Analysis of a Collatz Game and Other Variants of the $3n+1$ Problem

Abstract

Introduced and discussed are new variants of the $3n+1$ problem, in particular a 2-player game with moves $3n+1$ and $3n-1$ and subsequent halving, where the player is winner who first reaches the 1. Variants of the classical $3n+1$ problem are discussed with respect to a general convergence conjecture. Finally, we set up prizes for solutions of the Collatz problem and simpler variants.

A Random Walks with Drift

Given is the discrete strip of natural numbers, where each number $i > 0$ has neighbors $i-1$ and $i+1$, and $i=0$ is an absorbing state. Let there be a discrete set of times steps $t \in \mathbb {N}$. A particle starts at some number n and goes to a random neighbor in each discrete time step, independently of the former steps. In the case without drift the particle goes in time step t from $i>0$ to $i-1$ with probability 0.5 and to $i+1$ also with probability 0.5, independently of its history. It is well known that the particle will reach state 0 sooner or later with probability 1, and that the expected number of steps until absorption is infinite, even for starting state n = 1 [Woe00].

In a model with uniform drift, there is some $\epsilon \in [-\frac{1}{2},\frac{1}{2}]$, such that a particle standing on $i>0$ goes to $i-1$ with probability $0.5+\epsilon$ and to $i+1$ with probability $0.5-\epsilon$. In case of an $\epsilon >0$ the particle has a drift towards 0, and for $\epsilon <0$ a drift towards infinity.

Theory says [Woe00]: For $\epsilon > 0$ and any starting number the particle will reach 0 with probability 1 earlier or later in finitely many steps. For $\epsilon < 0$ and starting number $n_0$ there exists a value $q(\epsilon , n_0) < 1$, such that the particle will be absorbed at 0 with probability $q(\epsilon , n_0)$ and goes to infinity with the remaining probability $1 - q(\epsilon , n_0)$.

In a more general setting the particle can for instance jump from i to $i-3$, $i-2$, $i-1$, i, $i+1$, $i+2$, $i+3$, and $i+4$ with probabilities $p(-3)$, $p(-2)$, $p(-1)$, p(0), p(1), p(2), p(3), and p(4), respectively. If $3p(-3) + 2p(-2) + 1p(-1) > 1p(1) + 2p(2) + 3p(3) + 4p(4)$, the particle is drifting towards the left, i.e. towards zero. In this case it will be absorbed by 0 (or negative numbers) with probability 1 in the long run.

In the stochastic Collatz variant with logarithmic scale the particle is approximately jumping from its current state $n_t$ to $n_t + \log (\frac{3}{2})$, $n_t + \log (\frac{3}{4})$, $n_t + \log (\frac{3}{8})$, $n_t + \log (\frac{3}{16})$, $\ldots$ with probabilities $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, $\ldots$, respectively. Observe that $\log (\frac{3}{4})$, $\log (\frac{3}{8})$,... are all negative numbers. And $\frac{1}{2} \log (\frac{3}{2})+\frac{1}{4} \log (\frac{3}{4})+\frac{1}{4} \log (\frac{3}{8}) = \frac{1}{4} \log (\frac{3^4}{ 2^2\cdot 4\cdot 8}) = \frac{1}{4} \log (\frac{81}{128}) < 0$. The jumping widths are only approximately $\log (\frac{3}{2})$, $\log (\frac{3}{4})$, $\log (\frac{3}{8})$, $\ldots$, due to the ‘+1’ term in $3n+1$. However, for large values of $n_t$, this deviation is almost meaningless.