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Problem:

Given two convex polygons A, B, we can define Minkowski sum, as A + B = {a + b: a $\in$ A, b $\in$ B}, where a + b vector sum. Prove that:

for every external perpendicular u to an edge of A, there exists an external perpendicular to an edge of A + B, which will be parallel to u.


Attempt:

I know that the external perpendicular has maximum inner product for points that lie in that edge of the polygon, i.e. = max <=> p $\in$ edge.

enter image description here

So now, I assume I have an edge a of A and $\overline{b}$ = {b $\in$ B | $u^T$ to be max}.

Because B is convex, $\overline{b}$ will be a point or an edge. If it is a point a + $\overline{b}$ = a, u $\bot$ a and $u^T$(a + $\overline{b}$) = $u^Ta $+ $u^t$$\overline{b}$.

If a is not an edge, then there will be points outside of the polygon, which are going to maximize the inner product.


I am not sure if I am in the right track..any ideas? Please teach me how to fish, don't just hand me the fish! :)

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  • $\begingroup$ Do you know that, the Minkowski sum of two convex polygons is obtained by intertwinning the edges of these polygons taking them in the order of their normal angle ? See fig. 2.2 of the following document (in French). On the side of computer implementations www-i2.informatik.rwth-aachen.de/i2/fileadmin/user_upload/…. $\endgroup$ – Jean Marie Mar 23 '16 at 17:07
  • $\begingroup$ @JeanMarie thanks for the comment, but I am still unable to follow, I mean I am looking at the figure, but I can't get the point! Can you please expand on this? :) $\endgroup$ – gsamaras Mar 23 '16 at 22:01
  • $\begingroup$ I mean I understand that what I need to prove stands, from the Minkowski sum pictures (en.wikipedia.org/wiki/Minkowski_addition) and my intuition is clear on this. I just do not know how to prove it @JeanMarie. $\endgroup$ – gsamaras Mar 24 '16 at 0:10
  • $\begingroup$ What is an "external perpendicular" in this context? I can't seem to find the terminology used anywhere else. My first guess is that it's the same as a normal vector, but I don't understand the "maximum inner product" characterization. $\endgroup$ – Benjamin Bray Jun 10 '17 at 20:51
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First of all, a good reference (CGal s a very powerful library ) : http://doc.cgal.org/latest/Minkowski_sum_2/

I think the point is that you have to revert the definition of a convex polygon:

There is a perfect equivalence between two convex polygon definitions: through a list of points or through a set of vectors (sorted by their polar angle)

  • Being given a list of points $P_k$, the associated set of vectors is $P_{k+1}-P_k=\overrightarrow{P_kP_{k+1}}$ (it is a kind of derivative, denoted $\partial P$, part of a vast theory called "homology" ).

  • in a reverse way, being given a list of vectors $V_k$ (sorted by their polar angle), one takes an arbitrary origin point $P_1$, then $P_2=P_1+V_1$, $P_3=P_2+V_2$, etc.

The second way gives an immediate definition: the Minkowski sum of 2 polygons is the polygon associated with the (sorted) union of the list of vectors of the 2 polygons.

Philosophical note: It wouldn't be the first time in mathematics that a definition and a property take advantage to be interchanged...

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  • $\begingroup$ I know about CGAL, I have used it: gsamaras.wordpress.com/projects/#PIP, but now I approached the problem from a completely theoretical point of view, thankS! $\endgroup$ – gsamaras Mar 24 '16 at 21:35
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The proof you suggest is on the right track and I believe it works in general. Notice that $$\max_{a \in A , b \in B} \space u^\top ( a + b ) = \max_{a \in A} u^\top a + \max_{b \in B} u^\top b $$

In the general case where $u$ is perpendicular to some facet on $A$, the extremal set on $B$ will be a vertex, and that translated facet will be a facet of $A \bigoplus B$.

In the other cases where the extremal set on $B$ is not a vertex, the corresponding facet in $A \bigoplus B$ will be the Minkowski sum of the extremal facets, but that it is still perpendicular to $u$.

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