1 Introduction
Reconfiguration problems consist in finding stepbystep transformations between two feasible solutions of a problem such that all intermediate states are also feasible. Such problems model dynamic situations where a given solution already in place has to be modified for a more desirable one while maintaining some properties all along the transformation. Reconfiguration problems have been studied in various fields such as discrete geometry [BoseLPV18], optimization [BBRM18] or statistical physics [mohar4]. For a complete overview of the reconfiguration field, the reader is referred to the two recent surveys on the topic [Nishimura17, Heuvel13]. In this paper, our reference problem is graph colouring.
Let be a problem and be an instance of . The reconfiguration graph is the graph where vertices are solutions of and where there is an edge between two vertices if one can transform the first solution into the other in one step (for graph colouring one step means modifying the colour of a single vertex^{2}^{2}2Note that recolouring operations have been studied, for instance recolouring using Kempe chains (see e.g. [BonamyBFJ19]). In this article, we focus on single vertex recolourings.
). Given a reconfiguration problem, several questions may arise. (i) Is it possible to transform any solution into any other, i.e. is the reconfiguration graph connected? (ii) If yes, how many steps are needed to perform this transformation, i.e. what is the diameter of the reconfiguration graph? In this work, we will focus on the diameter of the reconfiguration graph. The diameter of the reconfiguration graph plays an important role, for instance in random sampling, since it provides a lower bound on the mixing time of the underlying Markov chain (and the connectivity of the reconfiguration graph ensures the ergodicity of the Markov chain
^{3}^{3}3Actually, it only gives the irreducibility of the chain. To get the ergodicity, we also need the chain to be aperiodic. For the chains associated to proper graph colourings, this property is usually straightforward.). Since proper colourings correspond to states of the antiferromagnetic Potts model at zero temperature, Markov chains related to graph colourings received a considerable attention in statistical physics and many questions related to the ergodicity or the mixing time of these chains remain widely open (see e.g. [ChenDMPP19, frieze2007survey]).Graph recolouring.
All along the paper denotes a graph, is the size of and is an integer. For standard definitions and notations on graphs, we refer the reader to [Diestel]. A (proper) colouring of is a function such that, for every edge , we have . Throughout the paper we will only consider proper colourings and will then omit the proper for brevity. The chromatic number of a graph is the smallest such that admits a colouring. Two colourings are adjacent if they differ on exactly one vertex. The reconfiguration graph of , denoted by and defined for any , is the graph whose vertices are colourings of , with the adjacency relation defined above. Cereceda, van den Heuvel and Johnson provided an algorithm to decide whether, given two colourings, one can be transformed into the other in polynomial time, and characterized graphs for which is connected [Cereceda09, CerecedaHJ11]. Given any two colourings of , it is complete to decide whether one can be transformed into the other for [BonsmaC07].
The recolouring diameter of a graph is the diameter of if is connected and is equal to otherwise. In other words, it is the minimum for which any colouring can be transformed into any other one through a sequence of at most adjacent colourings. Bonsma and Cereceda [BonsmaC07] proved that there exists a family of graphs and an integer such that, for every graph there exist two colourings whose distance in the reconfiguration graph is finite and superpolynomial in .
Cereceda conjectured that the situation is different for degenerate graphs. A graph is degenerate if any subgraph of admits a vertex of degree at most . In other words, there exists an ordering of the vertices such that for every , the vertex has at most neighbours in . It was shown independently by Dyer et al. [dyer2006randomly] and by Cereceda et al. [Cereceda09] that for any degenerate graph and every , is connected. However the (upper) bound on the recolouring diameter given by these constructive proofs is of order (where is a constant). Cereceda [Cereceda] conjectured that the the diameter of is of order as long as . If correct, the quadratic function is sharp, even for paths or chordal graphs as proved in [BonamyJ12]. The existence of a polynomial upper bound instead of a quadratic one also is open, even for or restricted to particular graph classes such as planar graphs. In what follows, we will call respectively the polynomial (resp. quadratic) Cereceda’s conjecture the question of proving that the recolouring diameter of degenerate graphs is polynomial (resp. quadratic) for .
The quadratic Cereceda’s conjecture is known to be true only for (for trees) [BonamyJ12] and with [FeghaliJP16] (where denotes the maximum degree of the graph). Cereceda [Cereceda] showed that for any degenerate graph and every , the diameter of is . Bousquet and Perarnau proved that when , the diameter of is linear [BousquetP16].
Even if the conjecture is open for general graphs, it has been proved for several graph classes such as chordal graphs [BonamyJ12] and bounded treewidth graphs [BonamyB18]. The polynomial Cereceda’s conjecture holds if we replace degeneracy by maximum average degree [BousquetP16, Feghali19]. In particular, it implies that the recolouring diameter of a planar graph is polynomial. The polynomial Cereceda’s conjecture asks for more since it states that the diameter should be polynomial when . So far, the best known upper bound on the recolouring diameter of planar graphs was subexponential [EibenF18].
Our result.
The main result of the paper is the following:
Theorem 1.
Let and be a degenerate graph. Then has diameter at most:

if (where is a constant independent of and ),

if and (where is a constant independent from and ),

for any and (where is a constant independent from and ).
recoloring diameter  Lower bound  Upper bound 

[BonamyJ12]  [Theorem 1]  
[Theorem 1]  
[Theorem 1]  
[Theorem 1]  
[BousquetP16] 
In particular, it implies that the recolouring diameter of planar graphs is polynomial (of order ), answering a question of [BousquetP16, Feghali19], and is quadratic if (improving the recent result of Feghali giving [Feghali19_2]). For general graphs, our result guarantees moreover that the diameter becomes a polynomial independent of as long as . We also obtain a quadratic diameter when the number of colours is at least , improving the result of Cereceda [Cereceda] who obtained a similar result for . Note moreover that Theorem 1 ensures that the diameter is polynomial as long as is a fixed constant, which was open even for (we get a diameter of order for ). The main known results on lower and upper bounds on the recolouring diameter are summarized in Figure 1.
In order to show Theorem 1, we need to prove a more general result that also holds for listcolourings. Indeed, we often need to consider induced subgraphs of our initial graph where the colours of some vertices are“frozen” (i.e. do not change). By considering the list colouring version, we can delete these vertices and remove their colours from the list of all their neighbours. This more general statement implying Theorem 1 is given in Section 2.
We complete Theorem 1 by proving the quadratic Cereceda’s conjecture for planar bipartite graphs. Euler’s formula ensures that planar bipartite graphs are degenerate. We prove the following in Section 3.
Theorem 2.
Let be a planar bipartite graph. The diameter of is at most .
Our proof is based on a discharging argument. It is, as far as we know, the first time such a method is used for reconfiguration.
The proofs of both Theorem 1 and Theorem 2 consist in showing that, given two colourings and of , there exists a transformation from to of the corresponding length. Since our proofs are algorithmic, they also provide polynomial time algorithms that, given two colourings and , output a transformation from to of length at most the diameter of the reconfiguration graph.
Further work and open problems.
Even if we obtain a polynomial bound on the diameter of the reconfiguration graph, the quadratic conjecture of Cereceda is still open, even for simple classes of graphs such as (trianglefree) planar graphs or degenerate graphs.
Conjecture 3 (Cereceda [Cereceda]).
For every , there exists such that for every degenerate graph , the diameter of is at most as long as .
The question of Cereceda in [Cereceda] is the following “For a graph with vertices and where is the degeneracy of , the diameter of is ”. We were not be able to determine if the coefficient in front of has to be a function of or not. As a consequence we decided to state the weaker possible version of the conjecture. Note that the constants of Theorem 2, for chordal graphs [BonamyJ12], and for bounded treewidth graphs [BonamyB18] do not depend on . A stronger conjecture would then be the following:
Question 4 (Stronger Conjecture).
There exists a constant such that, for every and every degenerate graph the diameter of is at most as long as .
Another interesting question is the following: what is the evolution of the diameter when the number of colours increases. With our proof technique, the power in the exponent decreases little by little and finally becomes quadratic when the number of colours is at least . Improving the factor may be another interesting way of tackling the general conjecture.
We also know that the diameter becomes linear when [BousquetP16]. Is it possible to improve this result? In particular, can we improve the recolouring diameter for ?
Question 5.
Is there and such that, for every and every degenerate graph the diameter of is at most when .
It was also proved in [BonamyB18] that the diameter becomes linear if is at least the grundy number plus , which in particular implies a linear diameter when . When , Feghali, Jonhson and Paulusma [FeghaliJP16] proved that the recolouring graph is composed of isolated vertices plus a unique component of diameter at most (and this nonisolated component of is exponentially larger than its number of isolated vertices [BonamyBP18]).
One can also notice in Figure 1 that, as long as , no nontrivial lower bound is known on the diameter of the reconfiguration graph. The subtle quadratic lower bound of [BonamyJ12] when seems to be hard to adapt if . Developing new techniques to find lower bounds on the recolouring diameter of a graph is a challenging open problem.
2 Polynomial Cereceda’s conjecture
Let be a degenerate graph and let be a degeneracy ordering. We denote by the outneighbours of (i.e., neighbours of which appear later in the ordering). Recall that a graph is degenerate if there is a degeneracy ordering such that for every vertex of the graph.
Let us first briefly discuss the main ideas of the proof before stating formally all the results. The main idea is to proceed recursively on the degeneracy. More precisely, we want to delete a subset of vertices in order to decrease by one both the degeneracy and the number of colours. In order to do this, observe that if at some point there is one colour such that every vertex of the graph is either coloured or has an outneighbour coloured , then by removing all the vertices coloured , we decrease the number of available colours by , but we also decrease the degeneracy of the graph by . If is the resulting graph, by applying induction, we can recolour however we want, and for example, we can remove completely one colour which we can then use to make the two colourings agree on a subset of vertices. If the colour satisfies the condition above, we will say that the colour is full. Our main objective will consist in finding a transformation from any colouring of to some colouring of which has a full colour (Note that any graph can have such a colouring, for example by applying the FirstFit algorithm in the reverse order of the elimination ordering). We will build the colouring (and the transformation) incrementally. However in order to do this, we will need to generalise the problem to list colouring.
A list assignment is a function which associates a list of colours to every vertex . An colouring is a (proper) colouring of such that for every vertex , . The total number of colours used by the assignment is . A list assignment is feasible if for every vertex . We just say that it is feasible if it is feasible. We denote by the reconfiguration graph of the colourings of . (One can easily prove by induction, that if a list assignment is feasible for , then is connected). We will prove a generalisation of Theorem 1 in the case of list colourings. Namely, we will prove that:
Theorem 6.
Let be a graph and . Let be an feasible list assignment and be the total number of colours. Then has diameter at most:

if .

if ( a constant independent of ),

if where is a constant and is independent of ,

, if .
The proof of Theorem 1 follows easily from this result. The only point that is not immediate is that the last point of Theorem 6 implies the last point of Theorem 1. In this case, we need a small trick to guarantee that the diameter does not increase if the number of colours increases. Note that the first point of Theorem 6 implies in terms of classical colouring that the recolouring diameter is linear when , which is an already known result [BousquetP16].
Proof of Theorem 1.
Note that given a degenerate graphs and colours, we can consider the list assignment where for every vertex . This list assignment is feasible with .
We start with the second point of Theorem 1. If then:
By applying the third case of Theorem 6, the result follows.
The first point follows immediately from the result above by taking .
Finally, in order to prove the last point, we need to prove that we can “replace” by . Let be a degenerate graph and let be a colouring of it. Let us prove that, if , any colouring can be transformed into within steps (and not as suggested by Theorem 6). Indeed, we simply simply “forget” the vertices coloured with colour in . Let be the graph without these vertices. The graph is degenerate and by Theorem 6, we can transform into within steps using colours in . We finally recolour the vertices of one by one with their colours in to obtain the colouring . ∎
The rest of this section is devoted to prove Theorem 6. In order to do it, we need to generalise the notion of full colour to the listcolouring setting. We also need to generalise it to sets of colours, to handle the case where . Given a colouring of , the set of colour is full if for every vertex and every colour one of the following holds:

[label= ()]

,

has at least one outneighbour coloured ,

or .
We have a property similar as previously: starting from an feasible list assignment with a colouring , if is full then by removing all the vertices with a colour from the graph, and removing all the colours from from all the lists, the resulting assignment is still feasible. Additionally, the total number of colours has decreased by (but the degeneracy of the graph might not have decreased as much if we had much larger than ).
In the following, we will denote by the maximum diameter of over all graphs with vertices, and all feasible assignments with total number of colours .
The proof of Theorem 6 is by induction on the total number of colours . The base case, which is the first point of Theorem 6, is the following lemma. This lemma ensures that if the number of excess colours is sufficient (at least half the number of colours), then the diameter is linear^{4}^{4}4The proof is similar to the linear diameter obtained in [BousquetP16] for colourings but adapted to list colourings..
Lemma 7.
Assume that , then .
Proof.
We show by induction on that if , then for any feasible list assignment on a graph on vertices, and any two colourings , there is a transformation from to such that every vertex is recoloured at most times.
The result is clearly true when , since in this case the unique vertex can be recoloured only once. Assume that the result holds for . Let be a graph on vertices with a degeneracy ordering , and an feasible list assignment using a total of colours. Let and be two colourings, be the subgraph obtained after removing , and define the number of neighbours of .
Using induction on , there is a transformation from to such that every vertex is recoloured at most times. Note that by assumption, the number of colours available to is at least , hence the total number of colours satisfies, , and in particular , and . Every time one of the neighbours of is recoloured in the sequence , we may have to recolour beforehand, so that the colouring remains proper. This happens if the neighbour wants to be recoloured with the current colour of .
Every time we have to recolour , we have the choice among colours for the new colour of . We can look ahead in to know which are the next modifications of colours of neighbours of in . One colour of does not appear in these modifications since and the first modified colour is not in (since we need to recolour at the first step, and then the target colour is the current colour of which is not in ). We recolour with . This way, we only need to recolour once out of every times its neighbours are recoloured. Finally, we may need to recolour one last time after the end of to colour with its target colour. Since by induction, the neighbours of are recoloured at most times, the total number of times is recoloured is at most:
where in the last inequality, we have used the fact that since . This concludes the induction step and proves the result. ∎
In the induction step of our proof, we will build a set of full colours. For a colouring and a set of vertices, we denote by . Before stating the main lemmas, let us make the following remark:
Remark 1.
Let be a graph, be a list assignment and be a list colouring. Let be an induced subgraph of with list assignment . Then any recolouring sequence from to some colouring also is a (valid) recolouring sequence from to where if and otherwise.
By abuse of notation and when no confusion is possible, we will then call both the recolouring sequence in and in .
The following lemma states that, if we already have a set of full colours, then changing it to an other given set can be done without too many additional recolouring steps.
Lemma 8.
Let be a colouring of , and a set of full colours for with . For any with , there exists a colouring such that is full for , and there is a transformation from to of length at most .
Proof.
Let be a set of colours with which is full for some colouring . Let be any set of colours of size . The main part of the proof consists in transforming into a colouring that does not use at all any colour of (such a colouring exists since the list assignment is feasible).
Let be the subgraph induced by the vertices not coloured in , and let be the list assignment for obtained from by removing from all the lists. Since is full in , is feasible for .
Consider the following preference ordering on the colours: an arbitrary ordering of , followed by an ordering of , and finally the colours from last. Let be the colouring of obtained by colouring greedily from to with this preference ordering. Since is feasible, and , no vertex is coloured with a colour in in . Indeed and only neighbours of have been coloured when is coloured. Since in no vertex has a colour in , is an colouring of . By induction hypothesis, there is a recolouring sequence that transforms the colouring of into within at most steps. By Remark 1, this recolouring sequence also is a recolouring sequence in . We can then recolour the vertices of to their target colour in in an additional steps. No conflict can happen at this step since is a proper colouring of .
One can easily check that, in , the set of colours is full. Let be the subgraph of induced by all the vertices coloured in , and let be a list assignment of these vertices where all the colours not in were removed. Then since is full, is feasible. We will recolour such that no vertex is coloured (such a colouring exists because is feasible, and ). Since the total number of colours used in is , this recolouring can be done in at most steps by Lemma 7. By Remark 1, this recolouring sequence also is a recolouring sequence in . The colouring of that we obtain is such that no vertex is coloured with . We can finally recolour the vertices of one by one, starting from , choosing a colour of if it is available, or leaving it with its current colour otherwise.
Let be the resulting colouring. By construction is full for . The total number of steps to reach is at most . ∎
Using Lemma 8, we show that we can incrementally construct a set of full colours.
Lemma 9.
Assume that . For any colouring , there exists a colouring containing a set of full colours with , and there is a transformation from to of length at most .
Proof.
Let be a degeneracy ordering of . A colouring is full up to step for a set of colour if , and all the vertices with satisfy the conditions for the set . If a colouring is full up to step for the set , then the set is full.
Note that for any colouring , any set of colours containing is full up to step . So we only need to show that given a colouring which is full up to step for some set , we can reach a colouring full up to step for some (potentially different) set in at most steps. Suppose now that is full up to step but not for some set . Let be the colours of the vertices . Up to adding arbitrary colours to , we can assume that . We will then recolour the graph in order to obtain a colouring where is full up to step .
Let be the graph induced by the vertices , and be the list assignment of the vertices of obtained from by fixing the colours of the vertices outside . In other words, for every vertex , we remove from the colours of all the vertices of . Note that is an feasible assignment of since was an feasible assignment of . Additionally, is a set of full colours for the colouring .
By Lemma 8, there is an colouring of which is full for such that we can transform into in at most steps. Let be the colouring which agrees with on the vertices with index larger than , and agrees with on . By Remark 1, can be obtained from into at most steps. By construction, is full up to step , and since the vertices are coloured with colours in , it is full up to step .
Finally, this procedure needs to be repeated at most times (the minus one coming from the fact that at the beginning we had for free a set of colours full for ). After this many steps, we obtain a colouring full up to step for some set with , which concludes the proof. ∎
Finally, we can use the two previous lemma to get the following recursive inequality.
Lemma 10.
Let , then .
Proof.
Let be an feasible list assignment of , and and be two colourings of . By Lemma 9, there exists a colouring and a set of colours with which is full for such that the colouring can be reached from in at most steps. Similarly, there exists a colouring and a set of colours with such that is full for such that the colouring can be reached from in at most steps. By Lemma 8, using an additional steps, we can get a colouring from such that the set of full colours in and is the same (namely ).
Let be some set of colours disjoint from with . Let be a colouring of that does not use any colour of (such a colouring exists since the list assignment of is feasible).
Let be the graph where the vertices coloured with colours in have been deleted and the colours in removed from the list assignment. Since is full for , the list assignment of is feasible. Note that is a proper colouring of . So by induction, it is possible to recolour into in at most steps. Since the vertices of are coloured with colours in , and since does not use any of these colours, one can finally recolour the vertices of one by one from their colours in to their target colours in .
Let be the graph where the vertices coloured with in have been deleted. One can similarly recolour into in at most steps. Since vertices of are coloured with colours in , we can also can recolour the vertices of one by one from their colours in to their target colours in .
The total number of steps to transform into is at most
∎
Lemma 11.
For all and , we have for some constant .
Proof.
Proof of Theorem 6.
The first point is the result from Lemma 7. The second and last points are consequences of Lemma 11 with the corresponding values of . Let us prove the third point. Since , we have . Consequently, the function given by Lemma 11 is . Note that for the second and third point, the constant does not depend on or since is a constant (but it depends on in the third point). ∎
3 Planar bipartite graphs
The rest of this article is devoted to prove Cereceda’s quadratic conjecture for planar bipartite graphs.
See 2
We start by defining the level of a vertex. Given a graph all the vertices with degree at most have level . For any , the vertices with level are the ones with degree at most after removal of all the vertices of level .
Let be a planar graph given with its representation. The size of a face is the number of vertices incident to the face. The proof of Theorem 2 is based on a discharging proof. We are assuming that the results does not hold. The proof then goes in two steps. First, we show that any planar bipartite graph must contain at least one ”configuration” of a list of (here two) configurations. Then, we prove that any counterexample cannot contain any other these configurations, which implies that this counterexample cannot exist. Let us first give the two configurations we will need.
Lemma 12.
Let be a planar bipartite graph. At least one of the following holds:

[label=()]

contains a vertex of degree at most or,

contains a vertex of degree incident to three vertices of level at most , and incident to a face of size four.
Proof.
We use a discharging argument to prove the result. Suppose by contradiction that there exists a planar bipartite graph satisfying none of the conditions above. Assign to each vertex the weight , and to each face the weight . The total weight assigned this way is:
where and are the number of vertices, edges, and faces of . The last equality comes from Euler’s formula for planar graphs.
Now the goal is to reallocate the weights on the faces and the vertices to obtain a total weight that is nonnegative, a contradiction with the equation above. Note that all the faces as well as vertices of degree at least have nonnegative weights. In order to get a contradiction, we apply the following procedure and prove that after applying it all the vertices and faces have nonnegative weights:

Every vertex of level at least gives weight to all its neighbours of degree .

Every face of size at least gives weight to all its incident vertices.
After applying this transformation, every face has a nonnegative weight. Indeed, the faces of lengt still have weight , and the faces of length at least gave a total of since .
Vertices of level initially have nonnegative weights since their degree is at least four, and they did not give away any of their weights. Hence their weights remain nonnegative. Vertices of level at least must have, by definition at least neighbours of degree larger than (otherwise they would be of level ), and thus keep a positive weight.
Finally, let be a vertex of degree . If is adjacent to a vertex of level at least , it receives weight during the first step, and its weight is nonnegative. Otherwise, is adjacent to three vertices of level at most , and then, by assumption on , the three faces incident to must have size at least . Then, received from each of the faces during the second step, and its weight is nonnegative.
Hence, after applying the the procedure, every face and every vertex has a nonnegative weight, a contradiction with the fact that the total weight is negative. ∎
We can now prove the Theorem 2. The proceeds by iteratively reducing the size of the graph, either by removing a vertex, or contracting two vertices together.
Proof of Theorem 2.
We will show by induction on the size of the graph that for any two colourings of , and , there exists a transformation from to such that each vertex is recoloured at most times, for some constant defined later.
If is reduced to a single vertex, the result is immediate. Otherwise, let . One of the two conditions of Lemma 12 must occur.
Case (i). The graph contains a vertex of degree at most .
Let be the subgraph obtained after removing . By induction, there exists a transformation from to such that each vertex is recoloured at most times. Given this transformation, we are going to produce a transformation for the whole graph . We follow the transformation steps of . If the same recolouring step is possible in we do it. Otherwise, it means that one of the (at most) two neighbours of is recoloured, and that the target colour is the current colour of . So we need to recolour to make the transition possible. To recolour , we have at least two possible new colour choices for distinct from . We can look forward at the sequence to see which colour will appear next in the neighbourhood of . We choose for a colour of distinct from . In particular, the choice of ensures that will not be recoloured next time a neighbour of is recoloured. At the end of the transformation , we may need one additional step to recolour to its target colour.
Thus in total, is recoloured at most times (which holds as long as ).
Case (ii). The graph contains a vertex of degree three such that:

all three neighbours of have level at most ,

is incident to a face of length .
Let be the face of length incident to , and let be the vertex opposite to on . Let be the vertex adjacent to , and not incident to . Let and be the two remaining vertices adjacent to (see Figure 2). In this case, we want to reduce the graph by merging the two vertices and (maintaining a bipartite planar graph). For this we will need the following result.
Claim 13.
Starting from , we can reach a colouring such that . This colouring can be reached by recolouring each vertex at most twice.
Proof.
Since and are adjacent to , their colour is different from . Hence, the only obstruction to recolour directly with is if is coloured with . Since has level at most , it has at most neighbours with degree larger than . Let be a colour different from these three neighbours, and different from . First, we recolour all the degree three neighbours of which are coloured with any other colour (note that is not recoloured since , but might be recoloured). Then, we recolour with , and finally can be recoloured with (since is bipartite, the recolouring sequence ensures that the only neighbor of that is recoloured is ). Every vertex is recoloured at most once, except which may be recoloured twice.
∎
By Claim 13 starting from and , we obtain two colourings and such that and . Let be the graph obtained after merging and into a single vertex . Clearly is still planar, and it is still bipartite since and are on the same side of the bipartition of . Moreover, and are proper colourings of . By induction, there is a transformation from to in such that each vertex is recoloured at most times. By applying each transformation on to both and , this gives a transformation in from to such that each vertex is recoloured at most times.
This gives a transformation from to where each vertex is recoloured at most times, taking . ∎