# Tag Info

## Hot answers tagged keys

11

An encryption algorithm does not need a keyspace. By definition, however, it has one. It sound to me like your confusion is mainly terminological. In cryptography, the "keyspace" of an encryption system is defined simply as the set of all possible (distinct) keys that the algorithm can accept. For example, let's say that we're back in the days of the ...

10

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

...wouldn't key still get repeated every few hours or so - i.e. you come to the end of the PRG(K)... This is where you are mistaken. Modern cryptographic PRGs simply do no repeat within any conceivable time frame. That is, starting from a seed, a well-constructed PRG (and this is true even when they are not so well constructed, like RC4) will simply ...

8

Most public key encryption schemes, such as PGP, support this. When you are encrypting a message to Bob, in fact you are encrypting the message with a random key using a symmetric cipher, then including the key encrypted to the public key of Bob. $$E_{\text{PK}}(\mathit{Bob}, \mathit{key}) \Vert E_{\text{Symmetric}}(\mathit{key}, \mathit{message})$$ ...

8

You're missing a piece in your understanding of modern encryption. AES is a symmetrical block encryption cipher. It describes how to use a key (which can be 128, 192 or 256 bits) long to encrypt and decrypt a single block of fixed size (128 bits) of data. That's it. In order to have a complete encryption/decryption system, you need to couple it with ...

7

If your software needs to decrypt the data and you want to prevent even those with physical access from decrypting without your software, you are basically out of luck. It is impossible to achieve purely in software, since even if a good white-box algorithm existed, an attacker could copy it into their software and be able to decrypt (without directly ...

7

The comments already have covered the two main points, but let me try to put it in the form of an answer. There are not (that we know) weak keys in AES, in the sense that you cannot formulate a routine $isWeak(key)$. However, there are weak ways to generate an AES key (i.e. bad randomness). An AES key is just a bit string of length $n$. That means that ...

7

How are these keys agreed upon/distributed? Practically speaking is asymmetric crypto a requirement to "bootstrap" and distribute keys? The answers to those questions are beyond the scope of the RFC. So, it depends on the context in which HMAC is being used. The keys can be agreed upon/distributed in any secure manner. The RFC doesn't care. It could be ...

6

In the first part of this answer, I consider the problem of decryption using leaked keys of a protocol not intended for that, which was my original reading of the question. I'll ignore that dominant industry practice is to use random symmetric session keys, leaving little opportunity to "hold a couple of secret keys" without knowing to what session they ...

6

SIV is a mode specially designed for this purpose. SIV-AES would be a good choice, but it has the same issues as AES-wrap; not many implementations. If you use a GCM you should make sure that the IV is unique (if your plaintext is ever not random you would otherwise be in problems). As for the password based key derivation function: yes, PBKDF2 is good, ...

6

What you are looking for is called white-box cryptography. In short white-box crypto aims to make an implementation of a cypher (for example AES) in such a way that it is impossible for an attacker to extract the key, even if the attacker (the user of the computer) has access to the source code and a debugger. Up till now all academic white-box ...

6

Yes, in any algorithm where keys are just random numbers, reading them from /dev/random is safe. However, /dev/random blocks if the kernel's entropy estimate goes to zero so it is often a good idea to use a user space CSPRNG seeded from /dev/random or /dev/urandom for session keys and other similar random numbers that are used in bulk. The newer getrandom ...

6

Yes a brute force key-guessing attack would be faster, but: It would be ridiculously slow for either. E.g. see this for 256-bit keys. There are faster attacks on both and those attacks break larger RSA sizes than ECC sizes. Related: Why can ECC key sizes be smaller than RSA keys for similar security?

5

There are two things here: Encryption uses mode of operation, and not "AES alone". Some of them are randomized by an initialization vector - that means the encryption of the same text under the same algorithm is still randomized and not deterministic. The encryption methods take care of that. You only need the correct key to decrypt. Passwords are not ...

5

I'm using it as a one way encryption on plaintext values such as SSN, names, dates, etc. I suggest rethinking your approach. None of these values have much entropy, so it would be straightforward to bruteforce the original plaintexts (just like cracking a password hashed with a fast hash function). If you're planning to use these values for ...

5

Many keys do not consist of a single value. Other key types, such as most symmetric keys, don't have a most significant bit or byte as the key value isn't interpreted as an integer. So this question would be different for each key type and encoding. In general though you would just decrease the "entropy" or key size. This is certainly the case for ECC, and ...

5

If we take some randomly generated key of AES-128 and we change any random 1 byte of that 16 byte key, will this make huge difference in the AES cipher text generated over same input string? Yes. The outputs with different keys differ greatly. If you pick two random keys the outputs must look completely uncorrelated, or an attacker could gain an ...

5

In two key 3DES two keys are equal so that key size is only 112 bits, compared to the 168 bits of full 3DES. The advantage is a smaller key size without a correspondingly large loss in security: both two and three key 3DES can be attacked in about $2^{112}$ time. With the encrypt-decrypt-encrypt construction it clearly must be the first and last key that ...

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 ...

4

There's actually an algorithm designed exactly for this purpose: generating a sequence of keys from one master key. It's called HKDF (HMAC-based Key Derivation Function, paper here). The algorithm essentially boils down to two steps: Extract and Expand. The Extract step accepts any type of "key material" as input, and outputs a pseudorandom key that will ...

4

Each 56-bit key has a unique 8-bit parity value. For this reason there are only $2^{56}$ keys.

4

For high security applications using 3DES, NIST recommends using keying option 1 (all keys are different). This is simply because it's the safest. For any application, keying option 1 should be used. If you set K1==K3, then you are reducing your key size to 112 bits, which is less than the smallest key size for AES. Worse still, due to cryptanalysis done on ...

4

yes,it is possible because in meet in the middle attack on 3DES,see below with Complementation Property of DES in red arrow,you can search $2^{55}$ key space instead of $2^{56}$,and for green arrow,you have $DEC_{K2}(ENC_{K1}(M))$ that without key Complementation Property,you need $2^{112}$ operations but with key Complementation Property of left ENC and ...

4

It is important and interesting to notice the different use cases you mention. You don't have millions of software updates you have to sign every day (or second for that matter). For SSL connections, however, you may have millions per second. Given the two use cases, it makes sense that the way you would protect the keys would be different. We don't know ...

4

Apologies if this is too basic but all the explanations about AES focussed on the details of the protocol, not these more basic concepts. In fact you are asking about general secret key management :) 1) Is the point of this (and other encryption techniques) to reuse the same private key for multiple messages? If we're using one-time keys, ...

3

The very fastest elliptic curve algorithms can generate a key-pair in about 40k cycles on a modern CPU. A high end laptop has four cores running at about 3 GHz, so it can generate about 300k key-pairs per second, or a billion per hour. (The cost of SHA-256 is negligible in comparison.) However, secp256k1 is nowhere near the fastest curve. It takes 10 times ...

3

I would just concatenate. Two 256-bit keys lead to a 512-bit key which is short enough for HMAC with common hash functions to use as is. XOR would allow the second party to easily choose a related key (and has worse behavior when neither key is perfectly random). Hashing and double HMAC use more resources without a clear benefit, unless you care about the ...

3

No it does not affect the parity bits. The plaintext or ciphertext doesn't have any parity bits specified. The application key is just a input to the cipher, which is a Pseudo Random Permutation. Permutation means a 1:1 mapping from plaintext - including parity bits - to the ciphertext. If you perform the permutation in the opposite direction the plaintext ...

3

If a cryptosystem is expecting a 16-byte key, making the first byte nonzero simply reduces the number of possible keys. It is better to allow all bytes to be all values.

3

The "obvious" (it really isn't that obvious) thing you are missing is: The same reasoning could be applied to literally any other private key! There is nothing special about $a=\lvert(\mathbb Z/p\mathbb Z)^\times\rvert=p-1$ (which would, by the way, more commonly be represented as $0$ modulo itself), except that checking for it is particularly easy. For ...

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