The Clarkkkkson vs. the xkcd Number

I just stumbled upon this webpage in which some kid (years ago, presumably in high school) who uses odd slang defined a huge number by putting various operators together and making recursive call after recursive call. He supposes that this is the biggest number anyone’s ever bothered to concisely define. My intuition is that A(g64, g64) (feel free to call it the xkcd number) is bigger. However, intuition is completely useless in this kind of question.

The Clarkkkkson defined:

(And the xkcd number is the result of the Ackermann function with Graham’s Number as both the arguments, as defined in this comic.)

Anyone want to take a crack at setting up some correspondences and demonstrating which is bigger — The Clarkkkkson or the xkcd number?

154 thoughts on “The Clarkkkkson vs. the xkcd Number

  1. I’m going to try to give my $0.02 as simply as possible.

    As has been previously mentioned, Clarkkkson got the hyper function wrong, making it left-associative instead of right-associative.
    This severely cripples it, because a^a^a = a^(a^a), but (a^a)^a = a^(a*a), which is obviously much smaller when using values of a greater than 2. Instead of a real power tower, it simply produces a large exponent.

    If we fix this and use the correct hyper function, then there is no contest.

    Graham’s number can be expressed in Knuth’s arrow notation as x ^y z,
    where x and z are 3, and y is determined through a 64-stage recursive process of y_n = 3 ^y_(n-1) 3,
    starting with y_1 = 4.
    Running it through the Ackermann function basically just adds a 65th stage,
    with x = 2, y = g64, z = g64.

    By comparison, Clarkkkson starts with x, y_1, and z all equal to k, which is at least 100.
    It then grows x and z along with y (whereas Graham keeps them at 3), and it uses k stages instead of 65.
    The repeats and the whole factorial aspect are just icing on the cake; Clarkkkson wins, hands down.

    If, on the other hand, we keep the broken hyper function, things are much less clear-cut.
    The explosive growth of x and z will certainly go a long way toward compensating,
    and the repeats and factorials will help too. After all, (100^100)^100 > 3^(3^3).
    But at higher values of y, the crippling really matters,
    and I wouldn’t be surprised if a better mathematician than myself declared xkcd the winner.

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