1. 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 & 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.

2. 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.

3. 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.

4. 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.

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

6. 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 :
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 :

7. Subsequences stored : 0, 1, 00
Data to be parsed :
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 :

8. 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

9. 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

10. 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

11. 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.

12. 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

13. 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,

14. 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.

15. 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)

16. 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)

17. Consider the numerical position 6 in figure
Consider the numerical position 6 in figure. It is partially reproduced below.
Row 1: Numerical position
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)

18. 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 :

19. 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

20. Example: Determine the Lempel ZIV code for the following bit steram 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

21. 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

22. 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

23. 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

24. 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