Lifting with Simple Gadgets and Applications to Circuit and Proof Complexity
Susanna F. de Rezende, Or Meir, Jakob Nordström, Toniann Pitassi, Robert Robere, Marc Vinyals
aa r X i v : . [ c s . CC ] J a n Lifting with Simple Gadgets andApplications to Circuit and Proof Complexity
Susanna F. de Rezende Or Meir Jakob Nordstr¨om
Mathematical Institute of the University of Haifa University of CopenhagenCzech Academy of Sciences KTH Royal Instituteof Technology
Toniann Pitassi Robert Robere Marc Vinyals
University of Toronto DIMACS TechnionInstitute for Advanced Study Institute for Advanced Study
Abstract
We significantly strengthen and generalize the theorem lifting Nullstellensatz degree to monotonespan program size by Pitassi and Robere (2018) so that it works for any gadget with high enough rank,in particular, for useful gadgets such as equality and greater-than . We apply our generalized theoremto solve two open problems: • We present the first result that demonstrates a separation in proof power for cutting planes withunbounded versus polynomially bounded coefficients. Specifically, we exhibit CNF formulasthat can be refuted in quadratic length and constant line space in cutting planes with unboundedcoefficients, but for which there are no refutations in subexponential length and subpolynomialline space if coefficients are restricted to be of polynomial magnitude. • We give the first explicit separation between monotone Boolean formulas and monotone realformulas. Specifically, we give an explicit family of functions that can be computed with mono-tone real formulas of nearly linear size but require monotone Boolean formulas of exponentialsize. Previously only a non-explicit separation was known.An important technical ingredient, which may be of independent interest, is that we show thatthe Nullstellensatz degree of refuting the pebbling formula over a DAG G over any field coincidesexactly with the reversible pebbling price of G . In particular, this implies that the standard decisiontree complexity and the parity decision tree complexity of the corresponding falsified clause searchproblem are equal. Introduction
Lifting theorems in complexity theory are a method of transferring lower bounds in a weak computationalmodel into lower bounds for a more powerful computational model, via function composition. Therehas been an explosion of lifting theorems in the last ten years, essentially reducing communication lowerbounds to query complexity lower bounds.Early papers that establish lifting theorems include Raz and McKenzie’s separation of the mono-tone NC hierarchy [RM99] (by lifting decision tree complexity to deterministic communication com-plexity), and Sherstov’s pattern matrix method [She11] which lifts (approximate) polynomial degreeto (approximate) matrix rank. Recent work has established query-to-communication lifting theorems ina variety of models, leading to the resolution of many longstanding open problems in many areas ofcomputer science. Some examples include the resolution of open questions in communication complex-ity [GPW15, GLM +
15, GKPW17, GJPW17, GPW18], monotone complexity [RPRC16, PR17, PR18],proof complexity [HN12, GP18, dRNV16, GGKS18], extension complexity of linear and semidefiniteprograms [KMR17, GJW18, LRS15], data structures [CKLM18] and finite model theory [BN16].Lifting theorems have the following form: given functions f : { , } n → { , } (the “outer function”)and g : X × Y → { , } (the “gadget”), a lower bound for f in a weak computational model implies alower bound on f ◦ g n in a stronger computational model. The most desirable lifting theorems are themost general ones. First, it should hold for any outer function, and ideally f should be allowed to be apartial function or a relation (i.e., a search problem). Indeed, nearly all of the applications mentionedabove require lifting where the outer function is a relation or a partial function. Secondly, it is oftendesirable that the gadget is as small as possible. The most general lifting theorems established so far,for example lifting theorems for deterministic and randomized communication complexity, require atleast logarithmically-sized gadgets; if these theorems could be improved generically to hold for constant-sized gadgets then many of the current theorems would be vastly improved. Some notable exampleswhere constant-sized gadgets are possible include Sherstov’s degree-to-rank lifting [She11], critical block-sensitivity lifting [GP18, HN12], and lifting for monotone span programs [PR17, PR18, Rob18]. In this paper, we generalize a lifting theorem of Pitassi and Robere [PR18] to use any gadget that hasnontrivial rank. This theorem takes a search problem associated with an unsatisfiable CNF, and liftsa lower bound on the Nullstellensatz degree of the CNF to a lower bound on a related communicationproblem.More specifically, let C be an unsatisfiable k -CNF formula. The search problem associated with C , Search ( C ) , takes as input an assignment to the underlying variables, and outputs a clause that is falsifiedby the assignments. [PR18] prove that for any unsatisfiable C , and for a sufficiently rich gadget g , de-terministic communication complexity lower bounds for the composed search problem Search ( C ) ◦ g n follow from Nullstellensatz degree lower bounds for C . We significantly improve this lifting theorem sothat it holds for any gadget of large enough rank.
Theorem 1.1.
Let C be a CNF over n variables, let F be any field, and let g be any gadget of rank at least r .Then the deterministic communication complexity of Search ( C ◦ g n ) is at least NS F ( C ) , the Nullstellensatzdegree of C , as long as r ≥ cn/ NS F ( C ) for some large enough constant c . An important special case of our generalized theorem is when the gadget g is the equality function. Inthis work, we apply our theorem to resolve two open problems in proof complexity and circuit complexity.Both solutions depend crucially on the ability to use the equality gadget.We note that lifting with the equality gadget has recently been the focus of another paper. Loff andMukhopadhyay [LM19] observed that a lifting theorem for total functions with the equality gadget can In fact the result is quite a bit stronger—it applies to Razborov’s rank measure [Raz90], which is a strict strengthening ofdeterministic communication complexity.
1e proven using a rank argument. Surprisingly, they also observed that it is not possible to lift querycomplexity to communication complexity for arbitrary relations! Concretely, [LM19] give an exampleof a relation with linear query complexity but whose composition with equality has only polylogarithmiccommunication complexity. Nonetheless, they are able to prove a lifting theorem for general relationsusing the equality gadget by replacing standard query complexity with a stronger complexity measure(namely, the -query complexity of the relation).Unfortunately, we cannot use either of the lifting theorems of [LM19] for our applications. Specifi-cally, in our applications we lift a search problem (and therefore cannot use their result for total functions),and this search problem has small -query complexity (and therefore we cannot use their lifting theoremfor general relations). Indeed, this shows that our lifting theorem is incomparable to the results of [LM19],even when specialized to the equality gadget. We note that our theorem, too, bypasses the impossibilityresult of [LM19] by using a stronger complexity measure, which in our case is the Nullstellensatz degree. The main application of our lifting theorem is the first separation in proof complexity between cuttingplanes proofs with high-weight versus low-weight coefficients. The cutting planes proof-system is a proofsystem that can be used to refute an unsatisfiable CNF by translating it into a system of integer inequalitiesand showing that this system has no integer solution. The latter is achieved by a sequence of steps thatderive new integer inequalities from old ones, until we derive the inequality ≥ (which clearly hasno solution). The efficiency of such a refutation is measured by its length (i.e., the number of steps)and its space (i.e., the maximal number of inequalities that have to be stored simultaneously during thederivation).The standard variant of the cutting planes proof system, commonly denoted by CP, allows the inequal-ities to use coefficients of arbitrary size. However, it is also interesting to consider the variant in whichthe coefficients are polynomially bounded, which is commonly denoted by CP ∗ . This gives rise to thenatural question of the relative power of CP vs. CP ∗ : are they polynomially equivalent or is there a super-polynomial length separation? This question appeared in [BC96] and remains stubbornly open to date.In this work we finally make progress by exhibiting a setting in which unbounded coefficients afford anexponential increase in proof power. Theorem 1.2.
There is a family of CNF formulas of size N that have cutting planes refutations of length ˜ O ( N ) and space O (1) , but for which any refutation in length L and space s with polynomially boundedcoefficients must satisfy s log L = ˜Ω( N ) . Our result is the first result in proof complexity demonstrating any situation where high-weight coeffi-cients are more powerful than low-weight coefficients. In comparison, for computing Boolean functions,the relative power of high-weight and low-weight linear threshold functions has been understood for along time. The greater-than function can be computed by high-weight threshold functions, but not bylow-weight threshold functions, and weights of length polynomial in n suffice [Mur71] for Boolean func-tions. For higher depth threshold formulas, it is known that depth- d threshold formulas of high-weightcan efficiently be computed by depth- ( d + 1) threshold formulas of low-weight [GHR92].In contrast to our near-complete knowledge of high versus low weights for functions, almost nothingis known about the relative power of high versus low weights in the context of proof complexity. Buss andClote [BC96], building on work by Cook, Coullard, and Tur´an [CCT87], proved an analog of Muroga’sresult for cutting planes, showing that weights of length polynomial in the length of the proof suffice.Quite remarkably, this result is not known to hold for other linear threshold proof systems: there is no nontrivial upper bound on the weights for more general linear threshold propositional proof systems (suchas stabbing planes [BFI + .3 A Separation in Circuit Complexity A second application of our lifting theorem relates to monotone real circuits, which were introduced byPudl´ak [Pud97]. A monotone real circuit is a generalization of monotone Boolean circuits where each gateis allowed to compute any non-decreasing real function of its inputs, but the inputs and output of the circuitare Boolean. A formula is a tree-like circuit, that is, every gate has fan-out one. The first (exponential)lower bound for monotone real circuits was proven already in [Pud97] by extending the lower bound forcomputing the clique-colouring function with monotone Boolean circuits [Raz85, AB87]. This lowerbound, together with a generalization of the interpolation technique [Kra97] which applied only to CP ∗ ,was used by Pudl´ak to obtain the first exponential lower bounds for CP.Shortly after monotone real circuits were introduced, there was an interest in understanding the powerof monotone real computation in comparison to monotone Boolean computation. By extending tech-niques in [RM99], Bonet et al. prove that there are functions with polynomial size monotone Booleancircuits that require monotone real formulas of exponential size [Joh98, BEGJ00]. This illustrates thepower of DAG-like computations in comparison to tree-like. In the other direction, we would like toknow whether monotone real circuits are exponentially stronger than monotone Boolean circuits. Rosen-bloom [Ros97] presented an elegant, simple proof that monotone real formulas are exponentially strongerthan (even non-monotone) Boolean circuits, since slice functions can be computed by linear-size mono-tone real formulas, whereas by a counting argument we know that most slice functions require exponentialsize Boolean circuits.The question of finding explicit functions that demonstrate that monotone real circuits are strongerthan general Boolean circuits is much more challenging since it involves proving explicit lower bounds forBoolean circuits—a task that seems currently completely out of reach. A more tractable problem is thatof finding explicit functions showing that monotone real circuits or formulas are stronger than monotone Boolean circuits or formulas, but prior to this work, no such separation was known either. We providean explicit separation for monotone formulas , that is, we provide a family of explicit functions that canbe computed with monotone real formulas of near-linear size but require exponential monotone Booleanformulas. This is the first explicit example that illustrates the strength of monotone real computation.
Theorem 1.3.
There is an explicit family of functions f n over O ( n polylog n ) variables that can be com-puted by monotone real formulas of size O ( n polylog n ) but for which every monotone Boolean formularequires size Ω( n/ log n ) . Another motivation for studying lifting theorems with simple gadgets, and in particular the equalitygadget, are connections with proving non-monotone formula size lower bounds. As noted earlier, liftingtheorems have been extremely successful in proving monotone circuit lower bounds, and it has also beenshown to be useful in some computational settings that are only “partially” monotone; notably monotonespan programs [RPRC16, PR17, PR18] and extended formulations [GJW18, KMR17].This raises the question of to what extent lifting techniques can help prove non-monotone lowerbounds. The beautiful work by Karchmer, Raz and Wigderson [KRW95] initiated such an approach forseparating P from NC —this opened up a line of research popularly known as the KRW conjecture . In-triguingly, steps towards resolving the KRW conjecture are closely connected to proving lifting theoremsfor the equality gadget. The first major progress was made in [EIRS01] where lower bounds for the uni-versal relation game are proven, which is an important special case of the KRW conjecture. Their resultwas recently improved in several papers [GMWW17, HW93, KM18], and Dinur and Meir [DM18] gavea new top-down proof of the state-of-the-art Ω( n ) formula-size lower bounds via the KRW approach.The connection to lifting using the equality gadget is obtained by observing that the KRW conjectureinvolves communication problems in which Alice and Bob are looking for a bit on which they differ—thisis exactly an equality problem. Close examination of the results in [EIRS01, HW93] show that they areequivalent to proving lower bounds for the search problem associated with the pebbling formula whenlifted with a -bit equality gadget on a particular graph [Pit16]. Our proof of Theorem 4.1 actuallyestablishes near-optimal lower bounds on the communication complexity of the pebbling formula liftedwith equality for any graph, but where the size of the equality is not 1. Thus if our main theorem could3e improved with one-bit equality gadgets this would imply the results of [EIRS01, HW93] as a directcorollary and with significantly better parameters. We conclude this section by giving a brief overview of our techniques, also trying to convey some of thesimplicity of the proofs which we believe is an extra virtue of these results.
Lifting theorem
In order to prove their lifting theorem, Pitassi and Robere [PR18] defined a notion ofa “good” gadget. They then showed that if we compose a polynomial p with a good gadget g , the rank ofthe resulting matrix p ◦ g n is determined exactly by the non-zero coefficients of p and the rank of g . Theirlifting theorem follows by using this correspondence to obtain bounds on the ranks of certain matrices,which in turn yield the required communication complexity lower bound.In this work, we observe that every gadget g can be turned into a good gadget using a simple trans-formation. This observation allows us to get an approximate bound on the rank of p ◦ g n for any g withnontrivial rank. While the correspondence we get in this way is only an approximation and not an exactcorrespondence as in [PR18], it turns out that this approximation is sufficient to prove the required lowerbounds. We thus get a lifting theorem that works for every gadget g with sufficiently large rank. Cutting planes separation
The crux of our separation between CP and CP ∗ is the following observa-tion: CP can encode a conjunction of linear equalities with a single equality, by using exponentially largecoefficients. This allows CP refutations to obtain a significant saving in space when working with linearequalities . This saving is not available to CP ∗ , and this difference between the proof systems allows theseparation.In order to exploit this observation, one of our main innovations is to concoct the separating formula.To do this, we must come up with a candidate formula that can only be refuted by reasoning about a largeconjunction of linear equalities, to show that cutting planes (CP) can efficiently refute it, and to show thatlow-weight cutting planes (CP ∗ ) cannot.To find such a candidate formula family we resort to pebbling formulas which have played a major rolein many proof complexity trade-off results. Interestingly, pebbling formulas have short resolution proofsthat reason in terms of large conjunctions of literals. When we lift such formulas with the equality gadgetthis proof can be simulated in cutting planes by using the large coefficients to encode many equalitieswith a single equality. This yields cutting planes refutation of any pebbling formula in quadratic lengthand constant space.On the other hand we prove our time-space lower bound showing that any CP ∗ refutation requires largelength or large space for the same formulas. To prove this lower bound, the first step is to instantiate theconnection in [HN12] linking time/space bounds for many proof systems to communication complexitylower bounds for lifted search problems. This connection means that we can obtain the desired CP ∗ -lower bounds for our formulas Peb G ◦ EQ n by proving communication complexity lower bounds for thecorresponding lifted search problem Search (Peb G ) ◦ EQ n .In order to prove the latter communication lower bounds, we prove lower bounds on the Nullstellensatzdegree of Search (Peb G ) , and then invoke our new lifting theorem to translate them into communicationlower bounds for Search (Peb G ) ◦ EQ n . To show the Nullstellensatz lower bounds, we prove the followinglemma, which establishes an equivalence between Nullstellsatz degree and the reversible pebbling price,and may be interesting in its own right. (We remark that connections between Nullstellensatz degree andpebbling were previously shown in [BCIP02]; however their result was not tight.) Lemma 1.4.
For any field F and any directed acyclic graph G the Nullstellensatz degree of Peb G is equalto the reversible pebbling price of G . We remark that due to known lower and upper bounds in query and proof complexity, this lemmaimmediately implies that Nullstellensatz degree coincides for (deterministic) decision tree and parity de-4ision tree complexity. We record this here as a corollary, as it may be of independent interest, and provideits proof in Appendix C.
Corollary 1.5.
For any field F and any directed acyclic graph G , the Nullstellensatz degree over F of Peb G , the decision tree depth of Search (Peb G ) , and the parity decision tree depth of Search (Peb G ) coincide and are equal to the reversible pebbling price of G . Using the above equivalence, we obtain near-linear Nullstellensatz degree refutations for a family ofgraphs with maximal pebbling price, which completes our time/space lower bound for CP ∗ . However,in order to separate CP and CP ∗ we require a very specific gadget and lifting theorem. Specifically, thegadget should be strong enough, so that lifting holds for deterministic communication complexity (whichcan efficiently simulate small time/space CP ∗ proofs), but on the other hand also weak enough, so thatlifting does not hold for stronger communication models (randomized, real) that can efficiently computehigh-weight inequalities. The reason that we are focusing on the equality gadget is that it hits this sweetspot—it requires large deterministic communication complexity, yet has short randomized protocols, andfurthermore equalities can be represented with a single pair of inequalities. Separation for monotone formulas
As was the case for the separation between CP and CP ∗ , to obtaina separation between monotone Boolean formulas and monotone real formulas we must find a functionthat has just the right level of hardness.To obtain a size lower bound for monotone Boolean formulas we invoke the characterization of for-mula depth by communication complexity of the Karchmer–Wigderson game [KW90]. By choosing afunction that has the same Karchmer–Wigderson game as the search problem of a lifted pebbling for-mula, we get a depth lower bound for monotone Boolean formulas from the communication lower boundof the search problem. Note that since monotone Boolean formulas can be balanced, a depth lower boundimplies a size lower bound.In the other direction, we would like to show that these functions are easy for real computation. Anal-ogously to the Karchmer–Wigderson relation, it was shown in [HP18] that there is a correspondence be-tween real DAG-like communication protocols (as defined in [Kra98]) and monotone real circuits. Usingthis relation, a small monotone real circuit can be extracted from a short CP proof of the lifted pebblingformula. However, we would like to establish a monotone real formula upper bound. One way to achievethis is by finding small tree-like CP refutations of lifted pebbling formulas. The problem is that for manygadgets lifted pebbling formulas require exponentially long tree-like proofs. Nevertheless, for pebblingformulas lifted with the equality gadget we are able to exhibit a short semantic tree-like CP refutation,which via real communication yields a small monotone real formula. Section 2 contains formal definitions of concepts discussed above and some useful facts. Our main liftingtheorem is proven in Section 3. Section 4 is devoted to proving our separation between high-weight andlow-weight cutting planes. In Section 5 we prove the separation between monotone real and Booleanformulas. We conclude in Section 6 with some open problems.
In this section we review some background material from communication complexity and proof complex-ity.
Given a function g : X × Y → I , we denote by g n : X n × Y n → I n the function that takes as input n independent instances of g and applies g to each of them separately. A total search problem is a relation5 ⊆ I × O such that for all z ∈ I there is an o ∈ O such that ( z, o ) ∈ S . Intuitively, S represents thecomputational task in which we are given an input z ∈ I and would like to find an output o ∈ O thatsatisfies ( z, o ) ∈ S .An important example of a search problem, which has proved to be very useful for proof complexityresults, comes from unsatisfiable k -CNF formulas. Given a k -CNF formula C over variables z , . . . , z n ,the search problem Search ( C ) ⊆ { , } n × C takes as input an assignment z ∈ { , } n and outputs aclause C ∈ C that is falsified by z .Given a search problem S ⊆ I n × O with a product input domain and a function g : X × Y → I ,we define the composition
S ◦ g n ⊆ X n × Y n × O in the natural way: ( x, y, o ) ∈ S ◦ g if and only if ( g n ( x, y ) , o ) ∈ S . We remark that this composition notation extends naturally to functions: for instance,if f : I n → F is a function taking values in some field F , for example, then the composition f ◦ g n is a X n × Y n matrix over F . Second, we remark that we will sometimes write S ◦ g instead of S ◦ g n if n isclear from context.A communication search problem is a search problem with a bipartite input domain I = A × B . Acommunication protocol for a search problem
S ⊆ A × B × O is a strategy for a collaborative gamewhere two players Alice and Bob hold x ∈ A , y ∈ B , respectively, and wish to output an o ∈ O suchthat (( x, y ) , o ) ∈ S while communicating as few bits as possible. Messages are sent sequentially untilone player announces the answer and only depend on the input of one player and past messages. The costof a protocol is the maximum number of bits sent over all inputs, and the communication complexity ofa search problem, which we denote by P cc ( S ) , is the minimum cost over all protocols that solve S . Formore details on communication complexity, see, for instance, [KN97].Given a CNF formula C on n variables z , z , . . . , z n and a Boolean function g : { , } q × { , } q →{ , } , we define a lifted formula C ◦ g n as follows. For each variable z i of C , we have q new variables x i, , . . . , x i,q , y i, , . . . , y i,q . For each clause C ∈ C we replace each literal z i or ¬ z i in C by a CNFencoding of either g ( x i, , . . . , x i,q , y i, , . . . , y i,q ) or ¬ g ( x i, , . . . , x i,q , y i, , . . . , y i,q ) according to the signof the literal. We then expand the resulting expression into a CNF, which we denote by C ◦ g , usingde Morgan’s rules. The substituted formula is C ◦ g = S C ∈C C ◦ g .For the sake of an example, consider the clause u ∨ v , and we will substitute with the equality gadgeton two bits. Formally, we replace u with x u, x u, = y u, y u, and v with x v, x v, = y v, y v, . We canencode a two-bit equality as the CNF formula ( x x = y y ) ≡ ( x ∨ y ) ∧ ( x ∨ y ) ∧ ( x ∨ y ) ∧ ( x ∨ y ) , and a two-bit disequality as the CNF formula ( x x = y y ) ≡ ( x ∨ x ∨ y ∨ y ) ∧ ( x ∨ x ∨ y ∨ y ) ∧ ( x ∨ x ∨ y ∨ y ) ∧ ( x ∨ x ∨ y ∨ y ) . So, in the clause u ∨ v , we would substitute u for the CNF encoding of x u, x u, = y u, y u, and v with theCNF encoding of x v, x v, = y v, y v, ; finally, we would convert the new formula to a CNF by distributingthe top ∨ over the ∧ s from the new CNF encodings.While Search ( C ) ◦ g n is not the same problem as Search ( C ◦ g n ) , we can reduce the former to the latter.Specifically, suppose we are given a protocol Π for Search ( C ◦ g n ) . Consider the following protocol Π ′ for Search ( C ) ◦ g n : Given an input ( x, y ) , the protocol Π ′ interprets ( x, y ) as an input to Π . Now, assumethat Π ′ outputs on ( x, y ) a clause D of C ◦ g n , which was obtained from a clause C of C . Then, the clause C is a valid Search ( C ) on ( x, y ) , so Π ′ outputs it. Let us record this observation. Observation 2.1. P cc ( Search ( C ◦ g )) ≥ P cc ( Search ( C ) ◦ g ) for any unsatisfiable CNF C and any Booleangadget g . As a proof system, Nullstellensatz allows verifying that a set of polynomials does not have a commonroot, and it can also be used to refute CNF formulas by converting them into polynomials. It plays animportant role in our lower bounds. 6et F be a field, and let P = { p = 0 , p = 0 , . . . , p m = 0 } be an unsatisfiable system of poly-nomial equations in F [ z , z , . . . , z n ] . A Nullstellensatz refutation of P is a sequence of polynomials q , q , . . . , q m ∈ F [ z , z , . . . , z n ] such that P mi =1 p i q i = 1 where the equality is syntactic. The degree of the refutation is max i deg( p i q i ) ; the Nullstellensatz degree of P , denoted NS F ( P ) , is the minimumdegree of any Nullstellensatz refutation of P .Let C = C ∧ C ∧ · · · ∧ C m be an unsatisfiable CNF formula over Boolean variables z , z , . . . , z n .We introduce a standard encoding of each clause C i as a polynomial equation. If C is a clause then let C + denote the set of variables occurring positively in C and C − denote the set of variables occurringnegatively in C ; with this notation we can write C = W z ∈ C + z ∨ W z ∈ C − z. From C define the polynomial E ( C ) ≡ Y z ∈ C + (1 − z ) Y z ∈ C − z, over formal variables z , z , . . . , z n . Observe that E ( C ) = 0 is satisfied (over / assignments to z i ) ifand only if the corresponding assignment satisfies C . We abuse notation and let E ( C ) = {E ( C ) : C ∈C} ∪ { z i − z i } i ∈ [ m ] , and note that the second set of polynomial equations restricts the z i inputs to { , } values. The F -Nullstellensatz degree of C , denoted NS F ( C ) , is the Nullstellensatz degree of refuting E ( C ) .How do we know that a Nullstellensatz refutation always exists? One can deduce this from Hilbert’sNullstellensatz, but for our purposes it is enough to use a simpler version proved by Buss et al. (Theorem5.2 in [BIK + P is a system of polynomial equations over F [ z , . . . , z n ] with no { , } solutions,then there exists a Nullstellensatz refutation of P ∪ { z i − z i = 0 } i ∈ [ n ] . The
Cutting planes (CP) proof system was introduced in [CCT87] as a formalization of the integer linearprogramming algorithm in [Gom63, Chv73]. Cutting planes proofs give a formal method to deducenew linear inequalities from old that are sound over integer solutions —that is, if some integral vector x ∗ satisfies a set of linear inequalities I , then x ∗ will also satisfy any inequality ax ≥ b deduced from I by a sequence of cutting planes deductions. The allowed deductions in a cutting planes proof are thefollowing:Linear combination P i a i x i ≥ A P i b i x i ≥ B P i ( ca i + db i ) x i ≥ cA + dB Division P i ca i x i ≥ A P i a i x i ≥ ⌈ A/c ⌉ where a i , b i , c , d , A , and B are all integers and c, d ≥ .In order to use cutting planes to refute unsatisfiable CNF formulas, we need to translate clauses toinequalities. It is easy to see how to do this by example: we translate the clause x ∨ y ∨¬ z to the inequality x + y + (1 − z ) ≥ , or, equivalently, x + y − z ≥ if we collect all constant terms on the right-hand side.For refuting CNF formulas we equip cutting planes proofs with the following additional rules ensuringall variables take { , } values: Variable axioms x ≥ − x ≥ − The goal, then, is to prove unsatisfiability by deriving the inequality ≥ . This is possible if andonly if there is no { , } -assignment satifying all constraints.As discussed in the introduction, we are interested in several natural parameters of cutting planesproof—length, space, and the sizes of the coefficients. So, we define a cutting planes refutation as asequence of configurations (this is also known as the blackboard model ). A configuration is a set of linearinequalities with integer coefficients, and a sequence of configurations C , . . . , C L is a cutting planesrefutation of a formula C if C = ∅ , C L contains the contradiction ≥ , and each configuration C t +1 follows from C t either by adding an inequality in C , by adding the result of one of the above inference ruleswhere all the premises are in C t , or by removing an inequality present in C t . The length of a refutation7s then defined to be the number of configurations L ; the space is max t ∈ [ L ] | C t | , the maximum numberof inequalities in a configuration; and the coefficient bit size is the maximum size in bits of a coefficientthat appears in the refutation.For any proof system, it is natural to ask what is the minimal amount of space needed to prove tau-tologies. Indeed, there has been much work in the literature studying this, and for proof systems suchas resolution (e.g. [ET01, ABRW02, BG03, BN08]) and polynomial calculus (e.g. [ABRW02, FLN + + ∗ to be cutting planes proofs with polynomially-bounded coefficients or, in other words, a cutting planes refutation Π of a formula C with n variablesis a CP ∗ refutation if the largest coefficient in Π has magnitude poly ( n, L ) .The question of how CP ∗ relates to unrestricted cutting planes has been raised in several papers, e.g.,[BPR97, BEGJ00]. This question was studied already in [BC96], where it was proven that any cuttingplanes refutation in length L can be transformed into a refutation with L O (1) lines having coefficents ofmagnitude exp( O ( L )) (here the asymptotic notation hides a mild dependence on the size of the coeffi-cients in the input). The authors write, however, that their original goal had been to show that coefficientsof only polynomial magnitude would be enough, i.e., that CP ∗ would be as powerful as cutting planes ex-cept possibly for a polynomial loss, but that they had to leave this as an open problem. To the best of ourknowledge, there has not been a single example of any unsatisfiable formula where CP ∗ could potentiallyperform much worse than general (high-weight) cutting planes.Finally, as observed in [BPS07, HN12], we can use an efficient cutting planes refutation of a formula C to solve Search ( C ) by an efficient communication protocol. Since the first configuration C is alwaystrue and the last configuration C L is always false, the players can simulate a binary search by evaluatingthe truth value of a configuration according to their joint assignment and find a true configuration followedby a false configuration. It is not hard to see that the inequality being added corresponds to a clause in C and it is a valid answer to Search ( C ) . Lemma 2.2 ([HN12]).
If there is a cutting planes refutation of C in length L , line space s , and coefficientbit size c , then there is a deterministic communication protocol for Search ( C ) of cost O ( s ( c +log n ) log L ) . In this section we discuss our new lifting theorem, restated next. Theorem 3.1.
Let C be any unsatisfiable k -CNF on n variables and let F be any field. For any Booleanvalued gadget g with rank( g ) ≥ enk/ NS F ( C ) we have P cc ( Search ( C ) ◦ g ) ≥ NS F ( C ) . This generalizes a recent lifting theorem from [PR18], which only allowed certain “good” gadgets.The main technical step of that proof showed that “good” gadgets can be used to lift the degree of multi-linear polynomials to the rank of matrices. In this section, we improve this, showing that any gadget withnon-trivial rank can be used to lift polynomial degree to rank. Given this result, Theorem 3.1 is proved byreproducing the proof of [PR18] with a tighter analysis. With this in mind, in this section we will prove Formally, this is known as the line space . In fact, we prove a somewhat more general theorem (see Theorem A.1 in Appendix A for details). We also remark that thistheorem in fact holds for a stronger communication measure (Razborov’s rank measure [Raz90]), and so implies lower boundsfor other models—see Appendix A for details.
Definition 3.2 (Definition 3.1 in [PR18]).
Let F be a field. A gadget g : X × Y → F is good if for anymatrices A, B of the same size we have rank( X , Y ⊗ A + g ⊗ B ) = rank( A ) + rank( g )rank( B ) where X , Y denotes the X × Y all- s matrix.In [PR18] it is shown that good gadgets are useful because they lift degree to rank when composedwith multilinear polynomials. Theorem 3.3 (Theorem 1.2 in [PR18]).
Let F be any field, and let p ∈ F [ z , z , . . . , z n ] be a multilinearpolynomial over F . For any good gadget g : X × Y → F we have rank( p ◦ g n ) = X S :ˆ p ( S ) =0 rank( g ) | S | . In the present work, we show that a gadget being good is not strictly necessary to obtain the abovelifting from degree to rank. In fact, composing with any gadget lifts degree to rank!
Theorem 3.4.
Let p ∈ F [ z , z , . . . , z n ] be any multilinear polynomial and let g : X × Y → F be anynon-zero gadget with rank( g ) ≥ . Then X S :ˆ p ( S ) =0 (rank( g ) − | S | ≤ rank( p ◦ g n ) ≤ X S :ˆ p ( S ) =0 rank( g ) | S | . We remark that the lower bound in the theorem can be sharpened to rank( g ) − if the gadget g isnot full rank. While the previous theorem does not require the gadget g to be good, the notion of a goodgadget will still play a key role in the proof. The general idea is that every gadget with non-trivial rankcan be transformed into a good gadget with a slight modification. With this in mind, en-route to provingTheorem 3.4 we give the following characterization of good gadgets which may be of independent interest. Lemma 3.5.
A gadget g is good if and only if the all- s vector is not in the row or column space of g . In the remainder of the section we prove Theorem 3.4 and Lemma 3.5.
We begin by proving Lemma 3.5, which is by a simple linear-algebraic argument. Given a matrix M over a field, let row ( M ) denote the row-space of M and let col ( M ) denote the column-space of M . Thefollowing characterization of when rank is additive will be crucial. Theorem 3.6 ([MS72]).
For any matrices
A, B of the same size over any field, rank( A + B ) = rank( A )+rank( B ) if and only if row ( A ) ∩ row ( B ) = col ( A ) ∩ col ( B ) = { } . The previous theorem formalizes the intuition that rank should be additive if and only if the corre-sponding linear operators act on disjoint parts of the vector space. Using the previous theorem we deducethe following general statement, from which Lemma 3.5 immediately follows.
Lemma 3.7.
Let f, g be matrices over any fixed field F of the same size. The following are equivalent:1. For all matrices A, B of the same size, rank( f ⊗ A + g ⊗ B ) = rank( f )rank( A )+rank( g )rank( B ) . rank( f + g ) = rank( f ) + rank( g ) . roof. By choosing A = B = (1) we instantly deduce (2) from (1). To prove the converse, we useTheorem 3.6. Let A, B be matrices such that rank( f ⊗ A + g ⊗ B ) = rank( f )rank( A )+rank( g )rank( B ) .Then by Theorem 3.6 it follows that there is a non-zero vector in the intersection of either the row- orcolumn-spaces of f ⊗ A and g ⊗ B . Suppose that there is a non-zero vector u ∈ col ( f ⊗ A ) ∩ col ( g ⊗ B ) ,and we prove that there is a non-zero vector in col ( f ) ∩ col ( g ) implying rank( f + g ) = rank( f )+rank( g ) .(A symmetric argument will apply to the row spaces.)Assume that f and g are a × b dimensional matrices, and that A and B are m × n dimensionalmatrices. Let u be the length am non-zero vector in the column spaces of both f ⊗ A and g ⊗ B , andsuppose without loss of generality that u = 0 . It follows that there are length bn vectors x, y such that ( f ⊗ A ) x = u = ( g ⊗ B ) y . Write x = ( x , x , . . . , x b ) ,y = ( y , y , . . . , y b ) where x i , y i are vectors of length n for each i .Let A denote the first row of A and B denote the first row of B ; note they are both vectors of length n . Define the length- b vectors x ′ = ( A x , A x , . . . , A x b ) ,y ′ = ( B y , B y , . . . , B y b ) . Then, by definition, for each i = 1 , , . . . , a we have ( f x ′ ) i = u ( i − m +1 = ( gy ′ ) i , and the vector isnon-zero since u = 0 by assumption. Thus f x ′ = gy ′ and the column spaces of f and g intersect at anon-zero vector.From Lemma 3.7 we can deduce Lemma 3.5 immediately. Proof of Lemma 3.5.
By the previous lemma, g is good if and only if rank( + g ) = rank( ) + rank( g ) .By Theorem 3.6 this is true iff the all- s vector is not in the row- or column-space of g . In this section we prove Theorem 3.4 using Lemma 3.5. The theorem follows by induction using thefollowing lemma, and the proof mimics the proof from [PR18, Rob18].
Lemma 3.8.
Let F be any field, and let g : X × Y → F be any gadget with rank( g ) ≥ . For anymatrices A, B of the same size we have rank( X , Y ⊗ A + g ⊗ B ) ≥ rank( A ) + (rank( g ) − B ) where X , Y is the X × Y all- s matrix.Proof. Assume without loss of generality that |X | ≥ |Y| and let = X , Y . Thinking of g as a matrix, let u be any column vector of g . If we zero the entries of u in g , then the remaining matrix cannot have fullrank, implying that some row-vector v of the remaining matrix will become linearly dependent. Let g be the X × Y matrix consisting of the u column and v row of g , and let g be the X × Y matrix obtainedby zeroing out u and v in g . Observe g = g + g , and also since g contains an all- row and an all- column it is good by Lemma 3.5 (as any linear combination of rows/columns of g must contain a zerocoordinate).Now, observe that rank( ⊗ A + g ⊗ B ) = rank( ⊗ A + g ⊗ B + g ⊗ B ) ≥ rank( ⊗ A + g ⊗ B ) − rank( g ⊗ B )= rank( ⊗ A + g ⊗ B ) − rank( g )rank( B ) R matrix can decrease the rank by at most R . Since g consists of a single non-zero row and column we have rank( g ) ≤ ; by the construction of g we have rank( g ) = rank( g ) − . Using these facts and the fact that g is good, we have rank( ⊗ A + g ⊗ B ) − rank( g )rank( B ) ≥ rank( A ) + rank( g )rank( B ) − B )= rank( A ) + (rank( g ) − B ) . With the lemma in hand we can prove Theorem 3.4.
Proof of Theorem 3.4.
We prove rank( p ◦ g n ) ≥ X S :ˆ p ( S ) =0 (rank( g ) − | S | by induction on n , the number of variables.Observe that the inequality is trivially true if n = 0 . Assume n > , and let = X , Y . Write p = q + z r for multilinear polynomials q, r ∈ F [ z , z , . . . , z n ] . Note that it clearly holds that p ◦ g n = ⊗ ( q ◦ g n − ) + g ⊗ ( r ◦ g n − ) . From the claim we have by induction that rank( p ◦ g n ) = rank( ⊗ ( q ◦ g n − ) + g ⊗ ( r ◦ g n − )) ≥ rank( q ◦ g n − ) + (rank( g ) − r ◦ g n − )= X S :ˆ q ( S ) =0 (rank( g ) − | S | + (rank( g ) − X T :ˆ r ( T ) =0 (rank( g ) − | T | = X S :ˆ p ( S ) =0 z S (rank( g ) − | S | + (rank( g ) − X T :ˆ p ( T ) =0 z ∈ T (rank( g ) − | T |− = X S :ˆ p ( S ) =0 (rank( g ) − | S | . For the upper bound, by subadditivity of rank we have rank( ⊗ A + g ⊗ B ) ≤ rank( ⊗ A ) + rank( g ⊗ B )= rank( )rank( A ) + rank( g )rank( B )= rank( A ) + rank( g )rank( B ) . Apply the above induction argument using this inequality mutatis mutandis . In this section we prove a new separation between high-weight and low-weight cutting planes proofs inthe bounded-space regime.
Theorem 4.1.
There is a family of O (log log n ) -CNF formulas over O ( n log log n ) variables and ˜ O ( n ) clauses that have CP refutations in length ˜ O ( n ) and line space O (1) , but for which any CP ∗ refutationin length L and line space s must satisfy s log L = Ω( n/ log n ) . By the results of [GPT15], any unsatisfiable CNF formula has a cutting planes refutation in constantline space, albeit with exponential length and exponentially large coefficients. In Theorem 4.1 we showthat the length of such a refutation can be reduced to polynomial for certain formulas, described next.At a high level, we prove Theorem 4.1 using the reversible pebble game. Given any DAG G with aunique sink node t , the reversible pebble game [Ben89] is a single-player game that is played with a setof pebbles on G . Initially the graph is empty, and at each step the player can either place or remove apebble on a vertex whose predecessors already have pebbles (in particular the player can always place or11emove a pebble on a source). The goal of the game is to place a pebble on the sink while using as fewpebbles as possible. The reversible pebbling price of a graph, denoted rpeb( G ) , is the minimum numberof pebbles required to place a pebble on the sink.The family of formulas witnessing Theorem 4.1 are pebbling formulas composed with the equalitygadget. Intuitively, the pebbling formula [BW01] Peb G associated with G is a formula that claims that itis impossible to place a pebble on the sink (using any number of pebbles). Since it is always possible toplace a pebble by using some amount of pebbles, this formula is clearly a contradiction.Formally, the pebbling formula Peb G is the following CNF formula. For each vertex u ∈ V there isa variable z u (intuitively, z u should take the value “true” if and only if it is possible to place a pebble on u using any number of pebbles). The variables are constrained by the following clauses. • a clause z s for each source vertex s (i.e., we can always place a pebble on any source), • a clause W u ∈ pred( v ) ¬ z u ∨ z v for each non-source vertex v with predecessors pred( v ) (i.e., if wecan place a pebble on the predecessors of v , then we can place a pebble on v ), and • a clause ¬ z t for the sink t (i.e., it is impossible to place a pebble on t ).Proving Theorem 4.1 factors into two tasks: a lower bound and an upper bound. By applying ourlifting theorem from the previous section, the lower bound will follow immediately from a good lowerbound on the Nullstellensatz degree of pebbling formulas. In order to prove lower bounds on the Nullstel-lensatz degree, we show in Section 4.1 that over every field, the Nullstellensatz degree required to refute Peb G is exactly the reversible pebbling price of G . We then use it together with our lifting theorem toprove the time-space tradeoff for bounded-coefficient cutting planes refutations of Peb G ◦ g in Section4.2 for any high-rank gadget g . Finally, in Section 4.3 we prove the upper bound by presenting a shortand constant-space refutation of Peb G ◦ EQ in cutting planes with unbounded coefficients. In this section we prove that the Nullstellensatz degree of the pebbling formula of a graph G equals thereversible pebbling price of G . Lemma 4.2.
For any field F and any graph G , NS F (Peb G ) = rpeb( G ) . We crucially use the following dual characterization of Nullstellensatz degree by designs [Bus98].
Definition 4.3.
Let F be a field, let d be a positive integer, and let P be an unsatisfiable system of poly-nomial equations over F [ z , z , . . . , z n ] . A d -design for P is a linear functional D on the space of poly-nomials satisfying the following axioms:1. D (1) = 1 .2. For all p ∈ P and all polynomials q such that deg( pq ) ≤ d , we have D ( pq ) = 0 .Clearly, if we have a candidate degree- d Nullstellensatz refutation P p i q i , then applying a d -design to both sides of the refutation yields , a contradiction. Thus, if a d -design exists for a systemof polynomials then there cannot be a Nullstellensatz refutation of degree d . Remarkably, a converseholds for systems of polynomials over { , } n . Theorem 4.4 (Theorems 3, 4 in [Bus98]).
Let F be a field and let P be a system of polynomial equationsover F [ z , z , . . . , z n ] containing the Boolean equations z i − z i = 0 for all i ∈ [ n ] . Then P does nothave a degree- d Nullstellensatz refutation if and only if it has a d -design. With this characterization in hand we prove Lemma 4.2.
Proof of Lemma 4.2.
Let G be a DAG, and consider the pebbling formula G . Following the standardtranslation of CNF formulas into unsatisfiable systems of polynomial equations, we express Peb G withthe following equations: 12 ource Equations. The equation (1 − z s ) = 0 for each source vertex s . Sink Equations.
The equation z t = 0 for the sink vertex t . Neighbour Equations.
The equation (1 − z v ) Q u ∈ pred( v ) z u = 0 for each internal vertex v . Boolean Equations.
The equation z v − z v = 0 for each vertex v .We prove that a d -design for the above system exists if and only if d < rpeb( G ) , and this implies thelemma. Let D be a d -design for the system. First, note that since the Boolean axioms are satisfiedand since D is linear, it follows that D is completely specified by its value on multilinear monomials z T := Q i ∈ T z i (with this notation note that z ∅ := 1 ). Moreover, D must satisfy the following properties: Empty Set Axiom. D ( z ∅ ) = 1 . Source Axioms. D ( z T ) = D ( z T z s ) for every source s and every T ⊆ [ n ] with | T ∪ { s }| ≤ d . Neighbour Axioms. D ( z T z pred( v ) ) = D ( z T z pred( v ) z v ) for every non-source vertex v and every T ⊆ [ n ] with | T ∪ pred( v ) ∪ { v }| ≤ d . Sink Axiom. D ( z T z t ) = 0 for the sink t and every T ⊆ [ n ] with | T ∪ { t }| ≤ d .We may assume without loss of generality that D ( z T ) = 0 for any set T with | T | > d .Given a set S of vertices of G , we think of S as the reversible pebbling configuration in which thereare pebbles on the vertices in S and there are no pebbles on any other vertex. We say that a configuration T is reachable from a configuration S if there is a sequence of legal reversible pebbling moves that changes S to T while using at most d pebbles at any given point.Now, we claim that the only way to satisfy the first three axioms is to set D ( x T ) = 1 for everyconfiguration T that is reachable from ∅ . To see why, observe that those axioms are satisfiable if and onlyif the empty configuration is assigned the value , any configuration containing the sink is labelled , and D ( z S ) = D ( z T ) for any two configurations S, T with at most d pebbles that are mutually reachable viaa single reversible pebbling move. Hence, this setting of D is the only one we need to consider.Finally, observe that this specification of a design D satisfies the sink axiom if and only if d < rpeb( G ) , since the sink is reachable from ∅ using rpeb( G ) pebbles but not with less (by the definition of rpeb( G ) ). Therefore, a d -design for Peb G exists if and only if d < rpeb( G ) , as required. In this section we prove the lower bound part of the time-space trade-off for CP ∗ . Lemma 4.5.
There is a family of graphs { G n } with n vertices and constant degree, such that every CP ∗ refutation of Peb G n ◦ EQ in length L and line space s must have s log L = Ω( n/ log n ) . Our plan is to lift a pebbling formula that is hard with respect to Nullstellensatz degree, and as we justproved it is enough to find a family of graphs whose reversible pebbling price is large. Paul et al. [PTC77]provide such a family (and in fact prove their hardness in the stronger standard pebbling model).
Theorem 4.6.
There is a family of graphs { G n } with n vertices, constant degree, and for which rpeb( G n ) =Ω( n/ log n ) . We combine these graphs with our lifting theorem as follows.
Lemma 4.7.
There is a family of graphs { G n } with n vertices and constant degree, such that P cc ( Search (Peb G ◦ EQ)) = Ω( n/ log n ) . roof. Let
Peb G be the pebbling formula of a graph G = G n from the family given by Theorem 4.6. ByLemma 4.2 the Nullstellensatz degree of the formula is NS F (Peb G ) = rpeb( G ) = Ω( n/ log n ) . (4.1)This allows us to use our lifting theorem, Theorem 3.1, with an equality gadget of arity q = O (log( n/ NS F (Peb G )) = O (log log n ) , and obtain that the lifted search problem Search (Peb G ) ◦ EQ requires deterministic com-munication P cc ( Search (Peb G ) ◦ EQ) ≥ N S F (Peb G ) = Ω( n/ log n ) . (4.2)As we noted in Observation 2.1, this implies that the search problem of the lifted formula also requiresdeterministic communication P cc ( Search (Peb G ◦ EQ)) ≥ P cc ( Search (Peb G ) ◦ EQ) = Ω( n/ log n ) . (4.3)Since we collected our last ingredient, let us finish the proof. Proof of Lemma 4.5.
Let S = Search (Peb G ◦ EQ) be the search problem given by Lemma 4.7. UsingLemma 2.2 we have that every cutting planes refutation of the lifted formula in length L , line space s ,and coefficient length c must satisfy s ( c + log n ) log L = Ω( P cc ( S )) = Ω( n/ log n ) . (4.4)Since the size of the lifted formula Peb G ◦ EQ is ˜ O ( n ) , the coefficients of a CP ∗ refutation arebounded by a polynomial of n in magnitude, and hence by O (log n ) in length. Substituting the value of c = O (log n ) in (4.4) we obtain that s log L = Ω( n/ log n ) (4.5)as we wanted to show. We now prove Theorem 4.8, showing that cutting planes proofs with large coefficients can efficientlyrefute pebbling formulas composed with equality gadgets in constant line space. Let EQ q denote theequality gadget on q bits. Theorem 4.8.
Let
Peb G be any constant-width pebbling formula. There is a cutting planes refutation of Peb G ◦ EQ log log n in length ˜ O ( n ) and space O (1) . We also use the following lemma, which is a “derivational” analogue of the recent result of [GPT15]showing that any set of unsatisfiable integer linear inequalities has a cutting planes refutation in constantspace. As the techniques are essentially the same we leave the proof to Appendix B.
Lemma 4.9 (Space Lemma).
Let C be a set of integer linear inequalities over n variables that implies aclause C . Then there is a cutting planes derivation of C from C in length O ( n n ) and space O (1) . Let us begin by outlining the high level proof idea. We would like to refute the lifted formula
Peb G ◦ EQ q using constant space. Consider first the unlifted formula Peb G . The natural way to refute it is thefollowing: Let v , . . . , v n be a topological ordering of the vertices of G . The refutation will go over thevertices in this order, each time deriving the equation that says that the variable z v i must take the value“true” by using the equations that were derived earlier for the predecessors of v i . Eventually, the refutationwill derive the equation that says that the sink must take the value “true”, which contradicts the axiomthat says that the sink must be false. 14oing back to the lifted formula Peb G ◦ EQ q , we construct a refutation using the same strategy, exceptthat now the equation of z v i is replaced with the equations x v i , = y v i , , . . . x v i ,q = y v i ,q . The main obstacle is that if we implement this refutation in the naive way, we will have to store all theequations simultaneously, yielding a refutation of space O ( q · n ) . The key idea of our proof is that CP canencode the conjunction of many equations using a single equation. We can therefore use this encodingin our refutation to store at any given point all the equations that were derived so far in a single equation.The implementation yields a refutation of constant space, as required.To see how we can encode multiple equations using a single equation, consider the following example.Suppose we wish to encode the equations x = y , x = y , x = y , where all the variables take values in { , } . Then, it is easy to see that those equations are equivalent tothe equation · x + 2 · x + x = 4 · y + 2 · y + y . This idea generalizes in a straightforward way to deal with more equations, as well as with arbitrary lineargadgets, to be discussed below.The rest of this section is devoted to the proof of Theorem 4.8. The following notion is central to theproof. Say a gadget g ( x, y ) : { , } q × { , } q → { , } is linear if there exists a linear expression withinteger coefficients L ( x, y ) = c + q X i =1 a i x i + b i y i such that g ( x, y ) = 1 if and only if L ( x, y ) = 0 . Note that the equality gadget is linear, as it correspondsto the linear expression P qi =1 i − ( x i − y i ) .Let g be any linear gadget with corresponding linear expression L . Let K = 1 + max x,y | L ( x, y ) | ,and let G be the underlying DAG of the composed pebbling formula Peb G ◦ g n . Note that for each vertex u of G the composed formula has corresponding variables x u , y u ∈ { , } q . Once and for all, fix anordering of the vertices of G and assume that all subsets are ordered accordingly. For a subset of vertices U ⊆ V define L ( U ) := X u i ∈ U K i L ( x u i , y u i ) . The following claim shows that L ( U ) encodes the truth of the conjunction V u i ∈ U g ( x u i , y u i ) . Claim 4.10.
For a set of vertices U and any x, y ∈ { , } qn , L ( U ) = 0 if and only if V u i ∈ U g ( x u i , y u i ) =1 . Proof.
Since g is linear, if g ( x u i , y u i ) = 1 for all u i ∈ U then it follows that L ( x u i , y u i ) = 0 for all u i ∈ U , which in turn implies L ( U ) = 0 . Conversely, suppose g ( x u i , y u i ) = 0 for some vertex u i , andlet i be the largest such index. It follows that L ( u i ) = 0 , and clearly (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X j From here on in the proof, we consider L ( U ) = 0 , or L ( U ) for short, as being syntactically repre-sented in cutting planes as the pair of inequalities L ( U ) ≥ , − L ( U ) ≥ . The bulk of the proof lies inthe following lemma, which shows how to “encode” and “decode” unit literals in the expressions L ( U ) . Lemma 4.11 (Coding Lemma).
Let U be any set of vertices. Then1. For any u ∈ U there is a cutting planes derivation of L ( u ) from L ( U ) in length O ( q | U | ) and space O (1) .2. Let C = ¬ z u ∨ ¬ z u ∨ · · · ∨ ¬ z u k − ∨ z u k be an axiom of Peb G with u , u , . . . , u k − ∈ U . Let ℓ g and s g be such that there exists a derivation of L ( u ) from a CNF encoding of g ( u ) in length ℓ g and space s g . From L ( u ) , L ( u ) , . . . , L ( u k − ) and C ◦ g n there is a cutting planes derivation of L ( u k ) in length O (2 kq ℓ g ) and space O ( s g ) .3. For any u U there is a cutting planes derivation of L ( U ∪ { u } ) from L ( U ) and L ( u ) in length O (1) and space O (1) . Let us first use the Coding Lemma to complete the proof. We show a more general statement fromwhich Theorem 4.8 follows immediately by setting k = 3 and g = EQ q , with q = O (log log n ) , andbounding ℓ EQ = O ( q ) and s EQ = O (1) . Lemma 4.12. If Peb G is a width- k pebbling formula on n variables and g is a linear gadget of arity q then there is a cutting planes refutation of Peb G ◦ g n in length O ( n ( kqn + 2 kq ℓ g )) and space O ( k + s g ) .Proof. We begin with L ( ∅ ) , which is represented as the pair of inequalities ≥ , ≥ . By combiningParts (2) and (3) of the Coding Lemma we can derive L ( S ) , where S is the set of sources of G . Wethen follow a unit-propagation proof of Peb G , deriving L ( u ) for each vertex of G in topological order.Suppose at some point during the derivation we have derived L ( U ) for some subset U of vertices. Forany axiom C of Peb G of the form C = ¬ z u ∨ ¬ z u ∨ · · · ∨ ¬ z u k − ∨ z u k with u , u , . . . , u k − ∈ U wedo the following: first apply Part (1) of the Coding Lemma to obtain L ( u i ) for each i ∈ [ k − . ApplyPart (2) to derive L ( u k ) , forget L ( u i ) for each i ∈ [ k − , and then apply Part (3) to L ( U ) and L ( u k ) toderive L ( U ∪ { u k } ) . Continue in this way until we derive L ( z ) where z is the sink vertex of G . Since { L ( z ) , ¬ z ◦ g } is an unsatisfiable set of linear inequalities, it follows by the Space Lemma (Lemma 4.9)that we can deduce a contradiction in length O ( q q ) and space O (1) .In the above proof we need to derive L ( u ) for each of the n vertices of the graph. Deriving L ( u ) requires at most O ( k ) applications of Part (1), one application of Part (2), and one application of Part (3).Thus, in total, we require length O ( n ( kqn + 2 kq ℓ g )) and space O ( k + s g ) .It remains to prove the Coding Lemma (Lemma 4.11). Proof of Coding Lemma.
Let U = { u , u , . . . , u t } be an arbitrary subset of vertices of size t . Recallthe definition L ( U ) = P ti =1 K i − L ( u i ) . For any u i ∈ U a term of L ( U ) will be one of the terms16 i − L ( u i ) , which is a sum of q variables itself. We begin by defining two auxiliary operations thatallow us to trim both the least and the most significant terms from L ( U ) .To trim the i least significant terms of an inequality we essentially divide by K i . More formally, forevery variable v with a positive coefficient a j less than K i we add the inequality − a j v ≥ − a j , and forevery variable with a negative coefficient greater than − K i we add the inequality a j v ≥ . This takeslength O ( qi ) , since each term contributes q coefficients, and space O (1) .At this point all the remaining coefficients on the LHS are divisible by K i , so we can apply the divisionrule. By construction the RHS is greater than − K i − P j ≥ i cK j , therefore when we divide by K i thecoefficient on the RHS becomes − P j ≥ i cK j − i .Finally, to restore the values of the coefficients to the values they had before dividing, we multiply by K i at the end to restore them.To trim the m − i most significant terms of an inequality with m terms we need to use the oppositeinequality, since the remaining part only has a semantic meaning when the most significant part vanishes.Hence we first trim the i least significant terms of the opposite inequality, keeping exactly the negation ofthe terms that we want to discard. Then we add both inequalities so that only the i least significant termsremain. This takes length O ( qi ) and space O (1) .Using the trimming operations we can prove items 1–3 in the lemma.1. We must show that for any u ∈ U there is a cutting planes derivation of L ( u ) from L ( U ) in length O ( qt ) and space O (1) . This is straightforward: begin by making copies of the pair of inequalities L ( U ) ≥ and − L ( U ) ≥ encoding L ( U ) = 0 . Trim the terms that are strictly more and strictlyless significant than L ( u ) from both of the inequalities, in length O ( qt ) and space O (1) .2. Recall that we assumed there is a derivation Π of L ( u k ) from the CNF formula z u k ◦ g in length ℓ g and space s g , so our goal is to produce the set of clauses z u k ◦ g . Any such clause D is impliedby the set of inequalities { L ( u i ) } k − i =1 together with the CNF encoding of C ◦ g n , therefore it hasa derivation Π D in length O (2 kq ) and space O (1) by the Space Lemma (Lemma 4.9). Replacingeach usage of a clause D ∈ z u k ◦ g in Π as an axiom by the corresponding derivation Π D we obtaina sound derivation L ( u t +1 ) in length O (2 tq ℓ g ) and space O ( s g ) .3. Simply add K t +1 L ( u ) ≥ to L ( U ) ≥ and − K t +1 L ( u ) ≥ to − L ( U ) ≥ ; this clearly usesbounded length and space.This completes the proof of Theorem 4.8.Note that the largest coefficient used in the refutation is bounded by K n . Indeed, the argument canbe generalized to give a continuous trade-off between the size of the largest coefficient and the numberof inequalities, simply by adding a new pair of empty inequalities once the coefficient required to add avertex to an existing pair would be too large. This means that if we allow up to ξ inequalities then we canuse coefficients of size bounded by K O ( n/ξ ) . In this section we exhibit an explicit function that exponentially separates the size of monotone Booleanformulas and monotone real formulas.
Theorem 5.1.
There is an explicit family of functions f n over O ( n polylog n ) variables that can be com-puted by monotone real formulas of size O ( n polylog n ) but for which every monotone Boolean formularequires size Ω( n/ log n ) . To prove the lower bound part of Theorem 5.1 we use the characterization of formula depth bycommunication complexity [KW90]. Given a monotone Boolean function f , the monotone Karchmer–Wigderson game of f is a search problem mKW( f ) : { , } n × { , } n → [ n ] defined as (( x, y ) , i ) ∈ mKW( f ) if f ( x ) = 1 , f ( y ) = 0 , x i = 1 , and y i = 0 . In other words, given a -input x and a -input17 for f , the task is to find an index i ∈ [ n ] such that x i = 1 , and y i = 0 . Such an index always existsbecause f is monotone.If we denote by mD( f ) the minimum depth of a monotone Boolean formula required to compute aBoolean function f , then we can write the characterization as Lemma 5.2 ([KW90]).
For every function f , it holds that mD( f ) = P cc (mKW( f )) . The analogue of this characterization for real circuits is in terms of DAG-like real protocols [Kra98,Sok17, HP18]. Since we are only interested in formulas rather than circuits we only consider tree-like pro-tocols, which we call locally real protocols to distinguish them from the stronger model of real protocols,also known as real games [Kra98].A locally real communication protocol, then, is a communication protocol where the set of inputscompatible with a node is defined by one half-space, as opposed to a real protocol where the set ofcompatible inputs is defined by the intersection of all half-spaces in the path leading to that node.Formally, a locally real protocol for a search problem
Search : X × Y → Z , where X and Y areBoolean hypercubes, is a tree where every internal node v is labelled with a half-space H v = { ( x, y ) ∈X × Y | h xy, c v i ≥ d v } , where c v ∈ R dim X +dim Y and d v ∈ R , and every leaf is additionally labelledwith an element z ∈ Z . The root is labelled with the full space X × Y , children are consistent in the sensethat if a node w has children u and v then H w ⊆ H u ∪ H v . Given an input ( x, y ) , the protocol produces anondeterministic output z as follows. We start at the root and at each internal node we nondeterministicallymove to a child that contains ( x, y ) , which exists by the consistency condition. The output of the protocolis the label of the resulting leaf. A protocol is correct if for any input ( x, y ) ∈ X × Y it holds that z ∈ Z .It is not hard to turn a real formula into a locally real protocol, and the converse also holds. Lemma 5.3 ([HP18]).
Given a locally real protocol for the monotone Karchmer–Wigderson game of apartial function f , there exists a monotone real formula with the same underlying graph that computes f . In order to obtain a function whose Karchmer–Wigderson game we can analyse we use the fact thatevery search problem can be interpreted as the Karchmer–Wigderson game of some function. To statethe result we need the notion of a nondeterministic communication protocol, which is a collection N ofdeterministic protocols such that (( x, y ) , z ) ∈ S if and only if there exists some protocol π ∈ N such that π ( x, y ) = z . The cost of a nondeterministic protocol is log | N | + max π ∈ N depth( π ) . Lemma 5.4 ([Raz90, G´al01], see also [Rob18]).
Let S be a two-party total search problem with nondeter-ministic communication complexity k . There exists a partial monotone Boolean function f : { , } k →{ , , ∗} such that S is exactly the monotone Karchmer–Wigderson game of f . We use as a search problem the falsified clause search problem of a hard pebbling formula composedwith equality given by Lemma 4.7. To exhibit a real formula for the function it induces, we first build atree-like cutting planes proof of small size of the composed pebbling formula.
Theorem 5.5. If C is the pebbling formula of a graph of indegree , then there is a tree-like semanticcutting planes refutation of C ◦ EQ log log n in length O ( n log n log log n ) . It is not hard to see that we can extract an efficient locally real protocol from a tree-like cutting planesrefutation of small size, but let us record this fact formally.
Lemma 5.6 (Folklore, see [Sok17]).
Given a semantic cutting planes refutation of a formula F , there isa locally real protocol for Search ( F ) with the same underlying graph. Before we move into the proof of Theorem 5.5, let us complete the proof of Theorem 5.1.
Proof of Theorem 5.1.
Let S = Search (Peb G ◦ EQ log log n ) be the search problem given by Lemma 4.7.The nondeterministic communication complexity of f is log( | Peb G ◦ EQ log log n | ) + 2 , since given acertificate consisting of a clause falsified by the inputs each party can independently verify that their partis falsified and communicate so to the other party. Therefore by Lemma 5.4 there is a partial monotone18unction f ∗ over O ( n polylog n ) variables whose monotone Karchmer–Wigderson game is equivalentto S . By Theorem 5.5 there is a semantic cutting planes refutation of the formula Peb G ◦ EQ log log n of length O ( n polylog n ) , which we convert into a locally real protocol for S of size O ( n polylog n ) using Lemma 5.6, and then into a monotone real formula for f ∗ of size O ( n polylog n ) using Lemma 5.3.Add a threshold gate on top of the formula to ensure that the output is always Boolean and let f be thetotal function that the formula computes. Since f extends f ∗ , by Lemma 5.2 and Lemma 4.7 f requiresmonotone Boolean formulas of depth Ω( n/ log n ) , and therefore size Ω( n/ log n ) . For simplicity in this section we reinterpret the pebbling formula of a graph G of indegree lifted withequality of q bits as the pebbling formula of a graph G ′ lifted with equality of bit or XNOR , where G ′ is the graph where we replace every vertex in G by a blob of q vertices and we replace every edge by abipartite complete graph between blobs, and with the difference that instead of having axioms assertingthat all sinks are false, the axioms assert that some sink is false.Without further ado, let us prove Theorem 5.5, which follows by setting q = log log n in the followingLemma. Lemma 5.7. If C is the pebbling formula of a graph of indegree , then there is a tree-like semanticcutting planes refutation of C ◦ EQ q in length O ( nq q ) . As in Section 4.3 we fix a topological order of G and we build a refutation by keeping two inequalities L ( W ) ≥ and − L ( W ) ≥ . The main difference is that we cannot use the Coding Lemma to isolatethe value of a single vertex, since then we would lose the information on the rest of vertices, therefore wehave to simulate the inference steps in place as we describe next.Let us set up some notation. If W is a set of vertices, let g ( W ) = V v ∈ W XNOR( v ) . We represent XNOR( v ) with L ( v ) = 0 , where L ( v ) = x v − y v , and g ( W ) with L ( W ) = 0 , where L ( W ) = P v j ∈ W j L ( v j ) . We begin with W = ∅ and with the trivial inequalities L ( ∅ ) = 0 . Let us showhow to derive each vertex. Lemma 5.8.
There is a tree-like semantic derivation of L ( W ∪ { w } ) ≥ from L ( W ) ≥ and theaxioms in q steps.Proof. If w is a source, then the inequality L ( w ) ≥ is already an axiom, hence it is enough to multiply L ( W ) ≥ by and add L ( w ) .The complex case is when w has predecessors u , . . . , u q . Let ℓ ( v, b ) = b + ( − b v be the literalover variable v and polarity − b . Consider the q axioms I b ≥ indexed by b ∈ { , } q and defined as J b = q X j =1 ℓ ( x u j , b j ) + ℓ ( y u j , b j ) (5.1) I b = L ( w ) + J b . (5.2)We start with an inequality L ( W ) ≥ . In order to have enough working space for the axioms wemultiply the inequality by q , and using weakening axioms we add a slack term defined as S = q X j =1 j − x u j (5.3)to obtain L +0 ≥ with L +0 = 2 q L ( W ) + S . (5.4)The coefficients for S are chosen so that if we evaluate S on a string b ∈ { , } q , the result is b interpretedas a binary number. We use it to keep track of which axioms we have processed so far, in a similar fashion19o how the space-efficient refutation of the complete tautology [GPT15] that we reproduce in Appendix Bkeeps track of processed truth value assignments.The next step is to add each axiom to L +0 , but for this to work we need to represent each intermediatestep with one inequality as follows. Claim 5.9.
We can represent the Boolean expression g + B = J L ( W ) ≥ K ∧ g ( W ) → ^ b ≤ B I b (5.5)with the inequality L + B ≥ defined as L + B = ( B + 1) L ( w ) + L +0 . (5.6) Proof.
Let us begin proving the claim by showing that g + B ⇒ L + B ≥ . First consider an assignment α that satisfies L ( W ) ≥ but not g ( W ) , that is an assignment where x u j = 1 and y u j = 0 for somepredecessor u j of w . Then L +0 ↾ α ≥ q , hence L + B ↾ α ≥ − ( B + 1) + 2 q ≥ .Now consider an assignment α that satisfies g ( W ) , hence L ( W ) = 0 . If α u ,...,u q = b ≤ B then,since α falsifies J b ≥ , α must satisfy L ( w ) ≥ , so both L +0 ≥ and L ( w ) ≥ . Otherwise if α u ,...,u q = b > B then S ↾ α = b ≥ B + 1 , and we have L + B ↾ α ≥ − ( B + 1) + b ≥ .Let us finish by showing that g + B ⇐ L + B ≥ . First consider an assignment α that falsifies L ( W ) ≥ .Then L + B ↾ α ≤ ( B + 1) − q < .Now consider an assignment α that satisfies g ( W ) but not an axiom I b with b ≤ B . Then in particular α falsifies L ( w ) ≥ , hence L + B ↾ α = ( B + 1) L ( w ) + S ↾ α = − ( B + 1) − b < . This concludes theproof of the claim.Since L + B +1 ≥ follows semantically from L + B ≥ and I B , we can derive g +2 q ≥ from L +0 ≥ and the set of axioms I b ≥ using q semantic inferences of arity . Also, L +2 q ≥ is semantically (butnot syntactically) equivalent to L ( W ∪ { w } ) ≥ , so we can be ready for the next step with a semanticinference of arity .We can derive the upper bound inequality − L ( W ∪ { w } ) ≥ similarly, the main differences beingthat we start with − L ( W ) ≥ and that we use the other half of the axioms, that is I b = − L ( w ) + J b .We handle the sinks in a slightly different way. Instead of using the pebbling axioms directly, wefirst use the pebbling axioms of all the sinks together with the axioms enforcing that some sink is falsein order to derive a set of inequalities similar to pebbling axioms but with − in place of L ( w ) . We thenuse the same derivation as in Lemma 5.8 using these inequalities in place of the axioms and we obtain L ( W ) − ≥ and analogously − L ( W ) − ≥ . Adding both inequalities leads to the contradiction − ≥ .To conclude the proof it is enough to observe that we do O (2 q ) inference steps for each vertex in G ′ ,which has order nq , hence the total length of the refutation is O ( nq q ) . In this paper, we show that the cutting planes proof system (CP) is stronger than its variant with polyno-mially bounded coefficients (CP ∗ ) with respect to simultaneous length and space. This is the first resultin proof complexity demonstrating any situation where high-weight coefficients are more powerful thanlow-weight coefficients. We also prove an explicit separation between monotone Boolean formulas andmonotone real formulas. Previously the result was only known to hold non-constructively. To obtain theseresults we strengthen a lifting theorem of [PR18] to allow the lifting to work with any gadget with suffi-ciently large rank, in particular with the equality gadget—a crucial ingredient for obtaining the separationsdiscussed above. 20his work raises a number of questions. Prior to our result, no explicit function was known separatingmonotone real circuits or formulas from monotone Boolean circuits or formula. Although we prove anexplicit formula separation, it remains open to obtain an explicit function that separates monotone realcircuits from monotone Boolean circuits.The most glaring open problem related to our cutting planes contribution is to strengthen our resultto a true length separation, without any assumption on the space complexity. It is natural to ask whethertechniques inspired by [Sok17, GGKS18] can be of use. Another thing to note about our trade-off result forCP ∗ is that it is not a “true trade-off”: we know that length and space cannot be optimised simultaneously,but we do not know if there in fact exist small space refutations. An interesting problem is, therefore, toexhibit formulas that present “true trade-offs” for CP ∗ but are easy with regard to space and length in CP.It follows from our results that standard decision tree complexity, parity decision tree complexity,and Nullstellensatz degree are equal for the falsified clause search problem of lifted pebbling formulas.In view of this we can ask ourselves what complexity measure we are actually lifting. We know thatfor general search problem decision tree complexity is not enough for a lifting result. How about paritydecision tree complexity? Or can we leverage the fact that we have “well-behaved” rectangle covers andsmall certificate complexity to lift weaker complexity models? It would be valuable to have a betterunderstanding of the relation between gadgets, outer functions/relations and complexity measures. Acknowledgements
Different subsets of the authors would like to acknowledge fruitful and enlightening conversations withdifferent subsets of Arkadev Chattopadhyay, Pavel Hrubeˇs, Christian Ikenmeyer, Bruno Loff, SagnikMukhopadhyay, Igor Carboni Oliveira, Pavel Pudl´ak, and Dmitry Sokolov. We are also grateful for dis-cussions regarding literature references with Albert Atserias, Paul Beame, and Massimo Lauria. We arethankful to the anonymous referees for their comments; in particular, indicating a simplified proof ofLemma 3.5.Part of this work was carried out while several of the authors were visiting the Simons Institute forthe Theory of Computing in association with the DIMACS/Simons Collaboration on Lower Bounds inComputational Complexity, which is conducted with support from the National Science Foundation.Or Meir was supported by the Israel Science Foundation (grant No. 1445/16). Toniann Pitassi andRobert Robere were supported by NSERC. Susanna F. de Rezende and Jakob Nordstr¨om were supportedby the European Research Council under the European Union’s Seventh Framework Programme (FP7/2007–2013) /ERC grant agreement no. 279611, as well as by the Knut and Alice Wallenberg grant KAW 2016.0066.Jakob Nordstr¨om also received funding from the Swedish Research Council grants 621-2012-5645 and2016-00782. Marc Vinyals was supported by the Prof. R Narasimhan post-doctoral award.
A Lifting Nullstellensatz Degree for All Gadgets
In this section we prove Theorem 3.1. In fact, we prove the following stronger result, which impliesTheorem 3.1 as a corollary.
Theorem A.1.
Let C be an unsatisfiable k -CNF on n variables and let F be any field. Let g be anyBoolean-valued gadget with rank( g ) ≥ . Then P cc ( Search ( C ) ◦ g n ) ≥ NS F ( C ) log (cid:18) NS F ( C )rank( g ) en (cid:19) − n log e rank( g ) − log k. The proof of the theorem follows the proof of a similar lifting theorem from [PR18]. As such, we willneed some notation from that paper. Let us begin by introducing a key notion: Razborov’s rank measure .Given sets U , V , a rectangle cover R of U × V is a covering of
U × V by combinatorial rectangles.21 efinition A.2.
Let U , V be sets and let R be any rectangle cover of U × V . Let A be any U × V matrixover a field F . The rank measure of R at A is the quantity µ F ( R , A ) = rank( A )max R ∈R rank( A ↾ R ) . Using the rank measure we can lower bound the deterministic communication complexity of com-posed CNF search problems as follows. The key observation is that any deterministic communicationprotocol outputs a rectangles that lie in a “structured” rectangle cover in the following sense. We notebelow that if A is a collection of tuples from some product set I n then we write A i to mean the projectionof A to the i th coordinate and A I for I ⊆ [ n ] to mean the projection onto the coordinates in I . Definition A.3.
Let C be an unsatisfiable k -CNF on n variables and let g : X × Y → { , } be a gadget.For a clause C ∈ C , a combinatorial rectangle R ⊆ X n × Y n is C -structured if g n ( x, y ) falsifies C forall ( x, y ) ∈ R and for all i Vars ( C ) we have R i = X × Y . A rectangle cover R of X n × Y n is C -structured if every R ∈ R is C -structured for some C ∈ C . Lemma A.4.
Let C be an unsatisfiable k -CNF on n variables and let g : X × Y → { , } be a gadget.Let F be any field and let A be any X n × Y n matrix over F . Then P cc ( Search ( C ) ◦ g ) ≥ min R log µ F ( R , A ) where the minimum is taken over C -structured rectangle covers of X n × Y n .Proof. Let Π be any communication protocol solving Search ( C ) ◦ g , and let T be the monochromaticrectangle partition corresponding to Π . Since Π solves the search problem, for every rectangle R ∈ T there is a clause C such that for all ( x, y ) ∈ R , g n ( x, y ) falsifies C . We can write R = A × B for somesets A ⊆ X Vars ( C ) × X [ n ] \ Vars ( C ) and B ⊆ Y Vars ( C ) × Y [ n ] \ Vars ( C ) . Consider R ′ = A ′ × B ′ where A ′ = A Vars ( C ) × X [ n ] \ Vars ( C ) and B ′ = B Vars ( C ) × X [ n ] \ Vars ( C ) . Since C only depends on indicesin Vars ( C ) we have that g n ( x, y ) falsifies C for all ( x, y ) ∈ R ′ and, moreover, R ′ i = X × Y for all i Vars (cid:0) C (cid:1) . It follows that R ′ is C -structured. Let R be the C -structured rectangle covering obtainedfrom T by relaxing all rectangles of T in this way.We now have an C -structured rectangle cover R such that every T ∈ T is contained in some rectangleof R . Razborov [Raz90] proved that if T is a rectangle partition and R is a rectangle cover such that foreach T ∈ T there is an R ∈ R such that T ⊆ R it holds that |T | ≥ µ F ( R , A ) for any matrix A . Since log |T | ≤ | Π | = P cc ( Search ( C ) ◦ g ) the lemma follows.We now introduce the notion of a certificate of an unsatisfiable CNF formula. Definition A.5.
Let C be an unsatisfiable Boolean formula on n variables in conjunctive normal form, andlet C be a clause in C . The certificate of C , denoted Cert( C ) , is the partial assignment π : [ n ] → { , , ∗} which falsifies C and sets the maximal number of variables to ∗ s. Let Cert( C ) denote the set of certificatesof clauses of C .We say that an assignment z ∈ { , } n agrees with a certificate π ∈ Cert( C ) if π ( i ) = z i for each i assigned to a { , } value by π . Since the CNF formula C is unsatisfiable, it follows that every assignmentin z ∈ { , } n agrees with some { , } -certificate of C . Next we introduce an alternative definition ofNullstellensatz degree called the algebraic gap complexity . Definition A.6.
Let F be a field. Let C be an unsatisfiable CNF on n variables. The F -algebraic gap com-plexity of C is the maximum positive integer gap F ( C ) ∈ N for which there exists a multilinear polynomial p ∈ F [ z , z , . . . , z n ] such that deg( p ) = n and ∀ π ∈ Cert( C ) : deg( p ↾ π ) ≤ n − gap F ( C ) . When the field is clear from context we will write gap( C ) .22n [PR18, Rob18] it was shown that the algebraic gap complexity is equal to Nullstellensatz degree. Theorem A.7.
For any unsatisfiable CNF formula C on n variables and any field F , gap F ( C ) = NS F ( C ) . We now prove a lifting theorem from Nullstellensatz degree to the rank measure, from which TheoremA.1 follows by applying Lemma A.4.
Theorem A.8.
Let C be an unsatisfiable k -CNF on n variables and let F be any field. Let g be anyBoolean-valued gadget with rank( g ) ≥ . There is a matrix A such that for any C -structured rectanglecover R we have µ F ( R , A ) ≥ k (cid:18) NS F ( C )rank( g ) en (cid:19) NS F ( C ) exp( − n/ rank( g )) . Proof.
Let p ∈ F [ z , z , . . . , z n ] be the polynomial witnessing the algebraic gap complexity gap( C ) , andlet A = p ◦ g n be the pattern matrix obtained by composing p and g . We need to analyze µ F ( R , p ◦ g n ) = rank F ( p ◦ g n )max R ∈R rank F ( p ◦ g n ↾ R ) . Let us first analyze the denominator. Let R be an arbitrary rectangle from the cover R , and supposethat R is C -structured for the clause C ∈ C . Let π = Cert( C ) . We want to show that rank F ( p ◦ g n ↾ R ) ≤ X S : [ p ↾ π ( S ) =0 rank( g ) | S | . (A.1)To prove this, we claim that p ◦ g n ↾ R is column-equivalent to the block matrix [( p ↾ π ) ◦ g [ n ] \ Vars ( C ) , ( p ↾ π ) ◦ g [ n ] \ Vars ( C ) , . . . , ( p ↾ π ) ◦ g [ n ] \ Vars ( C ) ] for some number of copies of the matrix ( p ↾ π ) ◦ g [ n ] \ Vars ( C ) . Indeed Equation A.1 immediately followsfrom this claim as rank F ( p ◦ g n ↾ R ) = rank F (( p ↾ π ) ◦ g [ n ] \ Vars ( C ) ) ≤ X S : [ p ↾ π ( S ) =0 rank( g ) | S | by Theorem 3.4. So, we now prove the claim.Write R = A × B . Fix assignments α ∈ A Vars ( C ) and β ∈ B Vars ( C ) , and note that since R is C -structured we have that g Vars ( C ) ( α, β ) = π and ( α, x ′ ) ∈ A and ( β, y ′ ) ∈ B for all x ′ , y ′ . Thus, byranging x [ n ] \ Vars ( C ) , y [ n ] \ Vars ( C ) over all values yields the matrix ( p ↾ π ) ◦ g [ n ] \ Vars ( C ) . Then, ranging x Vars ( C ) and y Vars ( C ) over all α, β such that g Vars ( C ) ( α, β ) = π yields the claim and Equation A.1.Now, consider the rank measure µ F ( R ) , which by Theorem 3.4 and Equation A.1 satisfies µ F ( R ) ≥ rank F ( p ◦ g n )max R ∈R rank F ( p ◦ g n ↾ R ) = X S :ˆ p ( S ) =0 (rank( g ) − | S | max π ∈ Cert( C ) X S : [ p ↾ π ( S ) =0 rank( g ) | S | By definition of gap( C ) we have deg p = n and thus the numerator is at least (rank( g ) − n . For thedenominator, since p witnesses the algebraic gap of C , we have that deg p ↾ π ≤ n − gap( C ) for all π ∈ Cert( C ) . We may assume that ˆ p ( S ) = 0 when | S | < n − gap( C ) as the definition of algebraicgaps depends only on the coefficients of monomials of p with degree larger than n − gap( C ) . So, for anyrestriction π : X S : [ p ↾ π ( S ) =0 rank( g ) | S | ≤ k X i =0 (cid:18) n gap( C ) − i (cid:19) rank( g ) n − gap( C ) − i ≤ k (cid:18) en gap( C ) (cid:19) gap( C ) rank( g ) n − gap( C ) rank( g ) ≥ en/ gap( C ) , we have µ F ( R , p ◦ g n ) ≥ (rank( g ) − n k ( en/ gap( C )) gap( C ) rank( g ) n − gap( C ) = 1 k (cid:18) gap( C )rank( g ) en (cid:19) gap( C ) · (cid:18) − g ) (cid:19) n ≥ k (cid:18) gap( C )rank( g ) en (cid:19) gap( C ) exp( − n/ rank( g )) . Since gap( C ) = NS ( C ) the theorem is proved.Theorem A.1 follows immediately from Theorem A.8 and Lemma A.4. B Proof of the Space Lemma
In this section we prove the Space Lemma (Lemma 4.9), restated next.
Lemma B.1.
Let C be a set of inequalities over n variables that implies a clause C . Then there is acutting planes derivation of C from C in length O ( n n ) and space O (1) . We do so by adapting the proof in [GPT15] that any formula has a cutting planes refutation in constantspace in order to show that, in fact, we can derive any clause that follows from a set of inequalities inconstant space.At a bird’s eye view, the proof in [GPT15] has two steps. The primary step is building a refutation ofthe complete tautology, the formula that contains all n clauses with n variables each forbiding one of thepossible n assignments, in constant space. The authors come up with an order and a way to encode thatthe first K clauses are all true in small space for an arbitrary K , and the rest of the primary step consistsof showing how to operate with this encoding also in small space, starting with no clause being true andadding clauses one by one until a contradiction arises. The secondary step is to transform the original setof linear inequalities into the complete tautology.If we do not start with an unsatisfiable set of linear inequalities we obviously cannot reach a contra-diction, but given a clause C that follows from C we can still encode that all the clauses that are a supersetof C must be true, and this expression is equivalent to C .Let us set up some notation. We number the variables from to n − . If α is a total assignment,we denote by C α the clause over n variables that is falsified exactly by α . We overload notation and alsodenote by C α the standard translation of the clause C α into an inequality. We say that an assignment is lessthan a natural number B and write α < B if α is lexicographically smaller than the binary representationof B , that is if P n − i =0 i α ( x i ) < B . We write T B to denote the inequality P n − i =0 i x i ≥ B that is falsifiedexactly by the assignments { α ∈ { , } n | α < B } .We can reuse the following two intermediate lemmas from [GPT15], corresponding to the primaryand the secondary steps. Lemma B.2 ([GPT15]).
There is a cutting planes derivation of T B from the set of clauses { C α | α < B } in length O ( nB ) and space O (1) . Lemma B.3 ([GPT15]).
If a total assignment α falsifies a set of inequalities C , then there is a cuttingplanes derivation of C α from C in length O ( n ) and space O (1) . Lemma B.2, which contains the core of the argument, follows from the proof of Lemma 3.2 in [GPT15].We repeat Claim 3 in that proof, which shows how to inductively derive T B ′ +1 from T B ′ and C B ′ , not n but B times.In turn Lemma B.3 follows from the proof of Theorem 3.4 in [GPT15]: since α must falsify someinequality I from C , we only need to reproduce the derivation of C α from I verbatim.24 roof of Lemma 4.9. Assume for now that C = x n − ∨ · · · ∨ x n − k . Consider the derivation Π of theinequality T n − k from { C α | α < B } , which is equivalent to C , given by Lemma B.2. We build a newderivation Π ′ extending Π as follows.Every time that we add an axiom C α to a configuration in Π , we replace that step by the derivationof C α from C given by Lemma B.3. Observe that we only add axioms C α with α < n − k , and since anysuch assignment falsifies C we meet the conditions to apply Lemma B.3.Finally we obtain C from T n − k by considering { T n − k } as a set of inequalities over the k variables x n − . . . x n − k and applying Lemma B.3 with α = 0 k being the only assignment over these variables thatfalsifies T n − k . The result is C = C .To derive a general clause C that contains k ′ negative literals, say C = ¬ x n − ∨ · · · ∨ ¬ x n − k ′ ∨ x n − k ′ − ∨ · · · ∨ x n − k , we build a derivation with the same structure as Π ′ , except that we replace everyoccurrence of x i by (1 − x i ) for n − k ′ ≤ i < n . To do so, we replace each derivation of an axiom C α witha derivation of the axiom C α +(0 n − k ′ k ′ ) and, for n − k ′ ≤ i < n , replace each use of x i ≥ and − x i ≥ − by − x i ≥ − and x i ≥ , respectively. Linear combination and division steps go through unchanged,we only observe that at a division step the coefficient on the right hand side differs by a multiple of thedivisor, so rounding is not affected. C Proof of Corollary 1.5
In this appendix, we provide the proof of Corollary 1.5, restated next.
Corollary C.1 (1.5, restated).
For any field F and any directed acyclic graph G , the Nullstellensatzdegree over F of Peb G , the decision tree depth of Search (Peb G ) , and the parity decision tree depth of Search (Peb G ) coincide and are equal to the reversible pebbling price of G . Our proof uses Lemma 4.2, restated next.
Lemma C.2 (4.2, restated).
For any field F and any graph G , NS F (Peb G ) = rpeb( G ) . Let F be the finite field of two elements. Given a search problem S , we denote the (deterministic)decision tree depth and parity decision tree depth of S by DT( S ) and PDT( S ) respectively. We use thefollowing lemma, which will be proved below. Lemma C.3 (Folklore).
Let C be an unsatisfiable CNF over n variables. Then, NS F ( C ) ≤ PDT(
Search ( C )) .Proof of Corollary 1.5 from Lemmas 4.2 and C.3. Let G be a directed acyclic graph. Note that it sufficesto prove the corollary for F = F , since Lemma 4.2 implies that NS F (Peb G ) is the same for every finitefield F . The corollary follows immediately from the following chain of inequalities: PDT(
Search (Peb G )) ≤ DT(
Search (Peb G )) = rpeb( G ) = NS F (Peb G ) ≤ PDT(
Search (Peb G )) . The first inequality is obvious, and the first equality was proved in the work of Chan [Cha13], but forcompleteness we provide a simplified proof in Appendix D. The second equality follows from Lemma 4.2,and the last inequality follows from Lemma C.3. Thus, the corollary is proved.In the remainder of this appendix, we prove Lemma C.3. Let C be an unsatisfiable CNF over variables z , . . . , z n , and let T be a parity decision tree of depth d that solves Search ( C ) . We prove that there existsa Nullstellensatz refutation over F for C of degree at most d , and this will imply the required result.Recall that the parity decision tree T takes as input an assignment α to z , . . . , z n , queries at most d parities of α , and then outputs a clause C of C that is violated by α . More formally, every internalnode v of T is associated with some linear polynomial p v in z , . . . , z n , and each outgoing edge e of v isassociated with bit b e ∈ { , } (so the edge e is taken if p v ( α ) = b e ). Every leaf ℓ of T is associated witha clause C ℓ ∈ C , such that every assignment α that leads T to ℓ violates the clasue C ℓ .We construct for each leaf ℓ of T a polynomial r ℓ ( z , . . . , z n ) of degree at most d that output on anassignment α if the tree T reaches the leaf ℓ when invoked on α , and outputs otherwise. We will use25hose polynomials later to construct the Nullstellensatz refutation of C . First, for every internal vertex v and an outgoing edge e of v , we associate with e the linear polynomial r e ( z , . . . , z n ) = p v ( z , . . . , z n ) + b e + 1 . Intuitively, the polynomial r e output on an assignment α if the query of v outputs b e on α , and otherwise. Now, to construct the polynomial r ℓ of a leaf ℓ , we multiply the polynomials r e for all theedges e on the path from the root to ℓ . Since there are at most d edges on that path, it follows that r ℓ isof degree at most d . Moreover, it is not hard to see that r ℓ ( α ) = 1 if α leads the tree T to the leaf ℓ , and r ℓ ( α ) = 0 otherwise.Let us denote by r ′ ℓ the “multilinearized” version of r ℓ , that is, r ′ ℓ is the polynomial obtained from r ℓ by reducing the degree of every variable to in each of its occurences in r ℓ . It is not hard to see that r ′ ℓ agrees with r ℓ on every assignment in { , } n . Our Nullstellensatz refutation of C is the polynomial r = X leaf ℓ of T r ′ ℓ . Clearly, this polynomial is of degree at most d . In order to prove that the polynomial r is a valid Nullstel-lensatz refutation of C , we need to prove that r equals , and that it can be derived from the axioms of C (in other words, that r belongs to ideal generated by E ( C ) ∪ (cid:8) z i − z i (cid:9) i ∈ [ n ] ). We start by proving that r equals . Claim C.4. r = 1 . Proof.
We use the well-known fact that a multilinear polynomial is determined by its values on { , } n :one way to see it is to observe that the multilinear monomials are a basis of the space of functions from { , } n to { , } , and therefore every such function has a unique representation as a multilinear polyno-mial. Since r is multilinear, it therefore suffices to prove that r outputs on every assignment in { , } n .Let α ∈ { , } n be an assignment to z , . . . , z n . Let ℓ be the leaf that T reaches when invoked on α .Then, by the construction of r ′ ℓ , it holds that r ′ ℓ ( α ) = 1 and that r ′ ℓ ′ ( α ) = 0 for every other leaf ℓ ′ of T . Itfollows that r ( α ) = 1 , as required.It remains to show that r can be derived from the axioms of C . To this end, we will show that eachof the polynomials r ′ ℓ can be derived from those axioms . It is not hard to show that for every leaf ℓ , thepolynomial r ℓ − r ′ ℓ can be derived from the boolean axioms (cid:8) z i − z i (cid:9) i ∈ [ n ] (in fact, this holds for anydifference of a polynomial and its multilinearized version). Thus, it suffices to prove that for every leaf ℓ ,the polynomial r ℓ can be derived from the axioms of C . We prove a stronger statement, namely, that forevery leaf ℓ , the polynomial r ℓ is divisible by the polynomial E ( C ℓ ) (i.e., the polynomial encoding of theclause C ℓ ). To this end, we prove the following result. Claim C.5.
Let p ( z , . . . , z n ) be a multilinear polynomial over F , and let i ∈ [ n ] . If p vanishes whenever z i = 0 , then z i divides p . Moreover, if p vanishes whenever z i = 1 , then (1 − z i ) divides p . Proof.
We can write p = z i · a + b , where a and b are polynomials that do not contain z i . Suppose firstthat p vanishes whenever z i = 0 . We would like to prove that b = 0 . Assume that this is not the case.Then, there is an assignment α ′ to the variables z , . . . , z n except for z i on which b does not vanish. Now,if we extend α ′ to an assignment α to z , . . . , z n by setting z i = 0 , it will follow that α is an assignmenton which z i = 0 but p does not vanish, which is a contradiction.Next, suppose that p vanishes whenever z i = 1 . Observe that we can write p = (1 − z i ) · a + ( b − a ) (here we use the fact that we are working over F ). We would like to prove that b − a = 0 . Assume thatthis is not the case. Then, there is an assignment α ′ to the variables z , . . . , z n except for z i on which b − a does not vanish. As before, if we extend α ′ to an assignment α to z , . . . , z n by setting z i = 1 , itwill follow that α is an assignment on which z i = 1 but p does not vanish, which is a contradiction.26e turn to prove that for every leaf ℓ , the polynomial r ℓ is divisible by the polynomial E ( C ℓ ) . Fix aleaf ℓ , and denote C = C ℓ . Recall that we denote by C + and C − the sets of variables that occur positivelyand negatively in C respectively, and that E ( C ) ≡ Y z ∈ C + (1 − z ) Y z ∈ C − z. Now, observe that for every variable z i ∈ C + , it holds that r ℓ vanishes whenever z i = 1 : to see it, observethat when z i = 1 , the assignment does not violate the clause C , and therefore the tree T cannot reach ℓ when invoked on that assignment. Thus, (1 − z i ) divides r ℓ for every z i ∈ C + . Similarly, the variable z i divides r ℓ for every z i ∈ C − . It follows that r ℓ is divisible by every factor of E ( C ) , and therefore it isdivisible by E ( C ) (here we used the fact that each factor occurs in E ( C ) at most once). This concludesthe proof. D Reversible Pebbling is Equivalent to Query Complexity
In this appendix we present a direct proof that the reversible pebbling price of a graph equals the querycomplexity of the search problem of the pebbling formula of that graph, originally proved by Chan [Cha13].
Theorem D.1 ([Cha13]).
For every DAG G with a single sink, it holds that DT(
Search (Peb G )) =rpeb( G ) . Let us introduce some notation to talk more formally about pebbling: A pebbling configuration is aset of vertices P . A (reversible) pebbling is a sequence of configurations P = P , . . . , P ℓ where P i +1 follows from P i by applying the pebbling rules. Its reverse R ( P ) = P ℓ , . . . , P is also a valid pebbling.Its cost is the maximal size of a configuration P i . Unless we call a pebbling partial , we assume that P = ∅ . A pebbling visits x if x ∈ P ℓ , and surrounds x if pred( x ) ⊆ P ℓ . The pebbling price of agraph G , denoted rpeb( G ) , is the minimum cost of all pebblings that visit the sink, and its surroundingprice , denoted speb( G ) , is the minimum cost of all pebblings that surround the sink.Given a decision tree for Search (Peb G ) , we associate each node with a state formed by a pair ( Q, Z ) of queried vertices and the vertices Z ⊆ Q whose queries were answered by . It is immediate toverify that at a leaf either the sink z belongs to Q \ Z , or there is a vertex in Z such that all of itspredecessors are in Q \ Z . It is useful to generalize the definition of the search problem Search (Peb G ) to start with intermediate states. Specifically, we associate a state ( Q, Z ) with the search problem inwhich we are given an assignment to Peb G that is promised to assign to the vertices in Z and to thevertices in Q \ Z , and would like to find a clause that is falsified by this assignment. We denote this searchproblem by Search G ( Q, Z ) and denote its query complexity by DT G ( Q, Z ) . We omit G from the latternotation when it is clear from the context. The crux of our proof is the following lemma, which impliesTheorem D.1. Lemma D.2.
For every DAG G with a single sink z , it holds that DT G ( { z } , { z } ) = speb( G ) . We claim that Lemma D.2 implies Theorem D.1. To see why, let G be a DAG with a single sink z ,and let G ′ be the DAG obtained from G by adding a new sink z ′ and an edge from z to z ′ . Then, it is nothard to see that Search ′ G ( { z ′ } , { z ′ } ) = Search (Peb G )) , and that every pebbling that surrounds the sinkof G ′ is pebbling that visits the sink of G , and vice versa. Hence, it holds that DT(
Search (Peb G )) = DT G ′ ( { z ′ } , { z ′ } ) = speb( G ′ ) = rpeb( G ) , where the second equality follows from Lemma D.2. In the rest of this appendix, we focus on provingLemma D.2. To this end, fix a DAG G with a single sink z .We first show that the states of an optimal decision tree for Search ( { z } , { z } ) are of a special form,which we call ”path-like”. Specifically, we say that a state ( Q, Z ) is path-like if there is a path P endingat the sink such that P ∩ Q = Z . Observe that in a path-like state there is a unique such path of maximal27ength that starts in a vertex in Z . We denote this path by P Z , and denote its first vertex by head( Z ) . Wesay that a vertex v is relevant to a path-like state ( Q, Z ) if there is a path P v from v to head( Z ) such that P v ∩ Q = { head( Z ) } . Observe that starting from a path-like state and querying a relevant vertex yieldsa path-like state where the new path is P Z ∪ P v if the answer is and P Z if it is .We now show that if ( Q, Z ) is a path-like state, then every optimal decision tree for Search ( Q, Z ) onlyqueries relevant vertices. Observe that this implies that all the nodes of the tree correspond to path-likestates. Moreover, this result holds in particular for ( { z } , { z } ) , since it is a path-like state. Lemma D.3. If ( Q, Z ) is a path-like state then every optimal decision tree for Search ( Q, Z ) only queriesvertices relevant to ( Q, Z ) .Proof. The proof is by induction on the query complexity p of Search ( Q, Z ) . The base case p = 0 holdsvacuously: the optimal decision tree does not make any queries. Assume that p > . Fix an optimaldecision tree T for ( Q, Z ) , and let v be the query made by the root of T . Observe that the children of theroot of T correspond to the states ( Q ∪ v, Z ∪ v ) and ( Q ∪ v, Z ) . Let T , T be the sub-trees rooted atthose children respectively, and note that these trees are optimal for the latter states and that their depthis at most p − .Suppose first that v is a relevant query to ( Q, Z ) . In this case, the states of the children of the root arepath-like. By the induction assumption, the trees T , T only make queries that are relevant to the states ( Q ∪ v, Z ∪ v ) and ( Q ∪ v, Z ) respectively. Observe that any such query is necessarily relevant to ( Q, Z ) as well. Therefore, every query of T is relevant to ( Q, Z ) .Next, suppose that v is irrelevant to ( Q, Z ) . We consider two separate cases: the case where v belongsto the path P Z , and the case where it does not belong to P Z . v belongs to P Z . Assume that v belongs to the path P Z . In this case, the state ( Q ∪ v, Z ∪ v ) ispath-like, and therefore by the induction assumption the tree T only makes queries that are relevant tothat state. We claim that T solves Search ( Q, Z ) , which contradicts the assumption that T is optimal for Search ( Q, Z ) .To see why, suppose for the sake of contradiction that T fails to solve Search ( Q, Z ) on some assign-ment α . Observe that the only case where this can happen is when α ( x v ) = 1 but T outputs that theviolated clause is C v = x v ∨ W u ∈ pred( v ) ¬ x u . Let u be the predecessor of v that lies on P Z (such u mustexist since v is queried by T and hence cannot be equal to head( Z ) ). Then, u cannot belong to Z (orotherwise T could not have output C v ) and therefore it cannot belong to Q (since P Z ∩ Q = Z ). Thismeans that T must have queried u in order to output C v . However, u is irrelevant to ( Q ∪ v, Z ∪ v ) , sowe reached a contradiction to the assumption that T does not make irrelevant queries. Hence, T solves Search ( Q, Z ) , as required. v does not belong to P Z . Next, assume that v does not belong to the path P Z . In this case, the state ( Q ∪ v, Z ) is path-like, and therefore by the induction assumption the tree T only makes queries that arerelevant to that state. We claim that T solves Search ( Q, Z ) , which contradicts the assumption that T isoptimal for ( Q, Z ) .To see why, suppose for the sake of contradiction that T fails to solve Search ( Q, Z ) on some assign-ment α . Observe that the only case where this can happen is when α ( x v ) = 0 but for some successor w of v the tree T outputs that the violated clause is C w = x w ∨ W u ∈ pred( w ) ¬ x u . The query w cannotbe relevant to ( Q, Z ) , since otherwise v would have been relevant for ( Q, Z ) . Since w is irrelevant to ( Q, Z ) , it cannot be queried by T , and therefore it must belong to Z in order for T to output C w . More-over, w cannot be equal to head( Z ) , since otherwise v would have been relevant for ( Q, Z ) . Thus, w hasa predecessor u in P Z .The vertex u cannot belong to Z (or otherwise T could not have output C w ) and therefore it cannotbelong to Q (since P Z ∩ Q = Z ). This means that T must have queried u in order to output C w .However, u is irrelevant to ( Q ∪ v, Z ) , so we reached a contradiction to the assumption that T does notmake irrelevant queries. Hence, T solves Search ( Q, Z ) pebbling assuming free pebbles on a set S is a partial pebbling P such that P ⊆ S , and its cost is the maximum size of P i \ S . Proposition D.4. If DT( { z } , { z } ) ≤ p , then speb( G ) ≤ p .Proof. We prove the following stronger claim: if for some path-like state ( Q, Z ) there is an optimaldecision tree T of depth p , then there is a pebbling that surrounds head( Z ) of cost p assuming freepebbles on Q \ Z . To see that this claim implies the proposition observe that ( { z } , { z } ) is path-like.Therefore, if DT( { z } , { z } ) ≤ p then the claim implies that there is a pebbling that surrounds z of cost atmost p without free pebbles.We prove the claim by induction on p . The base case is when p = 0 , so T consists of a single leaf.This means that there must be some vertex in Z that is surrounded by vertices in Q \ Z . This vertex mustbe head( Z ) , since any other vertex v ∈ Z has a predecessor in P Z , and this predecessor cannot belong to Q \ Z . Hence, { Q \ Z } is a surrounding pebbling of head( Z ) of cost assuming free pebbles on Q \ Z ,as required.We proceed to the induction step. Suppose that p > . Let v be the query made at the root of T .Let T , T be the subtrees rooted at the children of v that corresponds to the states ( Q ∪ v, Z ∪ v ) and ( Q ∪ v, Z ) respectively. By Lemma D. , the query v is relevant to ( Q, Z ) , and therefore the latter statesare path-like.Observe that head( Z ∪ v ) = v . Hence, by applying the induction assumption to T , we obtain apebbling P that surrounds v of cost at most p − assuming free pebbles on ( Q ∪ v ) \ ( Z ∪ v ) = Q \ Z. Furthermore, by applying the induction assumption to T , we obtain a pebbling P that surrounds head( Z ) of cost at most p − assuming free pebbles on ( Q ∪ v ) \ Z .We now construct a pebbling that surrounds head( Z ) assuming free pebbles on Q \ Z as follows: wefirst follow P , thus reaching a configuration that surrounds v . Then, we place a pebble on v (unless it isalready pebbled). Next, we follow { P ∪ v | P ∈ R ( P ) } to remove all the pebbles that were placed by P except for ( Q ∪ v ) \ Z . Finally, we follow P and reach a configuration that surrounds head( Z ) . It isnot hard to see that this pebbling has cost at most p assuming free pebbles on Q \ Z , as required.In the proof of the second proposition, we use the following notion: Given a pebbling P , we define static( P ) = ∩P to be the set of pebbles that are always present in P , and the non-static cost of P be themaximum size of P \ static( P ) for P ∈ P . We also denote the vertices reachable from v , including v itself, by desc( v ) . Proposition D.5. If speb( G ) ≤ p then DT( { z } , { z } ) ≤ p .Proof. We prove the following stronger claim: if for some state ( Q, Z ) there is a partial pebbling P thatsurrounds a vertex w ∈ Z of non-static cost p with static( P ) ⊆ Q \ Z , then there is a decision tree ofdepth p that solves Search ( Q, Z ) .To see that this claim implies the proposition, observe that a pebbling P that surrounds z of cost p starting from ∅ has static( P ) = ∅ . Thus, the claim implies that if such a pebbling exists, it holds that DT( { z } , { z } ) ≤ p .If static( P ) surrounds w then all the predecessors of w are in Q \ Z . Therefore, Search ( Q, Z ) canbe solved without making any queries: the decision tree can immediately output that the clause C w = x w ∨ W u ∈ pred( w ) ¬ x u is violated.Otherwise we prove the claim by induction on p . The base case p = 0 is a particular instance of static( P ) surrounding w , hence we turn to proving the induction step and suppose that p > . Having ℓ denote the length of P , let v be the earliest (in time) vertex to be placed at some time m > and notremoved until time ℓ . Note that v exists because w is not surrounded in some configuration of P .29et P ′ = P m , . . . , P ℓ be the subpebbling of P from time m to time ℓ . Observe that static( P ′ ) =static( P ) ∪ { v } by construction, and thus the non-static cost of P ′ is at most p − . By applying theinduction assumption to P ′ , it follows that there exists a decision tree T for Search ( Q ∪ v, Z ) of depth p − .Next, observe that R ( P ′ ) is a partial pebbling that surrounds v of non-static cost p − with static( R ( P ′ )) =static( P ) ∪ { v } , therefore P ′′ = { P \ desc( v ) | P ∈ R ( P ′ ) } is a partial pebbling that surrounds v ofnon-static cost p − with static( P ′′ ) ⊆ static( P ) . Hence, by applying the induction assumption to P ′′ ,it follows that there exists a decision tree T for Search ( Q ∪ v, Z ∪ v ) of depth p − .We now construct a decision tree T for Search ( Q, Z ) as follows: The tree T queries v . If the answeris , the tree proceeds by invoking T , and otherwise it invokes T . 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