So, for work I'm working on a robot controller that explores an arbitrary region. The region is defined by a series of vertices (it's a polygon). Here's an example:
The robot starts in the middle and tries to reach the outermost boundary and then follow it all the way around. However, due to the nature of terrain, it may be unable to reach certain areas, and can only explore a given region:
What I want to do is calculate all individual regions that have not been explored, and return the vertices that defines their boundaries, like this:
After this is computed, I should have a new array of polygons, containing the geometry for A, B, and C.
Unfortunately, I can't come up with a good, fast algorithm to do this. What's the best way to compute this?
One method is to define a predicate for a point p to be "touching" the boundary of the enclosing region, perhaps according to some tolerance ε > 0, e.g., T if and only if p is within a distance of ε of the boundary. Then traverse the boundary of the explored region, noting this predicate for each vertex: ..,T, T, T, F, F, F, F, F, T, T,... Then your regions are delimited by maximal strings of Fs, the two T vetices bounding those Fs, and the bounding region's boundary between. The string I just used as an example outlines your region B: five Fs.
ISTM that you're wanting the boolean difference of the outer bounding polygon minus the inner pink (explored) polygon. However there's no simple algorithm for this, so I suggest you look at and choose from the various polygon clipping libraries.
Here's a fairly recent comparison of a number of clipping libraries - http://rogue-modron.blogspot.com.au/2011/04/polygon-clipping-wrapper-benchmark.html
And also a shameless plug for my own open source freeware polygon clipper - http://angusj.com/delphi/clipper.php
Related
I'm a newbie to computer graphics so I apologize if some of my language is inexact or the question misses something basic.
Is it possible to calculate face normals correctly, given a list of vertices, and a list of faces like this:
v1: x_1, y_1, z_1
v2: x_2, y_2, z_2
...
v_n: x_n, y_n, z_n
f1: v1,v2,v3
f2: v4,v2,v5
...
f_m: v_j, v_k, v_l
Each x_i, y_i , z_i specifies the vertices position in 3d space (but isn't neccesarily a vector)
Each f_i contains the indices of the three vertices specifying it.
I understand that you can use the cross product of two sides of a face to get a normal, but the direction of that normal depends on the order and choice of sides (from what I understand).
Given this is the only data I have is it possible to correctly determine the direction of the normals? or is it possible to determine them consistently atleast? (all normals may be pointing in the wrong direction?)
In general there is no way to assign normal "consistently" all over a set of 3d faces... consider as an example the famous Möbius strip...
You will notice that if you start walking on it after one loop you get to the same point but on the opposite side. In other words this strip doesn't have two faces, but only one. If you build such a shape with a strip of triangles of course there's no way to assign normals in a consistent way and you'll necessarily end up having two adjacent triangles with normals pointing in opposite directions.
That said, if your collection of triangles is indeed orientable (i.e. there actually exist a consistent normal assignment) a solution is to start from one triangle and then propagate to neighbors like in a flood-fill algorithm. For example in Python it would look something like:
active = [triangles[0]]
oriented = set([triangles[0]])
while active:
next_active = []
for tri in active:
for other in neighbors(tri):
if other not in oriented:
if not agree(tri, other):
flip(other)
oriented.add(other)
next_active.append(other)
active = next_active
In CG its done by polygon winding rule. That means all the faces are defined so the points are in CW (or CCW) order when looked on the face directly. Then using cross product will lead to consistent normals.
However many meshes out there does not comply the winding rule (some faces are CW others CCW not all the same) and for those its a problem. There are two approaches I know of:
for simple shapes (not too much concave)
the sign of dot product of your face_normal and face_center-cube_center will tell you if the normal points inside or outside of the object.
if ( dot( face_normal , face_center-cube_center ) >= 0.0 ) normal_points_out
You can even use any point of face instead of the face center too. Anyway for more complex concave shapes this will not work correctly.
test if point above face is inside or not
simply displace center of face by some small distance (not too big) in normal direction and then test if the point is inside polygonal mesh or not:
if ( !inside( face_center+0.001*face_normal ) ) normal_points_out
to check if point is inside or not you can use hit test.
However if the normal is used just for lighting computations then its usage is usually inside a dot product. So we can use its abs value instead and that will solve all lighting problems regardless of the normal side. For example:
output_color = face_color * abs(dot(face_normal,light_direction))
some gfx apis have implemented this already (look for double sided materials or normals, turning them on usually use the abs value ...) For example in OpenGL:
glLightModeli(GL_LIGHT_MODEL_TWO_SIDE, GL_TRUE);
Assume you are given the equation of a line (in 2d), and the equations of lines that form a convex polygon (the polygon could be unbounded). How do I determine if the line intersects the polygon?
Furthermore, are there computational geometry libraries where such tasks are pre-defined? I ask because I'm interested not just in the 2D version but n-dimensional geometry.
For the 2D case, I think the problem simplifies a bit.
The line partitions the space into two regions.
If the polygon is present in only one of those regions, then the line does not intersect it.
If the polygon is present in both regions, then the line does intersect it.
So:
Take any perpendicular to the line, making the intersection with the
line the origin.
Project each vertex of the polytope onto the perpendicular.
If those projections occur with both signs, then the polygon
intersects the line.
[Update following elexhobby's comment.]
Forgot to include the handling of the unbounded case.
I meant to add that one could create a "virtual vertex" to represent the open area. What we really need is the "direction" of the open area. We can take this as the mean of the vectors for the bounding edges of the open area.
We then treat the dot product of that direction with the normal and add that to the set of vertex projections.
In geometry, typically see wikipedia a polygon is bounded.
What you are describing is usually called a polytope or a polyhedron see wikipedia
There are a few geometry libraries available, two that come to mind are boost (polygon) and CGAL. Generally, there is a distinct split between computational methods that deal with 2d,3d, and N-d - for obvious reasons.
For your problem, I would use a somewhat Binary Space Partitioning Tree approach. I would take the first line of your "poly" and trim the query line against it, creating a ray. The ray would start at the intersection of the two lines, and proceed in direction of the interior of the half-space generated by the first line of the "poly". Now I would repeat the procedure with the ray and the second line of the "poly". (this could generate a segment instead of ray) If at some point the ray (or now segment) origin lies on the outer side of a poly line currently considered and does not intersect it, then the answer is no - the line does not intersect your "poly". Otherwise it intersects. Take special care with various parallel edge cases. Fairly straight forward and works for multi-dimensional cases.
I am not fully sure, but I guess you can address this by use of duality. First normalize your line equations as a.x+b.y=1, and consider the set of points (a,b).
These must form a convex polygon, and my guess is that the new line may not correspond to a point inside the polygon. This is readily checked by verifying that the new point is on the same side of all the edges. (If you don't know the order of the lines, first construct the convex hull.)
Let's start from finite polygons.
To intersect polygon a line must intersect one of its edges. Intersection between line and an edge is possible only if two points lie on different sides from the line.
That can be easily checked with sign(cross_product(Ep-Lp,Ld)) for two points of the edge. Ep - edge point, Lp - some point on the line, Ld - direction vector of the line, cross_product(A,B)=Ax*By-Ay*Bx.
To deal with infinite polygons we may introduce "infinite points". If we have a half infinite edge with point E1 and direction Ed, its "second point" is something like E1+infinity*Ed, where infinity is "big enough number".
For "infinite points" the check will be slightly different:
cross_product(Ep-Lp,Ld)=
=cross_product(E1+infinity*Ed-Lp,Ld)=
=cross_product(E1-Lp+infinity*Ed,Ld)=
=cross_product(E1-Lp,Ld)+cross_product(infinity*Ed,Ld)=
=cross_product(E1-Lp,Ld)+infinity*cross_product(Ed,Ld)
If cross_product(Ed,Ld) is zero (the line is parallel to the edge), the sign will be determined by the first component. Otherwise the second component will dominate and determine the sign.
I have a project which takes a picture of topographic map and makes it a 3D object.
When I draw the 3D rectangles of the object, it works very slowly. I read about BSP trees and I didn't really understand it. Can someone please explain how to use BSP in 3D (maybe give an example)? and how to use it in my case, when some mountains in the map cover other parts so I need to organize the rectangles in order to draw them well?
In n-D a BSP tree is a spatial partitioning data structure that recursively splits the space into cells using splitting n-D hyperplanes (or even n-D hypersurfaces).
In 2D, the whole space is recursively split with 2D lines (into (possibly infinite) convex polygons).
In 3D, the whole space is recursively split with 3D planes (into (possibly infinite) convex polytopes).
How to build a BSP tree in 3D (from a model)
The model is made of a list of primitives (triangles or quads which is I believe what you call rectangles).
Start with an initial root node in the BSP tree that represents a cell covering the whole 3D space and initially holding all the primitives of your model.
Compute an optimal splitting plane for the considered primitives.
The goal of this step is to find a plane that will split the primitives into two groups of primitives of approximately the same size (either the same spatial extents or the same count of primitives).
A simple splitting strategy could be to chose a direction at random (which will be the normal of your plane) for the splitting. Then sort all the primitives spatially along this axis. And traverse the sorted list of primitives to find the position that will split the primitives into two groups of roughly equal size (i.e. this simply finds the median position from the primitives along this axis). With this direction and this position, the splitting plane is defined.
One typically used splitting strategy is however:
Compute the centroid of all the considered primitives.
Compute the covariance matrix of all the considered primitives.
The centroid gives the position of the splitting plane.
The eigenvector for the largest eigenvalue of the covariance matrix gives the normal of the splitting plane, which is the direction where the primitives are the most spread (and where the current cell should be split).
Split the current node, create two child nodes and assign primitives to each of them or to the current node.
Having found a suitable splitting plane in 1., the 3D space can be now be divided into two half-spaces: one positive, pointed to by the plane normal, and one negative (on the other side of the splitting plane). The goal of this step is to cut in half the considered primitives by assigning the primitives to the half-space where they belong.
Test each primitive of the current node against the splitting plane and assign it to either the left or right child node depending on whether it in the positive half-space or in the negative half-space.
Some primitives may intersect the splitting plane. They can be clipped by the plane into smaller primitives (and maybe also triangulated) so that these smaller primitives are fully inside one of the half-spaces and only belong to one of the cells corresponding to the child nodes. Another option is to simply attach the overlapping primitives to the current node.
Apply recursively this splitting strategy to the created child nodes (and their respective child nodes), until some criterion to stop splitting is met (typically not having enough primitives in the current node).
How to use a BSP tree in 3D
In all use cases, the hierarchical structure of the BSP tree is used to discard irrelevant part of the model for the query.
Locating a point
Traverse the BSP tree with your query point. At each node, go left or right depending on where the query point is located w.r.t. to the splitting plane of the node.
Compute a ray / model intersection
To find all the triangles of your model intersecting a ray (you may need this for picking your map), do something similar to 1.. Traverse the BSP tree with your query ray. At each node, compute the intersection of the ray with the splitting plane. Also check the primitives stored at the node (if any) and report the ones that intersect the ray. Continue traversing the children of this node that whose cell intersect your ray.
Discarding invisible data
Another possible use is to discard pieces of your model that lie outside the view frustum of your camera (that's probably what you are interested in here). The view frustum is exactly bounded by six planes and has 6 quad faces. Like in 1. and 2., you can traverse the BSP tree, check recursively which cell overlaps with the view frustum and completely discard the ones (and the corresponding pieces of your model) that don't. For the plane / view frustum intersection test, you could check whether any of the 6 quads of the view frustum intersect the plane, or you could conservatively approximate the view frustum with a bounding volume (sphere / axis-aligned bounding box / oriented bounding box) or even do a combination of both.
That being said, the solution to your slow rendering problem might be elsewhere (you may not be able to discard a lot of data with a 3D BSP tree for your model):
62K squares is not that big: if you're using OpenGL, you should however not draw these squares individually or continously stream the geometry to the GPU. You can put all the vertices in a single static vertex buffer and draw the quads by preparing a static index buffer containing the list of indices for the squares with either triangles or (better) triangle strips primitives to draw the corresponding squares in a single draw call.
Your data is highly structured (a regular grid with elevation). If you happen to have much larger data sets (that don't even fit in memory anymore), then you need not only spatial partitioning (that exploits the 2.5D structure of your data and its regularity, like a quadtree) but perhaps LOD techniques as well (to replace pieces of your data by a cheaper representation instead of simply discarding the data). You should then investigate LOD techniques for terrain rendering. This page lists a few resources (papers + implementations). A simplified Chunked LOD could be used as a starting point.
I'm using Unity, but the solution should be generic.
I will get user input from mouse clicks, which define the vertex list of a closed irregular polygon.
That vertices will define the outer edges of a flat 3D mesh.
To procedurally generate a mesh in Unity, I have to specify all the vertices and how they are connected to form triangles.
So, for convex polygons it's trivial, I'd just make triangles with vertices 1,2,3 then 1,3,4 etc. forming something like a Peacock tail.
But for concave polygons it's not so simple.
Is there an efficient algorithm to find the internal triangles?
You could make use of a constrained Delaunay triangulation (which is not trivial to implement!). Good library implementations are available within Triangle and CGAL, providing efficient O(n*log(n)) implementations.
If the vertex set is small, the ear-clipping algorithm is also a possibility, although it wont necessarily give you a Delaunay triangulation (it will typically produce sub-optimal triangles) and runs in O(n^2). It is pretty easy to implement yourself though.
Since the input vertices exist on a flat plane in 3d space, you could obtain a 2d problem by projecting onto the plane, computing the triangulation in 2d and then applying the same mesh topology to your 3d vertex set.
I've implemented the ear clipping algorithm as follows:
Iterate over the vertices until a convex vertex, v is found
Check whether any point on the polygon lies within the triangle (v-1,v,v+1). If there are, then you need to partition the polygon along the vertices v, and the point which is farthest away from the line (v-1, v+1). Recursively evaluate both partitions.
If the triangle around vertex v contains no other vertices, add the triangle to your output list and remove vertex v, repeat until done.
Notes:
This is inherently a 2D operation even when working on 3D faces. To consider the problem in 2D, simply ignore the vector coordinate of the face's normal which has the largest absolute value. (This is how you "project" the 3D face into 2D coordinates). For example, if the face had normal (0,1,0), you would ignore the y coordinate and work in the x,z plane.
To determine which vertices are convex, you first need to know the polygon's winding. You can determine this by finding the leftmost (smallest x coordinate) vertex in the polygon (break ties by finding the smallest y). Such a vertex is always convex, so the winding of this vertex gives you the winding of the polygon.
You determine winding and/or convexity with the signed triangle area equation. See: http://softsurfer.com/Archive/algorithm_0101/algorithm_0101.htm. Depending on your polygon's winding, all convex triangles with either have positive area (counterclockwise winding), or negative area (clockwise winding).
The point-in-triangle formula is constructed from the signed-triangle-area formula. See: How to determine if a point is in a 2D triangle?.
In step 2 where you need to determine which vertex (v) is farthest away from the line, you can do so by forming the triangles (L0, v, L1), and checking which one has the largest area (absolute value, unless you're assuming a specific winding direction)
This algorithm is not well defined for self-intersecting polygons, and due to the nature of floating point precision, you will likely encounter such a case. Some safeguards can be implemented for stability: - A point should not be considered to be inside your triangle unless it is a concave point. (Such a case indicates self-intersection and you should not partition your set along this vertex). You may encounter a situation where a partition is entirely concave (i.e. it's wound differently to the original polygon's winding). This partition should be discarded.
Because the algorithm is cyclic and involves partitioning the sets, it is highly efficient to use a bidirectional link list structure with an array for storage. You can then partition the sets in 0(1), however the algorithm still has an average O(n^2) runtime. The best case running time is actually a set where you need to partition many times, as this rapidly reduces the number of comparisons.
There is a community script for triangulating concave polygons but I've not personally used it. The author claims it works on 3D points as well as 2D.
One hack I've used in the past if I want to constrain the problem to 2D is to use principal component analysis to find the 2 axes of greatest change in my 3D data and making these my "X" and "Y".
Suppose there are a number of convex polygons on a plane, perhaps a map. These polygons can bump up against each other and share an edge, but cannot overlap.
To test if two polygons P and Q overlap, first I can test each edge in P to see if it intersects with any of the edges in Q. If an intersection is found, I declare that P and Q intersect. If none intersect, I then have to test for the case that P is completely contained by Q, and vice versa. Next, there's the case that P==Q. Finally, there's the case that share a few edges, but not all of them. (These last two cases can probably be thought of as the same general case, but that might not be important.)
I have an algorithm that detects where two line segments intersect. If the two segments are co-linear, they are not considered to intersect for my purposes.
Have I properly enumerated the cases? Any suggestions for testing for these cases?
Note that I'm not looking to find the new convex polygon that is the intersection, I just want to know if an intersection exists. There are many well documented algorithms for finding the intersection, but I don't need to go through all the effort.
You could use this collision algorithm:
To be able to decide whether two convex polygons are intersecting (touching each other) we can use the Separating Axis Theorem. Essentially:
If two convex polygons are not intersecting, there exists a line that passes between them.
Such a line only exists if one of the sides of one of the polygons forms such a line.
The first statement is easy. Since the polygons are both convex, you'll be able to draw a line with one polygon on one side and the other polygon on the other side unless they are intersecting. The second is slightly less intuitive. Look at figure 1. Unless the closest sided of the polygons are parallel to each other, the point where they get closest to each other is the point where a corner of one polygon gets closest to a side of the other polygon. This side will then form a separating axis between the polygons. If the sides are parallel, they both are separating axes.
So how does this concretely help us decide whether polygon A and B intersect? Well, we just go over each side of each polygon and check whether it forms a separating axis. To do this we'll be using some basic vector math to squash all the points of both polygons onto a line that is perpendicular to the potential separating line (see figure 2). Now the whole problem is conveniently 1-dimensional. We can determine a region in which the points for each polygon lie, and this line is a separating axis if these regions do not overlap.
If, after checking each line from both polygons, no separating axis was found, it has been proven that the polygons intersect and something has to be done about it.
There are several ways to detect intersection and / or containment between convex polygons. It all depends on how efficient you want the algorithm to be. Consider two convex polygons R and B with r and b vertices, respectively:
Sweep line based algorithm. As you mentioned you can perform a sweep line procedure and keep the intervals resulting from the intersection of the polygons with the sweeping line. If at any time the intervals overlap, then the polygons intersect. The complexity is O((r + b) log (r + b)) time.
Rotating Callipers based algorithm. See here and here for more details. The complexity is O(r + b) time.
The most efficient methods can be found here and here. These algorithms take O(log r + log b) time.
if the polygons are always convex, first calculate the angle of a line drawn from center to center of the polygons. you can then eliminate needing to test edge segments in the half of the polygon(s) 180 degrees away from the other polygon(s).
to eliminate the edges, Start with the polygon on the left. take the line segment from the center of the polygon that is perpendicular to the line segment from the previous bullet, and touches both sides of the polygon. call this line segment p, with vertexes p1 and p2. Then, for all vertexes if the x coordinate is less than p1.x and p2.x That vertex can go in the "safe bucket".
if it doesn't, you have to check to make sure it is on the "safe" side of the line (just check the y coordinates too)
-if a line segment in the polygon has all vertexes in the "safe bucket" you can ignore it.
-reverse the polarity so you are "right oriented" for the second polygon.
GJK collision detection should work.
Since altCognito already gave you a solution, I'll only point out an excellent book on computational geometry that might interest you.
Your test cases should work, since you're checking the case where the polygons don't intersect at all (completely outside or completely inside), as well as where there is any form of partial intersection (edges intersect always if there is overlap).
For testing, I would just make sure to test every potential combination. The one missing above from what I see is a single edge shared, but one poly contained in the other. I would also add tests for some more complex poly shapes, from tri -> many sided, just as a precaution.
Also, if you had a U shaped poly that completely surrounded the poly, but didn't overlap, I believe your case would handle that, but I would add that as a check, as well.
Here's an idea:
Find the center point of each polygon
Find the two points of each polygon closest to the center point of the other. They will be adjacent points in convex polygons. These define the nearest edge of each polygon, let's call the points {A, B} and {Y, Z}
Find the intersection of lines AB and YZ. If the line segments cross (the intersection on AB lies between A and B), your polygons intersect. If AB and XY are parallel ignore this condition, the next step will trap the problem.
There is one more case you need to check for, which is when the polygons intersect heavily enough that AB and XY are completely past each other and don't actually intersect.
To trap this case, calculate the perpendicular distances of AB and XY to each polygons center points. If either center point is closer to the opposite polygon's line segment your polygon overlap heavily.