Chapter 4 Variable–Length and Huffman Codes

Unique Decodability We must always be able to determine where one code word ends and the next one begins. Counterexample: Suppose: s 1 = 0; s 2 = 1; s 3 = 11; s 4 = = s 4 s 3 or s 1 s 1 s 3 Unique decodability means that any two distinct sequences of symbols (of possibly differing lengths) result in distinct code words. 4.1, 2

Instantaneous Codes No code word is the prefix of another. By reading a continuous sequence of code words, one can instantaneously determine the end of each code word. Consider the reverse: s 1 = 0; s 2 = 01; s 3 = 011; s 4 = ……111 is uniquely decodable, but the first symbol cannot be decoded without reading all the way to the end. s1s1 s2s2 s3s decoding tree s4s4 4.3 s 1 = 0 s 2 = 10 s 3 = 110 s 4 = 111

Constructing Instantaneous Codes comma code:s 1 = 0 s 2 = 10 s 3 = 110 s 4 = 1110 s 5 = 1111 modification:s 1 = 00 s 2 = 01 s 3 =10 s 4 = 110 s 5 = 111 s 1 = 00s 2 = 01s 3 = 10 s 4 = 110s 5 = Decoding tree Notice that every code word is located on the leaves 4.4

Kraft Inequality Basis: n = 1 s1s1 0,1 s1s1 0 s2s2 1 or Induction: n > 1 01 <n T0T0 T1T1 Prefixing one symbol at top of tree increases all the lengths by one, so Theorem:There exists an instantaneous code for S where each symbol s  S is encoded in radix r with length |s| if and only if Proof: (  ) By induction on the height (maximal length path) of the decoding tree, max{|s|: s  S}. For simplicity, pick r = 2 (the binary case). By IH, the leaves of T 0, T 1 satisfy the Kraft inequality. 4.5 Could use n = 0 here!

Same argument for radix r: Basis: n = 1 s1s1 0 s≤rs≤r ≤ r  1 …… ………… at most r Induction: n > 1 0 ≤ r  1 T0T0 T ≤r-1 at most r subtrees IH so adding at most r of these together gives ≤ 1  Inequality in the binary case implies that not all internal nodes have degree 2, but if a node has degree 1, then clearly that edge can be removed by contraction. 4.5 ……

Kraft Inequality (  ) Construct a code via decoding trees. Number the symbols s 1, …, s q so that l 1 ≤ … ≤ l q and assume K ≤ 1. Greedy method: proceed left-to-right, systematically assigning leaves to code words, so that you never pass through or land on a previous one. The only way this method could fail is if it runs out of nodes (tree is over-full), but that would mean K > 1. Exs:r = 21, 3, 3, 3r = 21, 2, 3, 3r = 21, 2, 2, 3 not used ½ + ⅛ + ⅛ + ⅛ < 1½ + ¼ + ⅛ + ⅛ = 1 ½ + ¼ + ¼ + ⅛ > 1 4.5

Shortened Block Codes With exactly 2 m symbols, we can form a set of code words each of length m : b 1 …… b m b i  {0,1}. This is a complete binary decoding tree of depth m. With < 2 m symbols, we can chop off branches to get modified (shortened) block codes s1s1 s2s2 s3s3 s4s4 s5s s1s1 s2s2 s3s3 s4s4 s5s5 0 1 Ex 1 Ex 2

McMillan Inequality Idea: Uniquely decodable codes satisfy the same bounds as instantaneous codes. Theorem: Suppose we have a uniquely decodable code in radix r of lengths of l 1 ≤ … ≤ l q. Then their Kraft sum is ≤ 1. Use a multinomial expansion to see that N k = the number of ways n l‘s can add up to k, which is the same as the number of different ways n symbols can form a coded message of length k. Because of uniqueness, this must be ≤ r k, the number of codewords. Conclusion: WLOG we can use only instantaneous codes. 4.7

Average code length Our goal is to minimize the average coded length. If p n > p m then l n ≤ l m. For if p m < p n with l m < l n, then interchanging the encodings for s m and s n we get So we can assume that if p 1 ≥ … ≥ p q then l 1 ≤ … ≤ l q, because if p i = p i+1 with l i > l i+1, we can just switch s i and s i old new >

Start with S = {s 1, …, s q } the source alphabet. And consider B = {0, 1} as our code alphabet (binary). First, observe that l q  1 = l q, since the code is instantaneous, s l q  1 ) won’t hurt. Huffman algorithm: So, we can combine s q  1 and s q into a “combo-symbol” (s q  1 +s q ) with probability (p q  1 +p q ) and get a code for the reduced alphabet. For q = 1, assign s 1 = ε. For q > 1, let s q-1 = (s q-1 +s q ) 0 and s q = (s q-1 +s q ) Example: ε N. B. the case for q = 1 does not produce a valid code. 4.8

Huffman is always of shortest average length Assume p 1 ≥ … ≥ p q Huffman Alternative  L avg  L L ≥ trying to show We know l 1 ≤ … ≤ l q Example: p 1 = 0.7; p 2 = p 3 = p 4 = 0.1 Compare L avg = 1.5 to log 2 q = 2. Base Case: For q = 2, no shorter code exists. Induction Step: For q > 2 take any instantaneous code for s 1, …, s q with minimal average length. s1s1 10 s2s2 4.8

Claim that l q  1 = l q = l q  1, q + 1 because So its reduced code will always satisfy: By IH, L′ avg ≤ L′. But more importantly the reduced Huffman code shares the same properties so it also satisfies the same equation L′ avg + (p q  1 + p q ) = L avg, hence L avg ≤ L. 4.8 reduced code 01 sq1 + sqsq1 + sq combined symbol total height = l q s 1, ………sq1sq1 sqsq

Code Extensions Takep 1 = ⅔ and p 2 = ⅓Huffman code gives s 1 = 0 s 2 = 1 L avg = 1 Square the symbol alphabet to get: S 2 : s 1,1 = s 1 s 1 ;s 1,2 = s 1 s 2 ; s 2,1 = s 2 s 1 ; s 2,2 = s 2 s 2 ; p 1,1 = 4 ⁄ 9 p 1,2 = 2 ⁄ 9 p 2,1 = 2 ⁄ 9 p 2,2 = 1 ⁄ 9 Apply Huffman to S 2 : s 1,1 = 1; s 1,2 = 01; s 2,1 = 000; s 2,2 = 001 But we are sending two symbols at a time! 4.10

Huffman Codes in radix r At each stage down, we merge the last (least probable) r states into 1, reducing the # of states by r  1. Since we end with one state, we must begin with no. of states  1 mod (r  1). We pad out states with probability 0 to get this. Example: r = 4; k =  4.11 pads