grhkm's blog

By grhkm, history, 4 weeks ago, In English

Lagrange Interpolation by grhkm

No polynomial stuff because I don't know $$$O(n\log{n})$$$ 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 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.

disclaimer
Why (n-1) degree?

Idea:

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:

$$$f(x)=(x-x_2)(x-x_3)\cdots (x-x_n)$$$

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.

$$$f_1(x)= \prod_{i=2}^n (x-x_i) \cdot \frac{y_1}{\prod_{i=2}^n (x_1-x_i)}=y_1\prod_{i=2}^n \frac{x-x_i}{x_1-x_i}$$$

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

$$$f_i(x) = y_i \prod_{j=1,j\neq i}^n \frac{x-x_i}{x_i-x_i}$$$

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:

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

$$$O(n^2)$$$ excluding inverse for division

Optimization:

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!

$$$f(x) = \sum_{i=1}^n f(i) \prod_{j=1,j\neq i}^n \frac{x-j}{i-j} = \sum_{i=1}^n f(i) \frac{\prod_{j=1,j\neq i}^n x-j}{\prod_{j=1,j\neq i}^n i-j}$$$

Let's consider the numerator and denominator separately

Numerator

$$$\prod_{j=1,j\neq i}^n x-j = [(x-1)(x-2)\cdots (x-(i-1))] \cdot [(x-(i+1))(x-(i+2))\cdots (x-n)]$$$

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

Denominator

$$$\prod_{j=1,j\neq i}^n i-j = [(i-1)(i-2)(i-3)\cdots(i-(i-1))] \cdot [(i-(i+1))(i-(i+2))\cdots (i-n)]$$$
$$$=(i-1)! \cdot (-1)^{n-(i+1)+1} \cdot (n-i)!$$$
$$$=(-1)^{n-i} (n-i)! (i-1)!$$$

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

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

Example 0

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 $$$deg(f)+1$$$ points)

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

Code can be found below

Example 1

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

Hint

Example 2 (ADVANCED)

(Luogu P4463 submit here): 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. MY code below

Hint
Hint2
Hint3
Hint4

Example 3 (ADVANCED TOO)

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

Hint

If you have any questions feel free to ask below

Practice question:

CF 622F (Example 0)

Luogu P4463 (Example 2)

Codechef XETF (Example 3)

Not in order of difficulty:

SPOJ ASUMEXTR

SPOJ ASUMHARD which is harder than EXTREME

Codechef COUNTIT

Atcoder ARC33D (Japanese)

Atcoder S8PC3G

  • Unfortunately you can't get full score, but you can get up to subtask 4

  • Still not trivial, do first 3 subtasks would give insight

  • For full solution please wait for my next blog (next year)

HDU 4059

SPOJ SOMESUMS very interesting

PE 487 Direct application

 
 
 
 
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4 weeks ago, # |
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Preview for next post that's coming next year or something:

Interpolate MULTIPLE values in O(nlog^2(n))

and more

Also if you know any more problems using this technique please tell me I'll put it in and credit you too thanks

EDIT 1: sorry if the 'and more' sounds depressing I don't have enough energy :(

EDIT 2: The O(n) optimization can also be done if the data points given are $$$f(a),f(a+1),\cdots,f(a+k-1)$$$ instead. Left as exercise for the reader :)

EDIT 3: After staring at the ceiling for a few seconds I believe that $$$O(n)$$$ is achievable for all arithmetic sequence input, as you can factor out the common difference and multiply/divide by the power of it directly. Left as an exercise for the reader :) I'll implement when I'm free

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    4 weeks ago, # ^ |
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    UPDATE: I believe for arithmetic sequence input, you can simply apply transformation g(x)=f(a+dx), then the input will be g(0), g(1), ..., g(k-1), which you can use the techniques in the blog to do. You can also directly find $$$f(x)=g(\frac{x-ad}{d})$$$ since you don't need to find the polynomial

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4 weeks ago, # |
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Great blog!

I learnt about this for the first time when I couldn't solve 622F.

https://codeforces.com/problemset/problem/622/F

This problem is exactly what is taught in this blog.

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    4 weeks ago, # ^ |
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    2600

    hmm

    Well I guess a lot of people will solve their first 2600 today :D

    Great that this blog helps

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4 weeks ago, # |
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Very Very Nice Blog.. Never thought of this much simple explanation of lagrange interpolation .

Thanks a lot .. I got this thing for the first time.

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    4 weeks ago, # ^ |
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    Glad that this is helpful

    (Tbh the explanation given is standard but I hope I explained it more detailedly and with clean code too :))

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      4 weeks ago, # ^ |
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      Yaa I know .. Even it seemed like very basic school maths to me , But the point is, u explained it using only basic terms without much technical terms .. and tbh i have always skipped this because of its name.

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4 weeks ago, # |
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Nice blog! but I have a question, on the idea part, shouldn't the degree of $$$f$$$ be $$$n - 1$$$? since you only had $$$n$$$ points at the start (from $$$(x_1,y_1),(x_2,y_2),...(x_n,y_n)$$$)

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    4 weeks ago, # ^ |
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    Oops, you're correct. n points -> (n-1) degree, n degree -> (n + 1) points required

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4 weeks ago, # |
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Just a note on spoj ASUMHARD. There are other methods with higher complexity that run faster than Lagrange interpolation for that particular problem (since k is small and number of test cases is large). For e.g. using a Newton divided difference table (O(k^2) pre-computation) with O(k) per test case runs super fast. See difference in run times here: https://www.spoj.com/status/ASUMHARD,suh_ash2008/ (only ID:26436058 uses Lagrange interpolation).

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    4 weeks ago, # ^ |
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    Interesting. Tbf I haven't Read them carefully so you're probably correct.

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      4 weeks ago, # ^ |
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      The complexity of the solution using Lagrange interpolation is actually O(k*log k) and not O(k) since the computation of f(0), f(1), f(2), ..., f(k) involves modular exponentiation. Even for the Lagrange interpolation, I've seen people do a modulo inverse (to divide by (x-i) instead of prefix and suffix products, which makes it O(k*log k). For ASUMHARD, let g[i][j] = i^j. We can write this as g[i][j] = g[i][j-1] * i and since k is small (321), this table can be pre-computed easily in O(k^2) and each test case can be answered in O(k) instead of O(k*log k). This seems to make a huge difference due to the large number of test cases. There are other methods (for e.g. http://oeis.org/A080779) which involve pre-computing the polynomial coefficients in O(k^2) or O(k^3) and answer each test case in O(k). :)

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        3 weeks ago, # ^ |
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        Shouldn't the k*log k be k*log p instead, if I understand properly? It is from exponentiation right?

        If so, I wrote a blog that explains how to optimise to O(k+log p) lol feel free to check it out

        Or maybe I'm misunderstanding what you mean

        Thank you for the information though :)

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          3 weeks ago, # ^ |
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          No, I did mean O(k*log k) due to modular exponentiation (and not due to inverse computation). In your Example 0, you mention "to find the data points, simply calculate f(0) = 0, f(x) = f(x-1) + x^k". This involves computation of x^k for x = 0 to k, each of which is O(log k). Hence, total complexity of computing f(0), f(1), ..., f(k) is O(k*log k) and not O(k).

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4 weeks ago, # |
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One more nice problem to try out if anyone wants to play around with their Lagrange interpolation code and try to optimise it: https://www.spoj.com/problems/PWSUMC/