Minimum degree conditions for the existence of cycles of all lengths modulo k in graphs
aa r X i v : . [ m a t h . C O ] A p r Minimum degree conditions for the existence of cycles of all lengthsmodulo k in graphs Shuya Chiba ∗ Tomoki Yamashita † Abstract
Thomassen, in 1983, conjectured that for a positive integer k , every 2-connected non-bipartitegraph of minimum degree at least k + 1 contains cycles of all lengths modulo k . In this paper, wesettle this conjecture affirmatively. Keywords : Cycles, Length modulo k , Minimum degree AMS Subject Classification : 05C38
All graphs considered in this paper are finite undirected graphs without loops or multiple edges. In[3], Thomassen conjectured the following.
Conjecture A (Thomassen [3])
For a positive integer k , every -connected non-bipartite graph ofminimum degree at least k + 1 contains cycles of all lengths modulo k . In 2018, Liu and Ma proved that this conjecture is true for all even integers k , see [2, Theorem 1.9](for the history and other related results to the conjecture, we also refer the reader to [2]). In thispaper, we settle Conjecture A by showing that it is also true for all odd integers k . For this purpose,we give the following result, which is our main theorem. Here E ( G ) denotes the edge set of a graph G . Theorem 1
For a positive integer k , every -connected graph of minimum degree at least k + 1 contains k cycles C , . . . , C k such that either (i) | E ( C i +1 ) | − | E ( C i ) | = 1 for ≤ i ≤ k − , or (ii) | E ( C i +1 ) | − | E ( C i ) | = 2 for ≤ i ≤ k − . We here prove Conjecture A assuming Theorem 1 for the case where k is odd. It follows from theproof that the condition “non-bipartite” in Conjecture A is not necessary if k is odd. Proof of Conjecture A for the case where k is odd. Let k be a positive odd integer, and let G be a2-connected graph of minimum degree at least k + 1 (the graph G may be bipartite). We will show ∗ Applied Mathematics, Faculty of Advanced Science and Technology, Kumamoto University, 2-39-1 Kurokami, Ku-mamoto 860-8555, Japan; E-mail address: [email protected] ; This work was supported by JSPS KAKENHIGrant Number 17K05347 † Department of Mathematics, Kindai University, 3-4-1 Kowakae, Higashi-Osaka, Osaka 577-8502, Japan; E-mailaddress: [email protected] ; This work was supported by JSPS KAKENHI Grant Number 16K05262 G contains cycles of all lengths modulo k . By Theorem 1, G contains k cycles satisfying (i) or(ii) in Theorem 1. If the k cycles satisfy (i), then the k cycles clearly have all lengths modulo k . So,suppose that the k cycles satisfy (ii). Since k is odd, we may assume that( l, l + 2 , . . . , l + k − , l + k − , l + k + 1 , . . . , l + 2 k − , l + 2 k − k cycles for some integer l ≥
3. Then it follows that l + 2 i + 1 ≡ l + k + 2 i + 1 (mod k ) for 0 ≤ i ≤ k − . Thus the k cycles have all lengths modulo k . (cid:3) Our proof of Theorem 1 is based on the technique of Liu and Ma [2]. In the next section, weintroduce results to prove Theorem 1. In particular, we give sharp degree conditions for the existenceof paths with specified end vertices whose lengths differ by one or two (see Theorems 2 and 3 for thedetail), which are our key results.
In this section, we introduce results for the proof of Theorem 1 according to the flowchart of Figure 1.
Step 1 ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴✤✤✤✤✤✤✤ ✤✤✤✤✤✤✤❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴✤✤✤✤✤ ✤✤✤✤✤
Theorem 2 (Proof: Secs.3,4) / / Theorem 3 (Proof: Secs.3,5)
Lemmas 5, 6 ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴ (cid:11) (cid:19) ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴ (cid:11) (cid:19) ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴✤✤✤✤✤✤ ✤✤✤✤✤✤
Theorem 5 (Proof: Sec.6) 3-connected,non-bipartite
Theorem B ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴ (cid:11) (cid:19)
Step 2 ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴✤✤✤✤✤✤ ✤✤✤✤✤✤ (I) 2-connected,but not 3-connected
Theorem 4 (Proof: Sec.6) (II) 3-connected,non-bipartite
Theorem 6 (III) bipartite
Theorem C ❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴❴ (cid:11) (cid:19)
Theorem 1
Figure 1: The overall flowchart of the proof of Theorem 12e require some terminology and notation. Let G be a graph. The vertex set of G is denotedby V ( G ). For a vertex v of G , deg G ( v ) denotes the degree of v in G , and let δ ( G ) denote theminimum degree of G . For S ⊆ V ( G ), G [ S ] denotes the subgraph of G induced by S , and let G − S = G [ V ( G ) \ S ]. For distinct vertices x and y of G , ( G, x, y ) is called a rooted graph . A rootedgraph (
G, x, y ) is 2 -connected if(R1) G is a connected graph of order at least 3 with at most two end blocks, and(R2) every end block of G contains at least one of x and y as a non-cut vertex.Note that ( G, x, y ) is 2-connected if and only if G + xy (i.e., the graph obtained from G by addingthe edge xy if xy / ∈ E ( G )) is 2-connected. For convenience, we say that a sequence of paths or cycles H , H , . . . , H k • have consecutive lengths if | E ( H ) | ≥ | E ( H i +1 ) | − | E ( H i ) | = 1 for 1 ≤ i ≤ k − • satisfy the length condition if | E ( H ) | ≥ | E ( H i +1 ) | − | E ( H i ) | = 2 for 1 ≤ i ≤ k − • satisfy the semi-length condition if | E ( H ) | ≥ j with 1 ≤ j ≤ k − switch , such that | E ( H i +1 ) | − | E ( H i ) | = 2 for 1 ≤ i ≤ j − , | E ( H j +1 ) | − | E ( H j ) | = 1 and | E ( H i +1 ) | − | E ( H i ) | = 2 for j + 1 ≤ i ≤ k − . The first step of the proof of Theorem 1 is to show the following two results concerning degreeconditions for the existence of paths satisfying the length condition or the semi-length condition.
Theorem 2
Let k be a positive integer, and let ( G, x, y ) be a -connected rooted graph. Suppose that deg G ( v ) ≥ k for any v ∈ V ( G ) \ { x, y } . Then G contains k paths from x to y satisfying the lengthcondition. Theorem 3
Let k be a positive integer, and let ( G, x, y ) be a -connected rooted graph. Suppose that deg G ( v ) ≥ k − for any v ∈ V ( G ) \ { x, y } . Then G contains k paths from x to y satisfying thelength condition or the semi-length condition. The complete graphs of orders 2 k and 2 k − heorem 4 Let k be a positive integer, and let G be a -connected but not -connected graph. If δ ( G ) ≥ k + 1 , then G contains k cycles satisfying the length condition. To show the case (II), in Section 6, we also prove the following theorem by using Theorems 2, 3and additional lemmas. Here, for a cycle C in a connected graph G , C is said to be non-separating if G − V ( C ) is connected. Theorem 5
Let k be a positive integer, and let G be a -connected graph containing a non-separatinginduced odd cycle. If δ ( G ) ≥ k + 1 , then G contains k cycles, which have consecutive lengths or satisfythe length condition. The following result is known for the existence of a non-separating induced odd cycle.
Theorem B (see the proof of Theorem 2 in [1])
Every -connected non-bipartite graph containsa non-separating induced odd cycle. Combining Theorems 5 and B, we can obtain the following theorem for the case (II).
Theorem 6
Let k be a positive integer, and let G be a -connected non-bipartite graph. If δ ( G ) ≥ k + 1 , then G contains k cycles, which have consecutive lengths or satisfy the length condition. On the other hand, for the case (III), the following theorem is proved by Liu and Ma.
Theorem C ([2, Theorem 1.2])
Let k be a positive integer, and let G be a bipartite graph. If δ ( G ) ≥ k + 1 , then G contains k cycles satisfying the length condition. Consequently, we can obtain Theorem 1 by Theorems 4, 6 and C. In Section 7, we give someremarks on the minimum degree and the connectivity conditions in Theorem 1.In the rest of this section, we prepare terminology and notation which will be used in the subse-quent sections. Let G be a graph. We denote by N G ( v ) the neighborhood of a vertex v in G . For S ⊆ V ( G ), we define N G ( S ) = (cid:0)S v ∈ S N G ( v ) (cid:1) \ S . For S, T ⊆ V ( G ) with S ∩ T = ∅ , E G ( S, T ) denotesthe set of edges of G between S and T , and let e G ( S, T ) = | E G ( S, T ) | . Furthermore, G [ S, T ] is thegraph defined by V ( G [ S, T ]) = S ∪ T and E ( G [ S, T ]) = E G ( S, T ). Note that G [ S, T ] is a bipartitesubgraph of G with partite sets S and T , and we always assume that G [ S, T ] is such a bipartitegraph. For a rooted graph (
G, x, y ), we define δ ( G, x, y ) = min { deg G ( v ) : v ∈ V ( G ) \ { x, y }} . In therest of this paper, we often denote the singleton set { v } by v , and we often identify a subgraph H of G with its vertex set V ( H ).For S ⊆ V ( G ), a path in G is an S - path if it begins and ends in S , and none of its internal verticesare contained in S . For S, T ⊆ V ( G ) with S ∩ T = ∅ , a path in G is an ( S, T )-path if one end vertexof the path belongs to S , another end vertex belongs to T , and the internal vertices do not belongto S ∪ T . We write a path or a cycle P with a given orientation as −→ P . If there exists no fear ofconfusion, we abbreviate −→ P by P . Let −→ P be an oriented path (or cycle). For u ∈ V ( P ), the h -thsuccessor and the h -th predecessor of u on −→ P (if exist) is denoted by u + h and u − h , respectively, and4e let u + = u +1 and u − = u − . For u, v ∈ V ( P ), the path from u to v along −→ P (if exist) is denotedby u −→ P v . The reverse sequence of u −→ P v is denoted by v ←− P u . In the rest of this paper, if −→ P is an( S, T )-path in G , we always assume that the orientation of P is given from the end vertex belongingto S to the end vertex belonging to T along the edges of P . In this section, we introduce the concept of a core which was used in the argument of [2] and givesome lemmas for the proofs of Theorems 2 and 3.Let x and y be two distinct vertices of a graph G . For a bipartite subgraph H = G [ S, T ] of G and an integer l , H is called an l - core with respect to ( x, y ) if(C1) H is complete bipartite and | T | ≥ | S | = l + 1 ≥ x ∈ S and y / ∈ V ( H ),(C3) e G ( v, S ) ≤ l for v ∈ V ( G ) \ ( V ( H ) ∪ { y } ), and(C4) e G ( v, T \ { v } ) ≤ l + 1 for v ∈ V ( G ) \ ( S ∪ { y } ).In the rest of this section, we fix the following notation. Let ( G, x, y ) be a 2-connected rooted graph,and let H = G [ S, T ] be an l -core of G with respect to ( x, y ). Furthermore, let C be the componentof G − V ( H ) such that y ∈ V ( C ). Since E G ( H − x, C ) = ∅ by (R2), the following two lemmas(Lemmas 1 and 2) easily follows from (C1). So, we omit the proof. Lemma 1
If either (i) l ≥ k or (ii) l = k − and E G ( T, C ) = ∅ , then G contains k ( x, y ) -pathssatisfying the length condition. Lemma 2 If l = k − , E G ( S \{ x } , C ) = ∅ and E ( G [ T ]) = ∅ , then G contains k ( x, y ) -paths satisfyingthe semi-length condition. The following two lemmas (Lemmas 3 and 4) are proved by Liu and Ma, see [2, Lemmas 2.7 and2.11]. Note that the argument in [2] can work for paths satisfying the length condition but also forpaths satisfying the semi-length condition.
Lemma 3
Let s be a vertex of S \ { x } such that E G ( s, C ) = ∅ . If one of the following (i)–(iii) holds,then G contains k ( x, y ) -paths satisfying the length condition (resp., the semi-length condition). (i) G − V ( C ) contains k − l + 1 T -paths internally disjoint from V ( H ) and satisfying the lengthcondition (resp., the semi-length condition) [2, Lemma 2.7-2] . (ii) G − V ( C ) contains k − l + 1 ( T, { x, s } ) -paths internally disjoint from V ( H ) and satisfying thelength condition (resp., the semi-length condition) [2, Lemma 2.7-4] . (iii) G − V ( C ) contains k − l + 2 ( T, S \ { x, s } ) -paths internally disjoint from V ( H ) and satisfyingthe length condition (resp., the semi-length condition) [2, Lemma 2.7-3] . emma 4 ([2, Lemma 2.11]) If one of the following (i) and (ii) holds, then G contains k ( x, y ) -paths satisfying the length condition (resp., the semi-length condition). (i) G contains k − l ( T, y ) -paths internally disjoint from V ( H ) and satisfying the length condition(resp., the semi-length condition). (ii) G contains k − l + 1 ( S \ { x } , y ) -paths internally disjoint from V ( H ) and satisfying the lengthcondition (resp., the semi-length condition). Proof of Theorem 2.
We prove it by induction on | V ( G ) | + | E ( G ) | . Let ( G, x, y ) be a minimumcounterexample with respect to | V ( G ) | + | E ( G ) | . If k = 1, then by (R1) and (R2), we can easily seethat G contains an ( x, y )-path of length at least 2, a contradiction. Thus k ≥
2. Since δ ( G, x, y ) ≥ k ,this implies that | G | ≥
5. By symmetry, we may assume that deg G ( x ) ≤ deg G ( y ). Claim 4.1 G is -connected.Proof. Suppose that G is not 2-connected. Then by (R1), G has a cut vertex c and G − c hasexactly two components C and C . Since | G | ≥ ≥ | C | ≥
2. Let G i = G [ V ( C i ) ∪ { c } ] for i ∈ { , } . Then by (R2), for some two distinct vertices x ′ , y ′ ∈ { x, y } ,(i) ( G , x ′ , c ) is a 2-connected rooted graph such that δ ( G , x ′ , c ) ≥ δ ( G, x, y ) ≥ k , and (ii) y ′ iscontained in a block of G . Hence by the induction hypothesis, G contains k ( x ′ , c )-paths −→ P , . . . , −→ P k satisfying the length condition. Then x ′ −→ P c −→ P y ′ , . . . , x ′ −→ P k c −→ P y ′ are k ( x, y )-paths in G satisfying thelength condition, where −→ P denotes the ( c, y ′ )-path in G , a contradiction. (cid:3) Claim 4.2 xy / ∈ E ( G ) .Proof. If xy ∈ E ( G ), then by Claim 4.1, ( G − xy, x, y ) is a 2-connected rooted graph such that δ ( G − xy, x, y ) = δ ( G, x, y ) ≥ k , and hence the induction hypothesis yields that G − xy (and also G ) contains k ( x, y )-paths satisfying the length condition, a contradiction. (cid:3) Case 1. G − y does not contain a cycle of length 4 passing through x .Since xy / ∈ E ( G ) by Claim 4.2, in this case, we have e G ( v, N G ( x ) \ { v } ) ≤ v ∈ V ( G ) \ { x, y } . (4.1)Let G ∗ be the graph obtained from G by contracting the subgraph induced by N G ( x ) ∪ { x } into asingle vertex x ∗ and then removing multiple edges. Then by (4.1),deg G ∗ ( v ) = deg G ( v ) for v ∈ V ( G ∗ ) \ { x ∗ , y } . (4.2)6y (4.1) and since δ ( G, x, y ) ≥ k ≥
4, we have | G ∗ | ≥
3. Hence by Claim 4.1, if G ∗ is not 2-connected, then x ∗ is the unique cut vertex of G ∗ and each block of G ∗ is an end block containing x ∗ .Now, let B ∗ be the block of G ∗ which contains y if G ∗ is not 2-connected; otherwise, let B ∗ = G ∗ .Assume that ( B ∗ , x ∗ , y ) is 2-connected. Since δ ( B ∗ , x ∗ , y ) ≥ δ ( G, x, y ) ≥ k by (4.2), it followsfrom the induction hypothesis that B ∗ contains k ( x ∗ , y )-paths −→ P , . . . , −→ P k satisfying the length con-dition. By the definition of G ∗ and since xy / ∈ E ( G ), we see that for each i with 1 ≤ i ≤ k , thereis a vertex u i of N G ( x ) such that u i u ′ i ∈ E ( G ), where u ′ i is the successor of x ∗ along −→ P i . Therefore, xu u ′ −→ P y, . . . , xu k u ′ k −→ P k y are k ( x, y )-paths in G satisfying the length condition, a contradiction. Thus( B ∗ , x ∗ , y ) is not 2-connected.Then we have V ( B ∗ ) = { x ∗ , y } . This together with the definition of G ∗ and Claim 4.2 impliesthat N G ( y ) ⊆ N G ( x ). Since deg G ( x ) ≤ deg G ( y ), this yields that N G ( x ) = N G ( y ). Since | G ∗ | ≥
3, thedefinition of G ∗ also implies that there is a component C of G − ( N G ( x ) ∪ { x, y } ). Then N G ( C ) ⊆ N G ( x ) and G [ V ( C ) ∪ N G ( C ) ∪ { x } ] is 2-connected, since G is 2-connected and N G ( x ) = N G ( y ). Hence N G ( C ) ( ⊆ N G ( x )) can be partitioned into two vertex-disjoint non-empty sets S and T so that thegraph C ∗ defined as follows is 2-connected: V ( C ∗ ) = V ( C ) ∪ { x, S, T } and E ( C ∗ ) = E ( C ) ∪ { xS, xT } ∪ { vS : v ∈ V ( C ) , E G ( v, S ) = ∅} ∪ { vT : v ∈ V ( C ) , E G ( v, T ) = ∅} . Then by the definition of C ∗ , it follows that ( C ∗ − x, S, T ) is a 2-connected rooted graph anddeg C ∗ − x ( v ) = deg G ( v ) for v ∈ V ( C ∗ ) \ { x, S, T } ; thus δ ( C ∗ − x, S, T ) ≥ δ ( G, x, y ) ≥ k . By theinduction hypothesis, C ∗ − x contains k ( S, T )-paths −→ P , . . . , −→ P k satisfying the length condition. Notethat each P i has order at least 3, and each P i − { S, T } is contained in C . For each i with 1 ≤ i ≤ k ,let s i and t i be vertices in S and T , respectively, such that s i s ′ i , t i t ′ i ∈ E ( G ), where s ′ i and t ′ i are thesuccessor of S and the predecessor of T along −→ P i , respectively. Then xs s ′ −→ P t ′ t y, . . . , xs k s ′ k −→ P k t ′ k t k y are k ( x, y )-paths in G satisfying the length condition, a contradiction.This completes the proof of Case 1. Case 2. G − y contains a cycle of length 4 passing through x .By the assumption of Case 2, G contains a bipartite subgraph H = G [ S, T ] such that H iscomplete bipartite, | T | ≥ | S | =: l + 1 ≥ x ∈ S , y / ∈ V ( H ) (i.e., H satisfies (C1) and (C2)). Let C be the component of G − V ( H ) such that y ∈ V ( C ). Choose H so that(a) | S | is maximum,(b) | T | is maximal, subject to (a),(c) | C | is maximum, subject to (a) and (b), and(d) | N G ( C ) ∩ S | is minimum, subject to (a), (b) and (c).Then by the choices (a) and (b), we can obtain the following. Claim 4.3 H is an l -core of G with respect to ( x, y ) . roof. Since H satisfies (C1) and (C2), it suffices to show that H also satisfies (C3) and (C4).We first show (C4). Suppose that there exists a vertex v ∈ V ( G ) \ ( S ∪ { y } ) such that e G ( v, T \{ v } ) ≥ l + 2. Let S ′ = S ∪ { v } and T ′ = { v ′ ∈ V ( G ) \ ( S ′ ∪ { y } ) : S ′ ⊆ N G ( v ′ ) } . Note that N G ( v ) ∩ T ⊆ T ′ . Hence G [ S ′ , T ′ ] is a complete bipartite subgraph of G such that | T ′ | ≥ e G ( v, T \{ v } ) ≥ l + 2 = | S ′ | > | S | , x ∈ S ′ and y / ∈ V ( G [ S ′ , T ′ ]), which contradicts the choice (a).We next show (C3). Suppose that there exists a vertex v ∈ V ( G ) \ ( V ( H ) ∪ { y } ) such that e G ( v, S ) ≥ l + 1. Since | S | = l + 1, this implies that S ⊆ N G ( v ). Hence, G [ S, T ∪ { v } ] is a completebipartite subgraph of G such that | T ∪ { v }| > | T | ≥ | S | , x ∈ S and y / ∈ V ( G [ S, T ∪ { v } ]) whichcontradicts the choice (b). (cid:3) By the choices (c) and (d), we can obtain the following.
Claim 4.4 If E G ( S \ { x } , C ) = ∅ , then (i) e G ( v, T ) ≤ l for v ∈ V ( G ) \ V ( H ∪ C ) , and (ii) e G ( v, T \{ v } ) ≤ l or E G ( v, C ) = ∅ for v ∈ T .Proof. Assume that E G ( S \ { x } , C ) = ∅ , and let s be a vertex of S \ { x } such that E G ( s, C ) = ∅ .We now suppose that there is a vertex v of V ( G ) \ ( S ∪ V ( C )) such that e G ( v, T \ { v } ) ≥ l + 1.Further, if v ∈ T , then we suppose that E G ( v, C ) = ∅ . Note that E G ( v, C ) = ∅ also holds in the caseof v ∈ V ( G ) \ V ( H ∪ C ), since C is a component of G − V ( H ).Let S ′ = ( S \ { s } ) ∪ { v } and T ′ = { v ′ ∈ V ( G ) \ ( S ′ ∪ { y } ) : S ′ ⊆ N G ( v ′ ) } , and let H ′ = G [ S ′ , T ′ ].Since e G ( v, T \ { v } ) ≥ l + 1, it follows from the definitions of S ′ , T ′ and H ′ that H ′ is a completebipartite subgraph of G such that | T ′ | ≥ e G ( v, T \ { v } ) ≥ l + 1 = | S ′ | = | S | , x ∈ S ′ and y / ∈ V ( H ′ ).In particular, H ′ satisfies (a) and (b).Let C ′ be the component of G − V ( H ′ ) such that y ∈ V ( C ′ ). Note that V ( C ) ⊆ V ( C ′ ) ⊆ V ( G ) \ V ( H ′ ) since S ′ = ( S \ { s } ) ∪ { v } and E G ( v, C ) = ∅ . Hence the choice (c) yields that C ′ = C .In particular, H ′ also satisfies (c). But then, since s ∈ N G ( C ) and v / ∈ N G ( C ), we have | N G ( C ′ ) ∩ S ′ | = | N G ( C ) ∩ (cid:0) ( S \ { s } ) ∪ { v } (cid:1) | = | N G ( C ) ∩ S | − , which contradicts the choice (d). Thus (i) and (ii) hold. (cid:3) Note that l ≤ k − Claim 4.5 If E G ( S \ { x } , C ) = ∅ , then N G ( T ) ⊆ V ( H ∪ C ) .Proof. Assume that E G ( S \ { x } , C ) = ∅ , and let s be a vertex of S \ { x } such that E G ( s, C ) = ∅ .Then we can take an ( s, y )-path −→ P in G [ V ( C ) ∪ { s } ].We now suppose that N G ( T ) V ( H ∪ C ), i.e., there exists a component D of G − V ( H ) suchthat D = C and E G ( T, D ) = ∅ . Subclaim 4.5.1
Let B be an end block of D , and let b be a cut vertex of D which is contained in B . Then E G ( B − b, T ∪ { x, s } ) = ∅ . roof. Suppose that E G ( B − b, T ∪{ x, s } ) = ∅ . Since G is 2-connected, we have E G ( B − b, S \{ x, s } ) = ∅ . In particular, S \ { x, s } 6 = ∅ , that is, l ≥
2. We define the graph B ∗ as follows: V ( B ∗ ) = V ( B ) ∪ { S ∗ } and E ( B ∗ ) = E ( B ) ∪ { vS ∗ : v ∈ V ( B ) , E G ( v, S \ { x, s } ) = ∅} . Then ( B ∗ , S ∗ , b ) is a 2-connected rooted graph. Since l ≥
2, it also follows that for each v ∈ V ( B ∗ ) \ { S ∗ , b } ,deg B ∗ ( v ) ≥ deg G ( v ) ≥ k (if E G ( v, S \ { x, s } ) = ∅ )deg G ( v ) − ( l −
1) + 1 ≥ k − l + 2) (if E G ( v, S \ { x, s } ) = ∅ ) , and thus δ ( B ∗ , S ∗ , b ) ≥ k − l + 2). By the induction hypothesis, B ∗ contains k − l + 2 ( S ∗ , b )-pathssatisfying the length condition. Then it follows from the definition of B ∗ that G − V ( C ) contains k − l + 2 ( S \ { x, s } , b )-paths internally disjoint from V ( H ) and satisfying the length condition. On theother hand, since E G ( T, D ) = ∅ and E G ( T, B − b ) = ∅ by the assumption, we can take a ( T, b )-pathin G [ V ( D − B ) ∪ { b } ∪ T ]. Combining the ( T, b )-path with the above k − l + 2 ( S \ { x, s } , b )-paths,we can get k − l + 2 ( T, S \ { x, s } )-paths internally disjoint from V ( H ) and satisfying the lengthcondition, which contradicts Lemma 3(iii). (cid:3) Subclaim 4.5.2 E G ( { x, s } , D ) = ∅ .Proof. Suppose that E G ( { x, s } , D ) = ∅ . We define the graph D ∗ as follows: V ( D ∗ ) = V ( D ) ∪ { x ∗ , T } and E ( D ∗ ) = E ( D ) ∪ { vx ∗ : v ∈ V ( D ) , E G ( v, { x, s } ) = ∅} ∪ { vT : v ∈ V ( D ) , E G ( v, T ) = ∅} . Then by Subclaim 4.5.1 and since E G ( T, D ) = ∅ , it follows that ( D ∗ , x ∗ , T ) is a 2-connected rootedgraph. By (C3), and since e G ( v, T ) ≤ l for v ∈ V ( D ) by Claim 4.4(i), it also follows that each v ∈ V ( D ∗ ) \ { x ∗ , T } satisfies the following:deg D ∗ ( v ) ≥ deg G ( v ) − (cid:0) ( l −
1) + l (cid:1) + 1 ≥ k − l + 1) (if E G ( v, { x, s } ) = ∅ , E G ( v, T ) = ∅ )deg G ( v ) − ( l + l ) + (1 + 1) ≥ k − l + 1) (if E G ( v, { x, s } ) = ∅ , E G ( v, T ) = ∅ )deg G ( v ) − ( l − ≥ k − l + 1) (if E G ( v, { x, s } ) = ∅ , E G ( v, T ) = ∅ )deg G ( v ) − l + 1 ≥ k − l + 1) (if E G ( v, { x, s } ) = ∅ , E G ( v, T ) = ∅ ) . Thus δ ( D ∗ , x ∗ , T ) ≥ k − l + 1). By the induction hypothesis, D ∗ contains k − l + 1 ( T, x ∗ )-pathssatisfying the length condition. Then it follows from the definition of D ∗ that G − V ( C ) contains k − l + 1 ( T, { x, s } )-paths internally disjoint from V ( H ) and satisfying the length condition, whichcontradicts Lemma 3(ii). (cid:3) We divide the rest of the proof of Claim 4.5 into two cases as follows.
Case (i) | N G ( D ) ∩ T | ≤ . E G ( T, D ) = ∅ , we have | N G ( D ) ∩ T | = 1, say N G ( D ) ∩ T = { t } . Since G is 2-connected, itfollows from Subclaim 4.5.2 that E G ( D, S \ { x, s } ) = ∅ . In particular, S \ { x, s } 6 = ∅ , that is, l ≥ D ∗ as follows: V ( D ∗ ) = V ( D ) ∪ { t, S ∗ } and E ( D ∗ ) = E ( D ) ∪ { vt : v ∈ V ( D ) , vt ∈ E ( G ) } ∪ { vS ∗ : v ∈ V ( D ) , E G ( v, S \ { x, s } ) = ∅} . Then by Subclaims 4.5.1 and 4.5.2, ( D ∗ , t, S ∗ ) is a 2-connected rooted graph. By Subclaim 4.5.2 andsince l ≥
2, it also follows that for each v ∈ V ( D ∗ ) \ { t, S ∗ } ,deg D ∗ ( v ) ≥ deg G ( v ) ≥ k (if E G ( v, S \ { x, s } ) = ∅ )deg G ( v ) − ( l −
1) + 1 ≥ (cid:0) k − l + 2) (if E G ( v, S \ { x, s } ) = ∅ ) , and thus δ ( D ∗ , t, S ∗ ) ≥ k − l + 2). By the induction hypothesis, D ∗ contains k − l + 2 ( t, S ∗ )-pathssatisfying the length condition. Then it follows from the definition of D ∗ that G − V ( C ) contains k − l + 2 ( t, S \ { x, s } )-paths internally disjoint from V ( H ) and satisfying the length condition, whichcontradicts Lemma 3(iii). Case (ii) | N G ( D ) ∩ T | ≥ . Fix t ∈ N G ( D ) ∩ T . Note that E G ( D, T \ { t } ) = ∅ . We define the graph D ∗ as follows: V ( D ∗ ) = V ( D ) ∪ { t, T ∗ } and E ( D ∗ ) = E ( D ) ∪ { vt : v ∈ V ( D ) , vt ∈ E ( G ) } ∪ { vT ∗ : v ∈ V ( D ) , E G ( v, T \ { t } ) = ∅} . Then by Subclaims 4.5.1 and 4.5.2, ( D ∗ , t, T ∗ ) is a 2-connected rooted graph. It also follows fromClaim 4.4(i) and Subclaim 4.5.2 that for each v ∈ V ( D ∗ ) \ { t, T ∗ } ,deg D ∗ ( v ) ≥ deg G ( v ) − ( l − ≥ k − l + 1) (if E G ( v, T \ { t } ) = ∅ )deg G ( v ) − (cid:0) ( l −
1) + l (cid:1) + 1 ≥ k − l + 1) (if E G ( v, T \ { t } ) = ∅ ) , and thus δ ( D ∗ , t, T ∗ ) ≥ k − l + 1). By the induction hypothesis, D ∗ contains k − l + 1 ( t, T ∗ )-pathssatisfying the length condition. Then it follows from the definition of D ∗ that G − V ( C ) contains k − l + 1 T -paths internally disjoint from V ( H ) and satisfying the length condition, which contradictsLemma 3(i).This completes the proof of Claim 4.5. (cid:3) Claim 4.6 If E G ( T, C − y ) = ∅ , then E G ( S \ { x } , C ) = ∅ .Proof. Assume that E G ( T, C − y ) = ∅ and E G ( S \ { x } , C ) = ∅ . Let t be a vertex of T , and let α = 1 if ty ∈ E ( G ); otherwise, let α = 0. Then by Claim 4.5 and since E G ( T, C − y ) = ∅ , it followsthat N G ( t ) ⊆ V ( H ) ∪ { y } . By the definition of α and Lemma 1(ii), we have l ≤ k − α −
1. Since N G ( t ) ∩ V ( C ) ⊆ { y } , by (C4) and Claim 4.4(ii), we also have e G ( t, T \ { t } ) ≤ l + α . Combining thiswith (C1), it follows from the inequality l ≤ k − α − k ≤ deg G ( t ) = | S | + e G ( t, T \ { t } ) + | E ( G ) ∩ { ty }| ≤ ( l + 1) + ( l + α ) + α ≤ ( k − α ) + ( k −
1) + α = 2 k − ,
10 contradiction. Thus the claim holds. (cid:3)
We divide the proof of Case 2 into two cases according as | C | = 1 or | C | ≥ Case 2.1. | C | = 1, i.e., V ( C ) = { y } . Claim 4.7 N G ( x ) = N G ( y ) = T .Proof. Since V ( C ) \ { y } = ∅ , it follows from Claim 4.6 that E G ( y, S \ { x } ) = ∅ . Since xy / ∈ E ( G )by Claim 4.2, we get E G ( y, S ) = ∅ . This together with deg G ( x ) ≤ deg G ( y ) implies that N G ( x ) = N G ( y ) = T . (cid:3) Claim 4.8 | T | ≥ .Proof. Suppose that | T | = 2, say V ( T ) = { t , t } . Then by (C1), | S | = 2 also holds, say S \{ x } = { s } . Let G ′ = G − { x, y } . By Claim 4.7, G ′ is connected and deg G ′ ( v ) = deg G ( v ) for v ∈ V ( G ′ ) \ { t , t } . If G ′ is 2-connected, then obviously ( G ′ , t , t ) is a 2-connected rooted graph. Onthe other hand, if G ′ is not 2-connected, then since G is 2-connected and N G ( x ) = N G ( y ) = { t , t } , t and t are in different end blocks of G ′ ; since { t , t } ⊆ N G ( s ), s is the cut vertex of G ′ containedin both of the two end blocks of G ′ , that is, ( G ′ , t , t ) is a 2-connected rooted graph. In eithercase, ( G ′ , t , t ) is a 2-connected rooted graph. Therefore, by the induction hypothesis, we can get k ( t , t )-paths −→ P , . . . , −→ P k in G ′ satisfying the length condition. Then xt −→ P t y, . . . , xt −→ P k t y are k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) Let s ∈ S \ { x } and t ∈ T , and let G ′ = G − { s, t } . Then deg G ′ ( v ) ≥ deg G ( v ) − ≥ k −
1) for v ∈ V ( G ′ ) \ { x, y } .We divide the proof of Case 2.1 into three cases according as the connectivity of G ′ . Case 2.1.1. G ′ is 2-connected.By applying the induction hypothesis to ( G ′ , x, y ), G ′ contains k − x, y )-paths satisfying thelength condition. Let −→ P be the longest path in the k − x, y )-paths, and then the k − x, y )-pathstogether with the ( x, y )-path x −→ P y − sty form k ( x, y )-paths in G satisfying the length condition, acontradiction. Case 2.1.2. G ′ is not connected.In this case, there exists a component D of G ′ such that V ( D ) ∩ ( V ( H ) ∪ { y } ) = ∅ , and let D ′ = G [ V ( D ) ∪ { s, t } ]. Since G is 2-connected, it follows that ( D ′ , s, t ) is a 2-connected rooted graphsuch that deg D ′ ( v ) = deg G ( v ) for v ∈ V ( D ′ ) \ { s, t } (= V ( D )). Then by the induction hypothesis, D ′ contains k ( s, t )-paths satisfying the length condition. By adding xt and st ′ y to each path, where t ′ isa vertex of T \ { t } , we can obtain k ( x, y )-paths in G satisfying the length condition, a contradiction. Case 2.1.3. G ′ is connected, but not 2-connected.11y Claims 4.7 and 4.8, G [( V ( H ) ∪ { y } ) \ { s, t } ] is 2-connected. Since G ′ is connected but itis not 2-connected, this implies that there is an end block B of G ′ with cut vertex b such that V ( B − b ) ∩ (cid:0) ( V ( H ) ∪ { y } ) \{ s, t } (cid:1) = ∅ . Let −→ P be a path from b to some vertex a ∈ ( V ( H ) ∪ { y } ) \{ s, t } in G ′ internally disjoint from (cid:0) V ( B ∪ H ) ∪ { y } (cid:1) \ { s, t } . Since N G ( x ) = N G ( y ) = T by Claim 4.7, itfollows that a / ∈ { x, y } and thus a ∈ V ( H ) \ { x, s, t } . Note that N G ( B − b ) ⊆ { s, t, b } .We now show that t ∈ N G ( B − b ). By way of contradiction, suppose t / ∈ N G ( B − b ), and let B ′ = G [ V ( B ) ∪ { s } ]. Then ( B ′ , s, b ) is a 2-connected rooted graph such that deg B ′ ( v ) = deg G ( v ) for v ∈ V ( B ′ ) \ { s, b } . Hence by the induction hypothesis, B ′ contains k ( s, b )-paths −→ P , . . . , −→ P k satisfyingthe length condition. If a ∈ T , then let −→ P ′ = b −→ P ay ; if a ∈ S , then we take a vertex t ′ of T \ { t } ,and let −→ P ′ = b −→ P at ′ y . Then xts −→ P b −→ P ′ y, . . . , xts −→ P k b −→ P ′ y form k ( x, y )-paths in G satisfying the lengthcondition, a contradiction. Thus t ∈ N G ( B − b ) is proved.Now let B ′′ = G [ V ( B ) ∪ { t } ]. Then ( B ′′ , t, b ) is a 2-connected rooted graph such that deg B ′′ ( v ) ≥ deg G ( v ) − > k −
1) for v ∈ V ( B ′′ ) \ { t, b } . By the induction hypothesis, B ′′ contains k − t, b )-paths −→ Q , . . . , −−−→ Q k − satisfying the length condition. If a ∈ T , then there exists a vertex t ′ ∈ T \ { t, a } and thus xt −→ Q b −→ P ay, . . . , xt −−−→ Q k − b −→ P ay and xt −−−→ Q k − b −→ P ast ′ y form k ( x, y )-paths in G satisfying thelength condition, a contradiction. Thus a ∈ S . Then there exist two distinct vertices t , t ∈ T \ { t } ,and hence xt −→ Q b −→ P at y, . . . , xt −−−→ Q k − b −→ P at y and xt −−−→ Q k − b −→ P at st y form k ( x, y )-paths in G satisfyingthe length condition, a contradiction.This completes the proof of Case 2.1. Case 2.2. | C | ≥ Claim 4.9 E G ( T, C − y ) = ∅ .Proof. By way of contradiction, suppose that E G ( T, C − y ) = ∅ . By Claim 4.6, E G ( S \ { x } , C ) = ∅ .Hence N G ( C − y ) ⊆ { x, y } . Let C ′ = G [ V ( C ) ∪ { x } ]. Since G is 2-connected, ( C ′ , x, y ) is a 2-connected rooted graph such that deg C ′ ( v ) = deg G ( v ) for v ∈ V ( C ′ ) \ { x, y } . By the inductionhypothesis, C ′ (and also G ) contains k ( x, y )-paths satisfying the length condition, a contradiction.Thus E G ( T, C − y ) = ∅ . (cid:3) By (C1), for a vertex t of T , H contains 2 ( x, t )-paths of lengths 1 and 3, respectiely. Since E G ( T, C − y ) = ∅ by Claim 4.9, this implies that k ≥ . (4.3)In the rest of this proof, we say that an end block B of C is feasible if y / ∈ V ( B ) \ { b } , where b isthe cut vertex of C contained in B . Claim 4.10 (i) C is not -connected, and (ii) if B is a feasible end block of C with cut vertex b , then E G ( T, B − b ) = ∅ .Proof. Suppose that either C is 2-connected or there is a feasible end block B of C with cut vertex b such that E G ( T, B − b ) = ∅ . In the former case, we define B ′ = C and b ′ = y . Note that, in this12ase, E G ( T, B ′ − b ′ ) = ∅ holds by Claim 4.9. In the latter case, we define B ′ = B and b ′ = b . Notethat, in the latter case, E G ( T, B ′ − b ′ ) = ∅ also holds by the assumption. Now we define the graph B ∗ as follows: V ( B ∗ ) = V ( B ′ ) ∪ { T } and E ( B ∗ ) = E ( B ′ ) ∪ { vT : v ∈ V ( B ′ ) , E G ( v, T ) = ∅} . Then ( B ∗ , T, b ′ ) is a 2-connected rooted graph. By (C3) and (C4), it also follows that for each v ∈ V ( B ∗ ) \ { T, b ′ } ,deg B ∗ ( v ) ≥ deg G ( v ) − l ≥ k − l ) (if E G ( v, T ) = ∅ )deg G ( v ) − (cid:0) l + ( l + 1) (cid:1) + 1 ≥ k − l ) (if E G ( v, T ) = ∅ ) , and thus δ ( B ∗ , T, b ′ ) ≥ k − l ). By the induction hypothesis, B ∗ contains k − l ( T, b ′ )-paths satisfyingthe length condition. Therefore, by the definition of B ∗ and by adding a ( b ′ , y )-path in C to each ofthe k − l paths, we can obtain k − l ( T, y )-paths in G internally disjoint from V ( H ) and satisfyingthe length condition, which contradicts Lemma 4(i). (cid:3) Claim 4.11 If B is a feasible end-block of C with cut vertex b , then E G ( S \ { x } , B − b ) = ∅ .Proof. Suppose that E G ( S \ { x } , B − b ) = ∅ . By Claim 4.10(ii) and the 2-connectivity of G , we have N G ( B − b ) = { x, b } . Let B ′ = G [ V ( B ) ∪ { x } ]. Then ( B ′ , x, b ) is a 2-connected rooted graph such thatdeg B ′ ( v ) = deg G ( v ) for v ∈ V ( B ′ ) \ { x, b } . By the induction hypothesis, B ′ contains k ( x, b )-pathssatisfying the length condition. Combining these k paths and a ( b, y )-path in C , we can obtain k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) Claim 4.12 (i) l = 1 , and (ii) if B is a feasible end block of C with cut vertex b , then N G ( B − b ) = S ∪ { b } .Proof. By Claim 4.10(i), C contains a feasible end block. Let B be an arbitrary feasible end blockof C with cut vertex b , and we show that l = 1 and N G ( B − b ) = S ∪ { b } . Suppose that either l ≥ l = 1 but N G ( B − b ) = S ∪ { b } . We define the graph B ∗ as follows: V ( B ∗ ) = V ( B ) ∪ { S ∗ } and E ( B ∗ ) = E ( B ) ∪ { vS ∗ : v ∈ V ( B ) , E G ( v, S \ { x } ) = ∅} . Then by Claim 4.11, ( B ∗ , S ∗ , b ) is a 2-connected rooted graph.We first consider the case of l ≥
2. Then by Claim 4.10(ii) and (C3), and since l ≥
2, it followsthat for each v ∈ V ( B ∗ ) \ { S ∗ , b } ,deg B ∗ ( v ) ≥ deg G ( v ) ≥ k − l + 1) (if E G ( v, S ) = ∅ )deg G ( v ) − ≥ k − l + 1) (if E G ( v, S \ { x } ) = ∅ and E G ( v, x ) = ∅ )deg G ( v ) − l + 1 ≥ k − l + 1) (if E G ( v, S \ { x } ) = ∅ and E G ( v, x ) = ∅ )deg G ( v ) − l + 1 − ≥ k − l + 1) (if E G ( v, S \ { x } ) = ∅ and E G ( v, x ) = ∅ ) , δ ( B ∗ , S ∗ , b ) ≥ k − l + 1). By the induction hypothesis, B ∗ contains k − l + 1 ( S ∗ , b )-pathssatisfying the length condition. Then by the definition of B ∗ and by adding a ( b, y )-path in C to eachof the k − l + 1 paths, we can obtain k − l + 1 ( S \ { x } , y )-paths in G internally disjoint from V ( H )and satisfying the length condition, which contradicts Lemma 4(ii).We next consider the case of l = 1. In this case, N G ( B − b ) = ( S \ { x } ) ∪ { b } since E G ( S \ { x } , B − b ) = ∅ (by Claim 4.11) and N G ( B − b ) = S ∪ { b } . Hence it follows that deg B ∗ ( v ) = deg G ( v ) for v ∈ V ( B ∗ ) \ { S ∗ , b } . By the induction hypothesis, B ∗ contains k ( S ∗ , b )-paths satisfying the lengthcondition. Therefore, by the definition of B ∗ , we can easily see that G contains k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) By Claim 4.12(i), | S | = 2, say S \ { x } = { s } . On the other hand, by Claim 4.12(ii), if B isa feasible end block of C with cut vertex b and v is a vertex of S , then ( G [ V ( B ) ∪ { v } ] , v, b ) is a2-connected rooted graph such that deg G [ V ( B ) ∪{ v } ] ( v ′ ) ≥ deg G ( v ′ ) − ≥ k − v ′ ∈ V ( B ) \ { b } .This together with the induction hypothesis implies that G [ V ( B ) ∪ { v } ] contains k − v, b )-paths satisfying the length conditionfor any feasible end block B of C with cut vertex b and any vertex v ∈ S . (4.4)Now let U = N G ( T ) ∩ (cid:0) V ( C ) \ { y } (cid:1) . Note that by Claim 4.9, U = ∅ . Let B , . . . , B h be all the feasible end blocks of C with cut vertices b , . . . , b h , respectively(note that by Claim 4.10(i), such blocks exist). We further let C ′ = C − [ ≤ i ≤ h ( V ( B i ) \ { b i } ) . Note that C ′ is connected and that by Claim 4.12(ii), C ′ contains all the vertices of U ∪{ b , b , . . . , b h } .In the rest of this proof, let −→ P and −→ P be 2 ( x, b )-paths in G [ V ( B ) ∪ { x } ] satisfying the lengthcondition (note that by (4.3) and (4.4), such two paths exist). Claim 4.13
There exists an end block B y of C with cut vertex b y such that y ∈ V ( B y ) \ { b y } .Proof. Suppose that B , . . . , B h are all the end blocks of C . Since y ∈ V ( C ′ ) and U ⊆ V ( C ′ ) \ { y } ,and since the block-cut tree of C has order at least 3, there exist two vertex-disjoint paths −→ P and −→ Q in C ′ such that −→ P is a path from b i to some vertex u ∈ U and −→ Q is a path from b j to y for some i, j with i = j , say i = 1 and j = 2.On the other hand, it follows from (4.4) that G [ V ( B ) ∪{ s } ] contains k − s, b )-paths −→ Q , . . . , −−−→ Q k − satisfying the length condition. By the definition of U , we can take a vertex t of T such that tu ∈ E ( G ).Then x −→ P b −→ P uts −→ Q b −→ Q y, x −→ P b −→ P uts −→ Q b −→ Q y, . . . , x −→ P b −→ P uts −−−→ Q k − b −→ Q y, x −→ P b −→ P uts −−−→ Q k − b −→ Q y k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) Let B y and b y be the same ones as in Claim 4.13. Then B , . . . , B h and B y are all the end blocksof C . Claim 4.14
For each v ∈ V ( C ) \ { y } , either e G ( v, H ) ≤ or v is a cut vertex of C separating y andall feasible end blocks of C .Proof. Suppose that there exist a vertex v of V ( C ) \ { y } such that e G ( v, H ) ≥ B i , say i = 1, such that C − v has a ( b , y )-path −→ Q ′ internally disjoint from B (note that v ∈ V ( C ′ ),since every vertex of S ≤ j ≤ h ( V ( B j ) \ { b j } ) is adjacent to at most two vertices of H by Claim 4.12).Since l = 1 by Claim 4.12(i), (C3) and (C4) ensure that v is adjacent to exactly two distinct verticesin T , say t and t . By (4.4), G [ V ( B ) ∪ { s } ] contains k − s, b )-paths −→ Q , . . . , −−−→ Q k − satisfyingthe length condition. Then the k − xt s −→ Q i b −→ Q ′ y (1 ≤ i ≤ k −
1) together with the path xt vt s −−−→ Q k − b −→ Q ′ y form k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) By adding a ( b , b y )-path in C ′ to each of −→ P and −→ P , we can get two ( x, b y )-paths −→ P ′ and −→ P ′ in G [( V ( C ) ∪ { x } ) \ V ( B y − b y )] satisfying the length condition. Claim 4.15 | B y | = 2 ( i.e., V ( B y ) = { y, b y } ) .Proof. Suppose that | B y | ≥
3, that is, B y is 2-connected. For each vertex v of V ( B y ) \ { y, b y } , v is not a cut vertex of C separating y and all feasible end blocks of C , and hence by Claim 4.14, wehave e G ( v, H ) ≤
2. This implies that ( B y , b y , y ) is a 2-connected rooted graph such that deg B y ( v ) ≥ deg G ( v ) − ≥ k −
1) for v ∈ V ( B y ) \ { y, b y } . By the induction hypothesis, B y contains k − b y , y )-paths satisfying the length condition. Concatenating these k − P ′ and P ′ , we canobtain k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) By Claim 4.15 and since G is 2-connected, E G ( y, H ) = ∅ . Since xy / ∈ E ( G ), there exists a vertex a of V ( H ) \ { x } such that ay ∈ E ( G ). Claim 4.16 h = 1 .Proof. Suppose that h ≥
2. By (4.4), G [ V ( B ) ∪ { s } ] contains k − b , s )-paths −→ Q , . . . −−−→ Q k − satisfying the length condition. Let −→ R be a ( b , b )-path in C ′ . Note that by Claim 4.13, y / ∈ V ( R ).We also let −→ R ′ be an ( s, a )-path in H − x . Then x −→ P b −→ R b −→ Q s −→ R ′ ay, x −→ P b −→ R b −→ Q s −→ R ′ ay, . . . , x −→ P b −→ R b −−−→ Q k − s −→ R ′ ay, x −→ P b −→ R b −−−→ Q k − s −→ R ′ ay are k ( x, y )-paths in G satisfying the length condition, a contradiction. (cid:3) By Claim 4.16, B and B y are all the end blocks of C . Therefore, C has a unique block W of C such that W = B y and b y ∈ V ( W ). Then by (C3), (C4), (4.3), Claims 4.12(i) and 4.15,deg W ( b y ) ≥ k − (2 l + 1) − |{ b y y }| ≥
2, which implies that W is 2-connected.15e show that W = B . Assume not. Then by the definition of U , Claims 4.12(ii), 4.15 and 4.16,we have U ⊆ { b y } . Then by the definition of U , we have N G ( t ) ∩ V ( C ) ⊆ { y, b y } for t ∈ T . Let t bean arbitrary vertex of T . Since E G ( S \ { x } , C ) = ∅ by Claim 4.12(ii), it follows from Claim 4.5 that N G ( t ) ⊆ V ( H ∪ C ). Combining this with the above, we have N G ( t ) ⊆ V ( H ) ∪ { y, b y } . Then by (C1),(C4), (4.3) and Claim 4.12(i),6 ≤ k ≤ deg G ( t ) = | S | + e G ( t, T \ { t } ) + | E ( G ) ∩ { ty, tb y }| ≤ l + 4 = 6 . Thus the equality holds. This yields that k = 3 and also that e G ( t, T \ { t } ) = l + 1 = 2 and ty, tb y ∈ E ( G ). Since t is an arbitrary vertex of T , it follows that G [ T ] contains an edge t t , t b y ∈ E ( G ) and t y ∈ E ( G ). Then x −→ P ′ b y y, x −→ P ′ b y y and x −→ P ′ b y t t y are 3 (= k ) ( x, y )-paths in G satisfying the length condition, a contradiction. Thus W = B . In short, W is 2-connected and W = B .Let w be a cut vertex of C which is contained in W such that w = b y . Note that w and b y are allthe cut vertices of C which is contained in W . Then each vertex v of V ( W ) \{ w, b y } is not a cut vertexof C separating y and all feasible end blocks of C , since W is 2-connected. Hence by Claim 4.14, wehave e G ( v, H ) ≤ v ∈ V ( W ) \ { w, b y } . This implies that ( W, w, b y ) is a 2-connected rooted graphsuch that deg W ( v ) ≥ deg G ( v ) − ≥ k −
1) for v ∈ V ( W ) \ { w, b y } . By the induction hypothesis, W contains k − w, b y )-paths satisfying the length condition. Concatenating these k − w, b y )-pathswith P and P , the edge b y y and a ( b , w )-path in C , we can obtain k ( x, y )-paths in G satisfyingthe length condition, a contradiction.This completes the proof of Theorem 2. (cid:3) In this section, we give the proof of Theorem 3. The direction is the same as the proof of Theorem 2and the argument is also similar. Therefore we mainly describe the difference from the proof ofTheorem 2. In the following proof of Theorem 3, the claims without proof are obtained by the samearguments as in the proof of Theorem 2 (note that the numberings of the claims correspond to theones of the claims in the proof of Theorem 2).
Proof of Theorem 3.
We prove it by induction on | V ( G ) | + | E ( G ) | . Let ( G, x, y ) be a minimumcounterexample with respect to | V ( G ) | + | E ( G ) | . If k = 1, then by (R1) and (R2), we can easily seethat G contains an ( x, y )-path of length at least 2, a contradiction. Thus k ≥
2. Since δ ( G, x, y ) ≥ k −
1, this implies that | G | ≥
4. By symmetry, we may assume that deg G ( x ) ≤ deg G ( y ). Claim 5.1 G is -connected. Claim 5.2 xy / ∈ E ( G ) . Case 1. G − y does not contain a cycle of length 4 passing through x .16ince xy / ∈ E ( G ) by Claim 5.2, in this case, we have e G ( v, N G ( x ) \ { v } ) ≤ v ∈ V ( G ) \ { x, y } . (5.1)Let G ∗ be the graph obtained from G by contracting the subgraph induced by N G ( x ) ∪ { x } into asingle vertex x ∗ and then removing multiple edges. Then by (5.1),deg G ∗ ( v ) = deg G ( v ) for v ∈ V ( G ∗ ) \ { x ∗ , y } . (5.2)Assume for the moment that | G ∗ | = 2. This implies that V ( G ) = { x, y } ∪ N G ( x ). Since e G ( v, N G ( x ) \ { v } ) ≤ v ∈ N G ( x ), each vertex v of N G ( x ) satisfies 3 ≤ k − ≤ deg G ( v ) ≤ e G ( v, N G ( x ) \ { v } ) + |{ xv }| + | E ( G ) ∩ { yv }| ≤
3. Thus the equality holds. This yields that k = 2and also that G [ N G ( x )] contains an edge v v and v y, v y ∈ ( G ). Therefore, xv y and xv v y are2 (= k ) ( x, y )-paths satisfying the semi-length condition, a contradiction. Thus | G ∗ | ≥ | G ∗ | ≥
3, we can prove the rest of Case 1 by the same way as in theparagraphs following (4.2) in Case 1 of the proof of Theorem 2.
Case 2. G − y contains a cycle of length 4 passing through x .By the assumption of Case 2, G contains a bipartite subgraph H = G [ S, T ] such that H iscomplete bipartite, | T | ≥ | S | =: l + 1 ≥ x ∈ S , y / ∈ V ( H ) (i.e., H satisfies (C1) and (C2)). Let C be the component of G − V ( H ) such that y ∈ V ( C ). Choose H so that(a) | S | is maximum,(b) | T | is maximal, subject to (a),(c) | C | is maximum, subject to (a) and (b), and(d) | N G ( C ) ∩ S | is minimum, subject to (a), (b) and (c). Claim 5.3 H is an l -core of G with respect to ( x, y ) . Claim 5.4 If E G ( S \ { x } , C ) = ∅ , then (i) e G ( v, T ) ≤ l for v ∈ V ( G ) \ V ( H ∪ C ) , and (ii) e G ( v, T \{ v } ) ≤ l or E G ( v, C ) = ∅ for v ∈ T . Note that l ≤ k − Claim 5.5 If E G ( S \ { x } , C ) = ∅ , then N G ( T ) ⊆ V ( H ∪ C ) . Claim 5.6 If E G ( T, C − y ) = ∅ , then E G ( S \ { x } , C ) = ∅ .Proof. Assume that E G ( T, C − y ) = ∅ and E G ( S \ { x } , C ) = ∅ . Let t be a vertex of T , and let α = 1 if ty ∈ E ( G ); otherwise, let α = 0. Then by Claim 5.5 and since E G ( T, C − y ) = ∅ , it followsthat N G ( t ) ⊆ V ( H ) ∪ { y } . By the definition of α and Lemma 1(ii), we have l ≤ k − α −
1. Since17 G ( t ) ∩ V ( C ) ⊆ { y } , by (C4) and Claim 5.4(ii), we also have e G ( t, T \ { t } ) ≤ l + α . Combining thiswith (C1), it follows from the inequality l ≤ k − α − k − ≤ deg G ( t ) = | S | + e G ( t, T \ { t } ) + α ≤ ( l + 1) + ( l + α ) + α ≤ ( k − α ) + ( k −
1) + α = 2 k − . Thus the equality holds in the above inequality. This implies that l = k − α − e G ( t, T \{ t } ) = l + α .Since e G ( t, T \ { t } ) = l + α ≥ l ≥
1, we have E ( G [ T ]) = ∅ . If α = 0, then l = k −
1, and so Lemma 2implies that G contains k ( x, y )-paths satisfying the semi-length condition, a contradiction. Therefore α = 1, i.e., ty ∈ E ( G ). Then the equality e G ( t, T \ { t } ) = l + α = l + 1 = k − | T | ≥ k . Now,since t is an arbitrary vertex of T , these arguments imply the following: G [ T ] contains an edge t t such that t y, t y ∈ E ( G ) and, | S | = k − | T | ≥ k .Therefore, we can easily find k ( x, y )-paths in G [ V ( H ) ∪ { y } ] satisfying the semi-length condition, acontradiction. Thus the claim holds. (cid:3) We divide the proof of Case 2 into two cases according as | C | = 1 or | C | ≥ Case 2.1. | C | = 1, i.e., V ( C ) = { y } . Claim 5.7 N G ( x ) = N G ( y ) = T . Claim 5.8 | T | ≥ . Note that if P , . . . , P k − are k − Q is anotherpath of length | E ( P k − ) | + 2, then P , . . . , P k − , Q are k paths satisfying the semi-length condition.Therefore we can prove the rest of Case 2.1 by the same way as in the paragraphs following Claim 4.8in Case 2.1 of the proof of Theorem 2. Case 2.2. | C | ≥ Claim 5.9 E G ( T, C − y ) = ∅ . By (C1), for a vertex t of T , H contains 2 ( x, t )-paths of lengths 1 and 3, respectively. Since E G ( T, C − y ) = ∅ by Claim 5.9, this implies that k ≥ . (5.3)In the rest of this proof, we say that an end-block B of C is feasible if y / ∈ V ( B ) \ { b } , where b isthe cut vertex of C contained in B . Claim 5.10 (i) C is not -connected, and (ii) if B is a feasible end block of C with cut vertex b , then E G ( T, B − b ) = ∅ . Claim 5.11 If B is a feasible end block of C with cut vertex b , then E G ( S \ { x } , B − b ) = ∅ . laim 5.12 (i) l = 1 , and (ii) if B is a feasible end block of C with cut vertex b , then N G ( B − b ) = S ∪ { b } . By Claim 5.12(i), | S | = 2, say S \ { x } = { s } . On the other hand, by Claim 5.12(ii), if B isa feasible end block of C with cut vertex b and v is a vertex of S , then ( G [ V ( B ) ∪ { v } ] , v, b ) is a2-connected rooted graph such that deg G [ V ( B ) ∪{ v } ] ( v ′ ) ≥ deg G ( v ′ ) − ≥ k −
1) for v ′ ∈ V ( B ) \ { b } .Therefore, by Theorem 2, G [ V ( B ) ∪ { v } ] contains k − v, b )-paths satisfying the length conditionfor any feasible end block B of C with cut vertex b and any vertex v ∈ S . (5.4)Notice that these paths satisfy the length condition, not the length condition or the semi-lengthcondition. Now let U = N G ( T ) ∩ (cid:0) V ( C ) \ { y } (cid:1) . Note that by Claim 5.9, U = ∅ . Let B , . . . , B h be all the feasible end blocks of C with cut vertices b , . . . , b h , respectively(note that by Claim 5.10(i), such blocks exist). We further let C ′ = C − [ ≤ i ≤ h ( V ( B i ) \ { b i } ) . Note that C ′ is connected and that by Claim 5.12(ii), C ′ contains all the vertices of U ∪{ b , b , . . . , b h } .In the rest of this proof, let −→ P , . . . , −−→ P k − be k − ≥
2) ( x, b )-paths in G [ V ( B ) ∪ { x } ] satisfyingthe length condition (note that by (5.3) and (5.4), such two paths exist). Claim 5.13
There exists an end block B y of C with cut vertex b y such that y ∈ V ( B y ) \ { b y } . Let B y and b y be the same ones as in Claim 5.13. Then B , . . . , B h and B y are all the end blocksof C . Claim 5.14
For each v ∈ V ( C ) \ { y } , either e G ( v, H ) ≤ or v is a cut vertex of C separating y andall feasible end blocks of C . By adding a ( b , b y )-path in C ′ to each −→ P i , we can get k − ≥
2) ( x, b y )-paths −→ P ′ , . . . , −−→ P ′ k − in G [( V ( C ) ∪ { x } ) \ V ( B y − b y )] satisfying the length condition. Note that if −→ Q , . . . , −−−→ Q k − are k − b y , y )-paths in B y satisfying the semi-length condition, then x −→ P ′ b y −→ Q y, x −→ P ′ b y −→ Q y, . . . , x −→ P ′ b y −−−→ Q k − y, x −→ P ′ b y −−−→ Q k − y are k ( x, y )-paths satisfying the semi-length condition. Therefore, the following claim is also obtainedby the same argument as in the proof of Claim 4.15. Claim 5.15 | B y | = 2 ( i.e., V ( B y ) = { y, b y } ) .
19y Claim 5.15 and since G is 2-connected, E G ( y, H ) = ∅ . Since xy / ∈ E ( G ),there exists a vertex a of V ( H ) \ { x } such that ay ∈ E ( G ).Furthermore, if N G ( y ) ∩ N G ( b y ) ∩ V ( H − x ) = ∅ , say v ∈ N G ( y ) ∩ N G ( b y ) ∩ V ( H − x ), then x −→ P ′ b y y, . . . , x −−→ P ′ k − b y y and x −−→ P ′ k − b y vy are k ( x, y )-paths satisfying the semi-length condition, a con-tradiction. Thus we also have N G ( y ) ∩ N G ( b y ) ∩ V ( H − x ) = ∅ . (5.5) Claim 5.16 h = 1 . By Claim 5.16, B and B y are all the end blocks of C . Therefore, C has a unique block W of C such that W = B y and b y ∈ V ( W ). Claim 5.17 W = B .Proof. Suppose that W = B . Then by the definition of U , Claims 5.12(ii), 5.15 and 5.16, wehave U ⊆ { b y } . By the definition of U , we have N G ( t ) ∩ V ( C ) ⊆ { y, b y } for t ∈ T . Let t be anarbitrary vertex of T . Since E G ( S \ { x } , C ) = ∅ by Claim 5.12(ii), it follows from Claim 5.5 that N G ( t ) ⊆ V ( H ∪ C ). Combining this with the above, we have N G ( t ) ⊆ V ( H ) ∪ { y, b y } . Moreover, by(5.5), we also have | E ( G ) ∩ { ty, tb y }| ≤
1. Therefore, by (C1), (C4), Claim 5.12(i) and (5.3),5 ≤ k − ≤ deg G ( t )= | S | + e G ( t, T \ { t } ) + | E ( G ) ∩ { ty, tb y }| ≤ ( l + 1) + ( l + 1) + 1 = 2 l + 3 = 5 . Thus the equality holds. This yields that k = 3 and also that e G ( t, T \ { t } ) = l + 1 = 2 and | E ( G ) ∩ { ty, tb y }| = 1. Since t is an arbitrary vertex of T , we have e G ( t, T \ { t } ) = 2 and | E ( G ) ∩ { ty, tb y }| = 1 for t ∈ T. (5.6)Assume first that N G ( y ) ∩ T = ∅ . We may assume that a ∈ N G ( y ) ∩ T (see the paragraph followingClaim 5.15). Let t , t ∈ N G ( a ) ∩ T with t = t (note that by (5.6), such two vertices exist). If t i b y ∈ E ( G ) for some i with i ∈ { , } , then x −→ P ′ b y y, x −→ P ′ b y y and x −→ P ′ b y t i ay are 3 (= k ) ( x, y )-pathssatisfying the length condition, a contradiction. Thus t i b y / ∈ E ( G ) for i ∈ { , } , and hence by (5.6), t i y ∈ E ( G ) for i ∈ { , } . Then xay, xat y and xat st y are 3 (= k ) ( x, y )-paths satisfying thesemi-length condition, a contradiction. Assume next that N G ( y ) ∩ T = ∅ . Then a ∈ S , i.e., a = s since a ∈ V ( H ) \ { x } . Since N G ( y ) ∩ T = ∅ , tb y ∈ E ( G ) for t ∈ T . Hence by taking a vertex t of T ,it follows that x −→ P ′ b y y, x −→ P ′ b y y and x −→ P ′ b y tsy are 3 (= k ) ( x, y )-paths satisfying the length condition,a contradiction. (cid:3) Moreover, we can show that the following holds.
Claim 5.18 W is -connected. roof. It suffices to show that deg W ( b y ) ≥
2. By way of contradiction, suppose that deg W ( b y ) ≤ ≤ k − ≤ deg G ( b y )= e G ( b y , S ) + e G ( b y , T ) + deg W ( b y ) + |{ b y y }| = l + ( l + 1) + 1 + 1 = 2 l + 3 = 5 . Thus the equality holds. This yields that k = 3 and also that e G ( b y , S ) = 1 and e G ( b y , T ) = 2.Let t , t ∈ N G ( b y ) ∩ T with t = t . If b y x ∈ E ( G ), then xb y y, xt b y y and xt st b y y are 3 (= k )( x, y )-paths satisfying the semi-length condition, a contradiction. Thus b y x / ∈ E ( G ), and hence theequality e G ( b y , S ) = 1 implies that b y s ∈ E ( G ). Then by (5.5), a ∈ T , and hence x −→ P ′ b y y, x −→ P ′ b y y and x −→ P ′ b y say are 3 (= k ) ( x, y )-paths satisfying the length condition, a contradiction. Thus deg W ( b y ) ≥ (cid:3) Since W = B and W is 2-connected by Claims 5.17 and 5.18, we can prove the rest of Case 2.2by the same way as in the last paragraph of Case 2.2 in the proof of Theorem 2.This completes the proof of Theorem 3. (cid:3) In this section, for a positive integer k , we let ϕ (= ϕ ( k )) = k is odd)1 (if k is even) . Now we first show Theorem 4.
Proof of Theorem 4.
By the definition of ϕ , we have k = 2 l − ϕ for some l ≥
1. Then by thedegree condition, δ ( G ) ≥ k + 1 = 2 l + ϕ = 2( l + ϕ ) − ϕ . Since G is 2-connected but not 3-connected,there exists a separation ( A, B ) of G of order two, say A ∩ B = { x, y } . Then it is easily seen thateach of ( G [ A ] , x, y ) and ( G [ B ] , x, y ) is a 2-connected rooted graph and, δ ( G [ A ] , x, y ) ≥ δ ( G ) ≥ l + ϕ ) − ϕ ≥ l + ϕ ) − δ ( G [ B ] , x, y ) ≥ l + ϕ ) − ϕ. Therefore, by applying Theorem 3 to each of ( G [ A ] , x, y ) and ( G [ B ] , x, y ), it follows that G [ A ] (resp., G [ B ]) contains l + ϕ ( x, y )-paths −→ P , . . . , −−→ P l + ϕ (resp., −→ Q , . . . , −−−→ Q l + ϕ ) satisfying the length condition orthe semi-length condition. In particular, Theorem 2 guarantees thatboth of P , . . . , P l + ϕ and Q , . . . , Q l + ϕ satisfy the length condition if ϕ = 0.Now, suppose that either P , . . . , P l + ϕ or Q , . . . , Q l + ϕ satisfy the length condition. By the sym-metry, we may assume that P , . . . , P l + ϕ satisfy the length condition. By applying Theorem 2to ( G [ B ] , x, y ), we can take other l ( x, y )-paths −→ Q ′ , . . . , −→ Q ′ l satisfying the length condition, since δ ( G [ B ] , x, y ) ≥ l + ϕ ) − ϕ ≥ l . Then Table 1 gives l + ( l + ϕ −
1) = 2 l − ϕ = k cycles in G Consider in order from the first column of the left in Table 1 P , . . . , P l + ϕ nor Q , . . . , Q l + ϕ satisfy the length condition, and then both of P , . . . , P l + ϕ and Q , . . . , Q l + ϕ satisfy the semi-lengthcondition. Note that ϕ = 1.Let p and q be the switches of P , . . . , P l + ϕ (= P l +1 ) and Q , . . . , Q l + ϕ (= Q l +1 ), respectively.Then Table 2 gives 2 l = 2 l − ϕ = k cycles in G satisfying the length condition. (cid:3) Both of P , . . . , P l + ϕ and Q ′ , . . . , Q ′ l satisfy the length condition x −→ P y ←− Q ′ x x −→ P y ←− Q ′ l xx −→ P y ←− Q ′ x x −→ P y ←− Q ′ l x ... ... x −→ P y ←− Q ′ l x x −−→ P l + ϕ y ←− Q ′ l xl cycles l + ϕ − Table 1:
Both of P , . . . , P l + ϕ and Q , . . . , Q l + ϕ satisfy the semi-length condition and ϕ = 1 x −→ P y ←− Q x x −→ P y ←− Q q x x −−−→ P p +1 y ←−−− Q q +2 x x −−−→ P p +2 y ←−− Q l +1 xx −→ P y ←− Q x x −→ P y ←− Q q x x −−−→ P p +1 y ←−−− Q q +1 x ... ...... ... x −−−→ P p +1 y ←− Q l x x −→ P l y ←−− Q l +1 xx −→ P y ←− Q q x x −→ P p y ←− Q q x x −−−→ P p +1 y ←−− Q l +1 x x −−→ P l +1 y ←−− Q l +1 xq cycles p − l − q cycles l − p cycles Table 2:We next show Theorem 5. In the proof, we also use the following two lemmas (Lemmas 5 and 6).
Lemma 5 ([2, Lemma 5.1])
Let G be a connected graph such that δ ( G ) ≥ . If G contains a non-separating induced odd cycle, then G contains a non-separating induced odd cycle −→ C satisfying oneof the following (1) and (2) . (1) | V ( C ) | = 3 , or (2) for every non-cut vertex v of G − V ( C ) , e G ( v, C ) ≤ , and the equality holds if and only if vu + , vu − ∈ E ( G ) for some u ∈ V ( C ) . Lemma 6
Let k and l be positive integers such that k = 2 l − ϕ ( k ) . Let G be a graph and −→ C be an odd cycle in G , say | V ( C ) | = 2 m + 1 for some m ≥ . We further let x ∈ V ( G ) \ V ( C ) and u ∈ V ( C ) . If one of the following (i)–(iii) holds, then G contains k cycles having consecutive lengths. ϕ ( k ) = 0 and G contains l ( u, { u + m , u − m } ) -paths internally disjoint from V ( C ) and satisfyingthe length condition or the semi-length condition. (ii) ϕ ( k ) = 1 and G contains l ( u, { u + m , u − m } ) -paths internally disjoint from V ( C ) and satisfyingthe length condition. (iii) m ≥ , xu + , xu − ∈ E ( G ) and G contains l − x, u + m ) -paths internally disjoint from V ( C ) ∪{ x } and satisfying the length condition.Proof of Lemma 6. To show (i) and (ii), suppose that G contains l ( u, { u + m , u − m } )-paths −→ P , . . . , −→ P l internally disjoint from V ( C ) and satisfying the length condition or the semi-length condition. Wefurther suppose that if ϕ = 1, then P , . . . , P l satisfy the length condition. For each i with 1 ≤ i ≤ l ,let v i be the end vertex of P i such that v i ∈ { u + m , u − m } , and let −→ Q i and −→ R i denote the paths in C from v i to u such that | E ( Q i ) | = m and | E ( R i ) | = m + 1. If P , . . . , P l satisfy the length condition, thenTable 3 gives 2 l ≥ l − ϕ = k cycles having consecutive lengths (see also Figure 2). Thus we mayassume that P , . . . , P l does not satisfy the length condition but satisfy the semi-length condition. Inparticular, by our assumption, we have ϕ = 1, i.e., ϕ = 0. Let j be the switch of P , . . . , P l , and thenTable 4 gives 2 l − l − ϕ = k cycles having consecutive lengths. Thus (i) and (ii) are proved.We next show (iii). Suppose that m ≥ xu + , xu − ∈ E ( G ) and G contains l − x, u + m )-paths −→ P , . . . , −−→ P l − internally disjoint from V ( C ) ∪ { x } and satisfying the length condition. Since | C | = 2 m + 1 ≥
5, we can let −−→ Q u + and −−→ R u + (resp., −−→ Q u − and −−→ R u − ) be the paths in C from u + m to u + (resp., from u + m to u − ) such that | E ( Q u + ) | = m − | E ( R u + ) | = m + 2 (resp., | E ( Q u − ) | = m and | E ( R u − ) | = m + 1). Hence Table 5 gives 2 l ≥ l − ϕ = k cycles having consecutive lengths(see also Figure 3). Thus (iii) is also proved. (cid:3) P , . . . , P l satisfy the length condition u −→ P v −→ Q u , u −→ P v −→ R uu −→ P v −→ Q u , u −→ P v −→ R u ... u −→ P l v l −→ Q l u , u −→ P l v l −→ R l u l cycles Table 3: u v i −→ P i −→ Q i C u −→ P i −→ R i C ( v i ∈ { u + m , u − m } ) v i Figure 2: The paths P i , Q i and R i , . . . , P l satisfy the semi-length condition and ϕ = 0 u −→ P v −→ Q u , u −→ P v −→ R u u −−→ P j +1 v j +1 −−−→ R j +1 uu −→ P v −→ Q u , u −→ P v −→ R u u −−→ P j +2 v j +2 −−−→ Q j +2 u , u −−→ P j +2 v j +2 −−−→ R j +2 u ... u −−→ P j +3 v j +3 −−−→ Q j +3 u , u −−→ P j +3 v j +3 −−−→ R j +3 uu −−→ P j − v j − −−−→ Q j − u , u −−→ P j − v j − −−−→ R j − u ... u −→ P j v j −→ Q j u , u −→ P j v j −→ R j u u −→ P l v l −→ Q l u , u −→ P l v l −→ R l u j cycles 2 l − j − Table 4: m ≥ xu + , xu − ∈ E ( G ) and P , . . . , P l − satisfy the semi-length condition x −→ P u + m −−→ Q u + u + x , x −→ P u + m −−→ Q u − u − xx −→ P u + m −−→ Q u + u + x , x −→ P u + m −−→ Q u − u − x x −−→ P l − u + m −−→ R u − u − x , x −−→ P l − u + m −−→ R u + u + x ... x −−→ P l − u + m −−→ Q u + u + x , x −−→ P l − u + m −−→ Q u − u − x l − Table 5: u u + m −→ P i C −−→ Q u + u + u − x u u + m −→ P i C −−→ Q u − u + u − x u u + m C −−→ R u − u + u − x u u + m −−→ P l − C −−→ R u + u + u − x −−→ P l − Figure 3: The paths P i , Q u + , Q u − , R u + and R u − We are now ready to prove Theorem 5.
Proof of Theorem 5. If k = 1, then the assertion clearly holds, since G is 2-connected. If k = 2, takean edge xy of G and find 2 ( x, y )-paths satisfying the length condition or the semi-length conditionby using Theorem 3, and then the edge xy and the 2 ( x, y )-paths induce 2 cycles which satisfy thelength condition or have consecutive lengths. Thus we may assume k ≥ k = 2 l − ϕ for some l ≥
2. Then by the degree condition, we have δ ( G ) ≥ k +1 (= 2 l + ϕ ) ≥ G contains a non-separating induced odd cycle −→ C in G , say | V ( C ) | = 2 m + 1for m ≥
1, such that C satisfies (1) or (2) in Lemma 5. Now, suppose that G does not contain24 cycles having consecutive lengths, and then we will show that G contains k cycles satisfying thelength condition. Claim 6.1 | V ( C ) | ≥ , that is, m ≥ .Proof. Suppose that | V ( C ) | = 3, that is, m = 1. Let u ∈ V ( C ). Consider the graph G ∗ obtainedfrom G by contracting u + and u − into a vertex u ∗ . Then G ∗ is a 2-connected graph and δ ( G ∗ ) ≥ δ ( G ) − ≥ k = 2 l − ϕ ( ≥ l − G ∗ contains l ( u, u ∗ )-paths satisfying thelength condition or the semi-length condition. In particular, if ϕ = 1, then by Theorem 2, we mayassume that the l ( u, u ∗ )-paths satisfy the length condition. Note that each of the l paths does notcontain the edge uu ∗ , since the length is at least 2. Then it follows from the definition of G ∗ that G contains l ( u, { u + , u − } )-paths internally disjoint from V ( C ) and satisfying the length condition orthe semi-length condition; in particular, they satisfy the former condition when ϕ = 1. This togetherwith Lemma 6(i) or (ii) leads to a contradiction. (cid:3) By Claim 6.1, C satisfies (2) in Lemma 5, i.e., every non-cut vertex v of G − V ( C ) satisfiesdeg G − V ( C ) ( v ) ≥ δ ( G ) − ≥ k − l −
1) + ϕ. (6.1) Fact 6.2 If B is an end block of G − V ( C ) , then B is -connected.Proof. Since k ≥
3, (6.1) implies that | B | ≥
3, and hence B is 2-connected. (cid:3) Claim 6.3
Let B be an end block of G − V ( C ) with cut vertex b when G − V ( C ) is not -connected;otherwise, let B = G − V ( C ) and b be an arbitrary vertex of G − V ( C ) . Further, let x ∈ V ( B − b ) and u ∈ V ( C ) . If one of the following (i) and (ii) holds, then we have E G ( { u + m , u − m } , V ( G − C ) \ V ( B − b )) = ∅ . (i) xu ∈ E ( G ) and every vertex of B − b is adjacent to at most one vertex of C . (ii) xu + , xu − ∈ E ( G ) .Proof. Suppose that (i) or (ii) holds and that E G ( { u + m , u − m } , V ( G − C ) \ V ( B − b )) = ∅ . By thesymmetry of u + m and u − m , we may assume that E G ( u + m , V ( G − C ) \ V ( B − b )) = ∅ . Let y be avertex of V ( G − C ) \ V ( B − b ) such that yu + m ∈ E ( G ).We first consider the case where (i) holds. Then by the assumption, every vertex v of B − b satisfies deg B ( v ) = deg G − V ( C ) ( v ) ≥ δ ( G ) − ≥ k = 2 l − ϕ ( ≥ l − . Then it follows from Fact 6.2 and Theorem 3 that B contains l ( x, b )-paths satisfying the lengthcondition or the semi-length condition. In particular, if ϕ = 1, then by Theorem 2, we may assumethat the l ( x, b )-paths satisfy the length condition. By adding a ( b, y )-path in G − ( V ( C ) ∪ V ( B − b ))to each of the l ( x, b )-paths, we can get l ( x, y )-paths in G − V ( C ) satisfying the length conditionor the semi-length condition; in particular, they satisfy the former condition when ϕ = 1. Since xu, yu + m ∈ E ( G ), this together with Lemma 6(i) or (ii) leads to a contradiction.25e next consider the case where (ii) holds. Then it follows from (6.1), Fact 6.2 and Theorem 2that B contains l − x, b )-paths satisfying the length condition. By adding a ( b, y )-path in G − ( V ( C ) ∪ V ( B − b )) to each of the l − x, b )-paths, we can get l − x, y )-paths in G − V ( C ) satisfyingthe length condition. Since xu + , xu − , yu + m ∈ E ( G ) and since m ≥ (cid:3) Claim 6.4 G − V ( C ) is not -connected.Proof. Suppose that B := G − V ( C ) is 2-connected. Choose a vertex x of B so that e G ( x, C ) isas large as possible. Then e G ( x, C ) = 1 or 2 and e G ( v, C ) ≤ v ∈ V ( B ) since G is 2-connectedand C satisfies (2) in Lemma 5. Furthermore, if e G ( x, C ) = 2, then there is a vertex u of C suchthat xu + , xu − ∈ E ( G ); otherwise, let u be the unique vertex of C such that xu ∈ E ( G ). Note thatif e G ( x, C ) = 2, then the choice of x implies that every vertex of B is adjacent to at most one vertexof C . Now, since δ ( G ) ≥ k + 1 ≥ C is an induced cycle, we have E G ( u + m , B ) = ∅ , and so let b be a vertex of B such that u + m b ∈ E ( G ). Since m ≥ u that b = x . These imply that (i) or (ii) of Claim 6.3 holds for the graph B and the vertices b, x, u ,but u + m b ∈ E G ( { u + m , u − m } , V ( G − C ) \ V ( B − b )), a contradiction. (cid:3) By Claim 6.4, G − V ( C ) has two end blocks B and B with cut vertices b and b , respectively. Claim 6.5
There is a vertex x i of B i − b i such that e G ( x i , C ) = 2 for i ∈ { , } .Proof. Let B = B i and b i = b and suppose that every vertex of B − b is adjacent to at most onevertex of C . Since G is 2-connected, there is a vertex u of C such that E G ( u, B − b ) = ∅ , i.e., somevertex x of B − b is adjacent to u . Hence (i) of Claim 6.3 holds for the graph B and the vertices b, x, u , and so we have E G ( u + m , V ( G − C ) \ V ( B − b )) = ∅ , that is, N G ( u + m ) ⊆ V ( C ) ∪ V ( B − b ).Since δ ( G ) ≥ C is an induced cycle, this in particular implies that E G ( u + m , B − b ) = ∅ . Thenagain by applying (i) of Claim 6.3 for the vertex u + m and a neighbor of u + m in B − b instead of u and x , respectively, we have E G ( u +2 m , V ( G − C ) \ V ( B − b )) = ∅ and thus N G ( u +2 m ) ⊆ V ( C ) ∪ V ( B − b ).Repeating this argument we get N G ( C ) ⊆ V ( B − b ) since | V ( C ) | is odd. This implies that b is a cutvertex of G , which contradicts the 2-connectivity of G . Thus e G ( x, C ) ≥ x ∈ V ( B − b ).Since C satisfies (2) in Lemma 5, we have e G ( x, C ) = 2. (cid:3) By Claim 6.5 and since C satisfies (2) in Lemma 5, each B i − b i contains a vertex x i such that x i u + i , x i u − i ∈ E ( G ) for some u i ∈ V ( C ).Assume for the moment that u = u . Then Claim 6.3 yields that E G ( u + m , V ( G − C ) \ V ( B − b )) = ∅ . On the other hand, since u + m = u + m , it also follows from Claim 6.3 that E G ( u + m , V ( G − C ) \ V ( B − b )) = ∅ . Therefore we get E G ( u + m , V ( G − C )) = ∅ . But, since C is an induced cycle,this implies that deg G ( u + m ) = deg C ( u + m ) = 2 <
4, a contradiction. Thus u = u .Assume next that u +1 = u . Then u − m = u + m . By using Claim 6.3 and arguing as in the above,we have E G ( u − m , V ( G − C )) = ∅ , which contradicts that deg G ( u − m ) ≥
4. Thus u +1 = u . Similarly,we have u − = u . Consequently, we get u ∈ V ( u +21 −→ C u − ) . | V ( C ) | ≥
5, without loss of generality, we may assume that u − = u +2 . We will show that there exist 2 l − ϕ ( x , x )-paths −→ P , . . . , −−−−−→ P l − ϕ in G − V ( C ) satisfying thelength condition. In order to show it, we divide the proof into two cases. Case 1. ϕ ( k ) = 1.Since δ ( B i , x i , b i ) ≥ l −
1) + ϕ = 2 l − i ∈ { , } , it follows from Fact 6.2 and Theorem 3 that B contains l ( x , b )-paths −→ Q , . . . , −→ Q l satisfying the length condition or the semi-length condition,and B contains l ( b , x )-paths −→ R , . . . , −→ R l satisfying the length condition or the semi-length condition.Now, suppose that either Q , . . . , Q l or R , . . . , R l satisfy the length condition. By the symmetry,we may assume that Q , . . . , Q l satisfy the length condition. By applying Theorem 2 to ( B , b , x ),we can take other l − b , x )-paths −→ R ′ , . . . , −−→ R ′ l − satisfying the length condition, since δ ( B , b , x ) ≥ l − Q , . . . , Q l and R ′ , . . . , R ′ l − with a ( b , b )-path in G − V ( C ), we can obtain2 l − l − ϕ ) ( x , x )-paths −→ P , . . . , −−−→ P l − in G − V ( C ) satisfying the length condition. Therefore,we may assume that neither Q , . . . , Q l nor R , . . . , R l satisfy the length condition, and then both of Q , . . . , Q l and R , . . . , R l satisfy the semi-length condition.Let q and r be the switches of Q , . . . , Q l and R , . . . , R l , respectively, and let P be a ( b , b )-pathin G − V ( C ). Then by considering the paths x −→ Q b −→ P b −→ R i x (1 ≤ i ≤ r ) , x −→ Q i b −→ P b −→ R r x (2 ≤ i ≤ q ) ,x −−−→ Q q +1 b −→ P b −−−→ R r +1 x , x −−−→ Q q +1 b −→ P b −→ R i x ( r + 2 ≤ i ≤ l ) , x −→ Q i b −→ P b −→ R l x ( q + 2 ≤ i ≤ l ) , we can obtain 2 l − l − ϕ ) ( x , x )-paths −→ P , . . . , −−−→ P l − in G − V ( C ) satisfying the lengthcondition. Case 2. ϕ ( k ) = 0.Since δ ( B i , x i , b i ) ≥ l −
1) for i ∈ { , } , it follows from Fact 6.2 and Theorem 2 that B contains l − x , b )-paths satisfying the length condition and B contains l − b , x )-pathssatisfying the length condition. Concatenating them with a ( b , b )-path in G − V ( C ), we can obtain2 l − l − ϕ ) ( x , x )-paths −→ P , . . . , −−−→ P l − in G − V ( C ) satisfying the length condition.Thus, in both cases, Table 6 gives 2 l − ϕ = k cycles satisfying the length condition (see alsoFigure 4).This completes the proof of Theorem 5. (cid:3) , . . . , P l − ϕ satisfy the lengthcondition x −→ P x u − ←− C u +1 x x −→ P x u − ←− C u +1 x ... x −−−−−→ P l − ϕ x u − ←− C u +1 x l − ϕ cycles x −−−−−→ P l − ϕ x u − ←− C u − x x −−−−−→ P l − ϕ x u +2 ←− C u − x Table 6: u −→ P i Cu +1 u − x x u u − u +2 u −−−−−→ P l − ϕ Cu +1 u − x x u u − u +2 u −−−−−→ P l − ϕ Cu +1 u − x x u u − u +2 Figure 4: The path P i In this paper, we have shown that the Thomassen’s conjecture on the existence of cycles of any lengthmodulo a given integer k (Conjecture A) is true by giving degree conditions for the existence of aspecified number of cycles whose lengths differ by one or two (Theorem 1).The complete graph of order k + 1, in a sense, shows the sharpness of the lower bound on theminimum degree condition in Theorem 1. On the other hand, we believe that the assumption of2-connectivity in Theorem 1 is not necessary. In fact, Liu and Ma conjectured that Theorem 1 alsoholds even if we drop the connectivity condition (see [2, Conjecture 6.2]). To approach the conjecture,the following improvements of Theorems 2 and 3 will be helpful. Problem 1
Let k be a positive integer, and let ( G, x, y ) be a -connected rooted graph such that | V ( G ) | ≥ , and z ∈ V ( G ) ( possibly z = x or z = y ) . Suppose that deg G ( v ) ≥ k for any v ∈ V ( G ) \ { x, y, z } . Then G contains k paths from x to y satisfying the length condition. Problem 2
Let k be a positive integer, and let ( G, x, y ) be a -connected rooted graph such that | V ( G ) | ≥ , and z ∈ V ( G ) ( possibly z = x or z = y ) . Suppose that deg G ( v ) ≥ k − for any v ∈ V ( G ) \ { x, y, z } . Then G contains k paths from x to y satisfying the length condition or thesemi-length condition. In fact, if Problems 1 and 2 are true, then by applying them to an end block B with cut vertex z in a given graph of minimum degree at least k + 1, and by arguing as in the proofs of Theorems 4 and5, we can show that B contains k cycles satisfying the length condition when B is 2-connected butnot 3-connected; B contains k cycles, which have consecutive lengths or satisfy the length conditionwhen B is 3-connected and non-bipartite. Therefore, Problems 1 and 2 leads to the improvement ofTheorem 1.The above problems also concern with another Thomassen’s conjecture on the existence cycles ofany even length modulo a given integer k . 28 onjecture D (Thomassen [3]) For a positive integer k , every graph of minimum degree at least k + 1 contains cycles of all even lengths modulo k . It is known that this conjecture is true for all even integers k (see [2, Theorem 1.9]). Theimprovement of Theorem 1 implies that it is also true for all odd integers k (note that Theorem 1implies that Conjecture D is true for the case where k is odd and a given graph is 2-connected).Problems 1 and 2 can be proven by arguing as in the proofs of Theorems 2 and 3 and by atedious case-by-case analysis (in fact, we have checked Problem 1 is true in a private discussion, butit is unpublished). Therefore, we think that giving a short proof for the above problems might beinteresting and helpful for future work of this research area. References [1] J.A. Bondy, A. Vince,
Cycles in a graph whose lengths differ by one or two , J. Graph Theory (1998) 11–15.[2] C-H. Liu, J. Ma, Cycle lengths and minimum degree of graphs , J. Combin. Theory Ser. B (2018) 66–95.[3] C. Thomassen,
Graph decomposition with applications to subdivisions and path systems modulo k , J. Graph Theory7