Optimally Reconfiguring
List and Correspondence Colourings
Abstract
The reconfiguration graph for the -colourings of a graph has a vertex for each proper -colouring of , and two vertices of are adjacent precisely when those -colourings differ on a single vertex of . Much work has focused on bounding the maximum value of over all -vertex graphs . We consider the analogous problems for list colourings and for correspondence colourings. We conjecture that if is a list-assignment for a graph with for all , then . We also conjecture that if is a correspondence cover for a graph with for all , then . (Here and denote the matching number and vertex cover number of .) For every graph , we give constructions showing that both conjectures are best possible, which also hints towards an exact form of Cereceda’s Conjecture for regular graphs. Our first main result proves the upper bounds (for the list and correspondence versions, respectively) and . Our second main result proves that both conjectured bounds hold, whenever all satisfy . We conclude by proving one or both conjectures for various classes of graphs such as complete bipartite graphs, subcubic graphs, cactuses, and graphs with bounded maximum average degree.
1 Introduction
In this paper we study questions of transforming one proper colouring of a graph into another, by a sequence of recolouring steps. Each step recolours a single vertex, and we require that each resulting intermediate colouring is also proper. Our work fits into the broader context of reconfiguration, in which some object (in our case a proper colouring) is transformed into another object of the same type, via a sequence of small changes, and we require that after each change we again have an object of the prescribed type (for us a proper colouring). It is natural to pose reconfiguration questions for a wide range of objects: proper colourings, independent sets, dominating sets, maximum matchings, spanning trees, and solutions to 3-SAT, to name a few. For each type of object, we must define allowable changes between successive objects in the sequence (in our case, recolouring a single vertex). Typically, we ask four types of questions. (1) Given objects and , is there a reconfiguration sequence from to ? (2) If the first question is answered yes, what is the length of a shortest such sequence? (3) Is the first question answered yes for every pair of objects and ? (4) If yes, what is the maximum value of over all and For an introduction to reconfiguration, we recommend surveys by van den Heuvel [16] and Nishimura [15]. Most of our definitions and notation are standard, but for completeness we include many of them at the start of Section 2.
For a graph and a positive integer , the -colouring reconfiguration graph of , denoted , has as its vertices all proper -colourings of and two vertices of are adjacent if their corresponding colourings differ on exactly one vertex of . Our goal here is to study the diameter of this reconfiguration graph, denoted . In general, may be disconnected, in which case its diameter is infinite. For example consists of two isolated vertices (here is a cycle of length ). More strongly, for each integer there exist -regular graphs such that has isolated vertices. The simplest example is the clique , but this is also true for every -regular graph with a -colouring such that all colours appear on each closed neighbourhood; such colourings are called frozen.
To avoid a colouring being frozen, some vertex must have some colour unused by on its closed neighbourhood, . And to avoid being disconnected, every induced subgraph must contain such a vertex with some colour unused by on . Thus, it is natural to consider the degeneracy of , denoted .
Our examples of frozen colourings above show that, if we aim to have connected, then in general it is not enough to require . However, an easy inductive argument shows that a slightly stronger condition is sufficient: If , then is connected. (Earlier, Jerrum [12] proved that suffices, but the arguments are similar.) This inductive proof only yields that . But Cereceda [7, Conjecture 5.21] conjectured something much stronger.
Cereceda’s Conjecture.
For an -vertex graph with , the diameter of is .
Bousquet and Heinrich [6] proved that , which is the current best known bound. When , Cereceda [7, Proposition 5.23] proved that . In particular, Cereceda’s conjecture is true for regular graphs. But, as we show here, if , then in fact we have the stronger bound , and we conjecture .
A folklorish result observes that when is large relative to (for example, see [14, Section 3.1]). But until now, it seems that no one has investigated exact values of the diameter (say, when ). This is the main goal of our paper. Before stating our two main conjectures, and our results supporting them, we present an easy lower bound on in terms of the matching number .
Proposition 1.
For a graph , if , then .
Proof.
Let be a maximum matching in . Form from as follows. If and , then add to . Note that . So . Let be a -colouring of . Form from by swapping colours on endpoints of each edge in and, for each not saturated by , picking outside of . To recolour from to , every vertex must be recoloured. Further, for each edge , the first endpoint of to be recoloured must initially receive a colour other than , so must be recoloured at least twice. Thus, , as desired. ∎
At the end of Section 2, we mention various special cases in which the hypothesis of Proposition 1 can be weakened. We pose it as an open question whether the bound in Proposition 1 holds for all .
We study this reconfiguration problem in the more general contexts of list colouring and correspondence colouring, both of which we define in Section 2. These more general contexts offer the added advantage of enabling us to naturally prescribe fewer allowed colours for vertices of lower degree. Analogous to , for a graph and a list-assignment or correspondence cover for , we define the -reconfiguration graph or -reconfiguration graph of . Generalizing our constructions of frozen -colourings above, it is easy to show that and can contain frozen -colourings (and thus be disconnected) if we require only that all satisfy . Thus, we adopt a slightly stronger hypothesis: all satisfy . Now we can state our two main conjectures.
Conjecture 1 (List Colouring Reconfiguration Conjecture).
For a graph , if is a list-assignment such that for every , then .
Conjecture 2 (Correspondence Colouring Reconfiguration Conjecture).
For a graph , if is a correspondence cover such that for every , then .
Here denotes the vertex cover number of . For brevity, we often call these conjectures the List Conjecture and the Correspondence Conjecture. Our aim in this paper is to provide significant evidence for both conjectures. Due to Proposition 1, the List Conjecture is best possible, when the lists are large enough. We will soon give easy constructions showing that both conjectures are best possible whenever all satisfy . But we defer these constructions until Section 2, where we formally define list and correspondence colourings.
Proposition 2.
For every graph (i) there exists a list-assignment such that for all for which and (ii) there exists a correspondence cover such that for all for which .
For both the List Conjecture and the Correspondence Conjecture, it is trivial to construct examples that need at least recolourings. We simply require that for all vertices . So we view these conjectured upper bounds as consisting of a “trivial” portion, recolourings, and a “non-trivial” portion, or recolourings. We give two partial results toward each conjecture. Our first result proves both conjectures up to a factor of 2 on the non-trivial portions of these upper bounds.
Theorem 1.
(i) For every graph and list-assignment with for all we have . (ii) For every graph and correspondence cover with for all we have .
Bousquet, Feuilloley, Heinrich, and Rabie [5, following Question 1.3] asked about the diameter of when . In this case, a result of Bonamy and Bousquet [3, Theorem 1] implies, for an -vertex graph and , that . The authors of [5] asked whether it is possible to remove this dependency on . We answer their question affirmatively. Always , so Theorem 1(i) implies that when .
To complement Theorem 1, we prove both conjectured upper bounds when is sufficiently large.
Theorem 2.
(i) For every graph and list-assignment with for all we have . (ii) For every graph and correspondence cover with for all we have .
In Section 2 we present definitions and notation, as well as some easy lower bounds on diameters of reconfiguration graphs. In Section 3 we prove the list colouring portions of Theorems 1 and 2; and in Section 4 we prove the correspondence colouring portions. In Section 5 we prove more precise results when is a tree (in this case the correspondence problem is identical to the list problem).
In Sections 6 and 7, we conclude by proving one or both conjectures exactly for various graph classes, such as complete bipartite graphs, subcubic graphs, cactuses, and graphs with low maximum average degree. These graphs are all sparse or have low chromatic number. On the other end of the sparsity spectrum, the List Conjecture is also true for all complete graphs, due to an argument of Bonamy and Bousquet [3, Lemma 5].
2 Definitions, Notation, and Easy Lower Bounds
For a graph , we denote its order by . A matching is a set of vertex disjoint edges in . A matching of is perfect if it saturates every vertex of , and is near-perfect if it saturates every vertex of but one. A vertex cover is a vertex subset such that every edge of has at least one endpoint in . The matching number of is the size of a largest matching. The vertex cover number of is the size of a smallest vertex cover. (Recall that if is bipartite, then .)
The distance between two vertices is the length of a shortest path in between and . The eccentricity of a vertex , denoted , is The diameter of is , which is equal to , while the radius of is . A diameter can also refer to a shortest path between and for which A graph is -degenerate if every non-empty subgraph contains a vertex such that . The degeneracy of is the minimum such that is -degenerate.
A directed graph or digraph is analogous to a graph, except that each edge is directed. The directed edges in are ordered pairs of vertices, and are called arcs.
Definition 3.
For a digraph , the vertex cover number and matching number are defined to be and , where is the undirected graph with vertex set whose edges correspond to the bidirected edges in .
Colourings are mappings from to , and we denote them by greek letters such as , and . A colouring of is proper if for every edge we have . Let . A -colouring is a colouring using at most colours, which are generally from the set . The chromatic number of is the smallest such that has a proper -colouring.
A list-assignment , for a graph , assigns to each a set of natural numbers (“list” of allowable colours). A proper -colouring is a proper colouring such that for every . The list chromatic number (or choice number) is the least such that admits a proper -colouring whenever every satisfies .
A correspondence cover for a graph assigns to each vertex a set of colours and to each edge a matching between and . (In this paper, we typically consider either or for all .) Given a correspondence cover , an -colouring of is a function such that for all and whenever the edge is not an edge of the matching assigned to . Here denotes the union of the matchings assigned to all edges of . (It is easy to check that correspondence colouring generalises list colouring.)
The reconfiguration graph has as its vertices the proper -colourings of , and two -colourings and are adjacent in if they differ on exactly one vertex of . In particular, if , then has no vertices. (The reconfiguration graph was first defined in [8], where it was called the -colour graph.) The distance between -colourings and is at most if we can form , starting from , by recolouring at most vertices, one at a time, so that after each recolouring the current -colouring of is proper. Similarly, for a graph and list-assignment , the reconfiguration graph , or reconfiguration graph of the -colourings of , is the graph whose vertices are the proper -colourings of . Again, two -colourings and are adjacent in if they differ on exactly one vertex. This extension was first defined in [4]. For a correspondence cover , the reconfiguration graph is defined analogously; its vertices are the correspondence colourings of and two vertices of are adjacent precisely when their colourings differ on exactly one vertex of . We will also use the associated colour-shift digraph for a graph and two proper colourings, which is defined as follows.
Definition 4.
Let be a graph and and be two proper colourings of . We define an associated colour-shift digraph where if and only if and .
2.1 Improved Lower Bounds
In this subsection we prove lower bounds which show that our upper bounds in the rest of the paper are sharp or nearly sharp. These will not be explicitly needed elsewhere, so the impatient reader should feel free to skip to Section 3, where we prove Theorems 1(i) and 2(i).
Observation 5.
If is a graph with list colourings and , then .
Proof.
Whenever , vertex must be recoloured. Further, if and are neighbours for which and , then we cannot recolour both and only once, since after every recolouring step the resulting colouring must be proper. So at least vertices must be recoloured, and at least vertices must be recoloured at least twice. ∎
Does the lower bound in Observation 5 always hold with equality? Our next example shows that it does not; see Figure 1. Specifically, this example exhibits, for the 4-cycle, colourings and such that but still .
Example 6.
Let and denote by . If and and the colour set is precisely , then transforming to uses at least 6 recolourings, i.e. . Part of this reconfiguration graph is shown in Figure 1.
Proof.
Starting from , each vertex has a single colour with which it can be recoloured. But each possible recolouring creates a colouring for which Observation 5 gives . ∎
Next we present the (previously promised) proof of Proposition 2. For easy reference, we restate it.
Proposition 2.
For every graph (i) there exists a list-assignment such that for all for which and (ii) there exists a correspondence cover such that for all for which .
Proof.
For a graph , fix a maximum matching . Assign to each vertex a list of colours such that if and otherwise . Pick and such that for each , we have and , and for each we have The lower bound holds by Observation 5. The upper bound is trivial, since for all . This proves (i).
Let be a correspondence cover of such that for every . For every , let the matching (from ) consist of the two edges and ; otherwise, and are unmatched. Let and for all . Starting from , we can recolour a vertex with only if all neighbours of have already been recoloured. Thus, the set of vertices recoloured only once must be an independent set; equivalently, the set of vertices recoloured at least twice must be a vertex cover. So we must use at least recolourings, which proves the lower bound. For the upper bound, fix a minimum vertex cover . First recolour each with ; afterward, recolour each vertex with , first the vertices in and then those in . So recolourings suffice. This proves (ii). ∎
It is helpful to note that the proof of Proposition 2 actually works (for both list colouring and correspondence colouring) whenever all satisfy .
Clearly, the graph constructed in Proposition 1 depends on our choice of a perfect matching . It is interesting to note that some choices of work quite well, while others work rather poorly.
Example 7.
Figure 2 shows a graph built from two cliques and a complete bipartite graph , with parts and , by adding a perfect matching between one copy of and the vertices of and a perfect matching between and the other copy of . Now if we choose our perfect matching to be . If is even, then we can instead choose a perfect matching that is the disjoint union of perfect matchings in the two copies of and in the so that the resulting instead satisfies .
Although Example 7 implies that our choice of can effect dramatically, for many graphs, any choice of is fine. To conclude this section, we present various cases in which (for all choices of ) we can weaken the hypothesis of Proposition 1.
Proposition 3.
For a graph and positive integer we have
whenever (a) or (b) and , where is as in Proposition 1. In particular, this is true in the following cases.
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(in particular, if is bipartite and ).
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and .
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and (in particular, if is outerplanar and ).
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and is triangle-free, if Reed’s Conjecture111Recall that Reed conjectured that for every graph . holds for every graph with .
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and for some function .
Proof.
Recall the definition of from Proposition 1. Let be a maximum matching in . Form from as follows. If and , then add to . To prove each case of the proposition, we construct the graph , let be a -colouring of (which is also a -colouring of ), and form a -colouring from by swapping the colours on the endpoints of each edge in the specified maximum matching ; for each not saturated by , we choose to avoid . The latter is possible because we assumed that is bounded from below by or by . By Observation 5, we have . Note that . So . Now we show that in each case listed above we have or (in the first case) .
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Fix a -colouring of . We assume that is a perfect matching; if not, then we add a pendent edge at each vertex that is unsaturated by and add all these new edges to . To colour , we give each vertex the colour , where , , and . It is easy to check that the resulting colouring of is proper.
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Suppose that . As noted above, we have . If , then as desired. So assume instead that . By Brooks’ Theorem it suffices to show that no component of is . Suppose, to the contrary, that there exists such that . It is easy to check that must be 3-regular, and every vertex of must be saturated by . Observe that if two edges of , say lie on a 4-cycle in , then , for each . Now we need only to check that such a configuration occurs for every 3-regular graph on 6 vertices and every perfect matching . This is straightforward to verify: is either a 6-cycle or two 3-cycles; in the first case, can be added in two (non-isomorphic) ways and in the second, can be added in only one way.
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If , then we are done by condition . If , then we are done by condition . If , then we are done by Proposition 1. And if , then we are done trivially.
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We show that if is triangle-free, then . Suppose, to the contrary, that . Consider such that , and colour each edge of with 1 if it appears in and with 2 otherwise. Since is triangle-free, has no triangle coloured 1. Let denote the diagonal Ramsey number for cliques of order . Since , we must have a triangle in that is coloured 2; denote its vertices by . Let denote the edges of incident with . Since , we conclude that . This contradicts that is triangle-free. Hence, , as claimed. Finally, by Reed’s Conjecture, .
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Let be a positive integer and consider any graph with . Note that can be decomposed into two (edge-disjoint) copies of ; here is formed from the two copies of by identifying, for each edge , the instance of in one copy of with the instance of in the other copy. This implies that . We also know that Now by e.g. [13, Theorem 2] we have
This is smaller than whenever is sufficiently large.
∎
We also consider the graph formed from by contracting each edge of a maximum matching .
Proposition 4.
For a graph and positive integer we have
whenever for some maximum matching of . In particular, this is
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true a.a.s. for , for fixed , and
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true if is -minor free and , provided Hadwiger’s Conjecture222Recall that Hadwiger conjectured that for every graph with no -minor. holds for -minor free graphs. The latter is true for .
Proof.
Let be a -colouring of . We construct a -colouring of such that if and arises from contracting edge and , then ; if does not arise from contracting an edge, then we choose . Form from by swapping the colours on the endpoints of each edge of and letting for each .
Hence by Observation 5. We now prove that in the two mentioned cases, we (a.a.s.) have , regardless of the maximum matching of we choose to form .
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Note that is also -minor free. So, by assumption, .
∎
Question 1.
Is it true, for every graph and every , that ?
If the answer to this question is yes, then the List Conjecture (if true) would imply that for every graph and every . In particular, this would constitute a precise version for Cereceda’s Conjecture in the case of regular graphs.
3 Proving the Main Results for Lists
In this section, we prove the list colouring portions of our main results: Theorems 1(i) and 2(i). Recall that a graph is factor-critical if is odd and has a perfect matching for each . Gallai showed that if every vertex is unsaturated by some maximum matching, i.e., for every vertex , then is factor-critical. So, in each proof we begin with a lemma to handle factor-critical graphs, and thereafter proceed to the general case. The following lemma serves as the base case in an inductive proof of Theorem 1(i). (We will only need it when is factor-critical, but we prove it for all .)
Lemma 8.
Let be graph and let be a list-assignment for such that for all . If and are -colourings of , then we can recolour from to in at most steps.
Proof.
We use induction on . The case , is trivial, since then is an independent set; starting from , we can recolour each vertex to . We use at most recolouring steps, which suffices since .
Now assume . For each colour , let and . Since , there exists such that . Starting from , for each , recolour to avoid colour . Now for each , recolour with . After at most steps, we have more vertices that are coloured to agree with . Now we delete these vertices, and delete the colour from whenever had some neighbours in , and finish by induction. ∎
Corollary 9.
For every graph and every , the diameter of is at most .
Now we prove Theorem 1(i). For easy reference, we restate it below.
Theorem 10.
Let be an -vertex graph and be a list-assignment with for all . If and are -colourings of , then we can recolour from to in at most steps.
Proof.
We use induction on . We assume is connected; otherwise, we handle each component separately. (This suffices because, for a graph with components , we have and .) If for all , then is factor-critical. This is an easy result of Gallai, but also follows from Theorem 12(2). Now , so we are done by Lemma 8.
Assume instead that some vertex is saturated by every maximum matching. Note that . By Pigeonhole there exists such that . (We handle the case when equality holds, since the other case is easier.) Assume there exists such that and . (If instead , then we swap the roles of and .) Form from by recolouring from and recolouring with . Let . Let for all and for all other . Denote the restrictions to of and by and . By induction, we can recolour from to in at most steps. After this, we can finish by recolouring with . Thus, . ∎
To prove Theorem 2(i), we will use the following lemma to handle the case when is factor-critical.
Lemma 11.
Let be a graph and let be a list-assignment for such that for all . If and are -colourings of , then we can recolour from to in at most steps.
Proof.
Let and be arbitrary -colourings of . (Recall that all colours are positive integers.) Let and be the subsets of , respectively, for which and . Let and . Since , we assume by symmetry that ; otherwise, we interchange and . We show how to recolour from to using at most steps.
First, iteratively for every we recolour from , where is the current colouring of . By definition this recolouring is proper (we actually do not need to recolour if for all ). Next, iteratively for from to , we recolour with all vertices for which . In these steps, we only have proper colourings, as we will now show. Consider a neighbour of . If has already been recoloured with some , then , by construction. If has not been recoloured, then . ∎
The following classical result is helpful in our proof of Theorem 2(i).
Theorem 12 (Edmonds–Gallai Decomposition Theorem).
For a graph , let
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such that some maximum matching avoids ,
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such that has a neighbour in , but ,
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.
Now the following statements hold.333We prefer the names , , over the standard terminology , , and , since the latter terms conflict with our use of for a digraph with vertex set and arc set .
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The subgraph induced by has a perfect matching.
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The components of the subgraph induced by are factor-critical.
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If is any maximum matching, then contains a perfect matching of each component of , and contains a near-perfect matching of each component of , and matches all vertices of with vertices in distinct components of .
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, where denotes the number of components of the graph spanned by .
Now we prove Theorem 2(i). For easy reference, we restate it below.
Theorem 13.
Let be a graph and be a list-assignment for with for all . If and are -colourings of , then we can recolour from to in at most steps.
Proof.
We refer to , , and , as defined in Theorem 12. Starting from , recolour each vertex to avoid each colour used currently on and also to avoid each colour used on in ; call the resulting colouring . Let . For each , let . Note that . Thus, for each component of , we can recolour from to in at most steps, by Lemma 11. Finally, we recolour each vertex of to its colouring . Let denote the set of components of . The number of recolouring steps used is at most . The final equality uses Theorem 12(4). The first equality uses that the vertices of a component are contained either in or in . If then is even by Theorem 12(3). On the other hand, if then is odd by Theorem 12(2). ∎
By the comment after Proposition 2, Theorem 13 is sharp. For (non-list) colouring, we get the following.
Corollary 14.
For every graph and every , we have .
4 Proving the Main Results for Correspondences
In this section, we prove the correspondence colouring portions of our main results: Theorems 1(ii) and 2(ii).
Theorem 15.
Let be a graph and be a correspondence cover with for all . If and are -colourings of , then can be recoloured from to in at most steps.
Proof.
Let be a minimum vertex cover. For every vertex , we recolour with a colour that, for every , conflicts (under cover ) with neither the current colour nor the final colour . This is possible because . Now we recolour every vertex with and then do this also for every vertex in . Note that we use at most recolourings. ∎
Theorem 16.
Let be a graph and be a correspondence cover for such that for all . If and are -colourings of , then we can recolour from to in at most steps.
Proof.
We use induction on . If , i.e., is an independent set, then we simply recolour every vertex to . So assume the theorem is true for all graphs with .
Let be a minimum vertex cover of and pick . Consider the union over all of the edges in from and to . By Pigeonhole, since , some colour has at most one incident edge in this union. (We handle the case when has exactly one incident edge, since the other case is easier.) Assume there exists such that is matched to and no other or is matched to . (If instead is matched to , then we swap the roles of and .) Form from by first recolouring with a colour different from , that still gives a proper -colouring, and afterward recolouring with . Let . For every , remove from the colour matched with ; call the resulting correspondence cover . Denote the restrictions to of and by and . By induction, since , we can recolour from to in at most steps. After this, we can finish by recolouring with . Thus, . This proves the theorem. ∎
5 Distances in Reconfiguration Graphs of Trees
For each tree , it is straightforward to check that reconfiguration of correspondence colouring is no harder than reconfiguration of list colouring. Given a tree and a correspondence cover , it is easy to construct a list-assignment for such that -colourings of are in bijection with -colourings of . We can do this by induction on , by deleting a leaf and extending the list-assignment , given by hypothesis, for . So, for simplicity, we phrase all results in this section only in terms of list colourings.
Theorem 17.
Fix a tree and a list-assignment with for every If and are proper -colourings of , then the distance between and in the reconfiguration graph is equal to
Proof.
The lower bound holds by Observation 5. Now we prove that this lower bound is also an upper bound.
We use induction on . The base case, , is trivial. Assume the theorem is true whenever has order at most . We will prove it for an arbitrary tree on vertices. If for some , then we use induction on the components of , with removed from the lists of the neighbours of .
Suppose instead there are neighbours and for which . Now has two components; so let and be the components, respectively, containing and . By deleting from , we can first recolour . Afterwards, we delete from and recolour . By using the induction hypothesis (twice), we get the desired upper bound, since
In the remaining case, we have two colour classes, say with colours and , which need to be swapped. We pick a smallest vertex cover , which has size Iteratively, we recolour every vertex in with a colour different from , , and all colours used to recolour neighbours of in . Note that has at most neighbours in , since otherwise would not be a minimal vertex cover. Since , we can recolour as desired. Next, for each , we recolour with Finally, we recolour each with . This proves the induction step, which finishes the proof. ∎
Corollary 18.
If is a tree and is a list-assignment with for every , then
Proof.
Theorem 17 implies that , and the bound holds trivially, since each edge of a matching saturates two vertices. ∎
Recall that denotes . Thus, we write -colouring to mean a -colouring from the colours .
Proposition 5.
For every tree and , we have
Proof.
The inequality holds by Corollary 18; and this inequality actually holds with equality, since the two -colourings of are at distance exactly . That is, . Now we consider . If we contract all edges of a maximum matching in , the result is also a tree, . When we 2-colour , one colour is used on at least half of the contracted edges. Denote the set of these contracted edges by . Note that is an induced matching in . Fix an arbitrary proper -colouring of . Now we construct a proper -colouring of by swapping the colours on the endpoints of each edge , i.e., let and for every edge For every vertex not belonging to an edge in , choose a colour in different from the colours already assigned (in ) to neighbours of and different from Now by Theorem 17. ∎
We construct trees for which certain colourings are more “central” in the reconfiguration graph than others. That is, we construct trees for which . We also study the maximum possible diameter and minimum possible radius of reconfiguration graphs of trees of given order, and the maximum possible difference of these quantities.
Proposition 6.
For every , the path satisfies
Furthermore, there exist -vertex trees , with maximum degree 3, such that for every we have
All such maximise, over all -vertex trees, the difference .
Proof.
Consider with vertex set . Fix , and let and be the colourings with ranges and , respectively, such that
Note that by Theorem 17, since switching colours and in requires recolouring steps. Furthermore, Corollary 18 implies that . On the other hand, for every proper colouring , we know that cannot have a pair of bidirected edges of the form and since So ; hence, , which implies that .
Next we prove the lower bound on . For every colouring of , we can choose at least disjoint edges such that swapping the colours in on the endpoints of each edge (and possibly recolouring vertices not in any of these edges) yields another proper colouring. To see this, first select edge Whenever has been selected, there exists for which . Now add edge to the set of selected edges. When our selection ends, the set contains at least edges. We can now construct a -colouring where and for every edge , and for every Theorem 17 gives .
Now let be the -vertex comb graph; see Figure 3. Here is formed from a path , where , by adding, for each (except when is odd), an additional neighbour . Since , by Proposition 5 the diameter of is . Let be the colouring shown and described in Figure 3.
For list colourings, the difference can be even larger.
Proposition 7.
For every tree , there exists a list-assignment , with for all , such that and . These values are, respectively, the maximum and minimum possible over all such list-assignments .
Proof.
Choose a maximum matching of and fix such that
Let and be -colourings such that, for every edge , we have and . For every vertex not saturated by , we pick and to be arbitrary distinct colours in . By Theorem 17, we have . Corollary 18 implies that . Further, this diameter is the maximum possible over all such list-assignments .
To show that , we construct an -colouring such that for every -colouring . Since for every vertex , we can form an -colouring such that every satisfies . For every proper colouring , by our construction of , we can recolour into greedily; hence, .
Now we show that . For every -colouring , there exists an -colouring such that for all . To see this, let . Since for all , we form by colouring greedily (in any order) from . Clearly, . Thus, . In fact, this lower bound does not depend on the specific choice of , but only uses that for all . Thus, this radius is the minimum over all such list-assignments . ∎
6 Proving the List Conjecture for Complete Bipartite Graphs and Cactuses
A cactus is a connected graph in which each edge lies on at most one cycle. In this section, we prove the List Conjecture for all complete bipartite graphs and for all cactuses. Both proofs are by induction, and rely on the following helpful lemma.
Lemma 19.
Fix a positive integer . Let be a graph for which there exists a list-assignment and two proper -colourings such that for every , and , but, for every proper induced subgraph of , no such list-assignment exists. Now for every partition into two non-empty subsets of vertices, has at least one arc from to and at least one arc from to i.e., is strongly connected.
Proof.
Assume the lemma is false; by symmetry, assume has no arcs from to . Let and . Let be the -colouring where when and when .
For every , let . Note that . Also note that still for every ; this is where we use that there is no arc from to . Since is a proper induced subgraph of , by hypothesis ; thus . Now for every , let . Similarly to before, , so ; thus . By the bounds above and the triangle inequality, we have
The last step uses that the union of a matching in and a matching in is a matching in . So is not a counterexample, and the lemma is true. ∎
It is helpful to note that an analogous statement holds for correspondence colouring. In fact, its proof is nearly identical to that given above. We use this observation in our proof of Theorem 27.
Using Lemma 19, we prove that the List Conjecture holds for all complete bipartite graphs.
Theorem 20.
The List Conjecture is true for all complete bipartite graphs.
Proof.
Suppose the theorem is false and choose a counterexample minimizing . By symmetry, we assume that . Denote the parts of by and , with and .
By Lemma 19, we have and . By swapping the roles of and , we also have and ; thus, and . Note that , because the graph is complete bipartite.
Case 1: . For each , we have . First recolour each from . Now recolour each with . Finally, recolour each with . The number of steps that we use is at most .
Case 2: . If and , then , which contradicts the case. So assume, by symmetry, that ; if not, then simply interchange the roles of and . Now, by Pigeonhole, there exists such that . So, recolour , if such exists, to avoid , and then recolour every with . Now we delete every such that , we delete from for every , and we finish on the resulting smaller graph (with an assignment of smaller lists) by the minimality of . In total, the number of recolouring steps we use is at most . ∎
Using Lemma 19, we prove that the List Conjecture holds for all cycles.
Lemma 21.
Let be a cycle and be a list-assignment such that, for all , we have . If and are proper -colourings of , then we can recolour from to in at most steps.
Proof.
By Lemma 19, with , if , then is strongly connected. This implies that contains a directed cycle as a subdigraph. (If contains a bidirected path , then . So, if is spanning, then its order must be even, to avoid a conflict between the colours of its endpoints. But then contains an additional arc, so contains a directed cycle, as claimed.)
If is a bidirected cycle, then must be even, as in the previous paragraph, and recolouring steps suffice. Suppose instead that is precisely a directed cyle. When , we recolour one vertex in a colour absent from , recolour the other two vertices in order (to match ), and finally recolour with . When , we recolour one vertex in a colour absent from . Now we can recolour correctly (in order) all vertices of except for and one of its neighbours. Correctly colouring these final two vertices takes at most 3 recolouring steps. Now we are done, since .
In the remaining case, we have and some directed edge is adjacent to a bidirected edge. Without loss of generality, we assume but . Recolour with a colour different from , , and . Now delete from and , and let . By Theorem 17, we can recolour from to (both restricted to ) using at most recolouring steps. Here we use that is not an arc, so . Finally, we recolour with . ∎
Using Lemmas 19 and 21, we can prove that the List Conjecture is true for every cactus.
Theorem 22.
The List Conjecture is true for every cactus.
Proof.
Instead assume the theorem is false. Let be a counterexample minimizing and let and be -colourings with . Every proper induced subgraph of is a disjoint union of cactuses, say ; since and , the List Conjecture must hold for , by the minimality of . Lemma 19 implies that is strongly connected. And Lemma 21 implies that is not a cycle. The following claim restricts the structure of , , and .
Claim 23.
If and , then . In particular, each block of containing gives rise to exactly two arcs in incident to , and has no adjacent leaf in .
Proof.
Instead assume there exists . Recolour with , let , let for all , and otherwise let . Denote by and the restrictions to of and . Since is minimal, we can recolour from to , using , in at most recolouring steps. We finish by recolouring with . This proves the first statement.
Since is strongly connected, each block of containing gives rise to at least two arcs in incident to , one in each direction. Since is a cactus, each such contains at most 2 neighbours of . So at least half of the neighbours of are coloured by , and also at least half of the neighbours of are coloured by . Therefore . If any such is either or gives rise to at least three arcs in incident with , then , which contradicts the first statement. This proves the second statement. ∎
It is easy to check that every neighbour of a leaf in satisfies . Thus, Claim 23 implies that has no leaf vertex. This implies that all endblocks of are cycles. Let be an endblock. Denote its vertices by , such that is the unique cut-vertex in the block.
Claim 24.
The endblock is not an even cycle.
Proof.
Assume instead that is an even cycle. It is easy to check that whenever is even. That is, every maximum matching saturates whenever is even. By Claim 23, every is incident in to exactly two arcs arising from . Since is strongly connected, this implies that is a directed cycle; by symmetry, we assume that it is oriented as .
We first recolour with a colour absent from . Now we recolour each to , with decreasing from to . Let , let , and otherwise let . Since is minimal, we can recolour from its current colouring to (restricted to ) using at most steps. Finally, recolour to . The total number of recolouring steps is at most . Since , this is at most . ∎
Claim 25.
The endblock is not an odd cycle.
Proof.
Assume instead that is an odd cycle. If contains fewer than disjoint digons (for example, this is true when ), then we can easily finish, as follows. We recolour in a colour different from . This is possible because is used by on at least two neighbours of , since is strongly connected. By Theorem 17, we recolour to in at most steps, and we then recolour to in at most steps. Thus, we recolour from to in at most steps.
Now we consider the other case, when is a digon of for every odd with . Since is strongly connected, we assume contains arc for every . Similar to the proof of Lemma 21, here cannot contain a spanning bidirected path, since is odd. This implies, for some even , that does not contain . Now we will recolour some set of ’s to reach a new colouring , such that is either acyclic or contains a single directed cycle, . From , we can recolour by the minimality of , and afterward finish on . The details follow.
First suppose that . For every such that either (a) and is odd or (b) and is even, recolour with a colour different from those in . Here we take the indices modulo , i.e., . If , then we recolour the same set of ’s, except for . Let . By the minimality of , we can recolour in at most steps. In the case that , we now recolour with a colour different from those in . For every odd such that , we recolour with . Next, for every even such that , we recolour with Finally, the remaining vertices in can also be recoloured with their colour in . This process uses at most steps. ∎
7 Proving the Correspondence Conjecture for Cactuses, Subcubic Graphs, and Graphs with Low Maximum Average Degree
The maximum average degree of a graph , denoted , is the maximum, taken over all subgraphs , of the average degree of . That is, . Let denote the number of neighbours of such that . In this section we prove the Correspondence Conjecture for all subcubic graphs, cactuses, and graphs with . To do so, we often implicitly use the following observation. We omit its proof, which is easy.
Observation 26.
If is a graph with a minimum vertex cover and , then .
Theorem 27.
Let be a graph and a correspondence cover for such that, for all we have . Now if at least one of the following holds.
-
(a)
.
-
(b)
is a cactus.
-
(c)
.
Proof.
Fix satisfying (a), (b), or (c) and as in the theorem. Fix arbitrary -colourings and . Our proof is by a double induction: primarily on and secondarily on . The base case, , is trivial, since is an independent set and we can greedily recolour from to . For the induction step, assume .
We first show that contains either (i) a vertex such that and lies in some minimum vertex cover or (ii) a vertex such that and . Next we show how to proceed by induction in each of cases (i) and (ii).
If satisfies (a), then clearly contains an instance of (i). Suppose satisfies (b), and consider an endblock of . If is a cycle, then contains adjacent non-cut vertices, and ; note that . Further, every vertex cover of contains or , so contains (i). Assume instead that every endblock of is an edge.
Form a graph with a vertex for every block in , where two vertices of are adjacent if their corresponding blocks share a vertex in . Now the endpoints of a diameter in correspond with pendent edges in . For the leaf corresponding to such a pendent edge, call its neighbour . Clearly lies in some minimum vertex cover of . If , then contains (i). Otherwise, , since by the choice of at most one block incident to is not an endblock; so contains (ii). This concludes the case that satisfies (b).
Assume instead that satisfies (c). Suppose, to reach a contradiction, that contains neither (i) nor (ii). Form from by deleting all vertices with degree at most 1. We will show that , a contradiction. Note that has minimum degree (at least) 2. If an arbitrary vertex satisfies , then , since otherwise is an instance of (i) or (ii) in . Further, if , then for all , since otherwise or is an instance of (i) in . Now we will reach a contradiction with discharging. Give each vertex in charge . We use a single discharging rule: Each 2-vertex in takes from each neighbour. Now each 2-vertex in finishes with charge . And each vertex with finishes with charge . This yields the desired contradiction, which proves that contains either (i) or (ii) if .
Now we prove the induction step in the cases that contains (i) or (ii).
Suppose contains (i). Suppose at most one neighbour, say , of has matched with . Now recolour with a colour not matched to , for every , and not matched to . Next, recolour with , and proceed on by induction (with the colour matched to deleted from the lists of all vertices in ). So instead assume there exist at least two such neighbours, say and . By interchanging the roles of and (and repeating this argument), we see that also there exist two neighbours, say and , such that and are matched to . The total number of colours in matched to or for some is at most , since . Hence, there exists that is not matched to or for all . Recolour with . Proceed on by induction, with that colour that is matched to deleted from the list for each . After finishing on , recolour with . The number of steps we use is at most .
Suppose instead contains (ii). The proof is almost the same as for (i). If there exists such that and is not matched to , then simply recolour with , and proceed by induction; here we use the secondary induction hypothesis, since possibly . So assume that no such exists. By interchanging the roles of and , we also assume that is matched to for each with . Further, if has a non-leaf neighbour , then at least one such neighbour has matched to and at least one (possibly the same one) has matched to . (This follows from the correspondence colouring analog of Lemma 19.) But now there exists such that for all , colour is not matched to either or . Recolour to , and let . Form from by deleting from , for every , the colour matched with . By induction, we can recolour from to (both restricted to ) using at most steps. Finally, recolour to . This uses at most steps, which finishes the proof. ∎
Corollary 28.
The List Conjecture holds for all bipartite graphs with .
Proof.
When is bipartite, recall that . ∎
8 Concluding Remarks
In this paper, we give evidence for both the List Conjecture and the Correspondence Conjecture. We also give evidence for an affirmative answer to Question 1. The List Conjecture and Question 1 would together determine the precise diameter for and, as such, a precise bound when Cereceda’s Conjecture is restricted to regular graphs. So we explicitly conjecture the following.
Conjecture 3 (Regular Cereceda’s Conjecture).
For a -regular graph , if , then .
In Theorem 10 we prove, when for all , that . In a similar vein, it would be interesting to show that is linear when The following conjecture can be viewed as a “balanced” version of the List Conjecture.
Conjecture 4 (Mad Colouring Reconfiguration Conjecture).
For a graph with , if is a list-assignment such that for every , then
Feghali [11] proved that if and , then . Conjecture 4 aims to prove a similar result for list colouring, with one more colour available for each vertex, and with a somewhat stronger bound on diameter. All planar graphs have , and all triangle-free planar graphs have , so we note that Conjecture 4 would imply stronger forms of [10, Conjecture 22]. If is regular, then , so Conjecture 4 is true by Theorem 10. Conjecture 4 is also related to444If is not an integer, then . But if has a regular subgraph of degree , then this inequality fails. a conjecture of Bartier et al. [1, Conjecture 1.6] that graphs with satisfy .
We do not know if Conjecture 4 might be true with a linear bound of the form instead of But we do note, for this version of the conjecture and every constant , that no bound of the form can hold. This is shown by the star and the colourings with , and . Here and
For a correspondence cover such that for all , in Theorem 1(ii) we proved that . We view this as modest evidence that Cereceda’s Conjecture might remain true in the more general context of correspondence colourings.
8.1 Open Problems
The three focuses of this paper are the List Conjecture, the Correspondence Conjecture, and (to a lesser extent) Question 1. All of these remain open. However, each of them seems rather hard. So, to motivate further research, below we identify some specific classes of graphs for which we believe that each conjecture may be approachable. We begin with some graph classes for which it would be particularly interesting to make further progress on the List Conjecture.
-
Complete -partite graphs for each . (We know the List Conjecture is true for both complete graphs and complete bipartite graphs, so this is a natural common generalization.)
-
Bipartite graphs, not necessarily complete.
-
Outerplanar graphs and, more generally, planar graphs.
-
Subcubic graphs. (We already proved the Correspondence Conjecture for this class; nonetheless, the List Conjecture remains open.)
Conversely, the Correspondence Conjecture remains open for the following basic graph classes.
-
Complete graphs. (The argument of Bonamy and Bousquet for complete graphs directly yields the List Conjecture, but does not yield the Correspondence Conjecture.)
-
Complete bipartite graphs.
-
Bipartite graphs, not necessarily complete. (This would imply the List Conjecture for the same class.)
-
Outerplanar graphs and, more generally, planar graphs.
Finally, we stress that it will be interesting to improve on Theorems 1 and 2. For example, find the smallest such that for every graph and list-assignment with for all , we have . We proved suffices, while would resolve the List Conjecture.
Acknowledgement
We thank the organisers of the online workshop Graph Reconfiguration of the Sparse Graphs Coalition555For more information, visit https://sparse-graphs.mimuw.edu.pl/doku.php., where this project started. We also thank the two referees for their careful reading and valuable feedback.
References
- [1] V. Bartier, N. Bousquet, C. Feghali, M. Heinrich, B. Moore, and T. Pierron. Recoloring planar graphs of girth at least five. SIAM J. Discrete Math., 37(1):332–350, 2023. doi:10.1137/21M1463598.
- [2] B. Bollobás, P. A. Catlin, and P. Erdős. Hadwiger’s conjecture is true for almost every graph. European J. Combin., 1(3):195–199, 1980. doi:10.1016/S0195-6698(80)80001-1.
- [3] M. Bonamy and N. Bousquet. Recoloring graphs via tree decompositions. European J. Combin., 69:200–213, 2018, arXiv:1403.6386.
- [4] P. Bonsma and L. Cereceda. Finding paths between graph colourings: PSPACE-completeness and superpolynomial distances. Theoret. Comput. Sci., 410(50):5215--5226, 2009. doi:10.1016/j.tcs.2009.08.023.
- [5] N. Bousquet, L. Feuilloley, M. Heinrich, and M. Rabie. Short and local transformations between (+1)-colorings. 2022, arXiv:2203.08885.
- [6] N. Bousquet and M. Heinrich. A polynomial version of Cereceda’s conjecture. J. Combin. Theory Ser. B, 155:1--16, 2022. doi:10.1016/j.jctb.2022.01.006.
- [7] L. Cereceda. Mixing graph colourings. PhD thesis, The London School of Economics and Political Science (LSE), 2007.
- [8] L. Cereceda, J. van den Heuvel, and M. Johnson. Connectedness of the graph of vertex-colourings. Discrete Math., 308(5-6):913--919, 2008. doi:10.1016/j.disc.2007.07.028.
- [9] M. Delcourt and L. Postle. Reducing linear Hadwiger’s conjecture to coloring small graphs. 2021, arXiv:2108.01633.
- [10] Z. Dvořák and C. Feghali. A Thomassen-type method for planar graph recoloring. European J. Combin., 95:Paper No. 103319, 12, 2021. doi:10.1016/j.ejc.2021.103319.
- [11] C. Feghali. Reconfiguring colorings of graphs with bounded maximum average degree. J. Combin. Theory Ser. B, 147:133--138, 2021. doi:10.1016/j.jctb.2020.11.001.
- [12] M. Jerrum. A very simple algorithm for estimating the number of -colorings of a low-degree graph. Random Structures Algorithms, 7(2):157--165, 1995. doi:10.1002/rsa.3240070205.
- [13] M. Molloy. The list chromatic number of graphs with small clique number. J. Combin. Theory Ser. B, 134:264--284, 2019. doi:10.1016/j.jctb.2018.06.007.
- [14] C. M. Mynhardt and S. Nasserasr. Reconfiguration of colourings and dominating sets in graphs. In 50 years of combinatorics, graph theory, and computing, Discrete Math. Appl. (Boca Raton), pages 171--191. CRC Press, Boca Raton, FL, 2020. doi:10.1201/9780429280092-10.
- [15] N. Nishimura. Introduction to reconfiguration. Algorithms (Basel), 11(4):Paper No. 52, 25, 2018.
- [16] J. van den Heuvel. The complexity of change. In Surveys in combinatorics 2013, volume 409 of London Math. Soc. Lecture Note Ser., pages 127--160. Cambridge Univ. Press, Cambridge, 2013, arXiv:1312.2816.