Lower Bounds for the Weak Pigeonhole. Principle Beyond Resolution. Departament de Llenguatges i Sistemes Informatics

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1 Lower Bounds for the Weak Pigeonhole inciple Beyond Resolution Albert Atserias?, Mara Luisa Bonet??, and Juan Luis Esteban??? Departament de Llenguatges i Sistemes Informatics Universitat Politecnica de Catalunya, Barcelona fatserias,bonet,estebang@lsi.upc.es y Abstract. We work with an extension of Resolution, called Res(), that allows clauses with conjunctions of two literals. In this system there are rules to introduce and eliminate such conjunctions. We prove that the weak pigeonhole principle PHP cn n and random unsatisable CNF formulas require exponential-size proofs in this system. This is the strongest system beyond Resolution for which such lower bounds are known. As a consequence to the result about the weak pigeonhole principle, Res(log) is exponentially more powerful than Res(). Also we prove that Resolution cannot polynomially simulate Res(), and that Res() does not have feasible monotone interpolation solving an open problem posed by Krajcek. 1 Introduction The pigeonhole principle, PHP n+1 n, expresses that it is not possible to have a one-to-one mapping from n+1 pigeons to n holes. This simple principle is central to many mathematical arguments. Since it can be formalized in propositional logic, it is natural to ask in which propositional proof systems such a principle can be proved in polynomial-size, with respect to the size of the encoding. A fair amount of information is known about sizes of proofs of PHP n+1 n in various proof systems. Haken [1] proved that this principle requires exponentialsize proofs in Resolution. His proof techniques were later extended and simpli- ed [4, 5]. Also Beame et al. [] proved that PHP n+1 n requires exponential-size proofs in bounded-depth Frege systems. Regarding upper bounds, Buss [8] gave polynomial-size proofs of PHP n+1 n in unrestricted Frege systems. The pigeonhole principle can be formulated in more general terms, allowing the number of pigeons to be greater than n + 1. We call this principle weak pigeonhole principle, or PHP m n, when the number of pigeons m is at least n. The proof techniques of Haken where extended in [9] to prove that PHPn n? requires exponential-size proofs in Resolution. A very intriguing and often studied? Supported by the CUR, Generalitat de Catalunya, through grant 1999FI 005.?? Partially supported by MEC through grant PB C04 (FRESCO project) and CICYT TIC C0-01.??? Partially supported by MEC through grant PB C04 (FRESCO project) y Partially supported by ALCOM-FT, IST

2 open problem is to prove exponential-size lower bounds for Resolution proofs of PHP nc n, if that can be done at all. As a contrast, the techniques of [] for proving lower bounds for the pigeonhole principle in bounded-depth Frege systems can only prove lower bounds for PHP n+c n, and it is again open whether lower bounds can be proved when the number of pigeons in greater than n + c. Regarding upper bounds, it is known that PHP n n has quasipolynomial-size proofs in bounded-depth Frege [15, 14]. We work with the proof system Res(), proposed by Krajcek [1], that can be viewed either as an extension of Resolution, or as a restriction of boundeddepth Frege. In this system the clauses do not only contain literals, but can also have conjunctions of two literals. The resolution rule gets modied to be able to eliminate a conjunction of two literals from a clause. We prove that PHP cn n requires exponential-size proofs in Res(). This is, to our knowledge, the rst lower bound proof for the weak pigeonhole principle in a subsystem of boundeddepth Frege that extends Resolution. We note that the quasipolynomial upper bound for bounded-depth Frege mentioned above can be carried over in depth-0:5 LK [14], which is equivalent to Res(log) (the analogue of Res() when we allow conjunctions of up to polylog literals). As a consequence of our lower bound, there is an exponential separation between Res() and Res(log). Combining our techniques with those of [], we also obtain an exponentialsize lower bound for Res()-refutations of random unsatisable k-cnf formulas with clause density near the threshold. Again, this is the strongest system beyond Resolution for which such a lower bound is known. This result may be considered as a rst step towards proving them hard for bounded-depth Frege. Another important question to ask is whether Res() is more powerful than Resolution. Here we prove that Resolution cannot polynomially simulate Res(), and therefore Res() is superpolynomially more ecient than Resolution. As a corollary, we see that Res() does not have the feasible monotone interpolation property, solving this way a conjecture of Krajcek [1]. Another motivation for working with the system Res() is to see how useful it can be in automated theorem proving. Given that it is more ecient than Resolution (at least there is a superpolynomial separation), it might be a good idea to try to nd good heuristics to nd proofs in Res() to be able to use it as a theorem prover. Denitions and Overview of the Lower Bound oof A k-term is a conjunction of up to k literals. A k-disjunction is an (unbounded fan-in) disjunction of k-terms. If F is a k-disjunction, a 1-term of F is also called a free-literal. The refutation system Res(k), dened by Krajcek [1], works with k-disjunctions. There are three inference rules in Res(k): Weakening, - Introduction, and Cut A A _ V ii l i A _ V ii l i B _ V ij l i A _ B _ V ii[j l i A _ V ii l i A _ B B _ W ii l i

3 where A and B are k-disjunctions, I; J are sets of indices such that ji [ Jj k, and the l i 's are literals. As usual, if l is a literal, l denotes its negation. Observe that Res(1) coincides with Resolution with the Weakening rule. The size of a Res(k)-refutation is the number of symbols in it. Mainly, we will work with Res(). Let G = (U [ V; E) be a bipartite graph on the sets U and V of cardinality m and n respectively, where m > n. The G-PHP m n, dened by Ben-Sasson and Wigderson [5], states that there is no matching of U into V. For every edge (u; v) E, let x u;v be a propositional variable meaning that u is mapped to v. The principle is then formalized as the conjunction of the following set of clauses: x u;v1 x u;vr u U; N G (u) = fv 1 ; : : : ; v r g (1) x u;v _ x u 0 ;v v V; u; u 0 N G (v); u 6= u 0 : () Here, N G (w) denotes the set of neighbors of w in G. Observe that if G is the complete bipartite graph Kn m, then G-PHP m n coincides with the usual pigeonhole principle PHP m n. It is easy to see that a lower bound for the size of Res()- refutations of G-PHP m n implies the same lower bound for the size of Res()- refutations of PHP m n. Ben-Sasson and Wigderson proved that whenever G is expanding in a sense dened next, every Resolution refutation of G-PHP m n must contain a clause with many literals. We observe that this result is not unique to Resolution and holds in a more general setting. Before we state the precise result, let us recall the denition of expansion: Denition 1. [5] Let G = (U [ V; E) be a bipartite graph where ju j = m, and jv j = n. For U 0 U, the boundary of U, denoted 0, is the set of vertices in V that have exactly one neighbor in U 0 ; that 0 = fv V : jn(v)\u 0 j = 1g. We say that G is (m; n; r; f)-expanding if every subset U 0 U of size at most r is such that j@u 0 j f ju 0 j. The proof of the following statement is the same as in [5] for Resolution. Theorem 1. [5] Let S be a sound refutation system with all rules having fan-in at most two. Then, if G is (m; n; r; f)-expanding, every S-refutation of G-PHP m n must contain a formula that involves at least rf= distinct literals. With these denitions, we are ready to outline the argument of the lower bound proof. In section, we will prove the existence of a bipartite graph G = (U [V; E) with ju j = cn 0 and jv j = n 0 such that if we remove a small random subset of nodes from V, and the corresponding edges, the resulting graph is (m; n; r; f)- expanding for certain m, n, r and f. Then we will argue that G-PHP cn0 n 0 requires exponential-size Res()-refutations as follows. Assume, for contradiction, that is a small refutation of G-PHP cn0 n 0. We say that a -disjunction in is large if it contains at least d = rf= distinct literals. We apply a random restriction 1 to the refutation such that for every large C, either Cj 1 contains many free literals, or the total number of literals in Cj 1 is less than d. Then we extend 1

4 to a new random restriction 1 that knocks out all those large C such that Cj 1 contains many free literals, ignoring those that are not free. After applying, we obtain a refutation of G()-PHP m n where all -disjunctions have less than rf= literals and G() is (m; n; r; f)-expanding. This contradicts Theorem 1. Random Graphs and Restrictions In this section we will prove the existence of a bipartite graph G as claimed in Section. The principle G-PHP m n will require exponential size Res()-proofs. Let G(m; n; p) denote the distribution on bipartite graphs on sets U and V of sizes m and n respectively, with edge probability p independently for each edge. Lemma 1. If G is drawn from G(m; n; p), then [8v V : mp= < deg G (v) < mp] 1? ne? mp 8. oof : Fix a vertex v V. Then, deg G (v) Bin(m; p), so that E[deg G (v)] = mp. By Cherno bounds, [deg G (v) mp] e?mp= and [deg G (v) mp=] e?mp=8. By a union bound, [9v V : deg G (v) mp= _ deg G (v) mp] ne?mp= + ne?mp=8 ne?mp=8, and so [8v V : mp= < deg G (v) < mp] 1? ne?mp=8. ut (lemma 1) Lemma. Let m = kn, p = 48k ln(m)=m, = 1=mp and f = np=6. Let G be drawn from G(m; n; p). Then, [G is (m; n; m; f)-expanding] 1=. oof : Fix U 0 U of size s m, and v V. Then, 0 ] = sp(1? p) s?1. Let q = 0 ]. Let X v P be the indicator random variable for the event that 0. Then, j@u 0 j = vv X v. Observe that X v and X v 0 are independent whenever v 6= v 0. Hence, j@u 0 j Bin(n; q), so that E[j@U 0 j] = nq. By Cherno bound, [j@u 0 j nq=] e?nq=8. On the other hand, nq = nsp(1? p) s?1 snp(1? p) m. Moreover, (1? p) m = (1? p) 1=p approaches 1=e for suciently large m. Therefore, nq snp=. It follows that nq= sf and e?nq=8 e?snp=4. We conclude that [j@u 0 j < f ju 0 j] [j@u 0 j nq=] e?nq=8 e?snp=4. Finally, we bound the probability that G is not (m; n; m; f)-expanding by mx m mx mx e?snp=4 m s e?snp=4 (me?np=4 ) s : () s s=1 s=1 Recall that p = 48k ln(m)=m and m = kn. P So me?np=4 me? ln(m) = m?1 < 1 1=4. Hence the sum in () is bounded by s=1. ut (lemma ) s=1 1 4 s 1 Let G be a xed bipartite graph on f1; : : : ; mg and f1; : : : ; ng. A restriction (for G) is a sequence of pairs = ((u 1 ; v 1 ); : : : ; (u r ; v r )) such that (u i ; v i ) E(G), and all v i 's are distinct. We let R r (G) be the set of restrictions of length r. We dene a distribution R r (G) on R r (G) as follows: Let V 0 = f1; : : : ; ng; for every i f1; : : : ; rg in increasing order, choose a hole v i uniformly at random in

5 V i?1, choose a pigeon u i uniformly at random in N G (v i ), and let V i = V i?1?fv i g. The nal restriction is ((u 1 ; v 1 ); : : : ; (u r ; v r )). We dene a distribution D(m; n; p; r) on the set of pairs (G; ) with R r (G): the graph G is drawn from G(m; n + r; p) rst, and then is drawn from R r (G). In other words, if (H; ) is a xed pair with R r (H), then [G = H = ] = p e(h) (1? p) m(n+r)?e(h) jr r (H)j?1 : If G is a bipartite graph on f1; : : : ; mg and f1; : : : ; n+rg, and a restriction ((u 1 ; v 1 ); : : : ; (u r ; v r )) R r (G), then G() denotes the graph that results from deleting v 1 ; : : : ; v r from G, and renaming nodes in an order-preserving way. With this denitions we are ready to prove: Lemma. Let m = kn, p = 48k ln(m)=m, = 1=mp and f = np=6. Let (G; ) be drawn from D(m; n; p; r). Then, [G() is (m; n; m; f)-expanding] 1=. oof : Let A be the event that G() is (m; P n; m; f)-expanding. Let S = fr f1; : : : ; n + rg : jrj = rg. Then, [A] = RS [A j ran() = R] [ran() = R]. The proof that [A j ran() = R] 1= is the same as the proof of Lemma replacing V by V? R. The result follows. ut (lemma ) Lemma 4. Let m = kn, p = 48k ln(m)=m, = 1=mp and f = np=6. For every r n, there exists a bipartite graph H on f1; : : : ; mg and f1; : : : ; n + rg such that the following two properties hold: (i) mp= deg H (v) mp for every v f1; : : : ; n + rg, (ii) [H() is (m; n; m; f)-expanding] 1=; when is drawn from R r (H). oof : Let (G; ) be drawn from D(m; n; p; r). According to Lemmas, we have We have that [G() is (m; n; m; f)-expanding] 1=. Moreover, [8v V : mp= < deg G (v) < mp] 1?(n+r)e?mp=9 5=6 by Lemma 1. Let E(G; ) be the event that G() is expanding and every right-node in G has degree between mp= and mp. Combining both equations P above we have that [E(G; )] 1=. On the other hand, [E(G; )] = H [E(G; ) j G = H] [G = H] where H ranges over all bipartite graphs on m and n + r nodes. Therefore, there exists some xed H such that [E(G; ) j G = H] 1=. Moreover, [E(G; ) j G = H] equals [E(H; )] when is drawn from R r (H). Finally, since this probability is strictly positive, it must be the case that H satises property (i) in the lemma since it is independent of. ut (lemma 4) 4 The lower bound argument Before we start the lower bound proof, we need a normal form for Res()- refutations of G-PHP m n. We claim that every Res()-refutation of G-PHP m n can be turned into a Res()-refutation of similar size in which no -term is of the form x u;v x u 0 ;v with u 6= u 0. To check this, observe that such a -term must have been introduced at some point by the rule of -introduction with clauses, say, A_x u;v and B _x u 0 ;v. Cutting them with the axiom x u;v _ x u 0 ;v we get A_B that can be used to continue the proof because it subsumes A_B _(x u;v x u 0 ;v).

6 Theorem. Let c > 1 be a constant. For all suciently large n, every Res()- refutation of PHP cn n=(log n has size at least e n)14. oof : Let k = c+1, r = n=c, n 0 = n+r, and m = kn = cn 0. Let G = (U [ V; E) with ju j = m and jv j = n + r be the bipartite graph of Lemma 4. We show that every Res()-refutation of G-PHP has size at least e n=(log n)14. This will imply the Theorem since a Res()-refutation of PHP cn0 n 0 gives a Res()-refutation of G-PHP of no bigger size. Let us assume, for contradiction, that G-PHP has a Res()-refutation of size S < e n=(log n)14. Let C be a -disjunction, and let (u; v) E(G). We let Cj (u;v) be the result of assigning x u;v = 1 and x u 0 ;v = 0 for every u 0 N G (v)? fug to C, and simplifying as much as possible. This includes replacing subformulas of the form l _ (l l 0 ) by l, and subformulas of the form l _ (l l 0 ) by l _ l 0 in some specied order; here l and l 0 are literals. Given a restriction = ((u 1 ; v 1 ); : : : ; (u r ; v r )), we let Cj be the result of applying (u 1 ; v 1 ); : : : ; (u r ; v r ) successively in this order. For every i f1; : : : ; rg, we let i = ((u 1 ; v 1 ); : : : ; (u i ; v i )). We say that C is large if it contains at least d = n=1 distinct literals; otherwise, C is small. We say that C is wide if it contains at least s = n=(log n) 5 free literals; otherwise, C is narrow. We say that a pair (u; v) E(G) hits C if either x u;v occurs positively in C, or x u 0 ;v occurs negatively in C for some u 0 N G (v)? fug. Equivalently, (u; v) hits C if it sets some literal of C to 1. If the literal is free, it knocks out the -disjunction. If the literal is part of a conjunction, it will locally create a free literal. In general, we say that (u; v) E(G) knocks C if Cj (u;v) 1. We say that (u; v) E(G) is a bad pair for C if it does not knock it and there exists u 0 N G (v)? fug such that (u 0 ; v) knocks C. A bad pair may or may not be a hit. In all probabilities that follow, is drawn from the distribution R r (G). Our main goal is to prove that the probability that a xed -disjunction C of remains large is exponentially small; that is, we aim for a proof that [Cj is large] e?n=(log n)1: (4) This will suce because then [9C : Cj is large] Se?n=(log n)1 < e n=(log n)14?n=(log n)1 e < 1=, and also [G() not (m; n; m; f)-expanding] = by Lemma 4. This means that there exists a restriction R r (G) such that G() is (m; n; m; f)-expanding and every -disjunction in j has less than d = mf= literals. This is a contradiction with Theorem 1. For i f1; : : : ; rg, let A i be the event that Cj i is large, and let B i be the event that Cj i is narrow. Recall that i = ((u 1 ; v 1 ); : : : ; (u i ; v i )). Then, [Cj is large] 4 Ar rx j=r= _ jr= Ar [A j ] + 4 Ar jr= jr= 5 5 :

7 We will show that every term in this expression is exponentially small. The bound on terms of the form [A j ] will be proven in Lemma 6. For the last term, we use an argument similar in spirit to the one by Beame and Pitassi [4]: h i Lemma 5. A r Vjr= e?n=(log n)8. oof : Let S i be the indicator random variable for the event that (u i ; v i ) knocks Cj i?1. Then, 4 Ar jr= 5 4 = Y i>r= Y i>r= Y i>r= i>r= S i = 0 4 Si = 0 jr= 5 = jr= 4 Si = 0 B i?1 4 Si = 0 B i?1 r=<j<i r=<j<i r=<j<i S j = 05 S j = 05 S j = 05 : Fix i fr= + 1; : : : ; rg and let H be the set of holes that occur in a free literal of Cj i?1. Given that B i?1 holds, Cj i?1 is wide which means that there are at least s free literals. Therefore jhj s=, where = mp is an upper bound on the right-degree of G. Moreover, every v H gives a possible knock, and dierent holes give dierent knocks. The reason is the following: if x u;v is a free literal, then (u; v) is a knock; and if x u;v is a free literal, then (u 0 ; v) is a knock for every u 0 N G (v)? fug, which is non-empty since the right-degree of G is at least two. Therefore, Therefore, 4 Ar 4 Si = 1 B i?1 jr= r=<j<i 5 1? s n S j = 0 r= e? 5 jhj (n + r? i + 1) sr s n : 6 n e?n=(log n)8 : ut (lemma 5) Lemma 6. Let j be such that r= j r. Then, [A j ] e?n=(log n)11. oof : Recall that A j is the event that Cj j is large, and is the event that Cj j is narrow. We let S i be the indicator random variable for the event P that j (u i ; v i ) hits Cj i?1, where i?1 = ((u 1 ; v 1 ); : : : ; (u i?1 ; v i?1 )). Let S = i=1 S i.

8 Then, for every h, [A j ] = [A j S < h] + [A j S h] [A j S < h] + [A j S h]: We show that each term in this expression is exponentially small. More precisely, we show that [A j S < h] e?n=(log n)?n=(log n)10 and [A j S h] e which is clearly enough to prove Lemma 6. Claim 1 Let h = n=(log n) 4. Then, [A j S < h] e?n=(log n). P oof : Let Y = f(a 1 ; : : : ; a j ) f0; 1g j j : i=1 a i < hg. Observe that A j implies A i for every i j because if Cj j is large, so is Cj i for every i j. Then, " j # X X " # j [A j S < h] = S i < h A j = S i = a i A j = i=1 ay i=1 = X ay X jy i=1 jy ay i=1 X jy ay i=1 i?1 "Si = ai Aj k=1 i?1 "Si = ai Ai?1 k=1 " i?1 S i = a i A i?1 S k = a k # k=1 # S k = a k # S k = a k : Fix i f1; : : : ; jg. Let H be the set of holes that occur in Cj i?1. We have jhj d= given that A i?1 holds. Again, = mp is an upper bound to the right-degree of G. Moreover, every v H gives a possible hit, and dierent holes give dierent hits (the reason is the same as in Lemma 5 for knocks). Therefore, " S i = 1 A i?1 i?1 k=1 S k = a k # jhj (n + r? i + 1) d n : Since there are at least j? h zeros in (a 1 ; : : : ; a j ), we obtain X [A j S < h] 1? d j?h X j e? d(j?h) n hj h e? d(j?h) n n i ay i<h exp? j? h 6 + h log(j) + log(h) e?n=(log n) :ut(claim 1) Claim [A j S h] e?n=(log n)10. oof : During this proof we will drop the subindex j in A j and since it will always be the same. For every i f1; : : : ; rg, let T i fk; b; ng be a random variable indicating whether (u i ; v i ) is a knock, a bad pair, or none of the previous

9 respectively for Cj i?1. For t fk; b; ng, let P Si t be the indicator random variable for the event that T i = t, and let S t j = i=1 St i. Thus, Sk is the number of knocks and S b is the number of bad pairs of j. Fix = ((u 1 ; v 1 ); : : : ; (u r ; v r )) such that A B S h holds under. Observe that (u i ; v i ) does not knock Cj i?1 for any i f1; : : : ; jg since Cj j must be large. Hence, S k = 0 under. Let b = (h? s)=( + 1). We now claim that S b b. Suppose for contradiction that the number of bad pairs is less than b. Every bad pair (u i ; v i ) removes at most free literals since at most those many literals about hole v i may appear. Moreover, since there are no knocks, every hit (u i ; v i ) that is not a bad pair increases the number of free literals by at least one. The reason is that such a hit turns a conjunction into a free literal whose hole component must be dierent from v i for otherwise a knock would be available; recall that the right-degree of G is at least three, and that conjunctions of the form x u;v x u 0 ;v with u 6= u 0 do not occur in the refutation. It follows that the number of free literals in Cj j is at least (S? S b )? S b > h? ( + 1)b = s, a contradiction with the fact that B holds under. We have proved that [A B S h] [S k = 0 S b b]. The intuition behind why this probability is small is that every bad pair could have been a knock. This makes unlikely that produces many bad pairs and no knocks. In what follows, we will prove this intuition using martingales. For t fk; b; ng and i f1; : : : ; jg, let Pi t denote the random variable [T i = t j 0 ; : : : ; i?1 ]. We dene a martingale X 0 ; : : : ; X j with respect to 0 ; : : : ; j as follows: Let X 0 = 0, and X i+1 = X i + S b i+1? P b i+1. Recall that S b is the indicator random variable for the event that i+1 T i+1 = b. So E[X i+1 j 0 ; : : : ; i ] = (X i + 1? P b i+1 ) P b i+1 + (X i? P b i+1 ) (1? P b i+1 ) = = (X i? P b i+1)(p b i+1 + 1? P b i+1) + P b i+1 = X i : Hence, fx i g i is a martingale with respect to f i g i. Observe also that X j = S b? P j i=1 P b i. Similarly, we dene Y 0 ; : : : ; Y j as follows: Let Y 0 = 0, and Y i+1 = Y i + S k i+1? P k i+1. It is also easy to see that fy ig i is a martingale with respect to f i g i. Again, Y j = S k? P j i=1 P k i. Subclaim 1 P k i () P b i ()= for every R r(g) and i f1; : : : ; jg. oof : Fix i f1; : : : ; jg and = ((u 1 ; v 1 ); : : : ; (u r ; v r )). We want to show that Pi k() P i b ()=. Dene three sets as follows: Q = f(u; v) E(G) : v 6 fv 1 ; : : : ; v i?1 gg, Q k = f(u; v) Q : (u; v) is a knock for Cj i?1 g, and Q b = f(u; v) Q : (u; v) is a bad pair for Cj i?1 g. Observe that Pi b() = jqb j jqj?1 and Pi k () = jq k j jqj?1. On the other hand, every bad pair (u; v) Q b gives a possible knock (u 0 ; v) Q k by denition. Moreover, bad pairs with dierent hole components give dierent possible knocks. Grouping Q b by holes, we have that jq k j jq b j=. Consequently, Pi k() P i b ()= as required. ut (subclaim 1) To complete the proof of claim we will need the following form of Azuma's Inequality: Let X 0 ; : : : ; X n be a martingale such that jx i? X i?1 j 1; then,

10 [jx n? X 0 j ] e? =n for every > 0 [11]. Now, [S k = 0 S b b] = [S k = 0 S b b X j b=] + + [S k = 0 S b b X j < b=]: The rst summand is bounded by [X j b=] e?b =4j by Azuma's Inequality. The second summand is bounded by h S k = 0 Pj i=1 P b i b= i h S k = 0 Pj i=1 P k i b= [Y j?b=] e?b =4 j : The rst inequality follows from Subclaim 1, and the third follows from Azuma's Inequality again. The addition of the two summands is then bounded by e?n=(log n)10 as required. ut (claim and lemma 6) We are ready to complete the proof of our goal: equation (4). We have shown that [Cj large] rx j=r=?n=(log n)11 e + e?n=(log n)8 e?n=(log n)1: ut (theorem ) The most general form of the previous theorem gives a lower bound for n 9 PHP m n of e (m log m) 8 log n, using r = n=8, h = n =((m log m) log n) and s = h=, therefore the best result is an exponential lower bound for PHPn n9=8?. Given that Res(log) and depth-0:5 LK are polynomially equivalent, and given that PHP n n has quasipolynomial-size proofs in depth-0:5 LK [14], we obtain: Corollary 1. There is an exponential separation between Res() and Res(log). The model of random k-cnf formulas that we use is the one considered in [10,]. The distribution is denoted Fm k;n and consists in choosing m clauses of exactly k literals independently with replacement. The following result can be obtained by an argument similar to that of Theorem. The proof can be found in the complete version of the paper. Theorem. Let F F ;n 5n. Almost surely, Res()-refutations of F require size (n1= =(log(n)) ). i 5 Separation between Res() and Resolution In this section we prove that Resolution cannot polynomially simulate Res(). More precisely, we prove that a certain Clique-Coclique principle, as dened by Bonet, Pitassi and Raz in [6], has polynomial-size Res()-refutations, but every Resolution refutation requires quasipolynomial size.

11 The Clique-Coclique principle that we use, CLIQUE n k;k0, is: x i;1 x i;n 1 l k x l;i _ x l;j 1 l k; 1 i; j n; i 6= j x l;i _ x l 0 ;i 1 l; l 0 k; 1 i n; l 6= l 0 y 1;i y k 0 ;i 1 i n y l;i _ y l 0 ;i 1 l; l 0 k 0 ; 1 i n; l 6= l 0 x l;i _ x l 0 ;j _ y t;i _ y t;j 1 l; l 0 k; 1 t k 0 ; 1 i; j n; l 6= l 0 ; i 6= j Theorem 4. Let k 0 < k n. If PHP k k0 has Resolution refutations of size S, then CLIQUE n k;k 0 has Res()-refutations of size Snc for some constant c > 0. The proof can be found in the complete version of the paper. We reduce the formula CLIQUE n k;k 0 to PHPk k 0 in Res(). The meaning of variable p i;j is that pigeon i sits in hole j. We perform the following substitutions: p i;j n_ l=1 (x i;l y i;l ) p i;j n_ l=1;j 0 6=j (x i;l y j 0 ;l) We will use the Monotone Interpolation Theorem for Resolution together with the following result of Alon and Boppana [1] establishing a lower bound to the size of monotone circuits that separate large cliques from small cocliques. In the following, F (m; k; k 0 ) is the set of monotone functions that separate k-cliques from k 0 -cocliques on m nodes. Let S + (f) be the monotone circuit size of f. Theorem 5. If f F (m; k; k 0 ), k 0 k and k p k 0 m=(8 log m), then p S + (f) 1 ( k m 0 +1)= 8 4k p k 0 : log m Theorem 6. Let k = p m and k 0 = (log m) =8 log log m. Then, (i) CLIQUE m k;k 0 has Res()-refutations of size polynomial in m, and (ii) every Resolution refutation of CLIQUE m k;k 0 has size at least exp(((log m) = p log log m)). oof : Regarding (i), we have that k 0 log k 0 1(log 4 m), and so p k 0 log k 0 m 1= = k. On the other hand, Buss and Pitassi [7] proved that PHP k k 0 has Resolution refutations of size polynomial in k whenever k p k 0 log k 0. Therefore, by Theorem 4, CLIQUE m k;k0 has Res()-refutations of size polynomial in m. Regarding (ii), we apply the feasible monotone interpolation theorem for Resolution. We have log m=( p log log m) p k 0 log m. Therefore, by Theorem 5, if f F (m; k; k 0 ) is a monotone interpolant, then S + (f) 1 log p m 6 log log m m 8 4 p 1 m log p m 6 log log m ; m(log m) 8 m =4 which is exp(((log m) = p log log m)). ut (theorem 6) As a corollary, we solve an open problem posed by Krajcek [1]. Corollary. Res() does not have the feasible monotone interpolation property.

12 6 Discussion and Open oblems We proved that there is a quasipolynomial separation between Resolution and Res(). It is an open question whether the separation could be exponential, or a quasipolynomial simulation of Res() by Resolution exists. It is important to notice, that our lower bound for PHP would not follow from such a simulation. Indeed, the lower bound that would follow from that would be of the form n. The previous separation was obtained using a lower bound for Resolution via the monotone interpolation theorem. It is open whether this separation (or a stronger one) could be obtained via the size-width trade-o [5]. It would also be interesting to see what would that mean in terms of possible size-width trade-os for Res(). We conjecture that Res() does not have a strong size-width trade-o. Notice that Res(log) does not have it. This is because (a) Res(log) is equivalent to depth-0:5 LK, (b) PHP n n has quasipolynomial-size proofs in depth-0:5 LK [14], and (c) PHP n n has (n) width lower bounds for Res(log). We extended the width lower bound technique beyond Resolution. A very interesting open question is whether the technique can be extended for Res(), etc. This is related to the optimality of the Res(log) upper bound for PHP n n. References 1. N. Alon and R. Boppana. The monotone circuit complexity of boolean functions. Combinatorica, 7(1):1{, P. Beame, R. Impagliazzo, J. Krajcek, T. Pitassi, P. Pudlak, and A. Woods. Exponential lower bounds for PHP. In STOC9, pages 00{0, P. Beame, R. Karp, T. Pitassi, and M. Saks. The eciency of resolution and Davis-Putnam procedures. Submitted. evious version in STOC'98, P. Beame and T. Pitassi. Simplied and improved resolution lower bounds. In FOCS96, pages 74{8, E. Ben-Sasson and A. Wigderson. Short proofs are narrow: Resolution made simple. In STOC99, pages 517{57, Revised version (000). 6. M. Bonet, T. Pitassi, and R. Raz. Lower bounds for cutting planes proofs with small coecients. The Journal of Symbolic Logic, 6():708{78, Sept S. Buss and T. Pitassi. Resolution and the weak pigeonhole principle. In CSL: 11th Workshop on Computer Science Logic. LNCS, Springer-Verlag, S. R. Buss. Polynomial size proofs of the propositional pigeonhole principle. The Journal of Symbolic Logic, 5(4):916{97, Dec S. R. Buss and G. Turan. Resolution proofs on generalized pigeonhole principles. Theoretical Computer Science, 6():11{17, Dec V. Chvatal and E. Szemeredi. Many hard examples for resolution. J. ACM, 5(4):759{768, G. Grimmet and D. Stirzaker. obability and Random ocesses. Oxford, A. Haken. The intractability of resolution. TCS, 9(-):97{08, Aug J. Krajcek. On the weak PHP. To appear in Fundamenta Mathematic, A. Maciel, T. Pitassi, and A. Woods. A new proof of the weak pigeonhole principle. In STOC00, pages 68{77, J. B. Paris, A. J. Wilkie, and A. R. Woods. ovability of the pigeonhole principle and the existence of innitely many primes. The Journal of Symbolic Logic, 5(4):15{144, 1988.