毒酒问题 Poison Wine


毒酒问题---1000桶酒,其中1桶有毒。而一旦吃了,毒性会在1周后发作。问最少需要多少只老鼠可在一周内找出毒酒
如对1000桶按1~1000编码,需要10位二进制数。

因此只需要取10只老鼠,每只老鼠只喝其对应位数为1的编号的酒。
即10只老鼠按以下编码:

第一只 00000 00001
第二只 00000 00010
第三只 00000 00100
第四只 00000 01000
...
第十只 10000 00000

每只老鼠只喝其编码与酒编码做位与运算非0的酒。如果毒酒的编码在某一位为1,则监控该位的老鼠必喝,结果为1.

即把10只老鼠的结果,按位填入一个10位二进制数中,其结果即为毒酒编号。

例如:编号为10001 00011的酒是毒酒。则对应的只有第一只,第二只,第六只,第十只死亡。其对应位数置1,即为10001 00011
http://www.matrix67.com/blog/archives/4361
这个问题的答案也堪称经典:把瓶子从 0 到 999 依次编号,然后全部转换为 10 位二进制数。让第一只老鼠喝掉所有二进制数右起第一位是 1 的瓶子,让第二只老鼠喝掉所有二进制数右起第二位是 1 的瓶子,等等。一星期后,如果第一只老鼠死了,就知道毒药瓶子的二进制编号中,右起第一位是 1 ;如果第二只老鼠没死,就知道毒药瓶子的二进制编号中,右起第二位是 0 ⋯⋯每只老鼠的死活都能确定出 10 位二进制数的其中一位,由此便可知道毒药瓶子的编号了。
    现在,有意思的问题来了:如果你有两个星期的时间(换句话说你可以做两轮实验),为了从 1000 个瓶子中找出毒药,你最少需要几只老鼠?注意,在第一轮实验中死掉的老鼠,就无法继续参与第二次实验了。




    答案:7 只老鼠就足够了。事实上,7 只老鼠足以从 37 = 2187 个瓶子中找出毒药来。首先,把所有瓶子从 0 到 2186 编号,然后全部转换为 7 位三进制数。现在,让第一只老鼠喝掉所有三进制数右起第一位是 2 的瓶子,让第二只老鼠喝掉所有三进制数右起第二位是 2 的瓶子,等等。一星期之后,如果第一只老鼠死了,就知道毒药瓶子的三进制编号中,右起第一位是 2 ;如果第二只老鼠没死,就知道毒药瓶子的三进制编号中,右起第二位不是 2,只可能是 0 或者 1 ⋯⋯也就是说,每只死掉的老鼠都用自己的生命确定出了,三进制编号中自己负责的那一位是 2 ;但每只活着的老鼠都只能确定,它所负责的那一位不是 2 。于是,问题就归约到了只剩一个星期时的情况。在第二轮实验里,让每只活着的老鼠继续自己未完成的任务,喝掉它负责的那一位是 1 的所有瓶子。再过一星期,毒药瓶子的三进制编号便能全部揭晓了。
    类似地,我们可以证明, n 只小白鼠 t 周的时间可以从 (t+1)n 个瓶子中检验出毒药来。
http://bbs.emath.ac.cn/thread-1511-1-1.html
国王为10天后的生日宴会准备了1000桶酒,不幸的是,其中两桶被下了毒。为了确定两桶毒酒,有人提议用死刑犯试毒。毒的潜伏期为10天。
问:至少需要多少个死刑犯才能确保找出毒酒?方案如何实行?
http://www.zhihu.com/question/19731686
http://www.cnblogs.com/graphics/archive/2010/03/28/1699048.html
1、现有1000个苹果,10个盒子,现在要你将1000个苹果装入10个盒子中,使得用户无论购买多少个苹果(1-1000),都能由若干个盒子拼装而成(卖的时候是整个盒子卖,不能拆盒子的包装)
解法:考虑1, 2, 4, 8这四个数,由这四个数可以组成1-15之间任意一个数,也即1,2, 4 ... 2n 可以组成1-2(n+1)-1之间所有的数。所以这十个盒子分别放入
1, 2, 4, 8, 16, 32, 64, 128, 256, 489个苹果,即可组成1-1000内所有的数。


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