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COMP313A ProgrammingLanguages

Logic Programming (2)

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Lecture Outline

Horn Clauses

Resolution Some Prolog

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A little bit of Prolog

Objects and relations between objects

Facts and rules

parent(pam, bob). parent(tom,bob).parent(tom, liz). parent(bob, ann).

parent(bob, pat). parent(pat, jim).

? parent(bob, pat).

? parent(bob, liz).

? parent(bob, ben).

? parent(bob, X).

? parent(X, Y).

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Prolog

grandparent (X,Y) :- parent(X, Z), parent(Z, Y).

For all X and Y

X is the grandparent of Y if

X is a parent of Z and

Z is a parent of Y

sister (X,Y) :- parent(Z, X), parent(Z, Y), female(X)

For all X and YX is the sister of Y if

Z is the parent of both X and Y and

X is a female

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Horn Clauses

A clause is either a single predicate called a fact

or

a rule of the formconditionconclusion

Conclusion is a single predicate

Condition is a conjunction of predicates

a1 a2 a3 an

parent(X,Z) parent(Z,Y) grandparent(X,Y)

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Horn Clauses

Horn clauses dont have any quantifiers

Assumes that the variables in the head areuniversally quantified and the variables in the

body are existentially quantified (if they dont

appear in the head)

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Horn Clauses

grandparent(X,Y) parent(X,Z) parent(Z,Y)

grandparent(X,Y) :-parent(X,Z),parent(Z,Y)

Equivalent to

"x : "y : $z parent(X,Z) parent(Z,Y) grandparent(X,Y)

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Horn clauses continued

A fact

mammal(human) A query or goal

mammal(human)

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Horn clause example

" x mammal(x) legs(x,2) legs(x,4)

mammal(x) legs(x,2) legs(x,4)

mammal(x) legs(x, 4) legs(x, 2)

Prolog

legs(x,4) :- mammal(x), not legs(x,2)

legs(x,2) :- mammal(x), not legs(x,4)

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legs(x,4) mammal(x) legs(x,2).

legs(x,2) mammal(x) legs(x,4).

mammal(human).

mammal(horse).

legs(horse, 2).

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Resolution

Inference rule for Horn clauses

Combine left and right hand sides of two horn

clauses Cancel those statements which match on

both sides

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Resolution example

Example 1Fact: mammal(human).

Query: mammal(human).

Resolution:

mammal(human).

mammal(human) mammal(human).

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Resolution example

Example 1Facts: see slide 10

Query: legs(horse,4).

Resolution: legs(horse,4).

legs(horse,4) legs(horse,4),

mammal(horse),

legs(horse, 2).

mammal(horse),

legs(horse, 2).

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Turn the following sentences into formulae in

first order predicate logic

John likes all kinds of food

Apples are food

Chicken is food Anything anyone eats and isnt killed by is food

Bill eats peanuts and is still alive

Sue eats everything Bill eats

Prove that John likes peanuts using backwardchaining

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"x : food(x) likes(John,x)

food(Apples)

food(Chicken)

"x :"x :eats(x,y) killed_by(x,y) food(x)eats(Bill, Peanuts)

alive (Bill)

"x:eats(Bill, x) eats(Sue,x)

We have no ors. Drop the qualifiers to get the horn

clauses

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1. likes(John,x) food(x).

2. food(Apples).

3. food(Chicken).

4. food(x) eats(x,y) killed_by(x,y).

5. eats(Bill, Peanuts).

6. alive (Bill).

7. eats(Sue,x) eats(Bill, x).

8. alive(x,y) killed_by(x,y).

9. killed_by(x,y) alive(x,y).

Proves that John likes Peanuts using resolution

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Horn clause example

The Euclidian algorithm to compute the gcdof two positive integers, u and v:

The algorithm

The gcd ofu and 0is u

The gcd of u and v, ifvis not 0, is the same asthe gcd ofvand the remainder of dividing vinto u.

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Horn clause example

greatest common divisor

"u gcd(u, 0, u)

" u, " v, " w,

zero(v) gcd(v, u mod v, w) gcd(u,v,w)

gcd(u,0,u)

gcd(u,v,w) zero(v) gcd(v, u mod v, w)

What is the greatest common denominator of 15, and 10. i.e. find

and instantiation for X ingcd(15,10,X)

What value of X makes gcd(15,10,X) True

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Lecture Outline

Some Prolog Unification

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Some more prolog

Recursive rules

example predecessor relation

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

And so on

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

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Some more prolog

parent(pam,bob).

parent(tom,bob).

parent(tom,liz).

parent(bob, ann).

parent(bob, pat).

parent(pat, jim).

? predecessor(pam,bob).? predecessor(pam, ann).

? predecessor(pam, liz).

? predecessor(pam, X).

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Data Objects

Atoms versus numbers

versus Variables Atoms

strings of letters, digits, and the underscore

characterbeginning with a lower case letter some strings of special characters

can enclose strings of characters in single quotes

e.g. Fred

Numbers integers and floats

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Data Objects

Atoms versus Variables Variables are strings of characters, digits and

underscores that begin with an uppercase letter or

an underscore

Can use the anonymous underscore

hasachild(X) :- parent(X,Y)

hasachild(X) :- parent(X,_)

? parent(X, _)

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Data Objects

Structured objects objects that have several components

location(X, Y, Orientation).location(156, 786, 25).

location(156, 786, Orientation).

location is the functor

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Structures

date(X,Y,Z).

date(20, june, 2005).

date(Day, june, 2005).

date(Day, Month, Year) :- Day > 0, Day

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data objects

simple objects structures

constants variables

atoms numbers

These are all terms

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Operations on lists

Membership

The member relation member(X,L)

? member(d, [a, b, h, d]).?

? member(d, [a,b,[d h], e]).

?

?member([d, h], [a, [d,h], e f]).

?

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Membership

X is a member of L if

X is the head of L, or

X is a member of the tail of L.

member(X, [X|Tail]).

member(X, [Head | Tail]) :- member(X,Tail).

Note two separate clauses

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Membership

X is a member of L if

X is the head of L, or

X is a member of the tail of L, or X is a member of a sublist of L

member(X, [X|Tail]).

member(X, [Head | Tail]) :- member(X,Tail).

plus one more clause

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Concatenation

The concatenation relation conc(L1, L2, L3)

? conc([a,b], [c,d], [a,b,c,d])

yes

? conc([a,b], [c,d], [a, b, [c,d]])

no

?conc([a,b], [c,d], X).

X = [a,b,c,d]

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Concatentation

conc(L1, L2, Result)

If L1 is the empty list

then L2 and Result must be equal

If L1 is nonempty

then have [X|L1tail]

recursively X becomes the head of Result and we

use conc to find the tail of result [X|TailResult] Eventually will have exhausted L1 and TailResult

will be L2.

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Concatentation

conc([], L, L).

conc([X | L1], L2, [X | L3]) :- conc(L1, L2, L3).

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Unification

Matching clauses with variables

Have to find the appropriate substitutions forvariables so that the clauses match

Process is called unification

Process by which variables are instantiated

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GCD example

gcd(u,0,u)

gcd(u,v,w) not zero(v), gcd(v, u mod v, w)

Using resolution the goal

gcd(15, 10, x)

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Unification in Prolog

A constant unifies only with itself

? me = me.

Yes

?me = you.

No

Gcd(5, 0, 5) Gcd(5, 0, 5)

Gcd(5, 10, w) Gcd(5, 0, w)

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A variable that is uninstantiated unifies with anythingand becomes instantiated with that thing

? me = X.

X = me

? f(a,X) = f(Y, b).X = b

Y = a

? f(X) = f(Y)

gcd(u,v,w) not zero(v), gcd(v, u mod v, w), gcd(15, 10, x).

gcd(15, 10, x) not zero(10), gcd(10, 15 mod 10, x), gcd(15, 10, x).

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A structured term (predicate or function applied to

arguments requires

Same predicate/function name

Same number of arguments

Arguments can be unified recursively

? f(X) = g(X)

? f(X) = f(a,b)? f(a, g(X)) = f(Y, b)

? f(a, g(X)) = f(Y, g(b))

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Unification examples

Unify the following :

p(X,Y) and p(a, Z)p(X,X) and p(a,b)

ancestor(X,Y) and ancestor(bill, W)

p(X,a,Y) and p(Z,Z,b)p(Marcus, g(X,Y)) and f(x, g(Caesar, Marcus))

g(X, X) and g(f(X), f(X))

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Lecture Outline

Some Prolog lists

Unification

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Concatentation

conc([], L, L).

conc([X | L1], L2, [X | L3]) :-

conc(L1, L2, L3).

Can use concat to decompose lists

How?

Can use concat to define the member predicate -

member2(X, L) :- conc(L1, [X|L2], L).

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Adding and deleting

How do you add an item to a list?

Deleting an item from a list del(X, L, Result)

1. If X is the head of L result is the tail of L

2. If X is contained in the tail of L then recursively split L

into its head and tail until X is at the head. Then 1 willapply.

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Deleting

del(X, [X|Tail], Tail]).

del(X, [Y|Tail], [Y|Tail1]) :- del(X, Tail, Tail1)

Deletes one occurrence from the list

What happens when:

del(a, [a, b, a, a], X).

What if we changed it to

del(X, [X|Tail], Tail]).

del(X, [Y|Tail], [Y|Tail1]) :- del(X, Tail, Tail1), X=\=Y.

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Delete

What if we want to delete every instance from

the list

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del (X, [], []).

del (X, [X|Tail], Tail1) :- del (X, Tail, Tail1).

del (X, [Y|Tail], [Y|Tail1]) :- del (X, Tail, Tail1).

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Relations/Terms/Queries

An important note

You can define different relations with the same

name but with different numbers of arguments

e.g. member/1, member/2, member/3

If you leave off an argument prolog thinks it is a

different relation

If it is undefined for that number of arguments you will

get an error message

And if there is such a relation predefined.

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Define two predicates evenlength(List) and

oddlength(List) so that they are true if their argument is

a list of even or odd length respectively. For example

? evenlength([a,b,c,d])

? yes

?oddlength([c, d, e])

?yes

The trick is to define them as mutually recursive clauses.

Start with []

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[] is even

A list is even if its tail is odd

A list is odd if its tail is even

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evenlength([]).

evenlength([First | Rest]) :- oddlength(Rest).

oddlength([First | Rest]) :- evenlength(Rest).

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Unification

Matching clauses with variables

Have to find the appropriate substitutions forvariables so that the clauses match

Process is called unification

Process by which variables are instantiated

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GCD example

gcd(u,0,u)

gcd(u,v,w) not zero(v), gcd(v, u mod v, w)

Using resolution the goal

gcd(15, 10, x)

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A constant unifies only with itself

? me = me.

Yes

?me = you.

No

Gcd(5, 0, 5) Gcd(5, 0, 5)

Gcd(5, 10, w) Gcd(5, 0, w)

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A variable that is uninstantiated unifies with anythingand becomes instantiated with that thing

? me = X.

X = me

? f(a,X) = f(Y, b).X = b

Y = a

? f(X) = f(Y)

gcd(u,v,w) not zero(v), gcd(v, u mod v, w), gcd(15, 10, x).

gcd(15, 10, x) not zero(10), gcd(10, 15 mod 10, x), gcd(15, 10, x).

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A structured term (predicate or function applied to

arguments requires

Same predicate/function name

Same number of arguments Arguments can be unified recursively

? f(X) = g(X)

? f(X) = f(a,b)? f(a, g(X)) = f(Y, b)

? f(a, g(X)) = f(Y, g(b))

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Unification examples

Unify the following :

p(X,Y) and p(a, Z)p(X,X) and p(a,b)

ancestor(X,Y) and ancestor(bill, W)

p(X,a,Y) and p(Z,Z,b)p(Marcus, g(X,Y)) and f(x, g(Caesar, Marcus))

g(X, X) and g(f(X), f(X))

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Lecture Outline

Cut operator

Negation as Failure

Control Information

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Backtracking

PROGRAMbig(bear). %clause 1

big(elephant). %clause 2

small(cat). %clause 3

brown(bear). %clause 4

black(cat). %clause 5

grey(elephant). %clause 6

dark(Z) :- black(Z). %clause 7dark(Z) :- brown(Z). %clause 8

?dark(X), big(X).

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Controlling Backtracking

1 2 3

4 5 6

7 8 9

Knights Tour Problem

move(1,8).move(1,6).

move(2,9).

move(2,7).

move(3,8).

move(3,4).move(4,3).

move(4,9).

move(6,1).

move(6,7).

move(7,2).

move(7,6).

move(8,1).

move(8,3).

move(9,2).

move(9,4).

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Find all the two-move paths

path2(X,Y) :- move(X,Z), move(Z,Y).

path2(1,W).

? W = 1

? W = 3? W = 1

? W = 7

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path2(X,Y) :- move(X,Z), !, move(Z,Y).

? path2(1, W).

? W = 1

? W = 3

! is the cut operator

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path2(X,Y) :- move(X,Z), move(Z,Y),!.

? path2(1, W).

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Controlling Recursive Calls

predicate path determines if there is a path between two nodes.

path1(Z,Z,L).

path1(X,Z,L) :- move(X,Y), not (member(Y,L)), path(Y, Z, [Y|L]).

path2((Z,Z,L).

path2(X,Z,L) :- move(X,Y), not (member(Y,L)), path(Y, Z, [Y|L]), ! .

path1(1,3, []).

path2(1,3, []).

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Negation as Failure

Closed world assumption

The goal not (X) succeeds whenever the goal X fails!

?mother(amy, bob).

?not(parent(amy, bob)).

Why did it fail?

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Negation as Failure

Nonmonotonic reasoning more information

sometimes means fewer things can be

proved

human(bob).

?human(X). X = bob

?not human(X). no

?not (not human(X)). X = X

why?

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Control Information

Prolog uses a depth first search strateggy

Therefore order is important

pred1(X,Z) :- parent(X,Z).

pred1(X,Z) :- parent(X,Y), pred1(Y,Z).

pred2(X,Z) :- pred2(Y,Z), parent(X,Y).

pred2(X,Z) :- parent(X,Z).

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parent(pam,bob).

parent(tom,bob).

parent(tom,liz).parent(bob, ann).

parent(bob, pat).

parent(pat, jim).

? pred(tom, pat).

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Some More Prolog

structures

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Tk li P l

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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/MyDocuments/prolog/fred.txt', read, Fred),read_string(Fred, "\r\n ", 10, X).

Fred = 31

X = "freddy"

Yes (0.00s cpu)

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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

Structures

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a 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

family

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

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

family(_,person(Name, Surname, _, _), [_, _, _, | _]).

% all married women

who have at

least three children

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

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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

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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)

??

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Selectors

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

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

Define relation which allow us to access particularcomponents of a structure without knowing the details

the structure

This is data abstraction

These selectors are useful when we want to create instances

of families, for example

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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).

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nth_member([X|_], 1, X).

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

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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.

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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?

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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

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

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

We do this using the member predicate

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member(X1, Robotslist),

X1 \= dalek,


X2 \= tinman,

X2 \= johnnyfive,


X3 = r2d2,member(X4, Robotslist),

X4 \= tinman,

dalek,

There are just two more things left to do

L = [movie(X1, wizardofoz),solution(L):-

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X1 \= dalek,


X2 \= tinman,

X2 \= johnnyfive,member(X3, Robotslist),

X3 = r2d2,


X4 \= tinman,

X2 \= X1, X2 \= X3, X2 \= X4, X3 \= X1, X3 \= X4, X4 \= X1print_movies(L).

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