Aufgaben:Exercise 1.1: For Labeling Books - Revision history

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	[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]
	[[de:Aufgaben:1.~~1_Zur_Kennzeichnung_aller_Bücher~~]]		[[de:Aufgaben:Aufgabe 1.1: Zur Kennzeichnung aller Bücher]]

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(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:		'''(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:
	:$$S \ = \ \hspace{-0.1cm} \sum_{i=1}^{13} \hspace{0.2cm} z_i \cdot 3^{(i+1) \hspace{-0.2cm} \cdot 2} = (9+8+8+7+6+8) \cdot 1 + (7+3+2+3+0+4) \cdot 3 = 110\hspace{0.3cm}		:$$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}.$$
	\Rightarrow \hspace{0.3cm} S \hspace{-0.2cm} \mod 10 \hspace{0.15cm}\underline {= 0} \hspace{0.05cm}.$$


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	'''(3)'''  One digit can be reconstructed  ⇒   <u>Yes</u>.  For the digit  $z_{\rm 8}$,  it must hold:		'''(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}.$$
	\Rightarrow \hspace{0.3cm} [108 + 3z_8] \hspace{-0.2cm} \mod 10 = 0 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_8 \hspace{0.15cm}\underline {= 4} \hspace{0.05cm}.$$


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	*The given ISBN satisfies this condition:		*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 is  <u>statement 2</u>,  since the check sum  $S = 0$  could result even with more than one error.		*Correct is  <u>statement 2</u>,  since the check sum  $S = 0$  could result even with more than one error.
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]
			[[de:Aufgaben:1.1_Zur_Kennzeichnung_aller_Bücher]]

Guenter at 11:53, 6 June 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Just by counting the ISBN digits, you can tell that <u>answer 2</u> is correct.		'''(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:
	:The weighted sum over all digits is a multiple of 10:
	:$$S \ = \ \hspace{-0.1cm} \sum_{i=1}^{13} \hspace{0.2cm} z_i \cdot 3^{(i+1) \hspace{-0.2cm} \cdot 2} = (9+8+8+7+6+8) \cdot 1 + (7+3+2+3+0+4) \cdot 3 = 110\hspace{0.3cm}		:$$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)'''  The answer is <u>No</u>. Only one cancellation can be reconstructed with a single check digit.		'''(2)'''  The answer is  <u>No</u>.  Only one cancellation can be reconstructed with a single check digit.


	'''(3)'''  One digit can be reconstructed  ⇒   <u>Yes</u>. For the digit $z_{\rm 8}$, it must hold:		'''(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)'''  By the modulo 11 operation $z_{10}$ can take the values $0, 1, \text{...} , 10$ ⇒ $\underline{M =11}$.		'''(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 $z_{10} = \rm X$.		*Since  "10"  is not a digit,  one makes do with  $z_{10} = \rm X$.
	*This corresponds to the Roman representation of the number "10".		*This corresponds to the Roman representation of the number  "10".


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	:$$3 \cdot 1 + 8 \cdot 2 + 2 \cdot 3 + 7 \cdot 4 + 3 \cdot 5 + 7 \cdot 6 + 0 \cdot 7 + 6 \cdot 8 + 4 \cdot 9 + 7 \cdot 10 = 264\hspace{0.3cm}		:$$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.		*Correct is  <u>statement 2</u>,  since the check sum  $S = 0$  could result even with more than one error.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 11:46, 6 June 2022

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	[[File:EN_KC_A_1_1.png\|right\|frame\|'''ISBN–10'''? Or '''ISBN–13'''?]]		[[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:		Since the 1960s,  all books are provided with a 10-digit  "International Standard Book Number"  $\rm (ISBN)$.  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:		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.		Some exemplary  "ISBNs"  are given opposite.  The following questions refer to these.


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	Hints:		Hints:  This exercise belongs to the chapter  [[Channel_Coding/Objective_of_Channel_Coding\|"Objective of Channel Coding"]]
	*This exercise belongs to the chapter  [[Channel_Coding/Objective_of_Channel_Coding\|~~Obective~~ of Channel Coding]]


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	What is the standard for  $\text{~~example~~ 1}$?		What is the standard for  $\text{Example 1}$?
	\|type="()"}		\|type="()"}
	- ISBN-10,		- ISBN-10,
	+ISBN-13.		+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.		{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,		- Yes,
	+No.		+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.		{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,		+Yes,
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	$M \ = \ $ { 11 3% } $\ \rm$		$M \ = \ $ { 11 3% } $\ \rm$

	{Assumed as ISBN-10 is 3-8273-7064-7. Which statement is true?		{Assumed as ISBN-10 is  "3-8273-7064-7".  Which statement is true?
	\|type="()"}		\|type="()"}
	- This is not a valid ISBN.		- This is not a valid ISBN.

Noah at 15:56, 4 June 2022

2022-06-04T15:56:05Z

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Guenter at 15:24, 2 June 2022

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	}}		}}

	[[EN_KC_A_1_1.png\|right\|frame\|'''ISBN–10'''? Oder '''ISBN–13'''?]]		[[File:EN_KC_A_1_1.png\|right\|frame\|'''ISBN–10'''? Oder '''ISBN–13'''?]]

	Seit den 1960er Jahren werden alle Bücher mit einer 10–stelligen  ''International Standard Book Number''  (ISBN) versehen. Die letzte Ziffer dieser so genannten  '''ISBN–10–Angabe'''  berechnet sich dabei entsprechend folgender Regel:		Seit den 1960er Jahren werden alle Bücher mit einer 10–stelligen  ''International Standard Book Number''  (ISBN) versehen. Die letzte Ziffer dieser so genannten  '''ISBN–10–Angabe'''  berechnet sich dabei entsprechend folgender Regel:

Guenter at 15:21, 2 June 2022

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	}}		}}

	[[~~File:P_ID2380__KC_A_1_1~~.png\|right\|frame\|'''ISBN–10'''? Oder '''ISBN–13'''?]]		[[EN_KC_A_1_1.png\|right\|frame\|'''ISBN–10'''? Oder '''ISBN–13'''?]]

	Seit den 1960er Jahren werden alle Bücher mit einer 10–stelligen  ''International Standard Book Number''  (ISBN) versehen. Die letzte Ziffer dieser so genannten  '''ISBN–10–Angabe'''  berechnet sich dabei entsprechend folgender Regel:		Seit den 1960er Jahren werden alle Bücher mit einer 10–stelligen  ''International Standard Book Number''  (ISBN) versehen. Die letzte Ziffer dieser so genannten  '''ISBN–10–Angabe'''  berechnet sich dabei entsprechend folgender Regel:

Guenter at 09:10, 11 May 2022

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	[[Category:Channel Coding: Exercises\|^1.1 Objective of ~~the~~ Channel Coding		[[Category:Channel Coding: Exercises\|^1.1 Objective of Channel Coding^]]

	^]]

Guenter at 09:09, 11 May 2022

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	[[Category:Channel Coding: Exercises\|^1.1 ~~Zielsetzung der Kanalcodierung~~		[[Category:Channel Coding: Exercises\|^1.1 Objective of the Channel Coding

	^]]		^]]