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2

Those numbers for the key sizes come from the bit length of the prime chosen for the finite field $F_p$. To clarify, there are no complex numbers used at all. The elliptic curve is over a quadratic extension of $F_p$, i.e. $K = F_{p^2}$, and a root $i$ of $x^2+1$ is chosen, so that elements of $K$ can be represented as uniquely as elements of $F_p \oplus i\...


2

Yes, PGP allows different-sized subkeys and subkeys are not derived of the main key, so it is possible to have the earlier key be a subkey of the new one. How to do it is off topic here, but this should get you started if you use GnuPG.


2

I am basing my answer on Cryptopals. The basic idea is that as {c0,c0+3,c0+6,…} have all been xor-ed with the same byte, the number of differing bits between c0 and c3 is the same as between p0 and p3. (this number is called the Hamming distance between two characters. Furthermore, the distance between [c0 c1 c2] and [c3 c4 c5] is the same as between [p0 ...


2

It will be uniformly random for such a simple statistical test. The problem is that you are treating the probability of bits having a particular state as independent of each other. You would need to look at the conditional probability distributions of certain bits being set given other bits. The entire joint distribution of output and input bits for a ...


6

I seriously doubt the standard for passwords these days being 8 characters. I would consider anything below 10 characters seriously flawed. I personally always use 20 character passwords. Often, encryption keys are used as such, while passwords are (or at least should be) strengthened with a KDF. The vulnerability of an n bits password or key depends on the ...


12

I know that humans would find it impossible to maintain a 128 bit password -- however, I wonder if there is some technical reason why a 52 bit password would not be as weak as a 52-bit encryption key for that matter. First, I would argue that 128 bits is not impossible to remember. My current password manager master password is almost 100 bits (6 words ...


0

The previous comment left a term out of the numerator. $$x = \frac{1.923 \times \sqrt[3]{L \times ln(2)} {\sqrt[3]{[ln(L \times ln(2))]^2}} - 4.69}{ln(2)}$$ I find this formula on page 92 of the NIST document that was previously mentioned: http://csrc.nist.gov/groups/STM/cmvp/documents/fips140-2/FIPS1402IG.pdf The GNU bc code to implement this equation ...


1

The security levels for RSA are based on the strongest known attacks against RSA compared to amount of processing that would be needed to break symmetric encryption algorithms. The equation NIST recommends to compute approximate length for key is found in FIPS 140-2 Implementation Guidance Question 7.5. It is: $x = \frac{1.923 \times \sqrt[3]{L \times ln(...


3

Short answer: Not enough. The AES algorithm is defined in the FIPS standard with keylenght of 128, 192 or 256 bits. So you cannot use directly a 56-bit key. One needs to have a key with the proper length to use the AES encryption algorithm. Data will be protected using AES-256 encryption with a 56-bit effective key length probably means that the key ...


3

If the effective key length is 56-bit, that means you have to enumerate only 56-bit. It's not secure at all. From Wikipedia: In February 1997, RSA Data Security ran a brute force competition with a $10,000 prize to demonstrate the weakness of 56-bit encryption; the contest was won four months later. In July 1998, a successful brute-force attack was ...


8

It would appear that this is as a consequence of the Wassenaar Arrangement (more detailed explanation). Basically, because cryptography falls under certain laws regarding munitions and arms trafficking there are restrictions on the strength of the cryptography you are allowed to sell (if the product is not a mass market product). It would appear that the ...


1

I have reformatted the above equation as a program for GNU bc (part of GNU coreutils, found on most Linux systems). GNU bc will be much easier to find than Mathematica (although it is quite eccentric). Here is the code: $ cat RSA-gnfs.bc #!/usr/bin/bc -l scale = 14 a = 1/3 b = 2/3 #print "RSA Key Length? " c = read() t = l( l(2 ^ c) ) # if b < 1, ...



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