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Surjective H-Colouring: New Hardness Results⋆ Petr A. Golovach1, Matthew Johnson2, Barnaby Martin2, Daniël Paulusma2, and Anthony Stewart2 7 1Department of Informatics, University of Bergen, Bergen, Norway 1 [email protected] 2 0 School of Engineering and Computing Sciences, Durham University, 2 South Road, Durham,DH13LE, U.K. {matthew.johnson2,barnaby.d.martin,daniel.paulusma,a.g.stewart}@durham.ac.uk r a M 6 Abstract. A homomorphism from a graph G to a graph H is a vertex 2 mapping f from the vertex set of G to the vertex set of H such that there is an edge between vertices f(u) and f(v) of H whenever there is ] C an edge between vertices u and v of G. The H-Colouring problem is to decide if a graph G allows a homomorphism to a fixed graph H. We C continue a study on a variant of this problem, namely the Surjective . s H-Colouringproblem,whichimposesthehomomorphismtobevertex- c surjective.Webuilduponpreviousresultsandshowthatthisproblemis [ NP-complete for every connected graph H that has exactly two vertices 2 with a self-loop as long as these two vertices are not adjacent. As a v result, we can classify the computational complexity of Surjective H- 8 Colouring for everygraph H on at most four vertices. 8 1 2 1 Introduction 0 . 1 0 Thewell-knownColouringproblemistodecideiftheverticesofagivengraph 7 can be properly coloured with at most k colours for some given integer k. If we 1 exclude k from the input and assume it is fixed, we obtain the k-Colouring : v problem.AhomomorphismfromagraphG=(VG,EG)toagraphH =(VH,EH) i is a vertex mapping f : V → V , such that there is an edge between f(u) X G H and f(v) in H whenever there is an edge between u and v in G. We observe ar that k-Colouring is equivalent to the problem of asking if a graph allows a homomorphism to the complete graph K on k vertices. Hence, a natural gen- k eralization of the k-Colouring problem is the H-Colouring problem, which asks if a given graph allows a homomorphism to an arbitrary fixed graph H. We call this fixed graph H the target graph. Throughout the paper we consider undirected graphs with no multiple edges. We assume that an input graph G contains no vertices with self-loops (we call such vertices reflexive), whereas a target graph H may contain such vertices. We call H reflexive if all its vertices are reflexive, and irreflexive if all its vertices are irreflexive. ⋆ Supported by the Research Council of Norway via the project “CLASSIS” and the LeverhulmeTrust (RPG-2016-258). For a survey on graph homomorphisms we refer the reader to the textbook of Hell and Neˇsetˇril [12]. Here, we will discuss the H-Colouring problem, a number of its variants and their relations to each other. In particular, we will focusonthesurjectivevariant:ahomomorphismf fromagraphGtoagraphH is (vertex-)surjective if f is surjective, that is, if for every vertex x ∈ V there H exists at least one vertex u∈V with f(u)=x. G The computational complexity of H-Colouring has been determined com- pletely. The problem is trivial if H contains a reflexive vertex u (we can map each vertex of the input graph to u). If H has no reflexive vertices, then the Hell-Neˇsetˇril dichotomy theorem [11] tells us that H-Colouring is solvable in polynomial time if H is bipartite and that it is NP-complete otherwise. The List H-Colouring problem takes as input a graph G and a function L that assigns to each u ∈ V a list L(u) ⊆ V . The question is whether G G H allows a homomorphism f to the target H with f(u) ∈ L(u) for every u ∈ V . G Feder, Hell and Huang [4] proved that List H-Colouring is polynomial-time solvable if H is a bi-arc graph and NP-complete otherwise (we refer to [4] for the definitionofabi-arcgraph).Ahomomorphismf fromGto aninducedsub- graph H of G is a retraction if f(x) = x for every x ∈ V , and we say that G H retracts toH.A retractionfromGto H canbe viewedasalist-homomorphism: chooseL(u)={u}if u∈V ,andL(u)=V if u∈V \V .The corresponding H H G H decision problem is called H-Retraction. The computational complexity of H-Retractionhasnotyetbeenclassified.Federetal.[5]determinedthe complexity of the H-Retraction problem whenever H is a pseudo-forest(a graph in which every connected component has at most one cycle). They also showed that H-Retraction is NP-complete if H contains a connected component in which the reflexive vertices induce a disconnected graph. As mentioned, we impose a (vertex-)surjectivity condition on the graph homomorphism.Suchaconditioncanbeimposedlocallyorglobally.Ifwerequirea homomorphismf froma graphGto a graphH to be surjective when restricted to the open neighbourhood of every vertex u of G, we say that f is an H-role assignment.ThecorrespondingdecisionproblemiscalledH-RoleAssignment and its computational complexity has been fully classified [8]. We refer to the survey of Fiala and Kratochv´ıl [7] for further details on locally constrained homomorphisms and from here on only consider global surjectivity. Ithas been shownthat deciding whether a givengraphG allowsa surjective homomorphismtoagivengraphH isNP-completeevenifGandH bothbelong to one of the following graph classes:disjoint unions of paths; disjoint unions of complete graphs; trees; connected cographs; connected proper interval graphs; andconnectedsplitgraphs[9].Henceitisnatural,justasbefore,tofixH,which yields the following problem: Surjective H-Colouring Instance: a graph G. Question: does there exist a surjective homomorphism from G to H? 2 Weemphasizethatweareconsideringvertex-surjectivityandthatbeingvertex- surjective is a different condition than being edge-surjective. A homomorphism from a graphG to a graphH is callededge-surjective or a compaction if for any edge xy ∈ E with x 6= y there exists an edge uv ∈ E with f(u) = x and H G f(v) = y. Note that the edge-surjectivity condition does not hold for any self- loopsxx∈E .Iff isacompactionfromGtoH,wesaythatGcompacts toH. H The corresponding decision problem is known as the H-Compaction problem. A full classification of this problem is still wide open. However partial results are known, for example when H is a reflexive cycle, an irreflexive cycle, or a graphonatmostfourvertices[16,18,19],orwhenGisrestrictedtosomespecial graphclass[15].VikasalsoshowedthatwheneverH-Retractionispolynomial- time solvable, then so is H-Compaction [18]. Whether the reverse implication holds is not known. A complete complexity classification of Surjective H- Colouring is also still open. Below we survey the known results. WefirstconsiderirreflexivetargetgraphsH.TheSurjectiveH-Colouring problemisNP-completeforeverysuchgraphH ifH isnon-bipartite,asobserved by Golovach et al. [10]. The straightforward reduction is from the corresponding H-Colouring problem, which is NP-complete due to the aforementioned Hell-Neˇsetˇril dichotomy theorem. However, the complexity classifications of H- Colouring and Surjective H-Colouring do not coincide: there exist bipartite graphs H for which Surjective H-Colouring is NP-complete, for instance when H is the graph obtained from a 6-vertex cycle to each of which verticesweaddapathoflength3 [1],orwhenH is the 6-vertexcycleitself [17]. We now consider target graphs with at least one reflexive vertex. Unlike the H-Colouring problem, the presence of a reflexive vertex does not make the Surjective H-Colouringproblemtrivialtosolve.Wecallaconnectedgraph loop-connectedifallitsreflexiveverticesinduceaconnectedsubgraph.Golovach, Paulusma andSong [10] showedthat if H is a tree (in this context, a connected graph with no cycles of length at least 3) then Surjective H-Colouring is polynomial-time solvableif H is loop-connectedand NP-completeotherwise.As such the following question is natural: Is Surjective H-Colouring NP-complete for every connected graph H that is not loop-connected? Thereversestatementisnottrue(ifP6=NP):SurjectiveH-ColouringisNP- completewhenH isthe 4-vertexcycleC∗ withaself-loopineachofitsvertices. 4 This result has been shown by Martin and Paulusma [13] and independently by Vikas, as announced in [15]. Recall also that Surjective H-Colouring is NP-complete if H is irreflexive (and thus loop-connected) and non-bipartite. It is known that Surjective H-Colouring is polynomial-time solvable whenever H-Compaction is [1]. Recall that H-Compaction is polynomial- time solvable whenever H-Retraction is [18]. Hence, for instance, the aforementioned result of Feder, Hell and Huang [4] implies that Surjective H- Colouring is polynomial-time solvable if H is a bi-arc graph. We also recall that H-Retraction is NP-complete whenever H is a connected graph that is not loop-connected [5]. Hence, an affirmative answer to the above question 3 wouldmeanthatforthesetargetgraphsH the complexitiesofH-Retraction, H-Compaction and Surjective H-Colouring coincide. In Figure 1 we display the relationships between the different problems dis- cussed.Inparticular,itisamajoropenproblemwhetherthecomputationalcom- plexities of H-Compaction, H-Retraction and Surjective H-Colouring coincide for each target graph H. Even showing this for specific cases, such as the case H =C∗, has been provento be non-trivial.If it is true, it would relate 4 the Surjective H-Colouring problem to a well-known conjecture of Feder andVardi[6],whichstatesthattheH-Constraint Satisfactionproblemhas a dichotomy when H is some fixed finite target structure and which is equivalent to conjecturing that H-Retraction has a dichotomy [6]. We refer to the survey of Bodirsky, Kara and Martin [1] for more details on the Surjective H-Colouring problem from a constraint satisfaction point of view. ListH-Colouring H-Retraction H-Compaction SurjH-Colouring H-Colouring Fig.1: Relations between H-Colouring and its variants. An arrow from one problemtoanotherindicatesthatthelatterproblemispolynomial-timesolvable for a target graph H if the former is polynomial-time solvable for H. Reverse arrows do not hold for the leftmost and rightmost arrows, as witnessed by the reflexive 4-vertex cycle for the rightmost arrow and by any reflexive tree that is not a reflexive interval graph for the leftmost arrow (Feder, Hell and Huang [4] showed that the only reflexive bi-arc graphs are reflexive interval graphs). It is not known if the reverse direction holds for the two middle arrows. 1.1 Our Results We present further progress on the research question of whether Surjective H-Colouring is NP-complete for every connected graph H that is not loop- connected. We first consider the case where the target graph H is a connected graph with exactly two reflexive vertices that are non-adjacent. In Section 2 we prove that Surjective H-Colouring is indeed NP-complete for every such targetgraphH.Inthesamesectionweslightlygeneralizethisresultbyshowing that it holds even if the reflexive vertices of H can be partitioned into two non-adjacent sets of twin vertices. This enables us to classify in Section 3 the computationalcomplexity ofSurjective H-Colouringfor everygraphH on at most four vertices, just as Vikas [19] did for the H-Compaction problem. A classification of Surjective H-Colouring for target graphs H on at most four vertices has also been announced by Vikas in [15]. As we will illustrate for one particular case, it is interesting to note that NP-hardness proofs for H- Compaction of [19] may lift to NP-hardness for Surjective H-Colouring. However, this is not true for the reflexive cycle C∗, where a totally new proof 4 was required. 4 1.2 Future Work To conjecture a dichotomy of Surjective H-Colouring between P and NP- complete seems still to be difficult. Our first goal is to prove that Surjec- tive H-Colouring is NP-complete for every connected graph H that is not loop-connected. However, doing this via using our current techniques does not seem straightforward and we may need new hardness reductions. Another way forward is to prove polynomial equivalence between the three problems Sur- jective H-Colouring, H-Compaction andH-Retraction.However,com- pletelyachievingthisgoalalsoseemsfarfromtrivial.Ourclassificationfortarget graphs H up to four vertices does show such an equivalence for these cases (see Section 3). 2 Two Non-Adjacent Reflexive Vertices We say that a graph is 2-reflexive if it contains exactly 2 reflexive vertices that are non-adjacent.InthissectionwewillprovethatSurjective H-Colouring isNP-completewheneverH is connectedand2-reflexive.The problemisreadily seen to be in NP. Our NP-hardness reduction uses similar ingredients as the reductionofGolovach,PaulusmaandSong[10]forprovingNP-hardnesswhenH isatreethatisnotloop-connected.Thereare,however,anumberofdifferences. Forinstance,wewillreducefromafactorcutprobleminsteadofthelessgeneral matchingcutproblemusedin[10].Wewillexplainthesetwoproblemsandprove NP-hardness for the former one in Section 2.1. Then in Section 2.2 we give our hardnessreduction,andinSection2.3weextendourresulttobevalidfortarget graphsH withmorethantworeflexiveverticesaslongasthesereflexivevertices can be partitioned into two non-adjacent sets of twin vertices. 2.1 Factor Cuts Let G = (V ,E ) be a connected graph. For v ∈ V and E ⊆ E , let d (v) G G G G E denote the number of edges of E incident with v. For a partition(V ,V ) of V , 1 2 G let E (V ,V ) denote the set of edges between V and V in G. G 1 2 1 2 Let i and j be positive integers, i≤j. Let (V ,V ) be a partition of V and 1 2 G let M = E (V ,V ). Then (V ,V ) is an (i,j)-factor cut of G if, for all v ∈ V , G 1 2 1 2 1 d (v) ≤ i, and, for all v ∈ V , d (v) ≤ j. Observe that if a vertex v exists M 2 M withdegreeatmostj,thenthere is atrivial(i,j)-factorcut(V \{v},{v}).Two distinct vertices s and t in V are (i,j)-factor roots of G if, for each(i,j)-factor G cut (V ,V ) of G, s and t belong to different parts of the partition and, if i<j, 1 2 s ∈ V and t ∈ V (of course, if i = j, we do not require the latter condition as 1 2 (V ,V )isalsoan(i,j)-factorcut).Wenotethatwhenno(i,j)-factorcutexists, 2 1 every pair of vertices is a pair of (i,j)-factor roots. We define the following decision problem. (i,j)-Factor Cut with Roots Instance: a connected graph G with roots s and t. Question: does G have an (i,j)-factor cut? 5 We emphasize thatthe (i,j)-factorrootsare givenas partofthe input. Thatis, the problemaskswhether ornotan(i,j)-factorcut(V ,V )exists,but weknow 1 2 alreadythatifitdoes,thensandtbelongtodifferentpartsofthepartition.That is,we actually define (i,j)-Factor Cut with Roots to be a promise problem in which we assume that if an (i,j)-factor cut exists then it has the property that s and t belong to different parts of the partition. The promise class may not itself be polynomially recognizablebut one may readily find a subclass of it thatis polynomiallyrecognizableandincludes allthe instanceswe needforNP- hardness.Infactthiswillbecomeclearwhenreadingourproofbutwereferalso to [10] where such a subclass is given for the case (i,j) = (1,1). A (1,1)-factor cut (V ,V ) ofG is also knownas a matching cut,as no two edgesin E (V ,V ) 1 2 G 1 2 have a common end-vertex, that is, E (V ,V ) is a matching. Similarly (1,1)- G 1 2 Factor Cut with Roots is known as Matching Cut with Roots and was proved NP-complete by Golovach, Paulusma and Song [10] (by making an observation about the proof of the result of Patrignani and Pizzonia [14] that deciding whether or not any given graph has a matching cut is NP-complete). WewillprovetheNP-completenessof(i,j)-Factor Cut with Rootsafter first presenting a helpful lemma (a clique is a subset of vertices of G that are pairwise adjacent to each other). Lemma 1. Let i, j and k be positive integers where i≤j and k >i+j. Let G be a graph that contains a clique K on k vertices. Then, for every (i,j)-factor cut (V ,V ) of G, either V ⊆V or V ⊆V . 1 2 K 1 K 2 Proof. Ifthelemmaisfalse,thenforsome(i,j)-factorcut(V ,V ),wecanchoose 1 2 v ∈V ∩V andv ∈V ∩V .LetM =E (V ,V ).SinceeveryvertexinV ∩V 1 1 K 2 2 K G 1 2 1 K is linkedby anedge ofM to v andeveryvertexin V ∩V is linked by anedge 2 2 K of M to v , we have d (v )+d (v ) ≥ k > i+j, contradicting the definition 1 M 1 M 2 of an (i,j)-factor cut. ⊓⊔ Theorem 1. Let i and j be positive integers, i ≤ j. Then (i,j)-Factor Cut with Roots is NP-complete. Proof. If i = j = 1, then the problem is Matching Cut with Roots which, as we noted, is known to be NP-complete [10]. We split the remaining cases in two according to whether or not i=1. In each case, we construct a polynomial time reduction from Matching Cut with Roots. In particular, we take an instance (G,s,t) of Matching Cut with Roots, and construct a graph G′ that is a supergraph of G=(V,E) and show that (1) (G′,s,t) is an instance of (i,j)-Factor Cut with Roots (that is, if G′ has an (i,j)-factor cut (V′,V′), then s ∈ V and t ∈ V or, possibly, vice 1 2 1 2 versa if i=j), (2) if G′ has an (i,j)-factor cut, then G has a matching cut, and (3) if G has a matching cut, then G′ has an (i,j)-factor cut. Wenotethat(1)isanatypicalfeatureofanNP-completenessproofas,unusually for(i,j)-Factor Cut with Roots,itisnotimmediatetorecognizeaproblem instance. We let n=|V|. 6 Case 1: i=1. Let k = max{(n − 1)(j − 1),1 + j}. Construct G′ from G by first adding a complete graph K on k vertices and adding edges from s to every vertex of V . K Then, for each v ∈ V \{s}, add edges from v to j−1 vertices of K in such a G way that no vertex of V has more than one neighbour in V \{s}. K G Let (V′,V′) be a (1,j)-factor cut of G′. The vertices of {s}∪V induce a 1 2 K cliqueon1+k>1+jvertices.So,byLemma1,{s}∪V ⊆V′ or{s}∪V ⊆V′. K 1 K 2 Suppose that {s}∪V ⊆ V′. Then V must contain vertices of both V′ K 2 G 1 (otherwise V′ wouldbe empty) andV′ (at leasts). Thus, as G is connected,we 1 2 can find a vertex v ∈ V′∩V that has a neighbour in V′∩V . But v also has 1 G 2 G j−1≥1neighboursinV andsohasatleast2neighboursinV′,contradicting K 2 the definition of a (1,j)-factor cut. So we must have that {s}∪V ⊆ V′. Let V = V′∩V and V = V′ be a K 1 1 1 G 2 2 partition of V , and let M =E (V ,V ) and M′ =E (V′,V′) and notice that G G 1 2 G 1 2 M′ istheunionofM and,foreachv ∈V ,thej−1edgesfromv toV .Foreach 2 K v ∈V1, dM(v)=dM′(v)≤1.For eachv ∈V2, dM(v)=dM′(v)−(j−1)≤1.So (V ,V ) is a matching cut of G; this proves (2). And as s∈V , we have, by the 1 2 1 definition of factor roots, t∈V ; this proves (1). 2 To prove (3), we note that if (V ,V ) is a matching cut of G, then we can 1 2 assume that s ∈ V and t ∈ V (else relabel them for the purpose of construct- 1 2 ing G′), and then (V ∪V ,V ) is a (1,j)-factor cut of G′. 1 K 2 Case 2: i≥2. Letk =max{(n−1)(j−1),i+j}.ConstructG′fromGbyfirstaddingacomplete graph Ks on k vertices and adding edges from s to every vertex of VKs, and thenaddingacompletegraphKt onk verticesandaddingedgesfromttoevery vertex of VKt. Then, for eachv ∈VG\{s},add edges fromv to j−1 vertices of Ks insuchawaythatnovertexofVKs hasmorethanoneneighbourinVG\{s}. Afterwards, for each v ∈ V \{t}, add edges from v to i−1 vertices of Kt in G such a way that no vertex of VKt has more than one neighbour in VG\{t}. Let (V1′,V2′) be an (i,j)-factor cut of G′. The vertices of {s}∪VKs induce a clique on at least 1+k > i+j vertices. So, by Lemma 1, {s}∪VKs ⊆ V1′ or {s}∪VKs ⊆V2′. Similarly {t}∪VKt ⊆V1′ or {t}∪VKt ⊆V2′. Suppose that{s}∪VKs and{t}∪VKt areboth subsetsofV1′.Then VG must contain vertices of both V′ (at least s and t) and V′ (else it would be empty). 1 2 Thus, as G is connected, we can find a vertex v ∈V′∩V that has a neighbour 2 G in V1′ ∩VG. But v also has j −1 neighbours in VKs and i−1 neighbours in VKt and so has at least 1+(i−1)+(j −1) = i+j −1 > j ≥ i neighbours in V′, contradicting the definition of an (i,j)-factor. By an analogousargument 2 {s}∪VKs and {t}∪VKt cannot both be subsets of V2′. Suppose that i<j and {s}∪VKs ⊆V2′. As G is connected and VG contains verticesofbothV′ andV′,wecanfindavertexv ∈V′∩V thathasaneighbour 1 2 1 G inV2′∩VG.But v alsohasj−1>i−1neighboursinVKs andsohas morethan i neighbours in V′, contradicting the definition of a (i,j)-factor. 2 Thus we have that {s}∪VKs and {t}∪VKt are subsets of separate parts and, moreover,either {s}∪VKs ⊆V1′ or i=j. Thus (1) is proved,and we have, 7 in either case, that each vertex in V′ ∩V is joined by i−1 edges to vertices 1 G in V′ \V , and each vertex in V′ ∩V is joined by j −1 edges to vertices in 2 G 2 G V′ \V . Therefore each vertex in V′ ∩V is joined to at most one vertex in 1 G 1 G V′∩V , and each vertex in V′∩V is joined to at most one vertex in V′∩V . 2 G 2 G 1 G Thus (V′∩V ,V′∩V ) is a matching cut of G. This proves (2). 1 G 2 G To prove (3), we note that if (V ,V ) is a matching cut of G, then we can 1 2 assume that s ∈ V and t ∈ V (else relabel them for the purpose of construct- 1 2 ing G′), and then (V1∪VKs,V2∪VKt) is an (i,j)-factor cut of G′. ⊓⊔ 2.2 The Hardness Reduction Let H be a connected 2-reflexive target graph. Let p and q be the two (non- adjacent) reflexive vertices of H. The length of a path is its number of edges. The distance between two vertices u and v in a graph G is the length of a shortest path between them and is denoted dist (u,v). We define two induced G subgraphs H and H of H whose vertex sets partition V . First H contains 1 2 H 1 those vertices of H that are closer to p than to q; and H contains those ver- 2 tices that are at least as close to q as to p (so contains any vertex equidistant to p and q). That is, V = {v ∈V :dist (v,p)<dist (v,q)} and V = H1 H H H H2 {v ∈V :dist (v,q)≤dist (v,p)}.See Figure 2 for anexample.The following H H H lemma follows immediately from our assumption that H is connected. Lemma 2. BothH andH areconnected.Moreover,dist (x,p)=dist (x,p) 1 2 H1 H for every x∈V and dist (x,q)=dist (x,q) for every x∈V . H1 H2 H H2 Let ω denote the size of a largest clique in H. From graphs H and H we 1 2 construct graphs F and F , respectively, in the following way: 1 2 1. for each x∈/ {p,q}, create a vertex t1; x 2. for p, create ω vertices t1,...,tω; p p 3. for q, create ω vertices t1,...,tω; q q 4. fori=1,2,addanedgeinF betweenanytwoverticesth andtj ifandonly i x y if xy is an edge of E . Hi We note that F is the graphobtained by taking H andreplacingp by a clique 1 1 of size ω. Similarly, F is the graphobtained by taking H and replacing q by a 2 2 clique ofsizeω.We saythat t1,...,tω arethe roots ofF andthatt1,...,tω are p p 1 q q the roots of F . Figure 3 shows an example of the graphs F and F obtained 2 1 2 from the graph H in Figure 2. Let ℓ= dist (p,q)≥2 denote the distance between p and q. Let N be the H p set of neighbours of p that are each on some shortest path (thus of length ℓ) from p to q in H. Let r be the size of a largest clique in N . We define N and p p q r similarly. We will reduce from (r ,r )-Factor Cut with Roots, which is q p q NP-complete due to Theorem 1. Hence, consider an instance (G,s,t) of (r ,r )- p q Factor Cut with Roots, where G is a connected graph and s and t form the (ordered) pair of (r ,r )-factor roots of G. Recall that we assume that G is p q irreflexive. 8 p q H2 H1 Fig.2: An example of the construction of graphs H and H from a connected 1 2 2-reflexive target graph H with ω =3. Fig.3:ThegraphsF (left)andF (right)resultingfromthegraphH inFigure2. 1 2 Wesaythatweidentifytwoverticesuandvofagraphwhenweremovethem from the graph and replace them with a single vertex that we make adjacent to every vertex that was adjacent to u or v. From F , F , and G we construct a 1 2 new graph G′ as follows: 1. For each edge e = uv ∈ E , we do as follows. We create four vertices, gr , G u,e gb , gr andgb .WealsocreatetwopathsP1 andP2,eachoflengthℓ−2, u,e v,e v,e e e between gr and gb , and between gr and gb , respectively. If ℓ = 2 we u,e v,e v,e u,e identify gr and gb and gr and gb to get paths of length 0. u,e v,e v,e u,e 2. For each vertex u ∈ V , we do as follows. First we construct a clique C G u on ω vertices. We denote these vertices by g1,...,gω. We then make every u u vertex in C adjacent to both gr and gb for every edge e incident to u; u u,e u,e we call gr and gb a red and blue neighbour of C , respectively; if ℓ = 2, u,e u,e u thenthevertexobtainedbyidentifyingtwoverticesgr andgb ,orgr and u,e v,e v,e gb is simultaneously a red neighbour of one clique and a blue neighbour u,e of another one. Finally, for every two edges e and e′ incident to u, we make gr andgr adjacent,thatis,thesetofredneighboursofC formaclique, u,e u,e′ u whereas the set of blue neighbours form an independent set. 3. We add F by identifying ti and gi for i = 1,...,ω, and we add F by 1 p s 2 identifying ti and gi for i = 1,...,ω. We denote the vertices in F and F q t 1 2 in G′ by their label ti in F or F . x 1 2 SeeFigure4foranexampleofagraphG′.Thenextlemmadescribesastraight- forward property of graph homomorphisms that will prove useful. Lemma 3. If there exists a homomorphism h : G′ → H then distG′(u,v) ≥ distH(h(u),h(v)) for every pair of vertices u,v∈VG′. 9 t F2 F1 s (a) An example of a graph G with (b)Thecorresponding graphG′ whereH isa a (1,2)-factor cut with (1,2)-factor 2-reflexivetargetgraphwithℓ=3andω=3. roots s and t. Fig.4: An example of a graph G and the corresponding graph G′. We now prove the key property of our construction. Lemma 4. For every homomorphism h from G′ to H, there exists at least one clique C with p∈h(C ) and at least one clique C with q ∈h(C ). a a b b Proof. Sinceforeachu∈V andanyedgeeincidenttou,everycliqueC ∪{gr } G u u,e inG′ isofsizeatleastω+1,wefindthathmustmapatleasttwoofitsvertices to a reflexive vertex,so either to p or q. Hence, for every u∈V , we find that h G maps at least one vertex of C to either p or q. u We prove the lemma by contradiction. We will assume that h does not map anyvertexofanyC toq,thus p∈h(C )forall u∈V .We willnote laterthat u u G if instead q ∈ h(C ) for all u ∈ V we can obtain a contradiction in the same u G way. We consider two vertices ti ∈ F and tj ∈ F such that h(ti) = h(tj) = p. p 1 q 2 p q Without loss of generality let i = j = 1. We shall refer to these vertices as t and t respectively. We now consider a vertex v ∈ V ∪V . By Lemma 3, p q F1 F2 distG′(v,tp)≥distH(h(v),p) and distG′(v,tq)≥distH(h(v),p). In other words: min(distG′(v,tp),distG′(v,tq))≥distH(h(v),p). InfactbyapplyingLemma3wecangeneralizethisfurthertoanyvertexmapped to p by h: min (distG′(v,w))≥distH(h(v),p). (1) w∈h−1(p) For every v ∈VG′ we define a value D(v) as follows: dist (v,t )if v ∈F  F1 p 1 D(v)= dist (v,t )if v ∈F  F2 q 2 ⌊ℓ/2⌋ otherwise  Claim 1 D(v)≥minw∈h−1(p)(distG′(v,w))≥distH(h(v),p) for all v ∈VG′. We prove Claim 1 by showing that D(v) ≥ minw∈h−1(p)(distG′(v,w)), which suffices due to (1). First suppose v ∈ V ∪ V . We may assume, without F1 F2 10