Deriving the Resolution of n-Bit Spacetime

Assumptions

The universe is a single finite bitstring. It is static and timeless. There is no external agent, clock, or law that acts on it from outside. Time, wherever it appears below, is the internal traversal of an observer through the space of configurations — never an externally imposed parameter. Anything that requires an outside "bit-flipper" to make a choice is disallowed and must be replaced by a combinatorial or probabilistic argument that needs no such choice.

1 — Definition of n

A bitstring of length $n$ is an ordered sequence of $n$ bits. $n$ is the bit count. The $n$ bits can be arranged into $2^n$ distinct configurations.

Examples of bitstring:

• byte: 8 bits
• double precision float sizeof(double): 64
• a computer with 8 GB of memory: 8 GB * 8

2 — Spatial Dimensions (Space)

Solution 1 — one dimensional rope

Each one of the $2^n$ possible configurations describes one complete spatial slice. The bitstring's $n$ bits are the spatial content of that slice.

• Spatial resolution: $n$ (bits per slice).
• Temporal resolution (naive upper bound): $2^n$ (number of distinct possible slices).
• Maximal resolution: $n \times 2^n$ — an absolute ceiling, not a claim that all $2^n$ configurations describe an observer, nor that the bound is achieved.

Example spacetime $n=3$:

	t=0	t=1	t=2	t=3	t=4	t=5	t=6	t=7
Bit 1	0	0	0	0	1	1	1	1
Bit 2	0	0	1	1	0	0	1	1
Bit 3	0	1	0	1	0	1	0	1

This is the minimal possible geometric interpretation of a bitstring. This is geometry agnostic resolution.

Solution 1 — 3-sphere

This is what the standard GR implies. A theory from first principles should not postulate it, but derive it. Why 3, why spherical symmetry?

Solution 2 — the hypercube

Instead of a rope, the emergent observer sees the bitstring as an $n$-cube. Every vertex of the $n$-cube is equidistant from the center (this is true for trivial reasons — vertex-transitivity of the hypercube graph — for every $n \ge 1$, not a discovery specific to $n=3$; it is a property of the topology choice, not evidence favoring it over the rope).

The full $n$-cube has exactly $n\cdot 2^{n-1}$ edges — this is a standard graph-theoretic fact.

Solution 3, 4, ...

Many different bitstring-geometry mappings were studied (Paper III); they fall between the rope and hypercube extremes. In order the universe to appear the same in all direction 3-sphere might be the most simple answer.

Conclusion on spatial mapping

Nothing internal to the bitstring dictates which topology it should be read under. The candidate resolution remains anthropic/Solomonoff-style: the emergent observer interprets the bitstring under whichever topology maximizes her probability of existence. The hypercube gives spherical symmetry "for free," which is attractive, but its naive edge-counting route to a spatial-size formula may not work

3 — Temporal Dimensions (Time)

The process, precisely

In static, finite timeless universe bits don't flip. There is no external clock flipping bits here because other bits flipped there. What the emergent observer see as bit-flip, is due to his random walk through $2^n$ configurations.

Observational fact: increasing entropy order. And the minimal difference between two consecutive configurations is one bit. We assume that determines the maximal time resolution (minimal time step).

It takes equally many bit-flips to drive an $m$-unit structure forward in time as it took units to encode it in space.

Start from the all-zero string. At each tick, pick one of the $n$ bit positions uniformly at random and flip it. No position, direction, or outcome is favored by any external rule — the only inputs are "how many positions are there" and "which ones are currently 0 vs 1," both of which are internal, countable facts about the current configuration, not externally imposed choices.

Tracking $k$ = the number of 1s, this is a Markov chain on ${0, \dots, n}$:

$$ P(k \to k+1) = \frac{n-k}{n}, \qquad P(k \to k-1) = \frac{k}{n}. $$

This is the Ehrenfest urn model (Ehrenfest, 1907) — a random walk on the hypercube graph, projected onto Hamming weight.

Constant flip rate, vanishing net drift. At every $k$, exactly one flip happens every tick: $P(k\to k+1) + P(k\to k-1) = 1$, always. What changes with $k$ is not whether a flip occurs but the net drift, $\mathbb{E}[\Delta k] = \frac{n-2k}{n}$, which is $+1$ at $k=0$, falls linearly, and is exactly $0$ at $k=n/2$. "Saturation" is a statement about vanishing net drift under constant flip rate, not about the system slowing down.

No hand-coded selection for "increasing entropy." Starting from $k=0$ (an extreme, atypical configuration), the drift above is a consequence of counting, not an imposed rule: with only 0s present, every flip must increase $k$, since $P(k=0\to 1)=1$ trivially. As $k$ grows toward $n/2$, the pool of available "either direction" flips balances out on its own. No selection of "increasing-entropy paths" needs to be added externally — it falls out of starting at an extreme point of a shape (the hypercube) whose overwhelming majority of volume lies near the middle.

Fuzziness of the saturation boundary

At equilibrium, $k$ is binomially distributed with mean $n/2$ and standard deviation $\sqrt n /2$. The relative width of the saturated region is therefore $\sim 1/\sqrt n$ — never exactly zero (the boundary is genuinely statistical, as it must be for a system with no hand-coded rules), but vanishingly small at macroscopic $n$:

$n$	relative spread (std/mean)
40	0.16
$10^6$	$1.0\times10^{-3}$
$1.5\times10^{77}$ (solar-mass BH entropy)	$2.6\times10^{-39}$

4 — Calibration Against Physical Systems

Any physical system that (a) starts at zero entropy, (b) evolves toward saturation along this process, and (c) has two independently known points $(t=0, r=0)$ and $(t_1, r_1)$ in Planck units, supplies enough information to fix $n$ and check the model's timescale against known physics.

Spatial size should compatible with the entropy/area relationship (Berkenstein-Hawking)

$$ n = \frac{\pi R^2}{l_P^2 \ln 2} \quad\Longleftrightarrow\quad R = l_P\sqrt{\frac{n \ln 2}{\pi}}. $$

Duration is "Own size"

$$ T_H(R) = \hbar c/(4\pi k_B R) $$