Show that it is possible for a directed graph with $n$ vertices and no directed cycles to have $n(n−1)/2$ edges .
I am approaching by saying $2$ vertices are required for $1$ edge, so total number of edges is ${n\choose 2}$. Is this correct?
Show that it is possible for a directed graph with $n$ vertices and no directed cycles to have $n(n−1)/2$ edges .
I am approaching by saying $2$ vertices are required for $1$ edge, so total number of edges is ${n\choose 2}$. Is this correct?
If you don't want cycles, look for a tree-like structure.
For instance, call the vertices $1,\ldots,n$ and form a directed edge $i\to j$ whenever $i<j$.