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Disclaimer : basic crypto knowledge

So this is a question specificly about veracrypt , but theoreticly applies to any other encryption software which supports cipher cascades .

So if you use a cipher cascade with 3 cipheres e.g : AES(TWOFISH(SERPENT)) , then the key is divided in 3 different keys (hashes from KDF ) with 1/3 of the original key strength . So like a 8-10 charakter long password would not be considered secure right ?

But what if you use a keyfile like some mp3 file or some random generated stuff like an RSA cert. Veracrypt uses the first 1024 byte of the file so that would make a 4096 bit master key and a 1365 bit key for every cipher in the cascade if I understood that right. Which would be more than enough to be secure right?

Please correct every misunderstanding in my previous claims.

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What you would do in such a case is take a master key (which could be derived from a password) of a certain length (in your cases probably 128 or 256 bit long) and use a key derivation function (like HKDF) to derive three distinct keys of equal length which can be used in one algorithm each.

This way, each of the encryption keys will have the same strength as the original masterkey and the leakage of one of the encryption keys won't affect the other ones.

Veracrypt uses the first 1024 byte of the file so that would make a 4096 bit master key and a 1365 bit key for every cipher in the cascade if I understood that right.

Just taking 1024 byte of some file will yield something with the length of 8192 bit, but the key won't have a security level of that size, since the data you are using is probably not uniformly distributed.

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  • $\begingroup$ 1024 bytes = 8192 bits, unless I misunderstand what you are doing here. $\endgroup$ – bmm6o Jul 26 '17 at 18:55

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