# Confused about the definition of convex drawing of plane graph

When I looked up the definition of convex drawing of planar graph, my confusion mainly focused on the outer face.

The following definition of convex drawing is from Wikipedia.

In graph drawing, a convex drawing of a planar graph is a drawing that represents the vertices of the graph as points in the Euclidean plane and the edges as straight line segments, in such a way that all of the faces of the drawing (including the outer face) have a convex boundary.

It gives two examples Surprisingly, the outer face is not convex in the second example. I 'd like to ask what is the convex boundary of outer face here? The outer face doesn't seem to be convex and I feel that the unbounded face is the complement of a convex set.

So I'm confused by the definition of convex drawing where boundary of the outer face is also convex. I don't know where I got it wrong.

I searched the Mathemaics Stack for a similar puzzle below. There seems to be no consensus. https://math.stackexchange.com/questions/940693/convex-planar-graphs

• I believe the operative phrase in the definition is “have a convex boundary”. The definition does not say that the faces must be convex sets, but rather that the boundary of each face must be convex (i.e., must bound a convex region in its interior). Sep 4, 2021 at 0:41
• @dwilburne I see. Thank you! I confused the polygons formed by the edges of the outer face with the outer face.
– lcz
Sep 4, 2021 at 0:46

Wikipedia does not state that the outerface is convex, but rather that it has a convex boundary. I assume this means that the set consisting of all points inside the boundary is convex. So for example, the outerface has a convex boundary in your two examples. Also, with this definition, it is true that every $$3$$-connected planar graph has a convex drawing (since it is the $$1$$-skeleton of a $$3$$-polytope).