Discussion
The FFT (a) returns DFT (y). To compute inverse of DFT, FFT algorithm can be modified to :
(i) Switch the roles of a and y
(ii) Replace \(w_{n}\) by \(w_{n}^{-1}\)
(iii) Divide the final result by n
Answer: D
All three modifications are needed, so the answer is (i), (ii) and (iii).
The DFT and its inverse are
\[ \begin{aligned} y_k &= \sum_{j=0}^{n-1} a_j\, \omega_n^{kj} \\ a_j &= \frac{1}{n} \sum_{k=0}^{n-1} y_k\, \omega_n^{-kj} \end{aligned} \]
Comparing the two, the inverse is the same computation with the roles of a and y switched (i), with \(\omega_n\) replaced by \(\omega_n^{-1}\) (ii), and with the result divided by n (iii). So the FFT computes the inverse DFT in \(\Theta(n \log n)\) time with all three changes; leaving out any one gives the wrong result.
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