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Inferencing

Inferencing. Finding solution to queries Propositionalization (For reference only, may leave slides 2 to 7) Unification Forward Chaining Backward Chaining Resolution. Luger’s Book: Chapter 2 and Chapter 12 (upto 12.2). Propositionalization. Brute Force Approach

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Inferencing

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  1. Inferencing Finding solution to queries • Propositionalization • (For reference only, may leave slides 2 to 7) • Unification • Forward Chaining • Backward Chaining • Resolution CS 331 Dr M M Awais

  2. Luger’s Book:Chapter 2andChapter 12(upto 12.2)

  3. Propositionalization • Brute Force Approach • Apply all possible bindings to the variables. • Add every clause in the sentence as a proposition to the knowledge base. • Search for the required proposition (Goal proposition) CS 331 Dr M M Awais

  4. Reduction to propositional inference Universal Instantiation Suppose the KB contains just the following: X wealthy(X)  greedy(X)  bad(X) wealthy(arif) greedy(arif) cousins(shahid,arif) • Instantiating the universal sentence in all possible ways, we have: wealthy(arif)  greedy(arif)  bad(arif) wealthy(shahid)  greedy(shahid)  bad(shahid) wealthy(arif) greedy(arif) cousins(shahid,arif) • The new KB is propositionalized: proposition symbols are • wealthy(arif), greedy(arif), bad(arif), wealthy(shahid), • greedy(shahid), bad(shahid), cousins(shahid,arif). CS 331 Dr M M Awais

  5. Reduction contd. • Every FOL KB can be propositionalized so as to preserve entailment (entailments are dependencies) • (A ground sentence is entailed by new KB iff entailed by original KB) • Idea: propositionalize KB and query, apply resolution, return result • Problem: with function symbols, there are infinitely many ground terms, • e.g., Father(Father(Father(John))) CS 331 Dr M M Awais

  6. Reduction contd. Theorem: Herbrand (1930). If a sentence αis entailed by an FOL KB, it is entailed by a finite subset of the propositionalized KB Idea: For n = 0 to ∞ do create a propositional KB by instantiating with depth-$n$ terms see if α is entailed by this KB Problem: works if α is entailed, loops if α is not entailed Theorem: Turing (1936), Church (1936) Entailment for FOL is semidecidable (algorithms exist that say yes to every entailed sentence, but no algorithm exists that also says no to every nonentailed sentence.) CS 331 Dr M M Awais

  7. Problems withpropositionalization • Propositionalization seems to generate lots of irrelevant sentences. • E.g., from: x King(x)  Greedy(x)  Evil(x) King(John) y Greedy(y) Brother(Richard,John) • it seems obvious that Evil(John), but propositionalization produces lots of facts such as Greedy(Richard) that are irrelevant • With p k-ary predicates and n constants, there are p·nk instantiations. CS 331 Dr M M Awais

  8. Unification “It is an algorithm for determining the substitutions needed to make two predicate calculus expressions match” The process of universal/existential instantiation require substitutions CS 331 Dr M M Awais

  9. Unification • Unify(α,β) = θ if αθ = βθ p q θ knows(ali,X) knows(ali,sana) knows(ali,X) knows(Y,hira) knows(ali,X) knows(Y,mother(Y)) knows(ali,X) knows(X,hira) • Standardizing aparteliminates overlap of variables, e.g., the fourth sentence can have knows(Z,hira) CS 331 Dr M M Awais

  10. Unification • Unify(α,β) = θ if αθ = βθ p q θ knows(ali,X) knows(ali,sana) {sana/X} knows(ali,X) knows(Y,hira) knows(ali,X) knows(Y,mother(Y)) knows(ali,X) knows(X,hira) • Standardizing aparteliminates overlap of variables, e.g., the fourth sentence can have knows(Z,hira) CS 331 Dr M M Awais

  11. Unification • Unify(α,β) = θ if αθ = βθ p q θ knows(ali,X) knows(ali,sana) {sana/X} knows(ali,X) knows(Y,hira) {hira/X,ali/Y} knows(ali,X) knows(Y,mother(Y)) knows(ali,X) knows(X,hira) • Standardizing aparteliminates overlap of variables, e.g., the fourth sentence can have knows(Z,hira) CS 331 Dr M M Awais

  12. Unification • Unify(α,β) = θ if αθ = βθ p q θ knows(ali,X) knows(ali,sana) {sana/X} knows(ali,X) knows(Y,hira) {hira/X,ali/Y} knows(ali,X) knows(Y,mother(Y)) {ali/Y,mother(ali)} knows(ali,X) knows(X,hira) • Standardizing aparteliminates overlap of variables, e.g., the fourth sentence can have knows(Z,hira) CS 331 Dr M M Awais

  13. Unification • Unify(α,β) = θ if αθ = βθ p q θ knows(ali,X) knows(ali,sana) {sana/X} knows(ali,X) knows(Y,hira) {hira/X,ali/Y} knows(ali,X) knows(Y,mother(Y)) {ali/Y,mother(ali)} knows(ali,X) knows(X,hira) {fail} • Standardizing aparteliminates overlap of variables, e.g., the fourth sentence can have knows(Z,hira) CS 331 Dr M M Awais

  14. Unification - MGU • To unify knows(ali,X) and knows(Y,Z) θ = {ali/Y, Z/X } or θ= {ali/Y, ali/X, ali/Z} (if only ali is present in the data base of facts) • The first unifier is more general than the second. • There is a single Most General Unifier (MGU) that is unique up to renaming of variables. MGU = { ali/Y, Z/X} CS 331 Dr M M Awais

  15. Implementation of Unification  Using“List format” PC SyntaxList Syntax p(a,b) (p a b) p(f(a), g(X,Y)) (p(f a)(g X Y)) p(x) ^ q(y) ((p x) ^ (q y)) CS 331 Dr M M Awais

  16. Unification Examples Unify ((parents X(father X) (mother bill)), (parents bill (father bill)Y)). Complete Substitution {bill/X, mother (bill)/Y} CS 331 Dr M M Awais

  17. CS 331 Dr M M Awais

  18. CS 331 Dr M M Awais

  19. CS 331 Dr M M Awais

  20. Logic – Based Financial Advisor • To help a user decide whether to invest in a saving A/C or the stock market orboth CS 331 Dr M M Awais

  21. Policy: Saving inadequate - should invest in saving A/c regardless of income Saving adequateand adequate income- More profitable option of stock investment Low Income& adequate savings- Split surplus between saving and stock investment CS 331 Dr M M Awais

  22. Defining Constraints Adequate Savings/Income: Income: Rs 15000/month extra Rs 4000/ dependent/month Saving: Rs 25,000/year extra Rs 5000/dependent/year CS 331 Dr M M Awais

  23. Calculating Adequacy Define Functions to calculate the minimum income and savings Minimum Savings: min_saving(X) min_saving(X)=25000+5000*X where X are the number of dependents Minimum Income:min_income(X) min_income(X)=15000+4000*X CS 331 Dr M M Awais

  24. Outputs Investment (savings) Investment (stocks) Investment (combination) How can we represent outputs? CS 331 Dr M M Awais

  25. Rules ·saving_acc (inadequate) investment (savings) ·saving_acc(adequate) ^ income(adequate) investment(stocks) ·saving_acc (adequate) ^ income (inadequate) investment (combination) CS 331 Dr M M Awais

  26. What next? • Define Saving adequate • Define saving inadequate • Define income adequate • Define income inadequate CS 331 Dr M M Awais

  27. Savings Check if saved amount is greater than the minimum saving required CS 331 Dr M M Awais

  28. Income Check if the income is greater than the minimum income CS 331 Dr M M Awais

  29. Try firing rules with 9, 10, 11 CS 331 Dr M M Awais

  30. Example knowledge base The law says that it is a crime for an American to sell weapons to hostile nations. The country Nono, an enemy of America, has some missiles, and all of its missiles were sold to it by Colonel West, who is American. Prove that Col. West is a criminal CS 331 Dr M M Awais

  31. Example knowledge base contd. ... it is a crime for an American to sell weapons to hostile nations: american(X)  weapon(Y)  sells(X,Y,z)  hostile(z)  criminal(X) Nono … has some missiles, i.e., X Owns(Nono,X)  Missile(X): owns(nono,m1) and missile(m1) … all of its missiles were sold to it by Colonel West missile(X)  owns(nono,X)  sells(west,X,nono) Missiles are weapons: missile(X)  weapon(X) An enemy of America counts as "hostile“: enemy(X,america)  hostile(X) West, who is American … american(west) The country Nono, an enemy of America … enemy(nono,america) CS 331 Dr M M Awais

  32. Example knowledge base contd. Rule Base: ... it is a crime for an American to sell weapons to hostile nations: R1: american(X)  weapon(Y)  sells(X,Y,z)  hostile(z)  criminal(X) Nono … has some missiles, i.e., R2: X owns(nono,X)  missile(X): … all of its missiles were sold to it by Colonel West R3: missile(X)  owns(nono,X)  sells(west,X,nono) Missiles are weapons: R4: missile(X)  weapon(X) An enemy of America counts as "hostile“: R5: enemy(X,america)  hostile(X) • Database of Facts: • owns(nono,m1) and missile(m1) • West, who is American … • american(west) • The country Nono, an enemy of America … • enemy(nono,america) CS 331 Dr M M Awais

  33. R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)  criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) • START FROM THE FACTS • THINK WHICH RULES APPLY Be careful about the convention constants start with capital letters and variables with small (applicable to next few slides) 1. FORWARD CHAINING CS 331 Dr M M Awais

  34. R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)  criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) R5 {nono/X} R4 {m1/X} R3 {m1/X} CS 331 Dr M M Awais

  35. R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)  criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) R1 {west/X , m1/Y , nono/Z} CS 331 Dr M M Awais

  36. 2. BACKWARD CHAINING R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) • START FROM THE CONCLUSION • THINK IS THE RULE SUPPORT AVAILABLE CS 331 Dr M M Awais

  37. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z) criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  38. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z) criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  39. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  40. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  41. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  42. Backward chaining example R1: american(X)  weapon(Y)  sells(X,Y,Z)  hostile(Z)criminal(X) R2: X owns(nono,X)  missile(X): R3: missile(X)  owns(nono,X)  sells(west,X,nono) R4: missile(X)  weapon(X) R5: enemy(X,america)  hostile(X) CS 331 Dr M M Awais

  43. 3. Modus Ponen Based Solution • X,Y,Z[american(X)^weapon(Y)^nation(Z)^hostile(Z)^sells(X,Z,Y)  criminal(X)] • X owns(nono,X)^missiles(X)  sells(west,nono,X) • X missiles(X)  weapons (X) • X enemy(X,america) hostile(X) • owns(nono,m1) • missiles(m1) • american(west) • nation(nono) • enemy(nono,america) • nation(america) • 6 with 3 suggests, through Modus Ponens: weapons(m1) :{m1/X} 2. From 4 and 9 hostile(nono) : {nono/X} • From 2, 5, 6, Modus Ponens sells(west,nono,m1):m1/X 4. From 2,3,4,5,6, and 9 criminal(west): {west/X, m1/Y, nono/Z} CS 331 Dr M M Awais

  44. Composition • Find Overall Substitutions • Given two or more sets of substitutions CS 331 Dr M M Awais

  45. Example • S1={A/K} • S2={a/A, H/J} What set would represent overall substitutions • S1.S2={a/K, H/J} • Algorithm? CS 331 Dr M M Awais

  46. Example • S1={A/K} S2={a/A, H/J} • S1.S2={a/K, H/J} • Algorithm: • Apply S2 to S1 (find overall substitution) • Add S2 to S1 • Drop all substitutions in S2 which have common denominators with S1 CS 331 Dr M M Awais

  47. Example • S1={A/K} S2={a/A, H/J} • Algorithm: Apply S2 to S1 (find overall substitution) Apply a/A to A/K would lead to a/K Apply H/J to A/K cannot be applied S1.S2={a/K} Add S2 to S1 S1.S2={a/K a/A,H/J} Drop all substitutions in S2 which have common denominators with S1 nothing gets dropped, hence the answer is {a/K,a/A,H/J} CS 331 Dr M M Awais

  48. Example • S1={D/A} • S2={a/D} • S12={a/A, a/D} • S1={D/A,S/G} • S2={x/X, f/D, q/S,w/A} • S1.S2={f/A,q/G, x/X, f/D, q/S,w/A} • S1.S2={f/A,q/G, x/X,q/S} (w/A gets dropped) S1.S2={f/A,q/G, x/X,q/S} CS 331 Dr M M Awais

  49. Is S1.S2 same as S2.S1 • S1={D/A} • S2={a/D} • S21={a/A, D/A}{a/A} where as S12={a/A, a/D} • S1={D/A,S/G} • S2={x/X, f/D, q/S,w/A} • S2.S1={x/X, f/D, q/S, w/A D/A,S/G} • S2.S1={x/X, f/D, q/S, w/A, S/G} (D/A gets dropped) Hence generally speaking S1.S2 not equal to S2.S1 CS 331 Dr M M Awais

  50. Is S12 same as S21 • S1={D/A} • S2={X/A} • S12={D/A, X/A} • S12={D/A} • S21={X/A,D/A} • S21={X/A} CS 331 Dr M M Awais

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