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The theorem states that

$$ H(P)\leq\mathrm{MinACL}(P)<H(P)+1 $$

where, $\mathrm{MinACL}$ means the minimum average code word length of a given information source, i.e. the average code word length of any Huffman coding and $H$ means the entropy of the probability distribution $P$.

Now, the problem is how to show that for any $\epsilon>0$, there is a probability distribution $P$ s.t. $\mathrm{MinACL}(P) - H(P)\geq1-\epsilon$?

(I was given a hint that I can start with a source s.t. $H(P)=\mathrm{MinACL}(P)$ and try to change the probabilities in order to skew the code.}

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Em…… I seem to have figured out how to construct a probability distribution to achieve $\mathrm{MinACL}$ as close to $H(P)+1$ as possible. Suppose, for the information source $(S,P)$ with $|S| = 2^l + 1$, $p_0 = 1 - \epsilon$, $p_i = \epsilon/2^l,1\leq i\leq2^l$, a valid Huffman tree could be constructed as follows: First build a full binary tree with $s_i,1\leq i\leq2^l$, as leaves and then make $s_0$ sibling of the root of this binary tree, and child of the Huffman tree root node. This scheme can be justified easily given $\epsilon\in(0,0.5)$. Then, $H(P) = -(1-\epsilon)\log(1-\epsilon)-(\epsilon/2^l)\log(\epsilon/2^l)$ and $\mathrm{MinACL}(P) = (l+1)\cdot\epsilon+1\cdot(1-\epsilon)$.

$$ \lim_{\epsilon\to0}\mathrm{MinACL}(P)-H(P) = 1 $$

(I haven't calculated the limit, but the graph below if of much evidence: enter image description here)

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  • $\begingroup$ Take $l=0$ for a simple example. $\endgroup$ – Yuval Filmus Feb 9 at 2:59
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You can take the distribution $P(0) = 1-\epsilon$, $P(1) = \epsilon$. This distribution has entropy $h(\epsilon) = O(\epsilon \log 1/\epsilon)$, but $\mathrm{MinACL}(P) = 1$. As $\epsilon$ tends to $0$, the gap tends to $1$, since $\lim_{\epsilon\to 0} h(\epsilon) = 0$.

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  • $\begingroup$ Thank you very much. I started out in a strange way so I didn't think of this simplest example. $\endgroup$ – Miangu Feb 14 at 2:35

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