Aufgaben:Exercise 2.10: Shannon-Fano Coding

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Tree diagram of the
Shannon-Fano coding

Another algorithm for entropy coding was given in 1949 by  Claude Elwood Shannon  and  Robert Fano , which is described in the theory section.

This special type of source coding will be described here using a simple example for the symbol range  $M = 4$  and the following symbol probabilities:

$$p_{\rm A} = 0.2 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm B}= 0.3 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm C}= 0.4 \hspace{0.05cm},\hspace{0.4cm} p_{\rm D}= 0.1 \hspace{0.05cm}. $$

The graph shows the corresponding tree diagram.  Proceed as follows:

1. Order the symbols according to decreasing probability of occurrence, here  $\rm C$ – $\rm B$ – $\rm A$ – $\rm D$.

2. Divide the symbols into two groups of approximately equal probability, here  $\rm C$  and  $\rm BAD$.

3. The binary symbol  0is assigned to the less probable group,  1  to the other group.

4. If there is more than one symbol in a group, the algorithm is to be applied recursively.

For this example, the following code assignment results (in the tree diagram, a red connection marks a  1  and a blue one a  0:

      $\rm A$   →   111,     $\rm B$   →   10,     $\rm C$   →   0,     $\rm D$   →   110.

This gives the following for the mean codeword length:

$$L_{\rm M} = 0.4 \cdot 1 + 0.3 \cdot 2 + (0.2 + 0.1) \cdot 3 = 1.9\,\,{\rm bit/source\:symbol}\hspace{0.05cm}.$$

The Huffman algorithm would produce a slightly different code here, but even in this case 

• $\rm C$  is coded with one bit, 
• $\rm B$  with two bits and  
• $\rm A$ and  $\rm D$  with three bits each. 

  This would also result in  $L_{\rm M} = 1.9 \ \rm bit/source\:symbol$.

In this task you are to calculate the Shannon-Fano code for  $M = 8$  and the probabilities

$$p_{\rm A} = 0.10 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm B}= 0.40 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm C}= 0.02 \hspace{0.05cm},\hspace{0.4cm} p_{\rm D}= 0.14 \hspace{0.05cm},\hspace{0.4cm} p_{\rm E} = 0.17 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm F}= 0.03 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm G}= 0.05 \hspace{0.05cm},\hspace{0.4cm}p_{\rm H}= 0.09$$

determine.  You will see that with these probabilities "Shannon-Fano" will also differ from "Huffman" in terms of efficiency. 

With the Huffman code, the following assignment results with the probabilities at hand:

      $\rm A$   →   100,     $\rm B$   →   0,     $\rm C$   →   111100,     $\rm D$   →   101,     $\rm E$   →   110,     $\rm F$   →   111101,     $\rm G$   →   11111,     $\rm H$   →   1110.

Hints:

• The assignment belongs to the chapter  Other source coding methods.
• In particular, reference is made to the page  The Shannon-Fano algorithm.
• To check your results, please refer to the interactive applet  Huffman and Shannon-Fano coding.

Questions

1

What is the mean codeword length  $L_{\rm M}$  for the  Huffman code?

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

2

What happens in the first step of  Shannon–Fano coding?  Hint:  All other symbols are grouped together in the second group.

	$\rm A$  and  $\rm B$  are combined into the first group.
	$\rm B$  and  $\rm E$  are combined into the first group.
	The first group consists only of symbol  $\rm B$.

3

Which assignments result for the  Shannon–Fano algorithm?

	The character  $\rm A$  is binary coded with  010 .
	The character  $\rm B$  is binary coded with  11 .
	The character  $\rm C$  is binary coded with  00110 .

4

What is the average codeword length  $L_{\rm M}$  for  Shannon–Fano code?

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

5

Which statements are true for arbitrary probabilities?

	$L_{\rm M}$  could be smaller for "Shannon–Fano" than for "Huffman".
	$L_{\rm M}$  could be larger at "Shannon–Fano" than at "Huffman".
	$L_{\rm M}$  could be the same for "Shannon–Fano" and "Huffman".

Solution

Solution

(1)  For the given Huffman code one obtains:

$$L_{\rm M} = 0.4 \cdot 1 + (0.17 + 0.14 + 0.10) \cdot 3 + 0.09 \cdot 4 + 0.05 \cdot 5 + (0.03 + 0.02) \cdot 6 =\underline{ 2.54 \,\,{\rm bit/source\:symbol}}\hspace{0.05cm}. $$

(2)  The correct answer is 2:

• Before applying the Shannon-Fano algorithm, the characters must first be sorted according to their occurrence probabilities.  Thus, answer 1 is incorrect.
• All sorted characters must be divided into two groups in such a way that the group probabilities are as equal as possible. For the first step:

$${\rm Pr}(\boldsymbol{\rm BE}) = 0.57\hspace{0.05cm}, \hspace{0.2cm}{\rm Pr}(\boldsymbol{\rm DAHGFC}) = 0.43 \hspace{0.05cm}.$$

Tree diagram of Shannon-Fano coding

• With the distribution according to solution suggestion 3, the equal distribution would be achieved even less:

$${\rm Pr}(\boldsymbol{\rm B}) = 0.40\hspace{0.05cm}, \hspace{0.2cm}{\rm Pr}(\boldsymbol{\rm EDAHGFC}) = 0.60 \hspace{0.05cm}.$$

(3)  All suggeted solutionss are correct:

• The graph shows the tree diagram of the Shannon-Fano coding.
• This results in the following assignment (a red connection indicates a  1 , a blue one indicates a  0):

$\underline{\rm A}$   →   010,     $\underline{\rm B}$   →   11,     $\underline{\rm C}$   →   00110,

${\rm D}$   →   011,     ${\rm E}$   →   10,     ${\rm F}$   →   00111,     ${\rm G}$   →   0010,     ${\rm H}$   →   000.

(4)  Using the result of sub-task  (3) , we get:

$$L_{\rm M}= (0.40 + 0.17) \cdot 2 + (0.14 + 0.10 + 0.09) \cdot 3 + 0.05 \cdot 4 + (0.03 + 0.02) \cdot 5 =\underline{ 2.58 \,\,{\rm bit/source\:symbol}}\hspace{0.05cm}. $$

(5)  Statements 2 and 3 are correct:

• In the present example, Shannon-Fano results in a less favourable value than Huffman.
• In most cases – including the example on the information page – Huffman and Shannon-Fano result in an equivalent code and thus also the same average code word length.
• Shannon-Fano, on the other hand, never delivers a more effective code than Huffman.