# Tag Info

## Hot answers tagged encryption

32

The answer is in the source, file sshrsag.c, line 9: #define RSA_EXPONENT 37 /* we like this prime */ This value $e=37$ matches the conditions for a reasonable fixed RSA public exponent: $e$ is odd, $e$ is at least $3$, $e$ is reasonably small. The later condition is good for speed of operations involving the public key (encryption, ...

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.

18

In short: You must authenticate the IV. Which particular attacks apply if you don't depends on the block cipher mode; I will give two common examples. In CTR mode, an attacker who fiddles with the IV can forge authenticated messages, but the content of the corresponding plaintext is beyond his control (since he doesn't know the key). Depending on the ...

15

If we want to compact an existing RSA private key expressed as $(N,e,d,p,q,d_p,d_q,q_\text{inv})$, we can reduce it to $(e,p,q)$ and easily recompute the rest as: \begin{align} N&=p\cdot q\\ d&=e^{-1}\bmod\operatorname{lcm}(p-1,q-1)\;\text{ or }\;d=e^{-1}\bmod((p-1)\cdot(q-1))\\ d_p&=d\bmod(p-1)\;\text{ or equivalently }\;d_p=e^{-1}\bmod(p-1)\\ ... 15 This question has many problems in the way it was asked, and clearly did not come after doing some investigation. However, since this seems to be a misconception that is spreading widely, I will relate to it. It is not true that the "crypto community" (whoever that is) believes that the NSA can break RSA. In fact, if Snowden taught us anything, it is that ... 14 Both of the other answers tackle the question of encryption in a particular format, but I would argue that neither of them is necessarily a good fit for your use case. You want to be able to generate 20 character codes that a server will be able to verify. A symmetric MAC is sufficient for this use case, if you don't need the codes to contain any secret ... 14 n is the exponent. So when n is doubled from 64 to 128 it doesn't mean that you have to try twice as many values. It means that you have to try2^{64}$times the amount you were already trying (as$2^{128} = 2^{2\times64} = 2 ^{64+64} = 2^{64}\times2^{64}$). It is required to only search half of the key space on average (if average is the correct term ... 14 Any$e$such that$\gcd(e, (p-1)(q-1)) = 1$will do. There is no need for it to be in the set$\{3,17,65537\}$; these last numbers are chosen for speed of encryption, mostly (two set bits leads to faster computation of modular exponentation), and these numbers happen to be prime, so the condiiton is easily checked. One often encounters other$e$, but many ... 13 Decrypt the ciphertext with every possible key and store the result:$2^{56}$decryptions. Now encrypt the (known) plaintext of the ciphertext with every possible key:$2^{56}$encryptions. Now you have to check every entry, which is in both lists and try it with another plaintext-ciphertext pair. If you can successfully decrypt that, you are very likely to ... 12 You can use a seed to start a PRNG. Then you can use that PRNG to generate the two (or more) primes required to generate the key pair. Now if you save that seed you can regenerate the key pair, which means you don't have the store the modulus, CRT components or private exponent. So yes, it is possible to reduce the size, but this approach does have ... 12 So I'm trying to find a method of encryption that not only obfuscates text, but also compresses the result. For example, if I encrypted ninechars, the ideal result would be less than nine characters. Even without encryption, it's not possible for a reversible data compression scheme to shorten all of its inputs. This can be easily proven using the ... 11 If you combine two affine ciphers, you obtain one affine cipher. Say the first cipher is$e_1(x) = a_1x+b_1$and the second is$e_2(x) = a_2x+b_2$. Then$e_1(e_2(x)) = a_1(a_2x+b_2)+b_1 = (a_1a_2)x+(a_1b_2+b_1)$. Note that if$a_1$and$a_2$are both relatively prime with the modulus, then so is$a_1a_2$, so the new cipher can also be deciphered. 11 Modern cryptographic algorithms are specified in terms of bytes or even bits, not characters. Whether the data you encrypt happens to represent latin or cyrillic letters or pictures or audio data or anything else does not matter at all to an encryption algorithm; all it ever sees is a bunch of bytes. What this means in practice is: You have to fix some ... 10 ECB leaks if blocks are identical. For uniformly random data identical blocks become likely when you encrypt about$2^{n/2}$blocks with an$n$bit block cipher. CBC and CTR mode develop similar weaknesses when they encrypt that much data. => As long as you encrypt reasonable amounts (up to a petabyte or so) of random data with a 128 bit block cipher, like ... 10 This is trivially true via the pigeonhole principle. SHA-2/512 has$2^{512}$possible outputs, but$2^{2^{128}} - 1$possible inputs. Trying$2^{512}+1$unique inputs is sufficient to produce at least one collision. That said, SHA-2/512 is designed to be collision resistant, which implies that it should be hard to find two inputs that hash to the same ... 9 You really don't want to use ChaCha20 alone in (nearly) any situation. What ChaCha20 does for you is to prevent attackers from (passively) reading your data, which is good. But ChaCha is a so-called stream cipher which works by XOR'ing a pseudorandom pad with the message (your file at rest). However it is for this very way of working that ChaCha doesn't ... 9 It is usually assumed that the length of the message is not secret. Even with padding the approximate length is usually leaked, and necessarily any encryption reveals a maximum length (or at least information content) because the ciphertext cannot in general be shorter than the message. NaCl secretbox does not use a block cipher, but a stream cipher ... 9 A Diffie-Hellman exchange needs not be synchronous. In DH, each party has a private key (x) and a public key (gx mod p). If the sender knows the recipient's public key ga, then he can build his own key pair (b and gb), compute the shared secret (gab), and send both his public key (gb) and the encrypted message (symmetric key derived from gab) to the ... 9 "Allowed" or "forbidden" are not the right terms; the Pope won't excommunicate you if you dare implement your own algorithms. The question is rather whether doing your own implementation is a smart move or not. For learning, doing your own implementation is a good idea. It will teach you a lot on how the said algorithms work. A lot of details on how ... 9 Let me first answer your actual question (and then I'll proceed to answer something slightly different that I think will be informative and helpful). Your question asks whether it's possible to use only a DSA key. Technically speaking, this is of course possible. The reason is that a DSA key has exactly the same format as an ElGamal key. No one forces you to ... 9 You've actually been trapped by the mindset, that OTP will hide all information about the underlying plaintext. This is not true as you have observed. The definition of perfect secrecy, given in Introduction to Modern Cryptography by Katz-Lindell reads like this: Definition 2.3 An encryption scheme$(\text{Gen, Enc, Dec})$with message space$\mathcal ...

9

CBC mode decryption for the first block is defined as: $$P_0 = IV \oplus D_k( C_0 )$$ where $P_0$ is the first plaintext block, $C_0$ is the first ciphertext block, and $D_k$ is the decryption by the block cipher using the key $k$. This can be rearranged as: $$IV = P_0 \oplus D_k( C_0 )$$ which allows you to reconstruct the IV, if you know the key, the ...

9

XORing a master key (presumably a long term key) with data is a very dangerous idea. If any data key is leaked, then the master key may be easily calculated, thus leaking all keys. ($m$ for the master key, $d_x$ for all data keys) $$c_x = d_x \oplus m$$ then somehow $d_4$ is leaked $$m = d_4 \oplus c_4$$ $$d_x = c_x \oplus m$$ You'd be better off applying a ...

8

The way this is usually done is to use a separate compression algorithm, then encrypt the compressed (shorter) message. However, compression has some disadvantages and nowadays its use is discouraged. Compression can leak information about the plaintext, like in CRIME and BREACH attacks on TLS. Arguably it is the protocol that combines the compression and ...

8

None of Twofish, Serpent and AES are currently known as broken, so as far as security is concerned, you can use any of them. AES has a slight advantage because it's very widely used, so if it gets broken you're more likely to hear about it and get relevant software updates quickly. The Snowden postings haven't changed much as far as cryptography usage is ...

8

It depends. Specifically, it depends on the type of cipher, and on the way it's used. For stream ciphers like RC4, and for block ciphers like AES in CTR and OFB modes, decryption is effectively identical to encryption, and thus takes the exact same time. (Minor exception: encryption may require generating a unique nonce / IV, which might take a small ...

8

Although there are already many answers here, I wanted to strongly advocate AGAINST MAC-then-encrypt. I fully agree with Thomas' first half of the answer, but completely disagree with the second half. The ciphertext is the ENTIRE ciphertext (including IV etc.), and this is what must be MACed. This is granted. However, if you MAC-then-encrypt in the ...

8

Question: Given $n$ values $v_1=\alpha \cdot r_1 \bmod p,..., v_n=\alpha \cdot r_n \bmod p$ for a large $n$ can the adversary learn the value $\alpha$? Answer: assuming that the $r_i$ values are random (that is, equidistributed and uncorrelated), then the attacker gets absolutely no information about $\alpha$ (other than whether or not it's 0). We can ...

8

You are looking at the ASN.1 encoding of private (and public) keys; the 00 values you see are an artifact of how ASN.1 encodes integers. ASN.1 is a method for describing data structures, and has ways to represents all sorts of data types. It wasn't designed with public keys (or cryptography) in mind; it was intended for more general use, initially ...

8

You're describing a form of three-pass protocol, which is a communication mechanism where neither party needs to know each other's secret key. Wikipedia describes a helpful metaphor using a box that can be locked by two padlocks: First, Alice puts the secret message in a box, and locks the box using a padlock to which only she has a key. She then sends ...

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