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 0∉S (where 0 is the identity). The Cayley graph G is defined with vertex set V=H and x∼y if x−y∈S. Since S is symmetric, y−x∈S iff x−y∈S 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 i∼j if Si∩Sj≠∅.

Example 2.4 (Delaunay graph).

P⊂ℝd,d≥1 - finite distinct set of points. For y∈ℝd,d⁢(y,P):=minx∈P⁡|x−y|. Define Cx:={y:d⁢(y,P)=|x−y|},x∈P. Delaunay graph is the intersection graph on P with intersecting sets Cx,x∈P. 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 ℤd as vertex set and B1/2⁢(z),z∈ℤd as the intersecting sets. Alternatively, z∼z′ if ∑i=1d|zi−zi′|=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=V1⊔V2 and E⊂V1×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 e1∼e2 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 vi∈V⁢(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 di≤1. But

∑i=1n(di2)=∑i=1ndi⁢(di−1)2=∑i=1n(di22−di2)

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

|E⁢(L⁢(G))|=12⁢∑i=1ndi2−m