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Ackermann function

From Wikipedia, the free encyclopedia
Quickly growing function
This article is about the mathematical function. For other uses, seeAckermann (disambiguation).

Incomputability theory, theAckermann function, named afterWilhelm Ackermann, is one of the simplest[1] and earliest-discovered examples of atotalcomputable function that is notprimitive recursive. All primitive recursive functions are total and computable, but the Ackermann function illustrates that not all total computable functions are primitive recursive.

After Ackermann's publication[2] of his function (which had three non-negative integer arguments), many authors modified it to suit various purposes, so that today "the Ackermann function" may refer to any of numerous variants of the original function. One common version is the two-argumentAckermann–Péter function developed byRózsa Péter andRaphael Robinson. This function is defined from therecurrence relationA(m+1,n+1)=A(m,A(m+1,n)){\displaystyle \operatorname {A} (m+1,n+1)=\operatorname {A} (m,\operatorname {A} (m+1,n))} with appropriatebase cases. Its value grows very rapidly; for example,A(4,2){\displaystyle \operatorname {A} (4,2)} results in2655363{\displaystyle 2^{65536}-3}, an integer with 19,729 decimal digits.[3]

History

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In the late 1920s, the mathematiciansGabriel Sudan andWilhelm Ackermann, students ofDavid Hilbert, were studying the foundations of computation. Both Sudan and Ackermann are credited[4] with discoveringtotalcomputable functions (termed simply "recursive" in some references) that are notprimitive recursive. Sudan published the lesser-knownSudan function, then shortly afterwards and independently, in 1928, Ackermann published his functionφ{\displaystyle \varphi } (from Greek, the letterphi). Ackermann's three-argument function,φ(m,n,p){\displaystyle \varphi (m,n,p)}, is defined such that forp=0,1,2{\displaystyle p=0,1,2}, it reproduces the basic operations ofaddition,multiplication, andexponentiation as

φ(m,n,0)=m+nφ(m,n,1)=m×nφ(m,n,2)=mn{\displaystyle {\begin{aligned}\varphi (m,n,0)&=m+n\\\varphi (m,n,1)&=m\times n\\\varphi (m,n,2)&=m^{n}\end{aligned}}}

and forp>2{\displaystyle p>2} it extends these basic operations in a way that can be compared to thehyperoperations:

φ(m,n,3)=m[4](n+1)φ(m,n,p)m[p+1](n+1)for p>3{\displaystyle {\begin{aligned}\varphi (m,n,3)&=m[4](n+1)\\\varphi (m,n,p)&\gtrapprox m[p+1](n+1)&&{\text{for }}p>3\end{aligned}}}

(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 asGoodstein'shyperoperation sequence.)

InOn 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 paperOn 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 almost all authors.

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

In 1963R. Creighton Buck based an intuitive two-variable[n 1] variantF{\displaystyle \operatorname {F} } on thehyperoperation sequence:[10][11]

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

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

F(0,n)=2[0]n=n+1F(1,n)=2[1]n=2+nF(2,n)=2[2]n=2×nF(3,n)=2[3]n=2nF(4,n)=2[4]n=222...2{\displaystyle {\begin{aligned}\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}}}}}}\\&\quad \vdots \end{aligned}}}

Many other versions of Ackermann function have been investigated.[12][13]

Definition

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Definition: asm-ary function

[edit]

Ackermann's original three-argument functionφ(m,n,p){\displaystyle \varphi (m,n,p)} is definedrecursively as follows for nonnegative integersm,n,{\displaystyle m,n,} andp{\displaystyle p}:

φ(m,n,0)=m+nφ(m,0,1)=0φ(m,0,2)=1φ(m,0,p)=mfor p>2φ(m,n,p)=φ(m,φ(m,n1,p),p1)for n,p>0{\displaystyle {\begin{aligned}\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,p>0\end{aligned}}}

Of the various two-argument versions, the one developed by Péter and Robinson (called "the" Ackermann function by most authors) is defined for nonnegative integersm{\displaystyle m} andn{\displaystyle n} as follows:

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

The Ackermann function has also been expressed in relation to thehyperoperation sequence:[14][15]

A(m,n)={n+1m=02[m](n+3)3m>0{\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2[m](n+3)-3&m>0\\\end{cases}}}

or, written inKnuth's up-arrow notation (extended to integer indices2{\displaystyle \geq -2}):

A(m,n)={n+1m=02m2(n+3)3m>0{\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2\uparrow ^{m-2}(n+3)-3&m>0\\\end{cases}}}

or, equivalently, in terms of Buck's function F:[10]

A(m,n)={n+1m=0F(m,n+3)3m>0{\displaystyle A(m,n)={\begin{cases}n+1&m=0\\F(m,n+3)-3&m>0\\\end{cases}}}

Byinduction onm{\displaystyle m}, one can show thatF(m,n)A(m,n){\displaystyle F(m,n)\leq A(m,n)} for allm,nN0{\displaystyle m,n\in \mathbb {N} _{0}}.

Definition: as iterated 1-ary function

[edit]

Definefn{\displaystyle f^{n}} as then-th iterate off{\displaystyle f}:

f0(x)=xfn+1(x)=f(fn(x)){\displaystyle {\begin{array}{rll}f^{0}(x)&=&x\\f^{n+1}(x)&=&f(f^{n}(x))\end{array}}}

Iteration is the process of composing a function with itself a certain number of times.Function composition is anassociative operation, sof(fn(x))=fn(f(x)){\displaystyle f(f^{n}(x))=f^{n}(f(x))}.

Conceiving the Ackermann function as a sequence of unary functions, one can setAm(n)=A(m,n){\displaystyle \operatorname {A} _{m}(n)=\operatorname {A} (m,n)}.

The function then becomes a sequenceA0,A1,A2,...{\displaystyle \operatorname {A} _{0},\operatorname {A} _{1},\operatorname {A} _{2},...} of unary[n 2] functions, defined fromiteration:

A0(n)=n+1Am+1(n)=Amn+1(1).{\displaystyle {\begin{array}{lcl}\operatorname {A} _{0}(n)&=&n+1\\\operatorname {A} _{m+1}(n)&=&\operatorname {A} _{m}^{n+1}(1)\,.\end{array}}}

Computation

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Computation by LOOP program

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The functionsAi{\displaystyle \operatorname {A} _{i}} fit into the (finite-level)fast-growing hierarchy (FGH) of functions[16]

FGH0(n)=n+1FGHm+1(n)=FGHmn(n).{\displaystyle {\begin{array}{lcl}\operatorname {FGH} _{0}(n)&=&n+1\\\operatorname {FGH} _{m+1}(n)&=&\operatorname {FGH} _{m}^{n}(n)\,.\end{array}}}

The following inequality holds:[17]m>1,n>1:Am(n)<FGHm(n){\displaystyle \forall m>1,\forall n>1:\operatorname {A} _{m}(n)<\operatorname {FGH} _{m}(n)}

For fixedk{\displaystyle k}, the functionFGHk(n){\displaystyle \operatorname {FGH} _{k}(n)} can be computed by aLOOP program of nesting depthk{\displaystyle k}:[18]

# INPUT (n)LOOPn:# nesting depth: 1LOOPn:# nesting depth: 2...# ...LOOPn:# nesting depth: kn+=1## OUTPUT (n)

The functionAk(n){\displaystyle \operatorname {A} _{k}(n)} can also be computed by a LOOP-k program.[19] (The program (schema) is not listed here.)

It is obvious thatA(m,n){\displaystyle \operatorname {A} (m,n)}, not being aprimitive recursive function —see below—, cannot be computed by a LOOP program.

Computation by term rewriting system, based on 2-ary function

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Therecursive definition of the Ackermann function can naturally be transposed to aterm rewriting system (TRS).

The definition of the2-ary Ackermann function leads to the obvious reduction rules[20][21]

(r1)A(0,n)S(n)(r2)A(S(m),0)A(m,S(0))(r3)A(S(m),S(n))A(m,A(S(m),n)){\displaystyle {\begin{array}{lll}{\text{(r1)}}&A(0,n)&\rightarrow &S(n)\\{\text{(r2)}}&A(S(m),0)&\rightarrow &A(m,S(0))\\{\text{(r3)}}&A(S(m),S(n))&\rightarrow &A(m,A(S(m),n))\end{array}}}

Example

ComputeA(1,2)4{\displaystyle A(1,2)\rightarrow _{*}4}

The reduction sequence is[n 3]

Leftmost-outermost (one-step) strategy:            Leftmost-innermost (one-step) strategy:
A(S(0),S(S(0)))_{\displaystyle {\underline {A(S(0),S(S(0)))}}}A(S(0),S(S(0)))_{\displaystyle {\underline {A(S(0),S(S(0)))}}}
    r3A(0,A(S(0),S(0))_){\displaystyle \rightarrow _{r3}{\underline {A(0,A(S(0),S(0))}})}    r3A(0,A(S(0),S(0))_){\displaystyle \rightarrow _{r3}A(0,{\underline {A(S(0),S(0))}})}
    r1S(A(S(0),S(0))_){\displaystyle \rightarrow _{r1}S({\underline {A(S(0),S(0))}})}    r3A(0,A(0,A(S(0),0)_)){\displaystyle \rightarrow _{r3}A(0,A(0,{\underline {A(S(0),0)}}))}
    r3S(A(0,A(S0,0))_){\displaystyle \rightarrow _{r3}S({\underline {A(0,A(S0,0))}})}    r2A(0,A(0,A(0,S(0))_)){\displaystyle \rightarrow _{r2}A(0,A(0,{\underline {A(0,S(0))}}))}
    r1S(S(A(S(0),0)_)){\displaystyle \rightarrow _{r1}S(S({\underline {A(S(0),0)}}))}    r1A(0,A(0,S(S(0)))_){\displaystyle \rightarrow _{r1}A(0,{\underline {A(0,S(S(0)))}})}
    r2S(S(A(0,S(0))_)){\displaystyle \rightarrow _{r2}S(S({\underline {A(0,S(0))}}))}    r1A(0,S(S(S(0))))_{\displaystyle \rightarrow _{r1}{\underline {A(0,S(S(S(0))))}}}
    r1S(S(S(S(0)))){\displaystyle \rightarrow _{r1}S(S(S(S(0))))}    r1S(S(S(S(0)))){\displaystyle \rightarrow _{r1}S(S(S(S(0))))}

To computeA(m,n){\displaystyle \operatorname {A} (m,n)} one can use astack, which initially contains the elementsm,n{\displaystyle \langle m,n\rangle }.

Then repeatedly the two top elements are replaced according to the rules[n 4]

(r1)0,n(n+1)(r2)(m+1),0m,1(r3)(m+1),(n+1)m,(m+1),n{\displaystyle {\begin{array}{lllllllll}{\text{(r1)}}&0&,&n&\rightarrow &(n+1)\\{\text{(r2)}}&(m+1)&,&0&\rightarrow &m&,&1\\{\text{(r3)}}&(m+1)&,&(n+1)&\rightarrow &m&,&(m+1)&,&n\end{array}}}

Schematically, starting fromm,n{\displaystyle \langle m,n\rangle }:

WHILE stackLength <> 1{POP 2 elements;PUSH 1 or 2 or 3 elements, applying the rules r1, r2, r3}

Thepseudocode is published inGrossman & Zeitman (1988).

For example, on input2,1{\displaystyle \langle 2,1\rangle },

the stack configurations    reflect the reduction[n 5]
2,1_{\displaystyle {\underline {2,1}}}A(2,1)_{\displaystyle {\underline {A(2,1)}}}
    1,2,0_{\displaystyle \rightarrow 1,{\underline {2,0}}}    r1A(1,A(2,0)_){\displaystyle \rightarrow _{r1}A(1,{\underline {A(2,0)}})}
    1,1,1_{\displaystyle \rightarrow 1,{\underline {1,1}}}    r2A(1,A(1,1)_){\displaystyle \rightarrow _{r2}A(1,{\underline {A(1,1)}})}
    1,0,1,0_{\displaystyle \rightarrow 1,0,{\underline {1,0}}}    r3A(1,A(0,A(1,0)_)){\displaystyle \rightarrow _{r3}A(1,A(0,{\underline {A(1,0)}}))}
    1,0,0,1_{\displaystyle \rightarrow 1,0,{\underline {0,1}}}    r2A(1,A(0,A(0,1)_)){\displaystyle \rightarrow _{r2}A(1,A(0,{\underline {A(0,1)}}))}
    1,0,2_{\displaystyle \rightarrow 1,{\underline {0,2}}}    r1A(1,A(0,2)_){\displaystyle \rightarrow _{r1}A(1,{\underline {A(0,2)}})}
    1,3_{\displaystyle \rightarrow {\underline {1,3}}}    r1A(1,3)_{\displaystyle \rightarrow _{r1}{\underline {A(1,3)}}}
    0,1,2_{\displaystyle \rightarrow 0,{\underline {1,2}}}    r3A(0,A(1,2)_){\displaystyle \rightarrow _{r3}A(0,{\underline {A(1,2)}})}
    0,0,1,1_{\displaystyle \rightarrow 0,0,{\underline {1,1}}}    r3A(0,A(0,A(1,1)_)){\displaystyle \rightarrow _{r3}A(0,A(0,{\underline {A(1,1)}}))}
    0,0,0,1,0_{\displaystyle \rightarrow 0,0,0,{\underline {1,0}}}    r3A(0,A(0,A(0,A(1,0)_))){\displaystyle \rightarrow _{r3}A(0,A(0,A(0,{\underline {A(1,0)}})))}
    0,0,0,0,1_{\displaystyle \rightarrow 0,0,0,{\underline {0,1}}}    r2A(0,A(0,A(0,A(0,1)_))){\displaystyle \rightarrow _{r2}A(0,A(0,A(0,{\underline {A(0,1)}})))}
    0,0,0,2_{\displaystyle \rightarrow 0,0,{\underline {0,2}}}    r1A(0,A(0,A(0,2)_)){\displaystyle \rightarrow _{r1}A(0,A(0,{\underline {A(0,2)}}))}
    0,0,3_{\displaystyle \rightarrow 0,{\underline {0,3}}}    r1A(0,A(0,3)_){\displaystyle \rightarrow _{r1}A(0,{\underline {A(0,3)}})}
    0,4_{\displaystyle \rightarrow {\underline {0,4}}}    r1A(0,4)_{\displaystyle \rightarrow _{r1}{\underline {A(0,4)}}}
    5{\displaystyle \rightarrow 5}    r15{\displaystyle \rightarrow _{r1}5}

Remarks

Computation by TRS, based on iterated 1-ary function

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The definition of the iterated1-ary Ackermann functions leads to different reduction rules

(r4)A(S(0),0,n)S(n)(r5)A(S(0),S(m),n)A(S(n),m,S(0))(r6)A(S(S(x)),m,n)A(S(0),m,A(S(x),m,n)){\displaystyle {\begin{array}{lll}{\text{(r4)}}&A(S(0),0,n)&\rightarrow &S(n)\\{\text{(r5)}}&A(S(0),S(m),n)&\rightarrow &A(S(n),m,S(0))\\{\text{(r6)}}&A(S(S(x)),m,n)&\rightarrow &A(S(0),m,A(S(x),m,n))\end{array}}}

As function composition is associative, instead of rule r6 one can define

(r7)A(S(S(x)),m,n)A(S(x),m,A(S(0),m,n)){\displaystyle {\begin{array}{lll}{\text{(r7)}}&A(S(S(x)),m,n)&\rightarrow &A(S(x),m,A(S(0),m,n))\end{array}}}

Like in the previous section the computation ofAm1(n){\displaystyle \operatorname {A} _{m}^{1}(n)} can be implemented with a stack.

Initially the stack contains the three elements1,m,n{\displaystyle \langle 1,m,n\rangle }.

Then repeatedly the three top elements are replaced according to the rules[n 4]

(r4)1,0,n(n+1)(r5)1,(m+1),n(n+1),m,1(r6)(x+2),m,n1,m,(x+1),m,n{\displaystyle {\begin{array}{lllllllll}{\text{(r4)}}&1&,0&,n&\rightarrow &(n+1)\\{\text{(r5)}}&1&,(m+1)&,n&\rightarrow &(n+1)&,m&,1\\{\text{(r6)}}&(x+2)&,m&,n&\rightarrow &1&,m&,(x+1)&,m&,n\\\end{array}}}

Schematically, starting from1,m,n{\displaystyle \langle 1,m,n\rangle }:

WHILE stackLength <> 1{POP 3 elements;PUSH 1 or 3 or 5 elements, applying the rules r4, r5, r6;}

Example

On input1,2,1{\displaystyle \langle 1,2,1\rangle } the successive stack configurations are

1,2,1_r52,1,1_r61,1,1,1,1_r51,1,2,0,1_r61,1,1,0,1,0,1_r41,1,1,0,2_r41,1,3_r54,0,1_r61,0,3,0,1_r61,0,1,0,2,0,1_r61,0,1,0,1,0,1,0,1_r41,0,1,0,1,0,2_r41,0,1,0,3_r41,0,4_r45{\displaystyle {\begin{aligned}&{\underline {1,2,1}}\rightarrow _{r5}{\underline {2,1,1}}\rightarrow _{r6}1,1,{\underline {1,1,1}}\rightarrow _{r5}1,1,{\underline {2,0,1}}\rightarrow _{r6}1,1,1,0,{\underline {1,0,1}}\\&\rightarrow _{r4}1,1,{\underline {1,0,2}}\rightarrow _{r4}{\underline {1,1,3}}\rightarrow _{r5}{\underline {4,0,1}}\rightarrow _{r6}1,0,{\underline {3,0,1}}\rightarrow _{r6}1,0,1,0,{\underline {2,0,1}}\\&\rightarrow _{r6}1,0,1,0,1,0,{\underline {1,0,1}}\rightarrow _{r4}1,0,1,0,{\underline {1,0,2}}\rightarrow _{r4}1,0,{\underline {1,0,3}}\rightarrow _{r4}{\underline {1,0,4}}\rightarrow _{r4}5\end{aligned}}}

The corresponding equalities are

A2(1)=A12(1)=A1(A1(1))=A1(A02(1))=A1(A0(A0(1)))=A1(A0(2))=A1(3)=A04(1)=A0(A03(1))=A0(A0(A02(1)))=A0(A0(A0(A0(1))))=A0(A0(A0(2)))=A0(A0(3))=A0(4)=5{\displaystyle {\begin{aligned}&A_{2}(1)=A_{1}^{2}(1)=A_{1}(A_{1}(1))=A_{1}(A_{0}^{2}(1))=A_{1}(A_{0}(A_{0}(1)))\\&=A_{1}(A_{0}(2))=A_{1}(3)=A_{0}^{4}(1)=A_{0}(A_{0}^{3}(1))=A_{0}(A_{0}(A_{0}^{2}(1)))\\&=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{aligned}}}

When reduction rule r7 is used instead of rule r6, the replacements in the stack will follow

(r7)(x+2),m,n(x+1),m,1,m,n{\displaystyle {\begin{array}{lllllllll}{\text{(r7)}}&(x+2)&,m&,n&\rightarrow &(x+1)&,m&,1&,m&,n\end{array}}}

The successive stack configurations will then be

1,2,1_r52,1,1_r71,1,1,1,1_r51,1,2,0,1_r71,1,1,0,1,0,1_r41,1,1,0,2_r41,1,3_r54,0,1_r73,0,1,0,1_r43,0,2_r72,0,1,0,2_r42,0,3_r71,0,1,0,3_r41,0,4_r45{\displaystyle {\begin{aligned}&{\underline {1,2,1}}\rightarrow _{r5}{\underline {2,1,1}}\rightarrow _{r7}1,1,{\underline {1,1,1}}\rightarrow _{r5}1,1,{\underline {2,0,1}}\rightarrow _{r7}1,1,1,0,{\underline {1,0,1}}\\&\rightarrow _{r4}1,1,{\underline {1,0,2}}\rightarrow _{r4}{\underline {1,1,3}}\rightarrow _{r5}{\underline {4,0,1}}\rightarrow _{r7}3,0,{\underline {1,0,1}}\rightarrow _{r4}{\underline {3,0,2}}\\&\rightarrow _{r7}2,0,{\underline {1,0,2}}\rightarrow _{r4}{\underline {2,0,3}}\rightarrow _{r7}1,0,{\underline {1,0,3}}\rightarrow _{r4}{\underline {1,0,4}}\rightarrow _{r4}5\end{aligned}}}

The corresponding equalities are

A2(1)=A12(1)=A1(A1(1))=A1(A02(1))=A1(A0(A0(1)))=A1(A0(2))=A1(3)=A04(1)=A03(A0(1))=A03(2)=A02(A0(2))=A02(3)=A0(A0(3))=A0(4)=5{\displaystyle {\begin{aligned}&A_{2}(1)=A_{1}^{2}(1)=A_{1}(A_{1}(1))=A_{1}(A_{0}^{2}(1))=A_{1}(A_{0}(A_{0}(1)))\\&=A_{1}(A_{0}(2))=A_{1}(3)=A_{0}^{4}(1)=A_{0}^{3}(A_{0}(1))=A_{0}^{3}(2)\\&=A_{0}^{2}(A_{0}(2))=A_{0}^{2}(3)=A_{0}(A_{0}(3))=A_{0}(4)=5\end{aligned}}}

Remarks

  • On any given input the TRSs presented so far converge in the same number of steps. They also use the same reduction rules (in this comparison the rules r1, r2, r3 are considered "the same as" the rules r4, r5, r6/r7 respectively). For example, the reduction ofA(2,1){\displaystyle A(2,1)} converges in 14 steps: 6 × r1, 3 × r2, 5 × r3. The reduction ofA2(1){\displaystyle A_{2}(1)} converges in the same 14 steps: 6 × r4, 3 × r5, 5 × r6/r7. The TRSs differ in the order in which the reduction rules are applied.
  • WhenAi(n){\displaystyle A_{i}(n)} is computed following the rules {r4, r5, r6}, the maximum length of the stack stays below2×A(i,n){\displaystyle 2\times A(i,n)}. When reduction rule r7 is used instead of rule r6, the maximum length of the stack is only2(i+2){\displaystyle 2(i+2)}. The length of the stack reflects the recursion depth. As the reduction according to the rules {r4, r5, r7} involves a smaller maximum depth of recursion,[n 6] this computation is more efficient in that respect.

Computation by TRS, based on hyperoperators

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AsSundblad (1971) — orPorto & Matos (1980) — showed explicitly, the Ackermann function can be expressed in terms of thehyperoperation sequence:

A(m,n)={n+1m=02[m](n+3)3m>0{\displaystyle A(m,n)={\begin{cases}n+1&m=0\\2[m](n+3)-3&m>0\\\end{cases}}}

or, after removal of the constant 2 from the parameter list, in terms of Buck's function

A(m,n)={n+1m=0F(m,n+3)3m>0{\displaystyle A(m,n)={\begin{cases}n+1&m=0\\F(m,n+3)-3&m>0\\\end{cases}}}

Buck's functionF(m,n)=2[m]n{\displaystyle \operatorname {F} (m,n)=2[m]n},[10] a variant of Ackermann function by itself, can be computed with the following reduction rules:

(b1)F(S(0),0,n)S(n)(b2)F(S(0),S(0),0)S(S(0))(b3)F(S(0),S(S(0)),0)0(b4)F(S(0),S(S(S(m))),0)S(0)(b5)F(S(0),S(m),S(n))F(S(n),m,F(S(0),S(m),0))(b6)F(S(S(x)),m,n)F(S(0),m,F(S(x),m,n)){\displaystyle {\begin{array}{lll}{\text{(b1)}}&F(S(0),0,n)&\rightarrow &S(n)\\{\text{(b2)}}&F(S(0),S(0),0)&\rightarrow &S(S(0))\\{\text{(b3)}}&F(S(0),S(S(0)),0)&\rightarrow &0\\{\text{(b4)}}&F(S(0),S(S(S(m))),0)&\rightarrow &S(0)\\{\text{(b5)}}&F(S(0),S(m),S(n))&\rightarrow &F(S(n),m,F(S(0),S(m),0))\\{\text{(b6)}}&F(S(S(x)),m,n)&\rightarrow &F(S(0),m,F(S(x),m,n))\end{array}}}Instead of rule b6 one can define the rule

(b7)F(S(S(x)),m,n)F(S(x),m,F(S(0),m,n)){\displaystyle {\begin{array}{lll}{\text{(b7)}}&F(S(S(x)),m,n)&\rightarrow &F(S(x),m,F(S(0),m,n))\end{array}}}To compute the Ackermann function it suffices to add three reduction rules

(r8)A(0,n)S(n)(r9)A(S(m),n)P(F(S(0),S(m),S(S(S(n)))))(r10)P(S(S(S(m))))m{\displaystyle {\begin{array}{lll}{\text{(r8)}}&A(0,n)&\rightarrow &S(n)\\{\text{(r9)}}&A(S(m),n)&\rightarrow &P(F(S(0),S(m),S(S(S(n)))))\\{\text{(r10)}}&P(S(S(S(m))))&\rightarrow &m\\\end{array}}}

These rules take care of the base case A(0,n), the alignment (n+3) and the fudge (-3).

Example

ComputeA(2,1)5{\displaystyle A(2,1)\rightarrow _{*}5}

using reduction ruleb7{\displaystyle {\text{b7}}}:[n 5]    using reduction ruleb6{\displaystyle {\text{b6}}}:[n 5]
A(2,1)_{\displaystyle {\underline {A(2,1)}}}A(2,1)_{\displaystyle {\underline {A(2,1)}}}
    r9P(F(1,2,4)_){\displaystyle \rightarrow _{r9}P({\underline {F(1,2,4)}})}    r9P(F(1,2,4)_){\displaystyle \rightarrow _{r9}P({\underline {F(1,2,4)}})}
    b5P(F(4,1,F(1,2,0)_)){\displaystyle \rightarrow _{b5}P(F(4,1,{\underline {F(1,2,0)}}))}    b5P(F(4,1,F(1,2,0)_)){\displaystyle \rightarrow _{b5}P(F(4,1,{\underline {F(1,2,0)}}))}
    b3P(F(4,1,0)_){\displaystyle \rightarrow _{b3}P({\underline {F(4,1,0)}})}    b3P(F(4,1,0)_){\displaystyle \rightarrow _{b3}P({\underline {F(4,1,0)}})}
    b7P(F(3,1,F(1,1,0)_)){\displaystyle \rightarrow _{b7}P(F(3,1,{\underline {F(1,1,0)}}))}    b6P(F(1,1,F(3,1,0)_)){\displaystyle \rightarrow _{b6}P(F(1,1,{\underline {F(3,1,0)}}))}
    b2P(F(3,1,2)_){\displaystyle \rightarrow _{b2}P({\underline {F(3,1,2)}})}    b6P(F(1,1,F(1,1,F(2,1,0)_))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,{\underline {F(2,1,0)}})))}
    b7P(F(2,1,F(1,1,2)_)){\displaystyle \rightarrow _{b7}P(F(2,1,{\underline {F(1,1,2)}}))}    b6P(F(1,1,F(1,1,F(1,1,F(1,1,0)_)))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,F(1,1,{\underline {F(1,1,0)}}))))}
    b5P(F(2,1,F(2,0,F(1,1,0)_))){\displaystyle \rightarrow _{b5}P(F(2,1,F(2,0,{\underline {F(1,1,0)}})))}              b2P(F(1,1,F(1,1,F(1,1,2)_))){\displaystyle \rightarrow _{b2}P(F(1,1,F(1,1,{\underline {F(1,1,2)}})))}
    b2P(F(2,1,F(2,0,2)_)){\displaystyle \rightarrow _{b2}P(F(2,1,{\underline {F(2,0,2)}}))}    b5P(F(1,1,F(1,1,F(2,0,F(1,1,0)_)))){\displaystyle \rightarrow _{b5}P(F(1,1,F(1,1,F(2,0,{\underline {F(1,1,0)}}))))}
    b7P(F(2,1,F(1,0,F(1,0,2)_))){\displaystyle \rightarrow _{b7}P(F(2,1,F(1,0,{\underline {F(1,0,2)}})))}    b2P(F(1,1,F(1,1,F(2,0,2)_))){\displaystyle \rightarrow _{b2}P(F(1,1,F(1,1,{\underline {F(2,0,2)}})))}
    b1P(F(2,1,F(1,0,3)_)){\displaystyle \rightarrow _{b1}P(F(2,1,{\underline {F(1,0,3)}}))}    b6P(F(1,1,F(1,1,F(1,0,F(1,0,2)_)))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,1,F(1,0,{\underline {F(1,0,2)}}))))}
    b1P(F(2,1,4)_){\displaystyle \rightarrow _{b1}P({\underline {F(2,1,4)}})}    b1P(F(1,1,F(1,1,F(1,0,3)_))){\displaystyle \rightarrow _{b1}P(F(1,1,F(1,1,{\underline {F(1,0,3)}})))}
    b7P(F(1,1,F(1,1,4)_)){\displaystyle \rightarrow _{b7}P(F(1,1,{\underline {F(1,1,4)}}))}    b1P(F(1,1,F(1,1,4)_)){\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,1,4)}}))}
    b5P(F(1,1,F(4,0,F(1,1,0)_))){\displaystyle \rightarrow _{b5}P(F(1,1,F(4,0,{\underline {F(1,1,0)}})))}    b5P(F(1,1,F(4,0,F(1,1,0)_))){\displaystyle \rightarrow _{b5}P(F(1,1,F(4,0,{\underline {F(1,1,0)}})))}
    b2P(F(1,1,F(4,0,2)_)){\displaystyle \rightarrow _{b2}P(F(1,1,{\underline {F(4,0,2)}}))}    b2P(F(1,1,F(4,0,2)_)){\displaystyle \rightarrow _{b2}P(F(1,1,{\underline {F(4,0,2)}}))}
    b7P(F(1,1,F(3,0,F(1,0,2)_))){\displaystyle \rightarrow _{b7}P(F(1,1,F(3,0,{\underline {F(1,0,2)}})))}    b6P(F(1,1,F(1,0,F(3,0,2)_))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,{\underline {F(3,0,2)}})))}
    b1P(F(1,1,F(3,0,3)_)){\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(3,0,3)}}))}    b6P(F(1,1,F(1,0,F(1,0,F(2,0,2)_)))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,F(1,0,{\underline {F(2,0,2)}}))))}
    b7P(F(1,1,F(2,0,F(1,0,3)_))){\displaystyle \rightarrow _{b7}P(F(1,1,F(2,0,{\underline {F(1,0,3)}})))}    b6P(F(1,1,F(1,0,F(1,0,F(1,0,F(1,0,2)_))))){\displaystyle \rightarrow _{b6}P(F(1,1,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,2)}})))))}
    b1P(F(1,1,F(2,0,4)_)){\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(2,0,4)}}))}    b1P(F(1,1,F(1,0,F(1,0,F(1,0,3)_)))){\displaystyle \rightarrow _{b1}P(F(1,1,F(1,0,F(1,0,{\underline {F(1,0,3)}}))))}
    b7P(F(1,1,F(1,0,F(1,0,4)_))){\displaystyle \rightarrow _{b7}P(F(1,1,F(1,0,{\underline {F(1,0,4)}})))}    b1P(F(1,1,F(1,0,F(1,0,4)_))){\displaystyle \rightarrow _{b1}P(F(1,1,F(1,0,{\underline {F(1,0,4)}})))}
    b1P(F(1,1,F(1,0,5)_)){\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,0,5)}}))}    b1P(F(1,1,F(1,0,5)_)){\displaystyle \rightarrow _{b1}P(F(1,1,{\underline {F(1,0,5)}}))}
    b1P(F(1,1,6)_){\displaystyle \rightarrow _{b1}P({\underline {F(1,1,6)}})}    b1P(F(1,1,6)_){\displaystyle \rightarrow _{b1}P({\underline {F(1,1,6)}})}
    b5P(F(6,0,F(1,1,0)_)){\displaystyle \rightarrow _{b5}P(F(6,0,{\underline {F(1,1,0)}}))}    b5P(F(6,0,F(1,1,0)_)){\displaystyle \rightarrow _{b5}P(F(6,0,{\underline {F(1,1,0)}}))}
    b2P(F(6,0,2)_){\displaystyle \rightarrow _{b2}P({\underline {F(6,0,2)}})}    b2P(F(6,0,2)_){\displaystyle \rightarrow _{b2}P({\underline {F(6,0,2)}})}
    b7P(F(5,0,F(1,0,2)_)){\displaystyle \rightarrow _{b7}P(F(5,0,{\underline {F(1,0,2)}}))}    b6P(F(1,0,F(5,0,2)_)){\displaystyle \rightarrow _{b6}P(F(1,0,{\underline {F(5,0,2)}}))}
    b1P(F(5,0,3)_){\displaystyle \rightarrow _{b1}P({\underline {F(5,0,3)}})}    b6P(F(1,0,F(1,0,F(4,0,2)_))){\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,{\underline {F(4,0,2)}})))}
    b7P(F(4,0,F(1,0,3)_)){\displaystyle \rightarrow _{b7}P(F(4,0,{\underline {F(1,0,3)}}))}    b6P(F(1,0,F(1,0,F(1,0,F(3,0,2)_)))){\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,{\underline {F(3,0,2)}}))))}
    b1P(F(4,0,4)_){\displaystyle \rightarrow _{b1}P({\underline {F(4,0,4)}})}    b6P(F(1,0,F(1,0,F(1,0,F(1,0,F(2,0,2)_))))){\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(2,0,2)}})))))}
    b7P(F(3,0,F(1,0,4)_)){\displaystyle \rightarrow _{b7}P(F(3,0,{\underline {F(1,0,4)}}))}    b6P(F(1,0,F(1,0,F(1,0,F(1,0,F(1,0,F(1,0,2)_)))))){\displaystyle \rightarrow _{b6}P(F(1,0,F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,2)}}))))))}
    b1P(F(3,0,5)_){\displaystyle \rightarrow _{b1}P({\underline {F(3,0,5)}})}    b1P(F(1,0,F(1,0,F(1,0,F(1,0,F(1,0,3)_))))){\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,F(1,0,F(1,0,{\underline {F(1,0,3)}})))))}
    b7P(F(2,0,F(1,0,5)_)){\displaystyle \rightarrow _{b7}P(F(2,0,{\underline {F(1,0,5)}}))}    b1P(F(1,0,F(1,0,F(1,0,F(1,0,4)_)))){\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,F(1,0,{\underline {F(1,0,4)}}))))}
    b1P(F(2,0,6)_){\displaystyle \rightarrow _{b1}P({\underline {F(2,0,6)}})}    b1P(F(1,0,F(1,0,F(1,0,5)_))){\displaystyle \rightarrow _{b1}P(F(1,0,F(1,0,{\underline {F(1,0,5)}})))}
    b7P(F(1,0,F(1,0,6)_)){\displaystyle \rightarrow _{b7}P(F(1,0,{\underline {F(1,0,6)}}))}    b1P(F(1,0,F(1,0,6)_)){\displaystyle \rightarrow _{b1}P(F(1,0,{\underline {F(1,0,6)}}))}
    b1P(F(1,0,7)_){\displaystyle \rightarrow _{b1}P({\underline {F(1,0,7)}})}    b1P(F(1,0,7)_){\displaystyle \rightarrow _{b1}P({\underline {F(1,0,7)}})}
    b1P(8)_{\displaystyle \rightarrow _{b1}{\underline {P(8)}}}    b1P(8)_{\displaystyle \rightarrow _{b1}{\underline {P(8)}}}
    r105{\displaystyle \rightarrow _{r10}5}    r105{\displaystyle \rightarrow _{r10}5}

The matching equalities are

A(2,1)+3=F(2,4)==F6(0,2)=F(0,F5(0,2))=F(0,F(0,F4(0,2)))=F(0,F(0,F(0,F3(0,2))))=F(0,F(0,F(0,F(0,F2(0,2)))))=F(0,F(0,F(0,F(0,F(0,F(0,2))))))=F(0,F(0,F(0,F(0,F(0,3)))))=F(0,F(0,F(0,F(0,4))))=F(0,F(0,F(0,5)))=F(0,F(0,6))=F(0,7)=8{\displaystyle {\begin{aligned}&A(2,1)+3=F(2,4)=\dots =F^{6}(0,2)=F(0,F^{5}(0,2))=F(0,F(0,F^{4}(0,2)))\\&=F(0,F(0,F(0,F^{3}(0,2))))=F(0,F(0,F(0,F(0,F^{2}(0,2)))))=F(0,F(0,F(0,F(0,F(0,F(0,2))))))\\&=F(0,F(0,F(0,F(0,F(0,3)))))=F(0,F(0,F(0,F(0,4))))=F(0,F(0,F(0,5)))=F(0,F(0,6))=F(0,7)=8\end{aligned}}}

A(2,1)+3=F(2,4)==F6(0,2)=F5(0,F(0,2))=F5(0,3)=F4(0,F(0,3))=F4(0,4)=F3(0,F(0,4))=F3(0,5)=F2(0,F(0,5))=F2(0,6)=F(0,F(0,6))=F(0,7)=8{\displaystyle {\begin{aligned}&A(2,1)+3=F(2,4)=\dots =F^{6}(0,2)=F^{5}(0,F(0,2))=F^{5}(0,3)=F^{4}(0,F(0,3))=F^{4}(0,4)\\&=F^{3}(0,F(0,4))=F^{3}(0,5)=F^{2}(0,F(0,5))=F^{2}(0,6)=F(0,F(0,6))=F(0,7)=8\end{aligned}}}Remarks

Huge numbers

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To demonstrate how the computation ofA(4,3){\displaystyle A(4,3)} results in many steps and in a large number:[n 5]

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))))A(3,A(3,A(3,13)))A(3,A(3,65533))A(3,2655363)22655363.{\displaystyle {\begin{aligned}A(4,3)&\rightarrow A(3,A(4,2))\\&\rightarrow A(3,A(3,A(4,1)))\\&\rightarrow A(3,A(3,A(3,A(4,0))))\\&\rightarrow A(3,A(3,A(3,A(3,1))))\\&\rightarrow A(3,A(3,A(3,A(2,A(3,0)))))\\&\rightarrow A(3,A(3,A(3,A(2,A(2,1)))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,A(2,0))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,A(1,1))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,A(0,A(1,0)))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,A(0,A(0,1)))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,A(0,2))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(1,3)))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(1,2))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(0,A(1,1)))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(0,A(0,A(1,0))))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(0,A(0,A(0,1))))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(0,A(0,2)))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,A(0,3))))))\\&\rightarrow A(3,A(3,A(3,A(2,A(0,4)))))\\&\rightarrow A(3,A(3,A(3,A(2,5))))\\&\qquad \vdots \\&\rightarrow A(3,A(3,A(3,13)))\\&\qquad \vdots \\&\rightarrow A(3,A(3,65533))\\&\qquad \vdots \\&\rightarrow A(3,2^{65536}-3)\\&\qquad \vdots \\&\rightarrow 2^{2^{65536}}-3.\\\end{aligned}}}

Table of values

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Computing the Ackermann function can be restated in terms of an infinite table. First, place thenatural 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 ofA(mn)
n
m
01234n
012345n+1{\displaystyle n+1}
123456n+2=2+(n+3)3{\displaystyle n+2=2+(n+3)-3}
23579112n+3=2(n+3)3{\displaystyle 2n+3=2\cdot (n+3)-3}
351329611252(n+3)3{\displaystyle 2^{(n+3)}-3}
41365533265536 – 322655363{\displaystyle {2^{2^{65536}}}-3}222655363{\displaystyle {2^{2^{2^{65536}}}}-3}222n+33{\displaystyle {\begin{matrix}\underbrace {{2^{2}}^{{\cdot }^{{\cdot }^{{\cdot }^{2}}}}} _{n+3}-3\end{matrix}}}
=2↑↑53{\displaystyle =2\uparrow \uparrow 5-3}
2.003531019728{\displaystyle \approx 2.00353\cdot {10^{19728}}}
=2↑↑63{\displaystyle =2\uparrow \uparrow 6-3}
=2↑↑73{\displaystyle =2\uparrow \uparrow 7-3}
=2↑↑(n+3)3{\displaystyle =2\uparrow \uparrow (n+3)-3}
5655332↑↑655363{\displaystyle 2\uparrow \uparrow 65536-3}2↑↑↑53{\displaystyle 2\uparrow \uparrow \uparrow 5-3}2↑↑↑63{\displaystyle 2\uparrow \uparrow \uparrow 6-3}2↑↑↑73{\displaystyle 2\uparrow \uparrow \uparrow 7-3}2↑↑↑(n+3)3{\displaystyle 2\uparrow \uparrow \uparrow (n+3)-3}
62↑↑655363{\displaystyle 2\uparrow \uparrow 65536-3}2↑↑↑↑43{\displaystyle 2\uparrow \uparrow \uparrow \uparrow 4-3}2↑↑↑↑53{\displaystyle 2\uparrow \uparrow \uparrow \uparrow 5-3}2↑↑↑↑63{\displaystyle 2\uparrow \uparrow \uparrow \uparrow 6-3}2↑↑↑↑73{\displaystyle 2\uparrow \uparrow \uparrow \uparrow 7-3}2↑↑↑↑(n+3)3{\displaystyle 2\uparrow \uparrow \uparrow \uparrow (n+3)-3}
m(2m23)3{\displaystyle (2\uparrow ^{m-2}3)-3}(2m24)3{\displaystyle (2\uparrow ^{m-2}4)-3}(2m25)3{\displaystyle (2\uparrow ^{m-2}5)-3}(2m26)3{\displaystyle (2\uparrow ^{m-2}6)-3}(2m27)3{\displaystyle (2\uparrow ^{m-2}7)-3}(2m2(n+3))3{\displaystyle (2\uparrow ^{m-2}(n+3))-3}

The numbers here that are only expressed with recursive exponentiation orKnuth 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 asGraham'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:

Values ofA(mn)
n
m
01234n
00+11+12+13+14+1n + 1
1A(0, 1)A(0,A(1, 0))
=A(0, 2)
A(0,A(1, 1))
=A(0, 3)
A(0,A(1, 2))
=A(0, 4)
A(0,A(1, 3))
=A(0, 5)
A(0,A(1,n−1))
2A(1, 1)A(1,A(2, 0))
=A(1, 3)
A(1,A(2, 1))
=A(1, 5)
A(1,A(2, 2))
=A(1, 7)
A(1,A(2, 3))
=A(1, 9)
A(1,A(2,n−1))
3A(2, 1)A(2,A(3, 0))
=A(2, 5)
A(2,A(3, 1))
=A(2, 13)
A(2,A(3, 2))
=A(2, 29)
A(2,A(3, 3))
=A(2, 61)
A(2,A(3,n−1))
4A(3, 1)A(3,A(4, 0))
=A(3, 13)
A(3,A(4, 1))
=A(3, 65533)
A(3,A(4, 2))A(3,A(4, 3))A(3,A(4,n−1))
5A(4, 1)A(4,A(5, 0))A(4,A(5, 1))A(4,A(5, 2))A(4,A(5, 3))A(4,A(5,n−1))
6A(5, 1)A(5,A(6, 0))A(5,A(6, 1))A(5,A(6, 2))A(5,A(6, 3))A(5,A(6,n−1))

Properties

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General remarks

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Not primitive recursive

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The Ackermann function grows faster than anyprimitive recursive function and therefore is not itself primitive recursive.

Proof sketch:

Primitive recursive functions are built from basic functions using composition and primitive recursion, and all grow within a certain rate. We define, constructively, a hierarchy of total functionsFGHk(n){\displaystyle \operatorname {FGH} _{k}(n)} by:

FGH0(n)=n+1,FGHk+1(n)=FGHkn(n){\displaystyle \operatorname {FGH} _{0}(n)=n+1,\quad \operatorname {FGH} _{k+1}(n)=\operatorname {FGH} _{k}^{n}(n)}

whereFGHkn{\displaystyle \operatorname {FGH} _{k}^{n}} denotesn{\displaystyle n}-fold iteration ofFGHk{\displaystyle \operatorname {FGH} _{k}} on inputn{\displaystyle n}.[23] This hierarchy grows strictly faster with increasingk{\displaystyle k}, and every primitive recursive function is eventually bounded above by someFGHk{\displaystyle \operatorname {FGH} _{k}}. This can be shown bystructural induction on the definitions of primitive recursive functions.

However, the Ackermann functionA(m,n){\displaystyle \operatorname {A} (m,n)} eventually exceeds everyFGHk{\displaystyle \operatorname {FGH} _{k}}; for everyk{\displaystyle k}, there existsm{\displaystyle m} such thatA(m,n)>FGHk(n){\displaystyle \operatorname {A} (m,n)>\operatorname {FGH} _{k}(n)} for all sufficiently largen{\displaystyle n}. Thus,A{\displaystyle \operatorname {A} } grows faster than any primitive recursive function and is therefore not primitive recursive.

Inverse

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Since the functionf(n) =A(n,n) considered above grows very rapidly, itsinverse function,f−1, grows very slowly. Thisinverse Ackermann functionf−1 is usually denoted byα. In fact,α(n) is less than 5 for any practical input sizen, sinceA(4, 4) is on the order of222216{\displaystyle 2^{2^{2^{2^{16}}}}}.

This inverse appears in thetime complexity of some algorithms, such as thedisjoint-set data structure andChazelle's algorithm forminimum 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, wherex{\displaystyle \lfloor x\rfloor } is thefloor function:

α(m,n)=min{i1:A(i,m/n)log2n}.{\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 whilen represents the number of elements; in the minimum spanning tree algorithm,m represents the number of edges whilen represents the number of vertices. Several slightly different definitions ofα(m,n) exist; for example,log2n is sometimes replaced byn, and the floor function is sometimes replaced by aceiling.

Other studies might define an inverse function of one wherem is set to a constant, such that the inverse applies to a particular row.[24]

The inverse of the Ackermann function is primitive recursive, since it is graph primitive recursive, and it is upper bounded by a primitive recursive function.[25]

Usage

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In computational complexity

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The Ackermann function appears in the timecomplexity of somealgorithms,[26] such asvector addition systems[27] andPetri net reachability, thus showing they are computationally infeasible for large instances.[28]

Theinverse of the Ackermann function appears in some time complexity results. For instance, thedisjoint-set data structure takesamortized time per operation proportional to the inverse Ackermann function,[29] and cannot be made faster within thecell-probe model of computational complexity.[30]

In discrete geometry

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Certain problems indiscrete geometry related toDavenport–Schinzel sequences have complexity bounds in which the inverse Ackermann functionα(n){\displaystyle \alpha (n)} appears. For instance, forn{\displaystyle n} line segments in the plane, the unbounded face of thearrangement of the segments has complexityO(nα(n)){\displaystyle O(n\alpha (n))}, and some systems ofn{\displaystyle n} line segments have an unbounded face of complexityΩ(nα(n)){\displaystyle \Omega (n\alpha (n))}.[31]

As a benchmark

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The Ackermann function, due to its definition in terms of extremely deep recursion, can be used as a benchmark of acompiler's ability to optimize recursion. The first published use of Ackermann's function in this way was in 1970 by Dragoș Vaida[32] and, almost simultaneously, in 1971, by Yngve Sundblad.[14]

Sundblad's seminal paper was taken up by Brian Wichmann (co-author of theWhetstone benchmark) in a trilogy of papers written between 1975 and 1982.[33][34][35]

See also

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Notes

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  1. ^with parameter order reversed
  2. ^'curried'
  3. ^In eachstep the underlinedredex is rewritten.
  4. ^abhere: leftmost-innermost strategy!
  5. ^abcdFor better readability
    S(0) is notated as 1,
    S(S(0)) is notated as 2,
    S(S(S(0))) is notated as 3,
    etc...
  6. ^The maximum depth of recursion refers to the number of levels of activation of a procedure that exist during the deepest call of the procedure.Cornelius & Kirby (1975)
  7. ^LOOP n+1TIMES DO F

References

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  1. ^Monin & Hinchey 2003, p. 61.
  2. ^abAckermann 1928.
  3. ^"Decimal expansion of A(4,2)".kosara.net. 27 August 2000. Archived fromthe original on 20 January 2010.
  4. ^Calude, Marcus & Tevy 1979.
  5. ^Hilbert 1926, p. 185.
  6. ^van Heijenoort 1977.
  7. ^Péter 1935.
  8. ^Robinson 1948.
  9. ^Ritchie 1965, p. 1028.
  10. ^abcBuck 1963.
  11. ^Meeussen & Zantema 1992, p. 6.
  12. ^Munafo 1999a.
  13. ^Ritchie 1965.
  14. ^abSundblad 1971.
  15. ^Porto & Matos 1980.
  16. ^Odifreddi 1999, p. 298.
  17. ^"The Ackermann hierarchy vs. the fast growing hierarchy".StackExchange.
  18. ^Indentation according to theoff-side rule (INDENT ... DEDENT), like inPython:
    for_inrange(n):n+=1
  19. ^Meyer & Ritchie 1967.
  20. ^Grossman & Zeitman 1988.
  21. ^Paulson 2021.
  22. ^Cohen 1987, p. 56, Proposition 3.16 (see in proof).
  23. ^Another sequence of functions,En{\displaystyle \operatorname {E} _{n}}, defining theGrzegorczyk hierarchy, is frequently used to partition the primitive recursive functions into "growth classes". However,FGHn{\displaystyle \operatorname {FGH} _{n}} (orAn{\displaystyle \operatorname {A} _{n}}) andEn{\displaystyle \operatorname {E} _{n}} do not align in their indexing.
  24. ^Pettie 2002.
  25. ^Matos 2014.
  26. ^Brubaker 2023.
  27. ^Czerwiński & Orlikowski 2022.
  28. ^Leroux 2022.
  29. ^Tarjan 1975.
  30. ^Fredman & Saks 1989.
  31. ^Wiernik & Sharir 1988.
  32. ^Vaida 1970.
  33. ^Wichmann 1976.
  34. ^Wichmann 1977.
  35. ^Wichmann 1982.

Bibliography

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External links

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Primary
Leftinverse
Rightinverse
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