How to use fftw Guru interface

问题

I used to use fftw_plan_dft for multi-dimensional Fourier transformation.

fftw_plan fftw_plan_dft(int rank, const int *n, fftw_complex *in,
                        fftw_complex *out, int sign, unsigned flags);

Now I want to pass 64 bit integer to fftw, it looks like I need to use fftw guru interface.

 fftw_plan fftw_plan_guru64_dft(
     int rank, const fftw_iodim64 *dims,
     int howmany_rank, const fftw_iodim64 *howmany_dims,
     fftw_complex *in, fftw_complex *out,
     int sign, unsigned flags);

But I do not understand what is howmany_rank and howmany_dims mean. The manual of fftw_plan_guru_dft says:

These two functions plan a complex-data, multi-dimensional DFT for the interleaved and split format, respectively. Transform dimensions are given by (rank, dims) over a multi-dimensional vector (loop) of dimensions (howmany_rank, howmany_dims). dims and howmany_dims should point to fftw_iodim arrays of length rank and howmany_rank, respectively.

I do know know what is "multi-dimensional vector (loop) of dimensions (howmany_rank, howmany_dims)" mean. Can you give me an example or explain how to use this guru interface?

回答1:

If the sizes and strides of your mulitdimensional arrays are larger than 2^32, the 64 bit guru interface becomes useful.

The prototype of the function creating complex to complex DTFs is:

fftw_plan fftw_plan_guru64_dft(
 int rank, const fftw_iodim64 *dims,
 int howmany_rank, const fftw_iodim64 *howmany_dims,
 fftw_complex *in, fftw_complex *out,
 int sign, unsigned flags);

where:

rank is the rank of the FFTW transform to be performed, that is the number of dimensions.
dims is an array of size rank. For each dimension i, dims[i].n is the size of the line, dims[i].is is the stride between lines of the input array and dims[i].os is the stride between lines of the output array. For instance, if the array is contiguous in memory, then the documentation of the guru interface suggests to use the recurrence dims[i].is = n[i+1] * dims[i+1].is. The number of transforms to be performed and the offsets between the starting points are given by howmany_rank and howmany_dims.
howmany_rank specifies the number of transforms featuring particular offsets.
howmany_dims is an array of size howmany_rank. For each transform i, howmany_dims[i].n is the number of transforms to be computed, each featuring offset between inputs howmany_dims[i].is and offset between output howmany_dims[i].os.

More detail about these arguments are provided in Confusion about FFTW3 guru interface: 3 simultaneous complex FFTs

The following code calls fftw_plan_guru64_dft() so that it performs the same thing as fftw_plan_dft_3d(). It can be compiled by gcc main.c -o main -lfftw3 -lm -Wall:

#include<stdlib.h>
#include<complex.h>
#include<math.h>
#include<fftw3.h>

int main(void){

    fftw_plan p;
    unsigned long int N = 10;
    unsigned long int M = 12;
    unsigned long int P = 14;
    fftw_complex *in=fftw_malloc(N*M*P*sizeof(fftw_complex));
    if(in==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    fftw_complex *out=fftw_malloc(N*M*P*sizeof(fftw_complex));
    if(out==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    unsigned int i,j,k;

    int rank=3;
    fftw_iodim64 *dims=malloc(rank*sizeof(fftw_iodim64));
    if(dims==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    dims[0].n=N;
    dims[0].is=P*M;
    dims[0].os=P*M;
    dims[1].n=M;
    dims[1].is=P;
    dims[1].os=P;
    dims[2].n=P;
    dims[2].is=1;
    dims[2].os=1;

    int howmany_rank=1;
    fftw_iodim64 *howmany_dims=malloc(howmany_rank*sizeof(fftw_iodim64));
    if(howmany_dims==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    howmany_dims[0].n=1;
    howmany_dims[0].is=1;
    howmany_dims[0].os=1;

    printf("sizeof fftw complex %ld\n",sizeof(fftw_complex));
    printf("sizeof fftw_iodim64 %ld\n",sizeof(fftw_iodim64));
    printf("creating the plan\n");
    p=fftw_plan_guru64_dft(rank, dims,howmany_rank, howmany_dims,in, out,FFTW_FORWARD, FFTW_ESTIMATE);
    if (p==NULL){fprintf(stderr,"plan creation failed\n");exit(1);}
    printf("created the plan\n");

    for(i=0;i<N;i++){
        for(j=0;j<M;j++){
            for(k=0;k<P;k++){
                //printf("ijk\n");
                in[(i*M+j)*P+k]=30.+12.*sin(2*3.1415926535*i/((double)N))*sin(2*3.1415926535*j/((double)M))*sin(2*3.1415926535*k/((double)P))*I;
            }
        }
    }

    fftw_execute(p);

    for (i = 0; i < N; i++){
        for (j = 0; j < M; j++){
            for (k = 0; k < P; k++){
                printf("result: %d %d %d %g %gI\n", i,j,k, creal(out[(i*M+j)*P+k]), cimag(out[(i*M+j)*P+k]));
            }
        }
    }


    fftw_destroy_plan(p);
    fftw_free(in);
    fftw_free(out);

    free(dims);
    free(howmany_dims);

    return(0);
}

For instance, the guru interface can be used to compute the DFT of a complex 3D electric field. At each point of the grid, the electric field is a vector of size 3. Hence, I can store the electric field as a 4D array, the last dimension specifying the component of the vector. Finally, the guru interface can be used to perform the three 3D DFTs at once:

#include<stdlib.h>
#include<complex.h>
#include<math.h>
#include<fftw3.h>

int main(void){

    fftw_plan p;
    unsigned long int N = 10;
    unsigned long int M = 12;
    unsigned long int P = 14;
    unsigned long int DOF = 3;
    fftw_complex *in=fftw_malloc(N*M*P*DOF*sizeof(fftw_complex));
    if(in==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    fftw_complex *out=fftw_malloc(N*M*P*DOF*sizeof(fftw_complex));
    if(out==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    unsigned int i,j,k;

    int rank=3;
    fftw_iodim64 *dims=malloc(rank*sizeof(fftw_iodim64));
    if(dims==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    dims[0].n=N;
    dims[0].is=P*M*DOF;
    dims[0].os=P*M*DOF;
    dims[1].n=M;
    dims[1].is=P*DOF;
    dims[1].os=P*DOF;
    dims[2].n=P;
    dims[2].is=DOF;
    dims[2].os=DOF;

    int howmany_rank=1;
    fftw_iodim64 *howmany_dims=malloc(howmany_rank*sizeof(fftw_iodim64));
    if(howmany_dims==NULL){fprintf(stderr,"malloc failed\n");exit(1);}
    howmany_dims[0].n=DOF;
    howmany_dims[0].is=1;
    howmany_dims[0].os=1;

    printf("sizeof fftw complex %ld\n",sizeof(fftw_complex));
    printf("sizeof fftw_iodim64 %ld\n",sizeof(fftw_iodim64));
    printf("creating the plan\n");
    p=fftw_plan_guru64_dft(rank, dims,howmany_rank, howmany_dims,in, out,FFTW_FORWARD, FFTW_ESTIMATE);
    if (p==NULL){fprintf(stderr,"plan creation failed\n");exit(1);}
    printf("created the plan\n");

    for(i=0;i<N;i++){
        for(j=0;j<M;j++){
            for(k=0;k<P;k++){
                //printf("ijk\n");
                in[((i*M+j)*P+k)*DOF]=30.+12.*sin(2*3.1415926535*i/((double)N))*sin(2*3.1415926535*j/((double)M))*sin(2*3.1415926535*k/((double)P))*I;
                in[((i*M+j)*P+k)*DOF+1]=42.0;
                in[((i*M+j)*P+k)*DOF+2]=1.0;
            }
        }
    }

    fftw_execute(p);

    for (i = 0; i < N; i++){
        for (j = 0; j < M; j++){
            for (k = 0; k < P; k++){
                printf("result: %d %d %d || %g %gI | %g %gI | %g %gI\n", i,j,k, creal(out[((i*M+j)*P+k)*DOF]), cimag(out[((i*M+j)*P+k)*DOF]),creal(out[((i*M+j)*P+k)*DOF+1]), cimag(out[((i*M+j)*P+k)*DOF+1]),creal(out[((i*M+j)*P+k)*DOF+2]), cimag(out[((i*M+j)*P+k)*DOF+2]));
            }
        }
    }


    fftw_destroy_plan(p);
    fftw_free(in);
    fftw_free(out);

    free(dims);
    free(howmany_dims);

    return(0);
}

来源：https://stackoverflow.com/questions/39938409/how-to-use-fftw-guru-interface

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