Theory

The theory associated with Experiment-7 is divided into three parts:

• Preliminaries
• Basics of MDS codes
• Construction of RS codes

1     Preliminaries

In this section, we will define a linearly independent set of vectors over the finite field $\mathbb{F}_{q}$ ($q = p^{m}$, $p$ prime, $m \geq 1$) and explain a method to determine whether a given set is linearly independent or not.

Definition 1. (Linearly Independent set): A set of vectors $S = \{ \mathbf{v}_1 , \mathbf{v}_2, \ldots, \mathbf{v}_l \} \subset \mathbb{F}_{q}^n$ is said to be a linearly independent set if and only if $a_1\mathbf{v}_1 + a_2\mathbf{v}_2+ \ldots + a_l\mathbf{v}_l = \mathbf{0}$ can happen only when $a_1 = a_2 = \ldots = a_l = 0$.

For example, suppose the set $S_1 = \bigl\{\mathbf{v}_1 = \begin{bmatrix} 3 & 4 & 2 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 6 & 1 & 5 \end{bmatrix}$, $\mathbf{v}_3 = \begin{bmatrix} 0 & 1 & 6 \end{bmatrix} \bigr\} \subset \mathbb{F}_{7}^{3}$. The set $S_{1}$ is linearly independent since $a_1\mathbf{v}_1 + a_2\mathbf{v}_2+ a_3\mathbf{v}_3$ gives an all-zero vector only when we choose $a_1=a_2=a_3=0$. Whereas $S_2 = \bigl\{\mathbf{v}_1 = \begin{bmatrix} 2 & 1 & 3 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 5 & 1 & 5 \end{bmatrix}$, $\mathbf{v}_3 = \begin{bmatrix} 1 & 2 & 4 \end{bmatrix} \bigr\} \subset \mathbb{F}_{7}^{3}$ is linearly dependent set since $4\mathbf{v}_1 + \mathbf{v}_2+ \mathbf{v}_3 = \mathbf{0}$.

Now we will provide a well-known procedure to determine the set is linearly independent or dependent.

• Write each vector of the set as a row of a matrix. Let's call this matrix A.
• Perform row operations on matrix A to convert it into its Reduced Row Echelon Form (RREF). Remember that arithmetic operations should be performed in the finite field $\mathbb{F}{q}$.
• Check if any row in the RREF of the matrix $A$ consists of all zeros. The set is linearly dependent if at least one row has all zeros.
• The set is linearly independent if no row contains all zeros.

Now with above-mentioned procedure we will check $S_{2} = \{\mathbf{v}_1 = \begin{bmatrix} 3 & 4 & 2 \end{bmatrix}, \mathbf{v}_2 = \begin{bmatrix} 6 & 0 & 5 \end{bmatrix}$, $\mathbf{v}3 = \begin{bmatrix} 0 & 1 & 6 \end{bmatrix} } \subset \mathbb{F}{7}^{3}$is dependent. While carrying out the row operations, we represent the matrix rows with$R_{i}$(where$i$ is the index of the row). $$\begin{align*} A = \begin{bmatrix} 2 & 1 & 3 \\ 5 & 1 & 5 \\ 1 & 2 & 4 \end{bmatrix} \xRightarrow{R_{1} \leftarrow 4R_{1}} \begin{bmatrix} 1 & 4 & 5 \\ 5 & 1 & 5 \\ 1 & 2 & 4 \end{bmatrix} & \xRightarrow{R_{2} \leftarrow R_{2}+2R_{1}} \begin{bmatrix} 1 & 4 & 5 \\ 0 & 2 & 1 \\ 1 & 2 & 4 \end{bmatrix} \\ & \xRightarrow{R_{3} \leftarrow R_{3}+6R_{1}} \begin{bmatrix} 1 & 4 & 5 \\ 0 & 2 & 1 \\ 0 & 5 & 6 \end{bmatrix} \\ & \xRightarrow{R_{2} \leftarrow 4R_{2}} \begin{bmatrix} 1 & 4 & 5 \\ 0 & 1 & 4 \\ 0 & 5 & 6 \end{bmatrix} \xRightarrow{R_{3} \leftarrow R_{3}+2R_{2}} \begin{bmatrix} 1 & 4 & 5 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix}. \end{align*}$$

The last row of the RREF form of the matrix $A$ has all zeros. Therefore, the set is linearly dependent, and the matrix's rank $A$ is 2.

2     Basics of MDS Codes

In this section, we define MDS codes and list a few important properties of these codes. MDS codes are a class of error-correcting codes that can correct a maximum number of errors for the given code rate (k/n). For any given code $\mathcal{C}(n,k)$, the upper bound on the minimum distance $d_{min}$ of the code is called a Singleton bound, which is as follows: $$\begin{equation} d_{min} \leq n-k+1. \end{equation}$$ Note that error-correcting capability depends upon the $d_{min}$, since the number of errors that the code can correct is equal to $\lfloor \frac{d_{min} -1}{2} \rfloor$ . So, the formal definition of MDS code is defined as follows:

Definition 2(MDS Code). A code $\mathcal{C}(n,k)$ is called MDS code if the minimum distance achieves the Singleton bound (1), i.e., $d_{min}=n-k+1$.

Now, We will describe a few properties of MDS codes without proof.

• A code $\mathcal{C}(n,k)$ is MDS, if and only if every set of $n-k$ columns of its parity check matrix $H$ are linearly independent.
• Dual code of an MDS code is MDS.
• Every set of $k$ columns of its generator matrix $G$ are linearly independent.

Below, we will provide generator matrices for a code with the parameters $(n=5, k=3)$, one of which generates an MDS code while the other one does not.

Example 1. Let $$A = \begin{bmatrix} 1 & 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 6 & 3 \end{bmatrix},$$ where the elements of the matrix $A$ belong to the finite field $\mathbb{F}_{7}$. We can easily verify that columns 1,2 and 3 are linearly independent, and columns 1,3 and 4 are linearly dependent. Therefore, $A$ is the generator matrix of the code $\mathcal{C}(5,3)$ but it is not MDS.

However, the matrix $$G = \begin{bmatrix} 1 & 4 & 2 & 2 & 1 \\ 1 & 1 & 6 & 1 & 6 \\ 1 & 2 & 4 & 4 & 1 \end{bmatrix}$$ generates the $\mathcal{C}(5,3)$ MDS code. Since every set of three columns of $G$ is linearly independent.

3     Construction of RS Codes

In this section, we will discuss the encoding procedure of RS codes.

Definition 3(Encoding of RS Codes): Let $\alpha$ be a primitive element in $\mathbb{F}_{p^{m}}$ and let $n \leq p^{m}-1$. Let $\mathbf{u} = \begin{bmatrix} u_0 & u_1 & \ldots & u_{k-1} \end{bmatrix}$ $\in \mathbb{F}_{p^{m}}^{k}$ be a message vector and let $\mathbf{u}(X) = u_{0} + u_{1}X+u_{2}X^{2}+ \ldots + u_{1}X^{k-1} \in \mathbb{F}_{p^{m}}[X]$ be its associated polynomial. Then the encoding is defined by the mapping $\sigma : \mathbf{u}(X) \mapsto \mathbf{v}$ by
$$\begin{equation} \mathbf{v} = \begin{bmatrix} v_0 & v_1 & \ldots & v_{n-1} \end{bmatrix} \triangleq \sigma (u(X)) = \begin{bmatrix} \mathbf{u}(\alpha_{1}) & \mathbf{u}(\alpha_{2}) & \ldots & \mathbf{u}(\alpha_{n}) \end{bmatrix}. \end{equation}$$ Here, $\alpha_1, \alpha_2, \ldots , \alpha_{n}$ are distinct elements chosen from $\mathbb{F}_{p^{m}}$ known as evaluation set. That is, $\sigma (u(X))$ evaluates $u(X)$ at all the elements of evaluation set $\{ \alpha_1, \alpha_2, \ldots , \alpha_{n} \} \subseteq$ $\mathbb{F}_{p^{m}}$.

Note 1. In the context of Reed-Solomon codes, an "evaluation set" is a set of distinct elements over the finite field used to evaluate the polynomial that represents the message to be encoded.

The corresponding generator matrix of the RS code is as follows: $$\begin{equation} \begin{aligned} G = \begin{bmatrix} 1 & 1 & \ldots & 1 \\ \alpha_{1} & \alpha_{2} & \ldots & \alpha_{n} \\ \alpha_{1}^{2} & \alpha_{2}^{2} & \ldots & \alpha_{n}^{2} \\ \vdots & & \ddots & \\ \alpha_{1}^{k-1} & \alpha_{2}^{k-1} & \ldots & \alpha_{n}^{k-1} \\ \end{bmatrix}. \end{aligned} \end{equation}$$ Now we will verify that an $(n, k)$ RS code is MDS. Let $\mathbf{u} = \begin{bmatrix} u_0 & u_1 & \ldots & u_{k-1} \end{bmatrix} \neq 0$ is the message vector. The message polynomial $\mathbf{u}(X) = u_{0} + u_{1}X+u_{2}X^{2}+ \ldots + u_{k-1}X^{k-1}$ is a non-zero polynomial of degree at most $k-1$. Therefore, $\mathbf{u}(X)$ has at most $k-1$ roots, which implies $\mathbf{u} = \begin{bmatrix} u_0 & u_1 & \ldots & u_{k-1} \end{bmatrix}$ has at most $k-1$ zeros. From this, it is clear that each nonzero codeword has at most $k-1$ zeros. Thus $d_{min} \geq n-(k-1) = n-k+1$. However, by the Singleton bound \eqref{eq:sigleton_bound}, we must have $d_{min} \leq n-k+1$. So $d_{min} =n-k+1$.

The following example illustrates the construction of $(6,3)$ RS code over the Galois field $\mathbb{F}_{2^{3}}$. The elements of $\mathbb{F}_{2^{3}}$ are given in Table 1.

Example 2. A $(6, 3)$ RS code can be obtained by listing all polynomials of degree $2$ with coefficients in $\mathbb{F}_{2^{3}}$, and subsequently evaluating these polynomials at the chosen evaluation points from the finite field $\mathbb{F}_{2^{3}}$. Due to the Maximum Distance Separable (MDS) property of RS codes, the minimum distance $d_{\text{min}}$ is equal to $4$. Therefore, the code corrects one error.

Now we will illustrate an encoding of bit stream $010011111$. Here $u_{0} = 010 = \alpha, u_{1} = 011 = \alpha^{4}, u_{2} = 111 = \alpha^{5}$ and the corresponding message polynomial is $\mathbf{u}(X) = \alpha + \alpha^{4} X+ \alpha^{5} X^{2} \in \mathbb{F}_{2^{3}}[X]$.

For this example, there are $7$ possible evaluation sets. Here we demonstrate the encoding of the above message $\mathbf{u} = \begin{bmatrix} \alpha & \alpha^{4} & \alpha^{5} \end{bmatrix}$ for two different evaluation sets.

1. For the evaluation set $\{\alpha_{1} = \alpha, \alpha_{2} = \alpha^{2}, \ldots , \alpha_{6} = \alpha^{6}$ }: Now the codeword of the given message $\mathbf{u} = \begin{bmatrix} \alpha & \alpha^{4} & \alpha^{5} \end{bmatrix}$ is obtained for the given evaluation set $\alpha_{1} = \alpha, \alpha_{2} = \alpha^{2}, \ldots , \alpha_{6} = \alpha^{6}$ as follows:

$$\begin{align*} \mathbf{v} = \begin{bmatrix} v_{0} & v_{1} & \ldots & v_{n-1} \end{bmatrix} & = \begin{bmatrix} \mathbf{u}(\alpha_{1})& \mathbf{u}(\alpha_{2}) & \ldots & \mathbf{u}(\alpha_{n}) \end{bmatrix}. \nonumber \\ & = \begin{bmatrix} \mathbf{u}(\alpha) & \mathbf{u}(\alpha^{2}) & \mathbf{u}(\alpha^{3}) & \mathbf{u}(\alpha^{4}) & \mathbf{u}(\alpha^{5}) & \mathbf{u}(\alpha^{6}) \end{bmatrix}. \end{align*}$$ Using Table 1 we can evaluate the message polynomial $\mathbf{u}(X)$ at all evaluation points. For understanding, evaluation of message polynomial $\mathbf{u}(X) = \alpha + \alpha^{4} X+ \alpha^{5} X^{2}$ is explained below for $i = 1, 2$: $$\begin{align*} \mathbf{u}(\alpha) & = \alpha + \alpha^{4} \alpha + \alpha^{5} \alpha^{2} \\ & = \alpha + \alpha^{5} + \alpha^{7} \\ & = \alpha + (1+\alpha +\alpha^{2}) + 1 \\ & = \alpha^{2}. \end{align*} \begin{align*} \mathbf{u}(\alpha^{2}) & = \alpha + \alpha^{4} \alpha^{2} + \alpha^{5} \alpha^{4} \\ & = \alpha + \alpha^{6} + \alpha^{9} \\ & = \alpha + (1+\alpha^{2}) + \alpha^{2} \\ & = 1 + \alpha = \alpha^{3}. \end{align*}$$ Similarly, we can find for $i = 3, 4, 5, 6$. The codeword $$\begin{align*} \mathbf{v} = \begin{bmatrix} v_{0} & v_{1} & \ldots & v_{n-1} \end{bmatrix} & = \begin{bmatrix} \mathbf{u}(\alpha_{1})& \mathbf{u}(\alpha_{2}) & \ldots & \mathbf{u}(\alpha_{n}) \end{bmatrix}. \nonumber \\ & = \begin{bmatrix} \mathbf{u}(\alpha) & \mathbf{u}(\alpha^{2}) & \mathbf{u}(\alpha^{3}) & \mathbf{u}(\alpha^{4}) & \mathbf{u}(\alpha^{5}) & \mathbf{u}(\alpha^{6}) \end{bmatrix} \\ & = \begin{bmatrix} \alpha^{2} & \alpha^{3} & \alpha^{6} &\alpha^{6} & \alpha^{2} & \alpha \end{bmatrix}. \end{align*}$$ Hence the encoded bit stream of $010011111$ is $001110101101001010$.

2. For the evaluation set $\{\alpha_{1} = 1, \alpha_{2} = \alpha^{}, \alpha_{3} = \alpha^{2} \ldots , \alpha_{6} = \alpha^{5}$ }: Now the codeword of the given message $u = [\alpha, \alpha^{4}, \alpha^{5} ]$ is obtained for the given evaluation set $\alpha_{1} = \alpha, \alpha_{2} = \alpha^{2}, \ldots , \alpha_{6} = \alpha^{5}$ as follows: $$\begin{align*} \mathbf{v} = \begin{bmatrix} v_{0} & v_{1} & \ldots & v_{n-1} \end{bmatrix} & = \begin{bmatrix} \mathbf{u}(\alpha_{1})& \mathbf{u}(\alpha_{2}) & \ldots & \mathbf{u}(\alpha_{n}) \end{bmatrix}. \nonumber \\ & = \begin{bmatrix} \mathbf{u}(1) & \mathbf{u}(\alpha) & \mathbf{u}(\alpha^{2}) & \mathbf{u}(\alpha^{3}) & \mathbf{u}(\alpha^{4}) & \mathbf{u}(\alpha^{5}) \end{bmatrix} \\ & = \begin{bmatrix} \alpha^{3} & \alpha^{2} & \alpha^{3} & \alpha^{6} &\alpha^{6} & \alpha^{2} \end{bmatrix}. \end{align*}$$ Hence the encoded bit stream of $010011111$ is $110001110101101001$.