Maximal Independent Sets in Bipartite Graphs with at Least One Cycle Shuchao Li, Huihui Zhang, Xiaoyan Zhang

Maximal independent sets in bipartite graphs with at least one cycle Shuchao Li, Huihui Zhang, Xiaoyan Zhang

To cite this version:

Shuchao Li, Huihui Zhang, Xiaoyan Zhang. Maximal independent sets in bipartite graphs with at least one cycle. Discrete Mathematics and Theoretical Computer Science, DMTCS, 2013, Vol. 15 no. 2 (2), pp.243–258. hal-00980771

HAL Id: hal-00980771 https://hal.inria.fr/hal-00980771 Submitted on 18 Apr 2014

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Discrete Mathematics and Theoretical Computer Science DMTCS vol. 15:2, 2013, 243–258

Maximal independent sets in bipartite graphs with at least one cycle

1 1 2 Shuchao Li Huihui Zhang Xiaoyan Zhang †

1Faculty of Mathematics and Statistics, Central China Normal University, Wuhan 430079, China 2School of Mathematical Science & Institute of Mathematics, Nanjing Normal University, Nanjing 210023, China received 18th Apr. 2013, revised 30th July 2013, accepted 21st Aug. 2013.

A maximal independent set is an independent set that is not a proper subset of any other independent set. Liu [J.Q. Liu, Maximal independent sets of bipartite graphs, J. Graph Theory, 17 (4) (1993) 495-507] determined the largest number of maximal independent sets among all n-vertex bipartite graphs. The corresponding extremal graphs are forests. It ′ is natural and interesting for us to consider this problem on bipartite graphs with cycles. Let Bn (resp. Bn) be the set of all n-vertex bipartite graphs with at least one cycle for even (resp. odd) n. In this paper, the largest number ′ of maximal independent sets of graphs in Bn (resp. Bn) is considered. Among Bn the disconnected graphs with n−2 the ﬁrst-, second-, ..., 2 -th largest number of maximal independent sets are characterized, while the connected graphs in Bn having the largest and the second largest number of maximal independent sets are determined. Among ′ Bn graphs having the largest number of maximal independent sets are identiﬁed.

Keywords: Maximal independent set, bipartite graph, cycle

1 Introduction

Given a graph G = (VG,EG), a set I VG 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 independentset in the complementgraph. The set of all maximal independentsets of a graph G is denoted by MI(G) and its cardinality by mi(G). Around 1960, Erd˝os 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 K3. 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

†Email: [email protected]. 1365–8050 c 2013 Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France 244 Shuchao Li, Huihui Zhang, Xiaoyan Zhang 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 well- known 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 n−2 first-, second-, ..., 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.

2 Preliminary Given a simple graph G = (V ,E ), the cardinality of V is called the order of G. And G v denotes G G G − the graph obtained from G by deleting vertex v VG (this notation is naturally extended if more than one vertex is deleted). For v V , let N (v) (or N∈(v) for short) denote the set of all the adjacent vertices of ∈ G G v in G and d(v) = NG(v) . For convenience, let NG[v] = NG(v) v .A leaf of G is a vertex of degree one, while an isolated| vertex| of G is a vertexof degreezero. Foranytwo∪{ } 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 connectedgraph H with maximum degree vertex x and a graph G = G1 G2 Gk with u being a maximum degree vertex in G , i = 1, 2, . . . , k. Deﬁne the graph H G to⊎ be⊎· the ··⊎ graph i i ∗ with vertex set VH∗G = VH VG and edge set EH∗G = EH EG xui : i = 1, 2, . . . , k . An odd (resp. even) component is a∪ component of odd (resp. even) order.∪ ∪ Through { out the text we} denote by Pn,Cn,K1,n−1 and Kn the path, cycle, star and complete graph on n vertices, respectively. Undeﬁned terminology and notation may be referred to [3]. Maximal independent sets in bipartite graphs 245

Throughout this paper, for simplicity, r denotes √2. We begin with some useful known results which are needed to prove our main results.

Lemma 1 ([7, 14]) For any vertex v in a graph G, mi(G) 6 mi(G v) + mi(G NG[v]). If v is a leaf adjacent to u, then mi(G) = mi(G N [v]) + mi(G N [u]). − − − G − G Lemma 2 ([14]) If G is the union of two disjoint graphs G1 and G2, then mi(G) = mi(G1) mi(G2). · Lemma 3 ([13]) If T is a tree with n > 1 vertices, then mi(T ) 6 t1(n), where

rn−2 + 1, if n is even; t n 1( ) = 1 ( rn− , if n is odd.

Furthermore, mi(T ) = t1(n) if and only if T T1(n), where ∈ n−2 n−4 B(2, 2 ) or B(4, 2 ), if n is even; T1(n) = n−1 ( B(1, 2 ), if n is odd, where B(i, j) is the set of batons, which are the graphs obtainedfrom a basic path Pi (i > 1) by attaching j > 0 paths of length two to the endpoints of Pi in any possible ways.

Lemma 4 ([13]) If F is a forest with n > 1 vertices, then mi(F ) 6 f1(n), where

rn, if n is even; f1(n) = 1 ( rn− , if n is odd.

Furthermore, mi(F ) = f1(n) if and only if F F1(n), where ∈ n 2 K2, if n is even; F1(n) = n−1−2s n−1 ( B(1, ) sK2 for some s with 0 6 s 6 , if n is odd. 2 ⊎ 2

Lemma 5 ([14]) If F is a forest with n > 4 vertices having F F1(n), then mi(F ) 6 f2(n), where 6∈ 3rn−4, if n > 4 is even;

f2(n) =  3, if n = 5;  7rn−7, if n > 7 is odd.  Furthermore, mi(F ) = f2(n) if and only if F F2(n), where ∈ n−4 P4 K2, if n > 4 is even; ⊎ 2 F2(n) = P1 P4 or,P1 P4, if n = 5;  ∗ ⊎  n−7  P7 K2, if n > 7 is odd. ⊎ 2  246 Shuchao Li, Huihui Zhang, Xiaoyan Zhang

For n > 2, 0 6 k 6 n , deﬁne ⌊ 2 ⌋ n−8 2P4 2 K2 or,T1(n 2k) kK2, if n is even; Bn,k = ⊎ − ⊎ ( T1(n 2k) kK2, if n is odd. − ⊎ k n−8 Then mi(B ) = 2 mi(T1(n 2k)), or 9r . n,k · − Lemma 6 ([17]) The maximum number of maximal independent sets among all bipartite graphs of order n n is 2⌊ 2 ⌋, and the only bipartite graphs of order n which have this many maximal independent sets are F1(n).

Lemma 7 ([17]) If G is an acyclic graph of order n and G ≇ Bn,k, then mi(G) < t1(n).

Lemma 8 If n > 6, then mi(Cn) = mi(Cn−2) + mi(Cn−3). Furthermore, one has rn−1, if n > 7 is odd; mi(Cn) < 2 ( rn− , if n > 12 is even.

Proof: The ﬁrst part is due to [13]; we show the second part by induction on n. It is routine to check that

6 8 10 mi(C7) = 7 < r , mi(C9) = 12 < r , mi(C11) = 22 < r , 10 12 12 mi(C12) = 29 < r , mi(C13) = 39 < r , mi(C14) = 51 < r . Assume the result holds for n 6 k. Now consider n = k + 1 > 15. By induction hypothesis, we obtain

rn−2−1 + rn−3−2 = 3rn−5 < rn−1, if n is odd; C C C < mi( n) = mi( n−2) + mi( n−3) 2 2 3 1 2 ( rn− − + rn− − = rn− , if n is even. This completes the proof. ✷

3 Sharp bounds and extremal graphs B B′ Let n (resp. n) be the set of all n-vertex bipartite graphs with at least one cycle for even (resp. odd) B n−2 n. In this section, among n the disconnected graphs with the first-, second-, ..., 2 -th largest number of maximal independent sets are characterized, while the connected graphs in Bn having the largest and B′ the second largest number of maximal independent sets are determined; among n the graphs having the largest number of maximal independent sets are identified. n n For even n and 3 6 k 6 2 , define Dn,k = B2k ( 2 k)K2, where B2k is depicted in Fig. 1. Then n −k n −k k−1 ⊎ n−−2 n−6 mi(D ) = 2 2 mi(B2 ) = 2 2 (2 + 1) 6 r + r . n,k · k 8 10 10 n n− n− n− Theorem 1 For graphs Dn,3,Dn,4,...,Dn, 2 ,C8 2 K2,C10 2 K2, (C8 K2) 2 K2, one has ⊎ ⊎ ∗ ⊎ n 8 n 10 mi(D 3) = mi(C8 − K2) > mi(D 4) > mi(D 5) = mi(C10 − K2) n, ⊎ 2 n, n, ⊎ 2 n 10 n−2 = mi((C8 K2) − K2) > mi(D 6) > mi(D 7) > > mi(D n ) = r + 1. ∗ ⊎ 2 n, n, ··· n, 2 Maximal independent sets in bipartite graphs 247

. x2 . . . . . . . .

Fig. 1: Graph B2k, where k > 3 and t > 2.

Proof: By direct computing, we have

n−2t n−2 n−2t n mi(D ) = 2 2 mi(B2 ) = 2 2 + 2 2 , t = 3, 4,..., , (1) n,t · t 2 n 8 n−8 mi(C8 − K2) = 10 2 2 , (2) ⊎ 2 · n 10 n−10 mi(C10 − K2) = 17 2 2 , (3) ⊎ 2 · n 10 n−10 mi((C8 K2) − K2) = 17 2 2 . (4) ∗ ⊎ 2 · n By (1), it is easy to see that mi(Dn,t) < mi(Dn,t−1) for 4 6 t 6 2 . Combining with (2)-(4), our results follow immediately. ✷

Let H0(6),H0(8),H0(10),H1(n),H2(n),H3(n) and O1(n) be the graphs depicted in Fig. 2.

. . . . . . . . ... . . . . . . . ... ... . . .

H 0(6) H 0(8) H 0(10) H 1(n) H 2(n) H 3(n) O 1(n)

Fig. 2: The extremal graphs

Theorem 2 Consider an n-vertex bipartite graph G containing cycles with n > 5.

n−8 n−10 n−10 (i) If G B D 3,D 4,...,D n ,C8 K2,C10 K2, (C8 K2) K2 , then ∈ n \{ n, n, n, 2 ⊎ 2 ⊎ 2 ∗ ⊎ 2 } mi(G) 6 rn−2. (5)

If G is connected, the equality holds in (5) if and only if G ∼= H0(n),H1(n),H2(n), or H3(n) for > n = 6, 8, 10; and G ∼= H1(n),H2(n), or H3(n) for n 12. n−4 n−6 If G is disconnected, the equality holds in (5) ifandonlyif G = C4 2 K2,H0(6) 2 K2,H0(8) n−8 n−10 ∼ ⊎ ⊎ ⊎ K2,H0(10) K2,H1(n 2k) kK2,H2(n 2k) kK2, or H3(n 2k) kK2. 2 ⊎ 2 − ⊎ − ⊎ − ⊎ 248 Shuchao Li, Huihui Zhang, Xiaoyan Zhang

B′ (ii) If G n, then ∈ mi(G) 6 3rn−5. (6)

The equality holds in (6) if and only if G = C6 K1, or O1(n) for connected G and G = O1(n n−7 ∼ ∗ ∼ − 2k) kK2, or (C6 K1) K2 for disconnected G. ⊎ ∗ ⊎ 2

B n Proof: We put the proofs of (i) and (ii) together as below. Choose G in n Dn,3,Dn,4,...,Dn, 2 , n−8 n−10 n−10 B′ \{ C8 2 K2,C10 2 K2, (C8 K2) 2 K2 (resp. n) such that mi(G) is as large as possible. We⊎ proceed by induction⊎ on n. ∗ ⊎ } For 5 6 n 6 6, it is straightforward to check that our results hold. Assume that our results hold for n 6 k. Now consider n = k + 1 > 7. First we consider that G is disconnected. Denote

G = G G G G G G , o1 ⊎ o2 ⊎ · · · ⊎ ol1 ⊎ e1 ⊎ e2 ⊎ · · · ⊎ el2

where Goi is an odd component for 1 6 i 6 l1 and Gej is an even component for 1 6 j 6 l2.

Case 1. n is even. In this case, we have that l1 is also even. If l1 > 2, then

l1 l2 mi(G) = mi(G ) mi(G ) (by Lemma 2) oi · ej i=1 j=1 Y Y l1 l2 |VG |−1 |VGe | 6 r oi r j (byLemma6) (7) · i=1 j=1 Y Y = rn−l1 6 rn−2. (8)

The equality in (7) holds if and only if Go = F1( VG ) for 1 6 i 6 l1, Ge = F1( VG ) for i ∼ | oi | j ∼ | ej | 1 6 j 6 l2; while the equality in (8) holds if and only if l1 = 2, which implies that G = F1( V ) ∼ | Go1 | ⊎ n−|VGo1 |−|VGo2 | n−2 F1( V ) K2. Obviously, G does not contain cycles, so mi(G) < r . | Go2 | ⊎ 2 Therefore we consider that l1 = 0, that is to say, each componentcontained in G is an even component. Without loss of generality, assume that VG > VG > > VG . Note that G contains cycles, | e1 | | e2 | ··· | el2 | hence V > 4. | Ge1 | If V = 4, then there exists a p 1, 2, . . . , l2 such that V = V = V = = | Ge1 | ∈ { } | Ge1 | | Ge2 | | Ge3 | ··· V = 4. As G contains cycles, there is at least one i 1, 2, . . . , p such that G = C4. We obtain Gep ei ∼ that| | ∈ { }

l2 mi(G) = mi(G ) mi(G ) mi(G ) (by Lemma 2) ei · ej · ek 16 = 6 = +1 Yj6 i p k Yp 6 2 3p−1 rn−4p (byLemmas3and6) (9) · · 3 = rn−2 ( )p−1 · 4 6 rn−2. (10) Maximal independent sets in bipartite graphs 249

The equality in (9) holds if and only if G C4, G P4 for 1 6 j = i 6 p, G (G G ei ∼= ej ∼= e1 e2 n−4p 6 − ⊎ ⊎ · · · ⊎ Gep ) = F1(n 4p) = 2 K2; while the equality in (10) holds if and only if p = 1, which implies that ∼ n−4− G = C4 K2, our result holds. ∼ ⊎ 2 If V > 6, then we consider the following two possible subcases. | Ge1 |

n−|VGe1 | VGe = 2. In this subcase, we have G Ge1 = K2. Note that G ≇ Dn,|V |/2,C8 2 ∼ 2 Ge1 •n− | 8 | n−10 n−−10 ⊎ K2,C10 K2 and (C8 K2) K2, hence Ge1 ≇ B|V |,C8,C10 and C8 K2. By 2 ⊎ 2 ∗ ⊎ 2 Ge1 ∗ induction hypothesis and Lemma 2, we have

mi(G) = mi(G ) mi(G G ) e1 · − e1 |V |−2 n−|V | 6 r Ge1 r Ge1 (11) · = rn−2.

The equality in (11) holds if and only if G 1 = H0(6),H0(8),H0(10),H1( V ),H2( V ), or e ∼ | Ge1 | | Ge1 | n−|VGe1 | H3( V ). Together with G G 1 = K2, our result follows immediately. | Ge1 | − e ∼ 2 V > 4. In this subcase, we have • | Ge2 |

l2 mi(G) = mi(G ) mi(G ) mi(G ) (by Lemma 2) e1 · e2 · ej j=3 Y |V |−2 |V |−2 n−|V |−|V | 6 (r Ge1 + 1) (r Ge2 + 1) r Ge1 Ge2 (by Lemmas 3, 6 and Theorem 1) · · 4 n−|V |−2 n−|V |−2 n−|V |−|V | = rn− + r Ge2 + r Ge1 + r Ge1 Ge2 6 rn−4 + rn−6 + rn−8 + rn−10 < rn−2.

Case 2. n is odd. In this case, l1 is also odd. If l1 > 3,byLemmas2and6,we get

l1 l2 l1 l2 |VG |−1 |VGe | n−l1 n−3 n−5 mi(G) = mi(G ) mi(G ) 6 r oi r j = r 6 r < 3r . oi · ej · i=1 j=1 i=1 j=1 Y Y Y Y

Hence, we consider that l1 = 1 in what follows.

If Go1 contains cycles, by induction hypothesis and Lemmas 2 and 6 , we obtain that

|V |−5 n−|V | 5 mi(G) = mi(G ) mi(G G ) 6 3r Go1 r Go1 = 3rn− , o1 · − o1 ·

n−|VGo1 | the equality holds if and only if G 1 = O1( V ) or C6 K1 and G G 1 = K2, which o ∼ | Go1 | ∗ − o 2 n−|VGo1 | n−7 implies that G = O1( V ) K2 or (C6 K1) K2, the result holds. ∼ | Go1 | ⊎ 2 ∗ ⊎ 2 If Go1 does not contain cycles, then there exists at least one even component Gej containing cycles 250 Shuchao Li, Huihui Zhang, Xiaoyan Zhang with VG > 4. If VG = 4, then Ge = C4. We obtain that | ej | | ej | j ∼ mi(G) = mi(G ) mi(G ) mi(G ) (by Lemma 2) o1 · ej · ei 16 = 6 2 iY6 j l |V |−1 n−4−|V | 6 r Go1 2 r Go1 (by Lemmas 3 and 6) · · < 3rn−5, the result holds. If VG > 6, we obtain that | ej | mi(G) = mi(G ) mi(G ) mi(G ) (by Lemma 2) o1 · ej · ei 16 = 6 2 iY6 j l |V |−1 |VG |−2 n−|VG |−|VG | 6 r Go1 (r ej + 1) r o1 ej (by Lemmas 3, 6 and Theorem 1) · · 3 n−|VG |−1 = rn− + r ej 6 rn−3 + rn−7 < 3rn−5.

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

mi(G) 6 mi(G v) + mi(G N [v]) (by Lemma 1) − − G 6 3rn−6 + rn−5−1 (byinductionhypothesisandLemma6) (12) = rn−2.

n−k−1 n−8 The equality in (12) holds if and only if G v = O1(k) K2 or (C6 K1) K2, G N [v] = − ∼ ⊎ 2 ∗ ⊎ 2 − G ∼ F1(n 5). But there is no such bipartite graph since G NG[v] is obtained from G v by deleting four independent− vertices, hence mi(G) < rn−2. − − Subcase 1.2. n is odd. n−3 n−7 In this subcase, we ﬁrst consider that G v = D 1 . Thenby Theorem1, mi(G v) 6 r +r . − ∼ n− ,k − If d(v) = 4, then G N [v] is either an acyclic bipartite graph which is not isomorphic to B 5 or a − G n− ,k bipartite graph with cycles which is not isomorphic to Dn−5,k. Hence by induction hypothesis or Lemma 7, mi(G N [v]) 6 rn−7, which gives − G mi(G) 6 mi(G v) + mi(G N [v]) (by Lemma 1) − − G 6 rn−3 + rn−7 + rn−7 (13) = 3rn−5.

G v C n−7 K G N v H n−11 K ,H The equalityin (13)holds if andonlyif ∼= 6 2 2 and G[ ] ∼= 0(6) 2 2 0(8) n−13 n−15 n−−k−5 ⊎ n−k−5− n⊎−k−5 ⊎ K2,H0(10) K2,H1(k) K2,H2(k) K2, or H3(k) K2. Note that 2 ⊎ 2 ⊎ 2 ⊎ 2 ⊎ 2 Maximal independent sets in bipartite graphs 251

G v C n−7 K G N v P n−7 K G N v ∼= 6 2 2, hence G[ ] is a subgraph of 5 2 1. It is easy to see that G[ ] contains− no cycles,⊎ a contradiction.− ⊎ − Now we consider that G v ≇ Dn−1,k. In order to use the induction hypothesis, we should show that n−9 − n−11 n−11 n−9 G v ≇ C8 K2,C10 K2, and (C8 K2) K2 ﬁrst. In fact, if G v = C8 K2, − ⊎ 2 ⊎ 2 ∗ ⊎ 2 − ∼ ⊎ 2 then 1 6 NG(v) NC8 (v) 6 4. | ∩ | n−9 If NG(v) NC8 (v) = 1, then n > 15 and G NG[v] = P7 2 K1. By Lemma 1, we get | ∩ | n−7 − n−5∼ ⊎ mi(G) 6 mi(G v) + mi(G NG[v]) = 5r + 7 < 3r . − − n−7 n−9 If NG(v) NC8 (v) = 2, then n > 13 and G NG[v] = P5 2 K1, or 2P3 2 K1. By Lemma | ∩ | − n−∼7 ⊎ n−5 ⊎ 1, we get mi(G) 6 mi(G v) + mi(G NG[v]) = 5r + 4 < 3r . − − n−5 If NG(v) NC8 (v) = 3, then n > 11 and G NG[v] = P3 2 K1. By Lemma 1, we get | ∩ | n−7 − n−5∼ ⊎ mi(G) 6 mi(G v) + mi(G NG[v]) = 5r + 2 < 3r . − − n−1 If NG(v) NC8 (v) = 4, then n > 9 and G NG[v] = 2 K1. By Lemma 1, we get mi(G) 6 | ∩ | n−7 n−−5 ∼ mi(G v) + mi(G NG[v]) = 5r + 1 < 3r . − − G v C n−11 K , C K n−11 K By a similar discussion as above, we may show that ∼= 10 2 2 ( 8 2) 2 2. Therefore, we have − 6 ⊎ ∗ ⊎ n 9 n 11 n 11 G v ≇ D 1 ,C8 − K2,C10 − K2 and (C8 K2) − K2. − n− ,k ⊎ 2 ⊎ 2 ∗ ⊎ 2 By induction hypothesis, we have mi(G v) 6 rn−3. Thus, − mi(G) 6 mi(G v) + mi(G N [v]) (by Lemma 1) − − G 6 rn−3 + rn−5 (byLemma6) (14) = 3rn−5.

G v H n−7 K ,H n−9 K ,H n−11 K , The equality in (14) holds if and only if ∼= 0(6) 2 2 0(8) 2 2 0(10) 2 2 n−k−1 n−k−1 − n⊎−k−1 ⊎ ⊎ H1(k) K2,H2(k) K2, or H3(k) K2 and G N [v] = F1(n 5). Since ⊎ 2 ⊎ 2 ⊎ 2 − G ∼ − G is connected and G N [v] has no isolated vertex, G v contains no K2 as a component, i.e., − G − G v ∼= H0(6),H0(8),H0(10),H1(n 1),H2(n 1), or H3(n 1). However, G NG[v] is obtained from− G v by deleting four independent− vertices which− is impossible.− Hence mi(G)−< 3rn−5. − Case 2. Every vertex of degree > 4 is in all cycles of G.

First we consider δ(G) = 1. Choose an edge uv E with d(v) = 1. Let G1 = G u, v and ∈ G − { } G2 = G N[u]. We distinguish the following two possible subcases to show our results. − Subcase 2.1. n is odd. n−3 If d(u) = 2, then G1 is a connected graph with cycles and G2 ≇ 2 K2. By induction hypothesis and n−7 n−7 n−5 n−9 n−7 Lemma5 andTheorem1, we have mi(G1) 6 3r and mi(G2) 6 max 3r , r +r = 3r . By Lemma 1, we get { }

n−7 n−7 n−5 mi(G) = mi(G1) + mi(G2) 6 3r + 3r = 3r ,

G O n C K G P n−7 K G O n the equality holds if and only if 1 ∼= 1( 2) or 6 1, 2 ∼= 4 2 2, i.e., ∼= 1( ). If d(u) > 3, by Lemmas 1 and 6 , we obtain− that ∗ ⊎

n−2−1 n−4−1 n−5 mi(G) = mi(G1) + mi(G2) 6 r + r = 3r , 252 Shuchao Li, Huihui Zhang, Xiaoyan Zhang the equality holds if and only if G1 = F1(n 2), G2 = F1(n 4) or F1(n 5). If G2 = F1(n 4), ∼ − ∼ − − ∼ − we have d(u) = 3. In this situation, we get G = C6 K1. If G2 = F1(n 5), we have d(u) = 4. G2 is ∼ ∗ ∼ − obtained from G1 by deleting three independent vertices which is impossible. Subcase 2.2. n is even.

d(u) = 2. In this subcase, we know that G1 is a connected graph with cycles. Note that G ≇ C8 K2, • n−4 ∗ hence G1 ≇ C8. By Theorem 1, we have mi(G1) 6 r + 1. On the other hand, notice that G ≇ Dn,k, hence if G2 = Bn−3,k, we obtain that G = H1(n),H2(n), or H3(n), our result holds. If G2 = Bn−3,k, ∼ ∼ n−4 n−8 n−4 6∼ by induction hypothesisor Lemma 7, we have mi(G2) 6 max r 1, 3r = r 1. Combining with Lemma 1, we have { − } −

n−4 n−4 n−2 mi(G) = mi(G1) + mi(G2) 6 r + 1 + r 1 = r , − n−4 the equality holds if and only if G1 ∼= Bn−2,C10 or C8 K2 and mi(G2) = r 1, but there is no such bipartite graph. ∗ − n−4 d(u) = 3. If G1 is disconnected, then G1 must have cycles and G2 ≇ 2 K2 since G have cycles. Thus, • n−4 n−8 n−8 n−8 n−6 byLemma5andTheorem1, weget mi(G1) 6 r +r = 5r and mi(G2) 6 max 3r , r + rn−10 = 3rn−8. By Lemma 1, we have { } n−8 n−8 n−2 mi(G) = mi(G1) + mi(G2) 6 5r + 3r = r , G C n−8 K G P n−8 K the equality holds if and only if 1 ∼= 6 2 2 and 2 ∼= 4 2 2, but there is no such bipartite G G n−4⊎K G ⊎H n G rn−2 graph. If 1 is connected and 2 ∼= 2 2, then we have ∼= 2( ) and mi( ) = , the result G G ≇ n−4 K n G K G H holds. If 1 is connected and 2 2 2, then for = 6, we get 2 = 2 1 and ∼= 0(6), the result holds. We assume n > 8. Obviously, G1 ≇ C8 since G ≇ C8 K2. By Lemmas 3, 5 and Theorem n−4 n−8 n−6 n−∗ 10 n−8 1, we have mi(G1) 6 r + 1, mi(G2) 6 max 3r , r + r = 3r . Thus, { } mi(G) = mi(G1) + mi(G2) (by Lemma 1) 6 rn−4 + 1 + 3rn−8 (15) = 7rn−8 + 1 6 rn−2. (16)

n−8 The equality in (15) holds if and only if G1 = T1(n 2), or B 2, G2 = P4 K2; while the equality ∼ − n− ∼ ⊎ 2 in (16)holds if and only if n = 8, i.e., G1 ∼= P4 K2, C6 or P6 and G2 ∼= P4, but there is no such bipartite graph. ∗ n−2 d(u) > 4. Note that G contains cycles, it is easy to see that G1 ≇ 2 K2. By Lemmas 5, 6 and • n−6 n−4 n−8 n−6 n−6 Theorem 1, we get mi(G1) 6 max 3r , r + r = 3r and mi(G2) 6 r . By Lemma 1, { } n−6 n−5−1 n−2 mi(G) = mi(G1) + mi(G2) 6 3r + r = r ,

n−6 the equality holds if and only if G1 = P4 K2, G2 = F1(n 5) or F1(n 6). Note that G2 is ∼ ⊎ 2 ∼ − − obtained from G1 by deleting three or four independent vertices, hence there is no such bipartite graph of order n. Hence mi(G) < rn−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). Maximal independent sets in bipartite graphs 253

Fact 1 Suppose n > 7 is odd and each vertex of degree > 4 is in all cycles of G, then mi(G) < 3rn−5. Fact 2 Suppose n > 6 is even and each vertex of degree > 4 is in all cycles of G, then mi(G) 6 rn−2. The equality holds if and only if G ∼= H1(n). Obviously, in this case, if δ(G) > 2, Theorem 2 holds directly from Facts 1 and 2. ✷

4 Concluding remark B n−2 In view of Theorems 1 and 2(i), the disconnected graphs among n with the first-, second-, ..., 2 -th largest number of maximal independent sets are characterized, while the connected graphs in Bn having the largest and the second largest number of maximal independent sets are determined; whereas in view of B′ Theorem 2(ii), graphs among n having the largest number of maximal independent sets are identified. Acknowledgements The authors would like to express their sincere gratitude to the referees for a very careful reading of the paper and for all of their insightful comments and valuable suggestions, which led to a number of improvements in this paper. This work was supported by the National Natural Science Foundation of China (Grant No. 11271149) and the Special Fund for Basic Scientific Research of Central Colleges (Grant No. CCNU13F020).

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[12] M.J. Jou, G. J. Chang, Algorithmic aspects of counting independent sets, Ars Combin. 65 (2002) 265-277. [13] M.J. Jou, G. J. Chang, Maximal independent sets in graphs with at most one cycle, Discrete Appl. Math. 79 (1997) 67-73 . [14] M.J. Jou, J.J. Lin, Trees with the second largest number of maximal independent sets, Discrete Math. 309 (2009) 4469-4474. [15] M.J. Jou, The second largest number of maximal independent sets in connected graphs with at most one cycle, J. Comb. Optim. 24 (2012) 192-201. [16] S.C. Li, W. Jing, Maximal independent sets in graphs with cyclomatic number at most two. Util. Math. 83 (2010) 107-120. [17] J.Q. Liu, Maximal independent sets in bipartite graphs, J. Graph Theory, 17 (1993) 495-507. [18] J.W. Moon, L. Moser, On Cliques in graphs, Israel J. Math. 3 (1965) 23-28. [19] T. Moscibroda, R. Wattenhofer, Efﬁcient computation of maximal independent sets in unstructured multi-hop radio network. In: Proc of 1-st international conference on mobile ad hoc and sensor systems, 2004, pp 51-59. [20] C. Ortiz, M. Villanueva, Maximal independent sets in caterpillar graphs, Discrete Appl. Math. 160 (3) (2012) 259-266. [21] N. Rhodes, P. Willett, A. Calvet, J.B. Dunbar, C. Humblet, CLIP: similarity searching of 3D databases using clique detection, J. Chem. Inf. Comput. Sci. 43 (2) (2003) 443-448. [22] G.C. Ying, K.K. Meng, B.E. Sagan, V.R. Vatter, Maximal independent sets in graphs with at most r cycles, J. Graph Theory 53 (4) (2006) 270-282.

Appendix In the appendix, we present the proofs for Facts 1 and 2.

Proof of Fact 1: We ﬁrst assume that G has a vertex v of degree 3 such that G − v has cycles. Note that G contains no odd cycles and δ(G) > 2, hence G − v (resp. G − NG[v]) does not contain K2 as a component. G v ≇ D G v ≇ C n−9 n−11 n−11 • − n−1,k. Then we are to show that − 8 ⊎ 2 K2,C10 ⊎ 2 K2 and (C8 ∗ K2) ⊎ 2 K2 by contradiction. ∼ n−9 ∼ ∼ If G − v = C8 ⊎ 2 K2, i.e., G − v = C8. Notice that d(v) = 3, we have G − NG[v] = 2K1 ⊎ P3. So we obtain that the graph G is depicted in Fig. 3(a). By Lemma 1, we get mi(G) 6 mi(G − u) + mi(G − N[u]) = 7 + 4 = 11 < 3r4 = 12. ∼ n−11 K , G v ∼ C . d v G N v ∼ K P K P If G − v = C10 ⊎ 2 2 i.e., − = 10 Note that ( ) = 3, we have − G[ ] = 2 1 ⊎ 5 or 1 ⊎ 2 3. By 6 Lemma 1, we get mi(G) 6 mi(G − v) + mi(G − NG[v]) = 17 + 4 = 21 < 3r = 24. G v ∼ C K n−11 K , G v ∼ C K . d v G N v ∼ K P ,K P , If − = ( 8 ∗ 2)⊎ 2 2 i.e., − = 8 ∗ 2 Note that ( ) = 3, we have − G[ ] = 2 1 ⊎ 5 1 ⊎ 2 3 6 K1 ⊎(K1 ∗(K2 ⊎P3)) or P3 ⊎K1,3. By Lemma 1, we get mi(G) 6 mi(G−v)+mi(G−NG[v]) = 17+4 = 21 < 3r = 24. Hence, we have n − 9 n − 11 n − 11 G − v ≇ Dn−1,k,C8 ⊎ K2,C10 ⊎ K2 and (C8 ∗ K2) ⊎ K2. 2 2 2 By induction hypothesis, mi(G − v) 6 rn−3. In view of Lemmas 1 and 6, we have

mi(G) 6 mi(G − v) + mi(G − NG[v]) 6 rn−3 + rn−4−1 (A.1) = 3rn−5.

The equality in (A.1) holds if and only if G − v =∼ H0(n − 1),H1(n − 1),H2(n − 1), or H3(n − 1) for n = 7, 9, 11 and G − v =∼ H1(n − 1),H2(n − 1), or H3(n − 1) for n > 13, G − NG[v] =∼ F1(n − 4), i.e., G − NG[v] =∼ T1(n − 4). Note that G − NG[v] is obtained from G − v by deleting three independent disjoint vertices, but there is no such bipartite graph of order n. Hence mi(G) < 3rn−5. Maximal independent sets in bipartite graphs 255

n−3 • G − v =∼ Dn−1,k, i.e., G − v =∼ Bn−1. In this case, we have mi(G − v) = r + 1. Furthermore, G − NG[v] is either an acyclic graph which is not isomorphic to Bn−4,k or a bipartite graph containing cycles. It follows from induction hypothesis or n−4−1 n−4−5 n−5 Lemma 7 that mi(G − NG[v]) 6 max{r − 1, 3r } = r − 1. Together with Lemma 1, we have

mi(G) 6 mi(G − v) + mi(G − NG[v]) 6 rn−3 + 1 + rn−5 − 1 (A.2) = 3rn−5.

n−5 The equality in (A.2) holds if and only if G − v =∼ Bn−1, mi(G − NG[v]) = r − 1. Note that G is a bipartite graph with δ(G) > 2, hence G−v must be the graph as depicted in Fig. 3(b). G−NG[v] is obtained from G−v by deleting three independent n−5 n−5 vertices. Elementary calculation yields mi(G − NG[v]) < r − 1. Hence, mi(G) < 3r . u 1 w 1 P n1 v1 ... v 2 ... u2 w 2 Pn2 . . . u v . . w u u . v . . . P n vt k... u uk w k (a ) (b) ( c) (d)

Fig. 3: Graphs used in the proof of Fact 1.

Now, we consider the case that each vertex of degree > 3 is in all cycles of G. Then G must be a graph with the structures as depicted in Figs. 3(c) or 3(d). Assume that the vertex u is in all cycles of G, then d(u) > 3. Note that n is odd and δ(G) > 2, n−3 hence G − u ≇ Bn−1,k and G − u contains no cycles. By Lemma 7, we get mi(G − u) 6 r . Thus, we have mi(G) 6 mi(G − u) + mi(G − N[u]) (by Lemma 1) 6 rn−3 + rn−4−1 (by Lemma 6 ) (A.3) = 3rn−5.

n−3 The equality in (A.3) holds if and only if mi(G − u) = r , G − N[u] =∼ F1(n − 4) or F1(n − 5), i.e., G − N[u] =∼ T1(n − 4) or T1(n − 5). Notice that δ(G) > 2 and u is in all cycles of G, it is straightforward to check that n ∈ {7, 9, 11} and G − N[u] ≇ T1(n − 5). So we get that G is a graph of the structure in Fig. 3(d). That is to say, G has exactly two vertices of degree > 3, say u and v. It follows that d(u) = d(v) = d and G − {u, v} = Pn1 ⊎ · · · ⊎ Pnk with n1 > n2 > ··· > nk. If n = 7, then n1 = 3, n2 = 1, n3 = 1. By elementary calculation, mi(G) = 5 < 6. If n = 9, then n1 = 5, n2 = 1, n3 = 1. By Lemma 1, mi(G) 6 mi(G − u) + mi(G − N[u]) = 4 + 7 = 11 < 12. If n = 11, then n1 = n2 = n3 = 3. By Lemma 1, mi(G) 6 mi(G − u) + mi(G − N[u]) = 13 + 8 = 21 < 24. Hence, we obtain mi(G) < 3rn−5. Thus, Fact 1 holds. ✷

Further on we need the following lemma to prove Fact 2.

Lemma A Suppose n > 6 is even. If G has a path P5 = u1u2u3u4u5 such that d(u3) = 2,G − {u1, u2, u3, u4, u5} has cycles ≇ n−12 n−14 n−14 n−2 and G − {u2, u3, u4, u5} Dn−4,k,C8 ⊎ 2 K2,C10 ⊎ 2 K2 and (C8 ∗ K2) ⊎ 2 K2, then mi(G) < r .

Proof: From the assumption it follows that G1 := G−{u1, u2, u3},G2 := G−{u2, u3, u4} and G3 := G−{u2, u3, u4, u5}, ≇ n−12 n−14 n−14 respectively, has cycles and G3 Dn−4,k,C8 ⊎ 2 K2,C10 ⊎ 2 K2 and (C8 ∗ K2) ⊎ 2 K2. Hence, by Lemma 1 we obtain

mi(G) 6 mi(G − u2 − u4) + mi(G − u2 − u4 − N(u4)) + mi(G − N[u2])

= mi(G2) + mi(G3) + mi(G1) 6 3rn−3−5 + rn−4−2 + 3rn−3−5 (by induction hypothesis) (A.4) = rn−2. 256 Shuchao Li, Huihui Zhang, Xiaoyan Zhang

∼ n−10 n−k−3 The equality in (A.4) holds if and only if d(u2) = d(u4) = 2, G1 = G2 = (C6 ∗ K1) ⊎ 2 K2 or O1(k) ⊎ 2 K2, and ∼ n−10 K ,H n−12 K ,H n−14 K ,H k n−k−4 K ,H k n−k−4 K , H k G3 = H0(6) ⊎ 2 2 0(8) ⊎ 2 2 0(10) ⊎ 2 2 1( ) ⊎ 2 2 2( ) ⊎ 2 2 or 3( ) ⊎ n−k−4 > ∼ ∼ 2 K2. Since δ(G) 2, G1,G2 contains no K2 as a component, i.e., G1 = G2 = C6 ∗ K1 or O1(n − 3). If G1 = G2 = C6 ∗ K1, we get G =∼ H0(10) or the graph depicted in Fig. 4(c). But this implies G − {u1, u2, u3, u4, u5} has no cycles, a contradiction. If G1 = G2 =∼ O1(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) < rn−2, as desired. ✷

Proof of Fact 2: We say a vertex v is good if d(v) = 3 and G − v has cycles. Since δ(G) > 2 and G is bipartite, then G − NG[v] ≇ Bn−4,k for any good vertex v. First we consider that G has good vertices by distinguishing the following two possible cases.

Case 1. For all good vertices v, G − NG[v] =∼ Dn−4,k.

For convenience, let G1 = G − NG[v]. Note that δ(G) > 2 and G is bipartite, hence G1 must be connected and G1 =∼ Bn−4. Furthermore, G1 must be a graph with the structure shown in Fig. 3(b). Clearly, if t > 1, then N(w) ∩ N(v) = ∅; if t = 0, then either N(w) ∩ N(v) = ∅ or N(u) ∩ N(v) = ∅. 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 6 d(x) 6 3 for x = w or x ∈ N(v). If d(u) 6 3, note that δ(G) > 2 and G − NG[v] =∼ Dn−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) 6 mi(G − v) + mi(G − NG[v]) = 8 + 5 = 13 < 16; For Fig. 4(b), mi(G) 6 mi(G − v) + mi(G − NG[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 = {uj : 1 6 j 6 k} ∪ {wj : 1 6 j 6 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] ≇ Dn−4,k, a contradiction. Similarly, we can conclude that each vertex in N(v) has degree 2. Now, let P5 = u1w1ww2u2. Obviously, d(w) = 2, G − {u1, w1, w, w2, u2} has cycles and G − {w1, w, w2, u2} ≇ n−12 n−14 n−14 n−2 Dn−4,k,C8 ⊎ 2 K2,C10 ⊎ 2 K2 and (C8 ∗ K2) ⊎ 2 K2. By Lemma A, we obtain that mi(G) < r .

u u . . . . . . v v (a) (b) (c) (d) (f)

Fig. 4: Graphs used in the proof of Fact 2.

Case 2. There exists a good vertex v such that G − NG[v] ≇ Dn−4,k. In this case, we are to show that G G N v ≇ C n−9 K ,C n−11 K and C K n−11 K . In 1 := − G[ ] 8 ⊎ 2 2 10 ⊎ 2 2 ( 8 ∗ 2) ⊎ 2 2 ∼ n−9 1 2 8 ∼ n−11 fact, if G1 = C8 ⊎ 2 K2, then there exist eight such graphs, i.e., G ,G ,...,G ; if G1 = C10 ⊎ 2 K2, then there 9 10 21 ∼ n−11 K exist thirteen such graphs, i.e., G ,G ,...,G ; if G1 = (C8 ∗ K2) ⊎ 2 2, then there exist twenty-ﬁve such graphs, i.e., G22,G23,...,G46, where G1,G2,...,G46 are depicted in Fig. 5. By direct computation, we obtain mi(G) < rn−2. Hence, ≇ n−9 K ,C n−11 K C K n−11 K . G G1 C8 ⊎ 2 2 10 ⊎ 2 2 and ( 8 ∗ 2) ⊎ 2 2 By Lemma 7 (if 1 is acyclic) or by induction hypothesis (if n−6 G1 contains cycles), we obtain that mi(G1) 6 r . Thus, we obtain that

mi(G) 6 mi(G − v) + mi(G1) (by Lemma 1) 6 3rn−6 + rn−6 (by induction hypothesis) (A.5) = rn−2.

∼ n−8 n−1−k n−6 The equality in (A.5) holds if and only if G − v = (C6 ∗ K1) ⊎ 2 K2 or O1(k) ⊎ 2 K2, mi(G1) = r . Note that δ(G) > 2 and G contains no odd cycles, G − v contains no K2 as a component, hence G − v =∼ C6 ∗ K1 or O1(n − 1). If 6 G−v =∼ C6 ∗K1, there are two such graphs. By a straightforward calculation, mi(G) = 6 or 7 < r = 8. If G−v =∼ O1(n−1), then 6 6 n 6 10 since d(v) = 3 and δ(G) > 2. For n = 6, there is one such graph. By a straightforward calculation, mi(G) = 3 < 4. For n = 8, there are two such graphs. By a straightforward calculation, mi(G) = 7 < 8. For n = 10, we get Maximal independent sets in bipartite graphs 257

v v v v v v

1 2 3 4 5 6 GG G G G G

21 22 22 24 21 22

v v v v v v

7 8 9 10 11 12 G GG G G G

23 24 36 39 36 38

v v v v v v

13 14 15 G 16 17 18 G G G GG

39 39 41 42 36 38 v v v v v v

19 20 21 22 23 24 G G GG G G

39 41 42 39 41 43 v v v v v v

25 26 27 28 29 30 G G GG GG

39 41 43 38 43 41 v v v v v v

31 32 33 34 35 36 G G G GG G

38 42 42 44 40 45 v v v v v v

37 38 39 40 41 42 GG G G G G

42 44 41 41 43 43 v v v v

43 44 45 46 G GG G

44 40 41 39

Fig. 5: Graphs G1,G2,...,G46 used in Case 2 of Fact 2, in which each number below the graph Gi is an upper bound of mi(Gi), i = 1, 2,..., 46. 258 Shuchao Li, Huihui Zhang, Xiaoyan Zhang

n−2 G1 =∼ C4 ∗ 2K1. By Lemma 1, mi(G) 6 mi(G − v) + mi(G1) = 12 + 3 = 15 < 16. Hence, mi(G) < r . The result holds. Now we consider that G has no good vertex, that is to say, each vertex of degree > 3 must be contained in all cycles of G. This implies that G must be the graph with structures shown in Figs. 3(c) or 3(d). By Lemma 8 and G ≇ C6,C8 and C10, we assume that G is not a cycle. If G is a graph of the structure in Fig. 3(c), then G has only one vertex u of degree > 3. Since G is a bipartite graph of even order, then d(u) > 6 and G − u consists of k > 3 disjoint paths of even length, say Pn1 ,Pn2 ,...,Pnk d(u) (see Fig. 3(c)), where k = 2 . By Lemmas 2 and 3, it follows that k k nj −1 n−4 mi(G − u) = Y mi(Pnj ) 6 Y r 6 r j=1 j=1 and so

mi(G) 6 mi(G − u) + mi(G − NG[u]) (by Lemma 1) 6 rn−4 + rn−8 (by Lemma 6 ) = 5rn−8 < rn−2. The result holds. Suppose G is a graph of the structure in Fig. 3(d), then G has exactly two vertices degree > 3, say u, v. It follows that d(u) = d(v) = d > 3 and G − {u, v} consists of k disjoint paths, say Pn1 ,Pn2 ,...,Pnk with n1 > n2 > ··· > nk, where k = d if u and v are not adjacent and k = d − 1 otherwise. Since G is a bipartite graph, either all nj ’s are odd or all nj ’s are even. If all nj ’s are odd, we have k = d > 4. By Lemmas 2 and 3, we get

k k nj −1 n−6 mi(G − u − v) = Y mi(Pnj ) 6 Y r 6 r . j=1 j=1 This gives

mi(G) 6 mi(G − u − v) + mi(G − u − v − NG(v)) + mi(G − NG[u]) (by applying Lemma 1 twice) 6 rn−6 + rn−6 + rn−5−1 (by Lemma 6) = 3rn−6 < rn−2.

Now we consider that all nj ’s are even. Let Pn1 = u1u2 . . . un1 . By Lemma A, we assume n1 6 4. If n1 = 4, then both L1 = G − {u1, u2, u3} and L2 = G − {u2, u3, u4} have cycles and L3 = G − {u1, u2, u3, u4, v1} is a tree. Furthermore, L3 ≇ T1(n − 4) unless G is one of the graphs shown in Figures 4(c), 4(d) and 4(f). If L3 ≇ T1(n − 4), by Lemma1 we get

mi(G) 6 mi(G − u1 − u3) + mi(G − u1 − u3 − N(u1)) + mi(G − N[u3])

= mi(L1) + mi(L3) + mi(L2) 6 3rn−8 + rn−6 + 3rn−8 (by induction hypothesis and Lemma 6) (A.6) = rn−2.

∼ ∼ n−10 K ,H The equality in (A.6) holds if and only if L1 = L2 = C6 ∗ K1 or O1(n − 3), L3 = H0(6) ⊎ 2 2 0(8) ⊎ n−12 n−14 n−k−4 n−k−4 n−k−4 ∼ 2 K2,H0(10)⊎ 2 K2,H1(k)⊎ 2 K2,H2(k)⊎ 2 K2, or H3(k)⊎ 2 K2. If L1 = L2 = C6 ∗K1, then n = 10 and G =∼ H0(10). So mi(G) = mi(H0(10)) = 16, hence we get the extremal graph H0(10). If L1 = L2 =∼ O1(n−3), then n = 8 and G =∼ H0(8). So mi(G) = mi(H0(8)) = 8, hence we get the extremal graph H0(8). Our result holds. If L3 =∼ T1(n − 4), G is one of the graphs shown in Figs. 4(c), 4(d) and 4(f). By direct calculation, we have, for Fig. 4(c), n−6 n−2 mi(G) = 13 < 16; for Fig. 4(d) (n > 8), mi(G) = 3r + 2 6 r , the equality holds if and only if n = 8, G =∼ H0(8), n−6 n−2 hence we get extremal graph H0(8); for Fig. 4(f) (n > 10), mi(G) = 3r + 4 6 r , the equality holds if and only if n = 10, G =∼ H0(10). Hence, we assume n1 6 2, which implies n1 = n2 = ··· = nk = 2. Since G ≇ Bn, we conclude that n−2 v1 must be adjacent to v2. So mi(G) = r and G =∼ H1(n). This completes the proof. ✷