Advances in Applied Mathematics 54 (2014) 105–120

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Advances in Applied Mathematics

www.elsevier.com/locate/yaama

Minimum degree conditions for vertex-disjoint even cycles inlarge graphs ✩

Shuya Chiba a,∗,1, Shinya Fujita b,2, Ken-ichi Kawarabayashi c,d,3,Tadashi Sakuma e,4

a Department of Mathematics and Engineering, Kumamoto University, 2-39-1, Kurokami,Kumamoto 860-8555, Japanb International College of Arts and Sciences, Yokohama City University, 22-2 Seto, Kanazawa-ku,Yokohama, Kanagawa 236-0027, Japanc National Institute of Informatics, 2-1-2, Hitotsubashi, Chiyoda-ku, Tokyo, Japand JST, ERATO, Kawarabayashi Large Graph Project, Japane Systems Science and Information Studies, Faculty of Education, Art and Science,Yamagata University, 1-4-12 Kojirakawa, Yamagata 990-8560, Japan

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

Article history:Received 31 December 2012Accepted 2 December 2013Available online 9 January 2014

MSC:05C70

Keywords:Vertex-disjoint cyclesEven cycleTheta graphMinimum degree

We prove a variant of a theorem of Corrádi and Hajnal (1963)[4] which says that if a graph G has at least 3k vertices andits minimum degree is at least 2k, then G contains k vertex-disjoint cycles. Specifically, our main result is the following.For any positive integer k, there is a constant ck such thatif G is a graph with at least ck vertices and the minimumdegree of G is at least 2k, then (i) G contains k vertex-disjointeven cycles, or (ii) (2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2(p � k � 2), or (iii) k = 1 and each block of G is either a K2or an odd cycle.

© 2013 Elsevier Inc. All rights reserved.

✩ An extended abstract has been published in: EuroComb 2009, Electr. Notes Discrete Math., vol. 34,2009, pp. 113–119.* Corresponding author.

E-mail addresses: schiba@kumamoto-u.ac.jp (S. Chiba), shinya.fujita.ph.d@gmail.com (S. Fujita),k_keniti@nii.ac.jp (K. Kawarabayashi), sakuma@e.yamagata-u.ac.jp (T. Sakuma).1 This work was supported by JSPS KAKENHI grant 23740087.2 This work was supported by JSPS KAKENHI grant 23740095.3 This work was supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific

Research, by C & C Foundation, by Kayamori Foundation and by Inoue Research Award for YoungScientists.

0196-8858/$ – see front matter © 2013 Elsevier Inc. All rights reserved.http://dx.doi.org/10.1016/j.aam.2013.12.001

106 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

1. Introduction

In this paper, we consider only finite simple graphs. For terminology and notationnot defined in this paper, we refer the readers to [5]. Let G be a graph. We denote byV (G), E(G), δ(G) and Δ(G) the vertex set, the edge set, the minimum degree and themaximum degree of G, respectively. We refer to the cardinality of V (G) as the order ofG and denote it by |G|. For a graph H, if H is a subgraph of G, then we write H ⊂ G.Subgraphs of G are said to be vertex-disjoint if no two of them have any common vertexin G. For X ⊆ V (G), we let G[X] denote the subgraph of G induced by X, and letG−X = G[V (G) \X]. For H ⊂ G, let G−H = G− V (H).

Packing and covering is one of the central areas in both graph theory and theoreticalcomputer science. The starting point of this research area goes back to the followingwell-known theorem due to Erdős and Pósa [7] in early 1960s.

Theorem A. (See Erdős and Pósa [7].) For any integer k with k � 1 and any graph G,either G contains k vertex-disjoint cycles or a vertex set X of order at most f(k) (forsome function f of k) such that G−X is a forest.

In fact, Theorem A gives rise to the well-known Erdős–Pósa property. A family F ofgraphs is said to have the Erdős–Pósa property, if for every integer k � 1, there is aninteger f(k,F) such that every graph G contains either k vertex-disjoint subgraphs eachisomorphic to a graph in F or a set X of at most f(k,F) vertices such that G−X hasno subgraph isomorphic to a graph in F . The term Erdős–Pósa property arose becauseof Theorem A which proves that the family of cycles has this property.

Theorem A concerns both “packing”, i.e., k vertex-disjoint cycles, and “covering”,i.e., a set of at most f(k) vertices that hits all the cycles in G. Starting with this result,there are a lot of the results in this direction. Packing appears almost everywhere inextremal graph theory, while covering leads to the well-known concept “feedback set”in theoretical computer science. Also, the cycle packing problem, which asks whether ornot there are k vertex-disjoint cycles in an input graph G, is a well-known problem, e.g.[12].

In graph theory, there are many results concerning packing cycles. The following isthe well-known theorem due to Corrádi and Hajnal [4] in 1960s.

Theorem B. (See Corrádi and Hajnal [4].) Let k be an integer with k � 1, and let G bea graph of order at least 3k. If δ(G) � 2k, then G contains k vertex-disjoint cycles.

Theorem B tells us that if we assumed that the minimum degree of a given graph is atleast 2k, then the covering result in Theorem A would not happen. In view of Theorems Aand B, we would like to discuss how a parity condition on the cycles affects Theorems A

4 This work was supported by JSPS KAKENHI.

S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120 107

and B. Let us first consider the odd cycle case. As observed by Lovász and Schrijver (see[14]), there is a graph that does not have two vertex-disjoint odd cycles but needs at least√n vertices to cover all odd cycles. Hence the Erdős–Pósa property does not hold for

odd cycles in general. Also, it is easy to see that in order to guarantee the existence of kvertex-disjoint odd cycles for a graph G, the minimum degree condition should be at least(|G| + k)/2 (just consider a graph obtained from K(n−k+1)/2,(n−k+1)/2 by adding k − 1vertices and joining each of these k − 1 vertices to all vertices in K(n−k+1)/2,(n−k+1)/2,where for m � 1 and n � 1, Km,n denotes the complete bipartite graph with partitesets of cardinalities m and n, respectively. This graph does not contain k vertex-disjointodd cycles, yet the minimum degree is n+k−1

2 ). On the other hand, the minimum degreerequirement in Theorem B is only 2k (which does not depend on the order of a graph G).Thus this parity condition makes a huge gap.

Somewhat surprisingly, the situation is quite different if we consider the even cyclecase. Thomassen [15] proved that the Erdős–Pósa property holds for even cycles. In viewof Theorems A, B and Thomassen’s result, one might expect that a result like Theorem Bfor even cycles would hold. In fact, we had conjectured that, if a graph has sufficientlylarge order, then we would be able to replace “vertex-disjoint cycles” to “vertex-disjointeven cycles” in Theorem B. We prove that this conjecture is true except for two singularcases, which is our main result (Theorem 1) in this paper.

Let Km denote the complete graph of order m. For two graphs G1, G2 with V (G1)∩V (G2) = ∅, let G1∪G2 denote the union of G1 and G2, and let G1∨G2 denote the join ofG1 and G2, i.e., the graph obtained from G1 ∪G2 by joining each vertex in V (G1) to allvertices in V (G2). For a graph G and s � 1, let sG denote the union of s vertex-disjointcopies of G.

Theorem 1. Let k be an integer with k � 1, and let ck = 4(3k+1)α3(k−1)(3(k−1))2(k−1)+(12 + α)(k − 1), where α := 9600k2 + 36k + 11. Let G be a graph of order at least ck. Ifδ(G) � 2k, then one of the following holds.

(i) G contains k vertex-disjoint even cycles.(ii) k � 2 and (2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2 for some integer p with p � k.(iii) k = 1 and each block of G is either a K2 or an odd cycle.

The minimum degree requirement in Theorem 1 is best possible precisely for thesame reason that the minimum degree in Theorem B is best possible. Furthermore, ifthe conclusion (ii) or (iii) holds, clearly there are no k vertex-disjoint even cycles. So ifthe minimum degree of a graph G is at least 2k + 1, and G has at least ck vertices, thenG always contains k vertex-disjoint even cycles.

We do not know what the best possible value of ck is. An obvious lower bound of ckis 4k, as we need at least 4k vertices in a packing of k even cycles. In fact, for the casewhere the graph has precisely 4k vertices, Theorem 1 is known to hold. A conjecture ofErdős and Faudree [6], recently proved by Wang, states that a minimum degree 2k in agraph with 4k vertices suffices for a packing with k vertex-disjoint copies of a 4-cycle.

108 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

Wang’s proof in [16] contains over 40 pages. This is because sharpening the bound on theorder of a graph requires us to consider many cases in the proof. For the sake of clarity inthe proof, we do not attempt to improve the bound on ck in this paper. Note in passingthat a conjecture of Alon and Yuster [1], which was proved by Komlós, Sárközy andSzemerédi [11], also implies a weaker version of the conjecture of Erdős and Faudree;that is, for sufficiently large k, if the minimum degree of a graph G with 4k vertices isat least 2k + 3, then G contains k vertex-disjoint copies of a 4-cycle.

In order to prove Theorem 1, we actually consider a stronger statement. A theta graphis a graph consisting of the union of three internally disjoint paths that have the sametwo distinct end vertices. We are interested in finding k vertex-disjoint theta graphs,because a theta graph always contains an even cycle, and hence k vertex-disjoint thetagraphs give rise to k vertex-disjoint even cycles. Note that, since every subdivision of atheta graph is again a theta graph and planar, from Robertson and Seymour [13], it hasthe Erdős–Pósa property.

The following result is our technical result, which immediately implies Theorem 1.

Theorem 2. Under the same conditions as Theorem 1, one of the following holds.

(i) G contains k vertex-disjoint theta graphs.(ii) k � 2 and (2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2 for some integer p with p � k.(iii) k = 1 and each block of G is either a K2 or a cycle.

(Note that δ(G) = 2k in the conclusions (ii) and (iii).) Concerning the above, let usnote some related results for the case of the smallest theta graph K−

4 , i.e., the graphobtained from K4 by deleting one edge. It is known that if the minimum degree of agraph G with at least 4k vertices is at least (|G| + k)/2, then G always contains k

vertex-disjoint copies of K−4 , and the minimum degree condition is best possible (see [9,

10]). On the other hand, if we have only to find theta graphs, then the weaker minimumdegree condition as in Theorem 2 guarantees the existence, provided that a given graphhas at least ck vertices. This is quite different from the case of K−

4 ’s.We mention about another special theta graph. A chorded cycle is a cycle with at

least one chord. In [2,3,8], it is shown that if the minimum degree of a given graph G

(with at least 4k vertices) is at least 3k, then there are k vertex-disjoint chorded cycles,and the minimum degree condition is best possible.

Now we prepare terminology and notation which we use in subsequent sections. LetG be a graph. For a vertex v of G, we denote by dG(v) and NG(v) the degree and theneighborhood of v in G, respectively. For an integer l, let Vl(G) = {v ∈ V (G): dG(v) = l},and let V�l(G) =

⋃m�l Vm(G) and V�l(G) =

⋃m�l Vm(G). For two subsets X and Y of

V (G) with X ∩Y = ∅, let EG(X,Y ) denote the set of edges between X and Y . We oftenidentify a subgraph H of G with its vertex set V (H). For example, we write EG(H,F )instead of EG(V (H), V (F )) for two vertex-disjoint subgraphs H and F of G. We write apath P with a given orientation by −→

P . If there exists no fear of confusion, we abbreviate

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−→P by P . Let −→

P be an oriented path. For u, v ∈ V (P ), we denote by u−→Pv a subpath of P

from u to v along −→P .

In the rest of this section, we give a very high level overview of our proof of Theorem 2.We prove Theorem 2 by induction on k. Let G be a graph with |G| � ck and δ(G) � 2k.

If k = 1, then we can easily see that (i) or (iii) of Theorem 2 holds. Thus we may assumethat k � 2. Suppose that G does not contain k vertex-disjoint theta graphs. Since|G| � ck � ck−1 and δ(G) � 2k > 2(k − 1), it follows from the induction hypothesisthat G contains k − 1 vertex-disjoint theta graphs T1, . . . , Tk−1. Let T ∗ =

⋃k−1i=1 Ti,

and let H = G − T ∗. We choose T1, . . . , Tk−1 so that |T ∗| is as small as possible. Letβ = max{|Ti|: 1 � i � k − 1}.

For a technical reason, we need to divide the proof into two cases.

Case 1. β � 9600k2 + 36k + 11.

In this case, we shall show that (2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2. Since H

does not contain a theta graph, it can be shown that |V�2(H)| > |H|/4 (see Lemma 4in Section 2). Therefore, since the order of G is sufficiently large compared with |T ∗|(� β(k − 1)), there are many vertices of H with degree at most 2 in H (there exist atleast (|G| − β(k − 1))/4 vertices with degree 2 in H), i.e., there are many vertices of Hsuch that each of them has at least 2k − 2 neighbors on T ∗. Then by the PigeonholePrinciple, we can find a complete bipartite subgraph with partite sets U and W in G

such that |U | = 2k−2 and |W | � 3k+2, see Claim 1 in Section 3. (In order to obtain thisstructure, we need the huge order.) This complete bipartite subgraph allows us to confirmthat (2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2, where (2k − 1)K1 ⊂ G[U ∪ {z}] ⊂ K2k−1for some vertex z in G− U .

Case 2. β � 9600k2 + 36k + 12.

Without loss of generality, we may assume that |Tk−1| = β. In this case, we willderive a contradiction by finding k − 1 vertex-disjoint theta graphs such that the sumof their order is less than |T ∗|. In order to show this, we take k− 2 other vertex-disjointtheta graphs R1, . . . , Rk−2 so that

∑k−2i=1 |Ri| is as small as possible. Let R∗ =

⋃k−2i=1 Ri.

By taking such k − 2 vertex-disjoint theta graphs, we can easily estimate the numberof edges between R∗ and G − R∗ (see Lemmas 5 and 6 in Section 2). Moreover, since|G − R∗| � |H| + |Tk−1| � β by the minimality of |R∗|, and since G − R∗ has no twovertex-disjoint theta graphs and β is sufficiently large compared with k, there are manyvertices of G − R∗ with degree at most 2 in G − R∗ (see Lemma 9 in Section 2 andClaim 8 in Section 3), i.e., there are many vertices of G − R∗ such that each of themhas at least 2k− 2 neighbors on R∗. Since R∗ =

⋃k−2i=1 Ri, it follows from the Pigeonhole

Principle that there exists a complete bipartite subgraph with partite sets U and W inG such that U ⊆ V (G − R∗), W ⊆ V (Ri) ∪ V (Rj) for some i, j with 1 � i � k − 2and 1 � j � k − 2, and if i = j, then |U | � 4 and |W | � 4; otherwise, |U | � 9,

110 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

|W ∩ V (Ri)| � 3 and |W ∩ V (Rj)| � 3. (In order to obtain this structure, we need thehuge order.) This together with the minimality of |R∗| implies that G[U ∪W ] containstwo or three vertex-disjoint small theta graphs. Hence we can find k − 1 vertex-disjointtheta graphs such that the sum of their orders is less than |T ∗|, a contradiction.

2. Technical lemmas

In this section, we give some lemmas which will be used in our proof of Theorem 2.

2.1. Graphs with no theta graph

In this subsection, let G be a graph, and suppose that G does not contain a thetagraph. Under this assumption, the following two lemmas hold. Lemma 1 is easy and weomit its proof.

Lemma 1. (i) Each block of G is a K1 or a K2 or a cycle.(ii) If δ(G) � 2 and |V�3(G)| � 2, then G contains two vertex-disjoint cycles.

The following lemma will be used to show the result concerning the number of verticeswith degree at most 2 in a graph which has no two vertex-disjoint theta graphs (seeLemma 9 in Subsection 2.4).

Lemma 2. Let S be a subset of V (G) with |S| � 3. If G is connected, then there exists aconnected subgraph H of G such that H satisfies the following:

(i) 3 � |V (H) ∩ S| � 4.(ii) |{v ∈ V (H): dH(v) = dG(v)}| � |H| − 2.

Proof. Clearly there exists a connected induced subgraph H of G such that |V (H)∩S| �3 and G−H is connected, or H = G. We choose such a graph H so that |H| is as smallas possible. Since G does not contain a theta graph, it follows from Lemma 1(i) thateach block of G is a K2 or a cycle. If H = G, then by the minimality of |H|, we caneasily see that |S| = 3. Hence G itself is the desired subgraph. Thus we may assume that|H| < |G|.

Let D = {v ∈ V (H): NG(v)\V (H) = ∅}. Note that 1 � |D| � 2 since G has no thetagraph and both H and G − H are connected. Hence if |V (H) ∩ S| � 4, then H is thedesired subgraph of G. Thus we may assume that |V (H) ∩ S| � 5. Let v1, v2 ∈ D withv1 = v2 if |D| = 2; otherwise, let v1 = v2 be the unique vertex in D. Let F1, F2, . . . , Fl

be components of H − v1 (by the minimality of |H|, H − v1 is not connected). Byagain the minimality of |H|, |V (Fi) ∩ S| � 2 for each 1 � i � l. Hence there existsa subset I of {1, 2, . . . , l} such that 3 � |S ∩ ({v1} ∪ (

⋃i∈I V (Fi)))| � 4. Then H ′ :=

G[{v1} ∪ (⋃

i∈I V (Fi))] is connected and EG(H ′ − {v1, v2}, G −H ′) = ∅. Therefore, H ′

is the desired subgraph of G. �

S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120 111

2.2. The number of vertices with degree at most 2 in a graph with no theta graph

In this subsection, let G be a graph, and suppose that G does not contain a thetagraph.

In Case 1 of the proof of Theorem 2, we shall show that G contains a complete bipartitegraph K2k−2,3k+2. In order to do that, we need to estimate the number of vertices withdegree at most 2 (see the first paragraph in Case 1 and Claim 1). More precisely, weneed the following lemmas.

Lemma 3. Let fG : V (G) → Z be a function such that for each v ∈ V (G), if dG(v) � 3,then fG(v) � −1; if dG(v) = 2, then fG(v) � 2; otherwise, fG(v) � 3. We have∑

v∈V (G) fG(v) > 0.

Proof. We may assume that G is connected. We prove Lemma 3 by induction on |G|. If|G| � 3, then V�3(G) = ∅. Hence by the assumption of fG,

∑v∈V (G) fG(v) > 0. Thus

we may assume that |G| � 4.Note that by Lemma 1(i), each block of G is a K2 or a cycle. If G has exactly one

block, then by the assumption of fG, we can easily see that the assertion holds. Thus wemay assume that G has two distinct end blocks. Let B be an end block of G, and let xB

be a unique cut vertex contained in B. Let G′ = G−V (B−{xB}). We define a functionfG′ : V (G′) → Z as follows. For each v′ ∈ V (G′), let

fG′(v′)

={∑

v∈V (B) fG(v) (if v′ = xB),fG(v′) (otherwise).

Then by the definitions of G′ and fG′ , we can easily see that the following hold (notethat if B ∼= K2 and dG′(xB) � 1, or B is a cycle, then

∑v∈V (B) fG(v) � 3; otherwise,∑

v∈V (B) fG(v) � 2).

(i) G′ is a connected graph such that |G′| < |G| and G′ does not contain a theta graph.(ii)

∑v′∈V (G′) fG′(v′) =

∑v∈V (G) fG(v).

(iii) For each v′ ∈ V (G′), fG′(v′) �

⎧⎨⎩

−1 (if dG′(v′) � 3),2 (if dG′(v′) = 2),3 (otherwise).

By (i) and (iii), we can apply the induction hypothesis to G′. Combining this with (ii),we obtain

∑v∈V (G) fG(v) =

∑v′∈V (G′) fG′(v′) > 0. �

Lemma 4. |V�2(G)| > |G|/4.

Proof. Suppose that |V�2(G)| � |G|/4, i.e., |V�3(G)| � 3|G|/4. For each v ∈ V (G), wedefine

112 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

fG(v) =

⎧⎪⎨⎪⎩

−1 (if dG(v) � 3),2 (if dG(v) = 2),3 (otherwise).

Then by Lemma 3 and since |V�3(G)| � 3|G|/4, we obtain

0 <∑

v∈V (G)

fG(v) =∑

v∈V�1(G)

fG(v) +∑

v∈V2(G)

fG(v) +∑

v∈V�3(G)

fG(v)

= 3∣∣V�1(G)

∣∣ + 2∣∣V2(G)

∣∣− ∣∣V�3(G)∣∣

= 3|G| −∣∣V2(G)

∣∣− 4∣∣V�3(G)

∣∣ � 3|G| − 4∣∣V�3(G)

∣∣ � 0,

a contradiction. �2.3. Upper bounds on the order of a minimal theta graph

In this and next subsection, we fix the following notation. Let G be a graph, andlet T be a theta graph in G. We let x and y be two distinct vertices in T such thatthere exist three internally disjoint paths connecting x and y, and let P1, P2 and P3 bethree internally disjoint paths in T such that V (Pi) ∩ V (Pj) = {x, y} for each i, j with1 � i < j � 3. For each Pi (1 � i � 3), we give an orientation from x to y along edgesof Pi.

Now we suppose that there exists no theta graph of order smaller than |T |. Underthis assumption, the following two lemmas hold.

Lemma 5. Let u ∈ V (G − T ), and let v1 and v2 be two distinct vertices in T such thatuvi ∈ E(G) for i ∈ {1, 2}. Then the following hold.

(i) If v1 ∈ V (Pi − {x, y}) and v2 ∈ V (Pj − {x, y}) for some i, j with 1 � i < j � 3,then |T | � 9.

(ii) If v1, v2 ∈ V (Pi) and v1 ∈ V (x−→Piv2) for some i with 1 � i � 3, then |v1

−→Piv2| � 3.

Proof. (i) We may assume that i = 1 and j = 2. Since G[V (x−−→P1v1 ∪ P2 ∪ P3) ∪ {u}]

contains a theta graph, it follows from the minimality of |T | that |v1−−→P1y| � 3. Similarly,

we have |x−−→P1v1| � 3, |v2

−−→P2y| � 3 and |x−−→

P2v2| � 3. Since G[V (T−P3)∪{x, y, u}] containsa theta graph, we have |P3| � 3. Thus |T | � 9.

(ii) We may assume that i = 1. Then G[V (P2 ∪ P3 ∪ x−−→P1v1 ∪ v2

−−→P2y) ∪ {u}] contains

a theta graph. Hence by the minimality of |T |, we have |v1−−→P1v2| � 3. �

Lemma 6. Let u ∈ V (G − T ). If |EG(u, T )| � 3, then |T | � 5. In particular, if|EG(u, T )| � 4, then |T | = 4.

Proof. Suppose that |EG(u, T )| � 3, and let v1, v2 and v3 be three distinct vertices in T

such that uvi ∈ E(G) for 1 � i � 3. By the symmetry of v1, v2 and v3, the symmetry of

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x and y and the symmetry of P1, P2 and P3, we may assume that one of the followingthree cases holds.

Case 1. v1, v2 and v3 appear in this order along −−→P1.

Then by Lemma 5(ii), |v1−→Pv3| = 3, i.e., v1v2v3 is a subpath of T . Since G[{u, v1, v2, v3}]

contains a theta graph of order 4, it follows from the minimality of |T | that |T | = 4.

Case 2. v1 and v2 appear in this order along −−→P1, v2 = y and v3 ∈ V (P2 − {x, y}).

Since G[V (x−−→P1v2∪x

−−→P2v3)∪{u}] contains a theta graph, it follows from the minimality

of |T | that |T − (x−−→P1v2 ∪ x

−−→P2v3)| = 1, in particular, |P3| = 2 and v2y ∈ E(T ). Since

G[V (P2∪P3)∪{v2, u}] contains a theta graph, it follows from the minimality of |T | that|x−−→P1v2| � 3. Moreover, since G[V (P1 ∪P3)∪{u}] contains a theta graph, |P2| = 3. Thus

|T | � 5.

Case 3. vi ∈ V (Pi − {x, y}) for 1 � i � 3.

Since G[V (x−−→P1v1∪x

−−→P2v2∪x

−−→P3v3)∪{u}] and G[V (v1

−−→P1y∪v2

−−→P2y∪v3

−−→P3y)∪{u}] contain

a theta graph, respectively, it follows from the minimality of |T | that vix, viy ∈ E(G)for 1 � i � 3. Thus |T | = 5.

By Cases 1–3, we have |T | � 5. In particular, if |EG(u, T )| � 4, then by the factthat |T | � 5, there exists a subpath P of T such that |P | = 3 and V (P ) ⊆ NG(u), i.e.,G[V (P )∪{u}] contains a theta graph of order 4, and hence it follows from the minimalityof |T | that |T | = 4. �2.4. The number of vertices with degree at most 2 in a graph with no twovertex-disjoint theta graphs

Now we suppose that there exists no theta graph of order smaller than |T |. We furthersuppose that |T | � 10 and G does not contain two vertex-disjoint theta graphs (notethat by the minimality of |T |, T is an induced subgraph of G), and we fix the followingnotation. Let F be the set of components of G − T . For each F ∈ F , we define SF ={v ∈ V (F ): NG(v) ∩ V (T ) = ∅}.

In this last subsection, we prepare the result concerning the number of vertices withdegree at most 2 in the graph G which satisfies the above assumptions, see Lemma 9.To show this, we use the following two lemmas (Lemmas 7 and 8).

Lemma 7. Let F ∈ F . If |SF | � 2, then V (F ) ∩ V�2(G) = ∅.

Proof. Since G has no two vertex-disjoint theta graphs, it follows from Lemma 1(i) thateach block of F is a K1 or a K2 or a cycle. Therefore, since |SF | � 2, if |F | � 3, thenwe can easily see that V (F ) ∩ V�2(G) = ∅. Thus we may assume that |F | � 2. Suppose

114 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

that |F | = 1, say V (F ) = {v}. Then by Lemma 6 and the minimality of |T | and since|T | � 10, we have |EG(v, T )| � 2, and hence v ∈ V (F ) ∩ V�2(G). Thus |F | = 2. Thenby Lemma 5 and the minimality of |T | and since |T | � 10, it is easy to check that thereexists a vertex v in F such that |EG(v, T )| � 1, and hence v ∈ V (F ) ∩ V�2(G). �Lemma 8. Let F ∈ F . If |SF | � 3, then |V (F ) ∩ V�2(G)| � |T |/12 − 7.

Proof. Since F has no theta subgraph and |SF | � 3, it follows from Lemma 2 thatthere exists a connected subgraph F ′ of F such that 3 � |V (F ′) ∩ SF | � 4 and |{v ∈V (F ′): dF ′(v) = dF (v)}| � |F ′| − 2.

We first show that

∣∣F ′∣∣ � |T |/3 − 1. (2.1)

Since |SF ∩V (F ′)| � 3, there exist three vertices v1, v2 and v3 in T such that NG(vi)∩V (F ′) = ∅ for 1 � i � 3 and |(

⋃3i=1 NG(vi)) ∩ V (F ′)| � 3. Assume that v1, v2 and v3

are not distinct vertices. Then there exists a connected subgraph Q of T such that{v1, v2, v3} ⊆ V (Q) and |Q| � |T |/2 + 1. Since G[V (F ′)∪ V (Q)] contains a theta graph,it follows from the minimality of |T | that |F ′| � |T |/2 − 1. Thus we may assume thatv1, v2 and v3 are three distinct vertices in T . By the symmetry of v1, v2 and v3, thesymmetry of x and y and the symmetry of P1, P2 and P3, it is enough to consider thefollowing three cases.

Case 1. v1, v2 and v3 appear in this order along −−→P1.

Let G1 = v1−−→P1v3, and let G2 = T [V (T −G1)∪{v1, v3}]. Then G[V (F ′ ∪Gi)] contains

a theta graph for i ∈ {1, 2}. Since |Gi| � |T |/2 + 1 for some i with i ∈ {1, 2}, it followsfrom the minimality of |T | that |F ′| � |T |/2 − 1.

Case 2. v1 and v2 appear in this order along −−→P1, v2 = y and v3 ∈ V (P2 − {x, y}).

Let G1 = v1−−→P1v2, G2 = x

−−→P1v1 ∪ x

−−→P2v3 and G3 = v2

−−→P1y ∪ v3

−−→P2y. Then G[V (F ′ ∪Gi ∪

Gj)] contains a theta graph for each i, j with 1 � i < j � 3. Since |Gi ∪Gj | � 2|T |/3+1for some i, j with 1 � i < j � 3, it follows from the minimality of |T | that |F ′| � |T |/3−1.

Case 3. vi ∈ V (Pi − {x, y}) for 1 � i � 3.

Let G1 =⋃3

i=1 x−→Pivi and G2 =

⋃3i=1 vi

−→Piy. Then G[V (F ′ ∪ Gi)] contains a theta

graph for i ∈ {1, 2}. Since |Gi| � |T |/2 + 1 for some i with i ∈ {1, 2}, it follows from theminimality of |T | that |F ′| � |T |/2 − 1.

By Cases 1–3, we have |F ′| � |T |/3 − 1, and thus (2.1) is proved.Since F ′ does not contain a theta graph, it follows from Lemma 4 and (2.1) that

|V�2(F ′)| � |F ′|/4 � |T |/12 − 1/4. Since |SF ∩ V (F ′)| � 4, |(V (F ′) ∩ V�2(F )) \ SF | �

S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120 115

|V (F ′) ∩ V�2(F )| − 4. Moreover, since |{v ∈ V (F ′) : dF ′(v) = dF (v)}| � |F ′| − 2,|V (F ′) ∩ V�2(F )| � |V�2(F ′)| − 2. Therefore, we obtain

∣∣V (F ) ∩ V�2(G)∣∣ � ∣∣V (

F ′) ∩ V�2(G)∣∣ � ∣∣(V (

F ′) ∩ V�2(F ))\ SF

∣∣�

∣∣V (F ′) ∩ V�2(F )

∣∣− 4 �∣∣V�2

(F ′)∣∣− 6 � |T |/12 − 7. �

Lemma 9. |V�2(G)| � |T |/12 − 7.

Proof. If |T |/12 − 7 � 0, then we clearly have |V�2(G)| � 0 � |T |/12 − 7. Thus we mayassume that |T |/12 − 7 > 0. By Lemma 8, we may assume that |SF | � 2 for all F ∈ F .Then by Lemma 7, we may assume that |F| < |T |/12 − 7 (otherwise, the assertionholds). Let FS = {F ∈ F : SF = ∅} and X = {v ∈ V (T ): NG(v) ∩ V (G − T ) = ∅}.Since |SF | � 2 for each F ∈ FS and |T | � 10, it follows from Lemma 6 and theminimality of |T | that |EG(F, T )| � 4 for each F ∈ FS . Hence |X| � |T | − 4|FS |. Notethat X \ {x, y} ⊆ V�2(G) because T is an induced subgraph of G. Thus |V�2(G)| �|X \ {x, y}| � |X| − 2 � |T | − 4|FS | − 2 � |T | − 4|F| − 2 > 2|T |/3 + 26 > |T |/12− 7. �3. Proof of Theorem 2

We prove Theorem 2 by induction on k. Let G be a graph with |G| � ck and δ(G) � 2k.If k = 1, then by Lemma 1(i), (i) or (iii) of Theorem 2 holds. Thus we may assume thatk � 2.

Suppose that G does not contain k vertex-disjoint theta graphs. Since |G| � ck � ck−1

and δ(G) � 2k > 2(k−1), it follows from the induction hypothesis that G contains k−1vertex-disjoint theta graphs T1, T2, . . . , Tk−1. Let T ∗ =

⋃k−1i=1 Ti, and let H = G − T ∗.

Then H does not contain a theta graph. We choose T1, . . . , Tk−1 so that |T ∗| is as smallas possible. By the minimality of |T ∗|, for any collection of k − 1 vertex-disjoint thetagraphs in G, the sum of their order is at least |T ∗|. Note that for each 1 � i � k − 1, if|Ti| � 5, then Ti is an induced subgraph of G, that is, G[V (Ti)] ∼= Ti (since otherwise,G[V (Ti)] contains a smaller theta graph, contradicting the minimality of |T ∗|). Let β =max{|Ti|: 1 � i � k − 1}.

Case 1. β � 9600k2 + 36k + 11.

Since H does not contain a theta graph, it follows from Lemma 4 that |V�2(H)| >|H|/4. Since |G| � ck = 4(3k+1)α3(k−1)(3(k−1))2(k−1) +(12+α)(k−1), β � 9600k2 +36k + 11 (= α) and |T ∗| � β(k − 1), we obtain

∣∣V�2(H)∣∣ > |H|/4 =

(|G| −

∣∣T ∗∣∣)/4� (3k + 1)β3(k−1)(3(k − 1)

)2(k−1) + 3(k − 1). (3.1)

116 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

Claim 1. There exist U ⊆ V (T ∗) and W0 ⊆ V�2(H) with |U | = 2k− 2 and |W0| � 3k+2such that U ⊆ NG(w) for all w ∈ W0.

Proof. Note that |EG(v, T ∗)| � 2k − 2 for all v ∈ V�2(H). Let W1 = {w ∈V�2(H): |EG(w, Ti)| � 4 for some i with 1 � i � k − 1}, and let W2 = V�2(H) −W1.

Assume for the moment that |W1| > 3(k − 1). Then there exist a subset W of W1and a theta graph Ti with 1 � i � k − 1 such that |W | = 4 and |EG(w, Ti)| � 4 for allw ∈ W . Then, since |Ti| = 4 by Lemma 6, G[V (Ti) ∪ W ] contains two vertex-disjointtheta graphs, and hence G contains k vertex-disjoint theta graphs, a contradiction. Thus|W1| � 3(k − 1).

By the definition of W1 and W2,∣∣EG(w, Ti)

∣∣ � 3 for each w ∈ W2 and each Ti with 1 � i � k − 1. (3.2)

Moreover, by (3.1),

|W2| =∣∣V�2(H)

∣∣− |W1| > (3k + 1)β3(k−1)(3(k − 1))2(k−1)

� (3k + 1)β3(k−1)(

3(k − 1)2(k − 1)

). (3.3)

Let U∗ = {⋃k−1

i=1 Ui: Ui ∈(V (Ti)

3)

for each i with 1 � i � k − 1}, where for a finiteset X and an integer s with s � 1,

(Xs

)denotes the family of all s-element subsets

of X. Since |Ti| � β for each 1 � i � k − 1, |(V (Ti)

3)| =

(|Ti|3)�

(β3)< β3 for each

1 � i � k − 1. Hence by the definition of U∗, |U∗| < β3(k−1). Let U =⋃

U∗∈U∗(

U∗

2k−2).

Since |(

U∗

2k−2)| =

( |U∗|2k−2

)=

(3k−32k−2

)for each U∗ ∈ U∗ by the definition of U∗, we obtain

|U| < β3(k−1)(3(k−1)2(k−1)

). On the other hand, by (3.2) and since |EG(v, T ∗)| � 2k − 2 for

all v ∈ V�2(H), we see that for each w ∈ W2, there exists U ∈ U such that U ⊆ NG(w).Since |U| < β3(k−1)(3(k−1)

2(k−1)), this together with (3.3) implies that there exist U ∈ U and

W0 ⊆ W2 with |W0| � 3k + 2 such that U ⊆ NG(w) for all w ∈ W0. Since U ⊆ V (T ∗)and |U | = 2k − 2, the desired conclusion holds. �

Let U and W0 be the same as in Claim 1. Let G′ = G − U , and let W = {w ∈V (G′): U ⊆ NG(w)}. Since |U | = 2k − 2, it follows from the definition of G′ thatδ(G′) � 2. Since W0 ⊆ W , we have |W | � |W0| � 3k + 2. Furthermore, for any subsetU ′ ⊆ U and W ′ ⊆ W with |U ′| = 2 and |W ′| = 3, G[U ′ ∪ W ′] contains a K2,3, i.e.,G[U ′ ∪W ′] contains a theta graph. Therefore we can easily see that the following factholds.

Fact 2. Let i be an integer with 1 � i � k. Then for any U ′ ⊆ U and W ′ ⊆ W with|U ′| � 2(k−i) and |W ′| � 3(k−i), G[U ′∪W ′] contains k−i vertex-disjoint theta graphs.

For a connected subgraph F of G′, we call F a partial theta graph of G′ if (i) |V (F )∩W | = 3, or (ii) |V (F )∩W | = 2 and F contains two distinct paths connecting two vertices

S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120 117

in V (F )∩W (note that two such distinct paths may have common edges or vertices otherthan end vertices). For a partial theta graph F of G′ and u ∈ U , G[V (F )∪{u}] containsa theta graph, and hence the following claim holds.

Claim 3. G′ does not contain two vertex-disjoint partial theta graphs.

Proof. Suppose that G′ contains two vertex-disjoint partial theta graphs F1 and F2. Letu1, u2 ∈ U with u1 = u2. Then G[V (Fi) ∪ {ui}] contains a theta graph for i ∈ {1, 2}.Note that by the definition of a partial theta graph, |V (Fi)∩W | � 3 for i ∈ {1, 2}. Since|U \{u1, u2}| = 2(k−2) and |W \V (F1∪F2)| � |W |−6 > 3(k−2), it follows from Fact 2that G[(U \ {u1, u2}) ∪ (W \ V (F1 ∪ F2))] contains k − 2 vertex-disjoint theta graphs,and hence G contains k vertex-disjoint theta graphs, a contradiction. �Claim 4. G′ does not contain a theta graph.

Proof. Suppose that G′ contains a theta graph T . Then by Claim 3, |V (T ) ∩ W | � 5(otherwise, T contains two vertex-disjoint partial theta graphs). Hence |W − V (T )| �(3k + 2)− 5 = 3(k− 1). Since |U | = 2(k− 1), it follows from Fact 2 that G− T containsk − 1 vertex-disjoint theta graphs, and hence G contains k vertex-disjoint theta graphs,a contradiction. �Claim 5. G′ is connected.

Proof. Suppose that G′ is not connected. Then by Claim 3, there exists a component F ofG′ such that |V (F )∩W | � 2. Hence by the definition of W , |V�2(F )| � 2. Since |F | � 3because δ(F ) � δ(G′) � 2, we see that V�3(F ) = V (F ) \ V�2(F ) = ∅. Since δ(F ) � 2and V�3(F ) = ∅, this together with Lemma 1(i) and Claim 4 implies that F has at leasttwo distinct end blocks which are cycles, and hence |V�2(F )| � 4, a contradiction. �Claim 6. Δ(G′) � 3.

Proof. Suppose that Δ(G′) � 2, that is, dG′(v) = 2 for all v ∈ V (G′). Then it followsfrom Claim 5 that G′ is a cycle. Moreover, since δ(G) � 2k, it follows that V (G′) = W ,and hence |W | = |G′| = |G|−|U | � ck−(2k−2) � 6. Thus G′ contains two vertex-disjointpartial theta graphs, which contradicts Claim 3. �Claim 7. |V�3(G′)| � 1.

Proof. Suppose that |V�3(G′)| � 2. Then by Lemma 1(ii) and Claim 4, G′ contains twovertex-disjoint cycles C1 and C2. By Claim 5, there exists a path Q connecting C1 andC2 in G′, i.e., Q is a path in G′ such that |V (Q) ∩ V (Ci)| = 1 for i ∈ {1, 2}.

Subclaim 7.1. Let C = Ci with i ∈ {1, 2} and v ∈ V (C −Q). Then there exists a path R

with an end vertex v such that |V (R) ∩W | = 1 and V (R) ∩ V (C1 ∪ C2 ∪Q) = {v}.

118 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

Proof. If v ∈ W , then v itself is the desired path. Thus we may assume that v /∈ W . Thenby the definition of W , dG′(v) � 3. Hence by Claim 4, NG′(v)∩V (G′−(C1∪C2∪Q)) = ∅.Therefore, there exists a path R with an end vertex v such that |R| � 2 and V (R) ∩V (C1 ∪C2 ∪Q) = {v}. We choose R so that |R| is as large as possible. Let v′ be an endvertex of R other than v. Suppose that V (R)∩W = ∅. Then dG′(v′) � 3 because v′ /∈ W .On the other hand, by the maximality of |R|, NG′(v′)∩ V (G′ − (C1 ∪C2 ∪Q∪R)) = ∅.Moreover, by Claim 4 and since |R| � 2, NG′(v′) ∩ (V (C1 ∪ C2 ∪ Q) − {v}) = ∅ and|NG′(v′)∩V (R)| � 2. Hence we obtain dG′(v) � 2, a contradiction. Thus V (R)∩W = ∅,and hence we can easily see that the assertion holds. �

Let v1, v2 ∈ V (C1 −Q) with v1 = v2 and v3, v4 ∈ V (C2 −Q) with v3 = v4. Then bySubclaim 7.1, there exist four distinct paths R1, R2, R3 and R4 in G′ such that for each i

with 1 � i � 4, vi is an end vertex of Ri, |V (Ri)∩W | = 1 and V (Ri)∩V (C1∪C2∪Q) ={vi}. By Claim 4, R1, R2, R3 and R4 are pairwise vertex-disjoint. Hence we can easilysee that G′[V (C1 ∪ C2 ∪ R1 ∪ R2 ∪ R3 ∪ R4)] contains two vertex-disjoint partial thetagraphs, which contradicts Claim 3. This completes the proof of Claim 7. �

By Claims 6 and 7, |V�3(G′)| = 1, say V�3(G′) = {z}. Since V2(G′) = V (G′) − {z},it follows from Claim 5 that there exist p distinct cycles C1, . . . , Cp in G′ such that⋃p

i=1 V (Ci) = V (G′) and V (Ci) ∩ V (Cj) = {z} for each i, j with 1 � i < j � p.Note that p � 2 and W = (

⋃pi=1 V (Ci)) \ {z}. Then each Ci contains a partial theta

graph, in particular, if |Ci| � 4, then Ci − {z} contains a partial theta graph. Henceby Claim 3, |Ci| = 3 for 1 � i � p. This implies that G′ ∼= K1 ∨ pK2. Therefore,(2k − 1)K1 ∨ pK2 ⊂ G ⊂ K2k−1 ∨ pK2, and thus (ii) of Theorem 2 holds.

Case 2. β � 9600k2 + 36k + 12.

Without loss of generality, we may assume that |Tk−1| = β. If k = 2, then by Lemma 9,the minimality of |T ∗| (= |T1|) and the assumption of Case 2, |V�2(G)| � |T1|/12−7 > 0,which contradicts the assumption that δ(G) � 2k > 2. Thus k � 3.

Let R1, R2, . . . , Rk−2 be k − 2 vertex-disjoint theta graphs in G. Let R∗ =⋃k−2

i=1 Ri,and let H ′ = G−R∗. We choose R1, . . . , Rk−2 so that |R∗| is as small as possible. Thenby the minimality of |R∗| and the definition of β and since |Tk−1| = β, it follows that|R∗| �

∑k−2i=1 |Ti| = |T ∗| − |Tk−1| = |T ∗| − β, and hence |H ′| � |H| + β � β. Note

that H ′ does not contain two vertex-disjoint theta graphs (since otherwise, G contains kvertex-disjoint theta graphs). Note also that H ′ may have no theta graph and that if H ′

has a theta graph, then |R| � |Tk−1| = β for every theta graph R in H ′ (since otherwise,R1, . . . , Rk−2, R are k − 1 vertex-disjoint theta graphs in G such that the sum of theirorder is less than |T ∗|, which contradicts the minimality of |T ∗|).

Claim 8. |V�2(H ′)| � β/12 − 7.

S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120 119

Proof. If H ′ has no theta graph, then by Lemma 4, |V�2(H ′)| � |H ′|/4 � β/4. Thus wemay assume that H ′ contains a theta graph. Choose a theta graph R in H ′ so that |R|is as small as possible. Then since |R| � β and H ′ does not contain two vertex-disjointtheta graphs, it follows from Lemma 9 that |V�2(H ′)| � |R|/12 − 7 � β/12 − 7. �

Since δ(G) � 2k and H ′ = G−R∗ = G−⋃k−2

i=1 Ri, it follows that for each v ∈ V�2(H ′),one of the following holds:

(I) there exists i with 1 � i � k − 2 such that |EG(v,Ri)| � 4;(II) there exist i, j with 1 � i < j � k−2 such that |EG(v,Ri)| � 3 and |EG(v,Rj)| � 3.

We define A1 = {v ∈ V�2(H ′): v satisfies (I)}, and let A2 = V�2(H ′) −A1. By Claim 8and the assumption of Case 2, |A1|+|A2| = |V�2(H ′)| � β/12−7 � (800k2+3k+1)−7 =800k2 + 3(k − 2).

We need to divide the proof of Case 2 into two cases.

Subcase 2.1. |A1| > 3(k − 2).

For each i with 1 � i � k − 2, let A(i)1 = {v ∈ A1: |EG(v,Ri)| � 4}. Then by the

assumption of Subcase 2.1, the definition of A(i)1 and the Pigeonhole Principle, there

exists i with 1 � i � k − 2 such that |A(i)1 | � 4. We may assume that i = 1. Since

|EG(v,R1)| � 4 for all v ∈ A(1)1 , it follows from Lemma 6 and the minimality of |R∗| that

|R1| = 4. Let U ⊆ A(1)1 with |U | = 4. Then G[U ∪ V (R1)] contains two vertex-disjoint

K−4 ’s S1 and S2 (actually there are two vertex-disjoint copies of K−

4 such that eachcontains exactly two vertices of U and two vertices of V (R1), respectively). Since (U ∪V (R1))∩ (

⋃k−2i=2 V (Ri)) = ∅, it follows that S1, S2, R2, . . . , Rk−2 are k− 1 vertex-disjoint

theta graphs in G. Moreover, by the assumption of Case 2 and since |R∗| � |T ∗|−|Tk−1|,

|S1| + |S2| +k−2∑i=2

|Ri| = 8 +(∣∣R∗∣∣− |R1|

)= 8 +

(∣∣R∗∣∣− 4)

=∣∣R∗∣∣ + 4 �

(∣∣T ∗∣∣− |Tk−1|)

+ 4 =(∣∣T ∗∣∣− β

)+ 4 <

∣∣T ∗∣∣.This contradicts the minimality of |T ∗|.

Subcase 2.2. |A1| � 3(k − 2).

Then |A2| = |V�2(H ′)| − |A1| � 800k2. For each i, j with 1 � i < j � k − 2,let A

(i,j)2 = {v ∈ A2: |EG(v,Ri)| � 3 and |EG(v,Rj)| � 3}. Then by the definition

of A(i,j)2 and the Pigeonhole Principle, there exist i, j with 1 � i < j � k − 2 such

that |A(i,j)2 | > 800. We may assume that i = 1 and j = 2. Since |EG(v,R1)| � 3 and

|EG(v,R2)| � 3 for all v ∈ A(1,2)2 , we see that for each v ∈ A

(1,2)2 , there exist W1 ⊆ V (R1)

and W2 ⊆ V (R2) such that |W1| = |W2| = 3 and W1∪W2 ⊆ NG(v). By Lemma 6, we also

120 S. Chiba et al. / Advances in Applied Mathematics 54 (2014) 105–120

have |Ri| � 5 for i ∈ {1, 2}. Since |A(1,2)2 | > 800 and |Ri| � 5 for i ∈ {1, 2} (and hence

|(V (Ri)

3)| �

(53)

= 10 for i ∈ {1, 2}), it follows from the Pigeonhole Principle that thereexist U ⊆ A

(1,2)2 , W1 ⊆ V (R1) and W2 ⊆ V (R2) with |U | = 9 and |W1| = |W2| = 3 such

that W1∪W2 ⊆ NG(v) for all v ∈ U . Then G[U ∪W1∪W2] contains three vertex-disjointK2,3’s S1, S2 and S3 (actually there are three vertex-disjoint copies of K2,3 such thateach contains exactly three vertices of U and two vertices of W1 ∪ W2, respectively).Since (U ∪ W1 ∪ W2) ∩ (

⋃k−2i=3 V (Ri)) = ∅, it follows that S1, S2, S3, R3, . . . , Rk−2 are

k−1 vertex-disjoint theta graphs in G. Moreover, by the assumption of Case 2 and since|R∗| � |T ∗| − |Tk−1|,

|S1| + |S2| + |S3| +k−2∑i=3

|Ri| = 15 +(∣∣R∗∣∣− |R1| − |R2|

)� 15 +

(∣∣R∗∣∣− 8)

=∣∣R∗∣∣ + 7 �

(∣∣T ∗∣∣− |Tk−1|)

+ 7 =(∣∣T ∗∣∣− β

)+ 7 <

∣∣T ∗∣∣,which contradicts the minimality of |T ∗|.

This completes the proof of Theorem 2. �Acknowledgments

The authors would like to thank the referee for valuable suggestions and comments.

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