如何在序言中使用递归绘制直角三角形? [英] How to draw right triangle using recursion in prolog?

问题描述

我得到了这个直角三角形的答案，如下所示:

I get the answer for this right triangle as shown below:

shape:- shape(0, 6).

shape(S, A) :- S < A, 
              count(0, S), 
              S1 is S+1, 
              shape(S1, A).

shape(S, X) :- S >= X.


count(A, B) :- A =< B, 
              write('*'), 
              A1 is A+1, 
              count(A1,B).

count(A, B) :- A > B, nl.

   *
   **
   ***
   ****
   *****
   ******

我应该修改什么来打印这种类型的直角三角形?

What should i modify to print the right triangle of this type?

         *
        **
       ***
      ****
     *****
    ******

推荐答案

shape(S, _) :- S =< 0.
shape(S, N) :- S > 0, 
              S1 is S-1, 
              count(0, S1, N), 
              shape(S1, N).

count(A, _, N) :- A >= N, nl.
count(A, B, N) :- A < N, 
    (   A >= B
    ->  write('*')
    ;   write('.')
    ),
    A1 is A+1, 
    count(A1,B, N).

?- shape(6, 6).
.....*
....**
...***
..****
.*****
******
true ;
false.

我在打印空间时遇到问题.SWI prolog 似乎有一个错误:它要么不打印任何内容，要么打印 ' '.因此，此代码可能与您的 prolog 解释器一起使用 write(' ') 而不是 write('.') 运行.

I have trouble printing the space. There seems to be a bug with SWI prolog: it either prints nothing or ' '. So maybe this code runs with your prolog interpreter with write(' ') instead of write('.').

代码与您的非常相似.主要区别在于计数有 3 个参数:迭代器 A、字符边界数 B 和写入的字符总数 (N>).迭代器 A 必须小于或等于 N.根据条件 A >= B，它会打印 '*' 或 '.'.另外我必须写它倒数，否则树会颠倒/从左到右.

The code is very similar to yours. The main difference is there are 3 parameters for the count: an iterator A, a character boundary number B and the total number of characters written (N). The iterator A has to be smaller or equal to N. Depending on the condition A >= B it either prints '*' or '.'. Also I had to write it counting backwards, otherwise the tree would have been upside down / left to right.

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