我试图理解这个位反转算法。我找到了很多来源，但它并没有真正解释伪代码是如何工作的。例如，我在http://www.briangough.com/fftalgorithms.pdf
下面找到了伪代码for i = 0 ... n − 2 do
k = n/2
if i < j then
swap g(i) and g(j)
end if
while k ≤ j do
j ⇐ j − k
k ⇐ k/2
end while
j ⇐ j + k
end for
从查看这个伪代码，我不明白你为什么会这样做
swap g(i) and g(j)
if语句为true。
另外：while循环有什么作用？如果有人能向我解释这个伪代码会很棒。
下面是我在网上找到的c ++代码。
void four1(double data[], int nn, int isign)
{
int n, mmax, m, j, istep, i;
double wtemp, wr, wpr, wpi, wi, theta;
double tempr, tempi;
n = nn << 1;
j = 1;
for (i = 1; i < n; i += 2) {
if (j > i) {
tempr = data[j]; data[j] = data[i]; data[i] = tempr;
tempr = data[j+1]; data[j+1] = data[i+1]; data[i+1] = tempr;
}
m = n >> 1;
while (m >= 2 && j > m) {
j -= m;
m >>= 1;
}
j += m;
}
以下是我发现的FFT代码的完整版本
/************************************************
* FFT code from the book Numerical Recipes in C *
* Visit www.nr.com for the licence. *
************************************************/
// The following line must be defined before including math.h to correctly define M_PI
#define _USE_MATH_DEFINES
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#define PI M_PI /* pi to machine precision, defined in math.h */
#define TWOPI (2.0*PI)
/*
FFT/IFFT routine. (see pages 507-508 of Numerical Recipes in C)
Inputs:
data[] : array of complex* data points of size 2*NFFT+1.
data[0] is unused,
* the n'th complex number x(n), for 0 <= n <= length(x)-1, is stored as:
data[2*n+1] = real(x(n))
data[2*n+2] = imag(x(n))
if length(Nx) < NFFT, the remainder of the array must be padded with zeros
nn : FFT order NFFT. This MUST be a power of 2 and >= length(x).
isign: if set to 1,
computes the forward FFT
if set to -1,
computes Inverse FFT - in this case the output values have
to be manually normalized by multiplying with 1/NFFT.
Outputs:
data[] : The FFT or IFFT results are stored in data, overwriting the input.
*/
void four1(double data[], int nn, int isign)
{
int n, mmax, m, j, istep, i;
double wtemp, wr, wpr, wpi, wi, theta;
double tempr, tempi;
n = nn << 1;
j = 1;
for (i = 1; i < n; i += 2) {
if (j > i) {
//swap the real part
tempr = data[j]; data[j] = data[i]; data[i] = tempr;
//swap the complex part
tempr = data[j+1]; data[j+1] = data[i+1]; data[i+1] = tempr;
}
m = n >> 1;
while (m >= 2 && j > m) {
j -= m;
m >>= 1;
}
j += m;
}
mmax = 2;
while (n > mmax) {
istep = 2*mmax;
theta = TWOPI/(isign*mmax);
wtemp = sin(0.5*theta);
wpr = -2.0*wtemp*wtemp;
wpi = sin(theta);
wr = 1.0;
wi = 0.0;
for (m = 1; m < mmax; m += 2) {
for (i = m; i <= n; i += istep) {
j =i + mmax;
tempr = wr*data[j] - wi*data[j+1];
tempi = wr*data[j+1] + wi*data[j];
data[j] = data[i] - tempr;
data[j+1] = data[i+1] - tempi;
data[i] += tempr;
data[i+1] += tempi;
}
wr = (wtemp = wr)*wpr - wi*wpi + wr;
wi = wi*wpr + wtemp*wpi + wi;
}
mmax = istep;
}
}
/********************************************************
* The following is a test routine that generates a ramp *
* with 10 elements, finds their FFT, and then finds the *
* original sequence using inverse FFT *
********************************************************/
int main(int argc, char * argv[])
{
int i;
int Nx;
int NFFT;
double *x;
double *X;
/* generate a ramp with 10 numbers */
Nx = 10;
printf("Nx = %d\n", Nx);
x = (double *) malloc(Nx * sizeof(double));
for(i=0; i<Nx; i++)
{
x[i] = i;
}
/* calculate NFFT as the next higher power of 2 >= Nx */
NFFT = (int)pow(2.0, ceil(log((double)Nx)/log(2.0)));
printf("NFFT = %d\n", NFFT);
/* allocate memory for NFFT complex numbers (note the +1) */
X = (double *) malloc((2*NFFT+1) * sizeof(double));
/* Storing x(n) in a complex array to make it work with four1.
This is needed even though x(n) is purely real in this case. */
for(i=0; i<Nx; i++)
{
X[2*i+1] = x[i];
X[2*i+2] = 0.0;
}
/* pad the remainder of the array with zeros (0 + 0 j) */
for(i=Nx; i<NFFT; i++)
{
X[2*i+1] = 0.0;
X[2*i+2] = 0.0;
}
printf("\nInput complex sequence (padded to next highest power of 2):\n");
for(i=0; i<NFFT; i++)
{
printf("x[%d] = (%.2f + j %.2f)\n", i, X[2*i+1], X[2*i+2]);
}
/* calculate FFT */
four1(X, NFFT, 1);
printf("\nFFT:\n");
for(i=0; i<NFFT; i++)
{
printf("X[%d] = (%.2f + j %.2f)\n", i, X[2*i+1], X[2*i+2]);
}
/* calculate IFFT */
four1(X, NFFT, -1);
/* normalize the IFFT */
for(i=0; i<NFFT; i++)
{
X[2*i+1] /= NFFT;
X[2*i+2] /= NFFT;
}
printf("\nComplex sequence reconstructed by IFFT:\n");
for(i=0; i<NFFT; i++)
{
printf("x[%d] = (%.2f + j %.2f)\n", i, X[2*i+1], X[2*i+2]);
}
getchar();
}
/*
Nx = 10
NFFT = 16
Input complex sequence (padded to next highest power of 2):
x[0] = (0.00 + j 0.00)
x[1] = (1.00 + j 0.00)
x[2] = (2.00 + j 0.00)
x[3] = (3.00 + j 0.00)
x[4] = (4.00 + j 0.00)
x[5] = (5.00 + j 0.00)
x[6] = (6.00 + j 0.00)
x[7] = (7.00 + j 0.00)
x[8] = (8.00 + j 0.00)
x[9] = (9.00 + j 0.00)
x[10] = (0.00 + j 0.00)
x[11] = (0.00 + j 0.00)
x[12] = (0.00 + j 0.00)
x[13] = (0.00 + j 0.00)
x[14] = (0.00 + j 0.00)
x[15] = (0.00 + j 0.00)
FFT:
X[0] = (45.00 + j 0.00)
X[1] = (-25.45 + j 16.67)
X[2] = (10.36 + j -3.29)
X[3] = (-9.06 + j -2.33)
X[4] = (4.00 + j 5.00)
X[5] = (-1.28 + j -5.64)
X[6] = (-2.36 + j 4.71)
X[7] = (3.80 + j -2.65)
X[8] = (-5.00 + j 0.00)
X[9] = (3.80 + j 2.65)
X[10] = (-2.36 + j -4.71)
X[11] = (-1.28 + j 5.64)
X[12] = (4.00 + j -5.00)
X[13] = (-9.06 + j 2.33)
X[14] = (10.36 + j 3.29)
X[15] = (-25.45 + j -16.67)
Complex sequence reconstructed by IFFT:
x[0] = (0.00 + j -0.00)
x[1] = (1.00 + j -0.00)
x[2] = (2.00 + j 0.00)
x[3] = (3.00 + j -0.00)
x[4] = (4.00 + j -0.00)
x[5] = (5.00 + j 0.00)
x[6] = (6.00 + j -0.00)
x[7] = (7.00 + j -0.00)
x[8] = (8.00 + j 0.00)
x[9] = (9.00 + j 0.00)
x[10] = (0.00 + j -0.00)
x[11] = (0.00 + j -0.00)
x[12] = (0.00 + j 0.00)
x[13] = (-0.00 + j -0.00)
x[14] = (0.00 + j 0.00)
x[15] = (0.00 + j 0.00)
*/
答案 0 :(得分：1)
位反转算法通过反转每个项的二进制地址来创建数据集的排列;所以例如在一个16项中设置地址：
0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111
将改为：
1000 0100 1100 0010 1010 0110 1110 0001 1001 0101 1101 0011 1011 0111 1111
然后将相应的项目移动到新地址。
或以十进制表示法：
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
成为
0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15
伪代码中的while循环是什么，将变量j设置为此序列。 （顺便说一下，j的初始值应为0）。
您会看到序列组成如下：
0
0 1
0 2 1 3
0 4 2 6 1 5 3 7
0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15
通过将先前版本乘以2来生成每个序列，然后在添加1的情况下重复它。或者以另一种方式看待它：通过重复前一个序列，与值+ n / 2隔行扫描（这更接近地描述了算法中发生的情况）。
0
0 1
0 2 1 3
0 4 2 6 1 5 3 7
0 8 4 12 2 10 6 14 1 9 5 13 3 11 7 15
然后在for循环的每次迭代中交换项目i和j，但仅当i＆lt;焦耳;否则每个项目都将被交换到新的位置（例如，当i = 3且j = 12时），然后再返回（当i = 12且j = 3时）。
function bitReversal(data) {
var n = data.length;
var j = 0;
for (i = 0; i < n - 1; i++) {
var k = n / 2;
if (i < j) {
var temp = data[i]; data[i] = data[j]; data[j] = temp;
}
while (k <= j) {
j -= k;
k /= 2;
}
j += k;
}
return(data);
}
console.log(bitReversal([0,1]));
console.log(bitReversal([0,1,2,3]));
console.log(bitReversal([0,1,2,3,4,5,6,7]));
console.log(bitReversal([0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]));
console.log(bitReversal(["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p"]));
您找到的C ++代码似乎使用序列的对称性以两步的方式遍历它。但它并没有产生正确的结果，所以要么是尝试失败，要么是它设计成完全不同的东西。这是一个使用两步理念的版本：
function bitReversal2(data) {
var n = data.length;
var j = 0;
for (i = 0; i < n; i += 2) {
if (i < j) {
var temp = data[i]; data[i] = data[j]; data[j] = temp;
}
else {
var temp = data[n-1 - i]; data[n-1 - i] = data[n-1 - j]; data[n-1 - j] = temp;
}
var k = n / 4;
while (k <= j) {
j -= k;
k /= 2;
}
j += k;
}
return(data);
}
console.log(bitReversal2([0,1]));
console.log(bitReversal2([0,1,2,3]));
console.log(bitReversal2([0,1,2,3,4,5,6,7]));
console.log(bitReversal2([0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]));
console.log(bitReversal2(["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p"]));
答案 1 :(得分：0)
首先，感谢大家帮忙回答我的问题。我正在和帮助我的人交谈，我想我现在理解我的C ++代码了。也许我的问题有点不清楚，但我想要做的是使用C ++实现FFT。我在问题中给出的C ++代码只是我在网上找到的FFT源代码的前半部分。本质上，这部分C ++代码将输入分类为实数和虚数，将实数存储到奇数索引中，将虚数存储到偶数索引中。 （real_0，imag_0，real_1，imag_1，real_2，imag_2，.....）索引从1开始，因为我们不需要交换第零索引。
下面的交换操作是交换实数和虚数。
tempr = data[j]; data[j] = data[i]; data[i] = tempr;
tempr = data[j+1]; data[j+1] = data[i+1]; data[i+1] = tempr;
例如，我们的数组长度为8（nn = 8），然后是2 * nn = 16，我们将输入分为实数和图像数，并将其存储到16的数组中。所以第一次去通过C ++代码的for循环，我的j=1，i=1，它将跳过if和while语句，现在在第二个for循环中， j=9，i=3，所以if语句为true，数据[9]，数据[3]将被交换，数据[9 + 1]和数据[3 + 1]将被交换。由于索引3和4具有第一个输入的实数和图像数，而索引9和10具有第四个的实数和图像数，因此执行位反转数。
结果，001 = 1（第一次输入）
100 = 4 (fourth input)
，因此，这是使用索引进行位反转。
我还没有真正了解while循环，但我从@ m69知道，它只是一种设置序列的方式，因此它可以进行位反转。对我有同样问题的人来说，对我自己的问题的解释很有帮助。再一次，谢谢大家。