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 Bird on a Wire (Posted on 2004-06-07)
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!

 No Solution Yet Submitted by Sam Rating: 3.7000 (10 votes)

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 re(3): Concrete Answer is (n-1)/(n+1) | Comment 27 of 42 |

What does this have to do with the problem at hand?  I gather it has to do with the range between the leftmost and rightmost bird.  But there are unpainted gaps between the birds. Again, each bird has zero width.  Even with 10,000,000 birds no length is taken up by the birds themselves. As birds cluster, the painted area concentrates in the clusters.

In the samples below, the left column shows bird positions as fraction of the length of the wire. There are 12 birds in each sample. The right column shows the distance between two birds if that stretch of wire represents either of the birds' nearest neighbor status. Stretches where the gap is longer than that on either side are not counted as they are not between birds one or both of whom is the nearest neighbor to the other. The sum of the counted segments is at the bottom of each:

`0.125988              0.1239570.2499450.460339              0.0666120.526951              0.0380890.5650400.625489              0.0498110.6753000.808922              0.0700270.878948              0.0127660.8917140.935140              0.0092060.944346`
`              0.370467`
`------------`
`0.008945              0.0859240.094869              0.0294720.124341              0.0080550.1323950.408642              0.1587470.567389              0.0602120.6276010.742005              0.0245260.7665310.892891              0.0262000.919090              0.0431710.962261`
`              0.436306`
`--------------`
`0.216360              0.0462820.262642              0.0343620.2970030.407908              0.0095120.417420              0.0053820.4228020.519325              0.0206960.540020              0.0705960.6106160.825474              0.0297740.855248              0.1384770.993725`
`              0.355081`

As you can see the total is not averaging 11/12. To average 11/12 you'd have to consider the whole distance from first bird to last, disregarding the unpainted gaps.

The differences are actually carried to more places and rounding may make them seem to disagree with the specified positions in the last decimal place.

 Posted by Charlie on 2004-06-09 11:16:34

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