Difference between revisions of "Aufgaben:Exercise 1.07Z: Classification of Block Codes"

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===Solution===
 
===Solution===
 
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'''(1)'''&nbsp; <u>Statements 1 and 2</u> are correct:
+
'''(1)'''&nbsp; <u>Statements 1 and 2</u>&nbsp; are correct:
* That is why there are $\rm 4 \ over \ 2 = 6$ codewords.  
+
* That is why there are&nbsp; "$\rm 4 \ over \ 2 = 6$"&nbsp; code words.  
* Statement 3 is false. For example, if the first bit is $0$, there is one codeword starting $00$ and two codewords starting $01$.
+
* Statement 3 is false.&nbsp; For example,&nbsp; if the first bit is&nbsp; "$0$",&nbsp; there is one code word starting&nbsp; "$00$"&nbsp; and two code words starting&nbsp; "$01$".
  
  
  
'''(2)'''&nbsp; <u>Statements 1 to 4</u> are correct:
+
'''(2)'''&nbsp; <u>Statements 1 to 4</u>&nbsp; are correct:
* All codes that can be described by a generator matrix $\boldsymbol {\rm G}$ and/or a parity-check matrix $\boldsymbol {\rm H}$ are linear.  
+
* All codes that can be described by a generator matrix&nbsp; $\boldsymbol {\rm G}$&nbsp; and/or a parity-check matrix&nbsp; $\boldsymbol {\rm H}$&nbsp; are linear.  
*In contrast, "Code 5" does not satisfy any of the conditions required for linear codes. For example
+
*In contrast,&nbsp; "code 5" does not satisfy any of the conditions required for linear codes.&nbsp; For example
  
:*is missing the all zero word,
+
:*is missing the all-zero word,
:*the code cardinality $|\mathcal{C}|$ is not a power of two,
+
:*the code size&nbsp; $|\mathcal{C}|$&nbsp; is not a power of two,
:*gives $(0, 1, 0, 1) \oplus (1, 0, 1, 0) = (1, 1, 1, 1)$ no valid codeword.
+
:*gives&nbsp; $(0, 1, 0, 1) \oplus (1, 0, 1, 0) = (1, 1, 1, 1)$&nbsp; no valid code word.
  
  
  
'''(3)'''&nbsp; <u>Statements 1 to 3</u> are correct:
+
'''(3)'''&nbsp; <u>Statements 1 to 3</u>&nbsp; are correct:
*In a systematic code, the first $k$ bits of each codeword $\underline{x}$ must always be equal to the information word $\underline{u}$.  
+
*In a systematic code,&nbsp; the first&nbsp; $k$&nbsp; bits of each code word&nbsp; $\underline{x}$&nbsp; must always be equal to the information word&nbsp; $\underline{u}$.  
*This is achieved if the beginning of the generator matrix $\boldsymbol {\rm G}$ is a unit matrix $\boldsymbol{\rm I}_{k}$.  
+
*This is achieved if the beginning of the generator matrix&nbsp; $\boldsymbol {\rm G}$&nbsp; is an identity matrix&nbsp; $\boldsymbol{\rm I}_{k}$.  
*This is true for "Code 1" (with dimension $k = 3$), "Code 2" (with $k = 1$) and "code 3" (with $k = 2$).
+
*This is true for&nbsp; "code 1"&nbsp; $($with dimension&nbsp; $k = 3)$,&nbsp; "code 2" $($with&nbsp; $k = 1)$&nbsp; and&nbsp; "code 3"&nbsp; $($with $k = 2)$.
*The generator matrix of "Code 2", however, is not explicitly stated. It is:
+
*The generator matrix of&nbsp; "code 2",&nbsp; however,&nbsp; is not explicitly stated.&nbsp; It is:
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  
  
  
'''(4)'''&nbsp; <u>Statement 1</u> is correct:
+
'''(4)'''&nbsp; <u>Statement 1</u>&nbsp; is correct:
*Dual codes are those where the parity-check matrix $\boldsymbol {\rm H}$ of one code is equal to the generator matrix $\boldsymbol {\rm G}$ of the other code.  
+
*Dual codes are those where the parity-check matrix&nbsp; $\boldsymbol {\rm H}$&nbsp; of one code is equal to the generator matrix&nbsp; $\boldsymbol {\rm G}$&nbsp; of the other code.  
*For example, this is true for "Code 1" and "Code 2."  
+
*For example,&nbsp; this is true for "code 1" and "code 2."  
*For the SPC (4, 3) holds:
+
*For the&nbsp; $\text{SPC (4, 3)}$&nbsp; holds:
  
 
:$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
 
:$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
  
:and for the repetition code RC (4, 1):
+
:and for the repetition code&nbsp; $\text{RC (4, 1)}$:
  
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  
*Statement 2 is certainly wrong, already for dimensional reasons: The generator matrix $\boldsymbol {\rm G}$ of "Code 3" is a $2×4$ matrix and the parity-check matrix $\boldsymbol {\rm H}$ of "Code 2" is a $3×4$ matrix.
+
*Statement 2 is certainly wrong,&nbsp; already for dimensional reasons:&nbsp; The&nbsp; $\boldsymbol {\rm G}$&nbsp; matrix  of&nbsp; "code 3"&nbsp; is a&nbsp; $2×4$&nbsp; matrix and the&nbsp; $\boldsymbol {\rm H}$&nbsp; matrix of&nbsp; "code 2"&nbsp; is a $3×4$&nbsp; matrix.
  
*"Code 3" and "Code 4" also do not satisfy the conditions of dual codes. The parity-check equations of
+
*"Code 3"&nbsp; and&nbsp; "code 4"&nbsp; also do not satisfy the conditions of dual codes.&nbsp; The parity-check equations of
  
 
:$${\rm Code}\hspace{0.15cm}3 = \{ (0, 0, 0, 0) \hspace{0.05cm},\hspace{0.1cm} (0, 1, 1, 0) \hspace{0.05cm},\hspace{0.1cm}(1, 0, 0, 1) \hspace{0.05cm},\hspace{0.1cm}(1, 1, 1, 1) \}$$
 
:$${\rm Code}\hspace{0.15cm}3 = \{ (0, 0, 0, 0) \hspace{0.05cm},\hspace{0.1cm} (0, 1, 1, 0) \hspace{0.05cm},\hspace{0.1cm}(1, 0, 0, 1) \hspace{0.05cm},\hspace{0.1cm}(1, 1, 1, 1) \}$$
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:$$x_1 \oplus x_4 = 0\hspace{0.05cm},\hspace{0.2cm}x_2 \oplus x_3 = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
 
:$$x_1 \oplus x_4 = 0\hspace{0.05cm},\hspace{0.2cm}x_2 \oplus x_3 = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
  
:In contrast, the generator matrix of "Code 4" is given as follows:
+
:In contrast,&nbsp; the generator matrix of&nbsp; "code 4"&nbsp; is given as follows:
  
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &0 &0\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
 
:$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &0 &0\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$

Revision as of 14:30, 10 July 2022

Block codes of length  $n = 4$ Korrektur code

We consider block codes of length  $n = 4$:

$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
  • the  repetition code  $\text{RC (4, 1)}$   ⇒   "code 2"   with the parity-check matrix
$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
  • the  $\text{(4, 2)}$  block code   ⇒   "code 3"   with the generator matrix
$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
  • the  $\text{(4, 2)}$  block code   ⇒   "code 4"   with the generator matrix
$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &0 &0\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
  • another "code 5"   with code size  $|\hspace{0.05cm}C\hspace{0.05cm}| = 6$.


The individual codes are explicitly indicated in the graphic.  The questions for these tasks are about the terms



Hints :



Questions

1

How can  "code 5"  be described?

There are exactly two zeros in each code word.
There are exactly two ones in each code word.
After each  "$0$",  the symbols  "$0$"  and  "$1$"  are equally likely.

2

Which of the following block codes are linear?

code 1,
code 2,
code 3,
code 4,
code 5.

3

Which of the following block codes are systematic?

code 1,
code 2,
code 3,
code 4,
code 5.

4

Which code pairs are dual to each other?

code 1 and code 2,
code 2 and code 3,
code 3 and code 4.


Solution

(1)  Statements 1 and 2  are correct:

  • That is why there are  "$\rm 4 \ over \ 2 = 6$"  code words.
  • Statement 3 is false.  For example,  if the first bit is  "$0$",  there is one code word starting  "$00$"  and two code words starting  "$01$".


(2)  Statements 1 to 4  are correct:

  • All codes that can be described by a generator matrix  $\boldsymbol {\rm G}$  and/or a parity-check matrix  $\boldsymbol {\rm H}$  are linear.
  • In contrast,  "code 5" does not satisfy any of the conditions required for linear codes.  For example
  • is missing the all-zero word,
  • the code size  $|\mathcal{C}|$  is not a power of two,
  • gives  $(0, 1, 0, 1) \oplus (1, 0, 1, 0) = (1, 1, 1, 1)$  no valid code word.


(3)  Statements 1 to 3  are correct:

  • In a systematic code,  the first  $k$  bits of each code word  $\underline{x}$  must always be equal to the information word  $\underline{u}$.
  • This is achieved if the beginning of the generator matrix  $\boldsymbol {\rm G}$  is an identity matrix  $\boldsymbol{\rm I}_{k}$.
  • This is true for  "code 1"  $($with dimension  $k = 3)$,  "code 2" $($with  $k = 1)$  and  "code 3"  $($with $k = 2)$.
  • The generator matrix of  "code 2",  however,  is not explicitly stated.  It is:
$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$


(4)  Statement 1  is correct:

  • Dual codes are those where the parity-check matrix  $\boldsymbol {\rm H}$  of one code is equal to the generator matrix  $\boldsymbol {\rm G}$  of the other code.
  • For example,  this is true for "code 1" and "code 2."
  • For the  $\text{SPC (4, 3)}$  holds:
$${ \boldsymbol{\rm H}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm G}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},$$
and for the repetition code  $\text{RC (4, 1)}$:
$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}, \hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &0 &1\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • Statement 2 is certainly wrong,  already for dimensional reasons:  The  $\boldsymbol {\rm G}$  matrix of  "code 3"  is a  $2×4$  matrix and the  $\boldsymbol {\rm H}$  matrix of  "code 2"  is a $3×4$  matrix.
  • "Code 3"  and  "code 4"  also do not satisfy the conditions of dual codes.  The parity-check equations of
$${\rm Code}\hspace{0.15cm}3 = \{ (0, 0, 0, 0) \hspace{0.05cm},\hspace{0.1cm} (0, 1, 1, 0) \hspace{0.05cm},\hspace{0.1cm}(1, 0, 0, 1) \hspace{0.05cm},\hspace{0.1cm}(1, 1, 1, 1) \}$$
are as follows:
$$x_1 \oplus x_4 = 0\hspace{0.05cm},\hspace{0.2cm}x_2 \oplus x_3 = 0 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} { \boldsymbol{\rm H}} = \begin{pmatrix} 1 &0 &0 &1\\ 0 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$
In contrast,  the generator matrix of  "code 4"  is given as follows:
$${ \boldsymbol{\rm G}} = \begin{pmatrix} 1 &1 &0 &0\\ 0 &0 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$