Exercise 1.09: Extended Hamming Code

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$\text{(7, 4)}$ Hamming code (yellow background)
and  $\text{(8, 4)}$ extension (green).

Two codes are to be compared, whose code tables are given on the right.

  • The first four bits of each code word  $\underline{x}$  are equal to the respective information word  $\underline{u}$  (black font).
  • Then follow  $m = n- k$  parity bit (red font).


The systematic  $\text{(7, 4)}$-Hamming code has already been discussed in  Exercise 1.6  and  Exercise 1.7 . The parity-check matrix and generator matrix of this code are given as follows:

$${ \boldsymbol{\rm H}}_1 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix}\hspace{0.05cm},$$
$${ \boldsymbol{\rm G}}_1 = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1\\ 0 &1 &0 &0 &1 &1 &0\\ 0 &0 &1 &0 &0 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 \end{pmatrix}\hspace{0.05cm}.$$

In the further course of the exercise this (yellow highlighted) code is called  $\mathcal{C}_{1}$ .


The right column in the above table specifies a block code with parameters  $n = 8$  and  $k = 4$ , usually referred to in the literature as the "Extended Hamming Code". We refer to this code (highlighted in green) in the following  $\mathcal{C}_{2}$  and denote its parity-check matrix by  ${ \boldsymbol{\rm H}}_{2}$  and the corresponding generator matrix by  ${ \boldsymbol{\rm G}}_{2}$ .

The questions for this exercise are related to




Hints:



Questions

1

Specify the code rates of  $\mathcal{C}_{1}$  and  $\mathcal{C}_{2}$ .

$\mathcal{C}_{1}\text{:}\hspace{0.4cm}R \ = \ $

$\mathcal{C}_{2}\text{:}\hspace{0.4cm}R \ = \ $

2

Give the minimum distances of  $\mathcal{C}_{1}$  and  $\mathcal{C}_{2}$  .

$\mathcal{C}_{1}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $

$\mathcal{C}_{2}\text{:}\hspace{0.4cm}d_{\rm min} \ = \ $

3

What is the format of the parity-check matrix  $\boldsymbol{\rm H}_{2}$  of  $\mathcal{C}_{2}$?

${\rm Spaltenzahl} \ = \ $

${\rm Zeilenzahl} \ = \ $

4

Derive the equation for the code bit  $x_ {8} (= p_{4})$  from the code table. Which specification is correct?

$x_{8} = 0.$
$x_{8} = x_{1}⊕x_{2}⊕x_{4}⊕x_{5}.$
$x_{8} = x_{1}⊕x_{2}⊕x_{3}⊕x_{4}⊕x_{5}⊕x_{6}⊕x_{7}.$

5

Which statements are true for  ${ \boldsymbol{\rm H}}_{2}$? Hint:  Correct are 3 out of 4 answers.

Row 1 reads:   $1 1 0 1 1 0 0 0$.
Row 2 reads:   $0 1 1 1 0 1 0 0$.
Row 3 reads:   $0 0 0 0 1 1 1 1$.
Row 4 reads:   $1 1 1 1 1 1 1 1$.

6

Which transformation is allowed for the last row of  ${ \boldsymbol{\rm H}}_{2}$ ?

$1 1 1 1 1 1 1 1 → 0 0 0 0 0 0 0 0,$
$1 1 1 1 1 1 1 1 → 1 1 1 0 0 0 0 1,$
$1 1 1 1 1 1 1 1 → 0 0 1 0 1 0 0 0.$

7

Give the corresponding generator matrix  ${ \boldsymbol{\rm G}}_{2}$ . Which statements are true?

${ \boldsymbol{\rm G}}_{2}$  has same format as matrix  ${ \boldsymbol{\rm G}}_{1}$ of a  $\text{(7, 4)}$ code.
${ \boldsymbol{\rm G}}_{2}$  starts like ${ \boldsymbol{\rm G}}_{1}$  with a diagonal matrix  ${ \boldsymbol{\rm I}}_{4}$ .
${ \boldsymbol{\rm G}}_{2}$  in the considered example has the same format as  ${ \boldsymbol{\rm H}}_{2}$ .


Solution

(1)  The corresponding equation for the code rate in both cases is $R = k/n\text{:}$

  • $\mathcal{C}_{1} \text{:} \ n = 7, k = 4\ ⇒ \ R = 4/7 \underline {= 0.571},$
  • $\mathcal{C}_{2} \text{:} \ n = 8, k = 4 \ ⇒ \ R = 4/8 \underline { =0.5}.$


(2)  The minimum distance of the (7, 4, 3)-Hamming code $\mathcal{C}_{1}$ is $d_{\rm min} \underline{= 3}$, which can be read from the naming alone.

  • From the table on the information page, it can be seen that for the extended Hamming code $d_{\rm min} \underline{= 4}$ holds.
  • $\mathcal{C}_{2}$ is therefore also called a (8, 4, 4) block code in the literature.


(3)  The parity-check matrix ${ \boldsymbol{\rm H}}$ generally consists of $n$ columns and $m = n - k$ rows, where $m$ indicates the number of parity-check equations.

  • For the (7, 4, 3)-Hamming code, ${ \boldsymbol{\rm H}}$ is a 3 × 7 matrix.
  • For the extended Hamming code   ⇒   code $\mathcal{C}_{2}$, on the other hand, $\underline{n = 8}$ (column number) and $\underline{m = 4}$ (row number) holds.


(4)  From the code table on the information page you can see that only answer 3 is correct.

  • The parity bit $p_{4}$ is to be determined in such a way that the modulo 2 sum over all bits of the code word results in the value $0$.


(5)  It should first be noted that the specification of the parity-check matrix is never unambiguous, if only because the order of the parity-check equations is interchangeable.

  • However, considering that only one of the given rows is wrong, ${ \boldsymbol{\rm H}}_{2}$ is uniquely determined:
$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &1 &1 &1 &1 &1 \end{pmatrix} \hspace{0.05cm}.$$
  • Correct are therefore the statements 1, 2 and 4. The rows of this parity-check matrix represent the four parity-check equations in this order:
$$ x_1\oplus x_2 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$
$$x_2 \oplus x_3 \oplus x_4 \oplus x_6 = 0 \hspace{0.05cm},$$
$$ x_1 \oplus x_3 \oplus x_4 \oplus x_7 = 0 \hspace{0.05cm},$$
$$ x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0 \hspace{0.05cm}.$$


(6)  Correct is the answer 2:

  • This result is obtained by replacing the last row with the modulo 2 sum over all four rows, which is allowed.
  • Proposition 1 does not represent a parity-check equation.
  • Proposal 3 represents the parity-check equation $x_{3}⊕x_{5} = 0$, which also does not correspond to the facts.


According to the correct solution suggestion 2, on the other hand, the parity-check equation becomes

$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 \oplus x_6 \oplus x_7 \oplus x_8 = 0$$

replaced by the following new parity-check equation:

$$x_1 \oplus x_2 \oplus x_3 \oplus x_8 = 0 \hspace{0.05cm}.$$

The modified parity-check matrix is now:

$${ \boldsymbol{\rm H}}_2 = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0 &0\\ 0 &1 &1 &1 &0 &1 &0 &0\\ 1 &0 &1 &1 &0 &0 &1 &0\\ 1 &1 &1 &0 &0 &0 &0 &1 \end{pmatrix} \hspace{0.05cm}.$$


(7)  After this matrix manipulation, ${ \boldsymbol{\rm H}}_{2}$ is in the form typical for systematic codes:

$${ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_m \right)\hspace{0.3cm} \Rightarrow\hspace{0.3cm} m = 4 {\rm :}\hspace{0.3cm}{ \boldsymbol{\rm H}}_2 =\left({ \boldsymbol{\rm P}}^{\rm T} \: ; \: { \boldsymbol{\rm I}}_4 \right) \hspace{0.05cm}.$$

Thus, the generator matrix is:

$${ \boldsymbol{\rm G_{2}}} =\left({ \boldsymbol{\rm I}}_4 \: ; \: { \boldsymbol{\rm P}}\right) = \begin{pmatrix} 1 &0 &0 &0 &1 &0 &1 &1\\ 0 &1 &0 &0 &1 &1 &0 &1\\ 0 &0 &1 &0 &0 &1 &1 &1\\ 0 &0 &0 &1 &1 &1 &1 &0 \end{pmatrix} \hspace{0.05cm}.$$

So the statements 2 and 3 are correct:

  • ${ \boldsymbol{\rm G}}_{2}$ starts like ${ \boldsymbol{\rm G}}_{1}$ (see specification sheet) with a diagonal matrix ${ \boldsymbol{\rm I}}_{4}$ , but unlike ${ \boldsymbol{\rm G}}_{1}$ now has 8 columns.
  • In the present case $n = 8, k = 4 \ ⇒ \ m = 4$ both ${ \boldsymbol{\rm G}}_{2}$ and ${ \boldsymbol{\rm H}}_{2}$ are 4×8 matrices respectively.