...MEAN-FIELD THEORY OF SPIN GLASSES AND BOOLEAN SATISFIABILITY...

Transcription

1 ...MEAN-FIELD THEORY OF SPIN GLASSES AND BOOLEAN SATISFIABILITY... JAESUNG SON THESIS ADVISOR: ROBIN PEMANTLE APRIL 16, 018 1

2 1 Introduction 1.1 The Ising model Suppose the Ising spins are localized the vertices of some region L = 1,, L} d of a d-dimensional cubic lattice of side L. On each site i L, there is an Ising spin σ i +1, 1}, which can be thought of as a spin that points either up +1 or down 1. A configuration of the system σ = σ 1,, σ N is an assignment of values of all the spins in the system. So, the space of configurations X N = X X = +1, 1} }} L, with X = +1, 1} and N = L d. N times Definition 1.1. We equip a configuration σ with an energy function, or Hamiltonian, as such: 1. Eσ = J i,j σ i σ j h i L σ i, where the sum over i, j is over all unordered pairs of sites i, j L that are nearest neighbors and h measures the applied external magnetic field. The variables J s are interactions of a bond ij and are assigned a value J > 0 when the neighboring spins are aligned σ i = σ j or or a value J < 0 when the neighboring spins are anti-aligned σ i σ j. The positive interaction J > 0 can lead to ferromagnetism macroscopic magnetism as all the pairs of spins in the system have the tendency to align in the same direction and thus is called a ferromagnetic interaction [N08]. The negative interaction J < 0 has the opposite effect in which spins tend to align in opposite directions and thus is called a anti-ferromagnetic interaction. Ising solved the one-dimensional Ising model in his 194 thesis, a problem given to him by Lenz, and showed the absence of any phase transitions [Is5]. Physical phase transitions can be understood as when a small temperature in some parameter such as temperature or pressure causes a drastic qualitative change in the state of the system. In the one-dimensional Ising model, Ising showed that for any positive, finite temperature, the system is always disordered. Although Ising erroneously concluded the absence of any phase transitions in higher dimensions, Peierls, in 1936, showed that phase transitions in fact do occur in higher dimensions using an argument based on the free energy of domain walls, a boundary separating magnetic domains [PB36]. In 1944, Lars Onsager solved the two-dimensional Ising model by using the transfer-matrix method and showed that the model undergoes a phase transition between an ordered phase and a disordered phase [On44]. The attempt to extend the

3 transfer-matrix method to the three-dimensional case was unsuccessful and the problem remains unsolved in the higher dimensional cases. Definition 1.3. In general, the probability for the system to be found in a specific configuration x from a configuration space X, µ β x, is given by the Boltzmann distribution: µ β x = e βex Zβ, Zβ = x X e βex. Here, the parameter β = 1/T is the inverse of temperature, with Boltzmann s constant k B = 1. Zβ is a normalization constant and is called the partition function. Note that at infinite temperature, β = 0, all configurations have equal weights according to the Boltzmann distribution and the energy function is irrelevant. In the case of the Ising model, all configurations have weight N and the Ising spins are completely independent. At zero temperature, β = +, the Boltzmann distribution has only weights at the ground states, where, in general, the configuration x 0 X is a ground state if Ex Ex 0 for any x X. The minimum value of the energy E 0 = Ex 0 is the ground state energy. If there is no magnetic field, h = 0, then there are two degenerate ground states: the configuration σ + with all spins σ i = +1 and the configuration σ with all spins σ i = Thermodynamic potentials Some of the properties of the Boltzmann distribution can be summarized through the thermodynamic potentials, which are functions of the temperature 1/β and the parameters that define the energy function Ex. Definition 1.6. Free energy is defined as 1.7 fβ = 1 ln Zβ. β Definition 1.8. It may be more convenient to work with free entropy 1.9 Φβ = βfβ = ln Zβ. 3

4 Definition From free entropy, we can derive the internal energy Uβ 1.11 Uβ = β βfβ. The thermodynamic potentials defined above are consolidated here for easier navigation and will appear later when we discuss the replica trick in detail. 1.3 Mean-field theory of spin glasses The interaction of localized magnetic moments can be ferromagnetic in which all neighboring spins align in the same direction or anti-ferromagnetic in which all neighboring spins point in opposite directions. Spin glasses are disordered magnets which have a random magnetic spin structure, with some ferromagnetic and anti-ferromagnetic interactions, while ferromagnetic solids have an ordered spin structure in which all magnetic spins align in the same direction. The simplest model for ferromagnetism is the Ising model, which was discussed above Sherrington-Kirkpatrick model In 1975, David Sherrington and Scott Kirkpatrick introduced an exactly solvable model of the spin glass called the Sherrington Kirkpatrick model or SK model [SK75]. Here, we will introduce the SK model and the idea of the replica calculation. Consider the space of N configurations of N Ising spins with values in ±1}. Definition 1.1. The energy function or Hamiltonian of the SK model is given by 1.13 Eσ = i<j J ij σ i σ j h i σ i, where the first sum is over all NN 1/ distinct pairs, the coupling random variables J ij, 1 i < j N are independent and identically distributed i.i.d Gaussian N J 0 /N, J /N, and h is an external magnetic field that interacts with the spins The replica trick A sample or instance of the SK model is given by one of the values of its N energy levels. In order to describe typical samples, we have to compute the average log-partition function E ln Z. However, this is a 4

5 rather difficult task and it is much easier to compute the integer moments of the partition function EZ n, with n N, since Z is the sum of a large number of simple terms [MM09]. Lemma The replica method utilizes the following identity ln EZ n lim EZ n E ln Z = lim = lim n 0 n n 0 n by preparing EZ n as if n were an integer and taking the limit as n 0 to apply the identity If we were to calculate EZ n with n as a non-integer real number, the computation would be not much easier than that of E ln Z. The heart of the replica trick is to extrapolate E ln Z by first assuming n to be an integer and later taking the limit as n 0. This bold justification has not yet been resolved; however, the replica method has been become standard practice as, when compared with other exact solutions, the method leads to the same results. We will explore the derivation of the replica calculation for the Sherrington-Kirkpatrick model, including the replica symmetric solution and the 1-step replica symmetry breaking solution in Section. We only state the result of the full replica symmetry breaking solution and suggest Appendix B for the Parisi equation [N08]. We will also discuss the complications regarding the assumptions of the replica method. 1.4 Computational complexity The theory of computational complexity deals with classifying computational problems by their difficulty, where an algorithm inputs some number N of variables and answers either yes or no. There are two classes of algorithms: polynomial and super-polynomial. Let T N be the number of operations required to solve, in the worst-case, an instance of size N. Definition The algorithm is polynomial if there exists a constant k such that T N = ON k. Definition In the big O notation, we say fx = Ogx if and only if there exists some positive number M and real number x 0 such that fx M gx for x x 0. If the algorithm is not polynomial, it is super-polynomial. Definition To compare two decision problems A and B, we say that B is polynomially reducible to A if the following conditions are satisfied [MM09]: 5

6 There exists a mapping R which takes any instance I of the decision problem B to the instance RI of A such that the solution of the instance RI of A is the same solution as that of I of B. The mapping R can be computed in polynomial time. Such a mapping R is a polynomial reduction. Some of the complexity classes of decision problems are as follows [MM09]: P: Polynomial problems can be solved by an algorithm running in polynomial time. NP: Non-deterministic polynomial problems can be solved by a non-deterministic algorithm running in polynomial time. NP-complete: A problem is NP-complete if it is in NP and if any other problem in NP can be polynomially reduced to it. NP-hard: Non-deterministic polynomial hard problems are not in NP but are as hard as NPcomplete problems. The following theorem by Stephen Cook and Leonid Levin is stated without proof: Theorem The satisfiability problem is NP-complete. The implications of the theorems are quite remarkable in that any problem in NP is polynomially reducible to an instance of the satisfiability problem. Thus, any problem in NP can be solved in polynomial time. Definition 1.0. We introduce the complexity class #P, which is the set of counting problems associated with decision problems in the complexity class NP. Definition 1.1. Parallel to the decision problems, a counting problem is #P-complete if and only if it is in #P. So, while an NP decision problem may ask whether there exists an assignment of variables that satisfies a given general Boolean formula, a #P counting problem would ask how many different variable assignments will satisfy that given general Boolean formula. Such a counting problem of counting the number of satisfying assignments of a given general Boolean formula is #SAT The satisfiability problem Definition 1.. A clause is a logical OR of some variables or their negations. 6

7 A variable x i takes values 1 TRUE or 0 FALSE and its negation x i := 1 x i. Definition 1.3. A variable or its negation is a literal, denoted by z i. A clause a involving K a variables is a constraint that forbids exactly one of the Ka possible assignments of these K a variables. When a clause is not satisfied, it is said to be violated. Definition 1.4. An instance of the satisfiability problem can be expressed by the conjunctive normal form CNF: 1.5 F = C 1 C M, where C a = z i a 1 z i a Ka, and i a 1, i a K a } 1,, N}. Given F, is there an assignment of the x i s in 0, 1} N such that F is true? If so, then F is SAT and otherwise, F is UNSAT. This question of satisfiability is known as the Boolean satisfiability problem. We have the K-satisfiability K-SAT problem if we require all the clauses to have the same length K a = K. The MAX-SAT problem is one in which we find the configuration that violates the fewest clauses. Note that if K a, then the problem is polynomial; however, if K a K with K 3, then the problem is NP-complete Random K-SAT threshold An instance of a random K-SAT has only clauses of length K. SAT N K, M denotes the random K-SAT ensemble with N as the number of variables and M as the number of clauses. A formula in SAT N K, M N with M clauses of size K is selected randomly from the K. K Definition 1.6. A parameter for the random K-SAT ensemble is the clause density α := M/N. N An instance of the ensemble SAT N K, α is generated by selecting each of the K possible clauses K N independently with probability αn K /. P N K, α is the probability that a randomly generated K formula is satisfiable. As α 0, the probability goes to 1 and as α, the probability goes to 0. Simulations indicate an existence of a phase transition at some finite value α sat K. So, for α < α sat K, a random K-SAT formula is SAT with P N K, α 1 as N, and for α > α sat K, the formula is UNSAT with P N K, α 1 as N. The existence of such a phase transition has been shown for the random -SAT at the critical clause density α sat = 1 and the proof by bicycles, found in [MM09], is a result of [Go96] and is presented below. 7

8 Theorem 1.7. Let P N K =, α be the probability for a SAT N K =, M random formula to be SAT. Then, 1.8 lim N P NK =, α = 1 if α < 1, 0 if α > 1. Proof. We first show that a random formula is SAT with high probability for α < 1. We define a directed graph DF of some formula F by associating each of the N literals with some vertex. Whenever there is a clause such as x 1 x, if x 1 = 1, then x = 1, and if x = 0, then x 1 = 0. We represent these cases graphically by drawing a directed edge from x 1 to x and an undirected edge from x to x 1. Then, F is UNSAT iff there exists some index i 1,, N} such that DF contains a directed path from x i to x i and from x i to x i. Define a bicycle of length s as a path u, w 1, w,, w s, v, where the w i s are literals on s distinct variables, and u, v w 1,, w s, w 1,, w s }. From above, if a formula is UNSAT, then there exists a cycle containing two literals x i and x i for some i 1,, N}. Since the probability that there exists a bicycle in DF, denoted PA, is bounded above by the expected number of bicycles, we have 1.9 PF is UNSAT PA N N s s s M s+1 where s is the size of the bicycle. The sum 1.9 is O1/N when α < 1. s= s+1 1, The proof for that the random formula is UNSAT with high probability for α > 1 follows from Theorem N Upper bounds on the satisfiability threshold can be obtained by using the first moment method: let UF be a random variable such that 0 if F is UNSAT, 1.30 UF = 1 otherwise. Thus, for any formula F, we have 1.31 PF is SAT EUF. The choice of UF = ZF, where ZF is the number of SAT assignments is ideal since EUF vanishes as N for α large enough. Since the probability that an assignment is SAT is uniform on all the possible 8

9 assignments, 1.3 EZF = N 1 K M = exp [ N ln + α ln1 K ]. So, if α > α K, where 1.33 α ln K := ln1 K, EZF is exponentially small at large N and so PF is SAT vanishes as N. We restate the result from [FP83], but found in [MM09], below as the following theorem. Theorem 1.34 FP83. If α > α K, then lim PF is SAT = 0, and so, αn sat K α K. N As simulations suggest that there exists a phase transition in the random K-SAT between the SAT and UNSAT phases for any K, we have the following conjecture. Conjecture 1.35 Satisfiability threshold conjecture. For any K, there exists a threshold α sat K with 1.36 lim N P N K, α = 1 if α < α sat K, 0 if α > α sat K. As mentioned previously, the conjecture has been proven for K =. The theorem below from Friedgut [Fr99] strongly supports the case for the conjecture and all that remains to prove the satisfiability threshold conjecture is that α N sat K α sat K as N. Theorem 1.37 Friedgut s Theorem. There exists a sequence of α N sat K such that, for any ε > 0, 1.38 lim P 1 if α N < α N sat K ε, N K, α N = N 0 if α N > α N sat K + ε. In section 3, we clarify the satisfiability threshold conjecture explore some of the recent developments on the bounds for the conjecture. 9

10 Mean-field Theory of Spin Glasses.1 Sherrington-Kirkpatrick model Suppose we have the space of N configurations of N Ising spins that take values ±1}. Definition.1. The energy function, or the Hamiltonian, of the SK model is given by. Eσ = N J ij σ i σ j h σ i, i<j N where the first sum is over all = NN 1/ distinct pairs with 1 i < j N, and h is an external magnetic field that interacts with the spins. Definition.3. The interactions J ij s are i.i.d. coupling random variable N J 0 /N, J /N with density function N.4 P J ij = πj exp N J J ij J } 0. N Some others let J ij N 0, 1/N but this is merely a technicality and the result is not affected. The SK model is an infinite-range interaction mean-field model as there is no notion of Euclidean distance between the positions of the spins and it is a fully connected model since each spin interacts directly with all the other spins [MM09]. The SK model is the closest to the original spin glass problem.. The replica derivation Note that a sample or instance of the SK model is given by one of the values of its N energy levels, E 1,, E N, and from 1.5, the partition function is.5 Z = N 10 exp βe i.

11 Assuming that n is an integer, the first step of the replica trick involves finding an expression for Z n.6.7 Z n = n N i exp βe ia = = n exp β E ia, σi a} N i 1=1 N i n=1 exp βe i1 + + E in which is the partition function of a new system with configuration given by the n-tuple i 1,, i n with i a 1,, N } and the sum σ a i } is over the Nn different configurations. This sum is will be shortened henceforth by the trace representation Tr. Using our expression for the energy function. gives us.8.9 n Z n = Tr exp β N J ij σi a σj a + h σi a i<j = Tr } [ n N n }] exp σi a σj a exp βh σi a. i<j βj ij..1 Replica average of the partition function To proceed with our replica calculation, we take the expectation of the new partition function.9: EZ n i = Tr E i<j [ exp ii βj 0 = Tr exp N i<j iii βj 0 = Tr exp N n βj ij }] N n } σi a σj a exp βh σi a n σi a σj a + n i<j [ βj σ a i σ a j + β J N n σi a σj a N i<j ] N n exp βh n N n σi a σj a + βh where i is from the linearity of expectation and the independence of the J ij s, ii is from the moment generating function of the normal distribution, and iii is from the property of exponentiation. With a little algebra, we have the following expression for the sum σ a i σ a i },.13 n n n σi a σj a = σi a σj a σi b σj, b b=1 σ a i σ a j σ b i σ b j = n + with the n from that σ a i σ b j = 1 and the sum over ordered pairs from the symmetry of the expression. 11

12 Using.13 and some manipulation gives us the following expression: β J N n σi a σj a = β J N i<j NN 1n = β J N 1n 4 = β J Nn 4 β J n 4 + β J N σi a σj a σi b σj b i<j N σi a σj a σi b σj b N σi a σi a σi b σ b i i,j=1 + β J N σi a σi b β J nn 1. N 4 + β J N Similarly, we have βj 0 N n i<j σ a i σ a j = βj 0 N = βj 0 N n i,j=1 n N N σi a σj a σ a i n βj 0n. N σi a σi a Since the constant terms those without N in.16 and.18 are irrelevant to the leading exponential order in the large N-limit, we discard them henceforth in our calculations, replacing =, the equal sign, with =, the notation for the leading exponential order. Thus, we have.19.0 EZ n βj 0 =Tr exp N n N σ a i β J Nn βj 0 = exp Tr exp 4 N Using the property of exponentiation, we have.1 β EZ n J Nn = exp Tr exp βh 4 N + β J Nn n σ a i 4 n N + β J N σ a i N N n σi a σi b + βh σ a i + β J N N n σi a σi b + βh N } n βj 0 N exp N σ a i β J exp N σ a i. N σi a σi b.. Reduction by Gaussian integral Lemma.. From the distribution of the Gaussian N µ, σ random variable,.3 1 π σ e x µ/σ dx, 1

13 we can derive the following identity:.4 From lemma., letting a = n βj 0 N exp.5 b e bx /+abx dx = exp π ba N σi a /N, b = βj 0 N, and the integration variables be m a, a 1,, n}, N σi a /N = n n = [ βj0 N π exp βj 0Nm ab βj0 N dm a exp βj 0N π. + βj 0 m a N n m a } exp σ a i } dm a ] where we instead use the integral notation dxfx instead of fxdx for convenience. In this unusual integral notation, we take dx i c fx i to be dx i cfx i, where c is some constant. Similarly, letting a = β J N exp.6 i i i βj 0 n N N σi a σi b /N, b = β J N, and the integration variables be Q ab, 1 a < b < n, N σi a σi b /N = [ β J N π = Substituting.5 and.6 into.1 gives us.7 exp β J NQ ab β J N dq ab exp β J N π β EZ n J Nn n βj0 N β J N = exp dm a dq ab 4 π π } βj 0 N n exp m a + β J N Q ab n N Tr exp β + J 0 m a σi h a + β J } N σi a σi b Q ab. Lemma.8. If g is some arbitrary function, we have that N }.9 Tr exp gσi a = exp N ln tr exp gσ a }, where the trace tr = σ a } is over all states of a single replicated spin σ a [FH93]. σ a i m a } } ] N + β J σi a σi b Q ab dq ab Q ab } exp β J, } N σi a σi b Q ab. 13

14 Using the relation.9, we have that.30 n EZ n = βj0 N β J N dm a dq ab exp NGQ, m }, π π where GQ, m is a function of nn 1/ + n variables Q ab, 1 a < b n, and m a, 1 a n, with.31 GQ, m := β J n 4 βj 0 n m a β J n Q ab + ln tr exp β + J 0 m a σ h a + β J } σ a σ b Q ab...3 Method of steepest descent The method of steepest descent or the saddle-point method approximates an integral in roughly the direction of the steepest descent. The integral to be approximated in.30 is in the form of.3 b a e λgz dz, where a and b are some limits of integration and λ is large. Because the integral will be dominated by the highest saddle point, a < z 0 < b, when the function gz is near z 0, we have, from the Taylor series expansion,.33 gz gz g z 0 z z 0, where the first derivative term is zero since we choose z 0 such that g z 0 = 0. Lemma.34. The integral in.3 can be approximated as such.35 b as λ if g z 0 < 0 and h is positive. a hze λgz π dz λ g z 0 hz 0e λgz0, At the extremum of GQ, m, namely Q, m which we will determine later, we can approximate the integral.30 for large N using the saddle-point method EZ n Q,m exp NGQ, m } β J N 1 + n βj 0N 4 n n m a β J N n Q ab + 1 n ln tr e LQ,m} Q,m. 14

15 where the function LQ, m of Q and m is defined as n.38 LQ, m := β + J 0 m a σ h a + β J σ a σ b Q ab, and in taking the limit n 0 and letting N < be large, we obtain the last approximation.37 from the tangent line approximation of e λx near x = 0:.39 e λx 1 + λx. With the definition of free entropy 1.8 and the replica identity 1.15, we have the following expression for free energy EZ n 1 β[f] = lim n 0 nn β J = lim n 0 4 βj 0 n n m a β J n Q ab + 1 n ln tr e LQ,m} Q,m, where [f] = f N is the configurational average of the free energy. To estimate the integral.30 at large N via the saddle-point method, we differentiate the function GQ, m.31 with respect to its arguments and set it equal to zero. For all a < b, we have that.4 which implies G Q ab = β J Q ab + tr β J σ a σ b n exp β + J 0 m a σ h a + β J } σ a σ b Q ab n tr exp β + J 0 m a σ h a + β J } = 0, σ a σ b Q ab.43 Q ab = σ a σ b n. Definition.44. We introduce here the notation.45 fσ n := 1 tr fσ exp β ZQ ab, m a n h + J 0 m a σ a + β J σ a σ b Q ab } n.46 ZQ ab, m a := tr exp β + J 0 m a σ h a + β J } σ a σ b Q ab, where fσ = fσ 1,, σ n is any function., 15

16 Similarly, we have.47 which implies that G m a = βj 0 m a + n tr βj 0 σ a exp β + J 0 m a σ h a + β J } σ a σ b Q ab n tr exp β + J 0 m a σ h a + β J }, σ a σ b Q ab.48 m a = σ a n. We have the following second partial derivatives G Q ab = β J + Qab = σ a σ b n β J σ a σ b ZQ ab, m a β J σ a σ b n ZQ ab, m a n ZQ ab, m a Qab = σ a σ b n.49 = β J, G = βj 0 + ma= σ a n m a βj 0 σ a n ZQ ab, m a βj 0 σ a n ZQ ab, m a ZQ ab, m a ma= σ a n.50 = βj 0. We can also confirm from taking the partial derivatives of.41 our results in.43 and Replica symmetric solution It is natural but naïve to assume replica symmetry RS, Q ab = q for a b, Q ab = q for a = b, and m a = m, and derive a replica-symmetric solution. Changing the sums over a < b to sums over a b by symmetry and substituting our values for Q ab and m a into the free energy expression.41 give us, before the limit n 0,.51 β[f] RS = β J 4 βj 0 n nm β J 4n nn 1q + 1 n ln tr elqrs,mrs, where LQ RS, m RS is simply.5 LQ RS, m RS := β n h + J 0 mσ a + β J q σ a σ b. 16 a b

17 With a little algebra, we have that.53 n n σ a σ b = σ a σ a σ a, a b and so LQ RS, m RS can be rewritten as.54 LQ RS, m RS = β n h + J 0 mσ a + β J n q σ a β J qn The last term can be simplified using the Gaussian reduction identity.4 with the integration variable as n z, a = σ a, and b = β J q ln tr e LQRS,mRS = ln tr β J q π = ln dz = ln dz dz exp β J q z + e z / exp n } π β J q tr e z / exp n π β J q n σ a β J qz + β } n h + J 0 mσ a n β J q n exp βj qz + βh + J 0 m σ a} } n tr exp βj qz + βh + J 0 m σ a}, where in.55 we replaced β J qz with z, which does not change the expression as β J q > 0 and the limits of integration are from to and in.56 we used the property of exponentiation to interchange n the order of tr. Since the single replica spin can take one of two values σ a ±1}, we have.57 e z / = ln dz exp n π β J q = ln = ln 1 + n } n cosh βj qz + βh + J 0 m σ a} Dz exp n ln [ cosh βj qz + βh + βj 0 m ] n β J q Dz ln cosh β Hz n β J q + On, where the hyperbolic function coshz = e z + e z /, Dz = dz e z / / π is the Gaussian measure, Hz = J qz + h + J 0 m, and O is the big O notation. In the last equality.57, we used the Taylor expansion e z = 1 + z + Oz. Using our expression.57 in.51 and taking the limit n 0 give us.58 β[f] RS = β J 4 1 q βj 0m + Dz ln cosh β Hz, } 17

18 where in applying L Hôpital s rule, we have the limit ln1 + ax + bx.59 lim = lim x 0 x x 0 a + bx 1 + ax + bx 1 = a. Solving the extremization condition for free energy.58 with respect to q gives us = β J q 1 + sinhβ Dz Hz βjz coshβ Hz q / βj/ 8πq = β J z q 1 tanhβ Hze = β J q β J + Dzβ J sech β Hz/, + + dze z / βj sech β HzβJ q/ 8πq by integration by parts using u = tanhβ Hz and dv = dze z / βjz/ 8πq and from that e z / is an even function of z that decays to zero. This implies.63 q = 1 Dz sech β Hz = Dz tanh β Hz Solving the extremization condition for free energy.58 with respect to m gives us.64 0 = βj 0 m + sinhβ Dz Hz coshβ Hz βj 0, which implies.65 m = Dz tanh β Hz..3.1 Phase diagram The behavior of the solutions.63 and.65 depends on the parameters β and J 0. For simplicity, suppose that the external magnetic field h = 0 for the rest of the section. If the distribution of the coupling random variable J ij is symmetric, then J 0 = 0 and Hz = J qz. So, tanhβ HZ is an odd function of z. Thus, the magnetization m in.63 is 0 and there is no ferromagnetic phase. The free energy is now.66 β[f] RS = 1 4 β J 1 q + Dz ln cosh βj qz = 1 4 β J 1 β J q β J q + ln + 1 β J q 1 4 β4 J 4 q + Oq 3 = 1 4 β J + ln β J q 1 q + Oq 3. 18

19 Lemma.69. We used the following asymptotic expansion below in.67 to obtain an expression.70 1 π e z / ln coshβj qzdz = N n=1 B n n!n n 1 1 n βj q n + O q N+, where B n = 1n+1 n! π n ζn for n 1 is the Bernoulli numbers and ζn = zeta function. 1 k n is the Riemann The Landau theory implies that the critical point is determined by the condition of a vanishing coefficient of the second order term q and so the spin glass transition exists at T = J =: T f [N08]. Figure 1. Phase diagram of the SK model. The dashed line is the boundary between the ferromagnetic F and spin glass SG phases and exists only under the ansatz of replica symmetry. The replica-symmetric solution is unstable below the AT line, and a mixed phase M emerges between the spin glass and ferromagnetic phases. The system is in the paramagnetic phase P in the high-temperature region [N08]. The boundary between the spin glass and ferromagnetic phases is given numerically by solving.63 and.65 and we can depict the phase diagram as shown above in figure 1 [N08]..3. Negative entropy We see that the assumption for replica symmetry fails at low temperatures as the ground-state entropy for J 0 = 0 is the negative value 1/π. Since lim x tanh x = 1, we have q 1 as 1/β = T 0, and assume 19

20 that q = 1 at, with a > 0 for small T > 0. To check this assumption, we have, for β, Dz sech βjz = 1 Dz d βj dz tanhβjz 1 Dz δz} βj T = π J. Thus, the assumption that q = 1 at is justified with a = expression for free energy.66 gives us, for large β, 1 π J. Substituting q = 1 at into the β[f] RS = 1 4 β J a T + J 4 J π Dz 1 βj1 at/ + + π π Dz ln e βj qz 1 + e βj qz βj qz + ln 1 + e βj qz 0 Dz e βj qz, where we changed ln1 + z z for small z..77 Dz to β[f] RS 1 βj π + 0 Dz since coshz is an event function of z and used the approximation 1 π π T J + exp β J 1 } T π J.78 = 1 π + π βj. So, we have the following expression for the free energy in low temperature.79 [f] RS T π π J, with the ground-state entropy as 1 π and the ground-state energy π J. Although the interchange of the limits n 0 and N to derive the free energy.41 using the method of steepest descent may have been the source of the negative entropy, we see in section.6 that the problem lies in the replica method itself. 0

21 .4 Replica symmetry breaking The free energy expression derived through the replica symmetry ansatz RS ansatz led to an erroneous negative entropy at low temperatures. So, it seems logical to break the replica symmetry and proceed with an approach in which the order parameter Q ab possibly depends on the replica indices a and b and m a on the replica index a..4.1 The Parisi Ansatz The n n matrix Q ab } for the replica-symmetric solution had Q ab = q for a b and 0 along the diagonals as such 0 q q q 0 q.80 Q ab } = q q 0 To start the first step of the replica symmetry breaking 1RSB, let m 1 n be a positive integer that divides the replicas into n/m 1 blocks. A diagonal block, which contains the main diagonal entries, has 0 s along its diagonals and q 1 as its off-diagonal entries while an off-diagonal block, which does not contain the main diagonal entries, has q 0 as its entries as such.81 0 q 1 q 1 q 1 q 1 0 q 1 q 1 q 1 q 1 0 q 1 q 1 q 1 q 1 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 0 q 1 q 1 q 1 q 1 0 q 1 q 1 q 1 q 1 0 q 1 q 1 q 1 q 1 0 Figure. 8 8 Parisi matrix with one step RSB n = 8, m 1 = 4. Then, in the second step of the replica symmetry breaking, the off-diagonal blocks remain the same, but the diagonal blocks are further divided into m 1 /m blocks. Within the subdivided blocks, the off-diagonal blocks remain the same while the diagonal blocks have q instead of q 1 in their off-diagonal entries as such The numbers n m 1 m 1 are integers and we define the function qx as.83 qx = q i for m i+1 x m i. 1

22 .8 0 q q 1 q 1 q 0 q 1 q 1 q 1 q 1 0 q q 1 q 1 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 q 0 0 q q 1 q 1 q 0 q 1 q 1 q 1 q 1 0 q q 1 q 1 q 0 Figure Parisi matrix with two step RSB n = 8, m 1 = 4, m =. Following the replica method, we take the limit n 0 and now have.84 0 m 1 m 1 for 0 x 1, treating 0 qx 1 as a continuous function..4. One-step replica symmetry breaking With our construction of the first step replica symmetry breaking, we now have the following expression.85 Q ab σ a σ b = 1 n q 0 σ a + q 1 q 0 n/m 1 block m1 a block σ a nq 1, where the first sum fills all the entries of Q ab } as q 0, the second sum replaces all the entries q 0 in the diagonal block with q 1, and the last term removes the entries q from the main diagonal and makes the diagonal entries zero [N08]. With the same approach of adding and subtracting terms, the sum with the quadratic term of Q ab from.41 is 1.86 lim n 0 n a b Q 1 ab = lim n q0 n/m 1 m n n 1q1 q0 nq1 = m1 q1 q0 q1. Substituting our expression.85 in the function LQ, m.38, we have.87 LQ RSB, m RSB = β n h + J 0 mσ a + β J q 0 n σ a + q 1 q 0 n/m 1 block m1 a block σ a nq 1

23 We can simplify the last term using the Gaussian reduction identity.4 with the integration variable u, n m 1 a 1 = σ a, and b 1 = β J q 0 and the integration variable v, a = σ a, and b = β J q 1 q a block β ln tr e LQRSB,mRSB J q n/m 1 0 = ln tr π du β J q 1 q 0 dv π block n } n exp β J q 0 u / + σ a β J q 0 u + β h + J 0 mσ a n β J q 1 n/m 1 exp β J q 1 q 0 v / + block = n β J q 1 + ln + 1 m 1 Du ln m1 a block where the derivation utilizes the same methods as was used in the RS ansatz. σ a β J q 1 q 0 v } Dv cosh m1 ηu, v, Definition.90. We define the function.91 ηu, v = β J q 0 u + J q 1 q 0 v + J 0 m + h. Using our two expressions above.86 and.89 for free energy.41, we have.9 βf 1RSB = βj 0 m β J 4 where we let m a = m be replica symmetric. 1 m1 q1 q0 + q1 1 q 1 + ln + m 1 Du ln Dv cosh m1 ηu, v, The parameters m, m 1, q 0, and q 1 are all between 0 and 1. Solving the extremization conditions of free energy.9 with respect to m, q 0, and q 1 gives the following [N08].93 Dv cosh m 1 ηu, v tanh ηu, v m = Du Dv cosh m 1 ηu, v q 0 = q 1 = Du Du Dv cosh m 1 ηu, v tanh ηu, v Dv cosh m 1 ηu, v Dv cosh m 1 ηu, v tanh ηu, v Dv cosh m 1. ηu, v When J 0 = h = 0, νu, v = βj q 0 u+j q 1 q 0 v is an odd function of u and v, which implies m = 0 from.93. The order parameter q 1 can be positive when T < T f = J because the first term in the expansion.95 is β J q [N08]. Thus, the RS and 1RSB ansatz result in the same transition point. 3

24 .5 Full replica symmetry breaking solution We have that the entropy per spin with J 0 = 0 and T = 0 changes from in the RS solution to π approximately 0.01 in the 1RSB solution [N08]. This seems to indicate that the symmetry breaking ansatz leads us towards a stable solution and that we should make further steps in our replica symmetry breaking. Thus, we present the expression free energy.41 using a full replica symmetry breaking solution FRSB found in [N08] and the interested reader should refer to Appendix B of [N08] for a complete derivation. For simplicity, we will only consider the case J 0 = 0. We have the following expression for the K-step RSB K-RSB qab l = q0n l + q1 l q0m l 1 a b = n K m j m j+1 qj, l j=0 n m 1 + q l q l 1m m1 m n m 1 + q l K n where l is some integer, m 0 = n, and m K+1 = 1. As n 0, we can replace m j m j+1 dx, giving us.98 1 n qab l a b 0 q l xdx. The internal energy for J 0 = 0, h = 0 can be evaluated by the partial derivative with respect to β of U = βf = βj 1 β n Q ab = βj q xdx. The magnetic susceptibility can be evaluated by the second partial derivative with respect to h of.41 as.100 χ = β n a b Q ab β 1 0 qxdx. The expression for free energy [N08] is given by.101 βf = β J } qxdx q1 Duf 0 0, q0u, where f 0, with the initial condition f 0 1, h = ln coshβh satisfies the Parisi equation.10 f 0 x, h x = J dq f 0 dx h + x } f0. h 4

25 .6 Possible issues with the replica trick As we have discussed previously, to evaluate E ln Z, we employ the replica method We first take n to be integers to calculate the nth moment EZ n and proceed to take the limit as n 0. A natural complication arises as the two steps are not compatible with each other. For the replica trick to hold, a sufficient condition of uniqueness of the moment problem is given by Carleman stated without proof. Theorem.103. Let Z be a real random variable with moments m n = EZ n such that.104 n=1 m 1/n n =. Then, any random variable with the same moments as Z is identically distributed to Z. So, if the sum in.104 converges, then the distribution of Z is not necessarily determined by its integer moments. However, the SK model does not satisfy Carleman s condition and there is yet to be a formal mathematical justification for the use of the replica trick in solving models in statistical mechanics like the SK model [Ta07]. In deriving the expression for free energy.41, we interchanged the limits n 0 and N as such E ln Z lim N N = lim lim N n 0 1 = lim n 0 n ln EZ n Nn lim N ln EZ n N to apply the method of steepest descent. Hemmen and Palmer prove the validity of interchanging the limits for the SK model and find that the complications with SK model lies only in taking the moments from integers to real n [HP78]. 3 Satisfiability From Section 1, we found that the threshold for the -SAT is α = 1 using the first moment method [FP83]. It is known that the K-SAT problem is NP-complete for K 3 [GJ79], which may explain the different 5

26 behavior of the K = and the K 3 problems. Following the construction of Monasson and Zecchina [MZ97], we minimize the energy of the random K-SAT with respect to the number of unsatisfied clauses. Let the Ising spin σ i take value +1 if the logical variable x i = TRUE and value 1 if the logical variable x i = FALSE. Let the coupling random variable C li be +1 if the clause C l includes x i, 1 if it includes x i, and 0 if the clause includes neither x i nor x i. Since we have a clause length of K, we choose K non-zero C li from C l1,, C ln } and assign ±1 randomly for these K random variables. So, we have that +1 if σ i = C li, 3.1 C li σ i = 0 if C li = 0, 1 if σ i C li. Thus, the K-SAT problem is satisfiable if there exists an assignment of x i s and thus a configuration σ such that there is at least one σ i for which C li σ i = 1 for every l 1,, M}. Since a clause contains only N K logical variables, the minimum value of C li σ i = K, which is coincidentally the only case that the clause C l is unsatisfied. So, if clause C l is satisfied. N C li σ i > K, then there exists at least one σ i such that C li σ i = 1 and the Definition 3.. The energy function or Hamiltonian of the K-SAT problem is given by M N 3.3 Eσ = δ C li σ i, K, l=1 where δi, j is the Kronecker delta function with δi, j = 1 if i = j and δi, j = 0 if i j. From this construction, we see that the ground-state energy level Eσ = 0 implies that the problem is satisfiable. 3.1 The replica derivation Note that an instance of the satisfiability problem is given by one of the values of its M N levels, E i, 1 i KM, and from 1.5, the partition function is K M N K energy K 3.4 Z = i exp βe i. 6

27 Assuming that n is an integer, the first step of the replica trick involves finding an expression for Z n n Z n = exp βe ia = exp βe i1 + + E in i a i 1 i n = n exp β E ia, σi a} which is the partition function of a new system with configuration given by the n-tuple i 1,, i n with } M N i a 1,, KM and the sum Mn N is over the KMn different configurations. This sum K K σi a} will henceforth shortened by the trace representation Tr. Using our expression for the energy function 3.3 gives us [ n M N ]} Z n = Tr exp β δ C li σi a, K l=1 l=1 [ M n N ]} = Tr exp β δ C li σi a, K Replica average of the partition function To proceed with our replica calculation, we take the expectation of the new partition function 3.8: 3.9 EZ n = Tr ζ K σ M where used the linearity of expectation and the independence of the l clauses and define ζ K σ below. Definition As we will be working with the expectation in 3.9, we define the expression as such 3.11 ζ K σ := E exp β [ n N ]} δ C i σi a, K. From 3.1, we can express the Kronecker delta function in 3.13 by a product of Kronecker delta functions N 3.1 δ C i σi a, K = N,C i 0 δ σ a i, C i, where the product is only over i such that C i takes a non-zero value. So, substituting 3.1 into 3.13 and taking the configurational average over the C i s give us 3.13 ζ K σ = 1 K Ci 1=±1 C ik =±1 1 N K N i 1=1 7 [ N n K exp β δ σi a ]} k, C ik, i K =1

28 where we ignore the term ON 1 as it goes to zero in the large N limit. Definition We can rewrite the expression for ζ K σ 3.13 in terms of cs} s, which is the set of the number of sites with a specific pattern in the replica space s = s 1, s n }: N n 3.15 Ncs := δ σi a, s a. ζ K σ 3.13 only depends on σ only through cs} since if we choose C ik σi a k = s a k. Note from 3.1 that s a k = C i k σ ik = 1 iff C ik σ ik and so δ s a k, 1 = δ σa k, C k over nonzero C k. So, 3.13 can be rewritten as 3.16 ζ K σ = ζ k c} = 1 K c C 1 s 1 c C K s K exp β s 1 C 1=±1 C K =±1 s K } n K δ s a k, 1. Since σi asa i = 1 iff δ σa i, sa i = 1 and σa i sa i = 1 iff δ σa i, sa i = 0, we have cs = 1 N = 1 n N n 1 N 1 + σ a i s a i N N n Q a s a + 1 Q ab s a s b N N < <n Q ab n s a s n. Definition We define the multioverlaps Q ab as 3.0 Q ab = 1 N N σi a σi b, where we assume that the multioverlaps with an odd number of indices Q a = Q abc = = 0 [MZ97]. Remark 3.1. We have the symmetry cs = c s because the multioverlaps with odd number of indices are zero. Thus, since C i takes only values ±1, we may remove C i from c C i s i in the expression Remark 3.. The average partition function 3.9 can now be simplified as EZ n = dcse NE0c} Tr } δ cs 1 N n 3.3 δ σi a, s a, 0 N s s n E 0 c} := α ln cs 1 cs K 1 + e β 1 K 3.4 δ s a k, 1, s 1,,s K where α = M/N is the clause density. 8

29 The first term in 3.3 is a straightforward result of algebraic manipulation n exp NE 0 c} = exp Nα ln cs 1 cs K 1 + e β 1 K δ s a k, 1 s 1,,s K n = cs 1 cs K 1 + e β 1 K M δ s a k, 1 s 1,,s K n K M = cs 1 cs K exp β δ s a k, 1, s 1,,s K where in, we use the multiplicative property of exponentiation and that 3.8 exp β K δ s a k, 1 = The second term in 3.3 applies the constraint e β when 1 when K δ s a k, 1 = 1, K δ s a k, 1 = cs = 1 s in the form of the Dirac delta, using the definition of cs in Remark We will now simplify the second term in 3.3. Applying the trace operation over the spin variables s gives the entropy 3.31 Tr s δ cs 1 N } N n δ σi a, s a N! = Ncs!. s 9

30 Using the following approximation by Stirling 3.3 lnn! n ln n n, yields ln N! Ncs! ln N N N s s = ln N N s [ ln Ncs Ncs] + s Ncs ln N + ln [cs Ncs] Ncs 3.35 = s ln[cs Ncs ], where in and we used that s cs = 1 since cs is the fraction of sites i, among the N possible indices, such that σ a i = sa for all a 1,, n}. With a little algebraic manipulation, we have that 3.36 s ln[cs Ncs ] = ln s cs Ncs 3.37 = N ln s cs cs 3.38 = N s cs ln cs. Thus, our average partition function is now 3.39 EZ n = dcs exp N E 0 c} + s 0 s cs ln cs Applying the method of steepest descent.35 to the integral above 3.39 EZ n exp N E 0 c} cs ln cs s NE 0 c} N s cs ln cs, where in we use the approximation e λx 1 + λx for small x. 30

31 With the definition of free entropy 1.8 and the replica identity 1.15, we have the following expression for configurational average of free entropy in the limit N, M β[f] = βf N = EZn 1 N = E 0 c} + s cs ln cs. 3. Replica-symmetric solution The free entropy in 3.43 is extremized with respect to cs and similarly to the RS solution in the SK model, we assume cs is symmetric under permutation of s 1,, s n to obtain the RS solution. We can express cs in terms of the distribution function of the local magnetization P m as such 3.44 cs = Such a cs is replica-symmetric. n 1 + ms a dmp m. 1 We show the following two remarks in the Appendix 4.1. Remark The extremization condition of the free energy 3.43 gives the following equation for P m P m = π1 m exp αk + αk A K 1 = 1 + e β 1 K 1 u du cos ln K m k. Remark The free energy is expressed in terms of P m as 1 + m 1 m dm k P m k cos u ln A K 1 }, 3.49 βf = ln + α1 K Nn + αk K K dm k P m k ln A k dm k P m k lna K dmp m ln 1 m.

32 3..1 The K = 1 case When K = 1, A 0 = e β and 3.46 gives 3.50 P m = 1 u 1 + m π1 m du cos ln e α+α cosuβ/. 1 m Lemma We can express the exponential cosine in 3.50 using the modified Bessel functions 3.5 e z cos θ = where the kth modified Bessel function k= I k z coskθ, z k I k z = n!γk + n + 1. which gives us 3.54 P m = n=0 z e α u 1 + m π1 m du cos ln I k α cos kuβ/. 1 m k= Lemma Using the exponential relation of cosine 3.56 cosθ = eiθ + e iθ, and the the Fourier transform of e iω0u n 3.57 e iuω ω0 du = πδω ω 0, along with the additivity of integrals, gives us 3

33 3.58 e α P m = π1 m [ 1 + m e iu/ ln 1 m k= +kβ I k α/4 du ] + e iu/ [ ln 1 + m 1 m kβ ] + e iu/ [ ln 1 + m 1 m kβ ] + e iu/ [ ln 1 + m 1 m +kβ ] = = e α π1 m e α 1 m k= k= I k α/ π [ I k α δ ln [ δ [ ln [ + δ ln ] [ 1 + m + kβ / + δ ln 1 m ] [ 1 + m kβ / + δ ln 1 m 1 + m + kβ + δ ln 1 m ] 1 + m kβ, 1 m where, in, we used the scaling property of the Dirac delta for non-zero scalar c 3.61 δcx = δx c. ] 1 + m kβ / 1 m ] ] 1 + m + kβ / 1 m Lemma 3.6. Composing the Dirac delta with a smooth function g, with g nowhere zero, gives the following identity 3.63 δ gx = δx x 0 g, x 0 where x 0 is the real root of the function g. Corollary A corollary of the above lemma is the following 3.65 δ x a = 1 δx + a + δx a a x Since tanh = ex 1 e x, using lemma 3.6 and 3.64 in equation 3.60 yields P m = e α k= I k αδ m tanh βk. 33

34 Lemma The summation of the kth modified Bessel function has the following identity 3.68 I k z = 1 ez I 0 z Since tanhx ±1 as x ±, as temperature T goes to zero β = 1/T, using lemma 3.67, the local magnetization 3.66 becomes 3.69 P m = e α I 0 αδm + 1 [ 1 e α I 0 α ] [δm 1 + δm + 1]. Substituting our expression for 3.66 into 3.49 for K = 1 gives us βf Nn = ln + α dmp m ln e β 1 1 = ln αβ 1 e α k= I k α 1 1 dm δ dmp m ln 1 m m tanh βk ln 1 m. Lemma 3.7. We use the identity δx a ln 1 x dx = H1 aha + 1 ln 1 a, where Hx = x δξdξ is the Heaviside function, to derive, βf αβ = ln Nn 1 e α = ln αβ 1 e α = ln αβ + e α k= k= k= I k α ln 1 tanh βk I k α ln 1 tanh βk I k α ln cosh βk, which, as β, yields Eα N = lim β ln β = α e α + α e α k= = α e α k= ln cosh βk I k α lim β β k I kα 34 k= 1 ln cosh βk I k α β k I kα,

35 where the limit can be evaluated by L Hôpital s rule 3.80 ln cosh βk lim β β = lim β k βk sinh cosh βk = ± k, where we have k/ when k 0 and k/ when k < 0. Lemma Using the definition of the Bessel function 3.53, we have for integer n that 3.8 I n z = I n z. Lemma We also have the following relationship involving the neighboring terms for any real ν 3.84 νi ν z = z I ν 1z I ν+1 z, Using lemmas 3.81 and 3.83 in 3.79, we have Eα N = α e α = α e α k I k kα + I kα α I k 1α I k+1 α 3.87 = α α e α I 0 α + I 1 α, where the telescoping sum 3.88 n z I k 1z I k+1 z = z I 0z + I 1 z I n z I n+1 z z I 0z + I 1 z for z sufficiently small. Since α > 0, the ground-state energy is always positive. So, for K = 1 the SAT problem is unsatisfiable with probability The K case As we found previously in Theorem 1.7, the critical point for K = is α N sat = 1. Comparing the results of numerical simulations to the RS solutions for K indicate that the RS theory is correct for α < α N sat but not for α > α N sat. A further discussion of the K case can be found in [MZ97]. 35

36 3.3 Satisfiability threshold for random K-SAT From above in Theorem 1.7, we have shown a proof by bicycles for the -SAT [Go96]. The -SAT problem is unsurprisingly much simpler to analyze than the K-SAT problem for K 3. Indeed, in terms of computational complexity, the -SAT problem is polynomial while the K-SAT problem is NP-complete for K 3. As discussed previously, numerical simulations suggest that the random K-SAT has a phase transition between the SAT and UNSAT phases for any K. Friedgut s theorem 3.91, restated below, states that there exists a sharp threshold sequence α N sat K for K. An important note is that the theorem does not state whether α N sat K converges to a unique limit, which is what Conjecture 3.89, restated below, requires. Conjecture 3.89 Satisfiability threshold conjecture. For any K, there exists a threshold α sat K with 3.90 lim N P N K, α = 1 if α < α sat K, 0 if α > α sat K. As mentioned previously, the conjecture has been proven for K =. The theorem below from Friedgut [Fr99] strongly supports the case for the conjecture and all that remains to prove the satisfiability threshold conjecture is that α N sat K α sat K as N. Theorem 3.91 Friedgut s Theorem. There exists a sequence of α N sat K such that, for any ε > 0, 3.9 lim P 1 if α N < α N sat K ε, N K, α N = N 0 if α N > α N sat K + ε Upper bound With the moment method 1.31, we have the upper bound 1.33 given by Theorem The upper bound from [FP83] is not sharp, but is actually quite close to the more precise upper bound found by [KKKS98] presented below. Theorem 3.93 KKKS98. Let ɛ K denote an error term that decays to zero as K. Then, 3.94 lim sup N α N sat K K ln ln + ɛ K. 36

37 The upper bound is achieved by truncating the first moment to locally maximal solution. The following is an outline of the result from [KKKS98]. Let A n be the class of all truth assignments and P n be the random class of truth assignments that satisfy a random formula φ. We define a class smaller than P n as follows. Definition For a random formula φ, P n is defined the random class of truth assignments A such that: i A φ and ii any assignment obtained from A by changing exactly one FALSE value of A to TRUE does not satisfy φ. Such a construction is known as single flips. Remark Here, is the symbol for entailment and we say that a formula φ is a semantic consequence within some system of a set of statements A A φ if and only if every model which makes members of A true makes φ true. Consider the lexicographic ordering among truth assignments in which the value FALSE is considered smaller than the value TRUE and the values of variables with higher index are of lower priority in establishing the way two assignments compare. We see that Pn is the set of elements of P n that are local maxima in the lexicographic ordering of assignments, where the neighborhood of determination of local maximality is the set of assignments that differ from A in at most one position [KKKS98]. We thus have the following method of moments upper bound: Lemma Let F be a random formula PF is SAT E P n. Proof. From definition 3.95, we see that if an instantiation φ of random formula is satisfiable, the P nφ. We straightforwardly have 3.99 PF is SAT = φ Pφ 1 φ, where the indicator random variable 1 if φ is satisfiable φ = 0 otherwise. 37

38 Also, E P n = Pφ P n φ. φ Since P nφ, P nφ 1 and the lemma follows. The single flip method sharpens the previous lowest upper bound of for the 3-SAT due to Kamath et al. [KMPS95] to [KKKS98]. By defining an even smaller subset of P n with a more general double flip method further improves the bound to a value between and [KKKS98] Lower bound More recent advances in finding a lower bound for α N sat K have used the second moment method below. Lemma 3.10 Second moment method. Let X 0 be a random variable with finite variance. Then, PX > 0 EX EX. Remark This sequence of development can be briefly summarized as such: K 1 ln d k [AM0]; lim inf N αn sat K K ln K + 1 ln 1 ɛ K [AP04]; K ln 3 ln + ɛ K [CP1], where d k 1 + ln / and, as before, ɛ K 0 as K. The second moment method brings two issues in the random K-SAT model: the first concerning the geometry of the solution space of possible assignments SOL ±1} N, and the second concerning the geometry of the underlying bipartite graph [DSS16]. Remark Coja-Oghlan and Panagiotou [CP16] addressed both issues simultaneously and showed that lim inf N αn sat K K ln ln ɛ K, which matches the upper bound in Theorem 3.93 up to the error term ɛ K. Remark Ding, Sly, and Sun [DSS16] resolve the satisfiability threshold conjecture 3.89 for large K: For K K 0, with K 0 an absolute constant, the random K-SAT has a sharp satisfiability threshold α SAT. 38