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The Prime Factor Algorithm

Module by: Douglas L. Jones

Note: Your browser may not currently support MathML. See our browser support page for additional details. You can always view the correct math in the PDF version.

General Index Maps

n= K 1 n 1 + K 2 n 2 modN n K 1 n 1 K 2 n 2 N n= K 3 k 1 + K 4 k 2 modN n K 3 k 1 K 4 k 2 N n 1 =01… N 1 −1 n 1 0 1 … N 1 1 k 1 =01… N 1 −1 k 1 0 1 … N 1 1 n 2 =01… N 2 −1 n 2 0 1 … N 2 1 k 2 =01… N 2 −1 k 2 0 1 … N 2 1

The basic ideas is to simply reorder the DFT computation to expose the redundancies in the DFT, and exploit these to reduce computation!

Three conditions must be satisfied to make this map serve our purposes

  1. Each map must be one-to-one from 00 to N−1 N 1 , because we want to do the same computation, just in a different order.
  2. The map must be cleverly chosen so that computation is reduced
  3. The map should be chosen to make the short-length transforms be DFTs. (Not essential, since fast algorithms for short-length DFT-like computations could be developed, but it makes our work easier.)

Conditions for one-to-oneness of general index map

Case I

N 1 N 1 , N 2 N 2 relatively prime (greatest common denominator =1 1 ) i.e. gcd N 1 N 2 =1 N 1 N 2 1

K 1 =a N 2 K 1 a N 2 and/or K 2 =b N 1 K 2 b N 1 and gcd K 1 N 1 =1 K 1 N 1 1 , gcd K 2 N 2 =1 K 2 N 2 1

Case II

N 1 N 1 , N 2 N 2 not relatively prime: gcd N 1 N 2 >1 N 1 N 2 1

K 1 =a N 2 K 1 a N 2 and K 2 ≠b N 1 K 2 b N 1 and gcda N 1 =1 a N 1 1 , gcd K 2 N 2 =1 K 2 N 2 1 or K 1 ≠a N 2 K 1 a N 2 and K 2 =b N 1 K 2 b N 1 and gcd K 1 N 1 =1 K 1 N 1 1 , gcdb N 2 =1 b N 2 1 where K 1 K 1 , K 2 K 2 , K 3 K 3 , K 4 K 4 , N 1 N 1 , N 2 N 2 , a a, b b integers

proof:

Requires number-theory/abstract-algebra concepts. Reference: C.S. Burrus

Note:

Conditions of one-to-oneness must apply to both kk and nn

Conditions for arithmetic savings

X k 1 k 2 =∑ n 1 =0 N 1 −1∑ n 2 =0 N 2 −1x n 1 n 2 W N ( K 1 n 1 + K 2 n 2 ) ( K 3 k 1 + K 4 k 2 =∑ n 1 =0 N 1 −1∑ n 2 =0 N 2 −1x n 1 n 2 W N K 1 K 3 n 1 k 1 W N K 1 K 4 n 1 k 2 W N K 2 K 3 n 2 k 1 W N K 2 K 4 n 2 k 2 X k 1 k 2 n 1 N 1 1 0 n 2 N 2 1 0 x n 1 n 2 W N ( K 1 n 1 + K 2 n 2 ) ( K 3 k 1 + K 4 k 2 n 1 N 1 1 0 n 2 N 2 1 0 x n 1 n 2 W N K 1 K 3 n 1 k 1 W N K 1 K 4 n 1 k 2 W N K 2 K 3 n 2 k 1 W N K 2 K 4 n 2 k 2 (1)
  • K 1 K 4 modN=0 K 1 K 4 N 0 exclusive or K 2 K 3 modN=0 K 2 K 3 N 0 ⇒ Common Factor Algorithm (CFA). Then Xk= DFT N i twiddle factors DFT N j x n 1 n 2 X k DFT N i twiddle factors DFT N j x n 1 n 2
  • K 1 K 4 modN K 1 K 4 N and K 2 K 3 modN=0 K 2 K 3 N 0 ⇒ Prime Factor Algorithm (PFA). Xk= DFT N i DFT N j X k DFT N i DFT N j No twiddle factors!

fact:

A PFA exists only and always for relatively prime N 1 N 1 , N 2 N 2

Conditions for short-length transforms to be DFTs

K 1 K 3 modN= N 2 K 1 K 3 N N 2 and K 2 K 4 modN= N 1 K 2 K 4 N N 1

Note:

Convenient choice giving a PFA
K 1 = N 2 K 1 N 2 , K 2 = N 1 K 2 N 1 , K 3 = N 2 N 2 -1mod N 1 mod N 1 K 3 N 2 N 2 1 N 1 N 1 , K 4 = N 1 N 1 -1mod N 2 mod N 2 K 4 N 1 N 1 1 N 2 N 2 where N 1 -1mod N 2 N 1 1 N 2 is an integer such that N 1 N 1 -1mod=1 N 1 N 1 1 1

Example 1

N 1 =3 N 1 3 , N 2 =5 N 2 5 N=15 N 15 n=5 n 1 +3 n 2 mod15 n 5 n 1 3 n 2 15 k=10 k 1 +6 k 2 mod15 k 10 k 1 6 k 2 15

  1. Checking Conditions for one-to-oneness: 5= K 1 =a N 2 =5a 5 K 1 a N 2 5 a 3= K 2 =b N 1 =3b 3 K 2 b N 1 3 b gcd53=1 5 3 1 gcd35=1 3 5 1 10= K 3 =a N 2 =5a 10 K 3 a N 2 5 a 6= K 4 =b N 1 =3b 6 K 4 b N 1 3 b gcd103=1 10 3 1 gcd65=1 6 5 1
  2. Checking conditions for reduced computation: K 1 K 4 mod15=5×6mod15=0 K 1 K 4 15 5 6 15 0 K 2 K 3 mod15=3×10mod15=0 K 2 K 3 15 3 10 15 0
  3. Checking Conditions for making the short-length transforms be DFTS: K 1 K 3 mod15=5×10mod15=5= N 2 K 1 K 3 15 5 10 15 5 N 2 K 2 K 4 mod15=3×6mod15=3= N 1 K 2 K 4 15 3 6 15 3 N 1
Therefore, this is a prime factor map.

Figure 1: n=5 n 1 +3 n 2 mod15 n 5 n 1 3 n 2 15 and k=10 k 1 +6 k 2 mod15 k 10 k 1 6 k 2 15
2-D map
2-D map (figures15.png)

Operation Counts

  • N 2 N 2 length- N 1 N 1 DFTs + N 1 N 1 length- N 2 N 2 DFTs N 2 N 1 2+ N 1 N 2 2=N N 1 + N 2 N 2 N 1 2 N 1 N 2 2 N N 1 N 2 complex multiplies
  • Suppose N= N 1 N 2 N 3 … N M N N 1 N 2 N 3 … N M N N 1 + N 2 +…+ N M N N 1 N 2 … N M Complex multiplies

Different Strategies:

radix-2, radix-4 eliminate all multiplies in short-length DFTs, but have twiddle factors: PFA eliminates all twiddle factors, but ends up with multiplies in short-length DFTs. Surprisingly, total operation counts end up being very similar for similar lengths.

References

  1. C.S. Burrus. (1977, June). Index Mappings for Multidimensional Formulation of the DFT and Convolution. ASSP, 25, 239-242.

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