A polynomial version of Cereceda's conjecture
AA polynomial version of Cereceda’s conjecture ∗ Nicolas Bousquet and Marc Heinrich Univ. Grenoble Alpes, CNRS, Laboratoire G-SCOP, Grenoble-INP, Grenoble, France. † LIRIS, Universit´e Claude Bernard, Lyon, France
Abstract
Let k and d be such that k ≥ d + 2. Consider two k -colourings of a d -degenerate graph G .Can we transform one into the other by recolouring one vertex at each step while maintaininga proper coloring at any step? Cereceda et al. answered that question in the affirmative, andexhibited a recolouring sequence of exponential length. However, Cereceda conjectured thatthere should exist one of quadratic length.The k -reconfiguration graph of G is the graph whose vertices are the proper k -colouringsof G , with an edge between two colourings if they differ on exactly one vertex. Cereceda’sconjecture can be reformulated as follows: the diameter of the ( d + 2)-reconfiguration graph ofany d -degenerate graph on n vertices is O ( n ). So far, the existence of a polynomial diameteris open even for d = 2.In this paper, we prove that the diameter of the k -reconfiguration graph of a d -degenerategraph is O ( n d +1 ) for k ≥ d + 2. Moreover, we prove that if k ≥ ( d + 1) then the diameterof the k -reconfiguration graph is quadratic, improving the previous bound of k ≥ d + 1. Wealso show that the 5-reconfiguration graph of planar bipartite graphs has quadratic diameter,confirming Cereceda’s conjecture for this class of graphs. Reconfiguration problems consist in finding step-by-step transformations between two feasible solu-tions of a problem such that all intermediate states are also feasible. Such problems model dynamicsituations where a given solution already in place has to be modified for a more desirable one whilemaintaining some properties all along the transformation. Reconfiguration problems have beenstudied in various fields such as discrete geometry [7], optimization [1] or statistical physics [20].For a complete overview of the reconfiguration field, the reader is referred to the two recent surveyson the topic [21, 22]. In this paper, our reference problem is graph colouring.Let Π be a problem and I be an instance of Π. The reconfiguration graph is the graph wherevertices are solutions of I and where there is an edge between two vertices if one can transformthe first solution into the other in one step (for graph colouring one step means modifying thecolour of a single vertex ). Given a reconfiguration problem, several questions may arise. (i) Is ∗ This work was supported by ANR project GrR (ANR-18-CE40-0032). † [email protected] Note that recolouring operations have been studied, for instance recolouring using Kempe chains (see e.g. [3]).In this article, we focus on single vertex recolourings. a r X i v : . [ c s . D M ] M a r t 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 diameterof the reconfiguration graph? In this work, we will focus on the diameter of the reconfigurationgraph. The diameter of the reconfiguration graph plays an important role, for instance in randomsampling, since it provides a lower bound on the mixing time of the underlying Markov chain (andthe connectivity of the reconfiguration graph ensures the ergodicity of the Markov chain ). Sinceproper colourings correspond to states of the antiferromagnetic Potts model at zero temperature,Markov chains related to graph colourings received a considerable attention in statistical physicsand many questions related to the ergodicity or the mixing time of these chains remain widely open(see e.g. [12, 19]). Graph recolouring.
All along the paper G = ( V, E ) denotes a graph, n is the size of V and k is an integer. For standard definitions and notations on graphs, we refer the reader to [13]. A (proper) k -colouring of G is a function f : V ( G ) → { , . . . , k } such that, for every edge xy ∈ E ,we have f ( x ) (cid:54) = f ( y ). Throughout the paper we will only consider proper colourings and will thenomit the proper for brevity. The chromatic number χ ( G ) of a graph G is the smallest k suchthat G admits a k -colouring. Two k -colourings are adjacent if they differ on exactly one vertex.The k -reconfiguration graph of G , denoted by G ( G, k ) and defined for any k ≥ χ ( G ), is the graphwhose vertices are k -colourings of G , with the adjacency relation defined above. Cereceda, vanden Heuvel and Johnson provided an algorithm to decide whether, given two 3-colourings, one canbe transformed into the other in polynomial time, and characterized graphs for which G ( G,
3) isconnected [10, 11]. Given any two k -colourings of G , it is PSPACE -complete to decide whetherone can be transformed into the other for k ≥ k -recolouring diameter of a graph G is the diameter of G ( G, k ) if G ( G, k ) is connected andis equal to + ∞ otherwise. In other words, it is the minimum D for which any k -colouring can betransformed into any other one through a sequence of at most D adjacent k -colourings. Bonsmaand Cereceda [6] proved that there exists a family G of graphs and an integer k such that, for everygraph G ∈ G there exist two k -colourings whose distance in the k -reconfiguration graph is finiteand super-polynomial in n .Cereceda conjectured that the situation is different for degenerate graphs. A graph G is d -degenerate if any subgraph of G admits a vertex of degree at most d . In other words, there existsan ordering v , . . . , v n of the vertices such that for every i ≤ n , the vertex v i has at most d neighboursin v i +1 , . . . , v n . It was shown independently by Dyer et al. [14] and by Cereceda et al. [10] thatfor any d -degenerate graph G and every k ≥ d + 2, G ( G, k ) is connected. However the (upper)bound on the k -recolouring diameter given by these constructive proofs is of order c n (where c is aconstant). Cereceda [9] conjectured that the the diameter of G ( G, k ) is of order O ( n ) as long as k ≥ d + 2. If correct, the quadratic function is sharp, even for paths or chordal graphs as provedin [5]. The existence of a polynomial upper bound instead of a quadratic one also is open, even for d = 2 or restricted to particular graph classes such as planar graphs. In what follows, we will callrespectively the polynomial (resp. quadratic) Cereceda’s conjecture the question of proving thatthe k -recolouring diameter of d -degenerate graphs is polynomial (resp. quadratic) for k ≥ d + 2.The quadratic Cereceda’s conjecture is known to be true only for d = 1 (for trees) [5] and d = 2with ∆ ≤ Actually, 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. -recoloring diameter Lower bound Upper bound k = d + 2 O ( n ) [5] O ( n d +1 ) [Theorem 1] k = d + 3 O ( n ) O ( n (cid:100) d +12 (cid:101) ) [Theorem 1] k ≥ (1 + ε )( d + 1) O ( n ) O ( n (cid:100) /ε (cid:101) ) [Theorem 1] k ≥ ( d + 1) O ( n ) O ( n ) [Theorem 1] k ≥ d + 1) O ( n ) O ( n ) [8]Figure 1: Known results on the k -recolouring diameter of d -degenerate graphs.for any d -degenerate graph G and every k ≥ d + 1, the diameter of G ( G, k ) is O ( n ). Bousquetand Perarnau proved that when k ≥ d + 2, the diameter of G ( G, k ) is linear [8].Even if the conjecture is open for general graphs, it has been proved for several graph classessuch as chordal graphs [5] and bounded treewidth graphs [2]. The polynomial Cereceda’s conjectureholds if we replace degeneracy by maximum average degree [8, 16]. In particular, it implies thatthe 8-recolouring diameter of a planar graph is polynomial. The polynomial Cereceda’s conjectureasks for more since it states that the diameter should be polynomial when k = 7. So far, the bestknown upper bound on the 7-recolouring diameter of planar graphs was subexponential [15]. Our result.
The main result of the paper is the following:
Theorem 1.
Let d, k ∈ N and G be a d -degenerate graph. Then G ( G, k ) has diameter at most: • Cn if k ≥ ( d + 1) (where C is a constant independent of k and d ), • C ε n (cid:100) /ε (cid:101) if k ≥ (1 + ε )( d + 2) and < ε < (where C ε is a constant independent from k and d ), • ( Cn ) d +1 for any d and k ≥ d + 2 (where C is a constant independent from k and d ). In particular, it implies that the 7-recolouring diameter of planar graphs is polynomial (of order O ( n )), answering a question of [8, 16], and is quadratic if k ≥ k ≥
10 [17]). For general graphs, our result guarantees moreover that the diameterbecomes a polynomial independent of d as long as k ≥ (1 + ε )( d + 2). We also obtain a quadraticdiameter when the number of colours is at least · ( d + 1), improving the result of Cereceda [9] whoobtained a similar result for k ≥ d + 1. Note moreover that Theorem 1 ensures that the diameteris polynomial as long as d is a fixed constant, which was open even for d = 2 (we get a diameter oforder O ( n ) for d = 2). The main known results on lower and upper bounds on the k -recolouringdiameter are summarized in Figure 1.In order to show Theorem 1, we need to prove a more general result that also holds for list-colourings. Indeed, we often need to consider induced subgraphs of our initial graph where thecolours 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. Thismore general statement implying Theorem 1 is given in Section 2.We complete Theorem 1 by proving the quadratic Cereceda’s conjecture for planar bipartitegraphs. Euler’s formula ensures that planar bipartite graphs are 3-degenerate. We prove thefollowing in Section 3. 3 heorem 2. Let G be a planar bipartite graph. The diameter of G ( G, is at most n . Our proof is based on a discharging argument. It is, as far as we know, the first time such amethod is used for reconfiguration.The proofs of both Theorem 1 and Theorem 2 consist in showing that, given two colourings α and β of G , there exists a transformation from α to β of the corresponding length. Since our proofsare algorithmic, they also provide polynomial time algorithms that, given two colourings α and β ,output a transformation from α to β 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 ofthe reconfiguration graph, the quadratic conjecture of Cereceda is still open, even for simple classesof graphs such as (triangle-free) planar graphs or 2-degenerate graphs.
Conjecture 3 (Cereceda [9]) . For every d , there exists C d such that for every d -degenerate graph G ,the diameter of G ( G, k ) is at most C d · n as long as k ≥ d + 2.The question of Cereceda in [9] is the following “ For a graph G with n vertices and k ≥ d + 2 where d is the degeneracy of G , the diameter of G ( G, k ) is O ( n )”. We were not be able to determineif the coefficient in front of n has to be a function of d or not. As a consequence we decided to statethe weaker possible version of the conjecture. Note that the constants of Theorem 2, for chordalgraphs [5], and for bounded treewidth graphs [2] do not depend on d . A stronger conjecture wouldthen be the following: Question 4 (Stronger Conjecture) . There exists a constant C such that, for every d and every d -degenerate graph G the diameter of G ( G, k ) is at most C · n as long as k ≥ d + 2.Another interesting question is the following: what is the evolution of the diameter when thenumber of colours increases. With our proof technique, the power in the exponent decreases littleby little and finally becomes quadratic when the number of colours is at least · ( d + 1). Improvingthe factor may be another interesting way of tackling the general conjecture.We also know that the diameter becomes linear when k ≥ d + 2 [8]. Is it possible to improvethis result? In particular, can we improve the k -recolouring diameter for ( d + 1) ≤ k ≤ d + 1? Question 5.
Is there α < β ∈ N such that, for every d and every d -degenerate graph G thediameter of G ( G, k ) is at most C d · n when k ≥ αd + β .It was also proved in [2] that the diameter becomes linear if k is at least the grundy numberplus 1, which in particular implies a linear diameter when k ≥ ∆ + 2. When k = ∆ + 1, Feghali,Jonhson and Paulusma [18] proved that the k -recolouring graph is composed of isolated vertices plusa unique component of diameter at most O ( n ) (and this non-isolated component of G ( G, ∆ + 1)is exponentially larger than its number of isolated vertices [4]).One can also notice in Figure 1 that, as long as k ≥ d +3, no non-trivial lower bound is known onthe diameter of the reconfiguration graph. The subtle quadratic lower bound of [5] when k = d + 2seems to be hard to adapt if k ≥ d + 3. Developing new techniques to find lower bounds on the k -recolouring diameter of a graph is a challenging open problem.4 Polynomial Cereceda’s conjecture
Let G be a d -degenerate graph and let v , . . . , v n be a degeneracy ordering. We denote by d + ( v )the out-neighbours of v (i.e., neighbours of v which appear later in the ordering). Recall that agraph is d -degenerate if there is a degeneracy ordering such that d + ( v ) ≤ d for every vertex v ofthe 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 asubset of vertices in order to decrease by one both the degeneracy and the number of colours. Inorder to do this, observe that if at some point there is one colour c such that every vertex of thegraph is either coloured c or has an out-neighbour coloured c , then by removing all the verticescoloured c , we decrease the number of available colours by 1, but we also decrease the degeneracyof the graph by 1. If H is the resulting graph, by applying induction, we can recolour H howeverwe want, and for example, we can remove completely one colour which we can then use to makethe two colourings agree on a subset of vertices. If the colour c satisfies the condition above, wewill say that the colour c is full . Our main objective will consist in finding a transformation fromany colouring α of G to some colouring α (cid:48) of G which has a full colour (Note that any graph canhave such a colouring, for example by applying the First-Fit algorithm in the reverse order ofthe elimination ordering). We will build the colouring α (cid:48) (and the transformation) incrementally.However in order to do this, we will need to generalise the problem to list colouring.A list assignment L is a function which associates a list of colours to every vertex v . An L -colouring is a (proper) colouring α of G such that for every vertex v , α ( v ) ∈ L ( v ). The totalnumber of colours used by the assignment is k = | (cid:83) v ∈ G L ( v ) | . A list assignment L is a -feasible if | L ( v ) | ≥ | d + ( v ) | + a + 1 for every vertex v ∈ G . We just say that it is feasible if it is 1-feasible.We denote by G ( G, L ) the reconfiguration graph of the L -colourings of G . (One can easily proveby induction, that if a list assignment is a -feasible for a ≥
1, then G ( G, L ) is connected). We willprove a generalisation of Theorem 1 in the case of list colourings. Namely, we will prove that:
Theorem 6.
Let G be a graph and a ∈ N . Let L be an a -feasible list assignment and k be the totalnumber of colours. Then G ( G, L ) has diameter at most: • kn if k ≤ a . • Cn if k ≤ a ( C a constant independent of k, a ), • C ε n (cid:100) /ε (cid:101) if k ≤ (1 + ε ) a where ε is a constant and C ε is independent of k, a , • ( Cn ) k − , if a ≥ . The proof of Theorem 1 follows easily from this result. The only point that is not immediate isthat the last point of Theorem 6 implies the last point of Theorem 1. In this case, we need a smalltrick to guarantee that the diameter does not increase if the number of colours increases. Note thatthe first point of Theorem 6 implies in terms of classical colouring that the k -recolouring diameteris linear when k ≥ d + 2, which is an already known result [8]. Proof of Theorem 1.
Note that given a d -degenerate graphs and k colours, we can consider thelist assignment L where L ( v ) = [ k ] for every vertex v . This list assignment is a -feasible with a = k − d −
1. 5e start with the second point of Theorem 1. If k ≥ (1 + ε )( d + 1) then: ka = kk − d − d + 1 k − d − ≤ ε . By applying the third case of Theorem 6, the result follows.The first point follows immediately from the result above by taking ε = .Finally, in order to prove the last point, we need to prove that we can “replace” k by d . Let G be a d -degenerate graph and let γ be a ( d + 1)-colouring of it. Let us prove that, if k > d + 2,any colouring α can be transformed into γ within O ( n d +1 ) steps (and not O ( n k − ) as suggested byTheorem 6). Indeed, we simply simply “forget” the vertices coloured with colour d + 3 , . . . , k in α .Let H be the graph without these vertices. The graph H is d -degenerate and by Theorem 6, wecan transform α | H into γ | H within O ( n d +1 ) steps using colours in 1 , . . . , d + 2. We finally recolourthe vertices of G \ H one by one with their colours in γ to obtain the colouring γ .The rest of this section is devoted to prove Theorem 6. In order to do it, we need to generalisethe notion of full colour to the list-colouring setting. We also need to generalise it to sets of colours,to handle the case where a >
1. Given a colouring α of G , the set of colour S is full if for everyvertex v and every colour c ∈ S one of the following holds:(i) α ( v ) ∈ S ,(ii) v has at least one out-neighbour coloured c ,(iii) or c (cid:54)∈ L ( v ).We have a property similar as previously: starting from an a -feasible list assignment with a colour-ing α , if S is full then by removing all the vertices v with a colour α ( v ) ∈ S from the graph,and removing all the colours from S from all the lists, the resulting assignment is still a -feasible.Additionally, the total number of colours has decreased by | S | (but the degeneracy of the graphmight not have decreased as much if we had k much larger than d + 2).In the following, we will denote by f a ( n, k ) the maximum diameter of G ( G, L ) over all graphs G with n vertices, and all a -feasible assignments L with total number of colours k .The proof of Theorem 6 is by induction on the total number of colours k . The base case, whichis the first point of Theorem 6, is the following lemma. This lemma ensures that if the number ofexcess colours is sufficient (at least half the number of colours), then the diameter is linear . Lemma 7.
Assume that k ≤ a , then f a ( n, k ) ≤ kn .Proof. We show by induction on n that if k ≤ a + 1, then for any a -feasible list assignment L ona graph on n vertices, and any two L -colourings α, β , there is a transformation from α to β suchthat every vertex is recoloured at most k times.The result is clearly true when n = 1, since in this case the unique vertex can be recoloured onlyonce. Assume that the result holds for n −
1. Let G be a graph on n vertices with a degeneracyordering v , . . . , v n , and an a -feasible list assignment L using a total of k colours. Let α and β be two L -colourings, H be the subgraph obtained after removing v n , and define d = | d + ( v ) | thenumber of neighbours of v . The proof is similar to the linear diameter obtained in [8] for colourings but adapted to list colourings. H , there is a transformation S from α | H to β | H such that every vertex isrecoloured at most k times. Note that by assumption, the number of colours available to v is atleast d + a + 1, hence the total number of colours k satisfies, d + a + 1 ≤ k ≤ a , and in particular a ≥ d + 1, and k ≥ d + a + 1 ≥ d + 2. Every time one of the neighbours of v is recoloured in thesequence S , we may have to recolour v beforehand, so that the colouring remains proper. Thishappens if the neighbour wants to be recoloured with the current colour of v .Every time we have to recolour v , we have the choice among a ≥ d + 1 colours C for the newcolour of v . We can look ahead in S to know which are the next d + 1 modifications of colours ofneighbours of v in S . One colour c of C does not appear in these modifications since | C | ≥ d + 1and the first modified colour is not in C (since we need to recolour v at the first step, and thenthe target colour is the current colour of v which is not in C ). We recolour v with c . This way,we only need to recolour v once out of every d + 1 times its neighbours are recoloured. Finally, wemay need to recolour v one last time after the end of S to colour v with its target colour. Sinceby induction, the neighbours of v are recoloured at most k times, the total number of times v isrecoloured is at most: (cid:24) dkd + 1 (cid:25) + 1 ≤ dd + 1 k + 2 ≤ k , where in the last inequality, we have used the fact that 2 ≤ kd +1 since k ≥ d + 2. This concludesthe induction step and proves the result.In the induction step of our proof, we will build a set of full colours. For a colouring α and aset X of vertices, we denote by α ( X ) = (cid:83) x ∈ X α ( x ). Before stating the main lemmas, let us makethe following remark: Remark 1.
Let G be a graph, L be a list assignment and α be a L -list colouring. Let H be aninduced subgraph of G with list assignment L (cid:48) ( v ) = L ( v ) \ α ( N ( v ) \ V ( H )). Then any recolouringsequence S from α | H to some colouring β | H also is a (valid) recolouring sequence from α to β where β ( v ) = β | H ( v ) if v ∈ H and α ( v ) otherwise.By abuse of notation and when no confusion is possible, we will then call S both the recolouringsequence in H and in G .The following lemma states that, if we already have a set of full colours, then changing it to another given set can be done without too many additional recolouring steps. Lemma 8.
Let α be a colouring of G , and S a set of full colours for α with | S | = a . For any S (cid:48) with | S (cid:48) | = a , there exists a colouring α (cid:48) such that S (cid:48) is full for α (cid:48) , and there is a transformationfrom α to α (cid:48) of length at most f a ( n, k − a ) + (2 a + 2) n .Proof. Let S be a set of colours with | S | = a which is full for some L -colouring α . Let S (cid:48) be anyset of colours of size a . The main part of the proof consists in transforming α into a colouring thatdoes not use at all any colour of S (cid:48) (such a colouring exists since the list assignment is a -feasible).Let H be the subgraph induced by the vertices not coloured S in α , and let L H be the listassignment for H obtained from L by removing S from all the lists. Since S is full in α , L H is a -feasible for H .Consider the following preference ordering on the colours: an arbitrary ordering of [ k ] \ ( S ∪ S (cid:48) ),followed by an ordering of S (cid:48) \ S , and finally the colours from S last. Let γ be the L -colouring of G G greedily from v n to v with this preference ordering. Since L is a -feasible,and | S | = a , no vertex is coloured with a colour in S in γ . Indeed | L ( v ) | ≥ | d + ( v ) | + a + 1 and only | d + ( v ) | neighbours of v have been coloured when v is coloured. Since in γ no vertex has a colourin S , γ | H is an L H -colouring of H . By induction hypothesis, there is a recolouring sequence thattransforms the colouring α | H of H into γ | H within at most f a ( n, k − a ) steps. By Remark 1, thisrecolouring sequence also is a recolouring sequence in G . We can then recolour the vertices of G \ H to their target colour in γ in an additional n steps. No conflict can happen at this step since γ isa proper colouring of G .One can easily check that, in γ , the set of colours [ k ] \ ( S ∪ S (cid:48) ) is full. Let K be the subgraph of G induced by all the vertices coloured S (cid:48) in γ , and let L K be a list assignment of these vertices whereall the colours not in S ∪ S (cid:48) were removed. Then since [ k ] \ ( S ∪ S (cid:48) ) is full, L K is a -feasible. We willrecolour K such that no vertex is coloured S (cid:48) (such a colouring exists because L K is a -feasible, and | S (cid:48) | = a ). Since the total number of colours used in L K is | S ∪ S (cid:48) | ≤ | S | + | S (cid:48) | = 2 a , this recolouringcan be done in at most f a ( n, a ) = 2 an steps by Lemma 7. By Remark 1, this recolouring sequencealso is a recolouring sequence in G . The colouring γ (cid:48) of G that we obtain is such that no vertexis coloured with c ∈ S (cid:48) . We can finally recolour the vertices of G one by one, starting from v n ,choosing a colour of S (cid:48) if it is available, or leaving it with its current colour otherwise.Let α (cid:48) be the resulting colouring. By construction α (cid:48) is full for S (cid:48) . The total number of stepsto reach α (cid:48) is at most f a ( n, k − a ) + n + 2 an + n = f a ( n, k − a ) + (2 a + 2) n .Using Lemma 8, we show that we can incrementally construct a set of full colours. Lemma 9.
Assume that k ≥ a . For any colouring α , there exists a colouring β containing aset of full colours S with | S | = a , and there is a transformation from α to β of length at most na f a ( n, k − a ) + 4 n .Proof. Let v , . . . , v n be a degeneracy ordering of G . A colouring is full up to step i for a set ofcolour S if | S | = a , and all the vertices v j with j ≤ i satisfy the conditions ( i ) , ( ii ) , ( iii ) for theset S . If a colouring γ is full up to step n for the set S , then the set S is full.Note that for any colouring α , any set of colours containing α ( v ) , . . . , α ( v a ) is full up to step a .So we only need to show that given a colouring α which is full up to step i for some set S , wecan reach a colouring α (cid:48) full up to step i + a for some (potentially different) set S (cid:48) in at most f a ( n, k − a ) + 4 an steps. Suppose now that α is full up to step i but not i + 1 for some set S . Let S (cid:48) be the colours of the vertices v i +1 , . . . , v i + a . Up to adding arbitrary colours to S (cid:48) , we can assumethat | S (cid:48) | = a . We will then recolour the graph in order to obtain a colouring where S (cid:48) is full up tostep i + a .Let H be the graph induced by the vertices v , . . . , v i , and L H be the list assignment of the ver-tices of H obtained from L by fixing the colours of the vertices outside H . In other words, for everyvertex v ∈ V ( H ), we remove from L ( v ) the colours of all the vertices of N ( v ) ∩ { v i +1 , . . . , v n } . Notethat L H is an a -feasible assignment of H since L was an a -feasible assignment of G . Additionally, S is a set of full colours for the colouring α | H .By Lemma 8, there is an L H -colouring α (cid:48) H of H which is full for S (cid:48) such that we can transform α | H into α | (cid:48) H in at most f a ( n, k − a ) + (2 a + 2) n steps. Let α (cid:48) be the colouring which agrees with α on the vertices with index larger than i , and agrees with α | (cid:48) H on H . By Remark 1, α (cid:48) can beobtained from α into at most f a ( n, k − a ) + (2 a + 2) n steps. By construction, S (cid:48) is full up to step i ,and since the vertices v i +1 , . . . v i + a are coloured with colours in S (cid:48) , it is full up to step i + a .8inally, this procedure needs to be repeated at most (cid:6) na (cid:7) − ≤ na times (the minus one comingfrom the fact that at the beginning we had for free a set of colours full for v , . . . , v a ). After thismany steps, we obtain a colouring full up to step n for some set S with | S | = a , which concludesthe proof.Finally, we can use the two previous lemma to get the following recursive inequality. Lemma 10.
Let k ≥ a , then f a ( n, k ) ≤ ( na + 3) f a ( n, k − a ) + 10 · n .Proof. Let L be an a -feasible list assignment of G , and α and β be two L -colourings of G . ByLemma 9, there exists a colouring α (cid:48) and a set of colours S α with | S α | = a which is full for α (cid:48) suchthat the colouring α (cid:48) can be reached from α in at most na f a ( n, k − a ) + 4 n steps. Similarly, thereexists a colouring β (cid:48) and a set of colours S β with | S β | = a such that S β is full for β (cid:48) such that thecolouring β (cid:48) can be reached from β in at most na f a ( n, k − a ) + 4 n steps. By Lemma 8, using anadditional f a ( n, k − a ) + 4 n steps, we can get a colouring β (cid:48)(cid:48) from β (cid:48) such that the set of full coloursin β (cid:48)(cid:48) and α (cid:48) is the same (namely S α ).Let S be some set of colours disjoint from S α with | S | = a . Let γ be a colouring of G that doesnot use any colour of S α (such a colouring exists since the list assignment of G is a -feasible).Let G α be the graph G where the vertices coloured with colours in S α have been deleted andthe colours in S α removed from the list assignment. Since S α is full for α (cid:48) , the list assignment of G α is a -feasible. Note that γ | G α is a proper colouring of G α . So by induction, it is possible torecolour α (cid:48) | G α into γ | G α in at most f a ( n, k − a ) steps. Since the vertices of W := V ( G ) \ V ( G α ) arecoloured with colours in S α , and since γ does not use any of these colours, one can finally recolourthe vertices of W one by one from their colours in S α to their target colours in γ .Let G β (cid:48)(cid:48) be the graph where the vertices coloured with S β (cid:48)(cid:48) = S α in β (cid:48)(cid:48) have been deleted.One can similarly recolour β (cid:48)(cid:48) | G β (cid:48)(cid:48) into γ | G β (cid:48)(cid:48) in at most f a ( n, k − a ) steps. Since vertices of W (cid:48) := V ( G ) \ V ( G β (cid:48)(cid:48) ) are coloured with colours in S α , we can also can recolour the vertices of W (cid:48) one by one from their colours in S α to their target colours in γ .The total number of steps to transform α into β is at most f ( n, k ) ≤ (cid:16) na f a ( n, k − a ) + 4 n (cid:17) + ( f a ( n, k − a ) + 4 n ) + 2 f a ( n, k − a ) + 2 n ≤ (cid:18) na + 3 (cid:19) f a ( n, k − a ) + 10 n Lemma 11.
For all k ≥ and a ≥ , we have f a ( n, k ) ≤ Ckn (cid:0) na + 3 (cid:1) (cid:100) ka (cid:101) − for some constant C > .Proof. We prove it by induction on the total number of colours k . The base case is when k ≤ a and simply is a consequence of Lemma 7 (note that (cid:6) ka (cid:7) = 2 since k > a ). The induction step isobtained using Lemma 10.We now have all the ingredients to prove Theorem 6, which completes the proof of Theorem 1.9 roof of Theorem 6. The first point is the result from Lemma 7. The second and last points areconsequences of Lemma 11 with the corresponding values of a . Let us prove the third point. Since k ≤ (1 + ε ) a , we have (cid:6) ka (cid:7) − ≤ (cid:100) ε (cid:101) − ≤ (cid:100) ε (cid:101) −
1. Consequently, the function given byLemma 11 is O ( n (cid:100) /ε (cid:101) ). Note that for the second and third point, the constant does not depend on k or a since k/a is a constant (but it depends on ε in the third point). The rest of this article is devoted to prove Cereceda’s quadratic conjecture for planar bipartitegraphs.
Theorem 2.
Let G be a planar bipartite graph. The diameter of G ( G, is at most n . We start by defining the level of a vertex. Given a graph G all the vertices with degree atmost 3 have level 1. For any i >
1, the vertices with level i are the ones with degree at most 3 afterremoval of all the vertices of level j < i .Let G be a planar graph given with its representation. The size of a face is the number of verticesincident to the face. The proof of Theorem 2 is based on a discharging proof. We are assumingthat the results does not hold. The proof then goes in two steps. First, we show that any planarbipartite graph must contain at least one ”configuration” of a list of (here two) configurations.Then, we prove that any counter-example cannot contain any other these configurations, whichimplies that this counter-example cannot exist. Let us first give the two configurations we willneed. Lemma 12.
Let G be a planar bipartite graph. At least one of the following holds:(i) G contains a vertex of degree at most or,(ii) G contains a vertex of degree incident to three vertices of level at most , and incident to aface of size four.Proof. We use a discharging argument to prove the result. Suppose by contradiction that thereexists a planar bipartite graph G satisfying none of the conditions above. Assign to each vertex v the weight deg( v ) −
4, and to each face f the weight deg( f ) −
4. The total weight assigned thisway is: (cid:88) v ∈ G (deg( v ) −
4) + (cid:88) f, face (deg( f ) −
4) = 2 | E | − n + 2 | E | − | F | = 4( | E | − n − | F | ) = − n, | E | and | F | are the number of vertices, edges, and faces of G . The last equality comesfrom Euler’s formula for planar graphs.Now the goal is to re-allocate the weights on the faces and the vertices to obtain a total weightthat is non-negative, a contradiction with the equation above. Note that all the faces as well asvertices of degree at least 4 have non-negative weights. In order to get a contradiction, we applythe following procedure and prove that after applying it all the vertices and faces have non-negativeweights: • Every vertex of level at least 3 gives weight 1 to all its neighbours of degree 3.10
Every face of size at least 6 gives weight to all its incident vertices.After applying this transformation, every face has a non-negative weight. Indeed, the faces of lengt4 still have weight 0, and the faces of length at least 6 gave a total of deg( f )3 ≤ deg( f ) − f ) ≥ v be a vertex of degree 3. If v is adjacent to a vertex of level at least 3, it receivesweight 1 during the first step, and its weight is non-negative. Otherwise, v is adjacent to threevertices of level at most 2, and then, by assumption on G , the three faces incident to v must havesize at least 6. Then, v received from each of the faces during the second step, and its weight isnon-negative.Hence, after applying the the procedure, every face and every vertex has a non-negative 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 G that for any two 5colourings of G , α and β , there exists a transformation from α to β such that each vertex isrecoloured at most cn times, for some constant c defined later.If G is reduced to a single vertex, the result is immediate. Otherwise, let n >
1. One of the twoconditions of Lemma 12 must occur.
Case (i).
The graph G contains a vertex v of degree at most 2.Let H be the subgraph obtained after removing v . By induction, there exists a transforma-tion S from α | H to β | H such that each vertex is recoloured at most c ( n −
1) times. Given thistransformation, we are going to produce a transformation S (cid:48) for the whole graph G . We followthe transformation steps of S . If the same recolouring step is possible in G we do it. Otherwise,it means that one of the (at most) two neighbours of v is recoloured, and that the target colour isthe current colour c of v . So we need to recolour v to make the transition possible. To recolour v ,we have at least two possible new colour choices X for v distinct from c . We can look forward atthe sequence S to see which colour c (cid:48) will appear next in the neighbourhood of v . We choose for v a colour of X distinct from c (cid:48) . In particular, the choice of c (cid:48) ensures that v will not be recolourednext time a neighbour of v is recoloured. At the end of the transformation S , we may need oneadditional step to recolour v to its target colour.Thus in total, v is recoloured at most c ( n − + 1 ≤ cn times (which holds as long as c ≥ Case (ii).
The graph G contains a vertex v of degree three such that: • all three neighbours of v have level at most 2, • v is incident to a face of length 4.Let f be the face of length 4 incident to v , and let w be the vertex opposite to v on f . Let u be thevertex adjacent to v , and not incident to f . Let v and v be the two remaining vertices adjacentto f (see Figure 2). In this case, we want to reduce the graph by merging the two vertices v and w (maintaining a bipartite planar graph). For this we will need the following result.11 uw v v f Figure 2: Reducible configuration in case ( ii ). Claim 13.
Starting from α , we can reach a colouring α (cid:48) such that α (cid:48) ( v ) = α (cid:48) ( w ). This colouringcan be reached by recolouring each vertex at most twice. Proof.
Since v and v are adjacent to w , their colour is different from α ( w ). Hence, the onlyobstruction to recolour directly v with α ( w ) is if u is coloured with α ( w ). Since u has level atmost 2, it has at most 3 neighbours with degree larger than 3. Let c be a colour different fromthese three neighbours, and different from α ( w ). First, we recolour all the degree three neighboursof u which are coloured c with any other colour (note that w is not recoloured since α ( w ) (cid:54) = c , but v might be recoloured). Then, we recolour u with c , and finally v can be recoloured with α ( w ) (since G is bipartite, the recolouring sequence ensures that the only neighbor of v that is recoloured is u ).Every vertex is recoloured at most once, except v which may be recoloured twice.By Claim 13 starting from α and β , we obtain two colourings α (cid:48) and β (cid:48) such that α (cid:48) ( v ) = α (cid:48) ( w )and β (cid:48) ( v ) = β (cid:48) ( w ). Let H be the graph obtained after merging v and w into a single vertex x .Clearly H is still planar, and it is still bipartite since v and w are on the same side of the bipartitionof G . Moreover, α (cid:48) and β (cid:48) are proper colourings of H . By induction, there is a transformationfrom α (cid:48) to β (cid:48) in H such that each vertex is recoloured at most c ( n −
1) times. By applying eachtransformation on x to both v and w , this gives a transformation in G from α (cid:48) to β (cid:48) such that eachvertex is recoloured at most c ( n −
1) times.This gives a transformation from α to β where each vertex is recoloured at most c ( n − ≤ cn times, taking c = 4. References [1] H.-J. B¨ockenhauer, E. Burjons, M. Raszyk, and P. Rossmanith. Reoptimization of Parame-terized Problems. arXiv e-prints , 2018.[2] M. Bonamy and N. Bousquet. Recoloring graphs via tree decompositions.
Eur. J. Comb. ,69:200–213, 2018.[3] M. Bonamy, N. Bousquet, C. Feghali, and M. Johnson. On a conjecture of mohar concerningkempe equivalence of regular graphs.
J. Comb. Theory, Ser. B , 135:179–199, 2019.[4] M. Bonamy, N. Bousquet, and G. Perarnau. Frozen (∆ + 1)-colourings of bounded degreegraphs. arXiv e-prints , 2018. 125] M. Bonamy, M. Johnson, I. Lignos, V. Patel, and D. Paulusma. Reconfiguration graphs for ver-tex colourings of chordal and chordal bipartite graphs.
Journal of Combinatorial Optimization ,pages 1–12, 2012.[6] P. Bonsma and L. Cereceda. Finding Paths Between Graph Colourings: PSPACE-Completeness and Superpolynomial Distances. In
MFCS , volume 4708 of
Lecture Notes inComputer Science , pages 738–749, 2007.[7] P. Bose, A. Lubiw, V. Pathak, and S. Verdonschot. Flipping edge-labelled triangulations.
Comput. Geom. , 68:309–326, 2018.[8] N. Bousquet and G. Perarnau. Fast recoloring of sparse graphs.
Eur. J. Comb. , 52:1–11, 2016.[9] L. Cereceda.
Mixing Graph Colourings . PhD thesis, London School of Economics and PoliticalScience, 2007.[10] L. Cereceda, J. van den Heuvel, and M. Johnson. Mixing 3-colourings in bipartite graphs.
Eur. J. Comb. , 30(7):1593–1606, 2009.[11] L. Cereceda, J. van den Heuvel, and M. Johnson. Finding paths between 3-colorings.
Journalof Graph Theory , 67(1):69–82, 2011.[12] S. Chen, M. Delcourt, A. Moitra, G. Perarnau, and L. Postle. Improved bounds for randomlysampling colorings via linear programming. In
Proceedings of the Thirtieth Annual ACM-SIAMSymposium on Discrete Algorithms, SODA 2019 , pages 2216–2234, 2019.[13] R. Diestel.
Graph Theory , volume 173 of
Graduate Texts in Mathematics . Springer-Verlag,Heidelberg, third edition, 2005.[14] M. Dyer, A. D. Flaxman, A. M. Frieze, and E. Vigoda. Randomly coloring sparse ran-dom graphs with fewer colors than the maximum degree.
Random Structures & Algorithms ,29(4):450–465, 2006.[15] E. Eiben and C. Feghali. Towards cereceda’s conjecture for planar graphs.
CoRR ,abs/1810.00731, 2018.[16] C. Feghali. Paths between colourings of sparse graphs.
Eur. J. Comb. , 75:169–171, 2019.[17] C. Feghali. Reconfiguring 10-colourings of planar graphs. arXiv e-prints , 2019.[18] C. Feghali, M. Johnson, and D. Paulusma. A reconfigurations analogue of brooks’ theoremand its consequences.
Journal of Graph Theory , 83(4):340–358, 2016.[19] A. Frieze and E. Vigoda. A survey on the use of markov chains to randomly sample colourings.
Oxford Lecture Series in Mathematics and its Applications , 34:53, 2007.[20] B. Mohar and J. Salas. On the non-ergodicity of the Swendsen–Wang–Koteck´y algo-rithm on the Kagom´e lattice.
Journal of Statistical Mechanics: Theory and Experiment ,2010(05):P05016, 2010.[21] N. Nishimura. Introduction to reconfiguration. preprint , 2017.1322] J. van den Heuvel.