**Simple pursuit**

Zombies moving towards you will always catch you, but due to their lack of intelligence, your survival time increases exponentially with your relative speed. In $k=O(1)$ dimensional space ($k=2$ in the problem), the expected survival time is $d⋅(1/Θ(μd^k))^{(1+1/v)(1±o(1))/(k-1)}$ if $v$ is bounded below 1 and $μd^k→0$. I conjecture that for fixed $v$, the above $o(1)$ is unnecessary. If $μ$ depended on the distance $r$ from the origin, the threshold value for survival (at large $r$ and constant $v<1$) is $μ(r) = r^{-(k-1)v/(1+v)±o(1)}$ (the $o(1)$ is likely negative and necessary here).

Consider first the continuous field version of the problem: The initial density is $μ$ and the player loses when the mass within distance $d$ of the player reaches (or exceeds) 1. Let $r(t)$ be the trajectory of the player; $r(0) = 0$. If $a$ and $b$ are trajectories of two possible zombies, we have:

- $a'(t) = v \frac{r(t)-a(t)}{|r(t)-a(t)|}$

- $|b(t)-a(t)|$ is nonincreasing

- For small $|b(t)-a(t)|$, $b'(t) = a'(t) - \frac{v}{|a(t)-r(t)|} (b(t)-a(t))_⊥ + O(\frac{|b(t)-a(t)|}{|r(t)-a(t)|})^2$ where the orthogonal projection $(b(t)-a(t))_⊥ = b(t)-a(t)-(b(t)-a(t))⋅(r(t)-a(t)) \frac{r(t)-a(t)}{|r(t)-a(t)|^2}$.

- If $f_T(a(0)) = a(T)$, then the density at time $T$ at $a(T)$ is $μ/\det J_{f_T}$, and by integrating the above $b'$ equation, we get $\log \det J_{f_T} = -(k-1)v \int_0^T \frac{dt}{|r(t)-a(t)|}$.

Now among all $a$ and $r$ with $|a(T)-r(T)|≤d$, the integral (and thus the density) is minimized if $|a(0)|=(1+v)T+d$, which requires moving in a straight line from the origin at maximum speed. Furthermore, in this case, the density at $a(T)$ matches the average density within distance $d$ up to a constant factor, and the precise bounds in the first paragraph follow.

*Lower bounds*

For the nonfield version, we can get the lower bounds by escaping in a nearly straight line while avoiding traps. This is possible here even if the player velocity is always within $ε$ (if $ε$ is $Θ(1)$) of moving with speed 1 in the positive $x$ direction. Intuitively, on a straight line, the average distance between traps of size $O(d)$ is $O(1/(μ_1 d^{k-1}))$ (where $μ_1$ is the relevant field density; $1-v=Ω(1)$), and using the (approximate) independence, the frequency of larger traps drops exponentially with trap size, with the lower bounds (in the first paragraph) reached with high probability unless the initial position is a trap.

However, formalization of traps is a bit tricky, so we instead observe that we can trace out a trajectory of speed $1+v$ of sufficient clearance against the initial configuration, and then evolve it the same way as the zombies to get the escape trajectory. As long as all tangents of the initial trajectory are at an angle $≤α$ (for $α≤45°$) to the $x$ axis, this property will hold throughout the trajectory evolution, allowing us to ensure that the clearance will not shrink too much. For fixed $v$ ($μ$ does not depend on $r$), we can remove the $o(1)$ from the lower bounds for survival time by choosing a trajectory with variable $α$ with, at each point, $1/α$ at least polynomial in the distance between the point and the final destination.

Also, for $v=1$ (and small $d$) and variable $μ(r) = r^{-k/2+1/6-ε}$, you can survive by making your trajectory increasingly smooth: Zombies following you at a small distance gain on you at a speed proportional to the square of the path curvature (and the square of the distance to you), so with curvature $r^{-0.5-ε}$ you can avoid $d→0$ as $r→∞$ for those zombies. In a straight line path and $d=Θ(1)$, you will encounter zombies at typical intervals $s = Θ(1/(μ(r) r^{(k-1)/2}))$. Avoiding an incoming zombie from a distance $s$ uses correction $O(\sqrt s)$, corresponding to curvature $O(s^{-1.5})$, which is $O(r^{-0.5-ε_2})$ for the above $μ(r)$.

Also, for $v≤1$ and $d=0$, you can survive indefinitely (i.e. not lose at finite time) by simply moving at speed 1 in a straight line in any unoccupied direction — or even, with probability 1, by moving with speed 1 along any curve with bounded curvature, with the curve chosen independently of zombie positions.

*Upper bound*

For the upper bound, it suffices to consider a single point and a linear approximation to the problem. To escape, for every $R>0$, the player would have to cross (at time $O(R/v)$) an $a(t)$ that starts at the sphere $|a(0)|=R$. A player cannot approach $a(t)$ to within distance $d$ without increasing the field density at $a(t)$ in $ρ =(R/d)^{(k-1)/(1+1/v)}$ times. Furthermore, as long as the increase in density at $a(t)$ is $o(ρ)$ times, the cumulative relative nonlinearity within distance $O(d)$ of $a(t)$ is $o(1)$. (Proof outline: If the remaining density increase is $ρ_1$ times, then the distance of the relevant points to $a$ is $D=O(ρ_1^{1/(k-1)} d)$, and with the player far enough compared to $D^2/d$, the nonlinearity is small enough.) From there, for large enough $R$, $\{b(0):|b(t)-a(t)|≤d|\}$ contains a volume $ω(\log R)$ ellipsoid (contained within distance $O(R)$ from the origin). By a counting argument, with probability $1-o(1)$ all such ellipsoids contain at least one relevant point, as required. For the bounds without $o(1)$, we are off by a factor of $\log(1/(μd^k))$ inside the $Θ(μd^k)$. However, I expect that the $\log$ factor can be eliminated by using the player path rather than a single point $a(t)$ and by analyzing the impact of nonlinearities.

**Intelligent pursuit**

If zombies (in $k = O(1)$ dimensional space) could strategize and cooperate, and letting $D=\frac{1}{μd^{k-1}v}$, they could surround you with gaps $<d$ in time $O(D)$ and capture you in time $O(D + \frac{\log^{1/k}(1+Dμ^{1/k})}{μ^{1/k}v})$ (and even do this with a strategy independent of your movement; also, the second summand is typically small and stems from random fluctuations in zombie density). You cannot escape for $μ(r) = ω(1/r^{k-1})$. I do not know whether you can escape if $μ(r) = O(1/r^{k-1})$ (and $d$ is small enough and $v<1$); Conway's angel problem has some of the similar subtleties.

At $v<1$, a fixed finite number of zombies cannot capture you at a small enough $d$ since you can perturb your path (with sufficient smoothness and clearance) to avoid the first zombie, use a smaller scale perturbation against the second one, and so on. For the naive implementation, you might have to avoid the $n$th zombie $2^{n-1}$ times, and your clearance drops exponentially with $n$. However, by considering groups of zombies, you can cut off a fraction at each length scale, allowing a polynomial clearance, and you can escape at density $μ(r) = r^{-k+ε}$ for a small enough $ε>0$ dependent on $v$ (and small enough $d$ dependent on $v,ε$).

At $v=1$, $k+1$ well-placed zombies (i.e. 3 for the plane) can win in finite time even at $d=0$. The reason is that a single pursuer can guard a half-space. They can even (deterministically) win if they have nonzero but small enough reaction time proportional to distance (i.e. the speed of light is finite). A single pursuer can guard a slowly receding half-space, allowing 1 (effectively 2) out of $k+1$ pursuers to advance.

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