"Algorithm topic" polynomial-related.

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"Fast Fourier Transform"

Reference: from polynomial multiplication to fast Fourier transform by Miskcoo

FFT Learning Note by Menci

Features: FFT for O (n log n) solves polynomial multiplication.

(i) Representation of the polynomial

Coefficient notation: F (x) =a[n-1]*x^ (n-1) +...+a[0], called n-1-quadratic polynomial.

Point value notation: a n-1 polynomial has n roots in a complex field, that is, n (x, Y) can uniquely determine a n-1 polynomial.

For a polynomial, the transformation from its coefficient representation to its point value representation is called a discrete Fourier transform (DFT) and, conversely, a Fourier inverse transformation (IDFT).

Simple discrete Fourier transform, the complexity of enumeration implementation is O (n^2).

Fast Fourier transform is an algorithm for implementing IDF and IDFT with the complexity of O (n log n), commonly used Cooley-tukey algorithm.

(ii) plural

The plural is a figure as A+bi, when B=0 is a real number, b≠0 is imaginary, and a=0 is pure imaginary.

Define a plane as a complex plane, then each point within the plane (A, B) uniquely corresponds to a complex a+bi,i that can be understood as the unit length on the y-axis, as 1 is the unit length on the x-axis.

The essence of I is to define the rotation transformation on the axis, I is 1 counterclockwise rotation 90 °, then i^2=-1.

The complex number is added and follows the parallelogram rule.

Multiply the plural, the modulus of appearance multiply, the amplitude angle adds.

(c) Unit root

Take the dot as the starting point, the complex plane units round the N equal point as the end point, as n vector, the resulting amplitude angle is the positive and the smallest vector corresponding to the complex number of Ω (1,n), that is, n-th unit root. (The brackets are superscript to the left and the subscript to the right).

Image source: The FFT by Zball in OI

Where the B point is the unit root ω (1,n), the counter-clockwise is Ω (2,n), ω (3,n) ..., ω (n,n) =ω (n,0) = 1.

Calculation formula: Ω (k,n) =cos (2kπ/n) + i*sin (2kπ/n)

The nature of the unit root:

(1) Elimination: Ω (2n,2k) =ω (n,k)

(2) Binary: Ω (N,K+N/2) =-ω (n,k)

The n unit root of Ω (n,0) ~ω (n,n-1) is used as the horizontal axis of n points representing the n-1 sub-polynomial, which can get good properties.

(d) Fast Fourier transform (FFT solution DFA)

This part because does not manipulate the mathematical formula, pastes the Menci blog Qaq directly.

I'm the figure. jpg

The coefficients of the n-1 polynomial a (x) are divided into two polynomial A1 (x) and A2 (x), then a (x) =a1 (x^2) +x*a2 (x^2).

For K<N/2, there is a (Ω (n,k)) =a1 (Ω (n/2,k)) +ω (n,k) *a2 (ω (n/2,k))

At the same time, there is a (Ω (N,K+N/2)) =a1 (Ω (n/2,k)) - ω (n,k) *a2 (ω (n/2,k))

For a K-polynomial, a two-k/2 polynomial is obtained by the odd-even sub-term, and then the values are calculated and then called to solve the K-th polynomial, that is, divide and conquer the solution.

(v) Fourier inverse transformation (IDFA)

For the n-1 polynomial, its n-1-sustain number vector {A0,a1...an-1} obtains the point value vector {b0,b1...bn-1} through the DFA, whereas the operation is called IDFA.

The point value vector is used as a coefficient, and the inverse of the unit root is FFT, and each number divided by n is the result of IDFA.

(vi) iterative implementation of FFT

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"Algorithm topic" polynomial-related.