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Obstructions to Injective Oriented Colourings
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Electronic Notes in Discrete Mathematics 38 (2011) 597–605 www.elsevier.com/locate/endm
Obstructions to Injective Oriented Colourings Gary MacGillivray a,1 Andr´e Raspaud b Jacobus Swarts c a b
Mathematics and Statistics, University of Victoria, PO BOX 3060 STN CSC, Victoria, B.C., V8W 3R4, Canada [email protected] LaBRI UMR CNRS 5800, Universit´e Bordeaux I, 33405 Talence Cedex, France [email protected] c Mathematics and Statistics, Vancouver Island University, 900 Fifth Street, Nanaimo, BC, Canada V9R 5S5 [email protected]
Abstract The question of deciding the existence of an oriented colouring which must also be locally injective on the in-neighbourhood of each vertex is polynomially solvable when the number of colours is at most three. In these cases, we describe efficient algorithms, based on colouring extension, which produce either the desired colouring or a forbidden substructure. Keywords: oriented colouring, injective colouring, homomorphism, colouring extension, colour critical
1
Introduction
Oriented colourings which are injective on in-neighbourhoods first arose in work of Courcelle [2], who used them as an example to demonstrate the expressive power of monadic second-order logic. These have also been considered by Raspaud and Sopena [13]. They improved a result of Courcelle by showing that every simple planar graph can be oriented to have such a colouring with at most 320 colours. It follows from Courcelle’s Theorem on the monadic second order logic of graphs that the problem of deciding if a given oriented graph admits an injective oriented colouring with a fixed number k of colours is polynomially solvable for digraphs of bounded treewidth. In the case where 1
Research supported by NSERC
1571-0653/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2011.10.001
598
G. MacGillivray et al. / Electronic Notes in Discrete Mathematics 38 (2011) 597–605
the colourings are required to assign different colours to adjacent vertices, the basic theory involving a homomorphism model, complexity, and bounds has recently been developed [10]. Our goal in this paper is to more closely examine the Polynomial cases. We describe an algorithm that either successfully colours the given oriented graph, or produces an “obstruction”, that is, a forbidden subdigraph. The complexity of injective homomorphisms, which generalize injective oriented colourings, has been studied by Swarts and MacGillivray [11,15]. They obtain a dichotomy theorem in the case when the target digraph is reflexive, and show that the situation when the target digraph is at least as rich as the collection of all digraph homomorphism problems (and therefore all constraint satisfaction problems [4]). Injective colourings and homomorphisms of undirected graphs also arose in the paper of Courcelle. These have been considered by many authors (e.g. see [5,6,7,8]). Oriented colourings are also well studied in the literature (e.g. see [3,14,16]).
2
Preliminaries
For basic results in graph theory we follow the text of Bondy and Murty [1]. An oriented graph is a directed graph with no loops such that, for every pair of vertices x and y, at most one of the arcs xy and yx exists. Let G and H be directed graphs. A homomorphism of G to H is a function f : V (G) → V (H) such that f (x)f (y) is an arc of H whenever xy is an arc of G. A homomorphism of G to H is locally injective if, for each vertex x of G, no two in-neighbours of x have the same image. Because homomorphisms generalize colourings, the problem of deciding if a given digraph has a homomorphism to a fixed digraph H has been called H-colouring. If the mapping is required to be injective, it is called locally injective H-colouring. The book by Hell and Neˇsetˇril [9] is an excellent introduction to the theory of homomorphisms of graphs and digraphs. In the remainder of this paper we shall drop the adjective “locally”. We shall refer to locally injective homomorphisms as injective homomorphisms, and locally injective colourings as injective colourings Let D be an oriented graph and k be a positive integer. An oriented kcolouring of D is an assignment of the colours 1, 2, . . . , k to the vertices of D so that adjacent vertices are assigned different colours, and if some arc of D joins a vertex of colour i to a vertex of colour j, then no arc of D joins a vertex of colour j to a vertex of colour i. Equivalently, an oriented k-colouring of D is a homomorphism to an oriented graph T on k vertices. Since arcs can
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be added joining non-adjacent vertices of T without destroying the existence of a homomorphism of D to T , the oriented graph T can be taken to be a tournament. An injective oriented k-colouring of an oriented graph D is an oriented colouring of D such that, for each vertex x of D, no two in-neighbours of x are assigned the same colour. The colouring which assigns every vertex of D a different colour is an injective oriented |V |-colouring of D. The injective oriented chromatic number of an oriented graph D, denoted χi (D), is the smallest k for which there is an injective oriented k-colouring of D. Equivalently, an injective oriented k-colouring of an oriented graph D is an injective homomorphism of D to a tournament T on k vertices, and χi (D) is the least k for which such a homomorphism exists. For each fixed positive integer k, the complexity of injective oriented k colouring – the problem of deciding if a given oriented graph has an injective oriented k-colouring – is described in the following theorem. Theorem 2.1 [10] Let k be a fixed integer. If k ≤ 3, then injective oriented k-colouring is Polynomial. If k ≤ 4, then injective oriented k-colouring is NP-complete. Since an oriented graph has an oriented injective 1-colouring if and only if it has no arcs, in the sequel we focus on the cases k = 2 and k = 3. An oriented graph has an oriented injective 2-colouring if and only if it has an injective homomorphism to T2 , the tournament on two vertices with one arc, and has an oriented injective 3-colouring if and only if it has an injective homomorphism to T3 , the transitive tournament on three vertices, or to the directed 3-cycle C3 . Because our colourings are injective on in-neighbourhoods, the directed graph that consists of two arcs meeting at a vertex plays an important role. The digraph with vertices v0 , s, v1 and arcs v0 s, v1 s will be denoted H3 (v0 , s, v1 ), or simply H3 when it is not necessary to emphasize the vertices, and called a hat. The vertex s is the point of the hat while v0 and v1 are its ends. All three vertices of a hat must be assigned different colours in an injective oriented colouring.
3
Obstructions to injective oriented 2- and 3-colouring
In this final section we describe the obstructions to injective oriented 2colouring and 3-colouring. To do so, we determine the obstructions for injective homomorphism to T2 , T3 and C3 . In two of these cases, the obstructions
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are very similar to the ones for (ordinary) homomorphism. Proposition 3.1 An oriented graph G has an injective oriented 2-colouring if and only if neither H3 nor P2 has an injective homomorphism to G. Proof. Suppose the oriented graph G admits an injective oriented 2-colouring, that is, an injective homomorphism to the transitive tournament on two vertices, T2 . If there is an injective homomorphism of H3 to G, then by composition there is an injective homomorphism of H3 to T2 , a contradiction. Similarly, there is no injective homomorphism of P2 to G. Suppose neither H3 nor P2 has an injective homomorphism to G. Since G is an oriented graph, any homomorphism of P2 to G is injective. Since P2 is the unique obstruction for (ordinary) homomorphism to T2 , it follows that there is a homomorphism of G to T2 . Since there is no injective homomorphism of H3 to G, no two vertices of G have a common out-neighbour. Hence the homomorphism of G to T2 is injective. 2 It is clear that the existence of an injective oriented 2-colouring can easily be decided by a greedy algorithm. Since each of H3 or P2 maps injectively to G if and only if it is a subgraph of G, Proposition 3.1 also gives a forbidden subgraph characterization of the digraphs that have an injective oriented 2-colouring. Corollary 3.2 An oriented graph G has an injective 2-colouring if and only if it is a star in which all arcs are directed away from the centre vertex. We now describe the obstructions for injective homomorphism to the transitive triple, T3 . Suppose T3 has vertices t0 , t1 , t2 and arcs t0 t1 , t1 t2 , t0 t2 . Since T3 is acyclic and has maximum in-degree two, both P3 and the digraph H4 with vertex set {h1 , h2 , h3 , h4 } and arcs h1 h4 , h2 h4 , h3 h4 are obstructions to injective homomorphism. Since the unique vertex of T3 with in-degree two has out-degree 0, the digraph A4 constructed from H3 by adding a new vertex x and the arc h3 x is also an obstruction. These are not all of the obstructions because, for example, the digraph Z2k+1 obtained from the undirected odd cycle C2k+1 by replacing each edge with a copy of a hat H3 – identifying the ends of the edge with the ends of the hat – admits a homomorphism from neither H4 nor P3 but does not map injectively to T3 . Hence these “zigzag odd cycles” form an infinite family of obstructions. There are other forbidden configurations as well. First, note that if x, y, z is a directed path of length two, then in any injective homomorphism to T3 these vertices must map to t0 , t1 and t2 in that order. Let Bn be the digraph obtained from an undirected path of length n by replacing each edge with a copy of the hat H3 , identifying
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the ends of the hat with the ends of the edge. By analogy with the zigzag odd cycles, each digraph Bn is a “zigzag path”. If the initial vertex of Bn maps to t0 , then the remaining vertices map to t2 , t1 , t2 , t0 , t2 , t1 , . . . in that order. If n is even, then the terminal vertex of Bn maps to t0 , and if n is odd it maps to t1 . Similarly, if the initial vertex maps to t1 , then the terminal vertex maps to t1 when n is even and t0 when n is odd. This leads to the following families of obstructions: (i) H0,0 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k+1 , where k is a non-negative integer, joining the initial vertices of the two paths; (ii) H1,1 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k+1 , where k is a non-negative integer, joining the middle vertices of the two paths; (iii) H0,1 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k , where k is a positive integer, joining the initial vertices of one path to the middle vertex of the other. Examples of these obstructions are shown in Figure 1. Notice that for the digraphs in H1,1 , for example, there is an injective homomorphism that maps the terminal vertex x of a directed path of length two to the point of the hat one of whose ends is an in-neighbour of x. Although it is tempting to define the family H1,0 , it is identical to H0,1 . Proposition 3.3 An oriented graph G has in injective homomorphism to T3 if and only if no digraph in {H4 , P3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G. Proof. The reasoning above demonstrates that none of the given digraphs have an injective homomorphism to T3 . Since injective homomorphisms compose (e.g. see [15]), none of them can have an injective homomorphism to G. Suppose that none of the given digraphs has an injective homomorphism to G. We will show that there is an injective homomorphism to T3 . Since different components can be treated separately, it can be assumed without loss of generality that the underlying undirected graph of G is connected. Suppose first that G has no directed path of length two. Then every vertex of G has either in-degree zero or out-degree zero. Let X be the set of vertices of out-degree zero. Map each vertex in X to t2 (this is possible because A4 does not map injectively to G). Construct an auxiliary graph L with vertex set V (G) − X and vertex u adjacent to vertex v whenever u and v have a common
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Fig. 1. Examples of the obstructions.
out-neighbour in G. Since no zigzag odd cycle Z2k+1 maps injectively to G, the undirected graph L is bipartite. A 2-colouring of its vertices using the colours t0 and t1 gives an injective homomorphism of G to T3 . Now suppose that G has one or more directed paths of length two. The vertices of each such path must necessarily be coloured t0 , t1 , t2 , in that order. Since A4 is forbidden, no vertex that is the origin of such a path is joined by an arc to a vertex that is the midpoint of another such path, and similarly no origin of a path of length two is joined to the midpoint of another path of length two. Since H4 is forbidden, no origin of a path of length two is joined to the terminus of another path of length two. And since P3 is forbidden, there can be no arc joining two vertices that are each the origin of a directed path of length two. Let S be the set of vertices of G that have out-degree zero. Because there are no forbidden configurations, S contains all vertices that have in-degree two or are the terminus of a directed path of length two. For the same reason, S is an independent set. Construct an auxiliary graph R with vertex set V (G) − S
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and vertex u adjacent to vertex v whenever u and v have a common outneighbour in G. As above, the undirected graph R is bipartite. Pre-colour some of its vertices by colouring vertices that are the origin of a directed path of length two with t0 and those that are the midpoint of such a path with t1 . Since no digraph in H0,0 ∪ H1,1 maps injectively to G, no two vertices pre-coloured with the same colour are joined by an odd path in R. Since no digraph in H0,1 has an injective homomorphism to G, no two vertices precoloured with different colours are joined by an even path in R. That is, the pre-colouring of R respects a bipartition. Therefore, the pre-colouring extends to a 2-colouring of R. An injective homomorphism of G to T3 can now be constructed by mapping the vertices in S to t2 , and map the vertices of R to the vertex of T3 corresponding to their colour. 2 The proof of the Proposition 3.3 implies a polynomial time algorithm that either finds an injective homomorphism of a given digraph to T3 or returns an obstruction. Proposition 3.4 An oriented graph G has in injective homomorphism to an C3 if and only if neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. Proof. Since neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to C3 , hence none of them can have an injective homomorphism to G. Suppose neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. The facts that G has maximum in-degree one, and any directed cycle that has a homomorphism to G is of length divisible by three, together imply that G has a homomorphism to C3 . This homomorphism is injective. 2 Note that if H3 does not map injectively to G (i.e. H3 is not a subgraph of G), then any injective homomorphism of a directed cycle to G is an isomorphism. Hence, Proposition 3.4 gives a forbidden subgraph characterization of the digraphs that map injectively to C3 . Its proof implies a polynomial time algorithm that either finds an injective homomorphism of a given digraph to T3 or returns an obstruction. The following results combine the two previous propositions. Theorem 3.5 Let G be an oriented graph. Then G has an injective oriented 3-colouring if and only if either no digraph in {H4 , P3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G, or neither H3
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nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. Define H0,0 to be the set of digraphs obtained from H0,0 be adding any digraph that can be formed from an injective homomorphism that fixes all vertices on the zigzag path. The sets H1,1 and H0,1 are defined similarly.
Corollary 3.6 Let G be an oriented graph. •
If the maximum in-degree of G is one, then G has an injective oriented 3colouring if and only if either there is no injective homomorphism of P3 to G, or G contains a directed cycle of length not divisible by three;
•
If the maximum in-degree of G is two, then G has an injective oriented 3-colouring if and only if it has no element of {P3 , C3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G;
•
If G has maximum in-degree at least three (that is, H4 maps injectively to G), then G does not have an injective oriented 3-colouring.
Proof. The first and third statements are easy to see, so we prove only the second statement. Suppose the oriented graph G has maximum in-degree two. Then there is an injective homomorphism of P3 to G if and only if P3 or C3 is a subgraph of G, and there is an injective homomorphism of A4 to G if and only if A4 is a subgraph of G. Essentially the same argument that shows that a homomorphic image of an undirected odd cycle (i.e. a closed walk of odd length) contains an odd cycle can be used to show that, when the target graph has maximum in-degree two, an injective homomorpic image of an odd zigzag cycle contains an odd zigzag cycle. If the oriented graph G contains no odd zigzag cycle, then an injective homomorphic image of a zigzag path of length n contains a zigzag path whose length is congruent to n modulo 2. The result now follows from Theorem 3.5 and the definition of H0,0 , H1,1 and H0,1 . 2
References [1] J. A. Bondy and U. S. R.Murty, Graph Theory. Springer, 2007. [2] B. Courcelle, The monadic second order logic of graphs VI: On several reresentations of graphs by relational structures, Discrete Appl. Math. 54 (1994), 117–149. (Erratum Discrete Appl. Math. 63 (1995), 199–200.) [3] L. Esperetand P. Ochem, Oriented colorings of 2-outerplanar graphs, Inform. Process. Lett. 101 (2007), 215–219.
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[4] T. Feder and M. Vardi, The computational structure of monotone monadic SNP and constraint satisfaction: A study through Datalog and group theory, SIAM J. Comput 28 (1998), 57–104. [5] G. Fertin, R. Rizzi, and S. Vialette, Finding exact and maximum occurrences of protein complexes in protein-protein interaction graphs, Mathematical foundations of computer science 2005, Lecture Notes in Comput. Sci., vol. 3618, Springer, Berlin, 2005, 328–339. [6] J. Fiala and J. Kratochvil, Locally injective graph homomorphism: lists guarantee dichotomy, Graph-theoretic concepts in computer science, Lecture Notes in Comput. Sci., vol. 4271, Springer, Berlin, 2006, pp. 15–26. [7] J. Fiala and J. Kratochvil, and A. Por, On the computational complexity of partial covers of theta graphs, Discrete Appl. Math. 156 (2008), 1143–1149. [8] G. Hahn, J. Kratochvil, J. S´ıˇran, and D. Sotteau, On the injective chromatic number of graphs, Discrete Math. 256 (2002), 179 –192. [9] P. Hell and J. Neˇsetˇril, Graphs and Homomorphisms, Oxford University Press, 2004. [10] G. MacGillivray, A. Raspaud and J. Swarts,Injective Oriented Colourings. In C. Paul and M. Habib (Eds.): WG 2009, Lecture Notes in Comput. Sci. vol. 5911, Springer, Berlin, 2010, (2010), 262–272. [11] G. MacGillivray and J. Swarts, The complexity of locally injective homomorphisms Discrete Math. 310 (2010), 2685–2696. [12] J. W. Moon, Topics on Tournaments, Holt, Rinehart and Winston, New York, 1968. [13] A. Raspaud and E. Sopena, Good and semi-strong colorings of oriented planar graphs, Information Processing Letters 51 (1994), 171–174. [14] E. Sopena, Oriented graph coloring. Combinatorics, graph theory, algorithms and applications. Discrete Math. 229 (2001), 359–369. [15] J. Swarts, The complexity of digraph homomorphisms: Local tournaments, Injective Homomorphisms and Polymorphisms, Ph.D. Thesis (2008), Department of Mathematics and Statistics, University of Victoria, Victoria, BC, Canada. [16] D. Wood, Acyclic, star and oriented colourings of graph subdivisions, Discrete Math. Theor. Comput. Sci. 7 (2005), 37–50
Electronic Notes in Discrete Mathematics 38 (2011) 597–605 www.elsevier.com/locate/endm
Obstructions to Injective Oriented Colourings Gary MacGillivray a,1 Andr´e Raspaud b Jacobus Swarts c a b
Mathematics and Statistics, University of Victoria, PO BOX 3060 STN CSC, Victoria, B.C., V8W 3R4, Canada [email protected] LaBRI UMR CNRS 5800, Universit´e Bordeaux I, 33405 Talence Cedex, France [email protected] c Mathematics and Statistics, Vancouver Island University, 900 Fifth Street, Nanaimo, BC, Canada V9R 5S5 [email protected]
Abstract The question of deciding the existence of an oriented colouring which must also be locally injective on the in-neighbourhood of each vertex is polynomially solvable when the number of colours is at most three. In these cases, we describe efficient algorithms, based on colouring extension, which produce either the desired colouring or a forbidden substructure. Keywords: oriented colouring, injective colouring, homomorphism, colouring extension, colour critical
1
Introduction
Oriented colourings which are injective on in-neighbourhoods first arose in work of Courcelle [2], who used them as an example to demonstrate the expressive power of monadic second-order logic. These have also been considered by Raspaud and Sopena [13]. They improved a result of Courcelle by showing that every simple planar graph can be oriented to have such a colouring with at most 320 colours. It follows from Courcelle’s Theorem on the monadic second order logic of graphs that the problem of deciding if a given oriented graph admits an injective oriented colouring with a fixed number k of colours is polynomially solvable for digraphs of bounded treewidth. In the case where 1
Research supported by NSERC
1571-0653/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2011.10.001
598
G. MacGillivray et al. / Electronic Notes in Discrete Mathematics 38 (2011) 597–605
the colourings are required to assign different colours to adjacent vertices, the basic theory involving a homomorphism model, complexity, and bounds has recently been developed [10]. Our goal in this paper is to more closely examine the Polynomial cases. We describe an algorithm that either successfully colours the given oriented graph, or produces an “obstruction”, that is, a forbidden subdigraph. The complexity of injective homomorphisms, which generalize injective oriented colourings, has been studied by Swarts and MacGillivray [11,15]. They obtain a dichotomy theorem in the case when the target digraph is reflexive, and show that the situation when the target digraph is at least as rich as the collection of all digraph homomorphism problems (and therefore all constraint satisfaction problems [4]). Injective colourings and homomorphisms of undirected graphs also arose in the paper of Courcelle. These have been considered by many authors (e.g. see [5,6,7,8]). Oriented colourings are also well studied in the literature (e.g. see [3,14,16]).
2
Preliminaries
For basic results in graph theory we follow the text of Bondy and Murty [1]. An oriented graph is a directed graph with no loops such that, for every pair of vertices x and y, at most one of the arcs xy and yx exists. Let G and H be directed graphs. A homomorphism of G to H is a function f : V (G) → V (H) such that f (x)f (y) is an arc of H whenever xy is an arc of G. A homomorphism of G to H is locally injective if, for each vertex x of G, no two in-neighbours of x have the same image. Because homomorphisms generalize colourings, the problem of deciding if a given digraph has a homomorphism to a fixed digraph H has been called H-colouring. If the mapping is required to be injective, it is called locally injective H-colouring. The book by Hell and Neˇsetˇril [9] is an excellent introduction to the theory of homomorphisms of graphs and digraphs. In the remainder of this paper we shall drop the adjective “locally”. We shall refer to locally injective homomorphisms as injective homomorphisms, and locally injective colourings as injective colourings Let D be an oriented graph and k be a positive integer. An oriented kcolouring of D is an assignment of the colours 1, 2, . . . , k to the vertices of D so that adjacent vertices are assigned different colours, and if some arc of D joins a vertex of colour i to a vertex of colour j, then no arc of D joins a vertex of colour j to a vertex of colour i. Equivalently, an oriented k-colouring of D is a homomorphism to an oriented graph T on k vertices. Since arcs can
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be added joining non-adjacent vertices of T without destroying the existence of a homomorphism of D to T , the oriented graph T can be taken to be a tournament. An injective oriented k-colouring of an oriented graph D is an oriented colouring of D such that, for each vertex x of D, no two in-neighbours of x are assigned the same colour. The colouring which assigns every vertex of D a different colour is an injective oriented |V |-colouring of D. The injective oriented chromatic number of an oriented graph D, denoted χi (D), is the smallest k for which there is an injective oriented k-colouring of D. Equivalently, an injective oriented k-colouring of an oriented graph D is an injective homomorphism of D to a tournament T on k vertices, and χi (D) is the least k for which such a homomorphism exists. For each fixed positive integer k, the complexity of injective oriented k colouring – the problem of deciding if a given oriented graph has an injective oriented k-colouring – is described in the following theorem. Theorem 2.1 [10] Let k be a fixed integer. If k ≤ 3, then injective oriented k-colouring is Polynomial. If k ≤ 4, then injective oriented k-colouring is NP-complete. Since an oriented graph has an oriented injective 1-colouring if and only if it has no arcs, in the sequel we focus on the cases k = 2 and k = 3. An oriented graph has an oriented injective 2-colouring if and only if it has an injective homomorphism to T2 , the tournament on two vertices with one arc, and has an oriented injective 3-colouring if and only if it has an injective homomorphism to T3 , the transitive tournament on three vertices, or to the directed 3-cycle C3 . Because our colourings are injective on in-neighbourhoods, the directed graph that consists of two arcs meeting at a vertex plays an important role. The digraph with vertices v0 , s, v1 and arcs v0 s, v1 s will be denoted H3 (v0 , s, v1 ), or simply H3 when it is not necessary to emphasize the vertices, and called a hat. The vertex s is the point of the hat while v0 and v1 are its ends. All three vertices of a hat must be assigned different colours in an injective oriented colouring.
3
Obstructions to injective oriented 2- and 3-colouring
In this final section we describe the obstructions to injective oriented 2colouring and 3-colouring. To do so, we determine the obstructions for injective homomorphism to T2 , T3 and C3 . In two of these cases, the obstructions
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are very similar to the ones for (ordinary) homomorphism. Proposition 3.1 An oriented graph G has an injective oriented 2-colouring if and only if neither H3 nor P2 has an injective homomorphism to G. Proof. Suppose the oriented graph G admits an injective oriented 2-colouring, that is, an injective homomorphism to the transitive tournament on two vertices, T2 . If there is an injective homomorphism of H3 to G, then by composition there is an injective homomorphism of H3 to T2 , a contradiction. Similarly, there is no injective homomorphism of P2 to G. Suppose neither H3 nor P2 has an injective homomorphism to G. Since G is an oriented graph, any homomorphism of P2 to G is injective. Since P2 is the unique obstruction for (ordinary) homomorphism to T2 , it follows that there is a homomorphism of G to T2 . Since there is no injective homomorphism of H3 to G, no two vertices of G have a common out-neighbour. Hence the homomorphism of G to T2 is injective. 2 It is clear that the existence of an injective oriented 2-colouring can easily be decided by a greedy algorithm. Since each of H3 or P2 maps injectively to G if and only if it is a subgraph of G, Proposition 3.1 also gives a forbidden subgraph characterization of the digraphs that have an injective oriented 2-colouring. Corollary 3.2 An oriented graph G has an injective 2-colouring if and only if it is a star in which all arcs are directed away from the centre vertex. We now describe the obstructions for injective homomorphism to the transitive triple, T3 . Suppose T3 has vertices t0 , t1 , t2 and arcs t0 t1 , t1 t2 , t0 t2 . Since T3 is acyclic and has maximum in-degree two, both P3 and the digraph H4 with vertex set {h1 , h2 , h3 , h4 } and arcs h1 h4 , h2 h4 , h3 h4 are obstructions to injective homomorphism. Since the unique vertex of T3 with in-degree two has out-degree 0, the digraph A4 constructed from H3 by adding a new vertex x and the arc h3 x is also an obstruction. These are not all of the obstructions because, for example, the digraph Z2k+1 obtained from the undirected odd cycle C2k+1 by replacing each edge with a copy of a hat H3 – identifying the ends of the edge with the ends of the hat – admits a homomorphism from neither H4 nor P3 but does not map injectively to T3 . Hence these “zigzag odd cycles” form an infinite family of obstructions. There are other forbidden configurations as well. First, note that if x, y, z is a directed path of length two, then in any injective homomorphism to T3 these vertices must map to t0 , t1 and t2 in that order. Let Bn be the digraph obtained from an undirected path of length n by replacing each edge with a copy of the hat H3 , identifying
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the ends of the hat with the ends of the edge. By analogy with the zigzag odd cycles, each digraph Bn is a “zigzag path”. If the initial vertex of Bn maps to t0 , then the remaining vertices map to t2 , t1 , t2 , t0 , t2 , t1 , . . . in that order. If n is even, then the terminal vertex of Bn maps to t0 , and if n is odd it maps to t1 . Similarly, if the initial vertex maps to t1 , then the terminal vertex maps to t1 when n is even and t0 when n is odd. This leads to the following families of obstructions: (i) H0,0 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k+1 , where k is a non-negative integer, joining the initial vertices of the two paths; (ii) H1,1 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k+1 , where k is a non-negative integer, joining the middle vertices of the two paths; (iii) H0,1 , the set of digraphs obtained from two directed paths of length two by adding a copy of B2k , where k is a positive integer, joining the initial vertices of one path to the middle vertex of the other. Examples of these obstructions are shown in Figure 1. Notice that for the digraphs in H1,1 , for example, there is an injective homomorphism that maps the terminal vertex x of a directed path of length two to the point of the hat one of whose ends is an in-neighbour of x. Although it is tempting to define the family H1,0 , it is identical to H0,1 . Proposition 3.3 An oriented graph G has in injective homomorphism to T3 if and only if no digraph in {H4 , P3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G. Proof. The reasoning above demonstrates that none of the given digraphs have an injective homomorphism to T3 . Since injective homomorphisms compose (e.g. see [15]), none of them can have an injective homomorphism to G. Suppose that none of the given digraphs has an injective homomorphism to G. We will show that there is an injective homomorphism to T3 . Since different components can be treated separately, it can be assumed without loss of generality that the underlying undirected graph of G is connected. Suppose first that G has no directed path of length two. Then every vertex of G has either in-degree zero or out-degree zero. Let X be the set of vertices of out-degree zero. Map each vertex in X to t2 (this is possible because A4 does not map injectively to G). Construct an auxiliary graph L with vertex set V (G) − X and vertex u adjacent to vertex v whenever u and v have a common
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Fig. 1. Examples of the obstructions.
out-neighbour in G. Since no zigzag odd cycle Z2k+1 maps injectively to G, the undirected graph L is bipartite. A 2-colouring of its vertices using the colours t0 and t1 gives an injective homomorphism of G to T3 . Now suppose that G has one or more directed paths of length two. The vertices of each such path must necessarily be coloured t0 , t1 , t2 , in that order. Since A4 is forbidden, no vertex that is the origin of such a path is joined by an arc to a vertex that is the midpoint of another such path, and similarly no origin of a path of length two is joined to the midpoint of another path of length two. Since H4 is forbidden, no origin of a path of length two is joined to the terminus of another path of length two. And since P3 is forbidden, there can be no arc joining two vertices that are each the origin of a directed path of length two. Let S be the set of vertices of G that have out-degree zero. Because there are no forbidden configurations, S contains all vertices that have in-degree two or are the terminus of a directed path of length two. For the same reason, S is an independent set. Construct an auxiliary graph R with vertex set V (G) − S
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and vertex u adjacent to vertex v whenever u and v have a common outneighbour in G. As above, the undirected graph R is bipartite. Pre-colour some of its vertices by colouring vertices that are the origin of a directed path of length two with t0 and those that are the midpoint of such a path with t1 . Since no digraph in H0,0 ∪ H1,1 maps injectively to G, no two vertices pre-coloured with the same colour are joined by an odd path in R. Since no digraph in H0,1 has an injective homomorphism to G, no two vertices precoloured with different colours are joined by an even path in R. That is, the pre-colouring of R respects a bipartition. Therefore, the pre-colouring extends to a 2-colouring of R. An injective homomorphism of G to T3 can now be constructed by mapping the vertices in S to t2 , and map the vertices of R to the vertex of T3 corresponding to their colour. 2 The proof of the Proposition 3.3 implies a polynomial time algorithm that either finds an injective homomorphism of a given digraph to T3 or returns an obstruction. Proposition 3.4 An oriented graph G has in injective homomorphism to an C3 if and only if neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. Proof. Since neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to C3 , hence none of them can have an injective homomorphism to G. Suppose neither H3 nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. The facts that G has maximum in-degree one, and any directed cycle that has a homomorphism to G is of length divisible by three, together imply that G has a homomorphism to C3 . This homomorphism is injective. 2 Note that if H3 does not map injectively to G (i.e. H3 is not a subgraph of G), then any injective homomorphism of a directed cycle to G is an isomorphism. Hence, Proposition 3.4 gives a forbidden subgraph characterization of the digraphs that map injectively to C3 . Its proof implies a polynomial time algorithm that either finds an injective homomorphism of a given digraph to T3 or returns an obstruction. The following results combine the two previous propositions. Theorem 3.5 Let G be an oriented graph. Then G has an injective oriented 3-colouring if and only if either no digraph in {H4 , P3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G, or neither H3
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nor any Cn where n ≡ 0 (mod 3) has an injective homomorphism to G. Define H0,0 to be the set of digraphs obtained from H0,0 be adding any digraph that can be formed from an injective homomorphism that fixes all vertices on the zigzag path. The sets H1,1 and H0,1 are defined similarly.
Corollary 3.6 Let G be an oriented graph. •
If the maximum in-degree of G is one, then G has an injective oriented 3colouring if and only if either there is no injective homomorphism of P3 to G, or G contains a directed cycle of length not divisible by three;
•
If the maximum in-degree of G is two, then G has an injective oriented 3-colouring if and only if it has no element of {P3 , C3 , A4 } ∪ {Z2k+1 : k ≥ 0} ∪ H0,0 ∪ H1,1 ∪ H0,1 has an injective homomorphism to G;
•
If G has maximum in-degree at least three (that is, H4 maps injectively to G), then G does not have an injective oriented 3-colouring.
Proof. The first and third statements are easy to see, so we prove only the second statement. Suppose the oriented graph G has maximum in-degree two. Then there is an injective homomorphism of P3 to G if and only if P3 or C3 is a subgraph of G, and there is an injective homomorphism of A4 to G if and only if A4 is a subgraph of G. Essentially the same argument that shows that a homomorphic image of an undirected odd cycle (i.e. a closed walk of odd length) contains an odd cycle can be used to show that, when the target graph has maximum in-degree two, an injective homomorpic image of an odd zigzag cycle contains an odd zigzag cycle. If the oriented graph G contains no odd zigzag cycle, then an injective homomorphic image of a zigzag path of length n contains a zigzag path whose length is congruent to n modulo 2. The result now follows from Theorem 3.5 and the definition of H0,0 , H1,1 and H0,1 . 2
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