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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"), ...


14

There is only one main difference between PKCS#5 and PKCS#7 padding is the block size. PKCS#5 padding is only defined for 8-byte block sizes. PKCS#7 padding would work for any block size from 2 to 255 bytes. This is the definition of PKCS#5 padding (6.2): The padding string PS shall consist of 8 - (||M|| mod 8) octets all having value 8 - (||M|| mod ...


11

A block cipher is an invertible transformation that maps an $n$ bit block of bits to an $n$ bit block of bits, under the control of a key (and where $n=128$ in the case of AES) Now, we most often need to do things other than mapping blocks of $n$ bits; how we do that is using the block cipher within a Mode of Operation. A mode of operation is just a way to ...


9

Such a block cipher would be completely unusable in any of the standard modes of operation, presuming you want at least IND-CPA security. Say, for instance, that you are using it in CTR mode. A chosen plain text adversary looking at a cipher text consisting of only two blocks (i.e. 2 times 32 bits), would have a non-negligible probability (higher than ...


9

Yes, they can be described as a multivariate polynomial over $GF(2)$ (or over $GF(2^8)$). See algebraic cryptanalysis. This expression does not seem to help cryptanalyze AES, so far as we know, but it can be done. For an example of how to write AES in this way, see the following paper: A simple algebraic representation of Rijndael. Niels Ferguson, ...


9

At the time of the competition (I can talk about it, I was there), there was a lot of discussion and various people showed arguments. However, there was never an official, publicly known "board of scores" with totals and definite rules, as the pictures you show seem to purport. It is possible that the NIST people did make something similar internally, but ...


8

Take $C_2$ and pick any $k_2$. Then decrypt using $k_2$ so that $M_2 = AES_{k_2}^{-1}(C_2)$. Now obviously we have $AES_{k_2}(M_2) = C_2 = C_1$. This extends to any blockcipher, because blockciphers are specifically designed to be reversible. In the comments you asked about the scenario where $M_2$ is also fixed. This is as hard as breaking AES. Consider ...


7

Analyzable in this case means "simple to study". If your cipher consists of a small function that mixes a few things together, and then you repeat that often then your cipher is more easily understood and more easily fit into previous research knowledge than a overclomplicated large design. This is the case with Simon and Speck. Being easy to analyze is a ...


7

CBC mode encryption is defined as: $C_i = E_k(P_i\oplus C_{i-1})$ (with $P_i$ being the $i$th plaintext block, and $C_{i-1}, C_i$ being the ciphertext blocks. What might happen if we have a lot of ciphertext encrypted with the same key is if two ciphertexts happen to be the same, that is: $C_i = C_j$ If we see that, we can then immediately deduce that: ...


6

The question makes a number of statements that are incorrect. It is not correct that a fixed point is guaranteed to exist. It is not correct that if you hold the plaintext constant and vary the key, then a fixed point is guaranteed to exist. Moreover, the existence of fixed points has only an extremely tenuous connection to security. Assume $E$ is a ...


6

There are many well known and studied ways of constructing a hash function from a block cipher. A thorough (but reasonably readable for a beginner) treatment of many of the classic approaches, and the security properties of the various constructions, can be found in Black-Box Analysis of the Block-Cipher-Based Hash-Function Constructions from PGV, which is ...


6

In their 2012 paper "The Security of Ciphertext Stealing", Phillip Rogaway, Mark Wooding and Haibin Zhang prove that all the NIST-approved ciphertext stealing modes provide the same level of security as ordinary CBC mode, i.e. ciphertext indistinguishability under a chosen-plaintext attack. To quote their abstract: "Abstract. We prove the security of ...


6

An Ideal Cipher with $k$-bit keys and a $b$-bit blocksize is a family of $2^k$ permutations on the set $\{0,1\}^b$ indexed by the set $\{0,1\}^k$, selected uniformly at random from the set of all such families of permutations. See e.g. http://eprint.iacr.org/2005/210.pdf. The IC model is primarily useful for proofs where you need to assume that the ...


6

First of all, we need to review what they mean by "parity of a permutation"; they don't mean whether the input block had a even number of 1's. Instead, they view the $n$ bit cipher (with a specific key) as a permutation on $2^n$ objects; that is, it can be review as a way of rearranging that set of $2^n$ objects onto itself. Now, permutations on a finite ...


5

Yes, we can build a hash function from a block cipher, and that's common, although with block ciphers designed for that purpose, when in the following I focus on AES, mentioned in the (different) question that motivated the present answer [which got moved here because said question was found to be a duplicate]. One classic method to obtain a hash function ...


5

Most vulnerabilities in block ciphers are related to key security. Successful attacks have not been practical against anything except smaller keysizes than 256 bits or fewer rounds of encryption. Since there are no variables to be selected for AES except the S-box and the P-box, the Holy Grail is key management. Lateral attacks against AES rely on bad ...


5

Well, whether $AES'$ is as secure as $AES$ depends on the length of $k_1, k_2$. If they are both 128 bit, then what you effectively have is a standard 128-bit AES, except that prior to round 6, you replace the running key with an independent key (and you tweaked the last round, but that's cryptographically harmless). Now, it is never a good idea to do ...


5

We can still safely use 64-bit-block ciphers when used in an otherwise sound protocol, and all of the following three conditions are met: The effective key size is made big enough; that disqualifies DES (55-bit), but not Blowfish (up to 448 bits), TEA (128 bits), 3DES (167 bits), and to some degree 2-keys-3DES (111 bits). Note: I computed the effective key ...


5

The exact information leaked depends on the mode of operation that is in use. The simplest case is ECB, where a duplicate ciphertext block means that the corresponding plaintext blocks are also equal. For the CBC mode, when two ciphertext block are equal, i.e. $C_i=C_j$, we know that there was equality before applying the block cipher. This in turn ...


5

A 6-byte nonce can expect to receive a collision 0.01% of the time after around 240,000 nonce generations (based on a birthday attack). After 100 such rotations (a little under 17 years, based on your 2-month rotation policy), that comes out to a likelihood of just under 1% of experiencing a collision. On the surface, to me that seems like a reasonable ...


5

Don't believe every claim ever made in any paper ever written, particularly when the paper provides little or no justification for the claim; not everything you read reflects the cryptographic consensus. This is particularly true for a paper written in 2002, which is a time our understanding of authenticated encryption and security engineering was still in ...


5

When cryptographers create algorithms, they usually provide some argument that the algorithm is secure. They need to start the argument with some set of assumptions. For example, the in public-key cryptography, they may begin with the assumption that factoring large numbers is hard. Many algorithms use use a block cipher as a building block. The arguments ...


5

A block cipher is (or tries to be) a pseudorandom permutation on a given space. Let $\mathcal{M}$ be the set of $n$-bit blocks for a given $n$. There are $2^n$ possible block values, and a permutation on $\mathcal{M}$ sends each block value to another value. There are $2^n!$ such permutations. A block cipher is a mapping from key values (in a given key space ...


5

A mode of operation is an explicit method by which we use a block cipher (eg AES) to do more than just encrypt one block of data. For example, it may allow us to encrypt multiple blocks of data (eg ECB,CBC etc), provide us with some authenticated encryption (eg GCM) or a method for encrypting disc storage (eg XTS). Rijndael,DES etc are block ciphers. That ...


5

Because CBC-MAC with inputs that are not prefix free is weak against existential forgery, meaning it is not a "secure" MAC. More precisely, CBC-MAC is easily distinguishable from a random function (i.e. not a PRF) when the input domain is not prefix-free. This is because an adversary can request the CBC-MAC of messages $M_0$ and $M_1$, and then xor the MAC ...


5

Did you try Wikipedia? DES consists of 16 rounds of the form: $$L_{i+1} = R_{i}, \quad R_{i+1} = L_i \oplus F(R_i, K_i),$$ which are identical except for the round subkeys $K_i$. (The last round is slightly different, in that the half-blocks $L$ and $R$ are not swapped as they are after all other rounds, but that makes no cryptanalytic difference.) The ...


5

Actually, for CFB mode, the IV is the same size as the block size, 16 bytes. As for your question "does keeping the IV secret help security", the answer is "not really". CFB mode processes the message in blocks, and for each block of plaintext, combines that with the previous block of ciphertext to generate the next block of ciphertext. What the IV is ...


5

The tweak is generally not secret, i.e. the adversary knows it (perhaps even chooses it, depending on the attack model) and thus doesn't have to try to guess it. So no, tweaks don't increase the effort to brute force a cipher ... unless of course the tweak is kept secret, in which case it may or may not increase the required effort, depending on how the ...


5

Quite apart from the correction that Reid made (it takes $2^{127}$ attempts to achieve a probability of 50% of finding the right key; with $2^{64}$ attempts, the probability of success is $2^{-64}$), with AES, there is no known way to take advantage of known (or even chosen) plaintext to speed up any brute force search; even with $2^{64}$ chosen ...


5

It can't be achieved under the assumptions you are making, because the attacker can distinguish it by selecting an arbitrary $k'$, and checking if $E(k')$ commutes with the permutation in question. That is, to check a permutation $P$, we pick an arbitrary $x$, and check if: $E(k', P(x)) = P(E(k',x))$ This equation always holds if $P = E(k)$ for some value ...



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