Hack Encrypted Dmg Password Could Be

Hack Encrypted Dmg Password Could Be

Because the AES encryption scrambles the data contained in a zip file, the password could be found by unscrambling that data correctly.

However, the condition for this type of cracking is that the zip file contains known file types which have a determined data format and places certain data in certain locations in the file.

For instance, a 3ds file always starts with a header containing the bytes 4D4D.

Hence a list of possible passwords to rearrange the sequence begin to be narrowed down as the cracking progresses.

So is this method plausible or is there something wrong with it, like maybe the actual address of the bytes in the zip is different from the address in the unzipped file.

In this scenario, it is not important to consider the expected number of AES operations for brute-force search of the AES key ( 2127 for AES-128, 2255 for AES-256) because thats not the best (nor a credible) attack strategy.

Users vary widely on that, and 123456 will be easier to find than MdSm2aZt5uZ5.

That varies widely with the context, and motivation of users.

A password-based key derivation function should be used here, parametrized so that it adds a sizable cost.

For example, if the password is 6 random decimal digits ( Elog2(106)approx19.9,textbit ), and the PBKDF plaintext testing uses 0.1 second of CPU time (300000 CPU cycles 3GHz), a competent attacker will use an average of 106times10-1250000,texts of CPU time to find a password, that is less than 2 hours of wall clock time with a single 8-core PC, minutes for a powerful adversary.

You dont even have enough storage to store even 1 of possibilities of results.

And usually password itself is weakest point (if there is password), not anything else.

So if it was encrypted using password, best bet is that hopefully password is weak.

See How do I detect a failed AES-256 decryption programmatically for a working example based on AES-256-CBC.

However AES isnt like that in that its nth byte of cipher text is not solely derived from the nth byte of zip file and a key.

How much else is input into a given byte depends on the mode of AES, but suffice to say with accepted modes its at least all bytes that have gone before, the iv, the key, and adjacent bytes in the block.

Oh, and to guess you (typically) have to redo the whole encryption, so guessing the remainder is relatively slow, even in the case of a single unknown appended log line on a known log.

Provide details and share your research But avoid Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

MathJax reference.

To learn more, see our tips on writing great answers.

Not the answer youre looking for Browse other questions tagged encryption aes passwords or ask your own question.