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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 TheoryLecture 16 Lw1w Summary and 0/1 Laws
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 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 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 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 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 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 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 • 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 • 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 • 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 • 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 ] • 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 • 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
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 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 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 Axiom tt,s = 8 x1.8 x2.8 x3. (t(x1, x2, x3) )9 z. s(x1, x2, x3, z))
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 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 • T has no finite models, hence it must have some infinite model • By Lowenheim-Skolem, it has a countable model
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 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 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 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 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 • 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 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 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 Theorem [Grandjean] The problem whether m(f) = 0 or 1 is PSPACE complete
Discussion • Old way to think about formulas and models: finite satsfiability/ validity FO f valid f unsatisfiable Undecidable
Discussion • New way to think about formulas and models: probability m(f)=1 FO m(f)=0 f valid f unsatisfiable PSPACE