I have a kind of cutting problem. There is an irregular polygon that doesn't have any holes and a list of standard sized of rectangular tiles and their values.
I want an efficient algorithm to find the single best valued tile that fit in this polygon; or an algorithm that just says if a single tile can fit inside the polygon. And it should run in deterministic time for irregular polygons with less than 100 vertices.
Please consider that you can rotate the polygon and tiles.
Answers/hints for both convex and non-convex polygons are appreciated.
Disclaimer: I've never read any literature on this, so there might be a better way of doing this. This solution is just what I've thought about after having read your question.
A rectangle has two important measurements - it's height and it's width
now if we start with a polygon and a rectangle:
1: go around the perimeter of the polygon and take note of all the places the height of the rectangle will fit in the polygon (you can store this as a polygon*):
2: go around the perimeter of the new polygon you just made and take note of all the places the width of the rectangle will fit in the polygon (again, you can store this as a polygon):
3: the rectangle should fit within this new polygon (just be careful that you position the rectangle inside the polygon correctly, as this is a polygon - not a rectangle. If you align the top left node of the rectangle with the top left node of this new polygon, you should be ok)
4: if no area can be found that the rectangle will fit in, rotate the polygon by a couple of degrees, and try again.
*Note: in some polygons, you will get more than one place a rectangle can be fitted:
After many hopeless searches, I think there isn't any specific algorithm for this problem. Until, I found this old paper about polygon containment problem.That mentioned article, present a really good algorithm to consider if a polygon with n points can fit a polygon with m points or not. The algorithm is of O(n^3 m^3(n+m)log(n+m)) in general for two transportable and rotatable 2D polygon.
I hope it can help you, if you are searching for such an irregular algorithm in computational geometry.
This might help. It comes with the source code written Java
http://cgm.cs.mcgill.ca/~athens/cs507/Projects/2003/DanielSud/
Related
I need an algorithm for splitting a (convex) polygon (let's call it base polygon). The polygon should be splitted into several smaller polygons by another polygon's edges (let's call this one the splitting polygon).
I know there exist algorithms for clipping a polygon (f.e. the Sutherland-Hodgman algorithm), but these algorithms discard the vertices that are lying outside of the splitting polygon instead of creating new polygons with them. I don't want to clip a polygon, i want to split it into several small parts.
I know the answer seems quite obvious because I would just have to extend the existing algorithms.
The problem is that I can't figure out a nice and performant way of doing this.
Are there existing algorithms that describe how to best split a polygon in a performant way?
There has to be a simple solution for this problem that I can't figure out at the moment.
You can see this problem as that of finding the intersection of the subject polygon and the complement of the window polygon. So you can use the standard Surtherland-Hodgman algorithm for that purpose, taking two precautions:
swap the roles of the subject and the window (actually your window is not convex as you consider its complement, only the subject is convex),
embed the window polygon in a large bounding box that covers both polygons, and consider the box as a polygon with a hole.
Example: the subject polygon is the rectangle and the window is the pentagon. Form the green polygon (larger rectangle with a hole) and clip it inside the rectangular (convex) window. The result of the clipping is in blue.
With some extra care, it should be possible to do without the large bounding box.
I am tyring to cover a simple concave polygon with a minimum rectangles. My rectangles can be any length, but they have maximum widths, and the polygon will never have an acute angle.
I thought about trying to decompose my concave polygon into triangles that produce a set of minimumally overlapping rectangles minimally bounding each triangle and then merging those rectangles into larger ones. However, I don't think this will work for small notches in the edges of the polygon. The triangles created by the reflex vertices on those notches will create the wrong rectangles. I am looking for rectangles that will span/ignore notches.
I don't really know anything about computational geometry, so I'm not really sure on how to begin asking the question.
I found other posts that were similar, but not what I need:
split polygon into minimum amount of rectangles and triangles
Covering an arbitrary polygon with minimum number of squares
Find $k$ rectangles so that they cover the maximum number of points
Algorithm for finding the fewest rectangles to cover a set of rectangles
Some examples: Black is the input. Red is the acceptable output.
Another exmaple: The second output is prefered. However, generating both outputs and using another factor to determine preference is probably necessary and not the responsibility of this algorithm.
Polygons that mimic curves are extremely rare. In this scenario much of the area of the rectangles is wasted. However, this is acceptable because each rectangle obeys the max width constraint.
Also, I found this article to be close to what I need:
Covering with rectangular pieces by Paul Iacob, Daniela Marinescu, and Cristina Luca
Maybe a better question is "How can I identify rectangular-like portions of a concave polygon?"
Here is an image showing the desired implementation:
The green is the actual material usage. The red rectangles are the layouts. The blue is the MBR of the entire polygon. I am thinking I should try to get little MBRs and fill them in. The 2-3 green rectangles in the upper left corner that terminate into the middle of the polygon are expensive. That is what I want to minimize. The green rectangles have a min and max width and height, but I can use as many rows and columns necessary to cover a region. Again, I must minimize the number of rectangles that do not span across the input. I can also modify the shape of the green rectangle to fit in small places that is also very expensive. In other words, getting as many rectangles as possible to span as much as possible is ideal.
Maybe I should simply be trying to identify rectangular regions like this:
Or, perhaps a better approach would be using largest-inscribed rectangles instead of MBRs. I could continually cut my polygon down using rectangles until I am left with regions were the largest-inscribed rectangle is not sharing an edge with the original polygon. The remaining regions would have to be handled with a heuristic approach.
I've been working with the engineering and manufacturing departments at my company to bring more clarificaiton to this problem. I am still waiting to confirm, but I am now thinking an algorithm that would return sets of largest inscribed rectangles would work. While it does not completely cover the shape, it would give preference to the orthognal regions while leaving the non-orthogonal regions to some heuristics. The only trick is to maxamize those orthogonal regions.
I have two 2D rectangles, defined as an origin (x,y) a size (height, width) and an angle of rotation (0-360°). I can guarantee that both rectangles are the same size.
I need to calculate the approximate area of intersection of these two rectangles.
The calculation does not need to be exact, although it can be. I will be comparing the result with other areas of intersection to determine the largest area of intersection in a set of rectangles, so it only needs to be accurate relative to other computations of the same algorithm.
I thought about using the area of the bounding box of the intersected region, but I'm having trouble getting the vertices of the intersected region because of all of the different possible cases:
I'm writing this program in Objective-C in the Cocoa framework, for what it's worth, so if anyone knows any shortcuts using NSBezierPath or something you're welcome to suggest that too.
To supplement the other answers, your problem is an instance of line clipping, a topic heavily studied in computer graphics, and for which there are many algorithms available.
If you rotate your coordinate system so that one rectangle has a horizontal edge, then the problem is exactly line clipping from there on.
You could start at the Wikipedia article on the topic, and investigate from there.
A simple algorithm that will give an approximate answer is sampling.
Divide one of your rectangles up into grids of small squares. For each intersection point, check if that point is inside the other rectangle. The number of points that lie inside the other rectangle will be a fairly good approximation to the area of the overlapping region. Increasing the density of points will increase the accuracy of the calculation, at the cost of performance.
In any case, computing the exact intersection polygon of two convex polygons is an easy task, since any convex polygon can be seen as an intersection of half-planes. "Sequential cutting" does the job.
Choose one rectangle (any) as the cutting rectangle. Iterate through the sides of the cutting rectangle, one by one. Cut the second rectangle by the line that contains the current side of the cutting rectangle and discard everything that lies in the "outer" half-plane.
Once you finish iterating through all cutting sides, what remains of the other rectangle is the result.
You can actually compute the exact area.
Make one polygon out of the two rectangles. See this question (especially this answer), or use the gpc library.
Find the area of this polygon. See here.
The shared area is
area of rectangle 1 + area of rectangle 2 - area of aggregated polygon
Take each line segment of each rectangle and see if they intersect. There will be several possibilities:
If none intersect - shared area is zero - unless all points of one are inside the other. In that case the shared area is the area of the smaller one.
a If two consecutive edges of one rectactangle intersect with a single edge of another rectangle, this forms a triangle. Compute its area.
b. If the edges are not consequtive, this forms a quadrilateral. Compute a line from two opposite corners of the quadrilateral, this makes two triangles. Compute the area of each and sum.
If two edges of one intersect with two edges of another, then you will have a quadrilateral. Compute as in 2b.
If each edge of one intersects with each edge of the other, you will have an octagon. Break it up into triangles ( e.g. draw a ray from one vertex to each other vertex to make 4 triangles )
#edit: I have a more general solution.
Check the special case in 1.
Then start with any intersecting vertex, and follow the edges from there to any other intersection point until you are back to the first intersecting vertex. This forms a convex polygon. draw a ray from the first vertex to each opposite vetex ( e.g. skip the vertex to the left and right. ) This will divide it into a bunch of triangles. compute the area for each and sum.
A brute-force-ish way:
take all points from the set of [corners of
rectangles] + [points of intersection of edges]
remove the points that are not inside or on the edge of both rectangles.
Now You have corners of intersection. Note that the intersection is convex.
sort the remaining points by angle between arbitrary point from the set, arbitrary other point, and the given point.
Now You have the points of intersection in order.
calculate area the usual way (by cross product)
.
I want to do the following: I have some faces in the 3D space as polygons. I have a projection direction and a projection plane. I have a convex clipping polygon in the projection plane. I wnat to get a polygon representing the shaddow of all the faces clipped on the plane.
What I do till now: I calculate the projections of the faces as polygons in the projection plane.
I could use the Sutherland–Hodgman algorithm to clip all the singe projected polygons to clip to the desired area.
Now my question: How can I combine the projected (maybe clipped) polygons together? Do I have to use algorithms like Margalit/Knott?
The algorithm should be quite efficient because it has to run quite often. So what algorithm do you suppose?
Is it maybe possible to modify the algorithm of Sutherland–Hodgman to solve the merging problem?
I'm currently implementing this algorithm (union of n concave polygons) using Bentley–Ottmann to find all edge intersections and meanwhile keeping track of the polygon nesting level on both sides of edge segments (how many overlapping polygons each side of the line is touching). Edges that have a nesting level of 0 on one side are output to the result polygon. It's fairly tricky to get done right. An existing solution with a different algorithm design can be found at:
http://sourceforge.net/projects/polyclipping/
I need to solve the following problem:
I have multiple rectangles of sizes: width height, width/2 height/2, width/4 height/4 , width/8 height/8 ... etc
I need to pack these rectangles in a big rectangle of size x*width y*height such that no rectangles overlap, the rectangles are distributed randomly in the packing and any rectangle should at least touch another rectangle. I tried a fairly basic greedy algorithm but it fails.
Can you give me some suggestions on how to solve the problem?
Thanks!
EDIT: You can have more than one rectangle of each size
This is not homework. I'm trying to create an effect similar to the effect on ted.com
By random I mean that there might exist more than one packing of the rectangles that satisfies the constraints. The algorithm should not produce the same packing at each run.
This sounds like a rectangle packing problem. There is a link there to an algorithm. That code packs the rectangles as tightly as possible. You said you want the rectangles to be distributed randomly, which I'm guessing means not all rectangles of one size next to each other and all rectangles spread out to fill the big rectangle. Maybe the code at the link above would be a good starting point to get some ideas.
You can use a spatial index or a quadtree to subdivide the 2d-plane. The idea is to reduce the 2d problem to a 1d-problem. Once you got the spatial index (or space-filling-curve) and you can discretize the 2d into 1d you can use the 1d to search for similarity or to sort from low to high or the reverse for example by the length. If you got this order you can then compute the index back to a 2d represenation and to pack them in most efficent way in your container. There are many ways to make a spatial index. Some of the best but difficult to make is the hilbert curve. Another one is the z-curve or morton-curve. It's different from zizag-curve because it's subdivide the plane into 4 squares (not rectangles).
EDIT: Here is a link for an Jquery-Plugin: http://www.fbtools.com/jquery/treemap/
Here with world poplulation: http://www.fbtools.com/jquery/treemap/population.html
EDIT: http://people.csail.mit.edu/konak/papers/socg_2008-circular_partitions_with_applications_to_visualization_and_embeddings.html
EDIT: http://lip.sourceforge.net/ctreemap.html
At each step you divide the surface of your new rectange by 4.
SUM(1/4n for n in [0,inf]) = 4/3**
So the best you can do is fit your rectangle in a rectangle of surface
4/3 (height*width)
(that's a lower bound)
#mloskot algorithm gives a possible solution that will be in a rectangle of surface 3/2*(height*width) : Here is an illustration:
I don't see how you can do better.
Assuming you have only one rectangle of each size, you can try to replicate the arrangement of paper sizes. Sort the rectangles by size from the biggest to the smallest, then
Take first rectangle and place it at the corner of the target plane.
Take next rectangle (assert it's smaller than the previous rectangle)
Rotate about 90 degrees
Place so
its shorter size is adjacent to the longer size of the last bigger neighbour
and its longer side is adjacent to the edge of the target plane or edge of neighbour of the same
size
Repeat 2 - 4
I realise the description might be unclear, so here is picture presenting the solution - it should help to grasp it:
This is a lot like MIP-mapping