Ackermann function

The Ackermann function \(A(x,y)\) is a recursive function invented by and defined as follows:

It grows at similar rates as much less known Sudan function.
 * \(y + 1\) if \(x = 0\),
 * \(A(x - 1,1)\) if \(y = 0\), or
 * \(A(x - 1,A(x,y - 1))\) otherwise.

So, for example,

\[ \begin{array}{cclclcl} A(2,2) &=& A(1,A(2,1))\\ &=& A(1,A(1,A(2,0)))&&\\ &=& A(1,A(1,A(1,1)))\\ &=& A(1,A(1,A(0,A(1,0))))\\ &=& A(1,A(1,A(0,A(0,1))))\\ &=& A(1,A(1,A(0,2)))\\ &=& A(1,A(1,2))\\ &=& A(1,A(0,A(1,1)))\\ &=& A(1,A(0,A(0,A(1,0))))\\ &=& A(1,A(0,A(0,1)))\\ &=& A(1,A(0,2))\\ &=& A(1,3)\\ &=& A(0,A(1,2))\\ &=& A(0,A(0,A(1,1)))\\ &=& A(0,A(0,A(0,A(1,0))))\\ &=& A(0,A(0,A(0,A(0,1))))\\ &=& A(0,A(0,A(0,2)))\\ &=& A(0,A(0,3))\\ &=& A(0,4))\\      &=& 5 \end{array} \]

Using Knuth's up-arrow notation \(A(x,y)=2\uparrow^{x-2}(y+3)-3\)

Friedman's definition
The definition of Ackermann function may slightly vary. For example, Harvey Friedman defines it like so:


 * \(A(x,y) = 2y\) (if \(x=1\))
 * \(A(x,y) = A(x-1,A(x,y-1))\) (otherwise)

The Ackermann function is related to the Ackermann numbers, because they exhibit equivalent growth rates.

"Ackermann function" often refers to the single-argument function \(A(n) = A(n, n)\). This is also known as the gag function.

The inverse of the single-argument Ackermann function \(\alpha(n)\) is called the inverse Ackermann function. Since the Ackermann function grows rapidly for small input values, the inverse Ackermann function grows slowly. It has applications in time complexity theory.

Goucher's definition
A.P. Goucher in his blog post proposed the following definition of Ackermann function:

\(f_1(n)=n+2\)

\(f_{m+1}(n) = f_m^n(2)\)

\(A(n) = f_n(n)\)

Also this post describes yet another variant of Ackermann function, which is related to the following problem:

Given a row of boxes, and some number coins in each. We can pick one box and operate by one of the following rules:


 * Remove one coin from this box and add two coins in the box n+1.
 * Remove one coin from this box and invert the number of coins in boxes n+1 and n+2.

We can choose a strategy of picking boxes and applying rules for them. Consider the situation when all boxes expect rightmost are empty. Now, given a row of n boxes with one coin in each, f(n) is the largest number of coins in this rightmost box. Computing exact values of this function can be tricky, but it is trivial to see that \(f(1) = 1\) and \(f(2) = 3\).