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Aug 25 at 14:30 comment added Joseph O'Rourke You might look at this for ideas when $k=2$: Bespamyatnikh, Sergei. "Packing two disks in a polygon." Computational Geometry 23, no. 1 (2002): 31-42.
Aug 25 at 11:55 comment added Joseph O'Rourke Each edge of a maximal rectangle $R$ must be flush with (pushing against) an edge of the polygon. Otherwise it could be expanded. So that accords with your $O(n^4)$. And if you select top and bottom polygon edges for $R$, then either $R$ goes exterior, or left/right edges of $R$ are determined. So this improves on brute force.
Aug 25 at 1:59 comment added user536106 For k = 1, we can solve in polynomial time via brute-force search. If I assume that my orthogonal convex polygon takes up at most $n$ horizontal units and $n$ vertical units, then a single rectangle is defined by an interval on the x axis and an interval on the y axis. There are $O(n^2)$ such intervals on the x axis and $O(n^2)$ on the y axis, for a total $O(n^4)$ runtime. I'm pretty sure the exponent can be reduced by being a little smarter with pruning intervals.
Aug 25 at 1:53 comment added user536106 Yes, the rectangles should be inscribed in the polygon. Every edge of the polygon and every edge of the k rectangles should be either horizontal or vertical.
Aug 24 at 23:02 comment added Joseph O'Rourke For $k=1$, perhaps a variant of this would work: Alt, Helmut, David Hsu, and Jack Snoeyink. "Computing the largest inscribed isothetic rectangle." In CCCG, pp. 67-72. 1995. See also this useful student project.
Aug 24 at 23:01 comment added Joseph O'Rourke Presumably you want your rectangles to be inside the polygon. Should the rectangles be "isothetic," oriented the same as the edges of the polygon?
Aug 24 at 6:22 history edited Daniele Tampieri CC BY-SA 4.0
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S Aug 24 at 4:26 review First questions
Aug 24 at 6:22
S Aug 24 at 4:26 history asked user536106 CC BY-SA 4.0