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50

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37

Are checksums basically toned-down versions of cryptographic hashes? As in: they are supposed to detect errors that occur naturally/randomly as opposed to being designed to prevent a knowledgeable attacker's meticulous engineering feat? That is one way to look at it. However, hash functions have many purposes. They are also meant to be one-way (an attacker ...


34

We don't ever know, in the information theory sense, that a crypto algorithm wont fail suddenly. If we ever knew that, we'd quit using it. However, it has been shown that crypto algorithm failing has a strong tendency to fail according to a two step process: Most crypto algorithms fail quickly in the initial analysis phase, as we apply a pile of known ...


30

Actually, that wikipedia article you mention in your question already answers your question: It is moderately common for companies and sometimes even standards bodies as in the case of the CSS encryption on DVDs – to keep the inner workings of a system secret. Some argue this "security by obscurity" makes the product safer and less vulnerable to attack. ...


28

Many cryptographic algorithms are expressed as iterative algorithms. E.g., when encrypting a message with a block cipher in CBC mode, each message "block" is first XORed with the previous encrypted block, and the result of the XOR is then encrypted. The first block has no "previous block" hence we must supply a conventional alternate "zero-th block" which we ...


28

Most encryption is based heavily on number theory, most of it being abstract algebra. Calculus and trigonometry isn't heavily used. Additionally, other subjects should be understood well; specifically probability (including basic combinatorics), information theory, and asymptotic analysis of algorithms. There's also more math that's worth knowing to be a ...


26

Yes, there are advantages to the attacker. Using a well vetted encryption algorithm provides a better assurance of security. There may be cryptographic algorithm flaws and/or coding mistakes. As noted, relying on the algorithm being private just adds a layer of false security.


22

I sent an email to Ron Rivest and got an answer back. The digits of $\pi$ are used as a sort of random number generator that is used in the Durstenfeld shuffle (see also Knuth vol 3, sec 3.4.2). Below is some pseudocode adapted from the description and code he sent me. S = [0, 1, ..., 255] digits_Pi = [3, 1, 4, 1, 5, 9, ...] # the digits of pi def rand(...


21

Observation: An individual 1-byte pearson hash behaves like an 8 bit block cipher, encrypting the initial state using the message as key. This means that given a fixed message, each possible initial state produces a different output. This implies that a combined hash will never contain duplicate bytes. Without this property a hash would forget about the ...


20

The standard Diffie-Hellman key exchange algorithm (or family of algorithms) works in an cyclic group with generator $g$, and relies on $$ {y_A}^{x_B} = (g^{x_A})^{x_B} = (g^{x_B})^{x_A} = {y_B}^{x_A}, $$ where $y_A$ and $y_B$ are publicly transmitted, while $x_A$ and $x_B$ remain private. With three parties, we still have $$((g^{x_A})^{x_B})^{x_C} = ((...


20

Yes, this is a widely-used cryptographic construction called a stream cipher. For more information about this and other encryption schemes, Coursera's cryptography class is a good resource.


20

It's a good question. As pg1989 said, this is the basis behind stream ciphers, which are very fast in practice. I thought I'd quickly expand upon your statement that "the one-time pad is the perfect cipher and impossible to crack." This is true, in a sense, but it's worth pointing out that sometimes an attacker wants to do something simpler than "cracking" ...


20

They are both linear, but in different algebraic Groups. Which is to say, xor is linear in any finite field of characteristic 2, while 'ordinary' addition is linear in the infinite field of the Real numbers. Addition modulo $n$ (which is more cryptologically significant than addition over the Reals) is also a linear operation, but in the ring of integers $\...


19

Well, the exact reason for an IV varies a bit between different modes that use IV. At a high level, what the IV does is act as a randomizer, so that each encrypted message appears to be encrypted to a random pattern, even if those messages are similar. In general, IVs disguise when you encrypt the same message twice (and more generally, when two messages ...


18

The really great thing about Diffie-Hellman is how light it is, network-wise: both parties send each other a single message; neither has to wait for the message from the peer before beginning to computing his own message. If you can tolerate something heavier, you can have a look at what @Paŭlo describes; with $n$ participants, it requires $n-1$ messaging ...


18

Not only we can turn block ciphers into hash functions, but we do. The usual hash functions (MD5, SHA-1, SHA-256...) use the Merkle-Damgård construction which relies on a block cipher E. A running state r is initialized to a conventional value. Then the input data is split into a number of chunks, each chunk being used as key for the block cipher: r is ...


18

CBC does not perform authentication This property makes it less suitable for places where authentication is required, basically any transport protocol. TLS uses CBC, but by default performs authentication over the plain text instead of the ciphertext, which opened up a host of attacks. CBC can be used here, but it is error prone and may require an ...


18

In practice, CRC operations are often started with a nonzero state. Because of this, the actual equation is usually of the form: $$crc(a) \oplus crc(b) = crc( a \oplus b ) \oplus c$$ for some constant $c$ (which depends on the length of $a$, $b$). An alternative way of expressing this is, for three any equal-length bitstrings $a, b, c$, we have: $$crc(a)...


17

As you suspected, there's a very close relationship between white-box cryptography and obfuscation. (Good instincts!) White-box cryptography is basically all about obfuscating an encryption implementation. White-box cryptography is obfuscation of crypto code. Imagine that you took an AES implementation, picked a random AES key, and then hardcoded that AES ...


17

People found MARS to be clunky and overly complex, leading to more effort for implementation and optimization, and also a less clear overall security picture. Assessments of "security" are, in fact, extremely subjective, because they rely on speculations about unknown future cryptanalytic attack, empiric traditions (e.g. "more rounds" = "more security"), ...


16

If a block cipher is linear with respect to some field, then, given a few known plaintext-ciphertext pairs, it is possible to recover the key using a simple Gaussian elimination. This clearly contradicts the security properties one expects from a secure block cipher.


16

You have clarified the question as asking about whether replacing ShiftRows with a random byte permutation would strengthen AES against differential attacks. It would not. ShiftRows and MixColumns were carefully selected to work in tandem, such that every byte affects every other byte in the state within just two rounds. MixColumns ensures that every ...


15

A block cipher is a deterministic and computable function of $k$-bit keys and $n$-bit (plaintext) blocks to $n$-bit (ciphertext) blocks. (More generally, the blocks don't have to be bit-sized, $n$-character-blocks would fit here, too). This means, when you encrypt the same plaintext block with the same key, you'll get the same result. (We normally also want ...


14

On software platforms, bytewise adding will not be faster than bitwise XORing. It may be a bit slower, though, also this will be negligible with regards to the process which generated the stream (and, for that matter, will probably also be negligible with regards to the memory bandwidth). On hardware platforms (FPGA, dedicated ASIC), addition is slower than ...


14

yyyyyyy's answer is the correct short version. There is only a single cryptographic algorithm that is mathematically proven secure: the one-time pad. It's hardly ever used because it's impractical: the key size is as large as the data to protect. You can prove that any algorithm that is secure against an adversary with infinite computational power is ...


13

The algorithm (now reasonably clear) is reminiscent of a block cipher in CFB mode, with $random$ as the IV (which can be public), $secret$ as the key, and MD5 used as keystream generator instead of the block cipher. Decryption works as in CFB: $$M_1 = C_1 \oplus \operatorname{MD5}( secret||random )$$ $$M_n = C_n \oplus \operatorname{MD5}( secret||C_{n-1} )$$...


13

I restrict to hash functions $H$ with an output of some fixed size $n\ge1$ bit(s), accepting as input some strings, including all $n$-bit strings; MD5 (resp. SHA-1, SHA-256) is an example of such function for $n=128$ (resp. $n=160$, $n=256$). Whether there exists a solution to $H(x)=x$ depends on the particular hash function. If $H$ is a random function (as ...


12

Timing attacks rely on operations which do not always take the same time to execute, depending on the processed data. For instance, on a typical software platform (say, a PC) implementing SHA-256, all operations are 32-bit additions or rotations or bitwise combinations which take a constant time to execute, regardless of the actual operand values. SHA-256 is ...


12

There are some approaches. In many algorithms it for the security doesn't really matter what constant is used, as long as it is not too simple, like initialization vectors for hash functions. (And of course, we need to use always the same number.) Then mathematical constants like binary expansions of irrational numbers like $\sqrt{2}$ (or roots of other ...


12

Sure. If you want a $b$-bit hash of the message $m$, then use the first $b$ bits of AES-CTR(SHA256($m$)). That'll do the trick. In other words, compute SHA256($m$) and treat the resulting 256-bit string as a 256-bit AES key. Next, use AES in counter mode (with this key) to generate an unending stream of pseudorandom bits. Take the first $b$ bits from ...



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