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.
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.