Given no of vertex & edges how to find no of Non Isomorphic graphs possible ?

Question

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 ______.

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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

Answer 1

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 k_3,3

Comments

That's the issue, if we are asked

Give no of isomorphic graphs for this many no of vertices & edges, how to find it ? That's real issue. (Updated the question !)

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how e = 2n-3??

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@Akash we can find how many graph we can draw with help of 6 Vertex and 9 Edge with different structure we can create e.g

k3,3,

graph 2 vertex with 5 degree and left 4 are with 2 degree.