Description edited after comments:
Let's say I have a 3d cuboid that I projected onto a 2d plane. Now I want to find the minimum-area parallelogram that contains all those projected vertices. How do I go about doing that?
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Sign up to join this communityDescription edited after comments:
Let's say I have a 3d cuboid that I projected onto a 2d plane. Now I want to find the minimum-area parallelogram that contains all those projected vertices. How do I go about doing that?
(I deleted my first answer, which misinterpreted the original question. Here I address the question in the comments, which is now incorporated in the revised question.)
The minimum area parallelogram enclosing a convex polygon can be found in linear time, linear in the number of vertices (for your case, a small constant). This result depends on proving that two adjacent edges of an optimal parallelogram must be flush with edges of the convex polygon. Once you have that lemma, it is easy to walk through the few possibilities.
Christian Schwarz, Jürgen Teich, Alek Vainshtein, Emo Welzl, and Brian L. Evans. Minimal enclosing parallelogram with application. In Proc. 11th Annu. ACM Sympos. Comput. Geom., pages C34–C35, 1995. ACM link.
^{ Fig.2. Illustration of proof of one lemma. }
Finding the minimum perimeter enclosing parallelogram (not what was asked) is a bit more involved, but can also be found in linear time.