Timeline for Annihilating random walkers
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2018 at 15:02 | comment | added | ofer zeitouni | For Q1, since after time $n$ the parity of the number of particles remains constant, and there is positive probability of odd parity, the answer can't be $0$. I believe that the answer is either 1 or 0 precisely depending on that (random) parity. | |
| Feb 5, 2018 at 21:00 | comment | added | Mateusz Kwaśnicki | @JosephO'Rourke: Yes, of course. Or simply throw in a new walker every other turn. | |
| Feb 5, 2018 at 19:58 | comment | added | Joseph O'Rourke | @MateuszKwaśnicki: I guess I could add in a "no-move" choice, which would then mix the populations. | |
| Feb 5, 2018 at 19:31 | answer | added | RaphaelB4 | timeline score: 8 | |
| Feb 5, 2018 at 19:24 | comment | added | Mateusz Kwaśnicki | This is kind of irrelevant, but, technically, there are two separate populations, introduced on odd and even generations, respectively. | |
| Feb 5, 2018 at 18:51 | comment | added | Robert Israel | On the large scale, I would expect this process to act like a nonlinear reaction-diffusion equation $$ \dfrac{\partial u}{\partial t} = a \Delta u - b u^2 + c \delta(x)$$ | |
| Feb 5, 2018 at 17:55 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |