Backbone colouring and algorithms for TDMA scheduling

We investigate graph colouring models for the purpose of optimizing TDMA link scheduling in Wireless Networks. Inspired by the BPRN-colouring model recently introduced by Rocha and Sasaki, we introduce a new colouring model, namely the BMRN-colouring model, which can be used to model link scheduling problems where particular types of collisions must be avoided during the node transmissions. In this paper, we initiate the study of the BMRN-colouring model by providing several bounds on the minimum number of colours needed to BMRN-colour digraphs, as well as several complexity results establishing the hardness of ﬁnding optimal colourings. We also give a special focus on these considerations for planar digraph topologies, for which we provide reﬁned results. Some of these results extend to the BPRN-colouring model as well.


Motivation
A Radio Network consists in a set of nodes distributed in space that communicate via broadcast radio waves, with all messages sent from a node transmitted to all nodes in its range.The range of a node is therefore the region in space within which it can communicate to others.Node a can transmit to node b if b is within the range of a, and we say there is a directional link ab.Typically, communication is done in a multi-hop fashion, with intermediate nodes forwarding data from a source to a destination that is distant.Since the channels are shared, transmissions are subject to collisions, that result in undesired effects such as loss of data and network efficiency.To avoid such collisions, medium access control techniques have been designed.
In the Time-Division Multiple Access (TDMA) method [10], time is divided into frames of fixed duration.Each frame is itself divided into a number of fixed duration time slots.A link schedule is then an assignment of time slots to the network transmissions.When the link ab is scheduled, it is required that b receives its message from a free of collision, although we do not require the same for other nodes receiving a message from a.The TDMA method is being used in standards such as IEEE 802.16 [18] and IEEE 802.11s [19], providing guaranteed Quality-of-Service (QoS).In particular, TDMA MAC protocols are being used for the increasingly popular Wireless Sensor and Mesh Networks [5].Although there are particularities in the distinct network scenarios, an efficient use of TDMA methods is related to an increase in network throughput, and to a reduction of delays and packet losses.
Given a network and an interference model, the challenge is to find a link schedule that avoids conflicts and minimizes the number of used time slots [6,9].Minimizing the number of time slots is important, since it affects network throughput and multi-hop transmission delays.That is, if there are N f different time slots in a frame, then, since a node or link is only active during its associated time slot, it becomes active only during a fraction 1  N f of time.For this reason, finding a broadcast or link scheduling that minimizes the number of slots needed in a frame can lead to an improvement in the network efficiency.
In a typical network, only a subset of the existing links become active.This is because the existence of a link ab does not necessarily imply that a ever communicates with b.As an example, consider a network formed by a Wi-Fi router that is connected to the internet by cable, and that is surrounded by a number of devices such as computers and smartphones, connected to the internet through it.The nearby devices may be in each other's ranges, but their transmissions are not intended to each other, so that their mutual links do not need to be scheduled.In the rest of this paper we refer to the subnetwork whose links are the active ones as the network backbone.

Modelling
Such problems on radio networks can be modelled as graph colouring problems, in the following way.A sensor network can be represented as a digraph D with vertex set V (D) and arc set A(D).The vertices of D correspond to the nodes, and there is an arc from a vertex u to a vertex v if the node corresponding to v is in the transmission range of the node corresponding to u.The backbone is represented by a spanning subdigraph B of D, whose arcs correspond exactly to the backbone links.The arcs in A(D) \ A(B) correspond to the networks links that are not used by the network.The pair (D, B) is called a backboned digraph throughout.Now, we consider the time slots as colours to be assigned to the arcs of B. Hence, a link schedule of k time slots is a k-arc-colouring of B that must satisfy some constraints, depicting the conflicts which must be avoided during transmissions.
In practice, transmission collisions may occur for various reasons (physical constraints, device constraints, etc.), such as the following four ones, which shall be considered throughout this paper.
• Type-1 constraint: During a time slot, a node cannot both transmit and receive messages along the backbone.
⇒ If uv and vw are two arcs of B, then they cannot be assigned the same colour.
• Type-2 constraint: During a time slot, a node receiving a message along the backbone must not receive a message from another transmitting node.⇒ If u 1 v 1 is an arc of B, and e is either an arc u 2 v 1 of B, or an arc u 2 v 2 of B such that u 2 v 1 ∈ A(D) \ A(B), then uv and e cannot be assigned the same colour.
• Type-3 constraint: During a time slot, a node transmits at most one message along the backbone.⇒ If uv 1 and uv 2 are two arcs of B, then they cannot be assigned the same colour.
• Type-3 * constraint: A node transmits all its messages along the backbone during one time slot only.⇒ If uv 1 and uv 2 are two arcs of B, then they must be assigned the same colour.
1.3 Colouring variants over the four types of constraints

BPRN-colouring
Arc-colourings of backboned digraphs fulfilling combinations of the constraints above were already considered in literature.In particular, an arc-colouring of a backboned digraph (D, B) verifying Type-1, Type-2 and Type-3 constraints is called a Backbone-Packet-Radio-Network-colouring (BPRN-colouring for short).Note that when B is an out-branching of D, i.e., a spanning oriented tree all arcs of which are oriented away from the root, BPRN-colourings model link schedules where, in the radio network, the root node wants to send a private message to all other nodes.The BPRN-chromatic index of (D, B), denoted by BPRN(D, B), is the minimum number of colours in a BPRN-colouring of (D, B).
The BPRN-colouring model is a generalization of the PRN-colouring model, first proposed by Arika [4], that appeared in other works about link scheduling [9,13,15].It was introduced by Rocha and Sasaki in [14], who, motivated by applications in Wireless Sensor Networks, studied the case where the backbone is an in-branching, i.e., an oriented tree where all arcs are oriented towards the root.Among several results on BPRN-colourings, they exhibited bounds on the BPRN-chromatic index of backboned digraphs (D, B) where D is complete or a cycle.They also proved that, in general, determining the BPRN-chromatic index is N P-hard, even when restricted to bipartite backboned digraphs.chromatic index of (D, B), denoted by BMRN(D, B), is the minimum number of colours in a BMRNcolouring of (D, B).To the best of our knowledge, BMRN-colourings were not studied in literature.According to the definitions, BMRN-colourings are subject to less constraints than BPRN-colourings, and can thus be regarded as a simpler colouring notion.In particular, results on BPRN-colourings hold for BMRNcolourings.Moreover, the particular case when B is an out-branching models the case when a root wants to send the same message to all other vertices via a multicast tree.

BMRN * -colouring
For a backboned digraph (D, B), an arc-colouring verifying Type-1, Type-2, and Type-3 * constraints is called a Star-Backbone-Multicast-Radio-Network-colouring or simply BMRN * -colouring of (D, B).The BMRN * -chromatic index of (D, B), denoted by BMRN * (D, B), is the minimum number of colours in a BMRN * -colouring of (D, B).A first main motivation for considering BMRN * -colourings is to derive results on the BMRN-colourings, since BMRN * -colourings are also BMRN-colourings.However, it should be pointed that BMRN * -colourings are also a model to broadcast scheduling, in which each node of the network is scheduled to a time slot, during which all of its neighbours must receive its transmission free of collisions.Other works such as [12,13] study broadcast scheduling by considering that the network links are symmetric, the network being modelled by an undirected graph and the scheduling problem corresponding to finding a distance-2 colouring that uses the smallest number of colours.This approach can be seen as a simplification of our problem, since a typical network may have devices with different transmission ranges, resulting in a directed graph.
The BMRN * -colouring model can therefore be seen as a more realistic model for broadcast scheduling.This motivates the investigation of more results on this model and its corresponding algorithms.Other works such as [8] already point that distance-2 colouring of graphs is not the most accurate model to broadcast scheduling in real wireless networks, and consider the directed version.But to the best of our knowledge, we are the first to model the problem as an arc-colouring problem, and to relate it to link scheduling.

Organization of the paper
We start off with Section 2, in which we introduce some definitions and terminology that are useful for understanding our investigations.
It should be intuitive to the readers that the three colouring variants presented in the previous section are related.From the definitions, BPRN-colourings and BMRN * -colourings are also BMRN-colourings.Section 3 is devoted to understanding some of these relations further.
In Section 4, we exhibit upper bounds on the BMRN-chromatic, BPRN-chromatic and BMRN * -chromatic indices of a spanned digraph, that is a backboned digraph (D, T ) where T is an out-branching in D. Firstly, in Subsection 4.1, we establish general bounds in terms of the maximum degree and maximum out-degree of the digraph and show that these bounds are tight up to a small constant factor.For example, we show that for every spanned digraph (D, T ) we have BMRN(D, T ) ≤ 2∆ + (D) + 1 (where ∆ + (D) denotes the maximum out-degree of a vertex of D) and BMRN(D, T ) ≤ 2∆(D) − 1 (where ∆(D) denotes the maximum degree of the underlying graph of D) while there are spanned digraphs (D, T ) for which BMRN(D, T ) ≥ 2∆ + (D) − 1 and BMRN(D, T ) ≥ 2∆(D) − 3. Secondly, in Subsection 4.2, we give more refined upper bounds in the realistic case where a planar topology is considered.We show that BMRN(D, T ) ≤ BMRN * (D, T ) ≤ 8 for every planar spanned digraph (D, T ), and that BMRN(D, T ) ≤ BMRN * (D, T ) ≤ 5 for every outerplanar spanned digraph (D, T ).
In Section 5, we show that determining any of the BMRN-, BPRN-and BMRN * -chromatic indices of a spanned digraph is N P-hard in general, and that this remains true if one is allowed to choose the outbranching of the input spanned digraph.We also show that given a planar spanned digraph (D, T ), deciding whether BMRN(D, T ) ≤ 3 is N P-complete.
Drawing conventions: In every figure depicting a backboned digraph (D, B), the bold arcs are the arcs of B, the dashed arcs are the arcs not in B, and the undirected edges are pairs of opposite arcs.Moreover, if B is an out-branching, then the white vertex is its root.

Definitions and notation
The square G 2 of an undirected graph G is the graph with same vertex set as G and in which two vertices are adjacent if and only if they are at distance at most 2 in G.
The notion of minor of a digraph that we use in this paper corresponds to the notion of minor in the underlying graph.That is, a digraph D is a minor of D if it can be obtained from D by a succession of vertex-deletions, arc-deletions, and arc-contractions.A family D of digraphs is minor-closed if for every D ∈ D, all minors of D are also in D.
The union of two digraphs D 1 and D 2 is the digraph ), then it is the disjoint union of D 1 and D 2 .A matching in a digraph D is a disjoint union of arcs.A matching M is perfect in D if every vertex of D is either the tail or the head of an arc of M .A matched digraph is a pair (D, M ) where D is a digraph and M is a perfect matching of D.
In a digraph, we say that a vertex u dominates a vertex v if uv is an arc.An out-star is a digraph consisting in a vertex, called the centre, dominating all the others, called spades.A galaxy is a disjoint union of outstars.For a digraph D, an out-generator of D is a vertex r such that D admits an out-branching rooted at r.In other words, there is a directed path from r to every other vertex of D.
If A is a directed path or cycle, and x, y are two vertices of A, then A[x, y] denotes the directed subpath of A with origin x and terminus y.Conversely, one can easily see that BPRN(D, B) cannot be bounded by a function of BMRN(D, B).To see this, consider, for example, the directed out-star with k spades S k , for which we clearly have BMRN( S k , S k ) = 1 and BPRN( S k , S k ) = k.
On the other hand, we show that BMRN * (D, T ) is bounded above by a function of BMRN(D, T ).The upper bound function we exhibit is actually best possible.Proof: Let φ be a BMRN-colouring of (D, B).For every vertex v ∈ V (D), let Now, let g be the arc-colouring of B defined by g(xy) = f (x) for every arc xy ∈ A(B).We claim that g is a BMRN * -colouring of (D, B): • For every two arcs uv and vw of B, we have φ(uv) / ∈ f (v) because φ is a BMRN-colouring of (D, B).Hence f (u) = f (v), and g(uv) = g(vw).In particular, g is a proper arc-colouring.Thus Type-1 constraints are satisfied.
• By definition, any two arcs of B with the same tail are assigned the same colour by g.Hence Type-3 * constraints are satisfied.
The conclusion is now obtained by noting that if φ takes values in a set S of k colours, then g takes values in a set of at most 2 k − 1 colours, namely the non-empty subsets of S.
We now prove that the upper bound in Theorem 3.1 is best possible.Proof: Let X be a set of k vertices.Let us construct a galaxy B as follows: for each non-empty subset U of X, we create an out-star S + (U ) with centre x U and spades y U (u) for all u ∈ U .Let now D be the digraph obtained from B by adding arcs as follows.For every two elements u = t of X, and every subset T such that t ∈ T , we add the arc x {u} y T (t).This ensures that, in any BMRN * -colouring of (D, B), the arcs x {u} y {u} (u) and x T y T (t) get different colours.Then, for every pair {U, T } of distinct subsets of X with |U | ≤ |T |, choose a vertex t in T \ U , and add the arc x U y T (t) (if not already present).Observe that this arc implies that, in any BMRN * -colouring of (D, B), the arcs of S + (U ) and the arcs of S + (T ) are assigned different colours, i.e., the two sets of colours are disjoint.Consequently, BMRN * (D, B) ≥ 2 k − 1.
Now consider the arc-colouring φ of B where every arc x U y U (u) is assigned colour u.One easily checks that φ is a k-BMRN-colouring of (D, B).Indeed, the arcs of A(D) \ A(B) are of the form x U y T (t) with t ∈ T \ U .According to how φ was defined, this means that the arcs with tail x U are assigned, by φ, a colour in U , while the arc x T y T (t) is assigned colour t.Hence all Type-2 constraints are satisfied, and, consequently, BMRN(D, B) ≤ k.Furthermore, we actually have equality, as every two arcs x {u} y {u} (u), x {t} y {t} (t) for u = t must receive distinct colours, due to the arc x {u} y {t} (t).So all k arcs of the form x {u} y {u} (u) (u ∈ U ) must be coloured differently.
The proof of Proposition 3.2 can be easily modified to hold for spanned digraphs (D, B).
Proof: We start from the pair (D, B) constructed in the proof of Proposition 3.2.We construct a spanned digraph (D, T ) as follows.We take an out-branching BPRN-, BMRN-and BMRN * -colourings of backboned digraphs can be viewed as particular cases of the classical colouring of graphs, where one aims at assigning colours to the vertices so that a proper colouring is attained, i.e., a colouring where no two adjacent vertices have the same colour.Let us now explain this in detail, for a backboned digraph (D, B).The BMRN-constraint graph of (D, B) (see Figure 1 for an example) is the undirected graph C BMRN (D, B) defined as follows: • There is an edge in C BMRN (D, B) between two vertices corresponding to arcs, say, By construction, there is a trivial one-to-one correspondence between the BMRN-colourings of (D, B) and the proper colourings of C BMRN (D, B).In particular, we have BMRN(D, B) = χ(C BMRN (D, B)), where χ is the usual chromatic number of undirected graphs.
The BPRN-constraint graph of (D, B) is the graph C BPRN (D, B) defined as follows: • There is an edge in C BPRN (D, B) between two vertices corresponding to arcs, say, By construction, there is a trivial one-to-one correspondence between the BPRN-colourings of (D, B) and the proper colourings of C BPRN (D, B), and BPRN(D, B) = χ(C BPRN (D, B)).
) by identifying all vertices corresponding to arcs of B with same tail.It can also be defined as follows: Again, by construction, there is a one-to-one correspondence between the BMRN * -colourings of (D, B) and the proper colourings of In

BMRN * -colouring
Observe that if u 1 u 2 is an edge of C BMRN * (D, B), then u 1 and u 2 are at distance at most 2 in D. This means that which provides a first upper bound on BMRN * (D, B) (and, hence, on BMRN(D, B)) in terms of ∆(D).
We now prove a better upper bound on BMRN * (D, T ) for spanned digraphs (D, T ).It is obtained from a particular orientation of C BMRN * (D, T ) and the following well-known lemma, whose short proof is given for sake of completeness.Proof: For every subdigraph H of D, we have Proof: Let C be the orientation of C BMRN * (D, T ) defined by , and For every u ∈ V (D), we have . Now since T is an out-branching, we have d − T (x) ≤ 1 for every vertex x, so

BPRN-colouring
As in the previous section, using an orientation of the BPRN-constraint graph C BPRN (D, T ) (for some spanned digraph (D, T )), we now establish upper bounds on BPRN(D, T ) in terms of ∆ + (D) and ∆(D).Proof: Let C be the orientation of C BPRN (D, T ) defined by For every arc uv ∈ A(T ), we have

Tightness of the bounds
The bounds of Theorems 4.2 and 4.3 and Corollary 4.4 are tight up to a small additive factor, as shown in the following proposition.
Proposition 4.5.For every k ≥ 2, there exists a spanned digraph Proof: Let (D k , T k ) be the spanned digraph defined as follows (see Figure 2): Clearly ∆ + (D k ) = k, and ∆(D k ) = k + 1.Now, in every BMRN-colouring of (D k , T k ), every two of the u i v i 's must receive distinct colours because they are involved in a Type-2 constraint.Consequently, BMRN(D k , T k ) ≥ 2k−1.It is easy to check that (2k−1)-BMRN-colourings of (D k , T k ) actually exist (by, e.g., generalizing the colouring scheme of (D 3 , T 3 ) depicted in Figure 2), so BMRN(D k , T k ) = 2k −1.
For larger values of ∆(D) and ∆ + (D), one could wonder whether the bounds of Theorems 4.2 and 4.3 and Corollary 4.4 can be improved.We thus address the following questions: Question 4.7.
• What is the maximum value • What is the maximum value • What is the maximum value P ∆ + (k) (resp.P ∆ (k)) of BPRN(D, T ) over all spanned digraphs (D, T ) with In the rest of this section, we make a first step towards these questions by studying M ∆ (3) and M * ∆ (3).Proof: Let D be a subcubic digraph and T an out-branching of D. Let D = D \ A(T ).We partition V (D) = V (T ) into four sets according to their in-and out-degrees in T .Recall that the root r is the unique vertex such that d − T (r) = 0, and that the leaves are the vertices with out-degree If u is a flat vertex, then u + denotes its out-neighbour in T .
Suppose for a contradiction that C * is not 3-degenerate.Then it has a subgraph H such that δ(H) ≥ 4. The graph H contains no leaves of T because they are isolated vertices in C * .
Consider a flat vertex u in V (H).The only possibility for it to have degree 4 in C * is that there exist three distinct vertices v, u 1 , u 2 in V (H) such that there exists w such that vw ∈ A(T ) and uw ∈ A(D ), u 1 u + ∈ A(D ), and either u 2 u + ∈ A(D ) or u 2 = u + .In the latter case, note that u 2 has degree at most 3 in C * ; so let us suppose the first situation occurs.Note that u 1 and u 2 are either flat vertices or the root.Therefore if u is a flat vertex in H, then u + is a leaf which is the tail of two arcs u 1 u + and u 2 u + of A(D ) (with u 1 , u 2 either flat vertices or the root).
Let U be the set of flat vertices in H and let U + be the set of leaves dominated by a vertex of U in T .We have |U | = |U + |.Moreover, in D each vertex of U + has two in-neighbours in U ∪ {r}, while every vertex of U has at most one out-neighbour of U + and r has at most two out-neighbours in and so u 1 (and u 2 as well) has degree 3 in C * , a contradiction; and U = {u} is not possible because u + must have an in-neighbour in U \ {u}.Therefore U = ∅.Now V (H) contains only branching vertices and possibly the root r.Thus all arcs in D originate from r.Let v be a vertex in V (H) such that its out-neighbours in T are not in H; recall that a such vertex exists since all leaves of T are not in H. Then v can only be adjacent in H to r, and its in-neighbour in T , a contradiction to the fact that δ(H) ≥ 4.
We believe that a similar result holds for BPRN-colourings of spanned digraphs with maximum degree 3. Note in particular that the spanned digraph (D, T ) in Figure 3  Let us end up this section with discussing the notion of proper edge-colourings, which seem legitimate to consider in our context as they are perhaps the most investigated type of edge-colourings.Recall that a proper edge-colouring of an undirected graph is an edge-colouring where no two adjacent edges receive the same colour.We note that a BPRN-colouring is always a proper edge-colouring because of Type-1 and Type-3 constraints.A BMRN * -colouring is not a proper edge-colouring as soon as the backbone has vertices with out-degree at least 2, because of Type-3 * constraints.A BMRN-colouring can be a proper edge-colouring, although this is not always the case as vertices are allowed to have several out-going arcs with the same colour.
A well-known result of Vizing [17] states that every undirected graph G with maximum degree ∆ has chromatic index χ (G) (smallest number of colours in a proper edge-colouring) ∆ or ∆ + 1.In light of the previous arguments, we wonder about a possible connection between BPRN(D, T ) and χ ( Ũ (D) for a spanned digraph (D, T ), for instance of the following form: Question 4.10.Is there a function f , such that BPRN(D, T ) ≤ f (χ ( Ũ (D)) for every spanned digraph (D, T )?
In case such a function f were to exist, it would also be interesting investigating the existence of such a function for BMRN-colourings and BMRN * -colourings as well.

Upper bounds for some families of digraphs 4.2.1 Minor-closed families of digraphs
We start off by pointing out the following obvious result for BMRN * -colouring.There exist planar spanned digraphs (D, T ) verifying BMRN(D, T ) = 7.One such example is given in Figure 4.One easily checks that the BMRN-constraint graph of this example is K 7 , the complete graph on seven vertices.However, we still do not know whether all planar spanned digraphs have BMRN-or BMRN * -chromatic index at most 7.In the next section, we manage to answer such questions for outerplanar spanned digraphs.

Upper bounds for outerplanar spanned digraphs
Since outerplanar graphs have chromatic number at most 3, Corollary 4.13 yields that BMRN * (D, T ) ≤ 6 for every outerplanar spanned digraph (D, T ).The aim of this section is to improve this bound and show the following theorem.The bound 5 of Theorem 4.16 is best possible as shown by the example depicted in Figure 5.One easily sees that, for this spanned digraph (D, T ), we have BMRN(D, T ) ≥ 5 because its BMRN-constraint graph contains a K 5 .Let (D, T ) be an outerplanar spanned digraph.A chord is an arc of D which is not incident to the outer face, and a T -chord is a chord in A(T ).The proof of Theorem 4.16 is by induction on the number of Tchords.We first prove the following which corresponds to the basis of the induction, that is the case when there is no T -chord.Lemma 4.17.Let (D, T ) be a 2-connected outerplanar spanned digraph such that all arcs of T are on the outer face of D. Then BMRN * (D, T ) ≤ 5.
Proof: By considering a minimum counterexample (i.e., with the minimum number of vertices).Free to add arcs, we may assume that D is outerplanar-maximal.Since D is a symmetric digraph, in what follows we sometimes regard it as an undirected graph G, in which every pair of arcs {uv, vu} is replaced with an edge uv.
Let us number the vertices of D by v 1 , . . ., v n so that The span of an arc v i v j of D or an edge v i v j of G is |j − i|.In particular, all arcs of T have span 1. T [i, j] denotes the subdipath of T induced by {v i , . . ., v j }.Let i 0 be the index of the root r of T (i.e., r = v i0 ).
The edge-tree of (D, T ) is the out-tree whose vertices are the edges of G and such that every edge v i v j (with i < j) of span at least 2 dominates the two edges v i v k and v k v j such that i < k < j and v i v k v j v i is a 3-cycle in G. Observe that the root of the edge-tree of (D, T ) is v 1 v n , and its leaves are the edges of span 1.Furthermore, the span of every edge that is not a leaf is the sum of the spans of the two edges it dominates.
If v i v j is an edge of G and T [i, j] is a directed path, then the v i v j -reduced spanned digraph is the spanned digraph (D i,j , T i,j ) defined as follows: • D i,j is obtained from D − {v i+1 , . . ., v j−1 } by adding a vertex x i,j and the arcs of the two directed cycles (v i , x i,j , v i ) and (v j , x i,j , v j ).
• If T [i, j] is a directed path from v i to v j (resp.from v j to v i ), then T i,j is obtained from T − {v i+1 , . . ., v j−1 } by adding a new vertex x i,j and the arcs v i x i,j and x i,j v j (resp.v j x i,j and x i,j v i ).
In this case, b = b i,j (resp.b = b i,j ) denotes the arc of T i,j between v i (resp.v j ) and x i,j .
If v i v j is an edge of G and T [i, j] is not a directed path, then the v i v j -reduced spanned digraph is the spanned digraph (D i,j , T i,j ) defined as follows: • D i,j is obtained from D − {v i+1 , . . ., v j−1 } by adding two vertices x i,j and y i,j and the arcs of the three directed cycles (v i , x i,j , v i ), (x i,j , y i,j , x i,j ), and (v j , y i,j , v j ).
• T i,j is obtained from T − {v i+1 , . . ., v j−1 } by adding the two vertices x i,j and y i,j and the arcs x i,j v i , x i,j y i,j , and y i,j v j .
In that case, b = b i,j (resp.b = b i,j ) denotes the arc x i,j v i (resp.y i,j v j ).
Observe moreover that if v i v j has at least 3 or span 2 and T [i, j] is a directed path, then the v i v jreduced spanned digraph has smaller order than (D, T ).Therefore by minimality, it admits a 5-BMRN *colouring φ.Moreover, the colours assigned to a i−1 , a j , b and b are all distinct, except possibly φ(a i−1 ) = φ(a j ) when The general idea of the proof is to show that there is an edge v i v j such that any 5-BMRN * -colouring of the v i v j -reduced spanned digraph can be modified to get a 5-BMRN * -colouring of (D, T ), which is a contradiction.

Proof of claim.
Assume for a contradiction that v i v i+4 , v i v i+2 and v i+2 v i+4 are edges and i ≤ i 0 ≤ i + 4.
By symmetry, we may assume i 0 ∈ {i, i + 1, i + 2}.By minimality of (D, T ), the v i v i+4 -reduced spanned digraph has a 5-BMRN * -colouring φ.Without loss of generality, φ(a 1 ) = 1 (if φ(a i+2 ) = 5 and φ(a i+3 ) = 3.This gives a 5-BMRN * -colouring of (D, T ), a contradiction.♦ Consider a deepest edge e of span 2 in the edge-tree of (D, T ), that is an edge with longest distance from v 1 v n in the edge-tree.Set e 1 = v i v i+2 .Let e 2 be the edge that dominates e 1 in the edge-tree, and let e 1 be the second edge that is dominated by e 2 .Since e 1 is a deepest edge of span 2, necessarily e 1 is an edge of span 1 or 2, for otherwise the branch of the edge-tree spanned at e 1 would contain an edge of span 2 deeper than e 1 .Therefore the span of e 2 , which is the sum of the spans of e 1 and e 1 , is either 3 or 4. Without loss of generality, either e 2 = v i v i+3 or e 2 = v i v i+4 and v i+2 v i+4 is an edge.
Case 1: e 2 = v i v i+3 .Then i 0 ∈ {i + 1, i + 2} by Claim 4.18.Let e 3 = v k v l be the edge that dominates e 2 in the edge-tree, and let e 2 be the second edge dominated by e 3 .Since e 1 was the deepest edge of span 2, e 2 has span at most 4. If e 2 has span 4, then swapping the names of e 2 and e 2 we are in Case 2. If e 2 has span 3, then e 2 contradicts Claim 4.18.Hence e 2 has span 1 or 2. Henceforth, we must be in one of the subcases below.For each of them, we take a 5-BMRN * -colouring φ of the e 3 -reduced spanned digraph, which exists by minimality of (D, T ).Without loss of generality, we may assume that φ(a k−1 ) = 1 (if a k−1 exists), φ(b) = 2, φ(b ) = 3, and φ(a l ) ∈ {1, 4} (if a l exists).We now show for each subcase how to derive a 5-BMRN * -colouring of (D, T ), which is a contradiction.
Case 2: e 2 = v i v i+4 and v i+2 v i+4 is an edge.By Claim 4.19, and by symmetry, we may assume that i 0 < i.Let e 3 = v k v l be the edge that dominates e 2 in the edge-tree, and let e 2 be the second edge dominated by e 3 .Since e 1 was the deepest edge of span 2, e 2 has span at most 4. Furthermore if e 2 has span 4, then it dominates two edges of span 2.
One of the subcases below must thus occur.For each of them, we take a 5-BMRN * -colouring φ of the e 3 -reduced spanned digraph, which exists by minimality of (D, T ).Without loss of generality, we may assume that φ( We now show for each subcase how to derive a 5-BMRN * -colouring of (D, T ), which is a contradiction.We first consider the subcases when k = i.In those subcases, φ(a l ) = 4 because i 0 < i.
We now consider the subcases when l = i + 4.
In that subcase, by Claim 4.18, We are in Subcase 2 (d), with e 2 and e 2 swapped.
This concludes the proof.
Proof of Theorem 4.16: By induction on the number of T -chords and next on the order of D. Without loss of generality, we may assume that D is outerplanar-maximal.Let C be the oriented cycle around the outer face.
If there is no T -chord, then we have the result by Lemma 4.17.Assume now that there is a T -chord uv.This chord divides D into two outerplanar digraphs, D 1 with outer face C 1 = C[u, v] ∪ {uv} and D 2 with outer face C 2 = C[v, u] ∪ {uv}.Without loss of generality, we may assume that the root r of T is in D 1 .For i = 1, 2, let D * i be the symmetric outerplanar graph obtained from D i by adding the arc vu (if necessary), and let T i = T ∩ D i .Observe that T i is an out-branching of D * i and that the number of T i -chords in (D * i , T i ) is less than the number of T -chords in (D, T ).We distinguish two major cases, each consisting of two subcases.
Case 1: There is an arc vw 1 in T 1 .
We distinguish two subcases depending on whether u is the root of T or not.
Subcase 1.1: u is the root of T .By the induction hypothesis, there is a 5-BMRN * -colouring φ i of each (D * i , T i ).Free to permute the colours, we may assume that φ 1 (uv) = φ 2 (uv) and that the arcs of T 2 with tail v (if some exist) are coloured (by φ 2 ) with φ 1 (vw 1 ).One can easily check that the colouring φ of A(T ), defined by φ(a) = φ 1 (a) if a ∈ A(T 1 ) and φ(a) = φ 2 (a) if a ∈ A(T 2 ), is a 5-BMRN * -colouring of (D, T ).Subcase 1.2: u is not the root of T .
In that subcase, u has a unique in-neighbour t in T , which must be in T 1 .Let D + 2 be the subdigraph of D induced by V (D 2 ) ∪ {t}, and T + 2 be the out-branching of D + 2 obtained from T by adding t and the arc tu.Observe that the number of T + 2 -chords in (D + 2 , T + 2 ) is not greater than the number of T -chords in (D, T ) and By the induction hypothesis, there are 5-BMRN *colourings φ 1 of (D 1 , T 1 ) and φ + 2 of (D 2 , T 2 ).Free to permute the colours, we may assume that φ 1 (uv) = φ + 2 (uv), φ 1 (tu) = φ + 2 (tu) and that the arcs of T 2 with tail v (if some exist) are coloured (by φ + 2 ) with φ 1 (vw 1 ).Note that this is possible because, in both colourings, tu and the arcs with tail v receive distinct colours because vu is an arc.One can easily check that the colouring φ of A(T ), defined by φ(a) = φ 1 (a) if a ∈ A(T 1 ) and φ(a) = φ + 2 (a) if a ∈ A(T + 2 ), is a 5-BMRN * -colouring of (D, T ).
Case 2: There is no arc with tail v in T 1 .
Since uv is a T -chord, |V (D 1 )| ≥ 3 and so u has a unique in-neighbour t in T 1 , which is also its unique in-neighbour in T .Subcase 2.1: We get a 5-BMRN * -colouring of (D, T ) exactly as in Subcase 1.2.
Subcase 2.2: We may assume that uv is the unique T -chord, for otherwise Case 1 or Subcase 2.1 would apply, and we would be done.Let v be the neighbour of u in C distinct from t.Let D be the digraph obtained from D by replacing the arcs tv and vt by the arcs tv and v t.Observe that (D, T ) and (D , T ) have the same BMRN * -chromatic index because those four arcs do not create any new constraint.If uv / ∈ A(T ), then, by Lemma 4.17, (D , T ) has a 5-BMRN * -colouring which is also a 5-BMRN *colouring of (D, T ).Henceforth, we may assume that uv ∈ A(T ).In particular, T − t is the union of two directed paths, P with first arc uv and P with first arc uv .
Assume u has at most five neighbours in D * 2 .Recall that D is symmetric, so every neighbour is both an in-and an out-neighbour.Two of these neighbours are v and v .By Lemma 4.17, (D * 2 , T 2 ) admits a 5-BMRN * -colouring φ 2 .Now at most four colours are forbidden for tu, namely φ 2 (uv) = φ 2 (uv ) and the colours assigned to arcs with tail a neighbour of u in D 2 distinct from v and v .Hence one can extend φ 2 to (D, T ) by assigning to tu a non-forbidden colour.Henceforth, we may assume that u has at least five neighbours in D * 2 .Free to consider (D , T ) instead of (D, T ), we may assume that u has at least three neighbours in P .Let z be the last neighbour of u along P .Let z − be the in-neighbour of z in P and let z + be the out-neighbour of z in P if it has one and z + = z otherwise, and let z * be the terminal vertex of P .Let D 3 be the subdigraph induced by the vertices of V (P [u, z + ]) ∪ {t}.By Lemma 4.17, (D 3 , T 3 ) admits a 5-BMRN * -colouring φ 3 .Observe moreover that the colours assigned to tu, uv, z − z and zz + (if z = z + ) are all distinct. .Thus, by minimality of (D, T ), (D 4 , T 4 ) admits a 5-BMRN *colouring.In addition, the colours assigned to tu, uv, z − z and zz + (if z = z + ) are all distinct.Hence, free to permute the colours, we may assume that φ 3 and φ 4 agree on those four arcs.Now one easily checks that the colouring φ of A(T ), defined by φ(a) = φ 3 (a) if a ∈ A(T 3 ) and φ(a) = φ 4 (a) if a ∈ A(T 4 ), is a 5-BMRN * -colouring of (D, T ).
This concludes the proof.

Determining the exact value of an index
In this section, we prove several results showing that the problems of determining BMRN(D, T ), BPRN(D, T ) and BMRN * (D, T ) are N P-hard, even when restricted to particular spanned digraphs (D, T ).We define these decision problems in the usual way: Recall that finding a k-BMRN-colouring (resp., k-BPRN-colouring, k-BMRN * -colouring) of (D, T ) is equivalent to finding a k-colouring of C BMRN (D, T ) (resp., C BPRN (D, T ), C BMRN * (D, T )).Furthermore, the usual k-COLOURING problem is well-known to be polynomial-time solvable when k = 2 and N Pcomplete for all k ≥ 3. Since the constraint graphs of (D, T ) can clearly be constructed in polynomial time, we directly get that 2-BMRN-COLOURING, 2-BPRN-COLOURING, and 2-BMRN * -COLOURING are polynomial-time solvable.Still from the previous colouring equivalence, we now establish the hardness of the three problems above for all k ≥ 3, using the following construction (see Figure 6 for an example).Let G be an oriented graph.
The matched digraph associated to G is the matched digraph (D G , M G ) defined by: Theorem 5.1.For every k ≥ 3, the problems k-BMRN-COLOURING, k-BPRN-COLOURING and k-BMRN * -COLOURING are N P-complete, even when restricted to spanned digraphs (D, T ) where T is a directed path.
Proof: Fix k ≥ 3. The problems are clearly in N P. Let us now prove that they are N P-hard.Observe that when T is a directed path, the Type-3 and Type-3 * constraints are vacuously fulfilled by any arc-colouring.Therefore, in that case, any BMRN-colouring of (D, T ) is also a BPRN-colouring and a BMRN * -colouring.It thus suffices to prove the N P-completeness of k-BMRN-COLOURING for such restricted instances.
The reduction is from k-COLOURING.Let G be a graph with vertex set {x 1 , . . ., x n }.Consider any orientation G of G. Let (D, T ) be the spanned digraph obtained from the associated matched digraph By construction, T is a directed path.One easily sees that each v xi u xi+1 is subject to at most two colouring constraints, or, in other words, that the associated vertex has degree 2 in the BMRN-constraint graph C BMRN (D, T ).Hence, since k ≥ 3, the graph We proved, in Theorem 4.8, that subcubic spanned digraphs have BMRN-chromatic index at most 4. As already pointed out, by looking at the BMRN-constraint graph, it can be decided in polynomial time whether the BMRN-chromatic index of a given subcubic spanned digraph is at most 2. It is not clear, however, whether it can easily be decided whether this index is 3 or 4; so we leave the following question open: Question 5.2.Is 3-BMRN-COLOURING N P-complete when restricted to subcubic spanned digraphs?

Finding a particular out-branching
We now consider problems where, for a given digraph D, one aims at finding an out-branching T (rooted at a given vertex r or not), such that BMRN(D, T ) (resp.BMRN(D, T ) and BPRN * (D, T )) is small.More precisely, we consider the following decision problems (which we define for BMRN-colouring only, but they can be derived for BPRN-colouring and BMRN * -colouring in an obvious way): • If φ(r i v i ) = φ(r i+1 v i+1 ), then add r i q i , r i p i , p i p i and p i r i+1 to A(T ) and set φ(r i q Now, it is simple matter to check that T is an out-branching of D and φ a 2-BMRN-colouring of (D, T ).
The exact same proof yields the following.
Proof: The proof of the N P-hardness of the two problems is by reduction from a restriction of the DI-RECTED HAMILTONIAN CYCLE problem, which asks whether a given digraph has a directed Hamiltonian cycle.In [11], Plesńik proved that DIRECTED HAMILTONIAN CYCLE remains N P-complete when restricted to small-degree digraphs, that are digraphs in which the in-degree and out-degree of each vertex are either 1 or 2.
A track is a vertex with in-degree 1 and out-degree 1, an out-switch is a vertex with in-degree 1 and outdegree 2, and an in-switch is a vertex with in-degree 2 and out-degree 1.A digraph is nice if every vertex is either a track, an out-switch or an in-switch.We now construct a digraph H such that H admits an out-branching T such that BPRN(H, T ) = 2 if and only if D has a directed Hamiltonian path, and so if and only if D has a directed Hamiltonian cycle.
To that end, we first associate a gadget in H to each vertex of D , in the following way: • For each of r and s, we add to H two gadgets that are actually exactly the vertices r and s.For the vertex gadget corresponding to r (resp.s), we call r (resp.s) its exit (resp.entry).
Our goal is now to extend (D G , M G ) into a planar spanned digraph (D, T ), preserving the colouring equivalence with G. From G, one can easily derive a plane straight-line embedding of D G in which all arcs u x v x , for x ∈ V ( G), are drawn horizontally and from left to right, of tiny length and whose middle is x.Towards getting (D, T ), we first add, to (D G , M G ), a vertex r below all other vertices (and thus on the outer face), which shall be the root of T .For every inner face F of G, let m(F ) be the lowest vertex of F (for the particular case of the outer face F , let m(F ) = r).For every face F of (D G , M G ), let M (F ) be the set of vertices x such that u x is inside F .Observe that the vertices of M (F ) are incident to F , and that m(F ) / ∈ M (F ).For each face F , we add a directed path P x (both in D and T ) of length 3 from u m(F ) to u x for each vertex x ∈ M (F ).Of course, we do this in such a way that the added paths do not cross.This results in the planar spanned digraph (D, T ) where D = D G ∪ x∈V (G) P x and T = M G ∪ x∈V (G) P x .
Let us now check that BMRN(D G , M G ) ≤ 3 if and only if BMRN(D, T ) ≤ 3. Since (D G , M G ) is the restriction of (D, T ) to V (D G ), every 3-BMRN-colouring of (D, T ) induces a 3-BMRN-colouring of (D G , M G ). Conversely, every 3-BMRN-colouring φ of (D G , M G ) can be easily extended in a 3-BMRNcolouring of (D, T ), for the following reasons.First, the added paths P x do not create any new constraint between the arcs of M G .Moreover, for every face F and every x ∈ M (F ), we can assign the colour φ(u m(F ) , v m(F ) ) to the first arc of P x , and then extend the colouring to the two other arcs, which are subject to at most two constraints in (D, T ).
Thus, (D, T ) is a planar spanned digraph that is 3-BMRN-colourable if and only if G is 3-colourable.
PLANAR k-COLOURING is trivial for all k ≥ 4, as the answer is always 'Yes' according to the Four-Colour Theorem.Henceforth, the above proof cannot be generalized to show that PLANAR k-BMRN-COLOURING is N P-complete for k larger than 3.

The
underlying simple graph Ũ (D) of a digraph D is the graph defined by V ( Ũ (D)) = V (D) and E( Ũ (D)) = {uv | uv ∈ A(D) or vu ∈ A(D)}.We call D symmetric if vu ∈ A(D) whenever uv ∈ A(D).Observe that if D is symmetric, then Ũ (D) is obtained by replacing each directed 2-cycle by an edge.The digraph obtained from D by contraction of the arc a = xy in A(D) is the digraph D/a obtained from D − {x, y} by adding a new vertex v a and the arc v a w (resp.wv a ) for every vertex u in V (D) \ {x, y} such that xw ∈ A(D) or yw ∈ A(D) (resp.wx ∈ A(D) or wy ∈ A(D)).

Proposition 3 . 2 .
For every k ≥ 1, there exists a digraph D and a spanning galaxy B of D such that BMRN(D, B) = k and BMRN * (D, B) = 2 k − 1.
and root z X .For every arc a U a T of B + , we add a directed path of length 3 with initial vertex a spade of S + (U ) (i.e., y U (u) for some u ∈ U ), terminal vertex x T , and new (private) internal vertices.This forms our out-branching T .One then easily checks thatBMRN(D, T ) = BMRN(D, B) = k and BMRN * (D, T ) = BMRN * (D, B) = 2 k − 1.4 Bounds on BPRN(D, B), BMRN(D, B) and BMRN * (D, B)

Lemma 4 . 11 .
Let D be a digraph and B 1 and B 2 be two subdigraphs of D.Then BMRN * (D, B 1 ∪ B 2 ) ≤ BMRN * (D, B 1 ) + BMRN * (D, B 2 ).In view of Lemma 4.11, to get upper bounds on BMRN * (D, T ) for spanned digraphs (D, T ), it might be interesting to get upper bounds on BMRN * (D, S) when S is a galaxy.We now use this approach for minor-closed families of backboned digraphs.For any class F of digraphs, we define χ(F) = max{χ(D) | D ∈ F}.

Theorem 4 . 12 .
Let F be a minor-closed family of digraphs.If D ∈ F and S is a galaxy in D, then BMRN * (D, S) ≤ χ(F).Proof: Consider the BMRN * -constraint graph C BMRN * (D, S).One easily sees that it is the minor of D obtained by contracting the arcs of S. Hence BMRN * (D, S) = χ(C BMRN * (D, S)) ≤ χ(F).

Claim 5 . 7 .
DIRECTED HAMILTONIAN CYCLE remains N P-complete for nice digraphs.Proof of claim.Reduction from DIRECTED HAMILTONIAN CYCLE restricted to small-degree digraphs.Consider a small-degree digraph D. Let D be the digraph obtained by "exploding" each vertex v to an arc v − v + .Formally,• V (D ) = v∈V (D)v − , v + , and• A(D ) = v − v + | v ∈ V (D) ∪ u + v − | uv ∈ A(D) .Clearly D is nice and it has a directed Hamiltonian cycle if and only if D has one.♦ We now give a reduction from DIRECTED HAMILTONIAN CYCLE restricted to nice digraphs to 2-BPRN-ROOT and 2-BPRN-BRANCHING.Let D be a nice digraph.Choose a vertex v of D with in-degree 1.Let u be its in-neighbour.Observe that every directed Hamiltonian cycle of D must contain the arc uv.Let D be the digraph obtained from D − uv by adding two vertices r and s and the arcs rv and us.Note that a directed Hamiltonian path in D necessarily starts in r and ends in s, so D has a directed Hamiltonian path if and only if D has a directed Hamiltonian cycle.

Figure 8 :
Figure 8: The out-gadget O v .A directed Hamiltonian path P of O v is displayed with bold arcs.The black and gray bold arcs form a 2-BPRN-colouring of (O v , P ).

Question 5 . 9 .
For every k ∈ {4, 5, 6, 7}, what is the complexity of PLANAR k-BMRN-COLOURING?Similarly, one can compute the chromatic number of an outerplanar graph in polynomial time.So we cannot establish the hardness of the restriction of k-BMRN-COLOURING to outerplanar digraphs via a proof similar to that of Theorem 5.8.We thus leave the following question open: Question 5.10.For any outerplanar spanned digraph (D, T ), can BMRN(D, T ) and BMRN * (D, T ) be determined in polynomial time?