Tutorial: A simple O(n log n) polynomial multiplication algorithm

Revision en4, by pajenegod, 2023-07-07 01:35:15

Hi Codeforces!

I have recently come up with a really neat and simple recursive algorithm for multiplying polynomials in $O(n \log n)$ time. It is so neat and simple that I think it might possibly revolutionize the way that fast polynomial multiplication is taught and coded. You don't need to know anything about FFT to understand and implement this algorithm.

I've split this blog up into two parts. The first part is intended for anyone to be able to read and understand. The second part is advanced and goes into a ton of interesting ideas and concepts related to this algorithm.

Prerequisite: Polynomial quotient and remainder, see Wiki article and Stackexchange example.

Given two polynomials $P$ and $Q$ and an integer $n$, where degree $P < n$ and degree $Q < n$. Your task is to calculate the polynomial $P(x) \, Q(x) \% (x^n - c)$ in $O(n \log n)$ time. You may assume that $n$ is a power of two.

### Solution:

We can create a divide and conquer algorithm for $P(x) \, Q(x) \% (x^n - c)$ based on the difference of squares formula. Assuming $n$ is even, then $(x^n - c) = (x^{n/2} - \sqrt{c}) (x^{n/2} + \sqrt{c})$. The idea behind the algorithm is to calculate $P(x) \, Q(x) \% (x^{n/2} - \sqrt{c})$ and $P(x) \, Q(x) \% (x^{n/2} + \sqrt{c})$ using 2 recursive calls, and then use the result of that to calculate $P(x) \, Q(x) \% (x^n - c)$.

So how do we actually calculate $P(x) \, Q(x) \% (x^n - c)$ using $P(x) \, Q(x) \% (x^{n/2} - \sqrt{c})$ and $P(x) \, Q(x) \% (x^{n/2} + \sqrt{c})$?

Well, we can use the following formula:

\begin{aligned} A(x) \% (x^n - c) = &\frac{1}{2} (1 + \frac{x^{n/2}}{\sqrt{c}}) (A(x) \% (x^{n/2} - \sqrt{c})) \, + \\ &\frac{1}{2} (1 - \frac{x^{n/2}}{\sqrt{c}}) (A(x) \% (x^{n/2} + \sqrt{c})). \end{aligned}
Proof of the formula

This formula is very useful. If we substitute $A(x)$ by $P(x) Q(x)$, then the formula tells us how to calculate $P(x) \, Q(x) \% (x^n - c)$ using $P(x) \, Q(x) \% (x^{n/2} - \sqrt{c})$ and $P(x) \, Q(x) \% (x^{n/2} + \sqrt{c})$. Additionally, if we substite $A(x)$ with $P(x)$, then the formula tells us how to calculate $P(x) \% (x^{n/2} - \sqrt{c})$ and $P(x) \% (x^{n/2} + \sqrt{c})$ using $P(x) \% (x^n - c)$. With this we have entire recipie for implementing a $O(n \log n)$ divide and conquer algorithm.

One final thing that I want to mention before going into the advanced section is that this algorithm can also be used to do fast unmodded polynomial multiplication, i.e. given polynomials $P(x)$ and $Q(x)$ calculate $P(x) \, Q(x)$. The trick is simply to pick $n$ large enough such that $P(x) \, Q(x) = P(x) \, Q(x) \% (x^n - c)$, and then call the algorithm from before.

Python implementation for general Fast polynomial multiplication

### (Advanced) Speeding up the algorithm

This section will be about tricks that can be used to speed up the algorithm. This will speed it up by a factor of between 2 and 4.

$n$ doesn't actually need to be a power of 2
Imaginary-cyclic convolution
Trick to go from $\% (x^n - c)$ to $\% (x^n - 1)$

### (Advanced) Connection between the algorithm and FFT

This algorithm is actually FFT in disguise. But it is not the same algorithm as Cooley–Tukey (standard FFT algorithm).

Using this algorithm to calculate FFT
Destinct difference between this algorithm and Cooley–Tukey

### (Advanced) Connection between the algorithm and NTT

Using modulo integers instead of complex numbers
What if $sqrt(c)$ doesn't exist?

#### History

Revisions

Rev. Lang. By When Δ Comment
en53 pajenegod 2023-07-21 20:29:32 631
en52 pajenegod 2023-07-21 19:29:42 0 (published)
en51 pajenegod 2023-07-21 19:29:23 120 (saved to drafts)
en50 pajenegod 2023-07-10 22:53:22 1131
en49 pajenegod 2023-07-10 20:27:39 204
en48 pajenegod 2023-07-10 20:18:13 0 (published)
en47 pajenegod 2023-07-10 20:14:48 435
en46 pajenegod 2023-07-10 19:55:32 288
en45 pajenegod 2023-07-10 19:51:08 13
en44 pajenegod 2023-07-10 19:48:33 7
en43 pajenegod 2023-07-10 19:41:29 1385
en42 pajenegod 2023-07-10 03:42:37 2274
en41 pajenegod 2023-07-09 22:40:43 571
en40 pajenegod 2023-07-09 22:00:53 39
en39 pajenegod 2023-07-09 21:57:38 1100
en38 pajenegod 2023-07-09 20:46:05 384
en37 pajenegod 2023-07-09 20:03:42 39
en36 pajenegod 2023-07-09 19:36:59 3745
en35 pajenegod 2023-07-09 17:00:26 121
en34 pajenegod 2023-07-09 16:53:59 2297
en33 pajenegod 2023-07-08 22:04:53 23
en32 pajenegod 2023-07-08 15:24:23 3
en31 pajenegod 2023-07-08 15:22:29 24
en30 pajenegod 2023-07-08 15:19:05 192
en29 pajenegod 2023-07-08 15:03:43 438
en28 pajenegod 2023-07-08 14:50:40 16
en27 pajenegod 2023-07-08 14:48:51 278
en26 pajenegod 2023-07-08 14:41:17 5
en25 pajenegod 2023-07-08 14:40:38 7540
en24 pajenegod 2023-07-08 13:38:20 1
en23 pajenegod 2023-07-08 13:27:45 24
en22 pajenegod 2023-07-08 13:26:39 844
en21 pajenegod 2023-07-08 13:08:19 1427
en20 pajenegod 2023-07-08 12:36:17 492
en19 pajenegod 2023-07-08 04:13:26 5
en18 pajenegod 2023-07-08 04:10:18 30
en17 pajenegod 2023-07-08 03:59:49 999
en16 pajenegod 2023-07-08 03:42:36 977
en15 pajenegod 2023-07-08 03:27:23 4195
en14 pajenegod 2023-07-07 20:07:54 1204
en13 pajenegod 2023-07-07 18:13:08 2
en12 pajenegod 2023-07-07 18:08:18 39
en11 pajenegod 2023-07-07 13:51:58 89
en10 pajenegod 2023-07-07 13:41:19 789
en9 pajenegod 2023-07-07 06:29:01 11
en8 pajenegod 2023-07-07 02:22:13 4
en7 pajenegod 2023-07-07 02:02:33 122 ffao,-is-this-fft-,meooow,ToxicPie9,algmyr,aryanc403,kostia244,nor,Spheniscine,magnus.hegdahl,jeroenodb
en6 pajenegod 2023-07-07 01:52:20 119
en5 pajenegod 2023-07-07 01:46:09 448
en4 pajenegod 2023-07-07 01:35:15 2157
en3 pajenegod 2023-07-07 00:53:13 1957
en2 pajenegod 2023-07-07 00:15:55 2037
en1 pajenegod 2023-07-06 23:46:15 3121 Initial revision (saved to drafts)