Hyperoperation

Definition

Definition, most common

The hyperoperation sequence H_n(a,b) \colon (\mathbb{N}_0)^3 \rightarrow \mathbb{N}_0 is the sequence of binary operations H_n \colon (\mathbb{N}_0)^2 \rightarrow \mathbb{N}_0, defined recursively as follows:

:H_n(a,b) = a[n]b = \begin{cases} b + 1 & \text{if } n = 0 \ a & \text{if } n = 1 \text{ and } b = 0 \ 0 & \text{if } n = 2 \text{ and } b = 0 \ 1 & \text{if } n \ge 3 \text{ and } b = 0 \ H_{n-1}(a, H_n(a, b - 1)) & \text{otherwise} \end{cases}

(Note that for n = 0, the binary operation essentially reduces to a unary operation (successor function) by ignoring the first argument.)

For n = 0, 1, 2, 3, this definition reproduces the basic arithmetic operations of successor (which is a unary operation), addition, multiplication, and exponentiation, respectively, as :\begin{align} H_0(a, b) &= b + 1, \ H_1(a, b) &= a + b, \ H_2(a, b) &= a \times b, \ H_3(a, b) &= a\uparrow{b} = a^{b}. \end{align}

The H_n operations for n ≥ 3 can be written in Knuth's up-arrow notation.

So what will be the next operation after exponentiation? We defined multiplication so that H_2(a, 3) = a[2]3 = a \times 3 = a + a + a, and defined exponentiation so that H_3(a, 3) = a[3]3 = a\uparrow 3 = a^3 = a \times a \times a, so it seems logical to define the next operation, tetration, so that H_4(a, 3) = a[4]3 = a\uparrow\uparrow 3 = \operatorname{tetration}(a, 3) = a^{a^a}, with a tower of three 'a'. Analogously, the pentation of (a, 3) will be tetration(a, tetration(a, a)), with three "a" in it.

:\begin{align} H_4(a,b) &= a\uparrow\uparrow{b}, \ H_5(a,b) &= a\uparrow\uparrow\uparrow{b}, \ \ldots & \ H_n(a,b) &= a\uparrow^{n-2}b \text{ for } n \ge 3, \ \ldots & \ \end{align}

Knuth's notation could be extended to negative indices ≥ −2 in such a way as to agree with the entire hyperoperation sequence, except for the lag in the indexing: :H_n(a, b) = a \uparrow^{n-2}b\text{ for } n \ge 0.

The hyperoperations can thus be seen as an answer to the question "what's next" in the sequence: successor, addition, multiplication, exponentiation, and so on. Noting that :\begin{align} a + b &= (a + (b - 1)) + 1 \ a \cdot b &= a + (a \cdot (b - 1)) \ a^b &= a \cdot \bigl(a^{(b - 1)}\bigr) \ a [4] b &= a^{a [4] (b - 1)} \end{align}

the relationship between basic arithmetic operations is illustrated, allowing the higher operations to be defined naturally as above. The parameters of the hyperoperation hierarchy are sometimes referred to by their analogous exponentiation term; so a is the base, b is the exponent (or hyperexponent), and n is the rank (or grade), and moreover, H_n(a, b) is read as "the bth n-ation of a", e.g. H_4(7,9) is read as "the 9th tetration of 7", and H_{123}(456,789) is read as "the 789th 123-ation of 456".

In common terms, the hyperoperations are ways of compounding numbers that increase in growth based on the iteration of the previous hyperoperation. The concepts of successor, addition, multiplication and exponentiation are all hyperoperations; the successor operation (producing x + 1 from x) is the most primitive, the addition operator specifies the number of times 1 is to be added to itself to produce a final value, multiplication specifies the number of times a number is to be added to itself, and exponentiation refers to the number of times a number is to be multiplied by itself.

Definition, using iteration

Define iteration of a function f of two variables as : f^{x}(a,b) = \begin{cases} f(a,b) & \text{if } x = 1 \ f(a, f^{x-1}(a,b)) & \text{if } x 1 \end{cases} The hyperoperation sequence can be defined in terms of iteration, as follows. For all integers x,n,a,b \geq 0, define : \begin{array}{lcl} H_{0}(a,b) & = & b+1 \ H_{1}(a,0) & = & a \ H_{2}(a,0) & = & 0 \ H_{n+3}(a,0) & = & 1 \ H_{n+1}(a,b+1) & = & H^{b+1}{n}(a,H{n+1}(a,0)) \ H^{x+2}{n}(a,b) & = & H{n}(a,H^{x+1}_{n}(a,b)) \end{array}

As iteration is associative, the last line can be replaced by : \begin{array}{lcl} H^{x+2}{n}(a,b) & = & H^{x+1}{n}(a,H_{n}(a,b)) \end{array}

Computation

The definitions of the hyperoperation sequence can naturally be transposed to term rewriting systems (TRS).

TRS based on definition sub 1.1

The basic definition of the hyperoperation sequence corresponds with the reduction rules : \begin{array}{lll} \text{(r1)} & H(0,a,b) & \rightarrow & S(b) \ \text{(r2)} & H(S(0),a,0) & \rightarrow & a \ \text{(r3)} & H(S(S(0)),a,0) & \rightarrow & 0 \ \text{(r4)} & H(S(S(S(n))),a,0) & \rightarrow & S(0) \ \text{(r5)} & H(S(n),a,S(b)) & \rightarrow & H(n,a,H(S(n),a,b)) \end{array}

To compute H_{n}(a, b) one can use a stack, which initially contains the elements \langle n,a,b \rangle.

Then, repeatedly until no longer possible, three elements are popped and replaced according to the rulesThis implements the leftmost-innermost (one-step) strategy. : \begin{array}{lllllllll} \text{(r1)} & 0 &,& a &,& b & \rightarrow & (b+1) \ \text{(r2)} & 1 &,& a &,& 0 & \rightarrow & a \ \text{(r3)} & 2 &,& a &,& 0 & \rightarrow & 0 \ \text{(r4)} & (n+3) &,& a &,& 0 & \rightarrow & 1 \ \text{(r5)} & (n+1) &,& a &,& (b+1) & \rightarrow & n &,& a &,& (n+1) &,& a &,& b \end{array}

Schematically, starting from \langle n,a,b \rangle:

WHILE stackLength 1 { POP 3 elements; PUSH 1 or 5 elements according to the rules r1, r2, r3, r4, r5; }

Example

Compute H_2(2,2) \rightarrow_{*} 4.

The reduction sequence is

When implemented using a stack, on input \langle 2,2,2 \rangle

TRS based on definition sub 1.2

The definition using iteration leads to a different set of reduction rules : \begin{array}{lll} \text{(r6)} & H(S(0),0,a,b) & \rightarrow & S(b) \ \text{(r7)} & H(S(0),S(0),a,0) & \rightarrow & a \ \text{(r8)} & H(S(0),S(S(0)),a,0) & \rightarrow & 0 \ \text{(r9)} & H(S(0),S(S(S(n))),a,0) & \rightarrow & S(0) \ \text{(r10)} & H(S(0),S(n),a,S(b)) & \rightarrow & H(S(b),n,a,H(S(0),S(n),a,0)) \ \text{(r11)} & H(S(S(x)),n,a,b) & \rightarrow & H(S(0),n,a,H(S(x),n,a,b)) \end{array}

As iteration is associative, instead of rule r11 one can define : \begin{array}{lll} \text{(r12)} & H(S(S(x)),n,a,b) & \rightarrow & H(S(x),n,a,H(S(0),n,a,b)) \end{array}

Like in the previous section the computation of H_n(a,b) = H^1_n(a,b) can be implemented using a stack.

Initially the stack contains the four elements \langle 1,n,a,b \rangle.

Then, until termination, four elements are popped and replaced according to the rules : \begin{array}{lllllllll} \text{(r6)} & 1 &, 0 &, a &, b & \rightarrow & (b+1) \ \text{(r7)} & 1 &, 1 &, a &, 0 & \rightarrow & a \ \text{(r8)} & 1 &, 2 &, a &, 0 & \rightarrow & 0 \ \text{(r9)} & 1 &, (n+3) &, a &, 0 & \rightarrow & 1 \ \text{(r10)} & 1 &, (n+1) &, a &, (b+1) & \rightarrow & (b+1) &, n &, a &, 1 &, (n+1) &, a &, 0 \ \text{(r11)} & (x+2) &, n &, a &, b & \rightarrow & 1 &, n &, a &, (x+1) &, n &, a &, b \end{array}

Schematically, starting from \langle 1,n,a,b \rangle: WHILE stackLength 1 { POP 4 elements; PUSH 1 or 7 elements according to the rules r6, r7, r8, r9, r10, r11; }

Example

Compute H_3(0,3) \rightarrow_{*} 0.

On input \langle 1,3,0,3 \rangle the successive stack configurations are :\begin{align} & \underline{1,3,0,3} \rightarrow_{r10} 3,2,0,\underline{1,3,0,0} \rightarrow_{r9} \underline{3,2,0,1} \rightarrow_{r11} 1,2,0,\underline{2,2,0,1} \rightarrow_{r11} 1,2,0,1,2,0,\underline{1,2,0,1} \ & \rightarrow_{r10} 1,2,0,1,2,0,1,1,0,\underline{1,2,0,0} \rightarrow_{r8} 1,2,0,1,2,0,\underline{1,1,0,0} \rightarrow_{r7} 1,2,0,\underline{1,2,0,0} \rightarrow_{r8} \underline{1,2,0,0} \rightarrow_{r8} 0. \end{align}

The corresponding equalities are :\begin{align} & H_3(0,3) = H^3_2(0,H_3(0,0)) = H^3_2(0,1) = H_2(0,H^2_2(0,1)) = H_2(0,H_2(0,H_2(0,1)) \ & = H_2(0,H_2(0,H_1(0,H_2(0,0)))) = H_2(0,H_2(0,H_1(0,0))) = H_2(0,H_2(0,0)) = H_2(0,0) = 0. \end{align}

When reduction rule r11 is replaced by rule r12, the stack is transformed according to :\begin{array}{lllllllll} \text{(r12)} & (x+2) &, n &, a &, b & \rightarrow & (x+1) &, n &, a &, 1 &, n &, a &, b \end{array}

The successive stack configurations will then be :\begin{align} & \underline{1,3,0,3} \rightarrow_{r10} 3,2,0,\underline{1,3,0,0} \rightarrow_{r9} \underline{3,2,0,1} \rightarrow_{r12} 2,2,0,\underline{1,2,0,1} \rightarrow_{r10} 2,2,0,1,1,0,\underline{1,2,0,0} \ & \rightarrow_{r8} 2,2,0,\underline{1,1,0,0} \rightarrow_{r7} \underline{2,2,0,0} \rightarrow_{r12} 1,2,0,\underline{1,2,0,0} \rightarrow_{r8} \underline{1,2,0,0} \rightarrow_{r8} 0 \end{align}

The corresponding equalities are :\begin{align} & H_3(0,3) = H^3_2(0,H_3(0,0)) = H^3_2(0,1) = H^2_2(0,H_2(0,1)) = H^2_2(0,H_1(0,H_2(0,0))) \ & = H^2_2(0,H_1(0,0)) = H^2_2(0,0) = H_2(0,H_2(0,0)) = H_2(0,0) = 0 \end{align}

Remarks

• H_3(0,3) = 0 is a special case. See below.
• The computation of H_{n}(a,b) according to the rules {r6 - r10, r11} is heavily recursive. The culprit is the order in which iteration is executed: H^{n}(a,b) = H(a, H^{n-1}(a,b)). The first H disappears only after the whole sequence is unfolded. For instance, H_4(2,4) converges to 65536 in 2863311767 steps, the maximum depth of recursion is 65534.
• The computation according to the rules {r6 - r10, r12} is more efficient in that respect. The implementation of iteration H^{n}(a,b) as H^{n-1}(a, H(a,b)) mimics the repeated execution of a procedure H. The depth of recursion, (n+1), matches the loop nesting. formalized this correspondence. The computation of H_4(2,4) according to the rules {r6-r10, r12} also needs 2863311767 steps to converge on 65536, but the maximum depth of recursion is only 5, as tetration is the 5th operator in the hyperoperation sequence.
• The considerations above concern the recursion depth only. Either way of iterating leads to the same number of reduction steps, involving the same rules (when the rules r11 and r12 are considered "the same"). As the example shows the reduction of H_3(0,3) converges in 9 steps: 1 X r7, 3 X r8, 1 X r9, 2 X r10, 2 X r11/r12. The modus iterandi only affects the order in which the reduction rules are applied.

Examples

Below is a list of the first seven (0th to 6th) hyperoperations (0⁰ is defined as 1).

Special cases

Hn(0, b) = :b + 1, when n = 0 :b, when n = 1 :0, when n = 2 :1, when n = 3 and b = 0 For more details, see Powers of zero or Zero to the power of zero. :0, when n = 3 and b 0 :1, when n 3 and b is even (including 0) :0, when n 3 and b is odd

Hn(1, b) = :b, when n = 2 :1, when n ≥ 3

Hn(a, 0) = :0, when n = 2 :1, when n = 0, or n ≥ 3 :a, when n = 1

Hn(a, 1) = :2, when n = 0 :a + 1, when n = 1 :a, when n ≥ 2

Hn(a, a) = :Hn+1(a, 2), when n ≥ 1

Hn(a, −1) = :0, when n = 0, or n ≥ 4 :a − 1, when n = 1 :−a, when n = 2 : , when n = 3

Hn(2, 2) = : 3, when n = 0 : 4, when n ≥ 1, easily demonstrable recursively.

History

One of the earliest discussions of hyperoperations was that of Albert Bennett in 1914, who developed some of the theory of commutative hyperoperations (see below). About 12 years later, Wilhelm Ackermann defined the function \phi(a, b, n), which somewhat resembles the hyperoperation sequence.

In his 1947 paper, Reuben Goodstein introduced the specific sequence of operations that are now called hyperoperations, and also suggested the Greek names tetration, pentation, etc., for the extended operations beyond exponentiation (because they correspond to the indices 4, 5, etc.). As a three-argument function, e.g., G(n, a, b) = H_n(a, b), the hyperoperation sequence as a whole is seen to be a version of the original Ackermann function \phi(a, b, n) — recursive but not primitive recursive — as modified by Goodstein to incorporate the primitive successor function together with the other three basic operations of arithmetic (addition, multiplication, exponentiation), and to make a more seamless extension of these beyond exponentiation.

The original three-argument Ackermann function \phi uses the same recursion rule as does Goodstein's version of it (i.e., the hyperoperation sequence), but differs from it in two ways. First, \phi(a, b, n) defines a sequence of operations starting from addition (n = 0) rather than the successor function, then multiplication (n = 1), exponentiation (n = 2), etc. Secondly, the initial conditions for \phi result in \phi(a, b, 3) = G(4,a,b+1) = a [4] (b + 1), thus differing from the hyperoperations beyond exponentiation. The significance of the b + 1 in the previous expression is that \phi(a, b, 3) = a^{a^{\cdot^{\cdot^{\cdot^a}}}}, where b counts the number of operators (exponentiations), rather than counting the number of operands ("a"s) as does the b in a [4] b, and so on for the higher-level operations. (See the Ackermann function article for details.)

Notations

This is a list of notations that have been used for hyperoperations.

Variant starting from ''a''

Main article: Ackermann function

In 1928, Wilhelm Ackermann defined a 3-argument function \phi(a, b, n) which gradually evolved into a 2-argument function known as the Ackermann function. The original Ackermann function \phi was less similar to modern hyperoperations, because his initial conditions start with \phi(a, 0, n) = a for all n 2. Also he assigned addition to n = 0, multiplication to n = 1 and exponentiation to n = 2, so the initial conditions produce very different operations for tetration and beyond.

Another initial condition that has been used is A(0, b) = 2b + 1 (where the base is constant a = 2), due to Rózsa Péter, which does not form a hyperoperation hierarchy.

Variant starting from 0

In 1984, C. W. Clenshaw and F. W. J. Olver began the discussion of using hyperoperations to prevent computer floating-point overflows. Since then, many other authors have renewed interest in the application of hyperoperations to floating-point representation. (Since Hn(a, b) are all defined for b = -1.) While discussing tetration, Clenshaw et al. assumed the initial condition F_n(a, 0) = 0, which makes yet another hyperoperation hierarchy. Just like in the previous variant, the fourth operation is very similar to tetration, but offset by one.

Lower hyperoperations

An alternative for these hyperoperations is obtained by evaluation from left to right. Since :\begin{align} a + b &= (a + (b - 1)) + 1 \ a \cdot b &= (a \cdot (b - 1)) + a \ a^b &= \left(a^{(b-1)}\right) \cdot a \end{align} define (with ° or subscript) :a_{(n+1)}b = \left(a_{(n + 1)}(b - 1)\right){(n)} a with :\begin{align} a{(1)} b &= a + b \ a_{(2)} 0 &= 0 \ a_{(n)} 1 &= a & \text{for } n2 \ \end{align}

This was extended to ordinal numbers by Doner and Tarski, by :

:\begin{align} \alpha O_0 \beta &= \alpha + \beta \ \alpha O_\nu \beta &= \sup\limits_{\delta \end{align}

It follows from Definition 1(i), Corollary 2(ii), and Theorem 9, that, for a ≥ 2 and b ≥ 1, that : a O_n b = a_{(n + 1)} b

But this suffers a kind of collapse, failing to form the "power tower" traditionally expected of hyperoperators:Ordinal addition is not commutative; see ordinal arithmetic for more information :\alpha_{(4)}(1 + \beta) = \alpha^{\left(\alpha^\beta\right)}.

If α ≥ 2 and γ ≥ 2,[Corollary 33(i)] :\alpha_{(1 + 2\nu + 1)}\beta \leq \alpha_{(1 + 2\nu)}(1 + 3\alpha\beta).

Commutative hyperoperations

Commutative hyperoperations were considered by Albert Bennett as early as 1914, which is possibly the earliest remark about any hyperoperation sequence. Commutative hyperoperations are defined by the recursion rule :F_{n+1}(a, b) = \exp(F_n(\ln(a), \ln(b))) which is symmetric in a and b, meaning all hyperoperations are commutative. This sequence does not contain exponentiation, and so does not form a hyperoperation hierarchy.

Numeration systems based on the hyperoperation sequence

R. L. Goodstein used the sequence of hyperoperators to create systems of numeration for the nonnegative integers. The so-called complete hereditary representation of integer n, at level k and base b, can be expressed as follows using only the first k hyperoperators and using as digits only 0, 1, ..., b − 1, together with the base b itself:

• For 0 ≤ n ≤ b − 1, n is represented simply by the corresponding digit.
• For n b − 1, the representation of n is found recursively, first representing n in the form :b [k] x**k [k − 1] x**k − 1 [k - 2] ... [2] x2 [1] x1 :where x**k, ..., x1 are the largest integers satisfying (in turn)

:b [k] x**k ≤ n

:b [k] x**k [k − 1] x**k − 1 ≤ n

:...

:b [k] x**k [k − 1] x**k − 1 [k - 2] ... [2] x2 [1] x1 ≤ n

:Any x**i exceeding b − 1 is then re-expressed in the same manner, and so on, repeating this procedure until the resulting form contains only the digits 0, 1, ..., b − 1, together with the base b.

Unnecessary parentheses can be avoided by giving higher-level operators higher precedence in the order of evaluation; thus,

: level-1 representations have the form b [1] X, with X also of this form;

: level-2 representations have the form b [2] X [1] Y, with X,Y also of this form;

: level-3 representations have the form b [3] X [2] Y [1] Z, with X,Y,Z also of this form;

: level-4 representations have the form b [4] X [3] Y [2] Z [1] W, with X,Y,Z,W also of this form;

and so on.

In this type of base-b hereditary representation, the base itself appears in the expressions, as well as "digits" from the set {0, 1, ..., b − 1}. This compares to ordinary base-2 representation when the latter is written out in terms of the base b; e.g., in ordinary base-2 notation, 6 = (110)2 = 2 [3] 2 [2] 1 [1] 2 [3] 1 [2] 1 [1] 2 [3] 0 [2] 0, whereas the level-3 base-2 hereditary representation is 6 = 2 [3] (2 [3] 1 [2] 1 [1] 0) [2] 1 [1] (2 [3] 1 [2] 1 [1] 0). The hereditary representations can be abbreviated by omitting any instances of [1] 0, [2] 1, [3] 1, [4] 1, etc.; for example, the above level-3 base-2 representation of 6 abbreviates to 2 [3] 2 [1] 2.

Examples: The unique base-2 representations of the number 266, at levels 1, 2, 3, 4, and 5 are as follows:

:Level 1: 266 = 2 [1] 2 [1] 2 [1] ... [1] 2 (with 133 2s) :Level 2: 266 = 2 [2] (2 [2] (2 [2] (2 [2] 2 [2] 2 [2] 2 [2] 2 [1] 1)) [1] 1) :Level 3: 266 = 2 [3] 2 [3] (2 [1] 1) [1] 2 [3] (2 [1] 1) [1] 2 :Level 4: 266 = 2 [4] (2 [1] 1) [3] 2 [1] 2 [4] 2 [2] 2 [1] 2 :Level 5: 266 = 2 [5] 2 [4] 2 [1] 2 [5] 2 [2] 2 [1] 2

Notes

References

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References

1. In each ''step'' the underlined ''redex'' is rewritten.
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3. '''LOOP''' ''n'' '''TIMES DO''' H.