Exercise 2.7: Huffman Application for Binary Two-Tuples
The application of the Huffman algorithm in its original form assumes a symbol set size $M > 2$ and is therefore useless for data compression of binary sources.
However, if one combines several consecutive binary characters of the source into a new symbol, one can usefully apply Huffman data compression to the new symbol set.
In this task we start from the source symbol set $\{$ $\rm X$, $\rm Y$ $\}$ ⇒ $M = 2$ and form two-tuples according to the table on the right with the new symbol set $\{$ $\rm A$, $\rm B$, $\rm C$, $\rm D$ $\}$ ⇒ $(M\hspace{0.03cm}' = M^2 = 4)$.
For example, the binary source symbol sequence $\rm XYXXYXXXYY$ thus becomes the quaternary sequence $\rm BACAD$.
Furthermore, three codes are given in the table, some of which have been created by the Huffman algorithm. The binary output sequences then result for our example as follows:
- $\text{Code 1}$: 1011011100,
- $\text{Code 2}$: 0110011000,
- $\text{Code 3}$: 10011001110.
Again for understanding:
- From the original symbol set $\{$ $\rm X$, $\rm Y$ $\}$ a quaternary set with symbol set $\{$ $\rm A$, $\rm B$, $\rm C$, $\rm D$ $\}$ is obtained by forming two-tuples.
- The sequence length $N$ is thereby halved to $N\hspace{0.03cm}' = N/2 $ .
- Huffman coding again results in a binary sequence whose symbol set is designated $\{$0, 1$\}$ for better differentiation.
- The application of Huffman coding makes sense exactly when the length $L$ of the initial sequence is smaller than $N$ (on statistical average).
This task is intended to clarify which of the given binary codes make sense under which boundary conditions.
- Let the binary message source $\{$ $\rm X$, $\rm Y$ $\}$ be memoryless and be described solely by the symbol probability $p_{\rm X}$ .
- The second probability is then always $p_{\rm Y} = 1 - p_{\rm X}$.
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.
- The average codeword length per two-tuple is $L_{\rm M}\hspace{0.03cm}' = p_{\rm A} \cdot L_{\rm A} +$ ... $ + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm}$. With respect to a source symbol, $L_{\rm M} = L_{\rm M}\hspace{0.03cm}'/2$.
- A comparable task with ternary input symbols is treated in Exercise 2.7Z .
- The idea for this task arose during a lecture by Prof. Robert Fischer from the University of Ulm at the Technical University of Munich on the topic of
"Der goldene Schnitt in der Nachrichtentechnik" ⇒ "The golden ratio in communications technology".
Questions
Solution
- $$L_{\rm M} = \big [ \hspace{0.05cm}p_{\rm A} \cdot L_{\rm A} + p_{\rm B} \cdot L_{\rm B} + p_{\rm C} \cdot L_{\rm C} + p_{\rm D} \cdot L_{\rm D} \hspace{0.05cm} \big ] / 2 = \big [ \hspace{0.05cm} L_{\rm A} + L_{\rm B} + L_{\rm C} + L_{\rm D}\hspace{0.05cm} \big ] / 8 \hspace{0.05cm}.$$
Taking into account the given assignments, we obtain for
- $\text{Code 1}$: $L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \hspace{0.15cm}symbol} }$,
- $\text{Code 2}$: $L_{\rm M} \hspace{0.15cm}\underline{= 1.125 \ {\rm bit/source \hspace{0.15cm}symbol} }$,
- $\text{Code 3}$: $L_{\rm M} \hspace{0.15cm}\underline{= 1.250 \ {\rm bit/source \hspace{0.15cm}symbol} }$.
In the course of the task it will become apparent that the first two codes are quite possible as a result of the Huffman algorithm (of course only with suitable symbol probabilities). $\text{Code 3}$ is also prefix-free, but never optimal in terms of average codeword length.
(2) The probabilities of the possible two-tuples are:
- $$p_{\rm A} = 0.6^2 = 0.36 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm B}= 0.6 \cdot 0.4 = 0.24 = p_{\rm C} \hspace{0.05cm},\hspace{0.4cm} p_{\rm D}= 0.4^2 = 0.16 \hspace{0.05cm}.$$
- This gives the tree diagram on the left (in the adjacent graph) and the following Huffman code:
- $\rm A$ → 11, $\rm B$ → 10, $\rm C$ → 01, $\rm D$ → 00.
- This is $\text{Code 1}$ ⇒ solution suggestion 1.
(3) Each two-tuple is represented by two bits. Thus
- $$L_{\rm M} \hspace{0.15cm}\underline{= 1.000 \ {\rm bit/source \hspace{0.15cm}symbol} }.$$
(4) Here the probabilities of each two-tuple are:
- $$p_{\rm A} = 0.8^2 = 0.64 \hspace{0.05cm}, \hspace{0.4cm}p_{\rm B}= 0.8 \cdot 0.2 = 0.16 \hspace{0.05cm}, $$
- $$p_{\rm C} = p_{\rm B}= 0.8 = 0.16 \hspace{0.05cm},\hspace{0.4cm} p_{\rm D}= 0.2^2 = 0.04 \hspace{0.05cm}. $$
According to the tree diagram on the right the result is now $\text{Code 2}$ ⇒ solution suggestion 2:
- $\rm A$ → 1, $\rm B$ → 01, $\rm C$ → 001, $\rm D$ → 010.
(5) For $\text{Code 2}$ the mean two-tuple length or the average codeword length is:
- $$L_{\rm M}\hspace{0.01cm}' = 0.64 \cdot 1 + 0.16 \cdot 2 + (0.16 + 0.04) \cdot 3 = 1.56\,{\rm bit/two-tuples}\hspace{0.3cm} \Rightarrow\hspace{0.3cm}L_{\rm M} = {L_{\rm M}\hspace{0.01cm}'}/{2}\hspace{0.15cm}\underline{ = 0.78\,{\rm bit/source \:symbol}}\hspace{0.05cm}.$$
(6) From the earlier subtasks it has been found:
- For $p_{\rm X} = 0.6$, $\text{Code 1}$ is optimal and the average codeword length of this code is $($independent of $p_{\rm X})$ $L_{\rm M} = 1$ bit/source symbol.
- For $p_{\rm X} = 0.8$ according to subtask (4), $\text{Code 2}$ is optimal and the average codeword length is $L_{\rm M} = 0.78$ bit/source symbol.
The sought maximum value $p_\text{X, max}$ will thus lie between $0.6$ and $0.8$ . The determining equation here is that for the limiting case $p_\text{X} = p_\text{X, max}$ both codes have exactly the same average codeword length $L_{\rm M} = 1$ bit/source symbol, or $L_{\rm M}\hspace{0.03cm}' = 2$ bit/two-tuple.
- With the abbreviation $p = p_\text{X, max}$ the corresponding determining equation is:
- $$L_{\rm M}\hspace{0.01cm}'\hspace{0.15cm}{\rm (Code \hspace{0.15cm}2)} = p^2 \cdot 1 + p \cdot (1-p) \cdot 2 + p \cdot (1-p) \cdot 3 + (1-p)^2 \cdot 3 \stackrel{!}{=} 2 \hspace{0.05cm}.$$
- This leads to the numerical result:
- $$p^2 + p - 1 \stackrel{!}{=} 0 \hspace{0.3cm}\Rightarrow\hspace{0.3cm} p_{\rm X,\hspace{0.05cm}max} = p = \frac{\sqrt{5}-1}{2} \hspace{0.15cm}\underline{ \approx 0.618} \hspace{0.05cm}.$$
- Since the basic Huffman structure does not change by swapping $\rm X$ and $\rm Y$ the lower bound is:
- $$p_{\rm X,\hspace{0.05cm}min} = 1 - p_{\rm X,\hspace{0.05cm}max}\hspace{0.15cm}\underline{ \approx 0.382} \hspace{0.05cm}.$$
The representation of the tuples of two by bit sequences of different lengths $\text{(Code 2)}$ therefore only makes sense if the symbol probabilities of $\rm X$ and $\rm Y$ differ significantly. If, on the other hand, they are between $0.382$ and $0.618$, $\text{Code 1}$ is to be applied.
Note: The division of a path of length $1$ into two sections of length $0.618$... and $0.382$... is called the Golden ratio, which is encountered again and again in the most diverse fields.
