科学与数学math.dartmouth.edu
七个你以为自己没听过的谜题（2006）
Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]
2009 — 2018，它一次次回到首页
黑点是上首页（≥ 50 分）的那次投稿，空心点是没冲上去的投稿，点越大分数越高。
导读
这份 PDF 给出七道数学与逻辑谜题及其解答，题目看似是听过的老题，实际答案往往反直觉。从评论看，其中涉及盒子里的名字、锁箱传情、蓝点自杀、盒子嵌套等经典设定，重点在严格策略与公共知识推理。
为什么经久不衰：它把经典谜题的反直觉点压缩成可证明的短解，既适合作为推理训练，也能反复提醒读者：直觉和题目隐含假设常常是错的。
HN 评论精选
补充了一个与“盒子里的名字”同类的单灯公共知识谜题及其策略设定。
我很喜欢这类谜题！再补一个与第一个（盒子里的名字）相似的：100 名囚犯，随机时刻随机选一人进入只有一盏灯的屋子；他可开灯、关灯或保持原状，其他东西不能改变。每次探访后，他会被问是否所有人都已来过：若答“是”且属实，所有人立即获释，否则全部处死；也可答“不知道”且无后果。他们只能靠灯通信，但可事先约定策略，且都是完美逻辑学家。目标是让某人有 100% 把握所有人都来过；笨办法是等几十亿年，优雅策略会结束得更早。由于探访时刻随机，囚犯无法知道上一位是谁，甚至可能是自己。
iliis查看英文原文
Ah, I love these kind of puzzles! Here's another one, similiar to the first one (Names in Boxes). Apologies for any incorrections in advance. There are 100 prisoners. At random times one prisoner is chosen uniformly at random and led into a room with a single lamp. The prisoner can choose to switch it on or off or leave it as the last visiting prisoner left it. Apart from the state of the lamp he must leave the room unchanged. After visiting the room, each prisoner is asked if every prisoner has visited the room by now. If he answers 'Yes' and indeed everyone has been to the room at least once, then everybody is freed immediately. Otherwise they are all executed ;) He can answer 'I don't know' without any consequences. Apart from the lamp being on or off the prisoners have no way of communication at all, but of course as is customary in such puzzles they can plot a strategy in advance and everybody is a perfect logician. So, to clarify: The goal is for one prisoner to be 100% sure that everybody has been to the room at least once. The "easiest" solution would simply be to wait a few billion years (it's an abstract puzzle, they are all immortal anyways ;). But of course there is a more elegant solution that terminates earlier. Also, as the time for a visit is chosen at random a prisoner has no way of knowing who the previous person in the room was. It might just as well have been himself! As there is some randomness involved, it is theoretically possible that the goal state never happens. Just assume that in the limit everybody will have visited the room infinitely often ;) In other words: Implement synchronization between 100 threads with only one bit of shared memory and completely random scheduling.
批评该题对锁箱结构的设定不够明确，可解性取决于隐含假设。
“Kleptopia 中的爱情”需要解释得更清楚。这个题只有在你能把两把挂锁都固定到同一个箱子上时才可解；我当时以为锁箱只有一个普通形状的锁扣，那样就不可能做到。很多“思想”问题都这样：我解不出来（有时还能证明解不出来），是因为题目规则没讲清。
deathanatos查看英文原文
Love in Kleptopia needs to be explained better. The problem can only be solved if you can afix two padlocks onto a box, and I was presuming the lock box had a single, normally shaped padlock eye, which would make such a thing impossible. I find this happens a lot with "thought" problems: I can't solve it (and can often prove that) because the rules of the problem are inadequately explained.
给出另一个反直觉的概率/博弈问题，问如何获得正期望收益。
另一个（对我来说）非常反直觉的问题：在下面这个游戏中，你怎样才能做到不亏本甚至更好？A 选择两个不同的整数，写在纸条上，分别握在左右拳中。你选一只手，让他展开一个数；然后你必须猜另一个数比看到的数更大还是更小，猜对赢 1 美元，猜错输 1 美元。
tomp查看英文原文
Another very counterintuitive (for me) problem: how do you do better that break even in the following game: 'A' chooses two distinct integers, writes them on slips of paper and holds one out in each hand in a fist. You choose a hand and reveal a number. You must then guess whether the other number is higher or lower than the revealed one, winning $1 if you guess right and losing $1 otherwise.
用体对角线平方加表面积比较，给出盒子嵌套问题的一个可视化几何证明。
我觉得盒子问题有一个更直观的解法。设 A 的尺寸为 (a,b,c)，B 为 (x,y,z)，并假设 A 能装进 B。注意 (a+b+c)^2 = a^2+b^2+c^2 + 2ab+2ac+2bc，这等于 A 的体对角线平方加上 A 的表面积；B 也同理。A 的体对角线至多是 B 的体对角线——它是 A 的最长轴，总得装进 B；而 A 的表面积小于 B 的表面积：把 A 放在 B 内，两者内部都是凸集，想象在 B 内“膨胀” A，取所有离 A 中某点距离不超过 t 且在 B 内的点，膨胀只会增大表面积，而最大不超过 B 的表面积。因此 (a+b+c)^2 = (A 体对角线)^2 + (A 表面积) <= (B 体对角线)^2 + (B 表面积) = (x+y+z)^2，命题得证。
jmblpati查看英文原文
I think there's an easier to visualize solution to the box problem. Let A have dimensions (a,b,c) and let B have dimensions (x,y,z). Assume A fits inside B. We have (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc. This is the sum of the A's hypotenuse squared plus its surface area. The same holds for B. Note that A's hypotenuse is at most that of B-- the hypotenuse of a is its longest axis, and it needs to fit in B somehow. Further, note that the surface area of A is less than that of B. To see this, consider the nesting of A inside B and realize that both boxes' interiors are convex sets. Imagine inflating A inside of B by taking the sets A_t consisting of all points within B that are within distance t of a point in A. It is not hard to see that this inflating operation can only increase the surface area of A, and since the maximum surface area we can get is that of B we have that A has smaller surface area than that of B. Thus, (a+b+c)^2 = (A hypotenuse)^2 + (A surface area) <= (B hypotenuse)^2 + (B surface area) = (x+y+z)^2. The claim follows.
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