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Ch. 9 – FOL Inference

Ch. 9 – FOL Inference. Supplemental slides for CSE 327 Prof. Jeff Heflin. EBNF Grammar for Prolog. <program>  <clause> { <clause> } <clause>  <fact> | <rule> <fact>  <compound> . <compound>  <atom> ( <termlist> ) <rule>  <head> :- <body> . <head>  <compound>

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Ch. 9 – FOL Inference

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  1. Ch. 9 – FOL Inference Supplemental slides for CSE 327 Prof. Jeff Heflin

  2. EBNF Grammar for Prolog <program>  <clause> { <clause> }<clause>  <fact> | <rule><fact>  <compound> .<compound>  <atom> ( <termlist> )<rule>  <head> :- <body> .<head>  <compound> <body>  <compound> { , <compound> } <termlist>  <term> { , <term> } <term>  <atom> | <num> | <var> | … • Notes: • { <a> } means 0, 1 or more <a>’s • This grammar is slightly more restrictive than the actual language

  3. Backward Chaining functionFOL-BC-Ask(KB, goals, ) returns a set of substitutionslocalanswers, a set of substitutions, initially empty if goals is empty then return { }q’ Subst( ,First(goals))for each sentence rin KB whereStandardize-Apart(r) = (p1… pn  q) and ’ Unify(q,q’) succeedsnew_goals  Prepend([p1,…,pn],Rest(goals))answers  FOL-BC-Ask(KB, new_goals, Compose(’, ))  answersreturnanswers Alternative description of Figure 9.6, p. 338

  4. Backward Chaining Example Set of sentences: S1: x1,y1 child(x1,y1)  parent(y1,x1) S2: x2,y2 parent(x2,y2)  female(x2)  mother(x2,y2) S3: child(Lisa,Homer) S4: child(Lisa,Marge) S5: female(Marge) Note: variables have already been standardized apart using subscripts Query: x mother(x,Lisa)

  5. Backward Chaining Search Tree mother(x0,Lisa) match rule S2’={x2/x0, y2/Lisa} parent(x0,Lisa), female(x0) match rule S1’={y1/x0, x1/Lisa} child(Lisa,x0), female(x0) match S4 ’={x0/Marge} match S3 ’={x0/Homer} female(homer) female(marge) matches S5 ’={ x0/Marge} no matches answers={} (FAIL!)  answers={ x0/Marge}

  6. Forward Chaining functionFOL-FC-Ask(KB,) returns a substitution or falselocalnew, the new sentences inferred on each iteration repeat until new is emptynew  {}for each sentence rin KBdo (p1… pn  q )  Standardize-Apart (r)for each  such thatSubst( , p1… pn)=Subst( , p1’ … pn’) for some p1’ ,…, pn’ in KBq’ Subst( ,q)if q’ does not unify with some sentence already in KB or newthen do add q’ to new Unify(q’, ) if  is not failthen return add new to KBreturnanswers From Figure 9.3, p. 332

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