Given two 3d objects, how can I find if one fits inside the second (and find the location of the object in the container).
The object should be translated and rotated to fit the container - but not modified otherwise.
Additional complications:
The same situation - but look for the best fit solution, even if it's not a proper match (minimize the volume of the object that doesn't fit in the container)
Support for elastic objects - find the best fit while minimizing the "distortion" in the objects
This is a pretty general question - and I don't expect a complete solution.
Any pointers to relevant papers \ articles \ libraries \ tools would be useful
Here is one perhaps less than ideal method.
You could try fixing the position (in 3D space) of 1 shape. Placing the other shape on top of that shape. Then create links that connect one point in shape to a point in the other shape. Then simulate what happens when the links are pulled equally tight. Causing the point that isn't fixed to rotate and translate until it's stable.
If the fit is loose enough, you could use only 3 links (the bare minimum number of links for 3D) and try every possible combination. However, for tighter fit fits, you'll need more links, perhaps enough to place them on every point of the shape with the least number of points. Which means you'll some method to determine how to place the links, which is not trivial.
This seems like quite hard problem. Probable approach is to have some heuristic to suggest transformation and than check is it good one. If transformation moves object only slightly out of interior (e.g. on one part) than make slightly adjust to transformation and test it. If object is 'lot' out (e.g. on same/all axis on both sides) than make new heuristic guess.
Just an general idea for a heuristic. Make a rasterisation of an objects with same pixel size. It can be octree of an object volume. Make connectivity graph between pixels. Check subgraph isomorphism between graphs. If there is a subgraph than that position is for a testing.
This approach also supports 90deg rotation(s).
Some tests can be done even on graphs. If all volume neighbours of a subgraph are in larger graph, than object is in.
In general this is 'refined' boundary box approach.
Another solution is to project equal number of points on both objects and do a least squares best fit on the point sets. The point sets probably will not be ordered the same so iterating between the least squares best fit and a reordering of points so that the points on both objects are close to same order. The equation development for this is a lot of algebra but not conceptually complicated.
Consider one polygon(triangle) in the target object. For this polygon, find the equivalent polygon in the other geometry (source), ie. the length of the sides, angle between the edges, area should all be the same. If there's just one match, find the rigid transform matrix, that alters the vertices that way : X' = M*X. Since X' AND X are known for all the points on the matched polygons, this should be doable with linear algebra.
If you want a one-one mapping between the vertices of the polygon, traverse the edges of the polygons in the same order, and make a lookup table that maps each vertex one one poly to a vertex in another. If you have a half edge data structure of your 3d object that'll simplify this process a great deal.
If you find more than one matching polygon, traverse the source polygon from both the points, and keep matching their neighbouring polygons with the target polygons. Continue until one of them breaks, after which you can do the same steps as the one-match version.
There're more serious solutions that're listed here, but I think the method above will work as well.
What a juicy problem !. As is typical in computational geometry this problem
can be very complicated with a mismatched geometric abstraction. With all kinds of if-else cases etc.
But pick the right abstraction and the solution becomes trivial with few sub-cases.
Compute the Distance Transform of your shapes and Voilà! Your solution is trivial.
Allow me to elaborate.
The distance map of a shape on a grid (pixels) encodes the distance of the closest point on the
shape's border to that pixel. It can be computed in both directions outwards or inwards into the shape.
In this problem, the outward distance map suffices.
Step 1: Compute the distance map of both shapes D_S1, D_S2
Step 2: Subtract the distance maps. Diff = D_S1-D_S2
Step 3: if Diff has only positive values. Then your shapes can be contained in each other(+ve => S1 bigger than S2 -ve => S2 bigger than S1)
If the Diff has both positive and negative values, the shapes intersect.
There you have it. Enjoy !
Related
We have some polylines (list of points, has start and end point, not cyclic) and polygons (list of points, cyclic, no such thing as endpoints).
We want to map each polyline to a new polyline and each polygon to a new polygon so the total number of edges is small enough.
Let's say the number of edges originally is N, and we want our result to have M edges. N is much larger than M.
Polylines need to keep their start and end points, so they contribute at least 1 edge, one less than their vertex count. Polygons need to still be polygons, so they contribute at least 3 edges, equal to their vertex count. M will be at least large enough for this requirement.
The outputs should be as close as possible to the inputs. This would end up being an optimization problem of minimizing some metric to within some small tolerance of the true optimal solution. Originally I'd have used the area of the symmetric difference of the original and result (area between), but if another metric makes this easier to do I'll gladly take that instead.
It's okay if the results only include vertices in the original, the fit will be a little worse but it might be necessary to keep the time complexity down.
Since I'm asking for an algorithm, it'd be nice to also see an implementation. I'll likely have to re-implement it for where I'll be using it anyway, so details like what language or what data structures won't matter too much.
As for how good the approximation needs to be, about what you'd expect from getting a vector image from a bitmap image. The actual use here is for a tool for a game though, there's some strange details for the specific game, that's why the output edge count is fixed rather than the tolerance.
It's pretty hard to find any information on this kind of thing, so without even providing a full workable algorithm, just some pointers would be very much appreciated.
Ramer–Douglas–Peucker algorithm (mentioned in the comment) is definitely good, but it has some disadvantages:
It requires open polyline on input, for closed polygon one has to fix an arbitrary point, which either decreases final quality or forces to test many points and decreases the performance.
The vertices of simplified polyline are a subset of original polyline vertices, other options are not considered. This permits very fast implementations, but again decreases the precision of simplified polyline.
Another alternative is to take well known algorithm for simplification of triangular meshes Surface Simplification Using Quadric Error Metrics and adapt it for polylines:
distances to planes containing triangles are replaced with distances to lines containing polyline segments,
quadratic forms lose one dimension if the polyline is two dimensional.
But the majority of the algorithm is kept including the queue of edge contraction (minimal heap) according to the estimated distortion such contraction produces in the polyline.
Here is an example of this algorithm application:
Red - original polyline, blue - simplified polyline, and one can see that all its vertices do not lie on the original polyline, while general shape is preserved as much as possible with so few line segments.
it'd be nice to also see an implementation.
One can find an implementation in MeshLib, see MRPolylineDecimate.h/.cpp
I'm using the centroid of polygons to attach a marker in a map application. This works definitely fine for convex polygons and quite good for many concave polygons.
However, some polygons (banana, donut) obviously don't produce the desired result: The centroid is in these cases outside the polygons area.
Does anybody know a better approach to find a suitable point within any polygons area (which may contain holes!) to attach a marker?
One approach would be to generate and refine a skeleton of the polygon, then use the midpoint of the skeleton to place your marker (and if it's text, to orient the text correctly). This works well for most shapes, including ones with holes, and banana-shaped or tadpole-shaped crescents.
The CGAL library has a 2D Straight Skeleton and Polygon Offsetting module, or you could use PostGIS, for example.
To rephrase comment of ChristopheRoussy we may look for the largest circle inside of the polygon.
The largest circle is the one which cannot grow anymore because it touches three vertices or edges (if it touches only two, it can become bigger or just moved until it touches third).
So if you have few vertices, you can just enumerate all possible triples of vertices/edges, find for each one a circle and then select the largest one.
But it will require creating four functions:
Circle(vertex,vertex,vertex)
Circle(vertex,vertex,edge)
Circle(vertex,edge,edge)
Circle(edge,edge,edge)
All of them are possible, but may require some effort.
Find the extreme ordinates and draw an horizontal line in the middle. It is guaranteed to cross the polygon.
Find the intersection with the sides and sort them by increasing abscissa. Pick a point in the middle of two intersections.
This is an O(N + K Log K) process where K is the number of intersections (usually a very small even number). Pretty straightforward to write.
To increase the chances of a nice placement, you can try three horizontals instead of one an pick the longest intersection segment.
I have no idea how to solve this for any possible shape (and not doing heavy computation), but maybe for simpler shapes like the ones you have shown:
https://en.wikipedia.org/wiki/Force-directed_graph_drawing
Heuristic: This could converge to a reasonable approximation after a while
transform shape border into many points (more = more precise)
start out with many random points inside the polygon
push them until they are furthest away from border points, or just compute distance ... (can be done in parallel)
take best point
Another way could be to use multiple algorithms depending on the nature of the shape (like another one for donuts ...). Also perhaps relying on measuring 'fattest' sections first ?
IMHO would ask this on a math forum.
Similar: Calculate Centroid WITHIN / INSIDE a SpatialPolygon
Similar: How to find two most distant points?
To get a point for a marker I would use Yves Daoust's method.
To get a point that is reliably "within any polygon with holes" I would split polygon into triangles with a reliable library (e.g. OpenGL's GLUtessellator), and then get centroid of triangle with largest area.
If I had time for developing and testing, and I wanted good performance, then I would use a hybrid method: First use Yves Daoust's method to get some candidate points and then test candidates to see if they are within polygon. If all candidates fail, then fall back to slower reliable method (e.g. GLUtesselator).
for (int i = 0; i < n; /*++i*/) {
p = RandomPointInsideConvexHull();
if (IsInsidePolygon(p)) {
++i;
d = DistanceToClosestEdge(p);
if (d > bestD) {
bestP = p;
}
}
}
After running this loop you will approximate solution by bestP. n is parameter to choose. If you want more accurate result you can restart search, but now instead of picking a point inside polygon's convex hull you can pick one in the neighborhood of bestP, say not farther than bestD / 5 (this time you don't need to check if random point is inside polygon).
I want to write a program to draw a picture which covers a plane with tiled irregular quadrangles, just like this one:
However, I don't know the relevant algorithms, for example, in which order should I draw the edges?
Could someone point a direction for me?
Apologies, in my previous answer, I misunderstood the question.
Here is one stab at an algorithm (not necessarily the most optimal way, but a way). All you need is the ability to render a polygon and a basic rotation.
If you don't want the labels to be flipped, draw them separately (the labels can be stored in the vertices, e.g., and rotated with the polygon points, but drawn upright as text).
Edit
I received a question about the "start with an arbitrary polygon" step. I didn't communicate that step very clearly, as I actually intended to merely suggest an arbitrary polygon from the provided diagram, and not any arbitrary polygon in the world.
However, this should work at least for arbitrary quads, including concave ones, like so:
I'm afraid I lack the proper background to provide a proof as to why this works, however. Perhaps more mathematically-savvy people can help there with the proof.
I think one way to tackle the proof is to first start with the notion that all tiled edges are manifold -- this is a given considering that we're generating a neighboring polygon at every edge in order to generate the tiled result. Then we might be able to prove that every 2-valence boundary vertex is going to become a 4-valence vertex as a result of this operation (since each of its two edges are going to become manifold, and that introduces two new vertex edges into the mix -- this seems like the hardest part to prove to me). Last step might be to prove that the sum of the angles at each 4-valence vertex will always add up to 360 degrees.
I'm working with a really slow renderer, and I need to approximate polygons so that they look almost the same when confined to a screen area containing very few pixels. That is, I'd need an algorithm to go through a polygon and subtract/move a bunch of vertices until the end polygon has a good combination of shape preservation and economy of vertice usage.
I don't know if there's a formal name for these kind of problems, but if anyone knows what it is it would help me get started with my research.
My untested plan is to remove the vertices that change the polygon area the least, and protect the vertices that touch the bounding box from removal, until the difference in area from the original polygon to the proposed approximate one exceeds a tolerance I specify.
This would all be done only once, not in real time.
Any other ideas?
Thanks!
You're thinking about the problem in a slightly off way. If your goal is to reduce the number of vertices with a minimum of distortion, you should be defining your distortion in terms of those same vertices, which define the shape. There's a very simple solution here, which I believe would solve your problem:
Calculate distance between adjacent vertices
Choose a tolerance between vertices, below which the vertices are resolved into a single vertex
Replace all pairs of vertices with distances lower than your cutoff with a single vertex halfway between the two.
Repeat until no vertices are removed.
Since your area is ultimately decided by the vertex placement, this method preserves shape and minimizes shape distortion. The one drawback is that distance between vertices might be slightly less intuitive than polygon area, but the two are proportional. If you really wish, you could run through the change in area that would result from vertex removal, but that's a lot more work for questionable benefit imo.
As mentioned by Angus, if you want a direct solution for the change in area, it's not actually super difficult. Was originally going to leave this as an exercise to the reader, but it's totally possible to solve this exactly, though you need to include vertices on either side.
Assume you're looking at a window of vertices [A, B, C, D] that are connected in that order. In this example we're determining the "cost" of combining B and C.
Calculate the angle offset from collinearity from A toward C. Basically you just want to see how far from collinear the two points are. This is |sin(|arctan(B - A)| - |arctan(C - A)|)| Where pipes are absolute value, and differences are the sensical notion of difference.
Calculate the total distance over which the angle change will effectively be applied, this is just the euclidean distance from A to B times the euclidean distance from B to C.
Multiply the terms from 2 and 3 to get your first term
To get your second term, repeat steps 2 - 4 replacing A with D, B with C, and C with B (just going in the opposite direction)
Calculate the geometric mean of the two terms obtained.
The number that results in step 6 presents the full-picture minus a couple constants.
I tried my own plan first: Protect the vertices touching the bounding box, then remove the rest in the order that changes the resultant area the least, until you can't find a vertice to remove that keeps the new polygon area within X% of the original one. This is the result with X = 5%:
When the user zooms out really far these shapes fit the bill well enough for me. I haven't tried any of the other suggestions. The savings are quite astonishing, sometimes from 80-100 vertices down to 4 or 5.
Here is a problem I am trying to solve:
I have an irregular shape. How would I go about evenly distributing 5 points on this shape so that the distance between each point is equal to each other?
David says this is impossible, but in fact there is an answer out of left field: just put all your points on top of each other! They'll all have the same distance to all the other points: zero.
In fact, that's the only algorithm that has a solution (i.e. all pairwise distances are the same) regardless of the input shape.
I know the question asks to put the points "evenly", but since that's not formally defined, I expect that was just an attempt to explain "all pairwise distances are the same", in which case my answer is "even".
this is mathematically impossible. It will only work for a small subset of base shapes.
There are however some solutions you might try:
Analytic approach. Start with a point P0, create a sphere around P0 and intersect it with the base shape, giving you a set of curves C0. Then create another point P1 somewhere on C0. Again, create a sphere around P1 and intersect it with C0, giving you a set of points C1, your third point P2 will be one of the points in C1. And so on and so forth. This approach guarantees distance constraints, but it also heavily depends on initial conditions.
Iterative approach. Essentially form-finding. You create some points on the object and you also create springs between the ones that share a distance constraint. Then you solve the spring forces and move your points accordingly. This will most likely push them away from the base shape, so you need to pull them back onto the base shape. Repeat until your points are no longer moving or until the distance constraint has been satisfied within tolerance.
Sampling approach. Convert your base geometry into a voxel space, and start scooping out all the voxels that are too close to a newly inserted point. This makes sure you never get two points too close together, but it also suffers from tolerance (and probably performance) issues.
If you can supply more information regarding the nature of your geometry and your constraints, a more specific answer becomes possible.
For folks stumbling across here in the future, check out Lloyd's algorithm.
The only way to position 5 points equally distant from one another (other than the trivial solution of putting them through the origin) is in the 4+ dimensional space. It is mathematically impossible to have 5 equally distanced object in 3D.
Four is the most you can have in 3D and that shape is a tetrahedron.