Prolog - Part 1

1. Prolog - Part 1 Alexis Raptarchis Patrick Rypalla raptarch@cs.uni-bonn.de rypalla@cs.uni-bonn.de

2. Table of Contents • Facts, Rules and Queries • Prolog examples • Prolog Syntax • Exercise 1 • Matching • Proof Search • Exercise 2 • Recursion • Exercise 3 • List • Arithmetic • Cuts and Negation • Meta-Predicates

3. Facts, Rules, and Queries • There are only three basic constructs in Prolog: facts, rules, and queries • Collection of facts and rules is called a knowledge base (KB) • Queries are questions about the information stored in the KB fact • Facts are used to state things that are unconditionally true of the domain of interest • We can ask Prolog whether Mia is a woman by posing a query • If we ask whether Jody is a woman Prolog will answer no, because Jody is not known to the KB woman(mia).?- woman(mia). query Prolog will answer: yes

4. Facts, Rules, and Queries • Rules state information that is conditionally true of the domain of interest playsAirGuitar(mia) :- listens2Music(mia). • Therule says that Mia plays air guitar if she listens to music • :- should be read as ``if'', or ``is implied by'‘ • In general rules say: if the body of the rule is true, then the head of the rule is true too • The body can contain more then one fact, for example:playsAirGuitar(mia):- listens2music(mia), happy(mia). rule head body

5. Facts, Rules, and Queries Lets add a fact to our KB, namely listens2Music(mia). playsAirGuitar(mia) :- listens2Music(mia). • We will ask Prolog whether Mia plays air guitar ?- playsAirGuitar(mia). • Remember playsAirGuitar(mia)Is not a fact in our KB • But Prolog will respond yes! Hence Prolog can use so called modus ponens to deduce facts from the KB • This new fact is not explicitly recorded in the knowledge base. It is only implicitly present fact rule

6. Facts, Rules, and Queries • Prolog answers this query by working from top to bottom through the KB, trying to match the expression woman(X) with the information KB contains. • Prolog instantiates X to mia, or that it binds X to mia • ; means or, so this query means: are there any more women? man(vincent). woman(mia). woman(jody). woman(yolanda). ?- woman(X). X = mia ?- ; X = jody; X = yolanda; no

7. Facts, Rules, and Queries The facts and rules contained in a KB are called clauses. In this case the KB contains 5 clauses, namely 2 facts and 3 rules happy(yolanda). listens2music(mia). listens2music(yolanda):- happy(yolanda). playsAirGuitar(mia):- listens2music(mia),happy(mia). playsAirGuitar(yolanda):- listens2music(yolanda). ?- playsAirGuitar(mia). no ?- playsAirGuitar(yolanda). yes

8. Facts, Rules, and Queries woman(mia). woman(jody). loves(vincent, mia). loves(marsellus, mia). loves(pumpkin, honey_bunny). loves(honey_bunny, pumpkin). jealous(X,Y):- loves(X,Z), loves(Y,Z). ?- loves(pumpkin,X), woman(X). no loves(marcellus,X),woman(X). X = mia ?- jealous(marsellus,W). W = vincent

9. Prolog Syntax • What exactly are facts, rules and queries built out of? Terms Terms Simple Terms Simple Terms Complex Terms Complex Terms Constants Constants Variables Variables Atoms Atoms Numbers Numbers

10. Prolog Syntax: Atoms and Variables • Atoms are sequence of characters of upper-case letters, lower-case letters, digits, or underscore, starting with a lowercase letter • Examples: butch, mia, playGuitar • An arbitrary sequence of characters enclosed in single quotes • Examples: 'Vincent', 'Five dollar shake', '@$%' • A sequence of special characters • Examples: : , ; . :- • Variables Same as Atoms, just starting with either an uppercase letter or an underscore • Examples: X, Y, Variable, Vincent, _tag

11. Prolog Syntax: Complex Terms • Operators • Implication :- • Conjunction , • Disjunction ; • Complex Terms • Atoms, numbers and variables are building blocks for complex terms • Complex terms are built out of a functor directly followed by a sequence of arguments • Arguments are put in round brackets, separated by commas • The functor must be an atom • Examples we have seen before: • playsAirGuitar(jody) • loves(vincent, mia) • jealous(marsellus, W) functor argument

12. Prolog Syntax: Arity • Complex terms inside complex terms: • hide(X,father(father(father(butch)))) • Functor is hide and it has two arguments: X and the complex term father(father(father(butch))) • The number of arguments a complex term has is called its arity • Examples:woman(mia) /1 is a term with arity 1 loves(vincent,mia) /2 has arity 2father(father(butch)) /1 arity 1 • In Prolog documentation arity of a predicate is usually indicated with the suffix "/" followed by a number to indicate the arity

13. Exercise 1 • Which of the following sequences of characters are atoms and which are variables? • vINCENT • Footmassage • variable23 • big_kahuna_burger • 'big kahuna burger' • Jules • ‘@Jules‘ • How many facts, rules and clausesare there in the following knowledge base? What are the heads of the rules, and what are the goals they contain?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).

14. Exercise 1 - Solution • Which of the following sequences of characters are atoms and which are variables? • vINCENT atom • Footmassage variable • Variable23 atom • big_kahuna_burger atom • 'big kahuna burger‘ atom • Jules variable • ‘@Jules‘ atom • How many facts, rules and clauses are there in the following knowledge base? What are the heads of the rules, and what are the goals they contain?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). 7 clauses 3 facts 4 rules

15. Matching • Two terms match, if they are equal or if they contain variables that can be instantiated in such a way that the resulting terms are equal Exampels: • =(mia,mia) match, because they are the same atom • woman(mia) = woman(mia) match, because they are the same complex term • mia = X match, because X can be instantiated to mia • How about loves(vincent, X) and loves(X, mia)? no match, because its impossible to find an instantiation of X • And does kill(shoot(gun), stab(knife)) = kill (X, stab(Y)) match? X = shoot(gun) Y = knife yes

16. Matching • Definition of Matching • If term1 and term2 are constants, then term1 and term2 match if and only if they are the same atom, or the same number. • If term1 is a variable and term2 is any type of term, then term1 and term2 match, and term1 is instantiated to term2. • If term1 and term2 are complex terms, then they match if and only if: • They have the same functor and arity. • All their corresponding arguments match • and the variable instantiations are compatible. (I.e. it is not possible to instantiate variable X to mia, when matching one pair of arguments, and to then instantiate X to vincent, when matching another pair of arguments.) • Two terms match if and only if it follows from the previous three clauses that they match.

17. Matching - occurs check • Consider the following query: father(X) = X. Let`s try and instantiate X to father(father(butch)):father(father(father(butch))) = father(father(butch)) • Prolog is desperately trying to match these terms, but it won't succeed.X = father(father(father(father(father(father(father(father(father(father(father(father(father(father(father(father(…. • newer versions of Prolog can detect cycles in termsX = father(father(father(father(father(father(...))))))))))yes • Now we know about matching • Next: we will learn how Prolog actually searches a KB to see if a query is satisfied • Proof search

18. Proof Search f(a). f(b). g(a). g(b). h(b). k(X):- f(X), g(X), h(X). ?- k(Y). Y=X ?- f(X), g(X), h(X). X=a X=b ?- g(a), h(a). ?- g(a), h(a). ?- g(b), h(b). ?- k(Y). Y=b; no ?- h(a). ?- h(a). ?- h(b). †

19. Proof Search loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). X=A Y=B ?- loves(A,C), loves(B,C). A=marsellus C=mia A=vincent C=mia ?- loves(B,mia). ?- loves(B,mia). ?- jealous(X,Y). X=marsellus Y=vincent; X=vincent Y=marsellus; no X = vincent Y = vincent; X = vincent Y = marsellus; B=vincent B=vincent B=marsellus B=marsellus

20. Exercise 2 • Which of the following pairs of terms match? Give the variable instantiations that lead to successful matching. • food(bread,X,beer) = food(Y,sausage,X) • food(bread,X,beer) = food(Y,kahuna_burger) • meal(food(bread),drink(beer)) = meal(X,Y) • ?- loves(X,X) = loves(marsellus,mia). • ?- k(s(g),Y) = k(X,t(k)). • ?- k(s(g),t(k)) = k(X,t(Y)).

21. Exercise 2 - Solution • food(bread,X,beer) = food(Y,sausage,X) no X = beer and X = sausage • food(bread,X,beer) = food(Y,kahuna_burger) no Because we have 3 arguments on the left side but only 2 on the right • meal(food(bread),drink(beer)) = meal(X,Y) X = food(bread) Y = drink(beer) Yes

22. Exercise 2 - Solution • ?- loves(X,X) = loves(marsellus,mia). no X = marsellus and X = mia • ?- k(s(g),Y) = k(X,t(k)). X=s(g) Y=t(k) yes • ?- k(s(g),t(k)) = k(X,t(Y)). X=s(g) Y=k yes

23. Recursion • Prolog predicates can be defined recursively • A predicate is recursively defined if one or more rules in its definition refers to itself • Let’s take a look on two rules: descend(X,Y):- child(X,Y). descend(X,Y):- child(X,Z), descend(Z,Y). • Whatdoesthissay? • if Y is a child of X, then Y is a descendant of X • if Z is a child of X, and Y is a descendant of Z, then Y is a descendant of X

24. Recursion - descend descend(martha,laura) child(martha, charlotte). child(charlotte, caroline).    child(caroline, laura).    child(laura, rose). descend(X,Y):- child(X,Y). descend(X,Y):- child(X,Z),descend(Z,Y). child(martha,laura) child (martha,Y) descend(Y,laura) † descend(charlotte,laura) child (charlotte,Y) descend(Y,laura) child(charlotte,laura) † ?- descend(martha, laura) yes descend(charlotte,laura) child(charlotte,laura)

25. Recursion - successor • Suppose we use the following way to write numerals: • 0 is a numeral. • If X is a numeral, then so is succ(X). • That is, succ(X) represents the number obtained by adding one to the number represented by X • It simply says that 0 is a numeral, and that all other numerals are built by stacking succ symbols in front. numeral(0). numeral(succ(X)) :- numeral(X). Exercise: What will happen if you ask: ?- numeral(X) X = 0 ;X = succ(0) ;X = succ(succ(0)) ;X = succ(succ(succ(0))) ;X = succ(succ(succ(succ(0)))) ;X = succ(succ(succ(succ(succ(0))))) ;X = succ(succ(succ(succ(succ(succ(0)))))) ;  

26. Exercise 3 descend(ron,harry) child(ron, hermione). child(hermione, draco). child(draco, harry).    descend(X,Y):- child(X,Y). descend(X,Y):- child(X,Z),descend(Z,Y). child(ron,harry) child (ron,Y) descend(Y,harry) † descend(hermione,harry) child (hermione,Y) descend(Y,harry) child(hermione,harry) † ?- descend(ron, harry) descend(draco,harry) child(draco,harry)