On sufficient spectral radius conditions for hamiltonicity of k-connected graphs

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Contents lists available atScienceDirect

Linear

Algebra

and

its

Applications

www.elsevier.com/locate/laa

On

suﬃcient

spectral

radius

conditions

for

hamiltonicity

of

k-connected

graphs

✩

Qiannan Zhoua,b_{, Hajo Broersma}b,∗_{, Ligong Wang}a_,_{Yong Lu}c a_School_of_Mathematics_and_Statistics,_Northwestern_{Polytechnical}_University,

Xi’an,Shaanxi710129,People’sRepublicofChina

b

FacultyofEEMCS,UniversityofTwente,P.O.Box217,7500AEEnschede, the Netherlands

c_School_of_Mathematics_and_Statistics,_Jiangsu_Normal_University,_Xuzhou,

Jiangsu221116,People’sRepublicofChina

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received20December2019 Accepted11June2020 Availableonline20June2020 Submittedby S.M.Cioaba MSC: 05C50 05C45 05C40 Keywords: k-Connected Hamiltonian Traceable Spectralradius

In this paper, we present two new sufficient conditions on thespectralradiusρ(G) thatguaranteethehamiltonicityand traceability of a k-connected graph G of sufficiently large order,respectively,unlessG isaspecifiedexceptionalgraph. Inparticular,ifk≥ 2,n≥ k3_+k+2,_and_ρ(G)_{> n}_−k−1₋1

n, then G is hamiltonian, unless G is a speciﬁed exceptional graph.Ifk_{≥ 1,}n_{≥ k}3_{+ k}2_{+ k + 3,}_and_ρ(G)_{> n}_{− k − 2}₋1

✩ _Research_supported_by_the _National_Natural_Science_Foundation_of _China_(No._11871398,_11901253),

theNatural ScienceBasicResearchPlaninShaanxiProvinceofChina(Program No.2018JM1032),the FundamentalResearch Funds forthe CentralUniversities (No.3102019ghjd003), the ChinaScholarship Council(No.201806290178)andtheNaturalScienceFoundationforCollegesandUniversitiesinJiangsu ProvinceofChina(No.19KJB110009).

* Correspondingauthor.

E-mailaddresses:qnzhoumath@163.com(Q. Zhou),h.j.broersma@utwente.nl(H. Broersma),

lgwangmath@163.com(L. Wang),luyong@jsnu.edu.cn(Y. Lu).

https://doi.org/10.1016/j.laa.2020.06.012

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1. Introduction

We usethe textbook of Bondyand Murty [3] for any terminologyand notation not deﬁnedhere.Throughoutthis paper,weconsideronlyﬁnite undirectedsimplegraphs.

Let G= (V (G),E(G)) beagraphwithvertex set V (G) andedge setE(G).We use

n(G) =|V (G)| to denote the order of G, and e(G) = |E(G)| to denote the size of G.

For X⊆ V (G),theinducedsubgraphG[X] isthesubgraphofG whose vertexset isX

and whoseedgeset consistsofalledgesofG whichhavebothends inX.Forv∈ V (G)

and subgraphs H and R of G, we let NH(v) and NH(R) denotethe neighbors of the vertex v and the subgraph R in H, thatis, NH(v) = {u ∈ V (H) | uv ∈ E(G)} and

NH(R)= (u∈V (R)NH(u))\ V (R).WhenH = G, |NH(v)| is called thedegree of the vertex v, simply written as d(v). Denote by δ(G) the minimum degree of G. For two subsets S and T of V (G), we say S isadjacent to T if every vertex of S is adjacent to every vertex of T . Let Puv and Pwz be two disjoint paths. Denote by Puv

Pwz a path obtainedfrom Puv and Pwz byjoiningv and w with anedge. Aconnected graph

G is saidto be k-connected (or k-vertex connected)if ithas morethan k verticesand

remains connectedwheneverfewer thank vertices areremoved.Theconnectivityκ(G)

ofG isthemaximumvalueofk forwhichG isk-connected.Theindependencenumberof

G,denotedbyα(G),isthecardinalityofalargestindependent(mutuallynonadjacent) set of vertices. Denote by ω(G) the clique number of G, that is, the cardinality of a largest clique, i.e., a set of mutually adjacent vertices. For two graphsG1 and G2, we useG1+ G2todenote thedisjoint unionofG1 andG2,andG1∨ G2 todenotethejoin of G1 and G2.

For a graph G with vertex set {v1,v2,. . . ,vn}, the adjacency matrix A(G) is the symmetric n× n matrix with entries A(i,j)= 1 if and onlyif vivj ∈ E(G) and zeros elsewhere.Thelargest eigenvalueofA(G) iscalled thespectral radiusofG,denotedby

ρ(G).

Agraphishamiltonian(traceable)ifitcontainsaHamiltoncycle(Hamiltonpath),i.e., acycle(path)containingallverticesofG.Decidingwhetheragivengraphishamiltonian (traceable) is an NP-complete problem.Many graph theorists have focused on ﬁnding suﬃcientconditionsfortheexistenceofaHamiltoncycle(orpath).In2010,Fiedlerand Nikiforov[6] gavesomeboundsonρ(G) thatimplytheexistenceofHamiltoncyclesand pathsinG.Thisworkmotivatedfurtherresearch,asin[1,10,12,13,15,16,18].

For graphs with minimum degree δ(G) ≥ k, Nikiforov [15] presented the following resultonthespectralradiusguaranteeingthehamiltonicityofG.

Theorem1.1([15]). LetG beagraph ofordern withδ(G)≥ k.Ifk≥ 2,n≥ k3_{+ k + 4}

and ρ(G) ≥ n− k − 1, then G is hamiltonian unless G = K1 ∨ (Kn−k−1 + Kk) or

G= Kk∨ (Kn−2k+ kK1).

Recently, Ge and Ning [7] showed that the statement in the abovetheorem due to Nikiforovalsoholdsfork≥ 1 andn≥ max{₂1k3+ k +5₂,6k + 5}.

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ForthetraceabilityofG,LiandNing[10] obtainedthefollowing resultinvolvingthe spectralradius.

Theorem1.2([10]). LetG beagraphof ordern withδ(G)≥ k.If k≥ 0,n≥ max{6k +

10,(k2_{+ 7k + 8)/2}_{} and} _ρ(G) _{≥ ρ(K}

k ∨ (Kn−2k−1+ (k + 1)K1)), then G is traceable

unless G= Kk∨ (Kn−2k−1+ (k + 1)K1).

Recently, for graphs with connectivity κ and minimum degree δ, Li [11] presented thefollowingsuﬃcientconditionsonthespectralradiusimplyingthehamiltonicityand traceabilityofG,respectively.

Theorem1.3([11]). LetG be agraphof order n≥ 3 withconnectivity κ≥ 2.If ρ(G)≤ δ

κ+1

n−κ−1,thenG ishamiltonian orG isKκ,κ+1.

Theorem 1.4 ([11]). Let G be a graph of order n ≥ 12 with connectivity κ ≥ 1. If

ρ(G)≤ δ κ+2

n−κ−2,thenG is traceableorG isKκ,κ+2.

In this paper, we present new suﬃcient conditions based onthe spectral radius for the hamiltonicity and traceability of k-connected graphs. Since being k-connected is a stronger condition than having minimum degree at least k, our motivation was to improve on the bounds for ρ(G) in Theorems 1.1 and 1.2. Obviously, we still haveto excludeKk∨ (Kn−2k+ kK1) andKk∨ (Kn−2k−1+ (k + 1)K1),respectively,sincethese graphsarek-connected.

Before stating our results, we introduce some families of graphs based on the two exceptionalgraphsKk∨(Kn−2k+ kK1) andKk∨(Kn−2k−1+ (k + 1)K1).Forn≥ 2k +1, wedeﬁne

G1_n,k= Kk∨ (Kn−2k+ kK1).

Forn ≥ 2k + 2,we startwith agraph consisting ofvertex-disjoint graphskK1 and

K_n−k−1,andanadditionalnewvertexv,andweletV (kK1)= X andV (Kn−k−1)= Y . WechooseY2⊆ Y with|Y2|= k.Now,weletG2n,k denotethegraphobtainedfrom this

kK1+ Kn−k−1+{v} byjoiningX toY2,andv toY2andanarbitraryvertexinX (see thegraphsketchedintheleft partofFig.1).Wealso deﬁne G3

n,k = Kk∨ (Kn−2k−1+ (k + 1)K1).

Forn≥ 2k+3,westartwithagraphconsistingofvertex-disjointgraphs(k+1)K1and

K_n−k−1,and weletV ((k + 1)K1)= X and V (Kn−k−1)= Y . WechooseX1⊆ X with

|X1|= k,andX2= X\X1,sowith|X2|= 1,andY1⊆ Y with|Y1|= k,andY2⊆ Y \Y1

with |Y2| = 1.Now, we let Gn,k4 denote thegraphobtained from (k + 1)K1+ Kn−k−1 byaddingedgesfrom X toY1andX2toY2(see thegraphsketchedintherightpartof Fig.1).WealsodeﬁneG_n,k5 = Kk+1∨ (Kn−2k−2+ (k + 1)K1).

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Fig. 1. Some of the graphs appearing in our proofs.

Additionally, we denote by G1_n,k,G2_n,k,G3_n,k,G4_n,k,G5_n,k the graphs obtained from

G1_n+1,k+1, G2_n+1,k+1,G3_n+1,k+1,G4_n+1,k+1,G5_n+1,k+1 by deleting onevertex of degree n,

respectively,i.e.,

G1_n,k = Kk∨ (Kn−2k−1+ (k + 1)K1),

G3_n,k = Kk∨ (Kn−2k−2+ (k + 2)K1),

G5_n,k = Kk+1∨ (Kn−2k−3+ (k + 2)K1).

One easilychecks thatG2_n,k is(again)the leftgraphinFig.1with |X|= k + 1,|Y | =

n−k−2,and|Y2|= k.Similarly,G4n,kis(again)therightgraphinFig.1with|X|= k+2,

|X1|= k + 1,|Y |= n− k − 2,and|Y1|= k.

InSection3,weprovethefollowingtwomain results.

Theorem 1.5.Let G be a k-connected graph of order n≥ k3_{+ k + 2,} _where _k _{≥ 2.} _If

ρ(G)> n− k − 1−_n1,then G ishamiltonian, unless G= G1

n,k. NotethattheexceptionalgraphG1

n,kintheabovetheoremispreciselythek-connected graphKk∨ (Kn−2k−1+ (k + 1)K1) thatwasexcludedintheconclusionofTheorem1.1. Theconditionρ(G)> n−k −1−1_n intheabovetheoremlooksbetterthanthecondition

ρ(G) > n− k − 1 in Theorem 1.1, but one should note the diﬀerent meaning of k:

Theorem 1.1involves graphswithminimum degreeat leastk,whereastheaboveresult dealswithk-connected graphs.Inthissense,thetworesultsareincomparable.

Theorem 1.6. LetG be ak-connected graph of ordern≥ k3_{+ k}2_{+ k + 3,} _where_k_{≥ 1.}

If ρ(G)> n− k − 2−_n1,then G istraceable, unlessG= G1_n,k.

Similar remarks pertain to the comparison of the abovetheorem and Theorem 1.2. The exceptional graph G1_n,k in the above theorem is precisely the k-connected graph

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Kk∨ (Kn−2k−1+ (k + 1)K1) thatwasexcludedintheconclusionof Theorem1.2. The conditionsonρ(G) inthetworesultsareincomparable,inthesensethatk inourresult refersto k-connectedgraphs,whereask inTheorem 1.2referstographswith minimum degreeatleastk.

Therest of the paper isorganized as follows. InSection 2, we will givesome useful techniquesand lemmasthatwillbe usedinourproofs. InSection3,wepresentseveral othernecessary lemmas,togetherwithourproofsofTheorems1.5 and1.6.

2. Preliminaries

We start this section by introducing a techniquebased on the concept of equitable partitions.Let M be asymmetric real n× n matrix. Therows and columns of M are

indexed byX = {1,. . . ,n}.Supposeπ = {X1,. . . ,Xm} is a partitionof X.Let M be partitionedaccordingto{X1,. . . ,Xm},i.e.,

M = ⎛ ⎜ ⎝ M11 . . . M1m .. . ... Mm1 . . . Mmm ⎞ ⎟ ⎠ ,

where Mij denotes theblock of M formed by therows in Xi and the columns in Xj. Let bij = 1

T

Mij1

|Xi| , i.e., the average row sum of Mij, where 1 is the column vector (of

thecorrectdimension)withallentries equalto 1.ThenthematrixM/π = (bij)m×m is called the quotientmatrix of M . If therow sumof eachblock Mij is aconstant, then thepartitioniscalled equitable.

The following lemma shows that equitable partitions can provide a simple way to calculateeigenvalues.

Lemma 2.1 ([8]).Let G be a graph, and let π be an equitable partition of G. Then ρ(G)= ρ(A(G))= ρ(A(G)/π).

TheconceptoftheclosureofagraphwasintroducedbyBondyandChvátal[2].The

k-closureofagraphG isdenotedbyclk(G),andistheuniquegraphthatcanbeobtained

from G by recursivelyjoiningnonadjacent vertices withdegree sumat leastk until no suchpairremainsinthecurrent graph.Whenk = n, weusecl(G) insteadof cln(G).A graphiscalled closedifG= cl(G).

Lemma2.2([2]).A graph G ishamiltonianif andonlyif cl(G) ishamiltonian.

Thenextlemmais oneof thegeneralizationsofDirac’stheorem dueto Chvátaland Erdős.

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Sinceweconsiderhamiltonicity(andtraceability)ofk-connectedgraphs,wecanmake use of Lemma2.3 (and its counterpart for traceable graphs) ifwe are ableto use the conditions on ρ(G) to obtain upper bounds on the independence number of G in our proofs. This is the main diﬀerence between our approachand the appliedmethods in [15] and[10].

Wewill alsofrequentlyusethefollowing knownlemmasinvolving ρ(G).

Lemma 2.4 ([4,8]). LetG be aconnected graph. If H is asubgraph of G, thenρ(H) ≤

ρ(G), withstrict inequalityin caseH isaproper subgraphofG.

Lemma 2.5([14]). LetG beagraph on n vertices andm edgeswith minimum degree δ.

Then ρ(G)≤ δ−1₂ +

2m− nδ +(δ+1)₄ 2.

Inconjunction withLemma2.5,we alsousethefollowing property.

Lemma 2.6 ([9,14]). Fornonnegative integers p and q with 2q≤ p(p− 1) and 0≤ x ≤

p− 1, thefunctionf (x)= x−1₂ +

2q− px +(x+1)₄ 2 isdecreasingwith respecttox.

3. Theproofsofourresults

In this section, we ﬁrst give two results involving alower bound on the numberof edges, the ﬁrst of which is known and implies a lower bound on the clique number. This is then used inthesecond result to characterizeexceptional, i.e., nonhamiltonian

k-connected graphsmeeting the conditions, which is oneof the key ingredients of our

proofofTheorem1.5.Theproofiscompletedbydeterminingupperboundsonρ(G) for

(subgraphsof)theexceptionalgraphs,thecontentofLemma3.3below.Forourproofof Theorem 1.6,we takeasimilarapproachandneedacounterpartofTheorem 3.2below ontraceability.

From Lemma 1in[10],wehavethefollowing lemma.

Lemma 3.1.Let G be a closed graph of order n ≥ 6k + 11, where k ≥ 0. If e(G) >

n− k − 2

2

+ (k + 2)2,thenω(G)≥ n− k − 1.

Using theabove,wecanprovethefollowingstructure theorem.

Theorem 3.2. Let G be a k-connected graph of order n ≥ 6k + 11, where k ≥

2. If e(G) >

n− k − 2

2

+ (k + 2)2_, _then _{G is} _hamiltonian _unless _cl(G) _∈

{G1

n,k,G2n,k,G3n,k,G4n,k,G5n,k}.

Proof. LetH = cl(G).IfH ishamiltonian,thensoisG byLemma2.2.Now,weassume thatH isnothamiltonian.NotethatH isk-connectedande(H)≥ e(G).ByLemma3.1,

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wehaveω(H)≥ n−k −1.Weclaimthatω(H)≤ n−k.Infact,ifω(H)≥ n−k +1,then

α(H)≤ k. Then, byLemma2.3,H is hamiltonian, whichcontradicts ourassumption.

Notethatδ(G)≥ κ(G)≥ k.Nextwe distinguishtwocases.

Case 1. ω(H)= n− k.

Inthis case,we have α(H)= k + 1. Set V (H) = X∪ Y , where X∩ Y = ∅, |X| = k,

|Y |= n− k,H[Y ]= Kn−k andX togetherwithonevertex inY ,sayw,isamaximum

independentset inH.Note thatδ(H)≥ κ(H)≥ k.From thefactthatH isclosed and

dH(w)= n− k − 1,wehavedH(x)= k foreveryx∈ X.LetF ⊆ Y betheneighborsof

X inY .Obviouslyw /∈ F .Theneveryvertex inF isadjacenttoeveryvertexinX,and hence|F |= k.SoH = Kk∨ (Kn−2k+ kK1)= G1n,k.

Case 2. ω(H)= n− k − 1.

Inthiscase,wehaveα(H)= k + 1 ork + 2.Wediscussthese twosubcasesseparately. Subcase 2.1 α(H)= k + 1.

Weﬁrstconsider thesituationthatV (H) canbedividedintotwosubsets,i.e.,V (H)=

X∪ Y , such thatX ∩ Y = ∅, |X| = k + 1, |Y | = n− k − 1, H[X] = (k + 1)K1 and

H[Y ]= K_n−k−1.Inthissituation,eachvertexofY mustbeadjacenttosomevertexin

X;otherwiseα(H)= k + 2.Set Y = Y1∪ Y2 suchthateachvertex of Y1 hasonlyone neighborinX andeach vertex ofY2 hasat leasttwo neighborsinX.Recall thatH is closed.IfY1=∅,thenH = Kn−k−1∨ (k + 1)K1, whichishamiltonian,acontradiction. IfY2=∅,thenobviouslyH hasaHamiltoncycle,acontradiction.Next,letY1= ∅ and

Y2= ∅.ThendH(y)= n− k − 1 fory∈ Y1,andY2isadjacenttoX sincedH(y)≥ n− k

fory∈ Y2.DenotebyX1 theneighborsetofY1 inX,andletX2= X\ X1.If|X1|≥ 2, thendH(x)= k forx∈ X1;otherwise Y1 will beadjacent toX1, andeachvertex ofY1 hasmorethanoneneighborinX.Then wehave|Y2|≤ k − 1 andhencedH(x)≤ k − 1 forx∈ X2,acontradiction.If|X1|= 1,thenthevertexx∈ X1isadjacentto Y ,which meansω(H)= n− k, acontradiction.

TheothersituationisthatV (H)= X∪ Y ∪ {v},suchthatX∩ Y ∩ {v}=∅,|X|= k,

|Y | = n− k − 1, H[Y ] = Kn−k−1, and X together with onevertex of Y , say w, is a

maximumindependentsetinH.Weusethesamenotationsasintheﬁrstsituation.We deducethat v is adjacentto Y2, andthatv must be adjacentto at least onevertex in

X;otherwiseα(H)= k + 2.

We ﬁrst assume v is also adjacent to Y \ {Y1∪ Y2}, that is to say, all possible w have degreen− k − 1. Then dH(x) = k for all x ∈ X. In this case, we have Y1 = ∅; otherwise v is adjacent to Y and ω(H) = n− k, and hence X1 = ∅. If X2 = ∅, then

|Y2|= k− 1,v isadjacenttoX2,andeachvertexofX1hasexactlyoneneighborinY1. Hence|Y1|=|X1|≤ k − 1.Suppose|X1|= a≥ 1 and|X2|= b≥ 1,wherea+ b= k.We useG1todenotethegraphwhenv hasnoneighborsinY\(Y1∪Y2) (see theleftgraphof Fig.2).Obviously,wehaveG1⊆ H.LabeltheverticesofG1asx11,. . . ,x1a;x21,. . . ,x2b;

y11,. . . ,y1a;y21,. . . ,y2a,y31,. . . ,y3,b−1;y41,. . . ,y4c (see theleftgraphofFig.2),where

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Fig. 2. The graphs G1and G2from the proof.

y31x21· · · y3,b−1x2,b−1vx2b and Q4 = y2,a−1y2ay14· · · y1ay41· · · y4cy11. Then

4i=1Qi is

aHamilton cycle.HenceH is alsohamiltonian,acontradiction.If X2=∅,then|Y2|≤

k− 2. Indeed, if |Y2| ≥ k − 1, then either v has no neighbors in X1 or the vertices in X1 have degree more than k, both leading to a contradiction. Suppose |Y2| = t. Then each x ∈ X1 \ NH[X1](v) has k− t neighbors in Y1, and each x ∈ NH[X1](v)

has k− t− 1 neighbors in Y1. When t = k− 2, we use G2 to denote the graph if v

has no neighborsin Y \ (Y1∪ Y2) (see the right graphof Fig. 2). Obviously, G2 ⊆ H. Label the vertices of G2 as x11,. . . ,x1a; x21,. . . ,x2b; y111 ,y112 ,. . . ,y11a,y1a2 ; y121,. . . ,y2b1;

y31,. . . ,y3,b−2,y41,. . . ,y4a;y51,. . . ,y5c,wherea≥ 1,b≥ 1,a+ b= k andc= n− 3k −

a+ 1.LetQ1= y111x11y112 · · · y11ax1ay1a2 ,Q2= y121x21y31· · · x2,b−2y3,b−2x2,b−1vx2by2b1 and

Q3 = y122· · · y2,b−11 y41· · · y4ay51· · · y5cy111 (see the right graphofFig.2).Then

3i=1Qi

isaHamiltoncycleofG2.HenceH isalsohamiltonian,acontradiction.Whent≤ k −3, we have thateach vertex of X1 has at least two neighbors in Y1. Obviously, H has a Hamilton cycle,acontradiction.

Next,weassumethatthereexistsavertexw suchthatitsdegreeisn− k − 2,i.e., v

is notadjacentto w.Inthiscase,dH(x)= k or k + 1 forx∈ X.

If Y1 = ∅, then X1 = ∅. If dH(x) = k + 1 for all x ∈ X2, then |Y2| = k, and v is adjacent to X2. Then we can easily ﬁnd a Hamilton cycle inH, a contradiction. If

dH(x)= k forallx∈ X2,then|Y2|= k− 1 (k ≥ 2) andagainv isadjacenttoX.Hence

dH(v) ≥ 2k − 1. We see that v must have at least one neighbor in Y \ Y2; otherwise

κ(H)= k− 1.NowdH(v)≥ 2k anddH(v)+ dH(w)≥ 2k + n− k − 2= n+ k− 2≥ n,

whichmeansv isadjacenttoY \ Y2,acontradiction.IfsomeverticesinX2havedegree

k and someverticesinX2 havedegreek + 1,then|Y2|= k,andtheverticesinX2with

degree k + 1 are adjacent to v. In this case, ifv has aneighbor inY \ Y2 or v hasat least twoneighborsinX,thenH isobviously hamiltonian,acontradiction.Ifv hasno neighborinY \ Y2and hasonlyoneneighborinX,thenH = G2n,k.

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IfY2=∅,itiseasytoseethatH ishamiltonian,acontradiction.

IfY1= ∅ andY2= ∅,when X2= ∅,then dH(x)= k forx∈ X2 sincetheverticesin

X2 have noneighborsinY1. If |Y2|= k, then ontheonehand, there isno neighborof

v inX2;otherwiseits neighborwillbe adjacentto Y1. Ontheother hand,v cannotbe adjacenttoanyx∈ X1;otherwisedH(x)= k + 2,acontradiction.Therefore,|Y2|= k−1 andhence,X2isadjacenttov.Inthiscase,everyvertexinX1hasaone-to-oneneighbor inY1.Also,v mustbeadjacenttoatleastonevertexinY\(Y1∪Y2);otherwiseY2willbe acutsetofH.NowweconcludethatG1⊆ H.Bytheformerproof,weobtainthatH is hamiltonian,acontradiction.WhenX2=∅,thenforx∈ X1,wehavedH(x)= k.Wecan obtainthat|Y2|≤ k − 2.Let|Y2|= t.DenotebyX11= NH[X1](v) andX12= X1\ X11.

Then each vertex of X11 has k− t− 1 neighbors in Y1, and each vertex of X12 has

k− t neighborsinY1.Obviously, whent≤ k − 3,everyvertexinX1 hasmorethantwo

neighbors in Y1, and it is easy to ﬁnd a Hamilton cycle inH, a contradiction. When

t= k− 2,weobtainthatG2⊆ H.Then,bytheformerproof, H isalso hamiltonian,a contradiction.

Subcase 2.2 α(H)= k + 2.

SetV (H)= X∪ Y ,where X∩ Y = ∅,|X|= k + 1, |Y |= n− k − 1,H[Y ]= K_n−k−1,

andX togetherwithonevertexinY ,sayw,isamaximumindependentsetinH.Since

dH(w)= n− k − 2,weknowthatdH(x)= k ork + 1 forx∈ X.LetX1⊆ X be theset,

eachvertexofwhichhasdegreek,andletX2= X\ X1 betheset,eachvertexofwhich hasdegreek + 1. DenotebyY1theneighborsofX1 inY , andY2= NH[Y ](X2)\ Y1.

Theﬁrstsituationweconsider isthatX1=∅.ThenX isadjacentto Y2, and|Y2|=

k + 1.HenceH = Kk+1∨ (Kn−2k−2+ (k + 1)K1)= G5n,k.

The second situation we consider is that X1 = ∅ and X2 = ∅. Firstly, Y1 is ad-jacent to X2 since dH(y) ≥ n− k − 1 for y ∈ Y1. So, now every vertex of Y1 has at least two neighbors in X, and hence dH(y) ≥ n− k for y ∈ Y1. Then every ver-tex inX1 is adjacent to every vertex in Y1, and hence |Y1| = k. Then every x ∈ X2 has a one-to-one neighbor in Y2, hence |X2| = |Y2|. When |X2| = |Y2| = 2, label thevertices of H asx1,x2,. . . ,xk+1;y0,y1,. . . ,yk+1;y11,y12,. . . ,y1,n−2k−3 (see Fig. 3 (1)).Let Q1= y0xkyky11y12. . . y1,n−2k−3yk+1xk+1,Q2 = y1x1y2x2. . . yk−1xk−1y0. Ob-viously P = Q1

Q2 is a Hamilton cycle, a contradiction. If |X2| = |Y2| ≥ 3, label the vertices of H as x11,. . . ,x1a; x21,. . . ,x2b; y11,. . . ,y1a, y1,a+1,. . . ,y1,a+b−1;

y21,. . . ,y2b;y31,. . . ,y3,n−2k−b−1(see Fig.3(2)),where |X1|= a, |X2|= b≥ 3,a+ b=

k + 1. Let Q1 = y11x11. . . y1ax1a. If b is even, let Q2i = y1,a+ix2,2i−1y2,2i−1y2,2ix2,2i

(1 ≤ i ≤ ₂b) and Q3 = y1,b

2+1. . . y1,a+b−1y31. . . y3,n−2k−b−1. Then we obtain a

Hamilton cycle Q1

(

b/2i=1Q2i)

Q3y11, a contradiction. If b is odd, let Q3 =

y_1,b−1

2 +1x2by2by1,b−12 +2. . . y1,a+b−1y31. . . y3,n−2k−b−1.Thenwe obtainaHamilton cycle

Q1

(

(bi=1−1)/2Q2i)

Q3y11, acontradiction.When|X2|=|Y2|= 1,H = G4n,k. The thirdsituation we consider is that X2 =∅. Then Y2 = ∅.Set Y1 = Y11∪ Y12, whereY11isthesetofverticeswithonlyoneneighborinX,andY12isthesetofvertices withat leasttwoneighborsinX.LetY3= Y \ Y1,i.e., Y3 istheset ofvertices withno

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Fig. 3. The graph H from the proof.

neighborinX.Obviouslyw∈ Y3,dH(y)= n− k − 1 fory∈ Y11 anddH(y)≥ n− k for

y ∈ Y12.

If Y11 = ∅, then it is easy to see that X is adjacent to Y12, |Y12| = k, and hence

H = Kk∨ (Kn−2k−1+ (k + 1)K1)= G3n,k.

IfY12=∅,thenobviouslyH ishamiltonian,acontradiction.

If Y11 = ∅ and Y12 = ∅, then |Y11|+|Y12| ≥ k + 1, and X is adjacent to Y12. Let

|Y11|= a.Then1≤ a≤ k+1.If1≤ a≤ k,then|Y12|= k andeachvertexinNH[X](Y11)

hasdegreek + 1,acontradiction.Soa= k + 1,i.e., everyvertex inX has aone-to-one neighborinY11, whichleadsto |Y12|= k− 1.Ifk + 1 is even,then labeltheverticesof

H as x1,x2,. . . ,xk+1; y11,y12,. . . ,y1,k+1; y21,. . . ,y_2,k+1

2 ,y2,

k+3

2 ,. . . ,y2,k−1;y31,. . . ,y3b

(see Fig.4(3)),whereb= n−3k−1.LetQi= y1,2i−1x2i−1y2,ix2iy1,2i(i= 1,2,. . . ,k+1₂ ) and let Q= y_2,k+3

2 . . . y2,k−1y31y32. . . y3b. Then (

k+1

2

i=1 Qi)

Qy11 isaHamilton cycle inH,acontradiction.

If k + 1 is odd, label the vertices of H as x1,x2,. . . ,xk+1;y11,y12,. . . ,y1,k+1;

y21,. . . ,y2,k

2,y2,

k

2+1,. . . ,y2,k−1;y31,. . . ,y3b (see Fig. 4 (4)), where b = n− 3k − 1.

Let Q = y1,k+1xk+1y2,k 2+1y2, k 2+2. . . y2,k−1y31y32. . . y3b. Then (

k 2 i=1Qi)

Qy11 is a Hamilton cycleinH,acontradiction.

TocompleteourproofofTheorem1.5,wealsoneedthefollowingresultonthespectral radius.

Lemma 3.3. Let G be a graph of order n with minimum degree δ(G) ≥ k, where n ≥ k3_{+ k + 2 and}_k_{≥ 2.}

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Fig. 4. The graph H from the proof.

(ii) IfG∈ {G2

n,k,G3n,k,G4n,k,G5n,k},thenρ(G)< n− k − 1−n1.

Proof. (i)ForG1_n,k = Kk∨(Kn−2k+ kK1),letX bethesetofverticeswithdegreek,let

Y bethesetofneighborsofX,andletZ betheremainingsetofn−2k vertices.Assume that G is a proper subgraph obtained from G1

n,k by deleting one edge uv ∈ E(G1n,k). According to the proof of Theorem 1.6 in [15], we know that there are three possible cases,and thatthecase withu,v∈ Z yields thelargest spectral radius. It is suﬃcient toproveρ(G)< n− k − 1−_n1 forthiscase.

Let us consider the following partition π of V (G) in X1 = X, X2 = Y , X3 = Z\

{u,v},andX4={u,v}.Itiseasytocheckthatthispartitionisequitable,andthatthe correspondingadjacency matrixofthequotientmatrixofG isasfollows:

A(G/π) = ⎛ ⎜ ⎜ ⎜ ⎝ 0 k 0 0 k k− 1 n − 2k − 2 2 0 k n− 2k − 3 2 0 k n− 2k − 2 0 ⎞ ⎟ ⎟ ⎟ ⎠.

ByusingtheLaplaceexpansionforthecalculationofthedeterminant,weobtainthe followingcharacteristicpolynomialofA(G/π).

f1(x) = x4+ (k− n + 4)x3+ (−k2+ 3k− 3n + 7)x2

+ (2k− 2n + k2n− 3k2− 2k3+ 4)x + 2k2n− 4k2− 4k3.

Then we canobtain the following derivatives of f1(x) with respect to x, by standard analysis:

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f1(x) = 4x3+ 3(k− n + 4)x2+ 2(−k2+ 3k− 3n + 7)x + 2k − 2n + k2n− 3k2− 2k3+ 4,

f₁(2)(x) = 12x2+ 6(k− n + 4)x + 2(−k2+ 3k− 3n + 7),

f₁(3)(x) = 24x + 6(k− n + 4),

f₁(4)(x) = 24.

Bysubstitutionand simplebuttediouscalculations,weobtain

f1(n− k − 1 − 1 n) = n 2_{− (k}3_{+ k + 2)n +}k3+ k2− 3k + 2 n + 2k2_{− 2} n2 + 3k n3 + 1 n4 + k4− k3− 2k2+ 2k > n2− (k3+ k + 2)n + k4− k3− 2k2+ 2k = g1(n)≥ 0,

whereg1(x)= x2− (k3+ k + 2)x+ k4− k3− 2k2+ 2k.Forthelastinequality,obviously, the functiong1(x) isstrictly increasingwhenx≥ k3+ k + 2.Usingn≥ k3+ k + 2,we obtaing1(n)≥ g1(k3+ k + 2)= k4− k3− 2k2+ 2k≥ 0. f₁(n− k − 1 − 1 n) = n 3_{− 3kn}2_{+ (2k}2_{− 3)n −} 4k2− 7 n − 9k n2 − 4 n3 − k 3_{− k}2_{+ 9k}_{− 2} = g2(n)≥ g2(k3+ k + 2) = k9+ 6k6− k5+ 8k3− 3k2+ 6k− 4k 2_{− 7} k3_{+ k + 2} − 9k (k3_{+ k + 2)}2 − 4 (k3_{+ k + 2)}3 ≥ k9_{+ (6k}6_{− k}5_{) + (8k}3_{− 3k}2_{) + 6k}₋3 4− 1 8− 1 432 > 0, where g2(x)= x3− 3kx2+ (2k2− 3)x−4k 2₋₇ x − 9k x2 −_x43− k3− k2+ 9k− 2.Fortheﬁrst

inequality,theﬁrstderivativeofg2(x) isg2(x)= 3x2− 6kx+ 2k2− 3+4k

2₋₇

x2 +18k_x3 +12_x4.

Sinceg₂(x)≥ g₂(k3_{+ k + 2)}_{= 3k}6_{+ 12k}3_−k2_{+ 9}₊ 4k2−7

(k3_+k+2)2+_(k3_+k+2)18k 3+_(k3_+k+2)12 4 > 0

forx≥ k3_{+ k + 2,}_we_obtain_that_g

2(x) isstrictlyincreasingwhenx≥ k3+ k + 2.Using

n≥ k3_{+ k + 2,}_we_have_g 2(n)≥ g2(k3+ k + 2). f₁(2)(n− k − 1 − 1 n) = 6n 2_{− 12kn +}18k n + 12 n2 + 4k 2_{− 16} > 6n2− 12kn + 4k2− 16 = g3(n) > 0,

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where g3(x) = 6x2 − 12kx+ 4k2− 16. For the last inequality, it is easy to see that

g3(x) is strictly increasing when x ≥ k3 + k + 2. Using n ≥ k3 + k + 2, we obtain

g3(n)≥ g3(k3+ k + 2)= 6k6+ 24k3− 2k2+ 8> 0. f₁(3)(n− k − 1 − 1 n) = 6(3n− 4 n− 3k) > 0. f₁(4)(n− k − 1 − 1 n) = 24 > 0.

Hence,bytheFourier-BudanTheorem(see,e.g.,[17]),thereisnorootoff1(x) inthe interval[n− k − 1−_n1,+∞).Then byLemma2.4,allsubgraphs ofG1

n,k havespectral radiuslessthann− k − 1−_n1.

(ii)Sincetheproofsareverysimilar,weomitthedetails.

Proof of Theorem1.5. Notethatδ(G)≥ κ(G)≥ k. ByLemmas 2.5and2.6,wehave

n− k − 1 − 1 n < ρ(G)≤ k− 1 2 + 2e(G)− nk + (k + 1) 2 4 .

Therefore,whenn≥ k3_{+ k + 2,}_we_obtain

e(G) > 1 2[n 2_{− (2k + 1)n +} 3k + 1 n + 1 n2 + 2k 2_{+ k}_{− 2]} > n 2_{− (2k + 1)n + 2k}2_{+ k}_{− 2} 2 > n− k − 2 2 + (k + 2)2. By Theorem 3.2, G is hamiltonian or cl(G) ∈ {G1 n,k,G2n,k,G3n,k,G4n,k,G5n,k}. Since K_n−k ⊆ G1

n,k, by Lemma 2.4, we have ρ(G1n,k) > ρ(Kn−k) = n− k − 1. Combining theabovewithLemmas 2.4and3.3,we concludethatG= G1

n,k.

Next,weturn totraceable graphsandthecounterparts oftheaboveresultsthatwe needforourproofofTheorem1.6.

Theorem 3.4. Let G be a k-connected graph of order n ≥ 6k + 16, where k ≥ 1.

If e(G) >

n− k − 3

2

+ (k + 2)(k + 3), then G is traceable unless cln−1(G) ∈

{G1

n,k,G2n,k,G3n,k,G4n,k,G5n,k}.

Proof. Let G = G ∨ K1. Then we have κ(G) = κ(G) + 1 ≥ k + 1, n(G) =

n+ 1 = 6(k + 1)+ 11 and e(G) = e(G)+ n >

n− k − 3 2 + (k + 2)(k + 3)+ n = n− k − 2 2

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cln+1(G)∈ {G1n+1,k+1,G2n+1,k+1,G3n+1,k+1,G4n+1,k+1,G5n+1,k+1}.Hence,G istraceable unless cln−1(G)∈ {G1_n,k,G2_n,k,G3_n,k,G4_n,k,G5_n,k}.

Lemma 3.5. Let G be a graph of order n with minimum degree δ(G) ≥ k, where n ≥ k3_{+ k}2_{+ k + 3 and} _k_{≥ 1.}

(i) IfG isaproper subgraphof G1_n,k,thenρ(G)< n− k − 2−_n1.

(ii) If G∈ {G2_n,k,G3_n,k,G4_n,k,G5_n,k}, thenρ(G)< n− k − 2−_n1.

Proof. (i)ForG1_n,k= Kk∨(Kn−2k−1+ (k + 1)K1),letX bethesetofdegreek,Y bethe set ofneighborsofX,andZ bethesetofremainingn− 2k − 1 vertices.AssumethatG

is apropersubgraph obtainedfrom G1_n,k bydeletingoneedge uv∈ E(G1_n,k). Similarly as inthe proof ofLemma3.3, thecase u,v ∈ Z yieldsthe largest spectral radius. Itis suﬃcient toproveρ(G)< n− k − 2−_n1 forthis case.

Let us consider the following partition π of V (G) in X1 = X, X2 = Y , X3 = Z\

{u,v},andX4={u,v}.Itiseasytocheckthatthispartitionisequitable,andthatthe corresponding adjacencymatrixofG/π isas follows:

A(G/π) = ⎛ ⎜ ⎜ ⎜ ⎝ 0 k 0 0 k + 1 k− 1 n − 2k − 3 2 0 k n− 2k − 4 2 0 k n− 2k − 3 0 ⎞ ⎟ ⎟ ⎟ ⎠.

ByusingtheLaplaceexpansionforthecalculationofthedeterminants,weobtainthe characteristic polynomialofA(G/π):

f2(x) = x4+ (k− n + 5)x3+ (−k2+ 2k− 3n + 10)x2

+ (kn− 2n − 2k + k2n− 6k2− 2k3+ 6)x + 2kn− 6k + 2k2n− 10k2− 4k3.

Then we can obtain the following derivatives of f2(x) with respect to x, by standard analysis: f₂(x) = 4x3+ 3(k− n + 5)x2+ 2(−k2+ 2k− 3n + 10)x + kn − 2n − 2k + k2n − 6k2_{− 2k}3_{+ 6,} f₂(2)(x) = 12x2+ 6(k− n + 5)x + 2(−k2+ 2k− 3n + 10), f₂(3)(x) = 24x + 6(k− n + 5), f₂(4)(x) = 24.

Bysubstitutionand simplebuttediouscalculations,weobtain

f2(n− k − 2 − 1 n) = n 2_{− (k}3_{+ k}2_{+ k + 3)n +}k3+ 3k2 n + 2k2_{+ 5k + 1} n2 + 3k + 3 n3 + 1 n4

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+ k4+ k3− 2k2− k + 1

> n2− (k3+ k2+ k + 3)n + k4+ k3− 2k2− k + 1

= h1(n)≥ 0,

whereh1(x)= x2−(k3+k2+k+3)x+k4+k3−2k2−k+1.Forthelastinequality,obviously, thefunctionh1(x) isstrictlyincreasingwhenx≥ k3+k2+k +3.Sincen≥ k3+k2+k +3, weobtainf2(n− k − 2−_n1)> h1(n)≥ h1(k3+ k2+ k + 3)= k4+ k3− 2k2− k + 1≥ 0. f₂(n− k − 2 − 1 n) = n 3_{− 3(k + 1)n}2_{+ (2k}2_{+ 5k)n}₋4k2+ 10k− 1 n − 9(k + 1) n2 − 4 n3 − k3_{− 3k}2_{+ 6k + 6} = h2(n)≥ h2(k3+ k2+ k + 3) = k9+ 3k8+ 3k7+ 7k6+ 11k5+ 4k4+ 7k3+ 5k2+ 3k + 6 − 4k2+ 10k− 1 k3_{+ k}2_{+ k + 3}− 9k + 9 (k3_{+ k}2_{+ k + 3)}2 − 4 (k3_{+ k}2_{+ k + 3)}3 ≥ k9_{+ 3k}8_{+ 3k}7_{+ 7k}6_{+ 11k}5_{+ 4k}4_{+ 7k}3_{+ 5k}2_{+ 3k + 6} −13 6 − 1 2− 1 54 > 0, whereh2(x)= x3−3(k +1)x2+ (2k2+ 5k)x−4k 2_+10k₋₁ x − 9(k+1) x2 −_x43−k3−3k2+ 6k + 6.

Fortheﬁrst inequality,the ﬁrstderivativeofh2(x) ish2(x)= 3x2− 6(k + 1)x+ (2k + 5)k +4k2+10k−1 x2 + 18(k+1) x3 +_x124.Since h2(x)≥ h2(k3+ k2+ k + 3) = 3k6+ 6k5+ 3k4+ 12k3+ 11k2− k + 9 + 4k 2_{+ 10k}_{− 1} (k3_{+ k}2_{+ k + 3)}2 + 18k + 18 (k3_{+ k}2_{+ k + 3)}3 + 12 (k3_{+ k}2_{+ k + 3)}4 > 0,

wehaveh2(x) isstrictlyincreasingwhenx≥ k3+ k2+ k + 3.Sincen≥ k3+ k2+ k + 3, weobtainh2(n)≥ h2(k3+ k2+ k + 3). f₂(2)(n− k − 2 − 1 n) = 6n 2_{− (12k + 12)n +} 18k + 18 n + 12 n2 + 4k 2_{+ 10k}_{− 10} > 6n2− (12k + 12)n + 4k2+ 10k− 10 = h3(n) > 0.

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Forthelast inequality,leth3(x)= 6x2− (12k + 12)x+ 4k2+ 10k− 10.Itiseasytosee thath3(x) isstrictlyincreasingwhenx≥ k3+ k2+ k + 3.Sincen≥ k3+ k2+ k + 3,we obtainh3(n)≥ h3(k3+ k2+ k + 3)= 6k6+ 12k5+ 6k4+ 24k3+ 22k2− 2k + 8> 0. f₂(3)(n− k − 2 − 1 n) = 18n− 24 n − 18k − 18 > 0. f₂(4)(n− k − 2 − 1 n) = 24 > 0.

Hence,bytheFourier-BudanTheorem(see,e.g.,[17]),thereisnorootoff2(x) inthe interval [n− k − 2−_n1,+∞).ThenbyLemma2.4, allsubgraphsof G1_n,k havespectral radius lessthann− k − 2− 1

n.

(ii)Since theproofs aresimilar,weomitthedetails.

Proof of Theorem1.6. Notethatδ(G)≥ κ(G)≥ k.ByLemmas2.5and 2.6,wehave

n− k − 2 − 1 n < ρ(G)≤ k− 1 2 + 2e(G)− nk +(k + 1) 2 4 .

Therefore, whenn≥ k3_{+ k}2_{+ k + 3,}_we_obtain

e(G) > 1 2[n 2_{− (2k + 3)n +}3k + 3 n + 1 n2 + 2k 2_{+ 4k]} >n 2_{− (2k + 3)n + 2k}2_{+ 4k} 2 > n− k − 3 2 + (k + 2)(k + 3).

By Theorem 3.4, G is traceable or cl_n−1(G) ∈ {G1_n,k,G2_n,k,G3_n,k,G4_n,k,G5_n,k}. Since

K_n−k−1 ⊆ G1_n,k, by Lemma 2.4, we have ρ(G1_n,k) > ρ(K_n−k−1) = n− k − 2. Com-biningthis withLemmas2.4and 3.5,wegetG= G1_n,k.

Declaration ofcompetinginterest

Theauthors declarethattheyhavenoconﬂictofinterest. Acknowledgements

We thankthe anonymous referees fortheir suggestionsthathaveimprovedthe pre-sentationofourresultsandproofs.

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