G is Hamiltonian graph. Hence a path exists which visits each vertex exactly once. If possible let G has a cycle of length more than n/2 Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once (except for the vertex that is both the start and end, which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph Consider the cycle path which has length more than n/2. We know each edge connects two vertices. And as the graph is hamilton and we have taken a cycle path, the path can visit each vertex exactly once except start point Hence if n is odd, (n-1)/2 edges will connect all the vertices Or if n is even n/2-1 edges will connect all the vertices. Hence cycle can not have length more than n/2 Thus proved Solution G is Hamiltonian graph. Hence a path exists which visits each vertex exactly once. If possible let G has a cycle of length more than n/2 Recall the definition A Hamiltonian cycle, Hamiltonian circuit, vertex tour or graph cycle is a cycle that visits each vertex exactly once (except for the vertex that is both the start and end, which is visited twice). A graph that contains a Hamiltonian cycle is called a Hamiltonian graph Consider the cycle path which has length more than n/2. We know each edge connects two vertices. And as the graph is hamilton and we have taken a cycle path, the path can visit each vertex exactly once except start point Hence if n is odd, (n-1)/2 edges will connect all the vertices Or if n is even n/2-1 edges will connect all the vertices. Hence cycle can not have length more than n/2 Thus proved.