9.5
Exercises
Exercise
9.1
Which of the following queries succeed, and which fail?
?-
12
is
2*6.
?-
14
=\=
2*6.
?-
14
=
2*7.
?-
14
==
2*7.
?-
14
\==
2*7.
?-
14
=:=
2*7.
?-
[1,2,3|[d,e]]
==
[1,2,3,d,e].
?-
2+3
==
3+2.
?-
2+3
=:=
3+2.
?-
7-2
=\=
9-2.
?-
p
==
’p’.
?-
p
=\=
’p’.
?-
vincent
==
VAR.
?-
vincent=VAR,
VAR==vincent.
Exercise
9.2
How does Prolog respond to the following queries?
?-
.(a,.(b,.(c,[])))
=
[a,b,c].
?-
.(a,.(b,.(c,[])))
=
[a,b|[c]].
?-
.(.(a,[]),.(.(b,[]),.(.(c,[]),[])))
=
X.
?-
.(a,.(b,.(.(c,[]),[])))
=
[a,b|[c]].
Exercise
9.3
Write a two-place predicate
termtype(Term,Type)
that takes a term
and gives back the type(s) of that term (atom, number, constant, variable, and so on).
The types should be given back in the order of their generality. The predicate should
behave in the following way.
?-
termtype(Vincent,variable).
yes
?-
termtype(mia,X).
X
=
atom
;
X
=
constant
;
X
=
simple_term
;
X
=
term
;
no
?-
termtype(dead(zed),X).
X
=
complex_term
;
X
=
term
;
no
Exercise
9.4
Write a Prolog program that defines the predicate
groundterm(Term)
which tests whether or not
Term
is a ground term. Ground terms are terms
that don’t contain variables. Here are examples of how the predicate should
behave:
?-
groundterm(X).
no
?-
groundterm(french(bic_mac,le_bic_mac)).
yes
?-
groundterm(french(whopper,X)).
no
Exercise
9.5
Assume that we have the following operator definitions.
:-
op(300,
xfx,
[are,
is_a]).
:-
op(300,
fx,
likes).
:-
op(200,
xfy,
and).
:-
op(100,
fy,
famous).
Which of the following are well-formed terms? What are the main operators? Give the
bracketings.
X
is_a
witch
harry
and
ron
and
hermione
are
friends
harry
is_a
wizard
and
likes
quidditch
dumbledore
is_a
famous
wizard
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