Can a 2 dimensional shape have more vertices than its edges, if there are no curved edges? For example, a hexagon has 6 edges, and 6 vertices. However, is there any shape, open or closed, where the number of vertices exceeds the number of edges?


Just draw a shape by its vertices. (I assumed a shape is a polygon here. Otherwise this is not true. For instance this does not hold when we draw a square with its diagonals.)

  • The first vertex will connect the first two edges. We now have one vertex and two edges.
  • Adding a new vertex will now add one edge (because one edge was already accounted for by the previous vertex). So for $m$ vertices we now have $m+1$ edges.
  • This will hold untill we draw the last edge. The last edge will connect two edges which are already acconted for.

Thus in the end we must have $n$ vertices and $n$ edges, for some $n$, by construction.


No, a 2d shape will always have the same number of vertices and edges.

In 3d, you have the Descartes-Euler formula $$s-a+f=2,$$ where $s$ is the number of edges, $a$ the number of vertices, and $f$ the number of faces.

  • $\begingroup$ A proof of some sort would be nice, I think... $\endgroup$ – gebruiker May 23 '16 at 8:15
  • $\begingroup$ How about 3 non-collinear points joined by 2 edges. And in general each planar graph that is a tree. $\endgroup$ – Raskolnikov May 23 '16 at 8:17

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