Programming Languages 2nd edition Tucker and Noonan

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1 Programming Languages 2nd edition Tucker and Noonan Chapter 15 Logic Programming Q: How many legs does a dog have if you call its tail a leg? A: Four. Calling a tail a leg doesn t make it one. Abraham Lincoln Introduction Logic programming languages, sometimes called declarative programming languages Express programs in a form of symbolic logic Use a logical inferencing process to produce results Declarative rather that procedural: Only specification of results are stated (not detailed procedures for producing them) Concepts of Copyright Programming Languages, 2006 The 9th McGraw-Hill ed., by Robert W. Companies, Sebesta. Addison Inc. Wesley,

2 Example: Sorting a List Describe the characteristics of a sorted list, not the process of rearranging a list sort(old_list, new_list) permute (old_list, new_list) sorted (new_list) sorted (list) j such that 1 j < n, list(j) list (j+1) Concepts of Copyright Programming Languages, 2006 The 9th McGraw-Hill ed., by Robert W. Companies, Sebesta. Addison Inc. Wesley, Logic and Horn Clauses A Horn clause has a head h, which is a predicate, and a body, which is a list of predicates p 1, p 2,, p n. It is written as: h p 1, p 2,, p n This means, h is true only if p 1, p 2,, and p n are simultaneously true. E.g., the Horn clause: snowing(c) precipitation(c), freezing(c) says, it is snowing in city C only if there is precipitation in city C and it is freezing in city C

3 Horn Clauses and Predicates Any Horn clause h p 1, p 2,, p n can be written as a predicate: p 1 p 2 p n h or equivalently: (p 1 p 2 p n ) h But not every predicate can be written as a Horn clause. E.g., literate(x) reads(x) writes(x) 16-6 Resolution and Unification If h is the head of a Horn clause h terms and it matches one of the terms of another Horn clause: t t 1, h, t 2 then that term can be replaced by h s terms to form: t t 1, terms, t 2 During resolution, assignment of variables to values is called instantiation. Unification is a pattern-matching process that determines what particular instantiations can be made to variables during a series of resolutions

4 Example The two clauses: speaks(mary, English) talkswith(x, Y) speaks(x, L), speaks(y, L), X Y can resolve to: talkswith(mary, Y) speaks(mary, English), speaks(y, English), Mary Y The assignment of values Mary and English to the variables X and L is an instantiation for which this resolution can be made Logic Programming in Prolog In logic programming the program declares the goals of the computation, not the method for achieving them. Logic programming has applications in AI and databases. Natural language processing (NLP) Automated reasoning and theorem proving Expert systems (e.g., MYCIN) Database searching, as in SQL (Structured Query Language) Prolog emerged in the 1970s. Distinguishing features: Nondeterminism Backtracking

5 Prolog Program Elements Prolog programs are made from terms, which can be: Variables Constants Structures Variables begin with a capital letter, like Bob. Constants are either integers, like 24, or atoms, like the, zebra, Bob, and.. Structures are predicates with arguments, like: n(zebra), speaks(y, English), and np(x, Y) The arity of a structure is its number of arguments (1, 2, and 2 for these examples) Facts, Rules, and Programs A Prolog fact is a Horn clause without a right-hand side. Its form is (note the required period.): term. A Prolog rule is a Horn clause with a right-hand side. Its form is (note :- represents and the required period.): term :- term 1, term 2, term n. A Prolog program is a collection of facts and rules

6 Inferencing Process of Prolog Queries are called goals If a goal is a compound proposition, each of the facts is a subgoal To prove a goal is true, must find a chain of inference rules and/or facts. For goal Q: A. B :- A. C :- B. Q :- P. Process of proving a subgoal called matching, satisfying, or resolution Concepts of Copyright Programming Languages, 2006 The 9th McGraw-Hill ed., by Robert W. Companies, Sebesta. Addison Inc. Wesley, Approaches Bottom-up resolution, forward chaining Begin with facts and rules of database and attempt to find sequence that leads to goal Works well with a large set of possibly correct answers A. {A} B :- A. {A, B} C :- B. {A, B, C} Q :- P. {A, B, C,, P, Q}? Q. true Copyright 2009 Addison-Wesley. All rights reserved

7 Approaches Top-down resolution, backward chaining Begin with goal and attempt to find sequence that leads to set of facts in database Works well with a small set of possibly correct answers A. {A, B, C,, P, Q} B :- A. C :- B. Q :- P. {A?, B?, C?,, P?, Q?} {B?, C?,, P?, Q?} {P?, Q?}? Q. {Q?} Prolog implementations use backward chaining Copyright 2009 Addison-Wesley. All rights reserved Subgoal Strategies When goal has more than one subgoal, can use either Depth-first search: find a complete proof for the first subgoal before working on others Breadth-first search: work on all subgoals in parallel Prolog uses depth-first search Can be done with fewer computer resources Copyright 2009 Addison-Wesley. All rights reserved

8 Backtracking With a goal with multiple subgoals, if fail to show truth of one of subgoals, reconsider previous subgoal to find an alternative solution: backtracking Begin search where previous search left off Can take lots of time and space because may find all possible proofs to every subgoal Copyright 2009 Addison-Wesley. All rights reserved Simple Arithmetic Prolog supports integer variables and integer arithmetic is operator: takes an arithmetic expression as right operand and variable as left operand A is B / 17 + C Not the same as an assignment statement! Copyright 2009 Addison-Wesley. All rights reserved

9 Prolog Example cat min.pro min(x, Y, X) :- X < Y. min(x, Y, Y) :- X >= Y. brb0164@faculty% gprolog GNU Prolog ?- consult( min ). compiling min.pro for byte code... min.pro compiled, 2 lines read bytes written, 17 ms?- min(0, 1, 0). true??- min(17, 45, Minimum). Minimum = 17??- min(65, 3, Minimum). Minimum = 3?- ^D Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc Prolog Example brb0164@faculty% cat factorial.pro factorial(0, 1). factorial(n, Factorial) :- M is N 1, factorial(m, M_Factorial), Factorial is N * M_Factorial. brb0164@faculty% gprolog?- consult( factorial ). compiling factorial.pro for byte code... factorial.pro compiled, 3 lines read bytes written, 58 ms?- factorial(5, Factorial). Factorial = 120??- factorial(10, Factorial). Factorial = ??- ^D Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

10 Example Program speaks(allen, russian). speaks(bob, english). speaks(mary, russian). speaks(mary, english). talkswith(x, Y) :- speaks(x, L), speaks(y, L), X \= Y. This program has four facts and one rule. The rule succeeds for any instantiation of its variables in which all the terms on the right of := are simultaneously true. E.g., this rule succeeds for the instantiation X=allen, Y=mary, and L=russian. For other instantiations, like X=allen and Y=bob, the rule fails Lists A list is a series of terms separated by commas and enclosed in brackets. The empty list is written []. The sentence The giraffe dreams can be written as a list: [the, giraffe, dreams] A don t care entry is signified by _, as in [_, X, Y] A list can also be written in the form: [Head Tail] The functions append joins two lists, and member tests for list membership

11 append Function append([], X, X). append([head Tail], Y, [Head Z]) :- append(tail, Y, Z). This definition says: 1. Appending the empty list to any list (X) returns an unchanged list (X again). 2. If Tail is appended to Y to get Z, then a list one element larger [Head Tail] can be appended to Y to get [Head Z]. Note: The last parameter designates the result of the function. So a variable must be passed as an argument member Function member(x, [X _]). member(x, [_ Y]) :- member(x, Y). The test for membership succeeds if either: 1. X is the head of the list [X _] 2. X is not the head of the list [_ Y], but X is a member of the list Y. Notes: pattern matching governs tests for equality. Don t care entries (_) mark parts of a list that aren t important to the rule

12 More List Functions X is a prefix of Z if there is a list Y that can be appended to X to make Z. That is: prefix(x, Z) :- append(x, Y, Z). Similarly, Y is a suffix of Z if there is a list X to which Y can be appended to make Z. That is: suffix(y, Z) :- append(x, Y, Z). So finding all the prefixes (suffixes) of a list is easy. E.g.:?- prefix(x, [my, dog, has, fleas]). X = []; X = [my]; X = [my, dog]; Prolog Example brb0164@faculty% cat length.pro length([], 0). length([x Xs], Length) :- length(xs, Xs_Length), Length is Xs_Length + 1. brb0164@faculty% gprolog?- consult( length ).?- length([a, b, c, d], Length). Length = 4?- length([a, [b, c]], Length). Length = 2 Yes?- length([], Length). Length = 0?- length([a, [b, c], d, [e, f, [g, h]]], Length). Length = 4 Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

13 Lisp Examples in Prolog cat lispfunctions.pro car([x Xs], X). cdr([x Xs], Xs). cons(x, Xs, [X Xs]). append([], Ys, Ys). append([x Xs], Ys, [X Zs]) :- append(xs, Ys, Zs). Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc Lisp Examples in Prolog gprolog?- consult( lispfunctions ).?- cons(x, [], A). A = [x]?- cons(y, [x], B). B = [y,x] Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

14 Lisp Examples in Prolog?- append([y, x], [z], D). D = [y,x,z]?- append([y, x], [z], D), cdr(d, Cdr_D), car(cdr_d, Cadr_D). Cadr_D = x Cdr_D = [x,z] D = [y,x,z] Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc Database Example in Prolog brb0164@faculty% cat ancestors.pro female(shelley). female(mary). female(lisa). female(joan). mother(mary, jake). mother(mary, shelley). mother(lisa, mary). mother(joan, bill). male(bill). male(jake). male(bob). male(frank). father(bill, jake). father(bill, shelley). father(bob, mary). father(frank, bill). Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

15 Database Example in Prolog parent(father, Child) :- father(father, Child). parent(mother, Child) :- mother(mother, Child). parents(father, Mother, Child) :- father(father, Child), mother (Mother, Child). Copyright Copyright 2006 The 2004 McGraw-Hill Barrett R. Companies, Bryant. Inc Database Example in Prolog gprolog?- consult( ancestors ).?- parents(father, Mother, jake). Father = bill Mother = mary??- parents(bill, mary, Child). Child = jake? ; Child = shelley Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

16 Database Example in Prolog?- parents(father, Mother, Child). Child = jake Father = bill Mother = mary? ; Child = shelley Father = bill Mother = mary? ; Child = mary Father = bob Mother = lisa? ; Child = bill Father = frank Mother = joan Copyright Copyright 2006 The 2007 McGraw-Hill Barrett R. Companies, Bryant. Inc Database Example in Prolog sibling(child1, Child2) :- father(father, Child1), father(father, Child2), mother(mother, Child1), mother(mother, Child2). Copyright Copyright 2006 The 2004 McGraw-Hill Barrett R. Companies, Bryant. Inc

17 Database Example in Prolog?- sibling(child1, Child2). Child1 = jake Child2 = jake? ; Child1 = jake Child2 = shelley? ; Child1 = shelley Child2 = jake? ; Child1 = shelley Child2 = shelley? ; Child1 = mary Child2 = mary? ; Child1 = bill Child2 = bill Copyright Copyright 2006 The 2007 McGraw-Hill Barrett R. Companies, Bryant. Inc Database Example in Prolog true_sibling(child1, Child2) :- sibling(child1, Child2), \+ Child1 = Child2. /* \+ is negation */?- true_sibling(child1, Child2). Child1 = jake Child2 = shelley? ; Child1 = shelley Child2 = jake? ; no Copyright Copyright 2006 The 2007 McGraw-Hill Barrett R. Companies, Bryant. Inc

18 Database Example in Prolog ancestor(ancestor, Descendant) :- parent(ancestor, Descendant). ancestor(ancestor, Descendant) :- parent(descendants_parent, Descendant), ancestor(ancestor, Descendants_Parent).?- ancestor(ancestor, jake). Ancestor = bill? ; Ancestor = mary? ; Ancestor = frank? ; Ancestor = joan? ; Ancestor = bob? ; Ancestor = lisa? ; no Copyright Copyright 2006 The 2007 McGraw-Hill Barrett R. Companies, Bryant. Inc Postfix Example in Prolog brb0164@faculty% cat postfix.pro /* This program converts a postfix expression, represented as a list of operands and operators, into a syntax tree. */ postfix(exp, Tree) :- postfix_stack(exp, [], [Tree]). Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

19 Postfix Example in Prolog postfix_stack([], Tree, Tree). postfix_stack([operator Rest_exp], [Top, Next_top Rest_stack], Tree) :- operator(operator), postfix_stack(rest_exp, [tree(operator, Next_top, Top) Rest_stack], Tree). postfix_stack([operand Rest_exp], Stack, Tree) :- postfix_stack(rest_exp, [Operand Stack], Tree). operator(+). operator(-). operator(*). operator(/). Copyright Copyright 2006 The 2004 McGraw-Hill Barrett R. Companies, Bryant. Inc Postfix Example in Prolog gprolog?- consult( postfix ).?- postfix([a,b,c,*,+],tree). Tree = tree(+,a,tree(*,b,c))??- postfix([a,b,+,c,*,d,/,e,-],tree). Tree = tree(-,tree(/,tree(*,tree(+,a,b),c),d),e)? Copyright Copyright 2006 The 2014 McGraw-Hill Barrett R. Companies, Bryant. Inc

20 Practical Aspects of Prolog Tracing The Cut Negation The is, not, and Other Operators The Assert Function Tracing To see the dynamics of a function call, the trace function can be used. E.g., if we want to trace a call to the following function: factorial(0, 1). factorial(n, Result) :- N > 0, M is N - 1, factorial(m, SubRes), Result is N * SubRes. we can activate trace and then call the function:?- trace(factorial/2).?- factorial(4, X). Note: the argument to trace may include the function s arity

21 Tracing Output?- factorial(4, X). Call: ( 7) factorial(4, _G173) Call: ( 8) factorial(3, _L131) Call: ( 9) factorial(2, _L144) Call: ( 10) factorial(1, _L157) Call: ( 11) factorial(0, _L170) Exit: ( 11) factorial(0, 1) Exit: ( 10) factorial(1, 1) Exit: ( 9) factorial(2, 2) Exit: ( 8) factorial(3, 6) Exit: ( 7) factorial(4, 24) X = 24 These are temporary variables These are levels in the search tree The Cut The cut is an operator (!) inserted on the right-hand side of a rule. semantics: the cut forces those subgoals not to be retried if the right-hand side succeeds once. E.g (bubble sort): bsort(l, S) :- append(u, [A, B V], L), B < A,!, append(u, [B, A V], M), bsort(m, S). bsort(l, L). So this code gives one answer rather than many

22 Bubble Sort Trace?- bsort([5,2,3,1], Ans). Call: ( 7) bsort([5, 2, 3, 1], _G221) Call: ( 8) bsort([2, 5, 3, 1], _G221) Call: ( 12) bsort([1, 2, 3, 5], _G221) Redo: ( 12) bsort([1, 2, 3, 5], _G221) Exit: ( 7) bsort([5, 2, 3, 1], [1, 2, 3, 5]) Ans = [1, 2, 3, 5] ; No Without the cut, this would have given some wrong answers The assert Function The assert function can update the facts and rules of a program dynamically. E.g., if we add the following to the foregoing database program:?- assert(mother(jane, joe)). Then the query:?- mother(jane, X). gives: X = ron ; X = joe; No F

23 Solving Word Problems A simple example: Baker, Cooper, Fletcher, Miller, and Smith live in a five-story building. Baker doesn't live on the 5th floor and Cooper doesn't live on the first. Fletcher doesn't live on the top or the bottom floor, and he is not on a floor adjacent to Smith or Cooper. Miller lives on some floor above Cooper. Who lives on what floors? We can set up the solution as a list of five entries: [floor(_,5), floor(,4), floor(_,3), floor(_,2), floor(_,1)] The don t care entries are placeholders for the five names Modeling the solution We can identify the variables B, C, F, M, and S with the five persons, and the structure floors(floors) as a function whose argument is the list to be solved. Here s the first constraint: member(floor(baker, B), Floors), B\=5 which says that Baker doesn't live on the 5th floor. The other four constraints are coded similarly, leading to the following program:

24 Prolog solution floors([floor(_,5),floor(_,4),floor(_,3),floor(_,2), floor(_,1)]). building(floors) :- floors(floors), member(floor(baker, B), Floors), B \= 5, member(floor(cooper, C), Floors), C \= 1, member(floor(fletcher, F), Floors), F \= 1, F \= 5, member(floor(miller, M), Floors), M > C, member(floor(smith, S), Floors), not(adjacent(s, F)), not(adjacent(f, C)), print_floors(floors) Auxiliary functions Floor adjacency: adjacent(x, Y) :- X =:= Y+1. adjacent(x, Y) :- X =:= Y-1. Note: =:= tests for numerical equality. Displaying the results: print_floors([a B]) :- write(a), nl, print_floors(b). print_floors([]). Note: write is a Prolog function and nl stands for new line. Solving the puzzle is done with the query:?- building(x). which finds an instantiation for X that satisfies all the constraints