4 ms·
> If you have the points in the convex hull and a set of points representing each hole, then rectangularization should be easy because the convex hull should be
by jibiki 17y ago
> If you have the points in the convex hull and a set of points representing each hole, then rectangularization should be easy because the convex hull should be rectilinear.
I don't follow.
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Convex hull is an octagon (with no right angles)?
- keefe 17y agoCan you just modify the gift wrapping algorithm such that the next point selected must be on a straight line from the previous point? http://en.wikipedia.org/wiki/Orthogonal_convex_hull http://en.wikipedia.org/wiki/Orthogonal_convex_hull Suggests it is doable in O(n log n) still, but totally overlooked this modification in my post my bad...
- jibiki 17y ago> http://en.wikipedia.org/wiki/Orthogonal_convex_hull http://en.wikipedia.org/wiki/Orthogonal_convex_hull Ah, that's what I was missing, thanks.