How do I go about having a cross-entropy layer in Caffe? - neural-network

I am working on a CNN where the last layer is supposed to be a cross-entropy layer where I compare 2 images. I am imposed to use Caffe. I checked on layer catalogue. And there isn't a cross-entropy layer, only a Sigmoid cross-entropy. I asked my supervisor who told that using wouldn't do.
So here is my question. Is vanilla cross-entropy hidden somewhere?

No, there is no vanilla cross-entropy in caffe till now. There are only some special instances of cross-entropy in caffe, for example, SigmoidCrossEntropyLossLayer and MultinomialLogisticLossLayer.
But you can get vanilla cross-entropy by simply modefiying the
MultinomialLogisticLossLayer. Because this layer computes cross-entropy assuming that the target probability distribution is a form of one-hot vector, so by modifying its computation formula by assumming the target probability distribution a general probability distribution you can get the vanilla cross-entropy. Hope that will help you.
Orignal MultinomialLogisticLossLayer in caffe:
template <typename Dtype>
void MultinomialLogisticLossLayer<Dtype>::Forward_cpu(
const vector<Blob<Dtype>*>& bottom, const vector<Blob<Dtype>*>& top) {
const Dtype* bottom_data = bottom[0]->cpu_data();
const Dtype* bottom_label = bottom[1]->cpu_data();
int num = bottom[0]->num();
int dim = bottom[0]->count() / bottom[0]->num();
Dtype loss = 0;
for (int i = 0; i < num; ++i) {
int label = static_cast<int>(bottom_label[i]);
Dtype prob = std::max(bottom_data[i * dim + label], Dtype(kLOG_THRESHOLD));
loss -= log(prob);
}
top[0]->mutable_cpu_data()[0] = loss / num;
}
Assume that bottom[1] contains target/true distribution, then Cross EntropyLossLayer should be like this now:
void CrossEntropyLossLayer<Dtype>::Forward_cpu(
const vector<Blob<Dtype>*>& bottom, const vector<Blob<Dtype>*>& top) {
const Dtype* bottom_data = bottom[0]->cpu_data();
const Dtype* bottom_label = bottom[1]->cpu_data();
int num = bottom[0]->num();
int dim = bottom[0]->count() / bottom[0]->num();
Dtype loss = 0;
for (int i = 0; i < num; ++i) {
for (int j = 0; j < dim; ++j){
Dtype prob = std::max(bottom_data[i * dim + j], Dtype(kLOG_THRESHOLD));
loss -= bottom_label[i * dim + j] * log(prob);
}
}
top[0]->mutable_cpu_data()[0] = loss / num;
}
Similarly, the corresponding backward_cpu function for back propagation will be:
template <typename Dtype>
void CrossEntropyLossLayer<Dtype>::Backward_cpu(
const vector<Blob<Dtype>*>& top, const vector<bool>& propagate_down,
const vector<Blob<Dtype>*>& bottom) {
if (propagate_down[1]) {
LOG(FATAL) << this->type()
<< " Layer cannot backpropagate to label inputs.";
}
if (propagate_down[0]) {
const Dtype* bottom_data = bottom[0]->cpu_data();
const Dtype* bottom_label = bottom[1]->cpu_data();
Dtype* bottom_diff = bottom[0]->mutable_cpu_diff();
int num = bottom[0]->num();
int dim = bottom[0]->count() / bottom[0]->num();
caffe_set(bottom[0]->count(), Dtype(0), bottom_diff);
const Dtype scale = -top[0]->cpu_diff()[0] / num;
for (int i = 0; i < num; ++i) {
for (int j = 0; j < dim; ++j){
Dtype prob = std::max(bottom_data[i * dim + j], Dtype(kLOG_THRESHOLD));
bottom_diff[i * dim + j] = scale * bottom_label[i * dim + j] / prob;
}
}
}
}

Related

Same C code different results TIv5.2.5 and gcc 5.4.1 c99 compiler

I am using MSP432P401R to do FFT of SAR ADC samples, did FFT in MATLAB and got results same as C compiler online but Code Composer Studio IDE is giving different output than MATLAB results, I thought that can be a compiler issue so tried reading same did some changes and tried but not getting results Like MATLAB.
Online C compiler was gcc 5.4.1 c99.
and in CCS TI v5.2.5 compiler is used.
float m;
float ur, ui, sr, si,tr, ti;
long double Temp_A[256],ArrayA[256]={2676,2840,2838,2832,2826,2818,2814,2808,
2804,2798,2790,2784,2778,2770,2764,2758,2752,2746,2740,2734,
2726,2720,2714,2706,2700,2692,2686,2680,2674,2668,2660,2654,
2646,2642,2634,2624,2618,2612,2604,2598,2590,2584,2576,2570,
2562,2556,2550,2542,2536,2530,2522,2512,2508,2498,2490,2484,
2478,2470,2462,2454,2448,2442,2432,2426,2420,2414,2404,2398,
2390,2382,2374,2368,2360,2352,2346,2338,2330,2322,2314,2306,
2300,2294,2286,2278,2272,2262,2258,2250,2238,2234,2228,2220,
2208,2202,2192,2186,2178,2170,2164,2156,2150,2142,2134,2126,
2116,2110,2104,2096,2088,2078,2070,2062,2054,2046,2040,2034,
2026,2018,2010,2002,1994,1986,1978,1970,1962,1954,1946,1936,
1930,1922,1914,1908,1902,1894,1886,1876,1868,1860,1852,1846,
1838,1830,1822,1814,1804,1796,1790,1784,1776,1768,1760,1754,
1746,1738,1728,1720,1714,1708,1698,1692,1684,1674,1668,1656,
1656,1644,1640,1628,1624,1612,1610,1598,1596,1584,1580,1570,
1564,1554,1546,1540,1532,1526,1520,1512,1504,1496,1490,1482,
1474,1468,1462,1454,1446,1438,1432,1424,1420,1410,1404,1398,
1392,1384,1376,1370,1364,1356,1348,1342,1336,1328,1322,1316,
1308,1300,1294,1286,1280,1276,1270,1262,1254,1248,1242,1236,
1230,1222,1216,1210,1206,1198,1192,1188,1178,1172,1168,1162,
1154,1148,1144,1138,1132,1126,1120,1114,1108,1102,1096,1090,
1084,1080,1074,1068,1062,1058,1052,1048},ArrayA_IMX[256]={0};
unsigned int jm1,i;
unsigned int ip,l;
void main(void)
{
WDT_A->CTL = WDT_A_CTL_PW |WDT_A_CTL_HOLD;
VCORE();
CLK();
P1DIR |= BIT5; //CLK--AD7352 OUTPUT DIRECTION
P1DIR |= BIT7; //CHIP SELECT--AD7352 OUTPUT DIRECTION
P5DIR &= ~BIT0; //SDATAA--AD7352 INPUT DIRECTION P5.0
P5DIR &= ~BIT2; //SDATAB--AD7352 INPUT DIRECTION P5.2
while(1)
{
bit_reversal(ArrayA);
fft(ArrayA,ArrayA_IMX);
}
}
void bit_reversal(long double REX[])
{
int i,i2,n,m;
int tx,k,j;
n = 1;
m=8;
for (i=0;i<m;i++)
{
n *= 2;
}
i2 = n >> 1;
j = 0;
for (i=0;i<n-1;i++)
{
if (i < j)
{
tx = REX[i];
//ty = IMX[i];
REX[i] = REX[j];
//IMX[i] = IMX[j];
REX[j] = tx;
//IMX[j] = ty;
}
k = i2;
while (k <= j)
{
j -= k;
k >>= 1;
}
j += k;
}
}
void fft(long double REX[],long double IMX[])
{
N = 256;
nm1 = N - 1;
nd2 = N / 2;
m = log10l(N) / log10l(2);
j = nd2;
for (l = 1; l <= m; l++)
{
le = powl(2, l);
le2 = le / 2;
ur = 1;
ui = 0;
// Calculate sine and cosine values
sr = cosl(M_PI/le2);
si = -sinl(M_PI/le2);
// Loop for each sub DFT
for (j = 1; j <= le2; j++)
{
jm1 = j - 1;
// Loop for each butterfly
for (i = jm1; i <= nm1; i += le)
{
ip = i + le2;
tr = REX[ip]*ur - IMX[ip]*ui;
ti = REX[ip]*ui + IMX[ip]*ur;
REX[ip] = REX[i] - tr;
IMX[ip] = IMX[i] - ti;
REX[i] = REX[i] + tr;
IMX[i] = IMX[i] + ti;
}
tr = ur;
ur = tr*sr - ui*si;
ui = tr*si + ui*sr;
}
}
}

finding local mean in an image using mex-cuda

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]));
}
}

Porting signal windowing code from Matlab to Java

This is part of a code from spectral subtraction algorithm,i'm trying to optimize it for android.please help me.
this is the matlab code:
function Seg=segment(signal,W,SP,Window)
% SEGMENT chops a signal to overlapping windowed segments
% A= SEGMENT(X,W,SP,WIN) returns a matrix which its columns are segmented
% and windowed frames of the input one dimentional signal, X. W is the
% number of samples per window, default value W=256. SP is the shift
% percentage, default value SP=0.4. WIN is the window that is multiplied by
% each segment and its length should be W. the default window is hamming
% window.
% 06-Sep-04
% Esfandiar Zavarehei
if nargin<3
SP=.4;
end
if nargin<2
W=256;
end
if nargin<4
Window=hamming(W);
end
Window=Window(:); %make it a column vector
L=length(signal);
SP=fix(W.*SP);
N=fix((L-W)/SP +1); %number of segments
Index=(repmat(1:W,N,1)+repmat((0:(N-1))'*SP,1,W))';
hw=repmat(Window,1,N);
Seg=signal(Index).*hw;
and this is our java code for this function:
public class MatrixAndSegments
{
public int numberOfSegments;
public double[][] res;
public MatrixAndSegments(int numberOfSegments,double[][] res)
{
this.numberOfSegments = numberOfSegments;
this.res = res;
}
}
public MatrixAndSegments segment (double[] signal_in,int samplesPerWindow, double shiftPercentage, double[] window)
{
//default shiftPercentage = 0.4
//default samplesPerWindow = 256 //W
//default window = hanning
int L = signal_in.length;
shiftPercentage = fix(samplesPerWindow * shiftPercentage); //SP
int numberOfSegments = fix ( (L - samplesPerWindow)/ shiftPercentage + 1); //N
double[][] reprowMatrix = reprowtrans(samplesPerWindow,numberOfSegments);
double[][] repcolMatrix = repcoltrans(numberOfSegments, shiftPercentage,samplesPerWindow );
//Index=(repmat(1:W,N,1)+repmat((0:(N-1))'*SP,1,W))';
double[][] index = new double[samplesPerWindow+1][numberOfSegments+1];
for (int x = 1; x < samplesPerWindow+1; x++ )
{
for (int y = 1 ; y < numberOfSegments + 1; y++) //numberOfSegments was 3
{
index[x][y] = reprowMatrix[x][y] + repcolMatrix[x][y];
}
}
//hamming window
double[] hammingWindow = this.HammingWindow(samplesPerWindow);
double[][] HW = repvector(hammingWindow, numberOfSegments);
double[][] seg = new double[samplesPerWindow][numberOfSegments];
for (int y = 1 ; y < numberOfSegments + 1; y++)
{
for (int x = 1; x < samplesPerWindow+1; x++)
{
seg[x-1][y-1] = signal_in[ (int)index[x][y]-1 ] * HW[x-1][y-1];
}
}
MatrixAndSegments Matrixseg = new MatrixAndSegments(numberOfSegments,seg);
return Matrixseg;
}
public int fix(double val) {
if (val < 0) {
return (int) Math.ceil(val);
}
return (int) Math.floor(val);
}
public double[][] repvector(double[] vec, int replications)
{
double[][] result = new double[vec.length][replications];
for (int x = 0; x < vec.length; x++) {
for (int y = 0; y < replications; y++) {
result[x][y] = vec[x];
}
}
return result;
}
public double[][] reprowtrans(int end, int replications)
{
double[][] result = new double[end +1][replications+1];
for (int x = 1; x <= end; x++) {
for (int y = 1; y <= replications; y++) {
result[x][y] = x ;
}
}
return result;
}
public double[][] repcoltrans(int end, double multiplier, int replications)
{
double[][] result = new double[replications+1][end+1];
for (int x = 1; x <= replications; x++) {
for (int y = 1; y <= end ; y++) {
result[x][y] = (y-1)*multiplier;
}
}
return result;
}
public double[] HammingWindow(int size)
{
double[] window = new double[size];
for (int i = 0; i < size; i++)
{
window[i] = 0.54-0.46 * (Math.cos(2.0 * Math.PI * i / (size-1)));
}
return window;
}
"Porting" Matlab code statement by statement to Java is a bad approach.
Data is rarely manipulated in Matlab using loops and addressing individual elements (because the Matlab interpreter/VM is rather slow), but rather through calls to block processing functions (which have been carefully written and optimized). This leads to a very idiosyncratic programming style in which repmat, reshape, find, fancy indexing et al. are used to do operations which would be much more naturally expressed through Java loops.
For example, to multiply each column of a matrix A by a vector v, you will write in matlab:
A = diag(v) * A
or
A = repmat(v', 1, size(A, 2)) .* A
This solution:
for i = 1:size(A, 2),
A(:, i) = A(:, i) .* v';
end;
is inefficient.
But it would be terribly foolish to try to do the same thing in Java and invoke a matrix product or to build a matrix with repeated copies of v. Instead, just do:
for (int i = 0; i < rows; i++) {
for (int j = 0; j < columns; j++) {
a[i][j] *= v[i]
}
}
I suggest you to try to understand what this matlab function is actually doing, instead of focusing on how it is doing it, and reimplement it from scratch in Java, forgetting all the matlab implementation except the specifications given in the comments. Half of the code you have written is useless, indeed. Actually, it seems to me that this function wouldn't be needed at all, and what it does could be efficiently integrated in the caller's code.

imregionalmax matlab function's equivalent in opencv

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.

Teaching a Neural Net: Bipolar XOR

I'm trying to to teach a neural net of 2 inputs, 4 hidden nodes (all in same layer) and 1 output node. The binary representation works fine, but I have problems with the Bipolar. I can't figure out why, but the total error will sometimes converge to the same number around 2.xx. My sigmoid is 2/(1+ exp(-x)) - 1. Perhaps I'm sigmoiding in the wrong place. For example to calculate the output error should I be comparing the sigmoided output with the expected value or with the sigmoided expected value?
I was following this website here: http://galaxy.agh.edu.pl/~vlsi/AI/backp_t_en/backprop.html , but they use different functions then I was instructed to use. Even when I did try to implement their functions I still ran into the same problem. Either way I get stuck about half the time at the same number (a different number for different implementations). Please tell me if I have made a mistake in my code somewhere or if this is normal (I don't see how it could be). Momentum is set to 0. Is this a common 0 momentum problem? The error functions we are supposed to be using are:
if ui is an output unit
Error(i) = (Ci - ui ) * f'(Si )
if ui is a hidden unit
Error(i) = Error(Output) * weight(i to output) * f'(Si)
public double sigmoid( double x ) {
double fBipolar, fBinary, temp;
temp = (1 + Math.exp(-x));
fBipolar = (2 / temp) - 1;
fBinary = 1 / temp;
if(bipolar){
return fBipolar;
}else{
return fBinary;
}
}
// Initialize the weights to random values.
private void initializeWeights(double neg, double pos) {
for(int i = 0; i < numInputs + 1; i++){
for(int j = 0; j < numHiddenNeurons; j++){
inputWeights[i][j] = Math.random() - pos;
if(inputWeights[i][j] < neg || inputWeights[i][j] > pos){
print("ERROR ");
print(inputWeights[i][j]);
}
}
}
for(int i = 0; i < numHiddenNeurons + 1; i++){
hiddenWeights[i] = Math.random() - pos;
if(hiddenWeights[i] < neg || hiddenWeights[i] > pos){
print("ERROR ");
print(hiddenWeights[i]);
}
}
}
// Computes output of the NN without training. I.e. a forward pass
public double outputFor ( double[] argInputVector ) {
for(int i = 0; i < numInputs; i++){
inputs[i] = argInputVector[i];
}
double weightedSum = 0;
for(int i = 0; i < numHiddenNeurons; i++){
weightedSum = 0;
for(int j = 0; j < numInputs + 1; j++){
weightedSum += inputWeights[j][i] * inputs[j];
}
hiddenActivation[i] = sigmoid(weightedSum);
}
weightedSum = 0;
for(int j = 0; j < numHiddenNeurons + 1; j++){
weightedSum += (hiddenActivation[j] * hiddenWeights[j]);
}
return sigmoid(weightedSum);
}
//Computes the derivative of f
public static double fPrime(double u){
double fBipolar, fBinary;
fBipolar = 0.5 * (1 - Math.pow(u,2));
fBinary = u * (1 - u);
if(bipolar){
return fBipolar;
}else{
return fBinary;
}
}
// This method is used to update the weights of the neural net.
public double train ( double [] argInputVector, double argTargetOutput ){
double output = outputFor(argInputVector);
double lastDelta;
double outputError = (argTargetOutput - output) * fPrime(output);
if(outputError != 0){
for(int i = 0; i < numHiddenNeurons + 1; i++){
hiddenError[i] = hiddenWeights[i] * outputError * fPrime(hiddenActivation[i]);
deltaHiddenWeights[i] = learningRate * outputError * hiddenActivation[i] + (momentum * lastDelta);
hiddenWeights[i] += deltaHiddenWeights[i];
}
for(int in = 0; in < numInputs + 1; in++){
for(int hid = 0; hid < numHiddenNeurons; hid++){
lastDelta = deltaInputWeights[in][hid];
deltaInputWeights[in][hid] = learningRate * hiddenError[hid] * inputs[in] + (momentum * lastDelta);
inputWeights[in][hid] += deltaInputWeights[in][hid];
}
}
}
return 0.5 * (argTargetOutput - output) * (argTargetOutput - output);
}
General coding comments:
initializeWeights(-1.0, 1.0);
may not actually get the initial values you were expecting.
initializeWeights should probably have:
inputWeights[i][j] = Math.random() * (pos - neg) + neg;
// ...
hiddenWeights[i] = (Math.random() * (pos - neg)) + neg;
instead of:
Math.random() - pos;
so that this works:
initializeWeights(0.0, 1.0);
and gives you initial values between 0.0 and 1.0 rather than between -1.0 and 0.0.
lastDelta is used before it is declared:
deltaHiddenWeights[i] = learningRate * outputError * hiddenActivation[i] + (momentum * lastDelta);
I'm not sure if the + 1 on numInputs + 1 and numHiddenNeurons + 1 are necessary.
Remember to watch out for rounding of ints: 5/2 = 2, not 2.5!
Use 5.0/2.0 instead. In general, add the .0 in your code when the output should be a double.
Most importantly, have you trained the NeuralNet long enough?
Try running it with numInputs = 2, numHiddenNeurons = 4, learningRate = 0.9, and train for 1,000 or 10,000 times.
Using numHiddenNeurons = 2 it sometimes get "stuck" when trying to solve the XOR problem.
See also XOR problem - simulation