Critical Case

The critical case is where $\alpha=1$ . Intuitively this is because the first-passage metric has no preference over the lengths of the edges hence there are many more paths to a specific vertex $\mathbf{x}$ of optimal length. This isn't the case when $\alpha \neq 1$ since for $\alpha<1$ we have that $k^\alpha$ is concave which means that longer edges are preferred therefore the only path to $\mathbf{x}$ which can be made in optimal time is the direct edge from the origin. Similarly, when $\alpha>1$ we have that $k^\alpha$ is convex and hence shorter edges are preferred. This means that the only paths to $\mathbf{x}$ which can be of optimal length consist of $||\mathbf{x}||$ edges of length $1$ .

In this case our simulations suggested that the limiting shape of $t^{-1}\mathcal{B}^{(\alpha)}_t$ should coincide precisely with the unit $l^1$ ball $B_1$ . The figure below shows one quadrant from a simulation where the edge weights have distribution

$f(x)=p\chi_{\{1\}}(x)+(1-p)e^{1-x}\chi_{\{(1,\infty)\}}(x)$

for $p=0.7$ . This is a distribution with an atom at $\{1\}$ and an exponential tail. The white region shows the unoccupied space, the black region shows the active vertices under the LRFPP model and the grey points are those which are active in both the nearest neighbour and long-range models. For this simulation the nearest neighbour model has been coupled with the long range one so that the nearest neighbour growth set cannot extend further than the corresponding long range one. The growth set for the long-range model has a flat piece which extends over the entire edge of the $l^1$ ball and the nearest neighbour growth set has a flat piece which coincides with the results of Marchand.

Our main result for the critical case is indeed exactly what the simulations have suggested in that there are only finitely many lattice points which are not reached in optimal time.

Theorem 1:

If $\alpha=1, p \in (0,1]$ we have that for $\mu \in \mathcal{M}_p$ that the limiting shape:

$A_\mu^{(\alpha)} :=\lim\limits_{t\rightarrow \infty} t^{-1}\mathcal{B}^{(\alpha)}_t$

exists a.s. and

$A_\mu^{(\alpha)} = B_1.$