Degree sums and graphs that are not covered by two cycles

Degree Sums and Graphsthat are not Covered byTwo Cycles

Akira SaitoDEPARTMENT OF MATHEMATICS

NIHON UNIVERSITYSAKURAJOSUI 3-25-40

SETAGAYA-KU, TOKYO 156JAPAN

FAX: +81-3-5317-9431E-mail: [email protected]

Received April 8, 1996; revised September 2, 1997

Abstract: For a graph G, let σ3(G) = min{degGx+degGy +degGz: {x, y, z}is an independent set in G}. Enomoto et al. [Enowoto et al., J Graph Theory 20(1995), 419–422] have proved that the vertex set of a 2-connected graph G of ordern with σ3(G) ≥ n is covered by two cycles, edges or vertices. Extending theirresult, we characterize the graphs of order n with σ3(G) ≥ n − 1 whose vertexset is not covered by two cycles, edges, or vertices. c© 1999 John Wiley & Sons, Inc. J Graph

Theory 32: 51–61, 1999

Keywords:degree sum; cycle covering

1. INTRODUCTION

We use [3] for terminology and notation not defined in this article and consideronly finite simple graphs.

Contract grant sponsor: Ministry of Education, Science and Culture of JapanContract grant no.: YSE(A) 07780286 (1995)

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/010051-11

52 JOURNAL OF GRAPH THEORY

For a graphG and a positive integerk we defineσk(G) by

σk(G) =

min {∑v∈S degGv:S is an independent

set of cardinalityk} if α(G) ≥ k+∞ if α(G) ≤ k − 1.

Ore [7] has proved that a graphG with σ2(G) ≥ |G| is Hamiltonian.

Theorem A ([7]). Let G be a graph of ordern ≥ 3. If σ2(G) ≥ n, thenG isHamiltonian.

This theorem is sharp in the sense that there exist non-Hamiltonian 2-connectedgraphsG with σ2(G) = |G| − 1. However, such graphs are characterized.

Theorem B ([1, 5, 6, 8]). LetG be a2-connected graph of ordern. If σ2(G) ≥n − 1, then one of the following holds:

(1) G is Hamiltonian.(2) Km,m+1 ⊂ G ⊂ Km + (m + 1)K1, wherem = n−1

2 andn ≥ 5.

In this article, we consider a Hamiltonian cycle as a cycle that covers the vertexset of a graph. LetH1, . . . , Hk be subgraphs ofG. Then we shall say thatGis covered byH1, . . . , Hk if V (G) =

⋃ki=1 V (Hi). Note that we do not require

V (H1), . . . , V (Hk) to be disjoint. Under this terminology, Theorem A says thata graphG of ordern ≥ 3 is covered by a cycle ifσ2(G) ≥ n, and Theorem Bcharacterizes the 2-connected graphsG with σ2(G) = n − 1 that are not coveredby one cycle.

Enomoto et al. [4] have recently given a sufficient condition for a 2-connectedgraphG to be covered by two cycles in terms ofσ3(G).

Theorem C ([4]). Let G be a graph of ordern. If σ3(G) ≥ n, thenG can becovered by two cycles, edges, or vertices.

Consequently, a 2-connected graphG of ordern with σ3(G) ≥ n can be coveredby two cycles. This is sharp in the sense that there exist 2-connected graphsG withσ3(G) = |G| − 1 such thatG cannot be covered by two cycles, edges, or vertices.However, if we see Theorem A and Theorem B, we may be able to characterizesuch graphs.

Motivated by the above observations, we extend Theorem C just as Theorem Ais extended to Theorem B, and characterize the graphsG with σ3(G) = |G| − 1such thatG cannot be covered by two cycles, edges, or vertices.

An endblock of a graphG is a block ofG, which contains precisely one cutvertexof G.

2. MAIN RESULT

Theorem 1. Let G be a graph of ordern. If σ3(G) ≥ n − 1, then one of thefollowing holds:

TWO CYCLES COVERING VERTICES 53

(1) G is covered by two cycles, edges, or vertices inG.

(2) Km,2m+1 ⊂ G ⊂ Km + (2m + 1)K1, wherem = n−13 ≥ 2.

(3) G = K1 +(Ka ∪Kb ∪Kc) for some positive integersa, b, c with a+ b+ c =n − 1.

(4) G has exactly two blocksH1 and H2 such thatKk,k+1 ⊂ H1 ⊂ Kk +(k + 1)K1 and H2 = Kl for somek and l with n = 2k + l, k ≥ 2, and2 ≤ l ≤ k + 1. The cutvertex inG is a vertex of degreek + 1 or more inH1.

(5) G has precisely two endblocksH1 and H2 and precisely one blockH oforder at least three, which is not an endblock such that bothH1 andH2 arecomplete andH is a complete graph possibly except for the edgec1c2, wherec1 andc2 are the cutvertices ofG.

Before we prove Theorem 1, we introduce some additional notation. For a graphG andv ∈ V (G), we denote byNG(v) the neighborhood ofv in G. If H is asubgraph ofG, then we sometimes writeNH(x) instead ofNG(x) ∩ V (H). ForS ⊂ V (G), let NG(S) = ∪v∈SNG(v). When we consider a path or a cycle, wealways assign to it an orientation. LetP = x0x1 · · ·xm be a path. We callx0 andxm thestarting vertexand theterminal vertexof P , respectively. Ifu, v ∈ V (P )andu precedesv onP , thenu

→Pv denotes the sequence of vertices ofP from u to

v. The same vertices in reverse order are given byv←Pu. We will consideru

→Pv and

v←Pu as both paths and vertex sets. Ifu ∈ V (P ), thenu+(P ) denotes the successor

of u on P andu−(P ) its predecessor. If we do not need to specify the pathP , wesimply writeu+ andu−, respectively. ForS ⊂ V (P ) let S+ = {v+: v ∈ S} andS− = {v−: v ∈ S}. We use the same notation for cycles. We denote the order of alongest cycle and a longest path in a graphG by c(G) andp(G), respectively.

Before we prove Theorem 1, we prove six lemmas.

Lemma 2. LetP be a longest path in a graphG, and letx andy be its startingvertex and terminal vertex, respectively. Suppose that we chooseP, x0 ∈ NG(x),y0 ∈ NG(y), so that|x→Px0|+ |y0

→Py| is as large as possible. If |x→Px0|+ |y0

→Py| ≤

|P | + 1, then the following conditions hold:

(1) NG(u) ⊂ x→Px0 for eachu ∈ NG(x)− and NG(v) ⊂ y0

→Py for eachv ∈

NG(y)+.

(2) If G is 2-connected, thenx→Px−

0 /⊂ NG(x)− andy+0→Py /⊂ NG(y)+.

Proof. (1) If NG(u) /⊂ V (P ), sayu′ ∈ NG(u) − V (P ), thenu′u←Pxu+→Py is

a path that is longer thanP , a contradiction. Thus,NG(u) ⊂ V (P ). If NG(u) ∩x+

0→Py /= ∅, sayu′ ∈ NG(u) ∩ x+

0→Py, then letP ′ = u

←Pxu+→Py. ThenP ′ is a

longest path, and

|u→P′u′| + |y0→P′y| = |x→Pu′| + |y0

→Py| > |x→Px0| + |y0

→Py|.

This contradicts the choice of(P, x0, y0). Therefore, we haveNG(u) ⊂ x→Px0

for eachu ∈ NG(x)−. By the same argument, we haveNG(v) ⊂ y0→Py for each

v ∈ NG(y)+.


(2) If x→Px−

0 ⊂ NG(x)−, thenNG(u) ⊂ x→Px0 for eachu ∈ x

→Px−

0 by (1). Thisimplies thatG − x0 is disconnected, which contradicts the 2-connectedness ofG.Hence,x

→Px−

0 /⊂ NG(x)−. We havey+0→Py /⊂ NG(y)+ by the same argument.

The next one is a key lemma.

Lemma 3. Let G be a2-connected graph of ordern with σ3(G) ≥ n − 1. IfG is not covered by two cycles, then eitherc(G) ≥ p(G) − 1 or G has a Hamil-tonian path.

Proof. Assume thatG has no Hamiltonian path andc(G) ≤ p(G) − 2. LetP be a longest path, and letx andy be its starting vertex and terminal vertex,respectively. SinceP is a longest path,NG(x)∪NG(y) ⊂ V (P ). Letx0 ∈ NG(x)andy0 ∈ NG(y). Now choose(P, x0, y0) so that|x→Px0| + |y0

→Py| is as large as

possible.By the assumptionV (P ) /= V (G). LetH = G−V (P ) andz ∈ V (H). SinceP

is a longest path,{x, y, z} is an independent set. Assume thatNG(x)− ∩NG(y)+ /=∅, sayv ∈ NG(x)− ∩ NG(y)+. Thenx

→Pv−y

←Pv+x is a cycle of order|P | − 1 =

p(G) − 1. This contradicts the assumption. Assume thatNG(x)− ∩ NG(z) /= ∅,sayv ∈ NG(x)− ∩ NG(z). Thenzv

←Pxv+→Py is a path of orderp(G) + 1, again a

contradiction. We have a similar contradiction ifNG(y)+ ∩NG(z) /= ∅. Therefore,NG(x)−, NG(y)+, andNG(z) are mutually disjoint.

Sincez /∈ NG(x)− ∪ NG(y)+ ∪ NG(z), NG(x)− ∪ NG(y)+ ∪ NG(z) ⊂V (G) − {z}, and

n − 1 ≥ |NG(x)− ∪ NG(y)+ ∪ NG(z)| = |NG(x)−| + |NG(y)+| + |NG(z)|= |NG(x)| + |NG(y)| + |NG(z)| = degGx + degGy + degGz ≥ n − 1.

Thus, we haveNG(x)− ∪ NG(y)+ ∪ NG(z) = V (G) − {z} for everyz ∈ V (H).In particular,H is a complete graph.

We consider two cases.

Case 1. |x→Px0| + |y0→Py| ≤ p(G) + 1.

In this case,y0 ∈ x0→Py. By the choice of(P, x0, y0), x0

→Py0 ∩ (NG(x)− ∪

NG(y)+) = ∅. Thus,x0→Py0 ⊂ NG(z) for eachz ∈ V (H). Sincez cannot be

adjacent to two consecutive vertices inP , we have|x0→Py0| = 1, or x0 = y0. By

Lemma 2,x→Px−

0 /⊂ NG(x)− andy+0→Py /⊂ NG(y)+. Let a ∈ x

→Px−

0 − NG(x)−

andb ∈ y+→Py − NG(y)+. Since{a, b} ∩ (NG(x)− ∪ NG(y)+) = ∅, V (H) ⊂NG(a) ∩ NG(b).

If |H| ≥ 2, choose distinct verticesz1, z2 in V (H). If |H| = 1, letz1 = z2 be theunique vertex inH. In either case,H has a Hamiltonian pathQ from z1 to z2, sinceH is complete. By the above arguments,a ∈ NG(z1) andb ∈ NG(z2). Now letC1 = xx0y

←Pbz2

←Qz1a

←Px andC2 = a

→Pbz2

←Qz1a. ThenV (C1) ∪ V (C2) = V (G).

This contradicts the assumption.

Case 2. |x→Px0| + |y0→Py| ≥ p(G) + 2.


In this case,x0 ∈ y+0→Py. Let P1 = x0x

→Py0, P2 = x0

←Py0 andP3 = x0

→Pyy0.

ThenP1, P2, andP3 are three internally disjoint paths fromx0 to y0 with V (P ) =V (P1) ∪ V (P2) ∪ V (P3).

SinceG is 2-connected, there exist a pair of independent edgesz1a andz2b withz1, z2 ∈ V (H) anda, b ∈ V (P ) if |H| ≥ 2, and if |H| = 1, sayV (H) = {z},there exist a pair of distinct edgesza andzb with a, b ∈ V (P ). If |H| = 1, we putz1 = z2 = z. In either case,H has a Hamiltonian pathQ from z1 to z2.

If {a, b} ⊂ V (Pi) for somei ∈ {1, 2, 3}, we may assumei = 1 and b ∈a+(P1)→P1y0. Let C1 = x0

→P1az1

→Qz2b

→Py0

←P2x0 and C2 = x0

→P1y0

←P3x0. Then

V (G) = V (C1) ∪ V (C2), a contradiction. Hence, we may assumea ∈ x+(P1)0→

P1y−(P1)0 , andb ∈ x

+(P2)0

→P2y

−(P2)0 . Let

C1 = x0→P1az1

→Qz2b

→P2y0

←P3x0 and

C2 = x0→P2bz2

←Qz1a

→P1y0

←P3x0.

ThenV (G) = V (C1) ∪ V (C2). This is the final contradiction, and the lemmafollows.

From the inequalityc(G) ≥ p(G) − 1, we obtain several useful facts.

Lemma 4. LetG be a2-connected graph withc(G) ≥ p(G) − 1. Then:

(1) Every longest cycle is a dominating cycle.(2) For a longest cycleC and for everyu, v ∈ V (G) − V (C), NC(u)+ ∩

NC(v) = ∅.

Proof. (1) Assume a longest cycleC is not dominating. SinceG is connected,there exists an edgeuv ∈ E(G − V (C)) with NC(v) /= ∅, sayw ∈ NC(v). Thenuvw

→Cw− is a path of orderc(G) + 2 ≥ p(G) + 1. This is a contradiction.

(2) Assume thatNC(u)+ ∩ NC(v) /= ∅, sayw ∈ NC(u)+ ∩ NC(v). SinceC is alongest cycle,u /= v. Thenvw

→Cw−u is a path of orderc(G) + 2 ≥ p(G) + 1, a

contradiction.

Lemma 5. LetG be a graph and letU be an independent set ofG. Suppose that|NG(u) ∩ NG(v)| ≥ |U | − 1 for everyu, v ∈ U. If G has no cycle containingU,then there exists a setT with |T | = |U | − 1 such thatNG(u) ∩ NG(v) = T forevery pair of distinct verticesu, v ∈ U.

Proof. Let U = {u0, u1, . . . , ul−1} and letxi = (ui, ui+1)(0 ≤ i ≤ l−1) (suf-fices are counted modulol). Let X = {x0, x1, . . . , xl−1} andY =

⋃l−1i=0 NG(ui).

Construct a bipartite graphH with partite setsX andY by joining xi ∈ Xandy ∈ Y , if y ∈ NG(ui) ∩ NG(ui+1). If H has a matching{x0y0, x1y1, . . . ,xl−1yl−1} from X to Y , thenG has a cycleu0y0u1y1u2 · · ·ul−1yl−1u0. This con-tradicts the assumption. Therefore,H has no matching fromX to Y . By Hall’sTheorem|NH(X ′)| ≤ |X ′| − 1 for someX ′ ⊂ X. In particular,X ′ /= ∅. On theother hand, forxi ∈ X ′, NG(ui) ∩ NG(ui+1) ⊂ NH(X ′) and, by the assumption,

l − 1 ≤ |NG(ui) ∩ NG(ui+1)| ≤ |NH(X ′)| ≤ |X ′| − 1 ≤ |X| − 1 = l − 1.


Therefore, we haveX ′ = X, |NH(X)| = l − 1, andNG(ui) ∩ NG(ui+1) =NH(X) for eachi (0 ≤ i ≤ l − 1). Let T = NH(X).

By the above arguments, the lemma follows ifl ≤ 3. Supposel ≥ 4 andj > i + 1. Let u′

0 = ui, u′1 = uj , andu′

2 = uj−1, and defineu′3, . . . , u

′l−1, so that

{u′3, . . . , u

′l−1} = U − {ui, uj−1, uj}. Applying the same arguments as above to

{u′0, u

′1, . . . , u

′l−1} instead ofU , we have

NG(ui) ∩ NG(uj) = NG(u′0) ∩ NG(u′

1) = NG(u′1) ∩ NG(u′

2)

= NG(uj) ∩ NG(uj−1) = T.

Therefore, the lemma follows.

Lemma 6. LetG be a2-connected graph of ordern, and letc be a vertex inG. IfdegGx+degGy ≥ n for every pair of nonadjacent verticesx andy in V (G)−{c},thenG is Hamiltonian.

Proof. Let G′ be the Bondy–Chv́atal closure [2] ofG. ThenG′ − c is complete.Since degGc ≥ 2 by the assumption, degG′c ≥ 2 and, hence,G′ is Hamiltonian.Therefore,G is also Hamiltonian.

Lemma 7. Let G be a2-connected graph of ordern ≥ 3 and letc be a vertexin G. If degGx + degGy ≥ n − 1 for every pair of nonadjacent verticesx, y inV (G) − {c}, then at least one of the following holds:

(1) G is Hamiltonian.(2) G − c is Hamiltonian.(3) Kk,k+1 ⊂ G ⊂ Kk + (k + 1)K1 with k = n−1

2 ≥ 2, anddegGc ≥ k + 1.

Proof. Choose a cycleC containingc so thatC is as long as possible. Supposethat C is not a Hamiltonian cycle. LetH be a largest component ofG − V (C).For u, v ∈ V (H), we denote byuHv a path inH that starts atu and ends atv.Let NC(H) = {y1, . . . , yk}. We may assume thaty1, . . . , yk appear in the con-secutive order alongC. SinceG is 2-connected,k ≥ 2. Let zi ∈ NH(yi)(1 ≤i ≤ k). If y+

i y+j ∈ E(G) for somei and j, C ′ = y+

j

→CyiziHzjyj

←Cy+

i y+j is a

cycle withV (C) ( V (C ′). This contradicts the assumption. Hence,NC(H)+ ={y+

1 , . . . , y+k } is independent. Similarly,{y−

1 , . . . , y−k } is also independent. Since

k ≥ 2, we may assume thatc /∈ {y+1 , y−

2 }. Let N1(y+1 ) = NG−V (C)(y

+1 ) ∪

NC(y+1 )− and N2(y−

2 ) = NG−V (C)(y−2 ) ∪ NC(y−

2 )+. SinceNG−V (C)(y+1 ) ∩

NC(y+1 )− = ∅,

|N1(y+1 )| = |NG−V (C)(y

+1 )| + |NC(y+

1 )−| = |NG−V (C)(y+1 )| + |NC(y+

1 )|= |NG(y+

1 )| = degGy+1 .

Similarly,|N2(y−2 )| = degGy−

2 . Letx ∈ V (H). If NH(y+1 ) /= ∅, sayv ∈ NH(y+

1 ),theny+

1→Cy1z1Hvy+

1 is a cycle that containsc and is longer thanC, a contradiction.


Therefore,NH(y+1 ) = ∅. In particular,xy+

1 /∈ E(G) andNG−V (C)(x)∩N1(y+1 ) =

∅. SinceNC(H)+ is independent,NC(x) ∩ N1(y+1 ) = ∅. Therefore,NG(x) ∩

N1(y+1 ) = ∅. SinceNG(x) ∪ N1(y+

1 ) ⊂ V (G) − {x} andc /∈ {x, y+1 },

n − 1 = |V (G) − {x}| ≥ |NG(x) ∪ N1(y+1 )| = |NG(x)| + |N1(y+

1 )|= degGx + degGy+ ≥ n − 1.

Therefore, we have

(A) V (G) − {x} = NG(x) ∪ N1(y+1 )(disjoint) for eachx ∈ V (H).

By a similar argument, we haveV (G) − {x} = NG(x) ∪ N2(y−2 ) (disjoint) for

eachx ∈ V (H). In particular,N1(y+1 ) = N2(y−

2 ). If {u−, u} ⊂ N1(y+1 ) for

someu ∈ y++2→Cy−

1 , then{u, u+} ⊂ NC(y+1 ) and{u−−, u−} ⊂ NC(y−

2 ). Thus,y+1 u→Cy1z1Hz2y2

→Cu−y−

2←Cy+

1 is a cycle that containsc and is longer thanC. Thisis a contradiction. Therefore,{u−, u} /⊂ N1(y+

1 ) for eachu ∈ y++2→Cy−

1 . Sincey+

i

→Cy−

i+1 ⊂ N1(y+1 ) by (A), we have|y+

i

→Cy−

i+1| = 1, or equivalentlyy+i =

y−i+1, for eachi ≥ 2, where we consideryk+1 = y1. SinceNH(y+

1 ) = ∅ and{y1, . . . , yk} ∩ N1(y+

1 ) = ∅, V (H) − {x} = NH(x) and{y1, . . . , yk} ⊂ NG(x)for eachx ∈ V (H) by (A). Hence,H is a complete graph andNH(yi) = V (H)for eachi with 1 ≤ i ≤ k.

Suppose thatk ≥ 3 ory+2 /= c. Theny+

i /= c for somei ≥ 2. If |H| ≥ 2, then wemay assume thatzi /= zi+1 andyi+1

→Cyizizi+1yi+1 is a cycle that containsc and

is longer thanC, a contradiction. Thus,|H| = 1. SinceH is a largest componentin G − V (C), G − V (C) contains no edges. Furthermore, we can apply the sameargument as in the previous paragraphs, usingy+

i = y−i+1 instead ofy+

1 andy−2 ,

and obtainy+1 = y−

2 . Therefore, we have|C| = 2k. If G − V (C) has two ormore vertices, sayx1 andx2, then, by the maximality ofC, we have degGx1 ≤ kand degGx2 ≤ k and, hence, degGx1 + degGx2 ≤ 2k. On the other hand, sincen ≥ |C| + |{x1, x2}| ≥ 2k + 2 andc /∈ {x1, x2}, we have degGx1 + degGx2 ≥n − 1 ≥ 2k + 1. This is a contradiction. Therefore,|V (G) − V (C)| = 1, sayV (G)−V (C) = {x}. LetA = {x, y+

1 , . . . , y+k } andB = {y1, . . . , yk}. ThenA is

an independent set, and, sinceNG(y+i ) ⊂ B, degGy+

i ≤ k (1 ≤ i ≤ k). If c ∈ A,sayc = y+

i , thenyi+1→Cyixyi+1 is a Hamiltonian cycle ofG − c. Hence, we may

assumec ∈ B. Then degGx + degGy+i ≥ n − 1 = 2k, and, since degGx = k, we

have degGy+i ≥ k. Therefore, degGy+

i = k andNG(y+i ) = B for eachy+

i ∈ A.This impliesKk,k+1 ⊂ G ⊂ Kk + (k + 1)K1.

Now we may assume thatk = 2 andy+2 = c. Assume thatG−V (C) has a com-

ponentH ′ that is different fromH. Letx ∈ V (H). SinceV (G)−{x} = NG(x)∪N1(y+

1 ) andV (H ′) ∩ NG(x) = ∅, we haveV (H ′) ⊂ NG(y+1 ). SinceN1(y+

1 ) =N2(y−

2 ), we also haveV (H ′) ⊂ NG(y−2 ). Let x′ ∈ V (H ′). If y1 ∈ NG(x′), then

y+1→Cy1x

′y+1 is a cycle that containsc and is longer thanC, a contradiction. There-

fore,y1 /∈ NG(x′). Similarly, y2 /∈ NG(x′). SinceNG(H) ∩ NG(H ′) ⊂ {y1, y2},


we haveNG(x) ∩ NG(x′) = ∅. Hence, we have

degGx + degGx′ = |NG(x)| + |NG(x′)| = |NG(x) ∪ NG(x′)| ≤ n − 2.

Sincexx′ /∈ E(G) andc /∈ {x, x′}, this contradicts the assumption. Therefore,G − V (C) = H. SinceH is a complete graph, we may choosez1, z2 andz1Hz2,so thatz1Hz2 is a Hamiltonian path ofH. Theny2

→Cy1z1Hz2y2 is a Hamiltonian

cycle ofG − c.Now we are ready to prove Theorem 1.

Proof of Theorem 1. Assume thatG cannot be covered by two cycles, edges,or vertices. We consider several cases based on the connectivity ofG.

Case 1. G is 2-connected.We further consider two subcases.

Subcase 1.1.G has no Hamiltonian path.By Lemma 3,c(G) ≥ p(G) − 1. Let C be a longest cycle. By the assumption

of the case,C is not a Hamiltonian cycle. LetU = V (G) − V (C). By Lemma 4(1), U is an independent set. Moreover, by assumption,U cannot be covered bya cycle. SinceG is 2-connected, it follows that|U | ≥ 3. Let u, v, andw be threedistinct vertices inU . Then degGu + degGv + degGw ≥ σ3(G) ≥ n − 1. Since(NG(u) ∪ NG(v)) ∩ NG(w)+ = ∅ by Lemma 4 (2),

|NG(u) ∪ NG(v) ∪ NG(w)+| = |NG(u) ∪ NG(v)| + |NG(w)+|= degGu + degGv − |NG(u) ∩ NG(v)| + degGw

≥ n − |NG(u) ∩ NG(v)| − 1.

SinceNG(u) ∪ NG(v) ∪ NG(w)+ ⊂ V (C), it follows that|C| ≥ n − |NG(u) ∩NG(v)| − 1. Since|U | = n − |C|, we have

(B) |NG(u) ∩ NG(v)| ≥ |U | − 1 for each pair of distinct verticesu andv in U.

SinceU cannot be covered by a cycle, by Lemma 5 there exists a set of verticesT ⊂ V (C) with |T | = |U | − 1 such thatNG(u) ∩ NG(v) = T for every pair ofdistinct verticesu, v ∈ U . Since|NG(u) ∩ NG(v)| = |U | − 1, the equality holdsin (B) and, hence,NG(w)+ = V (C) − (NG(u) ∪ NG(v)). Applying the samearguments toNG(u), NG(v), andNG(w)−, we also haveNG(w)− = V (C) −(NG(u) ∪ NG(v)). Thus,NG(w)+ = NG(w)−. Sinceu, v, andw are chosenarbitrarily fromU , by Lemma 4 (2) we haveNG(v) = T for eachv ∈ V (C) andV (C) = T ∪ T+ (disjoint). Let |T | = m. Then|U | = m + 1. SinceU ∪ T+ isan independent set of order2m + 1 by Lemma 4 (2), we haveKm,2m+1 ⊂ G ⊂Km + (2m + 1)K1.

Subcase 1.2.G has a Hamiltonian path.Let P be a Hamiltonian path and letx andy be its starting vertex and terminal

vertex, respectively. Letx0 ∈ NG(x) andy0 ∈ NG(y). We chooseP, x0, y0 so that|x→Px0| + |y0

→Py| is as large as possible.


If |x→Px0| + |y0→Py| ≥ |P | = n, thenx0 ∈ y−

0→Py and, hence, the cyclesx

→Px0x

andy0→Pyy0 coverV (P ) = V (G). This is a contradiction. Thus,|x→Px0|+|y0

→Py| ≤

n − 1. This impliesy0 ∈ x++0→Py.

Let v ∈ x+0→Py−

0 . By the choice ofx0 andy0, {x, y, v} is an independent set.Furthermore, by Lemma 2 (1),NG(x)−, NG(y)+, andNG(v) are mutually disjoint.Hence,

|NG(x)− ∪ NG(y)+ ∪ NG(v)| = |NG(x)−| + |NG(y)+| + |NG(v)|= degGx + degGy + degGv ≥ n − 1.

Sincev /∈ NG(x)− ∪ NG(y)+ ∪ NG(v), it follows that NG(x)− ∪ NG(y)+ ∪NG(v) = V (G) − {v}. Sincex0

→Py0 ∩ (NG(x)− ∪ NG(y)+) = ∅, we have

x0→Py0 ⊂ NG(v) for eachv ∈ x+

0→Py−

0 . In particular,x+0→Py−

0 induces a com-plete graph.

By Lemma 2 (2),x→Px−

0 /⊂ NG(x)− andy+0→Py /⊂ NG(y)+. Let a ∈ x

→Px−

0 −NG(x)− and b ∈ y+

0→Py − NG(y)+. Since{a, b} ∩ (NG(x)− ∪ NG(y)+) =

∅, x+0→Py−

0 ⊂ NG(a) ∩ NG(b).If x+

0 /= y−0 , then x++

0 ∈ NG(a) and x+0 ∈ NG(b). Then x

→Pax++

0→Py0y←

Pbx+0 x0x and a

→Px+

0 b←Px++

0 a are two cycles coveringV (G), a contradiction.Hence, we havex+

0 = y−0 .

SinceG−x+0 is connected, there exists an edgecd with c ∈ x

→Px0 andd ∈ y0

→Py.

If {c, d} ∩ {x0, y0} = ∅, thenx→Pcd

→Pyy0x

+0 x0x andc

→Pdc are two cycles covering

V (G), a contradiction. Hence, we have{c, d} ∩ {x0, y0} /= ∅. By symmetry, wemay assumec = x0. If d /= y0, x

→Pax+

0 y0y←Pdx0x anda

→Px0d

←Px+

0 a are two cyclescoveringV (G). If d = y0, thenx

→Pax+

0 b→Pyy0x0x anda

→Px0y0

→Pbx+

0 a are twocycles coveringV (G). Therefore, we have a contradiction in either case. This isthe final contradiction, and the theorem follows in this case.

Case 2. G is connected, but not 2-connected.SinceG is not covered by two edges,|G| ≥ 3. Therefore,G has a cutvertex and

at least two blocks. We consider three subcases.

Subcase 2.1.G has three or more endblocks.Let H1, H2, andH3 be three distinct endblocks ofG. Choosexi ∈ V (Hi) so

thatxi is not a cutvertex ofG (1 ≤ i ≤ 3). Then{x1, x2, x3} is an independent setand

3∑i=1

degGxi ≤3∑

i=1

(|Hi| − 1) =3∑

i=1

|Hi| − 3.

Therefore, by assumption we have∑3

i=1 |Hi| − 3 ≥ n − 1, or∑3

i=1 |Hi| ≥ n + 2.This is possible only ifH1, H2, andH3 share a common cutvertex, sayw, NG(xi) =V (Hi) − {xi}(1 ≤ i ≤ 3) and

⋃3i=1 V (Hi) = V (G). Sincexi is an arbitrary

vertex inHi, which is not a cutvertex, eachHi is complete(1 ≤ i ≤ 3). Therefore,G = K1 + (Ka ∪ Kb ∪ Kc) with a, b, c > 0 anda + b + c = n − 1.

Subcase 2.2.G has exactly two blocks.


LetH1 andH2 be the two blocks ofG. Let|Hi| = ni (i = 1, 2). Then bothH1 andH2 are endblocks. Letc be the unique cutvertex ofG. SinceG is not covered by twocycles, edges, or vertices, we may assume that|H1| ≥ 3 andH1 is not Hamiltonian.Choosez ∈ V (H2) − {c} so that degGz = degH2

z is as small as possible. LetdegGz = n2 − t (t ≥ 1). If x andy are a pair of nonadjacent vertices inH1 − c,then, by assumption, degGx + degGy + degGz ≥ n − 1. Sincen = n1 + n2 − 1and degGz = n2 − t, degH1x+degH2

y = degGx+degGy ≥ n1 + t− 2. If t ≥ 2,then degH1x+degH1

y ≥ n1 for each pair of nonadjacent verticesx andy in H1−c,and, hence,H1 is Hamiltonian by Lemma 6, a contradiction. Therefore,t = 1 anddegH1x + degH2

y ≥ n1 − 1 for each pair of nonadjacent verticesx, y of H1 − c.In particular,H2 is a complete graph, which implies that|H1| ≥ 4 and neitherH1norH1−c is Hamiltonian. Therefore, by Lemma 7Kk,k+1 ⊂ H1 ⊂ Kk+(k+1)K1for somek ≥ 2 and degH1c ≥ k + 1. The degree sum condition yields2 ≤ l ≤k + 1.

Subcase 2.3.G has three or more blocks, butG has exactly two endblocks.Let H1 and H2 be the endblocks ofG. Let ci be the unique cutvertex ofG

contained inHi (i = 1, 2). Because of the assumption of this subcasec1 /= c2.Choosexi ∈ V (Hi) − {ci} so that degHixi = degGxi is as small as possible(i = 1, 2), and let|Hi| = ni. Then degGxi = degHi

xi ≤ ni − 1 (i = 1, 2). Ifx is a vertex inV (G) − (V (H1) ∪ V (H2)), then{x1, x2, x} is an independentset and, hence, degGx1 + degGx2 + degGx ≥ n − 1. On the other hand, sinceNG(x) ∩ (V (H1) ∪ V (H2)) ⊂ {c1, c2},

degGx1 + degGx2 + degGx ≤ n1 − 1 + n2 − 1 + (n − n1 − n2 + 1) = n − 1.

Therefore, degGx1 = n1 − 1, degGx2 = n2 − 1 and degGx = n − n1 − n2 + 1.Because of the minimality of degGx1 and degGx2, bothH1 andH2 are completeandNG(x) = V (G) − (V (H1) ∪ V (H2) ∪ {x}) ∪ {c1, c2} for everyx ∈ V (G) −(V (H1) ∪ V (H2)). Therefore,G has precisely three blocksH1, H2, andH, andH is a complete graph, possibly except for the edgec1c2. If V (G) = V (H1) ∪V (H2), then the degree sum condition implies that bothH1 andH2 are Hamiltonian(possibly one or both of them are complete), andG is covered by two cycles. Thisis a contradiction.

Case 3. G is disconnected.If G has three or more components, we can choosex1, x2, x3, so that they belong

to different components. Then∑3

i=1 degGxi ≤ n−3, a contradiction. Thus,G hasexactly two components, sayG1 andG2. SinceG is not covered by two cycles,edges, or vertices, we may assumeG1 is not Hamiltonian and|G1| ≥ 3. By TheoremA, G1 has a pair of nonadjacent verticesx, y with degGx + degGy ≤ |G1| − 1.Take a vertexz from G2. Then{x, y, z} is an independent set and

degGx + degGy + degGz ≤ |G1| − 1 + |G2| − 1 = n − 2.

This is again a contradiction.Now all the cases are checked and the proof is complete.


ACKNOWLEDGMENTS

The author thanks Professor Hikoe Enomoto for stimulating discussions and impor-tant suggestions. He is also grateful to the referees for their helpful comments. Thiswork is partly supported by the Grant in Aid for Scientific Research of the Min-istry of Education, Science and Culture of Japan under the grant number: YSE(A)07780286 (1995).

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Degree sums and graphs that are not covered by two cycles

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