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Mar
17
reviewed Reject suggested edit on Randomness test question from FIPS 140-1 and comparison with 140-2
Mar
14
comment Is this a pseudo random function (PRF)? F(k,x) = f(k,x) - f(k,x-1)
@fgrieu This is getting way too long for the comments. But you have some basic misunderstanding about the proof technique here. IF the construction from the proof WOULD work against $f$, THEN the proof would lead to the conclusion that F'' is a PRF. But it does not.
Mar
14
comment Is this a pseudo random function (PRF)? F(k,x) = f(k,x) - f(k,x-1)
@fgrieu The reasoning (which while correct is missing a crucial argument about the distribution of sums of uniformly and independently distributed values) simply does not apply to your example. The reasoning goes as follows: To simulate the oracle for a dist. against $F$ we simply follow the construction of $F$, but use the oracle (containing either $f$ or a random function) instead of $f$. Now the following holds: (1) If the oracle is $f$, then we perfectly simulate $F$. (2) If the oracle is a random function then we perfectly simulate a random function. The second part fails for $F''$.
Mar
14
comment Is this a pseudo random function (PRF)? F(k,x) = f(k,x) - f(k,x-1)
@fgrieu Yes, the proof sketch is missing the argument, why it is that the reduction perfectly simulates a random function in the case that it itself is given access to a random function. But my point is that the proof does NOT need " profound change" besides that. The reduction trivially fails for the examples you give, because the distribution of outputs is incorrect for the random case. In particular your distinguisher would always output the same bit when run as a subroutine of the reduction, because the condition you check would hold in both cases.
Mar
13
comment Is this a pseudo random function (PRF)? F(k,x) = f(k,x) - f(k,x-1)
@fgrieu I don't see how the same argument would work for your examples. If we assume an analogous simulation of the oracle for $F'$, it is easy to see that the simulation of the random case simply fails. That is, in both cases (PRF and random function) the simulated oracle exhibits the bias you mention. The hard part in proving such PRF constructions secure is always to argue why the simulation does not fail in the random case. Here, such an argument can be made for the original function, but not for your examples.
Mar
3
reviewed Approve suggested edit on What RC4 key value will completely invert $S$ after initial permutation?
Feb
28
reviewed Approve suggested edit on Cryptographic Protocol using NaCl
Feb
21
revised Prove that two MACs with incremendal PRF application are not secure
fixed index of key
Feb
21
comment Prove that two MACs with incremendal PRF application are not secure
Well, to show that they are not secure, you have to present an adversary that is able to forge them. For the first one, think about how, given a message and a tag, you can find another message for which the same tag will verify. For the second one, think about how you might be able to compute a tag for a longer message without knowing the key.
Feb
20
reviewed Leave Closed Complexity class of an idealised version of Bitcoin's proof-of-work (hashcash)?
Feb
20
reviewed Reopen RSA vs El Gamal digital signature. Which is more secure?
Feb
20
reviewed Leave Closed Why does a one-time-pad key have to be at least as long as a message?
Feb
20
reviewed Close Finding if exponent share is present in dlog instance
Feb
20
reviewed Leave Open Signature based on public key cryptography and forgery
Feb
20
reviewed Approve suggested edit on What makes LSBit steganography detectable? And what would help in concealing it?
Feb
17
comment Proving that a function is not a OWF (One-way-function)
That information is exactly what your definition is missing. But in general when it comes to crypto we are interested in average-case one-way functions. Which means the probability is taken over the choice of x and therefore it is fine if the function is easy to invert for some inputs, as long as it is hard on average.
Feb
13
awarded  Civic Duty
Feb
10
comment Signature based on public key cryptography and forgery
It depends on what you mean. If the secret key is necessary to find the second document, then it is fine. If you know the secret key, you can already sign anything you want. However, if it is possible to find the second document given only the original message, signature and public key, then the scheme is trivially forgeable.
Feb
9
comment RSA assumption and cryptography
This sounds a lot like homework. But it sounds like you might want to check out random self-reducibility.
Feb
3
comment Going Blind, Group or Ring?
I still don't see how Alice would prove that she does NOT know the secret key. There is nothing stopping Alice from knowing a key for a different identity.