Aufgaben:Exercise 2.10: Reed-Solomon Error Detection - Revision history

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	[[Category:Channel Coding: Exercises\|^2.3 Reed–Solomon Codes^]]		[[Category:Channel Coding: Exercises\|^2.3 Reed–Solomon Codes^]]
	[[de:Aufgaben:2.~~10_Fehlererkennung_bei_RSC~~]]		[[de:Aufgaben:Aufgabe 2.10: Fehlererkennung bei Reed–Solomon]]

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	[[Category:Channel Coding: Exercises\|^2.3 Reed–Solomon Codes^]]		[[Category:Channel Coding: Exercises\|^2.3 Reed–Solomon Codes^]]
			[[de:Aufgaben:2.10_Fehlererkennung_bei_RSC]]

Hwang at 15:27, 23 January 2023

2023-01-23T15:27:53Z

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	:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}		:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$
	* However,  if the ~~corrupted~~ received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the ~~corrupted~~ block undetected.		* However,  if the falsified received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the falsified block undetected.


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	$W_3 \ = \ ${ 245 }		$W_3 \ = \ ${ 245 }

	{What is the probability that an incorrect block remains undetected? Let the ~~corruption~~ probability of each symbol be  $\varepsilon = 0.1$.		{What is the probability that an incorrect block remains undetected? Let the falsification probability of each symbol be  $\varepsilon = 0.1$.
	\|type="{}"}		\|type="{}"}
	$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$		$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$

Guenter at 15:27, 11 October 2022

2022-10-11T15:27:40Z

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	:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}		:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$
	* However,  if the corrupted received word  $(\underline{y} ≠ \underline{c})$  is a valid code word,    the decoding process will leave the corrupted block undetected.		* However,  if the corrupted received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the corrupted block undetected.

Guenter at 14:17, 11 October 2022

2022-10-11T14:17:38Z

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Guenter at 13:33, 11 October 2022

2022-10-11T13:33:56Z

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	*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].		*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].

	* The weights  $W_i$ marked in red in the above graph are to be calculated.		* The weights  $W_i$  marked in red in the above graph are to be calculated.

Guenter at 13:32, 11 October 2022

2022-10-11T13:32:40Z

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	{{quiz-Header\|Buchseite=Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes}}

	[[File: P_ID2524__KC_A_2_10.png\|right\|frame\|Distance spectra of two different Reed-Solomon codes]]		[[File: P_ID2524__KC_A_2_10.png\|right\|frame\|Distance spectra of two <br>different Reed-Solomon codes]]
	For a linear block code, up to  $e = d_{\rm min} - 1$  errors can be detected. For all Reed–Solomon codes the minimum distance is		For a linear block code,  up to   $e = d_{\rm min} - 1$   errors can be detected.  For all Reed–Solomon codes the minimum distance is
	:$$d_{\rm min} = n-k+1 \hspace{0.05cm}.$$		:$$d_{\rm min} = n-k+1 \hspace{0.05cm}.$$

	One must distinguish the following cases:		One must distinguish the following cases:
	* If no more than  $e = n - k$  symbol errors occur, the block is always detected as faulty.		* If no more than  $e = d_{\rm min} - 1= n - k$  symbol errors occur,  the block is always detected as faulty.
	* Error detection can still work even if more than  $n - k$  symbol errors occur~~, and this~~ is when the received word is not a valid ~~codeword~~ of the Reed–Solomon code:
			* Error detection can still work even if more than  $n - k$  symbol errors occur.  This is when the received word is not a valid code word of the Reed–Solomon code:
	:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}		:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$
	* However, if the corrupted received word  $(\underline{y} ≠ \underline{c})$  is a valid ~~codeword~~ &nbsp~~; &#8658~~;   ~~$\underline{y}$,~~ the decoding process will leave the corrupted block undetected.		* However,  if the corrupted received word  $(\underline{y} ≠ \underline{c})$  is a valid code word,    the decoding process will leave the corrupted block undetected.


	We define as block error probability:		We define as  '''block error probability''':
	:$${\rm Pr}({\rm ~~Block~~\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})		:$${\rm Pr}({\rm block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In this exercise, this probability is to be determined for the following codes:		In this exercise,  this probability is to be determined for the following codes:
	* Reed–Solomon code  $(7, \, 3, \, 5)_8 \ \Rightarrow \ d_{\rm min} = 5$,		* Reed–Solomon code  $(7, \, 3, \, 5)_8 \ \Rightarrow \ d_{\rm min} = 5$,

	* Reed–Solomon code  $(7, \, 5, \, 3)_8 \ \Rightarrow \ d_{\rm min} = 3$.		* Reed–Solomon code  $(7, \, 5, \, 3)_8 \ \Rightarrow \ d_{\rm min} = 3$.


	Furthermore, let it hold:		Furthermore,  let it hold:
	* Each symbol is corrupted into another symbol with probability  $\varepsilon_{\rm S} = 0.1$  and correctly transferred with probability  $1 - \varepsilon_{\rm S} = 0.9$~~ ~~ .		* Each symbol is corrupted into another symbol with probability  $\varepsilon_{\rm S} = 0.1$  and correctly transferred with probability  $1 - \varepsilon_{\rm S} = 0.9$.
	* For the distance spectrum of a Reed–Solomon code of length  $n$  holds with  $d = d_{\rm min}$:		* For the distance spectrum of a Reed–Solomon code of length  $n$  holds with  $d = d_{\rm min}$:
	:$$W_i = {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$		:$$W_i = {n \choose i} \cdot \sum_{j = 0}^{i-d}\hspace{0.15cm}(-1)^j \cdot {i \choose j} \cdot \big [\hspace{0.03cm}q^{i\hspace{0.03cm}-\hspace{0.03cm}j\hspace{0.03cm}-\hspace{0.03cm}d\hspace{0.03cm}+\hspace{0.03cm}1}-1 \hspace{0.03cm} \big ]\hspace{0.05cm}.$$

	Besides, two bounds for the block error probability will be considered and evaluated:		Besides,  two bounds for the block error probability will be considered and evaluated:
	* If the minimum distance ~~alone~~ is known, one can derive an ~~<i>~~upper bound~~</i>~~  from it. The weight enumerating factors  $W_i$  are to be chosen in such a way, that certainly (that is:   with all constellations) applies:		* If alone the minimum distance is known,  one can derive an  "upper bound"  from it.  The weight enumerating factors  $W_i$  are to be chosen in such a way,  that certainly (that is:   with all constellations)  applies:
	:$${\rm Pr}({\rm ~~Upper~~\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm ~~Block~~\hspace{0.15cm}error})		:$${\rm Pr}({\rm upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm block\hspace{0.15cm}error})
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$
	* A ~~<i>~~lower bound~~</i>~~  additionally requires knowledge of the weight function  $W_i$  for  $i = d_{\rm min}$. Thus, the following condition can be satisfied:		* A  "lower bound"  additionally requires knowledge of the weight function  $W_i$  for  $i = d_{\rm min}$.  Thus,  the following condition can be satisfied:
	:$${\rm Pr}({\rm ~~Upper~~\hspace{0.15cm} bound}) \le {\rm Pr}({\rm ~~Block~~\hspace{0.15cm}error})		:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) \le {\rm Pr}({\rm block\hspace{0.15cm}error})
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$





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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes\|"Definition and Properties of Reed-Solomon Codes"]].		*The exercise belongs to the chapter  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes\|"Definition and Properties of Reed-Solomon Codes"]].

	*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].		*Reference is made in particular to the page  [[Channel_Coding/Definition_and_Properties_of_Reed-Solomon_Codes#Singleton_bound_and_minimum_distance\|"Singleton Bound and Minimum Distance"]].
	* ~~To be calculated are the~~ weights marked in red in the above graph~~  $W_i$~~.
			* The weights  $W_i$ marked in red in the above graph are to be calculated.


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	$W_7 \ = \ ${ 217 }		$W_7 \ = \ ${ 217 }

	{What is the missing weight of the  $\rm RSC \, (7, \, 5, \, 3)_8$ in the graph?		{What is the missing weight of the  $\rm RSC \, (7, \, 5, \, 3)_8$  in the graph?
	\|type="{}"}		\|type="{}"}
	$W_3 \ = \ ${ 245 }		$W_3 \ = \ ${ 245 }

	{What is the probability that an incorrect block remains undetected? Let the corruption probability of each symbol be $\varepsilon = 0.1$.		{What is the probability that an incorrect block remains undetected? Let the corruption probability of each symbol be  $\varepsilon = 0.1$.
	\|type="{}"}		\|type="{}"}
	$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(~~Blockfehler~~) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$		$\rm RSC \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.263 3% } $\ \cdot 10^{-5}$
	$\rm RSC \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} Pr(~~Blockfehler~~) \ = \ ${ 0.942 3% } $\ \cdot 10^{-5}$		$\rm RSC \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} Pr(block error) \ = \ ${ 0.942 3% } $\ \cdot 10^{-5}$

	{Calculate and evaluate for both codes the upper bound proposed in the specification  $p_{\rm upper} = \rm Pr(~~Upper~~ \ bound)$.		{Calculate and evaluate for both codes the  "upper bound"  proposed in the specification:  $p_{\rm upper} = \rm Pr(upper \ bound)$.
	\|type="{}"}		\|type="{}"}
	${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$
	${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$

	{Calculate and evaluate for both codes the lower bound proposed in the specification  $p_{\rm lower} = \rm Pr(~~Lower~~ \ bound)$.		{Calculate and evaluate for both codes the  "lower bound"  proposed in the specification:  $p_{\rm lower} = \rm Pr(lower \ bound)$.
	\|type="{}"}		\|type="{}"}
	${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$

Guenter: Guenter moved page Aufgaben:Exercise 2.10: Reed-Solomon Fault Detection to Aufgaben:Exercise 2.10: Reed-Solomon Error Detection

2022-10-05T15:09:51Z

Guenter moved page Aufgaben:Exercise 2.10: Reed-Solomon Fault Detection to Aufgaben:Exercise 2.10: Reed-Solomon Error Detection

← Older revision	Revision as of 17:09, 5 October 2022
(No difference)

Noah at 17:57, 2 September 2022

2022-09-02T17:57:46Z

← Older revision		Revision as of 19:57, 2 September 2022
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	{Calculate and evaluate for both codes the upper bound proposed in the specification  $p_{\rm upper} = \rm Pr(Upper \ bound)$.		{Calculate and evaluate for both codes the upper bound proposed in the specification  $p_{\rm upper} = \rm Pr(Upper \ bound)$.
	\|type="{}"}		\|type="{}"}
	${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm ~~oben~~} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 0.81 3% } $\ \cdot 10^{-5}$
	${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm ~~oben~~} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm upper} \ = \ ${ 65.6 3% } $\ \cdot 10^{-5}$

	{Calculate and evaluate for both codes the lower bound proposed in the specification  $p_{\rm lower} = \rm Pr(Lower \ bound)$.		{Calculate and evaluate for both codes the lower bound proposed in the specification  $p_{\rm lower} = \rm Pr(Lower \ bound)$.
	\|type="{}"}		\|type="{}"}
	${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm ~~unten~~} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 3, \, 5) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.233 3% } $\ \cdot 10^{-5}$
	${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm ~~unten~~} \ = \ ${ 0.494 3% } $\ \cdot 10^{-5}$		${\rm RSC} \, (7, \, 5, \, 3) \text{:} \hspace{0.4cm} p_{\rm lower} \ = \ ${ 0.494 3% } $\ \cdot 10^{-5}$
	</quiz>		</quiz>

← Older revision		Revision as of 15:30, 16 March 2026
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	* Error detection can still work even if more than  $n - k$  symbol errors occur.  This is when the received word is not a valid code word of the Reed–Solomon code:		* Error detection can still work even if more than  $n - k$  symbol errors occur.  This is when the received word is not a valid code word of the Reed–Solomon code:
	:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}		:$$\underline {y} \notin C_{\rm RS} = \{ \underline {c}_{\hspace{0.05cm}0}, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_i, \hspace{0.05cm}\text{...} \hspace{0.1cm}, \underline {c}_{\hspace{0.05cm}n -1} \}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$
	* However,  if the falsified received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the falsified block undetected.		* However,  if the falsified received word  $(\underline{y} ≠ \underline{c})$  is a valid code word  $(\underline{y\in C_{\rm RS}})$,  the decoding process will leave the falsified block undetected.


	We define as  '''block error probability''':		We define as  '''block error probability''':
	:$${\rm Pr}({\rm block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})		:$${\rm Pr}({\rm block\hspace{0.15cm}error}) = {\rm Pr}(\underline {y} \ne \underline {c})\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In this exercise,  this probability is to be determined for the following codes:		In this exercise,  this probability is to be determined for the following codes:
Line 31:		Line 29:
	Besides,  two bounds for the block error probability will be considered and evaluated:		Besides,  two bounds for the block error probability will be considered and evaluated:
	* If alone the minimum distance is known,  one can derive an  "upper bound"  from it.  The weight enumerating factors  $W_i$  are to be chosen in such a way,  that certainly (that is:   with all constellations)  applies:		* If alone the minimum distance is known,  one can derive an  "upper bound"  from it.  The weight enumerating factors  $W_i$  are to be chosen in such a way,  that certainly (that is:   with all constellations)  applies:
	:$${\rm Pr}({\rm upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm block\hspace{0.15cm}error})		:$${\rm Pr}({\rm upper\hspace{0.15cm} bound}) \ge {\rm Pr}({\rm block\hspace{0.15cm}error})\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$
	* A  "lower bound"  additionally requires knowledge of the weight function  $W_i$  for  $i = d_{\rm min}$.  Thus,  the following condition can be satisfied:		* A  "lower bound"  additionally requires knowledge of the weight function  $W_i$  for  $i = d_{\rm min}$.  Thus,  the following condition can be satisfied:
	:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) \le {\rm Pr}({\rm block\hspace{0.15cm}error})		:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) \le {\rm Pr}({\rm block\hspace{0.15cm}error})\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	*Because of the minimum distance  $d_{\min} = 5$,  $W_3 \ \underline{= 0}$  and  $W_4 \ \underline{= 0}$.  The other weights result in		*Because of the minimum distance  $d_{\min} = 5$,  $W_3 \ \underline{= 0}$  and  $W_4 \ \underline{= 0}$.  The other weights result in
	:$$W_5 = {7 \choose 5} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 \cdot4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot4 \cdot 5} \cdot 7 = 21 \cdot 7		:$$W_5 = {7 \choose 5} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 \cdot4 \cdot 3}{1 \cdot 2 \cdot 3 \cdot4 \cdot 5} \cdot 7 = 21 \cdot 7\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
	\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$		:$$W_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 6} \cdot \sum_{j = 0}^{1}\hspace{0.15cm}(-1)^j \cdot {6 \choose j} \cdot \big (\hspace{0.03cm}8^{2\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )=7 \cdot \left [ (8^2 - 1) - 6 \cdot (8^1 - 1)\right ] = 7 \cdot (63-42)\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
	:$$W_6 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 6} \cdot \sum_{j = 0}^{1}\hspace{0.15cm}(-1)^j \cdot {6 \choose j} \cdot \big (\hspace{0.03cm}8^{2\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )=7 \cdot \left [ (8^2 - 1) - 6 \cdot (8^1 - 1)\right ] = 7 \cdot (63-42)		:$$W_7 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 7} \cdot \sum_{j = 0}^{2}\hspace{0.15cm}(-1)^j \cdot {7 \choose j} \cdot \big (\hspace{0.03cm}8^{3\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )= (8^3 - 1) - 7 \cdot (8^2 - 1) +21 \cdot (8^1 - 1) = 511 - 7 \cdot 63 + 21 \cdot 7\hspace{0.15cm}\underline {= 217}\hspace{0.05cm},$$
	\hspace{0.15cm}\underline {= 147}\hspace{0.05cm},$$
	:$$W_7 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {7 \choose 7} \cdot \sum_{j = 0}^{2}\hspace{0.15cm}(-1)^j \cdot {7 \choose j} \cdot \big (\hspace{0.03cm}8^{3\hspace{0.03cm}-\hspace{0.03cm}j} -1 \big )= (8^3 - 1) - 7 \cdot (8^2 - 1) +21 \cdot (8^1 - 1) = 511 - 7 \cdot 63 + 21 \cdot 7
	\hspace{0.15cm}\underline {= 217}\hspace{0.05cm},$$
	:$$\Rightarrow \hspace{0.3cm}W_0 + W_5 + W_6 + W_7 = 1 + 147 + 147 + 217 = 512 = 8^3 = q^k\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}W_0 + W_5 + W_6 + W_7 = 1 + 147 + 147 + 217 = 512 = 8^3 = q^k\hspace{0.05cm}.$$


	'''(2)'''  Analogous to subtask  '''(1)''',  we obtain:		'''(2)'''  Analogous to subtask  '''(1)''',  we obtain:
	:$$W_3 = {7 \choose 3} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 }{1 \cdot 2 \cdot 3 } \cdot 7 = 35 \cdot 7		:$$W_3 = {7 \choose 3} \cdot (8^1 - 1) = \frac {7 \cdot 6 \cdot 5 }{1 \cdot 2 \cdot 3 } \cdot 7 = 35 \cdot 7\hspace{0.15cm}\underline {= 245}\hspace{0.05cm}.$$
	\hspace{0.15cm}\underline {= 245}\hspace{0.05cm}.$$


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	:Then,  for the probability that in a code word with  $n = 7$  symbols		:Then,  for the probability that in a code word with  $n = 7$  symbols
	* exactly three symbols are falsified:		* exactly three symbols are falsified:
	:$$p_3 = 0.1^3 \cdot 0.9^4 = 0.6561 \cdot 10^{-3}		:$$p_3 = 0.1^3 \cdot 0.9^4 = 0.6561 \cdot 10^{-3}\hspace{0.05cm},$$
	\hspace{0.05cm},$$
	* exactly four symbols are falsified:		* exactly four symbols are falsified:
	:$$p_4 = 0.1^4 \cdot 0.9^3 = 0.729 \cdot 10^{-4}		:$$p_4 = 0.1^4 \cdot 0.9^3 = 0.729 \cdot 10^{-4}\hspace{0.05cm},$$
	\hspace{0.05cm},$$
	* exactly five symbols are falsified:		* exactly five symbols are falsified:
	:$$p_5 = 0.1^5 \cdot 0.9^2 = 0.81 \cdot 10^{-5}		:$$p_5 = 0.1^5 \cdot 0.9^2 = 0.81 \cdot 10^{-5}\hspace{0.05cm},$$
	\hspace{0.05cm},$$
	* exactly six symbols are falsified:		* exactly six symbols are falsified:
	:$$p_6 = 0.1^6 \cdot 0.9 = 0.9 \cdot 10^{-6}\hspace{0.05cm},$$		:$$p_6 = 0.1^6 \cdot 0.9 = 0.9 \cdot 10^{-6}\hspace{0.05cm},$$
Line 120:		Line 109:

	⇒   At the  $\rm RSC \, (7, \, 3, \, 5)_8$  the zero word can be falsified by symbol errors into one of  $q^k - 1 = 8^3 - 1 = 511$  other code words.  Thus,  using the weight enumerating functions of subtask  '''(1)''',  we obtain:		⇒   At the  $\rm RSC \, (7, \, 3, \, 5)_8$  the zero word can be falsified by symbol errors into one of  $q^k - 1 = 8^3 - 1 = 511$  other code words.  Thus,  using the weight enumerating functions of subtask  '''(1)''',  we obtain:
	:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{511} =\frac {147 \cdot 0.81 \cdot 10^{-5} + 147 \cdot 0.9 \cdot 10^{-6} + 217 \cdot 10^{-7}}{511}		:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{511} =\frac {147 \cdot 0.81 \cdot 10^{-5} + 147 \cdot 0.9 \cdot 10^{-6} + 217 \cdot 10^{-7}}{511}\hspace{0.15cm}\underline {\approx 0.263 \cdot 10^{-5}} \hspace{0.05cm}.$$
	\hspace{0.15cm}\underline {\approx 0.263 \cdot 10^{-5}} \hspace{0.05cm}.$$

	⇒   At the  $\rm RSC \, (7, \, 5, \, 3)_8$  we have to average over  $8^5 - 1 = 32767$  falsification probabilities because of  $k = 5$.  With  $W_3 = 245$  from subtask  '''(2)'''  and the weights		⇒   At the  $\rm RSC \, (7, \, 5, \, 3)_8$  we have to average over  $8^5 - 1 = 32767$  falsification probabilities because of  $k = 5$.  With  $W_3 = 245$  from subtask  '''(2)'''  and the weights
	:$$W_4 = 1224, \ W_5 = 5586, \ W_6 = 12838, \ W_7 = 12873$$		:$$W_4 = 1224, \ W_5 = 5586, \ W_6 = 12838, \ W_7 = 12873$$
	according to the specification sheet,  we obtain for this:		according to the specification sheet,  we obtain for this:
	:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_3 \cdot p_3 + W_4 \cdot p_4 + W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{32767} =\frac {245 \cdot 0.656 \cdot 10^{-3} + \hspace{0.15cm}... \hspace{0.15cm}+ 12873 \cdot 10^{-7}}{32767}		:$${\rm Pr}({\rm block\hspace{0.15cm}error}) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \frac {W_3 \cdot p_3 + W_4 \cdot p_4 + W_5 \cdot p_5 + W_6 \cdot p_6 + W_7 \cdot p_7}{32767} =\frac {245 \cdot 0.656 \cdot 10^{-3} + \hspace{0.15cm}... \hspace{0.15cm}+ 12873 \cdot 10^{-7}}{32767}\hspace{0.15cm}\underline {\approx 0.942 \cdot 10^{-5}} \hspace{0.05cm}. $$
	\hspace{0.15cm}\underline {\approx 0.942 \cdot 10^{-5}} \hspace{0.05cm}. $$


Line 146:		Line 133:

	'''(5)'''  Using the abbreviation  $d = d_{\rm min}$,  one obtains for the lower bound:		'''(5)'''  Using the abbreviation  $d = d_{\rm min}$,  one obtains for the lower bound:
	:$${\rm Pr}({\rm lower\hspace{0.15cm} bound})= \frac{W_d \cdot p_d}{q^k -1}		:$${\rm Pr}({\rm lower\hspace{0.15cm} bound})= \frac{W_d \cdot p_d}{q^k -1}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$
	* For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because of  $W_d = W_5$ and  $p_d = p_5$,  the lower bound is about  $11\%$  below the actual value $(0.263 \cdot 10^{-5})$:		* For the  $\rm RSC \, (7, \, 3, \, 5)_8$,  because of  $W_d = W_5$ and  $p_d = p_5$,  the lower bound is about  $11\%$  below the actual value $(0.263 \cdot 10^{-5})$:
	:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{147 \cdot 0.81 \cdot 10^{-5}}{511}		:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{147 \cdot 0.81 \cdot 10^{-5}}{511}\hspace{0.15cm}\underline {\approx 0.233 \cdot 10^{-5}}$$
	\hspace{0.15cm}\underline {\approx 0.233 \cdot 10^{-5}}$$
	* For the  $\rm RSC \, (7, \, 5, \, 3)_8$  holds with  $W_d = W_3$  and  $p_d = p_3$,  the lower bound here deviates from the actual value  $(0.942 \cdot 10^{-5})$  more strongly,  because in this code the contributions of the higher weights  $(W_4, \ W_5, \ W_6, \ W_7)$  are more relevant in relation to  $W_3$:		* For the  $\rm RSC \, (7, \, 5, \, 3)_8$  holds with  $W_d = W_3$  and  $p_d = p_3$,  the lower bound here deviates from the actual value  $(0.942 \cdot 10^{-5})$  more strongly,  because in this code the contributions of the higher weights  $(W_4, \ W_5, \ W_6, \ W_7)$  are more relevant in relation to  $W_3$:
	:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{245 \cdot 0.656 \cdot 10^{-3}}{32767}		:$${\rm Pr}({\rm lower\hspace{0.15cm} bound}) = \frac{245 \cdot 0.656 \cdot 10^{-3}}{32767}\hspace{0.15cm}\underline {\approx 0.494 \cdot 10^{-5}}\hspace{0.05cm}.$$
	\hspace{0.15cm}\underline {\approx 0.494 \cdot 10^{-5}}\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}