Obstructions for local tournament orientation completions
aa r X i v : . [ m a t h . C O ] M a r Obstructions for local tournament orientationcompletions
Kevin Hsu and Jing Huang ∗ Abstract
The orientation completion problem for a class of oriented graphs asks whethera given partially oriented graph can be completed to an oriented graph in the classby orienting the unoriented edges of the partially oriented graph. Orientation com-pletion problems have been studied recently for several classes of oriented graphs,yielding both polynomial time solutions as well as NP-completeness results.Local tournaments are a well-structured class of oriented graphs that generalizetournaments and their underlying graphs are intimately related to proper circular-arc graphs. According to Skrien, a connected graph can be oriented as a localtournament if and only if it is a proper circular-arc graph. Proper interval graphsare precisely the graphs which can be oriented as acyclic local tournaments. Ithas been proved that the orientation completion problems for the classes of localtournaments and acyclic local tournaments are both polynomial time solvable.In this paper we characterize the partially oriented graphs that can be completedto local tournaments by determining the complete list of obstructions. These arein a sense minimal partially oriented graphs that cannot be completed to localtournaments. The result may be viewed as an extension of the well-known forbiddensubgraph characterization of proper circular-arc graphs obtained by Tucker. Thecomplete list of obstructions for acyclic local tournament orientation completionshas been given in a companion paper.
We consider graphs, digraphs, and partially oriented graphs in this paper. For graphswe assume that they do not contain loops or multiple edges (i.e., they are simple ), and ∗ Department of Mathematics and Statistics, University of Victoria, Victoria, B.C., Canada V8W2Y2; [email protected]; Research supported by NSERC oriented graphs).A partially oriented graph is a mixed graph H obtained from some graph G by orientingthe edges in a subset of the edge set of G . The graph G is called the underlying graph of H . We denote H by ( V, E ∪ A ) where E is the set of (non-oriented) edges and A is theset of arcs in H . We use uv to denote an edge in E with endvertices u, v and use ( u, v )to denote an arc in A with tail u and head v . In either case we say that u, v are adjacent in H . The partially oriented graph H is connected if its underlying graph G is.A class C of graphs is called hereditary if it is closed under taking induced subgraphs,that is, if G ∈ C and G ′ is an induced subgraph of G then G ′ ∈ C . Similarly, a class ofdigraphs is hereditary if it is closed under taking induced subdigraphs. We extend thisconcept to partially oriented graphs.Let H = ( V, E ∪ A ) and H ′ = ( V ′ , E ′ ∪ A ′ ) be partially oriented graphs. We says that H critically contains H ′ (or H ′ is critically contained in H ) if V ′ ⊆ V and for all u, v ∈ V ′ , • u and v are adjacent in H ′ if and only if they are adjacent in H ; • if ( u, v ) ∈ A ′ then ( u, v ) ∈ A ; • if uv ∈ E ′ , then uv ∈ E , or ( u, v ) ∈ A , or ( v, u ) ∈ A .. Equivalently, H ′ is critically contained in H if and only if it is obtained from H bydeleting some vertices, followed by replacing some arcs ( u, v ) with edges uv .We note that, in case when H and H ′ are both graphs or digraphs, H critically contains H ′ if and only if H contains H ′ as an induced subgraph or as an induced subdigraph. Wecall a class C of partially oriented graphs hereditary if H ∈ C and H ′ is critically containedin H then H ′ ∈ C .Fix a hereditary class C of oriented graphs, the orientation completion problem for C asks whether a given partially oriented graph H = ( V, E ∪ A ) can be completed to anoriented graph in C by orienting the edges in E . The hereditary property of C ensuresthat if a partially oriented graph H can be completed to an oriented graph in C thenevery partially oriented graph that is critically contained in H can also be completed toan oriented graph in C . Therefore the partially oriented graphs which can be completedto oriented graphs in C form a hereditary class.Orientation completion problems have been studied for several classes of orientedgraphs, including local tournaments, local transitive tournaments, and acyclic local tour-naments, cf. [2, 9]. A local tournament is an oriented graph in which the in-neighbourhoodas well as the out-neighbourhood of each vertex induces a tournament. When the thein-neighbourhood as well as the out-neighbourhood of each vertex induces a transitivetournament, the oriented graph is called a local transitive tournament . If a local tourna-ment does not contain a directed cycle then it is called acyclic . These three classes oforiented graphs are nested; the class of local tournaments properly contains local transi-tive tournaments, which in turn as a class properly contains acyclic local tournaments. It2as been proved in [2] that the orientation completion problem is polynomial time solv-able for local tournaments and for acyclic local tournaments, but NP-complete for localtransitive tournaments.Any hereditary class of graphs or digraphs admits a characterization by forbiddensubgraphs or subdigraphs. The forbidden subgraphs or subdigraphs consists of minimalgraphs or digraphs which do not belong to the class and they are sometimes called ob-structions for the class. It turns out this is also the case for a hereditary class of partiallyoriented graphs and in particular for the class of partially oriented graphs which can becompleted to local tournaments.We call a partially oriented graph X = ( V, E ∪ A ) an obstruction for local tournamentorientation completions (or simply, an obstruction ) if the following three properties hold:1. X cannot be completed to a local tournament;2. For each v ∈ V , X − v can be completed to a local tournament;3. For each ( u, v ) ∈ A , the partially oriented graph obtained from X by replacing ( u, v )with the edge uv can be completed to a local tournament.Thus an obstruction X is a partially oriented graph which cannot be completed to alocal tournament and is minimal in the sense that if X ′ is critically contained in X and X ′ = X then X ′ can be completed to a local tournament.Clearly, every obstruction must be connected. If an obstruction contains no arc then itis a forbidden subgraph for the class of graphs which can be oriented as local tournaments(or equivalently, proper circular-arc graphs, cf. Theorem 2.1). We shall prove that if anobstruction contains arcs then it contains exactly two arcs. The dual of an obstruction X is the one obtained from X by reversing the arcs (if any) in X . If an obstruction X does not contain arcs then the dual of X is X itself. Clearly, the dual of an obstructionis again an obstruction. Proposition 1.1.
A partially oriented graph H cannot be completed to a local tourna-ment if and only if it critically contains an obstruction. Proof: If H can be completed to a local tournament, then every partially orientedgraph critically contained in H can also be completed to a local tournament so H doesnot critically contain an obstruction. On the other hand, suppose that H cannot becompleted to a local tournament. By deleting vertices and replacing arcs with edges in H as long as the resulting partially oriented graph still cannot be completed to a localtournament we obtain an obstruction that is critically contained in H .We will find in this paper all obstructions for local tournament orientation completions.In particular, we will prove the following: Theorem 1.2.
Let X be an obstruction for local tournament orientation completions.Then one of the following statements holds: X has no arcs and X is a C k with k ≥ , a C k +1 + K with k ≥ , or a graph inFigure 1; • X has exactly two arcs, and X or its dual is a graph in Figures 2–7; • X has exactly two arcs and U ( X ) is a graph in Figures 8–14 (with arcs in X or itsdual being specified in the figures). A proper circular-arc graph is the intersection graph of a family of circular-arcs on acircle where no circular-arc contains another. Skrien [11] proved that a connected graphis a proper circular-arc graph if and only if it can be oriented as a local tournament. Itfollows that if a partially oriented graph H can be completed to a local tournament theneach component of the underlying graph of H is a proper circular-arc graph.Tucker [12] found all minimal graphs which are not proper circular-arc graphs. Theorem 2.1. [12] A graph G is a proper circular-arc graph if and only if G does notcontain C k + K ( k ≥ ) or tent + K and G does not contain C k ( k ≥ ), C k +1 + K ( k ≥ ), or any of the graphs in Figure 1 as an induced subgraph. tent Figure 1
Corollary 2.2.
Let X be an obstruction for local tournament orientation completions.If X has no arc then X is a C k with k ≥ , a C k +1 + K with k ≥ , or a graph inFigure 1. Proof:
Since X is an obstruction for local tournament orientation completions thathas no arc, it is a forbidden subgraph for proper circular arc graphs. Since X is connected, X is not a C k + K ( k ≥
4) or tent + K . Hence, by Theorem 2.1, X is a C k with k ≥ C k +1 + K with k ≥
1, or a graph in Figure 1.4n view of Corollary 2.2 we only need to find obstructions that contain arcs. Bydefinition the underlying graph of any obstruction that contains arcs is a proper circular-arc graph and hence local tournament orientable.Let G = ( V, E ) be a graph and Z ( G ) = { ( u, v ) : uv ∈ E } be the set of all orderedpairs ( u, v ) such that uv ∈ E . Note that each edge uv ∈ E gives rise to two ordered pairs( u, v ) , ( v, u ) in Z ( G ). Suppose that ( u, v ) and ( x, y ) are two ordered pairs of Z ( G ). Wesay ( u, v ) forces ( x, y ) and write ( u, v )Γ( x, y ) if one of the following conditions is satisfied: • u = x and v = y ; • u = y , v = x , and vx / ∈ E ; • v = x , u = y , and uy / ∈ E .We say that ( u, v ) implies ( x, y ) and write ( u, v )Γ ∗ ( x, y ) if there exists a sequence of pairs( u , v ) , ( u , v ) , . . . , ( u k , v k ) ∈ Z ( G ) such that( u, v ) = ( u , v )Γ( u , v )Γ . . . Γ( u k , v k ) = ( x, y ) . We will call such a sequence a Γ- sequence from ( u, v ) to ( x, y ). It is easy to verify thatΓ ∗ is an equivalence relation on Z ( G ).We say a path P avoids a vertex u if P does not contain u or any neighbour of u . Proposition 2.3.
Let G be a graph and u, v, w be vertices Suppose that P is a path oflength k connecting v, w that avoids u in G . If k is even, then ( u, v )Γ ∗ ( u, w ) . Otherwise, ( u, v )Γ ∗ ( w, u ) . Proof:
Denote P : v v . . . v k where v = v and v k = w . Since P avoids u in G ,( u, p i )Γ( p i +1 , u ) for each 0 ≤ i ≤ k −
1. If k is even, then( u, v ) = ( u, p )Γ( p , u )Γ( u, p )Γ . . . Γ( u, p k ) = ( u, w ) . Otherwise, ( u, v ) = ( u, p )Γ( p , u )Γ( u, p )Γ . . . Γ( p k , u ) = ( w, u ) . Proposition 2.4. [8] Let G be a graph and D = ( V, A ) be a local tournament orien-tation of G . Suppose that ( u, v )Γ ∗ ( x, y ) for some ( u, v ) , ( x, y ) ∈ Z ( G ) . Then ( u, v ) ∈ A ifand only if ( x, y ) ∈ A . Regardless whether or not G is local tournament orientable, the relation Γ ∗ on Z ( G )induces a partition of the edge set of G into implication classes as follows: two edges uv, xy of G are in the same implication class if and only if ( u, v )Γ ∗ ( x, y ) or ( u, v )Γ ∗ ( y, x ).An implication class is called trivial if it has only one edge and non-trivial otherwise.An edge uv of G is called balanced if N [ u ] = N [ v ] and unbalanced otherwise. Clearly,any balanced edge forms a trivial implication class and the unique edge in any trivialimplication class is balanced.The following theorem characterizes the implication classes of a local tournamentorientable graph and describes all possible local tournament orientations of such a graph.5 heorem 2.5. [8] Let G = ( V, E ) be a connected graph and let H , H , . . . , H k be thecomponents of G . Suppose that G is local tournament orientable and F is an implicationclass of G . Then F is one of the following types: • F is trivial; • F consists of all unbalanced edges of G within H i for some i ; • F consists of all edges of G between H i and H j for some i = j .Moreover, suppose that F , F , . . . , F ℓ are the implication classes of G . For each ≤ i ≤ ℓ ,let A i be the equivalence class of Γ ∗ containing ( u, v ) for some uv ∈ F i and let A = ∪ ℓi =1 A i .Then D = ( V, A ) is a local tournament orientation of G . Let H = ( V, E ∪ A ) be a partially oriented graph and ( a, b ) , ( c, d ) be arcs of H . Wesay that the two arcs ( a, b ) , ( c, d ) are opposing in H if ( a, b )Γ ∗ ( d, c ). For convenience wealso call an arc of H balanced if the corresponding edge is balanced. Clearly, if ( a, b ) , ( c, d )are opposing then neither of them is balanced. Proposition 2.6.
Suppose that H is a partially oriented graph whose underlying graph U ( H ) is local tournament orientable. Then H can be completed to a local tournament ifand only if it does not contain opposing arcs. Proof: If H contains opposing arcs, then by Proposition 2.4 it cannot be completedto a local tournament. On the other hand, suppose that H does not contain opposingarcs. Let F , F , . . . , F ℓ be the implication classes of U ( H ). For each 1 ≤ i ≤ ℓ , if noedge in F i is oriented then let A i be an equivalence class of Γ ∗ containing ( u, v ) for some uv ∈ F i ; otherwise let A i be the equivalence class of Γ ∗ containing ( u, v ) where uv ∈ F i and ( u, v ) is an arc. With A = ∪ ℓi =1 A i , Theorem 2.5 ensures that D = ( V, A ) is a localtournament completion of H .The next theorem is fundamental in determining whether a partially oriented graphwhose underlying graph is local tournament orientable is an obstruction. Theorem 2.7.
Let X be a partially oriented graph whose underlying graph U ( X ) islocal tournament orientable. Then X is an obstruction if and only if X contains exactlytwo arcs ( a, b ) , ( c, d ) which are opposing and, for every vertex v ∈ V ( X ) \ { a, b, c, d } , thearcs ( a, b ) , ( c, d ) are not opposing in X − v (that is, the edges ab, cd belong to differentimplication classes in U ( X − v ) ). Moreover, any Γ -sequence connecting ( a, b ) and ( d, c ) must include all vertices of X . Proof:
For sufficiency, suppose that ( a, b ) , ( c, d ) are the only arcs and they are op-posing in X and that, for every vertex v ∈ V ( X ) \ { a, b, c, d } , the arcs ( a, b ) , ( c, d ) are notopposing in X − v . Since X contains opposing arcs, it cannot be completed to a localtournament by Proposition 2.6. Let v be a vertex in X . Since U ( X ) is local tournamentorientable, U ( X − v ) is also local tournament orientable. If v ∈ { a, b, c, d } , then X − v contains at most one arc and hence no opposing arcs. If v / ∈ { a, b, c, d } , then the only two6rcs in X − v are not opposing by assumption. Hence X − v can be completed to a localtournament by Proposition 2.6. Therefore X is an obstruction.Conversely, suppose that X is an obstruction. By Proposition 2.6 X must containopposing arcs. Let ( a, b ) , ( c, d ) be opposing arcs in X . If X contains an arc ( x, y ) thatis distinct from ( a, b ) , ( c, d ), then replacing the arc ( x, y ) by the edge xy gives a partiallyorientable graph in which ( a, b ) , ( c, d ) are still opposing and hence cannot be completedto a local tournament. This contradicts the assumption that X is an obstruction. So( a, b ) , ( c, d ) are the only arcs in X . Since X is an obstruction, for every v ∈ V ( X ), X − v can be completed to a local tournament and hence by Proposition 2.6 contains no opposingarcs. This implies in particular that if v ∈ V ( X ) \ { a, b, c, d } , the arcs ( a, b ) , ( c, d ) are notopposing in X − v .The second part of the theorem follows from the fact that deleting any vertex resultsin a graph that contains no Γ-sequence connecting ( a, b ) and ( d, c ).Let v be a vertex and ( x, y ) be an arc in a partially oriented graph H where v / ∈ { x, y } .We call v the ( x, y ) -balancing vertex if v is the only vertex adjacent to exactly one of x, y ;when the arc ( x, y ) does not need to be specified, we simply call v an arc-balancing vertex.Each obstruction has at most two arc-balancing vertices as it contains at most two arcs.A vertex of a graph G is called a cut-vertex of G if G − v has more components than G . For a partially oriented graph H , a cut-vertex of U ( H ) is also called a cut-vertex of H . Proposition 2.8.
Let X be an obstruction with opposing arcs ( a, b ) , ( c, d ) and let v / ∈ { a, b, c, d } . Then v is an arc-balancing vertex, or a cut-vertex of U ( X ) , or a cut-vertex of U ( X ) . Proof:
Assume that v is not a cut-vertex of U ( X ) or of U ( X ) as otherwise we aredone. We show that v must be an arc-balancing vertex. Since ab, cd are in the sameimplication class of U ( X ), by Theorem 2.5 ab, cd are unbalanced edges either containedin a component or between two components of U ( X ). Since v is not a cut-vertex of U ( X ), each component of U ( X − v ) is a component of U ( X ) except possibly missing v .It follows that ab, cd are contained in some component or between two components of U ( X − v ). Since v is not a cut-vertex of U ( X ), U ( X − v ) is connected. If ab, cd are bothunbalanced edges in U ( X − v ), then they remain in the same implication class of U ( X − v )and hence ( a, b ) , ( c, d ) are still opposing in X − v , which contradicts the assumption that X is an obstruction. So one of ab, cd is balanced in U ( X − v ), which means that v is( a, b )-balancing or ( c, d )-balancing.An arc-balancing triple in a partially oriented graph H is a set of three vertices inwhich one balances an arc between the other two. Corollary 2.9.
Let X be an obstruction with opposing arcs ( a, b ) , ( c, d ) . Suppose that U ( X ) has no cut-vertices. Then U ( X ) contains at most six non-cut-vertices. In thecase when U ( X ) has six non-cut-vertices, the six non-cut-vertices form two disjoint arc-balancing triples. roof: Let v be a non-cut-vertex of U ( X ). By assumption v is not a cut-vertex of U ( X ) and thus, by Proposition 2.8, it is either in { a, b, c, d } or an arc-balancing vertex.There are at most two arc-balancing vertices so U ( X ) contains at most six non-cut-vertices. When U ( X ) has six non-cut-vertices, among the six non-cut-vertices two arearc-balancing vertices and the other four are incident with arcs. Hence the six non-cut-vertices form two disjoint arc-balancing triples.A proper interval graph is the intersection graph of a family of intervals in a line whereno interval contains another. Proper interval graphs form a prominent subclass of propercircular-arc graphs and play an important role in the orientation completion problem forlocal tournaments. It is proved in [6] that a graph is a proper interval graph if and onlyif it can be oriented as an acyclic local tournament.A straight enumeration of a graph G is a vertex ordering ≺ such that for all u ≺ v ≺ w ,if uw is an edge of G , then both uv and vw are edges. This property is referred to as the umbrella property of the vertex ordering. Proposition 2.10. [8] A graph is a proper interval graph if and only if it has a straightenumeration.
Proposition 2.11.
Let G be a connected proper interval graph and let ≺ be a straightenumeration of G . Suppose that ( u, v )Γ ∗ ( x, y ) . Then u ≺ v if and only if x ≺ y . Proof:
It suffices to show that if u ≺ v and ( u, v )Γ( x, y ) then x ≺ y . So assume that( u, v )Γ( x, y ). Then one of the following holds: • u = x and v = y ; • u = y , v = x , and vx / ∈ E ( G ); • v = x , u = y , and uy / ∈ E ( G ).Clearly, x ≺ y when u = x and v = y . Suppose that u = y , v = x , and vx / ∈ E ( G ). If u ≺ x ≺ v , then it violates the umbrella property because uv ∈ E ( G ) but xv / ∈ E ( G ).If u ≺ v ≺ x , then it again violates the umbrella property because ux ∈ E ( G ) but vx / ∈ E ( G ). Hence we must have x ≺ u = y . The proof for the case when v = x , u = y ,and uy / ∈ E ( G ) is similar.Let H be a partially oriented graph whose underlying graph U ( H ) is a proper intervalgraph. Suppose that ≺ is a straight enumeration of U ( H ). We call an arc ( u, v ) of H positive (with respect to ≺ ) if u ≺ v and negative otherwise. If H does not containnegative arcs, then H can be completed to an acyclic local tournament by replacing alledges of H with positive arcs. Similarly, if H does not contain positive arcs then it canalso be completed to an acyclic local tournament. It follows that if X is an obstructionsuch that U ( X ) is a proper interval graph, then the two arcs in X must be opposite (i.e.,one is positive and the other is negative).A vertex in a graph is universal if it is adjacent to every other vertex.8 heorem 2.12. [8] Suppose that G = ( V, E ) is a connected proper interval graph thatis not a complete graph. Then G has a unique non-trivial component H . If F is animplication class of G , then F is one of the following types: • F is trivial; • F consists of all unbalanced edges within H ; • F consists of all edges of G between H and a universal vertex of G .In particular, if G contains no universal vertex, then G has a unique non-trivial implica-tion class. Proposition 2.13.
Let G be a connected proper interval graph and let v , v , . . . , v n bea straight enumeration of G . Suppose that v α is a cut-vertex of G . Then α ∈ { , n } and G − v α contains a vertex that is adjacent to every vertex except v α in G . Proof:
Since G has a cut-vertex, G is not a complete graph and by Theorem 2.12, G has a unique non-trivial component H . Thus the cut-vertex v α of G is in fact a cut-vertexof H . Again by Theorem 2.12, H − v α has at most one non-trivial component. Hence H contains a vertex v β that is only adjacent to v α in G , that is, in G it is adjacent to everyvertex except v α . If α < β , then α = 1 as otherwise we have 1 < α < β and v β is adjacentto v but not to v α , a contradiction to the umbrella property of the straight enumeration.Similarly, if β < α , then α = n as otherwise β < α < n and v β is adjacent to v n but not to v α , also a contradiction to the umbrella property of the straight enumeration. Therefore, α ∈ { , n } . Our goal is to find all obstructions for local tournament orientation completions thatcontains arcs. By Theorem 2.7 each of them contains exactly two arcs which are opposingand its underlying graph is a connected proper circular arc graph (i.e., local tournamentorientable). In this section, we examine such obstructions that contain cut-vertices. Sincethey contain cut-vertices, their underlying graphs are necessarily proper interval graphsand thus have straight enumerations according to Proposition 2.10.Let X be an obstruction that contains two arcs and let ≺ be a straight enumerationof U ( X ). Suppose that v is a cut-vertex of X . Then v is neither the first nor the lastvertex in ≺ and moreover, for all u, w with u ≺ v ≺ w , uw is not an edge in U ( X ). Wecall v a dividing cut-vertex if one of the two arcs in X is incident with a vertex preceding v and the other is incident with a vertex succeeding v in ≺ ; if v is not dividing then it iscalled non-dividing . 9 .1 Dividing cut-vertices In this subsection, we focus on the obstructions that contain dividing cut-vertices. Wewill show that they consist of the three infinite classes in Figure 2 and their duals. Ineach of these graphs, the dots in the middle represent a path of length ≥
0; when thelength of the path is 0 the two vertices beside the dots are the same vertex. . . . (i) . . . (ii) . . . (iii)Figure 2: Obstructions with dividing cut-vertices.
Lemma 3.1.
Let X be an obstruction that contains a dividing cut-vertex and let ≺ : v , v , . . . , v n be a straight enumeration of U ( X ) . Suppose that v c is the first dividing cut-vertex in ≺ . Then, either c = 2 and v , v are the endvertices of an arc, or c = 4 and v , v are the end vertices of an arc. In the case when c = 4 , v , v , v , v induce in U ( X ) the following graph: v v v v Proof:
By considering the dual of X if necessary we may assume that ( v j , v k ) and( v s , v t ) are the two arcs in X where j < k ≤ c ≤ t < s . By Theorem 2.7, there is aΓ-sequence from ( v t , v s ) to ( v j , v k ) that includes all vertices of X . Let( v t , v s ) = ( u , w )Γ( u , w )Γ . . . Γ( u q , w q ) = ( v j , v k )be a shortest such a sequence. Since v t ≺ v s , we have u i ≺ w i for each i by Proposition2.11. Let ℓ be the smallest subscript such that u ℓ +1 ≺ w ℓ +1 = v c = u ℓ ≺ w ℓ . Such ℓ exists because v c is a cut-vertex dividing ( v j , v k ) and ( v s , v t ). We distinguish two casesdepending on whether or not k = c . Suppose first k = c . Note that ( u ℓ , w ℓ )Γ( v j , v k ).Thus the choice of the Γ-sequence implies ( u ℓ +1 , w ℓ +1 ) = ( u q , w q ) = ( v j , v k ). Since theΓ-sequence includes all vertices of X , v j is the only vertex preceding v c in ≺ , that is, c = 2(and ( v , v ) is an arc in X ).Suppose now that k < c . Thus j < k < c . We claim that v j , v k , v c are consecutivevertices in ≺ (i.e., j + 1 = k = c − k > j + 1. Since v j , v k are adjacent, v j +1 cannot be a cut-vertex of U ( X ). Since v j +1 is not the first or the last vertex in ≺ , Proposition 2.13 ensures that v j +1 cannot be a cut-vertex of U ( X ). By Proposition2.8, v j +1 is an arc-balancing vertex. Clearly, v j +1 is not ( v j , v k )-balancing. So it must be10 v s , v t )-balancing. Since j +1 < c and v c is a cut-vertex, v j +1 has no neighbours succeeding v c . It follows that v t = v c . Since v j +1 v c is an edge and j + 1 < k < c , v k v c is an edge bythe umbrella property. Again, since v c is a cut-vertex, v k cannot be adjacent to v s . Thiscontradicts the fact that v j +1 is arc-balancing for the arc between v s , v t . Hence j + 1 = k ,i.e., v j and v k are consecutive vertices in ≺ .Suppose c > k + 1. Neither of v k , v k +1 can be a cut-vertex of U ( X ) as otherwise itwould be a dividing cut-vertex preceding v c , a contradiction to the choice of v c . Since v k +1 is not the first or the last vertex in ≺ , it is not a cut-vertex of U ( X ) according toProposition 2.13. By Proposition 2.8, v k +1 is an arc-balancing vertex. Since v k is not acut-vertex of U ( X ), v k − = v j is adjacent to v k +1 . So v k +1 is adjacent to both v j , v k andhence not arc-balancing for the the arc between them. So v k +1 is arc-balancing for thearc between v s , v t . Similarly as above we have v c = v t and v k +1 is adjacent to v t but notto v s . If c > k + 2, then v k +2 is adjacent to v c by the umbrella property and the fact v k +1 is adjacent to v c . Thus v k +2 is adjacent to v c = v t but not to v s , a contradiction to that v k +1 is arc-balancing to the arc between v s , v t . If c = k + 2, since v k +1 is not a cut-vertexof U ( X ), v k is adjacent to v k +2 = v c . Thus v k is adjacent to v t = v c but not to v s , acontradiction again to the fact that v k +1 is arc-balancing to the arc between v s , v t . Hence c = k + 1, i.e., v k , v c are consecutive vertices in ≺ . Therefore v j , v k , v c are consecutive in ≺ . Since v c is the first dividing cut-vertex in ≺ , v k cannot be a cut-vertex of U ( X ) andhence v j , v c are adjacent in X . We claim that there exists a vertex preceding v j in ≺ whichis adjacent to v j but not to v k . First, observe that if no vertex is adjacent to exactly oneof v j , v k , then v j and v k would share the same closed neighbourhood. In this case, the arcbetween v j and v k would be balanced, a contradiction. Hence, there is at least one vertexadjacent to exactly one of v j , v k . Clearly, such a vertex must precede v j in ≺ and henceis adjacent to v j but not to v k . Assume that v p is such a vertex closest to v j .We show that v p and v j are consecutive in ≺ , that is, p = j −
1. If p < j −
1, then v j − cannot be a cut-vertex of U ( X ) because v p is adjacent to v j . On the other hand, byProposition 2.13, v j − is not a cut-vertex of U ( X ). It follows from Theorem 2.8 that v j − is an arc-balancing vertex. The choice of v p implies that v j − is adjacent to both v j , v k soit does not balance the arc between v j and v k . Hence, v j − is an arc-balancing vertex forthe arc between v s and v t . By definition it is the unique vertex adjacent to exactly oneof v s and v t . This also implies v c = v t . But then v k is also a vertex adjacent to v t but notto v s , a contradiction. Hence p = j − u ℓ +1 , w ℓ +1 )Γ( u ℓ , w ℓ ) and u ℓ +1 ≺ w ℓ +1 = v c = u ℓ ≺ w ℓ (i.e., u ℓ +1 is a vertexpreceding and adjacent to v c but not adjacent to w ℓ ), u ℓ +1 can only be v c − or v c − , SinceΓ-sequence is chosen to be the shortest from ( v t , v s ) to ( v j , v k ), we must have u ℓ +1 = v c − .It follows that( v t , v s ) = ( u , w )Γ( u , w )Γ . . . ( u ℓ , w ℓ )Γ( v c − , v c )Γ( v c − , v c − )Γ( v c − , v c − ) = ( v j , v k )is a shortest Γ-sequence. The Γ-sequence must contain all vertices of X , which means v c − , v c − , v c − are all the vertices preceding v c . Therefore c = 4 and v , v , v , v inducein U ( X ) the graph in the statement. 11e can now apply Lemma 3.1 to prove the following: Theorem 3.2.
Let X be an obstruction that contains a dividing cut-vertex with respectto a straight enumeration. Then X or its dual belongs to one of the three infinite classesin Figure 2. Proof:
Let ≺ : v , v , . . . , v n be a straight enumeration of U ( X ) and let v c and v d be the first and last dividing cut-vertices respectively with respect to ≺ . By consideringthe dual of X if necessary we assume that ( v j , v k ) and ( v s , v t ) are the arcs in X where j < k ≤ c ≤ d ≤ t < s .Suppose c = 2 and d = n −
1. Since v c = v is a cut-vertex, v is the only neighbourof v . Similarly, v n − is the only neighbour of v n . If v p is adjacent to v q for some 2 ≤ p < q − ≤ n −
1, then it is easy to see that the partially oriented graph obtainedfrom X by deleting v p +1 , . . . , v q − cannot be completed to local tournament orientation,a contradiction to the assumption X is an obstruction. Hence X belongs to Figure 2(i).Suppose that c = 2. Then c = 4 by Lemma 3.1. If d = n −
1, then a similar argumentas above shows that X belongs to Figure 2(ii). On the other hand if d = n −
1, then againby Lemma 3.1 we must have d = n −
3. In this case X belongs to Figure 2(iii). In this subsection, we will determine the rest of obstructions that contain cut-vertices,i.e., those containing only non-dividing cut-vertices.
Lemma 3.3.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumeration of U ( X ) . Suppose that v c is a non-dividing cut-vertex. Then c = 2 or c = n − . Moreover,if v c is incident with both arcs then n = 4 . Proof:
Let ( v j , v k ) ( j < k ) and ( v s , v t ) ( s > t ) be the arcs in X . Since v c is non-dividing, either c ≤ min { j, t } or c ≥ max { k, s } . Suppose that c ≤ min { j, t } .Let ( v j , v k ) = ( u , w ) , . . . , ( u q , w q ) = ( v t , v s ) be a Γ-sequence of U ( X ) between ( v j , v k )and ( v t , v s ). By Theorem 2.7, the sequence must include all vertices of X . Let α bethe smallest subscript such that one of u α , w α precedes v c (and hence the other ver-tex is v c since v c is a cut-vertex). Similarly, let β be the largest subscript such thatone of u β , w β precedes v c (and hence the other vertex is v c ). Then it is easy to verifythat ( u , w ) , . . . , ( u α , w α ) , ( u β +1 , w β +1 ) , . . . , ( u q , w q ) is a Γ-sequence between ( v j , v k ) and( v t , v s ). Since this sequence contains a unique vertex preceding v c and includes all verticesof X , we must have c = 2. A similar argument shows that if c ≥ max { k, s } then c = n − v c is incident with both arcs. Then either c = j = t = 2 or c = k = s = n − c = j = t = 2, then ( v j , v k )Γ( v , v j )Γ( v t , v s ) and by Theorem 2.7, v , v j = v t , v k , v s areall the vertices of X so n = 4. A similar argument shows that X has exactly four verticesif c = k = s = n − v and v n − are both non-dividingcut-vertices of U ( X ). 12 heorem 3.4. Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v and v n − are the two cut-vertices of U ( X ) , both non-dividing.Then X or its dual is one of the two graphs in Figure 3. Figure 3: Obstructions with two non-dividing cut-vertices.
Proof:
Since both v and v n − are non-dividing cut-vertices, n ≥
5. Hence by Lemma3.3 each of v and v n − is incident with at most one arc.We show that v and v n are arc-balancing vertices. By symmetry we only prove that v is arc-balancing. Clearly v is not a cut-vertex of U ( X ) and is not incident with anarc. By Proposition 2.8, it can only be an arc-balancing vertex or a cut-vertex of U ( X ).Assume that v is a cut-vertex of U ( X ). By Proposition 2.13, some vertex v is adjacentto every vertex in X except v . Since v n − is the only neighbour of v n in U ( X ), v = v n − .Since the vertex v = v n − is adjacent to v , by the umbrella property, the vertices v i with2 ≤ i ≤ n − U ( X ). Thus the vertices v i with 3 ≤ i ≤ n − U ( X ) and hence cannot contain both endvertices ofany arc. It follows that each arc is incident with v or v n − . From the above we knowthat each of v and v n − is incident with at most one arc. It is not possible that v and v n − are incident with the same arc (as otherwise the endvertices of the other arc havethe same closed neighbourhood). Hence v and v n − are incident with different arcs. Wesee that v is an arc-balancing vertex.By taking the dual of X if necessary we assume ( v , v k ) and ( v n − , v t ) are the twoarcs in X where 3 ≤ k, t ≤ n −
2. Then v is the ( v , v k )-balancing vertex and v n is the( v n − , v t )-balancing vertex. No vertex v i with 2 < i < n − U ( X ) or U ( X ) and hence each must be incident with an arc of X by Proposition 2.8. Hence v k and v t are the only vertices between v and v n − in ≺ . It is now easy to verify that X isone of the two graphs in Figure 3.It remains to consider the case when X has only one cut-vertex and it is non-dividing.By Lemma 3.3 and reversing the straight enumeration ≺ if necessary we will assume v is this vertex. Lemma 3.5.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. Then, the followingstatements hold:(a) For each i ≥ , v i is an arc-balancing vertex or incident with an arc;(b) For some i ≥ , v i is adjacent to every vertex except for v . Moreover, there are atmost two such vertices, each incident with exactly one arc;(c) The number of vertices in X is between 4 and 8 (i.e., ≤ n ≤ ). roof: For (a), if each v i with i ≥ v i with i ≥ U ( X ). According to Proposition 2.13, v i = v n and there is a vertex adjacent to everyvertex except v n in U ( X ). Such a vertex can only be v . Since v n − is not a cut-vertexof U ( X ), v n is adjacent to v n − . Since v is not adjacent to v n , n − > n > v . This arc musthave endvertices strictly between v and v n in ≺ . Therefore v n is an arc-balancing vertex,which contradicts our assumption.Statement (b) holds if v is a cut-vertex of U ( X ). Indeed, by Proposition 2.13 there isa vertex v i which is adjacent to every vertex except v and it is clear that i ≥
3. So assume v is not a cut-vertex of U ( X ). Since v is the only cut-vertex and it is non-dividing, v isneither a cut-vertex of U ( X ) nor incident with an arc, and hence must be an arc-balancingvertex by Proposition 2.8. Without loss of generality, assume v balances an arc between v and v j for some j >
2. If v j = v n or v j is adjacent to v n , then v j is adjacent to everyvertex except v and we are done. Otherwise, j < n and v j is not adjacent to v n . Foreach j < k < n , v k is not a cut-vertex of U ( X ) by assumption so v k − must be adjacentto v k +1 . Since v j is not adjacent to v n , j < n − n > j + 2 >
4. By statement(a), each vertex v i with i ≥ v is arc-balancing and v is incident with an arc, there are at most four vertices v i with i ≥
3. Hence n ≤ n = 6. It is now easy to see that v is adjacent to everyvertex except v .Suppose v i with i ≥ v . Clearly v i isnot an arc-balancing vertex and hence by (a) it is incident with an arc. We show bycontradiction that v i is incident with exactly one arc. So suppose that v i is incident withboth arcs of X . Let v s and v t denote the other endvertices of the two arcs. We firstshow that either s = 2 or t = 2. By Theorem 2.7, the edges v i v s , v i v t belong to differentimplication classes in U ( X − v ). Since v i is an isolated vertex in U ( X − v ), each of v s , v t , v i belongs to a different component of U ( X − v ) by Theorem 2.12. In particular,one of v s , v t is an isolated vertex in U ( X − v ). Without loss of generality, assume v s issuch a vertex. Thus, v s is adjacent to every vertex except possibly v in X . If v s is notadjacent to v , then v s and v i share the same closed neighbourhood, so the arc between v s and v i is balanced, a contradiction. Hence, v s is adjacent to v and v s = v . Consider v t . Suppose t < i . Since 2 = s < t < i and v i is adjacent to v s , the umbrella propertyimplies v t is adjacent to v s . If v t is also adjacent to v n , then v i and v t have the sameclosed neighbourhood so the arc between them is balanced, a contradiction. Hence, v t is not adjacent to v n . Since s < t < n , the umbrella property implies that v s and v n are not adjacent. Thus ( v t , v i )Γ( v i , v n )Γ( v s , v i ) is a Γ-sequence between the arcs and notcontaining v , a contradiction by Theorem 2.7. It follows that i < t . If v t is non-adjacentto v s , then ( v i , v t )Γ( v s , v i ) is a Γ-sequence between the arcs and not containing v , acontradiction. Hence, v t is adjacent to v s = v . If t = n , then the arc between v i and v t is balanced by the umbrella property, a contradiction. If t < n , then v t is adjacent to v n because i < t < n and v i is adjacent to v n , leading to a similar contradiction. Therefore v i is incident with exactly one arc. Suppose v i , v j are two such vertices. By the above,each of them is incident with an arc. Moreover, they cannot be incident with the same14rc because they share the same neighbourhood. Hence, they are each incident with adifferent arc. Since X contains two arcs, there are at most two such vertices.Finally we prove (c). Clearly, n ≥
4. Since there are at most four vertices incidentwith arcs and at most two arc-balancing vertices in X , there can be at most six vertices v i with i ≥ n ≤ Theorem 3.6.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. If n = 4 or 5, then X or its dual is one of the graphs in Figure 4. (i) (ii) (iii) Figure 4: Obstructions with a unique non-dividing cut-vertex on 4 or 5 vertices.
Proof:
Suppose n = 4. Since v is not a cut-vertex, v and v are adjacent. Both v and v are adjacent to every vertex except for v and by Lemma 3.5(b) they are incidentwith different arcs. It is easy to see that X is Figure 4(i).Suppose n = 5. For each i = 3 , v i is not a cut-vertex, so v i − and v i +1 are adjacent.On the other hand if v is adjacent to v , then the umbrella property implies v , v , v are all adjacent to every vertex except for v , contradicting Lemma 3.5(b). So v and v are not adjacent. Each of v , v is adjacent to every vertex except v and by Lemma3.5(b) they are incident with different arcs. Since n = 4, v is not incident with botharcs according to Lemma 3.3. It follows that v must be incident with at least one arc. If v is incident with exactly one arc, then X is or its dual is Figure 4(ii). Otherwise v isincident with both arcs and X or its dual is Figure 4(iii). Lemma 3.7.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose v k is a ( v i , v j ) -balancing vertex. Then, either k < min { i, j } or k > max { i, j } . Moreover, • If k < min { i, j } , then no v p with p < k is adjacent to either one of v i , v j , and any v q with q > max { i, j } is adjacent to either both or neither of v i , v j ; • If k > max { i, j } , then no v p with p > k is adjacent to either one of v i , v j , and any v q with q < min { i, j } is adjacent to either both or neither of v i , v j . Proof:
First we show that either k < min { i, j } or k > max { i, j } . Otherwise, v k isbetween v i and v j . Since v i v j is an edge of U ( X ), the umbrella property implies thatboth v i and v j are adjacent to v k , a contradiction to the fact that v k is a ( v i , v j )-balancingvertex. Thus, either k < min { i, j } or k > max { i, j } .By symmetry, it suffices to consider the first case. Suppose k < min { i, j } . If v p with p < k is adjacent to either one of v i , v j , then it must also be adjacent to v k by theumbrella property. Since v k is the only vertex adjacent to exactly one of v i , v j , v p must be15djacent to both v i and v j . By the umbrella property, both v i and v j are adjacent to v k ,a contradiction. On the other hand, since v k is the only vertex adjacent to exactly oneof v i , v j , it is clear that any v q with q > max { i, j } is adjacent to either both or neither of v i , v j . Theorem 3.8.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. If n = 6 , then X or its dual is one of the graphs in Figure 5. (i) (ii)(iii) (iv)(v) (vi)(vii) (viii)(ix) (x)(xi) Figure 5: Obstructions with a unique non-dividing cut-vertex on 6 vertices.
Proof:
For each 3 ≤ i ≤ v i is not a cut-vertex, so v i − and v i +1 are adjacent. Now v v and v v cannot both be edges in U ( X ) as otherwise each of v , v and v is adjacentto every vertex except for v , contradicting Lemma 3.5(b).We claim that each of v and v is incident with an arc. Since v is adjacent to everyvertex except for v , it is incident with exactly one arc by Lemma 3.5(b). On the otherhand, suppose v is not incident with an arc. By Lemma 3.5(a), v is an arc-balancingvertex for some arc. Thus v is adjacent to exactly one endvertex of the arc. It is easyto see that the other endvertex can only be v . Since v is a cut-vertex, v is adjacent toexactly one endvertex (i.e., v ) of the arc, a contradiction to that v is arc-balancing forthe arc. Hence v is incident with an arc.Suppose v and v are also incident with arcs. Then v , v , v , v are endvertices ofthe two arcs. Suppose that the two arcs are between v and v and between v and v v v is not an edge of U ( X ) as otherwise the arc between v and v isbalanced, a contradiction. If v v is not an edge of U ( X ) then X or its dual is Figure 5(i);otherwise, X or its dual is Figure 5(ii). Suppose that the two arcs are between v and v and between v and v respectively. Then X or its dual is Figure 5(iii), (iv) or (v)depending whether or not v v and v v are edges of U ( X ). Suppose the two arcs arebetween v and v and between v and v respectively. Then X or its dual is againFigure 5(v) (with v and v being switched).Suppose v is not incident with an arc. By Lemma 3.5(a), v is an arc-balancingvertex. By Lemma 3.7, v balances an arc between v and v . Since v is incident with anarc and v is not, the arc incident with v has the other endvertex being v , v , v . Thesethree cases are represented by Figure 5(vi), (vii) and (viii).It follows from the above that at least one of v and v is incident with an arc. Thus itremains to consider the case that v is incident with an arc but v is not. By Lemma 3.5(a), v is an arc-balancing vertex for some arc. By Lemma 3.7, v cannot be an endvertex ofthis arc, so the arc must be between v and one of v , v . In particular, this implies v v is not an edge of U ( X ). It is now easy to verify that X or its dual is Figure 5(ix), (x) or(xi). Lemma 3.9.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. If n ≥ , then v is not incident with an arc and the subgraph of U ( X ) induced by the vertices v i with i ≥ cannot contain a copy of K . Proof:
Suppose that v is incident with an arc. Then v is adjacent to exactly oneendvertex of this arc so this arc cannot be balanced by any vertex v i with i ≥
3. It followsthat there is at most one arc-balancing vertex v i with i ≥
3. By Lemma 3.5(a) and theassumption n ≥ v i with i ≥ v is such a vertex.By Lemma 3.5(a) and (c), n ≤ v i with i ≥ v , v is incident with an arc, any set of five vertices v i with i ≥ K in U ( X ). Theorem 3.10.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. If n = 7 , then X or its dual is one of the graphs in Figure 6. i) (ii)(iii) (iv)(v) (vi)(vii) (viii) Figure 6: Obstructions with a unique non-dividing cut-vertex on 7 vertices.
Proof:
First, note that v v is not an edge in U ( X ) as otherwise the vertices v i with i ≥ K in U ( X ), a contradiction to Lemma 3.9. If v v and v v are both edgesof U ( X ), then each of v , v , v is adjacent to every vertex except for v , contradictingLemma 3.5(c). So, v v and v v cannot both be edges in U ( X ).By Lemma 3.5(b), there exists a vertex v i with i ≥ v . Since v v is not an edge in U ( X ), neither v nor v is such a vertex. It is easy tosee that if v is such a vertex, then v is also such a vertex. Hence, at least one of v , v is adjacent to every vertex except for v .Suppose v is adjacent to every vertex except for v . This implies in particular that v v is an edge of U ( X ) and thus v v is not an edge of U ( X ). By Lemma 3.5(b), v is incident with exactly one arc. Lemma 3.9 implies the other endvertex of this arc isone of v , v , v , and v . Suppose that the other endvertex is v . If no v i with i ≥ { v , v , v } must be an arc-balancing triple,a contradiction because these vertices induce a clique. Hence for some i ≥ v i is anarc-balancing vertex for the arc between v and v . By Lemma 3.7, it must be v . Since v balances the arc between v and v , we see that v must be adjacent to v . Both v and v are adjacent to v i for each i ≥ v and v are incident with arcs. This means there is an arcbetween v and v , which implies that v is adjacent to v (as otherwise v and v havethe same closed neighbourhood in U ( X )). Since v v is not an edge in U ( X ), X or itsdual is Figure 6(i). Suppose next that there is an arc between v and v . Since v and v cannot have the same closed neighbourhood in U ( X ), v v is not an edge in U ( X ).Clearly, the arc between v and v is not balanced by any of v , v , v , so { v , v , v } is anarc-balancing triple. By Lemma 3.7, the arc is between v and v . It follows that v v isan edge in U ( X ). Hence X or its dual is Figure 6(ii).18uppose next that there is an arc between v and v . By Lemma 3.7, the arc between v and v cannot be balanced by v , v , v . Similarly as above, { v , v , v } is an arc-balancingtriple. If v balances an arc between v and v , then v v and v v are edges in U ( X ), and X or its dual is Figure 6(i). Suppose that v balances an arc between v and v . Eachvertex except v is either adjacent to both v , v or neither. Since v is not adjacent to v , it is not adjacent to v . Hence, X or its dual is Figure 6(ii) or (iii) depending whetheror not v v is an edge of U ( X ). Finally, suppose there is an arc between v and v . ByLemma 3.7, none of v , v , v is an arc-balancing vertex for this arc. Hence, { v , v , v } isan arc-balancing triple. By Lemma 3.7, v balances an arc between v and v . It followsthat neither v v nor v v can be an edge in U ( X ). So X or its dual is Figure 6(iii).Suppose now v is not adjacent to every vertex except for v . From the above we knowthat v must be adjacent to every vertex except for v . So v is incident with exactly onearc, and the other endvertex of this arc is one of v , v , v , and v . We claim that itcannot be v . Suppose to the contrary that there is an arc between v and v . By Lemma3.7, none of v , v , v can be an arc-balancing vertex for this arc. Hence, { v , v , v } is anarc-balancing triple. Since neither v v nor v v is an edge of U ( X ), the second arc canonly be between v and v but it is not balanced by v , a contradiction. Hence, there isan arc between v and one of v , v , and v .Suppose first that there is an arc between v and v . Assume that this arc is balancedby a vertex. By Lemma 3.7, it is balanced by v . It follows that v v is an edge in U ( X ).Since v is adjacent to every vertex v i with i ≥
3, which are where all endvertices of arcsare, it cannot be an arc-balancing vertex. It follows that v is incident with an arc. Weclaim that the other endvertex of this arc is v . Indeed, if it is not v , then v would bethe arc-balancing vertex for this arc, a contradiction by Lemma 3.7. Thus, X or its dualis Figure 6(iv) or (v) depending whether or not v v is an edge of U ( X ). Assume nowthat the arc between v and v is not balanced by any vertex. In this case, { v , v , v } is an arc-balancing triple. By Lemma 3.7, either v balances an arc between v and v ,or v balances an arc between v and v . In the first case, v v cannot be an edge of U ( X ), as that would imply v v is also an edge, a contradiction. Hence, X or its dualis Figure 6(vi). In the second case, v v must be an edge of U ( X ) and X or its dual isFigure 6(v).Suppose there is an arc between v and v . We claim that v is not arc-balancingvertex. Indeed, if it is, then it must balance an arc between v and v . Thus, v v is anedge of U ( X ), a contradiction. Hence, v is incident with an arc. The other endvertexof this arc is v , v , or v . Clearly it cannot be v because that would imply v v is anedge of U ( X ), contradicting the fact that v is not adjacent to every vertex except for v .Suppose the second arc is between v and v . Then, v must be an arc-balancing vertex.Clearly, v cannot balance the arc between v and v , so it must balance the arc between v and v . It follows that v v is not an edge of U ( X ), so X or its dual is Figure 6(vii).On the other hand, suppose the second arc is between v and v . In this case, X or itsdual is Figure 6(iv) or (v) depending whether v v is an edge of U ( X ).Finally, suppose there is an arc between v and v . Clearly, none of v , v , v can be anarc-balancing vertex for this arc. Hence { v , v , v } is an arc-balancing triple. By Lemma19.7, v must balance an arc between v and v . It follows that v v is not an edge of U ( X ),so X or its dual is Figure 6(viii). Theorem 3.11.
Let X be an obstruction and ≺ : v , v , . . . , v n be a straight enumerationof U ( X ) . Suppose that v is the only cut-vertex and it is non-dividing. If n = 8 , then X or its dual is one of the graphs in Figure 7. (i) (ii) Figure 7: Obstructions with a unique non-dividing cut-vertex on 8 vertices.
Proof:
Since there are six vertices succeeding v , exactly two of them are arc-balancingvertices and the other four are incident with arcs by Lemma 3.5(a). By Lemma 3.5(b),there exists a vertex succeeding v that is adjacent to every vertex except for v . If anyof v , v , v , v is adjacent to every vertex except for v , then U ( X ) contains a copy of K among the vertices v i with i ≥
3, contradicting Lemma 3.9. Hence, only v and v can beadjacent to every vertex except for v .Suppose v is adjacent to every vertex except for v . By Lemma 3.5(b) again, v isincident with exactly one arc. By Lemma 3.7, v balances an arc between v and one of v , v . If v is an endvertex of this arc, then v v would be an edge in U ( X ), contradictingLemma 3.9. Hence, v balances an arc between v and v . It follows that v v is anedge of U ( X ). Moreover, there is an arc with both endvertices and arc-balancing vertexamong v , v , v . If v balances an arc between v and v , then v v is an edge of U ( X ),contradiction Lemma 3.5(b). Hence v balances an arc between v and v . It follows that X or its dual is Figure 7(i).On the other hand, suppose v is not adjacent to every vertex except for v . By theprevious discussion, v must be the unique such vertex. By Lemma 3.5(b), v is incidentwith an arc. By Lemma 3.7, v balances an arc between v and one of v , v , v . If v is anendvertex of this arc, then v v is an edge of U ( X ), contradicting Lemma 3.9. Suppose v balances an arc between v and v . Then, v v is an edge of U ( X ). Moreover, { v , v , v } is an arc-balancing triple. By Lemma 3.7, v balances an arc between v and v . If v v is an edge of U ( X ), then v v is also an edge, contradicting Lemma 3.9. Hence v v isnot an edge and so X or its dual is Figure 7(ii). Suppose instead that v balances an arcbetween v and v . In this case, { v , v , v } is an arc-balancing triple. By Lemma 3.7, v balances an arc between v and v . Since v v and v v are not edges of U ( X ), X or itsdual is Figure 7(ii). 20 Obstructions without cut-vertices
We now examine obstructions X that do not contain cut-vertices. We shall considerthe complements U ( X ) of the underlying graphs U ( X ). All theorems and proofs, includingthe drawings of obstructions X , in this section will be presented in terms of U ( X ) insteadof U ( X ). Lemma 4.1.
Suppose that X is an obstruction that contains no cut-vertices. Then,in U ( X ) , each vertex has at least two non-neighbours. Proof:
Note that X has at least three vertices. Since X has no cut-vertices and U ( X )is connected, in U ( X ) each vertex has at least two neighbours and hence in U ( X ) eachvertex has at least two non-neighbours.Recall from Corollary 2.9 that if an obstruction X has no cut-vertices then U ( X ) hasat most six non-cut-vertices. We show this holds for every connected subgraph of U ( X ). Lemma 4.2.
Let X be an obstruction that contains no cut-vertices and H be a con-nected subgraph of U ( X ) . Then U ( X ) contains at least as many non-cut-vertices as H .In particular, H has at most six non-cut-vertices. Proof:
Since adding edges does not decrease the number of non-cut-vertices, we mayassume H is an induced subgraph of U ( X ). Thus H can be obtained from U ( X ) bysuccessively deleting non-cut-vertices. Since each deletion of a non-cut-vertex does notincrease the number of non-cut-vertices, U ( X ) contains at least as many non-cut-verticesas H . By Corollary 2.9, U ( X ) has at most six non-cut-vertices. So H has at most sixnon-cut-vertices. Lemma 4.3. If X is an obstruction that contains no cut-vertices, then U ( X ) containsno induced cycle of length ≥ . Proof:
By Lemma 4.2, any connected subgraph of U ( X ) has at most six non-cut-vertices. Thus U ( X ) contains no induced cycle of length ≥
7. Theorem 2.1 ensures that U ( X ) does not contain an induced cycle of length 6. Therefore U ( X ) contains no inducedcycle of length ≥ U ( X ) has length 3, 4 or 5. We showthat U ( X ) contains at most one C and at most one induced C and moreover, if U ( X )contains an induced C , then it does not contain an induced C or C . Lemma 4.4.
Let X be an obstruction. Suppose C is an odd cycle (not necessarilyinduced) in U ( X ) . Then, in U ( X ) , each vertex is either in C or adjacent to a vertex of C . In particular, each cut-vertex of U ( X ) is in C . Proof:
Since C is an odd cycle, U ( X ) contains an induced odd cycle C k +1 on somevertices in C . By Theorem 2.1, U ( X ) does not contain C k +1 + K as an induced subgraph.Thus each vertex is either in C k +1 or adjacent to a vertex of C k +1 . Since the vertices of C k +1 are all in C , each vertex is either in C or adjacent to a vertex of C . Consequently,each cut-vertex of U ( X ) is in C . 21 emma 4.5. Suppose that X is an obstruction that contains no cut-vertices. Then U ( X ) contains at most one C . Proof:
Suppose that C and C ′ are two copies of C in U ( X ). If C and C ′ share nocommon vertex, then every vertex of U ( X ) is either not in C or not in C ′ and hence by byLemma 4.4 is a non-cut-vertex. But U ( X ) has at most six non-cut-vertices by Corollary2.9, so U ( X ) is a union of C and C ′ . According to Proposition 2.8 each vertex of U ( X )is an endvertex of an arc or an arc-balancing vertex. There are at most four endverticesof arcs and at most two arc-balancing vertices. So, among the six vertices of U ( X ), fourare the endvertices of arcs and the remaining two are arc-balancing vertices. Supposethat ( a, b ) is an arc (of X ) and u is its balancing vertex that is adjacent to a but not to b in U ( X ). Then each of the remaining three vertices is adjacent to a or b and thus toboth a, b . Hence b is the only non-neighbour of a in U ( X ), a contradiction to Lemma 4.1.Therefore any two copies of C in U ( X ) must share a common vertex.Suppose that C and C ′ share exactly one common vertex. Denote C : v v v and C ′ : v v v . Let u, w be two non-neighbours of v in U ( X ) guaranteed by Lemma 4.1.Each vertex except v is not in C or C ′ and hence by Lemma 4.4 is a non-cut-vertex.Since U ( X ) has at most six non-cut-vertices, it consists of C, C ′ and u, v . A similarargument as above shows that, among the six non-cut-vertices u, w, v , v , v , v , four arethe endvertices of arcs and the remaining two are arc-balancing vertices. We claim thatthe two arc-balancing vertices are u, w . Indeed, since v is not an arc-balancing vertex,there is no arc between u, w and v , v , v , v . Suppose that there is an arc between u and w . Assume without loss of generality that this arc is balanced by v which is adjacent to u but not w . By Lemma 4.4, w is adjacent to a vertex in C . Since w is not adjacent to v or v , it is adjacent to v . Since v does not balance the arc between u and w , v is adjacentto u . But then uv v and C ′ are vertex-disjoint copies of C , a contradiction. Henceneither of u, w is an endvertex of an arc so both are arc-balancing vertices. Without lossof generality assume that u balances an arc between v and v and is adjacent to v butnot v . Since v is adjacent to v , it must be adjacent to v . Similarly, v must be adjacentto v . By Lemma 4.4, u is adjacent to a vertex in C ′ which can only be v . Hence uv v and v v v are vertex-disjoint copies of C , a contradiction. Therefore any two copies of C in U ( X ) must share at least two common vertices.Suppose that C and C ′ share exactly two vertices. Denote C : v v v and C ′ : v v v .We claim that in U ( X ) any vertex v / ∈ C ∪ C ′ that is adjacent to one of v , v must beadjacent to both v , v and neither of v , v . Without loss of generality, suppose v / ∈ C ∪ C ′ is adjacent to v . If it is also adjacent to v , then v v v and v v v would be two distinctcopies of C in U ( X ) that share exactly one common vertex, a contradiction to the above.Hence, v is not adjacent to v . Similarly, v is not adjacent to v . By Lemma 4.4 v mustbe adjacent to a vertex in C ′ so it is adjacent to v .Now, we show that v and v are incident with different arcs. Suppose v balances anarc between a and b and is adjacent to a but not b . If a = v , then since v and v areadjacent to v , they must also be adjacent to b . So, bv v and C are two distinct copiesof C in U ( X ) that share exactly one common vertex, a contradiction. Thus, a = v .Similarly, a = v . Suppose a = v . Since v and v are adjacent to a , they must be22djacent to b as well. Thus v v b and v v v are two copies of C in U ( X ) that shareexactly one common vertex, a contradiction.It follows that a / ∈ C ∪ C ′ . Since a / ∈ C ∪ C ′ and a is adjacent to v , it is adjacent toboth of v , v by the above claim. Since v is adjacent to a , it must also be adjacent to b . Moreover, b / ∈ C ∪ C ′ because it is adjacent to v but not a . Since b / ∈ C ∪ C ′ and b is adjacent to v , the above claim implies b is adjacent to both of v , v , a contradictionbecause v balances the arc ( a, b ). Thus, v is not an arc-balancing vertex. Since v is notin C ′ , it is not a cut-vertex of U ( X ) by Lemma 4.4. By Proposition 2.8, v is incidentwith an arc. Similarly, v is incident with an arc. If v and v are incident with the samearc, then there must be a vertex u that is adjacent to exactly one of v , v because arcsin X are not balanced. Clearly, u / ∈ C ∪ C ′ . This is a contradiction because any vertexnot in C ∪ C ′ is adjacent to either both of v , v or neither, by above claim. Thus, v and v are each incident with a different arc.Suppose there exists a vertex v / ∈ C ∪ C ′ that is adjacent to either of v , v . By theabove claim, we know that v is adjacent to both of v , v and neither of v , v . Since v isadjacent to both of v , v , which are each incident with a different arc, v is not incidentwith an arc. Moreover, since v is not on the odd cycle C , Lemma 4.4 implies v is not acut-vertex of U ( X ). So by Proposition 2.8, v is an arc-balancing vertex. Since v and v are each incident with a different arc, we may assume without loss of generality that v balances an arc incident with v . Let w denote the other endvertex of this arc. Then, w is adjacent to v and v , so v v w is a triangle. By Lemma 4.4, v is adjacent to a vertexin v v w , which must be w , a contradiction because v balances the arc between v and w .It follows U ( X ) does not contain a vertex v / ∈ C ∪ C ′ that is adjacent to either of v , v .By Lemma 4.1, v has at least two non-neighbours, say u and w . Clearly, u, w / ∈ C ∪ C ′ .So by the above, neither of u, w is adjacent to either of v , v . By Lemma 4.4, each of u, w is adjacent to a vertex in C , which must be v . Similarly, v has at least two non-neighbours, say x and y , and each is adjacent to v . By Lemma 4.4, each of v , v , u, w, x, y is a non-cut-vertex of U ( X ) and so by Corollary 2.9 they form two disjoint arc-balancingtriples. Since v and v are each incident with a different arc, exactly two of u, w, x, y arearc-balancing vertices. Without loss of generality, assume u is an arc-balancing vertexfor an arc incident with v . Since v is adjacent to both v and v , the other endvertexmust also be adjacent to both v and v . This is a contradiction because none of w, x, y is adjacent to both v and v by assumption. It follows that C and C ′ cannot share twocommon vertices. Therefore U ( X ) contains at most one C . Lemma 4.6.
Suppose that X is an obstruction that contains no cut-vertices. Then U ( X ) contains at most one induced C . Proof:
Suppose that C and C ′ are induced copies of C contained in U ( X ). By Lemma4.4, any vertex not in C or C ′ is a non-cut-vertex of U ( X ) and hence by Corollary 2.9there can be at most six such vertices. Thus C and C ′ must share at least two commonvertices. If C and C ′ share less two or three common vertices, then the subgraph of U ( X )induced by C ∪ C ′ is connected and has at least seven non-cut-vertices, contradictingLemma 4.2. Hence, C and C ′ must share exactly four vertices.23enote C : v v v v v and C ′ : v v v v v . Then v v is not an edge in U ( X ) asotherwise v v v and v v v are two copies of C in U ( X ), a contradiction to Lemma 4.5.We claim that v , v are endvertices of arcs in X . By symmetry we only prove that v is an endvertex of an arc in X . We prove it by contradiction. So assume that v is notan endvertex of an arc in X . Since v is not in C ′ , by Lemma 4.4 it is not a cut-vertexof U ( X ). Hence v is an arc-balancing vertex for some arc according to Proposition 2.8.Suppose that v balances the arc between vertices a, b and is adjacent to a but not to b in U ( X ). If a is not in C ′ , then a must be adjacent to a vertex of C ′ by Lemma 4.4.But then the subgraph of U ( X ) induced by C ∪ C ′ ∪ { a } is connected and has sevennon-cut-vertices, contradicting Lemma 4.2. Hence a is a vertex of C ′ and therefore it is v or v . Assume by symmetry a = v . Since v balances the arc between a, b , every vertexnot in { v , a, b } is either adjacent to both a, b or neither. It follows that b cannot be in C ∪ C ′ . Thus the subgraph of U ( X ) induced by C ∪ C ′ ∪ { b } is connected and has sevennon-cut-vertices, a contradiction to Lemma 4.2. Therefore v , v are both endvertices ofarcs of X . We claim that there is no arc between v , v . Suppose not; there is an arcbetween v , v . Then there must exist a vertex u adjacent to exactly one of v , v . Asimilar argument as above shows that u is not in C ∪ C ′ but adjacent to a vertex in C ∪ C ′ . Thus the subgraph of U ( X ) induced by C ∪ C ′ ∪ { u } is connected and containsseven non-cut-vertices, a contradiction. Thus, v , v are endvertices of different arcs.The subgraph of U ( X ) induced by C ∪ C ′ contains six non-cut-vertices, so U ( X )contains six non-cut-vertices by Lemma 4.2. It follows from Proposition 2.8 that X contains exactly four vertices incident to arcs and exactly two arc-balancing vertices. Inparticular, both arcs have an arc-balancing vertex.By Proposition 2.8, v is a cut-vertex of U ( X ), an arc-balancing vertex, or is incidentwith an arc. We claim it must be a cut-vertex of U ( X ). Suppose instead v is an arc-balancing vertex. Without loss of generality, assume it balances the arc incident with v .Then, the other endvertex must be adjacent to each of v , v . Clearly, the subgraph of U ( X ) induced by C ∪ C ′ together with this endvertex contains seven non-cut-vertices,a contradiction. On the other hand, suppose v is incident with an arc. The otherendvertex is one of v , v . Without loss of generality, assume it is v . Then, v and v areboth vertices adjacent to exactly one of the endvertices of this arc, so the arc between v and v has no arc-balancing vertex, a contradiction. Thus, v is a cut-vertex of U ( X ).Let v be a neighbour of v belong to a different component of U ( X − v ) as thevertices in ( C ∪ C ′ ) \ { v } . By Lemma 4.4, v cannot be a cut-vertex of U ( X ). Onthe other hand, suppose v is incident with an arc. Without loss of generality, assumethe other endvertex is v . Since v and v are both vertices adjacent to exactly one of v , v , there is no corresponding arc-balancing vertex for this arc, a contradiction. Thus, v cannot be incident with an arc. By Proposition 2.8, v is an arc-balancing vertexfor either the arc incident with v or the arc incident with v . In either case, the otherendvertex must be adjacent to both v and v . Clearly, the subgraph of U ( X ) induced by C ∪ C ′ together with this endvertex contains seven non-cut-vertices, a contradiction. Lemma 4.7.
Let X be an obstruction that contains no cut-vertices. If U ( X ) containsan induced C , then it contains neither C nor induced C . roof: Let C : v v v v v be an induced C in U ( X ). We first show that U ( X ) doesnot contain C . Suppose otherwise and let C ′ be a C in U ( X ). A similar argument asthe one in Lemma 4.6 shows that C and C ′ have exactly two common vertices. Withoutloss of generality let C ′ : v v v . By Lemma 4.4, v must be adjacent to a vertex in C ′ ,which clearly must be v . The subgraph induced by C ∪ C ′ contains six non-cut-vertices,so U ( X ) contains six non-cut-vertices by Lemma 4.2. Each of these six non-cut-verticesis an arc-balancing vertex or incident with an arc by Proposition 2.8. Hence, each arc hasan arc-balancing vertex. If both endvertices of some arc are in C , then C contains twoother vertices which are both adjacent to exactly one of the endvertices, contradicting thefact that each arc has a unique arc-balancing vertex. It follows that each arc has at mostone endvertex in C . In particular, at most two vertices in C are incident with arcs. Onthe other hand, at most two vertices in C are arc-balancing. It follows from Proposition2.8 that C has a cut-vertex. By Lemma 4.4, each cut-vertex belongs to C ∩ C ′ , so only v and v can be cut-vertices.We claim that if v is a cut-vertex, then there exists a vertex v that is adjacent onlyto v and an arc between v and v that is balanced by v . Suppose v is a cut-vertex.Let v be a vertex adjacent to v that belongs to a different component of U ( X − v ) asthe vertices in ( C ∪ C ′ ) \ { v } . If v is adjacent to a vertex other than v , then that vertexmust be adjacent to a vertex in C ∪ C ′ by Lemma 4.4, contradicting the choice of v .Hence, v is adjacent only to v . By Lemma 4.4, v is not a cut-vertex. Suppose v is anarc-balancing vertex. Then, v is incident with an arc, and the other endvertex of thisarc must be adjacent to v , v , v . Clearly, this endvertex is none of the vertices in C ∪ C ′ ,so the subgraph of U ( X ) induced by C ∪ C ′ together with this endvertex contains sevennon-cut-vertices, contradicting Lemma 4.2. Hence, v is not an arc-balancing vertex. ByProposition 2.8, v is incident with an arc. Since v is a leaf, the other endvertex u of thearc incident with v has exactly two neighbours, one of which is v . Since any subgraphof U ( X ) contains at most six non-cut-vertices, u must be in C ∪ C ′ . Clearly, u = v andthe vertex which balances the arc between v and v is v . This proves our claim.Recall that at least one of v , v is a cut-vertex. Without loss of generality, assume v is a cut-vertex. By the above, there exist a vertex v that is adjacent only to v and anarc between v and v that is balanced by v . In particular, v is not an arc-balancingvertex for this arc. By Lemma 4.4, v is not a cut-vertex. Hence by Proposition 2.8, v isarc-balancing for the other arc or incident with it. By symmetry, if v is also a cut-vertex,then there exist a vertex v that is adjacent only to v and an arc between v and v , acontradiction because v is neither arc-balancing for this arc nor incident with this arc.Thus, v is not a cut-vertex. Since v balances an arc between v and v and v , v , v arenon-cut-vertices, { v , v , v } is an arc-balancing triple. It follows that v balances an arcbetween v and v , a contradiction. Thus, U ( X ) does not contain an induced C .It remains to show that U ( X ) does not contain an induced C . Suppose otherwise,and let C ′ be such a cycle. A similar argument as above shows that C and C ′ haveexactly three common vertices. Let C ′ : v v v v . If v is adjacent to neither v or v ,then v v v v v is an induced C , contradicting Lemma 4.6. Hence, v is adjacent to oneof v , v . It follows that U ( X ) contains an induced C , a contradiction.25 .1 U ( X ) is disconnected We first examine obstructions X that do not contain cut-vertices for which U ( X ) isdisconnected. These obstructions have a simple structure as described in the followingtheorem. Theorem 4.8.
Let X be an obstruction that does not contain cut-vertices. Supposethat U ( X ) is disconnected. Then the following statements hold: • U ( X ) is the union of two disjoint paths P : p , . . . , p k and Q : q , . . . , q ℓ ; • X or its dual contains the arcs ( p , q ) , ( q ℓ , p k ) if k + ℓ is even, and ( p , q ) , ( p k , q ℓ ) otherwise.That is, U ( X ) is one of the graphs in Figure 8 and X or its dual contains the dotted arcs. . . .. . .p p p k q q q ℓ (i) k + ℓ is even . . .. . .p p p k q q q ℓ (ii) k + ℓ is odd Figure 8: Obstructions X for which U ( X ) is disconnected. Proof:
Let ( a, b ) and ( c, d ) be the two arcs of X . Then ab and cd belong to the sameimplication class of U ( X ). By Theorem 2.5, either ab and cd are unbalanced edges of U ( X ) within a component of U ( X ) or they are edges between two components of U ( X ).Since U ( X ) is disconnected, it has at least two components. If some component of U ( X )does not contain any of a, b, c, d , then any non-cut-vertex of that component is not acut-vertex of U ( X ) by assumption and is not an arc-balancing vertex because it is notadjacent to any of a, b, c, d in U ( X ). This contradicts Proposition 2.8. Thus U ( X ) hasexactly two components and ab, cd are edges between them.Consider a component H of U ( X ) and let P be a shortest path in H between some twoof a, b, c, d . If H contains a vertex v that is not in P then it follows from Proposition 2.3that ab and cd are still in the same implication class of U ( X − v ), which is a contradictionto Theorem 2.7. This shows that each component of U ( X ) is a path connecting twovertices of a, b, c, d and U ( X ) is the union of two disjoint paths.Let P : p , . . . , p k and Q : q , . . . , q ℓ be the two paths in U ( X ). The two arcs arebetween p and q and between p k and q ℓ respectively. Without loss of generality, assume( p , q ) is an arc. Suppose k + ℓ is even. If k, ℓ are both even, then ( p , q )Γ ∗ ( q ℓ , p )and ( q ℓ , p )Γ ∗ ( p k , q ℓ ) by Proposition 2.3. Since the arcs must be opposing, the otherarc is ( q ℓ , p k ). Otherwise, k, ℓ are both odd. In this case, we have ( p , q )Γ ∗ ( p , q ℓ ) and( p , q ℓ )Γ ∗ ( p k , q ℓ ), so the other arc is ( q ℓ , p k ). Hence X or its dual is Figure 8(i). A similarproof shows that, when k + ℓ is odd, X or its dual is Figure 8(ii).26 orollary 4.9. If X is an obstruction that does not contain cut-vertices and for which U ( X ) is disconnected, then U ( X ) contains no cycle. U ( X ) is a tree We next examine obstructions X that do not contain cut-vertices and for which U ( X )is a tree. We begin with a useful lemma. Lemma 4.10.
Let X be an obstruction that contains no cut-vertices. If U ( X ) is a tree,then it is a caterpillar and has at most four leaves. Moreover, suppose P : p , p , . . . , p k isa longest path in U ( X ) . If p is an arc-balancing vertex, then p has only two neighbours(namely, p , p ) and p balances an arc between p and a leaf adjacent to p but not in P . Proof:
Since U ( X ) is a proper circular-arc graph, U ( X ) does not contain the fifthgraph in Figure 1 by Theorem 2.1 and hence is a caterpillar. If v is a leaf of U ( X ) that isnot incident with an arc of X , then by Proposition 2.8 v is an arc-balancing vertex andhence adjacent to a vertex that is incident with an arc. Clearly, the vertex adjacent to v cannot be adjacent to any other leaf. Since there are at most four vertices incident witharcs, U ( X ) has at most four leaves.Since P is a longest path, p is a leaf. If p is an arc-balancing vertex, then p isincident with an arc balanced by p . Let u be the other endvertex of the arc. Everyvertex other than p is adjacent either to both p , u or neither. Since p is adjacent to p , it is adjacent to u . Since U ( X ) is a tree, p is the only neighbour of p other than p and the only neighbour of u . It follows that p , p are the only neighbours of p . If u isin P then u = p and k = 4. Thus each vertex not in P can only be adjacent to p in U ( X ), which implies that p is a cut-vertex of U ( X ), a contradiction to the assumption.Therefore u is a leaf of U ( X ) adjacent to p but not in P . Theorem 4.11.
Let X be an obstruction that contains no cut-vertices and for which U ( X ) is a tree. Let P : p , p , . . . , p k be a longest path in U ( X ) . Then U ( X ) consists of P and u, v (possibly u = v ) where u is either a leaf adjacent to some p ℓ but not in P or u = p ℓ and v is either a leaf adjacent to some p j but not in P or v = p j , and one of thefollowing statements holds:(i) u is not in P and ℓ = 3 , v is not in P and j = k − , and X or its dual has arcs ( p , u ) , ( p k − , v ) (See Figure 9(i));(ii) u is not in P and ℓ = 3 , ≤ j ≤ k − with j > when v is not in P , and X or itsdual has arcs ( p , u ) , ( v, p k ) if either k + j is even and v is not in P or k + j is oddand v is in P ; otherwise X or its dual has arcs ( p , u ) , ( p k , v ) (See Figure 9(ii));(iii) u is not in P and ≤ ℓ ≤ k − , j = ℓ + 1 , X or its dual has arcs ( p , p k ) , ( v, u ) if either k is even and v is not in P or k is odd and v is in P ; otherwise X or itsdual has arcs ( p , p k ) , ( u, v ) (See Figure 9(iii)); . .p p p p k − p k − p k u v (i) . . . . . .p p p p j p k − p k u v (ii.a): The second arc is ( v, p k ) if k + j iseven and ( p k , v ) otherwise where 2 < j Suppose both p , p k are arc-balancing vertices. By Lemma 4.10, p balancesan arc between p and a leaf u adjacent to p but not in P , and p k balances an arc between p k − and a leaf v adjacent to p k − but not in P . In the tree U ( X ) the unique ( u, p k )-path avoids p and the unique ( p , v )-path avoids p k . The lengths of these two pathshave the same parity so by Proposition 2.3, we have either ( u, p )Γ ∗ ( p k , p )Γ ∗ ( p k , v ) or( u, p )Γ ∗ ( p , p k )Γ ∗ ( p k , v ). In both cases ( u, p )Γ ∗ ( p k , v ) so ( p , u )Γ( u, p )Γ ∗ ( p k , v )Γ( v, p k − ).Since the two arcs of X are opposing, X or its dual contains arcs ( p , u ) , ( p k − , v ). Theminimality of X ensures that U ( X ) consists of P and u, v and thus statement ( i ) holds.Suppose next that p is an arc-balancing vertex but p k is not. By Lemma 4.10 p balances an arc between p and a leaf u adjacent to p but not in P . Since p k is not anarc-balancing vertex, it is an endvertex of an arc. Let v be the other endvertex. Theneither v = p j for some 1 ≤ j ≤ k − P . Suppose that v is a leaf adjacent to p j . Then j / ∈ { , k } because P is the longest path in U ( X ), and j = 2 because p has no neighbour other than p , p according to Lemma 4.10. Moreover, j = k − p j and p k is balanced, which is not possible. So2 < j < k − 1. In the tree U ( X ) the unique ( u, p k )-path avoids p and the unique ( p , v )-path avoids p k . If k + j is even and v is not in P or k + j is odd and v is in P , then thelengths of these two paths have the same parity. By Proposition 2.3, ( u, p )Γ ∗ ( p k , v ) andso ( p , u )Γ( u, p )Γ ∗ ( p k , v ). Since the two arcs of X are opposing, X or its dual containsarcs ( p , u ) , ( v, p k ). Otherwise, the lengths of the two paths have the opposite parities andwe have ( p , u )Γ( u, p )Γ ∗ ( v, p k ). Hence X or its dual contains arcs ( p , u ) , ( p k , v ). Theminimality of X ensures that U ( X ) consists of P and u, v and thus statement ( ii ) holds.It remains to consider the case when neither of p , p k is an arc-balancing vertex. Sup-pose first that X contains an arc between p and p k . Let u, v be the endvertices of theother arc. We claim that at least one of u, v is not in P . Indeed, if they are both in P (say u = p i and v = p j where i < j ) then j > i + 1. It is easy to check that X − p i +1 cannotbe completed to a local tournament, which contradicts the minimality of X . So at leastone of u, v is not in P . Assume without loss of generality that u is not in P . Since U ( X )is a caterpillar, u is a leaf adjacent to some p ℓ in P , and v = p j or v is a leaf adjacent tosome p j in P . By reversing ≺ if needed we assume that j ≥ ℓ . Since uv is an unbalancededge of U ( X ), j = ℓ . In the tree U ( X ) the unique ( p , u )-path avoids p k and the unique( v, p k )-path avoids u . If k is even and v is not in P or k is odd and v is in P , thenthe lengths of these two paths have the same parity. By Proposition 2.3, ( p , p k )Γ ∗ ( u, v )and hence X or its dual contains arcs ( p , p k ) , ( v, u ). Otherwise, the lengths of the twopaths have opposite parities and we have ( p , p k )Γ ∗ ( v, u ) and X or its dual contains arcs( p , p k ) , ( u, v ). If j > ℓ +1, then the two arcs are still opposing in X − p ℓ +1 , a contradictionto the assumption that X is an obstruction. So j = ℓ + 1. The minimality of X ensures U ( X ) contains no other vertices. Therefore statement ( iii ) holds.Suppose now that X does not contain an arc between p , p k . Then p , p k are incidentwith different arcs. Let u, v be the other endvertices of the arcs incident with p , p k u = p ℓ or is a leaf adjacent to some p ℓ in P and v = p j or is a leafadjacent to some p j in P . Since X has no arc between p , p k and p u is an unbalancededge of U ( X ), 3 ≤ ℓ ≤ k − 1. Similarly, 2 ≤ j ≤ k − 2. In U ( X ) the unique ( p , v )-pathavoids p k and the unique ( u, p k )-path avoids p . If k + ℓ + j is even and P contains eitherboth u, v or neither, or k + ℓ + j is odd and P contains exactly one of u, v , then thelengths of these two paths have opposite parities. By Proposition 2.3, ( p , u )Γ ∗ ( v, p k ) andhence X or its dual contains arcs ( p , u ) , ( p k , v ). Otherwise, the lengths of the two pathshave the same parity and ( p , u )Γ ∗ ( p k , v ) and X or its dual contains arcs ( p , u ) , ( v, p k ).If j < ℓ − 1, then the two arcs are opposing in X − p ℓ − , contradicting that X is anobstruction. So j ≥ ℓ − 1. The minimality of X ensures that U ( X ) contains no othervertices. Therefore statement ( iv ) holds. U ( X ) contains a C but no induced C We now examine obstructions X that do not contain cut-vertices and for which U ( X )contains cycles. By Corollary 4.9, U ( X ) is connected. We know from Lemma 4.3 thatany induced cycle in U ( X ) is of length 3, 4 or 5, and also from Lemma 4.7 that if U ( X )contains an induced cycle of length 5 then it does not contain an induced cycle of length3 or 4.We divide our discussion into four cases: U ( X ) contains a C but no induced C ; U ( X ) contains an induced C but no C ; U ( X ) contains both a C and an induced C ;and U ( X ) contains an induced C . These four cases will be treated separately. Lemma 4.12. Let X be an obstruction that contains no cut-vertices. Suppose U ( X ) contains a C but no induced C . Then the C is the only cycle in U ( X ) and any vertexnot on the C is a leaf adjacent to a vertex on the C and incident with an arc. Moreover,any vertex on the C is adjacent to a vertex not on it. Proof: Since U ( X ) contains a C but no induced C , by Lemmas 4.3, 4.5, and 4.7,the C is the unique cycle in U ( X ). Let C : v v v the unique cycle. Consider a vertex u that is not on C . By Lemma 4.4, u is adjacent to a vertex on C . Since C is theunique cycle in U ( X ), u must be a leaf. Clearly, u is not a cut-vertex of U ( X ) and byassumption is not a cut-vertex of U ( X ). If u is an arc-balancing vertex, then u balancesan arc incident with a vertex on C . Thus the other two vertices of C must be adjacent toboth endvertices of the arc, a contradiction to the fact C is the unique cycle in U ( X ). So u is not an arc-balancing vertex and therefore by Lemma 2.8 it is incident with an arc.It remains to show that each vertex on C is adjacent to a vertex not on it. Suppose onthe contrary that v is not adjacent to a vertex not on C . By Lemma 4.1, v and v eachhas two non-neighbours. Clearly, the non-neighbours of v and of v are not on C . Weknow from the above they are endvertices of arcs. Since v is adjacent to none of them, v is not an arc-balancing vertex. By assumption v is not a cut-vertex of U ( X ). It cannotbe a cut-vertex of U ( X ) because it is adjacent only to v , v (which are adjacent). Thisis a contradiction to Lemma 2.8. 30 heorem 4.13. Let X be an obstruction that contains no cut-vertices. Suppose U ( X ) contains a C but no induced C . Then U ( X ) is one of the graphs in Figure 10 and X orits dual contains the dotted arcs. v v v uw v (i) v v v uzw v (ii) Figure 10: Obstructions X for which U ( X ) contains a C but no induced C . Proof: Suppose X is an obstruction. Let C : v v v be the unique C in U ( X ). ByLemma 4.12, each vertex of C is adjacent to a vertex not on it and each vertex not on C is a leaf adjacent to a vertex of C . Let u, v, w be vertices adjacent to v , v , v respectivelybut not on C . By Lemma 4.12, each of u, v, w is incident with an arc. Since X containsexactly two arcs, there must be an arc with both endvertices among u, v, w . Without lossof generality, assume there is an arc between u and v . By possibly considering the dualof X , let ( u, v ) be an arc. On the other hand, let z denote the other endvertex of the arcincident with w . First suppose z is on C . Then z ∈ { v , v } . Without loss of generality,assume z = v . Then ( u, v )Γ( v , u )Γ( u, v )Γ( w, u )Γ( v , w ) = ( z, w ). Since the two arcsin X are opposing, the second arc must be ( w, z ) = ( w, v ). Thus U ( X ) is Figure 10(i).Otherwise, z is not on the C . By Lemma 4.12, z is a leaf adjacent to a vertex on C .Clearly, z cannot be adjacent to v because otherwise the arc between w and z would bebalanced. Hence, assume without loss of generality that z is adjacent to v . If z = u ,then u, z belong to one component of U ( X − v ) and v, w belong to another, so uv and wz belong to the same implication class of U ( X − v ), contradicting Theorem 2.7. Hence, z = u . In this case, we have ( u, v )Γ ∗ ( w, u )Γ( v , w )Γ( w, z ). Hence, the second arc is ( z, w ).Thus U ( X ) is Figure 10(ii). U ( X ) contains an induced C but no C We consider next the case when U ( X ) contains an induced C but no C . Since U ( X )is a proper circular-arc graph, by Theorem 2.1 any induced C in U ( X ) contains at mosttwo cut-vertices of U ( X ). Theorem 4.14. Let X be an obstruction that contains no cut-vertices. Suppose U ( X ) contains a unique induced C but no C . Then U ( X ) is one of the graphs in Figure 11and X or its dual contains the dotted arcs. . .v v v p p p k (i): k is even . . .v v v p p p k (ii): k is odd . . . . . .v p i +1 p i p i − p p k (iii): k is even. . . . . . .v p i +1 p i p i − p p k (iv): k is odd. . . . . . .v v p i +1 p i p p k (v): k is even. . . . . . .v v p i +1 p i p p k (vi): k is odd. . . . . . .v v p i +1 p i p p k (vii): k is even. . . . . . .v v p i +1 p i p p k (viii): k is odd. . . .v v p p p p k − p k − p k x (ix): k > v v p p p p (x): k = 4 . . .v v p p p p k (xi): The second arc is between p k and any ver-tex x / ∈ { v , p , p k − , p k } . If k + d ( p , x ) is even,then ( x, p k ) is an arc. Otherwise, ( p k , x ) is anarc. Figure 11: Obstructions X for which U ( X ) contains a unique induced C but no C . Proof: Let C : v v v v be the unique induced C in U ( X ). Since U ( X ) and U ( X )are both connected, at least one vertex on C is adjacent to a vertex not on C . Moreover,since C is the unique cycle in U ( X ), any vertex on C that is adjacent to a vertex not on32 is a cut-vertex of U ( X ). So C contains at least one cut-vertex.Suppose that only one vertex on C is a cut-vertex of U ( X ). Without loss of generalityassume v is such a vertex. We claim that v , v are incident with different arcs. Indeed,since v , v are not cut-vertices, by Proposition 2.8 they are either incident with arcs orarc-balancing vertices. If v is an arc-balancing vertex, then it balances an arc incidentwith v or v . Note that v is adjacent to both v and v so v must be adjacent tothe other endvertex of the arc balanced by v , which is not possible. Hence v is not anarc-balancing vertex. By symmetry v is not an arc-balancing vertex either. Thereforeeach of v , v is incident with an arc. Since v , v have the same neighbourhood, therecannot be an arc between v , v , which implies that v , v are incident with different arcsas claimed.Let u, w denote the other endvertices of the arcs incident with v , v respectively.Clearly, u, w are not on C . Since v is the unique cut-vertex in C , each of u, w belongs toa component of U ( X − v ) that does not contain a vertex of C . According to Theorem2.7, v u and v w belong to different implication classes of U ( X − v ). Since v , v arein the same component of U ( X − v ), u, w are in different components of U ( X − v )by Theorem 2.5. The vertex v is not a cut-vertex so it is an arc-balancing vertex byProposition 2.8. Without loss of generality, assume v balances the arc between v and u . Then u must be a leaf adjacent to v , as otherwise there is a vertex adjacent to u but not to v , a contradiction to the fact v balances the arc between v and u . Let P : u = p , p , . . . , p k = w be a shortest ( u, w )-path. Such a path exists because U ( X ) isconnected. It is easy to see that p = v . By possibly considering the dual of X , assume( p , v ) is an arc of X . If k is even, then ( p , v )Γ( v , p )Γ( p , v )Γ ∗ ( p k , v )Γ( v , p k ) byProposition 2.3. The two arcs of X are opposing, so the second arc is ( p k , v ). Theminimality of X ensures U ( X ) is Figure 11(i) and X contains the dotted arcs. Otherwise, k is odd and the second arc is ( v , p k ), so U ( X ) is Figure 11(ii) and X contains the dottedarcs.Suppose that exactly two vertices of C are cut-vertices of U ( X ). We consider firstthe case when the two cut-vertices of U ( X ) on C are non-consecutive, say v and v .We claim that v , v are incident with different arcs. Since v , v are not cut-vertices,neither of them is adjacent to any vertex not on C . In particular, if v is an arc-balancingvertex, it must balance an arc incident with v or v , and the other endvertex is adjacentto v but not to v . Such a vertex does not exist, so v is not an arc-balancing vertex.Similarly, v is not an arc-balancing vertex. By Proposition 2.8, v , v are incident witharcs. Moreover, since v , v share the same neighbourhood, they must be incident withdifferent arcs as claimed. Let H denote a component of U ( X − v ) not containing verticeson C , and H denote a component of U ( X − v ) not containing vertices on C . Since C isthe unique cycle in U ( X ), H , H are vertex-disjoint trees. Let u, w be leaves of U ( X ) in H , H respectively. Clearly, neither u nor w can balance the arc incident with v becauseotherwise the other endvertex would be adjacent to both of v , v and thus would be v ,a contradiction to the fact that v , v are incident with different arcs. Similarly, neither u nor w can balance the arc incident with v . Hence each of u, w is incident with an arc byProposition 2.8. Without loss of generality, assume there is an arc between u, v and anarc between w, v . By the choice of u and w , there is a ( w, u )-path that contains v but33ot v . Let P : w = p , . . . , p k = u be a shortest ( w, u )-path where p i − = v , p i = v , and p i +1 = v for some i . By possibly considering the dual of X , assume ( w, v ) = ( p , v ) is anarc. Suppose k is even. If i is even, ( p , v )Γ ∗ ( p , p k )Γ ∗ ( p k , p i ) by Proposition 2.3. So thesecond arc is ( p i , p k ) = ( v , u ). If instead i is odd, then ( p , v )Γ ∗ ( p k , p )Γ ∗ ( p k , p i ). Thesecond arc is again ( p i , p k ) = ( v , u ). The minimality of X ensures U ( X ) is Figure 11(iii)and X contains the dotted arcs. Otherwise, k is odd and U ( X ) is Figure 11(iv) and X contains the dotted arcs.We now consider the case when the two cut-vertices of U ( X ) on C are consecutive,say v and v . First suppose both v and v are incident with arcs. Clearly, v , v areincident with different arcs. Let u, w be the other two endvertices of the arcs. By a similarargument as above, u, w are leaves in components of U ( X − p ) , U ( X − p ) respectively.Let P : w = p , . . . , p k = u be a shortest ( w, u )-path where p i = v and p i +1 = v forsome i . There are two possibilities: either w or u is the endvertex of the arc incident with v . Suppose there is an arc between w and v . By possibly considering the dual of X ,assume ( p , v ) = ( w, v ) is an arc in X . Suppose k is even. If i is odd, then Proposition2.3 implies ( p , v )Γ ∗ ( p , p k )Γ ∗ ( v , p k ). If i is even, then ( p , v )Γ ∗ ( p k , p )Γ ∗ ( v , p k ). Ineither case, the second arc is ( p k , v ), so U ( X ) is Figure 11(v) and X contains the dottedarcs. Otherwise, k is odd and U ( X ) is Figure 11(vi) and X contains the dotted arcs.On the other hand, suppose ( p , v ) is an arc in X . Suppose k is even. Then we have( p , v )Γ ∗ ( p , p k )Γ ∗ ( v , p k ) if i is even, and ( p , v )Γ ∗ ( p k , p )Γ ∗ ( v , p k ) if i is odd. In eithercase, the second arc is ( p k , v ), so U ( X ) is Figure 11(vii) and X contains the dotted arcs.Otherwise, k is odd and U ( X ) is Figure 11(viii) and X contains the dotted arcs.Suppose that one of v , v is not incident with an arc. Without loss of generality,assume it is v . Then v is an arc-balancing vertex by Proposition 2.8. Since v is acut-vertex, it is adjacent to a vertex x not on C . So, if v balances an arc incident with v , then the other endvertex must be adjacent to both v and x , contradicting thefact that C is the unique cycle. Hence v balances an arc incident with v . Since v isadjacent only to v and v , the other endvertex w is a leaf adjacent to v . Without lossof generality, assume ( w, v ) is an arc. Since v is a cut-vertex, there is a component H of U ( X − v ) not containing the vertices on C . Let u be a vertex of maximal distancefrom v in H , and let P : w = p , . . . , p k = u be a shortest ( w, u )-path in U ( X ). Clearly, p = v and p = v . Moreover, since C is the unique cycle and u is of maximal distancefrom v in H , u is a leaf. First suppose u balances an arc incident with p k − . Thereare two cases depending on whether or not k > 4. If k > 4, then p k − = v , so theother endvertex is a leaf x adjacent to p k − . If k = 4, then p k − = v , so the otherendvertex is v , because it must be adjacent to both v and v and C is the unique cycle.In either case, we have d ( v , p k ) + d ( p , x ) = 2 k − 3, so one of d ( v , p k ) , d ( p , x ) is evenand the other is odd. If d ( v , p k ) is even and d ( p , x ) is odd, then Proposition 2.3 implies( p , v )Γ ∗ ( p , p k )Γ ∗ ( p k , x )Γ( x, p k − ). On the other hand, if d ( v , p k ) is odd and d ( p , x )is even, then ( p , v )Γ ∗ ( p k , p )Γ ∗ ( p k , x )Γ( x, p k − ). In either case, the second arc must be( p k − , x ). Thus, U ( X ) is Figure 11(ix) if k > k = 4, and X contains the dotted arcs.Otherwise, u is incident with an arc by Proposition 2.8. Let x denote the other34ndvertex. Since v is not incident with an arc, x = v . Theorem 2.7 implies p v and p k x belong to different implication classes of U ( X − p ), so x = p by Theorem 2.5. Thus, x / ∈{ v , p , p k − , p k } . Suppose k + d ( p , x ) is even. Since d ( v , p k )+ d ( p , x ) = ( k − d ( p , x ),one of d ( v , p k ) and d ( p , x ) is even and the other is odd. If d ( v , p k ) is even and d ( p , x )is odd, then Proposition 2.3 implies ( p , v )Γ ∗ ( p , p k )Γ ∗ ( p k , x ). Otherwise if d ( v , p k ) isodd and d ( p , x ) is even, then ( p , v )Γ ∗ ( p k , p )Γ ∗ ( p k , x ). In either case, the second arcis ( x, p k ). Otherwise, k + d ( p , x ) is odd and the second arc is ( p k , x ). So, U ( X ) isFigure 11(xi) and X contains the dotted arcs. Theorem 4.15. Let X be an obstruction that has no cut-vertices. Suppose that U ( X ) contains two induced C ’s but no C . Then U ( X ) is one of the graphs in Figure 12 and X or its dual contains the dotted arcs. . . .v v v x v v v yp p k (i) v v v v v v x y (ii) v v v v v v (iii) . . .v v u v v p p k (iv): k is even. . . .v v u v v p p k (v): k ≥ v v u u zv v v (vi) v v u v v v u (vii) Figure 12: Obstructions X for which U ( X ) contains two induced C ’s but no C . Proof: Suppose there are two induced C ’s in U ( X ) which share at most one commonvertex. Let C and C ′ be such induced C ’s and let P : p , p , . . . , p k be a shortest pathbetween a vertex of C and a vertex of C ′ . By Lemma 4.2 any connected subgraph of U ( X ) has at most six non-cut-vertices. The (connected) subgraph of U ( X ) induced by C ∪ C ′ ∪ P has at least six non-cut-vertices and thus has exactly six non-cut-vertices.This implies that P is the unique path between C and C ′ and each p i of P is a cut-vertexof U ( X ). Since the subgraph of U ( X ) induced by C ∪ C ′ ∪ P has six non-cut-vertices, U ( X ) also has six non-cut-vertices according to Lemma 4.2. Thus by Corollary 2.9 the35ix non-cut-vertices of U ( X ) form two disjoint arc-balancing triples.Denote C : v v v p and C ′ : v v v p k . We first show that v is not incident with anarc. Suppose there is an arc between v and a vertex z . Since p is a cut-vertex of U ( X ),it does not balance the arc between v and z . Since p adjacent to v , it is adjacent to z . If z is not in C ∪ C ′ ∪ P , then the subgraph induced by C ∪ C ′ ∪ P ∪ { z } containsseven non-cut-vertices (i.e., v , v , v , v , v , v , z ), which contradicts Lemma 4.2. So z isin C ∪ C ′ ∪ P . Note that z is adjacent to p . If z = v , then v is adjacent to v but not z and there is another vertex in C ′ ∪ P adjacent to z but not v , a contradiction to thefact that v and z are in a an arc-balancing triple. Thus z = v . But then the vertex v which balances the arc between v and z can not be in C ∪ C ′ ∪ P . Assume without loss ofgenerality that v is adjacent to v but not to z . Since v is incident with an arc, it is not acut-vertex of U ( X ). So U ( X ) − v has a ( v, v )-path Q . The connected subgraph of U ( X )induced by C ∪ C ′ ∪ P ∪ Q contains seven non-cut-vertices (i.e., v , v , v , v , v , v , v ), acontradiction to Lemma 4.2. Therefore v is not incident with an arc. By symmetry, noneof v , v , v is incident with an arc.Since p is a cut-vertex and any induced C in U ( X ) contains at most two cut-verticesof U ( X ), v , v cannot both be cut-vertices of U ( X ). Moreover, we know from above thatneither of v , v is incident with an arc so Proposition 2.8 implies that one of v , v isan arc-balancing vertex. Similarly, one of v , v is an arc-balancing vertex. Hence oneof v , v is an arc-balancing vertex and the other is a cut-vertex of U ( X ). Without lossof generality, assume v is an arc-balancing vertex and v is a cut-vertex of U ( X ). Thevertex v is adjacent to exactly one endvertex u of the arc it balances. We claim that u = v . If u is not in C ∪ C ′ ∪ P , then U ( X ) contains a ( u, v )-path Q not containing v because v is a non-cut-vertex. So the connected subgraph induced by C ∪ C ′ ∪ P ∪ Q contains seven non-cut-vertices, contradicting Lemma 4.2. So u is in C ∪ C ′ ∪ P . Since p is a cut-vertex and no cut-vertex of U ( X ) is arc-balancing or incident with an arc, u = p .Hence u = v as claimed, and v balances an arc incident with v . The other endvertex x must therefore be outside of C ∪ C ′ ∪ P and adjacent to v .By symmetry, v balances an arc between v and a vertex y outside of C ∪ C ′ ∪ P andadjacent to v . By possibly taking the dual of X assume ( v , x ) is an arc in X . If k iseven, then Proposition 2.3 implies ( p , x )Γ ∗ ( x, p k ) and ( p , y )Γ ∗ ( y, p k ). Hence( v , x )Γ( x, v )Γ( p , x )Γ ∗ ( x, p k )Γ( v , x )Γ( x, y )Γ( y, v )Γ( p , y )Γ ∗ ( y, p k )Γ( v , y )Γ( y, v )and so the second arc is ( v , y ). Otherwise, k is odd and in this case, we have( v , x )Γ( x, v )Γ( p , x )Γ ∗ ( p k , x )Γ( x, v )Γ( y, x )Γ( v , y )Γ( y, p )Γ ∗ ( y, p k )Γ( v , y )Γ( y, v )and the second arc is again ( v , y ). The minimality of X ensure that U ( X ) is the graphin Figure 12(i) and X contains the dotted arcs.Suppose next there are two induced C ’s in U ( X ) which share two common verticesbut no two induced C ’s in U ( X ) share three common vertices. Then such two C ’s mustshare an edge. Let C : v v v v and C ′ : v v v v be such induced C ’s in U ( X ). Since U ( X ) contains no C and no two induced C ’s share three vertices, the subgraph inducedby C ∪ C ′ has exactly seven edges belonging to the two C ’s. We claim that no vertex36utside of C ∪ C ′ is adjacent to v or v . Indeed, if some vertex z outside of C ∪ C ′ isadjacent to v or v , then it must be adjacent to at least two vertices in C ∪ C ′ , becauseotherwise Theorem 2.1 would imply that U ( X ) is not a proper circular-arc graph. Butthen C ∪ C ′ ∪ { z } would induced a connected subgraph in U ( X ) having seven non-cut-vertices, a contradiction to Lemma 4.2.Suppose both v and v are arc-balancing vertices. If v balances an arc incident with v , then the other endvertex z of the arc is not in C ∪ C ′ that is adjacent to both v and v . Thus the (connected) subgraph of U ( X ) induced by C ∪ C ′ ∪ { z } has sevennon-cut-vertices, contradicting Lemma 4.2. Hence v does not balance an arc incidentwith v . Moreover, since no vertices outside of C ∪ C ′ is adjacent to v , v must balancean arc incident with v or v . Similarly, v must balance an arc incident with v or v .Without loss of generality, assume v balances an arc incident with v . Thus the otherendvertex x is a vertex whose only neighbour in C ∪ C ′ is v . We claim v balances an arcincident with v . Otherwise, v balances an arc incident with v , so the other endvertexof the arc has v as the only neighbour in C ∪ C ′ . Clearly, either v is a non-cut-vertexor U ( X ) has a non-cut-vertex in a component of U ( X − v ) not containing vertices in C ∪ C ′ . In either case, U ( X ) contains a non-cut-vertex that is neither an endvertex of anarc nor an arc-balancing vertex, a contradiction by Proposition 2.8. Hence v balancesan arc incident with v as claimed. The other endvertex y of the arc has v as the onlyneighbour in C ∪ C ′ . By possibly considering the dual of X , assume ( x, v ) is an arc in X .Since ( x, v )Γ( v , x )Γ( x, v )Γ( y, x )Γ( v , y )Γ( y, v )Γ( v , y ), the second arc is ( y, v ). Theminimality of X ensures that U ( X ) is Figure 12(ii) and X contains the dotted arcs.Suppose at least one of v , v is not an arc-balancing vertex. Without loss of generalityassume that v is not an arc-balancing vertex. Then by Proposition 2.8, v must beincident with an arc. The subgraph of U ( X ) induced by C ∪ C ′ has six non-cut-verticesso by Lemma 4.2 U ( X ) has six non-cut-vertices. Corollary 2.9 implies that the six non-cut-vertices of U ( X ) form two disjoint arc-balancing triples. Since v has three neighbours in C ∪ C ′ and the arc incident with v has an arc-balancing vertex, the other endvertex mustbe adjacent to at least two of the three neighbours of v in C ∪ C ′ . Since any connectedsubgraph of U ( X ) has at most six non-cut-vertices by Lemma 4.2, so any vertex not in C ∪ C ′ is adjacent to at most one vertex in C ∪ C ′ . It follows that the other endvertex ofthe arc incident with v is in C ∪ C ′ . Without loss of generality, assume ( v , v ) is an arcin X . Clearly, v is the ( v , v )-balancing vertex. We claim that v is incident with an arc.Otherwise, Proposition 2.8 would imply v is an arc-balancing vertex and hence balancesan arc incident with v . By the above, the other endvertex x of the arc incident with v is adjacent to v . Clearly, either v is a non-cut-vertex or U ( X ) has a non-cut-vertex in acomponent of U ( X − v ) not containing vertices in C ∪ C ′ . In either case, U ( X ) containsa non-cut-vertex that does not belong to either arc-balancing triple, a contradiction byProposition 2.8. Therefore v must be incident with an arc. By a similar argument asabove, the other endvertex of the arc incident with v is in C ∪ C ′ . Since U ( X ) contains twodisjoint arc-balancing triples and v is the ( v , v )-balancing vertex, the other endvertexcannot be v and hence must be v . Since ( v , v )Γ( v , v )Γ( v , v )Γ( v , v )Γ( v , v ), thesecond arc is ( v , v ). The minimality of X ensures that U ( X ) is Figure 12(iii) and X contains the dotted arcs. 37uppose now that there are two induced C ’s in U ( X ) which share three commonvertices. Let C : v v v v and C ′ : v v v v be such induced C ’s. Note that C ∪ C ′ induces a K , in U ( X ). Each v i with 1 ≤ i ≤ U ( X ). If v i is a cut-vertex of U ( X ), then U ( X − v i ) must contain a non-cut-vertex of U ( X ) that is not in C ∪ C ′ . Let u i be such a vertex in U ( X − v i ) when v i is a cut-vertex;otherwise let u i = v i for each 1 ≤ i ≤ 5. First note that u , u are non-adjacent andthat u , u , u are pairwise non-adjacent. Moreover, if u i = v i then u i is not adjacentto u j for all j = i . Since each u i is a non-cut-vertex, it is an endvertex of an arc oran arc-balancing vertex by Proposition 2.8. This implies that there is an arc-balancingtriple T contained in { u , u , . . . , u } . Since there is exactly one edge in T , we know fromthe above observation the only edge in T has one endvertex in { u , u } and the other in { u , u , u } . Without loss of generality assume that u u is the edge in T . Then we musthave u = v and u = v and thus neither v nor v is a cut-vertex of U ( X ). It is easy tosee that the third vertex of T is u and v balances the arc between u and v . Withoutloss of generality assume ( u , v ) is an arc in X . Clearly, u = v and v is a cut-vertex of U ( X ).Since C has at most two cut-vertices of U ( X ), at most one of v , v can be a cut-vertex. If neither of v , v is a cut-vertex, then one of them is an endvertex of an arcwhich is balanced by the other vertex. Since v is adjacent to both v , v , it is adjacentto the other endvertex of the arc, which implies U ( X ) contains a C , a contradiction tothe assumption. So exactly one of v , v is a cut-vertex of U ( X ) and we assume it is v .Suppose that there is an arc between v and u . Let P : v = p , . . . , p k = u be theshortest ( v , u )-path in U ( X ). Suppose k is even. Then by Proposition 2.3,( u , v )Γ( v , u )Γ( u , p )Γ ∗ ( p k , u )Γ( v , p k )Γ( p k , v )Γ( v , p k )Γ( p k , v )and hence the second arc is ( v , p k ). The minimality of X ensures U ( X ) is Figure 12(iv)and X contains the dotted arcs. Otherwise, k is odd and U ( X ) is Figure 12(v) and X contains the dotted arcs.Suppose that there is no arc between v and u . Then either u is incident with anarc balanced by v or v is incident with an arc balanced by u . Suppose it is the theformer. Let z denote the other endvertex. Since v is not adjacent to u , it is adjacentto z . Moreover, since u is not adjacent to v , z is also not adjacent to v . In particular, z / ∈ C ∪ C ′ . If a vertex other than v is adjacent to z , then it must also be adjacent to u .In particular, the choice of u implies that v is the only vertex that is possibly adjacentto u . Since U ( X ) is connected, v must be adjacent to u , so v is adjacent to z as well.Since ( u , v )Γ( v , u )Γ( u , v )Γ( u , u )Γ( v , u )Γ( u , v )Γ( z, u ), the second arc is ( u , z ).The minimality of X ensures U ( X ) is Figure 12(vi) and X contains the dotted arcs.Suppose instead that v is incident with an arc balanced by u . Since u is not adjacentto v , it is adjacent to the other endvertex. Moreover, since v is adjacent to v and v ,the other endvertex is also adjacent to v and v . By the choice of u , the only vertex thatcan be adjacent to all of u , v , v is v , so the other endvertex is v and u is adjacentto v . Since ( u , v )Γ( v , u )Γ( u , v )Γ( u , u )Γ( v , u )Γ( u , v )Γ( v , v ), the second arc is( v , v ). The minimality of X ensures U ( X ) is Figure 12(vii) and X contains the dottedarcs. 38 .5 U ( X ) contains a C and an induced C Lemma 4.16. Let X be an obstruction which has no cut-vertices. Suppose U ( X ) contains a C and an induced C . Then U ( X ) contains a unique C and a unique induced C , which share two common vertices. Moreover, each vertex not in any of the cycles isa leaf adjacent to a vertex on the C and is incident with an arc. Proof: By Lemmas 4.4 and 4.5, U ( X ) contains a unique C and each vertex not onthe C is adjacent to a vertex in the C . It follows that each vertex not on the C isadjacent to exactly one vertex on the C . We show that if C is an induced C in U ( X )then C shares exactly two vertices with the C . Clearly, C share at most two verticeswith the C . The fact that every vertex not on the C is adjacent to a vertex in the C implies that C cannot share exactly one vertex with the C . If C shares no vertex withthe C , then C ∪ C induces a connected subgraph in U ( X ) with seven non-cut-vertices,a contradiction to Lemma 4.2.Denote the unique C in U ( X ) by v v v and without loss of generality assume that v v v v is an induced C in the graph. Let u / ∈ { v , v , . . . , v } . From the above weknow that u is adjacent to exactly one vertex in the C . Suppose that u is adjacent to v . Then u cannot be adjacent to both v , v as otherwise uv v is another C in U ( X ), acontradiction. If u is adjacent to one of v , v then uv v v v or uv v v v is an induced C in U ( X ), which contradicts Lemma 4.7. If u is adjacent to a vertex w / ∈ { v , v , . . . , v } ,then w is not adjacent to v due to the uniqueness of the C and so is adjacent to v or v . But then { u, w, v , v , . . . , v } induces a connected subgraph of U ( X ) with sevennon-cut-vertices, a contradiction to Lemma 4.2. Hence u is a leaf in U ( X ). Suppose that u is not adjacent to v . Then it is adjacent to v or v . By symmetry we assume u isadjacent to v . It is not adjacent to v as otherwise uv v is another C in U ( X ). It is notadjacent to v as otherwise uv v v is an induced C which share just one vertex (namely, v ) with the C . Suppose that u is adjacent to a vertex w / ∈ { v , v , . . . , v } . Then w isnot adjacent to v due to the uniqueness of the C . It is not adjacent to v because fromthe above any such vertex is a leaf. So w is adjacent to v . But then { u, w, v , v , . . . , v } induces a connected subgraph of U ( X ) with seven non-cut-vertices, a contradiction toLemma 4.2. Therefore any vertex u / ∈ { v , v , . . . , v } is a leaf adjacent to a vertex inthe C . It follows that v v v v is the unique induced C in U ( X ). If such a vertex u is an arc-balancing vertex, then it balances an arc incident with one of v , v , v , so theother endvertex is adjacent to the remaining two vertices among v , v , v , contradictingthe uniqueness of the C . Moreover, such a u is not a cut-vertex of U ( X ) or of U ( X ).Therefore by Proposition 2.8 u is incident with an arc. Theorem 4.17. Let X be an obstruction that contains no cut-vertices. Suppose U ( X ) contains a C and an induced C . Then U ( X ) is one of the graphs in Figure 13 and X or its dual contains the dotted arcs. v v v v y x z (i) v v v v v y x (ii) v v v v v y z (iii) Figure 13: Obstructions X for which U ( X ) contains a C and an induced C . Proof: By Lemma 4.16, U ( X ) contains a unique C and a unique C sharing twovertices. Denote the C and the induced C by v v v and v v v v respectively. We alsoknow from the lemma that every vertex not in any of the cycles is a leaf adjacent toa vertex in the C and is incident with an arc. We claim that at least two vertices ofthe C are neighbours of leaves. Indeed, since U ( X ) does not contain cut-vertices and v , v , . . . , v induce a path in U ( X ), there must be at least one vertex not on the cycles,which implies at least one vertex on the C is adjacent to a leaf. If v is the only vertexin the C adjacent to a leaf. Then v is adjacent to every vertex except v in U ( X ), acontradiction to Lemma 4.1. So v cannot be the only vertex in the C adjacent to a leaf.By symmetry, v cannot be the only vertex in the C adjacent to a leaf.Suppose v is the only vertex in the C adjacent to a leaf. Let u be a leaf of U ( X )adjacent to v . By Lemma 4.16 u is incident with an arc. The other endvertex of thisarc cannot be another leaf v as otherwise uv is a balanced edge in U ( X ), a contradiction.So the other endvertex of this arc must be among v , v , v , v . Note first that none of v , v , v , v is a cut-vertex. If the arc is between u and v , then v does not balance thisarc as it is adjacent to neither of the endvertices. So v either balances or is incidentwith the second arc. If v is incident with the second arc then v and v are arc-balancingvertices, which is not possible. If v balances the second arc, then the second arc must beincident with exactly one of v , v . But then v is also a vertex adjacent to exactly one ofthe endvertices of the second arc, which is again impossible. This shows there is no arcbetween u and v . A similar argument shows that there is no arc between u and any of v , v , v . Therefore at least two vertices of the C are neighbours of leaves.Suppose that all three vertices of the C are neighbours of leaves. Let x, y, z be leavesadjacent to v , v , v respectively. By Lemma 4.16 each of x, y, z is incident with an arc.So there is an arc between two of x, y, z . Suppose there is an arc between y and z . ByLemma 4.4, neither of v , v can be a cut-vertex so each of them is an arc-balancingvertex or incident with an arc. At least one of v , v must be an arc-balancing vertexbecause otherwise x, y, z, v , v would be five vertices incident with arcs. Without loss ofgenerality, assume v is an arc-balancing vertex. Since v is adjacent to neither of y, z ,it cannot balance the arc between y and z . Thus v balances an arc between x and x or between x and v . This is a contradiction because v is another vertex adjacent toexactly one of endvertices of the arc balanced by v . Thus there is no arc between y and z . So there is an arc between x and y or between x and z . By symmetry and taking40he dual of X if necessary we may assume that ( x, z ) is an arc. Since v and v aretwo vertices adjacent to exactly one of x, z , there cannot be an ( x, z )-balancing vertex.By Lemma 4.4, v , v are non-cut-vertices of U ( X ) so each of them is arc-balancing orincident with an arc by Proposition 2.8. Clearly, none of them can be an ( x, z )-balancingvertex. So either v balances the arc between y and v or v balances the arc between y and v . The latter case is not possible because v is adjacent to v but not to y .Hence there is an arc between y and v . Since the two arcs of X are opposing and( x, z )Γ( z, v )Γ( v , z )Γ( z, y )Γ( y, v )Γ( v , y )Γ( y, v ), ( v , y ) is an arc. The minimality of X ensure that U ( X ) is the graph in Figure 13(i).Suppose now that exactly two of v , v , v are neighbours of leaves. First considerthe case when v and v are neighbours of leaves. Let x, y be leaves adjacent to v , v respectively. Clearly, v is not a cut-vertex. Since v , v are not the C , by Lemma 4.4they are not cut-vertices. Hence, each of v , v , v is an arc-balancing vertex or incidentwith an arc by Proposition 2.8. By Lemma 4.16, x, y are incident with arcs. So atleast one of v , v , v is an arc-balancing vertex. If v is an arc-balancing vertex, thenit must balance an arc incident with v or v . But then v is another vertex adjacentto exactly one endvertex of this arc, a contradiction. Hence v is not the arc-balancingvertex. For a similar reason, v is also not an arc-balancing vertex. Thus v is an arc-balancing vertex. It is easy check an arc balanced by v cannot be incident with v .So v balances an arc incident with v . The other endvertex of this arc cannot be aleaf adjacent to v . Hence v balances an arc between v and a leaf adjacent to v .Without loss of generality assume it is between v and y . By taking the dual of X ifnecessary we may assume ( v , y ) is an arc. Thus the second arc is between x and v .Since ( v , y )Γ( y, v )Γ( v , v )Γ( v , v )Γ( v , x )Γ( x, v ) and the two arcs of X are opposing,the second arc is ( v , x ). So U ( X ) is Figure 13(ii).The case when v and v are neighbours of leaves is symmetric to the case when v and v are cut-vertices. So we now consider the case where v and v are neighbours of leaves.Let y, z be leaves adjacent to v , v respectively. By assumption, v is not a cut-vertex.Since v , v are not on the C , they are not cut-vertices by Lemma 4.4. Thus each of v , v , v is an arc-balancing vertex or incident with an arc by Proposition 2.8. It followsthat at least one of them is an arc-balancing vertex. A similar proof as above shows that v balances an arc between v and y . Without loss of generality, assume ( v , y ) is an arc.It is easy to see that v is not an arc-balancing vertex so it is incident with an arc. So thesecond arc is between v and z . Since ( v , y )Γ( y, v )Γ( v , y )Γ( y, z )Γ( z, v )Γ( v , z ) and thetwo arcs are opposing, the second arc is ( z, v ). Hence U ( X ) is Figure 13(iii). U ( X ) contains an induced C Lemma 4.18. Let X be an obstruction which has no cut-vertices. If C is an induced C in U ( X ) , then the following statements hold:(a) C is the unique cycle in U ( X ) ;(b) Each vertex not in C is a leaf adjacent to a vertex in C and incident with an arc; c) C contains an arc-balancing vertex that is not a cut-vertex of U ( X ) ;(d) If v is an arc-balancing vertex in C , then v balances an arc between a neighbour of v in C and a leaf. Proof: Let C : v v v v v be an induced C in U ( X ). By Lemma 4.3, C is a longestinduced cycle in U ( X ). According to Lemmas 4.6 and 4.7, U ( X ) contains at most oneinduced C but neither C nor induced C . Thus C is the unique cycle in U ( X ).Suppose that u is a vertex not in C . Then by Lemma 4.4 u is adjacent to a vertex in C . Since C the unique cycle in U ( X ), u is a leaf and hence not a cut-vertex of U ( X ). Let v i be the neighbour of u . If u is an arc-balancing vertex, then it balances an arc between v i and some vertex w . Since v i is adjacent to both v i − , v i +1 which do not balance thearc between v i and w , w must be adjacent to both v i − , v i +1 . Thus v i v i +1 wv i − is a C , acontradiction to the fact C is the unique cycle in U ( X ). Hence u is incident with an arcby Proposition 2.8.Suppose there are k cut-vertices in C . We know from above that each such vertex isadjacent to a leaf that is incident with an arc. Among the 5 − k non-cut-vertices in C , atmost 4 − k can be incident with arcs because there are at most four vertices incident witharcs. It follows that C contains at least one vertex that is not a cut-vertex of U ( X ) andnot incident with an arc. Such a vertex must be an arc-balancing vertex by Proposition2.8. Hence C contains an arc-balancing vertex that is not a cut-vertex of U ( X ). Withoutloss of generality, assume v is an arc-balancing vertex and it balances an arc incidentwith v . Clearly the other endvertex cannot be in C so it is a leaf of U ( X ).Suppose that v a vertex C which balances an arc between u and w . If one of u, w isa leaf neighbour of v , then the other endvertex is an isolated vertex, contradicting thefact that U ( X ) is connected. So neither of u, w can be a leaf neighbour of v . Since v isadjacent to one of u, w , at least one of u, w is in C . If the other vertex is also in C , thenthere is a vertex in C which is not v but is adjacent to exactly one of u, w , a contradiction.Therefore exactly one of u, w is a neighbour of v in C and the other is a leaf. Theorem 4.19. Let X be an obstruction that has no cut-vertices. Suppose U ( X ) contains an induced C . Then U ( X ) is one of the graphs in Figure 14 and X or its dualcontains the dotted arcs. v v v v w u (i) v v v v v uw (ii) v v v v v w ux (iii) v v v v v u (iv) v v v v v uw (v) v v v v v w u (vi) Figure 14: Obstructions X for which U ( X ) contains an induced C . Proof: Let C : v v v v v be an induced C in U ( X ). By Lemma 4.18, C is theunique cycle in U ( X ) and has a vertex which is an arc-balancing but not a cut-vertex of U ( X ). Without loss of generality assume v is such a vertex. By Lemma 4.18, v balancesan arc between a neighbour of v in C and a leaf. Without loss of generality, assume v balances an arc between v and u and ( v , u ) is the arc. Since any vertex except v that isadjacent to u is also adjacent to v , u is adjacent to v . Suppose that neither v nor v isan arc-balancing vertex. Then by Proposition 2.8, each of v , v is a cut-vertex of U ( X )or incident with an arc. Since v , v are adjacent in U ( X ), there is no arc between them.So one of v , v is not incident with an arc and hence must be a cut-vertex of U ( X ).Suppose v is a cut-vertex and v is incident with an arc. By Lemma 4.18(b), thereis a leaf w adjacent to v and incident with an arc. Hence there is an arc between v and w . Since we have ( v , u )Γ( u, v )Γ( v , v )Γ( v , v )Γ( v , w )Γ( w, v ) and the two arcsare opposing in X , the second arc is ( v , w ). Thus U ( X ) is Figure 14(i). Suppose insteadthat v is incident with an arc and v is a cut-vertex. By Lemma 4.18, there is a leaf w adjacent to v and incident with an arc. Hence there is an arc between v and w .Since ( v , u )Γ( u, v )Γ( v , v )Γ( v , v )Γ( w, v )Γ( v , w ), the second arc is ( w, v ) and U ( X )is Figure 14(ii). Finally, suppose both v and v are cut-vertices. By Lemma 4.12, thereare leaves w, x adjacent to v , v , respectively, and incident with arcs. Hence there is anarc between w and x . Since we have ( v , u )Γ( u, v )Γ( v , u )Γ( u, x )Γ( x, v )Γ( v , x )Γ( x, w ),the second arc is ( w, x ). Thus U ( X ) is Figure 14(iii).Suppose exactly one of v , v is an arc-balancing vertex. Clearly they cannot bothbe arc-balancing vertices because v is such a vertex and there are at most two arc-balancing vertices. Consider first the case when v is an arc-balancing vertex. Then v v and a leaf adjacent to v or an arc between v anda leaf adjacent to v . However, the former is not possible, as otherwise v is not anarc-balancing vertex and not incident with an arc so it is a cut-vertex by Proposition2.8. But then a leaf adjacent to it is not incident with an arc, a contradiction to Lemma4.18. So v balances an arc between v and a leaf w adjacent to v . Then we have( v , u )Γ( u, v )Γ( v , v )Γ( w, v )Γ( v , w )Γ( w, v ). The second arc is ( v , w ). When w = u , U ( X ) is Figure 14(iv); otherwise U ( X ) is Figure 14(v).Consider now the case when v is an arc-balancing vertex. Then v either bal-ances an arc between v and a leaf adjacent to v or an arc between v and a leafadjacent to v . If v balances an arc between v and a leaf w adjacent to v , then( v , u )Γ( u, v )Γ( v , v )Γ( v , w )Γ( w, v )Γ( v , w ). The second arc is ( w, v ) and U ( X ) isFigure 14(vi). On the other hand, if v balances an arc between v and a leaf adjacentto v , then v is not an arc-balancing vertex and not incident with an arc so it is a cut-vertex by Proposition 2.8. But then a leaf adjacent to it is not incident with an arc, acontradiction to Lemma 4.18. Hence this is not possible. Theorem 1.2 follows immediately from Corollary 2.2 and Theorems 3.2, 3.4, 3.6, 3.8,3.10, 3.11, 4.8, 4.11, 4.13, 4.14, 4.15, 4.17, and 4.19.Obstructions for acyclic local tournament orientation completions can be defined anal-ogously. The complete list of obstructions for acyclic local tournament orientation com-pletions has been given in the companion paper [7].Partially oriented graphs which cannot be completed to local tournaments and areminimal with respect to only vertex deletions can be derived from Theorem 1.2. Indeed,suppose that Y is such a graph, that is, Y cannot be completed to a local tournament andfor each vertex v of Y , Y − v can be completed to a local tournament. Since Y cannot becompleted to a local tournament, by Proposition 1.1 it critically contains an obstruction.Since Y is minimal with respect to vertex deletion, Y is either an obstruction describedin Theorem 1.2 or is obtained from an obstruction by orienting some edges.An oriented graph D = ( V, A ) is transitive if, for any three vertices u, v, w ∈ V , uv ∈ A and vw ∈ A imply uw ∈ A ; it is quasi-transitive if, for any three vertices u, v, w ∈ V , uv ∈ A and vw ∈ A imply uw ∈ A or wu ∈ A , cf. [1]. The orientationcompletion problems for the classes of transitive and quasi-transitive oriented graphs arepolynomial time solvable, cf. [2]. The underlying graphs of transitive oriented graphscoincide with the underlying graphs of quasi-transitive oriented graphs, which are knownas comparability graphs , cf. [3, 4, 5]. It remains open problems to find obstructions fortransitive orientation completions and quasi-transitive orientation completions.44 eferences [1] J. Bang-Jensen and J. Huang, Quasi-transitive digraphs, J. Graph Theory 20 (1995)141 - 161.[2] J. Bang-Jensen, J. Huang, and X. Zhu, Completing orientations of partially orientedgraphs, J. Graph Theory 87 (2018) 285 - 304.[3] T. Gallai, Transitiv orientierbare graphen, Acta Mathematica Academiae Scien-tiarum Hungarica 18 (1967) 25 - 66.[4] A. Ghouila-Houri, Caract´erisation des graphes non orient´es dont on peut orienterles arˇetes de mani`ere `a obtenir le graphe d’une relation d’ordre, C. R. Acad. Sci.Paris 254 (1962) 1370–1371.[5] M.C. Golumbic, Algorithmic Graph Theory and Perfect Graphs , AcademicPress, 1980.[6] P. Hell and J. Huang, Lexicographic orientation and representation algorithms forcomparability graphs, proper circular-arc graphs, and proper interval graphs, J.Graph Theory 20 (1995) 361 - 374.[7] K. Hsu and J. Huang, Obstructions for acyclic local tournament orientation com-pletions, manuscript 2020.[8] J. Huang, On the structure of local tournaments, J. Combin. Theory B (1995) 200- 221.[9] J. Huang, Lexicographic orientation algorithms for orientation completion problems,Chapter in