CS 152: Programming Language Paradigms March 5 Class Meeting Department of Computer Science San Jose State University Spring 2014 Instructor: Ron Mak

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 2 Backtracking and the Cut in Prolog  Cut is a mechanism you can use in Prolog programs. Tell Prolog not to reconsider certain previous choices it made.  Your program will run faster. It won’t waste time attempting to satisfy goals that you know beforehand will never contribute to a solution.  Your program will use less memory. Fewer intermediate results.  Cut can be the difference between a program that runs and one that does not. _

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 3 Cut Example: Library Application  What library services can a client use? First version: service(Person, Service) :- book_overdue(Person, Book), basic_service(Service). service(Person, Service) :- general_service(Service). basic_service(reference). basic_service(inquiries). special_service(borrowing). special_service(interlibrary_loan). general_service(X) :- basic_service(X). general_service(X) :- special_service(X). book_overdue(jones, war_and_peace). book_overdue(jones, oliver_twist). ?- service(smith, S). ?- service(jones, S). library1.pl

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 4 Cut Example: Library Application, cont’d Second version: service(Person, Service) :- book_overdue(Person, Book), !, basic_service(Service). service(Person, Service) :- general_service(Service). basic_service(reference). basic_service(inquiries). special_service(borrowing). special_service(interlibrary_loan). general_service(X) :- basic_service(X). general_service(X) :- special_service(X). book_overdue(jones, war_and_peace). book_overdue(jones, oliver_twist). cut ?- service(smith, S). ?- service(jones, S). library2.pl

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 5 Cut Example: Library Application, cont’d  Match the first service rule. Subgoal: Any overdue books? Match book_overdue(jones, war_and_peace).  The next goal is the cut “ ! ”. Cut always succeeds. Cut commits Prolog to all its decisions since it matched the service rule. service(Person, Service) :- book_overdue(Person, Book), !, basic_service(Service). service(Person, Service) :- general_service(Service). basic_service(reference). basic_service(inquiries). special_service(borrowing). special_service(interlibrary_loan). general_service(X) :- basic_service(X). general_service(X) :- special_service(X). book_overdue(jones, war_and_peace). book_overdue(jones, oliver_twist). ?- service(jones, S).

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 6 Cut Example: Library Application, cont’d  The cut prevents any backtracking from going past (before) it.  Therefore, after satisfying the subgoal basic_service (Service) the query is done. Don’t look for more overdue books.  The choice of the rule that the cut appears in is also blocked from backtracking. Therefore, don’t consider any other service rules. service(Person, Service) :- book_overdue(Person, Book), !, basic_service(Service). service(Person, Service) :- general_service(Service). basic_service(reference). basic_service(inquiries). special_service(borrowing). special_service(interlibrary_loan). general_service(X) :- basic_service(X). general_service(X) :- special_service(X). book_overdue(jones, war_and_peace). book_overdue(jones, oliver_twist). ?- service(jones, S).

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 7 Cut-Failure Example: Tax Calculator  A foreigner is not an average tax payer.  If your spouse makes over $100,000, then you’re not an average taxpayer.  If you get less than $10,000 from a pension, then you’re considered not to have any gross income. average_taxpayer(X) :- foreigner(X), !, fail. average_taxpayer(X) :- spouse(X, Y), gross_income(Y, Inc), Inc > , !, fail. average_taxpayer(X) :- gross_income(Y, Inc), Inc > 20000, Inc < gross_income(X, Y) :- receives_pension(X, P), P < 10000, !, fail. gross_income(X, Y) :- gross_salary(X, Z), investment_income(X, W), Y is Z+W. gross_salary :-...

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 8 Prolog Lists  A list in Prolog is similar to a list in Lisp or Scheme, but with a different syntax.  Use square brackets and commas: [a, b, c, d, e]  Use the notation [H|T ] (where H and T are any variables) to specify the head and tail of a list. ?- [H|T] = [a, b, c, d, e]. H = a, T = [b, c, d, e]. ?- [W, X, Y|Z] = [a, b, c, d, e]. W = a, X = b, Y = c, Z = [d, e].

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 9 Prolog Lists: cons  What does this predicate do? cons(H, T, L) :- L = [H|T]. ?- cons(x, [1, 2, 3], L). L = [x, 1, 2, 3].

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 10 Prolog Lists: append  Rewrite the Scheme append procedure in Prolog: Appending the empty list to any list L2 returns the list L2. Appending a list whose head is H and tail is T to a list L2 returns a list whose head is also H and whose tail T2 is T appended to L2. (define append (lambda (lst1 lst2) (if (null? lst1) lst2 (cons (car lst1) (append (cdr lst1) lst2)) ))) append([], L2, L2). append([H|T], L2, [H|T2]) :- append(T, L2, T2). ?- append([a, b, c], [1, 2, 3], L). L = [a, b, c, 1, 2, 3].

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 11 Prolog Lists: append, cont’d  Run append backwards to find all the ways that two lists can be appended to get a specified list. ?- append(L1, L2, [a, b, c]). L1 = [], L2 = [a, b, c] ; L1 = [a], L2 = [b, c] ; L1 = [a, b], L2 = [c] ; L1 = [a, b, c], L2 = [] ; false.

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 12 Prolog Lists: reverse  Rewrite the Scheme reverse procedure in Prolog: Reversing the empty list returns the empty list. Reversing a list whose head is H and whose tail is T returns a list L which is the reverse of T (variable RT ) appended to H. (define reverse (lambda (lst) (if (null? lst) '() (append (reverse (cdr lst)) (list (car lst))) ))) reverse([], []). reverse([H|T], L) :- reverse(T, RT), append(RT, [H], L). ?- reverse([1, 2, 3, 4, 5], L). L = [5, 4, 3, 2, 1].

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 13 Prolog’s Search Strategy  Prolog does a depth-first search of the solution tree. Consider the database: Who is Bob’s ancestor? ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob) ?- ancestor(X, bob).

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 14 Prolog’s Search Strategy, cont’d  Can we understand this trace? ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob) ?- ancestor(X, bob). 1 ?- trace. true. [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep Call: (7) ancestor(bob, bob) ? creep Call: (8) parent(bob, _G2111) ? creep Fail: (8) parent(bob, _G2111) ? creep Redo: (7) ancestor(bob, bob) ? creep Exit: (7) ancestor(bob, bob) ? creep Exit: (6) ancestor(amy, bob) ? creep X = amy ; Redo: (6) ancestor(_G2037, bob) ? creep Exit: (6) ancestor(bob, bob) ? creep X = bob.

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 15 Prolog’s Search Strategy, cont’d ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob) ?- ancestor(X, bob). ancestor(X, bob) Y = bob 1 [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep parent(amy, bob) X = amy Z = bob 3 Success

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 16 Prolog’s Search Strategy, cont’d ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob) ?- ancestor(X, bob). ancestor(X, bob) Y = bob 1 parent(amy, bob) X = amy Z = bob [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep Call: (7) ancestor(bob, bob) ? creep 3 ancestor(bob, bob) Success ancestor(bob, bob) X = bob Y = bob 1

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 17 Prolog’s Search Strategy, cont’d ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob) ?- ancestor(X, bob). ancestor(X, bob) Y = bob 1 parent(amy, bob) X = amy Z = bob [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep Call: (7) ancestor(bob, bob) ? creep Call: (8) parent(bob, _G2111) ? creep Fail: (8) parent(bob, _G2111) ? creep Redo: (7) ancestor(bob, bob) ? creep 3 ancestor(bob, bob) X = bob Y = bob 1 parent(bob, Z) 3 Failure Success REDO

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 18 Prolog’s Search Strategy, cont’d ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob). 1 3 ?- ancestor(X, bob). ancestor(X, bob) Y = bob 1 parent(amy, bob) X = amy Z = bob 3 ancestor(bob, bob) [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep Call: (7) ancestor(bob, bob) ? creep Call: (8) parent(bob, _G2111) ? creep Fail: (8) parent(bob, _G2111) ? creep Redo: (7) ancestor(bob, bob) ? creep Exit: (7) ancestor(bob, bob) ? creep Exit: (6) ancestor(amy, bob) ? creep X = amy ; 2 ancestor(bob, bob) X = bob Success X = amy 2 REDO

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 19 Prolog’s Search Strategy, cont’d ancestor(X, Y) :- parent(X, Z), ancestor(Z, Y). ancestor(X, X). parent(amy, bob). 1 3 ?- ancestor(X, bob). [trace] 1 ?- ancestor(X, bob). Call: (6) ancestor(_G2037, bob) ? creep Call: (7) parent(_G2037, _G2111) ? creep Exit: (7) parent(amy, bob) ? creep Call: (7) ancestor(bob, bob) ? creep Call: (8) parent(bob, _G2111) ? creep Fail: (8) parent(bob, _G2111) ? creep Redo: (7) ancestor(bob, bob) ? creep Exit: (7) ancestor(bob, bob) ? creep Exit: (6) ancestor(amy, bob) ? creep X = amy ; Redo: (6) ancestor(_G2037, bob) ? creep Exit: (6) ancestor(bob, bob) ? creep X = bob. 2 ancestor(bob, bob) X = bob 2 Success X = bob REDO

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 20 Assignment #4: Prolog Murder Mystery  Use Prolog to create and solve a murder mystery. Construct a Prolog database that presents a good murder mystery that can be solved deductively.  Your database must have at least 20 facts and 20 rules.  You must make at least 15 queries.  At least half of your rules and half of your queries must have subgoals separated by commas.  Example facts:  Example rule: person(jenkins, butler). killed(alice, blunt_object). owns(john, baseball_bat). having_affair(Person1, Person2) :- married(Person1, Person3), seen_together(Person1, Person2), Person2 \= Person3.

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 21 Assignment #4, cont’d  Example query: If you were Sherlock Holmes at the scene of the crime, what questions would you ask?  a zip file to that A text file of your database of facts and rules that can be loaded into Prolog, for example: SuperCoders.pl A cut-and-paste text file or screen shots that show your queries and the responses.  Name the file after your team: SuperCoders.zip  Subject line: CS 152 Assignment #4, team name Don’t forget to CC all your team members.  Due: Friday, March 14. ?- having_affair(bill, X), killed(X, blunt_object), owns(Y, hockey_stick).

SJSU Dept. of Computer Science Spring 2014: March 5 CS 152: Programming Language Paradigms © R. Mak 22 Uses of Prolog  Expert systems  Natural language processing NASA’s Clarissa voice interface for the space station  Modelling weather prediction air pollution control  See: application-of-prolog/ http:// application-of-prolog/