I am a beginner in c++ and have coded a for loop to show a hollow circle when I run the code, however, I was wondering how I could achieve a filled-in circle using the distance formula (d = sqrt((ax-bx)^2 + (ay-by)^2). Here's what I have so far! Any help would be appreciated!
int MAX = 728;
for (float t = 0; t < 2 * 3.14; t += 0.01)
SetPixel(MAX / 4 + MAX / 6 * sin(t), MAX / 4 + MAX / 6 * cos(t), 255, 255, 0);
#include <windows.h>
#include <iostream>
using namespace std;
int main()
{
HWND consoleWindow = GetConsoleWindow(); // Get a console handle
HDC consoleDC = GetDC(consoleWindow); // Get a handle to device context
int max = 628;
float i = 0;
float t;
float doublePi = 6.29;
for (i = 0.0; i < max; i += 2.0) {
for (t = 0.0; t < doublePi; t += 0.01) {
SetPixel(consoleDC, max / 4 + (max - i) / 6 * sin(t), max / 4 + (max - i) / 6 * cos(t), RGB(255, 255, 0));
}
}
ReleaseDC(consoleWindow, consoleDC);
cin.ignore();
return 0;
}
Working almost well. Draw and fill in! A little slow...
Pffff... do not use sin and cos! instead use the sqrt(1-x^2) approach. You can view the formula rendering a circle in google for example: https://www.google.com/search?q=sqrt(1-x^2)
I edit this answer because it seems that is not clear:
float radius = 50.0f;
for (int x = -radius; x <= radius; ++x) {
int d = round(sqrt(1.0f - (x * x / radius / radius)) * radius);
for (int y = -d; y <= d; ++y) {
SetPixel(x, y, 255, 255, 0);
}
}
Note: each graphic library is different, so I assumed that you used rightfully the "SetPixel" function.
Now, for most people say the sqrt(1-x^2) approach should be enough, but it seem that some downvoters does not think the same XD.
Inefficient as can be, and probably the last way you really want to draw a circle ... but ...
Over the entire square encompassing your circle, calculate each pixel's distance from the center and set if under or equal the radius.
// Draw a circle centered at (Xcenter,Ycenter) with given radius using distance formula
void drawCircle(HDC dc, int XCenter, int YCenter, int radius, COLORREF c) {
double fRad = radius * 1.0; // Just a shortcut to avoid thrashing data types
for (int x = XCenter - radius; x<XCenter + radius; x++) {
for (int y = YCenter - radius; y<YCenter + radius; y++) {
double d = sqrt(((x - XCenter) * (x - XCenter)) + ((y - YCenter) * (y - YCenter)) );
if (d <= fRad) SetPixel(dc, x, y, c);
}
}
}
Caveat: No more caveat, used a C++ environment and tested it this time. :-)
Call thusly:
int main()
{
HWND consoleWindow = GetConsoleWindow();
HDC consoleDC = GetDC(consoleWindow);
drawCircle(consoleDC, 50, 50, 20, RGB(255, 0, 255));
ReleaseDC(consoleWindow, consoleDC);
return 0;
}
Issue
I'm trying to implement the Perlin Noise algorithm in 2D with a single octave with a size of 16x16. I'm using this as heightmap data for a terrain, however it only seems to work in one axis. Whenever the sample point moves to a new Y section in the Perlin Noise grid, the gradient is very different from what I expect (for example, it often flips from 0.98 to -0.97, which is a very sudden change).
This image shows the staggered terrain in the z direction (which is the y axis in the 2D Perlin Noise grid)
Code
I've put the code that calculates which sample point to use at the end since it's quite long and I believe it's not where the issue is, but essentially I scale down the terrain to match the Perlin Noise grid (16x16) and then sample through all the points.
Gradient At Point
So the code that calculates out the gradient at a sample point is the following:
// Find the gradient at a certain sample point
float PerlinNoise::gradientAt(Vector2 point)
{
// Decimal part of float
float relativeX = point.x - (int)point.x;
float relativeY = point.y - (int)point.y;
Vector2 relativePoint = Vector2(relativeX, relativeY);
vector<float> weights(4);
// Find the weights of the 4 surrounding points
weights = surroundingWeights(point);
float fadeX = fadeFunction(relativePoint.x);
float fadeY = fadeFunction(relativePoint.y);
float lerpA = MathUtils::lerp(weights[0], weights[1], fadeX);
float lerpB = MathUtils::lerp(weights[2], weights[3], fadeX);
float lerpC = MathUtils::lerp(lerpA, lerpB, fadeY);
return lerpC;
}
Surrounding Weights of Point
I believe the issue is somewhere here, in the function that calculates the weights for the 4 surrounding points of a sample point, but I can't seem to figure out what is wrong since all the values seem sensible in the function when stepping through it.
// Find the surrounding weight of a point
vector<float> PerlinNoise::surroundingWeights(Vector2 point){
// Produces correct values
vector<Vector2> surroundingPoints = surroundingPointsOf(point);
vector<float> weights;
for (unsigned i = 0; i < surroundingPoints.size(); ++i) {
// The corner to the sample point
Vector2 cornerToPoint = surroundingPoints[i].toVector(point);
// Getting the seeded vector from the grid
float x = surroundingPoints[i].x;
float y = surroundingPoints[i].y;
Vector2 seededVector = baseGrid[x][y];
// Dot product between the seededVector and corner to the sample point vector
float dotProduct = cornerToPoint.dot(seededVector);
weights.push_back(dotProduct);
}
return weights;
}
OpenGL Setup and Sample Point
Setting up the heightmap and getting the sample point. Variables 'wrongA' and 'wrongA' is an example of when the gradient flips and changes suddenly.
void HeightMap::GenerateRandomTerrain() {
int perlinGridSize = 16;
PerlinNoise perlin_noise = PerlinNoise(perlinGridSize, perlinGridSize);
numVertices = RAW_WIDTH * RAW_HEIGHT;
numIndices = (RAW_WIDTH - 1) * (RAW_HEIGHT - 1) * 6;
vertices = new Vector3[numVertices];
textureCoords = new Vector2[numVertices];
indices = new GLuint[numIndices];
float perlinScale = RAW_HEIGHT/ (float) (perlinGridSize -1);
float height = 50;
float wrongA = perlin_noise.gradientAt(Vector2(0, 68.0f / perlinScale));
float wrongB = perlin_noise.gradientAt(Vector2(0, 69.0f / perlinScale));
for (int x = 0; x < RAW_WIDTH; ++x) {
for (int z = 0; z < RAW_HEIGHT; ++z) {
int offset = (x* RAW_WIDTH) + z;
float xVal = (float)x / perlinScale;
float yVal = (float)z / perlinScale;
float noise = perlin_noise.gradientAt(Vector2( xVal , yVal));
vertices[offset] = Vector3(x * HEIGHTMAP_X, noise * height, z * HEIGHTMAP_Z);
textureCoords[offset] = Vector2(x * HEIGHTMAP_TEX_X, z * HEIGHTMAP_TEX_Z);
}
}
numIndices = 0;
for (int x = 0; x < RAW_WIDTH - 1; ++x) {
for (int z = 0; z < RAW_HEIGHT - 1; ++z) {
int a = (x * (RAW_WIDTH)) + z;
int b = ((x + 1)* (RAW_WIDTH)) + z;
int c = ((x + 1)* (RAW_WIDTH)) + (z + 1);
int d = (x * (RAW_WIDTH)) + (z + 1);
indices[numIndices++] = c;
indices[numIndices++] = b;
indices[numIndices++] = a;
indices[numIndices++] = a;
indices[numIndices++] = d;
indices[numIndices++] = c;
}
}
BufferData();
}
Turned out the issue was in the interpolation stage:
float lerpA = MathUtils::lerp(weights[0], weights[1], fadeX);
float lerpB = MathUtils::lerp(weights[2], weights[3], fadeX);
float lerpC = MathUtils::lerp(lerpA, lerpB, fadeY);
I had the interpolation in the y axis the wrong way around, so it should have been:
lerp(lerpB, lerpA, fadeY)
Instead of:
lerp(lerpA, lerpB, fadeY)
I've got a 3D terrain environment like so:
I'm trying to get the character (camera) to look up when climbing hills, and look down when descending, like climbing in real life.
This is what it's currently doing:
Right now the camera moves up and down the hills just fine, but I can't get the camera angle to work correctly. The only way I can think of aiming up or down depending on the terrain is getting the z-index of the cell my character is currently facing, and set that as the focus, but I really have no idea how to do that.
This is admittedly for an assignment, and we're intentionally not using objects so things are organized a little strangely.
Here's how I'm currently doing things:
const int M = 100; // width
const int N = 100; // height
double zHeights[M+1][N+1]; // 2D array containing the z-indexes of terrain cells
double gRX = 1.5; // x position of character
double gRY = 2.5; // y position of character
double gDirection = 45; // direction of character
double gRSpeed = 0.05; // move speed of character
double getZ(double x, double y) // returns the height of the current cell
{
double z = .5*sin(x*.25) + .4*sin(y*.15-.43);
z += sin(x*.45-.7) * cos(y*.315-.31)+.5;
z += sin(x*.15-.97) * sin(y*.35-8.31);
double amplitute = 5;
z *= amplitute;
return z;
}
void generateTerrain()
{
glBegin(GL_QUADS);
for (int i = 0; i <= M; i++)
{
for (int j = 0; j <= N; j++)
{
zHeights[i][j] = getZ(i,j);
}
}
}
void drawTerrain()
{
for (int i = 0; i < M; i++)
{
for (int j = 0; j < N; j++)
{
glColor3ub( (i*34525+j*5245)%256, (i*3456345+j*6757)%256, (i*98776+j*6554544)%256);
glVertex3d(i, j, getZ(i,j));
glVertex3d(i, j+1, getZ(i,j+1));
glVertex3d(i+1, j+1, getZ(i+1,j+1));
glVertex3d(i+1, j, getZ(i+1,j));
}
}
}
void display() // callback to glutDisplayFunc
{
glEnable(GL_DEPTH_TEST);
glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT);
glLoadIdentity();
double radians = gDirection /180.*3.141592654; // converts direction to radians
double z = getZ((int)gRX, (int)gRY); // casts as int to find z-index in zHeights[][]
double dx = cos(radians)*gRSpeed;
double dy = sin(radians)*gRSpeed;
double at_x = gRX + dx;
double at_y = gRY + dy;
double at_z = z; // source of problem, no idea what to do
gluLookAt(gRX, gRY, z + 2, // eye position
at_x, at_y, at_z + 2, // point to look at, also wrong
0, 0, 1); // up vector
drawTerrain();
glEnd();
}
void init()
{
generateTerrain();
}
Firstly, I don't see any reason to cast to int here:
double z = getZ((int)gRX, (int)gRY);
Just use the double values to get a smooth behavior.
Your basic approach is already pretty good. You take the current position (gRX, gRY), walk a bit in the viewing direction (dx, dy) and use that as the point to look at. There are just two small things that need adaptation:
double dx = cos(radians)*gRSpeed;
double dy = sin(radians)*gRSpeed;
Although multiplying by gRSpeed might be a good idea, in my opinion, this factor should not be related to the character's kinematics. Instead, this represents the smoothness of your view direction. Small values make the direction stick very closely to the terrain geometry, larger values smooth it out.
And finally, you need to evaluate the height at your look-at point:
double at_z = getZ(at_x, at_y);
I detected lines in an image and drew them in a separate image file in OpenCv C++ using HoughLinesP method. Following is a part of that resulting image. There are actually hundreds of small and thin lines which form a big single line.
But I want single few lines that represent all those number of lines. Closer lines should be merged together to form a single line. For example above set of lines should be represented by just 3 separate lines as below.
The expected output is as above. How to accomplish this task.
Up to now progress result from akarsakov's answer.
(separate classes of lines resulted are drawn in different colors). Note that this result is the original complete image I am working on, but not the sample section I had used in the question
If you don't know the number of lines in the image you can use the cv::partition function to split lines on equivalency group.
I suggest you the following procedure:
Split your lines using cv::partition. You need to specify a good predicate function. It really depends on lines which you extract from image, but I think it should check following conditions:
Angle between lines should be quite small (less 3 degrees, for example). Use dot product to calculate angle's cosine.
Distance between centers of segments should be less than half of maximum length of two segments.
For example, it can be implemented as follows:
bool isEqual(const Vec4i& _l1, const Vec4i& _l2)
{
Vec4i l1(_l1), l2(_l2);
float length1 = sqrtf((l1[2] - l1[0])*(l1[2] - l1[0]) + (l1[3] - l1[1])*(l1[3] - l1[1]));
float length2 = sqrtf((l2[2] - l2[0])*(l2[2] - l2[0]) + (l2[3] - l2[1])*(l2[3] - l2[1]));
float product = (l1[2] - l1[0])*(l2[2] - l2[0]) + (l1[3] - l1[1])*(l2[3] - l2[1]);
if (fabs(product / (length1 * length2)) < cos(CV_PI / 30))
return false;
float mx1 = (l1[0] + l1[2]) * 0.5f;
float mx2 = (l2[0] + l2[2]) * 0.5f;
float my1 = (l1[1] + l1[3]) * 0.5f;
float my2 = (l2[1] + l2[3]) * 0.5f;
float dist = sqrtf((mx1 - mx2)*(mx1 - mx2) + (my1 - my2)*(my1 - my2));
if (dist > std::max(length1, length2) * 0.5f)
return false;
return true;
}
Guess you have your lines in vector<Vec4i> lines;. Next, you should call cv::partition as follows:
vector<Vec4i> lines;
std::vector<int> labels;
int numberOfLines = cv::partition(lines, labels, isEqual);
You need to call cv::partition once and it will clusterize all lines. Vector labels will store for each line label of cluster to which it belongs. See documentation for cv::partition
After you get all groups of line you should merge them. I suggest calculating average angle of all lines in group and estimate "border" points. For example, if angle is zero (i.e. all lines are almost horizontal) it would be the left-most and right-most points. It remains only to draw a line between this points.
I noticed that all lines in your examples are horizontal or vertical. In such case you can calculate point which is average of all segment's centers and "border" points, and then just draw horizontal or vertical line limited by "border" points through center point.
Please note that cv::partition takes O(N^2) time, so if you process a huge number of lines it may take a lot of time.
I hope it will help. I used such approach for similar task.
First off I want to note that your original image is at a slight angle, so your expected output seems just a bit off to me. I'm assuming you are okay with lines that are not 100% vertical in your output because they are slightly off on your input.
Mat image;
Mat binary = image > 125; // Convert to binary image
// Combine similar lines
int size = 3;
Mat element = getStructuringElement( MORPH_ELLIPSE, Size( 2*size + 1, 2*size+1 ), Point( size, size ) );
morphologyEx( mask, mask, MORPH_CLOSE, element );
So far this yields this image:
These lines are not at 90 degree angles because the original image is not.
You can also choose to close the gap between the lines with:
Mat out = Mat::zeros(mask.size(), mask.type());
vector<Vec4i> lines;
HoughLinesP(mask, lines, 1, CV_PI/2, 50, 50, 75);
for( size_t i = 0; i < lines.size(); i++ )
{
Vec4i l = lines[i];
line( out, Point(l[0], l[1]), Point(l[2], l[3]), Scalar(255), 5, CV_AA);
}
If these lines are too fat, I've had success thinning them with:
size = 15;
Mat eroded;
cv::Mat erodeElement = getStructuringElement( MORPH_ELLIPSE, cv::Size( size, size ) );
erode( mask, eroded, erodeElement );
Here is a refinement build upon #akarsakov answer.
A basic issue with:
Distance between centers of segments should be less than half of
maximum length of two segments.
is that parallel long lines that are visually far might end up in same equivalence class (as demonstrated in OP's edit).
Therefore the approach that I found working reasonable for me:
Construct a window (bounding rectangle) around a line1.
line2 angle is close enough to line1's and at least one point of line2 is inside line1's bounding rectangle
Often a long linear feature in the image that is quite weak will end up recognized (HoughP, LSD) by a set of line segments with considerable gaps between them. To alleviate this, our bounding rectangle is constructed around line extended in both directions, where extension is defined by a fraction of original line width.
bool extendedBoundingRectangleLineEquivalence(const Vec4i& _l1, const Vec4i& _l2, float extensionLengthFraction, float maxAngleDiff, float boundingRectangleThickness){
Vec4i l1(_l1), l2(_l2);
// extend lines by percentage of line width
float len1 = sqrtf((l1[2] - l1[0])*(l1[2] - l1[0]) + (l1[3] - l1[1])*(l1[3] - l1[1]));
float len2 = sqrtf((l2[2] - l2[0])*(l2[2] - l2[0]) + (l2[3] - l2[1])*(l2[3] - l2[1]));
Vec4i el1 = extendedLine(l1, len1 * extensionLengthFraction);
Vec4i el2 = extendedLine(l2, len2 * extensionLengthFraction);
// reject the lines that have wide difference in angles
float a1 = atan(linearParameters(el1)[0]);
float a2 = atan(linearParameters(el2)[0]);
if(fabs(a1 - a2) > maxAngleDiff * M_PI / 180.0){
return false;
}
// calculate window around extended line
// at least one point needs to inside extended bounding rectangle of other line,
std::vector<Point2i> lineBoundingContour = boundingRectangleContour(el1, boundingRectangleThickness/2);
return
pointPolygonTest(lineBoundingContour, cv::Point(el2[0], el2[1]), false) == 1 ||
pointPolygonTest(lineBoundingContour, cv::Point(el2[2], el2[3]), false) == 1;
}
where linearParameters, extendedLine, boundingRectangleContour are following:
Vec2d linearParameters(Vec4i line){
Mat a = (Mat_<double>(2, 2) <<
line[0], 1,
line[2], 1);
Mat y = (Mat_<double>(2, 1) <<
line[1],
line[3]);
Vec2d mc; solve(a, y, mc);
return mc;
}
Vec4i extendedLine(Vec4i line, double d){
// oriented left-t-right
Vec4d _line = line[2] - line[0] < 0 ? Vec4d(line[2], line[3], line[0], line[1]) : Vec4d(line[0], line[1], line[2], line[3]);
double m = linearParameters(_line)[0];
// solution of pythagorean theorem and m = yd/xd
double xd = sqrt(d * d / (m * m + 1));
double yd = xd * m;
return Vec4d(_line[0] - xd, _line[1] - yd , _line[2] + xd, _line[3] + yd);
}
std::vector<Point2i> boundingRectangleContour(Vec4i line, float d){
// finds coordinates of perpendicular lines with length d in both line points
// https://math.stackexchange.com/a/2043065/183923
Vec2f mc = linearParameters(line);
float m = mc[0];
float factor = sqrtf(
(d * d) / (1 + (1 / (m * m)))
);
float x3, y3, x4, y4, x5, y5, x6, y6;
// special case(vertical perpendicular line) when -1/m -> -infinity
if(m == 0){
x3 = line[0]; y3 = line[1] + d;
x4 = line[0]; y4 = line[1] - d;
x5 = line[2]; y5 = line[3] + d;
x6 = line[2]; y6 = line[3] - d;
} else {
// slope of perpendicular lines
float m_per = - 1/m;
// y1 = m_per * x1 + c_per
float c_per1 = line[1] - m_per * line[0];
float c_per2 = line[3] - m_per * line[2];
// coordinates of perpendicular lines
x3 = line[0] + factor; y3 = m_per * x3 + c_per1;
x4 = line[0] - factor; y4 = m_per * x4 + c_per1;
x5 = line[2] + factor; y5 = m_per * x5 + c_per2;
x6 = line[2] - factor; y6 = m_per * x6 + c_per2;
}
return std::vector<Point2i> {
Point2i(x3, y3),
Point2i(x4, y4),
Point2i(x6, y6),
Point2i(x5, y5)
};
}
To partion, call:
std::vector<int> labels;
int equilavenceClassesCount = cv::partition(linesWithoutSmall, labels, [](const Vec4i l1, const Vec4i l2){
return extendedBoundingRectangleLineEquivalence(
l1, l2,
// line extension length - as fraction of original line width
0.2,
// maximum allowed angle difference for lines to be considered in same equivalence class
2.0,
// thickness of bounding rectangle around each line
10);
});
Now, in order to reduce each equivalence class to single line, we build a point cloud out of it and find a line fit:
// fit line to each equivalence class point cloud
std::vector<Vec4i> reducedLines = std::accumulate(pointClouds.begin(), pointClouds.end(), std::vector<Vec4i>{}, [](std::vector<Vec4i> target, const std::vector<Point2i>& _pointCloud){
std::vector<Point2i> pointCloud = _pointCloud;
//lineParams: [vx,vy, x0,y0]: (normalized vector, point on our contour)
// (x,y) = (x0,y0) + t*(vx,vy), t -> (-inf; inf)
Vec4f lineParams; fitLine(pointCloud, lineParams, CV_DIST_L2, 0, 0.01, 0.01);
// derive the bounding xs of point cloud
decltype(pointCloud)::iterator minXP, maxXP;
std::tie(minXP, maxXP) = std::minmax_element(pointCloud.begin(), pointCloud.end(), [](const Point2i& p1, const Point2i& p2){ return p1.x < p2.x; });
// derive y coords of fitted line
float m = lineParams[1] / lineParams[0];
int y1 = ((minXP->x - lineParams[2]) * m) + lineParams[3];
int y2 = ((maxXP->x - lineParams[2]) * m) + lineParams[3];
target.push_back(Vec4i(minXP->x, y1, maxXP->x, y2));
return target;
});
Demonstration:
Detected partitioned line (with small lines filtered out):
Reduced:
Demonstration code:
int main(int argc, const char* argv[]){
if(argc < 2){
std::cout << "img filepath should be present in args" << std::endl;
}
Mat image = imread(argv[1]);
Mat smallerImage; resize(image, smallerImage, cv::Size(), 0.5, 0.5, INTER_CUBIC);
Mat target = smallerImage.clone();
namedWindow("Detected Lines", WINDOW_NORMAL);
namedWindow("Reduced Lines", WINDOW_NORMAL);
Mat detectedLinesImg = Mat::zeros(target.rows, target.cols, CV_8UC3);
Mat reducedLinesImg = detectedLinesImg.clone();
// delect lines in any reasonable way
Mat grayscale; cvtColor(target, grayscale, CV_BGRA2GRAY);
Ptr<LineSegmentDetector> detector = createLineSegmentDetector(LSD_REFINE_NONE);
std::vector<Vec4i> lines; detector->detect(grayscale, lines);
// remove small lines
std::vector<Vec4i> linesWithoutSmall;
std::copy_if (lines.begin(), lines.end(), std::back_inserter(linesWithoutSmall), [](Vec4f line){
float length = sqrtf((line[2] - line[0]) * (line[2] - line[0])
+ (line[3] - line[1]) * (line[3] - line[1]));
return length > 30;
});
std::cout << "Detected: " << linesWithoutSmall.size() << std::endl;
// partition via our partitioning function
std::vector<int> labels;
int equilavenceClassesCount = cv::partition(linesWithoutSmall, labels, [](const Vec4i l1, const Vec4i l2){
return extendedBoundingRectangleLineEquivalence(
l1, l2,
// line extension length - as fraction of original line width
0.2,
// maximum allowed angle difference for lines to be considered in same equivalence class
2.0,
// thickness of bounding rectangle around each line
10);
});
std::cout << "Equivalence classes: " << equilavenceClassesCount << std::endl;
// grab a random colour for each equivalence class
RNG rng(215526);
std::vector<Scalar> colors(equilavenceClassesCount);
for (int i = 0; i < equilavenceClassesCount; i++){
colors[i] = Scalar(rng.uniform(30,255), rng.uniform(30, 255), rng.uniform(30, 255));;
}
// draw original detected lines
for (int i = 0; i < linesWithoutSmall.size(); i++){
Vec4i& detectedLine = linesWithoutSmall[i];
line(detectedLinesImg,
cv::Point(detectedLine[0], detectedLine[1]),
cv::Point(detectedLine[2], detectedLine[3]), colors[labels[i]], 2);
}
// build point clouds out of each equivalence classes
std::vector<std::vector<Point2i>> pointClouds(equilavenceClassesCount);
for (int i = 0; i < linesWithoutSmall.size(); i++){
Vec4i& detectedLine = linesWithoutSmall[i];
pointClouds[labels[i]].push_back(Point2i(detectedLine[0], detectedLine[1]));
pointClouds[labels[i]].push_back(Point2i(detectedLine[2], detectedLine[3]));
}
// fit line to each equivalence class point cloud
std::vector<Vec4i> reducedLines = std::accumulate(pointClouds.begin(), pointClouds.end(), std::vector<Vec4i>{}, [](std::vector<Vec4i> target, const std::vector<Point2i>& _pointCloud){
std::vector<Point2i> pointCloud = _pointCloud;
//lineParams: [vx,vy, x0,y0]: (normalized vector, point on our contour)
// (x,y) = (x0,y0) + t*(vx,vy), t -> (-inf; inf)
Vec4f lineParams; fitLine(pointCloud, lineParams, CV_DIST_L2, 0, 0.01, 0.01);
// derive the bounding xs of point cloud
decltype(pointCloud)::iterator minXP, maxXP;
std::tie(minXP, maxXP) = std::minmax_element(pointCloud.begin(), pointCloud.end(), [](const Point2i& p1, const Point2i& p2){ return p1.x < p2.x; });
// derive y coords of fitted line
float m = lineParams[1] / lineParams[0];
int y1 = ((minXP->x - lineParams[2]) * m) + lineParams[3];
int y2 = ((maxXP->x - lineParams[2]) * m) + lineParams[3];
target.push_back(Vec4i(minXP->x, y1, maxXP->x, y2));
return target;
});
for(Vec4i reduced: reducedLines){
line(reducedLinesImg, Point(reduced[0], reduced[1]), Point(reduced[2], reduced[3]), Scalar(255, 255, 255), 2);
}
imshow("Detected Lines", detectedLinesImg);
imshow("Reduced Lines", reducedLinesImg);
waitKey();
return 0;
}
I would recommend that you use HoughLines from OpenCV.
void HoughLines(InputArray image, OutputArray lines, double rho, double theta, int threshold, double srn=0, double stn=0 )
You can adjust with rho and theta the possible orientation and position of the lines you want to observe.
In your case, theta = 90° would be fine (only vertical and horizontal lines).
After this, you can get unique line equations with Plücker coordinates. And from there you could apply a K-mean with 3 centers that should fit approximately your 3 lines in the second image.
PS : I will see if i can test the whole process with your image
You can merge multiple close line into single line by clustering lines using rho and theta and finally taking average of rho and theta.
void contourLines(vector<cv::Vec2f> lines, const float rho_threshold, const float theta_threshold, vector< cv::Vec2f > &combinedLines)
{
vector< vector<int> > combineIndex(lines.size());
for (int i = 0; i < lines.size(); i++)
{
int index = i;
for (int j = i; j < lines.size(); j++)
{
float distanceI = lines[i][0], distanceJ = lines[j][0];
float slopeI = lines[i][1], slopeJ = lines[j][1];
float disDiff = abs(distanceI - distanceJ);
float slopeDiff = abs(slopeI - slopeJ);
if (slopeDiff < theta_max && disDiff < rho_max)
{
bool isCombined = false;
for (int w = 0; w < i; w++)
{
for (int u = 0; u < combineIndex[w].size(); u++)
{
if (combineIndex[w][u] == j)
{
isCombined = true;
break;
}
if (combineIndex[w][u] == i)
index = w;
}
if (isCombined)
break;
}
if (!isCombined)
combineIndex[index].push_back(j);
}
}
}
for (int i = 0; i < combineIndex.size(); i++)
{
if (combineIndex[i].size() == 0)
continue;
cv::Vec2f line_temp(0, 0);
for (int j = 0; j < combineIndex[i].size(); j++) {
line_temp[0] += lines[combineIndex[i][j]][0];
line_temp[1] += lines[combineIndex[i][j]][1];
}
line_temp[0] /= combineIndex[i].size();
line_temp[1] /= combineIndex[i].size();
combinedLines.push_back(line_temp);
}
}
function call
You can tune houghThreshold, rho_threshold and theta_threshold as per your application.
HoughLines(edge, lines_t, 1, CV_PI / 180, houghThreshold, 0, 0);
float rho_threshold= 15;
float theta_threshold = 3*DEGREES_TO_RADIANS;
vector< cv::Vec2f > lines;
contourCluster(lines_t, rho_max, theta_max, lines);
#C_Raj made a good point, for lines like this, i.e., most likely extracted from table/form-like images, you should make full use of the fact that many of the line segments captured by Hough transform from the same lines have very similar \rho and \theta.
After clustering these line segments based on their \rho and \theta, you can apply 2D line fitting to obtain estimate of the true lines in an image.
There is a paper describing this idea and it's making further assumptions of the lines in a page.
HTH.
I have line that is defined as two points.
start = (xs,ys)
end = (xe, ye)
Drawing function that I'm using Only accepts lines that are fully in screen coordinates.
Screen size is (xSize, ySize).
Top left corner is (0,0). Bottom right corner is (xSize, ySize).
Some other funcions gives me line that that is defined for example as start(-50, -15) end(5000, 200). So it's ends are outside of screen size.
In C++
struct Vec2
{
int x, y
};
Vec2 start, end //This is all little bit pseudo code
Vec2 screenSize;//You can access coordinates like start.x end.y
How can I calculate new start and endt that is at the screen edge, not outside screen.
I know how to do it on paper. But I can't transfer it to c++.
On paper I'm sershing for point that belongs to edge and line. But it is to much calculations for c++.
Can you help?
There are many line clipping algorithms like:
Cohen–Sutherland wikipedia page with implementation
Liang–Barsky wikipedia page
Nicholl–Lee–Nicholl (NLN)
and many more. see Line Clipping on wikipedia
[EDIT1]
See below figure:
there are 3 kinds of start point:
sx > 0 and sy < 0 (red line)
sx < 0 and sy > 0 (yellow line)
sx < 0 and sy < 0 (green and violet lines)
In situations 1 and 2 simply find Xintersect and Yintersect respectively and choose them as new start point.
As you can see, there are 2 kinds of lines in situation 3. In this situation find Xintersect and Yintersect and choose the intersect point near the end point which is the point that has minimum distance to endPoint.
min(distance(Xintersect, endPoint), distance(Yintersect, endPoint))
[EDIT2]
// Liang-Barsky function by Daniel White # http://www.skytopia.com/project/articles/compsci/clipping.html
// This function inputs 8 numbers, and outputs 4 new numbers (plus a boolean value to say whether the clipped line is drawn at all).
//
bool LiangBarsky (double edgeLeft, double edgeRight, double edgeBottom, double edgeTop, // Define the x/y clipping values for the border.
double x0src, double y0src, double x1src, double y1src, // Define the start and end points of the line.
double &x0clip, double &y0clip, double &x1clip, double &y1clip) // The output values, so declare these outside.
{
double t0 = 0.0; double t1 = 1.0;
double xdelta = x1src-x0src;
double ydelta = y1src-y0src;
double p,q,r;
for(int edge=0; edge<4; edge++) { // Traverse through left, right, bottom, top edges.
if (edge==0) { p = -xdelta; q = -(edgeLeft-x0src); }
if (edge==1) { p = xdelta; q = (edgeRight-x0src); }
if (edge==2) { p = -ydelta; q = -(edgeBottom-y0src);}
if (edge==3) { p = ydelta; q = (edgeTop-y0src); }
r = q/p;
if(p==0 && q<0) return false; // Don't draw line at all. (parallel line outside)
if(p<0) {
if(r>t1) return false; // Don't draw line at all.
else if(r>t0) t0=r; // Line is clipped!
} else if(p>0) {
if(r<t0) return false; // Don't draw line at all.
else if(r<t1) t1=r; // Line is clipped!
}
}
x0clip = x0src + t0*xdelta;
y0clip = y0src + t0*ydelta;
x1clip = x0src + t1*xdelta;
y1clip = y0src + t1*ydelta;
return true; // (clipped) line is drawn
}
Here is a function I wrote. It cycles through all 4 planes (left, top, right, bottom) and clips each point by the plane.
// Clips a line segment to an axis-aligned rectangle
// Returns true if clipping is successful
// Returns false if line segment lies outside the rectangle
bool clipLineToRect(int a[2], int b[2],
int xmin, int ymin, int xmax, int ymax)
{
int mins[2] = {xmin, ymin};
int maxs[2] = {xmax, ymax};
int normals[2] = {1, -1};
for (int axis=0; axis<2; axis++) {
for (int plane=0; plane<2; plane++) {
// Check both points
for (int pt=1; pt<=2; pt++) {
int* pt1 = pt==1 ? a : b;
int* pt2 = pt==1 ? b : a;
// If both points are outside the same plane, the line is
// outside the rectangle
if ( (a[0]<xmin && b[0]<xmin) || (a[0]>xmax && b[0]>xmax) ||
(a[1]<ymin && b[1]<ymin) || (a[1]>ymax && b[1]>ymax)) {
return false;
}
const int n = normals[plane];
if ( (n==1 && pt1[axis]<mins[axis]) || // check left/top plane
(n==-1 && pt1[axis]>maxs[axis]) ) { // check right/bottom plane
// Calculate interpolation factor t using ratio of signed distance
// of each point from the plane
const float p = (n==1) ? mins[axis] : maxs[axis];
const float q1 = pt1[axis];
const float q2 = pt2[axis];
const float d1 = n * (q1-p);
const float d2 = n * (q2-p);
const float t = d1 / (d1-d2);
// t should always be between 0 and 1
if (t<0 || t >1) {
return false;
}
// Interpolate to find the new point
pt1[0] = (int)(pt1[0] + (pt2[0] - pt1[0]) * t );
pt1[1] = (int)(pt1[1] + (pt2[1] - pt1[1]) * t );
}
}
}
}
return true;
}
Example Usage:
void testClipLineToRect()
{
int screenWidth = 320;
int screenHeight = 240;
int xmin=0;
int ymin=0;
int xmax=screenWidth-1;
int ymax=screenHeight-1;
int a[2] = {-10, 10};
int b[2] = {300, 250};
printf("Before clipping:\n\ta={%d, %d}\n\tb=[%d, %d]\n",
a[0], a[1], b[0], b[1]);
if (clipLineToRect(a, b, xmin, ymin, xmax, ymax)) {
printf("After clipping:\n\ta={%d, %d}\n\tb=[%d, %d]\n",
a[0], a[1], b[0], b[1]);
}
else {
printf("clipLineToRect returned false\n");
}
}
Output:
Before clipping:
a={-10, 10}
b=[300, 250]
After clipping:
a={0, 17}
b=[285, 239]