1. Agile Software Development Lab 2008 Dr. Günter Kniesel, Daniel Speicher, Tobias Rho, Pascal Bihler Spring 2008 R O O T S Prolog - Part 1 Alexis Raptarchis Patrick Rypalla raptarch@cs.uni-bonn.de rypalla@cs.uni-bonn.de

2. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)2 Table of Contents 1. Facts, Rules and Queries a.Prolog examples b.Prolog Syntax c.Exercise 1 2. Matching a.Proof Search b.Exercise 2 3. Recursion a.Exercise 3 4. List 5. Arithmetic 6. Cuts and Negation 7. Meta-Predicates

3. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)3 Facts, Rules, and Queries l There are only three basic constructs in Prolog: facts, rules, and queries l Collection of facts and rules is called a knowledge base (KB) l Queries are questions about the information stored in the KB l Facts are used to state things that are unconditionally true of the domain of interest l We can ask Prolog whether Mia is a woman by posing a query l If we ask whether Jody is a woman Prolog will answer no, because Jody is not known to the KB woman(mia). ?- woman(mia). fact query query Prolog will answer: yes

4. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)4 Facts, Rules, and Queries l Rules state information that is conditionally true of the domain of interest playsAirGuitar(mia) :- listens2Music(mia). l The rule says that Mia plays air guitar if she listens to music l :- should be read as ``if'', or ``is implied by'‘ l 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). head body rule

5. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)5 Facts, Rules, and Queries Lets add a fact to our KB, namely listens2Music(mia). playsAirGuitar(mia) :- listens2Music(mia). l We will ask Prolog whether Mia plays air guitar ?- playsAirGuitar(mia). Remember playsAirGuitar(mia) Is not a fact in our KB l But Prolog will respond yes! Hence Prolog can use so called modus ponens to deduce facts from the KB l This new fact is not explicitly recorded in the knowledge base. It is only implicitly present fact rule

6. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)6 Facts, Rules, and Queries man(vincent). woman(mia). woman(jody). woman(yolanda). ?- woman(X). X = mia ?- ; X = jody; X = yolanda; no 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?

7. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)7 Facts, Rules, and Queries happy(yolanda). listens2music(mia). listens2music(yolanda):- happy(yolanda). playsAirGuitar(mia):- listens2music(mia),happy(mia). playsAirGuitar(yolanda):- listens2music(yolanda). ?- playsAirGuitar(mia). no ?- playsAirGuitar(yolanda). yes 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 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

8. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)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. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)9 Prolog Syntax l What exactly are facts, rules and queries built out of? Terms Simple TermsComplex Terms ConstantsVariables AtomsNumbers Terms Simple TermsComplex Terms ConstantsVariables AtomsNumbers

10. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)10 Prolog Syntax: Atoms and Variables l Atoms are sequence of characters of upper-case letters, lower-case letters, digits, or underscore, starting with a lowercase letter Examples: butch, mia, playGuitar l An arbitrary sequence of characters enclosed in single quotes Examples: 'Vincent', 'Five dollar shake', '@$%' l A sequence of special characters Examples: :, ;. :- l Variables Same as Atoms, just starting with either an uppercase letter or an underscore l Examples: X, Y, Variable, Vincent, _tag

11. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)11 Prolog Syntax: Complex Terms l Operators uImplication :- uConjunction, uDisjunction ; l Complex Terms uAtoms, numbers and variables are building blocks for complex terms uComplex terms are built out of a functor directly followed by a sequence of arguments uArguments are put in round brackets, separated by commas uThe functor must be an atom l Examples we have seen before: uplaysAirGuitar(jody) uloves(vincent, mia) ujealous(marsellus, W) argument functor

12. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)12 Prolog Syntax: Arity l Complex terms inside complex terms: uhide(X,father(father(father(butch)))) l Functor is hide and it has two arguments: X and the complex term father(father(father(butch))) l The number of arguments a complex term has is called its arity uExamples: woman(mia) /1 is a term with arity 1 loves(vincent,mia) /2 has arity 2 father(father(butch)) /1 arity 1 l In Prolog documentation arity of a predicate is usually indicated with the suffix "/" followed by a number to indicate the arity

13. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)13 Exercise 1 l Which of the following sequences of characters are atoms and which are variables? uvINCENT uFootmassage uvariable23 ubig_kahuna_burger u'big kahuna burger' uJules u‘@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. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)14 Exercise 1 - Solution l Which of the following sequences of characters are atoms and which are variables? uvINCENTatom uFootmassagevariable uVariable23atom ubig_kahuna_burgeratom u'big kahuna burger‘atom uJulesvariable u‘@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). 3 facts 4 rules 7 clauses

15. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)15 Matching lTlTwo 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: l=l=(mia,mia) match, because they are the same atom lwlwoman(mia) = woman(mia)match, because they are the same complex term lmlmia = X match, because X can be instantiated to mia lHlHow about loves(vincent, X) and loves(X, mia)? no match, because its impossible to find an instantiation of X lAlAnd does kill(shoot(gun), stab(knife)) = kill (X, stab(Y)) match? X = shoot(gun) Y = knife yes

16. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)16 Matching l Definition of Matching 1. If term1 and term2 are constants, then term1 and term2 match if and only if they are the same atom, or the same number. 2. If term1 is a variable and term2 is any type of term, then term1 and term2 match, and term1 is instantiated to term2. 3. If term1 and term2 are complex terms, then they match if and only if: uThey have the same functor and arity. uAll their corresponding arguments match uand 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.) 4. Two terms match if and only if it follows from the previous three clauses that they match.

17. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)17 Matching - occurs check l Consider the following query: father(X) = X. Let`s try and instantiate X to father(father(butch)): father(father(father(butch))) = father(father(butch)) l 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(…. l newer versions of Prolog can detect cycles in terms X = father(father(father(father(father(father(...)))))))))) yes l Now we know about matching uNext: we will learn how Prolog actually searches a KB to see if a query is satisfied èProof search

18. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)18 Proof Search 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). ?- g(b), h(b). X=b ?- h(b). † Y=X ?- g(a), h(a). X=a ?- h(a).

19. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)19 Proof Search loves(vincent,mia). loves(marsellus,mia). jealous(A,B):- loves(A,C), loves(B,C). ?- jealous(X,Y). X=marsellus Y=vincent; X=vincent 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 X = vincent Y = vincent; X = vincent Y = marsellus;

20. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)20 Exercise 2 l Which of the following pairs of terms match? Give the variable instantiations that lead to successful matching. ufood(bread,X,beer) = food(Y,sausage,X) ufood(bread,X,beer) = food(Y,kahuna_burger) umeal(food(bread),drink(beer)) = meal(X,Y) u?- loves(X,X) = loves(marsellus,mia). u?- k(s(g),Y) = k(X,t(k)). u?- k(s(g),t(k)) = k(X,t(Y)).

21. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)21 Exercise 2 - Solution l food(bread,X,beer) = food(Y,sausage,X) no X = beer and X = sausage l food(bread,X,beer) = food(Y,kahuna_burger) no Because we have 3 arguments on the left side but only 2 on the right l meal(food(bread),drink(beer)) = meal(X,Y) X = food(bread) Y = drink(beer) Yes

22. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)22 Exercise 2 - Solution l ?- loves(X,X) = loves(marsellus,mia). no X = marsellus and X = mia l ?- k(s(g),Y) = k(X,t(k)). X=s(g) Y=t(k) yes l ?- k(s(g),t(k)) = k(X,t(Y)). X=s(g) Y=k yes

23. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)23 Recursion l Prolog predicates can be defined recursively l A predicate is recursively defined if one or more rules in its definition refers to itself l Let’s take a look on two rules: descend(X,Y):- child(X,Y). descend(X,Y):- child(X,Z), descend(Z,Y). l What does this say? 1.if Y is a child of X, then Y is a descendant of X 2.if Z is a child of X, and Y is a descendant of Z, then Y is a descendant of X

24. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)24 Recursion - descend 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). ?- descend(martha, laura) yes child(martha,laura) descend(martha,laura) child (martha,Y) descend(Y,laura) child (martha,Y) descend(Y,laura) descend(charlotte,laura) † child(charlotte,laura) † child (charlotte,Y) descend(Y,laura) child (charlotte,Y) descend(Y,laura) descend(charlotte,laura) child(charlotte,laura)

25. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)25 Recursion - successor lSlSuppose we use the following way to write numerals: 1. 0 is a numeral. 2. If X is a numeral, then so is succ(X). lTlThat is, succ(X) represents the number obtained by adding one to the number represented by X lIlIt 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. Agile Software Development Lab Spring 2008 R O O T S Vortragstitel (in „Ansicht -> Master änderbar)26 Exercise 3 child(ron, hermione). child(hermione, draco). child(draco, harry). descend(X,Y):- child(X,Y). descend(X,Y):- child(X,Z),descend(Z,Y). ?- descend(ron, harry) child(ron,harry) descend(ron,harry) child (ron,Y) descend(Y,harry) child (ron,Y) descend(Y,harry) descend(hermione,harry) † child(hermione,harry) † child (hermione,Y) descend(Y,harry) child (hermione,Y) descend(Y,harry) descend(draco,harry) child(draco,harry)