Prolog (cont d) Remark. Using multiple clauses. Intelligent Systems and HCI D7023E

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1 Intelligent Systems and HCI D703E Lecture : More Prolog Paweł Pietrzak Prolog (cont d) 1 Remark The recent version of SWI- Prolog displays true and false rather than and no Using multiple clauses Different clauses can be used to deal with different arguments.! greet(hamish):-!! write( How are you doin, pal? ).! greet(amelia):-!! write( Awfully nice to see you! ).! = Greet Hamish or Amelia = a disjunction of goals. -!?- greet(hamish).!?- greet(amelia). -! How are you doin, pal?! Awfully nice to see you! -!!!!! Clauses are tried in order from the top of the file. The first clause to match succeeds (= )

2 Variables in Questions We can call greet/1 with a variable in the question. A variable will match any head of greet/1.?- greet(anybody). How are you doin, pal? Anybody = hamish? The question first matches the clause closest to the top of the file. The variable is instantiated (i.e. bound) to the value hamish. As the variable was in the question it is passed back to the terminal (Anybody = hamish?). Re-trying Goals When a question is asked with a variable as an argument (e.g. greet(anybody).) we can ask the Prolog interpreter for multiple answers using: ;?- greet(anybody). How are you doin, pal? Anybody = hamish? ; Redo that match. Awfully nice to see you! Anybody = amelia? ; Redo that match. no Fail as no more matches. This fails the last clause used and searches down the program for another that matches. - RETURN = accept the match - ; = reject that match 6 Unification When two terms match we say that they unify. - Their structures and arguments are compatible. This can be checked using =/!?- loves(john,x) = loves(y,mary).! X = mary,! unification leads to instantiation! Y = john?!! Asking questions of the database We can ask about facts directly:?- mother(x,alan). X = jane? Yes Or we can define rules that prove if a property or relationship holds given the facts currently in the database. mother(jane,alan). father(john,alan). parent(mum,child):- mother(mum,child). parent(dad,child):- father(dad,child).?- parent(jane,x). X = alan? 7

3 Arithmetic Operators () Arithmetic Operators: +, -, *, / = Infix operators but can also be used as prefix. - Need to use is/ to access result of the arithmetic expression otherwise it is treated as a term:!!?- X = +4.!!?- X is +4.!!! X = +4?!! X = 9?!!!!! (Can X unify with +4?) (What is the result of +4?) Mathematical precedence is preserved: /, *, before +,- Can make compound sums using round brackets - Impose new precedence - Inner-most brackets first Satisfying Subgoals Most rules contain calls to other predicates in their body. These are known as Subgoals. These subgoals can match: - facts, - other rules, or - the same rule = a recursive call 1) drinks(alan,beer). ) likes(alan,coffee). 3) likes(heather,coffee). 4) likes(person,drink):-! drinks(person,drink). % a different subgoal ) likes(person,somebody):-! likes(person,drink), % recursive subgoals! likes(somebody,drink) Representing Proof using Trees To help us understand Prolog s proof strategy we can represent its behaviour using AND/OR trees. 1. Query is the top-most point (node) of the tree.. Tree grows downwards (looks more like roots!). Representing Proof using Trees () Using the tree we can see what happens when we ask for another match ( ; ) X/coffee 4 Backtracking 3. Each branch denotes a subgoal. 1. The branch is labelled with the number of the matching clause and. any variables instantiated when matching the clause head. 1 st match is failed and forgotten drinks(alan,x). 4. Each branch ends with either: 1. A successful match,. A failed match, or 3. Another subgoal. 11 X/coffee 1 st solution = Alan likes coffee. nd solution = Alan likes beer because Alan drinks beer. 1

4 Recursion using Trees When a predicate calls itself within its body we say the clause is recursing X/coffee 4 drinks(alan,x). likes(alan,drink) Drink/coffee Conjoined subgoals X/Somebody likes(somebody,drink) Drink = coffee Recursion using Trees () When a predicate calls itself within its body we say the clause is recursing X/coffee 4 drinks(alan,x). likes(alan,coffee) Drink/coffee X/Somebody likes(somebody,coffee) Drink = coffee Somebody /alan Somebody = alan 3 rd solution = Alan likes Alan because Alan likes coffee Recursion using Trees (3) When a predicate calls itself within its body we say the clause is recursing A question X/coffee 4 drinks(alan,x). likes(alan,coffee) likes(somebody,coffee) Drink/coffee Somebody/alan 3 Somebody/heather What is going to happen if we post a query?- likes(smb, coffee) 4 th solution = Alan likes Heather because Heather likes coffee. 1 Somebody = alan Somebody = heather 16

5 Prolog Data Objects (Terms) Atoms Constants Simple objects Integers Symbols a Strings bob l8r_day a Bob L8r day Signs <---> ==> Structured Objects Variables Structures Lists X A_var _Var date(4,10,04) [] person(bob,48) [a,b,g] [[a],[b]] [bit(a,d),a, Bob ] Structures To create a single data element from a collection of related terms we use a structure. A structure is constructed from a functor (a constant symbol) and one of more components. components functor somerelationship(a,b,c,[1,,3]) The components can be of any type: atoms, integers, variables, or structures. As functors are treated as data objects just like constants they can be unified with variables?- X = date(04,10,04). X = date(04,10,04)? Structure unification structures will unify if - the functors are the same, - they have the same number of components, - and all the components unify.?- person(nm,london,age) = person(bob,london,48). Nm = bob, Age = 48? More unification Terms that don t unify fred = jim. Hey you = Hey me. frou(frou) = f(frou). foo(bar) = foo(bar,bar). foo(n,n) = foo(bar,rab).?- person(someone,_,4) = person(harry,dundee,4). Someone = harry? (A plain underscore _ is not bound to any value. By using it you are telling Prolog to ignore this argument and not report it.) 19 0

6 More unification Terms that unify!! Outcome fred = fred.!!. Hey you = Hey you.!. fred=x.!!! X=fred. X=Y.!!! Y = X. foo(x) = foo(bar).!! X=bar. foo(n,n) = foo(bar,x). N=bar, X=bar. foo(foo(bar)) = foo(x) X = foo(bar) Structures = facts? The syntax of structures and facts is identical but: - Structures are not facts as they are not stored in the database as being true (followed by a period. ); - Structures are generally just used to group data; - Functors do not have to match predicate names. 1 A collection of ordered data. Lists Has zero or more elements enclosed by square brackets ( [ ] ) and separated by commas (, ). [a]!!! a list with one element []! an empty list [34,tom,[,3]]!! a list with 3 elements where the!!! 3 rd element is a list of elements. Like any object, a list can be unified with a variable?- [Any, list, of elements ] = X. X = [Any, list, of elements ]? List Unification Two lists unify if they are the same length and all their elements unify.?-[a,b,c,d]=[a,b,c,d]. A = a,!!! W = a, B = b,!!! X = c, C = c,!!! Y = d? D = d?!!?-[(a+x),(y+b)]=[(w+c),(d+b)].?- [[X,a]]=[b,Y].!?-[[a],[B,c],[]]=[X,[b,c],Y]. no!!!! B = b, Length 1 Length X = [a], Y = []? 3 4

7 Definition of a List Lists are recursively defined structures. An empty list, [], is a list. A structure of the form [X, ] is a list if X is a term and [ ] is a list, possibly empty. This recursiveness is made explicit by the bar notation [Head Tail] Head must unify with a single term. Tail unifies with a list of any length, including an empty list, []. - the bar notation turns everything after the Head into a list and unifies it with Tail. Head and Tail!?-[a,b,c,d]=[Head Tail].?-[a,b,c,d]=[X [Y Z]].! Head = a,!!! X = a,! Tail = [b,c,d]?!! Y = b,!!!!! Z = [c,d];!!!!!!!!!?-[a] = [H T].!!?-[a,b,c]=[W [X [Y Z]]].! H = a,!!! W = a,! T = [];!!! X = b,!!!!! Y = c, Z = []?!?-[] = [H T].!!?-[a [b [c []]]]= List.! no!!!! List = [a,b,c]?!!!!! 6