# non-planar graph question [closed]

The non-planar graph $G$ has degree sequence $$(2, 2, 3, 3, 3, 3, 4, 4).$$

Explain why $G$ cannot contain a subdivision of $K_5$, but must contain a subdivision of $K_{3,3}$.

Draw two such a graphs:

1. one in which $K_{3,3}$ is a subgraph, and
2. one in which there is a proper subdivision of $K_{3,3}$ as a subgraph.
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## closed as off-topic by Fundamental, Jyrki Lahtonen♦, RecklessReckoner, T. Bongers, Claude LeiboviciJul 14 '14 at 7:37

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Hi, what are your thoughts on this homework problem? –  gt6989b Feb 14 '14 at 19:48
What is the degree sequence of $K_5$? –  hbm Feb 14 '14 at 20:14

1. $K_5$ has $5$ vertices of degree $4$. The degree of these vertices remains unchanged after subdividing edges.
2. The assumption is that the graph is non-planar: that the graph has a $K_{3,3}$ subdivision follows from Kuratowski's Theorem.
3. By starting with $K_{3,3}$ we can add vertices and edges to achieve the desired degree sequence with a $K_{3,3}$ subgraph.
4. Start with $K_{3,3}$ and replace an edge with two $2$-edge paths.