Programming Praxis – The Playfair Cipher

Today’s Programming Praxis problem is about the Playfair Cipher, a way of encrypting messages that was used in the first and second World Wars. Our target is 44 lines, so let’s get cracking.

We’re going to need a bunch of imports on this one:

import Data.Char
import Data.List
import Data.List.HT
import Data.List.Split
import qualified Data.Map as M

And for convenience, we make a type definition for Keys.

type Key = M.Map (Int, Int) Char

In order for the algorithm to work, we’re going to have to filter all the input so that all we get is capital letters without a J.

process :: String -> String
process = replace "J" "I" . map toUpper . filter isLetter

Instead of working with a block, we turn the key into a map of of x-y coordinates to letters. This makes the rest of the algorithm simpler.

key :: String -> Key
key = M.fromList . concat .
      zipWith (\y -> zipWith (\x c -> ((x, y), c)) [0..]) [0..] .
      chunk 5 . (`union` delete 'J' ['A'..'Z']) . nub . process

bigram encodes or decodes two letters at a time according to the three given rules.

bigram :: Key -> Int -> Char -> Char -> String
bigram k dir c1 c2
    | y1 == y2  = get (x1 + dir, y1) : [get (x2 + dir, y2)]
    | x1 == x2  = get (x1, y1 + dir) : [get (x2, y2 + dir)]
    | otherwise = get (x2, y1)       : [get (x1, y2)]
    where (x1, y1) = head . M.keys $ M.filter (== c1) k
          (x2, y2) = head . M.keys $ M.filter (== c2) k
          get (x,y) = k M.! (mod x 5, mod y 5)

Encoding consists of inserting Xs in the correct places and then encoding all the bigrams.

encode' :: Key -> String -> String
encode' _ []       = []
encode' k [x]      = encode' k (x : "X")
encode' k (x:y:xs) | x == y    = encode' k [x] ++ encode' k (y:xs)
                   | otherwise = bigram k 1 x y ++ encode' k xs

Decoding is simpler, since there are no special cases there.

decode' :: Key -> String -> String
decode' k = concatMap (\[x,y] -> bigram k (-1) x y) . chunk 2

And finally, two convenience functions so we can work with strings directly.

encode :: String -> String -> String
encode k = encode' (key k) . process

decode :: String -> String -> String
decode k = decode' (key k) . process

All that remains is to test if everything works correctly:

main :: IO ()
main = do print $ encode "PLAYFAIR" "PROGRAMMING PRAXIS"
          print $ decode "PLAYFAIR" "LIVOBLKZEDOELIYWCN"

And we’re done. Interestingly, if you try to decode the 1943 message, you’ll notice two errors in the result: it says Blackess instead of Blackett and coce instead of cove. I wonder if the errors were made by the original sender or by the first person to type it on a computer.

Anyway, the solution is 24 lines, just over half the size of the Scheme version. That will do nicely.

About these ads

Tags: , , , , , ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 35 other followers

%d bloggers like this: