C]=1/10$.

For each $n$ we can construct prefix codes $C_n\colon \{A,B,C\}^n \to \{0,1\}^*$ such that $$ \lim_{n\to\infty} \frac{\operatorname*{\mathbb{E}}_{X_1,\ldots,X_n \sim \mathcal{D}}[C_n(X_1,\ldots,X_n)]}{n} = H(\mathcal{D}). $$

In words, if we encode a large number of independent samples from $\mathcal{D}$, then on average we need $H(\mathcal{D}) \approx 0.922$ bits per sample. Intuitively, the reason we can do with less than one bit is that each individual sample is quite likely to be $A$.

This is the real meaning of entropy, and it shows that computing the "entropy" of a string $A^8BC$ is a rather pointless exercise.


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