CS 403 - Programming Languages

CS 403 - Programming Languages Class 21 November 9, 2000

Today’s Agenda • Finish Chapter 15 • Assignment: • Read chapter 15 for today • Announcement: • Programming assignment.

Objectives • Introduction to using Prolog

Length of Lists I • The Problem: Define a relation llength(L,N) to mean that the length of the list L is N. • The Program: • llength([],0). • llength([X|Z],N):- llength(Z,M), N is M + 1.

Watch it work: • ?- [length]. • ?- llength([a,b,c,d],M). • M=4

Length of Lists II • The Program: • llength([],0). • llength([X|Z],N) :- N is M + 1, llength(Z,M).

Watch it work: • ?- [length2]. • ?- llength([a,b,c,d],M). • evaluation_error: [goal(_1629 is _1805+1),argument_index(2)]

Control in Prolog I • How Prolog tries to solve a query like: • <query1>, <query2>, .... , <queryN> • This is the control side of the equation: • Algorithm=Logic+Control • Step 1: Find Things that solve <query1>, if none then fail else goto Step 2 • Step 2: Do the things found from the previous step allow more things to be found that solve <query2>? If not then go back to step1 else goto step 3.......

Control in Prolog I • Prolog tries to solve the clauses from left to right If there is a database file around it will use it 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 II • 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; • SUCCEED

Control in Prolog II • Note 1: Thus the order of the queries is of paramount importance . • Note 2: The general paradigm in Prolog is Guess then Verify: Queries with the fewest solutions should come first, followed by those that filter or verify these few solutions

Binary Search Trees I • An example of user defined data structures. • The Problem: Recall that a binary search tree (with integer labels) is either : 1. the empty tree empty,or 2. a node labeled with an integer N, that has a left subtree and a right subtree, each of which is a binary search tree such that the nodes in the left subtree are labeled by integers strictly smaller than N, while those in the right subtree are strictly greater than N.

Data Types in Prolog • The primitive data types in prolog can be combined via structures,to form complex datatypes: • <structure>::= <functor>(<arg1>,<arg2>,...) • Example In the case of binary search trees we have: • <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))) • Draw this tree

Binary Search Trees II • The Problem: Define a unary predicate isbstree which is true only of those trees that are binary search trees . • The Program isbstree(empty). isbstree(node(N,L,R)):- number(N),isbstree(L),isbstree(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).

Watch it work: • ?- [btree]. • ?- isbstree(node(9,node(3,empty,empty),empty)). • true ? • yes

Binary Search Trees III • The Problem: 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. • Now you try it • The Program mymember(K,node(K,_,_)). mymember(K,node(N,S,_)) :- K < N,mymember(K,S). mymember(K,node(N,_,T)) :- K > N,mymember(K,T).

Watch it work: • ?- [btree]. • ?- [mymember]. • ?- mymember(3, node(10,node(9,node(3,empty,empty),empty), node(12,empty,empty))). • true ? • yes

Unification • Unification is a more general form of pattern matching. In that pattern variables can appear in both the pattern and the target. • The following summarizes how unification works: 1. a variable and any term unify 2. two atomic terms unify only if they are identical 3. two complex terms unify if they have the same functor and their arguments unify .

Prolog Search Trees Summary • 1. Goal Order affects solutions • 2. Rule Order affects Solutions • 3. Gaps in Goals can creep in • 4. More advanced Prolog programming manipulates the searching

Sublists (Goal Order) • Two definitions of S being a sublist of Z use: 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). Version 3 sublist3(S,Z) :- mysuffix(Y,Z), myprefix(S,Y).

Watch them work: | ?- [sublist]. consulting....sublist.pl yes | ?- 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: • Goal will generate an infinite number of guesses. • [e] [_,e] [_,_,e] [_,_,_,e] [_,_,_,_,e] [_,_,_,_,_,e] • .... • None of which pass the verify goal, so we never terminate

CS 403 - Programming Languages