Exercise 1.08Z: Equivalent Codes

Four $(6, 3)$ block codes

In the graph, the mappings $\underline{u} \rightarrow \underline{x}$ for different codes are given, each characterized below by the generator matrix $\boldsymbol{\rm G}$ and the parity-check matrix $\boldsymbol{\rm H}$ respectively:

${\boldsymbol{\rm Code \ A}}$:

$${ \boldsymbol{\rm G}}_{\rm A} = \begin{pmatrix} 1 &0 &0 &1 &1 &0\\ 0 &1 &0 &1 &0 &1\\ 0 &0 &1 &0 &1 &1 \end{pmatrix} \hspace{0.05cm},\hspace{0.5cm}{ \boldsymbol{\rm H}}_{\rm A} = \begin{pmatrix} 1 &1 &0 &1 &0 &0\\ 1 &0 &1 &0 &1 &0\\ 0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

${\boldsymbol{\rm Code \ B}}$:

$${ \boldsymbol{\rm G}}_{\rm B} = \begin{pmatrix} 0 &0 &1 &0 &1 &1\\ 1 &0 &0 &1 &1 &0\\ 0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm},\hspace{0.5cm} { \boldsymbol{\rm H}}_{\rm B} = \begin{pmatrix} 1 &0 &1 &0 &1 &0\\ 1 &1 &0 &1 &0 &0\\ 0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

${\boldsymbol{\rm Code \ C}}$:

$${ \boldsymbol{\rm G}}_{\rm C} = \begin{pmatrix} 1 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1\\ 0 &0 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm},\hspace{0.5cm}{ \boldsymbol{\rm H}}_{\rm C} = \begin{pmatrix} 1 &0 &1 &1 &0 &0\\ 0 &1 &1 &0 &1 &0\\ 1 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm},$$

${\boldsymbol{\rm Code \ D}}$:

$${ \boldsymbol{\rm G}}_{\rm D} = \begin{pmatrix} 1 &0 &0 &1 &0 &1\\ 0 &1 &0 &1 &0 &0\\ 0 &0 &1 &0 &1 &0 \end{pmatrix} \hspace{0.05cm},\hspace{0.5cm}{ \boldsymbol{\rm H}}_{\rm D} = \begin{pmatrix} 1 &1 &0 &1 &0 &0\\ 0 &0 &1 &0 &1 &0\\ 1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$

This task is to investigate which of these codes or code pairs are

are systematic,
are identical (that is: Different codes have same code words),
are equivalent (that is: Different codes have same code parameters).

Hints :

This exercise belongs to the chapter General Description of Linear Block Codes.
Reference is made in particular to the pages Systematic Codes and Identical Codes.
Note that the specification of a parity-check matrix $\boldsymbol{\rm H}$ is not unique.
If one changes the order of the parity-check equations, this corresponds to a swapping of rows.

Questions

Solution

(1) Correct are the answers 1, 3 and 4:

For a systematic (6, 3) block code, the following must hold:

$$\underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6) = ( u_1, u_2, u_3, p_1, p_2, p_{3}) \hspace{0.05cm}.$$

This condition is satisfied by code A, code C, and code D, but not by code B.

(2) Correct is only answer 1:

Only code A and code B are identical codes. They contain exactly the same code words and differ only by other assignments $\underline{u} \rightarrow \underline{x}$.
As indicated in the sample solution to Exercise 1.8 (3), one gets from the generator matrix ${ \boldsymbol{\rm G}}_{\rm B}$ to the generator matrix ${ \boldsymbol{\rm G}}_{\rm A}$

by swapping/permuting rows alone, or
by replacing a row with the linear combination between that row and another.

(3) Thus, the correct answer is answer 2 alone:

Code A and code B are more than equivalent, namely identical.
Code C and D also differ, for example, by the minimum Hamming distance $d_{\rm min} = 3$ and $d_{\rm min} = 2$, respectively, and are thus also not equivalent.
Code B and code C, on the other hand, show the same properties, for example $d_{\rm min} = 3$ holds for both. However, they contain different codewords.

(4) Correct is answer 3:

The last column of ${ \boldsymbol{\rm G}}_{\rm B}$ gives the first column of ${ \boldsymbol{\rm G}}_{\rm C}$.
The first column of ${ \boldsymbol{\rm G}}_{\rm B}$ gives the second column of ${ \boldsymbol{\rm G}}_{\rm C}$.
The second column of ${ \boldsymbol{\rm G}}_{\rm B}$ gives the third column of ${ \boldsymbol{\rm G}}_{\rm C}$, etc.

(5) All statements are true:

The condition ${ \boldsymbol{\rm H}} \cdot { \boldsymbol{\rm G}}^{\rm T} = \boldsymbol{0}$ holds for all linear codes.

	Code $\rm A$,
	Code $\rm B$,
	Code $\rm C$,
	Code $\rm D$.

	Code $\rm A$,
	Code $\rm B$,
	Code $\rm C$,
	Code $\rm D$.

	Code $\rm A$ and code $\rm B$,
	Code $\rm B$ and code $\rm C$,
	Code $\rm C$ and code $\rm D$.

	By different linear combinations of different rows.
	By cyclic shifting of rows by $1$ down.
	By cyclic shifting of columns by $1$ to the right.?