# Tag Info

6

Well, how resistant to attack would depend on what security properties you would need from it. There are three standard assumptions we can make about a hash function: Given a hash value, it is difficult to find an image that hashes to that value; this is known as preimage resistance Given a image that hashes to a specific value, it is difficult to find ...

0

Combining two hash functions is exactly what the PRF does in the original TLS 1.1 specification. Half of the message (secret) is put through MD5 while the other half is put through SHA-1, and the two outputs are XOR'd together: $PRF(secret, label, seed) = P_{MD5}(S1, label + seed) \; \otimes \; P_{SHA-1}(S2, label + seed);$ TLS's PRF is created by ...

0

Maybe. If the algorithms are individually broken, and your design is open (i.e. no “security” by obscurity), then chances are your scheme will also be broken more easily. In the case of hash algorithms, “broken” usually means finding collisions or second preimages, which will differ between algorithms, so yes, this will work. With the caveat above, of ...

3

At the time of answering, the question was can insecure algorithms be combined to form a secure algorithm? Yes: Think of any secure round-based block cipher. The independent rounds are not secure, but put together the overall cipher is. I think my favourite example is (currently) the Even Mansour cipher, which combines two xor operations and one unkeyed ...

0

Yes, that is sufficient. I realise this isn't the most helpful of answers, but I'm not quite sure how we're supposed to deal with questions that lend themselves so neatly to one word answers. Certainly it seems wrong that something that has been solves should continue to sit in the 'unanswered questions' column!

5

Computationally indistinguishable typically means that your adversary is computationally bounded and that because of this they cannot distingush between, for example, two messages. For example, say you encrypt (with proper padding) the messages $0$ and $1$ using RSA and send them to the adversary. We would not want the adversary to be able to distinguish ...

2

Can we state that $Adv_{G0} \le Adv_{G_1}$ ? Absolutely. Let us call $S0$ the strategy that $A$ can use in game $G0$ to achieve advantage $Adv_{G0}$; we notice that $A$ can also use strategy $S0$ to achieve that exact same advantage in $G1$, and hence we see that the maximum advantage he can achieve in $G1$ must be at least as large as in $G0$ Can ...

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