Question

您能告诉我如何在PROLOG中实现算法以生成N个平衡括号的所有组合吗？（方括号可正确关闭）

Answer 1

这是dcg的最佳选择。我们为圆括号定义了一种语法，然后对其进行了枚举。

:- set_prolog_flag(double_quotes, chars).

balanced --> "".
balanced --> "(", balanced, ")", balanced.

现在，在询问具体句子时，最好使用library(double_quotes)和this来查看更多信息。

我们可以要求固定长度的句子：

| ?- length(T, 6), phrase(balanced, T).
   T = "()()()"
;  T = "()(())"
;  T = "(())()"
;  T = "(()())"
;  T = "((()))"
;  false.

或者只增加一句话：

| ?- length(T, N), phrase(balanced, T).
   T = [], N = 0
;  T = "()", N = 2
;  T = "()()", N = 4
;  T = "(())", N = 4
;  T = "()()()", N = 6
;  T = "()(())", N = 6
;  T = "(())()", N = 6
;  T = "(()())", N = 6
;  T = "((()))", N = 6
;  T = "()()()()", N = 8
;  T = "()()(())", N = 8
;  T = "()(())()", N = 8
;  T = "()(()())", N = 8
;  T = "()((()))", N = 8
;  ...

Answer 2

这是获取加泰罗尼亚语数字的另一种解决方案。
它的括号类似于DCG解决方案：

catalan(L) :-
   append(['('|A], [')'|B], L),
   length(A, N), 0 =:= N mod 2,
   catalan(A),
   catalan(B).
catalan([]).

示例，验证A000108中的3个条目：

?- length(L, 18), findall(L, catalan(L), R), 
   length(R, N), write(N), nl, fail; true.
4862
?- length(L, 20), findall(L, catalan(L), R), 
   length(R, N), write(N), nl, fail; true.
16796
?- length(L, 22), findall(L, catalan(L), R), 
   length(R, N), write(N), nl, fail; true.
58786

Answer 3

#include<iostream>
using namespace std;
void parenthesis(int open, int close, string brak)
{
   if (open == 0 && close == 0)
      cout<<brak;
   if (open>close)
      return;
   if (open > 0)
      parenthesis(open - 1, close,brak + "(");
   if (close > 0)
      parenthesis(open, close-1 ,brak + ")");
}
void printPar(int n)
{
   parenthesis(n,n,"\n");
}
int main()
{
   int n = 3;
   printPar(n);
   return  0;

}

Prolog，生成n个平衡括号的所有组合

3 个答案: