I have a question,
Given a set of points , how do you place a point with the constraint that the distance to the farthest point is as small as possible?.
This is in reference to this problem. I do not know how to proceed. Some pointers anyone?
Thanks
Check out this page. It describes several methods to do this.
http://www.personal.kent.edu/~rmuhamma/Compgeometry/MyCG/CG-Applets/Center/centercli.htm
In case the link above ever dies, here's the relevant part that describes the most straight-forward method:
An O(n2)-time Algorithm
Draw a circle at center, c, such that points of given set lie within the circle. Clearly, this circle can be made smaller (or else we have the solution).
Make a circle smaller by finding the point A farthest from the center of circler, and drawing a new circle with the same center and passing through the point A. These two steps produce a smaller enclosing circle. The reason that the new circle is smaller is that this new circle still contains all the points of given set, except now it passes through farthest point, x, rather than enclosing it.
If the circle passes through 2 or more points, proceed to step 4. Otherwise, make the circle smaller by moving the center towards point A, until the circle makes contact with another point B from the set.
At this stage, we a circle, C, that passes through two or more points of a given set. If the circle contains an interval (point-free interval) of arc greater than half the circle's circumference on which no points lie, the circle can be made smaller. Let D and E be the points at the ends of this point-free interval. While keeping D and E on the circle's boundary, reduce the diameter of the circle until we have either case (a) or case (b).
Case (a) The diameter is the distance DE.
We are done!
Case (b) The circle C touches another point from the set, F.
Check whether there exits point-free arc intervals of length more than half the circumference of C.
IF no such point-free arc intervals exit THEN
We are done!
Else
Goto step 4.
In this case, three points must lie on an arc less than half the circumference in length. We repeat step 4 on the outer two of the three points on the arc.
Another page here, with a sample applet:
http://www.sunshine2k.de/stuff/Java/Welzl/Welzl.html
You need to use the Voronoi diagram, possibibly the Farthest-Point Voronoi diagram, where the plane is divided into regions, where points in the same region have the same fartherst point
Update
You need to build a Farthest-Point voronoi diagram first, which is O(nlogn) time, and find the center of the smallest circle among all the vertices(if the circle is defined by three points) and all the edges(if the circle is defined by two points). The total time complexity of this approach is O(nlogn)
I just saw the Smallest circle problem wiki page, seems like there is a O(n) time algorithm. You can check it out if you care about the speed, otherwise never mind.
Related
This question is an extension on some computation details of this question.
Suppose one has a set of (potentially overlapping) circles, and one wishes to compute the area this set of circles covers. (For simplicity, one can assume some precomputation steps have been made, such as getting rid of circles included entirely in other circles, as well as that the circles induce one connected component.)
One way to do this is mentioned in Ants Aasma's and Timothy's Shields' answers, being that the area of overlapping circles is just a collection of circle slices and polygons, both of which the area is easy to compute.
The trouble I'm encountering however is the computation of these polygons. The nodes of the polygons (consisting of circle centers and "outer" intersection points) are easy enough to compute:
And at first I thought a simple algorithm of picking a random node and visiting neighbors in clockwise order would be sufficient, but this can result in the following "outer" polygon to be constructed, which is not part of the correct polygons.
So I thought of different approaches. A Breadth First Search to compute minimal cycles, but I think the previous counterexample can easily be modified so that this approach results in the "inner" polygon containing the hole (and which is thus not a correct polygon).
I was thinking of maybe running a Las Vegas style algorithm, taking random points and if said point is in an intersection of circles, try to compute the corresponding polygon. If such a polygon exists, remove circle centers and intersection points composing said polygon. Repeat until no circle centers or intersection points remain.
This would avoid ending up computing the "outer" polygon or the "inner" polygon, but would introduce new problems (outside of the potentially high running time) e.g. more than 2 circles intersecting in a single intersection point could remove said intersection point when computing one polygon, but would be necessary still for the next.
Ultimately, my question is: How to compute such polygons?
PS: As a bonus question for after having computed the polygons, how to know which angle to consider when computing the area of some circle slice, between theta and 2PI - theta?
Once we have the points of the polygons in the right order, computing the area is a not too difficult.
The way to achieve that is by exploiting planar duality. See the Wikipedia article on the doubly connected edge list representation for diagrams, but the gist is, given an oriented edge whose right face is inside a polygon, the next oriented edge in that polygon is the reverse direction of the previous oriented edge with the same head in clockwise order.
Hence we've reduced the problem to finding the oriented edges of the polygonal union and determining the correct order with respect to each head. We actually solve the latter problem first. Each intersection of disks gives rise to a quadrilateral. Let's call the centers C and D and the intersections A and B. Assume without loss of generality that the disk centered at C is not smaller than the disk centered at D. The interior angle formed by A→C←B is less than 180 degrees, so the signed area of that triangle is negative if and only if A→C precedes B→C in clockwise order around C, in turn if and only if B→D precedes A→D in clockwise order around D.
Now we determine which edges are actually polygon boundaries. For a particular disk, we have a bunch of angle intervals around its center from before (each sweeping out the clockwise sector from the first endpoint to the second). What we need amounts to a more complicated version of the common interview question of computing the union of segments. The usual sweep line algorithm that increases the cover count whenever it scans an opening endpoint and decreases the cover count whenever it scans a closing endpoint can be made to work here, with the adjustment that we need to initialize the count not to 0 but to the proper cover count of the starting angle.
There's a way to do all of this with no trigonometry, just subtraction and determinants and comparisons.
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.
I am given a set of N points in 2D plane represented as (x,y) coordinate pairs. What is a fast algorithm to choose three points so that the triangle formed by these points has maximum perimeter?
This is preemptive in nature
Pick a point farthest from the flock, lets call it point A.
draw an imaginary straight line that cuts through A to rest of the flock.
pick another opposite point, that its deviation(from the imaginary
straight line) is highest to right.
pick another opposite point, that
is deviation(from the imaginary straight line) is highest to the
left.
Check if a triangle can be made?.
if no check another highest point in another axis
Here's a rough idea (I'm not too versed in computational geometry). A triangle with a fixed perimeter and base can generate an ellipse. For example, here B and C are fixed and any point, A, on the ellipse will keep the triangle perimeter the same:
For each segment connecting two points, pick a random third point in our set. Generate the relevant ellipse, then pick another random point from our set that's outside that ellipse. Each ellipse will exclude points that generate triangles of the same or smaller perimeter until we run out of points, having found the largest. Of course, we would need some efficient methods to find relevant points (perhaps using space partitioning?).
Suppose I have a set of points in the Cartesian plane, defined by an array/vector of (X,Y) coordinates. This set of points will be "contiguous" in the coordinate plane, if any set of discontinuous points can be contiguous. That is, these points originated as a rectangular grid in which regions of points were eliminated by a prior algorithm. The shape outlined by the points is arbitrary, but it will tend to have arcs for edges.
Suppose further that I can create circles of fixed radius r.
I would like an algorithm that will find me the center X,Y for a circle that will enclose as close to exactly half of the given points as possible.
OK, try this (sorry if I have very bad wording: I didn't learn my Maths in english)
Step 1: Find axis
For all pairs of points, that are less than 2r apart calculate how many points lie on either side of the connecting line
Chose the pair with the worst balance
Calculate the line, that bisects these two points as an axis ("Axis of biggest concavity")
Step 2: Find center
Start on the axis far (>2r) away on the side, that had the lower point count in step 1 (The concave side)
Move the center on the axis, until you reach the desired point. This can be done by moving up with a step of sqrt(delta), where delta is the smallest distance between 2 points in the set, if overreaching move back halfing the step, etc.
You might want to look into the algorithm for smallest enclosing circle of a point set.
A somewhat greedy algorithm would be to simply remove points 1 at a time until the circle radius is less or equal to r.
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