Finite Model Theory Lecture 16

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Finite Model TheoryLecture 16

L1 Summary

and 0/1 Laws

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Outline

• Summary on L1

– All you need to know in 5 slides !

• Start 0/1 Laws: Fagin’s theorem– Will continue next time

Infinitary Logics and 0-1 Laws, Kolaitis&Vardi, 1992

New paper:

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Summary on L1

Notation Comes from in classical logic

• L = formulas where:

– Conjunctions/disjunctions of ordinal < Çi 2 i, Æ2, where <

– Quantifier chains of ordinal < 9i 2 xi. , where <

• Hence, L1 = [ L

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Summary on L1

Motivation• Any algorithmic computation that applies FO formulas

is expressible in L1

• Relational machines• While-programs with statements R := • Fixpoint logics: LFP, IFP, PFP, etc, etc

Consequence: cannot express EVEN, HAMILTONEAN

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Summary on L1

Canonical Structure

Any algorithmic computation on A can be decomposed• Compute the ¼k equivalence relation on k-tuples, and order

the equivalence classes ) in LFP[how do we choose k ???]

• Then compute on ordered structure ) any complexity

Consequence: PTIME=PSPACE iff IFP=PFPBut note that DTC TC yet L ? NL [ why ?]

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Summary on L1

Pebble Games: with k pebbles• Notation: A 1

k B if duplicator wins

Theorem 1. For any two structures A, B:• A, B are Lk

1 equivalent iff• A 1

k BTheorem 2. If A, B are finite:• A, B are FOk equivalent iff• A, B are Lk

1 equivalent iff• A 1

k B

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Summary on L1

Definability of FOk types

• FOk types are the same as Lk1 types [ why ?]

Theorem [Dawar, Lindell, Weinstein] The type of A (or of (A, a)) can be expressed by some 2 FOk

B ² [b] iff Tpk(A,a) = Tpk(B,b)

Difficult result: was unknown to Kolaitis&Vardi

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0/1 Laws in Logic

Motivation: random graphs

• 0/1 law for FO proven by Glebskii et al., then rediscovered by Fagin (and with nicer proof) – Only for constant probability distribution

• Later extended to other logics, and other probability distributions

Why we care: applications in degrees of belief, probabilistic databases, etc.

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Definitions

• Let = a vocabulary

• Let n ¸ 0, and An µ STRUCT[] be all models over domain {0, 1, …, n-1}

• Uniform probability distribution on An

• Given sentence , denote n() its probability

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Definition

• Denote () = limn ! 1 n() if it exists

Definition A logic L has a convergence law if for every sentence , () exists

Definition A logic L has a 0/1 law if for every sentence , () exists and is 0 or 1

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Theorems

• Suppose has no constants

Theorem [Fagin 76, Glebskii et al. 69] FO admits a 0/1 law

Theorem [Kolaitis and Vardi 92] L

1 admits a 0/1 law

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Application

• What does this tell us for database query processing ?

• Don’t bother evaluating a query: it’s either true or false, with high probability

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Examples [ in class ]

• Compute n(), then ():

R(0,1) /* I’m using constants here */R(0,1) Æ R(0,3) Æ : R(1,3) 9 x.R(2,x) : (9 x.9 y.R(x,y)) 8 x.8 y.(9 z.R(x,z) Æ R(z,y))

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Types

• We only need rank-0 types (i.e. no quantifiers)

• Recall the definition

Definition A type t(x) over variables (x1, …, xm) is conjunction of a maximally consistent set of atomic formulas over x1, …, xm

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Types

The type t(x) says:

• For each i, j whether xi = xj or xi xj

• For each R and each xi1, …, xip

whether

R(xi1, …, xip

) or : R(xi1, …, xip

)

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Extension Axioms

Definition Type s(x, z) extends the type t(x) if {s, t} is consistent;

Equivalently: every conjunct in t occurs in s

Definition The extension axiom for types t, s is the formula

t,s = 8 x1…8 xk (t(x) ) 9 z.s(x, z))

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Example of Extension Axiom

t(x1, x2, x3) =x1 x2 Æ x2 x3 Æ x1 x3 ÆR(x1,x2) Æ R(x2,x3) Æ R(x2,x2) Æ : R(x1, x1) Æ : R(x2, x1) Æ …

x1

x2

x3

s(x1, x2, x3, z) =t(x1, x2, x3) Æ z x1 Æ z x2 Æ z x3 ÆR(z,x1) Æ R(x3,z) Æ R(z,z) Æ: R(x1, z) Æ : (z, x2) Æ …

z

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Example of Extension Axiom

8 x1.8 x2.8 x3. (t(x1, x2, x3) ) 9 z. s(x1, x2, x3, z))

t,s =

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The Theory T

• Let T be the set of all extension axioms– Studied by Gaifman

• Is T consistent ?– In a model of T the duplicator always wins [ why ? ]

• Does it have finite models ?

• Does it have infinite models ?

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The Theory T

• Let k be the conjunction of all extension axioms for types with up to k variables

• There exists a finite model for k [why ?]

• Hence any finite subset of T has a model

• Hence T has a model. [can it be finite ?]

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The Model(s) of T

• T has no finite models, hence it must have some infinite model

• By Lowenheim-Skolem, it has a countable model

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The Theory T

Theorem T is -categorical

Proof: let A, B be two countable model.

Idea: use a back-and-forth argument to find an isomorphism f : A ! B

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The Theory T

Theorem T is -categoricalProof: (cont’d)

A = {a1, a2, a3, ….} B = {b1, b2, b3, ….}

Build partial isomorphisms f1 µ f2 µ f3 µ …such that: 8 n.9 m. an 2 dom(fm)and 8 n.9 m. bn 2 rng(fm)

[in class]

Then f = ([m ¸ 1 fm) : A ! B is an isomorphism

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The Theory T

Corollary T has a unique countable model R

• R = the Rado graph = the “random” graph

Corollary The theory Th(T) is complete

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0/1 Law for FO

Lemma For every extension axiom , () = limn n() = 1

Proof: later

Corollary For any m extension axioms 1, …, m: (1 Æ … Æ m) = 1

Proof n(:(1 Æ … Æ m))

= n(: 1 Ç … Ç : m) · n(: 1) + … + n(: m) ! 0

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Fagin’s 0/1 Law for FO

Theorem For every 2 FO, either () = 0 or () = 1.

Proof. Case 1: R ² . Then there exists m extension

axioms s.t. 1, …, m ² . Then n() ¸ n(1 Æ … Æ m) ! 1

Case 2: R 2 . Then R ² : , hence (: ) = 1, and () = 0

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Proof for the Extension Axioms

• Let = 8 x. t(x) ) 9 z.s(x, z)• Assume wlog that t asserts xi xj forall i j.

Denote (x) the formula Æi < j xi xj

– Hence t(x) = (x) Æ t’(x)

• Similarly, s asserts z xi forall i.Denote (x, z) = Æi xi z– Hence s(x, z) = t(x) Æ (x, z) Æ s’(x, z)

where all atomic predicates in s’(x, z) contain z

• Hence: = 8 x.((x) Æ t’(x) ) 9 z. (x,z) Æ s’(x, z))

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Proof for the Extension Axioms

: = 9 x.((x) Æ t’(x) Æ 8 z.((x, z) ) : s’(x, z)))

n(: ) · n(9 x.((x) Æ 8 z.((x, z) ) : s’(x, z))))

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Proof for the Extension Axioms

n(: ) · n(9 x.((x) Æ 8 z.((x, z) ) :s’(x, z))))

· a1, ... , ak 2 {1, …, n} n(8 z. ((x, z) ) :s’(a1, …, ak, z)))

= n(n-1)…(n-k+1) n(8 z. (x, z) ) :s’(1, 2, …, k, z))

· nk n(8 z. (x, z) ) :s’(1, 2, …, k, z)) =

= nk z=k+1, n : s’(1,2,…,k,z) /* by independence !! */

= nk ( 1 - 1 / 22k+1 )n-k /* since s’ is about 2k+1 edges */! 0 when n ! 1

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Complexity

Theorem [Grandjean] The problem whether () = 0 or 1 is PSPACE complete

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Discussion

• Old way to think about formulas and models: finite satsfiability/ validity

FO

unsatisfiable

valid

Undecidable

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Discussion

• New way to think about formulas and models: probability

FO

unsatisfiable

valid

PSPACE

)=0)=1