To derive the FFT, we assume that the signal's duration is a
power of two:
N=2l
N
2
l
. Consider what happens to the even-numbered and odd-numbered
elements of the sequence in the DFT calculation.

Sk=s0+s2e(−j)2π2kN+…+sN−2e(−j)2π(N−2)kN+s1e(−j)2πkN+s3e(−j)2π×(2+1)kN+…+sN−1e(−j)2π(N−2+1)kN=s0+s2e(−j)2πkN2+…+sN−2e(−j)2π(N2−1)kN2+(s1+s3e(−j)2πkN2+…+sN−1e(−j)2π(N2−1)kN2)e−(j2πk)N
S
k
s
0
s
2
2
2
k
N
…
s
N
2
2
N
2
k
N
s
1
2
k
N
s
3
2
2
1
k
N
…
s
N
1
2
N
2
1
k
N
s
0
s
2
2
k
N
2
…
s
N
2
2
N
2
1
k
N
2
s
1
s
3
2
k
N
2
…
s
N
1
2
N
2
1
k
N
2
2
k
N

(1)
Each term in square brackets has the form of a
N2
N
2
-length DFT. The first one is a DFT of the even-numbered
elements, and the second of the odd-numbered elements. The
first DFT is combined with the second multiplied by the complex
exponential
e−(j2πk)N
2
k
N
. The half-length transforms are each evaluated at frequency indices
k∈0…N−1
k
0
…
N
1
. Normally, the number of frequency indices in a DFT
calculation range between zero and the transform length minus
one. The computational advantage of the FFT comes
from recognizing the periodic nature of the discrete Fourier
transform. The FFT simply reuses the computations made in the
half-length transforms and combines them through additions and
the multiplication by
e−(j2πk)N
2
k
N
, which is not periodic over
N2
N
2
, to rewrite the length-N DFT. Figure 1 illustrates this decomposition. As it stands, we
now compute two length-
N2
N
2
transforms (complexity
2ON24
2
O
N
2
4
), multiply one of them by the complex exponential (complexity
ON
O
N
), and add the results (complexity
ON
O
N
). At this point, the total complexity is still dominated by
the half-length DFT calculations, but the proportionality
coefficient has been reduced.

Now for the fun. Because
N=2l
N
2
l
, each of the half-length transforms can be reduced to two
quarter-length transforms, each of these to two eighth-length
ones, etc. This decomposition continues until we are left with
length-2 transforms. This transform is quite simple, involving
only additions. Thus, the first stage of the FFT has
N2
N
2
length-2 transforms (see the bottom part of Figure 1). Pairs of these transforms are
combined by adding one to the other multiplied by a complex
exponential. Each pair requires 4 additions and 4
multiplications, giving a total number of computations equaling
8N4=N2
8
N
4
N
2
. This number of computations does not change from stage to
stage. Because the number of stages, the number of times the
length can be divided by two, equals
log2N
2
N
, the complexity of the FFT is
ONlogN
O
N
N
.

Doing an example will make computational savings more obvious.
Let's look at the details of a length-8 DFT. As shown on Figure 1, we first decompose the DFT into
two length-4 DFTs, with the outputs added and subtracted
together in pairs. Considering Figure 1 as the frequency index goes from 0 through 7, we
recycle values from the length-4 DFTs into the final calculation
because of the periodicity of the DFT output. Examining how
pairs of outputs are collected together, we create the basic
computational element known as a butterfly (Figure 2).

By considering together the computations involving common output
frequencies from the two half-length DFTs, we see that the two
complex multiplies are related to each other, and we can reduce
our computational work even further. By further decomposing the
length-4 DFTs into two length-2 DFTs and combining their
outputs, we arrive at the diagram summarizing the length-8 fast
Fourier transform (Figure 1).
Although most of the complex multiplies are quite simple
(multiplying by
e−(jπ)
means negating real and imaginary parts), let's count those for
purposes of evaluating the complexity as full complex
multiplies. We have
N2=4
N
2
4
complex multiplies and
2N=16
2
N
16
additions for each stage and
log2N=3
2
N
3
stages, making the number of basic computations
3N2log2N
3
N
2
2
N
as predicted.

Note that the ordering of the input sequence in the two
parts of Figure 1 aren't quite
the same. Why not? How is the ordering determined?

The upper panel has not used the FFT algorithm to compute
the length-4 DFTs while the lower one has. The ordering is
determined by the algorithm.

Other "fast" algorithms were discovered, all of which make use
of how many common factors the transform length N has. In
number theory, the number of prime factors a given integer has
measures how composite it is. The numbers 16 and
81 are highly composite (equaling
24
2
4
and
34
3
4
respectively), the number 18 is less so (
2132
2
1
3
2
), and 17 not at all (it's prime). In over thirty years of
Fourier transform algorithm development, the original
Cooley-Tukey algorithm is far and away the most frequently
used. It is so computationally efficient that power-of-two
transform lengths are frequently used regardless of what the
actual length of the data.

Comments:"My introduction to signal processing course at Rice University."