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Cryptography coursera class exercise #4.
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{-# LANGUAGE UnicodeSyntax #-} | |
module Main where | |
import Control.Applicative ((<$>)) | |
import Control.Monad (foldM, join) | |
import qualified Data.Bits as Bits | |
import Data.List (genericReplicate, inits) | |
import Data.Monoid ((<>)) | |
import Data.Word (Word8) | |
import Text.Printf (printf) | |
import Data.ByteString (ByteString) | |
import qualified Data.ByteString as B | |
import qualified Network.Curl as C | |
randomByteString ∷ ByteString | |
randomByteString = B.replicate 16 57 | |
main ∷ IO () | |
main = | |
do cipher ← B.readFile "ciphertext.dat" | |
plainText ← decrypt cipher | |
print plainText | |
decrypt ∷ ByteString → IO ByteString | |
decrypt cipher = B.concat . map (B.pack . B.zipWith Bits.xor randomByteString) <$> mapM decryptitionRound (ciphers cipher) | |
where | |
ciphers = map B.concat . drop 2 . inits . chunk 16 | |
decryptitionRound ∷ ByteString → IO ByteString | |
decryptitionRound cipher = foldM (bruteforceByte cipher) (B.replicate 16 0) paddings | |
where | |
paddings = map (B.pack . reverse . take 16 . (<> repeat 0) . join genericReplicate) [1..16] | |
bruteforceByte ∷ ByteString → ByteString → ByteString → IO ByteString | |
bruteforceByte cipher acc padding = go 0 | |
where | |
paddedCipher = cipher `xor` padding `xor` acc `xor` randomByteString | |
go n = | |
do let guess = pretty $ paddedCipher `xor` fromWord8 (B.length $ B.dropWhile (== 0) acc) n | |
r ← C.withCurlDo $ C.curlGetResponse_ | |
("http://crypto-class.appspot.com/po?er=" <> guess) | |
[] ∷ IO (C.CurlResponse_ [(String, String)] ByteString) | |
case C.respStatus r of | |
403 → go (n + 1) | |
_ → return (addByte n acc) | |
addByte ∷ Word8 → ByteString → ByteString | |
addByte n xs = B.replicate (16 - t - 1) 0 <> B.cons n ys | |
where | |
ys = B.dropWhile (== 0) xs | |
t = B.length ys | |
xor ∷ ByteString → ByteString → ByteString | |
xor x y = B.concat as <> B.pack (B.zipWith Bits.xor c y) <> b | |
where | |
(as,c,b) = split $ chunk 16 x | |
fromWord8 ∷ Int → Word8 → ByteString | |
fromWord8 t n = B.pack $ replicate (16 - t - 1) 0 ++ n : replicate t 0 | |
pretty ∷ ByteString → String | |
pretty = concatMap (printf "%02x") . B.unpack | |
chunk ∷ Int → ByteString → [ByteString] | |
chunk n bs | |
| B.length bs <= n = [bs] | |
| otherwise = B.take n bs : chunk n (B.drop n bs) | |
split ∷ [α] → ([α], α, α) | |
split = go [] | |
where | |
go as [x,y] = (reverse as, x, y) | |
go as (x:xs) = go (x:as) xs | |
go _ _ = error "Main.split: [_]/empty list" |
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% time runhaskell Main.hs
"The Magic Words are Squeamish Ossifrage\t\t\t\t\t\t\t\t\t"
runhaskell Main.hs 17.64s user 4.23s system 2% cpu 15:31.76 total
ciphertext.dat is not very interesting since it just contains given ciphertext (f20bdba6ff29eed7b046d1df9fb7000058b1ffb4210a580f748b4ac714c001bd4a61044426fb515dad3f21f18aa577c0bdf302936266926ff37dbf7035d5eeb4) in binary.
You don't need to do any smart dictionary attacks or prepare sophisticated ciphertexts in file to solve the problem, bruteforce is enough. The key idea is to guess byte after byte given different oracle's responses on different kinds of problems. Simple example for one last byte is given in lecture slides (page 50-51).