At the moment I try to learn and understand geometric algorithms and I found the smallest-circle-problem quite interesting. I found many solutions and different algorithms including this one of which I'm not sure if it is Emo Welzl's.
However, I don't understand one specific (important) part:
You're given N points on a (XY)-Plane.
You order those points randomly
Choose 3 points and create the circle where they are on the circle boundary.
Get the next point and check if it is enclosed by the circle:
a) If it is enclosed, repeat 4 until there are no more points left.
b) If is not enclosed, create a new circle where the new point is on the circle boundary and still all other points are inside or on the circle.
Steps 1) to 4a) are simple, my problem is step 4b). How can I find this new circle? To me, it seems like this is the same problem just with a smaller (sub)set of points. (Divide-et-Impera)
I guessed I have to replace one of the 3 original points (the first 3 points, that made up the first circle) with the new point, but I'm not sure
if this really works...
From the 3 points A,B,C you can calculate the centre O of the circle: the point that is equidistant to those three. Its coordinates xO and yO are the means of xA,xB,xC and yA, yB, yC respectively.
Let's call D the 4th point, trace the circle of centre 0 and radius OD.
OD > OA (and OA=OB=OC) so A, B and C are in the circle.
EDIT
The solution I proposed above is not optimal.
I found a good explanation of Welzl's algorithm: see link
Of course, you can get his paper easily by looking on Google Scholar, but it is quite hard to read.
The basic principle is that in 4b) the circle is computed from all the possible circles having D on the boundary as well as two other points that were on the boundary before (or one point if that doesn't work and it will be diametrically opposed to D).
Related
I'm a high school student and I went to a coding competition recently and got this problem that I had no idea how to solve:
Given a maze enclosed in a 100x100 area, determine if a circle with a given radius could fit through the maze given the locations of all the walls. Walls will be defined as lines connecting two points within the space, and you will be given start and destination points for the circle. The circle must start with its center at the start point and touch the destination point for it to successfully fit through the maze. There will be a maximum of 20 walls. The radius of the circle and the locations of the walls can be "arbitrarily" precise. ("arbitrarily" for this case just means within far limits - let's say, up to a max of 10 digits after the decimal).
Here is an example. If this were the input:
Radius = 2.8
Start = (5,5), Destination = (95,95)
Walls (a wall connects each pair of points):
(20,0) to (27.5,22.6)
(27.5,22.6) to (55.1,35.5)
(55.1,35.5) to (80.3,80,4)
(80.3,80,4) to (95,63.9)
(1.7,25.8) to (17.5,53.2)
(17.5,53.2) to (56.4,69)
(56.4,69) to (67.9,90.6)
(85.6,98.94512) to (87.3,92.5)
then this (made on desmos) is what the maze would look like (the blue circle is just to show how big the circle is):
I would know how to solve the problem if it were in a quantized grid, but the exact locations of the walls and the radius of the circle can be arbitrarily precise. I've thought about using the "right-hand rule" to find a path, but I don't know how to implement that in a non-quantized space (nor am I very familiar with the method).
How would I go about solving this? Can someone point me to an algorithm, a link, some pseudocode, or just an intuition that could help me get an understanding of how I might solve this? Any help is appreciated. Thanks!
Thickennig/moving walls by r in each side like in the other answer (+1 btw) sounds simple but to code it its not trivial to do. For more info see
draw outline for some connected lines
The direction of normal in 2D is easy if dx,dy is line direction then (-dy,dx) and (dy,-dx) are normals to it ...
However I would encourage to do slower but safe and easier approach by computing the closest distance to wall for each vertex of the maze and close paths that are closer than 2r ...
Something like this:
So:
for each vertex
check all lines that does not belong to vertex path
compute perpendicular and min distance d to line and its vertexes
use the smallest d the distance can be computed easily see:
helix central axis angle
just look there for Perpendicular distance of any point P to AB so:
d = min
(
perpendicular_distance(line,vertex),
|line_vertex1-vertex|,
|line_vertex2-vertex|
)
if d<2r close the path. For example by adding a line that joins the wall its too close to tested vertex
ideally by joining tested vertex and the found closest point. Do not forget in such case to split the opposing wall line to two by the closest point so your graph algorithms still work...
As you can see this is O(n^2) instead of O(n) like in the other answer but its foul proof... enlarging polygons is not and in matters of fact its one of the hardest things in 2D geometry to code (IIRC even still open problem)...
It's quite a task and not easy to code, but here's a way that works:
Let r be the radius of the circle. That means that the center of the circle can't get within r of any obstacle.
Move the walls of your maze area in by r on each side.
Replace every wall endpoint with a circle of radius r.
Replace every wall with a rectangle of width 2r.
Now you don't need to worry about the circle -- only its center point, which must remain within the new boundaries and outside of any of the circles or rectangles you made from the walls.
Now, there is a path from the start to the end if they are in the same enclosed area. To find that out...
Cut the scene horizontally at every intersection and vertical maximum or minimum to create strips, with each strip divided into regions by a line or circular arc that passes all the way through it. A region does not connect direction to the regions to its left and right, but may connect to zero or more regions in the strips above and below it. The connections between regions form a graph.
Starting at the region containing the start point, run a BFS or DFS on this graph to see if you can reach the region containing the end point.
In
http://en.wikipedia.org/wiki/Closest_pair_of_points_problem
we can see that it mentions that is at most 6 points that is closest to the point on the other half, which can be represented as the graph below:
My question is for point P1 and Point P2, the distance to the red point will exceed sqrt(2)*d, why it is part of the solution? Why it is not at most 4 points that is closest to P rather than at most 6 points? Thanks.
P1 and P2 are not part of the solution, but they have to be examined on the way to the solution, because the algorithm examines all points in the box, and P1 and P2 are in the box.
Note that no such point as your Q can exist, because by hypothesis the minimum distance between points in the right-hand half of the diagram is d.
Edited to add: you seem to think that the Wikipedia article is making a claim like this:
There may be up to 6 points on the right side of the line that are within a distance d of P.
This claim would be false. But the article does not make such a claim. Instead, it makes two separate claims, both of which are true:
All the points on the right side of the line that are within a distance d of P are inside the box.
There may be up to 6 points in the box.
We are only counting the maximum number of points that can lie in the right d x 2d rectangle. Since any two points are constrained to have a minimum distance of d, we can place at most 6 points in the rectangle while satisfying this constraint, as shown in the figure.
Note that the points on the right side that are within d distance from P should all lie within a circular segment of a circle centered at P and whose radius is d. There can be at most 4 points in this segment. However, finding the number of points within a segment is more complicated than finding the number of points within a rectangle. So we use the rectangle instead and incur an extra cost of having to search for at most 2 additional points.
The bound is only important for complexity estimation. Code-wise, you may simply scan up and down within the distance dRmin. The bound here suggest that you'll at most see 6 points in each such scan, making this O(1).
I am trying to determine the distance from a point to a polygon in 2D space. The point can be inside or outside the polygon; The polygon can be convex or concave.
If the point is within the polygon or outside the polygon with a distance smaller than a user-defined constant d, the procedure should return True; False otherwise.
I have found a similar question: Distance from a point to a polyhedron or to a polygon. However, the space is 2D in my case and the polygon can be concave, so it's somehow different from that one.
I suppose there should be a method simpler than offsetting the polygon by d and determining it's inside or outside the polygon.
Any algorithm, code, or hints for me to google around would be appreciated.
Your best bet is to iterate over all the lines and find the minimum distance from a point to a line segment.
To find the distance from a point to a line segment, you first find the distance from a point to a line by picking arbitrary points P1 and P2 on the line (it might be wise to use your endpoints). Then take the vector from P1 to your point P0 and find (P2-P1) . (P0 - P1) where . is the dot product. Divide this value by ||P2-P1||^2 and get a value r.
Now if you picked P1 and P2 as your points, you can simply check if r is between 0 and 1. If r is greater than 1, then P2 is the closest point, so your distance is ||P0-P2||. If r is less than 0, then P1 is the closest point, so your distance is ||P0-P1||.
If 0<r<1, then your distance is sqrt(||P0-P1||^2 - (r * ||P2-P1||)^2)
The pseudocode is as follows:
for p1, p2 in vertices:
var r = dotProduct(vector(p2 - p1), vector(x - p1))
//x is the point you're looking for
r /= (magnitude(vector(p2 - p1)) ** 2)
if r < 0:
var dist = magnitude(vector(x - p1))
else if r > 1:
dist = magnitude(vector(p2 - x))
else:
dist = sqrt(magnitude(vector(x - p1)) ^ 2 - (r * magnitude(vector(p2-p1))) ^ 2)
minDist = min(dist,minDist)
In the event that this helps someone else, I reverse engineered doverbin's answer to understand why it worked showing graphically what the three cases are computing. (doverbin, feel free to incorporate this into your answer if you wish.)
If you have a working point to line segment distance function, you can use it to calculate the distance from the point to each of the edges of the polygon. Of course, you have to check if the point is inside the polygon first.
Do you need fast or simple?
Does it have to be always absolutely correct in edge cases or will good enough most of the time be OK?
Typical solution are to find the distance to each vertex and find the pair with the smallest values ( note that for a point outside a convex polygon these might not be adjacent) and then check point to line intersections for each segment.
For large complex shapes you can also store approx polygon bounding boxes (either rectangular or hexagons) and find the closest side before checking more detail.
You may also need code to handle the special case of exactly on a line.
I can help you with this pointers:
The distance can be calculated using Wikipedia entry, even with an untested snipped of code.
The inside/outside border can be solved with this Stack Post and links
and some remarks:
you should check only the nearest points, as Martin Beckett´s answer point outs, since another segment can "proyected" very near, but in reality don´t be close as need.
I do not know about the difference in performance with respect to the rest of answers, but in boost C++ libraries there is an generic implementation called distance. It has information about complexity in every case and in your problem case it is linear.
I was also looking for solutions to this problem some days ago and I want to share this finding. Hope it helps to someone.
I'm struggling to find a rock solid solution to detecting collisions between a circle and a circle segment. Imagine a Field of View cone for a game enemy, with the circles representing objects of interest.
The diagram at the bottom is something I drew to try and work out some possible cases, but i'm sure there are more.
I understand how to quickly exlude extreme cases, I discard any targets that don't collide with the entire circle, and any cases where the center of the main circle is within the target circle are automatically true (E in the diagram).
I'm struggling to find a good way to check the rest of the cases. I've tried comparing distances between circle centers and the end points of the segments outer lines, and i've tried working out the angle of the center of the target circle from the center of the main circle and determining whether that is within the segment, but neither way seems to catch all cases.
Specifically it seems to go funky if the target circle is close to the center but not touching it (somewhere between E and B below), or if the segment is narrower than the target circle (so that the center is within the segment but both edges are outside it).
Is there a reliable method for doing this?
Extra info: The segment is described by position P, orientation O (whose magnitude is the circle radius), and a view size, S.
My most successful attempt to date involved determining the angles of the vectors ca1 and ca2, and checking if either of them lies between the angles of vectors a1 and a2. This works for some cases as explained above, but not situations where the target circle is larger than the segment.
Edit 2
After implementing the best suggestion from below, there is still a false positive which I am unsure how best to eliminate. See the pink diagram below. The circle in the lower right is reporting as colliding with the segment because it's bounds overlap both half spaces and the main circle.
Final Edit
After discovering another edge case (4th image), i've settled on an approach which combines the two top answers from below and seems to cover all bases. I'll describe it here for the sake of those who follow.
First exclude anything that fails a quick circle-to-circle test.
Then test for collision between the circle and the two outer lines of the segment. If it touches either, return true.
Finally, do a couple of point-to-halfspace tests using the center of the circle and the two outer lines (as described by Gareth below). If it passes both of those it's in, otherwise return false.
A. Check if it is intersecting the whole cirlce.
B. Check if it is intersecting either of the straight segment lines.
C. If not, check if the angle between the circle centres lies in the angular range of the segment (dot product is good for this).
Intersection requires A && (B || C)
A circular segment (with central angle less than 180°) is the intersection of three figures: a circle, and two half-planes:
So a figure intersects the circular segment only if it intersects all three of these figures. [That's only if but not if; see below.]
Circle/circle intersection is easy (compare the distance between their centres with the sum of their radii).
For circle/half-plane intersection, represent the half-plane in the form p · n ≤ k (where p is the point to be tested, n is a unit vector that's normal to the line defining the half-plane, and k is a constant). Then a circle with centre x and radius r intersects the half-plane if x · n ≤ k + r.
(If you need to handle a circular segment with central angle greater than 180°, split it into two segments with central angle less than 180°. If I understand your problem description correctly, you won't need to do this, since your field of view will always be less than 180°, but it's worth mentioning.)
Edited to add: as pointed out by beeglebug, a circle can intersect all three figures without intersecting their intersection. Oops. But I believe that this can only happen when the circle is behind the centre of the segment, as in the diagram below, and in this case we can apply the separating axis test for convex figures.
The separating axis theorem says that two convex figures fail to intersect if there exists a line such that one figure falls entirely on one side of the line, and the other figure on the other.
If any separating axis exists in this case, then the axis that's perpendicular to the line between the centre of the circle and the centre of the segment is a separating axis (as shown).
Let the centre of the segment be at the origin, let the circle have centre x and radius r, and let the two half-planes have (outward) normals n₁ and n₂. The circle is "behind" the segment if
x · n₁ > 0 and x · n₂ > 0
and the axis separates it from the segment if
|x| > r
I am programming an algorithm where I have broken up the surface of a sphere into grid points (for simplicity I have the grid lines parallel and perpendicular to the meridians). Given a point A, I would like to be able to efficiently take any grid "square" and determine the point B in the square with the least spherical coordinate distance AB. In the degenerate case the "squares" are actually "triangles".
I am actually only using it to bound which squares I am searching, so I can also accept a lower bound if it is only a tiny bit off. For this reason, I need the algorithm to be extremely quick otherwise it would be better to just take the loss of accuracy and search a few more squares.
I decided to repost this question to Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates. More progress has been made here
For points on a sphere, the points closest in the full 3D space will also be closest when measured along the surface of the sphere. The actual distances will be different, but if you're just after the closest point it's probably easiest to minimize the 3D distance rather than worry about great circle arcs, etc.
To find the actual great-circle distance between two (latitidude, longitude) points on the sphere, you can use the first formula in this link.
A few points, for clarity.
Unless you specifically wish these squares to be square (and hence to not fit exactly in this parallel and perpendicular layout with regards to the meridians), these are not exactly squares. This is particularly visible if the dimensions of the square are big.
The question speaks of a [perfect] sphere. Matters would be somewhat different if we were considering the Earth (or other planets) with its flattened poles.
Following is a "algorithm" that would fit the bill, I doubt it is optimal, but could offer a good basis. EDIT: see Tom10's suggestion to work with the plain 3D distance between the points rather than the corresponding great cirle distance (i.e. that of the cord rather than the arc), as this will greatly reduce the complexity of the formulas.
Problem layout: (A, B and Sq as defined in the OP's question)
A : a given point the the surface of the sphere
Sq : a given "square" from the grid
B : solution to problem : point located within Sq which has the shortest
distance to A.
C : point at the center of Sq
Tentative algorithm:
Using the formulas associated with [Great Circle][1], we can:
- find the equation of the circle that includes A and C
- find the distance between A and C. See the [formula here][2] (kindly lifted
from Tom10's reply).
- find the intersect of the Great Circle arc between these points, with the
arcs of parallel or meridian defining the Sq.
There should be only one such point, unless this finds a "corner" of Sq,
or -a rarer case- if the two points are on the same diameter (see
'antipodes' below).
Then comes the more algorithmic part of this procedure (so far formulas...):
- find, by dichotomy, the point on Sq's arc/seqment which is the closest from
point A. We're at B! QED.
Optimization:
It is probably possible make a good "guess" as to the location
of B, based on the relative position of A and C, hence cutting the number of
iterations for the binary search.
Also, if the distance A and C is past a certain threshold the intersection
of the cicles' arcs is probably a good enough estimate of B. Only when A
and C are relatively close will B be found a bit further on the median or
parallel arc in these cases, projection errors between A and C (or B) are
smaller and it may be ok to work with orthogonal coordinates and their
simpler formulas.
Another approach is to calculate the distance between A and each of the 4
corners of the square and to work the dichotomic search from two of these
points (not quite sure which; could be on the meridian or parallel...)
( * ) *Antipodes case*: When points A and C happen to be diametrically
opposite to one another, all great circle lines between A and C have the same
length, that of 1/2 the circonference of the sphere, which is the maximum any
two points on the surface of a sphere may be. In this case, the point B will
be the "square"'s corner that is the furthest from C.
I hope this helps...
The lazy lower bound method is to find the distance to the center of the square, then subtract the half diagonal distance and bound using the triangle inequality. Given these aren't real squares, there will actually be two diagonal distances - we will use the greater. I suppose that it will be reasonably accurate as well.
See Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates for an exact solution