Collatz's Ant and Σ(n)
Correction (July 09, 2025): Updates on how $\tau$ was being computed through the count of frames (e.g. as in the following script). Now corrected to
τ = len(frames) - 1, not counting initial empty frame.
Relevant preceding posts here and here.
Consider the corresponding ant’s landscape development for $n = 500$:
with the last frame being:
Let’s also consider a score function $\Sigma(n)$ which returns the number of 1’s (or marked states) left by the ant on the corresponding landscape (regarding the last frame). With the prior example, we would have $\Sigma(500) = 54$. We might also want to normalize this by the corresponding stopping time $\tau_{n}$ characteristic of the collatz function dynamics for a given $n$. As such, given $\tau_{500} = 110$, we would have $\frac{\Sigma(500)}{\tau_{500}} ≈ 0.49$.
Additionally, let’s also consider some metrics relative to the ant’s (euclidean) distance to the origin. Say $α$ is the maximum distance reached relative to the origin over the whole development of the landscape, and $γ$ is the distance from the origin at the last frame (so where the ant ends up). For $n = 500$, we would have the ratio $\frac{\gamma}{\alpha} ≈ 0.87$.
The following is a representation of such metrics for numbers from $n = 2$ to $n = 50000$:
