Exercises on the Fundamentals of Prolog

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1 1 Introduction Exercises on the Fundamentals of Prolog These exercises are intended to help reinforce material taught in the lectures of CIS335 course in Prolog. They do not contribute any marks to the total for that module, and they are meant to be self-assessed. This work should take around 6 hours, including the self-assessment. No material is to be handed in. The work is not meant to be particularly difficult, but merely to remind you of what we have covered in the lectures don t be surprised if the answers are obvious! On the other hand, don t be downhearted if you find some of the material hard to follow there s plenty of time yet. If you have problems with the material, please feel free to discuss it with fellow students, or with me (g.wiggins@gold.ac.uk). Your labs are based around this material, so you will be able to discuss it there too. Your feedback is welcome, and if you feel the exercises could be made more useful, please let me know how. 2 Prolog Syntax Identify the errors in the following Prolog program syntax. If you are unsure, type your answers into a file to see if they load correctly. NB these are syntax errors, not logical program errors. % Problem 2.1 simple( program ) :-. % Problem 2.2 simple( Program ) :- Small( Program ). % Problem 2.3 simple( program1, X ) :- X = 2, simple( program2, X ) :- X = 1. % Problem 2.4 simple( simple program ). % Problem 2.5 simple( %program% ). % Problem 2.6 6thprogram( program ) :- too many( Errors ). 1

2 % Problem 2.7 simple program :- program(simple) % Problem 2.8 simple program :- program(simple). 3 List Processing The two example predicates append/3 and flatten/2 can be found in examples for practical2.pl on the Web in the course s prolog code directory. 3.1 Joining lists together Here is a program which takes two lists and sticks them end-to-end, to make a third: % append( List1, List2, BothLists ) append( [], List, List ). append( [Head Tail], Rest, [Head All] ) :- append( Tail, Rest, All ). Think about how the program works, and then answer the following questions (in your head not on the computer!). Remember that some queries may return more than one answer work out what all the possibilities are, and what order they appear in. Then check it out on the computer. % Problem 3.1.1?- append( [a], [b], [c] ). % Problem 3.1.2?- append( [a], [b], X ). % Problem 3.1.3?- append( [a], [], X ). % Problem 3.1.4?- append( [a], b, X ). % Problem 3.1.5?- append( [a], X, Y ). % Problem 3.1.6?- append( X, [b,c], Y ). 2

3 % Problem 3.1.7?- append( X, Y, [a,b,c] ). 3.2 Unifying lists What answers (including any substitutions) are produced by the following queries? % Problem 3.2.1?- [A B] = [a,a]. % Problem 3.2.2?- [A,b,C D] = [d,c,b,a]. % Problem 3.2.3?- [a,[a []]] = X. % Problem 3.2.4?- [A,g,d,A] = [x,d,g,z]. 3.3 Building an answer Here is a program with a programmer error in it. It is an error that appears commonly in code writen by beginning Prolog students. The program is supposed to take a list of pairs of items (eg pair( X, Y )) and flatten them out into a list of items. So the list should come out as [pair( a, b ),pair( c, d )] [a,b,c,d] The programmer types in the code, runs it with the query shown, and the answer is no. Why? (Hint: think about how unification works how many values can a variable be unified with at once?) 3

4 % Problem % append( List1, List2, BothLists ) append( [], List, List ). append( [Head Tail], Rest, [Head All] ) :- append( Tail, Rest, All ). % flatten( ListOfPairs, List ). flatten( [], [] ). % deal with empty lists flatten( [pair( A, B ) Pairs], Ans ) :- flatten( Pairs, Ans ), append( [A, B], Ans, Ans ). % And here s the query?- flatten( [pair( a, b ), pair( c, d )], Answer ). How would you correct the code? There is one choice of input for which the incorrect program would produce the correct answer. What is it? How many times will it be produced? To check, try running the program with the query?- flatten( A, B ). (Note: look carefully at how this program uses the list to control recursion. The two clauses represent two mutually exclusive cases where the input is an empty list and where it is a nonempty list.) 4 Unification This, the main section of the practical, is aimed understanding unification. The idea is to build a basic unification algorithm. This is really quite hard, so do not be dismayed if you find it tricky. Feel free to ask for help from me, the lab demonstrator or your fellow students. Those who don t have problems remember that explaining things to others is a valuable part of the learning process. We are going to use a special representation for variables, so that we don t have to worry about what is a Prolog variable and what is one of our own variables. So we use in an operator, +, to mark a variable, and we make its name a Prolog constant, like this: +variable note the lower case v. We will also use a special notation for terms, like this: functor-[term1,term2,...] so a term with no arguments is written noargs-[]; and we will restrict ourselves to symbols with only lower case letters in them. Choosing these notations will make life very much simpler. We want a predicate called unify/3 which will take two terms, and try to unify them. If it succeeds, the third argument will be a list of binary unifiers, which are pairs like this: u( +a, v-[] ) 4

5 which (in this example) means that variable a has been unified with term v. You can find a file called unification partial.pl on the CIS335 module web page. It contains some of the code needed for this assignment. You are required to add in the parts which are missing; to do this, you will need to understand all the code which is present. Missing parts are marked /* MISSING N */ where N is a number between 1 and 10 inclusive; fill in the missing parts in order of the numbers. Make up around 10 goals to test your program, deciding in advance what the answers should be. Remember to check that you don t get too many answers, and that non-unifiable terms don t unify! Here are some example questions and answers:?- unify( f-[+x,+y], f-[+y,+x], B ). B = [u(+(x),+(y))]??- unify( f-[g-[h-[],+y],+x], f-[+x,g-[+a,+a]], B ). B = [u(+(y),+(a)),u(+(a),h-[]),u(+(x),g-[h-[],+(y)])]??- unify( f-[], +x, B ). B = [u(+(x),f-[])]??- unify( f-[+x,+y], f-[g-[],+x], B ). B = [u(+(y),+(x)),u(+(x),g-[])]??- unify( f1-[], f2-[], B ). no?- Feel free to queries to me. 5

Advanced Prolog Programming

1 Introduction CIS335, Assignment 4 Advanced Prolog Programming Geraint Wiggins November 12, 2004 This practical is the second self-assessed exercise for MSc students on the Prolog module. It is intended