List circular backbone colouring

A natural generalization of graph colouring involves taking colours from a metric space and insisting that the end-points of an edge receive colours separated by a minimum distance dictated by properties of the edge. In the q - backbone colouring problem, these minimum distances are either q or 1 , depending on whether or not the edge is in the backbone . In this paper we consider the list version of this problem, with particular focus on colours in Z p – this problem is closely related to the problem of circular choosability. We ﬁrst prove that the list circular q -backbone chromatic number of a graph is bounded by a function of the list chromatic number. We then consider the more general problem in which each edge is assigned an individual distance between its endpoints, and provide bounds using the Combinatorial Nullstellensatz. Through this result and through structural approaches, we achieve good bounds when both the graph and the backbone belong to restricted families of graphs.


Introduction
All graphs considered in this paper are simple.Let G = (V, E) be a graph, and let H = (V, E(H)) be a subgraph of G, called the backbone.A k-colouring of G is a mapping f : V → {1, 2, . . ., k}.Let f be a k-colouring of G.It is a proper colouring if |f (u) − f (v)| ≥ 1.It is a q-backbone colouring for (G, H) if f is a proper colouring of G and |f (u) − f (v)| ≥ q for all edges uv ∈ E(H).The chromatic number χ(G) is the smallest integer k for which there exists a proper k-colouring of G.The q-backbone chromatic number BBC q (G, H) is the smallest integer k for which there exists a q-backbone k-colouring of (G, H).
If f is a proper k-colouring of G, then g defined by g(v) = q • f (v) − (q − 1) is a q-backbone (q • k − q + 1)-colouring of (G, H) for any spanning subgraph H of G. Hence, BBC q (G, H) ≤ q • χ(G) − q + 1. (1) In [4,5], Broersma et al. showed that for any integer k there is a graph G with χ(G) = k and a spanning tree T such that BBC 2 (G, T ) = 2k − 1.
One can generalize the notion of backbone colouring by allowing a more complicated structure of the colour space.A natural choice is to impose a circular metric on the colours.We can see Z k as a cycle of length k with vertex set {1, . . ., k} together with the graphical distance | • | k .Then |a − b| k ≥ q if and only if q ≤ |a − b| ≤ k − q.A circular q-backbone k-colouring of G or q-backbone Z k -colouring of (G, H) is a mapping f : V (G) → Z k such that c(v) = c(u) for each edge uv ∈ E(G) and |c(u) − c(v)| k ≥ q for each edge uv ∈ E(H).The circular q-backbone chromatic number of a graph pair (G, H), denoted CBC q (G, H), is the minimum k such that (G, H) admits a circular q-backbone k-colouring.
A backbone Z k -colouring is trivially a backbone k-colouring.On the other hand, a backbone kcolouring yields a circular backbone Z k+q−1 -colouring.Hence for every graph pair (G, H) (where H is a subgraph of G), we have BBC q (G, H) ≤ CBC q (G, H) ≤ BBC q (G, H) + q − 1. ( Inequalities 1 and 2 yield CBC q (G, H) ≤ q • χ(G) for every subgraph H of G. ( This bound is tight if H contains a complete graph of size χ(G).In particular, it is the case when G is bipartite, and H is non-empty.On the other hand, Broersma et al. [6] gave better upper bounds when the backbone is a matching.We explore similar restrictions shortly, but first we define a list colouring analogue.Such an analogue is motivated by the fact that backbone colouring models a special case of radio channel assignment: in such problem, because of technical reasons or dynamicity, the set of channels available (i.e.colours) very often varies from transmitter (i.e.vertex) to transmitter.

List backbone colourings
The notions of backbone and circular backbone colouring naturally generalize to list colouring.An S-list assignment of a graph G is a mapping L which assigns to each vertex v ∈ V (G) a prescribed list of colours L(v) ⊆ S, where S is a set of colours.For a list assignment L, an L-colouring is a colouring c such that c(v) ∈ L(v) for every vertex.The list chromatic number of a graph G, denoted χ (G), is the least integer k such that for any list assignment L such that |L(v)| ≥ k, there is a proper L-colouring.The graph pair (G, H) is said to be circularly q-backbone k-choosable if for any Z p -list assignment L such that |L(v)| ≥ k, there is an L-colouring that is a q-backbone Z p -colouring.The list circular q-backbone number of a pair (G, H), denoted CBC q (G, H) is the least integer k such that (G, H) is circularly qbackbone k-choosable.
The concept of circular choosability, introduced by Mohar [15] and Zhu [20], is closely related to list circular backbone colouring.Indeed, the circular list chromatic number or circular choice number of G may be defined as cch(G) := inf{CBC q (G, G)/q}.
Zhu [20] proved that cch(G) ≥ χ (G)−1 for every graph G.This implies that CBC q (G, G) ≥ q • χ (G)− q for every positive q.In view of this inequality and the fact that CBC q (G, G) = q • χ(G), it is natural to ask whether CBC q (G, G) can be bounded by a function of χ (G).We answer in the positive in Section 2.
However, the function we prove is exponential in χ (G) -we believe that it is far from optimal, and pose the following problem.
Problem 1 Let q be a positive integer.What is the minimum function m q such that for every k-choosable This is closely related to a conjecture of Zhu [20].
Conjecture 2 (Zhu [20]) There is a constant α such that, for every graph Note that if such an α exists then it is at least two, as for any positive integer m, χ m k as shown by Zhu [20].Conjecture 2 would be implied by the following.
Conjecture 3 Let q be a positive integer.There is a constant α q such that, for every graph As an evidence in support of the latter conjecture, we prove (Theorem 17) that for a graph G and a matching 1) is a function tending to 0 when χ (G) tends to infinity.These bounds generalize to any backbone H in an upper bound in terms of χ (G) and χ (H), the chromatic index of H: In Section 3, we consider list circular backbone colouring of 2-choosable graphs.In [7], Erdős, Rubin, and Taylor characterized such graphs as follows.The heart of a graph G is the maximal subgraph in which there is no vertex of degree one.The graph consisting of two vertices connected with three internally vertex-disjoint paths of length i, j and k is θ i,j,k .
Theorem 4 (Erdős et al. [7]) A connected graph is 2-choosable if and only if its heart is either a single vertex, an even cycle or θ 2m,2,2 for some integer m ≥ 1.
Conjecture 5 (Norine, Wong, and Zhu [17] This conjecture is trivial for trees and was proved by Norine [16] for even cycles (and so for all graphs whose heart is an even cycle).
Theorem 6 (Norine [16]) If C is an even cycle, then CBC q (C, C) = 2q for all q.
Note that Conjecture 5 implies that if G is 2-choosable, then CBC q (G, H) = 2q for any non-empty subgraph H of G.We prove that this is the case if H is a matching (Theorem 28).We also show that if H does not contain the heart of G, then CBC q (G, H) ≤ 2q + 1.

Edge-weighted backbones
In Section 4 we generalize the idea of a backbone colouring by giving every edge e a positive integer weight w(e) dictating the minimum distance between the colours of its endpoints.This corresponds to the general channel assignment problem (see e.g.[12,13]), in which the minimum distance between the channels assigned to two transmitters depends on these transmitters and how they can interfere with each other.In this paper, we apply the technique developed in this section to prove a result on list circular backbone colouring.However, as the technique is quite general, we believe that it could be used on other channel assignment problems.
Formally, a colouring c of there is a w-circular L-colouring.Clearly, every w-circular colouring is a w-coherent colouring, and every w-coherent k-colouring induces a w-circular (k + max e∈E(G) w(e) − 1)-circular colouring.
The w-semi-degree of a vertex v is dw (v) = e v w(e), and the maximum semi-degree of a graph Using the many-passes greedy algorithm, McDiarmid [14] proved that every graph G is w-coherent ( ∆w (G) + 1)-choosable.Therefore every graph G is w-circular ( ∆w (G) + max e∈E(G) w(e))-choosable.
For a digraph D, we denote by E + D (v) the set of arcs with v as their tails, and by E − D (v) the set of arcs with v as their heads.Similarly, if D is a digraph, its w-outdegree is d + w (v) = e∈E + D (v) (2w(e) − 1).Extending a result of Norine, Wong and Zhu [17] (Theorem 33), which itself extended a famous result of Alon and Tarsi [3] (Theorem 32), we use the Combinatorial Nullstellensatz to prove that if a graph G admits an orientation D with certain property, then for any Z p -list assignment L such that |L(v)| ≥ d + w (v) + 1 for each vertex v, there is a w-circular L-colouring (Lemma 34).In particular, if D is an orientation of G that contains no odd directed cycles, then it has the required property (Theorem 34).From this we deduce that if G is bipartite, then it is w-circular , where G i is the graph induced by the edges of weight i.

Restricted graph classes
Let G and H be two graph classes.For any parameter A among BBC q , CBC q and CBC q , we define A(G, H) as the maximum value of A(G, H) over all pairs (G, H) such that G ∈ G, H ∈ H and H is a subgraph of G. Let P, F and M be the classes of planar graphs, forests, and matchings respectively.
In Section 5, we consider the case when G is in P and H is in one of the three classes P, F and M. Inequality (1) and the Four-Colour Theorem imply BBC q (P, P) ≤ 3q + 1 and CBC q (P, P) ≤ 4q.Moreover, BBC q (G, G) = 3q + 1 and CBC q (G, G) = 4q, for any 4-chromatic planar graph G. Thus BBC q (P, P) = 3q + 1 and CBC q (P, P) = 4q.
But the exact value of CBC q (P, P) is still unknown.
Problem 7 What is the exact value of CBC q (P, P) ?
This problem is very closely related to the one posed by Mohar [15].
Problem 8 (Mohar [15]) What is the best upper bound on circular choosability for planar graphs?
Havet et al. [11] showed that Conjecture 10 implies Conjecture 9.They also proved that CBC q (P, F) ≤ 2q + 4 and conjectured that this upper bound can be reduced by at least one.
It might even be possible that the bound 2q + 3 is not optimal.
Regarding matching backbones, Broersma et al. [6] proved that for BBC q (P, M) = q + 3 for all q ≥ 3 and conjecture that the same holds for q = 2.
Proposition 13 (Broersma et al. [6]) CBC q (P, M) ≤ 2q + 2. They also show an example for which the bound is attained for q = 2, so CBC 2 (P, M) = 6.For larger value of q, it is still open if the bound 2q + 2 is best possible.Problem 14 Is it true that CBC q (P, M) ≤ 2q + 1, for all q ≥ 3 ?
We also believe that Conjecture 11 and Proposition 13 extend to list circular backbone colouring.
In support of both parts of this conjecture, we derive from our result of Section 4, that if G is a bipartite planar graph and F a forest in G, then CBC q (G, F ) ≤ 2q + 2. Observe that by (3), we have CBC q (G, F ) ≤ 2q.We also show that CBC q (P, M) ≤ 2q + 3 (Theorem 42).Our proof also uses the approach of Thomassen's 5-choosability proof [19].
2 Bounding CBC q (G, H) and CBC q (G, H) in terms of χ (H) and χ (G) We believe the bound M q (k) of Theorem 16 is far from being tight.In particular, when H is a matching, we can get an upper bound that is a lot smaller.
Theorem 17 Let G be a graph and M a matching in G.
(ii) for any q ≥ 3 and any α > 1, there is In order to prove this theorem, we need some definitions.Let A and B be two subsets of N (resp.Z p ).We say that A and B q-interfere if there exists a ∈ A and b ∈ B such that a = b and |a − b| < q (resp.|a − b| p < q).
For any pair of non-negative integers, we define f q (a, b) to be the smallest integer m such that for any two subsets A and B of N of cardinality at least m, there exist two non-q-interfering sets A ⊂ A and B ⊂ B such that |A | = a, |B | = b.Similarly, let g q (a, b) be the smallest integer m such that for any p and any two subsets A and B of Z p of cardinality at least m, there exist two non-q-interfering sets A ⊂ A and B ⊂ B such that |A | = a, |B | = b.These two functions are closely related.
Proof: If two sets q-interfere in N, then they also q-interfere in Z r (for r sufficiently large).So f q (a, b) ≤ g q (a, b).Now let C and D be two sets in Z p of cardinality f q (a, b) + q − 1, and consider A = C \ {0, . . ., q − 2} and B = D \ {0, . . ., q − 2}.Then |A|, |B| ≥ f q (a, b), so there exist two sets A ⊂ A and B ⊂ B of cardinality a and b respectively which do not q-interfere in N.But by the definitions of A and B, A ⊂ C, B ⊂ D and A and B cannot q-interfere in Z p . 2 Let us define h q by h q (k) = g q (k, k).The function h q is useful for our purpose as shown by the following lemma.
Lemma 19 Let G be a graph and M a matching in G. Then CBC q (G, M ) ≤ h q (χ (G)).
Proof: Let k = χ (G).Let L be an h q (k)-list assignment in Z p .By definition of h q , for every edge uv ∈ M , we can find two k-sets L (u) ⊂ L(u) and L (v) ⊂ L(v) that do not q-interfere.Since G is k-choosable, there is an L -colouring c of G.
We now show that c is a q-backbone colouring of (G, M ).First, for any edge xy ∈ E(G), c(x) = c(y).For any edge uv ∈ M , c(u) = c(v).In addition, since L (u) and L (v) do not q-interfere, |c(u)−c(v)| p ≥ q. 2 Notation For two integers a and b, we denote by [a, b] p , the set {a, a + 1, . . ., b}, where the numbers are modulo p.For every integer a, we denote by [a] p q the set [a − q + 1, a + q − 1] p .Very often p is either clear from the context or implicit and we omit the superscript p.
Lemma 20 (ii) For any integer q ≥ 3 and any real number α ≥ 1, we have f q (a, b) ≤ α(a + b) + R(q, α).(ii) We prove the result by induction on a + b.Let A and B be two sets of integers of size α(a + b) + R(q, α).
Assume a < ψ q,α (i * ).Then one can choose any a-subset A of A. It forbids at most a × (2q − 2) elements to be chosen in B. So since B is of cardinality at least b + a + R(q, α) ≥ b + (2q − 2)a, we can find a b-subset B that does not q-interfere with A .
We get the result similarly if b < ψ q,α (i * ).Henceforth we now assume that a and b are both greater or equal to ψ q,α (i * ).
Let x be the smallest element of A ∪ B. Without loss of generality, we may assume that x ∈ A.
0), then C 1 and D 1 are both of cardinality at least α(a + b − 1) + R(q, α).Hence by the induction hypothesis, we can find an (a − 1)-set C in C 1 and a b-set B in D 1 which do not q-interfere.Then, A = C ∪ {x} and B are the desired sets.Hence we may assume that at least ψ q,α (1) elements of B are in [x, x + q − 1].
We now extend the argument by removing the ψ q,α (1) smallest elements of B. For any k, let A(k) (resp.B(k)) be the set of the k smallest elements in A (resp.B).Consider D 2 = B \ B(ψ q,α (1)) and , then C 2 and D 2 are both of cardinality at least α(a+b−ψ q,α (1))+R(q).Hence by the induction hypothesis, we can find an a-set A in C 2 and a (b − ψ q,α (1))-set D in D 2 which do not q-interfere.Then, A and B = D ∪ B(ψ q,α (1)) are the desired sets.(They do not interfere because, as we proved in the last paragraph, B(ψ q,α (1)) ⊆ [x, x + q − 1]).Hence we may assume that A(ψ q,α (2)) ⊆ [x, x + 2(q − 1)].
We continue to extend the argument for ψ q,α (i), with 2 ≤ i ≤ i * , alternating the roles of A and B, as follows.If i is odd, consider C i = A \ A(ψ q,α (i − 1)) and , then C i and D i are both of cardinality at least α(a + b − ψ q,α (i − 1)) + R(q, α).Hence by the induction hypothesis, we can find an (a − ψ q,α (i − 1))-set C in C i and a b-set B in D i which do not q-interfere.Then, A = C ∪ A(ψ q,α (i − 1)) and B are the desired sets.Hence we may assume that B(ψ q,α (i , then C i and D i are both of cardinality at least α(a + b − ψ q,α (i − 1)) + R(q, α).Hence by the induction hypothesis, we can find an a-set A in C i and a (b − ψ q,α (i − 1))-set B in D i which do not q-interfere.Then, A and B = D ∪ B(ψ q,α (i)) are the desired sets.Hence we may assume that A(ψ q,α (i)) ⊂ [x, x + i(q − 1)].
Remark 22 a + b + 1 may not be replaced by a + b, even if a = b, in Corollary 21 (i).Just consider a = b and Z 4a with A the set of even numbers and B the set of odd numbers.One can check that any pair of two a-sets A ⊂ A and B ⊂ B 2-interfere.
Theorem 17 derives directly from Corollary 21 and Lemma 19.It can also be generalized to any subgraph H. Indeed in the very same way as Lemma 19, we can prove the following.
Lemma 23 Let G be a graph, H a subgraph of G and M a matching in H. Then CBC q (G, H) ≤ h q (CBC q (G, H − M )).
Hence by induction, we get the following theorem, where χ (H) denotes the chromatic index of H.
Theorem 24 Let G be a graph and H a subgraph of G and let q ≥ 2 be an integer.Then CBC q (G, H) ≤ 2 χ (H) (1 + o(1))χ (G).
Remark 25 In the case q = 2, one can get rid of the o(1) term and show that CBC 2 (G, H) ≤ 2 χ (H) χ (G)+ 1.It suffices to observe that we can remove one fixed colour so that the colours behave like on N. Then for each matching M of a proper edge-colouring, we must divide the list by 2 so that there is no 2-interference between vertices linked by an edge in M .

2-choosable graphs
In this section we study list backbone colouring of 2-choosable graphs.
Let t and q be two positive integers.The (t, q)-kernel of (G, H) is the maximal subset By convention, CBC q (N, N ) = 1 if N is the null graph.
Lemma 26 Let t and q be two positive integers, and let G be a graph and M a matching in G, and let K be its (t, q)-kernel.
Let L be an t-list assignment on G in Z r .It is easy to see that for any subgraph Now by definition of K, there exists a sequence v 1 , . . ., v p of V (G) \ K such that for all Lemma 26 and Theorem 6 immediately imply the following.
Corollary 27 Let G be a graph and H a nonempty subgraph of G.If the heart of G is a single vertex or an even cycle, then CBC q (G, H) = 2q.
Theorem 28 If G is a 2-choosable graph and M a matching, then CBC q (G, M ) = 2q.
Proof: Clearly, it is enough to prove the statement for connected graphs.So we may assume that G is connected.
Let H be the heart of G, and let Observe that if H is a θ 2m,2,2 for some integer m ≥ 1, with x and y its two vertices of degree three, then H −{x, y} has three connected components of odd order.Hence, there is a vertex v of H \ {x, y} that is not matched in N , and for which d By Theorem 4, it follows that the (2q, q)-kernel of (G, M ) is either the empty set, or a set K such that G[K] is an even cycle and M [K] a perfect matching of G[K].Hence, by Theorem 6, CBC q (G[K], M [K]) ≤ 2q, and so by Lemma 26, CBC q (G, M ) ≤ 2q. 2 Remark 29 In the above proof, Theorem 6 is not necessarily required as it is easy to see that CBC q (G, M ) ≤ 2q for an even cycle G and a matching M .
Theorem 30 If G is a 2-choosable graph and H is a subgraph of G not containing its heart, then CBC q (G, H) ≤ 2q + 1.
Proof: Again it is enough to prove the statement for G connected.If H does not contain the heart of G, then, using Theorem 4, one easily sees that the (2q + 1, q)-kernel of (G, H) is either the empty graph or an even cycle.Therefore Theorem 6 and Lemma 26 yield the result.2

Weighted circular list colouring via the Combinatorial Nullstellensatz
In this section we shall use the following theorem, called the Combinatorial Nullstellensatz, to establish some results on circular backbone colouring.
Theorem 31 Let F be an arbitrary field, and let P = P (x 1 , . . ., x n ) be a polynomial in Suppose that the degree deg(P ) of P is n i=1 t i , where each t i is a non-negative integer, and suppose the coefficient of Let G be a graph with vertex set {v 1 , . . ., Alon and Tarsi [3] applied the Combinatorial Nullstellensatz to the graph polynomial to obtain the following theorem on choosability.
Theorem 32 (Alon and Tarsi [3]) Let D be an orientation of a graph G, and let ee(D) and oe(D) respectively the numbers of even and odd eulerian spanning subdigraphs of D. If ee(D) = oe(D), then for any list assignment Let G be a graph with vertex set {v 1 , v 2 , . . ., v n } and let p and q be two positives integers.Its (p, q)circular polynomial of G is Let γ : {0, . . ., p − 1} → C be the function defined by γ(k) = exp(2πik/p).It is obvious that a mapping c : V (G) → {0, . . ., p − 1} is a (p, q)-colouring of G if and only if Applying the Combinatorial Nullstellensatz to the (p, q)-circular polynomials, Norine, Wong and Zhu [17] established a generalization of Theorem 32, from which they deduced the following.
Theorem 33 (Norine, Wong and Zhu [17]) Suppose G is a graph and D is an orientation of G which contains no odd directed cycles.Let L be a In this section, we shall extend Theorem 33 to w-circular colouring.We will prove the following theorem.
Theorem 34 Suppose G is a graph with a positive integer edge weighting w.Suppose moreover that D is an orientation of G which contains no odd directed cycles.If L is a Z p -list assignment for G such that Let G be a graph with vertex set {v 1 , v 2 , . . ., v n }, w an edge weight of G, and p a positive integer.The (p, w)-circular polynomial of G is Observe that when w(e) = q for all e ∈ E(G), then CP G p,w = CP G p,q .Let γ : Z p → C be defined as Thus, the graph G is w-circularly L-colourable if and only if there exist Thus by Theorem 31, if there is an orientation D such that the coefficient of the monomial φ(e).
A mapping that is consistent and eulerian is said to be nice.A mapping φ is called even (respectively, odd) if e∈E(D) φ(e) is even (respectively, odd).
For an arc (v j , v j ) of D of weight q and for 0 ≤ φ(e) ≤ 2q − 1, the coefficient of x It is easy to see that a w-consistent mapping φ makes a contribution to the monomial Observe now that each a l (p, q) is a real because it is equal to its conjugate.We can also see that a 0 (p, q) = a 2q−1 (p, q) = 1.A result of Evans and Montgomery [8] implies that if p ≥ 2q, then the values of a l (p, q), for fixed p and q, are unimodal.It follows that all values of a l (p, q) are at least 1 -Norine, Wong and Zhu proved this from first principles for certain values of p and q [17].
Lemma 36 Let p and q be two positive integers such that p ≥ 2q.For any 0 ≤ l ≤ 2q − 1, a l (p, q) ≥ 1.
From Lemmas 35 and 36, we now derive Theorem 34.For a nice mapping φ, let D φ be the multi-digraph obtained from D by replacing each arc (v j , v j ) by φ(e) parallel arcs from v j to v j .Since φ is eulerian, then D φ is an eulerian multi-digraph.Each directed cycle of D φ corresponds to a directed cycle of D. Since D has no directed cycle of odd length, D φ has no directed cycle of odd length.Thus |E(D φ )| is even, i.e. e∈E(D) φ(e) is even.So D has no odd nice mapping, and so φ nice and odd A(p, w, φ) = 0. Now Lemma 36 implies that all the A(p, w, φ) are positive.But there is at least one nice even mapping, namely the one for which φ(e) = 0 for all e ∈ E(D).So φ nice and even A(p, w, φ) > 0.

Proof
2 We shall now apply Theorem 34.We need the following simple lemma which was first proved by Hakimi [9].It also appears independently in several papers [18,2,3].
Lemma 37 A graph G has an orientation D with maximum outdegree at most ∆ + if and only if Mad(G)/2 ≤ ∆ + .
For a graph G and an edge weighting w, we define M (G, w) := i∈N * (2i − 1) Mad(G i )/2 , where G i is the subgraph of G induced by the edges of weight i, for all i ∈ N * .
Corollary 38 Let G be a graph and w an edge-weighting.Then G has an orientation D such that d + w (v) ≤ M (G, w) for all v ∈ V .
To prove this theorem, we need the following lemma.
Lemma 44 Let q be a positive integer and let S and T be two sets of size 2q and 2q + 3 respectively in Z p .Then there exist two elements a and b of S such that |T \ ([a] q ∪ [b] q )| ≥ 3.
Proof: The result holds trivially when q = 1.
Let us now prove the case q = 2.If two elements a and b of S are consecutive in Z p , then [a] 2 ∪[b] 2 = 4, and we trivially have the result.So we may assume that the elements of S are pairwise non-consecutive.Let 1 ≤ s 1 < s 2 < s 3 < s 4 ≤ p be the elements of S and set Because the elements of S are pairwise non-consecutive, |I ∩ J| ≤ 2. But, Hence either I ∩ T or J ∩ T has size at most 4. Therefore either s 1 and s 2 or s 3 and s 4 are the desired elements of S.
Suppose now that q ≥ 3. We prove the result by induction on p.If two elements a and b of S are consecutive (in particular if p ≤ 4q − 1), then [a] q ∪ [b] q = 2q, and we trivially have the result.Thus we may assume that the elements of S are pairwise non-consecutive.
Assume now that an element, say p, is not in S ∪ T .Then apply induction on Z p−1 with the same S and T .It is easy to see that the two elements a and b such that |T \ ( By the above arguments, each J i contains at least one vertex of T , so at most 3 vertices are in T ∩ S. Moreover a vertex of T is in exactly one I i if it is in T \ S and in exactly two I i if it is in T ∩ S. Hence if q ≥ 4, there must be an I i such that |T ∩ I i | = 1.Then s i and s i+1 satisfy |T \ ([s i ] q ∪ [s i+1 ] q )| ≥ 3.If q = 3, then either there is an I i such that |T ∩ I i | = 1 and we get the result, or S ∩ T is one of the two sets {s 1 , s 3 , s 5 } and {s 2 , s 4 , s 6 }.In that case, one can check that |T \ ( Proof Proof of Theorem 43: The proof is by induction on the number of vertices n.The result holds if G is a triangle since there is at least 3 − 2 = 1 choice to colour the last vertex v if it is not matched, and at least 2q + 1 − 2q = 1 choice to colour the last vertex v if it is matched.Assume now that the result is true for every near triangulation with at most n − 1 ≥ 3 vertices, and let G be a near triangulation with n vertices.We let u 1 u 2 . . .u k be the outer cycle of G, and u 1 and u 2 be the two precoloured vertices with respective colours c 1 and c 2 .
Case 1: G has a chord u i u j with i < j.We use the induction hypothesis on the near triangulation whose outer cycle is u 1 u 2 . . .u i u j u j+1 . . .u k u 1 .Next we use the induction hypothesis on the near triangulation whose outer cycle is u i u i+1 . . .u j u i , the two precoloured vertices being u i and u j .The result follows easily.
Case 2: G has no chord.Let v 1 , . . ., v d be the neighbours of u k that do not belong to C. Without loss of generality, we can assume that u k−1 v 1 v 2 . . .v d u 1 is a path.Set G = G − u k , noting that G is a near-triangulation.
We shall distinguish four cases: • Since this applies for any choice of S, for at least 2q − 1 distinct elements c of L(u k−1 ), there is a circular q-backbone L -colouring such that u k is assigned c.One of these values of c is not in [a] q ∩ [b] q .Hence we can extend the corresponding colouring by colouring u k with a if c / ∈ [a] q and with b otherwise, to obtain a circular q-backbone L-colouring of G.
Proof: (i) We prove the result by induction on a + b, the result holding trivially when a = 0 or b = 0. Let A and B be two sets of integers of size a + b, and let x be the minimum ofA ∪ B. Suppose first x ∈ A ∩ B. Let C = A \ {x, x + 1},and D = B \ {x, x + 1}.Then |C|, |D| ≥ a + b − 2, so by the induction hypothesis, there exist two non-2-interfering sets C ⊂ C and D ⊂ D such that |C | = a − 1, |D | = b − 1. Setting A = C ∪ {x} and B = D ∪ {x}, we obtain the desired sets.Suppose now x / ∈ A ∩ B. Then x is only in one of these two sets, say A. Let C = A \ {x}, and D = B \ {x + 1}.Then |C|, |D| ≥ a + b − 1, so by the induction hypothesis, there exist two non-2interfering sets C ⊂ C and D ⊂ D such that |C | = a − 1, |D | = b.Setting A = C ∪ {x} and B = D , we obtain the desired sets.
Proof of Theorem 34: By Lemma 35, it suffices to show that φ nice and even A(p, w, φ) = φ nice and odd A(p, w, φ).

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Subcase 3: u k is matched to v j , 1 ≤ j ≤ d.Since L(u k ) ≥ 2q + 1, by Lemma 44, there exist two colours a and b inL(u k ) \ {c 1 } such that |L(v j ) \ ([a] q ∪ [b] q )| ≥ 3. We define the list assignment L of G by L (v) := L(v) if v / ∈ {v 1 , v 2 , . . ., v d }, L (v j ) := L(v j ) \ ([a] q ∪ [b] q ) and and L (v) := L(v) \ {a, b} otherwise.Then |L (v i )| ≥ 2q + 3 − 2 = 2q + 1 for each i = j.Thus we can apply the induction hypothesis to G and L .Now we complete the colouring of G by colouring u k with a if c(u k−1 ) = a and with b otherwise.•Subcase 4: u k is not matched.Since L(u k ) ≥ 3, there exist two colours a and b in L(u k ) \ {c 1 }.We define the list assignmentL of G by L (v) := L(v) if v / ∈ {v 1 , v 2 ,. . ., v d } and L (v) := L(v) \ {a, b} otherwise.Then |L (v i )| ≥ 2q + 3 − 2 = 2q + 1 for each i.Thus we can apply the induction hypothesis to G and L .Now we complete the colouring of G by colouring u k with a if c(u k−1 ) = a and with b otherwise.