National Taipei University of Technology
Artificial Intelligence (spring, 2013)
Solution # 2
1. What kinds of object are the following Prolog objects (atom, number, variable,
structure)?
(a) Ted  variable
(b) teddy  atom
(c) ‘Tom’  atom
(d)
(e)
(f)
_bts  variable
‘Homework is easy’  atom
easy(homework)  structure
(g)
(h)
(i)
(j)
30  number
holiday(date(28, Feb, 2013))  structure
+(x, y)  structure
dangerous(car, ivy)  structure
2. Please add a new relation relatives(X,Y) based on hw1. (1-2.pl)
Two people are relatives if
(a) One is a predecessor of the other, or
(b) They have a common predecessor, or
(c) They have a common successor
And you must use the semicolon notation to shorten the program.
% 1-2.pl ignore
relatives(X,Y) :predecessor(X, Y) ;
predecessor(Y, X) ;
predecessor(Z, X) , predecessor(Z,Y) ;
%X and Y have a common predecessor
predecessor(X, Z) , predecessor(Y,Z).
%X and Y have a common successor
3. Define the following relation where the line exists in three-dimensional space.
(1) Define the relation parallelX(L) , parallelY(L) and parallelZ(L) where L is
parallel to the X-axis, Y-axis or Z-axis. For example:
parallelX( line( point( _, Y1, Z1), point( _, Y1, Z1))).
parallelY( line( point( X1, _, Z1), point( X1, _, Z1))).
parallelZ( line( point( X1, Y1, _), point( X1, Y1, _))).
(2) Define the relation parallelXY(L) , parallelYZ(L) and parallelXZ(L) where
L is parallel to the XY plane, YZ plane or XZ plane. For example:
parallelXY( line( point( _, _, Z1), point( _, _, Z1))).
parallelYZ( line( point( X1, _, _), point( X1, _, _))).
parallelXZ( line( point( _, Y1, _), point( _, Y1, _))).
4. According to clauses, answer the following question.
Clause 1：The bear is brown.
Clause 2：The cat is black.
Clause 3：The elephant is gray.
Clause 4：The dog is white.
Clause 5：The bear and the elephant are big.
Clause 6：The cat and the dog are small.
Clause 7：Gray and white is bright.
Clause 8：Black and brown is dark.
(1) Write a prolog program distinguishes animal. For example:
brown(bear).
%Clause 1
black(cat).
%Clause 2
gray(elephant).
%Clause 3
white(dog).
%Clause 4
big(bear).
big(elephant).
small(cat).
small(dog).
%Clause 5.1
%Clause 5.2
%Clause 6.1
%Clause 6.2
bright(X):-gray(X).
bright(X):-white(X).
%Clause 7.1
%Clause 7.2
dark(X):-black(X).
dark(X):-brown(X).
%Clause 8.1
%Clause 8.2
(2) According to problem4-1, if your program gets the right answer, please
explain the trace of procedure execute.
QUESTION
?-bright(X),small(X).
X = dog.
EXECUTION TRACE
(1) Initial goal list: bright(X), small(X).
(2) Scan the program from top to bottom looking for a clause whose head
matches the first goal bright(X). Clause 7.1 found: bright(X) :- gray(X).
Replace the first goal by the instantiated body of clause 7.1, giving a
new goal list: gray(X), small(X)
(3) Scan the program to find a match with gray(X). Clause 3 found:
gray(elephant). This clause has no body, so the goal list, properly
instantiated, shrinks to: small(elephant)
(4) Scan the program for the goal small(elephant). No clause found.
Therefore backtrack to step(3) and undo the instantiation X = elephant.
Now the goal list is again: gray(X), small(X)
(5) Continue scanning the program below clause 3. No clause found.
Therefore backtrack to step(2) and continue scanning below clause 7.1.
Clause 7.2 is found:bright(X):-white(X).
Replace the first goal by the instantiated body of clause 7.2, giving a
new goal list: white(X), small(X)
(6) Scan the program to find a match with white(X). Clause 4 found:
white(dog). This clause has no body, so the goal list, properly
instantiated, shrinks to: small(dog)
(7) Scan the program and find clause small(dog) . It has no body so the goal
list shrinks to empty. This indicates successful termination, and the
corresponding variable instantiation is: X = dog