Moller is a universal and fastest algorithm used for ray tracing triangles.
It based on barycentrics coordinates.
Even if Weber(2016) can be better than Moller.
What's the difference with Curless algorithm and why is-it always better?
Curless : https://courses.cs.washington.edu/courses/csep557/10au/lectures/triangle_intersection.pdf
Moller : https://www.tandfonline.com/doi/abs/10.1080/10867651.1997.10487468
Thank you
Curless needs to
compute the normal vector
do a triangle inside outside test (3 edge equation test)
compute barycentric coordiantes
Moeler does in one go compute barycentric coordiantes - boom! Less calculations.
Neverthless, checkout Watertight Ray/Triangle Intersection.
Related
I have a set of distinct 2D vectors (over real numbers), pointing in different directions. We are allowed to pick a pair of vectors and construct their linear combination, such that the coefficients are positive and their sum is 1.
In simple words we are allowed to take a "weighted average" of any two vectors.
My goal is for an arbitrary direction to pick a pair of vectors whose "weighted average" is in this direction and is maximized.
Speaking algebraically given vectors a and b and a direction vector n we are interested in maximizing this value:
[ a cross b ] / [ (a - b) cross n ]
i.e. pick a and b which maximize this value.
To be concrete the application of this problem is for sailing boats. For every apparent wind direction the boat will have a velocity given by a polar diagram. Here's an example of such a diagram:
(Each line in this diagram corresponds to a specific wind magnitude). Note the "impossible" front sector of about 30 degrees in each direction.
So that in some direction the velocity will be high, for some - low, and for some directions it's impossible to sail directly (for instance in the direction strictly opposite to the wind).
If we need to advance in a direction in which we can't sail directly (or the velocity isn't optimal) - it's possible to advance in zigzags. This is called tacking.
Now, my goal is to recalculate a new diagram which denotes the average advance velocity in any direction, either directly or indirectly. For instance for the above diagram the corrected diagram would be this:
Note that there are no more "impossible" directions. For some directions the diagram resembles the original one, where it's best to advance directly, and no maneuver is required. For others - it shows the maximum average advance velocity in this direction assuming the most optimal maneuver is periodically performed.
What would be the most optimal algorithm to calculate this? Assume the diagram is given as a discrete set of azimuth-velocity pairs, from which we can calculate the vectors.
So far I just check all the vector pairs to select the best. Well, there're cut-off criterias, such as picking only vectors with positive projection on the advance direction, and opposite perpendicular projections, but still the complexity is O(N^2).
I wonder if there's a more efficient algorithm.
EDIT
Many thanks to #mcdowella. For both computer-science and sailor answers!
I too thought in terms of convex polygon, figured out that it's only worth probing vectors on that hull (i.e. if you take a superposition of 2 vectors on this hull, and try to replace one of them by a vector which isn't on this hull, the result would be worse since new vector's projection on the needed direction is worse than of both source vectors).
However I didn't realize that any "weighted average" of 2 vectors is actually a straight line segment connecting those vectors, hence the final diagram is indeed this convex hull! And, as we can see, this is also in agreement with what I calculated by "brute-force" algorithm.
Now the computer science answer
A tacking strategy gives you the convex combination of the vectors from the legs that make up the tacks.
So consider the outline made by just one contour in your diagram. The set of all possible best speeds and directions is the convex polygon formed by taking all convex combinations of the vectors to the contour. So what you want to do is form the convex hull of your contour (https://en.wikipedia.org/wiki/Convex_hull). To find out how to go fast in any particular direction, intersect that vector with the convex hull, and use tacks with legs that correspond to the corners on either side of the edge of the convex hull that you intersect with.
Looking at your diagram, the contour is concave straight upwind and straight downwind, which is what you would expect. However there is also another concave section, somewhere between 4 and 5 O'Clock and also symmetrically between 7 and 8 O'Clock, which appears as a straight line in your corrected diagram - so I guess there is a third direction to tack in, using two reaches on the same side of the wind which I don't recognise from traditional sailing.
First the ex-laser sailor answer
At least for going straight upwind or downwind, the obvious guess is to tack so that each leg is of the same length and of the same bearing to the wind. If the polar diagram is symmetric around the upwind-downwind axis this is correct. Suppose upwind is the Y axis and possible legs are (A, B), (-A, B), (a, b) and (-a, b). Symmetrical tacking moves (A, B)/2 + (-A, B)/2 = (0, B) and the other symmetrical tack gives you (0, b). Asymmetrical tacking is (-A, B)a/(a+A) + (a, b)A/(a+A) = (0, (a/(a+A))B + (A/(a+A))b) and if b!=B lies between b and B and so is not as good as whichever of b or B is best.
For any direction which lies between the port and starboard tacks that you would take to work your way upwind, the obvious strategy is to change the length of those legs but not their direction so that the average vector traveled is in the required direction. Is this the best strategy? If not, the better strategy is making progress upwind faster that the port and starboard tacks that you would take to work your way upwind, which I think is a contradiction - so for any direction which lies between the port and starboard tacks made to go upwind I think the best strategy is indeed to make those tacks but alter the leg lengths to go in the required direction. The same thing should apply for tacking downwind, if you have a boat that makes that a good idea.
I am trying to develop an algorithm that performs the following :
Given a 2D polygon and a 3D polyhedron, determine if the 2D polygon is a projection of the 3D polyhedron (a perspective projection to be precise) without knowing which transformation matrix we may have possibly used for the projection.
input
{2D Polygon}
{3D Polyhedron}
output
{bool} whether or not it's a perspective projection
I am not asking for code, but I would simply like to know if this is feasible in polynomial time.
Any help will be greatly appreciated.
A 3D to 2D perspective projection has 7 degrees of freedom (6 for the relative motion of the scene with respect to the camera, 1 for the focal length).
Select four vertices in the 2D projection and consider all possible correspondences with polyhedron vertices (there is a polynomial number of such associations). Then form a system of 7 equations in the 7 unknown parameters (unfortunately a nonlinear one; maybe the eighth equation can be useful to select among multiple solutions).
Knowing the parameters, you can check a solution by re-projecting the polyhedron and comparing to the polygon (with further search for correspondences with vertices and edges).
All of this will take polynomial time (quartic if I am right), if one admits that the solver takes bounded time (hence bounded precision).
If the focal length is known, then a better approach is possible. Indeed, with only 6 unknowns, you can find the projection parameters from the projection of just three points. This problem is known to have an analytical solution (actually up to 4 of them), as described at length in "New Algorithms for the Perspective-Three-Point Problem, GAO Xiaoshan & CHEN Hangfei, Vol.16 No.3 J. Comput. Sci. & Technol."
This should lead to an O(N³) exact procedure.
More generally speaking, you form putative correspondences between N pairs of points, solve the corresponding Perspective-N-point problem, and check the hypothesis by reprojecting the polyhedron and comparing to the known projection to validate the hypothesis.
Just an idea for an algorithm:
Take a triangle of the projection made of three points next to each other not on the same line. Iterate through all corresponding triangles of the original. For all possible projections that solve the pair of triangles, check if the rest matches.
I must admit I am not sure right now if there could be infinite solutions for triangles (which would be hard to iterate)? If so, start with four points.
I think it is possible but you have to do a fair amount of reverse engineering. A 2D sketch that represents a 3D object is known as an Orthographic Projection. The link shows you the transformation matrices you need apply to transform the 3D point onto its 2D projection. Now, how do you go the opposite way? Inverse matrices with a mix of some inverse transformations (translation, scaling, rotation...)? I think this is a good lead to follow.
It is simple to fill rectangle: simply make some grid. But if polygon is unconditioned the task becomes not so trivial.
Probably "regularly" can be formulated as distance between each other point would be: R ± alpha. But I'm not sure about this.
Maybe there is some known algorithm to achieve this.
Added:
I need to generate net, where no large holes, and no big gathering of the points.
Have you though about using a force-directed layout of the points?
Scatter a number of points randomly over the bounding box of your polygon, then repeatedly apply two simple rules to adjust their location:
If a point is outside of the polygon, move it the minimum possible distance so that it lies within, i.e.: to the closest point on the polygon edge.
Points repel each other with a force inversely proportional to the distance between them, i.e.: for every point, consider every other point and compute a repulsion vector that will move the two points directly apart. The vector should be large for proximate points and small for distant points. Sum the vectors and add to the point's position.
After a number of iterations the points should settle into a steady state with an even distribution over the polygon area. How quickly this state is achieved depends on the geometry of the polygon and how you've scaled the repulsive forces between the points.
You can compute a Constrained Delaunay triangulation of the polygon and use a Delaunay refinement algorithm (search with this keyword).
I have recently implemented refinement
in the Fade2D library, http://www.geom.at/fade2d/html/. It takes an
arbitrary polygon without selfintersections as well as an upper bound on the radius of the circumcircle of each resulting triangle. This feature is not contained in the current release 1.02 yet, but I can compile the current development version for Linux or Win64 if you want to try that.
I've been scouring the internet for days, but have been unable to find a good answer (or at least one that made sense to me) to what seems like it should be a common question. How does one scale an arbitrary polygon? In particular, concave polygons. I need an algorithm which can handle concave (definitely) and self-intersecting (if possible) polygons. The obvious and simple algorithm I've been using to handle simple convex polygons is calculating the centroid of the polygon, translating that centroid to the origin, scaling all the vertices, and translating the polygon back to its original location.
This approach does not work for many (or maybe all) concave polygons as the centroid often falls outside the polygon, so the scaling operation also results in a translation and I need to be able to scale the polygon "in place" without the final result being translated.
Is anybody aware of a method for scaling concave polygons? Or maybe a way of finding the "visual center" which can be used as a frame of reference for the scaling operation?
Just to clarify, I'm working in 2D space and I would like to scale my polygons using the "visual center" as the frame of reference. So maybe another way to ask the question would be, how do I find the visual center of a concave and/or self-intersecting polygon?
Thanks!
I'm not sure what your problem is.
You're working in an affine space, and you're looking for an affine transformation to scale your polygon ?
If i'm right, just write the transformation matrix:
scaling matrix
homotethy
And transform your polygon with matrix
You can look up for affine transformation matrix.
hope it helps
EDIT
if you want to keep the same "center", you can just do an homotethy of parameter lambda with center G = barycenter of the polygon:
it verifies :
G won't move since it's the center of the homotethy.
It will still verify the relation below, so it will still be the barycenter. (you just multiply the relation by lambda)
in your case G is easy to determinate: G(x,y) : (average of x values of points, average of y values of points)
and it should do what you need
Perhaps Craig is looking for a "polygon offset" algorithm - where each edge in the polygon is offset by a given value. For example, given a clockwise oriented polygon, offsetting edges towards the left will increase the size of the polygon. If this is what Craig is looking for then this has been asked and answered before here - An algorithm for inflating/deflating (offsetting, buffering) polygons.
If you're looking for a ready made (opensource freeware) solution, I've also created a clipping library (Clipper) written in Delphi, C++ and C# which includes a rather simple polygon offsetting function.
The reason why you can't find a good answer is because you are being imprecise with your requirements. First explicitly define what you mean by "in-place". What is being kept constant?
Once you have figured that out, then translate the constant point to the origin, scale the polygon as usual, and translate back.
I am programming an algorithm where I have broken up the surface of a sphere into grid points (for simplicity I have the grid lines parallel and perpendicular to the meridians). Given a point A, I would like to be able to efficiently take any grid "square" and determine the point B in the square with the least spherical coordinate distance AB. In the degenerate case the "squares" are actually "triangles".
I am actually only using it to bound which squares I am searching, so I can also accept a lower bound if it is only a tiny bit off. For this reason, I need the algorithm to be extremely quick otherwise it would be better to just take the loss of accuracy and search a few more squares.
I decided to repost this question to Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates. More progress has been made here
For points on a sphere, the points closest in the full 3D space will also be closest when measured along the surface of the sphere. The actual distances will be different, but if you're just after the closest point it's probably easiest to minimize the 3D distance rather than worry about great circle arcs, etc.
To find the actual great-circle distance between two (latitidude, longitude) points on the sphere, you can use the first formula in this link.
A few points, for clarity.
Unless you specifically wish these squares to be square (and hence to not fit exactly in this parallel and perpendicular layout with regards to the meridians), these are not exactly squares. This is particularly visible if the dimensions of the square are big.
The question speaks of a [perfect] sphere. Matters would be somewhat different if we were considering the Earth (or other planets) with its flattened poles.
Following is a "algorithm" that would fit the bill, I doubt it is optimal, but could offer a good basis. EDIT: see Tom10's suggestion to work with the plain 3D distance between the points rather than the corresponding great cirle distance (i.e. that of the cord rather than the arc), as this will greatly reduce the complexity of the formulas.
Problem layout: (A, B and Sq as defined in the OP's question)
A : a given point the the surface of the sphere
Sq : a given "square" from the grid
B : solution to problem : point located within Sq which has the shortest
distance to A.
C : point at the center of Sq
Tentative algorithm:
Using the formulas associated with [Great Circle][1], we can:
- find the equation of the circle that includes A and C
- find the distance between A and C. See the [formula here][2] (kindly lifted
from Tom10's reply).
- find the intersect of the Great Circle arc between these points, with the
arcs of parallel or meridian defining the Sq.
There should be only one such point, unless this finds a "corner" of Sq,
or -a rarer case- if the two points are on the same diameter (see
'antipodes' below).
Then comes the more algorithmic part of this procedure (so far formulas...):
- find, by dichotomy, the point on Sq's arc/seqment which is the closest from
point A. We're at B! QED.
Optimization:
It is probably possible make a good "guess" as to the location
of B, based on the relative position of A and C, hence cutting the number of
iterations for the binary search.
Also, if the distance A and C is past a certain threshold the intersection
of the cicles' arcs is probably a good enough estimate of B. Only when A
and C are relatively close will B be found a bit further on the median or
parallel arc in these cases, projection errors between A and C (or B) are
smaller and it may be ok to work with orthogonal coordinates and their
simpler formulas.
Another approach is to calculate the distance between A and each of the 4
corners of the square and to work the dichotomic search from two of these
points (not quite sure which; could be on the meridian or parallel...)
( * ) *Antipodes case*: When points A and C happen to be diametrically
opposite to one another, all great circle lines between A and C have the same
length, that of 1/2 the circonference of the sphere, which is the maximum any
two points on the surface of a sphere may be. In this case, the point B will
be the "square"'s corner that is the furthest from C.
I hope this helps...
The lazy lower bound method is to find the distance to the center of the square, then subtract the half diagonal distance and bound using the triangle inequality. Given these aren't real squares, there will actually be two diagonal distances - we will use the greater. I suppose that it will be reasonably accurate as well.
See Math Overflow: https://mathoverflow.net/questions/854/closest-grid-square-to-a-point-in-spherical-coordinates for an exact solution