如何使用fftw Guru界面 [英] How to use fftw Guru interface

问题描述

我曾经使用fftw_plan_dft进行多维傅里叶变换.

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);

现在我想将64位整数传递给fftw，看来我需要使用fftw guru接口.

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);

但是我不明白什么是howmany_rank和howmany_dims的意思. fftw_plan_guru_dft的手册说:

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

这两个函数分别针对交错格式和分割格式计划复杂数据，多维DFT.变换尺寸是由尺寸(howmany_rank，howmany_dims)的多维矢量(循环)给出的(等级，暗淡). dims和howmany_dims应该分别指向fftw_iodim数组，它们的长度分别为rank和howmany_rank.

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.

我确实知道什么是维度(howmany_rank，howmany_dims)的多维向量(循环)"的意思.您能给我一个例子或解释一下如何使用此大师界面吗?

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?

推荐答案

如果多维度数组的大小和步幅大于2 ^ 32，则

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

创建复杂到复杂DTF的函数的原型为:

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);

其中:

• rank是要执行的FFTW变换的等级，即维数.
• dims是大小为rank的数组.对于每个尺寸i，dims[i].n是行的大小，dims[i].is是输入数组的行之间的跨度，而dims[i].os是输出数组的行之间的跨度.例如，如果数组在内存中是连续的，则

• 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.

有关这些参数的更多详细信息，请参见关于FFTW3专家界面的困惑:3个同时进行的复杂FFT

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

以下代码调用fftw_plan_guru64_dft()，以便它执行与fftw_plan_dft_3d()相同的操作.可以由gcc main.c -o main -lfftw3 -lm -Wall:

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);
}

例如，宗师接口可用于计算复杂3D电场的DFT.在网格的每个点处，电场都是大小为3的向量.因此，我可以将电场存储为4D数组，最后一个维指定向量的分量.最后，可以使用guru界面一次执行三个3D DFT:

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);
}

这篇关于如何使用fftw Guru界面的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！