Geometric properties of satisfying assignments of random $epsilon$-1-in-k SAT

Geometric prop erties of satisf ying assignmen ts of random ǫ -1-in- k SA T Gabriel Istrate, eAustria Researc h Institute Bd. V. Pˆ arv an 4, cam 045B, Timi ¸ soara, RO-300223, Romania, email: ga brielistrate@a cm.org No vem b e r 3, 2018 Abstract W e study the geometric structure of the set of solutions of rand o m ǫ -1-in- k SA T problem [2, 15]. F or l ≥ 1, tw o satisfying assignments A and B are l -c onne cte d if there exists a sequence of satisf ying assignments connecting them by c hanging at most l bits at a time. W e first pro ve that w.h.p. tw o assignments of a random ǫ -1-in- k SA T instance are O (log n )-connected, conditional on b eing satisfying assign- ments. Also, there exists ǫ 0 ∈ (0 , 1 k − 2 ) suc h that w.h.p. no tw o satisfying assignmen ts at distance at least ǫ 0 · n form a ”hole” in the set of assign- ments. W e b eliev e that this is true for all ǫ > 0, and thus satisfying assignmen ts of a random 1-in- k SA T instance form a single cluster. Keyw ords : ǫ -1-in- k SA T, o verlaps, random g raphs, phase transition. AMS Categories: Prima ry 68Q25, Seconda ry 82B27. A CM Class ificat ion: F.2.2, G.2. 1 In tro duction The geometric struc tur e of solutions of r a ndom constra in t sa tisfaction problems has lately b ecome a topic of sig nifi cant interest [1], [10], [9], [6 ]. The motiv ation is the study o f phase tr ansitions in c ombinatoria l op timization pr oblems [5, 14], particularly using metho ds fro m physics of Spin Glasses such as the so-ca lle d r eplic a metho d a nd c avity ap pr o ach . These metho ds, s o far without a complete rigoro us foundation, are large ly respo nsible for our substantially increased un- derstanding o f structural prop erties of constraint sa t isfaction problems. Of sp ecial interest are tw o s pecial cases when the r e plic a method [11] ap- plies, those characterized by so-calle d “ replica symmetry” or “one-step replica symmetry brea king”. T hes e assumptions mak e predictions on (and hav e impli- cations for) the typical geo met ry of the set of solutions o f a random instance. 1 Spec ific a lly , the t wo assumptions s eem to constrain the set of solutions in the following way: 1. F or problems displaying replica-symmetry , the set o f solutions forms a single clus t er. The t ypica l overlap is concentrated a round a single v alue, and the distr ibut ion o f ov er laps has contin uous suppor t . 2. In the presence of one- step replica-sy mm etry br e aking, the so lut ion space is no long er connected, but breaks into a n um ber of clusters . These clusters corres p ond to the emerge nc e o f Ω( n )-size mini-b ackb ones , sets of v ar iables taking the same v alue for a ll solutions in a cluster. The clusters do not p os- sess further g e ometrical s t ructure (hence the “o ne-step” qua lifier in ”one step RSB” ) , a nd are s eparated by Ω( n ) v a riable flips. The distribution of ov erla ps develops multiple p eaks and has discontin uous s upp ort. In this pap er w e study the geometric structure o f the s et of solutions o f the r andom 1-in- k SA T pr oblem [2] , and a generalizatio n of this pro ble m from [15], r andom ǫ - 1 -in- k satisfiability . This latter problem is parameterized by a real n umber ǫ ∈ [0 , 1 / 2], and essentially coincides with 1-in- k SA T for ǫ = 1 / 2. Results in the cited work sugge st that for ǫ ∈ ( ǫ c , 1 / 2], where ǫ c ∼ 0 . 272 6 is the solution of equation 2 x 3 − 2 x 2 + 3 x − 1 = 0, ǫ -1-in- k SA T b eha ves qualitatively “like 2- SA T”. In particular, for both pr oblems the threshold lo cation can b e predicted in b oth cases by a “p ercolation of contradictory cycles” argument, and the replica symmetry ansatz is correct. F or 2- SA T we hav e previously pr o ved [6] tw o r esults supp orting r eplica sym- metry: with hig h pro babilit y satisfying a ssignmen ts o f a random 2-CNF formula with claus e /v ar iable ratio c < 1 form a single cluster; also the ov erlap distribu- tion ha s contin uous supp ort. F rom the heuristic similar ity of the tw o pro blems, we exp ect similar r esults to also hold for ǫ -1-in- k SA T, ǫ ∈ ( ǫ c , 1 / 2]. Though the replica symmetric approach s eems c o rrect [1 5], we ca nn ot rig orously pr o ve such results. Instead we provide some evidence for them: • W e fir st note (Theor em 1) that the replica symmetric picture holds in the sub c ritical re g ime of the formu la hyperg raph. • W e show (Theorem 2) that for any tw o given assignments A, B a t suffi- ciently lar ge Hamming distance, with pro babilit y 1 − o (1) A, B are O (log n )- connected (conditional on b eing sa t isfying assignments). • W e show (Theorem 4 ) t hat with probability 1 − o (1) (as n → ∞ ) the set of satisfying assignments of a random instance o f 1-in- k SA T with clause/v ar iable ra t io λ < 1 ( k 2 ) do es not ha ve ho le s o f size > ǫ k n , for s ome ǫ k > 0. 2 Preliminaries Definition 1 L et ǫ ∈ [0 , 1 / 2] . A n instanc e of the ǫ -1- in- k SA T problem is a pr op ositional formula Φ in clausal form, with ex a ctly k liter als in e ach clause. 2 A satisfying a ssignmen t for instanc e Φ is a mapping of variables in Φ t o { 0 , 1 } such that in e ach clause of Φ e xactly one liter al is true. W e will use tw o rela ted mo dels to the c onstant pr ob ability mo del to genera te random insta nces of ǫ -1-in- k SA T. 1. The c ounting model is par ameterized by a rea l num b er r > 0 . A ra ndo m instance of ǫ -1-in- k SA T will hav e rn clauses, out of whic h r n · ǫ i (1 − ǫ ) j hav e i nega tiv e and j p ositiv e v ar iables (where i + j = k ). 2. The c onstant p r ob ability model is parameter ized by a pr obabilit y p . A random instance Φ is obtained by including indep enden tly with pr obabilit y pǫ i (1 − ǫ ) j each p o ssible clause with i neg ativ e and j p ositive v ariables (where i + j = k ). Using standa rd metho ds ([4], Chapter 2; s e e als o a similar issue in [1 2]) the tw o mo dels we describ ed ab o ve for ǫ -1-in- k SA T a r e equiv alent: Lemma 1 L et r > 0 and let p = p ( n ) b e such t h at p ·  n k  = rn . L et Φ 1 b e a r andom instanc e of ǫ -1-in- k SA T with rn clauses gener ate d ac c or ding to the c ounting mo del, and le t Φ 2 b e a r andom i nstanc e o f ǫ -1-in- k SA T, gener ate d ac c or ding to the c onstant pr ob ability mo del with pr ob ability p . L et B b e an arbitr ary monotone pr op ert y and µ ∈ { 0 , 1 } . Then: lim n →∞ P rob [Φ 1 | = B ] = µ, iff lim n →∞ P rob [Φ 2 | = B ] = µ. In the s e quel we will lib erally use o ne mo del or the o th er, dep ending on o ur goals. Results in [2] and [15] imply the fact that for ǫ ∈ ( ǫ c , 1 / 2 ] the threshold of satisfiability for the ǫ -1-in- k satisfiability (under the counting mo del) is lo cated at cr itical v alue 1 r k,ǫ = 1 4 ǫ (1 − ǫ ) · 1  k 2  The corresp onding threshold for ǫ -1-in- k SA T under the cons t ant probability mo del is p k,ǫ = ( k − 2)! 2 ǫ (1 − ǫ ) · n 1 − k Definition 2 The ov erlap of t w o assignments A and B for a formula Φ on n variables, denote d by ov erl ap ( A, B ) , is the fr action of va r iables on which the two assignments agr e e. F ormal ly ov erl ap ( A, B ) = |{ i : A ( x i )= B ( x i ) }| n . 1 in [15] the result i s only stated and prov ed f or k = 3, but the metho d outlined there works for any k ≥ 3 3 The distribution of ov erla ps is, indeed, the or ig inal order pa rameter that was originally used to study the phase transition in ra ndom k -SA T [1 3 ]. Definition 3 L et l ≥ 1 b e an inte ger and let A, B b e two satisfying assignments of an instanc e Φ of ǫ -1-in- k SA T. Pair ( A, B ) is c al le d l -connected if t her e ex ist s a se quenc e of satisfying assignments A 0 , A 1 , . . . A r , A 0 = A , A r = B , with A i and A i +1 at Hamming distanc e at most l . Definition 4 L et A, B b e arbitr ary assignments for t h e variables of an inst a nc e Φ of 1-in- k S A T. Pair ( A, B ) is c al le d a hole if : 1. A, B ar e satisfying assignments for Φ . 2. Ther e exists n o satisfying assignment C with d H ( A, C ) + d H ( C, B ) = d H ( A, B ) (wher e d H is the Hamming distanc e). The numb er λ = d H ( A, B ) is c al le d the size of hole ( A, B ) . 3 Results First, w e prov e that for lo w enough cla us e/v a riable ratios the set of satisfying assignments of a random instance of 1- in- k SA T b eha ves in the wa y pre dict ed by the replica symmetry ansatz: Theorem 1 L et k ≥ 3 and c < 1 / k ( k − 1) . Then ther e exists γ > 0 such that, with pr ob ability 1 − o (1 ) (as n → ∞ ), a r andom instanc e of 1-in- k SA T with n variables and cn cla uses has al l its satisfying assignments γ log( n ) -c onne cte d. W e b eliev e (and would like to prove) that the r esult in Theorem 1 is v alid for v alues of c up to 2 / k ( k − 1) (the s atisfiabilit y threshold of 1- in- k SA T [2]). W e cannot pr o ve this statemen t. Instead, we prov e a r esult that implies a weak er claim fo r 1-in- k SA T but is v alid, mo re generally , for ǫ -1-in- k SA T: Theorem 2 L et 0 ≤ ǫ ≤ 1 2 , let c < 1 , let Φ b e a r andom inst a nc e of ǫ -1-in- k SA T with clause/variable r atio 1 max[4 ǫ (1 − ǫ ) ,ǫ 2 +(1 − ǫ ) 2 ] · c ( k 2 ) , and let ( A n , B n ) ∈ { 0 , 1 } n × { 0 , 1 } n such that 2 · [ ov erl ap ( A n , B n )(1 − ǫ ) ] k − 2 ≤ 1 . (1) Then ther e exists λ c,ǫ > 0 such that Pr[( A n , B n ) ar e n o t λ c,ǫ · log( n ) -c onne cte d | A n , B n | = Φ } < 1 /n. for lar ge en ou gh n . In other w ords, every single p air of a s signmen ts is lik ely to be O (log n )- connected, co ndit ional on b eing a pair of satisfying assignments, and b eing far enough . The r emark able thing about co nd ition (1) is tha t it depends on ǫ and 4 k but not c . F or certain v a lues of ǫ (we sp ecifically believe this is the case in the region [0 , ǫ c )) it migh t simply signal the fa ct that there a re no satisfying assignments of a certain o verlap. This is not a problem for ǫ = 1 / 2 (i.e. for the 1-in- k SA T), since the condition (1) is triv ially satisfied for every overlap v alue. F or this problem, the results in Theo rems 1 and 2 are highly reminiscent of the results for 2- SA T in [6]. On the o ther hand for a n y c < 1 and all q ∈ (0 , 1) a r andom 2-CNF formula has w.h.p. t wo satisfying ass ignmen ts of ov erla p approximately q . Despite 2-SA T a nd 1-in- k SA T b eing similar in other wa ys (see e.g . [7]), the corre s ponding statement is not true for 1-in- k SA T: Theorem 3 F or any c > 0 ther e exists q c ∈ (0 , 1) su ch t h at w.h.p . a r andom instanc e of 1-in- k SA T with clause/va riable r atio c ha s, with pr ob ability 1 − o (1) no satisfying assignment s of overlap less t han q c . W e next consider an a lternativ e approach to characterizing the geometry of satisfying assignments of 1-in- k SA T by studying the existence of holes inbetw een such as s ignmen ts. F or other pro blems, e.g. k -SA T, k ≥ 9, that display clustering the set o f satisfying assignments has large holes. Indeed [10], for certain v a lue s of q 1 < q 2 < q 3 and c > 0, a ra ndo m instance of k -SA T of constraint density c will hav e, with high pr obabilit y , satisfying assignments of ov er lap q 3 , but no satisfying ass ignmen ts of ov er lap λ , q 1 ≤ λ ≤ q 2 . Consider A, B t wo s atisfying assignments of ov erla p q 3 . Then the set o f a ssignmen ts C b et ween A, B con ta ins a ho le of size at least ( q 2 − q 1 ) n . W e would like to stat that for any λ > 0 a r andom ins ta nce Φ o f 1-in- k SA T as in Theor em 2 has no hole of size a t leas t λ · n . W e canno t, how ever prov e this result (we leave it a s an intriguing op en pro blem) . Instea d we prov e a weak er result: Theorem 4 F or any k ≥ 3 ther e exists ǫ k ∈ (0 , 1 /k − 1) s u ch that with pr ob a- bility 1 − o (1) (as n → ∞ ) a r andom instanc e of 1-in- k S A T of clause/variable r atio c < 1 /  k 2  has no hol es of size ≥ ǫ k · n . 4 Pro ofs 4.1 Pro of of Theorem 1 First, note that lo cation c = 1 / k ( k − 1) in Theor em 1 is the phas e tra nsition lo cation for the random k -uniform h y pergra ph [16]. F o r smaller v alues of c , by results in [16] there ex is t s γ > 0 s uc h that w.h.p. the la rgest connected comp onen t of H has size no larger tha n γ log( n ). This argument immediately implies the desired result. Indeed, let P, Q b e t wo arbitrary satisfying assignmen ts, and let ( P 1 , Q 1 ), ( P 2 , Q 2 ), . . . , ( P v , Q v ) represent the restrictions of P and Q o n the connected comp onen ts o f Φ on which P 6 = Q . O ne can obtain a path from P to Q b y starting at P and then obtain the next satisfying a ssignmen ts by re pla cing P i by Q i for i = 1 , . . . , v . In this w ay w e are constructing satisfying assignments for Φ, s inc e we change 5 assignments c o nsisten tly on connected comp onents o f the formula hypergra ph. W e are changing at most γ log( n ) v alues a t a time, since this is the upp er b ound on the comp onen t size of H . 4.2 Pro of of Theorem 3 W e prov e the theorem b y a simple first momen t bo und . W e will work with the constant probability mo del. Definition 5 L et Φ b e a formula. A cov er of Φ is a set of varia bles W su ch that every clause of Φ c ontains at le ast one variable in W . The theorem now follows fr om the follo w ing t wo lemma s: Lemma 2 L et A , B b e satisfying assignments of an instanc e Φ of 1-in- k S A T. Then the set { x : A ( x ) = B ( x ) } is a c over of Φ . Pro of. Suppo se this was not the case, and ther e exists a clause C of Φ cons isting ent irely of v ariables in the set { x : A ( x ) 6 = B ( x ) } . Then clause C has t wo satisfying assignments at distance k . But this is not p ossible, since a ll s atisfying assignments of a given 1- in- k cla use hav e Ha m ming distance tw o . ♣ Lemma 3 F or any c > 0 ther e exists a q c > 0 such that a r andom instanc e of 1-in- k SA T of clause/vari able r atio c has, w.h. p. no c over of s i ze at m ost q c n . Pro of. Le t λ < 1 / 2 . The probabilit y that Φ has a cov er o f size i ≤ λn is a t most λn X i =1  n i  (1 − p ) ( n − i k ) ≤ λn · (1 − p ) ( n (1 − λ ) k ) · h λn X i =1  n i  i ≤ ≤ ( λn ) 2 · e − p ( n (1 − λ ) k ) ·  n λn  ≤ (1 + w ) · ( λn ) 2 · e − p ( n (1 − λ ) k ) ·  1 λ λ (1 − λ ) 1 − λ  n · · 1 p 2 π λ (1 − λ ) n . for some w > 0 (we ha ve applied the fact that λ < 1 / 2 and Stirling’s for mula) So the proba bilit y is at most (1 + w )( λn ) 2 p 2 π λ (1 − λ ) n · e − p ( n (1 − λ )) k /k !+ n [ λ ln(1 /λ )+(1 − λ ) ln(1 / (1 − λ ))] = = (1 + w )( λn ) 2 p 2 π λ (1 − λ ) n · e − n [ c (1 − λ )) k − λ ln(1 /λ ) − (1 − λ ) ln(1 / (1 − λ ))] Since c > 0 and lim λ → 0 λ ln(1 /λ ) − (1 − λ ) ln(1 / (1 − λ )) = 0, ther e exists q c > 0 such that for λ < q c , c (1 − λ )) k − λ ln(1 /λ ) + (1 − λ ) ln (1 / (1 − λ )) > 0. Thu s, fo r q < q c the pro babilit y that a random instanc e of 1-in- k SA T has a cov er o f size at most q n is exp onen tially small. ♣ 6 t yp e V 0 ( a ) V 3 ( d ) V 1 ( b ) V 2 ( c ) nu mber probability C 1 k − i − 2 [0] i [ i ] 1 [0 ] 1 [0 ]  a k − i − 2  d i  pǫ i (1 − ǫ ) k − i C 2 k − i − 2 [0] i [ i ] 1 [1 ] 1 [1 ]  a k − i − 2  d i  pǫ i +2 (1 − ǫ ) k − i − 2 C 3 k − i − 2 [0] i [ i ] 2 [1 ] 0 [0 ] 2  a k − i − 2  d i  pǫ i +1 (1 − ǫ ) k − i − 1 C 4 k − i − 2 [0] i [ i ] 0 [0 ] 2 [1 ] 2  a k − i − 2  d i  pǫ i +1 (1 − ǫ ) k − i − 1 Figure 1: The four type s of clauses leading to an edge ( x, y ) in graph H 4.3 Pro of of Theorem 2 F or a pa ir of assignments ( A, B ) define V 0 = { x : A ( x ) = B ( x ) = 0 } , V 1 = { x : A ( x ) = 0 , B ( x ) = 1 } , V 2 = { x : A ( x ) = 1 , B ( x ) = 0 } , V 3 = { x : A ( x ) = B ( x ) = 1 } . Pair ( A, B ) has typ e ( a, b, c, d ) if | V 0 | = a, | V 1 | = b, | V 2 | = c, | V 3 | = d . Also denote α = a /n, β = b/ n, γ = c/n, δ = d/n . Conditioning on A, B b eing satisfying as signmen ts, define a graph H on the set o f v ariables in A 6 = B as follo w s : x a nd y are connected if there exists a clause C of Φ consis ting of k − 2 literals whose v ariables are from V 0 ∪ V 3 and x, y . Since b oth A and B must be sa tisfying assig nmen ts, o nly four combinations ar e po ssible for the litera l combination present in C : 1. ( x, y ∈ C or x, y ∈ C ) and A ( x ) 6 = A ( y ), or 2. ( x, y ∈ C or x, y ∈ C ) and A ( x ) = A ( y ). W e can r ewrite conditions (1) a nd (2) as 1. ( x, y ∈ C or x, y ∈ C ) and ( x ∈ V 1 ∧ y ∈ V 2 ) ∨ ( x ∈ V 2 ∧ y ∈ V 1 ), or 2. ( x, y ∈ C or x, y ∈ C ) and ( x, y ∈ V 1 ) ∨ ( x, y ∈ V 2 ). T o s umma r ize this dis cussion, ther e ar e four types of cla uses that imply the existence of a n edge ( x, y ) in graph H . They are describ ed in the table fr o m Figure 1. The semantics of columns in the table is the following: first column (t ype ) lists the four types of clauses, lab eled C 1 to C 4 . Columns la beled V 0 to V 3 contain tw o num be r s. The first one is the num b er of literals of the giv en clause t yp e that are in the set V j . The second num ber (in square br a c kets) lists the nu mber of negated v ariables in the set V j . Column lab eled “num b er” co mput es the total n umber of cla uses of type C i . The column lab e led “Pr o babilit y” lists the probability that a fixed claus e of type C i be in Φ. The probability that an edge is present in graph H is the sa me fo r all pairs ( x, y ) suc h that A ( x ) = A ( y ). Similarly the pr obabilit y that an edge is pre sen t in g raph H is the same for all pair s ( x, y ) such that A ( x ) 6 = A ( y ). W e denote by µ = = µ = ( n, a, b, c, d ) a nd µ 6 = = µ 6 = ( n, a, b, c, d ) these tw o probabilities. 7 µ = ≤ p · k − 2 X i =0  a k − i − 2  d i  · h ǫ i (1 − ǫ ) k − i + ǫ i +2 (1 − ǫ ) k − i − 2 i µ 6 = ≤ p · n k − 2 X i =0  a k − i − 2  d i  · h 2 ǫ i +1 (1 − ǫ ) k − i − 1 + 2 ǫ i +1 (1 − ǫ ) k − i − 1 io Applying inequalit y  a i  ≤ a i i ! and r ewriting the s econd term o f the previous inequalities w e get µ = ≤ p [ ǫ 2 + (1 − ǫ ) 2 ] ( k − 2)! · n k − 2 X i =0  k − 2 i  · a k − i − 2 d i · ǫ i (1 − ǫ ) k − i − 2 o = = p [ ǫ 2 + (1 − ǫ ) 2 ] ( k − 2 ) ! · [ a ( 1 − ǫ ) + dǫ ] k − 2 µ 6 = ≤ 4 pǫ (1 − ǫ ) ( k − 2)! · n k − 2 X i =0  k − 2 i  · a k − i − 2 d i · ǫ i (1 − ǫ ) k − i − 2 o = = 4 pǫ (1 − ǫ ) ( k − 2 ) ! · [ a (1 − ǫ ) + dǫ ] k − 2 The equa tion ǫ 2 + (1 − ǫ ) 2 = 4 ǫ (1 − ǫ ) has a solution ǫ 0 = 3 − √ 3 6 ∼ 0 . 211 3 ... . F or ǫ ∈ ( ǫ 0 , 1 / 2 ] w e hav e ǫ 2 + (1 − ǫ ) 2 < 4 ǫ (1 − ǫ ). F or p = λ · k ! · n 1 − k , with λ = c 4 ǫ (1 − ǫ ) · 2 k ( k − 1) , with c < 1 we hav e max( µ = , µ 6 = ) = 2 c n [ α (1 − ǫ ) + δ ǫ ] k − 2 ≤ c n · 2 · [ ov erl ap ( A, B )(1 − ǫ )] k − 2 ≤ c n . It follows that the graph H has all its connected compo nen ts of size a t most λ c log n , where [8] λ c = 3 (1 − c ) 2 (2) By the discussion o f clause types in Figure 1, edges of t yp e (1) corresp ond to a constraint x 6 = y b et ween a v ar iable in V 1 and one in V 2 , while clauses of type (2) co rrespond to a constraint x = y betw een t wo v ariables, both in V 1 or both in V 2 . H does not contain con tr adictory cycles, since w e hav e conditioned on A, B being satisfying assignments. It is ea sy to see that s e t ting one v alue o f a given v ariable in H uniquely determines the v alues on its whole co nn ected co mponent. Similarly , different v alues of x lead to opp osite a ssignmen ts on the connected comp onen t of x . Given the small size of the connected co mponents, the statement o f the theorem immediately follows. ♣ 8 4.4 Pro of of Theorem 4 W e first prove a simple result ab out the connectivit y of a random graph that we w ill use in the sequel. Lemma 4 L et 0 < c < 1 and let G b e a r andom gr aph fr om G ( n, c/n ) . Then P r [ G is c onne cte d ] ≤ c n − 1 n . Pro of. Ther e a re u = n n − 2 lab eled trees on the s e t of v er tices of G . Denote by T 1 , . . . T u the edge sets of these tr e e s, a nd by W i the even t “ T i ⊆ E [ G ]”. It is ea sy to see that G is not connected if a nd only if V u i =1 W i . By Janson’s ineq ua lit y [3] Pr[ u ^ i =1 W i ] ≥ u Y i =1 Pr[ W i ] = (1 − ( c n ) n − 1 ) u . So, b y applying inequalit y 1 − (1 − x ) a ≤ ax w e get: Pr[ G connected ] ≤ 1 − (1 − ( c n ) n − 1 ) n n − 2 ≤ c n − 1 n , ♣ W e will work with the constant probability model. Ea c h clause w ill b e included in formula Φ with pro babilit y p , where p · 2 k ·  n k  = λ · 1 ( k 2 ) n , with λ < 1. Lemma 5 The pr ob ability that ther e exist two satisfying assignments A and B of overlap i t h at form a hole is at most  n i  · 2 i · [2 2 − k · ( i n ) k − 2 · (1 − i n )] n − i − 1 ( n − i ) · e − λn ( k 2 ) [1 − k ( i k ) +2 ( i k − 2 )( n − i 2 ) 2 k · ( n k ) ] Pro of. F or tw o as signmen ts A, B of ov erlap i to b e satisfying assignments of a formula Φ, all clauses of Φ m us t fall in to one of the following tw o categories: 1. Cla use C contains k − 1 literals from A = B = 0 and o ne literal from A = B = 1 . 2. Cla use C con ta ins k − 2 literals from A = B = 0 and t wo literals from A 6 = B , of o pp osite s igns in A . There are k ·  i k  clauses of the firs t type and 2  i k − 2  ·  n − i 2  clauses of the sec o nd t yp e. So the probability that all clauses of Φ fall into these t wo c a tegories is (1 − p ) 2 k · ( n k ) − k · ( i k ) − 2 ( i k − 2 )( n − i 2 ) ≤ e − p · [2 k ( n k ) − k · ( i k ) − 2 ( i k − 2 )( n − i 2 ) ] = = e − λn ( k 2 ) [1 − k ( i k ) +2 ( i k − 2 )( n − i 2 ) 2 k · ( n k ) ] . 9 The probability is at most  n i  · 2 i times the pro babilit y that giving s pecific v alues to i v ar iables we simplify the formula Φ to one for which the following graph H is connected: t wo v a riables x, y ∈ { λ : A ( λ ) 6 = B ( λ ) } are connected if there exists a clause C of Φ that contains b oth of them (and no other v ar iables in that set). This pro ba bilit y is a t most 2 ·  i k − 2  · p . So an uper b ound for the probability is 2 · i k − 2 ( k − 2)! · 2 − k · k ! n k · λ · 1  k 2  n = ( i 2 n ) k − 2 · λ 2 n = = 2 · ( i 2 n ) k − 2 · (1 − i n ) · λ n − i ≤ 2 2 − k · ( i n ) k − 2 · (1 − i n ) n − i . Since connectivity is an increasing prop ert y , applying Lemma 4, the proba- bilit y that H is co nnected is at mo st [2 2 − k · λ · ( i n ) k − 2 · (1 − i n )] n − i − 1 ( n − i ) . ♣ Let α = i/ n . T he n the upper b ound in the result a bov e reads:  n α · n  · 2 α · n · [2 2 − k · λ · ( α · n n ) k − 2 · (1 − α · n n )] n − α · n − 1 n (1 − α ) · e − λn ( k 2 ) · [1 − kα k + k ( k − 1) α k − 2 (1 − α ) 2 2 k ] Applying Stirling ’s formula for the fa c t orial, the ab o ve expressio n simplifies to ( n e ) n √ 2 π n · 2 α · n · [2 2 − k λ ( αn n ) k − 2 (1 − αn n )] n − αn − 1 · e − λn ( k 2 ) · [1 − kα k + k ( k − 1) α k − 2 (1 − α ) 2 2 k ] ( α · n e ) α · n √ 2 π αn · ( (1 − α ) · n e ) (1 − α ) · n p 2 π (1 − α ) n · n (1 − α ) = = θ (1 ) · 2 α · n · [2 2 − k λα k − 2 (1 − α )] n (1 − α ) · e − λn ( k 2 ) · [1 − kα k + k ( k − 1) α k − 2 (1 − α ) 2 2 k ] α αn √ α · (1 − α ) (1 − α ) n p 2 π (1 − α ) n · n (1 − α ) · ( α/ 2 ) k − 2 λ (1 − α ) = = θ (1 ) · n − 3 / 2 α k − 3 / 2 λ (1 − α ) 5 / 2 · · n 2 α [( α/ 2) k − 2 λ (1 − α )] (1 − α ) e − λ/ ( k 2 ) (1 − kα k + k ( k − 1) α k − 2 (1 − α ) 2 2 k ) α α · (1 − α ) 1 − α o n = = θ (1 ) · n − 3 / 2 α k − 3 / 2 λ (1 − α ) 5 / 2 · f k ( α ) n , where f k ( x ) = λ 1 − x · ( x/ 2 ) ( k − 2)(1 − x ) − x · e − λ (1 − kx k + k ( k − 1) x k − 2 (1 − x ) 2 2 k ) / ( k 2 ) and the θ (1) factor do es not depe nd on α o r λ . Let g k ( x ) = ln( f k ( x )) = = (1 − x ) ln λ + [( k − 2)(1 − x ) − x ] · ln( x/ 2) − − λ  k 2  (1 − k x k + k ( k − 1) x k − 2 (1 − x ) 2 2 k ) . 10 F or x ∈ (0 , k − 1 k − 2 ], s ince λ < 1, ln λ < 0 a nd 1 − x > 0 . Also ln( x/ 2) < 0 while ( k − 2)(1 − x ) − x > 0 . Finally , k ( k − 1) x k − 2 (1 − x ) 2 ≤ k ( k − 1 ) / 2 (since x < 1 and x (1 − x ) ≤ 1 / 4. Since k + k ( k − 1) / 2 = k ( k +1) 2 < 2 k (since k ≥ 3), w e infer that the las t ter m is positive. The conclusion o f this arg umen t is that x ∈ (0 , k − 1 k − 2 ] → g k ( x ) < 0. On the other hand g k (1) = ln 2 − λ ( k 2 )  1 − k 2 k  > ln 2 − 1 ( k 2 ) > 0, since k ≥ 3 . Thu s the equation g k ( x ) = 0 has a (smallest) ro ot x k ∈ ( k − 1 k − 2 , 1). F or α < x k , f ( α ) < 1 and the upp er bound is asymptotica lly equal to zer o. ♣ 5 Conclusions Theorem 1 connects the p ercolation of the g ian t comp onent in the random for - m ula hyper graph to the existence of a single cluster of satisfying assignments. Of course, since the phase transition in 1-in- k SA T is determined [2 ] by a “ gian t comp onen t“ phenomeno n in a dir e cted v er sion of the formula hyperg raph, the main op en question raised b y this work is to prov e that the sta tement of The- orem 1 ho lds up to critical threshold c = 2 k ( k − 1) . Theorem 2 provides further evidence that this might be true. W e b eliev e that it might b e po ssible to prov e this s t atement using a more robust gener alization o f the notion of ” ho le“ in the s et of sa t isfying assig nmen ts. Ac kno wledgmen ts I thank Romeo Negrea for useful discuss ions. This work has been supp orted by a Marie Curie International Rein teg ration Grant within the 6th Euro pean Communit y F r amew or k Pr ogramme and by a PN-I I/”P arteneriate” gr an t from the Romanian CNCSIS. References [1] D. Achlioptas and F. Ricci-T ersenghi, O n t h e solution sp ac e ge ometry of r andom c onstr aint s atisfaction pr oblems , in Pr o c e e dings of t h e 36th ACM Symp osium on The ory of Computing , 2006, pp. 130 –139. [2] D. Ac hlioptas , A. Chtc herba , G. Istrate, a nd C. Mo ore, The phase tr ansi- tion in 1-in- K SA T and NAE3SA T , in Pr o c e e dings of the 12th ACM-SIAM Symp osium on discr ete algorithms , 2001. [3] N. Alo n, P . Erd˝ os, and J. Sp encer, The pr ob abili stic metho d , 2nd ed., John Wiley and So ns (1992). [4] B. Bollo b´ as, R andom gr aphs , 2nd ed., Cambridge Universit y Press, 2001. 11 [5] A. Hartmann and M. W eigt, Phase tr ansitions in c ombinatorial optimiza- tion pr oblems , Wiley-VCH (200 5). [6] G. Istrate, S a tisfiability of b o ole an ra ndom c onstra int satisfaction: Clusters and overlaps , Journal of Universal Computer Science 13 (200 7), pp. 16 55– 1670. [7] G. I s t rate, A. Percus, and S. Bo ettc her, Spines of r andom c onstr aint satis- faction pr oblems: Definition and c onne ction with c omputational c omplexity , Annals of Mathematics and Artificial In tellige nce 44 (2005), pp. 353– 372. [8] S. Janson, T. Luczak, a nd A. Ruczinski, R andom Gr aphs , Jo hn Wiley & Sons (2000). [9] F. Kr zak a la, A. Montanari, F. Ricci-T ersenghi, G. Semerjian, and L. Zde- bo ro v a , Gibbs states and the set of solutions of r andom c onstra int satisfac- tion pr oblems , Pro ceedings of the Nationa l Academ y of Science s 104 (20 0 7), pp. 10318–1 0323. [10] M. M ´ e z ard, T. Mora, and R. Zecc hina, Clustering of solutions in t h e ra n - dom satisfiability pr oblem , Physical Review Letters 94 (2005). [11] M. M´ ezard, G. Parisi, and M. Vir asoro, Spin glass the ory and b eyond , W orld Scient ific (1987). [12] M. Molloy , Mo dels for r andom c onstr aint satisfaction pr oblems , in Pr o c e e d- ings of the 32nd ACM Symp osium on The ory of Computing , 2002. [13] R. Monasson and R. Zecchina, St a tistic al me chanics of the r andom k -SA T mo del , Physical Re v iew E 56 (199 7 ), p. 135 7. [14] A. Percus, G. Istrate, and C. Mo ore (eds.), Comp utational Complexity and Statistic al Physics , Oxfor d Universit y P ress (2006). [15] J. Ra y m ond, A. Spo rtiello, and L. Zdeb oro v´ a, The phase diagr am of r an- dom 1-in-3 satisfiabi lity , Phys. Rev. E 7 6 (20 07). [16] J. Schmidt-Pruznan and D. Sha mir, Comp onent structur e in the evolution of r andom hy p er gr aphs , Combinatorica 5 (1985), pp. 81 –94. 12