# Conway chained arrow notation

Conway chained arrow notation, created by mathematician John 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 integer (usually enormous) power.

 Contents

## Definition and overview

In the following definition, a chain is a sequence of one or more numbers connected by right-arrows "→", a sub-chain is a chain that is directly part of a larger chain (with no grouping), and a complete chain is a chain that is not a sub-chain.

If p and q are positive integers, and X stands for some chain (possibly of only one element), then the following rules apply to complete chains:

1. Xp→(q+1) is equivalent to X→(X→(...X→(X)→q...)→q)→q
(with p copies of X, p-1 copies of q, and p-1 pairs of parentheses).
2. X→1 is equivalent to X.
3. pq is equivalent to the exponential expression pq.

The last two rules do not conflict, since p→1 = p¹ = p

A chain of length 1 is simply that number: p.

A chain of length 2 is simply a power: pq.

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

[itex]\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}[itex]

A chain of length 4 or more is, generally, too large to comprehend.

## 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. [itex]2^{3^4}[itex] from right to left, that is not so with Conway's arrow.

For example:

• 2→3→2 = 2^^3 = 2^2^2 = 16
• 2→(3→2) = 2^3^2 = 512
• (2→3)→2 = (2^3)^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 are:

• a→b→2→2 = a→b→a^b
• a→b→3→2 = a→b→(a→b→a^b)

If, for any chain X, we define f(n) = Xn then Xp→2 = f p(1) (see functional powers).

## 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.

n

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

= pq (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 1number = 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: [itex]4 \uparrow \uparrow 3 = 4 \uparrow 4 \uparrow 4 = 4^{256}[itex]

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 2 which is q)
= 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)
= [itex]2^{2^{\dots^2}}[itex] (a tower with 216 = 65536 stories)

which is unimaginably large. With Knuth's arrows: [itex]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[itex].

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)
= mind bogglingly larger than previous number

With Knuth's arrows: [itex]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.[itex]

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: [itex]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.[itex]

## Graham's number

Graham's number [itex]G[itex] itself can not succinctly be expressed in Conway chained arrow notation, but by defining the intermediate function [itex]f(n) = 3 \rightarrow 3 \rightarrow n[itex], we have:

[itex]G = f^{64}(4)\, [itex] (see functional powers), and
[itex]3 \rightarrow 3 \rightarrow 64 \rightarrow 2 < G < 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\, [itex]


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

[itex]f^{64}(1)\, [itex]

[itex]= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 1))\dots ))\, [itex] (with 64 [itex]3 \rightarrow 3[itex]'s)
[itex]= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3) \rightarrow 1) \dots ) \rightarrow 1) \rightarrow 1\, [itex]
[itex]= 3 \rightarrow 3 \rightarrow 64 \rightarrow 2;\, [itex]

[itex]f^{64}(4) = G;\, [itex]

[itex]f^{64}(27)\, [itex]

[itex]= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27))\dots ))\, [itex] (with 64 [itex]3 \rightarrow 3[itex]'s)
[itex]= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\dots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (3 \rightarrow 3)))\dots ))\, [itex] (with 65 [itex]3 \rightarrow 3[itex]'s)
[itex]= 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\, [itex] (computing as above).

Since f is strictly increasing,

[itex]f^{64}(1) < f^{64}(4) < f^{64}(27)\, [itex]

which is the given inequality. Note that

[itex] 3 \rightarrow 3 \rightarrow 3 \rightarrow 3 = 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27 \rightarrow 2) \rightarrow 2\, [itex]

which is much greater than Graham's number.

## Ackermann function

The Ackermann function may be expressed using Conway chained arrow notation:

A(m, n) = (2 → (n+3) → (m − 2)) − 3 for m>2

hence

2 → nm = A(m+2,n-3) + 3 for n>2

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

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