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In the MP3 audio compression scheme, a sound signal is encoded in three steps 1 It is digitized by sampling at regular intervals, yielding a sequence of real numbers s1 , s2 , , sT For instance, at a rate of 44,100 samples per second, a 50-minute symphony would correspond to T = 50 60 44,100 130 million measurements 1
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2 Each real-valued sample st is quantized: approximated by a nearby number from a nite set This set is carefully chosen to exploit human perceptual limitations, with the intention that the approximating sequence is indistinguishable from s 1 , s2 , , sT by the human ear 3 The resulting string of length T over alphabet is encoded in binary It is in the last step that Huffman encoding is used To understand its role, let s look at a toy example in which T is 130 million and the alphabet consists of just four values, denoted by the symbols A, B, C, D What is the most economical way to write this long string in binary The obvious choice is to use 2 bits per symbol say codeword 00 for A, 01 for B, 10 for C, and 11 for D in which case 260 megabits are needed in total Can there possibly be a better encoding than this In search of inspiration, we take a closer look at our particular sequence and nd that the four symbols are not equally abundant Symbol A B C D Frequency 70 million 3 million 20 million 37 million
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Is there some sort of variable-length encoding, in which just one bit is used for the frequently occurring symbol A, possibly at the expense of needing three or more bits for less common symbols A danger with having codewords of different lengths is that the resulting encoding may not be uniquely decipherable For instance, if the codewords are {0, 01, 11, 001}, the decoding of strings like 001 is ambiguous We will avoid this problem by insisting on the pre x-free property: no codeword can be a pre x of another codeword Any pre x-free encoding can be represented by a full binary tree that is, a binary tree in which every node has either zero or two children where the symbols are at the leaves, and where each codeword is generated by a path from root to leaf, interpreting left as 0 and right as 1 (Exercise 528) Figure 510 shows an example of such an encoding for the four symbols A, B, C, D Decoding is unique: a string of bits is decrypted by starting at the root, reading the string from left to right to move downward, and, whenever a leaf is reached, outputting the corresponding symbol and returning to the root It is a simple scheme and pays off nicely for our toy example, where (under the codes of Figure 510) the total size of the binary string drops to 213 megabits, a 17% improvement
For stereo sound, two channels would be needed, doubling the number of samples
Figure 510 A pre x-free encoding Frequencies are shown in square brackets
0 1 [60] A [70] [23]
Symbol A B C D
Codeword 0 100 101 11
D [37]
B [3]
C [20]
In general, how do we nd the optimal coding tree, given the frequencies f 1 , f2 , , fn of n symbols To make the problem precise, we want a tree whose leaves each correspond to a symbol and which minimizes the overall length of the encoding, cost of tree =
n i=1
fi (depth of ith symbol in tree)
(the number of bits required for a symbol is exactly its depth in the tree) There is another way to write this cost function that is very helpful Although we are only given frequencies for the leaves, we can de ne the frequency of any internal node to be the sum of the frequencies of its descendant leaves; this is, after all, the number of times the internal node is visited during encoding or decoding During the encoding process, each time we move down the tree, one bit gets output for every nonroot node through which we pass So the total cost the total number of bits which are output can also be expressed thus: The cost of a tree is the sum of the frequencies of all leaves and internal nodes, except the root The rst formulation of the cost function tells us that the two symbols with the smallest frequencies must be at the bottom of the optimal tree, as children of the lowest internal node (this internal node has two children since the tree is full) Otherwise, swapping these two symbols with whatever is lowest in the tree would improve the encoding This suggests that we start constructing the tree greedily: nd the two symbols with the smallest frequencies, say i and j, and make them children of a new node, which then has frequency fi + fj To keep the notation simple, let s just assume these are f 1 and f2 By the second formulation of the cost function, any tree in which f 1 and f2 are sibling-leaves has cost f1 + f2 plus the cost for a tree with n 1 leaves of frequencies (f 1 + f2 ), f3 , f4 , , fn : 143
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