Aufgaben:Exercise 2.07Z: Reed-Solomon Code (15, 5, 11) to Base 16

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$\rm GF(2^4)$  in power, polynomial and coefficient notation

The task at hand is similar to the one in  "Exercise 2.7". However, we now refer here to the Galois field  $\rm GF(2^4)$, whose elements are given opposite both in powers and polynomial representations and by the coefficient vector. Further, in  $\rm GF(2^4)$:

$$\alpha^{16} = \alpha^{1}\hspace{0.05cm},\hspace{0.2cm} \alpha^{17} = \alpha^{2}\hspace{0.05cm},\hspace{0.2cm} \alpha^{18} = \alpha^{3}\hspace{0.05cm},\hspace{0.05cm}\text{...} $$

To encode the information block of length  $k = 5$,

$$\underline{u} = (u_0,\ u_1,\ u_2,\ u_3,\ u_4)\hspace{0.05cm},$$

we form the polynomial

$$u(x) = u_0 + u_1 \cdot x + u_2 \cdot x^2 + u_3 \cdot x^3 + u_4 \cdot x^4 $$

with  $u_0, \hspace{0.05cm}\text{...} \hspace{0.1cm} , u_4 ∈ \rm GF(2^4)$.

The  $n = 15$  codewords are then obtained by substituting into  $u(x)$  the elements of  $\rm GF(2^4) \ \backslash \ \{0\}$ :

$$c_0 = u(\alpha^{0})\hspace{0.05cm},\hspace{0.2cm} c_1 = u(\alpha^{1})\hspace{0.05cm}, \hspace{0.2cm} c_2 = u(\alpha^{2})\hspace{0.05cm}, \hspace{0.15cm} ... \hspace{0.15cm},\hspace{0.20cm} c_{14} = u(\alpha^{14})\hspace{0.05cm}.$$

Hints:

• This exercise belongs to the chapter  "Definition and Properties of Reed-Solomon Codes".

Questions

1

How many symbol errors can be corrected?

$t \ = \ $

2

What is the polynomial  $u(x)$  for  $\underline{u} = (\alpha^3, \, 0, \, 0, \, 1, \, \alpha^{10})$?

	$u(x) = \alpha^3 + x + \alpha^{10} \cdot x^2$,
	$u(x) = \alpha^3 + x^3 + \alpha^{10} \cdot x^4$,
	$u(x) = 1 + x + x^2 + x^3 + x^4$.

3

What is the symbol  $c_0$  of the associated codeword  $\underline{c}$?

	$c_0 = 1$,
	$c_0 = \alpha^5$,
	$c_0 = \alpha^{11}$,
	$c_0 = \alpha^{14}$.

4

What is the symbol  $c_1$  of the associated codeword  $\underline{c}$?

	$c_1 = 1$,
	$c_1 = \alpha^5$,
	$c_1 = \alpha^{11}$,
	$c_1 = \alpha^{14}$.

5

What is the symbol  $c_{13}$  of the associated codeword  $\underline{c}$?

	$c_{13} = 1$,
	$c_{13} = \alpha^5$,
	$c_{13} = \alpha^{11}$,
	$c_{13} = \alpha^{14}$.

6

What is the last symbol of the associated codeword  $\underline{c}$?

	$c_{15} = 1$,
	$c_{14} = 1$,
	$c_{14} = \alpha^7$,
	$c_{14} = \alpha^{14}$.

7

Which statements are true?

	The code symbol "$0$" is not possible for  $\rm RSC \, (15, \, 5, \, 11)_{16}$ .
	A code symbol "$0$" results only for  $\underline{u} = (0, \, 0, \, 0, \, 0)$.
	Also for  $\underline{u} ≠ (0, \, 0, \, 0, \, 0)$ there can be code symbols "$0$".

Solution

Solution

(1)  From $n = 15$ and $k = 5$ follows:

$$d_{\rm min} = n - k +1 = 15 - 5 + 1 = 11 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} t = \frac{d_{\rm min}-1}{2}\hspace{0.15cm}\underline {=5}\hspace{0.05cm}.$$

(2)  In general, for the sought polynomial $u(x)$ with $k = 5$:

$$u(x) = \sum_{i = 0}^{k-1} u_i \cdot x^{i}= u_0 + u_1 \cdot x + u_2 \cdot x^2 + u_3 \cdot x^3 + u_4 \cdot x^4 \hspace{0.05cm}.$$

For $u_0 = \alpha^3, \ u_1 = u_2 = 0, \ u_3 = 1$ and $u_4 = \alpha^{10}$ the proposed solution 2 turns out to be correct.

(3)  It holds $c_0 = u(\alpha^0) = u(1)$:

$$c_0 = \alpha^{3} + 1 \cdot 1^3 + \alpha^{10} \cdot 1^{4} = (1000) + (0001) + (0111) = (1110)= \alpha^{11} \hspace{0.05cm}.$$

The correct solution is therefore proposed solution 3.

(4)  From $c_1 = u(\alpha)$ one obtains the proposed solution 4.

$$c_1 = u(\alpha^{1}) =\alpha^{3} +1 \cdot \alpha^{3} + \alpha^{10} \cdot \alpha^{4} = \alpha^{14} \hspace{0.05cm}.$$

(5)  For the penultimate symbol $c_{13} = u(\alpha^{13})$ holds:

$$c_{13} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u(\alpha^{13}) =\alpha^{3} + 1 \cdot \alpha^{13 \hspace{0.05cm}\cdot \hspace{0.05cm}3} + \alpha^{10} \cdot \alpha^{13 \hspace{0.05cm}\cdot \hspace{0.05cm}4} = \alpha^{3} + \alpha^{39}+ \alpha^{62} =\alpha^{3} + \alpha^{15 \hspace{0.05cm}\cdot \hspace{0.05cm}2} \cdot \alpha^{9}+ \alpha^{15 \hspace{0.05cm}\cdot \hspace{0.05cm}4} \cdot \alpha^{2} = \alpha^{3} + \alpha^{9} + \alpha^{2}$$

$$\Rightarrow \hspace{0.3cm}c_{13} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1000) + (1010) + (0100) = (0110) = \alpha^{5} \hspace{0.05cm}.$$

The correct solution is therefore proposed solution 2.

(6)  The last code symbol is $c_{14} = u(\alpha^{14})$:

$$c_{14} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u(\alpha^{14}) =\alpha^{3} + 1 \cdot \alpha^{14 \hspace{0.05cm}\cdot \hspace{0.05cm}3} + \alpha^{10} \cdot \alpha^{14 \hspace{0.05cm}\cdot \hspace{0.05cm}4} = \alpha^{3} + \alpha^{42}+ \alpha^{66} =\alpha^{3} + \alpha^{15 \hspace{0.05cm}\cdot \hspace{0.05cm}2} \cdot \alpha^{12}+ \alpha^{15 \hspace{0.05cm}\cdot \hspace{0.05cm}4} \cdot \alpha^{6} = \alpha^{3} + \alpha^{12} + \alpha^{6} =$$

$$\Rightarrow \hspace{0.3cm}c_{14} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}(1000) + (1111) + (1100) = (1011) = \alpha^{7} \hspace{0.05cm}.$$

The correct solution is therefore proposed solution 3.

(7)  The code symbol "$0$" occurs just as often as all other symbols "$\alpha^i$"  ⇒  proposed solution 3.