Polynomial Multipoint Evaluation & Fast Interpolation

Polynomial Multipoint Evaluation

Description

Given a polynomial \(f\left(x\right)\) and \(n\) points \(x_{1},x_{2},\dots,x_{n}\), compute

\[ f\left(x_{1}\right),f\left(x_{2}\right),\dots,f\left(x_{n}\right) \]

Solution

Consider using divide and conquer to halve the problem size.

Divide the given points into two parts:

\[ \begin{aligned} X_{0}&=\left\{x_{1},x_{2},\dots,x_{\left\lfloor\frac{n}{2}\right\rfloor}\right\}\\ X_{1}&=\left\{x_{\left\lfloor\frac{n}{2}\right\rfloor+1},x_{\left\lfloor\frac{n}{2}\right\rfloor+2},\dots,x_{n}\right\} \end{aligned} \]

Construct the polynomial

\[ g_{0}\left(x\right)=\prod_{x_{i}\in X_{0}}\left(x-x_{i}\right) \]

Then \(\forall x\in X_{0}:g_{0}\left(x\right)=0\).

Consider expressing \(f\left(x\right)\) in the form \(g_{0}\left(x\right)Q\left(x\right)+f_{0}\left(x\right)\), i.e.:

\[ f_{0}\left(x\right)\equiv f\left(x\right)\pmod{g_{0}\left(x\right)} \]

Then \(\forall x\in X_{0}:f\left(x\right)=g_{0}\left(x\right)Q\left(x\right)+f_{0}\left(x\right)=f_{0}\left(x\right)\). The same applies to \(X_{1}\).

Thus, the problem size is halved, and can be solved using divide and conquer with polynomial modulo.

Time Complexity

\[ T\left(n\right)=2T\left(\frac{n}{2}\right)+O\left(n\log{n}\right)=O\left(n\log^{2}{n}\right) \]

Fast Polynomial Interpolation

Description

Given a set of \(n+1\) points

\[ X=\left\{\left(x_{0},y_{0}\right),\left(x_{1},y_{1}\right),\dots,\left(x_{n},y_{n}\right)\right\} \]

find an \(n\)-degree polynomial \(f\left(x\right)\) such that \(\forall\left(x,y\right)\in X:f\left(x\right)=y\).

Solution

Consider the Lagrange interpolation formula

\[ f(x) = \sum_{i=1}^{n} \prod_{j\neq i }\frac{x-x_j}{x_i-x_j} y_i \]

Let polynomial \(M(x) = \prod_{i=1}^n (x - x_i)\). By L'Hôpital's rule,

\[ \prod_{j\neq i} (x_i - x_j) = \lim_{x\rightarrow x_i} \frac{\prod_{j=1}^n (x - x_j)}{x - x_i} = M'(x_i) \]

Thus the polynomial can be expressed as

\[ f(x) = \sum_{i = 1}^n \frac{y_i}{M'(x_i)}\prod_{j \neq i}(x - x_j) \]

We first compute the coefficient representation of \(M(x)\) via divide and conquer. Then all \(M'(x_i)\) can be computed in \(O(n\log^2 n)\) time via multipoint evaluation.

Let \(v_i = \frac{y_i}{M'(x_i)}\). Next, consider computing \(f(x)\). For the case \(n = 1\), we have \(f(x) = v_1, M(x) = x - x_1\). Otherwise, let

\[ \begin{aligned} f_0(x) & = \sum_{i = 1}^{\left\lfloor \frac n2 \right \rfloor} v_i\prod_{j \neq i \wedge j \le \left\lfloor \frac n2 \right \rfloor}(x - x_j)\\ M_0(x) & = \prod_{i = 1}^{\left\lfloor \frac n2 \right \rfloor} (x - x_i)\\ f_1(x) & = \sum_{i = \left\lfloor \frac n2 \right \rfloor+1}^n v_i\prod_{j \neq i \wedge \left\lfloor \frac n2 \right \rfloor < j \le n}(x - x_j) \\ M_1(x) & = \prod_{i = \left\lfloor \frac n2 \right \rfloor+1}^n (x - x_i) \end{aligned} \]

We obtain \(f(x) = f_0(x)M_1(x) + f_1(x)M_0(x), M(x) = M_0(x)M_1(x)\). Therefore, it can be computed via divide and conquer, and this part also has complexity \(O(n\log^2 n)\).

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