这样做:
square2(0,_). % base case, recursion stops when X reaches 0, second argument is irrelevent
square2(X,Symbol):-
X1 is X - 1,
write(Symbol),
square2(X1,Symbol).
使用查询,结果为:
?- square2(5,'* ').
* * * * *
因此,我们需要另一个循环来使其写入 X 次。
square1(0,_,_). % base case, recursion stops when X reaches 0
square1(X,Y,Symbol):-
X1 is X - 1,
square2(Y,Symbol), % with the same Y passed in square2 predicate to print a line of symbols
nl, % creates a new line
square1(X1,Y,Symbol).
但是,问题是要求格式为square(5, '* ')
. 所以,
square(X,Symbol):-
square1(X,X,Symbol).
总结一下:
square(X,Symbol):-
square1(X,X,Symbol).
square1(0,_,_).
square1(X,Y,Symbol):-
X1 is X - 1,
square2(Y,Symbol),
nl,
square1(X1,Y,Symbol).
square2(0,_).
square2(X,Symbol):-
X1 is X - 1,
write(Symbol),
square2(X1,Symbol).