珍珠湾ART

标题: 囚犯问题新版 (自编 )[:D)] [打印本页]

作者: 野菜花 时间: 2005-4-22 23:10
标题: 囚犯问题新版 (自编 )[:D)]

一个小屋子里有 1000个开关，有些开着有些关着。每天进去一个囚犯，必须也只能动 15个开关。如果某一天囚犯能把所有的开关关了，他们可以全部获释。典狱长让他们在开始前讨论一下，但并不知道有几个开关是开的。

问他们能否商量出一个必胜的办法？

(就是以下开关的题目一般化，但我想大家对囚犯比较有同情心，所有会有动力帮他们想办法。 )

作者: fzy 时间: 2005-4-22 23:38
标题: 回复：囚犯问题新版 (自编 )[:D)]

Let n = number of current on switches, and a = n mod 15.If a is odd,turn on (15-a)/2 switches and turn off (15+a)/2 switches. If a = 0, turn off 15 switches. If a is even and > 0, turn on 7 and turn off 8 so it becomes odd.

Are there really bad prisoners that should not be released? If so, we change the rule to 14 switches every prizoner, but turn on an odd number of lights at the beginning. :)

作者: 乱弹 时间: 2005-4-22 23:52
标题: It should be OK.

In the first round, the prisoners enter the room and change the statuses of the lights following this simple rule: shut off a light whenever possible. Then everybody can predict the process, and after this round, everybody know the current number of lights that are on.

Then now we have simplified the problem. There are many algorithms to finish the work. The worst one might be this: in every round, the total effect is to shut off one light People of odd numbers shut off lights, and the others turn on lights). This method would keep the poor prisoners in the prison for aquite a long time (985*15 days, which is roughly 42 or 43 years).

作者: 野菜花 时间: 2005-4-23 00:50
标题: [:))]原来的题目的题目没人理，一变成囚犯题马上就有解了，看来大家对囚犯都很同情呀！

原来的题目的题目没人理，一变成囚犯题马上就有解了，看来大家对囚犯都很同情呀！

作者: 野菜花 时间: 2005-4-23 00:54
标题: You are really saviour of those poor prisoners! [:

Using you method, in at most 68 days, all of them can be released!!

作者: 野菜花 时间: 2005-4-23 01:04
标题: 回复：It should be OK.

Probabilty you think they deserve to stay there longer, you give them hope but let them be patient.

欢迎光临珍珠湾ART (http://zzwav.com/)