Aufgaben:Exercise 2.7Z: Huffman Coding for Two-Tuples of a Ternary Source

• Return to book

Huffman tree for
a ternary source

We consider the same situation as in  Exercise A2.7:  

• The Huffman algorithm leads to a better result, i.e. to a smaller average code word length  $L_{\rm M}$, if one does not apply it to individual symbols but forms  $k$–tuples beforehand.  
• This increases the symbol set size from  $M$  to  $M\hspace{0.03cm}' = M^k$.

For the message source considered here, the following applies:

• Symbol set size:   $M = 3$,
• Symbol set:   $\{$ $\rm X$,  $\rm Y$,  $\rm Z$ $\}$,
• Probabilities:   $p_{\rm X} = 0.7$,  $p_{\rm Y} = 0.2$,  $p_{\rm Z} = 0.1$,
• Entropy:   $H = 1.157 \ \rm bit/ternary\hspace{0.12cm}symbol$.

The graph shows the Huffman tree when the Huffman algorithm is applied to single symbols  $(k= 1)$.
In subtask  (2)  you are to give the corresponding Huffman code when two-tuples are formed beforehand  $(k=2)$.

Hints:

• The exercise belongs to the chapter  Entropy Coding according to Huffman.
• In particular, reference is made to the page  Application of Huffman coding to  $k$-tuples.
• A comparable task with binary input symbols is dealt with in  Exercise 2.7 .
• Designate the possible two-tuples with     $\rm XX = A$,   $\rm XY = B$,   $\rm XZ = C$,   $\rm YX = D$,   $\rm YY = E$,   $\rm YZ = F$,   $\rm ZX = G$,   $\rm ZY = H$,   $\rm ZZ = I$.

Questions

1

What is the average code word length when the Huffman algorithm is applied directly to the ternary source symbols  $\rm X$,  $\rm Y$  und  $\rm Z$ ?

$\underline{k=1}\text{:} \hspace{0.25cm}L_{\rm M} \ = \ $ $\ \rm bit/source\hspace{0.12cm}symbol$

2

What are the tuple probabilities here?  In particular:

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

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

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

3

What is the average code word length if you first form two-tuples and apply the Huffman algorithm to them?

$\underline{k=2}\text{:} \hspace{0.25cm}L_{\rm M} \ = \ $ $\ \rm bit/source\hspace{0.12cm}symbol$

4

Which of the following statements are true when more than two ternary symbols are combined  $(k>2)$?

	$L_{\rm M}$  decreases monotonically with increasing  $k$.
	$L_{\rm M}$  does not change when  $k$  is increased.
	Für  $k= 3$  you get  $L_{\rm M} = 1.05 \ \rm bit/source\hspace{0.12cm}symbol$.

Solution

Solution

(1)  The average code word length is with  $p_{\rm X} = 0.7$,  $L_{\rm X} = 1$,  $p_{\rm Y} = 0.2$,  $L_{\rm Y} = 2$,  $p_{\rm Z} = 0.1$,  $L_{\rm Z} = 2$:

$$L_{\rm M} = p_{\rm X} \cdot 1 + (p_{\rm Y} + p_{\rm Z}) \cdot 2 \hspace{0.15cm}\underline{= 1.3\,\,{\rm bit/source\:symbol}}\hspace{0.05cm}. $$

• This value is greater than the source entropy  $H = 1.157$  bit/source  symbol.

(2)  There are  $M\hspace{0.03cm}' = M^k = 3^2 = 9$  two-tuples with the following probabilities:

Huffman tree for ternary source and two-tuples.

$$p_{\rm A} = \rm Pr(XX) = 0.7 \cdot 0.7\hspace{0.15cm}\underline{= 0.49},$$

$$p_{\rm B} = \rm Pr(XY) = 0.7 \cdot 0.2\hspace{0.15cm}\underline{= 0.14},$$

$$p_{\rm C} = \rm Pr(XZ) = 0.7 \cdot 0.1\hspace{0.15cm}\underline{= 0.07},$$

$$p_{\rm D} = \rm Pr(YX) = 0.2 \cdot 0.7 = 0.14,$$

$$p_{\rm E} = \rm Pr(YY) = 0.2 \cdot 0.2 = 0.04,$$

$$p_{\rm F} = \rm Pr(YZ) = 0.2 \cdot 0.1 = 0.02,$$

$$p_{\rm G} = \rm Pr(ZX) = 0.1 \cdot 0.7 = 0.07,$$

$$p_{\rm H} = \rm Pr(ZY) = 0.1 \cdot 0.2 = 0.02,$$

$$p_{\rm I} = \rm Pr(ZZ) = 0.1 \cdot 0.1 = 0.01.$$

(3)  The graph shows the Huffman tree for the application with $k = 2$.  Thus we obtain

• for the individual two-tuples the following binary codings:
  

    $\rm XX = A$   →   0,     $\rm XY = B$   →   111,     $\rm XZ = C$   →   1011,

    $\rm YX = D$   →   110,     $\rm YY = E$   →   1000,     $\rm YZ = F$   →   10010,

    $\rm ZX = G$   →   1010,     $\rm ZY = H$   →   100111,     $\rm ZZ =I$   →   100110;

• for the average code word length:

$$L_{\rm M}\hspace{0.01cm}' =0.49 \cdot 1 + (0.14 + 0.14) \cdot 3 + (0.07 + 0.04 + 0.07) \cdot 4 + 0.02 \cdot 5 + (0.02 + 0.01) \cdot 6 = 2.33\,\,{\rm bit/two tuples}$$

$$\Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{ = 1.165\,\,{\rm bit/source\hspace{0.12cm}symbol}}\hspace{0.05cm}.$$

(4) Statement 1 is correct,  even if  $L_{\rm M}$  decreases very slowly as  $k$  increases.

• The last statement is false because  $L_{\rm M}$  cannot be smaller than  $H = 1.157$  bit/source  symbol even for   $k → ∞$  .
• But the second statement is not necessarily correct either:   Since  $L_{\rm M} > H$  still applies with  $k = 2$ ,  $k = 3$  can lead to a further improvement.