Recursive Ackermann's function - need help

This is a discussion on Recursive Ackermann's function - need help within the C++ Programming forums, part of the General Programming Boards category; Hello everyone. I'm new to this board. Just discovered it a short while ago while searching for help on homework. ...

  1. #1
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    Recursive Ackermann's function - need help

    Hello everyone. I'm new to this board. Just discovered it a short while ago while searching for help on homework. Was wondering if anybody here could help me.

    I need to write a recursive version of the Ackermann function/algorithm, which is:

    Ackermann(m,n) = { n+1 if m=0
    Ackermann(m-1, 1) if n = 0
    otherwise Ackermann(m-1), Ackermann(m, n-1))

    Now a direction coding of this function would be:
    int ackermann(int m, int n){
    if (m == 0) return n + 1;
    if (n == 0) return ackermann(m - 1, 1);
    return ackermann(m - 1, ackermann(m, n -1));
    }

    But this function is probably somewhere along the lines of O^3 as far as Big O is concerned. Which basically means any number halfway to 10 called by the function would take a lot of time. Forget double digit numbers. I need a more efficient version of function.

    What I have in mind is something that involves Dynamic Programming. Making a table through matrix and then storing already calculated integers in that table and then calling them whenever they come up when the function is checking for a base case. This would GREATLY increase the speed. So I have in mind, a for loop with the table. Anybody who could gimme a hand with this, I would greatly appreciate it. Thank you.

  2. #2
    ¡Amo fútbol!
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    Use iteration, i.e. loops. Can't help you right now cause I'm kinda busy. But if no one helps you by tomorrow, I'll help you then.

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    I need a valid response to this problem before 6:30 AM EST tomorrow.

  4. #4
    Señor Member
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    Ooooo being a little pushy?

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    Nope, just procrastination at its best.

  6. #6
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    This function is *much* slower than O(n^3) I think it's like O(n^n).

    It looks like you could maybe try to do a 2d
    static array and then assign all the elements to 0. Then if a[m][n] != 0, a[m][n] = the computed value otherwise a[m][n] = the recursive definition.

  7. #7
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    Thank you, Nick! For that last algorithm. It really put me back on the right track.

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