Estimating an Affine Transform between Two Images - image

I have a sample image:
I apply the affine transform with the following warp matrix:
[[ 1.25 0. -128 ]
[ 0. 2. -192 ]]
and crop a 128x128 part from the result to get an output image:
Now, I want to estimate the warp matrix and crop size/location from just comparing the sample and output image. I detect feature points using SURF, and match them by brute force:
There are many matches, of which I'm keeping the best three (by distance), since that is the number required to estimate the affine transform. I then use those 3 keypoints to estimate the affine transform using getAffineTransform. However, the transform it returns is completely wrong:
-0.00 1.87 -6959230028596648489132997794229911552.00
0.00 -1.76 -0.00
What am I doing wrong? Source code is below.
Perform affine transform (Python):
"""Apply an affine transform to an image."""
import cv
import sys
import numpy as np
if len(sys.argv) != 10:
print "usage: %s in.png out.png x1 y1 width height sx sy flip" % __file__
sys.exit(-1)
source = cv.LoadImage(sys.argv[1])
x1, y1, width, height, sx, sy, flip = map(float, sys.argv[3:])
X, Y = cv.GetSize(source)
Xn, Yn = int(sx*(X-1)), int(sy*(Y-1))
if flip:
arr = np.array([[-sx, 0, sx*(X-1)-x1], [0, sy, -y1]])
else:
arr = np.array([[sx, 0, -x1], [0, sy, -y1]])
print arr
warp = cv.fromarray(arr)
cv.ShowImage("source", source)
dest = cv.CreateImage((Xn, Yn), source.depth, source.nChannels)
cv.WarpAffine(source, dest, warp)
cv.SetImageROI(dest, (0, 0, int(width), int(height)))
cv.ShowImage("dest", dest)
cv.SaveImage(sys.argv[2], dest)
cv.WaitKey(0)
Estimate affine transform from two images (C++):
#include <stdio.h>
#include <iostream>
#include <opencv2/core/core.hpp>
#include <opencv2/features2d/features2d.hpp>
#include <opencv2/calib3d/calib3d.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/nonfree/nonfree.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <algorithm>
using namespace cv;
void readme();
bool cmpfun(DMatch a, DMatch b) { return a.distance < b.distance; }
/** #function main */
int main( int argc, char** argv )
{
if( argc != 3 )
{
return -1;
}
Mat img_1 = imread( argv[1], CV_LOAD_IMAGE_GRAYSCALE );
Mat img_2 = imread( argv[2], CV_LOAD_IMAGE_GRAYSCALE );
if( !img_1.data || !img_2.data )
{
return -1;
}
//-- Step 1: Detect the keypoints using SURF Detector
int minHessian = 400;
SurfFeatureDetector detector( minHessian );
std::vector<KeyPoint> keypoints_1, keypoints_2;
detector.detect( img_1, keypoints_1 );
detector.detect( img_2, keypoints_2 );
//-- Step 2: Calculate descriptors (feature vectors)
SurfDescriptorExtractor extractor;
Mat descriptors_1, descriptors_2;
extractor.compute( img_1, keypoints_1, descriptors_1 );
extractor.compute( img_2, keypoints_2, descriptors_2 );
//-- Step 3: Matching descriptor vectors with a brute force matcher
BFMatcher matcher(NORM_L2, false);
std::vector< DMatch > matches;
matcher.match( descriptors_1, descriptors_2, matches );
double max_dist = 0;
double min_dist = 100;
//-- Quick calculation of max and min distances between keypoints
for( int i = 0; i < descriptors_1.rows; i++ )
{ double dist = matches[i].distance;
if( dist < min_dist ) min_dist = dist;
if( dist > max_dist ) max_dist = dist;
}
printf("-- Max dist : %f \n", max_dist );
printf("-- Min dist : %f \n", min_dist );
//-- Draw only "good" matches (i.e. whose distance is less than 2*min_dist )
//-- PS.- radiusMatch can also be used here.
sort(matches.begin(), matches.end(), cmpfun);
std::vector< DMatch > good_matches;
vector<Point2f> match1, match2;
for (int i = 0; i < 3; ++i)
{
good_matches.push_back( matches[i]);
Point2f pt1 = keypoints_1[matches[i].queryIdx].pt;
Point2f pt2 = keypoints_2[matches[i].trainIdx].pt;
match1.push_back(pt1);
match2.push_back(pt2);
printf("%3d pt1: (%.2f, %.2f) pt2: (%.2f, %.2f)\n", i, pt1.x, pt1.y, pt2.x, pt2.y);
}
//-- Draw matches
Mat img_matches;
drawMatches( img_1, keypoints_1, img_2, keypoints_2, good_matches, img_matches,
Scalar::all(-1), Scalar::all(-1), vector<char>(), DrawMatchesFlags::NOT_DRAW_SINGLE_POINTS);
//-- Show detected matches
imshow("Matches", img_matches );
imwrite("matches.png", img_matches);
waitKey(0);
Mat fun = getAffineTransform(match1, match2);
for (int i = 0; i < fun.rows; ++i)
{
for (int j = 0; j < fun.cols; j++)
{
printf("%.2f ", fun.at<float>(i,j));
}
printf("\n");
}
return 0;
}
/** #function readme */
void readme()
{
std::cout << " Usage: ./SURF_descriptor <img1> <img2>" << std::endl;
}

The cv::Mat getAffineTransform returns is made of doubles, not of floats. The matrix you get probably is fine, you just have to change the printf command in your loops to
printf("%.2f ", fun.at<double>(i,j));
or even easier: Replace this manual output with
std::cout << fun << std::endl;
It's shorter and you don't have to care about data types yourself.

Related

Dealing with matrices in CUDA: understanding basic concepts

I'm building a CUDA kernel to compute the numerical N*N jacobian of a function, using finite differences; in the example I provided, it is the square function (each entry of the vector is squared). The host coded allocates in linear memory, while I'm using a 2-dimensional indexing in the kernel.
My issue is that I haven't found a way to sum on the diagonal of the matrices cudaMalloc'ed. My attempt has been to use the statement threadIdx.x == blockIdx.x as a condition for the diagonal, but instead it evaluates to true only for them both at 0.
Here is the kernel and EDIT: I posted the whole code as an answer, based on the suggestions in the comments (the main() is basically the same, while the kernel is not)
template <typename T>
__global__ void jacobian_kernel (
T * J,
const T t0,
const T tn,
const T h,
const T * u0,
const T * un,
const T * un_old)
{
T cgamma = 2 - sqrtf(2);
const unsigned int t = threadIdx.x;
const unsigned int b = blockIdx.x;
const unsigned int tid = t + b * blockDim.x;
/*__shared__*/ T temp_sx[BLOCK_SIZE][BLOCK_SIZE];
/*__shared__*/ T temp_dx[BLOCK_SIZE][BLOCK_SIZE];
__shared__ T sm_temp_du[BLOCK_SIZE];
T* temp_du = &sm_temp_du[0];
if (tid < N )
{
temp_sx[b][t] = un[t];
temp_dx[b][t] = un[t];
if ( t == b )
{
if ( tn == t0 )
{
temp_du[t] = u0[t]*0.001;
temp_sx[b][t] += temp_du[t]; //(*)
temp_dx[b][t] -= temp_du[t];
temp_sx[b][t] += ( abs( temp_sx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_dx[b][t] += ( abs( temp_dx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_sx[b][t] = ( temp_sx[b][t] == 0 ? 0.1 : temp_sx[b][t] );
temp_dx[b][t] = ( temp_dx[b][t] == 0 ? 0.1 : temp_dx[b][t] );
}
else
{
temp_du[t] = MAX( un[t] - un_old[t], 10e-6 );
temp_sx[b][t] += temp_du[t];
temp_dx[b][t] -= temp_du[t];
}
}
__syncthreads();
//J = f(tn, un + du)
d_func(tn, (temp_sx[b]), (temp_sx[b]), 1.f);
d_func(tn, (temp_dx[b]), (temp_dx[b]), 1.f);
__syncthreads();
J[tid] = (temp_sx[b][t] - temp_dx[b][t]) * powf((2 * temp_du[t]), -1);
//J[tid]*= - h*cgamma/2;
//J[tid]+= ( t == b ? 1 : 0);
//J[tid] = temp_J[tid];
}
}
The general procedure for computing the jacobian is
Copy un into every row of temp_sx and temp_dx
Compute du as a 0.01 magnitude from u0
Sum du to the diagonal of temp_sx, subtract du from the diagonal of temp_dx
Compute the square function on each entry of temp_sx and temp_dx
Subtract them and divide every entry by 2*du
This procedure can be summarized with (f(un + du*e_i) - f(un - du*e_i))/2*du.
My problem is to sum du to the diagonal of the matrices of temp_sx and temp_dx like I tried in (*). How can I achieve that?
EDIT: Now calling 1D blocks and threads; in fact, .y axis wasn't used at all in the kernel. I'm calling the kernel with a fixed amount of shared memory
Note that in int main() I'm calling the kernel with
#define REAL sizeof(float)
#define N 32
#define BLOCK_SIZE 16
#define NUM_BLOCKS ((N*N + BLOCK_SIZE - 1)/ BLOCK_SIZE)
...
dim3 dimGrid(NUM_BLOCKS,);
dim3 dimBlock(BLOCK_SIZE);
size_t shm_size = N*N*REAL;
jacobian_kernel <<< dimGrid, dimBlock, size_t shm_size >>> (...);
So that I attempt to deal with block-splitting the function calls. In the kernel to sum on the diagonal I used if(threadIdx.x == blockIdx.x){...}. Why isn't this correct? I'm asking it because while debugging and making the code print the statement, It only evaluates true if they both are 0. Thus du[0] is the only numerical value and the matrix becomes nan. Note that this approach worked with the first code I built, where instead I called the kernel with
jacobian_kernel <<< N, N >>> (...)
So that when threadIdx.x == blockIdx.x the element is on the diagonal. This approach doesn't fit anymore though, since now I need to deal with larger N (possibly larger than 1024, which is the maximum number of threads per block).
What statement should I put there that works even if the matrices are split into blocks and threads?
Let me know if I should share some other info.
Here is how I managed to solve my problem, based on the suggestion in the comments on the answer. The example is compilable, provided you put helper_cuda.h and helper_string.h in the same directory or you add -I directive to the CUDA examples include path, installed along with the CUDA toolkit. The relevant changes are only in the kernel; there's a minor change in the main() though, since I was calling double the resources to execute the kernel, but the .y axis of the grid of thread blocks wasn't even used at all, so it didn't generate any error.
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <math.h>
#include <assert.h>
#include <cuda.h>
#include <cuda_runtime.h>
#include "helper_cuda.h"
#include "helper_string.h"
#include <fstream>
#ifndef MAX
#define MAX(a,b) ((a > b) ? a : b)
#endif
#define REAL sizeof(float)
#define N 128
#define BLOCK_SIZE 128
#define NUM_BLOCKS ((N*N + BLOCK_SIZE - 1)/ BLOCK_SIZE)
template <typename T>
inline void printmatrix( T mat, int rows, int cols);
template <typename T>
__global__ void jacobian_kernel ( const T * A, T * J, const T t0, const T tn, const T h, const T * u0, const T * un, const T * un_old);
template<typename T>
__device__ void d_func(const T t, const T u[], T res[], const T h = 1);
template<typename T>
int main ()
{
float t0 = 0.; //float tn = 0.;
float h = 0.1;
float* u0 = (float*)malloc(REAL*N); for(int i = 0; i < N; ++i){u0[i] = i+1;}
float* un = (float*)malloc(REAL*N); memcpy(un, u0, REAL*N);
float* un_old = (float*)malloc(REAL*N); memcpy(un_old, u0, REAL*N);
float* J = (float*)malloc(REAL*N*N);
float* A = (float*)malloc(REAL*N*N); host_heat_matrix(A);
float *d_u0;
float *d_un;
float *d_un_old;
float *d_J;
float *d_A;
checkCudaErrors(cudaMalloc((void**)&d_u0, REAL*N)); //printf("1: %p\n", d_u0);
checkCudaErrors(cudaMalloc((void**)&d_un, REAL*N)); //printf("2: %p\n", d_un);
checkCudaErrors(cudaMalloc((void**)&d_un_old, REAL*N)); //printf("3: %p\n", d_un_old);
checkCudaErrors(cudaMalloc((void**)&d_J, REAL*N*N)); //printf("4: %p\n", d_J);
checkCudaErrors(cudaMalloc((void**)&d_A, REAL*N*N)); //printf("4: %p\n", d_J);
checkCudaErrors(cudaMemcpy(d_u0, u0, REAL*N, cudaMemcpyHostToDevice)); assert(d_u0 != NULL);
checkCudaErrors(cudaMemcpy(d_un, un, REAL*N, cudaMemcpyHostToDevice)); assert(d_un != NULL);
checkCudaErrors(cudaMemcpy(d_un_old, un_old, REAL*N, cudaMemcpyHostToDevice)); assert(d_un_old != NULL);
checkCudaErrors(cudaMemcpy(d_J, J, REAL*N*N, cudaMemcpyHostToDevice)); assert(d_J != NULL);
checkCudaErrors(cudaMemcpy(d_A, A, REAL*N*N, cudaMemcpyHostToDevice)); assert(d_A != NULL);
dim3 dimGrid(NUM_BLOCKS); std::cout << "NUM_BLOCKS \t" << dimGrid.x << "\n";
dim3 dimBlock(BLOCK_SIZE); std::cout << "BLOCK_SIZE \t" << dimBlock.x << "\n";
size_t shm_size = N*REAL; //std::cout << shm_size << "\n";
//HERE IS A RELEVANT CHANGE OF THE MAIN, SINCE I WAS CALLING
//THE KERNEL WITH A 2D GRID BUT WITHOUT USING THE .y AXIS,
//WHILE NOW THE GRID IS 1D
jacobian_kernel <<< dimGrid, dimBlock, shm_size >>> (d_A, d_J, t0, t0, h, d_u0, d_un, d_un_old);
checkCudaErrors(cudaMemcpy(J, d_J, REAL*N*N, cudaMemcpyDeviceToHost)); //printf("4: %p\n", d_J);
printmatrix( J, N, N);
checkCudaErrors(cudaDeviceReset());
free(u0);
free(un);
free(un_old);
free(J);
}
template <typename T>
__global__ void jacobian_kernel (
const T * A,
T * J,
const T t0,
const T tn,
const T h,
const T * u0,
const T * un,
const T * un_old)
{
T cgamma = 2 - sqrtf(2);
const unsigned int t = threadIdx.x;
const unsigned int b = blockIdx.x;
const unsigned int tid = t + b * blockDim.x;
/*__shared__*/ T temp_sx[BLOCK_SIZE][BLOCK_SIZE];
/*__shared__*/ T temp_dx[BLOCK_SIZE][BLOCK_SIZE];
__shared__ T sm_temp_du;
T* temp_du = &sm_temp_du;
//HERE IS A RELEVANT CHANGE (*)
if ( t < BLOCK_SIZE && b < NUM_BLOCKS )
{
temp_sx[b][t] = un[t]; //printf("temp_sx[%d] = %f\n", t,(temp_sx[b][t]));
temp_dx[b][t] = un[t];
//printf("t = %d, b = %d, t + b * blockDim.x = %d \n",t, b, tid);
//HERE IS A NOTE (**)
if ( t == b )
{
//printf("t = %d, b = %d \n",t, b);
if ( tn == t0 )
{
*temp_du = u0[t]*0.001;
temp_sx[b][t] += *temp_du;
temp_dx[b][t] -= *temp_du;
temp_sx[b][t] += ( abs( temp_sx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_dx[b][t] += ( abs( temp_dx[b][t] ) < 10e-6 ? 0.1 : 0 );
temp_sx[b][t] = ( temp_sx[b][t] == 0 ? 0.1 : temp_sx[b][t] );
temp_dx[b][t] = ( temp_dx[b][t] == 0 ? 0.1 : temp_dx[b][t] );
}
else
{
*temp_du = MAX( un[t] - un_old[t], 10e-6 );
temp_sx[b][t] += *temp_du;
temp_dx[b][t] -= *temp_du;
}
;
}
//printf("du[%d] %f\n", tid, (*temp_du));
__syncthreads();
//printf("temp_sx[%d][%d] = %f\n", b, t, temp_sx[b][t]);
//printf("temp_dx[%d][%d] = %f\n", b, t, temp_dx[b][t]);
//d_func(tn, (temp_sx[b]), (temp_sx[b]), 1.f);
//d_func(tn, (temp_dx[b]), (temp_dx[b]), 1.f);
matvec_dev( tn, A, (temp_sx[b]), (temp_sx[b]), N, N, 1.f );
matvec_dev( tn, A, (temp_dx[b]), (temp_dx[b]), N, N, 1.f );
__syncthreads();
//printf("temp_sx_later[%d][%d] = %f\n", b, t, (temp_sx[b][t]));
//printf("temp_sx_later[%d][%d] - temp_dx_later[%d][%d] = %f\n", b,t,b,t, (temp_sx[b][t] - temp_dx[b][t]) / 2 * *temp_du);
//if (t == b ) printf( "2du[%d]^-1 = %f\n",t, powf((2 * *temp_du), -1));
J[tid] = (temp_sx[b][t] - temp_dx[b][t]) / (2 * *temp_du);
}
}
template<typename T>
__device__ void d_func(const T t, const T u[], T res[], const T h )
{
__shared__ float temp_u;
temp_u = u[threadIdx.x];
res[threadIdx.x] = h*powf( (temp_u), 2);
}
template <typename T>
inline void printmatrix( T mat, int rows, int cols)
{
std::ofstream matrix_out;
matrix_out.open( "heat_matrix.txt", std::ofstream::out);
for( int i = 0; i < rows; i++)
{
for( int j = 0; j <cols; j++)
{
double next = mat[i + N*j];
matrix_out << ( (next >= 0) ? " " : "") << next << " ";
}
matrix_out << "\n";
}
}
The relevant change is on (*). Before I used if (tid < N) which has two downsides:
First, it is wrong, since it should be tid < N*N, as my data is 2D, while tid is a global index which tracks all the data.
Even if I wrote tid < N*N, since I'm splitting the function calls into blocks, the t < BLOCK_SIZE && b < NUM_BLOCKS seems clearer to me in how the indexing is arranged in the code.
Moreover, the statement t == b in (**) is actually the right one to operate on the diagonal elements of the matrix. The fact that it was evaluated true only on 0 was because of my error right above.
Thanks for the suggestions!

Comparing arrays for "similarity"?

I am trying to compare same size arrays.
Given the following array:
I am looking for an algorithm to tell me the most "similar" array to the input. I understand that the word "similar" is not very specific but I don't know how to be more specific.
For example the following is very similar to the input.
The following is somewhat similar.
The following is very different.
You could apply a smoothing kernel to the array, and then compute the L2 norm (Euclidean distance) on it.
This is often used to compare e.g. neural spike trains or other continuous signals.
http://www.cs.utah.edu/~suresh/papers/kerneld/kerneld.pdf
You didn't specify a language...I happen to have code in C++ (may not be the most efficient).
First, you do a smoothing of the vector based on your desired kernel width and parameterize it depending on the scale/desired amount of "blur", etc.. For example:
Output of code below (behaves as expected):
riveale#rv-mba:~/tmpdir$ g++ -std=c++11 test.cpp -o test.exe
riveale#rv-mba:~/tmpdir$ ./test.exe
Distance [1] to [2]: [31.488026] (should be far)
Distance [2] to [3]: [26.591297] (should be far)
Distance [1] to [3]: [12.468342] (should be closer)
And code (test.cpp):
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cmath>
double gauss_kernel_funct(const size_t& sourcetime, const size_t& thistime)
{
const double tauval = 5.0; //width of kernel
double dist = ((sourcetime-thistime)/tauval); //distance between the points in the vector
double retval = exp(-1 * dist*dist); //exponential decay away from center of that point, squared....
return retval;
}
std::vector<double> convolvegauss( const std::vector<double>& v1)
{
std::vector<double> convolved( v1.size(), 0.0 );
for(size_t t=0; t<v1.size(); ++t)
{
for(size_t u=0; u<v1.size(); ++u)
{
double coeff = gauss_kernel_funct(u, t);
convolved[t]+=v1[u] * coeff;
}
}
return (convolved);
}
double eucliddist( const std::vector<double>& v1, const std::vector<double>& v2 )
{
if(v1.size() != v2.size()) { fprintf(stderr, "ERROR v1!=v2 sizes\n"); exit(1); }
double sum=0.0;
for(size_t x=0; x<v1.size(); ++x)
{
double tmp = (v1[x] - v2[x]);
sum += tmp*tmp; //sum += distance of this dimension squared
}
return (sqrt( sum ));
}
double vectdist( const std::vector<double>& v1, const std::vector<double>& v2 )
{
std::vector<double> convolved1 = convolvegauss( v1 );
std::vector<double> convolved2 = convolvegauss( v2 );
return (eucliddist( convolved1, convolved2 ));
}
int main()
{
//Original 3 vectors. (1 and 3) are closer than (1 and 2) or (2 and 3)...like your example.
std::vector<double> myvector1 = {1.0, 32.0, 10.0, 5.0, 2.0};
std::vector<double> myvector2 = {2.0, 3.0, 10.0, 22.0, 2.0};
std::vector<double> myvector3 = {2.0, 20.0, 17.0, 1.0, 2.0};
//Now run the vectdist on each, which convolves each vector with the gaussian kernel, and takes the euclid distance between the convovled vectors)
fprintf(stdout, "Distance [%d] to [%d]: [%lf] (should be far)\n", 1, 2, vectdist(myvector1, myvector2) );
fprintf(stdout, "Distance [%d] to [%d]: [%lf] (should be far)\n", 2, 3, vectdist(myvector2, myvector3) );
fprintf(stdout, "Distance [%d] to [%d]: [%lf] (should be closer)\n", 1, 3, vectdist(myvector1, myvector3) );
return 0;
}

Function of Values

I am using the Mat_<float> aque(3,3);
aque << .240 , .640 , .450
,0,0,0
,0,0,0
I want to turn it in to light blue , how I can adjust these values ? I want to know the function of these three values .240 , .640 , .450 in matrix , I know these all are presenting blue color but what are their functionalities what they all 3 values present ?
just go and do your own experiments ?
int f[9] = {24,64,45,0,0,0,0,0,0}; // initial trackbar pos
Mat filt(3,3,CV_32F);
Mat img;
void onTrack(int,void*)
{
float *p = filt.ptr<float>(0);
for ( int i=0; i<9; i++ )
{
p[i] = float(f[i]) / 100; // get it back to [0..1] range for the transform matrix
}
Mat out;
cv::transform(img,out,filt);
imshow("result",out);
}
int main( int argc, const char** argv )
{
img = imread("lena.jpg");
namedWindow("result",0);
namedWindow("sliders",0);
createTrackbar("1","sliders",&f[0],100,onTrack);
createTrackbar("2","sliders",&f[1],100,onTrack);
createTrackbar("3","sliders",&f[2],100,onTrack);
createTrackbar("4","sliders",&f[3],100,onTrack);
createTrackbar("5","sliders",&f[4],100,onTrack);
createTrackbar("6","sliders",&f[5],100,onTrack);
createTrackbar("7","sliders",&f[6],100,onTrack);
createTrackbar("8","sliders",&f[7],100,onTrack);
createTrackbar("9","sliders",&f[8],100,onTrack);
onTrack(0,0);
waitKey();
cerr << filt << endl;
return 0;
}
if you 're happy with the results, just divide each slider value by 100, and compose a filter matrix from that.

Piecemeal processing of a matrix - CUDA

OK, so lets say I have an ( N x N ) matrix that I would like to process. This matrix is quite large for my computer, and if I try to send it to the device all at once I get a 'out of memory error.'
So is there a way to send sections of the matrix to the device? One way I can see to do it is copy portions of the matrix on the host, and then send these manageable copied portions from the host to the device, and then put them back together at the end.
Here is something I have tried, but the cudaMemcpy in the for loop returns error code 11, 'invalid argument.'
int h_N = 10000;
size_t h_size_m = h_N*sizeof(float);
h_A = (float*)malloc(h_size_m*h_size_m);
int d_N = 2500;
size_t d_size_m = d_N*sizeof(float);
InitializeMatrices(h_N);
int i;
int iterations = (h_N*h_N)/(d_N*d_N);
for( i = 0; i < iterations; i++ )
{
float* h_array_ref = h_A+(i*d_N*d_N);
cudasafe( cudaMemcpy(d_A, h_array_ref, d_size_m*d_size_m, cudaMemcpyHostToDevice), "cudaMemcpy");
cudasafe( cudaFree(d_A), "cudaFree(d_A)" );
}
What I'm trying to accomplish with the above code is this: instead of send the entire matrix to the device, I simply send a pointer to a place within that matrix and reserve enough space on the device to do the work, and then with the next iteration of the loop move the pointer forward within the matrix, etc. etc.
Not only can you do this (assuming your problem is easily decomposed this way into sub-arrays), it can be a very useful thing to do for performance; once you get the basic approach you've described working, you can start using asynchronous memory copies and double-buffering to overlap some of the memory transfer time with the time spent computing what is already on-card.
But first one gets the simple thing working. Below is a 1d example (multiplying a vector by a scalar and adding another scalar) but using a linearized 2d array would be the same; the key part is
CHK_CUDA( cudaMalloc(&xd, batchsize*sizeof(float)) );
CHK_CUDA( cudaMalloc(&yd, batchsize*sizeof(float)) );
tick(&gputimer);
int nbatches = 0;
for (int nstart=0; nstart < n; nstart+=batchsize) {
int size=batchsize;
if ((nstart + batchsize) > n) size = n - nstart;
CHK_CUDA( cudaMemcpy(xd, &(x[nstart]), size*sizeof(float), cudaMemcpyHostToDevice) );
blocksize = (size+nblocks-1)/nblocks;
cuda_saxpb<<<nblocks, blocksize>>>(xd, a, b, yd, size);
CHK_CUDA( cudaMemcpy(&(ycuda[nstart]), yd, size*sizeof(float), cudaMemcpyDeviceToHost) );
nbatches++;
}
gputime = tock(&gputimer);
CHK_CUDA( cudaFree(xd) );
CHK_CUDA( cudaFree(yd) );
You allocate the buffers at the start, and then loop through until you're done, each time doing the copy, starting the kernel, and then copying back. You free at the end.
The full code is
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
#include <cuda.h>
#include <sys/time.h>
#include <math.h>
#define CHK_CUDA(e) {if (e != cudaSuccess) {fprintf(stderr,"Error: %s\n", cudaGetErrorString(e)); exit(-1);}}
__global__ void cuda_saxpb(const float *xd, const float a, const float b,
float *yd, const int n) {
int i = threadIdx.x + blockIdx.x*blockDim.x;
if (i<n) {
yd[i] = a*xd[i]+b;
}
return;
}
void cpu_saxpb(const float *x, float a, float b, float *y, int n) {
int i;
for (i=0;i<n;i++) {
y[i] = a*x[i]+b;
}
return;
}
int get_options(int argc, char **argv, int *n, int *s, int *nb, float *a, float *b);
void tick(struct timeval *timer);
double tock(struct timeval *timer);
int main(int argc, char **argv) {
int n=1000;
int nblocks=10;
int batchsize=100;
float a = 5.;
float b = -1.;
int err;
float *x, *y, *ycuda;
float *xd, *yd;
double abserr;
int blocksize;
int i;
struct timeval cputimer;
struct timeval gputimer;
double cputime, gputime;
err = get_options(argc, argv, &n, &batchsize, &nblocks, &a, &b);
if (batchsize > n) {
fprintf(stderr, "Resetting batchsize to size of vector, %d\n", n);
batchsize = n;
}
if (err) return 0;
x = (float *)malloc(n*sizeof(float));
if (!x) return 1;
y = (float *)malloc(n*sizeof(float));
if (!y) {free(x); return 1;}
ycuda = (float *)malloc(n*sizeof(float));
if (!ycuda) {free(y); free(x); return 1;}
/* run CPU code */
tick(&cputimer);
cpu_saxpb(x, a, b, y, n);
cputime = tock(&cputimer);
/* run GPU code */
/* only have to allocate once */
CHK_CUDA( cudaMalloc(&xd, batchsize*sizeof(float)) );
CHK_CUDA( cudaMalloc(&yd, batchsize*sizeof(float)) );
tick(&gputimer);
int nbatches = 0;
for (int nstart=0; nstart < n; nstart+=batchsize) {
int size=batchsize;
if ((nstart + batchsize) > n) size = n - nstart;
CHK_CUDA( cudaMemcpy(xd, &(x[nstart]), size*sizeof(float), cudaMemcpyHostToDevice) );
blocksize = (size+nblocks-1)/nblocks;
cuda_saxpb<<<nblocks, blocksize>>>(xd, a, b, yd, size);
CHK_CUDA( cudaMemcpy(&(ycuda[nstart]), yd, size*sizeof(float), cudaMemcpyDeviceToHost) );
nbatches++;
}
gputime = tock(&gputimer);
CHK_CUDA( cudaFree(xd) );
CHK_CUDA( cudaFree(yd) );
abserr = 0.;
for (i=0;i<n;i++) {
abserr += fabs(ycuda[i] - y[i]);
}
printf("Y = a*X + b, problemsize = %d\n", n);
printf("CPU time = %lg millisec.\n", cputime*1000.);
printf("GPU time = %lg millisec (done with %d batches of %d).\n",
gputime*1000., nbatches, batchsize);
printf("CUDA and CPU results differ by %lf\n", abserr);
free(x);
free(y);
free(ycuda);
return 0;
}
int get_options(int argc, char **argv, int *n, int *s, int *nb, float *a, float *b) {
const struct option long_options[] = {
{"nvals" , required_argument, 0, 'n'},
{"nblocks" , required_argument, 0, 'B'},
{"batchsize" , required_argument, 0, 's'},
{"a", required_argument, 0, 'a'},
{"b", required_argument, 0, 'b'},
{"help", no_argument, 0, 'h'},
{0, 0, 0, 0}};
char c;
int option_index;
int tempint;
while (1) {
c = getopt_long(argc, argv, "n:B:a:b:s:h", long_options, &option_index);
if (c == -1) break;
switch(c) {
case 'n': tempint = atoi(optarg);
if (tempint < 1 || tempint > 500000) {
fprintf(stderr,"%s: Cannot use number of points %s;\n Using %d\n", argv[0], optarg, *n);
} else {
*n = tempint;
}
break;
case 's': tempint = atoi(optarg);
if (tempint < 1 || tempint > 50000) {
fprintf(stderr,"%s: Cannot use number of points %s;\n Using %d\n", argv[0], optarg, *s);
} else {
*s = tempint;
}
break;
case 'B': tempint = atoi(optarg);
if (tempint < 1 || tempint > 1000 || tempint > *n) {
fprintf(stderr,"%s: Cannot use number of blocks %s;\n Using %d\n", argv[0], optarg, *nb);
} else {
*nb = tempint;
}
break;
case 'a': *a = atof(optarg);
break;
case 'b': *b = atof(optarg);
break;
case 'h':
puts("Calculates y[i] = a*x[i] + b on the GPU.");
puts("Options: ");
puts(" --nvals=N (-n N): Set the number of values in y,x.");
puts(" --batchsize=N (-s N): Set the number of values to transfer at a time.");
puts(" --nblocks=N (-B N): Set the number of blocks used.");
puts(" --a=X (-a X): Set the parameter a.");
puts(" --b=X (-b X): Set the parameter b.");
puts(" --niters=N (-I X): Set number of iterations to calculate.");
puts("");
return +1;
}
}
return 0;
}
void tick(struct timeval *timer) {
gettimeofday(timer, NULL);
}
double tock(struct timeval *timer) {
struct timeval now;
gettimeofday(&now, NULL);
return (now.tv_usec-timer->tv_usec)/1.0e6 + (now.tv_sec - timer->tv_sec);
}
Running this one gets:
$ ./batched-saxpb --nvals=10240 --batchsize=10240 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.072 millisec.
GPU time = 0.117 millisec (done with 1 batches of 10240).
CUDA and CPU results differ by 0.000000
$ ./batched-saxpb --nvals=10240 --batchsize=5120 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.066 millisec.
GPU time = 0.133 millisec (done with 2 batches of 5120).
CUDA and CPU results differ by 0.000000
$ ./batched-saxpb --nvals=10240 --batchsize=2560 --nblocks=20
Y = a*X + b, problemsize = 10240
CPU time = 0.067 millisec.
GPU time = 0.167 millisec (done with 4 batches of 2560).
CUDA and CPU results differ by 0.000000
The GPU time goes up in this case (we're doing more memory copies) but the answers stay the same.
Edited: The original version of this code had an option for running multiple iterations of the kernel for timing purposes, but that's unnecessarily confusing in this context so it's removed.

Find local maxima in grayscale image using OpenCV

Does anybody know how to find the local maxima in a grayscale IPL_DEPTH_8U image using OpenCV? HarrisCorner mentions something like that but I'm actually not interested in corners ...
Thanks!
A pixel is considered a local maximum if it is equal to the maximum value in a 'local' neighborhood. The function below captures this property in two lines of code.
To deal with pixels on 'plateaus' (value equal to their neighborhood) one can use the local minimum property, since plateaus pixels are equal to their local minimum. The rest of the code filters out those pixels.
void non_maxima_suppression(const cv::Mat& image, cv::Mat& mask, bool remove_plateaus) {
// find pixels that are equal to the local neighborhood not maximum (including 'plateaus')
cv::dilate(image, mask, cv::Mat());
cv::compare(image, mask, mask, cv::CMP_GE);
// optionally filter out pixels that are equal to the local minimum ('plateaus')
if (remove_plateaus) {
cv::Mat non_plateau_mask;
cv::erode(image, non_plateau_mask, cv::Mat());
cv::compare(image, non_plateau_mask, non_plateau_mask, cv::CMP_GT);
cv::bitwise_and(mask, non_plateau_mask, mask);
}
}
Here's a simple trick. The idea is to dilate with a kernel that contains a hole in the center. After the dilate operation, each pixel is replaced with the maximum of it's neighbors (using a 5 by 5 neighborhood in this example), excluding the original pixel.
Mat1b kernelLM(Size(5, 5), 1u);
kernelLM.at<uchar>(2, 2) = 0u;
Mat imageLM;
dilate(image, imageLM, kernelLM);
Mat1b localMaxima = (image > imageLM);
Actually after I posted the code above I wrote a better and very very faster one ..
The code above suffers even for a 640x480 picture..
I optimized it and now it is very very fast even for 1600x1200 pic.
Here is the code :
void localMaxima(cv::Mat src,cv::Mat &dst,int squareSize)
{
if (squareSize==0)
{
dst = src.clone();
return;
}
Mat m0;
dst = src.clone();
Point maxLoc(0,0);
//1.Be sure to have at least 3x3 for at least looking at 1 pixel close neighbours
// Also the window must be <odd>x<odd>
SANITYCHECK(squareSize,3,1);
int sqrCenter = (squareSize-1)/2;
//2.Create the localWindow mask to get things done faster
// When we find a local maxima we will multiply the subwindow with this MASK
// So that we will not search for those 0 values again and again
Mat localWindowMask = Mat::zeros(Size(squareSize,squareSize),CV_8U);//boolean
localWindowMask.at<unsigned char>(sqrCenter,sqrCenter)=1;
//3.Find the threshold value to threshold the image
//this function here returns the peak of histogram of picture
//the picture is a thresholded picture it will have a lot of zero values in it
//so that the second boolean variable says :
// (boolean) ? "return peak even if it is at 0" : "return peak discarding 0"
int thrshld = maxUsedValInHistogramData(dst,false);
threshold(dst,m0,thrshld,1,THRESH_BINARY);
//4.Now delete all thresholded values from picture
dst = dst.mul(m0);
//put the src in the middle of the big array
for (int row=sqrCenter;row<dst.size().height-sqrCenter;row++)
for (int col=sqrCenter;col<dst.size().width-sqrCenter;col++)
{
//1.if the value is zero it can not be a local maxima
if (dst.at<unsigned char>(row,col)==0)
continue;
//2.the value at (row,col) is not 0 so it can be a local maxima point
m0 = dst.colRange(col-sqrCenter,col+sqrCenter+1).rowRange(row-sqrCenter,row+sqrCenter+1);
minMaxLoc(m0,NULL,NULL,NULL,&maxLoc);
//if the maximum location of this subWindow is at center
//it means we found the local maxima
//so we should delete the surrounding values which lies in the subWindow area
//hence we will not try to find if a point is at localMaxima when already found a neighbour was
if ((maxLoc.x==sqrCenter)&&(maxLoc.y==sqrCenter))
{
m0 = m0.mul(localWindowMask);
//we can skip the values that we already made 0 by the above function
col+=sqrCenter;
}
}
}
The following listing is a function similar to Matlab's "imregionalmax". It looks for at most nLocMax local maxima above threshold, where the found local maxima are at least minDistBtwLocMax pixels apart. It returns the actual number of local maxima found. Notice that it uses OpenCV's minMaxLoc to find global maxima. It is "opencv-self-contained" except for the (easy to implement) function vdist, which computes the (euclidian) distance between points (r,c) and (row,col).
input is one-channel CV_32F matrix, and locations is nLocMax (rows) by 2 (columns) CV_32S matrix.
int imregionalmax(Mat input, int nLocMax, float threshold, float minDistBtwLocMax, Mat locations)
{
Mat scratch = input.clone();
int nFoundLocMax = 0;
for (int i = 0; i < nLocMax; i++) {
Point location;
double maxVal;
minMaxLoc(scratch, NULL, &maxVal, NULL, &location);
if (maxVal > threshold) {
nFoundLocMax += 1;
int row = location.y;
int col = location.x;
locations.at<int>(i,0) = row;
locations.at<int>(i,1) = col;
int r0 = (row-minDistBtwLocMax > -1 ? row-minDistBtwLocMax : 0);
int r1 = (row+minDistBtwLocMax < scratch.rows ? row+minDistBtwLocMax : scratch.rows-1);
int c0 = (col-minDistBtwLocMax > -1 ? col-minDistBtwLocMax : 0);
int c1 = (col+minDistBtwLocMax < scratch.cols ? col+minDistBtwLocMax : scratch.cols-1);
for (int r = r0; r <= r1; r++) {
for (int c = c0; c <= c1; c++) {
if (vdist(Point2DMake(r, c),Point2DMake(row, col)) <= minDistBtwLocMax) {
scratch.at<float>(r,c) = 0.0;
}
}
}
} else {
break;
}
}
return nFoundLocMax;
}
The first question to answer would be what is "local" in your opinion. The answer may well be a square window (say 3x3 or 5x5) or circular window of a certain radius. You can then scan over the entire image with the window centered at each pixel and pick the highest value in the window.
See this for how to access pixel values in OpenCV.
This is very fast method. It stored founded maxima in a vector of
Points.
vector <Point> GetLocalMaxima(const cv::Mat Src,int MatchingSize, int Threshold, int GaussKernel )
{
vector <Point> vMaxLoc(0);
if ((MatchingSize % 2 == 0) || (GaussKernel % 2 == 0)) // MatchingSize and GaussKernel have to be "odd" and > 0
{
return vMaxLoc;
}
vMaxLoc.reserve(100); // Reserve place for fast access
Mat ProcessImg = Src.clone();
int W = Src.cols;
int H = Src.rows;
int SearchWidth = W - MatchingSize;
int SearchHeight = H - MatchingSize;
int MatchingSquareCenter = MatchingSize/2;
if(GaussKernel > 1) // If You need a smoothing
{
GaussianBlur(ProcessImg,ProcessImg,Size(GaussKernel,GaussKernel),0,0,4);
}
uchar* pProcess = (uchar *) ProcessImg.data; // The pointer to image Data
int Shift = MatchingSquareCenter * ( W + 1);
int k = 0;
for(int y=0; y < SearchHeight; ++y)
{
int m = k + Shift;
for(int x=0;x < SearchWidth ; ++x)
{
if (pProcess[m++] >= Threshold)
{
Point LocMax;
Mat mROI(ProcessImg, Rect(x,y,MatchingSize,MatchingSize));
minMaxLoc(mROI,NULL,NULL,NULL,&LocMax);
if (LocMax.x == MatchingSquareCenter && LocMax.y == MatchingSquareCenter)
{
vMaxLoc.push_back(Point( x+LocMax.x,y + LocMax.y ));
// imshow("W1",mROI);cvWaitKey(0); //For gebug
}
}
}
k += W;
}
return vMaxLoc;
}
Found a simple solution.
In this example, if you are trying to find 2 results of a matchTemplate function with a minimum distance from each other.
cv::Mat result;
matchTemplate(search, target, result, CV_TM_SQDIFF_NORMED);
float score1;
cv::Point displacement1 = MinMax(result, score1);
cv::circle(result, cv::Point(displacement1.x+result.cols/2 , displacement1.y+result.rows/2), 10, cv::Scalar(0), CV_FILLED, 8, 0);
float score2;
cv::Point displacement2 = MinMax(result, score2);
where
cv::Point MinMax(cv::Mat &result, float &score)
{
double minVal, maxVal;
cv::Point minLoc, maxLoc, matchLoc;
minMaxLoc(result, &minVal, &maxVal, &minLoc, &maxLoc, cv::Mat());
matchLoc.x = minLoc.x - result.cols/2;
matchLoc.y = minLoc.y - result.rows/2;
return minVal;
}
The process is:
Find global Minimum using minMaxLoc
Draw a filled white circle around global minimum using min distance between minima as radius
Find another minimum
The the scores can be compared to each other to determine, for example, the certainty of the match,
To find more than just the global minimum and maximum try using this function from skimage:
http://scikit-image.org/docs/dev/api/skimage.feature.html#skimage.feature.peak_local_max
You can parameterize the minimum distance between peaks, too. And more. To find minima, use negated values (take care of the array type though, 255-image could do the trick).
You can go over each pixel and test if it is a local maxima. Here is how I would do it.
The input is assumed to be type CV_32FC1
#include <vector>//std::vector
#include <algorithm>//std::sort
#include "opencv2/imgproc/imgproc.hpp"
#include "opencv2/core/core.hpp"
//structure for maximal values including position
struct SRegionalMaxPoint
{
SRegionalMaxPoint():
values(-FLT_MAX),
row(-1),
col(-1)
{}
float values;
int row;
int col;
//ascending order
bool operator()(const SRegionalMaxPoint& a, const SRegionalMaxPoint& b)
{
return a.values < b.values;
}
};
//checks if pixel is local max
bool isRegionalMax(const float* im_ptr, const int& cols )
{
float center = *im_ptr;
bool is_regional_max = true;
im_ptr -= (cols + 1);
for (int ii = 0; ii < 3; ++ii, im_ptr+= (cols-3))
{
for (int jj = 0; jj < 3; ++jj, im_ptr++)
{
if (ii != 1 || jj != 1)
{
is_regional_max &= (center > *im_ptr);
}
}
}
return is_regional_max;
}
void imregionalmax(
const cv::Mat& input,
std::vector<SRegionalMaxPoint>& buffer)
{
//find local max - top maxima
static const int margin = 1;
const int rows = input.rows;
const int cols = input.cols;
for (int i = margin; i < rows - margin; ++i)
{
const float* im_ptr = input.ptr<float>(i, margin);
for (int j = margin; j < cols - margin; ++j, im_ptr++)
{
//Check if pixel is local maximum
if ( isRegionalMax(im_ptr, cols ) )
{
cv::Rect roi = cv::Rect(j - margin, i - margin, 3, 3);
cv::Mat subMat = input(roi);
float val = *im_ptr;
//replace smallest value in buffer
if ( val > buffer[0].values )
{
buffer[0].values = val;
buffer[0].row = i;
buffer[0].col = j;
std::sort(buffer.begin(), buffer.end(), SRegionalMaxPoint());
}
}
}
}
}
For testing the code you can try this:
cv::Mat temp = cv::Mat::zeros(15, 15, CV_32FC1);
temp.at<float>(7, 7) = 1;
temp.at<float>(3, 5) = 6;
temp.at<float>(8, 10) = 4;
temp.at<float>(11, 13) = 7;
temp.at<float>(10, 3) = 8;
temp.at<float>(7, 13) = 3;
vector<SRegionalMaxPoint> buffer_(5);
imregionalmax(temp, buffer_);
cv::Mat debug;
cv::cvtColor(temp, debug, cv::COLOR_GRAY2BGR);
for (auto it = buffer_.begin(); it != buffer_.end(); ++it)
{
circle(debug, cv::Point(it->col, it->row), 1, cv::Scalar(0, 255, 0));
}
This solution does not take plateaus into account so it is not exactly the same as matlab's imregionalmax()
I think you want to use the
MinMaxLoc(arr, mask=NULL)-> (minVal, maxVal, minLoc, maxLoc)
Finds global minimum and maximum in array or subarray
function on you image

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