Expansion-Based QBF Solving Without Recursion

Transcription

1 1 Expnsion-Bsed QBF Solving Without Recursion Roderick Bloem, Nicols Brud-Sntoni, Vedd Hdzic, TU Grz Uwe Egly, Florin Lonsing, TU Wien Mrtin Seidl, JKU Linz rxiv: v1 [cs.lo] 24 Jul 218 Astrct In recent yers, expnsion-sed techniques hve een shown to e very powerful in theory nd prctice for solving quntified Boolen formuls (QBF), the extension of propositionl formuls with existentil nd universl quntifiers over Boolen vriles. Such pproches prtilly expnd one type of vrile (either existentil or universl) nd pss the otined formul to SAT solver for deciding the QBF. Stte-of-the-rt expnsionsed solvers process the given formul quntifier-lock wise nd recursively pply expnsion until solution is found. In this pper, we present novel lgorithm for expnsionsed QBF solving tht dels with the whole quntifier prefix t once. Hence recursive pplictions of the expnsion principle re voided. Experiments indicte tht the performnce of our simple pproch is comprle with the stte of the rt of QBF solving, especilly in comintion with other solving techniques. I. INTRODUCTION Efficient tools for deciding the stisfiility of Boolen formuls (SAT solvers) re the core technology in mny utomted verifiction nd synthesis pproches [45]. However, wide rnge of verifiction nd synthesis prolems re eyond the complexity clss NP s cptured y SAT, requiring more powerful formlisms for encoding nd solving such prolems. One promising formlism re quntified Boolen formuls (QBFs) tht extend propositionl formuls y universl nd existentil quntifiers over Boolen vriles [32] resulting in decision prolem tht is PSPACE-complete (see [5] for survey). Among mny others, pplictions like verifiction nd synthesis [8], [13], [14], [18], [2], [24], relizility checking [19], ounded model checking [16], [48], nd lso plnning [17], [41] motivte the quest for efficient QBF solvers. Unlike for SAT, where conflict-driven cluse lerning (CDCL) is the single dominnt solving pproch for prcticl prolems, two dominnt pproches exist for QBF solving. On one hnd, CDCL hs een successfully extended to QCDCL tht enles cluse nd cue lerning [21], [35], [47]. On the other hnd, vrile expnsion hs ecome populr solving prdigm (see Section II for short overview of relted work nd Section IV for detiled introduction). In short, expnsion-sed solvers eliminte one kind of vriles y ssigning them truth vlues nd solve the resulting propositionl formul with SAT solver. For QBFs with one quntifier lterntion (2QBF), nturl pproch is to use two SAT solvers: one tht dels with the existentilly quntified vriles nd nother one tht dels with the universlly quntified vriles. For generlising this SAT-sed pproch to QBFs with n ritrry numer of quntifier lterntions, This work hs een supported y the Austrin Science Fund (FWF) under projects W1255-N23, S1146-N23, S1148-N23, nd S1149-N23. expnsion is recursively pplied per quntifier lock, requiring multiple SAT solvers. As noted y Re nd Tentrup [39], these pproches show poor performnce for formuls with mny quntifier lterntions. In this pper, we present novel solving lgorithm sed on non-recursive expnsion for QBFs with ritrry quntifier prefixes using only two SAT solvers. Our pproch of non-recursive expnsion is theoreticlly (i.e., from proof complexity perspective) equivlent to pproches tht pply recursive expnsion since oth non-recursive nd recursive expnsion rely on the Exp+Res proof system [6]. However, the non-recursive expnsion hs prcticl implictions such s modified serch strtegy. Tht is, the use of recursive or nonrecursive expnsion results in different serch strtegies for the proof. With respect to proof serch, there is n nlogy to, e.g., implementtions of resolution-sed CDCL SAT solvers tht employ different serch heuristics. In our experimentl evlution, we compre the performnce of our pproch with the performnce of the stte-of-the-rt solvers. In ddition, we implemented hyrid pproch tht comines cluse lerning with non-recursive expnsion-sed solving for exploiting the power of QCDCL. Our experiments indicte tht this hyrid pproch performs very well, especilly on formuls with multiple quntifier lterntions. This pper is structured s follows. After review of relted work in the next section, we introduce the necessry preliminries in Section III. After short recpitultion of expnsion in Section IV, our novel non-recursive expnsionsed lgorithm is presented in Section V. Implementtion detils re discussed in Section VI together with short discussion of the hyrid pproch. In Section VII we compre our pproch to stte-of-the-rt solvers. II. RELATED WORK Alredy the erly QBF solvers Quos [2] nd Quntor [2] incorporte selective quntifier expnsion for eliminting one kind of quntifiction to reduce the given QBF to propositionl formul. The resulting propositionl formul is then solved y clling SAT solver once. Quos nd Quntor impressively demonstrted the power of expnding universl vriles ut lso showed its enormous memory consumption. As prgmtic compromise, ounded universl expnsion ws introduced for efficient preprocessing [11], [22], [23], [46]. The first pproch which uses two lternte SAT solvers A nd B for solving 2QBF, i.e., QBFs of the form U E.φ, ws presented in [4]. Solver A is initilised with φ, B with the empty formul. Both propositionl formuls re incrementlly refined with stisfying ssignments found y the other solver.

2 2 If A finds its formul unstisfile, then the QBF is flse. Otherwise, the negtion of the universl prt of the stisfying ssignment is pssed to solver B. If solver B finds its formul unstisfile, then the QBF is true. Otherwise, the existentil prt of the stisfying ssignment is pssed to solver A. Jnot nd Mrques-Silv generlised the ide of lternting SAT solvers [31] such tht one solver dels with the existentilly quntified vriles nd one solver dels with the universlly quntified vriles exclusively. Solver A gets instntitions of φ in which the universl vriles re ssigned, nd solver B gets instntitions of φ in which the existentil vriles re ssigned. The stisfying ssignment found y one solver is used to otin new instntition for the other. This loop is repeted until one solver returns unstisfile. This pproch relises nturl ppliction of the counter-exmple guided strction refinement (CEGAR) prdigm [15]. A detiled survey on 2QBF solving is given in [3]. A significnt dvncement of expnsion-sed solving for QBF with n ritrry numer of quntifier lterntions ws mde with the solver [26], [27], which recursively pplies the previously discussed 2QBF pproch [31] for ech quntifier lterntion. The pproch turned out to e highly competitive in competitions. 1 For formlising this solving pproch the clculus Exp+Res ws introduced [6], nd proof-theoreticl investigtions reveled the orthogonl strength of Exp+Res nd Q-resolution [33], the QBF vrint of the resolution clculus tht forms the sis for QCDCLsed solvers. Reserch on the proof complexity of QBF hs identified n exponentil seprtion etween Q-resolution nd the Exp+Res system. There re fmilies of QBFs for which ny Q-resolution proof hs exponentil size, in contrst to Exp+Res proofs of polynomil size, nd vice vers. Hence these two systems hve orthogonl strength. Recent work successfully comines mchine lerning with this CEGAR pproch [25]. Motivted y the success of expnsion-sed QBF solving, severl other pproches [1], [3], [39], [42] [44] hve een presented tht re sed on levelised SAT solving, i.e., one SAT solver is responsile for the vriles of one quntifier lock. In this pper, we lso introduce solving pproch tht is sed upon propositionl strction ut considers the whole quntifier prefix t once. III. PRELIMINARIES The QBFs considered in this pper re in prenex norml form Π.φ where Π is quntifier prefix Q 1 x 1 Q 2 x 2... Q n x n over the set of vriles X = {x 1,..., x n } with Q i {, } nd x i x j for i j. The propositionl formul φ contins only vriles from X. Unless stted otherwise, we do not mke ny ssumptions on the structure of φ. Sometimes Π.φ is in prenex conjunctive norml form (PCNF), i.e., Π is prefix s introduced efore nd φ is conjunction of cluses. A cluse is disjunction of literls, nd literl is vrile or the negtion of vrile. The prefix imposes the order < Π on the elements of X such tht x i < Π x j if i < j. By U Π (E Π ) we denote the set of universlly (existentilly) quntified vriles of the prefix Π. If cler from the context we omit 1 the suscript Π. We ssume the stndrd semntics of QBF. A QBF consisting of only the syntctic truth constnt ( ) is flse (true). A QBF xπ.φ is true if Π.φ[x ] nd Π.φ[x ] re oth true, where φ[x t] is the sustitution of x y t in φ. A QBF xπ.φ is true if Π.φ[x ] or Π.φ[x ] is true. Given set of vriles X, we cll function σ : X {,, ɛ} n ssignment for X. If there is n x X with σ(x) = ɛ then σ is prtil ssignment, otherwise σ is full ssignment of X. Informlly, σ(x) = ɛ mens tht σ does not ssign truth vlue to vrile x. A restriction σ Y : Y {,, ɛ} of ssignment σ : X {,, ɛ} to Y X is defined y σ Y (x) = σ(x) if x Y, otherwise σ Y (x) = ɛ. By Σ X we denote the set of ll full ssignments σ : X {, }. Let φ e propositionl formul over X. By σ(φ) we denote the ppliction of ssignment σ : X {,, ɛ} on φ, i.e., σ(φ) is the formul otined y replcing vriles x X y σ(x) if σ(x) {, } nd performing stndrd propositionl simplifictions. Let φ, ψ e propositionl formuls over the set of vriles X. If for every full ssignment σ Σ X, σ(φ) = σ(ψ) then φ nd ψ re equivlent. Let τ : X {,, ɛ} nd σ : Y {,, ɛ} e ssignments such tht for every x X Y, τ(x) = σ(x) if τ(x) ɛ nd σ(x) ɛ. Then the composite ssignment of σ nd τ is denoted y στ : X Y {,, ɛ} nd for every propositionl formul φ over X Y, it holds tht στ(φ) = τσ(φ) = σ(τ(φ)) = τ(σ(φ)). Furthermore, σσ = σ for ny ssignment σ. Exmple 1. Let σ : X {,, ɛ} e n ssignment over vriles {,, x, y} defined y σ() =, σ() = ɛ, σ(x) =, nd σ(y) = ɛ. The restriction τ = σ Y of σ to Y = {x, y} is given y τ() = ɛ, τ() = ɛ, τ(x) =, τ(y) = ɛ. For the propositionl formul φ = (x y) ( x y) ( y ), the ppliction of σ nd τ on φ gives us σ(φ) = y ( y ) nd τ(φ) = ( y) ( y ). IV. EXPANSION In the following, we introduce the nottion nd terminology used for descriing expnsion-sed QBF solving in generl, nd the lgorithm introduced in the next section in prticulr. We first define the notion of instntition tht is inspired y the xiom rule of the clculus Exp+Res [29]. Definition 1. Let Π.φ e QBF with prefix Π = Q 1 x 1... Q n x n over the set of vriles X = {x 1,..., x n } nd σ : Y {,, ɛ} with Y X n ssignment. If Y X, we extend the domin of σ to X y setting σ(x) = ɛ if x Y. The instntition of φ y σ, denoted y φ σ, is otined from φ s follows: 1) ll vriles x X with σ(x) ɛ re set to σ(x); 2) ll vriles x X with σ(x) = ɛ re replced y x ω where nnottion ω is uniquely defined y the sequence σ(x k1 )σ(x k2 )... σ(x km ) such tht the set formed from the vriles x ki contins ll vriles of X with x ki < Π x. Furthermore, x ki < Π x kj if k i < k j. If we instntite QBF Π.φ with the full ssignment σ : U Π {, } of the universl vriles, we otin propositionl formul tht contins only (possily nnotted)

3 3 vriles from E Π. The dul holds for the instntition y full ssignment σ : E Π {, }. Exmple 2. Given the QBF x y.φ with φ = ((x y) ( x y) ( y )). Then U = {, } nd E = {x, y}. Let σ : U {,, ɛ} e defined y σ() = nd σ() =. Then φ σ = ( x y ) y. Further, let τ : E {,, ɛ} with τ(x) = nd τ(y) =. Then φ τ =. Note tht is not nnotted ecuse it occurs in the first quntifier lock. Sometimes we wnt to remove the nnottions gin from n ssignment or n instntited formul. Therefore, we introduce the following nottion. Let φ σ e n instntition y ssignment σ : X {,, ɛ} nd X σ the set of nnotted vriles. If we hve n ssignment τ : X σ {,, ɛ}, then we define τ σ : X {, } y τ σ (x) = τ(x σ ) for x σ X σ. If we hve n instntited formul φ σ, the (φ σ ) σ is the formul otined y replcing every nnotted vrile x σ X σ y x. In generl, (φ σ ) σ φ. Lemm 1. Let Π.φ e QBF with vriles X nd σ : X {,, ɛ} e prtil ssignment. Then (φ σ ) σ nd σ(φ) re equivlent. Proof. By induction over the formul structure. For the se cse let φ = x with x X. If σ(x) = ɛ, σ(φ) = x nd φ σ = x. Then (φ σ ) σ = x. Otherwise, φ σ = σ(x). Oviously, σ(φ) = σ(x) = (σ(x)) σ {, }. The induction step nturlly follows from the semntics of the logicl connectives. Exmple 3. Reconsider the propositionl formul φ nd ssignments σ, τ from ove (Exmple 2). Then (φ σ ) σ = (( x y ) y ) σ = ( x y) y. Furthermore, (φ τ ) τ = () τ =. Finlly, we specify the semntics of QBF in terms of universl nd existentil expnsion on which expnsion-sed QBF solving is founded. Lemm 2. Let Φ = Π.φ e QBF with universl vriles U. Then Φ is flse if nd only if α Σ U φ α is unstisfile. The efficiency of expnsion-sed pproches comes from the following corollry tht sttes tht it is sufficient to consider only suset of Σ U for deciding the given QBF. Corollry 1. Let Φ = Π.φ e QBF with universl vriles U. There is set of ssignments A Σ U with α A φα is unstisfile if nd only if Φ is flse. The lemm nd the corollry ove lso hve dul version for true QBFs. This dulity will ply prominent role in our novel solving lgorithm. Lemm 3. Let Φ = Π.φ e QBF with existentil vriles E. Then Φ is true if nd only if σ Σ E φ σ is vlid. Corollry 2. Let Φ = Π.φ e QBF with existentil vriles E. There is set of ssignments S Σ E with σ S φσ is vlid if nd only if Φ is true. input : QBF Π.φ with universl vriles U nd existentil vriles E output: truth vlue of Π.φ 1 A := {α }, where α : U {, } is n ritrry ssignment 2 S := 3 i := 1 4 while true do 5 (isunst, τ) := SAT( α A i 1 φ α ) 6 if isunst then return flse; 7 S i := S i 1 {(τ E α) α α A i 1 } 8 (isunst, ρ) := SAT( σ S i φ σ ) 9 if isunst then return true; 1 A i := A i 1 {(ρ U σ) σ σ S i } 11 i++ 12 end Figure 1: Non-Recursive Expnsion-Bsed Algorithm V. A NON-RECURSIVE ALGORITHM FOR EXPANSION-BASED QBF SOLVING The pseudo-code in Figure 1 summrises the sic ide of our novel pproch for solving the QBF Π.φ with universl vriles U nd existentil vriles E. First, n ritrry ssignment α for the universl vriles is selected in Line 1. The instntition φ α is hnded over to SAT solver. If φ α is unstisfile, then Π.φ is flse nd the lgorithm returns. Otherwise, τ : E α {, } is stisfying ssignment of φ α. Let σ 1 denote the ssignment τ α. Then α σ 1 is stisfying ssignment of φ. Next, the propositionl formul φ σ1 is hnded over to SAT solver for checking the vlidity of φ σ1. If φ σ1 is unstisfile, then Π.φ is true nd the lgorithm returns. If φ σ1 is stisfile, then ρ: U σ1 {, } is stisfying ssignment of φ σ1. Let α 1 denote the ssignment ρ σ1. Then α 1 σ 1 is stisfying ssignment for φ. The following lemm shows tht α nd α 1 re different. Lemm 4. Let Π.φ e QBF with universl vriles U nd existentil vriles E. Further, let α: U {, } e n ssignment such tht the instntition φ α is stisfile nd hs the stisfying ssignment τ : E α {, }. Let σ : E {, } with σ = τ α. Then α flsifies ( φ σ ) σ. Proof. Since φ α is stisfied y τ, φ is stisfied y the composite ssignment ατ α = ασ, nd therefore φ is flsified y ασ. Then α flsifies σ( φ). According to Lemm 1 σ( φ) is equivlent to ( φ σ ) σ. Then α lso flsifies ( φ σ ) σ. In the next round of the lgorithm, the propositionl formul φ α φ α1 is hnded over to SAT solver. If this formul is unstisfile, Π.φ is flse nd the lgorithm returns. Otherwise, it is stisfile under some ssignment τ : E α E α1 {, }, then t lest one ssignment σ 2 : E {, } is otined. Note tht σ 2 σ 1. This ssignment is then used for otining

4 4 new propositionl formul φ σ1 φ σ2. To show the vlidity of this formul, its negtion is pssed to SAT solver. If this formul is unstisfile, Π.φ is true nd the lgorithm returns. Otherwise, it is stisfile under the ssignment ρ: U σ1 U σ2 {, }, from which α 2 : U {, } is otined. This ssignment is then used in the next round of the lgorithm. In this wy, the propositionl formuls α Σ U φ α nd σ Σ E φ σ re generted. If α A φα is unstisfile for some A Σ U, y Corollry 1 Π.φ is flse. Dully, if σ S φσ is vlid for some S Σ E, y Corollry 2 Π.φ is true. The lgorithm itertively extends the sets A nd S y dding prts of stisfying ssignments of φ to S nd prts of flsifying ssignments to A. In prticulr, A is extended y ssignments of the universl vriles nd S is extended y ssignments of the existentil vriles. The order in which ssignments re considered depends on the used SAT solver. Exmple 4. We show how to solve the QBF x y.φ with E = {x, y}, U = {, }, nd φ = (( x y) ( x y) ( y)) with the lgorithm presented ove. This formul cn e solved in two itertions: Init: We strt with some rndom ssignment α : U {, }, for exmple with α () = nd α () =. Itertion 1: The formul φ α = ( x y ) y is pssed to SAT solver nd found stisfile under the ssignment τ : E α {, } with τ(x ) = nd τ(y ) =. By removing the vrile nnottions we get ssignment σ 1 = (τ E α ) α, where σ 1 : E {, } with σ 1 (x) = nd σ 1 (y) =. Bsed on this ssignment we otin φ σ1 =. The formul φ σ1 is pssed to SAT solver. It is stisfile nd hs the stisfying ssignment ρ: U σ1 {, } with ρ() = nd ρ( ) =, which we then reduce to α 1 = (ρ U σ 1 ) σ1 where α 1 : U {, } with α 1 () = nd α 1 () =. Itertion 2: The formul φ α φ α1 = ( x y ) y (x y ) is pssed to SAT solver in the second itertion. It is stisfile nd one stisfying ssignment is τ : E α E α1 {, } with τ(x ) =, τ(x ) =, τ(y ) =, τ(y ) =. From τ, we cn extrct the ssignment σ 2 = (τ E α 1 ) α1 where σ 2 : E {, } with σ 2 (x) = nd σ 2 (y) =. Note tht σ 2 σ 1. Further note tht σ 2 σ 1 for ny stisfying ssignment different from τ. Next, we construct φ σ1 φ σ2 =. This formul is tutology, so its negtion tht is pssed to SAT solver is unstisfile, hence Π.φ is true. The soundness of our lgorithm immeditely follows from Corollries 1 nd 2: the lgorithm returns flse (true) if, in some itertion i, it finds tht the current prtil expnsion α A i 1 φ α (respectively σ S i φ σ ) is unstisfile. Theorem 1. The lgorithm shown in Figure 1 is sound. For showing tht the lgorithm lso termintes, we rgue tht sets A i nd S i increse in itertion i + 1. To this end, we hve to relte the vriles of the QBF, the nnotted vriles s well s their ssignments. Before we give the proof, we first consider nother exmple in which we illustrte how the different ssignments re relted. α 2σ 3 y y x x Models of φ y x α σ 1 σ 1(x) =, σ 1(y) = x α 1σ 2 α σ 1 σ 2(x) =, σ 2(y) = y x α 1σ 2 α σ 1 σ 3(x) =, σ 3(y) = y y x Itertion 1 y Itertion 2 y Itertion 3 y Itertion 4 y Counter-Models of φ y α 1σ 1 y α 2σ 2 α 2σ 1 y α 2σ 2 α 2σ 1 v v v x α 1() =, α 1() = x α 2() =, α 2() = x y α 3σ 3 α 3() =, α 3() = v set to v set to v unssigned Figure 2: Expnsion trees relting the ssignments found during solving the QBF x y.φ in Exmple 5, with initil ssignment α () =, α () =. The ssignments shown in the leves of the trees stisfy (left trees) or flsify (right trees) φ. Exmple 5. We show one possile run of the lgorithm presented ove for the QBF Φ := x y.φ with φ := ( x y) ( x (( y) ( y))) nd how it itertively genertes the sets Σ U nd Σ E. Figure 2 shows the expnsion trees tht re implicitly uilt during the serch. An expnsion tree reltes the vriles of the prtil expnsion of Φ constructed from A i (left column) nd S i (right column). Solid edges indicte tht the vrile on the top hs een set y n ssignment from A i or S i, nd dotted edges indicte tht the vrile hs to e ssigned vlue y the SAT solver. The order of the (nnotted) vriles in the expnsion tree respects the order of the (originl) vriles in the prefix.

5 5 Init: For the initilistion of A, n ritrry ssignment α : U {, } is chosen. Let α () = nd α () =. Itertion 1: The prtil expnsion φ α is found to e stisfile y the SAT solver, nd the ssignment σ 1 : E {, }, with σ 1 (x) = nd σ 1 (y) =, is extrcted from τ : E α1 {, } nd dded to S 1. Now the formul φ σ1 is checked for vlidity nd new ssignment α 1 : U {, }, with α 1 () = nd α 1 () =, is extrcted from the counterexmple ρ: U σ1 {, } nd dded to A 1. Itertion 2: The formul φ α φ α1 is checked for stisfiility. The SAT solver now returns n ssignment τ : E α E α1 {, } from which φ α gin gives σ 1, ut φ α1 forces new ssignment which is not in S 1 nd is dded to S 2. In prticulr, we get σ 2 : E {, } with σ 2 (x) = nd σ 2 (y) =. When the vlidity of φ σ1 φ σ2 is checked, we get counter-exmple ρ: U σ1 U σ2 {, }, from which the ssignment α 2 : U {, } cn e extrcted, with α 2 () = nd α 2 () =. This new ssignment is dded to A 2 nd results in new pth of the left expnsion tree of Itertion 3 in Figure 2. Itertion 3: We now check the stisfiility of φ α φ α1 φ α2, nd get n ssignment τ : E α E α1 E α2 {, } from which σ 3 : E {, } is extrcted, stisfying the suformul φ α2. This ssignment is different from oth σ 1 nd σ 2 y construction: σ 3 (x) = nd σ 3 (y) =. This gin results in new rnch of the expnsion tree, s shown y the left expnsion tree of Itertion 4 in Figure 2. The resulting formul φ σ1 φ σ2 φ σ3 is not vlid, nd from the counter-exmple ρ: U σ1 U σ2 U σ3 {, } we get new ssignment α 3 : U {, } with α 3 () = nd α 3 () =. Itertion 4: Lstly we check the full expnsion φ α φ α1 φ α2 φ α3 nd find tht it is not stisfile in this exmple, mening tht the originl formul x y.φ is flse. In the exmple ove we sw tht new ssignments re generted in ech itertion ecuse A i nd S i uild models nd counter-models of φ. The following definition formlises the reltionship etween A i nd S i. Definition 2. Let Π.φ e QBF over universlly quntified vriles U nd existentilly quntified vriles E. Further, let A {α α: U {, }} nd S {σ σ : E {, }}. If for every ssignment σ S, there exists n ssignment α A such tht ασ( φ) is true, then we sy tht A completes S. If for every ssignment α A, there exists n ssignment σ S such tht ασ(φ) is true, then we sy tht S completes A. We now show tht S i completes A i 1 nd A i completes S i if the lgorithm does not terminte in itertion i ecuse of the unstisfiility of the respective expnsion. Lemm 5. Let Π.φ e QBF over universlly quntified vriles U nd existentilly quntified vriles E. Further, let A i 1 nd A i with A i 1 A i e two sets of full ssignments to the universl vriles nd let S i e set of full ssignments to the existentil vriles otined y itertion i during n execution of the lgorithm shown in Figure 1. (1) If α A i 1 φ α is stisfile, then S i completes A i 1, i.e., for every µ A i 1, there is n ssignment ν S i such tht µν(φ) is true. (2) If σ S i φ σ is stisfile, then A i completes S i, i.e., for every ν S i, there is n ssignment µ A i such tht νµ( φ) is true. Proof. By contrdiction. For (1), ssume there is n ssignment µ A i 1 such tht there is no ssignment ν S i with µν(φ) is true. By ssumption α A i 1 φ α is stisfile, so there is stisfying ssignment τ with τ E µ(φ µ ) is true. Then lso µ(τ E µ) µ (φ) is true. But (τ E µ) µ S i. For (2), ssume tht there is n ssignment µ S i such tht there is no ν A i with µν( φ) is true. The rest of the rgument is similr s in (1). Next, we show tht the ddition of new ssignments A to set A of universl ssignments forces set S of existentil ssignments to increse if some completion criteri hold. Lemm 6. Let Φ = Π.φ e QBF over universlly quntified vriles U nd existentilly quntified vriles E. Further, let A A e set of universl ssignments such tht A A = nd A. Let S e set of existentil ssignments nd ssume tht σ S φσ hs the stisfying ssignment ρ, A {(ρ U σ) σ σ S}. If S completes A, nd A A completes S, nd α A A φα evlutes to true under ssignment τ, then there exists n ssignment ν {(τ E α) α α A A } with ν S. Proof. By induction over the numer of vriles in Π. Bse Cse. Assume tht Φ hs only one vrile, i.e., Π = Qx. Note tht A = 1 ecuse x is outermost in the prefix nd A is otined from su-ssignments of ρ. If Q =, then the elements of A re full ssignments of φ, nd S is either empty, or it contins the empty ssignment ω : {, }. Let A = {µ}. If S is empty, so is A (ecuse S hs to complete A). If τ is stisfying ssignment of φ µ, then ν = τ = ω is the empty ssignment nd ν S. Otherwise, ω S. If there is n ssignment α A, then φ α φ µ is full expnsion of Φ. If this full expnsion is true, then φ is unstisfile. Otherwise, φ α φ µ is unstisfile. In oth cses, the necessry preconditions for the lemm re not fulfilled. If A =, then µω( φ) is true. Then φ µ is unstisfile, gin violting precondition. If Q =, then µ = ω nd A =. If S = nd φ ω = φ hs the stisfying ssignment τ, then ν = τ nd ν S. Otherwise, if there is n ssignment σ S, then ωσ( φ) is true, ecuse A {µ} = {ω} completes S. Hence, if ssignment τ stisfies φ µ, then ν = τ, so ν S. Induction Step. Assume the lemm holds for QBFs with n vriles. We show tht it lso holds for QBFs with n + 1 vriles. Let Φ = QxΠ.φ e QBF over existentil vriles E nd universl vriles U with Π = Q 1 x 1... Q n x n nd A A nd S e s required (S completes A, A A completes S, α A A φα hs stisfying ssignment τ, nd σ S φσ hs stisfying ssignment ρ from which A is otined). If Q =, then ll ssignments α A ssign the sme vlue t to x, i.e., α(x) = t, ecuse these ssignments re extrcted from ssignment ρ nd since x is the outermost vrile of the prefix of Φ, ρ(x) = t. Further, let A t = {α A α(x) = t}. It is esy to rgue tht for Π.φ[x t] together with the ssignment sets A t A nd S

6 6 the induction hypothesis pplies, i.e., there is n ssignment ν S with ν {(τ E α) α α A t A } where τ is the prt of τ tht stisfies α A t A (φ[x t])α. Oviously, ν {(τ E α) α α A A }. If Q =, ssume tht τ(x) = t. Let {σ S σ(x) = t} S t S, nd let A t A such tht the induction hypothesis pplies to Π.φ[x t], A t A, nd S t. Let τ t e those su-ssignments of τ tht stisfy α A t φα. Then there is n ssignment ν tht cn e extrcted from τ t with ν S t. Since ν(x) = t, ν S. This concludes the proof. This property lso holds in the other direction, i.e., dding set S of new ssignments to S will force the set A to increse. Lemm 7. Let Φ = Π.φ e QBF over universlly quntified vriles U nd existentilly quntified vriles E. Further, let S S e set of existentil ssignments such tht S S =, S, let A e set of universl ssignments, α A φα hs the stisfying ssignment τ, S {(τ E α) α α A}. If A completes S nd S S completes A nd σ S S φσ evlutes to true under ssignment ρ, then there exists n ssignment ν {(ρ U σ) σ σ S S } with ν A. Proof. The proof is nlogous to the proof of Lemm 6. Now tht we hve identified the reltions etween the sets of universl nd existentil ssignments, we use them to show tht the lgorithm from Figure 1 termintes. Theorem 2. The lgorithm shown in Figure 1 termintes for ny QBF Φ = Π.φ. Proof. By induction over the numer of itertions i, we rgue tht sets A i 1 A i nd S i 1 S i. Bse Cse. Let i = 1 nd A = {α }. S S 1, ecuse S = nd σ 1 S 1 is stisfying ssignment of φ α (if φ α is unstisfile, the lgorithm termintes). A A 1 directly follows from Lemm 4. Induction Step. For i > 1, we rgue tht S i S i+1. By induction hypothesis the theorem holds for itertion i, i.e., A i = A i 1 A with A i 1 A = nd A nd S i = S i 1 S with S i 1 S = nd S. Becuse of Lemm 5, S i completes A i 1, nd A i completes S i. Furthermore, if σ S i φ σ is stisfile under some ssignment ρ (otherwise the lgorithm would terminte), y construction A {(ρ U σ) σ σ S i }. Hence, Lemm 6 pplies nd if α A i φ α is stisfile under some ssignment τ (otherwise the lgorithm would immeditely terminte), then there is n ssignment ν {(τ E α) α α A i } with ν S i. The rgument for A i A i+1 is similr nd uses the property shown in Lemm 7. Note tht the lgorithm presented ove does not mke ny ssumptions on the formul structure, i.e., for QBF Π.φ it is not required tht φ is in conjunctive norml form. Without ny modifiction, our lgorithm lso works on formuls in PCNF s SAT solvers typiclly process formuls in CNF only, we focus on this representtion for the rest of the pper. We conclude this section y rguing tht the Exp+Res [6], [28], [29] clculus yields the theoreticl foundtion of our lgorithm for refuting formul Π.φ in PCNF with universl vriles U. The Exp+Res clculus consists of two rules, the xiom rule C α where C is cluse of φ nd α: U {, } is universl ssignment s well s the resolution rule C 1 x ω C 2 x ω C 1 C 2 A derivtion in Exp+Res is sequence of cluses where ech cluse is either otined y the xiom or derived from previous cluses y the ppliction of the resolution rule. A refuttion of PCNF Π.φ is derivtion of the empty cluse. The ppliction of the xiom instntites the universl vriles of one cluse of φ. If enough of these instntitions cn e found in order to derive the empty cluse y the ppliction of the resolution rule, the QBF Π.φ is flse. Our lgorithm in Figure 1 does not instntite individul cluses, ut ll cluses of φ t once with prticulr ssignment of the universl vriles. Hence, when the SAT solver finds ψ = α A i φ α unstisfile for some A i, not necessrily ll cluses of ψ re required to derive the empty cluse vi resolution, ut only the miniml unstisfile core of ψ, i.e., suset of the cluses such tht the removl of ny cluse would mke this formul stisfile. Proposition 1. Let Π.φ e flse QBF. Further, let ψ = α A i φ α e otined y the ppliction of the lgorithm in Figure 1. Further, let ψ e n unstisfile core of ψ. Then there is Exp+Res refuttion such tht ll cluses tht re introduced y the xiom rule occur in ψ. VI. IMPLEMENTATION The lgorithm descried in Section V is relised in the solver Ijtihd 2 The most recent version of Ijtihd is ville t The solver is implemented in C++ nd currently processes formuls in PCNF ville in the QDIMACS formt. For ccessing SAT solvers, Ijtihd uses the IPASIR interfce [4], which mkes chnging the SAT solver very esy. The SAT solver used in ll of our experiments is Glucose [1]. The implementtion relises vrious optimiztions to mke Ijtihd vile in prctice. Some of them re discussed in the following. For solving QBF Π.φ, the sic lgorithm shown in Figure 1 dds instntitions of φ to ψ = α A i 1 φ α nd ψ = σ S i φ σ in ech itertion i until the formul is decided. The clls to the SAT solver in Line 5 nd Line 8 re done incrementlly, i.e., we crete two instnces of the SAT solver nd provide them with the cluses stemming from new instntitions of φ t ech itertion. For simplicity, we omit indices of sets A nd S nd refer to n ritrry itertion of the execution of the lgorithm in the following discussion. Figure 5 reltes set sizes of A nd S s well s the ccumulted time tht one SAT solver needs to solve ψ 2 The nme Ijtihd refers to the effort of solving cses in Islmic lw (for detils see

7 7 Size of universl expnsion set A Universl expnsion ψ solver time flse true Size of existentil expnsion set S Figure 3: Sizes of sets S nd A flse true Existentil expnsion ψ solver time Figure 4: Time used for solving ψ nd ψ Figure 5: Set sizes nd time consumed during SAT clls for solved instnces from QBFEVAL 17 preprocessed y Bloqqer with the time the other SAT solver needs to solve ψ for the formuls of the PCNF trck of QBFEVAL 17 (preprocessed with Bloqqer [7]). We lso distinguish etween true nd flse formuls. In Figure 3 we see tht for true formuls, set S tends to e lrger thn A, while for flse instnces the picture is less cler. Figure 4 shows the overll time needed for solving ψ (y-xis) nd ψ (x-xis). In lmost ll cses, the solver tht hndles ψ needs more time thn the solver tht hndles ψ. This my e founded on the oservtion tht mny QBFs hve considerly more existentil vriles thn universl vriles [37], hence the instntitions dded to ψ re much lrger thn the instntitions dded to ψ. In Line 1 of Fig. 1, the set of universl ssignments A is initilised with n ritrry ssignment α. In prctice, it turned out to e fvorle to initilize A with severl different ssignments t once. In the current implementtion, we initilize A with ssignments hving the vriles of one quntifier lock set to nd the vriles of ll other quntifier locks set to, respectively. This initiliztion heuristics seems to e very eneficil for some instnces. In Line 7 nd Line 1 our lgorithm increses the size of S nd A in ech itertion of the min loop, s rgued in Theorem 2. In the worst cse, this leds to n exponentil increse in spce consumption. However, some of the ssignments found in n erlier itertion could ecome osolete fter etter ssignments were found. It is therefore eneficil to empty either S or A nd then reconstruct them from ψ nd ψ respectively. We evluted severl heuristics for scheduling these set resets, nd we found tht periodic memory limit wre restrts work est. The regulr reset of one set hs similr effect s restrts in SAT solvers nd we oserved considerle improvement in performnce, nd in prticulr in terms of memory consumption. Our implementtion periodiclly resets the set A, since experiments indicte tht the resulting formul ψ is much hrder to solve thn ψ s seen in Figure 4. Besides the forementioned imlnce etween universl nd existentil vriles, it is lso likely due to the structure of ψ which is conjunction of formuls in disjunctive norml form. Note tht this reset of A does not ffect the termintion rgument presented in Theorem 2, since the sets A nd S still complete ech other. Finlly, we extended the presented pproch with orthogonl resoning techniques like QCDCL [21] for exploiting the different strengths of Exp+Res nd Q-resolution, yielding hyrid solver tht smoothly integrtes oth solving prdigms. To this end, we implemented the prototypicl solver clled which pursues the following ide: The min loop of the lgorithm shown in Figure 1 (Lines 4-12) is extended in sequentil portfolio style such tht QCDCL solver is periodiclly clled. After ech cll, ll cluses tht were lerned through QCDCL re dded to Π.Φ, mking them ville in further itertions. These new cluses potentilly exclude ssignments tht would otherwise e possile nd tht could result in more itertions of the min loop. The solver extends Ijtihd y dditionl invoctions of the QCDCL solver [36]. Aout every 3 seconds, is clled nd run for out 3 seconds. The lernt cluses re otined vi the API of. Leverging lerned cues is suject to future work. VII. EVALUATION We evlute non-recursive expnsion s implemented in our solvers Ijtihd nd its hyrid vrint on the enchmrks from the PCNF trck of the QBFEVAL 17 competition. All experiments were crried out on cluster of Intel Xeon CPUs (E5-265v4, 2.2 GHz) running Uuntu with CPU time limit of 1 seconds nd memory limit of 7 GB. We considered the following top-performing solvers from QBFEVAL 17: 3 [38], Rev-Qfun [25], [26], [39], [43], [12], [26], [34], [36], [3], nd [9], [1]. In the competition, mny of these solvers were coupled with the preprocessor Bloqqer [7], [23] s module pplied efore solving. 4 In our experiments, we rn the plin solvers without Bloqqer. However, to tke preprocessing into ccount, we first preprocessed the enchmrks using Bloqqer with timeout of two hours. For the empiricl evlution we used two enchmrk sets: originl enchmrks without preprocessing nd preprocessed ones. We included 76 formuls tht were solved lredy y Bloqqer in the preprocessed enchmrk set. 3 We excluded AIGSolve due to ssertion filures on certin instnces. 4 We refer to the ppendix for dditionl experiments with the preprocessor HQSpre [46].

8 8 Rev-Qfun K K K K K K K Ijtihd K K K K TABLE I: Originl instnces K K K Rev-Qfun K Ijtihd K K K K K K K TABLE II: Preprocessing y Bloqqer K K K K K Ijtihd K Rev-Qfun K K K K K TABLE III: 197 originl instnces with four or more quntifier locks. Tles I nd II show the totl numers of solved instnces (S), solved unstisfile ( ) nd stisfile ones ( ), nd totl CPU time including timeouts. 5 We focus on comprison of our solvers Ijtihd nd with. Like our solvers, relies on expnsion, however, it is sed on recursive implementtion. Rev-Qfun extends y mchine lerning techniques. In generl, preprocessing hs considerle impct on the numer of solved instnces. The difference in solved instnces etween Ijtihd nd is 17 on originl instnces (Tle I), nd ecomes lrger on preprocessed instnces (Tle II). Notly, despite its simple design, significntly outperforms Ijtihd on the two enchmrk sets. Moreover, is rnked third on preprocessed instnces (Tle II) nd thus cn e considered on pr with stte-of-the-rt solvers. On the two enchmrk sets, the gp in solved instnces etween nd is considerly smller thn the one etween nd Ijtihd. The strength of is emphsized y more detiled nlysis w.r.t. formuls tht hve four or more quntifier locks (i.e., three or more quntifier lterntions). As shown in Tle III, outperforms ll other solvers on respective originl instnces. We mde similr oservtion on preprocessed instnces. 6 Moreover, solves only four originl instnces less thn (Tle I), nd outperforms on preprocessed instnces (Tle II). These results indicte the potentil of comining the orthogonl proof systems Exp+Res s implemented in Ijtihd nd Q-resolution s implemented in in hyrid solver such s. R vs. I R vs. H I vs. H < = > < = > < = > No preprocessing Bloqqer Although outperforms oth Ijtihd nd on the two given enchmrk sets (Tles I nd II), filed to solve certin instnces tht were solved y Ijtihd nd. Tle IV shows relted sttistics. E.g., on preprocessed instnces (row Bloqqer ), 218 instnces were solved y oth nd (column R vs. H), 38 only y, nd 27 only y. Summing up these numers yields totl of 283 solved instnces (more thn ny individul solver on preprocessed instnces in Tle II) tht could hve een solved y hypotheticl solver comining nd. This oservtion underlines the strength of expnsion in generl nd, in prticulr, of the hyrid pproch implemented in. solved significnt mount of instnces not solved y nd clerly outperformed Ijtihd (column I vs. H ). VIII. CONCLUSION We presented novel non-recursive lgorithm for expnsion-sed QBF solving tht uses only two SAT solvers for incrementlly refining the propositionl strction nd the negted propositionl strction of QBF. We gve concise proof of termintion nd soundness nd demonstrted with severl experiments tht our prototype is competitive with the stte of the rt. In ddition to non-recursive expnsion, we lso studied the impct of effectively comining Q-resolution nd Exp+Res in hyrid pproch. To this end, we coupled QCDCL solver nd non-recursive expnsion to mke cluses derived y the QCDCL solver ville to the expnsion solver. Experimentl results indicted tht the hyrid pproch significntly outperforms our implementtion of non-recursive expnsion indicting the potentil of comining expnsionsed pproches with Q-resolution which gives rise to n exciting direction of future work. TABLE IV: Pirwise comprison of (R), Ijtihd (I), nd (H) y numers of instnces from QBFEVAL 17 with nd without preprocessing tht were solved y only one solver of the considered pir (<, >) or y oth (=). 5 Relted cctus plots of the runtimes sorted scendingly re shown in Figures 12 nd 13 in the ppendix. 6 Tle VIII nd Figures 11 nd 1 in the ppendix.

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10 1 APPENDIX A ADDITIONAL EXPERIMENTAL DATA Pir-wise comprison with Figure 6: Sctter plot of run times of nd on instnces preprocessed y Bloqqer (relted to Tle II) Figure 7: Sctter plot of run times of nd on originl instnces (relted to Tle I). Preprocessing with HQSpre R vs. I R vs. H I vs. H < = > < = > < = > No preprocessing Bloqqer HQSpre TABLE VII: Pirwise comprison of (R), Ijtihd (I), nd (H) y numers of instnces from QBFEVAL 17 with nd without preprocessing tht were solved y only one solver of the considered pir (<, >) or y oth (=) K K K Rev-Qfun K K K K K Ijtihd K K K TABLE V: Preprocessing y HQSpre. Ijtihd Figure 8: Plot relted to Tle V K K K K K K Rev-Qfun K K Ijtihd K K K TABLE VI: Preprocessing y HQSpre: 97 instnces with four or more quntifier locks. Ijtihd Figure 9: Plot relted to Tle VI.

11 11 Preprocessing with Bloqqer K K K K Ijtihd K Rev-Qfun K K K K K K TABLE VIII: 154 preprocessed instnces with four or more quntifier locks. Ijtihd Figure 1: Plot relted to Tle VIII. Additionl plots Ijtihd Figure 11: Plot relted to Tle III. Ijtihd Figure 12: Plot relted to Tle I. Ijtihd Figure 13: Plot relted to Tle II.