# Does the intersection of two concave polygons without holes result in a set of several polygons without holes?

I did not find a proof anywhere, so here is my assumption :

If I have 2 concave and simple polygons without holes, then the intersection yields a set of different polygons without any holes.

If it can result to a polygon with a hole inside, could you provide an example?

Edit: the polygons are simple. edges of one polygone don't intersect, except consecutive edges, which intersect in their common vertex. The polygons in input do not have holes. I consider the intersection of the interiors of the polygons. The resulting set is all the points that are included both in interior of polygon P1 and the interior of the polygon P2.

Thanks.

• The answer highly depends on the to be applied density rule. So an overlap of 2 inner regions either could be seen as density 2 region or, in a mod 2 sense, as a new empty region. - Similar for the sides. You could consider the former sides, when crossing, as still to be continued and the intersection point being not a vertex of the figure (like in a pentagram, for instance), or you could consider such intersection points to be added vertices and the former sides become broken up there. - Either way is fine, you just have to argue what you are considering. I.e. it does not go without telling! Apr 12, 2018 at 21:25
• . the polygons are simple. edges of one polygone don't intersect, except consecutive edges, which intersect in their common vertex. The polygons in input do not have holes. I consider the intersection of the interiors of the polygons. The resulting set is all the points that are included both in interior of polygon P1 and the interior of the polygon P2. Apr 12, 2018 at 22:03

Call the two polygons $$P_1$$ and $$P_2$$. Suppose that some point $$x \not \in P_1 \cap P_2$$, but there is a curve inside $$P_1 \cap P_2$$ encircling $$x$$. Since $$x \not \in P_1 \cap P_2$$ then $$x \not \in P_1$$ wlog. But there is a curve encircling $$x$$ which is inside $$P_1$$ and therefore $$P_1$$ has a hole.