On 18. mars 2018 14:52, Marius Kintel wrote:

> Nophead is right: Minkowski sums are commutative, but they’re implemented by accumulating binary sums as you describe.

>

> The exception is if all operators are convex. In that case we can exploit the fact that the "convex hulls of a minkowski sum” equals the “minkowski sum of convex hulls”. ..and simply collect vertices and calculate a single convex hull:

>

https://en.wikipedia.org/wiki/Minkowski_additionThanks, that makes it clearer. Perhaps the Wiki could mention minkowski

with more than 2 operators and that the order is not significant (i.e.

commutative).

Ok, I see that if all operators are convex then the minkowski sum is

convex as well, and the calculation becomes much simpler. Good point.

Carsten Arnholm

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