Tag Info

Hot answers tagged

47

We don't.${}{}{}{}{}{}{}{}{}{}$


28

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 ...


21

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 ...


16

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 ...


12

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

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 ...


9

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 ...


9

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 ...


8

There are several scenarios where you wouldn’t want to use AES in CBC mode. In CBC mode, each block is dependent on a previous one. As @fgrieu nicely hinted at in his comment, using CBC means that if you have a large, encrypted file and you only want to update/change/modify a tiny fraction of it, you would have to follow the decrypt-modify-encrypt path each ...


7

"In software" means programming the cipher on a multifunctional processor or smart chip. The normal instruction set is used to build the algorithm, and not available operations have to be emulated. "In hardware" means building the cipher in real hardware, like on a Field-programmable gate array (FPGA) or Application-specific integrated circuit (ASIC). ...


6

Using a MAC on the plaintext may potentially leak information about the plaintext (MAC algorithms do not necessarily ensure confidentiality of the data they are applied to, although some MAC algorithms like HMAC seem pretty safe). If you want to avoid this (theoretical) problem, then you should encrypt the MAC on the plaintext (i.e. MAC-then-encrypt, not ...


6

Actually, given that the MD2 S-box needs to be a permutation, using a Fisher–Yates shuffle does seem a fairly obvious choice to me. (It's pretty much the standard algorithm for generating a uniformly chosen random permutation.) The rand() function used to convert the digits of $\pi$ into a random number from a given interval looks complicated, but is ...


6

While it may be confusing, that Wikipedia article is actually correct! Let me try to explain it a bit better… Definition of key whitening Key whitening is an extremely simple technique to make block ciphers like DES much more resistant against brute-force attacks. Like you’ve already discovered yourself, this is the basic scheme: Or, defining it a bit ...


6

XSalsa20 uses the same cryptographic core as Salsa20 and comes with a security proof that it's secure if Salsa20 is secure. It doesn't use the core of ChaCha and thus has worse diffusion. The way XSalsa20 works is that it hashes its 256 bit key and the first 128 bits of the nonce using HSalsa down to a 256 bit key and then uses that key together with the ...


6

GF$(2^8)$ or $\mathbb F_{2^8}$ can also be viewed as the vector space $\mathbb F_2^8$ of $8$-bit vectors (or bytes) over GF$(2)$ or $\mathbb F_2$. Suppose $\{\beta_0, \beta_1, \cdots, \beta_7\}$ is a basis of $\mathbb F_2^8$ over $\mathbb F_2$, that is, the sum $$a_0\beta_0 \oplus a_1\beta_1 \oplus \cdots \oplus a_7\beta_7, ~ a_i \in \mathbb F_2$$ equals ...


5

Nobody can tell you not to "have fun with it" but I would strongly recommend you to first study attacks on other ciphers. Spritz (Rivest & Schuldt) fortunately mentions a lot of research on its predecessor, RC4. This makes it a rather good starting point in my opinion. It is necessary to understand the linguistics and mathematical constructs that are ...


5

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 ...


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

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 ...


4

You can get what you want programmatically. No special ciphers or modes needed. You say there is a single, continuous subset that needs to be encrypted. Thus, you could have a function where the programmer specifies the start of the portion of the plaintext that needs to be encrypted and the number of bytes to encrypt. The function could pull that part out, ...


4

The sum of PRPs is a secure PRF. That paper gives as a security bound for a sum of two independent PRPs $q^3/2^{2n-1}$, where $q$ is the number of queries and $n$ the block size (i.e. 128 for AES). That means that your construction, correctly used, is more secure than a single PRP, for which the bound is $q^2/2^n$. If you wanted to give an adversary an ...


4

No, this is safe. In fact, if you show a way of distinguish the stream $AES_{k_1}(C) \oplus AES_{k_2}(C)$ from a random stream with fewer than $2^{64}$ outputs, you have just demonstrated a way of distinguishing AES from a random permutation. Here is how this works: suppose we are given Oracle assess to a permutation $P$, which might be $AES_{k_1}$ for ...


4

Since you encrypt just a single letter, there are $26^2$ combinations of $p$ and $c$ where $c=E(p)$. This is because there are $26$ possible shift keys in the key space, an therefore each $p$ can be mapped to one of $26$ letters in the code space. Now, assuming that the key is distributed uniformly in the key space, each of those combinations of $(p,c)$ has ...


4

Yes, it would be more secure if they were used correctly. But as it would require a substantially different algorithm, you really would not be talking about DES anymore. Brute forcing usually scales exponentially with the size of the key. However, if the algorithm is substantially altered then it is required to analyze the algorithm again. Note that AES is ...


4

Where did SHAKE128 and SHAKE256 originate from? They follow from the general properties of the sponge construction. A sponge function can generate an arbitrary length of output. The submission of Keccak to the SHA-3 competition proposed a single "XOF" (extendable-output function) with a user defined length, which would have been essentially SHAKE-288. ...


4

Compared to fixed rotations, data-dependent rotations improve resistance to differential and linear cryptanalysis. A fixed rotation has no effect (beyond helping with diffusion) in the probability of a (xor-)differential characteristic, whereas a data-dependent rotation also introduces differences in the rotation amounts, which brings probabilities down. ...


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

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 ...


3

A Vigenère cypher can be unbreakable if, and only if: the key is random, the key is at least as long as the message being encrypted the key is used only once. In that case a Vigenère cypher is mathematically equivalent to a One Time Pad, and the same mathematical proof of unbreakability applies. You proposal is closer to being unbreakable than standard ...


3

It depends on what you think of as an alternative. If you think of the scheme where you do not use $M$ as a modulus, but the keys a picked as: $$ k \leftarrow \{1, \ldots, M-1\} $$ Encryption: $$ C = d + k $$ Decryption: $$ d = C - k $$ Then the scheme is insecure. One way to see this is to note that we have $C \geq d$. So the ciphertext communicates the ...



Only top voted, non community-wiki answers of a minimum length are eligible