Aufgaben:Exercise 2.8: Huffman Application for a Markov Source

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Binary symmetric Markov source

We consider the binary symmetric Markov source according to the graph, which is given by the single parameter

$$q = {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y})$$

completely.

• The given source symbol sequences apply to the conditional probabilities  $q = 0.2$  bzw.  $q = 0.8$, respectively.
• In subtask  (1)  it has to be clarified which symbol sequence – the red or the blue one – was generated with  $q = 0.2$  and which with  $q = 0.8$ .

The properties of Markov sources are described in detail in the chapter  Message Sources with Memory .  Due to the symmetry assumed here with regard to the binary symbols  $\rm X$  and  $\rm Y$ , some serious simplifications result, as is derived in  exercise 1.5Z :

• The symbols  $\rm X$  and  $\rm Y$  are equally probable, that is,  $p_{\rm X} = p_{\rm Y} = 0.5$ holds. 
  Thus the first entropy approximation is:   $H_1 = 1\,\,{\rm bit/source\:symbol}\hspace{0.05cm}. $
• The entropy of the Markov source for  $q = 0.2$  as well as for  $q = 0.8$  results in

$$H = q \cdot {\rm log_2}\hspace{0.15cm}\frac{1}{q} + (1-q) \cdot {\rm log_2}\hspace{0.15cm}\frac{1}{1-q} = 0.722\,\,{\rm bit/source\:symbol}\hspace{0.05cm}.$$

• For Markov sources, all entropy approximations  $H_k$  with order  $k \ge 2$  are determined by  $H_1$  and  $H = H_{k \to \infty}$ :

$$H_k = {1}/{k}\cdot \big [ H_1 + H \big ] \hspace{0.05cm}.$$

• The following numerical values again apply equally to  $q = 0.2$  and  $q = 0.8$ :

$$H_2 = {1}/{2}\cdot \big [ H_1 + H \big ] = 0.861\,\,{\rm bit/source\:symbol}\hspace{0.05cm},$$

$$H_3 = {1}/{3} \cdot \big [ H_1 + 2H \big ] = 0.815\,\,{\rm bit/source\:symbol}\hspace{0.05cm}.$$

In this exercise, the Huffman algorithm is to be applied to  $k$–tuples, where we restrict ourselves to  $k = 2$  and  $k = 3$ .

Hints:

• The task belongs to the chapter  Entropy coding according to Huffman.
• In particular, reference is made to the page  Application of Huffman coding to k-tuples .
• Useful information can also be found in the specification sheets for  exercise 2.7  and  exercise 2.7Z.
• To check your results, please refer to the interaction module  Huffman and Shannon–Fano––coding.

Questions

1

Which of the example sequences given at the front is true for  $q = 0.8$?

	the red source symbol sequence  1,
	the blue source symbol sequence  2.

2

Which of the following statements are true?

	The direct application of Huffman is also useful here.
	Huffman makes sense when forming two-tuples  $(k = 2)$  Sinn.
	Huffman makes sense when forming tuples of three  $(k = 3)$  Sinn.

3

What are the probabilities of two-tuples  $(k = 2)$  for  $\underline{q = 0.8}$?

$p_{\rm A} = \rm Pr(XX)\ = \ $

$p_{\rm B} = \rm Pr(XY)\ = \ $

$p_{\rm C} = \rm Pr(YX)\ = \ $

$p_{\rm D} = \rm Pr(YY)\ = \ $

Find the Huffman code for  $\underline{k = 2}$.  What is the mean codeword length in this case?

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$L_{\rm M} \ = \ $ $\ \rm bit/source\:symbol$

4

What is the bound on the mean codeword length when two-tuples are formed  $(k = 2)$? Interpretation.

	$L_{\rm M} \ge H_1 = 1.000$  $\ \rm bit/source\:symbol$
	$L_{\rm M} \ge H_2 \approx 0.861$  $\ \rm bit/source\:symbol$
	$L_{\rm M} \ge H_3 \approx 0.815$  $\ \rm bit/source\:symbol$
	$L_{\rm M} \ge H_{k \to \infty} \approx 0.722$  $\ \rm bit/source\:symbol$
	$L_{\rm M} \ge 0.5$  $\ \rm bit/source\:symbol$

5

Calculate the probabilities of the three-tuple  $(k = 3)$  for  $\underline{q = 0.8}$?

$p_{\rm A} = \rm Pr(XXX)\ = \ $

$p_{\rm B} = \rm Pr(XXY)\ = \ $

$p_{\rm C} = \rm Pr(XYX)\ = \ $

$p_{\rm D} = \rm Pr(XYY)\ = \ $

$p_{\rm E} = \rm Pr(YXX)\ = \ $

$p_{\rm F} = \rm Pr(YXY)\ = \ $

$p_{\rm G} = \rm Pr(YYX)\ = \ $

$p_{\rm H} = \rm Pr(YYY)\ = \ $

6

Find the Huffman code for $\underline{k = 3}$.  What is the mean codeword length in this case?

$L_{\rm M} \ = \ $ $\ \rm bit/source\:symbol$

Solution

Solution

(1)  Correct is the solution suggestion 2:

• In the blue source symbol sequence  2  one recognizes much less symbol changes than in the red sequence.
• The symbol sequence  2  was generated with the parameter  $q = {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.8$  and the red symbol sequence  1  with  $q = 0.2$.

(2) Answers 2 and 3 are correct.:

• Since here the source symbols  $\rm X$  and  $\rm Y$  were assumed to be equally probable, the direct application of Huffman makes no sense.
• In contrast, one can use the inner statistical bonds of the Markov source for data compression if one forms  $k$–tuples   $(k ≥ 2)$.
• The larger  $k$  is, the more the mean codeword length  $L_{\rm M}$  approaches the entropy  $H$.

(3)  The symbol probabilities are  $p_{\rm X} = p_{\rm Y} = 0.5$, which gives us for the two-tuples.: 

$$p_{\rm A} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XX}) = p_{\rm X} \cdot {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = 0.5 \cdot q = 0.5 \cdot 0.8 \hspace{0.15cm}\underline{ = 0.4} \hspace{0.05cm},$$

$$p_{\rm B} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XY}) = p_{\rm X} \cdot {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm X}) = 0.5 \cdot (1-q)= 0.5 \cdot 0.2 \hspace{0.15cm}\underline{ = 0.1} \hspace{0.05cm},$$

$$p_{\rm C} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm YX}) = p_{\rm Y} \cdot {\rm Pr}(\boldsymbol{\rm X}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.5 \cdot (1-q)= 0.5 \cdot 0.2 \hspace{0.15cm}\underline{ = 0.1} \hspace{0.05cm},$$

$$p_{\rm D} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm YY}) = p_{\rm Y} \cdot {\rm Pr}(\boldsymbol{\rm Y}\hspace{0.05cm}|\hspace{0.05cm}\boldsymbol{\rm Y}) = 0.5 \cdot q = 0.5 \cdot 0.8\hspace{0.15cm}\underline{ = 0.4} \hspace{0.05cm}.$$

For Huffman coding for $k = 2$

(4)  Opposite screen capture of the (earlier) SWF program  Shannon–Fano– and Huffman–coding  shows the construction of the Huffman code for  $k = 2$  with the probabilities just calculated.

• Thus, the mean codeword length is:

$$L_{\rm M}\hspace{0.01cm}' = 0.4 \cdot 1 + 0.4 \cdot 2 + (0.1 + 0.1) \cdot 3 = 1.8\,\,{\rm bit/two-tuple}$$

$$\Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{ = 0.9\,{\rm bit/source\:symbol}}\hspace{0.05cm}.$$

(5)  Correct is the suggested solution 2:

• According to the source coding theorem  $L_{\rm M} ≥ H$ holds.
• However, if we apply Huffman coding and disregard ties between non-adjacent symbols  $(k = 2)$, the lower bound of the codeword length is not  $H = 0.722$, but  $H_2 = 0.861$  (the addition bit/source symbol is omitted for the rest of the task).
• The result of subtask  (4)  was  $L_{\rm M} = 0.9.$
• If an asymmetric Markov chain were present and in such a way that for the probabilities  $p_{\rm A}$, ... , $p_{\rm D}$  the values  $50\%$,  $25\%$  and twice  $12.5\%$  would result, then one would come to the average code word length  $L_{\rm M} = 0.875$.
• How the exact parameters of this asymmetrical Markov source look, however, is not known even to the task creator (G. Söder).
• Nor how the value  $0.875$  could be reduced to  $0.861$ . In any case, the Huffman algorithm is unsuitable for this.

(6)  With  $q = 0.8$  and  $1 - q = 0.2$  we get:

$$p_{\rm A} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XXX}) = 0.5 \cdot q^2 \hspace{0.15cm}\underline{ = 0.32} = p_{\rm H} = {\rm Pr}(\boldsymbol{\rm YYY})\hspace{0.05cm},$$

$$p_{\rm B} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XXY}) = 0.5 \cdot q \cdot (1-q) \hspace{0.15cm}\underline{ = 0.08}= p_{\rm G} = {\rm Pr}(\boldsymbol{\rm YYX}) \hspace{0.05cm},$$

$$p_{\rm C} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XYX}) = 0.5 \cdot (1-q)^2\hspace{0.15cm}\underline{ = 0.02} = p_{\rm F}= {\rm Pr}(\boldsymbol{\rm YXY}) \hspace{0.05cm},$$

$$p_{\rm D} \hspace{0.2cm} = \hspace{0.2cm} {\rm Pr}(\boldsymbol{\rm XYY}) = 0.5 \cdot (1-q) \cdot q \hspace{0.15cm}\underline{ = 0.08} = p_{\rm E} = {\rm Pr}(\boldsymbol{\rm YXX})\hspace{0.05cm}.$$

(7)  The screen capture of the of the (earlier) SWF program  Shannon–Fano– and Huffman–coding  coding illustrates the constellation of the Huffman code for  $k = 3$.  This gives us for the mean codeword length:

$$L_{\rm M}\hspace{0.01cm}' = 0.64 \cdot 2 + 0.24 \cdot 3 + 0.04 \cdot 5 = 2.52\,\,{\rm bit/Dreiertupel}\hspace{0.3cm} \Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{3}\hspace{0.15cm}\underline{ = 0.84\,{\rm bit/Quellensymbol}}\hspace{0.05cm}.$$

Zur Huffman–Codierung für $k = 3$

• Man erkennt die Verbesserung gegenüber der Teilaufgabe  (4).
• Die für  $k = 2$  gültige Schranke  $H_2 = 0.861$  wird nun von der mittleren Codewortlänge  $L_{\rm M}$  unterschritten.
• Die neue Schranke für  $k = 3$  ist  $H_3 = 0.815$.
• Um die Quellenentropie  $H = 0.722$  zu erreichen  (besser gesagt:  diesem Endwert bis auf ein  $ε$  nahe zu kommen), müsste man allerdings unendlich lange Tupel bilden  $(k → ∞)$.