List-edge-colouring planar graphs with precoloured edges

# List-edge-colouring planar graphs with precoloured edges

Joshua Harrelson111jth0048@auburn.edu   Jessica McDonald222mcdonald@auburn.edu; Supported in part by NSF grant DMS-1600551.   Gregory J. Puleo333gjp0007@auburn.edu

Department of Mathematics and Statistics
Auburn University
Auburn, AL 36849
###### 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 -edge-coloured, then the edge-precolouring can be extended to an -edge-colouring 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 edge-colouring 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 -edge-colouring. The chromatic index of , denoted , is the minimum such that is -edge-colourable. It is obvious that , where is the maximum degree of , and Vizing’s Theorem [12] says that .

In this paper we are looking to edge-colour 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 edge-precolouring – if the edge-precoloured 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 edge-precolouring to a -edge-colouring of , then we will certainly need that is at least the maximum degree of , and that the edge-colouring of uses at most colours (i.e. is a -edge-colouring). 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 -edge-coloured, can the edge-precolouring of be extended to a -edge-colouring 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 NP-complete by Colbourn [4], and Marx [10] showed that this is true even when is a planar 3-regular bipartite graph. Since, as Holyer [7] showed, it is NP-complete to decide whether or not, the special case of Question 1 is also NP-complete 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 edge-precoloured subgraph. However, in order to extend the edge-precolouring to a -edge-colouring 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 edge-precolouring to , we need new colours, meaning that for a -edge-colouring of , we need .

If it happened that was edge-coloured 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 edge-colouring of with at most the following number of colours (according to Vizing’s Theorem):

 χ′(G−E(H))+χ′(H)≤χ′(G)+χ′(H)≤Δ+d+2. (1)

That is, when has been edge-coloured 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 edge-precoloured efficiently, and in general we have no control over the edge-precolouring on .

While edge-colouring is in general an NP-hard 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 -edge-colourable; 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 now-established 7). When focusing on planar graphs, there are additional techniques at one’s disposal, in particular the so-called discharging method, that make edge-colouring 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 -edge-coloured. If has maximum degree at most , then the edge-precolouring can be extended to a -edge-colouring of provided that either:

1. , or

2. and

 Δ≥⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩16+d,if d=t,9+d,if d=t−1,8+d,if d=t−2,7+d,if d=t−3.

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 edge-colouring 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 vertex-version of the precolouring extension problem received much more attention than Question 1. Edwards et al. [5] re-initiated 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 edge-precoloured matching in a planar graph where edges are at distance at least 3 from one another, then any -edge-colouring 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 -edge-colouring on can be extended to a -edge-colouring of .

As Edwards et al. [5] observed, extending an edge-colouring is closely related to list-edge-colouring. 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 -edge-colouring of is an edge-colouring of such that every edge is given a colour from . Note that a classical -edge-colouring of can be viewed as an -edge-colouring for the list assignment defined by for all . A graph is -list-edge-colourable if it is -edge-colourable for every edge list assignment such that for all . The notorious List-Edge-Colouring Conjecture (attributed to many sources, some as early as 1975; see [8]) asserts that every is -list-edge-colourable. If this conjecture is true, then given the above discussion on the chromatic index of planar graphs, should be -list-edge-colourable 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 -list-edge-colourable.

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 -list-edge-colourable.

In the present paper we have in fact proved the list-edge-colouring 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 -edge-coloured. If has maximum degree at most , then the edge-precolouring can be extended to an -edge-colouring of provided that either:

1. , or

2. and

 Δ≥⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩16+d,if d=t,9+d,if d=t−1,8+d,if d=t−2,7+d,if d=t−3.

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 “non-negative 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 -edge-colourable.

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 edge-precolouring, a list-edge-colouring 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 -edge-coloured. If has maximum degree at most , then the edge-precolouring can be extended to an -edge-colouring 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

 ∣∣L′(xy)∣∣≥|L(xy)|−degH(x)−degH(y)≥Δ+t−degH(x)−degH(y).

Since , this implies that

 ∣∣L′(xy)∣∣ ≥Δ−degH(x),and ∣∣L′(xy)∣∣ ≥Δ−degH(y).

On the other hand,

 degG′(x) =degG(x)−degH(x)≤Δ−degH(x),and degG′(y) =degG(y)−degH(y)≤Δ−degH(y).

Thus, , and this holds for all . By Theorem 6, it follows that is -edge-colourable, and any -edge-colouring of gives the desired -edge-colouring 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 -edge-coloured, and that this extends to an -edge-colouring 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

 degX(u)≥degG(u)−d,

then

 (t+1−d)|A|≤Δ∑i=b0(a+i−1−(Δ+t))|Vi|.

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 , ,

 degG(u)−t≥a0−t≥1 (2)

and

 a+degG(v)−(Δ+t)≥a+b0−(Δ+t)≥1, (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 edge-precolouring on extends to an -edge-colouring of . We derive a contradiction by showing we can further extend to an -edge-colouring 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

 ∣∣LJ(uv)∣∣≥Δ+t−degG(u)−degG(v)+degJ(u)+degJ(v).

Since is bad, we have , so that

 ∣∣LJ(uv)∣∣ ≥Δ−degG(v)+degJ(v)≥degJ(v),

and likewise so that

 ∣∣LJ(uv)∣∣ ≥a−degG(u)+degJ(u)≥degJ(u).

Hence, for every , we have . By Theorem 6, is -edge-colourable. Now any proper -edge-colouring of , combined with the -edge-colouring of , yields a proper -edge-colouring of that extends the edge-precolouring 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

 |E(X)| ≤∑u∈A[degG(u)−t−1]+∑v∈B[a+degG(v)−(Δ+t)−1] (4) ≤∑u∈A[(degX(u)+d)−t−1]+∑v∈B[a+degG(v)−(Δ+t)−1] =|E(X)|+∑u∈A[d−t−1]+Δ∑i=b0(a+i−(Δ+t)−1)|Vi|.

Rearranging the last inequality yields

 (t+1−d)|A|≤Δ∑i=b0(a+i−1−(Δ+t))|Vi|,

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 right-hand-side 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 edge-precolouring on can be extended to an -edge-colouring 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

 3∣∣E(G′)∣∣+∣∣V[2,t+1](G′)∣∣<3|E(G)|+∣∣V[2,t+1](G)∣∣.

Hence, by our choice of counterexample, the edge-precolouring of extends to an -edge-colouring of . ∎

If , then .

###### Proof of Claim.

By Claim 1, the edge-precolouring of can be extended to an -edge-colouring 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. ∎

.

###### 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 edge-precolouring of extends to if and only if the edge-precolouring 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 edge-precolouring of extends to , but this means that the edge-precolouring of extends to as well. ∎

###### Claim 5.

Every vertex of is either a leaf incident to an edge in , or of degree at least .

###### Proof.

This follows by combining Claim 3 and Claim 4. ∎

Let be the set of faces in with exactly vertices on its boundary having degree 3 or higher in .

.

###### 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:

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

2. Every vertex takes from its neighbor.

In the special case where for , we also add the following rules:

1. For every vertex , where :
takes from .

2. 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.

We get the desired inequality if and only if . If , then we have , so the hypothesis of Theorem 5 yields

 Δ+2ℓ=Δ≥16+d=16+t>8+t.

If we may rewrite hypothesis of Theorem 5 as

 Δ≥10+d−ℓ=10+(t−ℓ)−ℓ=10+t−2ℓ,\ so
 Δ+2ℓ≥10+t>8+t.

Using Euler’s formula for planar graphs, the sum of initial charges is at most :

 α(P)+∑v∈V(G)α(v)+∑f∈F(G)α(f) =0+∑v∈V(G)(3degG(v)−6)+∑f∈F(G)(−6) =6|E(G)|−6|V(G)|−6|F(G)|≤6(−2)=−12. (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 ),

 α′(f)=(−6)+m(6m)=0.

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

 t+5−ℓ∑i=t+2(t+6−ℓ−i)|Vi|<Δ∑j=Δ−3+ℓq(j)(q(j)+1)2(ℓ+1)|Vj|. (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

 a0 =t+2, a =t+5−ℓ−k, b0 =Δ−3+ℓ+k,

and hence to do so we must verify that (true) and that , which is true since

 (t+5−ℓ−k)+(Δ−3+ℓ+k)=t+2+Δ.

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

 degXk(u)≥degG(u)−d.

Consider such a vertex with incident edge in . Since , and by Claim 2, we know that

 degG(v)≥Δ+t+2−degG(u)≥Δ+t+2−(t+5−ℓ−k)=Δ−3+ℓ+k.

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

 (ℓ+1)|Ak|<Δ∑j=Δ−3+ℓ+k(q(j)−k)|Vj|, (7)

since by the hypothesis of Claim 8, and since, for our choices of parameters,

 a+j−1−(Δ+t) =(t+5−ℓ−k)+j−1−(Δ+t) =j−Δ+4−ℓ−k =q(j)−k.

Dividing (7) by and summing over all yields

 3−ℓ∑k=0|Ak|<(1ℓ+1)3−ℓ∑k=0Δ∑j=Δ−3+ℓ+k(q(j)−k)|Vj|. (8)

The left-hand-side of (8) is

 3−ℓ∑k=0|V[t+2,t+5−ℓ−k]| = |V[t+2,t+5−ℓ]|+|V[t+2,t+4−ℓ]|+⋯+|V[t+2,t+2]| = (4−ℓ)|Vt+2|+⋯+2|Vt+4−ℓ|+|Vt+5−ℓ| = t+5−ℓ∑i=t+2(t+6−ℓ−i)|Vi|,

matching the left-hand side of (6). It remains only to show that the right-hand-side of (8) equals the right-hand side of (6). To this end, note that

 j≥Δ−3+ℓ+k⟺k≤j−Δ+3−ℓ=q(j)−1,and so
 3−ℓ∑k=0Δ∑j=Δ−3+ℓ+k(q(j)−k)|Vj|=Δ∑j=Δ−3+ℓ⎛⎝q(j)−1∑k=0(q(j)−k)⎞⎠|Vj|.

Now the bracketed sum can be rewritten as

 q(j)−1∑k=0(q(j)−k)=q(j)+(q(j)−1)+(q(j)−2)+⋯+1=q(j)(q(j)+1)2,

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

 α′(v)=(−3)+3=0.

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

 2x+3y≤2degG(v)+d. (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

 3degG(v)−6−(2x+3y)≥degG(v)−6−d. (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),

 α′(v)≥degG(v)−6−d≥degG(v)−6−(t−4)=degG(v)−(t+2)≥0.

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

 α′(v)≥degG(v)−6−d+p. (11)

If neither discharging rule (c) nor (d) applies to , then we know that and therefore (11) says that

 α′(v)≥(t+5−ℓ+1)−6−d+(0)=(t−d)−ℓ=0,

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

 α′(v)≥degG(v)−6−d+(t+6−ℓ−degG(v))=0.

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),

 α′(v)≥degG(v)−6−d−((degG(v)−(Δ−4+ℓ))(degG(v)−(Δ−5+ℓ))2(ℓ+1)).

Writing as , where , we can rewrite this lower bound as

 α′(v) ≥Δ−h+ℓ−6−d−((4−h)(5−h)2(ℓ+1)) =Δ−d−(6+h−ℓ+(4−h)(5−h)2(ℓ+1)).

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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