Popular mathematics folklore provides some simple tools
enabling us compactly to describe some truly enormous
numbers. For example, the number $10^{100}$ is commonly
known as a googol,
and a googol
plex is
$10^{10^{100}}$. For any number $x$, we have the common
vernacular:
- $x$ bang is the factorial number $x!$
- $x$ plex is the exponential number $10^x$
- $x$ stack is the number obtained by iterated exponentiation
(associated upwards) in a tower of height $x$, also denoted $10\uparrow\uparrow x$,
$$10\uparrow\uparrow x = 10^{10^{10^{\cdot^{\cdot^{10}}}}}{\large\rbrace} x\text{ times}.$$
Thus, a googol bang is $(10^{100})!$, and a googol stack is
$10\uparrow\uparrow 10^{100}$. The vocabulary enables us to
name larger numbers with ease:
- googol bang plex stack. (This is the exponential tower $10^{10^{\cdot^{\cdot^{^{10}}}}}$ of height $10^{(10^{100})!}$)
- googol stack bang stack bang
- googol bang bang stack plex stack
- and so on…
Consider the collection of all numbers that can be named in
this scheme, by a term starting with googol and having
finitely many adjectival operands: bang, stack, plex, in
any finite pattern, repetitions allowed. (For the purposes
of this question, let us limit ourselves to these three
operations and please accept the base 10 presumption of the
stack and plex terminology simply as an artifact of its
origin in popular mathematics.)
My goal is to sort all such numbers nameable in this
vocabulary by size.
A few simple observations get us started. Once $x$ is large
enough (about 20), then the factors of $x!$ above $10$
compensate for the few below $10$, and so we see that
$10^x\lt x!$, or in other words, $x$ plex is less than $x$
bang. Similarly, $10^{10^{:^{10}}}x$ times is much larger
than $x!$, since $10^y\gt (y+1)y$ for large $y$, and so for
large values we have
- $x$ plex $\lt$ $x$ bang $\lt$ $x$ stack.
In particular, the order for names having at most one
adjective is:
googol
googol plex
googol bang
googol stack
And more generally, replacing plex with bang or bang with
stack in any of our names results in a strictly (and much)
larger number.
Continuing, since $x$ stack plex $= (x+1)$ stack, it
follows that
- $x$ stack plex $\lt x$ plex stack.
Similarly, for large values,
- $x$ plex bang $\lt x$ bang plex,
because $(10^x)!\lt (10^x)^{10^x}=10^{x10^x}\lt 10^{x!}$.
Also,
- $x$ stack bang $\lt x$ plex stack $\lt x$ bang stack,
because $(10\uparrow\uparrow x)!\lt (10\uparrow\uparrow
x)^{10\uparrow\uparrow x}\lt 10\uparrow\uparrow 2x\lt
10\uparrow\uparrow 10^x\lt 10\uparrow\uparrow x!$. It also
appears to be true for large values that
- $x$ bang bang $\lt x$ stack.
Indeed, one may subsume many more iterations of plex and
bang into a single stack. Note also for large values that
- $x$ bang $\lt x$ plex plex
since $x!\lt x^x$, and this is seen to be less than
$10^{10^x}$ by taking logarithms.
The observations above enable us to form the following
order of all names using at most two adjectives.
googol
googol plex
googol bang
googol plex plex
googol plex bang
googol bang plex
googol bang bang
googol stack
googol stack plex
googol stack bang
googol plex stack
googol bang stack
googol stack stack
My request is for any or all of the following:
-
Expand the list above to include numbers named using
more than two adjectives. (This will not be an
end-extension of the current list, since googol plex plex
plex and googol bang bang bang will still appear before
googol stack.) If people post partial progress, we can
assemble them into a master list later. -
Provide general comparison criteria that will assist
such an on-going effort. -
Provide a complete comparison algorithm that works for
any two expressions having the same number of adjectives. -
Provide a complete comparison algorithm that compares
any two expressions.
Of course, there is in principle a computable comparison
procedure, since we may program a Turing machine to
actually compute the two values and compare their size.
What is desired, however, is a simple, feasible algorithm.
For example, it would seem that we could hope for an
algorithm that would compare any two names in polynomial
time of the length of the names.
Best Answer
OK, let's attempt a sorting of the names having at most three operands. I'll make several observations, and then use them to assemble the order section by section, beginning with the part below googol stack.
googol bang bang bang $\lt$ googol stack. It seems clear that we shall be able to iterated bangs many times before exceeding googol stack. Since googol bang bang bang is the largest three-operand name using only plex and bang, this means that all such names will interact only with each below googol stack.
plex $\lt$ bang. This was established in the question.
plex bang $\lt$ bang plex. This was established in the question, and it allows us to make many comparisons in terms involving only plex and bang, but not quite all of them.
googol bang bang $\lt$ googol plex plex plex. This is because $g!!\lt (g^g)^{g^g}=g^{gg^g}=10^{100\cdot gg^g}$, which is less than $10^{10^{10^g}}$, since $100\cdot gg^g=10^{102\cdot 10^{100}}\lt 10^{10^g}$. Since googol bang bang is the largest two-operand name using only plex and bang and googol plex plex plex is the smallest three-operand name, this means that the two-operand names using only plex and bang will all come before all the three-operand names.
googol plex bang bang $\lt$ googol bang plex plex. This is because $(10^g)!!\lt ((10^g)^{10^g})!=(10^{g10^g})!=(10^{10^{g+100}})!\lt (10^{10^{g+100}})^{10^{10^{g+100}}}=10^{10^{g+100}10^{10^{g+100}}}= 10^{10^{(g+100)10^{g+100}}}\lt 10^{10^{g!}}$.
Combining the previous observations leads to the following order of the three-operand names below googol stack:
Perhaps someone can generalize the methods into a general comparison algorithm for larger smallish terms using only plex and bang? This is related to the topic of the Velleman article linked to by J. M. in the comments.
Meanwhile, let us now turn to the interaction with stack. Using the observations of the two-operand case in the question, we may continue as follows:
Now we use the following fact:
The order therefore continues with:
The order therefore continues with:
Thus, the order continues with:
This last item is clearly less than googol stack stack, and so, using all the pairwise operations we already know, we continue with:
Which seems to complete the list for three-operand names. If I have made any mistakes, please comment below.
Meanwhile, this answer is just partial progress, since we have the four-operand names, which will fit into the hierarchy, and I don't think the observations above are fully sufficient for the four-operand comparisons, although many of them will now be settled by these criteria. And of course, I am nowhere near a general comparison algorithm.
Sorry for the length of this answer. Please post comments if I've made any errors.