Prolog中的Asterisks Triangle

时间:2015-01-13 01:07:38

标签: prolog

我必须定义prolog金字塔(N),它打印出一个给定高度的星号金字塔,如下例所示。

pyramid(4).  
   * 
  *** 
 ***** 
******* 

true

这是我到目前为止所做的...... 我无法找到打印出每条线所需的其余星星的方法。 我还尝试定义支持谓词来处理程序的子部分。但是找不到它。

pyramid(0) :-
   nl.
pyramid(N) :-
   N > 0,
   N1 is N - 1,
   foreach(between(1,N1,_), write(' ')),
   write('*'), nl,
   pyramid(N1).

2 个答案:

答案 0 :(得分:0)

想想每个等级在N方面有多少个星星。假设你在i行,N = 4。

  • 第一行获得3个(实际上是N-1)个空格,一个星号和另外3个空格。
  • 第二行有3 - 1个空格,3个星,另有3到1个空格。
  • i行获得(N - 1) - (i - 1)个空格,1 + 2 * (i - 1)个星标和另一个(N - 1) - (i - 1)个空格。

所以这给了:

pyramid(N) :- pyramid(N, N-1).

pyramid(0, _) :- nl.
pyramid(N, K) :- N > 0, N1 is N - 1,
                 foreach(between(1, N1, _), write(' ')), 
                 Q is 2 * (K - N1) + 1,
                 foreach(between(1, Q, _), write('*')),
                 foreach(between(1, N1, _), write(' ')), 
                 nl, pyramid(N1, K).

我认为(但不确定)您也可以删除N > 0位,因为首先会检查案例pyramid(0, _)

答案 1 :(得分:0)

你应该这样做:

pyramid(N) :-         % to make an ASCII art pyramid...
  N > 0 ,             % - if first has to have a height,
  pyramid( N-1 , 1 ). % - then just invoke the helper predicate.
  .                   %

pyramid(I,_) :-         % If the indentation level has dropped below zero, we're done.
  I < 0 .               %
pyramid(I,C) :-         % otherwise...
  I >= 0 ,              % - if the indentation level is non-negative...
  repeat_write(I,' ') , % - write that many spaces,
  repeat_write(C,'*') , % - write the desired number of asterix characters
  nl ,                  % - a new line, 
  I1 is I-1 ,           % - decrement the indentation level
  C1 is C+2 ,           % - increment the asterix count
  pyramid(I1,C1).       % - and recurse down.

repeat_write(0,_) .   % writing zero characters is easy.
repeat_write(N,C) :-  % writing N characters is also easy:
  N > 0 ,             % - N must be positive
  write(C),           % - write a single character
  N1 is N-1 ,         % - decrement N
  repeat_write(N1,C). % - recurse down.