First off, let me apologize for joining a forums and posting a question right off the bat! I'm going to read the C++/C/Assembler/Perl sections right after this to see if I can offer my semy-professional advice :p
Now to my problem. I'm trying to find the algorithm to emulate this behavior. I don't need to know the key to this, though it should be easy to find once you know the algorithm, because I know the key changes with each packet. I just need to make an emulator for this.
42 02 98 FD 4D BE 39 ED 8D 3F 7A 37 92 FB D3 88 | B.˜ýM¾9í??z7’ûÓˆ
89 06 10 7F B2 D3 98 C9 AD 2D 3B 54 99 F8 B3 E2 | ‰..²Ó˜É*-;T™ø³â
86 06 10 7F B2 D3 98 C9 AD 2D 3B 54 99 F8 B3 E2 | †..²Ó˜É*-;T™ø³â
86 06 10 7F B2 D3 98 C9 AD 2D 3B 35 F8 86 CD 83 | †..²Ó˜É*-;5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 86 CD 83 | çxn.Ó*æ¨ÌSE5ø†Íƒ
E7 78 6E 1E D3 AD E6 A8 CC 53 45 35 F8 F8 B3 83 | çxn.Ó*æ¨ÌSE5øø³ƒ
E7 06 | ç.
Now what I know of this packet, it starts off as a header and I'm not exactly sure what information is in the header there. Second is a set of 20 chars of "a" and that's followed by 256 chars of "~" which I chose do to binary 111111110, hoping that might make it easier. There's a defenate pattern, but I see that 1 unencrypted byte equals 2 encrypted bytes. Not sure what this means, but I'm thinking it's an XOR operation and maybe a shift because it's extremely repeating so nothing tough like AES or DES or anything.
Sorry for the jumbled description of what I'm trying to do, but I'm not exactly great with cryptography so I don't exactly know where to start. Even any simple observations about it that I might miss would be greatly appreciated.
Sep 7th, 2005, 12:37 AM
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