# Tag Info

## Hot answers tagged algorithm-design

50

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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 ...

22

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.

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 ...

11

Contrary to your assumption, this is done, and it is secure: For instance, the hash functions SHA-224 and SHA-384 are basically the same algorithms as SHA-256 and SHA-512! The only differences are in the initial values for the Merkle-Damgård construction used internally and, of course, in that only the first $224$ or $384$ bits of the resulting hash are ...

11

What choice did they have? F1 is a bitwise function with three inputs and one output. There are $2^8 = 256$ such functions. Only 70 of them are "unbiased" (i.e. have as many 0 and 1 outputs in their image). If you further require that each input, as well as the order of inputs, matters for the output, you are left with only 36. However, those 36 are all ...

10

NO, we can't apply an hill-climbing algorithm to Diffie–Hellman. In order to break Diffie-Hellman key exchange, it is enough for Eve to reverse exponentiation modulo the public prime $p$; that is, given $g^x\bmod p$, find $x$. That's the Discrete Logarithm Problem. We do not know that hill-climbing can help for that (or the slightly less general DH ...

10

The simple answer is nobody can prove that an algorithm won't break in a given period of time. The achievable goal is to increase the probability that no effective attack will be developed without warning. There are a couple of characteristics that indicate a particular cipher may remain secure and if degraded will do so 'gracefully'. 1. Time. Time is the ...

8

Grover's algorithm treats the function it is evaluating as a black box and finds, with high probability, an input to the black box such that it outputs a specified value in $O(N^{1/2})$ evaluations of the function. Since Grover's algorithm works on the function as a black box, your modification does not hinder Grover's algorithm at all in finding the ...

8

The distinction is that ECDSA solves a problem that HMAC does not. If you need that problem solved, then you need to do ECDSA rather than HMAC; if you do not, then HMAC works just as well (and is a lot cheaper). With HMAC, here is what we have: we have an authenticator that has a secret key. It takes a message, and gives that (and the secret key) to the ...

8

The only advantage I can think of is that they're able to put "State of the art encryption" on their website. But even then, those with a trained eye may spot it as an issue, therefore rendering it as yet another disadvantage. But other than that pseudo-advantage, there are none. Chances are overwhelmingly good that this new cipher, having been ...

8

The main advantage is that using a proprietary algorithm gives you access to trade secrets like additional cryptographic attacks that other algorithms fall to but to which the proprietary algorithm is resistant. Whether this is important depends on the amount of trust you have in the vendor. As other answers have noted, usually the staff of any one ...

7

Your doubts are absolutely valid. Disguising the algorithm is not a valid argument for security. It also contradicts to Kerckhoffs Law. It (the algorithm) should not require secrecy, and it should not be a problem if it falls into enemy hands; Designing cryptographic algorithms (ciphers, hashfunctions, ...) is a long and complicated process. In ...

7

Most standard-use iterative hash functions (including SHA-512) are build in a way that these types of operation are not possible (without breaking the hash function). They work generally in this way: The message is split in same-size blocks (usually with some padding at the end to fill the last block): $pad(M) = M_0 || M_1 || M_2 ... || M_n$. There is ...

7

For any $x,y$ represented by $\{0, 1\}$, $x \lor y = 1 - (1-x)(1-y)$. It follows, any one-multiplication homomorphic scheme would do. It also follows, just additively homomorphic scheme would be not enough.

6

First, it's not said that AES is unbreakable, merely that none of the currently known attacks reduce the computational cost to a point where it's feasible. The current best attack on AES-128 takes 2^126.1 operations, if we had a computer (or cluster) several million times more efficient than any current computer and could operate at the thermodynamic ...

6

There are attacks on both blockciphers and hash functions that can exploit symmetry in the round functions. For example, completely identical round functions can permit Slide Attacks on Hash Functions, and rotational symmetries of the round function can permit rotational cryptanalysis. The round constant addition or 'iota' step of the Keccak Hash Function ...

6

The scheme you describe is essentially same as the "SIV construction"* introduced by Rogaway and Shrimpton in their 2007 paper "Deterministic Authenticated-Encryption: A Provable-Security Treatment of the Key-Wrap Problem". This construction takes a PRF (such as HMAC) and a conventional IV-based encryption scheme (such as, say, a block cipher in CTR mode), ...

5

Yes you could use a hash function as round function, but if you are using the "same key" over all rounds, you are vulnerable to slide attacks. Using a hash function is not a very good idea. Your round function should not introduce biases, should not lead to special differences (attack: differential cryptanalysis), and it should also not be writable as a ...

5

The numbers $n$ (coding source) and $m$ (coding index for a given source) can be combined into a single bitstring; e.g. with $0\le m<2^u$ and $0\le n<2^v$, as a bitstring of $\lceil u+v\rceil$ bits. Then, converting that single bitstring into a unique random-like number can be done by encryption with a secure block cipher with block width $w\ge u+v$ ...

5

As I noted in another answer, Auguste Kerckhoffs published his principles in the scientific/academic journal “Journal des sciences militaires, vol. IX, pp. 5–38 in his article "II. DESIDERATA DE LA CRYPTOGRAPHIE MILITAIRE.", Jan. 1883. So, when you ask since when academics and cryptographers “might” have been accepting (and even applying) those rules, the ...

5

The pseudocode has a serious issue: changing the value of nonce2 in an otherwise valid cryptogram is not detected, and results in invalid deciphered plaintext. That would be fixed by encrypt(password, string): nonce1 := generate_random_nonce() nonce2 := generate_random_nonce() key := derive_key(nonce1, password) encrypted := nonce2 || cipher(nonce2, ...

5

Does the value of the key array(T) have to be in this range [0-255] if yes could you please specify why? Yes. RC4 operates on bytes. There are 256 possible values for an 8 bit (1 byte) number, that range from 0 to 255. RC4 treats the key as an array of bytes, so every entry in the key array is by definition in the range 0 to 255. Why did they use ...

5

Before answering the actual question, I will offer some general advice. It is important to pay attention, both in class and to the textbook you are reading. If learning how to solve such exercises is a key goal of the course, such solutions have very probably been discussed at length in class. Moreover, your textbook also has proof examples, and in this ...

5

No, or at least, if you can, you have an Extremely Significant result; you've just shown that Paillier is a Fully Homomorphic system, and so it could perform any operation on encrypted data (and in a way that's significantly more efficient than any other known FHE system). Here's why: The $|a - b|$ operation is effectively an $XOR$; if the ciphertexts $a, ... 5 The key size is simply the amount of bits in the key. With AES, like most modern ciphers, the key size directly relates to the strength of the key / algorithm. The higher the stronger. AES is a bit different with respect to the key size in the sense that both the key schedule and the number of rounds are different for each key size. Because of this there ... 5 Custom crypto can be valuable when other aspects are more important than the confidentiality guarantee, and the well-known ciphers don't address those aspects. A custom cipher or custom application of a cipher would tend to offer a weaker guarantee of confidentiality than well-tested systems. But some users of encryption can handle an eventual breach so ... 4 There is a simpler way: implement a stream cipher using the hash function, and use that to encrypt the plaintext. Probably the most used stream mode is counter (CTR) mode, which is normally defined for block ciphers. CTR mode works equally well with a PRF (MAC) as with a PRP (block cipher). It only uses the function as a one-way function; with a block ... 4 You are essentially asserting that if$k \equiv 1 \pmod N$, then$a^k \equiv a \pmod N$. This is false in general. The correct assertion is the following:$a^k \equiv a^\ell \pmod N$if$k\equiv \ell \pmod{\phi(N)}$. In more general group-theoretic terms, if$a$is an element of order$n$in a group$G$, then$a^k = a^\ell$if and only if$k \equiv \ell ...

4

Ignore the integer overflow issue I mentioned in a comment, for a moment. I don't see how this adds any security. For all $n>2$, the function you are calling Fibonacci is one-to-one, and since $n<256$, you could easily build a lookup table without much memory to invert the function. Therefore, to break this, all one has to do is invert the Fibonacci ...

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