Berlekamp–Rabin algorithm

In number theory, Berlekamp's root finding algorithm (also Berlekamp-Rabin algorithm) is the probabilistic method of finding roots of polynomials over field $\mathbb {Z} _{p}$. The method was discovered by Berlekamp in 1970 as an auxiliary to the algorithm for polynomial factorization over finite field. The algorithm was later modified by Rabin for arbitrary finite field in 1979. The method was also independently discovered before Berlekamp by some other researchers.

History

The method was proposed by Elwyn Berlekamp in his work on polynomial factorization over finite fields. Original work lacked formal correctness proof and was later refined and modified for arbitrary finite fields by Michael Rabin. In 1986 René Peralta proposed similar algorithm for finding square roots in $\mathbb {Z} _{p}$. In 2000 Peralta's method was generalized for cubic equations.

Statement

Let $p$ be the odd prime number. Consider polynomial ${\textstyle f(x)=a_{0}+a_{1}x+\dots +a_{n}x^{n}}$ over field $\mathbb {Z} _{p}$ of remainders modulo $p$. One have to find all $\lambda _{1},\dots ,\lambda _{k}$ such that ${\textstyle f(\lambda _{i})\equiv 0{\pmod {p}}}$ for every possible $i$.

Algorithm

Randomization

Let ${\textstyle f(x)=(x-\lambda _{1})(x-\lambda _{2})\dots (x-\lambda _{n})}$. Finding all roots of such polynomial is equivalent to finding its factorization into linear factors. To find such factorization it's sufficient to split polynomial into any two non-trivial divisors and factorize them recursively. To do this consider the polynomial ${\textstyle f_{z}(x)=f(x-z)=(x-\lambda _{1}-z)(x-\lambda _{2}-z)\dots (x-\lambda _{n}-z)}$ where $z$ is some random element of $\mathbb {Z} _{p}$. If one can represent this polynomial as the product $f_{z}(x)=p_{0}(x)p_{1}(x)$then in terms of initial polynomial it means that $f(x)=p_{0}(x+z)p_{1}(x+z)$ which provides needed factorization of $f(x)$.

Classification of $\mathbb {Z} _{p}$ elements

Due to Euler's criterion for every monomial $(x-\lambda )$ exactly one of following properties holds:

1. Monomial is equal to $x$ if $\lambda =0$,
2. Monomial divides ${\textstyle g_{0}(x)=(x^{(p-1)/2}-1)}$ if $\lambda$ is quadratic residue modulo $p$,
3. Monomial divides ${\textstyle g_{1}(x)=(x^{(p-1)/2}+1)}$ if $\lambda$ is quadratic non-residual modulo $p$.

Thus if $f_{z}(x)$ is not divisible by $x$, which may be checked separately, then $f_{z}(x)$ is equal to the product of greatest common divisors $\gcd(f_{z}(x);g_{0}(x))$ and $\gcd(f_{z}(x);g_{1}(x))$.

Berlekamp's method

Property written above leads to the following algorithm:

1. Explicitly calculate coefficients of $f_{z}(x)=f(x-z)$,
2. Calculate remainders of ${\textstyle x,x^{2},x^{2^{2}},x^{2^{3}},x^{2^{4}},\dots ,x^{2^{\lfloor \log _{2}p\rfloor }}}$ modulo $f_{z}(x)$ by consequently squaring current polynomial and taking remainder modulo $f_{z}(x)$,
3. Using exponentiation by squaring and polynomials calculated on the previous steps calculate the remainder of ${\textstyle x^{(p-1)/2}}$ modulo ${\textstyle f_{z}(x)}$,
4. If ${\textstyle x^{(p-1)/2}\not \equiv \pm 1{\pmod {f_{z}(x)}}}$ then $\gcd$ mentioned above provide non-trivial factorization of $f_{z}(x)$,
5. Otherwise all roots of $f_{z}(x)$ are either residues or non-residues simultaneously and one has to choose another $z$.

If $f(x)$ is divisible by some non-linear primitive polynomial $g(x)$ over $\mathbb {Z} _{p}$ then when calculating $\gcd$ with $g_{0}(x)$ and $g_{1}(x)$ one will obtain non-trivial factorization of $f_{z}(x)/g_{z}(x)$, thus algorithm allows to find all roots of arbitrary polynomials over $\mathbb {Z} _{p}$.

Modular square root

Consider equation ${\textstyle x^{2}\equiv a{\pmod {p}}}$ having elements $\beta$ and $-\beta$ as its roots. Solution of this equation is equivalent to factorization of polynomial ${\textstyle f(x)=x^{2}-a=(x-\beta )(x+\beta )}$ over $\mathbb {Z} _{p}$. In this particular case problem is a bit simpler because it is sufficient to calculate only $\gcd(f_{z}(x);g_{0}(x))$. For this polynomial exactly one of the following properties will hold:

1. GCD is equal to $1$ which means that $z+\beta$ and $z-\beta$ are both quadratic non-residues,
2. GCD is equal to $f_{z}(x)$which means that both numbers are quadratic residues,
3. GCD is equal to $(x-t)$which means that exactly one of these numbers is quadratic residue.

In the third case GCD is equal to either $(x-z-\beta )$ or $(x-z+\beta )$. It allows to write the solution as ${\textstyle \beta =(t-z){\pmod {p}}}$.

Example

Assume we need to solve the equation ${\textstyle x^{2}\equiv 5{\pmod {11}}}$. For this we need to factorize $f(x)=x^{2}-5=(x-\beta )(x+\beta )$. Consider some possible values of $z$:

1. Let $z=3$. Then $f_{z}(x)=(x-3)^{2}-5=x^{2}-6x+4$, thus $\gcd(x^{2}-6x+4;x^{5}-1)=1$. Both numbers $3\pm \beta$ are quadratic non-residues, so we need to take some other $z$.

1. Let $z=2$. Then $f_{z}(x)=(x-2)^{2}-5=x^{2}-4x-1$, thus ${\textstyle \gcd(x^{2}-4x-1;x^{5}-1)\equiv x-9{\pmod {11}}}$. From this follows ${\textstyle x-9=x-2-\beta }$, so $\beta \equiv 7{\pmod {11}}$ and ${\textstyle -\beta \equiv -7\equiv 4{\pmod {11}}}$.

Manual check shows that, indeed, ${\textstyle 7^{2}\equiv 49\equiv 5{\pmod {11}}}$ and ${\textstyle 4^{2}\equiv 16\equiv 5{\pmod {11}}}$.

Correctness proof

Algorithm finds factorization of $f_{z}(x)$ in all cases except for ones when all numbers $z+\lambda _{1},z+\lambda _{2},\dots ,z+\lambda _{n}$ are quadratic residues or non-residues simultaneously. According to theory of cyclotomy probability of such event for the case when $\lambda _{1},\dots ,\lambda _{n}$ are all residues or non-residues simultaneously (that is, when $z=0$ would fail) may be estimated as $2^{-k}$ where $k$ is the number of different values among $\lambda _{1},\dots ,\lambda _{n}$. In this way even for the worst case of $k=1$ and $f(x)=(x-\lambda )^{n}$ probability of error may be estimated as $1/2$ and for modular square root case error probability is at most $1/4$.

Complexity

Assume that polynomial's degree is $n$. Here's estimation of algorithm's steps complexity:

1. Due to binomial theorem ${\textstyle (x-z)^{k}=\sum \limits _{i=0}^{k}{\binom {k}{i}}(-z)^{k-i}x^{i}}$, thus transition from $f(x)$ to $f(x-z)$ may be done in $O(n^{2})$,
2. Polynomial multiplication and taking remainder of one polynomial modulo another one may be done in ${\textstyle O(n^{2})}$, thus calculation of ${\textstyle x^{2^{k}}{\bmod {f}}_{z}(x)}$ is done in ${\textstyle O(n^{2}\log p)}$,
3. Binary exponentiation works in $O(n^{2}\log p)$ as well,
4. Taking $\gcd$ of two polynomials via Euclidean algorithm works in $O(n^{2})$.

Thus the whole procedure may be done in $O(n^{2}\log p)$. Using fast Fourier transform and Half-GCD algorithm complexity may be improved to $O(n\log n\log pn)$. For modular square root case the degree $n$ is equal to $2$, thus the whole complexity of algorithm in such case may be estimated as $O(\log p)$ per iteration.