Finding the "fitLine " of contours in OpenCV - c++

I'm a program that find contours in a stream, for example :
I want to find "set of points " that can describe this contours say like the red line :
the yellow part is the the moments of the contours ,
I tried to use fitLine function of opencv but the result was nonsense, any Idea how can I get the middle line of a contours, that should represent the key aspect of my Contours . by the way **I'm not asking for codes ! ** just a hint how can I do that ?
thanks in advance for any help

maybe try this approach with a distance transform and ridge detection:
cv::Mat input = cv::imread("fitLine.jpg");
cv::Mat gray;
cv::cvtColor(input,gray,CV_BGR2GRAY);
cv::Mat mask = gray>100;
cv::imshow("mask",mask);
cv::Mat dt;
cv::distanceTransform(mask,dt,CV_DIST_L1,CV_DIST_MASK_PRECISE);
cv::imshow("dt", dt/15.0f);
cv::imwrite("fitLineOut.png",255*dt/15.0f);
//care: this part doesn't work for diagonal lines, a ridge detection would be better!!
cv::Mat lines = cv::Mat::zeros(input.rows, input.cols, CV_8UC1);
//only take the maxDist of each row
for(unsigned int y=0; y<dt.rows; ++y)
{
float biggestDist = 0;
cv::Point2i biggestDistLoc(0,0);
for(unsigned int x=0; x<dt.cols; ++x)
{
cv::Point2i current(x,y);
if(dt.at<float>(current) > biggestDist)
{
biggestDist = dt.at<float>(current) ;
biggestDistLoc = current;
}
}
lines.at<unsigned char>(biggestDistLoc) = 255;
}
//and the maxDist of each row
for(unsigned int x=0; x<dt.cols; ++x)
{
float biggestDist = 0;
cv::Point2i biggestDistLoc(0,0);
for(unsigned int y=0; y<dt.rows; ++y)
{
cv::Point2i current(x,y);
if(dt.at<float>(current) > biggestDist)
{
biggestDist = dt.at<float>(current) ;
biggestDistLoc = current;
}
}
lines.at<unsigned char>(biggestDistLoc) = 255;
}
cv::imshow("max", lines);
cv::waitKey(-1);
the idea is to compute the distance transform and find the ridges within the contour.
this is how the distance transformed image looks like:
you can see that there is a local ridge maximum in the middle of the lines.
then I use a very simple method: just find the maximum distance in each row/column. That is very sloppy and should be changed for a real ridge detection or thinning method!!!!
edit: additional description:
The idea is to find all points that are in the "middle" of the contour. In mathematics/graphics, the medial axis in some kind of "middle" of an object and it's definition is to be all points that have the same minimum distance to at least two contour points at the same time.
A way to approximate the medial axis is to compute the distance transform. The distance transform is a matrix that holds for each pixel the distance to the next object point (for example a contour of an object)(see http://en.wikipedia.org/wiki/Distance_transform too). That's the first image. There you can see that the points in the middle of the lines are a little bit brighter than the points closer to the border, which means that the brightest points along the lines can be interpreted as the medial axis (approximation), since if you move away from it (orthogonal to the line direction) the distances become smaller, so the peak is the point where the distance to both borders in close to equality.
That way if you can find those "ridges" in the distance transform, you're done.
Ridge detection is normally done by a Harris Operator (see http://en.wikipedia.org/wiki/Ridge_detection ).
In the fast and dirty version that I've posted, I try to detect the ridge by accepting only the maximum value in each line and in each row. That's ok for most horizontal and vertical ridges, but will fail for diagonal ones. So maybe you really want to exchange the for loops with a real ridge detection.

Interesting task :) Here is my solution:
Here is code:
#include <iostream>
#include <vector>
#include <stdio.h>
#include <stdarg.h>
#include <set>
#include "opencv2/opencv.hpp"
#include "fstream"
#include "iostream"
using namespace std;
using namespace cv;
int Thinning(unsigned char * ucBinedImg, unsigned char * ucThinnedImage, long lWidth, long lHeight, long lIterativeLimit)
{
if(ucBinedImg == NULL)
return -1;
if(ucThinnedImage == NULL)
return -2;
if(lIterativeLimit == -1)
lIterativeLimit = 60000;
unsigned char x1, x2, x3, x4, x5, x6, x7, x8, xp;
unsigned char g1, g2, g3, g4;
unsigned char b1, b2, b3, b4;
unsigned char np1, np2, npm;
unsigned char *pUp, *pDown, *pImg;
long lDeletedPoints = 0;
// set border
memcpy(ucThinnedImage, ucBinedImg, lWidth*lHeight);
for(long it=0; it<lIterativeLimit; it++)
{
lDeletedPoints = 0;
for(long i=1; i<lHeight-1; i++)
{
// init neighborhood
pUp = ucBinedImg + (i-1)*lWidth;
pImg = ucBinedImg + i*lWidth ;
pDown = ucBinedImg + (i+1)*lWidth ;
for( long j=1; j<lWidth-1; j++)
{
pUp++;
pImg++;
pDown++;
if(!*pImg)
continue;
x6 = *(pUp-1);
x5 = *(pImg-1);
x4 = *(pDown-1);
x7 = *pUp;
xp = *pImg;
x3 = *pDown;
x8 = *(pUp+1);
x1 = *(pImg + 1);
x2 = *(pDown + 1);
b1 = !x1 && (x2 == 1 || x3 == 1);
b2 = !x3 && (x4 == 1 || x5 == 1);
b3 = !x5 && (x6 == 1 || x7 == 1);
b4 = !x7 && (x8 == 1 || x1 == 1);
g1 = (b1 + b2 + b3 + b4) == 1;
np1 = x1|| x2;
np1 += x3 || x4;
np1 += x5 || x6;
np1 += x7 || x8;
np2 = x2|| x3;
np2 += x4 || x5;
np2 += x6 || x7;
np2 += x8 || x1;
npm = np1>np2?np2:np1;
g2 = npm>=2 && npm<=3;
g3 = (x1 && (x2 || x3 || !x8)) == 0;
g4 = (x5 && (x6 || x7 || !x4)) == 0;
// first part
if(g1 && g2 && g3)
{
// delete this point
ucThinnedImage[lWidth*i + j] = 0;
++lDeletedPoints;
}
}
}
//syn
memcpy(ucBinedImg, ucThinnedImage, lWidth*lHeight);
for(long i=1; i<lHeight-1; i++)
{
// init neighborhood
pUp = ucBinedImg + (i-1)*lWidth;
pImg = ucBinedImg + i*lWidth ;
pDown = ucBinedImg + (i+1)*lWidth ;
for( long j=1; j<lWidth-1; j++)
{
pUp++;
pImg++;
pDown++;
if(!*pImg)
continue;
x6 = *(pUp-1);
x5 = *(pImg-1);
x4 = *(pDown-1);
x7 = *pUp;
xp = *pImg;
x3 = *pDown;
x8 = *(pUp+1);
x1 = *(pImg + 1);
x2 = *(pDown + 1);
b1 = !x1 && (x2 == 1 || x3 == 1);
b2 = !x3 && (x4 == 1 || x5 == 1);
b3 = !x5 && (x6 == 1 || x7 == 1);
b4 = !x7 && (x8 == 1 || x1 == 1);
g1 = (b1 + b2 + b3 + b4) == 1;
np1 = x1|| x2;
np1 += x3 || x4;
np1 += x5 || x6;
np1 += x7 || x8;
np2 = x2|| x3;
np2 += x4 || x5;
np2 += x6 || x7;
np2 += x8 || x1;
npm = np1>np2?np2:np1;
g2 = npm>=2 && npm<=3;
g3 = (x1 && (x2 || x3 || !x8)) == 0;
g4 = (x5 && (x6 || x7 || !x4)) == 0;
// second part
if(g1 && g2 && g4)
{
// delete this point
ucThinnedImage[lWidth*i + j] = 0;
++lDeletedPoints;
}
}
}
//syn
memcpy(ucBinedImg, ucThinnedImage, lWidth*lHeight);
// if no points to be deleted
if(lDeletedPoints == 0)
break;
}
// clear edge bar
for(long i=0; i<lHeight; i++)
{
for(long j=0; j<lWidth; j++)
{
if(i<16)
ucThinnedImage[i*lWidth+j] = 0;
else if(i>=lHeight-16)
ucThinnedImage[i*lWidth+j] = 0;
else if(j<16)
ucThinnedImage[i*lWidth+j] = 0;
else if(j>=lWidth-16)
ucThinnedImage[i*lWidth+j] = 0;
}
}
return 0;
}
void Thinning(Mat& src,Mat& dst,long IterativeLimit=-1)
{
Mat bin_img=src&1;
if(!dst.empty()){dst.release();}
dst=Mat::zeros(src.size(),CV_8UC1);
Thinning(bin_img.data,dst.data,bin_img.cols,bin_img.rows,IterativeLimit);
dst*=255;
}
int main(int argc, char* argv[])
{
namedWindow("source");
namedWindow("result");
Mat img=imread("raw_image.jpg",0);
cv::threshold(img,img,128,255,cv::THRESH_BINARY);
int erosion_size=5;
Mat element = getStructuringElement( cv::MORPH_ELLIPSE,Size( 2*erosion_size + 1, 2*erosion_size+1 ),Point( erosion_size, erosion_size ) );
cv::dilate(img,img,element);
Mat thinned;
Thinning(img,thinned);
vector<Vec2f> lines;
HoughLines(thinned, lines, 0.5, CV_PI/360, 50, 0, 0 );
float hist_theta[2]={0,0};
float hist_rho[2]={0,0};
float n[2]={0,0};
for( size_t i = 0; i < lines.size(); i++ )
{
float rho = lines[i][0], theta = lines[i][1];
if(fabs(theta-CV_PI/2)<CV_PI/4)
{
hist_theta[0]+=theta;
hist_rho[0]+=rho;
n[0]+=1;
}else
{
hist_theta[1]+=theta;
hist_rho[1]+=rho;
n[1]+=1;
}
}
for( size_t i = 0; i < 2; i++ )
{
float rho = hist_rho[i]/n[i], theta = hist_theta[i]/n[i];
Point pt1, pt2;
double a = cos(theta), b = sin(theta);
double x0 = a*rho, y0 = b*rho;
pt1.x = cvRound(x0 + 1000*(-b));
pt1.y = cvRound(y0 + 1000*(a));
pt2.x = cvRound(x0 - 1000*(-b));
pt2.y = cvRound(y0 - 1000*(a));
line( thinned, pt1, pt2, Scalar(255,255,255), 1, CV_AA);
}
imshow("source",img);
imshow("result",thinned);
cv::waitKey(0);
}
Here is a hack in this source it uses two 1D histograms for postprocessing. In real life application it should use 2D histogram for close lines averaging.

Related

What am I doing wrong with the Sutherland Hodgman algorithm?

Apologies for the wall of code below... but I'm not 100% sure of where the exact issue is. I'm using the Sutherland Hodgman clipping algorithm to find the contact points in my collision detection system (physics engine for my undergrad thesis). The vertices it returns are MOSTLY correct however I often get a lot of false positives on the InsideEdge function or incorrect intersections. If anyone can help I'd be so grateful! I have tried with many different input vertices and the result varies, however the X-axis does seem to be the most consistently correct.
Example IO with visual:
Input: clipping vertices [ (2.0, 2.0), (2.0, -2.0), (-2.0, -2.0), (-2.0, 2.0) ], poly vertices [ (2.5, 3.0), (2.5, 1.0), (0.5, 1.0), (0.5, 3.0) ]
Output: (2.0, 1.0), (0.5, 1.0), (0.5, 3.0 (wrong))
bool SAT::InsideEdge(double px, double py, double edgeMaxX, double edgeMaxY, double edgeMinX, double edgeMinY)
{
double one = (edgeMaxX - edgeMinX) * (py - edgeMinY);
double two = (edgeMaxY - edgeMinY) * (px - edgeMinX);
return (one - two) < 0;
}
void SAT::SutherlandHodgman(std::vector<Vector3>& _clipped, const Vector3& normal, const Vector3* polyVertices, const Vector3* clippingVertices)
{
const unsigned maxPoints = 16;
Vector3 newPoints[maxPoints];
for (unsigned i = 0; i < numVertices; i++)
{
newPoints[i] = polyVertices[i];
}
unsigned newSize = 0;
for (unsigned edge = 0; edge < numEdges; edge++) //for each clipping edge
{
//NOTE: axes is a 2D array to define which two axes to consider out of x, y, z
Vector3 edgeMin = clippingVertices[edge];
Vector3 edgeMax = clippingVertices[(edge + 1) % numEdges];
for (unsigned v = 0; v < numVertices; v++) //for each input vertex
{
Vector3 v1 = newPoints[v];
Vector3 v2 = newPoints[(v + 1) % numVertices];
bool insideEdge1 = InsideEdge(v1[axes[0]], v1[axes[1]], edgeMax[axes[0]], edgeMax[axes[1]], edgeMin[axes[0]], edgeMin[axes[1]]);
bool insideEdge2 = InsideEdge(v2[axes[0]], v2[axes[1]], edgeMax[axes[0]], edgeMax[axes[1]], edgeMin[axes[0]], edgeMin[axes[1]]);
if (insideEdge1 && insideEdge2)
{
newPoints[newSize] = v2;
newSize++;
}
else if (!insideEdge1 && insideEdge2)
{
newPoints[newSize] = CalculateIntersection(v1, v2, axes, edgeMin, edgeMax);
newSize++;
newPoints[newSize] = v2;
newSize++;
}
else if (insideEdge1 && !insideEdge2)
{
newPoints[newSize] = CalculateIntersection(v1, v2, axes, edgeMin, edgeMax);
newSize++;
}
}
numVertices = newSize;
if (numVertices >= maxPoints)
break;
}
for (unsigned i = 0; i < numVertices; i++)
{
//Removes duplicates before adding to the final list
VerifyVertex(_clipped, newPoints[i]);
}
}
And the calculate intersection code...
//x intersect
double num = (x1 * y2 - y1 * x2) * (x3 - x4) - (x1 - x2) * (x3 * y4 - y3 * x4);
double den = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);
double X = num / den;
//y intersect
num = (x1 * y2 - y1 * x2) * (y3 - y4) - (y1 - y2) * (x3 * y4 - y3 * x4);
den = (x1 - x2) * (y3 - y4) - (y1 - y2) * (x3 - x4);
double Y = num / den;
Vector3 intersection;
intersection[axes[0]] = X;
intersection[axes[1]] = Y;
//Get the value of the remaining axis
double percAcrossLine = 1;
if (x2 != 0)
percAcrossLine = (x1 + X) / x2;
else if (y2 != 0)
percAcrossLine = (y1 + Y) / y2;
unsigned i = 0;
if (axes[0] == 0)
{
if (axes[1] == 1)
i = 2;
else if (axes[1] == 2)
i = 1;
}
intersection[i] = v1[i] + (v2[i] - v1[i]) * percAcrossLine;
return intersection;

Unit testing not failing but hanging on subroutine call

I am unit testing the following C++ code using Visual Studio 2015's built in test framework. When I run the test below, no error is thrown (and the code compiles), but the test just hangs. It only does it when the line I have commented out which calls averageGradient is run. Why is this?
float averageGradient(int x1, int x2) {
int i = 0, y1 = 0, y2 = 0;
while (i < graph.size() && (y1 == 0 || y2 == 0)) { //if both y values haven't been solved then keep trying
if (x1 >= graph[i][0] && x1 < graph[i][1]) { // if x1 lies in the first straight line segment
y1 = (graph[i][2] * x1) + graph[i][2]; // then calculate it's y value (y1)
}
else if (x2 >= graph[i][0] && x2 < graph[i][1]) { //try the same thing for x2
y2 = (graph[i][2] * x2) + graph[i][3]; //calculate its y value (y2)
}
else { i++; } //otherwise incriment i to check the next straight line segment
}
float m = (y2 - y1) / (x2 - x1);
return m;
};
Unit Testing:
TEST_METHOD(Average_Gradient) {
int x1 = 683675;
int x2 = x1 + 86400;
//float gradient = averageGradient(x1, x2);
float answer = 0.0000895684639;
//Assert::AreEqual(answer, gradient);
}
There is an infinite loop in your code. This is not related to unit test.
float averageGradient(int x1, int x2) {
int i = 0, y1 = 0, y2 = 0;
while (i < graph.size() && (y1 == 0 || y2 == 0)) // 1
{
if (x1 >= graph[i][0] && x1 < graph[i][1]) // 2
{
y1 = (graph[i][2] * x1) + graph[i][2]; // 3
}
else if (x2 >= graph[i][0] && x2 < graph[i][1]) // 4
{
y2 = (graph[i][2] * x2) + graph[i][3];
}
else { i++; }
}
float m = (y2 - y1) / (x2 - x1);
return m;
};
In the first iteration of your loop, let's assume that condition in line marked with // 2 is true.
You enter line // 3, and change y1. i and y2 are not modified.
Now you go to next iteration.
i hasn't changed so i < graph.size() is still true.
y2 hasn't changed so (y1 == 0 || y2 == 0) is still true.
Thus, you go to line // 2: the condition is true again, as neither x1 nor graph[] values have changed.
Execution enters // 3, y1 is not modified
Execution continues at 3. (infinite loop)
If your first iteration enters // 4 instead of // 2, the same thing happens.
You should be able to easily analyze this just by running your code under debugger.

Delaunay triangulation : too many triangles

I'm trying to implement the Delaunay triangulation in C++. Currently it's working, but I'm not getting the correct amount of triangles.
I try it with 4 points in a square pattern : (0,0), (1,0), (0,1), (1,1).
Here's the algorithm I use :
std::vector<Triangle> Delaunay::triangulate(std::vector<Vec2f> &vertices) {
// Determinate the super triangle
float minX = vertices[0].getX();
float minY = vertices[0].getY();
float maxX = minX;
float maxY = minY;
for(std::size_t i = 0; i < vertices.size(); ++i) {
if (vertices[i].getX() < minX) minX = vertices[i].getX();
if (vertices[i].getY() < minY) minY = vertices[i].getY();
if (vertices[i].getX() > maxX) maxX = vertices[i].getX();
if (vertices[i].getY() > maxY) maxY = vertices[i].getY();
}
float dx = maxX - minX;
float dy = maxY - minY;
float deltaMax = std::max(dx, dy);
float midx = (minX + maxX) / 2.f;
float midy = (minY + maxY) / 2.f;
Vec2f p1(midx - 20 * deltaMax, midy - deltaMax);
Vec2f p2(midx, midy + 20 * deltaMax);
Vec2f p3(midx + 20 * deltaMax, midy - deltaMax);
// Add the super triangle vertices to the end of the vertex list
vertices.push_back(p1);
vertices.push_back(p2);
vertices.push_back(p3);
// Add the super triangle to the triangle list
std::vector<Triangle> triangleList = {Triangle(p1, p2, p3)};
// For each point in the vertex list
for(auto point = begin(vertices); point != end(vertices); point++)
{
// Initialize the edges buffer
std::vector<Edge> edgesBuff;
// For each triangles currently in the triangle list
for(auto triangle = begin(triangleList); triangle != end(triangleList);)
{
if(triangle->inCircumCircle(*point))
{
Edge tmp[3] = {triangle->getE1(), triangle->getE2(), triangle->getE3()};
edgesBuff.insert(end(edgesBuff), tmp, tmp + 3);
triangle = triangleList.erase(triangle);
}
else
{
triangle++;
}
}
// Delete all doubly specified edges from the edge buffer
// Black magic by https://github.com/MechaRage
auto ite = begin(edgesBuff), last = end(edgesBuff);
while(ite != last) {
// Search for at least one duplicate of the current element
auto twin = std::find(ite + 1, last, *ite);
if(twin != last)
// If one is found, push them all to the end.
last = std::partition(ite, last, [&ite](auto const &o){ return !(o == *ite); });
else
++ite;
}
// Remove all the duplicates, which have been shoved past "last".
edgesBuff.erase(last, end(edgesBuff));
// Add the triangle to the list
for(auto edge = begin(edgesBuff); edge != end(edgesBuff); edge++)
triangleList.push_back(Triangle(edge->getP1(), edge->getP2(), *point));
}
// Remove any triangles from the triangle list that use the supertriangle vertices
triangleList.erase(std::remove_if(begin(triangleList), end(triangleList), [p1, p2, p3](auto t){
return t.containsVertex(p1) || t.containsVertex(p2) || t.containsVertex(p3);
}), end(triangleList));
return triangleList;
}
And here's what I obtain :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 1 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
Triangle:
Point x: 0 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
While this would be the correct output :
Triangle:
Point x: 1 y: 0
Point x: 0 y: 0
Point x: 0 y: 1
Triangle:
Point x: 1 y: 0
Point x: 1 y: 1
Point x: 0 y: 1
I have no idea why there is a triangle with the (0, 0) and the (1, 1).
I need an outside eye to review the code and find out what's going wrong.
All the sources are on my Github repo. Feel free to fork it and to PR your code.
Thanks!
what about this implementation of Paul Bourke's Delaunay triangulation algorithm. Take a look at Triangulate() I have used this source many times without any complains
#include <iostream>
#include <stdlib.h> // for C qsort
#include <cmath>
#include <time.h> // for random
const int MaxVertices = 500;
const int MaxTriangles = 1000;
//const int n_MaxPoints = 10; // for the test programm
const double EPSILON = 0.000001;
struct ITRIANGLE{
int p1, p2, p3;
};
struct IEDGE{
int p1, p2;
};
struct XYZ{
double x, y, z;
};
int XYZCompare(const void *v1, const void *v2);
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri);
int CircumCircle(double, double, double, double, double, double, double, double, double&, double&, double&);
using namespace std;
////////////////////////////////////////////////////////////////////////
// CircumCircle() :
// Return true if a point (xp,yp) is inside the circumcircle made up
// of the points (x1,y1), (x2,y2), (x3,y3)
// The circumcircle centre is returned in (xc,yc) and the radius r
// Note : A point on the edge is inside the circumcircle
////////////////////////////////////////////////////////////////////////
int CircumCircle(double xp, double yp, double x1, double y1, double x2,
double y2, double x3, double y3, double &xc, double &yc, double &r){
double m1, m2, mx1, mx2, my1, my2;
double dx, dy, rsqr, drsqr;
/* Check for coincident points */
if(abs(y1 - y2) < EPSILON && abs(y2 - y3) < EPSILON)
return(false);
if(abs(y2-y1) < EPSILON){
m2 = - (x3 - x2) / (y3 - y2);
mx2 = (x2 + x3) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (x2 + x1) / 2.0;
yc = m2 * (xc - mx2) + my2;
}else if(abs(y3 - y2) < EPSILON){
m1 = - (x2 - x1) / (y2 - y1);
mx1 = (x1 + x2) / 2.0;
my1 = (y1 + y2) / 2.0;
xc = (x3 + x2) / 2.0;
yc = m1 * (xc - mx1) + my1;
}else{
m1 = - (x2 - x1) / (y2 - y1);
m2 = - (x3 - x2) / (y3 - y2);
mx1 = (x1 + x2) / 2.0;
mx2 = (x2 + x3) / 2.0;
my1 = (y1 + y2) / 2.0;
my2 = (y2 + y3) / 2.0;
xc = (m1 * mx1 - m2 * mx2 + my2 - my1) / (m1 - m2);
yc = m1 * (xc - mx1) + my1;
}
dx = x2 - xc;
dy = y2 - yc;
rsqr = dx * dx + dy * dy;
r = sqrt(rsqr);
dx = xp - xc;
dy = yp - yc;
drsqr = dx * dx + dy * dy;
return((drsqr <= rsqr) ? true : false);
}
///////////////////////////////////////////////////////////////////////////////
// Triangulate() :
// Triangulation subroutine
// Takes as input NV vertices in array pxyz
// Returned is a list of ntri triangular faces in the array v
// These triangles are arranged in a consistent clockwise order.
// The triangle array 'v' should be malloced to 3 * nv
// The vertex array pxyz must be big enough to hold 3 more points
// The vertex array must be sorted in increasing x values say
//
// qsort(p,nv,sizeof(XYZ),XYZCompare);
///////////////////////////////////////////////////////////////////////////////
int Triangulate(int nv, XYZ pxyz[], ITRIANGLE v[], int &ntri){
int *complete = NULL;
IEDGE *edges = NULL;
IEDGE *p_EdgeTemp;
int nedge = 0;
int trimax, emax = 200;
int status = 0;
int inside;
int i, j, k;
double xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r;
double xmin, xmax, ymin, ymax, xmid, ymid;
double dx, dy, dmax;
/* Allocate memory for the completeness list, flag for each triangle */
trimax = 4 * nv;
complete = new int[trimax];
/* Allocate memory for the edge list */
edges = new IEDGE[emax];
/*
Find the maximum and minimum vertex bounds.
This is to allow calculation of the bounding triangle
*/
xmin = pxyz[0].x;
ymin = pxyz[0].y;
xmax = xmin;
ymax = ymin;
for(i = 1; i < nv; i++){
if (pxyz[i].x < xmin) xmin = pxyz[i].x;
if (pxyz[i].x > xmax) xmax = pxyz[i].x;
if (pxyz[i].y < ymin) ymin = pxyz[i].y;
if (pxyz[i].y > ymax) ymax = pxyz[i].y;
}
dx = xmax - xmin;
dy = ymax - ymin;
dmax = (dx > dy) ? dx : dy;
xmid = (xmax + xmin) / 2.0;
ymid = (ymax + ymin) / 2.0;
/*
Set up the supertriangle
his is a triangle which encompasses all the sample points.
The supertriangle coordinates are added to the end of the
vertex list. The supertriangle is the first triangle in
the triangle list.
*/
pxyz[nv+0].x = xmid - 20 * dmax;
pxyz[nv+0].y = ymid - dmax;
pxyz[nv+1].x = xmid;
pxyz[nv+1].y = ymid + 20 * dmax;
pxyz[nv+2].x = xmid + 20 * dmax;
pxyz[nv+2].y = ymid - dmax;
v[0].p1 = nv;
v[0].p2 = nv+1;
v[0].p3 = nv+2;
complete[0] = false;
ntri = 1;
/*
Include each point one at a time into the existing mesh
*/
for(i = 0; i < nv; i++){
xp = pxyz[i].x;
yp = pxyz[i].y;
nedge = 0;
/*
Set up the edge buffer.
If the point (xp,yp) lies inside the circumcircle then the
three edges of that triangle are added to the edge buffer
and that triangle is removed.
*/
for(j = 0; j < ntri; j++){
if(complete[j])
continue;
x1 = pxyz[v[j].p1].x;
y1 = pxyz[v[j].p1].y;
x2 = pxyz[v[j].p2].x;
y2 = pxyz[v[j].p2].y;
x3 = pxyz[v[j].p3].x;
y3 = pxyz[v[j].p3].y;
inside = CircumCircle(xp, yp, x1, y1, x2, y2, x3, y3, xc, yc, r);
if (xc + r < xp)
// Suggested
// if (xc + r + EPSILON < xp)
complete[j] = true;
if(inside){
/* Check that we haven't exceeded the edge list size */
if(nedge + 3 >= emax){
emax += 100;
p_EdgeTemp = new IEDGE[emax];
for (int i = 0; i < nedge; i++) { // Fix by John Bowman
p_EdgeTemp[i] = edges[i];
}
delete []edges;
edges = p_EdgeTemp;
}
edges[nedge+0].p1 = v[j].p1;
edges[nedge+0].p2 = v[j].p2;
edges[nedge+1].p1 = v[j].p2;
edges[nedge+1].p2 = v[j].p3;
edges[nedge+2].p1 = v[j].p3;
edges[nedge+2].p2 = v[j].p1;
nedge += 3;
v[j] = v[ntri-1];
complete[j] = complete[ntri-1];
ntri--;
j--;
}
}
/*
Tag multiple edges
Note: if all triangles are specified anticlockwise then all
interior edges are opposite pointing in direction.
*/
for(j = 0; j < nedge - 1; j++){
for(k = j + 1; k < nedge; k++){
if((edges[j].p1 == edges[k].p2) && (edges[j].p2 == edges[k].p1)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
/* Shouldn't need the following, see note above */
if((edges[j].p1 == edges[k].p1) && (edges[j].p2 == edges[k].p2)){
edges[j].p1 = -1;
edges[j].p2 = -1;
edges[k].p1 = -1;
edges[k].p2 = -1;
}
}
}
/*
Form new triangles for the current point
Skipping over any tagged edges.
All edges are arranged in clockwise order.
*/
for(j = 0; j < nedge; j++) {
if(edges[j].p1 < 0 || edges[j].p2 < 0)
continue;
v[ntri].p1 = edges[j].p1;
v[ntri].p2 = edges[j].p2;
v[ntri].p3 = i;
complete[ntri] = false;
ntri++;
}
}
/*
Remove triangles with supertriangle vertices
These are triangles which have a vertex number greater than nv
*/
for(i = 0; i < ntri; i++) {
if(v[i].p1 >= nv || v[i].p2 >= nv || v[i].p3 >= nv) {
v[i] = v[ntri-1];
ntri--;
i--;
}
}
delete[] edges;
delete[] complete;
return 0;
}
int XYZCompare(const void *v1, const void *v2){
XYZ *p1, *p2;
p1 = (XYZ*)v1;
p2 = (XYZ*)v2;
if(p1->x < p2->x)
return(-1);
else if(p1->x > p2->x)
return(1);
else
return(0);
}
I didn't go with a debugger, but from the resulting triangles it seems that this is an accuracy/ambiguity problem.
When you are triangulating a square there are two ways to split it into triangles and both are OK from Delaunay criteria (circumscribed circle center is on border of triangle).
So if you evaluate every triangle independently you may sometimes get even 4 triangles (depending on implementation).
Normally in such cases I recommend to build algorithm as a series of questions which cannot produce contradicting answers. In this case the question is "to which point goes triangle based on edge (1,0)-(1,1)". But often this requires significant changes to the algorithm.
A quick fix usually involves adding some tolerances for comparisons and extra checks (like non-intersecting triangles). But usually it just makes problems rarer.
Most likely you didn't delete all the double edges, especially not the edges from same triangles but with vertices only in another order. The correct function is in the answer from #cMinor.

how to draw line in opencv

hello every ace i have some problem
Here is my code who can teach me how to draw a straight line in the video at center?
And how to reduce my green line in the video ,because i want to find the Vanishing Point.
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <iostream>e
using namespace cv;
using namespace std;
Point2f center(0,0);
Point2f computeIntersect(Vec4i a, Vec4i b)
{
int x1 = a[0], y1 = a[1], x2 = a[2], y2 = a[3], x3 = b[0], y3 = b[1], x4 = b[2], y4 = b[3];
float denom;
if (float d = ((float)(x1 - x2) * (y3 - y4)) - ((y1 - y2) * (x3 - x4)))
{
Point2f pt;
pt.x = ((x1 * y2 - y1 * x2) * (x3 - x4) - (x1 - x2) * (x3 * y4 - y3 * x4)) / d;
pt.y = ((x1 * y2 - y1 * x2) * (y3 - y4) - (y1 - y2) * (x3 * y4 - y3 * x4)) / d;
return pt;
}
else
return Point2f(-1, -1);
}
void sortCorners(std::vector<Point2f>& corners, Point2f center)
{
std::vector<Point2f> top, bot;
for (int i = 0; i < corners.size(); i++)
{
if (corners[i].y < center.y)
top.push_back(corners[i]);
else
bot.push_back(corners[i]);
}
Point2f tl = top[0].x > top[1].x ? top[1] : top[0];
Point2f tr = top[0].x > top[1].x ? top[0] : top[1];
Point2f bl = bot[0].x > bot[1].x ? bot[1] : bot[0];
Point2f br = bot[0].x > bot[1].x ? bot[0] : bot[1];
corners.clear();
corners.push_back(tl);
corners.push_back(tr);
corners.push_back(br);
corners.push_back(bl);
}
Mat src;
int main()
{
VideoCapture cap("C://pic/way.mp4");
while ( cap.isOpened() )
{
cap >> src;
//Mat src = imread("D:\\11.jpg");
if (src.empty())
return -1;
Mat bw;
cvtColor(src, bw, CV_BGR2GRAY);
blur(bw, bw, Size(3, 3));
Canny(bw, bw, 150, 100, 3);
std::vector<Vec4i> lines;
HoughLinesP(bw, lines, 1, CV_PI/180, 70, 30, 10);
// Expand the lines
for (int i = 0; i < lines.size(); i++)
{
Vec4i v = lines[i];
lines[i][0] = 0;
lines[i][1] = ((float)v[1] - v[3]) / (v[0] - v[2]) * -v[0] + v[1];
lines[i][2] = src.cols;
lines[i][3] = ((float)v[1] - v[3]) / (v[0] - v[2]) * (src.cols - v[2]) + v[3];
}
std::vector<Point2f> corners;
for (int i = 0; i < lines.size(); i++)
{
for (int j = i+1; j < lines.size(); j++)
{
Point2f pt = computeIntersect(lines[i], lines[j]);
if (pt.x >= 0 && pt.y >= 0)
corners.push_back(pt);
}
}
std::vector<Point2f> approx;
approxPolyDP(Mat(corners), approx, arcLength(Mat(corners), true) * 0.02, true);
// Get mass center
for (int i = 0; i < corners.size(); i++)
center += corners[i];
center *= (1. / corners.size());
sortCorners(corners, center);
Mat dst = src.clone();
// Draw lines
for (int i = 0; i < lines.size(); i++)
{
Vec4i v = lines[i];
line(dst, Point(v[0], v[1]), Point(v[2], v[3]), CV_RGB(0,255,0));
}
// Draw corner points
circle(dst, corners[3], 3, CV_RGB(255,255,255), 2);
cout << "Detected " << corners[3] << "\n";
// Draw mass center
circle(dst, center, 3, CV_RGB(255,255,0), 2);
Mat quad = Mat::zeros(300, 220, CV_8UC3);
std::vector<Point2f> quad_pts;
quad_pts.push_back(Point2f(0, 0));
quad_pts.push_back(Point2f(quad.cols, 0));
quad_pts.push_back(Point2f(quad.cols, quad.rows));
quad_pts.push_back(Point2f(0, quad.rows));
Mat transmtx = getPerspectiveTransform(corners, quad_pts);
warpPerspective(src, quad, transmtx, quad.size());
imshow("image", bw);
imshow("images", dst);
if(waitKey(27) >= 0) break; }
return 0;}
Check this out. Hope it will help, Let me know.

how to convert image in to matrix using opencv?

I am trying to make a program in OpenCV to convert an image into matrix form, with each value representing an image's pixel. I have converted the image into binary form and now I want to convert it's pixel values into a matrix.
If You need to use CvMat object, You may want to try to use cvCopy function. It takes CvArr* as its arguments, so both IPLImage and CvMat will fit. If You would leave the C API and go to something more modern, You can use cv::Mat object to load image into and use C++ threshold.
The question is why do You want to convert the format of matrix that you already have (IPLImage as well as all others are matrices already). If You want to have a matrix of bool type, use Matx or Mat_ template class for this.
First glance at your question raises more questions... try to specify a bit (I don't seem to be able to see your code example, I'm new to stackoverflow)
Such as, your open cv version and IDE (like codeblocks or Microsoft Visual Studio). But include it in your question. What I would also like to know, is what is the purpose of this? Why do you need a matrix and so forth :)
attempted answer
from what I can gather
"but I have installed OpenCV version 2.3.1 on Visual C++ 2010 – Ayesha Khan"
OpenCV uses the class called Mat, which you should have encountered a lot. This class is essentially a matrix already. If I remember correctly it is very similar to vectors, which I won't cover here.
so if you need to access any pixel value in, lets say.
Mat Img;
you would use a function in this instance of the class, as such
cout << Img.at<uchar>(x,y);
This will access and print the value of the pixel with the coordinates of x,y, to console. In this example I use uchar inside the pointy brackets <>. uchar is used for 8bit picures. You will have to change this if you work with images of more detail (more bits).
When using a binary picture, OpenCV will most likely will allocate the memory of 8bit, which means you need the example above.
I'd like to give more details, but not before you've specified what exactly it is that you are attempting to do.
Regards Scrub # Stackoverflow
Your code uses OpenCV version 1. I'll let someone else answer, since it's not my forte. In my opinion, the 2.0 template-based interface is much more intuitive, and it's my recommendation to use it for all new endeavors.
Have a look at the way I use imread() in this program...
Please inspect the type of value returned from imread()...
Also, search in the code for originalColor = imageArg(/*row*/chosenX, /*column*/chosenY); It's a way to index into the matrix returned from imread
// HW1 Intro to Digital Image Processing
// used OpenCV 2.3.1 and VS2010 SP1 to develop this solution
#include "opencv2/core/core.hpp"
#include "opencv2/highgui/highgui.hpp"
#include <iostream>
#include <cassert>
using namespace cv;
Mat_<Vec3b> image;
int discreteAngles = 512;
void on_mouse(int eventCode, int centerX, int centerY, int flags, void* params);
int str2int(const std::string &str);
int main(int argc, char* argv[])
{
// command itself is one element of argument array...
if(argc != 1 && argc != 3)
{
std::cout << "Expecting two arguments to the application: angular granularity as a whole number and a file name." << std::endl;
exit(0);
}
std::string discreteAnglesStr, fileName;
if(argc == 3)
{
discreteAnglesStr = argv[1];
fileName = argv[2];
}
else
{
discreteAnglesStr = "64";
fileName = "boats.tif";
}
try
{
discreteAngles = str2int(discreteAnglesStr);
auto image_ = imread(fileName);
int channels = image_.channels();
assert(channels == 3);
image = image_;
if(image.rows == 0)
throw new std::exception();
auto originalImageStr = "Original Image";
namedWindow(originalImageStr);
setMouseCallback(originalImageStr, on_mouse);
imshow(originalImageStr, image);
}
catch(std::exception e)
{
std::cout << "could not load image." << std::endl;
}
waitKey(0);
return -1;
}
// borrowed from http://stackoverflow.com/q/194465/90475, courtesy of Luka Marinko
int str2int(const std::string &str)
{
std::stringstream ss(str);
int num;
if((ss >> num).fail())
{
throw new std::exception("could not parse user input!");
}
return num;
}
double compute_max_madius(int imageRows, int imageCols, int centerX, int centerY)
{
auto otherX = imageCols - centerX;
auto otherY = imageRows - centerY;
auto a = sqrt((double)centerX * centerX + centerY * centerY);
auto b = sqrt((double)otherX * otherX + centerY * centerY);
auto c = sqrt((double)centerX * centerX + otherY * otherY);
auto d = sqrt((double)otherX * otherX + otherY * otherY);
return max(max(a,b), max(c,d));
}
Vec3b interpolate_with_nearest(const Mat_<Vec3b>& imageArg, double x, double y)
{
auto x0 = static_cast<int>(floor(x)); auto y0 = static_cast<int>(floor(y));
auto x1 = static_cast<int>(ceil(x)); auto y1 = static_cast<int>(ceil(y));
// Rolls over to the other side, esp. for angles
if(x0 < 0) x0 = imageArg.rows - 1;
if(y0 < 0) y0 = imageArg.cols - 1;
if (x1 == imageArg.rows) x1 = 0;
if (y1 == imageArg.cols) y1 = 0;
int chosenX, chosenY;
if (x - x0 < 0.5) chosenX = x0; else chosenX = x1;
if (y - y0 < 0.5) chosenY = y0; else chosenY = y1;
Vec3b originalColor = Vec3b(0, 0, 0);
if (chosenX >= 0 && chosenX < imageArg.rows &&
chosenY >= 0 && chosenY < imageArg.cols)
{
originalColor = imageArg(/*row*/chosenX, /*column*/chosenY);
}
return originalColor;
}
Vec3b interpolate_with_bilinear(const Mat_<Vec3b>& imageArg, double x, double y)
{
auto x0 = static_cast<int>(floor(x)); auto y0 = static_cast<int>(floor(y));
auto x1 = static_cast<int>(ceil(x)); auto y1 = static_cast<int>(ceil(y));
// Rolls over to the other side, esp. for angles
if(x0 < 0) x0 = imageArg.rows - 1;
if(y0 < 0) y0 = imageArg.cols - 1;
if (x1 == imageArg.rows) x1 = 0;
if (y1 == imageArg.cols) y1 = 0;
if (!(
x0 >= 0 && x0 < imageArg.rows &&
x1 >= 0 && x1 < imageArg.rows &&
y0 >= 0 && y0 < imageArg.cols &&
y1 >= 0 && y1 < imageArg.cols))
return Vec3b(0, 0, 0);
auto f00 = imageArg(x0, y0);
auto f01 = imageArg(x0, y1);
auto f10 = imageArg(x1, y0);
auto f11 = imageArg(x1, y1);
auto b1 = f00;
auto b2 = f10 - f00;
auto b3 = f01 - f00;
auto b4 = f00 + f11 - f01 - f10;
x = x - x0;
y = y - y0;
return b1 + b2 * x + b3 * y + b4 * x * y;
}
void on_mouse(int eventCode, int centerX, int centerY, int flags, void* params)
{
if(eventCode == 0)
return;
switch( eventCode )
{
case CV_EVENT_LBUTTONDOWN:
{
std::cout << "Center was (" << centerX << ", " << centerY << ")" << std::endl;
auto maxRadiusXY = compute_max_madius(image.rows, image.cols, centerX, centerY);
int discreteRadii = static_cast<int>(floor(maxRadiusXY));
Mat_<Vec3b> polarImg1;
polarImg1.create(/*rows*/discreteRadii, /*cols*/discreteAngles);
Mat_<Vec3b> polarImg2;
polarImg2.create(/*rows*/discreteRadii, /*cols*/discreteAngles);
for (int radius = 0; radius < discreteRadii; radius++) // radii
{
for (int discreteAngle = 0; discreteAngle < discreteAngles; discreteAngle++) // discreteAngles
{
// 3
auto angleRad = discreteAngle * 2.0 * CV_PI / discreteAngles;
// 2
auto xTranslated = cos(angleRad) * radius;
auto yTranslated = sin(angleRad) * radius;
// 1
auto x = centerX + xTranslated;
auto y = centerY - yTranslated;
polarImg1(/*row*/ radius, /*column*/ discreteAngle) = interpolate_with_nearest(image, /*row*/y, /*column*/x);
polarImg2(/*row*/ radius, /*column*/ discreteAngle) = interpolate_with_bilinear(image, /*row*/y, /*column*/x);
}
}
auto polarImage1Str = "Polar (nearest)";
namedWindow(polarImage1Str);
imshow(polarImage1Str, polarImg1);
auto polarImage2Str = "Polar (bilinear)";
namedWindow(polarImage2Str);
imshow(polarImage2Str, polarImg2);
Mat_<Vec3b> reprocessedImg1;
reprocessedImg1.create(Size(image.rows, image.cols));
Mat_<Vec3b> reprocessedImg2;
reprocessedImg2.create(Size(image.rows, image.cols));
for(int y = 0; y < image.rows; y++)
{
for(int x = 0; x < image.cols; x++)
{
// 1
auto xTranslated = x - centerX;
auto yTranslated = -(y - centerY);
// 2
auto radius = sqrt((double)xTranslated * xTranslated + yTranslated * yTranslated);
double angleRad;
if(xTranslated != 0)
{
angleRad = atan((double)abs(yTranslated) / abs(xTranslated));
// I Quadrant
if (xTranslated > 0 && yTranslated > 0)
angleRad = angleRad;
// II Quadrant
if (xTranslated < 0 && yTranslated > 0)
angleRad = CV_PI - angleRad;
// III Quadrant
if (xTranslated < 0 && yTranslated < 0)
angleRad = CV_PI + angleRad;
/// IV Quadrant
if (xTranslated > 0 && yTranslated < 0)
angleRad = 2 * CV_PI - angleRad;
if (yTranslated == 0)
if (xTranslated > 0) angleRad = 0;
else angleRad = CV_PI;
}
else
{
if (yTranslated > 0) angleRad = CV_PI / 2;
else angleRad = 3 * CV_PI / 2;
}
// 3
auto discreteAngle = angleRad * discreteAngles / (2.0 * CV_PI);
reprocessedImg1(/*row*/ y, /*column*/ x) = interpolate_with_nearest(polarImg1, /*row*/radius, /*column*/discreteAngle);
reprocessedImg2(/*row*/ y, /*column*/ x) = interpolate_with_bilinear(polarImg2, /*row*/radius, /*column*/discreteAngle);
}
}
auto reprocessedImg1Str = "Re-processed (nearest)";
namedWindow(reprocessedImg1Str);
imshow(reprocessedImg1Str, reprocessedImg1);
auto reprocessedImg2Str = "Re-processed (bilinear)";
namedWindow(reprocessedImg2Str);
imshow(reprocessedImg2Str, reprocessedImg2);
} break;
}
}