|
2
|
Hi, We know that for triangle-free graphs, if they are regular (i.e. same degree,d, for each vertex), then 2d <= n where n is the number of vertices. In words, the degree is less than or equal to the half of the number of vertices. (complete bipartite for 2d = n) My question is, for every graph with 2d <= n, can we always find a triangle-free graph? Do you know any related results in the literature? I'd be very glad if you could help me with it.. best, gizem |
|||
|
|
9
|
Yes, it is always possible to find regular triangle-free graphs of any degree up to half the number of vertices (as long as the number of vertices is even). To see this, first consider $K_{n,n}$. By Hall's Theorem, $K_{n,n}$ has a perfect matching $M$. Removing the edges of $M$ leaves a $(n-1)$-regular graph which is bipartite (and hence triangle-free). Repeat. It is obviously not true if the number of vertices is odd. If $n$ and $k$ are both odd, then there are no $k$-regular graphs on $n$ vertices and hence no triangle-free ones either. |
|||
|
