2,600 reputation
213
bio website
location
age
visits member for 1 year, 9 months
seen 18 hours ago

2d
comment Vigenère cipher: Security when key length and plaintext length are the same
Fixed. Something came up when I was writing the answer earlier.
2d
revised Vigenère cipher: Security when key length and plaintext length are the same
added 102 characters in body
2d
answered Vigenère cipher: Security when key length and plaintext length are the same
Jul
24
comment Protocol composition
You got a view recommendations already, so hopefully there was something useful for you. But anyway, reference recommendations are offtopic here.
Jul
24
comment Vigenère cipher: Security when key length and plaintext length are the same
If the key has the same length as the message and it used only once, then it is called a One-Time-Pad. And that is information theoretically secure. If you re-use the key, security is gone. And that's true for any key-length of Vigenere: Re-using a key means that security has left the building.
Jul
24
comment RSA with modulus n=p²q
That might be oversimplifying it. But either way you can get any modulus size you want, by picking primes in the correct size. As a rule of thumb: factoring $p^2q$ or $p'q'$ is only as hard as finding the smallest prime factor. Therefore, if you have two modulus of the same size, the normal RSA modulus is probably harder to factor than the one with one factor squared, because the lowest prime factor accounts only for $1/3$ of the total size.
Jul
23
comment RSA with modulus n=p²q
The "bigger numbers" argument is not valid, because you can set $p'$ and $q'$ to primes of arbitrary size. The exponentiation with $e$ is done in a group with $|n|$ bits, it doesn't matter if $n$ is a product of two primes, or the construction with one prime squared.
Jul
23
comment Advantages using Diffie-Hellman or Elgamal
One is a key agreement protocol, the other an encryption scheme. They are not interchangeable. However, they are two mechanisms based on the same computational problem, which is the CDH problem.
Jul
22
comment Ciphertext indistinguishability (IND-CPA) for symmetric cryptography
any encryption scheme which isn't IND-CPA can't be IND-CCA. The security game is basically the same, with additional access to a decryption oracle, which may be used on any ciphertext which isn't the challenge.
Jul
21
comment How to calculate complexity of ElGamal cryptosystem?
Your time complexity assumptions are correct for this algorithm, and space complexity is $O(1)$ in the number of variables or $O(n)$ if you consider the length of the numbers. However, the algorithm can be achieved faster, because of what poncho wrote in his answer. Basically, it is like calculating $a \cdot b$ as $a+a+a...+a$ with $b$ summands.
Jul
15
comment Leak-proof protocol: is such a thing possible?
There is no proof that no other communication channels exist. The only statement you can make about a system is "we don't know any other channel". And with an abstract model in mind, you might already have removed channels which can be used in reality. Timing, power, EM radiation, etc. are only some side channels. There are others like ordering of packets, length of messages, sound (for the air gaped example), choice of identifiers, etc. Even if you have e.g. something like a set of elements, in reality it will come as an ordered list of elements, and this can be used to transfer information.
Jul
15
answered Leak-proof protocol: is such a thing possible?
Jul
15
comment Determining if a stream of bits is statistically random
What e-sushi meant is, that there is no single test that is qualifies randomness as "good enough". Golomb tried that, which lead to LFSRs, which have some serious flaws. If you have some real randomness, you can test for the real statistical properties, uniform distribution, binomial distribution of runs, statistical independence, etc.
Jul
15
comment Protocol: Coin Flipping over phone
Since this looks like a homework question, I didn't want to spoil the answer right away. Even preimage resistance isn't needed, it just has to be a binding commitment scheme. Of course, preimage resistance provides some computational binding scheme.
Jul
15
comment Protocol: Coin Flipping over phone
What is "uncertainty" in this context? And to 2: No it is wrong. The function is not binding at all, even without knowledge of $n$.
Jul
15
comment A question about elliptic curves and finite fields in bilinear pairings
Formulas can be inserted here with common Latex notation, by using '$ $'.
Jul
1
comment Exactly two of the four roots must be greater than N/2
right... $p\neq q$. I wrote a comment, and not an answer, but I shouldn't have left it out. Anyway, you're making it more confusing anyone trying to learn something here, by throwing in other things. For example, $p \equiv q \equiv 3$ mod $4$ is not required for the theorem.
Jul
1
comment Exactly two of the four roots must be greater than N/2
This can easily be proven by the fact that with a square root $a$, there is also a square root $-a$. And if you look at this pair, one will be between 0 and $N/2$ and the other one between $N/2$ and $N$.
Jul
1
comment Exactly two of the four roots must be greater than N/2
It is correct for all primes greater than 2. There are (at least) 4 square roots if $\phi(N)$ is a multiple of 4. And for every uneven $p,q$, that's the case.
Jun
30
comment How to prove NIZK proof of knowledge?
The problem is, there is no $s$. Of course you can base $st_i$ on some value $s$ and reuse this. But you can not prove that to anyone. The protocol would have the very same statistical distribution as if you chose $st_i$ at random in every round. Assume e.g. the cyclic group $\mathbb{Z}_3^*=\{1,2\}$. Assume $s$ is from that group, too, but fixed, and $s'$ the other element (also fixed). Now it doesn't matter if I give you $t_i$, $s\cdot t_i$ or (s' \cdot t_i). In either case, if $t_i$ is chosen uniform, those 3 choices are uniform, too. But you can not prove that you used a fixed $s$ either.