cycles passing through k + 1 vertices in k-connected graphs

Cycles Passing Throughk + 1 Vertices ink-Connected Graphs

Jun Fujisawa1 and Tomoki Yamashita2

1DEPARTMENT OF COMPUTER SCIENCENIHON UNIVERSITY, SAKURAJOSUI 3-25-40

SETAGAYA-KU, TOKYO 156-8550, JAPANE-mail: [email protected] OF MATHEMATICS

ASAHI UNIVERSITY, GIFU 501-0296, JAPANE-mail: [email protected]

Received July 6, 2007; Revised January 21, 2008

Published online 10 April 2008 in Wiley InterScience(www.interscience.wiley.com).DOI 10.1002/jgt.20306

Abstract: In this article, we prove the following theorem. Let k ≥ 3 bean integer, G be a k-connected graph with minimum degree d and X bea set of k + 1 vertices on a cycle. Then G has a cycle of length at leastmin{2d, |V(G)|} passing through X. This result gives the positive answer tothe Question posed by Locke [8]. © 2008 Wiley Periodicals, Inc. J Graph Theory 58: 179–190,

2008

Keywords: minimum degree; long cycle; cyclable

1. INTRODUCTION

In this article, we consider only finite undirected graphs without loops or multipleedges. For standard graph-theoretic terminology not explained in this article, werefer the reader to [2].

Contract grant sponsor: JSPS Research Fellowships for Young Scientists (to J. F.);Contract grant sponsor: 21st Century COE Program.Journal of Graph Theory© 2008 Wiley Periodicals, Inc.

179

180 JOURNAL OF GRAPH THEORY

In 1952 and 1960, Dirac proved the following two results concerning cycles.

Theorem 1 (Dirac [3]). Let G be a 2-connected graph with minimum degree d.Then G has a cycle of length at least min{2d, |V (G)|}.Theorem 2 (Dirac [4]). Let k ≥ 2 be an integer. Let G be a k-connected graphand X ⊂ V (G) with |X| ≤ k. Then G has a cycle passing through X.

These two results are unified as the following theorem, by Egawa et al.

Theorem 3 (Egawa et al.[5]). Let k ≥ 2 be an integer. Let G be a k-connectedgraph with minimum degree d and X ⊂ V (G) with |X| ≤ k. Then G has a cycle oflength at least min{2d, |V (G)|} passing through X.

Here, one may ask the following question: What happens if we put more thank vertices on X of Theorem 3? In general, it is easy to see that Theorem 3 doesnot hold in the case |X| > k, since there exist k-connected graphs which do notcontain any cycle through given k + 1 vertices. (See Watkins and Mesner [9], inwhich a characterization of k-connected graphs, which do not contain any cyclethrough given k + 1 vertices, is shown.) However, if we restrict X to the vertex setwhich lies on a cycle, then the situation may change, and there is a hope to obtainlong cycle passing through X. Indeed, Locke and Zhang considered the followingquestion, and proved the case of k = 2 and |X| = 3.

Question 4 (Locke and Zhang [7]). Let k ≥ 2 and m > k be integers. Let G bea k-connected graph with minimum degree d. For any set X ⊆ V (G) of m verticeson a cycle, does G have a cycle C of length at least min{2d, |V (G)|} passingthrough X?

Theorem 5 (Locke and Zhang [7]). Let G be a 2-connected graph with minimumdegree d and X a set of three vertices on a cycle. Then G has a cycle of length atleast min{2d, |V (G)|} passing through X.

It seems difficult to find the maximum value m which gives the positive answerto Question 4. Even for the case m = k + 1, Question 4 has been open for k ≥ 3,and the following question was proposed by Locke in 2002.

Question 6 (Locke [8]). Let G be a 3-connected graph with minimum degree dand X a set of four vertices on a cycle. Does G have a cycle of length at leastmin{2d, |V (G)|} passing through X?

In this article, we give the positive answer to the above question, and solveQuestion 4 for the case k ≥ 3 and m = k + 1. Our main result is the following.

Theorem 7. Let k ≥ 3 be an integer. Let G be a k-connected graph with minimumdegree d and X a set of k + 1 vertices on a cycle. Then G has a cycle of length atleast min{2d, |V (G)|} passing through X.

Journal of Graph Theory DOI 10.1002/jgt

CYCLES PASSING THROUGH k + 1 VERTICES 181

2. NOTATION AND PRELIMINARIES

Now we prepare some terminology and notation used in this article. Let G be a graph.For a subgraph H of G and a vertex x ∈ V (G), we denote NH (x) := NG(x) ∩ V (H),dH (x) := |NH (x)|, δH (H) := min{dH (x) : x ∈ V (H)}, and δG(H) := min{dG(x) :x ∈ V (H)}. For X ⊂ V (G), ∪x∈XNG−X(x) is denoted by NG(X). Furthermore,for a subgraph H of G and X ⊂ V (G), we write NH (X) := NG(X) ∩ V (H).If there is no fear of confusion, we often identify a subgraph H of a graphG with its vertex set V (H). For example, G − V (H) is sometimes denotedby G − H .

Throughout this article, we consider that any cycle has a fixed orientation. Let Cbe a cycle of G. For x ∈ V (C), we denote the successor and the predecessor of x onC by x+ and x−, respectively. For X ⊂ V (C), we define X+ := {x+ : x ∈ X} andX− := {x− : x ∈ X}. For x, y ∈ V (C), we denote by C[x, y] a path from x to y alongthe orientation of C. The reverse sequence of C[x, y] is denoted by

←−C [y, x]. We

write C[x, y] − {x, y}, C[x, y] − {x}, C[x, y] − {y} by C(x, y), C(x, y] and C[x, y),respectively. A path with endvertices x and y is called an (x, y)-path. Let H be acomponent of G − C. For u, v ∈ NC(H), we write by PH [u, v] a longest (u, v)-path among ones whose internal vertices are contained in H. PH [u, v] − {u, v} isdenoted by PH (u, v). For a path P, we simply write |V (P)| by |P |. An endblock is ablock that has at most one cut vertex (For convenience, we call a 2-connected graphitself an endblock.) For an endblock B, we define IB := V (B)\{c} if B has a cutvertex c; otherwise IB := V (B). Throughout this article, for every graph H, we fixan endblock of H with maximum number of vertices, and denote it by BH . For u, v ∈NC(H), the pair of vertices (u, v) is called a good pair for H if u = v and there existx ∈ NH (u), y ∈ NH (v) such that x = y and {x, y} ∩ IBH

= ∅. If there is no fear ofconfusion, we sometimes omit “for H” and simply call (u, v) a good pair. We call avertex set S ⊆ V (C) a good set for H if every pair of vertices (u, v) in S is a good pairfor H.

In the rest of this section, we shall prove some lemmas. The following theoremis used to prove the next lemma.

Theorem 8 (Bondy and Jackson [1, Lemma 2], Locke [6, Lemma 3.2.2], Egawa etal.[5, Lemma 5]). Let G be a 2-connected graph with |V (G)| ≥ 4 and u, v, w ∈V (G). If d(x) ≥ d for any vertex x ∈ V (G)\{u, v, w}, then there exists a (u, v)-pathof length at least d.

In the next two lemmas, let G be a connected graph, C be a cycle of G, H be acomponent of G − C with at least two vertices, let B := BH and u, v ∈ NC(H).

Lemma 1. If (u, v) is a good pair for H, then PH (u, v) has at least δH (H) + 1vertices.

Proof. If |V (B)| ≤ 3, the assertion is obvious, and hence we assume that|V (B)| ≥ 4. Since (u, v) is a good pair, there exist x ∈ NH (u) and y ∈ NH (v)such that x = y and {x, y} ∩ IB = ∅. Without loss of generality, we may assume



that x ∈ IB. If y ∈ IB, then Theorem 8 implies that there exists an (x, y)-path P of length at least δB(IB) ≥ δH (H) whether B has a cutvertex or not.Hence, P has at least δH (H) + 1 vertices, which implies the assertion. If y /∈IB, then B has a cutvertex c. It follows from Theorem 8 that there existsan (x, c)-path of length at least δB(IB) ≥ δH (H). Take a (c, y)-path P ′ in H,then the path xPcP ′y has at least δH (H) + 1 vertices, and hence the assertionholds. �

Lemma 2. Suppose that (u, v) is a good pair for H. Then

(i) for any w ∈ NC(H), (u, w) or (v, w) is a good pair, and(ii) for any w, w′ ∈ NC(H), at least one of (u, w), (w, w′), or (w′, v) is a good

pair.

Proof. First we prove (i). If w = u or w = v, it is obvious. So we assume thatw = u, v. Since (u, v) is a good pair, there exist x ∈ NH (u) and y ∈ NH (v) such thatx = y and {x, y} ∩ IB = ∅. Without loss of generality, we may assume that x ∈ IB.If NH (w)\{x} = ∅, then (u, w) is a good pair. Otherwise, NH (w) = {x} holds, andhence (v, w) is a good pair. Next we prove (ii). Assume (u, w) is not a good pair,then by (i), (w, v) is a good pair. Therefore, also by (i), (w, w′) or (w′, v) is agood pair. �

Lemma 3. Let k ≥ 2 and G be a k-connected graph with δ(G) ≥ d. Let C be acycle of G with |V (C)| ≥ k + 1, let H be a component of G − C with |V (H)| ≥ 2,and let B := BH . If |NC(H)| < d, then for any vertex u ∈ NC(H), there exists agood set Y for H such that u ∈ Y and |Y | ≥ k.

Proof. Since G is k-connected and |V (C)| ≥ k + 1, we have |NC(H)| ≥ k.Suppose that |V (B)| = 2. Then, since B is an endblock of H with maximumnumber of vertices, there exist two vertices u1 ∈ V (IB) and u2 ∈ V (H) such thatdH (u1) = dH (u2) = 1. This implies dC(ui) = dG(ui) − 1 ≥ d − 1 for each i = 1, 2.With the fact |NC(H)| < d, we obtain NC(H) = NC(u1) = NC(u2). Therefore, it iseasy to see that NC(H) is a good set for H of cardinality at least k which containsu. Hence, we may assume that |V (B)| ≥ 3.

First, we consider the case where H is not 2-connected. Let W := {w ∈ V (C) :dIB

(w) ≥ 2}. Now we shall show that |V (B)| ≥ k − |W |. By the definition of W,∑v∈IB

dC−W (v) ≤ |NC(H)| − |W |. Hence, we deduce

∑

v∈IB

dG(v) ≤∑

v∈IB

(dB(v) + dW (v)) +∑

v∈IB

dC−W (v)

≤ (|V (B)| − 1) maxv∈IB

(dB(v) + dW (v)) + |NC(H)| − |W |

< (|V (B)| − 1)(|V (B)| − 1 + |W |) + d − |W |. (1)



On the other hand,

∑

v∈IB

dG(v) ≥ (|V (B)| − 1)d. (2)

It follows from (1) and (2) that (|V (B)| − 2)(|V (B)| + |W |) + 1 > (|V (B)| − 2)d.Since |V (B)| ≥ 3, |V (B)| + |W | > d − 1 ≥ k − 1 holds, and we obtain |V (B)| ≥k − |W |.

If |W | ≥ k − 1 and u ∈ W , then W ∪ {u} is a desired good set. If |W | ≥ k − 1 andu ∈ W , then W is a desired good set if |W | ≥ k; otherwise W ∪ {z} is a desired goodset, where z ∈ NC(H) \ W . Hence, we assume |W | ≤ k − 2. Let W ′ = NC(H)\W .Since G − W is (k − |W |)-connected and min{|W ′|, |V (B)|} ≥ k − |W |, there existk − |W | disjoint paths P1, . . . , Pk−|W |, where vi ∈ V (B) and w′

i ∈ W ′, and Pi is a(vi, w

′i)-path whose internal vertices are in V (H) \ V (B). Since at most one of

P1, . . . , Pk−|W | contains the cutvertex of B, we can find a (k − |W | − 1)-matchingbetween IB and W ′. Hence, {w′

1, . . . , w′k−|W |} is a good set of cardinality k − |W |.

Let W∗ = W ∪ {w′1, . . . , w

′k−|W |}. If W∗ contains u, then W∗ is a desired good

set. If u /∈ W∗ and there exists w∗ ∈ W∗, such that (u, w∗) is not a good pair,then by Lemma 2 (i), (u, w) is a good pair for every w ∈ W∗\{w∗}, and hence{u} ∪ W∗ \ {w∗} is a desired good set. If u /∈ W∗ and (u, w) is a good pair for everyw ∈ W∗, then {u} ∪ W∗ is a desired good set.

In the case H is 2-connected, let x ∈ V (H). By replacing IB and c with H\{x}and x in the above proof, we obtain a desired good set. This completes the proof ofLemma 3. �

3. PROOF OF THEOREM 7

Let C be a longest cycle passing through X. Then note that |V (C)| ≥ k + 1 and|NC(H)| ≥ k, since G is k-connected. Assume that |V (C)| < min{2d, |V (G)|}.Then there exists a component of G − C. Suppose that |NC(F )| ≥ d holds forsome component F of G − C. Since C is longest, NC(F ) ∩ NC(F )+ = ∅ and hence|V (C)| ≥ 2|NC(F )| ≥ 2d, a contradiction. Therefore, for each component F ofG − C, |NC(F )| < d holds, which implies |V (F )| ≥ 2. Choose a component H ofG − C so that

(i) NC(H) ∩ X = ∅ if possible, and(ii) δH (H) is as small as possible, subject to (i).

Let u0 be a vertex of V (H) which satisfies dH (u0) = δH (H). Let NC(H) ={u1, u2, . . . , um}, where u1, u2, . . . , um appear in this order along C, and letum+1 := u1. For 1 ≤ i ≤ m, C(ui, ui+1) is called a long segment if |C(ui, ui+1)| ≥δH (H) + 1.



Claim 1. If (v, w) is a good pair for H, such that C(v, w) ∩ X = ∅, then thereexists a long segment in C(v, w).

Proof. By Lemma 1 and the choice of C, it suffices to prove the existence ofui ∈ NC(H) such that (ui, ui+1) is a good pair for H and C(ui, ui+1) ⊆ C(v, w).Let v = uj and w = ul. Without loss of generality, we may assume j < l. Let i bethe largest integer such that j ≤ i < l and (ui, ul) is a good pair for H. Then, since(ui+1, ul) is not a good pair, Lemma 2 (ii) implies that (ui, ui+1) is a good pair.Hence, ui is the desired vertex. �Claim 2. For any U ⊆ NC(H), it follows that∑

ui∈U

|C(ui, ui+1)| < 2δH (H) + |U|.

Proof. Suppose∑

ui∈U |C(ui, ui+1)| ≥ 2δH (H) + |U| for some U ⊆ NC(H).By the choice of C, |C(ui, ui+1)| ≥ 1 for every ui ∈ NC(H) \ U. Hence,

|C| = |NC(H)| +∑

ui∈NC(H)

|C(ui, ui+1)|

= |NC(H)| +∑

ui∈NC(H)\U|C(ui, ui+1)| +

∑

ui∈U

|C(ui, ui+1)|

≥ |NC(H)| + |NC(H)| − |U| + 2δH (H) + |U|= 2|NC(H)| + 2dH (u0)

≥ 2dC(u0) + 2dH (u0)

≥ 2dG(u0) ≥ 2d,

a contradiction. �The following claim can easily be obtained from Claims 1 and 2.

Claim 3. Suppose that C(a1, b1) is a long segment, or (a1, b1) is a good pair forH and C(a1, b1) ∩ X = ∅. If C(a2, b2) ⊆ C(b1, a1) and C(a2, b2) ∩ X = ∅, then(a2, b2) is not a good pair for H.

By Lemma 3, there exists a good set Y for H of order k. Let Y be a good set chosensuch that |Y ∩ X| is maximum, let Y := {y1, y2, . . . , yk} where y1, y2, . . . , yk

appear in this order along C, and let yk+1 := y1. We denote by xi (or x′i) the first

(or last) vertex of X along C(yi, yi+1) for each i with 1 ≤ i ≤ k (Possibly xi = x′i

or xi does not exist.) For every i ∈ {1, . . . , k}, such that xi exists, let zi be a vertexin NC(H) ∩ C[yi, xi) such that NC(H) ∩ C(zi, xi) = ∅, and let z′

i be a vertex inNC(H) ∩ C(x′

i, yi+1] such that NC(H) ∩ C(x′i, z

′i) = ∅.

Claim 4. Let i, j be integers with 1 ≤ i < j ≤ k. Suppose that NG−C(xi) =NG−C(xj) = ∅ and (zi, zj) is a good pair. Then |C(zi, xi)| + |C(zj, xj)| ≥δH (H) + 1.



Proof. Let C1 := C(zi, xi), C2 := C[xi, zj], C3 := C(zj, xj) and C4 :=C[xj, zi]. Note that |PH (zi, zj)| ≥ δH (H) + 1 holds by Lemma 1. First, supposethat there exists a vertex v in NC3 (xi). Then xivC[v, zi]PH [zi, zj]

←−C [zj, xi] is a

cycle passing through X. By the choice of C, we have |C(zi, xi)| + |C(zj, xj)| ≥|C(zi, xi)| + |C(zj, v)| ≥ |PH (zi, zj)| ≥ δH (H) + 1, and hence the claim holds.Thus, we may assume that NC3 (xi) = ∅. By the symmetry of C, we may also assumethat NC1 (xj) = ∅. Next, suppose that there exists a vertex v in NC2 (xi)− ∩ NC2 (xj).Then the cycle xiv

+C[v+, zj]PH [zj, zi]←−C [zi, xj]xjv

←−C [v, xi] passes through X. By

the choice of C, we have |C(zi, xi)| + |C(zj, xj)| ≥ |PH (zi, zj)| ≥ δH (H) + 1, andhence the claim holds. Thus, we may assume that NC2 (xi)− ∩ NC2 (xj) = ∅. Bythe symmetry of C, we may also assume that NC4 (xi) ∩ NC4 (xj)− = ∅. Now, sinceNG−C(xi) = NG−C(xj) = ∅, we have

|V (C)| ≥ |C1| + |C2| + |C3| + |C4|≥ |NC1 (xi)| + |NC2 (xi)

− ∪ NC2 (xj)| + |NC3 (xj)| + |NC4 (xi) ∪ NC4 (xj)−|= |NC1 (xi)| + |NC1 (xj)| + |NC2 (xi)| + |NC2 (xj)|

+ |NC3 (xi)| + |NC3 (xj)| + |NC4 (xi)| + |NC4 (xj)|= dC(xi) + dC(xj)

= dG(xi) + dG(xj) ≥ 2d,

a contradiction. �Claim 5. Let i, j be integers with 1 ≤ i < j ≤ k. Suppose that NG−C(xi) =NG−C(xj) = ∅ and there exists a long segment C(a, b) in C(xi, yj) ∪ C(xj, yi).Then (zi, zj) is not a good pair.

Proof. Assume that (zi, zj) is a good pair. Then it follows from Claim 4 that|C(zi, xi)| + |C(zj, xj)| ≥ δH (H) + 1. Let zi = ur and zj = us. Since NG−C(xi) =NG−C(xj) = ∅, we have C(ur, xi] ⊆ C(ur, ur+1) and C(us, xj] ⊆ C(us, us+1).Therefore, |C(a, b)| + |C(ur, ur+1)| + |C(us, us+1)| ≥ |C(a, b)| + |C(ur, xi)| +1 + |C(us, xj)| + 1 ≥ δH (H) + 1 + δH (H) + 1 + 1 + 1 = 2δH (H) + 4, and so{a, ur, uj} contradicts Claim 2. �Claim 6. Suppose that |C(yi, yi+1) ∩ X| = |C(yj, yj+1) ∩ X| = 1 for some i, j,1 ≤ i < j ≤ k. Then NG−C(xi) = ∅ or NG−C(xj) = ∅ holds.

Proof. Suppose NG−C(xi) = ∅ and NG−C(xj) = ∅. Then C(zi, z′i) ∩ NC(H) =

C(zj, z′j) ∩ NC(H) = ∅ holds since |C(yi, yi+1) ∩ X| = |C(yj, yj+1) ∩ X| = 1.

First, assume that both of (zi, zj) and (z′i, z

′j) are good pairs. Then, by

Claim 4 and the symmetry of C, |C(zi, xi)| + |C(zj, xj)| ≥ δH (H) + 1 and|C(xi, z

′i)| + |C(xj, z

′j)| ≥ δH (H) + 1. Thus, |C(zi, z

′i)| + |C(zj, z

′j)| ≥ 2δH (H) +

2. Since C(zi, z′i) ∩ NC(H) = C(zj, z

′j) ∩ NC(H) = ∅, {zi, zj} contradicts Claim 2.

Next, assume that neither (zi, zj) nor (z′i, z

′j) is a good pair. Then it follows from



Lemma 2 (ii) that (yi, zi) or (yj, zj) is a good pair, and (z′i, yi+1) or (z′

j, yj+1) is agood pair. This contradicts Claim 3.

Hence, exactly one of (zi, zj) or (z′i, z

′j) is a good pair. By symmetry, we may

assume that (zi, zj) is a good pair and (z′i, z

′j) is not a good pair. Then it follows from

Lemma 2 (ii) that (z′i, yi+1) or (z′

j, yj+1) is a good pair. By Claim 1, there exists along segment C(a, b) in C(z′

i, yi+1) ∪ C(z′j, yj+1). This contradicts Claim 5, since

(zi, zj) is a good pair. �Claim 7. Y ∩ X = ∅.

Proof. Assume NC(H) ∩ X = ∅. Then by the choice of H, NC(H ′) ∩ X = ∅for every component H ′ of G − C. First, assume that C(yi, yi+1) ∩ X = ∅ holds forevery i with 1 ≤ i ≤ k. Since |X| = k + 1 and k ≥ 3, we can find p, q such that 1 ≤p < q ≤ k and |C(yp, yp+1) ∩ X| = |C(yq, yq+1) ∩ X| = 1. Then by Claim 6, thereexists a component H ′ of G − C such that NC(H ′) ∩ {xp, xq} = ∅, a contradiction.

Therefore, we may assume that C(yi, yi+1) ∩ X = ∅ holds for some i. Thenit follows from Claim 3 that C(yj, yj+1) ∩ X = ∅ for every j with j = i. Takep, q ∈ {1, . . . , k} \ {i}. Also by Claim 3, neither (yp, zp) nor (yq, zq) is a good pair.Then Lemma 2 (ii) implies that (zp, zq) is a good pair. This contradicts Claim 5.Therefore, NC(H) ∩ X = ∅. It follows from Lemma 3 and the choice of H and Y thatY ∩ X = ∅. �

In the next three claims, we assume that H ′ is a component of G − C such thatH ′ = H and NC(H ′) ∩ X = ∅, and (v, w) is a good pair for H ′. By Lemma 1 andthe choice of H, |PH ′(v, w)| ≥ δH ′(H ′) + 1 ≥ δH (H) + 1.

Claim 8. Suppose that v ∈ C(ui, uj) and w ∈ C(uj, ui) for some i, j withi = j. If

(i) there exists a long segment C(a, b) in C(v, uj) ∪ C(w, ui), or(ii) (ui, uj) is a good pair for H,

then C(ui, v) ∩ X = ∅ or C(uj, w) ∩ X = ∅.

Proof. Assume not. Take up ∈ NC(H) ∩ C[ui, v) and uq ∈ NC(H) ∩ C[uj, w)so that NC(H) ∩ C(up, v) = NC(H) ∩ C(uq, w) = ∅. Then, the cyclePH ′[v, w]C[w, up]PH [up, uq]

←−C [uq, v] passes through X. Hence, by the choice

of C,

|C(up, up+1)| + |C(uq, uq+1)| ≥ |C(up, v)| + |C(uq, w)|≥ |PH ′(v, w)| + |PH (up, uq)|≥ δH (H) + 1 + |PH (up, uq)|.

If (i) holds, then |C(a, b)| + |C(up, up+1)| + |C(uq, uq+1)| ≥ δH (H) + 1 +δH (H) + 1 + |PH (up, uq)| ≥ 2δH (H) + 3, and hence {a, up, uq} contradicts



Claim 2. Therefore, we may assume that (ii) holds, and it follows from Claim 2 andLemma 1 that neither (ui, up) nor (uj, uq) is a good pair. Now Lemma 2 (ii) impliesthat (up, uq) is a good pair. Then by Lemma 1, |C(up, up+1)| + |C(uq, uq+1)| ≥δH (H) + 1 + |PH (up, uq)| ≥ 2δH (H) + 2. Therefore, {up, uq} contradictsClaim 2. �Claim 9. Suppose that C(v, w) ∩ X = ∅ and there exists a long segment C(a, b)in C(w, v). Then NC(H) ∩ C(v, w) = ∅.

Proof. Suppose that NC(H) ∩ C(v, w) = ∅. Then C(v, w) ⊆ C(uj, uj+1) forsome j. By the choice of C, |C(a, b)| + |C(uj, uj+1)| ≥ |C(a, b)| + |C(v, w)| ≥|C(a, b)| + |PH ′(v, w)| ≥ 2δH (H) + 2, and hence {a, uj} contradicts Claim 2. �Claim 10. Suppose that C(v, w) ∩ X = ∅, C(v, w) ⊆ C(yi, yi+1) for some i, andthere exists a long segment C(a, b) in C(yi+1, yi). If C[v, w] ⊆ C(yi, yi+1](resp. C[v, w] ⊆ C[yi, yi+1)) for some i, then C(yi, v) ∩ X = ∅ (resp.C(w, yi+1) ∩ X = ∅).

Proof. By Claim 9, there exists a vertex of NC(H) ∩ C(v, w), say uj. Sincev ∈ C(yi+1, uj) and w ∈ C(uj, yi+1), by applying Claim 8 (i) as (v, w, ui, uj) =(v, w, yi, uj), we obtain C(yi, v) ∩ X = ∅. By the symmetry of C, we haveC(w, yi+1) ∩ X = ∅. �

Now we divide the rest of the proof.

Case 1. There exists a long segment C(a, b) in C.

Without loss of generality, we may assume that C(a, b) in C(y1, y2). Then itfollows from Claim 3 that

C(yi, yi+1) ∩ X = ∅ for every i with 2 ≤ i ≤ k. (3)

On the other hand, by Claim 7,

there are at most k vertices of X ink⋃

i=2

C(yi, yi+1). (4)

It follows from (3) and (4) that there exists an integer t = 1 such that |C(yi, yi+1) ∩X| = 1 for every i ∈ {2, . . . , k} \ {t} and 1 ≤ |C(yt, yt+1) ∩ X| ≤ 2. Note that|C(y1, y2) ∩ X| ≤ 2 − |C(yt, yt+1) ∩ X| ≤ 1.

Claim 11. Let p be an integer with |C(yp, yp+1) ∩ X| = 1. Then NG−C(xp) = ∅.

Proof. Assume NG−C(xp) = ∅. Let H∗ be a component of G − C withxp ∈ NC(H∗). If H∗ = H , then by Lemma 2 (i), (yp, xp) or (xp, yp+1) isa good pair. This contradicts Claim 3, and hence H∗ = H . By Lemma 3,there exists a good set Y∗ for H∗ such that xp ∈ Y∗ and |Y∗| = k. Note thatδH∗(H∗) ≥ δH (H) holds by the choice of H. If there exists v∗ ∈ Y∗ ∩ C(yi, xi]



for some i ∈ {2, . . . , k} \ {p}, then applying Claim 8 (ii) as (v, w, ui, uj) =(v∗, xp, yi, yp), we obtain a contradiction. Hence, for every i ∈ {2, . . . , k} \ {p}, itfollows that Y∗ ∩ C(yi, xi] = ∅, and by the symmetry Y∗ ∩ C[x′

i, yi+1) = ∅ holds.Therefore, Y∗ ∩ C(yi, yi+1) = ∅ for every i ∈ {2, . . . , k} \ {p, t}. Similarly, Y∗ ∩C(y1, y2) = ∅ (whether x1 exists or not). Moreover, by Claim 10, Y∗ ∩ C[yp, xp) =∅ and Y∗ ∩ C(xp, yp+1] = ∅ holds. Therefore, Y∗\{xp} ⊆ (Y\{yp, yp+1}) ∪C(xt, x

′t). This yields k − 1 = |Y∗\{xp}| ≤ |(Y\{yp, yp+1}) ∪ C(xt, x

′t)| = k − 2 +

|C(xt, x′t)|. Hence xt = x′

t , that is, |C(yt, yt+1) ∩ X| = 2 and |C(y1, y2) ∩ X| = 0.Now, if y2 ∈ Y∗, then applying Claim 8 (ii) as (v, w, ui, uj) = (y2, xp, y1, yp),we obtain a contradiction. Hence y2 /∈ Y∗, and by the symmetry y1 /∈ Y∗

holds. Therefore, k − 1 = |Y∗\{xp}| ≤ |Y\{y1, y2, yp, yp+1}| + |Y∗ ∩ C(xt, x′t)| ≤

k − 3 + |Y∗ ∩ C(xt, x′t)|, which implies |Y∗ ∩ C(xt, x

′t)| ≥ 2. Let v∗, w∗ ∈ Y∗ with

C(v∗, w∗) ⊆ C(xt, x′t). By Claim 9, there exists a vertex of NC(H) ∩ C(v∗, w∗),

say uq. Then applying Claim 8 (i) as (v, w, ui, uj) = (xp, w∗, yp, uq), we obtain acontradiction. �

Case 1.1. k ≥ 4 or k = 3 and |C(yt, yt+1) ∩ X| = 1.In this case, there exist distinct integers p, q in {2, . . . , k} such that |C(yp, yp+1) ∩

X| = 1 and |C(yq, yq+1) ∩ X| = 1. Then it follows from Claim 6 that NG−C(xp) =∅ or NG−C(xq) = ∅ holds. This contradicts Claim 11.

Case 1.2. k = 3 and |C(yt, yt+1) ∩ X| = 2.

By the symmetry of C, we may assume that t = 3. Then |C(y1, y2) ∩ X| = 0 and|C(y2, y3) ∩ X| = 1. By Claim 11, we have NG−C(x2) = ∅.

Assume that NG−C(x3) = ∅. Now it follows from Claim 3 that (y2, z2) is not agood pair. Similarly, (y3, z3) is not a good pair. Therefore, applying Lemma 2 (ii)on y2, z2, y3 and z3, we obtain that (z2, z3) is a good pair. This contradicts Claim5, and hence NG−C(x3) = ∅. By the similar argument (considering y3, z

′2, y1, and

z′3), we obtain NG−C(x′

3) = ∅.Let H3 and H ′

3 be components of G − C with x3 ∈ NC(H3) and x′3 ∈ NC(H ′

3).Assume H = H3. By Claim 3, (y3, x3) is not a good pair for H. By Lemma 2 (i), Y :={y1, y2, x3} is a good set for H. Now note that |C(y1, y2) ∩ X| = 0, |C(y2, x3) ∩X| ≤ 2, and C(x3, y1) ∩ X = {x′

3}. Since NG−C(x′3) = ∅, applying Claim 11 as Y =

Y and xp = x′3, we obtain a contradiction. Therefore, we may assume that H = H3.

By Lemma 3, there exists a good set Y∗ for H3 such that x3 ∈ Y∗ and |Y∗| = 3.By Claim 10, Y∗ ∩ (C(x3, x

′3] ∪ C(y3, x3)) = ∅. If there exists v∗ ∈ Y∗ ∩ C(y1, y2]

(resp. Y∗ ∩ C(y2, x2]), then applying Claim 8 (ii) as (v, w, ui, uj) = (x3, v∗, y3, y1)

(resp. (x3, v∗, y3, y2)), we obtain a contradiction. Hence Y∗ ∩ C(y1, x2] =

∅. Therefore, Y∗\{x3} ⊂ C(x2, y3] ∪ C(x′3, y1]. Let v∗, w∗ ∈ Y∗ ∩ (C(x2, y3] ∪

C(x′3, y1]). Suppose that v∗ ∈ Y∗ ∩ C(x2, y3] and w∗ ∈ Y∗ ∩ C(x′

3, y1]. By Claim9, C(v∗, x3) contains a vertex of NC(H), say v. By Claim 3, (v, y3) is not a good pairfor H. By Lemma 2 (i), (v, y2) is a good pair for H. Therefore, by applying Claim8 (ii) as (v, w, ui, uj) = (v∗, w∗, v, y2), we obtain a contradiction. Hence, we mayassume that v∗, w∗ ∈ Y∗ ∩ C(x2, y3] or v∗, w∗ ∈ Y∗ ∩ C(x′

3, y1] holds.



First, suppose that v∗, w∗ ∈ Y∗ ∩ C(x2, y3]. We may assume that x2, v∗ and

w∗ appear in this order along C. By Claim 9, each of C(v∗, w∗) and C(w∗, x3)contains a vertex of NC(H), say v and w, respectively. Then applying Claim 8 (i)as (v, w, ui, uj) = (v∗, w∗, v, w), we obtain a contradiction.

Next, suppose that v∗, w∗ ∈ Y∗ ∩ C(x′3, y1]. We may assume that x′

3, v∗ and w∗

appear in this order along C. By Claim 9, there exists a vertex in NC(H) ∩ C(v∗, w∗),say v. By Claim 3, (v, y1) is not a good pair for H. By Lemma 2 (i), (v, y2) is agood pair for H. Applying Claim 8 (ii) as (v, w, ui, uj) = (v∗, w∗, v, y2), we obtaina contradiction. This completes the proof of Cases 1.2 and 1.

Case 2. There exists no long segment in C.

In this case, it follows from Claim 1 that

for any a, b ∈ NC(H) with C(a, b) ∩ X = ∅, (a, b) is not a good pair for H. (5)

Also by Claim 1, C(yi, yi+1) ∩ X = ∅ for every i. By Claim 7, there are at most kvertices of X\Y , and hence |C(yi, yi+1) ∩ X| = 1 holds for every i. Then it followsfrom Claim 6 that NG−C(xp) = ∅ for some p. Let H∗ be a component of G − C

such that NH∗(xp) = ∅. Assume H∗ = H . By Lemma 2 (i), (yp, xp) or (xp, yp+1)is a good pair, which contradicts (5). Hence H∗ = H , and by Lemma 3, we cantake a good set Y∗ for H∗ such that xp ∈ Y∗ and |Y∗| = k.

It follows from Claim 8 (ii) that Y∗ ∩ C(yi, xi] = ∅ for every i ∈ {1, . . . , k} \ {p},and by the symmetry of C, Y∗ ∩ C[xi, yi+1) = ∅ for every i ∈ {1, . . . , k} \ {p}.Therefore, Y∗\{xp} ⊂ (Y\{yp, yp+1}) ∪ (C[yp, xp) ∪ C(xp, yp+1]). This impliesY∗ ∩ (C[yp, xp) ∪ C(xp, yp+1]) = ∅. By the symmetry of C, we may assume thatthere is a vertex y∗ in Y∗ ∩ C[yp, xp). Then |C(y∗, xp)| ≥ δH∗(H∗) + 1 ≥ δH (H) +1. By the assumption of Case 2, there exists a vertex of C(y∗, xp) ∩ NC(H), sayv. It follows from (5) that (yp, v) is not a good pair for H. Hence, by Lemma 2(i), (v, yp+1) is a good pair for H. Now applying Claim 8 (ii) as (v, w, ui, uj) =(y∗, xp, v, yp+1), we obtain a contradiction. This completes the proof of Case 2 andTheorem 7.

ACKNOWLEDGMENTS

This work was partially supported by the JSPS Research Fellowships for YoungScientists (to J. F.) and by the 21st Century COE Program.

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