I have a connected shape that consists of squares put together, e.g. take a squared paper and draw a line along the existing lines that ends at its beginning and does not cross itself.
The goal is now to find an algorithm (not brute-force) that fills this shape with as few, non-overlapping rectangles as possible.
I'm looking for the optimal solution. As can be seen in the images, the naive greedy approach (take the largest rectangle) does not work.
(Optimal)
(Greedy)
My scenario is vertices reduction, but I'm sure there are other use-cases as well.
Note: This problem seems basic, but I was not able to find a solution elsewhere. Also, is this problem NP-hard?
Edit: I just realized that, in my scenario, filling the shape with as few non-overlapping triangles as possible, would give an even better result.
I've spend a lot of time researching this, since I asked the initial question. For the first problem (optimally filling the shape with rectangles), I've written the solution here under the header "Optimal Greedy Meshing":
http://blackflux.wordpress.com/2014/03/01/meshing-in-voxel-engines-part-2/
The complexity is actually better (faster) than for optimally triangulating a polygon without holes. The slowest part is the Hopcroft-Karp algorithm.
Treating the shape as a polygon is also discussed in the linked blog post. Note that I'm also considering holes.
The first problem is harder than the one with triangles; for triangles, see the algorithms in
http://en.wikipedia.org/wiki/Polygon_triangulation
which can do it without any extra vertices.
Related
consider an image like this:
by grouping pixels by color into distinct rectangles, different configurations might be achieved, for example:
the goal is to find one of the best configurations, i.e. a configuration which has the least possible number of rectangles (rectangles sizes are not important).
any idea on how to design an efficient algorithm which is able to solve this problem?
EDIT:
i think the best answer is the one by #dshin, as they proved that this problem is a NP-HARD one so there probably isn't any efficient solution that is able to guarantee an optimal result.
other answers provide reasonable compromises to get an acceptable solution, but that won't always be the optimal one.
Each connected colored region is a rectilinear polygon that can be considered independently, and so your problem amounts to solving the minimum rectangle covering for rectilinear polygons. This is a well-studied problem that finds applications in some fields, like VLSI.
For convex rectilinear polygons, there is an algorithm that finds the optimal solution in polynomial time, described in this 1984 thesis.
The non-convex case is NP-hard (reference), so an efficient optimal solution likely does not exist. But there are several algorithms which produce good empirical results. This 1990 publication describes three separate algorithms, each of which are guaranteed to use at most twice as many rectangles as the optimal solution. This 2016 publication describes an algorithm that uses the common IP + LP relaxation technique, which apparently produces better results in real-life problem instances, although lacking in theoretical guarantees. Unfortunately, both publications are behind paywalls, and I haven't been able to find free resources that describe the algorithms.
If you are just looking for something reasonable, and your problem instances are not pathological in nature, then the algorithms described in other answers are probably good enough.
I don't have a proof but my feeling is a greedy approach should solve this problem:
Start on the upper left (or in whichever corner)
Expand rectangle 1px to the right as long as colors match
Expand rectangle 1px to the bottom as long as all colors in that row match
Line by line and column by column, find the next pixel that is not already part of a square (maybe keep track of visited pixels in a second array) and repeat 2 and 3.
You can switch lines and columns and go up and left or whatever instead and end up with different configurations, but from playing this through in my mind I think the number of rectangles should always be the same.
The idea here is based on the following links: Link 1 and Link 2.
In both the cases, the largest possible rectangle is computed within a given polygon/shape. Check both the above links for details.
We can extend the idea above to the problem at hand.
Steps:
Filter the image by color (say red)
Find the largest possible rectangle in the red region. After doing so mask it.
Repeat to find the next biggest rectangle until all the portions in red have been covered.
Repeat the above for every unique color.
Overview:
I want to programm a tool that can place objects on a rectangle with the minumum of waste, this problem is also known as the cutting problem.
So i looked around to find some algorithms and i found out there are a few for rectangles but not that much for n-edged polygones.
my first approach was to get a bounding box for the polygone, then run the normal rectangle algorithm. After that you cound slowly try to increase the number of edges but still have only isometric lines (only vertical and horizontal), to approximate the polygone.
I wonder if there is any good algorithm that implement such thing, but is more common than create my own stuff.
the other way ive come up with could be something with two dimensional knapsack and some sorting heuristics that sort the best fitting polygones and try to put them on the rectangle.
But all i come up with has some good detection of special polygones (such as a square or normal rectangle) but does not work on common polygones.
I have a final set of tiles in which every edge can have on of four colors.
The task is to find a maximal possible square build from a given set (finite) of this tiles. Tiles can be rotated.
I need to design 3 algorithms for finding a solution for this task. One complete and two aproximations.
Obviously it is my task for Algorithms class so Im not asking about complete solutions (as this would be unfair) but for some directions.
Im already designed a kind of complete algorithm (using backtracking - search for a square of size sqrt(n) - if it could not be found try finding smaller and so on) but I have no idea how to create aproximation algorithms. I think one will be kind of stupid which will find a good answer only in specific cases just to document that it is not a good aproach but still I need one much faster then backtracking and quite good one.
Also is this problem NP-hard one? My backtracking algorithm is exponential one but it doesnt mean that there cannot be a better one...
EDIT: I have complete algorithm with exponential time, could some one give me some hints how to build some kind of aproximation for this problem with polynomial time or something better then exponential?
EDIT2: I have the idea that this problem can be changed to a problem of reducting a graph to square grid graph ( http://mathworld.wolfram.com/GridGraph.html ). Still there is a problem if the tiles can be arranged in such a way to build a grid, but this could be a good point to start. Are there any, for example, greedy or any other aproximation algorithms for reducting graph to square-grid graph?
Suppose your backtracking algorithm constructs k-by-k squares for increasing values of k.
You can extend the backtracking algorithm with heuristics. So instead of choosing the next tile randomly, choose and attach a tile such that the colors of the free tiles "agree with" those on the square. The big problem is to find the "agreement" heuristics. One possible heuristics is to find the least common color on the free tiles and use it.
I have a detailed 2D polygon (representing a geographic area) that is defined by a very large set of vertices. I'm looking for an algorithm that will simplify and smooth the polygon, (reducing the number of vertices) with the constraint that the area of the resulting polygon must contain all the vertices of the detailed polygon.
For context, here's an example of the edge of one complex polygon:
My research:
I found the Ramer–Douglas–Peucker algorithm which will reduce the number of vertices - but the resulting polygon will not contain all of the original polygon's vertices. See this article Ramer-Douglas-Peucker on Wikipedia
I considered expanding the polygon (I believe this is also known as outward polygon offsetting). I found these questions: Expanding a polygon (convex only) and Inflating a polygon. But I don't think this will substantially reduce the detail of my polygon.
Thanks for any advice you can give me!
Edit
As of 2013, most links below are not functional anymore. However, I've found the cited paper, algorithm included, still available at this (very slow) server.
Here you can find a project dealing exactly with your issues. Although it works primarily with an area "filled" by points, you can set it to work with a "perimeter" type definition as yours.
It uses a k-nearest neighbors approach for calculating the region.
Samples:
Here you can request a copy of the paper.
Seemingly they planned to offer an online service for requesting calculations, but I didn't test it, and probably it isn't running.
HTH!
I think Visvalingam’s algorithm can be adapted for this purpose - by skipping removal of triangles that would reduce the area.
I had a very similar problem : I needed an inflating simplification of polygons.
I did a simple algorithm, by removing concav point (this will increase the polygon size) or removing convex edge (between 2 convex points) and prolongating adjacent edges. In any case, doing one of those 2 possibilities will remove one point on the polygon.
I choosed to removed the point or the edge that leads to smallest area variation. You can repeat this process, until the simplification is ok for you (for example no more than 200 points).
The 2 main difficulties were to obtain fast algorithm (by avoiding to compute vertex/edge removal variation twice and maintaining possibilities sorted) and to avoid inserting self-intersection in the process (not very easy to do and to explain but possible with limited computational complexity).
In fact, after looking more closely it is a similar idea than the one of Visvalingam with adaptation for edge removal.
That's an interesting problem! I never tried anything like this, but here's an idea off the top of my head... apologies if it makes no sense or wouldn't work :)
Calculate a convex hull, that might be way too big / imprecise
Divide the hull into N slices, for example joining each one of the hull's vertices to the center
Calculate the intersection of your object with each slice
Repeat recursively for each intersection (calculating the intersection's hull, etc)
Each level of recursion should give a better approximation.... when you reached a satisfying level, merge all the hulls from that level to get the final polygon.
Does that sound like it could do the job?
To some degree I'm not sure what you are trying to do but it seems you have two very good answers. One is Ramer–Douglas–Peucker (DP) and the other is computing the alpha shape (also called a Concave Hull, non-convex hull, etc.). I found a more recent paper describing alpha shapes and linked it below.
I personally think DP with polygon expansion is the way to go. I'm not sure why you think it won't substantially reduce the number of vertices. With DP you supply a factor and you can make it anything you want to the point where you end up with a triangle no matter what your input. Picking this factor can be hard but in your case I think it's the best method. You should be able to determine the factor based on the size of the largest bit of detail you want to go away. You can do this with direct testing or by calculating it from your source data.
http://www.it.uu.se/edu/course/homepage/projektTDB/ht13/project10/Project-10-report.pdf
I've written a simple modification of Douglas-Peucker that might be helpful to anyone having this problem in the future: https://github.com/prakol16/rdp-expansion-only
It's identical to DP except that it pushes a line segment outwards a bit if the points that it would remove are outside the polygon. This guarantees that the resulting simplified polygon contains all the original polygon, but it has almost the same number of line segments as the original DP algorithm and is usually reasonably good at approximating the original shape.
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...