最近，来自 wavegrower 的一张 gif 动画红遍了 reddit 。有人提出了这么一个问题：每个小点最后都会回到自己原来的位置上吗？注意，这些小点并不是沿着一个回路在运动，而是沿着三个交替出现的回路在运动。
答案是肯定的。 math 版上的 OmnipotentEntity 给出了一个简短的证明。假设某个地方的小点出发后永远不会回到原地。由于小点的运动规律是三步一个周期，因此每三步之后从此处出发的小点将会拥有完全相同的命运——永远不会回到原地。既然从这里出发的小点会不断地发生有去无回的情况，那么总有一个时候小点会被用光，此时就再也没有小点能从这里出发了。但这与我们看到的实际情况相矛盾：每个地方的小点都是用之不竭的。
熟悉群论的朋友会很快发现，这个结论几乎是显然的。小点的每一步运动都形成了一个置换，三个置换的复合本质上也还是一个置换，而这个置换的足够多次幂一定会变成单位置换。这意味着，不但每个点都能回到自己原来的位置，而且所有点能同时回到自己原来的位置（后者可能需要更长的时间）。事实上，有限群中的任意一个元素都有一个有限的阶，因而如果某类变换操作能构成一个有限群的话，不断地执行某一个操作，或者不断地循环执行某几个操作，最后总有一个时刻你会发现，一切又都重新变回了原样。拿出一副新的扑克牌，每次洗牌时都把牌分成两半并把它们完美地交叉在一起，那么不断这样洗下去之后，整副牌总会在某个时候重新变得有序。找一个复原好了的魔方，循环执行几个固定的操作，魔方很快就会被彻底打乱，但最终一定会奇迹般地再次复原。
证明可能性一般都会先试用反证法；但证明不可能性，反证法就不太好用啦。
漂亮
看了最后一节有感：把扑克牌和魔方看成宇宙的话，在达到最终复原之前的漫长时间里，其中的居民会总结出“事物总是往混乱方向演化”的“热力学定律”吧。
不对，整个宇宙的变化是非线性的。
刚才忘了一句，补充上：“而且宇宙中空间是连续的，不是离散的，即：物体的位置存在着无限个可能情况”
扑克牌和魔方都是有限群，并且博主说的是“循环执行几个固定的操作”。好比说拿一个复原好的魔方，重复执行RUR’U’连续六次，魔方又能还原。即使没有按照RUR’U’这样的固定操作，由于魔方是有限群，随机乱拧也存在还原的可能，尽管这个可能经计算近乎为零——1/(4.3×10^19)，以人脑和人手的能力是不可能在有生之年靠随机乱拧来还原魔方的。
宇宙中粒子的运动存在混沌和非线性，尽管受物理定律支配，很多无法用简单的周期性描述，也难以靠计算来预测
懒人的赖皮证明方法：
因为是GIF图像，所以一定是循环播放的。
（我自觉面壁
这个证明方法不对，图里每个点形状都一样，根本说明不了什么。
说得有理… 不过说不定这个gif并没有完整地循环, 反正你看不出来…
“动图循不循环”与“每一个小点能否回到原处”之间没有关系。可以明确地告诉你这个动画一个周期里只有三次运动，每个小点从出发到回到原处需要N多步，而要论证N是否有限与动图循不循环没有任何关系。注意q68257962回复中提到的“每个点形状都一样”的含义。
怎么能一下子看到最开始的博客呢？有人知道吗，谢谢
http://www.matrix67.com/blog/page/174
博主你好，我试图自己做一个这样的动图，但是在求小点的路径（使得可以循环起来）时遇到了困难，用普通深搜在8×8时就已经算不出来了。请问博主有没有见过类似的问题或算法。
后来用的方法是分块，比如用4块6×6拼出一个12×12，然后修改中心处的几个点的方向使四块相互联通，不过这样生成的回路的混乱度没有直接深搜高。
http://7xkk5t.com1.z0.glb.clouddn.com/dot_circulation_movement_result_25fps.gif
这是我自己写的，24×24点阵，动画效果没有原图好看。
目前只用了一个回路在循环，不像原图那样用了三个。
16*16=256个点
有256!种排序
动图是无限的，自然会有重复情况
这样的话，不仅证明了每个点都可能回到原处，还证明了所有点都会同时会到原处。假如给每个点都编一个号，所有的点产生的局面数量肯定是有限的，但动图是无限的，所以一定是循环的。
Dude, right on there brrheto.
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经证实该GIF总共75帧，每25帧为一次运动，故每一周期内只运动了三次。但GIF的循环周期与小点的运动周期显然不是一码事：由于每一个小点长得是一样的，GIF不需要制作出完整的周期就可以达到多步运动的效果。说明白点儿：GIF循环一次所显示的运动次数不等于一个小点从出发到回到原处所需要的步数。以上回复中凡是用“动图是循环的”来验证结论的都没理解题目意思。
为什么我的思考方式是 回路是完整的，进行数次置换后将回路完成，自然就回到了原点。感觉没用什么靠得住的理论- -。。
把每个点编号，每个点所在的位置和现在处于三个回路中的哪一个组成了现在的[状态]，每一个时刻的[状态]仅与前一个时刻的[状态]有关，[状态]只有有限种，所以必须有循环。(这个和递推数列不一样，不存在多对一的情况，所以一定能循环到当前状态)。。。其实还是有限群的思想
…
有意思
看得明白☺。
这里如此。但有理小数都是从某个位置开始进入循环节
“由于小点的运动规律是三步一个周期”
什么意思“由于小点的运动规律是三步一个周期”，，，为什么是三步一个周期
仔细看图，三步是有些区别的，但是第四步又会像第一步一样然后接着重复
还可以
有趣的证明
图画得可真有趣啊。证明也很有意思。
Grade A stuff. I’m unuqlstionabey in your debt.
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密恐啊啊啊啊啊啊啊啊
不能发图片啊
为什么评论不能发图片
图看着有点眼花
漂亮在盯着一个位置有没有点，而不是盯着一个点看它往什么轨迹动
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