Bramer M. Logic Programming with Prolog
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Built-in Operators 183
Name of Operator
>=
Type of Operator
infix
Syntax
Exp1>=Exp2
Description
Succeeds if the value of arithmetic expression Exp1 is greater than or equal to the
value of arithmetic expression Exp2.
Name of Operator
is/2
Type of Operator
infix
Syntax
Result is Expression
Description
Expression must be a valid arithmetic expression which is evaluated to give a
number. If Result is an unbound variable (the usual case) the variable is bound to
the value of the expression. If Result is a bound variable with a numerical value or
a number, the goal succeeds if the values of both sides of the is operator are the
same and fails otherwise.
Name of Operator
not/1
Type of Operator
prefix
Syntax
not Goal
Description
Succeeds if Goal fails, fails if Goal succeeds.
Appendix 3
Specimen Solutions to Practical Exercises
Practical Exercise 1
Question 2
A suitable program would comprise the five clauses:
animal(lion).
animal(tiger).
animal(cow).
carnivore(lion).
carnivore(tiger).
Suitable sequences of goals to test (a) to (d) are as follows.
?- animal(tiger).
yes
?- animal(cow),animal(tiger).
yes
?- animal(lion),carnivore(lion).
yes
?- animal(cow),carnivore(cow).
no
186 Logic Programming With Prolog
Question 3
Here is the output produced by one Prolog system for the specified goals, with
explanations inserted in italic.
?- write(hello).
helloyes
?- write(Hello).
_12544Hello = _
This obscure output is produced because Hello is not an atom (it begins with a
capital letter). To print Hello World with a capital H it must be enclosed in quotes.
?- write('Hello!').
Hello!yes
It is not (usually) possible to suppress the final 'yes', which indicates that the goal
has ben satisfied. However the output can be made more readable by using 'nl' to
generate a newline after 'Hello!'.
?- write('Hello!'),nl.
Hello!
yes
?- 100=100.
yes
?- 100=1000/10.
no
Using = is not the right way to do arithmetic – see Chapter 4.
?- 100 is 1000/10.
yes
?- 1000 is 100*10.
yes
?- 2 is (5+7)/6.
yes
?- 74 is (5+7)*6.
no
Specimen Solutions to Practical Exercises 187
Practical Exercise 2
Question 1
A suitable series of goals is as follows:
?- animal(mammal,A,_,_).
A = tiger ;
A = hyena ;
A = lion ;
A = zebra
?- animal(mammal,A,carnivore,_).
A = tiger ;
A = hyena ;
A = lion ;
no
?- animal(mammal,A,_,stripes).
A = tiger ;
A = zebra
?- animal(reptile,A,_,mane).
no
Question 2
A suitable additional rule would be
couple(X,Y):-person(X,male),person(Y,female).
Testing this gives the following output:
?- couple(X,Y).
X = bill ,
Y = carol ;
188 Logic Programming With Prolog
X = bill ,
Y = margaret ;
X = bill ,
Y = jane ;
X = george ,
Y = carol ;
X = george ,
Y = margaret ;
X = george ,
Y = jane ;
X = alfred ,
Y = carol ;
X = alfred ,
Y = margaret ;
X = alfred ,
Y = jane ;
no
Specimen Solutions to Practical Exercises 189
Practical Exercise 3
Question 1
Suitable definitions are given below:
child_of(A,B):-parent(B,A).
grandfather_of(A,B):-father(A,C),parent(C,B).
grandmother_of(A,B):-mother(A,C),parent(C,B).
great_grandfather_of(A,B):-
father(A,C),grandfather_of(C,B).
great_grandfather_of(A,B):-
father(A,C),grandmother_of(C,B).
?- child_of(X,ann).
X = henry ;
X = mary ;
no
?- grandfather_of(A,caroline).
A = francis ;
no
?- grandmother_of(B,caroline).
B = janice ;
no
?- great_grandfather_of(C,caroline).
C = john ;
no
Question 2
The system begins by matching the goal with the first clause defining the
ancestor/2 predicate, i.e. [A1].
190 Logic Programming With Prolog
?-ancestor(louise,Desc).
[A1] ancestor(louise,Y):-parent(louise,Y).
X is bound to louise. Variables Desc and Y are bound to each other.
The goal parent(louise,Y) is now matched with clause [P3], which is first
rewritten to replace X and Y by X1 and Y1, i.e.
parent(X1,Y1):-mother(X1,Y1).
?-ancestor(louise,Desc).
[A1] ancestor(louise,Y):-parent(louise,Y).
[P3] parent(louise,Y1):-mother(louise,Y1).
X is bound to louise. Variables Desc, Y and Y1 are bound to each other. X1 is
bound to louise.
The system now tries to satisfy the goal mother(louise,Y1). It matches it with
clause [M9].
?-ancestor(louise,Desc).
[A1] ancestor(louise,Y):-parent(louise,Y).
[P3] parent(louise,Y1):-mother(louise,Y1).
[M9] mother(louise,caroline).
X is bound to louise. Variables Desc, Y and Y1 are bound to each other and to
caroline. X1 is bound to louise.
This gives a solution to the user's goal, with Desc bound to caroline.
?- ancestor(louise,Desc).
Desc = caroline
If the user now forces the system to backtrack, the system will try to resatisfy the
goal mother(louise,Y1) and fail. This will cause the rule [P3] to be rejected.
Attempts to resatisfy parent(louise,Y) in the body of [A1] will also fail, so clause
[A1] will be rejected. This brings the system back to the original goal
ancestor(louise,Desc).
The system tries to resatisfy it by matching it with the second clause for
ancestor/2, i.e. [A2].
Specimen Solutions to Practical Exercises 191
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,Z),ancestor(Z,Y).
X is bound to louise. Variables Desc and Y are bound to each other. Variable Z is
unbound.
The system now tries to satisfy the goal parent(louise,Z). It matches it with [P3],
which is first rewritten as (say)
parent(X1,Y1):-mother(X1,Y1).
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,Z),ancestor(Z,Y).
[P3] parent(louise,Y1):-mother(louise,Y1).
X is bound to louise. Variables Desc and Y are bound to each other. Variables Y1
and Z are bound to each other. Variable X1 is bound to louise.
It now tries to satisfy the goal mother(louise,Y1). It matches it with clause [M9].
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,caroline),ancestor(caroline,Y).
[P3] parent(louise,caroline):-mother(louise,caroline).
[M9] mother(louise,caroline).
X is bound to louise. Variables Desc and Y are bound to each other. Variables Z
any Y1 are bound to each other and to caroline. Variable X1 is bound to louise.
The next step is to satisfy the goal ancestor(caroline,Y). It is matched with [A1],
which is first rewritten as (say)
ancestor(X2,Y2):-parent(X2,Y2).
192 Logic Programming With Prolog
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,caroline),ancestor(caroline,Y).
[P3] parent(louise,caroline):-mother(louise,caroline).
[M9] mother(louise,caroline).
[A1] ancestor(X2,Y2):-parent(caroline,Y2).
X is bound to louise. Variables Desc and Y are bound to each other. Variables Z
any Y1 are bound to each other and to caroline. Variable X1 is bound to louise.
Variable X2 is bound to caroline. Variables Y and Y2 are bound to each other.
Next the system tries to satisfy parent(caroline,Y2). It matches it with [P3], which
is first rewritten as (say)
parent(X3,Y3):-mother(X3,Y3).
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,caroline),ancestor(caroline,Y).
[P3] parent(louise,caroline):-mother(louise,caroline).
[M9] mother(louise,caroline).
[A1] ancestor(X2,Y2):-parent(caroline,Y2).
[P3] parent(caroline,Y3):-mother(caroline,Y3).
X is bound to louise. Variables Desc and Y are bound to each other. Variables Z
any Y1 are bound to each other and to caroline. Variable X1 is bound to louise.
Variable X2 is bound to caroline. Variables Y, Y2 and Y3 are bound to each other.
X3 is bound to caroline.
Specimen Solutions to Practical Exercises 193
Next the system tries to satisfy mother(caroline,Y3). It matches it with [M10].
?-ancestor(louise,Desc).
[A2] ancestor(louise,Y):-parent(louise,caroline),ancestor(caroline,Y).
[P3] parent(louise,caroline):-mother(louise,caroline).
[M9] mother(louise,caroline).
[A1] ancestor(X2,Y2):-parent(caroline,Y2).
[P3] parent(caroline,Y3):-mother(caroline,Y3).
[M10] mother(caroline,david).
X is bound to louise. Variables Desc and Y are bound to each other. Variables Z
any Y1 are bound to each other and to caroline. Variable X1 is bound to louise.
Variable X2 is bound to caroline. X3 is bound to caroline. Variables Y, Y2 and Y3
are bound to each other and to david.
Now all the goals in the body of [A2] have succeeded, so the original goal
ancestor(louise,Desc) succeeds with Desc bound to david.
Another solution with Desc bound to janet is available by backtracking.
?- ancestor(louise,Desc).
Desc = caroline ;
Desc = david ;
Desc = janet ;
no