03/01/2025
Imagine this hypothetical: You've just unexpectedly fallen into the Backrooms, the mysterious fictional dimension composed of labyrinths of bland empty rooms. For the sake of convenience, let's imagine that in this Backrooms you don't need to eat or drink and there are no monsters trying to kill you (a core feature of many interpretations), meaning your only goal is to walk around until you find an exit. How long would it take to find one? Let's take a look at a few different scenarios.
Let's start by assuming that there are a large number of exits (In Kane Pixels' Backrooms series this appears to be possible by randomly noclipping again, but in many other versions there are instead exits to further "levels"). To make the problem more approachable mathematically, we will also reduce the space to a two-dimensional square lattice: A series of points connected by paths in a square pattern.
This has an obvious correspondence under the more restrictive assumption that the Backrooms is a series of regular square rooms connected on all sides. Most interpretations of the Backrooms are more complex than this and can't be easily reduced to a mathematically workable form, but most are similar enough that any theorem here should still roughly apply. Regardless of the shape, if we treat the space as a series of points connected by paths, we are then free to place exits on those points randomly. This is where random walks come in: Our final assumption will be that you will be walking around completely randomly, including at times turning and walking through spaces you have just been through. This is most likely not exactly what someone would do in this position, since your chances of finding an exit dramatically increase if you avoid traversing places you've been before (mathematically, this is known as a "self-avoiding walk", a more complicated task to handle). However, on the large scale you would likely be walking effectively randomly without any obvious guides to go off, and it's not implausible that you will eventually forget which of the many places you've been through, making it closer to a true random walk.
Assuming that some proportion $p$ of the points on the lattice have an exit, the random distribution of these points and the memorylessness of the walk (the probability being unaffected by previous steps) means that the probability of finding an exit on each given step will always be $p$. This means that the walk acts as a series of Bernoulli trials, allowing us to model the time it would take to escape using a geometric distribution. Consider it this way: The probability of finding an exit after some number of steps ($n$) will be equivalent to the probability of not finding the exit on all the previous steps ($(1-p)$ $n-1$ times in a row, or $(1-p)^{n-1}$) and also finding an exit on this step ($p$), giving us a formula of $(1-p)^{n-1} p$. Let's graph this for different values of $p$:
However, this is just the probability of finding an exit after a certain specific number of steps. Let's instead graph the cumulative distribution to find the probability of finding an exit in $n$ steps or less:
This is promising! It means that if just 1 in 1000 rooms have an exit, you have nearly a 20% chance of escaping after having searched 200 rooms, and that's if you backtrack for a truly random walk. If each point (room) takes 10 seconds to explore, you'll have a 50/50 chance of having found an exit after about an hour and 55 minutes. If only 1 in 10,000 rooms have an exit, it'll take closer to 20 hours, and if just 1 in 100,000 rooms have an exit it'll take about 8 days.
The original Backrooms 4chan post claims it to have an area of 600 million square miles, equivalent to 1.5 quadrillion square meters, or three times the surface area of the earth. Probably the scariest case for us (besides no exits at all) is if there exists only a single exit in the entire space. If this is the case, we can find the average amount of time it would take to find it by estimating the total number of points (rooms) in the space. Let's give it 50 square meters based on the original image, which gives us a total of 30 trillion rooms. If the probability of finding the exit each step is 1 in 30 trillion, it will take approximately 6,595,000 years before there's a 50/50 chance you'll have found the exit, by which point you'll have undoubtedly gone mad.
Kane Pixels' Backrooms series centers around the "threshold", a portal between the real world and a point in the Backrooms created by the series' main mysterious corporation. One of the videos in the series follows a man who travels into the Backrooms in a group, experiences an inexplicable time travel event, and then becomes lost while trying to make his way back. This raises a question: If you were to travel into the backrooms through a portal (assume this is the only exit), travel away from the portal and then get lost, how long would it take to find it again knowing it's somewhere nearby? In fact, if you assume the Backrooms is infinite, are you guaranteed to eventually find the exit or is there a chance you could wander out into infinity forever?
Let's assume for simplicity that this is another random walk on a square lattice, this time starting at a specific point (the origin point) and aiming to return to it. We know that if we are guaranteed to return to the origin in a finite amount of time then we are also guaranteed to eventually find some point nearby, since A. we will eventually find the origin point an infinite number of times and B. from that point there is a finite probability of walking to some other point within a finite distance, meaning it will certainly eventually happen. A random walk which will certainly return to its starting position after a finite number of steps is said to be recurrent, and any random walk that is not recurrent is transient. Before we get to the square lattice, let's start by determining whether a random walk on a one-dimensional lattice (a line of points with two possible directions each step) is recurrent. This should be simpler.
Let $u_n$ be the probability that a random walk on a lattice is at 0 (the origin point) after $n$ steps. It is immediately apparent that $u_0 = 1$, and that $u_1 = u_3 = u_5 = ... = 0$, since it's impossible to be on an even-numbered point on the line after an odd number of steps. It is known that if the sum of $u_n$ for all values of $n$ from 0 is infinite, then the walk is recurrent, and if it is finite, then the walk is transient (Providing a proof here is unfortunately beyond my abilities, see [1]). Let's write this mathematically:
A random walk is recurrent if and only if \[ \sum_{n=0}^{\infty} u_n = \infty \] otherwise it is transient.
Now let's consider the one-dimensional lattice. First of all, since $u_n = 0$ for odd values of $n$ we know that $\sum_{n=0}^{\infty} u_n = \sum_{n=0}^{\infty} u_{2n}$. In order to be at 0 (the origin point) at step $2n$, you must have taken the same number of steps in both directions. Since at each step there are 2 options the total number of possibilities at step $2n$ is $2^{2n} = 4^n$, and of those there are ${2n} \choose n$ combinations that have the same number of steps in both directions, meaning the probability $u_{2n}$ is equal to $\frac{{2n} \choose n}{4^n}$. Thus, we must determine whether this is true or false: \[ \sum_{n=0}^{\infty} \frac{{2n} \choose n}{4^n} = \infty \]
Stirling's approximation tells us that $n! \sim \sqrt{2 \pi n} (\frac{n}{e})^n$. Here, the $\sim$ means that the two quantities are asymptotic; Their ratio converges towards 1 as $n$ approaches infinity, meaning if the sum of a formula is infinite then any asymtotic quantity will be as well. We can derive from this that ${{2n} \choose n} \sim \frac{4^n}{\sqrt{\pi n}}$ (See [1] for the proof), and thus \[ \frac{{2n} \choose n}{4^n} \sim \frac{1}{\sqrt{\pi n}} \]
$\sum_{n=1}^{\infty} \frac{1}{n}$ is the well-known harmonic series, which is known to diverge to infinity. Since $\frac{1}{n^x}$ converges slower than $\frac{1}{n}$ for all $0 \lt x \lt 1$, we can determine that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}}} = \infty$. For our case of $\sum_{n=1}^{\infty} \frac{1}{\sqrt{\pi n}}$ we can factor out the constant $\frac{1}{\sqrt{\pi}}$ to get $\frac{1}{\sqrt{\pi}} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$, which must also equal $\infty$. Note that we have switched from $n=0$ to $n=1$ for our formula since $\frac{1}{0}$ is undefined. This is no problem, since we know $u_0 = 1$ and thus $\sum_{n=1}^{\infty} u_{2n}$ = $\sum_{n=0}^{\infty} u_{2n} - 1$, meaning if one is infinite then the other will be as well. We can now build our final proof:
\[ \sum_{n=1}^{\infty} u_{n} = \sum_{n=1}^{\infty} u_{2n} = \sum_{n=1}^{\infty} \frac{{2n} \choose n}{4^n} \sim \sum_{n=1}^{\infty} \frac{1}{\sqrt{\pi n}} = \frac{1}{\sqrt{\pi}} \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \infty \]
Thus a random walk on a one-dimensional lattice is recurrent. An interesting fact that follows from this is that a gambler playing any game with fair odds (and finite money) will eventually hit zero and lose everything.
Determining the case for a two-dimensional square lattice (which we are assuming the Backrooms is) is not too difficult once we know this. We can use a trick for it: If we rotate our lattice by 45° and shrink it by a factor of $\sqrt{2}$ (see the light blue lattice below), we can show that our four movement directions are practically equivalent to if we were moving diagonally across the lattice.
If we think of it this way there are four movement options each step: (+1, +1), (+1, -1), (-1, +1), and (-1, -1). This is effectively the same as engaging in two one-dimensional random walks at the same time, where both walks have to return to 0 at the same time to be at the overall origin point. This means that for a two-dimensional random walk: \[ u_{2n} = (\frac{{2n} \choose n}{4^n})^2 \sim (\frac{1}{\sqrt{\pi n}})^2 = \frac{1}{\pi n} \] Hence, \[ \sum_{n=1}^{\infty} u_{2n} \sim \sum_{n=1}^{\infty} \frac{1}{\pi n} = \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} = \infty \] Thus a random walk on a two-dimensional lattice is also recurrent. This tells us that if you were to enter the Backrooms through the threshold, and then lose your way, you are guaranteed to eventually find the threshold again in a finite amount of time by walking around randomly. However, you might be surprised to learn that the average amount of time it would take to return to the origin is infinite. What this would mean in practice is although you are likely to find the exit in a relatively short amount of time, and you will certainly eventually find it, the off-chance of it taking an arbitrarily long amount of time means the average (or expected) amount of time is infinite, meaning in theory you should "expect" the walk to take forever. This is very counter-intuitive, but it really is the case for both one-dimensional and two-dimensional walks.
I have neglected to mention a crucial detail regarding Kane's Backrooms series: His interpretation of the Backrooms exists across multiple levels. This would, assuming there are an infinite number of layers, require comparison to a three-dimensional lattice, which is a more complicated problem to solve than the first two dimensions. But before we get there, I do admit that the random set of paths seen in the Backrooms series can't cleanly be reduced to any sort of regular lattice, and I can't prove that theorems that apply to these lattices would also apply to the "real" situation. These are merely pieces of evidence at this point, not proof. However, we can strengthen our evidence by considering multiple types of lattices (beyond the simple square lattice, cubic lattice, etc.) and looking at common points between them. The probability-generating function for finding the probability of returning to the origin point of a lattice is known as the lattice Green's function [2].
The values of $\sum_{n=1}^{\infty} u_{2n}$ are more complicated to determine for the other two types of regular two-dimensional lattices not yet covered: The triangular lattice and the hexagonal (or honeycomb) lattice. I believe (from my understanding of [2]) that in the case of the honeycomb lattice:
\[ u_{2n} = (\sum_{j=0}^{n} {n \choose j}^2 {2j \choose j}) \frac{1}{3^{2n}}\]
And for the triangular lattice:
\[ u_{n} = (\sum_{j=0}^{n} {n \choose j} (-3)^{n-j} b_j) \frac{1}{6^n}\]
Where $b_j = \sum_{j=0}^{n} {n \choose j}^2 {2j \choose j}$, the formula for the honeycomb lattice. Understanding these is not crucial; What matters is that just as with the square lattice, both of these diverge to infinity when summed.
Now let's consider three-dimensional lattices. The trick used for determining the probability $u_{2n}$ for a square lattice unfortunately does not extend to the simple cubic lattice. However, following the trend of the one and two-dimensional examples, $\sum_{n=1}^{\infty} u_{2n}$ is asymtotic with $C \sum_{n=1}^{\infty} \frac{1}{n^{\frac{3}{2}}}$ where $C$ is some constant (probably $3 \sqrt{3}$). This raises an issue: The sum $\sum_{n=1}^{\infty} \frac{1}{n^x}$ does not diverge to infinity for $x \gt 1$. This means that a random walk on a cubic lattice is actually transient; There is a possibility that it will wander off into infinity forever. Indeed, this is also the case for all regular three-dimensional lattices. However, the probability of eventually returning to the origin point differs based on the type of lattice. This can be determined if we know the average number of times a walk will intersect the origin, which for the simple cubic case is known as "Watson's integral" [4]. It is obtained using this unwieldy triple integral:
\[ \frac{3}{(2 \pi)^3} \int_{- \pi}^{\pi} \int_{- \pi}^{\pi} \int_{- \pi}^{\pi} \frac{dx dy dz}{3 - \cos x - \cos y - \cos z} \]
Which can be simplified to $\frac{1}{32 \pi^3} (\sqrt{3} - 1) (\Gamma (1/24) \Gamma (11/24))^2$. This evaluates to ~1.516386. This figure includes the initial step (which is guaranteed to be at the origin point), so the average number of times we can expect to return to the origin is ~0.516386. The probability of returning to the origin point is $1 - \frac{1}{1.516386...} = 0.340537...$, or just over 34%. This is not good! It means you have almost a two-thirds chance of wandering off forever, and it'll undoubtedly be higher if you start some distance from the intended destination. The other lattices aren't much more promising: 55.78% of returning for a diamond lattice, 28.22% for a body-centered cubic lattice, and just 25.63% for a face-centered cubic lattice.
There are undoubtedly flaws and simplifications in this work that hinder a more accurate picture of what would happen in this (highly specific and unrealistic) situation. Of course, if you know you are close to an exit you'll have a much higher chance of escaping using a systematic scanning/mapping approach than by wandering around randomly. It is also undetermined whether all infinite two-dimensional sets of paths (including random non-regular ones like Kane's Backrooms) experience recurrence, and whether all three-dimensional sets of paths experience transience.
A promising future path would be to attack this problem through simulation rather than mathematics (an easy task in the case of random walks). This might be something I will look into at some point, it could make for a good future post.
References:
[1] Tyler Zhu, Pólya's Recurrence Theorem. https://tylerzhu.com/assets/handouts/polya_rec.pdf
[2] Anthony J Guttmann (2010), Lattice Green's functions in all dimensions. DOI: 10.1088/1751-8113/43/30/305205
[3] W. H. McCrea, F. J. W. Whipple (1940), Random Paths in Two and Three Dimensions. DOI: 10.1017/S0370164600020265
[4] Eric W. Weisstein, Pólya's Random Walk Constants. https://mathworld.wolfram.com/PolyasRandomWalkConstants.html
https://oeis.org/A086230 and https://oeis.org/A086231