3 ms·

This is probably a solvable number, and these 1.4 numbers are suspiciously close to the square root of two 1.41421...

by koverda 6y ago

MarcelVos 6y ago
That was my thought as well, but I couldn't justify it being root 2 so I went with the number I got.
- crdrost 6y ago
  For the simplest case you have a system of transitions Start → Mid(1, ↑) prob: 1, time: 1? Mid(1,↑) → Start prob: ⅜, time: 3 Mid(1,↑) → End prob: ¼, time: 1 prob: ⅜, time: 3 (not 100% sure about those times, here they are in units of the timesteps that a person spends to go from one square to another, so the 3 is “go to the left square, go to the middle square but coming from the left, go from here up/down having the normal orientation again”). There seem to be two ways to do this. First you could determine the average time to leave the park by pumping 1 person into the park every timestep until it reached a steady-state where one person was leaving the park, then just count how many people there are in the park. Assuming steady state, you do not need to use travel times to calculate that exactly 1.6 people occupy Start and 1.6 occupy Mid, with 0.6 making the journey back and 0.4 making the fast journey out and 0.6 making the slow journey out. These two slow journeys taking time 3 can be viewed as having (3–1)•0.6 occupation, or 1.2 people in each, so the total is 1.2 + 1.2 + 1.6 + 1.6 = 5.6 people in the park, so this should be the number of timesteps. Or if you prefer a straight calculation it is S where S = 1 + T T = ⅜(3+S) + ¼(1) + ⅜(3) S being the average time from Start → End and T bring the average time from Mid → End. This can be solved to similarly find S = 28/5 = 5.6 timesteps. I think the former approach is going to be theoretically easier to understand when you are asking, “I want to convert End to Mid(n+1, ↑) and introduce a new End and also a node Mid(n, ↓), how does adding these three nodes change the system?” In fact I think to solve it you will want to always calculate three different flows. I have only given you one of them, a steady state where one person arrives in Start and then they leave out of End. The other one is that they arrive in Mid(n–1, ↑) at a steady rate and then leave out of End at that same rate. And finally there is the flow where they arrive in Mid(n–1, ↓) at a steady rate and then leave at the same rate. [I think Mid(0, _) is just Start here.] If I have those three populations/average travel times then I have a way to add those 3 nodes to the system at each stage and this gives me a recurrence which can calculate the thing exactly. I would need to think a bit about how to program all of that.
ironSkillet 6y ago
Seems like you could come up with a recursive formula to compute the expected number of steps to reach hedge N. Along the lines of how you compute the expected number of coin flips to get K heads in a row. I'm too lazy to figure it out right now though, so there may be some nuance I'm missing.
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This is probably a solvable number, and these 1.4 numbers are suspiciously close to the square root of two 1.41421...

MarcelVos 6y ago

That was my thought as well, but I couldn't justify it being root 2 so I went with the number I got.

crdrost 6y ago
For the simplest case you have a system of transitions Start → Mid(1, ↑) prob: 1, time: 1? Mid(1,↑) → Start prob: ⅜, time: 3 Mid(1,↑) → End prob: ¼, time: 1 prob: ⅜, time: 3 (not 100% sure about those times, here they are in units of the timesteps that a person spends to go from one square to another, so the 3 is “go to the left square, go to the middle square but coming from the left, go from here up/down having the normal orientation again”). There seem to be two ways to do this. First you could determine the average time to leave the park by pumping 1 person into the park every timestep until it reached a steady-state where one person was leaving the park, then just count how many people there are in the park. Assuming steady state, you do not need to use travel times to calculate that exactly 1.6 people occupy Start and 1.6 occupy Mid, with 0.6 making the journey back and 0.4 making the fast journey out and 0.6 making the slow journey out. These two slow journeys taking time 3 can be viewed as having (3–1)•0.6 occupation, or 1.2 people in each, so the total is 1.2 + 1.2 + 1.6 + 1.6 = 5.6 people in the park, so this should be the number of timesteps. Or if you prefer a straight calculation it is S where S = 1 + T T = ⅜(3+S) + ¼(1) + ⅜(3) S being the average time from Start → End and T bring the average time from Mid → End. This can be solved to similarly find S = 28/5 = 5.6 timesteps. I think the former approach is going to be theoretically easier to understand when you are asking, “I want to convert End to Mid(n+1, ↑) and introduce a new End and also a node Mid(n, ↓), how does adding these three nodes change the system?” In fact I think to solve it you will want to always calculate three different flows. I have only given you one of them, a steady state where one person arrives in Start and then they leave out of End. The other one is that they arrive in Mid(n–1, ↑) at a steady rate and then leave out of End at that same rate. And finally there is the flow where they arrive in Mid(n–1, ↓) at a steady rate and then leave at the same rate. [I think Mid(0, _) is just Start here.] If I have those three populations/average travel times then I have a way to add those 3 nodes to the system at each stage and this gives me a recurrence which can calculate the thing exactly. I would need to think a bit about how to program all of that.

ironSkillet 6y ago

Seems like you could come up with a recursive formula to compute the expected number of steps to reach hedge N. Along the lines of how you compute the expected number of coin flips to get K heads in a row. I'm too lazy to figure it out right now though, so there may be some nuance I'm missing.

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