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comment Why are there only positive value points on an elliptic curve?
@OlegGryb: when talking about $Z/p$, the values $x$ and $x+p$ are considered "the same" (no matter what $x$ is). In this case, the value $A = -3$ is considered the same as $A = p-3$ (and $b^2c = -27$ is considered the same as $b^2c = p-27"). Now, $-3$ is not the standard representation of that particular field element; however it can be considered a shorthand for that element.
6h
answered Use of Lamport's Signature as a primitive
1d
reviewed Leave Open Is it possible to transform an x-coordinate of ephemeral elliptic curve point?
1d
reviewed Close Decoding Base64 encoded password with internal hashing
1d
reviewed Leave Open Why cant I use ECB with some obfuscation for random access data stored in say a DDR
2d
comment Encryption of plain text using stream cipher algorithm
What makes you think that the output of Rabbit is limited to128 bits? It generates 128 bits per iteration; however you're not limited to doing only 1 iteration.
Apr
23
comment Common Modulus Attack in RSA
If you're looking for extra credit, answer these questions: 1) how could an employee decrypt a message to anyone, and 2) (harder) how could an employee factor n (and thus generate anyone's private key)
Apr
23
comment Double-and-add/Montgomery VS blinding
Actually, it's not at all true that blinding is slower; at least in the case of multiplication of the generator. If we assume 256-bit EC, then computing xG by double-and-add takes 256 doubles and 256 adds; in contrast (if we precompute the tables) a multiply by the generator might take 64 adds and no doubles (say, by radix-16; actually, there are smarter ways to do it). Even if we compute rG + (x-r)G, that brings us to 129 adds, which is still more than twice as fast as the double-and-add method.
Apr
22
revised SSL certificate / SSL configuration - FIPS-197 compliance
Corrected typo
Apr
21
reviewed Leave Closed suitable method for encryption and decryption
Apr
21
reviewed Reviewed Decrypting an encrypted text file
Apr
21
reviewed Close AES encryption only taking 32 bits
Apr
21
comment AES encryption only taking 32 bits
At first glance, this code appears to be an AES implementation; the AES block cipher handles 128 bits at once. Why do you think that it handles only 32 bits?
Apr
21
comment Why rogued certificate from hash collision is harmful
What makes you think that the attacker cannot acquire the private key? Depending on how to generates the collision, the public key can be either a value he selected (in which case he can know the private key), or a known random value (and a nontrivial fraction of random 2048 bit values can be factored, being a smooth number times a large prime).
Apr
20
comment DHKE choice of private keys
@CodesInChaos: oops, you're right -- I typo'ed that - I meant to say $[1,2^{256}]$.
Apr
20
comment DHKE choice of private keys
@CodesInChaos: probably. Now, when I do DH, I pick a secret randomly from a range such as $[1, 2^{128}]$; that keeps the DLOG problem just as hard, makes the computation cheaper (and incidentally avoids hard cases such as $(p-1)/2$
Apr
20
answered DHKE choice of private keys
Apr
20
comment Is it ever unsafe to compress an EC point?
@CodesInChaos: actually, there's a deterministic solution in that case as well; it's a tad more complex than the 3 mod 4 solution, but still works. What's tougher is the 1 mod 8 case...
Apr
19
comment Length of prime number used in Pedersen Commitment
One requirement would be that $g$ and $h$ be a part of the same subgroup (otherwise, the commitee can deduce some information on $x$). Given that it's also important that no one know the discrete log $\log_g h$, it would appear wise to select $(p-1)/2$ prime (and select $g$ and $h$ to be random quadratic residues); alternatively, work in an elliptic curve with a prime order.
Apr
19
comment Plaintext block chaining, bad idea why?
Hint: what happens if you encrypt extremely redundant plaintext, say, 3 blocks of all the same character?