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PrologProlog

CSE2303 Formal Methods I

Lecture 20

OverviewOverview

• Prolog

• Database

• List

• Arithmetic

• Implementing NFA

• Towers of Hanoi

• SWI-Prolog

PrologProlog

• PROgramming in LOGic• Used for solving problems between objects

and the relationships between objects.• A logic program is a finite set of program

clauses.• Programs clauses are:

– Facts.– Rules.

Parent RelationshipParent Relationship

• We represent the statement:

“bob is a parent of pat”

as

parent(bob, pat).

• The compound statement

“If X is a parent of Z, then X is a predecessor of Z”

is represented as:

predecessor(X, Z) :- parent(X,Z).

databasedatabaseparent(bob, ann).

parent(bob, pat).

parent(pat, jim).

predecessor(X, Z) :-

parent(X,Z).

predecessor(X, Z) :-

parent(X,Y),

predecessor(Y, Z).

Facts

Rules

Constants

Variables

Using Prolog Using Prolog

• To read in the file database. Type: [database].• To ask if bob is the parent of pat. Type: parent(bob, pat).• To ask who is the predecessors of jim. Type: predecessor(X, jim).• To quit. Type: ctrl-D

Example SessionExample Session?- [database].

?- parent(bob, pat).

yes

?- parent(bob, jim).

no

?- predecessor(X, jim).

X = pat ;

X = bob ;

no

no Nothing matches

no

prompt

Ask for another solution

Program ClausesProgram Clauses

• Rule– The head and the body are nonempty.– The body is the conditional part.– The head is the conclusion.

• Fact– The body is empty, and is written as:

A.

A :- B1, … , Bn.

Head BodyEnd of clause marker“ if ”

InterpretationsInterpretations

• Logic

X1,..Xm ((B1 … Bn) A)

• Procedural

– To execute A, first execute B1, then execute B2,...

• Process

– A, B1, … , Bn are considered processes.

– Shared variables are communication channels

A :- B1, … , Bn.

SyntaxSyntax• Constants

– Names: which always begin with a lowercase letter. likes mary parent predecessor– Numbers: 0 -23 3.4585 14092

• Variables– Always begin with either an capital letter or an

underline character. X Answer _ _Input

• Structures– Compound terms. owns(john, book) parent(X, parent(Y, jim))

ListsLists

• A list t1,…, tn can be represented as:

[t1,…, tn]

• For example:[a,b,c,d]

• Also represented as:[a | b,c,d]

• The empty list is [ ]

Head of the list

Tail of the list

Membership ProblemMembership Problem

Write a Prolog program which finds the members of a list.

Membership RelationshipsMembership Relationships

• X is always a member of a list whose head is X.

member(X, [X | T]).

• If X is a member of a list T, then X is a member of a list whose tail is T.

member(X, [H | T]) :-

member(X, T).

MemberMember

member(X, [X | T]).member(X, [H | T]) :- member(X, T).

?- member(john, [paul, john]).yes?- member(X, [paul, john]).X = paul ;X = john ;no

ArithmeticArithmetic

• Operators– X + Y, -X, X-Y, X*Y, X/Y, …

• Functions– abs(X), max(X, Y), sin(X), …

• Relations– X < Y, X > Y, X =< Y, X =:= Y, X =\= Y, ..– is “Evaluates arithmetic expression”

?- X is 2*3 + 4.X = 10yes

Factorial RelationshipsFactorial Relationships

• The factorial of 0 is 1.

factorial(0,F) :- F is 1.

• If N > 0, and N1 is N – 1, and the factorial of N1 is F1, and F is N*F1, then the factorial of N is F.

factorial(N, F) :- N > 0, N1 is N-1,

factorial(N1, F1), F is N*F1.

FactorialFactorial

factorial(0, F) :- F is 1.

factorial(N, F) :- N > 0, N1 is N-1,

factorial(N1, F1), F is N*F1.

?- factorial(3, F).

F = 6

yes

Size RelationshipsSize Relationships

• The size of an empty list is 0.

size([ ], 0).

• If the size of the list T is N1 and N is N1+1, then the size of the list with tail T is N.

size([H | T], N) :-

size(T, N1), N is N1+1.

SizeSizesize([ ], 0).size([H | T], N) :- size(T, N1), N is N1+1.

?- size([a,b,c], N).N = 3yes

Note: There is a predefined funtion in SWI-Prolog called length

accept(W) :- start(S), path(S, W).path(S, [ ]) :- final(S).path(S, [H | T]) :- arc(S, H, N), path(N, T).start(1).final(3).arc(1, a, 1).arc(1, b, 2).arc(1, b, 3).arc(2, b, 3).arc(3, a, 3).

a

a

b

b

b-

+

1

2

3

?- [nfa].

?- accept([a, b, a]).

yes

NFANFA

Towers of HanoiTowers of Hanoi

• The object is to move the disks, one at a time, from the left peg to the right peg.

• You are allowed to use the middle peg.

• At no stage are you allowed to place a bigger disk on top of a smaller one.

SolutionSolution

move(1, X, Y, Z) :- write(’ Move top disk from ’), write(X), write(’ to ’), write(Z), nl.move(N, X, Y, Z) :- N > 1, M is N-1, move(M, X, Z, Y), move(1, X, Y, Z), move(M, Y, X, Z).

?- move(3, left, middle, right).

More InformationMore Information

• Courseware web site.

• Links to:– Prolog tutorial, SWI-Prolog, etc.

• Books:– “The Art of Prolog: Advanced Programming

Techniques”, by L.Sterling and E. Shapiro.– “Prolog Programming for Artificial Intelligence”,

by I. Bratko.– “Programming in Prolog”, by W. Clocksin and D.

Mellish