# Aufgabe 1.7: Entropie natürlicher Texte

Text with erasures or errors

In the early 1950s,  Claude E. Shannon  estimated the entropy  $H$  of the English language at one bit per character.  A short time later,  Karl Küpfmüller  arrived at an entropy value of  $H =1.3\ \rm bit/character$, i.e. a somewhat larger value, in an empirical study of the German language.  Interestingly, the results of Shannon and Küpfmüller are based on two completely different methods.

The differing results cannot be explained by the small differences in the range of symbols  $M$ :

• Shannon assumed  $26$ letters and the space  ⇒  $M = 27$.
• Küpfmüller assumed only  $M = 26$  letters  (i.e. without the space).

Both made no distinction between upper and lower case.

This task is intended to show how

• Erasures   ⇒   one knows the location of an error, resp.
• Errors   ⇒   it is not clear to the reader what is wrong and what is right,

have an effect on the comprehensibility of a text.  Our text also contains the typical German letters "ä",  "ö",  "ü"  and  "ß"  as well as numbers and punctuation. In addition, a distinction is made between upper and lower case.

In the figure, a text dealing with Küpfmüller's approach is divided into six blocks of length between  $N = 197$  and  $N = 319$  aufgeteilt.  Described is the check of his  first analysis  in a completely different way, which led to the result  $H =1.51\ \rm bit/character$ .

• In the upper five blocks one recognises "Erasures" with different erasure probabilities between  $10\%$  and  $50\%$.
• In the last block, "character errors" with  $20$–error probability are inserted.

The influence of such character errors on the readability of a text is to be compared in subtask  (4)  with the second block (outlined in red), for which the probability of an erasure is also  $20\%$ .

Hints:

Entropy estimation according to Küpfmüller,  sowie
Another entropy estimation by Küpfmüller.
• For the  relative redundancy  of a sequence, with the decision content  $H_0$  and the entropy  $H$ , applies
$$r = \frac{H_0 - H}{H_0}\hspace{0.05cm}.$$

### Questions

1

What symbol range  $M$  did Küpfmüller assume?

 $M \ = \$

2

What relative redundancy  $r$  results from Küpfmüller's entropy value?

 $r \ = \$ $\ \%$

3

How can the result of subtask  (2)  be interpreted?  Assume a text file with  $M = 26$  different characters.

 Such a text file of sufficient length  $(N \to \infty)$  could be represented with  $1.3 \cdot N$  binary symbols. Such a text file with  $N= 100\hspace{0.1cm}000$  characters could be represented with  $130\hspace{0.1cm}000$  binary symbols. A reader can still understand (or at least guess) the text even if  $70\%$  of the characters are erased.

4

What makes a text more difficult to understand?

 $20\%$  Erasures, a  $20\%$ probability of character errors.

### Solution

#### Solution

(1)  The symbol range in Küpfmüller's investigations was  $\underline{M = 26}$,  because, in contrast to Shannon, he did not initially take the space into account.

In the given German text of this task, the symbol range is significantly larger,

• since the typical German characters "ä",  "ö",  "ü"  and  "ß"  also occur,
• there is a distinction between upper and lower case,
• and there are also numbers and punctuation marks.

(2)  With the decision content  $H_0 = \log_2 \ (31) \approx 4.7 \ \rm bit/character$  and the entropy  $H = 1.3\ \rm bit/character$ , one obtains for the relative redundancy:

$$r = \frac{H_0 - H}{H_0}= \frac{4.7 - 1.3}{4.7}\underline {\hspace{0.1cm}\approx 72.3\,\%}\hspace{0.05cm}.$$

(3)  Only the first suggested solution is correct:

• According to Küpfmüller, one only needs  $1.3$  binary characters per source character.
• With a file of length  $N$ ,  $1.3 \cdot N$  binary symbols would therefore be sufficient, but only if the source symbol sequence is infinitely long  $(N \to \infty)$  and this was encoded in the best possible way.
• In contrast, Küpfmüller's result and the relative redundancy of more than  $70\%$ calculated in subtask  (2)  do not mean that a reader can still understand the text if  $70\%$  of the characters have been erased.
• Such a text is neither infinitely long nor has it been optimally encoded beforehand.

(4)  Correct is statement 2:

• Test it yourself:   The second block of the graphic on the information page is easier to decode than the last block, because you know where there are errors.
• If you want to keep trying:   The exact same sequence of character errors was used for the bottom block  (F)  as for block  (B)  , that is, there are errors at characters  $6$,  $35$,  $37$, , and so on.

Original texts

Finally, the original text is given, which is only reproduced on the information page distorted by erasures or real character errors.