 # Towards Cereceda's conjecture for planar graphs

The reconfiguration graph R_k(G) of the k-colourings of a graph G has as vertex set the set of all possible k-colourings of G and two colourings are adjacent if they differ on the colour of exactly one vertex. Cereceda conjectured ten years ago that, for every k-degenerate graph G on n vertices, R_k+2(G) has diameter O(n^2). The conjecture is wide open, with a best known bound of O(k^n), even for planar graphs. We improve this bound for planar graphs to 2^O(√(n)). Our proof can be transformed into an algorithm that runs in 2^O(√(n)) time.

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## 1 Introduction

Let be a graph, and let be a non-negative integer. A -colouring of is a function such that whenever . The reconfiguration graph of the -colourings of has as vertex set the set of all -colourings of and two vertices of are adjacent if they differ on the colour of exactly one vertex. Let be a positive integer. Then is said to be -degenerate if every subgraph of contains a vertex of degree at most . Expressed differently, is -degenerate if there is an ordering of its vertices such that has at most neighbours with .

In the past decade, the study of reconfiguration graphs for graph colourings has been the subject of much attention. One typically asks whether the reconfiguration graph is connected. If so, what is its diameter and, in case it is not, what is the diameter of its connected components? See [4, 8, 9] for some examples. Computational work has focused on deciding whether there is a path in the reconfiguration graph between a given pair of colourings [5, 10, 14]. Other structural considerations of the reconfiguration graph have also been investigated in [1, 2]. Reconfiguration graphs have also been studied for many other decision problems; see  for a recent survey.

We remark that reconfiguration problems for graph colourings do not have known results for which the reconfiguration graph is connected but has a diameter that is not polynomial in the order of the graph. (In nearly all cases, the diameter turns out to be quadratic in the number of vertices.) On the other hand, the problem of deciding whether a pair of colourings are in the same component of the reconfiguration graph tends to be PSPACE-complete whenever the reconfiguration graph is disconnected. There are exceptions to this pattern such as, for example, deciding whether a pair of 3-colourings of a graph belong to the same component .

Given a -degenerate graph , it is not difficult to show that is connected . The foregoing pattern motivated Cereceda  to conjecture that has diameter that is quadratic in the order of .

###### Conjecture 1.

Let be a positive integer, and let be a -degenerate graph on vertices. Then has diameter .

Conjecture 1 has resisted several efforts and has only been verified (other than for trees) for graphs of bounded treewidth  and graphs with degeneracy at least where denotes the maximum degree of the graph .

In the expectation of the difficulty of Conjecture 1, Bousquet and Perarnau  have shown that for every and and every graph with maximum average degree , the diameter of is for some constant ; see  for a short proof. Their result in particular implied that the reconfiguration graph of -colourings for planar graphs has diameter that is polynomial in the order of the graph. Since planar graphs are -degenerate, the one outstanding case of Conjecture 1 restricted to planar graphs is thus (aside, of course, from improving the constant term in the exponent of the diameter); see also [3, Conjecture 16]. On the other hand, the best known upper bound on the diameter in Conjecture 1 is – even for planar graphs – and this follows from . In this note, we significantly improve this bound for planar graphs.

###### Theorem 1.

For every planar graph on vertices, has diameter at most .

## 2 Proof of Theorem 1

In this section, we prove Theorem 1. We begin with the following three lemmas. In the first lemma, we obtain a crude bound on the number of recolourings required for degenerate graphs to reduce the number of colours by one.

###### Lemma 1.

Let , and let be a -degenerate graph on vertices. Let be the set of vertices of of degree at least . If is a -colouring of , then we can recolour to some -colouring of by at most recolourings.

###### Proof.

We generalise the proof in  by describing an algorithm that finds a sequence of recolourings from to some -colouring of in time.

Let us fix a -degenerate ordering of and, without loss of generality, let appear before in whenever . In the following, we will describe an algorithm that given an index outputs a sequence of recolourings with the following properties:

• for , is not recoloured,

• for , is recoloured at most times, where is the first degree at least vertex with index at least in , and

• is recoloured once to a different colour.

Notice that the algorithm takes recolourings to recolour . Hence, by repeatedly using such a sequence on the lowest index of a vertex coloured by the colour , we can obtain the colouring in which colour does not appear using recolourings.

Given and a -degenerate ordering of , the algorithm recolour works as follows:

1. If there is a colour that is not used on or any of its neighbours, then recolour to and terminate.

2. If has degree exactly , let be the neighbour of that is latest in and let be the colour of . The algorithm first calls recolour and then recolours to .

3. If has degree at least , then

1. let be a colour not appearing on or any of its at most neighbours earlier in the ordering and

2. let be the neighbours of later in the ordering with .

3. For each in the ascending order: if colour of is at this point, then call recolour.

4. Recolour to .

We simultaneously prove the correctness and properties (i)–(iii) of the algorithm by induction on . Clearly, whenever Step 1. of recolour applies, we only recolour once to a colour not appearing on or any of its neighbours. Moreover, if , then, since is a -degenerate ordering, has degree at most and Step 1. again applies.

For the induction step, let us assume that , Step 1. of recolour does not apply and for all the algorithm recolour is correct and satisfies properties (i)–(iii). Since Step 1. does not apply, the degree of is at least . We distinguish two cases.

• has degree .

In this case recolour applies Step 2. Since Step 1. does not apply, each colour appears either on or on one of its neighbours. Since there are colours, it follows that each colour appears precisely once in the closed neighbourhood of . Note that since is a -degenerate ordering, the latest neighbour of is after in . Hence, properties (i) and (ii) follow from properties (i) and (ii) for recolour. Finally, the correctness and property (iii) follow from the fact that recolour is, by induction, both correct and recolours the unique neighbour of of colour before recolouring to colour .

• has degree at least .

In this case recolour applies Step 3. Again, the algorithm applies the recursive calls only on the vertices that are later in than and hence property (i) is satisfied. After the execution of Steps (a)–(c), colour no longer appears on or any of its neighbours, after which is recoloured to at Step (d). So the algorithm is correct and property (iii) holds. To prove (ii), notice that the algorithm calls recolour at most number of times and always for . Hence, each vertex will get recoloured at most , where , and property (ii) follows. ∎

The maximum average degree of a graph is defined as

By Euler’s formula, the maximum average degree of a planar graph is strictly less than six. By definition, if a graph has maximum average degree strictly less than for some positive integer , then this graph is also -degenerate.

In our next lemma, we show that we can reduce the number of colours by one using subexponentially many recolourings if we further assume our graph to have bounded maximum average degree.

###### Lemma 2.

Let , and let be a a graph on vertices and with . If is a -colouring of , then we can recolour to some -colouring of by recolourings.

###### Proof.

We shall prove by induction on the size of that we can recolour to a -colouring of such that each vertex in is recoloured at most times, which implies the lemma. We note that our inductive proof can be easily transformed to a recursive algorithm running in time .

As the base case, we show how to recolour by at most recolourings any graph with that contains at most vertices of degree at most . Let be a graph with that contains at most vertices of degree at most , let be a -colouring of , and let .

###### Claim 1.

We can recolour to some -colouring of by recolourings.

###### Proof of Claim..

Due to Lemma 1, we only need to show that , where is the set of vertices of degree at least in . Let be the set of vertices of degree less than or equal to in , and let be the set of vertices of degree precisely in . We can assume that is connected, because otherwise we can prove the claim for each connected component of .

Since is connected,

 (s∑i=1deg(ui))+(t∑i=1deg(wi)) ≥ (s∑i=1deg(ui))+t. (1)

On the other hand, since ,

 (s∑i=1deg(ui))+(t∑i=1deg(wi))+(h−s−t∑i=1deg(zi))<(k+1)h (2) ⟺ (s∑i=1deg(ui))+(t∑i=1deg(wi))+(k+1)(h−s−t)<(k+1)(h−s−t+s+t) ⟺ (s∑i=1deg(ui))+(t∑i=1deg(wi))<(k+1)(s+t).

Combine Inequalities (1) and (2):

 (s∑i=1deg(ui))+t<(k+1)(s+t)⟹(k+2)s+t<(k+1)(s+t)⟺s

Since, by assumption, , it follows that . Substituting these bounds into gives us

 s∑i=1deg(ui)

By the AM-GM inequality of arithmetic and geometric means, it holds that

 ∑si=1deg(ui)s≥(s∏i=1deg(ui))s−1. (4)

Combining Inequalities (3) and (4) we get

 as>(s∏i=1deg(ui))s−1,

or, since both sides of the inequality are positive, equivalently

 f(s)=(as)s>s∏i=1deg(ui).

It remains to find an upper bound for the expression when is between and . The derivative of with respect to is given by

 f′(s)=(as)s⋅(log(as)−1),

and since is positive for each , it follows that is maximized when . Therefore, we obtain

 (2(k+1))k(k+1)2√h>s∏i=1deg(ui),

finishing the proof of the claim. ∎

For the inductive step, suppose that contains more that vertices of degree at most and that we can recolour any subgraph of with vertices to some -colouring such that each vertex get recoloured at most times. Let be an independent set in containing only vertices of degree at most of size at least . Since can be greedily coloured with colours using its -degenerate ordering and contains more that vertices of degree at most , such a set exists and can be found in polynomial time. By the inductive hypothesis we can recolour the graph to some -colouring such that each vertex get recoloured at most times for some constant . We can extend this sequence of recolourings to a sequence in by recolouring a vertex whenever some neighbour of gets recoloured to its colour (this is possible because the number of colours is and has at most neighbours in ). At the end of the sequence, we can recolour each vertex of to a colour other than . It follows that the maxmimum number of times a vertex of is recoloured satisfies the inequality

 f(n)≤k⋅(kck2√h)+1≤k⋅(kck2√n−2√n)+1=kck2√n−2√n+1+1.

Since , to show that , it suffice to show that for each . Adding to both sides of the inequality and then squaring yields the result. ∎

In our next lemma, we adapt the proof method introduced in  to show that we can further reduce the number of colours by one for planar graphs.

###### Lemma 3.

Let be a planar graph, and let be a -colouring of . Then we can recolour to some -colouring of using seven colours by at most recolourings for some constant .

###### Proof.

Let be any subgraph of , and let . An independent set of is said to be special if it contains at least vertices and every vertex of has at most neighbours in . Let be the set of vertices of of degree at most . Then has at least vertices since otherwise

 ∑v∈V(H)deg(v)≥∑v∈V(H)−Sdeg(v)>7(h−h7)=6h,

which contradicts that . Let be a maximal independent subset of . Each vertex of has at most neighbours in and every vertex of has at least one neighbour in . Therefore, and so is a special independent set as needed.

Let us prove by induction on the order of that there is a sequence of recolourings from a -colouring of to some -colouring of . We will then argue that at most recolourings have been performed for some constant , thereby finishing the proof.

Let be a special independent set of , and let be the graph obtained from by

• removing all vertices of degree in from and

• for each vertex in of degree , deleting and identifying a pair of neighbours of that are coloured alike in (such a pair always exists since at most colours appear on and its neighbours).

Notice that is planar (one can think of some embedding of in the plane and then note that the neighbours of any vertex form part of the boundary of a face in ; thus, indentifying a pair of neighbours of inside the interior of in the graph does not break the planarity).

Let denote the colouring of that agrees with on and such that, for each , if is the vertex obtained by the identification of some vertices and of , then . Graph has less vertices than , so can we apply our induction hypothesis to find a sequence of recolourings from to some -colouring of .

We let be the -colouring of that agrees with on and such that, for each pair of vertices identified into a new vertex , . We can transform to by

• recolouring and using the same recolouring as for every pair identified into a vertex ;

• recolouring each using the same recolouring.

We can extend this sequence to by recolouring each vertex of to a colour from not appearing on it or its neighbours (this is possible since each vertex of either has degree at most or has degree but with at least two neighbours that are in some sense always coloured alike). At the end of this sequence, we recolour each vertex of of colour to another colour (this is again possible by the same reasoning). So our aim of transforming into a -colouring is achieved unless some vertex of has colour .

Suppose that there is a vertex of with colour . We emulate the proof of the -Colour Theorem to show that we can recolour to a colour from without introducing new vertices of colour or . By repeating the same procedure at most times, we can transform into a -colouring of , as needed. For this, we require some definitions.

Let and be two colours. Then a component of a subgraph of induced by colours and is called an -component. Suppose that  is an -component, . Then colours and are said to be swapped on if the vertices coloured are recoloured , then the vertices coloured are recoloured , and finally the vertices initially coloured are recoloured . Since no vertex coloured or in is adjacent to a vertex of colour , it is clear that each colouring is proper and that no new vertices of colour are introduced.

If for a vertex at least one colour in does not appear on its neighbour, we can immediately recolour . So we can assume that has either degree or with precisely five neighbours coloured distinctly. Suppose these neighbours appear in this order in a plane embedding of . Let us denote by the colour of (). If the -component that contains does not contain , we swap colours , on (this is possible since colour is not used on ), which in turn allows us to recolour to . So we can assume that contains both and . In the same vein, the vertices and must be contained in the same -component . By the Jordan Curve Theorem, this is impossible. Hence, either does not contain both and or does not contain both and and we are able to recolour , as required.

Let us now estimate the number of recolourings of a vertex

in terms of the number of recolourings of vertices of . When recolouring  to a -colouring that uses only colours to on , is recoloured at most five more times than any of its neighbours (this bound is achieved if is recoloured every time one of its neighbours is recoloured and these neighbours are recoloured the same number of times). Moreover, recolouring to a -colouring of has cost an additional recolourings per vertex. Therefore, the maximum number of recolourings per vertex satisfies the recurrence relation

 f(n)≤5⋅f(n−n49)+O(n),

and the theorem follows by the master theorem. ∎

We also require some auxiliary results whose algorithmic versions (running in polynomial time) is implicit in the respective papers.

###### Lemma 4 ().

Let and be positive integers, , and let be a -degenerate graph on vertices. Then has diameter .

###### Lemma 5 ().

Let be a planar graph. There is a partition such that is an independent set and is a -degenerate graph.

###### Lemma 6 ([15, 18]).

Let be a positive integer, and let be a -degenerate graph. There is a partition such that is an independent set and is a -degenerate graph.

We combine Lemmas 5 and 6 to obtain the following corollary.

###### Corollary 2.1.

Let be a planar graph. Then there is a partition such that and are independent sets and is a -degenerate graph.

We are now ready to prove Theorem 1.

###### Proof of Theorem 1.

Let and be two -colourings of . To prove the theorem, it suffices to show that we can recolour to by recolourings. Combining Lemmas 2 and 3, we can recolour to some -colouring of and to some -colourings by recolourings. We apply Corollary 2.1 to find a partition such that and are independent sets and is a -degenerate graph. From and we recolour the vertices in to colour and those in to colour (the colours that are not used in either or ). Let and denote, respectively, the restrictions of and to . We focus on and as long as we do not use colours and we can recolour to without worrying about adjacencies between and . Since is -degenerate, we can apply Lemma 4 with and to find by recolourings a recolouring sequence from to . This completes the proof of the theorem. ∎

## 3 Final remarks

The reader may have observed from the proof of Theorem 1 that, in order to settle [3, Conjecture 16], it would suffice to show that we can recolour any -colouring of a planar graph to some -colouring by polynomially many recolourings.

###### Problem 1.

Given a planar graph and a -colouring of , can we recolour to some -colouring of by recolourings for some constant ?

In order to obtain a sub-exponential bound on the diameter of reconfiguration graphs of colourings for graphs with any bounded maximum average degree, it would suffice to find a positive answer to the following problem. (The proof of this fact follows by combining Lemma 2 with an affirmative answer to Problem 2 in the same way that Lemmas 8, 9 and 10 in  are combined to obtain Theorem 6 in .)

###### Problem 2.

Let , and let be a graph with . Then there exists a partition of such that is an independent set and .

## Acknowledgements

Eduard Eiben was supported by Pareto-Optimal Parameterized Algorithms (ERC Starting Grant 715744). Carl Feghali was supported by the research Council of Norway via the project CLASSIS.

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