LAGRANGE INTERPOLATION by grhkm

No polynomial stuff because I don't know nlogn polynomial multiplication :(

MAIN RESULTS:

Given $$$f(x_1)=y_1, f(x_2)=y_2, ..., f(x_n)=y_n$$$ we can calculate f(X) where f is the unique (n+1)-degree polynomial

For general {$$$x_i$$$} we can do so in O(n^2)

For consecutive {$$$x_i$$$} i.e. $$$x_j=x_1+(j-1)$$$, we can do so in O(n) excluding binary exponentiation

WHY IS THIS USEFUL:

Calculate $$$1^n+2^n+...+X^n$$$ in O(n)

DP optimization

With polynomials you can do cool things but not gonna cover this blog

Let's get started :)

Assume all polynomials below are (n+1) degree unless specified.

To start let's deal with first problem:

Given $$$f(x_1)=y_1, f(x_2)=y_2, ..., f(x_n)=y_n$$$ and an integer X we can calculate f(X) where f is the unique (n+1)-degree polynomial

Let's consider an easier version:

Find polynomial satisfying $$$f(x_1)=y_i, f(x_2)=0, ..., f(x_n)=0$$$

As we learn in school, $$$f(r)=0 \implies (x-r)$$$ is divisor

We can write this down:

Now notice that $$$f(x_1)=\prod_{i=2}^n (x_1-x_i)$$$ (a constant), but we want it to be $$$y_1$$$

We can simply scale the polynomial by a correct factor i.e.

Similarly, we can find a similar polynomial such that $$$f_i(x_i)=y_i$$$ and $$$f_i(x_j)=0 \forall j \neq i$$$

Okay now finally to answer the original question, we can notice that $$$f(x)=\sum_{i=1}^n f_i(x)$$$ satisfies the constraints. So there we have it! Lagrange interpolation:

O(n^2) excluding inverse for division

Let's try to optimise it now! Assume that we're given $$$f(1), f(2), \cdots, f(n)$$$, how can we calculate f(X) quickly?

Notice that here, $$$x_i=i$$$ and $$$y_i=f(i)$$$. Substitute!

Let's consider the numerator and denominator separately

Numerator

We can pre-calculate prefix and suffix product of x, then we can calculate the numerator (for each $$$i$$$) in O(1)

Denominator

So we can precompute factorials (preferably their inverse directly) then O(1) calculation

So now we can calculate $$$f(X)$$$ in O(n)

Example problem: Find $$$\sum_{i=1}^N i^k$$$, $$$k\leq 1e6, N\leq 1e12$$$

Solution: Notice that the required sum will be a degree (k+1) polynomial, and thus we can interpolate the answer with (k+2) data points (Remember, we need deg+1 points)

To find the data points simply calculate $$$f(0)=0, f(x)=f(x-1)+x^k$$$

Here is my code: here

Example problem (BZOJ 3453): Find $$$\sum_{i=0}^n \sum_{j=1}^{a+id} \sum_{k=1}^j k^x \mod 1234567891, x \leq 123, a, n, d \leq 123456789$$$

ADVANCED EXAMPLE (Luogu P4463): Given $$$n \leq 500, k \leq 1e9$$$, we call a sequence $$${a_n}$$$ nice if $$$1\leq a_i \leq k \forall i$$$ and $$$(a_i\neq a_j) \forall i\neq j$$$. Find the sum of (products of elements in the sequence) over all nice sequences.

Problem https://www.codechef.com/problems/XETF : Find $$$\sum_{i=1,gcd(i,N)=1}^N i^k, N \leq 1e12, k \leq 256$$$

If you have any questions feel free to ask below