Given two convex polygons P, Q separated by a line, how can I find their common tangents?
There should be 4 total.
Geometry isn't my strong side so any help will be appreciated :)
Any tangent will be generated from (at least) one point on each polygon. The tangent forms a half-space for each polygon, containing that entire polygon; different tangents are generated by requiring that each polygon be above or below the tangent line.
Whenever you have a problem involving two convex polygons, the algorithm probably involves "pick a point on each one, then iterate." This algorithm is no different.
What you do, basically, is start with a random guess, and refine. Pick a vertex on each polygon and calculate the line through the two vertices. Look at polygon A and see whether either of the two neighboring vertices is on the wrong side of the line. If one is, replace the current vertex with that vertex. Then do the same check on polygon B. Then, if you updated either vertex, repeat. Eventually the lines will converge to tangents with each vertex the extremal one in the desired direction.
Related
I am working with several convex polygons that overlap each other and I need to combine them back together to form one single polygon that may be convex or concave.
The problem is always as follows:
1) The polygons that I need to merge together are always convex.
2) The vertices of each polygon are defined in clockwise order.
3) The polygons are never in any specific order.
4) The final polygon can only be simple convex or concave polygon, i.e. no self-intersection, no duplicate vertices or holes in the shape.
Here is an example of the kind of polygons that I am working with.
![overlapping convex polygons]"image removed")
My current approach is to start from the first polygon and vertex by vertex I loop through all vertices of all of the polygons to find overlap. If there is no overlap, I store the vertex for the final outline and continue.
Upon finding overlapping vertices, I determine which polygon to continue to by measuring the angles of the possible paths and by choosing the one that leads towards the outside of the shape.
This method works until I encounter polygons that do not have vertices overlapping each other, but instead one polygon's vertex is overlapping another polygon's side, as is the case with the rectangle in the image.
I am currently planning on solving these situations by running line intersect checks for all shapes that I have not yet processed, but I am convinced that this cannot be the easiest or the best method in terms of performance.
Does someone know how I should approach this problem in a more efficient manner and/or universal manner?
I solved this issue and I'm posting the answer here in case someone else runs into this issue as well.
My first step was to implement a pre-processing loop based on trincot's suggestions.
I calculated the minimum and maximum x and y bounds for each individual shape.
I used these values to determine all overlapping shapes and I stored a simple array for each shape that I could later use to only look at shapes that can overlap each other.
Then, for the actual loop that determines the outline of the final polygon:
I start from the first shape and simply compare its vertices to those of the nearby shapes. If there is at least one vertex that isn't shared by another vertex, it must be on the outer edge and the loop starts from there. If there are only overlapping vertices, then I add the first shape to a table for all checked shapes and repeat this process with another shape until I find a vertex that is on the outer edge.
Once the starting vertex is found, the main loop will check the vertices of the starting shape one by one and measure how far from the given vertex is from every nearby shapes' edges. If the distance is zero, then the vertex either overlaps with another shape's vertex or the vertex lies on the side of another shape.
Upon finding the aforementioned type of vertex, I add the previous shape's number to the table of checked shapes so that it isn't checked again. Then, I check if there are other shapes that share this particular vertex. If there are, then I determine the outermost shape and continue from there, starting back from step 2.
Once all shapes have been checked, I check that all non-overlapping vertices from the starting shape were indeed added to the outline. If they weren't, I add them at the end.
There may be computationally faster methods, but I found this one to be simple to write, that it meets all of my requirements and it is fast enough for my needs.
Given a vertex, you could speed up the search of an "overlapping" vertex or edge as follows:
Finding vertices
Assuming that the coordinates are exact, in the sense that if two vertices overlap, they have exactly the same x and y coordinates, without any "error" of imprecision, then it would be good to first create a hash by x-coordinate, and then for each x-entry you would have a hash by y-coordinate. The value of that inner hash would be a list of polygons that have that vertex.
That structure can be built in O(n) time, and will allow you to find a matching vertex in constant time.
Only if that gives no match, you would go to the next algorithm:
Finding edges
In a pre-processing step (only once), create a segment tree for these polygons where a "segment" corresponds to a min/max x-coordinate range for a particular polygon.
Given a vertex, use the segment tree to find the polygons that are in the right x-coordinate range, i.e. where the x-coordinate of the vertex is within the min/max range of x-coordinates of the polygon.
Iterate those polygons, and eliminate those that do not have an y-coordinate range that has the y-coordinate of the vertex.
If no polygons remain, the vertex does not participate in any edge of another polygon.
You cannot get more than one polygon here, since that would mean another polygon shares the vertex, which is a case already covered by the hash-based algorithm.
If you get just one polygon, then continue your search by going through the edges of that polygon to find a match -- which is what you already planned on doing (line intersect check), but now you would only need to do it for one polygon.
You could speed that line intersect check up a little bit by first filtering the edges to those that have the right x-range. For convex polygons you would end up with at most two edges. At most one of those two will have the right y-range. If you get such an edge, check whether the vertex is really on that edge.
Given set of 2D points find a triangle built from those points, that encloses the biggest number of points.
Brutal algorithm for this is just building triangles from every possible triad of points and checking how many points they enclose, but time complexity of this solution is O(n^4).
For the optimal solution I thought about first finding the convex hull of those points and arranging points inside this hull with some structure, but I can't figure it out.
Do you have any ideas about the optimal solution for this kind of problem?
In a set of n points, there are (n choose 3) triangles, and using brute force to check for each point whether it is contained in each triangle indeed has O(n4) complexity. To give a practical example of a few set sizes:
points: 100 1,000 10,000
triangles: 161,700 166,167,000 166,616,670,000
checks: 15,684,900 165,668,499,000 1,665,666,849,990,000
Below are a few geometrical ideas; they don't lead straight to a solution, but they can reduce the number of triangles that have to be checked.
Counter-example for convex hull
First of all, using only points on the convex hull is not guaranteed to give the optimal solution. Consider this counter-example:
The convex hull is the red rectangle. However, if we use two of its sides and a diagonal to form a triangle, the diagonal will cut through the central point cluster and leave out some of the points. And even if we only use 1 or 2 corners of the rectangle, combined with a point in the center, it will always cut through the blue triangle and leave out some points. The blue triangle, which has no points on the convex hull, is in fact the optimal solution.
Triangle contained in triangle
If you consider a triangle abc, and three points d, e and f contained within it, then the triangle def cannot be the triangle which contains the most points, because triangle abc contains at least three more points. Triangles made from a combination of points from abc and def, like abd, also contain fewer points than abc.
This means that finding a triangle and some points contained within it, allows you to discard a number of triangles. In the next paragraphs, we will use this idea to discard as many triangles as possible from having to be checked.
Expanding a triangle
If we consider a triangle made from three randomly chosen points a, b and c (named clock-wise), and then check whether all other points are on the left of right side of the lines |ab|, |bc| and |ca|, the points are partitioned into 7 zones:
If we replace a corner point of the triangle by a point in the adjacent coloured zone, e.g. zone LRL for point a, we get a larger triangle that contains triangle abc. If we randomly pick three points from zones LRL, LLR and RLL, we can expand the triangle like this:
We can then partition the points again using this new triangle a'b'c' (points already in zone RRR can be added to the new zone RRR without checking) and expand the triangle again as long as there is at least one point in the zones LRL, LLR or RLL.
If we have caught enough points inside the expanded triangle, we can now use the brute force algorithm, but skip any triangle which doesn't have a point outside of the expanded triangle a'b"c'.
If we haven't caught enough points to make that feasible, we can try again with another three random points. Note, however, that you should not use the union of the points contained within several triangles; three points which are each contained in another triangle, but not in the same triangle, can still be the triangle containing the most points.
Excluding triangles in multiple steps
We could repeatedly choose a random triangle, expand it maximally, and then mark the triangles made from three points on or inside the triangle, to then exclude these from the check. This would require storing a boolean for all possible triangles, e.g. in a 3D bit array, so it is only feasible for sets up to a few thousand points.
To simplify things, instead of expanding random triangles, we could do this with a number of randomly chosen triangles, or triangles made from points on the convex hull, or points far apart when sorted in the x or y-direction, or ... But any of these methods will only help us to find triangles which can be excluded, they will not give us optimal (or even good-enough) triangles by themselves.
Say I have a polygon. It can be a convex one or not, it doesn't matter, but it doesn't have holes. It also has "inner" vertices and edges, meaning that it is partitioned.
Is there any kind of popular/known algorithm or standard procedures for when I want to check if a point is inside that kind of polygon?
I'm asking because Winding Number and Ray Casting aren't accurate in this case
Thanks in advance
You need to clarify what you mean by 'inner vertices and edges'. Let's take a very general case and hope that you find relevance.
The ray casting (point in polygon) algorithm shoots off a ray counting the intersections with the sides of the POLYGON (Odd intersections = inside, Even = outside).
Hence it accurately gives the correct result regardless of whether you start from inside the disjoint trapezoidal hole or the triangular hole (inner edges?) or even if a part of the polygon is completely seperated and/or self intersecting.
However, in what order do you feed the vertices of the polygon such that all the points are evaluated correctly?
Though this is code specific, if you're using an implementation that is counting every intersection with the sides of the polygon then this approach will work -
- Break the master polygon into polygonal components. eg - trapezoidal hole is a polygonal component.
- Start with (0,0) vertex (doesn't matter whether (0,0) actually lies wrt your polygon) followed by the first component' vertices, repeating its first vertex after the last vertex.
- Include another (0,0) vertex.
- Include the next component , repeating its first vertex after the last vertex.
- Repeat the above two steps for each component.
- End with a final (0,0) vertex.
2 component eg- Let the vertices of the two components be (1x,1y), (2x,2y), (3x,3y) and (Ax,Ay), (Bx,By), (Cx,Cy). Where (Ax,Ay), (Bx,By), (Cx,Cy) could be anything from a disjoint triangular hole, intersecting triangle or separated triangle.
Hence , the vertices of a singular continous polygon which is mathematically equivalent to the 2 components is -
(0,0),(1x,1y),(2x,2y),(3x,3y),(1x,1y),(0,0),(Ax,Ay),(Bx,By),(Cx,Cy),(Ax,Ay),(0,0)
To understand how it works, try drawing this mathematically equivalent polygon on a scratch pad.-
1. Mark all the vertices but don't join them yet.
2. Mark the repeated vertices separately also. Do this by marking them close to the original points, but not on them. (at a distance e, where e->0 (tends to/approaches) ) (to help visualize)
3. Now join all the vertices in the right order (as in the example above)
You will notice that this forms a continuous polygon and only becomes disjoint at the e=0 limit.
You can now send this mathematically equivalent polygon to your ray casting function (and maybe even winding number function?) without any issues.
I'm facing the following problem: I'm given a set of coordinates on an integer grid that define the vertices of a polygon. The polygon is guaranteed to be convex. It's proven that such a polygon can always be cut into 4 equal area parts by 2 orthogonal lines. Let's call the point of these lines' intersection P.
Given that set, I should calculate the coordinates of P within the polygon and the angle the lines need to be turned on so that the lines cut the polygon into 4 equal parts.
I realise that, put generally, the cake cutting problem has no "good" solution. But this particular case of it should.
I've searched for an algorithm to solve that problem, but found nothing useful.
Where should I look?
My approach would be to calculate the coordinates of the centre of the polygon (that can be done more or less easily), place Pthere and then "wiggle" the lines until the areas of the parts match. But that sounds too inelegant.
UPD: that's the problem I'm dealing with. Perhaps this question should be suspended until I come up with actual code questions.
Here is a partial sketch of the solution:
Choose an arbitrary direction and find the line parallel to that direction that splits the polygon in two. To achieve this, draw a line by every vertex to decompose the polygon in slabs. The respective areas of the slabs will tell you what slab the desired line intersects. Simple linear interpolation will give the exact location of the line.
Now your polygon is split in two convex polygons. For each halve, repeat the above procedure using the perpendicular direction. In general, you will get two distinct splitters, and what remains to be done is to find the direction such that they do coincide.
In the given direction, the splitters intersect four specific edges of the polygon. If you slightly rotate, they still intersect the same four edges. You can decompose a full turn in angular ranges such that the four intersected edges remain the same.
Knowing the four intersected edges, you can establish the relation that tells you the distance between the two perpendicular splitters as a function of the angle. Then you can compute the angle at which the two splitters coincide, and check if this angle belongs to the range defined for these edges.
By trying all ranges in turn, you will find the solution.
Note: the limits of the angular ranges correspond to directions parallel or perpendicular to the lines joining two vertexes.
I need to compute the area of the region of overlap between two triangles in the 2D plane. Oddly, I have written up code for the triangle-circle problem, and that works quite well and robustly, but I have trouble with the triangle-triangle problem.
I already first check to see if one entirely contains the other, or if the other contains the first, as well as obtain all the edge-wise intersection points. These intersection points (up to 6, as in the star of David), combined with the triangle vertices that are contained within the other triangle, are the vertices of the intersection region. These points must form a convex polygon.
The solution I seek is the answer to either of these questions:
Given a set of points known to all lie on the convex hull of the point set, compute the area of the convex hull. Note that they are in random order.
Given a set of half-planes, determine the intersecting area. This is equivalent to describing both triangles as the intersection of three half-planes, and computing the solution as the direct intersection of this description.
I have considered for question 1 simply adding up all areas of all possible triangles, and then dividing by the multiplicity in counting, but that seems dumb, and I'm not sure if it is correct. I feel like there is some kind of sweep-line algorithm that would do the trick. However, the solution must also be relatively numerically robust.
I simply have no idea how to solve question 2, but a general answer would be very useful, and providing code would make my day. This would allow for direct computation of intersection areas of convex polygons instead of having to perform a triangle decomposition on them.
Edit: I am aware of this article which describes the general case for finding the intersection polygon of two convex polygons. It seems rather involved for just triangles, and furthermore, I don't really need the resulting polygon itself. So maybe this question is just asked in laziness at this point.
Question 1: why are the points in a random order? If they are, you have to order them so that connecting consecutive points with straight lines yields a convex polygon. How to order them -- for example, by running a convex hull algorithm (though there are probably also simpler methods). Once you have ordered them, compute the area as described here.
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Question 2 is simpler. Half-plane is defined by a single line having an implicit equation a*x+b*y+c=0 ; all points (x, y) for which a*x+b*y+c <= 0 (note the inequality) are "behind" the half-plane. Now, you need at least three planes so that the intersection of their negative half-spaces is closed (this is necessary, but not sufficient condition). If the intersection is closed, it will be a convex polygon.
I suggest that you maintain a linked list of vertices. The algorithm is initialized with THREE lines. Compute the three points (in general case) where the lines intersect; these are the starting vertices of your region (triangle). You must also check that each vertex is "behind" the half-plane defined by the line going through the other two vertices; this guarantees that the intersection actually IS a closed region.
These three vertices define also the the three edges of a triangle. When you intersect by a new half-plane, simply check for the intersection between the line defining the half-plane and each of the edges of the current region; in general you will get two intersection points, but you must watch out for degenerate cases where the line goes through a vertex of the region. (You can also end up with an empty set!)
The new intersection vertices define a line that splits the current region in TWO regions. Again, use orientation of the new half-plane to decide which of the two new regions to assign to the new "current region", and which one to discard.
The points in the list defining the edges of the current region will be correctly ordered so you can apply the formula in the above link to compute its area.
If this description is not detailed/understandable, the next-best advice I can give you is that you invest in a book on computational geometry and linear algebra.