Question about displaying combined variables - China DOS Union Forum

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...

Floor2　moniuming	Posted 2009-07-04 14:01
银牌会员　Posts 574　Credits 1,335　From 广西
How about looking here? http://www.cn-dos.net/forum/viewthread.php?tid=41243&fpage=0&highlight=&page=1

Floor3　Hanyeguxing	Posted 2009-07-04 17:59
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂
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...

Floor4　netbenton	Posted 2009-07-05 03:24
银牌会员　Posts 752　Credits 1,916　From 广西
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 ]

Floor5　Hanyeguxing	Posted 2009-07-05 03:50
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂
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

Floor6　Hanyeguxing	Posted 2009-07-05 04:08
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂
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)......32+ 　　　　　　　 n(n-1)(n-2)(n-3)(n-4)(n-5)......321 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 ]

Floor1　Hanyeguxing	Posted 2009-07-04 13:34
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂

Floor7　netbenton	Posted 2009-07-05 08:42
银牌会员　Posts 752　Credits 1,916　From 广西
Turns out a senior member had already solved this problem perfectly two years ago! http://www.cn-dos.net/forum/viewthread.php?tid=26180###

Floor8　Hanyeguxing	Posted 2009-07-05 09:05
银牌会员　Posts 897　Credits 1,039　From 在地狱中仰望天堂
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...