Implementing Ackermann Function by Java with BigInteger support

Question

I'm getting StackOverflowError (Exception in thread "main" java.lang.StackOverflowError) for the following code. But the program works fine for m=3, n=3 (or other lower values) but does not work for m=4 and n=2 or 3.

public class AckermannFunction
{
   static BigInteger One = BigInteger.ONE;

   static BigInteger Zero = BigInteger.ZERO;

   static BigInteger ackmnFun(BigInteger m, BigInteger n)
   {
      if (m.equals(Zero))
         return n.add(One);

      if (n.equals(Zero))
         return ackmnFun(m.subtract(One), One);

      return ackmnFun(m.subtract(One), ackmnFun(m, n.subtract(One)));
   }

   public static void main(String[] args)
   {
      BigInteger m = new BigInteger("4");
      BigInteger n = new BigInteger("3");

      System.out.println(ackmnFun(m, n));

   }
}

There are too many recursive calls I understand. Is there any way to get rid of this error?

Thanks.

You could rewrite the function using a `Stack` which will be stored on the heap. — marstran, Jun 12 '19 at 18:12
You might manage to compute n=2, but n=3 is probably going to be too much to compute. — Louis Wasserman, Jun 12 '19 at 18:13

score 1 · Answer 1 · answered Jun 12 '19 at 18:44

1

Rather than doing this recursively, you could approach it as a dynamic programming problem, and construct a table of values from the bottom up. Then rather than making recursive calls, you simply reference your table to create the next entry.

answered Jun 12 '19 at 18:44

ncbvs

88
8

1

Thanks for the idea. – User_67128 Jun 12 '19 at 18:48
Hi, I was able to find the value of Ackermann(4, 2) with a modified recursive function where I put some extra terminating condition by Running the function with some lower value (m=2, n= 1, 2, 3... and m=3, n=1 , 2, 3, ...) so that when I call ackmnFun(4, 2) there are less recursive calls. My program now finds the value of ackmnFun(4, 2) in less than a minute. – User_67128 Jun 19 '19 at 17:26

score 0 · Accepted Answer · answered Jun 19 '19 at 17:40

I've found a solution of my question by myself. I want to share it with you.

Since the Ackermann function calls itself for so many times even for a very low value of m & n, what I did, I added few more terminating condition up to m=3 and n=3 (to minimize recursive calls). And the tricks worked. I found initial values by running the original recursive function and then added all these as terminating condition.

The program finds the Ackermann(4,2) in a few seconds now. The answer is very long containing 19729 decimal digits.

Would you mind sharing this? – April Crude Jan 05 '22 at 22:55 — April Crude, Jan 05 '22 at 22:55

Implementing Ackermann Function by Java with BigInteger support

2 Answers2