[challenge] String Permutations

[challenge] String Permutations - Printable Version

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+--- Thread: [challenge] String Permutations (/Thread-challenge-String-Permutations)

[challenge] String Permutations - Inori - 06-03-2015

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:

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
[Image: 2f2Bd3d.png]

[/spoiler]

Some Encouragement:

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.

RE: [challenge] String Permutations - Lysergide - 06-21-2015

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)

RE: [challenge] String Permutations - phyrrus9 - 06-21-2015

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:
[Image: 3447WyE.png]

Where's my reward...

@"Chitoge" try this one: https://sinister.li/Thread-Dysfunctional-challenge-light-bulbs

RE: [challenge] String Permutations - Inori - 06-21-2015

(06-21-2015, 12:45 PM)phyrrus9 Wrote:

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

RE: [challenge] String Permutations - Eclipse - 06-21-2015

(06-21-2015, 05:27 PM)Chitoge Wrote: Is there a way to snip the comma off the end? If not, it doesn't follow the format. Soz D:

... srsly?

RE: [challenge] String Permutations - phyrrus9 - 06-21-2015

(06-21-2015, 05:27 PM)Chitoge Wrote: 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 Tongue

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

RE: [challenge] String Permutations - Inori - 06-21-2015

(06-21-2015, 05:56 PM)Eclipse Wrote: ... srsly?

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

(06-21-2015, 06:27 PM)phyrrus9 Wrote: 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

RE: [challenge] String Permutations - Shebang - 08-19-2015

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

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.

RE: [challenge] String Permutations - Inori - 08-19-2015

(08-19-2015, 03:16 AM)Shebang Wrote: 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

RE: [challenge] String Permutations - Eclipse - 08-19-2015

(08-19-2015, 03:16 AM)Shebang Wrote: 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.