Questions tagged [planar-graphs]

A planar graph is a graph (in the combinatorial sense) that can be embedded in a plane such that the edges only intersect at vertices. Consider tagging with [tag:combinatorics] and [tag:graph-theory]. For embeddings into higher-genus spaces, use [tag:graph-embeddings].

A planar graph is a graph (in the combinatorial sense) that can be embedded in a plane such that the edges only intersect at vertices. Consider tagging with and .

A finite graph $G$ is planar if and only if no minor of $G$ is the complete graph $K_5$, or the complete bipartite graph $K_{3,3}$ (Wagner's Theorem).

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Kuratowski's theorem on Planar graphs

I've got an exercise that I've been battling for 5 hours to solve. There is a picture of the graph below. I am sure that I can't find a K5 as a minor since every vertex has 3 edges leaving. I tried so long to find a K3,3 minor and failed. What is…
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Graph Theory Planar Graphs

A plane connected graph has 22 corners (eng. Vertices) and all surfaces the graph divides the plane (also the external, unlimited surface) has just 4 edges Find all possible values for the number of edges in the graph. I get all surfaces to r = 4*e…
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In graph theory, what's the difference of triangles and 3-faces?

I'm pretty sure that triangles and 3-faces are not the same but I cannot find their differences according to their definitions. Could you please help me with that? Thanks
arny
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G with n vertices is planar if it has most an vertices

“if a connected graph with $n$ vertices has at most $αn$ edges, then $G$ is planar.” For what real numbers $α$ is this statement always true? Prove your answer in both directions. I tried to use $f\le \frac23m$ but still can not find the value of…
Chloe
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If a graph with m nodes is complete and planar, all graphs with m nodes are planar.

How can I prove that if a there exists a complete graph with $m$ nodes that is planar, then all graphs with $m$ nodes must be planar. I know a complete planar graph has at most 4 nodes, and I think I could show it using that, but I assume there is a…
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