I have a bunch of markers placed in my scene as childNodes at fixed node positions in 3D world. When I move the phone around, I need to determine which marker node is the closest to the 2D screen center, so I can get the text description corresponding to that node and display it.
Right now, in a renderloop, I just determined the distance of each node from the screen center in a forEach loop, and decide if that distance is <150, if so get title and copy of that node. However, this doesn't solve my problem because there could be multiple nodes that satisfy that condition. I need to compare the distances from the center across all the nodes and get that one node that's is closest
func renderer(_ renderer: SCNSceneRenderer, willRenderScene scene: SCNScene, atTime time: TimeInterval){
scene.rootNode.childNodes.filter{ $0.name != nil }.forEach{ node in
guard let pointOfView = sceneView.pointOfView else { return }
let isVisible = sceneView.isNode(node, insideFrustumOf: pointOfView)
if isVisible {
let nodePos = sceneView.projectPoint(node.position)
let nodeScreenPos = CGPoint(x: CGFloat(nodePos.x), y: CGFloat(nodePos.y))
let distance = CGPointDistance(from: nodeScreenPos, to: view.center)
if distance < 150.0 {
print("display description of: \(node.name!)")
guard let title = ar360Experience?.threeSixtyHotspot?[Int(node.name!)!].title else { return }
guard let copy = ar360Experience?.threeSixtyHotspot?[Int(node.name!)!].copy else { return }
titleLabel.text = title
copyLabel.text = copy
cardView.isHidden = false
}else {
cardView.isHidden = true
}
}
}
}
There are various ways to do this. Like iterating all the nodes to find their distance. But like you said this becomes inefficient.
What you can do is to store your node data in a different format using something like GKQuadTree... https://developer.apple.com/documentation/gameplaykit/gkquadtree
This is a GameplayKit That will allow you to much more quickly iterate the data set so that you can find the node closest to the centre.
It works by breaking down the area into four (hence quad) sections and storing nodes into one of those sections storing the rect of the section too. It then breaks down each of those four sections into four more and so on.
So when you ask for the nodes closest to a given point it can quickly eliminate the majority of the nodes.
Can anybody tell me how i can format my array so i can load it to a QImage?
For now i have a 2d-char-array:
uchar dataArray[400][400];
time_t t;
srand(time(&t));
int x, y;
for(x=0; x< 400; x++)
{
for(y=0; y<400; y++)
{
dataArray[x][y] = rand()%1001;
}
}
QPainter MyPainter(this);
scene = new QGraphicsScene(this);
scene->addEllipse(200, 200, 20, 20);
ui.graphicsView->setScene(scene);
*image = new QImage(*dataArray, 400, 400, QImage::Format_Mono);
image->setColorCount(2);
image->setColor(1, qRgb(255, 0, 0));//red
image->setColor(0, Qt::transparent);
scene->addPixmap(QPixmap::fromImage(*image));
When the content of the array is 0 I want the one color and content > 0 the other color. So i want to load the array into a monochromatic QImage. Obviously this array does not work. How does the Array need to be formatted for my QImage to load properly?
The doc just says following, but i do not really get what that means...
data must be 32-bit aligned, and each scanline of data in the image must also be 32-bit aligned.
I want a QImage with Format_Mono like this Where "x" and "+" represent single pixels with different color (red and transparent):
xxxx+xxx++xxx
xxx++xx++xxxx
++x+x+xxxxxxx
+xxx+x+x+x+xx
For this i have an array (dataArray as in code above) with same pattern where x is above a specified value and + is under or equal (the value is 0 at the moment).
How do i convert this array to an array wich can be used by QImage with Format_Mono so i can see the correct pattern?
Assuming that dataArray is declared as uchar dataArray[400][400];
QImage tmpImg(dataArray[0], 400, 400, QImage::Format_Grayscale8);
// this image shares memory with dataArray and this is not safe
// for inexperienced developer
QImage result(tmpImg); // shallow copy
result.bits(); // enforce a deep copy of image
Found the solution to convert my dataArray into a working imageArray:
Its pretty easy as soon as you figure out how it is done (and it should be obviously i do not know why i dind't get it in first place...). I just needed to convert every data-point to the new array bit-wise, also I had to figure out, that the axis of the imageArray have to be in [y][x]-order not the other way round (who the hell would do that in this order?!).
Many time wasted to figure that out...
uchar dataArray[400][400]; //first index is x-axis, second one is y-axis
uchar imageArray[400][52]; //first index is y-axis, second one is x-axis
time_t t;
srand(time(&t));
int x, y;
for(y=0; y<400; y++)
{
for(x=0; x<400; x++)
{
dataArray[x][y] = rand()%1001;
}
}
for(y=0; y<400; y++)
{
for(x=0; x<400; x++)
{
if(dataArray[x][y] > 0)
imageArray[y][x/8] |= 1 << 7-x%8; //changing bit to 1
//7- because msb is left but we start counting left starting with 0
else
imageArray[y][x/8] &= ~(1 << 7-x%8); //changing bit to 0
}
}
scene = new QGraphicsScene(this);
ui.graphicsView->setScene(scene);
QImage *image = new QImage(*imageArray, 400, 400, QImage::Format_Mono);
Currently, I am using a convex hull algorithm to get the outer most points from a set of points randomly placed. What I aim to do is draw a polygon from the set of points returned by the convex hull however, when I try to draw the polygon it looks quite strange.
My question, how do I order the points so the polygon draws correctly?
Thanks.
EDIT:
Also, I have tried sorting using orderby(...).ThenBy(...) and I cant seem to get it working.
Have you tried the gift wrapping algorithm ( http://en.wikipedia.org/wiki/Gift_wrapping_algorithm)? This should return points in the correct order.
I had an issue where a random set of points were generated from which a wrapped elevation vector needed a base contour. Having read the link supplied by #user1149913 and found a sample of gift-wrapping a hull, the following is a sample of my implementation:
private static PointCollection CalculateContour (List<Point> points) {
// locate lower-leftmost point
int hull = 0;
int i;
for (i = 1 ; i < points.Count ; i++) {
if (ComparePoint(points[i], points[hull])) {
hull = i;
}
}
// wrap contour
var outIndices = new int[points.Count];
int endPt;
i = 0;
do {
outIndices[i++] = hull;
endPt = 0;
for (int j = 1 ; j < points.Count ; j++)
if (hull == endPt || IsLeft(points[hull], points[endPt], points[j]))
endPt = j;
hull = endPt;
} while (endPt != outIndices[0]);
// build countour points
var contourPoints = new PointCollection(points.Capacity);
int results = i;
for (i = 0 ; i < results ; i++)
contourPoints.Add(points[outIndices[i]]);
return contourPoints;
}
This is not a full solution but a guide in the right direction. I faced a very similar problem just recently and I found a reddit post with an answer (https://www.reddit.com/r/DnDBehindTheScreen/comments/8efeta/a_random_star_chart_generator/dxvlsyt/) suggesting to use Delaunay triangulation which basically returns a solution with all possible triangles made within the data points you have. Once you have all possible triangles, which by definition you know won't result on any overlapped lines, you can chose which lines you use which result on all nodes being connected.
I was coding my solution on python and fortunately there's lots of scientific libraries on python. I was working on a random sky chart generator which would draw constellations out of those stars. In order to get all possible triangles (and draw them, just for fun), before going into the algorithm to draw the actual constellations, all I had to do was this:
# 2D array of the coordinates of every star generated randomly before
points = list(points_dict.keys())
from scipy.spatial import Delaunay
tri = Delaunay(points)
# Draw the debug constellation with the full array of lines
debug_constellation = Constellation(quadrants = quadrants, name_display_style = config.constellation_name_display_style)
for star in available_stars:
debug_constellation.add_star(star)
for triangle in tri.simplices:
star_ids = []
for index in triangle:
star_ids.append(points_dict[points[index]].id)
debug_constellation.draw_segment(star_ids, is_closed = True)
# Code to generate the image follows below
You can see the full implementation here: fake_sky_chart_generator/fake_libs/constellation_algorithms/delaunay.py
This is the result:
I saw the below algorithm works to check if a point is in a given polygon from this link:
int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
int i, j, c = 0;
for (i = 0, j = nvert-1; i < nvert; j = i++) {
if ( ((verty[i]>testy) != (verty[j]>testy)) &&
(testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
c = !c;
}
return c;
}
I tried this algorithm and it actually works just perfect. But sadly I cannot understand it well after spending some time trying to get the idea of it.
So if someone is able to understand this algorithm, please explain it to me a little.
Thank you.
The algorithm is ray-casting to the right. Each iteration of the loop, the test point is checked against one of the polygon's edges. The first line of the if-test succeeds if the point's y-coord is within the edge's scope. The second line checks whether the test point is to the left of the line (I think - I haven't got any scrap paper to hand to check). If that is true the line drawn rightwards from the test point crosses that edge.
By repeatedly inverting the value of c, the algorithm counts how many times the rightward line crosses the polygon. If it crosses an odd number of times, then the point is inside; if an even number, the point is outside.
I would have concerns with a) the accuracy of floating-point arithmetic, and b) the effects of having a horizontal edge, or a test point with the same y-coord as a vertex, though.
Edit 1/30/2022: I wrote this answer 9 years ago when I was in college. People in the chat conversation are indicating it's not accurate. You should probably look elsewhere. 🤷♂️
Chowlett is correct in every way, shape, and form.
The algorithm assumes that if your point is on the line of the polygon, then that is outside - for some cases, this is false. Changing the two '>' operators to '>=' and changing '<' to '<=' will fix that.
bool PointInPolygon(Point point, Polygon polygon) {
vector<Point> points = polygon.getPoints();
int i, j, nvert = points.size();
bool c = false;
for(i = 0, j = nvert - 1; i < nvert; j = i++) {
if( ( (points[i].y >= point.y ) != (points[j].y >= point.y) ) &&
(point.x <= (points[j].x - points[i].x) * (point.y - points[i].y) / (points[j].y - points[i].y) + points[i].x)
)
c = !c;
}
return c;
}
I changed the original code to make it a little more readable (also this uses Eigen). The algorithm is identical.
// This uses the ray-casting algorithm to decide whether the point is inside
// the given polygon. See https://en.wikipedia.org/wiki/Point_in_polygon#Ray_casting_algorithm
bool pnpoly(const Eigen::MatrixX2d &poly, float x, float y)
{
// If we never cross any lines we're inside.
bool inside = false;
// Loop through all the edges.
for (int i = 0; i < poly.rows(); ++i)
{
// i is the index of the first vertex, j is the next one.
// The original code uses a too-clever trick for this.
int j = (i + 1) % poly.rows();
// The vertices of the edge we are checking.
double xp0 = poly(i, 0);
double yp0 = poly(i, 1);
double xp1 = poly(j, 0);
double yp1 = poly(j, 1);
// Check whether the edge intersects a line from (-inf,y) to (x,y).
// First check if the line crosses the horizontal line at y in either direction.
if ((yp0 <= y) && (yp1 > y) || (yp1 <= y) && (yp0 > y))
{
// If so, get the point where it crosses that line. This is a simple solution
// to a linear equation. Note that we can't get a division by zero here -
// if yp1 == yp0 then the above if will be false.
double cross = (xp1 - xp0) * (y - yp0) / (yp1 - yp0) + xp0;
// Finally check if it crosses to the left of our test point. You could equally
// do right and it should give the same result.
if (cross < x)
inside = !inside;
}
}
return inside;
}
To expand on the "too-clever trick". We want to iterate over all adjacent vertices, like this (imagine there are 4 vertices):
i
j
0
1
1
2
2
3
3
0
My code above does it the simple obvious way - j = (i + 1) % num_vertices. However this uses integer division which is much much slower than all other operations. So if this is performance critical (e.g. in an AAA game) you want to avoid it.
The original code changes the order of iteration a bit:
i
j
0
3
1
0
2
1
3
2
This is still totally valid since we're still iterating over every vertex pair and it doesn't really matter whether you go clockwise or anticlockwise, or where you start. However now it lets us avoid the integer division. In easy-to-understand form:
int i = 0;
int j = num_vertices - 1; // 3
while (i < num_vertices) { // 4
{body}
j = i;
++i;
}
Or in very terse C style:
for (int i = 0, j = num_vertices - 1; i < num_vertices; j = i++) {
{body}
}
This might be as detailed as it might get for explaining the ray-tracing algorithm in actual code. It might not be optimized but that must always come after a complete grasp of the system.
//method to check if a Coordinate is located in a polygon
public boolean checkIsInPolygon(ArrayList<Coordinate> poly){
//this method uses the ray tracing algorithm to determine if the point is in the polygon
int nPoints=poly.size();
int j=-999;
int i=-999;
boolean locatedInPolygon=false;
for(i=0;i<(nPoints);i++){
//repeat loop for all sets of points
if(i==(nPoints-1)){
//if i is the last vertex, let j be the first vertex
j= 0;
}else{
//for all-else, let j=(i+1)th vertex
j=i+1;
}
float vertY_i= (float)poly.get(i).getY();
float vertX_i= (float)poly.get(i).getX();
float vertY_j= (float)poly.get(j).getY();
float vertX_j= (float)poly.get(j).getX();
float testX = (float)this.getX();
float testY = (float)this.getY();
// following statement checks if testPoint.Y is below Y-coord of i-th vertex
boolean belowLowY=vertY_i>testY;
// following statement checks if testPoint.Y is below Y-coord of i+1-th vertex
boolean belowHighY=vertY_j>testY;
/* following statement is true if testPoint.Y satisfies either (only one is possible)
-->(i).Y < testPoint.Y < (i+1).Y OR
-->(i).Y > testPoint.Y > (i+1).Y
(Note)
Both of the conditions indicate that a point is located within the edges of the Y-th coordinate
of the (i)-th and the (i+1)- th vertices of the polygon. If neither of the above
conditions is satisfied, then it is assured that a semi-infinite horizontal line draw
to the right from the testpoint will NOT cross the line that connects vertices i and i+1
of the polygon
*/
boolean withinYsEdges= belowLowY != belowHighY;
if( withinYsEdges){
// this is the slope of the line that connects vertices i and i+1 of the polygon
float slopeOfLine = ( vertX_j-vertX_i )/ (vertY_j-vertY_i) ;
// this looks up the x-coord of a point lying on the above line, given its y-coord
float pointOnLine = ( slopeOfLine* (testY - vertY_i) )+vertX_i;
//checks to see if x-coord of testPoint is smaller than the point on the line with the same y-coord
boolean isLeftToLine= testX < pointOnLine;
if(isLeftToLine){
//this statement changes true to false (and vice-versa)
locatedInPolygon= !locatedInPolygon;
}//end if (isLeftToLine)
}//end if (withinYsEdges
}
return locatedInPolygon;
}
Just one word about optimization: It isn't true that the shortest (and/or the tersest) code is the fastest implemented. It is a much faster process to read and store an element from an array and use it (possibly) many times within the execution of the block of code than to access the array each time it is required. This is especially significant if the array is extremely large. In my opinion, by storing each term of an array in a well-named variable, it is also easier to assess its purpose and thus form a much more readable code. Just my two cents...
The algorithm is stripped down to the most necessary elements. After it was developed and tested all unnecessary stuff has been removed. As result you can't undertand it easily but it does the job and also in very good performance.
I took the liberty to translate it to ActionScript-3:
// not optimized yet (nvert could be left out)
public static function pnpoly(nvert: int, vertx: Array, verty: Array, x: Number, y: Number): Boolean
{
var i: int, j: int;
var c: Boolean = false;
for (i = 0, j = nvert - 1; i < nvert; j = i++)
{
if (((verty[i] > y) != (verty[j] > y)) && (x < (vertx[j] - vertx[i]) * (y - verty[i]) / (verty[j] - verty[i]) + vertx[i]))
c = !c;
}
return c;
}
This algorithm works in any closed polygon as long as the polygon's sides don't cross. Triangle, pentagon, square, even a very curvy piecewise-linear rubber band that doesn't cross itself.
1) Define your polygon as a directed group of vectors. By this it is meant that every side of the polygon is described by a vector that goes from vertex an to vertex an+1. The vectors are so directed so that the head of one touches the tail of the next until the last vector touches the tail of the first.
2) Select the point to test inside or outside of the polygon.
3) For each vector Vn along the perimeter of the polygon find vector Dn that starts on the test point and ends at the tail of Vn. Calculate the vector Cn defined as DnXVn/DN*VN (X indicates cross product; * indicates dot product). Call the magnitude of Cn by the name Mn.
4) Add all Mn and call this quantity K.
5) If K is zero, the point is outside the polygon.
6) If K is not zero, the point is inside the polygon.
Theoretically, a point lying ON the edge of the polygon will produce an undefined result.
The geometrical meaning of K is the total angle that the flea sitting on our test point "saw" the ant walking at the edge of the polygon walk to the left minus the angle walked to the right. In a closed circuit, the ant ends where it started.
Outside of the polygon, regardless of location, the answer is zero.
Inside of the polygon, regardless of location, the answer is "one time around the point".
This method check whether the ray from the point (testx, testy) to O (0,0) cut the sides of the polygon or not .
There's a well-known conclusion here: if a ray from 1 point and cut the sides of a polygon for a odd time, that point will belong to the polygon, otherwise that point will be outside the polygon.
To expand on #chowlette's answer where the second line checks if the point is to the left of the line,
No derivation is given but this is what I worked out:
First it helps to imagine 2 basic cases:
the point is left of the line . / or
the point is right of the line / .
If our point were to shoot a ray out horizontally where would it strike the line segment. Is our point to the left or right of it? Inside or out? We know its y coordinate because it's by definition the same as the point. What would the x coordinate be?
Take your traditional line formula y = mx + b. m is the rise over the run. Here, instead we are trying to find the x coordinate of the point on that line segment that has the same height (y) as our point.
So we solve for x: x = (y - b)/m. m is rise over run, so this becomes run over rise or (yj - yi)/(xj - xi) becomes (xj - xi)/(yj - yi). b is the offset from origin. If we assume yi as the base for our coordinate system, b becomes yi. Our point testy is our input, subtracting yi turns the whole formula into an offset from yi.
We now have (xj - xi)/(yj - yi) or 1/m times y or (testy - yi): (xj - xi)(testy - yi)/(yj - yi) but testx isn't based to yi so we add it back in order to compare the two ( or zero testx as well )
I think the basic idea is to calculate vectors from the point, one per edge of the polygon. If vector crosses one edge, then the point is within the polygon. By concave polygons if it crosses an odd number of edges it is inside as well (disclaimer: although not sure if it works for all concave polygons).
This is the algorithm I use, but I added a bit of preprocessing trickery to speed it up. My polygons have ~1000 edges and they don't change, but I need to look up whether the cursor is inside one on every mouse move.
I basically split the height of the bounding rectangle to equal length intervals and for each of these intervals I compile the list of edges that lie within/intersect with it.
When I need to look up a point, I can calculate - in O(1) time - which interval it is in and then I only need to test those edges that are in the interval's list.
I used 256 intervals and this reduced the number of edges I need to test to 2-10 instead of ~1000.
Here's a php implementation of this:
<?php
class Point2D {
public $x;
public $y;
function __construct($x, $y) {
$this->x = $x;
$this->y = $y;
}
function x() {
return $this->x;
}
function y() {
return $this->y;
}
}
class Point {
protected $vertices;
function __construct($vertices) {
$this->vertices = $vertices;
}
//Determines if the specified point is within the polygon.
function pointInPolygon($point) {
/* #var $point Point2D */
$poly_vertices = $this->vertices;
$num_of_vertices = count($poly_vertices);
$edge_error = 1.192092896e-07;
$r = false;
for ($i = 0, $j = $num_of_vertices - 1; $i < $num_of_vertices; $j = $i++) {
/* #var $current_vertex_i Point2D */
/* #var $current_vertex_j Point2D */
$current_vertex_i = $poly_vertices[$i];
$current_vertex_j = $poly_vertices[$j];
if (abs($current_vertex_i->y - $current_vertex_j->y) <= $edge_error && abs($current_vertex_j->y - $point->y) <= $edge_error && ($current_vertex_i->x >= $point->x) != ($current_vertex_j->x >= $point->x)) {
return true;
}
if ($current_vertex_i->y > $point->y != $current_vertex_j->y > $point->y) {
$c = ($current_vertex_j->x - $current_vertex_i->x) * ($point->y - $current_vertex_i->y) / ($current_vertex_j->y - $current_vertex_i->y) + $current_vertex_i->x;
if (abs($point->x - $c) <= $edge_error) {
return true;
}
if ($point->x < $c) {
$r = !$r;
}
}
}
return $r;
}
Test Run:
<?php
$vertices = array();
array_push($vertices, new Point2D(120, 40));
array_push($vertices, new Point2D(260, 40));
array_push($vertices, new Point2D(45, 170));
array_push($vertices, new Point2D(335, 170));
array_push($vertices, new Point2D(120, 300));
array_push($vertices, new Point2D(260, 300));
$Point = new Point($vertices);
$point_to_find = new Point2D(190, 170);
$isPointInPolygon = $Point->pointInPolygon($point_to_find);
echo $isPointInPolygon;
var_dump($isPointInPolygon);
I modified the code to check whether the point is in a polygon, including the point is on an edge.
bool point_in_polygon_check_edge(const vec<double, 2>& v, vec<double, 2> polygon[], int point_count, double edge_error = 1.192092896e-07f)
{
const static int x = 0;
const static int y = 1;
int i, j;
bool r = false;
for (i = 0, j = point_count - 1; i < point_count; j = i++)
{
const vec<double, 2>& pi = polygon[i);
const vec<double, 2>& pj = polygon[j];
if (fabs(pi[y] - pj[y]) <= edge_error && fabs(pj[y] - v[y]) <= edge_error && (pi[x] >= v[x]) != (pj[x] >= v[x]))
{
return true;
}
if ((pi[y] > v[y]) != (pj[y] > v[y]))
{
double c = (pj[x] - pi[x]) * (v[y] - pi[y]) / (pj[y] - pi[y]) + pi[x];
if (fabs(v[x] - c) <= edge_error)
{
return true;
}
if (v[x] < c)
{
r = !r;
}
}
}
return r;
}
I am using the GPC Polygon Clipping lib and want to create a polygon programatically. I only see code for how to create one from a file. How can I do the initialization in my code?
Read better from your link, find the doc page and read; in particular gpc_add_contour function is likely what you need. The struct gpc_vertex_list holds a pointer to gpc_vertex-s and the number of vertex, and is what you must fill in. Like
gpc_polygon p = {0, NULL, NULL}; // "void" polygon
gpc_vertex v[] = { {0.0, 0.0}, {10.0, 0.}, {10.0, 10.10}, {0.0, 10.0} };
gpc_vertex_list vl = {
4, v
};
//...
gpc_add_contour(&p, &vl, 0);
The doc is not too much clear, but you can deduce the use, and testing (try-error loops) is your friend (I won't install gpc to do it anyway, so my code could be wrong). The proposed code snippet should create a square. Several other gpc_add_countour with the same &p but different vertex list can be used to create a more complex polygon, and of course vl can be changed to have at the beginning a more complex polygon. The third parameter should be 1 if you want the defined contour to be a "hole" in the current (p) polygon.
gpc_polygon subject;
int w = 100, h = 100, verticesCnt = 30;
//setup a gpc_polygon container and fill it with random vertices ...
subject.num_contours = 1;
subject.hole = 0;
subject.contour = new gpc_vertex_list; //ie just a single polygon here
subject.contour->num_vertices = verticesCnt;
subject.contour->vertex = new gpc_vertex [verticesCnt];
for (i = 0; i < verticesCnt; i++){
subject.contour[0].vertex[i].x = random(w);
subject.contour[0].vertex[i].y = random(h);
}
//do stuff with it here, then ...
gpc_free_polygon(&subject);