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Basically, I have a set of up to 100 co-ordinates, along with the desired tangents to the curve at the first and last point.
I have looked into various methods of curve-fitting, by which I mean an algorithm with takes the inputted data points and tangents, and outputs the equation of the cure, such as the gaussian method and interpolation, but I really struggled understanding them.
I am not asking for code (If you choose to give it, thats acceptable though :) ), I am simply looking for help into this algorithm. It will eventually be converted to Objective-C for an iPhone app, if that changes anything..
EDIT:
I know the order of all of the points. They are not too close together, so passing through all points is necessary - aka interpolation (unless anyone can suggest something else). And as far as I know, an algebraic curve is what I'm looking for. This is all being done on a 2D plane by the way
I'd recommend to consider cubic splines. There is some explanation and code to calculate them in plain C in Numerical Recipes book (chapter 3.3)
Most interpolation methods originally work with functions: given a set of x and y values, they compute a function which computes a y value for every x value, meeting the specified constraints. As a function can only ever compute a single y value for every x value, such an curve cannot loop back on itself.
To turn this into a real 2D setup, you want two functions which compute x resp. y values based on some parameter that is conventionally called t. So the first step is computing t values for your input data. You can usually get a good approximation by summing over euclidean distances: think about a polyline connecting all your points with straight segments. Then the parameter would be the distance along this line for every input pair.
So now you have two interpolation problem: one to compute x from t and the other y from t. You can formulate this as a spline interpolation, e.g. using cubic splines. That gives you a large system of linear equations which you can solve iteratively up to the desired precision.
The result of a spline interpolation will be a piecewise description of a suitable curve. If you wanted a single equation, then a lagrange interpolation would fit that bill, but the result might have odd twists and turns for many sets of input data.
I'm trying to figure out the most efficient/fast way to add a large number of convex quads (four given x,y points) into an array/list and then to check against those quads if a point is within or on the border of those quads.
I originally tried using ray casting but thought that it was a little overkill since I know that all my polygons will be quads and that they are also all convex.
currently, I am splitting each quad into two triangles that share an edge and then checking if the point is on or in each of those two triangles using their areas.
for example
Triangle ABC and test point P.
if (areaPAB + areaPAC + areaPBC == areaABC) { return true; }
This seems like it may run a little slow since I need to calculate the area of 4 different triangles to run the check and if the first triangle of the quad returns false, I have to get 4 more areas. (I include a bit of an epsilon in the check to make up for floating point errors)
I'm hoping that there is an even faster way that might involve a single check of a point against a quad rather than splitting it into two triangles.
I've attempted to reduce the number of checks by putting the polygon's into an array[,]. When adding a polygon, it checks the minimum and maximum x and y values and then using those, places the same poly into the proper array positions. When checking a point against the available polygons, it retrieves the proper list from the array of lists.
I've been searching through similar questions and I think what I'm using now may be the fastest way to figure out if a point is in a triangle, but I'm hoping that there's a better method to test against a quad that is always convex. Every polygon test I've looked up seems to be testing against a polygon that has many sides or is an irregular shape.
Thanks for taking the time to read my long winded question to what's prolly a simple problem.
I believe that fastest methods are:
1: Find mutual orientation of all vector pairs (DirectedEdge-CheckedPoint) through cross product signs. If all four signs are the same, then point is inside
Addition: for every edge
EV[i] = V[i+1] - V[i], where V[] - vertices in order
PV[i] = P - V[i]
Cross[i] = CrossProduct(EV[i], PV[i]) = EV[i].X * PV[i].Y - EV[i].Y * PV[i].X
Cross[i] value is positive, if point P lies in left semi-plane relatively to i-th edge (V[i] - V[i+1]), and negative otherwise. If all the Cross[] values are positive, then point p is inside the quad, vertices are in counter-clockwise order. f all the Cross[] values are negative, then point p is inside the quad, vertices are in clockwise order. If values have different signs, then point is outside the quad.
If quad set is the same for many point queries, then dmuir suggests to precalculate uniform line equation for every edge. Uniform line equation is a * x + b * y + c = 0. (a, b) is normal vector to edge. This equation has important property: sign of expression
(a * P.x + b * Y + c) determines semi-plane, where point P lies (as for crossproducts)
2: Split quad to 2 triangles and use vector method for each: express CheckedPoint vector in terms of basis vectors.
P = a*V1+b*V2
point is inside when a,b>=0 and their sum <=1
Both methods require about 10-15 additions, 6-10 multiplications and 2-7 comparisons (I don't consider floating point error compensation)
If you could afford to store, with each quad, the equation of each of its edges then you could save a little time over MBo's answer.
For example if you have an inward pointing normal vector N for each edge of the quad, and a constant d (which is N.p for one of the vertcies p on the edge) then a point x is in the quad if and only if N.x >= d for each edge. So thats 2 multiplications, one addition and one comparison per edge, and you'll need to perform up to 4 tests per point.This technique works for any convex polygon.
Suppose you have a list of 2D points with an orientation assigned to them. Let the set S be defined as:
S={ (x,y,a) | (x,y) is a 2D point, a is an orientation (an angle) }.
Given an element s of S, we will indicate with s_p the point part and with s_a the angle part. I would like to know if there exist an efficient data structure such that, given a query point q, is able to return all the elements s in S such that
(dist(q_p, s_p) < threshold_1) AND (angle_diff(q_a, s_a) < threshold_2) (1)
where dist(p1,p2), with p1,p2 2D points, is the euclidean distance, and angle_diff(a1,a2), with a1,a2 angles, is the difference between angles (taken to be the smallest one). The data structure should be efficient w.r.t. insertion/deletion of elements and the search as defined above. The number of vectors can grow up to 10.000 and more, but take this with a grain of salt.
Now suppose to change the above requirement: instead of using the condition (1), let's request all the elements of S such that, given a distance function d, we want all elements of S such that d(q,s) < threshold. If i remember well, this last setup is called range-search. I don't know if the first case can be transformed in the second.
For the distance search I believe the accepted best method is a Binary Space Partition tree. This can be stored as a series of bits. Each two bits (for a 2D tree) or three bits (for a 3D tree) subdivides the space one more level, increasing resolution.
Using a BSP, locating a set of objects to compare distances with is pretty easy. Just find the smallest set of squares or cubes which contain the edges of your distance box.
For the angle, I don't know of anything. I suppose that you could store each object in a second list or tree sorted by its angle. Then you would find every object at the proper distance using the BSP, every object at the proper angles using the angle tree, then do a set intersection.
You have effectively described a "three dimensional cyclindrical space", ie. a space that is locally three dimensional but where one dimension is topologically cyclic. In other words, it is locally flat and may be modeled as the boundary of a four-dimensional object C4 in (x, y, z, w) defined by
z^2 + w^2 = 1
where
a = arctan(w/z)
With this model, the space defined by your constraints is a 2-dimensional cylinder wrapped "lengthwise" around a cross section wedge, where the wedge wraps around the 4-d cylindrical space with an angle of 2 * threshold_2. This can be modeled using a "modified k-d tree" approach (modified 3-d tree), where the data structure is not a tree but actually a graph (it has cycles). You can still partition this space into cells with hyperplane separation, but traveling along the curve defined by (z, w) in the positive direction may encounter a point encountered in the negative direction. The tree should be modified to actually lead to these nodes from both directions, so that the edges are bidirectional (in the z-w curve direction - the others are obviously still unidirectional).
These cycles do not change the effectiveness of the data structure in locating nearby points or allowing your constraint search. In fact, for the most part, those algorithms are only slightly modified (the simplest approach being to hold a visited node data structure to prevent cycles in the search - you test the next neighbors about to be searched).
This will work especially well for your criteria, since the region you define is effectively bounded by these axis-defined hyperplane-bounded cells of a k-d tree, and so the search termination will leave a region on average populated around pi / 4 percent of the area.
I have a set of vectors in multidimensional space (may be several thousands of dimensions). In this space, I can calculate distance between 2 vectors (as a cosine of the angle between them, if it matters). What I want is to visualize these vectors keeping the distance. That is, if vector a is closer to vector b than to vector c in multidimensional space, it also must be closer to it on 2-dimensional plot. Is there any kind of diagram that can clearly depict it?
I don't think so. Imagine any twodimensional picture of a tetrahedron. There is no way of depicting the four vertices in two dimensions with equal distances from each other. So you will have a hard time trying to depict more than three n-dimensional vectors in 2 dimensions conserving their mutual distances.
(But right now I can't think of a rigorous proof.)
Update:
Ok, second idea, maybe it's dumb: If you try and find clusters of closer associated objects/texts, then calculate the center or mean vector of each cluster. Then you can reduce the problem space. At first find a 2D composition of the clusters that preserves their relative distances. Then insert the primary vectors, only accounting for their relative distances within a cluster and their distance to the center of to two or three closest clusters.
This approach will be ok for a large number of vectors. But it will not be accurate in that there always will be somewhat similar vectors ending up at distant places.
My question is fairly simple. I have two tetrahedra, each with a current position, a linear speed in space, an angular velocity and a center of mass (center of rotation, actually).
Having this data, I am trying to find a (fast) algorithm which would precisely determine (1) whether they would collide at some point in time, and if it is the case, (2) after how much time they collided and (3) the point of collision.
Most people would solve this by doing triangle-triangle collision detection, but this would waste a few CPU cycles on redundant operations such as checking the same edge of one tetrahedron against the same edge of the other tetrahedron upon checking up different triangles. This only means I'll optimize things a bit. Nothing to worry about.
The problem is that I am not aware of any public CCD (continuous collision detection) triangle-triangle algorithm which takes self-rotation in account.
Therefore, I need an algorithm which would be inputted the following data:
vertex data for three triangles
position and center of rotation/mass
linear velocity and angular velocity
And would output the following:
Whether there is a collision
After how much time the collision occurred
In which point in space the collision occurred
Thanks in advance for your help.
The commonly used discrete collision detection would check the triangles of each shape for collision, over successive discrete points in time. While straightforward to compute, it could miss a fast moving object hitting another one, due to the collision happening between discrete points in time tested.
Continuous collision detection would first compute the volumes traced by each triangle over an infinity of time. For a triangle moving at constant speed and without rotation, this volume could look like a triangular prism. CCD would then check for collision between the volumes, and finally trace back if and at what time the triangles actually shared the same space.
When angular velocity is introduced, the volume traced by each triangle no longer looks like a prism. It might look more like the shape of a screw, like a strand of DNA, or some other non-trivial shapes you might get by rotating a triangle around some arbitrary axis while dragging it linearly. Computing the shape of such volume is no easy feat.
One approach might first compute the sphere that contains an entire tetrahedron when it is rotating at the given angular velocity vector, if it was not moving linearly. You can compute a rotation circle for each vertex, and derive the sphere from that. Given a sphere, we can now approximate the extruded CCD volume as a cylinder with the radius of the sphere and progressing along the linear velocity vector. Finding collisions of such cylinders gets us a first approximation for an area to search for collisions in.
A second, complementary approach might attempt to approximate the actual volume traced by each triangle by breaking it down into small, almost-prismatic sub-volumes. It would take the triangle positions at two increments of time, and add surfaces generated by tracing the triangle vertices at those moments. It's an approximation because it connects a straight line rather than an actual curve. For the approximation to avoid gross errors, the duration between each successive moments needs to be short enough such that the triangle only completes a small fraction of a rotation. The duration can be derived from the angular velocity.
The second approach creates many more polygons! You can use the first approach to limit the search volume, and then use the second to get higher precision.
If you're solving this for a game engine, you might find the precision of above sufficient (I would still shudder at the computational cost). If, rather, you're writing a CAD program or working on your thesis, you might find it less than satisfying. In the latter case, you might want to refine the second approach, perhaps by a better geometric description of the volume occupied by a turning, moving triangle -- when limited to a small turn angle.
I have spent quite a lot of time wondering about geometry problems like this one, and it seems like accurate solutions, despite their simple statements, are way too complicated to be practical, even for analogous 2D cases.
But intuitively I see that such solutions do exist when you consider linear translation velocities and linear angular velocities. Don't think you'll find the answer on the web or in any book because what we're talking about here are special, yet complex, cases. An iterative solution is probably what you want anyway -- the rest of the world is satisfied with those, so why shouldn't you be?
If you were trying to collide non-rotating tetrahedra, I'd suggest a taking the Minkowski sum and performing a ray check, but that won't work with rotation.
The best I can come up with is to perform swept-sphere collision using their bounding spheres to give you a range of times to check using bisection or what-have-you.
Here's an outline of a closed-form mathematical approach. Each element of this will be easy to express individually, and the final combination of these would be a closed form expression if one could ever write it out:
1) The equation of motion for each point of the tetrahedra is fairly simple in it's own coordinate system. The motion of the center of mass (CM) will just move smoothly along a straight line and the corner points will rotate around an axis through the CM, assumed to be the z-axis here, so the equation for each corner point (parameterized by time, t) is p = vt + x + r(sin(wt+s)i + cos(wt + s)j ), where v is the vector velocity of the center of mass; r is the radius of the projection onto the x-y plane; i, j, and k are the x, y and z unit vectors; and x and s account for the starting position and phase of rotation at t=0.
2) Note that each object has it's own coordinate system to easily represent the motion, but to compare them you'll need to rotate each into a common coordinate system, which may as well be the coordinate system of the screen. (Note though that the different coordinate systems are fixed in space and not traveling with the tetrahedra.) So determine the rotation matrices and apply them to each trajectory (i.e. the points and CM of each of the tetrahedra).
3) Now you have an equation for each trajectory all within the same coordinate system and you need to find the times of the intersections. This can be found by testing whether any of the line segments from the points to the CM of a tetrahedron intersects the any of the triangles of another. This also has a closed-form expression, as can be found here.
Layering these steps will make for terribly ugly equations, but it wouldn't be hard to solve them computationally (although with the rotation of the tetrahedra you need to be sure not to get stuck in a local minimum). Another option might be to plug it into something like Mathematica to do the cranking for you. (Not all problems have easy answers.)
Sorry I'm not a math boff and have no idea what the correct terminology is. Hope my poor terms don't hide my meaning too much.
Pick some arbitrary timestep.
Compute the bounds of each shape in two dimensions perpendicular to the axis it is moving on for the timestep.
For a timestep:
If the shaft of those bounds for any two objects intersect, half timestep and start recurse in.
A kind of binary search of increasingly fine precision to discover the point at which a finite intersection occurs.
Your problem can be cast into a linear programming problem and solved exactly.
First, suppose (p0,p1,p2,p3) are the vertexes at time t0, and (q0,q1,q2,q3) are the vertexes at time t1 for the first tetrahedron, then in 4d space-time, they fill the following 4d closed volume
V = { (r,t) | (r,t) = a0 (p0,t0) + … + a3 (p3,t0) + b0 (q0,t1) + … + b3 (q3,t1) }
Here the a0...a3 and b0…b3 parameters are in the interval [0,1] and sum to 1:
a0+a1+a2+a3+b0+b1+b2+b3=1
The second tetrahedron is similarly a convex polygon (add a ‘ to everything above to define V’ the 4d volume for that moving tetrahedron.
Now the intersection of two convex polygon is a convex polygon. The first time this happens would satisfy the following linear programming problem:
If (p0,p1,p2,p3) moves to (q0,q1,q2,q3)
and (p0’,p1’,p2’,p3’) moves to (q0’,q1’,q2’,q3’)
then the first time of intersection happens at points/times (r,t):
Minimize t0*(a0+a1+a2+a3)+t1*(b0+b1+b2+b3) subject to
0 <= ak <=1, 0<=bk <=1, 0 <= ak’ <=1, 0<=bk’ <=1, k=0..4
a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
= a0’*(p0’,t0) + … + a3’*(p3’,t0) + b0’*(q0’,t1) + … + b3’*(q3’,t1)
The last is actually 4 equations, one for each dimension of (r,t).
This is a total of 20 linear constraints of the 16 values ak,bk,ak', and bk'.
If there is a solution, then
(r,t)= a0*(p0,t0) + … + a3*(p3,t0) + b0*(q0,t1) + … + b3*(q3,t1)
Is a point of first intersection. Otherwise they do not intersect.
Thought about this in the past but lost interest... The best way to go about solving it would be to abstract out one object.
Make a coordinate system where the first tetrahedron is the center (barycentric coords or a skewed system with one point as the origin) and abstract out the rotation by making the other tetrahedron rotate around the center. This should give you parametric equations if you make the rotation times time.
Add the movement of the center of mass towards the first and its spin and you have a set of equations for movement relative to the first (distance).
Solve for t where the distance equals zero.
Obviously with this method the more effects you add (like wind resistance) the messier the equations get buts its still probably the simplest (almost every other collision technique uses this method of abstraction). The biggest problem is if you add any effects that have feedback with no analytical solution the whole equation becomes unsolvable.
Note: If you go the route of of a skewed system watch out for pitfalls with distance. You must be in the right octant! This method favors vectors and quaternions though, while the barycentric coords favors matrices. So pick whichever your system uses most effectively.