Semi-proper orientations of dense graphs

Semi-proper orientations of dense graphs Julio Araujo, Frédéric Havet, Claudia Linhares Sales, Nicolas Nisse, Karol Suchan To cite this version: Julio Araujo, Frédéric Havet, Claudia Linhares Sales, Nicolas Nisse, Karol Suchan. Semi-proper orientations of dense graphs. Inria & Université Cote d’Azur, CNRS, I3S, Sophia Antipolis, France. 2022. ฀hal-03907202v2฀ HAL Id: hal-03907202 https://hal.science/hal-03907202v2 Submitted on 13 Jan 2023 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L’archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d’enseignement et de recherche français ou étrangers, des laboratoires publics ou privés. Semi-proper orientations of dense graphs∗ Julio Araujo1 , Frédéric Havet2 , Claudia Linhares Sales1 , Nicolas Nisse2 , and Karol Suchan3,4 1 2 ParGO, Universidade Federal do Ceará, Fortaleza, Brazil Université Côte d’Azur, CNRS, Inria, I3S, Sophia Antipolis, France 3 Universidad Diego Portales, Santiago, Chile 4 AGH University of Science and Technology, Krakow, Poland January 13, 2023 Abstract An orientation D of a graph G is a digraph obtained from G by replacing each edge by exactly one of the two possible arcs with the same ends. An orientation D of a graph G is a k-orientation if the in-degree of each vertex in D is at most k. An orientation D of G is proper if any two adjacent vertices → have different in-degrees in D. The proper orientation number of a graph G, denoted by − χ (G), is the minimum k such that G has a proper k-orientation. A weighted orientation of a graph G is a pair (D, w), where D is an orientation of G and w is an arc-weighting A(D) → N \ {0}. A semi-proper orientation of G is a weighted orientation (D, w) of G such that for every two adjacent vertices u and v in G, we have that S(D,w) (v) 6= S(D,w) (u), where S(D,w) (v) is the sum of the weights of the arcs in (D, w) with head v. For a positive integer k, a semi-proper k-orientation (D, w) of a graph G is a semi-proper orientation of G such that maxv∈V (G) S(D,w) (v) ≤ k. →(G), is the least k such that G has a The semi-proper orientation number of a graph G, denoted by − χ s semi-proper k-orientation. →(G) ∈ {ω(G) − 1, ω(G)} for every split graph G, and that, given In this work, we first prove that − χ s →(G) = ω(G) − 1 is an NP-complete problem. We also show that, a split graph G, deciding whether − χ s − → for every k, there exists a (chordal) graph G and a split subgraph H of G such ≤ k and  3 that χ (G) − → − → p χ (H) = 2k − 2. In the sequel, we show that, for every n ≥ p(p + 1), χs (Pn ) = 2 p , where Pnp is the pth power of the path on n vertices. We investigate further unit interval graphs with no big clique: we show → that − χ (G) ≤ 3 for any unit interval graph G with ω(G) = 3, and present a complete characterization → →(G) = ω(G) of unit interval graphs with − χ (G) = ω(G) = 3. Then, we show that deciding whether − χ s can be solved in polynomial time in the class of co-bipartite graphs. Finally, we prove that computing − →(G) is FPT when parameterized by the minimum size of a vertex cover in G or by the treewidth of χ s → →(G), but also − χ χ (G), admits a polynomial kernel when G. We also prove that not only computing − s parameterized by the neighbourhood diversity plus the value of the solution. These results imply kernels 2 of size 4O(k ) and O(2k k2 ), in chordal graphs and split graphs, respectively, for the problem of deciding →(G) ≤ k parameterized by k. We also present exponential kernels for computing both − → whether − χ χ (G) s − → and χs (G) parameterized by the value of the solution when G is a cograph. On the other hand, we show →(G) does not admit a polynomial kernel parameterized by the value of the solution that computing − χ s when G is a chordal graph, unless NP ⊆ coNP/poly. 1 Introduction All graphs in this work are simple and finite. For terms not defined here, we refer to [9, 12]. ∗ This work was supported by the STIC-AmSud project GALOP (Programa Regional STICAMSUD 19-STIC-05). 1 A (proper) k-colouring of G is a mapping c : V (G) → {1, . . . , k} such that c(u) 6= c(v) for every edge uv ∈ E(G). The chromatic number of G, denoted by χ(G), is the minimum k such that G admits a k-colouring. Determining the chromatic number of a given graph G is one of the most studied problems in Graph Theory and it is one of the Karp’s NP-hard problems [18]. An orientation D of a graph G is a digraph obtained from G by replacing each edge by exactly one of the two possible arcs with the same ends. An orientation D of a graph G is a k-orientation if the in-degree of every vertex in D is at most k. An orientation D of G is proper if any two adjacent vertices have different − in-degrees in D. The proper orientation number of a graph G, denoted by → χ (G), is the minimum k such that G has a proper k-orientation. In other words, the values of the in-degrees in D define a proper colouring of G (with the first colour being zero). The existence of proper orientations was implicitely demonstrated − χ (G) ≤ ∆(G), where ∆(G) denotes the by Borowiecki, Grytczuk and Pilśniak in [10], who established that → maximum degree of G. Thus, − χ(G) − 1 ≤ → χ (G) ≤ ∆(G). (1) Afterwards, the proper orientation number was introduced in [1]. Since then, the proper orientation number has been studied by several authors and some parameters related to proper orientations were defined in the literature (for instance, see [1–7, 11, 13, 14, 19]). In [5], it is shown that the proper orientation number of a tree is at most 4, and that there exist trees whose proper orientation number equals 4. A natural question is to ask how it can be generalized. Problem 1.1. Which graph classes containing the trees have bounded proper orientation number? Araujo et al. [6] proved that this is the case for cacti (their proper orientation number is bounded by 7), but left open the question for the class of outerplanar graphs. It was partly answered by Ai et al. [3], − − who proved that → χ (G) ≤ 3 if G is a triangle-free 2-connected outerplanar graph, and → χ (G) ≤ 4 if G is a triangle-free bridgeless outerplanar graph. This question was already asked for the more general subclasses of planar graphs, like outerplanar graphs or (all) planar graphs. In [5], the authors posed the following − more general problem. Is → χ (G) upper bounded by a function of the treewidth or the maximum average degree of G? Recall that the maximum average degree of G, denoted by mad(G), is the maximum value of 2|E(H)|/|V (H)| over all subgraphs H of G. Note that mad(G) and χ(G) are upper bounded by the treewidth of G plus one, for every graph G, and that mad(G) and χ(G) are bounded by constants in (outer)planar graphs. The above problem has very recently been solved by Chen, Mohar and Wu [11] who proved the following theorem, which can be seen as a generalization of a result of Dehghan and Havet [14] on the semi-proper orientation number (Theorem 1.5). Theorem 1.2 (Chen, Mohar and Wu [11]). For all graph G, we have   χ(G) log χ(G) mad(G) → − χ (G) = . +O 2 log log χ(G) Dehghan and Havet [14] introduced the notion of semi-proper orientations. A weighted orientation of a graph G is a pair (D, w), where D is an orientation of G and w is an arc-weighting A(D) → N \ {0}. A semi-proper orientation of G is a weighted orientation (D, w) of G such that, for every two vertices u and v adjacent in G, we have that S(D,w) (v) 6= S(D,w) (u), where the in-weight of a vertex v, denoted S(D,w) (v), is the sum of the weights of the arcs in (D, w) for which v is the head, i.e.: X w(z, v), S(D,w) (v) = − z∈ND (v) − where ND (v) = {z | (z, v) ∈ A(D)} denotes the in-neighbourhood of v in D. For a positive integer k, a semi-proper k-orientation (D, w) of a graph G is a semi-proper orientation of G such that →(G), is maxv∈V (G) S(D,w) (v) ≤ k. The semi-proper orientation number of a graph G, denoted by − χ s the least k such that G has a semi-proper k-orientation. A semi-proper k-orientation (D, w) is optimum 2 →(G). In [14], it is shown that, for any graph G, there exists an optimum semi-proper orientation if k = − χ s (D, w) such that w : A(D) → {1, 2}. Note that every proper k-orientation D of a graph G gives a semi-proper k-orientation (D, w) of G such that w(e) = 1 for every e ∈ A(D). Consequently, by (1), we have →(G) ≤ → − χ(G) − 1 ≤ − χ χ (G) ≤ ∆(G). s (2) The degeneracy δ ∗ (G) of a graph G is defined as δ ∗ (G) = maxH⊆G δ(H), where δ(H) is the minimum degree of a vertex in H. Let ω(G) denote the maximum size of a clique in G. It is well known that: ω(G) ≤ χ(G) ≤ δ ∗ (G) ≤ ∆(G) + 1. (3) → − − → A considerable difference between the parameters χ and χs is that the latter is monotone. Recall that a graph parameter γ is monotone if γ(H) ≤ γ(G) for every subgraph H of G. As shown [6], the proper orientation number is not monotone, even when restricted to trees. In contrast, the semi-proper orientation number is. →(H) ≤ − →(G). Lemma 1.3 (Dehghan and Havet [14]). If H is a subgraph of G, then − χ χ s s When dealing with induced subgraphs, we have the following more precise statement. →(G − v) ≤ − →(G) ≤ − →(G − v) + 1. Proposition 1.4. Let G be a graph and v a vertex of G. Then, − χ χ χ s s s →(G − v) ≤ − →(G). Therefore it suffices to prove that − →(G) ≤ − →(G − v) + 1. Proof. By Lemma 1.3, − χ χ χ χ s s s s − → Let (D, w) be an optimum semi-proper orientation of G − v, i.e., with maximum in-weight χs (G − v). Let (D′ , w′ ) be the semi-proper orientation of G such that D′ agrees with D on G − v and v has in-degree 0. Moreover, for every u ∈ N (v), let w′ (v, u) = 1 (so the in-weight of u in G equals the in-weight of u in G − v plus one). Finally, for every u ∈ V (G) \ N [v] that has in-weight at least one in (D, w), choose one in-going arc (x, u) and let w′ (x, u) = w(x, u) + 1. For every other arc e, w′ (e) = w(e). Then, (D′ , w′ ) is a semi-proper →(G − v) + 1. orientation of G with maximum in-weight at most − χ s − →, since there exist a graph G and a Proposition 1.4 shows another major difference between → χ and − χ s → − → − vertex v ∈ V (G) such that χ (G − v) ≤ χ (G) − 2. Indeed, Araujo et al. exhibit a split graph G′ (a graph − whose vertex-set can be partitioned into a clique and a stable set) such that → χ (G′ ) = 2ω(G′ ) − 2 and they → − show that, for any split graph G, χ (G) ≤ 2ω(G) − 2. Therefore, for any vertex v of the clique of G′ , → − − χ (G′ − v) ≤ 2ω(G′ − v) − 2 = 2ω(G′ ) − 4 = → χ (G′ ) − 2 [4]. →(G) in χ Thanks to Lemma 1.3, Dehghan and Havet [14] succeeded to prove lower and upper bounds for − s ∗ function of mad(G) and δ (G): Theorem 1.5 (Dehghan and Havet [14]). For any graph G, we have      ∗  mad(G) →(G) ≤ mad(G) + χ(G) − 1 and δ (G) + 1 ≤ − →(G) ≤ 2δ ∗ (G). ≤− χ χ s s 2 2 2 Recall that graphs with treewidth at most k satisfy that δ ∗ (G) ≤ k, and thus previous bounds apply to the class of bounded treewidth graphs. Tight constant upper bounds for the classes of trees, cacti, outerplanar and planar graphs are presented in [14]. Theorem 1.2 (and Theorem 1.5) states that the proper orientation number (and thus semi-proper orientation number) of sparse graphs is bounded. It also has corollaries for classes of dense graphs. If every graph G of a class has maximum average degree bounded by a function of the clique number ω(G), then the proper and semi-proper orientation numbers are also bounded by a function of the ω(G). Split graphs, interval graphs and cobipartite graphs form such graph classes. For each of theses classes, a natural question − →(G) ≤ g(ω(G))) for is then to find the minimum function f (resp. g) such that → χ (G) ≤ f (ω(G)) (resp. − χ s every graph G in the class. 3 − Regarding the computational complexity, deciding if the proper orientation number → χ (G) of a graph G is at most 2 is an NP-complete problem even if G is planar and bipartite. Moreover, deciding whether → − χ (G) ≤ k is NP-complete for a graph G and a positive integer k given as inputs [4]. In [7], Araujo et al. 2 − showed that deciding whether → χ (G) ≤ k can be solved in O(2t · k 3t · t · n) time in the class of n-vertex graphs with treewidth at most t. In other words, the problem is FPT parameterized by the treewidth of G plus the value k of the solution. Moreover, they showed that the weighted version of this problem (when edges of the input graph G have weights given as input) is W[1]-hard when parameterized only by the treewidth. →(G) = → − For semi-proper orientation, Dehgan and Havet [14] proved that deciding whether − χ χ (G) is NPs − → − → complete, even if G is planar and χs (G) = 2, and that deciding whether χs (G) ≤ 2 is also NP-complete even if G is planar and bipartite. Our contributions. In Section 2, we present some definitions that we need along the text. Then, in →(G) ∈ {ω(G) − 1, ω(G)} for every split graph G, and that, given a split graph G, χ Section 3, we prove that − s − → − deciding whether χs (G) = ω(G) − 1 is an NP-complete problem. Note that, in contrast, → χ (G) ≤ 2ω(G) − 2 for every split graph G and that this bound is tight [4]. To conclude Section 3, we show that, for every k, − − there exists a chordal graph G and a split subgraph H of G such that → χ (G) ≤ k and → χ (H) = 2k − 2. In  − → 3 p p th Section 4, we show that χs (Pn ) = 2 p for every n ≥ p(p + 1), where Pn is the p power of the path on n → is monotone, this implies that − →(G) ≤ 3 (ω(G) − 1) for any unit interval graph G, and vertices. Since − χ χ s s 2 − that this bound is tight. On the other hand, since → χ is not monotone, we investigate further unit interval − graphs with small clique number: we show that → χ (G) ≤ 3 for any unit interval graph G with ω(G) = 3 − and, through a complete characterization of unit interval graphs with → χ (G) = ω(G) = 3, we design a linear→ − time algorithm that computes χ (G) in this class of graphs. In Section 5, we show that deciding whether − →(G) = ω(G) can be solved in polynomial time in the class of co-bipartite graphs. Finally, in Section 6, we χ s →(G) is FPT when parameterized by the minimum size of a vertex cover in G or by show that computing − χ s 2 the treewidth of G. As corollaries, we obtain kernels of size 4O(k ) and O(2k k 2 ), in chordal graphs and split →(G) ≤ k parameterized by k. We also show that graphs respectively, for the problem of deciding whether − χ s this problem does not admit a polynomial kernel in the class of chordal graphs, unless NP ⊆ coNP/poly. We →(G), but also → − also prove that not only computing − χ χ (G), admits a polynomial kernel when parameterized s − by the neighbourhood diversity plus the value of the solution. While the complexity of computing → χ and − → χs is open in cographs [4], we present exponential kernels for both problems parameterized by the value of the solution when restricted to cographs. We conclude with several open questions in Section 7. 2 Definitions We denote by [n] the set of first n positive integers {1, . . . , n}. Each graph G in this work is simple, i.e. there are no parallel edges nor loops. Thus, we consider edges as subsets of two distinct vertices, and we use uv to denote an edge, as a shorthand for {u, v}. Given a graph G, we use V (G) and E(G) to denote its vertex set and edge set, whose cardinalities are known as the order and size of G, and are denoted by n(G) and m(G), respectively. Given two subsets A and B of vertices, we denote by E(A, B) the set of edges having one end in A and the other in B. The neighbourhood of a vertex v ∈ V (G) in a graph G is denoted by NG (v) = {u | uv ∈ E(G)} and, since G is an undirected simple graph, the degree of v in G is dG (v) = |NG (v)|. Subscripts will be omitted when G is clear from the context. Two vertices u and v in a graph G are twins if NG (u) = NG (v). We denote by δ(G) and ∆(G) the minimum and maximum degree of a vertex in G, respectively. A graph H is a subgraph of G if V (H) ⊆ V (G) and E(H) ⊆ E(G), in which case we also say that G is a supergraph of H. When S ( V (G), the graph G − S is obtained from G by the removal of all vertices in S and all the edges that are incident to at least one vertex in S. A subgraph H of G is a subgraph induced by S ⊆ V (G), denoted as H = G[S], if H = G − (V (G) \ S). A clique in G is a set of pairwise adjacent vertices. The cardinality of a largest clique in G is called the clique number of G and denoted by ω(G). A stable (a.k.a. independent) set in G is a set of pairwise 4 non-adjacent vertices in G. The cardinality of a largest stable set in G is called the stability number of G and denoted by α(G). A graph G whose vertex set is itself a clique is called complete. A (u, v)-path on k vertices in a graph G is a sequence Pk = (u = v1 , e1 , v2 , e2 , . . . , ek−1 , vk = v) satisfying ei = vi vi+1 ∈ E(G) for every i ∈ {1, . . . , k − 1}, and vi 6= vj for every 1 ≤ i 6= j ≤ k. A (u, v)-path P in G is Hamiltonian if P traverses/contains all vertices of G. A graph G is connected if there exists a (u, v)-path in G for any u, v ∈ V (G). A cycle on k vertices in G is a sequence Ck = (u = v1 , e1 , v2 , e2 , . . . , ek−1 , vk = u) satisfying ei = vi vi+1 ∈ E(G) for every i ∈ {1, . . . , k − 1}, and vi 6= vj for every 1 ≤ i 6= j ≤ k − 1. A graph G is chordal if every cycle Ck in G such that k ≥ 4 has a chord, i.e., an edge linking two non-consecutive vertices of Ck . We do the usual abuse of notation of referring to the subgraph (and not the sequence) of G corresponding to a path or a cycle by the same nomenclature. A matching in G is a set of pairwise non-intersecting (i.e., with no end in common) edges. An antimatching is a set of pairwise non-intersecting pairs of non-adjacent vertices. In other words, it is a matching in the complement of G. Let M be a matching or an anti-matching. For every edge uv of M , we say that u is matched with v by M , and write u = M (v) and M (u) = v. We say that M saturates a set of vertices A if every vertex a ∈ A is matched by M . Similarly, each directed graph D = (V, A) in this work is simple, i.e., has no parallel arcs, no symmetric arcs, and no loops. Given a digraph D, we use V (D) and A(D) to denote its vertex set and arc set; and n(D) and m(D) to denote their cardinalities, respectively. If (u, v) ∈ A(D) is an arc in D, the tail of (u, v) is + u and its head is v. Let v be a vertex of D. The out-neighbourhood of v, denoted by ND (v), is the set of − vertices u such that (v, u) ∈ A(D). The in-neighbourhood of v, denoted by ND (v), is the set of vertices u − + − such that (u, v) ∈ A(D). The out-degree d+ D (v) (resp. the in-degree dD (v)) of v is |ND (v)| (resp. |ND (v)|). The notions of (Hamiltonian) path, cycle, (induced) subgraph, have their straightforward notions in digraphs called as (resp. Hamiltonian) directed path, directed cycle, (induced) subdigraph. For any W ⊆ V (D), we denote by DhW i the subdigraph induced by W in D, and we denote by D − W the digraph DhV (D) \ W i. A tournament T is an orientation of a complete graph G. A regular tournament T is a tournament where d− (v) = d+ (v) for every vertex v ∈ V (T ). A transitive tournament T is a tournament such that if (u, v), (v, w) ∈ A(T ), then (u, w) ∈ A(T ), for every u, v, w ∈ V (T ). 3 Split graphs A split graph is a graph whose vertex set can be partitioned into a clique and a stable set. A split partition of a split graph G is a bipartition (K, S) of V (G) where K is a clique and S is a stable set. →(G) ∈ {ω(G) − 1, ω(G)}. Theorem 3.1. If G is a connected split graph, then − χ s →(G) ≥ χ(G) − 1 = ω(G) − 1. Proof. Let G be a split graph. We have − χ s Let (K, S) be a split partition of G with K = {v1 , . . . , vk }. Let D be the orientation of G such that DhKi is a transitive tournament with Hamiltonian directed path (v1 , . . . , vk ) and all the edges between K and S are oriented towards S. Let w be an arc-weighting of D such that every arc of DhKi has weight 1, and every vertex in S has in-weight ω(G). Such an arc-weighting exists because d− D (v) ≤ |K| ≤ ω(G) for all v ∈ S. For every i ∈ [k], we have S(D,w) (vi ) = i − 1 < k = ω(G). Hence (D, w) is a semi-proper orientation →(G) ≤ ω(G). of G, so − χ s →(G) = A natural question that arises with Theorem 3.1 is whether one can decide in polynomial time if − χ s ω(G) − 1. We show in the sequel that this is not possible unless P = NP. Theorem 3.2. It is NP-complete to decide whether a given split graph G has semi-proper orientation number equal to ω(G) − 1. Proof. We reduce the well-known 3-SAT problem to ours. Let Φ be a 3-SAT formula with n variables x1 , . . . , xn and m clauses C1 , . . . , Cm . In addition, we assume that each clause has exactly three literals. 5 We construct S the split graph GΦ as follows: We create a clique K of size 4n − 3 which is the disjoint union of {x0 } ∪ i∈[n] {xi , x̄i } and two sets D and D′ of size n − 2. We create a stable set S of size 2n + m + 1 S which is the disjoint union of {y0 } ∪ i∈[n] {yi , yi′ } and {Cj | j ∈ [m]}. We then add edges between K and S as follows: • we link y0 to all vertices of K except x0 ; • for every i ∈ [n], we connect yi to all vertices of K except xi , x̄i and i − 1 vertices in D, and we connect yi′ to all vertices of K except xi , x̄i and i − 1 vertices in D′ ; • for every j ∈ [m], we connect the clause Cj to all vertices of K except its three literals and n − 2 vertices in D ∪ D′ . →(G) = ω − 1 By construction K is a maximum clique of G so ω = ω(G) = 4n − 3. We shall prove that − χ s if and only if Φ is satisfiable. Assume first that G admits a semi-proper weighted orientation (D, w) with maximum in-weight ω − 1. Since K is a clique of size ω, for every k ∈ {0, . . . , ω − 1} there is exactly one vertex of K with in-weight k. Thus DhKi is a transitive tournament, all the arcs of DhKi have weight 1, and all edges in E(K, S) are oriented towards their end in S. Consider now y0 . We have S(D,w) (y0 ) ≥ d− D (y0 ) = dG (y0 ) = ω − 1. Hence S(D,w) (y0 ) = ω − 1 and so x0 has also in-weight ω − 1, since D is proper. For every i ∈ [n], let zi be the vertex of K with in-weight ω − 1 − i. Claim 3.2.1. zi ∈ {xi , x̄i } for every i ∈ [n]. Proof: By induction on i ∈ [n]. Observe that both yi and yi′ are adjacent to {x0 , . . . , xi−1 }. Thus, by the induction hypothesis and because S(D,w) (x0 ) = ω − 1, they are adjacent to a vertex with in-degree ω − 1 − k for every k ∈ {0, . . . , i − 1}. But, by construction, yi and yi′ have in-weight ω − 1 − i in D. Therefore, S(D,w) (yi ) = S(D,w) (yi′ ) = ω − 1 − i. Now zi is neither adjacent to yi nor yi′ (in particular, zi ∈ / D ∪ D′ ). Hence, by construction, zi is either xi or x̄i .  Let ϕ be the truth assignment defined by ϕ(xi ) = true if xi = zi and ϕ(xi ) = f alse if zi = x̄i . This is well-defined by the above claim. Consider a clause Cj = ℓi1 ∨ ℓi2 ∨ ℓi3 , with ℓip ∈ {xip , x̄ip } for all p ∈ [3]. It has degree ω − 1 − n in G and so in-degree ω − 1 − n in D. Hence sj = S(D,w) (Cj ) ≥ ω − 1 − n. Moreover, Cj is adjacent to x0 . Thus, sj must be exactly the in-weight of one of the vertices corresponding to literals belonging to Cj and so, by our definition of ϕ, Cj is satisfied. Consequently, ϕ satisfies Φ. Reciprocally, assume that there is a truth assignment ϕ satisfying Φ. Let zi = xi if ϕ(xi ) = true and let zi = x̄i if ϕ(xi ) = f alse. Orient K in a transitive tournament such that x0 has in-degree ω − 1 and zi has in-degree ω − 1 − i for all i ∈ [n]. Orient all the edges between K and S from K to S. This results in the orientation D. Let us now assign weights on the arcs of D. All arcs with both end-vertices in K are weighted by 1. For each i ∈ [n], we assign weights to the arcs entering yi and yi′ so that they both have in-weight ω − 1 − i. For each j ∈ [m], the clause Cj has a literal which is true, so Cj is non-adjacent to zij for some ij ∈ [n]. Hence we assign weights to the arcs entering Cj so that Cj has in-weight ω − 1 − ij . The resulting weighted orientation (D, w) is a semi-proper orientation of G with maximum in-weight ω − 1. Let us now use Theorem 3.1 to argue that, for a given chordal graph G, one may find a subgraph H ⊆ G − − such that → χ (H) ≥ 2→ χ (G) − 2. − →(H). Proposition 3.3. Every graph H has a supergraph G such that → χ (G) ≤ − χ s Proof. Let H be a graph, and (D, w) be an optimal semi-proper orientation of H. Let G be the graph obtained from H by adding for each vertex v ∈ V (H) a set Sv of S(D,w) (v) − d− D (v) new vertices of degree 1 adjacent to v, and let D′ be the orientation of G obtained from D by orienting all the edges between the new vertices and V (H) towards V (H). Every new vertex is a source in D′ and, by the construction of G, ′ we have that S(D,w) (v) = d− D ′ (v). Therefore, D is a proper orientation of G with maximum in-degree equal − → to χs (H). 6 Observe that, if the graph H is chordal (resp. a tree), then the supergraph G given by the proof of Proposition 3.3 is also chordal (resp. a tree). Araujo et al. [4] proved that there exists a split graph H such − that → χ (H) = 2ω(H) − 2. Hence with Theorem 3.1, this implies the following. Corollary 3.4. For every k, there exists a (chordal) graph G and a (split) subgraph H of G such that → − − χ (G) ≤ k and → χ (H) = 2k − 2. − − Problem 3.5. Does there exists a function θ such that → χ (H) ≤ θ(→ χ (G)) for every graph G and every subgraph H of G? 4 Unit interval graphs A graph G is an interval graph if it can be obtained from a set of intervals on the real line, by adding a vertex for each interval, and adding an edge for each pair of vertices whose intervals intersect. It is the intersection graph of the intervals. A unit interval graph G is an interval graph that can be obtained from intervals on the real line of that are of the same length. If G is a unit interval graph, then ∆(G) ≤ 2ω(G) − 2, so by Equation (2), we get the following. →(G) ≤ → − Proposition 4.1. If G is a unit interval graph, then − χ χ (G) ≤ 2ω(G) − 2. s A unit interval graph with clique number 2 is the disjoint union of paths. It is easy to check that a path of order at least 4 has (semi-)proper orientation number equal to 2. Let Pnk be the k-th power of the path of order n, i.e., the graph with vertex set {v1 , . . . , vn } and edge set {vi vj | i − j ∈ [k]}. The following proposition is easy to prove and well known. Proposition 4.2. Every unit interval graph of order n with clique number at most ω is a subgraph of Pnω−1 . So, let us present our result on powers of paths. →(P ω−1 ) = Proposition 4.3. Let b and ω be positive integers such that b > ω, and let n = b × ω. Then − χ s n → − 3 ω−1 χ (Pn ) = 2 (ω − 1) . Proof. For every i ∈ [b], let Ci = {vj | (i − 1)ω + 1 ≤ j ≤ i · ω}. Observe that each Ci is a clique. Let (D, w) be a semi-proper weighted orientation of Pnω−1 and let k be its maximum in-weight. Therefore all vertices of Ci have different in-weights and so   X X X 1 S(D,w) (v) ≤ b(k + k − 1 + . . . + k − ω + 1) = b · ω k − (ω − 1) S(D,w) (v) = (4) 2 v∈V (G) i∈[b] v∈Ci But X v∈V (G) S(D,w) (v) ≥ X v∈V (G)   ω−1 1 . ω(ω − 1) = b · ω ω − 1 − d− (v) ≥ |E(G)| = n(ω − 1) − D 2 2b (5) Equations (4) and (5) (divided by b · ω) yield ω−1 3 1 3 (ω − 1) − > (ω − 1) − . 2 2b 2 2 3  Since k, b and ω are integers, we have k ≥ 2 (ω − 1) . →(P ω−1 ) ≥  3 (ω − 1). Consequently, − χ s n 2 k≥   − Let us now prove that → χ (Pnω−1 ) ≤ 32 (ω − 1) . We distinguish two cases depending on whether ω is odd or even. 7 • Assume first that ω is odd. Take an orientation D of G such that for every i ∈ [b − 1], all edges in E(Ci+1 , Ci ) are oriented towards Ci , DhCb i is a transitive tournament with Hamiltonian directed path (vn−ω+1 , . . . , vn ), and for every i ∈ [b − 1], DhCi i is a regular tournament. For every j ∈ [ω], we have d− D (v(b−1)ω+j ) = j − 1. For every i ∈ [b − 1] and every j ∈ [ω], the in-neighbours in Ci and j − 1 in-neighbours in Ci+1 . Hence d− vertex v(i−1)ω+j has w−1 D (v(i−1)ω+j ) = 2 3 ω−1 + j − 1. Consequently D is a proper orientation of D with maximum in-degree (ω − 1). 2 2 • Assume now that ω is even. Take an orientation D of G such that for every i ∈ [b − 1], all edges in E(Ci+1 , Ci ) are oriented towards Ci , DhCb i is a transitive tournament with Hamiltonian directed path (vn−ω+1 , . . . , vn ), and for every i ∈ [b − 1], DhCi i is a quasi-regular tournament such that, for every ω ω 1 ≤ j ≤ ω2 , v(i−1)ω+j has in-degree ω−2 2 and for every 2 + 1 ≤ j ≤ ω v(i−1)ω+j has in-degree 2 . For every j ∈ [ω], the vertex d− D (v(b−1)ω+j ) = j − 1. Let i ∈ [b − 1] and j ∈ [ω]. Since v(i−1)ω+j ω−2 ω has j − 1 in-neighbours in Ci+1 , it follows that and d− D (v(i−1)ω+j ) = 2 + j − 1 if 1 ≤ j ≤ 2 , and − ω ω dD (v(i−1)ω+j ) = 2 + j − 1 if  2 + 1 ≤ j ≤ ω. Consequently D is a proper orientation of D with maximum in-degree 32 ω − 1 = 23 (ω − 1) . 4.1 Unit interval graphs with clique number 3 In this subsection, we prove (in Lemma 4.4) that if G is an unit interval graph with clique number 3, then − →(G) ≤ → − χ χ (G) ≤ 3. Moreover, we prove that if ω(G) ≤ 3, then one can decide in linear time whether s → − − χ (G) = 2 (in Lemma 4.6). Overall, this gives a linear time algorithm that computes → χ (G) in the class of unit interval graph G with ω(G) = 3 (Theorem 4.7). It is easy to check that every connected unit interval graph of order n contains Pn as a subgraph. Moreover, every unit interval graph G with ω(G) ≤ 3 is a subgraph of Pn2 . Let us start with a result on proper 3-orientations of such graphs. Lemma 4.4. Let Pn = (v1 , . . . , vn ) be an n-node path and G be a graph such that Pn ⊆ G ⊆ Pn2 , for n ≥ 2. − − Then, G admits a proper 3-orientation D such that d− D (v1 ) ≤ 1, dD (v2 ) ≤ 2 and dD (vn ) ≤ 1. − Proof. A nice orientation of G is a proper 3-orientation of G such that d− D (v1 ) ≤ 1, dD (v2 ) ≤ 2 and − dD (vn ) ≤ 1. We shall prove by induction on the number n of vertices of G that G admits a nice orientation. − If n ≤ 4, then → χ (G) ≤ ∆(G) ≤ 3. So G admits a proper 3-orientation. If G is a path, then this orientation is necessarily nice. If G is not a path, then it is one of the graphs whose nice orientation is depicted in Figure 1. v1 v2 v3 v1 v2 v1 v3 v2 v4 v3 v1 v2 v3 v4 v4 Figure 1: Nice orientations when n ≤ 4. Thus, assume n ≥ 5 and any graph H such that Pk ⊆ H ⊆ Pk2 with 1 ≤ k < n has a nice orientation. Let G be a graph such that Pn ⊆ G ⊆ Pn2 . 8 We shall show a nice orientation of G. We distinguish several cases. In order to shorten the writing, in all cases the orientation is explicitly shown and thus one can verify that the constraints d− D (v1 ) ≤ 1 and − (v ) ≤ 1 will always be a consequence of the induction hypothesis. (v ) ≤ 2 are satisfied. The fact that d d− n 2 D D Therefore we shall only explicitly check that the orientation is proper. 1. d(v2 ) = 2 and d(v3 ) = 2. In this case, we extend a nice orientation D3 of G − {v1 , v2 , v3 } into a nice orientation D of G as − depicted in Figure 2. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 because D3 is nice. v1 v2 v3 v4 v5 Figure 2: The case d(v2 ) = 2 and d(v3 ) = 2. 2. d(v2 ) = 2 and d(v3 ) = 3. If v1 v3 ∈ E(G), then we extend a nice orientation D3 of G − {v1 , v2 , v3 } into a − nice orientation D of G as depicted in Figure 3. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 because D3 is nice. v1 v2 v3 v4 v5 Figure 3: The case d(v2 ) = 2, d(v3 ) = 3 and v1 v3 ∈ E(G). If v3 v5 ∈ E(G), then we extend a nice orientation D2 of G − {v1 , v2 } into a nice orientation D of G as − depicted in Figure 4. This orientation is proper since d− D (v3 ) = dD2 (v3 ) ≤ 1, because D2 is nice. v1 v2 v3 v4 v5 Figure 4: The case d(v2 ) = 2, d(v3 ) = 3 and v1 v3 ∈ E(G). 3. d(v2 ) = 2 and d(v3 ) = 4. In this case, we extend a nice orientation D3 of G − {v1 , v2 , v3 } into a − nice orientation D of G as in Figure 5. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 and − d− D (v5 ) = dD3 (v5 ) ≤ 2. v1 v2 v3 v4 v5 Figure 5: The case d(v2 ) = 2 and d(v3 ) = 4. 4. d(v2 ) = 3 and d(v3 ) = 2. Let D4 be a nice orientation of G − {v1 , v2 , v3 , v4 }. Depending on the degree of v4 in G, extend it into a nice orientation D of G as depicted in Figure 6, (d(v4 ) = 3 left and d(v4 ) = 4 9 v1 v2 v3 v4 v5 v1 v2 v3 v4 v5 v6 Figure 6: The case d(v2 ) = 3 and d(v3 ) = 2. right). In both cases, D is proper since d− D (v5 ) = dD4 (v5 ) ≤ 1 and dD (v6 ) = dD4 (v6 ) ≤ 2 (in case n ≥ 6) because D4 is nice. 5. d(v2 ) = 3 and d(v3 ) = 3. If v1 v3 ∈ E(G), then we extend a nice orientation D3 of G − {v1 , v2 , v3 } into a nice orientation D of G − as depicted in Figure 7. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 because D3 is nice. v1 v2 v3 v4 Figure 7: The case d(v2 ) = 3, d(v3 ) = 3 and v1 v3 ∈ E(G). If v3 v5 ∈ E(G), then we extend a nice orientation D3 of G−{v1 , v2 , v3 } into a nice orientation D of G as − − − depicted in Figure 8. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 and dD (v5 ) = dD3 (v5 ) ≤ 2 because D3 is nice. v1 v2 v3 v4 v5 Figure 8: The case d(v2 ) = 3, d(v3 ) = 3 and v3 v5 ∈ E(G). 6. d(v2 ) = 3 and d(v3 ) = 4. In this case, we extend a nice orientation D3 of G − {v1 , v2 , v3 } into a nice − orientation D of G as depicted in Figure 9. This orientation is proper since d− D (v4 ) = dD3 (v4 ) ≤ 1 and − − dD (v5 ) = dD3 (v5 ) ≤ 2 because D3 is nice. v1 v2 v3 v4 v5 Figure 9: The case d(v2 ) = 3 and d(v3 ) = 4. The proof of Lemma 4.4 is constructive. That is, given a unit interval graph G with clique number at most 3, it allows to build a proper orientation with proper orientation number at most 3. Recall that → − − χ (H) ≥ ω(H) − 1 for any graph H, and so → χ (G) ∈ {2, 3}. A simple dynamic programming algorithm would → − allow to decide whether χ (G) = 2, as the one presented in [7] for graphs of bounded treewidth. To conclude this section, we provide a full characterization of unit interval graphs with clique number 3 and with proper orientation number 2, which leads to a simpler algorithm. As noticed in [4]: 10 − Proposition 4.5. Let G be any graph with → χ (G) = ω(G) − 1. Then, for any optimal proper orientation D and any maximum clique K of G, we have: (a) the orientation D restricted to K induces a transitive tournament; (b) {d− D (v) | v ∈ V (K)} = {0, . . . , ω(G) − 1}; and (c) for every u ∈ V (K) and v ∈ / V (K), uv ∈ A(D). ′ ′ Thus, if K ′ is another maximum clique of G, {d− D (v) | v ∈ V (K) ∩ V (K )} = {0, . . . , |V (K) ∩ V (K )| − 1}. − Now, let us present a characterization of unit interval graphs with ω(G) = 3 and → χ (G) = 2. − Lemma 4.6. Let G = (V, E) be a connected unit interval graph with ω(G) = 3. Then, → χ (G) = 2 if and only if G has no induced subgraph isomorphic to one of the graphs depicted in Figure 10. (a) (b) (c) x (e) In addition to be an induced subgraph, in this case, the vertex x must have degree 1 in G. (d) (f) (g) (h) →(G) = 2. Figure 10: List of forbidden induced subgraphs for a unit interval graph G with ω(G) = 3 and − χ s − − Proof. Recall, by Lemma 4.4 and the fact that → χ (G) ≥ ω(G) − 1 for any graph, → χ (G) ∈ {2, 3}. − χ (G) = 3. Let us first prove that if one of the induced subgraphs in Figure 10 appears in G, then → → − Towards a contradiction, suppose that χ (G) = 2. Let D be a proper 2-orientation of G. Note that every triangle (clique of size three) in G contains exactly 1 vertex of in-degree 0 in D. For any three triangles K, K ′ , K ′′ in G such that K ∩ K ′ 6= ∅ and K ∩ K ′′ 6= ∅, by Proposition 4.5, there is K ∩ K ′ ∩ K ′′ 6= ∅. Therefore, G does not contain any of the graphs in Figures 10b and 10c. Moreover, it is easy to check that, since G is a unit interval graph and K ∩ K ′ ∩ K ′′ 6= ∅, we have that G contains the graph in Figure 10a, and the middle vertex of the path is of in-degree 0 in D. But, as it is easy to check, there exists no proper 2-orientation of the graph in Figure 10a with the middle vertex of in-degree 0. A contradiction. 11 By Proposition 4.5, in any proper 2-orientation of G, any edge xy with x in a triangle and y not in this triangle must be oriented from x to y. Therefore, if G contains the induced subgraph depicted in Figure 10d, there is no way to orient the edge between the two triangles in a proper 2-orientation of G. Now, if G contains an induced diamond H (two triangles intersecting on an edge) as subgraph, there − exists only one way (up to symmetry) to orient it in a proper way respecting that → χ (G) = 2. See Figure 11. In particular, the two vertices x and y that do not belong to the intersection of the triangles must have in-degree 2 and the edges between x (resp., y) and vertices not in V (H) must be oriented from x (resp., from y). It can be checked that there is no way to extend this orientation in the graphs of Figures 10e to 10h − while keeping → χ (G) = 2. y x Figure 11: Single proper orientation (up to symmetry of the central edge) of a diamond subgraph of a graph − G with → χ (G) = 2. Now, let us show that any unit interval graph G with ω(G) = 3 and with no induced subgraphs of − χ (G) = 2. Figure 10 satisfies → Let P = (v1 , . . . , vn ) be such that G is a subgraph of P 2 . Let X ⊆ V be the set of vertices not in any triangle. Let B1 , . . . , Br be the connected components (called blocks) of G − X, in the order they appear in P (i.e., for any vo ∈ Bi and vp ∈ Bj with i < j, then o < k). Note that, by the assumption, each Bi is isomorphic to some graph depicted in Figure 12. vi vi+1 vi+2 vi vi+1 vi vi+1 vi+2 vi+2 vi+3 vi+4 vi+3 Figure 12: Possible blocks. Let Q1 , . . . , Qℓ be the maximal subpaths with at least two vertices of P with internal vertices not in any triangle (note that the end of these paths may belong to some triangle) in the order they appear in P (note that r − 1 ≤ ℓ ≤ r + 1). If v1 ∈ V (B1 ) (in which case ℓ ≤ r), we set P1 = {v1 } and Pi = Qi−1 for every 2 ≤ i ≤ ℓ + 1, and otherwise, Pi = Qi for every 1 ≤ i ≤ ℓ. If vn ∈ Br (in which case ℓ ≤ r), we set Pr+1 = {vn }. Note that the Pi ’s are pairwise vertex-disjoint and that, for every 1 < i ≤ r, |V (Pi )| ≥ 3 6 3 (because (because of the forbidden induced subgraph depicted in Figure 10d), and |V (P1 )|, |V (Pr+1 )| = of the forbidden induced subgraph depicted in Figure 10e). Moreover, note that, for every 1 ≤ i ≤ r, Bi intersects Pi in one vertex, denoted by li (the “left” vertex of Bi and the “right” end of Pi ), and Pi+1 in one vertex, denoted by ri (the “right” vertex of Bi and the “left” end of Pi+1 ). Let r be the number of blocks. Note that r ≥ 1 since ω(G) = 3. We show by induction on 1 ≤ j ≤ r, that: S • if Bj is a diamond, then there exists a proper 2-orientation of Yj = G[Pi ∪ Bi ] such that rj has 1≤i≤j in-degree 2; 12 • otherwise, there are two proper 2-orientations of Yj one in which rj has in-degree 1 and another such that rj has in-degree 2. Furthermore, we also prove that when j = r, then one can extend a proper 2-orientation of Yr to G. If B1 is a diamond, let us orient P1 (if P1 6= {v1 }) in such a way that l1 has in-degree 0 (in P1 ), and the neighbour of l1 has in-degree different than 2 (looking at Figure 13, it is possible since the length of P1 is not 2 by assumption). Then, orient B1 as in Figure 14b, so that r1 has in-degree 2. If B1 is a single triangle or two triangles intersecting in one vertex, let us give to P1 (if P1 6= {v1 }) any orientation given in Figure 13 (so l1 has in-degree 0) and then: if B1 is one triangle, then orient B1 in such a way that l1 has in-degree 0 (Figure 14a, and, if B1 consists of two triangles intersecting in one vertex, then orient B1 in such a way that l1 has in-degree 2 (Figure 14c). In both cases, there are two possible such orientations, one leading r1 with in-degree 1 and the other with r1 with in-degree 2 (see Figures 14a and 14c). S G[Pi ∪ Bi ] Let us assume, by induction on j where 1 ≤ j ≤ r, that a proper 2-orientation of Yj = 1≤i≤j has been obtained for some 1 ≤ j ≤ r, such that: if Bj is a diamond, then rj has in-degree 2, and otherwise, we have both a proper 2-orientation of Yj such that rj has in-degree 1 and a proper 2-orientation of Yj such that rj has in-degree 2. Note that this is satisfied for j = 1 by the previous paragraph. If j = r and Pr+1 6= {vn } (as otherwise we already have a proper 2-orientation of G), then a proper 2-orientation of Pr+1 extending the one of Yr can be found according to Figure 13. Recall that, if Br is a diamond, then, by assumption, Pr+1 has length different than 2 and so the neighbour of rr in Pr+1 can be chosen with in-degree different than 2. Hence, one can obtain a proper 2-orientation of G. Suppose then that 1 ≤ j < r. There are several cases to consider depending whether Bj and Bj+1 are diamonds or not. • If Bj and Bj+1 are not diamonds. Consider a proper 2-orientation of Yj such that rj has in-degree 1, which exists by induction hypothesis. Give to Pj+1 any orientation with the neighbour of its left end rj+1 having in-degree 6= 1 (which exists as in Figure 13 since, by assumption, the length of Pj+1 is at least 2). Finally, – if Bj+1 is a triangle, there are two possible orientations of Bj+1 such that lj+1 has in-degree 0: one with rj+1 with in-degree 1 and the other one with rj+1 with in-degree 2 (see Figure 14a); – otherwise Bj+1 consists of two triangles intersecting in one vertex, and then there are two possible orientations of Bj+1 such that lj+1 has in-degree different from the left neighbour of lj+1 in Pj+1 : one with rj+1 with in-degree 1 and the other one with rj+1 with in-degree 2 (see Figure 14c). • If Bj is not a diamond and Bj+1 is a diamond. Orient Bj+1 as in Figure 14b and so rj+1 and lj+1 have both in-degree 2. Then, give to Pj+1 an orientation such that lj+1 has in-degree 2. As shown in Figure 13, here there exists such orientation in which the right neighbour of rj has in-degree 1 or, resp., 2 because Pj+1 has length 4 or larger than 6, by assumption. Then, it can be completed into a proper orientation of Yj+1 by using a proper 2-orientation of Yj such that rj has in-degree 2 or, resp., 1 that exists by induction. • If Bj is a diamond and Bj+1 is not a diamond. Then, by induction, there exists a proper 2-orientation of Yj such that rj has in-degree 2. Then, orient Pj+1 using an orientation such that the neighbour of rj in Pj+1 has in-degree 1. Note that the in-degree of the left neighbour of lj+1 is either 1 or 2 (see Figure 13). Then, – if Bj+1 is a triangle, there are two possible orientations of Bj+1 such that lj+1 has in-degree 0: one with rj+1 with in-degree 1 and the other one with rj+1 with in-degree 2 (see Figure 14a); – otherwise Bj+1 consists of two triangles intersecting in one vertex and there are two possible orientations of Bj+1 such that lj+1 has in-degree different from its left neighbour: one with rj+1 with in-degree 1 and the other one with rj+1 with in-degree 2 (see Figure 14c). 13 1 0 0 2 0 0 2 1 0 (a) Single possible orientations for paths of length at most 3 (up to left/right symmetry). 0 2 0 2 0 0 1 2 1 0 (b) Single possible orientations for paths of length 4. Note that no such orientation allows the two neighbours of the ends to have different in-degree. 0 2 0 2 1 0 0 2 0 1 2 0 (c) Single possible orientations for paths of length 5 (up to symmetry). Note that no such orientation allows the two neighbours of the ends to have both in-degree 1. 0 2 0 2 0 2 0 0 1 2 0 2 1 0 0 1 2 1 0 2 0 (d) Possible orientations for paths of length 2ℓ + 1, for ℓ ≥ 3. Note that, it is possible to ensure that the neighbours of the ends have in-degree 1 or 2. The case ℓ = 2 can be extended to any path of length 2ℓ + 1, ℓ ≥ 3, as follows. Let uv ∈ E such that the in-degree of u equals 0. Then subdivide twice uv into a path (u, x, y, v) with arcs ux, xy and vy. 0 2 0 2 0 1 2 0 0 1 2 1 0 2 1 0 0 1 2 1 0 1 2 0 (e) Possible orientations for paths of length 2ℓ, for ℓ ≥ 4. Note that, it is possible to ensure that the neighbours of the ends have in-degree 1 or 2. The case ℓ = 4 can be extended to any path of lenght 2ℓ, ℓ ≥ 4, as follows. Let uv ∈ E such that the in-degree of u equals 0. Then subdivide twice uv into a path (u, x, y, v) with arcs ux, xy and vy. Figure 13: Orientations of paths such that both ends are sources (but in the case of the path of length one). Integers in the nodes refer to their in-degree. • If both Bj and Bj+1 are diamonds. Then path Pj+1 between Bj and Bj+1 is of length 4 or at least 6 by assumption. It is possible to orient it in such a way that the neighbours of lj and rj+1 have in-degree different from 2 (see Figure 13). Therefore, we may orient Bj+1 as in Figure 14b such that we obtain a proper 2-orientation of Yj+1 with rj+1 with in-degree 2. So, we are ready to prove the main result of this section. − Theorem 4.7. For any unit interval graph G with ω(G) = 3, then → χ (G) ∈ {2, 3}. Moreover, there exists a → − linear-time algorithm to compute χ (G). 14 1 0 2 1 0 1 2 2 1 0 2 0 1 2 0 0 2 1 (a) Possible orientations of K3 (recall that the order of the vertices (from left to right) is the same as in P ). 2 1 0 2 (b) Single orientation of the diamond (up to reverse the central edge). 1 2 0 2 1 1 2 0 1 2 2 1 0 1 2 1 2 0 2 1 (c) Possible orientation of two triangles intersecting in one vertex (by Proposition 4.5, this common vertex must have in-degree 0. Figure 14: Possible proper 2-orientations of the blocks (where vertices are ordered from left to right as they appear in P ). − − Proof. The fact that → χ (G) ∈ {2, 3} follows from the facts that → χ (G) ≥ ω(G) − 1 for any graph G and from Lemma 4.4. The linear complexity comes from the fact that computing a Hamiltonian path P of G such that G is a subgraph of P 2 can be done in linear time [17], that the forbidden induced subgraphs and the blocks can be identified in linear time (by following the path P ) and by Lemma 4.6. − The above section gives an ad-hoc proof that → χ (G) ≤ ω(G) and an ad-hoc linear-time algorithm to → − compute χ (G) for unit interval graphs with ω(G) = 3. Similar results might be obtained by tedious case analysis for larger (but still constant) clique number. However, it would be interesting to obtain such general bound for all (unit) intervalgraphs. Recall that Proposition 4.3 provides a family of unit interval  − graphs with → χ (G) = 23 (ω(G) − 1) . Note that, in [11] (arXiv), it was stated that, for every graph G, log(χ(G)) mad(G) → − χ (G) = O( χ(G) . Supposing the provided proofs are correct, we obtain that, for any unit log log χ(G) ) + 2 ω(G) log(ω(G)) → − interval graph G, χ (G) = O( ) + ω(G) − 1, since, for any unit interval graph G, χ(G) = ω(G) log log ω(G) ω(G) (since G is a perfect graph) and mad(G) ≤ mad(Pn ) ≤ 2(ω(G) − 1). − Note that → χ (G) can be computed in linear time in interval graphs with bounded ω(G) by the dynamic programming for bounded treewidth graphs [7]. 5 Cobipartite graphs A graph is cobipartite if its vertex set has a partition (K1 , K2 ) such that both K1 and K2 are cliques. Such a partition is called a cobipartition of the graph. We assume for the remainder of the section that K1 is the largest clique of a cobipartite graph G with clique partition {K1 , K2 } of V (G). The aim of this section is to prove the following theorem. 15 Theorem 5.1. Given a cobipartite graph G with cobipartition (K1 , K2 ), one can decide in polynomial time →(G) = |K | − 1, assuming that |K | = max{|K |, |K |}. whether − χ s 1 1 1 2 →(G) = |K | − 1, note that any optimal weighted orientation when restricted to G[K ] To prove that − χ s 1 1 must be transitive with all weights equal to 1; and all edges from K1 to K2 must be oriented towards K2 . Thus, the goal in what follows it to decide which transitive orientation of K1 to pick, how to assign weights to the edges from K1 to K2 and, overall, which weighted orientation to use in K2 satisfying the upper bound of |K1 | − 1 and the constraints of being proper. → − We need some preliminary results. Let k1 and k2 be two positive integers. Let d = (d1 , . . . , dk2 ) be a → − sequence of non-negative integers. d is k1 -suitable if and only if there is a tournament T with vertex set {v1 , . . . , vk2 } such that, for every j ∈ [k2 ], there are at most j −1 indices of [k2 ] such that di +d− T (vi ) > k1 −j. →(G) = k − 1. Let us first characterize graphs with − χ s 1 Lemma 5.2. Let G be a cobipartite graph G with cobipartition (K1 , K2 ). Set k1 = |K1 | and k2 = |K2 |, and let K2 = {v1 , . . . , vk2 }. − →(G) = k − 1 if and only if the two following statements hold: χ s 1 (i) There exists an antimatching saturating K2 in the bipartite graph induced by E(K1 , K2 ); and (ii) There exists an ordering (v1 , . . . , vk2 ) of the vertices of K2 such that (dK1 (v1 ), . . . , dK1 (vk2 )) is a k1 suitable sequence. →(G) = k − 1. Let (D, w) be an optimal semi-proper orientation of G. Set T = DhK i. Proof. Assume − χ s 1 2 Since K1 is a clique, by Proposition 4.5, for each i ∈ {0, . . . , k1 }, there is exactly one vertex of K1 with in-weight i, and all the edges of E(K1 , K2 ) are oriented towards K2 . Now K2 is also a clique, so the in-weights of its vertices are all distinct. For each i ∈ [k2 ], there is a unique vertex ui ∈ K1 such that S(D,w) (ui ) = S(D,w) (vi ). Since (D, w) is semi-proper, then ui and vi are not adjacent. Hence M = {ui vi | i ∈ [k2 ]} is an antimatching saturating K2 in the bipartite graph induced by E(K1 , K2 ). So (i) holds. Furthermore, since the in-weight of the vi are all distinct, there are at most j − 1 vertices vi such that − S(D,w) (vi ) ≥ k1 − j + 1. But S(D,w) (vi ) ≥ d− D (vi ) = dK1 (vi ) + dT (vi ). So there are at most j − 1 vertices vi such that dK1 (vi ) + d− T (vi ) ≥ k1 − j + 1. Hence (dK1 (v1 ), . . . , dK1 (vk2 )) is a k1 -suitable sequence, so (ii) holds. Reciprocally, assume that (i) and (ii) hold. Let T be an orientation of K2 such that for every j ∈ [k2 ], there are at most j − 1 indices of [k2 ] such that di + d− T (vi ) > k1 − j. Free to relabel the vertices of K2 , we − − may assume that dK1 (v1 ) + d− (v ) + d (v ) ≥ d (v ) 1 K1 2 2 ≥ . . . ≥ dK1 (vk2 ) + dT (vk2 ). Consequently, for every T T − i ∈ [k2 ], dK1 (vi ) + dT (vi ) ≤ k1 − i. Let us construct a weighted orientation D of G as follows. First DhK2 i = T and all the edges of E(K1 , K2 ) are oriented towards K2 . Let i ∈ {2, . . . , k2 }. Choose an arc ei of A(T ) ∪ E(K1 , K2 ) entering vi ; − − − Such an arc exists since dK1 (vi ) + d− T (vi ) ≥ dK1 (vk2 ) + dT (vk2 ), and dK1 (vi ) + dT (vi ) + dK1 (vk2 ) + dT (vk2 ) ≥ |A(T h{vi , vk2 }i)| ≥ 1; assign to all arcs entering vi a weight of 1 except for ei to which we assign weight k1 − i − (dK1 (vi ) + d− T (vi )) + 1. Doing so, S(D,w) (vi ) = k1 − i for all i ∈ {2, . . . , k2 }. Assign to all arcs entering vk2 a weight of 1. Then S(D,w) (vk2 ) = dK1 (vk2 ) + d− T (vk2 ) ≤ k1 − k2 . Hence all the in-weights of the vertices of K2 are distinct. Now by (ii), there is an antimatching M saturating K2 . Take a transitive orientation T1 of K1 such that d− T1 (M (ui )) = S(D,w) (vi ) for all i ∈ [k2 ], and assign weight 1 to each of its arcs. One easily checks that the resulting weighted orientation of G is semi-proper. Let us now characterize k1 -suitable sequences. → − Lemma 5.3. Let k1 , k2 be two integers with k1 ≥ k2 , and let d = (d1 , . . . , dk2 ) be a non-increasing sequence → − of non-negative integers. Then d is k1 -suitable if and only if there is a tournament T with vertex set {v1 , . . . , vk2 } such that di + d− T (vi ) ≤ k1 − i for all i ∈ [k2 ]. 16 Proof. By definition if there is a tournament T with vertex set {v1 , . . . , vk2 } such that di + d− T (vi ) ≤ k1 − i for all i ∈ [k2 ], then D is k1 -suitable. → − Let us now prove the reciprocal. Assume that d is k1 -suitable. A tournament T with vertex set → − {v1 , . . . , vk2 } is a d -partner if for every j ∈ [k2 ], there are at most j −1 indices of [k2 ] such that di +d− T (vi ) > → − − k1 −j. Observe that if T is a d -partner, then there is a permutation σ of [k2 ] such that di +dT (vi ) ≤ k1 −σ(i) for every i ∈ [k2 ]. Such a permutation is an adjoint of T . The displacement index of T of a permutation σ of [k2 ] is the smallest index such that σ(i) 6= i or k2 + 1 of σ is the identity. → − → − → − Since d is k1 -suitable, by definition there is a d -partner. Consider the pair (T, σ) where T is a d -partner → − of d and σ is an adjoint of T that maximizes the displacement index of σ. We shall prove that σ is the identity, and thus di + d− T (vi ) ≤ k1 − i for all i ∈ [k2 ]. Suppose, for a contradiction, that σ is not the identity. Let i be the displacement index of σ, and let j = σ −1 (i). By definition, j, σ(i) > i, so di ≥ dj . We have di + d− T (vi ) − dj + dT (vj ) ≤ ≤ k1 − σ(i) ≤ k1 − i k1 − i ′ ′ ′ ′ If dj + d− T (vj ) ≤ k1 − σT (i), then the permutation σ defined by σ (i) = i, σ (j) = σ(j) and σ (k) = σ(k) for all k ∈ [k2 ] \ {i, j}, is also an adjoint of T and has a larger displacement index than σ, a contradiction. − Henceforth, we may assume that dj + d− T (vj ) > k1 − σ(i). Set α = dj + dT (vj ) − k1 + σ(i). Observe that vj has k1 − σ(i) − dj + α in-neighbours in T while vi has at most k1 − σ(i) − di ≤ k1 − σ(i) − dj in-neighbours in T . Therefore, there exists a set A of α in-neighbours of vj which are not in-neighbours of vi . Let T ′ be tournament obtained from T by reversing the direction of the arc between a and vi and vj if a ∈ A \ {vi } and reversing the arc vi vj if vi ∈ A. All vertices distinct from vi and vj have same in-degree in T ′ and T , − − − while d− T ′ (vi ) = dT (vi ) + α and dT ′ (vj ) = dT (vj ) − α. Hence di + d− T ′ (vi ) ≤ k1 − σ(i) + α = dj + d− T (vj ) ≤ k1 − i dj + d− T ′ (vj ) = dj + d− T ′ (vj ) − α = k1 − σ(i) Thus the permutation σ ′ defined by σ ′ (i) = i, σ ′ (j) = σ(j) and σ ′ (k) = σ(k) for all k ∈ [k2 ] \ {i, j}, is an adjoint of T ′ and has a larger displacement index than σ, a contradiction. A score sequence is a non-decreasing sequence of integers corresponding to the in-degrees of the vertices of a tournament. Let us recall a characterization of score sequences by Landau. − s = (s1 , . . . , sn ) be a non-decreasing sequence of non-negative integers. Theorem 5.4 (Landau [22]). Let →     k n X X k n − for all k ∈ [n]. si ≥ , and si = Then, → s is a score sequence if and only if 2 2 i=1 i=1 With the following two corollaries, we will get an algorithmic tool for checking if a sequence is k1 -suitable. − Corollary 5.5. Let → s = (s1 , . . . , sn ) be a non-decreasing sequence of non-negative integers. There is a tournament T with vertex set {v1 , . . . , vn } such that d− T (vi ) ≤ si for all i ∈ [n] if and only if   k X k for all k ∈ [n]. si ≥ 2 i=1 Proof. Assume first that there is a tournament T with vertex set {v1 , . . . , vn } such that d− T (vi ) ≤ si for all i ∈ [n]. Then for every k ∈ [n],   k k X X k = |A(T h{v1 , . . . , vk }i)| ≤ (v ) ≤ d− si . i T 2 i=1 i=1 17 
Pn Let us now prove the reciprocal by induction on Exc = i=1 si − n2 . If Exc = 0, then the results follows from Landau’s Theorem. Assume now that Exc > 0. Let r be the smallest integer such that sr = sn . Set s′r = sr − 1 and s′i = si for all i ∈ [n] \ {r}. We shall prove that
Pk k ′ i=1 si ≥ 2 for all k ∈ [n]. Thus, by the induction hypothesis, there is a tournament T with vertex set ′ ′ we get the result. {v1 , . . . , vn } such that d− T (vi ) ≤ si for all i ∈ [n]. Since si ≤ si for all i ∈ [n],
Pk k ′ Therefore assume for a contradiction that there exists k such that i=1 si < 2 . Choose such a k that is minimum. Clearly r ≤ k < n (in particular, k < n because Exc > 0), so sk = sn . Moreover

Pk Pk−1 = k. So sk ≤ s′k + 1 ≤ k, and thus si ≤ k for all k ≤ i ≤ n. Hence s′k = i=1 s′i − i=1 s′i < k2 − k−1 2 n X i=1 si = k X i=1 si + n X i=k+1 k X     n k + k(n − k) < si + k(n − k) ≤ si ≤ 2 2 i=1 because k < n. This a contradiction and completes the proof. Corollary 5.6. Checking if a sequence is k1 -suitable can be done in polynomial time. → − Proof. Let d = (d1 , . . . , dk2 ) be a sequence. Free to permute the indices, we may assume that the sequence → − → − is non-increasing. By Lemma 5.3, d is d is k1 -suitable if and only if there is a tournament T with vertex set {v1 , . . . , vk2 } such that d− T (vi ) ≤ k1 − i − di for all i ∈ [k2 ]. By Corollary 5.5, there is such a tournament   k X k k 1 − i − di ≥ if and only if for all k ∈ [k2 ]. Each of this k2 inequalities can easily be checked in 2 i=1 linear time. Now we are ready to complete the proof of Theorem 5.1. Proof of Theorem 5.1. Let G be a cobipartite graph with cobipartition (K1 , K2 ). Set k1 = |K1 |. By →(G) = k − 1 if and only if the following hold. χ Lemma 5.2, − s 1 (i) There is an antimatching saturating K2 in the bipartite graph induced by E(K1 , K2 ). (ii) (dK1 (v1 ), . . . , dK1 (vk2 )) is a k1 -suitable sequence. Checking if there is an antimatching saturating K2 can be done in polynomial time using any matching algorithm in the complement. Checking if a sequence is k1 -suitable can be done in polynomial time by Corollary 5.6. Nevertheless, the following problems remain open. →(G) = |K |−1+ℓ Problem 5.7. For any fixed positive integer ℓ, what is the complexity of deciding whether − χ s 1 for a given cobipartite graph with cobipartition (K1 , K2 )? − Problem 5.8. What is the complexity of deciding whether → χ (G) = |K | − 1 for a given cobipartite graph 1 with cobipartition (K1 , K2 )? Problem 5.9. Is it possible to characterize the cobipartite graphs G with cobipartition (K1 , K2 ) such that − →(G) ≤ |K | + |K | − 2 ? χ s 1 2 6 Parameterized Complexity We now study the parameterized complexity of determining if the semi-proper orientation number of a graph G is at most k when parameterized by the size of a minimum vertex cover in G, neighbourhood diversity of G plus k, treewidth of G, and only by the value of k. The result for neighbourhood diversity plus k, and the kernel for cographs also hold for the proper orientation number. Some results in this section use similar ideas to the ones presented in [4]. 18 6.1 Vertex Cover and Neighbourhood Diversity Given a graph G = (V, E), a subset S ⊆ V is a vertex cover of G if, for every uv ∈ E, we have that u ∈ S or v ∈ S. Then, define vc(G) = min{|S| | S is a vertex cover of G}. With these definitions, let us present the first parameterization. Semi-Proper Input: Parameter: Question: Orientation Parameterized by Vertex Cover A graph G and a positive integer k. vc(G). →(G) ≤ k? Is − χ s Let us prove that Semi-Proper Orientation Parameterized by Vertex Cover is FPT by showing that it admits an exponential kernel. Theorem 6.1. Semi-Proper Orientation Parameterized by Vertex Cover has a kernel of size O(4vc(G) vc(G)2 ). Proof. Let (G, k) be an instance of Semi-Proper Orientation Parameterized by Vertex Cover where G is a connected graph. We construct, in polynomial time, an equivalent instance (G′ , k) such that |V (G′ )| = O(4vc(G) vc(G)2 ). First, since G admits a vertex cover of size vc(G), then its vertex-set can be partitioned into two sets A and B such that |A| = vc(G) and B is an independent set. It follows that the degeneracy of G satisfies →(G) ≤ 2 vc(G) as it is always true that − →(G) ≤ 2δ ∗ (G) [14]. δ ∗ (G) ≤ vc(G). Thus, we know that − χ χ s s Therefore, if k ≥ 2 vc(G), by the previous argument, the answer to Semi-Proper Orientation Parameterized by Vertex Cover is “YES” (and one can just output G′ as a trivial vertex). Hence, let us assume that k < 2 vc(G). By using the well-known 2-approximation algorithm for Vertex Cover, one may obtain in polynomial time a subset C ⊆ V (G) such that C is a vertex cover of G and |C| = γ ≤ 2 vc(G). Thus, notice that V (G) can be partitioned into C and an independent set S. We partition the vertices of S into at most 2γ − 1 non-empty subsets Si of twins, for i ∈ {1, . . . , 2γ − 1} (i.e., two vertices are in a same part if and only if they are twins). Let M = kγ. Claim 6.1.1. If there exists i ∈ {1, . . . , 2γ − 1} such that |Si | ≥ M + 2, then removing (|Si | − (M + 1)) →(G) ≤ k if, and only if, − →(G′ ) ≤ k. vertices from Si builds a smaller graph G′ such that − χ χ s s →(G′ ) ≤ k, then χ Proof: Since the problem is monotone under subgraphs [14], we just need to prove that if − s − → χ (G) ≤ k. s Let Si′ be the M + 1 remaining vertices of Si in G′ . Note that, if (D′ , w′ ) is a semi-proper k-orientation of G′ such that exactly ℓ edges with one end in C and the other in Si′ are oriented towards C, then: P − ℓ v∈C dD ′ (v) ≥ . (v) ≥ k ≥ max d− ′ D v∈C γ γ Therefore, ℓ ≤ kγ = M ≤ 4 vc2 (G). So, for every i ∈ {1, . . . , 2γ − 1} and every semi-proper k-orientation, there are at most M edges oriented from Si′ to C. Since |Si′ | = M + 1, note that Si′ has a sink s in D′ . In order to build a semi-proper k-orientation (D, w) of G, it suffices to extend (D′ , w′ ) to G by making all the |Si | − (M + 1) removed vertices of Si as sinks and having the same in-weight as s.  By applying Claim 6.1.1 to each feasible Si , i ∈ {1, . . . , 2γ − 1}, we build a graph G′ as required. Note that |G′ | ≤ γ + (2γ − 1)(M + 1) = O(4vc(G) vc(G)2 ). The above result still leaves open the problem of existence of a polynomial kernel. Problem 6.2. Does there exist a polynomial kernel to Semi-Proper Orientation Parameterized by Vertex Cover? 19 Let us move to our second parameterization, based on the concept of neighbourhood diversity. The neighbourhood diversity of a graph G, denoted by nd(G), is the minimum k such that G can be partitioned into at most k subsets of twin vertices [21]. It is well known that if G has a vertex cover S satisfying |S| ≤ k, then G can be partitioned into at most 2k + k subsets of twin vertices (similarly to the proof of Theorem 6.1). →(G) ≤ k, resp., → − In the sequel, we prove that deciding whether − χ χ (G) ≤ k, is FPT parameterized by the s neighbourhood diversity of G plus k. Semi-Proper Input: Parameter: Question: Orientation Parameterized by Neighbourhood Diversity plus Solution Value A graph G and a positive integer k. nd(G) + k. →(G) ≤ k? Is − χ s Proper Orientation Parameterized by Neighbourhood Diversity plus Solution Value Input: A graph G and a positive integer k. Parameter: nd(G) + k. − Question: Is → χ (G) ≤ k? Let us first present a technical lemma and then prove the main result on parameterization by neighborhood diversity plus Solution value. − →(G) ≤ k and there are Lemma 6.3. Given a positive integer k, if G is a graph such that → χ (G) ≤ k or − χ s S1 , S2 ⊆ V (G) such that S1 ∩ S2 = ∅ and uv ∈ E(G) for every u ∈ S1 and v ∈ S2 , then min{|S1 |, |S2 |} ≤ 2k. Proof. For purpose of contradiction, let us assume that |S1 |, |S2 | > 2k. In any optimal proper orientation 2| > 2k 2 arcs oriented towards D (resp. semi-proper orientation (D, w)) of G, there are at least |S1 |·|S 2 Si , for some i ∈ {Si }, say S1 . Hence, one vertex of S1 has (resp. weighted) in-degree at least k + 1, a contradiction. Theorem 6.4. Semi-Proper Orientation Parameterized by Neighbourhood Diversity plus Solution Value and Proper Orientation Parameterized by Neighbourhood Diversity plus Solution Value have a kernel of size at most nd(G)(2k 2 + 1). Proof. Let (G, k) be an instance of Semi-Proper Orientation Parameterized by Neighbourhood Diversity plus Solution Value. We may assume that G is connected. Let nd(G) = d. In linear time, we can obtain a partition P = {N1 , . . . , Nd } such that each Ni is composed by twin vertices, for each i ∈ {1, . . . , d}. Let i ∈ {1, . . . , d}. By definition, note that the subgraph of G induced by Ni is either a clique or an independent set. In the case when G[Ni ] is a complete subgraph, if |Ni | ≥ k + 2, the instance (G, k) is trivially a “NO” instance and our kernelization algorithm just outputs the equivalent instance (Kk+2 , k). Let us consider a set Ni which induces a stable set. Since G is connected, there must be a set Nj , j 6= i, such that there are all edges between the vertices of Ni and the vertices of Nj . By Lemma 6.3, either |Ni | ≤ 2k or |Nj | ≤ 2k. Let us consider the latter case. If |Ni | > 2k 2 + 1, let us build an equivalent instance (G′ , k) by removing |Ni | − (2k 2 + 1) vertices from Ni . − − Let us show then that → χ (G) ≤ k if, and only if → χ (G′ ) ≤ k. Let D be a proper k-orientation of G. If 2 2 |Ni | > 2k , then at least |Ni | − 2k vertices of Ni must be sinks. Indeed, otherwise, at least 2k 2 + 1 arcs are oriented from Ni to Nj and, since |Nj | ≤ 2k, at least one vertex of Nj has in-degree at least k + 1, a contradiction. Let S ⊆ Ni be the set of sinks and let v ∈ S. Let D′ = D − (S \ {v}). Then D′ is a proper k-orientation of G′ . Reciprocally, let D′ be a proper k-orientation of G′ . By the same reasoning (since Ni has size 2k 2 + 1 in G′ ), there exists at least one sink v ∈ Ni in D′ . Let us build D by copying D′ and making the extra |Ni | − (2k 2 + 1) vertices of Ni as sinks. Then, D is a proper k-orientation of G. Repeating this process for all Ni , we build an equivalent instance G′ such that, for every 1 ≤ i ≤ d = nd(G), |Ni | ≤ 2k 2 + 1. Hence, |V (G′ )| ≤ nd(G)(2k 2 + 1). →(G′ ) ≤ k then, − →(G) ≤ k one can use →(G′ ) ≤ − →(G) by Lemma 1.3. To prove that if − χ χ Note that − χ χ s s s s analogous arguments as in the previous case. 20 The above result shows that parameterizing Semi-Proper Orientation by the neighborhood diversity plus the solution value not only leads to an FPT algorithm, but even yields a polynomial kernel. It leaves open the question whether adding the solution value is necessary for an FPT algorithm or a polynomial kernel. Problem 6.5. Id Semi-Proper Orientation is an FPT problem when parameterized by the neighbourhood diversity? 6.2 Treewidth Let us now present a parameterization based on the concept of tree-decomposition. Recall that a treedecomposition of a graph G = (V, E) is a pair (T, X ) where T is a tree and X = {Xv | v ∈ V (T )} is a family of subsets of V such that: S (a) v∈V (T ) Xv = V ; (b) for every e = uw ∈ E, there exists v ∈ V (T ) with u, w ∈ Xv ; (c) for every u ∈ V , {v ∈ V (T ) | u ∈ Xv } induces a subtree of T . The width w(D) of D = (T, X ) equals max |Xv | − 1, and the treewidth of G, denoted by tw(G), is v∈V (T ) the minimum width of its tree-decompositions. A tree-decomposition (T, X ) of G of width t is nice if T is a rooted tree with maximum degree 3, the root has at most two children and every node v ∈ V (T ) satisfies one of the following: Leaf node: v is a leaf (and not the root) and |Xv | = t + 1; Forget node: v has one child v ′ and, there exists u ∈ V such that Xv = Xv′ \ {u}; Introduce node: v has one child v ′ and, there exists u ∈ V such that Xv = Xv′ ∪ {u}; Join node: v has two children v ′ and v ′′ and Xv = Xv′ = Xv′′ . It is well known that, for any graph G on n vertices, a nice tree-decomposition D = (T, X ) of G with tw(G) ≤ w(D) ≤ t such that |V (T )| = O(n) can be computed in linear time from any tree-decomposition with width at most t of G [8]. In the sequel, we prove that the following problem is FPT: Semi-Proper Input: Parameter: Question: Orientation Parameterized by Treewidth A graph G and a positive integer k. tw(G). →(G) ≤ k? Is − χ s → − Let G = (V, E) be a graph, F ⊆ E and let w′ : F → {1, 2} and F be an orientation of the edges in F . A → − → − → − → − weighted orientation ( G , w : E → {1, 2}) of G extends (w′ , F ) if w ↾F = w′ and F agrees with G on F , → − → − i.e. if uv ∈ F , then (u, v) ∈ A( G ) if, and only if, (u, v) ∈ F . Theorem 6.6. Let t ≥ 1. For any k ≥ 0 and any graph G on n vertices with treewidth at most t, deciding →(G) ≤ k can be computed in 4O(t2 ) n time. whether − χ s →(G) ≤ 2δ ∗ (G) [14] Proof. Let k ≥ 0 and G be any graph on n vertices with treewidth at most t. Because − χ s − → ∗ and δ (G) ≤ tw(G) ≤ t, if k ≥ 2t, then χ (G) ≤ k. Hence, we may assume that k < 2t. s First, a nice tree-decomposition (T, X ) with O(n) nodes and of width at most 2t of G can be computed in 2O(t) n time [20]. Let r ∈ V (T ) be the root of TS. For every v ∈ V (T ), let Tv be the subtree of T rooted in v and let Gv be the subgraph of G induced by h∈V (Tv ) Xh . Let us design a bottom-up dynamic programming algorithm that computes, for every v ∈ V (T ), the set Ov of all the triples (L, C, O′ ) where L : Xv → 2{0,...,k} , C : Xv → {0, . . . , k} and O′ is a weighted orientation of G[Xv ] such that there exists a weighted orientation O = (D, w) of Gv satisfying: 21 • S(D,w) ≤ k; • O extends O′ ; and • for every vertex u ∈ Xv , SD,w (u) = C(u) and L(u) = {0, . . . , k} \ {SD,w (x) | x ∈ NGv (u) \ Xv }. Intuitively, C(u) represents the current colour of u in the weighted orientation O that extends O′ and L(u) corresponds to the set of available colours that u may use later (the ones that are not the colours of the neighbours of u in V (Gv ) \ Xv ). For every leaf ℓ ∈ V (T ), let SOℓ = {(D, w) | (D, w) is semi-proper k-orientation of Gℓ )}. For every v ∈ Xℓ and (D, w) ∈ SOℓ , define C(D,w) (v) = S(D,w) (v) and L(v) = {0, . . . , k}. Thus, define Oℓ = {(L, C(D,w) , O′ ) | O′ = (D, w) ∈ SOℓ }. Note that |Xℓ | = O(t), so |E(Gℓ )| = O(t2 ) and, since each edge can 2 2 take two values [14] and two orientations, |SOℓ | ≤ 4O(t ) , and so |Oℓ | ≤ 4O(t ) . Now, let v ∈ V (T ) be a non-leaf node and assume that Ov′ has been computed for every child v ′ of v. We explain how to compute Ov from the set(s) of its child(ren). There are three cases to be considered depending on the type of v. Forget node: Let v ′ be the unique child of v and let Xv = Xv′ \ {u}. The set Ov is obtained as follows. For every (L′ , C ′ , O′ ) ∈ Ov′ , let O be the weighted orientation of G[Xv ] such that O′ extends O to ′ G[Xv′ ]. For every x ∈ Xv ∩ N (u), let L(x) = L′ (x) \ {C ′ (u)}. Then, add (L, C↾X , O) to Ov , where v ′ C↾Xv is the restriction of C to Xv . Join node: Let v ′ and v ′′ be the two children of v. The set Ov is obtained as follows. For every (L′ , C ′ , O′ ) ∈ O′ = O′′ , let O = O′ = (D, w) and C = C ′ . Moreover, for every Ov′ and (L′′ , C ′′ , O′′ ) ∈ Ov′ such that P ′ ′′ x ∈ Xv , let C(x) = C (x) + C (x) − u∈Xv ,ux∈A(D) w(ux) and let L(x) = L′ (x) ∩ L′′ (x). If, for all u, x ∈ Xv , C(x) ∈ L(x) and, if ux ∈ A(D) or xu ∈ A(D), then C(x) 6= C(u), then add (L, C, O) to Ov . Introduced node: Let v ′ be the unique child of v and let Xv = Xv′ ∪{u}. The set Ov is obtained as follows. For every (L′ , C ′ , O′ ) ∈ Ov′ and for every weighted orientation O = (D, w) of Xv that extends O′ , let C(u) = S(D,w) (u) and let L(u) = {0, . . . , k}. Then, for every x ∈ Xv \ {u}, let C(x) = C ′ (x) + w(ux) if ux ∈ A(D) and let C(x) = C ′ (x) otherwise; and let L(x) = L′ (x). If, for all u, x ∈ Xv , C(x) ∈ L(x) and, if ux ∈ A(D) or xu ∈ A(D), then C(v) 6= C(u), then add (L, C, O) to Ot . →(G) ≤ k if and only if O 6= ∅, where r is the root of T . Moreover, the number of orientations Clearly, − χ s r 2 to be considered at any node v ∈ V (T ) is at most 4O(t ) and, for each vertex of Xv , the number of possible values of the function L is O(2k ) = O(4t ) since k ≤ 2t and the number of possible values of the function C 2 is k + 1 = O(t). This implies that |Ov | = 4O(t ) · O(tt ) · (4t )t , which gives the desired time complexity. 6.3 Natural parameter In this section, we combine the results of Sections 6.1 and 6.2 to obtain results with parameterization by the value of the solution. Some ideas are similar to the ones presented in [4]. Parameterized Proper Orientation Input: A graph G and a positive integer k. Parameter: k. − Question: Is → χ (G) ≤ k? Parameterized Semi-Proper Orientation Input: A graph G and a positive integer k. Parameter: k. →(G) ≤ k? Question: Is − χ s 22 It is important to recall that Parameterized Semi-Proper Orientation is para-NP-complete in the class of planar bipartite graphs [14]. In the sequel, we argue that a kernel of size polynomial in k is unlikely to exist even for the class of chordal graphs. For this, we need the following definition and result. Definition 6.7 (AND-cross-composition). Let L ⊆ Σ∗ be a language and Q ⊆ Σ∗ × N be a parameterized language. We say that L AND-cross-composes into Q if there exists a polynomial equivalence relation R and an algorithm A, called a cross-composition, satisfying the following conditions. The algorithm A takes as input P a sequence of strings x1 , x2 , . . . , xt ∈ Σ∗ that are equivalent with respect to R, runs in time t polynomial in i=1 |xi |, and outputs one instance (y, k) ∈ Σ∗ × N such that: 1. k ≤ p(maxti=1 |xi | + log t) for some polynomial function p(·), and 2. (y, k) ∈ Q if and only if xi ∈ L, for every i ∈ {1, . . . , t}. Theorem 6.8 (Drucker [15], see also [16]). Assume that an NP-hard language L AND-cross-composes into a parameterized language Q. Then Q does not admit a polynomial compression, unless NP ⊆ coNP/poly. Now, given graphs G1 , . . . , Gp with disjoint vertex sets, for some p ≥ 2, the disjoint union, or simply union, of G1 , . . . Gp is the graph G1 ∪ . . . ∪ Gp such that V (G1 ∪ . . . ∪ Gp ) = V (G1 ) ∪ . . . ∪ V (Gp ) and E(G1 ∪ . . . ∪ Gp ) = E(G1 ) ∪ . . . ∪ E(Gp ). Let us present the following series of results on chordal and split graphs. →(G) = max{− →(G ), . . . , − →(G )}, for any p ≥ 2. Proposition 6.9. If G = G ∪ . . . ∪ G , then − χ χ χ 1 p s s 1 s p Proposition 6.10. Parameterized Semi-Proper Orientation does not admit polynomial kernel for chordal graphs, unless NP ⊆ coNP/poly. Proof. By just taking the disjoint union of t instances of Semi-Proper Orientation in split graphs with the same k (this is the equivalence relation R), we have an AND-cross-composition to an instance of Parameterized Semi-Proper Orientation in chordal graphs as the disjoint union of split graphs is chordal and Proposition 6.9 holds. Thus, it follows, by Theorem 6.8. 2 Proposition 6.11. Parameterized Semi-Proper Orientation can be solved in 4O(k ) · n time for chordal graphs. →(G) ≤ k, when G has treewidth at most t, χ Proof. The algorithm presented Theorem 6.6 decides whether − s 2 in 4O(t ) · n time. Thus, given a chordal graph G, one may compute ω(G) in polynomial time, as they are perfect. In case ω(G) ≥ k + 2, then the answer to Parameterized Semi-Proper Orientation is trivially “NO”. Otherwise, note that a chordal graph with ω(G) ≤ k + 1 is a graph with treewidth at most k. Then, the result follows by Theorem 6.6. Proposition 6.12. Parameterized Semi-Proper Orientation admits a O(2k k 2 )-kernel for split graphs. Proof. Let {K, S} be a partition of V (G) such that K is a maximum clique of G and S is a stable set. Recall that |K| can be computed in linear time as G is chordal. By Proposition 6.9, we can suppose that G is connected. Note that if |K| ≥ k + 2 the answer is trivially “NO” and one can output a trivial equivalent instance as Kk+2 . Otherwise, note that K is a vertex cover of G and thus vc(G) ≤ |K| ≤ k + 1. Consequently, one may just apply the algorithm in Theorem 6.1. Note that, contrary to the algorithm in 6.1, here we directly get a vertex cover K of size at most k +1 (not at most 2k as in Theorem 6.1 that uses a 2-approximation algorithm for Vertex Cover. Therefore, the size of the kernel is slightly improved to O(2k k 2 ) in this case. In view of the above results, it is natural to ask the following. Problem 6.13. Does there exist a polynomial kernel to Parameterized Semi-Proper Orientation in split graphs? 23 Let us now address another dense-graph class admitting an exponential kernel for the natural parameter. Given p ≥ 2 graphs G1 , . . . , Gp with disjoint vertex sets, the join G1 + . . . + Gp of these graphs be the graph obtained from the disjoint union of G1 , . . . , Gp by adding an edge {x, y} for every x ∈ V (Gi ) and y ∈ V (Gj ), for every 1 ≤ i < j ≤ p. The class of cographs is defined recursively as follows: • a trivial graph (with one vertex and no edge) is a cograph; • the join of cographs is a cograph, and • the disjoint union of cographs is a cograph. Any cograph G admits a unique cotree T (G) defined recursively as follows. If G has a unique vertex, then its cotree T (G) is the rooted tree with a single node mapped to the single vertex of V (G). If G is the disjoint union of connected cographs G1 , . . . , Gp , then, the cotree T of G is obtained from T (G1 ), . . . , T (Gp ) by adding a root v (called a union node) linked to the roots of each T (Gi ), 1 ≤ i ≤ p. If G is the join of not connected and/or single vertex cographs G1 , . . . , Gp , then, the cotree T of G is obtained from T (G1 ), . . . , T (Gp ) by adding a root v (called a join node) linked to the roots of each T (Gi ), 1 ≤ i ≤ p. Note that the join nodes and union nodes of T (G) define a bipartition of the internal vertices of T (G). Note also that there is a one-to-one mapping between the leaves of T (G) and V (G). This implies the following: Proposition 6.14. Let G be a cograph with cotree T (G) and k ∈ N. If there exists f : N → N such that T (G) has depth at most f (k) and maximum degree f (k), then |V (G)| ≤ f (k)f (k) . As one may expect at this point, our kernelization algorithm uses the cotree T of a given cograph G to build an equivalent instance of bounded size. In the next lemmas, we show how to bound the depth and the degree of T as well as the number of vertices in G(vi ) for every child vi of an inner node v of T . For every t ∈ V (T (G)), let Gt be the cograph whose cotree is Tt , which is the subtree of T (G) rooted in t. Note that V (Gt ) is a module of G, i.e., for every u ∈ V (G) \ V (Gt ) either {u, w} ∈ E(G) for all w ∈ V (Gt ) or {u, w} ∈ / E(G) for all w ∈ V (Gt ). Given a rooted tree T with root r ∈ V (T ), the depth of T is the maximum length of a r, v-path in T , for some (leaf) v ∈ V (T ). − →(G) ≤ k, then Lemma 6.15. Let G be a connected cograph and k be a positive integer. If → χ (G) ≤ k or − χ s the depth of T (G) is at most 2k + 2. Proof. Let P = (r = t0 , t1 , . . . , tℓ ) be a path from the root r of T (G) to a leaf tℓ with length ℓ. Note that, since r is a join node (because G is connected), t2i is a join node for every integer i ≥ 0. Thus, for every, 0 ≤ i < ℓ/2, let t′i 6= t2i+1 be a child of t2i in T (G) not in P and let vi ∈ G(t′i ). Then G[{v0 , . . . , vℓ/2−1 }] induces a complete subgraph of G with ℓ/2 vertices. If ℓ > 2k + 2, then ℓ/2 − 1 > k, − →(G) ≥ ω(G) − 1 ≥ ℓ/2 − 1 > k, a contradiction. which implies that → χ (G) ≥ − χ s − →(G) ≤ k for some positive integer k. Then, for every Lemma 6.16. Let G be a cograph with → χ (G) ≤ k or − χ s join node t ∈ V (T (G)), t has at most k + 1 children and at most one child t′ of t is such that |V (Gt′ )| > 2k. Proof. If there is a join node t ∈ V (T (G)) with at least k + 2 children, by the definition of the join operation, − →(G) ≤ k. If t ∈ V (T (G)) we deduce that ω(G) ≥ k + 2. This would contradict the fact that → χ (G) ≤ k or − χ s →(G) > k and → − ′ ′′ χ χ (G) > k has at least two children t and t such that |V (Gt′ )| > 2k and |V (Gt′′ )| > 2k, then − s by Lemma 6.3, a contradiction. Let k ∈ N. A cograph G is k-nice if the depth of T (G) is at most 2k + 2 and every join node t ∈ V (T (G)) has at most k + 1 children, and at most one child t′ of t is such that |V (Gt′ )| > 2k. For every integer j ≥ 0, a node t ∈ V (T (G)) is called j-high if the depth of Tt equals j. For every 0 ≤ j ≤ 2k + 1, let fj : N → N defined recursively as follows. First, f0 (x) = 1 for every x ∈ N. Then, for every 0 < j ≤ 2k + 1 and x ∈ N, let gj (x) be the number of connected cographs with fj (x) vertices and fj (x) = (2x2 + 1)gj−1 (x)fj−1 (x). The cotree T (G) is (j, k)-thin if, for every 0 ≤ j ′ ≤ j, and for every j ′ -high node t ∈ V (T (G)), we have that |V (Gt )| ≤ fj ′ (k). Note that, if T (G) has depth at most d ≤ 2k + 2 and it is (d, k)-thin, then |V (G)| ≤ fd (k), i.e., the number of vertices of G is a function only on k. 24 Theorem 6.17. Parameterized Semi-Proper Orientation and Parameterized Proper Orientation admit a kernel of size f2k+3 (k) in cographs. Proof. Let k ∈ N. First let us consider a connected cograph G = (V, E) with cotree T (G). If G is not k-nice, − →(G) > k by Lemmas 6.15 and 6.16, and so we may return a trivial kernel (e.g., a then → χ (G) > k and − χ s clique of size k + 2). Hence, we may assume that G is k-nice. Claim 6.17.1. For every j ∈ {0, . . . , 2k + 1}, there exists a k-nice connected cograph G′ with T (G′ ) being − →(G) ≤ k) if and only if → − →(G′ ) ≤ k). (j, k)-thin such that → χ (G) ≤ k (resp., − χ χ (G′ ) ≤ k (resp., − χ s s Proof of Claim. We prove the claim by induction on j. Note that the leaves of T (G) correspond to vertices of G and so T (G) is (0, k)-thin, and so the induction hypothesis holds for j = 0. Let 0 ≤ j ≤ 2k + 2 and let us assume by induction that there exists a k-nice connected cograph G′ such − →(G) ≤ k) if and only if → − →(G′ ) ≤ k). If that T (G′ ) is (j, k)-thin and → χ (G) ≤ k (resp., − χ χ (G′ ) ≤ k (resp., − χ s s j = 2k + 2, we are done since T (G) has depth at most 2k + 2 because G is k-nice. Otherwise, let us show that the induction hypothesis also holds for j + 1. Let t ∈ V (T (G′ )) be a (j + 1)-high join node of T (G′ ). Then, because G′ is k-nice, t has at most k + 1 children, and for every child t′ of t but one t′′ , |V (Gt′ )| ≤ 2k and |V (Gt′′ )| ≤ fj (k) by the induction hypothesis. Hence, |V (Gt )| ≤ max{2(k + 1)k, 2k 2 + fj (k)} ≤ fj+1 (k). Let t ∈ V (T (G′ )) be a (j + 1)-high union node of T (G′ ). If |V (Gt )| ≤ 2k ≤ fj+1 (k), we are done. Otherwise, let X = NG (V (Gt )) (recall that Gt is a module and so, we have all edges between X and V (Gt )). Hence, by Lemma 6.3, |X| ≤ 2k. Let t′1 , . . . , t′p be the children of t in T (G′ ). By induction, |V (Gt′i )| ≤ fj (k) for every 1 ≤ i ≤ p. Let G′′ be the graph obtained from G by keeping at most 2k 2 + 1 copies of each subgraph Gt′i , for 1 ≤ i ≤ p. That is, if there exist I ⊆ {1, . . . , p} such that |I| > 2k 2 + 1 and all subgraphs Gt′j are isomorphic, for every j ∈ I, then remove |I| − (2k 2 + 1) children t′j , j ∈ I, of t in T (G′ ). Note that, after this operation, t has at most (2k 2 + 1)gj (k) children, where gj (k) is the number of connected cographs with at most fj (k) vertices, and so |V (G′′t )| ≤ (2k 2 + 1)gj (k)fj (k). →(G′′ ) ≤ − →(G′ ) since − → is closed under taking subgraphs. To show that, if → − Clearly, − χ χ χ χ (G′ ) ≤ k then s s s → − ′′ ′ ′ χ (G ) ≤ k, let D be a proper k-orientation of G . For every I ⊆ {1, . . . , p} such that |I| > 2k 2 + 1 and all subgraphs Gt′j are isomorphic to some graph H, for every j ∈ I, there must be at least |I| − (2k 2 + 1) copies C of H such that, in D′ , all edges are oriented from X to C (otherwise, at least one vertex of X would have in-degree at least k + 1 since |X| ≤ 2k). Let D′′ be the orientation of G′′ obtained from D′ by removing precisely these copies of H. Therefore, for every vertex of G′′ , its in-degree in D′′ is the same as its in-degree − in D′ . Hence, → χ (G′′ ) ≤ k. On the other hand, consider any proper k-orientation D′′ of G′′ . For every subgraph H isomorphic to some V (G′′t′ ) with t′ a child of t, there must be one copy of H where all arcs are oriented from X to H (otherwise there would be at least 2k 2 + 1 arcs to X and one vertex of X would have in-degree larger than k as |X| ≤ 2k). Consider the orientation D′ of G′ obtained from D′′ and by orienting all arcs from X toward the extra copies of H (and orienting the edges in the copies of H as they are oriented in H). This implies − →(G) ≤ → − →(G) ≤ k. that → χ (G) ≤ k. Since − χ χ (G), we also get that − χ s s After having considered all (j + 1)-high vertices, the induction hypothesis holds for j + 1. ⋄ Finally, we have to consider a disconnected cograph G, i.e., when we have a disjoint union of con− − →(G) = nected cographs G1 , . . . , Gp at the root of T (G). Note that → χ (G) = max1≤i≤p → χ (Gi ) (resp., − χ s − → ′ max1≤i≤p χs (Gi )). By Claim 6.17.1, for every 1 ≤ i ≤ p, we may find an instance Gi equivalent to Gi and with order at most f2k+2 (k). If p is larger than the number g2k+2 (k) of connected graphs with at most f2k+2 (k) vertices, then two G′i and G′j , i 6= j, must be isomorphic and we can remove one. Overall, we get a kernel of order at most g2k+2 (k)f2k+2 (k) ≤ f2k+3 (k). Although we present this exponential kernel for Parameterized Proper Orientation, the computational complexity for cographs remains open. This is an open problem posed by [4]. We have the same 25 situation for the semi-proper orientation number: →(G) ≤ k for a cograph G and Problem 6.18. What is the computational complexity of deciding whether − χ s an integer k given as inputs? 7 Further Research → restricted to split graphs also implies NP-hardness for chordal graphs, Our NP-hardness proof computing − χ s →(G) when G is an interval graph? In [4], the authors present an but what is the complexity of computing − χ s → − NP-hardness proof for computing χ of chordal graphs, but the complexity for split graphs is open. We have also shown that chordal graphs G may have split subgraphs whose proper orientation number − is at least 2→ χ (G) − 2. Can the same happen with respect to (unit) interval graphs? →(G) and It seems a challenging question to determine the computational complexity of computing both − χ s → − χ (G) when G is cobipartite or when G is a cograph. →(G) is NP-hard, but there is an With respect to a split graph G, we have shown that computing − χ s exponential kernelization algorithm when parameterized by the value of the solution. The straightforward question is whether this kernelization can be improved. − χ (G) ≤ k when G is a cobipartite The authors in [4] presented a straightforward linear kernel for deciding → →. graph parameterized by k, as both cliques must have at most k + 2 vertices. 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