Resolution Lower Bounds for the Weak PigeonholePrincipleRan Raz�Weizmann Instituteranraz@wisdom.weizmann.ac.ilAbstractWe prove that any Resolution proof for the weak pigeon hole principle, with n holesand any number of pigeons, is of length (2n�), (for some global constant � > 0). Onecorollary is that a certain propositional formulation of the statement NP 6� P=poly doesnot have short Resolution proofs.1 IntroductionThe Pigeon Hole Principle (PHP) is one of the most widely studied tautologies in propositionalproof theory. The tautology PHPn is a DNF encoding of the following statement: There is noone to one mapping from n+1 pigeons to n holes. The Weak Pigeon Hole Principle (WPHP)is a version of the pigeon hole principle that allows a larger number of pigeons. The tautologyWPHPmn (for m � n + 1) is a DNF encoding of the following statement: There is no one toone mapping from m pigeons to n holes. For m > n + 1, the weak pigeon hole principle is aweaker statement than the pigeon hole principle. Hence, it may have much shorter proofs incertain proof systems.The weak pigeon hole principle is one of the most fundamental combinatorial principles.In particular, it is used in most probabilistic counting arguments and hence in many combi-natorial proofs. Moreover, as observed by Razborov, there are certain connections betweenthe weak pigeon hole principle and the problem of P 6= NP [Razb3]. Indeed, the weak pi-geon hole principle (with a relatively large number of pigeons) can be interpreted as a certainencoding of the following statement: There are no small DNF formulas for SAT. Hence, (inmost proof systems, and in particular in Resolution), a short proof for certain formulations ofthe statement SAT 62 P=poly can be translated into a short proof for the weak pigeon holeprinciple. That is, a lower bound for the length of proofs for the weak pigeon hole principle�This work was done while the author was on sabattical year at the Institute for Advanced Study. Researchsupported by US-Israel BSF grant 98-00349, and NSF grant CCR-9987077.

usually implies a lower bound for the length of proofs for certain formulations of the statementNP 6� P=poly.Resolution is one of the most widely studied propositional proof systems. The Resolutionrule says that if C and D are two clauses and xi is a variable then any assignment that satis�esboth of the clauses, C _ xi and D _ :xi, also satis�es C _ D. The clause C _ D is calledthe resolvent of the clauses C _ xi and D _ :xi on the variable xi. A Resolution refutationfor a CNF formula F is a sequence of clauses C1; C2; : : : ; Cs, such that: (1) Each clause Cj iseither a clause of F or a resolvent of two previous clauses in the sequence. (2) The last clause,Cs, is the empty clause (and hence it has no satisfying assignments). We can represent aResolution refutation as an acyclic directed graph on vertices C1; : : : ; Cs, where each clause ofF has out-degree 0, and any other clause has two edges pointing to the two clauses that wereused to produce it. It is well known that Resolution is a sound and complete propositionalproof system, that is, a formula F is unsatis�able if and only if there exists a Resolutionrefutation for F . We think of a refutation for an unsatis�able formula F also as a proof forthe tautology :F . A well-known and widely studied restricted version of Resolution (thatis still complete) is called Regular Resolution. In a Regular Resolution refutation, along anypath in the directed acyclic graph, each variable is resolved upon at most once.There are trivial Resolution proofs (and Regular Resolution proofs) of length 2n � poly(n)for the pigeon hole principle and for the weak pigeon hole principle. In a seminal paper,Haken proved that for the pigeon hole principle, the trivial proof is (almost) the best possible[Hak]. More speci�cally, Haken proved that any Resolution proof for the tautology PHPnis of length 2(n). Haken's argument was further developed in several other papers (e.g.,[Urq, BeP, BSW]). In particular, it was shown that a similar argument gives lower boundsalso for the weak pigeon hole principle, but only for small values ofm. More speci�cally, super-polynomial lower bounds were proved for any Resolution proof for the tautology WPHPmn ,for m < c � n2= logn (for some constant c) [BT]. The weak pigeon hole principle with largervalues of m has attracted a lot of attention in recent years. However, the standard techniquesfor proving lower bounds for Resolution fail to give lower bounds for the weak pigeon holeprinciple. In particular, for m � n2, no non-trivial lower bound was known.For the weak pigeon hole principle with large values of m, there do exist Resolution proofs(and Regular Resolution proofs) which are much shorter than the trivial ones. In particular,it was proved by Buss and Pitassi that for m > cpn logn (for some constant c), there areResolution (and Regular Resolution) proofs of length poly(m) for the tautology WPHPmn[BuP]. Can this upper bound be further improved ? Can one prove a matching lower bound ?As mentioned above, for m � n2, no non-trivial lower bound was known. A partial progresswas made by Razborov, Wigderson and Yao, who proved exponential lower bounds for RegularResolution proofs, but only when the Regular Resolution proof is of a certain restricted form[RWY]. An exponential lower bound for any Regular Resolution proof was proved in [PR]. Inthis paper, we prove an exponential lower bound for any Resolution proof.More precisely, we prove that for any m, any Resolution proof for the weak pigeon holeprinciple, WPHPmn , is of length (2n�), (where � > 0 is some global constant).2

1.1 Lower Bounds for NP 6� P=polyAs mentioned above, our result implies that certain propositional formulations of the statementSAT 62 P=poly do not have short Resolution proofs.Let f : f0; 1gd ! f0; 1g be a Boolean function. For example, we can take f = SAT ,where SAT : f0; 1gd ! f0; 1g is the satis�ability function (or we can take any other NP -hardfunction). We assume that we are given the truth table of f . Let t � 2d be some integer. Wethink of t as a large polynomial in d, say t = d1000.In [Razb1] (see also [Razb2]), Razborov suggested to study propositional formulations ofthe following statement (in the variables ~Z):~Z is (an encoding of) a Boolean circuit of size t =)~Z does not compute the function f .Note that since the truth table of f is of length 2d, a propositional formulation of this statementwill be of length at least 2d, and it is not hard to see that there are ways to write this statementas a DNF formula of length 2O(d) (and hence, its negation is a CNF formula of that length).The standard way to do that is by including in ~Z both, the (topological) description of theBoolean circuit, as well as the value that each gate in the circuit gets on each input forthe circuit. The exact way to give the (topological) description of the circuit may also beimportant in some cases.In [Razb3], Razborov presented a lower bound for the degree of Polynomial Calculus proofsfor the weak pigeon hole principle, and used this result to prove a lower bound for the degreeof Polynomial Calculus proofs for a certain version of the above statement. Following thisline of research, we show (in a similar way) that if t is a large enough polynomial in d (sayt = d1000) then any Resolution proof for a certain version of the above statement is of lengthsuper-polynomial in 2d, that is, super-polynomial in the length of the statement.In particular, this can be interpreted as a super-polynomial lower bound for Resolutionproofs for certain formulations of the statement NP 6� P=poly.Our version of the above statement is slightly di�erent than the one used in [Razb3]. Themain di�erence is in the way we use the variables ~Z to encode the (topological) descriptionof the Boolean circuit. Here, we use ~Z to encode a Boolean circuit of unbounded fan-in,whereas [Razb3] considered Boolean circuits of fan-in 2. This di�erence is substantial, that is,our proof works only for the more general case of unbounded fan-in, and not for the weakercase of fan-in 2. Otherwise, our proof seems to be quite robust in the way the Boolean circuitis encoded. Following our result, the same result for the case of Boolean circuits of fan-in 2was recently proved in [Razb6]. This was done by proving a Resolution lower bound for theso called weak functional onto pigeon hole principle.3

1.2 Subsequent workAs mentioned above, our main result is a lower bound of (2n�) for any Resolution prooffor the weak pigeon hole principle. The constant � implicit in this paper is of the orderof 1=8 or 1=10. Following our result, Razborov came up with three related results. The�rst result [Razb4] presents a proof for an improved lower bound of (2n�) for � = 1=3. Thesecond result [Razb5] extends that proof for the so called weak functional pigeon hole principle,which is an important version of the weak pigeon hole principle. The above mentioned thirdresult [Razb6] extends the proof for the so called weak functional onto pigeon hole principle.2 Preliminaries2.1 Resolution as a Search ProblemA literal is either an atom (i.e., a variable xi) or the negation of an atom (i.e., :xi). A clauseis a disjunction of literals. If C and D are two clauses and xi is a variable then any assignmentthat satis�es both of the clauses, C _ xi and D _ :xi, obviously satis�es the clause C _ Das well. As mentioned in the introduction, the clause C _ D is called the resolvent of theclauses C _ xi and D _ :xi on the variable xi. A Resolution refutation for a CNF formulaF is a sequence of clauses C1; C2; : : : ; Cs, such that, each clause Cj is either a clause of F ora resolvent of two previous clauses in the sequence, and such that, the last clause, Cs, is theempty clause. We think of the empty clause as a clause that has no satisfying assignments. Wethink of a Resolution refutation for F also as a proof for :F . Without loss of generality, weassume that no clause in a Resolution proof contains both xi and :xi (such a clause is alwayssatis�ed and hence it can be removed from the proof). The length, or size, of a Resolutionproof is the number of clauses in it.As mentioned in the introduction, we represent a Resolution proof as an acyclic directedgraph G on the vertices C1; : : : ; Cs. In this graph, each clause Cj which is an original clauseof F has out-degree 0, and any other clause has two edges pointing to the two clauses thatwere used to produce it. We call the vertices of out-degree 0 (i.e., the clauses that are originalclauses of F ) the leaves of the graph. Without loss of generality, we can assume that the onlyclause with in-degree 0 is the last clause Cs (as we can just remove any other clause within-degree 0). We call the vertex Cs the root of the graph, and we denote it also by Root.We label each vertex Cj in the graph by the variable xi that was used to derive it (i.e., thevariable xi that was resolved upon), unless the clause Cj is an original clause of F (and thenCj is not labelled). If a clause Cj is labelled by a variable xi we label the two edges going outfrom Cj by 0 and 1, where the edge pointing to the clause that contains xi is labelled by 0,and the edge pointing to the clause that contains :xi is labelled by 1. That is, if the clauseC _D was derived from the two clauses C _ xi and D _:xi then the vertex C _D is labelledby xi, the edge from the vertex C _D to the vertex C _ xi is labelled by 0 and the edge fromthe vertex C _D to the vertex D _ :xi is labelled by 1. For a non-leaf node u of the graph4

G, de�ne,Label(u) = the variable labelling u.We think of Label(u) as a variable queried at the node u.Let p be a path on G, starting from the root. Note that along a path p, a variable ximay appear (as a label of a node u) more than once. We say that the path p evaluates xito 0 if xi = Label(u) for some node u on the path p, and after the last appearance of xi asLabel(u) (of a node u on the path) the path p continues on the edge labelled by 0 (i.e., if u isthe last node on p such that xi = Label(u) then p contains the edge labelled by 0 that goesout from u). In the same way, we say that the path p evaluates xi to 1 if xi = Label(u) forsome node u on the path p, and after the last appearance of xi as Label(u) (of a node u onthe path) the path p continues on the edge labelled by 1 (i.e., if u is the last node on p suchthat xi = Label(u) then p contains the edge labelled by 1 that goes out from u).For any node u of the graph G, we de�ne Zeros(u) to be the set of variables that the nodeu \remembers" to be 0, and Ones(u) to be the set of variables that the node u \remembers"to be 1, that is,Zeros(u) = the set of variables that are evaluated to 0 by every path p from theroot to u.Ones(u) = the set of variables that are evaluated to 1 by every path p from theroot to u.Note that for any u, the two sets Zeros(u) and Ones(u) are disjoint.The following proposition gives the connection between the sets Zeros(u), Ones(u) andthe literals appearing in the clause u. The proposition is particularly interesting when u is aleaf of the graph.Proposition 2.1 1 Let F be an unsatis�able CNF formula and let G be (the graph represen-tation of) a Resolution refutation for F . Then, for any node u of G and for any xi, if theliteral xi appears in the clause u then xi 2 Zeros(u), and if the literal :xi appears in theclause u then xi 2 Ones(u).Proof:Assume that the literal xi appears in the clause u. (The claim for the literal :xi is proved inthe same way). Let p be a path from the root Cs to u. We will show that the path p evaluatesxi to 0. If no node v < u on the path p satis�es Label(v) = xi then the literal xi appears inthe clause Cs, in contradiction to the fact that Cs is the empty clause. Hence, there exists anode v < u on the path p, such that, Label(v) = xi. Let v be the last (i.e., the largest) suchnode. Let w be the next node on p (i.e., the successor of v on the path p). Thus, the edge1We have learnt that a version of this proposition, presented as a game between two players, has alreadyappeared in [Pud]. 5

(v; w) is contained in the path p. Since v is the last node on p such that Label(v) = xi, nonode z on the path p from w to u satis�es Label(z) = xi. Hence, since the literal xi appears inu, it appears in w as well. Thus, (v; w) is the edge labelled by 0. That is, p evaluates xi to 0. 22.2 The Weak Pigeonhole PrincipleThe propositional weak pigeon hole principle, WPHPmn , states that there is no one-to-onemapping from m pigeons to n holes. The underlying Boolean variables, xi;j, for 1 � i � mand 1 � j � n, represent whether or not pigeon i is mapped to hole j. The negation ofthe pigeonhole principle, :WPHPmn , is expressed in conjunctive normal form (CNF) as theconjunction ofm pigeon clauses and �m2��n hole clauses. For every 1 � i � m, we have a pigeonclauses, (xi;1 _ : : :_ xi;n), stating that pigeon i maps to some hole. For every 1 � i1 < i2 � mand every 1 � j � n, we have a hole clauses, (:xi1;j _ :xi2;j), stating that pigeons i1 and i2do not both map to hole j. We refer to the pigeon clauses and the hole clauses also as pigeonaxioms and hole axioms.Let G be (the graph representation of) a Resolution refutation for :WPHPmn . Then, byProposition 2.1, for any leaf u of the graph G, one of the following is satis�ed:1. u is a pigeon axiom, and then for some 1 � i � m, the variables xi;1; : : : ; xi;n are allcontained in Zeros(u).2. u is a hole axiom, and then for some 1 � j � n, there exist two di�erent variablesxi1;j; xi2;j in Ones(u).2.3 Basic NotationsWe denote by n the number of holes, and by m the number of pigeons. We denote by Holesthe set of holes, and by Pigeons the set of pigeons. That is,Holes = f1; :::; ng.Pigeons = f1; :::; mg.We will usually denote a hole by j, and a pigeon by i. By xi;j we denote the variable corre-sponding to pigeon i and hole j. We denote by V ariables the set of all these variables, andby V ariablesi the set of variables corresponding to the ith pigeon. That is,Variables = fxi;jji 2 Pigeons; j 2 Holesg.Variablesi = fxi;jjj 2 Holesg. 6

We will consider (the graph representation of) Resolution proofs for the weak pigeon holeprinciple. We denote such a graph by G. By u we will usually denote a node in the graph.We say that u0 < u if there is a path in the graph from u0 to u. By p we will usually denotea path in the graph, starting from the root. Note that for any non-leaf node u of the graph,Label(u) is a variable xi;j. The sets Zeros(u) and Ones(u) are subsets of V ariables.We denote by � a small �xed constant (say � = 1=100). For simplicity, we assume that n islarge enough (say n� � 1000). For simplicity, we assume that expressions like n�; n1��; n1�8�=2,etc', are all integers.3 The Lower BoundIn this section, we prove our lower bound on the size of Resolution proofs for the weak pigeonhole principle. Fix � = 1=100, and assume for simplicity that n� � 1000. (We do not attempthere to optimize the value of �).Theorem 3.1 For any m � n + 1, any Resolution proof for the tautology WPHPmn is oflength larger than 2n�=100.In the rest of the section, we give the proof of Theorem 3.1. Let G be the graph represen-tation of a Resolution proof for WPHPmn , and assume for a contradiction that the size of G(i.e., the number of vertices in G) is at most 2n�=100. Note that since the size of G is at most2n�=100, we can assume w.l.o.g. that m < 2n�=100.3.1 Adding AxiomsFirst, de�ne for any integer 0 � k � n�,nk = k � n1��;mk = 2n��k:(Recall that we assume that n�; n1�� are integers). For any node u of the graph G and for anypigeon i, we de�ne Zerosi(u) to be the set of variables xi;j that the node u \remembers" tobe 0, that is, the set of variables xi;j that are evaluated to 0 by every path from the root tou. In other words,Zerosi(u) = Zeros(u) \ V ariablesi:We use the size of Zerosi(u) as a measure for the progress made on pigeon i along paths fromthe root. We compare this measure with the \mileposts" fnkg. We de�ne Overk(u) to be theset of pigeons that passed the kth milepost (for the node u), and we say that a node u is anaxiom of order k if for the node u at least mk pigeons passed the kth milepost. That is, forany integer 0 � k � n� and any node u of G, we de�ne,7

Overk(u) = the set of pigeons i such that jZerosi(u)j � nk.For 1 � k � n�, we say that a node u is a pigeon-axiom of order k ifjOverk(u)j � mk:Note that a pigeon-axiom of order k = n� is just a standard pigeon-axiom of the weak pigeonhole principle. We say that a node u is a hole-axiom if there exists a hole j and two di�erentpigeons i1; i2, such that xi1;j; xi2;j 2 Ones(u). Note that this is just a standard hole-axiom ofthe pigeon hole principle.In our lower bound proof, we allow the leaves of the graph G to be pigeon-axioms ofany order k (and not only pigeon-axioms of order k = n� as in the usual weak pigeon holeprinciple). That is, we say that G is a Resolution proof for the weak pigeon hole principle ifall its leaves are axioms (i.e., all the leaves of G are either hole-axioms or pigeon-axioms ofsome order). We assume w.l.o.g. that in G, a non-leaf node u is never an axiom (otherwise,we can just disconnect the two edges going out from u and hence convert u to a leaf). Inparticular, no non-leaf node is a pigeon-axiom of order k, for any k.One consequence of the assumption that no non-leaf node is a pigeon-axiom of order k isthat if a node u is a pigeon-axiom of order k then jOverk(u)j = mk. This is true because iffor some u we had jOverk(u)j > mk then any node v, such that there is an edge from v tou, would satisfy jOverk(v)j � mk, and hence the non-leaf node v would be a pigeon-axiom oforder k. Therefore, we can assume that for any node u in the graph,jOverk(u)j � mk:For our lower bound proof, we need to re�ne the scale fnkg. For any two integers 0 � k < n�and 0 � l � n� and any node u, de�ne,nk;l = k � n1�� + l � n1�2�:Overk;l(u) = the set of pigeons i such that jZerosi(u)j � nk;l.Note that nk;0 = nk, and nk;n� = nk+1.3.2 The Random AssignmentWe will de�ne a probabilistic assignment Ai;j to the variables xi;j. Unlike in previous lowerbound proofs, one should not interpret the assignment Ai;j as a \random restriction" of theResolution proof. The assignment Ai;j will be used in a di�erent way. The assignment Ai;jis chosen at random according to some speci�c probability distribution, de�ned below. First,de�ne,nHoleskon�k=1 = a random partition of Holes into n� sets of size n1�� each.8

That is, we partition Holes into n� sets of size n1�� each. The intuition is that the set of holesHolesk will be used \against" pigeon-axioms of order k. For each 1 � k � n�, de�ne,nHolesk;lon�l=1 = a random partition of Holesk into n� sets of size n1�2� each.That is, we further partition each set Holesk into n� sets of size n1�2� each. Altogether, theset Holes was partitioned into n2� sets of size n1�2� each. We denote by V ariablesk;li the setof variables corresponding to the ith pigeon and holes in Holesk;l. That is,Variablesk;li = fxi;jjj 2 Holesk;lg.Next, we would like to de�ne for every 1 � k � n�, a set of pigeons Pigeonsk. For mk � n�,we would like the set Pigeonsk to contain all pigeons. For larger values of mk, we would likeeach pigeon to be chosen (independently, at random) with a certain probability. For every1 � k � n�, de�ne,pk = min h1; n�mk i.Pigeonsk = a random subset of Pigeons, such that each pigeon is chosen (inde-pendently, at random) with probability pk.For every pigeon i, and every 1 � k � n� and 1 � l � n�, de�ne the subset AOnesk;li of theset V ariablesk;li , in the following way.AOnesk;li = ( a random subset of size n1�6� of V ariablesk;li if i 2 Pigeonsk; if i 62 PigeonskThe set AOnes is now de�ned to be the union of all the sets AOnesk;li , and the set AZerosis de�ned to be the complement of AOnes. The assignment Ai;j is de�ned by, Ai;j = 1 i�xi;j 2 AOnes. That is,AOnes = Si;k;lAOnesk;li .AZeros = V ariables n AOnes.Ai;j = ( 1 if xi;j 2 AOnes0 if xi;j 2 AZeros9

3.3 Properties of the AssignmentFor our lower bound proof, we do not need the assignment Ai;j, and the sets that wereinvolved in de�ning it, to be probabilistic. We just need them to satisfy certain properties.These properties are satis�ed (with high probability) by the probabilistic construction thatwe de�ned, but we will only need one assignment (and sets) that satisfy the properties. Theproperties that we will need are summarized in the following claim.Claim 3.1 With exponentially high probability, all the following are satis�ed, for every pigeoni, every hole j, every nodes u; v, and every 1 � k � n� and 1 � l � n�.1. If j 2 Holesk and i 62 Pigeonsk then Ai;j = 0:2. If i 2 Pigeonsk and jZerosi(u)j � jZerosi(v)j � n1�2� then���[Zerosi(u) n Zerosi(v)] \ AOnesk;li ��� > n1�8�=2:3. If i1 and i2 are two di�erent pigeons then���nj 0 2 Holesk;l j Ai1;j0 = 1 and Ai2;j0 = 1o��� < 2n1�10�:4. If jOnes(u)j � n� then Ones(u) \ AZeros 6= ;:5. If u is a pigeon-axiom of order k thenPigeonsk \ Overk(u) 6= ;:6. For any u, ���Pigeonsk \Overk�1(u)��� < 10n�:Proof:Recall that the number of pigeons and the number of nodes are both bounded by 2n�=100.The number of holes is n. Recall that we assume that � = 1=100, and n� � 1000. For theproof of the claim, we just have to verify that (for speci�c objects, i; j; k; l; u; v; i1; i2), therequirement in each one of the properties is falsi�ed with exponentially small probability (say,with probability smaller than 2�n�=25). This will usually follow by the standard Cherno�-Hoe�ding bounds or by other simple probabilistic arguments. Let us analyze the propertiesone by one.Property 1:By the de�nition of AOnesk;li , the requirement in this property is always satis�ed.10

Property 2:Zerosi(u) n Zerosi(v) is a �xed subset of V ariablesi of size � n1�2�. Assume w.l.o.g. thatthe size of Zerosi(u) n Zerosi(v) is exactly n1�2�. Since i 2 Pigeonsk, the set AOnesk;li is arandom subset of V ariablesi of size exactly n1�6�. Hence, the intersection[Zerosi(u) n Zerosi(v)] \ AOnesk;liis of expected size n1�8�, and by the standard Cherno�-Hoe�ding bounds the actual size ofthe intersection is very close to n1�8�, with high probability. In particular, the probabilitythat the size of the intersection is � n1�8�=2 is exponentially small (and in particular, smallerthan 2�n�=25).Property 3:Denote, Hi1 = nj 0 2 Holesk;l j Ai1;j0 = 1o ;and, Hi2 = nj 0 2 Holesk;l j Ai2;j0 = 1o :Then, nj 0 2 Holesk;l j Ai1;j0 = 1 and Ai2;j0 = 1o = Hi1 \Hi2 :If either i1 62 Pigeonsk or i2 62 Pigeonsk then Hi1 \ Hi2 is empty. If both i1; i2 2 Pigeonskthen Hi1 and Hi2 are both random subsets of Holesk;l of size n1�6� each. Recall that Holesk;lis a set of size n1�2�. Hence, the intersection Hi1 \ Hi2 is of expected size n1�10�, and bythe standard Cherno�-Hoe�ding bounds the actual size of the intersection is very close ton1�10�, with high probability. In particular, the probability that the size of the intersection is� 2n1�10� is exponentially small (and in particular, smaller than 2�n�=25).Property 4:Denote s = jOnes(u)j, and assume w.l.o.g. that s is exactly n�. Let xi1;j1; xi2;j2; :::; xis;js bethe s variables in Ones(u). It is easy to verify that for any 1 � t � s, the probability forAit;jt = 1 is smaler than 1=2, even under the condition that Ai1;j1; :::; Ait�1;jt�1 are all 1. Hence,the probability that Ai1;j1; :::; Ais;js are all 1 is smaller than 2�n�.Property 5:Overk(u) is a set of mk pigeons. If mk � n� then each one of these pigeons is in Pigeonskwith probability 1. Otherwise, the probability for each one of these pigeons to be in Pigeonskis n�=mk, and hence the probability that none of them is in Pigeonsk is�1� n�mk�mk < 2�n�:Property 6:As we have seen, Overk�1(u) is a set of at most mk�1 = 2mk pigeons. Assume w.l.o.g. thatOverk�1(u) is a set of exactly 2mk pigeons. If mk � n� then 2mk � 2n� and the requirement isobviously satis�ed. Otherwise, each one of these 2mk pigeons is in Pigeonsk with probabilityn�=mk. Hence, the intersection Pigeonsk \ Overk�1(u) is of expected size 2n�, and by the11

standard Cherno�-Hoe�ding bounds the actual size of the intersection is very close to 2n�,with high probability. In particular, the probability that the size of the intersection is � 10n�is exponentially small (and in particular, smaller than 2�n�=25). 23.4 The Adversary StrategyIn this subsection, we give the proof of Theorem 3.1, given one lemma (the main lemma).With high probability, all the properties in Claim 3.1 are satis�ed. Hence, we can �x theassignment Ai;j (and all the sets involved in de�ning it, such as, Pigeonsk, Holesk;l, etc') tosome �xed values that satisfy all these properties. Thus, from now on, we assume that theassignment Ai;j (and all the sets involved in de�ning it) are �xed (and are not probabilisticany more), and that all the properties in Claim 3.1 are satis�ed.For every non-leaf node u of the graph G, we de�ne a value Answer(u) 2 f0; 1g. We thinkof Answer(u) as an adversary \answer" for the \query" Label(u). The answer Answer(u)depends on the assignment Ai;j and the sets Holesk;l.Assume that Label(u) = xi;j, and j 2 Holesk;l. We de�ne Answer(u) in the fol-lowing way:1) If i 62 Overk�1;l�1(u) Answer(u) = 02) If 9i0 6= i s.t. xi0;j 2 Ones(u) Answer(u) = 03) Otherwise; Answer(u) = Ai;jThat is, the answer is automatically 0 if i 62 Overk�1;l�1(u), or if there exists i0 6= i suchthat xi0;j 2 Ones(u). Otherwise, the answer is the value of Ai;j. Given the values Answer(u)(for every non-leaf node u), we de�ne a path (called Path) on the graph G. The path startsfrom the root of G and in each step it follows the edge labelled by Answer(u), where u is thecurrent node. We denote by Leaf the leaf reached by the path Path. That is,Path = the path that starts from Root, and that satis�es that for every (non-leaf)node u on the path, the path contains the edge that goes out from u and is labelledby Answer(u).Leaf = the leaf reached by Path.Lemma 3.1 (Main Lemma) For any 1 � k � n�, and any node u on the path Path,Pigeonsk \ Overk(u) = ;:Lemma 3.1 is proved in the next subsection. Let us show how the proof of Theorem 3.1 followsfrom Lemma 3.1. 12

Proof of Theorem 3.1:By Lemma 3.1 and by Property 5 of Claim 3.1, no node u on Path is a pigeon-axiom (ofany order k). By the de�nition of Answer(u), if there exists i0 6= i such that xi0;j 2 Ones(u)then Answer(u) = 0. Hence, for no node u on Path we will have that both xi;j and xi0;j arein Ones(u). That is, no node u on Path is a hole-axiom. In particular, Leaf is neither apigeon-axiom (of any order k) nor a hole-axiom, in contradiction to the fact that all leaves ofthe graph G must be axioms. 23.5 Pigeon-SectionsIn this subsection, we give the proof of Lemma 3.1, given one claim (the main claim). For anynode u on Path, de�ne,u+ = the successor of u on Path.u� = the predecessor of u on Path.(u+ is unde�ned for u = Leaf , and u� is unde�ned for u = Root). For two nodes v � w onPath, denote by [v; w] the section of nodes (on Path) between them. That is,[v;w] = the set of nodes u on Path, such that, v � u � w.For a pigeon i 2 Pigeonsk, we will be interested in maximal sections on Path, such that, forevery node u in the section, i 2 Overk�1(u). For 1 � k � n�, we de�ne a pigeon-section oftype k, and the set PigSeck (of all these pigeon-sections), in the following way.(i; [v;w]) is a pigeon-section of type k if all the following are satis�ed:1. i 2 Pigeonsk, and v � w are nodes on Path.2. For any node u 2 [v; w], we have i 2 Overk�1(u).3. The section [v; w] is maximal with this property. That is, if v 6= Root theni 62 Overk�1(v�) and if w 6= Leaf then i 62 Overk�1(w+).PigSeck = the set of all pigeon-sections of type k.We will further re�ne the categorization of pigeon-sections into types. We say that a pigeon-section of type k is of type (k; l) if the section [v; w] contains a node u such that i 2Overk�1;l�1(u), and we de�ne the set PigSeck;l to be the set of all these pigeon-sections.That is, for 1 � k � n� and 1 � l � n� + 1,(i; [v;w]) is a pigeon-section of type (k; l) if all the following are satis�ed:13

1. (i; [v; w]) is a pigeon-section of type k.2. For some node u 2 [v; w], we have i 2 Overk�1;l�1(u).PigSeck;l = the set of all pigeon-sections of type (k; l).Note the asymmetric role of k and l in the de�nition of pigeon-section of type (k; l). Note alsothat PigSeck;1 = PigSeck.Claim 3.2 (Main Claim) For every 1 � k � n� and 1 � l � n�,���PigSeck;l+1��� � 12 � ���PigSeck;l��� :Claim 3.2 is proved in the next subsections. Let us show how the proof of Lemma 3.1 followsfrom Claim 3.2.Proof of Lemma 3.1:Since the number of pigeons and the number of nodes in the graph are both bounded by2n�=100, the number of pigeon-sections of type k is bounded by 2n�=50. That is,���PigSeck;1��� = ���PigSeck��� � 2n�=50:Hence, by n� applications of Claim 3.2,���PigSeck;n�+1��� � 2�n� � ���PigSeck;1��� � 2�n� � 2n�=50 < 1;and since ���PigSeck;n�+1��� is integer, ���PigSeck;n�+1��� = 0:That is, there are no pigeon-sections of type (k; n� + 1).Assume for a contradiction to the statement of the lemma that for some node u on Path,Pigeonsk \ Overk(u) 6= ;:Then, since Overk(u) = Overk�1;n�(u);there exists i 2 Pigeonsk, such that, i 2 Overk�1;n�(u):Denote by [v; w] the largest section (on Path) that contains u, and such that for everyu0 2 [v; w] we have i 2 Overk�1(u0) (such a section exists because i 2 Overk�1(u)). Then,(i; [v; w]) is a pigeon-section of type (k; n� + 1), in contradiction to the fact that there are nosuch pigeon-sections. 214

3.6 ForcingLet u be a node such that Label(u) = xi;j, and such that i 2 Pigeonsk and j 2 Holesk;l (forsome 1 � k � n� and 1 � l � n�). Recall that Answer(u) is 0 if there exists i0 6= i such thatxi0;j 2 Ones(u). If, in addition, i 2 Overk�1;l�1(u) and Ai;j = 1 we say that xi;j is forced to 0at the node u by xi0;j. (Recall that if i 62 Overk�1;l�1(u) or Ai;j = 0 then Answer(u) wouldbe 0 anyways, so we do not consider it as \forcing"). That is,Assume that Label(u) = xi;j, and j 2 Holesk;l. We say that xi;j is forced to 0at the node u by xi0;j if all the following are satis�ed:1. i 2 Pigeonsk and Ai;j = 1.2. i 2 Overk�1;l�1(u).3. xi0;j 2 Ones(u).Assume that xi;j is forced to 0 by xi0;j at a node u on Path. Then, since i 2 Pigeonsk andi 2 Overk�1;l�1(u), there exists a (unique) pigeon-section (i; [v; w]) of type (k; l) such thatu 2 [v; w]. (To see this, just denote by [v; w] the largest section on Path that contains u,and such that for every u 2 [v; w] we have i 2 Overk�1(u), such a section exists becausei 2 Overk�1(u). Then, (i; [v; w]) is a pigeon-section of type (k; l)).Consider the nodes on Path from the root to u, that is, the nodes in [Root; u]. Denote byu0 the last node in [Root; u], such that, Label(u0) = xi0;j. Since xi0;j 2 Ones(u), we know thatAnswer(u0) is 1. Therefore, by the de�nition ofAnswer(u0), we know that i0 2 Overk�1;l�1(u0),and by Property 1 of Claim 3.1 we know that i0 2 Pigeonsk (otherwise, Ai0;j would be 0, andhence Answer(u0) would be 0 as well). By the same argument as before, there exists a (unique)pigeon-section (i0; [v0; w0]) of type (k; l) such that u0 2 [v0; w0]. We categorize the \forcing" totypes according to the relations between the nodes u0; v; w; w0, as follows.Let u be a node on Path. Assume that Label(u) = xi;j, and j 2 Holesk;l. Assumethat xi;j is forced to 0 by xi0;j at the node u. Let u0 be the last node in [Root; u],such that Label(u0) = xi0;j. Let (i; [v; w]) be the pigeon-section of type (k; l) suchthat u 2 [v; w], and let (i0; [v0; w0]) be the pigeon-section of type (k; l) such thatu0 2 [v0; w0].1. We say that the forcing is a forcing of type 1 if u0 < v.2. We say that the forcing is a forcing of type 2 if u0 2 [v; w] and w0 � w.3. We say that the forcing is a forcing of type 3 if u0 2 [v; w] and w0 < w.Note that since u0 < u and u 2 [v; w], any forcing is a forcing of one of these three types. Forevery k; l, we would like to count the number of variables forced to 0 at pigeon-sections of type(k; l). In our counting, we would like to count a variable more than once if it is forced to 0 atmore than one pigeon-section. However, we would like to count a variable only once for eachpigeon-section, that is, if the variable is forced to 0 many times at the same pigeon-section wecount it only once. For every, 1 � k � n� and 1 � l � n�, de�ne,15

Forcedk;l = the set of all pairs (xi;j; [v; w]), such that all the following are satis�ed:1. (i; [v; w]) is a pigeon-section of type (k; l).2. j 2 Holesk;l.3. xi;j is forced to 0 at some node u 2 [v; w].Forcedk;l1 = the set of all pairs (xi;j; [v; w]), such that all the following are satis�ed:1. (i; [v; w]) is a pigeon-section of type (k; l).2. j 2 Holesk;l.3. xi;j is forced to 0 at some node u 2 [v; w], and the forcing is type 1.Forcedk;l2 = the set of all pairs (xi;j; [v; w]), such that all the following are satis�ed:1. (i; [v; w]) is a pigeon-section of type (k; l).2. j 2 Holesk;l.3. xi;j is forced to 0 at some node u 2 [v; w], and the forcing is type 2.Forcedk;l3 = the set of all pairs (xi;j; [v; w]), such that all the following are satis�ed:1. (i; [v; w]) is a pigeon-section of type (k; l).2. j 2 Holesk;l.3. xi;j is forced to 0 at some node u 2 [v; w], and the forcing is type 3.3.7 Bounding the Number of Forced VariablesIn this subsection, we give the proof of Claim 3.2. The proof will follow easily by the followingfour claims.Claim 3.3 For every 1 � k � n� and 1 � l � n�,���Forcedk;l1 ��� � ���PigSeck;l��� � n�:Claim 3.4 For every 1 � k � n� and 1 � l � n�,���Forcedk;l2 ��� � ���PigSeck;l��� � 20n1�9�:Claim 3.5 For every 1 � k � n� and 1 � l � n�,���Forcedk;l3 ��� � ���PigSeck;l��� � 20n1�9�:16

Claim 3.6 For every 1 � k � n� and 1 � l � n�,���Forcedk;l��� � ���PigSeck;l+1��� � n1�8�=2:Proof of Claim 3.2:Since any forcing is a forcing of type 1 or type 2 or type 3,���Forcedk;l��� � ���Forcedk;l1 ���+ ���Forcedk;l2 ���+ ���Forcedk;l3 ��� :Hence, the proof follows immediately from Claims 3.3, 3.4, 3.5, 3.6, using the assumptionsthat � = 1=100 and n� � 1000. 2Proof of Claim 3.3:Let (i; [v; w]) be a pigeon-section of type (k; l). Denote,F 1(i;[v;w]) = n(xi;j; [v; w]) 2 Forcedk;l1 o :We will show that for every such (i; [v; w]),���F 1(i;[v;w])��� � n�;(and hence the claim follows).Fix (i; [v; w]) to be a pigeon-section of type (k; l). For every (xi;j; [v; w]) 2 F 1(i;[v;w]), weknow that xi;j is forced to 0 at some node u 2 [v; w] by some xi0;j, and the forcing is type 1.Hence, the last node u0 2 [Root; u], such that Label(u0) = xi0;j, satis�es u0 < v. That is, xi0;jdoes not appear as Label(u) for any u 2 [v; u], and since xi0;j 2 Ones(u) we conclude thatxi0;j 2 Ones(v). Thus, for every (xi;j; [v; w]) 2 F 1(i;[v;w]), there is (at least one) correspondingxi0;j 2 Ones(v). Hence, ���F 1(i;[v;w])��� � jOnes(v)j:To �nish the proof of the claim, it is enough to show that for every node v on Path,jOnes(v)j � n�:Let v be a node such that jOnes(v)j > n�: We will show that v is not on Path. ByProperty 4 of Claim 3.1, Ones(v) \ AZeros 6= ;:Hence, there exists x~i;~j 2 Ones(v), such that A~i;~j = 0. Hence, for any node ~u such thatLabel(~u) = x~i;~j, we have Answer(~u) = 0. Since Path always follows the edge Answer(~u)(when ~u is the current node), it will never evaluate x~i;~j to 1. Since every path to v evaluatesx~i;~j to 1, we conclude that v is not on Path. 2Proof of Claim 3.4:Let (i; [v; w]) be a pigeon-section of type (k; l). Denote,F 2(i;[v;w]) = n(xi;j; [v; w]) 2 Forcedk;l2 o :17

We will show that for every such (i; [v; w]),���F 2(i;[v;w])��� � 20n1�9�;(and hence the claim follows).Fix (i; [v; w]) to be a pigeon-section of type (k; l). We will count the number of possibilitiesfor (xi;j; [v; w]) 2 F 2(i;[v;w]). For every (xi;j; [v; w]) 2 F 2(i;[v;w]), we know that xi;j is forced to 0 atsome node u 2 [v; w] by some xi0;j, and the forcing is type 2. Therefore, there exists a pigeon-section (i0; [v0; w0]) of type (k; l) such that v0 < u and w0 � w. Thus, w 2 [v0; w0]. Hence, since(i0; [v0; w0]) is a pigeon-section of type k, we know that i0 2 Pigeonsk and i0 2 Overk�1(w).Thus, for every (xi;j; [v; w]) 2 F 2(i;[v;w]), each corresponding xi0;j satis�es that i0 is inPigeonsk \ Overk�1(w):By Property 6 of Claim 3.1, ���Pigeonsk \ Overk�1(w)��� < 10n�;and hence for the pigeon-section (i; [v; w]), the number of possibilities for i0 is bounded by10n�.Since xi;j is forced to 0 at u by xi0;j, we know that Ai;j = 1 (by the de�nition of forcing),and Ai0;j = 1 (since xi0;j 2 Ones(u) and u is on Path, and as in the proof of Claim 3.3 Pathcannot evaluate xi0;j to 1 if Ai0;j = 0). Hence, j is innj 2 Holesk;l j Ai;j = 1 and Ai0;j = 1o :By Property 3 of Claim 3.1,���nj 2 Holesk;l j Ai;j = 1 and Ai0;j = 1o��� < 2n1�10�;and hence for every i0, the number of possibilities for j is bounded by 2n1�10�.Altogether, for the pigeon-section (i; [v; w]), the number of possibilities for i0 is boundedby 10n�, and for every i0 the number of possibilities for j is bounded by 2n1�10�. Hence,���F 2(i;[v;w])��� � 10n� � 2n1�10� = 20n1�9�: 2Proof of Claim 3.5:For every (xi;j; [v; w]) 2 Forcedk;l3 , we know that xi;j is forced to 0 at some node u 2 [v; w] bysome xi0;j, and the forcing is type 3. Let u0 be the last node in [Root; u], such that Label(u0) =xi0;j, and let (i0; [v0; w0]) be the pigeon-section of type (k; l) such that u0 2 [v0; w0]. We will say,in this case, that the pigeon-section (i0; [v0; w0]) is responsible for (xi;j; [v; w]) 2 Forcedk;l3 .18

Thus, for every (xi;j; [v; w]) 2 Forcedk;l3 , there is (at least one) pigeon-section (i0; [v0; w0]) oftype (k; l) responsible for it.Let (i0; [v0; w0]) be a pigeon-section of type (k; l). Denote by F 3(i0;[v0;w0]) the set of all(xi;j; [v; w]) 2 Forcedk;l3 that (i0; [v0; w0]) is responsible for. We will show that for every such(i0; [v0; w0]), ���F 3(i0;[v0;w0])��� � 20n1�9�;and hence, obviously, ���Forcedk;l3 ��� � ���PigSeck;l��� � 20n1�9�:The bound for jF 3(i0;[v0;w0])j is proved in a similar way to the proof of the bound for jF 2(i;[v;w])j,in Claim 3.4.Fix (i0; [v0; w0]) to be a pigeon-section of type (k; l). We will count the number of possi-bilities for (xi;j; [v; w]) 2 F 3(i0;[v0;w0]). For every (xi;j; [v; w]) 2 F 3(i0;[v0;w0]), we know that xi;j isforced to 0 at some node u 2 [v; w] by xi0;j, and the forcing is type 3. We also know thatif u0 is the last node in [Root; u] such that Label(u0) = xi0;j then u0 2 [v0; w0] (by the de�ni-tion of F 3(i0;[v0;w0])). Since the forcing is type 3, we know that u0 2 [v; w] and w0 < w. Thus,w0 2 [v; w]. Hence, since (i; [v; w]) is a pigeon-section of type k, we know that i 2 Pigeonskand i 2 Overk�1(w0). Thus, for every (xi;j; [v; w]) 2 F 3(i0;[v0;w0]), we know that i is inPigeonsk \Overk�1(w0):By Property 6 of Claim 3.1, ���Pigeonsk \Overk�1(w0)��� < 10n�;and hence for the pigeon-section (i0; [v0; w0]), the number of possibilities for i is bounded by10n�.Since xi;j is forced to 0 at u by xi0;j, we know that Ai;j = 1 (by the de�nition of forcing),and Ai0;j = 1 (as in the proof of Claim 3.4). Hence, j is innj 2 Holesk;l j Ai;j = 1 and Ai0;j = 1o :By Property 3 of Claim 3.1,���nj 2 Holesk;l j Ai;j = 1 and Ai0;j = 1o��� < 2n1�10�;and hence for every i, the number of possibilities for j is bounded by 2n1�10�.Altogether, for the pigeon-section (i0; [v0; w0]), the number of possibilities for i is boundedby 10n�, and for every i the number of possibilities for j is bounded by 2n1�10�. For every i, thenumber of possibilities for [v; w] is (at most) one, because there is (at most) one pigeon-section(i; [v; w]) of type (k; l) such that w0 2 [v; w]. Hence,���F 3(i0;[v0;w0])��� � 10n� � 2n1�10� = 20n1�9�:19

2Proof of Claim 3.6:Let (i; [v; w]) be a pigeon-section of type (k; l + 1). Then, obviously, (i; [v; w]) is a pigeon-section of type (k; l) as well. Denote,F(i;[v;w]) = n(xi;j; [v; w]) 2 Forcedk;lo :We will show that for every such (i; [v; w]),���F(i;[v;w])��� � n1�8�=2;(and hence the claim follows).Fix (i; [v; w]) to be a pigeon-section of type (k; l + 1). Then, for some node u 2 [v; w], wehave jZerosi(u)j � nk�1;l:For simplicity of the notations, assume that v is not the root, and hence v� exists. Lets be the last node in [v�; u], such that, i 62 Overk�1;l�1(s) (such an s exists because i 62Overk�1;l�1(v�)). Denote t = s+. Then,jZerosi(t)j = nk�1;l�1:(This is true because by the de�nition of s we know that i 2 Overk�1;l�1(t), and if we hadjZerosi(t)j > nk�1;l�1 then we would have had jZerosi(s)j � nk�1;l�1, in contradiction to thede�nition of s). Thus, jZerosi(u)j � jZerosi(t)j � n1�2�;and hence, by Property 2 of Claim 3.1,���[Zerosi(u) n Zerosi(t)] \ AOnesk;li ��� > n1�8�=2:To �nish the proof of the claim, it is enough to show thatxi;j 2 [Zerosi(u) n Zerosi(t)] \ AOnesk;li =) (xi;j; [v; w]) 2 F(i;[v;w]):Let xi;j 2 [Zerosi(u) n Zerosi(t)] \ AOnesk;li : First note that since xi;j 2 AOnesk;li , weknow that i 2 Pigeonsk and j 2 Holesk;l. Since xi;j 2 [Zerosi(u) nZerosi(t)], there is a nodez 2 [t; u], such that, Label(z) = xi;j and Answer(z) = 0. Since xi;j 2 AOnesk;li , we know thatAi;j = 1, and since z 2 [t; u], we know that i 2 Overk�1;l�1(z). Hence, Answer(z) is 0 only ifxi;j is forced to 0 at the node z. Thus, xi;j is forced to 0 at the node z, and since (i; [v; w]) isa pigeon-section of type (k; l), we conclude that (xi;j; [v; w]) 2 F(i;[v;w]). 220

4 Lower Bounds for NP 6� P=polyIn this section, we will show that a certain propositional formulation of the statement SAT 62P=poly does not have polynomial size Resolution proofs. Our proof is a version of the onegiven in [Razb3] and is included here for completeness and since this version of the argumenthas never appeared before. For the proof, we will need a lower bound for Resolution proofsfor the, so called, weak onto pigeon hole principle.4.1 The Weak Onto Pigeonhole PrincipleThe propositional weak onto pigeon hole principle,WOPHPmn , is a version of the weak pigeonhole principle that requires (in addition) that in each hole there is at least one pigeon. Theunderlying variables are, as before, xi;j, where 1 � i � m and 1 � j � n. The unsatis�ableformula :WOPHPmn is expressed in conjunctive normal form (CNF) as the conjunction ofthe m pigeon clauses and the �m2� � n hole clauses of the original weak pigeon hole principle,and n additional clauses that we call onto clauses. For every 1 � j � n, we have an ontoclauses, (x1;j _ : : :_ xm;j), stating that some pigeon is mapped to hole j. We refer to the ontoclauses also as onto axioms.The weak onto pigeon hole principle is a weaker principle than the weak pigeon holeprinciple. Hence, in some proof systems it may have shorter proofs. Nevertheless, it iswell known that in Resolution the two principles are equivalent up to a factor polynomialin m [BuP]. That is, any Resolution proof of length s for WOPHPmn can be converted intoa Resolution proof of length s � poly(m) for WPHPmn (where poly(m) is a small polynomialin m, say, smaller than 100 �m10).Corollary 4.1 For any m � n + 1, any Resolution proof for the tautology WOPHPmn is oflength 2(n�) (where � = 1=100).4.2 Propositional Formulation of SAT 62 P=polyLet f : f0; 1gd ! f0; 1g be a Boolean function. For example, we can take f = SAT , whereSAT : f0; 1gd ! f0; 1g is the satis�ability function. We assume that we are given the truthtable of f . Let t � 2d be some integer. We think of t as a large polynomial in d, say t = d1000.As mentioned in the introduction, we would like to formulate the following statement (inthe variables ~Z):~Z is (an encoding of) a Boolean circuit of size t =)~Z does not compute the function f .Our propositional formulation of the statement will be a DNF formula of length 2O(d) (i.e.,its negation is a CNF formula of that length). The variables ~Z will include the (topological)description of an unbounded fan-in Boolean circuit, as well as the value that each gate in the21

circuit gets on each input for the circuit. As mentioned in the introduction, our argumentseems to be robust and the exact encoding of the circuit seems to be unimportant (as longas the circuit is of unbounded fan-in). For simplicity of the notations, our circuit will nothave negation gates, and all negations will appear only on the input variables (note that anycircuit can be converted into such a circuit with only a constant factor increase in the sizeof the circuit). This is done here only for simplicity, and doesn't change the argument in asubstantial way.For simplicity of the notations, we partition the variables ~Z into groups, and give themnew names as follows:1. For every 1 � r � t, we have a variable GATE[r], saying whether the rth gate of thecircuit is and AND gate or an OR gate. We will use r = 0 to represent an AND gateand r = 1 to represent an OR gate.2. For every 1 � r � t and every 1 � q < r, we have a variable WIRE[r;q], sayingwhether or not the qth gate of the circuit is wired to the rth gate of the circuit.3. For every 1 � r � t and every 1 � k � d, we have a variable VAR:WIRE[r;k; 1],saying whether or not the kth input variable is wired to the rth gate of the circuit.4. For every 1 � r � t and every 1 � k � d, we have a variable VAR:WIRE[r;k; 0],saying whether or not the negation of the kth input variable is wired to the rth gate ofthe circuit.5. For every 1 � r � t and every w 2 f0; 1gd, we have a variable VALUE[r;w], giving thevalue that the rth gate of the circuit gets on the input w 2 f0; 1gd.For every such d; t; f , we will have a CNF formula CIRCUITd;t;f [~Z] that formulates thestatement:~Z is (an encoding of) a Boolean circuit of size t, and~Z computes the function f .The formula CIRCUITd;t;f [~Z] will be the conjunction of the following clauses:1. For every 1 � r � t and every 1 � q < r and every w 2 f0; 1gd, we will have a clauses,GATE[r] ^WIRE[r;q] ^VALUE[q;w]! VALUE[r;w]:2. For every 1 � r � t and every 1 � k � d and every w 2 f0; 1gd, we will have a clauses,GATE[r] ^VAR:WIRE[r;k;wk]! VALUE[r;w]:3. For every 1 � r � t and every w 2 f0; 1gd, we will have a clauses,22

GATE[r] ^VALUE[r;w]!Sdk=1VAR:WIRE[r;k;wk] _ Sr�1q=1(WIRE[r;q] ^VALUE[q;w]):4. For every 1 � r � t and every 1 � q < r and every w 2 f0; 1gd, we will have a clauses,:GATE[r] ^WIRE[r;q] ^ :VALUE[q;w]! :VALUE[r;w]:5. For every 1 � r � t and every 1 � k � d and every w 2 f0; 1gd, we will have a clauses,:GATE[r] ^VAR:WIRE[r;k; (1�wk)]! :VALUE[r;w]:6. For every 1 � r � t and every w 2 f0; 1gd, we will have a clauses,:GATE[r] ^ :VALUE[r;w]!Sdk=1VAR:WIRE[r;k; (1�wk)] _ Sr�1q=1(WIRE[r;q] ^ :VALUE[q;w]):7. For every w 2 f0; 1gd, we will have a clauses,VALUE[t;w] = f(w):4.3 Reduction to the Weak Onto Pigeonhole PrincipleWe will now show that for any d; f , if t is a large enough polynomial in d (say, t > d1000) thenthere are no short Resolution refutations for the formula CIRCUITd;t;f [~Z]. This will be doneby a reduction to the weak onto pigeon hole principle.Theorem 4.1 For any d; f and any t � 2d, any Resolution refutation for the formulaCIRCUITd;t;f [~Z] is of length larger than 2(t�) (where � = 1=100 is the constant from Corol-lary 4.1).(Hence, for (say) t > d1000, the length of the refutation is at least 2(d10), which is super-polynomial in 2O(d)).Proof:Let T be the set of all w 2 f0; 1gd such that f(w) = 1, and denote m = jT j. Let REFbe any Resolution refutation for the formula CIRCUITd;t;f [~Z]. We will convert REF into aResolution refutation for the formula :WOPHPmt�1.Let us �rst �x some of the variables in ~Z, as follows: We �x our Boolean circuit to be aDNF. That is, the top gate of the circuit is an OR gate and all the other t� 1 gates are ANDgates and are connected to the top gate. We �x each term in the DNF to be a \full-term",that is, it is a conjunction of d literals, where each one of the d input variables appears exactlyonce in each term (with or without negation). We �x the output of the circuit to be f . We�x the output of every gate to be 0 on every input not in T . For simplicity of the notations,we will also rename some of the variables by (the new notations) yr;k and xr;w. Formally, we�x (and rename) as follows:1. GATE[t] = 1. 23

2. For every 1 � q < t,WIRE[t; q] = 1.3. For every 1 � k � d,V AR:WIRE[t; k; 0] = V AR:WIRE[t; k; 1] = 0.4. For every 1 � r < t,GATE[r] = 0.5. For every 1 � r < t and every 1 � q < r,WIRE[r; q] = 0.6. For every 1 � r < t and every 1 � k � d,V AR:WIRE[r; k; 1] = :V AR:WIRE[r; k; 0] = yr;k(where yr;k is a new notation, introduced for simplicity).7. For every w 2 f0; 1gd,V ALUE[t; w] = f(w).8. For every 1 � r < t and every w 2 f0; 1gd n T ,V ALUE[r; w] = 0.9. For every 1 � r < t and every w 2 T ,V ALUE[r; w] = xw;r(where xw;r is a new notation, introduced for simplicity).The remained variables in REF are the variables xw;r (for every 1 � r < t and w 2 T ),and the variables yr;k (for every 1 � r < t and 1 � k � d). We will replace in REF each(positive) appearance of yr;k by [w2T;wk=1 xw;rand each appearance of :yr;k by [w2T;wk=0 xw;r:Our goal is to convert REF into a Resolution refutation for the formula :WOPHPmt�1 inthe variables fxw;rg. Let us �rst check what happened to the axioms of the original refutationREF (i.e., the clauses of CIRCUITd;t;f [~Z]). We say that an axiom (i.e., a clauses) becametrivial if one of its literals was �xed to 1 or if it contains a variable and its negation.All clauses of type 7 are now trivial. 24

For r = t, all clauses of types 1,2,4,5,6 are now trivial, as well as clauses of type 3 forw 2 f0; 1gd n T . Clauses of type 3 for r = t and w 2 T turned into the clauses,t�1[q=1 xw;q;which are just the pigeon axioms of WOPHPmt�1.For 1 � r < t, all clauses of types 1,2,3,4,5 are now trivial. Clauses of type 6 for 1 � r < t(and any w) turned into the clauses, [w2T xw;r;which are just the onto axioms of WOPHPmt�1.Let us now check what happened to the inferences of the original refutation REF . Eachtime that a variable other than yr;k was resolved upon, the inference is clearly still valid. Eachtime that a variable yr;k was resolved upon, we now have an inference of the form0@ [w2T;wk=1 xw;r1A _ A ; 0@ [w2T;wk=0 xw;r1A _ B �! A _B:This is not a valid Resolution inference. Nevertheless, it is well known and easy to show (seefor example [BuP], Section 3) that such an inference can be obtained as a sequence of poly(m)valid Resolution inferences, using the hole axioms of WOPHPmt�1 (where poly(m) is a smallpolynomial in m, say smaller than 5m5).Altogether, we obtained a Resolution refutation for :WOPHPmt�1, of length at most 5m5times the original length of REF . The proof of the theorem hence follows by Corollary 4.1. 2AcknowledgmentI would like to thank Toni Pitassi for years of collaboration that led to this work, and ToniPitassi and Sasha Razborov for helpful conversations about the content of Section 4. I wouldlike to thank Avi Wigderson and the other organizers and participants of the special complexityyear at the Institute for Advanced Study (2000-2001) for the great (and very special) yearthat we all had there. Finally, I would like to thank the Small World Caf�e, where most of thisresearch was done. Sometimes, the small things make the di�erence.References[BeP] Beame, P., and Pitassi, T., \Simpli�ed and improved resolution lower bounds," Foun-dations of Computer Science, 1996, pp. 274-282.25

[BuP] Buss, S., and Pitassi, T., \Resolution and the weak pigeonhole principle," Springer-Verlag Lecture Notes in Computer Science, Publications of selected papers presentedat Proceedings from Computer Science Logic 1997.[BSW] Ben-Sasson, E., andWigderson, A., \Short proofs are narrow{resolution made simple,"Symposium on Theory of Computing, 1999, pp. 517-526.[BT] Buss, S., and Turan, G., \Resolution proofs of generalized pigeonhole principles," The-oretical Computer Science, vol. 62, 1988, pp. 311-317.[Hak] Haken, A. \The intractability of resolution," Theoretical Computer Science, vol. 39,1985, pp. 297-308.[Pud] Pudlak, P. \Proofs as games," American Math. Monthly, June-July 2000, pp. 541-550.[PR] Pitassi, T., and Raz, R., \Regular resolution lower bounds for the weak pigeonholeprinciple," Symposium on Theory of Computing, 2001.[Razb1] Razborov, A., \Bounded arithmetic and lower bounds in Boolean complexity", Fea-sible Mathematics II. Progress in Computer Science and Applied Logic, vol. 13, 1995,pp. 344-386.[Razb2] Razborov, A., \Lower bounds for propositional proofs and independence results inBounded Arithmetic", Proceedings of the 23rd ICALP, Lecture Notes in ComputerScience, vol. 1099, 1996, pp. 48-62.[Razb3] Razborov, A., \Lower bounds for the polynomial calculus," Computational Complex-ity, vol. 7, 1998, pp. 291-324.[Razb4] Razborov, A., \Improved resolution lower bounds for the weak pigeonhole principle",manuscript 2001.[Razb5] Razborov, A., \Resolution lower bounds for the weak functional pigeonhole princi-ple", manuscript 2001.[Razb6] Razborov, A., \Resolution Lower Bounds for Perfect Matching Principles",manuscript 2001, to appear in Proc. of the 17th IEEE Conference on ComputationalComplexity.[RWY] Razborov, A., Wigderson, A., and Yao, A., \Read-once branching programs, rectan-gular proofs of the pigeonhole principle, and the transversal calculus," Symposium onTheory of Computing, 1997, pp. 739-748.[Urq] Urquhart, A., \Hard examples for resolution," Journal of ACM, vol. 34, 1987, pp.209-219.26