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1 Applicable Analysis and Discrete Mathematics available online at Appl. Anal. Discrete Math. 8 (2014), doi: /aadm z LAPLACIAN COEFFICIENTS OF UNICYCLIC GRAPHS WITH THE NUMBER OF LEAVES AND GIRTH Jie Zhang, Xiao-Dong Zhang Motivated by Ilić and Ilić s conjecture [A. Ilić, M. Ilić: Laplacian coefficients of trees with given number of leaves or vertices of degree two. Linear Algebra Appl., 431 (2009), ], we investigate properties of the minimal elements in the partial set (U g n,l, ) of the Laplacian coefficients, where U g n,l denote the set of n-vertex unicyclic graphs with the number of leaves l and girth g. These results are used to disprove their conjecture. Moreover, the graphs with minimum Laplacian-like energy in U g n,l are also studied. 1. INTRODUCTION Let G = (V,E) be a simple graph with n = V vertices and m edges and L(G) = D(G) A(G) be its Laplacian matrix, where A(G) and D(G) are its adjacency and degree diagonal matrices, respectively. The Laplacian polynomial L(G, λ) of G is the characteristic polynomial of its Laplacian matrix L(G), i.e., L(G,λ) = det(λi n L(G)) = n ( 1) k c k (G)λ n k. Then L(G) has nonnegative eigenvalues µ 1 µ 2 µ n 1 µ n = 0. From Viette s formula, c k = σ k (µ 1,µ 2 µ n 1 ) is a symmetric polynomial of order n 1. In particular, we have c 0 = 1,c n = 0,c 1 = 2 E(G),c n 1 = nτ(g), where τ(g) is the number of spanning trees of G (see [10]). If G is a tree, the Laplacian 2010 Mathematics Subject Classification. 05C50, 05C35. Keywords and Phrases. Unicyclic graph, Laplacian coefficients, Laplacian-like energy, balanced starlike tree. k=0 330

2 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 331 coefficient c n 2 is equal to its Wiener index, which is the sum of all distances between unordered pairs of vertices of G (see [1], [16]). c n 2 (T) = W(T) = d(u,v). u,v V In general, Laplacian coefficients c k can be expressed in terms of subtree structures of G. Theorem 1.1. [8] Let F k be the set of all spanning forests of G with exactly k components. The Laplacian coefficient c n k of a graph G is expressed by c n k (G) = γ(f), where F has k components T i with n i vertices, i = 1,2,...,k and γ(f) = F F k k n i. i=1 Recently, the study on the Laplacian coefficients has attracted much attention. Let G and H be two graphs of order n. We write G H if c k (G) c k (H) for all 0 k n, and write G H if G H and c k (G) < c k (H) for some k {0,1,...,n}. Mohar [11] first investigated properties of the poset (partially ordered set) of acyclic graphs with the partial order and proposed several questions. Later, Ilić [4] and Zhang et al. [17] investigated ordering trees by the Laplacian coefficients. Stevanović and Ilić [13] investigated and characterized the minimum and maximum elements in the poset of unicyclic graphs of order n with. Tan [15] proved that the poset of unicyclic graphs of order n and fixed matching number with has only one minimal element. He and Shan [3] studied the properties of the poset of bicyclic graphs of order n. The Laplacian-like energy in [9] of G, LEL for short, is defined as follows: n 1 LEL(G) = µk, since it has similar features as molecular graph energy defined by Gutman [2]. LEL describes well the properties which have a close relation with the molecular structures and was proved to be as good as the Randić index, better than Wiener index in some areas (see [14]). Further, Stevanović [12, Lemma 2] established a connection between LEL and Laplacian coefficients. Later, Ilić, Krtinić and Ilić [7, Theorem 1.1] provided a corrected proof of the following fact: Theorem 1.2. [12, 7] Let G and H be two graphs with n vertices. If G H, then LEL(G) LEL(H). Furthermore, if G H, then LEL(G) < LEL(H). Denote by k=1 U n,l = {G G is a n-vertex unicyclic graph with fixed l leaves }, U g n,l = {G G U n,l with fixed girth g }.

3 332 Jie Zhang, Xiao-Dong Zhang Let BST n,l be a balanced starlike tree of order n with l leaves which is obtained n 1 n 1 by identifying one end of each of the l paths of orders +1 or +1. l l Moreover, let U g,p n,l (see Fig. 1) be a balanced starlike unicyclic graph of order n with l leaves and girth g, which is obtained by identifying one end of a path P p+1 of order p+1 and one vertex of a cycle of order g, the other end of P p+1 and the center vertex of a balanced starlike tree BST n p g+1,l, respectively. Figure 1. Graph U g,p n,l Ilić and Ilić [6] proposed the following conjecture: Conjecture 1.3. [6] Among all n-vertex unicyclic graphs, the graph U 3,0 n,l has the minimum Laplacian coefficients c k, k = 0,...,n, i.e., U 3,0 n,l is the only one minimal element in the poset (U n,l, ). However, this conjecture is, in general, not true. Let G 1 and G 2 be the following two unicyclic graphs of orders 10 (see Fig. 2). Figure 2. Counter-example Then their Laplacian characteristic polynomials are L(G 1,x) = x 10 20x x 8 758x x x x x x 2 30x, L(G 2,x) = x 10 20x x 8 770x x x x x x 2 30x. Hence G 1 does not have minimal Laplacian coefficients c k, k = 0,...,10. So this conjecture is, in general, not true. But it is proven that G 1 is still a minimal element in the poset U 10,2. In fact, there are many minimal elements in the poset U n,l. This paperis organizedasfollows: in Section 2, we investigatesome properties of minimal elements in the poset U g n,l. In Sections 3 and 4, all minimal elements in

4 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 333 four special posets of U g n,l are characterized, respectively. Finally, in Section 5, we give some conjectures. 2. THE MINIMAL ELEMENTS IN (U G N,l, ) Let v be a vertex of a connected graph G and let N G (v) denote the set of the neighbors of v in G. Let d G (v) denote the degree of v in G, if d G (v) = 1, v is called a leaf or a pendent vertex. Say that P = vv 1 v k is a pendant path of length k attached at vertex v if its interval vertices v 1 v k 1 have degree two and v k is a leaf. If k = 1, then v 1 is a leaf and vv 1 is called a pendent edge. A branch vertex is a vertex having degree more than two. Moreover, let d(u,v) denote the distance between vertices u and v. For a n-vertex unicyclic graph G U g n,l, G can be obtained from a cycle C g = u 1 u g of order g by attaching trees T 1 T g rooted at u 1,...,u g, respectively. So G may be written to be C T1,...,T g. Lemma 2.4. [5] Let v be a vertex of a nontrivial connected graph G and for nonnegative integers p and q, let G(p,q) denote the graph obtained from G by adding two pendent paths of lengths p and q at v, respectively, p q 1. Then c k (G(p,q)) c k (G(p+1,q 1)),k = 0,1,...,n. Definition 2.5. Let G be an arbitrary connected graph with n vertices. If uv is a non-pendent cut edge of G with d G (v) 3 and d G (u) 2 such that there s at least one pendent path P t+1 = vv 1 v t attached at v, then the graph G = ξ(g,uv) obtained from G by changing all edges (except uv,vv 1 ) incident with v into new edges between u and N G (v)\{u,v 1 }. In other words, G = G {vx x N G (v)\{u,v 1 }}+{ux x N G (v)\{u,v 1 }}. We say that G is a ξ-transformation of G. (See Fig. 3). Figure 3. ξ-transformation Clearly, ξ-transformation preserves the number of leaves in G. Lemma 2.6. Let G be a connected graph of order n and G be obtained from G by ξ-transformation. If there exists a path P s+1 = uu p 1 up s 1 up s of order s+1 in the component of G uv and a pendent path P t+1 = vv 1 v t attached at v with s t, then G G, i.e, c k (G) c k (G ),k = 0,1,...,n, with equality if and only if k {0,1,n 1,n}.

5 334 Jie Zhang, Xiao-Dong Zhang Proof. Clearly, if k {0,1,n}, c k (G) = c k (G ). Since uv is a cut edge, then every spanning tree of G and G includes edge uv, which implies τ(g) = τ(g ). Hence c n 1 (G) = c n 1 (G ). Now assume that 2 k n 2 and consider the coefficients c n k (G). Let F (resp. F) be the set of all spanning forests of G (resp. G) with exactly k components. For an arbitrary spanning forest F F, denote by T the component of F containing u. Let f : F F,F = f(f ), where V(F) = V(F ) and E(F) = E(F ) {ux x N T (u) N G (v)}+{vx x N T (u) N G (v)}. Then f is injective. Let F = F (1) F (2), where F (1) = {F F uv E(F )} and F (2) = {F F uv / E(F )}. If F F (1), then F and F = f(f ) have the same components except T. Moreover, T and f(t ) have the same vertices. Hence γ(f) = γ(f ). If F F (2), let S be the component of F containingv. Then F and F = f(f ) have the same components except T and S in F. Assume T contains a vertices in the component of G uv containing v, V(T ) a vertices in the component of G uv containing u. Then F has two components f(t ) = T with V(T ) a vertices and f(s ) = S with a+ V(S ) vertices corresponding to T and S, respectively. Denote by N the product of the orders of all components of F except T and S. Then γ(f(f )) γ(f ) = [( V(T ) a)(a+ V(S ) ) V(T ) V(S ) ]N = ( V(T ) a V(S ) ) a N. Further let F (2) = F 20 F 21 F 22, where Hence it follows that F 20 = {F F (2) V(T ) a = V(S ) or a = 0}, F 21 = {F F (2) V(T ) a < V(S ),a > 0}, F 22 = {F F (2) V(T ) a > V(S ),a > 0}. F F 20,γ(f(F )) γ(f ) = 0, F F 21,γ(f(F )) γ(f ) < 0, F F 22,γ(f(F )) γ(f ) > 0. For every spanning forest F 1 F 21, let T and S be two components of F 1 containing u and v, respectively. Assume that u,u p 1,...,up r 1 V(T ) and u p r / V(T ). Moreover, let R be a component of F 1 containing u p r with b vertices. Thus let T be a tree obtained from T and R by joining u p r 1 and up r with the edge up r 1 up r, S be the path vv 1 v V (T ) a 1 and R be the path v V(T ) a v V(S ) 1. Then F 2 = (F 1 {T,S,R }) {T,S,R } is a spanning forest of G with exactly k components and F 2 F 22. Hence there exists an injective (not bijective) map from F 21 to F 22, i.e., ϕ : F 21 F 22 : F 1 F 2 = ϕ(f 1),

6 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 335 where F 2 = ϕ(f 1 ) = (F 1 {T,S,R }) {T,S,R }. Note that V(T ) = V(T ) +b, V(S ) = V(T ) a, and V(R ) = V(S ) V(T ) +a. It is easy to see that for F F 21, γ(f(ϕ(f ))) γ(ϕ(f )) = [ ( V(T ) +b a) ( V(T ) a+a) ( V(T ) +b) ( V(T ) a) ] Therefore, F F 21 F 22 + ( V(S ) V(T ) +a) N b = ( V(T ) a V(S ) ) an = (γ(f(f )) γ(f )) [γ(f(f )) γ(f )] = [γ(f(f )) γ(f )+γ(f(ϕ(f ))) γ(ϕ(f ))] F F 21 F F 22 \ϕ(f 21 ) [γ(f(f )) γ(f )] = It follows from Theorem 1.1 that c n k (G ) = F F γ(f ) < So the assertion holds. F F 22 \ϕ(f 21 ) [γ(f(f )) γ(f )] > 0. γ(f(f )) γ(f) = c n k (G). F F F F Now we are ready to present the main result in this section. Theorem 2.7. If C T1,...,T g U g n,l is obtained from the cycle C g = u 1 u g by attaching g trees T 1,...,T g at the roots u 1,...,u g, respectively, where V(T i ) = n i and the number of leaves in T i is l i, then C B1,...,B g C T1,...,T g, where B i is a tree with root u i obtained by identifying one end u i,pi of the path P pi : u i u i,2 u i,pi and the center of BST ni p i+1,l i for i = 1,...,g and p 1,...,p g are equal to 1 except at most one p j. Moreover, the equality holds if and only if C B1,...,B g = CT1,...,T g. Proof. Denote G = C T1,...,T g. We prove this result in two steps. Step 1. Let P be the longest path among all paths which start at u i in T i for i = 1,...,g. Without loss of generality, assume that P belongs to T 1. Let v be the farthest branch vertex from vertex u i in T i for i = 2,...,g. Thus there exists a cut edge uv in T i such that d(u i,u)+1 = d(u i,v). By ξ-transformation on uv and Lemma 2.6, we obtain G 1 = ξ(g,uv) = C T1,T 2,...,T i,...,tg G such that the number of branch vertices in T i is non-increasing. Hence after a series of ξ-transformations on the cut edge xy with y being a branch vertex and d(u i,x)+1 = d(u i,y) in T i,

7 336 Jie Zhang, Xiao-Dong Zhang G 2 = C T1,T 2,...,T, i,...,t g isobtained, wheret, i isastarliketreeofordern i andleaves l i with rootu i. Hence for i = 2,...,g, after performing aseries ofξ-transformations and repeatedly using Lemma 2.4, there exists a unicyclic graph G 3 = C T1,B 2,...,B g such that G 3 G 1 where B i is BST ni,l i with a center vertex u i for i = 2,...,g. Step 2. If T 1 is a path or a starlike tree of order n 1, then the assertion holds by using Lemma 2.4. Assume that T 1 has at least one branch vertex except the root u 1. Let v be the branch vertex which is nearest to some pendent vertices in T 1 (maybe v is not unique). Then there exists a cut edge uv, d(u 1,u)+1 = d(u 1,v). If there exists uv as defined above which satisfies the conditions of Lemma 2.6, then G 4 = ξ(g 3,uv) G 3. Moreover, the number of branch vertices in G 4 is no more than that in G 3. After performing a series of this type of ξ-transformations and repeatedly using Lemma 2.4, there exists an unicyclic graph G 5 = C T 1,B 2,...,B g such that G 5 G 4, where T 1 is a tree rooted at u 1 obtained by attaching l 1 1 pendent paths at some vertices of the longest path P of T 1. If all pendent paths are attached at the only one vertex u 1 or u 1,x of P, then the result holds. Otherwise, let v be the branch vertex which is nearest to u 1 in T 1 (when d G5 (u 1 ) > 3, v = u 1 ). Then there exists a cut edge u v which satisfies d(u 1,v )+ 1 = d(u 1,u ). By ξ-transformation on u v and Lemma 2.6, we obtain G 6 = ξ(g 5,u v ) G 5. Further, the number of branch vertices in G 6 is no more than that in G 5. Hence by performing a series of this type of ξ-transformations, G 7 = C B1,...,B g is obtained, where B 1 has exactly one branch vertex u 1 or B 1 has exactly one branch vertex u u 1 with d G7 (u 1 ) = 3. If u = u 1, then by Lemma 2.4, the assertionholds. If u u 1 andd G7 (u 1 ) = 3,then applyinglemma 2.4toall pendent paths in G 7 yields the result. Corollary 2.8. Let C T1,...,T g U g n,l be obtained from the cycle C g = u 1 u g by attaching g trees T 1,...,T g at the roots u 1,...,u g, respectively. If d(u 1 ) > 3 and V(T i ) = 1 for i = 2,...,g, then C B1,...,B g C T1,...,T g, where B 1 is BST n g+1,l with a center vertex u 1 and V(B i ) = 1 for i = 2,...,g. Moreover, the equality holds if and only if C B1,...,B g = CT1,...,T g. Proof. It is obvious that the assertion follows from the proof of Theorem THE MINIMAL ELEMENTS IN TWO SUBSETS OF U G N,l In this section, we characterize all extremal graphs which have minimal Laplacian coefficients in the following two special subsets of U g n,l. Denote Clearly, U g,p n,l U p. U g,1 n,l = {C T 1,...,T g V(T i ) = 1 for i = 2,...,g}, U g,2 n,l = {C T 1,...,T g V(T 1 ) > 1, V(T i ) > 1, V(T j ) = 1 for j 1,i}. is in Ug,1 n,l. For convenience, denote Ug,0 n,l = U0,U g,1 n,l = U1,...,U g,p n,l =

8 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 337 n g gl+l Lemma 3.9. For 1 p, U p and U p 1 are incomparable in the poset (U g n,l, ). (See Fig. 1). Proof. We first show that c n 2 (U p ) < c n 2 (U p 1 ). It is obvious that U p 1 can be regarded as U p 1 = ξ(u p,w p 1 w p ). For convenience, we denote U p and U p 1 by G and G, respectively. Let F 2 (resp. F 2 ) be the set of all spanning forests of G (resp. G ) with exactly 2 components. For an arbitrary spanning forest F F 2 (resp. F F 2), F (resp. F ) can be obtained by deleting two edges {e 1,e 2 } in E(G) (resp. E(G )) with e 1 belonging to the cycle in G (resp. G ). If e 2 w p 1 w p, then F and F have the same components, which implies γ(f) = γ(f ). If e 2 = w p 1 w p, then γ(f) γ(f ) ( n g )( p = (g +p 1)(n g p+1) +1 n l Therefore, c n 2 (G) c n 2 (G ) = F F 2 γ(f) F F 2 n g p l γ(f ) < 0. ) 1 < 0, We next show c m (G) > c m (G ) for 2p m 2(p + g) 3. Clearly P p+g 1 = w p 1 w p 2 w 1 u 1 u 2 u g (denote w i = u i+1,1 i p 1 for convenience) is the longest path of order p + g 1 in the component of G w p 1 w p. Let F (resp. F) be the set of all spanning forests of G (resp. G) with exactly n m components, in other words, F (resp. F) is the set of all spanning forests of G (resp. G) with exactly m edges. For an arbitrary spanning forest F F, denote by T the component of F containing w p 1. Let f : F F,F = f(f ), where V(F) = V(F ) and E(F) = E(F ) {w p 1 x x N T (w p 1 ) N G (w p )} +{w p x x N T (w p 1 ) N G (w p )}, thenf isinjective. LetF = F (1) F (2),whereF (1) = {F F w p 1 w p E(F )} and F (2) = {F F w p 1 w p E(F )}. If F F (1), then F and F = f(f ) have the same components except T. Moreover, T and f(t ) have the same vertices. Hence γ(f) = γ(f ). If F F (2), let S be the component of F containing w p. Then F and F = f(f ) have the same components except T and S in F. Assume that T contains a vertices in the component of G w p 1 w p containing w p, V(T ) a vertices in the component of G w p 1 w p containing w p 1. Then F has two components f(t ) = T with V(T ) a vertices and f(s ) = S with a+ V(S ) vertices corresponding to T and S, respectively. Denote by N the product of the orders of all components of F except T and S. Then γ(f(f )) γ(f ) = [ ( V(T ) a)(a+ V(S ) ) V(T ) V(S ) ] N = ( V(T ) a V(S ) ) a N.

9 338 Jie Zhang, Xiao-Dong Zhang Further, let F (2) = F 20 F 21 F 22, where Hence it follows that F 20 = {F F (2) V(T ) a = V(S ) or a = 0}, F 21 = {F F (2) V(T ) a < V(S ),a > 0}, F 22 = {F F (2) V(T ) a > V(S ),a > 0}. F F 20,γ(f(F )) γ(f ) = 0, F F 21,γ(f(F )) γ(f ) < 0, F F 22,γ(f(F )) γ(f ) > 0. For every spanning forest F 1 F 21, let T and S be two components of F 1 containing w p 1 and w p, respectively. Then T does not contain all the vertices of the path P p+g 1. In fact, if all the vertices of P p+g 1 belong to T, then by the definition of F 21, we have V(T ) = p + g 1 + a, E(T ) p + g 1, and V(S ) > V(T ) a = p + g 1, E(S ) > p + g 2, which implies m E(T ) + E(S ) > p+g 1+p+g 2 = 2(p+g) 3. It is a contradiction to m 2(p+g) 3. Therefore, assume that u (p 1)+1,u (p 2)+1,...,u r 1 V(T ) and u r / V(T ). Moreover, let R be a component of F 1 containing u r with b vertices. Thus let T be a tree obtained from T and R by joining u r 1 and u r with edge u r 1 u r, S be the path w p v 1 v V (T ) a 1 and R be the path v V(T ) a v V(S ) 1. Then F 2 = (F 1 {T,S,R }) {T,S,R } is a spanning forest of G with exactly m edges and F 2 F 22. Hence there exists an injective map from F 21 to F 22, i.e., ϕ : F 21 F 22 : F 1 F 2 = ϕ(f 1), where F 2 = ϕ(f 1 ) = (F 1 {T,S,R }) {T,S,R }. Note that V(T ) = V(T ) +b, V(S ) = V(T ) a, and V(R ) = V(S ) V(T ) +a. It is easy to see that for F F 21, γ(f(ϕ(f ))) γ(ϕ(f )) F F 21 F 22 = [( V(T ) +b a) ( V(T ) a+a) ( V(T ) +b) ( V(T ) a)] ( V(S ) V(T ) +a) N b = ( V(T ) a V(S ) ) an = (γ(f(f )) γ(f )) Therefore, [γ(f(f )) γ(f )] = [γ(f(f )) γ(f )+γ(f(ϕ(f ))) γ(ϕ(f ))] + F F 21 F F 22 \ϕ(f 21 ) [γ(f(f )) γ(f )] = F F 22 \ϕ(f 21 ) [γ(f(f )) γ(f )] 0.

10 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 339 It follows from Theorem 1.1 that for m 2(p+g) 3, c m (G ) = γ(f ) γ(f(f )) γ(f) = c m (G). F F F F Further we show that the above inequality is strict for m 2p. In other words, we show that ϕ is not a bijective map for 2p m 2(p+g) 3. Hence it is sufficient to find a spanning forest F 2 F 22, such that F 2 ϕ(f 21 ). Let T, S F 2, where T is the path: x 1 w p 1 w p 2 w 1 u 1 u g (x 1 N G (w p ) \ {w p 1,v 1 )}), S is the path: w p v 1 v p 1. The rest m 2p edges are chosen from the edge set E(G) \({w p 1 w p,u 1 u 2,u g u g 1,v p 1 v p } E( T ) E( S )) of order n 2p 4, it is obvious that n 2p 4 m 2p. Suppose that there exists F 1 F 21, and ϕ( F 1 ) = F 2. Let T be the component of F 1 which contains w p 1. By the definition ofϕ, V( S ) = V( T ) a,sou 2 isthefirstvertexonp p+g 1 whichdoesnotbelong to T, then u 1 u 2 ϕ( F 1 ). Since u 1u 2 F 2, then ϕ( F 1 ) F 2, a contradiction. This completes the proof of Lemma 3.9. n g gl+l Lemma For 1 p q the poset (U g n,l, ). F F, U p and U q are incomparable in Proof. Without loss of generality, assume p q = h 1. By using Lemma 3.9 repeatedly, we have c n 2 (U p ) < c n 2 (U p 1 ) < < c n 2 (U q ), c m (U p ) c m (U p 1 ) c m (U q+1 ) > c m (U q ), for 2(q+1) m 2(q+g+1) 3. Thus the assertion holds. Now we are ready to characterize all minimal elements in U g,1 n,l. Theorem There are exactly p + 1 minimal elements U 0,...,U p in U g,1 n g gl+l where p =. Proof. For any G = C T1,T 2,...,T g U g,1 n,l with V(T i) = 1 for i = 2,...,g, by Theorem 2.7, there exists a graph G G 1 = C B1,B 2,...,B g U g,1 n,l, where B 1 is a tree with root u 1 obtained by identifying one end u 1,p1 of the path P p1 : u 1 u 1,2 u 1,p1 and the center of BST n1 p 1+1,l 1 and V(B i ) = 1 for i = 2,...,g. n g gl+l If p 1 >, then by Lemma 2.6, there exists an cut edge uv such that G 1 G 2 = ξ(g 1,uv) = C B 1,B 2,...,B g U g,1 n,l, where B 1 is a tree with root u 1 obtained by identifying one end u 1,p1 1 of the path P p1 1 : u 1 u 1,2 u 1,p1 1 and the center of BST n1 p 1+2,l 1. After a series of ξ-transformations, there exists a U p such that G U p n g gl+l. If p 1, then G G 1 = U p1. On the other hand, by Lemma 3.10, U 0,...,U p are incomparable in the poset (U g,1 n,l U 0,...,U p are exactly all minimal elements in the poset (U g,1 n,l, ). n,l,, ). Hence

11 340 Jie Zhang, Xiao-Dong Zhang Theorem For any G = C T1,...,T g U g,2 n,l, U0 G. Proof. By Theorem 2.7, we may assume that G = C B1,...,B i,...,b g U g,2 n,l, where B 1 is a tree with root u 1 obtained by identifying one end u 1,p1 of the path P p1 : u 1 u 1,2 u 1,p1 and the center vertex of BST n1 p 1+1,l 1, B i is BST ni,l i with a center vertex u i, and V(B j ) = 1 for j 1,i. Let G = G {u i x x N G (u i )\{u i+1,u i 1 }}+{u 1 x x N G (u i )\{u i+1,u i 1 }}. We will prove c k (G) c k (G ) with at least one strict inequality. Clearly, when k {0,1,n 1,n}, c k (G) = c k (G ). For 2 k n 2, let F and F be the sets of all spanning forests of G and G with exactly k components, respectively. For an arbitrary spanning forest F F with T being the component of F containing u 1, let f : F F,F F = f(f ), where V(F) = V(F ), and E(F) = E(F ) {u 1 x x N G (u i ) N T (u 1 )\V(C g )} +{u i x x N G (u i ) N T (u 1 )\V(C g )}. Then f is injective from F to F. Denote by N the product of the orders of all components containing no u 1,u i. If u i T, then F and F have the same components except for T. Moreover, F has a component containing u 1 which corresponds to T in F. Clearly, the two components have the same orders. Hence γ(f) = γ(f ). If u i T, assume u i is in a component S of F. Moreover, there are b 0 vertices in the connected component containing u 2 in T u 1 u 2, and d 0 vertices in the connected component containing u g in T u 1 u g, e 1 1 vertices (including u 1 ) in the vertex set V(T 1 ) and e 2 0 vertices in the vertex set V(T i ) \ {u i }. Furthermore, the tree S contains a 1 vertices in the connected component containing u i in S u i u i+1 and c 0 vertices in the connected component containing u i+1 in S u i u i+1. Then F and F have the same components except for T and S. Moreover, F have two trees T containing u 1 and S containing u i which correspond to T and S in F, respectively. Hence γ(f(f )) γ(f ) = [(a+c+e 2 )(b+d+e 1 ) (a+c)(b+d+e 1 +e 2 )]N = e 2 (b+d+e 1 a c)n. Consider the subset F of those spanning forests F with k components which coincide on G \(T S ) with fixed values e 1,e 2. Since the two parts of F on the cycle between T and S may be translated, let a+b = M 1,c+d = M 2 be fixed. Then F F, a+b=m 1 c+d=m 2 c=0 ( γ(f(f )) γ(f ) ) = e 2 (b+d+e 1 a c)n a+b=m 1 c+d=m 2 M 2 M 1 1 M 2 = e 2 N (2b+M 2 +e 1 2c M 1 ) = e 2 NM 1 (e 1 +M 2 2c 1) b=0 = e 2 NM 1 (e 1 1)(M 2 +1) 0. c=0

12 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 341 Hence (γ(f(f )) γ(f )) = F F e 1 e 2 M 1 M 2 F F, a+b=m 1 c+d=m 2 (γ(f(f )) γ(f )) 0. Since V(T 1 ) > 1, V(T i ) > 1, there exists one forest F such that e 1 > 1 and e 2 > 0. Therefore c n k (G ) = γ(f ) < γ(f(f )) γ(f) = c n k (G),2 k n 2. F F F F Hence by Corollary 2.8, we have U 0 G with equality if and only if G = U 0. Therefore, U 0 G G, the assertion holds. F F It follows from Theorems 1.2, 3.11 and 3.12 that the following results hold. Corollary Let G = C T1,...,T g be an arbitrary unicyclic graph in U g,1 n,l. Then n g gl+l for p =, LEL(G) min{lel(u 0 ),LEL(U 1 ),...,LEL(U p )}. Corollary Let G = C T1,...,T g be an arbitrary unicyclic graph in U g,2 n,l. Then LEL(G) > LEL(U 0 ). 4. THE MINIMAL ELEMENTS IN U 3 N,l AND U4 N,l In this section, we determine all the minimal elements in the posets (Un,l 3, ) and (Un,l 4, ). Before stating our results, we need the following definitions. Definition For any G U 3 n,l, let G be the graph obtained from G by changing all the edges (except E(C 3 )) incident with u 2,u 3 into new edges between u 1 and N G (u 2 ) N G (u 3 )\V(C 3 ). In other words, G = G {u 2 x x N G (u 2 )\V(C 3 )} {u 3 x x N G (u 3 )\V(C 3 )} +{u 1 x x N G (u 2 ) N G (u 3 )\V(C 3 )}. We say G is a η-transformation of G. (See Fig. 4). Performing this transformation to the graph G can be seen as performing the transformation of Lemma 1 in [13] to G twice. Lemma For G U 3 n,l, if G is obtained from G by η-transformation, then G G, i.e., c k (G ) c k (G) with equality if and only if k {0,1,n 1,n}.

13 342 Jie Zhang, Xiao-Dong Zhang Figure 4. η-transformation Proof. Clearly, when k {0,1,n 1,n}, c k (G) = c k (G ). For 2 k n 2, let F (resp. F) be the set of all spanning forests of G (resp. G) with exactly n k components. Let F = F (1) F (2) F (3), where F (j),j = 1,2,3 is the set of all spanning forests of G in which u 1,u 2,u 3 belong to exactly j different components. Similarly, F (j),j = 1,2,3 can be defined. Let f : F F with F F = f(f ), where V(F) = V(F ) and E(F) = E(F ) {u 1 x x N R (u 1 ) N G (u 2 )\{u 3 }} {u 1 x x N R (u 1 ) N G (u 3 )\{u 2 }} +{u 2 x x N R (u 1 ) N G (u 2 )\{u 3 }} +{u 3 x x N R (u 1 ) N G (u 3 )\{u 2 }}, for R being a component of F containing u 1. Clearly f is injective and f(f (j) ) F (j) for j = 1,2,3. Denote V(T 1 ) V(R ) \ {u 1 } = a, V(T 2 ) V(R ) \{u 2 } = b, V(T 3 ) V(R )\{u 3 } = c, where a,b,c 0. Let N be the product of the orders of all components of F containing no {u 1,u 2,u 3 }. Now we consider the following three cases. Case 1. F F (1), u 1,u 2,u 3 belong to one component, then γ(f) = γ(f ). Thus F F (1) [γ(f) γ(f )] = 0. Case 2. F F (2), u 1,u 2,u 3 are in two components, then γ(f) γ(f ) = [(a+b+2)(c+1)+(a+c+2)(b+1)+(b+c+2)(a+1) 2(a+b+c+1) 2(a+b+c+2)]N = [(a+b)c+(a+c)b+(b+c)a]n 0. Thus F F (2) [γ(f) γ(f )] 0. Case 3. F F (3), u 1,u 2,u 3 are in three components, then γ(f) γ(f ) = [(a+1)(b+1)(c+1) (a+b+c+1)]n = (abc+ab+ac+bc)n 0. Thus F F (3) [γ(f) γ(f )] 0.

14 Laplacian coefficients of unicyclic graphs with the number of leaves and girth 343 Now the inequality c k (G ) < c k (G),k = 2,3,...,n 2 holds from Theorem 1.1 by summing over all possible subsets F of spanning forests F of G with n k components. Theorem There are exactly p +1 minimal elements U 3,0 the poset (Un,l 3, ), where p = n 3 2l. n,l,u3,1 n,l...,u3,p Proof. The assertion follows from Lemma 4.16 and Theorems 3.11 and n,l in Definition For G U 4 n,l, let G be the graph obtained from G by changing all the edges (except E(C 4 )) incident with u 2,u 3,u 4 into new edges between u 1 and 4 i=2 N G(u i )\V(C 4 ). In other words, G = G {u 2 x x N G (u 2 )\V(C 4 )} {u 3 x x N G (u 3 )\V(C 4 )} {u 4 x x N G (u 4 )\V(C 4 )}+{u 1 x x 4 i=2 N G(u i )\V(C 4 )}. We say G is a κ-transformation of G. Since for bipartite graphs, the Laplacian coefficients and the signless Laplacian coefficients are equal, from Lemma 3.1 in [18], we have the following lemma: Lemma For G U 4 n,l, if G is obtained from G by κ-transformation, then G G, i.e., c k (G ) c k (G) with equality if and only if k {0,1,n 1,n}. By combining Lemma 4.19 and Theorem 3.11, we have the following theorem. Theorem There are exactly p +1 minimal elements U 4,0 the poset (Un,l 4, ), where p = n 4 3l. From Theorem 1.2, we have the following corollary: n,l,u4,1 Corollary (1). Let G = C T1,T 2,T 3 U 3 n,l. Then for p = n 3 2l LEL(G) min{u 3,0 n,l,u3,1 n,l...,u3,p n,l }. (2). Let G = C T1,T 2,T 3,T 4 U 4 n,l. Then for p = n 4 3l LEL(G) min{u 4,0 n,l,u4,1 n,l...,u4,p n,l }., n,l...,u4,p, n,l in 5. REMARKS Although we showed that Ilić and Ilić s conjecture does not hold, it might be true in a slightly modified version. In fact, if there are at least two vertices in the cycle with degrees at least 3, then the conjecture is true for g = 3 and g = 4. Moreover, we compile a computer program with Matlab software and check that the conjecture is still true for all unicyclic graphs on 30 vertices with fixed l,g and having at least three vertices in the cycle with degrees at least 3. Hence their conjecture can be modified as follows:

15 344 Jie Zhang, Xiao-Dong Zhang Conjecture (1). For G U g n,l, if there are more than two vertices in the cycle having degrees 3, then U 0 G, with equality if and only if G = U 0. (2). There are exactly p+1 minimal elements U 3,0 n 3 2l. (U n,l, ), where p = n,l,u3,1 n,l,...,u3,p n,l in the poset Acknowledgements. This work is supported by National Natural Science Foundation of China (No ), innovation Program of Shanghai Municipal Education Commission (No.14ZZ016), the Ph.D. Programs Foundation of Ministry of Education of China (No ) and Funds of Innovation of Shanghai Jiao Tong University (No.Z071007). The authors would like to thank the referee for his/her suggestions and comments. REFERENCES 1. A. Dobrynin, R. Entringer, I. Gutman: Wiener index of trees: theory and applications. Acta Appl. Math., 66 (2001), I. Gutman: The energy of a graph. Ber. Math. Statist. Sekt. Forschungsz. Graz, 103 (1978), C. X. He, H. Y. Shan: On the Laplacian coefficients of bicyclic graphs. Discrete Math., 310 (2010), A. Ilić: On the ordering of trees by the Laplacian coefficients. Linear Algebra Appl., 431 (2009), A. Ilić: Trees with minimal Laplacian coefficients. Comput. Math. Appl., 59 (2010), A. Ilić, M. Ilić: Laplacian coefficients of trees with given number of leaves or vertices of degree two. Linear Algebra Appl., 431 (2009), A. Ilić, D. Krtinić, M. Ilić: On Laplacian like energy of trees. MATCH Commun. Math. Comput. Chem., 64 (2010), A. K. Kelmans, V. M. Chelnokov: A certain polynomial of a graph and graphs with extremal number of trees. J. Combin. Theory, Ser. B, 16 (1974), J. P. Liu, B. L. Liu: A Laplacian-energy-like invariant of a graph. MATCH Commun. Math. Comput. Chem., 59 (2008), R. Merris: A survey of graph Laplacians. Linear Multilinear Algebra, 39 (1995), B. Mohar: On the Laplacian coefficients of acyclic graphs. Linear Algebra Appl., 422 (2007), D. Stevanović: Laplacian-like energy of trees. MATCH Commun. Math. Comput. Chem., 61 (2009), D. Stevanović, A. Ilić: On the Laplacian coefficients of unicyclic graphs. Linear Algebra Appl., 430 (2009), D. Stevanović, A. Ilić, C. Onisor, M. Diudea: LEL-a newly designed molecular descriptor. Acta Chim. Slov., 56 (2009),

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