What is the maximum number of edges a planar subgraph that $K_n$ can have? Is there a simple way to calculation this if not, room there some worths of n for which this is easier to find out?


$egingroup$ correct it is well known, for instance it is top top the Wikipedia web page Planar graph; voted to close. $endgroup$
$egingroup$ No problem. Yes for n >= 3, it is 3(n-2); view in details the subsections "maximal planar graphs" and also "Eulers's formula" of the over mentioned page. $endgroup$
$egingroup$ since there is now likewise an price in the techncial sense, we can likewise leave it open up from my point of view (I currently voted, however have no solid feelings concerning this). $endgroup$
Suppose the graph has actually $n$ vertices, $m$ edges, and also $f$ faces. Through Euler"s formula we understand that$$n - m + f = 2$$Now assumed there space at the very least three vertices. Every challenge must be a triangle, otherwise you have the right to increase the variety of edges by separating a face with one edge. Since every edge borders two faces, $2m=3f$. Therefore$$n - m + frac23m = n - frac13m=2$$or$$m = 3n-6$$

This deserve to be achieved by triangulating the inside and also outside of one $n$-gon. There room $n$ edges that make the $n$-gon, $n-3$ that triangulate the inside, and also $n-3$ the triangulate the outside.

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Please delete. Ns didn"t watch Professor Elkies comment.

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reply Mar 10 "13 in ~ 1:36

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$egingroup$ ns am sure Yosef supposed the answer; however I execute not it have to be deleted. $endgroup$
Mar 10 "13 at 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 does not job-related 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 2 vertices with no intersectionsother the the end of the segments. This number has $|E|=5$ and also $|V|=2$, while the inequality $5 le 0$ is false.

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Of course $K_n$ is not a multigraph however it is an excellent to be aware of counter-examples.

We an initial assume that the encounters are all triangles and also show the inequality eginequation 2|E| ge 3|F| quad quad (1)endequation For example for one challenge $|E|=3$ and also $F=2$ therefore the equality $6=6$ is achieved. Yet for two deals with (for instance a rectangle v a diagonal) we have $|E|=5$ and $|F|=3$, right here the inequality $10 > 9$ is strict. We perform this by induction over $|F|$ . Let us present a brand-new face $|F|$ byadding one much more vertex and also two an ext edges ~ above the boundary of the graph (note that since all faces are triangular we can not include faces within the graph). We require to display that $2| |E| +2| ge 3| |F|+1|$. eginequation 2 | |E| + 2| = 2 | E| + 4 ge 3|F| + 4 ge 3|F| + 3 ge 3| |F|+1|.endequation

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

Now what happens if a face is not a triangle. We need to include edges until making it a triangle, use equation $|E"| le 3|V"| -6$ i beg your pardon is valid because that triangles then remove the edges and also find the for the brand-new graph $|E| le 3|V| - 6$ is a valid inequality. After adding edges to do all encounters triangles we have actually $|E"| le 3|V"| -6$ whereby $|E"|$ and also $|V"|$ are the variety of edges and vertices of the brand-new triangulated graph. Once we remove one edge which is usual to 2 triangular faces, we finish up with a quadrilateral. The graph has one less edge without removing any kind of vertex. In general, we remove edges yet no vertices to go from the triangulated to the original graph, therefore $|E|