A Lower Bound for the Circumference Involving Connectivity
aa r X i v : . [ m a t h . C O ] J u l A Lower Bound for the CircumferenceInvolving Connectivity ∗ Zh.G. Nikoghosyan † Institute for Informatics and Automation ProblemsNational Academy of SciencesP. Sevak 1, Yerevan 0014, ArmeniaE-mail: [email protected] 30, 2018
Abstract
Let G be a graph, C a longest cycle in G and p , c the lengths of a longestpath and a longest cycle in G \ C , respectively. Almost all lower bounds forthe circumference base on a standard procedure: choose an initial cycle C in G and try to enlarge it via structures of G \ C and connections between C and G \ C closely related to p , c and connectivity κ . Actually, eachlower bound obtained in result of this procedure, somehow or is related to κ , p , c but in forms of various particular values of κ , p , c and the majorproblem is to involve these invariants into such bounds as parameters. Inthis paper we present a lower bound for the circumference involving δ , κ and c and increasing with δ , κ and c . We consider only finite undirected graphs without loops or multiple edges. Agood reference for any undefined terms is [1]. The set of vertices of a graph G isdenoted by V ( G ); the set of edges by E ( G ). For S a subset of V ( G ), we denoteby G \ S the maximum subgraph of G with vertex set V ( G ) \ S . For a subgraph H of G we use G \ H short for G \ V ( H ).Paths and cycles in a graph G are considered as subgraphs of G . If Q is apath or a cycle, then the length of Q , denoted by | Q | , is | E ( Q ) | . For Q a path,we denote | Q | = − V ( Q ) = ∅ . Throughout the paper, each vertexand edge can be interpreted as cycles of lengths 1 and 2, respectively. ∗ The original version is preprinted in Transactions of the Institute for Informatics and Au-tomation Problems of the NAS (Republic of Armenia) and Yerevan State University, Mathe-matical Problems of Computer Science
21 (2000) 129–155. † G.G. Nicoghossian up to 1997 G be a graph and C a longest cycle in G . A cycle C is a Hamilton cycleif G \ C = ∅ and is a dominating cycle if G \ C is edgeless. We use n to denotethe order, δ the minimum degree and κ the connectivity in G . The length | C | ,denoted by c , is called a circumference. The lengths of a longest path and alongest cycle in G \ C , will be denoted by p and c , respectively.Almost all lower bounds for the circumference base on a standard procedure:choose an initial cycle C in a graph G and try to enlarge it via structures of G \ C and connections between C and G \ C closely related to connectivity κ and p , c . Actually, each lower bound obtained in result of this procedure,somehow or is related to κ , p , c but in forms of various particular values of κ , p , c and the major problem in long cycles theory is to involve these invariantsinto such bounds as parameters.The starting result in this area, due to Dirac [2], bases on the minimumdegree δ . Theorem A [2]. In every graph, c ≥ δ + 1.The second result in the same paper [2] shows that under 2-connectednessthe bound δ + 1 in Theorem A can be replaced by min { n, δ } . Theorem B [2]. In every 2-connected graph, c ≥ min { n, δ } .When G \ C has a simple structure, namely is edgeless, Voss and Zuluaga [6]obtained the following. Theorem C [6]. Let G be a 3-connected graph. Then either c ≥ δ − G is a dominating cycle.The first lower bound involving connectivity κ as a parameter has been ap-peared in 1981, by the author [3]. Theorem D [3]. Let G be a 2-connected graph. Then c ≥ min { n, δ − κ } .Further, the first two bounds involving p and c has been appeared in 1998and 2000, respectively, again by the author [4],[5]. Theorem E [4]. Let G be a graph and C a longest cycle in G . Then | C | ≥ ( p + 2)( δ − p ). Theorem F [5]. Let G be a graph and C a longest cycle in G . Then | C | ≥ ( c + 1)( δ − c + 1).As a defect, the bound in Theorem D decreases as κ increases. The boundsin Theorems E and F have the same defect for p ≥ ( δ − / c ≥ δ/ δ , κ and c and increasing with δ , κ and c . Theorem 1.
Let G be a graph and C a longest cycle in G . Then | C | ≥ ( c +1) κc + κ +1 ( δ + 2) if c ≥ κ and | C | ≥ ( c +1) c c +1 ( δ + 2) if c ≤ κ −
1. .The result is sharp, as can be seen from the following family of graphs.Take κ + 1 disjoint copies of the complete graph K δ − κ +1 and join each vertexin their union to every vertex of a disjoint complete graph K κ . This graph( κ + 1) K δ − κ +1 + K κ is clearly not hamiltonian. Moreover, c = κ ( δ − κ + 2) and c = δ − κ + 1, implying that c = ( c +1) κc + κ +1 ( δ + 2) . In view of Theorem 1, we belive the following is also true in terms of p . Conjecture 1 . Let G be a graph and C a longest cycle in G . Then | C | ≥ ( p +2) κp + κ +2 ( δ + 2) if p ≥ κ −
1, and | C | ≥ ( p +2) p p +2 ( δ + 2) if p ≤ κ − HC − extensions plays a central role in the sequel. In section 4 we investigate themain properties of HC − extensions and in the last section we prove our mainresult. An ( x, y )-path is a path with endvertices x and y . Given an ( x, y )-path L of G ,we denote by −→ L the path L with an orientation from x to y . If u, v ∈ V ( L ),then u −→ L v denotes the consequtive vertices on L from u to v in the directionspecified by −→ L . The same vertices, in reverse order, are given by v ←− L u . For −→ L = x −→ L y and u ∈ V ( L ), let u + ( −→ L ) (or just u + ) denotes the successor of u ( u = y ) on −→ L and u − denotes its predecessor ( u = x ). If A ⊆ V ( L ), then A + = { v + | v ∈ A \{ y }} and A − = { v − | v ∈ A \{ x }} . If Q is a cycle in G and A ⊆ V ( Q ), then −→ Q , A + and A − are analogously defined. For v ∈ V ( Q ), v −→ Q v will be interpreted as a vertex v. For v ∈ V , put N ( v ) = { u ∈ V | uv ∈ E } and d ( v ) = | N ( v ) | . We begin introducing some special definitions and convenient notations. Forthe remainder of this section let a longest cycle C in a graph G and a longestcycle H = u . . . u h u in G \ C with h = c be fixed. Definition 3.1. HC − extension; T ( u i ); o u ; ˆ u .3et T ( u ) , . . . , T ( u h ) are vertex-disjoint ( u i , ˆ u i )-paths in G \ C ( i = 1 , ..., h ).The union T = S hi =1 T ( u i ) is called HC − extension if N (ˆ u i ) ⊆ V ( T ) S V ( C ) foreach i ∈ { , ..., h } . An HC-extension T is called maximal if it is chosen so as tomaximize |{ u ∈ V ( H ) | u = ˆ u }| . If u = ˆ u for some u ∈ V ( H ), then we use o u todenote u + ( −→ T ( u )). Definition 3.2. ( A, B ) − path.Let A, B ⊂ V and A T B = ∅ . Let E is a path in G with all its inner ver-tices in V \ ( A S B ). Then E is called an ( A, B )-path if E starts at any vertexin A and terminates at any vertex in B . For subgraphs H and H of G , an( H , H )-path is analogously defined. Definition 3.3. Θ( −→ P , V neut , V fin ) = ( P , ..., P π ); P i = y i −→ P i z i ( i = 0 , . . . , π ).Let V ′ ⊂ V . A path with endvertices in V \ V ′ and all internal vertices in V ′ is called a V ′ − path. Let −→ P = v v . . . v n be a path in G of length n ≥ V neut , V fin be vertex-disjoint subsets in V \ V ( −→ P ). We define Θ( −→ P , V neut , V fin )as a sequence of paths P , . . . , P π as follows: For i = 0, put P = −−→ y z and X = V ( v −→ P z ), where y = v and z = v . Now let P i − = y i − −→ P i − z i − and X i − are defined for some integer i ≥
1. In order to define P i and X i wedistinguish three cases.( i ) If every V neut − path, starting in X i − − z i − , terminates in X i − , then X i = ∅ and P π = P i − (so P i is undefined).( ii ) If there is a V neut − path P ′ = v ′ −→ P ′ v ′′ with v ′ ∈ X i − − z i − and v ′′ ∈ V fin ,then X i = ∅ and P π = P i = y i −→ P ′ z i , where y i = v ′ and z i = v ′′ .( iii ) There is a V neut − path P ′′ = w ′ −→ P ′′ w ′′ with w ′ ∈ X i − − z i − and w ′′ ∈ V ( z + i − −→ P v n ) but there is no V neut − path satisfying (ii).Choose P ′′ so as to maximize | v −→ P w ′′ | . Then putting P i = y i −→ P ′′ z i , X i = V ( v −→ P z i ) , where y i = w ′ and z i = w ′′ , we complete the definition of P i and X i . Since X o ⊂ X ⊂ · · · , there must be some integer j ( j ≥
1) with P j = P π , which, infact, completes the definition of Θ( −→ P , V neut , V fin ). Definition 3.4. Φ u ; ϕ u ; Ψ u ; ψ u .Let T be a maximal HC-extension. For each u ∈ V ( H ), putΦ u = N (ˆ u ) \ V ( T ) , ϕ u = | Φ u | , Ψ u = N (ˆ u ) \ V ( C ) , ψ u = | Ψ u | . Definition 3.5. U ; U ; U ; U ; U ∗ .For T a maximal HC − extension, put U = { u ∈ V ( H ) | u = ˆ u } , U = V ( H ) \ U , U = { u ∈ U | Φ u V ( T ( u )) } . u ∈ V ( H ) \ ( U S U ) and Θ( ←− T ( u ) , V neut , V fin ) = ( P , . . . , P π ), where V neut = V \ ( V ( T ) [ V ( C )) , V fin = V ( T ) \ V ( T ( u )) . A vertex u is called special if P π starts and terminates in V ( T ( u )). The set ofall nonspecial vertices in V ( H ) \ ( U S U ) is denoted by U and the set of allspecial vertices by U ∗ . Definition 3.6. B u ; B ∗ u ; b u ; b ∗ u .Let T be a maximal HC-extension. For each u ∈ V ( H ), put B u = { v ∈ U | v o u ∈ E } . Clearly B u = ∅ if u ∈ U . Furthermore, for each u ∈ U , put B ∗ u = { v ∈ V ( H ) | u o v ∈ E . Clearly B ∗ u ⊆ U . Let b u = | B u | and b ∗ u = | B ∗ u | . Definition 3.7. A u ( v ); ρ u ( v ); ρ u ( v ); Λ u ; Λ u ( v, w ).Let T be a maximal HC-extension. For each u, v ∈ V ( H ), put A u ( v ) = (Φ u [ B u ) \ V ( T ( v )) . Let ρ u ( v ) denote the vertex in A u ( v ) maximizing | v −→ T ( v ) ρ u ( v ) | . In particular, ρ u ( u ) = ˆ u − . Put ρ u ( v ) = ˆ u if ρ u ( v ) ∈ Φ u and ρ u ( v ) = o u if ρ u ( v ) ∈ B u . Clearly ρ u ( u ) = ˆ u . Put Λ u = { v ∈ V ( H ) | A u ( v ) = ∅ } . For each v, w ∈ Λ u ( v = w ),put Λ u ( v, w ) = vT ( v ) ρ u ( v ) ρ u ( v ) T ( u ) ρ u ( w ) ρ u ( w ) T ( w ) w. Definition 3.8. ϕ ′ u ; γ u ; β u ; µ ( T ) . For T a maximal HC − extension, put ϕ ′ u = (cid:26) ϕ u if u ∈ V ( H ) \ U ∗ , u ∈ U ∗ , γ u = ( ϕ ′ u + b u if u ∈ U ,ϕ ′ u − b ∗ u if u ∈ U ,β u = ( γ u + γ u + )2 ( u ∈ V ( H )) , µ ( T ) = 1 h X u ∈ V ( H ) β u . Definition 3.9. T − transformation; T tr ( E , ..., E n ); T tr ( v , ..., v n ) . Let T be a maximal HC − extension and let E , ..., E n are vertex-disjoint( H, C ) − paths with E i = v i −→ E i w i ( i = 1 , ..., n ). Assume that the union of E , ..., E n intersect T ( z ) for some z ∈ V ( H ) \{ v , ..., v n } . Clearly z ∈ U . walk-ing along T ( z ) from z to ∧ z we stop at the first vertex w ∈ S ni =1 V ( E i ). Assumew.l.o.g. that w ∈ V ( E ). Replacing the segment v E w of a path E by zT ( z ) w we get a new path E instead of E . If the union of E , E , ..., E n intersect T ( z ′ ) for some z ′ ∈ V ( H ) \{ z, v , ..., v n } , then continue this procedure. In afinite number of steps we obtain | { v ∈ V ( H ) : (cid:0) n [ i =1 V ( E ′ i ) (cid:1) \ V ( T ( v )) = ∅} | = n H, C )-paths E ′ , ..., E ′ n . Let E ′ i = v ′ i E ′ i w i ( i = 1 , ..., n ) . By writing T tr ( E , ..., E n ) = ( E ′ , ..., E ′ n ) , T tr ( v , ..., v n ) = (cid:16) v ′ , ..., v ′ n (cid:17) , we say that E ′ , ..., E ′ n is a T − transformation of E , ..., E n . By the definition, v ′ i ∈ { v i } [ U ( i = 1 , ..., n ) , T tr ( w , ..., w n ) = ( w , ..., w n ) . Definition 3.10. O ( x, y ); O x ( x, y ); O ( y, o x ); O ( x, o x ); O y ( x, y ); O ( x, o y ); O ( y, o y ) . Let T be a maximal HC − extension. For each pair of distinct vertices x, y ∈ V ( H ) , put V = [ v x,y } V ( T ( v )) [ { x, y } , V = V [ { o x } . Let O ( x, y ) (resp. O x ( x, y ) , O ( y, o x ) , O ( x, o x )) be the longest ( x, y ) − path(resp. ( x, y )-path, ( y, o x ) − path, ( x, o x ) − path) in h V i (resp. h V i , h V i , h V i ). Thepaths O y ( x, y ) , O ( x, o y ) , and O ( y, o y ) are analogously defined. Definition 3.11.
Ω ( x, y ) ; Ω ( x, y, E, F ) ; Ω ( v, w, x, y, E, F ) . Let T be a maximal HC − extension and let E, F be a pair of vertex disjoint T − transformed ( H, C ) − paths with E = xEv and F = yF w. If | T ( x ) | − = 1,then we denote Ω x ( x, y, E, F ) = O ( x, y ) . Otherwise,Ω x ( x, y, E, F ) = O x ( x, y ) if o x V ( E ) S V ( F ) ,O ( o x, y ) if o x ∈ V ( E ) ,O ( o x, x ) if o x ∈ V ( F ) . Define Ω y ( x, y, E, F ) by the same way and denote by Ω ( x, y, E, F ) thelongest path among O ( x, y ) , Ω x ( x, y, E, F ) and Ω y ( x, y, E, F ) . Let Ω ( x, y ) bethe shortest path Ω ( x, y, E, F ) for fixed x, y and all possible
E, F.
By definition3.9, vEµ
Ω ( x, y, E, F ) νF w is a simple path for appropriate µ, ν ∈ { x, y, o x, o y } denoted by Ω ( v, w, x, y, E, F ) . Definition 3.12. ( v, L ) ∈ ∆ . Let L be a path in G with L = v ...v t − ( t ≥
1) and let v ∈ V \ V ( L ) . Wewill write ( v, L ) ∈ ∆ if vv i − ∈ E ( i = 1 , ..., t ) . For w ∈ V ( L ), we will write( w, L ) ∈ ∆ if wu ∈ E for each u ∈ V ( L ) \{ w } . Remarks.
If no ambiguity can arise, any notation of the type R u i in definitions2.4 and 2.6-2.8, having index u i (say Φ u i ), we abbreviate R u i = R i . Preliminaries
Throughout this section, let G be a graph, C be a longest cycle G and H = u ...u h u a longest cycle in G \ C with a maximal HC − extension T. Lemma 1.
Let G be a graph. (a1) For
E, F a pair of vertex-disjoint (
H, C ) − paths in G with E = xEv and F = yF w , if T tr ( E, F ) = ( E ′ , F ′ ) and T tr ( x, y ) = ( x ′ , y ′ ), then | v −→ C w | − ≥ | Ω( v, w, x ′ , y ′ , E ′ , F ′ ) | − ≥ a ( x ′ ) + a ( y ′ ) + | Ω( x ′ , y ′ ) | − , where a ( z ) = 1 if z U ∗ and a ( z ) = ϕ z + 1 if z ∈ U ∗ for each z ∈ { x ′ , y ′ } . (a2) Let u ∈ V ( H ) and Θ( ←− T ( u ) , V neut , V fin ) = ( P , ..., P π ) , where P i = y i −→ P i z i ( i = 0 , ..., π ) and V neut = V \ ( V ( T ) [ V ( C )) , V fin = V ( T ) \ V ( T ( u )) . If u ∈ U , then there is an ( u, z π ) − path L of length at least ϕ u + 1 with V ( L ) ⊆ V ( T ( u )) S V ∗ , where V ∗ = S πi =0 V ( P i ) . If u ∈ U ∗ , then for each vertex z ∈ ( V (ˆ u ←− T ( u ) z π ) [ V ∗ ) \{ z π } there is an ( u, z ) − path L of length at least ϕ u + 1 with V ( L ) ⊆ V ( T ( u )) S V ∗ . Lemma 2.
For each u ∈ V ( H ),( b1 ) if u ∈ U and ˆ u = o u, then Φ u T B u = ∅ . ( b2 ) P u U b u = P u ∈ U b ∗ u , P u ∈ V ( H ) γ u = P u ∈ V ( H ) ϕ ′ u , | Φ u S B u | = P v ∈ V ( H ) | A u ( v ) | . Lemma 3.
Let G be a graph, C be a longest cycle in G , Q be a path in G \ C and let P i = v i −→ P i w i ( i = 0 , ..., q ) are vertex-disjoint paths in G \ C having only v , ..., v q in common with Q. Then c ≥ q X i =0 | Z i | + | q [ i =0 Z i | , where Z i = N ( w i ) T V ( C ) ( i = 0 , ..., q ) . Lemma 4.
For each u ∈ V ( H ) , ( d1 ) if | T ( u ) | − ≥ , then h ≥ γ u . ( d2 ) if | T ( u ) | − , then h ≥ ϕ ′ u ≥ γ u + 1 . ( d3 ) h ≥ γ u + 1 . Lemma 5.
Let Λ u ⊆ V ( x −→ H y ) for some u, x, y ∈ V ( H ) . ( e1 ) if | T ( u ) | − ≥ , then | x −→ H y | − ≥ γ u . ( e2 ) if | T ( u ) | − , then | x −→ H y | − ≥ γ u − . ( e3 ) if | T ( u ) | − | x −→ H y | − γ u − , then(ˆ u, x −→ H y ) ∈ ∆ , B u = Λ u \{ u } ⊆ U and γ u − ϕ u − . emma 6. For each u ∈ U S U , let x −→ H y and x −→ H y be vertex-disjointsegments in H with { x , x , y , y } ⊆ Λ u ⊆ V ( x −→ H y ) S V ( x −→ H y ) and let v ∈ { x , y } . ( f1 ) If B u S { u } ⊆ V ( x −→ H y ) and Λ u \ ( B u S { u } ) ⊆ V ( x −→ H y ) , then | x −→ H y | − | x −→ H y | − | A u ( u ) | + | A u ( v ) | ≥ γ u − . Otherwise, | x −→ H y | − | x −→ H y | − | A u ( u ) | + | A u ( v ) | ≥ γ u − | A u ( u ) | ≥ γ u − . ( f2 ) If | x −→ H y | − | x −→ H y | − | A u ( u ) | + | A u ( v ) | = γ u − , then(ˆ u, x i −→ H y i ) ∈ ∆ ( i = 1 , , B u = Λ u \{ u } ⊆ U , γ u − ϕ u − . Lemma 7.
Let x, y be a pair of distinct vertices in H . For each u ∈ V ( H ) , ( g1 ) if u ∈ U ∗ , then | O ( x, y ) | − ≥ γ u + 1,( g2 ) if | T ( u ) | − ≥ , then | O ( x, y ) | − ≥ γ u ,( g3 ) if | T ( u ) | − , then | O u ( x, y ) | − ≥ γ u − g4 ) Let | T ( u ) | − | O u ( x, y ) | − γ u − . If either Λ u ⊆ V ( x −→ H y )and ( o u, H ) ∆ or Λ u ⊆ V ( y −→ H x ) and ( o u, H ) ∆ (say Λ u ⊆ V ( x −→ H y ) and( o u, H ) ∆), then( g4 . ) ( o u, x −→ H y ) ∈ ∆,( g4 . ) B u = Λ u \{ u } ⊆ U and | O u ( x, y ) | − | x −→ H y − | = γ u − ϕ u − , ( g4 . ) if z ∈ V ( x −→ H y ) \ Λ u , then either z ∈ U ∗ or z ∈ U andΛ z ⊆ Λ u S { z } , γ z ≤ ϕ u = ( γ u + 1) / , ( g4 . ) if z ∈ V ( x −→ H y ) \ { x, y } , then Λ z ⊆ V ( x −→ H y ) . Otherwise,( g4 . ) ( o u, H ) ∈ ∆ , ( g4 . ) B u = Λ u \{ u } ⊆ U and | O u ( x, y ) | − h − γ u − ϕ u − , ( g4 . ) if z ∈ V ( H ) \ Λ u , then either z ∈ U ∗ or z ∈ U , Λ z ⊆ Λ u S { z } and γ z ≤ ϕ u = ( γ u + 1) / h/ , ( g5 ) if | T ( x ) | − , then min { (cid:12)(cid:12)(cid:12) O ( o x, x ) (cid:12)(cid:12)(cid:12) − , (cid:12)(cid:12)(cid:12) O ( o x, y ) (cid:12)(cid:12)(cid:12) − } ≥ γ x . ( g6 ) if u ∈ { x + , x − , y + , y − } , then | O ( x, y ) | − ≥ γ u , ( g7 ) If u ∈ { x + , x − , y + , y − } (say u = x + ) and | O ( x, y ) | − γ u , then | T ( u ) | − ≤ u, v −→ H y ) ∈ ∆ for some v ∈ Λ u with Λ u ⊆ V ( v −→ H y ) , ( g8 ) If | T ( x ) | − | O x ( x, y ) | − | O x ( x, w ) | − γ x − w ∈ V ( H ) \ { x, y } , then for each z ∈ { x + , x − } , min {| O x ( x, y ) | − , | O x ( x, w ) | − } ≥ γ z + 1. Lemma 8.
Let x, y be a pair of distinct vertices in H and let a = min {| O x ( x, y ) | , | O ( o x, y ) | , | O ( o x, x ) |} − ,b = min {| O y ( x, y ) | , | O ( o y, x ) | , | O ( o y, y ) |} − . | Ω( x, y ) | − ≥ max {| O ( x, y ) | − , a, b } . Lemma 9.
Let x, y be a pair of distinct vertices in H. Then( i1 ) if { u i , u i +1 } T { x, y } = ∅ ( i ∈ { , ..., h } ) , then | Ω( x, y ) | − ≥ ( γ i + γ i +1 ) / β i , ( i2 ) if | T ( x ) | − ≥ z ∈ { x + , x − } , then | Ω( x, y ) | − ≥ ( γ x + γ z ) / i3 ) if x ∈ U ∗ and z ∈ { x + , x − } , then | Ω( x, y ) | − ≥ ( γ x + γ z + 1) / i4 ) If | T ( x ) | − w ∈ V ( H ) \ { x, y } and z ∈ { x + , x − } , max {| Ω( x, y ) | − , | Ω ( x, w ) | − } ≥ ( γ x + γ z ) / , ( i5 ) if z ∈ { x + , x − } and w ∈ V ( H ) \{ z } , then max {| Ω( x, y ) | − , | Ω ( z, w ) | − } ≥ ( γ x + γ z ) / , ( i6 ) If x ∈ U and h = 4, then | Ω( x, y ) | − ≥ ( γ x + γ z ) / z ∈ { x + , x − } ,( i7 ) if x, y ∈ U , then | Ω ( x, y ) | − ≥ max i β i , ( i8 ) if | x −→ H y | − , then | Ω ( x, y ) | − ≥ max i β i , ( i9 ) if | x −→ H y | − h = 4 , then | Ω( x, y ) | − ≥ ( γ x + γ x + ) / . Proof of lemma 1. (a1)
Following definition 3.11, we distinguish three cases.
Case 1. x, y U ∗ . Clearly, | v −→ C w | − ≥| Ω( v, w, x, y, E, F ) | − ≥ | Ω( x, y, E, F ) | − ≥| Ω( x, y ) | +1 . Case 2. x, y ∈ U ∗ . If | T ( x ) |− o x V (Ω ( x, y )) , since otherwise the segment of Ω ( x, y )between o x and y , contradict the fact that x ∈ U ∗ . Therefore, Ω ( x, y, E, F ) = O ( x, y ) . On the other hand, Ω x ( x, y, E, F ) = O ( x, y ) if | T ( x ) | − ≥ . Also,by the symmetric arguments, Ω y ( x, y, E, F ) = O ( x, y ) . Thus Ω ( x, y, E, F ) =Ω ( x, y ) and | v −→ C w | − ≥ | Ω ( v, w, x, y, E, F ) | − ≥ ( | E | −
1) + ( | F | −
1) + | Ω ( x, y ) | − ≥ ϕ x + ϕ y + | Ω ( x, y ) | + 1 . Case 3.
Either x U ∗ , y ∈ U ∗ or x ∈ U ∗ , y U ∗ . Apply the arguments in case 1 and case 2. ∆( a2 ) Suppose first that u ∈ U . By definition 3.3, z ∈ V ( T ( u )) and z π ∈ V fin . Let z π ∈ V ( T ( w )) for some w ∈ V ( H ) \{} u. Choose z ∈ V ( u −→ T ( u ) y − )such that z ˆ u ∈ E and | z −→ T ( u ) y | is minimum. Then we get the desiredresult putting together the following paths P , ..., P π , ˆ uz , ˆ u ←− T ( u ) y , z ←− T ( u ) y , z π − ←− T ( u ) u, z i ←− T ( u ) y i +2 , i = 2 , ..., π −
2. A similar proof holds for u ∈ U ∗ . ∆ Proof of lemma 2. (b1) . Case 1. u ∈ U . Suppose, to the contrary, that Φ u T B u = ∅ and let z ∈ Φ u T B u . Then, bydefinitions 3.4 and 3.1, the collection { T ( u ) , ..., T ( u h ) , u o u, z ˆ u }\ { T ( u ) , T ( z ) } generates another HC − extension, contradicting the maximality of T . Case 2. u ∈ U S U ∗ . By definition 3.5, Φ u ⊆ V ( T ( u )) and the result follows. ∆( b2 ) Immediately from definitions 3.6-3.8. ∆ Proof of lemma 3.
Assume first that v i = w i ( i = 0 , ..., q ). The result isimmediate if S qi =0 Z i = ∅ . Let S qi =0 Z i = ∅ and let ξ , ..., ξ m ( m ≥
1) be theelements of S qi =0 Z i occuring on −→ C in a consequtive order. Set F i = N ( ξ i ) \ { w , ..., w q } ( i = 1 , ..., m ) . Suppose that m = 1. If | F | = 1, then q = 0 and Z = Z q = { ξ } implying that c ≥ q X i =0 | Z i | + | q [ i =0 Z i | . If | F | ≥
2, then choosing u, v ∈ F ( u = v ) such that | u −→ Qv | is maximum,we get c ≥| ξ u −→ Q vξ |≥ q X i =0 | Z i | + 1 = q X i =0 | Z i | + | q [ i =0 Z i | . Thus, we may assume m ≥
2. It means, in particular, that c ≥
3. For i = 1 , ..., m , put f ( ξ i ) = | ξ i −→ C ξ i +1 | − m ). It is easy to see that c = P mi =1 f ( ξ i ) , P mi =1 | F i | = P qi =0 | Z i | , m = | S qi =0 Z i | . (1)For every i ∈ { , ..., m } , choose x i , y i ∈ F i S F i +1 such that | x i −→ Q y i | is maxi-mum (indices mod m ). Claim 3.1 f ( ξ i ) ≥ ( | F i | + | F i +1 | + 2) / i = 1 , ..., m ) . Proof of Claim 3.1.
We distingwish two cases.
Case 1.
Either x i ∈ F i , y i ∈ F i +1 or x i ∈ F i +1 , y i ∈ F i . If x i ∈ F i , y i ∈ F i +1 , then f ( ξ i ) ≥ | ξ i x i −→ Q y i ξ i +1 | − f ( ξ i ) ≥ max ( | F i | , | F i +1 | ) + 1 ≥ ( | F i | + | F i +1 | + 2) / . Otherwise, the result holds from f ( ξ i ) ≥| ξ i y i ←− Q x i ξ i +1 | − Case 2.
Either x i , y i ∈ F i or x i , y i ∈ F i +1 . x i , y i ∈ F i . We can assume also x i , y i F i +1 , since otherwisewe could argue as in case 1. Choose x ′ i , y ′ i ∈ F i +1 such that | x ′ i −→ Q y ′ i | ismaximum. If | x i −→ Q x ′ i | − ≥ ( | F i | − | F i +1 | ) /
2, then f ( ξ i ) ≥| ξ i x i −→ Q y ′ i ξ i +1 | − ≥ ( | F i | − | F i +1 | ) / | F i +1 | +1 = ( | F i | + | F i +1 | + 2) / . Otherwise, f ( ξ i ) ≥| ξ i y i ←− Q x ′ i ξ i +1 | − | x ′ i −→ Q y i | +1 = | x i −→ Q y i | − | x i −→ Q x ′ i | +2 ≥≥ | F i | − ( | F i | − | F i +1 | + 1) / > ( | F i | + | F i +1 | + 2) / . By symmetry, the case x i , y i ∈ F i +1 requires the same arguments. ∆By claim 3.1, m X i =1 f ( ξ i ) ≥ m X i =1 ( | F i | + | F i +1 | + 2) / m X i =1 | F i | + m, which by (1) gives the desired result. Finally, if v i = w i for some i ∈ { , ..., q } ,then we can argue exactly as in case v i = w i ( i = 0 , ..., q ). ∆ Proof of lemma 4. (d1) . Case1. u ∈ U . Let ξ , ..., ξ f be the elements of Λ u occuring on H in a consequtive orderwith u = ξ . For each integer i (1 ≤ i ≤ f ), let M i = ξ i −→ H ξ i +1 , ω i = | A u ( ξ i ) | + | A u ( ξ i +1 ) | (indices mod f ) . Since H is extreme, | M i | ≥ | Λ u ( ξ i , ξ i +1 ) | ( i = 1 , ..., f ) . (2)Let ξ r −→ H ξ s be the longest segment on H such that ξ ∈ V ( ξ r −→ H ξ s ) , { ξ r , ξ r +1 , ..., ξ s } ⊆ B u S { u } . Put Ω + = { M i ∈ { M , ..., M f − } | ρ u ( ξ i ) = ρ u ( ξ i +1 ) } , Ω − = { M i ∈ { M , M f } | ρ u ( ξ i ) = ρ u ( ξ i +1 ) } , Ω = { M , ..., M f } \ (Ω + S Ω − ) . Observe that | Ω − | ≤ | M i | − ≥ | Λ u ( ξ i , ξ i +1 ) | − i ∈ { , ..., f } . Then clearly if M i ∈ Ω + , then | M i | − ≥ ω i + | A u ( u ) | − , (3)if M i ∈ Ω − , then | M i | − ≥ ω i − | A u ( u ) | + 1 , (4) M i ∈ Ω = ⇒ | M i | − ≥ ω i . (5) Claim 4.1. If | Ω − | = 0 then | M i | − ≥ ω i ( i = 1 , ..., f ).11 roof of Claim 4.1. Immediate from (3), (4) and (5). ∆
Claim 4.2. ( k1 ) If | Ω − | = 1, say Ω − = { M } , then M s ∈ Ω + . ( k2 ) If Ω − = { M } and Ω + = { M s } , then B u S { u } ⊆ V ( ξ r −→ H ξ s ) , Λ u \ ( B u S { u } ) ⊆ V ( ξ s +1 −→ H ξ r − ) . Proof of Claim 4.2. ( k1 ) Let Ω − = { M } . By the definition, { ξ , ..., ξ s } ⊆ B u and ξ s +1 ∈ Λ u \ ( B u S { u } ) , implying that M s ∈ Ω + . ∆( k2 ) If follows that { ξ s +1 , ..., ξ f } ⊆ Λ u \ ( B u S { u } ) . On the other hand(by the definition), { ξ , ..., ξ s } ⊆ B u S { u } and the proof is complete. ∆ Claim 4.3. ( l1 ) If | Ω − | = 2 , i.e.Ω − = { M , M f } , then M s , M r − ∈ Ω + . ( l2 ) If Ω − = { M , M f } and Ω + = { M s , M r − } , then ξ , ξ r , ξ s are pairwisedifferent and B u S { u } ⊆ V ( ξ r −→ H ξ s ) , Λ u \ ( B u S { u } ) ⊆ V ( ξ s +1 −→ H ξ r − ) . Pr oof of claim 4.3. ( l1 ) By the definition, { ξ , ξ f , ξ s , ξ r } ⊆ B u and ξ s +1 , ξ r − ∈ Λ u \ ( B u S { u } ), which implies M s , M r − ∈ Ω + . ∆( l2 ) It follows that { M s +1 , ..., M r − } T Ω + = ∅ and hence { ξ s +1 , ..., ξ r − } ⊆ Λ u \ (cid:16) B u [ { u } (cid:17) . On the other hand (by the definition) { ξ r , ..., ξ s } ⊆ B u S { u } , which completesthe proof of claim 4.3. ∆The following three statements can be proved easely basing on (3), (4), (5)and claims 4.1, 4.2 and 4.3. Claim 4.4. P fi =1 ( | M i | − ≥ P fi =1 ω i . Claim 4.5. If t ∈ { , ..., f } , then P i = t ( | M i | − ≥ P i = t ω i −| A u ( u ) | +1 . Claim 4.6. If g, t ∈ { , ..., f } ( g = t ) , then P i g,t } ( | M i | − ≥ P i g,t } ω i − | A u ( u ) | + 2 . Using ( b
2) and claim 4.4, we get h = P fi =1 ( | M i | − ≥ P fi =1 ω i = P fi =1 ( | A u ( ξ i ) | + | A u ( ξ i +1 ) | )= 2 P fi =1 | A u ( ξ i ) | = 2 | Φ u S B u | . By ( b | Φ u S B u | = ϕ u + b u = γ u , implying that h ≥ γ u . ase 2. u ∈ U . Let Θ( ←− T ( u ) , V neut , V fin ) = ( P , ..., P π ), where P i = y i −→ P i z i ( i = 0 , ..., π ) . By ( a u, z π ) − path L of length at least ϕ u + 1 with V ( L ) ⊆ V ( T ( u )) S V ∗ . Let z π ∈ V ( T ( w )) for some w ∈ V ( H ). By denoting B u S { u, w } = { ξ , ..., ξ f } , we can argue exactly as in case 1. Case 3. u ∈ U ∗ . Clearly h ≥ b u + 1) = 2( ϕ ′ u + b u + 1) > γ u . ∆( d2 ) Since | T ( u ) | − u ∈ U S U ∗ . If u ∈ U ∗ , then b u = 0 and h ≥ ϕ ′ u + b u + 1) = 2 ( γ u + 1) ≥ γ u + 1. Let u ∈ U . Define ξ i , ω i , M i , ξ r −→ H ξ s , Ω + , Ω − , Ω (6)as in proof of ( d . It is easy to see that Ω + = Ω − = ∅ . By claim 4.1, P fi =1 ( | M i | − ≥ P fi =1 ω i and as in proof of ( d , h ≥ | Φ u S B u | = 2 ϕ u . Noting that ϕ u ≥ b u + |{ u }| = b u + 1, we obtain h ≥ ϕ u + b u + 1 = γ u + 1 . ∆( d3 ) It is easely checked that h ≥ γ u + 1 if u ∈ U . If u ∈ U , then by ( d d h ≥ min (2 γ u , γ u + 1) ≥ γ u + 1 . ∆ Proof of lemma 5.
Assume w.l.o.g. that x, y ∈ Λ u .( e1 ) Case 1 u ∈ U . Following (6) we let, in addition, y −→ H x = M t for some t (1 ≤ t ≤ f ). Byclaim 4.5 we can distingush the following two cases: Case 1.1. | x −→ H y | − ≥ P i = t ω i . By ( b , | Φ u S B u | = | Φ u | + | B u | = γ u , and using ( b , | x −→ H y | − ≥ P i = t ω i = P fi =1 ω i − ω t = 2 P fi =1 | A u ( ξ i ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | = P fi =1 | A u ( ξ i ) | + P i t,t +1 } | A u ( ξ i ) |≥ P fi =1 | A u ( ξ i ) | = | Φ u S B u | = γ u . Case 1.2. P i = t ω i − | A u ( u ) | + 1 ≤ (cid:12)(cid:12)(cid:12) x −→ H y − (cid:12)(cid:12)(cid:12) < P i = t ω i . If Ω − = ∅ , then by claim 4.1., | x −→ H y | − ≥ P i = t ω i , a contradition. LetΩ − = ∅ . Case 1.2.1. | Ω − | = 1 . Assume w.l.o.g Ω − = { M } . By claim 4.2, M s ∈ Ω + . If | Ω + | ≥
2, then by(3), (4) and (5), | x −→ H y | − ≥ P i = t ω i , a contradiction. Thus we can assumeΩ + = { M s } . If M t = M s , then again | x −→ H y | − ≥ P i = t ω i , a contradiction.Finally, if M t = M s , then A u ( ξ t ), A u ( ξ t +1 ) and A u ( u ) are pairwise differentand hence 13 x −→ H y | − ≥ P fi =1 | A u ( ξ i ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | − | A u ( u ) | + 1 = P fi =1 | A u ( ξ i ) | + P i ,t,t +1 } | A u ( ξ i ) | + 1 ≥ P fi =1 | A u ( ξ i ) | + ( f −
3) + 1 ≥ P fi =1 | A u ( ξ i ) | = | Φ u S B u | = γ u . Case 1.2.2. | Ω − | = 2 . By claim 4.3, M s , M r − ∈ Ω + . If | Ω + | ≥
3, then by (3), (4) and (5), | x −→ H y | − ≥ P i = t ω i , a contradiction. Let Ω + = { M s , M r − } . If M t Ω + ,then again | x −→ H y | − ≥ P i = t ω i , a contradiction. Finally, if M t ∈ Ω + , say M t = M s , then A u ( ξ t ), A u ( ξ t +1 ) , A u ( u ) are pairwise different and we canargue exactly as in case 1.2.1. Case 2. u ∈ U S U ∗ . Apply the arguments used in the proof of ( d
1) (case 2 and case 3) . ( e2 ) Clearly u ∈ U . Following (6) we see that Ω + = Ω − = ∅ . By claim4.1, | M i | − ≥ ω i ( i = 1 , ..., f ). Recalling that f ≥ b u + 1, | x −→ H y | − P i = t ( | M i | − ≥ P i = t ω i = P fi =1 | A u ( ξ i ) | + P i t,t +1 } | A u ( ξ i ) |≥ | Φ u S B u | + f − ϕ u + f − ϕ u + b u − γ u − . ( e3 ) It was shown in ( e
2) that | x −→ H y | − ≥ ϕ u + f − ≥ γ u −
1. Since | x −→ H y | − γ u −
1, we have | x −→ H y | − ϕ u + f − γ u −
1. This implies | B u | = b u = f − B u = Λ u \{ u } ⊆ U . But then ϕ u = b u + 1 and | x −→ H y | − γ u − ϕ u − u, x −→ H y ) ∈ ∆. ∆ Proof of lemma 6. ( f1 ) Case 1. u ∈ U . By symmetry, we can assume v = x . Following (6) we let, in addition, M g = y −→ H x and M t = y −→ H x for some integers g, t ∈ { , ..., f } . This meansthat y = ξ g , x = ξ g +1 , y = ξ t , x = ξ t +1 , v = x = ξ g +1 , A u ( ξ g +1 ) = A u ( v ) . Putting β = | x −→ H y | − | x −→ H y | − Case 1.1. β ≥ P i g,t } ω i + | A u ( u ) | − . Clearly β ≥ P fi =1 | A u ( ξ i ) | − | A u ( ξ g ) | − | A u ( ξ g +1 ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | + | A u ( u ) | − . (7)Observe that | A u ( u ) | ≥ f ≥ b u + 1. If x = y , then A u ( ξ g ), A u ( ξ t ) and A u ( ξ t +1 ) are pairwise different and by (7), β + | A u ( v ) | = β + | A u ( ξ g +1 ) | ≥ P fi =1 | A u ( ξ i ) | + P i g,t,t +1 } | A u ( ξ i ) |≥ P fi =1 | A u ( ξ i ) | + f − ≥ | Φ u S B u | + f − ≥ | Φ u | + f − ≥ ϕ u + b u − ≥ γ u − − | A u ( u ) | . x = y ) , A u ( ξ t +1 ) = A u ( u ) , and by (7), β + | A u ( v ) | = β + | A u ( ξ g +1 ) | ≥ P fi =1 | A u ( ξ i ) | − | A u ( ξ g ) | − | A u ( ξ t ) | − ≥ P fi =1 | A u ( ξ i ) | + P i g,t } | A u ( ξ i ) | − ≥ | Φ u S B u | + f − ≥ ϕ u + b u − ≥ γ u − − | A u ( u ) | . Case 1.2. P i g,t } ω i ≤ β < P i g,t } ω i + | A u ( u ) | − . Clearly β + | A u ( v ) | = β + | A u ( ξ g +1 ) |≥ f X i =1 | A u ( ξ i ) | − | A u ( ξ g ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | . If x = y , then we obtain the desired result as in case 1.1. Let x = y ,i.e. M g = M , M t = M f and ξ t +1 = ξ g = ξ = u . If Ω + = ∅ , then β ≥ P i ,t } ω i + | A u ( u ) | −
1, a contradiction. Let Ω + = ∅ . This implies M s = M and M r − = M f , and we deduce that B u [ { u } = { u } = V ( x −→ H y ) , Λ u \{ u } ⊆ V ( x −→ H y ) . Recalling that f ≥ b u + 1, we get β + | A u ( u ) | + | A u ( v ) | = β + | A u ( u ) | + | A u ( ξ g +1 ) |≥ P ti =1 | A u ( ξ i ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) |≥ | Φ u S B u | + f − ≥ ϕ u + b u − ≥ γ u − . Case 1.3. P i g,t } ω i − | A u ( u ) | +1 ≤ β < P i g,t } ω i . Case 1.3.1. ξ
6∈ { x , y } . It follows that A u ( u ) , A u ( ξ g ) , A u ( ξ t ) and A u ( ξ t +1 ) are pairwise different.Since f ≥ b u + 1 , we have β + | A u ( v ) | = β + | A u ( ξ g +1 ) |≥ P fi =1 | A u ( ξ i ) | − | A u ( u ) | − | A u ( ξ g ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | + 1 ≥ P fi =1 | A u ( ξ i ) | + f − ≥ ϕ u + b u − ≥ γ u − − | A u ( u ) | . Case 1.3.2. ξ ∈ { x , y } . Assume w.l.o.g. that ξ = x , i.e. M t = M f . If M Ω − , then β ≥ P i g,t } ω i , a contradiction. Let M ∈ Ω − . This implies ξ ∈ B u and M s ∈ Ω + .If M s = M g , then β ≥ P i g,t } ω i , a contradiction. So, assume M s = M g .Analogously, M r − = M f . If M j ∈ Ω + for some j ∈ { , ..., f } \ { g } , then again β ≥ P i g,t } ω i , a contradiction. Let i ∈ { , ..., f } \ { g } = ⇒ M i ∈ Ω [ Ω − . It follows that B u S { u } ⊆ V ( x −→ H y ) and Λ u \ ( B u S { u } ) ⊆ V ( x −→ H y ).Furthermore, noting that A u ( ξ g ) , A u ( ξ t ) , A u ( ξ t +1 ) are pairwise different and15 ≥ b u + 1, we get β + | A u ( u ) | + | A u ( v ) | = β + | A u ( u ) | + | A u ( ξ g +1 ) |≥ P fi =1 | A u ( ξ i ) | − | A u ( ξ g ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | + 1 ≥ P fi =1 | A u ( ξ i ) | + f − ≥ ϕ u + b u − ≥ γ u − . Case 1.4. P i g,t } ω i − | A u ( u ) | + 2 ≤ β < P i g,t } ω i − | A u ( u ) | + 1 . If | Ω − | ≤
1, then clearly β ≥ P i g,t } ω i − | A u ( u ) | + 1, a contradiction.Let | Ω − | = 2. This implies M , M f ∈ Ω − and M s , M r − ∈ Ω + . If | Ω + | ≥ β ≥ P i g,t } ω i − | A u ( u ) | + 1, a contradiction. Let | Ω + | = 2, i.e.Ω + = { M s , M r − } . By claim 4.3, B u [ { u } ⊆ V ( x −→ H y ) , Λ u \ ( B u [ { u } ) ⊆ V ( x −→ H y ) . Recalling that f ≥ b u + 1, we get β + | A u ( u ) | + | A u ( v ) | = β + | A u ( u ) | + | A u ( ξ g +1 ) |≥ P fi =1 | A u ( ξ i ) | − | A u ( u ) | − | A u ( ξ g ) | − | A u ( ξ t ) | − | A u ( ξ t +1 ) | + 2 ≥ P fi =1 | A u ( ξ i ) | + P i ,g,t,t +1 } | A u ( ξ i ) | + 2 ≥ | Φ u S B u | + ( f −
4) + 2 ≥ γ u − . Case 2. u ∈ U . Apply the arguments used in the proof of ( d
1) (see case 2 and case 3).( f2 ) Case 1. u ∈ U . As shown in the proof of ( f ,β + | A u ( u ) | + | A u ( v ) | ≥ (cid:12)(cid:12)(cid:12) Φ u [ B u (cid:12)(cid:12)(cid:12) + f − ≥ | Φ u | + f − ≥ ϕ u + b u − . Since β + | A u ( u ) | + | A u ( v ) | = ϕ u + b u − γ u − , we have equations β + | A u ( u ) | + | A u ( v ) | = (cid:12)(cid:12)(cid:12) Φ u [ B u (cid:12)(cid:12)(cid:12) + f − | Φ u | + f − ϕ u + b u − f = b u + 1. If | T ( u ) | − ≥
2, then Λ u \ ( B u S { u } ) = ∅ and hence f ≥ | B u | + |{ u }| + 1 = b u + 2, a contradiction. Otherwise ( | T ( u ) | − u = B u S { u } , | A u ( u ) | = | A u ( v ) | = 1 ,ϕ u = b u + 1 , β = ϕ u + b u − ϕ u − u, x i −→ H y i ) ∈ ∆ ( i = 1 ,
2) and γ u − ϕ u − Case 2. u ∈ U . Apply the arguments used in the proof of ( d Proof of lemma 7. ( g1 ) Clearly h ≥ b u + 1) = 2( ϕ ′ u + b u + 1) =2 ( γ u + 1) and therefore, | O ( x, y ) | − ≥ h/ ≥ γ u + 1. ∆16 g2 ) By (d1), h ≥ γ u which implies | O ( x, y ) | − ≥ h/ ≥ γ u . ∆( g3 ) , ( g5 ) If | Λ u | = 1, then γ u = 1 and there is nothing to prove. Let | Λ u | ≥
2, i.e. u ∈ U . Case 1. u
6∈ { x, y } . Assume w.l.o.g. that u ∈ V ( x + −→ H y − ) . We can assume also that Λ u V ( x −→ H y ), since otherwise the result holds by ( e x −→ H y be the longestsegment in x −→ H y − with x , y ∈ Λ u and x −→ H y be the longest segmnet in y −→ H x − with x , y ∈ Λ u . Putting β = | x −→ H y | − | x −→ H y | − β ≥ γ u − − | A u ( u ) | − | A u ( y ) | = γ u − − | A u ( y ) | and therefore | O u ( x, y ) | − ≥| x −→ H y Λ u ( y , y ) y ←− H y | − ≥ β + | A u ( y ) | + | A u ( y ) | ≥ γ u − | A u ( y ) | ≥ γ u − . Case 2. u ∈ { x, y } . Assume w.l.o.g. that u = x . Let x −→ H y be the longest segment in x + −→ H y with x , y ∈ Λ x and x −→ H y be the longest segmnet in y + −→ H x with y ∈ Λ x .Putting β = | x −→ H y | + | x −→ H x | − β ≥ γ x − − | A x ( x ) | − | A x ( x ) | = γ x − − | A x ( x ) | , and therefore | O x ( x, y ) | − ≥| x ←− H x Λ x ( x , x ) x −→ H y | − ≥ β + | A x ( x ) | + | A x ( x ) | ≥ γ x + | A x ( x ) | − ≥ γ x − . Also, by ( e
1) and ( e | x −→ H x | − ≥ γ x − , | x −→ H y | − ≥ γ x − | O ( o x, x ) | − ≥| o x ρ x ( x ) T ( x ) x −→ H x | − ≥| x −→ H x |≥ γ x , | O ( o x, y ) | − ≥| o x ρ x ( x ) T ( x ) x −→ H y | − ≥| x −→ H y |≥ γ x . ∆( g4 ) We can suppose u
6∈ { x, y } , since otherwise the arguments are thesame. Assume w.l.o.g. that u ∈ V ( x + −→ H y − ). Clearly | O u ( x, y ) | = | O ( x, y ) | . Inorder to prove ( g . − ( g . u ⊆ V ( x −→ H y )and ( u, H ) ∆.( g4 . ) , ( g4 . ) By ( e γ u − | O ( x, y ) | − ≥| x −→ H y | − ≥ γ u − | O ( x, y ) | − | x −→ H y | − γ u −
1. Using ( e o u, x −→ H y ) ∈ ∆, γ u − ϕ u −
1) and B u = Λ u \{ u } ⊆ U .( g4 . ) Case 1. z ∈ U . Case 1.1. Λ z V ( x −→ H y ) . w ∈ Λ z T V ( y + −→ H x − ) . If z = x + , then | O ( x, y ) | − ≥| y ←− H z Λ z ( z, w ) w −→ H x | − ≥| x −→ H y | , a contradiction. Otherwise we reach a contradiction by the following way | O ( x, y ) | − ≥| y ←− H z + 0 u x ++ −→ H z Λ z ( z, w ) w −→ H x | − ≥| x −→ H y | . Case 1.2. Λ z ⊆ V ( x −→ H y ) . Choose w ∈ Λ z \{ z } . Assume w.l.o.g. that w ∈ V ( x −→ H z − ). Since z ∈ U , wehave | Λ z ( z, w ) | − ≥ | O ( x, y ) | − ≥| y ←− H z Λ z ( z, w ) w −→ H z − o u w ′ ←− H x | − ≥| x −→ H y | for some w ′ ∈ { w − , w −− } , a contradiction. Case 2. z ∈ U S U . If z ∈ U , then apply the arguments used in the proof of ( d
1) (see case 2 and3). Let z ∈ U . If there exists a vertex w ∈ (Λ z \{ z } ) \ Λ u , then we can reach acontradiction as in case 1. Otherwise, Λ z ⊆ Λ u S { z } and γ z ≤ ϕ u = ( γ u + 1) / . ( g4 . ) Suppose, to the contrary, that Λ z V ( x −→ H y ). If | y −→ H x | − o u, H ) ∈ ∆, a contradiction. Let | y −→ H x | − ≥
3. Choose w ∈ Λ z T V ( y + −→ H x − ). Assume w.l.o.g. that | w −→ H x | − ≥
2. If z Λ u , then by( g . | O ( x, y ) | − ≥| y ←− H z ++ o u x ++ −→ H z Λ z ( z, w ) w −→ H x | − ≥| x −→ H y | , a contradiction. So, ( g . − ( g .
4) are proved. A similar proof holds for ( g . − ( g .
7) when Λ u V ( x −→ H y ) and Λ u V ( y −→ H x ). So, the proof of ( g
4) is com-pleted. ∆( g6 ) , ( g7 ) Let u = x + . Choose v ∈ Λ u so as to maximize | v −→ H y | . Clearly, v ∈ V ( y −→ H u ). Case 1. v = u. Case 1.1 | T ( u ) | − ≥ . By ( e
1) and ( e | O ( x, y ) | − ≥| u −→ H y |≥ γ u . If | O ( x, y ) | − γ u , thenby ( e , | T ( u ) | − ≤ | u −→ H y | = γ u − e
3) holds (ˆ u, u −→ H y ) ∈ ∆ . Case 1.2 | T ( u ) | − . Clearly, | O ( x, y ) | − ≥| x −→ H y | − ≥ γ u . If | O ( x, y ) | − γ u , then | x −→ H y | − γ u implying that uw ∈ E for each w ∈ V ( x −→ H y ) \{ u } , i.e.( u, x −→ H y ) ∈ ∆. Case 2. v = u. Case 2.1 | T ( u ) | − ≥ . By ( e
1) and ( e | v −→ H y | − ≥ γ u − | O ( x, y ) |− ≥| y ←− H u Λ u ( u, v ) v −→ H x | − ≥| v −→ H y | − | T ( u ) |− ≥ γ u + | T ( u ) |− . | O ( x, y ) | − γ u , then | T ( u ) | − | v −→ H y | − γ u − e u, v −→ H y ) ∈ ∆. Case 2.2 | T ( u ) | − . Clearly, | O ( x, y ) | − ≥| y ←− H u Λ u ( u, v ) v −→ H x | − ≥| v −→ H y | − ≥ γ u . If | O ( x, y ) | − γ u , then | v −→ H y | − γ u implying that uw ∈ E for each w ∈ V ( v −→ H y ) \{ u } , i.e. ( u, v −→ H y ) ∈ ∆. ∆( g8 ) By ( g o x, H ) ∈ ∆. Since { w, y } ⊆ Λ x , we have h ≥
6. If | O x ( x, y ) | − ≤ γ z , then by ( g .
6) and ( g . ,h − | O x ( x, y ) | − ≤ γ z ≤ h/ h ≤
4, a contradiction. So, | O x ( x, y ) | − ≥ γ z + 1. By symmetry, | O x ( x, w ) | − ≥ γ z + 1 and the result follows. ∆ Proof of lemma 8.
Immediate from definition 3.11. ∆
Proof of lemma 9.
By ( d , h ≥ γ i + 1 and h ≥ γ i +1 + 1 for each i ∈{ , ..., h } . In other words, h − ≥ ( γ i + γ i +1 ) / i = 1 , ..., h ) . (8)( i1 ) By lemma 8, it sufficies to prove | O ( x, y ) | − ≥ β i . Assume w.l.o.g.that i = 1 and u , u ∈ V ( x + −→ H y − ). Case 1. u , u ∈ U . Putting Γ i = Φ i T V ( H ) ( i = 1 ,
2) we see that | Γ i | = ϕ i − b ∗ i = γ i ( i = 1 , Case 1.1. Γ S Γ ⊆ V ( x −→ H y ) . Clearly | O ( x, y ) | − ≥| x −→ H y | − ≥ max ( | Γ | , | Γ | ) ≥ ( γ + γ ) / . Case 1.2. Γ S Γ V ( x −→ H y ) . Assume w.l.o.g. that Γ T ( V ( y + −→ H x − )) = ∅ . Let z −→ H z be the longestsegment in y + −→ H x − with z , z ∈ Γ . Case 1.2.1. Γ T V ( y + −→ H x − ) = ∅ . Choose w ∈ V ( x + −→ H u ) such that u w ∈ E and | x −→ H w | is minimum. Then | O ( x, y ) | − ≥| x −→ H y | − ≥| w −→ H y | + | x −→ H w | − ≥ γ + | x −→ H w | − , | O ( x, y ) | − ≥| x ←− H z u ←− H wu −→ H y | − ≥ γ − | x −→ H w | +2 . Combining these two inequalities yields | O ( x, y ) | − ≥ ( γ + γ ) / . Case 1.2.2. Γ T V ( y + −→ H x − ) = ∅ . Let w −→ H w be the longest segment in y + −→ H x − with w , w ∈ Γ . Case 1.2.2.1. z , w ∈ V ( w −→ H z ) . It follows that | O ( x, y ) | − ≥| x −→ H u z ←− H w u −→ H y | − ≥ max ( γ , γ ) ≥ ( γ + γ ) / . Case 1.2.2.2. z , w ∈ V ( z −→ H w ) . Clearly | O ( x, y ) | − ≥| x −→ H u z −→ H w u −→ H y | − ≥ max ( γ , γ ) ≥ ( γ + γ ) / . Case 1.2.2.3.
Either w , w ∈ V ( z −→ H z ) or z , z ∈ V ( w −→ H w ).19ssume w.l.o.g. that w , w ∈ V ( z −→ H z ). If w = z (resp. w = z ),then we can argue as in case 1.2.2.1. (resp. 1.2.2.2.). Otherwise ( w = z and w = z ), | O ( x, y ) | − ≥| x −→ H u z −→ H w u −→ H y | − ≥ γ + | z −→ H w | − , | O ( x, y ) | − ≥| x −→ H u z ←− H w u −→ H y | − ≥ γ − | z −→ H w | +1 . Combining these two inequalities yields | O ( x, y ) | − ≥ ( γ + γ ) / . Case 2. u , u ∈ U . By ( g
2) and ( g | O ( x, y ) | − ≥ γ i − i = 1 , u ∈ U ∗ or u ∈ U ∗ , then by ( g
1) we are done. Let u , u ∈ U S U . Case 2.1.
Either | O ( x, y ) | − γ − | O ( x, y ) | − γ − | O ( x, y ) |− γ − . Using ( g
2) we see that | T ( u ) |− g .
1) and ( g . u ∈ U S U ∗ , a contradiction. Case 2.2. | O ( x, y ) | − ≥ γ and | O ( x, y ) | − ≥ γ . Clearly, | O ( x, y ) | − ≥ max ( γ , γ ) ≥ ( γ + γ ) / . Case 3. u ∈ U , u ∈ U . By ( g
2) and ( g | O ( x, y ) | − ≥ γ − . (9) Case 3.1. Φ \ B ∗ ⊆ V ( x −→ H y ) . Clearly | O ( x, y ) | − ≥| x −→ H y | − ≥ γ . The result is immediate if either | O ( x, y ) | − > γ or | O ( x, y ) | − > γ −
1. Thus we can assume | O ( x, y ) | − | x −→ H y | − γ = γ − . Since | x −→ H y | − γ , we have Λ = V ( x −→ H y ) \{ u } .On the other hand, by ( g . ⊆ Λ S { u } , a contradiction. Case 3.2. Φ \ B ∗ V ( x −→ H y ) . Let z −→ H z be the maximal segment in y + −→ H x − with z , z ∈ N ( u ). Case 3.2.1. Λ ⊆ V ( x −→ H y ) . For each v ∈ Λ T V ( x −→ H u ), put P v = x ←− H z u −→ H y if v = u and P v = x ←− H z u ←− H v Λ ( v, u ) u −→ H y if v = u . Choose w ∈ Λ T V ( x −→ H u ) so as to maximize | w −→ H u | . By ( g g | w −→ H y | − ≥ γ − . If w = x , then clearly | z −→ H y | − ≥ γ andhence | O ( x, y ) | − ≥ | P w | − ≥| z −→ H y | − | x −→ H w | +2 ≥ γ − | x −→ H w | +3 , | O ( x, y ) | − ≥| x −→ H y | − ≥| w −→ H y | + | x −→ H w | − ≥ γ + | x −→ H w | − . Combining these two inequalities yields the results. Now let w = x . Choose w ∈ Λ T V ( x + −→ H u ) so as to maximize | w −→ H u | . Since H is extreme, h ≥| u Λ ( u , x ) x −→ H u z ←− H u | − ρ ( x ) = ˆ u = ⇒ | z −→ H x | − ≥ | A ( x ) | + | A ( u ) | , (10)20 ( x ) = u = ⇒ | z −→ H x | − ≥ | A ( x ) | + 1 ≥ . (11)Observe also, that | O ( x, y ) | − ≥| x −→ H y | − ≥ γ − | z −→ H z | . (12) Claim 9.1. | O ( x, y ) | − ≥ γ + | z −→ H z | . Proof of claim 9.1.
Clearly, | O ( x, y ) | − ≥ | P w | − ≥| w −→ H y | + | z −→ H x | + | z −→ H z | − . (13)By ( f | w −→ H y | − ≥ γ − − | A ( x ) | if ρ ( x ) = o u and | w −→ H y | − ≥ γ − − | A ( u ) | − | A ( x ) | if ρ ( x ) = ˆ u . Using also (10) and (11), we obtain | w −→ H y | + | z −→ H x | − ≥ γ −
1, which by (13) implies | O ( x, y ) | − ≥ γ + | z −→ H z | . ∆Claim 9.1 with together (12) implies the result. Case 3.2.2. Λ V ( x −→ H y ) . Let y −→ H y be the maximal segment in y + −→ H z − with y , y ∈ Λ . Case 3.2.2.1.
Either z , y ∈ V ( y −→ H z ) or z , y ∈ V ( z −→ H y ).Assume w.l.o.g that z , y ∈ V ( y −→ H z ). Then | O ( x, y ) | − ≥| y ←− H u Λ ( u , y ) y −→ H z u ←− H x | − ≥ γ + 1 , and the result follows by (9). Case 3.2.2.2. z , z ∈ V ( y −→ H y ) . Apply the arguments in case 3.2.2.1.
Case 3.2.2.3. y , y ∈ V ( z −→ H z ) . Putting β = | x −→ H y | + | y −→ H y | − P = y ←− H u Λ ( u , y ) y ←− H z u ←− H x,P = y ←− H u Λ ( u , y ) y −→ H z u ←− H x, we obtain | O ( x, y ) | − ≥ | P | − ≥| x −→ H y | + | y −→ H z |≥ γ − | z −→ H y | +2 . (14) Claim 9.2. | O ( x, y ) | − ≥ γ + | z −→ H y | − . Proof of claim 9.2. If ρ ( y ) = ˆ u , then by ( f β ≥ γ − − | A ( u ) | −| A ( y ) | and | O ( x, y ) | − ≥ | P | − ≥ β + | A ( u ) | + | A ( y ) | + | z −→ H y | − ≥ γ + | z −→ H y | − . Otherwise ( ρ ( y ) = o u ), β ≥ γ − − | A ( y ) | = γ − | O ( x, y ) | − ≥ | P | − ≥ β + | z −→ H y | +1 ≥ γ + | z −→ H y | − . ∆21laim 9.2 together with (14) implies the result. ∆( i2 ) By ( d h ≥ γ x imlying that | O ( x, y ) | − ≥ h/ ≥ γ x . Also, | O ( x, y ) | − ≥ γ z by ( g | Ω ( x, y ) | − ≥ | O ( x, y ) | − ≥ ( γ x + γ z ) / . ∆( i3 ) By ( g | O ( x, y ) | − ≥ γ x + 1 and by ( g | O ( x, y ) | − ≥ γ z . Usinglemma 8, we obtain the result immediately. ∆( i4 ) Claim 9.3. max ( | O x ( x, y ) | − , | O x ( x, w ) | − ≥ ( γ x + γ z ) / . Proof of claim 9.3.
By ( g , min( | O ( x, y ) | − , | O ( x, w ) | − ≥ γ z . Ifeither | O x ( x, y ) | − ≥ γ x or | O x ( x, w ) | − ≥ γ x , then clearly we are done.Otherwise, by ( g , | O x ( x, y ) | − | O x ( x, w ) | − γ x − g
8) and lemma 8. ∆
Claim 9.4. min( | O ( o x, y ) | − , | O ( o x, w ) | − ≥ ( γ x + γ z + 1) / . Proof of claim 9.4.
By ( g
5) and ( g , | O ( o x, y ) | − ≥ γ x and | O ( x, y ) | − ≥ γ z , respectively. Since V ( O ( x, y )) T { o x } = ∅ (by definition 2.10), (cid:12)(cid:12)(cid:12) O ( o x, y ) (cid:12)(cid:12)(cid:12) − ≥ | O ( x, y ) | ≥ γ z + 1, implying that | O ( o x, y ) | − ≥ ( γ x + γ z + 1) /
2. Analo-gously, | O ( o x, w ) | − ≥ ( γ x + γ z + 1) /
2. ∆
Claim 9.5. | O ( o x, x ) | − ≥ ( γ x + γ z + 1) / Proof of claim 9.5.
Let v ∈ Λ x \{ x } . By ( g
5) and ( g | O ( o x, x ) | − ≥ γ x and | O ( x, v ) | − ≥ γ z , respectively. Hence | O ( o x, x ) | − ≥ | O ( x, v ) | + | vT ( v ) ∧ v o x | − ≥ γ z + 1which implies | O ( o x, x ) | − ≥ ( γ x + γ z + 1) / . ∆The result holds from claims 9.3-9.5 and lemma 8. ∆( i5 ) By ( g | O ( x, y ) |− ≥ γ z and | O ( z, w ) |− ≥ γ x and the result followsfrom lemma 8. ∆( i6 ) By ( g | O ( x, y ) | − ≥ max ( γ x + , γ x − ) . If | T ( x ) | − ≥
2, then by( g , | O ( x, y ) | − ≥ γ x and the result holds immediately. Thus we can assume | T ( x ) | − . Put z = x + and w = x − . By ( g
3) and ( g , | O x ( x, y ) | − ≥ max ( γ x − , γ z , γ w ) . If | O x ( x, y ) | − ≥ min ( γ x , γ z + 1 , γ w + 1), then clearlywe are done. Now let | O x ( x, y ) | − γ x − γ z = γ w . Since u U ∗ (by( g g .
3) and ( g . ,γ z ≤ ( γ x + 1) / γ z + 2) / γ z ≤ | O x ( x, y ) | − ≤
2. It means that h ≤
4. Recalling also ( g .
1) and( g . h = 4, a contradiction. ∆( i7 ) If either | O x ( x, y ) | − γ x − | O y ( x, y ) | − γ y −
1, say | O x ( x, y ) | − γ x −
1, then by ( g − ( g , | T ( x ) | − g o x, x −→ H y ) ∈ ∆ or ( o x, y −→ H x ) ∈ ∆ or ( o x, H ) ∈ ∆. This implies by ( g .
2) and ( g . y ∈ U , a contradiction. Thus | O x ( x, y ) | − ≥ γ x and | O y ( x, y ) | − ≥ γ y .Using ( g
6) with lemma 8, we obtain | Ω( x, y ) | − ≥ ( γ x + γ z ) / z ∈ { x + , x − } and | Ω( x, y ) | − ≥ ( γ y + γ w ) / w ∈ { y + , y − } . Thenthe result follows from ( i i8 ) Observing that | O ( x, y ) | − ≥| y −→ H x | − h −
1, we obtain the resultfrom (8) immediately. ∆( i9 ) Put z = x + . We can assume h ≥
4, since otherwise the result holdsfrom ( i d , | O ( x, y ) | − ≥ h − ≥ γ x −
1. If | O ( x, y ) | − ≥ γ z + 1,then clearly we are done. Let | O ( x, y ) | − γ z . By ( g , | T ( z ) | − ≤
1. IfΛ z T V ( y + −→ H x − ) = ∅ , then by ( g7 ) , ∧ z x − ∈ E . Hence | O ( x, y ) | − ≥ h − z ⊆ V ( x −→ H y ). It means that γ z = 2 and | O ( x, y ) | − h = 4, a contradiction. ∆ Proof of the theorem.
Let G be a graph, C be a longest cycle in G and H = u ...u h u a longest cycle in G \ C with a maximal HC − extension T . Putting U ∗ = { v ∗ , ..., v ∗ r } and using definition 3.3, we let for each i ∈ { , ..., r } ,Θ( ←− T ( v ∗ i ) , V neut , V ( i ) fin ) = ( P ( i )0 , ..., P ( i ) π ( i ) ) ,R i = < ( V ( ∧ v ∗ i ←− T ( v ∗ i ) z ( i ) π ( i ) ) S S π ( i ) j =0 V ( P ( i ) j )) \{ z ( i ) π ( i ) } >, where V neut = V \ ( V ( C ) S V ( T )) and V ( i ) fin = V ( T ) \ V ( T ( v ∗ i )) . Since c ≥ δ + 1 ≥ κ + 1, for each i ∈ { , ..., r } there are κ − E (1) i , ..., E ( κ − i in ( κ − − connected graph G \{ z ( i ) π ( i ) } , starting at R i , pass-ing through V neut and terminating on C at κ − E ( a ) j has a vertex v in common with E ( b ) e for some a, b ∈ { , ..., κ − } and j, e ∈{ , ..., r } ( j = e ). If v V ( C ), then there is a path starting in R j , passingthrough V neut and terminating in R e , contradicting the fact that v ∗ j , v ∗ e ∈ U ∗ . So, v ∈ V ( C ) . Choose vertex-disjoint paths E ( i )1 , ..., E ( i t ) t ( i j ∈ { , ..., κ − } ) foreach j ∈ { , ..., t } so as to maximize t and put E ( i j ) j = x j −→ E ( i j ) j w ∗ j ( j = 1 , ..., t ) , where x j ∈ V ( R j ) and w ∗ j ∈ V ( C ). It is easy to see that t ≥ min ( r, κ − a j ∈ { , ..., t } there is an (cid:0) x j , v ∗ j (cid:1) − path F ( i j ) j passing through V ( R j ) S V ( T ( v ∗ j )) and having length at least ϕ v ∗ j . Denoting E ∗ j = v ∗ j F ( i j ) j x j E ( i j ) j w ∗ j ( j = 1 , ..., t ) , we see that E ∗ , ..., E ∗ t are vertex disjoint ( H, C ) − paths with | E ∗ i | − ≥ ϕ v ∗ i +1 ( i = 1 , ..., t ). Case 1. κ ≥ , h ≥ . Case 1.1. r ≥ κ. It follows that t ≥ κ −
1. Let ξ , ..., ξ t be the elements of { w ∗ , ..., w ∗ t } occuringon C in a consequtive order. 23 ase 1.1.1. t ≥ κ. Assume w.l.o.g. that ϕ v ∗ ≥ ... ≥ ϕ v ∗ r . Since r ≥ κ , we have1 κ κ X i =1 ϕ v ∗ i ≥ r r X i =1 ϕ v ∗ i implying that κ X i =1 ϕ v ∗ i ≥ κr r X i =1 ϕ v ∗ i ≥ κh r X i =1 ϕ v ∗ i . (15)By ( i
1) and ( i | Ω ( v ∗ a , v ∗ b ) | − ≥ β i for each a, b ∈ { , ..., t } and i ∈{ , ..., h } . Hence | Ω ( v ∗ a , v ∗ b ) | − ≥ h h X i =1 β i = µ ( T ) . Then for each i, j ∈ { , ..., t } , | w ∗ i −→ C w ∗ j | − ≥ | E ∗ i |− (cid:12)(cid:12) E ∗ j (cid:12)(cid:12) − (cid:12)(cid:12) Ω (cid:0) v ∗ i , v ∗ j (cid:1)(cid:12)(cid:12) − ≥ ϕ v ∗ i + ϕ v ∗ j +2+ µ ( T ) . (16)Using (15),(16) and recalling that t ≥ κ , we obtain c = P ti =1 ( | ξ i −→ C ξ i +1 | − ≥ P ti =1 ϕ v ∗ i + 2 t + tµ ( T ) ≥ P ti =1 ϕ v ∗ i + 2 t + tµ ( T ) ≥ κh P ri =1 ϕ v ∗ i + κµ ( T ) + 2 κ ≥ κh ( P ri =1 ϕ v ∗ i + P hi =1 ϕ ′ i + 2 h ) = κh ( P hi =1 ϕ i + 2 h ) , where ξ t +1 = ξ . It follows that P hi =1 ϕ i ≤ h ( c/κ − . Since ϕ i + ψ i = d ( u i ) ≥ δ ( i = 1 , ..., h ), we have h X i =1 ψ i ≥ hδ − h X i =1 ϕ i ≥ hδ − ch/κ + 2 h. In particular, max i ψ i ≥ δ − c/κ + 2. Using lemma 3, we obtain c ≥ h X i =1 ψ i + max i ψ i ≥ hδ − ch/κ + 2 h + δ − c/κ + 2 , and the result follows immediately. Case 1.1.2. t = κ − . Observe that E ( i ) κ terminates in (cid:8) w ∗ , ..., w ∗ κ − (cid:9) for each i ∈ { , ..., κ − } ,since otherwise E ∗ , ..., E ∗ κ − , E ( j ) κ contradict the maximality of t for some j ∈{ , ..., κ − } . By the same arguments, E ( i ) j terminates in (cid:8) w ∗ , ..., w ∗ κ − (cid:9) foreach i ∈ { , ..., κ − } and j ∈ { , ..., κ } . Then there is a path E = vEξ t +1 starting in < ( V ( T ) [ κ [ i =1 R i [ κ [ j =1 κ − [ i =1 V ( E ( i ) j )) \ (cid:8) w ∗ , ..., w ∗ κ − (cid:9) > C \{ w ∗ , ..., w ∗ κ − } . Assume w.l.o.g. that ξ , ..., ξ t +1 occurson −→ C in a consequtive order. Then it is easy to see that c = t +1 X i =1 ( | ξ i −→ C ξ i +1 | − ≥ κ X i =1 ϕ v ∗ i + 2 κ + κµ ( T )where ξ t +2 = ξ . Further, we can argue exactly as in case 1.1.1. Case 1.2. r ≤ κ − . It follows that t = r . There are κ vertex-disjoint ( H, C ) − paths E i = v i E i w i ( i = 1 , ..., κ ). Assume w.l.o.g. that w , ..., w κ occurs on −→ C in a conse-qutive order. Put W = { w , ..., w κ } , W ∗ = { w ∗ , ..., w ∗ r } . Let a, b ∈ { , ..., κ } . Denote W ∗ ( a, b ) = W ∗ \ V ( w a −→ C w b )We will say that w a −→ C w b is a suitable segment if | w a −→ C w b | − ≥ X v ∗ i ∈ W ∗ ( a,b ) ϕ v ∗ i + 2 ( b − a ) + b − a X i =1 (cid:0)(cid:12)(cid:12) Ω (cid:0) v a + i − , v a + i (cid:1)(cid:12)(cid:12) − (cid:1) , where v j , v j ∈ { v j } S U ( j = 1 , ..., κ ). Claim 1.
Let i ∈ { , ..., κ } . If either | W ∗ ( i, i + 1) | 6 = 1 or | W ∗ ( i, i + 1) | =1 and W ∗ ( i, i + 1) T { w i , w i +1 } = ∅ , then w i −→ C w i +1 is suitable. Proof of claim 1. Case a1 | W ∗ ( i, i + 1) | = 0 . Let T tr ( E i , E i +1 ) = ( E ′ i , E ′ i +1 ) and T tr ( v i , v i +1 ) = (cid:0) v i , v i +1 (cid:1) . Then w i −→ C w i +1 is suitable, since by ( a , | w i −→ C w i +1 | − ≥ (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) . Case a2. | W ∗ ( i, i + 1) | ≥ . Let
E, F be any two elements of { E ∗ , ..., E ∗ r } with E = xEv, F = yF w for some v, w ∈ W ∗ i . Since T tr ( E, F ) = (
E, F ) and { x, y } ⊆ U , we have by( a , | v −→ C w | − ≥ ϕ x + ϕ y + 2 + ( | Ω ( x, y ) | −
1) implying that w i −→ C w i +1 issuitable. Case a3. | W ∗ ( i, i + 1) | = 1 . Assume w.l.o.g that W ∗ ( i, i + 1) = { w ∗ } . If either E i or E i +1 (say E i )has no vertex in common with E ∗ , then using transformation T tr ( E i , E ∗ ) =( E ′ i , E ∗ ), we obtain by ( a | w i −→ C w i +1 | − ≥| w i −→ C w ∗ | − ≥ ϕ v ∗ + 2 + (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) v i ∈ { v i } S U and v i +1 = v ∗ . It means that w i −→ C w i +1 issuitable. Now let both E i and E i +1 have vertices in common with E ∗ . Walkingalong E ∗ from w ∗ to v ∗ we stop at the first vertex v ∈ V ( E i ) S V ( E i +1 ) . Assume.w.l.o.g. that v ∈ V ( E i +1 ). Putting E ′ i +1 = w ∗ E ∗ vE i +1 v i +1 and T tr ( E i , E ′ i +1 ) = ( E ′ i , E ′′ i +1 ), we see by ( a
1) that for some appropriate v i ∈{ v i } S U and v i +1 ∈ { v i +1 } S U , | w i −→ C w i +1 | − ≥| w i −→ C w ∗ | − ≥ ϕ v ∗ + (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) . So, again w i −→ C w i +1 is suitable. ∆ Claim 2. If w a −→ C w b and w b −→ C w e are suitable segments, then | w a −→ C w e | issuitable as well. Proof of claim 2.
Immediate from the definition. ∆
Claim 3.
Let w a −→ C w b is a suitable segment. If w b −→ C w b +1 is not suitable and W ∗ ( b, b + 1) = { w b } , then w a −→ C w b +1 is suitable. Proof of claim 3.
Immediate from the definition. ∆
Claim 4.
Let i ∈ { , ..., κ } . If W ∗ ( i, i + 1) ⊆ { w i , w i +1 } and | W ∗ ( i, i + 1) | = 1 (say W ∗ ( i, i + 1) = { w i } ), then w i − −→ C w i +1 is suitable. Proof of claim 4.
Assume w.l.o.g. that W ∗ ( i, i + 1) = { w ∗ } , i.e. w ∗ = w i . If | W ∗ ( i − , i ) | ≥
2, then by claims 1 and 3, w i − −→ C w i +1 is suitable. Let | W ∗ ( i − , i ) | = 1, i.e W ∗ ( i − , i ) = { w ∗ } . If either E i − or E i +1 (say E i − )has no vertices in common with E ∗ , then using transformations T tr ( E i − , E ∗ )and T tr ( E i , E i +1 ), we see that w i − −→ C w i is suitable and by claim 3, w i − −→ C w i +1 is suitable as well. Now let both E i − and E i +1 have vertices in commonwith E ∗ . Walking along E ∗ from w ∗ to v ∗ we stop at the first vertex v ∈ V ( E i − ) S V ( E i +1 ) . Assume w.l.o.g. that v ∈ V ( E i +1 ). If v = o v ∗ , i.e. v i − = v ∗ , then using T tr ( E i − , w i E ∗ vE i +1 v i +1 ) and T tr ( E i , w i +1 E i +1 vv ∗ ) we see that w i − −→ C w i is suitable. By claim 3, w i − −→ C w i +1 is suitable as well. Finally, if v = o v ∗ i (i. e. v i +1 U ), then using T tr ( E i − , w i E ∗ vE i v i +1 ) and T tr ( E i , E i +1 ),we see that w i − −→ C w i is suitable, implying by claim 3 that w i − −→ C w i +1 is suit-able as well. ∆ Claim 5.
For appropriate v i , v i ∈ { v i } S U , c = κ X i =1 ( | w i −→ C w i +1 | − ≥ r X i =1 ϕ v ∗ i + 2 κ + κ X i =1 (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) . Proof of claim 5.
Suppose not. Let i ∈ { , ..., κ } . If w i −→ C w i +1 is notsuitable, then by claims 1 and 4, either w i − −→ C w i +1 or w i −→ C w i +2 is suitable.Thus there exist some suitable segment on C and let w a −→ C w b be the longestone for some a, b ∈ { , ..., κ } ( a = b ). If w b −→ C w b +1 is suitable, then by claim2, w a −→ C w b +1 is suitable as well, a contradiction. Otherwise, by claims 3 and26, w b −→ C w b +2 is suitable and hence (by claim 2) w a −→ C w b +2 is suitable as well, acontradiction. ∆ Claim 6. If κ ≥ h ≥
5, then for appropriate v i , v i ∈ { v i } S U ( i = 1 , ..., κ ), κ X i =1 (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) ≥ κµ ( T ) . Proof of claim 6.
Assume w.l.o.g that β = max i { β i } . Put A = (cid:8) i (cid:12)(cid:12)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β (cid:9) , A = { , ..., κ }\ A ,A j = { i ∈ A (cid:12)(cid:12) u j ∈ (cid:8) v i , v i +1 (cid:9) } ( j = 1 , . We can assume that A = ∅ , since otherwise by ( i κ X i =1 (cid:0)(cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:1) ≥ κβ ≥ κµ ( T ) . If (cid:8) v i , v i +1 (cid:9) = { u , u } , then by ( i , i ∈ A . It means that A T A = ∅ . On the other hand, by ( i A = ∅ or A = ∅ . Assume w.l.o.g. that A = ∅ , i.e. A = A . Case b1. | A | ≥ . Recalling definition 3.9, it is not hard to see that there are at least twopaths among E , ..., E κ having vertices in common with V ( T ( u )) \{ u } , i.e. | T ( u ) | − ≥
2. By ( i A = ∅ , a contradiction. Case b2. | A | = 3 . If follows that at least one of the paths E , ..., E κ has a vertex in commonwith V ( T ( u )) \{ u } , i.e. | T ( u ) | − ≥
1. Clearly | T ( u ) | − , sinceotherwise | A | ≥
4. Assume w.l.o.g. that A = { , , } and v = v = v = u . If v = v = v , then clearly v , v ∈ U and by ( i , (cid:12)(cid:12) Ω (cid:0) v , v (cid:1)(cid:12)(cid:12) − ≥ β , acontradiction. Otherwise, by ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β for some i ∈ A , againa contradiction. So, 1 ≤ | A | ≤ . Case b3. | A | = 2 . Case b3.1. h ≥ . Let i ∈ A . Assume w.l.o.g. that v i = u , v i +1 = u s for some s ∈ { , ..., h } .By ( i (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β j for each j ∈ { , ..., h }\ { , h, s − , s } . Since h ≥
8, there are at least four pairwise different integers f , f , f , f in { , ..., h }\{ , h, s − , s } . By ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β f j ( j = 1 , , , . So,if i ∈ A , then (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( h X i =1 β i − β − β h − β s − − β s + X i =1 β f i ) . On the other hand,if i ∈ A , then (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β = 1 h ( h X i =1 β i − h X i =1 β i + hβ ) . (17)27ince hβ − β − β h − β s − − β s ≥ P hi =1 β i − P i =1 β f i , we haveif i ∈ A , j ∈ A , then (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − (cid:12)(cid:12) Ω( v j , v j +1 ) (cid:12)(cid:12) − ≥ µ ( T ) . Observing that | A | ≥ | A | , we obtain κ X i =1 ( | Ω( v i , v i +1 ) | −
1) = X i ∈ A ( | Ω( v i , v i +1 ) | −
1) + X i ∈ A ( | Ω( v i , v i +1 ) | − ≥ ( | A | − | A | ) µ ( T ) + 2 | A | µ ( T ) = ( | A | + | A | ) µ ( T ) = κµ ( T ) . Case b3.2. ≤ h ≤ . Let i ∈ A . Assume w.l.o.g that v i = u , v i +1 = u s for some s ∈ { , ..., h } .We will write i ∈ A ∗ if and only if (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β j for some j ∈{ , h, s − , s } . Case b3.2.1. A = A ∗ . Let i ∈ A ∗ and let v i = u , v i +1 = u s ( s ∈ { , ..., h } ). By the defini-tion, (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β j for some j ∈ { h, s − , s } , say j = s . Since6 ≤ h ≤ , there are at least three pairwise different integers f , f , f in { , ..., h }\ { , h, s − } . By ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ max ( β f , β f , β f ) . So,if i ∈ A , then (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( h X i =1 β i − β − β h − β s − + β f + β f + β f )and hence we can argue exactly as in case h ≥ . Case b3.2.2. A = A ∗ . Let A = { i, j } , where i A ∗ and let v i = v j = u , v i +1 = u s , v j +1 = u r for some s, r ∈ { , ..., h } ( s ≤ r ). By ( i
8) and ( i , ≤ s ≤ r ≤ h −
1. If s = r ,then it is easy to see (by definition 3.9) that either u ∈ U or u s ∈ U implyingby ( i
6) that i ∈ A ∗ , a contradiction. So, assume s = r , i. e. 4 ≤ s < r ≤ h − Case b3.2.2.1. h = 7 . Case b3.2.2.1.1. s = 4 and r = 5 . By ( i , either (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ β or (cid:12)(cid:12) Ω (cid:0) v j , v j +1 (cid:1)(cid:12)(cid:12) − ≥ β . Since i A ∗ ,we have (cid:12)(cid:12) Ω (cid:0) v j , v j +1 (cid:1)(cid:12)(cid:12) − ≥ β . Using ( i (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( X i =1 β i − β − β − β − β + 2 β + β + β ) , (cid:12)(cid:12) Ω (cid:0) v j , v j +1 (cid:1)(cid:12)(cid:12) − ≥ h ( X i =1 β i − β − β − β + 2 β + β ) . Using also all κ − h ≥ Case b3.2.2.1.2. s = 4 and r = 6 .
28y ( i
1) and ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( P i =1 β i − β − β − β − β + 2 β + β + β ) , (cid:12)(cid:12) Ω( v j , v j +1 ) (cid:12)(cid:12) − ≥ h ( P i =1 β i − β − β + β + β ) . Apply the arguments in case b3.2.2.1.1.
Case b3.2.2.1.3. s = 5 and r = 6 . By ( i , ( i
5) and ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( P i =1 β i − β − β − β − β + 2 β + β + β ) , (cid:12)(cid:12) Ω (cid:0) v j , v j +1 (cid:1)(cid:12)(cid:12) − ≥ h ( P i =1 β i − β − β + β + β ) . Apply the arguments in case b3.2.2.1.1.
Case b3.2.2.2. h = 6 . Clearly s = 4 , r = 5. By ( i , ( i
5) and ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( P i =1 β i − β − β − β − β + 2 β + 2 β ) , (cid:12)(cid:12) Ω( v j , v j +1 ) (cid:12)(cid:12) − ≥ h ( P i =1 β i − β + β ) . Apply the arguments in case b3.2.2.1.1.
Case b3.3. h = 5 . Let A = { i, j } and v i = v j = u , v i +1 = u s , v j +1 = u r for some s, r ∈{ , ..., h } ( s ≤ r ). By ( i
8) and ( i , s = r = 4 and we can reach a contradictionas in case b3.2.2. Case b4. | A | = 1 . Let A = { i } and v i = u , v i +1 = u s for some s ∈ { , ..., h } . Case b4.1. h = 5 . By ( i
8) and ( i , s = 4 . Also, by ( i
1) and ( i (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥
15 ( X i =1 β i − β − β + β + β ) . Apply the arguments in case b3.2.2.1.1.
Case b4.2. h ≥ . There are at least two distinct integers f , f in { , ..., h } − { , h, s − , s } .By ( i , (cid:12)(cid:12) Ω (cid:0) v i , v i +1 (cid:1)(cid:12)(cid:12) − ≥ h ( h X i =1 β i − β − β h − β s − − β s + 2 β f + 2 β f ) . Since | A |≥ − | A | = 3, we have at least two inequalities of type (17).So, we can argue as in case b3.2.2.1.1. ∆By claims 5 and 6, c ≥ P ri =1 ϕ v ∗ i + 2 κ + κµ ( T ) and the result follows as incase 1.1.1.1. 29 ase 2. κ ≥ , h ≤ . Since h ≥ κ , we have h = κ = 4. Then there are four vertex-disjoint( H, C ) − paths. It can be easily cheeked that c ≥
18. If δ ≤
6, then c ≥ ≥
20 ( δ + 2) / h + 1) κ ( δ + 2) / ( h + κ + 1). Let δ ≥
7. Using ( d
3) we canshow that ϕ i ≤ i ∈ { , , , } , i.e. max i ψ i ≥ δ −
3. Then bylemma 3, c ≥ P i =1 ( δ − ϕ i ) + δ − δ − − P i =1 ϕ i . If P i =1 ϕ i ≤ c ≥ δ − ≥ ( h + 1) κ ( δ + 2) / ( h + κ + 1). So, it sufficies to prove P i =1 ϕ i ≤ Case 2.1.
Either | U | = 0 or | U | = 4 . It follows that ϕ i ≤ i ∈ { , , , } . Case 2.2 . | U | = 3 . Assume w.l.o.g. that U = { u } . If u o u E , then it is easy to see that ϕ i ≤ i ∈ { , , , } . Otherwise ( u o u ∈ E ) , u u E and hence ϕ ≤ , ϕ ≤ , ϕ ≤ ϕ ≤ Case 2.3. | U | = 2 . By symmetry, we can distinguish the following two cases.
Case 2.3.1. U = { u , u } . If u u E and u u E , then clearly ϕ i ≤ i ∈ { , , , } .Assume w.l.o.g. that u u ∈ E. We can assume also u u E , since otherwisethe cycle u u u u u u u is larger than H , which is impossible. Then clearly ϕ ≤ , ϕ ≤ , ϕ ≤ ϕ ≤ Case 2.3.2. U = { u , u } . It is easy to see that ϕ i ≤ i ∈ { , , , } . Case 2.4. | U | = 1 . Returning to the proof of lemma 3, we can see that in this special case thelower bound in lemma 3 can be improved by a unit. So, it suffices to show P i =1 ϕ i ≤
13. Denoting U = { u } , we see that ϕ ≤ , ϕ ≤ , ϕ ≤ , ϕ ≤ Case 3. κ ≤ . Claim 7.
Let κ ∈ { , } and h ≥ κ . If there are no κ + 1 vertex-disjoint( H, C ) − paths, then c ≥ min ( κ ( h + 1) , κ ( δ − κ + 4)) . Proof of claim 7. Case d1. κ = 3 . Assume w.l.o.g. that E , E and E are T − transformed. We now prove that | w −→ C w | − ≥ min ( h + 1 , δ + 1). If v = v +1 , then clearly | w −→ C w | − ≥| w E v ←− H v E w | − ≥ h + 1 . Let v = v +1 . Walking along −→ H from v to v − we stop at the first vertex z with either ∧ z w ∈ E or ∧ z w E or z = v − . If ∧ z w ∈ E or z = v − , thenclearly | w −→ C w | − ≥ h + 1. Let ∧ z w ∈ E and ∧ z w E . If ∧ z w ∈ E forsome w ∈ V ( C ) \ { w , w , w } , then there are 4 vertex-disjoint ( H, C ) − paths,contradicting our assumption. So, N ( ∧ z ) T V ( C ) ⊆ { w } , i.e. ϕ z ≥ δ − h ≥ ϕ z + 1 ≥ δ . By ( g | O ( z − , v ) | − ≥ γ z ≥ ϕ z ≥ δ − w −→ C w | − ≥ δ +2. Thus we have proved | w −→ C w | − ≥ min ( h + 1 , δ + 1) . By symmetry, we have similar inequalities for segments w −→ C w and w −→ C w andthe result holds from h + 1 ≥ δ + 1. Case d2. κ = 2 . Apply the arguments in case 1. Claim 7 is proved. ∆
Case 3.1. κ = 3 . We can assume that there are no 4 vertex-disjoint (
H, C ) − paths, since oth-erwise c ≥ ( h + 1) 4 h + 4 + 1 ( δ + 2) > ( h + 1) κh + κ + 1 ( δ + 2) . Then by claim 7 we can distinguish the following two cases.
Case 3.1.1. c ≥ h + 1) . If h ≥ δ −
2, then c ≥ h + 1) ≥ h + 1) ( δ + 2) / ( h + 4). Otherwise, theresult holds from c ≥ δ −
1) (see [ ? ]). Case 3.1.2. c ≥ δ + 1) . If h ≤ δ + 2, then c ≥ δ + 1) ≥ h + 1) ( δ + 2) / ( h + 4). Let h ≥ δ + 1). Observing that c ≥ h/ Case 3.2. κ = 2 . Apply the arguments in case 1.Thus we have proved the theorem for h the length of a longest cycle in G \ C .Observing that c ≥ ( h + 1) κh + κ + 1 ( δ + 2) > (cid:16) h ′ + 1 (cid:17) κh ′ + κ + 1 ( δ + 2)for any h ′ < h , we complete the proof of the theorem. ∆ Acknowledgment
The author is very grateful to N. K. Khachatrian for his careful reading and hiscorrections for the manuscript.