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Finite Model TheoryLecture 18

Extended 0/1 LawsOr “Getting Real”

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Outline

• A better probabilistic model• Probabilities of conjunctive queries• Probabilities for FO

• Based on work done with N. Dalvi and G.Miklau, and on papers by Lynch, Shelah and Spencer

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Annomalies 0/1 Laws

Database schema:Employee(name, city, occupation)

We are not given the instance.

• Any person belongs to Employee with = 1/2 !• The expected size E[Employee] = n3/2 ! 1 !!• In practice need conditional probabilities,

( | ), but they often don’t exists [ why ?]

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A Better Model

• Postulate that for each R 2 E[R] = cR (a constant)

• This leads to: for each tuple t:Pr[t 2 R] = cR / na where a = arity(R)

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A Better Model

No more anomalies:• For a given person, the probability of it

belonging to Employee is ! 0

• The expected size is E[R] = cR

• Asymptotic conditional probabilities always exists for conjunctive queries

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Conjunctive Queries

• Have the form:

9 x1…9 xk.(C1 Æ … Æ Cm)

• Where each Ci is R(…) or xi=xj or xi xj

Empolyee(x,Seattle,-),Employee(x,y,Clerk),Employee(-,y,Lawer)Empolyee(x,Seattle,-),Employee(x,y,Clerk),Employee(-,y,Lawer)

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Conjunctive Queries

TheoremFor every Q there are numbers E, C s.t: Pr[Q] =C / nE + O(1/NE+1)

Corollary Pr[Q1 | Q2] always has a limit

• Will show next how to compute C, E

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Subgraph Properties

• Consider R(x,y);

• For every edge, Pr(R(u,v)) = c/n2

• Given Q, let H = Q obtained by adding all predicates of the form xi xj

• H checks for the presence of a subgraph

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Subgraph Properties

Example 1:

• Q = R(x,y),R(y,z),R(z,x)H=Q = R(x,y),R(y,z),R(z,x),x y,y z,z x

H =

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Subgraph Properties

Pr(H) = Pr(Çu,v,w H(u,v,w))

· u,v,w Pr(H(u,v,w))

= n(n-1)(n-2) * 1/3 * c3 / n6

= 1/3 c3 / n3 + O(1/n4)

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Subgraph Properties

Example 2:

Q = R(x,y),R(y,a),R(b,x)

H=Q=R(x,y),R(y,z),R(z,x),x y,ya,ax,xb, bx

ab

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Subgraph Properties

Pr(H) = Pr(Çu,v H(u,v))

· u,v Pr(H(u,v))

= n(n-1) * 1/1 * c3 / n6

= c3 / n4 + O(1/n5)

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Subgraph Properties

Let Q = G1, G2, …, Gm

Lemma Pr(Q) · C/H * 1/nE

V = number of variables in QA = arity(Q) = arity(G1) + … + arity(Gm)E = A - V = “the exponent of Q”

H = number of automorphisms Q ! QC = c1 * c2 * … * cm = “the coefficient of Q”

V = number of variables in QA = arity(Q) = arity(G1) + … + arity(Gm)E = A - V = “the exponent of Q”

H = number of automorphisms Q ! QC = c1 * c2 * … * cm = “the coefficient of Q”

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Subgraph Properties

Lower bound, for the triangle:

Pr(H) = Pr(Çu,v,w H(u,v,w))

¸ Pr(H(u,v,w)) – Pr(H(u,v,w)Æ H(u’,v’,w’)= 1/3 c3/n3 + O(1/n4) - Pr(HH)

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Subgraph Properties

• What is Pr(H) ? Each term belongs to one of the following cases:

E = 12 – 6 = 6

E = 12 – 5 = 7

E = 10 – 4 = 6

A few others…. But all have E > 3 ! Hence Pr(HH) is neglijible

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Subgraph Properties

• Hence, for the triangle:

Pr(H) ¼ 1/3 c3/n3

• This generalizes easily to any subgraph property

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Subgraphs with E = 0

H = R(x,y) E = 2-2 = 0; what is Pr(H) ?

H = R(x,y)R(u,v) E = 4–4 = 0what is Pr(H) ?

H = R(x,y)R(y,z)R(z,x), R(u,v) E(H) = E(triangle);

Exponent in the theorem is always correct, but need to adjust the coefficient

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Conjunctive Queries

• Consider the query:R(x,y),R(y,z),R(z,x)

• Any of the variables x,y,z may be equal: results in the following subgraphs:H1 = R(x,y)R(y,z)R(z,x) E=6-3=3H2 = R(x,x)R(x,z)R(z,x) E=6-2=4H3 = R(x,x)R(x,x)R(x,x) = R(x,x) E=2

• Hence Pr(Q) = Pr(H3) = cR/n2

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Conjunctive Queries

• Now considerQ = R(a,x),R(y,b)

• Two graphs:H1 = R(a,x)R(y,b) E = 4-2=2H2 = R(a,b) E = 2

• One can prove:Pr(Q) = Pr(H1) + Pr(H2) = (c + c2)/n2

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More General Distributions

[Shelah&Spencer, Lynch]

• Pr(tuple) = / n

• Example: H = triangle

• Pr(H) ¼ n3 * 1/3 * 3 / n3= C / nE

• Simply redefine E(H) to use

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More General Distributions

• But, problem here; let \alpha = 3/2:

E( ) = 3 – 3 = 3/2

E( ) = 3 – 3 + – 2 = 1

Hence the more complex graph is more likely !

Solution: adjust E(H) to be the max of E(H0) for H0 µ H

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Threshold Functions for Subgraphs

[Erdos and Reny]Edge probability Pr(t) = p(n) = some function

Main theorem of random graphs:For any monotone property C there exists a threshold function t(n) s.t.– If p(n) ¿ t(n) then limn Pr(C) = 0– If p(n) À t(n) then limn Pr(C) = 1

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Threshold Functions

[Erdos and Reny]

The threshold function for subgraph property H is the following:

Let = maxH0 µ H |nodes(H0)| / |edges(H0)|

Then t(n) = 1/n

Can derive it from the exponent [ show in class ]

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Extended 0/1 Laws

• Shelah and Spencer, and Lynch consider the following general case:

• Pr(t) = / n, for > 0

• Lynch: a logic admits an extended 0/1 law if for each one of the following holds:Pr() ¼ C/nE, orPr() < 1/nE for every E >0