Lets say you have each Planck length in the observable universe represent a googolxgoogolxgoogol Rubik's cube, and create a cube with a total volume of 4.6 x 10^185 of these cubes, each move on any individual cube counts as one permutation, also you can switch each individual cube with another cube in the larger cube, this too counts as another permutation.

The purpose of this exercise is because I am of course unable to imagine Graham's number {at least by any methods known by myself} and was wondering if there was any way to even come close to doing so.

I wanted to use measurements that are already in the real world such as "Planck length" and "observable universe" while also avoiding using variables that are in themselves unimaginable for the number of squares for each face.

In closing, how would I write this on paper, and is this even comparable to grahams number {the variable "googol" can be replaced with another number that can be somewhat imagined}?

  • $\begingroup$ I suspect that Ron Graham's number is rather bigger than this. $\endgroup$ – Lord Shark the Unknown Aug 9 at 8:18
  • $\begingroup$ Ron Graham himself on the size of Graham's number. $\endgroup$ – Vepir Aug 9 at 8:34
  • $\begingroup$ It is way too big for that approach. Try this simpler problem first. Imagine a circle with radius equal to the limit of the observable universe and express the radius in units of Planck length. How many digits of $\pi$ do you need to calculate its circumference to Planck length precision? (Assuming a flat universe) $\endgroup$ – badjohn Aug 9 at 8:44

A tl;dr of the below is basically what was echoed in the comments: it's not even remotely close to Graham's number, nor even close to the "building block" of it that is $3 \uparrow \uparrow \uparrow \uparrow 3$. Truthfully, pretty much any analogy grounded in reality would only serve to show "yeah Graham's number is way bigger."

Still, for its own sake and because I found this a little fun, I felt like demonstrating this properly.

Let's take a result from here as granted: the number of legal positions of a Rubik's cube of arbitrary dimension. I don't know why it's called "combinations" there; in my opinion it's misleading, but it matches up with the number of legal positions and in turn symmetries of a $3\times 3 \times 3$ cube if you plug in $n=3$, well-known to be $43,252,003,274,489,856,000$:

I feel it is fairly self-evident that the dimension of the resulting cube is an even number, let it be $k$; this allows for a little simplifying. Then $k \equiv 0 \pmod 2$ and the resulting formula for the number of symmetries for a $k \times k \times k$ Rubik's cube is:

$$7! \cdot 3^6 \cdot \frac{24! ^{\; \text{floor}[(k^2 - 2k)/4]}}{4!^{ \; 6 \cdot \text{floor}[(k-2)^2/4]}}$$

So what is $k$?

You have $4.6 \times 10^{185}$ cubes, in each of which you have $10^{300}$ (a googol cubed) cubes. Let's make the math simpler (and $k$ larger) by basically just assuming each cube is unique from the rest - don't put them together in one cube. This prevents nonsense about "well the inside faces of the smaller cubes won't affect the overall cube," simplifying it all greatly. Also, this way, each symmetry of one cube counts as one for the overall "universe-cube" and ultimately the "universe-cube" need not even be a cube. We now just have a bunch of Rubik's cubes.

Anyhow we see $k = 4.6 \times 10^{485}$ with these considerations.

Plugging this into WolframAlpha gives us that the number of symmetries of the universe-cube is roughly $10 \uparrow 10 \uparrow 973$ if we round up. Thus, the number of the symmetries of the universe-cube has roughly $10^{972}$ digits ... so far.

This neglects the cube-switching symmetry. Let's assume the worst-case value again, and basically line up each cube in order, and consider the overall configuration as one single symmetry. Thus if cube $1$ and cube $2$ individually show the exact same symmetry (i.e. look identical aside from the numbering), swapping them still counts as a new symmetry for the overall "universe-cube" you propose, even if it looks identical. Again, all in the interest of simplifying the math (even if it hugely overestimates the actual number).

Thus we have $(4.6 \times 10^{185})!$ more symmetries to account for, this being the number of ways to arrange that many objects. Ironically, that doesn't matter in the end: Wolfram gives that factor to be $10 \uparrow 10 \uparrow 188$, approximately. So it doesn't even meaningfully affect the overall number (though it is still calculated here if interested).

Still, a number with $10^{972}$ digits, pretty hefty, right?

Okay, but how many digits does Graham's number have? I'll largely follow the explanation given by him on Numberphile.

Let us recall its construction with the up-arrow notation. $3 \uparrow 3$ is basically $3^3$, and $3 \uparrow \uparrow 3 = 3 \uparrow (3 \uparrow 3)$. Proceeding numbers of arrows "reduce" similarly to the two-arrow case:

  • $3 \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow (3 \uparrow \uparrow 3)$
  • $3 \uparrow \uparrow \uparrow \uparrow 3 = 3 \uparrow \uparrow \uparrow (3 \uparrow \uparrow \uparrow 3)$

To see the escalation:

  • $3 \uparrow 3 = 3^3 = 27$
  • $3 \uparrow \uparrow 3 = 3^{27} > 7$ trillion
  • $3 \uparrow \uparrow \uparrow 3 > 10^{\text{3.6 trillion}}$
  • $3 \uparrow \uparrow \uparrow \uparrow 3$ amounts to amounts to a "power tower" of $3$ to its own power $7,625,597,484,987$ times per Wikipedia.

Starting at this last one, let's try to even approach that with the universe cube. After all, Wolfram can at least approximate fairly high. Again, not even in the realm of Graham's number, just a building block of it, but it's a good place to start.

Sadly, we're already well beyond the number of symmetries of your universe-cube at just calculating $3 \uparrow 3 \uparrow 3 \uparrow 3 \uparrow 3$ (a power tower of five $3$'s), which I chose as our random starting point. Wolfram claims this number by itself to simply have about $6×10^{3,638,334,640,023}$ decimal digits. We're well, well past the universe-cube already.

Its suffices to say, if this isn't even in the realm of $3 \uparrow \uparrow \uparrow \uparrow 3$, and $3 \uparrow \uparrow \uparrow \uparrow 3$ isn't even in the realm of Graham's number, the universe-cube is just not enough. Graham's number pretty much defies analogies based in reality. Its status as one of the biggest (meaningful) numbers is not given lightly.

For fun, just to continue seeing just how bad off we are, I continued approximating the power tower of $n$ threes for higher $n$ than $5$. For simplicity I keep using $x \uparrow y$ as a substitute for $x^y$ for readability - that I even have to do so shows just how nuts this gets.

  • $n=10$ yielded about $\underbrace{10 \uparrow \cdots \uparrow 10}_{\text{seven 10's}} \uparrow 12$ digits (Wolfram link)
  • $n=15$ yielded about $\underbrace{10 \uparrow \cdots \uparrow 10}_{\text{twelve 10's}} \uparrow 12$ digits (Wolfram link)

Wolfram wouldn't even return anything in a meaningful timeframe for $n=20$ but I imagine the idea is the number of $10$'s in these is roughly $n-3$.

And then bear in mind what Graham's number is. Let $G_1 = 3 \uparrow^4 3$ (where the $4$ means "put four arrows here"). Then Graham's number is defined recursively by

$$G_{n+1} = 3 \uparrow^{G_n} 3$$

Each time a number is giving the number of arrows in the next, and $G_{64}$ is Graham's number. And remember just how going from one to four arrows blew up to the point of $G_1$ being incalculable for all intents and purposes?

To say it is big, large, huge, monstrous, or any single adjective in the English language doesn't do it justice.

It just straight up defies reality and practicality; good luck at even trying to approach it with an analogy grounded in reality and not something else abstract and mathematical in nature like the problem it was designed for. At best all the analogy would indicate is just how massive Graham's number is by the sheer way in which Graham's number dwarfs it.

  • $\begingroup$ And yet as a set cardinaliy it is smaller than the set of natural numbers. $\endgroup$ – Oscar Lanzi Aug 9 at 12:40
  • 1
    $\begingroup$ Of similar interest: Magnitude of Graham's Number? $\endgroup$ – Dave L. Renfro Aug 9 at 13:33
  • $\begingroup$ This is exactly what I was looking for, Thank you! $\endgroup$ – Mephistophilus Aug 9 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.