1 COMP313A Programming Languages Logic Programming (6)

1

COMP313A Programming Languages

Logic Programming (6)

2

Some More Prolog

• structures

3

I/O in Prolog

• Files associated with input and output streams• I/O from terminal – user • In some Prologs only two files/streams are

active at any one time– current input stream– current output stream

• see switches input streams• tell switches output streams• In prolog I/O is a side effect of the predicate

4

Tkeclipse Prolog• stdin, stdout

– standard input and output streams

?- write("fred").Yes (0.00s cpu)

• open a stream for input or output– open(SourceSink, Mode, Stream)

? open('//C/Documents and Settings/mjeff/My Documents/prolog/fred.txt', read, Fred), read_string(Fred, "\r\n ", 10, X).

Fred = 31X = "freddy"Yes (0.00s cpu)

5

Bits and pieces

• Anonymous variables

family(person(tom, fox, date(7,may, 1950), employed),

person(ann, fox, date(9, may, 1951), employed),

[person(pat, fox, date(5, may, 1973), unemployed),

person(jim, fox, date(5, may, 1973), unemployed)]).

husband(X) :- family(X,Y, Z).

Y and Z are singleton variables

husband(X) :- family(X, _, _)

6

More bits and pieces

• Comparison operators– X > Y– X < Y– X >= Y– X =< Y– X = Y 1 + A = B + 2 1 + A = 2 + B– X \= Y– X =:= Y 1 + 2 =:= 2 + 1 1 + A =:=

B + 2– X =\= Y

7

Structuresa family database

family(person(tom, fox, date(7,may, 1950), employed), person(ann, armstrong, date(9, may, 1951), employed), [person(pat, fox, date(5, may, 1973, unemployed), person(jim, fox, date(5, may, 1973), unemployed)]).

family(_,_,_). % any familyfamily(person(_, fox, _, _), _, _). % the fox

familyfamily(_, person(_,armstrong, _, _), _) % or the armstrong family

family(_, _, [_, _, _]). % any three child family

family(_,person(Name, Surname, _, _), [_, _, _, | _]). % all married women

who have at least three children

8

husband(X) :- family(X, _, _).wife(X) :- f amily(_,X,_).

?- husband(X).X = person(tom, fox, date(7, may, 1950), employed)Yes (0.00s cpu)

?- wife(X).X = person(ann, armstrong, date(9, may, 1951), employed)Yes (0.00s cpu)

Problem – this may be too simplistic

9

husband2(X,Y) :- family(person(X,Y,_,_), _, _).

married(X,Y) :- family(person(X, _, _, _), person(Y,_, _, _),_).married(X,Y) :- married(Y,X).

child(X) :- family(_, _, Children), member(X, Children).

?? child(X)??

10

child(X) :- family(_, _, Children), mem_children (X, Children).

mem_children (X, [person(X, _, _, _) | Tail]).

mem_children (X, [person(Y, _, _, _) | Tail]) :- mem_children1(X,

Tail).

but

?? child(pat)

11

Selectors

husband(family(Husband, _, _), Husband).

wife(family(_, Wife, _), Wife).

children(family(_, _, Childlist), Childlist).

Define relation which allow us to access particular components of a structure without knowing the detailsthe structure

This is data abstraction

These selectors are useful when we want to create instances of families, for example

12

Selectors…

firstchild(Family, First) :- children(Family, [First | _]).

secondchild(Family, Second) :- children(Family, [_, Second | _]).

nthchild(N, Family, Child) :- children(Family, ChildList),

nth_member(ChildList, N, Child).

firstname(person(Name, _, _,_), Name).

surname(person(_, Surname, _, _), Surname).

13

nth_member([X|_], 1, X).nth_member([_| L], N, Child) :- N1 is N-1, nth_member(L, N1, Child).

14

Using them..

• Tom Fox and Jim Fox belong to the same family and Jim is the second child of Tom

firstname(Person1, tom), surname(Person1, fox), % person1 is Tom Fox

firstname(Person2, jim), surname(Person2, fox),%person2 is Jim Fox

husband(Family, Person1), secondchild(Family, Person2).

Person1 = person(tom, fox, _, _)Person2 = person(jim, fox, _, _)Family = family(person(tom, fox, _, _), _, [_, person(jim, fox, _,_) |

_])

15

Logic Puzzles

• Use the following clues to write a prolog program to determine the movies the robots tinman, r2d2, johnny five, and a dalek starred in.– Neither Tinman nor Johnny five were one of

the daleks in Dr Who and the Daleks– The movie Short Circuit did not star Tinman.– R2d2 wowed the audiences in Star Wars.– A dalek did not star in the Wizard of Oz.

16

Structure is important

• Solution(L) :-

We want a binding for L which will contain the result we are after

• What is the result we want?

17

L = [movie(X1, wizardofoz), movie(X2, drwhoandtheDaleks), movie(X3, starwars), movie(X4, shortcircuit)],

Now we have to supply a mechanism for instantiating X1..X4

We need a way of selecting a value and then checking it against some constraints

18


19


Robotslist = [tinman, dalek, r2d2, johnnyfive],

We will draw the values for X1..X2, from Robotslist

We do this using the member predicate

20


Robotslist = [tinman, dalek, r2d2, johnnyfive],

21


Robotslist = [tinman, dalek, r2d2, johnnyfive], member(X1, Robotslist), X1 \= dalek, member(X2, Robotslist), X2 \= tinman, X2 \= johnnyfive, member(X3, Robotslist), X3 = r2d2, member(X4, Robotslist), X4 \= tinman,

X1 \= dalek,

There are just two more things left to do

22


Robotslist = [tinman, dalek, r2d2, johnnyfive], member(X1, Robotslist), X1 \= dalek, member(X2, Robotslist), X2 \= tinman, X2 \= johnnyfive, member(X3, Robotslist), X3 = r2d2, member(X4, Robotslist), X4 \= tinman, X2 \= X1, X2 \= X3, X2 \= X4, X3 \= X1, X3 \= X4, X4 \= X1, print_movies(L).

solution(L):-

23

print_movies([A|B]) :- !, write(A), nl, print_movies(B).print_movies([]).