Ackermann function

In the late 1920s, the mathematicians Gabriel Sudan and Wilhelm Ackermann, students of David Hilbert, were studying the foundations of computation. Both Sudan and Ackermann are credited[4] with discovering total computable functions (termed simply "recursive" in some references) that are not primitive recursive. Sudan published the lesser-known Sudan function, then shortly afterwards and independently, in 1928, Ackermann published his function ${\displaystyle \varphi }$ (the Greek letter phi). Ackermann's three-argument function, ${\displaystyle \varphi (m,n,p)}$, is defined such that for p = 0, 1, 2, it reproduces the basic operations of addition, multiplication, and exponentiation as

${\displaystyle \varphi (m,n,0)=m+n,}$

${\displaystyle \varphi (m,n,1)=m\times n,}$

${\displaystyle \varphi (m,n,2)=m^{n},}$

and for p > 2 it extends these basic operations in a way that can be compared to the hyperoperations:

${\displaystyle \varphi (m,n,3)=m[4](n+1),\,\!}$

${\displaystyle \varphi (m,n,p)\gtrapprox m[p+1](n+1){\text{, for }}p>3.\,\!}$

(Aside from its historic role as a total-computable-but-not-primitive-recursive function, Ackermann's original function is seen to extend the basic arithmetic operations beyond exponentiation, although not as seamlessly as do variants of Ackermann's function that are specifically designed for that purpose—such as Goodstein's hyperoperation sequence.)

In On the Infinite,[5] David Hilbert hypothesized that the Ackermann function was not primitive recursive, but it was Ackermann, Hilbert's personal secretary and former student, who actually proved the hypothesis in his paper On Hilbert's Construction of the Real Numbers.[2][6]

Rózsa Péter[7] and Raphael Robinson[8] later developed a two-variable version of the Ackermann function that became preferred by many authors.

The generalized hyperoperation sequence, e.g. ${\displaystyle \operatorname {G} (m,a,b)=a[m]b}$, is a version of Ackermann function as well.[9]

In 1963 R.C. Buck based an intuitive two-variable variant (F[10]) on the hyperoperation sequence:[11]

${\displaystyle \operatorname {F} (m,n)=2[m]n}$.

Compared to most other versions Buck's function has no unessential offsets:

${\displaystyle {\begin{array}{lr}\operatorname {F} (0,n)=2[0]n=n+1\\\operatorname {F} (1,n)=2[1]n=2+n\\\operatorname {F} (2,n)=2[2]n=2\times n\\\operatorname {F} (3,n)=2[3]n=2^{n}\\\operatorname {F} (4,n)=2[4]n=2^{2^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}}\\...\end{array}}}$

Ackermann's original three-argument function ${\displaystyle \varphi (m,n,p)}$ is defined recursively as follows for nonnegative integers m, n, and p:

${\displaystyle {\begin{array}{lr}\varphi (m,n,0)=m+n\\\varphi (m,0,1)=0\\\varphi (m,0,2)=1\\\varphi (m,0,p)=m{\text{ for }}p>2\\\varphi (m,n,p)=\varphi (m,\varphi (m,n-1,p),p-1){\text{ for }}n>0{\text{ and }}p>0.\end{array}}}$

Of the various two-argument versions, the one developed by Péter and Robinson (called "the" Ackermann function by some authors) is defined for nonnegative integers m and n as follows:

${\displaystyle {\begin{array}{lcl}\operatorname {A} (0,n)&=&n+1\\\operatorname {A} (m,0)&=&\operatorname {A} (m-1,1)\\\operatorname {A} (m,n)&=&\operatorname {A} (m-1,A(m,n-1))\end{array}}}$

It may not be immediately obvious that the evaluation of ${\displaystyle A(m,n)}$ always terminates. However, the recursion is bounded because in each recursive application either m decreases, or m remains the same and n decreases. Each time that n reaches zero, m decreases, so m eventually reaches zero as well. (Expressed more technically, in each case the pair (m, n) decreases in the lexicographic order on pairs, which is a well-ordering, just like the ordering of single non-negative integers; this means one cannot go down in the ordering infinitely many times in succession.) However, when m decreases there is no upper bound on how much n can increase—and it will often increase greatly.

The Péter-Ackermann function can also be expressed in relation to various other versions of the Ackermann function:

• hyperoperation:

${\displaystyle A(m,n)={\begin{cases}0[m]n&{\text{, if }}m=0\\2[m](n+3)-3&{\text{, if }}m>0\\\end{cases}}}$

• hyperoperation, represented in Knuth's up-arrow notation (extended to integer indices ≥ −2):

${\displaystyle A(m,n)={\begin{cases}0\uparrow ^{m-2}n&{\text{, if }}m=0\\2\uparrow ^{m-2}(n+3)-3&{\text{, if }}m>0\\\end{cases}}}$

• hyperoperation, represented in Conway chained arrow notation:

${\displaystyle A(m,n)=(2\rightarrow (n+3)\rightarrow (m-2))-3}$ for ${\displaystyle m\geq 3}$

hence

${\displaystyle 2\rightarrow n\rightarrow m=A(m+2,n-3)+3}$ for ${\displaystyle n>2}$.

(n = 1 and n = 2 would correspond with A(m, −2) = −1 and A(m, −1) = 1, which could logically be added.)

For small values of m like 1, 2, or 3, the Ackermann function grows relatively slowly with respect to n (at most exponentially). For m ≥ 4, however, it grows much more quickly; even A(4, 2) is about 2×10¹⁹⁷²⁸, and the decimal expansion of A(4, 3) is very large by any typical measure.

An interesting aspect of the (Péter-)Ackermann function is that the only arithmetic operation it ever uses is addition of 1. Its fast growing power is based solely on nested recursion. This also implies that its running time is at least proportional to its output, and so is also extremely huge. In actuality, for most cases the running time is far larger than the output; see below.

A single-argument version f(n) = A(n, n) that increases both m and n at the same time dwarfs every primitive recursive function, including very fast-growing functions such as the exponential function, the factorial function, multi- and superfactorial functions, and even functions defined using Knuth's up-arrow notation (except when the indexed up-arrow is used). It can be seen that f(n) is roughly comparable to f_ω(n) in the fast-growing hierarchy. This extreme growth can be exploited to show that f, which is obviously computable on a machine with infinite memory such as a Turing machine and so is a computable function, grows faster than any primitive recursive function and is therefore not primitive recursive.

In a category with exponentials, using the isomorphism ${\displaystyle ((X\times Y)\rightarrow Z)\cong (X\rightarrow (Y\rightarrow Z))}$ (in computer science, this is called currying), the Ackermann function may be defined via primitive recursion over higher-order functionals as follows:

${\displaystyle {\begin{array}{lcl}\operatorname {Ack} (0)&=&\operatorname {S} \\\operatorname {Ack} (m+1)&=&\operatorname {Iter} (\operatorname {Ack} (m))\end{array}}}$

where S(n) = n + 1 is the usual successor function and Iter denotes the functional power operator, defined by primitive recursion as well:

${\displaystyle {\begin{array}{lcl}\operatorname {Iter} (f)(0)&=&f(1)\\\operatorname {Iter} (f)(n+1)&=&f(\operatorname {Iter} (f)(n)).\end{array}}}$

The function ${\displaystyle \mathrm {Ack} }$ defined in this way agrees with the Ackermann function ${\displaystyle A}$ defined above: ${\displaystyle \mathrm {Ack} (m)(n)=A(m,n)}$.

Number of recursions before the return of Ackerman(3,3)

To see how the Ackermann function grows so quickly, it helps to expand out some simple expressions using the rules in the original definition. For example, one can fully evaluate ${\displaystyle A(1,2)}$ in the following way:

${\displaystyle {\begin{aligned}A(1,2)&=A(0,A(1,1))\\&=A(0,A(0,A(1,0)))\\&=A(0,A(0,A(0,1)))\\&=A(0,A(0,2))\\&=A(0,3)\\&=4.\end{aligned}}}$

To demonstrate how ${\displaystyle A(4,3)}$'s computation results in many steps and in a large number:

${\displaystyle {\begin{aligned}A(4,3)&=A(3,A(4,2))\\&=A(3,A(3,A(4,1)))\\&=A(3,A(3,A(3,A(4,0))))\\&=A(3,A(3,A(3,A(3,1))))\\&=A(3,A(3,A(3,A(2,A(3,0)))))\\&=A(3,A(3,A(3,A(2,A(2,1)))))\\&=A(3,A(3,A(3,A(2,A(1,A(2,0))))))\\&=A(3,A(3,A(3,A(2,A(1,A(1,1))))))\\&=A(3,A(3,A(3,A(2,A(1,A(0,A(1,0)))))))\\&=A(3,A(3,A(3,A(2,A(1,A(0,A(0,1)))))))\\&=A(3,A(3,A(3,A(2,A(1,A(0,2))))))\\&=A(3,A(3,A(3,A(2,A(1,3)))))\\&=A(3,A(3,A(3,A(2,A(0,A(1,2))))))\\&=A(3,A(3,A(3,A(2,A(0,A(0,A(1,1)))))))\\&=A(3,A(3,A(3,A(2,A(0,A(0,A(0,A(1,0))))))))\\&=A(3,A(3,A(3,A(2,A(0,A(0,A(0,A(0,1))))))))\\&=A(3,A(3,A(3,A(2,A(0,A(0,A(0,2)))))))\\&=A(3,A(3,A(3,A(2,A(0,A(0,3))))))\\&=A(3,A(3,A(3,A(2,A(0,4)))))\\&=A(3,A(3,A(3,A(2,5))))\\&=\ldots \\&=A(3,A(3,A(3,13)))\\&=\ldots \\&=A(3,A(3,65533))\\&=\ldots \\&=A(3,2^{65536}-3)\\&=\ldots \\&=2^{2^{\overset {65536}{}}}-3.\\\end{aligned}}}$

Computing the Ackermann function can be restated in terms of an infinite table. First, place the natural numbers along the top row. To determine a number in the table, take the number immediately to the left. Then use that number to look up the required number in the column given by that number and one row up. If there is no number to its left, simply look at the column headed "1" in the previous row. Here is a small upper-left portion of the table:

Values of A(m, n)

n m	0	1	2	3	4	n
0	1	2	3	4	5	${\displaystyle n+1}$
1	2	3	4	5	6	${\displaystyle n+2=2+(n+3)-3}$
2	3	5	7	9	11	${\displaystyle 2n+3=2\cdot (n+3)-3}$
3	5	13	29	61	125	${\displaystyle 2^{(n+3)}-3}$
4	13 ${\displaystyle ={2^{2^{2}}}-3}$ ${\displaystyle =2\uparrow \uparrow 3-3}$	65533 ${\displaystyle ={2^{2^{2^{2}}}}-3}$ ${\displaystyle =2\uparrow \uparrow 4-3}$	265536 − 3 ${\displaystyle ={2^{2^{2^{2^{2}}}}}-3}$ ${\displaystyle =2\uparrow \uparrow 5-3}$	${\displaystyle {2^{2^{65536}}}-3}$ ${\displaystyle ={2^{2^{2^{2^{2^{2}}}}}}-3}$ ${\displaystyle =2\uparrow \uparrow 6-3}$	${\displaystyle {2^{2^{2^{65536}}}}-3}$ ${\displaystyle ={2^{2^{2^{2^{2^{2^{2}}}}}}}-3}$ ${\displaystyle =2\uparrow \uparrow 7-3}$	${\displaystyle {\begin{matrix}\underbrace {{2^{2}}^{{\cdot }^{{\cdot }^{{\cdot }^{2}}}}} _{n+3}-3\end{matrix}}}$ ${\displaystyle =2\uparrow \uparrow (n+3)-3}$
5	65533 ${\displaystyle =2\uparrow \uparrow (2\uparrow \uparrow 2)-3}$ ${\displaystyle =2\uparrow \uparrow \uparrow 3-3}$	${\displaystyle 2\uparrow \uparrow \uparrow 4-3}$	${\displaystyle 2\uparrow \uparrow \uparrow 5-3}$	${\displaystyle 2\uparrow \uparrow \uparrow 6-3}$	${\displaystyle 2\uparrow \uparrow \uparrow 7-3}$	${\displaystyle 2\uparrow \uparrow \uparrow (n+3)-3}$
6	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow 3-3}$	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow 4-3}$	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow 5-3}$	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow 6-3}$	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow 7-3}$	${\displaystyle 2\uparrow \uparrow \uparrow \uparrow (n+3)-3}$
m	${\displaystyle (2\rightarrow (3)\rightarrow (m-2))-3}$	${\displaystyle (2\rightarrow (4)\rightarrow (m-2))-3}$	${\displaystyle (2\rightarrow (5)\rightarrow (m-2))-3}$	${\displaystyle (2\rightarrow (6)\rightarrow (m-2))-3}$	${\displaystyle (2\rightarrow (7)\rightarrow (m-2))-3}$	${\displaystyle (2\rightarrow (n+3)\rightarrow (m-2))-3}$

The numbers here which are only expressed with recursive exponentiation or Knuth arrows are very large and would take up too much space to notate in plain decimal digits.

Despite the large values occurring in this early section of the table, some even larger numbers have been defined, such as Graham's number, which cannot be written with any small number of Knuth arrows. This number is constructed with a technique similar to applying the Ackermann function to itself recursively.

This is a repeat of the above table, but with the values replaced by the relevant expression from the function definition to show the pattern clearly:

In a sense, the Ackermann function grows faster than any primitive recursive function and therefore is not itself primitive recursive.

Specifically, one shows that to every primitive recursive function ${\displaystyle f(x_{1},\ldots ,x_{n})}$ there exists a non-negative integer ${\displaystyle t}$ such that for all non-negative integers ${\displaystyle x_{1},\ldots ,x_{n}}$,

${\displaystyle f(x_{1},\ldots ,x_{n})<A(t,\max _{i}x_{i}).}$

Once this is established, it follows that ${\displaystyle A}$ itself is not primitive recursive, since otherwise putting ${\displaystyle x_{1}=x_{2}=t}$ would lead to the contradiction ${\displaystyle A(t,t)<A(t,t)}$.

The proof[12] proceeds as follows: define the class ${\displaystyle {\mathcal {A}}}$ of all functions that grow slower than the Ackermann function

${\displaystyle {\mathcal {A}}=\left\{f\,{\bigg |}\,\exists t\ \forall x_{1}\cdots \forall x_{n}:\ f(x_{1},\ldots ,x_{n})<A(t,\max _{i}x_{i})\right\}}$

and show that ${\displaystyle {\mathcal {A}}}$ contains all primitive recursive functions. The latter is achieved by showing that ${\displaystyle {\mathcal {A}}}$ contains the constant functions, the successor function, the projection functions and that it is closed under the operations of function composition and primitive recursion.

Since the function  f(n) = A(n, n) considered above grows very rapidly, its inverse function, f⁻¹, grows very slowly. This inverse Ackermann function f⁻¹ is usually denoted by α. In fact, α(n) is less than 5 for any practical input size n, since A(4, 4) is on the order of ${\displaystyle 2^{2^{2^{2^{16}}}}}$.

This inverse appears in the time complexity of some algorithms, such as the disjoint-set data structure and Chazelle's algorithm for minimum spanning trees. Sometimes Ackermann's original function or other variations are used in these settings, but they all grow at similarly high rates. In particular, some modified functions simplify the expression by eliminating the −3 and similar terms.

A two-parameter variation of the inverse Ackermann function can be defined as follows, where ${\displaystyle \lfloor x\rfloor }$ is the floor function:

${\displaystyle \alpha (m,n)=\min\{i\geq 1:A(i,\lfloor m/n\rfloor )\geq \log _{2}n\}.}$

This function arises in more precise analyses of the algorithms mentioned above, and gives a more refined time bound. In the disjoint-set data structure, m represents the number of operations while n represents the number of elements; in the minimum spanning tree algorithm, m represents the number of edges while n represents the number of vertices. Several slightly different definitions of α(m, n) exist; for example, log₂ n is sometimes replaced by n, and the floor function is sometimes replaced by a ceiling.

Other studies might define an inverse function of one where m is set to a constant, such that the inverse applies to a particular row. [13]

The inverse of the Ackermann function is primitive recursive.[14]

The Ackermann function, due to its definition in terms of extremely deep recursion, can be used as a benchmark of a compiler's ability to optimize recursion. The first published use of Ackermann's function in this way was in 1970 by Dragoș Vaida[15] and, almost simultaneously, in 1971, by Yngve Sundblad.[16]

Sundblad's seminal paper was taken up by Brian Wichmann (co-author of the Whetstone benchmark) in a trilogy of papers written between 1975 and 1982.[17][18][19]

• Computability theory
• Double recursion
• Fast-growing hierarchy
• Goodstein function
• Primitive recursive function
• Recursion (computer science)

• "Ackermann function". Encyclopedia of Mathematics. EMS Press. 2001 [1994].
• Weisstein, Eric W. "Ackermann function". MathWorld.
•  This article incorporates public domain material from the NIST document: Black, Paul E. "Ackermann's function". Dictionary of Algorithms and Data Structures.
• An animated Ackermann function calculator
• Scott Aaronson, Who can name the biggest number? (1999)
• Ackermann functions. Includes a table of some values.
• Hyper-operations: Ackermann's Function and New Arithmetical Operation
• Robert Munafo's Large Numbers describes several variations on the definition of A.
• Gabriel Nivasch, Inverse Ackermann without pain on the inverse Ackermann function.
• Raimund Seidel, Understanding the inverse Ackermann function (PDF presentation).
• The Ackermann function written in different programming languages, (on Rosetta Code)
• Ackermann's Function (Archived 2009-10-24)—Some study and programming by Harry J. Smith.