Outer region of a polygon

Questions
,
1
Open in Wolfram Cloud ā†—
2

This also works:

pts = {{1, 1}, {-1, 1}, {-1, 0}, {-1/2, 0},
   {0, -1/2}, {0, -1}, {1, -1}};
RegionDifference[Rectangle[{-2, -2}, {2, 2}], Polygon[pts]]
3

Your method has the boundary region going from -2 to 2 in the x and y directions. My method has the boundary region going from -Infinity to Infinity in the x and y directions.

6061plot.jpg
6061plot.jpg691Ɨ266 13.2 KB

4

How about this?:

pts = {{1, 1}, {-1, 1}, {-1, 0}, {-1/2, 
    0}, {0, -1/2}, {0, -1}, {1, -1}};
exterior = RegionDifference[FullRegion[2], Polygon[pts]]
5

That also works. Thanks

6

Thanks for the different approaches ā€” this clarifies the distinction between bounding with a finite rectangle and using FullRegion.

I have a related question: if the goal is to work symbolically with the true exterior region (unbounded) ā€” for example to integrate over it or apply further region operations ā€” is RegionDifference[FullRegion[2], [\[Polygon\]][1] ][pts]] always the preferred formulation?

In particular:

  • Does using FullRegion[2] have any performance or simplification implications compared to a large bounding rectangle?
  • Are there cases where Reduce[!Element[{x,y}, poly], ā€¦] would produce a more suitable representation for downstream symbolic manipulation?

Iā€™m trying to understand which representation is more robust when the exterior region is used in subsequent analytic computations rather than just plotting.

Appreciate any insight.

7

Playing around with FullRegion[2] I discovered, to my surprise, that we can DiscretizeRegion the whole plane, without any warning, resulting in a square with finite measure:

DiscretizeRegion[FullRegion[2]]
RegionMeasure[%]

or even this:

BoundaryDiscretizeRegion[FullRegion[2]]
RegionMeasure[%]