J. Comp. & Math. Sci. Vol.4 (5), 322-324 (2013)
Set Independence Number of A Graph V. R. KULLI Department of Mathematics, Gulbarga University, Gulbarga, INDIA. (Received on: June 25, 2013) ABSTRACT The set independence number β s(G) of a graph G in the minimum number of subsets into which the vertex set of G should be partitioned so that each subset is independent. In this paper, the exact values of β s(G) for some standard graphs are found and some bounds are obtained. Also a Nordhaus-Gaddum type result is established. Keywords: independent independence number.
set,
independence
number,
set
Mathematics Subject Classification: 05C.
1. INTRODUCTION The graphs considered here are finite, undirected without loops and multiple edges. A set S of vertices in a graph G is called an independent set if no two vertices in S are adjacent. The independence number β0(G) is the maximum number of vertices in an independent set. In this paper, we define a new parameter set independence number βs(G) as follows: The set independence number βs(G) of a graph G is the minimum number of
subsets into which the vertex set of G should be partitioned so that each subset is independent. Consider the graph G as shown in Figure 1. The only maximum independent set of G is S = {1, 2, 4}. Therefore β0(G) = | S | = 3. The set S ' = {{1, 2, 4}, {3, 5}} is the minimum independent partition of vertex set of G. Thus βs(G) = | S ' | = 2.
1 3
G:
4
2 Figure 1
Journal of Computer and Mathematical Sciences Vol. 4, Issue 5, 31 October, 2013 Pages (322-402)
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V. R. Kulli, J. Comp. & Math. Sci. Vol.4 (5), 322-324 (2013)
2. RESULTS We obtain the exact values of βs(G) for some standard graphs. Proposition 1. For any complete graph Kp with p vertices, βs(Kp) = p. Proposition 2. For any path Pp with p ≥ 2 vertices, βs(Pp) = 2. Proposition 3. For any cycle Cp with p ≥ 3 vertices, if p is even, βs(Cp) = 2, = 3, if p is odd. Proposition 4. For any complete bipartite graph G, βs(G) = 2. Proposition 5. For any wheel Wp with p ≥ 4 vertices, βs(Wp) = 3, if p is odd, = 4, if p is even. Theorem 6. For any subgraph H of a graph G, βs(H) ≤ βs(G). Proof: Let S = {S1, S2, ... , Sn} be a minimum independent partition of vertex set of G. Then {S1', S2', ..., Sm'}, where Si' ⊆ Si, i ≤ m ≤ n, is a partition vertex set of H into independent sets. Thus |S '| ≤ | S |. Hence βs(H) ≤ βs(G). We obtain a lower bound for βs(G).
Theorem 7. For any graph G, ω(G) ≤ βs(G) where ω(G) is the clique number of G. Proof: Let H be a maximum order of a complete subgraph of G. Then ω(G) = ω(H). Also βs(H) = ω(H). Therefore βs(H) = ω(G). Since βs(H) ≤ βs(G), ω(G) ≤ βs(G). βs(G).
Now we obtain upper bounds for
Theorem 8. For any graph G, (i) βs(G) ≤ p – β0(G) + 1. (ii)
βs(G) ≤ α0(G) + 1.
Furthermore, the first upper bound is attained if and only if for some vertex u∈S, V − S ∪ {u} is a complete graph where S is a maximum independent set in G. Proof: Let S be a maximum independent set in G. Then V should be partitioned into at most |V – S| + 1 independent sets. Hence βs(G) ≤ |V – S| + 1 or βs(G) ≤ p – β0(G) + 1.
Since α0(G) + β0(G) = p, βs(G) ≤ α0(G) + 1.
Suppose for some vertex u ∈ S, V − S ∪ {u} is a complete graph. Then βs(G) = | (V – S) ∪ {u} | or
βs(G) = p – β0(G) + 1.
Conversely suppose βs(G) = p – β0(G) + 1. We now show that for some
Journal of Computer and Mathematical Sciences Vol. 4, Issue 5, 31 October, 2013 Pages (322-402)
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V. R. Kulli, J. Comp. & Math. Sci. Vol.4 (5), 322-324 (2013)
vertex u ∈ S, V − S ∪ {u} is complete. On the contrary, assume for every vertex u ∈ S, V − S ∪ {u} is not a complete graph. Then we consider the following two cases: Case 1. If every ui ∈ S is not adjacent to at
least one vertex vj ∈ V – S, then the vertices ui, vj belong to some independent set in G. Thus V should be partitioned into at most |V – S | independent sets. Thus βs(G) ≤ | V – S | or βs(G) ≤ p – β0(G) which is a contradiction. Case 2. Suppose there is a vertex u ∈ S adjacent to every other vertex of S. Then there exist two nonadjacent vertices v and w in V – S and hence they belong to some independent set in G. Thus βs(G) ≤ V – S or βs(G) ≤ p – β0(G), which is a contradiction. Hence from Cases 1 and 2, for some vertex u in S, V − S ∪ {u} is a complete graph. Now we establish an another upper bound for βs(G). Theorem 9. For any graph G, βs(G) ≤ ∆(G) + 1. Furthermore, the bound is attained if G = Kp or C2p+1. Proof: Let u be a vertex of maximum degree
∆(G)=r and u1, u2, ... , ur be the r vertices adjacent to u. Then V is partitioned into at most r +1 independent sets. Hence βs(G) ≤ r + 1 or βs(G) ≤ ∆(G) + 1. Suppose G = Kp. Then βs(Kp) = p
and ∆(Kp) = p – 1. Thus
βs(Kp) = ∆(Kp) + 1.
Suppose G = C2p+1. Then βs(C2p+1) =
3 and ∆(C2p+1) = 2. Thus
βs(C2p+1) = ∆( C2p+1) + 1.
Next we obtain a Nordhaus Gaddum type result. Theorem 10. For any graph G,
( )
(i) βs ( G ) + β s G ≤ p + 2 + ∆ ( G ) – δ(G).
( )
(ii) βs ( G ) .βs G ≤ ( ∆ ( G ) + 1) ( p − δ ( G ) + 1) . Proof: (i) Since βs(G) ≤ ∆(G)+1,
( )
βs G ≤ ∆ ( G ) + 1 = p − δ (G ) + 1.
Thus
( )
β s ( G ) + β s G ≤ ∆ ( G ) + 1 + p − δ (G ) + 1
≤ p + 2 + ∆(G) – δ(G).
( ) (
)( ( )
)
(ii) βs ( G ) .βs G ≤ ∆ ( G ) + 1 ∆ G + 1 ≤ (∆(G) + 1) (p – δ(G) + 1). REFERENCE
1. F. Harary, Graph Theory, Addison Wesley, Reading Mass. (1969).
Journal of Computer and Mathematical Sciences Vol. 4, Issue 5, 31 October, 2013 Pages (322-402)