原链接https://wiki.xkcd.com/irc/Puzzles 因为安全问题无法访问，好在我存有一些备份。
先随便贴几个我觉得很有意思的：
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Ants: 100 ants (zero-length points) walk on a meter stick (a line) at 1 cm/second. When two ants collide, they both reverse direction. If an ant reaches the end of the stick, it falls off. What arrangement of ants maximizes the time before all ants have fallen off? How long can they last? 蚂蚁：有100只蚂蚁（假设长度为0）在一个一米长的棍子上以1厘米每秒的速度行走。如果两只蚂蚁相撞，它们会各自调头向反方向走。如果有一只蚂蚁碰到了棍子的一端，它会掉下去。怎样安排蚂蚁的初始位置，使所有蚂蚁掉落前的时间最大化？它们最长能在棍子上待多久？
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Smurfs and Gargamel：Gargamel has captured 100 Smurfs. Feeling confident he proposes the following game: He will exchange the white hats of an undefined number of smurfs with red hats. No smurf knows his own color. All smurfs can see each other, but they have no other means of communication. Then, each Smurf (order selected by Gargamel) should say the color of his hat (loudly so that everyone can hear it). If it is correct, then he lives, otherwise he dies. After a short discussion the smurfs accept. They know that the first Smurf to be chosen might die, but all 99 other smurfs will survive. How? Note: as usual no trick. Nothing time-related... 蓝精灵和格格巫：格格巫抓住了100个蓝精灵。出于自信，他建议了如下游戏：他会将一部分蓝精灵的白帽子换成红帽子。没有蓝精灵知道自己帽子的颜色。蓝精灵可以看到其他蓝精灵，但（在戴上帽子之后）他们没有其他沟通的办法。然后，蓝精灵按照格格巫选择的顺序，一个接一个说出自己帽子的颜色（说得足够响，让所有人可以听到）。如果正确，他就能活下来，否则他就要死去。 在短暂的讨论后，蓝精灵们接受了提议。他们知道，第一个被选中说话的蓝精灵可能死去，但是其他的99个蓝精灵会活下来。他们的计划是什么呢？
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Smurfs and Gargamel Alternative 1: Suppose that instead of just red and white hats, Gargamel randomly distributes n different hat colors (including white) among the smurfs, where the number of colors n is less than 100, the number of smurfs. Assume the smurfs know all the possible hat colors. How many smurfs must be put at risk of death before the rest are guaranteed to survive? 蓝精灵和格格巫版本2：假设格格巫不仅仅分配红色和白色帽子，而是在蓝精灵中随机分配n种不同颜色的帽子（包括白色），n小于100（即蓝精灵的数量）。假如蓝精灵们知道所有可能的帽子的颜色。在余下的蓝精灵保证存活前，有几个蓝精灵要冒可能送命的危险？
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Smurfs and Gargamel Alternative 2: what if the Smurfs are positioned in a long queue such that each Smurf can only see the hats of the Smurfs in front of him? Assume Gargamel always starts at the back of the queue and works forward. 蓝精灵和格格巫版本2：（依然是红、白帽子）假设蓝精灵站在一个长长的队伍里，每个蓝精灵只能看到他前面的帽子颜色。会怎么样呢？这里假设，格格巫总是让队伍最后的蓝精灵先发言，然后一个个从后往前。
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Smurfs and Gargamel Alternative 3: Suppose there are infinitely many Smurfs (perhaps uncountably many) and all at once each must say the color of his own hat. Again, the Smurfs formulate a plan (perhaps using a bit of Smurf magic) and accept, knowing that a finite number may die but the rest will live. How? 蓝精灵和格格巫版本3：（依然是红、白帽子）假设现在有无限数量的蓝精灵（数不清的数量），所有蓝精灵必须同时说出自己帽子的颜色。蓝精灵们（也许是通过魔法）再一次形成了一个计划，接受了提议。他们知道有限数量的蓝精灵可能死去，但余下的都会存活。如何达到呢？
