Ok, I found myself with a nice little problem:

you will get from stdin a string. In this string you will find some hex colors (like #ff00dd). You must return the string with the colors inverted (#ff00dd inverted is #00ff11).

A sample input/output could be:

ascscasad #ff00ff csdcas dcs c#001122 #qq5500
ascscasad #00ff00 csdcas dcs c#ffeedd #qq5500

As I am proficient in Haskell, I tackled the problem with it. Came up with this code, and I think it's the best I can do. Do you see any way to improve it / a completely different way to make it shorter?

import Text.Regex.Posix ( (=~) )

main = getLine >>= putStrLn . f

f [] = []
f x = let (b, q, p) = x =~ "#[0-f]{6}"
          c = '#' : ['0'..'f'] ++ "#"
          a = zip c (reverse c)
          r = map ( (maybe '@' id) . (flip lookup $ a) ) q
      in b ++ r ++ f p

I am also interested to compare other languages to Haskell, so post in other languages are welcome, too!

  • 9
    \$\begingroup\$ Shouldn't the inverse of #ff00dd be #00ff22? \$\endgroup\$ Mar 8, 2013 at 23:21
  • 1
    \$\begingroup\$ main = interact f \$\endgroup\$ Mar 11, 2013 at 10:13
  • 1
    \$\begingroup\$ Can we assume #ff00 is equivalent to #00ff00 (since it is)? \$\endgroup\$
    – Griffin
    Mar 14, 2013 at 9:33

3 Answers 3


Perl, 44 chars

Two different tweaks can get it to this length, one suggested by Primo:

s/#\K[\da-f]{6}/sprintf"%06x",~hex ff.$&/eg

Another one by playing around with the /ee modifier:


Requires -p command line switch (1 char included in count).

Works for lower-case hex numbers (as in example). An i switch added to the regex would make it work for upper-case as well (though not preserving the case of the original number).

  • 1
    \$\begingroup\$ Very nice solution. I see only one minor improvement: (~0>>8)-hex$& can be replaced by ~hex ff.$&. \$\endgroup\$
    – primo
    Mar 12, 2013 at 4:31

APL 86

Index origin zero. Takes either a single colour hex string or multiple space separated colour strings as screen input via ←⍞

,' ',h[⊃16,¨,¨⍉¨(⊂16 16)⊤¨255-16⊥¨⍉¨(⊂3 2)⍴¨(⊂h←'0123456789abcdef#')⍳¨(c≠' ')⊂c←⍞~'#']

input: #ff00dd  output: #00ff22

input: #ff00dd #00ff22 output: #00ff22 #ff00dd
  • \$\begingroup\$ APL strikes again... :) Nice solution though! \$\endgroup\$
    – tomsmeding
    Mar 9, 2013 at 22:04

Ruby, 46 bytes


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