Splitting any $(m+n)$-step <self-avoiding walk> after $m$ steps and translating its remaining segment to the origin injects it into an ordered pair of an $m$-step and an $n$-step self-avoiding walk. Hence $b_{m+n}\leq b_mb_n$. The sequence $a_n=\log b_n$ is subadditive, so the <Fekete lemma> gives
$$
\lim_{n\to\infty}\frac{\log b_n}{n}=\inf_{n\geq1}\frac{\log b_n}{n}.
$$
Exponentiating proves existence of the <connective constant> $\kappa=\lim_nb_n^{1/n}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 204 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-204/1/a/solution