Finding an induced subdivision of a digraph
FFinding an induced subdivision of a digraph ∗ Jørgen Bang-Jensen † Fr´ed´eric Havet ‡ Nicolas Trotignon § April 17, 2012
Abstract
We consider the following problem for oriented graphs and digraphs: Given an oriented graph(digraph) G , does it contain an induced subdivision of a prescribed digraph D ? The complexityof this problem depends on D and on whether G must be an oriented graph or is allowed tocontain 2-cycles. We give a number of examples of polynomial instances as well as several NP-completeness proofs. Keywords:
NP-completeness, induced paths and cycles, linkings, 3-SAT.
Many interesting classes of graphs are defined by forbidding induced subgraphs, see [4] for a survey.This is why the detection of several kinds of induced subgraphs is interesting, see [7] where severalsuch problems are surveyed. In particular, the problem of deciding whether a graph G contains, as aninduced subgraph, some graph obtained after possibly subdividing prescribed edges of a prescribedgraph H has been studied. This problem can be polynomial or NP-complete depending on H andto the set of edges that can be subdivided. The aim of the present work is to investigate varioussimilar problems in digraphs, focusing only on the following problem: given a digraph H , is there apolynomial algorithm to decide whether an input digraph G contains a subdivision of H ?Of course the answer depends heavily on what we mean by “contain”. Let us illustrate thisby surveying what happens in the realm of non-oriented graphs. If the containment relation is thesubgraph containment, then for any fixed H , detecting a subdivision of H in an input graph G canbe performed in polynomial time by the Robertson and Seymour linkage algorithm [9] (for a shortexplanation of this see e.g. [2]). But if we want to detect an induced subdivision of H then theanswer depends on H (assuming P (cid:54) = NP). It is proved in [7] that detecting an induced subdivisionof K is NP-complete, and the argument can be reproduced for any H whose minimum degree is atleast 4. Polynomial-time solvable instances trivially exist, such as detecting an induced subdivisionof H when H is a path, or a graph on at most 3 vertices. But non-trivial polynomial-time solvableinstances also exist, such as detecting an induced subdivision of K , that can be performed in time ∗ This work was done while the first author was on sabbatical at Team Mascotte, INRIA, Sophia Antipolis France whosehospitality is gratefully acknowledged. Financial support from the Danish National Science research council (FNU) (undergrant no. 09-066741) is gratefully acknowledged. † Department of Mathematics and Computer Science, University of Southern Denmark, Odense DK-5230, Denmark(email: [email protected]). ‡ Projet Mascotte, I3S (CNRS, UNSA) and INRIA, Sophia Antipolis, France. Partly supported by ANR Blanc AGAPEANR-09-BLAN-0159. (email:[email protected]). § CNRS, LIP – ENS Lyon, France. Partially supported by the French
Agence Nationale de la Recherche under reference
ANR JCJC a r X i v : . [ c s . D M ] S e p ( n ) by the Chudnovsky and Seymour’s three-in-a-tree algorithm, see [5]. Note that for manygraphs H , nothing is known about the complexity of detecting an induced subdivision of H : when H is cubic (in particular when H = K ) or when H is a disjoint union of 2 triangles, and in many othercases.When we move to digraphs, the situation becomes more complicated, even for the subdigraphcontainment relation. All the digraphs we will consider here are simple , i.e. they have no loops normultiple arcs. We rely on [1] for classical notation and concepts. A subdivision of a digraph D , alsocalled a D-subdivision , is a digraph obtained from D by replacing each arc ab of D by a directed ( a , b ) -path. From the NP-completeness of the 2-linkage problem, proved by Fortune, Hopcroft andWyllie [6], it is straightforward to construct an oriented graph H such that deciding whether a givenoriented graph G contains a subdivision of H as a subgraph is NP-complete. See Theorem 33.Let us now think about the induced subdigraph relation. An induced subdigraph of a digraph G which is a subdivision of D is called an induced subdivision of D . When D is a digraph, we define:P ROBLEM Π D Input: A digraph G .Question: Does G contain an induced subdivision of D ?In Π D , the instance digraph G may have (directed) 2-cycles, where the 2 -cycle is the digraph C on 2 vertices a , b with 2 arcs ab and ba . Because of these 2-cycles, NP-completeness resultsare often quite easy to obtain, because no induced directed path can go through a 2-cycle (whichby itself contains a chord). Hence 2-cycles are very convenient to force an induced directed pathto go through many places of a large digraph that models an instance of 3-SAT. This yields NP-completeness results that cover large classes of detection problems. See Section 4. In fact, it can beeasily shown (see Section 2) that if D is the disjoint union of spiders (trees obtained from disjointdirected paths by identifying one end of each path into a vertex) and at most one 2-cycle, then Π D is polynomial-time solvable. However, except from those digraphs, we are not aware of any D forwhich Π D is polynomial time solvable. We indeed conjecture that there are none. As an evidence, weshow that if D is an oriented graph , i.e. a digraph with no 2-cycles, then Π D is NP-complete unless itis the disjoint union of spiders (see Corollary 13).It seems that allowing or not allowing 2-cycles is an essential distinction. Hence we also considerthe restricted problem Π (cid:48) D in which the input graph G is an oriented graph.P ROBLEM Π (cid:48) D Input: An oriented graph G .Question: Does G contain an induced subdivision of D ?Observe that if Π D is polynomial-time solvable then Π (cid:48) D is also polynomial-time solvable. Con-versely, if Π (cid:48) D is NP-complete then Π D is also NP-complete. Hence, NP-completeness results coverless cases for Π (cid:48) D .Similarly to Π D , for several D ’s, Π (cid:48) D is solvable by very simple polynomial-time algorithms (SeeSection 2). However, in this case they are not the only ones. We could obtain several digraphs forwhich Π (cid:48) D is solvable in polynomial time with non-trivial algorithms.We denote by T T the transitive tournament on 3 vertices a , b , c and arcs ab , ac , bc . In Subsec-tion 5.1, we use a variant of Breadth First Search that computes only induced trees to solve Π (cid:48) T T inpolynomial time.We also study oriented paths in Subsection 5.2. An oriented path is an orientation of a path. The length of an oriented path P is its number of arcs and is denoted l ( P ) . Its first vertex is called its origin and its last vertex its terminus . The blocks of an oriented paths are its maximal directed subpaths. We2enote by A − k the path on vertices s , s , . . . , s k , s k + and arcs s s , s s , s s , s s , . . . and A + k the pathon vertices s , s , . . . , s k , s k + and arcs s s , s s , s s , s s , . . . . These two paths are the antidirectedpaths of length k −
1. Observe that A − k is the converse of A + k (i.e. it is obtained from A + k by reversingall the arcs); if k is odd they are isomorphic but the origin and terminus are exchanged. Clearly, anoriented path with k -blocks can be seen as a subdivision of A − k or A + k . In particular, paths with oneblock are the directed paths. We show that if P is an oriented path with three blocks such that thelast one has length one then Π P is polynomial-time solvable. We also use classical flow algorithms toprove that Π (cid:48) A − is polynomial-time solvable.If D is any of the two tournaments on 3 vertices, namely the directed 3-cycle C and the transitivetournament T T , then Π (cid:48) D is polynomial time solvable. Hence it is natural to study the complexityof larger tournaments. In Section 6, it is shown that if D is a transitive tournament on more than 3vertices or the strong tournament on 4 vertices, then Π (cid:48) D is NP-complete.Finally, in Section 7, we point out several open questions. There are digraphs D for which Π D or Π (cid:48) D can be easily proved to be polynomial-time solvable. Forexample, it is the case for the directed k -path P k on k vertices. Indeed, a P k -subdivision is a directedpath of length at least k − k − P k . Hence a digraph has a P k -subdivision if and only if it has P k as an induced subdigraph. This canbe checked in time O ( n k ) by checking for every set of k vertices whether or not it induces a P k .A vertex of a digraph is a leaf if its degree is one, a node if its out-degree or its in-degree is at least2, and a continuity otherwise, that is if both its out- and in-degree equal 1. A spider is a tree havingat most one node. Proposition 1.
If D is the disjoint union of spiders then Π D is polynomial-time solvable.Proof. A digraph G contains an induced D -subdivision if and only if it contains D as an inducedsubdigraph. This can be checked in time O ( n | V ( D ) | ) .It is also not difficult to see that Π C is polynomial-time solvable. Proposition 2. Π C is polynomial-time solvable.Proof. A subdivision of the directed 2-cycle is a directed cycle. In a digraph, a shortest cycle isnecessarily induced, hence a digraph has a C -subdivision if and only if it is not acyclic. Sinceone can check in linear time if a digraph is acyclic or not [1, Section 2.1], Π C is polynomial-timesolvable.Since an oriented graph contains no 2-cycle, then Π (cid:48) C = Π (cid:48) C . Similarly to Π C , this problem ispolynomial-time solvable. Proposition 3. Π (cid:48) C is polynomial-time solvable.Proof. An oriented graph contains an induced subdivision of C if and only if it is not acyclic.Moreover, the following is polynomial-time solvable. Proposition 4.
If D is the disjoint union of spiders and a C then Π D is polynomial-time solvable.Proof. D (cid:48) = D − C is a collection of spiders. Let p be its order. For each set A of p vertices, wecheck if the digraph G (cid:104) A (cid:105) induced by A is D (cid:48) and if yes we check if G − ( A ∪ N ( A )) has a directedcycle. 3imilarly, Proposition 5.
If D is the disjoint union of spiders and a C then Π (cid:48) D is polynomial-time solvable. In all proofs below it should be clear that the reductions can be performed in polynomial time andhence we omit saying this anymore. Before starting with the NP-completeness proofs, we state aproposition.
Proposition 6.
Let D be a digraph and C a connected component of D. If Π C is NP-complete then Π D is NP-complete. Similarly, if Π (cid:48) C is NP-complete then Π (cid:48) D is NP-complete.Proof. Let D , . . . , D k be the components of D and assume that Π D is NP-complete. To show that Π D is NP-complete, we will give a reduction from Π D to Π D .Let G be an instance of Π D and G be the digraph obtained from D by replacing D by G . Weclaim that G has an induced D -subdivision if and only if G has an induced D -subdivision.Clearly, if G has an induced D -subdivision S then the disjoint union of S and the D i , 2 ≤ i ≤ k is an induced D -subdivision in G .Reciprocally, assume that G contains an induced D -subdivision S . Let S i , 1 ≤ i ≤ k be the con-nected components of S such that S i is an induced D i -subdivision. Set G i = D i if i ≥
2. Then the G i ’s are the connected components of G . Thus S is contained in one of the G i ’s. If it is G thenwe have the result. Otherwise, it is contained in some other component say G = D . In turn, S is contained in some G j . Hence G j contains a D -subdivision because S contains a D -subdivisionsince D contains S . Thus G j cannot be G since G already contains D and | S | ≥ | G | . If j = j =
3. And so on, for every i ≥
3, applying thesame reasoning, we show that one of the following occurs: • S i is contained in G and thus G contains a D -subdivision because S i did. • S i is contained in G j which cannot be any of the G i , 1 ≤ l ≤ i , for cardinality reasons. Hencewe may assume that G j = G i + and that G i + and hence S i + contains a D -subdivision.Since the number of components is finite, the process must stop, so G contains an induced D -subdivision. ( a , b ) -path in an oriented graph Our first result is an easy modification of Bienstock’s proof [3] that finding an induced cycle throughtwo given vertices is NP-complete for undirected graphs.
Lemma 7.
It is NP-complete to decide whether an oriented graph contains an induced ( a , b ) -pathfor prescribed vertices a and b.Proof. Given an instance I of 3-SAT with variables x , x , . . . , x n and clauses C , C , . . . , C m we firstcreate a variable gadget V i for each variable x i , i = , , . . . , n and a clause gadget C j for each clause C j , j = , , . . . , m as shown in Figure 1. Then we form the digraph G ( I ) as follows (see Figure 2):Form a chain U of variable gadgets by adding the arcs b i a i + for i = , , . . . , n − W of clause gadgets by adding the arcs d j c j + , j = , , . . . , m −
1. Add the arcs aa , b n c , c m b to geta chain from a to b . For each clause C , we connect the three literal vertices of the gadget for C tothe variable gadgets for variables occuring as literals in C in the way indicated in the figure. To beprecise, suppose C p = ( x i ∨ ¯ x j ∨ x k ) , then we add the following three 3-cycles l p x i v i l p , l p ¯ x j ¯ v j l p and l p x k v k l p . This concludes the construction of G ( I ) . A connected component of a digraph H is a connected component in the underlying undirected graph of H . j v i ¯ x i a i x i d j c j b i ¯ v i l j l j Figure 1: The variable gadget V i (left) and the clause gadget C j (right). b d x ∨ ¯ x ∨ x ¯ x ∨ x ∨ ¯ x ba ¯ x ¯ x ¯ x x x x c Figure 2: The digraph G ( I ) when I has variables x , x , x and three clauses C , C , C where C =( ¯ x ∨ x ∨ ¯ x ) and C = ( x ∨ ¯ x ∨ x ) (for clarity we do not show the arcs corresponding to C )5e claim that there is an induced directed ( a , b ) -path in G ( I ) if and only if I is satisfiable.Suppose first that I is satisfiable and consider a truth assignment T which satisfies I . Now form adirected ( a , b ) -path P by taking the arcs aa , c m b and the following subpaths: for each variable x i takethe subpath a i ¯ x i ¯ v i b i if T sets x i true and otherwise take the subpath a i x i v i b i . For each clause C j we fixa litteral l (cid:48) j which is satisfied by T and take the subpath c j l (cid:48) j d j . It is easy to check that P is inducedas we navigate it to avoid each of the arcs between the variable chain U and the clause chain W .Suppose now that Q is an induced directed ( a , b ) -path in G ( I ) . It follows from the construction that Q starts by a directed ( a , b n ) -path through all variable gadgets which contains no vertices from W and continues with a directed ( c , d m ) -path through all clause gadgets which contains no vertices from U . This follows from the presence of the directed 3-cycles that prevent Q from using any of the arcsgoing from a variable gadget to a clause gadget other than the arc b n c . Similarly there is no induceddirected ( c , d m ) -path which contains any vertex from U . Now form a truth assignment by setting x i true if and only if Q uses the subpath a i ¯ x i ¯ v i b i and false otherwise. Since Q is induced, for eachclause C j if Q uses the subpath c j l (cid:48) j d j , then we claim that l (cid:48) j will be true with the truth assignment justdescribed: if l (cid:48) j = x k for some k then since Q is induced the presence of the arc l (cid:48) j x k implies that Q usesthe path a k ¯ x k ¯ v k b k and similarly, if l (cid:48) j = ¯ x k then Q uses the path a k x k v k b k and again C j is satisfied. We first show that for any k ≥
4, the problem Π (cid:48) C k is NP-complete. Theorem 8.
It is NP-complete to decide whether an oriented graph contains an induced subdivisionof a fixed directed cycle of length at least 4.Proof.
Given an instance I of 3-SAT with variables x , x , . . . , x n and clauses C , C , . . . , C m we formthe digraph G ∗ ( I ) from G ( I ) which we defined above by adding the arc ba .Let C be an induced cycle of G ∗ ( I ) . Since the variable chain U and the clause chain W are bothacyclic, C must contain an arc with tail l in W and head y in U . If ly (cid:54) = ba , then there exists i such that y ∈ { x i , ¯ x i } and so C = lx i v i l or C = l ¯ x i ¯ v i l by construction of G ∗ ( I ) . Hence every induced directedcycle of length at least 4 contains the arc ba . Thus G ∗ ( I ) has an induced cycle of length at least 4 ifand only if G ( I ) has an induced directed ( a , b ) -path. As shown in the proof of Lemma 7 this is ifand only if I is satisfiable. Theorem 9.
Let D be an oriented graph containing an induced directed cycle of length at least 4with a vertex of degree
2. It is NP-complete to decide whether a given oriented graph contains aninduced subdivision of D.Proof.
Let D be given and let I be an arbitrary instance of 3-SAT. Fix an induced directed cycle C oflength at least 4 in D and fix an arc uv on C such that u is of degree 2. Let G (cid:48) ( I ) be the oriented graphthat we obtain by replacing the arc uv by a copy of G ( I ) and the arcs ua , bv . We claim that G (cid:48) ( I ) contains an induced subdivision of D if and only if I is satisfiable (which is if and only if G ( I ) contains an induced directed ( a , b ) -path).Clearly, if G ( I ) has an induced directed ( a , b ) -path, then we may use the concatention of thispath with ua and bv instead of the deleted arc uv to obtain an induced D -subdivision in G (cid:48) ( I ) (theonly subdivided arc will be uv ).Conversely, suppose that G (cid:48) ( I ) contains an induced subdivision D (cid:48) of D . Clearly D (cid:48) has at leastas many vertices as D and thus must contain at least one vertex z of V ( G ( I )) . Since u is of degree2, the digraph D \ uv has fewer induced directed cycles of length at least 4 than D . (Note that the fact The degree of a vertex v in a digraph is the number of arcs with one end in v , that is, the sum of the in- and out-degreeof v . u is of degree 2 is important: if u has degree more than 2, deleting uv could create new induceddirected cycles. ) Thus z must be on a cycle of length at least 4 in D (cid:48) . But this and the fact that G ( I ) has no induced directed cycle of length at least 4 implies that G (cid:48) ( I ) contains an induced directed ( a , b ) -path (which passes through z ).We move now to the detection of induced subdivisions of digraphs H when H is the disjointunion of one or more directed cycles, all of length 3. If there is just one cycle in H , the problemis polynomial-time solvable by Proposition 3. But from two on, it becomes NP-complete. We needresults on the following problem.P ROBLEM
DIDPPInput: An acyclic digraph G and two vertex pairs ( s , t ) , ( s , t ) . Moreover, there is no directed pathfrom { s , t } to { s , t } .Question: Does G have two paths P , P such that P i is a directed ( s i , t i ) -path, i = ,
2, and G (cid:104) V ( P ) ∪ V ( P ) (cid:105) is the disjoint union of P and P ?Problem k -DIDPP was shown to be NP-complete by Kobayashi [8] using a proof similar to Bien-stock’s proof in [3]. Theorem 10.
Let D be the disjoint union of two directed cycles with no arcs between them. Then Π (cid:48) D is NP-complete.Proof. Let G be an instance of DIDPP and H the oriented graph obtained from it by adding newvertices u , u and the arcs t u , u s , t u and u s . Since G was acyclic it is not difficult to see that H is a yes-instance of Π (cid:48) D if and only if G is a yes-instance of DIDPP. l ¯ x i a i x i d j l l c j b i Figure 3: The variable gadget V i (left) and the clause gadget C j (right). Unoriented bold edgesrepresent 2-cycles. Theorem 11.
Let k ≥ be an integer. Then Π C k is NP-complete.Proof. Reduction from 3-SAT. Let I be an instance of 3-SAT with variables x , x , . . . , x n and clauses C , C , . . . , C m . We first create a variable gadget V i for each variable x i , i = , , . . . , n and a clausegadget C j for each clause C j , j = , , . . . , m as shown in Figure 3. Then we form the digraph G ( I ) as follows (see Figure 4): Form a chain U of variable gadgets by adding the arcs b i a i + for i = , , . . . , n − W of clause gadgets by adding the arcs d j c j + , j = , , . . . , m −
1. Add thearcs aa , b n c , c m b to get a chain from a to b . For each clause C , we connect the three literal verticesof the gadget for C to the variable gadgets for variables occuring as literals in C in the following way.Suppose C p = ( x i ∨ ¯ x j ∨ x k ) , then we add the following three 2-cycles l p x i l p , l p ¯ x j l p and l p x k l p . Thisconcludes the construction of G ( I ) . See Figure 4. b d x ∨ ¯ x ∨ x ¯ x ∨ x ∨ ¯ x ba ¯ x ¯ x ¯ x x x x c Figure 4: The digraph G ( I ) when I has variables x , x , x and three clauses C , C , C where C =( ¯ x ∨ x ∨ ¯ x ) and C = ( x ∨ ¯ x ∨ x ) (for clarity we do not show the arcs corresponding to C )Similarly to the proof of Lemma 7, one can show that there is an induced directed ( a , b ) -path in G ( I ) if and only if I is satisfiable.Let G k ( I ) be the digraph obtained from C k by replacing one arc ab by G ( I ) . It is easy to checkthat G ( I ) has no induced cycle of length at least 3. Hence G k ( I ) has an induced directed cycle oflength k if and only if G ( I ) has an induced directed ( a , b ) -path. Hence by Lemma 7, G k ( I ) has aninduced D -subdivision if and only if I is satisfiable.A branch is a directed walk such that all the vertices are distinct except possibly its ends, its endsare nodes or leaves and all its internal vertices are continuities. A branch is central if its two ends arenodes.The skeleton of a multidigraph D is the digraph whose vertices are the nodes and leaves in D andin which ab is an arc if and only if there is a directed ( a , b ) -branch in D . Observe that a skeleton mayhave loops and multiple arcs. Clearly, any subdivision of D has the same skeleton as D . Theorem 12.
Let D be an oriented graph. If D contains a central branch, then Π D is NP-complete. roof. Reduction from 3-SAT. Let I be an instance of 3-SAT. Let B be a central branch with origin a and terminus c . Let G D ( I ) be the digraph obtained from D by replacing the first arc ab of B by G ( I ) .Clearly if G ( I ) has an induced directed ( a , b ) -path P , then the union of P and D \ ab is a D -subdivision (in which only ab is subdivided) in G D ( I ) .Conversely, assume that G D ( I ) contains an induced D -subdivision S . It is easy to check that novertex in V ( G ( I )) \ { a , b } can be a node of S (the 2-cycles prevent this). Then since S has the sameskeleton as D , a and b are nodes of S . In addition, since the number of central branches in D \ ab isone less than the number of central branches in D , one central branch of D must use vertices of G ( I ) .Thus, there is an induced directed ( a , b ) -path in G ( I ) .Hence G D ( I ) has an induced D -subdivision if and only if G ( I ) has an induced directed ( a , b ) -path and thus if and only if I is satisfiable. Corollary 13.
Let D be an oriented graph. Then Π D is NP-complete unless D is the disjoint union ofspiders.Proof. Let D be an oriented graph. If one of its connected components is neither a directed cycle nora spider, then it contains at least one central branch. So Π D is NP-complete by Theorem 12.If one of the components is directed cycle of length at least 3, then Π D is NP-complete by Theo-rem 11 and Proposition 6.Finally, if all its connected components are spiders then Π D is polynomial-time solvable accordingto Theorem 5.We believe that Corollary 13 can be generalized to digraphs. Conjecture 14.
Let D be a digraph. Then Π D is NP-complete unless D is the disjoint union of spidersand at most one 2-cycle.As support for this conjecture, we give some other digraphs D (which are not oriented graphs),for which Π D is NP-complete. In particular, when D is the lollipop , that is the digraph L with vertexset { x , y , z } and arc set { xy , yz , zy } . Note that the lollipop seems to be the simplest digraph that is notan oriented graph nor a C . So it should be an obvious candidate for a further polynomial case if oneexisted. Theorem 15.
Deciding if a digraph contains an induced subdivision of the lollipop is NP-complete.Proof.
Reduction from 3-SAT. Let I be an instance of 3-SAT with variables x , x , . . . , x n and clauses C , C , . . . , C m . We first create a variable gadget V i for each variable x i , i = , , . . . , n and a clausegadget C j for each clause C j , j = , , . . . , m as shown in Figure 5. Then we form the digraph G ( I ) as follows: Form a chain U of variable gadgets by adding the arcs b i a i + for i = , , . . . , n − W of clause gadgets by adding the arcs d j c j + , j = , , . . . , m −
1. Add the arcs aa , b n c , c m b to get a chain from a to b . For each clause C , we connect the three literal vertices of the gadget for C to the variable gadgets for variables occuring as literals in C in the way indicated in the figure.Similarly to the proof of Lemma 7, one can check that there is an induced directed ( a , b ) -path in G ( I ) if and only if I is satisfiable.The digraph G L ( I ) is obtained from L and G ( I ) by deleting the arc yz and adding the arcs ya and bz .It is easy to see that G ( I ) has no induced directed cycle of length 3 and that no 2-cycle iscontained in an induced lollipop. Hence if G L ( I ) contains an L -subdivision, the induced directedcycle in it is the concatenation of the path bzya and a induced directed ( a , b ) -path in G ( I ) . Thus I issatisfiable. The other direction is (as usual) clear. 9igure 5: The variable gadget V i , (top left), the clause gadget C i (bottom left) and the way to connectthem in G ( I ) (right). Bold unoriented edges represent 2-cycles. Only the connection for one variablegadget and one clause gadget is shown and the general strategy for connecting variable and clausegadgets is the same as in G ( I ) (Figure 2). Remark 16.
The cone is the digraph C with vertex set { x , y , z } and arc set { xy , xz , yz , zy } . In the verysame way as Theorem 15, one can show that finding an induced subdivision of the cone in a digraphis NP-complete. According to Conjecture 14, the only digraphs for which Π D is polynomial-time solvable are disjointunions of spiders and possibly one 2-cycle. For such digraphs, easy polynomial-time algorithms exist(See Section 2).In this section, we show that the picture is more complicated for Π (cid:48) D than for Π D . We show someoriented graphs D for which Π (cid:48) D is polynomial-time solvable. For all these oriented graphs, Π D isNP-complete by Corollary 13. Let s , u , v be three vertices such that s (cid:54) = v and u (cid:54) = v (so s = u is possible). A cherry on ( s , u , v ) is anyoriented graph made of three induced directed paths P , Q , R such that: • P is directed from s to u (so when s = u it has length 0); • Q and R are both directed from u to v (so they both have length at least 1 and since we do notallow parallel edges, at least one of them has length at least 2); • u , v are the only vertices in more than one of P , Q , R ; • there are no other arcs than those from P , Q , R .The cherry is rooted at s .An induced cherry contains an induced T T -subdivision (made of Q and R ) and a T T -subdivisionis a cherry (with u = s ). Hence detecting an induced cherry is equivalent to detecting an induced T T -subdivision. 10n order to give an algorithm that detects a cherry rooted at a given vertex, we use a modificationof the well-known Breadth First Search algorithm (BFS), see e.g. [1, Section 3.3]. Given an orientedgraph G and a vertex s ∈ V ( D ) , BFS returns an out-tree rooted at s and spanning all the verticesreachable from s . It proceeds as follows: BFS( G , s ) Create a queue Q consiting of s ; Intialize T = ( { s } , /0 ) while Q is not empty do Consider the head u of Q and visit u , that is foreach out-neighbour v of u in D doif v / ∈ V ( T ) then V ( T ) : = V ( T ) ∪ { v } and A ( T ) : = A ( T ) ∪ { uv } Put v to the end of Q Delete u from Q Note that the arc-set of the out-branching produced by BFS depends on the order in which the verticesare visited, but the vertex-set is always the same: it is the set of the vertices reachable from s . See [1]p. 92 for more details on BFS. We need the following variant: IBFS( G , s ) Create a queue Q consisting of s ; Intialize T = ( { s } , /0 ) while Q is not empty do Consider the head u of Q and visit u , that is foreach out-neighbour v of u in G doif N G ( v ) ∩ V ( T ) = { u } then V ( T ) : = V ( T ) ∪ { v } and A ( T ) : = A ( T ) ∪ { uv } Put v to the end of Q Delete u from Q Observe that IBFS (which we also call induced-BFS ) is the same as BFS except that we add theout-neighbour v of u to T only if it has no other neighbour already in T , hence ensuring that theresulting out-tree is an induced subdigraph of G . Contrary to BFS, the vertex-set of a tree obtainedafter IBFS may depend on the order in which the vertices are visited.IBFS can easily be implemented to run in time O ( n ) . When T is an oriented tree, we denote by T [ x , y ] the unique oriented path from x to y in T . Theorem 17.
Let G be an oriented graph, s a vertex and T a tree obtained after running IBFS ( G , s ) .Then exactly one of the following outcomes is true:(i) D contains an induced subdigraph that is a cherry rooted at s;(ii) for every vertex x of T , any out-neighbour of x not in T has an out-neighbour that is an ancestorof x in T .This is algorithmic in the sense that there is an O ( n ) algorithm that either outputs the cherry of(i) or checks that (ii) holds.Proof. Suppose that T does not satisify (ii). Then some vertex x of T has an out-neighbour y not in T and no out-neighbour of y is an ancestor of x . Without loss of generality, we assume that x is thefirst vertex added to T when running IBFS with such a property. In particular, T [ s , x ] y is an induceddirected path because a chord would contradict (ii) or the choice of x . Let v be the neighbour of y in T , different from x , that was first added to T when running IBFS. Note that v exists for otherwise y would have been added to T when visiting x . If x is the parent of v in T then T [ s , x ] y together with v s (whatever the orientation of the arc between y and v ). So we may assumethat x is not the parent of v . When visiting x , vertex y was not added to T , hence v was already visited(because x is not the parent of v ). In addition, when v was visited, it was the unique neighbour of y in the current out-tree, so y is an in-neighbour of v , for otherwise it would have been added to T . Let u be the common ancestor of x and v in T , chosen closest to x . Since T does not satisfy (ii) by thechoice of x and y , u (cid:54) = v . Now the directed paths sTu , T [ u , x ] yv and T [ u , v ] form an induced cherryrooted at s . Indeed since T is an induced out-tree, it suffices to prove that y has no neighbour in thesethree paths except x and v . By definition of v , there is no neighbour of y in T [ s , u ] and T [ u , v ] except v . Moreover, y has no out-neighbour in T [ u , x ] by the assumption that (ii) does not hold for y and x and it has no in-neighbour in T [ u , x ] except x by the choice of x .Conversely, let us assume that T satisfies (ii) and suppose by contradiction that G contains aninduced cherry C rooted at s . Since T is an induced out-branching, some vertices of C are not in T .So, let y be a vertex of V ( C ) \ V ( T ) as close to s as possible in the cherry. Let x be an in-neighbour of y in C ∪ T . From the choice of y , x and all its ancestors along the cherry are in T . Since T is induced,the ancestors of x along the cherry are in fact the ancestors of x along T . Hence, x is a vertex of T with an out-neighbour y not in T having no out-neighbour among the ancestors of x along T . Thiscontradicts T satisifying (ii).All this may be turned in an O ( n ) -algorithm that finds a cherry rooted at s if it exists or answer nootherwise. Indeed we first run IBFS and then check in time O ( n ) if the obtained tree T satisfies (ii).If not, then we can find the cherry following the first paragraph of the proof. Remark 18.
Since a digraph contains an induced
T T -subdivision if and only if it contains an in-duced cherry, Theorem 17 implies directly that Π (cid:48) T T is solvable in time O ( n ) (because we need toenumerate all potential roots).We can slightly extend our result. A tiny cherry is a cherry such that the path Q and R as in thedefinition form a T T . Corollary 19.
For any tiny cherry D, the problem Π (cid:48) D is solvable in time O ( n | V ( D ) | ) .Proof. Let P be the path of D as in the definition of cherry. Let G be the input oriented graph. Enu-merate by brute force all induced directed paths of order | P | by checking all the possible subdigraphsof order | P | . For each such path P (cid:48) with terminus x , look for a cherry rooted at x in the graph G (cid:48) obtained by deleting all the vertices of P − x and their neighbourhoods except x . If there is such acherry C then the union of P and C is an induced D -subdivision.Similarly to Propositions 4 and 5, we have the following. Corollary 20.
If D is the disjoint union of spiders and a tiny cherry then Π (cid:48) D is polynomial-timesolvable. By Proposition 1, for any oriented path P with at most two blocks Π P and thus Π (cid:48) P are polynomial-time solvable. In this section, we shall prove that Π (cid:48) P is polynomial-time solvable for some orientedpaths with three or four blocks. In contrast, Π P is NP-complete for every oriented path with at leastthree blocks as shown in Corollary 13. There exists an algorithm of complexity O ( m ) that given a connected oriented graphon n vertices and m arcs with a specified vertex s returns an induced A + -subdivision with origin s ifone exists, and answer ‘no’ if not. roof. Observe that any induced A + -subdivision with origin s contains an induced A + -subdivisionwith origin s such that the directed path corresponding to the arc s s is some arc f . Such a subdivisionis called f -leaded .Given an oriented graph G , we enumerate all arcs f = s (cid:48) s (cid:48) . For each arc in turn we either showthat there is no f -leaded induced A + -subdivision with origin s or give an induced subdivision of A + with origin s , (but not necessarily f -leaded). This will detect the A + -subdivision since if some exists,it is f -leaded for some f .We do this as follows. We delete all in-neighbours of s and all neighbours of s (cid:48) except s (cid:48) . Let usdenote by G (cid:48) the resulting graph. Then we compute by BFS a shortest directed path P from s to s (cid:48) . Ifit is induced, together with s (cid:48) s (cid:48) , it forms the desired A + subdivision. So, as P has no forward chord(since it is a shortest path), there is an arc uv in G (cid:48) (cid:104) V ( P ) (cid:105) such that u occurs after v on P . Take such anarc b b such that b is as close as possible to s (in P ). Observe that since we deleted all in-neighboursof s we have b (cid:54) = s . Now, P [ s , b ] together with b b forms the desired A + -subdivision.There are O ( m ) arcs and for each of them we must find a shortest path in G (cid:48) which can be donein O ( m ) . Hence the complexity of the algorithm is O ( m ) .From this theorem, one can show that finding an induced A − -subdivision is polynomial-timesolvable. It is enough to enumerate all arcs s (cid:48) s (cid:48) , to delete s (cid:48) and its neighbours except s (cid:48) , and todecide whether there exists in what remains an A + -subdivision with origin s . One can also de-rive polynomial-time algorithms for finding induced subdivisions of other oriented paths with threeblocks. Corollary 22.
Let P be a path with three blocks such that the last one has length . One can check intime O ( n | P |− m ) whether a given oriented graph contains an induced P-subdivision.Proof. By directional duality, we may assume that P is an A − -subdivision. Let Q be the subdigraphof P formed by the first block of P and the second block of P minus the last arc. Let s be the terminusof Q . For each induced oriented path Q (cid:48) in the instance graph, isomorphic to Q (there are at most O ( n | P |− ) of them), we delete Q (cid:48) − s and all vertices that have neighbours in Q − s except s . We thendetect an A + -subdivision rooted at s in the resulting graph. This will detect a P -subdivision if there isone. A − in an oriented graph We show how to check the presence of an induced copy of A − by using flows (for definitions andalgorithms for flows see e.g. [1, Chapter 4]). Theorem 23.
There exists an algorithm of complexity O ( nm ) that given an oriented graph on nvertices and m arcs with a specified vertex s returns an induced A + -subdivision rooted at s, if oneexists, and answer ‘no’ if not.Proof. The general idea is close to the one of the proof of Theorem 21. Observe that any induced A + -subdivision with origin s = a contains an induced subdivision of A + with origin s = a such thatthe directed path corresponding to the arc s s is some arc f . If, in addition, the vertex correspondingto s is v , such a subdivision is called ( v , f ) -leaded .Given an oriented graph G , we enumerate all pairs ( a , a a ) such that a , a , a are distinct ver-tices and a a ∈ E ( G ) . For each such pair in turn we either show that there is no ( a , a a ) -leadedinduced A + -subdivision with origin a or give an induced subdivision of A + with origin a (but notnecessarily ( a , a a ) -leaded).We do this as follows. We first delete all the neighbours of a except a , all in-neighbours of a and a and finally all out-neighbours of a . If this results in one or more of the vertices a , . . . , a to13e deleted, then there cannot be any ( a , a a ) -leaded induced A + -subdivision with origin a becausethere is an arc in G (cid:104){ a , . . . , a }(cid:105) which is not in { a a , a a , a a } . So we skip this pair and proceedto the next one. Otherwise we delete a and we use a flow algorithm to check in the resulting digraph G (cid:48) the existence of two internally-disjoint directed paths P , Q such that the origin of P and Q are a and a respectively and such that a is the terminus of both P and Q . Moreover, we suppose thatthese two paths have no forward chord (this can easily be ensured by running BFS on the graphsinduced by each of them). If no such paths exist , then we proceed to the next pair because there isno ( a , a a ) -leaded induced A + -subdivision. If we find such a pair of directed paths P , Q , then weshall provide an induced subdivision of A + with origin a . If P and Q are induced and have no arcsbetween them, then these paths together with the arc a a form the desired induced subdivision of A + .Suppose that P is not induced. As P has no forward chord, there is an arc uv in G (cid:48) (cid:104) V ( P ) (cid:105) such that u occurs after v on P . Take such an arc b b such that b is as close as possible to a (in P ), and subjectto this, such that b is as close as possible to a . Observe that since we deleted all in-neighbours of a and all out-neighbours of a before, we must have b (cid:54) = a and b (cid:54) = a . Let b be the successor of b on P . Now P [ a , b ] and the arcs b b , b b form the desired induced subdivision of A + . From hereon, we suppose that P is induced.Suppose now that there is an arc e with an end x ∈ V ( P ) and the other y ∈ V ( Q ) . Choose such anarc so that the sum of the lengths of P [ a , x ] and Q [ a , y ] is as small as possible. If e is from x to y wehave y (cid:54) = a because we removed all the in-neighbours of a , else e is from y to x and we have x (cid:54) = a because we removed all the in-neighbours of a . In all cases, we get an induced subdivision of A + bytaking the paths P [ a , x ] and Q [ a , y ] and the arcs a a , e . From here on, we suppose that there are noarcs with an end in V ( P ) and the other in V ( Q ) .The last case is when Q is not induced. Since Q has no forward chord, there is an arc uv in G (cid:48) (cid:104) V ( Q ) (cid:105) such that u occurs after v on Q . Take such an arc b b such that b is as close as possible to a (in Q ). Observe that since we deleted all out-neighbours of a before, we must have b (cid:54) = a . Now P , Q [ b , a ] and the arc b b form the desired induced subdivision of A + .There are O ( nm ) pairs ( a , a a ) and for each of them, we run an O ( m ) flow algorithm (we justneed to find a flow of value 2, say, by the Ford-Fulkerson method [1, Section 4.5.1]) and do somelinear-time operations. Hence the complexity of the algorithm is O ( nm ) .One can check in polynomial time if there is an induced A − -subdivision: it is enough to enumerateall arcs t t , to delete t and its neighbours except t , and to decide whether there exists in what remainsan A + subdivision with origin t . One can also derive polynomial-time algorithm for finding inducedsubdivision of other oriented paths with four blocks. Corollary 24.
Let P be an oriented path that can be obtained from A − by subdividing the first arcand the second arc. One can check in time O ( n | P |− m ) whether a given oriented graph contains aninduced subdivision of P.Proof. Let R be the subdigraph of P formed by the first block of P and its second block minus the lastarc. Let s be the last vertex of R . For each induced oriented path Q in the instance graph, isomorphicto R (there are O ( n | P |− ) of them), we delete Q − s , all vertices that have neighbours in Q − s except s and detect an A − -subdivision with origin s . This will detect a P -subdivision if there is one.14 j c j d j d j c j l j l j l j l j l j Figure 6: Left: clause gadget of G ( I ) . Right: clause gadget of G ( I ) . The transitive tournament on k vertices is denoted T T k . We saw in Section 5.1 that Π (cid:48) T T is polynomial.The next result shows that Π (cid:48) T T k is NP-complete for all k ≥ Theorem 25.
For all k ≥ , Π (cid:48) T T k is NP-completeProof. For a given instance I of 3-SAT, let G ( I ) be the digraph we obtain from G ( I ) by replacingeach clause gadget C j by the modified one C j from Figure 6. Also for each variable, modify thegadget V i as follows: replace the path a i x i v i b i by a path a i x i v i x i v i . . . x mi v mi b i , and similarly the path a i ¯ x i ¯ v i b i by a path a i ¯ x i ¯ v i ¯ x i ¯ v i . . . ¯ x mi ¯ v mi b i . Then in G ( I ) the links representing a variable x i and a clause C j that uses this variable are represented by arcs between vertices from the variable gadget withsuperscript j (as in Figure 2).Recall that G ( I ) has an induced directed ( a , b ) -path if and only if I is satisfiable. It is easy tosee that the same holds for G ( I ) . Note that in G ( I ) no vertex has in- or out-degree larger than 2.Given an instance I of 3-SAT we form the digraph G k ( I ) from G ( I ) and a copy of T T k (withvertices v , v , . . . , v k and arcs v i v j , 1 ≤ i < j ≤ k ) by deleting the arc v v k and adding the arcs v a , bv k .We claim that G k ( I ) contains an induced subdivision of T T k if and only if G ( I ) has an induceddirected ( a , b ) -path which is if and only if I is satisfiable.Clearly, if I is satisfiable, we may use the concatenation of an induced directed ( a , b ) -path in G ( I ) with v a and bv k in place of v v k to obtain an induced T T -subdivision in G k ( I ) .Conversely, suppose that G k ( I ) contains an induced subdivision of T T k and let h ( v i ) , 1 ≤ i ≤ k ,denote the image of v i in some fixed induced subdivision H of T T k . Then we must have h ( v ) = v and h ( v k ) = v k , because G ( I ) does not contain any vertex of out-degree k − k − k ≥
4. For all i , 1 < i < k , the vertex h ( v i ) could not be in V ( G ( I )) since otherwise theremust be either two disjoint directed ( v i , v k ) -paths to v k or two disjoint directed ( v , v i ) -paths. Thisis impossible because there is no directed ( v i , v k ) -path in G k ( I ) \ bv k and no directed ( v , v i ) -path in G k ( I ) \ v a . Hence h ( v i ) = v i for all 1 ≤ i ≤ k and so it is clear that we have an induced directed ( a , b ) -path in G ( I ) , implying that I is satisfiable.In the proof above we used that the two vertices v , v k cannot be mapped to vertices of G ( I ) ,the fact that the connectivity between these and the other vertices is too high to allow any of these15o be mapped to vertices of G ( I ) and finally we could appeal to the fact that G ( I ) has an induceddirected ( a , b ) -path if and only if I is satisfiable. Refining this argument it is not difficult to seethat the following holds where a ( z , X ) -path is a path whose initial vertex is z and whose last vertexbelongs to X . Theorem 26.
Let D = ( V , A ) be a digraph and let X (resp. Y ) be the subset of vertices with out-degree (resp. in-degree) at least 3 and let Z = V \ ( X ∪ Y ) (note that X ∩ Y (cid:54) = /0 is possible and alsoZ = /0 is possible). Suppose that for every z ∈ Z the digraph D contains either two internally disjoint ( X , z ) -paths or two internally disjoint ( z , Y ) -paths. Then Π (cid:48) D is NP-complete.Proof. (Sketch) For a given instance I of 3-SAT we form the digraph G (cid:48) ( I ) from D by replacingone arc uv with at least one of its endvertices in X ∪ Y by G ( I ) and the arcs ua , bv . Again it isclear how to obtain an induced subdivision of D in G (cid:48) ( I ) when I is satisfiable. Let us now assumethat G (cid:48) ( I ) contains an induced subdivision D (cid:48) of D . Let { h ( v ) | v ∈ V } be the vertices correspondingto the vertices of V in the subdivision. For degree reasons, none of the vertices in X ∪ Y can have h ( v ) ∈ V ( G ( I )) and because of connectivity, none of the vertices of Z can have h ( z ) ∈ V ( G ( I )) because there is only one arc entering and leaving V ( G ( I )) in G (cid:48) ( I ) . Thus { h ( v ) | v ∈ V } = V (possibly with h ( v ) (cid:54) = v for several vertices). However, since we deleted the arc uv and replaced it by G ( I ) and the arcs ua , bv , it follows that G (cid:48) ( I ) and so G ( I ) contains an induced directed ( a , b ) -path,implying that I is satisfiable. vertices Let ST be the unique strong tournament of order 4. It can be seen has a directed cycle αγβδα togetherwith two chords αβ and γδ . The aim of this section is to show that Π (cid:48) ST is NP-complete.An ( x , y , y ) -switch is the digraph with vertex set { x , z , y , y } and edge set { xz , xy , zy , zy , y y } .See Figure 7. zx y y Figure 7: An ( x , y , y ) -switch.A good ( x , y , y ) -switch in a digraph D is an induced switch Y such that all the arcs entering Y have head x and all arcs leaving Y have tail in { y , y } . Lemma 27.
Let Y be a good ( x , y , y ) -switch in a digraph D. Then every induced subdivision S ofST in D intersects Y on either the path ( x , y ) , the path ( x , z , y ) , or the empty set.Proof. Suppose for a contradiction, that y y ∈ E ( S ) . Then S must contain the unique in-neighbour z of y and the unique in-neighbour x of z . Hence y has in-degree 3 in S , a contradiction.Suppose for a contradiction, that zy ∈ E ( S ) . Then S must contain x the unique in-neighbour of z .Hence xy is a chord of S and so z must have degree 3 in S . Thus y ∈ V ( S ) and y has in-degree 3 in S , a contradiction. Theorem 28. Π (cid:48) ST is NP-complete. roof. Reduction from 3-SAT. Let I be an instance of 3-SAT with variables x , x , . . . , x n and clauses C , C , . . . , C m . We first create a variable gadget V i for each variable x i , i = , , . . . , n and a clausegadget C j for each clause C j , j = , , . . . , m as shown in Figure 8. Then we form the digraph G ( I ) asfollows: Form a chain U of variable gadgets by adding the arcs b i a i + for i = , , . . . , n − W of clause gadgets by adding the arcs d j c j + , j = , , . . . , m −
1. Add the arcs aa , b n b , cc , t m d .For each clause C , we connect the three literal vertices of the gadget for C to the variable gadgets forvariables occuring as literals in C in the way indicated in Figure 9. x i d j q j q j q j q j p j p j p j p j p j q j c j q (cid:48) j r j r j r j r j a (cid:48) i x i a i ¯ x i ¯ x i ¯ x i b (cid:48) i b i ¯ x i ¯ x i x i x i x i Figure 8: The variable gadget V i (left) and the clause gadget C j (right).Figure 9: Connections between a clause gadget and a variable gadget in G ( I ) . Only the connec-tion for one variable gadget and one clause gadget is shown and the general strategy for connectingvariable and clause gadgets is the same as in G ( I ) (Figure 2).We denote by X i the path a i a (cid:48) i x i x i x i x i x i b i , by ¯ X i the path a i ¯ x i ¯ x i ¯ x i ¯ x i ¯ x i b (cid:48) i b i , by P j the path c j p j p j p j p j p j d j ,by Q j the path c j q j q (cid:48) j q j q j q j q j d j , and by R j the path c j q j r j r j r j r j d j .Similarly to the proof of Lemma 7, one can check that I is satisfiable if and only if there are twoinduced disjoint directed ( a , b ) - and ( c , d ) -paths in G ( I ) .Let G ∗ ( I ) be the digraph obtained from G ( I ) by adding the edges ac , cb , bd and da . Observethat G ∗ ( I ) \ da is acyclic.Let us prove that G ∗ ( I ) contains an induced ST -subdivision if and only if I is satisfiable.17f I is satisfiable, then in G ( I ) there are two induced disjoint directed ( a , b ) - and ( c , d ) -paths.The union of these paths and the directed cycle acbd is an induced ST -subdivision in G ∗ ( I ) .Conversely, assume that G ∗ ( I ) contains an induced subdivision S of ST . For sake of simplicity(and with a slight abuse of notation), we will denote the vertices of S corresponding to α , β , γ and δ by the same names. Let T and T be the paths corresponding to the chord αβ and γδ respectively in S and let C be the directed cycle corresponding to αγβδα . Observe that the ends of T and T mustalternate on C .Notice that the subdigraphs induced by the sets { a i , a (cid:48) i , x i , ¯ x i } , 1 ≤ i ≤ n , { c j , p j , p j , q j } and { q j , q (cid:48) j , q j , r j } are good switches. In addition, the subdigraphs induced by the sets { b i , b (cid:48) i , x i , ¯ x i } ,1 ≤ i ≤ n , are the converse of good switches. Hence Lemma 27 (and its converse) imply the followingproposition. Claim 28.1. (i) For ≤ i ≤ n, if a i ∈ V ( S ) , then exactly one of the two paths ( a i , a (cid:48) i , x i ) and ( a i , ¯ x i ) is in S.(ii) For ≤ i ≤ n, if b i ∈ V ( S ) , then exactly one of the two paths ( ¯ x i , b (cid:48) i , b i ) and ( x i , b i ) is in S.(iii) For ≤ j ≤ m, if c j ∈ V ( S ) , then exactly one of the three paths ( c j , p j , p j ) , ( c j , q j , q (cid:48) j , q j ) and ( c j , q j , r j ) is in S. Since G ∗ ( I ) \ da is acyclic, C must contain the arc da . Moreover since there is no arc with tailin some clause gadget and head in some variable gadget, C contains at most one arc with tail in somevariable gadget and head in some clause gadget. Claim 28.2.
For any ≤ i ≤ n and any ≤ j ≤ m, the cycle C contains no arc with tail in { x i , ¯ x i } and head in { p j , q j , r j } .Proof. Assume for a contradiction that C contains such an arc y i l j . Then since l j and l j have out-degree 1 then C must also contain l j and l j . Thus, in S , y i has out-degree 3 in S , a contradiction. Claim 28.3.
For any ≤ i ≤ n and any ≤ j ≤ m the cycle C contains no arc with tail in { x i , ¯ x i } and head in { p j , q j , r j } .Proof. Assume for a contradiction that C contains such an arc y i l j . Then since y i and y i have in-degree 1 then C must also contain y i and y i . Thus, in S , l j has in-degree 3 in S , a contradiction. Claim 28.4.
For any ≤ i ≤ n and any ≤ j ≤ m then C contains no arc with tail in { x i , ¯ x i } andhead in { p j , q j , r j } .Proof. Assume for a contradiction that C contains such an arc y i l j . The vertex l j has a unique out-neighbour l j which must be in C . It follows that y i l j corresponds to one of the chords αβ or γδ .Thus l j must have degree 3 in S . It follows that l j is in V ( S ) and so y i has out-degree 3 in S , acontradiction. Claim 28.5.
For any ≤ i ≤ n and any ≤ j ≤ m the cycle C contains no arc with tail in { x i , ¯ x i } and head in { p j , q j , r j } .Proof. Assume for a contradiction that C contains such an arc y i l j . The vertex y i has a unique in-neighbour y i which must be in C . It follows that y i l j corresponds to one of the chords αβ or γδ .Thus y i must have degree 3 in S . It follows that y i is in V ( S ) and so l j has in-degree 3 in S , acontradiction. 18 laim 28.6. For any ≤ i ≤ n and any ≤ j ≤ m the cycle C contains no arc with tail in { x i , ¯ x i } and head in { p j , q j , r j } .Proof. Assume for a contradiction that C contains such an arc y i l j . Without loss of generality y i = x i .By the remark after Claim 28.1 this is the only arc from a variable gadget to a clause gadget.Furthermore, we have that b is not on C .Thus, by Claim 28.1, for every 1 ≤ k < i , the intersection of C and V k is either X k or ¯ X k , and forevery j < l ≤ m , the intersection of C and C j is either P j , Q j or R j .Consider y ∈ { α , β } . It is on C and has outdegree 2. On the other hand, applying Claim 28.1 wesee that the following must hold as none of these vertices can belong to S and at the same time havetwo of their out-neighbours in S : • y (cid:54)∈ ∪ ≤ j ≤ m { c j , p j , q j , q (cid:48) j , q j } , • y (cid:54)∈ ∪ k (cid:54) = i { a k , a (cid:48) k , x k , x k } and • y (cid:54)∈ { a i , a (cid:48) i , x i , x i } .By Claims 28.2-28.5, we have y (cid:54)∈ { x i , x i } and since b is not on C we also have y (cid:54) = b . If y = x i ,then using that yl j is and arc of C we get a contradiction because x i l j is an arc (so we cannot obtainan induced copy of S using both arcs yl j , x i l j ). Hence (as y was any of α , β ) we have a = α = β , acontradiction. Claim 28.7. C = acbda.Proof. Suppose not. Then by the above claims, C either does not intersect the clause gadget andintersect all the variable ones or does not intersect the variable gadget and intersect all the gadget ones.In both cases, similarly to the proof of Claim 28.5, one shows that a = α = β , a contradiction.Since C = acbda and by construction of G ∗ ( I ) , T and T are two induced disjoint path in G ( I ) and so I is satisfiable. It would be nice to have results proving a full dichotomy between the digraphs D for which Π D (resp. Π (cid:48) D ) is NP -complete and the ones for which it is polynomial-time solvable. Regarding Π D ,Conjecture 14 gives us what the dichotomy should be. But for Π (cid:48) D we do not know yet.A useful tool to prove such a dichotomy would be the following conjecture. Conjecture 29. If D is a digraph such that Π D (resp. Π (cid:48) D ) is NP-complete, then for any digraph D (cid:48) that contains D as an induced subdigraph, Π D (cid:48) (resp. Π (cid:48) D (cid:48) ) is NP-complete.We were able to settle the complexity of Π (cid:48) D when D is a directed cycle, a directed path, or somepaths with at most four blocks. The following problems are perhaps the natural next steps. Problem 30.
What is the complexity of Π (cid:48) D when D is an oriented cycle which is not directed? Problem 31.
What is the complexity of Π (cid:48) D when D is an oriented path which is not directed?19ote that the approach used above to find an induced subdivision of A − relied on the fact thatone can check in polynomial time (using flows) whether a digraph contains internally disjoint ( x , z ) -, ( y , z ) -paths for prescribed distinct vertices x , y , z . If we want to apply a similar approach for A − , thenfor prescribed vertices x , y , z , w we need to be able to check the existence of internally disjoint paths P , Q , R such that P is an ( x , y ) -path, Q is a ( z , y ) -path and R is a ( z , w ) -path such that these paths areinduced and have no arcs between them. However, the problem of deciding just the existence ofinternally disjoint paths P , Q , R with these prescribed ends is NP-complete by the result of Fortune etal. [6]. Thus we need another approach to obtain a polynomial-time algorithm (if one exists). v H u H y H x H Figure 10: The digraph H with specified vertices u H , v H , x H , y H .It seems that little is known about detecting a subdivision of some given digraph D as a subgraph(possibly non-induced). This leads us to the following problem: Problem 32.
When D is fixed directed graph, what is the complexity of deciding whether a givendigraph G contains a D -subdivision as a subgraph?The following shows that the problem above can be NP-complete. Theorem 33.
Let H be the digraph in Figure 10. It is NP-complete to decide whether a given digraphG contains an H-subdivision.Proof.
By the classical result of Fortune, Hopcroft and Wyllie [6], the so-called 2-linkage problem(given a digraph and four distinct vertices u , v , x , y ; does G contain a pair of vertex-disjoint paths P , Q so that P is a directed ( u , v ) -path and Q is a directed ( x , y ) -path?) is NP-complete. By inspectingthe proof (see [1, Section 10.2]) it can be seen that the problem is NP-complete even when G hasmaximum in- and out-degree at most 3. Given an instance G of the 2-linkage problem with maximumin- and out-degree at most 3 and a copy of H we form a new digraph G H by identifying the vertices { u , v , x , y } with { u H , v H , x H , y H } in that order. Clearly, if G has disjoint directed ( u , v ) , ( x , y ) -paths,then we can use these to realize the needed paths from u H to v H and from x H to y H (and all otherpaths are the original arcs of H ). Conversely, suppose there is a subdivision H ∗ of H in G H . Forevery v ∈ { u H , v H , x H , y H } , let us denote by v ∗ the vertex corresponding to v in H ∗ . Since d − ( u H ) = , d + ( v H ) = , d − ( x H ) = , d + ( y H ) = G H , we have u ∗ H = u H , v ∗ H = v H , x ∗ H = x H , and y ∗ H = y H .Thus the two disjoint directed ( u ∗ H , v ∗ H ) - x ∗ H , y ∗ H ) -paths in H ∗ are disjoint directed ( u , v ) , ( x , y ) -paths in G implying that G is a ’yes’-instance.Finally, we would like to point out that in all detection problems about induced digraphs, back-ward arcs of paths play an important role, especially in NP-completeness proofs. Also, these back-ward arcs make all “connectivity-flavoured” arguments fail: when two vertices x , y are given, it is not20ossible to decide whether x can be linked to y . So, maybe another notion of induced subdigraphcontainment would make sense: chords should be kept forbidden between the different directed pathsthat arise from subdividing arcs, but backward arcs inside the paths should be allowed. Acknowledgement
The authors would like to thank Joseph Yu for stimulating discussions.
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