introduction:
Last week one of my friend the N second confession was denied to girls, as a good friend, I think we should do something in addition to sympathy. Before a chat inspired brothers, plus interest for mathematics, I girls Policies "and select Denied" try to do a simple model, and draw more meaningful conclusions.
Summary:
Each of the girls are eager to find Prince Charming in my mind, to find his life's happiness. But the face of suitors, girls should be selected or rejected, how can we most likely to find their Mr. Right it? In this article we use knowledge of mathematical probability theory for girls who choose to pursue this process mathematical modeling, optimal policy selection of girls, and finally the results are briefly discussed.
Key words:
Cannon fodder ranked-choice model
Model assumptions:
As we all know things in life involves feelings are complex, all the factors that may affect are taken into account is almost impossible. To this end we first simplify reality, and make some reasonable assumptions, consider the relatively simple case.
Suppose a girl is willing for a time in a boy and start a relationship, and there are N boys and girls in the pursuit of this time. Description: N here is not determined in advance, each according to its own conditions girls, and combined with previous experience and experience to guess determine this number N. For example each of the other two aspects of the same female, in general, should PP girls than girls relatively large value of N is not some PP. In the sense that for this girl, assuming suitors in any two of the guys they were comparable, but not identical situation. The N so that we boys are numbered from 1 to N, where the larger numbers indicate more suitable for this girl. So during this time, the girl's Mr. Right is N the boys. Now the question becomes the face of the N suitor in what should be the policy to make it the first time to accept the possibility of the boys is greatest N, N noticed that boys in a different order to pursue the girls of.
In order to simplify the complex practical problems, we make the following few reasonable assumptions:
. 1 , N boys girls in different order to declare that at any one time do not exist two or more of M
Student in the case of the girls confession, and any kind of order is completely equal probability.
2 , the face of the confession boys and girls can only be made to accept and reject two options, ambiguous or other options do not exist.
3 , any time, a boy and a girl can only love, the situation is more than a foot boat does not exist.
4 , has been denied the boys will not pursue the girls again.
Based on the above assumptions, we want to find a strategy that women with the greatest probability for the first time opt-in
That boy is N, ie Mr. Right.
First consider the simplest strategy, if there are boys to tell the truth once the girls, the girls would choose to accept. This strategy is clearly to women the probability of 1 / N to find their Mr. Right. When N relatively large, this probability is very small, and obviously this strategy is not optimal.
Based on these assumptions and models above, we propose such a strategy: For the first M individual confession, no matter how the girls feel have chosen to reject; after the encounter case of boys to girls confession, as long as the number of boys than the previous M boys the numbers are large, that this is more appropriate for girls than boys in front of M boys, then the girls choose to accept, or opt out.
Below an example N = 3:
Three boys and girls to pursue, a total of six arrangement:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
If the above girls simplest strategy, then only the last two arrangements choose to Mr. Right, the probability is 2/3! = 1/3.
If our girls adopt strategies presented above, here we take M = 1, that is, whether the first person is excellent, the girls have chosen to reject. Then for suitors after, as long as he is more appropriate for girls than boys first choice to accept, otherwise reject. Based on this strategy , "132", "213", "231" under the order of these three women will encounter "3" in the first to make an acceptable option, so we put this probability increases to 3/3! = 1/2.
Now the question comes down to, for general N, M what will make this maximum probability of it? (In this model, the previous M boys are called "cannon fodder", no matter how good they have been rejected)
Model:
In this section, according to the above model assumptions, we first find (1 <M <N), the probability of selecting Mr. Right female of expression for a given M and N.
1 are arranged to the total numbers N N! Possible. When the number N P appears in the first position (M <P <= N), so that the policy if encountered in the first selection acceptance is N, the arrangement needs to meet the following two conditions:
. 1 , N at the position P
2 , the number M + 1 to the maximum number of P-1 than in the previous position of the location is smaller M
The use of knowledge in mathematics of permutations and combinations, it is easy to know meet two conditions above arrangement total
Thus for a given M and N, P can be obtained from the simple summation of M + 1 to N to a given change in total N and M
Kind of sequence to meet the requirements.
Thus the probability of getting met Mr. Right while girls opt for the
Model Solution :( not interested, then you can skip this section is derived)
Solving this section we make this expression to obtain maximum value when the M's.
Hutchison function 
, The argument and the set value M, the function to obtain the maximum value.
therefore:
So should meet M
We know that when x> 0, In (1 + x) <x;
当x-->0, In(1+x) ~ x 。
So from left inequality
and so:
当N比较大时，同理由右不等式可得M≈N/e， 以上e为自然对数。
若记[x]为不大于x的最大整数，由以上推导我们可猜测当M取[N/e]或[N/e]+1时，该表达式取得最大值。
用MATLAB仿真，上述结论正确。
结果分析：
由上述分析可以得到如下结论：为了使一个女生以最大的概率在第一次选择接受男生时遇到的正是Mr. Right，女生应该采用以下的策略：
拒绝前M=[N/e]或者[N/e]+1个追求者，当其后的追求者比前M个追求者更适合则接受，否则拒绝。
“打战的时候，很多士兵身先士卒，跑到前线勇往直前。通常来说，走在最前面的，都会给大炮打中（古代的大炮像象个球一样滚过来的）成为灰烬。而后来的士兵，就踏着炮灰走到胜利，所以成为别人利益的牺牲品的人就叫炮灰.。”--------百度上关于炮灰的解释
在本篇文章中介绍的“炮灰模型”中，前M个男生就成了炮灰的角色，无论其有多么优秀，都会被拒绝。
朋友，如果你追求一个女生而遭到拒绝，看完这篇文章后你会突然发现，也许这不是你的的错，也许你真的很优秀，只是很不幸，你成了“炮灰”。
这几天在校内上看到很多朋友都因为拒绝或失恋而苦恼。希望上面这些看似复杂的推导和模型对你能有所启发。不要因为一次的拒绝而伤心、失落，振作起来，你的Miss Right is waiting for you somewhere!
谨以此篇文章献给所有为爱而战的猛士们！
附：
将这个策略的最优性简证如下（限于篇幅，不借助复杂的数学公式了）：
1.作为“策略”，可以认为应该类似于算法，对于确定的输入有确定的输出。因此对第M号追求者是否同意仅取决于之前M-1个人与该人的状况比较，以及M的大小；进一步地，显然与前M-1个人的好坏顺序无关（因为前M-1个人的顺序与第M个人及以后无关）。
2.如果仅考虑选中N号，那么答应某个人的必要条件是此人比之前的都好（否则一定不是No.N）
3.综1、2，所有可能的策略都有相同形式：对于第K1,K2,...,Kt号人，如果比以前的都好，OK；如果不符合条件，“还是做朋友吧”
4.进一步，如果Km + 1<K(m+1),将Km替换为Km + 1。简单计算可以发现（其实是我不想写了）在这一步答应且选对的概率不变（始终是1/n*前面没有答应的概率），但这一步答应的概率减小，后面答应且选对的概率相应增大（如果替换的是Kt，概率不变，但可以接着换K(t-1)使概率增大）。由此可以得出K1到Kt应该是连续整数且Kt=n
5.（从楼主的文章继续）
再由作者的理论小推论一下：
设女性最为灿烂的青春为18-28岁，在这段时间中将会遇到一生中几乎全部的追求者（之前之后的忽略不计），且追求者均匀分布（则女性从18+10/e=21.7即22岁左右开始接受追求……这告诉我们，想谈恋爱找大四的……
看完之后，我又简单想了一下，在文章中我只考虑了N个男生表白的先后顺序是完全随机的，并没有考虑相邻两次之间的时间隔。如果把时间因素也考虑进去的话，在一个相对较短的时间中，可以近似的假设为齐次泊松过程，这样不仅可以得出女生应该选择上面的第M个男生的结论，而且找到男生表白的最佳时间在t=T/e时刻。 例如如果取时间段为大学四年的话，则T/e=1.4715。 也就是说，在大学四年里，男生表白的最佳时刻在第三个学期的期末或寒假（大二的ddmm们现在要把握机会哟）
如果这个时间段较长的话，那么男生追求可近似假设为了一个非齐次泊松过程，或者分段齐次泊松过程，具体建模中对各段参数lamma的估计就比较困难了，而且每个人以后的经历都会不同，不太可能找到一个统一的参数集，我就不再进一步考虑了，欢迎大家继续提出改进意见~~
本文纯属在家无聊系列，勿真信。最后祝大家中新年快乐！各位小可爱要保重身体哦，有空来踩踩我的主页～谢谢啦^_^
追MM的各种算法（转）
1.动态规划，基本上就是说：
你追一个MM的时候，需要对该MM身边的各闺中密友都好，这样你追MM这个问题
就分解为对其MM朋友的问题，只有把这些问题都解决了，最终你才能追到MM。
因此，该问题适用于聪明的MM，懂得“看一个人，不是看他如何对你，而是看
他如何对他人。”的道理，并且对付这样的MM总能得到最优解。但确定是开销
较大，因为每个子问题都要好好对待。。。。
2.贪心法，基本上就是：
你追一个MM的时候，从相识到相知，每次都采用最aggressive的方式，进攻进攻
再进攻！从不采用迂回战术或是欲擒故纵之法！目标是以最快的速度确立两人
关系。
该法优点是代价小，速度快，但缺点是不是每次都能得到最优解。。。。。
3.回溯算法，基本上就是： :
追一个MM，但也许你还是情窦初开的新手，不知道如何才能讨得MM的欢心，于是你只好一条路一条路的试，MM不开心了，你就回溯回去换另一种方式。当然其间你也许会从某些途径得到一些经验，能够判断哪些路径不好，会剪枝（这就是分支估界了）。你也可以随机选择一些路径来实施，说不定能立杆见影（这就是回溯的优化了）但总的来说，你都需要一场持久战。。。。该算法一般也能得到最优解，因为大多数MM会感动滴！！但其缺点是开销大除非你是非要谈一场恋爱不可，否则不推荐使用。特别是你可能还有许多其他的事情要做，比如学习，比如事业。。。。
4.NP完全问题：
呵呵，那你为什么那么贱，非要去追呢？记住：“天涯何处无芳草！” . 不过如果你“非如此不可”的话，建议升级你的硬件，好好学习，好好工作，加强实力，人到中年的时候也许你能解开NP难。。。。
5.网络流:
追MM的时候总避免不了送礼物，但是你老是直接送礼物就会给MM造成很大的压力
，于是你就想到了通过朋友来转送的方法。你希望送给MM尽可能多的礼物，所以
就是需要找到一中配送方案，就是最大流了。然而你请别人帮忙并不是不要开销
的，你让A同学拿去给B同学可能需要一些花费，自然你不是一个大款，想最小化
这个花费，那么就是最小费用最大流了……
6.NP：
在你追了若干美女都失败告终后，你发现有一批美女追起来是一样困难的，
如果你能追到其中任何一个就能追到其他所有的美女，你把这样的女人叫作
NP-Complete。P=NP：这是一个美好的猜想，追美女和恐龙的难度其实一样。
APX与Random：NP的美女难追，你无法完全占有她。你只好随机的去靠近她
装作若无其事；或者用一种策略，追到她的一个approximation ratio，
例如50%。APX-hard：这样的女人，连一个固定的百分比都不给你，还是另谋高就吧。
7.匹配：从初中到高中到大学大家追来追去，就是个二分图匹配的过程....
"和谐社会"应该就一个最大匹配...
可是后来有某些MM同时跟>1个人发展，违背了匹配的基本原则...大家都很BS之...
然后最近断背山很火，人们惊奇得发现原来还可以是 任意图匹配...
8.深度优先和广度优先
深度优先就是追一个mm追到底，直到失败然后换个mm继续追……
广度优先就是同时追多个mm，一起发展……
9.前序遍历就是直接搞定MM，然后搞定她爸妈（左）和你自己爸妈（右）
10.中序遍历就是先搞定未来岳父岳母，然后搞定她，最后告诉你爸妈
11.后序遍历就是，让未来的岳父岳母和自己爸妈都觉得你们合适之后，才对MM下手，这个时候就没有障碍了啊
12.STL：
某位贝尔实验室的大牛在追了N个MM后，为了造福后来人，总结了自己的经验，
出了本《 追MM求爱秘笈大全》，英文名叫Standard courTing Library，
/* court : vt.向...献殷勤, 追求; vi.求爱)
缩写为 STL. 广大同学在使用STL后，惊喜地发现追MM变得异常方便，大大缩短了时间和精力...
用数学模型来表述，就是在众多约束条件下，一个非线性最优化求解问题，极大值点很难有显示解，要求解，得找个合适的数值算法来逼近，把把全国的男人比做自变量，这个函数是个多峰的，很难求全局最优解，比较理想的就是找个局部最优解了，那个局部最优就是聪明漂亮女生的老公了。