I know there are lots of posts about collision detection generally for sprites moving about a 2D plane, but my question is slightly different.
I'm inserting circles into a 2D plane. The circles have variable radii. I'm trying to optimize my method of finding a random position within the plane where I can insert a new circle without it colliding with any other circles already on the plane. Right now I'm using a very "un-optimized" approach that simply generates a random point within the plane and then checks it against all the other circles on the plane.
Are there ways to optimize this? For this particular app, the bounds of the plane can only hold 20-25 circles at a time and typically there are between 5-10 present. As you would expect, when the number of circles approaches the max that can fit, you have to test many points before finding one that works. It gets very slow.
Note: safeDistance is the radius of the circle I want to add to the plane.
Here's the code:
- (CGPoint)getSafePosition:(float)safeDistance {
// Point must be far enough from edges
// Point must be far enough from other sprites
CGPoint thePoint;
BOOL pointIsSafe = NO;
int sd = ceil(safeDistance);
while(!pointIsSafe) {
self.pointsTested++; // DEBUG
// generate a random point inside the plane boundaries to test
thePoint = CGPointMake((arc4random() % ((int)self.manager.gameView.frame.size.width - sd*2)) + sd,
(arc4random() % ((int)self.manager.gameView.frame.size.height - sd*2)) + sd);
if(self.manager.gameView.sprites.count > 0) {
for(BasicSprite *theSprite in self.manager.gameView.sprites) {
// get distance between test point and the sprite position
float distance = [BasicSprite distanceBetweenPoints:thePoint b:theSprite.position];
// check if distance is less than the sum of the min safe distances of the two entities
if(distance < (safeDistance + [theSprite minSafeDistance])) {
// point not safe
pointIsSafe = NO;
break;
}
// if we get here, the point did not collide with the last tested point
pointIsSafe = YES;
}
}
else {
pointIsSafe = YES;
}
}
return thePoint;
}
Subdivide your window into w by h blocks. You'll be maintaining a w by h array, dist. dist[x][y] contains the size of the largest circle that can be centred at (x, y). (You can use pixels as blocks, although we'll be updating the entire array with each circle placed, so you may want to choose larger blocks for improved speed, at the cost of slightly reduced packing densities.)
Initialisation
Initially, set all dist[x][y] to min(x, y, w - x, h - y). This encodes the limits given by the bounding box that is the window.
Update procedure
Every time you add a circle to the window, say one positioned at (a, b) with radius r, you need to update all elements of dist.
The update required for each position (x, y) is:
dist[x][y] = min(dist[x][y], sqrt((x - a)^2 + (y - b)^2) - r);
(Obviously, ^2 here means squaring, not XOR.) Basically, we are saying: "Set dist[x][y] to the minimum distance to the circle just placed, unless the situation is already worse than that." dist values for points inside the circle just placed will be negative, but that doesn't matter.
Finding the next location
Then, when you want to insert the next circle of radius q, just scan through dist looking for a location with dist value >= q. (If you want to randomly choose such a location, find the complete list of valid locations and then randomly choose one.)
Honestly, with only 20-25 circles, you're not going to get much of a speed boost by using a fancier algorithm or data structure (e.g. a quadtree or a kd-tree). Everything is fast for small n.
Are you absolutely sure this is the bottleneck in your application? Have you profiled? If yes, then the way you're going to speed this up is through microoptimization, not through advanced algorithms. Are you making lots of iterations through the while loop because most of the plane is unsafe?
You could split your plane in lots of little rectangles (slightly quadtree-related) and save which rectangles are hit by at least one of the circles.
When you look for a insertion-point, you'll just have to look for some "empty" ones (which doesn't need any random jumps and is possible in constant time).
The number and constellation of rectangles can be computed by the radius.
Just an outline, since this solution is fairly involved.
If you want to guarantee you always find a place to put a circle if it's possible, you can do the following. Consider each existing circle C. We will try to find a location where we can place the new circle so that it is touching C. For each circle D (other than C) that is sufficiently close to C, there will be a range of angles where placing a new circle at one of those angles around C will make it intersect with D. Some geometry will give you that range. Similarly, for each of the four boundaries that are close enough to C, there will be a range of angles where placing a new circle at one of those angles will make it intersect with the boundary. If all these intervals cover all 360 degrees around C, then you cannot place a circle adjacent to C, and you will have to try the next circle, until there are no more candidates for C. If you find a place to put the new circle, you can move it some random distance away from C so that all your new circles do not have to be adjacent to an existing circle if that is not necessary.
Related
Given a list of points forming a polygonal line, and both height and width of a rectangle, how can I find the number and positions of all rectangles needed to cover all the points?
The rectangles should be rotated and may overlap, but must follow the path of the polyline (A rectangle may contain multiple segments of the line, but each rectangle must contain a segment that is contiguous with the previous one.)
Do the intersections on the smallest side of the rectangle, when it is possible, would be much appreciated.
All the solutions I found so far were not clean, here is the result I get:
You should see that it gives a good render in near-flat cases, but overlaps too much in big curbs. One rectangle could clearly be removed if the previous were offset.
Actually, I put a rectangle centered at width/2 along the line and rotate it using convex hull and modified rotating calipers algorithms, and reiterate starting at the intersection point of the previous rectangle and the line.
You may observe that I took inspiration from the minimum oriented rectangle bounding box algorithm, for the orientation, but it doesn't include the cutting aspect, nor the fixed size.
Thanks for your help!
I modified k-means to solve this. It's not fast, it's not optimal, it's not guaranteed, but (IMHO) it's a good start.
There are two important modifications:
1- The distance measure
I used a Chebyshev-distance-inspired measure to see how far points are from each rectangle. To find distance from points to each rectangle, first I transformed all points to a new coordinate system, shifted to center of rectangle and rotated to its direction:
Then I used these transformed points to calculate distance:
d = max(2*abs(X)/w, 2*abs(Y)/h);
It will give equal values for all points that have same distance from each side of rectangle. The result will be less than 1.0 for points that lie inside rectangle. Now we can classify points to their closest rectangle.
2- Strategy for updating cluster centers
Each cluster center is a combination of C, center of rectangle, and a, its rotation angle. At each iteration, new set of points are assigned to a cluster. Here we have to find C and a so that rectangle covers maximum possible number of points. I don’t now if there is an analytical solution for that, but I used a statistical approach. I updated the C using weighted average of points, and used direction of first principal component of points to update a. I used results of proposed distance, powered by 500, as weight of each point in weighted average. It moves rectangle towards points that are located outside of it.
How to Find K
Initiate it with 1 and increase it till all distances from points to their corresponding rectangles become less than 1.0, meaning all points are inside a rectangle.
The results
Iterations 0, 10, 20, 30, 40, and 50 of updating cluster centers (rectangles):
Solution for test case 1:
Trying Ks: 2, 4, 6, 8, 10, and 12 for complete coverage:
Solution for test case 2:
P.M: I used parts of Chalous Road as data. It was fun downloading it from Google Maps. The I used technique described here to sample a set of equally spaced points.
It’s a little late and you’ve probably figured this out. But, I was free today and worked on the constraint reflected in your last edit (continuity of segments). As I said before in the comments, I suggest using a greedy algorithm. It’s composed of two parts:
A search algorithm that looks for furthermost point from an initial point (I used binary search algorithm), so that all points between them lie inside a rectangle of given w and h.
A repeated loop that finds best rectangle at each step and advances the initial point.
The pseudo code of them are like these respectively:
function getBestMBR( P, iFirst, w, h )
nP = length(P);
iStart = iFirst;
iEnd = nP;
while iStart <= iEnd
m = floor((iStart + iEnd) / 2);
MBR = getMBR(P[iFirst->m]);
if (MBR.w < w) & (MBR.h < h) {*}
iStart = m + 1;
iLast = m;
bestMBR = MBR;
else
iEnd = m - 1;
end
end
return bestMBR, iLast;
end
function getRectList( P, w, h )
nP = length(P);
rects = [];
iFirst = 1;
iLast = iFirst;
while iLast < nP
[bestMBR, iLast] = getBestMBR(P, iFirst, w, h);
rects.add(bestMBR.x, bestMBR.y, bestMBR.a];
iFirst = iLast;
end
return rects;
Solution for test case 1:
Solution for test case 2:
Just keep in mind that it’s not meant to find the optimal solution, but finds a sub-optimal one in a reasonable time. It’s greedy after all.
Another point is that you can improve this a little in order to decrease number of rectangles. As you can see in the line marked with (*), I kept resulting rectangle in direction of MBR (Minimum Bounding Rectangle), even though you can cover larger MBRs with rectangles of same w and h if you rotate the rectangle. (1) (2)
I'm currently developing an application that will alert users of incoming rain. To do this I want to check certain area around user location for rainfall (different pixel colours for intensity on rainfall radar image). I would like the checked area to be a circle but I don't know how to do this efficiently.
Let's say I want to check radius of 50km. My current idea is to take subset of image with size 100kmx100km (user+50km west, user+50km east, user+50km north, user+50km south) and then check for each pixel in this subset if it's closer to user than 50km.
My question here is, is there a better solution that is used for this type of problems?
If the occurrence of the event you are searching for (rain or anything) is relatively rare, then there's nothing wrong with scanning a square or pixels and then, only after detecting rain in that square, checking whether that rain is within the desired 50km circle. Note that the key point here is that you don't need to check each pixel of the square for being inside the circle (that would be very inefficient), you have to search for your event (rain) first and only when you found it, check whether it falls into the 50km circle. To implement this efficiently you also have to develop some smart strategy for handling multi-pixel "stains" of rain on your image.
However, since you are scanning a raster image, you can easily implement the well-known Bresenham circle algorithm to find the starting and the ending point of the circle for each scan line. That way you can easily limit your scan to the desired 50km radius.
On the second thought, you don't even need the Bresenham algorithm for that. For each row of pixels in your square, calculate the points of intersection of that row with the 50km circle (using the usual schoolbook formula with square root), and then check all pixels that fall between these intersection points. Process all rows in the same fashion and you are done.
P.S. Unfortunately, the Wikipedia page I linked does not present Bresenham algorithm at all. It has code for Michener circle algorithm instead. Michener algorithm will also work for circle rasterization purposes, but it is less precise than Bresenham algorithm. If you care for precision, find a true Bresenham on somewhere. It is actually surprisingly diffcult to find on the net: most search hits erroneously present Michener as Bresenham.
There is, you can modify the midpoint circle algorithm to give you an array of for each y, the x coordinate where the circle starts (and ends, that's the same thing because of symmetry). This array is easy to compute, pseudocode below.
Then you can just iterate over exactly the right part, without checking anything.
Pseudo code:
data = new int[radius];
int f = 1 - radius, ddF_x = 1;
int ddF_y = -2 * radius;
int x = 0, y = radius;
while (x < y)
{
if (f >= 0)
{
y--;
ddF_y += 2; f += ddF_y;
}
x++;
ddF_x += 2; f += ddF_x;
data[radius - y] = x; data[radius - x] = y;
}
Maybe you can try something that will speed up your algorithm.
In brute force algorithm you will probably use equation:
(x-p)^2 + (y-q)^2 < r^2
(p,q) - center of the circle, user position
r - radius (50km)
If you want to find all pixels (x,y) that satisfy above condition and check them, your algorithm goes to O(n^2)
Instead of scanning all pixels in this circle I will check only only pixels that are on border of the circle.
In that case, you can use some more clever way to define circle.
x = p+r*cos(a)
y = q*r*sin(a)
a - angle measured in radians [0-2pi]
Now you can sample some angles, for example twenty of them, iterate and find all pairs (x,y) that are border for radius 50km. Now check are they on the rain zone and alert user.
For more safety I recommend you to use multiple radians (smaller than 50km), because your whole rain cloud can be inside circle, and your app will not recognize him. For example use 3 incircles (r = 5km, 15km, 30km) and do same thing. Efficiency of this algorithm only depends on number of angles and number of incircles.
Pseudocode will be:
checkRainDanger()
p,q <- position
radius[] <- array of radii
for c = 1 to length(radius)
a=0
while(a<2*pi)
x = p + radius[c]*cos(a)
y = q + radius[c]*sin(a)
if rainZone(x,y)
return true
else
a+=pi/10
end_while
end_for
return false //no danger
r2=r*r
for x in range(-r, +r):
max_y=sqrt(r2-x*x)
for y in range(-max_y, +max_y):
# x,y is in range - check for rain
How to rotate an object, so that its vertices never overlap with any of the other rotations? With a predefined number of rotations.
Idea:
It can be achieved with relaxation. (Idea comes from Greg Turk's paper: Generating Textures on Arbitrary Surfaces Using Reaction-Diffusion)
Steps:
Generate x dodecahedrons or any object symmetric to its centre
point.
These objects should be identical in position, orientation
and size. (so we can create an easy relation between vertices =>
ones that overlap at the beginning are related)
Create a function that calculates the distance between the related points.
Maximize the average distance between related points. (every point has x-1
related points)
Problems:
This is not a simple relaxation problem with points. Here, due to the dodecahedron constraint, I cannot just translate around. Rotation matrixes/quaternions are needed.
Possible solution with Brute force
Summary:Rotate randomly until desired average distance is achieved.
Explanation:Every dodecahedron is rotated until it's vertices do not overlap any vertices of the other dodecahedrons.
Then average distance is calculated and checked against the best (minimal) so far. Save all vertex positions, and the rotation quaternion, that will turn the base dodecahedron into the rotated one.
float minThreshold <- user defined
int iterationThreshold <- user defined
float minAveDistance = Infinity;
while (minAveDistance > minThreshold || maxIteration > iterationThreshold) {
Foreach (dodecahedron) { // except the first one, that can stay as is
// rotate randomly until there are no overlapping positions with the other dodecahedrons
while (checkOverlappingWithOtherDodecahedrons(dodecahedron)) {
rotateRandomly(dodecahedron);
}
}
float aveDistance = CalculateAverageDistanceBetweenAllPointsOfDodecahedrons();
if (aveDistance < minAveDistance) {
minAveDistance = aveDistance;
SaveAllPositions(); // e.g.: to a file
SaveAllRotationQuaternionsFromStartOrientation(); // e.g.: to a file
}
}
In 3-D space I have an unordered set of, say, 6 points; something like this:
(A)*
(C)*
(E)*
(F)*
(B)*
(D)*
The points form a 3-D contour but they are unordered. For unordered I mean that they are stored in an
unorderedList = [A - B - C - D - E - F]
I just want to reorganize this list starting from an arbitrary location (let's say point A) and traversing the points clockwise or counter-clockwise. Something like this:
orderedList = [A - E - B - D - F - C]
or
orderedList = [A - C - F - D - B - E]
I'm trying to implement an algorithm as simple as possible, since the set of points in mention corresponds to a N-ring neighborhood of each vertex on a mesh of ~420000 points, and I have to do this for each point on the mesh.
Some time ago there was a similar discussion regarding points in 2-D, but for now it's not clear for me how to go from this approach to my 3-D scenario.
The notion of "clockwise" or "counterclockwise" is not well-defined without an axis and orientation! (proof: What if you looked at those points from the other side of your monitor screen, or flipped them, for example!)
You must define an axis and orientation, and specify it as an additional input. Ways to specify it include:
a line (1x=2y=3z), using the right-hand rule
a (unit) vector (A_x, A_y, A_z), using the right-hand rule; this is the preferred way to do so
In order to determine the orientation, you have to look deeper at your problem: You must define a "up" and "down" size of the mesh. Then for each set of points, you must take the centroid (or another "inside" point) and construct a unit vector pointing "up" which is normal to the surface. (One way to do this would be to find the least-squares-fit plane, then find the two perpendicular vectors through that point, picking the one in the "up" direction.)
You will need to use any of the above suggestions to determine your axis. This will allow you to reformulate your problem as follows:
Inputs:
the set of points {P_i}
an axis, which we shall call "the z-axis" and treat as a unit vector centered on the centroid (or somewhere "inside") of the points
an orientation (e.g. counterclockwise) chosen by one of the above methods
Setup:
For all points, pick two mutually-orthogonal unit vectors to the axis, which we shall call "the y-axis" and "the x-axis". (Just rotate the z-axis unit-vector 90 degrees in two directions, http://en.wikipedia.org/wiki/Rotation_matrix#Basic_rotations )
Algorithm:
For each point P, project P onto the x-axis and y-axis (using the dot product), then use http://en.wikipedia.org/wiki/Atan2
Once you have the angles, you can just sort them.
I can't attest for the efficiency of this code, but it works, and you can optimize parts of it as needed, I'm just not good at it.
Code is in C#, using system collection classes, and linq.
Vector3 is a class with floats x, y, z, and static vector math functions.
Node is a class with Vector3 variable called pos
//Sort nodes with positions in 3d space.
//Assuming the points form a convex shape.
//Assuming points are on a single plain (or close to it).
public List<Node> sortVerticies( Vector3 normal, List<Node> nodes ) {
Vector3 first = nodes[0].pos;
//Sort by distance from random point to get 2 adjacent points.
List<Node> temp = nodes.OrderBy(n => Vector3.Distance(n.pos, first ) ).ToList();
//Create a vector from the 2 adjacent points,
//this will be used to sort all points, except the first, by the angle to this vector.
//Since the shape is convex, angle will not exceed 180 degrees, resulting in a proper sort.
Vector3 refrenceVec = (temp[1].pos - first);
//Sort by angle to reference, but we are still missing the first one.
List<Node> results = temp.Skip(1).OrderBy(n => Vector3.Angle(refrenceVec,n.pos - first)).ToList();
//insert the first one, at index 0.
results.Insert(0,nodes[0]);
//Now that it is sorted, we check if we got the direction right, if we didn't we reverse the list.
//We compare the given normal and the cross product of the first 3 point.
//If the magnitude of the sum of the normal and cross product is less than Sqrt(2) then then there is more than 90 between them.
if ( (Vector3.Cross( results[1].pos-results[0].pos, results[2].pos - results[0].pos ).normalized + normal.normalized).magnitude < 1.414f ) {
results.Reverse();
}
return results;
}
Given a set of n points on plane, I want to preprocess these points somehow faster than O(n^2) (O(nlog(n)) preferably), and then be able to answer on queries of the following kind "How many of n points lie inside a circle with given center and radius?" faster than O(n) (O(log(n) preferably).
Can you suggest some data structure or algorithm I can use for this problem?
I know that such types of problems are often solved using Voronoi diagrams, but I don't know how to apply it here.
Build a spatial subdivision structure such as a quadtree or KD-tree of the points. At each node store the amount of points covered by that node. Then when you need to count the points covered by the lookup circle, traverse the tree and for each subdivision in a node check if it is fully outside the circle, then ignore it, if it is fully inside the circle then add its count to the total if it intersects with the circle, recurse, when you get to the leaf, check the point(s) inside the leaf for containment.
This is still O(n) worst case (for instance if all the points lie on the circle perimeter) but average case is O(log(n)).
Build a KD-tree of the points, this should give you much better complexity than O(n), on average O(log(n)) I think.
You can use a 2D tree since the points are constrained to a plane.
Assuming that we have transformed the problem into 2D, we'll have something like this for the points:
struct Node {
Pos2 point;
enum {
X,
Y
} splitaxis;
Node* greater;
Node* less;
};
greater and less contains points with greater and lesser coordinates respectively along the splitaxis.
void
findPoints(Node* node, std::vector<Pos2>& result, const Pos2& origin, float radius) {
if (squareDist(origin - node->point) < radius * radius) {
result.push_back(node->point);
}
if (!node->greater) { //No children
return;
}
if (node->splitaxis == X) {
if (node->point.x - origin.x > radius) {
findPoints(node->greater, result, origin radius);
return;
}
if (node->point.x - origin.x < -radius) {
findPoints(node->less, result, origin radius);
return;
}
findPoints(node->greater, result, origin radius);
findPoints(node->less, result, origin radius);
} else {
//Same for Y
}
}
Then you call this function with the root of the KD-tree
If my goal is speed, and the number of points weren't huge (millions,) I'd focus on memory footprint as much as algorithmic complexity.
An unbalanced k-d tree is best on paper, but it requires pointers, which can expand memory footprint by 3x+, so it is out.
A balanced k-d tree requires no storage, other than for an array with one scalar for each point. But it too has a flaw: the scalars can not be quantized - they must be the same 32 bit floats as in the original points. If they are quantized, it is no longer possible to guarantee that a point which appears earlier in the array is either on the splitting plane, or to its left AND that a point which appears later in the array is either on the splitting plane, or to its right.
There is a data structure I developed that addresses this problem. The Synergetics folks tell us that volume is experientially four-directional. Let's say that a plane is likewise experientially three-directional.
We're accustomed to traversing a plane by the four directions -x, +x, -y, and +y, but it's simpler to use the three directions a, b, and c, which point at the vertices of an equilateral triangle.
When building the balanced k-d tree, project each point onto the a, b, and c axes. Sort the points by increasing a. For the median point, round down, quantize and store a. Then, for the sub-arrays to the left and right of the median, sort by increasing b, and for the median points, round down, quantize, and store b. Recurse and repeat until each point has stored a value.
Then, when testing a circle (or whatever) against the structure, first calculate the maximum a, b, and c coordinates of the circle. This describes a triangle. In the data structure we made in the last paragraph, compare the median point's a coordinate to the circle's maximum a coordinate. If the point's a is larger than the circle's a, we can disqualify all points after the median. Then, for the sub-arrays to the left and right (if not disqualified) of the median, compare the circle's b to the median point's b coordinate. Recurse and repeat until there are no more points to visit.
This is similar in theme to the BIH data structure, but requires no intervals of -x and +x and -y and +y, because a, b, and c are just as good at traversing the plane, and require one fewer direction to do it.
Assuming you have a set of points S in a cartesian plane with coordinates (xi,yi), given an arbitrary circle with center (xc,yc) and radius r you want to find all the points contained within that circle.
I will also assume that the points and the circle may move so certain static structures that can speed this up won't necessarily be appropriate.
Three things spring to mind that can speed this up:
Firstly, you can check:
(xi-xc)^2 + (yi-yc)^2 <= r^2
instead of
sqrt((xi-xc)^2 + (yi-yc)^2) <= r
Secondly, you can cull the list of points somewhat by remembering that a point can only be within the circle if:
xi is in the range [xc-r,xc+r]; and
yi is in the range [yc-r,yc+r]; and
This is known as a bounding box. You can use it as either an approximation or to cut down your list of points to a smaller subset to check accurately with the first equation.
Lastly, sort your points in either x or y order and then you can do a bisection search to find the set of points that are possibly within your bounding box, further cutting down on unnecessary checks.
I used Andreas's code but it contains a bug. For example I had two points on the plane [13, 2], [13, -1] and my origin point was [0, 0] with a radius of 100. It finds only 1 point. This is my fix:
void findPoints(Node * root, vector<Node*> & result, Node * origin, double radius, int currAxis = 0) {
if (root) {
if (pow((root->coords[0] - origin->coords[0]), 2.0) + pow((root->coords[1] - origin->coords[1]), 2.0) < radius * radius) {
result.push_back(root);
}
if (root->coords[currAxis] - origin->coords[currAxis] > radius) {
findPoints(root->right, result, origin, radius, (currAxis + 1) % 2);
return;
}
if (origin->coords[currAxis] - root->coords[currAxis] > radius) {
findPoints(root->left, result, origin, radius, (currAxis + 1) % 2);
return;
}
findPoints(root->right, result, origin, radius, (currAxis + 1) % 2);
findPoints(root->left, result, origin, radius, (currAxis + 1) % 2);
}
}
The differnce is Andreas checked for now children with just if (!root->greater) which is not complete. I on the other hand don't do that check, I simply check if the root is valid.
Let me know if you find a bug.
depending on how much precomputing time you have, you could build a tree like this:
first node branches are x-values, below them are y-value nodes, and below them are radius value nodes. at each leaf is a hashset of points.
when you want to compute the points at x,y,r: go through your tree and go down the branch that matches your x,y values the closest. when you get down to the root level, you need to do a little match (constant time stuff), but you can find a radius such that all the points in that circle (defined by the path in the tree) are inside the circle specified by x,y,r, and another circle (same x_tree,y_tree in the tree as before, but different r_tree) such that all of the points in the original circle (specified by x,y,r) are in that circle.
from there, go through all the points in the larger of the two tree circles: if a point is in the smaller circle add it to the results, if not, run the distance check on it.
only problem is that it takes a very long time to precompute the tree. although, you can specify the amount of time you want to spend by changing how many x,y, and r branches you want to have in your tree.