CoCalc -- 18.783 Lecture 3 Root finding in finite fields.ipynb

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Path: 18.783 Lecture 3 Root finding in finite fields.ipynb

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Problem:	Given a polynomial $f\in \mathbb F_q[x]$ , determine the $\mathbb F_q$ -rational roots of $f$ .
Solution:	Rabin's algorithm!

In [1]:

p=61; F=GF(p); R.<x>=PolynomialRing(F); f = x^8-2*x+5

In [2]:

f.roots()

[(44, 1), (39, 1), (15, 1)]

In [3]:

f.factor()

(x + 17) * (x + 22) * (x + 46) * (x^2 + 46*x + 1) * (x^3 + 52*x^2 + 41*x + 33)

In [4]:

f.change_ring(GF(p^6)).roots()

[(44, 1),
 (39, 1),
 (15, 1),
 (46*z6^5 + 32*z6^4 + 56*z6^3 + 22*z6^2 + 48*z6 + 44, 1),
 (7*z6^5 + 35*z6^4 + 2*z6^3 + 4*z6 + 37, 1),
 (15*z6^5 + 29*z6^4 + 5*z6^3 + 39*z6^2 + 13*z6 + 32, 1),
 (28*z6^5 + 30*z6^4 + 20*z6^3 + 26*z6^2 + 21*z6 + 30, 1),
 (26*z6^5 + 57*z6^4 + 39*z6^3 + 35*z6^2 + 36*z6 + 3, 1)]

Theorem:	$\mathbb F_{p^n} = \{\alpha\in \overline{\mathbb F}_p: \alpha^{p^n}=\alpha\}$ , in other words, we have $x^{p^n}-x=\prod_{\alpha\in \mathbb F_{p^n}}(x-\alpha)$
Proof:	For $n=1$ this follows from Fermat's little theorem. If we identify $\mathbb F_p\simeq \mathbb Z/p\mathbb Z$ , we have $a^{p-1}=1\bmod p$ for $a\in [0,p-1]$ , so the $p$ elements of $\mathbb Z/p\mathbb Z$ are precisely the roots of $x^p-x$ . For $n>1$ this is true by definition: $\mathbb F_{p^n}:=\overline{\mathbb{F}}_{p}^{\langle \alpha \mapsto \alpha^{p^n}\rangle}$ is the fixed field of the $p^n$ -power Frobenius automorphism of $\overline{\mathbb F}_p$ , equivalently, the splitting field of $x^{p^n}-x$ over $\mathbb F_p$ .
Corollary:	For any prime power $q$ we have $\mathbb F_{q^n} = \{\alpha\in \overline{\mathbb F}_q: \alpha^{q^n}=\alpha\}$ and $x^{q^n}-x=\prod_{\alpha\in \mathbb F_{q^n}}(x-\alpha)$ .

In [5]:

product([x-i for i in F]) == x^p-x

True

In [6]:

product([x.change_ring(GF(p^2))-i for i in GF(p^2)]) == x^(p^2)-x

True

Corollary:

Let

f\in \mathbb F_q[x]

be nonzero and let

S:=\{\alpha\in \mathbb F_{q^n}:f(\alpha)=0\}

. Then

\gcd(f(x),x^{q^n}-x)=\prod_{\alpha\in S}(x-\alpha).

In particular,

\#S=\deg(\gcd(f(x),x^{q^n}-x)

As usual,

\gcd(a,b)

denotes the unique monic representative of the nonzero principal ideal

(a,b)

\mathbb F_q[x]

In [7]:

%time gcd(f,x^p-x).degree()

CPU times: user 57 µs, sys: 0 ns, total: 57 µs
Wall time: 60.6 µs

3

In [8]:

%time gcd(f,x^(p^2)-x).degree()

CPU times: user 465 µs, sys: 90 µs, total: 555 µs
Wall time: 562 µs

5

In [9]:

%time gcd(f,x^(p^3)-x).degree()

CPU times: user 42.7 ms, sys: 330 µs, total: 43 ms
Wall time: 43 ms

6

In [0]:

%time gcd(f,x^(p^6)-x).degree()

Key point:

Make sure you exponentiate in the right ring!

The first step in computing

\gcd(f(x),x^q-x)

is to compute

x^q-x\bmod f

(note that

q

may be exponentially large, but

\deg f

certainly won't be).
If

\pi_f\colon \mathbb{F}_q[x]\to\mathbb{F}_q[x]/(f)

is the ring homomorphism defined by

x\mapsto x\bmod f

, then

\pi_f(x^q-x)=\pi_f(x^q)-\pi_f(x)=\pi_f(x)^q-\pi_f(x),

so the order of operations makes no mathematical difference, but the RHS can be computed exponentially faster than either the middle or the LHS!

The complexity of computing

x^q\bmod f

with Euclidean division is quasi-linear in

q

,
The complexity of computing

\pi_f(x)^q

using binary exponentiation is quasi-linear in $\log q$ .

In [0]:

pif=R.quotient(f)

In [0]:

%time print(pif(x)^(p^6))

In [0]:

p=2^255-19;  F=GF(p); R.<x>=PolynomialRing(F); f=x^8-2*x+5; pif=R.quotient(f);
%time print(f.factor())

In [0]:

%time print(pif(x)^p)
%time print(pif(x)^(p^2))
%time print(pif(x)^(p^3))
%time print(pif(x)^(p^4))
%time print(pif(x)^(p^10))

In [0]:

def distinct_root_poly(f):
    q = f.base_ring().cardinality()
    pif = f.parent().quotient(f)
    return gcd(f,(pif(x)^q).lift()-x)

def count_distinct_roots(f):
    return distinct_root_poly(f).degree()

In [0]:

p=61;  F=GF(p); R.<x>=PolynomialRing(F); f=x^8-2*x+5; pif=R.quotient(f)
for n in range(1,7):
    print("%d distinct roots over F_{%d^%d}" % (count_distinct_roots(f.change_ring(GF(p^n))),p,n))

Okay, we can count (distinct) roots, but what if we actually want to know what they are?

In [0]:

gcd(f,(pif(x)^p-x).lift())

In [0]:

s = (p-1) // 2 # we assume p (or more generally q) is odd
gcd(f,(pif(x)^s-1).lift())

In [0]:

print(gcd(f,(pif(x)^p-x).lift())/gcd(f,(pif(x)^s-1).lift()))

Excellent, we managed to split the roots of $f$ by picking out two that are roots of $x^s-1$ , where $s=(q-1)/2$ (we are assuming $q$ is odd).
What are the roots of $x^s-1$ anyway?

In [0]:

print(sorted((x^s-1).roots()))

My keen sense of pattern recognition (or a recollection of Euler's criterion) tells me these are the roots of $x^s-1$ are precisely the squares (quadratic residues) in $\mathbb F_p^\times.$

In [0]:

product([x-a^2 for a in F if a < p-a]) == (x^s-1)

But if $f$ has multiple roots that are quadratic residues (or maybe none), then what?

Rabin: Instead of $x^s-1$ , use $(x+\delta)^s-1$ for a random $\delta\in \mathbb F_q$ .

Definition:	Let us say that $\alpha,\beta\in \mathbb F_q$ are of different type if both are nonzero and $\alpha^s\ne \beta^s$ , in other words, one is a quadratic residue and one is not.
Theorem:	For every pair of distinct $\alpha,\beta\in \mathbb F_q$ we have $\#\{\delta\in \mathbb F_q:\alpha+\delta \text{ and } \beta+\delta \text{ are of different type }\} = \frac{q-1}{2}.$
Proof:	See Section 3.11 in the lecture notes for a short easy proof.
Corollary:	Let $f\in \mathbb F_q[x]$ , let $g(x)=\gcd(f,x^q-x)$ , let $s=(q-1)/2\in \Z$ , and let $\delta$ be a uniformly random element of $\mathbb F_q$ . If $g$ has more than one nonzero root, then $\Pr[0<\deg\gcd(g,(x+\delta)^s-1)<\deg g] \ge \frac{1}{2}.$

In [0]:

def rational_root(f):
    """Returns a rational root of f or None if f has no rational roots."""
    q = f.base_ring().cardinality()
    assert q % 2 == 1
    s = (q-1) // 2 # Note // yields an integer (whereas / would yield a rational)
    x = f.parent().gen()
    if f[0]==0:
        return f.base_ring()(0)
    g = distinct_root_poly(f)
    print("Initial g = gcd(f,x^q-x) = %s" % g)
    pig = g.parent().quotient(g)
    while g.degree() > 1:
        delta = g.base_ring().random_element()
        print("delta = %s" % delta)
        h = gcd(g, ((pig(x)-delta)^s-1).lift())
        if 0 < h.degree() < g.degree():
            g = h if 2*h.degree() <= g.degree() else g // h # as above, we use // to get a polynomial
            pig = g.parent().quotient(g)
            print("Better g = %s" % g)
    return -g[0] if g.degree() == 1 else None

Theorem:

Let

n=\log q

and

d=\deg f

. The expected running time of

\texttt{rational\_root}

\tilde O(n^2d)

bit operations.

Here the soft-O notation means up to a poly-logarithmic factors. See the lecture notes for a more precise bound and a proof.

In [0]:

for i in range(5):
    r = rational_root(f)
    print("Found root %s\n" % r)
    assert f(r) == 0

In [0]:

h = product([x-i for i in range(1,20)])
for i in range(5):
    r = rational_root(h)
    print("Found root %s\n" % r)
    assert h(r) == 0

In [0]:

p=2^61-1;  F=GF(p); R.<x>=PolynomialRing(F);  f=x^8-2*x+5; h = product([x-i for i in range(1,20)])
for i in range(3):
    %time print("Found root %s\n" % rational_root(h))

What if we want all the rational roots?

In [0]:

def distinct_rational_roots(f,recurse=False):
    """Returns a list of all the distinct rational roots of f."""
    q = f.base_ring().cardinality()
    assert q % 2 == 1
    s = (q-1) // 2
    x = f.parent().gen()
    roots = []
    if recurse:
        g = f
    else:
        g = distinct_root_poly(f)
        if g(0) == 0:
            roots = [g.base_ring()(0)]
            g = g // x
    if g.degree() <= 1:
        return [-g[0]] if g.degree() == 1 else []
    while True:
        delta = g.base_ring().random_element()
        pig = g.parent().quotient(g)
        h = gcd(g, ((pig(x)-delta)^s-1).lift())
        if 0 < h.degree() < g.degree():
            roots += distinct_rational_roots(h, True)
            roots += distinct_rational_roots(g//h, True)
            return roots

Theorem:

Let

n=\log q

and

d=\deg f

. The expected running time of

\texttt{distinct\_rational\_root}

\tilde O(n^2d)

bit operations.

This looks the same as the bound for $\texttt{rational\_root}$ , but in fact there is an extra factor of $\log d$ hidden in the soft-O notation.

In [0]:

%time distinct_rational_roots(f)

In [0]:

%time distinct_rational_roots(product([x-F.random_element() for i in range(10)]))

What if we want all the rational roots and their multiplicities?

Yun: Compute the squarefree factorization!

Definition:	The squarefree factorization of a monic $f\in \mathbb F_q[x]$ is the unique list of monic squarefree polynomials $g_1,\ldots,g_m$ with $g_m\ne 1$ for which $f=g_1g_2^2\cdots g_m^m$ .
Key fact:	If $f=f_1\cdots f_n$ is the factorization of a monic $f\in \mathbb F_q[x]$ into irreducibles then $f_i=f_j$ for some $i\ne j$ if and only if $f_i\|\gcd(f,f')$ . This holds in any perfect field.

In [0]:

def squarefree_factorization(f):
    """Yun's algorithm to compute a list of polynomials g_1,..,g_m such that f=lc(f)g_1g_2^2...g_m^m with g_m != 1, assuming deg(f) < char(Fq).  We include g_0=1 for convenience."""
    assert f.degree() < f.base_ring().characteristic()
    if not f.is_monic:
        f = f.monic()
    df = f.derivative()
    u = gcd(f,df)
    v, w = f//u, df//u
    gs =[f.parent()(1)]
    while True:
        g = gcd(v, w-v.derivative()); gs.append(g)
        if v.degree() == g.degree():
            return gs
        v, w = v//g, (w-v.derivative())//g

Theorem:

Let

n=\log q

and

d=\deg f

. The running time of

\texttt{squarefree\_factorization}

\tilde O(nd)

bit operations.

This is negligible compared to the complexity of $\texttt{distinct\_rational\_roots}$ , so we might as well call $\texttt{squarefree\_factorization}$ first.
Even if we only want distinct roots, this will actually save time if $f$ is not squarefree.

In [0]:

h = product([(x^2+F.random_element()*x+F.random_element())^i for i in range(1,10)])
%time gs = squarefree_factorization(h)
assert h == product([gs[i]^i for i in range(len(gs))])
gs

In [0]:

def rational_roots(f):
    """Returns a list of tuples identifying the rational roots of f and their multiplicities."""
    gs = squarefree_factorization(f)
    return reduce(lambda x, y: x+y,[[(a,i) for a in distinct_rational_roots(gs[i])] for i in range(1,len(gs))])

In [0]:

h = product([(x-F.random_element())^i for i in range(1,6)])
%time rational_roots(h)

What if we want to factor $f$ into irreducibles?

Rabin: You can do that with my root-finding algorithm.

Cantor-Zassenhaus: Yes, but that requires working in extension fields, and with our algorithm there is no need!

The first step is to compute the distinct degree factorization

Definition:	The distinct degree factorization of a monic $f\in \mathbb F_q[x]$ is the unique list of polynomials $g_1,\ldots,g_d$ with $f=g_1\cdots g_d$ such that each $g_i$ is a (possibly empty) product of distinct monic irreducible polynomials of degree $i$ , for $1\le i\le d=\deg f$ .
Key point:	This can be done deterministically by taking succesive gcds with $x^{q^i}-x$ .

In [0]:

def distinct_degree_poly(f,i):
    q = f.base_ring().cardinality()
    pif = f.parent().quotient(f)
    return gcd(f,(pif(x)^(q^i)).lift()-x)

def distinct_degree_factorization(f, squarefree=False):
    """
        Computes the distinct degree factorization of f as list gs with gs[0]=1 and gs[i]=g_i such that f=prod_i gs[i]^i.
        The optional parameter squarefree inidicates whether f is known to be squarefree (usually one computes the squarefree factorization first).
    """
    if not squarefree:
        return naive_distinct_degree_factorization(f)
    d = f.degree()
    gs = [f.parent()(1) for i in range(d+1)]
    for i in range(1,d+1):
        if 2*i > f.degree():
            gs[f.degree()] = f
            break
        gs[i] = distinct_degree_poly(f,i)
        f = f // gs[i]
    return gs

def naive_distinct_degree_factorization(f):
    """
        Computed distinct degree factorization of arbitrary f by combining distinct degree factorization of polys in square free factorization.
        There is no reason to ever use this, it is provided just for the sake of completeness
    """
    d = f.degree()
    gs = [f.parent()(1) for i in range(d+1)]
    hs = [distinct_degree_factorization(g, squarefree=True) for g in squarefree_factorization(f)]
    # combine results (this is silly, we are undoing work here)
    for i in range(1,len(hs)):
        for j in range(1,len(hs[i])):
            gs[j] *= hs[i][j]^i
    return gs

def facpat(f):
    """Returns a dictionary {d:i} whose keys are the degrees of the irreducible factors of f and whose values are the number of irreducible factors of that degree."""
    X = {}

    gs = squarefree_factorization(f)
    for i in range(1,len(gs)):
        gis = distinct_degree_factorization(gs[i],squarefree=True)
        for j in range(1,len(gis)):
            if gis[j].degree() > 0:
                X[j] = X.get(j,0) + i * (gis[j].degree() // j)
    return X

In [0]:

%time facpat(f)

In [0]:

h = (x^9-1)^2*(x^32-1)
assert h == product(distinct_degree_factorization(h))
%time print(facpat(h))
print([(r[0].degree(),r[1]) for r in h.factor()])

The key innovation introduced by Cantor-Zassenhaus is that when searching for irreducible factors $g$ of $f$ of degree $j$ , rather than computing $\gcd(f,(x+\delta)^s-1))$ using a random $\delta\in \mathbb F_{q^j}$ and computing the gcd in $\mathbb F_{q^j}[x]$ (which is how one would go about finding a root of $g$ using Rabin's algorithm), it is better to compute $\gcd(f,u^s-1)$ in $\mathbb F_q[x]$ using a random $u\in \mathbb F_q[x]$ coprime to $f$ with $\deg u < \deg f$ . This faster because we are working the ring $\mathbb F_q[x]/(f)$ rather than $\mathbb F_{q^j}[x]/(f)$ (so the bit-size of each element is smaller by a factor of $j$ ), and the probability of success is the same.

Theorem:	Let $f\in \mathbb F_q[x]$ be the product of $r\ge 2$ distinct monic irreducible polynomials of degree $j$ and let $u\in \mathbb F_q[j]$ be a random polynomial coprime to $f$ with $\deg u < \deg f$ . Then $\Pr[0<\deg\gcd(f,u^s-1)<\deg f] = 2^{1-r} \ge \frac{1}{2}.$
Proof:	This follows from applying the Chinese Remainder Theorem to $\mathbb F_q[x]/(f)\simeq \mathbb F_{q^j}^r$ . See Section 3.12 of the lecture notes for details.

In [0]:

def equal_degree_factorization(f,j):
    """Completely factors a monic polynomial f in Fq[x] that is known to be the product of distinct irreducible polynomials of degree j"""
    d = f.degree()
    assert d % j == 0
    if d == 0:
        return []
    elif d == j:
        return [f]
    Fq = f.base_ring(); q = Fq.cardinality()
    assert q%2 == 1
    s = (q^j-1) // 2
    x = f.parent().gen()
    pif = f.parent().quotient(f)
    while True:
        # generate a random polynomial of degree < deg(f), note f.parent()(v) converts a vector of Fq coefficients to a polynomial
        u = f.parent()([Fq.random_element() for i in range(d)])
        h = gcd(f,u)
        if 0 < h.degree() < d:
            return equal_degree_factorization(h,j) + equal_degree_factorization(f//h,j)
        # at this point we know u is coprime to f
        h = gcd(f,(pif(u)^s-1).lift())
        if 0 < h.degree() < d:
            return equal_degree_factorization(h,j) + equal_degree_factorization(f//h,j)

def factorization(f):
    """Computes the complete factorization of a non-constant f in Fq[x] using the Cantor-Zassenhaaus algorithm (assumes q is odd)."""
    assert f.degree() > 0
    if not f.is_monic:
        f = f.monic()
    x = f.parent().gen()
    res = []
    gs = squarefree_factorization(f)
    for i in range(1,len(gs)):
        gis = distinct_degree_factorization(gs[i],squarefree=True)
        for j in range(1,len(gis)):
            res += [(g,i) for g in equal_degree_factorization(gis[j],j)]
    # sort output to make it canonical (independent of random choices)
    return sorted(res)

In [0]:

assert factorization(f) == sorted(list(f.factor()))
print(factorization(f))
%timeit -n 10 factorization(f)
%timeit -n 10 f.factor()

In [0]:

assert factorization(h) == sorted(list((h).factor()))
print(factorization(h))
%timeit -n 1 -r 5 factorization(h)
%timeit -n 1 -r 5 h.factor()

Theorem:

Let

n=\log q

and

d=\deg f

. The expected running time of

\texttt{factorization}

\tilde O(n^2d^2)

bit operations.

There are other factorization methods based on linear algebra and modular composition that improve the dependence on $d$ .
Currently the best asymptotic bound is achieved by the Kedlaya-Umans algorithm with an expected running time of $\tilde O(d^{3/2}n + dn^2)$ .

In [0]:

p=2^255-19;  F=GF(p); R.<x>=PolynomialRing(F);  f=x^8-2*x+5; h = product([x-i for i in range(1,20)])
assert factorization(f^3+f^2+1) == sorted(list((f^3+f^2+1).factor()))
%time print(factorization(f))

In [0]:

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