About The Second Neighborhood Problem in Tournaments Missing Disjoint Stars
aa r X i v : . [ m a t h . C O ] S e p About the Second Neighborhood Problem in Tournaments Missing Disjoint Stars
Salman Ghazal , Abstract
Let D be a digraph without digons. Seymours second neighborhood conjecturestates that D has a vertex v such that d + ( v ) ≤ d ++ ( v ) . Under some conditions, weprove this conjecture for digraphs missing n disjoint stars. Weaker conditions arerequired when n = In this paper, a digraph D is a pair of two disjoint finite sets ( V , E ) such that E ⊆ V × V . E is the arc set and V is the vertex set and they are denoted by E ( D ) and V ( D ) respec-tively. An oriented graph is a digraph without loop and digon (directed cycles of lengthtwo). If K ⊆ V ( D ) then the induced restriction of D to K is denoted by D [ K ] . As usual, N + D ( v ) (resp. N − D ( v ) ) denotes the (first) out-neighborhood (resp. in-neighborhood) ofa vertex v ∈ V . N ++ D ( v ) (resp. N −− D ( v ) ) denotes the second out-neighborhood ( in-neighborhood ) of v , which is the set of vertices that are at distance 2 from v (resp.to v ). We also denote d + D ( v ) = | N + D ( v ) | , d ++ D ( v ) = | N ++ D ( v ) | , d − D ( v ) = | N − D ( v ) | and d −− D ( v ) = | N −− D ( v ) | . We omit the subscript if the digraph is clear from the context. Forshort, we write x → y if the arc ( x , y ) ∈ E . A vertex v ∈ V ( D ) is called whole if it isadjacent to every vertex in V ( D ) − { v } . A sink v is a vertex with d + ( v ) =
0, while asource v is a vertex with d − ( v ) =
0. For x , y ∈ V ( D ) , we say xy is a missing edge of D if neither ( x , y ) nor ( y , x ) are in E ( D ) . The missing graph G of D is the graph whoseedges are the missing edges of D and whose vertices are the non whole vertices of D .In this case, we say that D is missing G . So, a tournament does not have any missingedge. A star of center x is a graph whose edge set has the form { a i x ; i = , ..., k } . In thispaper, n stars are said to be disjoint if any two of them do not share a common vertex.A vertex v of D is said to have the second neighborhood property (SNP) if d + D ( v ) ≤ d ++ D ( v ) . In 1990, Seymour conjectured the following: Conjecture 1. (Seymour’s Second Neighborhood Conjecture (SNC)) [1] Every ori-ented graph has a vertex with the SNP.
In 1996, Fisher [3] solved the SNC for tournaments by using a certain probabil-ity distribution on the vertices. Another proof of Dean’s conjecture was established in2000 by Havet and Thomass´e [2]. Their short proof uses a tool called median orders.Furthermore, they have proved that if a tournament has no sink vertex then there areat least two vertices with the SNP. In 2007 Fidler and Yuster [5] proved, using medianorders and dependency digraphs, that SNC holds for digraphs missing a matching, a Department of Mathematics, Faculty of Sciences I, Lebanese University, Hadath, Beirut, Lebanon.E-mail: [email protected] Institute Camille Jordan, D´epartement de Math´ematiques, Universit´e Claude Bernard Lyon 1, France.
Let L = v v ... v n be an ordering of the vertices of a digraph D . An arc e = ( v i , v j ) is forward with respect to L if i < j . Otherwise e is a backward arc. The weight of L is w ( L ) = |{ ( v i , v j ) ∈ E ( D ) ; i < j }| . L is called a median order of D if w ( L ) = max { w ( L ′ ) ; L ′ is an ordering of the vertices of D } ; that is L maximizes the numberof forward arcs. In fact, the median order L satisfies the feedback property : For all1 ≤ i ≤ j ≤ n : d + D [ i , j ] ( v i ) ≥ d − D [ i , j ] ( v i ) and d − D [ i , j ] ( v j ) ≥ d + D [ i , j ] ( v j ) where [ i , j ] : = { v i , v i + , ..., v j } (See [2]).It is also known that if we reverse the orientation of a backward arc e = ( v i , v j ) of D with respect to L , then L is again a weighted median order of the new digraph D ′ = D − ( v i , v j ) + ( v j , v i ) (See [7]).Let L = v v ... v n be a median order. Among the vertices not in N + ( v n ) two typesare distinguished: A vertex v j is good if there is i ≤ j such that v n → v i → v j , otherwise v j is a bad vertex . The set of good vertices of L is denoted by G DL [2] ( or G L if thereis no confusion ). Clearly, G L ⊆ N ++ ( v n ) . The last vertex v n is called a feed vertex of D .We say that a missing edge x y loses to a missing edge x y if: x → x , y / ∈ N + ( x ) ∪ N ++ ( x ) , y → y and x / ∈ N + ( y ) ∪ N ++ ( y ) . The dependency digraph D of D is defined as follows: Its vertex set consists of all the missing edges and ( ab , cd ) ∈ E ( D ) if ab loses to cd [5, 7]. Note that D may contain digons. Definition 1. [6] In a digraph D, a missing edge ab is called a good missing edge if: ( i ) ( ∀ v ∈ V \{ a , b } )[( v → a ) ⇒ ( b ∈ N + ( v ) ∪ N ++ ( v ))] or ( ii ) ( ∀ v ∈ V \{ a , b } )[( v → b ) ⇒ ( a ∈ N + ( v ) ∪ N ++ ( v ))] .If ab satisfies ( i ) we say that ( a , b ) is a convenient orientation of ab.If ab satisfies ( ii ) we say that ( b , a ) is a convenient orientation of ab. We will need the following observation:
Lemma 1. ([5], [7]) Let D be an oriented graph and let D denote its dependencydigraph. A missing edge ab is good if and only if its in-degree in D is zero. Let D be a digraph and let D denote its dependency digraph. Let C be a connectedcomponent of D . Set K ( C ) = { u ∈ V ( D ) ; there is a vertex v of D such that uv is a2issing edge and belongs to C } . The interval graph of D , denoted by I D is definedas follows. Its vertex set consists of the connected components of D and two vertices C and C are adjacent if K ( C ) ∩ K ( C ) = f . So I D is the intersection graph of thefamily { K ( C ) ; C is a connected component of D } . Let x be a connected component of I D . We set K ( x ) = ∪ C ∈ x K ( C ) . Clearly, if uv is a missing edge in D then there is aunique connected component x of I D such that u and v belong to K ( x ) . For f ∈ V ( D ) ,we set J ( f ) = { f } if f is a whole vertex, otherwise J ( f ) = K ( x ) , where x is the uniqueconnected component of I D such that f ∈ K ( x ) . Clearly, if x ∈ J ( f ) then J ( f ) = J ( x ) and if x / ∈ J ( f ) then x is adjacent to every vertex in J ( f ) .Let L = x · · · x n be a median order of a digraph D . For i < j , the sets [ i , j ] : =[ x i , x j ] : = { x i , x i + , ..., x j } and ] i , j [= [ i , j ] \{ x i , x j } are called intervals of L . We recallthat K ⊆ V ( D ) is an interval of D if for every u , v ∈ K we have: N + ( u ) \ K = N + ( v ) \ K and N − ( u ) \ K = N − ( v ) \ K . The following shows a relation between the intervals of D and the intervals of L . Proposition 1. [8] Let I = { I , ..., I r } be a set of pairwise disjoint intervals of D.Then for every median order L of D, there is a weighted median order L ′ of D suchthat: L and L ′ have the same feed vertex and every interval in I is an interval of L ′ . We say that D is good digraph if the sets K ( x ) ’s are intervals of D . By the previousproposition, every good digraph has a median order L such that the K ( x ) ’s form inter-vals of L . Such an enumeration is called a good median order of the good digraph D [8]. Theorem 1. [8] Let D be a good oriented graph and let L be a good median order ofD, with feed vertex f. Then for every x ∈ J ( f ) , we have | N + ( x ) \ J ( f ) | ≤ | G L \ J ( f ) | . Soif x has the SNP in D [ J ( f )] , then it has the SNP in D. Corollary 1. ([2]) Let L be a median order of a tournament with feed vertex f. Then | N + ( f )) | ≤ | G L | . Let L be a good median order of a good oriented graph D and let f denote its feedvertex. By theorem 1, for every x ∈ J ( f ) , | N + ( x ) \ J ( f ) | ≤ | G L \ J ( f ) | . Let b , · · · , b r denote the bad vertices of L not in J ( f ) and v , · · · , v s denote the non bad vertices of L not in J ( f ) , both enumerated in increasing order with respect to their index in L .If | N + ( x ) \ J ( f ) | < | G L \ J ( f ) | , we set Sed ( L ) = L . If | N + ( x ) \ J ( f ) | = | G L \ J ( f ) | , we set sed ( L ) = b · · · b r J ( f ) v · · · v s . This new order is called the sedimentation of L . Lemma 2. [8] Let L be a good median order of a good oriented graph D. Then Sed ( L ) is a good median order of D. In the rest of this section, D is an oriented graph missing a matching and D denotesits dependency digraph. We begin by the following lemma: Lemma 3. [5] The maximum out-degree of D is one and the maximum in-degree of D is one. Thus D is composed of vertex disjoint directed paths and directed cycles. roof. Assume that a b loses to a b and a b loses to a ′ b ′ , with a → a and a → a ′ . The edge a ′ b is not a missing edge of D . If a ′ → b then b → a ′ → b , acontradiction. If b → a ′ then b → b → a ′ , a contradiction. Thus, the maximumout-degree of D is one. Similarly, the maximum in-degree is one.In the following, C = a b , ..., a k b k denotes a directed cycle of D , namely a i → a i + , b i + / ∈ N ++ ( a i ) ∪ N + ( a i ) , b i → b i + and a i + / ∈ N ++ ( b i ) ∪ N + ( b i ) , for all i < k . Lemma 4. ([5]) If k is odd then a k → a , b / ∈ N ++ ( a k ) ∪ N + ( a k ) , b k → b and a / ∈ N ++ ( b k ) ∪ N + ( b k ) . If k is even then a k → b , a / ∈ N ++ ( a k ) ∪ N + ( a k ) , b k → a andb / ∈ N ++ ( b k ) ∪ N + ( b k ) . Lemma 5. [5] K ( C ) is an interval of D.Proof. Let f / ∈ K ( C ) . Then f is adjacent to every vertex in K ( C ) . If a → f then b → f , since otherwise b ∈ N ++ ( a ) ∪ N + ( a ) which is a contradiction. So N + ( a ) \ K ( C ) ⊆ N + ( b ) \ K ( C ) . Applying this to every losing relation of C yields N + ( a ) \ K ( C ) ⊆ N + ( b ) \ K ( C ) ⊆ N + ( a ) \ K ( C ) ... ⊆ N + ( b k ) \ K ( C ) ⊆ N + ( b ) \ K ( C ) ⊆ N + ( a ) \ K ( C ) ... ⊆ N + ( a k ) \ K ( C ) ⊆ N + ( a ) \ K ( C ) if k is even. So these inclusion are equalities. An anal-ogous argument proves the same result for odd cycles. We recall that a vertex x in a tournament T is a king if { x } ∪ N + ( x ) ∪ N ++ ( x ) = V ( T ) .It is well known that every tournament has a king. However, for every natural number n / ∈ { , } , there is a tournament T n on n vertices, such that every vertex is a king forthis tournament.A digraph is called non trivial if it has at least one arc. Proposition 2.
Let D be a digraph missing disjoint stars. If the connected componentsof its dependency digraph are non-trivial strongly connected, then D is a good digraph.Proof.
Let x be a connected component of D . First, suppose that K ( x ) = K ( C ) forsome directed cycle C = a b , a b , ..., a n b n in D , namely a i → a i + and b i + / ∈ N + ( a i ) ∪ N ++ ( a i ) . If the set of the missing edges { a i b i ; i = , ..., n } forms a matching, then bylemma 5, K ( C ) is an interval of D .So we will suppose that a center x of a missing star appears twice in the list a , b , a , b , ..., a n , b n and assume without loss of generality that x = a . Suppose that n is even. Set K = { a , b , ..., a n − , b n } and K = K ( C ) \ K .Suppose that a n → b and a / ∈ N + ( a n ) ∪ N ++ ( a n ) . Then by following the proof oflemma 5 we get the desired result. 4uppose a n → a and b / ∈ N + ( a n ) ∪ N ++ ( a n ) . Then by following the proof oflemma 5 we get that K and K are intervals of D . Assume, for contradiction that K ∩ K = f and let i > x is incident to a i b i . Clearly i >
2. However, b / ∈ K and x = a → a → a implies that i >
3. Suppose that x = a i . Note that i must be odd by definition of K . Since b → a = x = a i and a / ∈ N + ( x ) ∪ N ++ ( x ) then a → x . Similarly b , a , ..., b i − are in-neighbors of x .However, b i − is an out-neighbor of a i = x , a contradiction. Suppose that x = b i . Sim-ilarly, a , b , ..., a i − are in-neighbors of x . However, a i − is an out-neighbor of b i = x ,a contradiction. Thus K ∩ K = f . Whence, K = K ∪ K is an interval of D . Similarargument is used to prove it when n is odd.This result can be easily extended to the case when K ( x ) = K ( C ) and C is a nontrivial strongly connected component of D , because between any two missing edges uv and zt there is directed path from uv to zt and a directed path from zt to uv . These twodirected paths creat many directed cycles that are used to prove the desired result.This also is extended to the case when K ( x ) = ∪ C ∈ x K ( C ) : Let u and u ′ be twovertices of K ( x ) . There are two non trivial strongly connected components of D suchthat u ∈ K ( C ) and u ′ ∈ K ( C ′ ) . Since x is a connected component of I D , there isa path C = C C ... C n = C ′ . For all i >
0, there is u i ∈ K ( C i − ) ∩ K ( C i ) , by definitionof edges in I D . Therefore, N + ( u ) \ K ( x ) = N + ( u ) \ K ( x ) = ... = N + ( u i ) \ K ( x ) = ... = N + ( u n ) \ K ( x ) = N + ( u ′ ) \ K ( x ) and N − ( u ) \ K ( x ) = N − ( u ) \ K ( x ) = ... = N − ( u i ) \ K ( x ) = ... = N − ( u n ) \ K ( x ) = N − ( u ′ ) \ K ( x ) . Theorem 2.
Let D be a digraph obtained from a tournament by deleting the edges ofdisjoint stars. Suppose that, in the induced tournament by the centers of the missingstars, every vertex is a king. If d − D > then D satisfies SNC.Proof. Orient every missing edge of D towards the center of its star. Let L be a medianorder of the obtained tournament T and let f be its feed vertex. Then f has the SNP in T . We prove that f has the SNP in D as well.First, suppose that f is a whole vertex. Then N + ( f ) = N + T ( f ) . Let v ∈ N ++ T ( f ) .Then there ∃ u ∈ V ( T ) = V ( D ) such that f → u → v → f in T . Since f is whole, then ( f , u ) and ( v , f ) ∈ D . If ( u , v ) ∈ D then v ∈ N ++ ( f ) . Otherwise, uv is a missing edgeand hence, ∃ ab that loses to uv , say b → v and u / ∈ N + ( b ) ∪ N ++ ( b ) . But f b is nota missing edge, since f is whole. Then ( f , b ) ∈ D , since otherwise, b → f → u in D which is a contradiction. Therefore, f → b → v in D . Whence, v ∈ N ++ ( f ) . So N ++ T ( f ) ⊆ N ++ ( f ) . Therefore, d + ( f ) = d + T ( f ) ≤ d ++ T ( f ) ≤ d ++ ( f ) .Now suppose that f is the center of a missing star. Then N + ( f ) = N + T ( f ) . Let v ∈ N ++ T ( f ) . Then there ∃ u ∈ V ( T ) = V ( D ) such that f → u → v → f in T . Then ( f , u ) ∈ D while ( f , v ) / ∈ D . If ( u , v ) ∈ D then v ∈ N ++ ( f ) . Otherwise, uv is a missingedge and v is the center of a missing star. Then v ∈ N + ( f ) ∪ N ++ ( f ) , because f is aking for the centers of the missing stars. Note that v / ∈ N + ( f ) . So N ++ T ( f ) ⊆ N ++ ( f ) .5herefor, f has the SNP in D .Finally, suppose that f is not whole and not the center of a missing star. Then ∃ x a center of a missing star such that f x is a missing edge. We distinguish between twocases.In the first case, we suppose that f x does not lose to any missing edge. We reori-ent f x as ( x , f ) . Since ( f , x ) ∈ T is a backward arc with respect to L , the again L isa median order of the new tournament T ′ obtained by reversing the orientation of f x .Moreover, N + ( f ) = N + T ′ ( f ) and f has the SNP in T ′ . Let v ∈ N ++ T ′ ( f ) . Then there ∃ u ∈ V ( T ) = V ( D ) such that f → u → v → f in T ′ . Then ( f , u ) ∈ D while ( f , v ) / ∈ D .If ( u , v ) inD then v ∈ N ++ ( f ) . Otherwise uv is a missing edge and v is the center of amissing star.Since D has no source, there is a missing edge that loses to uv . Suppose thatthis edge is of the form ax . Then we must have x → v and u / ∈ N + ( x ) ∪ N ++ ( x ) , by def-inition of losing relation and due to the fact that v ∈ N + ( x ) ∪ N ++ ( x ) ( x is a king for thecenters of the missing stars). If v / ∈ N ++ ( f ) , then f x loses uv which is a contradiction tothe supposition of this case. Hence, v / ∈ N ++ ( f ) . Now, suppose that the missing edgethat loses to uv is of the form by with x / ∈ { b , y } . Suppose without loss of generality that y is the center of a missing star containing by . Then y → v and u / ∈ N + ( y ) ∪ N ++ ( y ) ,by definition of losing relation and due to the fact that v ∈ N + ( y ) ∪ N ++ ( y ) ( y is a kingfor the centers of the missing stars). But ( f , u ) ∈ D and f y is not a missing edge, then ( f , y ) ∈ D . Thus f → y → v . Whence, v ∈ N + ( f ) ∪ N ++ ( f ) . So N ++ T ′ ( f ) ⊆ N ++ ( f ) .Therefor, f has the SNP in D as well.In the second case, we suppose that f x loses to some missing edge by . We mayassume without loss of generality that y is the center of a missing star containing by .Then we must have x → y and b / ∈ N + ( x ) ∪ N ++ ( x ) . Clearly, N + ( f ) ∪{ y } = N + T ( f ) . Weprove that N ++ T ( f ) ⊆ N ++ ( f ) ∪{ y } . Let v ∈ N ++ T ( f ) \ y . Then there ∃ u ∈ V ( T ) = V ( D ) such that f → u → v → f in T . Suppose that u = x . Since bv is not a missing edge, x = u → v and b / ∈ N + ( x ) ∪ N ++ ( x ) then we must have ( b , v ) ∈ D . Whence, f → b → v in D . Therefore v ∈ N ++ ( f ) . Now suppose that u = x . Then ( f , u ) ∈ D . If ( u , v ) ∈ D then v ∈ N ++ ( f ) . Otherwise, uv is a missing edge. Hence there is a missing edge pq that loses to uv , namely, q → v and u / ∈ N + ( q ) ∪ N ++ ( q ) . If q = x , then we have f → x → v → f in T , which is the same as the case when u = x . So we may supposethat q = x . Note that q must be the center of a missing star. So f , x / ∈ { p , q } . Thus f q is not a missing edge, u / ∈ N + ( q ) ∪ N ++ ( q ) and ( f , u ) ∈ D . Then we must have ( f , q ) ∈ D , since otherwise we get q → f → u in D which is a contradiction. Thus f → q → v in D . Whence v ∈ N ++ ( f ) . So N ++ T ( f ) ⊆ N ++ ( f ) ∪ { y } . Therefore d + ( f ) + = d + T ( f ) ≤ d ++ T ( f ) ≤ d ++ ( f ) +
1. Whence f has the SNP in D . A more general statement to the following theorem is proved in [6] . Here we introduceanother prove that uses the sedimentation technique of a median order.6 heorem 3. [5] Let D be an oriented graph missing a star. Then D satisfies SNC.Proof.
Orient all the missing edges of D towards the center x of the missing star. Theobtained digraph is a tournament T . Let L be a median order of T that maximizes a , the index of x in L , and let f denote its feed vertex. Reorient the missing edgesincident to f towards f (if any). L is also a median order of the new tournament T ′ .Note that N + ( f ) = N + T ′ ( f ) and we have d + T ′ ( f ) ≤ | G T ′ L | . If x ∈ G T ′ L and d + T ′ ( f ) = | G T ′ L | then sed ( L ) is a median order of T ′ in which the index of x is greater than a , andalso greater than the index of f . So we can give the missing edge incident to f (if itexists then it is x f ) its initial orientation (as in T ) such that sed ( L ) is a median orderof T , a contradiction to the fact that L maximizes a . So x / ∈ G T ′ L or d + T ′ ( f ) < | G T ′ L | .If f = x then, clearly, d + ( f ) = d + T ′ ( f ) ≤ | G T ′ L | ≤ d ++ T ′ ( f ) = d ++ ( f ) . Now supposethat f = x . We have that x is the only possible gained second out-neighbor vertex for f . If x / ∈ G T ′ L then G T ′ L ⊆ N ++ ( f ) , whence the result follows. If d + T ′ ( f ) < | G T ′ L | then d + ( f ) = d + T ′ ( f ) ≤ | G T ′ L | − ≤ d ++ ( f ) . So f has the SNP in D . In this section, let D be a digraph obtained from a tournament by deleting the edgesof 2 disjoint stars and let D denote its dependency digraph. Let S x and S y be the twomissing disjoint stars with centers x and y respectively, A = V ( S x ) \ x , B = V ( S y ) \ y , K = V ( S x ) ∪ V ( S y ) (the set of non whole vertices) and assume without loss of generalitythat x → y . In [6] it is proved that if the dependency digraph of any digraph consists ofisolated vertices only then it satisfies SNC. Here we consider the case when the D hasno isolated vertices. Theorem 4.
Let D be an oriented graph missing 2 disjoint stars. If D has no isolatedvertex, then D satisfies SNC.Proof. Assume without loss of generality that x → y . We note that the condition D hasno isolated vertex, implies that for every a ∈ A and y ∈ B we have y → a and b → x .We shall orient all the missing edges of D . First, we give every good edge a convenientorientation. For the other missing edges, let the orientation be towards the center of the2 missing stars S x or S y . The obtained digraph is a tournament T . Let L be a medianorder of T such that the index k of x is maximum and let f denote its feed vertex. Weknow that f has the SNP in T . We have only 5 cases:Suppose that f is a whole vertex. In this case N + ( f ) = N + T ( f ) . Suppose f → u → v in T . Clearly ( f , u ) ∈ D . If ( u , v ) ∈ D or is a convenient orientation then v ∈ N + ( f ) ∪ N ++ ( f ) . Otherwise there is a missing edge zt that loses to uv with t → v and u / ∈ N + ( f ) ∪ N ++ ( f ) . But f → u , then f → t , whence f → t → v in D. Therefore, N ++ ( f ) = N ++ T ( f ) and f has the SNP in D as well.Suppose f = x . Orient all the edges of S x towards the center x . L is a median orderof the modified completion T ′ of D . We have N + ( f ) = N + T ′ ( f ) . Suppose f → u → v in T ′ . If ( u , v ) ∈ D or is a convenient orientation then v ∈ N + ( f ) ∪ N ++ ( f ) . Otherwise ( u , v ) = ( b , y ) for some b ∈ B , but f = x → y . Thus, N ++ ( f ) = N ++ T ′ ( f ) and f has the7NP in T ′ and D .Suppose f = b ∈ B . Orient the missing edge by towards b . Again, L is a me-dian order of the modified tournament T ′ and N + ( f ) = N + T ′ ( f ) . Suppose f → u → v in T ′ . If ( u , v ) ∈ D or is a convenient orientation then v ∈ N + ( f ) ∪ N ++ ( f ) . Oth-erwise ( u , v ) = ( b ′ , y ) for some b ′ ∈ B or ( u , v ) = ( a , x ) for some a ∈ A , however x , y ∈ N ++ ( f ) ∪ N + ( f ) because f = b → x → y in D . Thus, N ++ ( f ) = N ++ T ′ ( f ) and f has the SNP in T ′ and D .Suppose f = y . Orient the missing edges towards y and let T ′ denote the new tour-nament. We note that B ⊆ N ++ ( y ) ∩ N ++ T ′ ( y ) due to the condition d D >
0. Also, x is theonly possible new second neighbor of y in T ′ . If B ∪ { x } " G L or d + T ′ ( y ) < d ++ T ′ ( y ) , then d + ( y ) = d + T ′ ( y ) ≤ d ++ T ′ ( y ) − ≤ d ++ ( y ) . Otherwise, B ∪ { x } * G L and d + T ′ ( y ) = | G L | .In this case we consider the median order Sed ( L ) of T ′ . Now the feed vertex of sed ( L ) is different from y , the index of x had increased, and the index of y became less thanthe index of any vertex of B which makes Sed ( L ) a median order of T also, in whichthe index of x is greater than k , a contradiction.Suppose f = a ∈ A . Orient the missing edge ax as ( x , a ) and let T ′ denote thenew tournament. Note that y is the only possible new second neighbor of a in T ′ andnot in D . Also x ∈ N ++ T ( a ) ∩ N ++ ( a ) . If d + T ′ ( a ) < d ++ T ′ ( a ) , then d + ( a ) = d + T ′ ( a ) ≤ d ++ T ′ ( a ) − ≤ d ++ ( a ) , hence a has the SNP in D . Otherwise, d + T ′ ( a ) = | G L | = d ++ T ′ ( a ) and in particular x ∈ G L . In this case we consider sed ( L ) which is a median order of T ′ . Note that the feed vertex of Sed ( L ) is different from a and the index of a is lessthan the index of x in the new order Sed ( L ) . Hence Sed ( L ) is a median of T as well, inwhich the index of x is greater than k , a contradiction.So in all cases f has the SNP in D . Therefore D satisfies SNC. Theorem 5.
Let D be a digraph obtained from a tournament by deleting the edges of2 disjoint stars. If D has neither a source nor a sink and D has no sink, then D has atleast two vertices with the SNP.Proof. claim 1: Suppose K = V ( D ) . If D has no isolated vertex, then D has at least two verticeswith the SNP. Proof of claim 1:
The condition D has no isolated vertex implies that for every a ∈ A and b ∈ B we have y → a and b → x . Clearly, N + ( x ) = { y } , N + ( y ) = A , d + ( x ) ≤ ≤| A | ≤ d ++ ( x ) , thus x has the SNP. Let H be the tournament D − { x , y } . Then H hasa vertex v with the SNP in H . If v ∈ A , then d + ( v ) = d + H ( v ) ≤ d ++ H ( v ) = d ++ ( v ) . If v ∈ B , then d + ( v ) = d + H ( v ) + ≤ d ++ H ( v ) + = d ++ ( v ) . Whence, v also has the SNP in D . Claim 2: D is a good digraph. Proof of claim 2:
Let I D be the interval graph of D . Let C and C be two distinctconnected components of D . Then the centers x and y appear in each of the these twoconnected components, whence K ( C ) ∩ K ( C ) = f . Therefore, I D is a connected8raph, having only one connected component x . Then, K = K ( x ) .So if D is composed of non trivial strongly connected components, the result holds bylemma 2.Due to the condition D has neither a source nor a sink, D has a non trivial stronglyconnected component, hence N + ( x ) \ K = N + ( y ) \ K . Now let v ∈ K and assume with-out loss of generality that xv is a missing edge. Due to the condition D has nei-ther a source nor a sink, we have that xv belongs to a non trivial strongly connectedcomponent of D , and in this case v ∈ R and N + ( v ) \ K = N + ( x ) \ K = N + ( y ) \ K , or xv belongs to a directed path P = xa , yb , · · · , xa p , yb p joining 2 non trivial stronglyconnected components C and C with xa ∈ C and yb p ∈ C . There is i > v = a i . L = xa i − , yb i − , xa i , yb i is a path in D . By the definition of losing cy-cles we have N + ( x ) \ K ⊆ N + ( b i − ) \ K ⊆ N + ( a i ) \ K ⊆ N + ( y ) \ K = N + ( x ) \ K . Hence N + ( x ) \ K = N + ( v ) \ K for all v ∈ K . Since every vertex outside K is adjacent to everyvertex in K we also have N − ( x ) \ K = N − ( v ) \ K for all v ∈ K . This proves the secondclaim.Since D is a good digraph, then it has a good median order L = x x ... x n . If J ( x n ) = K , then the result follows by claim 1 and theorem 1. Otherwise, x n is whole,that is J ( x n ) = { x n } . By theorem 1, x n has the SNP in D . So we need to find an-other vertex with the SNP in D . Consider the good median order L ′ = x x ... x n − ofthe good digraph D ′ = D [ { x , ..., x n − } ] . Suppose first that L ′ is stable. There is q forwhich Sed q ( L ′ ) = y ... y n − and | N + ( y n − ) \ J ( y n − ) | < | G Sed q ( L ′ ) \ J ( y n − ) | ( ∗ ) . Notethat y ... y n − x n is also a good median order of D . By theorem 1 and claim 1, thereis y ∈ J ( y n − ) that has the SNP in D ′ , more precisely | N + ( y ) | < | N ++ ( y ) | due to ( ∗ ) .Since y ∈ J ( y n − ) and y n − → x n then y → x n . So | N + ( y ) | = | N + D ′ ( y ) | + ≤| N ++ ( y ) | .Now suppose that L ′ is periodic. Since D has no sink then x n has an out-neighbor x j . Choose j to be the greatest (so that it is the last vertex of its corresponding inter-val). Note that for every q , x n is an out-neighbor of the feed vertex of Sed q ( L ′ ) . So x j is not the feed vertex of any Sed q ( L ′ ) . Since L ′ is periodic, x j must be a bad vertex of Sed q ( L ′ ) for some integer q , otherwise the index of x j would always increase duringthe sedimentation process. Let q be such an integer and set Sed q ( L ′ ) = y ... y n − . Bytheorem 1 and claim 1, there is y ∈ J ( y n − ) that has the SNP in D ′ , more precisely | N + D ′ ( y ) \ J ( y n − ) | < | G Sed q ( L ′ ) \ J ( y n − ) | due to ( ∗ ) . Since y ∈ J ( y n − ) and y n − → x n then y → x n . Note that y → x n → x j , G Sed q ( L ′ ) ∪ { x j }\ J ( y n − ) ⊆ N ++ ( y ) \ J ( y n − ) and | N + D ′ ( y ) \ J ( y n − ) | = | G Sed q ( L ′ ) \ J ( y n − ) | .Therefore | N + ( y ) | = | N + D ′ ( y ) | + = | N + D ′ ( y ) \ J ( y n − ) | + + | N + D ′ ( y ) ∩ J ( y n − ) | = | G Sed q ( L ′ ) \ J ( y n − ) | + + | N + D ′ ( y ) ∩ J ( y n − ) \ J ( y n − ) | = | G Sed q ( L ′ ) ∪{ x j }\ J ( y n − ) | + | N + D ′ ( y ) ∩ J ( y n − ) | ≤ | N ++ D ( y ) \ J ( y n − ) | + | N ++ D ( y ) ∩ J ( y n − ) | ≤ | N ++ ( y ) | . In this section, D is an oriented graph missing three disjoint stars S x , S y and S z withcenters x , y and z respectively. Set A = V ( S x ) − x , B = V ( S y ) − x , C = V ( S z ) − z and9 = A ∪ B ∪ C ∪ { x , y , z } . Let D denote the dependency digraph of D . The triangleinduced by the vertices x , y and z is either a transitive triangle or a directed triangle.First we will deal with the case when this triangle is directed, and assume without lossof generality that x → y → z → x . This is a particular case of the case when the missinggraph is a disjoint union of stars such that, in the induced tournament by the centers ofthe missing stars, every vertex is a king. Theorem 6.
Let D be an oriented graph missing 3 disjoint stars whose centers form adirected triangle. If D has no isolated vertices, then D satisfies SNC.Proof. Claim:
The only possible arcs in D have the forms xa → yb or yb → zc or zc → xa ,where a ∈ A , b ∈ B and c ∈ C . Proof of the claim: xa can not lose to zc because z → x and z ∈ N ++ ( x ) . Similarly yb can not lose to xa and zc can not lose to yb .Orient the good missing edges in a convenient way and orient the other edges to-ward the centers. The obtained digraph T is a tournament. Let L be a median orderof T such that the sum of the indices of x , y and z is maximum. Let f denote the feedvertex of L . Due to symmetry, we may assume that f is a whole vertex or f = x or f = a ∈ A .Suppose f is a whole vertex. Clearly, N + ( f ) = N + T ( f ) . Suppose f → u → v in T . If ( u , v ) ∈ E ( D ) or uv is a good missing edge then v ∈ N + ( f ) ∪ N ++ ( f ) . Otherwise, thereis missing edge rs that loses to uv with r → v and u / ∈ N ++ ( r ) ∪ N + ( r ) . But f → u ,then f → r , whence f → r → v and v ∈ N + ( f ) ∪ N ++ ( f ) . Thus, N ++ T ( f ) = N ++ ( f ) and f has the SNP in D .Suppose f = x . Reorient all the missing edges incident to x toward x . In the newtournament T ′ we have N + ( x ) = N + T ′ ( x ) and x has the SNP in T ′ . Since y ∈ N + ( x ) and z ∈ N ++ ( x ) we have that N ++ ( x ) = N ++ T ′ ( x ) . Thus x has the SNP in D .Suppose that f = a ∈ A . Reorient ax toward a . Suppose a → u → v in the newtournament T ′ with v = y . If ( u , v ) ∈ E ( D ) or uv is a good missing edge then v ∈ N + ( a ) ∪ N ++ ( a ) . Otherwise, there is b ∈ B and c ∈ C such that ( u , v ) = ( c , z ) and by loses to cz , then f → c implies that a → y , but y → z , whence z ∈ N ++ ( a ) ∪ N + ( a ) . Sothe only possible new second out-neighbor of a is y , hence if y / ∈ N ++ T ′ ( a ) then a has theSNP in D . Suppose y ∈ N ++ T ′ ( a ) . If d + T ′ ( a ) < d ++ T ′ ( a ) then d + ( a ) = d + T ′ ( a ) ≤ d ++ T ′ ( a ) = d ++ ( a ) , hence a has the SNP in D . Otherwise, d + T ′ ( a ) = | G L | and G L = N ++ T ′ ( a ) . So x , y and z are not bad vertices, hence the index of each increases in the median order Sed ( L ) of T ′ . But the index of a is less than the index of x , then we can give ax itsinitial orientation as in T nd the same order Sed ( L ) is a median order of T . However,the sum of indices of x , y and z has increased. A contradiction. Thus f has the SNP in D and D satisfies SNC . Theorem 7.
Let D be an oriented graph missing 3 disjoint stars whose centers form a irected triangle. If D has neither a source nor a sink and D has no sink, then D has atleast two vertices with the SNP.Proof. Claim 1:
For every a ∈ A , b ∈ B and c ∈ C we have: b → x → c → y → a → z → b . Proof of Claim 1:
This is due to the claim in the previous proof and the condition that D has neither a source nor a sink. Claim 2: If K = V ( D ) then D has at least 3 vertices with the SNP. Proof of Claim 2:
Let H = D − { x , y , z } . H is a tournament with no sink (dominatedvertex). Then H has 2 vertices u and v with SNP in H . Without loss of generality wemay assume that u ∈ A . But y → u → z , the adding the vertices x , y and z makes u gainsonly one vertex to its first out-neighborhood and x to its second out-neighborhood.Thus, also u has the SNP in D . Similarly, v has the SNP in D . Suppose, without lossof generality, that | A | ≥ | C | . We have C ∪ { y } = N + ( x ) and A ∪ { z } = N ++ ( x ) . Hence, d + ( x ) = | C | + ≤ | A | + ≤ d ++ ( x ) , whence, x has the SNP in D . Claim 3: D is a good oriented graph. Proof of Claim 3:
Let I D be the interval graph of D . Let C and C be two distinctconnected components of D . The three centers of the missing disjoint stars must appearin each of the these two connected components, whence K ( C ) ∩ K ( C ) = f . Therefore, I D is a connected graph, having only one connected component x . Then, K = K ( x ) .So if D is composed of non trivial strongly connected components, the result holds byproposition 2.Due to the condition that D has neither a source nor a sink, D has a non trivial stronglyconnected component C .Since x , y and z must appear in C , we have N + ( x ) \ K = N + ( y ) \ K = N + ( z ) \ K .Now let v ∈ K . If v appears in a non trivial strongly connected component of D then N + ( v ) \ K = N + ( x ) \ K = N + ( y ) \ K = N + ( z ) \ K .Otherwise,due to the condition that D has neither a source nor a sink, v appears in adirected path P of D joining two non trivial strongly connected components C and C of D . By the definition of losing relations we can prove easily that for all a ∈ K ( C ) , b ∈ K ( P ) and c ∈ K ( C ) we have N + ( a ) \ K ( x ) ⊆ N + ( b ) \ K ( x ) ⊆ N + ( c ) \ K ( x ) . Inparticular, for a = x = c and b = v . So N + ( v ) \ K = N + ( x ) \ K . Similarly, N − ( v ) \ K = N − ( x ) \ K . This proves claim 3.To conclude we apply the same argument of the proof of theorem 5. The author thanks Pr. Amine El-Sahili for many useful discussions.11 eferences [1] N. Dean and B. J. Latka, squaring the tournament: an open problem , CongressNumerantium 37-80.[2] F. Havet and S. Thomase´e,
Median Orders of Tournaments: A Tool for the SecondNeighborhood Problem and Sumner’s Conjecture , J. Graph Theory 35 (2000),244-256.[3] D. Fisher, squaring a tournament: a proof of Dean’s conjecture , J. Graph Theory23 (1996), 43-48.[4] Y. Kaneko and S.C. Locke,
The minimum degree approach for Paul Seymour’sdistance 2 conjecture , Congressus Numerantium 148 (2001), 201-206.[5] D. Fidler and R. Yuster,
Remarks on the Second Neighborhood Problem , J. GraphTheory 55 (2007), 208-220.[6] S. Ghazal: Seymour’s second neighborhood conjecture in tournaments missing ageneralized star,
J. Graph Theory , (2012), 89-94.[7] S. Ghazal: A contribution to the second neighborhood problem, Graphs and Com-binatorics ,29