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Joseph O'Rourke
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Yes, $K_{3,3}$ is such an example. $K_{3,3}$ is cubic, i.e., $3$-regular, but not planar.


         
          Image from [Wikipedia](https://en.wikipedia.org/wiki/Cubic_graph).
[Steinitz's theorem](https://en.wikipedia.org/wiki/Steinitz%27s_theorem) says that the polyhedral graphs are ($1$-skeletons of convex polyhedra) are exactly the $3$-connected planar graphs (with at least $4$ vertices).

Yes, $K_{3,3}$ is such an example. $K_{3,3}$ is cubic, i.e., $3$-regular, but not planar.


         
          Image from [Wikipedia](https://en.wikipedia.org/wiki/Cubic_graph).
[Steinitz's theorem](https://en.wikipedia.org/wiki/Steinitz%27s_theorem) says that the polyhedral graphs are exactly the $3$-connected planar graphs.

Yes, $K_{3,3}$ is such an example. $K_{3,3}$ is cubic, i.e., $3$-regular, but not planar.


         
          Image from [Wikipedia](https://en.wikipedia.org/wiki/Cubic_graph).
[Steinitz's theorem](https://en.wikipedia.org/wiki/Steinitz%27s_theorem) says that the polyhedral graphs ($1$-skeletons of convex polyhedra) are exactly the $3$-connected planar graphs (with at least $4$ vertices).
Source Link
Joseph O'Rourke
  • 150.9k
  • 36
  • 358
  • 958

Yes, $K_{3,3}$ is such an example. $K_{3,3}$ is cubic, i.e., $3$-regular, but not planar.


         
          Image from [Wikipedia](https://en.wikipedia.org/wiki/Cubic_graph).
[Steinitz's theorem](https://en.wikipedia.org/wiki/Steinitz%27s_theorem) says that the polyhedral graphs are exactly the $3$-connected planar graphs.