Minimum degree, independence number and pseudo [2,b]-factors in graphs
aa r X i v : . [ c s . D M ] A p r Minimum Degree, Independence Number andPseudo [2 , b ]-Factors in Graphs
Siham BEKKAI ∗ USTHB, Faculty of Mathematics, PO Box 32 El-Alia Bab Ezzouar16111 Algiers, Algeria
Abstract
A pseudo [2 , b ]-factor of a graph G is a spanning subgraph in whicheach component C on at least three vertices verifies 2 ≤ d C ( x ) ≤ b ,for every vertex x in C . The main contibution of this paper, is togive an upper bound to the number of components that are edgesor vertices in a pseudo [2 , b ]-factor of a graph G . Given an integer b ≥
4, we show that a graph G with minimum degree δ , independencenumber α > b ( δ − and without isolated vertices possesses a pseudo[2 , b ]-factor with at most α − ⌊ b ( δ − ⌋ edges or vertices. This boundis sharp. Key words:
Pseudo [2 , b ]-Factor; Independence Number; Minimum De-gree.
Throughout this paper, graphs are assumed to be finite and simple. Forunexplained concepts and notations, the reader could refer to [2].Given a graph G , we let V ( G ) be its vertex set, E ( G ) its edge set and n its order. The neighborhood of a vertex x in G is denoted by N G ( x ) anddefined to be the set of vertices of G adjacent to x ; the cardinality of this setis called the degree of x in G . For convenience, we denote by d ( x ) the degreeof a vertex x in G ; by δ the minimum degree of G and by α its independencenumber. However, if H is a subgraph of G then we write d H ( x ); δ H and α ( H ) respectively for the degree of x in H ; the minimum degree and the ∗ e-mail address: [email protected] H . We denote by d G ( x, y ) the distance between x and y in the graph G .A factor of G is a spanning subgraph of G , that is a subgraph obtainedby edge deletions only. If S is the set of deleted edges, then this subgraphis denoted G − S . If H is a subgraph of G , then G − H stands for thesubgraph induced by V ( G ) − V ( H ) in G . By starting with a disjoint unionof two graphs G and G and adding edges joining every vertex of G toevery vertex of G , we obtain the join of G and G , denoted G + G . For apositive integer p , the graph pG consists of p vertex-disjoint copies of G . Inall what follows, we use disjoint to stand for vertex-disjoint.In [1], we defined a pseudo 2-factor of a graph G to be a factor eachcomponent of which is a cycle, an edge or a vertex. It can also be seen asa graph partition by a family of vertices, edges and cycles. Graph partitionproblems have been studied in lots of papers. They consist in partitioning thevertex set of G by disjoint subgraphs chosen to have some specific properties.In [3], Enomoto listed a variety of results dealing with partitions into pathsand cycles. The emphasis is generally on the existence of a given partitionhowever, in our study of pseudo-factors, we take interest in the number ofcomponents that are edges or vertices in a pseudo-factor of G . In [1], weproved that every graph with minimum degree δ ≥ α ≥ δ possesses a pseudo 2-factor with at most α − δ + 1 edges orvertices and that this bound is best possible. Motivated by the desire toknow what happens in general cases, we define a pseudo [ a, b ] -factor (where a and b are two integers such that b ≥ a ≥
2) as a factor of G in whicheach component C on at least three vertices verifies a ≤ d C ( x ) ≤ b , for every x ∈ C . Clearly, a pseudo [ a, b ]-factor with no component that is an edge ora vertex is nothing but an [ a, b ]-factor. Surveys on factors and specifically[ a, b ]-factors and connected factors can be found in [6, 5]. In the presentwork, we study pseudo [2 , b ]-factors, we consider the case b ≥ δ , α and b ) for the number of componentsthat are edges or vertices in a pseudo [2 , b ]-factor of G . Note that, from aresult by Kouider and Lonc ([4]), we deduce that if α ≤ b ( δ − then G hasa [2 , b ]-factor. Laying down the condition α > b ( δ − , the main result of thispaper reads as follows: Theorem 1
Let b be an integer such that b ≥ and G a graph of minimumdegree δ ≥ and independence number α with α > b ( δ − . Then G possessesa pseudo [2 , b ] -factor with at most α − ⌊ b ( δ − ⌋ components that are edgesor vertices. The bound given in Theorem 1 is best possible. Indeed, let b be aninteger such that b ≥ H be a nonempty set of vertices. The graph2 = H + pK , where p > b | H | , has minimum degree δ = | H | + 1 andindependence number α = p . We can easily verify that G possesses a pseudo[2 , b ]-factor with α − ⌊ b ( δ − ⌋ edges and we can not do better. Also, asimple example reaching the bound of Theorem 1, is a graph G obtained bytaking a graph H on n vertices in which every vertex is of degree between2 and b ( b ≥ n additional independent vertices and joiningexactly one isolated vertex to exactly one vertex of H . The graph G hasminimum degree δ = 1, independence number α = n and can be partitionedinto one component that is H and n = α − ⌊ b ( δ − ⌋ vertices (or simply n edges) and we can not do better.Combining Theorem 1 with the results of [1] and [4], we obtain Corollary 1
Let b ≥ be an integer such that b = 3 . Let G be a graph ofminimum degree δ and independence number α and without isolated vertices.Then G possesses a pseudo [2 , b ] -factor with at most max(0 , α − ⌊ b ( δ − ⌋ ) edges or vertices. [2 , b ] -factors First of all, we put aside the case δ = 1 for which we know that we have in G a pseudo [2 , b ]-factor with at most α edges or vertices. Indeed, if we regarda cycle as a component each vertex of which is of degree between 2 and b ,then we know that any graph G can be covered by at most α cycles, edges orvertices (see for instance [7]). So the bound α − ⌊ b ( δ − ⌋ holds for δ = 1.From now on, we assume that G has minimum degree δ ≥
2. Let F bea subgraph of G such that 2 ≤ d F ( x ) ≤ b for all x ∈ V ( F ). For the sake ofsimplifying the writing, such a subgraph F will be called a [2 , b ]-subgraph of G . Denote by D a smallest component of G − F , set W = G − ( D ∪ F ) andchoose F in such a manner that:( a ) α ( G − F ) is as small as possible;( b ) subject to ( a ), the number of vertices of D is as small as possible;( c ) subject to ( a ) and ( b ), the number of vertices in F is as small aspossible.Notice that a subgraph F satisfying the conditions above exists since δ ≥
2. Indeed, let us consider a longest path in G and let u be one of itsendpoints. Let v be the farthest neighbor of u on this path and P uv thesegment of P joining u and v . The cycle C formed by the path P uv and theedge uv contains u and all its neighbors so α ( G − C ) < α . Hence F is notempty. 3e shall show the following theorem which yields Theorem 1: Theorem 2
Let b be an integer such that b ≥ . Let G be a graph of min-imum degree δ ≥ and independence number α such that α > b ( δ − . Thenthere exists a pseudo [2 , b ] -factor of G such that F is the [2 , b ] -subgraph ofthis pseudo [2 , b ] -factor and F gives α ( G − F ) ≤ α − ⌊ b ( δ − ⌋ . Proof of Theorem2.
Let F be a [2 , b ]-subgraph of G satisfying theconditions ( a ), ( b ) and ( c ). Denote by u , ..., u m ( m ≥
1) the neighbors of D on F and by P ij a path with internal vertices in D joining two vertices u i and u j with 1 ≤ i, j ≤ m and i = j . The proof of Theorem 2 will be dividedinto several claims. The following one which will be intensively used remindsLemma 1 in [1]. Claim 1
Let F ′ be a [2 , b ] -subgraph of G which contains the neighbors of D in F and at least one vertex of D . Setting W ′ = G − ( F ′ ∪ D ) , we have α ( W ′ ) > α ( W ) .Proof of Claim 1. Set D ′ = D − F ′ .(1) If D ′ = ∅ then by the choice of F , we have α ( G − F ) ≤ α ( G − F ′ ). But α ( G − F ) = α ( W ) + α ( D ) ≥ α ( W ) + 1 and α ( G − F ′ ) = α ( W ′ ), so α ( W ) <α ( W ′ ).(2) If D ′ = ∅ then F ′ gives a component D ′ smaller than D , so again by thechoice of F , we have α ( W )+ α ( D ) = α ( G − F ) < α ( G − F ′ ) = α ( W ′ )+ α ( D ′ ).But as α ( D ′ ) ≤ α ( D ) then we obtain α ( W ) < α ( W ′ ). (cid:3) In the next claims, we try to learn more about the degrees in F of itsvertices. Claim 2
For every i , ≤ i ≤ m , we have N F ( u i ) ∩ { u , . . . , u m } = ∅ .Proof of Claim 2. Suppose that for some i , N F ( u i ) ∩ { u , ..., u m } 6 = ∅ ,then there exists a vertex u j (1 ≤ j ≤ m and j = i ) such that u i u j ∈ E ( F ).Put e = u i u j , then ( F − e ) ∪ P ij is a [2 , b ]-subgraph. Indeed, none of thevertices of F changes its degree in ( F − e ) ∪ P ij and the internal verticesof P ij are of degree 2. So taking F ′ = ( F − e ) ∪ P ij in Claim 1 we obtain α ( W ) > α ( W ), which is absurd. (cid:3) Claim 3 d F ( u i ) ≤ b − for at most one vertex u i , i = 1 , ..., m .Proof of Claim 3. Suppose to the contrary that there exist at least twodistinct vertices u k and u l such that d F ( u k ) ≤ b − d F ( u l ) ≤ b − F ′ = F ∪ P kl in Claim 1 (notice that in F ′ , d F ( u k ) and d F ( u l )are at most b , and the internal vertices of P kl are of degree 2 in F ′ so F ′ is a[2 , b ]-subgraph of G ), we obtain α ( W ) < α ( W ) which is absurd. (cid:3) Let S be the set of vertices x in ∪ mi =1 N F ( u i ) such that x is a commonneighbor of at least two vertices in { u , ..., u m } . We have: Claim 4 d F ( x ) ≤ for every x ∈ S .2. If S contains a vertex x such that d F ( x ) = 3 , then(a) For every k, ≤ k ≤ m , we have d F ( u k ) = b .(b) For every y ∈ ∪ mi =1 N F ( u i ) − { x } we have d F ( y ) = 2 .Proof of Claim 4.
1. Suppose that d F ( x ) ≥ x ∈ S . By definition, x is the neighborin F of at least two vertices say u i and u j with 1 ≤ i, j ≤ m, i = j . Put e = xu i and e ′ = xu j . Then in F ′ = ( F − e − e ′ ) ∪ P ij only x changesits degree but it remains at least 2. So F ′ is a [2 , b ]-subgraph whichleads to a contradiction by Claim 1.2. Let x be in N F ( u i ) ∩ N F ( u j ) (1 ≤ i, j ≤ m and i = j ) such that d F ( x ) = 3. Suppose that there exists u k (which will be the only one byClaim 3) such that d F ( u k ) ≤ b −
1, we can always assume that k = i .Then taking F ′ = ( F − e ) ∪ P ik , where e = xu i , in Claim 1 gives acontradiction.Furthermore, if we suppose that there exists y ∈ N F ( u k ) − { x } , with1 ≤ k ≤ m (we can suppose without loss of generality that k = i )such that d F ( y ) ≥
3. Then setting e = xu i , e ′ = yu k and taking F ′ = ( F − e − e ′ ) ∪ P ik in Claim 1 gives a contradiction. Notice that F ′ is a [2 , b ]-subgraph: indeed, only x and y lose 1 in their degree butthey remain of degree at least 2 in F ′ and the internal vertices of P ik are of degree 2 in F ′ . (cid:3) Claim 4 implies that S is an independent set in F and we will deducelater that it is also independent in G . But before that, we take a look at theneighbors of { u , ..., u m } which are not in S . For each u i ( i = 1 , ..., m ), set N ∗ F ( u i ) = { x ∈ N F ( u i ); x / ∈ S } . Claim 5
1. If there exist vertices x in ∪ mi =1 N ∗ F ( u i ) such that d F ( x ) ≥ ,then these vertices are in the neighborhood of a same u k , ≤ k ≤ m . . If there exist k, ≤ k ≤ m , such that d F ( u k ) ≤ b − and x ∈∪ mi =1 N F ( u i ) such that d F ( x ) ≥ , then x ∈ N ∗ F ( u k ) .Proof of Claim 5.
1. Suppose that there exist x ∈ N ∗ F ( u k ) such that d F ( x ) ≥ x ′ ∈ N ∗ F ( u j )such that d F ( x ′ ) ≥ ≤ j, k ≤ m, j = k . Then the subgraph( F − e − e ′ ) ∪ P kj , where e = u k x and e ′ = u j x ′ is a [2 , b ]-subgraph of G .Taking F ′ = ( F − e − e ′ ) ∪ P kj in Claim 1, we obtain a contradiction.2. Suppose that there exist k, ≤ k ≤ m , such that d F ( u k ) ≤ b − x ∈ ∪ mi =1 N F ( u i ) with d F ( x ) ≥
3. By Claim 4(2), x / ∈ S . Suppose that x ∈ N ∗ F ( u i ), with i = k . Notice that the fact that d F ( u k ) ≤ b − d F ( u i ), by Claim 3, to be equal to b . Taking F ′ = ( F − e ) ∪ P ik , where e = xu i , in Claim 1, we obtain α ( W ) > α ( W ) which is absurd. (cid:3) Looking more closely at the structure of D , we can say more about thedegrees of the vertices in ∪ mi =1 N F [ u i ], where N F [ u i ] = N F ( u i ) ∪ { u i } is theclosed neighborhood of u i . First, we remark that D has minimum degree atmost 1. Remark 1 δ D ≤ .Proof. Suppose, by contradiction, that δ D ≥ D provides a cycle C which verifies α ( D − C ) < α ( D ). Put F ′ = F ∪ C , then F ′ is a [2 , b ]-subgraph of G . Moreover, α ( G − F ′ ) = α ( D − C ) + α ( W ) <α ( D ) + α ( W ) = α ( G − F ) and this contradicts the choice of F . (cid:3) Two cases are to consider, the case where D is a tree (a single vertex isa trivial tree) and the case where D contains a cycle. The following claimdeals with this latter case. Claim 6
Suppose that D contains a cycle. Then1. d F ( x ) = 2 for all x ∈ ∪ mi =1 N F ( u i ) .2. d F ( u i ) = b for all i, ≤ i ≤ m . roof of Claim 6.
1. Suppose that there exists a vertex x ∈ N F ( u i ) such that d F ( x ) ≥ Q be an edge or a path with internal vertices in D − C joining u i and C . Then taking F ′ = ( F − e ) ∪ Q ∪ C , where e = xu i , in Claim 1 givesa contradiction. Notice that d F ′ ( x ) ≥ u i does not changeits degree (nor do the other vertices of F ) then F ′ is a [2 , b ]-subgraphof G .2. Suppose that d F ( u k ) ≤ b − k, ≤ k ≤ m and let Q be anedge or a path with internal vertices in D − C joining u k and C . Then,taking F ′ = F ∪ Q ∪ C in Claim 1 gives a contradiction. (cid:3) If D is a tree and δ D = 0, then D has at least two leaves, say x and y .We relabel u , ..., u m , with m ≤ m , the vertices in N F ( x ) ∪ N F ( y ). Claim 7
Suppose that D is a tree and that there exist two vertices x and y in D with d D ( x ) = d D ( y ) = 1 such that N F ( x ) = N F ( y ) . Then1. For all k, ≤ k ≤ m , d F ( u k ) ≥ b − .2. If there exists a vertex x ∈ ∪ m i =1 N F ( u i ) such that d F ( x ) ≥ then it isthe only one.3. If there exists a vertex u k (with ≤ k ≤ m ) such that d F ( u k ) = b − then for every vertex x ∈ ∪ m i =1 N F ( u i ) we have d F ( x ) = 2 .Proof of Claim 7. Let P be a path in D joining x to y .1. If there exists a vertex u i (1 ≤ i ≤ m ) such that d F ( u i ) ≤ b −
2. Thentaking F ′ = F ∪ u i x P y u i in Claim 1 gives a contradiction.2. Suppose that there exists a vertex x ∈ ∪ m i =1 N F ( u i ) such that d F ( x ) ≥ x ∈ S then Claim 4 gives what desired. If x / ∈ S , then x ∈ N ∗ F ( u k )for some 1 ≤ k ≤ m . Suppose that there exists y ∈ ∪ m i =1 N F ( u i ) suchthat d F ( y ) ≥ y ∈ N ∗ F ( u j ), for some 1 ≤ j ≤ m . By Claim 5(1), j = k . Taking F ′ = ( F − e − e ′ ) ∪ u k x P y u k , where e = xu k and e ′ = yu k , in Claim 1, we obtain a contradiction.3. Finally, suppose that there exist a vertex u k (1 ≤ k ≤ m ) such that d F ( u k ) = b − x ∈ ∪ m i =1 N F ( u i ) such that d F ( x ) ≥
3. ByClaim 5(2), x ∈ N ∗ F ( u k ). Put e = xu k . Then taking F ′ = ( F − e ) ∪ u k x P y in Claim 1 gives a contradiction. (cid:3) I in F with V ( I ) ⊂ V ( F ), E ( I ) ⊂ E ( F ) and such that everyinternal vertex x of I has d F ( x ) = 2 is called an interval (or a segment) of F . We say that two disjoint intervals I (1) and I ( ) in F are path-independent if there exists no path internally disjoint from F ∪ D joining a vertex in I (1) to a vertex in I (2) . We say that t intervals I (1) , I (2) , . . . , I ( t ) ( t ≥
2) in F are path-independent if they are pairwise path-independent. The followingclaim will be very useful. It is a shorter version of Lemma 2 in [1] with ashort proof. Claim 8
Let I (1) , I (2) , . . . , I ( t ) ( t ≥ ) be t disjoint intervals in F , containingno neighbor of D and such that α ( W ∪ I ( i ) ) = α ( W ) for every i = 1 , ..., t . If I (1) , I (2) , . . . , I ( t ) are path-independent, then α ( D ∪ W ∪ I (1) ∪ I (2) ∪ . . . ∪ I ( t ) ) = α ( W ∪ D ) .Proof of Claim 8. Let W i be the union of components of W with neighbors in I ( i ) ( i =1 , . . . , t ). By hypothesis, the intervals I ( i ) are pairwise path-independent so W i ∩ W j = ∅ , for all 1 ≤ i, j ≤ t, i = j . Hence W −∪ ti =1 W i , W ∪ I (1) , . . . , W t ∪ I ( t ) form a partition of W ∪ I (1) ∪ . . . ∪ I ( t ) and it follows that α ( W ∪ I (1) ∪ . . . ∪ I ( t ) ) = α ( W ∪ I (1) ) + · · · + α ( W t ∪ I ( t ) ) + α ( W − ∪ ti =1 W i ). On the other hand,as α ( W ∪ I ( i ) ) = α ( W ), for every i = 1 , . . . , t then α ( W i ∪ I ( i ) ) = α ( W i ).This yields α ( W ∪ I (1) ∪ . . . ∪ I ( t ) ) = P ti =1 α ( W i ) + α ( W − ∪ ti =1 W i ) = α ( W ).We finally get α ( W ∪ D ∪ I (1) ∪ . . . ∪ I ( t ) ) = α ( W ∪ D ) because the intervals I ( i ) (with i = 1 , . . . , t ) do contain no neighbor of D . (cid:3) Let s be the vertex of S (if it exists) such that d F ( s ) = 3. We put s asidebefore applying the procedure described hereafter. Provided always that s exists, we set N ′ F ( u i ) = N F ( u i ) − { s } if s ∈ N F ( u i ) and N ′ F ( u i ) = N F ( u i )otherwise ( i = 1 , ..., m ).For u k , 1 ≤ k ≤ m , denote by x ki ( i = 1 , ..., | N ′ F ( u k ) | ) its neighborsthat belong to N ′ F ( u k ). Using this notation, we can have x ki = x lj , for some1 ≤ i ≤ | N ′ F ( u k ) | and 1 ≤ j ≤ | N ′ F ( u l ) | (1 ≤ k, l ≤ m, k = l ), in case x ki ∈ N ′ F ( u k ) ∩ N ′ F ( u l ) ⊂ S .From now on, let m ′ = m if D is a tree with at least two leaves and m ′ = m otherwise. We choose the sense u k → x ki , as a sense of ”orientation”.Let u k (1 ≤ k ≤ m ′ ) be such that d F ( u k ) = b and all its neighbors thatare in N ′ F ( u k ) are of degree 2 in F . Starting at x k and following the chosenorientation we go over from a vertex to its neighbor until meeting a vertexwhich we call y k , such that d F ( y k, +1 ) ≥ y k, +1 = u j for some j, ≤ j ≤ m (where y k, +1 is the successor of y k following the chosen orientation). Thisgives an interval x k ...y k which we denote by P k = [ x k , y k ] F .8e repeat the process using the other neighbors of u k that are in N ′ F ( u k ).At the p th step, we consider a vertex x kp ∈ X p = N ′ F ( u k ) − ( ∪ p − i =1 V ( P ki ))and construct a path P kp = x kp ...y kp containing x kp and such that d F ( y k, + p ) ≥ y k, + p = u j for some j, ≤ j ≤ m . When X r becomes empty at the r th step ( r ≥ p ), then we consider another vertex u l ( l = k ). We choose aslong as possible, u l such that d F ( u l ) = b and its neighborhood that are in N ′ F ( u l ) are all of degree 2 in F . We Choose a vertex in N ′ F ( u l ) − ∪ r − i =1 V ( P ki ),and we do the same construction, until the vertices in N ′ F ( u l ) are all in( ∪ r − i =1 V ( P ki )) ∪ ( ∪ r ′ − i =1 V ( P li )). Denote by P the set of paths obtained so far.When it is no more possible to choose a vertex u p , 1 ≤ p ≤ m ′ , such that d F ( u p ) = b and with all its neighbors that are in N ′ F ( u p ) having degree 2in F , then we take the vertex u q of degree at most b − N ′ F ( u q ) vertices of degree at least 3 in F . Notice that u q existsonly if s does not (see Claim 4(2)) and if both a vertex u q of degree at most b − x ji of degreeat least 3 exist, then these vertices are in the neighborhood of u q (see Claim5). Put N q = { x qi ∈ N ′ F ( u q ) − V ( P ) such that d F ( x qi ) = 2 } . Starting at avertex x qi ∈ N q , we repeat the construction described above until N q becomesempty. We update the set P at each step.By construction all the vertices of P ki are of degree 2 in F so V ( P ki ) ∩ V ( P lj ) = ∅ for every couple P ki , P lj of paths in P (they are disjoint), moreoverno vertex in P ki is adjacent in F to a vertex in P lj , for all P ki , P lj in P .We divide the set P into three subsets, each containing the paths P ki =[ x ki , y ki ] F of Type 1, Type 2 or Type 3, defined as follows: Type 1 If y k, + i = u j , with j = k . Type 2 If y k, + i = u j , with j = k . Type 3 If y k, + i = u j for every j , 1 ≤ j ≤ m .For technical reasons, in case D is a trivial tree or a tree having no coupleof leaves with the same neighborhood in F , we stop the procedure describedabove when it remains no vertex u p (1 ≤ p ≤ m ′ ) such that d F ( u p ) = b , orwhen the remaining vertex u p (1 ≤ p ≤ m ′ ) has in its neighborhood N ′ F ( u p ) avertex of degree at least 3. We consider Q the subset of P , of paths obtainedtill then. Let P = Q in this case and P = P in the others.We show in what follows that the addition of a path of P to W ∪ D augments α ( W ∪ D ) by at least 1. Claim 9
For each P ki ∈ P , we have α ( W ∪ D ∪ P ki ) > α ( W ∪ D ) . roof of Claim 9. Let P ki be a path in P .1. If D contains a cycle, then taking F ′ = F − P ki gives what desired.Indeed, in this case all the vertices u i are of degree b (by Claim 6),as b ≥ P ki , the degree of the vertices u i (1 ≤ i ≤ m ′ ) remains at least 2. Moreover, by construction of P ki ,the degree of no vertex in F becomes smaller than 2, after deletion of P ki . So F ′ is a [2 , b ]-subgraph of G . F ′ contradicts Condition ( c ) inthe choice of F (because | V ( F ′ ) | < | V ( F ) | ) so α ( G − F ′ ) > α ( G − F )which yields α ( W ∪ D ∪ P ki ) > α ( W ∪ D ).2. If D is a tree possessing two vertices x and y of degree 1 in D , havingthe same neighborhood in F ( N F ( x ) = N F ( y )). Then if P ki is ofType 1 or 3, then we reason as in (1) and we obtain what desired. If P ki is of Type 2, then (1) is no more efficient if d F ( u k ) = b − u k may become smaller than 2 when P ki is deleted). Sowe take F ′ = ( F − P ki ) ∪ u k x P y u k , where P is a path with internalvertices in D joining x to y . The subgraph F ′ is a [2 , b ]-subgraph of G (we have d F ( u k ) = d F ′ ( u k )) which gives by Claim 1, what desired.3. In the other cases, as b ≥ P , thedeletion of any path P ki ∈ P , gives a [2 , b ]-subgraph. Reasoning as in(1), we get what desired. (cid:3) Notice that as D is independent from P ki (by construction) and from W then α ( W ∪ D ∪ P ki ) = α ( W ∪ P ki ) + α ( D ). Hence the conclusion in Claim 9is equivalent to α ( W ∪ P ki ) > α ( W ). For each path P ki = [ x ki , y ki ] F in P andfollowing the chosen orientation, let v ki be the first vertex of P ki such that α ( W ∪ [ x ki , v ki ] F ) > α ( W ). Notice that v ki is well defined by Claim 9. Denoteby P ′ ki the interval [ x ki , v ki ] F of P ki and by P ′ the set of the intervals P ′ ki . Inwhat follows, we take interest in the path-independence of the intervals of P ′ . Claim 10
Let P ′ ki and P ′ lj be two distinct intervals [ x ki , v ki ] F and [ x lj , v lj ] F in P ′ such that ≤ k, l ≤ m ′ , k = l . Then, P ′ ki and P ′ lj are path-independent.Proof of Claim 10. By way of contradiction, suppose that there exist twovertices a ki ∈ P ′ ki and a lj ∈ P ′ lj such that a ki and a lj are joined by Q whichis an edge in G or a path with internal vertices in W . Choose a ki and a lj soas to minimize the sum d F ( x ki , a ki ) + d F ( x lj , a lj ). Recall that by construction xy / ∈ E ( F ) for every x ∈ P ki and y ∈ P lj .The segments [ x ki , a ki [ F and [ x lj , a lj [ F verify the hypothesis of Claim 8. Indeed,10y the choice of a ki and a lj , they are path-independent. Furthermore, as[ x ki , a ki [ F ⊂ [ x ki , v ki [ F ; [ x lj , a lj [ F ⊂ [ x lj , v lj [ F and by the choice of v ki and v lj , wehave α ( W ∪ [ x ki , a ki [ F ) = α ( W ) and α ( W ∪ [ x lj , a lj [ F ) = α ( W ). So by Claim8, we obtain α ( W ∪ [ x ki , a ki [ F ∪ [ x lj , a lj [ F ) = α ( W ). ( ⋆ )Also, taking the [2 , b ]-subgraph F ′ = ( F − ([ x ki , a ki [ F ∪ [ x lj , a lj [ F )) ∪ Q ∪ P kl ,in Claim 1, gives α (( W − Q ) ∪ [ x ki , a ki [ F ∪ [ x lj , a lj [ F ) > α ( W ). But as α ( W ∪ [ x ki , a ki [ F ∪ [ x lj , a lj [ F ) ≥ α (( W − Q ) ∪ [ x ki , a ki [ F ∪ [ x lj , a lj [ F ) hence we get α ( W ∪ [ x ki , a ki [ F ∪ [ x lj , a lj [ F ) > α ( W ) which contradicts ( ⋆ ). (cid:3) When k = l in the previous claim, then we consider the structure of D .If D contains a cycle or D is a tree with two leaves x and y such that N F ( x ) = N F ( y ), then the following claim gives the path-independence ofany couple of segments P ′ ki and P ′ kj in P ′ . Claim 11
Let P ′ ki and P ′ kj be two distinct segments of P ′ . Suppose that D contains a cycle or D is a tree with two leaves x and y such that N F ( x ) = N F ( y ) . Then P ′ ki and P ′ kj are path-independent, for every k, ≤ k ≤ m ′ .Proof of Claim 11. Let P ′ ki and P ′ kj (with 1 ≤ i, j ≤ m ′ , i = j ) be twosegments in P ′ . By way of contradiction, suppose that there is a path Q internally disjoint from F ∪ D joining a vertex a ki ∈ P ′ ki to a vertex a kj ∈ P ′ kj and choose a ki and a kj so that the sum d F ( x ki , a ki ) + d F ( x kj , a kj ) is minimum.The segments [ x ki , a ki [ F and [ x kj , a kj [ F verify the hypothesis of Claim 8. Indeed,they are path-independent, by the choice of a ki and a kj . Furthermore, as[ x ki , a ki [ F ⊂ [ x ki , v ki [ F and [ x kj , a kj [ F ⊂ [ x kj , v kj [ F and by the choice of v ki and v kj we have α ( W ∪ [ x ki , a ki [ F ) = α ( W ) and α ( W ∪ [ x kj , a kj [ F ) = α ( W ). So byClaim 8, we obtain α ( W ∪ [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) = α ( W ). ( ⋆⋆ )On the other hand, if D contains a cycle C , then let Q ′ be a path with internalvertices in D − C joining u k to a vertex on C . If D is a tree with two leaves x and y such that N F ( x ) = N F ( y ). Then let P be a path with internalvertices in D joining x to y . Taking F ′ = ( F − ([ x ki , a ki [ F ∪ [ x kj , a kj [ F )) ∪ Q ∪ Q ′ ∪ C in the first case and F ′ = ( F − ([ x ki , a ki [ F ∪ [ x kj , a kj [ F )) ∪ Q ∪ u k x P y u k in the second one and using Claim 1, we obtain in both cases α ( W ∪ [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) ≥ α (( W − Q ) ∪ [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) > α ( W ) whichcontradicts ( ⋆⋆ ). (cid:3) Suppose now that D is either a trivial tree or D has no leaves with thesame neighborhood in F . • If all couples of distinct segments ( P ′ ki , P ′ kj ) ( k, ≤ k ≤ m ′ , 1 ≤ i, j ≤| N ′ F ( u k ) | ) in P ′ are path-independent then we have finished. It is par-11icularly the case if s exists. Indeed, if we suppose to the contrarythat there exist two distinct segments P ′ ki and P ′ kj ( k, ≤ k ≤ m ′ ,1 ≤ i, j ≤ | N ′ F ( u k ) | ) in P ′ that are path-dependent, that is there isa path internally disjoint from D ∪ F joining a vertex in a ki ∈ P ′ ki toa vertex in a kj ∈ P ′ kj . We choose these vertices so as to minimize thesum d F ( x ki , a ki ) + d F ( x kj , a kj ). Reasoning as in the previous claims usingClaim 8 and taking in Claim 1 F ′ = ( F − [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) ∪ P kr − u r s where u r is a neighbor of s such that r = k , we get a contradiction.It is also the case if there exists a vertex u r (1 ≤ r ≤ m ) that is ofdegree at most b − F or that has in its neighborhood N F ( u r ) avertex x such that d F ( x ) ≥
3. Recall that in our case, this vertex issupposed to be put apart in the procedure we have used. So, if we sup-pose that there is a path internally disjoint from D ∪ F joining a vertexin a ki ∈ P ′ ki to a vertex in a kj ∈ P ′ kj ( k = r ). We choose these verticesso as to minimize the sum d F ( x ki , a ki ) + d F ( x kj , a kj ). Here again, usingClaim 8 and taking in Claim 1, F ′ = ( F − [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) ∪ P kr if d F ( u r ) ≤ b − F ′ = ( F − [ x ki , a ki [ F ∪ [ x kj , a kj [ F ) ∪ P kr − u r x if d F ( u r ) = b and d F ( x ) ≥ x ∈ N ′ F ( u r ), we get a contradiction. • If not, then this case is treated in following claim.
Claim 12
Suppose that D is a trivial tree or a tree with no leaves havingthe same neighborhood in F . Suppose moreover that there exist two distinctsegments P ′ ki and P ′ kj ( k , ≤ k ≤ m ′ , ≤ i, j ≤ | N ′ F ( u k ) | ) in P ′ thatare path-dependent. Then there exists no other couple of segments ( P ′ lp , P ′ lq ) ( l = k , ≤ l ≤ m ′ , ≤ p, q ≤ | N ′ F ( u l ) | , p = q ) in P ′ that are path-dependent.Proof of Claim 12. The proof is basically the same as the previous. Let a ki and a kj be two vertices in P ′ ki and P ′ kj respectively that are joined by apath internally disjoint from F ∪ D and chosen so as to minimize the sum d F ( x ki , a ki ) + d F ( x kj , a kj ). Suppose to the contrary that there exist two dis-tinct segments P ′ lp , P ′ lq ( l = k , 1 ≤ l ≤ m ′ , 1 ≤ p, q ≤ | N ′ F ( u l ) | ) and apath internally disjoint from F ∪ D joining a vertex a lp ∈ P ′ lp to a vertex a lq ∈ P ′ lq and choose these vertices in such a way that d F ( x lp , a lp ) + d F ( x lq , a lq )is minimum. Then using Claim 8 with four intervals and taking F ′ =( F − ([ x ki , a ki [ F ∪ [ x kj , a kj [ F ) − ([ x lp , a lp [ F ∪ [ x lq , a lq [ F )) ∪ P kl in Claim 1 yields acontradiction. As k = l , then Claim 10 guarantees the path-independence ofthe segments P ′ kr , P ′ lt for r ∈ { i, j } , t ∈ { p, q } . (cid:3) By the claims above, we have that P ′ contains several segments that arepath-independent. Furthermore, the following remark claims that an addi-tional segment can be considered when needed, particularly when S containsa vertex of degree 3. 12 emark 2 If there exists a vertex s ∈ S such that d F ( s ) = 3 ( s is uniqueby Claim 4(2)). We consider two cases:(i) If s is in the neighborhood of three vertices u k , u l and u p , with ≤ k, l, p ≤ m and k, l, p pairwise distinct. Then setting P ∗ = { s } we have that P ∗ ispath-independent from any path in P ′ and α ( W ∪ P ∗ ) > α ( W ) .(ii) If s is in the neighborhood of exactly two vertices, say u k and u l , k = l, ≤ l, k ≤ m . If furthermore P does not contain paths of Type 3, then thereexists a path P ∗ that is path-independent from any segment in P ′ included ina path of Type 1 or Type 2. Furthermore α ( W ∪ P ∗ ) > α ( W ) .Proof. (i) First, taking F ′ = ( F −{ s } ) ∪ P kl in Claim 1 we obtain α ( W ∪{ s } ) > α ( W ).Of course, since by Claim 4(2) d F ( u i ) = b for all i = 1 , ..., m , then we havethat F ′ is a [2 , b ]-subgarph of G . As s ∈ N F ( u k ) ∩ N F ( u l ) ∩ N F ( u p ) ⊂ N F ( u k ) ∩ N F ( u l ) then we can write { s } = P ki or { s } = P li as suitable toapply Claim 10 and show the path-independence of { s } from any segment in P ′ .(ii) Let us start from s and go forward following the chosen orientation froma vertex of degree 2 in F to a vertex of degree 2 in F , until coming acrossa vertex y whose successor y + is of degree at least 3. We have that y + / ∈{ u , . . . , u m } , otherwise, going in the opposite direction, the segment [ y, s [ F (where s is not taken) is a path of Type 3. Moreover, y + / ∈ ∪ mi =1 N F ( u i )because since s exists then by Claim 4 every vertex in ∪ mi =1 N F ( u i ) is of degree2. So y + ∈ V ( F ) − ∪ mi =1 N F [ u i ]. The path P = s . . . y can be consideredas a path deriving from u k ( P = P ′ ki ) or deriving from u l ( P = P ′ li ) andhence reasoning as in Claim 9, taking F ′ = F − P , we obtain α ( W ∪ P ) >α ( W ). Let v be the first vertex of P following the chosen orientation suchthat α ( W ∪ [ s, v ] F ) > α ( W ). Setting P ∗ = [ s, v ] F , we can show its path-independence with any segment in P ′ included in a path of Type 1 or Type2, like in Claim10. (cid:3) Finally, to count the number of pairwise path-independent segments in P ′ , those whose independence is guaranteed by Claims 10, 11, and 12, wedistinguish different cases according to the structure of D and get in any case,at least ⌊ b ( δ − ⌋ (recall that m ′ ≥ δ −
1) path-independent segments, addingwhen necessary the path P ∗ (in particular when s exists). Notice that when P contains paths of Type 3, then in these paths one vertex in ∪ mi =1 N F ( u i )is used at once, so the bound ⌊ b ( δ − ⌋ holds, otherwise P ∗ is added.The segments in P ′ ∪ { P ∗ } , when added to W ∪ D augment α ( W ∪ D ).Put L = P ′ ∪ { P ∗ } . Recall that the segments of L are independent from D by construction. For each P ∈ L , let W P be the union of components of W P .We have that α ( W S ∪ P ∈ L P ) = α ( W − ∪ P ∈ L W P ) + P P ∈ L α ( W P ∪ P ) ≥ α ( W − ∪ P ∈ L W P ) + P P ∈ L α ( W P ) + | L |≥ α ( W ) + ⌊ b ( δ − ⌋ .Hence α = α ( G ) = α ( W ∪ D ∪ F ) ≥ α ( W ∪ D S ∪ P ∈ L P ) ≥ α ( D ) + α ( W S ∪ P ∈ L P ) ≥ α ( D ) + α ( W ) + ⌊ b ( δ − ⌋ = α ( W ∪ D ) + ⌊ b ( δ − ⌋ .So α ( W ∪ D ) = α ( G − F ) ≤ α − ⌊ b ( δ − ⌋ and the proof of Theorem 2 isachieved. (cid:4) Proof of Theorem 1.
Since by Theorem 2, α ( G − F ) ≤ α − ⌊ b ( δ − ⌋ ,then the subgraph of G induced by V ( W ∪ D ) = V ( G − F ) can be covered byat most α − ⌊ b ( δ − ⌋ cycles, edges or vertices (see for instance [7]). Denoteby E the set of cycles, edges or vertices covering G − F . The graph F ∪ E is a pseudo [2 , b ]-factor of G with at most α − ⌊ b ( δ − ⌋ edges or vertices.This completes the proof of Theorem 1. (cid:4) References [1] S. Bekkai and M. Kouider.
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