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Find the set of largest contiguous rectangles to cover multiple areas [duplicate]
(3 answers)
Closed 9 years ago.
Suppose I have a grid containing 1's and 0's... How can I find the shortest list of rectangles (really, boundaries) that, together, cover all of the 1's and don't cover any 0's?
Currently, I'm repeatedly finding the largest rectangle in the grid, storing it's dimensions, and setting all of the cells in the area to 0, until they are all set to 0.
My two issues with this are: it seems very efficient, and I can't seem to prove that it actually covers the area in the fewest rectangles but I can't find any scenarios to disprove it either.
Update: I found a grid that disproves the method I described...
0000000
0000010
0111110
0000010
0000000
The horizontal string of 1's is the largest existing rectangle, but once removed, it splits the vertical string of 1's into two small rectangles... Clearly, this could be solved in just 2 rectangles.
A generic depth first search of the solution space may not be the most efficient, but will find the answer. You can even have it output every solution (for example the grid you show has two solutions where every '1' is covered using just two triangles.) This should be easy to implement and then you can use it as a baseline to compare with other methods, both to test them for correctness and to compare times.
A slight modification of your current method would solve that example grid: Find the 'largest' grid in terms of how many currently uncovered squares it covers, and instead of changing squares to '0' have a separate state that means '1' but already covered. This way later rectangles can span these squares but already covered squares don't count toward making rectangles 'larger'. (Of course this only works if you're allowed to have overlapping rectangles.)
Related
I have any number of points on an imaginary 2D surface. I also have a grid on the same surface with points at regular intervals along the X and Y access. My task is to map each point to the nearest grid point.
The code is straight forward enough until there are a shortage of grid points. The code I've been developing finds the closest grid point, displaying an already mapped point if the distance will be shorter for the current point.
I then added a second step that compares each mapped point to another and, if swapping the mapping with another point produces a smaller sum of the total mapped distance of both points, I swap them.
This last step seems important as it reduces the number crossed map lines. (This would be used to map points on a plate to a grid on another plate, with pins connecting the two, and lines that don't cross seem to have a higher chance that the pins would not make contact.)
Questions:
Can anyone comment on my thinking that if the image above were truly optimized, (that is, the mapped points--overall--would have the smallest total distance), then none of the lines were cross?
And has anyone seen any existing algorithms to help with this. I've searched but came up with nothing.
The problem could be approached as a variation of the Assignment Problem, with the "agents" being the grid squares and the points being the "tasks", (or vice versa) with the distance between them being the "cost" for that agent-task combination. You could solve with the Hungarian algorithm.
To handle the fact that there are more grid squares than points, find a bounding box for the possible grid squares you want to consider and add dummy points that have a cost of 0 associated with all grid squares.
The Hungarian algorithm is O(n3), perhaps your approach is already good enough.
See also:
How to find the optimal mapping between two sets?
How to optimize assignment of tasks to agents with these constraints?
If I understand your main concern correctly, minimising total length of line segments, the algorithm you used does not find the best mapping and it is clear in your image. e.g. when two line segments cross each other, simple mathematic says that if you rearrange their endpoints such that they do not cross, it provides a better total sum. You can use this simple approach (rearranging crossed items) to get better approximation to the optimum, you should apply swapping for more somehow many iterations.
In the following picture you can see why crossing has longer length than non crossing (first question) and also why by swapping once there still exists crossing edges (second question and w.r.t. Comments), I just drew one sample, in fact one may need many iterations of swapping to get non crossed result.
This is a heuristic algorithm certainly not optimum but I expect to be very good and efficient and simple to implement.
For example we have two rectangles and they overlap. I want to get the exact range of the union of them. What is a good way to compute this?
These are the two overlapping rectangles. Suppose the cords of vertices are all known:
How can I compute the cords of the vertices of their union polygon? And what if I have more than two rectangles?
There exists a Line Sweep Algorithm to calculate area of union of n rectangles. Refer the link for details of the algorithm.
As said in article, there exist a boolean array implementation in O(N^2) time. Using the right data structure (balanced binary search tree), it can be reduced to O(NlogN) time.
Above algorithm can be extended to determine vertices as well.
Details:
Modify the event handling as follows:
When you add/remove the edge to the active set, note the starting point and ending point of the edge. If any point lies inside the already existing active set, then it doesn't constitute a vertex, otherwise it does.
This way you are able to find all the vertices of resultant polygon.
Note that above method can be extended to general polygon but it is more involved.
For a relatively simple and reliable way, you can work as follows:
sort all abscissas (of the vertical sides) and ordinates (of the horizontal sides) independently, and discard any duplicate.
this establishes mappings between the coordinates and integer indexes.
create a binary image of size NxN, filled with black.
for every rectangle, fill the image in white between the corresponding indexes.
then scan the image to find the corners, by contour tracing, and revert to the original coordinates.
This process isn't efficient as it takes time proportional to N² plus the sum of the (logical) areas of the rectangles, but it can be useful for a moderate amount of rectangles. It easily deals with coincidences.
In the case of two rectangles, there aren't so many different configurations possible and you can precompute all vertex sequences for the possible configuration (a small subset of the 2^9 possible images).
There is no need to explicitly create the image, just associate vertex sequences to the possible permutations of the input X and Y.
Look into binary space partitioning (BSP).
https://en.wikipedia.org/wiki/Binary_space_partitioning
If you had just two rectangles then a bit of hacking could yield some result, but for finding intersections and unions of multiple polygons you'll want to implement BSP.
Chapter 13 of Geometric Tools for Computer Graphics by Schneider and Eberly covers BSP. Be sure to download the errata for the book!
Eberly, one of the co-authors, has a wonderful website with PDFs and code samples for individual topics:
https://www.geometrictools.com/
http://www.geometrictools.com/Books/Books.html
Personally I believe this problem should be solved just as all other geometry problems are solved in engineering programs/languages, meshing.
So first convert your vertices into rectangular grids of fixed size, using for example:
MatLab meshgrid
Then go through all of your grid elements and remove any with duplicate edge elements. Now sum the number of remaining meshes and times it by the area of the mesh you have chosen.
I have different dimension of small rectangles (1cm x 2xm, 2cmx3cm, 4cm*6cm etc). The number of different type rectangles may vary depending on case. Each type of different rectangles may have different number of counts.
I need to create a big rectangle with all these small rectangles which these small rectangles can only be placed on the edges. no rotations. The final outer rectangle should ideally be smiliar to a square shape. X ~Y. Not all edges need to be filled up. There can be gaps in between smaller rectangles. Picture Example:
http://i.stack.imgur.com/GqI5z.png
I am trying to write a code that finds out the minimum possible area that can be formed.
I have an algorithm that loop through all possible placement to find out the minimum area possible. But that takes a long run time as number of different type rectangles and number of rectangles increase. i.e. 2 type of rectangles, each has 100 + rectangles. 8 for loops. That will be ~100^8 iterations
Any ideas on better and faster algorithm to calculate the minimum possible area? code is in python, but any algorithm concept is fine.
for rectange_1_top_count in (range(0,all_rectangles[1]["count"]+1)):
for rectange_1_bottom_count in range(0,all_rectangles[1]["count"]-rectange_1_top_count+1):
for rectange_1_left_count in (range(0,all_rectangles[1]["count"]-rectange_1_top_count-rectange_1_bottom_count+1)):
for rectange_1_right_count in ([all_rectangles[1]["count"]-rectange_1_top_count-rectange_1_bottom_count-rectange_1_left_count]):
for rectange_2_top_count in (range(0,all_rectangles[2]["count"]+1)):
for rectange_2_bottom_count in (range(0,all_rectangles[2]["count"]-rectange_2_top_count+1)):
for rectange_2_left_count in (range(0,all_rectangles[2]["count"]-rectange_2_bottom_count-rectange_2_top_count+1)):
for rectange_2_right_count in [(all_rectangles[2]["count"]-rectange_2_bottom_count-rectange_2_left_count-rectange_2_top_count)]:
area=calculate_minimum_area()
if area< minimum_area:
minimum_area=area
This looks like an NP-hard problem, so there exists no simple and efficient algorithm. It doesn't mean that there is no good heuristic that you can use, but if you have many small rectangles, you won't find the optimal solution fast.
Why is it NP-hard? Let's assume all your rectangles have height 1 and you have on rectangle of height 2, then it would make sense to look for a solution with total height 2 (basically, you try to form two horizontal lines of height-1 rectangles with the same length). To figure out if such a solution exists, you would have to form two subsets of your small rectangles, both adding up to the same total width. This is called the partition problem and it is NP-complete. Even if there may be gaps and the total widths are not required to be the same, this is still an NP-hard problem. You can reduce the partition problem to your rectangle problem by converting the elements to partition into rectangles of height 1 as outlined above.
I'll wait for the answer to the questions I posted in the comments to your question and then think about it again.
I've got a binary matrix and I'm trying to find all the largest rectangles that can be formed by adjoining elements in the matrix. By largest rectangles I mean, all the rectangles which are unique, non subsets of any other rectangles. For example, the following matrix contains six such rectangles.
This is related to the set cover problem, though here I'm interested in the maximum number of rectangles, not the minimum. An approach I've tried is to find all the rectangles regardless of size, then compare rectangles and remove them if they are a subset of another rectangle. This is not an optimal approach. It seems like this case of the set cover problem shouldn't be too hard.
I've had a look and not found anything similar to this problem. There is this paper, which has some good ideas, but still wide of the mark. Is there another name for this particular problem? Are there any existing algorithms for finding all possible rectangles in a set cover problem?
After a bit more work, I've realised this is not really related to the set cover problem. It's actually the 'finding the unique rectangles that are not contained within in any other rectangle in a binary matrix problem'.
I have come up with something that works well, but I have no idea about it's complexity.
Basically, line sweep across the matrix horizontally and vertically. In each case, look for contiguous groups of 1's that can form rectangles with the next line. This results in a number of rectangles, some of which are duplicates or sub rectangles of others. These rectangles are reduced to a unique set where no rectangle is a sub rectangle of another. Then you have all the rectangles.
Here is a diagram that relates to the image in the original post:
i dont speek english very wheel but, maximal rectangles overlapping is a problem that resolve many pb, if you somme maximal rectangles, geometrique sommation you find optimal volume that you derive to find queleton etc
This question already has answers here:
Algorithm for finding the fewest rectangles to cover a set of rectangles without overlapping
(2 answers)
Closed 5 years ago.
Simplified, I have to solve the following problem:
You have a 2-dimensional array filled with 0s and 1s. Find the minimum number of rectangles such that they cover all 1s. Rectangles should not overlap.
The function signature might look like:
List<Rectangle> FindCoveringRectangles(bool[,] array)
I already have a solution that is "good enough" but doesn't always find the minimum number of rectangles. I'd like to know if there is some well known and efficient algorithm that can be applied to solve this problem?
Example:
An input array of:
..........
.1.....11.
.......11.
...111....
...111....
...111....
....1111..
....1111..
......11..
..........
(0s replaced by dots for readability)
Might result in the following rectangles:
(2,2,2,2),
(2,8,3,9),
(4,4,6,6),
(7,5,8,8),
(9,7,9,8)
(top, left, bottom, right), 1-based
There can be more than one solution, but one is enough.
This is a matching problem of a kind, that could easily have shown NP-hard. However it seems there is actually a very fast solution!
Using a bfs flood-fill you can find each connected component, O(n). Hence wlog we can assume that we just have to fill one connected area.
If the area doesn't have holes in it, you can use the algorithm described in this paper (or here on google scholar.)
An O(n log log n) algorithm is proposed for minimally rectangular partitioning a simple rectilinear polygon. For any simple rectilinear polygon P, a vertex-edge visible pair is a vertex and an edge that can be connected by a horizontal or vertical line segment that lies entirely inside P. It is shown that, if the vertex-edge visible pairs are found, the maximum matching and the maximum independent set of the bipartite graph derived from the chords of a simple rectilinear polygon can be found in linear time without constructing the bipartite graph. Using this algorithm, the minimum partition problem for convex rectilinear polygons and vertically (horizontally) convex rectilinear polygons can be solved in O(n) time
Some of the papers referred to also cover the case of an area with holes. These run in O(n^(3/2)logn) though, but they still seam pretty good.
Alternatively, one thing you might do is solving the problem without holes, solving the problem for each hole, and then subtract. This might not give an optimal solution though, but would keep the runtime.
You could also try and split the shape in its different topological parts, but that would likely run exponentially in the number of holes.
Thirdly you could try to adapt the proposed algorithm for the more general case, but it might be hard.