Non-separating spanning trees and out-branchings in digraphsof independence number 2
aa r X i v : . [ c s . D M ] J u l Non-separating spanning trees and out-branchings in digraphsof independence number 2
J. Bang-Jensen ∗ S. Bessy † A. Yeo ‡ July 7, 2020
Abstract
A subgraph H = ( V, E ′ ) of a graph G = ( V, E ) is non-separating if G \ E ′ , that is, thegraph obtained from G by deleting the edges in E ′ , is connected. Analogously we say thata subdigraph X = ( V, A ′ ) of a digraph D = ( V, A ) is non-separating if D \ A ′ is stronglyconnected. We study non-separating spanning trees and out-branchings in digraphs ofindependence number 2. Our main results are that every 2-arc-strong digraph D ofindependence number α ( D ) = 2 and minimum in-degree at least 5 and every 2-arc-strongoriented graph with α ( D ) = 2 and minimum in-degree at least 3 has a non-separatingout-branching and minimum in-degree 2 is not enough. We also prove a number of otherresults, including that every 2-arc-strong digraph D with α ( D ) ≤ G with δ ( G ) ≥ α ( G ) = 2has a non-separating hamiltonian path. Keywords: non-separating branching; spanning trees; digraphs of independence number2; strongly connected; hamiltonian path. An out-tree in a digraph D = ( V, A ) is a connected subdigraph T + s of D in which everyvertex of V ( T + s ), except one vertex s (called the root ) has exactly one arc entering. This isequivalent to saying that s can reach every other vertex of V ( T + s ) by a directed path usingonly arcs of T + s . An out-branching in a digraph D = ( V, A ) is a spanning out-tree, that is,every vertex of V is in the tree. We use the notation B + s for an out-branching rooted at thevertex s . An in-branching , B − t , rooted at the vertex t is defined analogously. The followingclassical result due to Edmonds and the algorithmic proof due to Lov´asz [14] implies thatone can check the existence of k arc-disjoint out-branchings in polynomial time. Theorem 1 (Edmonds) . [12] Let D = ( V, A ) be a digraph and let s ∈ V . Then D contains k arc-disjoint out-branchings, all rooted at s , if and only if there are k arc-disjoint ( s, v ) -pathsin D for every v ∈ V . ∗ Department of Mathematics and Computer Science, University of Southern Denmark, Odense, Denmark(email: [email protected]). Part of this work was done while the author was visiting LIRMM, Universit´ede Montpellier as well as INRIA Sophia Antipolis. Hospitality and financial support by both is gratefullyacknowledged. Ce travail a b´en´efici´e d’une aide du gouvernement franais, g´er´ee par l’Agence Nationale de laRecherche au titre du projet Investissements d’Avenir UCAJEDI portant la r´ef´erence no ANR-15-IDEX-01. † LIRMM, CNRS, Universit´e de Montpellier, Montpellier, France (email:[email protected]), financialsuports: PICS CNRS DISCO and ANR DIGRAPHS n.194718. ‡ Department of Mathematics and Computer Science, University of Southern Denmark, Odense, Denmarkand Department of Mathematics, University of Johannesburg, Auckland Park, 2006 South Africa (email:[email protected]).
Theorem 2.
It is NP-complete to decide whether a digraph contains arc-disjoint branchings B + s , B − t for given vertices s, t . It was shown in [6] that the problem remains NP-complete even for 2-arc-strong 2-regulardigraphs.Thomassen conjectured that sufficiently high arc-connectivity will guarantee the existenceof arc-disjoint in- and out-branchings with prescribed roots. As defined in Section 2, λ ( D )denotes is the arc-connectivity of the digraph D . Conjecture 3. [16] There exists a natural number K such that every digraph D with λ ( D ) ≥ K contains arc-disjoint branchings B + s , B − t for every choice of s, t ∈ V .It was pointed out in [2] that Conjecture 3 is equivalent to the following (the same valueof K would work for both conjectures). Conjecture 4.
There exists a natural number K such that every digraph D with λ ( D ) ≥ K contains an out-branching which is arc-disjoint from some in-branching.In this paper we study digraphs of independence number 2. The structure of digraphswith independence number 2 is not well understood and there are numerous interesting openproblems. For instance it is an open problem whether the existence of vertex disjoint paths P , P such that P i is an ( s i , t i )-path for i = 1 , Theorem 5. [2] Every digraph D = ( V, A ) with α ( D ) = 2 and λ ( D ) ≥ contains arc-disjointbranchings B + s , B − t for some choice of s, t ∈ V . Conjecture 6. [2] Every 2-arc-strong digraph D = ( V, A ) with α ( D ) = 2 has a pair ofarc-disjoint branchings B + s , B − s for every choice of s ∈ V . Conjecture 7. [2] Every 3-arc-strong digraph D = ( V, A ) with α ( D ) = 2 has a pair ofarc-disjoint branchings B + s , B − t for every choice of s, t ∈ V .In the present paper we are interested in the existence an out-branching B + in a stronglyconnected digraph D of independence number 2 such that the digraph D \ A ( B + ) that weobtain by deleting all arcs of B + is still strongly connected. Clearly if D has such an out-branching, then it also has arc-disjoint in- and out-branchings B + s , B − s from some vertex s (namely the root of B + ). The main result of the paper is the following which, besides beingof interest in connection with Conjecture 11 below, also provides support for Conjecture 6. Theorem 8.
Let D be a -arc-strong digraph with α ( D ) ≤ . If either of the followingstatements hold then there exists an out-branching, B + , in D , such that D \ A ( B + ) is stronglyconnected. i): δ − ( D ) ≥ , or (ii): δ − ( D ) ≥ and D is an oriented graph (no cycles of length 2). Theorem 8 (ii) is best possible in the sense that there exists a digraph ˜ D , with α ( ˜ D ) = 2and λ ( ˜ D ) ≥ δ − ( ˜ D ) ≥
2) which does not contain a non-separating out-branching. See Figure 6 and Proposition 24.In Section 2 we provide some preliminary results. In particular, we show in Proposition12 that there are infinitely many 2-arc-strong digraphs with independence number 2 and highin- and out-degrees that do not have an arc-partition into two spanning strong subdigraphs,implying that we cannot replace B + in Theorem 8 by some spanning strong subdigraph.We also describe some structural results on semicomplete digraphs that will be used in latersections. Finally we prove a structural result on strong spanning subdigraphs with few arcsin digraphs with α ( D ) = 2 ≤ λ ( D ).In Section 3 we characterize semicomplete digraphs with non-separating out-branchingsand prove a more general result which will be used in the proof of Theorem 8.In Section 4 we prove Theorem 8 and in Section 5 we study non-separating spanningtrees in digraphs of independence number 2. The main result here is Theorem 23 which saysthat every 2-arc-strong digraph D with α ( D ) = 2 and n ≥
14 vertices has a non-separatingspanning tree. We conjecture that already n ≥ D has a hamiltonian cycle and no cycle of length 2. In Section 6 weconstruct an infinite family of 2-arc-strong digraphs with α = 2 for which every hamiltonianpath is separating and in Section 7 we show that for undirected graphs with independencenumber 2 a non-separating hamiltonian path always exists, provided the minimum degree isat least 4. Finally, in Section 8 we pose a number of open problems. Terminology not defined here or above is consistent with [3]. Let D = ( V, A ) be a digraph.The underlying graph of D is the graph U G ( D ) = ( V, E ) where uv ∈ E if and only ifthere is at least one arc between u and v in D . For a non-empty subset X ⊂ V we denote by d + D ( X ) (resp. d − D ( X )) the number of arcs with tail (resp. head) in X and head (resp. tail)in V \ X . We call d + D ( X ) (resp. d − D ( X )) the out-degree (resp. in-degree ) of the set X .Note that X may be just a vertex. We will drop the subscript when the digraph is clear fromthe context. We denote by δ ( D ) the minimum over all in- and out-degrees of vertices of D .This is also called the minimum semi-degree of a vertex in D . The arc-connectivity of D ,denoted by λ ( D ), is the minimum out-degree of a proper subset of V . A digraph is stronglyconnected (or just strong ) if λ ( D ) ≥
1. An arc a of a strong digraph D is a cut-arc if D \ { a } is not strong.When X is a subset of the vertices of a digraph D , we denote by D [ X ] the subdigraph induced by X , that is, the vertex set of D [ X ] is X and the arc set consists of those arcs of D which have both end vertices in X .The independence number , denoted α ( D ), of a digraph D = ( V, A ) is the size of alargest subset X ⊆ V such that the subdigraph of D induced by X has no arcs.A strong component of a digraph D is a maximal (with respect to inclusion) subdigraph D ′ which is strongly connected. The strong components of D are vertex disjoint and their3ertex sets form a partition of V ( D ). If D has more than one strong component, then wecan order these as D , . . . , D k such that there is no arc from a vertex in V ( D j ) to a vertexin V ( D i ) when j > i . A strong component D i is initial ( terminal ) if there is no arc of D which enters (leaves) V ( D i ).The following result is well-known and easy to show. Proposition 9.
A digraph D has an out-branching if and only it it has precisely one initialstrong component. In that case every vertex of the initial strong component can be the rootof an out-branching in D . A digraph is semicomplete if it has no pair of nonadjacent vertices. A tournament is a semicomplete digraph with no directed cycle of length 2. A digraph D = ( V, A ) is co-bipartite is it has a vertex-partition V , V such that D [ V i ] is semicomplete for i ∈ [2].We shall make use of the following classical result due Camion. He formulated it only fortournaments but it is easy to see that it holds for semicomplete digraphs also. Theorem 10. [9] Every strongly connected semicomplete digraph is hamiltonian.
The following conjecture, which would clearly imply Conjecture 3, has been verified forsemicomplete digraphs (see Theorem 15 below).
Conjecture 11. [7] There exists a natural number K such that every digraph D with λ ( D ) ≥ K contains arc-disjoint spanning strong subdigraphs D , D .The infinite family of digraphs described below shows that no condition on semi-degree isenough to imply the conclusion of Conjecture 11 for 2-arc-strong digraphs, even for digraphswith independence number 2. Proposition 12.
For every natural number K there are infinitely many 2-arc-strong di-graphs D with α ( D ) = 2 and δ ( D ) ≥ K that have no pair of arc-disjoint spanning strongsubdigraphs. x x x x T T T T Figure 1: A 2-arc-strong digraph D with α ( D ) = 2 and no decomposition into 2 arc-disjointspanning subdigraphs. 4 roof. Let T be a 2-arc-strong tournament with δ ( T ) ≥ K and let x be a vertex of T . Let D = ( V, A ) be the digraph that we obtain from 4 disjoint copies T i , i ∈ [4], of T by addingthe arcs of the 4-cycle x x x x x , the arcs x x , x x , all possible arcs from V ( T ) to V ( T )and from V ( T ) to V ( T ). Here x i is the copy of x in T i . See Figure 1. Then D is co-bipartiteand 2-arc-strong and we claim that D ′ does not contain a pair of arc-disjoint spanning strongsubdigraphs.Indeed, suppose there is a partition A = A ∪ A such that D i = ( V, A i ) is strong for i = 1 ,
2. There are exactly two arcs in D in both directions between V ( T ) ∪ V ( T ) and V ( T ) ∪ V ( T ). Without loss of generality we have x x ∈ A and x x ∈ A . As thereare only two arcs entering V ( T ), this implies that the arc x x must be in A (in order toreach the vertices in V ( T )) and as there are only two arcs leaving V ( T ) we have x x ∈ A .We must also have x x ∈ A , since the only other arc leaving V ( T ) is in A . This impliesthat the arc x x must be in A now we see that there is no path from V ( T ) ∪ V ( T ) to V ( T ) ∪ V ( T ) in D , contradiction. Let D be a digraph. A decomposition of D is a partition ( S , . . . , S p ), p ≥
1, of its vertexset. The index of vertex v in the decomposition, denoted by ind( v ), is the integer i suchthat v ∈ S i . An arc uv is forward if ind( u ) < ind( v ), backward if ind( u ) > ind( v ), and flat if ind( u ) = ind( v ).A decomposition ( S , . . . , S p ) is strong if D h S i i is strong for all 1 ≤ i ≤ p . The followingproposition is well-known (just consider an acyclic ordering of the strong components of D ). Proposition 13.
Every digraph has a strong decomposition with no backward arcs. A nice decomposition of a strong digraph D is a strong decomposition such that theset of cut-arcs of D is exactly the set of backward arcs. Note that if D has no cut-arc, thatis, λ ( D ) ≥
2, then the strong decomposition with just one set S = V ( D ) is nice. Proposition 14. [5] Every strong semicomplete digraph of order at least admits a nicedecomposition. Given a semicomplete digraph and a nice decomposition of it, the natural ordering ofits backward arcs is the ordering of these arcs in decreasing order according to the index oftheir tail. Note that this ordering is unique [5].Denote by S the semicomplete digraph on 4 vertices that we obtain from a directed4-cycle v v v v v by adding the arcs v v , v v , v v , v v . The following result shows thatConjecture 11 holds for semicomplete digraphs. Theorem 15. [7] Let D = ( V, A ) be a 2-arc-strong semicomplete digraph which is not iso-morphic to S . Then D contains two arc disjoint strong spanning subdigraphs D , D . d ca b S , d ca b S , d ca b S , Figure 2: The digraphs S , , S , , S , Theorem 16. [4] Let D be a 2-arc-strong semicomplete directed multigraph. Then D has apair of arc-disjoint strong spanning subdigraphs if and only if D is not isomorphic to S orone of three directed multigraphs shown in Figure 2 that can be obtained from S by addingone or two extra arcs parallel to existing ones. Furthermore, if D is not one of those fourdigraphs, then we can find a pair of arc-disjoint strong spanning subdigraphs in polynomialtime. α = 2 . Theorem 17 (Chen-Manalastras) . [10] Let D be a strongly connected digraph with α ( D ) = 2 .Then either D has a hamiltonian cycle or it has two cycles C , C that cover V ( D ) and whoseintersection is a (possibly empty) subpath of both cycles. Corollary 18.
Let D = ( V, A ) be a strong digraph with α ( D ) = 2 . Then (A) or (B) belowholds. (A): V ( D ) can be partitioned into V and V , such that D [ V i ] are strong semicomplete di-graphs for i ∈ [2] and there exists u i ∈ V i that is not adjacent to any vertex in V − i . (B): D has a strong spanning subdigraph S with one of the following properties. (B1) S is a hamiltonian cycle of D . (B2) There are two vertices x, y of S such that d + S ( x ) = d − S ( y ) = 1 , d − S ( x ) = d + S ( y ) = 2 and d + S ( z ) = d − S ( z ) = 1 for all z ∈ V − { x, y } .Furthermore N − S ( x ) and N + S ( y ) are independent sets in D . (B3) There exists a vertex x ∈ V such that d + S ( x ) = d − S ( x ) = 2 and d + S ( v ) = d − S ( v ) = 1 for all v = x .Furthermore N + S ( x ) and N − S ( x ) are independent sets in D .In particular when one of (B1)-(B3) holds, the sum of the degrees of any two distinct verticesof S is at most 6.Proof. Let D have α ( D ) = 2. By Theorem 17 D has a hamiltonian cycle or it has two cycles C , C that cover V ( D ) and whose intersection is a (possibly empty) subpath of both cycles.If D has a hamiltonian cycle we take S to be that cycle and we are done as (B1) holds. So nowassume that D contains no hamiltonian cycle, which by Theorem 17 implies that D containstwo cycles C , C that cover V ( D ) and whose intersection is a (possibly empty) subpath ofboth cycles. Let such C and C be chosen such that | V ( C ) ∩ V ( C ) | is maximum possible.We now consider the case when | V ( C ) ∩ V ( C ) | >
0. If | V ( C ) ∩ V ( C ) | = 1, then let x be the vertex in V ( C ) ∩ V ( C ) and let A ( S ) to be the union of A ( C ) and A ( C ). Now thefirst part of (B3) holds. If N + S ( x ) is not an independent set, then without loss of generalityassume that xu ∈ A ( C ) and xv ∈ A ( C ) and uv ∈ A ( D ). Now remove the arc xv from C and add the path xuv , in order to obtain a new cycle C ′ , with | V ( C ) ∩ V ( C ′ ) | = 2 > | V ( C ) ∩ V ( C ) | , and thereby contradicting the maximality of | V ( C ) ∩ V ( C ) | . Therefore N + S ( x ) is an independent set. We can analogously show that N − S ( x ) is an independent set,and therefore part (B3) holds. This completes the case when | V ( C ) ∩ V ( C ) | = 1. We maytherefore assume that | V ( C ) ∩ V ( C ) | ≥
2. That is, there are vertices x, y ∈ V ( C ) ∩ V ( C )6uch that the path common to C and C is P and P = C i [ x, y ] for i = 1 ,
2. Now the firstpart of (B2) holds. If N + S ( y ) is not an independent set then analogously to above we get acontradiction to the maximality of | V ( C ) ∩ V ( C ) | (or to D not being hamiltonian). And,again analogously to above, we can show that N − S ( x ) is also an independent set. Therefore(B2) holds in this case. This completes the case when | V ( C ) ∩ V ( C ) | > | V ( C ) ∩ V ( C ) | = 0 and therefore C and C are vertex disjoint.As D is strongly connected there exists a ( C , C )-arc, say x x ∈ A ( D ). Let x +1 be thesuccessor of x on C . If there is any ( C , x +1 )-arc, y x +1 , in D , then considering the cycle C [ x +1 , x ] C [ x , y ] x +1 instead of C , would contradict the maximality of | V ( C ) ∩ V ( C ) | .So there is no ( C , x +1 )-arc in D . If there is an ( x +1 , C )-arc in D , then consider x +1 insteadof x . Continuing this process either gives us a vertex which is not adjacent to any vertex in C or there is no arc from C to x , x +1 , (cid:0) x +1 (cid:1) + , etc., a contradiction to D being strong. Sothere must be a vertex u ∈ V ( C ) which is not adjacent to any vertex in C .Analogously we can show that there must be a vertex u ∈ V ( C ) which is not adjacentto any vertex in C . This implies that D [ V ( C i )] is semicomplete, as if two vertices, x i , y i ,in D [ V ( C i )] are non-adjacent then { x i , y i , u − i } is an independent set, a contradiction to α ( D ) = 2. p p p p W r t r t W Figure 3: The semicomplete digraphs W and W . Theorem 19.
Let D be a strong semicomplete digraph. Then the following holds.(a) If D has at least two vertices with in-degree one, then D contains no non-separatingbranching. Furthermore if D contains exactly two vertices with in-degree one and is notisomorphic to W (see Figure 3), then there exists an out-tree T + rooted at r , such that V ( T + ) = V ( D ) − r and D \ A ( T + ) is strong, where d − D ( r ) = d − D ( r ) = 1 .(b) If D is isomorphic to W (see Figure 3), then D contains no non-separating branching.(c) If D is not isomorphic to W and contains exactly one vertex, r , of in-degree one,then D contains a non-separating branching, rooted at r .(d) If δ − ( D ) ≥ and | V ( D ) | ≤ , then for every r ∈ V ( D ) the digraph D contains anon-separating branching, rooted at r .(e) If δ − ( D ) ≥ and | V ( D ) | ≥ , then D admits a nice decomposition ( S , S , . . . , S p ) ,and for every r ∈ S the digraph D contains a non-separating branching, rooted at r . roof. Recall that in an out-branching every vertex except the root has one arc entering it.Hence if a vertex has in-degree one in D , it must be the root of any non-separating out-branching. This shows that if D admits a non-separating out-branching, it has at most onevertex with in-degree one. This proves the first part of (a).Now assume that D contains exactly two vertices, r and r , with in-degree one and let H be a hamiltonian cycle in D ( H exists by Theorem 10) and let D ′ = D \ A ( H ). If there isonly one initial strong component in D ′ − r , then letting T + be an out-branching in D ′ − r gives us the desired out-tree. So assume that there are at least two initial strong componentsin D ′ − r , { r } and S . If | V ( S ) | ≥ r is non-adjacent to { r } ∪ V ( S ) in D ′ and |{ r } ∪ V ( S ) | ≥ H was a hamiltonian cycle, meaningthat we removed only 2 arcs incident to r when we obtained D ′ from D ). So | S | = 1 andwe let V ( S ) = { t } . As d − D ( t ) ≥
2, we have d − D ′ ( t ) = 1 and N − D ′ ( t ) = { r } . Analogouslyconsidering D ′ − r instead of D ′ − r we obtain an initial strong component S in D ′ − r where V ( S ) = { t } and N − D ′ ( t ) = { r } . Note that t = t (as r t , r t ∈ A ( D ′ )). Furthermore t and t are not adjacent in D ′ , as if t t ∈ A ( D ′ ) then r t t is a path in D ′ − r and S isnot an initial strong component in D ′ − r . So in D ′ , r is non-adjacent to { r , t } and t isnon-adjacent to { r , t } . Therefore | V ( D ) | = 4 and D is isomorphic to W . This proves thesecond part of (a).It is easy to check that if D is the semicomplete digraph W in Figure 3, then everyout-branching is separating (the vertex p with in-degree one must be the root of all out-branchings), which proves part (b).Now suppose that D = ( V, A ) is different from W and has exactly one vertex of in-degreeone, which implies that n = | V ( D ) | ≥
3. Let H = p p . . . p n p be a hamiltonian cyclein D and let D ′ = D \ A ( H ). Without loss of generality assume that d − D ( p ) = 1, whichimplies that d − D ′ ( p ) = 0. If D ′ only has one initial strong component, then D ′ contains anout-branching, B + p , rooted at p , which implies that B + p is a non-separating out-branchingin D . Therefore we may assume that D ′ contains at least two initial strong components,one of which is just the vertex p . As d − D ′ ( x ) ≥ x ∈ V ( D ′ ) \ { p } we note that anyother initial strong component, S , in D ′ must contain at least two vertices. Furthermoreas there is no arc between p and the vertices in S in D ′ , we must have V ( S ) = { p , p n } and p p n , p n p ∈ A ( D ′ ). Therefore n ≥
4, as otherwise p p ∈ A ( H ) and p p ∈ A ( D ′ ), acontradiction.If n = 4, then we note that D = W , a contradiction (as H = p p p p p and A ( D ′ ) = { p p , p p , p p } ). So assume that n ≥
5. Let T be obtained from H by deleting the arc p p and adding the arcs p n p and p p . Note that T is a strongly connected spanningsubgraph of D . Let D ∗ = D \ A ( T ), and note that p p i ∈ A ( D ∗ ) for all i ∈ [ n ] \ { n, } and p p , p p n , p n p ∈ A ( D ∗ ) (as the vertex set of the initial component, S , in D ′ was { p , p n } ).Therefore the only initial component in D ∗ is { p } and there exists an out-branching, B + p in D ∗ rooted at p , which is therefore a non-separating branching in D . This proves part (c).We now consider the case when δ − ( D ) ≥
2. If n ≤
3, then D is the complete digraph onthree vertices and part (d) holds. So assume that n ≥
4. By Proposition 14, D admits a nicedecomposition ( S , S , . . . , S p ).First consider the case when p = 1. That is D is 2-arc-strong. If D is isomorphic to S (see Theorem 15), then as can be seen in Figure 4, S has a non-separating branching B + r for each r ∈ V ( S ). So we may assume that D is not isomorphic to S , which by Theorem 15implies that D contains two arc disjoint strong spanning subdigraphs D and D . For every r ∈ V ( D ) we note that D contains an out-branching rooted at r and therefore D contains anon-separating branching, rooted at r . This proves part (e) when p = 1.8 bcd Figure 4: Decomposing S into an out-branching B + c in red and a strong spanning subdigraphin blue.We now consider the case when p ≥
2. Let r ∈ S be arbitrary. Let st be the ( V ( D ) \ V ( S ) , V ( S ))-arc in D . That is, st , is the cut-arc entering S .Construct a new digraph H r from S by adding a vertex x to S and adding the two arcs xt and xr (if t = r we add two parallel arcs). We will first show that x has two arc-disjointpaths to every other vertex in H r . As S is strong we note that x can reach all other verticesin H r if we delete xt or xr . Furthermore if we delete any arc e ∈ A ( S ) then x can still reachall other vertices in S by starting with the arc xt , as t can reach all other vertices in S evenafter deleting one arc (by the definition of a nice decomposition). By Theorem 1 this impliesthat there exists two arc-disjoint out-branchings both rooted in x in H r . Deleting x fromthese gives us two arc-disjoint out-branchings B + t , B + r in S , rooted at t and r , respectively.We will now show that we may assume that B + t is not just an out-star rooted at t .Assume that B + t is an out-star rooted at t . We first consider the case when | S | ≥
4. Let l , l , . . . , l | S |− be the leaves of B + t and note that { l , l , . . . , l | S |− } is not independent in D \ A ( B + r ) as the underlying graph of B + r is acyclic. So without loss of generality we mayassume that l l ∈ D \ A ( B + r ). Now delete the arc tl from B + t and add the arc l l instead.We have then obtained a B + t (arc disjoint from B + r ) that is not an out-star, as desired. So wemay now consider the case when | S | ≤
3. Notice that | S | = 1 is impossible as δ − ( D ) ≥ | S | ≥
2. However if | S | = 2, denoting S by { t, y } , we have d − D ( y ) = 1, acontradiction. So we must have | S | = 3. Let S = { t, x, y } . As d − D ( y ) , d − D ( x ) ≥ xy, yx, tx, ty ∈ A ( D ). As S is strong we note that xt ∈ A ( D ) or yt ∈ A ( D ) (or both).Without loss of generality assume that xt ∈ A ( D ). We now obtain the desired B + t and B + r as follows. • If r = t , then B + t = { tx, xy } and B + r = { ty, yx } . • If r = x , then B + t = { ty, yx } and B + r = { xt, xy } . • If r = y , then B + t = { tx, xy } and B + r = { yx, xt } .Now, as B + t is not just an out-star it contains a vertex q which is neither the root or leaf.As D is a strong semicomplete digraph it contains a hamiltonian cycle, H = p p p . . . p n p .Without loss of generality assume that the cut-arc st is the arc p n p . Then there must beexactly one arc in H leaving S , say p i p i +1 . Note that p p . . . p i is a hamiltonian path in S and p i +1 p i +2 . . . p n is a hamiltonian path in D − V ( S ). Let Q be the union of B + t and thepath p i +1 p i +2 . . . p n p where we add an arc from every leaf of B + t to p i +1 (which exists bythe definition of the nice decomposition and the fact that t is not a leaf of B + t ). Note that Q is a strong spanning subdigraph of D .Now construct B + by starting with B + r and adding an arc from q (the vertex that was notthe root or a leaf of B + t ) to every vertex in V ( D ) \ V ( S ). Note that B + is an out-branching9n D rooted at r and D − A ( B + ) contains all arcs of Q and is therefore strongly connected.This completes the proof of part (e) and therefore also of the theorem.As the digraph S has a non-separating out-branching B + v for each of its 4 vertices, thesame holds for any digraph obtained from S by adding arcs parallel to existing ones. Thuswe one can prove the following corollary of Theorem 16. Corollary 20.
Every 2-arc-strong semicomplete directed multigraph D = ( V, A ) has a non-separating out-branching B + v for every choice of v ∈ V . Before we prove Theorem 8 we need the following lemma.
Lemma 21.
Let D have α ( D ) = 2 ≤ λ ( D ) and assume that δ − ( D ) ≥ . If D satisfies (A)in Corollary 18, then D has a non-separating out-branching.Proof. Let D be a digraph with α ( D ) = 2 ≤ λ ( D ) which consists of vertex disjoint strongsemicomplete digraphs D , D , such that there exists u i ∈ V ( D i ) that is not adjacent toany vertex in D − i for i = 1 ,
2. As δ − ( D ) ≥ d − D i ( u i ) ≥
3. Therefore | V ( D ) | , | V ( D ) | ≥ D nor D is isomorphic with W or W (see Figure 3). Wewill now construct in D an out-branching, B + , and a spanning strong subdigraph, Q , whichare arc-disjoint, as follows. To start this construction, consider the following three cases for i = 1 , Case 1. There are at least two vertices of in-degree one in D i . As | V ( D i ) | ≥ D i is a strong semicomplete digraph, we note that there are exactlytwo vertices, r i and r i , of in-degree one (there can be at most 3 vertices of in-degree one ina semicomplete digraph and if there were 3 such vertices in D i , then it would not be strong).By Theorem 19 and the fact that D i is not isomorphic to W , there exists an out-tree T + r i rooted at r i and spanning V ( D i ) \ { r i } such that D i \ A ( T + r i ) is strongly connected. Add thearcs of T + r i to B + and add the arcs of D i \ A ( T + r i ) to Q . As δ − ( D ) ≥ D − i , r i )-arcs and at least two ( D − i , r i )-arcs in D . Case 2. There is exactly one vertex, r i , of in-degree one in D i . As D i is not isomorphic to W , Theorem 19 implies that there is a non-separating out-branching, B + r i , in D i , rooted at r i . In this case add the arcs of B + r i to B + and the remainingarcs of D i to Q . As δ − ( D ) ≥ D − i , r i )-arcs in D . Case 3. δ − ( D i ) ≥ . As | V ( D i ) | ≥
4, then Theorem 19 (e) implies that D i admits a nice decomposition( S i , S i , . . . , S ip i ), and for every r ′ ∈ S i the digraph D i contains a non-separating branch-ing, rooted at r ′ . As λ ( D ) ≥
2, there must be a ( D − i , S i )-arc, ur i , in D . Let B + r i be anon-separating branching, rooted at r i in D i . Add the arcs of B + r i to B + and the remainingarcs of D i to Q .This completes our three cases. Note that Q contains a strong spanning subdigraph of D and of D . Furthermore in cases 2 and 3, B + contains an out-branching of D i rooted at avertex r i , such that there exists a ( D − i , r i )-arc in D . In Case 1, B + consists of an out-tree,10ooted at r i , containing all vertices of D i except r i , such that both r i and r i have at leasttwo arcs into them from D − i . We now consider the following possibilities.We were in Case 2 or 3 for both D and D . Add an arc from D to the root of the out-branching of D to B + . As λ ( D ) ≥
2, we can add a further ( D , D )-arc and a ( D , D )-arcto Q , in order for B + and Q to fulfill the desired properties.We were in Case 2 or 3 for D and Case 1 for D . Add an arc from D to r and to r .As there were at least two ( D , r )-arcs in D we can add a further ( D , r )-arc to Q and as λ ( D ) ≥
2, we can add a ( D , D )-arc to Q . Now B + and Q fulfill the desired properties.We were in Case 2 or 3 for D and Case 1 for D . This case is analogous to the previouscase.We were in Case 1 for both D and D . Add an arc from V ( D ) \ { r } to r and to r to B + (which is possible as r and r have at least two arcs into them from D ). Also addan arc from V ( D ) \ { r } to r to B + . Now B + is an out-branching rooted at r in D . Asthere are at least two arcs into r i and into r i from D − i we note that there are at least four( D , D )-arcs and at least four ( D , D )-arcs in D . We can therefore add a ( D , D )-arc anda ( D , D )-arc to Q such that Q and B + are arc-disjoint. Now B + and Q fulfill the desiredproperties, completing the proof of the theorem.Let us recall Theorem 8. Theorem 8.
Let D be a -arc-strong digraph with α ( D ) ≤ . If either of the followingstatements hold then there exists an out-branching, B + , in D , such that D \ A ( B + ) is stronglyconnected. (i): δ − ( D ) ≥ , or (ii): δ − ( D ) ≥ and D is oriented (has no 2-cycle).Proof. Let D be a 2-arc-strong digraph with α ( D ) ≤
2. By Theorem 19 we may assumethat α ( D ) = 2. By Lemma 21, we may assume that D has a strong spanning subdigraph H satisfying one of the conditions (B1)-(B3) in Case B of Corollary 18. Let D ′ = D \ A ( H ).If D ′ has only one initial strong component, then, by Proposition 9 there is an out-branching in D ′ and the theorem is proved. So we may assume that R ′ , R ′ , . . . , R ′ t are theinitial strong components in D ′ and t ≥
2. For all i ∈ [ t ], let R i = D h V ( R ′ i ) i . We will nowprove the following claims. Claim A: t = 2 and | V ( R ) | , | V ( R ) | ≥ . Furthermore, all in-degrees in D ′ are atleast two, except possibly for one vertex whose indegree is at least one. That is, there exists r ∈ V ( D ′ ) , such that d − D ′ ( r ) ≥ and d − D ′ ( x ) ≥ for all x ∈ V ( D ′ ) \ { r } .In Case (i) we actually have δ − ( D ′ ) ≥ . Proof of Claim A:
First consider Case (i) (when δ − ( D ) ≥ − ( H ) ≤ δ − ( D ′ ) ≥
3, and therefore also δ − D ′ ( R ) ≥
3. This further implies that | V ( R i ) | = | V ( R ′ i ) | ≥
4, since N − D ′ [ x ] ⊆ V ( R ′ i ), for all x ∈ V ( R ′ i ) and i ∈ [ t ]. Now noting that there is atmost one vertex of H whose in-degree is more than 1, we see that R ′ i contains a vertex withat least 4 in-neighbours inside R ′ i so | V ( R i ) | = | V ( R ′ i ) | ≥ D ′ are at least two,except possibly for one vertex whose indegree is at least one. Let n i = | V ( R i ) | and note thatthe number of arcs in R i is at least 2 n i − R i belong to R i ). As D is oriented this implies that (cid:0) n i (cid:1) ≥ | E ( V i ) | ≥ n i −
1. As (cid:0) n i (cid:1) < n i − n i ∈ [4], wemust have n i ≥
5, which implies that | V ( R i ) | ≥ t ≥
3. Let x ∈ V ( R ) be an arbitrary vertexwith d + H ( x ) = d − H ( x ) = 1 and let N be the set of the two neighbours of x in H . As | V ( R ) | ≥ V ( R ) that are not in N . By part (B) inCorollary 18 we note that at most two vertices in H have degree more than two, so we cancan choose a vertex x ∈ V ( R ) \ N such that d + H ( x ) = d − H ( x ) = 1. Let N be the set of thetwo neighbours of x in H . Now there exists a vertex x ∈ V ( R ) \ ( N ∪ N ) which impliesthat { x , x , x } is an independent set in D , contradicting α ( D ) ≤
2. Therefore t = 2, whichcompletes the proof of Claim A. Claim B: R and R are semicomplete digraphs. Furthermore, for all z ∈ V ( D ) thefollowing holds, | N + H ( z ) ∩ V ( R ) | , | N + H ( z ) ∩ V ( R ) | , | N − H ( z ) ∩ V ( R ) | , | N − H ( z ) ∩ V ( R ) | ≤ Proof of Claim B:
For the sake of contradiction assume that x , y ∈ V ( R ) and x and y are non-adjacent in D . Let N = ( N + H ( x ) ∪ N − H ( x ) ∪ N + H ( y ) ∪ N − H ( y )) ∩ V ( R ).If V ( R ) N , then let z ∈ V ( R ) \ N and note that { x , y , z } is independent in D ,contradicting that α ( D ) ≤
2. So, V ( R ) ⊆ N , which implies that | N | ≥ | V ( R ) | ≥
5, byClaim A. By Corollary 18, we note that d H ( x ) + d H ( y ) ≤
6, which implies the following, ≥ | N + H ( x ) ∩ V ( R ) | + | N − H ( x ) ∩ V ( R ) | + | N + H ( y ) ∩ V ( R ) | + | N − H ( y ) ∩ V ( R ) | ≥ | N | ≥ First consider the case when | N + H ( x ) ∩ V ( R ) | ≥
2. By the construction of H we notethat | N + H ( x ) ∩ V ( R ) | = 2, so let N + H ( x ) ∩ V ( R ) = { x , y } . By Corollary 18, x and y are non-adjacent in D . As α ( D ) = 2 we note that either x or y has to be adjacent to y .Without loss of generality assume that y is adjacent to y . As y is adjacent to both x and y in D we note that we must have d H ( x ) + d H ( y ) = 6, as H satisfies one of (B2), (B3) inCorollary 18 (and | N | = | V ( R ) | = 5).If | N − H ( x ) ∩ V ( R ) | ≥ | N + H ( y ) ∩ V ( R ) | ≥ | N − H ( y ) ∩ V ( R ) | ≥ d H ( x ) + d H ( y ) = 6 and there exists two non-adjacent vertices x and y in R .If we had considered { x , y } instead of { x , y } then we would analogously have obtain d H ( x ) + d H ( y ) = 6. However, by part (B) in Corollary 18 we note that it is not pos-sible to have vertex-disjoint sets, { x , y } and { x , y } , such that d H ( x ) + d H ( y ) = 6 = d H ( x ) + d H ( y ). Therefore R is semicomplete. Analogously we can show that R is also asemicomplete digraphs.Let z ∈ V ( D ) be arbitrary. For the sake of contradiction assume that | N + H ( z ) ∩ V ( R ) | ≥ | N + H ( z ) ∩ V ( R ) | = 2, so let N + H ( z ) ∩ V ( R ) = { x , y } . ByCorollary 18 x and y are non-adjacent in D . This contradicts the fact that R is semicom-plete. Therefore | N + H ( z ) ∩ V ( R ) | ≤
1. Analogously | N + H ( z ) ∩ R | , | N − H ( z ) ∩ R | , | N − H ( z ) ∩ R | ≤
1, which completes the proof of Claim B.
Claim C:
Let y ∈ V ( D ) \ ( V ( R ) ∪ V ( R )) be arbitrary. If y has at most one arc enteringit from V ( R ) in D , then y is adjacent to all vertices in V ( R ) and furthermore has at leastfour arcs entering it from V ( R ) .Analogously, if y has at most one arc entering it from V ( R ) in D , then y is adjacent toall vertices in V ( R ) and furthermore has at least four arcs entering it from V ( R ) . he above implies that every vertex in V ( D ) \ ( V ( R ) ∪ V ( R )) has at least four arcsentering it from V ( R ) ∪ V ( R ) in D . Proof of Claim C:
Assume that y ∈ V ( D ) \ ( V ( R ) ∪ V ( R )) and y has at most one arcentering it from V ( R ) in D . For the sake of contradiction assume that there exists r ∈ V ( R )which is non-adjacent to y (in D ). By Claim B, we note that | N + H ( r ) ∩ V ( R ) | , | N − H ( r ) ∩ V ( R ) | , | N + H ( y ) ∩ V ( R ) | , | N − H ( y ) ∩ V ( R ) | ≤
1. As R ′ and R ′ are initial strong componentsin D ′ , and therefore all arcs between r and R in D belong to H , we note that r is adjacentto at most two vertices in R (in D ). As y has at most one arc entering it from V ( R ) in D and at most one arc from y to R in H (and therefore also in D , as R ′ is an initial strongcomponents in D ′ ), we note that y is adjacent to at most two vertices in R (in D ). As | V ( R ) | ≥
5, by Claim A, this implies that there exists a r ∈ V ( R ) which is not adjacentto r or y in D , contradicting α ( D ) = 2. Therefore y is adjacent to every vertex in R .As R ′ was an initial strong component in D ′ and y V ( R ′ ) we note that every arc from y to R in D belongs to H . By Claim A and Claim B, we note that | N + H ( y ) ∩ V ( R ) | ≤ | V ( R ) | ≥
5, which implies that there are at least four arcs from V ( R ) to y in D . Thiscompletes the first part of the proof of Claim C. The second part is proven analogously (byswapping the names of R and R ).Let x ∈ V ( D ) \ ( V ( R ) ∪ V ( R )) be arbitrary. If x has less than two arcs entering itfrom R i then it has four arcs entering it from R − i ( i ∈ [2]). And if x has at least two arcsentering it from both R and from R , then it also has at least four arcs entering it from V ( R ) ∪ V ( R ). This completes the proof of Claim C. Construction of R ∗ . Initially let R ∗ = R . Now for every u ∈ V ( D ) \ ( V ( R ∗ ) ∪ V ( R ))with at least one arc into R ∗ in D and at most one arc from R to u , add u to R ∗ . Continuethis process until no further vertex can be added. R R v v v l R ∗ · · · At most one arc from R to v i for each i Figure 5: An illustration of the construction of R ∗ , where R ∗ is constructed from R by addingthe vertices v , v , . . . , v l in that order. Every v i has at least one arc into R ∪{ v , v , . . . , v i − } .By Claim C there are at least four arcs from R into v i for each i ∈ [ l ]. Claim D: R ∗ is a strong semicomplete digraph. Proof of Claim D:
Let Q = V ( R ∗ ) \ V ( R ). That is, Q denotes the set of verticesadded to R in the construction of R ∗ . By construction, when it was added to the current R ∗ each such vertex had at least one arc into the current set V ( R ∗ ) and at most one arcentering it from V ( R ) in D . We will first show that R ∗ is semicomplete. Assume for the13ake of contradiction that q , q ∈ V ( R ∗ ) are non-adjacent in D . By Claim B we note that q and q cannot both belong to V ( R ). By part 2 of Claim C we note that we cannothave q ∈ V ( R ) and q ∈ Q (or vice versa), which implies that we must have q , q ∈ Q .This implies that q and q both have at most one arc into them from V ( R ). By Claim B, | N + H ( q ) ∩ V ( R ) | , | N + H ( q ) ∩ V ( R ) | ≤
1, which implies that | N + D ( q ) ∩ V ( R ) | , | N + D ( q ) ∩ V ( R ) | ≤
1. Therefore each of q and q are adjacent to at most two vertices in R . As | V ( R ) | ≥
5, there is a vertex r ∈ V ( R ) which is non-adjacent to both q and q in D ,contradicting that α ( D ) ≤
2. Therefore R ∗ is semicomplete.As R is strongly connected and every vertex we add in the process of building R ∗ hasan arc into the current set R ∗ and an arc (actually at least 4 arcs) entering it from R (and R ⊆ R ∗ ), by Claim C, we note that the current set R ∗ is strongly connected in every step ofthe construction. Therefore the final R ∗ is also strongly connected. This completes the proofof Claim D. Definitions.
By Claim A and D and Proposition 14 we note that R ∗ has a nice decom-position ( S , . . . , S p ). If p ≥ S in R ∗ so let uv be anarc entering S in D , which does not belong to R ∗ . Such an arc exists as D is 2-arc-strong.If p = 1 then R ∗ is 2-arc-strong. In this case let uv be any arc entering R ∗ in D . Let D ∗ = D ′ − uv (that is, delete the arc uv from D ′ ). Claim E:
There exists an out-branching B +1 in R ∗ rooted at v , such that R ∗ − A ( B +1 ) isstrongly connected.Furthermore there exists an out-branching B +2 in R , such that R − A ( B +2 ) is stronglyconnected. Proof of Claim E: As R ∗ is strongly connected by Claim D, We can apply Theorem 19to R ∗ . By Claim A we note that | V ( R ∗ ) | ≥ R ∗ is not isomorphic to W .Also, by Claim A we note that all vertices of R , except possibly one, have in-degree atleast two in R . As every vertex we add to R in the construction of R ∗ have in-degree atleast four (from R , by Claim C), we note that all vertices of R ∗ , except possibly one, havein-degree at least two in R ∗ . Therefore we are in case (c)-(e) of Theorem 19.Note that if there is a vertex of in-degree one in R ∗ , then v is that vertex. Furthermore v ∈ S (where S was defined just above Claim E, as part of the nice decomposition on R ∗ ).Therefore by Theorem 19 we note that R ∗ contains a non-separating out-branching rootedat v , which completes the first part of the proof of Claim E.The second part of Claim E, follows analogously, by Theorem 19. Completion of the proof.
Let B + = B +1 ∪ B +2 (defined in Claim E) and let Q =( R ∗ − A ( B +1 )) ∪ ( R − A ( B +2 )) and let V ∗ = V ( R ∗ ) ∪ V ( R ). Note that B + consists oftwo vertex-disjoint out-trees (whose union span V ∗ ) and Q of two vertex-disjoint strongcomponents (whose union also span V ∗ ). Let P be a path from V ( R ∗ ) to V ( R ) in D ∗ and let P be a path from V ( R ) to V ( R ∗ ) in D ∗ Add the arcs of P and P to Q . Forevery vertex p ∈ ( V ( P ) ∪ V ( P )) \ V ∗ do the following. Add an arc, which is not in A ( P ) ∪ A ( P ), from V ( R ∗ ) ∪ V ( R ) to p to B + . This is possible by Claim C and the factthat there are at most two arcs in A ( P ) ∪ A ( P ) leaving V ( R ∗ ) ∪ V ( R ) (at most one leaves V ( R ∗ ) and at most one leaves V ( R )). Furthermore if p = u (recall that the arc uv wasdefined above Claim E), then make sure the added arc leaves V ( R ) (and not V ( R ∗ )), whichis possible as u was not added to R ∗ and therefore has at least two arcs into it from V ( R )(here we used that the arc uv enters R ∗ ). Finally add p to V ∗ .When this process is completed V ∗ = V ( R ∗ ) ∪ V ( R ) ∪ V ( P ) ∪ V ( P ), Q induces astrong subgraph on the vertex set V ∗ and B + still consist of two out-trees also spanning V ∗ ,14ne of which is rooted at v . Furthermore the arc uv is not used above and if u ∈ V ∗ then itbelongs to the out-tree not rooted at v . We now add the remaining vertices as follows. While V ∗ = V ( D ) let P ′ be any ( V ∗ , V ∗ )-path in D ∗ with at least one internal vertex. We canconstruct P ′ by letting p p be any arc out of V ∗ in D ∗ and then taking any path from p back to V ∗ in D ∗ . Now add A ( P ′ ) to Q and for every vertex p ∈ V ( P ′ ) \ V ∗ do the following(analogously to above). Add an arc, which is not in A ( P ′ ), from V ( R ) ∪ V ( R ) to p to B + ,which is possible by Claim C and the fact that there is at most one arc in A ( P ′ ) leaving V ( R ) ∪ V ( R ). Furthermore if p = u (recall that the arc uv was defined above Claim E),then make sure the added arc leaves V ( R ) (and not V ( R ∗ )), which is possible as u was notadded to R ∗ and therefore has at least two arcs into it from V ( R ) (here we used that thearc uv enters R ∗ ). Finally add p to V ∗ .We continue the above process until V ∗ = V ( D ). Now Q is a strong spanning subgraphof D , which does not include any arcs from B + and also does not include the arc uv . B + consists of two out-trees, one rooted at v and u belonging to the out-tree which was notrooted at v . Therefore by adding the arc uv to B + we obtain a spanning out-branching of D which is arc-disjoint to Q , thereby completing the proof. Let us recall that for a subdigraph H of D , we denote by D − H the sudigraph of D obtainedfrom removing the vertices of H from D , that is D − H = D [ V ( D ) \ V ( H )]. Furthermore wedenote by D \ A ( H ) the sudigraph of D obtained from removing the arcs of H from D , thatis V ( D − H ) = V ( D ) and A ( D − H ) = A ( D ) \ A ( H ).A spanning tree T of a connected digraph D is safe if for every pair of distinct vertices x and y of D , there exists an oriented path from x to y in D if and only if there exists also anoriented path from x to y in D \ A ( T ). In particular, a safe spanning tree of a strong digraphis a non separating spanning tree.At several places, we use the following fact. Assume that H is an induced subdigraph of D such that H admits a safe spanning tree T , and assume also that there exist u, v ∈ H and x ∈ D \ H such that ux, vx ∈ A ( D ) and that there exist a path from u to v in H . Then D [ V ( H ) ∪ x ] admits the safe spanning tree T + ux . Indeed, there exists also a path from u to v in H \ A ( T ) and thus a path from u to x in D [ V ( H ) ∪ x ] \ ( A ( T ) ∪ { ux } ).First, we derive some results on safe spanning trees of semicomplete digraphs. Lemma 22.
Every semicomplete digraph on at least five vertices admits a safe spanning tree.Proof.
Let D be a semicomplete digraph on at least five vertices. If D is strong, as D containsat least five vertices, then we find a spanning tree in the complement of a Hamiltonian cycleof D . This spanning tree is clearly safe.So, assume that D is not strong and denote by C , C , . . . , C t the strongly connectedcomponents of D such that there is no arc from C i to C j if i > j . We denote by K thesubdigraph of D containing the vertices V ( D ) and the set of arcs of D connecting its strongcomponents (that is the arcs uv with u ∈ C i , v ∈ C j and i = j ). Moreover, for every i = 1 , . . . , t let x i be a vertex of C i and P be the path x . . . x t . If there exists a spanningtree T of D all of whose arcs are in the subdigraph K ′ = K \ A ( P ), then T is a safe spanningtree of D . Thus we have to check that K ′ is a connected subdigraph of D .First, if there exist i and j such that | C i | ≥ | C j | ≥
2, then we pick a vertex y i in C i different from x i and a vertex y j in C j different from x j . In K ′ , every vertex not in C i is15djacent to y i and every vertex in C i is adjacent to y j . So, K ′ is connected in this case.Moreover, if t ≥ K ′ every vertex of D \ C ∪ C is adjacent to x and everyvertex of D \ C t − ∪ C t is adjacent to x t . So, K ′ is connected.Thus we may assume that t ≤ i ∈ { , . . . , t } such that for every i = i , the component C i has size 1 exactly. If t = 3, then, as D contains at least 5 vertices,we have | V ( C i ) | ≥
3. If i = 1, then K ′ contains all the arcs from V ( C ) − x to { x , x } and the arc x x . So K ′ is connected. The case i = 3 is symmetrical.So we may assume that i = 2. Let y ∈ V ( C ) \ { x } be arbitrary and let K ∗ = K \ { x x , y x } . As every vertex in V ( C ) is adjacent to x or x in K ∗ and x x ∈ A ( K ∗ )we note that K ∗ is a spanning connected subgraph of D and we can therefore in K ∗ find asafe spanning tree of D .Finally, if t = 2, by symmetry again we can assume that i = 1. We have | V ( C ) | ≥ C contains an arc xy such that C \ xy is strongly connected. In this case the arcsfrom V ( C ) \ { x } to x plus the arc xy form the arcs of a safe spanning tree of D .The following claim will be useful in the proof of the main theorem of the section. Claim 22.1.
Let D be a digraph with λ ( D ) ≥ . If D contains a tree T such that D \ A ( T ) is strongly connected and V ( D ) \ V ( T ) has size at most two and induces a semicompletedigraph, then D admits a non-separating spanning tree.Proof. Let D and T as stated, and let C be a terminal strong component of D [ V ( T )] \ A ( T ).Suppose first that V ( D ) \ V ( T ) = { x } . As λ ( D ) ≥
2, there are at least two arcs from C to x . Let be u an in-neighbour of x in C . The digraph D \ ( A ( T ) ∪ { ux } ) is still strong andso T + ux is a non-separating spanning tree of D .Now, assume that V ( D ) \ V ( T ) = { x, y } . If there are two arcs ux and vx from C to x , then T ′ = T + ux is a tree on n − D \ A ( T ′ ) is strongly connectedand we can conclude from the previous case that D admits a separating spanning tree. So,as λ ( D ) ≥
2, there are at least two arcs from C to { x, y } and we can assume that one, say ux , has head x and the other, say vy , has head y . Similarly, we can assume by the previouscase, that if C ′ is an initial strong component of D [ V ( T )] \ A ( T ), then there exist two arcs xu ′ and yv ′ with u ′ , v ′ ∈ C ′ . As D [ { x, y } ] is semicomplete, we can assume without loss ofgenerality that xy is an arc of D . Now it is easy to check that D \ ( A ( T ) ∪ { vy, xu ′ } ) isstrongly connected and that D admits the non-separating spanning tree T + vy + xu ′ .Now we can prove the following. Theorem 23.
Every digraph D = ( V, A ) with α ( D ) ≤ ≤ λ ( D ) such that D contains asemicomplete digraph on at least 5 vertices has a non separating spanning tree. In particular,every digraph D = ( V, A ) with α ( D ) ≤ ≤ λ ( D ) such that | V | ≥ has a non-separatingspanning tree.Proof. If D is semicomplete, then the result follows from Lemma 22. So we may assumethat α ( D ) = 2. As the Ramsey number R (3 ,
5) is 14 [15] and α ( D ) = 2, it follows that if | V | ≥
14, then D contains a semicomplete subdigraph on five vertices. Hence we may assumebelow that D is a semicomplete digraphs of D on 5 vertices. By Lemma 22, D contains asafe spanning tree. So let R be a maximal induced subdigraph of D containing V ( D ) andadmitting a safe spanning tree. We now show that R = D . Suppose for a contradiction thatthis is not the case.Let T be a safe spanning tree of R and consider a vertex x of S = D [ V \ V ( R )]. Thevertex x has at most one in-neighbour in D . Indeed, otherwise, assume that y and z are two16n-neighbours of x in D with yz being an arc of D (recall that D is semicomplete). Butthen, T + yx would be a safe spanning tree of D [ V ( R ) ∪ x ], a contradiction to the maximalityof R . Similarly, x has at most one out-neighbour in D . So we can conclude that S isa semicomplete subdigraph of D . Indeed, otherwise, S would contain an independent set { u, v } of size two. But as u and v have each at most two neighbours in D , there would existin D a vertex not adjacent to any of u or v , contradicting α ( D ) = 2.First, assume that S contains a safe spanning tree T ′ and denote by C a strong terminalcomponent of R . As λ ( D ) ≥
2, there exist at least two arcs xu and yv from C to S (with x, y ∈ C and u, v ∈ S ). If u = v , then there is an arc between u and v as S is semicomplete.Without loss of generality assume that uv is an arc of S and let e be the arc yv . If u = v then we can choose arbitrarily e = xu or e = yv . In both cases T + T ′ + e is a safe spanningtree of D , contradicting the maximality of R .So, S has no safe spanning tree and as S is semicomplete, it follows from Lemma 22, that | V ( S ) | ≤
4. We also have | V ( S ) | >
1, as a unique vertex always has a safe spanning tree.Thus, to conclude the proof of the Lemma, we have three cases to handle: | V ( S ) | ∈ { , , } .Assume first that S contains two vertices. Then, D \ A ( T ) is strong, D \ T has size twoand is semicomplete. So, by Claim 22.1, D has a separating spanning tree, a contradictionagain to the maximality of R .Now assume that S contains three vertices, and denote by C a strong terminal componentof R . As previously, as λ ( D ) ≥
2, there exist at least two arcs xu and yv from C to S (with x, y ∈ C and u, v ∈ S ). If u = v , as S is semicomplete there is an arc between u and v .Without loss of generality assume that uv is an arc of S and let e be the arc yv . If u = v thenwe can choose arbitrarily e = xu or e = yv . To conclude, denote T + e by T ′ and notice that D \ A ( T ′ ) is strongly connected. As D \ T ′ has size two and is semicomplete, by Claim 22.1, D has a separating spanning tree, a contradiction again to the maximality of R .Finally, assume that S contains four vertices. As in the previous case, we can find an arc e = zw with z ∈ R and w ∈ S such that D \ ( A ( T ) ∪ { e } ) is strongly connected. As S as fourvertices, w has in-degree or out-degree at least 2 in S . Assume that w has out-degree at least2 and denote by u and v two out-neighbours of w in S such that uv is an arc of D . So, if weremove the arc wv from the digraph D \ ( A ( T ) ∪ { e } ), the resulting digraph still contains thepath wuv from w to u and so is still strongly connected. That is, if we denote by T ′ the tree T + e + wv , the digraph D \ A ( T ′ ) is strongly connected and D − T ′ is semicomplete and hassize two. Thus by Claim 22.1, D has a separating spanning tree, a contradiction again to themaximality of R . The case when w has in-degree at least 2 in S is analogous. abc xyz a bc x yz Figure 6: Two different drawings of the same 2-arc-strong co-bipartite digraph ˜ D in whichevery spanning tree is separating. 17 roposition 24. The digraph ˜ D in Figure 6 has no non-separating spanning tree.Proof. Note that H has 12 arcs and 6 vertices so if H would have a pair of arc-disjoint sub-digraphs T, S where T is a spanning tree and S a strong spanning digraph, then | A ( S ) | ≤ S is either a hamiltonian cycle of H or it consist of a cycle C and a ( C, C )-path P which picks up the remaining vertices of V . Now note that the onlycycle lengths of D are 3 and 6. We now use that H has a number of automorphisms: thereare 4 pairs of 3-cycles joined by a hamiltonian cycle on their vertices, namely ( abca, xyzx ),( ayza, bcxb ), ( abza, cxyc ), ( ayca, bzxb ). Hence up to automorphisms there is only one hamil-tonian cycle, namely C = abcxyza . It is easy to check that D \ A ( C ) is not connected.Hence, if T, S exist then we must have | A ( S ) | = 7 and S must consist of a 3-cycle C anda ( C, C )-path P which picks up the remaining 3 vertices of V . By the symmetries above,we may assume that C = abca . Again, by permuting the vertices a, b, c if necessary, we canassume that P starts with the arc cx . This implies that P = cxyza (as P picks up all thevertices x, y, z ). Now we see that S contains the hamiltonian cycle abcxyza and we saw abovethat removing the arcs of this cycle we disconnect the graph. v v v v v v v v Figure 7: A 2-arc-strong digraph ˆ D with α ( ˆ D ) = 2 in which every spanning tree is separating. Proposition 25.
The digraph ˆ D in Figure 7 has no non-separating spanning tree.Proof. As every vertex of ˆ D has in- and out-degree 2 we see that if T, S is a pair of arc-disjoint spanning subdigraphs of ˆ D such that T is connected and S is strongly connected,then T must be a hamiltonian path in ˆ D (as d + S ( x ) , d − S ( x ) ≥ x ∈ V ( ˆ D ) andtherefore d + T ( x ) , d − T ( x ) ≤ T = p p . . . p . Let C denote the arcs on the hamiltoniancycle, v v . . . v v and let ¯ C = A ( ˆ D ) \ C . We first prove the following statement. (i) p i p i +1 , p i +1 p i +2 ∈ C is not possible for any i ∈ [6].For the sake of contradiction, assume the above is true and without loss of generalitythat p i = v . That is, v v v is a subpath of T . Continuing along T out of v andinto v we note that all arcs of T belong to C (i.e. we cannot use the arc v v in T , sothe only possible arc out of v is v v and the only possible arc into v is v v , etc.).However S would then contain two disjoint 4-cycles plus an extra arc, meaning it is notstrongly connected, a contradiction.As T is a hamiltonian path we note that p i p i +1 ∈ C for some i ∈ { , , , } (otherwise T contains a 4-cycle). By (i) we must have p i − p i , p i +1 p i +2 ∈ ¯ C . Without loss of generalityassume that p i = v , which implies that v v v v is a subpath of T . As T is a path (and i ≤
5) we must have that v v v v v is a subpath of T (as v v A ( T )). This implies that18he arcs v v , v v , v v ∈ A ( S ) (as d + S ( v ) , d − S ( v ) , d + S ( v ) , d − S ( v ) ≥ S has to containarcs into and out of { v , v , v } we must have v v , v v ∈ A ( S ). But now all arcs incidentwith v are in S , a contradiction.By Proposition 25, the following Conjecture would be best possible in terms of the numberof vertices. Conjecture 26.
Every digraph D on at least 9 vertices with λ ( D ) = 2 and α ( D ) ≤ Theorem 27.
Every hamiltonian oriented graph D = ( V, A ) on at least 9 vertices with λ ( D ) ≥ and α ( D ) = 2 has a non-separating spanning treeProof. Let C be a hamiltonian cycle of D and let X , X , . . . , X k be the vertex sets of theconnected components of H = U G ( D ) \ A ( C ). If k = 1 we are done, so assume that k ≥ d H ( v ) ≥
2, and D contains no 2-cycle.Suppose first that k ≥ v ∈ X i for some i ∈ [ k ]. In U G ( D ) thevertex v has at most 2 neighbours outside X i . If v has no neighbours in some X q , q = i , thenlet w be a non-neighbour of v in X j , j
6∈ { i, q } and let z be an arbitrary non-neighbour of w in X q . Then { v, w, z } is an independent set, contradiction. Therefore k = 3 and every vertexin X i has a neighbour in each of the other sets X j , implying that we can pick x i ∈ X i , i ∈ [3]such that { x , x , x } is an independent set, contradiction.Hence k = 2 and we may assume that | X | ≥ | X | . As | V | ≥ | X | ≥ X and X belongs to C this implies that X inducesa complete subgraph of U G ( D ) (every vertex of X must be adjacent at least one of thevertices x, y in a pair of non-adjacent vertices x, y ∈ X so if such a pair existed, at leastone of x, y would be incident to 3 arcs of C , contradiction). Hence, by Theorem 23, we canassume that | X | ≤ D [ X ] is not semicomplete. Note that for every pair of vertices u, v ∈ X such that D has no arc between these, every vertex of X has at least one edge to { u, v } in A ( C ). Also note that δ ( D [ X i ]) ≥ i ∈ [2] as only the arcs of C go between X and X .First suppose that | X | = 4. If X contains vertices u , u , v , v all distinct exceptpossibly u = v so that there is no arc in D between u i and v i for i = 1 ,
2, then we getthe contradiction that the undirected graph induced by A ( C ) contains either a 4-cycle oran 8-cycle, contradicting the C is a hamiltonian cycle of D . Thus X contains exactly onepair u, v of non-adjacent vertices and, by Theorem 23, we may assume that | X | = 5 so D has 9 vertices. As all 4 vertices of X are adjacent to either u or v , exactly two of themare adjacent to u and the other two are adjacent to v . Now it is easy to see that C has atmost one arc inside X (otherwise C would contain a 3-cycle or a 6-cycle as a subdigraph).Consider first the case when C uses no arc inside X . Then we can label the vertices of V such that X = { v , v , v , v , v } , X = { v , v , v , v } and C = v v . . . v v .19 v v v v v v v v X X (a) v v v v v v v v v X X (b)Figure 8: Illustrating two cases in the proof when | X | = 4. In (a) we illustrate the solutionwhen v v is an arc. The blue arcs form a strong spanning subdigraph S and the red edges,together with a spanning tree T ′ in D [ X ], avoiding v v and the blue arc into v form aspanning tree T which is edge-disjoint from S . In (b) we indicate a solution when D containsthe directed path v v v v .Suppose first that D contains the arc v v . Then let wv be an arbitrary arc entering v in D [ X ] (this exists as δ ( D [ X ]) ≥
1) and let T ′ be a spanning tree avoiding the arcs wv , v v in G ′ = U G ( D [ X ]). This tree exists as G ′ \ { wv , v v } has 5 vertices and 7 edgesand hence is connected. Then we obtain a strong spanning subdigraph S of D from C bydeleting the arc v v and adding the arcs v v , wv and note that S is arc-disjoint from thespanning tree formed by T ′ and the edges of the path v v v v v in U G ( D ), see Figure 8(a).Hence we can assume that v v ∈ A ( D ) and by an analogous argument we can assumethat v v , v v ∈ A ( D ). Now let v w be an arbitrary arc leaving v in D [ V ( X )], let T ′′ be aspanning tree avoiding the arcs v v , v w in G ′ and let S ′ be the strong spanning subdigraphof D obtained from C by deleting the arc v v and adding the arcs of the directed path v v v v and the arc v w and note that S ′ is arc-disjoint from the spanning tree formed by T ′′ and the edges of the path v v v v (in U G ( D )) and the arc v v . See Figure 8(b).Next we consider the case when C contains one arc of D [ V ( X )]. In this case, wemay assume that v and v are the two vertices in X that are non-adjacent in D and X = { x , x , x , x } such that x x x and x x x are subpaths of C . We may furthermoreassume without loss of generality that x x ∈ A ( C ) (the case when x x ∈ A ( C ) is iden-tical, by renaming vertices). This implies that we can label C as C = v v . . . v v , and X = { v , v , v , v , v } and X = { v , v , v , v } (and v and v are non-adjacent in D ).If v v is an arc of D [ V ( X )], then let j ∈ [3] be chosen such that v j is an out-neighbourof v in D [ V ( X )] and let j ′ ∈ [3] \ { j } be arbitrary. Now the strong spanning subdigraph S consisting of the cycle v v . . . v v v and the path v v v j is arc-disjoint from the spanningtree using the edges v v , v v , v v , v v , v v j ′ , v v , v v , v v . Hence we can assume that v v is an arc of D [ V ( X )]. A similar argument shows that we may assume that v v is anarc of D [ V ( X )]. Now using that δ ( D ) ≥ D [ V ( X )]are v v , v v and v v . See Figure 9(a). 20 v v v v v v v v X X (a) v v v v v v v v v X X (b)Figure 9: Illustrating the two last cases in the proof when | X | = 4.Now choose p ∈ [3] such that v p v is an arc of D [ V ( X )] and let S ′ be the strong spanningsubdigraph of D formed by the arcs of the cycle v v v v v v and the path v p v v v v v . Let q ∈ [3] \ { p } be arbitrary and note that D \ A ( S ′ ) is connected as it contains the spanningtree on the edges v v , v v , v v , v v , v v , v v , v v , v v q , see Figure 9(b). This completesthe case when | X | = 4.Consider now the case when | X | = 3. Recall that X induces a 3-cycle in U G ( D ). Let G be the complement of U G [ X ]. That is, V ( G ) = V ( X ) and uv ∈ E ( G ) if and only if u and v are non-adjacent in D . By Theorem 23 we may assume that α ( G ) ≤ α ( D ) = 2 wemay assume that G contains no 3-cycle. As | V ( G ) | ≥ G . Let uv and u ′ v ′ be two edges in a matching in G .Note that at least three arcs between X and { u, v } belong to C (as otherwise there wouldbe an independent set of size 3 containing u and v ). There are also at least three arcs between X and { u ′ , v ′ } in C . As | X | = 3 these 6 arcs are all the arcs between X and X . Withoutloss of generality assume that u is incident to two arcs between X and X and u ′ is alsoincident to two arcs between X and X , which implies that both v and v ′ are incident withexactly one arc between X and X . As α ( D ) = 2 and α ( G ) ≤ | X | = 6and E ( G ) = { uv, u ′ v ′ } or E ( G ) = { uv, u ′ v ′ , uu ′ } .Let u = u , u = v , u = u ′ and u = v ′ . We can now without loss of generality, labelthe vertices of V by u , . . . , u such that X = { u , u , u , u , u , u } , X = { u , u , u } , C = u u . . . u u and there is no arc between u and u and no arc between u and u .There may or may not be an arc between u and u . See Figure 10.21 u u u u u u u u X X Figure 10: The hamiltonian cycle C in D when | X | = 3. The dotted edges indicate the twopairs of non-adjacent vertices in D [ X ].Moreover, as δ + ( D ) ≥ D is oriented, we know that D [ X ] is a directed 3-cycle. If D contains the arc u u , then as above we can find the desired pair S, T , see Figure 11(a).Otherwise, it means that D [ X ] is the directed 3-cycle u u u u , and then, as above wecan find the desired pair S, T , see Figure 11(b). u u u u u u u u u X X u u u u u u u u u X X (a) (b)Figure 11: In (a): strong spanning subdigraph (in blue) and spanning tree (in red) when D contains the arc u u . In (b): strong spanning subdigraph (in blue) and spanning tree (inred) when D contains the 3-cycle u u u u Note that Theorem 15 implies that every 2-arc-strong semicomplete digraph D different from S has an out-branching B + such that D \ A ( B + ) is strong. It is easy to check that S alsohas such an out-branching. The purpose of this section is to prove that there exists 2-arc-strong digraphs with independence number 2 for which no hamiltonian path is non-separating.22or every natural number r ≥ T r = ( V, A ) be the tournament with vertex set { u , u , . . . , u r +1 , v , v , . . . , v r +1 } and arc set { u i − u i | i ∈ [ r ] } ∪ { v i v i +1 | i ∈ [ r ] } ∪ { u i v i | i ∈ [ r ] } ∪ { v v , v u , v u , u v } ∪ { u r +1 u r , v r +1 u r +1 , u r v r +1 , v r u r +1 } and for all remaining pairsnot mentioned above the arcs goes from the vertex of higher index to the one with the lowerindex. See Figure 12. u v u v u r v r u r +1 v r +1 u v u r − v r − Figure 12: The tournament T r . The fat arc in the middle indicates that all arcs not shownin the figure go from right to left. Lemma 28.
For every r ≥ the tournament T r is 2-arc-strong. Furthermore if P is ahamiltonian path in T r starting in v then v cannot reach v r in T r \ A ( P ) .Proof. It is easy to check that v has two arc-disjoint paths to every other vertex and thatevery vertex different from v has two arc-disjoint paths to v . This implies that T r is 2-arc-strong. For the sake of contradiction assume that there is a hamiltonian path, P , in T r , andthat v can reach v r in S = T r \ A ( P ). As for every i ∈ [ r −
1] the two arcs u i u i +1 , v i v i +1 form a 2-arc-cut of T r seperating v from v r , one of these arcs must belong to S and theother to P . Similarly, as for every i ∈ [ r −
2] the two arcs u i u i +1 , v i +1 v i +2 form a 2-arc-cutof T r seperating v from v r one of these arcs must belong to S and the other to P . Let A = { u u , u u , . . . , u r − u r } and let A = { v v , v v , . . . , v r − v r } , and note that by theprevious argument we must have A i ⊆ A ( S ) and A − i ⊆ A ( P ) for some i ∈ [2]. Without lossof generality assume that A ⊆ A ( S ) and A ⊆ A ( P ), which implies that P cannot containboth u and u , a contradiction to the existence of P and S .The following corollary follows immediatly from Lemma 28. Corollary 29.
For every r ≥ the tournament T r is 2-arc-strong and for every hamiltonianpath P starting in the vertex v the digraph D \ A ( P ) is not strongly connected. Theorem 30.
There exist infinitely many 2-arc-strong digraphs D with α ( D ) = 2 such thatdeleting the arcs of any hamiltonian path leaves a non-strong digraph.Proof. For each r ≥ T r be the 2-arc-strong tournament defined in Lemma 28 and formthe digraph D r from two copies T r , T r of T r (whose vertices are superscripted) by addingtwo arbitrary arcs from V ( T ir ) to v − i for i = 1 ,
2. Since each T r is a tournament, we have α ( D r ) = 2. Moreover, as T r and T r arc 2-arc strong, D r is 2-arc strongly connected also.Suppose that D r has a hamiltonian path P such that D \ A ( P ) is strong. Without loss ofgenerality P starts in V ( T r ) and thus the restriction of P to V ( T r ) is a hamiltonian path P ′ starting at v . By Lemma 28 we note that v cannot reach v r in T r \ A ( P ′ ), which implies thatno vertex in T r can reach v r in D r \ A ( P ). So D r \ A ( P ) is not strong, a contradiction.23 Non-separating hamiltonian paths in graphs with indepen-dence number 2
In contrast to the result in Theorem 30 above, for the case of undirected graphs of indepen-dence number 2 we have the following positive result on non-separating hamiltonian paths.
Theorem 31.
Let G be a 2-edge-connected graph with δ ( G ) ≥ and α ( G ) ≤ . Then, G contains a spanning tree and a Hamiltonian path which are edge-disjoint.Proof. Let G be a 2-edge connected graph with δ ( G ) ≥ α ( G ) ≤
2. It is easy to seethat every connected graph with independence number at least 2 has a spanning tree witha number of leaves at most its independence number. Hence G contains a Hamiltonian path P . If AY G \ E ( P ) is connected, we are done. Otherwise let X , X , . . . , X p be the connectedcomponents of G \ E ( P ). Notice that as δ ( G ) ≥ P is a path, we have δ ( G \ E ( P )) ≥ | X i | ≥ i = 1 , . . . , p . If p ≥
3, consider u an extremityof P and assume without loss of generality that u ∈ X and that its neighbour v in P is in X ∪ X . It means that all the vertices of X \ { v } and X are non neighbours of u and hencemust form a complete subgraph of G . In particular, all the edges between X \ { v } and X are edges of P , which is not possible as | X \ { v }| ≥ | X | ≥
3, implying that P wouldcontain a cycle. So, G \ E ( P ) contains exactly the two connected components X and X .Notice that the case | X | = | X | = 3 is not possible, as in this case, as δ ( G ) ≥
4, everyvertex of X would have at least two neighbours in X and every vertex of X would haveat least two neighbours in X , and so P would contain a cycle, which is not possible. Hencemax {| X | , | X |} ≥ | X | ≥ α ( G [ X ]) = α ( G [ X ]) = 1, that is, they are both complete graphs. Inthis case, we will show how to build a Hamiltonian path and a spanning tree of G whichare edge-disjoint. If x is a vertex of any complete graph K on at least 4 vertices, it is easyto find a Hamiltonian path which starts in x and a spanning tree which are edge-disjoint.Suppose first that | X | ≥
4. Then it follows from the fact that G is 2-edge connected, thatthere exist two distinct edges x x and y y with x , y ∈ X and x , y ∈ X . Now considerfor i = 1 , P i of G [ X i ] starting in x i and a spanning tree T i of G [ X i ]edge-disjoint from P i . We conclude by considering the Hamiltonian path ( P ∪ P ) + x x of G edge-disjoint from the spanning tree ( T ∪ T ) + y y of G . Hence we may assumethat | X | = 3 and denote the vertices in X by { x , y , z } . As δ ( G ) ≥
4, there exist threedistinct edges of
G x x , y y , z z such that x , y and z belong to X . So, we consider aHamiltonian path P of G [ X ] starting in x and a spanning tree T of G [ X ] edge-disjointfrom P . We conclude then with the Hamiltonian path P ′ = ( P ∪ x y z ) + x x of G andthe spanning tree T + x z + y y + z z of G edge-disjoint from P .If α ( G [ X ]) = 1 and α ( G [ X ]) = 2, then as δ ( G ) ≥
4, we must have | X | ≥
4. In thiscase we may swap the names of X and X , which implies that we may assume without lossof generality that α ( G [ X ]) = 2. Let x and y be two vertices of X which are not adjacentin G . As every vertex of X is adjacent to x or y in G and as the corresponding edges mustbe edges of P , we have | X | ≤
4. Suppose | X | = 4. Then we will prove that G [ X ] is almostcomplete, that is it contains all the possible edges except x y . Denote the vertices of X by { x , y , z , t } , and as α ( G ) = 2, we can assume that x and y are adjacent to x and that z and t are adjacent to y . Assume first that G [ X ] contains two non-adjacent vertices z and t both distinct from x and y . As α ( G ) = 2, the vertex x has to be adjacent to z or t . We can assume that z x is an edge of G , but then y has to be adjacent to t as thereis no cycle induced by the edges of P . Similarly, we can assume that z z is an edge of P ,24ut then t cannot be adjacent to any of z , t without creating a cycle induced by the edgesof P . So, { z , t , t } is an independent set, contradicting α ( G ) = 2. Now assume that X contains a vertex z = y which is non-adjacent to x . Then z has to be adjacent to x or z and as x and y are already adjacent to x in P , z must be adjacent to z . Similarly, t is adjacent to z , but then z z y t would form a cycle with the edges of P , a contradiction.Similarly we can prove that every vertex of X except x is adjacent to y so G [ X ] is thegraph K | X | − x y . As | X | ≥
4, it is easy to find then in G [ X ] two vertex-disjoint paths P and P ′ such that V ( P ) ∪ V ( P ′ ) = X , the path P ends in x and the path P ′ ends in y and G [ X ] − P − P is connected and so contains a spanning tree T . On the other hand, as δ ( G [ X ] − P ) ≥
2, the graph G [ X ] contains a 4-cycle. One of the edges of this 4-cycle goesfrom { x , y } to { z , t } . So, denote this 4-cycle by abcda such that a is adjacent to x and d is adjacent to y . Now, we consider the Hamiltonian path P ′ = ( P ∪ P ′ ) + ab + bc + cd of G . As b and c have at least one neighbour each in X , we know that G \ E ( P ′ ) has at most2 connected components, one containing X ∪ { b, c } , the other one containing { a, d } . Butas δ ( G ) ≥ a has a neighbour in X ∪ { b, c } different from x , and so, G \ E ( P ′ ) is connected.The only remaining case is when | X | = 3. Denote the vertices of X by { x , y , z } andassume without loss of generality that x and y are adjacent to x and that z is adjacentto y . As z as degree at least 4 in G , there exist a vertex z ∈ X distinct from x and y such that z is a neighbour of z in P . More generally, as δ ( G ) ≥
4, every vertex of X hasexactly two neighbours in X , which are then neighbours in P , G [ X ] is a complete graphand finally every vertex of X has degree exactly 4. In particular, { x , z } is an independentset of size 2, and every vertex of X \ { x , y , z } is adjacent to x . Now, let us focus on theextremities of the path P . Both cannot lie in Y = { x , y , z , x , y , z } as the only verticesof Y with degree less than 2 in P are y and z , and they cannot be both extremities of P as otherwise, P would be the path y z z . So, denote by p an extremity of P not lying in Y and recall that x is adjacent to every vertex of V \ Y so x p is an edge of G . Now considerthe Hamiltonian path P ′ of G defined by P ′ = ( P − x x − x y ) + x y + x p . To conclude,let us prove that G \ E ( P ′ ) is connected. Indeed, every vertex of V \ { y , z , z } is linked to x , and all the corresponding edges except px are edges of G \ E ( P ′ ). So, G \ E ( P ′ ) inducesa connected graph on Z = V \ { y , z , z , p } . Moreover, z is adjacent to x and x z is notan edge of P ′ . And by choice, p is not adjacent to z and so has a neighbour in Z differentfrom x . Finally, y and z have both at least one neighbour in G \ E ( P ′ ) which belongs to V \ { y , z } . Thus, G \ E ( P ′ ) is connected and the proof is complete.Notice that we cannot replace δ ( G ) ≥ δ ( G ) = 3 (even if λ ( G ) = 3) as shown by thegraph built from two 3-cycles linked by a perfect matching. Also, any 3-regular graph, G , has | E ( G ) | = 3 | V ( G ) | /
2, so cannot contain two edge-disjoint spanning trees when | V ( G ) | > All proofs in this paper are constructive and it is not difficult to derive polynomial algorithmsfor finding the desired objects in case they exist. We leave the details to the interested reader.
Problem 32.
Determine the complexity of deciding whether a strong digraph of indepen-dence number 2 has a non separating out-branching.
Problem 33.
Determine the complexity of deciding whether a strong digraph of indepen-dence number 2 has a non separating spanning tree.25his problem is NP-complete for general digraphs as shown in [8].Theorem 23 suggests that perhaps we can get rid of the requirement on the minimumin-degree in Theorem 8 when the digraph has enough vertices.
Conjecture 34.
There exists an integer K such that every digraph D on at least K verticeswith λ ( D ) ≥ α ( D ) = 2 has a non-separating out-branching.It is not difficult to check that every member of the infinite class of digraphs that we usedin Proposition 12 has a non-separating branching from every vertex. Conjecture 35.
There exists an integer L such that every digraph D on at least L verticeswith λ ( D ) ≥ α ( D ) = 2 has a non-separating out-branching B + s for every choice of s ∈ V . Question 36.
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