Lampel ZIV (LZ) code The Lempel-Ziv algorithm is a variable-to-fixed length code Basically, there are two versions of the algorithm LZ77 and LZ78 are the two lossless data compression algorithms published by Abraham Lempel and Jacob Ziv in 1977 & 1978. They are also known as LZ1 and LZ2 respectively. These two algorithms form the basis for many variations including LZW, LZSS, LZMA and others. Besides their academic influence, these algorithms formed the basis of several ubiquitous compression schemes, including GIF and the DEFLATE algorithm used in PNG.

They are both theoretically dictionary coders. LZ77 keeps track of last n-bytes seen of data & when a phrase is encountered that already has been seen, it outputs a pair of values corresponding to the position of pharase in previously seen buffer data & it moves a fixed size window over data. It does so by maintaining a sliding window during compression. This was later shown to be equivalent to the explicit dictionary constructed by LZ78—however, they are only equivalent when the entire data is intended to be decompressed. LZ78 decompression allows random access to the input as long as the entire dictionary is available, while LZ77 decompression must always start at the beginning of the input.

If we compare it Huffman code then we find the major disadvantage of the Huffman code is that the symbol probabilities must be known or estimated if they are unknown. In addition to this, the encoder and Decoder must know the coding tree. Moreover in the modeling text, the storage requirement prevent the Huffman code from capturing the higher order relationship between words and phrases.

So we have to compromise the efficiency of code. These practical limitation of Huffman code can be overcome by using the lampel ZIV algorithm. It is adaptive and simpler to implement as compared to Huffman coding.

Principal of Lampel ZIV algorithm The illustrate this principle let us consider the example of an input binary sequence specified as : 000101110010 The encoding in this algorithm is accomplished by parsing the source data stream into segments that are the shortest substances not encountered previously.

We assume that the binary symbols 0 and 1 are already stored in this order in the code book. Hence we write, subsequences stored : 0, 1 Data to be parsed : 000101110010 Now examine the data in above equation from LHS and find the shortest subsequence which is not encountered previously. It is 00. so we include 00 as the next entry in the subsequence and move00 from data to subsequence as follow :

Subsequences stored : 0, 1, 00 Data to be parsed : 010110010 The next shortest Subsequences which is not previously repeated is 01. In above equation Note that we are examining from LHS. Hence we write, Subsequences stored : 0, 1, 00, 01 Data to be parsed : 01110010

The next shortest Subsequences which is previously not encountered is 011. so we write, Subsequences stored : 0, 1, 00, 01,011 Data to be parsed : 10010 Similarly we can continue until the data stream has been completely parsed. The code book of binary Subsequences gets ready as shown in figure

Code book of Sequence The first row in the codebook shows the numerical position of various subsequence in the codebook. Numerical Position 1 2 3 4 5 6 7 Subsequences 00 01 001 10 010

Numerical representation : Let us now add third row to figure. This row is called as numerical representation as shown in figure Numerical Position 1 2 3 4 5 6 7 Subsequences 00 01 011 10 010 Numerical representation 11 12 42 21 41

The sequences 0 and 1 are originally stored The sequences 0 and 1 are originally stored. So consider the third Subsequences i.e. 00. this is the first Subsequences in the data stream and it is made up of concatenation of the first Subsequences i.e. 0 with itself. Hence it is represented by 11 in the row of numerical representation in above figure Similarly, subsequences 01 obtained by concatenation of first and second subsequences so we enter 12 below that. The remaining subsequences are treated accordingly.

Binary Encoded Representation : The last (4th ) row added as shown in figure, is the binary encoded representation of each subsequence. Numerical Position 1 2 3 4 5 6 7 Subsequences 00 01 011 10 010 Numerical representation 11 12 42 21 41 Binary encoded blocks 0010 0011 1001 0100 1000

The question is how to obtain binary encoded blocks. the last symbol of each subsequence in the second row of above figure (called as codebook) is called as an innovation symbol. So the last bit in each binary encoded block (4th row) is the innovation symbol of the corresponding subsequence,

The remaining bits provide the equivalent binary representation of the “pointer” to the “root subsequence” that matches the one in question except for the innovation symbol. This can be explained as follow. Consider Numerical position 3 in figure. The binary encoded block is 0010. Consider Numerical position 5 in figure. It is partially reproduced below.

Row 1: Numerical position 3 Row 2: Subsequence Row 4: Binary encoded Block Innovation number This is the first subsequence Take as it is 001 Binary equivalent of 1(this is called pointer)

Row 1: Numerical position 5 Row 2: Subsequence Row 4: Binary encoded Block Innovation number 01 1 This is the 4th subsequence Take as it is 100 1 Binary equivalent of 4. (this is called pointer)

Consider the numerical position 6 in figure Consider the numerical position 6 in figure. It is partially reproduced below. Row 1: Numerical position 6 Row 2: Subsequence Row 4: Binary encoded Block Similarly the other entries in the fourth row are made. Innovation number 1 This is the 2nd subsequence Take as it is 010 Binary equivalent of 2. (this is called pointer)

Decoder The decoding is as simple encoding. The steps followed at the time of decoding are as follows : Step 1 : Take the binary encoded block. For example consider the binary encoded block in position 5 i.e. 1001 Step 2 : use the pointer to identify the root subsequence :

Append the innovation symbol to the subsequence in step 2: Binary encoded block Append the innovation symbol to the subsequence in step 2: Append the innovation number i.e. 1 to the root subsequence of 01 to get the subsequence 011 corresponding to position 5. 100 1 Innovation number Pointer = 4 Pointer value 4 corresponds to 4th subsequence i.e. 01

Example: Determine the Lempel ZIV code for the following bit steram 01001111100101000001010101100110000 Recover the original sequence from the encoded stream Soln. Part 1 : Encoding We assume that the binary symbols 0 and1 are already stored in the code book. Subsequences stored : 0, 1

Encoding is accomplished by parsing the source data stream into segment that are shortest substances, not encountered previously. The given stream of bits can be parsed into subsequence as shown below : 0, 1, 00, 11, 111, 001, 01, 000, 0010, 10, 101, 100, 110, 000 The encoding table is as shown in table

Numerical representation Part II Decoding Consider the code for example Numerical Position 1 2 3 4 5 6 7 8 9 10 11 12 Subsequences 00 111 001 01 000 0010 101 100 Numerical representation - 22 42 32 31 61 21 102 code 0101 1001 0111 0011 0110 1100 0100 10101 10100

The decoding table is shown in table Corre Ss corresponding subsequence is 00 The decoding table is shown in table 001 Innovation number (do not change) Pointer = 1 This value corresponds to 1st subsequence is 0

Thus we get the original sequence back Decoding table. Thus we get the original sequence back Code 0010 0101 1001 0111 0011 0110 1100 0100 10101 10100 Innovation bit 1 Pointer 001 010 100 011 110 1010 Decoded subsequence 0 0 1 1 1 1 1 0 0 1 0 1 0 0 0 0 010 1 0 1 0 1 1 0 0