{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE ViewPatterns #-}
module MD5 (
Dictionary,
hashcatWord, hashcatDict, readMD5, showMD5, md5Round
) where
import Numeric
import Control.Monad
import Control.Applicative
import Data.Maybe
import Data.List ( foldl' )
import Data.ByteString.Lex.Integral ( readHexadecimal )
import qualified Data.Serialize as S
import qualified Data.ByteString as B
import qualified Data.ByteString.Lazy as L
import Prelude as P hiding ( Eq(..), (&&) )
import Data.Array.Accelerate hiding ( Ord(..) )
import Data.Array.Accelerate.Data.Bits as A
import qualified Data.Array.Accelerate as A
-- Generate an MD5 hash for every word in the dictionary, and if an entry
-- matches the given unknown md5, returns the index into the dictionary of said
-- match. If not found, this returns (-1).
--
hashcatDict
:: Acc Dictionary
-> Acc (Scalar MD5)
-> Acc (Scalar Int)
hashcatDict dict passwd
= permute const res (\ix -> crypt A.! ix `cmp` the passwd ? (Just_ Z_, Nothing_))
(enumFromN (index1 n) 0)
where
n = A.size crypt
res = fill Z_ (-1)
crypt = md5 dict
cmp x y = let (x1,x2,x3,x4) = unlift x
(y1,y2,y3,y4) = unlift y
in x1 == y1 && x2 == y2 && x3 == y3 && x4 == y4
-- Generate an MD5 hash for a single word, and if an entry matches the
-- given unknown md5, returns the given index. If not matched, this
-- returns (-1).
--
hashcatWord
:: Acc (Scalar MD5)
-> Acc (Vector Word32)
-> Acc (Scalar Int)
-> Acc (Scalar Int)
hashcatWord passwd word ix
= unit (crypt `cmp` the passwd ? (the ix, -1))
where
crypt = md5Round (\i -> word A.! index1 i)
cmp x y = let (x1,x2,x3,x4) = unlift x
(y1,y2,y3,y4) = unlift y
in x1 == y1 && x2 == y2 && x3 == y3 && x4 == y4
-- An MD5 round processes 512 bits of the input, as 16 x 32-bit values. We
-- require that the input dictionary of words to hash is of dimension
-- (Z :. 16 :. n), which will then output n MD5 hashes values.
--
-- NOTE: All values are unsigned 32-bit integers and wrap modulo 2^32 when
-- calculating. All values are little endian.
--
-- TODO: How to apply salt to passwords?
--
type MD5 = (Word32, Word32, Word32, Word32)
type Dictionary = Array DIM2 Word32
md5 :: Acc Dictionary -> Acc (Vector MD5)
md5 dict
= let n = A.snd . unindex2 $ A.shape dict
in
A.generate (index1 n) (\ (unindex1 -> ix) -> md5Round (\i -> dict A.! index2 i ix))
md5Round :: (Exp Int -> Exp Word32) -> Exp MD5
md5Round fetch
= lift
$ P.foldl step (a0,b0,c0,d0) [0..64]
where
step (a,b,c,d) i
| i < 16 = shfl ((b .&. c) .|. ((complement b) .&. d))
| i < 32 = shfl ((b .&. d) .|. (c .&. (complement d)))
| i < 48 = shfl ((b `xor` c `xor` d))
| i < 64 = shfl (c `xor` (b .|. (complement d)))
| otherwise = (a+a0,b+b0,c+c0,d+d0)
where
shfl f = (d, b + ((a + f + k i + get i) `rotateL` r i), b, c)
get :: Int -> Exp Word32
get i
| i < 16 = fetch (constant i)
| i < 32 = fetch (constant ((5*i + 1) `rem` 16))
| i < 48 = fetch (constant ((3*i + 5) `rem` 16))
| otherwise = fetch (constant ((7*i) `rem` 16))
-- Initial values. For a multi-round implementation we would initialise the
-- context with these values, and then after applying the round update the
-- context with the addition of the new result. Since we only ever apply one
-- round, we begin with these values and add them again at the end, which
-- seems odd but is correct.
--
a0, b0, c0, d0 :: Exp Word32
a0 = 0x67452301
b0 = 0xefcdab89
c0 = 0x98badcfe
d0 = 0x10325476
-- The binary integer part of sines and cosines (in radians) as constants
--
-- K[i] := floor(abs(sin(i + 1)) × (2 pow 32))
--
k :: Int -> Exp Word32
k i = constant (ks P.!! i)
where
ks = [ 0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee
, 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501
, 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be
, 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821
, 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa
, 0xd62f105d, 0x02441453, 0xd8a1e681, 0xe7d3fbc8
, 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed
, 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a
, 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c
, 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70
, 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x04881d05
, 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665
, 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039
, 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1
, 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1
, 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391 ]
-- Specifies the per-round shift amounts
--
r :: Int -> Exp Int
r i = constant (rs P.!! i)
where
rs = [ 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22, 7, 12, 17, 22
, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20, 5, 9, 14, 20
, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23, 4, 11, 16, 23
, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21, 6, 10, 15, 21 ]
-- Display the result in big endian hexadecimal format, which is consistent with
-- other MD5 hash implementations.
--
showMD5 :: MD5 -> String
showMD5 (a,b,c,d) =
let bs = S.runPutLazy $ S.putWord32be d >> S.putWord32be c >> S.putWord32be b >> S.putWord32be a
showsHex str w = let x = showHex w str
in if P.length x < P.length str + 2
then '0':x
else x
in
foldl' showsHex "" (L.unpack bs)
-- Read an input hexadecimal string into our MD5 representation. This string is
-- the same as that displayed by the standard md5 utility, and also produced by
-- 'showMD5'.
--
readMD5 :: L.ByteString -> MD5
readMD5 =
let get = (,,,) <$> readHex32be <*> readHex32be <*> readHex32be <*> readHex32be
readHex32be = do
s <- B.concat . P.reverse <$> replicateM 4 (S.getBytes 2)
return . P.fst $ fromMaybe (error "readHex32be: parse failure") (readHexadecimal s)
in
either error id . S.runGetLazy get