© Patrick Blackburn, Johan Bos & Kristina Striegnitz Lecture 2 Theory –Unification –Unification in Prolog –Proof search Exercises –Correction exercises.

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Presentation transcript:

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Lecture 2 Theory –Unification –Unification in Prolog –Proof search Exercises –Correction exercises of LPN chapter 1 –Exercises of LPN chapter 2

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Aim of this lecture Discuss unification in Prolog –Show how Prolog unification differs from standard unification Explain the search strategy that Prolog uses when it tries to deduce new information from old, using modus ponens

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification Recall previous example, where we said that Prolog unifies woman(X) with woman(mia) thereby instantiating the variable X with the atom mia.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Terms Simple TermsComplex Terms ConstantsVariables AtomsNumbers Recall Prolog Terms

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification Working definition: Two terms unify if they are the same term or if they contain variables that can be uniformly instantiated with terms in such a way that the resulting terms are equal

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification This means that: mia and mia unify 42 and 42 unify woman(mia) and woman(mia) unify This also means that: vincent and mia do not unify woman(mia) and woman(jody) do not unify

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification What about the terms: mia and X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification What about the terms: mia and X woman(Z) and woman(mia)

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Unification What about the terms: mia and X woman(Z) and woman(mia) loves(mia,X) and loves(X,vincent)

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Instantiations When Prolog unifies two terms it performs all the necessary instantiations, so that the terms are equal afterwards This makes unification a powerful programming mechanism

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Revised Definition 1/3 1.If T 1 and T 2 are constants, then T 1 and T 2 unify if they are the same atom, or the same number.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Revised Definition 2/3 1.If T 1 and T 2 are constants, then T 1 and T 2 unify if they are the same atom, or the same number. 2.If T 1 is a variable and T 2 is any type of term, then T 1 and T 2 unify, and T 1 is instantiated to T 2. (and vice versa)

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Revised Definition 3/3 1.If T 1 and T 2 are constants, then T 1 and T 2 unify if they are the same atom, or the same number. 2.If T 1 is a variable and T 2 is any type of term, then T 1 and T 2 unify, and T 1 is instantiated to T 2. (and vice versa) 3.If T 1 and T 2 are complex terms then they unify if: a)They have the same functor and arity, and b)all their corresponding arguments unify, and c)the variable instantiations are compatible.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Prolog unification: =/2 ?- mia = mia. yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Prolog unification: =/2 ?- mia = mia. yes ?- mia = vincent. no ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Prolog unification: =/2 ?- mia = X. X=mia yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz How will Prolog respond? ?- X=mia, X=vincent.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz How will Prolog respond? ?- X=mia, X=vincent. no ?- Why? After working through the first goal, Prolog has instantiated X with mia, so that it cannot unify it with vincent anymore. Hence the second goal fails.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example with complex terms ?- k(s(g),Y) = k(X,t(k)).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example with complex terms ?- k(s(g),Y) = k(X,t(k)). X=s(g) Y=t(k) yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example with complex terms ?- k(s(g),t(k)) = k(X,t(Y)).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example with complex terms ?- k(s(g),t(k)) = k(X,t(Y)). X=s(g) Y=k yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz One last example ?- loves(X,X) = loves(marsellus,mia).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Prolog and unification Prolog does not use a standard unification algorithm Consider the following query: ?- father(X) = X. Do these terms unify or not?

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Infinite terms ?- father(X) = X. X=father(father(father(father(father(father (father(father(father(father(father(father( father(father(father(father(father(father(f ather(father(father(father(father(father(f ather(father(father(father(father(father(f ather(father(father(father(father(father(f ather(father(father(father(father(father(f ather(father(father(father(father(father(

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Infinite terms ?- father(X) = X. X=father(father(father(…)))) yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Occurs Check A standard unification algorithm carries out an occurs check If it is asked to unify a variable with another term it checks whether the variable occurs in the term In Prolog: ?- unify_with_occurs_check(father(X), X). no

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))). ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))). ?- vertical(line(point(1,1),point(1,3))). yes ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))). ?- vertical(line(point(1,1),point(1,3))). yes ?- vertical(line(point(1,1),point(3,2))). no ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))). ?- horizontal(line(point(1,1),point(1,Y))). Y = 1; no ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Programming with Unification vertical( line(point(X,Y), point(X,Z))). horizontal( line(point(X,Y), point(Z,Y))). ?- horizontal(line(point(2,3),Point)). Point = point(_554,3); no ?-

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Proof Search Now that we know about unification, we are in a position to learn how Prolog searches a knowledge base to see if a query is satisfied. In other words: we are ready to learn about proof search

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). X=a Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a † Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a ?- g(b), h(b). X=b † Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a ?- g(b), h(b). X=b ?- h(b). † Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). Y=b ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a ?- g(b), h(b). X=b ?- h(b). † Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Example: search tree f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). Y=b; no ?- ?- k(Y). ?- f(X), g(X), h(X). ?- g(a), h(a). ?- h(a). X=a ?- g(b), h(b). X=b ?- h(b). † Y=X

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y).

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). ?- loves(A,C), loves(B,C). X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). X=vincent Y=vincent ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia B=vincent X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). X=vincent Y=vincent; X=vincent Y=marsellus ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia B=vincent B=marsellus X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). X=vincent Y=vincent; X=vincent Y=marsellus; ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia ?- loves(B,mia). A=marsellus C=mia B=vincent B=marsellus X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). …. X=vincent Y=marsellus; X=marsellus Y=vincent ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia ?- loves(B,mia). A=marsellus C=mia B=vincent B=marsellus X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). …. X=marsellus Y=vincent; X=marsellus Y=marsellus ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia ?- loves(B,mia). A=marsellus C=mia B=vincent B=marsellus X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Another example loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). …. X=marsellus Y=vincent; X=marsellus Y=marsellus; no ?- jealous(X,Y). ?- loves(A,C), loves(B,C). ?- loves(B,mia). A=vincent C=mia ?- loves(B,mia). A=marsellus C=mia B=vincent B=marsellus X=AY=B

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Solution Exercise vINCENT Atom 2. Footmassage Variable 3. variable23 Atom 4. Variable2000 Variable 5. big_kahuna_burger Atom 6. 'big kahuna burger' Atom 7. big kahuna burger neither: spaces without single quotes 8. 'Jules' Atom 9. _Jules Variable: starts with an underscore 10. '_Jules’ Atom: in single quotes Which of the following sequences of characters are atoms, which are variables, and which are neither?

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Solution Exercise loves(Vincent,mia) complex: loves/2 2. 'loves(Vincent,mia)' atom (in single quotes) 3. Butch(boxer) not a term (the functor must be an atom, and it is here a variable) 4. boxer(Butch) complex: boxer/1 5. and(big(burger),kahuna(burger)) complex: and/2, big/1, kahuna/1 6. and(big(X),kahuna(X)) complex: and/2, big/1, kahuna/1 7. _and(big(X),kahuna(X)) not a term (the functor must be an atom, and it is here a variable) 8. (Butch kills Vincent) not a term (the character sequence ‘(Butch’ does not start with a lower-case letter or with a single quote, so it is not an atom; it does not start with an upper-case letter or with an underscore, so it is not a variable; and it is obviously not a number or a complex term.) 9. kills(Butch Vincent) not a term (an operator is missing for the variables ‘Butch’ and ‘Vincent’, or a comma is missing for them to be arguments of the functor ‘kills’) 10. kills(Butch,Vincent not a term (closing parenthesis for it to be a predicate; and the opening parenthesis is not allowed in atoms and is not an operator) Which of the following sequences of characters are atoms, which are variables, which are complex terms, and which are not terms at all? Give the functor and arity of each complex term.

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Solution Exercise 1.3 woman(vincent). woman(mia). man(jules). person(X) :- man(X); woman(X). loves(X,Y) :- knows(Y,X). father(Y,Z) :- man(Y), son(Z,Y). father(Y,Z) :- man(Y), daughter(Z,Y). goal head Total 3 facts (clauses 1-3), 4 rules (clauses 4-7), 7 clauses and 8 predicates (woman/1, man/1, person/1, loves/2, knows/2, father/2, son/2, daughter/2) How many facts, rules, clauses, and predicates are there in the following knowledge base? What are the heads of the rules, and what are the goals they contain?

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Exercises Chapter 2 2.1, 2.2, 2.4

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Summary of this lecture In this lecture we have –defined unification –looked at the difference between standard unification and Prolog unification –introduced search trees

© Patrick Blackburn, Johan Bos & Kristina Striegnitz Next lecture Discuss recursion in Prolog –Introduce recursive definitions in Prolog –Show that there can be mismatches between the declarative meaning of a Prolog program, and its procedural meaning.