Long paths, long cycles, and their relative length

Long Paths, Long Cycles,and Their Relative Length

Akira SaitoDEPARTMENT OF MATHEMATICS

NIHON UNIVERSITYSAKURAJOSUI 3-25-40

SETAGAYA-KU, TOKYO 156JAPAN

E-mail: [email protected]

Received March 12, 1996; revised August 19, 1998

Abstract: Let p(G) and c(G) be the order of a longest path and a longest cyclein a graph G, respectively. Let σ3(G) = min{degG x + degG y + degG z : {x,y, z} is an independent set of vertices of G}. Extending the result by Enomotoet al. (J Graph Th 20 (1995), 213–225) on the difference p(G) − c(G), we provethat a 2-connected graph G of order n satisfies (1) p(G) − c(G) ≤ 1 or p(G) ≥σ3(G) − 1, and (2) if σ3(G) ≤ n, then p(G) − c(G) ≤ n − σ3(G) + 1 or p(G) ≥σ3(G). Then, using the above result, we give a new lower bound for p(G). Thisbound corresponds to the bound on c(G) given by Bauer et al. (Discrete Math 79(1989/90), 59–70). c© 1999 John Wiley & Sons, Inc. J Graph Theory 30: 91–99, 1999

Keywords: longest path, longest cycle

1. INTRODUCTION

For graph-theoretic notation, we refer the reader to [6]. For a graph G with α(G) ≥3, we define σ3(G) by σ3(G) = min{degG x + degG y + degG z: {x, y, z} is anindependent set of vertices in G}. If α(G) ≤ 2, we define σ3(G) = 3|G| − 8.(Here 3|G| − 8 is the smallest number that σ3(G) with α(G) ≥ 3 cannot attain.)Let p(G) and c(G) be the order of a longest path and the order of a longest cycle

Contract grant sponsor: Grant in Aid for Scientific Research, Ministry of Education,Science, and Culture of Japan. Contract grant number: YSE(A) 07780286 (1995)

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/020091-09

92 JOURNAL OF GRAPH THEORY

in G, respectively. In this article, we study the difference p(G)− c(G). In general,this difference can be arbitrarily large. On the other hand, a graph G with smallp(G) − c(G) tends to have a long cycle. In particular, for a connected graph Ghaving a cycle, G is hamiltonian if and only if p(G) − c(G) = 0.

A 2-connected graph G with large σ3(G) satisfies p(G) − c(G) = 0.

Theorem A (Bondy [3]). Let G be a 2-connected graph of order n. If σ3(G) >32(n − 1), then G is hamiltonian.

The lower bound on σ3(G) in the above theorem is sharp. A complete bipartitegraph G = Km,m+1(m ≥ 2) has n = 2m + 1 and σ3(G) ≥ 3m = 3

2(n − 1), butG is not hamiltonian.

Recently Enomoto et al. [7] proved the following theorem.

Theorem B. Let G be a 2-connected graph of order n with σ3(G) ≥ s.(1) If s ≥ n + 2, then c(G) ≥ p(G) − 1.(2) If s ≥ n, then either c(G) ≥ p(G) − 1, or G is traceable.

According to the above theorem, a 2-connected graph G with σ3(G) ≥ |G| + 2may not be hamiltonian, but p(G)− c(G) is still small and G is ‘‘almost’’ hamilto-nian in this sense. Actually, the class of 2-connected graphs G with σ3(G) ≥ |G|+2has several nice properties, and a number of lower bounds on c(G) are given forthis class (e.g., see [1], [2], and [5]).

A 2-connected graph G with σ3(G) ≥ |G| may not satisfy p(G)− c(G) ≤ 1. Infact, we can easily construct infinitely many graphs G with |G|+1 ≥ σ3(G) ≥ |G|with p(G) − c(G) arbitrarily large (see [7]). However, (2) of the above theoremsays such graphs are traceable.

The purpose of this article is to continue the investigation along this line forgraphs G with σ3(G) < |G|. We prove the following theorem.

Theorem 1. Let G be a 2-connected graph of order n. Then(1) p(G) − c(G) ≤ 1 or p(G) ≥ σ3(G) − 1, and(2) if σ3(G) ≤ n, then p(G) − c(G) ≤ n − σ3(G) + 1 or p(G) ≥ σ3(G).

Theorem B (2) corresponds to the case σ3(G) = n in (2) of Theorem 1.According to Theorem 1, roughly speaking, for a 2-connected graph G with

large σ3(G) (but not exceeding |G|), either p(G) − c(G) is small or p(G) is large.If p(G) − c(G) is small, G tends to have a long cycle and, hence, G tends to havea long path. However, possibly p(G) − c(G) is large. If this happens, we can atleast say that p(G) is large, relative to σ3(G). We present our results on long pathsin Section 3.

2. PROOF OF THEOREM 1

First we give the notation that we use in the subsequent arguments. For a graphG and x ∈ V (G), we denote by NG(x) the neighborhood of x in G. Thus,degG x = |NG(x)|. For S ⊂ V (G), let NG(S) = ∪v∈SNG(v). We sometimesidentify a graph and its vertex set if there is no fear of confusion. For example, if H

LONG PATHS AND CYCLES 93

is a subgraph of G, we sometimes write NG(H) instead of NG(V (H)). When weconsider a path or a cycle, we always assign an orientation. Let P = x0x1 · · ·xm

be a path. Then we call x0 and xm the starting vertex and the terminal vertexof P , respectively, and define x+

i = xi+1 and x−i = xi−1. Furthermore, we

define x++i = xi+2. More generally, for a positive integer k we define x

(k)+i by

x(k)+i = xi+k. For S ⊂ V (P ), let S+ = {v+ : v ∈ S}, and for xi, xj ∈ V (P )

with i ≤ j we denote the subpath xixi+1 · · ·xj by xiP→

xj . The same path traversedin the opposite direction is denoted by xjP

←xi. We use the same notation for cycles.

The following two lemmas are used in the proof of Theorem 1.

Lemma 1. Let G be a graph and let P be a longest path in G. Let x and y be thestarting vertex and the terminal vertex of P, respectively. If c(G) ≤ p(G)−2, then{x, y, z} is an independent set of vertices, p(G) ≥ |NG(x)− ∪NG(y)+∪(NG(z)∩V (P ))|, and |NG(x)− ∪NG(y)+ ∪NG(z)| ≥ σ3(G) for each z ∈ V (G)−V (P ).

Proof. First note that NG(x)∪NG(y) ⊂ V (P ), since P is a longest path. If xy ∈E(G), then xP

→yx is a cycle of order p(G), which contradicts c(G) ≤ p(G) − 2.

Hence, {x, y, z} is an independent set. Since NG(x)− ∪ NG(y)+ ∪ (NG(z) ∩V (P )) ⊂ V (P ), we have p(G) ≥ |NG(x)− ∪ NG(y)+ ∪ (NG(z) ∩ V (P ))|.

If NG(x)− ∩ NG(y)+ 6= ∅, say u ∈ NG(x)− ∩ NG(y)+, then xu+P→

yu−P←

xis a cycle of order |V (P )| − 1 = p(G) − 1, a contradiction. Hence, we haveNG(x)− ∩ NG(y)+ = ∅. If NG(x)− ∩ NG(z) 6= ∅, say v ∈ NG(x)− ∩ NG(z),then zvP

←xv+P

→y is a path of order p(G)+1, again a contradiction. Thus, NG(x)−

∩ NG(z) = ∅. Similarly, we have NG(y)+ ∩ NG(z) = ∅. Therefore, NG(x)−,NG(y)+, and NG(z) are mutually disjoint, and

|NG(x)− ∪ NG(y)+ ∪ NG(z)| = |NG(x)−| + |NG(y)+| + |NG(z)|= |NG(x)| + |NG(y)| + |NG(z)|= deg x + deg y + deg z ≥ σ3(G).

We use another lemma, which appears in [9, Sect. 10, Problem 19].

Lemma C. Let G be a 2-connected graph and let x and y be distinct vertices inG. If degG v ≥ d for each v ∈ V (G) − {x, y}, then x and y are joined by a pathof order at least d + 1.

Proof of Theorem 1. Let P be the set of all longest paths in G. In this proof,whenever P is chosen from P , we always assume x is the starting vertex of P andy is the terminal vertex of P .

Assume that c(G) ≤ p(G) − 2. For P ∈ P and z ∈ V (G) − V (P ), defineR(z) = V (G)−(NG(x)−∪NG(y)+∪NG(z)), R0(z) = R(z)−V (P ) = (V (G)−V (P )) − NG(z), R1(z) = R(z) ∩ V (P ), and ri(z) = |Ri(z)| (i = 0, 1). ByLemma 1, n − |R(z)| ≥ σ3(G). If NG(z) ⊂ V (P ) for some P ∈ P and z ∈V (G)−V (P ), then Lemma 1 gives p(G) ≥ |NG(x)−∪NG(y)+∪NG(z)| ≥ σ3(G)


and, hence, we are done. So we assume that G−V (P ) contains no isolated vertices.Let H = G − V (P ).

First we prove (1). We consider two cases.Case 1. NG(H) ∩ xP

→x−

0 6= ∅ for some P ∈ P and x0 ∈ NG(x).In this case, choose P ∈ P, x0 ∈ NG(x), and v ∈ NG(H) ∩ xP

→x−

0 so thatvP→

x0 is as short as possible. By the choice of (P, x0, v), v 6= x−0 , and NG(H)∩v+

P→

x−0 = ∅.

Assume NG(y) ∩ v+P→

x−0 6= ∅, say u ∈ NG(y) ∩ v+P

→x−

0 . Let Q = u+P→

x0

xP→

uyP→

x+0 . Then Q ∈ P , and vQ

→u = vP

→u is shorter than vP

→x0. This contr-

adicts the choice of P . Therefore, NG(y) ∩ v+P→

x−0 = ∅.

Let a ∈ NG(v) ∩ V (H), and take a path Q′ in H starting from a so that Q′ is as

long as possible. Let b be the terminal vertex of Q′, and let P ′ = bQ←′

avP←

xx0P→

sy.Then NG(b) ⊂ V (P ′) − {b}, NG(x)− ⊂ (V (P ′) ∪ {x−

0 }) − {b}, and NG(y)+ ⊂(V (P ′) ∪ {v+}) − {b}. Using Lemma 1, with b substituted for z, we get

p(G) + 1 ≥ |(V (P ′) ∪ {x−0 , v+}) − {b}|

≥ |NG(x)− ∪ NG(y)+ ∪ NG(b)| ≥ σ3(G).

Case 2. NG(H) ∩ xP→

x−0 = ∅ for every P ∈ P and x0 ∈ NG(x).

By symmetry, we may assume that NG(H) ∩ y+0 P→

y = ∅ for every y0 ∈ NG(y).Choose P, x0 and y0 so that |xP

→x0| + |y0P

→y| is as large as possible. By the

assumption of the case, |xP→

x0| + |y0P→

y| ≤ |P | + 1 = p(G) + 1, NG(H) ∩V (P ) ⊂ x0P

→y0, and since G is 2-connected, |x0P

→y0| ≥ |NG(H) ∩ V (P )| ≥ 2.

Let H0 be a component of H . Then |H0| ≥ 2, since G − V (P ) contains noisolated vertices. Let z be a vertex in H0 with degH z = δ(H0). First supposeH0 is a block. Since G is 2-connected, there exist two independent edges au, bvwith a, b ∈ V (H0), u ∈ x0P

→y0, and v ∈ u+P

→y0. Choose such edges so that uP

→v

is as short as possible, noting that v = u+ implies the existence of a longer paththan P . By Lemma C there exists a path Q0 in H0 from a to b of order at leastδ(H0) + 1 = degH z + 1. (This is true even if H0 = K2.) By the choice of theedges ua and vb, u+P

→v− ⊂ R1(z) and, hence, |u+P

→v−| ≤ r1(z). On the other

hand,

r0(z) = |V (H) − NH(z)| = |V (H)| − degH z = n − p(G) − degH z.

Therefore, |Q0| ≥ n − p(G) − r0(z) + 1. Let P0 = xP→

uaQ→

0bvP→

y. Then

p(G) ≥ |P0| = |P | − |u+P→

v−| + |Q0|≥ p(G) − r1(z) + n − p(G) − r0(z) + 1= n − |R(z)| + 1 ≥ σ3(G) + 1.

Next, suppose that H0 is not a block. Let B0 be an endblock of H0, and let c0be the unique cutvertex of H0 contained in B0. Since G is 2-connected, thereexist two independent edges ua and vb with u, v ∈ x0P

→y0, a ∈ B0 − c0, and


b ∈ (V (H0) − B0) ∪ {c0}. We may assume v ∈ u+P→

y0. Choose such edges sothat uP

→v is as short as possible. Take a path Q1 in H0 from a to b so that Q1 is as

long as possible. By Lemma C,

|Q1| ≥ |aQ→

1c0|≥

{min{degB0

w : w ∈ B0 − {a, c0}} + 1 if V (B0) 6= {a, c0}2 if V (B0) = {a, c0},

and, hence, we have |Q1| ≥ δ(H0) + 1 = degH z + 1 in either case. Bythe choice of ua and vb, u+P

→v− ⊂ R1(z). Let P1 = xP

→uaQ→

1bvP→

y. SincedegH z = n − p(G) − r0(z),

p(G) ≥ |P1| = |P | − |u+P→

v−| + |Q1|≥ p(G) − r1(z) + n − p(G) − r0(z) + 1 ≥ n − |R(z)| + 1 ≥ σ3(G) + 1.

Next, we prove (2). Assume that c(G) ≤ p(G) + σ3(G) − n − 2. In Case 2 ofthe proof of (1), we always conclude that p(G) ≥ σ3(G). Therefore, we have onlyto deal with Case 1. Furthermore, in Case 1, if |v+P

→x−

0 | = 1 or v+ 6∈ NG(y)+,then |P ′| ≥ σ3(G). Hence, we may assume that |v+P

→x−

0 | ≥ 2 and v ∈ NG(y).Therefore, we may assume that x1 ∈ y+

1 P→

y for some P ∈ P, x1 ∈ NG(x), andy1 ∈ NG(y).

Now choose such P ∈ P, x1 ∈ NG(x), and y1 ∈ NG(y) so that y1P→

x1

is as short as possible. Since C ′ = xx1P→

yy1P←

x is a cycle, |C ′| ≤ p(G) +σ3(G) − n − 2. Therefore, |y+

1 P→

x−1 | ≥ n − σ3(G) + 2. Let z ∈ V (H). Since

r0(z) + r1(z) = |R(z)| ≤ n − σ3(G), and z ∈ R0(z), 0 ≤ r1(z) ≤ n −σ3(G) − 1. This implies n − σ3(G) ≥ 1 and |y+

1 P→

x−1 | ≥ 3. Hence, y++

1 P→

x−−1 is

well defined and |y++1 P

→x−−

1 | ≥ 1. By the choice of P, x1 and y1, y++1

P→

x−−1 ∩ (NG(x)− ∪ NG(y)+) = ∅. Let A = {y++

1 , y(3)+1 , . . . , y

(n−σ3(G)+1)+1 } ⊂

y++1 P

→x−−

1 . Since r1(z) ≤ n − σ3(G) − 1 < |A|, A 6⊂ R(z). And sinceA ∩ (NG(x)− ∪ NG(y)+) = ∅, A ∩ NG(z) 6= ∅.

Now choose y(k)+1 ∈ A ∩ NG(H) so that k is as small as possible. Let a ∈

NG(y(k)+1 ) ∩ V (H). Take a path Q2 in H starting from a so that Q2 is as long

as possible. Let b be its terminal vertex. Let P2 = bQ←

2ay(k)+1 P

→yy1P

←x. Then

NG(b) ⊂ V (P2) − {b}, NG(x)− ⊂ V (P2) − {b} (note that y(k)+1 6∈ NG(x)), and

NG(y)+ ⊂ (V (P2) ⊂ {y+1 }) − {b}. Again using Lemma 1, we obtain

p(G) ≥ |(V (P2) ∪ {y+1 }) − {b}|

≥ |NG(x)− ∪ NG(y)+ ∪ NG(b)| ≥ σ3(G).


3. LONG PATHS IN GRAPHS WITH LARGEDEGREE SUMS

In [2], Bauer et al. proved the following theorem.

Theorem D [2]. Let G be a 2-connected graph of order n. If σ3(G) ≥ n + 2,then c(G) ≥ min{n + 1

3σ3(G) − α(G), n}.Since p(G) ≥ c(G) + 1 for a connected nonhamiltonian graph G, we make the

following observation.

Corollary 1. Let G be a 2-connected graph of order n. If σ3(G) ≥ n + 2, thenp(G) ≥ min{n + 1

3σ3(G) − α(G) + 1, n}.A cycleC in a graphG is said to be dominating, if V (G)−V (C) is an independent

set in G. The essential part of the proof of Theorem D is that every longest cyclein a graph G satisfying the hypothesis of the theorem is a dominating cycle. Sincethis property does not hold for graphs G with smaller σ3(G), we can no longergive the same lower bound to c(G). (Actually, Theorem D is best possible withrespect to the condition on σ3(G). See [2].) On the other hand, we will show thatin Corollary 1 we can relax the condition on σ3(G) by two without changing theconclusion.

Theorem 2. Let G be a 2-connected graph of order n. If σ3(G) ≥ n, thenp(G) ≥ min{n + 1

3σ3(G) − α(G) + 1, n}.Using Theorem 1, we give a similar bound to p(G), which can be applied to

graphs G with σ3(G) < |G| and almost contains Theorem 2.

Theorem 3. Let G be a 2-connected graph of order n. Then p(G) ≥ min{n +13σ3(G) − α(G) + 1, σ3(G) − 1, n}.

Note that Theorem 3 is weaker than Theorem 2 only when σ3(G) = n.Theorems 2 and 3 are best possible in the following sense. Let G = Kp,q with

p ≥ 2 and q ≥ p + 1. Then |G| = p + q, σ3(G) = 3p and α(G) = q. Therefore,

min{|G| +

13σ3(G) − α(G) + 1, σ3(G) − 1, |G|

}= min{2p + 1, 3p − 1, p + q}= 2p + 1 = p(G).

Thus, the bound n+ 13σ3(G)−α(G)+1 in Theorems 2 and 3 cannot be improved.

Next, let G = Kp + (p + 2)K2. Then |G| = 3p + 4, σ3(G) = 3(p + 1), andα(G) = p + 2. Hence,

min{

|G| +13σ3(G) − α(G) + 1, σ3(G) − 1, |G|

}

= min{3p + 4, 3p + 2, 3p + 4} = 3p + 2 = p(G).

Thus, the bound σ3(G) − 1 in Theorem 3 cannot be improved.Before proving Theorems 2 and 3, we need two lemmas.

Lemma 2. Let G be a 2-connected graph with c(G) ≥ p(G) − 1. Then


(1) Every longest cycle is dominating.(2) For a longest cycle C and v ∈ V (G) − V (C), NC(v)+ ∪ (V (G) − V (C)) isindependent.

Proof. (1) Let C be a longest cycle in G and assume that C is not dominating.Then G−V (C) has an edge uv with NG(u)∩V (C) 6= ∅, say w ∈ NG(u)∩V (C).Then vuwC

→w− is a path of order c(G) + 2 ≥ p(G) + 1, a contradiction.

(2) By (1), V (G) − V (C) is independent. Assume that vw ∈ E(G) for someu, v ∈ V (G) − V (C) and w ∈ NC(u)+. Since C is a longest cycle, u 6= v. ThenvwC→

w−u is a path of order c(G) + 2 ≥ p(G) + 1, again a contradiction.

Lemma 3. Let G be a 2-connected graph of order n with σ3(G) ≥ n. If n−2 ≥c(G) ≥ p(G) − 1, then G has a longest cycle C and a vertex v ∈ V (G) − V (C)with degG v ≥ 1

3σ3(G).Although we give a proof of the above lemma, it essentially appears in [2, Outline

proof of Theorem 9]. Note that in Theorem 9 of [2] the assumption that a graphunder consideration is 1-tough is used. However, this assumption is used when thecase c(G) = |G| − 1 is considered, and it is not necessary here.

Proof of Lemma 3. For a longest cycle C in G, let m(C) = max{degG v : v ∈V (G) − V (C)}. Choose C so that m(C) is as large as possible.

Since c(G) ≥ p(G)−1, C is a dominating cycle by Lemma 2 (1). Therefore, thelemma easily follows if |G − V (C)| ≥ 3. Thus, we may assume |G − V (C)| = 2,say V (G) − V (C) = {x1, x2} with degG x1 ≥ degG x2. Assume that m(C) =degG x1 < 1

3σ3(G), and let y ∈ NC(x1)+. By Lemma 2 (2), NG(y) ⊂ V (C).Since C is a longest cycle, NG(y) ∩ NC(x1)+ = ∅.

Assume that NG(y) ∩ NC(x1)++ − {y+} 6= ∅, say v ∈ NG(y) ∩ NC(x1)++ −{y+}. Let C ′ = vC

→y−x1v

−−C←

yv. Then C ′ is a longest cycle. By Lemma 2 (2),{x1, x2, v

−} is an independent set, and, hence, degG x1 + degG x2 + degG v− ≥σ3(G). Since degG(x1) ≥ degG(x2), and degG(x1) ≥ degG(v−) by the choiceof C, we have degG(x1) ≥ 1

3σ3(G), a contradiction. Thus, we have NG(y) ∩NG(x1)++ − {y+} = ∅, and, therefore, degG y ≤ |C| − 2 degG x1 + 1. However,this implies that

σ3(G) ≤ degG x1 +degG x2 +degG y ≤ 2 degG x1 +degG y ≤ |C|+1 = n−1.

This is a contradiction, and the lemma follows.Now we prove Theorems 2 and 3.

Proof of Theorem 2. Assume that p(G) ≤ n − 1. Then by Theorem B (2),c(G) ≥ p(G) − 1. If c(G) = p(G), then G is hamiltonian and p(G) = n, acontradiction. Hence, c(G) = p(G) − 1 ≤ n − 2. By Lemma 3, G has a longestcycle C and a vertex v ∈ V (G) − V (C) with degG v ≥ 1

3σ3(G). By Lemma2 (2), NG(v)+ ∪ (V (G) − V (C)) is an independent set of order degG v + n −c(G) ≥ n − c(G) + 1

3σ3(G). Therefore, we have n − c(G) + 13σ3(G) ≤ α(G), or

c(G) ≥ n + 13σ3(G) − α(G). This implies p(G) ≥ n + 1

3σ3(G) − α(G) + 1.


Proof of Theorem 3. By Theorem 2, we may assume that σ3(G) ≤ n − 1.Assume that p(G) ≤ σ3(G) − 2. Then by Theorem 1, c(G) ≥ p(G) − 1. Thus, bythe same arguments as in the proof of Theorem 2, we have p(G) ≥ n + 1

3σ3(G) −α(G) + 1.

One of the referees points out the natural correspondence between the set oflongest paths in a graph G and the set of longest cycles in G+K1. If G is a connectedgraph of order at least two, then c(G+K1) = p(G)+1. This observation suggestsanother approach when we investigate the relation between c(G) and p(G). Forexample, we can give an alternative proof to Theorem 2, using Theorem D.

Alternative Proof of Theorem 2. Let G∗ = G + K1. Then c(G∗) = p(G) + 1.Furthermore, G is 2-connected, α(G∗) = α(G), |G∗| = n + 1, and σ3(G∗) =σ3(G) + 3 ≥ |G∗| + 2. Therefore, by Theorem D,

c(G∗) ≥ min{

|G∗| +13σ3(G∗) − α(G∗), |G∗|

}

= min{n +13σ3(G) − α(G) + 2, n + 1}.

Hence,

p(G) = c(G∗) − 1 ≥= min{n +13σ3(G) − α(G) + 1, n}.

ACKNOWLEDGMENTS

The author thanks Professor Hikoe Enomoto and Professor Yoshimi Egawa forstimulating discussions and important suggestions. He also thanks the referees fora number of comments and suggestions.

References

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[3] J. A. Bondy, Longest paths and cycles in graphs of high degree. ResearchReport CORR 80–16, University of Waterloo, Waterloo, Ontario, 1980.

[4] J. A. Bondy and S. C. Locke, Relative length of paths and cycles in 3-connected graphs. Discrete Math 33 (1981), 111–122.

[5] H. J. Broersma, J. van den Heuvel, and H. J. Veldman, Long cycles, degreesums and neighborhood unions. Discrete Math 121 (1993), 25–35.


[6] G. Chartrand and L. Lesniak, Graphs & digraphs, 2nd ed., Wadsworth &Brooks/Cole, Monterey, CA, 1986.

[7] H. Enomoto, J. van den Heuvel, A. Kaneko, and A. Saito, Relative length oflong paths and cycles in graphs with large degree sums. J Graph Theory 20(1995), 213–225.

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Long paths, long cycles, and their relative length