By S. A. Chodum
An introdctory booklet approximately graph concept at undergraduate point.
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Extra resources for Graph Theory: A NPTEL Course
Example text
2) Suppose e belongs to cycle C in G. 2. Connected graphs So, C − e is a (u, v)-path in G − e. Hence, u and v are in the same components of G − e, that is G − e is connected. Therefore, e is not a cut-edge. 16. Every connected graph G with n ≥ 2, contains at least 2 vertices which are not cut-vertices. Proof. Let P be a path of maximum length in G; let P = P (x, y). Our claim is that neither x nor y is a cut-vertex of G. 9. It contains at least 2 components say C and D where y ∈ D. Let v be any vertex in C.
Vn . Let G − vi have mi edges for i = 1, 2, . . , n. Show the following: n 1 (i) m = n−2 i=1 mi , (ii) deg(vi ) = 1 n−2 n j=1 mj − mi , i = 1, 2, . . , n. 13. Give an example of a simple graph on 9 vertices and 20 edges which contains no K3 as a subgraph. 14. Give an example of a graph G on 8 vertices such that neither G contains a K3 nor Gc contains K4 . 15. (a) Draw a simple graph on 7 vertices with maximum number of edges which contains no complete subgraph on 4 vertices. (b) Draw a simple graph on n vertices with maximum number of edges which contains no complete subgraph on p, 2 ≤ p ≤ n vertices.
17: Application of 2-switch. We delete the edges (v1 , vj ), (vi , vp ) and add the edges (v1 , vi ), (vj , vp ) (and retain all other edges of G). The resultant graph H is simple and degG (vk ) = degH (vk ) for every k, 1 ≤ k ≤ n. So, H is also a realization of S in which v1 is adjacent with one more vertex in {v2 , v3 , . . , vd1 +1 } than G does. If v1 is not adjacent to some vertex in {v2 , v3 , . . , vd1 +1 } in H, then we can continue the above procedure to eventually get a realization G∗ of S such that v1 is adjacent to all the vertices in {v2 , v3 , .