 GeeksforGeeks App
Open App Browser
Continue

# Fast Fourier Transformation for polynomial multiplication

Given two polynomial A(x) and B(x), find the product C(x) = A(x)*B(x). There is already an O( ) naive approach to solve this problem. here. This approach uses the coefficient form of the polynomial to calculate the product.
A coefficient representation of a polynomial is a = a0, a1, …, an-1.

Example-  Coefficient representation of A(x) = (9, -10, 7, 6)
Coefficient representation of B(x) = (-5, 4, 0, -2)

Input :
A[] = {9, -10, 7, 6}
B[] = {-5, 4, 0, -2}
Output : We can do better, if we represent the polynomial in another form.

yes

Idea is to represent polynomial in point-value form and then compute the product. A point-value representation of a polynomial A(x) of degree-bound n is a set of n point-value pairs is{ (x0, y0), (x1, y1), …, (xn-1, yn-1)} such that all of the xi are distinct and yi = A(xi) for i = 0, 1, …, n-1.

Example xi    -- 0, 1, 2, 3 A(xi) -- 1, 0, 5, 22

Point-value representation of above polynomial is { (0, 1), (1, 0), (2, 5), (3, 22) }. Using Horner’s method, (discussed here), n-point evaluation takes time O( ). It’s just calculation of values of A(x) at some x for n different points, so time complexity is O( ). Now that the polynomial is converted into point value, it can be easily calculated C(x) = A(x)*B(x) again using horner’s method. This takes O(n) time. An important point here is C(x) has degree bound 2n, then n points will give only n points of C(x), so for that case we need 2n different values of x to calculate 2n different values of y. Now that the product is calculated, the answer can to be converted back into coefficient vector form. To get back to coefficient vector form we use inverse of this evaluation. The inverse of evaluation is called interpolation. Interpolation using Lagrange’s formula gives point value-form to coefficient vector form of the polynomial.Lagrange’s formula is – So far we discussed, .
This idea still solves the problem in O( ) time complexity. We can use any points we want as evaluation points, but by choosing the evaluation points carefully, we can convert between representations in only O(n log n) time. If we choose “complex roots of unity” as the evaluation points, we can produce a point-value representation by taking the discrete Fourier transform (DFT) of a coefficient vector. We can perform the inverse operation, interpolation, by taking the “inverse DFT” of point-value pairs, yielding a coefficient vector. Fast Fourier Transform (FFT) can perform DFT and inverse DFT in time O(nlogn).

DFT
DFT is evaluating values of polynomial at n complex nth roots of unity . So, for k = 0, 1, 2, …, n-1, y = (y0, y1, y2, …, yn-1) is Discrete fourier Transformation (DFT) of given polynomial.
The product of two polynomials of degree-bound n is a polynomial of degree-bound 2n. Before evaluating the input polynomials A and B, therefore, we first double their degree-bounds to 2n by adding n high-order coefficients of 0. Because the vectors have 2n elements, we use “complex 2nth roots of unity, ” which are denoted by the W2n (omega 2n). We assume that n is a power of 2; we can always meet this requirement by adding high-order zero coefficients.

FFT
Here is the Divide-and-conquer strategy to solve this problem.
Define two new polynomials of degree-bound n/2, using even-index and odd-index coefficients of A(x) separately   The problem of evaluating A(x) at reduces to evaluating the degree-bound n/2 polynomials A0(x) and A1(x) at the points Now combining the results by The list does not contain n distinct values, but n/2 complex n/2th roots of unity. Polynomials A0 and A1 are recursively evaluated at the n complex nth roots of unity. Subproblems have exactly the same form as the original problem, but are half the size. So recurrence formed is T(n) = 2T(n/2) + O(n), i.e complexity O(nlogn).

Algorithm
1. Add n higher-order zero coefficients to A(x) and B(x)
2. Evaluate A(x) and B(x) using FFT for 2n points
3. Pointwise multiplication of point-value forms
4. Interpolate C(x) using FFT to compute inverse DFT

Pseudo code of recursive FFT

Recursive_FFT(a){
n = length(a) // a is the input coefficient vector
if n = 1
then return a

// wn is principle complex nth root of unity.
wn = e^(2*pi*i/n)
w = 1

// even indexed coefficients
A0 = (a0, a2, ..., an-2 )

// odd indexed coefficients
A1 = (a1, a3, ..., an-1 )

y0 = Recursive_FFT(A0) // local array
y1 = Recursive-FFT(A1) // local array

for k = 0 to n/2 - 1

// y array stores values of the DFT
// of given polynomial.
do y[k] = y0[k] + w*y1[k]
y[k+(n/2)] = y0[k] - w*y1[k]
w = w*wn
return y
}
Recursion Tree of Above Execution- Why does this work? since, Thus, the vector y returned by Recursive-FFT is indeed the DFT of the input
vector a.

## C++

 #include using namespace std; // For storing complex values of nth roots// of unity we use complextypedef complex<double> cd; // Recursive function of FFTvector fft(vector& a){    int n = a.size();     // if input contains just one element    if (n == 1)        return vector(1, a);     // For storing n complex nth roots of unity    vector w(n);    for (int i = 0; i < n; i++) {        double alpha = -2 * M_PI * i / n;        w[i] = cd(cos(alpha), sin(alpha));    }     vector A0(n / 2), A1(n / 2);    for (int i = 0; i < n / 2; i++) {         // even indexed coefficients        A0[i] = a[i * 2];         // odd indexed coefficients        A1[i] = a[i * 2 + 1];    }     // Recursive call for even indexed coefficients    vector y0 = fft(A0);     // Recursive call for odd indexed coefficients    vector y1 = fft(A1);     // for storing values of y0, y1, y2, ..., yn-1.    vector y(n);     for (int k = 0; k < n / 2; k++) {        y[k] = y0[k] + w[k] * y1[k];        y[k + n / 2] = y0[k] - w[k] * y1[k];    }    return y;} // Driver codeint main(){    vector a{1, 2, 3, 4};    vector b = fft(a);    for (int i = 0; i < 4; i++)        cout << b[i] << endl;     return 0;}

## Python3

 from math import sin,cos,pi # Recursive function of FFTdef fft(a):     n = len(a)     # if input contains just one element    if n == 1:        return [a]     # For storing n complex nth roots of unity    theta = -2*pi/n    w = list( complex(cos(theta*i), sin(theta*i)) for i in range(n) )         # Separe coefficients    Aeven = a[0::2]    Aodd  = a[1::2]     # Recursive call for even indexed coefficients    Yeven = fft(Aeven)     # Recursive call for odd indexed coefficients    Yodd = fft(Aodd)     # for storing values of y0, y1, y2, ..., yn-1.    Y = *n        middle = n//2    for k in range(n//2):        w_yodd_k  = w[k] * Yodd[k]        yeven_k   =  Yeven[k]                 Y[k]          =  yeven_k  +  w_yodd_k        Y[k + middle] =  yeven_k  -  w_yodd_k         return Y  # Driver codeif __name__ == '__main__':     a = [1, 2, 3, 4]    b = fft(a)    for B in b:        print(B)

## Javascript

 // JavaScript program to implement// the approach // For storing complex values of nth roots// of unity we use complexclass complex{    constructor(a, b = 0)    {        this.x = a;        this.y = b;    }     } function product(c1, c2){    let c =new complex(0, 0);    c.x = c1.x * c2.x - c1.y * c2.y    c.y = c1.x * c2.y + c2.x * c1.y    return c} function sum(c1, c2){    let c = new complex(0, 0);    c.x = c1.x + c2.x    c.y = c1.y + c2.y    return c} function difference(c1, c2){    let c =new complex(0, 0);    c.x = c1.x - c2.x    c.y = c1.y - c2.y    return c}   // Recursive function of FFTfunction fft(a){    let n = a.length;     // if input contains just one element    if (n == 1)        return [a]     // For storing n complex nth roots of unity    let w = new Array(n);    let alpha = -2 * Math.PI / n;    for (var i = 0; i < n; i++) {        w[i] = new complex(Math.cos(alpha * i), Math.sin(alpha * i));    }     let A0 = new Array(Math.floor(n / 2));    let A1 = new Array(Math.floor(n / 2));    for (var i = 0; i < Math.floor(n / 2); i++) {         // even indexed coefficients        A0[i] = a[i * 2];         // odd indexed coefficients        A1[i] = a[i * 2 + 1];    }     // Recursive call for even indexed coefficients    let y0 = fft(A0);     // Recursive call for odd indexed coefficients    let y1 = fft(A1);     // for storing values of y0, y1, y2, ..., yn-1.    let y = new Array(n);     for (var k = 0; k < Math.floor(n / 2); k++) {                 y[k] =  sum(y0[k], product(w[k], y1[k]));        y[k + Math.floor(n / 2)] = difference(y0[k], product(w[k], y1[k]));    }    return y;} // Driver code let a = [ new complex(1, 0), new complex(2, 0), new complex(3, 0), new complex(4, 0)];let b = fft(a);for (var b0 of b)    console.log("(", b0.x, ",", b0.y, ")")  // This code is contributed by phasing17

## Java

 import java.util.*; // For storing complex values of nth roots// of unity we use complexclass cd {    double re;    double im;     cd(double r, double i) {        re = r;        im = i;    }     cd add(cd b) {        return new cd(re + b.re, im + b.im);    }     cd sub(cd b) {        return new cd(re - b.re, im - b.im);    }     cd mul(cd b) {        return new cd(re * b.re - im * b.im, re * b.im + im * b.re);    }} // Recursive function of FFTclass FFT {    int n;    cd[] a, w;     FFT(int n) {        this.n = n;        a = new cd[n];        w = new cd[n];    }     void fft() {        // if input contains just one element        if (n == 1)            return;         for (int i = 0; i < n; i++) {            double alpha = -2 * Math.PI * i / n;            w[i] = new cd(Math.cos(alpha), Math.sin(alpha));        }         cd[] A0 = new cd[n / 2];        cd[] A1 = new cd[n / 2];        for (int i = 0; i < n / 2; i++) {            // even indexed coefficients            A0[i] = a[i * 2];            // odd indexed coefficients            A1[i] = a[i * 2 + 1];        }         // Recursive call for even indexed coefficients        FFT y0 = new FFT(n / 2);        y0.a = A0;        y0.fft();         // Recursive call for odd indexed coefficients        FFT y1 = new FFT(n / 2);        y1.a = A1;        y1.fft();         // for storing values of y0, y1, y2, ..., yn-1.        cd[] y = new cd[n];        for (int k = 0; k < n / 2; k++) {            y[k] = y0.a[k].add(w[k].mul(y1.a[k]));            y[k + n / 2] = y0.a[k].sub(w[k].mul(y1.a[k]));        }        System.arraycopy(y, 0, a, 0, n);    }// } // Driver code// public class Main {    public static void main(String[] args) {        FFT fft = new FFT(4);        fft.a = new cd(1, 0);        fft.a = new cd(2, 0);        fft.a = new cd(3, 0);        fft.a = new cd(4, 0);        fft.fft();         for (int i = 0; i < 4; i++)            System.out.println(fft.a[i].re + " + " + fft.a[i].im + "i");    }}

## C#

 using System; // For storing complex values of nth roots// of unity we use a struct named Complexstruct Complex{    public double Re;    public double Im;     public Complex(double r, double i)    {        Re = r;        Im = i;    }     public Complex Add(Complex b)    {        return new Complex(Re + b.Re, Im + b.Im);    }     public Complex Sub(Complex b)    {        return new Complex(Re - b.Re, Im - b.Im);    }     public Complex Mul(Complex b)    {        return new Complex(Re * b.Re - Im * b.Im,                           Re * b.Im + Im * b.Re);    }} // Recursive function of FFTclass FFT {    int n;    Complex[] a, w;       public FFT(int n)    {        this.n = n;        a = new Complex[n];        w = new Complex[n];    }     public void Fft()    {        // if input contains just one element        if (n == 1)            return;         for (int i = 0; i < n; i++) {            double alpha = -2 * Math.PI * i / n;            w[i] = new Complex(Math.Cos(alpha),                               Math.Sin(alpha));        }         Complex[] A0 = new Complex[n / 2];        Complex[] A1 = new Complex[n / 2];        for (int i = 0; i < n / 2; i++) {            // even indexed coefficients            A0[i] = a[i * 2];            // odd indexed coefficients            A1[i] = a[i * 2 + 1];        }         // Recursive call for even indexed coefficients        FFT y0 = new FFT(n / 2);        y0.a = A0;        y0.Fft();         // Recursive call for odd indexed coefficients        FFT y1 = new FFT(n / 2);        y1.a = A1;        y1.Fft();         // for storing values of y0, y1, y2, ..., yn-1.        Complex[] y = new Complex[n];        for (int k = 0; k < n / 2; k++) {            y[k] = y0.a[k].Add(w[k].Mul(y1.a[k]));            y[k + n / 2] = y0.a[k].Sub(w[k].Mul(y1.a[k]));        }        Array.Copy(y, a, n);    }     public static void Main()    {        FFT fft = new FFT(4);        fft.a = new Complex(1, 0);        fft.a = new Complex(2, 0);        fft.a = new Complex(3, 0);        fft.a = new Complex(4, 0);        fft.Fft();         // css        // Copy code        for (int i = 0; i < 4; i++)            Console.WriteLine(fft.a[i].Re + " + "                              + fft.a[i].Im + "i");    }}

Input:  1 2 3 4
Output:
(10, 0)
(-2, 2)
(-2, 0)
(-2,-2)

Interpolation
Switch the roles of a and y.
Replace wn by wn^-1.
Divide each element of the result by n.
Time Complexity: O(nlogn).

My Personal Notes arrow_drop_up