A Lower Bound on DNNF Encodings of Pseudo-Boolean Constraints
aa r X i v : . [ c s . A I] J a n A Lower Bound on DNNF Encodings ofPseudo-Boolean Constraints
Alexis de ColnetCRIL, CNRS & Univ Artois
Abstract
Two major considerations when encoding pseudo-Boolean (PB) constraintsinto SAT are the size of the encoding and its propagation strength, that is,the guarantee that it has a good behaviour under unit propagation. Severalencodings with propagation strength guarantees rely upon prior compilation ofthe constraints into
DNNF (decomposable negation normal form),
BDD (binarydecision diagram), or some other sub-variants. However it has been shown thatthere exist PB-constraints whose ordered
BDD ( OBDD ) representations, and thusthe inferred
CNF encodings, all have exponential size. Since
DNNF s are moresuccinct than
OBDD s, preferring encodings via
DNNF to avoid size explosionseems a legitimate choice. Yet in this paper, we prove the existence of PB-constraints whose
DNNF s all require exponential size.
Pseudo-Boolean constraints (or PB-constraints) are Boolean functions over 0 / x , . . . , x n of the form P ni =1 w i x i ’op’ θ where the w i are integer weights, θ isan integer threshold and ’op’ is a comparison operator < , ≤ , > or ≥ . PB-constraintshave been studied extensively under different names (e.g. threshold functions [14],Knapsack constraints [13]) due to their omnipresence in many domains of AI and theirwide range of practical applications [3, 7, 9, 15, 21].One way to handle PB-constraints in a constraint satisfaction problem is to trans-late them into a CNF formula and feed it to a SAT solver. The general idea is togenerate a
CNF , possibly introducing auxiliary Boolean variables, whose restriction tovariables of the constraint is equivalent to the constraint. Two major considerationshere are the size of the
CNF encoding and its propagation strength. One wants, onthe one hand, to avoid the size of the encoding to explode, and on the other hand, toguarantee a good behaviour of the SAT instance under unit propagation – a techniqueat the very core of SAT solving. Desired propagation strength properties are, for in-stance, generalised arc consistency (GAC) [4] or propagation completeness (PC) [6].Several encodings to
CNF follow the same two-steps method: first, each constraint isrepresented in a compact form such as
BDD (Binary Decision Diagram) or
DNNF (De-composable Negation Normal Form). Second, the compact forms are turned into
CNF s1sing Tseitin or other transformations. The SAT instance is the conjunction of all ob-tained
CNF s. It is worth mentioning that there are GAC encodings of PB-constraintsinto polynomial size
CNF s that do not follow this two-steps method [5]. However nosimilar result is known for PC encodings. PC encodings are more restrictive that GACencodings and may be obtained via techniques requiring compilation to
DNNF [17].Thus the first step is a knowledge compilation task.Knowledge compilation studies different representations for knowledge [10, 19] un-der the general idea that some representations are more suitable than others whensolving specific reasoning problems. One observation that has been made is that themore reasoning tasks can be solved efficiently with particular representations, the largerthese representations get in size. In the context of constraint encodings to SAT, theconversion of compiled forms to
CNF s does not reduce the size of the SAT instance,therefore it is essential to control the size of the representations obtained by knowledgecompilation.Several representations have been studied with respect to different encoding tech-niques with the purpose of determining which properties of representations are suffi-cient to ensure propagation strength [1,2,11,12,16,17]. Popular representations in thiscontext are
DNNF and
BDD and their many variants: deterministic
DNNF , smooth
DNNF , ordered
BDD ( OBDD ) . . . As mentioned above, a problem occurring when com-piling a constraint into such representations is that exponential space may be required.Most notably, it has been shown in [2,14] that some PB-constraints can only be repre-sented by
OBDD s whose size is exponential in √ n , where n is the number of variables.Our contribution is the proof of the following theorem where we lift the statementfrom OBDD to DNNF . Theorem 1.
There is a class of PB-constraints F such that for any constraint f ∈ F on n variables, any DNNF representation of f has size Ω( n ) . Since
DNNF s are exponentially more succinct than
OBDD s [10], our result is ageneralisation of the result in [2, 14]. The class F is similar to that used in [2, 14],actually the only difference is the choice of the threshold for the PB-constraints. Yet,adapting proofs given in [2, 14] for OBDD to DNNF is not straightforward, thus ourproof of Theorem 1 bears very little resemblance.It has been shown in [18] that there exist sets of PB-constraints such that the whole set (so a conjunction of PB-constraints) requires exponential size
DNNF to represent.Our result is a generalisation to single
PB-constraints.
Conventions of notation.
Boolean variables are seen as variables over { , } , where0 and 1 represent f alse and true respectively. Via this 0 / Z . For notational convenience, wekeep the usual operators ¬ , ∨ and ∧ to denote, respectively, the negation, disjunctionand conjunction of Boolean variables or functions. Given X a set of n Boolean vari-ables, assignments to X are seen as vectors in { , } n . Single Boolean variables arewritten in plain text ( x ) while assignments to several variables are written in bold ( x ).2e write x ≤ y when the vector y dominates x element-wise. We write x < y when x ≤ y and x = y . In this framework, a Boolean function f over X is a mapping from { , } n to { , } . f is said to accept an assignment x when f ( x ) = 1, then x is calleda model of f . The function is monotone if for any model x of f , all y ≥ x are modelsof f as well. The set of models of f is denoted f -1 (1). Given f and g two Booleanfunctions over X , we write f ≤ g when f -1 (1) ⊆ g -1 (1). We write f < g when theinclusion is strict. Pseudo-Boolean constraints.
Pseudo-Boolean (PB) constraints are inequalitiesthe form P ni =1 w i x i ’op’ θ where the x i are 0 / w i and θ are integers, and ’op’ is one of the comparison operator < , ≤ , > or ≥ . A PB-constraint is associated with a Boolean function whose models are exactly the assign-ments to { x , . . . , x n } that satisfy the inequality. For simplicity we directly considerPB-constraints as Boolean functions – although the same function may represent dif-ferent constraints – while keeping the term “constraints” when referring to them. Inthis paper, we restrict our attention to PB-constraints where ’op’ is ≥ and all weightsare positive integers. Note that such PB-constraints are monotone Boolean functions.Given a sequence of positive integer weights W = ( w , . . . , w n ) and an integer thresh-old θ , we define the function w : { , } n → N that maps any assignment to its weightby w ( x ) = P ni =1 w i x i . With these notations, a PB-constraint over X for a given pair( W, θ ) is a Boolean function whose models are exactly the x such that w ( x ) ≥ θ . Example 1.
Let n = 5 , W = (1 , , , , and θ = 9 . The PB-constraint for ( W, θ ) isthe Boolean function whose models are the assignments such that P i =1 ix i ≥ . E.g. x = (0 , , , , is a model of weight w ( x ) = 10 . For notational clarity, given any subset Y ⊆ X and denoting x | Y the restriction of x to variables of Y , we overload w so that w ( x | Y ) is the sum of weights activated byvariables of Y set to 1 in x . Decomposable
NNF . A circuit in negation normal form ( NNF ) is a single outputBoolean circuit whose inputs are Boolean variables and their complements, and whosegates are fanin-2
AND and OR gates. The size of the circuit is the number of its gates.We say that an NNF is decomposable ( DNNF ) if for any
AND gate, the two sub-circuitsrooted at that gate share no input variable, i.e., if x or ¬ x is an input of the circuitrooted at the left input of the AND gate, then neither x nor ¬ x is an input of thecircuit rooted at the right input, and vice versa. A Boolean function f is encoded bya DNNF D if the assignments of variables for which the output of D is 1 ( true ) areexactly the models of f . Rectangle covers.
Let X be a finite set of Boolean variables and let Π = ( X , X )be a partition of X (i.e., X ∪ X = X and X ∩ X = ∅ ). A rectangle r with respect toΠ is a Boolean function over X defined as the conjunction of two functions ρ and ρ over X and X respectively. Π is called the partition of r . We say that the partitionand the rectangle are balanced when | X | ≤ | X | ≤ | X | (thus the same holds for X ).Whenever considering a partition ( X , X ), we use for any assignment x to X the3otations x := x | X and x := x | X . And for any two assignments x and x to X and X , we note ( x , x ) the assignment to X whose restrictions to X and X are x and x . Given f a Boolean function over X , a rectangle cover of f is a disjunctionof rectangles over X , possibly with different partitions, equivalent to f . The size ofa rectangle cover is the number of its rectangles. A cover is called balanced if all itsrectangles are balanced. Example 2.
Going back to Example 1, consider the partition X := { x , x , x } , X := { x , x } and define ρ := x ∧ x and ρ := x ∨ x . Then r := ρ ∧ ρ isa rectangle w.r.t. this partition that accepts only models of the PB-constraint fromExample 1. Thus it can be part of a rectangle cover for this constraint. Any function f has at least one balanced rectangle cover as one can create a balancedrectangle accepting exactly one chosen model of f . We denote by C ( f ) the size of thesmallest balanced rectangle cover of f . The following result from [8] links C ( f ) to thesize of any DNNF encoding f . Theorem 2.
Let D be a DNNF encoding a Boolean function f . Then f has a balancedrectangle cover of size at most the size of D . Theorem 2 reduces the problem of finding lower bounds on the size of
DNNF s encoding f to that of finding lower bounds on C ( f ). The strategy to prove Theorem 1 is to find a PB-constraint f over n variables such that C ( f ) is exponential in √ n and then use Theorem 2. We first show that we can restrictour attention to covering particular models of f with rectangles rather than the wholefunction. In this section X is a set of n Boolean variables and f is a PB-constraintover X . Recall that we only consider constraints of the form P ni =1 w i x i ≥ θ where the w i and θ are positive integers. Definition 1.
The threshold models of f are the models x such that w ( x ) = θ . Threshold models should not be confused with minimal models (or minimals).
Definition 2. A minimal of f is a model x such that no y < x is a model of f . For a monotone PB-constraint, a minimal model is such that its sum of weights dropsbelow the threshold if we remove any element from it. Any threshold model is mini-mal, but not all minimals are threshold models. There even exist constraints with nothreshold models (e.g. take even weights and an odd threshold) while there always areminimals for satisfiable constraints.
Example 3.
The minimal models of the PB-constraint from Example 1 are (0 , , , , , (0 , , , , , (1 , , , , and (0 , , , , . The first three are threshold models. Let f ∗ be the Boolean function whose models are exactly the threshold models of f .In the next lemma, we prove that the smallest rectangle cover of f ∗ has size at most C ( f ). Thus, lower bounds on C ( f ∗ ) are also lower bounds on C ( f ).4 emma 1. Let f ∗ be the Boolean function whose models are exactly the thresholdmodels of f . Then C ( f ) ≥ C ( f ∗ ) .Proof. Let r := ρ ∧ ρ be a balanced rectangle with r ≤ f and assume r accepts somethreshold models. Let Π := ( X , X ) be the partition of r . We claim that there existtwo integers θ and θ such that θ + θ = θ and, for any threshold model x acceptedby r , there is w ( x ) = θ and w ( x ) = θ . To see this, assume by contradiction thatthere exists another partition θ = θ ′ + θ ′ of θ such that some other threshold model y with w ( y ) = θ ′ and w ( y ) = θ ′ is accepted by r . Then either w ( x ) + w ( y ) < θ or w ( y ) + w ( x ) < θ , but since ( x , y ) and ( y , x ) are also models of r , r wouldaccept a non-model of f , which is forbidden. Now let ρ ∗ (resp. ρ ∗ ) be the functionwhose models are exactly the models of ρ (resp. ρ ) of weight θ (resp. θ ). Then r ∗ := ρ ∗ ∧ ρ ∗ is a balanced rectangle whose models are exactly the threshold modelsaccepted by r .Now consider a balanced rectangle cover of f of size C ( f ). For each rectangle r ofthe cover, if r accepts no threshold model then discard it, otherwise construct r ∗ . Thedisjunction of these new rectangles is a balanced rectangle cover of f ∗ of size at most C ( f ). Therefore C ( f ) ≥ C ( f ∗ ). K n,n We define the class of hard PB-constraints for Theorem 1 in this section. Recall thatfor a hard constraint f , our aim is to find an exponential lower bound on C ( f ). We willshow, using Lemma 1, that the problem can be reduced to that of covering all maximalmatchings of the complete n × n bipartite graph K n,n with rectangles. In this section, X is a set of n Boolean variables. For presentability reasons, assignments to X arewritten as n × n matrices. Each variable x i,j has the weight w i,j := (2 i + 2 j + n ) / W := ( w i,j : 1 ≤ i, j ≤ n ) and the threshold θ := 2 n − f for the pair ( W, θ ) is such that f ( x ) = 1 if and only if x satisfies X ≤ i,j ≤ n (cid:18) i + 2 j + n (cid:19) x i,j ≥ n − f writing the weights and threshold as binary numbers of2 n bits. Bits of indices 1 to n form the lower part of the number and those of indices n + 1 to 2 n form the upper part . The weight w i,j is the binary number where the onlybits set to 1 are the i th bit of the lower part and the j th bit of the upper part. Thuswhen a variable x i,j is set to 1, exactly one bit of value 1 is added to each part of thebinary number of the sum.Assignments to X uniquely encode subgraphs of K n,n . We denote U = { u , . . . , u n } the nodes of the left side and V = { v , . . . , v n } those of the right side of K n,n . Thebipartite graph encoded by x is such that there is an edge between the u i and v j ifand only if x i,j is set to 1 in x . 5 xample 4. Take n = 4 . The assignment x = encodes u u u u v v v v Definition 3. A maximal matching assignment (or maximal matching model) is anassignment x to X such that • for any i ∈ [ n ] , there is exactly one k such that x i,k is set to in x , • for any j ∈ [ n ] , there is exactly one k such that x k,j is set to in x . As the name suggests, the maximal matching assignments are those encodinggraphs whose edges form a maximal matching of K n,n (i.e., a maximum cardinalitymatching). One can also see them as encodings for permutations of [ n ]. Example 5.
The maximal matching model x = encodes u u u u v v v v For a given x , define var k ( x ) by var k ( x ) := { j | x k,j is set to 1 in x } when 1 ≤ k ≤ n and by var k ( x ) := { i | x i,k − n is set to 1 in x } when n + 1 ≤ k ≤ n . var k ( x )stores the index of variables in x that directly add 1 to the k th bit of w ( x ). Note thata maximal matching model is an assignment x such that | var k ( x ) | = 1 for all k . It iseasy to see that maximal matching models are threshold models of f : seeing weights asbinary numbers of 2 n bits, for every bit of the sum the value 1 is added exactly once,so exactly the first 2 n bits of the sum are set to 1, which gives us θ . Note that notall threshold models of f are maximal matching models, for instance the assignmentfrom Example 4 does not encode a maximal matching but one can verify that it is athreshold model. Recall that f ∗ is the function whose models are the threshold modelsof f . In the next lemmas, we prove that lower bounds on the size of rectangle coversof the maximal matching models are lower bounds on C ( f ∗ ), and a fortiori on C ( f ). Lemma 2.
Let
Π := ( X , X ) be a partition of X . Let x := ( x , x ) and y := ( y , y ) be maximal matching assignments. If ( x , y ) and ( y , x ) both have weight θ := 2 n − then both are maximal matching assignments.Proof. It is sufficient to show that | var k ( x , y ) | = 1 and | var k ( y , x ) | = 1 forall 1 ≤ k ≤ n . We prove it for ( x , y ) by induction on k . First observe thatsince | var k ( x ) | = 1 and | var k ( y ) | = 1 for all 1 ≤ k ≤ n , the only possibilities for | var k ( x , y ) | are 0, 1 or 2. • For the base case k = 1, if | var ( x , y ) | is even then the first bit of w ( x )+ w ( y )is 0 and the weight of ( x , y ) is not θ . So | var ( x , y ) | = 1. • For the general case 1 < k ≤ n , assume it holds that | var ( x , y ) | = · · · = | var k − ( x , y ) | = 1. So the k th bit of w ( x ) + w ( y ) depends only on theparity of | var k ( x , y ) | : the k th bit is 0 if | var k ( x , y ) | is even and 1 otherwise.( x , y ) has weight θ so | var k ( x , y ) | = 1.The argument applies to ( y , x ) analogously.6 emma 3. Let f be the PB-constraint (1) and let ˆ f be the function whose models areexactly the maximal matching assignments. Then C ( f ) ≥ C ( ˆ f ) .Proof. By Lemma 1, it is sufficient to prove that C ( f ∗ ) ≥ C ( ˆ f ). We already knowthat ˆ f ≤ f ∗ . Let r := ρ ∧ ρ be a balanced rectangle of partition Π := ( X , X ) with r ≤ f ∗ , and assume r accepts some maximal matching assignment. Let ˆ ρ (resp. ˆ ρ )be the Boolean function over X (resp. X ) whose models are the x (resp. x ) suchthat there is a maximal matching assignment ( x , x ) accepted by r . We claim thatthe balanced rectangle ˆ r := ˆ ρ ∧ ˆ ρ accepts exactly the maximal matching models of r . On the one hand, it is clear that all maximal matching models of r are modelsof ˆ r . On the other hand, all models of ˆ r are threshold models of the form ( x , y ),where ( x , x ) and ( y , y ) encode maximal matchings, so by Lemma 2, ˆ r accepts onlymaximal matching models of r .Now consider a balanced rectangle cover of f ∗ of size C ( f ∗ ). For each rectangle r of the cover, if r accepts no maximal matching assignment then discard it, otherwiseconstruct ˆ r . The disjunction of these new rectangles is a balanced rectangle cover ofˆ f of size at most C ( f ∗ ). Therefore C ( f ∗ ) ≥ C ( ˆ f ). Theorem 1.
There is a class of PB-constraints F such that for any constraint f ∈ F on n variables, any DNNF encoding f has size Ω( n ) . F is the class of constraints defined in (1). Thanks to Theorem 2 and Lemma 3, theproof boils down to finding exponential lower bounds on C ( ˆ f ), where ˆ f is the Booleanfunction on n variables whose models encode exactly the maximal matchings of K n,n (or equivalently, the permutations of [ n ]). ˆ f has n ! models. The idea is now to provethat rectangles covering ˆ f must be relatively small, so that covering the whole functionrequires many of them. Lemma 4.
Let
Π = ( X , X ) be a balanced partition of X . Let r be a rectangle withrespect to Π with r ≤ ˆ f . Then | r -1 (1) | ≤ n ! / (cid:0) nn √ / (cid:1) . The function ˆ f has already been studied extensively in the literature, often under thename PERM n (for permutations on [ n ]), see for instance Chapter 4 of [22] or section6.2 of [20] where a statement similar to Lemma 4 is established. With Lemma 4 wecan give the proof of Theorem 1. Theorem 1.
Let W C ( ˆ f ) k =1 r k be a balanced rectangle cover of ˆ f . We have P C ( ˆ f ) k =1 | r -1 k (1) | ≥| ˆ f -1 (1) | = n !. Lemma 4 gives us ( C ( ˆ f ) n !) / (cid:0) nn √ / (cid:1) ≥ n !, thus C ( ˆ f ) ≥ (cid:18) nn p / (cid:19) ≥ nn p / ! n √ / = (cid:18) (cid:19) n √ / ≥ n √ / = 2 Ω( n ) where we have used (cid:0) ab (cid:1) ≥ ( a/b ) b and 3 / ≥ √
2. Using Lemma 3 we get that C ( f ) ≥ C ( ˆ f ) ≥ Ω( n ) . Theorem 2 allows us to conclude.7 u u u v v v v (a) Balanced partition Π of K , u u u u v v v v x x U V U V (b) Partition of a maximal matching w.r.t. Π Figure 1: Partition of maximal matchingAll that is left is to prove Lemma 4.
Lemma 4.
Let r := ρ ∧ ρ and Π := ( X , X ). Recall that U := { u , . . . , u n } and V := { v , . . . , v n } are the nodes from the left and right part of K n,n respectively.Define U := { u i | there exists x i,l ∈ X such that a model of ρ has x i,l set to 1 } and V := { v j | there exists x l,j ∈ X such that a model of ρ has x l,j set to 1 } . Define U and V analogously (this time using X and ρ ). Figure 1 illustrates the constructionof these sets: Figure 1a shows a partition Π of the edges of K , (full edges in X ,dotted edges in X ) and Figure 1b shows the contribution of a model of r to U , V , U , and V after partition according to Π.Models of ρ are clearly matchings of K n,n . Actually they are matchings between U and V by construction of these sets. We claim that they are maximal. To verifythis, observe that U ∩ U = ∅ and V ∩ V = ∅ since otherwise r has a model that isnot a matching. Thus if ρ were to accept a non-maximal matching between U and V then r would accept a non-maximal matching between U and V . So ρ acceptsonly maximal matchings between U and V , consequently | U | = | V | . The argumentapplies symmetrically for V and U . We note k := | U | . It stands that U ∪ U = U and V ∪ V = V as otherwise r accepts matchings that are not maximal. So | U | = | V | = n − k . We now have | ρ -1 (1) | ≤ k ! and | ρ -1 (1) | ≤ ( n − k )!, leading to | r -1 (1) | ≤ k !( n − k )! = n ! / (cid:0) nk (cid:1) .Up to k edges may be used to build matchings between U and V . Since r isbalanced we obtain k ≤ n /
3. Applying the same argument to U and V gives us( n − k ) ≤ n /
3, so n (1 − p / ≤ k ≤ n p /
3. Finally, the function k n ! / (cid:0) nk (cid:1) ,when restricted to some interval [[ n (1 − α ) , αn ]], reaches its maximum at k = αn , hencethe upper bound | r -1 (1) | ≤ n ! / (cid:0) nn √ / (cid:1) . Acknowledgments
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