Thresholds in Random Graphs and k-sat

Emmanuel Abbe

December 5, 2006

1 Introduction

In their seminal work [6],[7], Erdos and Renyi invented the notion of randomgraphs and made the fundamental observation that when the parameter con-trolling the edge probability varies, the random system undergoes a dramaticand swift qualitative change. “Much like water that freezes abruptly as itstemperature drops below zero the structure of the random graph quicklychanges from a primordial soup of small tree-like connected components to agraph in which the giant component, one of linear size, captures a constantproportion of the vertices.” — Friedgut [10].In this report we wish to present several results dealing with the existenceand sharpness of thresholds in random graphs, having in mind the purposeof understanding the sharp threshold of random k-sat. As mentioned is [15],two main categories of questions can be addressed. One type is algorithmic(e.g., finding an algorithm that decides whether an instance is SAT or UN-SAT, or that tries to minimize the number of violated constraints). Anotherone is more theoretical and deals with random instances for which one wantsto predict the typical behavior (e.g., phase transitions and structure of thesolution space).Beside being fascinating in its own right, the randomized version of the satis-fiability problem, contributes to the evaluation of the average case complexityof this problem, and helps understanding the behavior and performance ofalgorithms.A random k-CNF (conjunctive normal form) formula with parameter M , is

a boolean formula having the following structure:

(x1 ∨ x2 ∨ x4) ∧ (x1 ∨ x3 ∨ x5),

each bracket component is called a k-clause, which is the disjunctions (OR)of k variables possibly negated, and the conjunction (AND) of the clauses

1

Figure 1: Probability that a random 3-sat problem is satisfiable as a functionof α for n = 50 variables (taken from [11]).

is taken to form the formula. In previous example, k = 3, M = 2 and thenumber of variables is n = 5.A satisfying assignment, is an assignment of “true” or “false” to each vari-ables such that the global formula is true (each clauses has to be true, i.e.has to contain at least one true value).A random k-CNF formula is a random variable taking values in the set of k-CNF formulas. There are various model defining the probability distributionof the random formula, we will focus on the Binomial one. For a numbern of variables, let N =

(nk

)2k be the number of all possible k-clauses (we

consider here a model with no repetitions of the clauses and no repetition ofthe literals in the clauses, but those types of hypotheses are not influent inthe results), we then pick up each of these clauses with probability p. Theexpected number of clauses is then Np, so if p = M

N, the formula contains M

clauses in average.

Computer simulations suggest that this problem exhibits a sharp thresh-old behavior with respect to the M/n, i.e. defining α = M/n, the probabilityof having a satisfying assignment tends to 1 (with n) as long as α is belowsome critical value αc, and tends to 0 if it s above (cf. figure 1).

In [8], this result is proven and the sharpness of the swift is analyzed.General conditions are given to characterize what kind of graph propertiesenjoys sharp versus coarse thresholds. Before presenting this result, will gothrough several preliminary results.

2

2 Random graphs preliminaries

We will consider undirected graphs on n vertices. We denote by v(G) ande(G) the vertices and edges of a graph G. A graph property is a propertythat depends on the isomorphism type only, i.e. the property holds or not, nomatter what the labeling of the vertices is. Examples of graph properties are“being connected”, “containing a triangle”, “having 5 isolated vertices”, butnot “vertex number 1 is isolated”. A graph property is increasing if wheneverthe property holds for a graph, it holds for all supergraph containing it (onthe same number of vertices).

All these definitions can be interpreted as set properties. Let us define N =N(n) =

(n2

), then the set of graphs on n vertices G(n) may be thought

as {0, 1}N , where each component represents an edges of the graph. Wewill sometimes think of {0, 1}N as a group with component wise modulo 2addition, or with the △ operation (exclusive union) on all subsets of [N ].In this setting, a graph property is a subset of {0, 1}N , say A, which issymmetric, i.e. independent under graph automorphisms. Equivalently, onecan define 1A : {0, 1}N → {0, 1} and a graph property has to satisfy 1A(x) =1A(y) if x ≃ y. Similarly, monotone means

A ∈ A, B ⊆ [n], A ⊆ B =⇒ B ⊆ Aor x ∈ A, y ∈ {0, 1}n, x ≤ y =⇒ 1A(x) ≤ 1A(y).

Given p ∈ [0, 1], a binomial random graph on n vertices — G(n, p) — is arandom graph whose law is given by

P(G) = p|e(G)|(1 − p)N−|e(G)|

µp(x) = pP

xi(1 − p)N−P

xi

≡ N independent coin flippings.

3 Thresholds

Form now on, A = An is a sequence of increasing graph properties in {0, 1}N

(we assume that the property is natural, meaning its statement do not dependon n, such as “being connected”, “containing a triangle”, “having a perfectmatching” but not “containing a triangle for n even and being connected forn odd”).Claim: p 7→ µp(A) is continuous nondecreasing.

3

Proof: µp(A) =∑

m P{A||G| = m}P{|G| = m}, with P{|G| = m} =(Nm

)pm(1−p)N−m, which implies the continuity. The nondecreasing property

can be seen from a coupling argument. If q > p, drawing according to G(n, q),can be expressed as drawing according to G(n, p), and independently addinga second round of drawing according to G(n, p

′

), with q = p + p′ − pp

′

.

Definition 1. Threshold functionWe say that p(n) is a threshold function for A if

µp(n)(A) →{

0 if p ≪ p

1 if p ≫ p.

where p ≪ q means that p/q → 0.

Note: this definition of the threshold does not imply uniqueness of it, in factif p is a threshold and p

′ ≍ p, then p′

is a threshold too.

Examples:

1. Connectivity: p = log(n)n

.

2. Triangle containment: p = 1n

(see below for a specific proof).

The first and second moment methods:To prove subgraph containment thresholds, one can use the so-called firstand second moment methods, which are consequences of Markov inequality.For a non-negative integer valued random variable X:

P{X > 0} ≤ EX

and

P{X = 0} ≤ Var X

(EX)2.

Let us consider the triangle containment, we denote by Xn the number of tri-angles in G(n, p(n)), from previous inequalities, we have a threshold functionif

EXn → 0Var Xn

(EXn)2→ 0.

But clearly EXn =(

n3

)p3, so if p(n) = o( 1

n), the first condition is verified.

Moreover, VarXn ≤ EXn +∑

S≃T P{AS ∩ AT}, and P{AS|AT} = p2, for

4

3(n− 3) possibilities, so by symmetry VarXn = O(EXnnp2), which will stillbe dominated by (EXn)2 if p ≫ 1

n.

→ When do we have existence of a threshold?

Theorem 1. Bollobas, Thomason (weakened version).Every Monotone graph property has a threshold.

→ How “sharp” are those thresholds?

Definition 2. Let ǫ ∈ (0, 1), we define p(ǫ) such that µp(ǫ)(A) = ǫ (which iswell define by continuous increasing properties) and δ(ǫ) = p(1 − ǫ) − p(ǫ).

In terms of thresholds width, previous theorem tells us that

δ(ǫ) = O(p).

There are refinement of this:

Theorem 2. For any ǫ ∈ (0, 1)

δ(ǫ) = O(log(1/ǫ)/ log(n)), Friedgut, Kalai (1)

∀ν > 0, δ(ǫ) = O(1/ log2−ν(n)), Bourgain, Kalai (2)

δ(ǫ) = O(1/ log2(n)), Conjectured! (3)

But in view of δ(ǫ) = O(p), previous results are mainly interesting forconstant or very slowly decaying thresholds.The first result above can be refined as follows:

∃Cǫ s.t. δ(ǫ, n) ≤ Cǫp(ǫ, n) log(2/p(ǫ, n))

log(n). (4)

Last result tells us that if a threshold behaves like log(1/p) = o(log(n)), thenthe threshold width tends to zero even with respect to the threshold value,i.e. δ(n)/p(n) → 0, which ensures a “sharp” transition. If the thresholddecays faster, as some power of n, then we are back to our O(p) estimate(as it is the case for both previous examples). This discussion leads to thefollowing definition:

Definition 3. Sharp VS Coarse thresholdIf for every ǫ, we have δ(ǫ) = o(p), the threshold is called sharp, if insteadδ(ǫ) = Θ(p), the threshold is called coarse.

5

Note: a coarse threshold means that δ(ǫ, n)/p(1/2, n) is bounded away fromzero, whereas a sharp thresholds guarantees that δ(ǫ, n)/p(1/2, n) → 0, whichis equivalent to ask for the existence of p such that for any ǫ > 0,

µp(n)(A) →{

0 if p ≤ (1 − ǫ)p

1 if p ≥ (1 + ǫ)p.

and this determines p uniquely up to ∼.Examples:

1. Connectivity: p = log(n)n

and δ(ǫ) = Θ(1/n), thus the threshold is sharp.

2. Triangle containment: p = 1n

and δ(ǫ) = Θ(1/n), thus the threshold iscoarse.

One way to see that the triangle containment has a coarse threshold is tosimply observe that ∀c > 0, µc/n{contain triangle} → 1 − e−c3/6, thus thethreshold cannot be sharp. We will see in next section why connectivity hasa sharp threshold.

→ When is a threshold sharp or coarse?

It turns out that previous examples are typical, in the sense that roughlyspeaking, coarse thresholds happen only for properties of the type “containinga certain graph as a subgraph”. The proof of this result, as well as some ofthe previous ones, relies on the Fourier analysis of boolean functions. Forexample, in order to prove theorem 2, one can use the following theoremwhich uses strongly Fourier analysis (in particular the famous result in KKL[13]).

Theorem 3. Bourgain, Kahn, Kalai, Katznelson, LinialFor every graph property A

maxi

Ii > c(p)log n

nµp(A)(1 − µp(A)).

We now give a short introduction to this subject.

4 Fourier Analysis on Zn2

Measure: We equipped {0, 1}n with the product measure µp = ×ni=1µ

ip,

where p ∈ [0, 1] and µip(1) = p for any i, hence

∀x ∈ {0, 1}n, µp(x) = pP

xi(1 − p)n−P

xi

∀A ⊆ {0, 1}n, µp(A) =∑

x∈A

µp(x).

6

Space, inner product and norms: We equipped the space {0, 1}n → C ≡L2({0, 1}n, R) with the following inner product:

〈f, g〉 =

∫fg dµp = Eµp

fg.

For 1 ≤ q < ∞, we define

||f ||q = (Eµp|f |q)1/q,

such that ||f ||22 = 〈f, f〉.Basis: The canonical orthonormal basis is {ey}y∈{0,1}n

ey(x) = µp(x)−1/2δy(x) =

{µp(x)−1/2 if x = y

0 o.w..

leading to the expansion f =∑

y f(y)δy. But this is not the basis we areinterested in.Fourier basis: we start by considering p = 1/2. We will often switch tothe set notation, on behalf of {0, 1}n ∼= P([n]) (x ↔ S = {i|xi = 1}). Let usdefine the “characters” as

∀S, x ⊆ [n], χS(x) = (−1)|S∩x|,

with the multiplicationχS · χT = χS△T .

Claim:∀S, T ⊆ [n], 〈χS, χT 〉 = δST.

Proof: 〈χS, χT 〉 = EχSχT = 2−n∑

x(−1)|R∩x|, where R = S△T . But bysymmetry, last expression is 0, unless |R ∩ x| = 0 for all x ∈ {0, 1}n, i.e.unless R = ∅ (⇔ S = T ), in which case we get 1.So we have a new orthonormal basis {χS|S ⊆ [n]} (since we showed it was anorthonormal set, and its dimension is 2n) and we have the Fourier expansion

f =∑

S⊆[n]

f(S)χs,

wheref(S) = 〈f, χS〉

are called the Fourier coefficient.If p 6= 1/2, we define for i ∈ [n]

χi(x) =

−

√1−p

pif x(i) = 1√

p1−p

o.w.

7

and∀S ⊆ [n], χS =

∏

i∈S

χi.

Comment: motivations for the use of this basis will appear throughoutthe next sections. The justification of the “Fourier” terminology comes fromthe following fact. We can look at Zn

2 as an abelian group with compo-nent wise modulo 2 addition, or equivalently, in the set notation, the group2[n] = {S}S⊆[n] with operation S△T . Then, the function χR are homomor-phisms of this group into C, with χ(S) ∈ ±1, defining the characters of thisgroup, and {χS}S⊆[n], with the operation χS ·χT = χS△T , forms a group whichis isomorphic to 2[n]. The same role is played by the characters x 7→ e2πix ofR, with analogy to the usual Fourier transform on L2(R).

Properties of the Fourier coefficient: most of the usual properties knownfor the case of L2(R), hold in a similar way. One of the most important isthe Parseval’s identity:

〈f, g〉 =∑

S

f(S)g(S),

||f ||22 =∑

S

f 2(S).

We also have:

• f + g(S) = f(S) + g(S),

• f⊕y(S) = χS(y)f(S),

• f ⋆ g(S) = f g(S) .

Interpretation of the Fourier coefficient:

• f(∅) = Ef = P{f = 1}.

Assume now p = 1/2

• f({i}) = 1/2 − P{f(X) = Xi}.So f({i}) is a measure of the correlation between the value of f andits ith coordinate.

• f(S) = 1/2 − P{f(X) = ⊕i∈SXi}.

• P{f(X) 6= f(X ⊕ ei)} = 4∑

S∋i f2(S).

8

The last property is a crucial one. We already went through this this quantityin one of the previous theorem. Let us now expand on this.

Definition 4. Let f be a boolean function. We define the influence of itsi-th variable Ii and its total influence I by

fi = f − f⊕ei

Ii(f) = P{fi 6= 0} = ||fi||22

I(f) =

n∑

i=1

Ii(f)

In order to prove

P{f(X) 6= f(X ⊕ ei)} = 4∑

S∋i

f 2(S),

note that

fi(S) =

{2f(S) if i ∈ S

0 o.w. .

and recalling that from Parseval

||fi||22 =∑

S

f 2i (S),

we get a proof of the last property and of the following identity

I(f) = 4∑

S

f 2(S)|S|.

With similar arguments, we get the following lemma for any value of p.

Lemma 1.

p(1 − p)Ii(f) =∑

S∋i

f 2(S),

p(1 − p)I(f) =∑

S

|S|f 2(S).

At that point, one may agree on the fact that the Fourier transform isuseful to get estimate on the influence of a boolean function. But how dowe relate this to our initial problem of thresholds in random graphs? Weinvestigate this question in the next section, before that we give a quick hint:

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Lemma 2. Russo, Margulis

∂µp(A)

∂p= I(1A)

Proof. Let f = 1A and think of each component being independent butwith a different probability of having a one, namely pi for component i andp = (p1, . . . , pn). Let

X = {x = (x1, . . . , xi−1, xi+1, . . . , xn) ∈ {0, 1}n−1}|f(x1, . . . , xi−1, 0, xi+1, . . . , xn) = 1 ∩ f(x1, . . . , xi−1, 1, xi+1, . . . , xn) = 1}

Y = {x = (x1, . . . , xi−1, xi+1, . . . , xn) ∈ {0, 1}n−1}|f(x1, . . . , xi−1, 0, xi+1, . . . , xn) = 0 ∩ f(x1, . . . , xi−1, 1, xi+1, . . . , xn) = 1}.

Thenµp(A) = µ(X) + piµ(Y ),

where the only dependence in pi is exhibited. Also note that

∂µp(A)

∂pi

= µ(Y ) = Ii(f).

Now, consider p = p1(p) = . . . = pn(p), and

∂µp(A)

∂p=

n∑

i=1

∂µp(A)

∂pi

∂pi

∂p=

n∑

i=1

Ii(f) = I(f).

5 Sharp thresholds characterization

In this section, we will give a necessary and sufficient condition that coarsethresholds have. Therefore, if this condition does not hold for a monotonesymmetric property, it implies that its threshold is sharp. First start bynoting that for a property to have a coarse threshold, there must be pointswithin critical interval for which the derivative of the function µp with re-spect to p is small. More precisely:

Remark 1. if A is a property with a coarse threshold, i.e. for any ǫ > 0, thereexists C > 0 such that for any n, δ(ǫ, n)/p(1/2, n) > C. Then, there existsp∗(n) in the critical interval for each An with p∗ ∂µ

∂p|p=p∗ < 1/C. In view of

our formula∂µp(A)

∂p= I(1A),

any coarse threshold must have pI bounded for a critical p.

10

Theorem 4. FriedgutFor any ǫ, c > 0, there exists k(ǫ, c) such that for any n and any monotoneproperty A for which pI ≤ c (with p in the critical interval), there exists amonotone property A′

such that

||A′|| ≤ k

andµp(A△A′

) ≤ ǫ.

Remark: ||A′|| denotes the number of edges of the largest minimal graph inA′

, the above condition basically says that A′

is the property of containinga subgraph in the list G1, G2, . . . , Gm, where m is finite and independent ofn, and |e(Gi)| ≤ k, for any i. Also, from lemma 2 one gets a converse of thetheorem, i.e. if ||A|| is small, then so is pI.

A similar result which is more suitable for applications is the following.

Theorem 5. FriedgutLet 0 < α < 1, ǫ, c > 0. There exist B = B(ǫ, c), b1 = b1(ǫ, c) and b2 =b2(ǫ, c) such that for any n and any monotone symmetric property A suchthat pI ≤ c and α < µp(A) < 1−α, there exists a graph property G satisfying:

• G is balanced

• b1 < EG < b2

• |e(G)| ≤ B

• P{A|G} ≥ 1 − ǫ, where P{A|G} denotes the probability that a randomgraph belongs to A conditioned on the appearance of G, a specific copyof G.

Finally, the following result gives a characterization of the possible valuesof critical probabilities for coarse thresholds.

Theorem 6. FriedgutLet 0 < α < 1/2 and c > 0. There exist b1, b2, L > 0, such that for anymonotone symmetric property A, if pI ≤ c and α < µp(A) < 1 − α, thenb1n

α ≤ p ≤ b2nα with α rational, namely α = −k/l, k and l positive integer

with l ≤ L.

11

Proof. A graph H is said to be modest with respect to the 3-uple L, c1, c2,if |H| ≤ L, c1 ≤ EH ≤ c2 and H is balanced. Then, an important result isthat when pI ≤ c, for any ǫ, there exists L, c1, c2 such that for large n

∑

S not modest

f 2(S) ≤ ǫ.

Thus, being in the critical interval, we can conclude that there exists graphwith less than L edges and c1 ≤ ES ≤ c2, call the number of edges andvertices of such a graph e and v, we then have

ES = Θ(S)p|S| =

(n

v

)me ∼ 1,

hence p ∼ n−v/e, implying the statement about the rationality of the expo-nent.

Corollary 1. Connectivity has a sharp threshold (proof: its threshold islog(n)/n).

We now try to to outline some steps required to prove theorems 4 and 5.The starting point is the following formula:

∑

H:|H|>L

f 2(H) ≤ p(1 − p)I/L,

which we proved in section 4. So, roughly speaking, from remark (1), usingthe hypothesis pI ≤ c, above inequality tells us that the Fourier transformenergy concentrates on graphs of relative small size. We shall see how thefact that the Fourier transform concentrates on small graphs characterizesthe graph property we are dealing with. Note that

f(S) =∑

x

f(x)χS(x)µp(x)

and in general

|f(S)| ≤ ||f ||∞||χS||1,

but ||f ||∞ = 1 and ||χS||1 = ||χ1|||S|1 =, with ||χ1||1 =√

p1−p

(1−p)+√

1−pp

p =

2√

p(1 − p). So, in general

|f(S)| ≤ (4p(1 − p))|S|/2,

12

it turns out that because the Fourier transforms concentrate on graph ofsmall size and as we assume to be in the critical interval, we indeed have

|f(S)| ∼ p|S|/2.

Let us now look at an example. We consider the property of containinga triangle C3. Choose p such that E = E#C3 =

(n3

)p3 = log 2. Then,

E1G⊇C3= µp(G ⊆ C3) = 1− (1−p3)(

n

k) → 1− e(− log 2) = 1/2. Using relation(4), this tells us that

f 2(∅) → 1/4,

where f denotes the characteristic function of the family of all graph con-taining a copy of C3. Furthermore, using similar as previous estimates onthe Fourier coefficients, extracting the correct constants, one can show that

∑

S≃C3

f 2(S) → 1

4E.

Generally, we get ∑

S≃kC3

f 2(S) → 1

4Ek/k!,

so that ∑

S≃Z+C3

f 2(S) → 1

4eE = 1/2.

But as shown before||f ||22 = 1/2,

thus, from Parseval, all the energy of f is concentrated on copies of triangles,in other words, the Fourier transform announces that f is a function thatdeals with triangles.

In the general setting, the proof of theorem (4) works as follows:1. First, truncate f to sets which are not large, i.e. define g1 by

g1(S) =

{0 if S is large

f(S) o.w. .

From Lemma 1 and the hypothesis, this leaves us with a close L2 approxi-mation of f . The notion of “large” is captured as follows:A graph H is said to be modest with respect to the 3-uple L, c1, c2, if |H| ≤ L,

13

Figure 2: Graph representing a 3-CNF formula on 5 variables (taken from[18]).

c1 ≤ EH ≤ c2 and H is balanced. Then, one can show that for any ǫ, thereexists L, c1, c2 such that for large n

∑

S not modest

f 2(S) ≤ ǫ.

2. So we now have a list S1, . . . , Sl of modest graphs. The next step is toshow that one can approximate g1 with g2, by leaving the support of g2 onlyon a small number of “nice” graphs, which are balanced and whose expectednumber of copies in G(n, p) is bounded.3. Finally, one shows that the value of g2 on a graph H can be (w.h.p.)approximated very closely just by knowing the number of subgraphs of Hisomorphic to our previous nice graphs. But g2 is not necessarily boolean,however, since it is a close L2 approximation of f which is boolean, one canapproximate g2 by g3, which “counts” appearances of those nice graphs, e.g.g3(H) = 1 iff H has as a subgraph, a triangle or at least two copies of C4.

6 Sharp threshold of k-sat

The k-SAT problem can be formulated in terms of hyper graphs as follows:the vertices are the variables and the edges are replaced by hyper edgesconsisting of a selection of k vertices each of them plain or dashed for thenegation; this corresponds to a k-clause. One can represent this with smallgraphs having a squared center and blue/red colored edges connecting thek vertices. A formula becomes then the graph composed of all such hyperedges (see figure 2). A random instance can then be generated in differentways. According to the uniform model, M of the possible

(nk

)2k clauses are

14

picked up uniformly at random and one takes the conjunction of them to getthe formula. The Binomial formulation picks each of the

(nk

)2k clauses with

probability p = M/N , where N =(

nk

)2k, so that the expected number of

clauses in the final formula is M . Let us define α = M/n, thus

p =αn

N,

and we denote such random k-CNF formulas by f(α, n).We consider the property that a formula is satisfiable, i.e. that there exists amapping (assignment) of the vertices to true/false such that each hyper edgecontains at least one vertex leading to the value true, and we denote by Athe family of all such graphs. We will work with the Binomial model, butthe question of sharp threshold is equivalent with the uniform model.

Theorem 7. Friedgut The satisfiability property for random k-CNF has asharp threshold, i.e. there exists p such that for any ǫ > 0,

µp(A) →{

1 if p ≤ (1 − ǫ)p

0 if p ≥ (1 + ǫ)p

where

p =αc(n)n

N.

In order to prove this result, we use the equivalence of theorem 4 in oursetting:

Theorem 8. FriedgutLet ǫ, c > 0. There exist B = B(ǫ, c), b1 = b1(ǫ, c) and b2 = b2(ǫ, c) such thatfor any n and any monotone symmetric family T of k-DNF formulas with nvariables such that pI ≤ c, there exists a formula F satisfying:

• F is balanced

• b1 < EF < b2

• |F | ≤ B

• P{T |F} ≥ 1 − ǫ, where P{T |G} denotes the probability that a randomformula belongs to T conditioned on the appearance of F , a specificcopy of F .

15

Note: 1. the notion of “symmetric family” refers to the equivalent notion of“graph property” for the hyper-graphs we are considering now now, namelyinvariant under permutation of the clauses and flipping of the negations.2. We are now dealing with a decreasing property, to keep the analogywith previous sections, we will then analyze the complement property ofsatisfiability, i.e unsatisfiability, and we define

Pn(α) = µp(n)(Ac).

Proof of theorem 7 (attempt of modifications from original versionin [8]):We prove that for a p in the critical interval, the approximation described inthe theorem cannot take place, it will be actually sufficient to assume thatproperties 1 and 4 (from the theorem) holds in order to get a contradiction.

Let αc be such that for any n,

P{F unsat} = 1/2.

Let G be a satisfiable formula with a finite number of clauses and, say, rvariables. And let us assume (w.l.o.g.) that G contains only the literalsx1, . . . , xr, and thus is satisfied when all variables are set to true. We thenhave

P{F unsat|F ⊃ G} = P{F ∧ G unsat}≤ P{F (n, αc)x1=...=xr=T unsat}.

The random formula F (n, αc)|x1=...=xr=T has now a different structure andprobability distribution. Let us denote by F∗ an equivalently distributedrandom formula, which contains clauses of size 1 to k and has only n − rvariables. More precisely,

• the k-clauses are selected independently in Ck(n− r), with probabilityp = αcn

(n

k)2k

• the (k − 1)-clauses are selected independently in Ck−1(n − r), withprobability 1 − (1 − p)r

• the (k − 2)-clauses are selected independently in Ck−2(n − r), with

probability 1 − (1 − p)(r

2)

• . . .

16

• the 1-clauses are selected independently in C1(n− r), with probability

1 − (1 − p)(r

k−1)

Recall that

|Ck(n)| =

(n

k

)2k,

thus, denoting by Sl the number of l-clauses present in F∗, we have

ESk =

(n − r

k

)2k αn(

nk

)2k

= O(n), (5)

ESk−1 =

(n − r

k − 1

)2k−1 αn(

nk−1

)2k−1

= O(1), (6)

and more generally

ESk−i = O(n1−i), ∀1 ≤ i ≤ k − 1. (7)

Then, for any d ≥ 1

P{F∗ unsat} ≤ P{F∗ unsat|Sk−1 ≤ d, max2≤i≤k−1

Sk−i ≤ 1/2}+ 2P{Sk−1 > d}+ P{ max

2≤i≤k−1Sk−i > 1/2}.

Now, let τ > 0. From (6) and (7) (and Markov’s inequality), we can choosen and d large enough, such that

P{F∗ unsat} ≤ P{F∗ unsat|Sk−1 ≤ d, max2≤i≤k−1

Sk−i ≤ 1/2} + τ/4

≤ P{F∗ unsat|Sk−1 = d, max2≤i≤k−1

Sk−i = 0} + τ/4.

So we can define a random formula F∗∗, containing only k-clauses and (k−1)-clauses, for which

• the k-clauses are selected independently in Ck(n− r), with probabilityp = αcn

(n

k)2k

• exactly d (k − 1)-clauses are uniformly selected among Ck−1(n − r).

So we get

P{F (n, αc) unsat|F ⊃ G} ≤ P{F∗ unsat}≤ P{F∗∗ unsat} + τ/4.

17

Because 1-clauses are more constraining than (k − 1)-clauses, we can upperbound our estimate by replacing each k-clause by the disjunction of its kliterals, but note that by doing so, the number of 1-clauses that we get isnot exactly D ≡ kd, as some literals can be repeated in the different (k− 1)-clauses. But as it is at most D, considering F∗∗ with D 1-clauses lead us toan upper bound.So F∗∗ is a conjunction of ∧D

i=1C1i , where the C1

i ’s are 1-clauses, i.e. literals,chosen uniformly within all

(2(n−r)

D

)possible ones, and F , which denotes a k-

CNF random formula with k-clauses selected independently in Ck(n−r) withprobability p(n, αc). As p(·, αc) is decreasing, we can consider the k-clausesto be drawn from p(n − r, αc) instead of p(n, αc), obtaining the followingupper bound,

P{F (n, αc) unsat|F ⊃ G} ≤ P{F (n − r, αc) ∧Di=1 C

(1)i unsat} +

τ

5. (8)

Claim: Let F a random formula on n variables, C(1)i be l-clauses uniformly

selected in C1(n). Then for any unbounded increasing sequence f(n) and forany ǫ > 0, we can take n large enough such that,

P{F ∧Di=1 C

(1)i unsat} ≤ P{F ∧f(n)

i=1 C(k)i unsat} + ǫ.

To see this, we first check that for n large enough,

P{F ∧ C(1)1 unsat} ≤ P{F ∧f(n)/D

i=1 C(k)i unsat} + ǫ/D, (9)

and iterating this inequality will prove the claim. So we need to check (9).But for any g ≥ 1

P{F ∧gi=1 C

(k)i unsat} ≥ P{∃1 ≤ i ≤ g, F ∧ C

(k)i unsat}

and

P{∃1 ≤ i ≤ g, F ∧ C(k)i unsat} = 1 − P{∀1 ≤ i ≤ g, F ∧ C

(k)i sat}

= 1 − P{F ∧ C(k)1 sat}g

≥ 1 − (1 − P{F ∧ C(k)1 unsat})g

so that

P{F ∧gi=1 C

(k)i unsat} ≥ 1 − (1 − P{F ∧ C

(k)1 unsat})g. (10)

But C(k)1 = ∨k

i=1C(1)i and

P{F ∧ C(k)1 unsat} = P{F ∧ ∨k

i=1C(1)i unsat}

= P{∨ki=1F ∧ C

(1)i unsat}

≥ P{∀1 ≤ i ≤ k, F ∧ C(1)i unsat}

≥ P{F ∧ C(1)1 unsat}k. (11)

18

Therefore, putting (10) and (11) together, we find

P{F ∧gi=1 C

(k)i unsat} ≥ 1 − (1 − P{F ∧ C

(1)1 unsat}k)g,

but this implies that we can take g large enough, such that

P{F ∧ C(1)1 unsat} ≤ P{F ∧g

i=1 C(k)i unsat} + ǫ/D,

thus, since f(n) is increasing and not bounded, for n large enough,

P{F ∧ C(1)1 unsat} ≤ P{F ∧f(n)/D

i=1 C(k)i unsat} + ǫ/D, (12)

which proves the claim.

We now can use our claim, to upper bound our estimate in (8), getting

P{F (n, αc) unsat|F ⊃ G} ≤ P{F (n − r, αc) ∧f(n)i=1 C

(k)i unsat} + 2τ/4.

But we can take t large enough such that

P{F (n − r, αc) ∧f(n)i=1 C

(k)i unsat} ≤ P{F (n − r, αc + tf(n)/(n − r)) unsat} + τ/4,

redefining m ≡ n − r and f(m) ≡ tf(n + r),

P{F (n, αc) unsat|F ⊃ G} ≤ Pm(αc + f(m)/m) + 3τ/4.

From now on, we keep n instead of m. So, taking f(n) = τ√

n4C

, and doing aTaylor expansion, we get

P{F (n, αc) unsat|F ⊃ G} ≤ Pn(αc +τ

4C√

n) + 3τ/4

= Pn(αc) +τ

4C√

n

∂

∂αPn(α)|αc

+ 3τ/4

In order to conclude we need the following claim:

∂

∂αPn(α) ≤ C

√n.

Proof of the claim: by definition

Pn(α) =∑

0≤m≤N

P{F (n, α) unsat||F | = m}P{|F | = m}

=∑

0≤m≤N

(N

m

)pm(1 − p)N−mP{F (n, α) unsat||F | = m},

19

and p = αnN

, thus

∂

∂αpm(1 − p)N−m =

nm

Npm−1(1 − p)N−m − n(N − m)

Npm(1 − p)N−m−1

=n

Npm−1(1 − p)N−m−1(m − Np),

so,

∂

∂αPn(α) =

n

Np−1(1 − p)−1

∑

0≤m≤N

(N

m

)pm(1 − p)N−m(m − Np)P{F (n, α) unsat||F | = m}

≤ n

Np−1(1 − p)−1

∑

m≥Np

(N

m

)pm(1 − p)N−m(m − Np)

=n

Np−1(1 − p)−1E[(Np − |F |), |F | ≥ Np]

≤ n

Np−1(1 − p)−1(E(Np − |F |)2)1/2

=n

Np−1(1 − p)−1(Np(1 − p))1/2

≤ C√

n,

which conclude the proof of the claim.Applying it to previous expression, we find

P{F (n, αc) unsat|F ⊃ G} ≤ Pn(αc) + τ

= 1/2 + τ.

Therefore, we showed that for a probability p(n, αc) in the critical interval,conditioning on any formula with a finite number of variables cannot lead toan arbitrarily high probability. Thus, using theorem (8), we get the sharpthreshold property, concluding the proof of theorem (7). �

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