I was looking around a lot but only found how to check collision for 2d objects.. for my current project I want to check colisions in 3d (I'm using obj models) - I could probably figure out something myself the problem is that I only know the center point of each object..
Is there a way to get the boarders of the object so I can check if it touches the boarders of another object? What would be the best way to get this information?
Edit: Some more information that might help:
I'm using lwjgl 2.8,
my objects are obj files,
I can get position, scale and the rotation of an object
EDIT: This is what I found on youtube:
https://www.youtube.com/watch?v=Iu6nAXFm2Wo&list=PLEETnX-uPtBXm1KEr_2zQ6K_0hoGH6JJ0&index=4
What you have is called a triangle soup[1]: you got vertex position (, texture and normal) information couppled with triangle information. What you can do is intersect each triangle from one mesh with another mesh. You can do this either brute-force (by testing all the other triangles each time) or build up a space partitioning data structure to speed up your intersections.
E.g. build an Octree per mesh and iterate over the leaves in one of them: for each of those leaves, test its bounding box against the leaves in the other tree and for each of those collision pairs brute-force test each triangles within either of those pairs against each other (or only test those from mesh A against those in mesh B, if you don't care about self-intersections).
There are libraries for these sorts of algorithms like OpenMesh or Bullet for example. But I only know of one port to Java: JBullet.
EDIT: If you're only interested in approximate collisions you can throw away all information about triangles and build a bounding volume out of your vertices (e.g. an axis aligned box is just the min and the max for all vertex positions, an oriented box is built similarly, but you have to find a good enough orientation first, a sphere is a bit more involved and finally you have a convex mesh, which uses the same sorts of intersection tests as a normal mesh, but is smaller than the original and convex, allowing for some optimisations of the general intersection test).
[1]: there are other kinds of 3D representations that record information about how the triangles are connected. You might imagine finding only two intersecting triangles in mesh A and mesh B and then start searching for intersecting triangles only in the neighbourhood of the initial intersection... these algorithms are much more involved, but for meshes that deform have the advantage, that you don't have to rebuild the space partitioning data structure each time as you have to do with triangle soups.
Related
Say you have a collection of points with coordinates on a Cartesian coordinate system.
You want to plot another point, and you know its coordinates in the same Cartesian coordinate system.
However, the plot you're drawing on is distorted from the original. Imagine taking the original plane, printing it on a rubber sheet, and stretching it in some places and pinching it in others, in an asymmetrical way (no overlapping or anything complex).
(source)
You know the stretched and unstretched coordinates of each of your set of points, but not the underlying stretch function. You know the unstretched coordinates of a new point.
How can you estimate where to plot the new point in the stretched coordinates based on the stretched positions of nearby points? It doesn't need to be exact, since you can't determine the actual stretch function from a set of remapped points unless you have more information.
other possible keywords: warped distorted grid mesh plane coordinate unwarp
Ok, so this sounds like image warping. This is what you should do:
Create a Delaunay triangulation of your unwarped grid and use your knowledge of the correspondences between the warped and unwarped grid to create the triangulation for the warped grid. Now you know the corresponding triangles in each image and since there is no overlapping, you should be able to perform the next step without much difficulty.
Now, to find the corresponding point A, in the warped image:
Find the triangle A lies in and use the transformation between the triangle in the unwarped grid and the warped grid to figure out the new position.
This is explained explicitly in detail here.
Another (much more complicated) method is the Thin Plate Spline (which is also explained in the slides above).
I understood that you have one-to-one correspondence between the wrapped and unwrapped grid points. And I assume that the deformation is not so extreme that you might have intersecting grid lines (like the image you show).
The strategy is exactly what Jacob suggests: Triangulate the two grids such that there is a one-to-one correspondence between triangles, locate the point to be mapped in the triangulation and then use barycentric coordinates in the corresponding triangle to compute the new point location.
Preprocess
Generate the Delaunay triangulation of the points of the wrapped grid, let's call it WT.
For every triangle in WT add a triangle between the corresponding vertices in the unwrapped grid. This gives a triangulation UWT of the unwrapped points.
Map a point p into the wrapped grid
Find the triangle T(p1,p2,p3) in the UWT which contains p.
Compute the barycentric coordinates (b1,b2,b3) of p in T(p1,p2,p3)
Let Tw(q1,q2,q3) be the triangle in WT corresponding to T(p1,p2,p3). The new position is b1 * q1 + b2 * q2 + b3 * q3.
Remarks
This gives a deformation function as a linear spline. For smoother behavior one could use the same triangulation but do higher order approximation which would lead to a bit more complicated computation instead of the barycentric coordinates.
The other answers are great. The only thing I'd add is that you might want to take a look at Free form deformation as a way of describing the deformations.
If that's useful, then it's quite possible to fit a deformation grid/lattice to your known pairs, and then you have a very fast method of deforming future points.
A lot depends on how many existing points you have. If you have only one, there's not really much you can do with it -- you can offset the second point by the same amount in the same direction, but you don't have enough data to really do any better than that.
If you have a fair number of existing points, you can do a surface fit through those points, and use that to approximate the proper position of the new point. Given N points, you can always get a perfect fit using an order N polynomial, but you rarely want to do that -- instead, you usually guess that the stretch function is a fairly low-order function (e.g. quadratic or cubic) and fit a surface to the points on that basis. You then place your new point based on the function for your fitted surface.
i want to find a circular object(Iris of eye, i have used Haar Cascase with viola Jones algorithm). so i found that hough circle would be the correct way to do it. can anybody explain me how to implement Hough circle in Java or any other easy implementation to find iris with Java.
Thanks,
Duda and Hart (1971) has a pretty clear explanation of the Hough transform and a worked example. It's not difficult to produce an implementation directly from that paper, so it's a good place for you to start.
ImageJ provides a Hough Circle plugin. I've been playing around with it several times in the past.
You could take a look at the source code if you want or need to modify it.
If you want to find an iris you should be straightforward about this. The part of the iris you are after is actually called a limbus. Also note that the contrast of the limbus is much lower than the one of the pupil so if image resolution permits pupil is a better target. Java is not a good option as programming language here since 1. It is slow while processing is intense; 2. Since classic Hough circle requires 3D accumulator and Java probably means using a cell phone the memory requirements will be tough.
What you can do is to use a fact that there is probably a single (or only a few) Limbuses in the image. First thing to do is to reduce the dimensionality of the problem from 3 to 2 by using oriented edges: extract horizontal and vertical edges that together represent edge orientation (they can be considered as horizontal and vertical components of edge vector). The simple idea is that the dominant intersection of edge vectors is the center of your limbus. To find the intersection you only need two oriented edges instead of three points that define a circle. Hence dimensionality reduction from 3 to 2.
You also don’t need to use a classical Hough circle transform with a huge accumulator and numerous calculations to find this intersection. A Randomized Hough will be much faster. Here is how it works (~ to RANSAC): you select a minimum number of oriented edges at random (in your case 2), find the intersection, then find all the edges that intersect at approximately the same location. These are inliers. You just iterate 10-30 times choosing a different random sample of 2 edges to settle in a set with maximum number of inliers. Hopefully, these inliers lie on the limbus. The median of inlier ray intersections will give you the center of the circle and the median distance to the inliers from the center is the radius.
In the picture below bright colors correspond to inliers and orientation is shown with little line segment. The set of original edges is shown in the middle (horizontal only). While original edges lie along an ellipse, Hough edges were transformed by an Affine transform to make those belonging to limbus to lie on a circle. Also note that edge orientations are pretty noisy.
I need to generate a Voronoi diagram around a concave (non-convex) inside polygon. I have looked for methods online, but I haven't been able to figure out how to do this. Basically, I generate the convex hull of the points, calculate the dual points and build an edge network between these points. However, when meeting the edges of the inside polygon, it has to look like the edge of the shape, just like the convex hull. So, by doing this and clipping all the edges at the borders, I should end up with a Voronoi diagram that has nice edges to the borders of the inside polygon and no cells that are on both sides of the inside polygon.
Let me give you an example:
The problem with this is that the cells cross the inside polygon edges and there is no visual relation between the cell structure and the polygon shape.
Does anybody know how to approach this problem? Is there some algorithm that already does this or gets close to what I'm trying to achieve?
Thank you so much for any kind of input!
You might be able to build a conforming Delaunay triangulation (i.e. a triangulation that includes the polygon edges as constraints) and then form the Voronoi diagram as the dual. A conforming triangulation will ensure that no edge in the triangulation intersects with a constraint edge - all constraint edges will be an edge in the triangulation.
Have a look at the Triangle package here, as a reference for this type of approach. In my experience it's a fast and robust library, although it's written in c not java.
I'm not sure I understand at this stage how the points (the Voronoi centres) are generated in your diagram. If you're actually looking to do mesh generation in a polygonal domain, then there may be other approaches to consider, although the Triangle package supports (conforming) Delaunay refinement mesh generation.
EDIT: It looks like you can also directly form the Voronoi diagram for general line segments, check out the VRONI library, here. Addressing your comment - I'm not sure that you can always expect to have a uniform Voronoi diagram that also conforms to a general polygonal boundary. I would expect that the shape of the polygonal boundary would impose a maximum dimension on the boundary Voronoi cells.
Hope this helps.
Clearly you need to generate your Voronoi diagram to the constraints of the greater polygon. Although you refer to it as a polygon, I notice that your example diagram has spline-based edges. Let's forget that for now.
What you want to do is to ensure that you start out with the containing polygon (whether generated by you or from another source) having edges of fairly equal length; a variance factor would make this look more natural. I would probably go for a variance of 10-20%.
Now that you have your containing polygon bounded by lines segments of approximately equal length, you have a basis from which to begin generating your Voronoi diagram. For each edge on your container:
Determine the edge normal (perp line jutting inward from centre of that segment).
Use the edge normal as a sliding scale on which to place a new Voronoi node centre. The distance away from the edge itself would be determined by what you want your average Voronoi cell "diameter" to be, if they were all taken as circles. In your example that looks like maybe 30 pixels (or whatever the equivalent in your world units would be). Again, you should apply a variance factor to this so that not every cell centre is placed equidistant from its source edge.
Generate the Voronoi cell for your newly placed centre.
Store your Voronoi cell source point in a list.
As you incrementally generate each point, you should begin to see that the algorithm subdivides each convex "constituent area" of your concave container in a radial fashion.
You may be wondering what the list is for. Well, obviously, you're not done yet, you've only generated a fraction of the total Voronoi tesselation you want. Once you have created these "boundary" cells of your concave space, you don't want new cells to be generated closer to the boundary than the boundary cells already are, you only want them inside that area. By maintaining a list of the boundary cell source points, you can then ensure that any further points you create are inside that area. It's a little bit like taking an internal Minkowski sum to ensure you have a buffer zone. Now you can randomise the rest of your cells in this derived concave space, to completion.
(Caveat emptor: You will have to be careful with this previous step. If any "passage" areas are too narrow, then the boundaries of this derived space will overlap, you will have a non-simple polygon, and you may find yourself placing points in the wrong places in spite of your efforts. The solution is to ensure that either your maximum placement distance from edges is never more than half of your minimum passage width... or use some other geometric means, including Minkowski summation as one possibility, to ensure you do not wind up with a degenerate derived polygon. It is quite possible that you will end with a multipolygon, i.e. fragments.)
I've not applied this method myself yet, but although there will certainly be bugs to work out, I think the general idea will get you started in the right direction.
Look for a paper called:
"Efficient computation of continuous skeletons" by Kirkpatrick, David G, written in 1979.
Here's the abstract:
An O(n lgn) algorithm is presented for the construction of skeletons
of arbitrary n-line polygonal figures. This algorithm is based on an
O(n lgn) algorithm for the construction of generalized Voronoi
diagrams (our generalization replaces point sets by sets of line
segments constrained to intersect only at end points). The generalized
Voronoi diagram algorithm employs a linear time algorithm for the
merging of two arbitrary (standard) Voronoi diagrams.
"Sets of line segments is the constrained to intersect only at end points" is the concave polygon you describe.
i would like to build a dynamic data structure that can hold a list of polygons and return a list of polygons that overlaps a specified rectangle.
i looked into bst trees (and quad trees) but these dont seem to work too well when the polygons overlap heavily.
any good ideas i should check out before i roll my own nonsense?
edit
lets assume all the polygons are normal non rotated rectangles. im willing to take the hit (point in polygon test) during point tests (i might be doing it anyway), and during a region test getting their bounding boxes is just as good. only a small percentage of them will actually not overlap the region in question.
I would look at 2-d segment delaunay graphs. Look also at Nef polygons. CGAL has a lot of set operations on polygons. Answers to this question may also be of value
Edit If your polygons are non rotated rectangles see R-Trees
Why do you write that yourself? Java offers complex intersection tests. You can convert your polygon data structures and your rectangle to Java.awt.geom.Area and then call the Area.intersect() method which does all the math for you.
It also takes care of all the rarely occurring (but still important) special cases which are really nasty to catch.
i just wrote a regular quadtree, that allowed each leaf node to hold unlimited polys, if the intersection of the bounds of the leaf and the bounds of each poly in the bucket were equivalent. otherwise leaf nodes are limited to 8 polys, before splitting.
This is sort of a homework question, however no expectations for code or whatever just an idea or hint towards the following problem.
I have a set of cubes in 3D world coordinates and i have to display them using two projections in two separate areas, parallel and perspective. The parallel went fine, no problems there, however displaying the same scene using perspective projection is becoming a nuisance for me.
The world to screen coordinates seemed like a good idea, but i don't know on which coordinates to apply them to, the original real coordinates, the new coordinates.
Thank you for your time.
PS: we are only allowed Java2D Api.
To perform a perspective projection, you need two additional things: the perspective point (where the "eye" is) and the projection plane. With a parallel projection, the perspective point/eye and plane can be any arbitrary distance from the objects (e.g., the cubes). But it is a little more complex with perspective projection.
Once you establish your eye and projection plane, you will need to iterate over your cubes. Ideally, you would iterate over them from the farthest cube to the eye to the nearest - that way the nearer cubes will overwrite the farther ones.
For each cube, determine the distance from the eye for each point. Then for each face (again in order of decreasing distance), calculate the projected points for each vertex. You can skip those faces with occluded points (the farthest vertex for each cube).
To calculate the projected point for a particular vertex, you need to find the point on the projection plane. This point will be where the line from the eye to the vertex intersects the projection plane. This will require some math, but should not be too difficult.