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Is there a theorem using the number of edges and vertices, or something about the max degree a vertex can have? I know that you can use $e ≤ 3v−6$. But what if that condition holds, and it still might not be planar? I'm having a hard time finding a planar representation for the graph I'm looking at.

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    There is a characterization due to Kuratowski, see https://en.m.wikipedia.org/wiki/Kuratowski%27s_theorem – Mariano Suárez-Álvarez Feb 18 '16 at 02:19
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    There are many ways to show a graph is non-planar. We have that every non-planar graph has a subgraph that is a subdivision of $K_5$ or $K_{3,3}$. We also have that if the minimum degree is greater than 5, the graph is non-planar. You can also look at induced subgraphs and try to find one where $e>3v-6$. If you provide more information about the graph you are looking at that would help. – Sean English Feb 18 '16 at 02:20
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    There are a good number of Related Questions linked to the right on your post. Have a look at (say) three of them to see if you can make your Question more precise. – hardmath Feb 18 '16 at 02:29
  • Thanks. I found out that the graph I was looking at had a homeomorphic K3,3. –  Feb 18 '16 at 02:40

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As stated above, a variation of Kuratowski's and Wagner's Theorems are popular for proving non-planarity of a graph. Another method could be to simply refer to Euler's Formula for planar graphs i.e. $$v - e + f = 2,$$ Where $v$ is the number of vertices, $e$ is the number of edges, and $f$ is the number of faces. Any connected planar graph has this equality holding, and often we can prove a graph is non-planar by assuming it is planar, and showing that it contradicts the above formula. Another variation of this formula for planar graphs with multiple connected components is $$v - e + f = k + 1,$$ where $k$ is the number of connected components within the graph.

Leanne
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