What is the maximum variety of edges a planar subgraph the $K_n$ deserve to have? Is there a simple method to calculation this if not, are there some values of n because that which this is less complicated to uncover out?


$\begingroup$ correct it is well known, for example it is top top the Wikipedia page Planar graph; voted come close. $\endgroup$
$\begingroup$ No problem. Yes because that n >= 3, it is 3(n-2); view in details the subsections "maximal planar graphs" and "Eulers's formula" the the above mentioned page. $\endgroup$
$\begingroup$ due to the fact that there is now likewise an answer in the techncial sense, we can additionally leave it open up from my point of check out (I already voted, yet have no solid feelings concerning this). $\endgroup$
Suppose the graph has actually $n$ vertices, $m$ edges, and also $f$ faces. By Euler"s formula we know that$$n - m + f = 2$$Now suspect there are at the very least three vertices. Every face must be a triangle, otherwise you can increase the variety of edges by splitting a confront with an edge. Due to the fact that every edge borders two faces, $2m=3f$. Therefore$$n - m + \frac23m = n - \frac13m=2$$or$$m = 3n-6$$

This have the right to be achieved by triangulating the inside and also outside of an $n$-gon. There space $n$ edges the make the $n$-gon, $n-3$ the triangulate the inside, and also $n-3$ that triangulate the outside.

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Please delete. I didn"t see Professor Elkies comment.

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answered Mar 10 "13 in ~ 1:36

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$\begingroup$ i am sure Yosef expected the answer; however I perform not it need to be deleted. $\endgroup$
Mar 10 "13 in ~ 2:20
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Proposition: If $\Gamma(E,V)$, is a planar graph (no multigraph) then$|E| \le 3 |V| - 6$.

Proof: let us note that this go not occupational for a multigraph where much more than one edge might be attached come the same two vertices. Imagine a number (below)of two vertices and 5 segment attached come the two vertices through no intersectionsother that the end of the segments. This number has $|E|=5$ and also $|V|=2$, when the inequality $5 \le 0$ is false.

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Of food $K_n$ is no a multigraph however it is good to be aware of counter-examples.

We first assume the the encounters are every triangles and show the inequality \beginequation 2|E| \ge 3|F| \quad \quad (1)\endequation For example for one confront $|E|=3$ and also $F=2$ therefore the equality $6=6$ is achieved. But for two deals with (for instance a rectangle v a diagonal) we have $|E|=5$ and also $|F|=3$, below the inequality $10 > 9$ is strict. We execute this by induction end $|F|$ . Let us present a brand-new face $|F|$ byadding one more vertex and two more edges ~ above the boundary of the graph (note that because all encounters are triangle we can not include faces inside the graph). We require to show that $2| |E| +2| \ge 3| |F|+1|$. \beginequation 2 | |E| + 2| = 2 | E| + 4 \ge 3|F| + 4 \ge 3|F| + 3 \ge 3| |F|+1|.\endequation

Now we usage Eulers equation $|V|-|E|+|F|=2$, indigenous which $|F|=2+|E|-|V|$ and also replacing $|F|$ in equation (1) \beginequation 2|E| \ge 3 (2 + |E| - |V|) \endequation climate \beginequation |E| \le 3|V| - 6.\endequation

Now what happens if a confront is no a triangle. We require to include edges till making the a triangle, use equation $|E"| \le 3|V"| -6$ i m sorry is valid because that triangles then eliminate the edges and also find the for the new graph $|E| \le 3|V| - 6$ is a precious inequality. After including edges to make all faces triangles we have $|E"| \le 3|V"| -6$ wherein $|E"|$ and $|V"|$ space the number of edges and vertices the the new triangulated graph. Once we eliminate one sheet which is usual to 2 triangular faces, we finish up through a quadrilateral. The graph has one less edge there is no removing any vertex. In general, we eliminate edges but no vertices to go from the triangulated to the initial graph, for this reason $|E|