84

CHAPTER 6

b-Chromatic Number of Total Graph of Some Graphs

In this Chapter, some special properties of Total graph of Cycle and Path are discussed

along with its b-Chromatic number. Also the b-Chromatic number of Total graph of Star graph,

Double Star graph, Fan graph, Bistar, Complete Bipartite graph, Crown graph are obtained along

with its structural properties.

6.1 Introduction [2, 11, 84]

Let G be a graph with vertex set V(G) and edge set E(G). Total graphs are generalizations

of Line graphs. The Total Graph of graph G, denoted by T(G) is defined as follows. The vertex

set of T(G) is V(G)∪ E(G). Two vertices x, y in the vertex set of T(G) are adjacent in T(G) in case

one of the following condition holds:

• x, y are in V(G) and x is adjacent to y in G.

• x, y are in E(G) and x, y are adjacent in G

• x is in V(G), y is in E(G) and x, y are incident in G.

Example

Figure 1(a): C4 Figure 1(b): Total Graph of C4

85

The following observations on Total Graph are made in [85]

• For any graph G with v>1 and e ≥1, the Total graph has atleast one 3 cycle.

• For all n-regular graph, Total graph is a 2n regular graph.

• For any graph G, and any vertex v in G, the degree of v in T(G) is twice the degree of

v in G.

• The number of edges in T(G) is equal to n times the number of vertices in T(G) if G

is a n-regular graph.

6.2 b-Chromatic Number of Total Graph of Cycle and its Structural Properties

6.2.1 Theorem

The b-Chromatic number of Total graph of every Cycle is 5 for every n≥5.

i.e. φ[T(Cn)] =5 for n ≥ 5.

Proof

Total graph of every Cycle is a 4-regular graph i.e. every vertex in the Total graph of

Cycle is incident with four edges. So we can assign more than or equal to five colours to the

Total graph of Cycle (for n≥5) for producing a b-chromatic colouring. Suppose if we assign

more than five colours, it contradicts the definition of b-colouring because in T(Cn) each vertex is

adjacent only with four vertices. Thus by the colouring procedure, the b-chromatic number of

Total graph of every Cycle is 5.

Example

Figure 2 : φ[ T(C5)] = 5

86

6.2.1.1 Remark

The Number of vertices in T(Cn) is twice the number of vertices in Cycle Cn.

6.2.1.2 Remark

The Number of edges in T(Cn) is four times the number of edges in Cycle Cn.

6.2.2 Theorem

The Total graph of every Cycle Cn (n>4) has 2n times 3-cycle and twice n cycles.

Proof

In T(Cn), by definition each edge vi,vi+1 is subdivided by the new vertex vi′ for i=1,2,3..n-1.

Consider the following cases to prove the above statement.

Case 1

Consider any arbitrary vertex vi. Here the vertex vi is adjacent vi+1, vi+1′,vn and vn′ for

i =2,3,4,..n-1 and vn is adjacent with vi-1, vi-1, v1 and v1′ and v1 is adjacent with v1′,v2,vn and vn′.

Here for i=2,3..n-1, vi along with vi′ and vi-1′ forms a 3 cycle and vi,vi+1 and vi’ forms another

3 cycle. Similarly the vertex v1 along with the vertices v1′ and vn′ forms a three cycle and the

vertex v1 along with v2 and v5 forms another 3 cycle, vertex vn with vn′ and v1 forms a three cycle

and the vertex Vn′ along with vn-1 and vn-1′ forms another 3 cycle. Thus, there are 2n times 3

cycles.

Case 2

Under observation , the vertices vi′ for i=1,2,3..n forms a n cycle and vi for i=1,2,3..n

forms another n cycle. Clearly there are two n cycles.

Example

Figure 3 : T (Cn)

87

6.2.2.1 Observation

The Total graph of Cycle T(C3) has (2n+1) times 3-cycle.

6.2.2.2 Observation

The Total graph of Cycle T(C2) has unique 3 cycle.

6.2.2.3 Remark

The Total graph of Cycle Cn is Eulerian and Hamiltonian.

6.3. b-Chromatic Number of Total Graph of Path and its Properties

6.3.1 Theorem

The b-Chromatic number of Total graph of every Path is 5 for every n ≥ 5.

Proof

Let Pn be a Path of length n-1 with vertices v1,v2,v3…vn. Let ui = vivi+1 for 1≤ i ≤ n-1 be

the edges of the Path Pn. By the definition of Total graph, V[T(Pn)]= V(Pn)∪E(Pn) i.e.

V[T(Pn)] = {vi,ui: 1≤ i ≤ n} and E[T(Pn)] = {vi, ui, uivi+1, vivi+1 ,uiui+1 : 1≤ i ≤ n-1}.

Consider any internal vertex ui or vi for i=2,3,4..n-1 in T(Pn) . Here the vertex vi or ui is

adjacent only with four vertices, so there is a possibility of assigning only five colours to the

Total graph of Path graph for every n ≥ 5, which produces a b-chromatic colouring. Suppose if

we assign more than five colours, it contradicts the definition of b-chromatic colouring. Thus by

the colouring procedure the above said colouring produces a maximum and b-chromatic

colouring.

Example

Figure 4: T(Pn)

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Figure 5: φ [ T(P6)] = 5.

6.3.2 Structural Properties of Total Graph of Path

• Number of vertices in T(Pn)= 2n-1

• Number of edges in T(Pn)= 4n-5

• Maximum degree of T(Pn) is ∆ = 4

• Minimum degree of T(Pn) is δ = 2

• n+1 vertices is of degree 4 and two vertices of degree 2 and remaining vertices is of

degree 3

6.3.3 Theorem

The Graph G is a Cycle if and only if G is T(P2).

6.4 b-Chromatic Number of Total Graph of Star Graph

6.4.1 Theorem

For n ≥ 3, ( )1, 1nT k nϕ = +

Proof

Consider the Star graph K1,n with V(K1,n) = v1, v2, v3, …., v where v is the root vertex. In

T(K1,n) by the definition of Total graph, each edge vvi for 1 ≤ i ≤ n of K1,n is subdivided by the

vertex v1′, v2′, …., vn′. The vertex set of Total graph of Star graph is defined as follows:

V [T(K1,n)] ={ vi / 1≤ i ≤ n } ∪{ vi′/ 1≤ i ≤ n }∪{v}.

Here the root vertex v along with v1′, v2′, …., vn′ induces a clique of order n+1 in T (K1,n)

Now assign a proper colouring to these vertices as follows. First assign the colours ci to the

vertex vi′ for 1≤ i ≤ n and assign the colour cn+1 to the root vertex v. As mentioned above T(K1,n)

contains a clique of order n + 1, so by proper colouring procedure it requires minimum of n+1

89

colours, which produces a b-chromatic colouring. Next if we assign the colour cn+1 to the vertex

vi for i = 1, 2, …… n, it will not produce a b-colouring because none of the vertices vi′s are

mutually adjacent to each other. So assigning any new colour to the vertex vi is not possible.

Thus by the colouring procedure the above said colouring produces the maximum and

b-chromatic colouring.

( )1, 1nT k nϕ ∴ = + , n ≥ 3

Example

v

c4

v1

v1'

c1

v2c3

c2

v2'

v3c2

v3'c3

v4'c4

c1v4

c5

Figure 6: ϕ [T(K1,4)] = 5

6.4.2 Structural Properties of Total Graph of Star Graph

• Number of vertices in T(K1,n) = 2n+1

• Number of edges in T(K1,n) = �(���)

�

• Maximum degree of T(K1,n) is ∆ = 2n

• Minimum degree of T(K1,n) is δ = 2

• The number of vertices having maximum degree ∆ in T(K1,n) is n(p∆)=1.

• The number of vertices having minimum degree δ in T(K1,n) is n(pδ)= n

• n vertices with degree n+1.

90

6.5 b-Chromatic Number of Total Graph of Double Star Graph

6.5.1 Theorem

For any integer n≥ 2, the b-Chromatic number of Total graph of Double Star graph is n+1

i.e. φ[T(K1,n,n)] = n+1.

Proof

Consider the Double Star graph K1,n,n i.e. V(K1,n,n) = {v}∪ {vi / 1≤ i≤ n}∪{vi′ / 1≤ i≤ n}.

In T(K1,n,n) by the definition of Total graph, each edge vvi is subdivided by the new vertex ui and

the edge vivi′ is subdivided by the another new vertex ui′ for i=1,2,3..n.

i.e. V [T(K1,n,n)]= {v}∪{vi / 1≤ i≤ n}∪{vi′ / 1≤ i≤ n}∪{ui / 1≤ i≤ n}∪{ui′ / 1≤ i≤ n}

We see that the vertices ui (1≤ i ≤n) are mutually adjacent with each other and the vertices

ui′ are adjacent with vi,vi′ and ui for i=1,2,3..n. Also the vertices u1,u2,u3..un along with root

vertex v induces a clique of order n+1 in T(K1,n,n). Therefore we say that φ[T (K1,n,n)] ≥ n+1.

Now we will prove the other side φ[T(K1,n,n)] ≤ n+1, for this consider a proper colouring of

T(K1,n,n) as follows.

Consider the colour class C={c1,c2,c3…cn,cn+1}. Assign the colour ci to the vertex ui for

i=1,2,3….n and cn+1 to the root vertex v. Here the vertices v,u1,u2,u3…….un realizes its own

colour class, which produces a b-chromatic colouring.

Next, if we assign the colour cn+2 to the vertices vi′ for i=1,2,3..n, the vertices

v,u1,u2,u3..un realizes its own colour class but the vertex vi′ (1≤ i ≤ n) does not realize its own

colour class due to the above mentioned non adjacency condition. So we should assign only the

existing colours to the remaining vertices i.e. assign the colour ci+1 to the vertices vi′ and vi for

i=1,2,3…n-1 and assign the colour c1 to the vertices vn and vn′ then for i=1,2,3..n , ui′

are

assigned with the colour cn. Thus there is no possibility of assigning more than n+1 colours to

every T(K1,n,n) i.e. φ[T (K1,n,n)] ≤ n+1. Therefore we have φ[T (K1,n,n)] = n+1.

Thus by the colouring procedure the above said colouring is maximum and b-chromatic.

91

Example

Figure 7: φ[T(K1,4,4)] = 5

6.5.2 Structural Properties of Total Graph of Double Star Graph

• Number of vertices in T(K1,n,n) = 4n+1

• Number of edges in T(K1,n,n) =�(��)

�

• Maximum degree of T(K1,n,n) is ∆= 2n

• Minimum degree of T(K1,n,n) is δ = 2

• The number of vertices having maximum degree ∆ in T(K1,n,n) is n(p∆)=1

• The number of vertices having minimum degree δ in T(K1,n,n) is n(pδ)= n

6.5.3 Theorem

For any n>2, q[T(K1,n,n)]= �(��)

�

Proof

q[T(K1,n,n)]= Number of edges in Kn + Number of edges not in Kn

= q(Kn)+ Number of edges not in Kn

= ����+7n

= �(� )

� +7n

92

= �� ����

�

= ����

�

= �(��)

�

Therefore q[T(K1,n,n)] = �(��)

�

6.6 b-Chromatic Number of Total Graph of Fan Graph

6.6.1 Theorem

For any integer n>2, the b-Chromatic number of Total graph of Fan graph is n+1.

i.e. φ[T(F1,n)] = n+1.

Proof

Let (X,Y) be a bipartition of F1,n with |x| = 1 and |y| = n. Let X={v} and Y={u1,u2,…un}.

In T(F1,n) by the definition of Total graph each edge vui for 1≤ i ≤ n of F1,n is subdivided by new

vertex vi′ and for i=1,2,3..n-1, ui,ui+1 is subdivided by vertex wi where T={wi / 1≤ i≤ n-1}.

Clearly T is an independent set. The vertex set of Total graph of F1,n is defined as

V [T(F1,n)] = {v}∪{vi ′ / 1≤ i ≤ n}∪{wi / 1≤ i ≤ n-1}∪{ui / 1≤ i ≤ n}

Here the vertices v,v1′,v2′,v3′…vn′ induces a clique of order n+1 in T(F1,n). So that we say

the b-chromatic number of Total graph of Fan graph will have more than or equal to n+1 colours

i.e. φ[T(F1,n)] ≥ n+1. Now we will prove for φ[T(F1,n)] ≤ n+1, for this assign a proper colouring

to these vertices as follows. Consider the colour class C={c1,c2,c3…cn,cn+1}. Assign the colour ci

to the vertex vi′ for i=1,2,3,….n and cn+1 to the vertex v. Here the vertices v,vi′ for i=1,2,3..n

realizes its own colour, which produces a b-chromatic colouring.

Suppose if we assign any new colour to the vertices ui for i=1,2..….n and wi for

i=1,2,3..n-1, the vertices ui and wi does not realizes the new colour due to the non-adjacency

condition. Note that the rearrangement of colours also fails to accommodate any new colour

93

class. Thus there is a possibility of assigning not more than n+1 colours to every total graph of

Fan graph i.e. φ[T(F1,n)] ≤ n+1. Therefore we have φ [T(F1,n)] = n+1.

Thus by the colouring procedure the above said colouring produces a maximal and

b-chromatic colouring.

Example

Figure 8: φ [ T(F1,4)] = 5

6.6.2 Structural Properties of Total Graph of Fan Graph

• Number of vertices in T(F1,n) = 3n

• Number of edges in T(F1,n)= ����� �

�

• Maximum degree of T(F1,n) is ∆= 2n

• Minimum degree of T(F1,n) is δ = 4

6.6.3 Theorem

For any integer n>2, q[ T(F1,n)]= ����� �

�

Proof

q[ T(F1,n)] = Number of edges in Kn+1 + Number of outer edges in T(F1,n) + Number of inner

edges in T(F1,n) + Remaining edges

94

=��(��)� �+2n-1+5n-4+n-2

= n2+n+4n-2+10n-8+2n-4

= ����� �

�

Therefore q[ T(F1,n)]= ����� �

�

6.7 b-Chromatic Number of Total Graph of Bistar

6.7.1 Theorem

The b-Chromatic number of Total graph of Bistar has n +3 colours for every n≥2.

Proof

Consider the Bistar Bn,n. By definition of Bistar, let u1,u2,….un be the n pendant edges

attached to the vertex u and v1,v2,….vn be another n pendant edges attached to the vertex v.

Consider the Total graph of Bistar, by the definition of Total graph each edge uui and vvi is

subdivided by the newly introduced vertices ui′ and vi′ for i=1,2,3…n. Let S be the newly

introduced vertex in between the vertices u and v. In T(Bn,n), both vertex set and edge set of Bn,n

corresponds to the vertex set of T(Bn,n).

i.e. V [T(Bn,n)] = {v}∪{u}∪{ui / 1≤ i≤ n}∪{vi / 1≤ i≤ n}∪{S}∪{ui′ / 1≤ i≤ n}∪{vi′ / 1≤ i≤ n}

In T(Bn,n), u1′,u2

′,u3

′..un

′ along with the vertex u and s induces a clique of order n+2. Also

v1′,v2

′…vn

′ along with vertex v and s induces another clique of order n+2.

Now assign a proper colouring to these vertices as follows. Consider the colour class

C={c1,c2,c3…cn,cn+1,cn+2,cn+3}. First assign the colour ci to ui′ for i=1,2,3..n and assign the colour

cn+1 to vi and cn+2 to S . Here the above assignment of colouring produces a b-chromatic

colouring. Next assign the colour cn+3 to the vertex v and cn+3+i to the vertices ui for

i=1,2,3…n, it does not produce a b-chromatic colouring because none of the vertices vi,vi′ for

i=1,2,3..n are adjacent with ui and ui′. Due to this non-adjacency condition the vertices ui does

not realizes the colour cn+3+i for i=1,2,….n. To make the above colouring as b-chromatic one,

95

assign the colour ci to ui′, cn+2 to ui and cn+1 to u for i=1,2,3..n. Now the vertices v1′,v2′..vn′

along with the vertex u,v and s realizes its own colours, which produces a b-chromatic

colouring. Thus by the colouring procedure the above said colouring produces maximum and

b-chromatic colouring.

Example

Figure 9: φ [T(B5,5)] = 8

6.7.2 Result

The Total graph of Bi-star (n ≥ 2) is a separable graph.

From Figure 10, it is clear that the vertex connectivity of Total graph of Bistar is one. By

definition, a graph is said to be separable if its vertex connectivity is one. Hence Total graph of

any Bistar is a separable graph.

6.7.3 Structural Properties of Total Graph of Bi Star

• Number of vertices in T(Bn,n) = 4n+3

• Number of edges in T(Bn,n) = n2+7n+3

• Maximum degree of T(Bn,n) is ∆= 2n+2

• Minimum degree of T(Bn,n) is δ = 2

96

6.7.4 Theorem

For any n≥2, q[T(Bn,n)] = n2+7n+3

Proof

q[T(Bn,n)] = 2[ Number of edges in Kn+1] + Number of edges in K3 + Remaining edges

= 2��(��)� � + 3+6n

= n(n+1)+3+6n

= n2+n+6n+3

= n2+7n+3

Therefore q[T(Bn,n)]= n2+7n+3

6.8 b-Chromatic Number of Total Graph of Complete Bipartite Graph

6.8.1 Theorem

Let Km,n be a Complete Bipartite graph on m and n vertices respectively. Then

φ[T(Km,n)] = �� + 2if� = �� + 1if� < �

Proof

Consider the Complete Bipartite graph Km,n with bipartition X={v1,v2,v3…vm} and

Y={u1,u2,u3….un}. By the definition of Total graph, let vij be the newly introduced vertex in the

edge connecting vi and uj in T(Km,n).

The vertex set of V [T(Km,n)]={vi / 1≤ i≤ m}∪{uj / 1≤ j ≤ n}∪{vij / 1≤ i ≤ m,1≤ j ≤ n} .

Here every vertex vi:1≤ i ≤ m and uj:1 ≤ j ≤ n are mutually adjacent with each other.

Also every vertex vi is incident with n edges and every vertex uj is incident with m edges.

Therefore degree of vertices in X is of degree 2n in T(Km,n) and degree of vertices in Y is of

degree 2m in T(Km,n). Also we find n disjoint cliques of order n in every T(Km,n).

97

Case 1 when m = n

Clearly u1,u2,u3….un and v1,v2,v3…vm are vertices in vi and uj i.e. |vi| = m and |uj| = n.

Consider the colour class C={c1,c2,c3…cn,cn+1,cn+2}. Now assign a proper colouring to T(Km,n) as

follows. Assign the colour c1 to the vertex vi for i=1,2,3…m , assign the colour ci+j to vij’s when

i+j ≤ n+1 and assign the colour ci+j-n to vij’s when i+j > n+1.

Here in T(Km,n) the vertex vi along with edges incident with vi forms a clique of order

n+1, also each vi(1≤ i ≤ m) is adjacent with uj for 1≤ j ≤ n . Due to this, the above colouring

produces a b-chromatic colouring. Also we see that in T(Km,n) each vi receives one colour

different from the colour class assigned to the clique introduced by vij’s for 1≤ i ≤ n , 1 ≤ j ≤ m.

Next we assign the colour cn+2 to uj for j=1,2,3..n, the vertices vi (1≤ i≤ m) and uj ( 1≤ j ≤ n)

realizes its own colour, which produces a b-chromatic colouring. Thus by the colouring

procedure, the above said colouring is maximum and b-chromatic.

Therefore φ[T(Km,n)]= n+2. When m=n.

Example

Figure 10: φ[T( K2,2)]= 4

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Case 2 when m < n

Assign the colour c1 to the vertex vi for i=1,2,3…n , assign the colour ci+j to vij’s when

i+j ≤ n+1 and assign the colour ci+j-n to remaining vij when i+j > n+1. Suppose if we assign the

colour cn+2 to the uj for j=1,2,3…n (as in case 1), it contradicts the definition of b-chromatic

colouring because all uj for j=1,2,3..n are adjacent with vi for i=1,2,3…..m but m<n , due to this

condition the vertex uj does not realize the new colour, which does not produce a b-chromatic

colouring. Hence to make the colouring as b-chromatic, assign the colour cn+1 to u1 and cj to uj

for j=2,3,…n such a way that all vi(1≤ i ≤ m) and uj(1≤ j ≤ n) realizes its own colours. Note that

rearrangement of colours also does not accommodate any new colour. Thus by the colouring

procedure, the above said colouring is maximum and b-chromatic.

Therefore φ[T(Km,n)] = n+1

Example

Figure 11: φ[T(K2,3)] =4

6.8.2 Corollary

The Total graph of every Kn,n is a 2n regular graph.

99

6.8.3 Structural Properties of Total Graph of Complete Bipartite Graph

• Number of vertices in T(Km,n) = m+n+mn

• Number of edges in T(Km,n) = �������

� +∑ �(� − )�!"

• Maximum degree of T(Km,n) is ∆ = 2n

• Minimum degree of T(Km,n) is δ = Min(m+n, 2m)

6.8.4 Theorem

For any integer m,n>2, q[T(Km,n)] = �������

� +∑ �(� − )�!"

Proof

q[T(Km,n)] = m( number of edges in X) + n( number of edges in Y but not included in X)

+ m(Number of edges in Kn) + Remaining edges

= m(2n)+n(m)+mq(Kn) + Remaining edges

= 2mn+nm+ m����+ Remaining edges

= 3mn+ m��(� )� �+∑ �(� − )�

!"

= #������ ��

� +∑ �(� − )�!"

=�������

� + ∑ �(� − )�!"

Therefore q[T(Km,n)] = �������

� +∑ �(� − )�!"

6.9 Structural Properties of Total Graph of Crown Graph

• Number of vertices in Total graph of Crown graph = n(n+1)

• Number of edges in Total graph of Crown Graph = n(n+1)(n-1)

• Maximum degree of Total graph of Crown Graph is ∆ = 2n-2

• Minimum degree of Total graph of Crown Graph is δ = 2n-2

100

6.9.1 Theorem

If G is a Crown graph then Total graph of Crown graph is a (2n-2) regular graph.

Proof

Let G= Sn be the Crown graph. Consider the Bipartition of Sn namely X= {v1,v2,….vn}

and Y = {u1,u2,….um} respectively. Here the vertex vi(i=1,2.3..n) is not adjacent with vertex uj

for j=1,2…m.

Consider the Total graph of Crown graph, by the definition of the Total graph , let vij for

i=1,2,3..n, j=1,2,3..m , i ≠j be the newly introduced vertex in the edge joining the vertex vi and

uj. By the definition of Total graph, each vi for i=1,2,3…n is adjacent with n-1 vertices in Y and

each vi is incident with n-1 edges. So each vi(i=1,2,3..n) has (n-1+n-1) degree in T(Sn) i.e.

vi(i=1,2,3..n) is of degree 2n-2 in T(Sn). Similarly let uj for j=1,2,3..m is adjacent with n-1

vertices in X and each uj is incident with n-1 edges. Also each uj ( j=1,2,3..n) has (n-1+n-1)

degree in T(Sn) i.e uj ( j=1,2,3..n) is of degree 2n-2 in T(Sn). Also here all vij’s are with degree

2n-2. Thus every vertices in T(Sn) is of degree 2n-2 .Therefore ,the Total graph of every Crown

graph is a 2n-2 regular graph.

Example

Figure 12: Total graph of Crown graph