I'm looking for a packing algorithm which will reduce a regular polygon into rectangles and right triangles. The algorithm should attempt to use as few such shapes as possible and should be relatively easy to implement (given the difficulty of the challenge).
If possible, the answer to this question should explain the general heuristics used in the suggested algorithm.
I think the answer is fairly simple for regular polygons.
Find an axis of symmetry, and draw a line between each vertex and its mirror. This divides the polygon into trapezoids. Each trapezoid can be turned into a rectangle and two right triangles.
It's not specifically rectangles + right triangles, but a good research point for looking into tesselating polygons is Voronoi Diagrams and Delaunay Triangulations and here and here.
In fact, if "just right triangles" is good enough, these are guaranteed to triangulate for you, and you can always split any triangle into two right triangles, if you really need those. Or you can chop off "tips" of triangles to make more right triangles and some rectangles out of the right-triangles.
You can also try ear-clipping, either by sweeping radially, if you know you have fairly regular polygons, or by "clipping the biggest convex chunk" off. Then, split each remaining triangle into two to create right triangles.
(source: eruciform.com)
You could try to make less breaks by sweeping one way and then the other to make a trapezoid and split it differently, but you then have to do a check to make sure that your sweep-line hasn't crossed another line someplace. You can always ear-clip, even with something practically fractal.
However, this sometimes creates very slim triangles. You can perform heuristics, like "take the biggest", instead of clipping continuously along the edge, but that takes more time, approaching O(n^2). Delaunay/Vornoi will do it more quickly in most cases, with less slim triangles.
You can try "cuting out" the largest rectangle that can fit in the polygon, leaving behind some leftovers. Keep repeating the cutting out of rectangles on the leftovers, until you end up with triangular pieces. Then, you can split them into two right triangles if necessary. I do not know if this will always yield solutions that will give you the least amount rectangles and right triangles.
Related
I have a big rectangle filled with circles. The circles might overlap each other, but they all have the same diameter. I need to find the "borderline" circles. If there are gaps between these borderline circles - and the gap is bigger than a circle diameter - the one inside should also be included.
Here are some examples:
What I need to do in is to make these outer circles immovable, so that when the inner circles move - they never exit the rectangle. How can it be made, are there any known algorithms for such a thing? I'm doing it in TypeScript, but I guess, any imperative language solution can be applied
This is perhaps too conservative but certainly won't let any disks out.
Compute a Delaunay triangulation on the centers of the circles. A high-quality library is best because the degenerate cases and floating-point tests are tricky to get right.
Using depth-first search on the planar dual, find all of the faces that are reachable from the infinite face crossing only edges longer than (or equal to? depends on whether these are closed or open disks) two times the diameter.
All of the points incident to these faces correspond to the exterior disks.
I'm not quite sure that I have the answer to your question, but I put together two quick examples using the Tinfour Delaunay Triangulation library (which is written in Java). I had to digitize your points by hand, so I didn't quite hit the center in all cases.
The first picture shows that the boundary of a Delaunay Triangulation is a convex polygon. This is easy to build, simply add the vertices (circle centers) to Tinfour's IncrementalTin class and then ask it for the bounding polygon. Pretty much any Delaunay library will support this. So you wouldn't necessarily need Tinfour.
The problem is that this may include areas that are not valid for your interior circles. I played a little with ways to introduce concavities to the bounding polygon. As you can see below, the vertices in the lower-right corner had to be lopped off entirely (if I understand what you are looking for). I then iterated over the perimeter edges and introduced concavities where the vertex opposite each exterior edge was a "guard" vertex.
The code I wrote to do this is pretty messy. But if this is what you are looking for, I'll try to get it cleaned up and post it here.
Is there an optimized way to get the convex hull, if I know my points are always arranged into two rectangles?
I programmed the classic convex hull algorithm (just by enumerating all points), but since I have a bunch of pairs of rectangles I was wandering if there might be a more efficient way to do it for this special case.
This is what I am talking about, to clarify:
I tried sorting the points in various ways, but I just cannot find a general rule to optimize it. Is the basic convex hull algorithm the most efficient way to do this case also?
Update
To clarify my final goal, I have ~100 rectangles already grouped into pairs of two, and thousands of points for which I have to check whether they lie within each of these convex hulls, in real time. Now that I've given some thought to it, I guess the convex hull part won't be the bottleneck in the entire operation (but there is still ~100 of them, and I am aiming for realtime 60fps processing), so I might as well use a plain ol' algorithm as #BartKiers suggested and then get back to this after profiling.
I will leave the question open for a while, perhaps someone has an idea for an optimization which might be useful anyway.
If I am right, you can enumerate all relevant configurations by noting that if the left side of a rectangle is more to the left than the left side of the other, then its two left vertices are on the convex hull.
With the same reasoning in the four cardinal directions, there are 16 distinct cases that you can hard-code.
Another way to look at it is to observe that the convex hull is the tightest bounding box of the two rectangles, with 0, 2 or 4 corners "cut off". Finding the bounding box is trivial, and you decide if a corner is to be cut when it does not belong to any of the rectangles.
You easily derive a point inclusion test from this rule. If you already have a bounding box test, it suffices to add the corner tests.
What is the best method to detect whether the red rectangle overlaps the black polygon? Please refer to this image:
There are four cases.
Rect is outside of Poly
Rect intersects Poly
Rect is inside of Poly
Poly is inside of Rect
First: check an arbitrary point in your Rect against the Poly (see Point in Polygon). If it's inside you are done, because it's either case 3 or 2.
If it's outside case 3 is ruled out.
Second: check an arbitrary point of your Poly against the Rect to validate/rule out case 4.
Third: check the lines of your Rect against the Poly for intersection to validate/rule out case 2.
This should also work for Polygon vs. Polygon (convex and concave) but this way it's more readable.
If your polygon is not convex, you can use tessellation to subdivide it into convex subparts. Since you are looking for methods to detect a possible collision, I think you could have a look at the GJK algorithm too. Even if you do not need something that powerful (it provides information on the minimum distance between two convex shapes and the associated witness points), it could prove to be useful if you decide to handle more different convex shapes.
Christer Ericson made a nice Powerpoint presentation if you want to know more about this algorithm. You could also take a look at his book, Real-Time Collision Detection, which is both complete and accessible for anyone discovering collision detection algorithms.
If you know for a fact that the red rectangle is always axis-aligned and that the black region consists of several axis-aligned rectangles (I'm not sure if this is just a coincidence or if it's inherent to the problem), then you can use the rectangle-on-rectangle intersection algorithm to very efficiently compute whether the two shapes overlap and, if so, where they overlap.
If you use axis-aligned rectangles and polygons consist of rectangles only, templatetypedef's answer is what you need.
If you use arbitrary polygons, it's a much more complex problem.
First, you need to subdivide polygons into convex parts, then perform collision detection using, for example, the SAT algorithm
Simply to find whether there is an intersection, I think you may be able to combine two algorithms.
1) The ray casting algorithm. Using the vertices of each polygon, determine if one of the vertices is in the other. Assuming you aren't worried about the actual intersection region, but just the existence of it. http://en.wikipedia.org/wiki/Point_in_polygon
2) Line intersection. If step 1 produces nothing, check line intersection.
I'm not certain this is 100% correct or optimal.
If you actually need to determine the region of the intersection, that is more complex, see previous SO answer:
A simple algorithm for polygon intersection
I've been searching far and wide on the seven internets, and have come to no avail. The closest to what I need seems to be The cutting stock problem, only in 2D (which is disappointing since Wikipedia doesn't provide any directions on how to solve that one). Another look-alike problem would be UV unwrapping. There are solutions there, but only those that you get from add-ons on various 3D software.
Cutting the long talk short - what I want is this: given a rectangle of known width and height, I have to find out how many shapes (polygons) of known sizes (which may be rotated at will) may I fit inside that rectangle.
For example, I could choose a T-shaped piece and in the same rectangle I could pack it both in an efficient way, resulting in 4 shapes per rectangle
as well as tiling them based on their bounding boxes, case in which I could only fit 3
But of course, this is only an example... and I don't think it would be much use to solving on this particular case. The only approaches I can think of right now are either like backtracking in their complexity or solve only particular cases of this problem. So... any ideas?
Anybody up for a game of Tetris (a subset of your problem)?
This is known as the packing problem. Without knowing what kind of shapes you are likely to face ahead of time, it can be very difficult if not impossible to come up with an algorithm that will give you the best answer. More than likely unless your polygons are "nice" polygons (circles, squares, equilateral triangles, etc.) you will probably have to settle for a heuristic that gives you the approximate best solution most of the time.
One general heuristic (though far from optimal depending on the shape of the input polygon) would be to simplify the problem by drawing a rectangle around the polygon so that the rectangle would be just big enough to cover the polygon. (As an example in the diagram below we draw a red rectangle around a blue polygon.)
Once we have done this, we can then take that rectangle and try to fit as many of that rectangle into the large rectangle as possible. This simplfies the problem into a rectangle packing problem which is easier to solve and wrap your head around. An example of an algorithm for this is at the following link:
An Effective Recursive Partitioning Approach for the Packing of Identical Rectangles in a Rectangle.
Now obviously this heuristic is not optimal when the polygon in question is not close to being the same shape as a rectangle, but it does give you a minimum baseline to work with especially if you don't have much knowledge of what your polygon will look like (or there is high variance in what the polygon will look like). Using this algorithm, it would fill up a large rectangle like so:
Here is the same image without the intermediate rectangles:
For the case of these T-shaped polygons, the heuristic is not the best it could be (in fact it may be almost a worst case scenario for this proposed approximation), but it would work very well for other types of polygons.
consider what the other answer said by placing the t's into a square, but instead of just leaving it as a square set the shapes up in a list. Then use True and False to fill the nested list as the shape i.e. [[True,True,True],[False,True,False]] for your T shape. Then use a function to place the shapes on the grid. To optimize the results, create a tracker which will pay attention to how many false in a new shape overlap with trues that are already on the grid from previous shapes. The function will place the shape in the place with the most overlaps. There will have to be modifications to create higher and higher optimizations, but that is the general premise which you are looking for.
I'm looking for a packing algorithm which will reduce an irregular polygon into rectangles and right triangles. The algorithm should attempt to use as few such shapes as possible and should be relatively easy to implement (given the difficulty of the challenge). It should also prefer rectangles over triangles where possible.
If possible, the answer to this question should explain the general heuristics used in the suggested algorithm.
This should run in deterministic time for irregular polygons with less than 100 vertices.
The goal is to produce a "sensible" breakdown of the irregular polygon for a layman.
The first heuristic applied to the solution will determine if the polygon is regular or irregular. In the case of a regular polygon, we will use the approach outlined in my similar post about regular polys: Efficient Packing Algorithm for Regular Polygons
alt text http://img401.imageshack.us/img401/6551/samplebj.jpg
I don't know if this would give the optimal answer, but it would at least give an answer:
Compute a Delaunay triangulation for the given polygon. There are standard algorithms to do this which will run very quickly for 100 vertices or fewer (see, for example, this library here.) Using a Delaunay triangulation should ensure that you don't have too many long, thin triangles.
Divide any non-right triangles into two right triangles by dropping an altitude from the largest angle to the opposite side.
Search for triangles that you can combine into rectangles: any two congruent right triangles (not mirror images) which share a hypotenuse. I suspect there won't be too many of these in the general case unless your irregular polygon had a lot of right angles to begin with.
I realize that's a lot of detail to fill in, but I think starting with a Delaunay triangulation is probably the way to go. Delaunay triangulations in the plane can be computed efficiently and they generally look quite "natural".
EDITED TO ADD: since we're in ad-hoc heuristicville, in addition to the greedy algorithms being discussed in other answers you should also consider some kind of divide and conquer strategy. If the shape is non-convex like your example, divide it into convex shapes by repeatedly cutting from a reflex vertex to another vertex in a way that comes as close to bisecting the reflex angle as possible. Once you've divided the shape into convex pieces, I'd consider next dividing the convex pieces into pieces with nice "bases", pieces with at least one side having two acute or right angles at its ends. If any piece doesn't have such a "base" you should be able to divide it in two along a diameter of the piece, and get two new pieces which each have a "base" (I think). This should reduce the problem to dealing with convex polygons which are kinda-sorta trapezoidal, and from there a greedy algorithm should do well. I think this algorithm will subdivide the original shape in a fairly natural way until you get to the kinda-sorta trapezoidal pieces.
I wish I had time to play with this, because it sounds like a really fun problem!
My first thought (from looking at your diagram above) would be to look for 2 adjacent right angles turning the same direction. I'm sure that won't catch every case where a rectangle will help, but from a user's point of view, it's an obvious case (square corners on the outside = this ought to be a rectangle).
Once you've found an adjacent pair of right angles, take the length of the shorter leg, and there's one rectangle. Subtract this from the polygon left to tile, and repeat. When there's no more obvious external rectangles to remove, then do your normal tiling thing (Peter's answer sounds great) on that.
Disclaimer: I'm no expert on this, and I haven't even tried it...