\lecture{10: Distinct Degree Factoring}{CS681}{Piyush P Kurur}{Ramprasad Saptharishi}

\section*{Overview}

Last class we left of with a glimpse into distant degree
factorization. This class, we shall look at the details of it and also
how it can be used as an irreducibility test. 

\section{DDF: The Problem}

We are given a polynomial $f$ over a finite field $\F_p$ of degree say
$n.$ We want to factor them in to degree $1$ factors, degree $2$
factors etc. To make this more explicit, let us assume that $f$
factorizes as
$$
f = g_{11}g_{12}\cdots g_{1m_1}g_{21}\cdots g_{dm_d}
$$
where each $g_{ij}$ is an irreducible factor of $f$ of degree $i.$
Hence given $f$, we want to return $\prod_j g_{ij}$ for each $i.$ That
is, return the factor of $f$ that is the product of all degree $i$
irreducible factors of $f.$ And we want to do this for all $i$. This
is called {\em distinct degree factorization.} We want an efficient
algorithm for this.

But before we start thinking of algorithms, what do we mean by
efficient? It is the usual 'polynomial time in input length' but what
is the input length? We are looking at $f(X) = \F_p[X]$ and each
coeffiecient of the polynomial is from $\F_p.$ And since $\F_p$
contains just $p$ elements, we can encode them in binary using $\log
p$ bits. Hence the input size is about $n\log p.$ Thus we are
interested in algorithms that have running time of $(n\log p)^c$ for
some constant $c.$

\section{Extracting Square-free Parts}

When we said that $f$ factorizes as $g_{11}\cdots g_{dm_d}$, it is
very much possible that there are some $g{ij} = g_{ik}$ or in other
words the square of $g_{ij}$ divides $f.$ The first step of any
factoring algorithm is to remove such multiple factors and make $f$
{\em square free} (make sure that $f$ is not divisible by any square).\\

Suppose we were looking at polynomials over $\Z[X]$ or something. Then
we have a very nice way of checking for such multiple factors. We know
that if $f$ has a repeated root $\alpha$ , then the derivative $f'$
has $\alpha$ as a root as well. Infact, if $(x-\alpha)^m$ divided $f$
then $(x-\alpha)^{m-1}$ will divide $f'.$ And hence, the gcd of $f$
and $f'$ will have $(x-\alpha)^{m-1}$ as a factor. Thus, we can divide
$f$ by this gcd and get rid of higher multiplicity terms. 

But in the case of $\Z[X]$, we had an interpretation of real numbers,
limits and hence we could define what a derivative is. But when it
comes to finite fields, what does differentiation mean? How can we use
this technique to get extract the square-free part?\\

Note that we don't need the notion of a derivative as a tangent or
something. That is where the limits come in. What we need is a
condition where if some $g^m$ divides $f$ then $g^{m-1}$ should divide
$f'$ for the $f'$ that we are going to define. Thus, since we are just
in the realms of polynomials, we shall use the rules of
differentiation as just a formula. 

Thus let $D$ be a map that sends $X^m$ to $mX^{m-1}.$ Now extend this
linearly to all polynomials to get 
$$
D\inparen{a_0 + a_1X + a_2X^2 + \cdots a_mX^m} = a_1 + 2a_2X + \cdots ma_mX^{m-1}
$$

Now, we leave the following things to be proven as an exercise.

\subsection*{Exercise}
\begin{enumerate}
\item $D(f+g) = D(f)+D(g)$
\item $D(fg) = fD(g) + gD(f)$
\item $D(f^m) = mf^{m-1}D(f)$
\item {\bf Theorem: } If $h$ is a factor of $f$ such that $h^m\mid f$ then $h^{m-1}\mid g.$
And further, if $h^m$ is the highest power of $h$ that divides $f$
then $h^{m-1}$ is the highest power of $h$ that divides $f'.$
\end{enumerate}

Once we have these properties, we can extract the square-free part of
$f$ easily. Given an $f$ construct the formal derivative $f'.$ Let $g
= gcd(f,f').$ The polynomial $f/g$ consists is the square free
extraction of $f.$\\

There is, however, a small catch here. But we shall get back to it in
the end of the class and discuss it in the next class in detail. For
the moment, let us proceed with distinct degree factorization. 

\section{Distinct Degree Factorization}

Given $f\in \F_p[X]$ of degree $n$, we want to get the
distant degree factorization of $f.$ That is, we want to get
$g_1,g_2,\cdots ,g_n$ such that $f = g_1g_2\cdots g_n$ where each $g$
is the product of all irreducible factors of $f$ of degree $i.$ And we
want to do this efficiently. \\

The key idea is the formula we proved last class:
$$
X^{p^m} - X = \prod_{\substack{f\in \text{Irr}(\F_p,d)\\d\mid m}} f(X)
$$

Firstly, let us look at the case when $m=1.$ Then $X^p - 1$ is the
product of all irreducible polynomials over $\F_p$ of degree $1.$ And
thus, in particular, it contains the degree $1$ irreducible factors of
$f$ within it.

Thus, $g_1 = gcd(f,X^p - X)$ will be the product of all degree $1$
irreducible factors of $f$ as $X^p - X$ has just degree $1$
irreducible factors and all those that divide $f$ will be a part of
the gcd. Thus, we have obtained the required $g_1.$ Now, call $f_2 =
f/g_1$ and now let $g_2 = gcd(f_2,X^{p^2} - X)$ and this will have
precisely all degree $2$ irreducible factors of $f$ (degree $1$
factors won't appear since we have removed all degree $1$ factors by
dividing by $g_1$). Thus, we can repeat this procedure. 

This is a naive algorithm and has a pretty serious problem. The
trouble is that the algorithm would have to compute
$gcd(f_i,X^{p^i}-X)$ which is a HUGE degree polynomial. We want our
running time as $(n\log p)^c$ but this naive algorithm takes about
$O(p)$ time! And this is definitely not acceptable since even finding
all linear irreducible factors might take $O(p)$ time! We definitely
need to change this. But the polynomial $X^{p^i} - X$ is a nice
polynomial and hence allows a nice simple trick to make the algorithm
polynomial time. \\

Let us look at the gcd algorithm. The first step is to compute
$X^{p^i} -X \bmod{f_i}$ and the problem with this is that the degree of
the polynomial is too large to apply the naive algorithm. But we can
do far better in the case of this special polynomial. 

Note that $X^{p^i} - X \bmod{f_i} = X^{p^i}\bmod{f_i} - X\bmod{f_i}$
and the difficulty was in computing $X^{p^i} \bmod{f_i}.$ This can be
done quite efficiently using the technique of repeated squaring.

\subsection{Repeated Squaring}

In general, we have a polynomial $f$ of degree $n$ and we want to
calculate $X^M\bmod{f}$ in time $(n\log M)^c$ for some constant $c.$
The idea is pretty simple. 

Firstly assume $M$ to be a power of $2.$ Then we can do the
following. Start with $X$, and square it. And square it again and so
on. The moment the degree goes beyond $n$, take it modulo $f$ to
reduce it's degree back to less than $n.$ And continue this squaring
process, for $\log M$ steps. Thus, we would have computed
$X^M\bmod{f}$ in time $(n\log M)^c.$

Now what do we do for $M$ that is not a power of two? The answer is
simple again. Just look at the binary representation of $M$. Say $m =
b_0 + b_12 + b_22^2 + \cdots b_{l}2^l.$ Then 
$$
X^M = X^{b_0}(X^{b_1})^{2^1}(X^{b_2})^{2^2}\cdots (X^{b_l})^{2^l}
$$
And since each $b_i$ is either $1$ or $0$, $X^{b_i}$ is either $1$ or
$X.$ Thus, we can just compute $X^{2^i}\bmod{f}$ for $0\leq i\leq \log
M$ and multiply all those residues that have $b_i$ as $1.$

\begin{algorithm}
\caption{\sc{Evaluate $X^M \bmod{f}$ using repeated squaring}}
\begin{algorithmic}[1]
\STATE Let $M = b_0 + b_12 + \cdots b_l2^l$, the binary representation
of $M,$ where $l =\floor{\log M}$.
\STATE $g_0 = 1$ and $g = b_0X.$
\FOR{$i = 1$ to $l$}
\STATE $g_i = (g_{i-1})^2\bmod{f}$
\IF{$b_i = 1$}
\STATE $g = g\cdot g_i\bmod{f}$
\ENDIF
\ENDFOR
\STATE {\bf return} $g$
\end{algorithmic}
\end{algorithm}

Now, with this repeated squaring algorithm, we can do distinct degree
factorization. The algorithm is given at the end of the lecture. 

\section{A Catch and a Hint}

First let us get to the catch that we had mentioned earlier. Our
method for extracting the square free part of $f$ involved taking the
formal derivative and then the gcd of $f$ with $f'.$ But strange
things can happen in a finite field. For example, let us take the
polynomial $f(X) = X^{2p} - 3X^p + 5.$ The formal derivative is
$2pX^{2p - 1} - 3pX^{p-1}$ which is $0$ in $\F_p$! What do we do now?

It is easy to see that the derivative can become $0$ if the only
surviving terms of the polynomial have degree that's a multiple of a
prime. Then this polynomial $f(X)$ can be thought of as $g(X^p)$ and
we can go on to do our algorithm for $g$ instead of $f.$ In our
example, $g(X) = X^2 - 3X + 5$ which is well behaved. All we need to
do is get the factors of $g$ and replace every occurance of $X$ by
$X^p.$ And more over, since we are in $\F_p$, it is easy to check that
$f(X) = g(X^p) = (g(X))^p.$ \\

So much for the small catch. Now here is a hint. The algorithm for
distinct degree factorization immediately gives us an algorithm to
test if a given polynomial over $\F_p[X]$ is irreducible. The
reduction is very straightforward and the students are encouraged to
think about it. We shall discuss this in the next class. 

\begin{algorithm}
\caption{\sc{Distinct Degree Factorization}}
\begin{algorithmic}[1]
\REQUIRE $f(X) \in \F_p[X]$ of degree $n.$
\STATE $f_0 = f.$
\FOR{$i=1$ to $n$}
\STATE Using repeated squaring, compute $s_i = X^{p^i} \bmod{f_{i-1}}$
\STATE Compute $g_i = gcd(f_{i-1},s_i - X).$
\STATE $f_i = f_{i-1}/g_i$
\ENDFOR
\STATE {\bf return} $\inbrace{g_1,g_2,\cdots, g_n}.$
\end{algorithmic}
\end{algorithm}

