2 2 votes Let $G$ be a simple graph on $n$ vertices. Prove that if $G$ has more than $\binom{n-1}{2}$ edges then $G$ is connected. For every $n>2$, find a graph $G_{n}$ which has exactly $n$ vertices and $\binom{n-1}{2}$ edges, and is not connected. Graph Theory cmi2018 graph-theory graph-connectivity descriptive + – gatecse 1.2k views answer comment Share Follow Print See 1 comment 1 1 comment reply Shaik Masthan commented Sep 16, 2019 reply Follow flag https://gateoverflow.in/237427/graph-theory 0 0 replyShare Please log in or register to add a comment.
2 2 votes Part (a) Let's assume the graph has $n$ vertices. We take $n-1$ vertices and form a complete graph with it . So the situation can be imagined as only one vertex is not touched by any edge and other $n-1$ vertices are connected in the best possible way . If we add one more edge , this edge cannot be within the chosen $n-1$ vertices otherwise the graph won't be simple anymore , and adding an edge to the isolated vertex will connect it. Thus we get a connected connected graph here. Part (b) We take $n=3$ A-B C and we get a disconnected graph here. prashant jha 1 answered Sep 18, 2019 prashant jha 1 comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Answer: soujanyareddy13 answered May 4, 2021 soujanyareddy13 comment Share Follow 0 reply Please log in or register to add a comment.