1 / 24

Finite Model Theory Lecture 18

Finite Model Theory Lecture 18. Extended 0/1 Laws Or “Getting Real”. 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. Annomalies 0/1 Laws.

lel
Download Presentation

Finite Model Theory Lecture 18

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Finite Model TheoryLecture 18 Extended 0/1 LawsOr “Getting Real”

  2. 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

  3. Annomalies 0/1 Laws Database schema:Employee(name, city, occupation) We are not given the instance. • Any person belongs to Employee with m = 1/2 ! • The expected size E[Employee] = n3/2 !1 !! • In practice need conditional probabilities, m(f | y), but they often don’t exists [ why ?]

  4. A Better Model • Postulate that for each R 2sE[R] = cR (a constant) • This leads to: for each tuple t:Pr[t 2 R] = cR / na where a = arity(R)

  5. 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

  6. Conjunctive Queries • Have the form:9 x1…9 xk.(C1Æ … Æ Cm) • Where each Ci is R(…) or xi=xj or xi¹ xj Empolyee(x,Seattle,-),Employee(x,y,Clerk),Employee(-,y,Lawer)

  7. 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

  8. Subgraph Properties • Consider R(x,y); • For every edge, Pr(R(u,v)) = c/n2 • Given Q, let H = Q¹ obtained by adding all predicates of the form xi¹ xj • H checks for the presence of a subgraph

  9. Subgraph Properties Example 1: • Q = R(x,y),R(y,z),R(z,x)H=Q¹ = R(x,y),R(y,z),R(z,x),x¹ y,y¹ z,z¹ x H =

  10. Subgraph Properties Pr(H) = Pr(Çu,v,w H(u,v,w)) ·åu,v,w Pr(H(u,v,w)) = n(n-1)(n-2) * 1/3 * c3 / n6 = 1/3 c3 / n3 + O(1/n4)

  11. Subgraph Properties Example 2: Q = R(x,y),R(y,a),R(b,x) H=Q¹=R(x,y),R(y,z),R(z,x),x¹ y,y¹a,a¹x,x¹b, b¹x b a

  12. Subgraph Properties Pr(H) = Pr(Çu,v H(u,v)) ·åu,v Pr(H(u,v)) = n(n-1) * 1/1 * c3 / n6 = c3 / n4 + O(1/n5)

  13. Subgraph Properties Let Q = G1, G2, …, Gm Lemma Pr(Q) · C/H * 1/nE V = number of variables in Q A = arity(Q) = arity(G1) + … + arity(Gm) E = A - V = “the exponent of Q” H = number of automorphisms Q ! Q C = c1 * c2 * … * cm = “the coefficient of Q”

  14. Subgraph Properties Lower bound, for the triangle: Pr(H) = Pr(Çu,v,w H(u,v,w)) ¸åPr(H(u,v,w)) – åPr(H(u,v,w)Æ H(u’,v’,w’)= 1/3 c3/n3 + O(1/n4) - å Pr(HH)

  15. 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

  16. Subgraph Properties • Hence, for the triangle: Pr(H) ¼ 1/3 c3/n3 • This generalizes easily to any subgraph property

  17. 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

  18. 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

  19. 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

  20. More General Distributions [Shelah&Spencer, Lynch] • Pr(tuple) = b / na • Example: H = triangle • Pr(H) ¼ n3 * 1/3 * b3 / n3a = C / nE • Simply redefine E(H) to use a

  21. More General Distributions • But, problem here; let \alpha = 3/2: E( ) = 3a – 3 = 3/2 E( ) = 3a – 3 + a – 2 = 1 Hence the more complex graph is more likely ! Solution: adjust E(H) to be the max of E(H0) for H0µ H

  22. 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

  23. Threshold Functions [Erdos and Reny] The threshold function for subgraph property H is the following: Let a = maxH0µ H |nodes(H0)| / |edges(H0)| Then t(n) = 1/na Can derive it from the exponent [ show in class ]

  24. Extended 0/1 Laws • Shelah and Spencer, and Lynch consider the following general case: • Pr(t) = b / na, for a > 0 • Lynch: a logic admits an extended 0/1 law if for each f one of the following holds:Pr(f) ¼ C/nE, orPr(f) < 1/nE for every E >0

More Related