On a generalization of extended resolution

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1 Discrete Applied Mathematics (1999) On a generalization of extended resolution O. Kullmann Johann Wolfgang Goethe-Universitat, Fachbereich Mathematik D Frankfurt, Germany Received 1 November 1996; revised 1 May 1997; accepted 5 October 1998 Abstract Motivated by improved SAT algorithms ((O. Kullmann, DIMACS Series, vol. 35, Amer. Math. Soc., Providence, RI, 1997; O. Kullmann, Theoret. Comput. Sci. (1999); O. Kullmann, Inform. Comput., submitted); yielding new worst-case upper bounds) a natural parameterized generalization GER of Extended Resolution (ER) is introduced. ER can simulate polynomially GER, but GER allows special cases for which exponential lower bounds can be proven.? 1999 Elsevier Science B.V. All rights reserved. Keywords: Propositional logic; Extended resolution; Blocked clauses; Generalized extended resolution; Lower bounds 1. Introduction 1.1. Extended resolution G. Tseitin introduced in [21] the Extension Rule for the Resolution Calculus: F F {{v; a; b}; {v; a}; {v; b}} for arbitrary variables a; b and a new variable v (new relative to the set F of premises and to a; b). Thereby the clause-set {{v; a; b}; {v; a}; {v; b}} is the Conjunctive Normal Form of the formula v (a b). An Extended Resolution Proof (for short: ER proof) of the empty clause from the clause-set F is an ordinary resolution proof of from F, where F F is obtained by repeated applications of the Extension Rule. The length of an ER proof is the (total) number of (dierent) clauses in it. We denote by Comp ER (F) the minimal length of an ER proof of from F. Till today no super polynomial lower bound for Comp ER (F) is known. For all (concrete) examples of dicult formulas we know short ER proofs, because the Extension This work was supported by DFG-Leibniz-Programm Schn 143=5-1. address: kullmann@mi.informatik.uni-frankfurt.de (O. Kullmann) X/99/$ - see front matter? 1999 Elsevier Science B.V. All rights reserved PII: S X(99)

2 150 O. Kullmann / Discrete Applied Mathematics (1999) Rule enables one to mimic the (informal) proofs of unsatisability for the (concrete) examples. In [4 6,9] it is proved that ER has the same power (up to polynomial transformations) as the most powerful (known) proof systems, Extended Frege Systems or Frege Systems with the substitution rule Blocked clauses In this note 1 I make some remarks on a (natural) generalization of the concept of ER proofs. This generalization is based on the concept of Blocked Clauses. Blocked Clauses are special cases of redundant clauses, i.e., they can be satisability-equivalently added to or eliminated from the given clause-set. They are dened by the condition that there is a literal l in them such that every candidate for a resolution on this literal, i.e., every clause in the given clause-set containing l, also contains another complementary literal and hence the resolvent is tautological. Note that for the above Extension Rule all three new clauses are blocked for the literal v respectively v (in any order of addition), and thus the addition of Blocked Clauses covers the Extension Rule. The concept of blocked clauses has been developed with the aim to improve worstcase upper bounds for SAT-algorithms. In [14,15] (see also [13]) the addition and elimination of blocked clauses under various circumstances is an important tool for improving the bound for 3-SAT-decision (no clause has more than three literals) to 1:5045 n (n= number of variables) and for improving the bound for SAT-decision to 2 l=g ( = number of literal occurrences). In all these applications we use blocked clauses without new variables. This restriction is important to obtain control on the process of introducing new clauses: If a clause is blocked for a literal which is not new, then this clauses reects a certain structure of the clause-set in consideration, while the Extension Rule does not depend on the structure of the set of premises. A predecessor of these methods (in a more general context) is the concept of Complement Search in [18] Generalized extended resolution Although the addition of blocked clauses already generalizes the above Extension Rule in a symmetrical manner, it is still not fully satisfactory because of the dependence on the order of additions. For example, consider the following 2 3 = 6 clauses coming from two applications of the Extension Rule: C 1 = {v; a; b}; C 2 = {v; a}; C 3 = {v; b}; C 4 = {w; v; c}; C 5 = {w; v}; C 6 = {w; c}: As mentioned, for i =1;:::;6 the clause C i is blocked w.r.t. F {C 1 ;:::;C i 1 }, but if we add one of the clauses C 4 ;C 5 ;C 6 rst, then one of the clauses C 1 ;C 2 ;C 3 is thereafter no longer blocked. 1 A rst version appeared in [12].

3 O. Kullmann / Discrete Applied Mathematics (1999) To overcome this dependence on the order of added clauses, we generalize the addition of blocked clauses in the following natural way: For a clause-set F we dene the kernel K(F) as the (uniquely dened) subset of F obtained by repeated eliminations of blocked clauses until no blocked clause is left. Now a Blocked Extension F of F is any clause-set F with the property: K(F F )=K(F). A Generalized ER proof (for short: GER proof) for F is a resolution proof of from F F, where F is a blocked extension of F (whether F is a superset of F, or only contains the new clauses, does not matter) Results The addition of one blocked clause C to F can speed up resolution proofs for F at most by a factor C (while, due to hidden additions of blocked clauses, an extension by one clause in the GER calculus can already cause an exponential speed up). The concept of GER proofs is a generalization of ER proofs (an extension F by ER is also a blocked extension of F), which eliminates the special form of the additional clauses as well as the special ordering of introduction. However, the strength of the calculus is not increased: We show that the ER calculus can polynomially simulate the GER calculus. GER allows to study the eect of various restrictions on the blocked extension F. We obtain the following results: 1: 1-clauses in F containing new variables (relative to F) can be eliminated without aecting the shortest proof length. In minimal unsatisable F we can eliminate all 1-clauses from F. 2: If F contains only clauses of length less than or equal to 2, then we can eliminate all new variables from F. Thus new variables play a role only if F contains at least one clause of length at least 3. 3: (i +1; 0)-GER cannot be simulated polynomially by (i; 0)-GER for i =0; 1; 2, where (i; 0)-GER stands for the GER calculus with the restrictions: 3:1: the blocked extensions F contain only clauses of length less or equal i, 3:2: F does not contain new variables, (thus (0; 0)-GER is the ordinary resolution calculus). 4: We give an (sub)exponential lower bound for those GER proofs, which only use blocked extensions F without new variables (but with arbitrary clause length). The hard formulas here are the Pigeonhole formulas. Hence, although the addition of clauses without new variables can cause an exponential speed up (see result 3 above), in general the introduction of new variables is necessary (ER admits proofs of polynomial size for the Pigeonhole formulas; see [2] or [6]). Considering the simulation of GER by ER mentioned above, the exponential lower bound in 4 shows that there are some harmless applications of the Extension Rule.

4 152 O. Kullmann / Discrete Applied Mathematics (1999) Prevailing backtracking algorithms for SAT-decision can be simulated by ordinary resolution and thus the exponential lower bounds carries over. See [10,11] and also [13] for a generalization of the (well-known) simulation by (regular) resolution of SAT decision algorithms using (only) backtracking in its simplest form (i.e., semantic trees, see [22]). However, for algorithms working with blocked clauses (without new variables) the resolution calculus has to be generalized. As a rst step by the lower bound for the GER calculus without new variables we obtain a lower bound for a general class of DPLL-like algorithms, whose reduction component allows the addition of resolvents and of blocked clauses without new variables (but does not eliminate clauses). See [11] Organization of the paper Section 2 introduces the notations used in this paper. Section 3 gives the background for the Resolution Calculus. The notion of Blocked Clauses is introduced in Section 4, and examples for its use for SAT decision are given. The maximal speed up achieved by adding one blocked clause is estimated in Section 5. Blocked Extensions and the GER calculus are introduced in Section 6. Section 7 is devoted to the simulation of GER by ER. 1- and 2-clauses in blocked extensions are discussed in Section 8. The exponential lower bound for GER without new variables is proven in Section 9. Finally some open problems are given in Section Notation By VA we denote the set of variables and by LIT := {v: v VA} {v: v VA} the set of literals (l is the complement of l, l = l). A clause is a nite and complement-free (i.e., non-tautological) set of literals. We denote by CL the set of all clauses: CL := {C LIT: C nite and C C = }, where for L LIT: L := {l: l L} (i.e., a set of literals (clauses for example) is complemented elementwise). A clause-set is a nite set of clauses, the set of all clause-sets is CLS := {F CL: F nite}. A special clause is the empty clause := CL, a special clause-set is the empty clause-set := CLS. By Var(l) VA for l LIT we denote the variable of l (l = Var(l) orl = Var(l)), and we use Var(C):={Var(l): l C}, Var(F):= C F Var(C). Furthermore, Lit(F):= C F C. A literal l is pure for F CLS i l Lit(F). A partial assignment is a complement-preserving mapping : L {0; 1} for L LIT closed under complement: L = L. We dene Var( ) := Var(L).

5 O. Kullmann / Discrete Applied Mathematics (1999) (C)=1 holds for a clause C i l C: (l)=1, and (C)=0 holds i l C : (l)= 0; otherwise (i.e., no literal of C is mapped to 1 by, and is not dened for at least one literal of C) the term (C) is undened. (F) = 1 for a clause-set F i C F: (C)=1; (F)=0 i C F: (C)=0; otherwise, (F) is undened. For C CL with (C) 1 we denote by C := C\{l C: (l)=0} CL the result of substituting truth-values via in C ( (l) = 0 implies that is dened on l), and for F CLS: F := { C: C F (C) 1} CLS: By l 1 j 1 ;:::;l n j n we denote the partial assignment with Var( )=Var({l 1 ;:::; l n }) and (l i )=j i {0; 1} for i {1;:::;n}. A substitution is a complement-preserving mapping : LIT LIT. For C CL we dene (C):={(l): l C}. Note that (C) CL is possible since clauses must be complement-free. And for F CLS we dene (F):={(C): C F (C) CL} CLS: A renaming is a bijective substitution. We write : L 1 L 2 for substitutions and subsets L 1 ;L 2 LIT, i(l 1 ) (L 2 ) holds and is the identity on VA\Var(L 1 ). 3. The resolution calculus Denition 3.1. A clause R CL is the Resolvent of clauses C 1 ;C 2 CL (C 1 ;C 2 R) i there is l C 1 with l C 2 such that R =(C 1 \{l}) (C 2 \{l}). (Note that, because R is a clause which are dened to be complement-free, the resolution literal l is uniquely determined: C 1 C 2 = {l}.) A Resolution Proof P of F 2 CLS from F 1 CLS is a sequence P=(C 1 ;:::;C n ) of clauses (C i CL, n 0), such that the following holds: 1: For each C F 2 there is i {1;:::;n} with C i C. 2: For i {1;:::;n} we have C i F 1, or there are j; k {1;:::;i 1} with C j ;C k C i. The length of P is n. AResolution Proof for F CLS is a resolution proof of { } from F. For F; F 1 ;F 2 CLS and C CL we dene 2 Comp R (F 1 ;F 2 ):=inf{n: P resolution proof of F 2 from F 1 with length n}; Comp R (F; C) := Comp R (F; {C}); Comp R (F) := Comp R (F; ): 2 inf =+.

6 154 O. Kullmann / Discrete Applied Mathematics (1999) Remark 1: 1.1. Comp R (F 1 ;F 2 )=0 F 2 =, 1.2. Comp R (F)= F SAT. 2: Resolution proofs have the structure of a forest. 3: Resolution proofs (C 1 ;:::;C n )of can be restricted w.l.o.g. to proofs containing less than n 2 literal occurrences ( n i=1 C i n 2 ), because the length of a clause decreases at most by one by a resolution step, and thus clauses C i with C i n i can be eliminated. Lemma 3.1. For F; F 1 ;F 1 ;F 2;F 2 CLS and a partial assignment we have 1: If the two conditions 1:1: C F 1 C F 1 : C C; 1:2: C F 2 C F 2: C C hold; then: Comp R (F 1 ;F 2 )6Comp R (F 1;F 2 ). 2: Comp R ( F 1 ; F 2 )6Comp R (F 1 ;F 2 ). 3: If C F 2 : (C) 1; and furthermore one of the following two conditions hold: 3:1: F 1 F 1 ; or 3:2: C F 2 : Var( ) Var(C) then Comp R ( F 1 ; F 2 ) = Comp R (F 1 ;F 2 ). Proof. 1. For a resolution proof (C 1 ;:::;C n )off 2 from F 1 one constructs inductively (in a straight forward manner) a resolution proof (C 1 ;:::;C n) of F 2 from F 1 with C i C i for i =1;:::;n. 2. By part 1 we have Comp R (F 1 ;F 2 ) Comp R (( F 1 ) {C F 1 : (C)=1};F 2 ): Let P =(C 1 ;:::;C n ) be a proof of F 2 from ( F 1 ) {C F 1 : (C)=1}, and dene G := {C i : C i F 1 \( F)}. Prove inductively that for C j G we have (C j ) = 1, where G is the set of all successors in G in the underlying forest structure of P. Thus after elimination of G from P we still have a resolution proof of F 2 \{C F 2 : (C)=1} from F 1, which in fact must be a proof of F 2 from F Because of part 2 we only have to prove Comp R ( F 1 ; F 2 ) Comp R (F 1 ;F 2 ): 3:1: Here ( ) follows from part 1. 3:2: Consider a proof (C 1 ;:::;C n ) of F 2 from F 1. Replace all axioms C for C F 1 (with (C) 1)byC and obtain a proof of F 1 from F 2.

7 O. Kullmann / Discrete Applied Mathematics (1999) Denition 3.2. For a clause C we dene the corresponding partial assignment C by C := l 0: l C : Corollary 3.2. For F CLS and C CL we have Comp R (F; C) = Comp R ( C F): Lemma 3.3 (Generalized splitting lemma). For clause-sets F; T CLS such that T is minimally unsatisable (i.e.; T SAT; and C T : T \{C} SAT) we have Comp R (F) 6 Comp R (F; T ) + Comp R (T ) T 6 Comp R ( C F) + Comp R (T ) T : C T Proof. For the rst inequality note that in a proof for T all clauses of T must occur (because of the minimality condition), and thus when combining the proofs of T from F with the proof for T we can subtract T clauses. The second inequality follows by Comp R (F; T )6 C T Comp R (F; C) and Corollary 3.2. Lemma 3.4. For F CLS and a substitution we have Comp R ((F))6Comp R (F). Proof. Consider a resolution proof (C 1 ;:::;C n ) for F. Replace axioms C i F by (C i ) and obtain a pre -resolution proof for {(C): C F}, containing possibly tautological clauses, which can be eliminated by the following observations: if the resolvent is non-tautological, then at least one of its parent clauses is also non-tautological; a non-tautological resolvent R of a non-tautological clause C 1 with a tautological clause C 2 properly contains C 1 : C 1 R Lower bounds For a natural number n 2 consider variables v i; j with i {1;:::;n} and j {1;:::; n 1}. By PHP we denote the Pigeonhole Principle: PHP := PHP p n PHP n n; PHP p n := {{v i; j } j {1;:::;n 1} } i {1;:::;n} ; PHP n n := {{v i1;j; v i2;j}} i 1 ;i 2 {1;:::;n};i 1 i 2 : j {1;:::;n 1} Variable v i; j has the meaning pigeon i in hole j. The positive part PHP p n states that every pigeon i is in some hole j. And the negative part PHP p n states that no hole contains two pigeons. Thus, because there are more pigeons than holes, we have PHP SAT. The size of PHP is: Var(PHP) =O(n 2 ), and c(php); (PHP) = O(n 3 )

8 156 O. Kullmann / Discrete Applied Mathematics (1999) where c(f):= F is the number of clauses, and (F):= C F C is the number of literal occurrences. Theorem 1 (Haken [8]). For all n 1676 using c := 2 1=20 =1:03526:: we have Comp R (PHP) c n : The value for c is taken from [1]. Later on we will strengthen this result and then we give an outline of the proof. 4. Blocked Clauses and their use for SAT-decision Denition 4.1. A clause C CL is called blocked for l LIT with respect to F CLS i all (envisaged) resolvents of C with C F for l C are tautological, i.e., i l C C F[l C x C [x l x C]] holds. C is called blocked w.r.t. F i there is a literal l such that C is blocked for l w.r.t. F. For brevity we use for F CLS and l LIT: B l (F):={C CL: C blocked for l w:r:t: F}; B(F):={C CL: C blocked w:r:t: F} = B in (F):=B(F) F: l LIT B l (F) Remark (for F 1 ;F 2 ;F CLS, C 1 ;C 2 ;C CL and l LIT) 1: As a rst example consider F := {{a; b}; {a; b}; {a; b}} B in (F)={{a; b}; {a; b}}: 2: Smaller clause-sets have more blocked clauses: F 1 F 2 B(F 2 ) B(F 1 ). 3: Superclauses of blocked clauses are also blocked: C 1 C 2 C 1 B(F) C 2 B(F). 4: Clauses with pure literals are blocked: C * Lit(F) C B(F) (especially clauses with new variables are blocked). 5: Cis blocked w.r.t. F C is blocked w.r.t. F\{C} C is blocked w.r.t. F {C}. 6: C B l (F) l C C\{l} ({C F : l C})=1. 7: If the clauses C 1 ;:::;C r are the only occurrences of l in F, then every clause C, containing l and from every C i another complementary literal (C C i \{l} ), is blocked for l w.r.t. F. For example: 7:1: If {l; x 1 ;:::;x m } is the only occurrence of l in F, then the clauses {l; x 1 };:::; {l; x m } are blocked for l w.r.t. F. 7:2: And if {l; x 1 ;:::;x m } and {l; y 1 ;:::;y n } are the only occurrences of l in F, then the clauses {l; x i ; y j } (i {1;:::;m}, j {1;:::;n}) are blocked for l w.r.t. F.

9 O. Kullmann / Discrete Applied Mathematics (1999) (If x i = y j, then {l; x i ; y j } in fact is a 2-clause, and if x i = y j, then it is no clause at all (in our notation).) 8: Generalizing the example from 1 the following holds: 3 C F C F[Var(C) = Var(C )] (B in (F)= F =2 Var(F) ): Lemma 4.1. For F CLS and C B(F) we have F\{C} sat F sat F {C}; where sat sat denotes equivalence w.r.t. satisability (F 1 F 2 i both F 1 ;F 2 are satisable or both F 1 ;F 2 are unsatisable). Proof. It is enough to show F sat F\{C} for C F. The direction F SAT F\{C} SAT is obvious. Consider a satisfying assignment for F: (F\{C}) =1. W.l.o.g.: Var( ) =Var(F). If (C) = 1 then immediately (F) = 1 also. Otherwise, let C be blocked for l w.r.t. F. Thus we have (l) = 0. Dene by ipping the value of l: (l):=1 ( (l):=0) and (x):= (x) else. Now (F) = 1 holds, since on the one hand we have (l)=1 (C)=1 and on the other hand we have for C F\{C} : if l C, then (C ) = 1 because of (C )=1 and (l)=1; if l C then there is another a C\{l} with a C (because of the blocking condition) and by (a)= (a) = 0 we have (C ) = 1 as well. Another possibility for a proof is to use satisability equivalence of F and DP l (F), where DP l (F) denotes the result of substituting all clauses of F containing l or l by their non-tautological resolvents (on l): If C is blocked for l w.r.t. F, then addition of C to F has no eect on DP l. Or one uses completeness of (non-tautological) SL-resolution for any start clause from a minimal unsatisable clause-set and for any selected resolution literal from that clause. To become familiar with the concept of blocked clauses, and for later use, we determine the blocked clauses without new variables for PHP: 3 Proof. For G CLS we dene the Resolution Graph R(G) as the undirected graph (without parallel edges and loops) whose vertices are the clauses of G, and an edge joins clauses C and C i C C =1 holds (i.e., C and C have exactly one clashing literal). Now consider R(F) for the special F here, where all clauses contain the same variables. R(F) is a sub-graph of R(F ) where F is that clause-set containing all 2 n clauses C with Var(C) = Var(F) (n = Var(F) ). If for 0 F 2 n no clause of F would be blocked w.r.t. F, then F would be a connected component of R(F ), since for every clause of F and every literal in it there is exactly one resolution partner (in F ). But if G is unsatisable (for any G), then at least one connected component of R(G) is also unsatisable. (The proof for that is not completely trivial since in the def. of R(G) not all complementary literal pairs correspond to edges.) Now we obtain a contradiction since F is minimally unsatisable.

10 158 O. Kullmann / Discrete Applied Mathematics (1999) Lemma 4.2. For C CL and i {1;:::;n}; j {1;:::;n 1} the following holds: 1: C B vi; j (PHP) {v 1;j ;:::;v n;j } C. 2: C B vi; j (PHP) k {1;:::;n 1}\{j} : {v i; j ; v i;k } C Blocked Clauses for SAT-decision Two new features are invented for the improved 3-SAT-algorithm in [14] (see also [13]): Generalized Autarkness and Blocked Clauses 4. The use of blocked clauses (without new variables) in [14] for 3-SAT-decision can be summarized as follows: 5 1: Testing of a not-blocked 2-clause {a; b}, i.e., branching via ( a 0;b 1 ; a 1 ), has a greater impact on the formula than testing a blocked 2-clause. The existence of a not blocked 2-clause is established by a combination of autarkness with the blocking concept (called Br-Autarkness in [14]). 2: The main idea in [14] is to consider not only the decrease in the number of variables in the course of the algorithm, but also to consider the alteration in the number of 2-clauses: An increase in the number of 2-clauses can shorten the computation. Now in the situation of point 7.2 in Remark (given after Denition 4.1) one can add blocked 3-clauses, which become new 2-clauses after applying l 1. 3: And also the situation of point 7.1 in the Remarks (given after Denition 4.1) is applied, but with a dierent eect: Here, by applying l 1, more variables vanish because of 1-clause eliminations. 4: Blocked 3-clauses are eliminated to establish some normal form. (Blocked 2-clauses are not eliminated since they count for the analysis.) The application of point 7 of the Remark to SAT-decision is already prescribed in [18], called Complement Search there (in a more general setting). The elimination of blocked clauses and the (implicit) addition of blocked 2-clauses is also part of the L -algorithm from [15] with the improved bound 2 1=10 for SAT-decision. 5. An upper bound for the speed up achieved by adding one blocked clause Lemma 5.1. For F CLS; a clause C CL; a literal l and a partial assignment fullling (C) 1and Var(l) Var( ) we have C B l (F) C B l ( F): (The opposite direction is true under the additional assumption Var( ) Var(C):) 4 The Autarkness Principle has been introduced in [17] to ensure the existence of a 2-clause in the course of the (recursive) decision procedure, and also [16] used a similar scheme. Generalized Autarkness is a branching scheme which gives an alternative branching in the case of an arbitrary number of new 2-clauses. See [14]. 5 Ref. [19] applies our concept of blocked clauses in a manner similar to 2 and 3 below.

11 O. Kullmann / Discrete Applied Mathematics (1999) Now we are able to prove an upper bound for the speed up achieved by adding one blocked clause. Lemma 5.2. For F CLS and C B(F) we have Comp R (F)6 C Comp R (F {C})+ C 1: Proof. (By induction on C ) C = 1: Thus C = {l}, where l is pure for F Comp R (F) = Comp R (F {C}). C 1: Let C B l (F) and choose x C\{l}. x 0 C =C\{x} B l ( x 0 F) by the previous lemma. Thus by the induction hypothesis (and Lemma 3.1, part 2): Comp R ( x 0 F) 6 ( C 1) Comp R (( x 0 F) ( x 0 {C})) + C 2 =( C 1) Comp R ( x 0 (F {C})) + C 2 6 ( C 1) Comp R (F {C})+ C 2: On the other hand we have Comp R ( x 1 F) = Comp R ( x 1 (F {C}))6Comp R (F {C}): Together by Lemma 3.3 Comp R (F) 6 Comp R ( x 0 F) + Comp R ( x 1 F) C Comp R (F {C})+ C 1: 6. Blocked extensions and generalized extended resolution Denition 6.1. A sequence C 1 ;:::;C n (n 0) of clauses is called a blocking sequence for the clause-set F CLS i i {1;:::;n}[C i B in (F\{C 1 ;:::;C i 1 })] holds. Literals l 1 ;:::;l n are called blocking literals for C 1 ;:::;C n i all C i are blocked for l i w.r.t. F\{C 1 ;:::;C i 1 }. A blocking sequence C 1 ;:::;C n is called maximal i B in (F\{C 1 ;:::;C n })=. Note that a blocked clause w.r.t. F may or may not be in F, while a blocking sequence for F is always contained in F. Lemma 6.1. Two maximal blocking sequences for F CLS dier only by a permutation of the clauses. Proof. Consider two maximal blocking sequences C 1 ;:::;C n and D 1 ;:::;D m for F. We prove by induction that for i {1;:::;n} we have C i {D 1 ;:::;D m }

12 160 O. Kullmann / Discrete Applied Mathematics (1999) and thus by symmetry the assertion follows. Suppose {C 1 ;:::;C i 1 } {D 1 ;:::;D m }. Assume C i {D 1 ;:::;D m }. Then C i F\{D 1 ;:::;D m }. Furthermore we have F\{D 1 ;:::;D m } F\{C 1 ;:::;C i 1 }, and thus by Remark 2 after Denition 4.1 also D 1 ;:::;D m ;C i would be a blocking sequence for F contradicting the maximality of D 1 ;:::;D m. Denition 6.2. By the previous lemma we are justied to dene the kernel K(F) for F CLS: K(F):=F\{C 1 ;:::;C n } where C 1 ;:::;C n is a maximal blocking sequence for F: K is a kernel operator : F 1 F 2 K(F 1 ) K(F 2 ), K(K(F)) = K(F), K(F) F. Lemma 6.2. For F CLS we have K(F) sat F. Denition 6.3. F CLS is called a Blocked Extension for F CLS i K(F F )=K(F) holds. Lemma 6.3. F blocked extension for F F sat F F. Proof. F sat K(F)=K(F F ) sat F F. Remark 1: The property F is a blocked extension for F is polynomially decidable. 2: If C is blocked w.r.t. F, then {C} is a blocked extension for F. If K(F) =F holds, then also the opposite direction is true (for example: F minimal unsatisable K(F)=F). 3: F 1 F 2, F 1 F 2, F 2 blocked extension for F 2 F 1 blocked extension for F 1. 4: Consider F CLS, an arbitrary propositional formula A (over VA), and a variable v VA\(Var(F) Var(A)). Suppose E is a CNF of v A (i.e., the formula C E l C l is (logically) equivalent to v A). Then E is a blocked extension for F: Every clause C E is blocked for v or v w.r.t. E (otherwise a clause C with v Var(C ) would follow from v A). 5: Hence the Extension Rule (and its obvious generalization) is covered by the use of blocked extensions. 6: Unlike extensions by the Extension Rule, blocked extensions are not conservative extensions since for example {{a; b}} is a blocked extension for {{a; b}} while {{a; b}} does not contain new variables and does not follow from {{a; b}}. 7: If F is a blocked extension for F with F F =, then F SAT holds, because F is also a blocked extension for SAT. More generally it holds for any blocked extension F for F CLS: S F[F\S SAT (F F )\S SAT]:

13 O. Kullmann / Discrete Applied Mathematics (1999) The next lemma provides an alternative characterization of blocked extensions by iterated addition of single clauses. Lemma 6.4. A clause-set F CLS is a blocked extension for F CLS i there exists an order F = {C 1 ;:::;C n } such that i {1;:::;n}[{C i } is a blocked extension for F {C 1 ;:::;C i 1 }] ( ) holds; and also i for all orders F ={C 1 ;:::;C n } condition ( ) holds. Thus the notion of Blocked Extension overcomes the special order inherent to the Extension Rule. Proof. (a) Consider a blocked extension F for F and an (arbitrary) order F = {C 1 ;:::;C n }. Then ( ) is a simple conclusion from Remark 2 of Section 4. (b) Consider F CLS, F = {C 1 ;:::;C n } such that ( ) holds. W.l.o.g. F F =. Suppose that D 1 ;:::;D m is a maximal blocking sequence for F {C 1 ;:::;C n }.We have to show {C 1 ;:::;C n } {D 1 ;:::;D m }. Assume that there is a maximal index i {1;:::;n} with C i {D 1 ;:::;D m }. We know that {C i } is a blocked extension for F {C 1 ;:::;C i 1 }. Now, since {C i+1 ;:::;C n } {D 1 ;:::;D m } holds, by Lemma 6.1 we conclude C i {D 1 ;:::;D m } contradicting our assumption. Lemma 6.5. Call clause-set F CLS a simple blocked extension for F CLS i there is an order F \F = {C 1 ;:::;C n } such that i {1;:::;n}[C i is blocked w:r:t: F {C 1 ;:::;C i 1 }] holds. Now for F; F CLS the following assertions are equivalent: 1: F is a blocked extension for F. 2: There exists F 0 F such that F (F\F 0 ) is a simple blocked extension for F 0. 3: F (F\K(F)) is a simple blocked extension for K(F). Proof. W.l.o.g. F F =. (i) (iii): Reverse the order of a maximal blocking sequence for F F and obtain a sequence required in the denition of simple blocked extensions. (iii) (ii): F 0 := K(F). (ii) (i): Under the assumption (ii) we have K(F F ) F 0 and thus F is a blocked extension for F. Denition 6.4. A Generalized Extended Resolution Proof P (for short: GER proof) for F CLS is a pair P =(F ; (C 1 ;:::;C n )), such that F is a blocked extension for F, and (C 1 ;:::;C n ) is a resolution proof for F F. The length of P is n. For F CLS we dene: Comp GER (F):=inf{n N: P GER proof for F of length n}: A GER proof P =(F ; (C 1 ;:::;C n )) is, more specically, an (a; b)-resolution Proof for functions a; b : CLS N 0 {+ }, if is an (a; b)-extension for F, which

14 162 O. Kullmann / Discrete Applied Mathematics (1999) means that the length of the new clauses is bounded by a: C F [ C 6a(F)], and the number of new variables is bounded by b: Var(F )\Var(F) 6b(F). Remark Comp (a;b) (F):=inf{n N: P (a; b)-resolution proof for F of length n}: 1: 1.1. Comp (0;0) = Comp (0; ) = Comp R ; 1.2. Comp (3; ) 6Comp ER ; 1.3. Comp ( ; ) = Comp GER. 2: Comp (a;b) (F) = min{comp R (F F ): F blocked (a; b)-extension for F}. 3: a a, b b Comp (a;b) 6Comp (a ;b ). 4: To obtain more general concepts of proofs, the concept of a kernel K(F) could be generalized to any polynomially computable K : CLS CLS such that F CLS: K(F) sat F holds. 7. Polynomial simulation of GER by ER Denition 7.1. For clause-sets F 1 ;F 2 CLS we dene F 1 = F2 i there is a renaming with (F 1 )=F 2. Denition 7.2. F CLS is called a Normal Extension for F CLS i there is m 0 with F =3m and there is an order F ={C 0 ;:::;C 3m 1 } such that for i {0;:::; m 1} the following holds: 1. C 3i = {v i ; a i ; b i }, a i ;b i VA;a i b i ;v i VA\(Var(F {C 0 ;:::;C 3i 1 }) {a i ;b i }). 2. C 3i+1 = {v i ;a i };C 3i+2 = {v i ;b i }. An Extended Resolution proof P (for short: ER proof )off 2 CLS from F 1 CLS is a pair P =(F 1 ; (C 1;:::;C n )) such that F 1 is a normal extension for F 1, and (C 1 ;:::;C n ) is a resolution proof of F 2 from F 1 F 1. The length of P is n. AnER proof for F CLS is an ER proof of { } from F. For F CLS we dene Comp ER (F):=inf {n: P ER proof for F of length n}: Additionally we dene for F 1 ;F 2 CLS: Comp ER (F 1 ;F 2 ):=inf {n: F 2 = F 2 P ER proof of F 2 from F 1 with length n}: Remark 1. The restrictions of signs in the new clauses is due to [21]. Every other distribution of signs has the same eect (e.g. we could have introduced the new clauses {v; a; b}; {v; a}; {v; b}, which are the CNF of v a b). 2. Every normal extension is a blocked extension, every ER proof is a GER proof, but not vice versa.

15 O. Kullmann / Discrete Applied Mathematics (1999) If F is a normal extension for F, then F is also a normal extension for every F + with Var(F + ) Var(F). This is not the case for blocked extension, if they contain blocked clauses which are blocked only for literals whose variables are already in F: These blocked clauses depend on the special shape of F. 4. Comp ER (F 1 ;F 2 ) is introduced for the purpose of simulation: F 2 shall contain (in renamed form) F 1 together with the blocked extension F 1 (which shall be simulated). 5. Comp ER (F) = Comp ER (F; { }). Lemma 7.1. For every F CLS and every blocked extension F for F we have Comp ER (F; F F )6O( (F F ) 5 ); where (F):= C F C. Corollary 7.2. For F CLS we have Comp ER (F)6O(Comp GER (F) 10 ). Proof. By Lemma 7.1 and point 3 of Remark that follows Denition 3.1. Proof of Lemma 7.1. We proceed by showing in I IV how to handle F s of increasing generality. In order to increase the strength of the induction hypothesis in fact we derive exactly F F (i.e., for condition 1 of Denition 3.1 here we have = instead of ). I. F = {{v; a; b}; {v; a}; {v; b}}; v VA\(Var(F) Var({a; b})), a; b LIT. (If a = b holds, then actually we have F = {{v; a}; {v; a}}.) Here we have Comp ER (F; F )6O(1). Proof. Case A: a b A.1: a; b VA: A.2: a VA;b VA. Introduce {w; a; b}; {w; a}; {w; b}; {v; w; a}; {v; w}; {v; a}: Infer {v; w; a}; {w; b} {v; a; b}; {w; a; b}; {v; a} {v; w; b}; {v; w; b}; {v; w} {v; b}: A.3: a VA;b VA. Use A.2. A.4: a; b VA. Introduce (by A.2) {w; a; b}; {w; a}; {w; b}; {v; w; b}; {v; w}; {v; b}:

16 164 O. Kullmann / Discrete Applied Mathematics (1999) Infer {v; w; b}; {w; a} {v; a; b}; {w; a; b}; {v; w} {v; a; b}; {v; a; b}; {v; b} {v; a}: Case B: a = b Introduce (by A) {w; a; x}; {w; a}; {w; x}; {v; w; a}; {v; w}; {v; a}: Infer {v; w; a}; {w; a} {v; a}: II. F ={{v; a 1 ;:::;a n }; {v; a 1 };:::;{v; a n }}, n 0, v VA\(Var(F) Var({a 1 ;:::;a n })), a 1 ;:::;a n LIT. Here we have Comp ER (F; F )6O(n 2 ). Proof (By induction). n = 0 Introduce (by I) {w; x}; {w; x}; {u; w; x}; {u; w}; {u; x}; {v; u}; {v; u}: Infer {w; x}; {u; x} {w; u}; {w; u}; {w; u} {u}; {u}; {v; u} {v}: n 1: Introduce (by induction hypothesis) {w; a 1 ;:::;a n 1 }; {w; a 1 };:::;{w; a n 1 } {v; w; a n }; {v; w}; {v; a n } Infer {w; a 1 ;:::;a n 1 }; {v; w; a n } {v; a 1 ;:::;a n }; for i {1;:::;n 1}: {v; w}; {w; a i } {v; a i }: III. F = {C 0 };C 0 blocked w.r.t. F. Here we have Comp ER (F; F F )6O( (F) 3 + C 0 2 ). Proof. Let C 0 = {x; x 1 ;:::;x m } ( C 0 = m +1;m 0);C 0 blocked for x w.r.t. F. Dene p := {C F: x C} ; n := {C F: x C} : Let C 1 ;:::;C p be the x-occurrences in F ({C 1 ;:::;C p } = {C F: x C}), and let D 1 ;:::;D n be the x-occurrences in F ({D 1 ;:::;D n } = {C F: x C}). Furthermore we use C i \{x} = {c i; j } j {1;:::; Ci 1} (i {1;:::;p}); D i \{x} = {d i; j } j {1;:::; Di 1} (i {1;:::;n}):

17 O. Kullmann / Discrete Applied Mathematics (1999) Case A: n =0 By II introduce {v i ;c i;1 ;:::;c i; Ci 1}; {v i ; c i;1 };:::; {v i ; c i; Ci 1} for i {1;:::;p}; {v p+1 ;x 1 ;:::;x m }; {v p+1 ; x 1 };:::; {v p+1 ; x m }; {v; v 1 ;:::;v p+1 }; {v; v 1 };:::; {v; v p+1 }: Infer for i {1;:::;p} {v i ;c i;1 ;:::;c i; Ci 1}; {v; v i } {v; c i;1 ;:::;c i; Ci 1}: And {v p+1 ;x 1 ;:::;x m }; {v; v p+1 } {v; x 1 ;:::;x m }: Thus we inferred x v ({C 1 ;:::;C p ;C 0 }), where x v denotes the renaming : {v; x} {v; x} with (v)=x and (x)=v. Case B: n 0 Step (a): For i {1;:::;n} introduce {v i ;d i;1 ;:::;d i; Di 1}; {v i ; d i;1 };:::;{v i ; d i; Di 1}: And infer for i {1;:::;n}: {x; d i;1 ;:::;d i; Di 1}; {v i ; d i;1 };:::;{v i ; d i; Di 1} {x; v i }: Step (b): Because C 0 is blocked for x w.r.t. F, for every i {1;:::;n} there is z i {x 1 ;:::;x m } such that z i {d i;1 ;:::;d i; Di 1} holds. Introduce {v; v 1 ;:::;v n }; {v; v 1 };:::;{v; v n }: Infer {v; v 1 ;:::;v n }; {v 1 ;z 1 };:::;{v n ;z n } {v; z 1 ;:::;z n }: And for i {1;:::;n} {v i ;d i;1 ;:::;d i; Di 1}; {v; v i } {v; d i;1 ;:::;d i; Di 1}: Step (c): Introduce {w; v; x 1 ;:::;x m }; {w; v}; {w; x 1 };:::;{w; x m }. Infer (remember: {z 1 ;:::;z n } {x 1 ;:::;x m }): {v; z 1 ;:::;z n }; {w; z 1 };:::;{w; z n } {v; w}: And {v; w}; {w; v; x 1 ;:::;x m } {v; x 1 ;:::;x m }: Step (d): Infer (using (a) and (b)) {v; v 1 ;:::;v n }; {x; v 1 };:::;{x; v n } {v; x}: And for i {1;:::;p}: {v; x}; {x; c i;1 ;:::;c i; Ci 1} {v; c i;1 ;:::;c i; Ci 1}: Altogether we inferred x v ({C 1 ;:::;C p ;D 1 ;:::;D n ;C 0 }):

18 166 O. Kullmann / Discrete Applied Mathematics (1999) IV. F = {C}; {C} blocked extension for F. Here we have Comp ER (F; F F )6 O( (F) 4 + C 2 ). Proof. Let C 1 ;:::;C m be a maximal blocking sequence for F {C}. There is i {1;:::;m} with C i = C. Now by III add subsequently C j to (F\{C m ;:::;C i+1 }) {C i ;C i 1 ;:::;C j+1 } for j = i; i 1;:::;1: Finally Lemma 7.1 is an immediate consequence of IV and 2-clauses in Blocked Extensions Lemma 8.1. Assume that F is a blocked extension for F with {l} F such that l is pure for F. Then also l 1 F is a blocked extension for F; and we have Proof. Comp R (F ( l 1 F ))6Comp R (F F ): Comp R (F F ) Comp R ( l 1 (F F )) (by Lemma 3:1 part 2) =Comp R ( l 1 F l 1 F ) =Comp R ( l 1 F {C F: C} l 1 F ) (by Lemma 3:1 part 3(a)) =Comp R (F l 1 F ) (since l is pure for F): To show that l 1 F is a blocked extension for F, consider a maximal blocking sequence C 1 ;:::;C n for F F with blocking literals l 1 ;:::;l n. Because of {l} F we have i {1;:::;n}: l i l: (Before {l} is eliminated, no clause can be blocked for l, and for {l} being blocked l must be pure.) Dene I:={i {1;:::;n}: l C i }, and for i I: C i := l 1 C i. Now (C i ) i I is a blocking sequence for l 1 (F F ) because of ( ) (with blocking literals l i ). By denition l 1 F {C i } i I holds, and therefore l 1 F is a blocked extension for l 1 F, and thus also for F (clauses with pure literals are blocked). Corollary : Comp (1; ) = Comp (1;0). 2: For F CLS with K(F)=F( B in (F)= ) we only have to consider blocked extension without any 1-clause: Comp (a;b) (F) = min{comp R (F F ): F blocked (a; b)-extension for F and C F [ C 2]}: Proof. For part 2 note that if {l} is blocked for F, then l must be pure for F. ( )

19 O. Kullmann / Discrete Applied Mathematics (1999) The assumption K(F) = F is fullled for example, if F is minimally unsatisable. But for F with K(F) F in general even (1; 0)-resolution proofs can cause exponential speed ups (compared to (0; 0)-resolution). To prove this we need the following lemma: Lemma 8.3 (Cook [2] or Cook and Reckhow [6]). Comp ER (PHP)6O(n 4 ). Proof. We give the proof in some detail because later we can make use of it. The idea is simple: Suppose an assignment for the variables v i; j (i {1;:::;n}; j {1;:::;n 1}) is given fullling PHP. We want to derive a fullling assignment for PHP n 1 from this. Therefore, we introduce new variables v 1 i; j for i {1;:::;n 1}; j {1;:::;n 2} and project the assignment from domain (v i; j ) to domain (v 1 i; j) by v 1 i; j v i; j (v n;j v i;n 1 ) ( ) (critical are only the pigeons in the nth row or the (n 1)th column: either there is a pigeon at (n; n 1), and then we simply forget this pigeon, or there are uniquely determined pigeons at (n; j) and (i; n 1) for i {1;:::;n 1}; j {1;:::;n 2}, and in this case these two pigeons are collapsed into a new one at (i; j)). In this way one gets a fullling assignment for PHP n 1 (but with variables (v 1 i; j) instead of (v i; j )), and by iteration of this process we eventually reach a contradiction. This idea is put into work as follows: Consider new variables v r i; j; x r i; j for r {1;:::;n 2}; i {1;:::;n r}; j {1;:::; n r 1}. Dene Ci; r j(1) := {xi; r j; v r 1 n r+1;j ; vr 1 i;n r }; Ci; r j(2) := {xi; r j ;vr 1 i;n r };Cr i; j(3):={xi; r j ;vr 1 n r+1;j }; Ci; r j(4) := {vi; r j ;vr 1 i; j ;xi; r j}; Ci; r j(5) := {vi; r j; vi; r 1 j }; Ci; r j(6):={vi; r j; xi; r j } where vi; 0 j :=v i; j. And n 2 E n := En; r r=1 En r := {Ci; r j(1);:::;ci; r j(6)} i {1;:::; n r} : j {1;:::; n r 1} E n is a normal extension for PHP (the order of introduction is: levels En;:::;E 1 n n 2 ; within the levels the order of the 6-clause blocks is arbitrary, but within the blocks choose numerical order), and we want to show Comp R (PHP E n )6O(n 4 ). To that end rst derive for r {1;:::;n 2}; i {1;:::;n r}; j {1;:::;n r 1} the following four clauses from the corresponding 6-clause block in En r (using three resolution steps for each triple (r; i; j)): Di; r j(1):={vi; r j ;vr 1 i; j ;v r 1 n r+1;j }; Dr i; j(2):={v r i; j ;vr 1 i; j ;v r 1 i;n r }

20 168 O. Kullmann / Discrete Applied Mathematics (1999) D r i; j(3):={v r i; j; v r 1 n r+1;j ; vr 1 i;n r }; Dr i; j(4):={v r i; j; v r 1 i; j } (these four clauses are a CNF of the generalization of ( )). Because of the completeness of resolution we have for r {1;:::;n 2}: r 1 (PHP n r+1 ) En r r (PHP n r ); ( ) where E r n :={D r i; j(1);:::;d r i; j(4)} i {1;:::; n r} j {1;:::; n r 1} and the renamings r are dened by r := v i; j v r i; j i {1;:::; n r} j {1;:::; n r 1} ( 0 is the identity). Fortunately in ( ) each single resolution proof of a long (positive) clause is of length O(n), and of a short (negative) clause is of length O(1), and thus altogether we obtain Comp R (PHP E n )6O(1)O(n 3 ) + (O(n)O(n) + O(1)O(n 3 ))O(n)=O(n 4 ): Lemma 8.4. There is a sequence (F n ) n N of (unsatisable) clause-sets (lim n (F n ) = ) with Comp (1;0) (F n )6O(n 4 ); but Comp R (F n ) = Comp R (PHP) c n (c as dened in Theorem 1). Proof. Consider E n from the proof of the previous lemma. Choose a new variable v VA\Var(PHP E n ). Dene E n := {C {v}: C E n }; F n := PHP E n: With the help of Lemma 3.1 part 3(a) we get Comp R (F n ) = Comp R (PHP): But {{v}} is a blocked extension for F n, and thus Comp (1;0) (F n )6O(n 4 ): (We used the simple fact, that if F is a blocked extension for F and v VA\Var(F F ), then also {C {v}: C F } {{v}} is a blocked extension for F.) clauses in blocked extensions Lemma 8.5. Assume that {C 0 } is a blocked extension for F CLS and that C 1 ;:::; C n is a maximal blocking sequence for F {C 0 } with blocking literals l 1 ;:::;l n. There is i {1;:::;n} with C i = C 0. If now there is x C 0 \{l i } with C F: {l i ; x} * C; then also {C 0 \{x}} is a blocked extension for F.

21 O. Kullmann / Discrete Applied Mathematics (1999) Proof. C 1 ;:::;C i 1 ;C 0 \{x} is a blocking sequence for F {C 0 \{x}} with blocking literals l 1 ;:::;l i (the assumption {l i ; x} * C for C F ensures that no original blockade has been destroyed). Corollary 8.6. If {{a; b}} is a blocked extension for F CLS such that C F: {a; b} * C holds; then at least one of {{a}} or {{b}} is also a blocked extension for F. Corollary 8.7. Assume a blocked extension F for F CLS with {a; b} F such that C F F : {a; b} * C holds. Then for x = a or x = b also F :=(F \{a; b}) {x} is a blocked extension for F with Comp R (F F )6Comp R (F F ). Proof. Use Lemmas 8:6 and 6:4 and part 1 of Lemma 3.1. Hence we can restrict (a; b)-resolution proofs for F CLS without aecting Comp (a;b) (F) to such blocked extensions F which fulll: 1: Lit (F ) Lit (F F ), 2: {l} F l Lit (F), 3: {a; b} F C F F : {a; b} C. Now consider a (2; )-extension F for F fullling restrictions 1 3. Suppose {l; x} F with Var (l) Var (F). Then by 3 also {l; x} F holds, and furthermore (also by 3 and by F SAT): {l; x}; {l; x} F. The next lemma shows how to eliminate the new variable Var (l) in this situation. Lemma 8.8. Assume F CLS; n 1; C i CL; C i = {l; x i } for i {1;:::;n}; Var(l) Var (F); Var (x i ) Var (x j ) for i j. Assume that {C 1 ;:::;C n ; C 1 ;:::;C n } is a blocked extension for F. Then also l x 1 ({C 1 ;:::;C n ; C 1 ;:::;C n })={{x 1 ;x 2 };:::;{x 1 ;x n }; {x 1 ; x 2 };:::;{x 1 ; x n }} is a blocked extension for F. Proof. Consider a maximal blocking sequence D 1 ;:::;D m for F {C 1 ;:::;C n ; C 1 ;:::;C n } with blocking literals l 1 ;:::;l m.weuse:= l x 1. There are p; q {1;:::;m} with D p ={l; x 1 };D q ={l; x 1 }. Dene I:={1;:::;m}\{p; q}. In the following we prove that (D i );i I is a blocking sequence for F ({C 1 ;:::;C n ; C 1 ;:::;C n })=F {(C 2 );:::;(C m );(C 2 );:::;(C m )} with blocking literals (l i ) (and thus the assertion follows). Consider i I. We have to show that (D i ) is blocked for (l i ) w.r.t. (F {(C 2 );:::;(C m );(C 2 );:::;(C m )})\{(D j )} j I; j i :

22 170 O. Kullmann / Discrete Applied Mathematics (1999) I. D i F( (D i )=D i ): We have two critical cases: l i = x 1 or l i = x 1. W.l.o.g. l i = x 1. Because of Var (l) Var (F) we have q6i 1. Thus l q = l holds, and hence {C 2 ;:::;C n } {D 1 ;:::;D q 1 }, yielding {(C 2 );:::;(C n )} {(D 1 );:::;(D q 1 )}. Therefore, all additional x 1 -clauses have been deleted before step i: II. D i = C j for one j {2;:::;n}: Here only l i = l is critical ((l i )=x 1 ). Now {C 2 ;:::;C n }\{C j } {D 1 ;:::;D i 1 } holds, and hence, as in I, it follows that all additional x 1 -clauses have been eliminated before step i. Furthermore we have q6i 1, and thus here l q = x 1 must hold. Hence x 1 is pure for F\{D 1 ;:::;D i 1 } = F\({D 1 ;:::;D i 1 }): III. D i = C j for one j {2;:::;n}: Analogous to II. Lemma 8.9. Comp (2; ) = Comp (2;0). Proof. By the above argumentation and by Lemma 3.4. We conclude this section by showing that (2; 0)-resolution cannot be bounded polynomially by (1; 0)-resolution: Lemma There is a sequence (F n ) n N of (unsatisable) clause-sets (lim n F n = ) with Comp (2;0) (F n )6O(n 4 ); but Comp (1;0) (F n ) = Comp R (PHP) c n. Proof. Consider E n from the proof of Lemma 8.3 and dene n 2 F n :=PHP {C En: 1 C =3} En: r r=2 Now {C E 1 n: C =2} isa(2; 0)-extension for F n, and thus we can estimate Comp (2;0) (F n )6Comp R (PHP E n )6O(n 4 ): On the other hand consider a (1; 0)-extension F for F n. Since the clauses C 1 i; j = {x 1 i; j; v n;j ; v i;n+1 } are not blocked for v n;j or v i;n+1 w.r.t. PHP, and Var (x 1 i; j) Var ( n 2 r=2 Er n), the literals x 1 i; j are pure in F n F. Thus, using Lemma 3.1, part 3(a), we get ) n 2 Comp R (F n F ) = Comp R (PHP En r F = Comp R (PHP): r=2 9. An exponential lower bound for GER without new variables In order to obtain a lower bound for Comp ( ;0) we have to strengthen Theorem 1 by adding such clauses to PHP which are fullled by every critical assignment :

23 O. Kullmann / Discrete Applied Mathematics (1999) Denition 9.1. Let V n :={v i; j } i {1;:::; n} j {1;:::; n 1} denote the set of variables of PHP (see Section 3). An i-critical assignment (i {1;:::;n}) for the pigeonhole formula PHP is a partial assignment with variables Var ( ) =V n such that for L i :={v i; j } j {1;:::;n 1} PHP p n we have (L i )=0 and (PHP\{L i })=1: is simply called a critical assignment if is i-critical for any i. ByCASS n we denote the set of all critical assignments for PHP. Critical assignments correspond to the dierent ways of distributing n 1 pigeons on n 1 holes such that no hole contains two pigeons. Theorem 2. Let E n be the set of clauses C CL with Var (C) V n such that for all CASS n we have (C)=1. Then for c:=2 1=20 and n 1676 the lower bound Comp R (PHP E n ) c n holds. 6 Proof. The point here is, that the lower bound of [8] for PHP (see also [3,20,1]) immediately can be generalized to PHP E n, since the proof follows only the trace of long clauses, and this by means of critical assignments, i.e., only clauses which are falsied by some critical assignment are interesting here. For readers not familiar with that proof we give a survey in the following, using the improved version of [1] Outline of proof We use for abbreviation: N:={1;:::;n}; N :={1;:::;n 1}. In this subsection we consider only clause-sets F; G and clauses C; D with variables from V n. The next notions reect the concentration on critical assignments. Denition 9.2. For clause-sets F and clauses C we dene: F = cn C : CASS n : (F)=1 (C)=1: A sequence (C 1 ;:::;C k ) of clauses is a = cn -proof of from F i C k = holds, and for each i {1;:::;k} either we have C i F or there is G {C 1 ;:::;C i 1 } with G 62 and G = cn C i (or both). Finally by n (C) we denote the minimal number of clauses from PHP p n implying C with respect to = cn : n (C):=min{ F : F PHP p n F = cn C}: 6 Note PHP E n = PHP p n E n.

24 172 O. Kullmann / Discrete Applied Mathematics (1999) Note that E n = {C: = cn C}, and thus = cn -proofs can arbitrarily introduce clauses from E n. Every resolution proof of from PHP is a = cn -proof of from PHP p n. Lemma 9.1. Consider a = cn -proof (C 1 ;:::;C k ) of from PHP p n ; where all C i are positive (i.e.; C i V n ): Then there is r {1;:::;k} with C r 2=9n 2. Proof. Consider r {1;:::;k} and F PHP p n with F = cn C r ; F =(C r ). Let I r :={i {1;:::;n}: L i F}. Consider i 1 I r. Due to the minimality of F there is i1 CASS n such that i1 is i 1 -critical and i1 (C r )=0. For i 2 N\I r consider the (uniquely determined) k(i 1 ;i 2 ) N with i1 (v i2;k(i 1;i 2))=1. Note that for i 2 i 2 we have k(i 1;i 2 ) k(i 1 ;i 2 ), and that i 1 (v i1;k(i 1;i 2)) = 0 holds. Obtain i2 i 1 from i1 by ipping the values for v i1;k(i 1;i 2) and v i2;k(i 1;i 2). i2 i 1 is an i 2 -critical assignment, thus i2 i 1 (F) = 1 holds, and we can infer i2 i 1 (C r )=1. Since C r is a positive clause, we conclude v i1;k(i 1;i 2) C r. It follows that C r I r N\I r =(C r )(n (C r )). Due to C PHP p n : (C)=1; ( )= n, and G = cn C (C)6 D G (D); there is r {1;:::;k} with 1=3n6(C r )62=3n. Using elementary calculus we get C r n=3(n n=3)=2=9n 2. The restriction to positive clauses in the previous lemma is justied by the next denition (and lemma). Denition 9.3. For a clause C we dene C + procedure: C + :=C; FOR j N DO IF =1 i N : v i; j C + THEN C + :=(C + \{v i; j }) {v i ;j} i N \{i} ELSE IF 2 i N : v i; j C + THEN C + :=(C + \{v i ;j} i N ) {v i ;j} i N END FOR. as the outcome of the following Lemma 9.2. For a clause C and any critical assignment we have (C) = (C + ). Thus; if (C 1 ;:::;C k ) is a = cn -proof of from PHP p n ; then also (C + 1 ;:::;C+ k ). By the next denition (and lemma) we are enabled to reduce PHP p n to PHP p n 1. Denition 9.4. For i N; j N let i; j be the partial assignment with i; j(v i; j )=1; i; j (v i; j )= i; j (v i ;j)=0 for i N\{i}; j N \{j}; and undened else.

25 O. Kullmann / Discrete Applied Mathematics (1999) Lemma 9.3. There is a renaming i; j :(V n \Var ( i; j )) V n 1 with i; j ( i; j PHP p n ) = PHP p n 1 : For CASS n 1 we have ( i; j ) i; j CASS n and ( i; j ( i; j {C})) = (( i; j ) i; j )(C) for all clauses C. Thus in case of G = cn C we either have i; j (C) =1or i; j ( i; j G) = cn 1 i; j ( i; j C). Now we are ready to prove Theorem 2. Assume there is a resolution proof P = (C 1 ;:::;C k )of from PHP E n with k c n. Consider P + =(C + 1 ;:::;C+ k ). By Lemma 9.2 the sequence P + is a = cn -proof of from PHP p n. Dene the number of large clauses in P + by #lc(p + ):= {i {1;:::;k}: C q V n } ; where 0 q 1 is a parameter. Trivially #lc(p + ) c n. There must be a variable v i; j which appears in at least q #lc(p + )-many large clauses, and thus i; j makes at least q #lc(p + )-many large clauses come true. Obtain P + from P + by deleting clauses from P + which become true by i; j, and applying rst i; j and then i; j to the rest of the clauses (see Lemma 9.3). By Lemma 9.3 P + is a = cn 1 -proof of from PHP p n 1. For the number of large clauses (which still refers to the size of V n, and not of V n 1 ) we know #lc(p + )6(1 q)#lc(p + )6(1 q)k: By repeating this process at most 1 + log (1 q) 1k times we are sure that no large clause is left, and we obtain a = c n -proof P of from PHP p n with n n 1 + log (1 q) 1k n 1+(1=20)nlog (1 q) 12 : Lemma 9.1 yields the existence of a clause of length at least (2=9)n 2 in P, but on the other hand, since all large clauses have been eliminated, every clause in P has length strictly less than qn(n 1), which yields a contradiction for suciently large n, when we choose q = 1=10 for example. By using q = 0: numerical calculations show that for all n 1676 we have 2=9n 2 qn(n 1) (using ( )) Applications to GER Lemma 9.4. n 1676 : Comp ( ;0) (PHP) c n. Proof. By Theorem 2 and Lemma 4.2. Lemma 9.5. There is a sequence (F n ) n N of (unsatisable) clause-sets (lim n F n = ) with Comp (3;0) (F n )6O(n 4 ); but Comp (2;0) (F n ) c n.