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醉鬼走路问题
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GhostBrain  
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 More options Jun 4 2005, 11:21 am
From: "GhostBrain" <ghostbr...@sina100.com>
Date: Sat, 04 Jun 2005 08:21:33 -0700
Local: Sat, Jun 4 2005 11:21 am
Subject: 醉鬼走路问题
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醉鬼走路问题:一个醉鬼从点A向点B行走,由于醉鬼喝醉了酒,所以他行走的方向是随机

的。先假设AB之间的距离为 D ,醉鬼每步的步长为
s,问醉鬼需要多少步可以走到B?
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  解答:这个问题可以结合向量和概率理论来解答。
首先,由于醉鬼行走的方向是随机的,故我们可以认为,当醉鬼从某一点走时,方向角α~

[0,2π]。所以,醉鬼在第j步行走的可表示成向量
Vi = s(cosαi + isinαi),i
为虚单位。不妨设A点为原点(0,0),则到第j步时醉鬼所在位

置为
   Dj = ∑Vi - 0,where i = 1,1,...,j,
αi~[0,2π]

   Dj = ∑s(cosαk + isinαk)
      = s∑(cosαk + isinαk), αk~[0,2π]
离开A点的距离为Dj的模,即、|Dj|,假设醉鬼用了n步走到了B,则,
        D = |Dn|
          = |s∑cosαk + isinαk)| , where k = 1,2,...,k
,αk~[0,2π]
现在,问题归结到了:
       已知α~[0,2π],求
cosα和sinα的分布的问题了。这留给大家求解吧,如果解出来了,别忘了拿出来与大家分享啊

   哦,忘了公布答案了,步数n = (D/s)^2


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