成绩公布前的小娱乐

路过，看一下

如果一开始确实有三顶黑帽，那么每个人会至少看到两顶黑帽，这样大家就会预测到第一次关灯不会有人打耳光，这样关灯就无法增加信息

有意思

这道题出错啦~
题目最后一句应该是：
一直到第三次关灯，才有“啪啪啪”打耳光的声音响起。问有多少人戴着黑帽子？

然后标准答案是，因为啪啪啪打了三声，所以有三个人戴着黑帽子~

对了，说点正经的，上次那个称球的游戏，其实题目应该是3次从13个球里找出重量不一样的。
不过解法跟12个几乎没有区别，一开始也是分成4 跟 4，唯一的区别是平的话用剩下的两次从5个球里找出不一样的来。

Case (i): only 1 black hat
The one with black hat should notice by himself (because it is given at least 1 black hat, and he find nobody wearing black hats). Then, he must show up during the 1st light off, which is a contradiction to the problem. Hence, there should be at least 2 black hats.

Case (ii): only 2 black hats
Each of the one who wears black hat will notice the other one wearing a black hat, and they will think that the other one is the only one who wear black hat. Hence, during the 1st light off, nobody will show up. Then, both of them will realize that themselves also wear black hat (because if there is only 1 black hat, it become case (i) and the other one must show up during 1st light off). Thus, they will show up during the 2nd light off, which is also a contradiction to the problem. Hence, there should be at least 3 black hats.

Case (iii): only 3 black hats
Each of the one who wears black hat will notice 2 people wearing a black hat. So, the immediate thought to them is that there are only 2 people wearing black hats. Hence, both of them will not show up for the 1st and 2nd light off. Then, both of them will realize that themselve also wear a black hat (because if there are only 2 black hats, it become case (ii) and they must already show up in 2nd light off). Thus, they will show up during the 3rd light off, which is the situation of this problem.

Similarly, we can deduce that when there are n people wearing black hats, it need n round of light off for them to realize and show off.

Case (i): only 1 black hat
The one with black hat should notice by himself (because it is given at least 1 black hat, and he find nobody wearing black hats). Then, he must show up during the 1st light off, which is a contradiction to the problem. Hence, there should be at least 2 black hats.

Case (ii): only 2 black hats
Each of the one who wears black hat will notice the other one wearing a black hat, and they will think that the other one is the only one who wear black hat. Hence, during the 1st light off, nobody will show up. Then, both of them will realize that themselves also wear black hat (because if there is only 1 black hat, it become case (i) and the other one must show up during 1st light off). Thus, they will show up during the 2nd light off, which is also a contradiction to the problem. Hence, there should be at least 3 black hats.

Case (iii): only 3 black hats
Each of the one who wears black hat will notice 2 people wearing a black hat. So, the immediate thought to them is that there are only 2 people wearing black hats. Hence, both of them will not show up for the 1st and 2nd light off. Then, both of them will realize that themselve also wear a black hat (because if there are only 2 black hats, it become case (ii) and they must already show up in 2nd light off). Thus, they will show up during the 3rd light off, which is the situation of this problem.

Similarly, we can deduce that when there are n people wearing black hats, it need n round of light off for them to realize and show off.

喂，不知道轻重的话，27个是不可能的啦，13个就是极限了。
如果事先知道不一样的比别的轻或者重的话，才能做到27个 （9-9 -> 3-3 -> 1-1)三次这样

maomaochongz 发表于 2013-11-30 11:15
喂，不知道轻重的话，27个是不可能的啦，13个就是极限了。
如果事先知道不一样的比别的轻或者重的话，才能 ...

这个算吗？
还是现在公布正解算了，无谓夜长梦多……

先设15个球编号为00,01,02,03,04,05,06,07,08,09,10,11,12,13,14（可以画一张图表，记录什么编号对应什么颜色），00号球为白色小球（肯定是好球）。

现在作如下三次的称量：

第一次：00,06,07,10,11 vs 05,08,09,12,13
第二次：00,02,03,05,11 vs 04,06,07,12,13
第三次：00,05,07,08,10 vs 01,02,04,11,13

下面是三次称量结果和哪个是坏球的对照表：

>代表左重、=代表平衡、<代表右重
+代表坏球比较重、-代表坏球比较轻、?代表坏球不知轻重

>,>,>: 13-
>,>,=: 12-
>,>,<: 11+
>,=,>: 10+
>,=,=: 09-
>,=,<: 08-
>,<,>: 07+
>,<,=: 06+
>,<,<: 05-

=,>,>: 04-
=,>,=: 03+
=,>,<: 02+
=,=,>: 01-
=,=,=: 14?
=,=,<: 01+
=,<,>: 02-
=,<,=: 03-
=,<,<: 04+

<,>,>: 05+
<,>,=: 06-
<,>,<: 07-
<,=,>: 08+
<,=,=: 09+
<,=,<: 10-
<,<,>: 11-
<,<,=: 12+
<,<,<: 13+
哈哈哈，虽然有个质量不能判断

詹姆斯 发表于 2013-11-30 11:22
这个算吗？
还是现在公布正解算了，无谓夜长梦多……

你这个解答假设了一个00号标准球，这个是本来题目没有的。
严格的看，题目应该是，你只有13个球和一个天平，你没有其他的标准球来做参考。
不过你只要把上面00号跟13号同时拿掉，你这个解答就还是有效的（1-12号等同于上面的解答，三次平的话就是13号）