Aufgaben:Exercise 1.1: For Labeling Books: Difference between revisions
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{{quiz-Header|Buchseite= | {{quiz-Header|Buchseite=Channel_Coding/Objective_of_Channel_Coding | ||
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[[File:EN_KC_A_1_1.png|right|frame|'''ISBN–10'''? | [[File:EN_KC_A_1_1.png|right|frame|'''ISBN–10'''? Or '''ISBN–13'''?]] | ||
Since the 1960s, all books are provided with a 10-digit ''International Standard Book Number'' (ISBN). The last digit of this so-called '''ISBN-10 specification''' is calculated according to the following rule: | |||
:$$ z_{10}= \left ( \sum_{i=1}^{9} \hspace{0.2cm} i \cdot z_i \right ) \hspace{-0.2cm} \mod 11 \hspace{0.05cm}.$$ | :$$ z_{10}= \left ( \sum_{i=1}^{9} \hspace{0.2cm} i \cdot z_i \right ) \hspace{-0.2cm} \mod 11 \hspace{0.05cm}.$$ | ||
Since 2007, the specification according to the standard '''ISBN-13''' is additionally mandatory, whereby the check digit $z_{\rm 13}$ then results as follows: | |||
:$$z_{13} = 10 - \left ( \sum_{i=1}^{12} \hspace{0.2cm} z_i \cdot 3^{(i+1)\mod 2} \right ) \hspace{-0.2cm} \mod 10 \hspace{0.05cm}.$$ | :$$z_{13} = 10 - \left ( \sum_{i=1}^{12} \hspace{0.2cm} z_i \cdot 3^{(i+1)\mod 2} \right ) \hspace{-0.2cm} \mod 10 \hspace{0.05cm}.$$ | ||
Some exemplary "ISBNs" are given opposite. The following questions refer to these. | |||
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Hints: | |||
* | *This exercise belongs to the chapter [[Channel_Coding/Objective_of_Channel_Coding|Obective of Channel Coding]] | ||
=== | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
What is the standard for $\text{example 1}$? | |||
|type="()"} | |type="()"} | ||
- | - ISBN-10, | ||
+ | +ISBN-13. | ||
{ | {Accordingly $\text{example 2}$ two digits of an ISBN-13 are deleted. Is it possible to reconstruct the ISBN? If yes: Specify the ISBN-13. | ||
|type="()"} | |type="()"} | ||
- | - Yes, | ||
+ | +No. | ||
{ | {According to $\text{example 3}$ one digit of an ISBN-13 is erased. Can the ISBN be reconstructed? If Yes: Specify the ISBN-13. | ||
|type="()"} | |type="()"} | ||
+ | +Yes, | ||
- | -No. | ||
{ | {How many different values $(M)$ can the check digit $z_{\rm 10}$ take for ISBN-10? | ||
|type="{}"} | |type="{}"} | ||
$M \ = \ $ { 11 3% } $\ | $M \ = \ $ { 11 3% } $\ \rm$ | ||
{ | {Assumed as ISBN-10 is 3-8273-7064-7. Which statement is true? | ||
|type="()"} | |type="()"} | ||
- | - This is not a valid ISBN. | ||
+ | + The ISBN could be correct. | ||
- | - The ISBN is certainly correct. | ||
</quiz> | </quiz> | ||
=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
'''(1)''' | '''(1)''' Just by counting the ISBN digits, you can tell that <u>answer 2</u> is correct. | ||
: | :The weighted sum over all digits is a multiple of 10: | ||
:$$S \ = \ \hspace{-0.1cm} \sum_{i=1}^{13} \hspace{0.2cm} z_i \cdot 3^{(i+1) \hspace{-0.2cm} \ | :$$S \ = \ \hspace{-0.1cm} \sum_{i=1}^{13} \hspace{0.2cm} z_i \cdot 3^{(i+1) \hspace{-0.2cm} \cdot 2} = (9+8+8+7+6+8) \cdot 1 + (7+3+2+3+0+4) \cdot 3 = 110\hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm} S \hspace{-0.2cm} \mod 10 \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$ | \Rightarrow \hspace{0.3cm} S \hspace{-0.2cm} \mod 10 \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$ | ||
'''(2)''' | '''(2)''' The answer is <u>No</u>. Only one cancellation can be reconstructed with a single check digit. | ||
'''(3)''' | '''(3)''' One digit can be reconstructed ⇒ <u>Yes</u>. For the digit $z_{\rm 8}$, it must hold: | ||
:$$[(9+8+4+3+0+1+2) \cdot 1 + (7+3+5+z_8+7+5) \cdot 3] \hspace{-0.2cm} \mod 10 = 0\hspace{0.3cm} | :$$[(9+8+4+3+0+1+2) \cdot 1 + (7+3+5+z_8+7+5) \cdot 3] \hspace{-0.2cm} \mod 10 = 0\hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm} [108 + 3z_8] \hspace{-0.2cm} \mod 10 = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_8 \hspace{0.15cm}\underline {= 4} \hspace{0.05cm}.$$ | \Rightarrow \hspace{0.3cm} [108 + 3z_8] \hspace{-0.2cm} \mod 10 = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_8 \hspace{0.15cm}\underline {= 4} \hspace{0.05cm}.$$ | ||
'''(4)''' | '''(4)''' By the modulo 11 operation $z_{10}$ can take the values $0, 1, \text{...} , 10$ ⇒ $\underline{M =11}$. | ||
* | *Since "10" is not a digit, one makes do with $z_{10} = \rm X$. | ||
* | *This corresponds to the Roman representation of the number "10". | ||
'''(5)''' | '''(5)''' The test condition is: | ||
:$$ | :$$\ \ S= \left ( \sum_{i=1}^{10} \hspace{0.2cm} i \cdot z_i \right ) \hspace{-0.2cm} \mod 11 = 0 \hspace{0.05cm}.$$ | ||
* | *The given ISBN satisfies this condition: | ||
:$$3 \cdot 1 + 8 \cdot 2 + 2 \cdot 3 + 7 \cdot 4 + 3 \cdot 5 + 7 \cdot 6 + 0 \cdot 7 + 6 \cdot 8 + 4 \cdot 9 + 7 \cdot 10 = 264\hspace{0.3cm} | :$$3 \cdot 1 + 8 \cdot 2 + 2 \cdot 3 + 7 \cdot 4 + 3 \cdot 5 + 7 \cdot 6 + 0 \cdot 7 + 6 \cdot 8 + 4 \cdot 9 + 7 \cdot 10 = 264\hspace{0.3cm} | ||
⇒\hspace{0.3cm} S= 264 \hspace{-0.3cm} \mod 11 = 0 \hspace{0.05cm}.$$ | ⇒\hspace{0.3cm} S= 264 \hspace{-0.3cm} \mod 11 = 0 \hspace{0.05cm}.$$ | ||
* | *Correct <u>statement 2</u>, since the checksum $S = 0$ could result even with more than one error. | ||
{{ML-Fuß}} | {{ML-Fuß}} | ||
Revision as of 17:56, 4 June 2022

Since the 1960s, all books are provided with a 10-digit International Standard Book Number (ISBN). The last digit of this so-called ISBN-10 specification is calculated according to the following rule:
- $$ z_{10}= \left ( \sum_{i=1}^{9} \hspace{0.2cm} i \cdot z_i \right ) \hspace{-0.2cm} \mod 11 \hspace{0.05cm}.$$
Since 2007, the specification according to the standard ISBN-13 is additionally mandatory, whereby the check digit $z_{\rm 13}$ then results as follows:
- $$z_{13} = 10 - \left ( \sum_{i=1}^{12} \hspace{0.2cm} z_i \cdot 3^{(i+1)\mod 2} \right ) \hspace{-0.2cm} \mod 10 \hspace{0.05cm}.$$
Some exemplary "ISBNs" are given opposite. The following questions refer to these.
Hints:
- This exercise belongs to the chapter Obective of Channel Coding
Questions
Solution
- The weighted sum over all digits is a multiple of 10:
- $$S \ = \ \hspace{-0.1cm} \sum_{i=1}^{13} \hspace{0.2cm} z_i \cdot 3^{(i+1) \hspace{-0.2cm} \cdot 2} = (9+8+8+7+6+8) \cdot 1 + (7+3+2+3+0+4) \cdot 3 = 110\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} S \hspace{-0.2cm} \mod 10 \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$
(2) The answer is No. Only one cancellation can be reconstructed with a single check digit.
(3) One digit can be reconstructed ⇒ Yes. For the digit $z_{\rm 8}$, it must hold:
- $$[(9+8+4+3+0+1+2) \cdot 1 + (7+3+5+z_8+7+5) \cdot 3] \hspace{-0.2cm} \mod 10 = 0\hspace{0.3cm}
\Rightarrow \hspace{0.3cm} [108 + 3z_8] \hspace{-0.2cm} \mod 10 = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_8 \hspace{0.15cm}\underline {= 4} \hspace{0.05cm}.$$
(4) By the modulo 11 operation $z_{10}$ can take the values $0, 1, \text{...} , 10$ ⇒ $\underline{M =11}$.
- Since "10" is not a digit, one makes do with $z_{10} = \rm X$.
- This corresponds to the Roman representation of the number "10".
(5) The test condition is:
- $$\ \ S= \left ( \sum_{i=1}^{10} \hspace{0.2cm} i \cdot z_i \right ) \hspace{-0.2cm} \mod 11 = 0 \hspace{0.05cm}.$$
- The given ISBN satisfies this condition:
- $$3 \cdot 1 + 8 \cdot 2 + 2 \cdot 3 + 7 \cdot 4 + 3 \cdot 5 + 7 \cdot 6 + 0 \cdot 7 + 6 \cdot 8 + 4 \cdot 9 + 7 \cdot 10 = 264\hspace{0.3cm}
⇒\hspace{0.3cm} S= 264 \hspace{-0.3cm} \mod 11 = 0 \hspace{0.05cm}.$$
- Correct statement 2, since the checksum $S = 0$ could result even with more than one error.