Multivalued morphism gadgets

The main technical tool in the hardness proofs are the gadgets defined in this section. Intuitively, assuming that D={0,1, . . . ,∆}, we want a gadget consisting of variablesv₀,v₁,. . .,v_∆ that has only two possible satisfying assignments: (1) either 0 appears on everyvi, or (2) valueiappears on v_i for everyi. Such gadgets would allow us to use a counterexample to weak separability to prove hardness in similar way as hardness is proved in the Boolean case (see Example 6.9). However, due to the endomorphisms of the constraint language, such a gadget is not always possible to create: a nontrivial endomorphism can be used to transform satisfying assignments into new ones.

Therefore, our goal is more modest: we want a gadget that is either fully zero or represents an endomorphism in every satisfying assignment. That is, if variable vi gets value ci, then the mappingh defined by h(i) =c_i is an endomorphism. We enforce this requirement by introducing, for every R ∈ Γ and (a₁, . . . , a_r) ∈ R, a constraint h(v_a1, . . . , v_ar), Ri. Such a constraint ensures that applying the mapping h given by an assignment to the tuple (a1, . . . , ar) gives a tuple inR.

The gadgets we use in the reductions are more general than the one described in the previous paragraph: instead of a single variable v_i representing value i, we have a bag B_i of a variables representing this value. Setting the size of these bags and the cardinality/size constraint is an essential and delicate part of the reduction. The requirement that we want to enforce now is that if ψ(i) is the set of values appearing on the variables of B_i in a satisfying assignment, then ψ is a multivalued morphism of Γ. This can be ensured in a way similar to the construction in the previous paragraph (see below for details).

A minor technical detail is that we defined morphisms in such a way that 0 is always mapped to 0, thus there is no need to introduce variables representing what 0 is mapped to; it is more convenient to use constant 0’s instead. We need the following definition to formulate this conveniently. For a relationRand a tuplet∈R, we denote bysupp(t) the set of coordinate positions oftoccupied by nonzero elements. Letsupp_t(R) denote the relation obtained by substituting 0 into all coordinates of R except for supp(t), i.e. if R is r-ary and supp(t) ={1, . . . , r} \ {i₁, . . . , iq}, then supp_t(R) = R^{|i1,...,iq;0,...,0}.

For a cc0-language Γ and some 0∈D⁰ ⊆dom(Γ), amultivalued morphism gadget MVM(Γ, D⁰) consists of |D⁰| −1 bags of vertices B_d, d ∈ D⁰ \ {0}. The number of variables in each bag will be specified every time we use the gadget in a proof. The gadget is equipped with the following set of constraints. For every R ∈ Γ and every tuple t = (a1, . . . , ar) ∈ R|D⁰, we add all possible constraints hs,supp_t(R)i where supp(t) = {i₁, . . . , iq}, s = (vi1, . . . , viq), and vij ∈ Ba_ij for every 1 ≤ j ≤ q. The standard assignment of a gadget assigns a to every variable in bag B_a. Observe that the standard assignment satisfies all the constraints of the gadget. We say that bag B_a and the variables in bag Ba represent a. When we say that a gadget is fully nonzero, then we mean that all the variables are assigned nonzero values.

Proposition 3.20. Let 0∈D⁰⊆dom(Γ). Consider a satisfying assignment f of anMVM(Γ, D⁰) gadget. If h_f :D⁰ →2^dom(Γ) is a mapping such that h_f(a) is the set of values appearing in bag B_a of the gadget andhf(0) ={0}, then hf is an inner multivalued morphism ofΓ fromD⁰ todom(Γ).

Proof. Let R be a relation of Γ and let t = (a₁, . . . , a_r) ∈ R_|D⁰. Let (b₁, . . . , b_r) ∈ h_f(t). By the definition of hf, for every 1≤i≤r, either ai =bi = 0 or ai 6= 0 and there is a variable vi in bag Bai having valuebi. Letsupp_t={i₁, . . . , i_r⁰}. The gadget is defined such that there is a constraint

h(v_i1, . . . , v_ir0),supp_t(R)i, implying that (b_i1, . . . , b_ir0) ∈ supp_t(R) and hence (b₁, . . . , b_r) ∈ R. It follows thath_f(t)⊆R.

Note that if D⁰ = dom(Γ), then h_f is a multivalued morphism of Γ. Mapping h_f, or any mapping h⁰ with h⁰(a)⊆ hf(a) (for a∈ D⁰), is said to be an (inner) multivalued morphism given by the gadget and assignment f. If |h⁰(a)|= 1 for all a∈D⁰, we call it an endomorphism (inner homomorphism)givenby the gadget and assignment f.

We define two types of gadgets connecting MVM gadgets. The gadget NAND(G₁, G₂) on MVM(Γ, D⁰) gadgets G1, G2 consists of the following constraints. For every R ∈ Γ and disjoint tuples t₁ = (a₁, . . . , a_r), t₂ = (b₁, . . . , b_r) in R_|D⁰, we add a constraint hs,supp_t1+t2(R)i, for every s= (v_i1, . . . , v_iq) with {i₁, . . . i_q} =supp(t₁+t₂), such thatv_ij is in bagB_aij of G₁ ifa_ij 6= 0 and vij is in bag Bb_ij of G2 ifbij 6= 0.

It is not difficult to see (Lemma 3.21 below) that if one ofG1,G2 has the standard assignment and the other is fully zero, then all the constraints in NAND(G1, G2) are satisfied. On the other hand, if bothG₁ and G₂ have the standard assignment and there is a union counterexample, then NAND(G1, G2) is not satisfied. For the reductions, we need this second conclusion not only if both G1 and G2 have the standard assignment, but also assignments that “behave well” in some sense.

The right notion for our purposes is the following: An inner homomorphism h : D⁰ → dom(Γ) of Γ is t-recoverable for some tuple t if h is invertible on elements of t in the following sense: Γ has a multivalued morphism φ such that t ∈ (h◦φ)(t). In particular, this is true if there is an endomorphismh⁰ of Γ with (h◦h⁰)(t) =t. We say that a MVM(Γ, D⁰) gadget is t-recoverable in a given assignment if at least one of the inner homomorphisms given by it is t-recoverable.

Lemma 3.21. Let 0 ∈ D⁰ ⊆ dom(Γ) and let there be a NAND(G₁, G₂) gadget on MVM(Γ, D⁰) gadgets G1, G2.

1. If one of G₁ and G₂ has the standard assignment and the other gadget is fully zero, then all constraints of NAND(G1, G2) are satisfied.

2. Letf be a satisfying assignment of NAND(G1, G2). Ifhi is an inner homomorphism given by gadget Gi and assignment f for i= 1,2 and t1,t2 are disjoint tuples inR_|D⁰, then h1(t1) + h₂(t₂)∈R.

3. If there is a union counterexample (R,t1,t2) in Γ_|D⁰ and gadget Gi is ti-recoverable in as-signment f (for i= 1,2), then some constraint of NAND(G₁, G₂) is not satisfied by f. Proof. 1. Suppose without loss of generality that G₁ has the standard assignment and G₂ is fully zero. Consider a relationR∈Γ and disjoint tuplest₁ = (a₁, . . . , a_r), t₂ = (b₁, . . . , b_r)∈R_|D⁰. In a corresponding constrainth(v₁, . . . , vr),supp_t1+t2(R)i, ifai 6= 0, then variablevi has value ai (since it is in bagB_ai of G₁) and has value 0 otherwise. Thus (a₁, . . . , a_r)∈R implies that the constraint is satisfied.

2. If the NAND(G1, G2) instance is satisfied, then one of the constraints corresponding to t1

and t2 ensures that h1(t1) +h2(t2)∈R.

3. For i = 1,2, let h_i be a t_i-recoverable inner homomorphism given by G_i and let φ_i be a multivalued morphism such that ti ∈ (hi◦φi)(ti). Since hi is an inner homomorphism, we have that h1(t1), h2(t2) ∈R. Statement 2 implies that h1(t1) +h2(t2) ∈R. By Lemma 3.12, we have thath₁(t₁) +t⁰₂ ∈R for any t⁰₂ ∈φ₂(h₂(t₂)); in particular, this means that h₁(t₁) +t₂ is inR. As t2 ∈ R, we can apply Lemma 3.12 once more to get that t⁰₁+t2 ∈ R for any t⁰₁ ∈φ1(h1(t1)); in particular,t1+t2∈R, a contradiction.

The IMP(G₁, G₂) gadget is defined similarly, but instead oft₁,t₂ ∈R_|D⁰, we requiret₂,t₁+t₂ ∈ R_|D⁰.

Lemma 3.22. Let 0 ∈ D⁰ ⊆ dom(Γ) and let there be an IMP(G₁, G₂) gadget on MVM(Γ, D⁰) gadgets G1, G2.

1. If G₂ has the standard assignment and G₁ either has the standard assignment or fully zero, then all constraints of IMP(G1, G2) are satisfied.

2. Let f be a satisfying assignment of IMP(G₁, G₂). If h_i is an inner endomorphism given by gadget Gi and assignment f for i= 1,2 and t1,t2 are disjoint tuples such witht2,t1+t2 ∈ R_|D⁰, then h₁(t₁) +h₂(t₂)∈R.

3. If there is a difference counterexample (R,t1,t2) in Γ_|D⁰, G1 is t1-recoverable in assign-ment f and G₂ gives an inner homomorphism h₂ with h₂(t₂) = 0, then some constraint of IMP(G₁, G₂) is not satisfied.

Proof. 1. Similarly to the proof of Lemma 3.21, the constraint corresponding to (R,t₁,t₂) is satisfied, since t2,t1+t2 ∈R.

2. As in Lemma 3.21, follows from the definition of IMP(G1, G2).

3. Let h₁ be a t₁-recoverable inner homomorphism given by G₁. Let φ₁ be a multivalued morphism such thatt1∈(h1◦φ1)(t1). One of the constraints of IMP(G1, G2) ensure thath1(t1) + h2(t2) =h1(t1) is inR. Nowt1 ∈φ1(h1(t1)) is also inR, a contradiction.

When the multivalued morphism gadgets are used in the reductions, it will be essential that the bags of the gadgets have very specific sizes. We will ensure somehow that in a solution each bag is either fully zero or fully nonzero. Our aim is to choose the sizes of the bags in such a way that if the sum of the sizes of a collection of bags add up to a certain integer, then this is only possible if the collection contains exactly one bag of each size.

In most of the reductions, the MVM(Γ, D⁰) gadgets are arranged intgroups (corresponding to the t groups of vertices in the Multicolored Independent Set or Multicolored Implica-tionsinstance we are reducing from). For a gadget in group i, the bag representing d∈D⁰\ {0}

has size Z_i,d^t,D0, which is defined as follows. Fix an integer t and a set 0 ∈ D⁰ ⊆ D. It will be convenient to assume that D⁰ ={0,1, . . . ,∆}. For 1≤i≤tand 1≤d≤∆, we define

Z_i,d^t,D0 := (4t∆)^{2t∆+(i∆+d)}+ (4t∆)^{5t∆−(i∆+d)}.

By Z^t,D0 we denote the set of integersZ_i,d^t,D0 for 1≤i≤t and 1≤d≤∆. Note that these integers have exactly two nonzero digits if written in base-(4t∆) expansion; the positions of these two digits depend oni∆ +d. Furthermore, the “larger nonzero digit” of any number inZ^t,D0 is always larger than the “smaller nonzero digit” of any other number inZ^t,D0. We will use the following property of these integers:

Lemma 3.23. Let us fixt and D⁰={0,1, . . . ,∆}. IfA is a subset of Z^t,D0 and B is a multiset of values from Z^t,D0 such that |P

S∈AS−P

S∈BS|<(4t∆)^2t∆, then B is a set and B=A.

Proof. It can be assumed that A∩ B = ∅, since removing integers from both A and B does not change the difference of the sums. LetT be the largest integer inA∪B. Assume first thatT ∈ B \A.

This means that A contains only integers strictly smaller than T, and (as A is a set) an integer appears at most once inA. Since there aret∆ integers inZ^t,D0 and every integer smaller thanT is

at most 2T /(4t∆), we have that the sum of the integers inA is at most T /2. Thus the difference of the sums is at least (4t∆)^2t∆, a contradiction.

Assume now that T ∈A\ B. Suppose thatT =Z_i,d^t,D0 and letX:= (4t∆)^{2t∆+(i∆+d)}. SinceT is the largest integer inA∪ B (i.e, i∆ +dis as small as possible), every integer inA∪ B other thanT is divisible by (4t∆)2t∆+(i∆+d+1) =X·4t∆, whileT is equal toX modulo X·4t∆. Thus P

S∈AS is X modulo X·4t∆, while P

S∈BS is divisible by X·4t∆. Therefore, |P

S∈AS−P

S∈BS| is at least min{X,4t∆·X−X}>(4t∆)^2t∆, a contradiction.