Kraft's Inequality - Proof of The General Case

Proof of The General Case

Consider the generating function in inverse of x for the code S

in which —the coefficient in front of —is the number of distinct codewords of length . Here min is the length of the shortest codeword in S, and max is the length of the longest codeword in S.

For any positive integer m consider the m-fold product Sm, which consists of all the words of the form, where are indices between 1 and n. Note that, since S was assumed to uniquely decodable, if, then . In other words, every word in comes from a unique sequence of codewords in . Because of this property, one can compute the generating function for from the generating function as

G(x) = \left( F(x) \right)^m = \left( \sum_{i=1}^n x^{-|s_i|} \right)^m
= \sum_{i_1=1}^n \sum_{i_2=1}^n \cdots \sum_{i_m=1}^n x^{-\left(|s_{i_1}| + |s_{i_2}| + \cdots + |s_{i_m}|\right)} = \sum_{i_1=1}^n \sum_{i_2=1}^n \cdots \sum_{i_m=1}^n x^{-|s_{i_1} s_{i_2}\cdots s_{i_m}|}
= \sum_{\ell=m \cdot \min}^{m \cdot \max} q_\ell \, x^{-\ell} \; .

Here, similarly as before, —the coefficient in front of in —is the number of words of length in . Clearly, cannot exceed . Hence for any positive x


\left( F(x) \right)^m \le \sum_{\ell=m \cdot \min}^{m \cdot \max} r^\ell \, x^{-\ell} \; .

Substituting the value x = r we have


\left( F(r) \right)^m \le m \cdot (\max-\min)+1

for any positive integer . The left side of the inequality grows exponentially in and the right side only linearly. The only possibility for the inequality to be valid for all is that . Looking back on the definition of we finally get the inequality.


\sum_{i=1}^n r^{-\ell_i} = \sum_{i=1}^n r^{-|s_i|} = F(r) \le 1 \; .

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