Almost but not quite, a repetitive block of plain-text is encoded with the previous cipher-text as the IV (initialization vector) which sets the 'state' of the cipher. Then it is encoded with the same key for each block. Quote from RFC 3602.. http://www.faqs.org/rfcs/rfc3602.html
The IV is XOR'd with the first plaintext block before it is encrypted. Then for successive blocks, the previous ciphertext block is XOR'd with the current plaintext, before it is encrypted. About memory dumps, only the first block (16-bytes) is required to recover the key. Some aes implementations have the key and the iv. Some references always set iv as 0x0000000000000000 but it has the ability to be set as any 128-bit value. Even if we did have the plain-text we would need the key and the iv (if its used). -----Original Message----- From: MsTiFtS [mailto:[EMAIL PROTECTED] Sent: Wednesday, October 03, 2007 8:05 AM To: Hardware and developpement mailing list. Subject: Re: [Linux4nano-dev] Concerning a block cipher. Jeremy Prater schrieb: > > The 240byte position split in the various versions in the aupd verify > the block size is 16byte (128-bit). It is odd that the aupd and osos > filesizes are multiples of 2048 also, I thought it could be just a > coincidence in osos, but both files is odd. To use decrypting > functions in winhex goto edit | convert (ctrl + r) and it brings a box > up, also hackman hexeditor can do decryption. I use them both for > testing stuff. I don't really know where to go from here, aes-cbc is > difficult to get the plain-text from the cipher-text, I was looking at > 1g nano firmware to see about strings and stuff, but I doubt any of > that data made it into the 2g firmware. Later -- Jeremy > I guess these files are just padded to fill the 2K blocksize of the flash. Well, there are definitely a lot of strings and stuff we know must be in there, but we don't know where they are. I would have also guessed that there would be at least some zero padding, but I couldn't find any repetitive ciphertext blocks, no matter how hard I search. If I understood the workings of a CBC cipher correctly, repetitive blocks of plaintext must yield repetitive blocks of ciphertext, right? What ways are there to crack the key if you know plaintext AND ciphertext? Does that need a brute force attack? Thanks for your good and hard work ;) > > ------------------------------------------------------------------------ > > _______________________________________________ > Linux4nano-dev mailing list > [email protected] > https://mail.gna.org/listinfo/linux4nano-dev > http://www.linux4nano.org _______________________________________________ Linux4nano-dev mailing list [email protected] https://mail.gna.org/listinfo/linux4nano-dev http://www.linux4nano.org
