aa r X i v : . [ m a t h . C O ] J un A contribution to the second neighborhoodproblem
Salman GHAZAL Abstract
Seymour’s Second Neighborhood Conjecture asserts that every digraph(without digons) has a vertex whose first out-neighborhood is at mostas large as its second out-neighborhood. It is proved for tournaments,tournaments missing a matching and tournaments missing a generalizedstar. We prove this conjecture for classes of digraphs whose missing graphis a comb, a complete graph minus 2 independent edges, or a completegraph minus the edges of a cycle of length 5.
In this paper, graphs are finite and simple. Directed graphs (digraphs) areorientations of graphs, so they do not contain loops, parallel arcs, or digons(directed cycles of length 2). Let G = ( V, E ) be a graph. The neighborhoodof a vertex v in G is denoted by N G ( v ) and its degree is d G ( v ) = | N G ( v ) | .For A ⊆ V , N G ( A ) denotes the set of negihbors outside A of the elementsof A . Let D = ( V, E ) denote a digraph with vertex set V and arc set E .As usual, N + D ( v ) (resp. N − D ( v )) denotes the (first) out-neighborhood (resp.in-neighborhood) of a vertex v ∈ V . N ++ D ( v ) (resp. N −− D ( v )) denotes thesecond out-neighborhood (in-neighborhood) of v , which is the set of verticesthat 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 thesubscript if the digraph (resp. graph) is clear from the context. For short, wewrite x → y if the arc ( x, y ) ∈ E . We say that a vertex v has the second neig-borhood property (SNP) if d + ( v ) ≤ d ++ ( v ).In 1990, P. Seymour conjectured [2] the following statement: Conjecture 1. (The Second Neighborhood Conjecture (SNC))
Every digraph has a vertex with the SNP.
Seymour’s conjecture restricted to tournaments is known as Dean’s conjec-ture [2]. In 1996, Fisher [5] proved Dean’s conjecture, thus asserting the SNCfor tournaments. Another proof of Dean’s conjecture was given by Thomass´eand Havet [3], in 2000, using a tool called (local) median order. In 2007, Fidlerand Yuster used also median orders to prove SNC for tournaments missing amatching. Ghazal proved the weighted version of SNC for tournaments missing Department of Mathematics, Faculty of Sciences I, Lebanese University, Hadath, Beirut,Lebanon.E-mail: [email protected] Camille Jordan, D´epartement de Math´ematiques, Universit´e Claude Bernard Lyon1, France.E-mail: [email protected] L = v v ...v n of a digraph D is an order of the vertices of D the maximizes the size of the set of forward arcs of D , i.e., the set { ( v i , v j ) ∈ E ; i < j } . In fact, L satisfies the feedback property: For all 1 ≤ i ≤ j ≤ n : d +[ i,j ] ( v i ) ≥ d − [ i,j ] ( v i )and d − [ i,j ] ( v j ) ≥ ωd +[ i,j ] ( v j )where [ i, j ] := D [ v i , v i +1 , ..., v j ].An order L = v v ...v n satisfying the feedback property is called a local medianorder . The last vertex v n of a weighted local median order L = v v ...v n of D is called a feed vertex of the digraph D [3]. Theorem 1. [3] Every feed vertex of a tournament has the SNP.
Let D = ( V, E ) be a digraph. For 2 vertices x and y , we call xy a missingedge if ( x, y ) / ∈ E and ( y, x ) / ∈ E . The missing graph G of D is the graph formedby the missing edges, formally, E ( G ) is the set of all the missing edge and V ( G )is the set of non whole vertices (vertices incident to some missing edges). Inthis case, we say that D is missing G .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 dependencydigraph ∆ of D is defined as follows: Its vertex set consists of all the missingedges and ( ab, cd ) ∈ E (∆) if ab loses to cd . Note that ∆ may contain digons.These digraphs were used in [4] to prove SNC for tournaments missing amatching. However, our defintion is general and is suitable for any digraph. Definition 1. [1] A missing edge ab is called good 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 . The following holds by the definition of good missing edges and losing rela-tion between them.
Lemma 1.
Let D be a digraph and let ∆ denote its dependency digraph. Amissing edge ab is good if and only if its in-degree in ∆ is zero. Let H be a family of digraphs (digons are allowed) and let G be a givengraph. We say that G is H -forcing if the dependency digraph of every digraphmissing G is a member of H . The set of all H -forcing graphs is denoted by F ( H ).A digraph is trivial if it has no arc.2 roposition 1. Let H be a family of digraphs. Then F ( H ) is nonempty if andonly if H has a trivial digraph.Proof. Let G be a graph and let D be any digraph missing it. Suppose xy → uv in ∆, the dependency digraph of D , namely v / ∈ N + ( x ) ∪ N ++ ( x ). We add to D an extra whole vertex α such that x → α → v . This breaks the arc ( xy, uv ).Hence, by adding a sufficient number of such vertices, one obtains a digraphwhose missing graph is G and such that its dependency digraph is trivial. Thisestablishes the necessary condition.The converse holds, by absorving that the dependency digraph of any digraphmissing a star ( edges sharing only one endpoint) is trivial.Let S denote the class of all trivial digraphs. In [1] Ghazal showed thatthe only S -forcing graphs are generalized stars and proved that every digraphmissing a generalized star satisfies Seymour’s Second Neighborhood Conjecture.In fact, a weighted version the following statement is proved in [1]. Lemma 2.
Let D be a digraph. If all the missing edges of D are good then ithas a vertex with the SNP. Problem 1.
Let ~ P be the family of all digraphs composed of vertex disjointdirected paths only. Characterize F ( ~ P ) . In the next section we present some classes of graphs contained in F ( ~ P ) andprove that every digraph missing a member of these classes, satisfies SNC. F ( ~ P ) A comb G is a graph defined as follows: V ( G ) is disjoint union of three set A , X and Y . G [ X ∪ Y ] is a complete graph. A is stable set with N ( A ) = X and N ( a ) ∩ N ( b ) = φ for any distinct vertices a, b ∈ A . For every a ∈ A , d ( a ) = 1.Observe that the edges with an end in A form a matching, say M. Proposition 2.
Combs are ~ P -forcing.Proof. Let D be a digraph missing a comb G . We follow the previous notations.The only possible arcs of ∆ occurs between the edges in M . For i = 1 , , a i x i ∈ M with a i ∈ A and x i ∈ X . Suppose a x loses to the 2 others. Then wehave a → x , x → a , a / ∈ N ++ ( a ) ∪ N + ( a ) and x / ∈ N ++ ( x ) ∪ N + ( x ).Since a x is not a missing edge then either a → x or a ← x . Whence,either x ∈ N ++ ( x ) ∪ N + ( x ) or a ∈ N ++ ( a ) ∪ N + ( a ). A contradiction.Therefore, the maximimum out-degree in ∆ is 1. Similarly, the maximum in-degree is 1. Thus ∆ is composed of at most vertex disjoint directed pathsand directed cycles. Now it is enough to prove that it has no directed cycles.Suppose that C = a x , a x , ..., a n x n is a cycle. Then we have a i +1 / ∈ N ++ ( a i )3nd a i ← a i +1 for all i < n . We prove, by induction on i , that a i → a n for all i < n . In particular, a n − → a n , a contradiction. The case i = 1 holds since a n x n loses to a x . Now let 1 < i < n . By induction hypothesis, a i − → a n .Since a i / ∈ N ++ ( a i − ) and a i a n is not a missing edge we must have ( a i , a n ) ∈ D . Theorem 2.
Every digraph missing a comb satisfies SNC.Proof.
Let D be a digraph missing a comb G . We follow the previous notations.Let P = a x , a x , ... be a maximal directed path in ∆ ( a i ∈ A ). By lemma 1, a x has a convenient orientation. Suppose ( a , x ) is a convenient orientation.In this case add ( a i , x i ) and ( x i +1 , a i +1 ) to D . Otherwise, we orient in thereverse direction. We do this for all such paths of ∆. The obtained digraph D ′ is missing the complete graph G [ X ∪ Y ]. Clearly, all the missing edges of D ′ are good (in D ′ ), so we give each one a convenient orientation and add it to D ′ .The obtained digraph T is a tournament. Let L be a local median order of T and let f denote its feed vertex. By theorem 1, f has the SNP in T . We claimthat f has the SNP in D as well.Suppose f is a whole vertex. We show that f gains no vertex in its secondout-neighborhood and hence our claim holds. Assume f → u → v → f in T .Since f is whole, f → u in D . If u → v in D ′ − D , then it is either a convenientorientation and hence v ∈ N ++ ( f ) or there is a missing edge rs that loses to uv , namely s → v and u / ∈ N + ( s ) ∪ N ++ ( s ). However, f s is not a missing edge,then we must have f → s . Whence v ∈ N ++ ( f ). Now, if u → v in T − D ′ then v ∈ N ++ D ′ ( f ). But this case is already discussed. This argument is usedimplicitly in the rest of the proof.Suppose f ∈ A . There is a maximal directed path P = a x , ..., a i x i , ..., a k x k with f = a i . If ( x i , a i ) ∈ D ′ then d + ( f ) = d + T ( f ) ≤ d ++ T ( f ) = d ++ ( f ). In fact f gains no new first nor second out-neighbor. Otherwise ( a i , x i ) ∈ D ′ . If i < k , f gains only x i (resp. a i +1 ) as a first (resp. second ) out-neighbor. If i = k , wereorient a k x k as ( x k , a k ). The same order L is also a local median order of T ′ the modified tournament. Now f gains no vertex in its second out-neighborhood.Suppose f ∈ X . There is a maximal directed path P = a x , ..., a i x i , ..., a k x k with f = x i . If ( a i , x i ) ∈ D ′ we reorient all the missing edges incident to x i towards x i . In this case f gains no new first nor second out-neighbor in themodified tournament. Otherwise ( x i , a i ) ∈ D ′ . If i = k , we reorient all themissing edges incident to x i towards x i . In this case f gains no new first norsecond out-neighbor in the modified tournament. If i < k , we reorient all themissing edges incident to x i towards x i except ( x i , a i ). In this case f gainsonly a i (resp. x i +1 ) as a first (resp. second ) out-neighbor in the modifiedtournament.Suppose f ∈ Y . Reoreient all the missing edges incident to y towards y . Inthe modified tournament f gains no vertex in its second out-neighborhood.Therefore D satisfies SNC. 4 ˜ K is a graph obtained from the complete graph by removing 2 nonadjacent edges. If xy and uv are the removed edges then ˜ K restricted to { x, y, u, v } is a cycle of length 4. Proposition 3.
The graphs ˜ K are ~ P -forcing.Proof. This is clear because the dependency digraph can have at most onearc.
Theorem 3.
Every digraph whose missing graph is a ˜ K satisfies SNC.Proof. Let D be a digraph missing a ˜ K . If ∆ has no arc then D satisfies SNCby lemma 2. Otherwise, it has exactly one arc, say xy → uv with x → u and v / ∈ N ++ ( v ). Note that the cycle C = xyuv is an induced cycle in the missinggraph. We may suppose that ( x, y ) is a convenient orientation. Add ( x, y ) and( u, v ) to D . The rest of the missing edges are good missing edges. So we givethem a convenient orientation and add to D . The obtained digraph T is a tour-nament. Let L be a local median order of T and let f denote its feed vertex.Now f has the SNP in T . We discuss according to f .Suppose f is a whole vertex. Then f gains no vertex in its second out-neighborhood.Suppose f = x . Reorient all the missing edges incident to x towards x except( x, y ). The same order L is a local median order of the modified tournament T ′ . The only new first (resp. second) out-neighbor of f is y (resp. v ).Suppose f = y , u , v or a non whole vertex that does not belong C . Reorientall the missing edges incident to f towards f . In the modified tournament, f gains no vertex in its second out-neighborhood.A ˜ K is a graph obtained from the complete graph by removing a cycle oflength 5. Note that ˜ K restricted to the vertices of the removed cycle is also acycle of length 5.In the following ab → cd means ab loses to cd , namely, a → c and b → d (theorder of the endpoints is considered). Let D be a digraph missing ˜ K and let ∆denote its dependency digraph. Let C = xyzuv be the induced cyle of length 5in ˜ K . Checking by cases, we find that ∆ has at most 3 arcs. If ∆ has exactly3 arcs then its arcs are (isomorphic to) uv → xy → zu → vx or uv → xy → zu and xv → zy .If ∆ has exactly 2 arcs then they are (isomorphic to) uv → xy → zu or uv → xy and vx → yz .If ∆ has exactly 1 arc then it is (isomorphic to) uv → xy . So we have thefollowing. Proposition 4.
The graphs ˜ K are ~ P -forcing. Theorem 4.
Every digraph whose missing graph is a ˜ K satisfies SNC. roof. Let D be a digraph missing a ˜ K . Let C = xyzuv be the induced cyleof length 5 in ˜ K . If ∆ has no arcs then D satisfies SNC by lemma 2.Suppose ∆ has exactly one arc uv → xy . Without loss of generality, we mayassume that ( u, v ) is a convenient orientation. Add ( u, v ) and ( x, y ) to D . Wegive the rest of the missing edges (they are good) convenient orientatins andthem add to D . Let L be a local median order of the obtained tournament T and let f denote its feed vertex. f has the SNP in T . If f = u the only newfirst (resp. second) out-neighbor of u is v (resp. y ). Whence f has the SNP in D . Otherwise, we reorient all the missing edges incident to f towards f , if anyexist. The same order L is a local median order of the new tournament T ′ and f has the SNP in T ′ . However, f gains neither a new first out-neighbor nor anew second out-neighbor. So f has the SNP in D .Suppose ∆ has exactly 2 arcs, say uv → xy and vx → yz . We may assumethat ( u, v ) is a convenient orientation. Add ( u, v ) and ( x, y ) to D . If ( v, x )is a convenient orientation, we add ( v, x ) and ( y, z ) to D , otherwise we addtheir reverse. We give the rest of the missing edges (they are good) convenientorientatins and add them to D . Let L be a local median order of the obtainedtournament H and let f denote its feed vertex. We reorient every missing edgeincident to f , whose other endpoint is not in { u, v, x } , towards f if any exists. The same L is a local median order of the new tournament T and f has theSNP in T .If f / ∈ { u, v, x } then it gains neither a new first out-neighbor nor a newsecond out-neighbor. So f has the SNP in D .If f = u , then the only new first (resp. second ) out-neighbor of f is v (resp. y ), whence f has the SNP in D .If f = v either v → x in T and in this case the only new first (resp. second) out-neighbor of v is x (resp. z ) or x → v and in this case f gains neither anew first out-neighbor nor a new second out-neighbor. Whence f has the SNPin D .If f = x we reorient xy as ( y, x ). The same L is a local median order of thenew tournament T ′ and f has the SNP in T ′ . If v → x in T ′ then f gains neithera new first out-neighbor nor a new second out-neighbor. Otherwis, x → v in T ′ then the only new first (resp. second ) out-neighbor of f is v (resp. y ). Whence f has the SNP in D .Suppose ∆ has exactly 2 arcs with uv → xy → zu . We may assume that( u, v ) is a convenient orientation. Add ( u, v ), ( x, y ) and ( z, u ) to D . We give therest of the missing edges (they are good) convenient orientatins and add themto D . Let L be a local median order of the obtained tournament H and let f denote its feed vertex. We reorient every missing edge incident to f , whoseother endpoint is not in { u, v, x, y, z } , towards f if any exists . The same L isa local median order of the new tournament T and f has the SNP in T .If f / ∈ { u, v, x, y, z } then it gains neither a new first out-neighbor nor a newsecond out-neighbor. So f has the SNP in D .If f = u , then the only new first (resp. second ) out-neighbor of f is v (resp. y ), whence f has the SNP in D .If f = v we orient xv as ( x, v ). The same L is a local median order of thenew tournament T ′ and f has the SNP in T ′ .6f f = x we orient xv as ( v, x ). The same L is a local median order of thenew tournament T ′ and f has the SNP in T ′ . The only new first (resp. second)out-neighbor of f is y (resp. u ). Whence f has the SNP in D .If f = y we orient yz as ( z, y ). The same L is a local median order ofthe new tournament T ′ and f has the SNP in T ′ . f gains neither a new firstout-neighbor nor a new second out-neighbor. So f has the SNP in D .If f = z we orient yz and zu towards z . The same L is a local median orderof the new tournament T ′ and f has the SNP in T ′ . f gains neither a new firstout-neighbor nor a new second out-neighbor. So f has the SNP in D .Suppose ∆ has exactly 3 arcs with uv → xy → zu → vx . We may assumethat ( u, v ) is a convenient orientation. Add ( u, v ), ( x, y ), ( z, u ) and ( v, x ) to D .We give the rest of the missing edges (they are good) convenient orientatins andthen add to D . Let L be a local median order of the obtained tournament H and let f denote its feeed vertex. We reorient every missing edge incident to f ,whose other endpoint is not in { u, v, x, y, z } , towards f if any exists . The same L is a local median order of the new tournament T and f has the SNP in T .If f / ∈ { u, v, x, y, z } then it gains neither a new first out-neighbor nor a newsecond out-neighbor. So f has the SNP in D .If f = u , then the only new first (resp. second ) out-neighbor of f is v (resp. y ), whence f has the SNP in D .If f = v we orient xv as ( x, v ). The same L is a local median order of thenew tournament T ′ and f has the SNP in T ′ . In this case f gains neither a newfirst out-neighbor nor a new second out-neighbor. So f has the SNP in D .If f = x , the only new first (resp. second ) out-neighbor of f is y (resp. u ).Whence f has the SNP in D .If f = y we orient yz as ( z, y ). The same L is a local median order ofthe new tournament T ′ and f has the SNP in T ′ . f gains neither a new firstout-neighbor nor a new second out-neighbor. So f has the SNP in D .If f = z we orient yz towards z . The same L is a local median order of thenew tournament T ′ and f has the SNP in T ′ . The only new first (resp. second)out-neighbor of f is u (resp. x ). Whence f has the SNP in D .Finally, suppose ∆ has exactly 3 arcs with uv → xy → zu and xv → zy . Wemay assume that ( u, v ) is a convenient orientation. Add ( u, v ), ( x, y ) and ( z, u )to D . Note that xv is a good missing edge. If ( x, v ) is a convenient orientationadd it with ( z, y ), otherwise we add the reverse of these arcs. We give the restof the missing edges (they are good) convenient orientatins and add them to D . Let L be a local median order of the obtained tournament H and let f denote its feeed vertex. We reorient every missing edge incident to f , whoseother endpoint is not in { u, v, x, y, z } , towards f if any exists. The same L is alocal median order of the new tournament T and f has the SNP in T .If f / ∈ { u, v, x, y, z } then it gains neither a new first out-neighbor nor a newsecond out-neighbor. So f has the SNP in D .If f = u , then the only new first (resp. second ) out-neighbor of f is v (resp. y ), whence f has the SNP in D .If f = v either x → f = v in T and in this case it gains neither a new firstout-neighbor nor a new second out-neighbor or f = v → x and in this case theonly new first (resp. second ) out-neighbor of f is x (resp. z ). Whence f hasthe SNP in D . 7f f = x either v → f = x in T and in this case the only new first (resp.second ) out-neighbor of f is y (resp. z ) or f = x → v and in this case the onlynew first (resp. second ) out-neighbor of f are y and v (resp. z ). Whence f hasthe SNP in D .If f = y or z , we orient the missing edges incident to f towards f . The same L is a local median order of the new tournament T ′ and f has the SNP in T ′ . f gains neither a new first out-neighbor nor a new second out-neighbor. So f has the SNP in D .Digraphs missing a matching are the digraphs with minimum degree | V ( D ) |−
2. These digraphs satisfies SNC [4]. A more general class of digraphs is the classof digraphs with minimum degree at least | V ( D ) | −
3. The missing graph ofsuch a digraph is composed of vertex disjoint directed paths and directed cycles. P is the path of length 3 and C , C and C are the cycles of length 3, 4 and5 respectively. Theorems 2, 3 and 4 implies the following. Corollary 1.
Every digraph whose missing graph is P , C , C or a C satisfiesSNC. References [1] S. Ghazal,
Seymour’s second neighborhood conjecture in tournaments miss-ing a generalized star , J. Graph Theory ( accepted, 24 June 2011).[2] N. Dean and B. J. Latka, squaring the tournament: an open problem ,Congress Numerantium 109 (1995), 73-80.[3] F. Havet and S. Thomass´e,
Median Orders of Tournaments: A Tool for theSecond Neighborhood Problem and Sumner’s Conjecture , J. Graph Theory35 (2000), 244-256.[4] D. Fidler and R. Yuster,
Remarks on the second neighborhood problem , J.Graph Theory 55 (2007), 208-220.[5] D. Fisher, squaring a tournament: a proof of Dean’s conjecturesquaring a tournament: a proof of Dean’s conjecture