(AutoIt) Tips on Matching Encryption with another System

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

Chilkat ActiveX Downloads ActiveX for 32-bit and 64-bit Windows

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

$oCrypt = ObjCreate("Chilkat.Crypt2")

; Let's examine 256-bit AES encryption in CBC mode.
; CBC mode is Cipher Block Chaining, and it uses an IV (initialization vector)
$oCrypt.CryptAlgorithm = "aes"
$oCrypt.CipherMode = "cbc"
$oCrypt.KeyLength = 256
$oCrypt.PaddingScheme = 0
Local $sIvHex1 = "000102030405060708090A0B0C0D0E0F"
Local $sIvHex2 = "FF0102030405060708090A0B0C0D0E0F"
$oCrypt.SetEncodedIV $sIvHex1,"hex"
Local $sKeyHex = "000102030405060708090A0B0C0D0E0F101112131415161718191A1B1C1D1E1F"
$oCrypt.SetEncodedKey $sKeyHex,"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.
$oCrypt.EncodingMode = "hex"

; Encrypt something small:
ConsoleWrite($oCrypt.EncryptStringENC("Hello") & @CRLF)
; The result is 5B827AB3B4F9F2292C2B74C8A6C99A3D
; This 16 bytes -- exactly one AES encryption block.

; Let's change only the padding scheme.
$oCrypt.PaddingScheme = 3

; Encrypt again:
ConsoleWrite($oCrypt.EncryptStringENC("Hello") & @CRLF)
; 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:
ConsoleWrite("-- Only the padding scheme differs --" & @CRLF)
$oCrypt.PaddingScheme = 0
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)
$oCrypt.PaddingScheme = 3
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)

; 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:
$oCrypt.PaddingScheme = 0
$oCrypt.CipherMode = "cbc"
ConsoleWrite("-- Only the IV differs, CBC mode produces different output. --" & @CRLF)
$oCrypt.SetEncodedIV $sIvHex1,"hex"
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)
$oCrypt.SetEncodedIV $sIvHex2,"hex"
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)

$oCrypt.CipherMode = "ecb"
ConsoleWrite("-- Only the IV differs, ECB does not use the IV.  The outputs are the same. --" & @CRLF)
$oCrypt.SetEncodedIV $sIvHex1,"hex"
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)
$oCrypt.SetEncodedIV $sIvHex2,"hex"
ConsoleWrite($oCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello") & @CRLF)

; 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.