Aufgaben:Exercise 4.7Z: Principle of Syndrome Decoding - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:58:27Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:58, 16 March 2026
Line 116:		Line 116:

	[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]		[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]
	[[de:Aufgaben:4.~~7Z_Syndromdecodierung_–_Prinzip~~]]		[[de:Aufgaben:Aufgabe 4.7Z: Zum Prinzip der Syndromdecodierung]]

Maintenance script: Fix 3 split formula blocks

2026-03-16T15:23:10Z

Fix 3 split formula blocks

← Older revision		Revision as of 17:23, 16 March 2026
Line 94:		Line 94:

	*Since  $\mathcal{C}_2$  is also a systematic code,  the generator matrix  $\mathbf{G}$  can be represented  in the following form:		*Since  $\mathcal{C}_2$  is also a systematic code,  the generator matrix  $\mathbf{G}$  can be represented  in the following form:
	:$$		:$${ \boldsymbol{\rm G}}= \begin{pmatrix}1 &0 &0 &1 &1 &0 \\0 &1 &0 &1 &0 &1 \\0 &0 &1 &0 &1 &1\end{pmatrix}= \left ( { \boldsymbol{\rm I}}_3 ; \hspace{0.15cm} { \boldsymbol{\rm P}} \right )\hspace{0.5cm}{\rm mit }\hspace{0.5cm}{ \boldsymbol{\rm P}}= \begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1\end{pmatrix}\hspace{0.05cm}.$$
	{ \boldsymbol{\rm G}}= \begin{pmatrix}1 &0 &0 &1 &1 &0 \\0 &1 &0 &1 &0 &1 \\0 &0 &1 &0 &1 &1\end{pmatrix}= \left ( { \boldsymbol{\rm I}}_3 ; \hspace{0.15cm} { \boldsymbol{\rm P}} \right )\hspace{0.5cm}{\rm mit }\hspace{0.5cm}{ \boldsymbol{\rm P}}= \begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1\end{pmatrix}\hspace{0.05cm}.$$		*So it can be written for the parity-check matrix::$${ \boldsymbol{\rm H}}= \left ( { \boldsymbol{\rm P}}^{\rm T} ; \hspace{0.15cm} { \boldsymbol{\rm I}}_3 \right )= \begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\end{pmatrix}\hspace{0.05cm}.$$
	*So it can be written for the parity-check matrix::$$		Here  $\mathbf{I}_3$  denotes a  $3 &times 3$  diagonal matrix typical of the systematic code.<u>Proposed solutions 2, 3, and 4</u>  are therefore correct::* Row 1:   $110100$,:* Row 2:   $101010$,:* Row 3:   $011001$.'''(3)'''  Correct is the  <u>proposed solution 1</u>:According to the statements in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]]   ⇒   $\underline{s} = \underline{e} \cdot \mathbf{H}^{\rm T}$  can be written.Thus,  for the error-free case   ⇒   $\underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 0)$::$$\underline{s}= \left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right ) = \underline{s}_0.$$
	{ \boldsymbol{\rm H}}= \left ( { \boldsymbol{\rm P}}^{\rm T} ; \hspace{0.15cm} { \boldsymbol{\rm I}}_3 \right )= \begin{pmatrix}1 &1 &0 &1 &0 &0\\1 &0 &1 &0 &1 &0\\0 &1 &1 &0 &0 &1\end{pmatrix}\hspace{0.05cm}.$$
	Here  $\mathbf{I}_3$  denotes a  $3 &times 3$  diagonal matrix typical of the systematic code.<u>Proposed solutions 2, 3, and 4</u>  are therefore correct::* Row 1:   $110100$,:* Row 2:   $101010$,:* Row 3:   $011001$.'''(3)'''  Correct is the  <u>proposed solution 1</u>:According to the statements in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]]   ⇒   $\underline{s} = \underline{e} \cdot \mathbf{H}^{\rm T}$  can be written.Thus,  for the error-free case   ⇒   $\underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 0)$::$$
	\underline{s}= \left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right ) = \underline{s}_0.$$

Maintenance script: Fix 1 split formula blocks

2026-03-16T14:24:56Z

Fix 1 split formula blocks

@@ Line 94: / Line 94: @@
 *Since&nbsp; $\mathcal{C}_2$&nbsp; is also a systematic code,&nbsp; the generator matrix&nbsp; $\mathbf{G}$&nbsp; can be represented&nbsp; in the following form:
-:$${ \boldsymbol{\rm G}}
+:$$
-=  \begin{pmatrix}
+{ \boldsymbol{\rm G}}=  \begin{pmatrix}1 &0 &0 &1 &1 &0  \\0 &1 &0 &1 &0 &1  \\0 &0 &1 &0 &1 &1\end{pmatrix}= \left (  { \boldsymbol{\rm I}}_3 ;  \hspace{0.15cm} { \boldsymbol{\rm P}} \right )\hspace{0.5cm}{\rm mit }\hspace{0.5cm}{ \boldsymbol{\rm P}}=  \begin{pmatrix}1 &1 &0   \\1 &0 &1   \\0 &1 &1\end{pmatrix}\hspace{0.05cm}.$$
-&0 &0 &1 &1 &0  \\
+*So it can be written for the parity-check matrix::$$
-&1 &0 &1 &0 &1  \\
+{ \boldsymbol{\rm H}}=   \left ( { \boldsymbol{\rm P}}^{\rm T} ;  \hspace{0.15cm} { \boldsymbol{\rm I}}_3  \right )=  \begin{pmatrix}1 &1 &0  &1 &0  &0\\1 &0 &1  &0 &1  &0\\0 &1 &1  &0 &0  &1\end{pmatrix}\hspace{0.05cm}.$$
-&0 &1 &0 &1 &1
+*Here&nbsp; $\mathbf{I}_3$&nbsp; denotes a&nbsp; $3 &times 3$&nbsp; diagonal matrix typical of the systematic code.*<u>Proposed solutions 2, 3, and 4</u>&nbsp; are therefore correct::* Row 1: &nbsp; $110100$,:* Row 2: &nbsp; $101010$,:* Row 3: &nbsp; $011001$.'''(3)'''&nbsp; Correct is the&nbsp; <u>proposed solution 1</u>:*According to the statements in the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes| "Decoding of Linear Block Codes"]] &nbsp; &rArr; &nbsp;  $\underline{s} = \underline{e} \cdot \mathbf{H}^{\rm T}$&nbsp; can be written.*Thus,&nbsp; for the error-free case &nbsp; &#8658; &nbsp; $\underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 0)$::$$
-	\end{pmatrix}
+\underline{s}= \left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot\begin{pmatrix}1 &1 &0 \\1 &0 &1 \\0 &1 &1 \\1 &0 &0 \\0 &1 &0 \\0 &0 &1\end{pmatrix}=\left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0 \right ) = \underline{s}_0.$$

Maintenance script: Join 1 split formula blocks

2026-03-16T14:17:00Z

Join 1 split formula blocks

@@ Line 115: / Line 115: @@
 &1 &1  &0 &0  &1
 	\end{pmatrix}
-\hspace{0.05cm}.$$
+\hspace{0.05cm}.$$*Here&nbsp; $\mathbf{I}_3$&nbsp; denotes a&nbsp; $3 &times 3$&nbsp; diagonal matrix typical of the systematic code.*<u>Proposed solutions 2, 3, and 4</u>&nbsp; are therefore correct::* Row 1: &nbsp; $110100$,:* Row 2: &nbsp; $101010$,:* Row 3: &nbsp; $011001$.'''(3)'''&nbsp; Correct is the&nbsp; <u>proposed solution 1</u>:*According to the statements in the chapter&nbsp; [[Channel_Coding/Decoding_of_Linear_Block_Codes| "Decoding of Linear Block Codes"]] &nbsp; &rArr; &nbsp;  $\underline{s} = \underline{e} \cdot \mathbf{H}^{\rm T}$&nbsp; can be written.*Thus,&nbsp; for the error-free case &nbsp; &#8658; &nbsp; $\underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 0)$::$$\underline{s}= \left ( 0, \hspace{0.03cm} 0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0, \hspace{0.03cm}0 \right ) \cdot
    \begin{pmatrix}
 &1 &0 \\

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:45Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 8:		Line 8:
	For syndrome decoding one proceeds as follows:		For syndrome decoding one proceeds as follows:
	* One forms the syndrome from the received vector  $\underline{y}$  of length  $n$  and the parity-check matrix  $\mathbf{H}$:		* One forms the syndrome from the received vector  $\underline{y}$  of length  $n$  and the parity-check matrix  $\mathbf{H}$:
	:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m)		:$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m)\hspace{0.05cm}, \hspace{0.5cm}{\rm Note\hspace{-0.10cm}:} \hspace{0.15cm}m = n-k \hspace{0.05cm}. $$
	\hspace{0.05cm}, \hspace{0.5cm}{\rm Note\hspace{-0.10cm}:} \hspace{0.15cm}m = n-k \hspace{0.05cm}. $$

	* The received word  $\underline{y} = \underline{x} \ {\rm (code\:word)} + \underline{e} \ {\rm (error\:vector)}$  is not necessarily an element of  ${\rm GF}(2^m)$,  but certainly an element of  ${\rm GF}(2^n)$  and it holds because of   $\underline{x} \cdot \mathbf{H}^{\rm T} = \underline{0}$   equally:		* The received word  $\underline{y} = \underline{x} \ {\rm (code\:word)} + \underline{e} \ {\rm (error\:vector)}$  is not necessarily an element of  ${\rm GF}(2^m)$,  but certainly an element of  ${\rm GF}(2^n)$  and it holds because of   $\underline{x} \cdot \mathbf{H}^{\rm T} = \underline{0}$   equally:
Line 22:		Line 21:

	A similar table is to be created for the truncated Hamming code  $\rm HC \ (6, 3, 3)$.  This has already been used in the  [[Aufgaben:Exercise_4.6:_Product_Code_Generation\|$\text{Exercise 4.6}$]]  and the  [[Aufgaben:Exercise_4.6Z:_Basics_of_Product_Codes\| $\text{Exercise 4.6Z}$]]  and is given by its generator matrix:		A similar table is to be created for the truncated Hamming code  $\rm HC \ (6, 3, 3)$.  This has already been used in the  [[Aufgaben:Exercise_4.6:_Product_Code_Generation\|$\text{Exercise 4.6}$]]  and the  [[Aufgaben:Exercise_4.6Z:_Basics_of_Product_Codes\| $\text{Exercise 4.6Z}$]]  and is given by its generator matrix:
	:$${ \boldsymbol{\rm G}}		:$${ \boldsymbol{\rm G}}= \begin{pmatrix}1 &0 &0 &1 &1 &0 \\0 &1 &0 &1 &0 &1 \\0 &0 &1 &0 &1 &1\end{pmatrix} \hspace{0.05cm}.$$
	= \begin{pmatrix}
	1 &0 &0 &1 &1 &0 \\
	0 &1 &0 &1 &0 &1 \\
	0 &0 &1 &0 &1 &1
	\end{pmatrix} \hspace{0.05cm}.$$

	Unlike the original  $\rm (7, 4, 3)$  Hamming code,  the shortened  $\rm (6, 3, 3)$  Hamming code is not perfect, so a single-error coset leader  $\underline{s}_{\mu}$  cannot be found for all possible  $\underline{e}_{\mu}$.		Unlike the original  $\rm (7, 4, 3)$  Hamming code,  the shortened  $\rm (6, 3, 3)$  Hamming code is not perfect, so a single-error coset leader  $\underline{s}_{\mu}$  cannot be found for all possible  $\underline{e}_{\mu}$.
Line 88:		Line 82:
	'''(1)'''  Correct is the  <u>proposed solution 2</u>:		'''(1)'''  Correct is the  <u>proposed solution 2</u>:
	*From the  [[Aufgaben:Exercise_4.7Z: _Principle_of_Syndrome_Decoding\| "syndrome table"]]  on the information page – valid for the  $\rm (7, 4, 3)$  Hamming code – it can be read that the syndrome   $\underline{s} = \underline{s}_3 = (0, \, 1, \, 1)$  corresponds to the error pattern   $\underline{e} = (0, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0)$. Thus the code word		*From the  [[Aufgaben:Exercise_4.7Z: _Principle_of_Syndrome_Decoding\| "syndrome table"]]  on the information page – valid for the  $\rm (7, 4, 3)$  Hamming code – it can be read that the syndrome   $\underline{s} = \underline{s}_3 = (0, \, 1, \, 1)$  corresponds to the error pattern   $\underline{e} = (0, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0)$. Thus the code word
	:$$\underline{x} = \underline{y} \hspace{0.15cm}+ \hspace{0.15cm} \underline{y} = (0, 1, 1, 0, 1, 1, 0) \hspace{0.15cm}+ \hspace{0.15cm}(0, 0, 1, 0, 0, 0, 0)		:$$\underline{x} = \underline{y} \hspace{0.15cm}+ \hspace{0.15cm} \underline{y} = (0, 1, 1, 0, 1, 1, 0) \hspace{0.15cm}+ \hspace{0.15cm}(0, 0, 1, 0, 0, 0, 0)\hspace{0.15cm}= \hspace{0.15cm}(0, 1, 0, 0, 1, 1, 0)$$
	\hspace{0.15cm}= \hspace{0.15cm}(0, 1, 0, 0, 1, 1, 0)$$

	is most likely and the syndrome decoder will output this as the result.		is most likely and the syndrome decoder will output this as the result.
Line 115:		Line 108:
	0 &1 &1		0 &1 &1
	\end{pmatrix}		\end{pmatrix}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$*So it can be written for the parity-check matrix::$${ \boldsymbol{\rm H}}

	*So it can be written for the parity-check matrix:
	:$${ \boldsymbol{\rm H}}
	= \left ( { \boldsymbol{\rm P}}^{\rm T} ; \hspace{0.15cm} { \boldsymbol{\rm I}}_3 \right )		= \left ( { \boldsymbol{\rm P}}^{\rm T} ; \hspace{0.15cm} { \boldsymbol{\rm I}}_3 \right )
	= \begin{pmatrix}		= \begin{pmatrix}

Maintenance script: Fix spaces in LaTeX braces to prevent Quiz misparse

2026-03-16T13:21:29Z

Fix spaces in LaTeX braces to prevent Quiz misparse

← Older revision		Revision as of 15:21, 16 March 2026
Line 153:		Line 153:

	'''(4)'''  <u>All statements</u>  are true,  as can be seen from the sample solution to the last subtask:		'''(4)'''  <u>All statements</u>  are true,  as can be seen from the sample solution to the last subtask:
	*The rows of the transposed parity-check matrix,  read from top to bottom,  give the respective syndromes for the error patterns $\underline{e} = (1, \, 0, \, 0, \, 0, \, 0, \, 0), \hspace{0.05cm} \text{ ... } \hspace{0.05cm} , \ \underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 1)$.		*The rows of the transposed parity-check matrix,  read from top to bottom,  give the respective syndromes for the error patterns $\underline{e} = (1, \, 0, \, 0, \, 0, \, 0, \, 0), \hspace{0.05cm} \text{...} \hspace{0.05cm} , \ \underline{e} = (0, \, 0, \, 0, \, 0, \, 0, \, 1)$.

Maintenance script: Add German interlanguage link

2026-03-16T12:33:49Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 168:		Line 168:

	[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]		[[Category:Channel Coding: Exercises\|^4.2 About the Product Codes^]]
			[[de:Aufgaben:4.7Z_Syndromdecodierung_–_Prinzip]]

Guenter at 14:53, 22 December 2022

2022-12-22T14:53:56Z

← Older revision		Revision as of 16:53, 22 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	~~[[File:~~[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code $\mathcal{C}_1~~$]] \|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)~~$]]		[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code  $\mathcal{C}_1$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].

Guenter at 11:44, 21 December 2022

2022-12-21T11:44:58Z

← Older revision		Revision as of 13:44, 21 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	[[File:~~EN_KC_T_4_2_S2b_v2~~.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]		[[File:[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code $\mathcal{C}_1$]] \|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].

Guenter at 11:41, 21 December 2022

2022-12-21T11:41:27Z

← Older revision		Revision as of 13:41, 21 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	[[File:~~EN_KC_Z_4_7_neu~~.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]		[[File:EN_KC_T_4_2_S2b_v2.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].

← Older revision		Revision as of 16:53, 22 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	~~[[File:~~[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code $\mathcal{C}_1~~$]] \|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)~~$]]		[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code  $\mathcal{C}_1$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].

← Older revision		Revision as of 13:44, 21 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	[[File:~~EN_KC_T_4_2_S2b_v2~~.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]		[[File:[[File:EN KC T 4 2 S2b v2 neu.png\|right\|frame\|Syndrome table for code $\mathcal{C}_1$]] \|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].

← Older revision		Revision as of 13:41, 21 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}		{{quiz-Header\|Buchseite=Channel_Coding/The_Basics_of_Product_Codes}}

	[[File:~~EN_KC_Z_4_7_neu~~.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]		[[File:EN_KC_T_4_2_S2b_v2.png\|right\|frame\|Coset leader for the code under consideration  $\rm HC \ (7, \ 4, \ 3)$]]
	The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].		The syndrome decoding was already treated in detail in the chapter  [[Channel_Coding/Decoding_of_Linear_Block_Codes\| "Decoding of Linear Block Codes"]].