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Muslix64 Returns: Blu-ray Disc This Time

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  • Muslix64 Returns: Blu-ray Disc This Time

    Link to Doom9 thread: http://tinyurl.com/2cftqs

    In less that 24 hours, without any Blu-Ray equipment, but with the help of Janvitos, I managed to decrypt and play a Blu-Ray media file using my known-plaintext attack...

    The file from the movie "Lord of war", play well with VideoLan.

    Janvitos gave me few files on the BD disc and a memory dump...
    Jerry Jones

  • #2
    (cue drums)...another one bites the dust....
    Dr. Mordrid
    ----------------------------
    An elephant is a mouse built to government specifications.

    I carry a gun because I can't throw a rock 1,250 fps

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    • #3
      More from muslix64:

      Many people ask me more details about the known-plaintext attack. This is a very basic, but powerfull crypto attack that I have used to decrypt both format.

      After reading posts of people trying to get the keys in memory, I realized, I have a different way of looking into the problem.

      A lot of people try to attack the software, I'm attacking the data!

      So I spent more time analysing the data, to look for patterns or something special to mount my known-plaintext attack. Because I know the keys are unprotected in memory, I can skip all the painfull process of code reversal.

      I don't have any Blu-Ray equipment but I was able to recover the keys anyways... because I had access to a memory dump file and a media file.


      To give you an example, let's take the Blu-Ray case.

      First, I had to read the documentation about the media file format.

      In the case of Blu-Ray, the media files are divided in blocks called "Aligned unit". Let's simply call them "Unit" for short. A Unit is a block of 6144 bytes. The first 16 bytes are unencrypted, and the rest are encrypted using AES in CBC mode.

      A unit is composed of 32 blocks called "MPEG source packet". Each packet is 192 bytes long. The first 16 bytes of the first MPEG source packet of a Unit are decrypted.

      Just to see the decrypted part of the packet, I have printed a few. Have a look:

      D13BF428474000100000B0110000C100
      D13C5DE84710111C6E3468D1861B8D1A
      D13CC7A84710111CE3468D1861B8D1A3
      D13D31684710111C1A346186E3468D18
      D13D9B284710111C6186E3468D1861B8
      D13E04E84710111C8D1861B8D1A34618
      D13E6EA84710111CD1861B8D1A346186
      D13ED8684710111C186E3468D1861B8D
      D14D57924710111CFCC810FE80107F08
      D14DC1524710111C1007647E401C002E
      D14E2B124710111C8001880350400300
      D14E94D24710111C007690DE581426A3
      D14EFE924710111C80800E8081F9E081
      D14F68524710111CA01300C007408C00
      D14FD2124710111C005200B002E00D49

      Do you see something special? Do you see any pattern?

      The first byte is always D1 and the 5th byte is always 47. Can we use that to mount the known-plaintext attack? Of course!

      Because we know we have multiple MPEG source packet inside a Unit, we know the decrypted version of the unit at position 192 will probably look like the sequences shown above.

      In most cases, the know-plaintext attack is in fact a guessed-plaintext attack. We "assume" the data will look like something we "guessed" when decrypted. Most of the time, it works!

      Knowing that, all you have to do, is to write a small program that scan a memory dump file, that comes from of a software player while it was playing the movie. The key is in that file, you have to locate it.

      You just have to decrypt the first 2 MPEG source packets of the first unit until, you find a key that decrypt to something like:

      D1??????47?????????????????????? at position 192.

      That's it!

      I also do something similar for the HD-DVD format.

      Once you know the value and the position of the key in memory, you can do like people are doing here. Use "memory landmark" to locate the key.

      Any questions?
      Link: http://tinyurl.com/2labaq

      Amazing.

      Jerry Jones

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