Ackermann’s Function

May 25, 2012

Our function is a simple transliteration of the mathematical definition into code; note that in the second clause of the cond it is already known that m is zero:

(define (a m n)
  (cond ((zero? m) (+ n 1))
        ((zero? n) (a (- m 1) 1))
        (else (a (- m 1) (a m (- n 1))))))

Here’s an example:

> (a 3 4)
125

You can run the program at http://programmingpraxis.codepad.org/lLpK9rFw.

Ackermann’s function was important in math, and is also important in computer science, being part of the computation of the time complexity of algorithms such as the disjoint-set data structure and others. It is also used as a test of recursion for compilers.

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10 Responses to “Ackermann’s Function”

  1. Graham said

    Given how heavily recursive this function is, I went with Haskell instead of Python:

    a :: Integer -> Integer -> Integer
    a 0 n = n + 1
    a m 0 = a (m - 1) 1
    a m n = a (m - 1) $ a m (n - 1)
    
    main :: IO ()
    main = print $ a 3 4
    
  2. [...] Programming News: Ackermann’s Function    In the 1920s, Wilhelm Ackermann demonstrated a computable function that was not primitive-recursive, settling an important argument in the run-up to the theory of computation.     Read full story => Programming Praxis [...]

  3. slabounty said
    def a(m, n)
      if m == 0
        n+1
      elsif m > 0 && n == 0
        a(m-1, 1)
      else
        a(m-1, a(m, n-1))
      end
    end
    
    puts "ackerman 3, 4 = #{a(3, 4)}"
    
  4. toecutter said

    Feel free to laugh at me. I tried to use visual C# for this. Stack overflow or there is some arbitrary stack limit where visual C# in its infinite wisdom decided I didn’t know what I was doing. Fun times

  5. programmingpraxis said

    Toecutter: If your language doesn’t provide a sufficient stack, you must create your own stack; that’s the point of the exercise. A(4,1)=65533 should be within range of your function.

  6. iapain said

    Ha! Ackermann’s function. At least this can computer A(4,2) in python


    def ackermann(m, n):
    while m >= 4:
    if n == 0:
    n = 1
    else:
    n = ackermann(m, n - 1)
    m = m - 1
    if m == 3:
    return (1 << n + 3) - 3
    elif m == 2:
    return (n << 1) + 3
    elif m == 1:
    return n + 2
    else:
    return n + 1

    print ackermann(4,2)

    http://codepad.org/QdneSl1Q

  7. function A(m,n) {
    	return (m == 0) ?
    			n + 1 :
    		( m > 0 && n ==0 ) ?
    			A(m - 1,1) :
    		( m > 0 && n > 0) ?
    			A(m - 1, A(m,n-1)) :
    			undefined;
    }
    

    http://jsfiddle.net/vcKdF/3248/

  8. igorii said

    How about some q


    A:{[m; n]
    if [m ~ 0; : n+1];
    $ [n ~ 0;
    if [0 < m;
    : A[m-1; 1]];
    if [0 < n;
    : A[m-1; A[m; n-1]]]];
    };

  9. Christian said

    My solution written in Go lang: https://gist.github.com/2942073

  10. David Gottner said

    Using the closed form, recursion only is necessary for Ackerman(m, n) with m >= 4.

    def hyper(n, a, b)
        return b+1  if n == 0
        return a+b  if n == 1
        return a*b  if n == 2
        return a**b if n == 3
    
        return 1 if b == 0
    
        x = a
        (b-1).times do |_|
            x = hyper(n-1, a, x)
        end
        return x
    end
    
    
    def ackerman(m, n)
        hyper(m, 2, n+3) - 3
    end
    

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