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soler
Member since May-13-05
May-14-05, 07:10 AM (EST)   "Birds on a wire puzzle"

 I may be a little thick but I could not understand the problem Description. If the birds are wider than points then obviously you cannot fit an infinitely large number of them. If they are just points then the answer is the length of the wire. What am I missing?

alexb
Charter Member
1604 posts
May-16-05, 11:49 AM (EST)    1. "RE: Birds on a wire puzzle"
In response to message #0

 >If the birds are wider than points then >obviously you cannot fit an infinitely large number of them. Right. They are just points.>If they are just points then the answer is the length of the >wire. Why?>What am I missing? I do not know.

Mark Huber guest
May-17-05, 01:32 PM (EST)

2. "RE: Birds on a wire puzzle"
In response to message #0

 Let the wire be the closed interval from 0 to 1, and suppose that there are three birds on the wire, points a = .1, b = .2, and c = .35.Then bird a is closest to bird b, so interval (.1, .2) is colored yellow. Bird b is closest to bird c, so interval (.2, .35) is colored yellow. Bird c is closes to bird b, so interval (.2, .35) is colored (of course it already was colored earlier so this interval is redundant.)So from (.1, .2) and (.2,.35) is colored yellow and the rest of the wire is clear.Here's another example. If the birds sat at .2, .3, .7, and .8, thenthe interval (.2,.3) and (.7,.8) would be colored yellow, and the rest of the wire would be clear.Hope that helps!-mark happyusers
Member since Jul-29-05
Jul-29-05, 09:28 AM (EST)    3. "RE: Birds on a wire puzzle"
In response to message #2

 I understand the question quite well but am not totally convinced by the solutions proposed. In particular, the 4th solution rely on computation of conditionnal probability without saying it.When you suppose three consecutive intervals, the intermediate having a length of x, the length of the other two is governed by a conditionnal law. To be convinced of that is very simple, Let us suppose a wire of length 1. f(x) has <0,1> for support. Let us say that x=.5. clearly the probability that it is smaller that its two neighbours is null ; nevertheless, the formulation proposed exhibit a positive probability (F(x)*F(x)*f(x) dx).Furthermore, I don't clearly visualize what is the F(x) law. Why should all intervals would have the same law? What is easy to speak of is the joint law of the xi intervals F(x1,..xi,..xn+1) if there are n birds. One can asset that sum(xi)=1. Another clue that variables xi are not independant.Hope that helps. Mark Huber guest
Jul-30-05, 02:47 PM (EST)

4. "RE: Birds on a wire puzzle"
In response to message #3

 Well, happyusers, none of the 4 solutions proposed are rigorous in the sense that they at the level of mathematical proof needed for a journal article, but each contains the essential idea for what could be turned into a rigorous proof.In the case of the 4th solution by Stuart Anderson, he begins by essentially considering an infinite length wire where the distance between successive birds are independent random variables each with the same distribution.As you point out, this is not the actual problem of N birds on the wire from 0 to 1. However, as the number of birds grows to infinity, the wire "looks" like it is growing relative to the distance between any two birds, since that distance is going to zero. So fixed wire length and shrinking distance between birds is the same essentially as infinite wire length and fixed distribution of distance between birds.You are right that this does not give a formal proof of the result unless these two related problems are shown to be equivalent, although it does contain the essential mathematical intuition needed to write out such a proof.Mark Huber mr_homm
Member since May-22-05
Jul-31-05, 01:05 AM (EST)    5. "RE: Birds on a wire puzzle"
In response to message #4

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