Find the set of largest contiguous rectangles to cover multiple areas [duplicate]

Answer 1

I found the paper Fast Algorithms To Partition Simple Rectilinear Polygons from San-Yuan Wu and Sartaj Sahni, which could be of interest to you. In your example the region with character 'd' forms a rectilinear polygon, also the regions with 'c' and '.'. This paper includes algorithms for hole-free simple rectilinear polygons.

If a polygon includes holes, there are algorithms running with O(n^3/2 log n) time, as JM Keil in the paper Polygon Decomposition on page 11 states.

If minimizing the total length of the line segments introduced in the partitioning process is the other optimization criterion, the problem becomes NP-complete if the polygon contains holes (page 12). For these problems approximation algorithms exist (the paper refers to papers with such algorithms). If the polygon doesn't contain holes there is an O(n^4) time algorithm.

Answer 2

This is not really an answer but using a naive search you can get

. 1 . 2 3 3
4 1 5 2 3 3
. 1 5 2 . 6
7 1 5 2 8 6
. 1 . 2 8 6

Basically you start from the top left corner and use it as the top left corner of your next rectangle, then you check how far you can extend it to the right and down, then find the topmost and leftmost cell of the remaining bits and so on.

This is probably very ineffective in some cases but it's quick as you don't have to recalculate anything..

Answer 3

In my view, all solutions that find a set of rectangles that cover the original area is a correct one. Finding a smaller set of rectangles is better because it compresses/performs better.

So I would not advise trying to find the optimal solution. (I would guess it is NP-hard as well).

For a faster running solution, you could tile the matrix into groups of 4 cells initially, and try to merge them if they are the same. After that, you can merge groups of 4 groups, if they are the same. And do so recursively if you are done.

This won't find the optimal solution but will be very fast. If your matrixes are large, with large contiguous areas, the difference with the optimal won't be that great.