Aufgaben:Exercise 1.2Z: Three-dimensional Representation of Codes

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Space $\rm GF(2^3)$ and
Code of length $n = 3$

Codes for error detection or error correction can be represented very clearly in  $n$-dimensional space. We restrict ourselves here to binary codes of length  $n = 3$:

$$\underline{x} = (x_{1}, x_{2}, x_{3}) \hspace{0.1cm} \in \hspace{0.1cm}{\rm GF}(2^3) \hspace{0.05cm},\hspace{0.5cm} x_i = \{0, 1 \}\hspace{0.05cm},\hspace{0.2cm} i = 1, 2, 3\hspace{0.05cm}.$$

In general, for block coding:

• The information word  $\underline{u} = (u_{1}, u_{2}, \ \text{...} , \ u_{k})$  is uniquely transformed into the codeword  $\underline{x} =(x_{1}, x_{2}, \ \text{...} , \ , x_{n})$  .
• The code rate is  $R = k/n$.
• The Hamming distance  $d_{\rm H}(x, \hspace{0.05cm}x\hspace{0.05cm}')$  between two codewords  $x ∈ \mathcal{C}$  and  $x\hspace{0.05cm}' ∈ \mathcal{C}$  indicates the number of bit positions in which  $x$  and  $x\hspace{0.05cm}'$  differ.
• The minimum distance  $d_{\rm min} = {\rm min}\big[d_{\rm H}(x, \hspace{0.05cm}x\hspace{0.05cm}')\big]$  is a measure of the correctability of a code.
• It can  $e =d_{\rm min} - 1$  detect errors and  $t =(d_{\rm min} - 1)/2$  correct errors.
• The last statement, however, is valid only for odd  $d_{\rm min}$.

Hints:

• This exercise belongs to the chapter  Obective of Channel Coding
• In addition, some simple questions about the chapter  Examples of binary block codes  are anticipated.

Questions

1

Which statements hold if all points in  $\rm GF(2^3)$  are occupied?

	The assignment  $\underline{u} = (u_{1}, u_{2}, u_{3})$   →   $\underline{x} = (x_{1}, x_{2},x_{3})$ holds.
	The identity  $\underline{x} = \underline{u}$ holds.
	The code rate is  $R = 1$.
	The minimum distance between two codewords is  $d_{\rm min} = 2$.

2

Which statements are true for a (3, 2, 2)-block code?

	code  $\mathcal{C}_{1} = \{(0, 0, 0), (0, 1, 1), (1, 0, 1), (1, 1, 0)\}$  is possible.
	Code  $\mathcal{C}_{2} = \{(0, 0, 1), (0, 1, 0), (1, 0, 0), (1, 1, 1)\}$  is possible.
	Code  $\mathcal{C}_{3} = \{(0, 0, 0), (0, 1, 1), (1, 0, 0), (1, 1, 1)\}$  is possible.

3

What properties does the code defined in subtask (2) show  $\mathcal{C}_{1}$?

	A bit error can be detected.
	A bit error can be corrected.

4

What properties does the code show  $\mathcal{C}_{4}= \{(0, 0, 0), (1, 1, 1)\}$?

	The code rate is  $R = 1/4$.
	The code rate is  $R = 1/3$.
	A bit error can be detected.
	A bit error can be corrected.

Solution

Solution

(1)  Correct statements 1 and 3:

• In this assignment, $k = 3$ information bits are mapped to $n = 3$ code bits   ⇒   $R = k/n = 1$.
• The statement $\underline{x} = \underline{u} $ would only hold in the case of systematic coding.
• For example, in principle, $(0, 0, 0)$   →   $(0, 1, 1)$ would also be possible.
• The last statement is certainly false: from the graph one can see the minimum distance $d_{\rm min} = 1$.

Two (3, 2, 2)-block codes

(2)  Correct statements 1 and 3:

• $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ actually describe codes with rate $R = 2/3$ and minimum distance $d_{\rm min} = 2$.
• In the graph, the green dots mark the code $\mathcal{C}_{1}$ and the blue dots mark the code $\mathcal{C}_{2}$.
• For the code $\mathcal{C}_{3}$ - also with rate $R = 2/3$ - the minimum distance between two codewords is $d_{\rm min} = 1$, for example between $(0, 0, 0)$ and $(1, 0, 0)$ or between $(0, 1, 1)$ and $(1, 1, 1)$.

(3)  Correct is only statement 1:

• Only a bit error can be detected with the minimum distance $d_{\rm min} = 2$.
• In the upper graph, the green dots indicate allowed codewords of $\mathcal{C}_{1}$. If a blue dot is received, this indicates a transmission error.
• On the other hand, error correction is not possible with $d_{\rm min} = 2$.
• The code $\mathcal{C}_{1}$ corresponds to the Single Parity-check Code (3, 2, 2).

(3, 1, 3) block code

(4)  Correct answers 2, 3, and 4:

• $C_{4}$ describes the (3, 1, 3) repetition code.
• In this code, two of the total eight possible points are occupied, from which one could incorrectly conclude the code rate $R = 1/4$. However, the code rate is calculated according to $R = k/n = 1/3$.
• From the lower diagram one recognizes that because of $d_{\rm min} = 3$ now also a bit error can be corrected.
• During decoding, all light green points (with black outline) are transferred to the green point $(0, 0, 0)$ and all light blue points are transferred to the blue point $(1, 1, 1)$.
• Up to two bit errors can be detected at the same time (one of course).