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Hot answers tagged multiple-encryption

10

The usual method to do this is to turn the block cipher into a stream cipher. In that way the ciphertext is generated by XOR'ing the plaintext with a generated key stream. This key stream in turn is generated by the mode of operation that turns the block cipher into a key stream. There are several of these modes, but CTR mode of operation is most often used ...

6

I don't know about computing things in parallel, so I will ignore that part of the question. First, please note that the encryption algorithm is rarely the the weak point of the security. It is far more likely that you will have problems with the implementation, some spyware installed on your computer, a weak password (If you use qwerty as your password, ...

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

There is a very simple, completely generic solution, that unlike the other solution doesn't assume anything about how the two encryption schemes work internally (e.g., that they are built from block ciphers or have pseudorandom ciphertexts): given a message $m$, choose a uniformly random $m_1$ of the same length and let $m_2 = m \oplus m_1$. Then encrypt ...

3

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.

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

It seems the result is specifically for multiple encryption with a single cipher (like in 3DES). It probably applies for different ciphers as well, but key and block sizes would need to be equal. You might get a lower bound by using the minimum key size, but don't quote me on that. However, I don't think this is really relevant for a practical ...

2

EDIT: I think what I do is provide a little more than a hint so don't read further if that is an issue. That is: Resulting in: Alternatively: and

2

At its simplest, to encrypt a message for $n$ different recipients, you could just make $n$ copies of the message, encrypt each one with a different key, and join the encrypted messages together into a single long ciphertext. Of course, the disadvantage of this scheme is that the ciphertext length grows linearly with the number of recipients. To avoid ...

2

They rely on problems not so different as you might think. They are based either in the factoring problem or in the discrete logarithm problem, which have a deep connection between each other. Once you have an algorithm that can efficiently solve one, you most likely would be able to adapt it to reproduce an answer for the other in polynomial time. Thus ...

1

If the ciphers are different, with independent keys, you can say that it is at least as strong as the first cipher. If the ciphers commute, like with stream ciphers, you can even say that it is at least as strong as the strongest. See Cascade Ciphers: The Importance of Being First. That's really all you can say in general. In practice, the combinations you ...

1

In general, all you can say is it can be as weak as the weakest encryption layer, if you're lucky. Edit: It can also be even weaker, for badly chosen components that cancel out some mathematically desirable properties, as pointed out in the comment.

1

For a crypto algorithm that acts like a group, the first thing that comes to mind is Pohlig-Hellman. In this method, we have a large prime $p$, and define: $$E_A(Data) = Data^A \bmod p$$ (with $A$ relatively prime to $p-1$) This has the property that $E_B(E_A(Data)) = E_{A \times B \bmod p-1}(Data)$; however it has the security properties you're looking ...

1

This technique is known as Ciphertext stealing. Ciphertext stealing avoids padding, but only works if the total message size is bigger than one block. Ciphertext stealing secure in principle, but as @mikeazo already pointed out using ECB and using 64 bit block ciphers is generally a bad idea. There are fancier length preserving encryption schemes. FFX mode ...

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