Translation from Complex-FFT to Finite-Field-FFT

Question

Good afternoon!

I am trying to develop an NTT algorithm based on the naive recursive FFT implementation I already have.

Consider the following code (co

Answer 1

I recently wanted to implement NTT for fast multiplication instead of DFFT too. Read a lot of confusing things, different letters everywhere and no simple solution, and also my finite fields knowledge is rusty , but today i finally got it right (after 2 days of trying and analog-ing with DFT coefficients) so here are my insights for NTT:

1. Computation

   X(i) = sum(j=0..n-1) of ( Wn^(i*j)*x(i) );
   

   where X[] is NTT transformed x[] of size n where Wn is the NTT basis. All computations are on integer modulo arithmetics mod p no complex numbers anywhere.

2. Important values

   
   Wn = r ^ L mod p is basis for NTT
   Wn = r ^ (p-1-L) mod p is basis for INTT
   Rn = n ^ (p-2) mod p is scaling multiplicative constant for INTT ~(1/n)
p is prime that p mod n == 1 and p>max'
max is max value of x[i] for NTT or X[i] for INTT
   r = <1,p)
L = <1,p) and also divides p-1
r,L must be combined so r^(L*i) mod p == 1 if i=0 or i=n
r,L must be combined so r^(L*i) mod p != 1 if 0 < i < n
max' is the sub-result max value and depends on n and type of computation. For single (I)NTT it is max' = n*max but for convolution of two n sized vectors it is max' = n*max*max etc. See Implementing FFT over finite fields for more info about it.

3. functional combination of r,L,p is different for different n

   this is important, you have to recompute or select parameters from table before each NTT layer (n is always half of the previous recursion).

Here is my C++ code that finds the r,L,p parameters (needs modular arithmetics which is not included, you can replace it with (a+b)%c,(a-b)%c,(a*b)%c,... but in that case beware of overflows especial for modpow and modmul) The code is not optimized yet there are ways to speed it up considerably. Also prime table is fairly limited so either use SoE or any other algo to obtain primes up to max' in order to work safely.

DWORD _arithmetics_primes[]=
    {
    2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,
    179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,
    419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,
    661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,
    947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,
    0}; // end of table is 0, the more primes are there the bigger numbers and n can be used
// compute NTT consts W=r^L%p for n
int i,j,k,n=16;
long w,W,iW,p,r,L,l,e;
long max=81*n;  // edit1: max num for NTT for my multiplication purposses
for (e=1,j=0;e;j++)             // find prime p that p%n=1 AND p>max ... 9*9=81
    {
    p=_arithmetics_primes[j];
    if (!p) break;
    if ((p>max)&&(p%n==1))
     for (r=2;r0)&&(i

and here is my slow NTT and INTT implementations (i havent got to fast NTT,INTT yet) they are both tested with Schönhage–Strassen multiplication successfully.

//---------------------------------------------------------------------------
void NTT(long *dst,long *src,long n,long m,long w)
    {
    long i,j,wj,wi,a,n2=n>>1;
    for (wj=1,j=0;j>1;
    rN=modpow(n,m-2,m);
    for (wj=1,j=0;j

dst is destination array
src is source array
n is array size
m is modulus (p)
w is basis (Wn)

hope this helps to someone. If i forgot something please write ...

[edit1: fast NTT/INTT]

Finally I manage to get fast NTT/INTT to work. Was little bit more tricky than normal FFT:

//---------------------------------------------------------------------------
void _NFTT(long *dst,long *src,long n,long m,long w)
    {
    if (n<=1) { if (n==1) dst[0]=src[0]; return; }
    long i,j,a0,a1,n2=n>>1,w2=modmul(w,w,m);
    // reorder even,odd
    for (i=0,j=0;i

[edit3]

I have optimized my code (3x times faster than code above),but still i am not satisfied with it so i started new question with it. There I have optimized my code even further (about 40x times faster than code above) so its almost the same speed as FFT on floating point of the same bit size. Link to it is here:

• Modular arithmetics and NTT (finite field DFT) optimizations