B4X

Tips on Matching Encryption with another System

See more Encryption Examples

This example provides tips on matching encryption results produced by another system. B4X

'  This example assumes the Chilkat API to have been previously unlocked.
'  See Global Unlock Sample for sample code.

Dim crypt As ChilkatCrypt2
crypt.Initialize("crypt")

'  Let's examine 256-bit AES encryption in CBC mode.
'  CBC mode is Cipher Block Chaining, and it uses an IV (initialization vector)
crypt.CryptAlgorithm = "aes"
crypt.CipherMode = "cbc"
crypt.KeyLength = 256
crypt.PaddingScheme = 0
Dim ivHex1 As String = "000102030405060708090A0B0C0D0E0F"
Dim ivHex2 As String = "FF0102030405060708090A0B0C0D0E0F"
crypt.SetEncodedIV(ivHex1, "hex")
Dim keyHex As String = "000102030405060708090A0B0C0D0E0F101112131415161718191A1B1C1D1E1F"
crypt.SetEncodedKey(keyHex, "hex")

'  Matching encryption requires all of the above settings to be matched exactly.
'  Let's get our output in hex format so we can easily see the values of the encrypted bytes.
crypt.EncodingMode = "hex"

'  Encrypt something small:
Log(crypt.EncryptStringENC("Hello"))
'  The result is 5B827AB3B4F9F2292C2B74C8A6C99A3D
'  This 16 bytes -- exactly one AES encryption block.

'  Let's change only the padding scheme.
crypt.PaddingScheme = 3

'  Encrypt again:
Log(crypt.EncryptStringENC("Hello"))
'  The result is entirely different: 469C28CC576069F807891FEE2DE76D68

'  The padding scheme only affects the very last block of output.  Therefore,
'  if all settings match except for the padding scheme, we're unable to
'  know if we encrypt a very small amount of data. However, if we encrypt
'  a larger amount of data, the single difference becomes apparent:
Log("-- Only the padding scheme differs --")
crypt.PaddingScheme = 0
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.PaddingScheme = 3
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))

'  Now examine the outputs:
'  F6A201F8E0B6595FA20E4A212A2AD9A5046DAF29E8B35AD15CEE56A1A69F2A3A7B347A7C15E26E7A6760533C7A8E0D44
'  F6A201F8E0B6595FA20E4A212A2AD9A5046DAF29E8B35AD15CEE56A1A69F2A3A292CA61D03A85E1AC39B50D4DA71691E
'  We can see the output matches except for the last block, which is affected by the padding scheme.

'  If we are able to easily use ECB mode w/ the other system
'  we are trying to match, then eliminate the IV from the picture.
'  If the encryption matches in ECB mode, but not in CBC mode,
'  then we know all correct except for the IV.
'  For example, you can see how the IV changes everything with CBC mode,
'  but it's not used in ECB mode:
crypt.PaddingScheme = 0
crypt.CipherMode = "cbc"
Log("-- Only the IV differs, CBC mode produces different output. --")
crypt.SetEncodedIV(ivHex1, "hex")
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.SetEncodedIV(ivHex2, "hex")
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))

crypt.CipherMode = "ecb"
Log("-- Only the IV differs, ECB does not use the IV.  The outputs are the same. --")
crypt.SetEncodedIV(ivHex1, "hex")
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.SetEncodedIV(ivHex2, "hex")
Log(crypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello"))


'  If we can eliminate the padding scheme and IV from the degrees of freedom,
'  then the only remaining likely differences are (1) the secret key,
'  and (2) the input data itself.

'  The secret key is composed of binary bytes of exactly KeyLength bits.
'  For 256-bit AES encrytion, the key length is 256, and therefore the 
'  secret key is exactly 32 bytes.  (32 * 8 bits/byte = 256 bits)
'  If the secret key is derived from an arbitrary password string, then one must
'  exactly duplicate the derivation scheme (such as PBKDF2, for example)
'  The input bytes to the derivation scheme must also match.  For example,
'  is it the utf-8 byte representation of the password string that is used
'  as the starting point for the derivation, or perhaps utf-16, or ANSI (1 byte per char)?

'  Likewise, if the data being encrypted is a string, what byte representation of
'  the string is being encrypted?  If the bytes presented to the encryptor are different,
'  then the output is different.