China DOS Union

How about looking here?
http://www.cn-dos.net/forum/viewthread.php?tid=41243&fpage=0&highlight=&page=1

After messing with it all night, it actually gave me:
****** B A T C H R E C U R S I O N exceeds STACK limits ******
Recursion Count=495, Stack Usage=90 percent
****** B A T C H PROCESSING IS A B O R T E D ******
Going to sleep now...

Should it be made like this?
Based on the link in post #2, I learned from it and then adapted it.

@echo off&setlocal enabledelayedexpansion

set in=寒夜孤星

set /p in=输入字符串

set m=0

set svr=abcdefghijklmnopqrstuvwxyz

:rep

if "!in:~%m%,1!" neq "" (set s0=!s0! !in:~%m%,1! &set/a m+=1&goto :rep)

set n=0

for %%a in (!s0!) do (

	for %%b in (!n!) do (

		set /a n+=1

		set "coe1=!coe1!for %%!svr:~%%b,1! in (^!s%%b^!) do (set s!n!=^!s%%b:%%!svr:~%%b,1!=^!^&"

		set "coe2=!coe2!%%!svr:~%%b,1!"

		set "coe3=!coe3!)"

		set doe1%%b=!coe1:~,-16!

		set doe2%%b=!coe2!

		set doe3%%b=!coe3!

	)

)

set n=0

:loop

set coe1=!doe1%n%!

set coe2=!doe2%n%!

set coe3=!doe3%n%!



::dynamic code

%coe1%

echo %coe2%

%coe3%

::dynamic code



set/a n+=1

pause

if !n! neq !m! goto loop

[ Last edited by netbenton on 2009-7-5 at 01:27 ]

Thanks to the friend in the previous post.
What the link given in post #2 produces is permutations, but what I need is combinations.
For combinations, “寒夜孤星” and “寒夜星孤” are the same case, although for permutations they are different. So after generating permutations, they still need to be filtered... that is, filter out all cases that use the same Chinese characters (the same elements).

By the way, celebrating a bit: I can finally give other people 4 points now, haha

The general idea would be to list all possible permutations first, then filter them to get the combinations.

According to the requirements I gave in the first post, the possible numbers of permutations and combinations are: (n is the number of characters in the variable)

For combinations: Σ(n)=2(n-1)+2(1)+2(2)+2(3)+......+2(n) Note: in this formula, the values in () are exponents of 2.
　　　That is: Σ(n)=2(n)-1 Note: in this formula, the values in () are exponents of 2.

For permutations: Σ(n)=n+
　　　　　　　 n*(n-1)+
　　　　　　　 n*(n-1)*(n-2)+
　　　　　　　 n*(n-1)*(n-2)*(n-3)+
　　　　　　　 n*(n-1)*(n-2)*(n-3)*(n-4)+
　　　　　　　 n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)+
　　　　　　　 ......
　　　　　　　 n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*......*3+
　　　　　　　 n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*......*3*2+
　　　　　　　 n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*......*3*2*1

When there are only four Chinese characters, that is, n=4, then the number of permutations is 64, and the number of combinations is 15. As n increases, the number of combinations grows algebraically, while the number of permutations grows geometrically.

For example, when n=10, the number of combinations is 1023, while the number of permutations is 9183700, an astronomical figure close to ten million. Then having to filter through 9183700 permutation cases to get the 1023 valid combination cases is a pretty terrifying thing...

So I had this idea: is it possible not to use the “permute first, then filter” method?

[ Last edited by Hanyeguxing on 2009-7-5 at 08:15 ]

Turns out a senior member had already solved this problem perfectly two years ago!
http://www.cn-dos.net/forum/viewthread.php?tid=26180###

I'm embarrassed...
I've been figuring it out myself for two days and still didn't get it done...
After reading the code in post #15, I really feel ashamed. That's the gap, and also the direction I need to work toward...