| Copyright | (c) Austin Seipp |
|---|---|
| License | BSD3 |
| Maintainer | erkokl@gmail.com |
| Stability | experimental |
| Safe Haskell | None |
| Language | Haskell2010 |
Documentation.SBV.Examples.Crypto.RC4
Description
An implementation of RC4 (AKA Rivest Cipher 4 or Alleged RC4/ARC4), using SBV. For information on RC4, see: http://en.wikipedia.org/wiki/RC4.
We make no effort to optimize the code, and instead focus on a clear implementation. In fact, the RC4 algorithm relies on in-place update of its state heavily for efficiency, and is therefore unsuitable for a purely functional implementation.
Synopsis
- type S = STree Word8 Word8
- initS :: S
- type Key = [SWord8]
- type RC4 = (S, SWord8, SWord8)
- swap :: SWord8 -> SWord8 -> S -> S
- prga :: RC4 -> (SWord8, RC4)
- initRC4 :: Key -> S
- keySchedule :: Key -> [SWord8]
- keyScheduleString :: String -> [SWord8]
- encrypt :: String -> String -> [SWord8]
- decrypt :: String -> [SWord8] -> String
- rc4IsCorrect :: IO ThmResult
- hex2 :: (SymVal a, Show a, Integral a) => SBV a -> String
Types
type S = STree Word8 Word8 Source #
RC4 State contains 256 8-bit values. We use the symbolically accessible
full-binary type STree to represent the state, since RC4 needs
access to the array via a symbolic index and it's important to minimize access time.
Construct the fully balanced initial tree, where the leaves are simply the numbers 0 through 255.
type RC4 = (S, SWord8, SWord8) Source #
Represents the current state of the RC4 stream: it is the S array
along with the i and j index values used by the PRGA.
The PRGA
prga :: RC4 -> (SWord8, RC4) Source #
Implements the PRGA used in RC4. We return the new state and the next key value generated.
Key schedule
keySchedule :: Key -> [SWord8] Source #
The key-schedule. Note that this function returns an infinite list.
keyScheduleString :: String -> [SWord8] Source #
Generate a key-schedule from a given key-string.
Encryption and Decryption
encrypt :: String -> String -> [SWord8] Source #
RC4 encryption. We generate key-words and xor it with the input. The following test-vectors are from Wikipedia http://en.wikipedia.org/wiki/RC4:
>>>concatMap hex2 $ encrypt "Key" "Plaintext""bbf316e8d940af0ad3"
>>>concatMap hex2 $ encrypt "Wiki" "pedia""1021bf0420"
>>>concatMap hex2 $ encrypt "Secret" "Attack at dawn""45a01f645fc35b383552544b9bf5"
decrypt :: String -> [SWord8] -> String Source #
RC4 decryption. Essentially the same as decryption. For the above test vectors we have:
>>>decrypt "Key" [0xbb, 0xf3, 0x16, 0xe8, 0xd9, 0x40, 0xaf, 0x0a, 0xd3]"Plaintext"
>>>decrypt "Wiki" [0x10, 0x21, 0xbf, 0x04, 0x20]"pedia"
>>>decrypt "Secret" [0x45, 0xa0, 0x1f, 0x64, 0x5f, 0xc3, 0x5b, 0x38, 0x35, 0x52, 0x54, 0x4b, 0x9b, 0xf5]"Attack at dawn"
Verification
rc4IsCorrect :: IO ThmResult Source #
Prove that round-trip encryption/decryption leaves the plain-text unchanged. The theorem is stated parametrically over key and plain-text sizes. The expression performs the proof for a 40-bit key (5 bytes) and 40-bit plaintext (again 5 bytes).
Note that this theorem is trivial to prove, since it is essentially establishing
xor'in the same value twice leaves a word unchanged (i.e., x ).
However, the proof takes quite a while to complete, as it gives rise to a fairly
large symbolic trace.xor y xor y = x