meta user
network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 2 months |
| seen | 4 hours ago | |
| stats | profile views | 3 |
|
May 8 |
awarded | Informed |
|
Apr 19 |
comment |
PKC McEliece + $S$ + $P$ As per your own Wikipedia link: $S$ is simply any invertible binary $k \times k$ matrix, and $P$ is any $n \times n$ permutation matrix. What is it that you find unclear? |
|
Mar 17 |
awarded | Yearling |
|
Mar 16 |
accepted | Proof of the standard pseudorandom generator + XOR encryption scheme in Goldreich |
|
Mar 16 |
comment |
Proof of the standard pseudorandom generator + XOR encryption scheme in Goldreich Exactly the answer I was looking for, thx! It's funny though: I had done exactly the same thing as you, including the consideration of the "triangle" inequality. But, when I got to your last equation, I couldn't figure out where that $1/2$ factor should come from. However, its good too see that I wasn't to far of :) |
|
Mar 16 |
asked | Proof of the standard pseudorandom generator + XOR encryption scheme in Goldreich |
|
Feb 28 |
awarded | Commentator |
|
Feb 28 |
comment |
Is SHA-1 still practical secure under specific scenarios? I think it should be mentioned that the security guarantees given by that HMAC-paper is disputed. See Another Look at HMAC and the youtube presentation Another look at provable security. While controversial, they do bring up relevant points to the practical (security) merit of the Bellare paper. |
|
Feb 16 |
comment |
How can an S-Box be reversed? I think you should make it clear that S-boxes often ARE invertible. For instance, the S-box in AES is invertible. @Liam Inverting S-boxes can be very easy: you simply create a lookup table that reverse all the possible substitutions of the S-box. E.g. if the S-box maps 0xA5 to 0x3F (this would be an 8x8 S-box), then the inverse transformation would map 0x3F to 0xA5. Thus, you simply enumerate all the possible values the S-box can have, and create an inverse table that "undoes" all those transformations (this effectively limits how large the S-boxes can be in practice). |
|
Feb 15 |
comment |
Encrypting a key with the same key @madhukar2k2 "I tried searching for an answer for the above, but couldnt not find one." - try looking at circular security, it might give you some answers. Additionally, I've added some things in my answer below. |
|
Feb 15 |
answered | Encrypting a key with the same key |
|
Feb 1 |
revised |
Questions on rank-attacks in Multivariate Cryptography Added another question and changed the title of the question |
|
Jan 29 |
revised |
Questions on rank-attacks in Multivariate Cryptography added 1 characters in body |
|
Jan 28 |
asked | Questions on rank-attacks in Multivariate Cryptography |
|
Jan 23 |
comment |
Why is MixColumns omitted from the last round of AES? Ah, apparently my vote has been locked in, so I'm unfortunately not able to change my downvote. Sorry about that :( |
|
Jan 23 |
comment |
Why is MixColumns omitted from the last round of AES? Ok, fair enough. I will remove my downvote when the timeout has passed. Still, since many modes of operation allows you get away with only the encryption direction, I don't understand why an equal RoundFunction wouldn't be preferable for all rounds (Ref. Fixee's comment about the hassle's of the last round omission, in his answer to PulpSpy's answer) |
|
Jan 22 |
awarded | Critic |
|
Jan 22 |
comment |
Why is MixColumns omitted from the last round of AES? I feel this answer is not answering the question (even if accepted). Let me reformulate what I would like to know: If our goal is to make all rounds in the encryption/decryption equal (I stress that the round function will be different in enc/dec), what is the problem of adding MixColumns in the last round? As far as I can see, AES-ENC would then simply become: 1. AddKey 2. repeat RoundFunc 10 times. And for AES-DEC we get: 1. AddKey (last RndKey) 2. repeat InvRoundFunc' 10 times, where InvRoundFunc' = 1. InvMixColumns 2. InvShift 3. InvSubBytes 4. AddRndKey. Why is this not possible? |
|
Jan 7 |
comment |
Why WPA-PSK do not uses Diffie-Hellman key exchange? @Michael Yes, if you know the 4-w-h then you can do that. However, I think you are confusing the situation you are in: in WPA-PSK, the pre-shared key is the ONLY (true) secret in the scheme. It is not obvious how you could allow all the participants to know this key, while at the same time not being able to read each others messages. A stronger variant is WPA-EAP, where all participants use their own individual secret. Here the 4-w-h will not reveal anything. However, this solution is more complex to set up, and usually not used for home-networks. |
|
Jan 7 |
comment |
Why WPA-PSK do not uses Diffie-Hellman key exchange? @Steel City Hacker Yes, I'm aware of that, but I considered this to be details not very relevant for the question at hand. Thus I only provided the link, where those details are included. |
