Prove that the average codeword length in a Huffman tree is $\Omega(\log n)$, where $n$ is the number of characters.

My try:

I think that the worst case is when the tree is full and all the characters are in the highest level.

Therefore: $n=2^h \to h=\log n$, and the average codeword length is $\Omega(\log n)$.

Am I missing something?

  • $\begingroup$ You did not use the probabilities of the characters? Most probable characters have the shortest code. $\endgroup$ Commented Jun 1, 2019 at 12:11

2 Answers 2


This answer assumes that by average you mean just that – the sum of all codeword lengths divided by $n$.

Let us show that any prefix code satisfies your property. Consider any prefix code whose codewords have lengths $\ell_1,\ldots,\ell_n$. Kraft's inequality shows that $$ 2^{-\ell_1} + \cdots + 2^{-\ell_n} \leq 1. $$ The function $x \mapsto 2^{-x}$ is convex, hence $$ \frac{1}{n} \geq \frac{2^{-\ell_1} + \cdots + 2^{-\ell_n}}{n} \geq 2^{-(\ell_1+\cdots+\ell_n)/n}. $$ It follows that $$ \frac{\ell_1 + \cdots + \ell_n}{n} \geq \log_2 n. $$ Since $x \mapsto 2^{-x}$ is strictly convex, there can be equality only if $\ell_1 = \cdots = \ell_n$, that is, only if all codewords are of length exactly $\log_2 n$. That can only happen if $n$ is a power of 2. This leaves open the following question:

What is the minimum average codeword length of a prefix code on $n$ symbols?

(We note in passing that every prefix code satisfying Kraft's inequality tightly is a minimum redundancy code for some distribution, namely the one in which the probability of the $i$th symbol is $2^{-\ell_i}$.)

To answer this, let us note that convexity implies that if $a < b$ then $$ 2^{-a} + 2^{-b} \geq 2^{-(a+1)} + 2^{-(b-1)}. $$ (We can also see this directly: without loss of generality $a=0$, and then we need to prove $1 + 2^{-b} \geq 1/2 + 2 \cdot 2^{-b}$, that is, $1/2 \geq 2^{-b}$, which follows from $b \geq 1$.)

Applying this operation repeatedly starting at some arbitrary integer solution of Kraft's equation, we eventually reach a solution which involves at most two different values $\ell,\ell+1$. If $n = 2^k + m$, where $0 \leq m < 2^k$, then this solution must consist of $2^k - m$ many $k$'s and $2m$ many $k+1$'s. Hence the optimal average codeword length is $$ \frac{k(2^k-m)+(k+1)2m}{2^k+m} = k + \frac{2m}{2^k+m}. $$

  • 1
    $\begingroup$ You can describe any prefix code as a binary tree, where the root-to-leaf path of element $x$ spells the code of $x$. The binary tree corresponding to a Huffman code is a Huffman tree. $\endgroup$ Commented Jan 3, 2021 at 20:44

If by average you mean the average length of the n codewords, then if the k-th symbol has probability $2^{-k}$, the length of the code words ranges from 1 to n-1, with average about n/2.

If by average you mean the average length of a codeword in a compressed message, then if the first symbol has probability 1-eps, and the others all have probability (1 - eps) / (n - 1), then the average length of a codeword in a compressed message is just a bit more than 1.


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