2c03-review - 00068 - 13[20 Suppose T is a Huffman tree and...

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11 13.[20] Suppose T is a Huffman tree, and that the leaf for symbol a has greater depth than the leaf for symbol b . Prove that the probability of symbol b is no less than that of a . It can be proven in many ways by induction. The simplest seems to be the following. Let a 1 , … , a n , be a sequence of characters. Without loss of generality we may assume that prob(a i ) prob(a j ) if i<j, so a 1 , a 2 are characters with the smallest probabilities. First we prove the following lemma: “a 1 , a 2 are siblings nodes whose depth is at least as big as any other leaf in the tree”. Suppose that a 1 and a 2 are not the deepest nodes in the tree. In this case, the Huffman tree must either look as the below one, or be symmetrical to this. v u a 1 a 2 x For this situation to occur, the parent of a 1 and a 2 , labeled v, must have greater combined probability than the node labeled x. Otherwise we would have selected v in place of node x as the child of u. However, this is impossible since a 1 and a 2 are the symbols with the smallest probabilities. Hence our lemma is true.
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