A Dichotomy Theorem for Circular Colouring

Reconfiguration

Richard C. Brewstera,1,∗, Sean McGuinnessa,1, Benjamin Moorea,1,Jonathan A. Noelb,1

aDepartment of Mathematics and Statistics, Thompson Rivers University, Kamloops,Canada

bMathematical Institute, University of Oxford, Oxford, United Kingdom

Abstract

The “reconfiguration problem” for circular colourings asks, given two (p, q)-colourings f and g of a graph G, is it possible to transform f into g bychanging the colour of one vertex at a time such that every intermediatemapping is a (p, q)-colouring? We show that this problem can be solvedin polynomial time for 2 ≤ p/q < 4 and is PSPACE-complete for p/q ≥ 4.This generalizes a known dichotomy theorem for reconfiguring classical graphcolourings.

Keywords: reconfiguration, recolouring, dichotomy, PSPACE, circularcolouring, homomorphism

1. Introduction

In recent years, a large body of research has emerged concerning so called“reconfiguration” variants of combinatorial problems (see, e.g., the survey ofvan den Heuvel [16] and the references therein, and well as [13, 14]). Theseproblems are typically of the following form: given two solutions to a fixedcombinatorial problem (e.g. two cliques of order k in a graph or two satisfying

∗Corresponding authorEmail addresses: rbrewster@tru.ca (Richard C. Brewster), smcguinness@tru.ca

(Sean McGuinness), mooreb104@mytru.ca (Benjamin Moore), noel@maths.ox.ac.uk(Jonathan A. Noel)

1Research supported by the Natural Sciences and Engineering Research Council ofCanada.

Preprint submitted to Elsevier August 21, 2015

assignments of a specific 3-SAT instance) is it possible to transform one ofthese solutions into the other by applying a sequence of allowed modificationssuch that every intermediate object is also a solution to the problem?

As a specific example, for a fixed integer k and a graph G, one may askthe following: given two (proper) k-colourings f and g of G, is it possibleto transform f into g by changing the colour of one vertex at a time suchthat every intermediate mapping is a k-colouring?2 In the affirmative we saythat f reconfigures to g. This problem is clearly solvable in polynomial timefor k ≤ 2. Rather surprisingly, Cereceda, van den Heuvel and Johnson [11]proved that it is also solvable in polynomial time for k = 3, despite the factthat determining if a graph admits a 3-colouring is NP-complete. On theother hand, Bonsma and Cereceda [5] proved that, when k ≥ 4, the problemis PSPACE-complete. (As pointed out in [8], a similar result was proved byJakob [15], but where the integer k is part of the input.) Combining thesetwo results yields the following dichotomy theorem:

Theorem 1 (Cereceda, van den Heuvel and Johnson [11]; Bonsma and Cere-ceda [5]). The reconfiguration problem for k-colourings is solvable in polyno-mial time for k ≤ 3 and is PSPACE-complete for k ≥ 4.

In this paper, we study the complexity of the reconfiguration problem forcircular colourings. Given a graph G and positive integers p and q with p/q ≥2, a (circular) (p, q)-colouring of G is a mapping f : V (G)→ {0, . . . , p− 1}such that

if uv ∈ E(G), then q ≤ |f(u)− f(v)| ≤ p− q. (1)

Clearly, a (p, 1)-colouring is nothing more than a p-colouring and so (p, q)-colourings generalize classical graph colourings. Circular colourings wereintroduced by Vince [17], and have been studied extensively; see the survey ofZhu [21]. Analogous to that of classical graph colourings, the reconfigurationproblem for circular colourings asks, given (p, q)-colourings f and g of G,whether it is possible to reconfigure f into g by recolouring one vertex at atime while maintaining (1) throughout.

Classical graph colourings and circular colourings are both special casesof graph homomorphisms. Recall, a homomorphism from a graph G to agraph H (also called an H-colouring of G) is a mapping f : V (G) → V (H)

2Throughout the paper, the term k-colouring will refer to a proper k-colouring.

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such that f(u)f(v) ∈ E(H) whenever uv ∈ E(G). In this language, a k-colouring is simply a homomorphism to a complete graph on k vertices. A(p, q)-colouring of G is equivalent to a homomorphism from G to the graphGp,q which has vertex set {0, . . . , p−1} and edge set {ij : q ≤ |i−j| ≤ p−q}.The graph Gp,q is called a circular clique.

Given two homomorphisms f, g : G → H, we say f reconfigures to g ifthere a sequence (f = f0), f1, f2, . . . , (fn = g) of homomorphisms from G toH such that fi and fi+1 differ on only one vertex. The sequence is referredto as a reconfiguration sequence. Clearly the existence of a reconfigurationsequence from f to g can be determined independently for each componentof G, so we may assume that G is connected. We define the general homo-morphism reconfiguration problem as follows. Let H be a fixed graph.

H-Recolouring

Instance: A connected graph G, and two homomorphisms f, g : G→ H.

Question: Does f reconfigure to g?

When H = Kk or H = Gp,q we will call the problem k-Recolouring and(p, q)-Recolouring respectively. Thus, Theorem 1 is a dichotomy theoremfor k-Recolouring. Our main result is a dichotomy theorem for (p, q)-Recolouring:

Theorem 2. Let p, q be fixed positive integers with p/q ≥ 2. Then the (p, q)-Recolouring problem is solvable in polynomial time for 2 ≤ p/q < 4 andis PSPACE-complete for p/q ≥ 4.

The complexity of H-Recolouring is only known for a handful of fam-ilies of targets. Theorem 1 is a dichotomy theorem for the family of com-plete graphs and Theorem 2 is a dichotomy theorem for the family of circularcliques. Recently, Wrochna [19] (see also [20]) proved that H-Recolouringis polynomial whenever H does not contain a 4-cycle. In contrast, one canobserve that Gp,q contains 4-cycles whenever p > 2q + 1 and so the polyno-mial side of Theorem 2 does not follow directly from the result of Wrochna.In a follow-up paper [7], we determine the complexity of H-Recolouringfor several additional classes of graphs including, for example, odd wheels.

The rest of the paper is outlined as follows. In Section 2, we provide an ex-plicit polynomial-time algorithm for deciding the (p, q)-Recolouring prob-lem when 2 ≤ p/q < 4. In addition, we provide an argument of Wrochna [18]which uses the algorithm of [11] (on which ours is based) to show that graphs

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with no cycle of length 0 mod 3 are 3-colourable; this result was originallyproved by Chudnovsky, Plumettaz, Scott and Seymour in [12]. We thengeneralize Wrochna’s argument to show that graphs with chromatic numbergreater than k must contain at least (k−1)!

2distinct cycles of length 0 mod k,

and conjecture a stronger bound. In Section 3, we show that, when p/q ≥ 4,the reconfiguration problem for bp/qc-colourings can be reduced to the re-configuration problem for (p, q)-colourings, thereby completing the proof ofTheorem 2 (via Theorem 1). We close the paper in Section 4 by stating anopen problem regarding reconfiguring colourings of the Erdos-Renyi randomgraph G(n, p).

2. The Polynomial Cases: 2 ≤ p/q < 4

We now extend the ideas of [11] to study the complexity of (p, q)-Recolouringfor 2 ≤ p/q < 4. Given two colourings f and g of a graph G, the algorithmin [11] consists of two phases. The first phase tests whether the so-called“fixed vertices” (vertices that cannot be recoloured under any sequence ofrecolourings) are assigned to the same colours under f and g. If not, thenwe know that f does not reconfigure to g and the algorithm terminates.

If the algorithm does not terminate at this point, then it enters the secondphase. In this phase, the algorithm obtains two edge labelleings of G fromthe vertex colourings f and g. The key property of these labellings is that,given that the algorithm did not terminate after the first phase, one canshow that f reconfigures to g if and only if every cycle of G has the same“weight” under both edge labellings. After a polynomial number of steps,the algorithm either produces a reconfiguration sequence or discovers a cyclewith different weights under f and g. We follow a similar approach, but webegin by examining the edge relabelling problem which turns out to dependonly on cycle weights.

2.1. Edge relabellings

Let f : G→ Gp,q be a (p, q)-colouring of a graph G. Let−→G be an oriented

graph obtained by arbitrarily orienting each edge of G. The edge-labellinginduced by f is defined as ϕf : E(G) → Zp by ϕf (e) = f(v) − f(u) wheree = −→uv and the subtraction is done in Zp (i.e. reduced modulo p). Let Cbe a cycle of G and arbitrarily choose a direction of traversal. Let E+(C)

be those edges of C whose orientation in−→G agrees with the direction of

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traversal (forward arcs) and E−(C) be those edges of C whose orientation in−→G is reversed to the direction of traversal (backward arcs). Define

ϕf (C) =∑

e∈E+(C)

ϕf (e) +∑

e∈E−(C)

p− ϕf (e). (2)

Note that the summation above is in Z. The following properties of ϕf areimmediate from the fact that f is a (p, q)-colouring and the sum ϕf (C) is atelescoping series in the values of f on C.

P1 q ≤ ϕf (e) ≤ p− q for all edges e.

P2 ϕf (C) ≡ 0 (mod p) for all cycles C.

Given a graph G, an edge-labelling ψ : E(G) → Zp is a (p, q)-labelling if itsatisfies properties P1 and P2 above. (Clearly ψ is a group labelling of E(G)with the group Zp. The discussions here are easily generalized to arbitraryabelian groups.) Similar to the weighting function for cycles in (2), we definea weighting for paths: ϕ(P ) is the sum of ϕ(e) and p−ϕ(e) for forward andbackward arcs in P respectively.

We now introduce the reconfiguration process for edge-labellings. Let G

be a graph and−→G an orientation of G. For ∅ 6= X ⊂ V (G), the edge cut

∂(X) is the set of edges with one end in X and the other in X. Let ∂+(X)

(resp. ∂−(X)) be those edges oriented from X to X (resp. X to X) in−→G .

Given a (p, q)-labelling ϕ : E(G) → Zp, let α ∈ Zp where ϕ(e) ≥ q + α fore ∈ ∂+(X) and ϕ(e) ≤ p− q−α for e ∈ ∂−(X). The labelling ϕ′ is obtainedfrom ϕ by relabelling on ∂(X) by α is defined by:

ϕ′(e) =

{ϕ(e)− α if e ∈ ∂+(X),ϕ(e) + α if e ∈ ∂−(X)

This process is called an edge cut relabelling. Clearly if ϕ is a (p, q)-labelling,then ϕ′ is as well.

Throughout this section the weighting of cycles, paths, and the orienta-tion of edge cuts all require that an orientation has been applied to G. Wewill explicitly assume this throughout the remainder of this section.

Theorem 3. Let G be a connected graph and ϕ, ψ be two (p, q)-edge-labellingsof G. Then ϕ reconfigures to ψ through a sequence of edge cut relabellings ifand only if ϕ(C) = ψ(C) for all cycles C in G.

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Proof. Relabelling on an edge cut does not change the weight of any cycles.Thus, if ϕ reconfigures to ψ, then ϕ(C) = ψ(C) for all cycles C is a necessarycondition. To prove sufficiency, assume ϕ(C) = ψ(C) for each cycle C. Wemay assume ϕ 6= ψ. Let −→wu be an arc such that ϕ(wu) < ψ(wu). The casewhere ϕ(wu) > ψ(wu) is analogous. Let T be a spanning tree of G rootedat u. For each v ∈ V (G) define wt(v, ϕ) = ϕ(P ) where P is the (u, v) pathin T . Similarly define wt(v, ψ).

Let X = {v : wt(v, ϕ) ≤ wt(v, ψ)}. Thus, X = {v : wt(v, ϕ) > wt(v, ψ)}.Clearly u ∈ X and it is easy to see w ∈ X. Suppose to the contrary w ∈ X.Then uw is not an edge of T and thus the uw path in T plus the edge wuis a cycle C. Moreover this cycle must have ϕ(C) < ψ(C) contrary to ourassumption. Hence ∅ 6= X ⊂ V (G) and ∂(X) is a well defined edge cut.

We claim that ∂+(X) must only contain arcs where ϕ(e) > ψ(e) andarcs in ∂−(X) must have ϕ(e) < ψ(e). Suppose to the contrary there is anedge e = −→xy, x ∈ X, y ∈ X such that ϕ(e) ≤ ψ(e). (The proof for arcsin ∂−(X) analogous.) By construction there is a (u, x)-path P1 such thatϕ(P1) ≤ ψ(P1). Thus P1+y is an (u, y)-path such that ϕ(P1+y) ≤ ψ(P1+y).On the other hand, since y ∈ X, the path (u, y) path in T , say P2, hasthe property that ϕ(P2) > ψ(P2). Hence, reverse of the path is a (y, u)-path, PR

2 such that ϕ(PR2 ) < ψ(PR

2 ). The concatenation of P1 + y andPR2 is a closed (u, u)-walk with less weight under ϕ than ψ. In particular,

it must contain a cycle C such that ϕ(C) 6= ψ(C), a contradiction. Letα = mine∈∂(X){|ϕ(e) − ψ(e)|}. Relabel ∂(X) by α to obtain ϕ′. Now ϕ′

agrees with ψ on more edges than ϕ. The result follows.

Corollary 4. Let G be a graph together with two (p, q)-edge-labellings ϕ andψ. Either ϕ reconfigures to ψ through a sequence of edge cut relabellings orthere is a cycle C such that ϕ(C) 6= ψ(C). Determining which of the twocases (exclusively) holds can be determined in polynomial time.

2.2. Vertex recolouring

We now return to recolouring vertices. We have already observed thatgiven a (p, q)-colouring f of a graph, we can construct a (p, q)-edge labellingϕf . Conversely given ϕ, a (p, q)-edge labelling, using the ideas from the proofof Theorem 3, we can generate a weight for each vertex. Reducing theseweights modulo p yields a (p, q)-colouring f where ϕ = ϕf . The followingproposition is immediate.

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Proposition 5. Given a connected graph G, an edge-labelling ψ : E(G)→ Zpis a (p, q)-labelling if and only if ψ = ϕf for some f : G → Gp,q. Moreover,if ϕf = ϕg for f, g : G → Gp,q, then f(v) = g(v) + k for all vertices v andsome constant k.

To connect our work on edge cut relabellings back to vertex recolourings,we must realize edge cut relabelling through vertex recolourings where werecolour one vertex at a time. First, we observe that relabelling ϕ on ∂(X)by α corresponds to the following vertex recolouring where we recolour anentire set of vertices (simultaneously). Let f ′ be the colouring obtained fromf by:

f ′(v) =

{f(v) + α if v ∈ X,f(v) otherwise.

Then ϕf ′ is the edge-labelling obtained from ϕf by relabelling on ∂(X) by α.We call this process recolouring the vertex set X by α. Provided ϕ(e) ≥ q+αfor all e ∈ ∂+(X) and ϕ(e) ≤ p− q − α for all e ∈ ∂−(X), the new colouringf ′ is a proper (p, q)-colouring. The following corollary is immediate fromProposition 5 and Theorem 3.

Corollary 6. Let f and g be two (p, q)-colourings of a graph G. Then thereis a sequence of colourings f, f1, f2, . . . , fn each obtained from its predecessorby a vertex set recolouring where fn(v) = g(v) +k for all vertices v and someconstant k if and only if ϕf (C) = ϕg(C) for all cycles C in G.

To complete our algorithm, we need to determine when recolouring avertex set X by α can be realized by a sequence of single vertex recolourings.We begin proving for p/q < 4, recolouring by α can always be acheivedthrough recolouring X by 1 (α times). Note this not the case when p/q ≥ 4.For example, consider a (4, 1)-colouring of C4 where the vertices are coloured0, 1, 2, 3 (in their cyclic order). The vertex with colour 0 can be recolouredto 2. However, the recolouring cannot be done by increasing the colour by 1(twice).

Given a pair a, b ∈ {0, . . . , p−1}, the interval [a, b] is defined to be the set{a, a+ 1, . . . , b− 1, b} (where addition is modulo p). An important propertyof Gp,q used in the proof below is that, when 2 ≤ p/q < 4, the commonneighbours of any two vertices is an interval. This fact, and its proof, appearin [6].

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Proposition 7. Let f, g be two (p, q)-colourings of a graph G where g isobtained from f by recolouring the vertex set X by α. Then there is a se-quence of colourings (f = f0), f1, f2, . . . , (fα = g) where each colouring isobtained from its predecessor by recolouring X by 1 (for the entire sequence)or each is obtained from its predecessor by recolouring X by −1 (for the entiresequence).

Proof. Let v ∈ X and vu be an edge. If u ∈ X, then (f(v) ± 1, f(u) ± 1)is an edge of Gp,q. Consequently for each vertex in X we can increment ordecrement the colour by 1 and maintain a proper (p, q)-colouring on X.

On the other hand, consider a neighbour u of v such that u 6∈ X. Notef(u)f(v) and f(u)(f(v) + α) are both edges of Gp,q. If α = p/2, then f(u)is distance at least q from both f(v) and f(v) + p/2 which implies p/2 ≥ 2q,contrary to our assumption. If α < p/2, then f(u) must belong to theinterval [f(v) + α, f(v)]. (The interval [f(v), f(v) + α] has length strictlyless than 2q and f(u) is adjacent to both endpoints.) Thus f(u) is adjacentto every vertex of [f(v), f(v) + α] and in particular f(v) + 1. Note thisargument depends only on α < p/2. Hence the colouring f1 obtained from fby increasing the colour of each vertex in X by 1 is a (p, q)-colouring of G.

A similar argument shows for α > p/2, the colouring f1 obtained from fby decreasing the colour of each vertex in X by 1 is a (p, q)-colouring. Theresult follows.

Given G, an orientation−→G , and an (p, q)-edge-labelling, say ϕ, let Dϕ be

the subdigraph of−→G induced by edges with label q or p− q. Without loss of

generality we may assumed Dϕ is oriented so that all edges have label q. (Tosimplify notation we shall write Df instead of Dϕf

when the edge-labellingϕf comes from some vertex colouring f .)

Lemma 8. Let f be a (p, q)-colouring of a graph G. Let f ′ be the (p, q)-colouring obtained from f by recolouring some vertex set X by 1. Then f ′

can obtain by a sequence of recolourings of single vertices if and only if Df [X]contains no directed cycles.

Proof. Suppose Df [X] contains no directed cycles. We topologically sort thevertices of X to obtain an ordering x1, x2, . . . , xt where e = −−→xixj ∈ E(Df [X])implies i < j. For each i = t, t− 1, . . . , 1 increase f(xi) by 1.

On the other hand, suppose Df [X] contains a directed cycle. If one couldsequentially increase the colour of each vertex in X by 1, then there would

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be a first vertex ci in the cycle to have its colour changed. However, ci hasan out neighbour ci+1 (in Df [X]) meaning f(ci+1)− f(ci) = q. The resultingcolouring gives the edge cici+1 a weight of q − 1, i.e. the resulting colouringis a not a proper (p, q)-colouring.

2.3. Polynomial Time Algorithm

Let f and g be two (p, q)-colourings of a graph G. Following the ideasof [11] and using the results above, we identify (the only) three obstructionspreventing the recolouring of f to g: cycle weights, fixed vertices, and weightsof paths between fixed vertices.

First, Corollary 6 shows if some cycle has different weight under ϕf andϕg, then f does not recolour to g.

Secondly by Lemma 8, the relabelling of an edge cut can be realized byrecolouring vertices one at a time provided the vertices do not belong toa directed cycle in Df . Vertices in a directed cycle can never have theircolour changed under any recolouring and are called fixed in [11]. Clearlyif f recolours to g, then the set of fixed vertices under each colouring mustbe equal and f(v) = g(v) for each fixed vertex v. A vertex belongs toa directed cycle in Df if and only if it belongs to a non-trivial stronglyconnected component of Df . The strongly connected components can befound in linear time; hence, we can find the fixed vertices and verify equalityof f and g on the fixed vertices in linear time. (See, e.g., [1].)

Finally, observe in the case we wish to increase the vertex set X by 1 andDf [X] contains a directed cycle, we can achieve the same change to the edgelabelling on ∂(X) by decreasing X by 1 provided Df [X] does not contain adirected cycle. In the event both subgraphs contain a directed cycle, we havevertices v ∈ X and u ∈ X which are fixed and a (u, v)-path P which musthave different weights under ϕf and ϕg. Recolouring an internal vertex ofthis path will not change the weight, and thus f cannot recolour to g. Ourfinal necessary condition for f to recolour to g is that all paths P whose endpoints are fixed must have ϕf (P ) = ϕg(P ).

In [11], fixed vertices are found by successively deleting sources and sinksfrom Df . Thus vertices belonging to a directed cycle will be fixed, but alsovertices on a direct path between two directed cycles are also fixed. In ourwork, we identify fixed vertices of the former type using strongly connectedcomponents of Df and vertices of the latter type by the weight of pathsbetween strongly connected components (our third obstruction). Directedpaths in Df have the smallest possible weight over all edge labellings.

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We now present our algorithm.

Recolouring Algorithm:Let 2 ≤ p/q < 4.

Input: A graph G and two (p, q)-colourings f, g.

Output: A recolouring sequence from f to g or an obstruction to recolour-ing.

1. Find the strongly connected components of Df . For each vertex vbelonging to a nontrivial strongly connected component, verify f(v) =g(v). If for some such v, f(v) 6= g(v), then answer NO, return thedirected cycle to which v belongs, and STOP.

2. Construct a spanning tree T rooted at a vertex u. Partition V (G) intothree sets: Xu, Xe, X` consisting of those vertices v whose (u, v)-pathin T has weight larger, equal, smaller under ϕf versus ϕg respectively.Repeat until Xu ∪X` = ∅.

(a) If Xu 6= ∅, then let X = Xe ∪ X` and X = Xu. (If Xu = ∅and X` 6= ∅ apply an analogous process reversing their roles.)For each e ∈ ∂+(X) (resp. ∂−(X)), verify ϕf (e) > ϕg(e) (resp.ϕf (e) < ϕg(e)). If some edge fails this test, then answer NO,return a cycle with different weights under ϕf and ϕg, and STOP.

(b) Let α = mine∈∂(X) |ϕf (e) − ϕg(e)|. If Df [X] and Df [X] bothcontain nontrivial strongly connected components, then answerNO, return a path P between two fixed vertices with ϕf (P ) 6=ϕg(P ), and STOP. Otherwise, recolour the vertices of X (or X)by 1 using a sequence of single vertex recolourings. Repeat thisrecolouring by X (by 1) α times.

(c) Update the path weights in T and the sets Xu, Xe, X`.

3. At this point ϕf = ϕg. If f(u) = g(u), then answer YES, return thesequence of recolourings, and STOP. Otherwise, we know G containsno directed cycles and we can topologically sort V (G) : v1, v2, . . . vn.Increase the colour of all the vertices by 1 in the order vn to v1. Repeatthis until f(u) = g(u). Answer YES, return the recolouring sequence,and STOP.

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The following theorem is immediate from the algorithm.

Theorem 9. Let 2 ≤ p/q < 4. Suppose f and g are (p, q)-colourings of a

graph G,−→G is an orientation of G and ϕf , ϕg are the edge labellings of G

obtained from f and g. Then f recolours to g if and only if:

1. the same set of vertices are fixed under f and g and f(v) = g(v) for allfixed vertices;

2. for each cycle C of G, ϕf (C) = ϕg(C); and

3. for each path P whose end points are fixed, ϕf (P ) = ϕg(P ).

Moreover, one can find a recolouring sequence from f to g or an obstructionof the above type in polynomial time.

At each iteration of the algorithm we recolour O(|V (G)|) vertices andincrease Xe by at least one vertex. Once a vertex is in Xe it never leaves.Thus we do O(|V (G)|) recolouring steps.

Corollary 10. Let 2 ≤ p/q < 4. Suppose f and g are (p, q)-colourings ofa graph G. If f recolours to g, then there is a reconfiguration sequence oflength O(|V (G)|2) which certifies this.

2.4. Cycles in Graphs of Large Chromatic Number

To close this section, we present an elegant argument of Wrochna [18]which uses the recolouring algorithm of [11] to prove the following theo-rem, which was originally proved by Chudnovsky, Plumettaz, Scott and Sey-mour [12].

Theorem 11 (Chudnovsky, Plumettaz, Scott and Seymour [12]). If G con-tains no cycle of length 0 mod 3, then G is 3-colourable.

Proof (Wrochna [18]). Suppose that the statement is false and let G be acounterexample for which |E(G)| is minimum. Define G′ := G − e wheree = uv is an edge of G. By hypothesis, G′ admits a 3-colouring, say f :V (G′)→ {0, 1, 2}. Clearly f must map u and v to the same colour; otherwisewe are done. So, without loss of generality, we have f(u) = f(v) = 0.

Now, let g : V (G) → {0, 1, 2} be defined by g(v) ≡ f(v) + 1 mod 3 forall v ∈ V (G). It is clear that ϕf (C) = ϕg(C) for all cycles C in G′. Also,since G′ has no cycles of length 0 mod 3, there are no directed cycles in Df

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or Dg, and therefore no fixed vertices either. Thus, by Theorem 9, f can bereconfigured to g. Let h be the first colouring of the reconfiguration sequencesuch that h(u) 6= 0 or h(v) 6= 0. Since the reconfiguration sequence coloursonly one vertex in each step, it is clear that h is a 3-colouring of G, whichcompletes the proof.

A closer look at the proof shows that it actually yields something slightlystronger: If G contains an edge e such that G − e has no cycle of length0 mod 3, then χ(G) ≤ 3. Using a similar strategy, we can generalize this tok-colourings.

Theorem 12. If G contains an edge e such that G− e contains fewer than(k−1)!

2cycles of length 0 mod k, then χ(G) ≤ k.

Proof. We proceed by induction on |E(G)|, where the case |E(G)| = 0 istrivial. Let e = uv be an edge of G such that G′ := G − e contains fewerthan (k−1)!

2cycles of length 0 mod k. Then, by the inductive hypothesis, G′

admits a k-colouring f : V (G′)→ {0, . . . , k− 1}. Clearly f must map u andv to the same colour so, without loss of generality, f(u) = f(v) = 0.

Let π be a permutation of {0, . . . , k − 1} with π(0) = 0. For each i andeach edge xy of G′ with f(x) = π(i) and f(y) = π(i + 1) (where additionis modulo k), orient the edge from x to y. The resulting oriented, spanningsubgraph of G′ is called Fπ. If Fπ is acyclic, we can topologically sort thevertices as above, successively increase the colour of each vertex from sayπ(j) to π(j+ 1), and eventually increase the colour of u or v to produce a k-colouring of G′ where u and v receive different colours. Thus we are finishedif we can find a permutation π such that Fπ is acyclic.

A directed cycle in Fπ must have vertex colours π(i), π(i + 1), π(i +2), . . . , π(i − 1), π(i), and thus have length 0 mod k. On the other hand,let C be a cycle of length 0 mod k in G′. Suppose the colours of C underf have the following property: there is a vertex of colour 0, starting at thatvertex and traversing the cycle we see all k colours in the first k vertices, andeach successive block of k vertices is coloured with the same colours in thesame order. In other words, the first k vertices define a permutation π suchthat π(0) = 0, and C is a directed cycle in Fπ. Observe if we traverse C isthe opposite direction we obtain a second permutation π′ where C is directedin Fπ′ . Moreover if C is a directed cycle in Fσ with σ(0) = 0, then σ must beπ or π′. Since G′ has fewer than (k − 1)!/2 cycles of length 0 mod k, theremust be some permutation π such that Fπ is acyclic. The result follows.

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Corollary 13. If χ(G) > k, then G contains at least (k−1)!2

cycles of length0 mod k.

The complete graph of order k+1 has precisely (k+1)(k−1)!2

cycles of length0 mod k, and so Corollary 13 is within a factor k + 1 of being tight. Wewonder whether Kk+1 contains the fewest cycles of length 0 mod k among allnon-k-colourable graphs.

Conjecture 14. If χ(G) > k, then G contains at least (k+1)(k−1)!2

cycles oflength 0 mod k.

The case k = 3 may be particularly instructive: Is it true that everygraph G with χ(G) ≥ 4 contains at least 4 cycles of length 0 mod 3? IfConjecture 14 turns out to be false, then it would still be interesting todetermine the minimum number of cycles of length 0 mod k in a graph ofchromatic number greater than k.

3. The PSPACE-complete Cases: p/q ≥ 4

3.1. Overview

As in the previous section, all addition and subtraction involving elementsof {0, . . . , p − 1} is viewed modulo p. We say that i, j ∈ {0, . . . , p − 1} arecompatible if they are adjacent in Gp,q. For notational convenience, let usdefine k := bp/qc and r := p− kq.

Given an instance (G, f, g) of the k-Recolouring problem, we will con-struct an instance (G′, α, β) of the (p, q)-Recolouring problem in polyno-mial time such that:

• |V (G′)| = O(|V (G)|+ |E(G)|) and

• f reconfigures to g if and only if α reconfigures to β.

It will follow from Theorem 1 that the reconfiguration problem for (p, q)-colourings is PSPACE-complete, thereby completing the proof of Theo-rem 2.

We give a brief outline of the ideas behind the construction before movinginto the finer details. The first step is to divide the set {0, . . . , p− 1} into kintervals which we will, in some sense, treat as k separate colours. Define

S0 := [0, q + r − 1], and

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Si := [iq + r, (i+ 1)q + r − 1] for 1 ≤ i ≤ k − 1.

It is clear that Si and Sj are disjoint for i 6= j and that⋃k−1i=0 Si = {0, . . . , p−

1}. Let γ : {0, . . . , k − 1} → {0, . . . , p − 1} be the function which maps ito the left endpoint of Si for 0 ≤ i ≤ k − 1. For i 6= j, it is easily observedthat the left endpoint of Si is adjacent to the left endpoint of Sj in Gp,q.Therefore, given any k-colouring f : V (G)→ {0, . . . , k−1}, the compositionγf is a (p, q)-colouring.

This observation guides part of our construction. The graph G′ will con-tain G as an induced subgraph and the (p, q)-colourings α and β will bedefined so that their restrictions to G will be equal to γf and γg, respec-tively. Now, if it is possible to reconfigure α to β in such a way that noneof the intermediate colourings map a pair of adjacent vertices of G to thesame set Si, then we immediately obtain a reconfiguration sequence takingf to g. That is, to obtain a sequence of k-colourings taking f to g, onecould simply compose each (p, q)-colouring of the sequence with the functionϕ : {0, . . . , p − 1} → {0, . . . , k − 1} which maps every vertex of Si to i for0 ≤ i ≤ k− 1. Notice that, for i 6= 0, the set Si is an independent set in Gp,q

and therefore no adjacent pair is ever mapped to Si. Thus, all that we needto worry about is that some of the intermediate colourings may map twoadjacent vertices of G to S0. To remedy this, we will add some structures(or “gadgets”) to G′ which will forbid adjacent vertices of G from mappingto S0.

To this end, we start by adding a copy of Gp,q disjoint from G with vertexset {y0, . . . , yp−1}, where yiyj is an edge whenever q ≤ |i − j| ≤ p − q. Weextend the (p, q)-colourings α and β to {y0, . . . , yp−1} by setting α(yi) =β(yi) = i for all i. It is clear that each vertex yi is fixed in α and β and,moreover, in any (p, q)-colouring which reconfigures to α or β.

Now, for each edge uv of G, we will add two paths Puv and Pvu fromu to v in G′ which are internally disjoint from V (G) ∪ {y0, . . . , yp−1} andone another, as well as from every other such path. We want these pathsto somehow restrict the colour pairs which can appear on u and v during areconfiguration process. We achieve this by joining the internal vertices of thepaths to specific subsets of {y0, . . . , yp−1}, which has the effect of restrictingtheir colours to lie within specified lists. For example, if we want a vertexx to always obtain a colour in the set [3q, p − 1], then it suffices to join xto the vertices {yq−1, . . . , y2q}. As we will show, given that the paths are ofthe correct length, there exists an assignment of lists to the internal vertices

14

which forbids u and v from mapping to S0 and can be enforced by joiningthe internal vertices to subsets of {y0, . . . , yp−1}. Also, once we describe ourconstruction in detail, it will be clear that the length of the forbidding pathsdepends only on p and q, and so |V (G′)| = O (|V (G)|+ |E(G)|).

The difficulty now comes in proving that if f reconfigures to g, then αreconfigures to β. Specifically, given a reconfiguration sequence (hi)

si=1 taking

f to g, we need that the lists assigned to the internal vertices of the forbiddingpaths are flexible enough that we can use (hi)

si=1 to obtain a reconfiguration

sequence taking α to β. This will be proved using a moderate amount of caseanalysis at the end of the section. Before moving on, we remark that thegeneral strategy of using some sort of forbidding paths in which the internalvertices are confined to lists was also used by Bonsma and Cereceda [5] (ina somewhat different manner) to prove that the reconfiguration problem fork-colourings is PSPACE-complete for k ≥ 4.

3.2. Defining the Forbidding Paths

Let uv be an edge of G. We will now describe our construction of theforbidding path Puv = uxuv0 x

uv1 . . . xuvt v from u to v, including the definition of

the lists assigned to the internal vertices xuv0 , . . . , xuvt . As mentioned before,

the construction of G′ also includes a path Pvu from v to u which is definedsimilarly.

u

xuv0

xvut

v

xuvt

xvu0

. . .

. . .

Figure 1: Forbidding paths attached to each edge uv

First, define t to be the smallest positive integer such that (t + 1)q ≡r mod p. Note that (p− k)q ≡ r mod p and so such a t exists. Now, the listsfor the internal vertices of Puv are intervals of Gp,q and are defined as follows:

L (xuv0 ) = L (xuvt ) := [p− 1, 2q − 1],

L (xuvi ) := [iq, (i+ 2)q − 1] for 1 ≤ i ≤ t.

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As stated in the outline, in order to enforce these lists we add a copy of Gp,q

on vertex set {y0, . . . , yp−1} and join the vertices of Puv \ {u, v} to the ap-propriate vertices of this set. Specifically, we join the vertices xuv0 and xuvt to{yi : i ∈ [3q − 1, p− q − 1]} and join xuvi to {yi : i ∈ [(i+ 3)q − 1, (i− 1)q]}for 1 ≤ i ≤ t− 1.

Figure 2 contains a useful tabular representation of the lists, and a specificcase (p, q) = (18, 4) is depicted in Figure 3. In both of these figures, thecolours are laid out so that below colour c in the table is the colour c + q.Consequently, when colouring the path xuv0 , x

uv1 , . . . , x

uvt the colours used (one

per row) must be directly below or to the right of the preceding vertex, andany such sequence of colours from the table give a good colouring of the path.Finally the vertical bar in the table provides the following information: ifvertex u receives colour 0, then only colours to the right of the vertical barcan be used to colour the path Puv.

Vertex Listsxuv0 p− 1 0 . . . q − 1 q . . . 2q − 1xuv1 q . . . 2q − 1 2q . . . 3q − 1xuv2 2q . . . 3q − 1 3q . . . 4q − 1

......

...xuvt−1 r − 2q . . . r − q − 1 r − q . . . r − 1xuvt p− 1 . . . r . . . q + r − 1 . . . 2q − 1

Figure 2: Assignment of lists to the internal vertices of the forbidding path.

Vertex Listsxuv0 17 0 1 2 3 4 5 6 7xuv1 4 5 6 7 8 9 10 11xuv2 8 9 10 11 12 13 14 15xuv3 12 13 14 15 16 17 0 1xuv4 17 0 1 2 3 4 5 6 7

Figure 3: Assignment of lists to the internal vertices of the forbidding path for the (p, q) =(18, 4) case.

Before moving on, let us check that the paths Puv and Pvu actually do

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the job that they are meant to do; namely, that they forbid u and v fromboth being mapped to S0.

Proposition 15. Let uv be an edge of G. If ψ is a (p, q)-colouring of Puv∪Pvusuch that the internal vertices of each path are coloured from their lists, thenψ(u) and ψ(v) cannot both be contained in S0.

Proof. Suppose to the contrary that ψ(u), ψ(v) ∈ S0. In Gp,q, the only edgescontained in S0 are between vertices in [0, r− 1] and vertices in [q, q+ r− 1].So, without loss of generality, we can assume that

ψ(u) ∈ [0, r − 1] and ψ(v) ∈ [q, q + r − 1]. (3)

However, the fact that ψ(u) ∈ [0, r − 1] implies that ψ (xuv0 ) ∈ [q, 2q − 1](Note that r − 1 < q − 1 and thus (r − 1, p− 1) is not an edge of Gp,q). Byour observations above, we see that only colours to the right of the verticalline in Figure 2 can be used to colour Puv. In particular ψ (xuvt ) ∈ [r, 2q− 1].This implies that ψ(v) 6∈ [q, q + r − 1], which contradicts (3) and completesthe proof.

3.3. Extending α and β to the Forbidding Paths

As we have already mentioned, α and β are defined in such a way thattheir restrictions to G are equal to γf and γg, respectively, and α(yi) =β(yi) = i for all i. Next, we will describe the way in which we extend α andβ to the vertices of the forbidding paths.

Recall that the vertices of G each receive a colour from {0, q + r, 2q +r, . . . , (k − 1)q + r} under α. Given an edge uv of G, the internal verticesof Puv are coloured as follows. If α(u) 6= 0, then we set α(xuvi ) = iq fori = 0, . . . , t − 1. The list of colours for xuvt is sufficiently long to contain aneighbour of each of {0, q + r, 2q + r, . . . , (k − 1)q + r}, and each colour onthis list is compatible with the colour of xuvt−1, namely r − 2q. Thus there isa colour for xuvt compatible with both the colour of xuvt−1 and the colour of v.If α(u) = 0, then we set α(xuvi ) = (i + 1)q for i = 0, . . . , t − 1. The coloursbelow and to the right of the colour of xuvt−1, namely r − q, in Figure 2 are[r, 2q − 1]. Again these allowed colours for xuvt contain a neighbour of anycolour that could be assigned to v.

We similarly define β using g. By Proposition 15, given any sequenceof (p, q)-colourings which reconfigures α to β, we can compose each of thesecolourings with ϕ to obtain a sequence of k-colourings taking f to g. Thisproves the following proposition.

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Proposition 16. Suppose (G′, α, β) is an instance of (p, q)-Recolouringobtained from (G, f, g), an instance of k-Recolouring, as described above.If α reconfigures to β, then f reconfigures to g.

3.4. Recolouring G′

To complete the reduction we need to prove that if f reconfigures to g,then α reconfigures to β. Let (hi)

si=1 be any reconfiguration sequence taking

f to g. We show that there is a sequence (ηi)s+1i=1 of (p, q)-colourings of G′

such that

• η1 = α and ηs+1 = β,

• the restriction of ηi to G is γhi for 1 ≤ i ≤ s, and

• ηi reconfigures to ηi+1 for 1 ≤ i ≤ s.

Clearly this will prove that α reconfigures to β and complete the proof ofTheorem 2. Also, notice that the colouring ηs is the same as β except on thevertices of the forbidding paths. In order to reconfigure ηs to β, we start bychanging the colour of xuv0 to β (xuv0 ), which is guaranteed to be compatiblewith β(u) and to the left of the colour assigned to xuv1 by ηs. We then shiftthe colour of each vertex xuvi for 1 ≤ i ≤ t− 1 to it smallest available colour(which will be β (xuvi )). Finally, we are free to change the colour of xuvt toβ(xuvt ). Therefore, it suffices to construct the sequence (ηi)

si=1 as above.

Before tackling the general case we illustrate our method with the examplein Figure 4, in which (p, q) = (18, 4). In this case, the colourings ηi map tothe colours {0, 6, 10, 14}. Let 1 ≤ j ≤ s− 1 be fixed and suppose that (ηi)

ji=1

have been constructed to satisfy the conditions above; our goal is to constructthe colouring ηj+1 and a reconfiguration sequence taking ηj to ηj+1.

By definition, hj and hj+1 differ on at most one vertex, say u ∈ V (G).Suppose, for example that hj(u) = 0 and hj+1(u) = 1; this requires us changethe colour of u from 0 (its current colour under ηj) to 6 (its desired colourunder ηj+1) without changing the colour of any other vertex of G. We remarkthat this is the most involved case, and corresponds to Case (c) in Lemma 17below. The other cases use similar ideas but are more straightforward. Foreach vertex v ∈ V (G) to which u is adjacent, v must receive a colour from{2, 3} under hj to facilitate the recolouring of u, which implies that ηj(v) ∈{10, 14}. Each step below is applied for each neighbour v of u, one by one,before moving on to the subsequent step. The recoloured vertices at each

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u

xuv0

xvu4

xuv1

xvu3

xuv2

xvu2

xuv3

xvu1

xuv4

xvu0

v

0

48 12 0

6

10612 8 4

0

ηj

0

711 15 1

5

10712 8 4

0

(i)

3

711 15 1

5

10712 8 4

0

(ii)

3

1711 15 1

5

101712 8 4

0

(iii)

6

1711 15 1

5

101712 8 4

0

ηj+1

(iv)

Figure 4: Recolouring u from 0 to 6 in the (18, 4) case.

step are shown as white vertices in Figure 4 with the new colours indicatedin boldface.

(i) First, recolour xvu4 to 7. Then, recolour Puv by recolouring xuv4 to 5,xuv3 to 1, xuv2 to 15, xuv1 to 11 and xuv0 to 7. At this point all vertices of{xuv0 , xvut : v ∈ NG(u)} have colour 7.

(ii) Now, change the colour of u to 3.

(iii) Next, change the colour of xuv0 to 17. Then, recolour Pvu by recolouringxvu0 to 0, xvu1 to 4, xvu2 to 8, xvu3 to 12 and xvu4 to 17. At this point allvertices of {xuv0 , xvut : v ∈ NG(u)} have colour 17.

(iv) Finally, recolour u to 6 to obtain the colouring ηj+1.

We now describe the recolouring technique in general.

Lemma 17. Let h and h′ be k-colourings of G which differ on a uniquevertex u and let η be a (p, q)-colouring of G′ such that

• η(yi) = i for all i, and

• the restriction of η to G is γh.

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Then there exists a (p, q)-colouring η′ of G′ such that the restriction of η′ toG is γh′ and η reconfigures to η′.

Proof. We provide a reconfiguration sequence which takes η to a colouringη′ with the desired properties. The proof is divided into cases. In each case,each of the reconfiguration steps is done for every neighbour v of u, one byone, before moving on to the subsequent step.

Case (a): η(u), η′(u) ∈ {q + r, 2q + r, . . . , (k − 1)q + r}

(i) Change the colour of xuv0 to 0. This is possible because in thetable in Figure 2 the colour of xuv1 is below and possibly to theright of η(xuv0 ).

(ii) If η(v) 6= 0, recolour Pvu by changing the colour of xvui to iq for0 ≤ i ≤ t−1 and then changing the colour of xvut to 0. If η(v) = 0,then recolour Pvu by changing the colour of xvui to (i + 1)q for0 ≤ i ≤ t.

(iii) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have coloursfrom {0, r}. This allows the colour of u to change to any value in{q + r, 2q + r, . . . , (k − 1)q + r}.

Case (b): η(u) = 0, η′(u) ∈ {2q + r, 3q + r, . . . , (k − 1)q + r}

(i) We have η(xuv0 ) ∈ {q, . . . , 2q − 1} since η(u) = 0. Change thecolour of xuv0 to q.

(ii) Change the colour of xvut to q + r. This colour is compatible withthe colours of u and xvut−1.

(iii) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have coloursfrom {q, q+ r}. This allows the colour of u to change to any valuein {2q + r, 3q + r, . . . , (k − 1)q + r}.

Case (c): η(u) = 0, η′(u) = q + r

(i) We begin observing that h(u) = 0 and h′(u) = 1, which impliesthat h(v) ∈ {2, . . . , k − 1}. Thus, η(v) ∈ {2q + r, 3q + r, . . . , (k −1)q + r}. For i = t, . . . , 0, in order, recolour xuvi to (i+ 2)q − 1.

(ii) Change the colour of xvut to 2q−1. This colour is compatible withthe colour of xvut−1.

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(iii) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have colour 2q−1. Recolour u to q − 1 (temporarily). Note that this colour iscompatible with all neighbours v of u in G since η(v) ∈ {2q +r, 3q + r, . . . , (k − 1)q + r}.

(iv) Recolour xuv0 to p− 1.

(v) For i = 0, . . . , t − 1 recolour xvui to iq, and then recolour xvut top− 1.

(vi) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have colour p−1. Change the colour of u to q + r.

Case (d): η(u) ∈ {2q + r, 3q + r, . . . (k − 1)q + r}, η′(u) = 0

(i) Change the colour of xvut to q + r.

(ii) If η(v) 6= q+ r, then for i = t, . . . , 1 recolour xuvi with (i+ 2)q− 1and recolour xuv0 to q. If η(v) = q+ r, then η(xuvt ) ∈ [p− 1, r]. Fori = t, . . . , 0 recolour xuvi to (i+ 1)q.

(iii) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have colours inin {q, q + r}. Now change the colour of u to 0.

Case (e): η(u) = q + r, η′(u) = 0

(i) For i = 0, . . . , t − 1 recolour xvui to iq and then recolour xvut top− 1.

(ii) Recolour xuv0 to p− 1.

(iii) At this point all vertices of {xuv0 , xvut : v ∈ NG(u)} have colour p−1. Change the colour of u to q − 1 (temporarily). Note thatthis colour is compatible with all neighbours v of u in G sinceη(v) ∈ {2q + r, 3q + r, . . . , (k − 1)q + r}.

(iv) Since η(v) ∈ {2q + r, 3q + r, . . . , (k − 1)q + r}, for i = t, . . . , 0 wecan recolour xuvi to (i+ 2)q − 1.

(v) Change the colour of xvut to 2q−1. This colour is compatible withthe colours of u and xvut−1.

(vi) Change the colour of u to 0.

This completes the proof of the lemma, and of Theorem 2.

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4. An Open Problem

We conclude the paper by stating a reconfiguration problem of a some-what different flavour. Given a graph G and an integer k ≥ χ(G), one canask whether every pair of k-colourings of G can be reconfigured to one an-other. When this is the case, we say that G is k-mixing. The problem ofdetermining whether a graph is k-mixing is investigated in [9, 10], and theanalogous problem for (p, q)-colourings is studed in [6]. The mixing thresholdof a graph G is the minimum integer r such that G is k-mixing for all k ≥ r.To our knowledge, the mixing threshold of the Erdos-Renyi random graphhas not been studied.

Question 18. What is the expected mixing threshold of the random graphG(n, 1/2)? Moreover, what is the distribution of this random variable? Inparticular, is it concentrated around its expected value?

The same questions can be asked for the mixing number which, for ageneral graph G, is defined to be the minimum integer k ≥ χ(G) such thatG is k-mixing. Clearly the mixing number is bounded above by the mixingthreshold and, for general graphs, the mixing threshold is not bounded aboveby any function of the mixing number (see the examples of [9]). In the case ofG(n, 1/2), it is not clear whether these values should be close to one anotheror very far apart.

A celebrated result of Bollobas [2] says that the chromatic number ofG(n, 1/2) is (1+o(1))n/2 log(n) asymptotically almost surely (a.a.s.), and sothis gives a lower bound for both quantities. For an upper bound, Bonamyand Bousquet [4] proved that the mixing threshold of a graph is boundedabove by its Grundy number plus one; the Grundy number of G is maximumnumber of colours over all permutations of V (G) used by greedy colouringalgorithm (sometimes called the First Fit algorithm). The Grundy num-ber of G(n, 1/2) is known to be (1 + o(1))n/ log(n) a.a.s. (see Bollobas [3,Section 11.3]). So, with probability tending to one as n → ∞, the mix-ing threshold and mixing number of G(n, 1/2) are both somewhere between(1 + o(1))n/2 log(n) and (1 + o(1))n/ log(n). It goes without saying that itwould also be interesting to study the mixing threshold and number of therandom graph G(n, p) for p 6= 1/2, including when p is a function of n.

Acknowledgements. The fourth author would like to thank Marthe Bonamyand Guillem Perarnau for stimulating discussions about reconfiguration prob-lems during a workshop at the Bellairs Institute of McGill University in

22

Holetown, Barbados in 2015. In particular, we thank Marthe for sharingWrochna’s beautiful proof of Theorem 11 with us.

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