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Logic Programming

Logic Programming. Some "declarative" slides on logic programming and Prolog. James Brucker. Introduction to Logic Programming. Declarative programming describes what is desired from the program, not how it should be done

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Logic Programming

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  1. Logic Programming Some "declarative" slides on logic programming and Prolog. James Brucker

  2. Introduction to Logic Programming • Declarative programming describes what is desired from the program, not how it should be done • Declarative language: statements of facts and propositions that must be satisfied by a solution to the program real(x). proposition: x is a real number. x > 0. proposition: x is greater than 0.

  3. Declarative Languages • what is a "declarative language"? • give another example (not Prolog) of a declarative language. SELECT * FROM COUNTRY WHERE CONTINENT = 'Asia';

  4. Facts, Rules, ... • What is a proposition? • What are facts? • What are rules? • What is a predicate? • What is a compound term?

  5. Facts: fish(salmon). likes(cat, tuna). Predicates: fish, likes Compound terms: likes(cat, X), fish(X) Atoms: cat, salmon, tuna Rule: eats(cat,X)  likes(cat,X), fish(X).

  6. A Really Simple Directed Graph (1) edge(a, b). (2) edge(a, c). (3) edge(c, d). (4) path(X, X). (5) path(X, Y)  edge(X, N), path(N, Y). a c b Question: What are the... • atoms • facts • rules d

  7. Clausal Form • Problem: There are too many ways to express propositions. • difficult for a machine to parse or understand • Clausal form: standard form for expressing propositions Consequent Antecedent Example: path(X, Y)  edge(X, N)  path(N, Y).

  8. Clausal Form Example Meaning: if there is an edge from X to N and there a path from N to Y, then there is a path from X to Y. The above is also called a "headed Horn clause". In Prolog this is written as a proposition or rule: path(X, Y)  edge(X, N)  path(N, Y). path(X, Y) :- edge(X, N) , path(N, Y).

  9. Query • A query or goalis an input proposition that we want Prolog to "prove" or disprove. • A query may or may not require that Prolog give us a value that satisfies the query (instantiation). 1 ?- edge(a,b). Yes 2 ?- path(c,b). No 3 ?- path(c,X). X = c ; X = d ; No

  10. Logical Operations on Propositions • What are the two operations that a logic programming language performs on propositions to establish a query? That is, how does it satisfy a query, such as:

  11. Unification Unification is a process of finding values of variables (instantiation) to match terms. Uses facts. (1-3) edge(a,b). edge(a,c). edge(c,d). (Facts) (4) path(X,X). (Rule) (5) path(X,Y) := edge(X,N), path(N,Y). (Rule) ?- path(a,d). This is the query (goal). Instantiate { X=a, Y=d }, and unifypath(a,d) with Rule 5. After doing this, Prolog must satisfy: edge(a,N). This is a subgoal. path(N,d). This is a subgoal.

  12. Unification in plain English Compare two atoms and see if there is a substitution which will make them the same. 1. edge(a, b). (Fact) 5. path(X, Y) :- edge(X, N) , path(N, Y). 6. path(a, Z). (Query) How can we unify 6 with 5? Let X := a Let Y := Z

  13. Resolution Resolution is an inference rule that allows propositions to be combined. • Idea: match the consequent (LHS) of one proposition with the antecedent (RHS term) of another. • Examples are in the textbook and tutorials.

  14. Resolution Example 1. edge(a, b). (Fact) 5. path(X, Y) :- edge(X, N) , path(N, Y). 6. path(a, Z). (Query) How can we unify 6 with 5? Let X := a Let Y := Z Resolution:

  15. Resolution Resolution is an inference rule that allows propositions to be combined. • Idea: match the consequent (LHS) of one proposition with the antecedent (RHS term) of another. • Examples are in the textbook and tutorials.

  16. How to handle failures • Prolog can work backwards towards the facts using resolution, instantiation, and unification. • As it works, Prolog must try each of several choices. • These choices can be stored as a tree. ?- path(a,d). The goal. Unify:unify path(a,d)with Rule 5 by instantiate { X=a,Y=d } Subgoal: edge(a,N). Instantiate: N=b which is true by Fact 1. Subgoal: path(b,d). Unify: path(b,d)with Rule 5: path(b,d) :- edge(b,N),path(N,d) Failure: can't instantiate edge(b,N) using any propositions.

  17. How to handle failures (2) • When a solution process fails, Prolog must undo some of the decisions it has made. • This is called backtracking. • same as backtracking you use in recursion. • Marks a branch of the tree as failed.

  18. How it Works (1) There are 2 search/execution strategies that can be used by declarative languages based on a database of facts. • Forward Chaining • Backward Chaining • what are the meanings of these terms?

  19. How it Works (2) • Forward Chaining • Backward Chaining • Which strategy does Prolog use? • Under what circumstances is one strategy more effective than the other? Consider two cases: • large number of rules, small number of facts • small number of rules, large number of facts

  20. PROLOG: PROgramming in LOGic The only "logic" programming language in common use.

  21. 3 Parts of a Prolog Program • A database contains two kinds of information. • What information is in a database? • A command to read or load the database. • in Scheme you can use load("filename") • in Prolog use consult('filename') • A query or goal to solve.

  22. Ancestors File: ancestors.pl ancestor(X,Y) :- parent(X,Y). ancestor(X,Y) :- ancestor(X,Z), ancestor(Z,Y). parent(X,Y) :- mother(X,Y). parent(X,Y) :- father(X,Y). father(bill, jill). mother(jill, sam). mother(jill, sally).

  23. Query the Ancestors ?- consult('/pathname/ancestors.pl'). ancestor(bill,sam). Yes ?- ancestor(bill,X). X = jill ; X = sam ; ERROR: Out of local stack ?- ancestor(X,bob). ERROR: Out of local stack

  24. Understanding the Problem • You need to understand how Prolog finds a solution. ancestor(X,Y) :- parent(X,Y). ancestor(X,Y) :- ancestor(X,Z), ancestor(Z,Y). parent(X,Y) :- mother(X,Y). parent(X,Y) :- father(X,Y). father(bill,jill). mother(jill,sam). father(bob,sam). Depth-first search causes immediate recursion

  25. Factorial File: factorial1.pl factorial(0,1). factorial(N,N*M) :- factorial(N-1,M). • The factorial of 0 is 1. • The factorial of N is N*M if the the factorial of N-1 is M ?- consult('/path/factorial1.pl'). ?- factorial(0,X). X = 1 Yes ?- factorial(1,Y). ERROR: Out of global stack

  26. Query Factorial ?- consult('/path/factorial1.pl'). ?- factorial(2,2). No ?- factorial(1,X). ERROR: Out of global stack Problem: Arithmetic is not performed automatically. ?- 2*3 = 6.No?- 2*3 = 2*3.Yes ?- 6 is 2*3.Yes?- 2*3 is 6.No is(6,2*3). l-value = r-value ?

  27. Arithmetic via Instantiation: is • "=" simply means comparison for identity. factorial(N, 1) :- N=0. • "is" performs instantiation if the left side doesn't have avalue yet. product(X,Y,Z) :- Z is X*Y. • this rule can answer the query: product(3,4,N). Answer: N = 12. • but it can't answer: product(3,Y,12).

  28. is does not mean assignment! • This always fails: N is N - 1. % sumto(N, Total): compute Total = 1 + 2 + ... + N. sumto(N, 0) :- N =< 0. sumto(N, Total) := Total is Subtotal + N, N is N-1, alwaysfails sumto(N, Subtotal). ?- sumto(0, Sum). Sum = 0. Yes ?- sumto(1, Sum). No

  29. is : how to fix? • How would you fix this problem? % sumto(N, Total): compute Total = 1 + 2 + ... + N. sumto(N, 0) :- N =< 0. sumto(N, Total) := N1 is N-1, always fails sumto(N1, Subtotal), Total is Subtotal + N. ?= sumto(5, X).

  30. Factorial revised File: factorial2.pl factorial(0,1). factorial(N,P) :- N1 is N-1, factorial(N1,M), P is M*N. Meaning: • The factorial of 0 is 1. • factorial of N is P if N1 = N-1 and factorial of N1 is M and P is M*N.

  31. Query Revised Factorial ?- consult('/path/factorial2.pl'). ?- factorial(2,2). Yes ?- factorial(5,X). X = 120 Yes but still has some problems... ?- factorial(5,X). X = 120 ;ERROR: Out of local stack ?- factorial(X,120). request another solution

  32. Factorial revised again Makes the rules mutually exclusive. File: factorial3.pl factorial(0,1). factorial(N,P) :- not(N=0), N1 is N-1, factorial(N1,M), P is M*N. ?- factorial(5,X). X = 120 ; No ?-

  33. Readability: one clause per line factorial(0,1). factorial(N,P) :- not(N=0), N1 is N-1, factorial(N1,M), P is M*N. factorial(0,1). factorial(N,P) :- not(N=0), N1 is N-1, factorial(N1,M), P is M*N. Better

  34. a b c d e f Finding a Path through a Graph edge(a, b). edge(b, c). edge(b, d). edge(d, e). edge(d, f). path(X, X). path(X, Y) :- edge(X, Z), path(Z, Y). ?- edge(a, b). Yes ?- path(a, a). Yes ?- path(a, e). Yes ?- path(e, a). No

  35. a b c d e f How To Define an Undirected Graph? edge(a, b). edge(b, c). edge(b, d). edge(d, e). edge(d, f). edge(X, Y) := not(X=Y), edge(Y, X). path(X, X). path(X, Y) :- edge(X, Z), path(Z, Y). ?- edge(b, a). Yes ?- path(a, b). Yes ?- path(b, e). No

  36. Queries and Answers • When you issue a query in Prolog, what are the possible responses from Prolog? % Suppose "likes" is already in the database :- likes(jomzaap, 219212).% Programming Languages. Yes. :- likes(papon, 403111).% Chemistry. No. :- likes(Who, 204219).% Theory of Computing? Who = pattarin • Does this mean Papon doesn't like Chemistry?

  37. Closed World Assumption • What is the Closed World Assumption? • How does this affect the interpretation of results from Prolog?

  38. List Processing • [Head | Tail] works like "car" and "cdr" in Scheme. • Example: ?- [H | T ] = [a,b,c,d,e]. returns: H = a T = [b,c,d,e] • This can be used to build lists and decompose lists. • Can use [H|T] on the left side to de/construct a list: path(X, Y, [X|P]) :- edge(X, Node), path(Node, Y, P).

  39. member Predicate • Test whether something is a member of a list ?- member(a, [b,c,d]). No. • can be used to have Prolog try all values of a list as values of a variable. ?-member(X, [a1,b2,c3,d4] ). X = a1 X = b2 X = c3

  40. member Predicate example • Use member to try all values of a list. • Useful for problems like • Queen safety • enumerating possible rows and columns in a game. % dumb function to find square root of 9 squareroot9(N) :- member(N,[1,2,3,4,5,5,6,7,8,9]), 9 is N*N.

  41. appending Lists ?- append([a,b],[c,d,e],L). L = [a,b,c,d,e] ?- append([],[a,b,c],L). L = [a,b,c] • append can resolve other parameters, too: ?- append(X, [b,c,d], [a,b,c,d] ). X = a

  42. Defining your own 'append' append([], List, List). append([Head|Tail], X, [Head|NewTail]) :- append(Tail, X, NewTail).

  43. Type Determination • Prolog is a weakly typed language. • It provides propositions for testing the type of a variable PREDICATE SATISFIED (TRUE) IF var(X) X is a variable nonvar(X) X is not a variable atom(A) A is an atom integer(K) K is an integer real(R) R is a floating point number number(N) N is an integer or real atomic(A) A is an atom or a number or a string functor(T,F,A) T is a term with functor F, arity A T =..L T is a term, L is a list. clause(H,T) H :- T is a rule in the program

  44. Tracing the Solution ?- trace. [trace] ?- path(a,d). Call: (8) path(a, d) ? creep Call: (9) edge(a, _L169) ? creep Exit: (9) edge(a, b) ? creep Call: (9) path(b, d) ? creep Call: (10) edge(b, _L204) ? creep Exit: (10) edge(b, c) ? creep Call: (10) path(c, d) ? creep Call: (11) edge(c, _L239) ? creep ^ Call: (12) not(c=_G471) ? creep ^ Fail: (12) not(c=_G471) ? creep Fail: (11) edge(c, _L239) ? creep Fail: (10) path(c, d) ? creep Redo: (10) edge(b, _L204) ? creep

  45. Solution Process (1) path(X,X). (2) path(X,Y) := edge(X,Z), path(Z,Y). ?- path(a,c). StackSubstitution (Instantiation) [path(a,c), path(X,X)] [path(a,c), path(a,a)] X = a Undo. [path(a,c), path(X,X)] [path(a,c), path(c,c)] X = c Undo.

  46. Solution Process (2) (1) path(X,X). (2) path(X,Y) := edge(X,Z), path(Z,Y). ?- path(a,c). StackSubstitution (Instantiation) [path(a,c), path(X,Y)] (Rule 2) [path(a,c), path(a,Y)] X = a X = a, Y = c edge(a,Z), path(Z,c) new subgoals edge(a,b), path(b,c) X = a, Y = c, Z = b path(b,c) edge(a,b) is a fact - pop it.

  47. What does this do? % what does this do? sub([], List). sub([H|T], List) :- member(H, List), sub(T, List).

  48. What does this do? Underscore (_) means "don't care". It accepts any value. % what does this do? foo([], _, []). foo([H|T], List, [H|P]) :- member(H, List), foo(T, List, P). foo([H|T], List, P) :- not( member(H, List) ), foo(T, List, P).

  49. Max Function • Write a Prolog program to find the max of a list of numbers: • max( List, X). • max( [3, 5, 8, -4, 6], X). X = 8. • Strategy: • use recursion • divide the list into a Head and Tail. • compare X to Head and Tail. Two cases: • Head = max( Tail ). in this case answer is X is Head. • X = max( Tail ) and Head < X. • what is the base case?

  50. Max Function % max(List, X) : X is max of List members max([X], X). base case max([H|Tail], H) :- 1st element is max max(Tail, X), H >= X. max([H|Tail], X) :- 1st element not max complete this case.

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