Conway chained arrow notation

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Conway chained arrow notation, created by mathematician John Horton Conway, is a means of expressing certain extremely large numbers. It is simply a finite sequence of positive integers separated by rightward arrows, e.g. 2→3→4→5→6.

As with most combinatorial symbologies, the definition is recursive. In this case the notation eventually resolves to being the leftmost number raised to some (usually enormous) integer power.

1 Definition and overview
2 Interpretation
3 Examples
4 Graham's number
5 Ackermann function
6 See also
7 External links

[edit] Definition and overview

A conway chain (or chain for short) is defined as follows:

Any positive integer is a chain of length 1.
A chain of length n, followed by a right-arrow → and a positive integer, together form a chain of length $n + 1$ .

A chain Y not contained inside a larger chain in one of the forms $X \to Y$ , $Y \to Z$ , or $X \to Y \to Z$ , where X and Z are also chains, is a complete chain.

If p and q are positive integers, and X stands for some chain, then the following rules apply to complete chains:

$X \to p \to (q + 1)$ is equivalent to $X \to ( X \to ( \dots (X \to ( X ) \to q)\dots ) \to q ) \to q$
(with p copies of X, p -1 copies of q, and p -1 pairs of parentheses).
$X \to 1$ is equivalent to X.
$p \to q$ is equivalent to the exponential expression $p q$ .

Some observations for longer chains:

A chain of length 3 corresponds to Knuth's up-arrow notation and hyper operators:

$\begin{matrix} p \to q \to r = \mbox{hyper}(p,r+2,q) = p \!\!\! & \underbrace{ \uparrow \dots \uparrow } & \!\!\! q = p\uparrow^r q.\\ & \!\!\! r \mbox{ arrows} \!\!\! \end{matrix}$

A chain of length 4 or more has a value that is, generally, too large to comprehend.

[edit] Interpretation

One must be careful to treat an arrow chain as a whole. Whereas chains of other infixed symbols (e.g. 3+4+5+6+7) can often be considered in fragments (e.g. (3+4)+5+(6+7)) without a change of meaning (see associativity), or at least can be evaluated step by step in a prescribed order, e.g. $2^{3^4}$ from right to left, that is not so with Conway's arrow.

For example:

$2\rightarrow3\rightarrow2 = 2\uparrow\uparrow3 = 2^{2^2} = 16$
$2\rightarrow\left(3\rightarrow2\right) = 2^{3^2} = 512$
$\left(2\rightarrow3\right)\rightarrow2 = \left(2^3\right)^2 = 64$

Note that in the second case no parentheses are needed in the power notation, since right-to-left evaluation is implied, while the Conway chain needs them because otherwise the meaning is as in the first case.

The first rule is the core: A chain of 3 or more elements ending with 2 or higher becomes a chain of the same length with a (usually vastly) increased penultimate element. But its ultimate element is decremented, eventually permitting the second rule to shorten the chain. After, to paraphrase Knuth, "much detail," the chain is reduced to two elements and the third rule terminates the recursion.

Note that X→p→q is (eventually) of the form X→p'.

Therefore:

a chain starting with a is a power of a
a chain starting with 1 is equal to 1
a chain starting with 2→2 is (eventually) of the form 2→2→p' and therefore equal to 4 (see below)

The simplest cases with four arrows (containing no integers less than 2) are:

$a \to b \to 2 \to 2 = a \to b \to 2 \to (1 + 1) = a \to b \to (a \to b) \to 1 = a \to b \to a^b$
$a \to b \to 3 \to 2 = a \to b \to 3 \to (1 + 1) = a \to b \to (a \to b \to (a \to b) \to 1) \to 1 = a \to b \to (a \to b \to a^b)$

Any more complicated and it gets large:

$a \to b \to 2 \to 3 = a \to b \to 2 \to (2 + 1) = a \to b \to (a \to b) \to 2 = a \to b \to a^b \to 2$
$a \to b \to 4 \to 2 = a \to b \to (a \to b \to (a \to b \to a^b))$

If, for any chain X, we define $f(n) = X \to n$ then $X \to p \to 2 = f^p(1)\!$ (see functional powers).

[edit] Examples

It is impossible to give a fully worked-out interesting example since at least 4 elements are required. However 1-, 2- and 3-length chains, which are subsumed in other notations, are expanded here as illustrated examples.

any single integer n is just the value n, e.g. 7 = 7. This does not conflict with the rules, since combining rule 2 (backwards) with rule 3 we have 7 = 7→1 = 7¹ = 7.

p→q

= p^q (by rule 3)

Thus 3→4 = 3⁴ = 81

Also 123456→1 = 123456¹ = 123456 (by both rules 2 and 3)

1→(any arrowed expression)

= 1 since the entire expression eventually reduces to 1^number = 1. (Indeed, any chain containing a 1 can be truncated just before that 1; e.g. X→1→Y=X for any (embedded) chains X,Y.)

4→3→2

= 4→(4→(4)→1)→1 (by 1) and then, working from the inner parentheses outwards,

= 4→(4→4→1)→1 (remove redundant parentheses [rrp])

= 4→(4→4)→1 (2)

= 4→(256)→1 (3)

= 4→256→1 (rrp)

= 4→256 (2)

= 1.34078079299e+154 approximately (3)

4→3→2 alternatively analysed

= 4→(4→(4)→1)→1 (by 1) and then, removing trailing "→1",

= 4→(4→(4)→1) (2)

= 4→(4→(4)) (2)

= 4→(256) (rrp, 3)

= 1.34078079299e+154 approximately (rrp, 3)

With Knuth's arrows: $4 \uparrow \uparrow 3 = 4 \uparrow 4 \uparrow 4 = 4^{256}$

2→2→4

= 2→(2)→3 (by 1)

= 2→2→3 (rrp)

= 2→2→2 (1, rrp)

= 2→2→1 (1, rrp)

= 2→2 (2)

= 4 (3) (In fact any chain beginning with two 2s stands for 4.)

2→4→3

= 2→(2→(2→(2)→2)→2)→2 (by 1) The four copies of X (which is 2 here) are in bold to distinguish them from the three copies of q (which is also 2)

= 2→(2→(2→2→2)→2)→2 (rrp)

= 2→(2→(4)→2)→2 (previous example)

= 2→(2→4→2)→2 (rrp) (expression expanded in next equation shown in bold on both lines)

= 2→(2→(2→(2→(2)→1)→1)→1)→2 (1)

= 2→(2→(2→(2→2→1)→1)→1)→2 (rrp)

= 2→(2→(2→(2→2)))→2 (2 repeatedly)

= 2→(2→(2→(4)))→2 (3)

= 2→(2→(16))→2 (3)

= 2→65536→2 (3,rrp)

= 2→(2→(2→(...2→(2→(2)→1)→1...)→1)→1)→1 (1) with 65535 sets of parentheses

= 2→(2→(2→(...2→(2→(2))...)))) (2 repeatedly)

= 2→(2→(2→(...2→(4))...)))) (3)

= 2→(2→(2→(...16...)))) (3)

= $2^{2^{\dots^2}}$ (a tower with 2¹⁶ = 65536 stories)

which is unimaginably large. With Knuth's arrows: $2 \uparrow \uparrow \uparrow 4 = 2 \uparrow \uparrow 2\uparrow \uparrow 2 \uparrow \uparrow 2=2 \uparrow \uparrow 2 \uparrow \uparrow 2 \uparrow 2=2\uparrow \uparrow 2 \uparrow \uparrow 4=2 \uparrow \uparrow 2 \uparrow 2 \uparrow 2 \uparrow 2 = 2 \uparrow \uparrow 65536$ .

2→3→2→2

= 2→3→(2→3)→1 (by 1)

= 2→3→8 (2 and 3)

= 2→(2→2→7)→7 (1)

= 2→4→7 (two initial 2's give 4 [ttgf])

= 2→(2→(2→2→6)→6)→6 (1)

= 2→(2→4→6)→6 (ttgf)

= 2→(2→(2→(2→2→5)→5)→5)→6 (1)

= 2→(2→(2→4→5)→5)→6 (ttgf)

= 2→(2→(2→(2→(2→2→4)→4)→4)→5)→6 (1)

= 2→(2→(2→(2→4→4)→4)→5)→6 (ttgf)

= 2→(2→(2→(2→(2→(2→2→3)→3)→3)→4) →5)→6 (1)

= 2→(2→(2→(2→(2→4→3)→3)→4)→5)→6 (ttgf)

= 2→(2→(2→(2→(2→65536→2)→3)→4)→5)→6 (previous example)

= still much larger than previous number

With Knuth's arrows: $2 \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow 2 \uparrow \uparrow \uparrow \uparrow \uparrow 2 \uparrow \uparrow \uparrow \uparrow 2 \uparrow \uparrow \uparrow 2 \uparrow \uparrow 65536.$

3→2→2→2

= 3→2→(3→2)→1 (1)

= 3→2→9 (2 and 3)

= 3→3→8 (1)

= huge

With Knuth's arrows: $3 \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow 3 \uparrow \uparrow \uparrow \uparrow \uparrow \uparrow 3 \uparrow \uparrow \uparrow \uparrow \uparrow 3 \uparrow \uparrow \uparrow \uparrow 3 \uparrow \uparrow \uparrow 3 \uparrow \uparrow 7.6e12.$

[edit] Graham's number

Graham's number $G \!$ itself can not succinctly be expressed in Conway chained arrow notation, but by defining the intermediate function $f(n) = 3 \rightarrow 3 \rightarrow n \!$ , we have: $G = f^{64}(4)\,$ (see functional powers), and $3 \rightarrow 3 \rightarrow 64 \rightarrow 2 < G < 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\,$

Proof: Applying in order the definition, rule 2, and rule 1, we have:

$f^{64}(1)\,$

$= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 1))\dots ))\,$ (with 64 $3 \rightarrow 3$ 's)

$= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3) \rightarrow 1) \dots ) \rightarrow 1) \rightarrow 1\,$

$= 3 \rightarrow 3 \rightarrow 64 \rightarrow 2;\,$

$f^{64}(4) = G;\,$

$f^{64}(27)\,$

$= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27))\dots ))\,$ (with 64 $3 \rightarrow 3$ 's)

$= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (3 \rightarrow 3)))\dots ))\,$ (with 65 $3 \rightarrow 3$ 's)

$= 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\,$ (computing as above).