New answers tagged

2

Around and about one hundred years ago, your idea would surely have made sense… but nowadays, modern technology and evolved cryptanalytic techniques are too smart to have a real problem coping with something like that. (Also see my related answer to “Why was the Navajo code not broken by the Japanese in WWII?”) Even when we completely ignore Kerckhoffs’ ...


8

Historically, there did exist a benefit to using a language that the adversary was not familiar with. The name for this is code talkers, and the most famous ones (at least in the USA) are the Navajo code talkers of World War II. The idea was to defeat attacks that relied on statistics about the language used in the plaintext. In modern cryptography, ...


4

If you're referring to a classical cipher, it might complicate frequency analysis and other such techniques. For a modern cipher, it makes no difference. Modern ciphers operate on arbitrary patterns of information. Ideally, the ciphertext of a modern cipher should have no relation of any kind to the associated plaintext, other then the key.


3

The actual "encryption" is done on this line: mysecretmessage[i] ^= ((mysecretvalue>>(8*(i%4)))&255); Clearly, this line XORs every byte (or at least, every element; but it makes sense to assume that this is indeed a byte array) of mysecretmessage with some value derived from mysecretvalue and the byte counter i. So what does the expression ((...


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 ...


1

"Guessed ID" means ID that the oracle guesses the attacker algorithm $A$ will attack.


4

Disclaimer: I'm answering without any knowledge on the content of the paper in question. Why there are numbers in this Figure 1, that are placed in vertical form? Because otherwise the figure would not fit into the page width (or the authors would have to use unreadably small font). How to interpret this axis X? I imagine that are interval times ...


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 ...


0

I think I can spot an implicit assumption that you're making that could easily trip this up: Assumption: None of the ciphers is its own inverse. Counterexample: Stream ciphers are a popular class of algorithms where encryption and decryption are the same function: encyrpt(key, encrypt(key, plaintext) = decrypt(key, encrypt(key, plaintext)) = plaintext. ...


0

I am no expert in cryptography, so this method of encryption sounded pretty good until you explained how the password was the only thing that seemed to matter. While this does mean that if someone gets the encrypted password it will be difficult to decrypt. It also means that this method would be very prone to brute force attacks. This is why RSA is so ...


0

TL;DR: As of 2015 quantum computing is not an immediate threat to RSA. But it is a long-term threat (decades away). Whether you should be worried depends whether you are interested in keeping the secrets for so long (2035, if you want a pessimistic guesstimate). As the other answers already say, D-Wave machines ARE not generic quantum computers and will ...


2

This is not very secure. You directly leak the symbol distribution, because only the order of symbols changes. For short enough messages this allows easy decryption – e.g. "dr olllWeoH" is quite clearly "Hello World". Even for long messages or binary values, the fact that you leak e.g. a crucial byte may be enough. You also have not defined how the same key ...


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(...


0

Just for simplicity lets assume that your short period "password" was just XORed with plaintext. So we have encryption procedure like: for(int i = 0; i < plaintext_len; i++){ ciphertext[i] = plaintext[i] ^ password[i % password_len]; } When you shift your ciphertext by password_len and XOR it with original ciphertext, you'll cancel out your ...


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, ...


0

First of all, you cannot uniquely determine the keyword of a Playfair cipher, or even the key table constructed from it, simply because there are multiple equivalent key tables that will produce the same ciphertext (and multiple keywords that will produce each table). In particular, the following key tables are all equivalent: Original: Row shift: ...


2

Let $n = pq$. By assumption, $3$ divides $\varphi(n) = (p-1)(q-1)$. Without loss of generality, I assume that $3$ divides $(p-1)$ or, equivalently, that $p \equiv 1 \pmod {3}$. Fact Let $p$ be a prime such that $p \equiv 1 \pmod 3$. Let also $c$ be a cubic residue modulo $p$. If $y$ is a cubic root of $c$ then so are $y\cdot \omega \pmod p$ and $y \cdot \...


3

Assuming your differential uniformity and non-linearity figures are correct, then yes, your s-box is slightly stronger against basic differential and linear cryptanalysis. Although Anubis already was essentially immune to basic differential and linear cryptanalysis. However your s-box would need to be evaluated against other forms of cryptanalysis (e.g. ...


1

To prove an encryption scheme to be perfectly secure, we need to prove: $$P[M=m|C=c]=P[M=m]$$ where $c$ is a cipher text and $m$ is a plain text. From Bayes theorem, we have: $$P[M=m|C=c]=\frac{P[C=c|M=m] \cdot P[M=m]}{P[C=c]}$$ It is noteworthy that: $$P[C=c|M=m]=P[K=k]$$ where $K$ is the key space and $k$ is a particular key. Now: $$P[C=c]=P[K=k]=\frac{...


1

I am wondering why people are using RSA keys when some types of double substitution ciphers seem to be just as secure if not better off. First of all, RSA is an asymmetric cipher while a substitution cipher is a symmetric cipher. Asymmetric ciphers are used to achieve different security needs, e.g. TLS authentication or non-repudiation of documents. Or, ...


5

Is there a reason that expanded messages are much more difficult to crack? When looking purely at encryption, message expansion does not really tell you anything about the difficulty to crack it. Message expansion is often a feature of asymmetric cryptosystems. Those are not inherently more difficult to crack than symmetric systems. Block ciphers also ...


1

a) The question seems to be about a comparison of the size of the key spaces. The hint already shows the key spaces for Vigenère with a 10-letter key ($26^{10}$) and for simple substitution ($26!$). Simple substitution has the much bigger key space. b) Using frequency analysis simple substitution is much easier to solve. Basic frequency analysis does not ...


0

It is easily seen that the value of $q$ is invariant modulo $e$. As a consequence, we always have $q \equiv p\cdot (p-1)^{-1} \pmod e$ and thus $q(p-1) \equiv p \pmod{e}$. So, the resulting modulus $N = pq$ is such that $N \equiv (p+q) \pmod e$. Example: With $p=10598342506117936057$ and $e=5004898192290387253$, the proposed construction yields the prime $...


1

As SEJPM already hinted at in his comment, a 7 bit key is short enough to allow a plain and simple brute-force attack because there are merely 128 different keys to be tested. Hint: 7 bit = 1111111 = 127, + 1 for the 0000000 key = 128 possible keys. Even when using a low-resource device, testing 128 different keys to find the correct one should be a pretty ...


0

This is called homophonic encryption, and has been around for a long time. In terms of cryptanalysis of such ciphers, there is a nice thesis from SJSU on this topic which is available here. The attacks tested in that cipher were based on hill climbing and local optimization techniques. The conclusion states: We designed and implemented an efficient ...



Top 50 recent answers are included