Implement Oplop

Question

Oplop is an algorithm to generate website specific passwords based on a master password and a keyword.

It is described here: http://code.google.com/p/oplop/wiki/HowItWorks

There's an online implementation here: https://oplop.appspot.com/

My attempt in Q is 204 200 167 165 139 characters ignoring whitespace:

{d:.Q.A,.Q.a,(n:"0123456789"),"-_";
b:d 2 sv'6 cut(,/)0b vs'(md5 x,y),2#0x00;
i:b in n;j:i?1b;k:(j _i)?0b;
8#$[0=0+/i;"1";8>j;();b j+(!)k],b}

EDIT: I achieved significant character savings by removing some redundant code from my Hex->Base64 conversion.

2012.03.07 - Cut out 2 more characters

2012.03.11 - Removed duplication and got accurate char count

2012.03.20 - Down to 141

Criteria

Candidate functions/implementations should take two strings as arguments and return an 8 character password as defined in the algorithm.

Judging

Code golf so shortest code takes the prize.

Answer 1

Python - 198 166 152 150

import base64,md5,sys,re
b=base64.urlsafe_b64encode(md5.md5(sys.argv[1]+sys.argv[2]).digest())
r=re.findall('\d+',b)if all(not c.isdigit()for c in b[:8])else['']
print((r[0]if len(r)else'1')+b)[:8]

import base64,md5,sys,re
b=base64.urlsafe_b64encode(md5.md5("".join(sys.argv[1:])).digest())
print((''if re.search('\d',b[:8])else re.findall('\d+',b+'1')[0])+b)[:8]

import base64,md5,sys,re
b=base64.urlsafe_b64encode(md5.md5("".join(sys.argv[1:])).digest())
print(re.findall('(?:[^\d]{8,}(\d*)|)',b+' 1')[0]+b)[:8]

This was a fun one. Could be a lot shorter if not for the verbosity of the md5 and base64 modules. The third and fourth lines are the interesting ones. My favourite trick was using a list containing only an empty string to avoid having to use any real if statements.

EDIT: Had to change from checking c.isalpha() to not c.isdigit() (added 4 characters)

EDIT: Shaved off 32 characters. Many thanks to Ugoren!

EDIT: Shaved off 14 more chars again many thanks to Ugoren. Used some regex trickiness

print(re.findall('(?:[^\d]{8,}(\d*)|)',b+' 1')[0]+b)[:8]

instead of

print((re.findall('^[^\d]{8,}(\d*)',b+' 1')+[''])[0]+b)[:8]

To save three chars.

Answer 2

Shell - 116

echo -n $1$2|md5sum|xxd -r -p|base64|sed -re'y!+/!-_!;/^.{0,7}[0-9]/!{s/^([^0-9]*)([0-9]+)/\2\1/;t;s/^/1/}'|cut -c-8

Depends on cut, md5sum and base64 from coreutils, xxd and GNU sed.

Answer 3

Ruby (113 110 chars)

require 'digest'
b=Digest::MD5.base64digest(ARGV.join)
puts "#{b} 1".sub(/(^\D{8,}(\d*))/,'\2\1')[0..7].tr'+/','-_'

require'digest'
b=Digest::MD5.base64digest(ARGV.join)
puts"#{b} 1".sub(/(^\D{8,}(\d*))/,'\2\1')[0,8].tr'+/','-_'

I notice that all the non-Python languages have to do a translation of the + and / to be - and _, respectively. It seems that Python's base64 is different/wrong?

Answer 4

PHP (126)

function f($m,$n){return substr(preg_replace('/^\D{8,}(\d+)/','$1$0',strtr(base64_encode(md5($m.$n,1)),'+/','-_').' 1'),0,8);}

Beautified:

function oplop( $master, $nick ) {
    return substr( preg_replace(
        '/^\D{8,}(\d+)/', '$1$0',
        strtr( base64_encode( md5( $master . $nick, true ) ), '+/', '-_' ) . ' 1'
    ), 0, 8 );
}