Complexity of Correspondence Homomorphisms
CCOMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS
TOM ´AS FEDER AND PAVOL HELL
Abstract.
Correspondence homomorphisms are both a generalization of standard homo-morphisms and a generalization of correspondence colourings. For a fixed target graph H ,the problem is to decide whether an input graph G , with each edge labeled by a pair of per-mutations of V ( H ), admits a homomorphism to H ‘corresponding’ to the labels, in a senseexplained below.We classify the complexity of this problem as a function of the fixed graph H . It turns outthat there is dichotomy – each of the problems is polynomial-time solvable or NP-complete.While most graphs H yield NP-complete problems, there are interesting cases of graphs H for which the problem is solved by Gaussian elimination.We also classify the complexity of the analogous correspondence list homomorphism prob-lems, and also the complexity of a bipartite version of both problems. We emphasize theproofs for the case when H is reflexive, but, for the record, we include a rough sketch of theremaining proofs in an Appendix. Introduction
Let H be a fixed graph, and let P ( H ) denote the set of all permutations of V ( H ). Let G be a graph with an edge-labeling (cid:96) : E ( G ) → P ( H ) × P ( H ). Each edge xy of G isassumed to have a label ( π, ρ ) where π is viewed as associated with x and ρ associated with y . An (cid:96) - correspondence homomorphism of G to H of G ) is a mapping f : V ( G ) → V ( H )such that xy ∈ E ( G ) with (cid:96) ( xy ) = ( π, ρ ) implies π ( f ( x )) ρ ( f ( y )) ∈ E ( H ). When H is theirreflexive (i.e., loopless) complete graph on k vertices, correspondence homomorphisms to H have been called correspondence k -colourings , and applied to answer a question of Borodinon choosability of planar graphs with certain excluded cycles [2]. These colorings have alsobeen called DP-colourings, in honour of the authors of [2], and have proved quite interestingfrom a graph theoretic point of view. cf. [1, 11] and the references therein. To emphasizethis connection, we sometimes call correspondence homomorphisms to H also correspondence H -homomorphisms , or correspondence H -colourings .The correspondence H -homomorphism problem takes as input a graph G with labeling (cid:96) andasks whether or not an (cid:96) -correspondence homomorphism to G exists. In the correspondence H -list-homomorphism problem , G is also equipped with lists L ( x ) , x ∈ V ( G ), each a subset of V ( H ), and the (cid:96) -correspondence homomorphism f also has to satisfy f ( x ) ∈ L ( x ) , x ∈ V ( G ).Clearly, these problems are generalizations of the well-known H -homomorphism (i.e., H -colouring) and list H -homomorphism (list H -colouring) problems respectively. Those prob-lems are obtained as special cases when (cid:96) chooses two identity permutations on each edge of G . They have been studied in [4, 6, 7, 8], cf. [9]. a r X i v : . [ c s . D M ] M a r T. FEDER AND P. HELL
There is another way to think of the correspondence homomorphism problem, one that isoften helpful in the proofs and illustrations. Let
G, (cid:96) be an instance of the correspondence H -homomorphism problem. Construct a new graph G ∗ by replacing each vertex x of G with its own separate copy V x of the set V ( H ), with the following edges. If xy is an edgeof G labelled by (cid:96) ( xy ) = ( π, ρ ), we join V x and V y with the edges from π ( x )( u ) ∈ V x to ρ ( y )( v ) ∈ V y for all edges uv of H . (Recall that π ( x ) is a permutation of V ( H ), and we view itas also a permutation of V x , and similarly for ρ ( y ).) Then an (cid:96) -correspondence homomorphismcorresponds precisely to a transversal of the sets V x , x ∈ V ( G ), (a choice of exactly one vertexfrom each V x ) which induces in G ∗ an isomorphic copy of G . Consider the edges between twoadjacent sets V x , V y , xy ∈ E ( G ). Recall the bipartite graph H (cid:48) associated with H , in whicheach v ∈ V ( H ) yields two vertices v , v in H (cid:48) and each edge uv ∈ E ( H ) yields two edges u v , u v in E ( H (cid:48) ). The edges in G ∗ between adjacent sets V x , V y form an isomorphic copy of H (cid:48) , with the part V x permuted according to π , and the part in V y according to ρ . The label (cid:96) ( xy ) specifies the way H (cid:48) is layed out between the sets V x and V y .In this note we focus on the reflexive case, i.e., we assume that H has a loop at everyvertex. While the standard H -homomorphism problem is trivial for reflexive graphs, thecorrespondence H -homomorphism problem turns out to be more interesting. Moreover, itmakes sense to consider inputs G that may have loops and parallel edges, as the permutationconstraints on these may introduce significant restrictions. We will use this freedom in Section4, to simplify NP-completeness proofs. We will also show, in Section 2, that the complexityof the problem for graphs with loops and parallel edges allowed or forbidden are the same.A reflexive clique is a complete graph with all loops, a reflexive co-clique is a set of discon-nected loops with no other edges. A disjoint union of cliques K p and K q will be denoted by K p ∪ K q or 2 K p if p = q , and similarly for more than two cliques. We use the same symbolsfor reflexive cliques and irreflexive cliques, and the right context will always be specified orclear from the context.Our main result is the following dichotomy classification of both the correspondence homo-morphism problem and the correspondence list homomorphism problem. Theorem 1.1.
Suppose H is a reflexive graph.If H is a reflexive clique, a reflexive co-clique, or a reflexive K , then the correspondence H -homomorphism problem is polynomial-time solvable. In all other cases, the correspondence H -homomorphism problem is NP-complete.If H is a reflexive clique or a reflexive co-clique, then the correspondence H -list-homomorphismproblem is polynomial-time solvable. In all other cases the correspondence H -list-homomorphismproblem is NP-complete. In the last section we state the analogous result for general graphs (with possible loops)and for bipartite graphs, and in the Appendix we provide a rough sketch of the proofs.2.
Loops and Parallel Edges
Suppose H is a fixed reflexive graph, and G is a graph with loops and parallel edges allowed,with an edge-labeling (cid:96) . Thus there are labels on loops, and parallel edges may have differentlabels. We will construct a modified simple graph G (cid:48) (without loops and parallel edges), and OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 3 a modified edge-labeling (cid:96) (cid:48) on G (cid:48) such that G has an (cid:96) -correspondence homomorphism to H ifand only if G (cid:48) has an (cid:96) (cid:48) -correspondence homomorphism to H . The changes from G, (cid:96) to G (cid:48) , (cid:96) (cid:48) proceed one loop and one pair of parallel edges at a time.Suppose G has a loop xx with label (cid:96) ( xx ) = ( π ( x ) , ρ ( x )). Replace x by a clique with ver-tices x , x , x , . . . , x n where n = | V ( H ) | . Each edge x i x j will have the same label (cid:96) ( x i x j ) =( π ( x ) , ρ ( x )). Each vertex x i will have the same adjacencies, with the same labels, as x did.Call the resulting graph G and the resulting labeling (cid:96) . Then we claim that G has an (cid:96) -correspondence homomorphism to H if and only if G has an (cid:96) -correspondence homomor-phism to H . The one direction is obvious, if f is an (cid:96) -correspondence homomorphism of G to H , then the same mapping, extended to all copies x i of x is an (cid:96) -correspondence homomor-phism of G to H . Conversely, suppose that f is an (cid:96) -correspondence homomorphism of G to H . This gives n + 1 values f ( x i ) among the n possible images V ( H ). Thus the majorityvalue must appear on at least two distinct vertices v i , v j , and therefore π ( v i ) ρ ( v j ) is an edgeof H . Thus the mapping F which assigns to x the majority value amongst f ( x i ) and equalsto f on all other vertices is an (cid:96) - correspondence homomorphism of G to H .Parallel edges are removed by a similar trick using expanders instead of cliques. To bespecific, assume that ( xy ) , ( xy ) (cid:48) are two different parallel edges of G joining the same vertices x and y , with labels (cid:96) (( xy )) = ( π ( x ) , ρ ( y )) and (cid:96) (cid:48) (( xy ) (cid:48) ) = ( π (cid:48) ( x ) , ρ (cid:48) ( y )). Replace x and y by alarge number N of vertices x i , y j , each having the same adjacencies and labels to other verticesas x, y respectively, and all edges x i y j for all i and j . Some of the edges x i y j are labelled by (cid:96) (( xy )) and the others by (cid:96) (cid:48) (( xy ) (cid:48) ). The set of edges labelled by (cid:96) (( xy )) defines a bipartitegraph B and those labelled by (cid:96) (cid:48) (( xy ) (cid:48) ) form its bipartite complement B . With the rightchoice of B we will be able to conclude that between any large sets of x i ’s and y j ’s there is atleast one edge of B and at least one edge of B . This allows the above idea of using majorityvalues to work, namely if f is a correspondence homomorphism on the replaced graph, wemay define F ( x ) to be the majority value of f ( x i ) and similarly for F ( y ). Then the two valuesappear at both an edge labelled by (cid:96) (( xy )) and an edge labelled by (cid:96) (cid:48) (( xy ) (cid:48) ), and so F is acorrespondence homomorphism on the original graph G .There are many known proofs that such expanders B do exist [10]. It is also not hard tosee directly. For instance, if | V ( H ) | = n , let N > n log n and take for B a random bipartitegraph on N versus N vertices. The number of ways of choosing sets of size t = (cid:100) N/n (cid:101) > n for both sides is A = (cid:0) Nt (cid:1) ≤ ( eNt ) t ≤ tlog N . The probability that the sets will not be joinedby a random choice of edges is bounded by t − < /A , giving a positive probability to theexistence of a suitable bipartite graph B .3. Polynomial Cases for Reflexive H If H is a reflexive clique, the correspondence H -homomorphism problem can be triviallysolved. Each vertex x of the input graph G can be assigned any image, and regardless of thepermutation labels (cid:96) , the mapping is a perm-homomorphism. In fact, this observation alsosolves the correspondence H -list-homomorphism problem. T. FEDER AND P. HELL If H is a reflexive co-clique, the correspondence H -homomorphism problem also has an easysolution, since every choice of an image for a vertex x of G implies a unique image of anyadjacent vertex y . Thus for each component of G we may try all images of a particular vertex x , and the component can be mapped to H if and only if one of these images produces an (cid:96) -consistent homomorphism. This also works for the correspondence list H -homomorphismproblem problem.The most interesting case occurs when H = 2 K . We name the vertices of H by binary2-strings, 00 , , ,
11. In addition to the loops 00 − , − , − , −
11, the twoedges of H are, say, 00 −
01 and 10 −
11. Note that because of symmetry, there are only threedifferent permutations of H , this one, with edges 00 −
01 and 10 −
11, and two more, withedges 00 −
10 and 01 −
11, or with edges 00 −
11 and 01 −
10. Consider the alternate viewvia the auxiliary graph G ∗ discussed earlier. For each edge xy of an input graph G , the edgesbetween V x and V y depend only on which of the above possibilities apply to each of V x and V y , because of the nature of the adjacencies, in the form of two copies of K , . Moreover, itdoes not matter which vertex on each side of the K , is chosen. (Here we use K , to denotethe irreflexive complete bipartite graph with two vertices on each side; see Figure 1.) Let usassociate with each vertex x of G , two { , } variables x a , x b , to be understood as describingthe two coordinates of the name of the chosen vertex for V x . Each of the above partitions of V ( H ) into two edges can now be described by linear equations modulo two: • −
01 and 10 −
11 correspond to the equations x a = 0 and x a = 1 • −
10 and 01 −
11 correspond to the equations x b = 0 and x b = 1 • −
11 and 01 −
10 correspond to the equations x a + x b = 0 and x a + x b = 1In Figure 1, we have the partition on V x corresponding to the first bullet, and the partitionof V y corresponding to the second bullet, while the edges of 2 K , joint the first part of thepartition on V x with the second part of the partition on V y and vice versa. To satisfy theconstraints of correspondence, we have to make sure that if x selects a vertex in the partwith x a = 0, then y selects a vertex in the part with y a + y b = 1, and if x selects in x a = 1then y selects in y a + y b = 0. These constraints can be expressed by the linear equation x a + y a + y b = 1. In all the other cases it is also easy to check that there is a linear equationmodulo two, which describes the constraints. Thus we have reduced the problem of existenceof a correspondence 2 K -homomorphism to a system of linear equations modulo two, whichcan be solved in polynomial-time by Gaussian elimination.It turns out, see the next section, that the list version of the correspondence 2 K -homomorphismproblem is NP-complete.4. NP-complete Cases for Reflexive H We begin with the following simple example. Suppose H is ( K ∪ K ), the disjoint union ofa reflexive cliques K and K . Specifically, let ( K ∪ K ) have a loop on a , and two adjacentloops on b and c . Proposition 4.1.
The correspondence ( K ∪ K ) -homomorphism problem is NP-complete.Proof. We give a reduction from 1-IN-3-SAT (without negated variables). Consider an in-stance, with variables x , x , . . . , x n and triples (of variables) T , T , . . . , T m . Construct the OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 5 x = y + y =x = aa a b a y + y = b Figure 1.
The edges are described by the equation x a + y a + y b = 1 modulo twocorresponding instance of the correspondence H -homomorphism problem as follows. The in-stance G will contain vertices v , . . . , v n , as well as T , . . . , T m , and all edges are of the form v i T j with v i appearing in the triple T j . The variables in the triples are arbitrarily ordered,each T j having a first, second, and third variable. The edge between T j and its first variable v p is labeled so that the special vertex a in V v p is adjacent to a in V ( T j ), the edge between T j and its second variable v q is labeled so that the special vertex a in V v q is adjacent to b in V ( T j ), and similarly for the third variable and c in V ( T j ). (See Figure 2.) Because of theadjacencies, if any one vertex x is chosen in V ( T j ), there is exactly one of its variables v p , v q , v r that has its special vertex a adjacent to x .We claim that this instance has a correspondence homomorphism if and only if the originalinstance of 1-IN-3-SAT was satisfiable. Indeed, any satisfying truth assignment sets as true aset of v i ’s such that exactly one appears in each T j ; hence we can set the value of the vertex v i to be a whenever the variable v i was true in the truth assignment (and, say, b otherwise).The value of the vertex T j will be a if its first variable was true, b if its second variable wastrue, and c if its third variable was true. It is easy to see that this defines a correspondence( K ∪ K )-homomorphism. For the converse, observe that a correspondence homomorphismselects one vertex from each V ( T j ), which forces exactly one of its variable to select the value a , thus defining a satisfying truth assignment. (cid:3) We note that this result implies that the list version of the correspondence 2 K -homomorphismproblem is also NP-complete, since the correspondence K + K -homomorphism problem re-duces to the correspondence 2 K -list-homomorphism problem. Indeed, lists may be used torestrict the input vertices never to use one of the four vertices of 2 K .A similar reduction from 1-IN-t-SAT shows the following fact. Proposition 4.2.
The correspondence ( K ∪ K t ) -homomorphism problem is NP-complete, forall t ≥ . The method we use most often is described in the following proposition.
T. FEDER AND P. HELL j abc abc vabc vabc va b c H pqrT Figure 2.
The triple T j with variables v p , v q , v r , considered in that order Proposition 4.3.
The correspondence ( K p ∪ K q ∪ · · · ∪ K z ) -homomorphism problem reducesto the correspondence ( K p +1 ∪ K q +1 ∪ · · · ∪ K z ) -homomorphism problem.Proof. Given an instance G of the ( K p ∪ K q ∪ · · · ∪ K z )-homomorphism problem, we create anew graph G (cid:48) by adding at each vertex v of G a loop e v (even if v may already have loops),labelled by two permutations ( π ( e v ) , ρ ( e v )) that forbid one vertex of K p and one vertex of K q .Specifically, suppose we want to forbid a vertex a of K p and a vertex b of K q . Choose π ( e v ) tobe the identity and ρ ( e v ) to be the involution exchanging a and b . Now we claim that G (cid:48) hasa correspondence ( K p ∪ K q ∪ · · · ∪ K z )-homomorphism if and only if the original graph G hasa correspondence ( K p +1 ∪ K q +1 ∪ · · · ∪ K z )-homomorphism. Indeed, if f is a correspondence( K p ∪ K q ∪ · · · ∪ K z )-homomorphism for the target graph has neither a nor b , and so, theadded loops create no problem, i.e., f remains a correspondence ( K p +1 ∪ K q +1 ∪ · · · ∪ K z )-homomorphism. On the other hand, if f is a ( K p +1 ∪ K q +1 ∪ · · · ∪ K z )-homomorphism of G (cid:48) ,the label on e v ensures that v cannot map to a or b as neither π ( e v )( a ) = a is adjacent to ρ ( e v )( a ) = b nor π ( e v )( b ) = b is adjacent to ρ ( e v )( b ) = a . Thus f is also a ( K p ∪ K q ∪ · · · ∪ K z )-homomorphism. (cid:3) Proposition 4.3 allows us to prove the NP-completeness of the correspondence ( K p +1 ∪ K q +1 ∪ · · · ∪ K z )-homomorphism problem from the NP-completeness of the correspondence( K p ∪ K q ∪ · · · ∪ K z )-homomorphism problem. Let us call the operation of removing a vertexfrom each of two distinct cliques a pair-deletion . Thus any H that is a union of reflexive cliquesthat can be transformed to K ∪ K by a sequence of pair-deletions yields an NP-completecorrespondence H -homomorphism problem. OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 7
We now consider two additional important special cases, starting with H = 2 K consistingof two reflexive triangles. Note that any pair-deletion of 2 K produces a 2 K which has apolynomial-time solvable problem. Proposition 4.4.
The correspondence K -homomorphism problem is NP-complete.Proof. Assume H = 2 K has triangles abc and a (cid:48) b (cid:48) c (cid:48) . We add to each vertex v of G two separateloops e v , e (cid:48) v with labels designed to make it impossible to map v to a or a (cid:48) or b (cid:48) . This reducesthe NP-complete correspondence ( K ∪ K )-homomorphism problem to the correspondence2 K -homomorphism problem. The first loop e v will have the label ( π ( e v ) , ρ ( e v ) where π ( e v ) isthe identity and ρ ( e v ) is the involution exchanging a and a (cid:48) . The second loop e (cid:48) v will have thelabel ( π ( e (cid:48) v ) , ρ ( e (cid:48) v )) where π ( e (cid:48) v ) is the identity and ρ ( e (cid:48) v ) is the involution exchanging a and b (cid:48) .Now we claim that the resulting graph G (cid:48) has a correspondence ( K ∪ K )-homomorphism forthe target graph on b, c, c (cid:48) (isomorphic to K ∪ K ) if and only if the original graph G has acorrespondence 2 K -homomorphism. If f is a correspondence ( K ∪ K )-homomorphism forthe target graph on b, c, c (cid:48) , neither of loops creates a problem, and f remains correspondenceto H . The converse follows from the fact that the first loop forbids a and a (cid:48) for any vertex v , since neither π ( e v )( a ) = a is adjacent to ρ ( e v )( a ) = a (cid:48) nor π ( e v )( a (cid:48) ) = a (cid:48) is adjacent to ρ ( e v )( a (cid:48) ) = a , and similarly the second loop e (cid:48) v forbids a and b (cid:48) . (cid:3) The second example is the union of two isolated loops a, b and two adjacent loops c, d , i.e.,the graph H = K ∪ K ∪ K . It also does not admit a pair-deletion that results in K ∪ K . Proposition 4.5.
The correspondence ( K ∪ K ∪ K ) -homomorphism problem is NP-complete.Proof. We reduce the NP-complete correspondence ( K ∪ K )-homomorphism problem to thecorrespondence ( K ∪ K ∪ K )-homomorphism problem. Suppose G, (cid:96) is an instance of thecorrespondence ( K ∪ K )-homomorphism problem with the target graph consisting of thefollowing vertices inducing K ∪ K : an isolated loop at a and a reflexive triangle bcd . Weform a new graph G (cid:48) and labeling (cid:96) (cid:48) by replacing each edge xy of G by a path xx (cid:48) , x (cid:48) y (cid:48) , y (cid:48) y so that if (cid:96) ( xy ) = ( π, ρ ) then (cid:96) (cid:48) ( xx (cid:48) ) = ( π, π (cid:48) ), (cid:96) (cid:48) ( x (cid:48) y (cid:48) ) = ( π, π (cid:48) ), and (cid:96) (cid:48) ( y (cid:48) y ) = ( π, ρ (cid:48) ) where π (cid:48) is obtained by composing π with the involution exchanging b and c , and ρ (cid:48) by composing ρ with the same involution of b and c . (See Figure 3). We observe that the path xx (cid:48) , x (cid:48) y (cid:48) , y (cid:48) y admits a corresponding homomorphism to K ∪ K ∪ K taking x and y to any of the pairs aa, bb, cc, dd, bc, cb, bd, db, cd, dc , but never a with another vertex. It is easy to see that thisimplies that G admits an (cid:96) -correspondence homomorphism to K ∪ K on a, b, c, d if and onlyif G (cid:48) admits an (cid:96) (cid:48) -correspondence homomorphism to K ∪ K ∪ K with edge cd (and all loops).(This is a correspondence version of the ’indicator construction’ from [8, 9], where the methodis discussed in more detail.)It is now easy to check that any union of reflexive cliques other than a single clique, a unionof disjoint K ’s (i.e., a co-clique), or 2 K , can be transformed by pair-deletions to one of theNP-complete cases K ∪ K , K ∪ K ∪ K , K , and hence they are all NP-complete. (cid:3) Now consider a reflexive target graph H that is not a union of cliques. The square H of agraph H has the same vertex-set V ( H ) = V ( H ), and two vertices are adjacent in H if andonly if they have distance at most two in H . The following observation is useful. T. FEDER AND P. HELL y a’b’c’d’ a’b’c’d’ a’b’c’d’ a’’b’’c’’d’’V V V V x x’ y’ Figure 3.
A replacement for the edge xy labeled by π = a (cid:48) b (cid:48) c (cid:48) d (cid:48) and ρ = a (cid:48)(cid:48) b (cid:48)(cid:48) c (cid:48)(cid:48) d (cid:48)(cid:48) Proposition 4.6.
The correspondence H -homomorphism problem reduces to the correspon-dence H -homomorphism problem.Proof. Let G (cid:48) be obtained from G by subdividing each edge xy to be two edges xz, zy . If (cid:96) ( xy ) = ( π, ρ ) then the label of xz is ( π, zy is (1 , ρ ). Now it can beseen that G has a correspondence H -homomorphism if and only if G (cid:48) has a correspondence H -homomorphism. (We again apply the logic of the indicator construction [9], since the path xz, zy can map x and y to any edge of H .) (cid:3) It follows that if H is not connected, we can deduce the NP-completeness of the corre-spondence H -homomorphism problem from the above results on the union of reflexive cliques(with the only exceptions of 2 K and tK ). For connected H , we can apply Proposition 4.6to a sufficiently high power of H that has diameter two. Proposition 4.7. If H has diameter two but is not the reflexive path of length two, then hecorrespondence H -homomorphism problem is also NP-complete.Proof. There must, in H , be a path ab, bc where a and c are not adjacent. Suppose firstthat H − a is not a clique. Then the correspondence ( H − a )-homomorphism problem canbe assumed NP-complete by induction (on | V ( H ) | ), and it reduces to the correspondence H -homomorphism problem as follows. Suppose G is an instance of the ( H − a )-homomorphismproblem, and form G (cid:48) by adding at each vertex of G a loop labelled ( π, ρ ) where π is theidentity and ρ is the cyclic permutation ( a, b, c ). The effect of these loops is to prevent anyvertex from mapping to a , as aa is not equal to π ( u ) ρ ( v ) for any edge uv ∈ E ( H ) (but bb and cc are, and even though cb is not an edge bc is an edge, which is sufficient for a loop). If H − c is not a clique we proceed analogously. Otherwise there exists a vertex d adjacent to a, b, c .In this case we can add a loop to each vertex of G that effectively deletes b and we can againapply induction on | V ( H ) | . These loops will have labels ( π, ρ ) where π is the involution of a and b and ρ is the involution of b and c . (See Figure 4.) The argument is similar, noting thatonly bb is missing, because ca is also equal to ac for a loop. (cid:3) OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 9 (b) aacba b b bc c caV V V V x x x x (a)
Figure 4. (a) Labelling of a loop to remove a ; (b) labelling of a loop to remove b Proposition 4.8. If H is the reflexive path of length two, then he correspondence H -homomorphismproblem is NP-complete.Proof. Suppose H is the reflexive path ab, bc . We reduce from 3-colourability. Thus suppose G is an instance of 3-colourability, and form G (cid:48) by replacing each edge xy of G by two paralleledges ( xy ) , ( xy ) with permutations ( π , ρ ) on ( xy ) and ( π , ρ ) on ( xy ) . Both π and ρ are identity permutations, both π and ρ are the involutions exchanging a and b . They areillustrated in Figure 5. The effect of ( xy ) is that if x maps to a , then y maps to either b or c but not a , and if x maps to c then y maps to either a or b but not c . The effect of the secondedge ( xy ) does not preclude any of these possible images, but also restricts b to map to a or c but not b . Thus a correspondence homomorphism of G (cid:48) is a 3-colouring of G with the colours a, b, c . The converse is also easy to see. (cid:3) ac caacbb bV V V y yx (xy ) (xy ) Figure 5.
Labelling of parallel edges ( xy ) and ( xy ) This completes the proof of Theorem 1.1.5.
The General Results
We have similar results for general graphs H , where some vertices may have loops and othersnot. Note that in this case isolated loopless vertices can be removed from H without affecting the complexity of the correspondence H -homomorphism problem or list H -homomorphismproblem. Theorem 5.1.
Let H be a graph with possible loops. Suppose moreover that if H has both avertex with a loop and a vertex without a loop, then it has no isolated loopless vertices.The following cases of the correspondence H -homomorphism problem are polynomial-timesolvable. (1) H is a reflexive clique (2) H is a reflexive co-clique (3) H is a reflexive K (4) H is an irreflexive pK ∪ qK (5) H is an irreflexive K , (6) H is a star in which the center has a loop and the other vertices do not (7) H is an irreflexive pK together with a disjoint reflexive qK .Otherwise, the correspondence H -homomorphism problem is NP-complete. For the correspondence H -list-homomorphism problem, the classification is the same, exceptthe cases (3) and (5) are NP-complete. Theorem 5.2.
Let H be a graph with possible loops. Suppose moreover that if H has both avertex with a loop and a vertex without a loop, then it has no isolated loopless vertices.Then the correspondence H -list homomorphism problem is polynomial-time solvable in cases(1, 2, 4, 6, 7), and is NP-complete otherwise. Note that the graphs in cases (1-3) are reflexive, in cases (4-5) irreflexive, and cases (6-7)mix loops and non-loops.In the process of proving Theorem 5.1, we also classified the complexity of a bipartiteversion of the correspondence H -homomorphism problems and (list)- H -homomorphism prob-lems. Specifically, we assume that H is a bipartite graph with a set of black vertices anda disjoint set of white vertices . The by-side correspondence H -homomorphism problem askswhether an input bipartite graph G (with edge-labeling (cid:96) ) admits an (cid:96) -correspondence homo-morphism to H taking black vertices of G to black vertices of H and white vertices of G towhite vertices of H . The by-side correspondence H -list homomorphism problem asks whetheran input bipartite graph G (with edge-labeling (cid:96) and lists L ( x ) , x ∈ V ( G ), such that for blackvertices x the lists L ( x ) contain only black vertices, and similarly for white vertices) admitsan (cid:96) -correspondence list homomorphism to H . Theorem 5.3.
Let H be a bipartite graph. Then the by-side correspondence H homomorphismproblem is polynomial-time solvable in case (4) above, as well as (8) H is a complete bipartite graph plus any number of isolated vertices, (9) H is a tree of diameter plus any number of isolated vertices, (10) H consists of two disjoint copies of K , with white leaves, plus any number of blackisolated vertices, (11) H consists of two disjoint copies of K , .In all other cases it is NP-complete. OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 11
For the list version we have the following result.
Theorem 5.4.
Let H be a bipartite graph. Then the by-side correspondence H -list homo-morphism problem is polynomial-time solvable in cases (4, 8, 9) above, and is NP-completeotherwise. In the Appendix, we sketch the proofs for Theorems 5.1, 5.2, 5.3, and 5.4.
Acknowledgement
The second author wishes to acknowledge the research support of NSERC Canada, througha Discovery Grant.
References [1] A. Bernshteyn, A. Kostochka, and X. Zhu, Fractional DP-colorings of sparse graphs, arXiv:1801.07307v1[math.CO][2] Z. Dvoˇr´ak and L. Postle, Correspondence coloring and its application to list-coloring planar graphswithout cycles of lengths 4 to 8, J. Combin. Theory B 129 (2017).[3] T. Feder and M. Y. Vardi, The computational structure of monotone monadic SNP and constraintsatisfaction: a study through Datalog and group theory, SIAM J. Computing 28 (1998) 57–104.[4] T. Feder and P. Hell, List homomorphisms to reflexive graphs, J. Combinatorial Theory B 72 (1998)236–250.[5] T. Feder and P. Hell, Complexity of correspondence homomorphisms, arXiv:1703.05881v2 [cs.DM][6] T. Feder, P. Hell, and J. Huang, List homomorphisms and circular arc graphs, Combinatorica 19 (1999)487–505.[7] T. Feder, P. Hell, and J. Huang, Bi-arc graphs and the complexity of list homomorphisms, J. GraphTheory 42 (2003) 61–80.[8] P. Hell and J. Neˇsetˇril, On the complexity of H –colouring, J. Combin. Theory B 48 (1990) 92–110.[9] P. Hell and J. Neˇsetˇril, Graphs and Homomorphisms , Oxford University Press 2004.[10] S. Hoory, N. Linial, and A. Wigderson, Expander graphs and their applications, Bull. Amer. Math. Soc.43 (2006) 439–561.[11] T. Wang, Every toroidal graph without triangles adjacent to 5-cycles is DP-4-colorable, manuscript 2018. Appendix
We sketch the proofs of all statements in Theorems 5.1, 5.2, 5.3, and 5.4, by case analysis.
Suppose H is a bipartite graph. If H contains a connected component C with | C ∩ U | ≥ | C ∩ V | ≥
3, say the former. Then H is reflexive and contains at least the two componentson C ∩ U and C ∩ V . By a reduction from Theorem 1.1 the correspondence H -homomorphismproblem is NP-complete. If H contains at least two components with at least one edge each,then H has at least four components which must all be loops or isolated vertices, else thecorrespondence H -homomorphism problem is again NP-complete by Theorem 1.1. This coversall H except case (4), which is easily solved by a trivial algorithm (similar to the case of areflexive co-clique in Theorem 1.1), even with lists.If H contains only one component C other than an isolated vertex, then the remaining casesare (a) C is a path uvu (cid:48) , (b) C is a path uvu (cid:48) v (cid:48) , and (c) C is K , . In case (a) H contains loopedcomponents of size one and two, which has NP-complete correspondence H -homomorphismproblem by Theorem 1.1. In case (b) we introduce a loop with the permutations ( uvu (cid:48) v (cid:48) )and identity, which eliminate only v (cid:48) , since v (cid:48) u is not an edge of H , leaving us with the path uvu (cid:48) which has NP-complete correspondence H -homomorphism problem by case (a). (SeeProposition 4.3 for a similar technique of loop addition.) In case (c) H has two reflexivecomponents of size two plus isolated vertices. An isolated vertex can simulate lists in C , soeither with lists or with at least one isolated vertex, case (c) is NP-complete by Theorem 1.1.Finally, the case (5) H = K , the correspondence H -homomorphism problem is polynomialagain by Gaussian elimination. If we view H as K , with parts { , } , { , } ), then wehave the adjacency ( xy )( zt ) ifand only if z = x + 1 modulo 2, and for a permutation ρ wehave ρ ( xy ) = zt if and only if zt = ρ (00) + ( ρ (10) − ρ (00))( xx ) + ( ρ (01) − ρ (00))( yy ) modulo2. Thus the problem again reduces to linear equations modulo 2. We now consider the by-side problems . Suppose H contains an induced path uvu (cid:48) v (cid:48) u (cid:48)(cid:48) .We show that the by-side correspondence H -homomorphism problem is NP-complete. If uu (cid:48)(cid:48) is not an edge in H , then the corresponding component H ∩ ( U × U ) is reflexive but notcomplete, so the problem is NP-complete by Theorem 1.1; otherwise there is a path u (cid:48)(cid:48) v (cid:48)(cid:48) u ,giving a cycle uvu (cid:48) v (cid:48) u (cid:48)(cid:48) v (cid:48)(cid:48) plus possibly the edge u (cid:48) v (cid:48)(cid:48) . Consider the permutation ρ = ( uu (cid:48) ) andjoin two variables x, y by an edge labelled by two identity permutations, and another paralleledge labelled by ( ρ, ρ ). If the edge u (cid:48) v (cid:48)(cid:48) was present, then the edge u (cid:48) v (cid:48) will be removed, andwe still have the induced path v (cid:48) u (cid:48)(cid:48) w (cid:48)(cid:48) uv , and we can repeat the process. Otherwise u (cid:48) v (cid:48)(cid:48) isabsent. If there is a vertex w (cid:54) = u (cid:48) and a path vwv (cid:48) , there will again remain the induced path uvwv (cid:48) u (cid:48)(cid:48) , and we can repeat. Otherwise no vertex is adjacent to the two vertices of the 6-cycle.If there is an edge vw , with w not in the 6-cycle, then the edge wu (cid:48)(cid:48) is in H , so there is apath vww (cid:48) u (cid:48)(cid:48) , and there will remain the induced path vww (cid:48) u (cid:48)(cid:48) v (cid:48) , again we can repeat. Finallyif the component is just the 6-cycle, there will remain at least components uvu (cid:48) , v (cid:48) u (cid:48)(cid:48) v (cid:48)(cid:48) , whichfor H ∩ ( U × U ) gives components of uu (cid:48) , u (cid:48)(cid:48) for which the correspondence H -homomorphismproblem is NP-complete from Theorem 1.1.In the remaining case, for each component C , if u, u (cid:48) ∈ U ∩ C and v, v (cid:48) ∈ V ∩ C , thenone of u, u (cid:48) dominates the other (similarly one of v, v (cid:48) dominates the other). Otherwise thereexists an induced subgraph uv, u (cid:48) v (cid:48) and a path uwu (cid:48) , giving an induced path vuwu (cid:48) v (cid:48) . Orderthe vertices u , u , . . . , u k and v , v , . . . , v (cid:96) so that u i dominates u i +1 , and v j dominates v j +1 . OMPLEXITY OF CORRESPONDENCE HOMOMORPHISMS 13
Let a i , b j be the degrees of u i , v j respectively. If a > a k , the by-side correspondence H -homomorphism problem is NP-complete. If H has more that one component, then H hasa component of size three or more, so NP-completeness follows as in the reflexive case. Wethus assume H is connected. We may assume a = a and a = a k . Indeed, if we let π be theidentity and ρ = ( u i u i (cid:48) ) with i < i (cid:48) and in parallel another edge with both πρ the identity, thisreduces the neighbors of u i to those of u i (cid:48) . Note that b > b (cid:96) , and we may similarly assume b = b (cid:96) . Then xy ∈ E ( H ) if and only if x ∈ { a , a } ∨ y = b i . By permutation, we also get x ∈ { a , a } ∨ y (cid:48) = b i (cid:48) and x ∈ { a , a } ∨ y (cid:48) = b i (cid:48)(cid:48) . This gives y = b i ∨ y (cid:48) = b i (cid:48) ∨ y (cid:48)(cid:48) = b i (cid:48)(cid:48) with i, i (cid:48) , i (cid:48)(cid:48) ∈ { , } . Thus there is a natural reduction from 3-satisfiability, and the by-sidecorrespondence H -homomorphism problem is NP-complete.Assume H is connected. Then the remaining cases are (3) a = a k and b = b (cid:96) , and (4) a > a = a k and b > b = b (cid:96) . In both these cases the by-side correspondence H -list-homomorphism problem is polynomial. For (3), any choice from the lists solves the problem.For (4), xy ∈ E ( H ) iff x = a ∨ y = b . By permutation we get boolean clauses x i ∨ y j where x i stands for x = a i and y j stands for y = b j . We also add clauses x i ∨ x i (cid:48) for i (cid:54) = i (cid:48) y j ∨ y j (cid:48) for j (cid:54) = j (cid:48) . Solve this instance of 2-satisfiability. This gives a partial assignment x = a i , y = b i tovariables x, y ; the remaining variables may be assigned arbitrarily.If H is not connected, in case (4) the by-side problem follows from the bipartite versionabove, which allows lists. Cases (8),(9) are still polynomial with lists; adding an isolatedvertex to either or both U, V just simulates lists on the appropriate side. By Theorem 1.1applied to H , these three are the only polynomial by-side cases with lists or or those withoutlists but with only one component having more than one vertex.The remaining by-side cases have H having two components plus isolated vertices, wherethe components are (*) either K , , or a path P on four vertices, with no isolated vertices; and(10) two K , (with white leaves) plus isolated vertices on the black side. Any P can have oneside permuted in parallel with the P , giving K , plus one isolated vertex, making this case of(*) NP-complete. Suppose H is the disjoint union of H , and H obtained from H by flippingthe two sides as a by-side correspondence H -homomorphism problem. Then H can be viewedas a standard (not by-side) correspondence H -homomorphism problem, which if polynomialmakes the by-side correspondence H -homomorphism problem polynomial. Thus the remainingcase (11) of two K , , is polynomial for the by-side correspondence H -homomorphism problem.Finally (10) is polynomial with one boolean variable for the black vertices and two booleanvariables for the white vertices. For graphs H that have both loops and non-loops, we may assume H has no isolatedloopless vertices. We create a bipartite graph H (cid:48) = ( U, V, E ) that has for each edge ww (cid:48) in H ,two edges uv (cid:48) and vu (cid:48) . If the by-side problem (with or without lists) for H (cid:48) is NP-complete,then so is the corresponding problem for H . The only cases where both loops and non-loopsare present are (6, 7), both easily checked to be polynomial even with lists.
268 Waverley Street, Palo Alto, CA 94301, USA
E-mail address : [email protected] School of Computing Science, Simon Fraser University, Burnaby, B.C., Canada V5A 1S6
E-mail address ::