The algebraic connectivity of a graph is defined as the second smallest eigenvalue of the Laplacian matrix of the graph, which is a parameter to measure how well a graph is connected. In this paper, we present two unique graphs whose algebraic connectivity attain the minimum among all graphs whose complements are trees, but not stars, and among all graphs whose complements are unicyclic graphs, but not stars adding one edge, respectively.
1. Introduction
Let G=(V(G),E(G)) be a simple graph with vertex set V(G)={v1,v2,…,vn}. The complement of G is denoted by Gc∶=(V(G),E′(G)), where E′(G)={xy:x,y∈V(G),xy∉E(G)}. The adjacency matrix of G is defined to be a matrix A(G)=[aij] of order n, where aij=1 if vi is adjacent to vj, and aij=0 otherwise. The degree matrix of G is denoted by D(G)=diagdG(v1),dG(v2),…,dG(vn), where dG(v) denotes the degree of a vertex v in the graph G. The matrix Q(G)=D(G)+A(G) is the signless Laplacian matrix of G. The Laplacian matrix of G is defined to be L(G)=D(G)-A(G). Since A(G), Q(G), and L(G) are real and symmetric, their eigenvalues are real and can be arranged, respectively. We simply call the eigenvalues of A(G) as eigenvalues of G, the largest eigenvalue of A(G) as spectral radius of G, the eigenvalues of Q(G) as the signless Laplacian eigenvalues of G, and the eigenvalues of L(G) as the Laplacian eigenvalues of G. The second smallest Laplacian eigenvalue of G is called the algebraic connectivity ofG, denoted by a(G). The eigenvectors corresponding to a(G) are called Fiedler vectors ofG. a(G) is a good parameter to measure how well a graph is connected and plays an important role in control theory and communications, and so forth. In particular, it is related to the synchronization ability of complex network [1, 2]. On the other hand it is related to the convergence speed in networks; one important topic is to increase a(G) as much as possible. There are several techniques to enhance this metric; see [3].
When the structures of graphs are very complex, but the structures of their complements are simple, we naturally think whether we can study the graphs by studying their complements. Recently, there are some works about this subject. A connected graph G with n vertices and m edges is called k-cyclic graph if m=n-1+k. Specially, if k=0,1,2, or 3, then G is called a tree, unicyclic graph, bicyclic graph, or tricyclic graph, respectively. Let Tn be the set of trees on n vertices, and let Sn be the set of unicyclic graphs of order n. Denote by K1,n-1 the star graph on n vertices, and K1,n-1+e the graph obtained from K1,n-1 by adding one edge e. Fan et al. [4] have got the unique minimizing graph whose least eigenvalue attains the minimum among all graphs G for Gc∈Tn∖{K1,n-1}. Li and Wang [5] have researched the unique minimizing graph which whose signless Laplacian eigenvalue attains the minimum among all graphs G for Gc∈Tn∖{K1,n-1}. Liu and Zhang [6] have studied the unique maximizing graph whose spectral radius attains the maximum among all graphs G for Gc∈Sn. Yin and Guo [7, 8] have characterized two maximizing graphs whose spectral radius attains the maximum among all the complements of bicyclic graphs of order n and among all the complements of tricyclic graphs of order n, respectively. Yu and Fan [9, 10] have studied two unique minimizing graphs whose least eigenvalue attains the minimum among all graphs of order n whose complements are connected and whose complements are connected and have no cut edges or cut vertices, that is, 2-edge connected or 2-vertex connected graphs, respectively.
We note that the least Laplacian eigenvalue of a graph is zero, and Fiedler [11] proves that a graph G is connected if and only if a(G)>0. In fact, K1,n-1 is the unique graph whose complement is not connected in Tn, and K1,n-1+e is the unique graph whose complement is not connected in Sn. One may naturally ask what is the minimizing graph whose second smallest Laplacian eigenvalue (or the algebraic connectivity) attains the minimum among all graphs G for Gc∈Tn∖{K1,n-1} or for Gc∈Sn∖{K1,n-1+e}? In this paper, we study those questions and obtain two unique graphs which have the least algebraic connectivity among all graphs G for Gc∈Tn∖{K1,n-1} and for Gc∈Sn∖{K1,n-1+e}, in Sections 3 and 4, respectively.
2. Preliminaries
We begin with some definitions. Given a graph G of order n, a vector X∈Rn is called to be defined on G, if there is a 1-1 map φ from V(G) to the entries of X, simply written Xu=φ(u) for each u∈V(G). If X is an eigenvector of L(G), then X is defined on G naturally; Xu is the entry of X corresponding to the vertex u. One can find that, for an arbitrary vector X∈Rn, (1)XTLGX=∑uv∈EGXu-Xv2,and when μ is a Laplacian eigenvalue of G corresponding to the eigenvector X if and only if X≠0, (2)dGv-μXv=∑u∈NGvXu,foreachvertexv∈VG,where NG(v) denotes the neighborhood of v in the graph G. Equation (2) is called the Laplacian eigenvalue-equation for the graph G. In addition, by the well-known Courant-Fisher Theorem [11], for an arbitrary unit vector X∈Rn, when X≠0, X⊥1,(3)aG≤XTLGX,with equality if and only if X is a Fiedler vector of G, where 0 is null vector and 1 is the vector such that each of its coordinates is equal to 1.
Let Gc be the complement of a graph G of order n. It is easily seen that L(Gc)=nI-J-L(G), where J,I, respectively, denote the all-ones square matrix and the identity matrix both of suitable sizes. So for an arbitrary vector X∈Rn, (4)XTLGcX=XTnI-JX-XTLGX.
Lemma 1 (see [11]).
Let G be a simple graph. Then a(G)≤δ(G), where δ(G)=min{dG(v),v∈V(G)}.
Lemma 2.
If {Xi,i=1,2,…,n} is a nonincreasing sequence, then, for any 1≤i, j≤n, (Xi-Xj)2≤max{(Xi-X1)2,(Xi-Xn)2}≤(X1-Xn)2.
Proof.
For any 1≤i, j≤n, if Xi-Xj≥0, by the monotone of {Xi,i=1,2,…,n}, we have (5)0≤Xi-Xj≤Xi-Xn≤X1-Xn.
Then(6)Xi-Xj2≤Xi-Xn2≤X1-Xn2.
Similarly, if Xi-Xj≤0, we have(7)0≥Xi-Xj≥Xi-X1≥Xn-X1.
Then(8)Xi-Xj2≤Xi-X12≤X1-Xn2.
The result follows.
3. The least Algebraic Connectivity of the Complements of Trees
Given a graph G with u,v∈V(G), let dG(u,v) be the distance between u and v in G. Denote by Kp,q a complete bipartite graph whose bipartition has p vertices and q vertices, respectively. Denote by T(p,q) the special tree, which is obtained from two disjoint stars K1,p (p≥1) and K1,q (q≥1) by joining the center of K1,p and K1,q by a path of length 1; see Figure 1.
The graph T(p,q).
Lemma 3.
Given a positive integer n (n≥6), for any positive integers p,q with p+q=n-2, p≥q≥2, one has(9)aTp,qc>aTp+1,q-1c.
Proof.
Let T(p,q) be the graph as depicted in Figure 1 with some vertices labeled. Let X be a unit Fiedler vector of T(p,q)c. By Lemma 1, we have a(T(p,q)c)≠dT(p,q)c(v)+1 for any vertex v∈V(T(p,q)c). Then by the Laplacian eigen-equations (2), all vertices which are the pendant vertices of v2 have the same values given by X, say X1; all vertices which are the pendant vertices of v3 have the same values given by X, say X4. Write X2∶=Xv2, X3∶=Xv3. Now considering the Laplacian eigen-equations of X for T(p,q)c, and simply writing a:=a(T(p,q)c), we have(10)p+q-aX1=p-1X1+X3+qX4,q-aX2=qX4p-aX3=pX1p+q-aX4=pX1+X2+q-1X4.Transform (10) into a matrix equation B-aIX′=0, where X′=(X1,X2,X3,X4)T and (11)B=q+10-1-q0q0-q-p0p0-p-10p+1.
Let f(μ;p,q)∶=det(μI-B); then we have(12)fμ;p,q=μ4-2p+q+1μ3+p2+3pq+2p+q2+2q+1μ2-pqp+q+2μ.
Because 0 and a are roots of the polynomial f(μ;p,q), then a is the second smallest root of the polynomial f(μ;p,q). Observe that(13)fμ;p+1,q-1-fμ;p,q=μq-p-1μ-p-q-2=μq-p-1μ-n.
By Lemma 1, we have a=a(T(p,q)c)<n. Then by p≥q, a>0, and f(a;p,q)=0, we have that (14)fa;p+1,q-1=aq-p-1a-n>0,which implies that (15)aTp+1,q-1c<aTp,qc.
The result now follows from the above discussion.
Corollary 4.
Given a positive integer n (n≥6), for any positive integers p,q with p+q=n-2, p≥q≥2, one has(16)aTp,qc>aTp+1,q-1c>⋯>aTn-3,1c.
Lemma 5.
Given a positive integer n (n≥4), for any tree T∈Tn∖{K1,n-1}, there exist some integers p,q with 1≤p, q≤n-3, p+q=n-2, such that (17)aTc≥aTp,qc.
Proof.
Let X be a unit Fiedler vector of Tc; then X≠0 and X⊥1. Thus we can get a sequence {Xvi,i=1,2,…,n} such that (18)Xv1≥Xv2≥⋯≥Xvn.
If dT(v1,vn)>1, we can let the path v1Tvn=v1u1⋯u2vn, where u1=u2 when dT(v1,vn)=2. Add the edge v1vn, and delete the edge v1u1 or u2vn such that the result tree T∗ is not star. Then, by (1) and Lemma 2, we have(19)XTLTX=∑vivj∈ETXvi-Xvj2≤∑vivj∈ET∗Xvi-Xvj2=XTLT∗X.
If T∗≇T(p,q) and there exists a pendant vertex v, whose neighbor u is neither v1 nor vn, satisfying(20)Xv-Xv12≥Xv-Xvn2,then delete uv and add vv1; otherwise delete uv and add vvn. Repeat this rearranging until the resulting tree T′≅T(p,q). By (1) and Lemma 2, we have(21)XTLT∗X=∑vivj∈ET∗Xvi-Xvj2≤∑vivj∈ET′Xvi-Xvj2=XTLTp,qX.
By (3), (4), (19), and (21), we have(22)aTc=XTLTcX=XTnI-JX-XTLTX≥XTnI-JX-XTLTp,qX=XTLTp,qcX≥aTp,qc.The result follows.
By Corollary 4 and Lemma 5, we now obtain one main result of this paper.
Theorem 6.
For n≥4, T∈Tn∖{K1,n-1}, one has(23)aTc≥aTn-3,1c,with equality if and only if T≅T(n-3,1).
4. The Least Algebraic Connectivity of the Complements of Unicyclic Graphs
Denote by G1(p,q) the special unicyclic graph, which is obtained from C3 by identifying two vertices with the center of K1,p and the center of K1,q, respectively; see Figure 2.
The graph G1(p,q).
Lemma 7.
Given a positive integer n (n≥7), for any positive integers p,q with p+q=n-3, p≥q≥2, we have (24)aG1p,qc>aG1p+1,q-1c.
Proof.
Let G1(p,q) be the graph as depicted in Figure 2 with some vertices labeled. Let X be a unit Fiedler vector of G1(p,q)c. By Lemma 1, we have a(G1(p,q)c)≠dG1(p,q)c(v)+1 for any vertex v∈V(G1(p,q)c). Then, by the Laplacian eigen-equations (2), all vertices which are the pendant vertices of v2 have the same values given by X, say X4; all vertices which are the pendant vertices of v3 have the same values given by X, say X5. Write Xi∶=Xvii=1,2,3. Now considering the Laplacian eigen-equations of X for G1(p,q)c, and simply writing a∶=a(G1(p,q)c), we have(25)p+q-aX1=pX4+qX5,q-aX2=qX5p-aX3=pX4p+q+1-aX4=X1+X3+p-1X4+qX5p+q+1-aX5=X1+X2+pX4+q-1X5.Transform (25) into a matrix equation B-aIX′=0, where X′=(X1,X2,X3,X4,X5)T and (26)B=p+q00-p-q0q00-q00p-p0-10-1q+2-q-1-10-pp+2.
Let f1(μ;p,q)∶=det(μI-B); then we have(27)f1μ;p,q=μ5-3p+3q+4μ4+3p2+7pq+8p+3q2+8q+4μ3-p+q+2p2+4pq+2p+q2+2qμ2+pqp+q+1p+q+3μ.
Because 0 and a are roots of the polynomial f1(μ;p,q), then a is the second smallest root of the polynomial f1(μ;p,q). Observe that(28)f1μ;p+1,q-1-f1μ;p,q=-μp-q+1μ-p-q-1μ-p-q-3.
By Lemma 1, we have a=a(G1(p,q)c)≤q<p+q+1. Then, by p≥q≥2, a>0, and f1(a;p,q)=0, we have(29)f1a;p+1,q-1=-ap-q+1a-p-q-1a-p-q-3<0,which implies that(30)aG1p+1,q-1c<aG1p,qc.
The result now follows from the above discussion.
Corollary 8.
Given a positive integer n (n≥7), for any positive integers p,q with p+q=n-3, p≥q≥2, one has(31)aG1p,qc>aG1p+1,q-1c>⋯>aG1n-4,1c.
Denote by G2(p,q) the special unicyclic graph, which is obtained from C3 by identifying one vertex with the center of K1,p and the pendent vertex of K1,q+1; see Figure 3.
The graph G2(p,q).
Lemma 9.
Given a positive integer n (n≥8), for any positive integers p,q with p+q=n-4, one has(32)aG2p,qc>aG2p+1,q-1c,ifp+3>q≥2aG2p,qc=aG2p+1,q-1c,ifp+3=qaG2p,qc<aG2p+1,q-1c,ifp+3<q.
Proof.
Let G2(p,q) be the graph as depicted in Figure 3 with some vertices labeled. Let X be a unit Fiedler vector of G2(p,q)c. By Lemma 1, we have a(G2(p,q)c)≠dG2(p,q)c(v)+1 for any vertex v∈V(G2(p,q)c) and a(G2(p,q)c)≠dG2(p,q)c(v1). Write Xi∶=Xvi, i=1,2,3,4,5,6. Then by the Laplacian eigen-equations (2), X1=X6, all vertices which are the pendant vertices of v2 have the same values given by X, namely, X4; all vertices which are the pendant vertices of v3 have the same values given by X, namely, X5. Now considering the Laplacian eigen-equations of X for G2(p,q)c, and simply writing a∶=a(G2(p,q)c), we have(33)p+q+1-aX1=X3+pX4+qX5,q-aX2=qX5,p+2-aX3=2X1+pX4,p+q+2-aX4=2X1+X3+p-1X4+qX5,p+q+2-aX5=2X1+X2+pX4+q-1X5.Transform (33) into a matrix equation B-aIX′=0, where X′=(X1,X2,X3,X4,X5)T and (34)B=p+q+10-1-p-q0q00-q-20p+2-p0-20-1q+3-q-2-10-pp+3.
Let f2(μ;p,q)∶=det(μI-B); then we have(35)f2μ;p,q=μ5-3p+3q+9μ4+3p2+7pq+18p+3q2+20q+27μ3-p3+5p2q+9p2+5pq2+29pq+27p+q3+13q2+41q+27μ2+qp+2p+q+3p+q+4μ.
Because 0 and a are roots of the polynomial f2(μ;p,q), then a is the second smallest root of the polynomial f2(μ;p,q). Observe that(36)f2μ;p+1,q-1-f2μ;p,q=-μp-q+3p+q-μ+3p+q-μ+4.
Because f2(a;p,q)=0, we have(37)f2a;p+1,q-1=-ap-q+3p+q-a+3p+q-a+4.
By Lemma 1, we note that 0<a=a(G2(p,q)c)≤p+q+2.
If p+3>q≥2, we have f2(a;p+1,q-1)<0, which implies that (38)aG2p+1,q-1c<aG2p,qc.
If p+3=q, we have f2(a;p+1,q-1)=0, which implies that(39)aG2p+1,q-1c=aG2p,qc.
If p+3<q, we have(40)f2a;p+1,q-1>0,and 0<a=a(G2(p,q)c)≤p+2 by Lemma 1.
Let(41)f2μ;p,q=μp+q-μ+3gμ.
Then(42)gμ=-μ3+2p+2q+6μ2-p2+3pq+6p+q2+8q+9μ+p2q+pq2+6pq+2q2+8q,and a is the least root of the polynomial g(μ) by 0<a≤p+2. We calculate that(43)g′μ=-3μ2+4p+4q+12μ-p2-3pq-6p-q2-8q-9,g′′μ=-6μ+4p+4q+12.
Thus when 0<μ≤p+2, p+3<q, g′′(μ)≥g′′(p+2)=4q-2p>0, which implies that g′(μ) is monotonously increasing for μ when 0<μ≤p+2,p+3<q.
So, when 0<μ≤p+2, p+3<q, g′(μ)≤g′(p+2)=-q2+pq+2p+3<-q-3<0, which implies that g(μ) is monotonously decreasing for μ when 0<μ≤p+2, p+3<q.
Then, if p+3<q, by f2(a;p+1,q-1)>0, we get(44)aG2p+1,q-1c>aG2p,qc.
The result now follows from the above discussion.
Lemma 10.
Given a positive integer n(n≥8), for any positive integers p,q with p+q=n-4, q≥1, one has(45)aG2p,qc≥aG2n-5,1c,with equality if and only if G2(p,q)=G2(n-5,1).
Proof.
From the proof of Lemma 9, we note that a1′∶=a(G2(n-5,1)c), a2′∶=a(G2(0,n-4)c) are the second smallest roots of the following two polynomials, respectively:(46)f2n-5,1=μμ-n+1μ3+2-2nμ2=μμ-nwi+n2-n-2μ+3n-n2,f20,n-4=μμ-n+1μ3+2-2nμ2=μμ-n+1ee+n2-7μ+8n-2n2.
By Lemma 1 and n≥8, we have 0<a1′, a2′≤2<n-1. Thus, a1′, a2′ are the least roots of the following two polynomials, respectively:(47)hμ=μ3+2-2nμ2+n2-n-2μ+3n-n2,kμ=μ3+2-2nμ2+n2-7μ+8n-2n2.
We observe that(48)hμ-kμ=n-μn-5>0,for 0<μ≤2 and n≥8.
Because k(a2′)=0, n≥8, and 0<a2′≤2, we have(49)ha2′>0,which imply that a1′<a2′; namely, (50)aG2n-5,1c<aG20,n-4c.
By Lemma 9, the result follows.
Denote by G3(p,q) the special unicyclic graph, which is obtained from C4 by identifying two adjacent vertices with the center of K1,p and the center of K1,q, respectively; see Figure 4.
The graph G3(p,q).
Lemma 11.
Given a positive integer n (n≥6), for any positive integers p,q with p+q=n-4, p≥q≥1, one has(51)aG3p,qc>aG3p+1,q-1c.
Proof.
Let G3(p,q) be the graph as depicted in Figure 4 with some vertices labeled. Let X be a unit Fiedler vector of G3(p,q)c. By Lemma 1, we have a(G3(p,q)c)≠dG3(p,q)c(v)+1 for any vertex v∈V(G3(p,q)c). Write Xi∶=Xvi, i=1,2,3,4,5,6. Then by the Laplacian eigen-equations (2), all vertices which are the pendant vertices of v3 have the same values given by X, namely, X5; all vertices which are the pendant vertices of v4 have the same values given by X, namely, X6. Now considering the Laplacian eigen-equations of X for G3(p,q)c, and simply writing a:=a(G3(p,q)c), we have(52)p+q+1-aX1=X3+qX5+pX6,p+q+1-aX2=X4+qX5+pX6,p+1-aX3=X1+pX6,q+1-aX4=X2+qX5,p+q+2-aX5=X1+X2+X4+q-1X5+pX6,p+q+2-aX6=X1+X2+X3+qX5+p-1X6.
Transform (52) into a matrix equation B-aIX′=0, where X′=(X1,X2,X3,X4,X5,X6)T and(53)B=p+q+10-10-q-p0p+q+10-1-q-p-10p+100-p0-10q+1-q0-1-10-1p+3-p-1-1-10-qq+3.Let f3(μ;p,q)∶=det(μI-B); then we have(54)f3μ;p,q=μ6-4p+4q+10μ5+6p2+13pq+31p+6q2+31q+37μ4-4p3+15p2q+33p2+15pq2+74pqww+83p+4q3+33q2+83q+60μ3+p4+7p3q+13p3+12p2q2+55p2qwee+55p2+7pq3+55pq2+129pq+86p+q4wee+13q3+55q2+86q+36μ2-p+q+3p+q+4·p2q+p2+pq2+3pq+2p+q2+2qμ.
Because 0 and a are roots of the polynomial f3(μ;p,q), then a is the second smallest root of the polynomial f3(μ;p,q). Observe that(55)f3μ;p+1,q-1-f3μ;p,q=-μp-q+1μ-p-q-1μ-p-q-4·μ-p-q-3.
By Lemma 1, we have a=a(G3(p,q)c)≤q+1. Then by p≥q≥1, a>0, and f3(a;p,q)=0, we have(56)f3a;p+1,q-1=-ap-q+1a-p-q-1a-p-q-4·a-p-q-3>0,which implies that (57)aG3p+1,q-1c<aG3p,qc.
The result now follows from the above discussion.
Corollary 12.
Given a positive integer n (n≥6), for any positive integers p,q with p+q=n-4, p≥q≥1, one has(58)aG3p,qc>aG3p+1,q-1c>⋯>aG3n-4,0c.
Lemma 13.
Given a positive integer n (n≥6), for any unicyclic graph G∈Sn∖{K1,n-1+e}, there exist some integers p,q with p,q≥1, such that (59)aGc≥aGip,qc,i=1or2or3.
Proof.
Let G be obtained from l-cycle Cl (the cycle which has length l) by attaching some trees to it, and let X be a unit Fiedler vector of Gc. Then X≠0 and X⊥1. Thus we can get a sequence {Xvn} such that(60)Xv1≥Xv2≥⋯≥Xvn.
If dG(v1,vn)>1, we can let the path v1Gvn=v1u1⋯u2vn, where u1=u2 when dG(v1,vn)=2. Add the edge v1vn, and delete the edge v1u1 or u2vn such that the result unicyclic graph G∗ is not K1,n-1+e. Then, by (1) and Lemma 2, we have(61)XTLGX=∑vivj∈EGXvi-Xvj2≤∑vivj∈EG∗Xvi-Xvj2=XTLG∗X,where G∗ is obtained from Cl′ by attaching some trees to it and dG∗(v1,vn)=1.
If G∗∉{G1(p,q),G2(p,q),G3(p,q)}, we will discuss G∗ in three cases in the following.
Case 1. Both v1 and vn are on the cycle Cl′.
If l′≥5, we can let the cycle Cl′=v1vnu1u2u3⋯v1. In this time, if (Xu2-Xv1)2≥(Xu2-Xvn)2, we delete the edge u2u3 and add the edge v1u2; otherwise, we delete the edge u2u3 and add the edge u2vn. Then the result unicyclic graph G∗∗ is obtained from C3 (vn is on the cycle) or C4 (both v1 and vn are on the cycle) by attaching some trees to it.
Case 2. One of v1,vn is on the cycle Cl′.
We may let v1 be on the cycle Cl′, and it has same discussion when vn is on the cycle Cl′. If l′≥4, we can let the cycle Cl′=v1u1u2u3⋯v1. In this time, if (Xv1-Xu2)2≥(Xu2-Xvn)2, we delete the edge u2u3 and add the edge v1u2; otherwise, we delete the edge u2u3 and add the edge u2vn. Then the result unicyclic graph G∗∗ is obtained from C3 (one of v1,vn is on the cycle) or C4 (both v1 and vn are on the cycle) by attaching some trees to it.
Case 3. Neither v1 nor vn is on the cycle Cl′.
We can let the cycle Cl′=u1u2u3⋯u1, and neither v1 nor vn is on the tree which is attached to u2. In this time, if (Xv1-Xu2)2≥(Xu2-Xvn)2, we delete the edge u2u3 and add the edge v1u2; otherwise, we delete the edge u2u3 and add the edge u2vn. Then the result unicyclic graph G∗∗ is in Case 1 or Case 2.
Then, by (1) and Lemma 2, we have(62)XTLG∗X=∑vivj∈EG∗Xvi-Xvj2≤∑vivj∈EG∗∗Xvi-Xvj2=XTLG∗∗X,where dG∗∗(v1,vn)=1 and G∗∗ is obtained from C3 by attaching some trees to it, where at least one of v1,vn is on the cycle, or G∗∗ is obtained from C4 by attaching some trees to it, where both v1 and vn are on the cycle.
If G∗∗∉{G1(p,q),G2(p,q),G3(p,q)} and there exists a pendant vertex v, whose neighbor u is neither v1 nor vn, satisfying (63)Xv-Xv12≥Xv-Xvn2,then delete uv and add vv1; otherwise delete uv and add vvn. Repeat this rearranging until the result graph G∗∗∗∈{G1(p,q),G2(p,q),G3(p,q)}.
Then by (1) and Lemma 2, we have(64)XTLG∗∗X=∑vivj∈EG∗∗Xvi-Xvj2≤∑vivj∈EG∗∗∗Xvi-Xvj2=XTLG∗∗∗X.
According to (3), (4), (61), (62), and (64), we have(65)aGc=XTLGcX=XTnI-JX-XTLGX≥XTnI-JX-XTLG∗X≥XTnI-JX-XTLG∗∗X≥XTnI-JX-XTLG∗∗∗X=XTLG∗∗∗cX≥aG∗∗∗c.The result follows.
Theorem 14.
For n≥8, G∈Sn∖{K1,n-1+e}, one has(66)aGc≥aG3n-4,0c,with equality if and only if G≅G3(n-4,0).
Proof.
From the proofs of Lemmas 7, 9, and 11, we note that a1∶=a(G1(n-4,1)c), a2∶=a(G2(n-5,1)c), and a3∶=a(G3(n-4,0)c) are the second smallest roots of the following three polynomials, respectively:(67)f1μ;n-4,1=μ5+5-3nμ4+3n2-9n+3μ3+-n3+3n2+3n-5μ2+n3-6n2+8nμ=μμ4-3n-5μ3+3n2-9n+3μ2w=-n3-3n2-3n+5μ+n3-6n2+8n,f2μ;n-5,1=μ5+3-3nμ4+3n2-5nμ3+-n3+n2+4n-2μ2+nn-1n-3μ=μμ-n+1·μ3+2-2nμ2+n2-n-2μ+3n-n2,f3μ;n-4,0=μ6+6-4nμ5+6n2-17n+9μ4+-4n3+15n2-11nμ3+n4-3n3-5n2+14n-4μ2+-n4+7n3-14n2+8nμ=μμ-n+1μ-n+2·μ3+3-2nμ2+n2-2n-2μ+4n-n2.
By Lemma 1 and n≥8, we have 0<a1,a2,a3≤1<n-2. Thus, a1,a2,a3 are the least roots of the following three polynomials, respectively:(68)g1μ=μ4-3n-5μ3+3n2-9n+3μ2-n3-3n2-3n+5μ+n3-6n2+8n,g2μ=μ3+2-2nμ2+n2-n-2μ+3n-n2,g3μ=μ3+3-2nμ2+n2-2n-2μ+4n-n2.
Claim 1. Consider a(G3(n-4,0)c)<a(G1(n-4,1)c).
We observe that(69)f1μ;n-4,1-μμ-n+2g3μ=μ2n-μ-1>0,for 0<μ≤1 and n≥8.
Because f1(a1;n-4,1)=0, n≥8, and 0<a1≤1, we have (70)g3a1>0,which imply that a3<a1; namely,(71)aG3n-4,0c<aG1n-4,1c.
Claim 2. a(G3(n-4,0)c)<a(G2(n-5,1)c).
We observe that(72)g2μ-g3μ=-μ2+nμ-n<0,for 0<μ≤1.
Because 0<a3≤1 and g3(a3)=0, we have g2(a3)<0.
We note that(73)g2′μ=3μ2+22-2nμ+n2-n-2>22-2n+n2-n-2=n2-5n+2>0for n≥8 and 0<μ≤1.
Then g2(μ) is monotonously increasing for u when n≥8 and 0<μ≤1.
Thus by g2(a3)<0, g2(a2)=0, and 0<a2, a3≤1, we gain a3<a2; that is, (74)aG3n-4,0c<aG2n-5,1c.
According to Corollary 8, Lemma 10, Corollary 12, Lemma 13, Claim 1, and Claim 2, the result follows.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Acknowledgments
This work is jointly supported by the National Natural Science Foundation of China under Grant nos. 11071001 and 11071002, the Natural Science Foundation of Anhui Province of China under Grant no. 11040606M14, and the Natural Science Foundation of Department of Education of Anhui Province of China under Grant nos. KJ2013A196 and KJ2015ZD27.
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