The Gateway to Computer Science Excellence

+2 votes

Best answer

+1 vote

since , all conditions like

no if points , no of edges, no of in degree and out degree sequences , no of cycle length all are same still they are necessary and but not sufficient condition so we cant say at this point

for this if we map each vertex as one to one correspondence of both the graphs then they are isomorphic to each other

i.e. f(x5)=y4 , f(x4)=y5, f(x1)=(y2), f(x2)=y2, f(x3)=y3 , these one to one correspondence is done on the basis of thier no of in degree and no of out degree , since all vertices of both graphs matched so its isomorphic graph

no if points , no of edges, no of in degree and out degree sequences , no of cycle length all are same still they are necessary and but not sufficient condition so we cant say at this point

for this if we map each vertex as one to one correspondence of both the graphs then they are isomorphic to each other

i.e. f(x5)=y4 , f(x4)=y5, f(x1)=(y2), f(x2)=y2, f(x3)=y3 , these one to one correspondence is done on the basis of thier no of in degree and no of out degree , since all vertices of both graphs matched so its isomorphic graph

52,345 questions

60,513 answers

201,931 comments

95,360 users