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Finite Model Theory Lecture 16. L w 1 w Summary and 0/1 Laws. Outline. Summary on L w 1 w All you need to know in 5 slides ! Start 0/1 Laws: Fagin’s theorem Will continue next time. New paper:. Infinitary Logics and 0-1 Laws , Kolaitis&Vardi, 1992. Summary on L w 1 w.

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finite model theory lecture 16

Finite Model TheoryLecture 16

Lw1w Summary

and 0/1 Laws

outline
Outline
  • Summary on Lw1w
    • All you need to know in 5 slides !
  • Start 0/1 Laws: Fagin’s theorem
    • Will continue next time

New paper:

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

summary on l w 1 w
Summary on Lw1w

Notation Comes from in classical logic

  • Lab = formulas where:
    • Conjunctions/disjunctions of ordinal < aÇi 2gfi, Æi 2 g, where g < a
    • Quantifier chains of ordinal < b 9i 2g xi. f, where g < b
  • Hence, L1w = [a Law
summary on l w 1 w1
Summary on Lw1w

Motivation

  • Any algorithmic computation that applies FO formulas is expressible in Lw1w
  • Relational machines
  • While-programs with statements R := f
  • Fixpoint logics: LFP, IFP, PFP, etc, etc

Consequence: cannot express EVEN, HAMILTONEAN

summary on l w 1 w2
Summary on Lw1w

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=PFP

But note that DTC ¹ TC yet L ¹? NL [ why ?]

summary on l w 1 w3
Summary on Lw1w

Pebble Games: with k pebbles

  • Notation: A 1wk B if duplicator wins

Theorem 1. For any two structures A, B:

  • A, B are Lk1w equivalent iff
  • A 1wk B

Theorem 2. If A, B are finite:

  • A, B are FOk equivalent iff
  • A, B are Lk1w equivalent iff
  • A 1wk B
summary on l w 1 w4
Summary on Lw1w

Definability of FOk types

  • FOk types are the same as Lk1w types [ why ?]

Theorem [Dawar, Lindell, Weinstein] The type of A (or of (A, a)) can be expressed by some f2 FOk

B ²f[b] iff Tpk(A,a) = Tpk(B,b)

Difficult result: was unknown to Kolaitis&Vardi

0 1 laws in logic
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.

definitions
Definitions
  • Let s = a vocabulary
  • Let n ¸ 0, and Anµ STRUCT[s] be all models over domain {0, 1, …, n-1}
  • Uniform probability distribution on An
  • Given sentence f, denote mn(f) its probability
definition
Definition
  • Denote m(f) = limn !1mn(f) if it exists

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

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

theorems
Theorems
  • Suppose s has no constants

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

Theorem [Kolaitis and Vardi 92] Lw1w admits a 0/1 law

application
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 
examples in class
Examples [ in class ]
  • Compute mn(f), then m(f):

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))

types
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

types1
Types

The type t(x) says:

  • For each i, j whether xi = xj or xi¹ xj
  • For each R and each xi1, …, xip whether R(xi1, …, xip) or : R(xi1, …, xip)
extension axioms
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 tt,s = 8 x1…8 xk (t(x) )9 z.s(x, z))

example of extension axiom
Example of Extension Axiom

t(x1, x2, x3) = x1¹ x2Æ x2¹ x3Æ x1¹ x3Æ R(x1,x2) Æ R(x2,x3) Æ R(x2,x2) Æ: R(x1, x1) Æ: R(x2, x1) Æ …

x1

x2

z

s(x1, x2, x3, z) = t(x1, x2, x3) Æ z ¹ x1Æ z ¹ x2Æ z ¹ x3Æ R(z,x1) Æ R(x3,z) Æ R(z,z) Æ: R(x1, z) Æ: (z, x2) Æ …

x3

example of extension axiom1
Example of Extension Axiom

tt,s =

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

the theory t
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 ?
the theory t1
The Theory T
  • Let qk be the conjunction of all extension axioms for types with up to k variables
  • There exists a finite model for qk [why ?]
  • Hence any finite subset of T has a model
  • Hence T has a model. [can it be finite ?]
the model s of t
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
the theory t2
The Theory T

Theorem T is w-categorical

Proof: let A, B be two countable model.

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

the theory t3
The Theory T

Theorem T is w-categorical

Proof: (cont’d)

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

Build partial isomorphisms f1µ f2µ f3µ …such that: 8 n.9 m. an2 dom(fm)and 8 n.9 m. bn2 rng(fm)

[in class]

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

the theory t4
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

0 1 law for fo
0/1 Law for FO

LemmaFor every extension axiom t, m(t) = limnmn(t) = 1

Proof: later

Corollary For any m extension axioms t1, …, tm: m(t1Æ … Ætm) = 1

Proofmn(:(t1Æ … Ætm)) = mn(:t1Ç … Ç:tm) ·mn(:t1) + … + mn(:tm) ! 0

fagin s 0 1 law for fo
Fagin’s 0/1 Law for FO

Theorem For every f2 FO, either m(f) = 0 or m(f) = 1.

Proof.

Case 1: R²f. Then there exists m extension axioms s.t. t1, …, tm²f. Then mn(f) ¸mn(t1Æ … Ætm) ! 1

Case 2: R2f. Then R²:f, hence m(:f) = 1, and m(f) = 0

proof for the extension axioms
Proof for the Extension Axioms
  • Let t = 8x. t(x) )9 z.s(x, z)
  • Assume wlog that t asserts xi¹ xj forall i ¹ j. Denote ¹(x) the formula Æi < j xi¹ xj
    • Hence t(x) = ¹(x) Æ t’(x)
  • Similarly, s asserts z ¹ xi forall i.Denote ¹(x, z) = Æi xi¹ z
    • Hence s(x, z) = t(x) ƹ(x, z) Æ s’(x, z)where all atomic predicates in s’(x, z) contain z
  • Hence:t = 8x.(¹(x) Æ t’(x) ) 9 z. ¹(x,z) Æ s’(x, z))
proof for the extension axioms1
Proof for the Extension Axioms

:t = 9x.(¹(x) Æ t’(x) Æ8 z.(¹(x, z) ): s’(x, z)))

mn(:t) ·mn(9x.(¹(x) Æ8 z.(¹(x, z) ): s’(x, z))))

proof for the extension axioms2
Proof for the Extension Axioms

mn(:t) ·mn(9x.(¹(x) Æ8 z.(¹(x, z) ):s’(x, z))))

·åa1, ... , ak2 {1, …, n}mn(8 z. (¹(x, z) ):s’(a1, …, ak, z)))

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

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

= nkÕz=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

complexity
Complexity

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

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

FO

f valid

f unsatisfiable

Undecidable

discussion1
Discussion
  • New way to think about formulas and models: probability

m(f)=1

FO

m(f)=0

f valid

f unsatisfiable

PSPACE

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