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7

EEE and EDE are effectively the same in terms of security. EDE is used because it is "backwards compatible:" by setting all three keys to be the same, it becomes equivalent to just single encryption (E) with that key.


5

I won't say someone would be able to break it 'easily'; however it won't be anywhere as difficult as with a true 128 bit cipher (or even 120 bit cipher; your construction ignores 8 of the key bits). Here's an outline of how the attack would work: we assume we know the plaintext and the ciphertext, and are trying to recover the key. When we do is encrypt ...


3

Start with “Shamir's Secret Sharing” concepts… Abstract. In this paper we show how to divide data D into n pieces in such a way that D is easily reconstructable from any k pieces, but even complete knowledge of k - 1 pieces reveals absolutely no information about D. This technique enables the construction of robust key management schemes for ...


3

Does re-encrypting the same value with multiple keys reduce security? The answer is "it depends"; there are some attack models and encryption methods where the security is reduced, there are other cases where there appears to be no security reduction. Let us go through some models where we actually see a security reduction: Plaintext guessing attack ...


3

What you are asking appears to be 'is AES commutative'? The short answer to which is no: encrypting with AES with key 1 then key 2 will not (generally) give the same output as encrypting with key 2 then key 1, which is what would be required for naive implementation. However, there are modes in which AES can be used which would be commutative. For example, ...


3

In general (especially without knowledge what encryption you consider), it's not possible to detect "correct decryption of one layer", if that's all you have and this "middle ciphertext" is not in a specific format. However, from today's point of view this is almost entirely irrelevant, because stronger attacks are considered: Kerckhoff's principle states ...


3

The actual security would probably be about 65 bits. A meet-in-the-middle attack can be used to find the keys of both ciphers in less time than naive brute force. The attack would decrypt the ciphertext with all the 64 bit l keys of the outer cipher, encrypt the plaintext with all the 56 bit keys of the inner cipher, then look for matches. It only requires ...


2

What you propose is called Double Encryption. With two independent keys, it is vulnerable to meet-in-the-middle attacks as described in another comment. I just add that this attack can be performed almost memoryless. Details are in the answer to similar question about Double-DES.


2

To begin with 4: Remember Kerckhoff's principle. You should always assume that the attacker knows which algorithm is used to encrypt your data. All the algorithms used in practice are designed to be secure under this assumption, so you should consider that hiding the algorithm from the attacker is superfluous. But as a hypothetical... I can't think of any ...


2

In short: does storing these encrypted files in an encrypted partition/folder create the same potential weaknesses as cascade encryption? If not, what of encrypting a .tar of encrypted documents? An encrypted file inside an encrypted container is a cascade of ciphers, almost by definition. Is that a problem? Not really. A cascade of ciphers shouldn't ...


1

Two things to consider: encryption and authentication. In general you can only say that a cascade of ciphers is as secure as its weakest link. If the encryption in NaCL had a side-channel attack, it might leak information about the plaintext, whether or not the ciphertext is sent through TLS. Authentication, on the other hand, is additive. If you can ...


1

If you have a secret key (256-bits) shared between the two systems that see the entity identifiers, you can use HMAC-SHA256 to map entity identifiers to a random string. Under the assumption that HMAC-SHA256 security is good (which is widely believed to be a reasonable one), this is just as secure as having generated a truly random mapping, but requires ...


1

If by "encrypted" you mean generating a keystream, then what you propose is to use in the CTR mode $$ C_i = P_i \oplus F_K(IV||i) $$ the following function $F$: $$ F_{K_1||K_2||K_3} = E_{K_1}\oplus E_{K_2} \oplus E_{K_3}. $$ This is secure as long as you ensure that for each key all the used IVs are different (i.e. are nonces). As mentioned in another ...



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