edited by
2,273 views
2 votes
2 votes

Assume that ‘e’ is the number of edges and n is the number of vertices. The number of non-isomorphic graphs possible with n-vertices such that graph is 3-regular graph and e = 2n – 3 are ______.

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Here I got as No of vertexes = 6

No of Edges = 9

From here on, Given no of vertex & edges how to find no of Non Isomorphic graphs possible ? , this is real question ! Is there any algorithm for this ?

From Made Easy FLT 6-Practice Test 14

edited by

1 Answer

2 votes
2 votes

since it is regular graph so max no. of edge

3n= 2e

as we know e = 2n – 3

equat both

3n/2 = 2n -4

n= 6 // no. of vertex

 so no. of edges= 2n-3= 2*6-3= 12-3= 9 

so using 6 vertex and 9 edges with every vertex has tree degree only two graph possible. one is k3,3

Related questions

1 votes
1 votes
2 answers
1
0 votes
0 votes
0 answers
3
SKP asked Dec 1, 2016
1,216 views
The Number of Non-Isomorphic simple graphs upto 5 Nodes is _______