A modification of the CSP algorithm for infinite languages
AA modification of the CSP algorithm for infinitelanguages
Dmitriy ZhukDepartment of Mechanics and MathematicsMoscow State UniversityMoscow, Russia
Abstract
Constraint Satisfaction Problem on finite sets is known to be NP-complete in gen-eral but certain restrictions on the constraint language can ensure tractability. It wasproved [4, 18] that if a constraint language has a weak near unanimity polymorphismthen the corresponding constraint satisfaction problem is tractable, otherwise it is NP-complete. In the paper we present a modification of the algorithm from [18] that worksin polynomial time even for infinite constraint languages.
Formally, the
Constraint Satisfaction Problem (CSP) is defined as a triple (cid:104) X , D , C (cid:105) , where • X = { x , . . . , x n } is a set of variables, • D = { D , . . . , D n } is a set of the respective domains, • C = { C , . . . , C m } is a set of constraints,where each variable x i can take on values in the nonempty domain D i , every constraint C j ∈ C is a pair ( t j , ρ j ) where t j is a tuple of variables of length m j , called the constraint scope , and ρ j is an m j -ary relation on the corresponding domains, called the constraint relation .The question is whether there exists a solution to (cid:104) X , D , C (cid:105) , that is a mapping thatassigns a value from D i to every variable x i such that for each constraints C j the image of theconstraint scope is a member of the constraint relation.In this paper we consider only CSP over finite domains. The general CSP is known to beNP-complete [11, 13]; however, certain restrictions on the allowed form of constraints involvedmay ensure tractability (solvability in polynomial time) [6, 8, 9, 10, 3, 5]. Below we providea formalization to this idea.To simplify the presentation we assume that all the domains D , . . . , D n are subsets of afinite set A . By R A we denote the set of all finitary relations on A , that is, subsets of A m forsome m . Then all constraint relations can be viewed as relations from R A .For a set of relations Γ ⊆ R A by CSP(Γ) we denote the Constraint Satisfaction Problemwhere all the constraint relations are from Γ. The set Γ is called a constraint language .Another way to formalize the Constraint Satisfaction Problem is via conjunctive formulas.Every h -ary relation on A can be viewed as a predicate, that is, a mapping A h → { , } .Suppose Γ ⊆ R A , then CSP(Γ) is the following decision problem: given a formula ρ ( x , , . . . , x ,n ) ∧ · · · ∧ ρ s ( x s, , . . . , x ,n s )1 a r X i v : . [ c s . CC ] M a r here ρ i ∈ Γ for every i ; decide whether this formula is satisfiable.It is well known that many combinatorial problems can be expressed as CSP(Γ) for someconstraint language Γ. Moreover, for some sets Γ the corresponding decision problem can besolved in polynomial time; while for others it is NP-complete. It was conjectured that CSP(Γ)is either in P, or NP-complete [7].An operation f is called idempotent if f ( x, x, . . . , x ) = x . An operation f is called a weaknear-unanimity operation (WNU) if f ( y, x, . . . , x ) = f ( x, y, x, . . . , x ) = · · · = f ( x, x, . . . , x, y ) . In the paper we present a modification of the algorithm from [18] that works in polynomialtime for infinite constraint languages and therefore prove CSP dichotomy conjecture for infiniteconstraint languages.
Theorem 1.1.
Suppose Γ ⊆ R A is a set of relations. Then CSP(Γ) can be solved in polynomialtime if there exists a WNU preserving Γ ; CSP (Γ) is NP-complete otherwise.
Note that the algorithm presented in [4] also works for infinite constraint languages.The paper is organized as follows. In Section 2 we give all necessary definitions, in Section 3we explain the algorithm starting with the new ideas. In Section 4 we prove statements thatshow the correctness of the algorithm.
A set of operations is called a clone if it is closed under composition and contains all projec-tions. For a set of operations M by Clo( M ) we denote the clone generated by M .An idempotent WNU w is called special if x ◦ ( x ◦ y ) = x ◦ y , where x ◦ y = w ( x, . . . , x, y ).It is not hard to show that for any idempotent WNU w on a finite set there exists a specialWNU w (cid:48) ∈ Clo( w ) (see Lemma 4.7 in [12]).A relation ρ ⊆ A × · · · × A n is called subdirect if for every i the projection of ρ onto the i -th coordinate is A i . For a relation ρ by pr i ,...,i s ( ρ ) we denote the projection of ρ onto thecoordinates i , . . . , i s . Algebras.
An algebra is a pair A := ( A ; F ), where A is a finite set, called universe , and F is a family of operations on A , called basic operations of A . In the paper we always assumethat we have a special WNU preserving all constraint relations. Therefore, every domain D can be viewed as an algebra ( D ; w ). By Clo( A ) we denote the clone generated by all basicoperations of A . Congruences.
An equivalence relation σ on the universe of an algebra A is called acongruence if it is preserved by every operation of the algebra. A congruence (an equivalencerelation) is called proper , if it is not equal to the full relation A × A . We use standard universalalgebraic notions of term operation, subalgebra, factor algebra, product of algebras, see [2].We say that a subalgebra R = ( R ; F R ) is a subdirect subalgebra of A × B if R is a subdirectrelation in A × B .We say that the i -th variable of a relation ρ is compatible with an equivalence relation σ if ( a , . . . , a n ) ∈ ρ and ( a i , b i ) ∈ σ implies ( a , . . . , a i − , b i , a i +1 , . . . , a n ) ∈ ρ . We say that arelation is compatible with σ if every variable of this relation is compatible with σ .For a relation ρ by Con( ρ, i ) we denote the binary relation σ ( y, y (cid:48) ) defined by ∃ x . . . ∃ x i − ∃ x i +1 . . . ∃ x n ρ ( x , . . . , x i − , y, x i +1 , . . . , x n ) ∧ ρ ( x , . . . , x i − , y (cid:48) , x i +1 , . . . , x n ) . For a constraint C = (( x , . . . , x n ) , ρ ), by Con( C, x i ) we denote Con( ρ, i ). Essential and critical relations.
A relation ρ is called essential if it cannot be repre-sented as a conjunction of relations with smaller arities. It is easy to see that any relation ρ ρ ⊆ A ×· · ·× A n is called critical if it cannot be represented as an intersection of other subalgebras of A × · · · × A n and it has no dummy variables.A tuple ( a , a , . . . , a n ) is called essential for an n -ary relation ρ if ( a , a , . . . , a n ) / ∈ ρ andthere exist b , b , . . . , b n such that ( a , . . . , a i − , b i , a i +1 , . . . , a n ) ∈ ρ for every i ∈ { , , . . . , n } .It is not hard to check the following lemma. Lemma 2.1. [15, 16, 17] Suppose ρ ⊆ A n , where n (cid:62) . Then the following conditions areequivalent:1. ρ is an essential relation;2. there exists an essential tuple for ρ . Parallelogram property.
We say that a relation ρ has the parallelogram property if anypermutation of its variables gives a relation ρ (cid:48) satisfying ∀ α , β , α , β : ( α β , β α , β β ∈ ρ (cid:48) ⇒ α α ∈ ρ (cid:48) ) . We say that the i -th variable of a relation ρ is rectangular if for every ( a i , b i ) ∈ Con( ρ, i )and ( a , . . . , a n ) ∈ ρ we have ( a , . . . , a i − , b i , a i +1 , . . . , a n ) ∈ ρ . We say that a relation is rectangular if all of its variables are rectangular. The following facts can be easily seen:if the i -th variable of ρ is rectangular then Con( ρ, i ) is a congruence; if a relation has theparallelogram property then it is rectangular. Polynomially complete algebras.
An algebra ( A ; F A ) is called polynomially complete(PC) if the clone generated by F A and all constants on A is the clone of all operations on A . Linear algebra.
A finite algebra ( A ; w A ) is called linear if it is isomorphic to ( Z p × · · · × Z p s ; x + . . . + x n ) for prime numbers p , . . . , p s . It is not hard to show that for every algebra( B ; w B ) there exists a minimal congruence σ , called the minimal linear congruence , such that( B ; w B ) /σ is linear. Absorption.
Let B = ( B ; F B ) be a subalgebra of A = ( A ; F A ). We say that B absorbs A if there exists t ∈ Clo( A ) such that t ( B, B, . . . , B, A, B, . . . , B ) ⊆ B for any position of A .In this case we also say that B is an absorbing subuniverse of A . If the operation t can bechosen binary or ternary then B is called a binary or ternary absorbing subuniverse of A . Center.
Suppose A = ( A ; w A ) is a finite algebra with a special WNU operation. C ⊆ A is called a center if there exists an algebra B = ( B ; w B ) with a special WNU operation of thesame arity and a subdirect subalgebra ( R ; w R ) of A × B such that there is no binary absorbingsubuniverse in B and C = { a ∈ A | ∀ b ∈ B : ( a, b ) ∈ R } . CSP instance.
An instance of the constraint satisfaction problem is called a CSP in-stance . Sometimes we use the same letter for a CSP instance and for the set of all constraintsof this instance. For a variable z by D z we denote the domain of the variable z .We say that z − C − z − · · · − C l − − z l is a path in Θ if z i , z i +1 are in the scope of C i for every i . We say that a path z − C − z − . . . C l − − z l connects b and c if there exists a i ∈ D z i for every i such that a = b , a l = c , and the projection of C i onto z i , z i +1 containsthe tuple ( a i , a i +1 ).A CSP instance is called if every constraint of the instance is subdirect. ACSP instance is called cycle-consistent if for every variable z and a ∈ D z any path startingand ending with z in Θ connects a and a . A CSP instance Θ is called linked if for everyvariable z appearing in Θ and every a, b ∈ D z there exists a path starting and ending with z in Θ that connects a and b .Suppose X (cid:48) ⊆ X . Then we can define a projection of Θ onto X (cid:48) , that is a CSP instancewhere variables are elements of X (cid:48) and constraints are projections of the constraints of Θ3nto X (cid:48) . We say that an instance Θ is fragmented if the set of variables X can be dividedinto 2 nonempty disjoint sets X and X such that the constraint scope of any constraint ofΘ either has variables only from X , or only from X .A CSP instance Θ is called irreducible if any instance Θ (cid:48) such that every constraint of Θ (cid:48) is a projection of a constraint from Θ on some set of variables is fragmented, linked, or itssolution set is subdirect. Weaker constraints.
We say that a constraint (( y , . . . , y t ) , ρ ) is weaker than a con-straint (( z , . . . , z s ) , ρ ) if { y , . . . , y t } ⊆ { z , . . . , z s } , ρ ( z , . . . , z s ) → ρ ( y , . . . , y t ), and, addi-tionally, ρ ( y , . . . , y t ) (cid:54)→ ρ ( z , . . . , z s ) or { y , . . . , y t } (cid:54) = { z , . . . , z s } . Suppose (( y , . . . , y t ) , ρ )is a constraint and ρ (cid:48) ( y , . . . , y t ) = ∃ z ρ ( z, y , . . . , y t ) ∧ σ ( z, y ) , where σ is a minimal con-gruence such that σ (cid:41) Con( ρ, y , . . . , y t ) , ρ (cid:48) ) is called a congruence-weakenedconstraint . Minimal linear reduction.
Suppose the domain set of the instance Θ is D = ( D , . . . , D n ).The domain set D (1) = ( D (1)1 , . . . , D (1) n ) is called a minimal linear reduction if D (1) i is anequivalence class of the minimal linear congruence of D i for every i . The reduction D (1) =( D (1)1 , . . . , D (1) n ) is called if the instance obtained after the reduction of everydomain is 1-consistent. Crucial instances.
Let D (1) i ⊆ D i for every i . A constraint C of Θ is called crucial in ( D (1)1 , . . . , D (1) n ) if Θ has no solutions in ( D (1)1 , . . . , D (1) n ) but the replacement of C ∈ Θ by allweaker constraints gives an instance with a solution in ( D (1)1 , . . . , D (1) n ). A CSP instance Θis called crucial in ( D (1)1 , . . . , D (1) n ) if every constraint of Θ is crucial in ( D (1)1 , . . . , D (1) n ). Tosimplify, instead of “crucial in ( D , . . . , D n )” we say “crucial” In this section we present a modified algorithm from [18]. The main problem that does notallow to use the original algorithm for infinite languages is in Steps 3 and 11, where we replace aconstraint by all weaker constraints. If Γ is infinite, we do not have a polynomial upper boundon the number of such replacements. To fix this problem, instead of replacing a constraintby all weaker constraints we replace it by all congruence-weakened constraints. Moreover, torestrict the depth of the recursion we additionally transform our instance to ensure that allthe constraint relations are essential relations with the parallelogram property.We start with the new procedures, then we explain auxiliary procedures from [18], andfinish with the modified main part of the algorithm.We have made only the following modifications of the algorithm.1. We added Step 3 to work only with essential relations. This property is important be-cause, by Lemma 4.1, any essential relation preserved by an idempotent WNU operationhas exponentially many tuples.2. We added Step 4 to ensure that every relation has the parallelogram property.3. We added Step 5 to ensure that Con( ρ,
1) is an irreducible congruence for every con-straint relation ρ .4. We changed Step 6 (Step 3 in [18]). Instead of replacing every constraint by all weakerconstraints we replace it by all congruence-weakened constraints, and therefore we in-crease the congruence Con( ρ,
1) for every constraint relation ρ .5. Similarly, we replaced Step 11 in [18] by Steps 13 and 14. In Step 13, instead of re-placing a constraint by all weaker constraints we replace it by all congruence-weakened4onstraints. In Step 14, we try to replace a constraint by all projections onto all variablesbut one appearing in the constraint.6. Instead of considering a center we consider a ternary absorption, thus we replace Steps4 and 5 in [18] by Step 7. Finding an appropriate projection.
Suppose ρ ⊆ A × · · · × A n , α = ( a , . . . , a n ) / ∈ ρ .Here we explain how to find a minimal subset I ⊆ { , . . . , n } such that pr I ( α ) / ∈ pr I ( ρ ). Notethat pr I ( ρ ) is always an essential relation.1. Put I := ∅ .2. Put k := 1.3. While pr { ,...,k }∪ I ( α ) ∈ pr { ,...,k }∪ I ( ρ ) do k := k + 1.4. Put I := I ∪ { k }
5. If pr I ( α ) ∈ pr I ( ρ ), go to step 2. Essential Representation.
Suppose ρ ⊆ A × · · · × A n . An essential representation of ρ is the following formula ρ ( x , . . . , x n ) = δ ( z , , . . . , z ,n ) ∧ · · · ∧ δ s ( z s, , . . . , z s,n s ) , where δ i is an essential relation from Γ, z i,j ∈ { x , . . . , x n } , z i,j (cid:54) = z i,k for every i and j (cid:54) = k .Below we explain how to find a set G = { I , . . . , I s } , where I i ⊆ { , . . . , n } for every i , suchthat pr I ( ρ ) , . . . , pr I s ( ρ ) form an essential representation of ρ . To guarantee this property werequire that every pr I i ( ρ ) is essential and for every α ∈ ( A × · · · × A n ) \ ρ there exists i suchthat α / ∈ pr I i ( ρ ).1. Put G := ∅ .2. Choose a tuple α = ( a , . . . , a n ) / ∈ ρ such that ( a , . . . , a n − , b n ) ∈ ρ for some b n (wehave at most | ρ | · | A n | such tuples).3. Find a minimal subset I ⊆ { , . . . , n } such that pr I ( α ) / ∈ pr I ( ρ ). Put G := G ∪ { I } .4. Go to the next tuple in 2).5. Put ρ (cid:48) = pr { ,...,n − } ( ρ ).6. By recursive call we calculate the essential representation G (cid:48) corresponding to ρ (cid:48) .7. Put G := G (cid:48) ∪ G .8. Remove from G all sets that are not maximal in G by inclusion.Note that the obtained essential representation of an essential relation consists of theoriginal relation. Therefore, the above procedure can also be used to check whether a relationis essential. Providing the Parallelogram Property.
In this section we explain how to find theminimal relation ρ (cid:48) ⊇ ρ having the parallelogram property. We say that tuples α , α , α , α orm a rectangle if there exists I ⊆ { , . . . , n } such that pr I ( α ) = pr I ( α ), pr I ( α ) = pr I ( α ),pr { ,...,n }\ I ( α ) = pr { ,...,n }\ I ( α ), pr { ,...,n }\ I ( α ) = pr { ,...,n }\ I ( α ). To find the minimal relationwith the parallelogram property it is sufficient to close the relation under adding the forthtuple of a rectangle. This can be done in the following way.For each α , α , α ∈ ρ .1. Put I := { i | α ( i ) = α ( i ) } , I := { i | α ( i ) = α ( i ) } .2. If I ∪ I = { , . . . , n } then find the forth tuple α of the corresponding rectangle.3. If α / ∈ ρ , add α to ρ .By Lemma 4.1, every essential relation has exponentially many tuples, therefore, thisprocedure works in polynomial time on the size of ρ if ρ is an essential. Provide cycle-consistency.
To provide cycle-consistency it is sufficient to use constraintpropagation providing (2,3)-consistency. Formally, it can be done in the following way.First, for every pair of variables ( x i , x j ) we consider the intersections of projections of allconstraints onto these variables. The corresponding relation we denote by ρ i,j . For every i, j, k ∈ { , , . . . , n } we replace ρ i,j by ρ (cid:48) i,j where ρ (cid:48) i,j ( x, y ) = ∃ z ρ i,j ( x, y ) ∧ ρ i,k ( x, z ) ∧ ρ k,j ( z, y ) . It is not hard to see that this replacement does not change the solution set.We repeat this procedure while we can change some ρ i,j . If at some moment we get arelation ρ i,j that is not subdirect in D i × D j , then we can either reduce D i or D j , or, if ρ i,j isempty, state that there are no solutions. If we cannot change any relation ρ i,j and every ρ i,j is subdirect in D i × D j , then the original CSP instance is cycle-consistent. Solve the instance that is not linked.
Suppose the instance Θ is not linked and notfragmented, then it can be solved in the following way. We say that an element d i ∈ D i andan element d j ∈ D j are linked if there exists a path that connects d i and d j . Let P be theset of pairs ( i ; a ) such that i ∈ { , , . . . , n } , a ∈ D i . Then P can be divided into the linkedcomponents.It is easy to see that it is sufficient to solve the problem for every linked component andjoin the results. Precisely, for a linked component by D (cid:48) i we denote the set of all elements d such that ( i, d ) is in the component. It is easy to see that ∅ (cid:40) D (cid:48) i (cid:40) D i for every i . Therefore,the reduction to ( D (cid:48) , . . . , D (cid:48) n ) is a CSP instance on smaller domains. Check irreducibility.
For every k ∈ { , , . . . , n } and every maximal congruence σ k on D k we do the following.1. Put I = { k } .2. Choose a constraint C having the variable x i in the scope for some i ∈ I , choose anothervariable x j from the scope such that j / ∈ I .3. Denote the projection of C onto ( x i , x j ) by δ .4. Put σ j ( x, y ) = ∃ x (cid:48) ∃ y (cid:48) δ ( x (cid:48) , x ) ∧ δ ( y (cid:48) , y ) ∧ σ i ( x (cid:48) , y (cid:48) ). If σ j is a proper equivalence relation,then add j to I .5. go to the next C , x i , and x j in 2. 6s a result we get a set I and a congruence σ i on D i for every i ∈ I . Put X (cid:48) = { x i | i ∈ I } . Itfollows from the construction that for every equivalence class E k of σ k and every i ∈ I thereexists a unique equivalence class E i of σ i such that there can be a solution with x k ∈ E k and x i ∈ E i . Thus, for every equivalence class of σ k we have a reduction to the instance on smallerdomains. Then for every i and a ∈ E i we consider the corresponding reduction and checkwhether there exists a solution with x i = a .Thus, we can check whether the solution set of the projection of the instance onto X (cid:48) is subdirect or empty. If it is empty then we state that there are no solutions. If it is notsubdirect, then we can reduce the corresponding domain. If it is subdirect, then we go to thenext k ∈ { , , . . . , n } and next maximal congruence σ k on D k , and repeat the procedure. In this section we provide an algorithm that solves CSP(Γ) in polynomial time for constraintlanguages Γ (finite or infinite) that are preserved by an idempotent WNU operation. We knowthat Γ is also preserved by a special WNU operation w . We extend Γ to the set of all relationspreserved by w . Let the arity of the WNU w be equal to m . Suppose we have a CSP instanceΘ = (cid:104) X , D , C (cid:105) , where X = { x , . . . , x n } is a set of variables, D = { D , . . . , D n } is a set of therespective domains, C = { C , . . . , C q } is a set of constraints.The algorithm is recursive, the list of all possible recursive calls is given in the end of thissubsection. One of the recursive calls is the reduction of a subuniverse D i to D (cid:48) i such thateither Θ has a solution with x i ∈ D (cid:48) i , or it has no solutions at all. Step 1.
Check whether Θ is cycle-consistent. If not then we reduce a domain D i for some i or state that there are no solutions. Step 2.
Check whether Θ is irreducible. If not then we reduce a domain D i for some i orstate that there are no solutions. Step 3.
Replace every constraint by its essential representation.
By Theorem 4.13, if Θ has no solutions then we cannot get a solution while doing thefollowing step.
Step 4.
Replace every constraint relation by the corresponding constraint relation having theparallelogram property. If one of the obtained constraint relation is not essential, go to Step 3.
By Lemma 4.18, if Θ has no solutions then we cannot get a solution in the following step.
Step 5.
If the congruence
Con( ρ, is not irreducible for some constraint relation ρ , thenreplace the constraint by the corresponding congruence-weakened constraints and go to Step 3. At the moment all constraint relations ρ have the parallelogram property and Con( ρ, Step 6.
Replace every constraint of Θ by the corresponding congruence-weakened constraint,then replace every constraint relation by the corresponding constraint relation having the par-allelogram property. Recursively calling the algorithm, check that the obtained instance has asolution with x i = b for every i ∈ { , , . . . , n } and b ∈ D i . If not, reduce D i to the projectiononto x i of the solution set of the obtained instance. By Theorems 4.8 and 4.11 we cannot loose the only solution while doing the followingstep. 7 tep 7. If D i has a binary or ternary absorbing subuniverse B i (cid:40) D i for some i , then wereduce D i to B i . By Theorem 4.10 we can do the following step.
Step 8.
If there exists a congruence σ on D i such that the algebra ( D i ; w ) /σ is polynomiallycomplete, then we reduce D i to any equivalence class of σ . By Theorem 4.5, it remains to consider the case when for every domain D i there exists acongruence σ i on D i such that ( D i ; w ) /σ i is linear, i.e. it is isomorphic to ( Z p × · · · × Z p l ; x + · · · + x m ) for prime numbers p , . . . , p l . Moreover, σ i is proper if | D i | > D i /σ i by L i . We define a new CSP instance Θ L with domains L , . . . , L n .To every constraint (( x i , . . . , x i s ) , ρ ) ∈ Θ we assign a constraint (( x (cid:48) i , . . . , x (cid:48) i s ) , ρ (cid:48) ), where ρ (cid:48) ⊆ L i × · · · × L i s and ( E , . . . , E s ) ∈ ρ (cid:48) ⇔ ( E × · · · × E s ) ∩ ρ (cid:54) = ∅ . The constraints of Θ L are all constraints that are assigned to the constraints of Θ.Since every relation on Z p ×· · ·× Z p l preserved by x + . . . + x m is known to be a conjunctionof linear equations, the instance Θ L can be viewed as a system of linear equations in Z p fordifferent p .Our general idea is to add some linear equations to Θ L so that for any solution of Θ L thereexists the corresponding solution of Θ. We start with the empty set of equations Eq , whichis a set of constraints on L , . . . , L n . Step 9.
Put Eq := ∅ . Step 10.
Solve the system of linear equations Θ L ∪ Eq and choose independent variables y , . . . , y k . If it has no solutions then Θ has no solutions. If it has just one solution, then,recursively calling the algorithm, solve the reduction of Θ to this solution. Either we get asolution of Θ , or Θ has no solutions. Then there exist Z = Z q × · · · × Z q k and a linear mapping φ : Z → L × · · · × L n suchthat any solution of Θ L ∪ Eq can be obtained as φ ( a , . . . , a k ) for some ( a , . . . , a k ) ∈ Z .Note that for any tuple ( a , . . . , a k ) ∈ Z we can check recursively whether Θ has a solutionin φ ( a , . . . , a k ). To do this, we just need to solve an easier CSP instance (on smaller domains).Similarly, we can check whether Θ has a solution in φ ( a , . . . , a k ) for every ( a , . . . , a k ) ∈ Z .To do this, we just need to check the existence of a solution in φ (0 , . . . , , , , . . . ,
0) and φ (0 , . . . ,
0) for any position of 1.
Step 11.
Check whether Θ has a solution in φ (0 , . . . , . If it has then stop the algorithm. Step 12.
Put Θ (cid:48) := Θ . Iteratively remove from Θ (cid:48) all constraints that are weaker than someother constraints of Θ (cid:48) . In the following two steps we try to weaken the instance so that it still does not have asolution in φ ( a , . . . , a k ) for some ( a , . . . , a k ) ∈ Z . Step 13.
For every constraint C ∈ Θ (cid:48)
1. Let Ω be obtained from Θ (cid:48) by replacing the constraint C ∈ Θ (cid:48) by the correspondingcongruence-weakened constraints.2. Replace every new constraint of Ω by its essential representation and remove from Ω allconstraints that are weaker than some other constraints of Ω .3. If Ω has no solutions in φ ( a , . . . , a k ) for some ( a , . . . , a k ) ∈ Z , then put Θ (cid:48) := Ω andrepeat Step 13. tep 14. For every constraint C ∈ Θ (cid:48)
1. Let Ω be obtained from Θ (cid:48) by replacing the constraint C ∈ Θ (cid:48) by its projections onto allvariables but one appearing in C .2. Replace the new constraints by its essential representation and remove from Ω all con-straints that are weaker than some other constraints of Ω .3. If Ω has no solutions in φ ( a , . . . , a k ) for some ( a , . . . , a k ) ∈ Z , then put Θ (cid:48) := Ω andgo to Step 13. At this moment, the CSP instance Θ (cid:48) has the following property. Θ (cid:48) has no solutionsin φ ( b , . . . , b k ) for some ( b , . . . , b k ) ∈ Z but if we replace any constraint C ∈ Θ (cid:48) by thecorresponding congruence-weakened constraints then we get an instance that has a solutionin φ ( a , . . . , a k ) for every ( a , . . . , a k ) ∈ Z . Unlike the original algorithm, we cannot claimthat Θ (cid:48) is crucial in φ ( b , . . . , b k ). Nevertheless, Theorem 4.19 proves that we can finish thealgorithm in the same way as in the original paper.In the remaining steps we will find a new linear equation that can be added to Θ L . Suppose V is an affine subspace of Z hp of dimension h −
1, thus V is the solution set of a linear equation c x + · · · + c h x h = c . Then the coefficients c , c , . . . , c h can be learned (up to a multiplicativeconstant) by ( p · h + 1) queries of the form “( a , . . . , a h ) ∈ V ?” as follows. First, we need atmost ( h + 1) queries to find a tuple ( d , . . . , d h ) / ∈ V . Then, to find this equation it is sufficientto check for every a and every i whether the tuple ( d , . . . , d i − , a, d i +1 , . . . , d h ) satisfies thisequation. Step 15.
Suppose Θ (cid:48) is not linked. For each i from to k
1. Check that for every ( a , . . . , a i ) ∈ Z q × · · · × Z q i there exist ( a i +1 , . . . , a k ) ∈ Z q i +1 ×· · · × Z q k and a solution of Θ (cid:48) in φ ( a , . . . , a k ) .2. If yes, go to the next i .3. If no, then find an equation c y + · · · + c i y i = c such that for every ( a , . . . , a i ) ∈ Z q × · · · × Z q i satisfying c a + · · · + c i a i = c there exist ( a i +1 , . . . , a k ) ∈ Z q i +1 × · · · × Z q k and a solution of Θ (cid:48) in φ ( a , . . . , a k ) .4. Add the equation c y + · · · + c i y i = c to Eq.5. Go to Step 10. It is not hard to see that Θ (cid:48) satisfies the conditions of Theorem 4.19. Then there existsa constraint (( x i , . . . , x i s ) , ρ ) in Θ (cid:48) and a relation ξ ⊆ D i × D i × Z p such that ( x , x , ∈ ξ ⇔ ( x , x ) ∈ Con( ρ, , ( ξ ) (cid:41) Con( ρ, ρ,
1) is a congruence. We add a newvariable z with domain Z p and a variable x (cid:48) i with the same domain as x i . Then we replace(( x i , . . . , x i s ) , ρ ) by (( x (cid:48) i , x i , . . . , x i s ) , ρ ) and add the constraint (( x i , x (cid:48) i , z ) , ξ ). We denotethe obtained instance by Υ. Let L be the set of all tuples ( a , . . . , a k , b ) ∈ Z q × · · · × Z q k × Z p such that Υ has a solution with z = b in φ ( a , . . . , a k ). By Theorem 4.14, if we replace theconstraint relation ρ in Θ (cid:48) by the minimal relation ρ (cid:48) ⊇ ρ having the parallelogram propertythen a minimal linear reduction cannot get a solution after the replacement. Therefore, L hasno tuple ( b , . . . , b k , ρ in Θ (cid:48) by ρ (cid:48)(cid:48) defined by ρ (cid:48)(cid:48) ( x i , x i , . . . , x i s ) = ∃ x (cid:48) i ρ ( x (cid:48) i , x i , . . . , x i s ) ∧ σ ( x i , x (cid:48) i ) , where σ = pr , ( ξ ), then we get a solution in φ ( b , . . . , b k ). Otherwise, Theorem 4.14 impliesthat the replacement of ρ (cid:48)(cid:48) by the minimal relation ρ (cid:48)(cid:48)(cid:48) ⊇ ρ (cid:48)(cid:48) having the parallelogram property9till does not give a solution in φ ( b , . . . , b k ). This contradicts the fact that if we replace anyconstraint C ∈ Θ (cid:48) by the corresponding congruence-weakened constraints then we get aninstance that has a solution in φ ( a , . . . , a k ) for every ( a , . . . , a k ) ∈ Z . Thus, we know thatthe projection of L onto the first k coordinates is a full relation.Therefore, L is defined by one linear equation. If this equation is z = b for some b (cid:54) = 0,then both Θ (cid:48) and Θ have no solutions. Otherwise, we put z = 0 in this equation and get anequation that describes all ( a , . . . , a k ) such that Θ (cid:48) has a solution in φ ( a , . . . , a k ). It remainsto find this equation. Step 16.
Suppose Θ (cid:48) is linked.1. Find an equation c y + · · · + c k y k = c such that for every ( a , . . . , a k ) ∈ ( Z q × · · · × Z q k ) satisfying c a + · · · + c k a k = c there exists a solution of Θ (cid:48) in φ ( a , . . . , a k ) .2. If the equation was not found then Θ has no solutions.3. Add the equation c a + · · · + c k a k = c to Eq.4. Go to Step 10. Note that every time we reduce our domains, we get constraint relations that are stillfrom Γ.We have four types of recursive calls of the algorithm:1. we reduce one domain D i , for example to a binary absorbing subuniverse (Steps 1, 7,8).2. we solve an instance that is not linked. In this case we divide the instance into thelinked parts and solve each of them independently (Steps 2, 15).3. we replace every constraint by the corresponding congruence-weakened constraint andsolve an easier CSP instance (Step 6).4. we reduce every domain D i such that | D i | > | A | . It is easyto see that the depth of the recursive calls of type 2 and 4 is at most | A | . Lemma 4.1.
Suppose ρ ⊆ A n is an essential relation preserved by a special WNU w of arity m < n. Then | ρ | (cid:62) n − m +1 Proof.
By Lemma 2.1, there exists an essential tuple ( a , a , . . . , a n ) for ρ . For every i choose b i such that ( a , . . . , a i − , b i , a i +1 , . . . , a n ) ∈ ρ. Put b (cid:48) i = w ( b i , a i , . . . , a i ) for every i . Without loss of generality assume that b (cid:48) i = a i for every i (cid:54) k and b (cid:48) i (cid:54) = a i for every i > k . Since w preserves ρ , we can show that ( b (cid:48) , . . . , b (cid:48) m , a m +1 , . . . , a n ) ∈ ρ . Hence, k < m . Consider the projection of ρ onto the last n − m +1 variables, which we denoteby ρ (cid:48) . It is not hard to check that α = ( a m , . . . , a n ) ∈ ρ (cid:48) and ( a m , . . . , a i − , b (cid:48) i , a i +1 . . . , a n ) ∈ ρ (cid:48) for every i ∈ { m, . . . , n } . 10or any subset I ⊆ { m, . . . , n } put α I = ( c m , . . . , c n ) ∈ ρ , where c i = b (cid:48) i if i ∈ I and c i = a i otherwise. We know that α { i } ∈ ρ (cid:48) for every i ∈ { m, . . . , n } . Since w is a special WNU, w ( b (cid:48) i , a i , . . . , a i ) = b (cid:48) i . Then we can check that for any disjoint subsets I , I ⊆ { m, . . . , n } wehave w ( α I , α I , α, . . . , α ) = α I ∪ I . Thus, ρ (cid:48) contains the tuple α I for any I ⊆ { m, . . . , n } .Therefore, both ρ and ρ (cid:48) contain at least 2 n − m +1 tuples. Lemma 4.2.
Suppose ρ is an essential relation with the parallelogram property, Con( ρ, isan irreducible congruence, σ = Con( ρ, ∗ , ρ (cid:48) ( y , . . . , y t ) = ∃ z ρ ( z, y , . . . , y t ) ∧ σ ( z, y ) . Then
Con( ρ (cid:48) , i ) (cid:41) Con( ρ, i ) for every i .Proof. Put σ i ( y , y (cid:48) ) = ∃ y (cid:48) i ∃ y . . . ∃ y t ρ ( y (cid:48) , y , . . . , y t ) ∧ ρ ( y , . . . , y i − , y (cid:48) i , y i +1 , . . . , y t ) . Since ρ is an essential relation with the parallelogram property, we have σ i (cid:41) Con( ρ, ρ,
1) is irreducible, σ i ⊇ σ . Therefore, for any ( a , a (cid:48) ) ∈ σ \ Con( ρ,
1) there exist a (cid:48) i , a , a , . . . , a t such that ( a (cid:48) , a , . . . , a t ) , ( a , . . . , a i − , a (cid:48) i , a i +1 , . . . , a t ) ∈ ρ. Hence ( a i , a (cid:48) i ) ∈ Con( ρ (cid:48) , i ) \ Con( ρ, i ) and Con( ρ (cid:48) , i ) (cid:41) Con( ρ, i ). Lemma 4.3.
The depth of the recursive calls of type 3 in the algorithm is less than | A | .Proof. First, we introduce a partial order on variables of instances. For a constraint C anda variable x by M axComp ( C, x ) we denote the maximal equivalence relation σ such that thevariable x of C is compatible with σ . We say that x in Θ is weaker than x (cid:48) in Θ (cid:48) if one of thefollowing conditions holds:1. the domain of x is a proper subset of the domain of x (cid:48) ;2. the domain of x is equal to the domain of x (cid:48) , for every C ∈ Θ there exists C (cid:48) ∈ Θ (cid:48) suchthat M axComp ( C, x ) ⊇ M axComp ( C (cid:48) , x (cid:48) ).Since every constraint relation ρ we have in Step 6 is an essential relation with the parallel-ogram property and Con( ρ,
1) is an irreducible congruence, it follows from Lemma 4.2 thatCon( C (cid:48) , x ) (cid:41) Con(
C, x ) for every new constraint C (cid:48) we generate from C in Step 6. Since weprovide the parallelogram property after the replacement in Step 6, we have M axComp ( C (cid:48) , x ) (cid:41) M axComp ( C, x ). Thus, every variable is getting weaker. It is easy to see that any other re-duction makes all variables weaker or does not change them. Therefore, the depth of therecursive calls of type 3 is less than the number of binary relations on the set A , that is2 | A | . In [18] we proved the following theorem.
Theorem 4.4. [18] Suppose A = ( A ; w ) is an algebra, w is a special WNU of arity m . Thenone of the following conditions hold:1. there exists a binary absorbing set B (cid:40) A ,2. there exists a center C (cid:40) A ,3. there exists a proper congruence σ on A such that ( A ; w ) /σ is polynomially complete,4. there exists a proper congruence σ on A such that ( A ; w ) /σ is isomorphic to ( Z p ; x + · · · + x m ) . Using Corollary 7.9.2 from [18], this theorem can be rewritten in the following form.11 heorem 4.5.
Suppose A = ( A ; w ) is an algebra, w is a special WNU of arity m . Then oneof the following conditions hold:1. there exists a binary or ternary absorbing set B (cid:40) A ,2. there exists a proper congruence σ on A such that ( A ; w ) /σ is polynomially complete,3. there exists a proper congruence σ on A such that ( A ; w ) /σ is isomorphic to ( Z p ; x + · · · + x m ) . An operation is called cyclic if f ( x , x , . . . , x n ) = f ( x , x , . . . , x n , x ). We know from [1](see Theorem 4.1) that the existence of a cyclic term is equivalent to the existence of a WNUterm. Theorem 4.6. [1] An idempotent algebra ( A ; F ) has a WNU term operation if and only if ithas a cyclic term operation of arity p for every prime number p > | A | . Lemma 4.7.
Suppose B absorbs D with a ternary idempotent operation f , u is an idempotentcyclic operation. Then there exists a cyclic operation v ∈ Clo( { u, f } ) preserving the relation ( B × D ) ∪ ( D × B ) . Moreover, B is a center for the algebra ( D ; v ) .Proof. By Theorem 4.6 there exists a cyclic operation u (cid:48) ∈ Clo( u ) of an odd arity. Let n bethe arity of u (cid:48) . Let us consider a ternary majority operation m on 2-element set. We knowfrom [14] that the n -ary majority operation belongs to Clo( m ). Consider a term t over m that defines the n -ary majority operation. Replace every operation m in it by f to definean operation g of arity n . It is not hard to see that the operation g ( x , . . . , x n ) satisfies thefollowing property: if more than half of the variables are from B then the result is from B .Put v ( x , . . . , x n ) = u (cid:48) ( g ( x , . . . , x n ) , g ( x , . . . , x n , x ) , . . . , g ( x n , x , . . . , x n − )) . It is not hard to see that v is a cyclic operation preserving the relation ( B × D ) ∪ ( D × B ).Let us show that B is a center for the algebra ( D ; v ). Put E = { , } . Let v be definedon E as a majority operation. Put R = ( B × { , } ) ∪ ( D × { } ). It is not hard to see that v preserves R , hence B is a center.The following three theorems are proved in [18]. Theorem 4.8. [18] Suppose Θ is a cycle-consistent irreducible CSP instance, B is a binaryabsorbing set of D i . Then Θ has a solution if and only if Θ has a solution with x i ∈ B . Theorem 4.9. [18] Suppose Θ is a cycle-consistent irreducible CSP instance, C is a centerof D i . Then Θ has a solution if and only if Θ has a solution with x i ∈ C . Theorem 4.10. [18] Suppose Θ is a cycle-consistent irreducible CSP instance, there does notexist a binary absorbing subuniverse or a center on D j for every j , ( D i ; w ) /σ is a polynomiallycomplete algebra, E is an equivalence class of σ . Then Θ has a solution if and only if Θ hasa solution with x i ∈ E . Combining Theorem 4.9 and Lemma 4.7 we obtain the following theorem.
Theorem 4.11.
Suppose Θ is a cycle-consistent irreducible CSP instance, B is a ternaryabsorbing set of D i . Then Θ has a solution if and only if Θ has a solution with x i ∈ B . .3 Adding a new linear variable To prove the main result of this section we will need the following definitions from [18].For an instance Ω by ExpCov(Ω) (
Expanded Coverings ) we denote the set of all instancesΩ (cid:48) such that there exists a mapping S from the set of all variables of Ω (cid:48) to the set of allvariables of Ω satisfying the following conditions:1. for every constraint (( x , . . . , x n ) , ρ ) of Ω (cid:48) either the variables S ( x ) , . . . , S ( x n ) are differ-ent and the constraint (( S ( x ) , . . . , S ( x n )) , ρ ) is weaker than or equal to some constraintof Ω, or ρ is a binary reflexive relation and S ( x ) = S ( x );2. if a variable x appears in Ω and Ω (cid:48) then S ( x ) = x .We say that a congruence σ is irreducible if it cannot be represented as an intersection ofother binary relations δ , . . . , δ s compatible with σ . For an irreducible congruence σ on a set A by σ ∗ we denote the minimal binary relation δ (cid:41) σ compatible with σ .Suppose σ and σ are congruences on D and D , correspondingly. A relation ρ ⊆ D × D is called a bridge from σ to σ if the first two variables of ρ are compatible with σ , the lasttwo variables of ρ are compatible with σ , pr , ( ρ ) (cid:41) σ , pr , ( ρ ) (cid:41) σ , and ( a , a , a , a ) ∈ ρ implies ( a , a ) ∈ σ ⇔ ( a , a ) ∈ σ . A bridge ρ ⊆ D is called reflexive if ( a, a, a, a ) ∈ ρ forevery a ∈ D .We say that two congruences σ and σ on a set D are adjacent if there exists a reflexivebridge from σ to σ . We say that two constraints C and C are adjacent in a commonvariable x if Con( C , x ) and Con( C , x ) are adjacent. An instance is called connected if everyconstraint in it is rectangular and for every two constraints there exists a path that connectsthem.We will need the following statements from [18]. Lemma 4.12. [Lemma 8.3 in [18]] Suppose ρ is a critical subdirect relation, the i -th variableof ρ is rectangular. Then Con( ρ, i ) is an irreducible congruence. Theorem 4.13. [Theorem 9.5 in [18]] Suppose Θ is a cycle-consistent irreducible CSP in-stance, its constraint (( x , . . . , x n ) , ρ ) is crucial. Then ρ is a critical relation with the paral-lelogram property. Theorem 4.14. [Theorem 9.5 in [18]] Suppose D (1) is a 1-consistent minimal linear reductionfor a cycle-consistent irreducible CSP instance Θ , the constraint (( x , . . . , x n ) , ρ ) is crucial in D (1) . Then ρ is a critical relation with the parallelogram property. Theorem 4.15. [Theorem 9.8 in [18]] Suppose D (1) is a 1-consistent minimal linear reductionof a cycle-consistent irreducible CSP instance Θ , Θ is crucial in D (1) and not connected. Thenthere exists an instance Θ (cid:48) ∈ ExpCov(Θ) that is crucial in D (1) and contains a linked connectedcomponent whose solution set is not subdirect. Lemma 4.16. [Corollary 8.15.1 in [18]] Suppose Θ is a cycle-consistent linked connectedinstance whose constraint relations are critical rectangular relations. Then for every constraint C and its variable x there exists a bridge δ from Con(
C, x ) to Con(
C, x ) such that δ ( x, x, y, y ) defines a full relation. Lemma 4.17. [Corollary 8.10.1 in [18]] Suppose σ ⊆ A is an irreducible congruence, ρ ( x , x , y , y ) is a bridge from σ to σ such that ρ ( x, x, y, y ) defines a full relation. Then thereexists a prime number p and a relation ζ ⊆ A × A × Z p such that ( x , x , ∈ ζ ⇔ ( x , x ) ∈ σ and pr , ζ = σ ∗ . emma 4.18. Suppose Θ is a cycle-consistent irreducible CSP instance without a solution,every constraint relation of Θ has the parallelogram property, the congruence Con( ρ, is notirreducible for some constraint relation (( x i , . . . , x i t ) , ρ ) , Θ (cid:48) is obtained from Θ by replacementof (( x i , . . . , x i t ) , ρ ) by the corresponding congruence-weakened constraints. Then Θ (cid:48) has nosolutions.Proof. Consider binary relations δ , . . . , δ s compatible with Con( ρ,
1) such that δ ∩ · · · ∩ δ s =Con( ρ, ρ i ( x i , . . . , x i t ) = ∃ z ρ ( z, x i , . . . , x i t ) ∧ δ i ( z, x i ) , and replace the constraint(( x i , . . . , x i t ) , ρ ) by the constraints (( x i , . . . , x i t ) , ρ ) , . . . , (( x i , . . . , x i t ) , ρ s ). Since ρ has theparallelogram property, the obtained instance still does not have a solution. By Theorem 4.13,if we replace every new constraint relation by the corresponding relation having the parallel-ogram property, then we cannot get a solution. Therefore, Θ (cid:48) has no solutions.Similarly to Theorem 9.8 from [18], we prove the following theorem. Theorem 4.19.
Suppose the following conditions hold:1. Θ is a linked cycle-consistent irreducible CSP instance with domain set ( D , . . . , D n ) ;2. there does not exist a binary absorbing subuniverse or a center on D j for every j ;3. suppose we replace every constraint of Θ by the corresponding congruence-weakened con-straints, then replace every constraint relation ρ by the minimal constraint relation ρ (cid:48) ⊇ ρ having the parallelogram property; then the obtained instance has a solution with x i = b for every i and b ∈ D i ;4. D (1) = ( D (1)1 , . . . , D (1) n ) is a minimal linear reduction of Θ ;5. Θ has no solutions in D (1) ;6. if we replace any constraint by the corresponding congruence-weakened constraints thenthe obtained instance has a solution in D (1) ;7. if we replace any constraint by its projections onto all variables but one appearing in theconstraint then the obtained instance has a solution in D (1) .Then there exists a constraint (( x i , . . . , x i s ) , ρ ) of Θ and a relation ξ ⊆ D i × D i × Z p such that ( x , x , ∈ ξ ⇔ ( x , x ) ∈ Con( ρ, , pr , ( ξ ) (cid:41) Con( ρ, , and Con( ρ, is a congruence.Proof. Assume the contrary. First, we want to make our instance crucial in D (1) . To do this wereplace our constraints by all weaker constraints while we still do not have a solution in D (1) .The obtained instance we denote by Θ (cid:48) . Since Θ is linked, condition 7 guarantees that Θ (cid:48) isalso linked. By Theorem 4.14, every constraint in Θ (cid:48) has the parallelogram property. SupposeΘ (cid:48) is not connected. Then by Theorem 4.15 there exists an instance Θ (cid:48)(cid:48) ∈ ExpCov(Θ (cid:48) ) thatis crucial in D (1) and contains a linked connected component Ω such that the solution set ofΩ is not subdirect. By condition 3, there exists a constraint relation ρ (cid:48) from Ω such that fortheir ancestor ρ from Θ we have Con( ρ,
1) = Con( ρ (cid:48) , (cid:48) is connected, then Θ (cid:48) is a linked connected component itself and we choose a constraintrelation ρ (cid:48) from Θ (cid:48) such that Con( ρ,
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