# Polynomial Algorithms

** Published:**

This post introduces some tricks on polynomials widely used in ICPC. I will try to practice algebraic knowledge as well.

# Polynomial Inverse

**Prop**: Let $R$ be a commutative ring and $I$ be an ideal. Then $[p ]_I$ is a unit in $R/I$ if and only if $\langle p\rangle$ is coprime with $I$.

*Proof*: $1\in Rp+I$ is equivalent to $\langle p\rangle+I = R$.

**Cor**: For $R$ UFD, $f\in R[ X]/\langle X^n\rangle$ is a unit if and only if $f(0)$ is a unit in $R$.

*Proof*: $R$ being a domain implies $X^a$ ($a< n$) are the only factors of $X^n$. Suppose $f(0)$ is a unit, then $f$ is coprime with $X^n$ in $K[ X]$, where $K$ is the fractional field of $R$. Suppose $fg = 1$ in $K[ X]$. Since the content of $f$ is associated to $1$, by Gauß’s lemma, $C(f\hat{g})$ is also associated to $1$. Since $f\hat{g} = h\cdot X^n + f(0)\hat{g}(0)$, $[ f(0)\hat{g}(0)]^{-1}\hat{g}$ is the inverse of $f$ modulo $X^n$. The “only if” part is trivial.

**Prop**: If $fg = h\cdot X^n + 1$, then $(2-fg)g\cdot f = -h^2\cdot X^{2n} +1$.

This proposition gives a binary lifting algorithm to calculate $f^{-1}$ modulo $X^n$, in $O(n\log n)$.

# Newton’s Method

Let $R$ be a UFD and $f\in R[[ X]]$ be a power series. Suppose $f(a)\in Rp$ for some $a,p\in R$, i.e. $a$ is a root to $f$ modulo $p$. Then, consider $x = a+rp$. We have $f(a) + f’(a)rp \in Rp^2$. Since $p\mid f(a)$, if $f’(a)$ is invertible in $R/p^2R$, we have $p\mid f(a)f’(a)^{-1}$. Thus, $x = a - f(a)f’(a)^{-1}$ is a root to $f$ modulo $p^2$ if $f’(a)$ is invertible.

Now, substitute $R$ with $R[ X]$ and $p$ with $X^m$. Consider $g\in R[ X, Y]$, but it can have infinite terms on $Y$. If there is some $a\in R$ s.t. $g(0, a) = 0, \frac{\partial g}{\partial Y}(0, a)\in R^{\times}$, then we can calculate the “root” of $g(X, f(X))$ in $R[ X]/\langle X^n\rangle$ for any $n$:

- Let $f_0(X) = a$ and $n_0 = 1$
- Let $f_i(X) = a - g(X, f_{i-1}) \left(\frac{\partial g}{\partial Y}(0, f_{i-1})\right)^{-1}$ and $n_i = 2n_{i-1}$.
- Note that $X\mid g(X, f_{i-1})$. Thus, if the first derivative is invertible, all derivatives are invertible.

## Examples

- Inverse $h(X)$: let $g(X, Y) = \frac{1}{Y} - h(X)$. Get $f_{i+1} = 2f_{i} - f_i^2h$.
- Condition: $h(0)$ is invertible.

- Sqrt of $h(X)$: let $g(X, Y) = Y^2 - h(X)$. Get $f_{i+1} = \frac{1}{2}(f_i^2+h)f_i^{-1}$.
- Condition: there exists $a$ s.t. $a^2 = h(0)$ and $2a\in R^{\times}$.
- For exmaple, let $R = \mathbb{Z}/9\mathbb{Z}$ and $h(X) = X+1$. Starting from $f_0 = 1$ and $2\times 5 = 1$, we can get $f_2(X)= -5X^3+X^2+5X+1$ is the solution modulo $X^4$.
- Well, $R$ is not a domain, but it works.

- Exp of $h(X)$: let $g(X, Y) = \ln Y - h(X)$. Get $f_{i+1} = f_i(X)(1 - \ln {f_i(X)} + h(X))$.
- Log can be calculated by integral on $\frac{f’(X)}{f(X)}$.
- Condition: $h(0)=0$

# Polynomial Modulus

Let $K$ be a field. For $f\in K[ X]$ of degree $d$, let $f^T(X) := X^df(1/X)$ be the reverse of coefficients of $f$. Suppose $f = gq + r$, where $f, g$ are of degree $n, m$ resp. Then, $f^T(X) = g^T(X)q^T(X) + X^nr(1/X)$. Since $r$ is of degree at most $m-1$, $X^{n-m+1}\mid X^nr(1/X)$. Also, $q(X)$ is of degree $n-m < n-m+1$. Thus, $q = [ f^T\cdot (g^T)^{-1}]^T (\text{mod }X^{n-m+1})$ and $r = f - gq$.

# FFT

Let $A$ be an $R$-algebra, and $g\in A^{\times}$ of order $n$. Suppose $\sum_{i=0}^{n-1}g^{ik} = 0$ for all $1\leq k < n$ (generally true if $g^k-1$ is not a zero divisor). For a sequence $a = (a_0, \ldots, a_{n-1})\in R^n$, its discrete Fourier transform is $F(a) = (F_0(a), \ldots, F_{n-1}(a))\in A^n$ where

\[F_k(a) = \sum_{i=0}^{n-1}a_ig^{ik}\]If $n$ is invertible in $A$, the inverse of DFT is

\[a_i = \frac{1}{n}\sum_{k=0}^{n-1} F_k(a)g^{-ik}\]This is given by the zero sum condition.

The DFT of the cyclic convolution $a * b$ is the pointwise product of their DFT:

\[\begin{eqnarray} F_k(a*b) &=& F_k(a)F_k(b) \nonumber\\ (a*b)_i &=& \sum_{j=0}^{n-1} a_j b_{(i-j)\mod n} \end{eqnarray}\]Thus, DFT can be used to compute convolution. However, directly computing the sum does not reduce the time complexity. Now we try to accelerate this. Suppose $n = 2^m$.

\[\begin{eqnarray} F_k(a) &=& \sum_{i=0}^{2^m-1}a_ig^{ik} \nonumber\\ &=& \left(\sum_{i=0}^{2^{m-1}-1}a_{2i}g^{i\cdot 2k}\right) + g^k\left(\sum_{i=0}^{2^{m-1}-1}a_{2i+1}g^{i\cdot 2k}\right) \nonumber\\ &=& F_k'(a_{even}) + F_k'(a_{odd}) \end{eqnarray}\]where $F_k’$ is the DFT of length $n/2$ using $g^2$ as the primitive root instead of $g$. Repeating this, we get $O(n\log n)$ fast Fourier transform (FFT) algorithm.

## Common settings

- $R=\mathbb{Z}$, $A=\mathbb{C}$, $g = e^{\frac{2\pi i}{n}}$
- $R=A=\mathbb{Z}/p$, for some prime number $p = c\cdot 2^m + 1$.
- $p = 0\text{x}\text{C}0000001$, $g = 5$, $m = 30$
- $p = 0\text{x}78000001$, $g = 22$, $m = 27$
- $p = 0\text{x}3\text{B}800001$, $g = 3$, $m = 23$
- $p = 0\text{x}0\text{A}000001$, $g = 3$, $m = 25$

# Lagrange polynomial

Let $K$ be a field and $f\in K[ X]$ is of degree $d$. If we know its value on $d+1$ different points $x_0, \ldots, x_d\in K$ and $f(x_0), \ldots, f(x_d)\in K$, then we can recover the coefficients of $f$.

Let $l_i\in K[ X]$ be defined as:

\[l_i(X) := \prod_{j\neq i}\frac{X - x_j}{x_i - x_j}\]Then, $l_i$ is $1$ at $x_i$ and $0$ at other points $x_j\neq x_i$. Let $l := \sum_{i=0}^{d} y_il_i$, then $f = l$. Because otherwise, $f-l$ will be a polynomial of degree $d$ but have $d+1$ disjoint roots.

To evaluate $f(x)$, we can directly substitute $x$ into $l_i$, and get an $O(d^2)$ algorithm.

## Example: sum of powers

Let $f_k(n) = \sum_{i=1}^{n} i^k$. By induction we can prove $f_k$ is a polynomial of degree $k+1$. To calculate $f_k(n)$, it suffices to compute $f_k(0), \ldots, f_k(k+1)$ and then interpolate.

# Multipoint Evaluation

Let $R$ be an commutative ring, $f\in R[ X]$ be a polynomial with degree less than $n$, $x_1,\ldots, x_n\in R$ be points. Compute $f(x_1), \ldots, f(x_n)$.

Let $p_0(X) = \prod_{i=1}^{n/2} (X - x_i)$ and $p_1(X) = \prod_{i=n/2+1}^{n} (X - x_i)$. Define $r_0(X) = f\mod p_0$ and $r_1(X) = f\mod p_1$. Then it suffices to evaluate $r_0$ on $x_1,\ldots, x_{n/2}$ and $r_1$ on $x_{n/2+1},\ldots, x_{n}$. This divide-and-conquer algorithm works in $O(n\log^2 n)$.

In special case there are algorithms that run faster, e.g. $O(n\log n)$, but need more memory.

## Example: factorial

Let $f(X) := \prod_{i=1}^{n}(X+i)$. Then, $n! = f(0)$. Let $g(X) := \prod_{i=1}^{v}(X+i)$, where $v = \lfloor\sqrt{n}\rfloor$. Then,

\[n! = \left( \prod_{i=0}^{v-1}g(vi) \right)\cdot \prod_{i=v^2+1}^{n}i\]Thus, it is enough to calculate $g(X)$ on $0, v, \ldots, v(v-1)$.

# Recursive Sequence

Suppose ${a_i}_ {\infty}$ is a sequence s.t. $a_i = \sum_{j=1}^{d} c_ja_{i-j}$. Then, the array is decided by its first $n$ elements.

## Matrix Exponentiation

Clearly, we have

\[\begin{pmatrix} a_{n+d-1} \\ a_{n+d-2} \\ a_{n+d-3} \\ \vdots \\ a_n \end{pmatrix} = \begin{pmatrix} c_1 & c_2 & \cdots & c_{d-1} & c_d \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{pmatrix}^n \cdot \begin{pmatrix} a_{d-1} \\ a_{d-2} \\ a_{d-3} \\ \vdots \\ a_0 \end{pmatrix}\]Thus, we can let $M$ denote the recurrenc matrix and calculate $M^n$ in $O(d^3\log n)$:

- $M^{2m} = (M^m)^2$
- $M^{2m+1} = (M^m)^2\cdot M$

## Polynomial Modulus

By Hamilton-Cayley theorem, the characteristic polynomial of $M$ is an annihilator of $M$. Let $p(X) := X^d - \sum_{i=1}^{d}c_iX^{d-i}$ and $r_n(X) := X^n\mod p$. Then, $A^n = r_n(A)$. Thus, if $r_n(X) = \sum_{i=0}^{d-1} s_iX^i$, then $a_n = \sum_{i=0}^{d-1} s_ia_i$.

However, there is a much cleaner way to view this. Recall that given $c\in R$, we have $R[ X]/\langle X-c \rangle\cong R$ via $X\mapsto c$. This somehow gives an solution for $a_i = ca_{i-1}$ as $a_i = c^i$. We want to generalize this, but $a_1 \times a_1 \neq a_2$. This inspires us that we should **forget the multiplication**.

Consider $R[ X]$ as a abelian group. Then, define $f: R[ X]\to R$ as $f(uX^i) = u\cdot a_i$ for all $u\in R$ and $i\in\mathbb{N}$. By the distributivity of $R$, $f$ is a well-defined *group* homomorphism. Let $I = \langle p \rangle$ be the ideal generated by $p$. We claim that $I$ is in the kernel of $f$. Since $f$ preserves the addition, so it suffices to show that $f(uX^ip) = 0$ for all $u\in R$ and $i\in\mathbb{N}$. This is clearly true since

Observe that now $R$ becomes a cocone of the diagram $1\leftarrow I \hookrightarrow R$. And $R/I$ is exactly the coproduct of this diagram, with canonical map $\pi: R\to R/I$ via $f\mapsto (f\mod p)$. By the universality, $f = f\restriction_{R/I}\circ \pi$. Hence, $a_n = f(X^n\mod p) = \sum_{i=0}^{d-1} s_ia_i$, with $s_i$ defined above.

## Berlekamp-Massey Algorithm

This is the inverse problem: given a recurrence sequence $\{a_i\}_ {\infty}$ (first $2d$ elements are enough), can we compute its recurrence relation $a_i = \sum_{j=1}^{d} c_ja_{i-j}$?

Again we use a polynomial to represent the recurrence relation. Suppose $p_i\in R[ X]$ s.t. $p_i$ works for $a_j$, $0\neq j < i$. Formally, this means $f(X^j\mod p_i) = a_j$ and $f(X^{j-\deg p_i}p_i) = 0$. Let the error

\[e_i = a_i + \sum_{j=1}^{\deg p_i} [ X^{\deg p_i-j}]p_i\cdot a_{i-j} = f(X^{i-\deg p_i}p_i)\]- If $e_i = 0$, we can keep it $p_{i+1} = p_i$.
- If $e_i \neq 0$, we need to fix it.
- Suppose $p_k\neq p_i$ is
*some*polynomial we have different from $p_i$. Let $p_{i+1} = p_i + \frac{e_i}{e_k}p_kX^{i-k}$.

- Suppose $p_k\neq p_i$ is

Clearly, $f(X^{i-\deg p_{i+1}} p_{i+1})=0$, so $p_{i+1}$ works for $a_{i}$. One can verify it works for all $a_j$, $0\leq j\leq i$.

Now we want to minimize the degree of final $p$. The flexibility we have is $p_k$ when fixing the error. Massey argues that we can use the last $p_k$ s.t. $\deg p_{k+1} > \deg p_k$.

# References

- OI Wiki. Polynomials.
- Wikipedia. Lagrange polynomial.
- Factorial modulo Prime Numbers.
- Chandan Saha. Computational Number Theory and Algebra. Lecture 6.