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An outerplanar graph is a graph that can be drawn as a planar graph where every vertex is incident to the outer region. A maximal outerplanar graph can be drawn such that every vertex is part of a simple cycle where every vertex is part of the outer region and every region inside the cycle is a triangle.

I have to show with proof by induction that for all $n\geq 2$ A graph with that drawing has exactly $2n-3$ edges.

The base case is easy: $n=2$: $v$---$u$ is the only graph with that drawing. $1=2\cdot2 -3 =1$ holds.

The induction hypothesis is stated above.

I can wrap my head around the induction step. I've seen that many induction proofs on graphs deconstruct a graph, use the induction hypothesis on the smaller graph and rebuild the graph to proof that the hypothesis holds. Here I don't see how I can do it. I thought about using rules such as that every new vertex needs to be on the circle and every inner region has to be a triangle which leads to the only possibility of adding 2 edges but I don't use the hypothesis there so it's no proof by induction right?

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  • $\begingroup$ Shouldn't $n=3$ be the base case, since for $n=2$ there are no triangles? In the induction step, pick one triangle. This triangle has one edge on the cycle. If $\{u,v,w\}$ are its vertices, and $uv$ is the edge on the cycle, let $z$ be its midpoint. Now add the edges $zu$, $zv$ and $zw$ and delete $uv$. You've deleted 1 and added 3 new edges so in total you have $2n-3+2=2(n+1)-3$ edges. $\endgroup$ Oct 21 '20 at 21:07
  • $\begingroup$ @RandyMarsh In order for that induction argument to work, you also need to prove every Maxouterplanar graph on $n+1$ vertices arises from one on $n$ vertices by that edge-splitting construction. Which is not necessarily wrong, but it's very much more difficult than necessary, $\endgroup$ Oct 21 '20 at 21:20
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    $\begingroup$ @BrandonduPreez you are correct, and now that I've though more carefully about it, those maxouterplanar graphs that don't have a vertex of degree $3$ can't be obtained in this way. I thought it was obvious that they'll always have a vertex of degree $3$, but they do not. $\endgroup$ Oct 21 '20 at 22:37
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Hint: Every maximal outerplanar graph of order at least $3$ has at least one vertex of degree $2$ (if you aren't convinced, draw some examples and then try prove this first), and if you remove this vertex, you still have a maximal outerplanar graph.

![enter image description here

The proof from there works in much the same way as Randy Marsh's comment. You remove the vertex of degree $2$ to get a maximal outerplanar graph with $2$ fewer edges and $1$ less vertex.

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  • $\begingroup$ Thank you! So I can finish the proof by removing the vertex of degree 2 to get a smaller maximal outerplanar graph (as Randy suggested). By the induction hypothesis this graph has 2(n-1) - 3 edges. If I put the vertex back I get two edges more which would be 2(n-1) - 3 + 2 = 2n-3 edges. Is this reasoning correct? $\endgroup$
    – Ahiru
    Oct 21 '20 at 21:46
  • $\begingroup$ @Ahiru Yeah, that works ^^ $\endgroup$ Oct 22 '20 at 9:06

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