3
$\begingroup$

We went over that the maximum number of edges in a graph with no even cycles is $3(n-1)/2$ by using a spanning tree and making triangles. Can we do the same thing for this but instead of adding one edge, we would have to add two each time? But then I am getting stuck, because we also have the theorem that says if it has more then $(n^2)/4$ edges it will have a triangle or odd cycle. Is the professor just trying to trick us? Doesn't this mean most edges is the $(n^2)/4$, because if more then has the odd cycle? And why doesn't this following instead of the $3(n-1)/2$

  • 1
    Hint, not posted as an answer because I haven't checked it out. With no odd cycles the graph is bipartite. Check out the complete bipartite graph. – 2017-02-13
  • 0
    @EthanBolker Yes, a graph is bipartite if and only if it has no odd cycles. You should post an aswer. andemw01: All you need to do is find the maximum of $(n-m)m$ for $1 \leq m < n$. Your professor didn't trick you. – 2017-02-13

1 Answers 1

2

Hint: With no odd cycles the graph is bipartite. Check out the complete bipartite graphs with a fixed total number of vertices.