# Exercise 2.10: 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 set size  $M = 4$  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  0  is 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 average codeword length:

$$L_{\rm M} = 0.4 \cdot 1 + 0.3 \cdot 2 + (0.2 + 0.1) \cdot 3 = 1.9\,\,{\rm bit/source\hspace{0.15cm}symbol}\hspace{0.05cm}.$$

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

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

This would also result in  $L_{\rm M} = 1.9 \ \rm bit/source\hspace{0.15cm}symbol$.

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

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

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:

### Questions

1

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

 $L_{\rm M}\ = \$ $\ \rm bit/source\hspace{0.15cm}symbol$

2

What happens in the first step of  Shannon–Fano codingHint:  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 encoded with  010 . The character  $\rm B$  is binary encoded with  11 . The character  $\rm C$  is binary encoded with  00110 .

4

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

 $L_{\rm M}\ = \$ $\ \rm bit/source\hspace{0.15cm}symbol$

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:

$$L_{\rm M} = 0.4 \cdot 1 + (0.17 + 0.14 + 0.10) \cdot 3 + 0.09 \cdot 4 + 0.05 \cdot 5 + (0.03 + 0.02) \cdot 6 =\underline{ 2.54 \,\,{\rm bit/source\hspace{0.15cm}symbol}}\hspace{0.05cm}.$$

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

$$L_{\rm M}= (0.40 + 0.17) \cdot 2 + (0.14 + 0.10 + 0.09) \cdot 3 + 0.05 \cdot 4 + (0.03 + 0.02) \cdot 5 =\underline{ 2.58 \,\,{\rm bit/source\hspace{0.15cm}symbol}}\hspace{0.05cm}.$$

(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 to the same average code word length.
• Shannon-Fano, on the other hand, never delivers a more effective code than Huffman.