compile error in shader ourside of code - unity3d

I have the following attempt at translating the perlin noise to the GPU in
unity compute shader:
#pragma kernel CSMain
RWTexture2D<float4> Result;
[numthreads(8,8,1)]
//based on http://mrl.nyu.edu/~perlin/noise/
void CSMain (uint3 id : SV_DispatchThreadID)
{ float res = noise((float)id.x,(float)id.y,0.0f);
Result[id.xy] = float4(res,res,res,res);
}
int p[256]= {151,160,137,91,90,15,
131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,
190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,
88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,
77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,
102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,
135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,
5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,
223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,
129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,
251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,
49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,
138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180};
double fade(double t) { return t * t * t * (t * (t * 6 - 15) + 10); }
double lerp(double t, double a, double b) { return a + t * (b - a); }
double grad(int hash, double x, double y, double z) {
int h = hash & 15;
double u = h<8 ? x : y,
v = h<4 ? y : h==12||h==14 ? x : z;
return ((h&1) == 0 ? u : -u) + ((h&2) == 0 ? v : -v);
}
double noise(float xx, float yy, float zz){
int X = (int)floor(xx) & 255;
int Y = (int)floor(yy) & 255;
int Z = (int)floor(zz) & 255;
int x = -floor(x);
int y = -floor(y);
int z = -floor(z);
double u = fade(x);
double v = fade(y);
double w = fade(z);
}
However this gives me an error on line 57 (the code is 47 lines long)
Shader error in "PerlinClouds.ompute": noise(floatM|halfM|min10floatM|min16floatM) at line 57 (on)
Does anybody know what that means? It's not even in my code so I don't quite know where to look.


Make sure you define functions earlier in the file than you first try and use them.
In your noise function:
You don't return a value.
You initialise 'x' on the same line as you use it. You might have meant:
int x = -floor(X);
But maybe you meant:
int x = -floor(xx);
Don't use different case of the same variables (x, xx and X); you appear to have confused yourself and probably anyone reading your code.

Related

Color conversion RGB to HSL using Core Image Kernel Language

I'm trying to create image filter that will shift color of the image. In order to do this I need to convert rgb color to hsl and after shift, convert hsl back to rgb. I make some researches and found formulas that can help me with this task.
I implement them in my playground using Swift just to test if they are reliable and they are. I won't post Swift code here just to keep things clean, but I'll show my test results:
input: rgb (61, 117,237) or (0.24,0.46,0.93)
result:
rgb2hsl [0.613527 0.831325 0.585] or (221, 83, 58.5) //hsl
hsl2rgb [0.24 0.46 0.93] //back to rgb
Great! So far so good.
Now we need to convert our Swift code to Core Image Kernel Language (CIKL).
And here it is:
float hue2rgb(float f1, float f2, float hue) {
if (hue < 0) {
hue += 1.0;
}
else if (hue > 1) {
hue -= 1.0;
}
float res;
if (6*hue<1) {
res = f1 + (f2 - f1) * 6 * hue;
}
else if (2*hue<1) {
res = f2;
}
else if (3*hue<2) {
res = f1 + (f2 - f1) * (2.0/3.0 - hue) * 6;
}
else {
res = f1;
}
return res;
}
vec3 hsl2rgb(vec3 hsl) {
vec3 rgb;
if (hsl.y == 0) {
rgb = vec3(hsl.z,hsl.z,hsl.z);
}
else {
float f2;
if (hsl.z < 0.5) {
f2 = hsl.z * (1.0 + hsl.y);
}
else {
f2 = hsl.z + hsl.y - hsl.y * hsl.z;
}
float f1 = 2 * hsl.z - f2;
float r = hue2rgb(f1, f2, hsl.x + 1.0/3.0);
float g = hue2rgb(f1, f2, hsl.x);
float b = hue2rgb(f1, f2, hsl.x - 1.0/3.0);
rgb = vec3(r,g,b);
}
return rgb;
}
vec3 rgb2hsl(vec3 rgb) {
float maxC = max(rgb.x, max(rgb.y,rgb.z));
float minC = min(rgb.x, min(rgb.y,rgb.z));
float l = (maxC + maxC)/2.0;
float h = 0;
float s = 0;
if (maxC != minC) {
float d = maxC - minC;
s = l > 0.5 ? d / (2.0 - maxC - minC) : d / (maxC + minC);
if (maxC == rgb.x) {
h = (rgb.y - rgb.z) / d + (rgb.y < rgb.z ? 6.0 : 0);
} else if (maxC == rgb.y) {
h = (rgb.z - rgb.x) / d + 2.0;
}
else {
h = (rgb.x - rgb.y) / d + 4.0;
}
h /= 6.0;
}
return vec3(h,s,l);
}
And here comes the problem. I'm not able to get right values using this functions in my filter. To check everything I made a Quartz Composer Patch.
Since I didn't find any print/log option in CIKL, I made this to check if my conversions work right:
The logic of this patch: my filter takes color as an input, convert it to hsl and back to rgb and returns it; image input ignored for now.
Kernel func of my filter:
kernel vec4 kernelFunc(__sample pixel, __color color) {
vec3 vec = color.rgb;
vec3 hsl = rgb2hsl(vec);
return vec4(hsl2rgb(hsl), 1);
}
Filter includes functions listed above.
The result I see in the viewer is:
Image on the right is cropped constant color image from the input color.
The left image is the output from our filter.
Digital color picker returns rgb (237, 239.7, 252) for left image.
I have no more ideas how to debug this thing and find a problem. Any help will be highly appreciated. Thanks.

I found the problem. It was me, converting code from Swift to CIKL I made a stupid mistake that was very hard to find, because you have no print / log tools in CIKL or I don't know about it.
Anyway, the problem was in the rgb2hsl function:
float l = (maxC + maxC)/2.0; // WRONG
it should be:
float l = (maxC + minC)/2.0;
Hope it'll help someone in the future.

How to get the points (coordinates) on 2D Line?

When I plot point1(p1) and point2(p2), the line between p1 and p2 is drawn. I wanna know a set of the points making the line.
For example, I wanna get x, y coordinates (as array type: x[], y[]). Is there any algorithms or code?

Here's what I have come up with:
It is fair to say that we need to use the slope formula, y = m*x + b to find the slope so we can plot our points along that line. We need the following:
(x1, y1)
(x2, y2)
to find the following:
m = (y2 - y1) / (x2 - x1)
b = y1 - (m * x1)
minX = min(x1, x2) used for limiting our lower bound
maxX = max(x1, x2) used for limiting our upper bound
Now that everything is set, we can plot our line pixel by pixel and obtain all (x,y) coordinates we need. The logic is simple:
let x loop from minX to maxX and plug it in y = m*x + b (we already have all the variables except y). Then, store the (x,y) pair.
I have used Java for coding this logically and visually. Also, I used LinkedList instead of arrays (because I we can't know the number of points we will obtain).
I have also drawn what Java would draw (in blue) and my approach (in red). They are almost perfectly the exact output and coordinates. The image below is zoomed 5x the original size.
Note! The above explanation is what you would use if the line is not vertical (because the slope would be undefined, division by zero). If it is, then you will plug y (instead of x) values and find the x (instead of y) value from the following formula x = (y - b) / m (instead of y = m*x + b). Though, the code takes care of vertical lines.
import java.awt.Canvas;
import java.awt.Color;
import java.awt.Graphics;
import java.awt.Point;
import java.util.LinkedList;
import javax.swing.JFrame;
public class LineDrawing extends Canvas {
int x1 = 5;
int y1 = 10;
int x2 = 105;
int y2 = 100;
double m = ((double) (y2 - y1)) / ((double) (x2 - x1));//slope
double b = y1 - (m * ((double) x1));//vertical shift
//Takes care of the domain we will loop between.
//min and max will be assigned minX and maxX if the line is not vertical.
//minY and maxY are assigned to min and max otherwise.
int minX = Math.min(x1, x2);//minimum x value we should consider
int maxX = Math.max(x1, x2);//maximum x value we should consider
int minY = Math.min(y1, y2);//minimum y value we should consider
int maxY = Math.max(y1, y2);//maximum y value we should consider
int min = 0;
int max = 0;
boolean plugX = true;//if true, the line is not vertical.
LinkedList<Point> points = new LinkedList<>();//Store all points here
public LineDrawing() {
if (x1 == x2) {//plug the y value instead the x, this is a vertical line.
plugX = false;
min = minY;
max = maxY;
} else {//dont change and plug x values.
min = minX;
max = maxX;
}
}
#Override
public void paint(Graphics g) {
super.paint(g);
//Draw the line, using default java drawLine in blue.
g.setColor(Color.BLUE);
g.drawLine(x1, y1, x2, y2);
//change the color to red, it will draw our verison.
g.setColor(Color.RED);
//Draw the points, point by point on screen.
//Plug m, x, and b in the formula y = m*x + b
//to obtain the y value.
//OR
//Plug m, y, and b in the formula x = (y - b) / m
//to obtain the x value if vertical line.
//Then plot (x,y) coordinate on screen and add the point to our linkedList.
for (int i = min; i <= max; i++) {
int obtained = 0;
if (plugX) {//not a vertical line
obtained = (int) Math.round((m * i + b));
System.out.println("x = " + i + " , y = " + obtained);
points.add(new Point(i, obtained));
//Uncomment to see the full blue line.
g.drawLine(i, obtained, i, obtained);
} else {//vertical line
obtained = (int) Math.round((double) (i - b) / (double) m);
System.out.println("x = " + x1 + " , y = " + i);
g.drawLine(x1, i, x1, i);//Uncomment to see the full blue line.
points.add(new Point(x1, i));
}
}
//Print out the number of points as well as the coordinates themselves.
System.out.println("Total points: " + points.size());
for (int i = 0; i < points.size(); i++) {
System.out.println(i + " ( " + points.get(i).x
+ ", " + points.get(i).y + " )");
}
}
public static void main(String[] args) {
JFrame frame = new JFrame();
frame.setSize(120, 150);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.add(new LineDrawing());
frame.setVisible(true);
}
}

Best Way to Add 3 Numbers (or 4, or N) in Java - Kahan Sums?

I found a completely different answer to this question, the whole original question makes no sense anymore. However, the answer way be useful, so I modify it a bit...
I want to sum up three double numbers, say a, b, and c, in the most numerically stable way possible.
I think using a Kahan Sum would be the way to go.
However, a strange thought occured to me: Would it make sense to:
First sum up a, b, and c and remember the (absolute value of the) compensation.
Then sum up a, c, b
If the (absolute value of the) compensation of the second sum is smaller, use this sum instead.
Proceed similar with b, a, c and other permutations of the numbers.
Return the sum with the smallest associated absolute compensation.
Would I get a more "stable" Addition of three numbers this way? Or does the order of numbers in the sum have no (use-able) impact on the compensation left at the end of the Summation? With (use-able) I mean to ask whether the compensation value itself is stable enough to contain Information that I can use?
(I am using the Java programming language, although I think this does not matter here.)
Many thanks,
Thomas.

I think I found a much more reliable way to solve the "Add 3" (or "Add 4" or "Add N" numbers problem.
First of all, I implemented my idea from the original post. It resulted into quite some big code which seemed, initially, to work. However, it failed in the following case: add Double.MAX_VALUE, 1, and -Double.MAX_VALUE. The result was 0.
#njuffa's comments inspired me dig somewhat deeper and at http://code.activestate.com/recipes/393090-binary-floating-point-summation-accurate-to-full-p/, I found that in Python, this problem has been solved quite nicely. To see the full code, I downloaded the Python source (Python 3.5.1rc1 - 2015-11-23) from https://www.python.org/getit/source/, where we can find the following method (under PYTHON SOFTWARE FOUNDATION LICENSE VERSION 2):
static PyObject*
math_fsum(PyObject *self, PyObject *seq)
{
PyObject *item, *iter, *sum = NULL;
Py_ssize_t i, j, n = 0, m = NUM_PARTIALS;
double x, y, t, ps[NUM_PARTIALS], *p = ps;
double xsave, special_sum = 0.0, inf_sum = 0.0;
volatile double hi, yr, lo;
iter = PyObject_GetIter(seq);
if (iter == NULL)
return NULL;
PyFPE_START_PROTECT("fsum", Py_DECREF(iter); return NULL)
for(;;) { /* for x in iterable */
assert(0 <= n && n <= m);
assert((m == NUM_PARTIALS && p == ps) ||
(m > NUM_PARTIALS && p != NULL));
item = PyIter_Next(iter);
if (item == NULL) {
if (PyErr_Occurred())
goto _fsum_error;
break;
}
x = PyFloat_AsDouble(item);
Py_DECREF(item);
if (PyErr_Occurred())
goto _fsum_error;
xsave = x;
for (i = j = 0; j < n; j++) { /* for y in partials */
y = p[j];
if (fabs(x) < fabs(y)) {
t = x; x = y; y = t;
}
hi = x + y;
yr = hi - x;
lo = y - yr;
if (lo != 0.0)
p[i++] = lo;
x = hi;
}
n = i; /* ps[i:] = [x] */
if (x != 0.0) {
if (! Py_IS_FINITE(x)) {
/* a nonfinite x could arise either as
a result of intermediate overflow, or
as a result of a nan or inf in the
summands */
if (Py_IS_FINITE(xsave)) {
PyErr_SetString(PyExc_OverflowError,
"intermediate overflow in fsum");
goto _fsum_error;
}
if (Py_IS_INFINITY(xsave))
inf_sum += xsave;
special_sum += xsave;
/* reset partials */
n = 0;
}
else if (n >= m && _fsum_realloc(&p, n, ps, &m))
goto _fsum_error;
else
p[n++] = x;
}
}
if (special_sum != 0.0) {
if (Py_IS_NAN(inf_sum))
PyErr_SetString(PyExc_ValueError,
"-inf + inf in fsum");
else
sum = PyFloat_FromDouble(special_sum);
goto _fsum_error;
}
hi = 0.0;
if (n > 0) {
hi = p[--n];
/* sum_exact(ps, hi) from the top, stop when the sum becomes
inexact. */
while (n > 0) {
x = hi;
y = p[--n];
assert(fabs(y) < fabs(x));
hi = x + y;
yr = hi - x;
lo = y - yr;
if (lo != 0.0)
break;
}
/* Make half-even rounding work across multiple partials.
Needed so that sum([1e-16, 1, 1e16]) will round-up the last
digit to two instead of down to zero (the 1e-16 makes the 1
slightly closer to two). With a potential 1 ULP rounding
error fixed-up, math.fsum() can guarantee commutativity. */
if (n > 0 && ((lo < 0.0 && p[n-1] < 0.0) ||
(lo > 0.0 && p[n-1] > 0.0))) {
y = lo * 2.0;
x = hi + y;
yr = x - hi;
if (y == yr)
hi = x;
}
}
sum = PyFloat_FromDouble(hi);
_fsum_error:
PyFPE_END_PROTECT(hi)
Py_DECREF(iter);
if (p != ps)
PyMem_Free(p);
return sum;
}
This summation method is different from Kahan's method, it uses a variable number of compensation variables. When adding the ith number, at most i additional compensation variables (stored in the array p) get used. This means if I want to add 3 numbers, I may need 3 additional variables. For 4 numbers, I may need 4 additional variables. Since the number of used variables may increase from n to n+1 only after the nth summand is loaded, I can translate the above code to Java as follows:
/**
* Compute the exact sum of the values in the given array
* {#code summands} while destroying the contents of said array.
*
* #param summands
* the summand array – will be summed up and destroyed
* #return the accurate sum of the elements of {#code summands}
*/
private static final double __destructiveSum(final double[] summands) {
int i, j, n;
double x, y, t, xsave, hi, yr, lo;
boolean ninf, pinf;
n = 0;
lo = 0d;
ninf = pinf = false;
for (double summand : summands) {
xsave = summand;
for (i = j = 0; j < n; j++) {
y = summands[j];
if (Math.abs(summand) < Math.abs(y)) {
t = summand;
summand = y;
y = t;
}
hi = summand + y;
yr = hi - summand;
lo = y - yr;
if (lo != 0.0) {
summands[i++] = lo;
}
summand = hi;
}
n = i; /* ps[i:] = [summand] */
if (summand != 0d) {
if ((summand > Double.NEGATIVE_INFINITY)
&& (summand < Double.POSITIVE_INFINITY)) {
summands[n++] = summand;// all finite, good, continue
} else {
if (xsave <= Double.NEGATIVE_INFINITY) {
if (pinf) {
return Double.NaN;
}
ninf = true;
} else {
if (xsave >= Double.POSITIVE_INFINITY) {
if (ninf) {
return Double.NaN;
}
pinf = true;
} else {
return Double.NaN;
}
}
n = 0;
}
}
}
if (pinf) {
return Double.POSITIVE_INFINITY;
}
if (ninf) {
return Double.NEGATIVE_INFINITY;
}
hi = 0d;
if (n > 0) {
hi = summands[--n];
/*
* sum_exact(ps, hi) from the top, stop when the sum becomes inexact.
*/
while (n > 0) {
x = hi;
y = summands[--n];
hi = x + y;
yr = hi - x;
lo = y - yr;
if (lo != 0d) {
break;
}
}
/*
* Make half-even rounding work across multiple partials. Needed so
* that sum([1e-16, 1, 1e16]) will round-up the last digit to two
* instead of down to zero (the 1e-16 makes the 1 slightly closer to
* two). With a potential 1 ULP rounding error fixed-up, math.fsum()
* can guarantee commutativity.
*/
if ((n > 0) && (((lo < 0d) && (summands[n - 1] < 0d)) || //
((lo > 0d) && (summands[n - 1] > 0d)))) {
y = lo * 2d;
x = hi + y;
yr = x - hi;
if (y == yr) {
hi = x;
}
}
}
return hi;
}
This function will take the array summands and add up the elements while simultaneously using it to store the compensation variables. Since we load the summand at index i before the array element at said index may become used for compensation, this will work.
Since the array will be small if the number of variables to add is small and won't escape the scope of our method, I think there is a decent chance that it will be allocated directly on the stack by the JIT, which may make the code quite fast.
I admit that I did not fully understand why the authors of the original code handled infinities, overflows, and NaNs the way they did. Here my code deviates from the original. (I hope I did not mess it up.)
Either way, I can now sum up 3, 4, or n double numbers by doing:
public static final double add3(final double x0, final double x1,
final double x2) {
return __destructiveSum(new double[] { x0, x1, x2 });
}
public static final double add4(final double x0, final double x1,
final double x2, final double x3) {
return __destructiveSum(new double[] { x0, x1, x2, x3 });
}
If I want to sum up 3 or 4 long numbers and obtain the precise result as double, I will have to deal with the fact that doubles can only represent longs in -9007199254740992..9007199254740992L. But this can easily be done by splitting each long into two parts:
public static final long add3(final long x0, final long x1,
final long x2) {
double lx;
return __destructiveSum(new long[] {new double[] { //
lx = x0, //
(x0 - ((long) lx)), //
lx = x1, //
(x1 - ((long) lx)), //
lx = x2, //
(x2 - ((long) lx)), //
});
}
public static final long add4(final long x0, final long x1,
final long x2, final long x3) {
double lx;
return __destructiveSum(new long[] {new double[] { //
lx = x0, //
(x0 - ((long) lx)), //
lx = x1, //
(x1 - ((long) lx)), //
lx = x2, //
(x2 - ((long) lx)), //
lx = x3, //
(x3 - ((long) lx)), //
});
}
I think this should be about right. At least I can now add Double.MAX_VALUE, 1, and -Double.MAX_VALUE and get 1 as result.

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.

Transformation of coordinates, Concept

i want to convert x,y,z coordinates to polar coordinates. I am getting (-) in y coordiantes. Can someone explain me why I am getting it. It would be great help.
I am reading these values (xyz , az_elev_r) from a software and can't be changed.I am just not sure of the order of angles( az & elev). Using my code I get -y instead of y. It means there is 180 rotation.My code is:
xyz=[-0.564 3.689 -0.735;
2.011 5.067 -1.031;
-1.181 3.943 -1.825; % Reference values
];
%% az_elev_r-->xyz
az_elev_r=[ 261.30 -11.24 3.80;
291.65 -10.692 5.548;
253.34 -23.897 4.50]; % Also Reference (degree)
az_elev_r(:,1:2)=deg2rad(az_elev_r(:,1:2));
r=az_elev_r(:,3);
az=az_elev_r(:,1);
elev=az_elev_r(:,2);
x=r.*cos(az).*cos(elev)
y=r.*sin(az).*cos(elev)
z=r.*sin(elev)

Your az_elev_r matrix is not consistent with your xyz reference.
>> [az, el, r] = cart2sph(xyz(:,1), xyz(:,2), xyz(:,3));
>> rad2deg(az)
ans =
98.6924675475501
68.3527736950233
106.673911589314
Your answers are consistent with the values returned by the sph2cart function. (Example starts with your original input, before the dec2rad replacement.
>> [x, y, z] = sph2cart(deg2rad(az_elev_r(:,1)), deg2rad(az_elev_r(:,2)), az_elev_r(:,3))
x =
-0.563766229670505
2.01131973806906
-1.17951822049783
y =
-3.68422880893852
-5.06709019311118
-3.94153436658676
z =
-0.740692730942158
-1.02931719412937
-1.82292172199717
Incidentally, you're code will be more readable if you just use the sph2cart function, and work in radians, unless you are trying to understand the conversions for their own sake.

OpenCV has the code for conversion to polar coordinates and back. This conversion is useful for finding object rotation through correlation or otherwise creating object-centred 'rotation-independent' representation of objects. It is useful to visualize each of the polar coordinates as well as their joint image. The images below should be self_explanatory. The polar plot has angle as a horizontal axis and Radius as a vertical axis, so that 4 peaks correspond to the 4 corners of the input image. The code (C++ with OpenCV) is attached.
//================================
// Name : PolarCoord.cpp
// Author : V.Ivanchenko cudassimo#gmail.com
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//======================================
#include <iostream>
#include "opencv.hpp"
using namespace std;
using namespace cv;
#define VALID(x, y, w, h) ((x)>=0 && (y)>=0 && (x)<(w) && (y)<(h)) // validates index
/*
* 1. Original binary image HxW CV_8U
* |
* |
* V
* 2. Two coordinate Mats HxW CV_32F
* |
* |
* V
* 3. Visualization CV_8U
* a. gray HxW for a single coordinate image
* b. binary Rx360 for two coordinate images
*/
// convert a binary 2D image into two Mats with float coordiantes
void imageToCoord(const Mat& img, Mat& X, Mat& Y, bool centered = true) {
if (img.empty())
return;
int h = img.rows;
int w = img.cols;
X.create(h, w, CV_32F);
Y.create(h, w, CV_32F);
float Cx = w/2.0f;
float Cy = h/2.0f;
for (int i=0; i<h; ++i){
const uchar* img_row = img.ptr<uchar>(i);
float* x_row = X.ptr<float>(i);
float* y_row = Y.ptr<float>(i);
for (int j=0; j<w; ++j) {
if (img_row[j]>0) {
float x = j;
float y = i;
if (centered) {
x-=Cx;
y-=Cy;
}
x_row[j] = x;
y_row[j] = y;
}
} // j
} // i
} //imageToCoord()
// convert a single float ploar coord Mat to a gray image
void polarToImg(const Mat& PolarCoord, Mat& img) {
if (PolarCoord.empty())
return;
int h = PolarCoord.rows;
int w = PolarCoord.cols;
img.create(h, w, CV_8U);
float maxVal = std::numeric_limits<float>::min();
// find maxVal
for (int i=0; i<h; ++i){
const float* x_row = PolarCoord.ptr<float>(i);
for (int j=0; j<w; ++j) {
if (maxVal < x_row[j])
maxVal = x_row[j];
} // j
} // i
// create an image
if (maxVal>0) {
float k = 255.0/maxVal;
for (int i=0; i<h; ++i){
uchar* img_row = img.ptr<uchar>(i);
const float* x_row = PolarCoord.ptr<float>(i);
for (int j=0; j<w; ++j) {
img_row[j] = saturate_cast<uchar>(k*x_row[j]);
}// j
} // i
} // if
} // plarToImg()
// convert two polar coord Mats to a binary image
void polarToImg(const Mat& radius, const Mat& angle, Mat& img) {
if (angle.empty() || radius.empty())
return;
int h = angle.rows;
int w = angle.cols;
assert(radius.cols==w && radius.rows==h);
const int imgH = sqrt(h*h+w*w)+0.5f; // radius
const int imgW = 360; // angle, deg
img.create(imgH, imgW, CV_8U);
// create an image
for (int i=0; i<h; ++i){
const float* ang_row = angle.ptr<float>(i);
const float* r_row = radius.ptr<float>(i);
for (int j=0; j<w; ++j) {
int x = ang_row[j] + 0.5f;
int y = r_row[j] + 0.5f;
if (x>0) {
cout<<x<<endl;
}
if (VALID(x, y, imgW, imgH))
img.at<uchar>(y, x) = 255;
else {
cout<<"Invalid x, y: "<<x<<", "<<y<<endl;
}
}// j
} // i
} // plarToImg()
int main() {
cout << "Cartesian to polar" << endl; // prints "Syntax training in openCV"
const int W=400, H=400;
Mat Minput(H, W, CV_8U);
Minput(Rect(W/4, H/4, W/2, H/2)) = 255;
Mat X, Y, Angle, Radius, Mr, Mang, Mpolar;
// processing
imageToCoord(Minput, X, Y); // extract coordinates
cartToPolar(X, Y, Radius, Angle, true);// convert coordiantes
// visualize
polarToImg(Radius, Mr);
polarToImg(Angle, Mang);
polarToImg(Radius, Angle, Mpolar);
// debug
//cout<<Mpolar<<endl;
namedWindow("input", 0);
namedWindow("angle", 0);
namedWindow("radius", 0);
namedWindow("Polar", 0);
const int winw=200, winh=200;
resizeWindow("input", winw, winh);
resizeWindow("angle", winw, winh);
resizeWindow("radius", winw, winh);
resizeWindow("Polar", 360, (int)sqrt(H*H + W*W));
moveWindow("input", 0, 0);
moveWindow("angle", winw, 0);
moveWindow("radius", 2*winw, 0);
moveWindow("Polar", 3*winw, 0);
imshow("input", Minput);
imshow("angle", Mang);
imshow("radius", Mr);
imshow("Polar", Mpolar);
waitKey(-1);
return 0;
}