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Minimizing a class of unicyclic graphs by means of Hosoya index
Mathematical and Computer Modelling 48 (2008) 940–948 www.elsevier.com/locate/mcm
Minimizing a class of unicyclic graphs by means of Hosoya indexI Hongbo Hua Department of Computing Science, Huaiyin Institute of Technology, Huaian, Jiangsu 223000, PR China Received 18 August 2007; received in revised form 9 November 2007; accepted 5 December 2007
Abstract The Hosoya index, denoted by z(G), of a (molecular) graph G is defined as the total number of independent-edge sets of G. Let Un be the set of unicyclic graphs with n vertices. A fully loaded unicyclic graph is a unicyclic graph with the property that there is no vertex with degree less than 3 in its unique cycle. Denote by Un1 the set of fully loaded unicyclic graphs. In this paper, graphs in Un1 with minimal, second-minimal and third-minimal Hosoya indices are uniquely determined, respectively. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Hosoya index; Unicyclic graph; Permanent; Matching
1. Introduction The Hosoya index, proposed by Hosoya [2] in 1971, acts as one of the important topological parameters in the study of the relation between molecular structure and the physical and chemical properties of a number of hydrocarbon compounds. Let G be a graph with n vertices. Hosoya index, denoted by z(G), is defined to be the total number of matchings, namely, n
z(G) =
b2c X
m(G; s),
s=0
where m(G; s) is the number of s-matchings of G. An s-matching of a graph G is a subset M of its edge set with the property that |M| = s and M contains no two edges sharing a common vertex. For convenience and consistence, it will be always assumed that m(G; 0) = 1. It is both interesting and significant to determine the graph with extremal (maximal or minimal) Hosoya index. Along these lines, many results have been put forward in recent years. Gutman in [3] proved that the linear hexagonal chain is the unique chain with minimal Hosoya index among all hexagonal chains. Zhang [4] showed that the zig-zag hexagonal chain is the unique chain with maximal Hosoya index among all hexagonal chains. Zhang and Tian [5] gave new proofs of Gutman’s results in [3] and Zhang’s results in [4]. In [6], Zhang and Tian determined the graphs with I Partially supported by SRF of Huaiyin Institute of Technology (HGQ 0611, HGQN0726, HGQN0727).
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.12.003
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H. Hua / Mathematical and Computer Modelling 48 (2008) 940–948
Fig. 1. (a). P52 (1, 1, 1, 1); (b). Plk (1, 1, . . . , 1).
4 (1, 2, 2, 3); (b). C 3 (2, 2, 2). Fig. 2. (a). C12 9
minimal and second-minimal Hosoya indices among all catacondensed systems. In [7], the path and star have been shown to have the maximal and minimal Hosoya indices resp., among all trees on n vertices. Hou [8] characterized the trees having minimal and second-minimal Hosoya indices among all trees with a given size of matching. Yu and Tian [9] investigated the graph having minimal Hosoya index among all graphs with given edge-independence number and cyclomatic number. In [10], Yu and Lv investigated the tree having minimal Hosoya index among all trees with k pendent vertices. More recently, Ou [11] determined, respectively, the unique unicyclic graphs with the first and second largest Hosoya indices among all unicyclic graphs with n vertices. He also determined the unicyclic graphs with the first and second smallest Hosoya indices in the successive paper [12]. All graphs considered here are both connected and simple if not stated in particular. For a vertex v in G, let d(v) be the number of edges incident with v. If d(v) ≥ 3, then we call v a branched vertex. By N G (v) we denote the set of vertices in G adjacent to v. In order to state and prove our main results, we need some further notations. Let Un be the set of unicyclic graphs with n vertices. A fully loaded unicyclic graph is a unicyclic graph with the property that there exists no vertex with degree less than 3 in its unique cycle. Let Un1 be the set of fully loaded unicyclic graphs. Let Un (l) and Un1 (l) denote resp. the subset of Un and Un1 in which every graph has a unique cycle of length l. Let Pn be the path on n vertices and its vertices be ordered successively as x1 , x2 , . . . , xn . For 1 ≤ k ≤ n, we denote, by Pnk (1, 1, . . . , 1), the graph obtained from Pn by attaching exactly one pendent edge to each of the vertices xk , xk+1 , . . . , xn , respectively. Here Pnn (1, 1, . . . , 1) = Pn+1 . Let Cl be the cycle of length l and its vertices be ordered successively as y1 , y2 , . . . , yl . We denote, by Cnl (t1 , t2 , . . . , tl ), the graph obtained from Cl by attaching P exactly ti pendent edges to the vertex yi for i = 1, 2, . . . l, where ti ≥ 1 and li=1 ti = n − l. As examples, we 4 (1, 2, 2, 3) and C 3 (2, 2, 2) in Fig. 2, respectively. Other illustrate P52 (1, 11, 1) and Plk (1, 1, . . . , 1) in Fig. 1 and C12 9 notations and terminology not defined here will conform to those in [1]. In this paper, we shall investigate the Hosoya index of fully loaded unicyclic graphs. Graphs in Un1 with minimal, second-minimal and third-minimal Hosoya indices are determined. 2. Graph in Un1 with minimal Hosoya index Let G be an n-vertex graph with adjacency matrix A(G). Define the neighbor matrix B(G) of G as B(G) = A(G) + I , where I is the unit matrix of order n. We immediately have Lemma 1. Let G be a graph with m (≥1) components G 1 , G 2 , . . . G m . Then Per B(G) = Proof. By a suitable labeling vertices of G, we obtain B(G 1 ) B(G 2 ) B(G) = . . .. B(G m ) The result follows by a direct expansion of Per B(G).
Qm
i=1 Per
B(G i ).
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Fig. 3. Transformation A.
Lemma 2. Let G be a graph on n ≥ 2 vertices. If uv is a pendent edge of G with pendent vertex v, then Per B(G) = Per B(G − v) + Per B(G − v − u). Proof. Ordering vertices of G as v, u, . . . , we have 1 1 0 B(G) = 1 1 x . 0 x D By an expansion of Per B(G) along its first row, we obtain 1 x 1 x 1 x Per B(G) = Per + Per = Per + Per D. x D 0 D x D Note that B(G − v) = x1 Dx and B(G − v − u) = D. Thus the result follows.
For convenience, we set Per B(∅) = 1. Let G ∈ Un (l) and we denote, by T (G), the forest obtained from G by deleting all vertices in V (Cl ) as well as edges incident with each vertex in V (Cl ). Thus, we have Theorem 3. Let G be a unicyclic graph in Un (l) with n ≥ 3 vertices. Then z(G) = Per B(G) − 2Per B(T (G)). Proof. From the definition of Permanent [13], we obtain Per B(G) =
n XY σ i=1
bi,σ (i) ,
(1)
Qn where σ goes over all symmetric group of order n. In the above Eq. (1), term i=1 bi,σ (i) = 0 if and only if either vertex i is not adjacent to vertex σ (i) for some i 6= σ (i) or, σ contains a cycle with length greater than 2 but not equal to l since G ∈ Un (l) contains exactly one cycle of length l. Consequently, every non-zero term of Per B(G) corresponds to a matching of G with the exception that σ is one of two cycles of length l. So z(G) = Per B(G) − 2Per B(T (G)) and then the theorem follows. In what follows we shall introduce a transformation on a graph G which will decrease the total number of matchings, namely, Hosoya index of G. 2.1. Transformation A Lemma 4. Let G 1 and G 2 be graphs as shown in Fig. 3, then z(G 1 ) > z(G 2 ). Proof. It can be easily seen that every k-matching (k ≥ 1) of G 2 is also a k-matching of G 1 . On the other hand, every k-matching (k ≥ 2) of G 1 containing two edges uu 1 and vv1 is not a k-matching (k ≥ 2) of G 2 , where u 1 ∈ N G 0 (u). This completes the proof. We denote the number of pendent vertices and branched vertices in a graph G as P(G) and B(G), respectively. For any graph G in Un1 (l), let Cl be the unique cycle in it. If there exists a branched vertex outside the cycle Cl , we can employ one step of transformation A on it, and then we get a graph G 0 . By Lemma 4, z(G) > z(G 0 ). Moreover, B(G 0 ) = B(G) − 1 and P(G 0 ) = P(G) + 1. Continue this process, we have
H. Hua / Mathematical and Computer Modelling 48 (2008) 940–948
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Fig. 4. Transformation B.
Lemma 5. Let G be a unicyclic graph in Un1 (l) with l ≥ 3. If G Cnl (k1 , k2 , . . . , kl ), then z(G) > z(Cnl (k1 , k2 , . . . , kl )), where ki ≥ 1. This lemma can be proved by repeatedly using transformation A. 2.2. Transformation B Lemma 6. Let Gˆ1 , Gˆ2 and Gˆ3 be graphs as shown in Fig. 4. Then Per B(Gˆ1 ) > Per B(Gˆ2 ) or Per B(Gˆ1 ) > Per B(Gˆ3 ). Proof. Note that Per B(t K 1 ) = 1 for any t ≥ 1, where t K 1 is the graph consisting of t isolated vertices. Using Lemmas 1 and 2, we obtain Per B(Gˆ1 ) = Per B(Gˆ1 − v1 ) + Per B(Gˆ1 − v1 − v) = Per B(Gˆ1 − v1 − v2 ) + 2Per B(Gˆ1 − v) = ··· = Per B(Gˆ1 − v1 − v2 − · · · − vr ) + r Per B(Gˆ1 − v) = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 ) + r Per B(Gˆ1 − v) + Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u) = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u 2 ) + r Per B(Gˆ1 − v) + 2Per B(Gˆ0 − u) = ··· = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u 2 − · · · − u s ) + r Per B(Gˆ1 − v) + sPer B(Gˆ0 − u) = Per B(Gˆ0 ) + r Per B(Gˆ1 − v) + sPer B(Gˆ0 − u).
(2)
It can be easily seen that Per B(Gˆ1 − v) = Per B(Gˆ1 − v − u 1 ) + Per B(Gˆ0 − v − u) = ··· = Per B(Gˆ0 − v) + sPer B(Gˆ0 − v − u).
(3)
In view of Eqs. (2) and (3), we arrive at Per B(Gˆ1 ) = Per B(Gˆ0 )+r Per B(Gˆ0 −v)+sPer B(Gˆ0 −u)+r sPer B(Gˆ0 − v−u). Similarly, we have Per B(Gˆ2 ) = Per B(Gˆ0 )+Per B(Gˆ0 −u)+(r +s −1)Per B(Gˆ0 −v)+(r +s −1)Per B(Gˆ0 − v − u) and Per B(Gˆ3 ) = Per B(Gˆ0 ) + Per B(Gˆ0 − v) + (r + s − 1)Per B(Gˆ0 − u) + (r + s − 1)Per B(Gˆ0 − v − u). Thus, we obtain Per B(Gˆ1 ) − Per B(Gˆ2 ) = (s − 1)[Per B(Gˆ0 − u) + (r − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − v)],
(4)
Per B(Gˆ1 ) − Per B(Gˆ3 ) = (r − 1)[Per B(Gˆ0 − v) + (s − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − u)].
(5)
Note that the two terms Per B(Gˆ0 − u) + (r − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − v) and Per B(Gˆ0 − v) + (s − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − u) in Eqs. (4) and (5) are of opposite parities. Consequently, the result follows.
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Theorem 7. Let G be a graph in Un1 (l) with l ≥ 3. If G Cnl (1, 1, . . . , 1, n − 2l + 1), then z(G) > z(Cnl (1, 1, . . . , 1, n − 2l + 1)). Proof. Let G be a graph in Un1 (l). If G Cnl (k1 , k2 , . . . , kl ), then z(G) > z(Cnl (k1 , k2 , . . . , kl )) by Lemma 5, where ki ≥ 1. So we need only to verify that z(Cnl (k1 , k2 , . . . , kl )) > z(Cnl (1, 1, . . . , 1, n − 2l + 1)) for (k1 , k2 , . . . , kl ) 6= (1, 1, . . . , 1, n − 2l + 1). Since Cnl (k1 , k2 , . . . , kl ) Cnl (1, 1, . . . , 1, n − 2l + 1), there exists 1 ≤ i < j ≤ l such that ki , k j ≥ 2 in Cnl (k1 , k2 , . . . , kl ). We repeatedly employ transformation B on Cnl (k1 , k2 , . . . , kl ) and we finally get the graph Cnl (1, 1, . . . , 1, n − 2l + 1). From Lemma 6, we have Per B(Cnl (k1 , k2 , . . . , kl )) > Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)). According to Theorem 3, we have z(Cnl (k1 , k2 , . . . , kl )) = Per B(Cnl (k1 , k2 , . . . , kl ))−2 > Per B(Cnl (1, 1, . . . , 1, n− 2l + 1)) − 2 = z(Cnl (1, 1, . . . , 1, n − 2l + 1)). This completes the proof. It is easy to obtain the following result. Lemma 8. Let G 0 be a subgraph of G. Then Per B(G) > Per B(G 0 ). Lemma 9 ([7]). Let T be an n-vertex tree. Then n = z(Sn ) ≤ z(T ) ≤ z(Pn ) = Fn+1 , z(Sn ) < z(T ) if and only if T Sn and z(T ) < z(Pn ) if and only if T Pn . Lemma 10. For l ≥ 5, we have Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) > Per B(Cnl−1 (1, 1, . . . , 1, n − 2l + 3)). Proof. In view of Lemma 2, we obtain l 1 Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) = Per B(Cn−1 (1, 1, . . . , 1, n − 2l)) + Per B(Pl−1 (1, 1, . . . , 1)) l 1 = Per B(Cn−2 (1, 1, . . . , 1, n − 2l − 1)) + 2Per B(Pl−1 (1, 1, . . . , 1)) = ··· l 1 = Per B(C2l (1, 1, . . . , 1, 1)) + (n − 2l)Per B(Pl−1 (1, 1, . . . , 1)).
(6)
Similarly, we have l−1 Per B(Cnl−1 (1, 1, . . . , 1, n − 2l + 3)) = Per B(C2l−2 (1, 1, . . . , 1, 1)) 1 + (n − 2l + 2)Per B(Pl−2 (1, 1, . . . , 1)).
(7)
Once again, by Lemma 2, we have l Per B(C2l (1, 1, . . . , 1, 1)) = Per B(Cl ) +
l−2 X
p
Per B(Pl−1 (1, 1, . . . , 1)) + Per B(Pl ) + Per B(Pl−1 )
(8)
p=1
and l−1 Per B(C2l−2 (1, 1, . . . , 1, 1)) = Per B(Cl−1 ) +
l−3 X
q
Per B(Pl−2 (1, 1, . . . , 1))
q=1
+ Per B(Pl−1 ) + Per B(Pl−2 ). Since
1 (1, 1, . . . , 1) Pl−2
is a proper subgraph of
1 (1, 1, . . . , 1), Pl−1
(9) then
1 1 Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1))
(10)
by Lemma 8. From Lemma 2 it follows easily that 2 1 Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1)).
(11)
In what follows, we will prove that k+2 k Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1)) k (1, 1, . . . , 1) and P k+2 (1, 1, . . . , 1). for k = 1, 2, . . . , l − 4. See Fig. 5 for Pl−2 l−1
(12)
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k+2 k (1, 1, . . . , 1). Fig. 5. (a). Pl−1 (1, 1, . . . , 1); (b). Pl−2
Fig. 6. The graph Cn3 (1, 1, n − 5).
k+1 k+2 k+2 (1, 1, . . . , 1) and Pl−1 (1, 1, . . . , 1) − s1 − s2 ∼ Note from Fig. 5 that Pl−1 (1, 1, . . . , 1) − s1 ∼ = = Pl−2 k+1 k k k ∼ Pl−3 (1, 1, . . . , 1). Note from Fig. 5 also that Pl−2 (1, 1, . . . , 1) − yk = Pl−2 (1, 1, . . . , 1) and Pl−2 (1, 1, . . . , 1) − S 1 yk − xk = Pk−1 Pl−2−k (1, 1, . . . , 1). S 1 k (1, 1, . . . , 1)) > Per B(P From Lemma 2, it suffices to prove that Per B(Pl−3 Pl−2−k (1, 1, . . . , 1)). k−1 S k (1, 1, . . . , 1) contains P 1 It is not difficult to see that Pl−3 Pl−2−k (1, 1, . . . , 1) as its proper subgraph, then k−1 S 1 k (1, 1, . . . , 1)) > Per B(P Per B(Pl−3 Pl−2−k (1, 1, . . . , 1)) by Lemma 8. This proves Eq. (12). Moreover, we k−1 l−3 can directly get Per B(Pl−2 (1, 1, . . . , 1)) < Per B(Pl ) by Lemma 2. So what remains is to verify that
Per B(Cl ) > Per B(Cl−1 ) + Per B(Pl−2 ).
(13)
Bear in mind that z(Tn ) = Per B(Tn ) [8] for any n-vertex tree. Combining this fact with Lemma 9, we have Per B(Pl−2 ) = Fl−1 . Directly expanding Per B(Cm ) or according to lemma 2.4 [11], we arrive at Per B(Cl ) = Fl−1 + Fl+1 + 2. By an elementary calculation, it is easy to check that Eq. (13) holds. From the combination of Eqs. (6)–(13) the lemma follows as expected. Lemma 11. For n ≥ 6, we have z(Cn3 (1, 1, n − 5)) = 5n − 16, where Cn3 (1, 1, n − 5) is the graph as depicted in Fig. 6. Proof. By a suitable labeling of the vertices of Cn3 (1, 1, n − 5), we obtain 1 1 1 1 · · · 1 1 1 0 0 1 1 1 . . . B(Cn3 (1, 1, n − 5)) = 1 1 1 0 0
1 0 0 .. .
0 1 0 .. .
0 0 1 .. .
··· ··· ··· .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
··· ··· ··· ··· ···
1 0 0 0 0
0 1 1 1 0
0 1 1 0 1
0 1 0 1 0
0 0 0 .. . . 0 0 1 0 1 n
Directly expanding Per B(Cn3 (1, 1, n − 5)) along its first n − 4 rows gives Per B(Cn3 (1, 1, n − 5)) = 5n − 14. From Theorem 3, we have z(Cn3 (1, 1, n − 5)) = 5n − 16. This completes the proof. The following theorem determines the minimum cardinality of z(G) for G ∈ Un1 as well as the corresponding extremal graph. Theorem 12. Let G ∈ Un1 with n ≥ 8 vertices. Then z(G) ≥ 5n − 16 with the equality if and only if G∼ = Cn3 (1, 1, n − 5).
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Fig. 7. (a). Cn3 (1, 2, n − 6); (b). Cn3 (1, 3, n − 7).
Fig. 8. The graph G( p).
Proof. From the combination of Lemma 10 and Theorems 3 and 7 it follows that z(G) ≥ z(Cn4 (1, 1, 1, n−7)). Similar to Lemma 11, we can obtain z(Cn4 (1, 1, 1, n − 7)) = 10n − 48. Thus, for n ≥ 8, z(G) ≥ z(Cn4 (1, 1, 1, n − 7)) > z(Cn3 (1, 1, n − 5)) and then the result follows. 3. Graphs in Un1 with second- and third-minimal Hosoya indices For brevity, we will denote transformations A or B simply as T.A. or T.B., respectively, in what follows. Theorem 13. Let n ≥ 8 and G ∈ Un1 (3) − Cn3 (1, 1, n − 5). Then z(G) ≥ 7n − 30 with equality if and only if G∼ = Cn3 (1, 2, n − 6), where Cn3 (1, 2, n − 6) is the graph as indicated in Fig. 7(a). Proof. We distinguish between two cases. Case 1. G ∼ = Cn3 (k1 , k2 , k3 ). Since G Cn3 (1, 1, n − 5), then (k1 , k2 , k3 ) 6= (1, 1, n − 5). Assume that (k1 , k2 , k3 ) 6= (1, 2, n − 6) and k1 ≤ k2 ≤ k3 . We consider the following two subcases. Subcase 1.1. k1 = 1. Note that k2 ≥ 3 in this case. Take the graph Cn3 (1, 1, 1) as Gˆ0 as depicted in Fig. 4. Using T.B. and by Lemma 6, we obtain Per B(Cn3 (1, k2 , k3 )) > Per B(Cn3 (1, 2, k3 + (k2 − 2))) = Per B(Cn3 (1, 2, n − 6)) or Per B(Cn3 (1, k2 , k3 )) > Per B(Cn3 (1, 2, k2 + (k3 − 2))) = Per B(Cn3 (1, 2, n − 6)). By Theorem 3, z(Cn3 (1, k2 , k3 )) = Per B(Cn3 (1, k2 , k3 )) − 2 > Per B(Cn3 (1, 2, n − 6)) − 2 = z(Cn3 (1, 2, n − 6)). Subcase 1.2. k1 ≥ 2. Using T.B. and by Lemma 6, we obtain Per B(Cn3 (k1 , k2 , k3 )) > Per B(Cn3 (1, k2 , k3 + (k1 − 1))). If (1, k2 , k3 + (k1 − 1)) = (1, 2, n − 6), the theorem holds. Otherwise, we have k2 ≥ 3. After a further step of T.B., we obtain Per B(Cn3 (1, k2 , k3 + (k1 − 1))) > Per B(Cn3 (1, 2, k3 + (k1 − 1) + (k2 − 2))) = Per B(Cn3 (1, 2, n − 6)). Consequently, z(Cn3 (k1 , k2 , k3 )) > z(Cn3 (1, 2, k3 + (k1 − 1) + (k2 − 2))) = z(Cn3 (1, 2, n − 6)). Case 2. G Cn3 (k1 , k2 , k3 ). Subcase 2.1. Cn3 (1, 1, n − 5) can be obtained from G by a sequence of T.A.s. Since G Cnl (1, 1, n − 5), then Cn3 (1, 1, n − 5) can be obtained from G by at least one step of T.A. Let G( p) be the graph in Un1 (3) with the property that G is reduced to Cn3 (1, 1, n − 5) by exactly one step of T.A. Then G( p) must have the form as indicated in Fig. 8. It is easy to obtain that z(Cn3 (1, 2, n − 6)) = 7n − 30 and z(G( p)) = −5 p 2 + (5n − 23) p + 3n − 6. So z(G( p)) − z(Cn3 (1, 2, n − 6)) = −5 p 2 + (5n − 23) p + 24 − 4n. Set f ( p) = −5 p 2 + (5n − 23) p + 24 − 4n (1 ≤ p ≤ n − 6). Then
d f ( p) dp
= (5n − 23) − 10 p.
d f ( p) From this it is deduced that if 1 ≤ p ≤ then > 0 and if d 5n−23 10 e ≤ p ≤ n − 6, then d p < 0. So we have z(G( p)) − z(Cn3 (1, 2, n − 6)) = f ( p) ≥ f (1) = n − 4 > 0 for 1 ≤ p ≤ b 5n−23 10 c whereas 5n−23 3 z(G( p)) − z(Cn (1, 2, n − 6)) = f ( p) ≥ f (n − 6) = 3n − 18 > 0 for d 10 e ≤ p ≤ n − 6. Therefore z(G( p)) > z(Cn3 (1, 2, n − 6)).
b 5n−23 10 c,
d f ( p) dp
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˘ Fig. 9. (a). H ( p); (b). G.
If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 2, n − 6) by at least two steps of T.A., then by Lemma 4, we have z(G) > z(G( p 0 )) for some positive integer p 0 . Similarly, we can prove z(G( p 0 )) > z(Cn3 (1, 2, n − 6)) and then the theorem holds. Subcase 2.2. Cn3 (1, 1, n − 5) cannot be obtained from G by a sequence of T.A.s. Repeatedly using T.A. on G, we finally obtain the graph Cn3 (k1 , k2 , k3 ) and z(G) > z(Cn3 (k1 , k2 , k3 )) by Lemma 5, where (k1 , k2 , k3 ) 6= (1, 1, n − 5). What remains is similar to case 1, we can show that the theorem holds. Combining cases 1 and 2, we complete the proof. Theorem 14. Let n ≥ 8. Then Cn3 (1, 2, n − 6) has the second-minimal Hosoya index among all graphs in Un1 . Proof. By repeatedly using Lemma 10, we have Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) > Per B(Cn4 (1, 1, 1, n − 7)) for l ≥ 5. Then by Theorem 3, z(Cnl (1, 1, . . . , 1, n −2l +1)) > z(Cn4 (1, 1, 1, n −7)). When n ≥ 8, z(Cn4 (1, 1, 1, n −7)) = 10n −48 > 7n −30 = z(Cn3 (1, 2, n −6)). Combining this fact with Lemma 5 and Theorem 13, the result follows. Theorem 15. Let n ≥ 10 and G ∈ Un1 (3) − {Cn3 (1, 1, n − 5), Cn3 (1, 2, n − 6)}. Then 9n − 48, 10 ≤ n ≤ 12 (a) z(G) ≥ 8n − 35, n ≥ 13 (b). Equality holds in (a) if and only if G ∼ = Cn3 (1, 3, n − 7); Equality holds in (b) if and only if G ∼ = G˘ for n ≥ 14 3 3 ∼ ˘ ˘ and G = G or Cn (1, 3, n − 7) for n = 13. Here Cn (1, 3, n − 7) and G are the graphs as illustrated in Fig. 7(b) and Fig. 9(b), respectively. Proof. We consider the following two cases. Case 1. G ∼ = Cnl (k1 , k2 , k3 ). Since G Cn3 (1, 1, n − 5), Cn3 (1, 2, n − 6), then (k1 , k2 , k3 ) 6= (1, 1, n − 5), (1, 2, n − 6). Assume that (k1 , k2 , k3 ) 6= (1, 3, n − 7) and k1 ≤ k2 ≤ k3 . Similar to the proof of case 1 in Theorem 13, we can show that z(G) > z(Cn3 (1, 3, n − 7)) in this case. Case 2. G Cn3 (k1 , k2 , k3 ). We divide this case further into three subcases. Subcase 2.1. Cn3 (1, 1, n − 5) can be obtained from G by a sequence of T.A.s. Since G Cnl (1, 1, n − 5), then Cn3 (1, 1, n − 5) can be obtained from G by at least one step of T.A. We consider only the case when G is reduced to Cn3 (1, 1, n − 5) by exactly one step of T.A. Let G( p) be the graph in Un1 (3) with the above property. As obtained in Theorem 13, G( p) = −5 p 2 + (5n − 23) p + 3n − 6. Similar to Lemma 11, we obtain z(Cn3 (1, 3, n − 7)) = 9n − 48. Consequently, z(G( p)) − z(Cn3 (1, 3, n − 7)) = −5 p 2 + (5n − 23) p − 6n + 42. Set g( p) = −5 p 2 + (5n − 23) p − 6n + 42 (1 ≤ p ≤ n − 6). Then From this it is deduced that if 1 ≤ p ≤ b 5n−23 10 c, then z(Cn3 (1, 3, n
dg( p) dp
> 0 and
dg( p) d p = (5n − 23) − 10 p. dg( p) if d 5n−23 10 e ≤ p ≤ n − 6, then d p
< 0.
So we have z(G( p)) − − 7)) = g( p) ≥ g(2) = 4n − 24 > 0 for 2 ≤ p ≤ b 5n−24 10 c 5n−24 3 whereas z(G( p)) − z(Cn (1, 32, n − 7)) = g( p) ≥ g(n − 6) = n > 0 for d 10 e ≤ p ≤ n − 6. Therefore z(G( p)) > z(Cn3 (1, 3, n − 7)) for p ≥ 2. When p = 1, z(G(1)) − z(Cn3 (1, 3, n − 7)) = 14 − n. If 10 ≤ n ≤ 13, then z(G(1)) > z(Cn3 (1, 3, n − 7)), and if n ≥ 15, z(G(1)) < z(Cn3 (1, 3, n − 7)).
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If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 1, n − 5) by at least two steps of T.A. We can prove that z(G) > z(G( p 00 )) for some positive integer p 00 . Similarly as above, we obtain z(G( p 00 )) > z(Cn3 (1, 3, n − 7)) for p 00 ≥ 2 and z(G( p 00 )) > z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 13, while z(G( p 00 )) < z(Cn3 (1, 3, n − 7)) when n ≥ 15 for p 00 = 1. In conclusion, z(G) ≥ z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 13, whereas z(G) ≥ z(G(1)) when n ≥ 15. As for n = 14, z(G) ≥ z(G(1)) = z(Cn3 (1, 3, n − 7)). Subcase 2.2. Cn3 (1, 2, n − 6) can be obtained from G by a sequence of T.A.s. ˘ We consider only the case when G is reduced to Cn3 (1, 2, n − 6) by exactly one step of T.A. Then G ∼ = H ( p)or G. If G ∼ = H ( p), by an elementary calculation, we get z(H ( p)) = −7 p 2 + (7n − 40) p + 5n − 18. Hence z(H ( p)) − z(Cn3 (1, 3, n − 7)) = −7 p 2 + (7n − 40) p − 4n + 30. Set t ( p) = −7 p 2 + (7n − 40) p − 4n + 30 (1 ≤ p ≤ n − 7). It is easily seen that if 1 ≤ p ≤ b 7n−40 14 c, then z(H ( p)) − z(Cn3 (1, 3, n − 7)) = t ( p) ≥ t (1) = 3n − 17 > 0 and if d 7n−40 e ≤ p ≤ n − 7, then 14 z(H ( p)) − z(Cn3 (1, 3, n − 7)) = t ( p) ≥ t (n − 7) = 5n − 33 > 0. ˘ by an elementary calculation, we have z(G) ˘ = 8n − 35. When 10 ≤ n ≤ 12, we have If G ∼ = G, ˘ when n ≥ 14, z(Cn3 (1, 3, n − 7)) > z(G). ˘ z(Cn3 (1, 3, n − 7)) < z(G); If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 2, n − 6) by at least two steps of T.A. We can use the same method as employed in subcase 2.1. ˘ when n ≥ 14. As for n = 13, In conclusion, z(G) ≥ z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 12, whereas z(G) ≥ z(G) 3 ˘ = z(Cn (1, 3, n − 7)). z(G) ≥ z(G) Case 2.3. Neither Cnl (1, 1, n − 5) nor Cn3 (1, 2, n − 6) can be obtained from G by T.A. Repeatedly using T.A. on G, we finally obtain the graph Cn3 (k1 , k2 , k3 ) and z(G) > z(Cn3 (k1 , k2 , k3 )), where (k1 , k2 , k3 ) 6= (1, 1, n − 5), (1, 2, n − 6). What remains is similar to case 1 in Theorem 13, we can show that z(G) > z(Cn3 (1, 3, n − 7)). ˘ Combining cases 1, 2 and 3, we have z(G) ≥ z(Cn3 (1, 3, n − 7)) Note that z(G(1)) = 8n − 34 > 8n − 35 = z(G). ˘ ˘ for 10 ≤ n ≤ 12, while z(G) ≥ z(G) for n ≥ 14. when n = 13, z(G) ≥ z(Cn3 (1, 3, n − 7)) = z(G). We thus completed the proof here. Theorem 16. Among all graphs in Un1 with n ≥ 10, Cn3 (1, 3, n − 7) has the third-minimal Hosoya index when 10 ≤ n ≤ 12, while G˘ has the third-minimal Hosoya index for n ≥ 13. Proof. When n ≥ 8, z(Cn4 (1, 1, 1, n − 7)) = 10n − 48 > 9n − 48 = z(Cn3 (1, 3, n − 7)) and z(Cn4 (1, 1, 1, n − 7)) = ˘ From Lemmas 5 and 10, and Theorems 3 and 15, the result follows as expected. 10n − 48 > 8n − 34 = z(G). Acknowledgements The author is indebted to the anonymous referees for their helpful suggestions and useful comments, which improved considerably the presentation of this paper. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, Amsterdam, 1976. [2] H. Hosoya, Topological index, a newly proposed quantity characterizing the topological nature of structural isomers of saturated hydrocarbons, Bull. Chem. Soc. Jpn. 44 (1971) 2332–2339. [3] I. Gutman, Extremal hexagonal chains, J. Math. Chem. 12 (1993) 197–210. [4] L.Z. Zhang, The Z -transformation graph of perfect matchings of a catacondensed hexagonal system, J. Math. Study 31 (1998) 437–441 (in Chinese). [5] L.Z. Zhang, F. Tian, Extremal hexagonal chains concerning largest eigenvalue, Sci. China Ser. A 44 (2001) 1089–1097. [6] L.Z. Zhang, F. Tian, Extremal catacondensed benzenoids, J. Math. Chem. 34 (2003) 111–122. [7] I. Gutman, O.E. Polansky, Mathematical Concepts in Organic Chemistry, Springer, Berlin, 1986. [8] Y. Hou, On acyclic systems with minimal Hosoya index, Discrete Appl. Math. 119 (2002) 251–257. [9] A. Yu, F. Tian, A kind of graphs with minimal Hosoya indices and maximal Merrifield-Simmons indices, MATCH Commun. Math. Comput. Chem. 55 (1) (2006) 103–118. [10] A. Yu, X. Lv, The Merrifield–Simmons indices and Hosoya indices of trees with k pendent vertices, J. Math. Chem. 41 (1) (2007) 33–43. [11] J. Ou, On extremal unicyclic molecular graphs with maximal Hosoya index (submitted for publication). [12] J. Ou, On extremal unicyclic molecular graphs with prescribed girth and minimal Hosoya index, J. Math. Chem. 42 (3) (2007) 423–432. [13] H. Minc, Permanents, Addison-Wesley, Reading, MA, 1978.
Minimizing a class of unicyclic graphs by means of Hosoya indexI Hongbo Hua Department of Computing Science, Huaiyin Institute of Technology, Huaian, Jiangsu 223000, PR China Received 18 August 2007; received in revised form 9 November 2007; accepted 5 December 2007
Abstract The Hosoya index, denoted by z(G), of a (molecular) graph G is defined as the total number of independent-edge sets of G. Let Un be the set of unicyclic graphs with n vertices. A fully loaded unicyclic graph is a unicyclic graph with the property that there is no vertex with degree less than 3 in its unique cycle. Denote by Un1 the set of fully loaded unicyclic graphs. In this paper, graphs in Un1 with minimal, second-minimal and third-minimal Hosoya indices are uniquely determined, respectively. c 2007 Elsevier Ltd. All rights reserved.
Keywords: Hosoya index; Unicyclic graph; Permanent; Matching
1. Introduction The Hosoya index, proposed by Hosoya [2] in 1971, acts as one of the important topological parameters in the study of the relation between molecular structure and the physical and chemical properties of a number of hydrocarbon compounds. Let G be a graph with n vertices. Hosoya index, denoted by z(G), is defined to be the total number of matchings, namely, n
z(G) =
b2c X
m(G; s),
s=0
where m(G; s) is the number of s-matchings of G. An s-matching of a graph G is a subset M of its edge set with the property that |M| = s and M contains no two edges sharing a common vertex. For convenience and consistence, it will be always assumed that m(G; 0) = 1. It is both interesting and significant to determine the graph with extremal (maximal or minimal) Hosoya index. Along these lines, many results have been put forward in recent years. Gutman in [3] proved that the linear hexagonal chain is the unique chain with minimal Hosoya index among all hexagonal chains. Zhang [4] showed that the zig-zag hexagonal chain is the unique chain with maximal Hosoya index among all hexagonal chains. Zhang and Tian [5] gave new proofs of Gutman’s results in [3] and Zhang’s results in [4]. In [6], Zhang and Tian determined the graphs with I Partially supported by SRF of Huaiyin Institute of Technology (HGQ 0611, HGQN0726, HGQN0727).
E-mail address: [email protected]. c 2007 Elsevier Ltd. All rights reserved. 0895-7177/$ - see front matter doi:10.1016/j.mcm.2007.12.003
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H. Hua / Mathematical and Computer Modelling 48 (2008) 940–948
Fig. 1. (a). P52 (1, 1, 1, 1); (b). Plk (1, 1, . . . , 1).
4 (1, 2, 2, 3); (b). C 3 (2, 2, 2). Fig. 2. (a). C12 9
minimal and second-minimal Hosoya indices among all catacondensed systems. In [7], the path and star have been shown to have the maximal and minimal Hosoya indices resp., among all trees on n vertices. Hou [8] characterized the trees having minimal and second-minimal Hosoya indices among all trees with a given size of matching. Yu and Tian [9] investigated the graph having minimal Hosoya index among all graphs with given edge-independence number and cyclomatic number. In [10], Yu and Lv investigated the tree having minimal Hosoya index among all trees with k pendent vertices. More recently, Ou [11] determined, respectively, the unique unicyclic graphs with the first and second largest Hosoya indices among all unicyclic graphs with n vertices. He also determined the unicyclic graphs with the first and second smallest Hosoya indices in the successive paper [12]. All graphs considered here are both connected and simple if not stated in particular. For a vertex v in G, let d(v) be the number of edges incident with v. If d(v) ≥ 3, then we call v a branched vertex. By N G (v) we denote the set of vertices in G adjacent to v. In order to state and prove our main results, we need some further notations. Let Un be the set of unicyclic graphs with n vertices. A fully loaded unicyclic graph is a unicyclic graph with the property that there exists no vertex with degree less than 3 in its unique cycle. Let Un1 be the set of fully loaded unicyclic graphs. Let Un (l) and Un1 (l) denote resp. the subset of Un and Un1 in which every graph has a unique cycle of length l. Let Pn be the path on n vertices and its vertices be ordered successively as x1 , x2 , . . . , xn . For 1 ≤ k ≤ n, we denote, by Pnk (1, 1, . . . , 1), the graph obtained from Pn by attaching exactly one pendent edge to each of the vertices xk , xk+1 , . . . , xn , respectively. Here Pnn (1, 1, . . . , 1) = Pn+1 . Let Cl be the cycle of length l and its vertices be ordered successively as y1 , y2 , . . . , yl . We denote, by Cnl (t1 , t2 , . . . , tl ), the graph obtained from Cl by attaching P exactly ti pendent edges to the vertex yi for i = 1, 2, . . . l, where ti ≥ 1 and li=1 ti = n − l. As examples, we 4 (1, 2, 2, 3) and C 3 (2, 2, 2) in Fig. 2, respectively. Other illustrate P52 (1, 11, 1) and Plk (1, 1, . . . , 1) in Fig. 1 and C12 9 notations and terminology not defined here will conform to those in [1]. In this paper, we shall investigate the Hosoya index of fully loaded unicyclic graphs. Graphs in Un1 with minimal, second-minimal and third-minimal Hosoya indices are determined. 2. Graph in Un1 with minimal Hosoya index Let G be an n-vertex graph with adjacency matrix A(G). Define the neighbor matrix B(G) of G as B(G) = A(G) + I , where I is the unit matrix of order n. We immediately have Lemma 1. Let G be a graph with m (≥1) components G 1 , G 2 , . . . G m . Then Per B(G) = Proof. By a suitable labeling vertices of G, we obtain B(G 1 ) B(G 2 ) B(G) = . . .. B(G m ) The result follows by a direct expansion of Per B(G).
Qm
i=1 Per
B(G i ).
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Fig. 3. Transformation A.
Lemma 2. Let G be a graph on n ≥ 2 vertices. If uv is a pendent edge of G with pendent vertex v, then Per B(G) = Per B(G − v) + Per B(G − v − u). Proof. Ordering vertices of G as v, u, . . . , we have 1 1 0 B(G) = 1 1 x . 0 x D By an expansion of Per B(G) along its first row, we obtain 1 x 1 x 1 x Per B(G) = Per + Per = Per + Per D. x D 0 D x D Note that B(G − v) = x1 Dx and B(G − v − u) = D. Thus the result follows.
For convenience, we set Per B(∅) = 1. Let G ∈ Un (l) and we denote, by T (G), the forest obtained from G by deleting all vertices in V (Cl ) as well as edges incident with each vertex in V (Cl ). Thus, we have Theorem 3. Let G be a unicyclic graph in Un (l) with n ≥ 3 vertices. Then z(G) = Per B(G) − 2Per B(T (G)). Proof. From the definition of Permanent [13], we obtain Per B(G) =
n XY σ i=1
bi,σ (i) ,
(1)
Qn where σ goes over all symmetric group of order n. In the above Eq. (1), term i=1 bi,σ (i) = 0 if and only if either vertex i is not adjacent to vertex σ (i) for some i 6= σ (i) or, σ contains a cycle with length greater than 2 but not equal to l since G ∈ Un (l) contains exactly one cycle of length l. Consequently, every non-zero term of Per B(G) corresponds to a matching of G with the exception that σ is one of two cycles of length l. So z(G) = Per B(G) − 2Per B(T (G)) and then the theorem follows. In what follows we shall introduce a transformation on a graph G which will decrease the total number of matchings, namely, Hosoya index of G. 2.1. Transformation A Lemma 4. Let G 1 and G 2 be graphs as shown in Fig. 3, then z(G 1 ) > z(G 2 ). Proof. It can be easily seen that every k-matching (k ≥ 1) of G 2 is also a k-matching of G 1 . On the other hand, every k-matching (k ≥ 2) of G 1 containing two edges uu 1 and vv1 is not a k-matching (k ≥ 2) of G 2 , where u 1 ∈ N G 0 (u). This completes the proof. We denote the number of pendent vertices and branched vertices in a graph G as P(G) and B(G), respectively. For any graph G in Un1 (l), let Cl be the unique cycle in it. If there exists a branched vertex outside the cycle Cl , we can employ one step of transformation A on it, and then we get a graph G 0 . By Lemma 4, z(G) > z(G 0 ). Moreover, B(G 0 ) = B(G) − 1 and P(G 0 ) = P(G) + 1. Continue this process, we have
H. Hua / Mathematical and Computer Modelling 48 (2008) 940–948
943
Fig. 4. Transformation B.
Lemma 5. Let G be a unicyclic graph in Un1 (l) with l ≥ 3. If G Cnl (k1 , k2 , . . . , kl ), then z(G) > z(Cnl (k1 , k2 , . . . , kl )), where ki ≥ 1. This lemma can be proved by repeatedly using transformation A. 2.2. Transformation B Lemma 6. Let Gˆ1 , Gˆ2 and Gˆ3 be graphs as shown in Fig. 4. Then Per B(Gˆ1 ) > Per B(Gˆ2 ) or Per B(Gˆ1 ) > Per B(Gˆ3 ). Proof. Note that Per B(t K 1 ) = 1 for any t ≥ 1, where t K 1 is the graph consisting of t isolated vertices. Using Lemmas 1 and 2, we obtain Per B(Gˆ1 ) = Per B(Gˆ1 − v1 ) + Per B(Gˆ1 − v1 − v) = Per B(Gˆ1 − v1 − v2 ) + 2Per B(Gˆ1 − v) = ··· = Per B(Gˆ1 − v1 − v2 − · · · − vr ) + r Per B(Gˆ1 − v) = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 ) + r Per B(Gˆ1 − v) + Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u) = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u 2 ) + r Per B(Gˆ1 − v) + 2Per B(Gˆ0 − u) = ··· = Per B(Gˆ1 − v1 − v2 − · · · − vr − u 1 − u 2 − · · · − u s ) + r Per B(Gˆ1 − v) + sPer B(Gˆ0 − u) = Per B(Gˆ0 ) + r Per B(Gˆ1 − v) + sPer B(Gˆ0 − u).
(2)
It can be easily seen that Per B(Gˆ1 − v) = Per B(Gˆ1 − v − u 1 ) + Per B(Gˆ0 − v − u) = ··· = Per B(Gˆ0 − v) + sPer B(Gˆ0 − v − u).
(3)
In view of Eqs. (2) and (3), we arrive at Per B(Gˆ1 ) = Per B(Gˆ0 )+r Per B(Gˆ0 −v)+sPer B(Gˆ0 −u)+r sPer B(Gˆ0 − v−u). Similarly, we have Per B(Gˆ2 ) = Per B(Gˆ0 )+Per B(Gˆ0 −u)+(r +s −1)Per B(Gˆ0 −v)+(r +s −1)Per B(Gˆ0 − v − u) and Per B(Gˆ3 ) = Per B(Gˆ0 ) + Per B(Gˆ0 − v) + (r + s − 1)Per B(Gˆ0 − u) + (r + s − 1)Per B(Gˆ0 − v − u). Thus, we obtain Per B(Gˆ1 ) − Per B(Gˆ2 ) = (s − 1)[Per B(Gˆ0 − u) + (r − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − v)],
(4)
Per B(Gˆ1 ) − Per B(Gˆ3 ) = (r − 1)[Per B(Gˆ0 − v) + (s − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − u)].
(5)
Note that the two terms Per B(Gˆ0 − u) + (r − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − v) and Per B(Gˆ0 − v) + (s − 1)Per B(Gˆ0 − v − u) − Per B(Gˆ0 − u) in Eqs. (4) and (5) are of opposite parities. Consequently, the result follows.
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Theorem 7. Let G be a graph in Un1 (l) with l ≥ 3. If G Cnl (1, 1, . . . , 1, n − 2l + 1), then z(G) > z(Cnl (1, 1, . . . , 1, n − 2l + 1)). Proof. Let G be a graph in Un1 (l). If G Cnl (k1 , k2 , . . . , kl ), then z(G) > z(Cnl (k1 , k2 , . . . , kl )) by Lemma 5, where ki ≥ 1. So we need only to verify that z(Cnl (k1 , k2 , . . . , kl )) > z(Cnl (1, 1, . . . , 1, n − 2l + 1)) for (k1 , k2 , . . . , kl ) 6= (1, 1, . . . , 1, n − 2l + 1). Since Cnl (k1 , k2 , . . . , kl ) Cnl (1, 1, . . . , 1, n − 2l + 1), there exists 1 ≤ i < j ≤ l such that ki , k j ≥ 2 in Cnl (k1 , k2 , . . . , kl ). We repeatedly employ transformation B on Cnl (k1 , k2 , . . . , kl ) and we finally get the graph Cnl (1, 1, . . . , 1, n − 2l + 1). From Lemma 6, we have Per B(Cnl (k1 , k2 , . . . , kl )) > Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)). According to Theorem 3, we have z(Cnl (k1 , k2 , . . . , kl )) = Per B(Cnl (k1 , k2 , . . . , kl ))−2 > Per B(Cnl (1, 1, . . . , 1, n− 2l + 1)) − 2 = z(Cnl (1, 1, . . . , 1, n − 2l + 1)). This completes the proof. It is easy to obtain the following result. Lemma 8. Let G 0 be a subgraph of G. Then Per B(G) > Per B(G 0 ). Lemma 9 ([7]). Let T be an n-vertex tree. Then n = z(Sn ) ≤ z(T ) ≤ z(Pn ) = Fn+1 , z(Sn ) < z(T ) if and only if T Sn and z(T ) < z(Pn ) if and only if T Pn . Lemma 10. For l ≥ 5, we have Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) > Per B(Cnl−1 (1, 1, . . . , 1, n − 2l + 3)). Proof. In view of Lemma 2, we obtain l 1 Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) = Per B(Cn−1 (1, 1, . . . , 1, n − 2l)) + Per B(Pl−1 (1, 1, . . . , 1)) l 1 = Per B(Cn−2 (1, 1, . . . , 1, n − 2l − 1)) + 2Per B(Pl−1 (1, 1, . . . , 1)) = ··· l 1 = Per B(C2l (1, 1, . . . , 1, 1)) + (n − 2l)Per B(Pl−1 (1, 1, . . . , 1)).
(6)
Similarly, we have l−1 Per B(Cnl−1 (1, 1, . . . , 1, n − 2l + 3)) = Per B(C2l−2 (1, 1, . . . , 1, 1)) 1 + (n − 2l + 2)Per B(Pl−2 (1, 1, . . . , 1)).
(7)
Once again, by Lemma 2, we have l Per B(C2l (1, 1, . . . , 1, 1)) = Per B(Cl ) +
l−2 X
p
Per B(Pl−1 (1, 1, . . . , 1)) + Per B(Pl ) + Per B(Pl−1 )
(8)
p=1
and l−1 Per B(C2l−2 (1, 1, . . . , 1, 1)) = Per B(Cl−1 ) +
l−3 X
q
Per B(Pl−2 (1, 1, . . . , 1))
q=1
+ Per B(Pl−1 ) + Per B(Pl−2 ). Since
1 (1, 1, . . . , 1) Pl−2
is a proper subgraph of
1 (1, 1, . . . , 1), Pl−1
(9) then
1 1 Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1))
(10)
by Lemma 8. From Lemma 2 it follows easily that 2 1 Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1)).
(11)
In what follows, we will prove that k+2 k Per B(Pl−1 (1, 1, . . . , 1)) > Per B(Pl−2 (1, 1, . . . , 1)) k (1, 1, . . . , 1) and P k+2 (1, 1, . . . , 1). for k = 1, 2, . . . , l − 4. See Fig. 5 for Pl−2 l−1
(12)
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945
k+2 k (1, 1, . . . , 1). Fig. 5. (a). Pl−1 (1, 1, . . . , 1); (b). Pl−2
Fig. 6. The graph Cn3 (1, 1, n − 5).
k+1 k+2 k+2 (1, 1, . . . , 1) and Pl−1 (1, 1, . . . , 1) − s1 − s2 ∼ Note from Fig. 5 that Pl−1 (1, 1, . . . , 1) − s1 ∼ = = Pl−2 k+1 k k k ∼ Pl−3 (1, 1, . . . , 1). Note from Fig. 5 also that Pl−2 (1, 1, . . . , 1) − yk = Pl−2 (1, 1, . . . , 1) and Pl−2 (1, 1, . . . , 1) − S 1 yk − xk = Pk−1 Pl−2−k (1, 1, . . . , 1). S 1 k (1, 1, . . . , 1)) > Per B(P From Lemma 2, it suffices to prove that Per B(Pl−3 Pl−2−k (1, 1, . . . , 1)). k−1 S k (1, 1, . . . , 1) contains P 1 It is not difficult to see that Pl−3 Pl−2−k (1, 1, . . . , 1) as its proper subgraph, then k−1 S 1 k (1, 1, . . . , 1)) > Per B(P Per B(Pl−3 Pl−2−k (1, 1, . . . , 1)) by Lemma 8. This proves Eq. (12). Moreover, we k−1 l−3 can directly get Per B(Pl−2 (1, 1, . . . , 1)) < Per B(Pl ) by Lemma 2. So what remains is to verify that
Per B(Cl ) > Per B(Cl−1 ) + Per B(Pl−2 ).
(13)
Bear in mind that z(Tn ) = Per B(Tn ) [8] for any n-vertex tree. Combining this fact with Lemma 9, we have Per B(Pl−2 ) = Fl−1 . Directly expanding Per B(Cm ) or according to lemma 2.4 [11], we arrive at Per B(Cl ) = Fl−1 + Fl+1 + 2. By an elementary calculation, it is easy to check that Eq. (13) holds. From the combination of Eqs. (6)–(13) the lemma follows as expected. Lemma 11. For n ≥ 6, we have z(Cn3 (1, 1, n − 5)) = 5n − 16, where Cn3 (1, 1, n − 5) is the graph as depicted in Fig. 6. Proof. By a suitable labeling of the vertices of Cn3 (1, 1, n − 5), we obtain 1 1 1 1 · · · 1 1 1 0 0 1 1 1 . . . B(Cn3 (1, 1, n − 5)) = 1 1 1 0 0
1 0 0 .. .
0 1 0 .. .
0 0 1 .. .
··· ··· ··· .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 .. .
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
··· ··· ··· ··· ···
1 0 0 0 0
0 1 1 1 0
0 1 1 0 1
0 1 0 1 0
0 0 0 .. . . 0 0 1 0 1 n
Directly expanding Per B(Cn3 (1, 1, n − 5)) along its first n − 4 rows gives Per B(Cn3 (1, 1, n − 5)) = 5n − 14. From Theorem 3, we have z(Cn3 (1, 1, n − 5)) = 5n − 16. This completes the proof. The following theorem determines the minimum cardinality of z(G) for G ∈ Un1 as well as the corresponding extremal graph. Theorem 12. Let G ∈ Un1 with n ≥ 8 vertices. Then z(G) ≥ 5n − 16 with the equality if and only if G∼ = Cn3 (1, 1, n − 5).
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Fig. 7. (a). Cn3 (1, 2, n − 6); (b). Cn3 (1, 3, n − 7).
Fig. 8. The graph G( p).
Proof. From the combination of Lemma 10 and Theorems 3 and 7 it follows that z(G) ≥ z(Cn4 (1, 1, 1, n−7)). Similar to Lemma 11, we can obtain z(Cn4 (1, 1, 1, n − 7)) = 10n − 48. Thus, for n ≥ 8, z(G) ≥ z(Cn4 (1, 1, 1, n − 7)) > z(Cn3 (1, 1, n − 5)) and then the result follows. 3. Graphs in Un1 with second- and third-minimal Hosoya indices For brevity, we will denote transformations A or B simply as T.A. or T.B., respectively, in what follows. Theorem 13. Let n ≥ 8 and G ∈ Un1 (3) − Cn3 (1, 1, n − 5). Then z(G) ≥ 7n − 30 with equality if and only if G∼ = Cn3 (1, 2, n − 6), where Cn3 (1, 2, n − 6) is the graph as indicated in Fig. 7(a). Proof. We distinguish between two cases. Case 1. G ∼ = Cn3 (k1 , k2 , k3 ). Since G Cn3 (1, 1, n − 5), then (k1 , k2 , k3 ) 6= (1, 1, n − 5). Assume that (k1 , k2 , k3 ) 6= (1, 2, n − 6) and k1 ≤ k2 ≤ k3 . We consider the following two subcases. Subcase 1.1. k1 = 1. Note that k2 ≥ 3 in this case. Take the graph Cn3 (1, 1, 1) as Gˆ0 as depicted in Fig. 4. Using T.B. and by Lemma 6, we obtain Per B(Cn3 (1, k2 , k3 )) > Per B(Cn3 (1, 2, k3 + (k2 − 2))) = Per B(Cn3 (1, 2, n − 6)) or Per B(Cn3 (1, k2 , k3 )) > Per B(Cn3 (1, 2, k2 + (k3 − 2))) = Per B(Cn3 (1, 2, n − 6)). By Theorem 3, z(Cn3 (1, k2 , k3 )) = Per B(Cn3 (1, k2 , k3 )) − 2 > Per B(Cn3 (1, 2, n − 6)) − 2 = z(Cn3 (1, 2, n − 6)). Subcase 1.2. k1 ≥ 2. Using T.B. and by Lemma 6, we obtain Per B(Cn3 (k1 , k2 , k3 )) > Per B(Cn3 (1, k2 , k3 + (k1 − 1))). If (1, k2 , k3 + (k1 − 1)) = (1, 2, n − 6), the theorem holds. Otherwise, we have k2 ≥ 3. After a further step of T.B., we obtain Per B(Cn3 (1, k2 , k3 + (k1 − 1))) > Per B(Cn3 (1, 2, k3 + (k1 − 1) + (k2 − 2))) = Per B(Cn3 (1, 2, n − 6)). Consequently, z(Cn3 (k1 , k2 , k3 )) > z(Cn3 (1, 2, k3 + (k1 − 1) + (k2 − 2))) = z(Cn3 (1, 2, n − 6)). Case 2. G Cn3 (k1 , k2 , k3 ). Subcase 2.1. Cn3 (1, 1, n − 5) can be obtained from G by a sequence of T.A.s. Since G Cnl (1, 1, n − 5), then Cn3 (1, 1, n − 5) can be obtained from G by at least one step of T.A. Let G( p) be the graph in Un1 (3) with the property that G is reduced to Cn3 (1, 1, n − 5) by exactly one step of T.A. Then G( p) must have the form as indicated in Fig. 8. It is easy to obtain that z(Cn3 (1, 2, n − 6)) = 7n − 30 and z(G( p)) = −5 p 2 + (5n − 23) p + 3n − 6. So z(G( p)) − z(Cn3 (1, 2, n − 6)) = −5 p 2 + (5n − 23) p + 24 − 4n. Set f ( p) = −5 p 2 + (5n − 23) p + 24 − 4n (1 ≤ p ≤ n − 6). Then
d f ( p) dp
= (5n − 23) − 10 p.
d f ( p) From this it is deduced that if 1 ≤ p ≤ then > 0 and if d 5n−23 10 e ≤ p ≤ n − 6, then d p < 0. So we have z(G( p)) − z(Cn3 (1, 2, n − 6)) = f ( p) ≥ f (1) = n − 4 > 0 for 1 ≤ p ≤ b 5n−23 10 c whereas 5n−23 3 z(G( p)) − z(Cn (1, 2, n − 6)) = f ( p) ≥ f (n − 6) = 3n − 18 > 0 for d 10 e ≤ p ≤ n − 6. Therefore z(G( p)) > z(Cn3 (1, 2, n − 6)).
b 5n−23 10 c,
d f ( p) dp
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˘ Fig. 9. (a). H ( p); (b). G.
If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 2, n − 6) by at least two steps of T.A., then by Lemma 4, we have z(G) > z(G( p 0 )) for some positive integer p 0 . Similarly, we can prove z(G( p 0 )) > z(Cn3 (1, 2, n − 6)) and then the theorem holds. Subcase 2.2. Cn3 (1, 1, n − 5) cannot be obtained from G by a sequence of T.A.s. Repeatedly using T.A. on G, we finally obtain the graph Cn3 (k1 , k2 , k3 ) and z(G) > z(Cn3 (k1 , k2 , k3 )) by Lemma 5, where (k1 , k2 , k3 ) 6= (1, 1, n − 5). What remains is similar to case 1, we can show that the theorem holds. Combining cases 1 and 2, we complete the proof. Theorem 14. Let n ≥ 8. Then Cn3 (1, 2, n − 6) has the second-minimal Hosoya index among all graphs in Un1 . Proof. By repeatedly using Lemma 10, we have Per B(Cnl (1, 1, . . . , 1, n − 2l + 1)) > Per B(Cn4 (1, 1, 1, n − 7)) for l ≥ 5. Then by Theorem 3, z(Cnl (1, 1, . . . , 1, n −2l +1)) > z(Cn4 (1, 1, 1, n −7)). When n ≥ 8, z(Cn4 (1, 1, 1, n −7)) = 10n −48 > 7n −30 = z(Cn3 (1, 2, n −6)). Combining this fact with Lemma 5 and Theorem 13, the result follows. Theorem 15. Let n ≥ 10 and G ∈ Un1 (3) − {Cn3 (1, 1, n − 5), Cn3 (1, 2, n − 6)}. Then 9n − 48, 10 ≤ n ≤ 12 (a) z(G) ≥ 8n − 35, n ≥ 13 (b). Equality holds in (a) if and only if G ∼ = Cn3 (1, 3, n − 7); Equality holds in (b) if and only if G ∼ = G˘ for n ≥ 14 3 3 ∼ ˘ ˘ and G = G or Cn (1, 3, n − 7) for n = 13. Here Cn (1, 3, n − 7) and G are the graphs as illustrated in Fig. 7(b) and Fig. 9(b), respectively. Proof. We consider the following two cases. Case 1. G ∼ = Cnl (k1 , k2 , k3 ). Since G Cn3 (1, 1, n − 5), Cn3 (1, 2, n − 6), then (k1 , k2 , k3 ) 6= (1, 1, n − 5), (1, 2, n − 6). Assume that (k1 , k2 , k3 ) 6= (1, 3, n − 7) and k1 ≤ k2 ≤ k3 . Similar to the proof of case 1 in Theorem 13, we can show that z(G) > z(Cn3 (1, 3, n − 7)) in this case. Case 2. G Cn3 (k1 , k2 , k3 ). We divide this case further into three subcases. Subcase 2.1. Cn3 (1, 1, n − 5) can be obtained from G by a sequence of T.A.s. Since G Cnl (1, 1, n − 5), then Cn3 (1, 1, n − 5) can be obtained from G by at least one step of T.A. We consider only the case when G is reduced to Cn3 (1, 1, n − 5) by exactly one step of T.A. Let G( p) be the graph in Un1 (3) with the above property. As obtained in Theorem 13, G( p) = −5 p 2 + (5n − 23) p + 3n − 6. Similar to Lemma 11, we obtain z(Cn3 (1, 3, n − 7)) = 9n − 48. Consequently, z(G( p)) − z(Cn3 (1, 3, n − 7)) = −5 p 2 + (5n − 23) p − 6n + 42. Set g( p) = −5 p 2 + (5n − 23) p − 6n + 42 (1 ≤ p ≤ n − 6). Then From this it is deduced that if 1 ≤ p ≤ b 5n−23 10 c, then z(Cn3 (1, 3, n
dg( p) dp
> 0 and
dg( p) d p = (5n − 23) − 10 p. dg( p) if d 5n−23 10 e ≤ p ≤ n − 6, then d p
< 0.
So we have z(G( p)) − − 7)) = g( p) ≥ g(2) = 4n − 24 > 0 for 2 ≤ p ≤ b 5n−24 10 c 5n−24 3 whereas z(G( p)) − z(Cn (1, 32, n − 7)) = g( p) ≥ g(n − 6) = n > 0 for d 10 e ≤ p ≤ n − 6. Therefore z(G( p)) > z(Cn3 (1, 3, n − 7)) for p ≥ 2. When p = 1, z(G(1)) − z(Cn3 (1, 3, n − 7)) = 14 − n. If 10 ≤ n ≤ 13, then z(G(1)) > z(Cn3 (1, 3, n − 7)), and if n ≥ 15, z(G(1)) < z(Cn3 (1, 3, n − 7)).
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If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 1, n − 5) by at least two steps of T.A. We can prove that z(G) > z(G( p 00 )) for some positive integer p 00 . Similarly as above, we obtain z(G( p 00 )) > z(Cn3 (1, 3, n − 7)) for p 00 ≥ 2 and z(G( p 00 )) > z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 13, while z(G( p 00 )) < z(Cn3 (1, 3, n − 7)) when n ≥ 15 for p 00 = 1. In conclusion, z(G) ≥ z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 13, whereas z(G) ≥ z(G(1)) when n ≥ 15. As for n = 14, z(G) ≥ z(G(1)) = z(Cn3 (1, 3, n − 7)). Subcase 2.2. Cn3 (1, 2, n − 6) can be obtained from G by a sequence of T.A.s. ˘ We consider only the case when G is reduced to Cn3 (1, 2, n − 6) by exactly one step of T.A. Then G ∼ = H ( p)or G. If G ∼ = H ( p), by an elementary calculation, we get z(H ( p)) = −7 p 2 + (7n − 40) p + 5n − 18. Hence z(H ( p)) − z(Cn3 (1, 3, n − 7)) = −7 p 2 + (7n − 40) p − 4n + 30. Set t ( p) = −7 p 2 + (7n − 40) p − 4n + 30 (1 ≤ p ≤ n − 7). It is easily seen that if 1 ≤ p ≤ b 7n−40 14 c, then z(H ( p)) − z(Cn3 (1, 3, n − 7)) = t ( p) ≥ t (1) = 3n − 17 > 0 and if d 7n−40 e ≤ p ≤ n − 7, then 14 z(H ( p)) − z(Cn3 (1, 3, n − 7)) = t ( p) ≥ t (n − 7) = 5n − 33 > 0. ˘ by an elementary calculation, we have z(G) ˘ = 8n − 35. When 10 ≤ n ≤ 12, we have If G ∼ = G, ˘ when n ≥ 14, z(Cn3 (1, 3, n − 7)) > z(G). ˘ z(Cn3 (1, 3, n − 7)) < z(G); If G is a graph in Un1 (3) with the property that it is reduced to Cn3 (1, 2, n − 6) by at least two steps of T.A. We can use the same method as employed in subcase 2.1. ˘ when n ≥ 14. As for n = 13, In conclusion, z(G) ≥ z(Cn3 (1, 3, n − 7)) when 10 ≤ n ≤ 12, whereas z(G) ≥ z(G) 3 ˘ = z(Cn (1, 3, n − 7)). z(G) ≥ z(G) Case 2.3. Neither Cnl (1, 1, n − 5) nor Cn3 (1, 2, n − 6) can be obtained from G by T.A. Repeatedly using T.A. on G, we finally obtain the graph Cn3 (k1 , k2 , k3 ) and z(G) > z(Cn3 (k1 , k2 , k3 )), where (k1 , k2 , k3 ) 6= (1, 1, n − 5), (1, 2, n − 6). What remains is similar to case 1 in Theorem 13, we can show that z(G) > z(Cn3 (1, 3, n − 7)). ˘ Combining cases 1, 2 and 3, we have z(G) ≥ z(Cn3 (1, 3, n − 7)) Note that z(G(1)) = 8n − 34 > 8n − 35 = z(G). ˘ ˘ for 10 ≤ n ≤ 12, while z(G) ≥ z(G) for n ≥ 14. when n = 13, z(G) ≥ z(Cn3 (1, 3, n − 7)) = z(G). We thus completed the proof here. Theorem 16. Among all graphs in Un1 with n ≥ 10, Cn3 (1, 3, n − 7) has the third-minimal Hosoya index when 10 ≤ n ≤ 12, while G˘ has the third-minimal Hosoya index for n ≥ 13. Proof. When n ≥ 8, z(Cn4 (1, 1, 1, n − 7)) = 10n − 48 > 9n − 48 = z(Cn3 (1, 3, n − 7)) and z(Cn4 (1, 1, 1, n − 7)) = ˘ From Lemmas 5 and 10, and Theorems 3 and 15, the result follows as expected. 10n − 48 > 8n − 34 = z(G). Acknowledgements The author is indebted to the anonymous referees for their helpful suggestions and useful comments, which improved considerably the presentation of this paper. References [1] J.A. Bondy, U.S.R. Murty, Graph Theory with Applications, North-Holland, Amsterdam, 1976. [2] H. Hosoya, Topological index, a newly proposed quantity characterizing the topological nature of structural isomers of saturated hydrocarbons, Bull. Chem. Soc. Jpn. 44 (1971) 2332–2339. [3] I. Gutman, Extremal hexagonal chains, J. Math. Chem. 12 (1993) 197–210. [4] L.Z. Zhang, The Z -transformation graph of perfect matchings of a catacondensed hexagonal system, J. Math. Study 31 (1998) 437–441 (in Chinese). [5] L.Z. Zhang, F. Tian, Extremal hexagonal chains concerning largest eigenvalue, Sci. China Ser. A 44 (2001) 1089–1097. [6] L.Z. Zhang, F. Tian, Extremal catacondensed benzenoids, J. Math. Chem. 34 (2003) 111–122. [7] I. Gutman, O.E. Polansky, Mathematical Concepts in Organic Chemistry, Springer, Berlin, 1986. [8] Y. Hou, On acyclic systems with minimal Hosoya index, Discrete Appl. Math. 119 (2002) 251–257. [9] A. Yu, F. Tian, A kind of graphs with minimal Hosoya indices and maximal Merrifield-Simmons indices, MATCH Commun. Math. Comput. Chem. 55 (1) (2006) 103–118. [10] A. Yu, X. Lv, The Merrifield–Simmons indices and Hosoya indices of trees with k pendent vertices, J. Math. Chem. 41 (1) (2007) 33–43. [11] J. Ou, On extremal unicyclic molecular graphs with maximal Hosoya index (submitted for publication). [12] J. Ou, On extremal unicyclic molecular graphs with prescribed girth and minimal Hosoya index, J. Math. Chem. 42 (3) (2007) 423–432. [13] H. Minc, Permanents, Addison-Wesley, Reading, MA, 1978.