A telephone wire stretched tight between two poles placed ten meters apart is a favorite resting spot for a flock of crows.

Suppose one morning two crows land on the wire, each at a random spot (the probability is uniformly distributed). With a bucket of paint and a brush you mark the stretch of wire between them. A certain length of wire will have been painted.

On average, what length of wire would you expect to have painted? Assume that each bird is a single point along the line, and so has no width.

Suppose instead that a dozen crows landed on the wire, each at an independent, random location, and you painted the stretch of wire between each bird and its nearest neighbor. On average, what total length of wire would you expect to have painted now?

And if a thousand crows landed?

A computer-generated solution could be found, but bonus points will be awarded for a formal proof!

2 Birds

I'll set the origin at the end of the wire closest to the first bird landing. let x (<1/2) be the landing coordinate of the first bird. If y marks the position of the second bird lands, then the |y-x| has value x for y = 0, decreases with slope -1 until its value reaches 0 at y=x and thereafter increases with slope 1 until 1-x at y=1.

Integrating |y-x| over the two intervals (0,x) and (x,1) gives the result (1+x^2)/(2*(1+x)). (This function has value 1/2 at x=0, and decreases to 5/12 at x=5). We want to find the average of this function over the range x=0 to ,5

To perform the integration I found it useful to first write

(1+x^2)/(1+x) = x - 1 + 2/(1+x). This integrates easily enough to

x^2/2 - x + 2ln(1+x), yielding the answer listed in the subject.

(Maybe I'll try the 1000 bird case someday too, but for now I'd like to welcome feedback)

 

 

Posted by FrankM on 2008-01-18 00:14:53