Maximal independent sets in bipartite graphs with at least one cycle

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Introduction
Given a graph G = (V G , E G ), a set I ⊆ V G is independent if there is no edge of G between any two vertices of I.A maximal independent set is an independent set that is not a proper subset of any other independent set.The dual of an independent set is a clique, in the sense that a clique corresponds to an independent set in the complement graph.The set of all maximal independent sets of a graph G is denoted by MI(G) and its cardinality by mi(G).
Around 1960, Erdős and Moser proposed the problem to determine the maximum value of mi(G) when G runs over all n-vertex graphs and to characterize the graphs attaining this maximum, both of which were answered by Moon and Moser [18].It is interesting to see that the extremal graphs turn to have most components isomorphic to the complete graph K 3 .On the other hand, the theory on maximal independent set has some applications in other research field.For example, in chemistry, a Clar structure is defined to be a maximal independent set of vertices of the Clar graph of the corresponding benzenoid hydrocarbons [5].Clar structures recently are used as basis-set to compute resonance energies.The theory of maximal independent set is also applied to the areas of managements and networks.For the compatibility graph G, its vertices denote tasks, and an edge denotes a resource sharing constraint between the two tasks linked by it.A maximal independent set of a compatibility graph represents a maximal set of tasks that can be executed concurrently.Basagni [2] and Alzoubi et al. [21] pointed out the importance of a maximal independent set in the wireless network.Moscibroda and Watenhofer [19] obtained the wellknown Maximal Independent Set problem when they modeled the unstructured radio network as a graph.
Along the line on the study of the maximal independent set in mathematical literature, mathematicians focused on determining the largest number of mi(G) in various interesting classes of graphs.Ying et al. [22] determined the maximum number of maximal independent sets in graphs of order n with at most r cycles and in connected graphs of order n 3r with at most r cycles.Ortiz [20] established a sharp upper bound for the number of maximal independent sets in caterpillar graphs.Arumugam et al. [1] studied the maximal independent set of graphs with minimum coloring.Duffus, Frankl and Rödl [4] studied maximal independent sets in graphs obtained from Boolean lattices.Jing and one of the present authors [16] studied the maximal independent set of graphs with cyclomatic number at most two.Jin and Li [8] determined the second largest number of maximal independent sets among all graphs of order n.Hua and Hou [6] determined the n-vertex graph having the third largest number of maximal independent sets.Jou and Lin [10,13] settled the problem for trees and forests.Jin and Yan [9] settled the problem for the second and third largest number of maximal independent sets of trees.
This paper is motivated directly from [17], in which the author completely characterized the n-vertex bipartite graphs having the largest number of maximal independent sets.The corresponding extremal graphs are forests.Furthermore, for the bipartite graph of order n that contains cycles, the author in [17] determined the upper bound on the number of maximal independent sets for odd n.Unfortunately, the graphs which attain this value were not characterized.It is natural and interesting to determine the corresponding extremal graphs, which is settled in this paper.On the other hand, it is necessary and interesting to consider this problem on the n-vertex bipartite graphs with cycles for even n.In this paper, among the set of all disconnected bipartite graphs with even order, the extremal graphs which have the first-, second-, . . ., n−2 2 -th largest number of maximal independent sets are characterized respectively, while among the set of all connected bipartite graphs with even order, the extremal graphs having the largest and the second largest number of maximal independent sets are identified.

Given a simple graph
A leaf of G is a vertex of degree one, while an isolated vertex of G is a vertex of degree zero.For any two graphs G and H, let G⊎H denote the disjoint union of G and H, and for any nonnegative integer t, let tG stand for the disjoint union of t copies of G.For a connected graph H with maximum degree vertex x and a graph An odd (resp.even) component is a component of odd (resp.even) order.Throughout the text we denote by P n , C n , K 1,n−1 and K n the path, cycle, star and complete graph on n vertices, respectively.Undefined terminology and notation may be referred to [3].
Throughout this paper, for simplicity, r denotes √ 2. We begin with some useful known results which are needed to prove our main results.

Lemma 3 ([13])
If T is a tree with n 1 vertices, then mi(T ) t 1 (n), where where where B(i, j) is the set of batons, which are the graphs obtained from a basic path P i (i 1) by attaching j 0 paths of length two to the endpoints of P i in any possible ways.

Lemma 4 ([13])
If F is a forest with n 1 vertices, then mi(F ) f 1 (n), where where if n 4 is even;

Lemma 6 ([17])
The maximum number of maximal independent sets among all bipartite graphs of order n is 2 ⌊ n 2 ⌋ , and the only bipartite graphs of order n which have this many maximal independent sets are F 1 (n).

Lemma 7 ([17]) If G is an acyclic graph of order n and
Proof: The first part is due to [13]; we show the second part by induction on n.It is routine to check that Assume the result holds for n k.Now consider n = k + 1 15.By induction hypothesis, we obtain This completes the proof.✷

Sharp bounds and extremal graphs
Let B n (resp.B ′ n ) be the set of all n-vertex bipartite graphs with at least one cycle for even (resp.odd) n.In this section, among B n the disconnected graphs with the first-, second-, . . ., n−2 2 -th largest number of maximal independent sets are characterized, while the connected graphs in B n having the largest and the second largest number of maximal independent sets are determined; among B ′ n the graphs having the largest number of maximal independent sets are identified.
For even n and 3 Proof: By direct computing, we have By (1), it is easy to see that mi(D n,t ) < mi(D n,t−1 ) for 4 t n 2 .Combining with ( 2)-( 4), our results follow immediately. ✷ If G is disconnected, the equality holds in (5) if and only if The equality holds in (6) if and only if Proof: We put the proofs of (i) and (ii) together as below.Choose is as large as possible.We proceed by induction on n.
For 5 n 6, it is straightforward to check that our results hold.Assume that our results hold for n k.Now consider n = k + 1 7.
First we consider that G is disconnected.Denote where G oi is an odd component for 1 i l 1 and G ej is an even component for 1 j l 2 .
Case 1. n is even.In this case, we have that l 1 is also even.If The equality in (7) holds if and only if ) for 1 j l 2 ; while the equality in (8) holds if and only if l 1 = 2, which implies that Obviously, G does not contain cycles, so mi(G) < r n−2 .Therefore we consider that l 1 = 0, that is to say, each component contained in G is an even component.Without loss of generality, assume that (by Lemma 2) The equality in (9) holds if and only if 2 K 2 ; while the equality in (10) holds if and only if p = 1, which implies that 2 K 2 , our result holds.If |V Ge 1 | 6, then we consider the following two possible subcases.
The equality in (11) holds if and only if Case 2. n is odd.In this case, l 1 is also odd.If l 1 3, by Lemmas 2 and 6 , we get Hence, we consider that l 1 = 1 in what follows.
If G o1 contains cycles, by induction hypothesis and Lemmas 2 and 6 , we obtain that the equality holds if and only if , the result holds.If G o1 does not contain cycles, then there exists at least one even component G ej containing cycles (by Lemmas 3 and 6) < 3r n−5 , the result holds.If |V Ge j | 6, we obtain that (by Lemmas 3, 6 and Theorem 1) The result holds.Now we consider that G is connected.It suffices to consider the following two possible cases.
Case 1. G has a vertex v of degree 4 such that G − v has cycles.
Subcase 1.1.n is even.In this subcase, we have (by induction hypothesis and Lemma 6)

The equality in (12) holds if and only if
In this subcase, we first consider that is either an acyclic bipartite graph which is not isomorphic to B n−5,k or a bipartite graph with cycles which is not isomorphic to D n−5,k .Hence by induction hypothesis or Lemma

The equality in (13) holds if and only if
In order to use the induction hypothesis, we should show that By a similar discussion as above, we may show that By induction hypothesis, we have mi(G − v) r n−3 .Thus, (by Lemma 6) (14) = 3r n−5 .

The equality in (14) holds if and only if
by deleting four independent vertices which is impossible.Hence mi(G) < 3r n−5 .
Case 2. Every vertex of degree 4 is in all cycles of G.

First we consider δ(
. We distinguish the following two possible subcases to show our results. By induction hypothesis and Lemma 5 and Theorem 1, we have mi(G 1 ) 3r n−7 and mi(G 2 ) max{3r n−7 , r n−5 +r n−9 } = 3r n−7 .By Lemma 1, we get

the equality holds if and only if
obtained from G 1 by deleting three independent vertices which is impossible.Subcase 2.2.n is even.
• d(u) = 2.In this subcase, we know that G 1 is a connected graph with cycles.Note that G ≇ C 8 * K 2 , hence G 1 ≇ C 8 .By Theorem 1, we have mi(G 1 ) r n−4 + 1.On the other hand, notice that the equality holds if and only if 2 K 2 since G have cycles.Thus, by Lemma 5 and Theorem 1, we get mi(G 1 ) r n−4 +r n−8 = 5r n−8 and mi(G 2 ) max{3r n−8 , r n−6 + r n−10 } = 3r n−8 .By Lemma 1, we have

• d(u)
4. Note that G contains cycles, it is easy to see that G 1 ≇ n−2 2 K 2 .By Lemmas 5, 6 and Theorem 1, we get mi(G 1 ) max{3r n−6 , r n−4 + r n−8 } = 3r n−6 and mi(G 2 ) r n−6 .By Lemma 1, the equality holds if and only if Note that G 2 is obtained from G 1 by deleting three or four independent vertices, hence there is no such bipartite graph of order n.Hence mi(G) < r n−2 .Now we consider δ(G) 2. In this subcase, we used the following two facts (for their proofs one may be referred to the Appendix).
Fact 1 Suppose n 7 is odd and each vertex of degree 4 is in all cycles of G, then mi(G) < 3r n−5 .Fact 2 Suppose n 6 is even and each vertex of degree 4 is in all cycles of G, then mi(G) r n−2 .

The equality holds if and only if
Obviously, in this case, if δ(G) 2, Theorem 2 holds directly from Facts 1 and 2. ✷

Concluding remark
In view of Theorems 1 and 2(i), the disconnected graphs among B n with the first-, second-, . . ., n−2 2 -th largest number of maximal independent sets are characterized, while the connected graphs in B n having the largest and the second largest number of maximal independent sets are determined; whereas in view of Theorem 2(ii), graphs among B ′ n having the largest number of maximal independent sets are identified.
is either an acyclic graph which is not isomorphic to B n−4,k or a bipartite graph containing cycles.It follows from induction hypothesis or Lemma 7 that mi(G The equality in (A.2) holds if and only if Note that G is a bipartite graph with δ(G) 2, hence G−v must be the graph as depicted in Fig. 3  Further on we need the following lemma to prove Fact 2.
Lemma A Suppose n 6 is even.If G has a path Proof: From the assumption it follows that G The equality in (A.4) holds if and only if d(u (10) or the graph depicted in Fig. 4(c).But this implies G − {u 1 , u 2 , u 3 , u 4 , u 5 } has no cycles, a contradiction.If G 1 = G 2 ∼ = O 1 (n − 3), we get two such graphs which contain odd cycles, a contradiction.That is to say, the equality in (A.4) does not hold.Hence, mi(G) < r n−2 , as desired.
First we consider that G has good vertices by distinguishing the following two possible cases.

Case 1. For all good vertices
Note that δ(G) 2 and G is bipartite, hence G 1 must be connected and G 1 ∼ = B n−4 .Furthermore, G 1 must be a graph with the structure shown in Fig. 3

Fig. 3 : 3 . 1 (= 3r n− 5 .
Fig. 3: Graphs used in the proof of Fact 1. Hence, without loss of generality, assume N (w) ∩ N (v) = ∅.Note that d(v) = 3 and δ(G) 2, hence G − w has cycles.Thus, by the assumption that each vertex of degree 4 is in all cycles of G, it follows that 2 d(x)3 for x = w or x ∈ N (v).If d(u) 3, note that δ(G) 2 and G − N G [v] ∼ = D n−4,k for every good vertex v, G must be the graph shown in Figs.4(a) or 4(b).For Fig. 4(a), mi(G) mi(G − v) + mi(G − N G [v]) = 8 + 5 = 13 < 16; For Fig. 4(b), mi(G) mi(G − v) + mi(G − N G [v]) = 6 + 5 = 11 < 16, they are not the extremal graphs.If d(u) 4, then G has at most one edge between N (v) and A = {u j : 1 j k} ∪ {w j : 1 j k}.Consequently, we have d(w) = 2. Otherwise, w is a good vertex of G, so in G − N [w], there is at least one vertex y ∈ N (u) such that d(y) = 1.Obviously, G − N [w] ≇ D n−4,k , a contradiction.Similarly, we can conclude that each vertex in N (v) has degree 2. Now, let P

Fig. 4 :
Fig. 4: Graphs used in the proof of Fact 2.
C 8 , C 10 and C 8 * K 2 .By induction hypothesis and Lemma 2, we have