Planar first-passage percolation fluctuations

Imagine a square street grid with a random travel time on each edge. A longer route can arrive first if its edges are quicker. With independent exponential times averaging one, how widely do fastest arrival times vary across fresh grids as the destination moves farther east? The conjectured variance grows as distance to the two-thirds power.[1] [2]

Independent random delays define a shortest-path metric on a square grid. Does optimizing over detours produce the conjectured two-thirds variance power? Let $T_n=\inf_\gamma\sum_{e\in\gamma}\tau_e$, with $\gamma$ ranging over nearest-neighbor paths from $0$ to $(n,0)$ and independent $\tau_e\sim\mathrm{Exp}(1)$. The question is $\mathrm{Var}(T_n)=\Theta(n^{2/3})$, with fixed positive comparison constants as $n\to\infty$.[1] [2]

A logarithmic variance lower bound shows that the arrival-time spread cannot stay bounded. The predicted two-thirds growth is a much sharper claim. A newer improvement applies to a smooth random landscape with rotational symmetry, rather than this grid of independent edge times.[1] [2]

Lattice variance estimates leave the predicted exponent undetermined. The reported Riemannian improvement uses a smooth positive bounded field, rotational invariance and positive-correlation assumptions. Its bound is not a theorem about the independent exponential lattice model; neither are exact exponents in solvable last-passage settings.[1] [2]

The fastest route depends on the very delays being measured, so its uncertainty differs from adding delays along one fixed route. The problem tests how route choice shapes fluctuations in random spreading, and whether this simple model shares the scaling found in other growth models.[2]

Optimizing over paths couples a route to the random environment, so independent edge times do not make the chosen route’s cost a sum of independently selected delays. Quantifying the resulting fluctuations connects random metric geometry to the proposed universality of planar growth, beyond models with an exact solution.[2]