0
$\begingroup$

Thanks firstly! : )

Now we have n persons $n_1,n_2,n_3,n_4...$. for each person, he have written 2 kinds of articles: A B.
For example, for person $n_2$, there are some articles A and B. We will compare each A of person $n_2$ with each B of person $n_1$,$n_3$,$n_4$,$n_5$.... For $n_2$ himself, A articles and B articles (we call them $A_2[_1]$,$A_2[_2]$,$A_2[_3]$...,$B_2[_1]$,$B_2[_2]$..., first subscript means person number, $A_2$ means all A articles for person $n_2$. second means article number) will have some similarity (we have evaluation standard for this, you can call it one standard value $S$).

We say it's one $F$ event if the similarity of $A_2[_1]$ and $B_1[_2]$ is bigger than $S$ (note that $A_2$ and $B_1$ are different persons). If it's smaller than S, we call it as $T$ event.

We use $M(n_1,n_2)$ to represent number of $F$ event between $A_1$ with $B_2$, use $N(n_1,n_2)$ to represent number of comparing between $A_1$ with $B_2$ (we know that it's number of $A_1$ multiply number of $B_2$).

We use $M(n_2,n_1)$ to represent number of $F$ event between $A_2$ with $B_1$. use $N(n_2,n_1)$ to represent number of comparing between $A_2$ with $B_1$ (we know that it's number of $A_2$ multiply number of $B_1$).

I can get all $M(n_x,n_y)$ values and $N(n_x,n_y)$. Now, I need select 5 persons from n persons randomly (use $C^5_n$ to enumerate all possible situations) and compare all 5 persons' A articles and all other persons' B articles.

I want compute that :

for each combination in $C^5_n$ , I need to get $M(n_yn_xn_dn_en_u, n_z)$/$N(n_yn_xn_dn_en_u, n_z)$. here $y,x,d,e,u$ will construct $C^5_n$ and $z$ means other person.

ps: I can enumerate all $C^5_n$ situations, and computate all $M$ and $N$, but when number of persons are big, it's so big for compuater to simulate.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.