使用内置 Numpy.fft 在 Python 中进行 FFT 多项式乘法

Question

I want to multiply two polynomials fast in python. As my polynomials are rather large (> 100000) elements and I have to multiply lots of them. Below, you will find my approach,

from numpy.random import seed, randint
from numpy import polymul, pad 
from numpy.fft import fft, ifft
from timeit import default_timer as timer

length=100

def test_mul(arr_a,arr_b):  #inbuilt python multiplication

    c=polymul(arr_a,arr_b)

    return c    

def sb_mul(arr_a,arr_b):    #my schoolbook multiplication

    c=[0]*(len(arr_a) + len(arr_b) - 1 )
    for i in range( len(arr_a) ):
        for j in range( len(arr_b) ):
            k=i+j
            c[k]=c[k]+arr_a[i]*arr_b[j]
    return c    


def fft_test(arr_a,arr_b):  #fft based polynomial multuplication

    arr_a1=pad(arr_a,(0,length),'constant')
    arr_b1=pad(arr_b,(0,length),'constant')
    a_f=fft(arr_a1)
    b_f=fft(arr_b1)

    c_f=[0]*(2*length)

    for i in range( len(a_f) ):
        c_f[i]=a_f[i]*b_f[i]

    return c_f


if __name__ == '__main__':

    seed(int(timer()))
    random=1
    if(random==1):
        x=randint(1,1000,length)
        y=randint(1,1000,length)
    else:
        x=[1]*length
        y=[1]*length

    start=timer()
    res=test_mul(x,y)
    end=timer()
    print("time for built in pol_mul", end-start)

    start=timer()
    res1=sb_mul(x,y)
    end=timer()
    print("time for schoolbook mult", end-start)

    res2=fft_test(x,y)

    print(res2)

    #########check############
    if( len(res)!=len(res1) ):
        print("ERROR");

    for i in range( len(res) ):
        if( res[i]!=res1[i] ):
            print("ERROR at pos ",i,"res[i]:",res[i],"res1[i]:",res1[i])

Now, here are my approach in detail, 1. First, I tried myself with a naive implementation of Schoolbook with complexity O(n^2). But as you may expect it turned out to be very slow.

2. Second, I came to know polymul in the Numpy library. This function is a lot faster than the previous one. But I realized this is also a O(n^2) complexity. You can see, if you increase the length k the time increases by k^2 times.

3. My third approach is to try a FFT based multiplication using the inbuilt FFT functions. I followed the the well known approach also described here but Iam not able to get it work.

Now my questions are,

1. Where am I going wrong in my FFT based approach? Can you please tell me how can I fix it?

2. Is my observation that polymul function has O(n^2) complexity correct?

Please, let me know if you have any question. Thanks in advance.

Answer 1

1. Where am I going wrong in my FFT based approach? Can you please tell me how can I fix it?

The main problem is that in the FFT based approach, you should be taking the inverse transform after the multiplication, but that step is missing from your code. With this missing step your code should look like the following:

def fft_test(arr_a,arr_b):  #fft based polynomial multiplication

    arr_a1=pad(arr_a,(0,length),'constant')
    arr_b1=pad(arr_b,(0,length),'constant')
    a_f=fft(arr_a1)
    b_f=fft(arr_b1)

    c_f=[0]*(2*length)

    for i in range( len(a_f) ):
        c_f[i]=a_f[i]*b_f[i]

    return ifft(c_f)

Note that there may also a few opportunities for improvements:

• The zero padding can be handled directly by passing the required FFT length as the second argument (eg a_f = fft(arr_a, length) )
• The coefficient multiplication in your for loop may be directly handled by numpy.multiply .
• If the polynomial coefficients are real-valued, then you can use numpy.fft.rfft and numpy.fft.irfft (instead of numpy.fft.fft and numpy.fft.ifft ) for some extra performance boost.

So an implementation for real-valued inputs may look like:

from numpy.fft import rfft, irfft
def fftrealpolymul(arr_a, arr_b):  #fft based real-valued polynomial multiplication
    L = len(arr_a) + len(arr_b)
    a_f = rfft(arr_a, L)
    b_f = rfft(arr_b, L)
    return irfft(a_f * b_f)

2. Is my observation that polymul function has O(n ² ) complexity correct?

That also seem to be the performance I am observing, and matches the available code in my numpy installation (version 1.15.4, and there doesn't seem any change in that part in the more recent 1.16.1 version).