# Tag Info

50

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35

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 ...

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.

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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 ...

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: ...

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

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 ...

12

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 ...

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 ...

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 ...

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

Because it is not secure enough. Hash functions rely a lot on diffusion (a single bit change must change half of the other bits) and confusion (the value of a bit should depend on the value of other bits). This is also known as the avalanche effect. Because it lacks a permutation, my first intuition: it lacks diffusion and has weakness to differential ...

9

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 ...

9

I assume that you mean the S-box. The answer is NO! Randomly chosen S-boxes are not good choices for differential and linear cryptanalysis. When Biham and Shamir presented differential attacks on DES, one of the things that they showed was that if you replace the S-boxes in DES with randomly chosen ones, then the differential attack becomes much more ...

8

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 ...

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 ...

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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

I think it's more helpful to think of checksums as toned-down versions of message authentication codes (not hashes). Message authentication codes (MACs) are designed to detect any modification to a message, while it is in transit. They are secure against even adversarially-chosen modifications. Checksums are designed to detect some modifications to a ...

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.

7

Thanks for the question! You are correct that there is a bug here. Indeed, the sentence "choose $\mathbf{b}_i$ s.t. $\ldots$" makes no sense: the LHS is in $H$, but the RHS may not be. Fortunately, there is a simple fix which guarantees $\mathbf{y}'_i \in H$. (This must have been what we intended in the first place, based on how the rest of the proof ...

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), ...

6

Many properties of boolean functions are used in stream and block cipher design, e.g., when they are used as filtering and combining functions. Some important examples are: Nonlinearity (minimal Hamming distance of the truth table of the boolean function from affine functions), must be high for resisting linear/affine approximation attacks. Correlation ...

5

This question is based on opinion. At least kind-of. But the variants from which one can choose are quite a few. As for general construction the sponge construction (like Keccak / SHA-3 uses) are very versatile and can be used for many purposes, for example hashing, authenticating (= "MAC'ing"), authenticated encryption (see “General Overview of the ...

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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, ...

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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 ...

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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 ...

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