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67

As usual, journalism talking about technical subjects tends to be fuzzy about details... Assuming that a true Quantum Computer can be built, then: RSA, and other algorithms which rely on the hardness of integer factorization (e.g. Rabin), are toast. Shor's algorithm factors big integers very efficiently. DSA, Diffie-Hellman ElGamal, and other algorithms ...


55

Your example string (WeJcFMQ/8+8QJ/w0hHh+0g==) is Base64 encoding for a sequence of 16 bytes, which do not look like meaningful ASCII or UTF-8. If this is a value stored for password verification (i.e. not really an "encrypted" password, rather a "hashed" password) then this is probably the result of a hash function computed over the password; the one ...


48

An unknown "encryption" algorithm has been historically achieved at least once. I am speaking of Minoan Linear B script, a writing method which was used in Crete around 1300 BC. The method was lost a few centuries later, with the death of all practitioners and the overall collapse of civilization during the so-called Greek Dark Ages. When archaeologists ...


45

See this site for a summary of the key strength estimates used by various researchers and organizations. Your "512-bits in 12μs" is completely bogus. Let's see from where it comes. 1999 was the year when the first 512-bit general factorization was performed, on a challenge published by RSA (the company) and called RSA-155 (because the number consisted in ...


35

Edit: I just noticed a very cool script named 'Hash Identifier" http://code.google.com/p/hash-identifier/ The name pretty much describes it. ~~~ Generally speaking, using experience to make educated guesses is how these things are done. Here is a list with a very big number of hash outputs so that you know how each one looks and create signatures/patters ...


33

(LZMA is a compression algorithm, not cryptographic.) For the purpose of implementing cryptographic algorithms, the generic method is getting the relevant descriptive standard, grabbing your keyboard, and trying. Most standards include "test vectors", i.e. sample values which let you know whether your implementation returns the correct answers. At that ...


31

There is no known weakness for any short or long public exponent for RSA, as long as the public exponent is "correct" (i.e. relatively prime to p-1 for all primes p which divide the modulus). If you use a small exponent and you do not use any padding for encryption and you encrypt the exact same message with several distinct public keys, then your message ...


29

To attack a cryptographic protocol, you have the following attack methods Known plaintext: Trying to find correlations between the plaintext you have and the corresponding ciphertext. Chosen plaintext: Encrypting specific plaintext and studying the changes to the ciphertext as the plaintext changes. Choosen ciphertext: Decrypting specific ciphertext and ...


24

They wouldn't. The problem you described is well-addressed in the Wikipedia article on Unicity Distance. That article also links to one by Bruce Schneier that may be more accessible.


22

I am not aware of any published cryptanalysis on MySQL OLD_PASSWORD(), but it is so weak that it is kind of a joke. It could be given as an exercise during a cryptography course. Update: a cryptanalysis similar to the meet-in-the-middle described below was published in F. Muller and T. Peyrin "Cryptanalysis of T-Function-Based Hash Functions" in ...


16

Yes. There are extremely efficient ways to break a linear congruential generator. A linear congruential generator is defined by sn+1 = a sn + b mod m, where m is the modulus. In its simplest form, the generator just outputs sn as the nth pseudorandom number. If m is known to the attacker and a, b are not known, then Thomas described how to break it. If ...


15

Short answer: there is no issue in encrypting 104 files with the same key. Longer answer: it is called public encryption for a reason: the public key is, well, public. This means that anybody can encrypt data with the public key (it is the decrypting part which is reserved for the private key owner). In particular, the attacker can encrypt millions of files ...


15

I think nobody has said it aloud here, so I will. If a cryptographer is given only one ciphertext with no means to get more, the ciphertext is short and no knowledge of the plaintext is given, it is near impossible to decrypt the text. The only way this is still possible is if the cipher is around the difficulty level of a substitution cipher. Given the ...


14

The developers are simply incorrect. There's nothing wrong with the exponent 0x4451 (decimal 17489); it doesn't create any security problems. Long ago people used to think that small exponents were a problem, because of an attack that Thomas Pornin explained with sending the same message to multiple recipients. But today we understand that the exponents ...


14

There are two generic ways to produce a "sufficiently unbiased" random digit. First method is to loop if the byte was not in the right range. I.e.: Get next random byte b. If b is in the 0..249 range, return b mod 10. Loop. This method may consume an unbounded number of random bytes, but it is perfectly unbiased and it is very unlikely to require ...


13

Just don't: http://xkcd.com/153/ Bruce Schneier likes to say, "Anyone can invent an encryption algorithm they themselves can't break; it's much harder to invent one that no one else can break" The current algorithms RSA, AES etc has been through rigorous scrutiny over years by tens of thousands. Why not just use of these? Also: Lessons learned and ...


13

This could be encrypted with any key length that is equal or longer than 28 characters (sum of lengths of ciphertext you provided) and as such unsolvable. The character variation between the plaintext CANDY VERY CRANBERRYhttphttp and its ciphertext TXOtWjYhVk 8&O$4AmSAcZf.r5Hz is: ...


12

Those five numbers are Fermat primes. Since they are of the form 2k + 1, encryption is me = m·((m2)2...k times...)2, which is simpler and faster than exponentiation with an exponent of a similar size in the general case. Since they are primes, the test that e is coprime to (p − 1)(q − 1) is just a test that e doesn't divide it. So this is more likely ...


12

If you're going to distribute a secret algorithm, why not just distribute one-time pads instead? It's more secure. If you don't like the idea of one-time pads because too much data is moving over the wire, then why are you assuming that the attacker only has one cyphertext? Assuming somebody only has one cyphertext, and doesn't have the algorithm, (two ...


12

Nope Generally speaking: No. Hashing is not encryption. Hashing is not reversible. At all. It always generates a fixed length output. So with an output fixed to say 32 characters, and an input of 33 characters, there is no possible way to reverse this. The information of that one character is irretrievably lost. -- And along with it all other characters. ...


11

What you have posted is 16 bytes (128 bits) of base 64 encoded data. The fact that it is base 64 encoded doesn't tell us much because base 64 is not an encryption/hashing algorithm it is a way to encode binary data into text. This means that this block includes one useful piece of information, namely that the output is 16 bytes long. We can compare this ...


11

http://www.cryptool.org/ - free, and have all that you need for educational purposes. I use it in different cryptography trainings. From their site: " Numerous classic and modern cryptographic algorithms (encryption and decryption, key generation, secure passwords, authentication, secure protocols, etc.) Visualization of several algorithms (Caesar, Enigma, ...


11

Isn't this a Vigenère cipher ? Lookup the article, I'm pretty sure it talks about cryptanalysis of this code (find the ley length, frequency analysis etc) And if the key is of the same length that the plaintext, you have a One-time pad that is virtually unbreakable without the key (provided this key has good random properties and is never reused).


11

The short answer is "no". The longer answer is that SSL is specifically designed to make this impossible. If this was possible, SSL would be useless for its primary intended purpose -- to secure things like credit card numbers sent to secure web sites. The short (and technically incorrect) answer for how SSL does this is as follows: The server presents a ...


11

A cryptanalyst would probably just do this by hand; This is a text file. All printable characters. Frequency analysis shows 2 * '<' and 2 * '>' which is true of all plaintexts, whereas all other character frequencies change, which probably means something interesting is going on. I would say less than an hour to figure out your password scheme. ...


10

I can't give you an answer that is going to leave you perfectly satisfied, because there is no such answer. How do we know that our algorithms are secure? Strictly speaking, we don't. We have no proof that SHA256, or AES, or RSA are secure -- it is widely believed that they are secure, but I could not give you a mathematical proof of that fact, and who ...


10

Generally people do not encrypt random garbage. Assuming they did, a ciphertext only attack would be impossible. Consider, however, that common cryptographic schemes add to the message non random data, such as padding data and message authentication codes. In some schemes, these can be used to check whether a guessed key is correct.


10

I assume you mean encryption when you say "encoding". This model of attack is known as the known-plaintext attack. It is a situation where an attacker has samples of both the encrypted ciphertext and the corresponding plaintext. All commonly used encryption schemes thought to be secure are resistant against this form of attack (resistant meaning there is no ...



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