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Topic: Lost in the Woods, Again (Read 414 times) |
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SMQ
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Lost in the Woods, Again
« on: Aug 15th, 2007, 8:13am » |
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We've seen the setup (and similar ones) before: you find yourself lost in the woods. You know that you are exactly halfway between two long, parallel roads which are two miles apart, but you don't know what direction you're facing. Due to darkness and fog you will only be able to find a road when you're very near it.
You know from your previous experience that there's a pattern you can follow to ensure that you come to a road before walking more than about 31/4 miles, but you can't remember the details so you settle on a simpler plan: you'll walk a certain distance, then, if you haven't found a road, make a right-angle turn and continue walking until you're out.
What distance should you plan to walk before turning in order to minimize 1) the worst-case distance, or 2) the expected distance you'll walk before finding one of the roads?
--SMQ
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| « Last Edit: Aug 15th, 2007, 8:20am by SMQ » |
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mikedagr8
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Re: Lost in the Woods, Again
« Reply #1 on: Aug 15th, 2007, 5:12pm » |
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Yay, nice puzzle. Since it links over to the previous puzzle, for a start 1.1 miles 
Seriously though, sqrt(2) miles (45 degree angle, so just go straight. I am thinking about phythagoras, and i'm not very good at explaining my thought progress in words), then continue untill you find the road. Also wait until it's day time so you can have better visability.
That's my 10 second mathematics there.
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| « Last Edit: Aug 15th, 2007, 11:08pm by mikedagr8 » |
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Grimbal
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Re: Lost in the Woods, Again
« Reply #2 on: Aug 16th, 2007, 12:54am » |
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For the worst case, I would walk 1.608540 before turning,
for a worst-case distance of 4.161938
But it is a numerical optimization. I don't have an exact formula.
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| « Last Edit: Aug 16th, 2007, 7:57am by Grimbal » |
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SMQ
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Re: Lost in the Woods, Again
« Reply #3 on: Aug 16th, 2007, 6:58am » |
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I should maybe bump this over to Meduim. I put it in Easy because I thought it was a simple optimization problem -- a bit of number crunching but no real insight required -- but looking a little deeper it seems that mostly by luck I chose a particularly simplifying parameterization. So there's a hint: with the right parameterization the first part is almost trivial.
The second part is definitely Medium either way. I have a simple equation of which the optimal distance is a solution, but I haven't yet been able to wrangle a closed-form solution out of it (although I do know the numerical answer).
--SMQ
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Grimbal
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Re: Lost in the Woods, Again
« Reply #4 on: Aug 16th, 2007, 8:50am » |
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OK, I got it
You should turn after (2^(2/3)+1)^(1/2)
for a worst distance of (2^(2/3)+1)^(3/2)
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SMQ
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Re: Lost in the Woods, Again
« Reply #5 on: Aug 17th, 2007, 10:07am » |
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Aye, that's what I have as well. Any thoughts on the expected distance version? 
--SMQ
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Grimbal
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Re: Lost in the Woods, Again
« Reply #6 on: Aug 17th, 2007, 3:32pm » |
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Ouch!. That would involve an integral of trigonometric functions. I always hated these...
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