A Lower Bound for the Pigeonhole Principle in Tree-like Resolution by Asymmetric Prover-Delayer Games

Transcription

1 A Lower Boud for the Pigeohole Priciple i Tree-like Resolutio by Asymmetric Prover-Delayer Games Olaf Beyersdorff Nicola Galesi 2 Massimo Lauria 2 Istitut für Theoretische Iformatik, Leibiz Uiversität Haover, Germay 2 Dipartimeto di Iformatica, Sapieza Uiversità di Roma, Italy Abstract I this ote we show that the asymmetric Prover-Delayer game developed i [3] for Parameterized Resolutio is also applicable to other tree-like proof systems. I particular, Ω( log ) we use this asymmetric Prover-Delayer game to show a lower boud of the form 2 for the pigeohole priciple i tree-like Resolutio. This gives a ew ad simpler proof of the same lower boud established by Datchev ad Riis [5]. Itroductio Provig lower bouds by games is a very fruitful techique i proof complexity [, 8 0]. I particular, the Prover-Delayer game of Pudlák ad Impagliazzo [0] is oe of the caoical tools to study lower bouds i tree-like Resolutio [2,0] ad tree-like Res(k) [6]. The Prover-Delayer game of Pudlák ad Impagliazzo arises from the well-kow fact [7] that a tree-like Resolutio proof for a formula F ca be viewed as a decisio tree which solves the search problem of fidig a clause of F falsified by a give assigmet. I the game, Prover queries a variable ad Delayer either gives it a value or leaves the decisio to Prover ad receives oe poit. The umber of Delayer s poits at the ed of the game is the proportioal to the height of the proof tree. It is easy to argue that showig lower bouds by this game oly works if (the graph of) every tree-like Resolutio refutatio cotais a balaced sub-tree as a mior, ad the height of that sub-tree the gives the size lower boud. I [3] we developed a ew asymmetric Prover-Delayer game which exteds the game of Pudlák ad Impagliazzo to make it applicable to obtai lower bouds to tree-like proofs whe the proof trees are very ubalaced. I [3] we used the ew asymmetric game to obtai lower bouds i tree-like Parameterized Resolutio, a proof system i the cotext of parameterized proof complexity recetly itroduced by Datchev, Marti, ad Szeider [4]. The lower bouds we obtai i [3] for tree-like Parameterized Resolutio are of the form Ω( k ) ( is the formula size ad k the parameter), but the tree-like Parameterized Resolutio refutatios of the formulas i questio oly cotai balaced sub-trees of height k. The aim of this ote is to show that the asymmetric Prover-Delayer game is also applicable to other (o-parameterized) tree-like proof systems. Oe of the best studied priciples is the pigeohole priciple. Datchev ad Riis [5] show that the pigeohole priciple requires treelike Resolutio refutatios of size roughly! while its tree-like Resolutio proofs oly cotai This work was doe while the first author was visitig Sapieza Uiversity of Rome uder support of DFG grat KO 053/5 2. Supported by grat Limiti di compressioe i combiatoria e complessità computazioale by Sapieza Uiversity Rome.

2 balaced sub-trees of height. Therefore the game of Pudlák ad Impagliazzo oly yields a 2 Ω() lower boud which is weaker tha the optimal boud 2 Ω( log ) established by Datchev ad Riis. Here we provide a ew ad easier proof of this lower boud by our asymmetric Prover-Delayer game. 2 Prelimiaries A literal is a positive or egated propositioal variable ad a clause is a set of literals. A clause is iterpreted as the disjuctios of its literals ad a set of clauses as the cojuctio of the clauses. Hece clause sets correspod to formulas i CNF. The Resolutio system is a refutatio system for the set of all usatisfiable CNF. Resolutio uses as its oly rule the Resolutio rule {x} C { x} D C D for clauses C, D ad a variable x. The aim i Resolutio is to demostrate usatisfiability of a clause set by derivig the empty clause. If i a derivatio every derived clause is used at most oce as a prerequisite of the Resolutio rule, the the derivatio is called tree-like, otherwise it is dag-like. The size of a Resolutio proof is the umber of its clauses. Udoubtedly, Resolutio is the most studied ad best-uderstood propositioal proof system (cf. []). It is well kow (cf. [7]) that a tree-like refutatio of F ca equivaletly be described as a boolea decisio tree. A boolea decisio tree for F is a biary tree where ier odes are labeled with variables from F ad leafs are labeled with clauses from F. Each path i the tree correspods to a partial assigmet where a variable x gets value 0 or accordig to whether the path braches left or right at the ode labeled with x. The coditio o the decisio tree is that each path α must lead to a clause which is falsified by the assigmet correspodig to α. Therefore, a boolea decisio tree solves the search problem for F which, give a assigmet α, asks for a clause from F falsified by α. It is easy to verify that each tree-like Resolutio refutatio of F yields a boolea decisio tree for F ad vice versa, where the size of the Resolutio proof equals the umber of odes i the decisio tree. I the sequel, we will therefore cocetrate o boolea decisio trees to prove our lower boud to tree-like Resolutio. 3 Tree-like Lower Bouds via Asymmetric Prover-Delayer Games We review the asymmetric Prover-Delayer game from [3]. Let F be a set of clauses i variables x,..., x. I the asymmetric game, Prover ad Delayer build a (partial) assigmet to x,..., x. The game is over as soo as the partial assigmet falsifies a clause from F. The game proceeds i rouds. I each roud, Prover suggests a variable x i, ad Delayer either chooses a value 0 or for x i or leaves the choice to the Prover. I this last case, if the Prover sets the value, the the Delayer gets some poits. The umber of poits Delayer ears depeds o the variable x i, the assigmet α costructed so far i the game, ad two fuctios c 0 (x i, α) ad c (x i, α). More precisely, the umber of poits that Delayer will get is 0 if Delayer chooses the value, log c 0 (x i, α) if Prover sets x i to 0, ad log c (x i, α) if Prover sets x i to. Moreover, the fuctios c 0 (x, α) ad c (x, α) are chose i such a way that for each variable x ad assigmet α c 0 (x, α) + c (x, α) = () 2

3 holds. Let us call this game the (c 0, c )-game o F. The coectio of this game to size of proofs i tree-like Resolutio is give by Theorem. The theorem is essetially cotaied i [3], but for completeess we iclude the full proof. Theorem ( [3]). Let F be usatisfiable formula i CNF ad let c 0 ad c be two fuctios satisfyig () for all partial assigmets α to the variables of F. If F has a tree-like Resolutio refutatio of size at most S, the the Delayer gets at most log S poits i each (c 0, c )-game played o F. Proof. Let F be a usatisfiable CNF i variables x,..., x ad let Π be a tree-like Resolutio refutatio of F. Assume ow that Prover ad Delayer play a game o F where they successively costruct a assigmet α. Let α i be the partial assigmet costructed after i rouds of the game, i. e., α i assigs i variables a value 0 or. By p i we deote the umber of poits that Delayer has eared after i rouds, ad by Π αi we deote the sub-tree of the decisio tree of Π which has as its root the ode reached i Π alog the path specified by α i. We use iductio o the umber of rouds i the game to prove the followig claim: Π αi Π 2 p i for ay roud i. To see that the theorem follows from this claim, let α be a assigmet costructed durig the game yieldig p α poits to the Delayer. As a cotradictio has bee reached i the game, the size of Π α is, ad therefore by the iductive claim Π 2 pα, yieldig p α log Π as desired. I the begiig of the game, Π α0 is the full tree ad the Delayer has 0 poits. Therefore the claim holds. For the iductive step, assume that the claim holds after i rouds ad Prover asks for a value of the variable x i roud i+. If the Delayer chooses the value, the p i+ = p i ad hece Π αi+ Π αi Π 2 p i = Π 2 p i+ If the Delayer defers the choice to the Prover, the the Prover uses the followig strategy to set the value of x. Let αi x=0 be the assigmet extedig α i by settig x to 0, ad let αi x= be the assigmet extedig α i by settig x to. Now, Prover sets x = 0 if Π α x=0 i c 0 (x,α i ) Π α i, otherwise he sets x =. Because c 0 (x,α i ) + c (x,α i ) =, we kow that if Prover sets x =, the Π α x= i c (x,α i ) Π α i. Thus, if Prover s choice is x = j with j {0, }, the we get. Π αi+ = Π α x=j i Π α i c j (x, α i ) Π c j (x, α i )2 p = Π i 2 p i+log c j (x,α i ) = Π 2 p i+. This completes the proof of the iductio. As remarked i [3] we get the game of Pudlák ad Impagliazzo [0] by settig c 0 (x, α) = c (x, α) = 2 for all variables x ad partial assigmets α. 3

4 4 Tree-like Resolutio Lower Bouds for the Pigeohole Priciple The weak pigeohole priciple PHP m with m > uses variables x i,j with i [m] ad j [], idicatig that pigeo i goes ito hole j. PHP m cosists of the clauses for all pigeos i [m] j [] x i,j ad x i,j x i2,j for all choices of distict pigeos i, i 2 [m] ad holes j []. We prove that PHP m is hard for tree-like Resolutio. Showig the lower boud by the asymmetric game from the last sectio, requires a suitable choice of the fuctios c 0 ad c ad the the defiitio of the Delayer-strategy for the (c 0, c )-game. Theorem 2. Ay tree-like Resolutio refutatio of PHP m has size 2 Ω( log ). Proof. Let α be a partial assigmet to the variables { x i,j i [m], j [] }. Let p i (α) = { j [] α(x i,j ) = 0 ad α(x i,j) = for all i [m] }. Ituitively, p i (α) correspods to the umber of holes which are still free but are explicitely excluded for pigeo i by α (we do ot cout the holes which are excluded because some other pigeo is sittig there). We defie c 0 (x i,j, α) = 2 + p i(α) 2 p i(α) ad c (x i,j, α) = 2 + p i(α). For simplicity we assume that is divisible by 2. Durig the game it will ever be the case that Prover gets the choice whe p i (α) 2. Therefore the fuctios c 0 ad c are always greater tha zero whe the Delayer gets poits, thus the score fuctio is always well defied. Furthermore otice that this defiitio satisfies (). Let us try to provide a ituitive explaatio of why we choose the fuctios c 0 ad c ad thus the poits for Delayer as above. As a first observatio, Delayer always ears more whe Prover is settig a variable x i,j to istead of settig it to 0. This is ituitively correct as the amout of freedom for Delayer to cotiue the game is by far more dimiished by sedig pigeo i to some hole j tha by just excludig that hole j for pigeo i. Thus, whe Prover sets x i,j to, Delayer should ear a umber of poits approximately equal to the quatity of the iformatio that is ot aymore available for that pigeo i after the choice of Prover. How much is this iformatio? We take as measure of iformatio (approximately hece the 2 ) the umber of possible holes where Prover ca choose to sed pigeo i ad hece are still free for mappig that pigeo. We ow describe Delayer s strategy i a (c 0, c )-game played o PHP m. If Prover asks for a value of x i,j, the Delayer decides as follows: set α(x i,j ) = 0 set α(x i,j ) = let Prover decide if there exists i [m] {i} such that α(x i,j) = or if there exists j [] {j} such that α(x i,j ) = ; if p i (α) 2 ad there is o i [m] with α(x i,j) =, ad otherwise. Ituitively, Delayer leaves the choice to Prover as log as pigeo i does ot already sit i a hole, hole j is still free, ad there are at most 2 excluded free holes for pigeo i. 4

5 If Delayer uses this strategy, the the small clauses x i,j x i2,j from PHP m will ot be violated i the game. Therefore, a cotradictio will always be reached o oe of the big clauses j [] x i,j. Let us assume ow that the game eds by violatig j [] x i,j, i. e., for pigeo i all variables x i,j with j [] have bee set to 0. As soo as the umber p i (α) of excluded free holes for pigeo i reaches the threshold 2, Delayer will ot leave the choice to Prover. Istead, Delayer will try to place pigeo i ito some hole. If Delayer still aswers 0 to x i,j eve after p i (α) > 2, it must be the case that some other pigeo already sits i hole j, i. e., for some i = i, α(x i,j) =. Therefore, at the ed of the game at least 2 variables have bee set to. W. l. o. g. we assume that these are the variables x i,ji for i =,..., 2. Let us check how may poits Delayer ears i this game. We calculate the poits separately for each pigeo i =,..., 2 ad distiguish two cases: whether x i,j i was set to by Delayer or Prover. Let us first assume that Delayer sets the variable x i,ji to. The pigeo i was ot assiged to a hole yet ad, moreover, there must be 2 uoccupied holes which are already excluded for pigeo i by α, i. e., there is some J [] with J = 2, α(x i,j ) = for i [m], j J, ad α(x i,j ) = 0 for all j J. All of these 0 s have bee assiged by Prover, as Delayer has oly assiged a 0 to x i,j whe some other pigeo was already sittig i hole j, ad this is ot the case for the holes from J (at the momet whe Delayer assigs the to x i,ji ). Thus, before Delayer sets α(x i,ji ) =, she has already eared poits for all 2 variables x i,j, j J, yieldig 2 2 log + p 2 2 p=0 2 p = log + p ( ) p=0 2 p = log 2 + poits for the Delayer. Let us ote that because Delayer ever allows a pigeo to go ito more tha oe hole, she will really get the umber of poits calculated above for every of the variables which she set to. If, coversely, Prover sets variable x i,ji to, the Delayer gets log( 2 + p i(α)) poits for this, but she also received poits for the p i (α) variables set to 0 before by Prover. Thus, i this case Delayer ears o pigeo i log( p i (α) 2 + p i(α)) + log p=0 = log( 2 + p i(α)) + log ( ) = log p 2 p p i(α) + poits. I total, Delayer gets at least ( ) 2 log 2 + poits i the game. Applyig Theorem, we obtai 2 2 log( 2 +) as a lower boud to the size of each tree-like Resolutio refutatio of PHP m. By ispectio of the above Delayer strategy it becomes clear that the lower boud from Theorem 2 also holds for the fuctioal pigeohole priciple where i additio to the clauses from PHP m we also iclude x i,j x i,j2 for all pigeos i [m] ad distict holes j, j 2 []. 5

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