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48 Pages

### lec07

Course: ICOM 4036, Fall 2009
School: UPR Mayagüez
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Word Count: 1920

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4036: ICOM PROGRAMMING LANGUAGES Lecture 5 Logic Programming 05/17/09 What is Prolog Prolog is a `typeless' language with a very simple syntax. Prolog is declarative: you describe the relationship between input and output, not how to construct the output from the input (&quot;specify what you want, not how to compute it&quot;) Prolog uses a subset of first-order logic First-Order Logic Simplest form...

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4036: ICOM PROGRAMMING LANGUAGES Lecture 5 Logic Programming 05/17/09 What is Prolog Prolog is a `typeless' language with a very simple syntax. Prolog is declarative: you describe the relationship between input and output, not how to construct the output from the input ("specify what you want, not how to compute it") Prolog uses a subset of first-order logic First-Order Logic Simplest form of logical statement is an atomic formula. An assertion about objects. Examples: is-man(tom) is-woman(mary) married-to(tom,mary) mother-of(mary,john) First Order Logic More complex formulas can be built up using logical connectives: Men and Women are humans X [is-men(X) v is-woman(x) is-human(x)] Somebody is married to Tom X married-to(tom,X) Some woman is married to Tom X [married-to(tom,X) is-woman(X)] John has a mother X mother-of(X, john) Two offspring of the same mother are siblings X Y Z [mother-of(Z, X) mother-of(Z, Y) siblings(X, Y)] is the Existential quantifier is the Universal quantifier Logical Inference Example 2: Given these facts: is-man(carlos) is-man(pedro) and this rule: X [is-mortal(X) is-man(X)] derive: is-mortal(carlos), is-mortal(pedro). Logical Inference Logic programming is based on a simple idea: From facts and inferences try to prove more facts or inferences. Prolog Notation A rule: X [p(X) (q(X) r(X))] is written as p(X) q(X), r(X). Prolog conventions: variables begin with upper case (A, B, X, Y, Big, Small, ACE) constants begin with lower case (a, b, x, y, plato, aristotle) Prolog Assertions /* list of facts in prolog, stored in an ascii file, `family.pl'*/ mother-of(mary, ann). mother-of(mary, joe). mother-of(sue, mary). father-of(mike, ann). father-of(mike, joe). grandparent-of(sue, ann). /* reading the facts from a file */ ?- consult ( `family.pl' ). family.pl compiled, 0.00 sec, 828 bytes Prolog Evaluation ?- mother-of(sue, mary). Yes ?- mother-of(sue, ann). no ?- father-of( X, Y ). X = mike; Y = joe ; no % Prolog returns these solutions one at a time, in the order it finds them. You can press semicolon (;) to repeat the query and find the next solution. Prolog responds "no" when no more valid variable bindings of the variables can be found. Prolog Inference Rulesc /* Rules */ parent-of( X , Y ) : mother-of( X , Y ). % if mother(X,Y) then parent(X,Y) parent-of( X , Y ) : father-of( X , Y ). % if father(X,Y) then parent(X,Y) grandparent( X , Z ) : parent-of( X , Y ), parent-of(Y, Z ). % if parent(X,Y) and parent(Y,Z) then grandparent(X,Z) := means Prolog Inference Rule Evaluation ?- parent-of( X , ann), parent-of( X , joe). X = mary; X = mike; no ?- grandparent-of(sue, Y ). Y = ann; Y = joe; no Factorial in Prolog /* specification of factorial n! */ factorial(0,1). factorial(N, M) : N1 is N 1, factorial (N1, M1), M is N*M1. Takes 1 assertion and 1 inference Factorial in Prolog - Evaluation Lists in Prolog mylength( [ ], 0). mylength( [X | Y], N): mylength(Y, Nx), N is Nx+1. ? mylength( [1, 7, 9], X ). X=3 ? - mylength(jim, X ). No ? - mylength(Jim, X ). Jim = [ ] X=0 List Membership mymember( X , [X | Y] ). mymember( X , [W | Z ] ) : mymember( X , Z ). ?mymember(a, [b, c, 6] ). no ? mymember(a, [b, a, 6] ). yes ? mymember( X , [b, c, 6] ). X = b; X = c; X = 6; no Appending Lists The Problem: Define a relation append(X,Y,Z) as X appended to Y yields Z Appending Lists The Problem: Define a relation append(X,Y,Z) to mean that X appended to Y yields Z append([], Y, Y). append([H|X], Y, [H|Z]) :append(X,Y,Z). Appending Lists ?- append([1,2,3,4,5],[a,b,c,d],Z). Z = [1,2,3,4,5,a,b,c,d]; no ?- append(X,Y,[1,2,3]). X = [] Y = [1,2,3]; X = [1] Y = [2,3]; X = [1,2] Y = [3]; X = [1,2,3] Y = []; no Prolog Computes ALL Possible Bindings! Control in Prolog Prolog tries to solve the clauses from left to right. If there is a database file around, it will be used in a similarly sequential fashion. 1. Goal Order: Solve goals from left to right. 2. Rule Order: Select the first applicable rule, where first refers to their order of appearance in the program/file/database Control in Prolog The actual search algorithm is: 1. start with a query as the current goal. 2. WHILE the current goal is non-empty DO choose the leftmost subgoal ; IF a rule applies to the subgoal THEN select the first applicable rule; form a new current goal; ELSE backtrack; ENDWHILE SUCCEED Control in Prolog Thus the order of the queries is of paramount importance . The general paradigm in Prolog is Guess then Verify: Clauses with the fewest solutions should come first, followed by those that filter or verify these few solutions Fibonacci in Prolog fib1(1, 1). fib1(2, 1). fib1(N1, F1) :N1 > 2, N2 is N1 - 1, N3 is N1 - 2, fib1(N2, F2), fib1(N3, F3), F1 is F2 + F3. More List Processing remove(X, L1, L2) ~ sets L2 to the list obtained by removing the first occurrence of X from list L1 remove(X, [X|Rest], Rest). remove(X, [Y|Rest], [Y|Rest2]) :X \ == Y, remove(X, Rest, Rest2). More List Processing replace(X, Y, L1, L2) ~ sets L2 to the list obtained by replacing all occurrences of X in list L1 with Y replace(_, _, [], []). replace(X, Y, [X|Rest], [Y|Rest2]) :replace(X, Y, Rest, Rest2). replace(X, Y, [Z|Rest], [Z|Rest2]) :Z \== X, replace(X, Y, Rest, Rest2). More List Processing Write a predicate insert(X, Y, Z) that can be used to generate in Z all of the ways of inserting the value X into the list Y. insert(X, [], insert(X, [X]). [Y|Rest], [X,Y|Rest]). insert(X, [Y|Rest], [Y|Rest2]) :insert(X,Rest, Rest2). More List Processing Write a predicate permutation(X, Y) that can be used to generate in Y all of the permutations of list X permutation([], []). permutation([X|Rest], Y) :permutation(Rest, Z), insert(X, Z, Y). Graphs in Prolog b c path(a,b). path(b,c). path(c,d). path(d,b). path(a,c). Route(X,X). Route(X,Y):- path(X,Z), route(Z,Y). d Write a predicate route(X,Y) that success if there is a connection between X and Y a Binary Search Trees in Prolog <bstree>::= empty node(<number>, <bstree>, <bstree>) node(15,node(2,node(0,empty,empty), node(10,node(9,node(3,empty,empty), empty), node(12,empty,empty))), node(16,empty,node(19,empty,empty))) Binary Search Trees isbtree(empty). isbtree(node(N,L,R)):- number(N),isbtree(L),isbtree(R), smaller(N,R),bigger(N,L). smaller(N,empty). smaller(N, node(M,L,R)):- N < M, smaller(N,L), smaller(N,R). bigger(N, empty). bigger(N, node(M,L,R)):- N > M, bigger(N,L), bigger(N,R). Binary Search Trees ?- [btree]. ?isbtree(node(6,node(9,empty,empty),empty)). no ?isbtree(node(9,node(6,empty,empty),empty)). yes Binary Search Trees Define a relation which tells whether a particular number is in a binary search tree . mymember(N,T) should be true if the number N is in the tree T. mymember(K,node(K,_,_)). mymember(K,node(N,S,_)) :K <N, mymember(K,S). mymember(K,node(N,_,T)) :K >T, mymember(K,T). Binary search Trees ?mymember(3,node(10,node(9,node(3,empty,empty),empty) , node(12,empty,empty))). yes Sublists (Goal Order) myappend([], Y, Y). myappend([H|X], Y, [H|Z]) :myappend(X,Y,Z). myprefix(X,Z) :- myappend(X,Y,Z). mysuffix(Y,Z) :- myappend(X,Y,Z). Version 1 sublist1(S,Z) :myprefix(X,Z), mysuffix(S,X). Version 2 sublist2(S,Z) :mysuffix(S,X), myprefix(X,Z). Sublists ?- [sublist]. ?- sublist1([e], [a,b,c]). no ?- sublist2([e], [a,b,c]). Fatal Error: global stack overflow ... Version 1 So what's happening? If we ask the question: sublist1([e], [a,b,c]). this becomes prefix(X,[a,b,c]), suffix([e],X). and using the guess-query idea we see that the first goal will generate four guesses: [] [a] [a,b] [a,b,c] none of which pass the verify goal, so we fail. Version 2 On the other hand, if we ask the question: sublist2([e], [a,b,c]). this becomes suffix([e],X),prefix(X,[a,b,c]). using the guess-query idea note that the goal will generate an infinite number of guesses. [e] [_,e] [_,_,e] [_,_,_,e] [_,_,_,_,e] None of which pass the verify goal, so we never terminate!! Towers of Hanoi A B C You can move N disks from A to C in three general recursive steps. Move N-1 disks from A pole to the B pole using C as auxiliary. Move the last (Nth) disk directly over to the C pole. Move N-1 disks from the B pole to the C pole using A as auxiliary. Towers of Hanoi loc := right;middle;left hanoi(integer) move(integer,loc,loc,loc) inform(loc,loc) inform(Loc1, Loc2):write("\nMove a disk from ", Loc1, " to ", Loc2). Towers of Hanoi hanoi(N):move(N,left,middle,right). move(1,A,_,C) :- inform(A,C),!. move(N,A,B,C):N1 is N-1, move(N1,A,C,B), inform(A,C), move(N1,B,A,C). Logic Circuits construct an exclusive OR circuit from AND, OR, and NOT circuits, and then che...

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vti_encoding:SR|utf8-nl vti_timelastmodified:TR|02 Oct 2001 15:15:56 -0000 vti_extenderversion:SR|5.0.2.4803 vti_title:SR| vti_backlinkinfo:VX|courses/fall2001/icom4015/icom4015.htm vti_syncwith_ece.uprm.edu\:21/public_html:TX|02 Oct 2001 15:15:56 -0
UPR Mayagüez - ICOM - 4015
vti_encoding:SR|utf8-nl vti_author:SR|Michael vti_timecreated:TR|29 Feb 2000 02:30:38 -0000 vti_timelastmodified:TR|27 Mar 2000 14:32:00 -0000 vti_filesize:IR|50176 vti_title:SR|ICOM 4015 Advanced Programming vti_assignedto:SR| vti_approvallevel:SR|
UPR Mayagüez - ICOM - 4015
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UPR Mayagüez - ICOM - 4029
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UPR Mayagüez - ICOM - 4029
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