Chapter 6* The PPSZ Algorithm

Transcription

1 Chapter 6* The PPSZ Algorithm In this chapter we present an improvement of the pp algorithm. As pp is called after its inventors, Paturi, Pudlak, and Zane [PPZ99], the improved version, pps, is called after Paturi, Pudlak, Saks, and Zane [PPSZ5]. The idea behind the improvement is ver beautiful and simple; its analsis is challenging, but ver elegant uses a variet of interesting concepts. In this chapter, we anale the algorithm for the case that the input formula has a unique satisfing assignment. This analsis is from [PPSZ5]. If there are multiple satisfing assignments, the analsis surprisingl becomes much more dicult. In fact, it has been open for thirteen ears whether one generalie the analsis for the unique case to the general case { without getting a worse running time. This was achieved in 211 b Timon Hertli [Her11]. The basic idea of the pp-algorithm is ver simple: Given a CNF formula F, choose a variable 2 vbl(f) uniforml at random. If {} 2 F or {} 2 F, set the obvious wa. Otherwise, set randoml. Then iterate. The analsis of this simple algorithm shows surprising connections to encodings of satisfing assignments and lower bounds on depth-3 boolean circuits. From an algorithmic point of view, here is a simple idea for improvement: Instead of checking whether {} 2 F or {} 2 F, check whether or can be derived from F in some wa. Of course, checking this is again NPcomplete and no big help. However, we can check whether or follows from F in some simple, polnomial-time checkable wa. Indeed, {} being a unit clause in F is probabl the simplest wa how can follow from F. Definition 6*.1 Let F be a CNF formula, let D 2 N, and let u be a literal. We sa that F implies u if an assignment α satisfing F also satises 1

2 2 CHAPTER 6*. THE PPSZ ALGORITHM u. We sa F D-implies u if there is some G F with G D that implies u. In this terminolog, in the pp-algorithm we check whether is 1- implied in F. The pps-algorithm applies an obvious improvement: Check whether is D-implied for some (large) constant D. This can be done in polnomial time. Simple as this sounds, the analsis of this algorithm turns out to be quite sophisticated. To state the pps-algorithm formall, we think of it choosing a random permutation π of vbl(f) and an assignment β 2 {, 1} n in advance and then processing the variables according to π and assigning them values according to β, unless the turn out to be D-implied. F a 3-CNF formula, π = ( 1,..., n ), a permutation on vbl(f), β 2 {, 1} n, D 2 N. Postcondition: returns a satisfing assignment or failure function pps(f, π, β, D) α the empt assignment ; F 1 F; for i 1 to vbl(f) do if F i D-implies i then α() 1; elseif F i D-implies i then α() ; else α() β(); F i+1 F [7 α()] i ; if α satises F then return α; else return failure; When does this algorithm succeed in nding a satisfing assignment? For this need the concept of forced and guessed variables. Definition 6*.2 Let F be a CNF formula over n variables, α 2 {, 1} n a satisfing assignment, and π = ( 1,..., n ) a permutation of vbl(f), and D. A variable i is called forced (with respect to F, α, π, and D) if F [ 1 7 α( 1),..., i 1 7 α( i 1 )] D-implies i or i. Otherwise the variable is called guessed. We denote the set of forced (guessed) variables b Forced(F, α, π, D) (Guessed(F, α, π, D)). When F, α, π and D are understood from the contet, we drop these parameters and just write Forced and Guessed.

3 6*.1. HAVING A UNIQUE SATISFYING ASSIGNMENT 3 Observation 6*.3 Let F be a CNF formula and α be a satisfing assignment. Then pps(f, π, β, D) returns α if and onl if α() = β() for all 2 Guessed(F, α, π, D). 6*.1 Having a Unique Satisfing Assignment In this section we anale the special case that F has a unique satisfing assignment α. This case is easier to anale and at the same time we can introduce all the concepts we will need for the general case. Without loss of generalit, assume α = (1,..., 1) sets all variables to 1. B the above observation, pps returns α if and onl if β agrees with α on all variables from Guessed(F, α, π, D). This set is a random object, since it depends on π, which is a random permutation. Pr β,π [pps returns α] = E π h 2 Guessed(F,α,π,D) This epression is rather dicult to work with. It would be easier to bound E [ Guessed(F, α, π, D) ], for eample. Luckil, we have Jensen's inequalit, which in this case states that for a random variable X, it holds that E h 2 Xi 2 E[X], (6*.1) since the function 7 2 is a conve function. With this, we obtain Pr [pps returns α] β,π 2 Eπ[ Guessed(F,α,π,D) ] (6*.2) The right-hand side here is a much nicer epression, since b linearit of epectation it equals the sum of the probabilities 6*.1.1 Building a Tree Pr[ 2 Guessed(F, α, π, D)]. (6*.3) We will now tr to bound Pr[ 2 Guessed(F, α, π, D)] from above. In this section, we conne ourselves to k-cnf formulas F having a unique satisfing assignment, which we assume to be α = (1,..., 1). Fi some 2 vbl(f). We construct a binar tree T, called the critical clause tree of that in some wa represents the multiple was can become D-implied in pps. i

4 4 CHAPTER 6*. THE PPSZ ALGORITHM Definition 6*.4 Let F be a ( k)-cnf formula. A clause tree T is a tree with the following properties: Ever node has at most k 1 children. Ever node u 2 V(T) is labeled with a variable from vbl(()f), which we denote b var-label(u), and additionall with a clause C 2 F, denoted b clause-label(u). If u is a proper ancestor of v, then var-label(u) 6= var-label(v). We will dene a clause tree T for each 2 vbl(f). See Figures 6*.1{ 6*.5 for an illustration. 1. Start with T consisting of a single root. This root has variable label, and does not have a clause label et. 2. As long as there is a leaf u 2 V(T) that does not et have a clause label, do the following: (a) Dene U := {var-label(v) v 2 V(T) is an ancestor of u in T}, where ancestor includes u itself and the root. (b) Dene the assignment β as β : vbl(f) {, 1}, 7 { if 2 U, 1 otherwise. (c) Let C 2 F be a clause not satised b β. Since β 6= α and α is the unique satisfing assignment, such a clause eists. Set clause-label(u) := C. (d) For each negative literal 2 C, create a new leaf, label it with, and attach it to u as a child. The new leaf does not et have a clause label. Note that ever clause in F has at most k 1 negative literals, and thus in step (d) we append at most k 1 children. Suppose v is an ancestor of u and var-label(v) =. Since β() =, we conclude that 62 C. Therefore, no node has the same label as one of its ancestors. This also implies that the height of the tree cannot eceed n, thus the process terminates. We denote the resulting clause tree b T.

5 6*.1. HAVING A UNIQUE SATISFYING ASSIGNMENT 5 Figure 6*.1: Consider the formula F = {{,, }, {, v, w}, {,, v}, {,, a},... }. We begin the construction of T with a tree consisting of a root. It is labeled with, but does not et have a clause label. We consider the assignment α[ 7 ] and observe that this leaves {,, } unsatised. We attach two new leaves to the root, labeled with and, respectivel. {, ȳ, } Figure 6*.2: We take the leaf labeled with. The assignment α[, 7 ] leaves the clause {,, a} unsatised. Thus, we attach one new child to and label it with a. {, ȳ, } {,, ā} a Figure 6*.3: We take the leaf labeled with. The assignment α[, 7 ] leaves the clause {, v, w} unsatised. Thus, we attach two new children to. {, ȳ, } {, v, w} {,, ā} v w a Figure 6*.4: We pick the leftmost leaf. The assignment α[,, v 7 ] leaves the clause {,, v} unsatised. We do not attach further leaves here. {, ȳ, } v {,, v} {, v, w} w {,, ā} a Figure 6*.5: The tree now has 6 nodes. Two leaves are still do not have a clause label, thus the process is not nished et.

6 6 CHAPTER 6*. THE PPSZ ALGORITHM π() =.5 {, v, w} π(v) =.9 v {,, v} π(w) =.2 w π() =.6 {, ȳ, } π() =.8 {,, ā} π(a) =.4 a Figure 6*.6: Illustration of Lemma 6*.5. Nodes u with π(u) < π() are shown gra. Two nodes are reachable, those labeled with and. Their clause labels ield the subformula {{,, }, {,, a}}. This formula implies the clause {,, a} (which can be seen b resolution, for eample). Therefore, ever assignment satisfing this formula and setting all Gra variables to 1 must set to 1. From now on, we do not think about π as a permutation of vbl(f), but rather as a function vbl(f) [, 1]. If the values π() are chosen independentl and uniforml at random from [, 1] for each 2 vbl(f), then with probabilit 1, π is injective, and b sorting the values π() we obtain a uniforml distributed permutation of vbl(f). Note that π also denes a function V(T) [, 1] via π(u) := π(var-label(u)). Let γ 2 [, 1] and T be a clause tree. We call a node reachable if there eists a path root = v, v 1,..., v m = u in T such that π(v i ) γ for all 1 i m. We let Reachable(T, γ, π) be the set of all reachable nodes. Lemma 6*.5 (Clause Tree Lemma) If Reachable(T, π(), π) D, then 2 Forced(F, α, π, D). Proof. Let α be the restriction of α to the variables with π() < π(). B denition, is forced if there is a formula F F [α ] with F D that implies. Let G := clause-label(reachable(t, π(), π)), i.e., the subformula of F consisting of all clause labels of reachable nodes in T. Clearl G D. We claim that G [α ] implies. Suppose, for the sake of contradiction, that G [α ] does not impl. Then there is some assignment β : vbl(g) {, 1} such that (i) β satises G, (ii)

7 6*.1. HAVING A UNIQUE SATISFYING ASSIGNMENT 7 β() = 1 for all 2 vbl(g) for which π() < π(), (iii) β() =. Choose a maimal path in T starting at the root and containing onl nodes v such that β(var-labelv) =. Since β() =, this path is nonempt. Let u be its endpoint. The node u is reachable. All children of u are labeled with some variable for which β() = 1, and all ancestors with some variable for which β() =. One checks that clause-label(u) is unsatised b β, a contradiction. 6*.1.2 How to Bound Pr[ 2 Forced(F, α, π, D)] For this section, a variable B Lemma 6*.5 we know that Pr[ 2 Forced(F, α, π, D)] Pr[ Reachable(T, π(), π) D]. From now on we can forget the clause labels in our tree and focus onl on its structure and variable labels. For a tree T and γ 2 [, 1] let h(t, γ) be the number of nodes in a longest path in T starting at the node and containing onl reachable nodes. Hence h(t, γ) = if π(root(t)) < γ. Observation 6*.6 Reachable(T, γ, π) (k 1) h(t,γ), and therefore, for d := log k 1 D, Pr[ Reachable(T, γ, π) D] Pr[h(T, γ) d]. Definition 6*.7 For a clause tree T, γ 2 [, 1] and d 2 we dene P γ (T, d) := Pr[h(T, γ) d π(root(t)) γ]. For a tree T, let T be the tree we obtain b replacing each variable label in T b a new distinct variable. Lemma 6*.8 (Distinct Label Lemma) Let T be a clause tree, γ 2 [, 1], and d 2 N, and γ 2 [, 1]. Then P γ (T, d) P γ (T, d). Proof. The proof consists of two crucial facts, plus a straightforward induction. The one fact is that if u is a proper ancestor in T of v, then var-label(u) 6= var-label(v), and thus the event [π(u) < γ] is independent of everthing happening in u's subtree. The second fact is that two monotone boolean functions are positivel correlated:

8 8 CHAPTER 6*. THE PPSZ ALGORITHM 1 u u u 1 2 u 2 Figure 6*.7: Illustration of Lemma 6*.8: The left tree T, with non-unique labels, is at least as good as the right tree T with distinct labels. Theorem 6*.9 (FKG Inequalit [FKG71]) Let f, g : {, 1} n R be two monotonicall increasing boolean functions. Sample 2 {, 1} n b setting each i to 1 with some probabilit p i 2 [, 1], independentl. Then Pr[f() = 1 g() = 1] Pr[f() = 1] Pr[g() = 1]. We are read to prove the lemma b induction on d. Let u be the root of T and u the root of T. If d =, then P(T, ) =, since π(u) γ. The same holds for T. For the induction step, let v 1,..., v l be the children of u in T, and v1,..., v l the children of u in T. Let T 1,..., T l and T1,..., T l be the respective subtrees. Clearl h(t, γ) d if and onl if for each 1 i l, either (i) π(v i ) < γ or (ii) π(v i ) γ and h(t i ) d 1. We calculate " # l^ P(T, d) = Pr h(t i, γ) d 1 i=1 l Pr [h(t i, γ) d 1] (6*.4) = i=1 l (γ + (1 γ) P(T i, d 1)) (6*.5) i=1 l (γ + (1 γ) P(Ti, d 1)) (6*.6) i=1 " # l^ = Pr h(ti, γ) d 1 (6*.7) i=1 = Pr[h(T, γ) d]. (6*.8)

9 6*.1. HAVING A UNIQUE SATISFYING ASSIGNMENT u 1 2 u 2 T u 1 2 T T T T u 2 Figure 6*.8: Making the tree innite can onl increase Reachable(T, γ, π). Here, (6*.4) follows from the FKG Inequalit, since the events [h(t i, γ) d 1] are monotone boolean functions in the events [π(w) < γ]. Inequalit (6*.6) follows from induction, and (6*.7) holds since the events [h(ti, γ) d 1] are independent, since all variable labels in Ti are distinct. Let us condition the event [ 2 Forced(F, α, π, D)] on π() = γ. For d = log k 1 D it holds that Pr[ 2 Forced(F, α, π, D) π() = γ] Pr[h(T, γ) d π() = γ] = Pr[h(T, γ) d π() γ] = P(T, d) P(T, d). Net, we etend T to an innite tree T : Attach several innite (k 1)-ar tree to ever node in T having fewer than k 1 children. Label each new node with a new variable. This gives an innite (k 1)-ar tree T, whose root is labeled with. Observation 6*.1 Reachable(T, γ, π) Reachable(T, γ, π) and therefore P(T, d) P(T, d). Note that P(T, d) is some number depending on γ, k, and d, but not depending on the input formula F. Since T is innite, there is a possibilit that h(t, γ) is unbounded. Eercise 6*.1 Show that if h(t, γ) d for all d 2 N, i.e., T contains arbitraril long paths of reachable nodes starting at the root, then T contains an innite path of reachable nodes starting at the root.

10 1 CHAPTER 6*. THE PPSZ ALGORITHM We write h(t, γ) = as a shorthand for \there eist arbitraril long paths of reachable nodes starting at the root", and h(t, γ) < if there are no such paths. We dene P γ (T ) := Pr[h(T, γ) < ]. The following lemma is sounds deceptivel obvious, but is not entirel trivial to prove. Lemma 6*.11 For each γ 2 [, 1], it holds that lim d P γ(t, d) = P γ (T ). This means that there are numbers ɛ γ,d such that for ever ed γ 2 [, 1], it holds that P γ (T, d) P γ (T ) ɛ γ,d and lim d ɛ γ,d =. Let us combine everthing we have learned so far: Pr[ 2 Forced(F, α, π, D) π() = γ] P γ (T, d) P γ (T ) ɛ γ,d, and b the law of total probabilit, we conclude that Pr[ 2 Forced(F, α, π, D)] 1 P γ (T )dγ 1 ɛ γ,d dγ. If D = D(n) is some slowl growing function, then d = log 2 D grows, too. One has to check that if ɛ γ,d goes to for ever ed γ, then also 1 ɛ γ,d dγ goes to. We write R k := 1 P γ(t )dγ. This is a number that onl depends on k. We have Pr[ 2 Forced(F, α, π, D)] R k o(1), Pr[pps(F, β, π, D) = α] 2 (1 R k)(n+o(n)). We can nd an equation for P γ (T ), much like in the proof of Lemma 6*.8. Let u be the root of T and v 1,..., v k 1 be its children, and let T 1,..., T k 1 be the corresponding subtrees. The crucial observation is that all T i are

11 6*.1. HAVING A UNIQUE SATISFYING ASSIGNMENT 11 isomorphic to T itself, and therefore P γ (T ) = P γ (T i ). On the other hand, observe that h(t, γ) < if and onl if for ever 1 i k 1 either (i) π(v i ) < γ or (ii) π(v i ) γ and h(t i, γ) <. Therefore P γ (T ) = (γ + (1 γ)p γ (T )) k 1. (6*.9) For k = 3, (6*.9) has the two solutions P γ (T ) = 1 and P γ (T ) = γ 2 /(1 γ) 2. Without knowing which one is the true value, we can at least sa that P γ (T ) min(1, γ 2 /(1 γ) 2 (in fact, we will show later that this holds with equalit) and thus Thus 1 P γ (T ) = 1 min 1,! γ 2 = 2 2 ln 2 (1 γ) Pr[pps(F, β, π, D) = α] n o(n) Ω (1.371 n ). Theorem 6*.12 Let D be a slowl growing function. Then for an 3-CNF formula F over n variables that has a unique satisfing assignment α, pps(f, π, β, D) returns α with probabilit Ω(1.371 n ).

12 12 CHAPTER 6*. THE PPSZ ALGORITHM

13 Bibliograph [FKG71] [Her11] C. M. Fortuin, P. W. Kastelen, and J. Ginibre. Correlation inequalities on some partiall ordered sets. Comm. Math. Phs., 22:89{13, Timon Hertli. 3-sat faster and simpler - unique-sat bounds for pps hold in general. In Rafail Ostrovsk, editor, FOCS, pages 277{284. IEEE, 211. [PPSZ5] Ramamohan Paturi, Pavel Pudlak, Michael E. Saks, and Francis Zane. An improved eponential-time algorithm for k-sat. J. ACM, 52(3):337{364, 25. [PPZ99] Ramamohan Paturi, Pavel Pudlak, and Francis Zane. Satisabilit coding lemma. Chicago J. Theoret. Comput. Sci., pages Article 11, 19 pp. (electronic),