Non-homogenizable classes of finite structures
A relational structure with a countable domain is called homogeneous if it is highly symmetric in the precise technical sense that any isomorphism between any two of its finite induced substructures extends to an automorphism of the whole structure. In many areas of combinatorics, logic, discrete geometry, and computer science, homogeneous structures abound, often in the form of nicely behaved limit objects for classes of finite structures. Typical examples include the Rado graph , which can be seen as the limit of the class of all finite graphs; the linear order of the rational numbers , seen as the limit of all finite linear orders; or the countable Urysohn space , the limit of all rational metric spaces. The literature on the subject is very extensive; we refer the reader to  for a recent survey.
Homogeneous structures arise as limits of well-behaved classes of finite structures in a way made precise by Fraïssé’s theorem, which describes them combinatorially in a finitary manner. The theorem states that a homogeneous structure is characterized, up to isomorphism, by its age, i.e., the class of its finite induced substructures. Moreover, classes of finite structures arising as ages of homogeneous structures are precisely Fraïssé classes, i.e., classes closed under taking induced substructures and under amalgamation – a form of glueing pairs of structures along a common induced substructure (see  and Section 2 for precise definitions).
Thanks to Fraïssé’s theorem, combinatorial arguments involving finite structures can often be replaced by, or aided by, arguments involving highly symmetric, infinite structures. In combinatorics, for example, homogeneous structures appear unavoidably in structural Ramsey theory . At the intersection between combinatorics and computer science, homogeneous structures appear in the theory of logical limit laws for various models of random graphs . In computer science proper, homogeneous structures appear in the theory of constraint satisfaction problems , and automata theory , and verification .
One of the advantages of working with homogeneous structures, rather than classes of finite structures, is that their automorphism groups are very rich. For example, over a finite relational signature, the homogeneity of the structure immediately implies that, up to automorphism, it has finitely many elements, pairs of elements, triples, etc. In model theoretic terms, this means that the structure is -categorical by the classical Ryll-Nardzewski theorem, and its first-order theory admits elimination of quantifiers. In turn, since in any such structure there are only finitely many first-order definable relations of each arity, homogeneous structures over finite relational signatures are, in a strong technical way, close to being finite.
Thus, with Fraïssé’s theorem in hand and the many applications of homogeneous structures in mind, it becomes quite important a task to identify more Fraïssé classes. More generally, one would like to identify classes of finite structures that are perhaps not Fraïssé classes themselves, but appear as reducts of some Fraïssé class over a richer yet finite signature. Such classes of finite structures are called homogenizable . The point in case is that the lifted Fraïssé class can be thought of as taming its reduct by providing a homogeneous structure that plays the role of limit object for it. Many of the application examples mentioned above do actually go through lifted Fraïssé classes and their corresponding homogeneous Fraïssé limits. See  and the references therein for a discussion on this.
A noticeable amount of work has gone into providing sufficient conditions for a class of finite structures to be homogenizable. Instances include the model-theoretic methods of Covington , and the combinatorial explicit constructions of Hubička and Nešetřil. Here we provide a combinatorial necessary condition for homogenizability (Theorem ? in Section 3). This allows us to prove that certain natural classes of finite structures previously considered in the literature are not homogenizable.
Our first example of a non-homogenizable class comes from the theory of constraint satisfaction problems (CSPs). We show that the class of locally consistent systems of linear equations over the two-element field is not homogenizable. More generally, the result holds for systems of equations over any finite Abelian group. This answers a question first raised by the first author of this paper in . Precisely, by a locally consistent system of equations we mean one whose satisfiability cannot be refuted by the -consistency algorithm for small and , which is a well-studied heuristic algorithm for solving CSPs. Moreover, in combination with the resolution of the Bounded Width Conjecture by Barto and Kozik , this shows that the constraint languages whose classes of locally consistent instances are homogenizable are, up to pp-interpretability, precisely those that are solvable by local consistency methods. All this is worked out in Section 4.
In Section 5 we give a second example of a non-homogenizable class that, in this case, is motivated by the works of Hubička and Nešetřil , and Erdös, Tardif, and Tardos . It was shown in  that every class of finite structures that is of the form , where is a regular class of connected finite structures, is homogenizable. In words, is the class of finite structures that do not admit homomorphisms from any structure in . The notion of regularity considered in  is closely related to the notion of regularity in automata theory, and agrees with it on coloured paths and trees. However, our second example shows that even if is MSO-definable and has maximum treewidth two, the class need not be homogenizable. This shows that for MSO-definable classes, treewidth one versus two of the forbidden structures in is the dividing line to homogenizability.
Signatures, structures, reducts, and expansions. A relational signature is a set of relation symbols , each with an associated natural number called its arity. In this paper, we consider only finite relational signatures. A -structure is composed of a set , called its domain, and a relation on for each in , where is the arity of . We say that is the interpretation of in . We write to denote the cardinality of the domain of . A -structure is sometimes referred to as a structure over the signature . If is a signature that contains and is a -structure, then the -reduct of is the structure obtained from by forgetting all relations from . In this case, we also say that is an expansion of . Expansions and reducts are also called lifts and shadows, respectively.
Substructures, homomorphisms, and embeddings. If is a -structure and is a subset of its domain , we write for the substructure of induced by , that is, the -structure with domain in which each relation symbol in is interpreted by , where is the arity of .
Let and be structures over the same relational signature . Let and denote their domains. A homomorphism from to is a mapping for which the inclusion holds for every in . The homomorphism is strong if in addition the inclusion holds for every in , where is the arity of . A monomorphism from to is an injective homomorphism. Whenever is a subset of and the inclusion mapping is a monomorphism, we say that is a substructure of . An embedding from to is an injective strong homomorphism. Whenever is a subset of and the inclusion mapping is an embedding, we say that is an induced substructure of . An isomorphism from to is a surjective embedding. If there is an isomorphism from to we say that the two structures are isomorphic. If is a partial mapping with domain and image , we say that is a partial homomorphism from to if it is a homomorphism from to . We write to denote the set of all embeddings from to . Sometimes we write to mean that is a mapping from the domain of to the domain of .
Amalgamation. If and are -structures with domains and , we write for their union, i.e. the -structure with domain and relations for every in . Let and be embeddings from the same structure into structures and , respectively. The structure is an amalgam of and through and if there exist embeddings and from to and to , respectively, such that the diagram in Figure 1 commutes, i.e., .
We say that is a strong amalgam if , where , and denote the domains of , and , respectively. We say that is a free amalgam if it is strong and, additionally, . We also say that is the union of and amalgamated along through and via and . Note that the free amalgam of and through and is uniquely defined up to isomorphism, and is isomorphic to the disjoint union of and , quotiented by the equivalence relation identifying with , for in . We denote this free amalgam . When and are implicit, we denote it . We also say that and are glued along .
Classes of structures. All our structures will have finite or countably infinite domain. Moreover we assume that all structures have a domain that is a subset of a common background countable set, say . For a fixed signature , a class of structures is a set of structures that is closed under isomorphisms, i.e. if and are isomorphic structures and belongs to the class, then also belongs to the class. A class of structures is closed under amalgamation if for every two embeddings and from the same structure in into structures and in , there exists in an amalgam of and through and . A class of finite structures is an amalgamation class, also called a Fraïssé class, if it is closed under taking induced substructures and amalgamation. For example, the class of all finite graphs is an amalgamation class – in fact, it is closed under free amalgamation – so is the class of all finite digraphs. The class of all finite linear orders is also an amalgamation class, although it is not closed under free amalgamation. Fraïssé’s theorem states that a class is Fraïssé if and only if it is the class of finite induced substructures of a homogeneous structure.
For two signatures and with the second containing the first, if and are classes of -structures and -structures, respectively, then we say that is the -reduct of if is the class of -reducts of the structures in .
Homogenizable classes. We say that a class of -structures is homogenizable if there is a signature extending , and an amalgamation class of -structures, such that is the -reduct of . For a class of -structures , let denote the class of all finite -structures such that for no in there is a homomorphism from to . Hubička and Nešetřil define a notion of regularity, which we call HN-regularity (we omit its technical definition), and prove in Theorem 3.1 from  that if is a HN-regular class of finite connected structures, then is homogenizable. In particular, if is finite, then is homogenizable.
3Necessary condition for homogenizability
Fix a finite relational signature . Except for the examples, in this section all structures are over this signature, or over a signature that extends . Before we state the necessary condition for homogenizability we need some notation and terminology.
Let be a class of finite structures. If , and are structures in , and and are embeddings such that no amalgam of and through and is in , then we say that is a diagram that witnesses failure of amalgamation of . We illustrate the definitions with a running example.
Let be a diagram that witnesses failure of amalgamation of . For a structure and a partial mapping , let be the structure that is obtained by glueing to , for each in , a fresh copy of either or depending on whether or . More formally, is defined by induction on the cardinality of the domain of : if , then ; otherwise, if , where and , then define , where is composed with the identity embedding from to .
For a natural number and a -structure with domain , let denote the structure with domain in which the interpretation of a relation in of arity is the set of all tuples where and . Observe that every function induces an embedding , defined by . Let denote the set of all embeddings of the form for . In particular, is a subset of containing exactly embeddings.
A diagram is confusing for if the following conditions hold:
it witnesses failure of amalgamation of , and
for every natural number , if , then for every coloring the structure belongs to the class .
Its order is the cardinality of the domain of .
This theorem is the main technical result of this paper. Before we prove it, we illustrate it by applying it to our running example.
Theorem ? follows easily from Lemma ? stated below.
Let witness failure of amalgamation of . An -confusion for is a structure in , together with a set , such that is in for every coloring . For and a natural number bounded by the cardinality of the domain of , let denote the set of all restrictions of in , where ranges over all -element subsets of the domain of .
Before we prove Lemma ? we show how Theorem ? follows from it.
Suppose that has confusing diagrams of arbitrarily large order. For every two fixed natural numbers and , we apply Lemma ? to conclude that is not a reduct of an amalgamation class over a signature with symbols of arity at most . Let and be as in the statement of the lemma. Consider a confusing diagram and let be its order. Fix a natural number , and let and . Then is an -confusion for , by the definition of confusing diagram, and and . Since the order of the diagram can be chosen arbitrarily large, we can assume . Taking large enough, so that and , we get:
which gives condition in Lemma ?. Since and were arbitrary, this proves that is not the reduct of an amalgamation class.
It remains to prove the lemma.
Fix natural numbers and . In anticipation of the proof, let be the maximum number of atomic types of -tuples over any signature with at most predicates of arity at most , and let . Suppose that is a class of -structures as in the lemma, with a diagram that witnesses its failure of amalgamation, and an -confusion satisfying condition from Lemma ?.
Let be a -structure with domain and let be a function from some set to . Define the pullback as the -structure with universe in which the interpretation of a relation symbol of of arity is , i.e., the inverse image of the interpretation of in under the mapping . By definition, is the unique -structure on for which is a strong homomorphism into .
By definition of the structure , there is a distinguished embedding of into . Therefore, by composition, any embedding in defines an embedding of into , denoted . Note that for any expansion of , the pullback is an expansion of , which is isomorphic (via ) to an induced substructure of .
We show how the claim yields the lemma. Figure ? illustrates the proof.
Assume that is the class of -reducts of a class of -structure . To reach a contradiction, suppose that is closed under induced substructures and amalgamation. Let be as in the claim. Since belongs to by the definition of confusion, there exists an expansion of in . Let and be as in the conclusion of the claim, and suppose without loss of generality that and . By the definition of , the embeddings and induce embeddings , such that the diagram to the left below commutes:
Let be the pullback structure; this structure is an expansion of . Moreover, belongs to the class , since it is a pullback under an injective mapping, and hence is isomorphic to an induced substructure of , which is in .
Similarly, the embeddings and induce embeddings such that the diagram to the right above commutes. Let be the pullback structure, which is an expansion of , isomorphic to an induced substructure of , hence belongs to the class .
Let be the pullback , which, by the claim, is the same as the pullback . Note that by commutativity of the diagram to the left above, the pullback is the same as the pullback . Similarly, is the same as . In other words, is an embedding of into , and is an embedding of into . Since is closed under amalgamation, there exists an amalgamation of the diagram and , which consists of a structure in and two embeddings and . Taking -reducts, we obtain an amalgamation in of and . But the pair of embeddings and were supposed to witness failure of amalgamation in – a contradiction proving that cannot be closed under amalgamation.
Next we show how to prove Claim ? and hence Lemma ?. Call any embedding in a spot, and any restriction of a spot to an -element subset of the domain of a partial spot. For each coloring of the spots, and each two spots and , define if and only if . For each coloring of the partial spots, and each two spots and , define if and only if for every -element subset of the domain of . Both are equivalence relations on spots.
For this proof, let be the cardinality of the domain of and assume without loss that the domain of is . Let be the number of spots and let be the number of partial spots. With this notation, condition reads as follows:
Color the spots independently at random with either or , each with probability . Let be the random variable describing this process. In particular, is a random variable taking as values strings of length over alphabet , each with the same probability. Thus the binary entropy of the random variable is equal to .
Suppose for contradiction that the opposite of what the claim states holds. Then there is a random variable taking as values colorings of the partial spots using colors such that the inclusion holds with probability . The relation has at most equivalence classes; for each fixed spot , there are at most choices of colors for each of the restrictions to -element subsets , and any two spots sharing these choices are equivalent. In particular, there is a random variable taking as values strings of length over alphabet such that and determine . That is, , or equivalently,
We will show that this is impossible by proving that
Indeed, is determined by random variables , where the random variable takes as values the colorings of the restrictions of the spots to the subset . If denotes the number of such restrictions, the random variable takes values in , and therefore
Noting that is the sum of as ranges over all -element subsets of , it follows that
Moreover since takes as values strings of length over alphabet . Hence
However states that this quantity is strictly smaller than , as required.
Finally we use Claim ? to prove Claim ?. Let be the coloring of Claim ? with the two colors interpreted as the embeddings and . For each expansion of , let be the coloring of partial spots defined as follows: for each spot and each -element subset of the domain of , let be the atomic type of in . This is a coloring of partial spots using at most colors. By Claim ?, there is a pair of spots and such that but . From and the fact that is at least as large as the maximum arity of any new predicate in , it follows that the pullbacks and are equal. On the other hand, from we get by definition. This proves Claim ? and Lemma ?.
4Classes of consistent structures
In this section we work out the first of our two examples of non-homogenizable classes. We start by defining some basic notions from the theory of constraint satisfaction problems as described, for example, in Chapter 6 of the monograph . Recall that, for a structure , we write for the class of all finite structures over the same signature as for which there is a homomorphism from to . The ’s are called instances, the ’s are called templates.
Let be a relational signature, let and be -structures, and let and be integers such that . A -consistent family on and is a non-empty family of partial homomorphisms from to , such that the following three conditions hold for each in :
if is a subset of , then is in ,
if and is a subset of such that and , then there exists in such that and .
If there is a -consistent family on and , then we say that is -consistent with respect to . Note for later use that the class of structures that are -consistent with respect to is closed under inverse homomorphisms: if there is a homomorphism from to , and is -consistent with respect to , then is also -consistent with respect to . To see this, it suffices to compose the homomorphism from to with each partial homomorphism in the -consistent family for to get a -consistent family for .
We describe the special case of -consistency in terms of a pebble game. The game is played between spoiler and duplicator, each having three pebbles, numbered , and . Spoiler can place his pebbles on the nodes of , while duplicator can place his pebbles on the nodes of . They can also keep the pebbles in their pockets, in which they have all pebbles at the beginning of the game. The game proceeds in rounds as follows. In each round, spoiler places some of the pebbles from his pocket on the nodes of and duplicator replies by placing his corresponding pebbles on the nodes of . If the partial mapping defined by the pebble placement is not a partial homomorphism from to , then duplicator loses. Otherwise, spoiler puts back some of the pebbles into his pocket, and duplicator removes the corresponding pebbles, and the game continues to the next round. It is not hard to see that is -consistent with respect to if and only if duplicator can avoid losing forever.
4.2Systems of linear equations over
We define a finite template that can be used to represent the solvability of systems of linear equations over the 2-element field. Let us note that our definition of the template will not be the standard one as it can be found, for example, in the original Feder-Vardi paper . The main difference is that we want to have a signature of smallest possible arity, in this case two. We achieve this by letting be the natural encoding of the standard template as its incidence structure. Concretely, is defined as follows. Its domain is , where
The elements of are called values, and those of are called triples. The signature includes three partial functions , , and that map triples in to values in , and four unary relations value, triple, and . Formally, in order to have a relational structure, has binary relations that correspond to the graphs of the partial functions , and . The interpretations of the symbols in are as follows:
, and map in to , and , respectively,
holds of all elements in ,
holds of all elements in ,
holds of in , and
holds of in .
The purpose of triple is to encode equations of the type , and the purposes of and are to encode equations of the type and , respectively. Note that even though the language does not allow writing more complicated equations, such as or , such equations can be simulated in the language of with the help of auxiliary variables.
In the following, we fix the template , and when we refer to consistency, we mean -consistency with respect to . Finite structures on the signature of are called instances. Homomorphisms from an instance to are called solutions. By denote the class of consistent instances. Observe that, as noted earlier, the class of consistent instances is closed under inverse homomorphisms.
The plan is to apply Theorem ? to , and for that we need to find a confusing diagram with arbitrarily large .
Let be an exact power of two. Let be a rooted, ordered tree with leaves at depth ; in particular, no node at depth is a leaf, and no node at depth is a root. Let be the instance obtained from , with elements of two types: nodes, which correspond to the nodes of , and triples, which correspond to triples , where is an internal node in , and and are its left and right sons, respectively. Nodes are labeled by the unary predicate value and triples are labeled by the unary predicate triple. We say that the triple is the triple below node , and is adjacent to, or contains , , and . For each such triple, we declare:
We call a structure of this kind simply a tree. Since we will work with -structures that are made of trees, for the sake of intuition from now on we use the names , , and in place of , , and . If is a value in , then the -marking of is the -structure obtained from by marking the root by the predicate . Observe that in any solution of , the sum of the values of the leaves is equal to modulo . Conversely, any mapping from the leaves of to such that the sum of the values of the leaves is equal to modulo extends uniquely to a solution .
The structures and are the markings and of the tree , respectively. The structure is the substructure of induced by the leaves of the tree. Note that consists of isolated points, labeled by the unary relation value. The unary relations triple, and , as well as the binary relations , , and , are empty in . Note that and share as an induced substructure. Let and be the corresponding embeddings.
When spoiler has only two pebbles on the board, we allow him to perform a move we call a slide, in which he moves one pebble from a node to a triple adjacent to it, or from a triple to a node belonging to this triple. Duplicator has to respond accordingly: if spoiler slides a pebble from a node to a triple containing on the -th coordinate, then duplicator must move his corresponding pebble from a value in to a triple in containing on the -th coordinate. Symmetrically, in the case when spoiler slides his pebble from a triple to the node in the -th coordinate, duplicator must move his corresponding pebble from the corresponding triple to the value on its -th coordinate. The slide moves can be simulated in the original game, using a third pebble.
Denote the two (overlapping) trees and , respectively; they have common leaves in the free amalgam . Here is the strategy for spoiler; it consists of several steps. In the beginning of the -th step, spoiler has two pebbles placed on corresponding nodes and of and , at depth of the tree. In particular, in the beginning of the first step, two pebbles are placed on the roots of and , respectively. For a node on which spoiler has his pebble, denote by the value of the corresponding pebble of duplicator. The invariant is that . This invariant is clearly satisfied in the beginning of the first step, since has its root labeled with and has its root labeled with .
In the -th step, spoiler slides his pebble from node to the triple below in , and then slides his pebble from node to the triple below in . Duplicator’s responses have to satisfy and . In particular, , by the invariant. It follows that , so either or (or both). Since the cases are symmetric, suppose without loss of generality that the first case occurs. Then spoiler slides the pebble from to and then slides the pebble from to , and continues the game from these two nodes playing the role of and . The invariant is satisfied.
Since in each step the depth of increases by , at some point, must be a leaf of , and is the corresponding leaf in . But then and are the same element in , and by the invariant . In other words, spoiler has two pebbles placed at the same node of , but the corresponding pebbles of duplicator are not placed on the same element of . So duplicator loses.
This follows at once from the previous lemma and the fact that is closed under inverse homomorphisms. Indeed, the free amalgam through and maps homomorphically to any amalgam of and through and .
Let be a natural number, and let and .
We modify the game, by giving more power to spoiler. We show that even in this game, duplicator wins. In the modified game, the pebbles of spoiler can be placed only on triples of , and the pebbles of duplicator can be placed only on triples of . If the pebbles of spoiler are placed on triples , with , then duplicator must have his corresponding pebbles placed on triples in , so that the following conditions are satisfied:
Whenever is a triple containing a node with unary predicate on some coordinate, then the same coordinate of is equal to .
Whenever and agree on some coordinate, then and also agree on the same coordinate.
We show how spoiler can copy a strategy which is winning in the original game to win in the modified game.
Suppose that in the original game spoiler places a pebble on a node . We copy this move in the modified game by placing a pebble on any triple containing on some coordinate, say, the -th coordinate, and await the response of duplicator. If in the modified game duplicator places his corresponding pebble on a triple in , then we pretend that the duplicator in the original game places his pebble on the -th coordinate of , and the game continues. At some point, duplicator loses in the original game. This means that one of two cases occurred in the original game:
Spoiler has placed a pebble on a node with unary predicate and duplicator replied by placing his corresponding pebble on a value with .
One pebble of spoiler is placed on a node and another pebble of spoiler is placed on a triple containing on the -th coordinate, and the corresponding pebbles of duplicator are placed on a value and a triple that, however, do not satisfy the condition that the -th coordinate of equals .
Since duplicator is only copying his strategy from the modified game, it must be the case that duplicator must have lost as well in the modified game. In particular, if spoiler wins in the original game, then he wins in the modified game.
We show a winning strategy for duplicator in the modified game on . By the claim above, this means that duplicator also has a winning strategy in the original game.
The arena on which spoiler places his pebbles is a union of trees of the form glued along the leaves. Therefore, it is meaningful to talk about roots, children (or sons), brothers, and leaves, and parents in the case of nodes from trees which are not roots nor leaves (leaves have very many parents). Every triple in is of the form , where is an internal node of some tree, and and are its left and right son.
Call two tree nodes and in congruent, and write , if the following conditions hold:
The nodes correspond to the same node in ,
The leaves below coincide with the leaves below .
We lift this notion to triples: two triples and are congruent, also written , if , , and . Observe that two distinct roots in are not congruent, since by construction, not all their leaves are identified.
During the game, let , with , denote the triples on which the pebbles of spoiler are placed. Let denote the set of nodes that are congruent to some component of some pebbled triple, and let denote the union of with the roots. We say that a function is nice if it satisfies the following conditions.
For every triple in , if , then .
For every root , if is marked with unary predicate , then .
Whenever are congruent, then .
We show that duplicator has a strategy which satisfies the following invariant at each moment of the game:
There is a nice function such that for each pebble of spoiler occupying a triple , duplicator’s corresponding pebble occupies the triple .
At the beginning of the game, the invariant is satisfied: since consists only of roots, we can define for a root with unary predicate , yielding a nice function – the last condition of nicety holds since no two distinct roots are congruent.
Suppose that at some moment during the game there is a function satisfying the above conditions, and spoiler performs a move. If in this move he removes a pebble from some triple, then duplicator responds by removing the corresponding pebble from , and it is easy to see that the restriction of to the resulting set satisfies the above conditions.
Suppose now that spoiler makes his move by placing a new pebble on the board. In particular, before the move he had pebbles on triples , and a new pebble is placed on the triple , which we denote for simplicity. Below, unless indicated, when we speak about , , or , we refer to their values just before spoiler placed the new pebble on . The case that is a triple with is trivial: duplicator just responds with . This response is not loosing thanks to the invariant and the first two conditions of the nicety of . Moreover, the values of and after duplicator’s response are unmodified, so the same function can be used in the invariant. From now on we assume that at least one of the coordinates of is not in .
Note that after spoiler’s move, the new includes the congruence classes of the three components of . We say that a triple is completed by spoiler’s move if not all three components of the triple are in before spoiler’s move, but the addition of these congruence classes to makes all three components of the triple belong to the new . In particular, and its congruents are completed by spoiler’s move. The new after spoiler’s move will be defined to extend the old by assigning values to the components of and its congruents in such a way that the conditions of nicety are satisfied for the new . We need to distinguish several cases:
Case 1: is a triple in which and are not leaves, and is already in . Let and be the left and right sons of , and let and be those of . We need the following claim:
Since not all three , and are in but is in , at most one among and is in . It follows that not all four , , , and can be in . To argue for this, note that at most two pebbles occupy at most two triples and before spoiler’s move, but it cannot be the case that and if not both and are in . Moreover, for the same reason, if both and are in , then is not in , and if both and are in , then is not in . We use this to choose and by cases.
Case (i): both and are in . First choose to satisfy condition 4 and then choose to satisfy condition 3. Note that in case condition 1 also applies, then the only choice of that makes condition 4 hold is guaranteed to satisfy condition 1 too by the first condition of nicety of . Note also that in this case conditions 2 and 5 do not apply.
Case (ii): both and are in . First choose to satisfy condition 5 and then choose to satisfy condition 3. Again, note that in case condition 2 also applies, then the only choice of that makes condition 5 hold is guaranteed to satisfy condition 2 too by the first condition of nicety of . Note also that in this case conditions 1 and 4 do not apply.
Case (iii): otherwise. In this case the only conditions that can apply are 1, 2, and 3, and among 1 and 2 at most one can apply. In case 1 applies and is in , first choose to satisfy condition 1 and then choose to satisfy condition 3. In case 2 applies and is in , first choose to satisfy condition 2 and then choose to satisfy condition 3.
Case 2: is a triple in which and are not leaves, and is not yet in . Let and be the left and right sons of , and let and be those of . Since is not in , it is not a root. Let then be the sibbling of , and let be their parent. We need the following claim:
If both and are in , we argue that and are not in . To see this, note that at most two pebbles occupy at most two triples before spoiler’s move. But if both and are in , then these triples must contain nodes that are congruent to and , and be different and hence different from any triple that contains a node congruent to or , since all triples that contain both and are congruent to . Thus, in case both and are in , we choose and , and to satisfy condition 3. Conditions 1, 2 and 3 are then true by construction, condition 4 does not apply, and conditions 5 and 6 hold because is nice with respect to .
Assume then that not both and are in . In such a case we argue that not all four , , , and can be in . To see this, note again that at most two pebbles occupy at most two triples and , and it cannot be that and if not both and are in . Moreover, for the same reason, if both and are in , then is not in , and if both and are in , then is not in . We use this to choose and by cases. In all cases we first choose to satisfy condition 4.
Case (i): both and are in . First choose to satisfy condition 5 and then choose to satisfy condition 3. Note that in case condition 1 also applies, then the only choice of that makes condition 5 hold is guaranteed to satisfy condition 1 too by the first condition of nicety of . Note also that in this case conditions 2 and 6 do not apply.
Case (ii): both and are in . First choose to satisfy condition 6 and then choose to satisfy condition 3. Again, note that in case condition 2 also applies, then the only choice of that makes condition 6 hold is guaranteed to satisfy condition 2 too by the first condition of nicety of . Note also that in this case conditions 1 and 5 do not apply.
Case (iii): otherwise. In this case the only conditions that can apply are 1, 2, and 3 (and 4), and among 1 and 2 at most one can apply. In case 1 applies and is in , first choose to satisfy condition 1 and then choose to satisfy condition 3. In case 2 applies and is in , first choose to satisfy condition 2 and then choose to satisfy condition 3.
Case 3 (and last): is a triple in which and are leaves. Since is not a root, let be its sibling, and let be their parent.
As in the previous case, if both and are in , then and are not in , but the argument to show why this is the case is slightly different. First note that if both and are in then is not in because not all three components of are in by assumption. Second, at most two pebbles occupy at most two triples before spoiler’s move. If both and are in , then these triples must contain and , which are congruent only to themselves, and be different and hence different from any triple that contains a node congruent to or , since all the triples that contains both and are congruent to . Thus, in case both and are in , we choose and , and to satisfy condition 4. Conditions 1 and 5 just do not apply.
Assume then that not both and are in . In such a case, first choose to satisfy condition 5. Note that if condition 1 also applies, then the unique choice that satisfies 5 also satisfies 1 by the first condition of the nicety of . Once is chosen, choose either or to satisfy whichever condition among 2 or 3 applies, if any, and then choose the other to satisfy condition 4.
This completes the cases analysis over . Now, fix , , and as in the claim in whichever of the three cases applies. We claim that can be extended to a function that is defined on , , and so that , , and , and that is nice with respect to the new . Indeed, let , , and denote the sets of nodes that are congruent to , , and , respectively. We define the extension of by setting for all , for all , and for all . By the choices of , and in the claims, and the third condition of nicety of , this is well defined for those on which was already defined. Note that the domain of is precisely the value of after spoiler’s move. Let us argue that is nice with respect to this new .
First we note that on all triples that are congruent to , its three components get the same three values . This shows that satisfies the third condition of nicety with respect to the new . The second condition of nicety is also satisfied since extends and was nice with respect to the old , which contained all roots already. Finally, in order to argue that satisfies the first condition of nicety we need to argue which triples are completed by spoiler’s move. The triple and its congruents are definitely completed and, for these, the condition from the claims guarantees the first condition of nicety. The addition of , , and to can complete the triples , , and , when they exist, and their congruents, but no other triples. And for these, the conditions of the claims guarantee that the choices of , , and satisfy the first condition of nicety.
Lemma ? and Lemma ? show that the diagram is confusing for the class of consistent structures. Since can be taken arbitrarily large, Theorem ? follows immediately from Theorem ?.
4.3Other finite Abelian groups
The template for systems of equations over the 2-element field can be generalized to all finite Abelian groups. Let be a finite Abelian group; we write for the group operation and for its neutral element. Let be the structure with domain , where
The elements of are called values, and those of are called triples. As in , the signature of has three binary relations , , and , two unary relations value and triple, and one unary relation for each value in . The interpretations of all relation symbols are as in ; in particular, the unary relation symbol is interpreted by the singleton set . It is straightforward to check that can be used to encode arbitrary systems of equations over . As in the 2-element field case, equations more complex than the basic or can be encoded with the help of auxiliary variables.
The proof of Theorem ? does not rely in any way on the fact that the group is addition mod 2, except for it being Abelian and having at least two different values in it.
It is known that, for any non-trivial finite Abelian group, the constraint satisfaction problem of the template has unbounded width, i.e. for every two natural numbers and there exist instances that do not have homomorphisms to , but are nonetheless -consistent with respect to . We also say that