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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 1999, Jyrki Lahtonen, RecklessReckoner, T. Bongers, Claude Leibovici Jul 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 Answer 1


  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.

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