# 3.4 The prime factor algorithm

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## General index maps

$n=({K}_{1}{n}_{1}+{K}_{2}{n}_{2})\mod N$ $n=({K}_{3}{k}_{1}+{K}_{4}{k}_{2})\mod N$ ${n}_{1}$

0 1 N 1 1
${k}_{1}$
0 1 N 1 1
${n}_{2}$
0 1 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

• Each map must be one-to-one from $0$ to $N-1$ , because we want to do the same computation, just in a different order.
• The map must be cleverly chosen so that computation is reduced
• 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.)

## Case i

${N}_{1}$ , ${N}_{2}$ relatively prime (greatest common denominator $1$ ) i.e. $\gcd ({N}_{1}, {N}_{2})=1$

${K}_{1}=a{N}_{2}$ and/or ${K}_{2}=b{N}_{1}$ and $\gcd ({K}_{1}, {N}_{1})=1$ , $\gcd ({K}_{2}, {N}_{2})=1$

## Case ii

${N}_{1}$ , ${N}_{2}$ not relatively prime: $\gcd ({N}_{1}, {N}_{2})> 1$

${K}_{1}=a{N}_{2}$ and ${K}_{2}\neq b{N}_{1}$ and $\gcd (a, {N}_{1})=1$ , $\gcd ({K}_{2}, {N}_{2})=1$ or ${K}_{1}\neq a{N}_{2}$ and ${K}_{2}=b{N}_{1}$ and $\gcd ({K}_{1}, {N}_{1})=1$ , $\gcd (b, {N}_{2})=1$ where ${K}_{1}$ , ${K}_{2}$ , ${K}_{3}$ , ${K}_{4}$ , ${N}_{1}$ , ${N}_{2}$ , $a$ , $b$ integers

Requires number-theory/abstract-algebra concepts. Reference: C.S. Burrus
Conditions of one-to-oneness must apply to both $k$ and $n$

## Conditions for arithmetic savings

$X({k}_{1}, {k}_{2})=\sum_{{n}_{1}=0}^{{N}_{1}-1} \sum_{{n}_{2}=0}^{{N}_{2}-1} x({n}_{1}, {n}_{2}){W}_{N}^{\left({K}_{1}{n}_{1}+{K}_{2}{n}_{2}\right)\left({K}_{3}{k}_{1}+{K}_{4}{k}_{2}}=\sum_{{n}_{1}=0}^{{N}_{1}-1} \sum_{{n}_{2}=0}^{{N}_{2}-1} x({n}_{1}, {n}_{2}){W}_{N}^{{K}_{1}{K}_{3}{n}_{1}{k}_{1}}{W}_{N}^{{K}_{1}{K}_{4}{n}_{1}{k}_{2}}{W}_{N}^{{K}_{2}{K}_{3}{n}_{2}{k}_{1}}{W}_{N}^{{K}_{2}{K}_{4}{n}_{2}{k}_{2}}$
• ${K}_{1}{K}_{4}\mod N=0$ exclusive or ${K}_{2}{K}_{3}\mod N=0$ Common Factor Algorithm (CFA). Then $X(k)={\mathrm{DFT}}_{Ni}(\text{twiddle factors}{\mathrm{DFT}}_{Nj}(x({n}_{1}, {n}_{2})))$
• ${K}_{1}{K}_{4}\mod N$ and ${K}_{2}{K}_{3}\mod N=0$ Prime Factor Algorithm (PFA). $X(k)={\mathrm{DFT}}_{Ni}({\mathrm{DFT}}_{Nj})$ No twiddle factors!
A PFA exists only and always for relatively prime ${N}_{1}()$ , ${N}_{2}()$

## Conditions for short-length transforms to be dfts

${K}_{1}{K}_{3}\mod N={N}_{2}$ and ${K}_{2}{K}_{4}\mod N={N}_{1}$

Convenient choice giving a PFA
${K}_{1}={N}_{2}$ , ${K}_{2}={N}_{1}$ , ${K}_{3}={N}_{2}{N}_{2}^{-1}\mod {N}_{1}\mod {N}_{1}$ , ${K}_{4}={N}_{1}{N}_{1}^{-1}\mod {N}_{2}\mod {N}_{2}$ where ${N}_{1}^{-1}\mod {N}_{2}$ is an integer such that ${N}_{1}{N}_{1}^{-1}\mod =1$

${N}_{1}=3$ , ${N}_{2}=5$ $N=15$ $n=(5{n}_{1}+3{n}_{2})\mod 15$ $k=(10{k}_{1}+6{k}_{2})\mod 15$

• ## Checking conditions for one-to-oneness

$5={K}_{1}=a{N}_{2}=5a$ $3={K}_{2}=b{N}_{1}=3b$ $\gcd (5, 3)=1$ $\gcd (3, 5)=1$ $10={K}_{3}=a{N}_{2}=5a$ $6={K}_{4}=b{N}_{1}=3b$ $\gcd (10, 3)=1$ $\gcd (6, 5)=1$
• ## Checking conditions for reduced computation

${K}_{1}{K}_{4}\mod 15=5\times 6\mod 15=0$ ${K}_{2}{K}_{3}\mod 15=3\times 10\mod 15=0$
• ## Checking conditions for making the short-length transforms be dfts

${K}_{1}{K}_{3}\mod 15=5\times 10\mod 15=5={N}_{2}$ ${K}_{2}{K}_{4}\mod 15=3\times 6\mod 15=3={N}_{1}$
Therefore, this is a prime factor map.

## Operation counts

• ${N}_{2}$ length- ${N}_{1}$ DFTs ${N}_{1}$ length- ${N}_{2}$ DFTs ${N}_{2}{N}_{1}^{2}+{N}_{1}{N}_{2}^{2}=N({N}_{1}+{N}_{2})$ complex multiplies
• Suppose $N={N}_{1}{N}_{2}{N}_{3}{N}_{M}$ $N({N}_{1}+{N}_{2}++{N}_{M})$ Complex multiplies
radix-2 , radix-4 eliminate all multiplies in short-length DFTs, but have twiddle factors: PFA eliminates all twiddlefactors, but ends up with multiplies in short-length DFTs . Surprisingly, total operation counts end up being very similarfor similar lengths.

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