| bio | website | |
|---|---|---|
| location | Wellington, New Zealand | |
| age | ||
| visits | member for | 1 year, 3 months |
| seen | 5 mins ago | |
| stats | profile views | 53 |
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Sep 19 |
answered | Which is more secure using a CSPRNG for a One-time pad, or AES? |
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Sep 15 |
awarded | Nice Answer |
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Sep 14 |
revised |
Why RSA can't handle numbers above 76? typo |
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Sep 14 |
revised |
Why RSA can't handle numbers above 76? added practical demonstration |
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Sep 14 |
answered | Why RSA can't handle numbers above 76? |
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Sep 13 |
comment |
fixing WEP by treating all messages as one long stream Note you need to be careful with this approach: if the key is compromised, you lose both forward and backward secrecy (which are conserved to some extent if you change keys, perhaps using a KDF). Of course, it depends how long the stream is - you never set up a new key to send three bytes, but you do change the key every few dozen MB or so. But I am unsure how this "fixes WEP"... |
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Sep 13 |
answered | Implementing AES encryption for firmware distribution system |
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Sep 11 |
revised |
modulus condition post latex'd |
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Sep 11 |
comment |
modulus condition By the way, did you mean 8k - 1 instead of 8(k - 1) in your inequality? Because the former requires the modulus to be 8k bits long, whereas the latter allows any bit length between 8k - 7 and 8k. |
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Sep 11 |
suggested | suggested edit on modulus condition |
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Sep 11 |
comment |
modulus condition I don't understand the question. These conditions are a mathematical way to describe the statement "the modulus n is 8k bits long". Try representing the numbers in binary, you'll see the conditions imply something about the binary representation of $n$! (hint: it forces one bit of $n$ to be 1...) |
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Sep 8 |
comment |
Can I combine two of SHA-3 candidates cryptography hash functions and obtain more secure Algorithm? What makes you believe Skein is not secure enough for your purposes? More isn't always better in terms of security (and can, in specific cases, actually be worse), and in general there is a practical limit, in that throwing more iterations/cascades essentially changes nothing. Note, though, that if done properly, you can combine hash functions into something "better", but it's often not really elegant nor efficient. Also, if CPU cycles do not matter at all, I suggest iterating Skein an infinite number of times :-) |
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Sep 6 |
comment |
Password crackers in CUDA Not appropriate to Crypto.SE (this is a website about theoretical cryptography), and furthermore product lists/requests are not on-topic on Stack Exchange. |
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Sep 5 |
comment |
Randomized stream cipher using multivariant quadratic equations @SimonJohnson Oh, right, I see, my bad - I will delete my previous comment as it serves no purpose. |
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Sep 5 |
comment |
Why RC4 used constant 24bits for initial vector Also note 40-bit keys are way too short to be secure nowadays, I could break one in a couple hours on the computer I'm on right now. You want 80 bits at the very least and 128-bit to be comfortable. |
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Sep 5 |
comment |
Why RC4 used constant 24bits for initial vector We must not be looking at the same stream cipher, to my knowledge RC4 uses no constant value other than the identity permutation (0, 1, 2, ..., 255), and has a key up to 2048 bits long. If this is a custom implementation which uses an IV, we need to have references. You might be talking about the WEP protocol, which uses a 24-bit IV, if so, mention it. |
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Sep 2 |
comment |
How to encrypt data and know it will be secure for at least a few decades? "Serpent alone should be safe for a few decades" That's one bold claim, and past history doesn't really support this conclusion. Care for some references? As for CSPRNG's, it would be better to hash their outputs together, as XOR'ing them can destroy entropy and in general do more harm than good. |
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Aug 31 |
comment |
No SHA-1 Collision? Yet SHA1 is broken? As for alternatives, SHA-3 is just around the corner, so hang in there! |
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Aug 29 |
comment |
Random Sequence Generator function No closed form function can describe a truly random permutation by definition. But if $N = 2^k$, then block ciphers come a significant fraction of the way. |
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Aug 28 |
comment |
Secure Hash Function based on AES Would adding a feedforward step to the AES permutation function like so $\mathrm{Encrypt}(X) = E(X) \oplus X$ fix this vulnerability? |