\lecture{14: BCH Codes}{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

\section*{Overview}

We shall look a a special form of linear codes called cyclic
codes. These have very nice structures underlying them and we shall
study BCH codes. 

\section{General Codes}

Recall that a linear code $\mathcal{C}$ is just a subspace of
$\F_q^n.$ We saw last time that by picking a basis of $\mathcal{C}$ we
can construct what is known as a parity check matrix $H$ that is zero
precisely at $\mathcal{C}.$

Let us understand how the procedure works. Alice has a message, of
length $k$ and she wishes to transmit across the channel. The channel
is unreliable and therefore both Alice and Bob first agree on some
code $\mathcal{C}.$ Now how does Alice convert her message into a code
word in $\mathcal{C}$? If Alice's message could be written as $(x_1,
x_2, \cdots, x_k)$ where each $x_i\in \F_q$, then Alice simply sends
$\sum_{i=1}^k x_ib_i$ which is a codeword. 

Bob receives some $y$ and he checks if $Hy = 0.$ Assuming that they
choose a good distance code (the channel cannot alter one code into
another), if Bob finds that $Hy = 0$, then he knows that the message
he received was untampered with. 

But what sort of errors can the channel give? Let us say that the
channel can change atmost $t$ positions of the codeword. If $x$ was
sent and $x'$ was received with atmost $t$ changes between them, then
the vector $e=x' - x$ can be thought of as the error vector. And since
we assumed that the channel changed atmost $t$ positions, the error
vector can have weight atmost $t.$

This then means that Alice sent an $x$ and Bob received $x+e.$ Bob
runs the parity check matrix on $x+e$ to get $H(x+e) = Hx + He = He.$
The quantity $He$ is called the {\em syndrome,} which is just the
evaluation of Bob's received message by the parity check matrix. If
the syndrome is zero, Bob knows that the received word is a valid
codeword. \\

Of course, in order to determine what the actual message was, Bob
needs to figure out what $e$ is (for then he knows the message was $x'
- e$) but recovering $e$ from $He$ is still a hard thing. It is not
clear how this can be done efficiently on a general setting. 

\section{Cyclic Codes}

\begin{definition}
A cyclic code $\mathcal{C}$ is a linear code such that if
$(c_0,c_1, \cdots, c_{n-1})$ is a codeword, then so is $(c_{n-1},c_0,
c_1, \cdots, c_{n-2}).$ To put it algebraically, the space of
codewords is invariant under cyclic shifts.
\end{definition}

If course any codeword that is shifted by $i$ places, to the left or the
right, will also be a codeword. In order to be able to see the strong
structure behind them, we need a different perspective on $\F_q^n.$ 

\subsection{Codewords as Polynomials}

Given a vector $(c_0,c_1, \cdots, c_{n-1})$, we can associate a
polynomial naturally which is $c(X) = c_0 + c_1X + \cdots +
c_{n-1}X^n.$ This is just interpretting the vector space $\F_q^n$ as
the additive group of the ring $\F_q[X]/(f(X))$ where $f$ is a
polynomial of degree $n,$ since they are both isomorphic.

The ring picture has the extra multiplicative structure which is very
useful here. Suppose we have a codeword $c = (c_0, \cdots, c_{n-1})$,
what can we say about the codeword $c' = (c_{n-1},c_0, \cdots,
c_{n-2})$? As a polynomial, $c = c_0 + c_1X + \cdots + c_{n-1}X^{n-1}$
and $c' = c_{n-1} + c_0X + \cdots + c_{n-2}X^{n-1}.$ So essentially we
just took the polynomial $c$ and multiplied by $X.$ The last term
$c_{n-1}X^n$, however, was changed to $c_{n-1}.$ How do we achieve
this? Do the multiplication modulo $X^n - 1$ which is just identifying
$X^n$ by $1.$

Thus, cyclic shifts is just multiplication of polynomials in
$\F_q[X]/(X^n - 1)$ by powers of $X.$ With this observation, the
following theorem summarizes the strong underlying structure in cyclic
codes. 

\begin{theorem}
Any cyclic code $\mathcal{C}$ is an ideal of $R = \F_q[X]/(X^n - 1).$ And
conversely, every ideal is a cyclic code
\end{theorem}
\begin{proof}
Let us prove the easier converse first. Let $f(X)\in R$ be an element
of the ideal $\mathcal{C}.$ Then it follows that for any polynomial $a(X)$,
$a(X)f(X)\in \mathcal{C}$ and in particular $X^if(X) \in \mathcal{C}.$
But we already say that multiplying by powers of $X$ was just shifting
and therefore our code is also invariant under shifts. \\

The other direction is straightforward too. We want to show that given
a cyclic code $\mathcal{C}$, for any code word $f(X)$ and any polynomial $a(X)$,
$a(X)f(X) \in \mathcal{C}.$ 

\bea{
a(X)f(X) & = & (a_0 + a_1X + \cdots + a_{n-1}X^{n-1})f(X)\\
         & = & a_0f(X) + a_1(Xf(X)) + \cdots + a_{n-1}(X^{n-1}f(X))\\
         & = & a_0f_0(X) + a_1f_1(X) + \cdots +
         a_{n-1}f_{n-1}X \qquad\text{$X^if(X)$ is shifting}\\
         & = & f'(X) \in \mathcal{C}
}
\end{proof}

Suppose $X^n-1$ factorizes into irreducible polynomials over $\F_q$,
say
$$
X^n - 1 = g_1g_2\cdots g_k
$$
Then it easy to check that infact all ideals of $R$ are principle, of
the form $g(X)R$ where $g(X)$ is a factor of $X^n - 1.$ And hence, we
have a simple corollary to above theorem. 

\begin{corollary}
Every cyclic code $\mathcal{C}$ is just the set of multiples of a
single polynomial $g(X) \in R.$
\end{corollary}

This polynomial is called the generator polynomial. Let us say we pick
a factor $g(X)$ of $X^n - 1$ and let its degree be $d.$ What can we
say about the dimension of the code $(g(X))$? For this, we will need
the rank-nullity theorem.

\begin{theorem}[Rank-Nullity]
If $T$ is a linear map from a between two vector spaces $V$ and $W$,
then $\text{rank}(T) + \text{nullity}(T) = \dim{V}$ where
$\text{rank}(T)$ is defined to be the dimension of the image of $V$
and nullity the dimension of the kernel. 
\end{theorem}

Now look at the map $\phi: R\longrightarrow R/(g(X)).$ This, being a
homomorphism of rings will also be a linear map on the additive groups
which are vector spaces. The dimension of $R$ is $n$ and the dimension
of the image, which is $R/(g(X))$, is $d$. And therefore, the
dimension of the kernel which is $\mathcal{C} = (g(X))$ is $n-d.$\\

What about the parity check matrix? That is extremely simple
here. Since the ideal is generated by a single polynomial $g(X)$, we
just need to check if any given polynomial is in the code or not by
just checking if $g$ divides it. Thus, just the modulo operation is
the parity check. This can be written as a matrix as well but the idea
is clear. 

\section{BCH Codes}

BCH\footnote{Bose, Ray-Chaudhuri, Hocquenghem} codes is an example of
a cyclic code that is widely studied in coding theory. In order to
get a cyclic code, we just need to get the generating polynomial of
that code. 

Instead of asking for the polynomial in terms of the coefficient, what
if we identify the polynomial by the roots instead? This is the
general idea of a BCH code. \\

We are working in a vector space of dimension $n$ over $\F_q$ and
identifying cyclic codes as ideals of $R = \F_q[X]/(X^n - 1).$ Let us
further impose the constraint that the roots of $X^n - 1$ are distinct
by making sure $gcd(n,q)=1$ so that the derivative is non-zero. 

Let $\zeta$ be a primitive $n$-th root of unity in $R$ and look at the
set $\inbrace{\zeta, \zeta^2, \cdots, \zeta^d}$ where $d < \phi(n)$
(to prevent some $\zeta^i = \zeta^j$)\footnote{why won't $d<n$
  suffice?}. Now we ask for the smallest degree polynomial $g$ that
has $\zeta^i$ as a root for $1\leq i\leq d.$ This polynomial is going
to be our generating polynomial for the cyclic code. \\

The parity check matrix of a BCH code is pretty simple. Note that if
$c(X) \in \mathcal{C}$, then $c(X)$ is a multiple of $g(X)$ and in
particular $c(X)$ will also have the $\zeta^i$ as roots. And
therefore, all we need to check is if $c(\zeta^i)=0$ for all $1\leq
i\leq d.$ Now interpretting $c(X)$ as a vector $(c_0,c_1, \cdots,
c_{n-1})$ of coefficients, the parity check reduces to the following
matrix multiplication. 

$$
\insquar{\begin{array}{ccccc}
1 &\zeta & \zeta^2 & \cdots & \zeta^{n-1}\\
1 &\zeta^2 & (\zeta^2)^2 & \cdots & {\zeta^2}^{n-1}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
1 & \zeta^d & (\zeta^d)^2 & \cdots & (\zeta^d)^{n-1}
\end{array}
}
\insquar{\begin{array}{c}
c_0\\
c_1\\
\vdots\\
c_{n-1}
\end{array}
} = 
\insquar{\begin{array}{c}
0\\
0\\
\vdots\\
0
\end{array}
}
$$
Note that the parity check matrix $H$ is a $(n-d)\times d$ matrix.  

\subsection{Distance of a BCH Code}

Suppose $g(X)$ is the generating polynomial for the set being the
first $d$ powers of $\zeta$, what can we say about the distance of the
cyclic code $(g(X))?$ 

\begin{theorem} A BCH code obtained by considering the first $d$
  powers of $\zeta$ has distance $d+1.$
\end{theorem}
\begin{proof}
  We would like to show that the minimum weight of the code
  $\mathcal{C} = (g(X))$ has to be atleast $d+1.$ Suppose not, then
  there is a codeword $c$ such that the weight of $c$ is less than or
  equal to $d.$ Then this polynomial has atmost $d$ positions with
  non-zero entries. Let us denote those coefficients by
  $\inbrace{c_{i_1},c_{i_2}, \cdots, c_{i_d}}$ and say in
  increasing order of indices. 

  We just need to check that for each $1\leq k \leq d$
  $$
  \sum_{j=1}^{d} c_{i_j} (\zeta^k)^{i_j} = 0
  $$
  
  But the above equation corresponds to the following matrix product
  $$
  \insquar{\begin{array}{cccc}
  \zeta^{i_1} & \zeta^{i_2} & \cdots & \zeta^{i_d}\\
  (\zeta^{i_1})^2 & (\zeta^{i_2})^2 & \cdots & (\zeta^{i_d})^2\\
  \vdots & \vdots & \ddots & \vdots \\
  (\zeta^{i_1})^d & (\zeta^{i_2})^d & \cdots & (\zeta^{i_d})^d  
  \end{array}}
  \insquar{\begin{array}{c}
  c_{i_1}\\c_{i_2}\\\vdots\\c_{i_d}\end{array}} = 
  \insquar{\begin{array}{c}
  0\\0\\\vdots\\0\end{array}}   
  $$
  Note that the $d\times d$ matrix is essentially in the form of a
  vandermonde matrix:
  $$
  \insquar{\begin{array}{cccc}
  1 & 1 & \cdots &1\\
  x_1 & x_2 & \cdots & x_n\\
  x_1^2 & x_2^2 & \cdots & x_n^2\\
  \vdots & \vdots & \ddots & \vdots \\
  x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1}
  \end{array}}
  $$
  and it is well known that the determinant of this matrix is
  $\prod_{i<j} (x_i - x_j)$ and therefore non-zero if each $x_i$ is
  distinct as in our case of $\zeta^{i_j}$. Therefore, $Hc = 0$ and
  $H$ being invertible forces that $c$ has to be the zero vector as
  well!

  Therefore, the only codeword that can have weight less than or equal
  to $d$ is the zero vector. And therefore the minweight of the BCH
  code is atleast $d+1.$
\end{proof}

Now that we have this, we can use
$\zeta, \zeta^2, \cdots, \zeta^{d-1}$ to get a guarantee that our code
has distance atleast $d.$ This is called the {\em designed distance}
of the BCH code. Note that the actual distance you could be larger
than $d.$ We just have a guarantee that it is atleast $d$ but the
could potentially give you codes of larger distance. There are
examples of BCH codes with the actual distance larger than the
designed distance. 
