2.1 Some useful classes of graphs :

Example 2.1 (Complete graph).

Kn : V=[n],E=(V2).

Figure 2.1: Complete graph on 5 vertices
Example 2.2 (Cayley graphs).

Let (H,+) be a finite group with a finite set of generators S such that S=S (symmetric) and 0S (where 0 is the identity). The Cayley graph G is defined with vertex set V=H and xy if xyS. Since S is symmetric, yxS iff xyS and so abelianness of the group does not matter.

Refer to caption
Figure 2.2: Cayley graph for free group generated by a and b
Example 2.3 (Intersection graph).

Let S1,,Sn be subsets of a set S. Define G with V=[n] and ij if SiSj.

Example 2.4 (Delaunay graph).

Pd,d1 - finite distinct set of points. For yd,d(y,P):=minxP|xy|. Define Cx:={y:d(y,P)=|xy|},xP. Delaunay graph is the intersection graph on P with intersecting sets Cx,xP. Cx is called as the voronoi cell of x.

Refer to caption
Figure 2.3: Delaunay graph
Example 2.5 (Euclidean Lattices).

Let Br(x) be the closed ball of radius r centered at x. The d-dimensional integer lattice is the intersection graph formed with d as vertex set and B1/2(z),zd as the intersecting sets. Alternatively, zz if i=1d|zizi|=1. Show that Euclidean lattices are Cayley graphs. Find the generators of S.

Exercise 2.6.

Show that every graph is an intersection graph.

2.1.1 Some graph constructions

Example 2.7 (Bi-partite graph).

Graphs (V,E) such that V=V1V2 and EV1×V2.

Figure 2.4: Bipartite graph
Example 2.8 (Complementary graph).

Let G=(V,E) be a graph. The complementary graph is Gc=(V,([V]2)E).

Example 2.9 (Line graph).

Let G=(V,E) be a graph. The line graph is L(G) with vertex set V=E and e1e2 is they are adjacent in G.

Example 2.10 (Petersen graph).

The vertex set of the graph is ([5]2) and {i,j}{k,l} if {i,j}{k,l}=.

Figure 2.5: Petersen graph
Exercise(A) 2.11.

Show that the Petersen graph is the complement of the line graph of K5.

Exercise* 2.12.

Can you find the number of edges in a Line graph L(G) in terms of the number of edges in G ?

Let G have n vertices and m edges. For each vertex viV(G) we assume its degree to be di. In L(G), two vertices share an edge iff the two edges in G share a common vertex. So for each choice of two distinct edges coming out of a vertex, we get a distinct edge in L(G). Thus, |E(L(G))|=i=1n(di2) where (di2)=0 if di1. But

i=1n(di2)=i=1ndi(di1)2=i=1n(di22di2)

Now sum of degrees of vertices is a graph is basically counting the edges twice, so finally we have,

|E(L(G))|=12i=1ndi2m