Long time no see, saweet. Nice to see you again on ePvP's 9D section. You told it's easy to figure out the encryption keys. Can you tell - how?Quote:
Done. Searches tables for rows matching the Name column. Find Next wraps around when it reaches the end. Find Next table search order can be altered by manually selecting a table.
US Cipher Keys is 17 & 08Quote:
By the way - thanks for great update.
VN is: 07 & 11Quote:
KR Cipher Keys is 27 & 09
DE same as US
Dont know about VN yet
imma downloading it now anyway ><Quote:
Thank you very much :handsdown: it's hella easier now.Quote:
The 2nd key is obvious because it's only used in 2 byte characters and only ever used in the 2nd byte position. They use XOR encryption so it becomes very obvious when null is XOR'd, because 0 XOR N = N. Thus N is the 2nd key.Quote:
By the way - thanks for great update.
>>> [hex(i) for i in range(256) if 0x4d ^ i == 0x5a] ['0x17']
>>> ciphertext = bytearray([0x5A, 0x78, 0x75, 0x5A, 0x78, 0x73, 0x72, 0x7B, 0x48, 0x4F, 0x5A, 0x44]) >>> bytearray([ciphertext[i] ^ 0x17 for i in range(len(ciphertext))]) bytearray(b'MobModel_XMS') >>>
greate work brb ... but can you explain how to program that 2 cipher XOR encrytion ? :DQuote:
The 2nd key is obvious because it's only used in 2 byte characters and only ever used in the 2nd byte position. They use XOR encryption so it becomes very obvious when null is XOR'd, because 0 XOR N = N. Thus N is the 2nd key.
The 1st key can be learned by knowing the most common character in a string, with that you can guess the possible keys. What I did was grab all the table names in both an unencrypted and an encrypted XSD and compared string length to find matches as possible candidates for the real value. MobModel_XMS was a good choice because we have 3 Ms in a single short string. With that I found ZxuZxsr{HOZD of the same string length in the encrypted file. Notice the pattern, 3 Zs in the same position?
If you know Z (5A) in a string is really M (4D), all we have to do is figure out what possible XORs of 4D will equal 5A to get a list of possible keys. I did this in python and had the key in a fraction of a second:
Lets see what happens (ciphertext is "ZxuZxsr{HOZD" in hex)Code:>>> [hex(i) for i in range(256) if 0x4d ^ i == 0x5a] ['0x17']
ViolaCode:>>> ciphertext = bytearray([0x5A, 0x78, 0x75, 0x5A, 0x78, 0x73, 0x72, 0x7B, 0x48, 0x4F, 0x5A, 0x44]) >>> bytearray([ciphertext[i] ^ 0x17 for i in range(len(ciphertext))]) bytearray(b'MobModel_XMS') >>>
Wow, great job. That's the one from Rage Zone? Or VN one?Quote:
it's working great !
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VN .Quote:
well that's CuuLongViet... handling 9D for VietnameseQuote:
---> VN . ---> YesQuote:
Arale
Can you upload the XSD so I can take a look?Quote:
Here it is:Quote: