I have an image named HSIImage, of size is 565x585, in which I have find the local mean and standard deviation at every pixel. For this I am using a window W of size 9x9, if we a re finding the mean of x(i,j) we need values in the W where x(i,j) is at its center.
For working on the corner and edge pixels, I am padding the HSIImage and naming it as HSIImage2.
MATLAB code
[m,n,~] = size(HSIImage);
HSIImage2=padarray(HSIImage,[4,4],'symmetric');
mean1 = zeros(m,n);
sd = zeros(m,n);
phi_x=zeros(m,n);
for i=5:m+4
for j=5:n+4
mean1(i-4,j-4) = mean( mean(HSIImage2(i-4:i+4, j-4:j+4, 3) )); %sum / (4*4);
sd(i-4,j-4) = std( std(HSIImage2(i-4:i+4, j-4:j+4, 3), 1));
end
end
[phi_x2,mean2,sd2] = getPhi(HSIImage(:,:,3)',HSIImage2(:,:,3)',m,n);
Serial mean displayed as image.
My cuda code for finding mean and sd is
__global__ void phi(double *d_HSIImage,double *d_HSIImage2, int row, int col, double *d_phi_x, double *d_mean, double *d_std)
{
int X = blockDim.x * blockIdx.x + threadIdx.x;
int Y = blockDim.y * blockIdx.y + threadIdx.y;
int i,j;
double sum = 0;
if(Y>3 && X>3 && Y<row+4 && X<col+4)
{
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
sum= sum + d_HSIImage2[i*col+j];
}
}
d_mean[(Y-4)*col+X-4] = sum/81;
double mean = sum/81;
sum = 0;
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
int index = i*col+j;
sum= sum + (d_HSIImage2[index]-mean) * (d_HSIImage2[index]-mean);
}
}
d_std[(Y-4)*col+X-4] = sqrt(sum/81);
}
void mexFunction( int nlhs, mxArray *plhs[],int nrhs, const mxArray *prhs[])
{
double* HSIImage;
double* d_HSIImage;
double* HSIImage2;
double* d_HSIImage2;
double row;
double col;
double* phi_x;
double* d_phi_x;
double* mean2;
double* d_mean;
double* d_std;
double* sd2;
HSIImage = (double*)mxGetPr(prhs[0]);
HSIImage2 = (double*)mxGetPr(prhs[1]);
row = mxGetScalar(prhs[2]);
col = mxGetScalar(prhs[3]);
plhs[0] = mxCreateDoubleMatrix(row,col,mxREAL);
phi_x = mxGetPr(plhs[0]);
plhs[1] = mxCreateDoubleMatrix(row,col,mxREAL);
mean2 = mxGetPr(plhs[1]);
plhs[2] = mxCreateDoubleMatrix(row,col,mxREAL);
sd2 = mxGetPr(plhs[2]);
dim3 grid(((col+8)/TILE_WIDTH)+1,((row+8)/TILE_WIDTH)+1,1);
dim3 block(TILE_WIDTH,TILE_WIDTH,1);
if ( cudaMalloc(&d_HSIImage,sizeof(double)*row*col)!= cudaSuccess )
mexErrMsgTxt("Memory allocating failure on the GPU");
if ( cudaMalloc(&d_HSIImage2,sizeof(double)*(row+8)*(col+8))!= cudaSuccess )
mexErrMsgTxt("Memory allocating failure on the GPU");
if ( cudaMalloc(&d_phi_x,sizeof(double)*row*col)!= cudaSuccess )
mexErrMsgTxt("Memory allocating failure on the GPU");
if ( cudaMalloc(&d_mean,sizeof(double)*row*col)!= cudaSuccess )
mexErrMsgTxt("Memory allocating failure on the GPU");
if ( cudaMalloc(&d_std,sizeof(double)*row*col)!= cudaSuccess )
mexErrMsgTxt("Memory allocating failure on the GPU");
cudaMemcpy(d_HSIImage,HSIImage,sizeof(double)*row*col,cudaMemcpyHostToDevice);
cudaMemcpy(d_HSIImage2,HSIImage2,sizeof(double)*(row+8)*(col+8),cudaMemcpyHostToDevice);
phi <<< grid,block >>> (d_HSIImage,d_HSIImage2,row,col,d_phi_x,d_mean,d_std);
cudaMemcpy(phi_x,d_phi_x,sizeof(double)*row*col,cudaMemcpyDeviceToHost);
cudaMemcpy(mean2,d_mean,sizeof(double)*row*col,cudaMemcpyDeviceToHost);
cudaMemcpy(sd2,d_std,sizeof(double)*row*col,cudaMemcpyDeviceToHost);
cudaFree(d_HSIImage);
cudaFree(d_HSIImage2);
cudaFree(d_phi_x);
}
its working fine when image is full of ones. but when I give regular image, there is lot of difference in serial(MATLAB) and parallel(CUDA) outputs(When mean1 and mean2 are compared). Please tell me the error.
I am launching with
dim3 grid(((col+8)/TILE_WIDTH)+1,((row+8)/TILE_WIDTH)+1,1);
dim3 block(TILE_WIDTH,TILE_WIDTH,1);
TILEWIDTH is 32. row=565, col=584.
Parallel mean displayed as image
It is important to note Matlab's c api is column-major ordered, however as mentioned in the comments OP has made sure of the consistency. The problem is that the stride used to access the data did not include the pads of the image. Going from one row to another requires a stride of col+8 (8 being padding of 4 on each side.
changing
__global__ void phi(double *d_HSIImage,double *d_HSIImage2, int row, int col, double *d_phi_x, double *d_mean, double *d_std)
{
int X = blockDim.x * blockIdx.x + threadIdx.x;
int Y = blockDim.y * blockIdx.y + threadIdx.y;
int i,j;
double sum = 0;
if(Y>3 && X>3 && Y<row+4 && X<col+4)
{
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
sum= sum + d_HSIImage2[i*col+j];
}
}
d_mean[(Y-4)*col+X-4] = sum/81;
double mean = sum/81;
sum = 0;
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
int index = i*col+j;
sum= sum + (d_HSIImage2[index]-mean) * (d_HSIImage2[index]-mean);
}
}
d_std[(Y-4)*col+X-4] = sqrt(sum/81);
}
to
__global__ void phi(double *d_HSIImage,double *d_HSIImage2, int row, int col, double *d_phi_x, double *d_mean, double *d_std)
{
int X = blockDim.x * blockIdx.x + threadIdx.x;
int Y = blockDim.y * blockIdx.y + threadIdx.y;
int i,j;
double sum = 0;
if(Y>3 && X>3 && Y<row+4 && X<col+4)
{
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
sum= sum + d_HSIImage2[i*(col+8)+j];
}
}
d_mean[(Y-4)*col+X-4] = sum/81;
double mean = sum/81;
sum = 0;
for(i=Y-4;i<=Y+4;i++){
for(j=X-4;j<=X+4;j++){
int index = i*(col+8)+j;
sum= sum + (d_HSIImage2[index]-mean) * (d_HSIImage2[index]-mean);
}
}
d_std[(Y-4)*col+X-4] = sqrt(sum/81);
}
Should work, however, I have included a compilable example that I validated on a small sample, that should be easy to expand.
It is not optimized, but that wasn't part of your question. Optimization using shared memory would give a large performance boost.
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <cuda.h>
using namespace std;
__global__ void phi(double *img, int row, int col, double *d_mean){
int X=blockDim.x*blockIdx.x+threadIdx.x+4;
int Y=blockDim.y*blockIdx.y+threadIdx.y+4;
double sum = 0;
if(Y<row+4 && X<col+4){
for(int i=-4; i<=4; ++i){
for(int j=-4; j<=4; ++j){
sum+=img[ (Y+j)*(col+8)+X+i];
}
}
sum/=81;
d_mean[(Y-4)*col+X-4]=sum;
}
}
int main(int argc, char * argv[]) {
int width=10, height=10;
double *h_img=new double[(width+8)*(height+8)];
for(int i=0; i<height+8; i++){
for(int j=0; j<width+8; j++){
h_img[i*(width+8)+j]=0.0;
}
}
for(int i=0; i<height; i++){
for(int j=0; j<width; j++){
int index = (i+4)*(width+8)+j+4;
h_img[index]=i*width+j;
}
}
for(int i=0; i<height+8; i++){
for(int j=0; j<width+8; j++){
cout<<h_img[i*(width+8)+j]<<" ";
}cout<<endl;
}
double *d_img;
size_t size=sizeof(double)*(height+8)*(width*8);
cudaMalloc(&d_img, size);
cudaMemcpy(d_img, h_img, size, cudaMemcpyHostToDevice);
size = sizeof(double)*height*width;
double *d_avg;
cudaMalloc(&d_avg, size);
dim3 block(32, 32, 1);
dim3 grid(width/32+1, height/32+1, 1);
phi<<<grid, block>>>(d_img, height, width, d_avg);
cudaDeviceSynchronize();
double *h_avg=new double[width*height];
cudaMemcpy(h_avg, d_avg, size, cudaMemcpyDeviceToHost);
for(int i=0; i<height; i++){
for(int j=0; j<width; j++){
cout<<h_avg[i*width+j]<<" ";
}cout<<endl;
}
return 0;
}
Here's my 2 cents regarding local mean and local std.
You should check whether using matlab's optimized built-in functions (conv2 and stdfilt , with their gpu support) gives you better performance than a "simple" mex version. For example, to take the local mean, the fastest will be to use conv2 as follows:
local_mean_image=conv2(image,normalized_window,'same');
where in your case normalized_window=ones(9)./9^2;
For local std use stdfilt :
local_std_image = stdfilt(image, ones(9));
Both options are available for faster GPU performance, I use conv2 with Jacket routinely, and I saw the stdfilt supports gpuarray variables.
By observing the answers of #Christian Sarofeen and of #bla, I made some changes to my code and now I am able to find the mean exactly same as MATLAB. I posting this thinking that some one may use it in future(I am sending the image as is from MATLAB). Still finding standard deviation is little problem.
__global__ void phi(double *d_HSIImage,double *d_HSIImage2, int row, int col, double *d_phi_x, double *d_mean, double *d_std)
{
int X = blockDim.x * blockIdx.x + threadIdx.x;
int Y = blockDim.y * blockIdx.y + threadIdx.y;
int i,j;
double sum = 0;
if(Y>3 && X>3 && Y<row+4 && X<col+4)
{
int index = (X-4)*row+Y-4;
for(i=-4;i<=4;i++){
for(j=-4;j<=4;j++){
sum= sum + d_HSIImage2[(X+j)*(row+8)+(Y+i)];
}
}
d_mean[index] = sum/81;
double mean = 0;
double temp_std[9] = {0} ;
for(j=-4;j<=4;j++){
sum = 0;
for(i=-4;i<=4;i++){
sum = sum + d_HSIImage2[(X+j)*(row+8)+(Y+i)];//vector mean
}
mean = sum/9;
sum =0 ;
for(i=-4;i<=4;i++){
int index = (X+j)*(row+8)+(Y+i);
sum= sum + (d_HSIImage2[index]-mean) * (d_HSIImage2[index]-mean);
}
temp_std[j+4] = (sqrt(sum/9));//vector std
}
sum =0 ;
for(j=-4;j<=4;j++){
sum = sum + temp_std[j+4];//mean of vectors
}
mean = sum/9;
sum = 0 ;
for(j=-4;j<=4;j++){
sum = sum + (temp_std[j+4]-mean) * (temp_std[j+4]-mean);
}
d_std[index] = sqrt(sum);//std of vectors
d_phi_x[index] = 1.0/(1.0+exp((d_mean[index]-d_HSIImage[index])/d_std[index]));
}
}
Related
For a matrix such as
A = [...
12 34 67;
90 78 15;
10 71 24];
how could we determine efficiently if it is subset of a larger matrix?
B = [...
12 34 67; % found
89 67 45;
90 78 15; % found
10 71 24; % found, so A is subset of B.
54 34 11];
Here are conditions:
all numbers are integers
matrices are so large, i.e., row# > 100000, column# may vary from 1 to 10 (same for A and B).
Edit:
It seems that ismember for the case of this question, when called only few times works just fine. My initial impression was due to previous experiences where ismember was being invoked many times inside a nested loop resulting in the worst performance.
clear all; clc
n = 200000;
k = 10;
B = randi(n,n,k);
f = randperm(n);
A = B(f(1:1000),:);
tic
assert(sum(ismember(A,B,'rows')) == size(A,1));
toc
tic
assert(all(any(all(bsxfun(#eq,B,permute(A,[3,2,1])),2),1))); %user2999345
toc
which results in:
Elapsed time is 1.088552 seconds.
Elapsed time is 12.154969 seconds.
Here are more benchmarks:
clear all; clc
n = 20000;
f = randperm(n);
k = 10;
t1 = 0;
t2 = 0;
t3 = 0;
for i=1:7
B = randi(n,n,k);
A = B(f(1:n/10),:);
%A(100,2) = 0; % to make A not submat of B
tic
b = sum(ismember(A,B,'rows')) == size(A,1);
t1 = t1+toc;
assert(b);
tic
b = ismember_mex(A,sortrows(B));
t2 = t2+toc;
assert(b);
tic
b = issubmat(A,B);
t3 = t3+toc;
assert(b);
end
George's skm's
ismember | ismember_mex | issubmat
n=20000,k=10 0.6326 0.1064 11.6899
n=1000,k=100 0.2652 0.0155 0.0577
n=1000,k=1000 1.1705 0.1582 0.2202
n=1000,k=10000 13.2470 2.0033 2.6367
*issubmat eats RAM when n or k is over 10000!
*issubmat(A,B), A is being checked as submat of B.
It seems that ismember is hard to beat, at least using MATLAB code. I created a C implementation which can be used using the MEX compiler.
#include "mex.h"
#if MX_API_VER < 0x07030000
typedef int mwIndex;
typedef int mwSize;
#endif /* MX_API_VER */
#include <math.h>
#include <stdlib.h>
#include <string.h>
int ismember(const double *y, const double *x, int yrow, int xrow, int ncol);
void mexFunction(int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[])
{
mwSize xcol, ycol, xrow, yrow;
/* output data */
int* result;
/* arguments */
const mxArray* y;
const mxArray* x;
if (nrhs != 2)
{
mexErrMsgTxt("2 input required.");
}
y = prhs[0];
x = prhs[1];
ycol = mxGetN(y);
yrow = mxGetM(y);
xcol = mxGetN(x);
xrow = mxGetM(x);
/* The first input must be a sparse matrix. */
if (!mxIsDouble(y) || !mxIsDouble(x))
{
mexErrMsgTxt("Input must be of type 'double'.");
}
if (xcol != ycol)
{
mexErrMsgTxt("Inputs must have the same number of columns");
}
plhs[0] = mxCreateLogicalMatrix(1, 1);
result = mxGetPr(plhs[0]);
*result = ismember(mxGetPr(y), mxGetPr(x), yrow, xrow, ycol);
}
int ismemberinner(const double *y, int idx, const double *x, int yrow, int xrow, int ncol) {
int from, to, i;
from = 0;
to = xrow-1;
for(i = 0; i < ncol; ++i) {
// Perform binary search
double yi = *(y + i * yrow + idx);
double *curx = x + i * xrow;
int l = from;
int u = to;
while(l <= u) {
int mididx = l + (u-l)/2;
if(yi < curx[mididx]) {
u = mididx-1;
}
else if(yi > curx[mididx]) {
l = mididx+1;
}
else {
// This can be further optimized by performing additional binary searches
for(from = mididx; from > l && curx[from-1] == yi; --from);
for(to = mididx; to < u && curx[to+1] == yi; ++to);
break;
}
}
if(l > u) {
return 0;
}
}
return 1;
}
int ismember(const double *y, const double *x, int yrow, int xrow, int ncol) {
int i;
for(i = 0; i < yrow; ++i) {
if(!ismemberinner(y, i, x, yrow, xrow, ncol)) {
return 0;
}
}
return 1;
}
Compile it using:
mex -O ismember_mex.c
It can be called as follows:
ismember_mex(x, sortrows(x))
First of all, it assumes that the columns of the matrices have the same size. It works by first sorting the rows of the larger matrix (x in this case, the second argument to the function). Then, a type of binary search is employed to identify whether the rows of the smaller matrix (y hereafter) are contained in x. This is done for each row of y separately (see ismember C function).
For a given row of y, it starts from the first entry and finds the range of indices (using the from and to variables) that match with the first column of x using binary search. This is repeated for the remaining entries, unless some value is not found, in which case it terminates and returns 0.
I tried implementing it this idea in MATLAB, but it didn't work that well. Regarding performance, I found that: (a) in case there are mismatches, it is usually much faster than ismember (b) in case the range of values in x and y is large, it is again faster than ismember, and (c) in case everything matches and the number of possible values in x and y is small (e.g. less than 1000), then ismember may be faster in some situations.
Finally, I want to point out that some parts of the C implementation may be further optimized.
EDIT 1
I fixed the warnings and further improved the function.
#include "mex.h"
#include <math.h>
#include <stdlib.h>
#include <string.h>
int ismember(const double *y, const double *x, unsigned int nrowy, unsigned int nrowx, unsigned int ncol);
void mexFunction(int nlhs, mxArray *plhs[],
int nrhs, const mxArray *prhs[])
{
unsigned int xcol, ycol, nrowx, nrowy;
/* arguments */
const mxArray* y;
const mxArray* x;
if (nrhs != 2)
{
mexErrMsgTxt("2 inputs required.");
}
y = prhs[0];
x = prhs[1];
ycol = (unsigned int) mxGetN(y);
nrowy = (unsigned int) mxGetM(y);
xcol = (unsigned int) mxGetN(x);
nrowx = (unsigned int) mxGetM(x);
/* The first input must be a sparse matrix. */
if (!mxIsDouble(y) || !mxIsDouble(x))
{
mexErrMsgTxt("Input must be of type 'double'.");
}
if (xcol != ycol)
{
mexErrMsgTxt("Inputs must have the same number of columns");
}
plhs[0] = mxCreateLogicalScalar(ismember(mxGetPr(y), mxGetPr(x), nrowy, nrowx, ycol));
}
int ismemberinner(const double *y, const double *x, unsigned int nrowy, unsigned int nrowx, unsigned int ncol) {
unsigned int from = 0, to = nrowx-1, i;
for(i = 0; i < ncol; ++i) {
// Perform binary search
const double yi = *(y + i * nrowy);
const double *curx = x + i * nrowx;
unsigned int l = from;
unsigned int u = to;
while(l <= u) {
const unsigned int mididx = l + (u-l)/2;
const double midx = curx[mididx];
if(yi < midx) {
u = mididx-1;
}
else if(yi > midx) {
l = mididx+1;
}
else {
{
// Binary search to identify smallest index of x that equals yi
// Equivalent to for(from = mididx; from > l && curx[from-1] == yi; --from)
unsigned int limit = mididx;
while(curx[from] != yi) {
const unsigned int mididx = from + (limit-from)/2;
if(curx[mididx] < yi) {
from = mididx+1;
}
else {
limit = mididx-1;
}
}
}
{
// Binary search to identify largest index of x that equals yi
// Equivalent to for(to = mididx; to < u && curx[to+1] == yi; ++to);
unsigned int limit = mididx;
while(curx[to] != yi) {
const unsigned int mididx = limit + (to-limit)/2;
if(curx[mididx] > yi) {
to = mididx-1;
}
else {
limit = mididx+1;
}
}
}
break;
}
}
if(l > u) {
return 0;
}
}
return 1;
}
int ismember(const double *y, const double *x, unsigned int nrowy, unsigned int nrowx, unsigned int ncol) {
unsigned int i;
for(i = 0; i < nrowy; ++i) {
if(!ismemberinner(y + i, x, nrowy, nrowx, ncol)) {
return 0;
}
}
return 1;
}
Using this version I wasn't able to identify any case where ismember is faster. Also, I noticed that one reason ismember is hard to beat is that it uses all cores of the machine! Of course, the function I provided can be optimized to do this too, but this requires much more effort.
Finally, before using my implementation I would advise you to do extensive testing. I did some testing and it seems to work, but I suggest you also do some additional testing.
For small matrices ismember should be enough, probably.
Usage: ismember(B,A,'rows')
ans =
1
0
1
1
0
I put this answer here, emphasizing on a need to solutions with higher performance. I will accept this answer only if there was no better solution.
Using ismember, if a row of A appears twice in B while another one is missing, might wrongly indicate that A is a member of B. The following solution is suitable if the rows of A and B doesn't need to be in the same order. However, I haven't tested its performance for large matrices.
A = [...
34 12 67;
90 78 15;
10 71 24];
B = [...
34 12 67; % found
89 67 45;
90 78 15; % found
10 71 24; % found, so A is subset of B.
54 34 11];
A = permute(A,[3 2 1]);
rowIdx = all(bsxfun(#eq,B,A),2);
colIdx = any(rowIdx,1);
isAMemberB = all(colIdx);
You have said number of columns <= 10. In addition, if the matrix elements are all integers representable as bytes, you could code each row into a two 64 bit integers. That would reduce the number of comparisons by a factor of 64.
For the general case, the following may not be all that much better for thin matrices, but scales very well as the matrices get fat due to the level 3 multiplication:
function yes = is_submat(A,B)
ma = size(A, 1);
mb = size(B, 1);
n = size(B, 2);
yes = false;
if ma >= mb
a = A(:,1);
b = B(:,1);
D = (0 == bsxfun(#minus, a, b'));
q = any(D, 2);
yes = all(any(D,1));
if yes && (n > 1)
A = A(q, :);
C = B*A';
za = sum(A.*A, 2);
zb = sum(B.*B, 2);
Z = sqrt(zb)*sqrt(za');
[~, ix] = max(C./Z, [], 2);
A = A(ix,:);
yes = all(A(:) == B(:));
end
end
end
In the above, I use the fact that the dot product is maximized when two unit vectors are equal.
For fat matrices (say 5000+ columns) with large numbers of unique elements the performance beats ismember quite handily, but otherwise, it is slower than ismember. For thin matrices ismember is faster by an order of magnitude.
Best case test for this function:
A = randi(50000, [10000, 10000]);
B = A(2:3:end, :);
B = B(randperm(size(B,1)),:);
fprintf('%s: %u\n', 'Number of columns', size(A,2));
fprintf('%s: %u\n', 'Element spread', 50000);
tic; is_submat(A,B); toc;
tic; all(ismember(B,A,'rows')); toc;
fprintf('________\n\n');
is_submat_test;
Number of columns: 10000
Element spread: 50000
Elapsed time is 10.713310 seconds (is_submat).
Elapsed time is 17.446682 seconds (ismember).
So I have to admit, all round ismember seems to be much better.
Edits: Edited to correct bug when there is only one column - fixing this also results in more efficient code. Also previous version did not distinguish between positive and negative numbers. Added timing tests.
I have implemented a CUDA version of inverse discrete cosine transform (IDCT), by "translating" the MATLAB built-in function idct.m into CUDA:
My implementation is cuIDCT.cu, works when m = n and both m and n are even numbers.
cuIDCT.cu
#include <stdio.h>
#include <stdlib.h>
#include <cuda.h>
#include <cufft.h>
#include <cuComplex.h>
// round up n/m
inline int iDivUp(int n, int m)
{
return (n + m - 1) / m;
}
typedef cufftComplex complex;
#define PI 3.1415926535897932384626433832795028841971693993751
__global__
void idct_ComputeWeightsKernel(const int n, complex *ww)
{
const int pos = threadIdx.x + blockIdx.x * blockDim.x;
if (pos >= n) return;
ww[pos].x = sqrtf(2*n) * cosf(pos*PI/(2*n));
ww[pos].y = sqrtf(2*n) * sinf(pos*PI/(2*n));
}
__global__
void idct_ComputeEvenKernel(const float *b, const int n, const int m, complex *ww, complex *y)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
// Compute precorrection factor
ww[0].x = ww[0].x / sqrtf(2);
ww[0].y = ww[0].y / sqrtf(2);
y[iy + ix*m].x = ww[iy].x * b[pos];
y[iy + ix*m].y = ww[iy].y * b[pos];
}
__global__
void Reordering_a0_Kernel(complex *y, const int n, const int m, complex *yy)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
yy[iy + ix*n].x = y[pos].x / (float) n;
yy[iy + ix*n].y = y[pos].y / (float) n;
}
__global__
void Reordering_a_Kernel(complex *yy, const int n, const int m, float *a)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
// Re-order elements of each column according to equations (5.93) and (5.94) in Jain
if (iy < n/2)
{
a[ix + 2*iy*n] = yy[pos].x;
a[ix + (2*iy+1)*n] = yy[ix + (m-iy-1)*n].x;
}
}
/**
* a = idct(b), where a is of size [n m].
* #param b, input array
* #param n, first dimension of a
* #param m, second dimension of a
* #param a, output array
*/
void cuIDCT(float *h_in, int n, int m, float *h_out) // a is of size [n m]
{
const int data_size = n * m * sizeof(float);
// device memory allocation
float *d_in, *d_out;
cudaMalloc(&d_in, data_size);
cudaMalloc(&d_out, data_size);
// transfer data from host to device
cudaMemcpy(d_in, h_in, data_size, cudaMemcpyHostToDevice);
// compute IDCT using CUDA
// begin============================================
// Compute weights
complex *ww;
cudaMalloc(&ww, n*sizeof(complex));
dim3 threads(256);
dim3 blocks(iDivUp(n, threads.x));
idct_ComputeWeightsKernel<<<blocks, threads>>>(n, ww);
complex *y;
complex *yy;
cufftHandle plan;
dim3 threads1(32, 6);
dim3 blocks2(iDivUp(n, threads1.x), iDivUp(m, threads1.y)); // for even case
int Length[1] = {m}; // for each IFFT, the length is m
cudaMalloc(&y, n*m*sizeof(complex));
idct_ComputeEvenKernel<<<blocks2, threads1>>>(d_in, n, m, ww, y);
cufftPlanMany(&plan, 1, Length,
Length, 1, m,
Length, 1, m, CUFFT_C2C, n);
cufftExecC2C(plan, y, y, CUFFT_INVERSE); // y is of size [n m]
cudaMalloc(&yy, n*m*sizeof(complex));
Reordering_a0_Kernel<<<blocks2, threads1>>>(y, n, m, yy);
Reordering_a_Kernel<<<blocks2, threads1>>>(yy, n, m, d_out);
// end============================================
// transfer result from device to host
cudaMemcpy(h_out, d_out, data_size, cudaMemcpyDeviceToHost);
// cleanup
cufftDestroy(plan);
cudaFree(ww);
cudaFree(y);
cudaFree(yy);
cudaFree(d_in);
cudaFree(d_out);
}
Then I compared the result of my CUDA IDCT (i.e. cuIDCT.cu) against MATLAB idct.m using following code:
a test main.cpp function, and
a MATLAB main function main.m to read result from CUDA and compare it against MATLAB.
main.cpp
#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <helper_functions.h>
#include <stdlib.h>
#include <stdio.h>
// N must equal to M, and both must be even numbers
#define N 256
#define M 256
void WriteDataFile(const char *name, int w, int h, const float *in, const float *out)
{
FILE *stream;
stream = fopen(name, "wb");
float data = 202021.25f;
fwrite(&data, sizeof(float), 1, stream);
fwrite(&w, sizeof(w), 1, stream);
fwrite(&h, sizeof(h), 1, stream);
for (int i = 0; i < h; i++)
for (int j = 0; j < w; j++)
{
const int pos = j + i * h;
fwrite(in + pos, sizeof(float), 1, stream);
fwrite(out + pos, sizeof(float), 1, stream);
}
fclose(stream);
}
void cuIDCT(float *b, int n, int m, float *a);
int main()
{
// host memory allocation
float *h_in = new float [N * M];
float *h_out = new float [N * M];
float *h_temp = new float [N * M];
// input data initialization
for (int i = 0; i < N * M; i++)
{
h_in[i] = (float)rand()/(float)RAND_MAX;
h_out[i] = h_in[i];
h_temp[i] = h_in[i];
}
// please comment either one case for testing
// test case 1: use cuIDCT.cu once
// cuIDCT(h_in, N, M, h_out);
// test case 2: iteratively use cuIDCT.cu
for (int i = 0; i < 4; i++)
{
if (i % 2 == 0)
cuIDCT(h_out, N, M, h_temp);
else
cuIDCT(h_temp, N, M, h_out);
}
// write data, for further visualization using MATLAB
WriteDataFile("test.flo", N, M, h_in, h_out);
// cleanup
delete [] h_in;
delete [] h_out;
delete [] h_temp;
cudaDeviceReset();
}
main.m
clc;clear;
% read
[h_in, h_out] = read_data('test.flo');
% MATLAB result, for test case 1, comment the for-loop
matlab_out = h_in;
for i = 1:4
matlab_out = idct(matlab_out);
end
% compare
err = matlab_out - h_out;
% show
figure(1);
subplot(221); imshow(h_in, []); title('h\_in'); colorbar
subplot(222); imshow(h_out, []); title('h\_out'); colorbar
subplot(223); imshow(matlab_out, []); title('matlab\_out'); colorbar
subplot(224); imshow(err, []); title('error map'); colorbar
disp(['maximum error between CUDA and MATLAB is ' ...
num2str(max(max(abs(err))))])
I ran the code on Visual Studio 11 (i.e. VS2012) in Windows 7 with Nvidia GPU Tesla K20c, using CUDA Toolkit version 7.5, and my MATLAB version is R2015b.
My test steps:
For test case 1. Un-comment test case 1 and comment test case 2.
Run main.cpp.
Run main.m in MATLAB.
Repeat step 1 and step 2 (without any change, just re-run the code).
I repeated step 3 for 20 times. The output result is unchanged, and results in main.m are:
results of test case 1
The maximum error is 7.7152e-07.
For test case 2. Un-comment test case 2 and comment test case 1.
Run main.cpp.
Run main.m in MATLAB.
Repeat step 1 and step 2 (without any change, just re-run the code).
I repeated step 3 for 20 times. The output result is changed, and results in main.m are (not enough reputation to put all images, only wrong case is shown below):
one situation (the wrong one) of test case 2
The maximum error is 0.45341 (2 times), 0.44898 (1 time), 0.26186 (1 time), 0.26301 (1 time), and 9.5716e-07 (15 times).
From the test results, my conclusion is:
From test case 1: cuIDCT.cu is numerically correct (error ~10^-7) to idct.m.
From test case 2: recursively use of cuIDCT.cu leads to unstable result (i.e. the output changes every time when re-run the code and may sometimes be numerically wrong, error ~0.1)
My question:
From test case 1 we know cuIDCT.cu is numerically correct to idct.m. But why recursiviely use of cuIDCT.cu leads to different output result each time when re-run the code?
Any helps or suggestions are highly appreciated.
I believe the variability in your results is coming about due to this code in your idct_ComputeEvenKernel:
// Compute precorrection factor
ww[0].x = ww[0].x / sqrtf(2);
ww[0].y = ww[0].y / sqrtf(2);
It's not entirely clear what your intent is here, but it's doubtful that this code could be doing what you want. You may be confused about the CUDA execution model.
The above code will be executed by every CUDA thread that you launch for that kernel that passes the thread check:
if (ix >= n || iy >= m) return;
I believe this means 65536 threads will all execute this code in that kernel. Furthermore, the threads will execute that code in more-or-less any order (not all CUDA threads execute in lock-step). They may even step on each other as they are trying to write out their values to the location ww[0]. So the final result in ww[0] will be quite unpredictable.
When I comment out those lines of code, the results become stable for me (albeit different from what they were with those lines in place), unchanging from run to run.
I'd like to point something else out. Wherever you are calculating the .x and .y values of a complex quantity, my suggestion would be to rework the code from this (for example):
y[iy + ix*m].x = ww[iy].x * b[pos];
y[iy + ix*m].y = ww[iy].y * b[pos];
to this:
complex temp1, temp2;
temp1 = ww[iy];
temp2.x = temp1.x * b[pos];
temp2.y = temp2.y * b[pos];
y[iy + ix*m] = temp2;
At least according to my testing, the compiler doesn't seem to be making this optimization for you, and one side-effect benefit is that it's much easier to test your code with cuda-memcheck --tool initcheck .... In the first realization, the compiler will load y[iy + ix*m] as an 8 byte quantity, modify either 4 or 8 bytes of it, then store y[iy + ix*m] as an 8 byte quantity. The second realization should be more efficient (it eliminates the load of y[]), and eliminates the load of an uninitialized quantity (y[]), which the cuda-memcheck tool will report as a hazard.
This variability I'm describing should be possible whether you run either the 1-pass version of your code or the 4-pass version of your code. Therefore I think your statements about the 1-pass version being correct are suspect. I think if you run the 1-pass version enough, you will eventually see variability (although it may require varying initial memory conditions, or running on different GPU types). Even in your own results, we see that 15 out of 20 runs of the 4 pass code produce "correct" results, i.e. the residual error is ~1e-7
Here's my modified cuIDCT.cu file, modified from the version you posted here. The assumption I'm making below is that you wanted to compute the scaling on ww[0] only once, in which case we can easily handle that arithmetic as an addendum to the previous idct_ComputeWeightsKernel:
#include <stdio.h>
#include <stdlib.h>
#include <cuda.h>
#include <cufft.h>
#include <cuComplex.h>
#include <helper_cuda.h>
#include "assert.h"
// round up n/m
inline int iDivUp(int n, int m)
{
return (n + m - 1) / m;
}
typedef cufftComplex complex;
#define PI 3.1415926535897932384626433832795028841971693993751
#define cufftSafeCall(err) __cufftSafeCall(err, __FILE__, __LINE__)
inline void __cufftSafeCall(cufftResult err, const char *file, const int line)
{
if( CUFFT_SUCCESS != err) {
fprintf(stderr, "CUFFT error in file '%s', line %d\n %s\nerror %d: %s\nterminating!\n",__FILE__, __LINE__,err, \
_cudaGetErrorEnum(err)); \
cudaDeviceReset(); assert(0); \
}
}
__global__
void idct_ComputeWeightsKernel(const int n, complex *ww)
{
const int pos = threadIdx.x + blockIdx.x * blockDim.x;
if (pos >= n) return;
complex temp;
temp.x = sqrtf(2*n) * cosf(pos*PI/(2*n));
temp.y = sqrtf(2*n) * sinf(pos*PI/(2*n));
if (pos == 0) {
temp.x /= sqrtf(2);
temp.y /= sqrtf(2);}
ww[pos] = temp;
}
__global__
void idct_ComputeEvenKernel(const float *b, const int n, const int m, complex *ww, complex *y)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
/* handle this in idct_ComputeWeightsKernel
// Compute precorrection factor
ww[0].x = ww[0].x / sqrtf(2);
ww[0].y = ww[0].y / sqrtf(2);
*/
complex temp1, temp2;
temp1 = ww[iy];
temp2.x = temp1.x * b[pos];
temp2.y = temp1.y * b[pos];
y[iy + ix*m] = temp2;
}
__global__
void Reordering_a0_Kernel(complex *y, const int n, const int m, complex *yy)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
complex temp1, temp2;
temp1 = y[pos];
temp2.x = temp1.x / (float) n;
temp2.y = temp1.y / (float) n;
yy[iy + ix*n] = temp2;
}
__global__
void Reordering_a_Kernel(complex *yy, const int n, const int m, float *a)
{
const int ix = threadIdx.x + blockIdx.x * blockDim.x;
const int iy = threadIdx.y + blockIdx.y * blockDim.y;
if (ix >= n || iy >= m) return;
const int pos = ix + iy*n;
// Re-order elements of each column according to equations (5.93) and (5.94) in Jain
if (iy < n/2)
{
a[ix + 2*iy*n] = yy[pos].x;
a[ix + (2*iy+1)*n] = yy[ix + (m-iy-1)*n].x;
}
}
/**
* a = idct(b), where a is of size [n m].
* #param b, input array
* #param n, first dimension of a
* #param m, second dimension of a
* #param a, output array
*/
void cuIDCT(float *h_in, int n, int m, float *h_out) // a is of size [n m]
{
const int data_size = n * m * sizeof(float);
// device memory allocation
float *d_in, *d_out;
checkCudaErrors(cudaMalloc(&d_in, data_size));
checkCudaErrors(cudaMalloc(&d_out, data_size));
// transfer data from host to device
checkCudaErrors(cudaMemcpy(d_in, h_in, data_size, cudaMemcpyHostToDevice));
// compute IDCT using CUDA
// begin============================================
// Compute weights
complex *ww;
checkCudaErrors(cudaMalloc(&ww, n*sizeof(complex)));
dim3 threads(256);
dim3 blocks(iDivUp(n, threads.x));
idct_ComputeWeightsKernel<<<blocks, threads>>>(n, ww);
complex *y;
complex *yy;
cufftHandle plan;
dim3 threads1(32, 6);
dim3 blocks2(iDivUp(n, threads1.x), iDivUp(m, threads1.y)); // for even case
int Length[1] = {m}; // for each IFFT, the length is m
checkCudaErrors(cudaMalloc(&y, n*m*sizeof(complex)));
idct_ComputeEvenKernel<<<blocks2, threads1>>>(d_in, n, m, ww, y);
cufftSafeCall(cufftPlanMany(&plan, 1, Length,
Length, 1, m,
Length, 1, m, CUFFT_C2C, n));
cufftSafeCall(cufftExecC2C(plan, y, y, CUFFT_INVERSE)); // y is of size [n m]
checkCudaErrors(cudaMalloc(&yy, n*m*sizeof(complex)));
Reordering_a0_Kernel<<<blocks2, threads1>>>(y, n, m, yy);
cudaMemset(d_out, 0, data_size);
Reordering_a_Kernel<<<blocks2, threads1>>>(yy, n, m, d_out);
// end============================================
// transfer result from device to host
checkCudaErrors(cudaMemcpy(h_out, d_out, data_size, cudaMemcpyDeviceToHost));
// cleanup
cufftDestroy(plan);
checkCudaErrors(cudaFree(ww));
checkCudaErrors(cudaFree(y));
checkCudaErrors(cudaFree(yy));
checkCudaErrors(cudaFree(d_in));
checkCudaErrors(cudaFree(d_out));
}
You'll note I threw an extra cudaMemset on d_out in there, because it helped me clean up an issue with cuda-memcheck --tool initcheck .... It shouldn't be necessary, you can delete it if you want.
Its been two days and I am still cant figure it out why my implementation of CUDA matrix multiplication differs from the results produced in MATLAB.
CUDA kernel: A(200x60000) = W(200x784) * Data(784x6000)
__global__ void CalculateA(Matrix W, Matrix Data, Matrix A)
{
int Row = blockIdx.y * blockDim.y + threadIdx.y;
int Col = blockIdx.x * blockDim.x + threadIdx.x;
if ((Row < W.row) && (Col < Data.col)){
float Cvalue = 0.0;
for (int i = 0; i < W.col; ++i){
Cvalue += W.elements[Row*W.col+i] * Data.elements[i*Data.col+Col];
}
A.elements[Row*A.col+Col] = Cvalue;
}
}
And calling the kernel:
void myFunc(Matrix W1, Matrix data){
Matrix d_W1, d_data, d_a2, a2;
size_t size;
a2.row = W1.row; d_a2.row = a2.row;
a2.col = data.col; d_a2.col = a2.col;
size = a2.col*a2.row*sizeof(float);
cudaMalloc(&d_a2.elements,size);
d_W1.row = W1.row; d_W1.col = W1.col;
size = W1.col*W1.row*sizeof(float);
cudaMalloc(&d_W1.elements,size);
cudaMemcpy(d_W1.elements,W1.elements,size,cudaMemcpyHostToDevice);
d_data.col = data.col; d_data.row = data.row;
size = data.row*data.col*sizeof(float);
cudaMalloc(&d_data.elements,size);
cudaMemcpy(d_data.elements,data.elements,size,cudaMemcpyHostToDevice);
dim3 dimGrid(data.col/32 + 1, W1.row/32 + 1, 1);
dim3 dimBlock(32, 32, 1);
CalculateA<<<dimGrid, dimBlock>>>(d_W1, d_data, d_a2);
a2.elements = new float [a2.row*a2.col];
cudaMemcpy(a2.elements,d_a2.elements,sizeof(float)*a2.row*a2.col,cudaMemcpyDeviceToHost);
printf("\nA2 first and last member %f - %f\n",a2.elements[0],a2.elements[a2.row*a2.col-1]);
}
Results difference is not low for example first and last elements of CUDA code is 0.011322 and -0.179534 but multiplying in MATLAB results in 0.4280 and 0.0056.
this is how I do it in MATLAB:
>> size(W1) ans = 200 784
>> size(data) ans = 784 60000
>> z2=W1*data;
>> size(z2) ans = 200 60000
>> z2 = z2(:);
>> z2(1) ans = 0.4280
>> z2(200*60000)ans = 0.0056
There is nothing wrong with the code you posted. If I expand your kernel and function into a complete running example like this:
#include <iostream>
struct Matrix
{
int row;
int col;
float *elements;
__device__ __host__
float& operator()(int r, int c) { return elements[r*col + c]; };
};
__global__ void CalculateA(Matrix W, Matrix Data, Matrix A)
{
int Row = blockIdx.y * blockDim.y + threadIdx.y;
int Col = blockIdx.x * blockDim.x + threadIdx.x;
if ((Row < W.row) && (Col < Data.col)){
float Cvalue = 0.0;
for (int i = 0; i < W.col; ++i){
Cvalue += W.elements[Row*W.col+i] * Data.elements[i*Data.col+Col];
}
A.elements[Row*A.col+Col] = Cvalue;
}
}
void myFunc(Matrix W1, Matrix data)
{
Matrix d_W1, d_data, d_a2, a2;
size_t size;
a2.row = W1.row; d_a2.row = a2.row;
a2.col = data.col; d_a2.col = a2.col;
size = a2.col*a2.row*sizeof(float);
cudaMalloc(&d_a2.elements,size);
d_W1.row = W1.row; d_W1.col = W1.col;
size = W1.col*W1.row*sizeof(float);
cudaMalloc(&d_W1.elements,size);
cudaMemcpy(d_W1.elements,W1.elements,size,cudaMemcpyHostToDevice);
d_data.col = data.col; d_data.row = data.row;
size = data.row*data.col*sizeof(float);
cudaMalloc(&d_data.elements,size);
cudaMemcpy(d_data.elements,data.elements,size,cudaMemcpyHostToDevice);
dim3 dimGrid(data.col/32 + 1, W1.row/32 + 1, 1);
dim3 dimBlock(32, 32, 1);
CalculateA<<<dimGrid, dimBlock>>>(d_W1, d_data, d_a2);
a2.elements = new float [a2.row*a2.col];
cudaMemcpy(a2.elements,d_a2.elements,sizeof(float)*a2.row*a2.col,cudaMemcpyDeviceToHost);
for(int j=0; j<a2.col; ++j) {
for(int i=0; i<a2.row; ++i) {
std::cout << a2(i,j) << " ";
}
std::cout << std::endl;
}
}
int main(void)
{
float a[6] = { 1.0f, 2.0f, 3.0f, 4.0f, 5.0f, 6.0f };
float b[6] = { 0.1f, 0.2f, 0.3f, 0.4f, 0.5f, 0.6f};
Matrix W1; W1.row=2; W1.col=3; W1.elements = &a[0];
Matrix Data; Data.row=3; Data.col=2; Data.elements = &b[0];
myFunc(W1, Data);
return 0;
}
and run it, I get this:
>nvcc -arch=sm_21 -Xptxas="-v" -m32 matrix.cu
matrix.cu
tmpxft_000014f4_00000000-5_matrix.cudafe1.gpu
tmpxft_000014f4_00000000-10_matrix.cudafe2.gpu
matrix.cu
ptxas : info : 132 bytes gmem, 28 bytes cmem[14]
ptxas : info : Compiling entry function '_Z10CalculateA6MatrixS_S_' for 'sm_21'
ptxas : info : Function properties for _Z10CalculateA6MatrixS_S_
0 bytes stack frame, 0 bytes spill stores, 0 bytes spill loads
ptxas : info : Used 14 registers, 68 bytes cmem[0]
tmpxft_000014f4_00000000-5_matrix.cudafe1.cpp
tmpxft_000014f4_00000000-15_matrix.ii
>cuda-memcheck a.exe
========= CUDA-MEMCHECK
2.2 4.9
2.8 6.4
========= ERROR SUMMARY: 0 errors
which is the correct answer for the dot product assuming column major ordering (which is the Matlab convention).
So if your results don't agree, it is because of something you haven't shown us. One likelihood is that your test problem is so large (and kernel so inefficient) that if you are running this on a display GPU, your program is hitting the display driver watchdog timer limit and being killed before the kernel finishes running. Also note that you have no CUDA API error checking whatsoever, so it is possible that you are getting runtime errors which is either stopping your kernel from finishing or even running at all, but you simply don't notice because of the lack of error checking.
I have to read a matrix of double, handling its values and insert them in a new matrix, whose one of its dimension is unkwown at the beginning.
In a static memory allocation, my code is:
#include <mex.h>
void mexFunction( int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[] )
{
const mxArray *I = prhs[0];
double *indata = mxGetPr(I);
double *submtx = mxGetPr(prhs[1]);
const int *size = mxGetDimensions(I);
int xp = (int)submtx[0], yp = (int)submtx[1], zp = (int)submtx[2];
int xi = size[0], yi = size[1], zi = size[2];
int numsubmtx = (xi - xp + 1)*(yi - yp + 1)*(zi - zp + 1);
int out_rows = xp*yp*zp;
int out_cols = numsubmtx;
mxArray *out = mxCreateDoubleMatrix( out_rows, out_cols, mxREAL );
double *outdata = mxGetPr(out);
int submtx_counter = 0;
for( int z_offset = 0; ...; z_offset++ ){
for( int y_offset = 0; ...; y_offset++ ){
for( int x_offset = 0; ...; x_offset++ ){
int row = 0;
for( int z_counter = 0; ...; z_counter++ ){
for( int y_counter = 0; ...; y_counter++ ){
for( int x_counter = 0; ...; x_counter++ ){
outdata[submtx_counter*out_rows + row] =
indata[ (x_offset+x_counter) + (y_offset+y_counter)*xi + (z_offset+z_counter)*xi*yi ];
++row;
}}}
++submtx_counter;
}}}
plhs[0] = out;
}
In a dynamic version, I do not know the value of out_cols, so I have to reallocate *out when a condition on the values of indata is satisfied.
My idea is something like this:
#include <mex.h>
void mexFunction( int nlhs, mxArray *plhs[], int nrhs, const mxArray *prhs[] )
{
const mxArray *I = prhs[0];
double *indata = mxGetPr(I);
double *submtx = mxGetPr(prhs[1]);
const int *size = mxGetDimensions(I);
int xp = (int)submtx[0], yp = (int)submtx[1], zp = (int)submtx[2];
int xi = size[0], yi = size[1], zi = size[2];
int numsubmtx = (xi - xp + 1)*(yi - yp + 1)*(zi - zp + 1);
int out_rows = xp*yp*zp;
//int out_cols = numsubmtx; NOW UNKNOWN!
mxArray *out = NULL;
double *outdata = mxGetPr(out);
int submtx_counter = 0;
for( int z_offset = 0; ...; z_offset++ ){
for( int y_offset = 0; ...; y_offset++ ){
for( int x_offset = 0; ...; x_offset++ ){
int row = 0;
double condition=0;
for( int z_counter = 0; ...; z_counter++ ){
for( int y_counter = 0; ...; y_counter++ ){
for( int x_counter = 0; ...; x_counter++ ){
condition += indata[ (x_offset+x_counter) + (y_offset+y_counter)*xi + (z_offset+z_counter)*xi*yi ]/(xp*yp*zp);
++row;
}}}
if(coundition>0.5){
out = mxRealloc(out, (submtx_counter+1)*out_rows*sizeof(double));
double *outdata = mxGetPr(out);
int row = 0;
for( int z_counter = 0; ...; z_counter++ ){
for( int y_counter = 0; ...; y_counter++ ){
for( int x_counter = 0; ...; x_counter++ ){
outdata[submtx_counter*out_rows + row] = indata[ (x_offset+x_counter) + (y_offset+y_counter)*xi + (z_offset+z_counter)*xi*yi ];
++row;
}}}
++submtx_counter;
}
}}}
plhs[0] = out;
}
What is wrong?
I haven't looked closely at the logic of your code, but I can see the following problems:
mxArray *out = NULL;
double *outdata = mxGetPr(out);
and
out = mxRealloc(out, (submtx_counter+1)*out_rows*sizeof(double));
will have strange behavior. An mxArray is a structure, and you want to reallocate memory for the data pointed to by the structure rather than the structure itself. Instead, try
double *outdata = mxMalloc(out_rows*sizeof(double));
and
outdata = mxRealloc(outdata, (submtx_counter+1)*out_rows*sizeof(double));
and then at the very end, create your output matrix:
mxArray *out = mxCreateDoubleMatrix(out_rows, submtx_counter, mxREAL);
mxSetPr(out, outdata);
plhs[0] = out;
This will guarantee that the metadata in out gives you a matrix of the correct size, etc.
I have an image of connected components(circles filled).If i want to segment them i can use watershed algorithm.I prefer writing my own function for watershed instead of using the inbuilt function in OPENCV.I have successfu How do i find the regionalmax of objects using opencv?
I wrote a function myself. My results were quite similar to MATLAB, although not exact. This function is implemented for CV_32F but it can easily be modified for other types.
I mark all the points that are not part of a minimum region by checking all the neighbors. The remaining regions are either minima, maxima or areas of inflection.
I use connected components to label each region.
I check each region for any point belonging to a maxima, if yes then I push that label into a vector.
Finally I sort the bad labels, erase all duplicates and then mark all the points in the output as not minima.
All that remains are the regions of minima.
Here is the code:
// output is a binary image
// 1: not a min region
// 0: part of a min region
// 2: not sure if min or not
// 3: uninitialized
void imregionalmin(cv::Mat& img, cv::Mat& out_img)
{
// pad the border of img with 1 and copy to img_pad
cv::Mat img_pad;
cv::copyMakeBorder(img, img_pad, 1, 1, 1, 1, IPL_BORDER_CONSTANT, 1);
// initialize binary output to 2, unknown if min
out_img = cv::Mat::ones(img.rows, img.cols, CV_8U)+2;
// initialize pointers to matrices
float* in = (float *)(img_pad.data);
uchar* out = (uchar *)(out_img.data);
// size of matrix
int in_size = img_pad.cols*img_pad.rows;
int out_size = img.cols*img.rows;
int x, y;
for (int i = 0; i < out_size; i++) {
// find x, y indexes
y = i % img.cols;
x = i / img.cols;
neighborCheck(in, out, i, x, y, img_pad.cols); // all regions are either min or max
}
cv::Mat label;
cv::connectedComponents(out_img, label);
int* lab = (int *)(label.data);
in = (float *)(img.data);
in_size = img.cols*img.rows;
std::vector<int> bad_labels;
for (int i = 0; i < out_size; i++) {
// find x, y indexes
y = i % img.cols;
x = i / img.cols;
if (lab[i] != 0) {
if (neighborCleanup(in, out, i, x, y, img.rows, img.cols) == 1) {
bad_labels.push_back(lab[i]);
}
}
}
std::sort(bad_labels.begin(), bad_labels.end());
bad_labels.erase(std::unique(bad_labels.begin(), bad_labels.end()), bad_labels.end());
for (int i = 0; i < out_size; ++i) {
if (lab[i] != 0) {
if (std::find(bad_labels.begin(), bad_labels.end(), lab[i]) != bad_labels.end()) {
out[i] = 0;
}
}
}
}
int inline neighborCleanup(float* in, uchar* out, int i, int x, int y, int x_lim, int y_lim)
{
int index;
for (int xx = x - 1; xx < x + 2; ++xx) {
for (int yy = y - 1; yy < y + 2; ++yy) {
if (((xx == x) && (yy==y)) || xx < 0 || yy < 0 || xx >= x_lim || yy >= y_lim)
continue;
index = xx*y_lim + yy;
if ((in[i] == in[index]) && (out[index] == 0))
return 1;
}
}
return 0;
}
void inline neighborCheck(float* in, uchar* out, int i, int x, int y, int x_lim)
{
int indexes[8], cur_index;
indexes[0] = x*x_lim + y;
indexes[1] = x*x_lim + y+1;
indexes[2] = x*x_lim + y+2;
indexes[3] = (x+1)*x_lim + y+2;
indexes[4] = (x + 2)*x_lim + y+2;
indexes[5] = (x + 2)*x_lim + y + 1;
indexes[6] = (x + 2)*x_lim + y;
indexes[7] = (x + 1)*x_lim + y;
cur_index = (x + 1)*x_lim + y+1;
for (int t = 0; t < 8; t++) {
if (in[indexes[t]] < in[cur_index]) {
out[i] = 0;
break;
}
}
if (out[i] == 3)
out[i] = 1;
}
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;
}
I do not know if it is what you want, but in my answer to this post, I gave some code to find local maxima (peaks) in a grayscale image (resulting from distance transform).
The approach relies on subtracting the original image from the dilated image and finding the zero pixels).
I hope it helps,
Good luck
I had the same problem some time ago, and the solution was to reimplement the imregionalmax algorithm in OpenCV/Cpp. It is not that complicated, because you can find the C++ source code of the function in the Matlab distribution. (somewhere in toolbox). All you have to do is to read carefully and understand the algorithm described there. Then rewrite it or remove the matlab-specific checks and you'll have it.