Baby/giant step strategy

I wrote in my previous post that Cantor and Zassenhaus' algorithm of polynomial factorisation over finite fields can be divided into three stages:

• squarefree factorisation
• distinct-degree factorisation
• equal-degree factorisation

(in the standard Cantor-Zassenhaus algorithm the squarefree factorisation step is merged with the algorithm itself)

According to Modern Computer Algebra, "the second stage consumes the bulk of the computing time". In 1998 Erich Kaltofen and Victor Shoup designed a new approach to distinct-degree factorisation which allows to decrease it's cost. This strategy is called "baby/giant step" and exploits the fact from number theory:

Lemma. For nonnegative integers $i$ and $j$, the polynomial $x^{q^i} - x^{q^j} \in F_q [x]$ is divisible by precisely those irreducible polynomials in $F_q [x]$ whose degree divides $i−j$.

The proof is easy: if $i \geq j$ then  $x^{q^i} - x^{q^j} = (x^{q^{i-j}} - x)^{q^j}$. From the Theorem from the previous post we know that the factorisation of $x^{q^k} - x$ consists of all irreducible factors whose degree is a divisor of $k$. 

Now it's possible to formulate the algorithm:

Fast distinct-degree factorisation (DDF)

Input and output

This algorithm takes as input a square-free polynomial $f \in F_q [x]$ of degree $n$. 

The output is $f ^{[1]} , . . . , f^{[n]} \in F_q [x]$ such that for $1 \leq d \leq n$, $f ^{[d]}$ is the product of the monic irreducible factors of f of degree $d$. The algorithm is parameterized by a constant $\beta$, with $0 \leq \beta \leq 1$.

Step 1 (compute baby steps)

Let $l = \lceil n\beta \rceil$ . For $0\leq i\leq l$, compute $h_i = x^{q^i} \mod f$.
Step 2 (compute giant steps) 

Let $m = \lceil \frac{n}{2l} \rceil$ . For $1\leq j\leq m$, compute $H_j = x^{q^{lj}} \mod f$. 

Step 3 (compute interval polynomials) 

For $1\leq j\leq m$, compute $I_j = \prod\limits_{0\leq i\leq l}(H_j - h_i) \mod f$ .

(by Lemma the polynomial $I_j$ is divisible by those irreducible factors of $f$ whose degree divides an integer $k$ with  $(j-1)l < k \leq jl$)

Step 4 (compute coarse DDF) 

In this step, we compute polynomials $F_1, ... , F_m $ where $F_j = f^{[(j-1)l+1]} f^{[(j-1)l+2]} ... f^{[jl]}$.  This is done as follows.
$s\leftarrow f;$
for $j \leftarrow 1$ to $m$ do
$\left\{F_j \leftarrow \gcd(s, I_j); s\leftarrow \frac{s}{F_j}\right\}$
Step 5 (compute fine DDF) 

In this step, we compute the output polynomials $f^{[1]}, ... , f ^{[n]}$. 

First, initialize $f^{[1]}, ... , f^{[n]}$ to 1. Then do the following.

for $j \leftarrow1$ to $m$ do

{

$g\leftarrow F_j ;$

for $i\leftarrow l - 1$ down to 0 do
    $\left\{ f^{[lj-i]} \leftarrow \gcd(g, H_j - h_i); g\leftarrow \frac{g}{f^{[lj-i]}} \right\}$

}

if $s \neq 1$ then $f^{[\deg(s)]}\leftarrow s$

Some technical remarks

1.  Lemma. For any positive integer $r$ if we are given $h = x^{q^r} \mod f \in F_q [x]$, then for any $g \in F_q [x]$, we can compute  $g^{q^r} \mod f$ as $g(h) \mod f$.

Now to perform Step 2 we set $H_1 = h_l$ and then compute each $H_j$ as $H_{j−1} (H_1) \mod f$.

2. Computing $f(g) \mod h$ is a so-called "modular composition" problem. For solving this problem one can use e.g. Horner scheme or Brent and Kung's algorithm (this also requires fast matrix multiplication).

By now I implemented the simplest version of the fast CZ algorithm: it uses Horner scheme in Step 2.

The next step might be to implement matrix multiplication and Brent and Kung's algorithm.