Unsatisfiable CNF Formulas need many Conflicts

Unsatisfiable CNF F orm ulas need man y Conflicts ∗ Dominik Sc heder ETH Z ¨ uric h dscheder@in f.ethz.ch Philipp Zumstein ETH Z ¨ uric h zuphilip@in f.ethz.ch No v em b er 11, 2021 Abstract A pair of clauses in a CNF f ormula constitutes a conflict if there is a v ariable that o ccurs po sitiv ely in o ne cla use and negatively in the other. A CNF formula without any conflicts is satisfiable. The Lov´ asz Lo cal Lemma implies that a k -CNF for m ula is satisfiable if eac h clause conflicts with at most 2 k e − 1 clauses. It do es not, how ever, give any go o d b ound on how many conflicts an unsatisfiable form ula has globally . W e show here that every unsatisfiable k -CNF formula requir es Ω(2 . 69 k ) conflicts and there exist unsatisfiable k -CNF form ulas with O (3 . 51 k ) conflicts. 1 In tro duction A b oolean form ula in conjunctiv e normal f orm, a CNF formula for sh ort, is a conjunction (AND) of clauses , wh ic h are disj unctions (OR) of literals. A liter al is either a b o olean v ariable x or its negation ¯ x . W e assume that a clause do es neither conta in the same literal t wice nor a v ariable and its negation. A CNF form ula where eac h clause conta ins exactly k literals is called a k -CNF form ula. Satisfiabilit y , the problem o f deciding whether a CNF form ula is s atisfiable, plays a ma jor role in compu ter science. Ho w can a k -CNF form ula b e uns atisfiable? If k is large, ea c h clause is extremely easy to satisfy individually . Ho w ev er, it can b e that there are conflicts b etw een the clauses, m aking it imp ossible to satisfy all of them sim u ltaneously . If a k -CNF formula is unsatisfiable, then we exp ect that there are many conflicts. T o giv e a formal setup, we sa y t w o clauses c onflict if there is at least one v ariable that app ears p ositiv ely in one clause and negativ ely in the other. F or example, the tw o clauses ( x ∨ y ) and ( ¯ x ∨ u ) conflict, as w ell as ( x ∨ y ) and ( ¯ x ∨ ¯ y ) do. An y CNF f orm u la without the empt y clause and without an y confl icts is satisfiable. F or a formula F we defin e th e c onflict gr aph C G ( F ), whose v ertices are the clauses of F , and tw o clauses are connected b y an edge if they confl ict. ∆( F ) d enotes th e maxim um degree of C G ( F ), and e ( F ) the n umb er of conflicts in F , i.e., th e num b er of edges in C G ( F ). In fact, any k -CNF formula is satisfiable unless ∆( F ) and e ( F ) are large. A quan titativ e result follo ws fr om the lop- sided Lo v´ asz Lo cal Lemma [4, 1, 8]: A k -CNF formula F is satisfiable unless some clause conflicts with 2 k e or m ore clauses, i.e., unless ∆ ( F ) ≥ 2 k e . Up to a constan t factor, this is tigh t: Consider the form ula conta ining all 2 k clauses o v er the v ariables x 1 , . . . , x k , the c omplete k -CNF formula which w e denote by K k . It is u nsatisfiable, and ∆( K k ) = 2 k − 1. ∗ Researc h is supp orted by th e S NF Gran t 200 021-118001/1 1 As its n ame su gge sts, the lopsided Lov´ asz Lo cal L emma implies a lo c al result. O ur goal is to obtain a glob al result: F is satisfiable unless the total num b er of confl icts is very large. W e defin e t wo fu n ctions lc ( k ) := max { d ∈ N 0 | ev ery k -CNF form ula F w ith ∆( F ) ≤ d is satisfiable } , g c ( k ) := max { e ∈ N 0 | ev ery k -CNF form ula F w ith e ( F ) ≤ e is satisfiable } . The abbreviations lc and g c stand for lo c al c onflicts and glob al c onflicts , resp ectiv ely . F rom the ab ov e discussion, 2 k e − 1 ≤ lc ( k ) ≤ 2 k − 2, hence we kno w lc ( k ) up to a constan t factor. In con trast, it d oes not seem to b e easy to p ro v e n ontrivial upp er and lo w er b ounds on g c ( k ). Certainly , g c ( k ) ≥ l c ( k ) ≥ 2 k e − 1 and g c ( k ) ≤ e ( K k ) − 1 =  2 k 2  − 1. Ignoring constan t factors, g c ( k ) lies somewher e b et w een 2 k and 4 k . T his lea ves m uc h space for impro ve ment. In [10], w e prov ed that g c ( k ) ∈ Ω(2 . 27 k ) and g c ( k ) ≤ 4 k log 3 k k . In this pap er, w e improv e up on these b ounds. Surp risingly , g c ( k ) is exp onential ly sm alle r than 4 k . Theorem 1. A ny unsatisfiable k -CNF f ormula c ontains Ω  2 . 69 k  c onflicts. Ther e ar e unsatisfiable k -CNF formulas with O  3 . 51 k  c onflicts. W e obtain the lo we r b ound by a more sophisticated app lication of the idea w e us ed in [10]. The u pp er b ound follo ws from a construction that is p artially probabilistic, and inspired in parts by Erd˝ os’ construction in [3] of spars e k -uniform h yp ergraphs that are not 2-colo rable. T o simplify notation, we view formulas as sets of clauses, and clauses as sets of literals. Hence, | F | den ote s th e num b er of clauses in F . Still, we will sometimes find it con v enien t to use the more tr ad itional logic n otation. Related W ork Let F b e a CNF form ula and u b e a literal. W e define o cc F ( u ) := |{ C ∈ F | u ∈ C }| . F or a v ariable x , w e write d F ( x ) = o cc F ( x ) + o cc F ( ¯ x ) and call it the de gr e e of x . W e w rite d ( F ) = m ax x d F ( x ). It is easy to see that for a k -C NF formula, ∆( F ) ≤ k ( d ( F ) − 1). Define f ( k ) := m ax { d ∈ N 0 | ev ery k -CNF form ula F w ith d ( F ) ≤ d is satisfiable } . By an application of Hall’s T heorem, T o v ey [11] sho we d that ev ery k -CNF form ula F with d ( F ) ≤ k is satisfiable, h ence f ( k ) ≥ k . Later, Krato c h v ´ ıl, Sa vic k´ y and T u za [7] sh o wed that f ( k ) ≥ 2 k ek and f ( k ) ≤ 2 k − 1 − 2 k − 4 − 1. The upp er b ound w as impr ov ed by Sa vic k´ y and Sgall [9] to f ( k ) ∈ O ( k − 0 . 26 2 k ), by Ho ory and Szeider [6] to f ( k ) ∈ O  log( k )2 k k  , and recen tly by Gebauer [5] to f ( k ) ≤ 2 k +2 k − 1, closing th e gap b et w een lo we r and upp er b ound on f ( k ) up to a constan t factor. Actually , we used the formulas constructed in [6] to pro v e the upp er b oun d g c ( k ) ≤ 4 k log 3 k k in [10]. 2 A First Atte mpt W e sk etc h a first attempt on pr o ving a non trivial lo we r b ound on g c ( k ). Th ough this attempt d oes not succeed, it lea ds us to other in teresting questions, resu lts, and finally pro of metho ds which can b e u sed to pro v e a lo w er b ound on gc ( k ). Let F b e a k -CNF 2 form ula and x a v ariable. Every clause con taining x conflicts with ev ery clause cont aining ¯ x , thus e ( F ) ≥ o cc F ( x ) · o cc F ( ¯ x ). F urthermore, e ( F ) ≥ 1 k X x o cc F ( x ) · o cc F ( ¯ x ) , (1) where the 1 k comes from the fact that eac h conflict might b e counted up to k times, if t w o clauses conta in several complemen tary literals. Ev ery unsatisfiable k -CNF form ula F con tains a v ariable x with d F ( x ) ≥ 2 k ek . I f this v ariable is b alanc e d , i.e., o cc F ( x ) and o cc F ( ¯ x ) differ only in a p olynomial factor in k , then e ( F ) ≥ 4 k poly( k ) . Indeed, in the form ulas constructed in [5], all v ariables are balanced. The same holds for the complete k -CNF form ula K k . It follo ws that when trying to obtain an up p er b ound on g c ( k ) that is exp onentiall y smaller than 4 k , we should construct a v ery unb alanc e d formula. W e ask the follo w ing qu estion: Question: Is there a n umb er a > 1 suc h th at for ev ery un s atisfiable k -CNF form ula F there is a v ariable w ith o cc F ( x ) ≥ a k and o cc F ( ¯ x ) ≥ a k ? The answer is a v ery strong no: In [10] w e ga v e a simple inductiv e construction of a k -CNF f ormula F with occ F ( ¯ x ) ≤ 1 for ev ery v ariable x . H o we ve r, in this formula one has o cc F ( x ) ≈ k !. Allo wing o cc F ( ¯ x ) to b e a small exp on ential in k , we h a ve the follo wing result: Theorem 2. (i) F or every a > 1 , b ≥ a a − 1 ther e is a c onstant c suc h that for al l sufficiently lar ge k , ther e is an unsatisfiable k -CNF f ormula F with o cc F ( ¯ x ) ≤ ck 2 a k and o cc F ( x ) ≤ ck 2 b k , for al l x . (ii) L et 1 < a < √ 2 and b = q a 4 a 2 − 1 . Then every k -CNF formula F with o cc F ( x ) ≤ b k 8 k and o cc F ( ¯ x ) ≤ a k 8 k is satisfiable. Of cour s e, w e can interc hange the roles of x and ¯ x , but it is conv enient to assume that o cc F ( ¯ x ) ≤ o cc F ( x ) for ev ery x . In the spirit of these results, w e m ight susp ect that if F is u nsatisfiable, then for some v ariable x , the pro duct o cc F ( x ) · o cc F ( ¯ x ) is large. Question: Is there a n umb er a > 2 suc h that ev ery unsatisfiable k -CNF form ula con tains a v ariable x with o cc F ( x ) · o cc F ( ¯ x ) ≥ a k ? Clearly , gc ( k ) ≥ a k for any su c h n umber a . The complete k -CNF form ula witnesses that a cannot b e greater than 4, and it is not at all easy to come up w ith an un satisfiable k -CNF formula where o cc F ( x ) · occ F ( ¯ x ) is exp onential ly smaller than 4 k for ev ery x . W e cannot answer the ab o v e question, but we susp ect that the answe r is yes. W e prov e an upp er b ound on the p ossib le v alue of a : Theorem 3. Ther e ar e unsatisfiable k -CNF formulas with o cc F ( x ) · o cc F ( ¯ x ) ∈ O (3 . 01 k ) for al l variables x . 3 Pro ofs F or a tru th assignmen t α and a clause C , we will write α | = C if α satisfies C , and α 6| = C if it does not. S im ilarly , if α satisfies a form ula F , w e write α | = F . W e b egin b y stating a v ersion of the Lopsided Lov´ asz Lo cal Lemma form ulated in terms of satisfiabilit y . See [10] for a d eriv ation of this ve rsion. 3 Lemma 4 (SA T v ersion of the L op s ided Lo v´ asz Lo cal Lemma) . L et F b e a CN F formula not c ontaining the empty clause. Sample a truth assignment α by indep endently setting e ach variable x to true with some pr ob ability p ( x ) ∈ [0 , 1] . If f or any clause C ∈ F , it holds that X D ∈ F : C and D conflict Pr[ α 6| = D ] ≤ 1 4 , (2) then F is satisfiable. It is not p ossible to apply Lemma 4 directly to a formula F which w e w an t to pro v e b eing satisfiable. Instead, w e apply it to a form ula F ′ w e obtain from F in th e follo wing w a y: Definition 5. L et F b e a CNF f ormula. A truncation of F is a CNF formula F ′ that is obtaine d fr om F by deleting some liter als fr om some clauses. F or example, ( x ∨ y ) ∧ ( ¯ y ∨ z ) is a trun cat ion of ( x ∨ y ∨ ¯ z ) ∧ ( ¯ x ∨ ¯ y ∨ z ). A trun cat ion of a k -CNF form ula is n ot necessarily a k -CNF f ormula anymore. Any truth assignmen t satisfying a truncation F ′ of F also satisfies F . In our pro ofs, w e will often find it easier to app ly Lemma 4 to a sp ecial trun cati on of F than to F itself. W e n eed a tec h nical lemma on the b inomial co efficien t. Lemma 6. L et a, b ∈ N with b/a ≤ 0 . 75 . Then a b b ! ≥  a b  > a b b ! e − b 2 /a . Pr o of. The up p er b oun d is trivial and true f or all a, b . Th e lo wer b ound follo ws lik e this.  a b  = a ( a − 1) · · · 1 b ! = a b b ! b − 1 Y j =0 a − j a > a b b ! e − 2 /a P b − 1 j =0 j ≥ a b b ! e − b 2 /a , where w e used th e fact that 1 − x > e − 2 x for 0 ≤ x ≤ 0 . 75. 3.1 Pro of of Theorem 2 and 3 As w e h av e argued in Section 2, in order to impr o ve significan tly u p on the up p er b ound g c ( k ) ≤ 4 k , w e m ust construct a formula that is v ery unbalanced, i.e. o cc F ( x ) is exp o- nen tially larger than occ F ( ¯ x ). First, we will construct an unsatisfiable CNF form ula with k -clauses and some smaller clauses. In a second step, we expand all clauses to size k . Definition 7. L et F b e a CNF formula with clauses of size at most k . F or e ach k ′ -clause C with k ′ < k , c onstruct a c omplete ( k − k ′ ) -CNF formula K k − k ′ over k − k ′ new variables y C 1 , . . . , y C k − k ′ . We r eplac e C by C ∨ K k − k ′ . Using distributivi ty, we exp and it into a k -CN F formula G c al le d a k -CNFification of F . F or example, a 3-CNFification of ( x ∨ y ) ∧ ( ¯ x ∨ y ∨ z ) is ( x ∨ y ∨ y 1 ) ∧ ( x ∨ y ∨ ¯ y 1 ) ∧ ( ¯ x ∨ y ∨ z ). A tru th assignment satisfies F if and only if it satisfies its k -CNFification G . Definition 8. L et ℓ, k ∈ N 0 . An ( ℓ, k ) -CNF formula is a formula c onsisting of ℓ -clauses c ontaining only p ositive liter als, and k -clauses c ontaining only ne gative liter als. 4 If F is an ( ℓ, k )-CNF formula, we write F = F + ∧ F − , where F + consists of the p ositiv e ℓ -clauses and F − of the n ega tiv e k -clauses. Prop osition 9. L et ℓ ≤ k and let F = F + ∧ F − b e an ( ℓ, k ) -CN F formula. L et G b e the k -CNFific ation of F . Then (i) e ( G ) ≤ 4 k − ℓ | F + | + 2 k − ℓ | F + | · | F − | , (ii) o cc G ( x ) · o cc G ( ¯ x ) ≤ max { 4 k − ℓ , 2 k − ℓ | F + | · | F − |} . Pr o of. T o pro v e (i), note that ev ery edge in C G ( F ) runs b et we en a p ositiv e ℓ -clause C and a negativ e k -clause D . Thus, e ( F ) ≤ | F + | · | F − | . In G , this edge is replaced b y 2 k − ℓ edges, since C is replaced b y 2 k − ℓ copies. Th is explains the term 2 k − ℓ | F + | · | F − | . Replacing C b y 2 k − ℓ man y k -cla uses in tro duces at most 4 k − ℓ new conflicts. Th is exp lains the term 4 k − ℓ | F + | , and pr o ves (i). T o pro v e ( ii ) , ther e are t wo cases. First, if x app ears in F , then o cc G ( ¯ x ) = o cc F ( ¯ x ) and o cc G ( x ) = o cc F ( x )2 k − ℓ , thus o cc G ( x )o cc G ( ¯ x ) ≤ 2 k − ℓ | F + | · | F − | . Second, if x app ears in G , bu t not in F , then o cc G ( x ) = o cc G ( ¯ x ) = 2 k − ℓ − 1 , and o cc G ( x ) · o cc G ( ¯ x ) ≤ 4 k − ℓ . W e should explore f or wh ic h v alues of | F + | and | F − | th ere are unsatisfiable ( ℓ, k )-CNF form ulas. W e can then use Pr op osition 9 to deriv e upp er b ounds. Lemma 10. F or any ρ ∈ (0 , 1) , ther e is a c onstant c such that for al l k ∈ N 0 and ℓ ≤ k , ther e exists an unsatisfiable ( ℓ, k ) -CNF formula F = F + ∧ F − with | F − | ≤ ck 2 ρ − k and | F + | ≤ ck 2 (1 − ρ ) − ℓ . Pr o of. W e c ho ose a set v ariables V = { x 1 , . . . , x n } of n = k 2 v ariables. Th ere are  n k  k - clauses ov er V con taining only negativ e literals. W e form F − b y sampling ck 2 ρ − k of them, uniformly with replacemen t, and similarly , we f orm F + b y sampling ck 2 (1 − ρ ) − ℓ purely p ositiv e ℓ -clauses, wh er e c is some suitable constan t determined later. Set F = F − ∧ F + . W e claim that w ith high probabilit y , F is u nsatisfiable. Let α b e any truth assignmen t. There are t w o cases. Case 1. α sets at least ρn v ariables to true . F or a rand om negativ e clause C , Pr[ α 6| = C ] ≥  ρn k   n k  ≥ c ′ ρ k , The last inequalit y follo ws from Lemma 6. S in ce we select the clauses of F − indep endently of eac h other, w e obtain Pr[ α | = F − ] ≤ (1 − c ′ ρ k ) ck 2 ρ − k < e − cc ′ k 2 = e − k 2 , pro vided we c hose c large enough, i.e., c ≥ 1 c ′ . Case 2 : α sets at most n a v ariables to true . Now a similar calculation shows that α satisfies F + with probabilit y at most e − k 2 . In any case, Pr[ α | = F ] ≤ e − k 2 . The exp ected num b er of satisfying assignments of F is at most 2 k 2 e − k 2 ≪ 1 and with high probability F is unsatisfiable. The b ound in Lemma 10 is tight u p to a p olynomial factor in k : Lemma 11. L et F = F + ∧ F − b e an ( ℓ, k ) -CNF formula. If ther e is a ρ ∈ (0 , 1) such that | F + | < 1 2 (1 − ρ ) − ℓ and | F − | < 1 2 ρ − k , then F is satisfiable. 5 Pr o of. Sample a tru th assignmen t α by setting eac h v ariable indep endently to t rue with probabilit y ρ . F or a negativ e k -clause C , it holds th at Pr[ α 6| = C ] = ρ k . Similarly , for a p ositiv e ℓ -clause D , Pr[ α 6| = D ] = (1 − ρ ) ℓ . Hence the exp ect ed n umber of clauses in F that are un satisfied by α is ρ k | F − | + (1 − ρ ) ℓ | F + | < 1 2 + 1 2 = 1. Th er efore, with p ositiv e probabilit y α satisfies F . Pr o of of The or em 2. ( i ) Apply Lemma 10 with ℓ = k and ρ = 1 a . ( ii ) W e fix some probabilit y p := 1 a 2 ≥ 1 2 , and set every v ariable of F to tru e with probabilit y p , indep endent of eac h other. This give s a rand om truth assignmen t α . W e define a truncation F ′ of F as follo ws: F or eac h clause C ∈ F , if at least half the literals of C are negativ e, we remo v e all p ositiv e literal s from C and insert the truncated clause in to F ′ , otherw ise w e insert C in to F ′ without tru ncating it. W e write F ′ = F k ∧ F − , w here F − consists of purely negat iv e clauses of size at least k 2 , and F k consists of k -cla uses, eac h con taining at least k 2 p ositiv e lite rals. A clause in F − is u nsatisfied w ith probability at most p k 2 , and a clause in F k with probabilit y at most p k 2 (1 − p ) k 2 . This is b ecause in the w orst case, half of all literals are negativ e: Since p ≥ 1 2 , n ega tiv e literals are more lik ely to b e unsatisfied than p ositiv e ones. Let C ∈ F ′ b e an y clause. A p ositiv e literal x ∈ C causes conflicts b et we en C and the o cc F ′ ( ¯ x ) ≤ a k 8 k clauses of F ′ con taining ¯ x . S im ilarly , a negativ e literal ¯ y ∈ C causes conflicts w ith the at most b k 8 k clauses of F k con taining y . Therefore X D ∈ F : C and D conflict Pr[ α 6| = D ] ≤ a k 8 p k 2 + b k 8 p k 2 (1 − p ) k 2 = 1 4 , since p = 1 a 2 and b = q a 4 a 2 − 1 . By Lemma 4, F ′ is satisfiable. P art ( ii ) of Theorem 2 can easily b e impro ve d by d efining a more careful truncation pro cedure: W e r emo ve all p ositiv e literals from a clause C if C con tains less than λk of them, for some λ ∈ [0 , 1]. Cho osing λ and p optimally , we obtain a b etter result, b ut the calculations b ecome messy , and it offers n o additional insight. The crucial part of the pro of is that by remo ving p ositive literals from a clause, we can use the fact that o cc F ( ¯ x ) is small to b ound the num b er of clauses D that conflict with C and hav e a large probabilit y of b eing unsatisfied. This is also the main idea in our p roof of the lo we r b ound of Th eorem 1 . I t should b e p ointed out that for k = ℓ , an ( ℓ, k )-CNF form ula is just a monotone k -CNF formula . The size of a smallest unsatisfiable monotone k -CNF form ula is the same—up to a factor of at most 2—as the minimum n um b er of h yp eredges in a k -uniform hyp ergraph that is not 2-colo rable. In 1963, Erd˝ os [2] raised th e question what this num b er is, and pr o ved lo w er b oun d of 2 k − 1 (this is easy , simple c ho ose a random 2-colo ring). On e year later, he [3] ga v e a pr obabilistic construction of a n on-2-co lorable k -uniform hyp ergraph using ck 2 2 k h yp eredges. F or ℓ = k and ρ = 1 2 , th e statemen t and pro of of Lemma 10 are basically th e same in [3]. Pr o of of The or em 3. Combining Lemma 10 and Prop osition 9, we conclud e that for any ρ ∈ (0 , 1) and 0 ≤ ℓ ≤ k , there is an unsatisfiable k -CNF f ormula F with o cc F ( x ) · o cc F ( ¯ x ) ≤ max { 4 k − ℓ , 2 k − ℓ c 2 k 4 ρ − k (1 − ρ ) − ℓ } , for ev ery v ariable x . The constant c dep ends on ρ , b ut n ot on k or ℓ . F or fixed k , ℓ > 1, the term ρ − k (1 − ρ ) − ℓ is minimized for ρ = k k + ℓ . Cho osing ℓ = ⌈ 0 . 2055 k ⌉ , w e get ρ ≈ 0 . 8 3 and o cc F ( x ) · o cc F ( ¯ x ) ∈ O (3 . 01 k ). 6 3.2 P o of of t he Main Theorem Pr o of of the upp er b ound of The or em 1. As in the previous pro of, Prop osition 9 together with Lemma 10 yield an uns atisfiable k -CNF f orm ula F with e ( F ) ≤ 4 k − ℓ ck 2 (1 − ρ ) − ℓ + 2 k − ℓ c 2 k 4 ρ − k (1 − ρ ) − ℓ . F or ρ ≈ 0 . 6298 and ℓ = ⌈ 0 . 333 k ⌉ , we obtain e ( F ) ∈ O (3 . 51 k ). Pr o of of the lower b ound i n The or e m 1 . Let F b e an unsatisfiable k -CNF and let e ( F ) b e the n umber of conflicts in F . W e will sh o w that e ( F ) ∈ Ω  2 . 69 k  . In the p roof, x denotes a v ariable and u a p ositive or negativ e literal. W e assume o cc F ( ¯ x ) ≤ o cc F ( x ) for all v ariables x . W e can do so since otherwise w e ju st replace x b y ¯ x and vice v ersa. This c hanges neither e ( F ), nor satisfiabilit y of F . Also w e can assume that o cc F ( x ) and o cc F ( ¯ x ) are b oth at least 1, if x o ccurs in F at all. F or x , we defin e p ( x ) := max ( 1 2 , k s o cc F ( x ) 16 e ( F ) ) , and set x to true with probabilit y p ( x ) indep endently of all other v ariables yielding a random assignmen t α . Since o cc F ( u ) ≤ e ( F ), we hav e p ( x ) ≤ 1. W e set p ( ¯ x ) = 1 − p ( x ). By definition, p ( x ) ≥ p ( ¯ x ). Let us list some prop erties of this d istribution. First, if p ( u ) < 1 2 for some literal u , then u is a negativ e literal ¯ x , and p ( x ) = k q occ F ( x ) 16 e ( F ) > 1 2 . Second, if p ( u ) = 1 2 , then b oth k q occ F ( u ) 16 e ( F ) ≤ 1 2 and k q occ F ( ¯ u ) 16 e ( F ) ≤ 1 2 hold. W e distinguish t w o typ es of clauses: Bad clauses, which con tain at least one literal u with p ( u ) < 1 2 , and go o d clauses, whic h contai n only literals u with p ( u ) ≥ 1 2 . Let B ⊆ F denote the set of bad clauses and G ⊆ F the set of go od clauses. Lemma 12. P C ∈B Pr [ α 6| = C ] ≤ 1 8 . Pr o of. F or eac h clause C ∈ B , let u C b e th e literal in C m in imizing p ( u ), breaking ties arbitrarily . T his means Pr[ α 6| = C ] ≤ p ( ¯ u C ) k . Since C is a bad clause, p ( u C ) < 1 2 , u C is a negativ e literal ¯ x C , and p ( x C ) = k q occ F ( x C ) 16 e ( F ) . T hus X C ∈B Pr[ α 6| = C ] ≤ X C ∈B p ( x C ) k = X C ∈B o cc F ( x C ) 16 e ( F ) . (3) Since clause C con tains ¯ x C , it conflicts with all o cc F ( x C ) clauses con taining x C , thus P C ∈B o cc F ( x C ) ≤ 2 e ( F ). The factor 2 arises since we count eac h conflict p ossibly t wice, once from eac h side. Combining this with (3) prov es the lemma. W e cannot directly apply Lemm a 4 to F . Therefore w e apply the b elo w sparsifi cation pro cess to F . Lemma 13. L e t F ′ b e the r esult of the sp arsific ation pr o c e ss. If F ′ do es not c ontain the empty clause, then F is satisfiable. Pr o of. W e will sho w that (2) applies to F ′ . Fix a clause C ∈ F ′ . After th e sparsifi cation pro cess, ev ery literal u fu lfills P D ∈G ′ : u ∈ D Pr[ α 6| = D ] ≤ 1 8 k . T herefore, the terms Pr[ α 6| = D ], for all go o d clauses D conflicting with C , sum up to at most 1 8 . By L emm a 12, the terms Pr[ α | = D ] for all b ad clauses D also sum u p to at most 1 8 . Hence (2) holds, and b y Lemma 4, F ′ is satisfiable, and clearly F as wel l. 7 Algorithm: Sparsification Pro cess Let G ′ = { D ∈ F | p ( u ) ≥ 1 2 , ∀ u ∈ D } b e the set of go o d clauses in F . while ∃ a literal u : P D ∈G ′ : u ∈ D Pr[ α 6| = D ] > 1 8 k do Let C b e some clause maximizing Pr[ α 6| = D ] among all clauses D ∈ G ′ : u ∈ D . C ′ := C \ { u } G ′ := ( G ′ \ { C } ) ∪ { C ′ } end return F ′ := G ′ ∪ B Con trary , if F is unsatisfiable, the sparsification pr ocess pro du ces the empt y clause. W e will sho w that e ( F ) is large. Th ere is some C ∈ G all whose literal s are b eing deleted during the sparsification pro cess. W rite C = { u 1 , u 2 , . . . , u k } , and ord er th e u i suc h that o cc F ( u 1 ) ≤ o cc F ( u 2 ) ≤ · · · ≤ o cc F ( u k ). One c hec ks that this implies that p ( u 1 ) ≤ p ( u 2 ) ≤ · · · ≤ p ( u k ). Fix an y ℓ ∈ { 1 , . . . , k } and let u j b e the first literal among u 1 , . . . , u ℓ that is deleted from C . Let C ′ denote wh at is left of C just b efore that deletion, and consider the s et G ′ at this p oin t of time. Then { u 1 , . . . , u ℓ } ⊆ C ′ ∈ G ′ . By the definition of the p rocess, 1 8 k < X D ∈G ′ : u j ∈ D Pr[ α 6| = D ] ≤ X D ∈G ′ : u j ∈ D Pr[ α 6| = C ′ ] ≤ ≤ o cc F ( u j ) Pr[ α 6| = C ′ ] ≤ ≤ o cc F ( u ℓ ) ℓ Y i =1 (1 − p ( u i )) . Since p ( u ) ≥ k q occ F ( u ) 16 e ( F ) for all literals u in a go o d clause, it follo ws that 1 128 ke ( F ) ≤ p ( u ℓ ) k Q ℓ i =1 (1 − p ( u i )), for every 1 ≤ ℓ ≤ k . Let ( q 1 , . . . , q k ) ∈ [ 1 2 , 1] k b e any sequence satisfying the k inequalities 1 128 ke ( F ) ≤ q k ℓ Q ℓ i =1 (1 − q i ) for all 1 ≤ ℓ ≤ k , f or example, the p ( u i ) are s uc h a sequence. W e w ant to mak e the q ℓ as s m all as p ossible: If (i) q ℓ > 1 2 and (ii) 1 128 ke ( F ) < q k ℓ Q ℓ i =1 (1 − q i ), w e can d ecrease q ℓ unt il one of (i) and (ii) b ecomes an equalit y . The other k − 1 inequ alit ies sta y s atisfied. In the end we get a sequence q 1 , . . . , q k satisfying 1 128 ke ( F ) = q k ℓ Q ℓ i =1 (1 − q i ) whenev er q ℓ > 1 2 . T his sequence is n on-decreasing: If q ℓ > q ℓ +1 , then q ℓ > 1 2 , and 1 128 ke ( F ) ≤ q k ℓ +1 Q ℓ +1 i =1 (1 − q i ) < q k ℓ Q ℓ i =1 (1 − q i ) = 1 128 ke ( F ) , a con tradiction. If all q i are 1 2 , then th e k th inequalit y yields 128 k e ( F ) ≥ 4 k , and w e are d one. Oth- erwise, there is some ℓ ∗ = min { i | q i > 1 2 } . F or ℓ ∗ ≤ j < k b oth q j and q j +1 are greater than 1 2 , th us q k j +1 Q j +1 i =1 (1 − q i ) = q k j Q j i =1 (1 − q i ), and q j = q j +1 k p 1 − q j +1 . W e define f k ( t ) := t k √ 1 − t , th us q j = f k ( q j +1 ). By f ( j ) k ( t ) we d enote f k ( f k ( . . . ( f k ( t )) . . . )), the j -fold iterated app li- cation of f k ( t ), with f (0) k ( t ) = t . W e obtain q j = f ( k − j ) k ( q k ) > 1 2 for ℓ ∗ ≤ j ≤ k . By P art (v) of Prop osition 15, f ( k − 1) k ( q k ) ≤ 1 2 , thus ℓ ∗ ≥ 2. Therefore q 1 = · · · = q ℓ ∗ − 1 = 1 2 , and 8 the ( l ∗ − 1) st inequalit y reads as 1 128 ke ( F ) ≤ q k ℓ ∗ − 1 ℓ ∗ − 1 Y i =1 (1 − q i ) = 2 − k − ℓ ∗ +1 . W e obtain e ( F ) ≥ 2 k + ℓ ∗ − 1 128 k . Ho w large is ℓ ∗ ? Define S k := min { ℓ ∈ N 0 | f ( ℓ ) k ( t ) ≤ 1 2 ∀ t ∈ [0 , 1] } . By Pa rt (v) of Prop osition 15 (see app endix), S k is finite. Since f ( k − ℓ ∗ ) k ( q 1 ) = q ℓ ∗ > 1 2 , w e conclud e that k − ℓ ∗ ≤ S k − 1, thus e ( F ) ≥ 2 2 k − S k 128 k . Lemma 14. The se qu e nc e S k k c onver ges to lim k →∞ S k k = − R 1 1 2 1 x ln(1 − x ) dx < 0 . 572 . The pro of of this lemma is tec hnical and n ot related to satisfiabilit y . W e pro ve it in the app endix. W e conclude that e ( F ) ≥ 2 (2 − 0 . 572) k 128 k ∈ Ω  2 . 69 k  . 4 Conclusion W e w an t to giv e some hind sigh t why a sparsification pro cedure is necessary in b oth lo w er b ound pro ofs in this pap er. The prob ab ility distribution we d efine is not a un iform one, but biased to wards setting x to true if o cc F ( x ) ≫ o cc F ( ¯ x ). Let C b e a clause conta ining ¯ x . It conflicts w ith all clauses con taining ¯ x . It could happ en that in all those clauses, x is the on ly literal with p ( x ) > 1 2 . In this case, eac h suc h clause is un satisfied with p robabilit y not m uc h smaller than 2 − k , and the sum (2) is great er th an 1 4 . By remo ving x fr om these clauses, we r educe the n umb er of clauses conflicting with C , making the sum (2) m u c h smaller. Ho w ev er, for other clauses C ′ , this sum migh t increase b y remo ving x . W e think that one will n ot b e able to p ro ve a tigh t lo w er b ound using just a smarter sp arsification pro cess. W e state some op en p r oblems and questions. Question: Do es lim k →∞ k p g c ( k ) exist? If it do es, it lies b et w een 2 . 69 and 3 . 51. One wa y to pr ov e existence wo uld b e to define “pro duct” taking a k -CNF formula F and an ℓ -CNF formula G to a ( k + ℓ )-CNF form ula F ◦ G that is unsatisfiable if F and G are, and e ( F ◦ G ) = e ( F ) e ( G ). With 2 and 4 ru led out, there seems to b e no ob vious guess f or the v alue of th e limit. What abou t √ 8 ≈ 2 . 828, the geometric mean of 2 and 4? Question: Is there an a > 2 suc h that every unsatisfiable k -CNF formula con tains a v ariable x with o cc F ( x ) · o cc F ( ¯ x ) ≥ a k ? Where do our m ethods f ail to pro ve this? T he part in the pro of of the lo wer b ound of Theorem 1 that fails is Lemma 12. On the other hand, Lemma 12 pr o ves more than w e need for Theorem 1: It pro ve s that Pr[ α | = D ], summed up o v er al l bad clauses giv es at most 1 8 . W e only need that th e b ad clauses conflicting with a sp ecific clause sum up to at most 1 8 . Still, we do not see how to apply or extend our metho ds to pro v e that such an a > 2 exists. W e discuss ed lo w er and u pp er b ounds on the minimum of several parameters of un- satisfiable k -CNF f orm u las. The f ollo wing table lists them w h ere b ounds lab eled with an asterisk are from this pap er and un lab eled b ound s are not attribu ted to an y sp ecific pap er. 9 parameter notation lo w er b ound upp er b ound o ccurrences of a literal o cc( x ) 1 1 o ccurrences of a v ariable f ( k ) 2 k ek [7] 2 k +3 k [5] lo cal confl ict num b er lc ( k ) 2 k e [7] 2 k − 1 conflicts caused b y a v ariable o cc( x )occ( ¯ x ) 2 k ek [7] O (3 . 01 k ) ∗ global conflict n umber g c ( k ) Ω(2 . 69 k ) ∗ O (3 . 51 k ) ∗ References [1] N. Alon an d J. Sp encer. The pr ob abilistic metho d . Interscie nce Series in Discrete Mathematics and O ptimizatio n. John Wiley , second edition, 2000. [2] P . Er d˝ os. On a com binatorial problem. Nor disk Mat. Tidskr. , 11:5–10, 40, 1963. [3] P . E r d˝ os. On a com binatorial problem. I I. A cta Math. A c ad. Sci. Hungar , 15:44 5–447, 1964. [4] P . Er d˝ os and J. S p encer. Lopsided Lov´ asz Lo cal Lemma and Latin trans v ers als. Discr ete Appl. Math. , 30(2- 3):151–15 4, 1991. ARID AM I I I (New Brunsw ic k, NJ, 1988) . [5] H. Gebauer. Dispro of of the n eigh b orh oo d conjecture and its implications to SA T, 2009. sub mitted. [6] S. Ho ory and S. S zeider. A note on un satisfiable k -CNF formulas with few o ccurences p er v ariable. SIAM Journal on D iscr e te M athem atics , 20(2):523– 528, 200 6. [7] J. Krato c hv ´ ıl, P . Savic k´ y, and Z . T uza. O n e more o ccurren ce of v ariables makes satis- fiabilit y jump fr om tr ivial to NP-complete. SIAM Journal of Computing , 22(1):2 03– 210, 1993 . [8] L. Lu and L. Sz ´ ek ely . Using Lo v´ asz Lo cal Lemma in the sp ace of random injections. Ele ctr on. J. Combin. , 14(1):Rese arc h P ap er 63, 13 p p. (electronic), 2007. [9] P . S a vic k ´ y and J. Sgall. DNF tautologies with a limited num b er of o ccurren ces of ev ery v ariable. The or et. Comput. Sci. , 238(1–2):4 95–498, 2000. [10] D. Sc heder and P . Zumstein. Ho w man y confl icts do es it need to b e un satisfiable? In Eleventh Internationa l Confer enc e on The ory and Applic ations of Satisfiability T esting (SA T), L e ctur e Notes in Computer Sci enc e, V ol. 4996 , pages 246–2 56, 2008. [11] C . A. T o v ey . A simplified NP-complete satisfiabilit y problem. Discr ete Appl. Math. , 8(1):8 5–89, 1984. 10 A Pro of of Lemma 14 Prop osition 15. L et k ∈ N and f k : [0 , 1] → [0 , 1] with f k ( t ) = t k √ 1 − t . F or t ∈ [0 , 1] , the fol lowing statements hold. (i) f k ( t ) attains its unique maximum at t = t ∗ k := k k +1 . (ii) f k ( t ) ≤ t , and f k ( t ) = t if and only i f t = 0 . (iii) F or ℓ ≥ 1 , f ( ℓ ) k ( t ) ≤ f ( ℓ ) k  k k +1  . (iv) F or ℓ ≥ 0 and t ∈ [0 , 1] , (1 − t ) ℓ/k t ≤ f ( ℓ ) k ( t ) ≤ (1 − f ( ℓ ) k ( t )) ℓ/k t . (v) F or k ≥ 2 and any t ∈ [0 , 1] , f ( k − 1) k ( t ) ≤ 1 2 . Pr o of. ( i ) follo w s from elementa ry cal culus. ( ii ) h olds since k √ 1 − t is less than 1 for all t > 0. F or ℓ = 1, ( iii ) follo ws from ( i ), and for greater ℓ , it follo ws from ( ii ) and induction on ℓ . ( iv ) holds b ecause eac h of the ℓ applications of f k m ultiplies its argument w ith a factor that is at least k √ 1 − t and at most k q 1 − f ( ℓ ) t . Su pp ose ( v ) d oes not h old. Then b y (iii) w e get f ( k − 1) k  k k +1  ≥ f ( k − 1) k ( t ) > 1 2 , and by ( iv ), we ha ve 1 2 < f ( k − 1) k  k k + 1  ≤  1 2  k − 1 k k k + 1 . An elemen tary calculation sho ws that this do es n ot hold for an y k ≥ 1. T o pro ve Lemma 14, we compute lim k →∞ S k k (and sho w that th e limit exists). Recall the definition S k = min { ℓ ∈ N 0 | f ( ℓ ) k ( t ) ≤ 1 2 ∀ t ∈ [0 , 1] } , where f k ( t ) = t k √ 1 − t . By Pa rt ( iii ) of Pr op ositio n 15, S k = min { ℓ | f ( ℓ ) k ( t ∗ k ) ≤ 1 2 } , for t ∗ k := k k +1 . W e generalize the definition of S k b y defining for t ∈ (0 , 1], S k ( t ) := min { ℓ | f ( ℓ ) k ( t ∗ k ) ≤ t } . F urther, w e set s k ( t ) := S k ( t ) k . Let 0 < t 2 < t 1 < t ∗ k . W e w an t to estimate s k ( t 2 ) − s k ( t 1 ). This should b e small if | t 1 − t 2 | is small. F or br evit y , w e write a := S k ( t 1 ), b := S k ( t 2 ). Clearly a ≤ b . W e calculate t 2 ≥ f ( b ) k ( t ∗ k ) = f ( b − a +1) k ( f ( a − 1) k ( t ∗ k )) ≥ f ( b − a +1) k ( t 1 ) ≥ (1 − t 1 ) (( b − a +1) / k ) t 1 , t 2 < f ( b − 1) k ( t ∗ k ) = f ( b − a − 1) k ( f ( a ) k ( t ∗ k )) ≤ f ( b − a − 1) k ( t 1 ) ≤ (1 − t 2 ) (( b − a − 1) / k ) t 1 . Where we used p art ( iv ) of Prop osition 15. In fact, these in equ alit ies also hold if t 1 ≥ t ∗ k , when a = 0: t 2 ≥ f ( b ) k ( t ∗ k ) ≥ (1 − t ∗ k ) b/k t 1 ≥ (1 − t 1 ) ( b +1) /k t 1 , t 2 < f ( b − 1) k ( t ∗ k ) = (1 − t 2 ) (( b − 1) /k ) t 1 . 11 One c hec ks that the inequalities even hold if t ∗ ≤ t 2 < t 1 ≤ 1. Note that b − a k = s k ( t 2 ) − s k ( t 1 ). Solving for b − a k , the ab ov e in equaliti es yield log t 2 − log t 1 log(1 − t 1 ) − 1 k ≤ s k ( t 2 ) − s k ( t 1 ) ≤ log t 2 − log t 1 log(1 − t 2 ) + 1 k , (4) for all 0 < t 2 < t 1 < 1. The r igh t inequalit y also holds for 0 < t 2 < t 1 ≤ 1. Multiplying with − 1, w e see that it also holds if t 2 > t 1 . I f t 2 = t 1 , it is trivially true. Hence this inequalit y is true for all t 1 , t 2 ∈ (0 , 1). Supp ose s ( t ) = lim k →∞ s k ( t ) exists, for ev ery fixed t . Inequalit y (4) also h olds in the limit. W riting t 1 = t and t 2 = t + h and d ividing (4) by h giv es log( t + h ) − log t h log(1 − t ) ≤ s ( t + h ) − s ( t ) h ≤ log( t + h ) − log t h log(1 − t − h ) , Letting h go to 0, we obtain s ′ ( t ) = 1 t log(1 − t ) , th u s s ( t ) = s (1) − R 1 t 1 x log(1 − x ) dx . O bserving that S k k = s k ( 1 2 ) and s k (1) = 0 for all k pro v es the Lemma. The ab o ve argument shows that if s k ( t ) conv erges p oint wise, then it con v erges to a conti nuous fu n ction s ( t ) on (0 , 1). W e ha ve to show that lim k →∞ s k ( t ) d o es in fact exist. First p lug in t 1 = 1 into the righ t inequalit y of (4) to observe that for eac h fixed t 2 , the sequence ( s k ( t 2 )) k ∈ N is b oun ded fr om ab o ve. Clearly it is b ounded from b elo w b y 0. Hence there exist s ( t ) := lim su p s k ( t ) and similarly s ( t ) := lim inf s k ( t ). W e write s h orthand L ( t 1 , t 2 ) := log t 2 − log t 1 log(1 − t 1 ) and U ( t 1 , t 2 ) := log t 2 − log t 1 log(1 − t 2 ) . No w (4) r eads as L ( t 1 , t 2 ) − 1 k ≤ s k ( t 2 ) − s k ( t 1 ) ≤ U ( t 1 , t 2 ) + 1 k . W e claim that L ( t 1 , t 2 ) ≤ s ( t 2 ) − s ( t 1 ) ≤ U ( t 1 , t 2 ) , (5) L ( t 1 , t 2 ) ≤ s ( t 2 ) − s ( t 1 ) ≤ U ( t 1 , t 2 ) . (6) F or sequences ( a k ) k ∈ N , ( b k ) k ∈ N , lim sup a k − lim su p b k = lim sup( a k − b k ) d oes not hold in general, hence the claim is now completely trivial. W e will p ro of that s ( t 2 ) − s ( t 1 ) ≤ U ( t 1 , t 2 ). This will p r o ve one claimed inequalit y . Th e other thr ee inequalities can b e pro v en similarly . Fix some small ǫ > 0. F or all sufficientl y large k , 1 k ≤ ǫ . W e ha ve s k ( t 2 ) ≥ s ( t 2 ) − ǫ for infi nitely man y k , thus s k ( t 1 ) ≥ s k ( t 2 ) − U ( t 1 , t 2 ) − 1 k ≥ s ( t 2 ) − U ( t 1 , t 2 ) − 2 ǫ for infinitely man y k . Therefore s ( t 1 ) ≥ s ( t 2 ) − U ( t 1 , t 2 ) − 2 ǫ . By making ǫ arb itrarily small, the claimed in equalit y follo w s. W e can no w app ly our non-rigorous argumen t from ab o ve, this time rigorously . W rite t = t 1 , t 2 = t + h , and divide (5) and (6) b y h , send h to 0, and we obtain s ′ ( t ) = s ′ ( t ) = 1 t log(1 − t ) . S ince s (1) = s (1) = 0, we obtain s ( t ) = s ( t ) = Z 1 t − 1 x log (1 − x ) dx . 12