[Inori]

Warning: this challenge is a HUGE pain in the dick and took me 2+ hours to complete

The purpose of said pain in the dick challenge to write a program which prints all the permutations of a string in alphabetical order separated by a comma, considering that digits < upper case letters < lower case letters.

In case you're confused, a permutation is simply a way of arranging characters in a given set of data. This is done a lot in grade school math and science.

I/O example:

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Input:

Code:

acb fde igh

Output:

Code:

abc,acb,bac,bca,cab,cba def,dfe,edf,efd,fde,fed ghi,gih,hgi,hig,igh,ihg

Guidelines:

---

The code you will need to permute (one set per line) is:

Code:

AV7 A2La faHd 7peP eAg y68B HgAf mt06p V2c4 FfNh

Rules:
- No imports unless required
- No use of semicolons when not absolutely necessary
- Code < 6 lines AND 250 characters or less = 20 NSP prize
- Don't bother copying all the text to show output, just post a screenshot

[spoiler=My Code:]
Code (omitted user from filepath):
7 lines, 364 characters

Code:

File.open("C:/Users/Kagetane/Desktop/Code/Ruby/test.txt").each_line { |line| perms = [] v = line.include?("\n") ? line.split("")[0..-2] : line.split("")[0..-1] v.permutation.map { |str| perm = [] str.include?("\n") and str != v[-1] ? str = nil : nil str.each { |u| perm << u } if str != nil perms << perm.join("") } puts perms.sort.join(",") }

Output:
fits perfectly in a default-sized windows cmd window, which is nice

[/spoiler]

Some Encouragement:

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This challenge is listed under "hard" on CodeEval, which, believe me, is really, really, fucking hard.
The only reason I got this done was because I had a few hours to kill during school, which is currently in moratorium.

[Lysergide]

God that was hard.
Thank you for your challenge, it made me learn how to use recursive functions. I almost gave up!
I also would like to thank you for letting me know CodeEval.
I am having tons of fun: 680/6k, but the best thing is that i am learning a lot.

Here is the code (I didn't manage to get under 6 lines, still thinking about it tho):

Spoiler:

6 lines, 228 characters

Code:

def perm(l,a): if len(l)==1: return l[0] for x in range(len(l)): for b in perm(l[:x]+l[x+1:],[]): a.append(l[x]+b) return a for a in open('ha.txt', 'r').read().split(): print ','.join(sorted(perm(a,[])))

And here is the screenshot:
http://i.imgur.com/ZlRmUnX.png
(not posting through img tags because it is big)

[phyrrus9]

Hey, that isn't really fair to us C programmers...anyways, here is my 6 line, 250 character solution.

Spoiler:

Code:

#define y int #define x char void s(x*a,x*b){x t=*a;*a=*b;*b=t;} void p(x*a,y i,y n){if(i==n)printf("%s,",a);else for(y j=i;j<=n;j++){s(a+i,a+j);p(a,i+1,n);s(a+i,a+j);}} void main(y c,x**d){for(y i=1;i<c;i++){p(d[i],0,strlen(d[i])-1);printf("\n");}}

and a screenshot:


Where's my reward...

---

@"Chitoge" try this one: https://sinister.li/Thread-Dysfuncti...ight-bulbs

[Inori]

Is there a way to snip the comma off the end? If not, it doesn't follow the format. Soz D:

[Eclipse]

Is there a way to snip the comma off the end? If not, it doesn't follow the format. Soz D:

... srsly?

[phyrrus9]

Is there a way to snip the comma off the end? If not, it doesn't follow the format. Soz D:

Actually, the initial problem says nothing about format

And yes, you could trim it, but it would add characters. C makes it near impossible to do in under 250.

[Inori]

... srsly?

Hey, the challenge on CodeEval gave me 10/100 if I didn't follow the format

Actually, the initial problem says nothing about format

And yes, you could trim it, but it would add characters. C makes it near impossible to do in under 250.

Hmm.. I might still give you a good chunk of the prize

[Shebang]

I know I'm months late, but here's something that blows the other answers out of the water

148 Bytes / 2 Lines

Code:

p=lambda s:[[s[k]+m for k in range(len(s))for m in p(s[:k]+s[-~k:])],[s]][not s] print'\n'.join(','.join(p(x))for x in open('x','r').read().split())

If I could take the strings space-separated through raw_input() this would be down to 139 chars.

[Inori]

I know I'm months late, but here's something that blows the other answers out of the water

148 Bytes / 2 Lines

Code:

p=lambda s:[[s[k]+m for k in range(len(s))for m in p(s[:k]+s[-~k:])],[s]][not s] print'\n'.join(','.join(p(x))for x in open('x','r').read().split())

If I could take the strings space-separated through raw_input() this would be down to 139 chars.

Jesus, well done

[Eclipse]

I know I'm months late, but here's something that blows the other answers out of the water

148 Bytes / 2 Lines

Code:

p=lambda s:[[s[k]+m for k in range(len(s))for m in p(s[:k]+s[-~k:])],[s]][not s] print'\n'.join(','.join(p(x))for x in open('x','r').read().split())

If I could take the strings space-separated through raw_input() this would be down to 139 chars.

#golfingpro

OT: Wow, I forgot all about this challenge. I might yet attempt it.