Planar graph has an euler cycle iff its faces can be properly colored with 2 colors (such way the colors of two faces that has the common edge are different).

I have an idea to consider the dual graph (turn faces into vertexes and make edge when the two faces have a common edge), but I am stucked with the following proof.

  • 1
    $\begingroup$ For one direction: If there isn't an euler cycle in the graph, what kind of vertex do you know must exist? What happens when you try to colour the faces around that vertex? $\endgroup$ – Arthur Nov 24 '16 at 0:47

Hint: The faces of the dual graph have exactly $\deg(v)$ edges, where $v$ is the corresponding vertex in the actual graph.... Use this to conclude that the original graph is Eulerian if and only if all the faces of the dual graph are even cycles.

Hint 2: Prove that a planar graph has an odd cycle if and only if it has a face which is an odd cycle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.