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Assumptions:

  • The other person is constantly and randomly roaming
  • Foot traffic concentration is the same at all points of the park
  • Field of vision is always the same and unobstructed
  • Same walking speed for both parties
  • The other person is NOT looking for you. They are wandering around having the time of their life without you.
  • You could also assume that you and the other person are the only two people in the park to eliminate issues like others obstructing view etc.

Bottom line: the theme park is just used to personify a general statistics problem. So things like popular rides, central locations, and crowds can be overlooked. I know this can be simulated, but can this be calculated analytically?

Dargscisyhp
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  • I don't think there would be a difference. If you move, you could find him faster, but you could also find him slower than if you stood in place. – SalmonKiller May 14 '15 at 01:41
  • Saw this on reddit, huh? – Spooky May 14 '15 at 01:46
  • I did. Though I could not find an analytic answer. – Dargscisyhp May 14 '15 at 02:05
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    There's a difference. If you can run fast compared to how your friend is moving at least, then you should do better by running around. For example, in the limit where you're running really fast, you'll nearly instantly run around the whole park before your friend has time to move far. This doesn't solve the you-move-as-slow-as-your-friend case, but it shows it isn't just a simple equivalence. Also, in that case, boundary effects are going to matter. – aes May 14 '15 at 02:34
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    Another comment: Even with random walks on an infinite plane with both people moving the same speed, the two situations are not the same. In the one with both moving, the random walk step size is larger on average. – aes May 14 '15 at 02:44

1 Answers1

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Suppose your friend moves around taking a step in a uniformly random direction each time step and suppose you can either stand still or move by the same rule as your friend.

Then, if the step size is small compared to the visibility radius and the park is large (so boundary effects are to be ignored), I claim you'll find each other roughly twice as fast by both moving.

Why is this? You each taking a step is equivalent to your friend taking two steps, because we're ignoring the boundary of the park and we're assuming your friend's direction is uniformly random.

Then if the step size is small compared to the visibility radius, we can (approximately) ignore the possibility that you step into view just for one step and then out of view. So only checking if you can see each other every other step should be approximately equivalent.

(This argument can be formalized further in the framework of brownian motion, where [if we put you on an infinite plane] it will work exactly because brownian motion corresponds to a limit in which the discretized step size goes to zero.)

aes
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