Lumifer comments on The correct response to uncertainty is *not* half-speed - Less Wrong

77 Post author: AnnaSalamon 15 January 2016 10:55PM

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Comment author: IlyaShpitser 17 January 2016 05:32:17AM 3 points [-]

Imagine an undirected graph where each node has a left and right neighbor (so it's an infinitely long chain). You are on a node in this graph, and somewhere to the left or right of you is a hotel (50/50 chance to be in either direction). You don't know how far -- k steps for an arbitrarily large k that an adversary picks after learning how you will look for a hotel.

The solution that takes 1 step left, 2 steps right, 3 steps left, etc. will find the hotel in O(k^2) steps. Is it possible to do better?

Comment author: Lumifer 19 January 2016 05:53:05PM 0 points [-]

Does the solution have to be deterministic? If an adversary places the hotel after looking at my algorithm, my first inclination is introduce uncertainty into my search.

Comment author: IlyaShpitser 19 January 2016 06:02:39PM 0 points [-]

No. But you can't do better than O(k) anyways, and you can do O(k) deterministically.

Comment author: RichardKennaway 20 January 2016 11:47:00AM 0 points [-]

Randomising the length of the first step will improve on the constant factor by about 2. Similar analysis to the non-adversarial case, and with the same ETA I just added to my earlier comment.

Comment author: gjm 20 January 2016 12:18:10PM 0 points [-]

If you make the run lengths increase exponentially instead of linearly then you get O(k) unconditionally.

Comment author: RichardKennaway 20 January 2016 02:14:25PM 0 points [-]

I know. I was improving the constant factor.

Comment author: gjm 20 January 2016 05:18:45PM 0 points [-]

Ah yes, OK, essentially the same randomization helps against an adversary when you're growing exponentially as when you're growing only linearly. Fair enough.