Imagine a city that has 15 public phones. Is it possible to connect them to each other with some cables in case that every phone must connect to exactly 5 another phone. I tried to draw this graph with 15 vertices but

I could just fill 14 vertices and the last vertex 's degree was 4.

Is there a strong way to prove that it's possible?


Hint In any graph, the number of vertices of odd degree must be even.

  • $\begingroup$ Is there always a graph with $n$ nodes, such that each node is connected to $k<n$ other nodes (for any choices of $n$ and $k$ that meet the criterion you stated)? $\endgroup$ – DreiCleaner Jun 16 '20 at 15:13
  • $\begingroup$ @DreiCleaner It is possible if and only if $n \cdot k$ is even. $\endgroup$ – N. S. Jun 16 '20 at 15:56
  • 1
    $\begingroup$ @Integrand What do you mean by "It does not provide an answer"? 15 is ODD. $\endgroup$ – N. S. Jun 16 '20 at 15:57

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