Timeline for Escape the zombie apocalypse
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2021 at 15:04 | comment | added | Dmytro Taranovsky | @Eric For the escape from a very thin ring, most density increase manifests by ring layers becoming closer together, and so each extra random layer increases the final density. | |
| Jun 11, 2021 at 16:18 | comment | added | Eric | If you can escape a circle, it's straight forward to prove you can escape a thin ring with only several layers of zombies. Doesn't that contradict what you argued in the field version and upper bound case? For the density of zombies at distance R in the middle of the ring is 1. By your argument aforementioned, the density of zombies will be too large to pass through for sufficiently large R, when you reach the zombies there. That can't be the case if you can escape. | |
| May 26, 2021 at 21:18 | comment | added | Dmytro Taranovsky | @Eric For the circle (note: I previously misread the circle as a disk), I think you can escape ($v<1$, small $d$) by moving in one direction to make the zombie circle sufficiently nonconvex, and then approach the zombie circle tangentially, and swerve near the end. | |
| May 26, 2021 at 5:13 | comment | added | Eric | But of course, a straight line move is suboptimal. If there were only 2 zombies, who're at distance 1 from each other, then you always find a path to slip through between them, for small enough d and v (v=d=1/100 is more than enough). This is true regardless of the pair's distance and relative positions to you. | |
| May 26, 2021 at 2:23 | comment | added | Eric | Let's forget about randomness. Let v=d=1/100. In my previous comment the threshold R is about $50^{100}$ by direct calculation, if you move in a straight line towards the midpoint of two zombies. This is too small relative to your estimation. Did I miss something? | |
| May 25, 2021 at 0:38 | comment | added | Eric | In the simple pursuit, given v and d>0, suppose we only have zombies on a circle of radius R, equally spaced with distance 1 from each other. Does your argument imply we get caught for sufficiently large R? | |
| May 15, 2021 at 22:35 | comment | added | Dmytro Taranovsky | @Eric That would only work for two specific zombies that are already nearly aligned. | |
| May 15, 2021 at 2:53 | comment | added | Eric | Suppose $a(0)$ and $b(0)$ are $(1, c)$ and $(1, c+\delta)$ respectively. With $c$ sufficiently large, isn't it better to first move locally to aline $a, b$ and $r$, and then approach $a$ in a straight line? $\vert a(T)-b(T)\vert$ will be (far) greater this way than if you move in a straight line from the start. | |
| May 14, 2021 at 14:57 | comment | added | Dmytro Taranovsky | @Eric It comes from $(b(t)-a(t))_⊥$ (which accounts for zombies moving closer to each other); $k-1$ is the dimension of the space orthogonal to $a(t)-r(t)$. | |
| May 14, 2021 at 2:29 | comment | added | Eric | "by integrating the above $b'$ equation, we get $\log \det J_{f_T} = -(k-1)v \int_0^T \frac{dt}{|r(t)-a(t)|}$" May I ask why $\det J_{f_T}$ appears on the left hand side? There's no $\det J$ in the $b'$ equation. | |
| Apr 25, 2021 at 1:37 | comment | added | Dmytro Taranovsky | @Eric If an intelligent zombie is too close, you can move sideways to recover distance, and then continue in the original direction. | |
| Apr 23, 2021 at 12:18 | comment | added | Eric | What do you mean by “perturb your path”? How does that work? | |
| Apr 24, 2019 at 3:14 | history | edited | Dmytro Taranovsky | CC BY-SA 4.0 |
improved lead; fixed one formula
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| Apr 16, 2019 at 17:54 | history | answered | Dmytro Taranovsky | CC BY-SA 4.0 |