Principles of Programming Languages, Spring 2016 Assignment 5 Logic Programming

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1 Principles of Programming Languages, Spring 2016 Assignment 5 Logic Programming Submission instructions: a. Submit an archive file named id1_id2.zip where id1 and id2 are the IDs of the students responsible for the submission (or id1.zip for one student in the group). b. A contract must be included for every implemented procedure (Signature, Purpose, possibly Examples). E.g.: % Signature: member(x,l) % Purpose: holds when item X occurs within list L. % Examples: % member(2, [1, 2, 3]) true % member(x, [1,2,3]) X = 1 c. Use exact procedure and file names, as your code will be tested automatically. d. Answer theoretical questions in ex5.pdf. You can scan hand-drawn trees and graphs and add them to the submitted work. e. Answer coding questions in their specified files. f. Your implementation will be held correct if it returns answers in any order given a query, as long as all the correct answers are returned, and only the correct answers are returned, and infinite computations are avoided. Question 1: Comprehension Questions Answer in file ex5.pdf. a. Specify two differences and a similarity between composite terms and atomic formulas in LP. b. According to the answer-query algorithm given in class, is every proof tree in Relational Logic Programming finite? Explain. c. Suggest how the answer-query algorithm can be modified, so that it always finishes the computation. That is, infinite branches are cropped at some point in which it is obvious that the branch is infinite. You can assume that there exists a given value N(P, Q), which equals the number of all atomic formulas possible, with respect to a given program P and a given query Q, in RLP. List the changes that should be done to the answer-query algorithm given in the lecture notes (p.348, repeated here for convenience). Tree operations: make_node(label): Creates a node labeled label. add_branch(tree, edge_label, branch): Adds branch as a right branch to the root node of tree, with edge labeled by edge_label. label(node) selects the label of node. -1-

2 answer-query algorithm: Signature: answer-query(q,p,gsel,rsel) Input: A query: Q =?- G1,..., Gn., where Gi, i = 1..n is an atomic formula A program P Output: A set of substitutions for variables of Q (not necessarily for all variables). Method: 1. answers = {} 2. PT := proof-tree(make_node(q)). 3. Return answers/q the restriction of the substitutions in answers to the variables of Q. proof-tree: Signature: proof-tree(node). (node is a tree node with label query edges are labelled with substitutions) Input: A tree node node Output: A proof tree rooted in node. Method: if label(node) is?- true,..., true. then 1. Mark node as a Success node (leaf). 2. answers = answers {s1 s2... sn}, where s1,..., sn are the substitution labels of the path from the tree-root to node. Mark node with this substitution. else 1. Goal selection: G = Gsel(label(node)) 2. Rename variables in every rule and fact of P 3. Rule selection: rules = Rsel(G, P) = ( R, σ ) R P, G σ=head(r) σ 4. apply-rules(node, G, rules) d. Define and give examples for a program with (1) success finite proof tree, (2) success infinite proof tree, (3) failure proof tree, and (4) infinite failure proof tree. Question 2. Unify Answer in file ex5.pdf. a. What is the result of the following operations? Provide the computation steps according to the algorithm presented in the Course book p.345. Explain in case of failure. 1. unify[n(d(d),d,d,k,n(n),k),n(d(d),d,d,k,n(n),d)] 2. unify[k(n,d,k,k,n,k,a(d)),k(n,a(n),k,a(d),a(k),k,a(d))] 3. unify[d(k,d),k(k,d)] 4. unify[p(.(.(a,d),k)),p(.(a,.(d,k)))] 5. unify[p([[a D] K]),p([a [D K]])] b. In chapter 5 we used unification of type expressions and of typing statements, in order to infer a type for a language expression. For example, we computed (in Ex.3, question 5), the mgu of the expressions: {f:[t1 -> [T2 -> T1]], x:t1} - (f x):t1 {f:t3, x:t4} - (f x):[t1 -> T2] Compare the two kinds of unification processes. Suggest two differences and two similarities. -2-

3 Question 3: Relational Logic Programming (with SQL operations) Answer in file ex5.pl. Answer the question using SQL operations, as defined in Section in the book. The procedure not_member, which is provided in file ex5-aux.pl can be used. Relational databases are useful for managing large amounts of data. Each table in the database represents a relation. Elementary query operations, such as select, project, join and Cartesian-product enable data access. A relational database for Book-Library management is given in the file ex5-aux.pl. The database consists of four tables, represented as fact-based procedures: 1) The table book(id, Name, Author, Category, Local)/5 contains information about books available in the library. Each book has an id (that identifies it uniquely), a name, a single (for simplicity) author id (from the author table), a category (fiction, children, and so on), and boolean that defines whether this book is written locally or abroad. 2) The table editor(id, Name)/2 is the editors list. 3) The table author(id, Name)/2 is the authors list. 4) The table editorial(book, Editor)/2 contains an entry with two ids for each editorial role. a. Write a procedure local_children_book_authoring(book, Author)/2 that defines the relation between a book name and its author name, for local children books. Which SQL operations are performed by this procedure? Mark in the contract. b. Write a procedure busy_editor(name)/1 that identifies editors that edit at least two books. Which SQL operations are performed by this procedure? Mark in the contract. c. Write a procedure people(book, List)/2 that defines the relation between a book name and a list of names of all people participated in the book preparation - author and editors (in any order). This question involves lists (non-rlp elements). Question 4: Lists management. Answer in file ex5.pl. a. sub procedure Write a procedure sub(sublist, List)/2 that defines the following relation between two lists: % Signature: sub(sublist, List)/2 % Purpose: All elements in Sublist appear in List in the same order. % Precondition: List is fully instantiated (queries do not include variables in their first argument). % Example: %?- sub(x, [1, 2, 3]). % X = [1, 2, 3]; % X = [1, 2]; % X = [1, 3]; % X = [2, 3]; -3-

4 % X = [1]; % X = [2]; % X = [3]; % X = []; % false The following Scheme predicate implements one of the modes of this relation: ; Signature: sub?(sub, long) ; Type: [List(T) * List(T) -> Boolean] ; Purpose: Determine whether we can obtain sub by only removing items from long. ; Pre-conditions: - ; Examples: ; (sub? empty empty) -> true ; (sub? empty '(1)) -> true ; (sub? '(1) empty) -> false ; (sub? '(1) '(1 2 3)) -> true ; (sub? '(1 3) '(1 2 3)) -> true ; (sub? '(3 1) '(1 2 3)) -> false ; (sub? '(4) '(1 2 3)) -> false (define sub? (lambda (sub long) (cond ((empty? sub) #t) ((empty? long) #f) (else (let ((suffix (member (car sub) long))) (if suffix (sub? (cdr sub) (cdr suffix)) #f)))))) The following Scheme procedure computes another mode of the same relation: ; Signature: all-subs(long) ; Type: [List(T) -> List(List(T))] ; Purpose: compute all lists that can be obtained from long by removing items from it. ; Pre-conditions: - ; Tests: ; (all-subs '(1 2 3)) -> ; '(() (3) (2) (2 3) (1) (1 3) (1 2) (1 2 3)) (define all-subs (lambda (long) (if (empty? long) (list empty)) (else (let ((all-rests (all-subs (cdr long)))) (append all-rests (map (lambda (sub) (cons first sub)) all-rests)))))))) b. splitlist procedure Recall the Scheme interpreter for LP. The file auxiliary.rkt has a function split-list: ; Signature: split-list(l, elt) ; Type: [List(T) * T -> Pair(List(T), List(T)) -4-

5 ; Purpose: split a list of items into a pair (prefix. suffix) s.t. (prefix elt suffix) = l ; Pre-conditions: (member elt l) ; Tests: (split '( ) 3) ==> ((1 2) 4 5) (define split-list (lambda (l elt) (letrec ((iter (lambda (prefix suffix) (if (equal? (car suffix) elt) (cons prefix (cdr suffix)) (iter (append prefix (list (car suffix))) (cdr suffix)))))) (iter empty l)))) Implement a predicate splitlist(l, Elt, LeftList, RightList)/4 in Prolog. Question 5: Grammars Answer in file ex5.pl. a. Compound Nouns English Form The following context-free grammar describes a very partial yet infinite subset of English. S is the start nonterminal (variable). Other non-terminals represent syntax components such as nouns, verbs, prepositions, adjectives and determiners. S --> Np Vp Np --> Det Adjs N Vp --> V Np Pp Adjs --> ε Adj Adjs Pp --> ε P Np Adj --> beautiful funny tall big Det --> a the N --> cat dog mouse house table V --> saw ate P --> in from on with The file ex5.pl includes a program that implements the above grammar. Read the contract and run the examples. Extend the grammar above to support the form of compound nouns in English. Compound nouns are formed by a sequence of nouns for example: Science book, tooth paste, bus stop, school bus, school bus stop Modify the derivation rules accordingly you will need to add a new nonterminal. Implement the change in the Prolog program. b. subcfg procedure Write a procedure subcfg(subtext, Text)/2 that defines the relation between a sentence Text and a sequence of words SubText, such that SubText can contain only words from a sentence Text (but -5-

6 not all of them), exclusively in that order. Both SubText and Text must fit the new grammar, and Text is fully instantiated. For example: subcfg([a, mouse, ate, a, table], [a, beautiful, mouse, dog, ate, a, cat, table, in, the, house]) true subcfg(x, [a, beautiful, mouse, dog, ate, a, cat, table, in, the, house]) X = [a, dog, ate, a, house]; % more outputs Question 6: Answer-query algorithm, Answer in ex5.pdf file. Church numbers (numerals) provide symbolic representation for the natural numbers. They are defined inductively: 1. zero is a Church number, 2. For a Church number N, s(n) is a Church number. The following program defines addition: ; Signature: natural_number(n)/1 Purpose: N is a natural number. natural_number(zero). %1 natural_number(s(x)) :- natural_number(x). %2 Signature: plus(x, Y, Z)/3 Purpose: Z is the sum of X and Y. plus(x, zero, X) :- natural_number(x). %1 plus(x, s(y), s(z)) :- plus(x, Y, Z). %2 a. Draw the proof tree for the query below and the given program. For success leaves, compute the substitution composition and report the answer.?- plus(s(x), s(x), s(s(s(zero)))). b. What are the answers of the answer-query algorithm for this query? c. Is this a success or a failure proof tree? Explain. d. Is this tree finite or infinite? Explain. Question 7: Scheme LP interpreter This question refers to the LP-interpreter that is presented in the course site. The architecture of the interpreter is also given in the course site, and a general description was given in the practice session. a. The substitution-adt.rkt module Answer in ex5.pdf file. Two main operations of substitutions are substitution-combination, and substitution-application to expressions in the language. The substitution-combination operation is implemented by the sub-combine procedure, -6-

7 while substitution-application is implemented by four procedures: sub-apply, sub-apply-atomicformula, sub-apply-query, sub-apply-rule. Explain why substitution-application is split into four procedures, and describe the dependencies between these four procedures you can draw a graph, in which a directed arrow describes dependency (as seen in practices about scheme evaluators). b. Gsel procedure Answer in folder interpreter_gsel. Modify the Gsel procedure to select the goal which is not true and has the minimal number of variables, instead of the first non-true goal on the left. Examples: > (Gsel (make-query (list '(member (var Y) (cons (var X) empty)) '(member (var X) (cons 2 empty)) '(member 1 (cons 2 empty))))) (member 1 (cons 2 empty)) c. Rule tracing Answer in folder interpreter_trace. 1. Nodes of the proof-tree that are constructed by answer-query carry a query and a substitution that combines all the substitutions on the edges of the proof tree. This way, the substitution in a success leaf is already an answer to the query (after restriction to the query variables). In order to trace the rules that lead to the answer, we can extend a PT-node to carry also the list of rules that lead to this node (from the tree root). Extend PT-node to include also a list of rules, where a rule is represented by its predicate and number, as in: '( (apply 3) Rule 0). In the extended version, a PY-node is a triplet: <query, substitution, Rule-list> The new kind of node will be called PT-node-rules, and has the operations: Constructor: make-pt-node-rules/3 Getters: PT-node-rules->query/1, PT-node-rules->sub/1, PT-node-rules->rules/1 2. Write a procedure LP-node-expander-rules/2, which is a variant of LP-node-expander, that expands a PT-node-rules in a way that the list of rules in child nodes is the list of rules of the parent node, extended with the rule, whose application yields the child node. Signature: LP-node-expander(PT-node, program) Type: [PT-Node-rules * Program -> List(PT-Node-rules)] The following drawing explains the operation of LP-node-expander-rules/2: < Q, s, (V1,, Vn) > s1, R1 sm, Rm < Q, s s1, (V1,, Vn,R1) > < Q, s sm, (V1,, Vn, Rm) > -7-

8 3. Write a procedure answer-query-trace which extends answer-query to return pairs of a substitution and the list of rules that create it. Signature: answer-query(query, program) Type: [Query * Program -> List(Record (Sub,List(Rule)) )] where Record(T1,, Tn) stands for a list of n elements of types T1,, Tn, respectively. Example: > (answer-query-trace (make-query '(append (cons 1 empty) (cons 2 empty) Z))) program) '({Z = (cons 1 (cons 2 empty))} ((append/3 Rule 1). ) -8-

Principles of Programming Languages

Principles of Programming Languages Lesson Collaboration 22 An Introduction and Management to Logic Programming Dana Fisman Logic Programming ( Pure Prolog ) www.cs.bgu.ac.il/~ppl172 1 Review of Last Lecture