HEORY T OF C OMPUTING, Volume 3 (2007), pp. 2543 http://theoryofcomputing.org
Easily refutable subformulas of large random 3CNF formulas
Uriel Feige Eran Ofek
Received: May 2, 2006; published: February 9, 2007.
Abstract: A simple nonconstructive argument shows that most 3CNF formulas with cn clauses (where c is a sufciently large constant) are not satisable. It is an open question whether there is an efcient refutation algorithm that for most formulas with cn clauses proves that they are not satisable. We present a polynomial time algorithm that for most 3CNF formulas with cn3/2 clauses (where c is a sufciently large constant) nds a subformula with (c2 n) clauses and then uses spectral techniques to prove that this subformula is not satisable (and hence that the original formula is not satisable). Previously, it was only known how to efciently certify the unsatisability of random 3CNF formulas with at least poly(log(n)) n3/2 clauses. Our algorithm is simple enough to run in practice. We present some experimental results. ACM Classication: F.2.2 AMS Classication: 68Q17,68Q25 Key words and phrases: proof complexity, average case analysis, Boolean formula, 3CNF, refutation, spectral methods
1
Introduction
A 3CNF formula over n variables is a set of m clauses, each one contains exactly 3 literals of three different variables. A formula is satisable if there exists an assignment to its n variables such that in each clause there is at least one literal whose value is true. The problem of deciding whether an input 3CNF formula is satisable is NP-hard, but this does not rule out the possibility of designing a good
Authors retain copyright to their work and grant Theory of Computing unlimited rights to publish the work electronically and in hard copy. Use of the work is permitted as long as the author(s) and the journal are properly acknowledged. For the detailed copyright statement, see http://theoryofcomputing.org/copyright.html.
c 2007 Uriel Feige and Eran Ofek
DOI: 10.4086/toc.2007.v003a002
U RIEL F EIGE AND E RAN O FEK
heuristic for it. A heuristic for satisability may try to nd a satisfying assignment for an input formula , in case one exists. A refutation heuristic may try to prove that no satisfying assignment exists. In this paper we present an algorithm which tries to refute an input formula . The algorithm has one sided error, in the sense that it will never say unsatisable on a satisable formula, but for some unsatisable formulas it will fail to output unsatisable. It then follows that for a formula on which the algorithm outputs unsatisable, its execution on is a witness for the unsatisability of . How does one measure the quality of a refutation heuristic? A possible test may be to check how good the heuristic is on a random input. But then, how do we generate a random unsatisable formula? To answer this question we review some known properties of random 3CNF formulas. The satisability property has the following interesting threshold behavior. Let be a random 3CNF formula with n variables and cn clauses (each new clause is chosen independently and uniformly from the set of all possible clauses). As the parameter c is increased, it becomes less likely that is satisable, as there are more constraints to satisfy. In [8] it is shown that there exists cn such that for c < cn (1 ) almost surely is satisable, and for c > cn (1 + ), is almost surely unsatisable (for some which tends to zero as n increases). It is also known that 3.52 < cn < 4.596 [14, 12, 13]. We will use random formulas with cn clauses (for c > cn (1 + )) to measure the performance of a refutation heuristic. Specically, the refutation heuristic is considered good if for some c > (1 + )cn it almost surely proves that a random formula with cn clauses is unsatisable. Notice that for any n, as c is increased (for c > cn (1 + )), the algorithmic problem of refutation becomes less difcult since we can always ignore a xed fraction of the clauses. The following question is still open: how small can c be so that there is still a polynomial time algorithm which almost surely refutes random 3CNF formulas with cn clauses (c may also be an increasing function of n). A possible approach for refuting a formula is to nd a resolution proof for the unsatisability of . In this approach one derives new clauses implied by by combining pairs of clauses in which one clause contains a variable and the other clause contains the negation of this variable. Any satisfying assignment must satisfy at least one of the remaining literals contained in the two clauses, and hence the collection of these literals is a CNF clause implied by . A sequence of iterations of this resolution step that eventually generates the empty clause is a proof that is not satisable. Chv tal and Szemer di [3] a e proved that a resolution proof of a random 3CNF formula with linear number of clauses is almost surely of exponential size. A result of a similar avor for denser formulas was given by Ben-Sasson and Wigderson [2] who showed that a random formula with n3/2 clauses almost surely requires a resolution /(1) ) proof of size 2(n . These lower bounds imply that nding a resolution proof for a random formula is computationally inefcient. A simple refutation algorithm can be used to refute random formula with n2 clauses. This is done by considering only those clauses that contain a particular variable x. Fixing x to be true leaves about half of the selected clauses as a random 2CNF formula with roughly 3n/2 clauses. A 2CNF formula with this number of clauses is almost surely not satisable. Moreover, any polynomial time algorithm for 2SAT can be used to certify that this particular sub-formula is not satisable. The same can be done when xing x to be false, implying that no matter how x is assigned, the formula cannot be satised. A new approach, introduced by Goerdt and Krivelevich in [10], gave a signicant improvement and reduced the bound to log7 n nk clauses for efcient refutation of 2kCNF formulas. This approach was later extended in [9] and [11] to handle also random 3CNF formulas with n3/2+ and poly(log n) n3/2
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clauses, respectively. In [5, 7] it is shown how to efciently refute a random 2kCNF instances with at least cnk clauses. The difculty of nding refutation algorithms for formulas with linearly many clauses may lead one to assume that no such algorithm exists. It is shown in [6] that this assumption (that there is no polynomial time refutation heuristic that works for most 3CNF formulas with cn clauses, where c is an arbitrarily large constant) implies that certain combinatorial optimization problems (like minimum graph bisection, the 2-catalog segmentation problem, and others) have no polynomial time approximation schemes. It is an open question whether it is NP-hard to approximate these problems arbitrarily well, though further evidence that these problems are indeed hard to approximate is given in [15]. Our refutation algorithm is based on techniques similar to those used in earlier work, such as [9, 5, 6]. We use these techniques in a different way, resulting in an algorithm that is easier to implement, easier to analyze, and works at lower densities than previous algorithms. For example, both the algorithms in [9, 11] and our algorithm perform eigenvalue computations on some random matrices derived from the random input formula . However, our matrices are much smaller (of order n rather than n2 ), making the computational task easier. Moreover, the structure of our matrices is simpler, making the analysis of our algorithm simpler, and easier to apply also to formulas with fewer clauses than those in [9, 11]. As a result of this simplicity, we can show that our algorithm refutes formulas with cn3/2 clauses, whereas the algorithms given in [9] and [11] are claimed only to refute formulas with (n3/2+ ) and (poly(log n) n3/2 ) clauses, respectively. An implementation of our algorithm refuted a random formula with n = 50000 variables and 27335932 < 3n3/2 clauses (see details in Section 5). In some other respects, our algorithm is more limited then the algorithms in [9, 11]. An algorithm is said to provide strong refutation if it shows not only that the input 3CNF formula is not satisable, but also that every assignment to the variables fails to satisfy a constant fraction of the clauses. Our refutation algorithm does not provide a strong refutation. The problem of strong refutation was addressed in [4], where it was shown that variations of the algorithms of [9, 11] can strongly refute random 3CNF formulas with at least log6 (n) n3/2 clauses. The ability to perform strong refutation is an important issue, and its relation to approximability is discussed in [6].
2
2.1
Preliminaries
The random model
Denition 2.1. A clause is a 3-tuple of literals that belong to three different variables. The set Cn is the set of all possible clauses over n xed variables (there are 2n 2(n 1) 2(n 2) such clauses). We use the following model for generating the random formula . Denition 2.2. A 3CNF formula is generated by choosing m clauses from Cn independently at random m with repetitions. Such a random formula is denoted by R Cn . Although we concentrate on a specic random model for generating random formulas, our algorithm succeeds also on other related random models. For example the formula can be a random set of cn3/2 distinct clauses.
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2.2
Notation
a(n) We write a(n) b(n) if limn b(n) 1. We use the term w.h.p. (with high probability) to denote a sequence of probabilities that converges to 1 as n increases.
2.3
Efcient certication of a property
An important concept which will be frequently used is the concept of efcient certication. Let P be some property of graphs/formulas or any other combinatorial object. An algorithm A certies the property P if the following holds: 1. On any input instance the algorithm returns either has P or dont know. 2. Soundness: The algorithm never outputs has P on an instance which does not have the property P. The algorithm may output dont know on an instance which has the property P (the algorithm has one sided error). We will use certication algorithms on random instances of formulas/graphs taken from some probability space C. We shall consider properties that are almost surely true for the random object taken from C. A certication algorithm is complete with respect to the probability space C and a property P if it almost surely outputs has P on an input taken from C. The computationally heavy part of our algorithm is certifying that two different graphs derived from the random formula do not have a large cut. One of these two graphs is random, and the other is a multigraph that is the union of 6 graphs, where each of these graphs by itself is essentially random, but there are correlations among the graphs. A cut in a graph is a partition of its vertices into two sets. The size of the cut is the number of edges with one endpoint in each part. A certication algorithm for verifying that an input graph with m edges has no cut signicantly larger than m/2 is implicit in [16]. This algorithm is based on semi-denite programming; if the maximum cut in the input graph is of size at most m(1/2 + ), then the algorithm outputs a certicate that the maximum cut in G is bounded by m(1/2 + ()), where () 0 as 0. A computationally simpler algorithm can be applied if the graph is random. In [7] it is shown how to certify that in a random graph taken from Gn,d/n the size of the maximum cut is bounded by dn 1/4 + (1/ d) , thus bounding the maximum cut by m(1/2 + ) when d is large enough. This is done by removing the vertices of highest degree from G, and then computing the most negative eigenvalue of the adjacency matrix of the resulting graph.
2.4
An overview of our refutation algorithm
Our algorithm builds on ideas taken from earlier work ([10, 9, 6, 7, 5, 11]). This section gives an informal overview of the algorithm at a fairly detailed level. Other sections of this manuscript ll in the formal details. The input is an arbitrary 3CNF formula with n variables and m = cn3/2 clauses, where c is a large enough constant. Below we describe the expected behavior of our algorithm when the input formula is random. The algorithm rst greedily extracts from a subformula . This is done as follows. We say that two clauses match if they differ in their rst literal, but agree on their second literal and on their third
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literal. For example, the clauses (x1 x2 x3 ) and (x4 x2 x3 ) match. The subformula is constructed by greedily putting into pairs of clauses that match, until no further matches are found in (we allow each clause of to participate in at most one matched pair of ). Let m be the number of clauses in . A simple probabilistic argument shows that we can expect m = (m2 /n2 ) = (c2 n). Moreover, is essentially a union of two random (but correlated) formulas 1 and 2 (each containing one clause from every pair of clauses that are matched in ). Our algorithm will now ignore the rest of , and refute . As explained, is a union of two random formulas. Here we use an observation that is made in [6], that we shall call the 3XOR principle. The 3XOR principle: In order to show that a random 3CNF formula is not satisable, it sufces to strongly refute it as a 3XOR formula. Let us explain the terms used in the 3XOR principle. A clause in a 3XOR formula is satised if either one or three of its literals are satised. A strong refutation algorithm is one that shows that every assignment to the variables leaves at least a constant fraction of the clauses not satised (as 3XOR clauses, in our case). A proof of the 3XOR principle is given in [6]. We sketch it here, and give it in more details in Section 3. Observe that in a random formula every literal is expected to appear the same number of times, and if the number of clauses is large enough, then things behave pretty much like their expectation. As a consequence, every assignment to the variables sets roughly half the occurrences of literals to true, and roughly half to false. Hence every assignment satises on average 3/2 literals per clause. Moreover, this property is easily certiable, by summing up the number of occurrences of the n most popular literals. Given that every assignment satises on average 3/2 literals per clause, let us consider properties of satisfying assignments (if such assignments exist). The good option is that they satisfy one literal in roughly 3/4 of the clauses, three literals in roughly 1/4 of the clauses, and 2 literals in a negligible fraction of the clauses. This keeps the average roughly at 3/2, and indeed nearly satises the formula also as a 3XOR formula, as postulated by the 3XOR principle. The bad option (which also keeps the average at 3/2) is that the fraction of clauses that are satised three times drops signicantly below 1/4, implying that signicantly more than 3/4 of the clauses are satised either once or twice, or in other words, as a 3NAE (3-not all equal SAT) formula. But here, let us combine two facts. One is that for a random large enough formula, every assignment satises roughly 3/4 of the clauses as a 3NAE formula. The other (to be explained below) is that there are known efcient algorithms for certifying that no assignment satises more than 3/4 + fraction of the clauses of a 3NAE formula. Hence for a random 3CNF formula, one can efciently certify that the bad option mentioned above does not occur. Having established the 3XOR principle, the next step of our algorithm makes one round of Gaussian elimination. That is, under the assumption that we are looking for near satisability as 3XOR (which is simply a linear equation modulo 2), we can add clauses modulo 2. Adding (modulo 2) two matched clauses, the common literals drop out, and we get a clause with only two literals whose XOR is expected to be 0, namely, a 2EQ clause (EQ for equality). For example, from the clauses (x1 x2 x3 ) and (x4 x2 x3 ) one gets the clause (x1 = x4 ). Doing this for all pairs of matched clauses in , we get a random 2EQ formula 2eq . If was nearly satisable as 3XOR, then 2eq must be nearly satisable as 2EQ. But if is random, then 2eq is essentially a random 2EQ formula. For such formulas, every assignment satises roughly half the clauses. Moreover, there are known algorithms that certify this (to be explained shortly). Hence we can strongly refute 2eq as 2EQ, implying strong refutation of as
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3XOR, implying strong refutation of as 3SAT, implying refutation (though not strong refutation) of as 3SAT. Let us briey explain here the major part that we skipped over in the description of our algorithm, namely, how to certify that a random 2EQ formula is not 1/2 + satisable, and how to certify that a random 3NAE formula is not 3/4 + satisable. In both cases, we reduce the certication problem to certifying that certain random graphs do not have large cuts, and then use the certication algorithms mentioned in Section 2.3. (The principle of refuting random formulas by reduction to random graph problems was introduced in [10].) To strongly refute random 2EQ formulas, we negate the rst literal in every clause, getting a 2XOR formula. Now we construct a graph whose vertices are the literals, and whose edges are the clauses. A nearly satisfying 2XOR assignment naturally partitions the vertices into two sides (those literals set to true by the assignment versus those that are set to false), giving a cut containing nearly all the edges. On the other hand, if the original 2EQ formula was random, then so is the graph, and it does not have any large cut. As explained in Section 2.3, we can efciently certify that the graph does not have a large cut, thus strongly refuting the 2XOR formula, and hence also strongly refuting the original 2EQ formula. To strongly refute random 3NAE formulas, we again consider a max-cut problem on a graph (in fact, a multigraph, as there may be parallel edges) whose vertices are the literals. From each 3NAE clause we derive three edges, one for every pair of literals. For example, from the 3NAE clause (x1 , x2 , x3 ) we get the edges (x1 , x2 ), (x2 , x3 ) and (x3 , x1 ). It is not hard to see that if a 3/4 + fraction of the 3NAE clauses 2 are satised as 3NAE, than a 3 ( 3 + ) = 1 + 2 fraction of the edges of the graph are cut by the partition 4 2 3 induced by the corresponding assignment. Note that in our case (of that is the union of random 1 and random 2 ) this graph is essentially a union of 6 random graphs: 3 graphs derived from the clauses of 1 (one with edges derived from the rst two literals in every clause, one with edges derived from the last two literals, and one from the rst and third literal), and 3 graphs derived from 2 . Hence it is not expected to have a cut containing signicantly more than half the edges. One may certify that this is indeed the case either by using the algorithm of [16] on the whole graph, or by using the algorithm of [7] on each of the 6 random graphs separately. Summarizing, our refutation algorithm extracts from a subformula (composed of matched pairs of clauses), checks that in almost all literals appear roughly the same number of times, derives from certain graphs on 2n vertices and certies that they do not have large cuts (e. g., by computing the most negative eigenvalue of their adjacency matrices). The combination of all this evidence forms a proof that is not satisable. If is random and large enough (cn3/2 clauses), then almost surely the algorithm will indeed manage to collect all the desired evidence.
3
The refutation algorithm
3/2
cn The input formula is taken from Cn . We will use (?, w, ) to denote a clause in which the second and the third literals are w and , respectively, and the rst literal can be any literal. The following algorithm is used to extract from .
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Algorithm Extract( ) Set 1 = 2 = 0. / For every ordered pair of literals (w, ): 1. Count the number of clauses in of the form (?, w, ) and store it in N(w, ) . 2. If N(w, ) 2 add to 1 the rst appearance of a (?, w, ) clause and add to 2 the second appearance of a (?, w, ) clause. Return = (1 , 2 ). Each of 1 , 2 is a random formula, though clauses in 1 (and 2 ) are not completely independent of each other: if a clause (x, y, z) appears, then the clause (t, y, z) cannot appear. From now on we will concentrate on refuting , ignoring the rest of . The number of matched pairs in is denoted by m (in Section 2.4 we used m to denote the number of clauses in ; from here on, m will denote the number of matched pairs in ).
cn Lemma 3.1. Let be the formula returned by Extract( ), where R Cn . W.h.p. the number of c2 n matched pairs in is 8 .
3/2
The proof of Lemma 3.1 is deferred to Section 4. Before specifying the algorithm, we introduce additional notation and denitions which will ease the description of the algorithm. Denition 3.2. Let be any 3CNF formula over n variables. The graph induced by has 2n vertices (corresponding to all possible literals) and the following edges. Each clause of induces three edges by taking all (unordered) pairs of literals from it (e. g., the clause (x, y, z) induces the edges (x, y), (x, z), (y, z)). We denote the (multi) graph induced by by G . Denition 3.3. Let = (1 , 2 ) be a 3CNF formula with m pairs of matched clauses. The graph G is the graph induced by (as in Denition 3.2). The graph G2eq is a graph with 2n vertices (corresponding to all literals). Its edges are as follows: each matched pair from 1 , 2 , say (x, w, ), (y, w, ), induces exactly one edge (x, y). Note that we negate the literal which corresponds to the clause of 1 . Denition 3.4. Let be a formula with n variables and m clauses. The imbalance of a variable i (denoted by Imi ) is the difference in absolute value between the number of times it appears with positive polarity and the number of times it appears with negative polarity. The total imbalance of is n Imi and the i=1 normalized imbalance of is (1/3m) n Imi . i=1 If the normalized imbalance of is bounded by , then is -balanced. Denition 3.5. A 3CNF formula has the (1 ) 3XOR property if for every assignment A, if A satises as 3CNF, then at least 1 fraction of the clauses are satised as 3XOR. Denition 3.6. A graph is said to have a -cut if there is a partition of its vertices into two disjoint sets such that at least a -fraction of the edges cross this partition.
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Algorithm Refute( ) 1. Certify that has the (1 ) 3XOR property. Specically: (a) Find the normalized imbalance of and denote it by . (b) Certify that G has no ( 1 + )-cut ( is returned by a subroutine). 2 Set = 3 ( + 2 ). 2 2. Certify that G2eq has no ( 1 + )-cut ( is returned by a subroutine). 2 3. If + 2 <
1 2
return unsatisable, otherwise return dont know.
In steps 1(b) and 2 of Refute( ) we use as a blackbox a subroutine for bounding the maximum cuts in G and G2eq . This subroutine is explained in Theorem 3.10. We rst show that Refute( ) is sound, i. e., whenever it returns unsatisable it holds that can not be satised. This follows from Lemma 3.8 and Theorem 3.7. Theorem 3.7 (Soundness). Let = (1 , 2 ) be a 3CNF formula composed of pairs of matched clauses. Denote by G2eq the graph induced by as described in Denition 3.3. The formula is not satisable if all the following conditions hold: 1. has the (1 )3XOR property,
1 2. G2eq has no ( 2 + )-cut,
3. + 2 < 1 . 2 Proof. We shall show that if is satisable and has the (1 )3XOR property, then G2eq has a (1 2) cut. Combined with the fact that G2eq has no 1/2 + cut we derive a contradiction (since + 2 < 1/2). Consider the cut induced on the vertices of G2eq by a satisfying assignment A (where in one side there are all the literals whose value is true and in the other side there are all the literals whose value is false). A(x) denotes the value of the literal x induced by the assignment A. By the 3XOR property of , all but fraction of the clauses of are satised as 3XOR clauses. Hence at least a (1 2) fraction of the pairs of matched clauses have both clauses in the pair satised by A as 3XOR. Let (x, w, ), (y, w, ) be a pair such that both (x, w, ) and (y, w, ) are satised as 3XOR. It holds that: A(x) + A(y) + 2(A(w) + A( )) = 0 mod 2. Thus exactly one of the literals x, y is true and the other is false (under the assignment A), and the edge (x, y) induced by this pair of clauses crosses the cut. It then follows that at least 1 2 fraction of the edges in G2eq are cut edges. Lemma 3.8 (The 3XOR lemma). Let be a 3CNF formula. If the following hold: 1. is -balanced, and 2. the graph (induced by ) G has no ( 1 + )-cut, 2
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3 then has the (1 2 ( + 2))3XOR property.
Proof. The proof of this lemma appears in [6]; also a similar version of this lemma (for denser random 2kCNF formulas) appears in [5]. We repeat the proof for the sake of self-containment. Denote by m the number of clauses in and let A be a satisfying assignment. We bound from above the number of satised appearances of literals. The assignment which maximizes the number of satised appearances of literals is the majority vote: a variable x is assigned true iff it appears more times with positive polarity than with negative polarity. Using this assignment the total number 1 of satised appearances is 2 3m + n Imi (where Imi denotes the imbalance of variable i). It follows i=1 1 3 that on average each clause is satised at most 3 + 2m n Imi = 2 (1 + ) times. i=1 2 A clause is satised as 3AND by A if all its literal are by satised A. We next show that the fraction of clauses satised by A as 3AND is at least 1/4 3/2. Equivalently it is enough to show that the fraction of clauses satised as 3NAE, denoted by , is at most 3/4 + 3/2 (since A is a satisfying assignment). Consider the graph G induced by . We remind the reader that each clause of contributes a triangle of 3 edges to G (e. g., the clause (x, y, z) contributes the edges (x, y), (y, z), (x, z)). Consider the cut induced by the satisfying assignment A. Each clause satised as 3NAE contributes exactly 2 edges to the cut, thus this cut has at least 2 m edges. But G has no (1/2 + )-cut. It follows that 2 m (1/2 + )3m, giving 3/4 + 3/2 as needed. It remains to show that all but a small fraction of the clauses are satised as 3XOR by A. Denote by 1 , 2 , 3 the fraction of clauses which are satised exactly once, exactly twice and exactly 3 times respectively (3 i = 1). We already know that 3 1/4 3/2 and that each clause is satised at i=1 most 3/2 + times on average, thus 3(1 + ) 3 3 + 2 2 + 1 1 . 2
3 3 1 Substituting 1 = (1 3 2) and 3 with 4 2 yields that 2 2 ( + 2). cn The following two Theorems show that our refutation algorithm refutes most formulas R Cn . cn Theorem 3.9. For any > 0 there is a constant c = c() such that w.h.p. over the choice of R Cn :
3/2 3/2
(a) the subformula is -balanced, and (b) each of the graphs G and G2eq has no ( 1 + )-cut. 2 It is well known and follows from standard probabilistic calculations that a random graph (with large enough average degree) has no (1/2+)-cut, and that a random formula is -balanced. The distributions of , G are close enough to the standard models of random formulas/graphs respectively, so that the proof techniques used for the random cases can be applied also in our case. The proof of Theorem 3.9 is deferred to Section 4.2. Theorem 3.10. There is a polynomial time algorithm that nds the imbalance of a 3CNF formula. There is a polynomial time algorithm that for every graph G that has no ( 1 + )-cut, certies that G has no 2 1 ( 2 + )-cut, where () 0 as 0.
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Proof. Finding the normalized imbalance of a formula is done by counting positive and negative appearances for every variable, computing its imbalance and averaging. An algorithm for certifying a bound on the maximum cut of a graph is given in [16]. Given a graph with m edges whose maximum cut is bounded by m(1/2 + ) this algorithm produces a proof that the input graph has no cut of cardinality m(1/2 + ). This algorithm has the property that 0 as 0. Corollary 3.11 (Almost-completeness). For sufciently large constant c, the refutation algorithm w.h.p. cn3/2 returns unsatisable for a random formula R Cn . Proof. Using Theorems 3.9, 3.10 we will show that by taking c to be a sufciently large constant, the terms , from Refute( ) can be made arbitrary small (and thus + 2 < 1/2). The term equals (3/2)( + 2 ) where is the imbalance of and (1/2 + ) is the bound returned by the algorithm of [16] when applied to G . By Theorem 3.9, 0 as c increases. Furthermore, the value of the maximum cut in both G , G2eq approaches 1/2 as c increases. It then follows, using Theorem 3.10, that the bounds 1/2 + and 1/2 + returned by the certication algorithm (of [16]) applied to G and G2eq , respectively, can be made arbitrary close to 1/2 by setting c to be a sufciently large constant.
4
4.1
Proofs of Lemma 3.1 and Theorem 3.9
Proof of Lemma 3.1
Proof of Lemma 3.1. For two random clauses the probability that they induce a matched pair of clauses is 1 1 1 p 2 . 2n 2n 2 4n Thus, the expected number of pairs that match is = c2 n cn3/2 c2 n3 1 . p 2 4n2 8 2 (4.1)
Let X denote the number of pairs of clauses in that match. We use the second moment to show that w.h.p. X . By Chebyshevs inequality Pr[|X | > ] < Var(X) . 22 (4.2)
For every two clause locations in (e.g., rst clause and third clause) we set an indicator random variable 3/2 Xi (i = 1, 2, ..., cn2 ) to be 1 if the respective two clauses match and otherwise 0. For i = j we say that i j if the pair i and pair j share one clause location, and otherwise i j. (Note that if pair i and pair j share two clause locations, then i = j. Note also that i j might share the same clause without sharing a clause location, if a certain clause happened to appear twice in .) For any xed i we let =
j: ji
Pr[X j | Xi ] .
(4.3)
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From symmetry, does not depend on i. Var(X) = E X 2 2 = E ( Xi )2 2 = E
i
Xi2 + Xi X j
i i= j j: j i
2
= 2 + Pr[Xi X j ] = 2 + Pr[Xi ]
i= j i
Pr[X j ] +
j: ji
Pr[X j | Xi ]
2 + Pr[Xi ] Pr[X j ] + Pr[Xi ]
i j i 2
j: ji
Pr[X j | Xi ]
= + Pr[Xi ] = (1 + ) .
i
(4.4)
Substituting Var(X) with (1 + ) in inequality (4.2) we derive Pr[|X | > ] < (1 + ) 1 + 2 . 22 (4.5)
Thus, since = (1), it sufces to show that = o(). It holds that 2(cn3/2 2)p = o() . (4.6)
So far we showed that w.h.p. X . Note that X may over-count the number of matched pairs in . The reason is that in there are expected to be sets of three or more clauses in which any two clauses match. From each such set, Extract( ) takes to only the rst two clauses of the set. For i 3, we call a set of i clauses bad if each two clauses of the set match. Let Yi be the number of i bad sets of size i in . The number of matched clauses in is at least X i3 2 Yi . Thus, in order to show that the number of matched pairs in is it is enough to prove that E
i3
i Yi = o() . 2
i 2
Then, using Markovs inequality we derive that w.h.p. i3 E i i Yi = 2 2 cn3/2 i1 i2 p i 2
Yi = o(). It holds
i
cn3/2 e i
1 4n2 (1 1/n)
i1
.
(4.7)
i Thus, the sum i3 E 2 Yi is bounded by the sum of a geometric sequence whose rst term (i = 3) is i o(). It follows that E i3 2 Yi = o().
4.2
Proof of Theorem 3.9
Let P Cn Cn be the set of all possible matched pairs of clauses, and let Pm be the set of all m-tuples of matched pairs of clauses. For a pair of matched clauses (c1 , c2 ) P, the inducing pair is the pair formed
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U RIEL F EIGE AND E RAN O FEK
by the second and third literals in each of the two matched clauses c1 , c2 . Let m denote the number of matched pairs of clauses in . Note that Pm , however not all the elements of Pm are in the support P of . We denote by m Pm the support of , i. e., the collection of all (ordered) m matched pairs for which every matched pair of clauses has a distinct inducing pair. We claim that is a random element P of m . Lemma 4.1. Given that has m matched clauses, the set of inducing pairs is a random set of m different ordered pairs of literals (each pair has two distinct literals). Proof. Let C denote the set of clauses that have a matching clause (from the original formula ). Denote by L the set of inducing pairs, i. e., pairs that participate as the second and third literal in one of the clauses of C. The set L may be any set of distinct ordered pairs of size m. By symmetry, any such set is equally likely to be L. The explanation is as follows. Assume we expose the indices of the clauses in C and also the partition of C into equivalent classes (each equivalent class is a maximal set of clauses that have the same second and third literals). Given this information, for each choice of L, the probability that L is the set of inducing pairs is the same (for any L the number of ways to match the pairs of L with the equivalent classes is the same; additionally, the probability for all other clauses not in C to avoid all the pairs of L in the second and third literals is the same). Proof of Theorem 3.9. From here on we will assume that m is a xed number and that m c2 n/8 (this P is justied by Lemma 3.1). The formula is a random element of m . Let R Pm (i. e., is composed of m random and independent samples from P). Denote by the event that every matched pair of clauses in has a distinct inducing pair. Conditioned on , has the distribution of . As Lemma 4.2 shows, the event is not too rare (the proof of Lemma 4.2 is deferred to the end of this section). Lemma 4.2. Let m =
c2 n 8 .
For R Pm it holds that Pr[ ] ec
4 /128
Furthermore, as Lemma 4.3 states, G is unlikely to have a large cut. Lemma 4.3. Let m
c2 n 8 . 1 For R Pm with probability 1 o(1) it holds that G has no ( 2 + 4 )-cut. c
Combining Lemmas 4.2, 4.3 we now show that G is unlikely to have a large cut. The reasoning is as follows:
P R (m)
Pr [G has ( 1 + 4 ) cut] = 2 c
R Pm
1 Pr [G has ( 2 + 4 ) cut | ] c
(4.8) (4.9)
Pr [G has ( 1 + 4 ) cut] o(1) 2 c c2 /128 = o(1) , Pr [ ] e
where the last equality is because c is a xed constant. A similar argument shows that if w.h.p. G2eq has no (1/2 + )-cut and is -balanced, then w.h.p. G2eq has no (1/2 + )-cut and is -balanced. Hence, we only need to prove the following lemmas. Lemma 4.4. Let m
c2 n 8 . 1 For R Pm with probability 1 o(1) it holds that G2eq has no ( 2 + 5 )-cut. c
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Lemma 4.5. Let m
c2 n 8 .
For R Pm , with probability 1 o(1) it holds that is 3 -balanced. c
To complete the proof of Theorem 3.9 we now give the proofs of Lemmas 4.3, 4.4, 4.5, and 4.2. Proof of Lemma 4.3. Fix a partition (V1 ,V2 ) of the vertices of G . We denote by W (V1 ,V2 ) the number of edges crossing the cut (V1 ,V2 ) in G . For R Pm the expectation of W (V1 ,V2 ) is at most 6m(1/2 + 1/n). For any formula Pm we let f ( ) be equal to W (V1 ,V2 ). Let Xi be the expected value of f after exposing the rst i pairs of (for i = 0, 1, . . . , m). The sequence X0 , X1 , . . . , Xm is a martingale. The following two facts: 1. for any Pm , changing one matched pair (an element of P) can change the value of f by at most 4 (each clause forms a triangle that contributes at most 2 edges to the cut), and 2. is taken from a product measure Pm , imply that for every i it holds that |Xi Xi+1 | 4 (see Theorem 7.4.1 from [1]). Azumas inequality implies that for any > 0 Pr[ f ( ) 6m( 1 + 1 ) > ] < e 2m42 . 2 n Setting =
17m c
2
and using m
c2 n 8
we derive
R Pm
Pr [W (V1 ,V2 ) > 6m( 1 + 4 )] < 21.1n . 2 c
Using the union bound over all possible cuts we derive that w.h.p. (for R Pm ) the graph G has no 1 ( 2 + 4 )-cut. c The proof of Lemma 4.4 is very similar to the proof of Lemma 4.3, details are omitted. Proof of Lemma 4.5. We rst bound the expected imbalance of . The total imbalance of is bounded by the sum of the imbalances of 1 , 2 (where 1 /2 are formed by taking the rst/second clause from each matched pair of ). Since 2 has the same distribution of 1 , it is enough to bound the expected total imbalance of 1 (and then multiply by two). The total imbalance of 1 is the sum of the imbalances of all variables in 1 , i. e., n Imi (the imbalance of xi in 1 is denoted by Imi ). i=1 For any variable xi we denote by di the number of appearances of x in 1 . For R Pm it holds that n E[di ] = 3m. By symmetry, for every i it holds that E[di ] = 3m/n. We denote d 3m/n. Given i=1 that di = k the polarities of the appearances of xi are still random. Given that di = k, the imbalance of xi is the absolute value of the sum of k independent random variables, where each random variable has probability 1/2 of being 1 and probability 1/2 of being 1. Hence E[Im2 | di = k] = k. It then follows i that E[Im2 ] = Pr[di = k] E[Im2 | di = k] = k Pr[di = k] = E[di ] = d , i i
k k
(4.10)
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U RIEL F EIGE AND E RAN O FEK
where all probabilities and expectations are taken over R Pm . Using the convexity of the square function E[Imi ] E[Im2 ] i d . (4.11)
So far we showed that E R Pm n Imi n d. Thus, for R Pm the total imbalance of is i=1 expected to be less than 2n d. We will now show that the total imbalance of is not likely to be too large relative to 2n d. For any Pm we let f ( ) be the total imbalance of . Changing one matched pair of clauses in changes the total imbalance of by at most 12. Azumas inequality implies that for any > 0 2 Pr m f ( ) 2n d > < e 2m122 .
R P
Setting = n d and using d = 3m/n we derive
R P
n Pr m [the total imbalance of > 3n d] < e 96 .
It then follows that the normalized imbalance of is w.h.p. bounded by 3n d/6m 3/c (using m c2 n/8 and d = 3m/n). Proof of Lemma 4.2. We generate iteratively by choosing each time (independently) a random element of P. For each new random element of P, the probability for it to have an inducing pair which is different from all previous inducing pairs is 1 m/N, where N = 2n (2n 2) is the number of possible inducing pairs. It then follows that with probability of at least 1 m 2n(2n 2)
m (1)
exp m
m 2n2 (11/n)
(2)
ec
4 /128
,
each of the matched pairs in has a distinct inducing pair. Inequality (1) is because 1 x e2x holds for every x [0, 1/2]. The constant in the exponent following Inequality (2) is derived by taking m = c2 n/8.
5
Practical considerations for the refutation algorithm
Recall that our refutation algorithm extracts from a subformula that contains matched pairs of clauses, and then refutes . The longer is, the easier it is to refute it. For simplicity, we matched a pair of clauses only if they agreed on their last two literals. Moreover, every clause of participated in at most one pair of matched clauses in , even though a clause may be eligible to participate in more than one matched pair. In practical implementations, it is advantageous not to have these restrictions, and thus get a longer formula . In particular, we may allow the same clause to participate in several pairs of matched clauses, by duplicating it. More importantly, we may match any two clauses that share two variables (regardless of the polarity of the variables, and of their location within the clauses). For example, the two clauses (x, w, ) and (w, , y) can be matched. If is satised by an assignment A that
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has the 3XOR property, then one step of Gaussian elimination gives in this case A(x) + A(w) + A( ) + A(y) + A(w) + A( ) = 0 mod 2, thus A(x) + A(y) = 1 mod 2. Hence, we will associate the edge (x, y) with this pair of matched clauses so that the edge induced by this pair in G2eq will cross the cut which corresponds to the assignment A. Using the principles above, the number of pairs of matched clauses 3/2 that one expects to extract from a random formula of length cn3/2 is roughly 3cn / n 9c2 n. 2 2 We used the principles above to implement the algorithm in practice. In the current implementation, the problem of refuting is reduced to strong refutation of two 2XOR formulas. We performed 2 eigenvalue computations on matrices of size n n, whereas the original refutation algorithm performed eigenvalue computations on matrices of size 2n 2n. Our implementation uses the conditions of Theorem 5.2 to refute . Before stating Theorem 5.2 we need the following denition. Denition 5.1. Let be a 2XOR formula with m clauses and n variables. A is the following n n symmetric matrix associated with . Initially A is the zero matrix. For each clause of the forms (x, y) or (x, y) we add +1 to positions A(x, y) and A(y, x). For each clause of the forms (x, y) or (x, y) we add 1 to positions A(x, y) and A(y, x). A similar matrix can be dened for a 2EQ formula, just by reducing the 2EQ formula into a 2...