Self-avoiding random walk on $\mathbb{Z}^2$ getting stuck

Asked Jun 23 '14 at 13:42

Active Jun 23 '14 at 13:42

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Let $W_n$ be a self-avoiding random walk (SAW) on $\mathbb{Z}^2$, starting at the origin. Formally, $W_0=0$ and for $n\ge 0$, $W_{n+1}$ is chosen uniformly from the neighbours of $W_n$ which were not visited earlier: $$W_{n+1}\sim U\left(\{u\in\mathbb{Z}^2\mid u\sim W_n,\ \neg\exists0\le m<n:\ u=W_m\}\right)$$ (the inner $\sim$ sign means "neighbour of").

However, for a specific roll and for a specific $n$, it is possible that the set of unvisited neighbours from which $W_{n+1}$ should be chosen uniformly is empty. In that case, we say the walk is stuck, and leave it determined as is.

So my question is as follows:

what is the probability that such a random walk will be stuck?
if the probability is 1, what is the expectation of the hitting time of being stuck?

asked Jun 23 '14 at 13:42

Bach

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experiments show that it will always get stuck and that the expectation is around 70 steps. – mercio Jun 23 '14 at 14:21
Experiments you have made, or published ones? – Bach Jun 23 '14 at 14:26
that I have made. – mercio Jun 23 '14 at 14:27
1

@mercio https://aip.scitation.org/doi/abs/10.1063/1.447349?journalCode=jcp – Gabriel Romon May 23 '18 at 18:30

Self-avoiding random walk on $\mathbb{Z}^2$ getting stuck

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