Aufgaben:Exercise 2.3Z: About the LZ77 Coding

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Sliding–Window with  $G = 4$  and  $G = 5$
(valid for "LZ77")

In  Exercise 2.3,  you were supposed to compress  "BARBARA–BAR"  (string of length  $11$, four different characters)  using the LZ78 algorithm. 

In this exercise we use the same text to demonstrate LZ77 compression.  To note:

• While the successor  "LZ78"  successively builds a global dictionary,  "LZ77"  uses a local dictionary.
• The LZ77 method works with a  "Sliding Window"  that is gradually shifted over the input text.
• This  "sliding window"  is divided into the  "preview buffer"  (highlighted in blue in the graphic)  and the  "search buffer"  (highlighted in red).  Both buffers have a size of  $G$  memory locations each.
• Each coding step  $i$  is characterised by a triple  $(P,\ L,\ Z)$.  Here,  $P$  and  $L$  are integers and  $Z$  is a character.  Transmitted are the binary representations of  $P$,  $L$  and  $Z$.
• After transmission, the  sliding window  is shifted one or more positions to the right and the next coding step  $i + 1$ begins.

The upper diagram shows the initial allocation with the buffer size  $G = 4$  at the times  $i = 1$  as well as  $i = 4$.

• At time  $i = 1$ , the search buffer is empty, so the coder output is  $(0, 0,$ B$)$ .
• After the shift by one position, the search buffer contains a  B, but no string that begins with  A.  The second number triple is therefore  $(0, 0,$ A$)$.
• The output for  $i = 3$  is  $(0, 0,$ R$)$, since there is no string beginning with  R  in the search buffer even now.

The snapshot at time  $i = 4$  is also shown in the graphic.  The search is now for the character string in the search buffer that best matches the preview text  BARA .  A number triple  $(P,\ L,\ Z)$, is transmitted again, but now with the following meaning:

• $P$  indicates the position in the (red) search buffer where the match starts.  The  $P$  values of the individual memory locations can be seen in the graphic.
• $L$  denotes the number of characters in the search buffer that, starting at  $P$,  match the current string in the preview buffer.
• Finally $Z$  denotes the first character in the preview buffer that differs from the matching string found in the search buffer.

The larger the LZ77 parameter  $G$ , the easier it is to find the longest possible match.  In subtask  (4)  you will notice that the LZ77 encoding with  $G = 5$  gives a better result than the one with  $G = 4$.

• However, because of the later binary representation of  $P$,  one will always choose  $G$  as a power of two, so that  $G$  can be represented with  $\log_2 \ P$  bits  $(G = 8$   ⇒   three-digit binary number  $P)$.
• This means that a  "Sliding Window"  with  $G = 5$  has rather little practical relevance.

Hints:

• The task belongs to the chapter  Compression according to Lempel, Ziv and Welch.
• In particular, reference is made to the page  LZ77 – the basic form of the Lempel-Ziv algorithms .
• The original literature  [LZ77]  on this method is:
        Ziv, J.; Lempel, A.:  A Universal Algorithm for Sequential Data Compression.  In: IEEE Transactions on Information Theory, no. 3, vol. 23, 1977, p. 337–343.

•   Exercise 2.3  and  Exercise 2.4  deal with other Lempel–Ziv methods in a similar way.

Questions

1

What is the LZ77 output with  $G = 4$  at step  $i = 4$?

	$(0, 0,$ B$)$,
	$(2, 1,$ A$)$,
	$(2, 3,$ A$)$.

2

Which statement is true for the same buffer size  $G = 4$  at step  $i = 5$?

	The search buffer contains  "BARA".
	The preview buffer says  "–BAR".
	The output is  $(0, 0,$ A$)$.

3

After which step  $i_{\rm end}$  is the coding with  $G = 4$  finished?

$i_{\rm end} \ = \ $

4

Now let  $G = 5$.  After which step  $i_{\rm end}$  is the coding finished?

$i_{\rm end} \ = \ $

5

What advantages does LZ78 have over LZ77 for "very large" files?

	One finds more often already stored phrases in the dictionary.
	Fewer bits have to be transferred per coding step.

Solution

Solution

Example of the LZ77 algorithm with  $G = 4$

(1) Solution suggestion 3 is correct.

• The preview buffer contains the character string  "BARA" at the time  $i = 4$ .
• The search buffer contains  "BAR" in the last three digits:

$$P = 2\hspace{0.05cm},\hspace{0.2cm}L = 3\hspace{0.05cm},\hspace{0.2cm}Z = \boldsymbol{\rm A}\hspace{0.05cm}.$$

(2) The first two solutions are correct:

• The hyphen is not found in the search buffer at time  $i = 5$.
• The output is  $(0, 0,$ –$)$.

(3)  The upper graphic shows the  "Sliding Window"  and the coder output at times  $i>5$.

• After  $i = 9$  coding steps the coding process is finished considering  eof    ⇒   $\underline{i_{\rm end} = 9}$.

Example of the LZ77 algorithm with  $G = 5$

(4)  With a larger buffer size  $(G = 5$  instead of  $G = 4)$  the coding is already completed after the 6th coding step   ⇒   $\underline{i_{\rm end} = 6}$.

• A comparison of the two graphs shows that nothing changes for  $G = 5$  compared to  $G = 4$  up to and including  $i = 5$ .
• Due to the larger buffer, however,  "BAR"  can now be coded together with  "eof"  (end-of-file) in a single step, whereas with  $G = 4$  four steps were necessary for this.

(5) Only statement 1 is correct.

• A disadvantage of LZ77 is the local dictionary.  Phrases that are actually already known cannot be used for data compression if they occurred more than  $G$  characters earlier in the text.  In contrast, with LZ78 all phrases are stored in the global dictionary.
• It is true that with LZ78 only pairs  $(I, \ Z)$  have to be transmitted, whereas with LZ77 each coding step is identified by a triple  $(P, \ L, \ Z)$ .  However, this does not mean that fewer bits have to be transmitted per coding step.
• Consider, for example, the buffer size  $G = 8$.  With LZ77, one must then represent  $P$  with three bits and  $L$  with four bits.  Please note that the match found between the preview buffer and the search buffer may also end in the preview buffer.
• The new character  $Z$  requires exactly the same number of bits in LZ78 as in LZ77 (namely two bits), if one assumes the symbol set size  $M = 4$  as here.
• Statement 2 would only be correct if  $N_{\rm I}$  were smaller than  $N_{\rm P}+ N_{\rm L}$, for example  $N_{\rm I} = 6$.  But this would mean that one would have to limit the dictionary size to  $I = 2^6 = 64$ .  This is not sufficient for large files.
• Our rough calculation, however, is based on a uniform number of bits for the index  $I$.  With a variable number of bits for the index, you can save quite a few bits by transmitting  $I$  only with as many bits as are necessary for the coding step  $i$ .
• In principle, however, this does not change the limitation of the dictionary size, which will always lead to problems with large files.