2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.