Research In Computer Applications

Published Date: 02 Nov 2017

1 Department of Computer Science, Alagappa Government Arts College,

Karaikudi − 630003, India. [email protected]

2 Research Department of Mathematics, St. Xavier’s College (Autonomous)

Palayamkottai − 627002, India. [email protected]

3Ananda College, Devakottai, India. [email protected]

Abstract

In this paper, we introduce algorithms to find the clique-to-vertex (or (ζ, V ))-distance d(C, v) between a clique C and a vertex v in a graph G, (ζ, V )-eccentricity e2 (C ) of a clique C , and (ζ, V )-center Z2(G) of a graph G using BC -representation. Moreover, the algorithms are proved for their correctness and analyzed for their time complexity.

Keywords: clique, distance, eccentricity, radius, center, binary count.

1. Introduction

By a graph G = (V, E) we mean a finite undirected connected simple graph. |V | and |E| denote the order and size of a graph G respectively. A clique of a graph G is a maximal complete subgraph of G. For other basic definitions not mentioned in this paper, we refer [2, 3].

For vertices u and v in a graph G, the distance d(u, v) between u and v is the length of a shortest u − v path. For subsets A and B of the vertex set V of G, the distance between A and B is defined as d(A, B) = min{d(x, y) : x  A, y  B}. For any vertex v of G, the eccentricity of v is e(v) = max{d(v, u) : u  V }. The radius of G is r = min{e(v) : v  V }. The center of G is Z(G) = {v  V : e(v) = r}. A vertex in Z(G) is called a central vertex. The distance matrix D(G) = [dij ] of G is a n × n matrix, where n is the order of G, and dij = d(vi, vj ) the distance between vertex vi and the vertex vj in G (1 ≤ i ≤ n, 1 ≤ j ≤ n).

In [4] Santhakumaran and Arumugam introduced and studied the following central structures: Let G be a connected graph and ζ = {C : C is a clique in G}. Let G be a connected graph and ζ = {C : C is a clique in G}. For a clique C and a vertex v in G, the clique-to-vertex (or (ζ, V ))-distance d(C, v) from a clique C to a vertex v is defined as d(C, v) = min{d(u, v) : u  C, v  C}. For a clique C of G, (ζ, V )-eccentricity e2(C) of a clique C is defined as e2 (C) = max{d(C, v) : v  V }. The (ζ, V )-radius of graph G is, r2 = min{e2 (C) : C  ζ }. A clique C for which e2 (C) = r2 is called a (ζ, V )-central clique of G and set of all (ζ, V )-central cliques is called the (ζ, V )-center of G and is denoted by Z2(G).

In [1] Ashok, Athisayanathan and Antonysamy introduced a method to represent a subset of a set which is called binary count (or BC) representation. That is, if X = {1, 2, 3, 4} is a set, then the binary count (or BC) representation of the subsets {Ф}, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4} of X are (0000), (1000), (0100), (0010), (0001), (1100), (1010), (1001), (0110), (0101), (0011), (1110), (1101), (1011), (0111), (1111) respectively. Using this BC-representation, given a graph G with the vertex set V = {1, 2, 3, . . . , n} and a subset A of V, they introduced an algorithm to verify whether the subgraph <A> induced by the set A in G is a clique or not. Moreover, a general algorithm is introduced to generate all cliques in G and proved the correctness of these algorithms and analyzed their time complexities.

Example 1.1 Consider the graph G given in Figure 1.1 with the vertex set V = {1, 2, 3, 4, 5, 6}. Then the distance matrix D(G) of G is

Moreover, the set of all cliques in graph G is ζ = {{1, 2}, {1, 3}, {2, 4, 5}, {3, 4}, {4, 6}}. Now using the algorithms discussed in [1], it is easy to verify that the set ζ of all cliques in G in BC representation is ζ = {(110000), (101000), (010110), (001100), (000101)}. Note that if C is the clique {3, 4}, then the BC representation of C is BC(C) = (001100), and further BC(C(1)) = BC(C(2)) = BC(C(5)) = BC(C(6)) = 0, and BC(C(3)) = BC(C(4)) = 1. That is, BC(C(i)) (1 ≤ i ≤ n) denotes the integer ( 1 or 0 ) in the ith place in the BC representation of the clique C in the graph G.

In this paper we introduce algorithms to find (ζ, V )-distance, (ζ, V )-eccentricity and (ζ, V )-center in a connected graph G of order n(> 1) using BC representation.

2. Clique-to-Vertex Center

First, we introduce an algorithm to find the (ζ, V )-distance d(C, i) between a clique C and a vertex i in a graph G using BC representation.

Algorithm 2.1 Let G be a graph with V = {1, 2, 3, . . . , n} and ζ = {C : C is a clique in BC representation in G}.

Let D = [dij ] be the distance matrix of G.

Let i ∈ V .

Let C ∈ ζ .

4. if BC(C(i)) = 1 then d(C, i) = 0; goto step 10

for j = 1 to n

Let d(j, i) = n

if BC(C(j)) = 1 then d(j, i) = dji

next j

Find d(C, i) = min1≤j≤n{d(j, i)}

return d(C, i)

stop

Theorem 2.2 For any clique C and a vertex i in a graph G, the Algorithm 2.1 finds the (ζ, V )-distance d(C, i) from the clique C to the vertex i.

Proof. Let G be a graph with V = {1, 2, 3, . . . , n} , ζ = {C : C is a clique in BC representation in G} and D(G) the distance matrix of G. Let i ∈ V and C ∈ ζ . If the vertex i is a vertex of the clique C, then BC(C(i)) = 1 so that the (ζ, V )- distance d(C, i) = 0. If the vertex i is not a vertex of the clique C, then BC(C(i)) = 0, then the steps 5 to 8 of the Algorithm 2.1 find the distance d(j, i ) from the vertices j(1 ≤ j ≤ n) to the vertex i of G as follows: If j is a vertex of the clique C then BC(C(j)) = 1 otherwise BC(C(j)) = 0. Hence d(j, i) = n if BC(C(j)) = 0 and d(j, i) = dji if BC(C(j)) = 1(1 ≤ j ≤ n). Then the step 9 of Algorithm 2.1 finds the (ζ, V )-distance d(C, i) = min{d(j, i) : 1 ≤ j ≤ n} from the clique C to the vertex i.

Theorem 2.3 The distance between clique C and a vertex i in a graph G can be found in O(n) time using Algorithm 2.1.

Proof. It follows from the fact that the step 4 is executed in O(1) time, the steps 5 to 8 are executed in O(n) time and step 9 is executed in O(n) time in the Algorithm 2.1.

Example 2.4 Consider the graph G of order n(= 6) given in Figure 1.1 and the distance matrix D(G) of G as in Example 1.1. Now using Algorithm 2.1, let us find the ((ζ, V)-distance between the clique C = {1, 2} and the vertex i = 1. Clearly BC(C) = (110000). Since BC(C(i)) = 1 , the Algorithm 2.1 returns (ζ, V )-distance d(C, i) = 0. Again using the Algorithm 2.1, let us find the (ζ, V )-distance d(C, i) between the clique C = {2, 4, 5} and the vertex i = 1. Clearly BC(C) = (010110). Since BC(C(i)) = 0, the Algorithm 2.1 finds the (ζ, V )-distance d(C, i) = min{d(j, i) : 1 ≤ j ≤ n}. Since BC(C(j)) = 0, d(j, i) = n for j = 1, 3, 6 and since BC(C(j)) = 1, for j = 2, 4, 5, d(2, i) = d2i = 1, d(4, i) = d4i = 2 and d(5, i) = d5i = 2. Hence the Algorithm 2.1 returns (ζ, V )-distance d(C, i) = min{d(j, i) : 1 ≤ j ≤ n} = min{d(1, 1), d(2, 1), d(3, 1), d(4, 1), d(5, 1), d(6, 1)} = min{6, 1, 6, 1, 2, 6} = 1.

Next, we introduce an algorithm to find the (ζ, V )-eccentricity e2(C) of a clique C in a graph G of order n using BC representation.

Algorithm 2.5 Let G be a graph with V = {1, 2, 3, . . . , n} and ζ = {C : C is a clique in BC representation in G}.

Let C ∈ ζ

for i = 1 to n

Find d(C, i), by calling Algorithm 2.1

next i

Find e2 (C) = max1≤i≤n {d(C, i)}

return e2 (C)

stop

Theorem 2.6 For a clique C and set of all vertices V in G, the Algorithm 2.5 finds (ζ, V )-eccentricity e2 (C).

Proof. Let G be a graph with V = {1, 2, 3, . . . , n} and ζ = {C1, C2, . . . , Cm } be the set of all cliques in their BC representation in G. Let i ∈ V . Step 3 of Algorithm 2.5, finds the (ζ, V ) distance d(C, i) between a clique C and every vertex i (1 ≤ i ≤ n) in G. Then e2 (C) = max1≤i≤n {d(C, i)}. Hence the theorem.

Theorem 2.7 The Algorithm 2.5 finds (ζ, V )-eccentricity e2 (C) of a clique C in a graph G is O(n2) time.

Proof. By Theorem 2.3, the time complexity of the step 3 in the Algorithm 2.5 is O(n), so that steps 2 to 4 in the Algorithm 2.5 are executed in O(n2) time. The time complexity of the step 5 in the Algorithm 2.5 is O(n). Hence the theorem.

Example 2.8 Consider the graph G given in Figure 1.1 with the vertex set V and the clique set ζ a in the Example 1.1. Clearly the order of n of G is 6 and the number of cliques m in G is 5. Let V = {1, 2, 3, 4, 5, 6}, and C = (101000) ∈ ζ . Now we find the (ζ, V )-eccentricity e2(C). By calling Algorithm 2.1 n times, the step 3 of Algorithm 2.5 finds the (ζ, V )- distances d(C, 1) = 0,d(C, 2) = 1, d(C, 3) = 0, d(C, 4) = 1, d(C, 5) = 2, d(C, 6) = 2. Then the step 5 of Algorithm 2.5 finds the (ζ, V )-eccentricity e2(C) = max{0, 1, 0, 1, 2, 2} = 2

Finally, we introduce an algorithm to find the (ζ, V )-center Z2 (G) of a graph G of order n using BC representation.

Algorithm 2.9 Let G be a graph with V = {1, 2, 3, . . . , n} and ζ = {C : C is a clique in BC representation in G}.

Let Adj(G) be the adjacency matrix of G.

Let D = [dij ] be the distance matrix of G.

Let ζ = {C1, C2, . . . , Cm}.

for i = 1 to m

Find e2 (Ci), by calling Algorithm 2.5

next i

Find the r2 = min1≤i≤m {e2(Ci)}

for i = 1 to m

if r2 = e2 (Ci) then list Ci.

next i

stop

Theorem 2.10 For a graph G, the Algorithm 2.9 finds (ζ, V )-center Z2(G) of G.

Proof. Let G be a graph with V = {1, 2, . . . , n} and ζ = {C1, C2, . . . , Cm } be the set of all cliques in their BC representation in G. The step 5 of Algorithm 2.9, finds (ζ, V )-eccentricity e2 (Ci) for all clique Ci ∈ ζ (1 ≤ i ≤ m). Then the step 7 finds (ζ, V )- radius r2 = min1≤i≤m {e2(Ci) ∈ ζ } of G and the steps 8 to 10 find (ζ, V )-center Z2(G) = {Ci ∈ ζ : e2 (Ci) = r2 }. Thus the Algorithm 2.9 finds (ζ, V )-center Z2 (G) of G.

Theorem 2.11 The (ζ, V )-center Z2(G) of a graph G can be obtained in O(mn2) time using Algorithm 2.9.

Proof. By Theorem 2.7, the computing time for step 5 of the Algorithm 2.9 is O(n2), so that time complexity for the steps 4 to 6 of the Algorithm 2.9 is O(mn2). The step 7 of the Algorithm 2.9 finds r2 in O(m) time and the steps 8 to 10 of the Algorithm 2.9 finds Z2(G) of G in O(m) time. Hence the theorem.

Example 2.12 Consider the graph G given in Figure 1.1 as in the Example 1.1. Clearly the vertex set of G is V = {1, 2, 3, 4, 5, 6} and the set of all cliques in G is ζ = (110000), (101000), (010110), (001100), (000101)}. Now we find the (ζ, V )-center Z2(G). By calling Algorithm 2.5 m times, the step 5 of Algorithm 2.9 finds the (ζ, V )-eccentricities e2 (C1) = 2, e2 (C2) = 2, e2(C3 ) = 1, e2(C4) = 1 and e2 (C5) = 2. The step 7 of Algorithm 2.9 finds the (ζ, V )-radius r2 = min1≤i≤m {e2(Ci)} = 1. Finally, the step 9 of Algorithm 2.9 finds the (ζ, V )-center Z2(G) = {Ci ∈ ζ : r2 = e2} = {C3 , C4}.

3 Conclusion

In this paper we have developed sequential algorithms to find the (ζ, V )- central structures in a graph G and these algorithms may be used in networking, data mining , distributed computing and cluster analysis.