Listedgecolouring planar graphs with precoloured edges
Abstract
Let be a simple planar graph of maximum degree , let be a positive integer, and let be an edge list assignment on with for all . We prove that if is a subgraph of that has been edgecoloured, then the edgeprecolouring can be extended to an edgecolouring of , provided that has maximum degree and either or is large enough ( suffices). If , there are examples for any choice of where the extension is impossible.
1 Introduction
In this paper all graphs are simple.
An edgecolouring of is an assignment of colours to the edges of so that adjacent edges receive different colours; if at most colours are used we say it is a edgecolouring. The chromatic index of , denoted , is the minimum such that is edgecolourable. It is obvious that , where is the maximum degree of , and Vizing’s Theorem [12] says that .
In this paper we are looking to edgecolour a graph , but with the constraint that some edges have already been coloured and cannot be changed. In this scenario we have no control over the edgeprecolouring – if the edgeprecoloured subgraph is , then it will certainly have at least colours, but it could have many more, perhaps even more than colours. If we are looking to extend the edgeprecolouring to a edgecolouring of , then we will certainly need that is at least the maximum degree of , and that the edgecolouring of uses at most colours (i.e. is a edgecolouring). In general we consider the following question, first posed by Marcotte and Seymour [9]:
Question 1.
Given a graph with maximum degree and a subgraph of that has been edgecoloured, can the edgeprecolouring of be extended to a edgecolouring of ?
Marcotte and Seymour’s main result in [9] is a necessary condition for the answer to Question 1 to be “yes”; they prove that this condition is also sufficient when is a multiforest (the condition is rather technical, so we do not state it here). Question 1 was shown to be NPcomplete by Colbourn [4], and Marx [10] showed that this is true even when is a planar 3regular bipartite graph. Since, as Holyer [7] showed, it is NPcomplete to decide whether or not, the special case of Question 1 is also NPcomplete for general graphs. In this paper we focus on Question 1 for planar graphs. Before saying more about planar graphs in particular however, let us make several quick observations about Question 1 in general.
Firstly, if is huge – say at least – then the answer is yes, and moreover, the extension can be done greedily. This is because an edge in sees at most other edges, and when , this value is at most . If the maximum degree of is then this threshold for is actually sharp. To see this, consider the graph shown in Figure 1, formed by taking a copy of with one edge coloured and the rest uncoloured, and joining each leaf to distinct new vertices via edges coloured . Then has maximum degree , as does its edgeprecoloured subgraph. However, in order to extend the edgeprecolouring to a edgecolouring of , we need new colours, which forces .
Given the above paragraph, Question 1 is only interesting when the maximum degree of , say , is strictly less than . Here, we get a natural barrier to extension when , via nearly the same example as above. Let be the graph shown in Figure 2, formed by taking an (uncoloured) copy of and joining each leaf to distinct new vertices, via edges coloured . The resulting graph has maximum degree , and contains a precoloured subgraph with maximum degree . However, in order to extend the edgeprecolouring to , we need new colours, meaning that for a edgecolouring of , we need .
If it happened that was edgecoloured efficiently (i.e. using at most colours), then our problem would be significantly reduced. In this special situation, one could use a completely new set of colours to extend to an edgecolouring of with at most the following number of colours (according to Vizing’s Theorem):
(1) 
That is, when has been edgecoloured efficiently, the answer to Question 1 is yes whenever . Since extension can be impossible when (according to the above paragraph), this makes the only interesting values in this case, with further restrictions if any of the inequalities in (1) are strict. For example, if both and have chromatic index equal to their maximum degrees, then the colouring described above works whenever , and hence we get a sharp threshold. Of course, this only works when has been edgeprecoloured efficiently, and in general we have no control over the edgeprecolouring on .
While edgecolouring is in general an NPhard problem, the situation is somewhat simpler for planar graphs. For there are examples of planar graphs with chromatic index and . However, every planar with is edgecolourable; the case was proved independently by Sanders and Zhao [11] and Zhang [13], and the case was proved by Vizing, who conjectured it should also hold for 6 (as well as the nowestablished 7). When focusing on planar graphs, there are additional techniques at one’s disposal, in particular the socalled discharging method, that make edgecolouring easier.
We make progress on Question 1 in this paper by focusing on planar graphs. In particular, we prove that the answer to Question 1 is yes whenever , provided is small enough or is large enough. As discussed above, the assumption is sharp. (In fact, we actually prove a stronger result involving list edge colouring, stated at the end of this section as Theorem 5, but more exposition is required to properly state and contextualize the stronger result.)
Theorem 2.
Let be a planar graph of maximum degree at most , let be a positive integer, and let be a subgraph of that has been edgecoloured. If has maximum degree at most , then the edgeprecolouring can be extended to a edgecolouring of provided that either:

, or

and
Theorem 2 does not include the case , however the requirement of means that would correspond to being edgeless. Then the problem is not about precolouring at all, but simply about edgecolouring planar graphs as discussed above.
The case of Theorem 5 was previously established by Edwards, Girão, van den Heuvel, Kang, Sereni and the third author [5], with the slightly stronger assumption of . (Note that the restriction of our proof for Theorem 2 to this case provides a somewhat new proof; both arguments use global discharging, but we discharge in a different way). After the seminal work of Marcotte and Seymour [9], the vertexversion of the precolouring extension problem received much more attention than Question 1. Edwards et al. [5] reinitiated this study in their paper, with planar graphs being only one of the many families they considered. The main concern in [5] however is when is a matching, and in order to guarantee extensions they often impose distance conditions on the edges in the precoloured matching. In particular, this means avoiding the issues with being too small as exhibited in Figures 1 and 2. Specifically, in addition to the aforementioned result for , they showed that if is an edgeprecoloured matching in a planar graph where edges are at distance at least 3 from one another, then any edgecolouring on can be extended to provided . More recently, Girão and Kang [6] studied extension from precoloured matchings in general graphs, proving that if is a matching in a (not necessarily planar) graph where edges are distance at least from each other, then any edgecolouring on can be extended to a edgecolouring of .
As Edwards et al. [5] observed, extending an edgecolouring is closely related to listedgecolouring. An edge list assignment on a graph is a function that assigns to each edge a list of colours . If is an edge list assignment on a graph , an edgecolouring of is an edgecolouring of such that every edge is given a colour from . Note that a classical edgecolouring of can be viewed as an edgecolouring for the list assignment defined by for all . A graph is listedgecolourable if it is edgecolourable for every edge list assignment such that for all . The notorious ListEdgeColouring Conjecture (attributed to many sources, some as early as 1975; see [8]) asserts that every is listedgecolourable. If this conjecture is true, then given the above discussion on the chromatic index of planar graphs, should be listedgecolourable whenever (or perhaps 6). This has been verified when .
Theorem 3 (Borodin, Kostochka, and Woodall [2]).
If is a planar graph with maximum degree , then is listedgecolourable.
Borodin [1] proved a similar result; a short proof of this result was later obtained by Cohen and Havet [3].
Theorem 4 (Borodin [1]).
If is a planar graph with maximum degree , then is listedgecolourable.
In the present paper we have in fact proved the listedgecolouring analog of Theorem 2. This stronger result is as follows.
Theorem 5.
Let be a planar graph of maximum degree at most , let be an edge list assignment on with for all , where is a positive integer, and let be a subgraph of that has been edgecoloured. If has maximum degree at most , then the edgeprecolouring can be extended to an edgecolouring of provided that either:

, or

and
We again omit the case , however the required condition means that is edgeless and hence the best result is that of Theorem 3 above. Theorem 5 does have something meaningful to say when edgeless however: the case and gives Theorem 4 precisely.
The following section contains some technical results needed for our proof of Theorem 5, which comprises Section 3. The final section of this paper, Section 4, is about pushing Theorem 5 beyond planar graphs. We show that requiring to be planar is sufficient, and in fact “planar” can be replaced by “nonnegative Euler characteristic”.
2 Technical Lemmas
In this section, we gather some technical lemmas that will be needed for the proof of Theorem 5.
Theorem 6 (Borodin, Kostochka, Woodall [2]).
Let be a bipartite graph and let be an edge list assignment on . If for every edge , then is edgecolourable.
Edwards et al. [5] applied Theorem 6 to obtain a precolouring extension result for bipartite graphs (Theorem 15 of [5]), which we will use as part of our proof. While the result as stated in [5] only applies to classical edgeprecolouring, a listedgecolouring version can be obtained using essentially the same proof:
Theorem 7.
Let be a bipartite multigraph, and let be an edge list assignment on with for all . Let be a subgraph of that has been edgecoloured. If has maximum degree at most , then the edgeprecolouring can be extended to an edgecolouring of provided that .
Proof.
Let . For each edge , let be obtained from by removing all colours used on the edges of incident to . Let be an arbitrary edge of . Now
Since , this implies that
On the other hand,
Thus, , and this holds for all . By Theorem 6, it follows that is edgecolourable, and any edgecolouring of gives the desired edgecolouring of . ∎
In what follows and in the main argument, given a graph , we define as the set of all vertices with , and we define as .
Lemma 8.
Let be a graph of maximum degree at most , and let be an edge list assignment on with for all . Let be a subgraph of with maximum degree at most . Suppose that has been edgecoloured, and that this extends to an edgecolouring of for all , but not to .
Let and , where are positive integers with , , and . Let be the bipartite subgraph of induced by the bipartition . If every vertex has the property that
then
Moreover, if and then the above inequality is strict.
Proof.
Say that an induced subgraph is bad if

for all , and

for all .
Notice that for all , ,
(2) 
and
(3) 
so that if a bad induced subgraph exists, it has no isolated vertices, and in particular has at least one edge. We will first show that has no bad induced subgraph, and then show that this implies the desired claim.
Suppose that has a bad induced subgraph . Let . Since is nonempty, is a proper subgraph of , so by assumption, the edgeprecolouring on extends to an edgecolouring of . We derive a contradiction by showing we can further extend to an edgecolouring of . To this end, let be the edge list assignment on defined as follows: for each edge , is the set of colours from that do not appear on any edge adjacent to . Observe that for each , we have
Since is bad, we have , so that
and likewise so that
Hence, for every , we have . By Theorem 6, is edgecolourable. Now any proper edgecolouring of , combined with the edgecolouring of , yields a proper edgecolouring of that extends the edgeprecolouring of as desired; contradiction.
Hence, contains no bad induced subgraph, and so every induced subgraph of contains a vertex violating the definition of a “bad” subgraph. By iteratively removing these vertices and counting the edges removed when each vertex is deleted, we see that
(4)  
Rearranging the last inequality yields
which is the desired conclusion. If we additionally know that and , then inequalities (2) and (3) become strict. Hence each and is contributing a positive amount to the righthandside of (4). Since the last vertex removed is isolated, this is an overcount, and hence we get a strict inequality. ∎
3 Proof of Theorem 5
For fixed values of , we choose a counterexample where the quantity is as small as possible.
Claim 1.
The edgeprecolouring on can be extended to an edgecolouring of for any .
Proof of Claim.
Let any be given, and let . Note that satisfies the hypotheses of the theorem with . Exactly two vertices in have lower degrees than in , so may be as large as . However, since has one edge less than , we still get that
Hence, by our choice of counterexample, the edgeprecolouring of extends to an edgecolouring of . ∎
Claim 2.
If , then .
Proof of Claim.
By Claim 1, the edgeprecolouring of can be extended to an edgecolouring of . The edge sees at most different colours in , so since is a counterexample, it must be that . ∎
Claim 3.
If , then every edge incident to in is also in .
Proof of Claim.
Assume for contradiction that and is incident to an edge not in , say . By Claim 2, we know that . However, since , this implies that , a contradiction. ∎
Claim 4.
.
Proof of Claim.
Suppose not, and take . By Claim 3, every edge incident to must lie in .
Let and be the graphs obtained from and , respectively, by deleting and, for each , adding a new vertex adjacent only to . We precolour each edge with the same colour received by the edge in the precolouring of . See Figure 3. Observe that the edgeprecolouring of extends to if and only if the edgeprecolouring of extends to .
Now has the same number of edges as , and has one fewer vertex in . As and , our choice of counterexample implies that the edgeprecolouring of extends to , but this means that the edgeprecolouring of extends to as well. ∎
Claim 5.
Every vertex of is either a leaf incident to an edge in , or of degree at least .
Let be the set of faces in with exactly vertices on its boundary having degree 3 or higher in .
Claim 6.
.
Proof of Claim.
Suppose that ; we will show a contradiction. We know that by Claim 5, since . So, if the boundary of contains a cycle, then it contains at least three vertices of degree at least three, yielding a contradiction. Thus, the boundary of contains no cycle. This means that is a forest, and is its one face. In particular, is bipartite. By Theorem 7, this implies that the precolouring of extends to all of , contradicting our choice of as a counterexample. ∎
We now introduce a discharging argument. To each vertex in assign an initial charge of . To each face in assign an initial charge of . We also define an additional structure (a “global pot”) and assign to it an initial charge of . We discharge along the following rules:

For each , every face takes from each vertex of degree 3 or higher on its boundary.

Every vertex takes from its neighbor.
In the special case where for , we also add the following rules:

For every vertex , where :
takes from . 
For every vertex , where :
gives to , where .
While it is not immediately obvious, discharging rules (c) and (d) never apply to the same vertex, due to the following claim.
Claim 7.
If for some , then .
Proof of Claim.
Using Euler’s formula for planar graphs, the sum of initial charges is at most :
(5) 
For each graph element (either a vertex, a face, or the global pot), let denote the final charge of . Since each discharging rule conserves the total charge, we see that . We will achieve our desired contradiction by showing that the final charge of each element is nonnegative.
First consider a face . By Claim 6, for . So according to discharging rule (a) (the only rule affecting ),
Now consider the global pot . We know and that the charge of is unaffected when , so the following claim precisely amounts to showing showing that when .
Claim 8.
If for some , then
(6) 
Proof of Claim.
For each , define and and let be the bipartite subgraph of induced by the partition . We will show we can apply Lemma 8 for each value of , and then we will sum the resulting inequalities to get our desired result. For fixed , this means we want to apply Lemma 8 with parameter choices
and hence to do so we must verify that (true) and that , which is true since
In fact, since both these inequalities hold strictly, we will apply the strict version of Lemma 8. Of course, there are several other hypotheses we must check. In particular, we must verify that , which is equivalent to showing that . Since , we get this inequality by Claim 7. By Claim 1, we can therefore apply Lemma 8 for provided that every vertex has the property that
Consider such a vertex with incident edge in . Since , and by Claim 2, we know that
This means, by definition of , that the edge is in . So we get , as desired.
For any fixed , we can now apply Lemma 8 to get
(7) 
since by the hypothesis of Claim 8, and since, for our choices of parameters,
Dividing (7) by and summing over all yields
(8) 
The lefthandside of (8) is
matching the lefthand side of (6). It remains only to show that the righthandside of (8) equals the righthand side of (6). To this end, note that
Now the bracketed sum can be rewritten as
which is precisely what we needed to prove. ∎
We have now shown , so it remains only to consider the final charge of an arbitrary vertex . If , then only discharging rule (b) affects , and we get
By Claim 5, we may now assume that .
Suppose lies on the boundary of distinct faces and is incident to leaves. We know that is no more than , so . We also know that , by Claim 5 and by definition of . By doubling the first inequality and adding the result to the second inequality we get
(9) 
Since by Claim 6, each of the distinct faces incident to has at least vertices of degree at least 3 on their boundary. This means that each of these faces takes charge at most 2 from , according to discharging rule (a). Each of the leaves incident to takes exactly 3 from , according to discharging rule (b). Hence by inequality (9), after applying discharging rules (a) and (b) (but before considering discharging rules (c) or (d)), the charge of is at least
(10) 
Note that since , the additional discharging rules (c) and (d) are applied precisely when . If , then we do not apply them, and by inequality (10),
We may now assume that for . Let denote the total charge transferred from to according to discharging rules (c) and (d); note that may be positive, negative, or zero. In all cases, by inequality (10), we have that
(11) 
If neither discharging rule (c) nor (d) applies to , then we know that and therefore (11) says that
as desired.
Now suppose that discharging rule (c) applies to (and hence (d) does not, according to Claim 7). In this situation, (11) implies that
Finally, we may assume that discharging rule (d) applies to (and hence (c) does not, according to Claim 7). In this case, we have , where , andF . By (11),
Writing as , where , we can rewrite this lower bound as
Table 1 computes the bracketed quantity for each permissible combination of and . For each possible value of , the hypothesis of Theorem 5 ensures that this lower bound is always nonnegative.
We have proved that for every graph element , and this completes the proof of Theorem 5.
4 Beyond planarity
In the proof of Theorem 5 our initial charges sum to at most , and after discharging the vertices, faces, and global pot all have nonnegative charge. In fact, when we examine inequality (5), we see that the sum of initial charges is at most , where is the Euler characteristic of the plane. Hence our argument works for any surface of positive Euler characteristic; namely may be embedded on the plane or projective plane. Moreover, this embedding requirement need not concern the edges of the precoloured : imagine applying Theorem 5 to the graph obtained by replacing every edge in with a pair of edges and where are new leaves, and and retain the precolouring (and lists) of . Given these observations, we can strengthen Theorem 5 by removing the assumption that “ is planar” and replacing it by the somewhat milder “ can be embedded in a surface of positive Euler characteristic”.
Acknowledgements
We are indebted to an anonymous referee whose insightful comments strengthened our main result.
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