Navigation
  • Home
  • Recent
  • Most Active
  • Popular
  • Blog
  • Credits
  • RSS
  •   Interaction
  • Register
  • Statistics
  •   Help
  • Suggestions
  • Contact Us
  • How to Edit
  • Help



  • [Edit]


    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 integer (usually enormous) power.


        Conway chained arrow notation
            Definition and overview
            Interpretation
            Examples
            Grahams number
            Ackermann function
            See also

    top

    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 o Y, Y o Z, or X o Y o 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 o p o (q + 1) is equivalent to X o ( X o ( dots X o ( X ) o qdots ) o q ) o q
      (with p copies of X, p -1 copies of q, and p -1 pairs of parentheses).
      X o 1 is equivalent to X.
      p o q is equivalent to the exponential expression p^q.

    Some observations for longer chains:

    egin

    p o q o r = mbox(p,r+2,q) = p !!! & underbrace & !!! q = puparrow^r q.\
    & !!! r mbox !!!
    end
      A chain of length 4 or more has a value that is, generally, too large to comprehend.

    top

    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^ from right to left, that is not so with Conway's arrow.

    For example:
      2
    ightarrow3
    ightarrow2 = 2uparrowuparrow3 = 2^ = 16
      2
    ightarrowleft(3
    ightarrow2
    ight) = 2^ = 512
      left(2
    ightarrow3
    ight)
    ightarrow2 = left(2^3
    ight)^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 o b o 2 o 2 = a o b o a^b
      a o b o 3 o 2 = a o b o (a o b o a^b)

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

    top

    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: 4 uparrow uparrow 3 = 4 uparrow 4 uparrow 4 = 4^

    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)

    = 2^ (a tower with 216 = 65536 stories)

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

    top

    Grahams 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
    ightarrow 3
    ightarrow n !, we have:
    G = f^(4), (see functional powers), and
    3
    ightarrow 3
    ightarrow 64
    ightarrow 2 < G < 3
    ightarrow 3
    ightarrow 65
    ightarrow 2,

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

    f^(1),
    = 3

    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow (dots (3
    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow 1))dots )), (with 64 3
    ightarrow 3's)
    = 3

    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow (dots (3
    ightarrow 3
    ightarrow (3
    ightarrow 3)
    ightarrow 1) dots )
    ightarrow 1)
    ightarrow 1,
    = 3

    ightarrow 3
    ightarrow 64
    ightarrow 2;,

    f^(4) = G;,

    f^(27),
    = 3

    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow (dots (3
    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow 27))dots )), (with 64 3
    ightarrow 3's)
    = 3

    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow (dots (3
    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow (3
    ightarrow 3)))dots )), (with 65 3
    ightarrow 3's)
    = 3

    ightarrow 3
    ightarrow 65
    ightarrow 2, (computing as above).

    Since f is strictly increasing,
    f^(1) < f^(4) < f^(27),

    which is the given inequality. Note that
    3

    ightarrow 3
    ightarrow 3
    ightarrow 3 = 3
    ightarrow 3
    ightarrow (3
    ightarrow 3
    ightarrow 27
    ightarrow 2)
    ightarrow 2,
    which is much greater than Graham's number.

    top

    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).

    top

    See also
     
    Search more:
     

       
    Source Privacy License Download Contact Us Atlas
    Scientus.org Dictionary (Yet Another Wiki) RC : 1.39
    This article is licensed under the GNU Free Documentation License [copyleft]. It uses material from the Wikipedia article "Conway chained arrow notation". link