Question

Please answer the following question in c++ language

IS INTENDED TO BE DONE BY HAND WITH POSSIBLE USE OF A CALCULATOR.   YOU NEED TO SHOW SOME WORK BECAUSE THE OBECTIVE IS KNOWING WHAT A PROGRAM IS DOING, NOT RUNNING A PROGRAM. SHOWING HOW YOU GOT ANSWERS IS REQUIRED.

The divide and conquer Fast Fourier Transform (FFT) has run time of order   

a) O(logn)                      

b) O(n logn)  

c) O( )  

d) O(n)

Answer 1

Answer is

B)O(nlogn)

It is fast, and has running time in Θ(n log n) • There are many variations, often to take advantage of the numerical structure of n. Most commonly n is a power of 2. It can also be computed optically using lenses.

A Recursive Fast Fourier Transform #basis case Initialize results to 0. • Divide - even subproblem . - odd subproblem # FOCUSive call det FFT (f): n lan () if na 1 return [ ] F = [0] aven - [0::2] _odd [1:2] F_aven - FT (_even) F_odd - PIT (£_odd) n2 - int (n/2) for in sange (n2): twiddle - exp(-2*pi 13°/n) oddTerm - Todd [1] . twiddle [1] - Taven (1) + oddTerm [i+n2) - Tavan (1) - oddTerm return • Prepare to combine results • These could be precomputed . Odd terms need an adjustment • Compute a new term • Compute one more new term

An N log N algorithm Like in merge sort, in each recursive call, we divide the number of elements in half. The number of levels of recursive calls is therefore log, N. When we combine subproblem results, we spend linear time. Total time is bounded by CN log N.

Recursive FFT FFT(n,(a, a, .... a -1]): if n=1: return ao Feven = FFT(n/2, (a, a, ..., an-2]) Fodd = FFT(n/2, [a, a3, ..., an-1]) for k = 0 to n/2 - 1: wk = emrik/n yk = Feven k + wk Fodd k yk+n/2 = Feven k-wk Fodd k return [yo: 41: .... Yn-1]