Exercise 2.08Z: Addition and Multiplication in GF(2 power 3)

$\rm GF(2^3)$: Incomplete addition and multiplication tables

The graph shows the addition and multiplication table for the finite field $\rm GF(2^3)$. The tables are not complete. Some fields $($highlighted in color$)$ should be completed.

The elements are given both

in the exponent representation $($with red lettering, left and above$)$ and

in the coefficient representation (gray lettering, right and below).

From this assignment one can already recognize the underlying irreducible polynomial $p(\alpha)$.

Additions $($and subtractions$)$ are best done in the coefficient representation $($or with polynomials firmly linked to it$)$.

For multiplications, however, the exponential representation is more convenient.

Hints:

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

However, reference is also made to the chapter "Extension Field".

Questions

Solution

(1) Adding any element of an extension field based on $\rm GF(2)$ to itself always yields $0$, as can be easily seen from the coefficient representation, for example:

$$\alpha^3 + \alpha^3 = (011) + (011) = (000) = 0\hspace{0.05cm}.$$

That is: $\rm A$ stands for the zero element ⇒ Solution 1.

(2) $\rm B$ is the result of adding $\alpha^5$ and $\alpha^6$ ⇒ Solution 3:

$$\alpha^5 + \alpha^6 = (111) + (101) = (010) = \alpha^1 \hspace{0.05cm}.$$

One could have found this result more simply, since in each row and column each element occurs exactly once.

After $\rm A = 0$ is fixed, exactly only the element $\alpha^1$ is missing in the last row and the last column.

(3) $\rm C$ is the result of the sum $\alpha^1 +\alpha^2$ ⇒ Solution 3:

$$\alpha^1 + \alpha^2 = (010) + (100) = (110) = \alpha^4 \hspace{0.05cm}.$$

(4) $\rm D$ is the result of $\alpha^3$ and $\alpha^5$ ⇒ Solution 1:

$$\alpha^3 + \alpha^5 = (011) + (111) = (100) = \alpha^2\hspace{0.05cm}.$$

(5) All proposed solutions are correct, as can be seen from row 2 (multiplication with the "identity element"):

$\rm GF(2^3)$: Complete addition and multiplication tables

The complete tables for addition and multiplication are shown opposite.

Because of the validity of $\alpha^i \cdot \alpha^j = \alpha^{(i+j)\hspace{0.1cm} {\rm mod}\hspace{0.1cm} 7} $, multiplication yields a symmetry that could be used to solve.

(6) Correct here is the proposed solution 3:

All polynomials are indeed irreducible. However, one needs a degree-3 polynomial for $\rm GF(2^3)$.

The third proposed solution results from the relation

$$\alpha^3 = \alpha + 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}p(\alpha) = \alpha^3 + \alpha + 1 = 0 \hspace{0.05cm}.$$

	$p(\alpha) = \alpha^2 + \alpha + 1$,
	$p(\alpha) = \alpha^3 + \alpha^2 + 1$,
	$p(\alpha) = \alpha^3 + \alpha + 1$.

	$\rm C = \alpha^2$,
	$\rm C = \alpha^3$,
	$\rm C = \alpha^4$.

	$\rm D = \alpha^2$,
	$\rm D = \alpha^3$,
	$\rm D = \alpha^4$.

	$\rm E = \alpha^5$,
	$\rm F = \alpha^1$,
	$\rm G = \alpha^6$.