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Topic: Car parking (Read 4998 times) |
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towr
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Car parking
« on: Jan 31st, 2008, 8:39am » |
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You have a circular parking lot that would fit N cars, if only they would park them neatly next to each other. However, there are no individual spaces marked out for each car, so everyone just parks somewhere on the circle where there's room.
How many cars would you expect there to be parked when the lot is full?
(The usual mathematical abstractions apply: the lot is a circle with circumference N, and each car is an arc of length 1 on such a circle with, cars are placed according to a uniform distribution over the available space.
You can generalize for non-integer N, if you prefer.)
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| « Last Edit: Jan 31st, 2008, 8:41am by towr » |
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Joe Fendel
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Re: Car parking
« Reply #1 on: Jan 31st, 2008, 9:55am » |
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My inclination would be:
1) Yes, only work in the generalization for non-integer N.
2) Work on the problem for a linear parking lot instead, since then:
2a) The circular problem with N cars is equivalent to the 1 + linear problem with N-1 cars, since after the first car is parked that's what's left
2b) We can recurse - parking a car in a lot of size N leaves two linear lots of combined size N-1.
But I'm not sure how to work out the details of the recursion at first glance.
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Eigenray
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Re: Car parking
« Reply #2 on: Jan 31st, 2008, 5:01pm » |
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I think if E(x) is for a linear lot of length x, then for 0  t 1,
E(t) = 0
E(1+t) = 1
E(2+t) = [ 3t + 1 ] / [ 1 + t ]
E(3+t) = [ 4 + 7t - 4log(1+t) ] / [ 2 + t ]
E(4+t) = [ 11 - 2 2/3 + 16log(2) + 15t - 4(5+2log(1+t))log(2+t) + (2log(2+t))2 + 8PolyLog[2,(1+t)/(2+t)] ] / [ 3 + t ]
Laplace would probably work. (Maybe not.)
If we write E(x) = m(x)(x+1) - 1, i.e., F(x+1) = m(x)(x+1), where F(x) is for a circular lot of length x, then
m(1) = 0.5
m(2) = 0.67...
m(3) = 0.75
m(4) = 0.7485...
m(5) = 0.74751...
m(6) = 0.7475982...
The limiting value is actually 0.7475979202...
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| « Last Edit: Jan 31st, 2008, 8:32pm by Eigenray » |
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