To take a simple example, say there is 2 bounding boxes (not necessarily axis aligned), each defined by 6 planes.
Is there a good way to determine if the volumes defined by each set of planes overlap?
(Only true/false, no need for the intersecting volume).
A solution to this problem, if its general should be able to scale up to many sets of planes too.
So far the solutions I've come up with basically rely on converting each set of planes into geometry - (vertices & polygons), then performing the intersection as you would if you have to intersect any 2 regular meshes. However I was wondering if there was a more elegant method that doesn't rely on this.
The intersection volume (if any) is the set of all points on the right side of all planes (combined, from both volumes). So, if you can select 3 planes whose intersection is on the right side of all the remaining planes, then the two volumes have an intersection.
This is a linear programming problem. In your case, you only need to find if there is a feasible solution or not; there are standard techniques for doing this.
You can determine the vertices of one of your bodies by mutually intersecting all possible triples that its planes form, and then check whether each of the resulting vertices lies on the good side of the planes defining the second body. When each of the second body's planes is given as base vertex p and normal v, this involves checking whether (x-p).v>=0 .
Assume that your planes are each given as base vertices (p,q,r) and normals (u,v,w) respectively, where the normals form the columns of a matrix M, the intersection is x = inv(M).(p.u, q.v, r.w).
Depending on how regular your two bodies are (e.g. parallelepipeds), many of the dot products and matrix inverses can be precomputed and reused. Perhaps you can share some of your prerequisites.
Posting this answer since this is one possible solution (just from thinking about the problem).
first calculate a point on each plane set (using 3 planes), and simply check if either of these points is inside the other plane-set.This covers cases where one volume is completely inside another, but won't work for partially overlapping volumes of course.
The following method can check for partial intersections.
for one of the sets, calculate the ray defined by each plane-plane pair.
clip the each of these rays by the other planes in the set, (storing a minimum and maximum value per ray).
discard any rays that have a minimum value greater then their maximum.The resulting rays represent all 'edges' for the volume.
So far all these calculations have been done on a single set of planes, so this information can be calculated once and stored for re-use.
Now continue clipping the rays but this time use the other set of planes, (again, discarding rays with a min greater then the maximum).
If there are one or more rays remaining, then there is an intersection.
Note 0): This isn't going to be efficient for any number of planes, (too many On^2 checks going on). In that case converting to polygons and then using more typical geometry tree structures makes more sense.
Note 1): Discarding rays can be done as the plane-pairs are iterated over to avoid first having to store all possible edges, only to discard many.
Note 2): Before clipping all rays with the second set of planes, a quick check could be made by doing a point-inside test between the plane-sets (the point can be calculated using a ray and its min/max). This will work if one shape is inside another, however clipping the rays is still needed for a final result.
Related
i have two concentric circles and three points are given for each circle that are on circumference.
I need a optimized method to check if a given random point exist inbetween these circles or not.
You can compute (x²+y²), x, y, 1 for each point. The last entry is simply the constant one. Put these terms for four given points into a matrix and compute its determinant. The determinant will be zero if the points are cocircular. Otherwise the sign will tell you which point is on which side with respect to the circle defined by the other three. Use a simple example to check which sign corresponds to which direction. Be prepared for the fact that the three circle-defining points being oriented in a clockwise or counter-clockwise orientation will affect this sign, too.
Computing a 4×4 determinant can be done horribly inefficiently, too. I'd suggest you compute all the 2×2 minors from the first two rows, and all the 2×2 minors from the last two, then you can combine them to form the full determinant. See this Math SE post for details. If you need further mathematical help (as opposed to programming help), you might find more suitable answers there.
Nothe that the above works for each circle independently. Check whether the point is inside the one, then check whether it is outside the other. It does not make use of the fact that the circles are assumed to be cocircular.
Given some points in plane (upto 500 points), no 3 collinear. We have to determine the number of triangles whose vertices are from the given points and that contain exactly N points inside them. How to efficiently solve this problem? The naive O(n^4) algorithm is too slow. Any better approach?
You could try thinking of the triangle as the intersection of three half-spaces. To find the number of points inside a triangle A, B, C first consider the set of points on one side of the infinite line in direction AB. Let these sets L(AB) and R(AB) for points of the left and right. Similarly you the same with other two edges and build sets L(AC) and R(AC) and sets L(BC) and R(BC).
So the number of points in ABC will be the number of points in the intersection of L(AB), L(AC) and L(BC). (You might want to consider R(AB) instead depending on the orientation of the triangle).
Now if we want to consider the full set of 500 points. First take all pairs of points AB and construct the sets L(AB) and R(AB). This will take O(n^3) operations.
Next we test all triangles and find the intersections of the three sets. If we use some hash table structure for the sets then to find the intersection points is like a hashtable lookups. If L(AB) has l elements, L(AC) has m elements and L(BC) n elements. Say l > m > n. For each point in L(BC) we need to do a lookup in L(AC) and L(BC) so thats a maximum of 2n hashtable lookups.
It might be faster to consider a geometric lookup table.
Divide your whole domain into a coarse grid say a 10 by 10 grid. We can then put each point into a set G(i,j). We can then split the sets L(AB) into each grid cell. Say call these sets L(AB,i,j) and R(AB,i,j). In testing for intersections first workout which grid cells lie in the intersection. This dramatically reduces the search space and as each set L(AB,i,j) contain fewer members there will be fewer hashtable lookups.
Actually I happened to encounter similar problem recently but the only difference was that there were around 300 pts and I solved it using bitset (C++ STL). For every pair of points, say (x[i],y[i]) and (x[j],y[j]), I formed a bitset<302>B[i][j] and B[i][j][k] stores 1 if k-th point is above line segment from point i to point j else I would store 0.
Now in a brute force manner I get three points so as to form a triangle, lets say (x[i],y[i]), (x[j],y[j]) and (x[k],y[k]), then a point,say z-th point ,would be inside triangle if B[i][j][z]==B[i][j][k] && B[j][k][z]==B[j][k][i] && B[k][i][z]==B[k][i][j] because a point inside triangle would show similar sign w.r.t. a side of triangle as the third point of triangle(one which is not on this side).
So i get three bitset variables P=B[i][j], Q=B[j][k] and R=B[k][i] and there taking there bitwise AND then applying count() function to give me the active number of bits and hence the number of points within the triangle. But make sure you change variable P such that it gives B[i][j][k]=1 if not then take bitwise not (~) of this variable.
Though the above solution is problem specific, i hope it helps. This is the problem link: http://usaco.org/current/index.php?page=viewproblem&cpid=660
I am trying to find sphere that surly encompasses given list of points.
Points will have x, y and z co-ordinate[Points are in 3D].
Actually I am trying to find new three points based on given list of points by some calculations like find MinX,MaxX ,MinY,MaxY,and MinZ and MaxZ and do some operation and find new three points
And I will draw sphere from these three points.
And I will also taking all these three points on the diameter of sphere so I have a unique sphere.
Is there any standard way for finding encompassing sphere of given list of points?
Yes, the standard algorithm is Welzl's algorithm (assuming you want the minimal sphere around your points). Particularly the improved version of Gaertner is very useful, robust and numerically stable! It handles all the degenerate cases well too.
At its core, the algorithm permutes the points (randomly) to find the 1-4 points that lie on the boundary of the sphere. It's basically a clever trial-and-error algorithm. From these points, you can find the center by finding a point that has the same distance to all those points. Gärtner's version uses an improved numerical device to find the center. Also, it employs an extra pivoting step that presumably makes the algorithm work better for a large number of input points.
If all you want is a sphere around three points, I suggest you still use Gärtners "device" to compute the circumsphere of the triangle. Otherwise, the method will probably degenerate easily (i.e. when the triangle is very flat).
Do you need 3 points, or any number of points?
If you only need the answer for 3 points, each pair of points defines a line segment. Take the longest line segment. Take a sphere centered at the middle of that line segment, whose radius is half the length of the line segment. There are two cases.
The third point is inside of that initial sphere. If so, then you have the smallest sphere.
The third point is outside of that initial sphere. Then the solution at Find Circum Center of Three point of Triangle [Not using Compass] will give you the center of the smallest sphere containing those 3 points.
If you need an arbitrary number of points, I'd do some sort of iterative approximation algorithm. Since you don't seem like you need that, I won't work out the details.
I am taking Computer Graphics course.In 3D,I have a point and a polygon and I want to determine this point is located above or below my polygon.Thanks for your replies,in advance.
If above or below the plane on which the polygon is resting will do, you can compare the dot product of the point onto the plane normal and that of any point on the plane. Or look at the sign of the dot product between the normal and a vector from a point on the plane to the point, if you prefer.
To check whether it is actually 'above' or 'below' in the sense of being directly above or below (ie, not off to the side somewhere) then do a point in polygon by projecting the whole thing into 2d along the normal and then a distance along normal test.
It depends on your definition of above and below, let me first talk about the easy case:
If you think of above/below in terms of a global direction (typically the y-axis or z-axis), just compare the values on that axis.
Ok, now the more difficult interpretation: On which side of the polygon is the point.
Unless it is complanar you cannot decide it for the polygon at once. So if it is non-complanar you have to tesselate it into triangles and decide for each of them.
For a triangle you can decide whether a point is above or below it (in 3D), first calculate the cross product of 2 vectors that make up the sides of the triangle; this will define a direction (= the definition of "above" and "below"), this depends on the order in which you use those 2 vectors so be careful. Then calculate the dot-product of the new vector (which is called the perpendicular of that triangle) and the difference-vector of the point-to-test and the triangle-base.
If I have 5 Vertices in 3D coordinate space how can I determined the ordering of those Vertices. i.e clockwise or anticlockwise.
If I elaborate more on this,
I have a 3D model which consists of set of polygons. Each polygon is collection of vertices and I want to calculate the norm of the polygon surface. To calculate the norm I have to consider the vertices in counter clockwise order . My question is given set of vertices how can I determine whether it is ordered in clockwise or counter clockwise?
This is for navigation mesh generation where I want to remove the polygons which cannot be walked by the agent. To do so my approach is to calculate the surface norm(perpendicular vector of the polygon) and remove the polygon based on the angle with 2D plane. To calculate the norm I should know in which order points are arranged. So for given set of points in polygon how can I determine the order of the arrangement of points.
Ex.
polygon1 consist of Vertex1 = [-21.847065 -2.492895 19.569759], Vertex2 [-22.279873 1.588395 16.017160], Vertex3 [-17.234818 7.132950 7.453146] these 3 points and how can I determine the order of them
As others have noted, your question isn't entirely clear. Is the for something like a 3D backface culling test? If so, you need a point to determine the winding direction relative to. Viewed from one side of the polygon the vertices will appear to wind clockwise. From the other side they'll appear to wind counter clockwise.
But suppose your polygon is convex and properly planar. Take any three consecutive vertices A, B, and C. Then you can find the surface normal vector using the cross product:
N = (B - A) x (C - A)
Taking the dot product of the normal with a vector from the given view point, V, to one of the vertices will give you a value whose sign indicates which way the vertices appear to wind when viewed from V:
w = N . (A - V)
Whether this is positive for clockwise and negative for anticlockwise or the opposite will depend on the handedness of your coordinate system.
Your question is too poorly defined to give a complete answer, but here's the skeleton of one.
The missing part (the meat if you will), is a function that takes any two coordinates and tells you which one is 'greater' than the other. Without a solid definition for this, you won't be able to make anything work.
The rest, the skeleton, is pretty simple. Sort your list of vectors using your comparison function. For five vectors, a simple bubble sort will be all you need, although if the number of vertices increases considerably you may want to look into a faster sorting algorithm (ie. Quicksort).
If your chosen language / libraries provide sorting for you, you've already got your skeleton.
EDIT
After re-reading your question, it also occurred to me that since these n vertices define a polygon, you can probably make the assumption that all of them lie on the same plane (if they don't, then good luck rendering that).
So, if you can map the vector coordinates to 2d positions on that plane, you can reduce your problem to ordering them clockwise or counterclockwise in a two dimensional space.
I think your confusion comes from the fact that methods for computing cross products are sometimes taught in terms of clockwiseness, with a check of the clockwiseness of 3 points A,B,C determining the sign of:
(B-A) X (C - A)
However a better definition actually determines this for you.
In general 5 arbitrary points in 3 dimensions can't be said to have a clockwise ordering but 3 can since 3 points always lie in a plane.