I have two concave polygons on input represented as two vectors of points. I want to do some polygon operation on it - union, intersection and difference. I found intersection points between these polygons and insert them into the right place in each polygon. Then I give an information about its position (Inner - it is inside the other polygon, Outer - it is outside the other polygon, Intersection - point, where two edges of polygons intersects) to each vertex. Now I know which points create the union of these polygons (Outer and Intersection) etc., but I need to know how to sort them to the right order. In case of the intersection operation I need to divide these sorted points into the right number of sets, because the result of intersection could be more than one polygon.
I am using C++, but I don't need necessarily the code, I only want to need how to sort these final polygon points. And I don't want to use any library for these operations because I already have my own functions and want to use them.
I looked at this question How to intersect two polygons? and also some others but none of them is solving final sorting of points.
I also read this article http://www.gvu.gatech.edu/~jarek/graphics/papers/04PolygonBooleansMargalit.pdf , but I probably don't get it.
Any help would be appreciated.
If you follow all my recommendations from my comments:
Distinguish between intersection and touching points
Keep a link between the two equivalents of the intersection points in the two polygons
The solution for the union will be:
Consider only outer, touching and intersection points
Make sure the points in the two polygons are ordered in different direction (one of the sets is in clockwise direction, the other one in counter-clockwise)
Start from random point in any of the two polygons.
For every vertex in any of the two polygons keep if you have visited it
Every time you encounter an intersection point keep on traversing from the next to follow point in the other polygon after the equivalent of the intersection point.
If you come back to the point you started from this closes one of the components of the join of the two polygons. If not all vertices were traversed repeat the whole of it from any unvisited vertex.
In the end calculate the area of all the polygons you have found. The largest in area will be the real union. The rest will be holes in the union.
The solution for the join will be:
Consider only inner and intersection points
Make sure the points in the two polygons are ordered in the same direction
Start from random point in any of the two polygons.
For every vertex in any of the two polygons keep if you have visited it
Every time you encounter an intersection point keep on traversing from the next to follow point in the other polygon after the equivalent of the intersection point.
If you come back to the point you started from this closes one of the components of the join of the two polygons. If not all vertices were traversed repeat the whole of it from any unvisited vertex.
EDIT: As I already mentioned, I have the god feeling my approach with the polygon orientation needs to be revised. However, when searching through the web I found a description of algorithm that might do the work for you: The Vatti clipping algorithm
EDIT2 One more article describing such clipping algorithm.
Related
I'm trying to write an algorithm for cutting tessellated mesh with the given plane (plane defined with the point on the plane and unit normal vector). Also, this algorithm should triangulate all polygons and fill the hole after split.
I faced with a problem to find a polygon that lies on the plane (like the orange plane on the image)
I tried to process all edges of all triangles and find those that lies on the plane and stored them in an array. After that, I formed an array of vertices by searching next suitable edge.
Can someone explain an easier and faster way to find this polygon?
All vertices must be stored in CCW order.
Identify all edges that you cut by the indexes (or labels) of the endpoints. Make sure that every edge belongs to exactly two faces and is cut twice. Also make sure to orient the edge that results from the intersection consistently with the direction of the face normal.
Now the intersection edges form a chain that you can reconstruct by sorting: store the indexes of the endpoints separately, each with a link to the originating edge. After sorting on the indexes, the common vertices will appear in pairs in the sorted array. Using this structure, you can trace the polygon(s).
In the example below, from the faces aebf, bcgf, cdgh and dhea, you generate the edges ae-dh, bf-ae, cg-bf and dh-cg in some order. After splitting the endpoints and sorting, ae-, -ae, dh-, -dh, cg-, -cg, bf-, -bf, which generate the cycle ae-dh, dh-cg, cg-bf, bf-ae.
Given a set of points S (x, y, z). How to find the convex hull of those points ?
I tried understanding the algorithm from here, but could not get much.
It says:
First project all of the points onto the xy-plane, and find an edge that is definitely on the hull by selecting the point with highest y-coordinate and then doing one iteration of gift wrapping to determine the other endpoint of the edge. This is the first part of the incomplete hull. We then build the hull iteratively. Consider this first edge; now find another point in order to form the first triangular face of the hull. We do this by picking the point such that all the other points lie to the right of this triangle, when viewed appropriately (just as in the gift-wrapping algorithm, in which we picked an edge such that all other points lay to the right of that edge). Now there are three edges in the hull; to continue, we pick one of them arbitrarily, and again scan through all the points to find another point to build a new triangle with this edge, and repeat this until there are no edges left. (When we create a new triangular face, we add two edges to the pool; however, we have to first check if they have already been added to the hull, in which case we ignore them.) There are O(n) faces, and each iteration takes O(n) time since we must scan all of the remaining points, giving O(n2).
Can anyone explain it in a more clearer way or suggest a simpler alternative approach.
Implementing the 3D convex hull is not easy, but many algorithms have been implemented, and code is widely available. At the high end of quality and time investment to use is CGAL. At the lower end on both measures is my own C code:
In between there is code all over the web, including this implementation of QuickHull.
I would suggest first try an easier approach like quick hull. (Btw, the order for gift wrapping is O(nh) not O(n2), where h is points on hull and order of quick hull is O(n log n)).
Under average circumstances quick hull works quite well, but processing usually becomes slow in cases of high symmetry or points lying on the circumference of a circle. Quick hull can be broken down to the following steps:
Find the points with minimum and maximum x coordinates, those are
bound to be part of the convex.
Use the line formed by the two points to divide the set in two
subsets of points, which will be processed recursively.
Determine the point, on one side of the line, with the maximum
distance from the line. The two points found before along with this
one form a triangle.
The points lying inside of that triangle cannot be part of the
convex hull and can therefore be ignored in the next steps.
Repeat the previous two steps on the two lines formed by the
triangle (not the initial line).
Keep on doing so on until no more points are left, the recursion has
come to an end and the points selected constitute the convex hull.
See this impementaion and explanation for 3d convex hull using quick hull algorithm.
Gift wrapping algorithm:
Jarvis's match algorithm is like wrapping a piece of string around the points. It starts by computing the leftmost point l, since we know that the left most point must be a convex hull vertex.This process will take linear time.Then the algorithm does a series of pivoting steps to find each successive convex hull vertex untill the next vertex is the original leftmost point again.
The algorithm find the successive convex hull vertex like this: the vertex immediately following a point p is the point that appears to be furthest to the right to someone standing at p and looking at the other points. In other words, if q is the vertex following p, and r is any other input point, then the triple p, q, r is in counter-clockwise order. We can find each successive vertex in linear time by performing a series of O(n) counter-clockwise tests.
Since the algorithm spends O(n) time for each convex hull vertex, the worst-case running time is O(n2). However, if the convex hull has very few vertices, Jarvis's march is extremely fast. A better way to write the running time is O(nh), where h is the number of convex hull vertices. In the worst case, h = n, and we get our old O(n2) time bound, but in the best case h = 3, and the algorithm only needs O(n) time. This is a so called output-sensitive algorithm, the smaller the output, the faster the algorithm.
The following image should give you more idea
GPL C++ code for finding 3D convex hulls is available at http://www.newtonapples.net/code/NewtonAppleWrapper_11Feb2016.tar.gz and a description of the O(n log(n)) algorithm at http://www.newtonapples.net/NewtonAppleWrapper.html
One of the simplest algorithms for convex hull computation in 3D was presented in the paper The QuickHull algorithm for Convex Hulls by Barber, etc from 1995. Unfortunately the original paper lacks any figures to simplify its understanding.
The algorithm works iteratively by storing boundary faces of some convex set with the vertices from the subset of original points. The remaining points are divided on the ones already inside the current convex set and the points outside it. And each step consists in enlarging the convex set by including one of outside points in it until no one remains.
The authors propose to start the algorithm in 3D from any tetrahedron with 4 vertices in original points. If these vertices are selected so that they are on the boundary of convex hull then it will accelerate the algorithm (they will not be removed from boundary during the following steps). Also the algorithm can start from the boundary surface containing just 2 oppositely oriented triangles with 3 vertices in original points. Such points can be selected as follows.
The first point has with the minimal (x,y,z) coordinates, if compare coordinates lexicographically.
The second point is the most distant from the first one.
The third point is the most distant from the line through the first two points.
The next figure presents initial points and the starting 2 oppositely oriented triangles:
The remaining points are subdivided in two sets:
Black points - above the plane containing the triangles - are associated with the triangle having normal oriented upward.
Red points - below the plane containing the triangles - are associated with the triangle having normal oriented downward.
On the following steps, the algorithm always associates each point currently outside the convex set with one of the boundary triangles that is "visible" from the point (point is within positive half-space of that triangle). More precisely each outside point is associated with the triangle, for which the distance between the point and the plane containing the triangle is the largest.
On each step of algorithm the furthest outside point is selected, then all faces of the current convex set visible from it are identified, these faces are removed from the convex set and replaced with the triangles having one vertex in furthest point and two other points on the horizon ridge (boundary of removed visible faces).
On the next figure the furthest point is pointed by green arrow and three visible triangles are highlighted in red:
Visible triangles deleted, back faces and inside points can be seen in the hole, horizon ridge is shown with red color:
5 new triangles (joining at the added point) patch the hole in the surface:
The points previously associated with the removed triangles are either become inner for the updated convex set or redistributed among new triangles.
The last figure also presents the final result of convex hull computation without any remaining outside points. (The figures were prepared in MeshInspector application, having this algorithm implemented.)
I am running into another problem. I have an algorithm that implements the winding number algorithm to detect if a point lies within a polygon. This algorithm requires that the edges of the polygon are oriented in a counter clockwise fashion. I am currently doing this by checking if the center of the polygon is to the left of the edge. If not, then the line is fashioned clockwise and the sign of the result of the winding number algorithm is changed.
For most cases, this method works great. However, I am running into a case where the center is outside of the polygon. At this point, my algorithm breaks because some edges are being recorded as counter clockwise when they are in fact clockwise.
I have been looking at some resources to gain some inspiration:
How to determine if a list of polygon points are in clockwise order?
Ordering CONCAVE polygon vertices in (counter)clockwise?
https://math.stackexchange.com/questions/340830/clockwise-or-anticlockwise-edges-in-a-polygon
http://jeffe.cs.illinois.edu/teaching/373/notes/x05-convexhull.pdf
However, these resources deal mainly with convex polygons. I would need to develop a generic algorithm that can handle both convex and concave polygons. Although, if this doesn't exist (or can't be done), then I will settle with creating two separate algorithms and detecting if the polygon is concave/convex.
The ideal algorithm would simply detect if a particular edge is oriented clockwise or counter clockwise for a choosen polygon. but I am open to algorithms that will resort the edges and vertices in a counter clockwise flow for the polygon.
I have been considering an algorithm that not need to use the center point of the polygon as the result will be dependent on whether the center is inside/outside the polygon.
If you have any questions, please feel free to as in the comments and I will answer them as quickly as possible. Thank you!
Edit:
I should note that the orientation of the lines are random. Some will be counter clockwise. Others will be clockwise
Assuming that you have an unordered list of lines, and each one has a start vertex and an end vertex:
Create an empty list for your ordered lines.
Choose a line from the unordered list at random, move it from the unordered to the ordered list and make it your current line.
While there are lines remaining in the unordered list:
Find a line in the unordered list that has a start or end vertex equal to the end vertex of the current line and remove it from the list. Call this the next line.
If the start vertex of the next line is equal to the end vertex of the current line, add it to the end of the ordered list as is
Else if the end vertex of the next line is equal to the end vertex of the current line, swap the start and end vertices of the next line and add it to the end of the ordered list
Make the next line the new current line
Once complete, check the ordering of the entire list using the shoelace formula as described in Yves Daoust's answer, if it's negative then swap the start and end vertices of all the lines
You could use a hash table (using the vertex values as keys) to make finding the matching lines faster. If the lines are already in order (but some are around the wrong way) then it's just a matter of checking each line in turn against the previous line and swapping start and end if the start vertex of the current is not equal to the end vertex of the previous.
If you have more than one line sharing a single vertex then this algorithm won't work.
It suffices to compute the polygon area using the shoelace formula (without the absolute value). If the area is negative, reverse the order.
The basics of Weiler-Atherton Polygon Clipping algorithm are:
Start from the first edge which is going inside the clipping area.
When an edge of a candidate/subject polygon enters the clipping area, save the intersection point.
When an edge of a candidate/subject polygon exits the clipping area, save the intersection point and follow the clipping polygon.
How to distinguish between an inbound and an outbound edge of a polygon?
It seems like finding inbound edges invole another huge algorithm and thereby affects the efficiency of the algorithm.
Another question is, how can I find the first inbound intersection?
This answer seems to be shedding some light on the problem. But, sadly it doesn't work.
For example, if I reverse the direction of vectors, the angle is not negated.
https://www.wolframalpha.com/input/?i=angle+between+vector+%7B0%2C180%7D+%7B180%2C0%7D
https://www.wolframalpha.com/input/?i=angle+between+vector+%7B0%2C180%7D+%7B-180%2C0%7D
First, a reminder that the Weiler–Atherton algorithm uses polygons defined by vertices in a specific order, clockwise. In short, you test for edges going in or out by traversing the polygon clockwise. The first edge going in (and therefore the first inbound intersection) is simply the first edge you traverse which started outside the clipping area (see below).
Also, the algorithm is typically run in two phases. First find all intersections, these are added to a list of vertices for your polygons, inserted at the correct position. During this phase you would typically mark whether each vertex is within the other polygon. For the second phase, traverse the vertices to determine clipping polygons.
Lets try some examples. Take a triangle defined by vertices A,B,C, and a rectangle w,x,y,z. The triangle will be the clipping area, rectangle is the subject.
The list of points we have generated for the subject is therefore w,x,R,Q,y,z. The triangle list is now A,B,Q,C,R.
Starting at w, R is the first intersection, it is inbound because the previous point (x) is outside. The traversal of the area will be R,Q,C, and back to R(done).
The intersections are unlabeled here, but they will still be R and Q. The list of points we have generated for the subject is therefore w,x,R,y,Q,z. The triangle list is now A,B,C,Q,R.
The clipping traversal is R,y,Q, and R(done)
Let P and Q be two polygons. One can pick any vertex v of P in order to determine the position of v with respect to Q (i.e inside or outside it) via the ray casting algorithm (or any other algorithm that suits all the requirements of the problem).
You only need to determine the position of one such vertex v of P with respect to Q in this manner because the position of the other vertices of P can be inferred by iterating over the ordered set of vertices and intersection points of P.
Lets say v is outside Q. Then, by iterating over the ordered set of vertices and intersection points of P, the first intersection point one finds is laying on an entering edge. If v is inside Q, the first intersection point one finds is laying on an exiting edge. Keep in mind that one edge can be both entering and exiting, depending on the number of intersection points laying on it.
The idea behind the ray casting algorithm is simple, but one should pick vertex v of P if |V(P)|>=|V(Q)| and v of Q otherwise (in order to lower the impact the ray casting algorithm has on the overall performance, though not significantly).
You do not necessarily need to start at the first inbound intersection, this is fine when you are looking at the polygons drawn on a piece of paper and can drop your pen wherever you want, but as you noted would require more effort to find when coding it.
You just need to make sure you get all the intersections calculated for your two polygons first walking around the source polygons line segments checking for intersections with the clipping polygons line segments. At this point it does not matter whether it is inside or outside.
Once you have all the intersections and your two polygons points in order (I think I had two lists that could link to each other), walk around your source polygon point by point. If your first source polygon point is inside the clip polygon that is the first point of your solution polygon, if not the first point of your solution polygon is the first intersection with the clip polygon.
Once you have your first solution point each point from there is the next solution point. As you hit intersections you switch to the other polygon and carry on until you return back to your first solution point.
It has been a while since I have coded this, but if I remember correctly points that can catch you out are when polygons are entirely inside each other (in which case the contained one is your solution) and make sure you are prepared for more than one solution polygon if you have some odd polygon shapes.
I have a set of non-overlapping polygons. These polygons can share nodes, edges, but strictly no overlapping.
Now, I am going to mesh them using Constrainted Delaunay Triangulation (CDT) technique. I can get the mesh without problem.
My problem is, after the mesh, I want to know which mesh element belongs to which original polygon. MY current approach is to compute the centroid for each mesh element, and check which of the original polygon this centroid falls into. But I don't like this approach as it is very computationally intensive.
Is there any efficient ways to do this ( in terms of Big O the runtime)? My projects involve tens of thousands of polygons and I don't want the speed to slow down.
Edit: Make sure that all the vertices in a mesh element share a common face is not going to work, because there are cases where the all the vertices can have more than one common face, as below ( the dotted line forms a mesh element whose vertices have 2 common faces):
I can think of two options, both somehow mentioned :
Maintain the information in your points/vertices. See this other related question.
Recompute the information the way you did, locating each mesh element centroid in the original polygon, but this can be optimized by using a spatial_sort, and locating them sequentially in your input polygon (using the previous result as hint for starting the next point location).
What about labeling each of your original vertices with a polygon id (or several, I guess, since polys can share vertices). Then, if I understand DT correctly, you can look at the three verts in a given triangle in the mesh and see if they share a common label, if so, that mesh came from the labeled polygon.
As Mikeb says label all your original vertices with a polygon id.
Since you want the one that's inside the polygon, just make sure you only go clockwise around the polygons, this makes sure that if the points overlap for two polygons you get the one facing the correct direction.
I would expect this approach to remain close to O(n) where n represents number of points as each triangle can at only have one or two polygons that overlap all three points.
Create a new graph G(V,E) in the following way. For every mesh create a node in V. For every dashed edge create an edge in E that connects the two corresponding meshes. Don't map solid edges into edges in E.
Run ConnectedComponents(G).
Every mesh will be labeled with a label (with 1-to-1 correspondence to polygons.)
Maybe you can call CDT separately for each polygon, and label the triangles with their polygon after each call.