For some recent supervision work on my Security lectures, I was given the task of decoding a string encrypted with a simple shift cipher. This cipher, given a key, simply moves each letter on in the alphabet by an amount given by the key, wrapping at the end. Being a good functional programmer, I decided to implement a brute force solver for this in Haskell, and I’m rather pleased by the terseness of the result, so I felt compelled to share it.
import List(elemIndex)
cipherText = “LUXDZNUAMNDODJUDTUZDGYQDLUXDGOJDCKDTKKJDOZ”shift i x = (cycle ['A'..'Z'])!!(maybe (error “Bad character ” ++ (show x)) (+i) (elemIndex x ['A'..'Z']))
main = sequence_ (map (\i -> putStrLn $ map (shift i) cipherText) [0..25])
If you ignore the boilerplate, there’s only two lines of actual code there Image may be NSFW.
Clik here to view.
. I happen to think it’s rather understandable, though I’d be interested to get the opinion of someone who isn’t a Haskell hacker on the veracity of that statement. A hint you may need understanding it is that “cycle a_list” just returns the list which is “a_list” concatenated infinite times, and that “!!” is Haskells list indexing operator.
On this note, I saw an interesting blog post at retrospections the other day which compared the LOC count for a number of open source source control programs: Darcs (Haskell) and Mercurial (Python) were tied at 20KLOC, with the next best being Bazaar-NG (Python) at 47KLOC. Obviously this isn’t a strictly apples-to-apples comparison, though, so take it with a grain of salt!
Image may be NSFW.
Clik here to view.
Image may be NSFW.Clik here to view.
Image may be NSFW.Clik here to view.
Image may be NSFW.Clik here to view.
