(CkPython) Tips on Matching Encryption with another System

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

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import chilkat

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

crypt = chilkat.CkCrypt2()

# Let's examine 256-bit AES encryption in CBC mode.
# CBC mode is Cipher Block Chaining, and it uses an IV (initialization vector)
crypt.put_CryptAlgorithm("aes")
crypt.put_CipherMode("cbc")
crypt.put_KeyLength(256)
crypt.put_PaddingScheme(0)
ivHex1 = "000102030405060708090A0B0C0D0E0F"
ivHex2 = "FF0102030405060708090A0B0C0D0E0F"
crypt.SetEncodedIV(ivHex1,"hex")
keyHex = "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.put_EncodingMode("hex")

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

# Let's change only the padding scheme.
crypt.put_PaddingScheme(3)

# Encrypt again:
print(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:
print("-- Only the padding scheme differs --")
crypt.put_PaddingScheme(0)
print(crypt.encryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.put_PaddingScheme(3)
print(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.put_PaddingScheme(0)
crypt.put_CipherMode("cbc")
print("-- Only the IV differs, CBC mode produces different output. --")
crypt.SetEncodedIV(ivHex1,"hex")
print(crypt.encryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.SetEncodedIV(ivHex2,"hex")
print(crypt.encryptStringENC("HelloHelloHelloHelloHelloHelloHello"))

crypt.put_CipherMode("ecb")
print("-- Only the IV differs, ECB does not use the IV.  The outputs are the same. --")
crypt.SetEncodedIV(ivHex1,"hex")
print(crypt.encryptStringENC("HelloHelloHelloHelloHelloHelloHello"))
crypt.SetEncodedIV(ivHex2,"hex")
print(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.