Intersections of Cycles in $$k$$ k -Connected Graphs

Graphs and CombinatoricsDOI 10.1007/s00373-014-1427-z

ORIGINAL PAPER

Intersections of Cycles in k-Connected Graphs

Jessica Chen · Le Chen · Derrick Liu

Received: 19 July 2008 / Revised: 30 August 2012© Springer Japan 2014

Abstract Let k ≥ 2 be a positive integer and G be a k-connected graph. We provethat for any two cycles C and D in G there exist two cycles C∗ and D∗ such thatV (C) ∪ V (D) ⊆ V (C∗) ∪ V (D∗) and |V (C∗) ∩ V (D∗)| ≥ 1(

3√256+3)3/5 · k3/5.

Moreover, we conjecture that 1(3√256+3

)3/5 · k3/5 can be replaced by k and verify this

conjecture for k ≤ 6. For α > k, Fouquet and Jolivet conjectured that if a k-connectedgraph G with an independent number α has a cycle of length at least k(n + α − k)/α.Based on the proof techniques used by Manoussakis and by Chen, Hu and Wu, webelieved that resolving our conjecture will provide some help on tackling the Fouquet–Jolivet conjecture.

Keywords Longest cycle · k-connected graph · Common vertices

1 Introduction

The notation and terminology not defined in this paper can be found in the textbookof Bondy and Murty [2]. All graphs considered in this paper are simple graphs. Let

J. Chen (B)University of California, Berkeley, CA 94720, USAe-mail: [email protected]

L. Chen814 Highland Ridge Ave., Gaithersburg, MD 20878, USAe-mail: [email protected]

D. Liu9 Dogwood Drive, Plainsboro, NJ 08536, USAe-mail: [email protected]

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Graphs and Combinatorics

G be a graph. We use V (G) to denote the vertex set of G, E(G) the edge set ofG, |G| = |V (G)| the order of G, and ||G|| = |E(G)| the size of G. The connectivityκ(G) is the minimum number of the vertices whose removal disconnect G if G is notcomplete and k if G = Kk+1. A graph is k-connected if κ(G) ≥ k. Evidently, every 2-connected graph contains a cycle. The length �(C) of a cycle C is the number of verticesof C or equivalently the number of edges of C . A cycle C of G is called the longest if�(C) is maximum. Every two longest cycles in a 2-connected graph share at least twovertices. We expect that any two longest cycles share more vertices if the connectivityis higher. More specifically, the following conjecture is due to Scott Smith.

Conjecture 1.1 (Smith [1,9]) Let k ≥ 2 be a positive integer and let G be ak-connected graph. Then any two longest cycles in G meet in at least k vertices.

Grötchel [9] pointed out progress on Conjecture 1.1 are of interest, for instance, indesigning recursive algorithms (e.g. in conbinatorial optimization) in which in everystep a longest cycle is shrunk or deleted and one wants to know a bound on the lengthof the longest cycle in the resulting graph. The conjecture may have been verified up tok = 10 as indicated by Grötchel in [9]. Theorem 1.2(a) of [9] showed the conjecture istrue up to k = 6. Grötchel and Nemhauser [10] studied the properties of 2-connectedgraphs in which there are two longest cycles sharing exactly 2 vertices. Grötchel [9]studied the properties of two longest cycles sharing exactly k vertices for k = 3, 4,5. For general k, S. Burr and T. Zamfirescu proved that in a k-connected graph everytwo longest cycles share at least

√k − 1 vertices. Chen et al. [3] improved the lower

bound to k3/5

(3√256+3)3/5

.

In the paper, we propose the following conjecture and obtain a few results similarto these related to Smith’s Conjecture.

Conjecture 1.2 Let G be a k-connected graph, k ≥ 2, and let C and D be two cyclesof G. Then there exist two cycles C∗ and D∗ such that V (C∗)∪V (D∗) ⊇ V (C)∪V (D)

and |V (C∗) ∩ V (D∗)| ≥ k.

In Sect. 2, we give a short proof that Conjecture 1.2 implies Smith’s Conjecture.We note that it is not a mere straightforward generalization; for example, the newconjecture is related to the following result of Suil et al. [12], which was known as theFournier–Jolivet Conjecture and was open for more than 30 years.

Theorem 1.3 (Solution of the Fournier–Jolivet Conjecture) A k-connected graph oforder n with independence number α contains a cycle of length at least k(n+α−k)/α

if α > k.

In the previous version of this paper, we speculated that Conjecture 1.2 gives a proofof the Fournier–Jolivet Conjecture based on partial results obtained by Fournier [7,8],by Manoussakis [11], and by Chen et al. [5]. Suil et al. [12] completely confirmed theFournier–Jolivet Conjecture using an elegant proof bypassing Conjecture 1.2. Chen etal. [4] have recently proved that Conjecture 1.2 actually implies the Fournier–JolivetConjecture. Therefore, Conjecture 1.2 still remains open and meaningful.

In this paper, we denote by A := B to define B as A. We will use the notationG ⊆ Km,n to signify a bipartite graph in which one part has m vertices and the other

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part has n vertices. Let X be a path or a cycle in a graph. We will usually give anorientation to X . In this case, for any pair of vertices u and v ∈ V (X), we will usethe notation X [u, v] to signify the segment of X from u to v along the orientation (ifsuch a segment exists). The same segment with reversed orientation will be denotedby X−[v, u].

2 Related to the Smith Conjecture

Theorem 2.1 Conjecture 1.2 contains Conjecture 1.1.

Proof Assume that Conjecture 1.2 is true, and let G be a k-connected graph and Cand D be two longest cycles of G. We will show |V (C) ∩ V (D)| ≥ k.

By Conjecture 1.2, there are two cycles C∗ and D∗ such that V (C∗) ∪ V (D∗) ⊇V (C) ∪ V (D) and |V (C∗) ∩ V (D∗)| ≥ k. Since C and D are two longest cycles inG, then |V (C)| + |V (D)| ≥ |V (C∗)| + |V (D∗)|, that is

|V (C) ∪ V (D)| + |V (C) ∩ V (D)| ≥ |V (C∗) ∪ V (D∗)| + |V (C∗) ∩ V (D∗)|.

Since V (C) ∪ V (D) ⊆ V (C∗) ∪ V (D∗), then

|V (C) ∩ V (D)| ≥ |V (C∗) ∩ V (D∗)| ≥ k.

��

3 A Lower Bound

Theorem 3.1 Let G be a k-connected graph, k ≥ 2, and let C and D be two cycles ofG. Then there exist two cycles C∗ and D∗ such that V (C∗)∪ V (D∗) ⊇ V (C)∪ V (D)

and |V (C∗) ∩ V (D∗)| ≥ k3/5

(3√256+3)3/5

.

We will basically follow the proof strategy of [3]. In particular, we use the followingtwo lemmas which were proved in [3].

Lemma 3.2 Let G ⊆ Kn,n. Then G contains K3,257 if ||G|| ≥ 3√

256(n−2)n2/3 +2n.

Lemma 3.3 Let � be a set of n permutations of a sequence of S of 22n + 1 elements.Then there is a subsequence (a, b, c) of S on which each permutation σ ∈ � ismonotonic (that is, either σ(a) < σ(b) < σ(c) or σ(a) > σ(b) > σ(c) ).

Lemma 3.4 Let k be a positive integer and G be a nonhamiltonian k-connected graph.Then, for any cycle C of G, there exists a cycle C∗ such that V (C∗) ⊇ V (C) and|C∗| ≥ 2k.

Proof Since each cycle contains at least 3 vertices, Lemma 3.4 holds for k = 1. Weassume k ≥ 2. Among all cycles C∗ such that V (C∗) ⊇ V (C) (which includes Citself), let C∗ be one of them such that |C∗| is maximum. We will show that |C∗| ≥2k. Suppose the contrary: |C∗| < 2k. Since G is not hamiltonian, there is a vertex

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v ∈ V (G) − V (C∗). Using the connectivity of G, we obtain two paths P1 and P2from v to two consecutive vertices x and y of C∗. Then P1 ∪ P2 ∪ C∗ contains a cycleC∗∗ such that V (C∗∗) ⊃ V (C∗) ⊇ V (C) and |C∗∗| > |C∗|, a contradiction to themaximality of |C∗|. ��Lemma 3.5 (Dirac [6]) Let G be a 2-connected graph of minimum degree δ on nvertices, where n ≥ 3. Then G contains either a cycle of length at least 2δ or ahamiltonian cycle.

3.1 The Proof of Theorem 3.1

Let C∗ and D∗ be two cycles in G such that (1). V (C∗)∪V (D∗) ⊇ V (C)∪V (D), and(2)., subject to (1), |C∗|+|D∗| is maximum. Note that such two cycles C∗ and D∗ existbecause C∗ = C and D∗ = D satisfying (1). We will show that |V (C∗) ∩ V (D∗)| ≥ηk

35 where η = 1

(3√256+3)3/5

. Suppose the contrary: m := |V (C∗) ∩ V (D∗)| < ηk35 .

We will lead to a contradiction.Let A := {a1, a2, . . . , am} = {b1, b2, . . . , bm} = V (C∗) ∩ V (D∗), where

a1, a2, . . . , am are listed in the order along the orientation of C∗ while b1, b2, . . . , bm

are listed in the order along the orientation of D∗. Clearly, m < ηk3/5. For eachi = 1, 2, . . . , m, let Xi := C∗(ai , ai+1) and Yi := D∗(bi , bi+1) be the segments ofC∗ between ai and ai+1 and the segment of D∗ between bi and bi+1, respectively.Note that if ai and ai+1 (resp. bi and bi+1) are consecutive vertices along C∗ (resp.D∗), then Xi (resp. Yi ) is empty. We note C∗ − A = X1 ∪ X2 ∪ · · · ∪ Xm andD∗ − A = Y1 ∪ Y2 ∪ · · · ∪ Ym (Fig. 1).

We note that G is not hamiltonian, otherwise we reach a contradiction by takingC∗ = D∗ be a hamiltonian cycle of G. By Lemma 3.4, we have |C∗| ≥ 2k and|D∗| ≥ 2k. Thus, |C∗−A| ≥ 2k−m ≥ k > k−m and |D∗−A| ≥ 2k−m ≥ k > k−m.Since G is k-connected, G− A is (k−m)-connected so there are k−m pairwise vertex-disjoint paths, P1 := P1[x1, y1], P2 := P2[x2, y2], . . . , Pk−m := Pk−m[xk−m, yk−m]from V (C∗) − A to V (D∗) − A. The following claim is needed in our proof.

Claim 3.6 For any two segments Xs and Yt there do not exist two paths Pi [xi , yi ] andPj [x j , y j ] with xi , x j ∈ Xs and yi , y j ∈ Yt .

Proof Suppose the contrary: such two paths Pi [xi , yi ] and Pj [x j , y j ] exist. Let C∗∗ :=C∗[xi , x j ]Pj D∗[y j , yi ]Pi and D∗∗ := C∗[x j , xi ]Pi D∗[yi , y j ]Pj . Then, we have

Fig. 1 Partitioning segments

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Fig. 2 No two paths between Xs and Yt

V (C∗∗) ∪ V (D∗∗) ⊇ V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D),

and

|C∗∗| + |D∗∗| = |C∗| + |D∗| + 2�(Pi ) + 2�(Pj ) > |C∗| + |D∗|

which gives a contradiction to the maximality of |C∗| + |D∗|. ��

We now create an auxiliary graph H with vertex set (Fig. 2)V (H) = {X1, X2, . . . , Xm, Y1, Y2, . . . , Ym} such that two vertices Xs and Yt are

adjacent if there is a path Pi [xs, yt ] with xs ∈ Xs and yt ∈ Yt . By Claim 3.6, Hcontains no multiple edges. Therefore, H is a simple bipartite graph in which eachpartite set contains m vertices. In particular, from our construction of H , we have||H || = k −m. Since m ≤ 1

(3√256+3)3/5

k3/5, we have k −m ≥ (3√

256+3)m5/3 −m ≥(

3√

256 + 2)m5/3. Using Corollary 3.2, we obtain H ⊇ K3,257. Relabeling the seg-ments, we may assume the vertex set of this K3,257 is {X1, X2, X3, Y1, Y2, . . . , Y257}.We will also let Pi, j denote the path from Xi to Y j where 1 ≤ i ≤ 3 and1 ≤ j ≤ 257.

Starting at vertex a1 along the orientation of the cycle C∗, we obtain a linearorder of V (C∗) such that, for any pair of vertices x1 and x2 ∈ V (C∗), x1 ≺ x2 ifx1 ∈ C∗[a1, x2]. Furthermore, if S and T are two disjoint segments of C∗ − a1, wedefine S ≺ T if s ≺ t for every s ∈ S and t ∈ T . Similarly, we obtain a linear orderof V (D∗) and the disjoint segments of D∗ − b1.

By the above definition, we lose no generality by assuming X1 ≺ X2 ≺ X3 andY1 ≺ Y2 ≺ · · · ≺ Y257. We also assume that Pi, j := Pi, j [xi, j , yi, j ] where xi, j ∈ Xi

and yi, j ∈ Y j .In each Xi , xi,1, xi,2, ldots, xi,257 appear as distinct vertices. Let σi be the permu-

tation on {1, 2, . . . , 257} defined by xi,σ (1) ≺ xi,σ (2) ≺ · · · ≺ xi,σ (257). Using Lemma3.3 with n = 3, we know there are 3 integers a, b, c ∈ {1, 2, . . . , 257} on whicheach of these three permutations is monotonic. By the Pigeonhole Principle, we mayassume the following two cases: σi (a) > σi (b) > σi (c) or σi (a) < σi (b) < σi (c) fori = 1, 2

To avoid cumbersome notations, we may assume, without loss of generality, thefollowing

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x1,1 ≺ x1,2 ≺ x1,3

x2,1 ≺ x2,2 ≺ x2,3, and

either x3,1 ≺ x3,2 ≺ x3,3 or x3,3 ≺ x3,2 ≺ x3,1

Remember that the end vertex of Pi, j at Y is yi, j .

Claim 3.7 There are i and j with 1 ≤ i < j ≤ 3 such that either y1,i ≺ y2,i andy1, j ≺ y2, j or y2,i ≺ y1,i and y2, j ≺ y1, j .

Proof Suppose y1,1 ≺ y2,1. If y1,2 ≺ y2,2 or y1,3 ≺ y2,3, the first case follows with(i, j) = (1, 2) or (i, j) = (1, 3), respectively. Otherwise, y2,2 ≺ y1,2 and y2,3 ≺ y1,3and the second case holds with (i, j) = (2, 3).

Next suppose y2,1 ≺ y1,1. If y2,2 ≺ y1,2 or y2,3 ≺ y1,3, then we have the secondcase. Otherwise, y1,2 ≺ y2,2 and y1,3 ≺ y2,3 and the first case occurs. ��

By Claim 3.7, we assume that, without loss of generality, the following two cases(1) occur. y1,1 ≺ y2,1 and y1,2 ≺ y2,2 or (2). y2,1 ≺ y1,1 and y2,2 ≺ y1,2.

Case 1: y1,1 ≺ y2,1 and y1,2 ≺ y2,2. Let

C∗∗ := P1,1 D∗[y1,1, y2,1]P−2,1C∗−[x2,1, x1,2]P1,2 D∗[y1,2, y2,2]P−

2,2C∗[x2,2, x1,1],D∗∗ := P−

1,1C∗[x1,1, x1,2]P1,2 D∗−[y1,2, y2,1]P−2,1C∗[x2,1, x2,2]P2,2 D∗[y2,2, y1,1].

Since C∗[x1,2, x2,1] and C∗[x2,2, x1,1] are two disjoint segments of C andD∗[y1,1, y2,1] ⊆ Y1 and D∗[y1,2, y2,2] ⊆ Y2, C∗∗ is a cycle. Similarly, D∗∗ is acycle. It is readily seen that

V (C∗∗) ∪ V (D∗∗) ⊇ V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D)

and

|C∗∗|+|D∗∗|=|C∗|+|D∗| + 2(�(P1,1)+�(P1,2)+�(P2,1)+�(P2,2)) > |C∗|+|D∗|,

which leads to a contradiction to the maximality of |C∗| + |D∗| (Fig. 3).

Fig. 3 Construction of larger cycles I

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Fig. 4 Construction of larger cycles II

Case 2: y2,1 ≺ y1,1 and y2,2 ≺ y1,2. Let

C∗∗ := P1,1 D∗−[y1,1, y2,1]P−2,1C∗−[x2,1, x1,2]P1,2 D∗−[y1,2, y2,2]P−

2,2C∗[x2,2, x1,1],D∗∗ := P−

2,1C∗[x2,1, x2,2]P2,2 D∗−[y2,2, y1,1]P−1,1C∗[x1,1, x1,2]P1,2 D∗[y1,2, y2,1].

Similarly, we can check that both C∗∗ and D∗∗ are cycles and that

V (C∗∗) ∪ V (D∗∗) ⊇ V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D)

and

|C∗∗| + |D∗∗|=|C∗|+|D∗|+2(�(P1,1)+�(P1,2)+�(P2,1)+�(P2,2)) > |C∗| + |D∗|,

which leads to a contradiction to the maximality of |C∗| + |D∗| (Fig. 4).

4 Small k for Conjecture 1.2

We verify Conjecture 1.2 for 2 ≤ k ≤ 6 in this section. In order not to carry cumber-some notations C∗ and D∗ throughout the whole proof, let G be a graph and C and Dbe two cycles of G such that there do not exist two cycles C∗ and D∗ in G such thatV (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|. We will recursivelyshow that |V (C) ∩ V (D)| ≥ k if G is k-connected for k = 2, 3, 4, 5, and 6. Suppose,to the contrary, |V (C)∩ V (D)| < k. Clearly, we may assume |V (C)∩ V (D)| = k −1if G is k-connected for k ≥ 3. The following claim helps us reduce the number ofcases.

Claim 4.1 Let v ∈ V (C) ∩ V (D) and x ∈ V (C) and y ∈ V (D) such thatV (C(v, x)) ∩ V (D) = ∅ and V (D(v, y)) ∩ V (C) = ∅, respectively. Then thereis no path P[x, y] that is internally vertex-disjoint from V (C) ∪ V (D).

Proof Suppose, to the contrary, such a path P[x, y] exists. Let C∗ := D[v, y]P−[y, x]C[x, v] and D∗ := C[v, x]P[x, y]D[y, v]. Clearly, V (C∗) ∪ V (D∗) = V (C) ∪V (D) ∪ V (P) and |C∗| + |D∗| = |C | + |D| + 2�(P) > |C | + |D|, a contradiction. ��

For any two vertices u, v ∈ V (C) ∩ V (D), a segment C[u, v] of C and a segmentD[u, v] of D are called parallel if V (C(u, v))∩V (D) = ∅ and V (D(u, v))∩V (C) =∅.

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Claim 4.2 Let P and Q be segments of C and D, respectively. Suppose P and Q areparallel, then P = uv if and only if Q = uv.

Proof Suppose, to the contrary and without loss of generality, P = uv and |Q| ≥ 3.Let C∗ := Q[u, v] ∪ (C − P[u, v]) and D∗ := D. Then, |C∗| + |D∗| > |C | + |D|, acontradiction. ��

4.1 k = 2

In this case, we have |V (C) ∩ V (D)| ≤ 1. If V (C) ∩ V (D) = ∅, letP[u, v] and Q[x, y] be two disjoint paths from V (C) to V (D). Then, C∗ :=C[u, x]Q[x, y]D[y, v]P−[v, u] and D∗ := C−[u, x]Q[x, y]D−[u, v]P−[v, u] aretwo cycles such that V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|,a contradiction. So, |V (C) ∩ V (D)| = 1. Let {v} := V (C) ∩ V (D). By Claim 4.1,there is no path between C − v and D − v. So G is not 2-connected, a contradiction.

4.2 k = 3

In this case, let {u, v} := V (C) ∩ V (D). By Claim 4.1, there is no path betweenC − {u, v} and D − {u, v}. So G is not 3-connected, a contradiction.

4.3 k = 4

In this case, let {u, v, w} := V (C) ∩ V (D). By Claim 4.1, there is no path betweenC−{u, v, w} and D−{u, v, w}. Since G is 4-connected, we have either C−{u, v, w} =∅ or D − {u, v, w} = ∅. Assume, without loss of generality, D = uvwu. Since Gis 4-connected, we may assume |C | ≥ 4. Let C∗ = D∗ := C . Then, |C∗| + |D∗| >

|C | + |D|, a contradiction.

4.4 k = 5

In this case, let I := {v1, v2, v3, v4} := V (C) ∩ V (D). We choose the notation suchthat v1, v2, v3 and v4 are in the order along the orientation of C . Let P1 := C[v1, v2],P2 := C[v2, v3], P3 := C[v3, v4], and P4 := C[v4, v1]. Let Q1, Q2, Q3 and Q4 befour segments of D divided by v1, v2, v3, and v4. We may assume, without loss ofgenerality, the following two cases as illustrated in Fig. 5:

1. For each i = 1, 2, 3, 4, Pi and Qi are parallel each other;2. P1 and Q1 are parallel, P3 and Q3 are parallel, Q2 is a path from v2 to v4 and Q4

is a path from v3 to v1.

Case 1: There are four pairs of parallel segments.

Assume along the orientation of D the vertices of I are in the order v1, v2, v3, v4,v1, and (Pi , Qi ) is a pair of parallel segments between vi and vi+1 for each i = 1, 2,3, 4 (where v5 = v1).

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Fig. 5 Two cases for k = 5

For each i = 1, 2, 3, 4, swapping Pi and Qi does not change our assumptionsthat there are not C∗ and D∗ such V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| +|D∗| > |C |+|D|. By swapping parallel segments, we obtain pairs of cycles satisfyingthe above properties. By Claim 4.1, there is no path in G − {v1, v2, v3, v4} fromP1 ∪ P3 ∪ Q1 ∪ Q3 − I to P2 ∪ P4 ∪ Q2 ∪ Q4 − I . Since G is 5-connected, we mayassume and without loss of generality, P1 = Q1 = v1v2 and P3 = Q3 = v3v4.

If P2 �= v2v3 and P4 �= v4v1, there is a path R[x, y] in G − I from P2 ∪ Q2 − Ito P4 ∪ Q4 − I . Assume, without loss of generality, x ∈ V (P2) − {v2, v3} andy ∈ V (P4) − {v4, v1}. Let

C∗ := P−2 [x, v2]D[v2, v1]P−

4 [v1, y]R−[y, x], and

D∗ := P2[x, v3]D−[v3, v4]P4[v4, y]R−[y, x].

Clearly, V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|, acontradiction. So, we may assume P4 = v4v1. By Claim 4.2, Q4 = v4v1. So,C[v3, v2] = D[v3, v2] = v3v4v1v2. Since G is 5-connected, every set of 4 vertices iscontained in a cycle of length at least 5. So, |P2| ≥ 3 and |Q2| ≥ 3. Using the factthat G is 5-connected again, we obtain a path in G − I from P2 − I to Q2 − I , whichleads to a contradiction of Claim 4.1.

Case 2: There are exact two pairs of parallel segments.

Assume, along the orientation of D, vertices of I follow the order v1, v2, v4, v3,v1, that is, we assume that (P1, Q1) and (P3, Q3) are two pairs of parallel segments.

By swapping Pi and Qi for i = 1 and/or 3 and applying Claim 4.1 these resultedin cycles, we show that there are no paths in G − I from P1 ∪ Q1 ∪ P3 ∪ Q3 − I toP2 ∪ Q2 ∪ P4 ∪ Q4 − I . Since G is 5-connected, either all four paths P1, Q1, P3 andQ3 are single edges or all four paths P2, Q2, P4 and Q4 are single edges.

Case 2.1: P1 = Q1 = v1v2 and P3 = Q3 = v3v4.

In this case, the union of P2, Q2, P4 and Q4 forms a cycle C∗ such that V (C∗) =V (C) ∪ V (D). By setting D∗ := C∗, we obtain a contradiction to the maximality of|C | + |D|.

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Case 2.2: P2 = v2v3, Q2 = v3v1, P4 = v4v1 and Q4 = v4v2.

Assume, without loss of generality, |P1| ≥ 3. Following Claim 4.2, |Q1| ≥ 3. If|P3| ≥ 3, then |Q3| ≥ 3. Since G is 5-connected, there is a path R[x, y] in G − Ifrom P1 ∪ Q1 − I to P3 ∪ Q3 − I . Assume, without loss of generality, x ∈ P1(v1, v2)

and y ∈ P3(v3, v4). Let

C∗ := P1[x, v2]D−[v2, v1]D−[v4, v3]P3[v3, y]R−[y, x], and

D∗ := P−1 [x, v1]D[v1, v2]D[v3, v4]P−

3 [v4, y]R−[y, x].

Clearly, V (C∗)∪V (D∗) ⊇ V (C)∪V (D) and |C∗|+|D∗| > |C |+|D|, a contradiction.Thus, P3 = Q3 = v3v4. Since G is 5-connected, there is a path in G −{v1, v2, v3, v4}from P1 to Q1, which in turn gives a contradiction to Claim 4.1.

4.5 k = 6

Let I := {v1, v2, v3, v4, v5} = V (C) ∩ V (D). We assume that v1, v2, v3, v4, v5, v1are labeled in the order along the orientation of C . Let Pi := C[vi , vi+1] for eachi = 1, 2, 3, 4, 5 where v6 = v1. Let Q1, Q2, Q3, Q4 and Q5 be five segments of Ddivided by {v1, v2, v3, v4, v5} including the end-vertices. By eliminating symmetriccases, we may assume the following four cases as illustrated in Fig. 6.

Case 1: There are five parallel segment pairs

In this case, we can assume along the orientation of D the vertices of I are in theorder of v1, v2, v3, v4, v5, v1. Label the five segments of D as Qi := D[vi , vi+1]for each 1 ≤ i ≤ 5. Consequently, (Pi , Qi ) for 1 ≤ i ≤ 5 are five pairs of parallelsegments. By Claim 4.2, Pi = vivi+1 if and only if Qi = vivi+1. We consider a fewsubcases according to the number of these paths are edges.

Case 1.1: |Pi | ≥ 3 and |Qi | ≥ 3 for all i = 1, 2, 3, 4 with the possibilityP5 = Q5 = v5v1

Since G is 6-connected, there is a path R in G − I from P1 ∪ Q1 − I to ∪5i=2(Pi ∪

Qi ) − I . By swapping Pi and Qi for i = 1, 2, 5, if necessary, and applying Claim4.1, we can show that there is no path from P1 ∪ Q1 − I to P2 ∪ Q2 ∪ P5 ∪ Q5 − I .By Claim 4.1, we may assume that R is from P1 to P3. For the same reason, we mayassume that there is a path S in G − I from P2 to P4. By Claim 4.1, V (R)∩ V (S) = ∅.Moreover, we assume that R = R[x1, x3] and S = S[x2, x4]. Let (Fig. 7)

C∗ := D[v1, v2]C−[v2, x1]RC[x3, v4]D−[v4, v3]C−[v3, x2]SC[x4, v1]D∗ := C[v1, x1]RC−[x3, v3]D−[v3, v2]C[v2, x2]SC−[x4, v4]D[v4, v1].

Clearly, V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|, acontradiction.

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Fig. 6 Four Cases for k = 6

Fig. 7 Cycles C∗ and D∗ inCase 1.1

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Case 1.2: Pi = vivi+1 holds for at least two i = 1, 2, 3, 4, 5

If Pi = vivi+1 for two consecutive segments, assume P4 = v4v5 and P5 = v5v1.By Claim 4.2, Q4 = v4v5 and Q5 = v5v1. By swapping Pi and Qi for i = 1, 2, 3, ifnecessary, and applying Claim 4.1, we can show that there is no path in G − I fromP2 ∪ Q2 − I to P1 ∪ Q1 ∪ P3 ∪ Q3 − I . Since G is 6-connected, we have eitherP2 = Q2 = v2v3 or P1 = Q1 = v1v2 and P3 = Q3 = v3v4. If P2 = Q2 = v2v3,there is no path in G − I from P1 ∪ Q1 to P3 ∪ Q3, which in turn shows that eitherP1 = Q1 = v1v3 or P3 = Q3 = v3v4. So, in either case, we obtain four outof five segments Pi (1 ≤ i ≤ 5) are edges. Assume, without loss of generality,Pi = Qi = vivi+1 for i = 2, 3, 4, 5. Since G is 6-connected, there is a path in G − Ifrom P1 to Q1, which gives a contradiction to Claim 4.1.

So exactly two of P1, P2, P3, P4, P5 are edges and they are not two consecutiveedges. We assume, without loss of generality, P1 = Q1 = v1v2 and P3 = Q3 = v3v4.By the maximality of |C | + |D|, there is no path in G − I from P2 ∪ Q2 − I toP4 ∪ Q4 ∪ P5 ∪ Q5 − I , which gives a contradiction to that G is 6-connected.

Case 2: There are exactly three pairs of (Pi , Qi ) that are parallel

Assume along the orientation of D the vertices of I are in the order of v1, v3,v4, v5, v2, v1. Label the five segments of D as Q1 := D[v2, v1], Q2 := D[v1, v3],Q3 := D[v3, v4], Q4 := D[v4, v5] and Q5 := D[v5, v2]. Consequently, (P1, Q1),(P3, Q3) and (P4, Q4) are pairs of parallel segments.

Statement 2.1: P1 = v1v2 and Q1 = v2v1.Otherwise, by Claim 4.2, we may assume there is a path R[x, y] in G − I from P1 to

C ∪ D− P1. Assume x ∈ V (P1). By Claim 4.1, y ∈ V (P3)∪V (Q3)∪V (P3)∪V (P4).Assume, without loss of generality, y ∈ V (P3). Let (Fig. 8)

C∗ := C[x, v2]D[v2, v1]C−[v1, v4]D−[v4, v3]C[v3, y]R−[y, x]D∗ := C−[x, v1]D[v1, v3]C−[v3, v2]D−[v2, v4]C−[v4, y]R−[y, x]

Clearly, V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|, acontradiction.

Statement 2.2: P5 = v5v1. Similarly, P2 = v2v3, Q2 = v1v3 and Q5 = v5v2.Otherwise, since G is 6-connected, there is a path R[x, y] in G − I from P5 to

V (C)∪ V (D)− V (P5). Assume x ∈ V (P5)− I . By Claim 4.1, y ∈ V (P2)∪ V (P3)∪V (Q3) − I .

If y ∈ V (P2) − I , let (Fig. 9)

C∗ := C[x, v2]D−[v2, v3]C−[v3, y]R−[y, x]D∗ := C−[x, v3]D−[v3, v1]C[v1, y]R[y, x].

Clearly, V (C∗) ∪ V (D∗) ⊇ V (C) ∪ V (D) and |C∗| + |D∗| > |C | + |D|, acontradiction. So, y ∈ V (P3) ∪ V (Q3) − I . Assume, without loss of generality,y ∈ V (P3) − I .

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Fig. 8 Cycles C∗ and D∗ inStatement 2.1

Fig. 9 Cycles C∗ and D∗ inStatement 2.2

If P2 �= v2v3, similarly, we may assume that there is a path S[x∗, y∗] internallyvertex-disjoint from V (C) ∪ V (D) with x∗ ∈ V (P2) − I and y∗ ∈ V (P4) − I . ByClaim 4.1, R and S are disjoint. The following two cycles (Fig. 10) C∗∗ and D∗∗ givea contradiction:

C∗∗ := C[x, x∗]S[x∗, y∗]C[y∗, v5]D−[v5, v3]C[v3, y]R−[y, x],D∗∗ := C−[x, v5]D[v5, v2]D[v2, v3]C−[v3, x∗]S[x∗, y∗]C−[y∗, y]R−[y, x].

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Fig. 10 Cycles C∗∗ and D∗∗ inCase 2.2

Fig. 11 Cycles C∗∗∗ and D∗∗∗in Statement 2.2

So, P2 = v2v3. Let (Fig. 11)

C∗∗∗ := C[x, v2]D−[v2, v3]C[v3, y]R−[y, x]D∗∗∗ := D[v1, v3]C[v3, v1].

Clearly, V (C∗∗∗) ∪ V (D∗∗∗) = V (C) ∪ V (D) ∪ V (R) and |C∗∗∗| + |D∗∗∗| =|C |+|D|−�(v1v2v3)+�(R)+�(C[x, v1])+�(C[v3, y]) > |C |+|D|, a contradiction.

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Fig. 12 Cycles C∗ and D∗ in3.1

Since G is 6-connected, V (C) ∪ V (D) − I �= ∅. So, V (P3) ∪ V (P4) ∪ V (Q3) ∪V (Q4) − I �= ∅. By Claim 4.1 (swapping segments if necessary), there is no path inG − I connecting any two of P3, P4, Q3, and Q4, which gives a contradiction to thatG is 6-connected.

Case 3. There are exactly two pairs of parallel segments.

We assume that along the orientation of D the vertices of I are in the order v1, v4,v2, v3, v5, v1. Label the five segments of D as Q1 := D[v3, v5], Q2 := D[v2, v3],Q3 := D[v4, v2], Q4 := D[v1, v4], and Q5 := D[v5, v1]. Consequently, (P2, Q2)

and (P5, Q5) are two pairs of parallel segments.

Statement 3.1: Either V (P2) ∪ V (Q2) ∪ V (P5) ∪ V (Q5) − I = ∅ or V (P1) ∪V (Q1) ∪ V (P3) ∪ V (Q3) ∪ V (P4) ∪ V (Q4) − I = ∅.

Otherwise, by Claim 4.1 and 6-connectivity, we may assume that there is a pathR[x, y] in G − I internally vertex-disjoint from V (C) ∪ V (D) with x ∈ V (P5) − Iand y ∈ V (P3) − I . The following two cycles (Fig. 12) lead a contradiction to themaximality of |C | + |D|.

C∗ := C[x, v3]D[v3, v5]C−[v5, y]R−[y, x]D∗ := C−[x, v5]D[v5, v3]C[v3, y]R−[y, x]

First, suppose V (P1) ∪ V (Q1) ∪ V (P3) ∪ V (Q3) ∪ V (P4) ∪ V (Q4) − I = ∅.Since G is 6-connected, there is a path R in G − I connecting two segments P2,

Q2, P5, and Q5. Assume, without loss of generality, R := R[x, y] with x ∈ V (P2)

and y ∈ V (P5). Then, D∗ := C[x, v3]D−[v3, v5]C[v5, y]R[y, x] (Fig. 13) contains

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Fig. 13 Cycle D∗ in Subcase3.1.1

all vertices of V (D) and more, which gives a contradiction to the maximality of|C | + |D|.

Next, suppose V (P2) ∪ V (Q2) ∪ V (P5) ∪ V (Q5) − I = ∅.In this case, P2 = Q2 = v2v3 and P5 = Q5 = v5v1. We have V (C) �= I , otherwise

we can use (D, D) instead of (C, D) to obtain a contradiction to the maximality of|C |+|D|. Similarly, we have V (D) �= I . Since G is 6-connected, there is a path R[x, y]in G − I from C − I to D− I . By Claim 4.1, x ∈ V (C(v1, v2)) and y ∈ V (D(v3, v5)).Let (Fig. 14)

Fig. 14 Cycle C∗ and D∗ inSubcase 3.1.2

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Fig. 15 Cycle C∗ and D∗ inCase 4

C∗ := C[x, v5]D−[v5, y]R−[y, x],D∗ := C−[x, v1]D[v1, y]R−[y, x].

Clearly, V (C∗)∪ V (D∗) = V (C)∪ V (D)∪ V (R) and |C∗|+ |D∗| = (|C | − 1)+(|D| − 1) + 2�(R) + �(C[v2, v3]) ≥ |C | + |D| + 1, a contradiction.

Case 4. There is no parallel segment.

In this case, we assume that along the orientation of D the vertices of I are inthe order of v1, v3, v5, v2, v4, v1. By the maximality of |C | + |D|, V (C) − I �= ∅and V (D) − I �= ∅. Since G is 6-connected, there is a path R[x, y] in G − I fromV (C) − I to V (D) − I . Using Claim 4.1, we may assume that x ∈ V (C(v1, v2)) andy ∈ V (D(v3, v5)). Let (Fig. 15)

C∗ := C[x, v2]D−[v2, v5]C[v5, v1]D−[v1, v4]C−[v4, v3]D[v3, y]R−[y, x]D∗ := C−[x, v1]D[v1, v3]C−[v3, v2]D[v2, v4]C[v4, v5]D−[v5, y]R−[y, x]

Clearly, V (C∗)∪ V (D∗) = V (C)∪ V (D)∪ V (R) and |C∗|+ |D∗| = |C |+ |D|+2�(R) > |C | + |D|, a contradiction. ��Acknowledgments We wish to acknowledge the valuable comments provided by the anonymous referee.

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