学习Java递归,Ackerman函数

Question

我正在研究Java中的递归Ackermann函数.我在可能的递归线上得到一个错误,23.

return Ack(m - 1, Ack(m, n - 1));

非常感谢,如果有人能指出什么是错的.

-Kyle

   /*enter code here

     Ackerman's function, A(m, n) is defined:
     A(0 , n) = n + 1 for n  >= 0 
     A(m , 0) = A(m – 1 , 1) for m > 0
     A(m , n) = A(m – 1 , A(m , n - 1)) for n >= 0

    */

    public class AckFun {

     public static int Ack(int m, int n) {

      if (m == 0) {
       return 2 * n;
      } else if (m >= 1) {
       if (n == 0) {
        return 0;
       } else if (n == 1) {
        return 2;
       } else {
        return Ack(m - 1, Ack(m, n - 1));
       }
      }
      return n; // Not sure what to return here, Eclipse suggested this.
     }

     public static void main(String args[]) {
      System.out.println(Ack(3, 4));
     }
    }

Answer 1

你需要让你的堆叠更大:

http://thilinamb.wordpress.com/2008/12/22/how-to-increase-the-java-stack-size/

使用较大的堆栈,它运行时没有stackoverflow,但给出0.

编辑:你的代码是错误的,这就是它给出错误的原因.尝试完全按照定义所说的重写代码:

    //I assume that you check that n and m are non-negative before you run this
    if (m == 0) {
         return n + 1;
    } else if (n == 0) {
        return Ack(m - 1, 1);
    } else {
        return Ack(m - 1, Ack(m, n - 1));
    }

PS.不要责怪我发布家庭作业问题的源代码.我相信学习编程的最好方法是阅读和理解别人的代码.