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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
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This note was uploaded on 12/10/2009 for the course CAS 2c03 taught by Professor Janicki during the Spring '03 term at McMaster University.
 Spring '03
 janicki

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