FFT實(shí)現(xiàn)的C語言代碼- -（基2FFT及IFFT算法C語言實(shí)現(xiàn)）

Given two images A and B, use image B to cover image A. Where would we put B on A, so that the overlapping part of A and B has the most likelihood? To simplify the problem, we assume that A and B only contain numbers between 0 and 255. The difference between A and B is defined as the square sum of the differences of corresponding elements in the overlapped parts of A and B.

For example, we have

When B is placed on position a5, the difference of them is ((b1-a5)^2 + (b2-a6)^2 + (b4-a8)^2 + (b5-a9)^2). Now we hope to have the position of the top left corner of B that gives the minimum difference. (B must completely reside on A)

It is clear that a simple solution will appear with very low efficiency when A and B have too many elements. But we can use 1-dimensional repeat convolution, which can be computed by Fast Fourier Transform (FFT), to improve the performance.

A program with explanation of FFT is given below:

/**
* Given two sequences {a1, a2, a3.. an} and {b1, b2, b3... bn},
* their repeat convolution means:
* r1 = a1*b1 + a2*b2 + a3*b3 + ... + an*bn
* r2 = a1*bn + a2*b1 + a3*b2 + ... + an*bn-1
* r3 = a1*bn-1 + a2*bn + a3*b1 + ... + an*bn-2
* ...
* rn = a1*b2 + a2*b3 + a3*b4 + ... + an-1*bn + an*b1
* Notice n >= 2 and n must be power of 2.
*/
#include 
#include 
#include 
#define for if (0); else for
using namespace std;
const int MaxFastBits = 16;
int **gFFTBitTable = 0;
int NumberOfBitsNeeded(int PowerOfTwo) {
for (int i = 0;; ++i) {
if (PowerOfTwo & (1 << i)) {
return i;
}
}
}
int ReverseBits(int index, int NumBits) {
int ret = 0;
for (int i = 0; i < NumBits; ++i, index >>= 1) {
ret = (ret << 1) | (index & 1);
}
return ret;
}
void InitFFT() {
gFFTBitTable = new int *[MaxFastBits];
for (int i = 1, length = 2; i <= MaxFastBits; ++i, length <<= 1) {
gFFTBitTable[i - 1] = new int[length];
for (int j = 0; j < length; ++j) {
gFFTBitTable[i - 1][j] = ReverseBits(j, i);
}
}
}
inline int FastReverseBits(int i, int NumBits) {
return NumBits <= MaxFastBits ? gFFTBitTable[NumBits - 1][i] : ReverseBits(i, NumBits);
}
void FFT(bool InverseTransform, vector >& In, vector >& Out) {
if (!gFFTBitTable) { InitFFT(); }
// simultaneous data copy and bit-reversal ordering into outputs
int NumSamples = In.size();
int NumBits = NumberOfBitsNeeded(NumSamples);
for (int i = 0; i < NumSamples; ++i) {
Out[FastReverseBits(i, NumBits)] = In[i];
}
// the FFT process
double angle_numerator = acos(-1.) * (InverseTransform ? -2 : 2);
for (int BlockEnd = 1, BlockSize = 2; BlockSize <= NumSamples; BlockSize <<= 1) {
double delta_angle = angle_numerator / BlockSize;
double sin1 = sin(-delta_angle);
double cos1 = cos(-delta_angle);
double sin2 = sin(-delta_angle * 2);
double cos2 = cos(-delta_angle * 2);
for (int i = 0; i < NumSamples; i += BlockSize) {
complex a1(cos1, sin1), a2(cos2, sin2);
for (int j = i, n = 0; n < BlockEnd; ++j, ++n) {
complex a0(2 * cos1 * a1.real() - a2.real(), 2 * cos1 * a1.imag() - a2.imag());
a2 = a1;
a1 = a0;
complex a = a0 * Out[j + BlockEnd];
Out[j + BlockEnd] = Out[j] - a;
Out[j] += a;
}
}
BlockEnd = BlockSize;
}
// normalize if inverse transform
if (InverseTransform) {
for (int i = 0; i < NumSamples; ++i) {
Out[i] /= NumSamples;
}
}
}
vector convolution(vector a, vector b) {
int n = a.size();
vector > s(n), d1(n), d2(n), y(n);
for (int i = 0; i < n; ++i) {
s[i] = complex(a[i], 0);
}
FFT(false, s, d1);
s[0] = complex(b[0], 0);
for (int i = 1; i < n; ++i) {
s[i] = complex(b[n - i], 0);
}
FFT(false, s, d2);
for (int i = 0; i < n; ++i) {
y[i] = d1[i] * d2[i];
}
FFT(true, y, s);
vector ret(n);
for (int i = 0; i < n; ++i) {
ret[i] = s[i].real();
}
return ret;
}
int main() {
double a[4] = {1, 2, 3, 4}, b[4] = {1, 2, 3, 4};
vector r = convolution(vector(a, a + 4), vector(b, b + 4));
// r[0] = 30 (1*1 + 2*2 + 3*3 + 4*4)
// r[1] = 24 (1*4 + 2*1 + 3*2 + 4*3)
// r[2] = 22 (1*3 + 2*4 + 3*1 + 4*2)
// r[3] = 24 (1*2 + 2*3 + 3*4 + 4*1)
return 0;
}

Input

The first line contains n (1 <= n <= 10), the number of test cases.

For each test case, the first line contains four integers m, n, p and q, where A is a matrix of m * n, B is a matrix of p * q (2 <= m, n, p, q <= 500, m >= p, n >= q). The following m lines are the elements of A and p lines are the elements of B.

Output

For each case, print the position that gives the minimum difference (the top left corner of A is (1, 1)). You can assume that each test case has a unique solution.

Sample Input

2
2 2 2 2
1 2
3 4
2 3
1 4
3 3 2 2
0 5 5
0 5 5
0 0 0
5 5
5 5

Sample Output

1 1
1 2

Author: DU, Peng