# Tag Info

12

Brute forcing the key would hardly be an issue: 128-bit keys (assuming they have been properly generated) are in a space which is way too large to be successfully explored by brute force; and 256-bit keys (the kind you put in AES-256) are even more larger. Whether AES is "faster" than HMAC or not does not make such brute force more feasible: even if each key ...

9

The crucial difference between plain encryption and authenticated encryption (AE) is that AE additionally provides authenticity, while plain encryption provides only confidentiality. Let's investigate in detail these two notions. In the further text, we assume $K$ to be a secret key, which is known to authorized parties, but unknown to attackers. Goals ...

7

The GCM authentication tag doesn't need to be encrypted. Just attach it to the ciphertext in the clear. A very quick intuitive justification: It's an authentication tag derived from the ciphertext, it doesn't contain any sensitive information itself. The security of the GCM model assumes the tag is left in the open. (The GCM spec, SP 800-38D, shows the ...

6

Clearly, if you had been using AES-256-CBC for confidentiality and AES-256-CBC-MAC for authentication, it would not be secure to use the same key for both confidentiality and authentication. Hence, using the same key for confidentiality and authentication cannot generally be secure; you need additional premises to arrive at that conclusion. In your case it ...

6

CRAM-MD5 is a protocol to demonstrate knowledge of a password. In the context of email, it is sometime used by an email client to authenticate to a POP, IMAP, or/and SMTP server. Basically, the password is used as the key of HMAC-MD5 in a challenge-response protocol. Among positive things there are to say about CRAM-MD5: The password is not exchanged in ...

6

The article mentions that 3-DES was used to encrypt these passwords in ECB mode. DES has a 64-bit/8-byte block. So let's say you use ECB to encrypt a nine byte password. The first 8-bytes are encrypted using ECB. So far so good. But what happens when we come to the ninth byte? Well we're now in a new block but only the first byte is populated with any ...

6

A is acting as a square-root oracle in that protocol. We can use that oracle to factor $n$ and break the scheme. Suppose you are an attacker that wants to impersonate A. You: Pick a random $m$; Send $m^2$ to A; Compute $p = \gcd(m_1 - m, n)$, thus factoring $n$. This works with probability $1/2$ for each attempt.

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

All looks pretty secure except for your auth key derivation. You should use a better key derivation method like HKDF instead of just SHA-512. I don't think your random nonce is doing anything in this scenario - an attacker who wants to brute-force a weak password wouldn't be slowed down by a nonce transmitted in the clear. Why not just use a ...

5

GCM mode already incorporates any params that could affect the outcome of the decryption. The associated authenticated data is there to allow you to rely on context for your decryption. For example, say you are encrypting some records associated with a user. You may want to include the user's database ID as the authenticated data. If a user found a way to ...

5

Using EAX with a 64-bit block cipher is problematic, because the short block size causes some weaknesses due to internal collisions. I do not recommend it. Use a 128-bit block cipher. Indeed, the world has moved away from 3DES and towards AES exactly because of these fundamental problems with a 64-bit block size: the internal collision effect means that, ...

5

The same key is indeed used in EAX to key both the CTR mode and the underlying OMAC (which is actually used in 3 distinct phases: randomising the CTR nonce, authenticating the Additional Authenticated Data, and authenticating the Ciphertext). This is explicitly acknowledged in the security proof. Where EAX differs from a naive reuse of the key is that it ...

4

"Efﬁcient, Compromise Resilient and Append-only Cryptographic Schemes for Secure Audit Logging" (PDF) gives a publicly verifiable approach that allows fine-grained verification, but it is in the Random Oracle Model. The Simple Method: The verifier and logger start with a seed for a forward-secure pseudo-random number generator. To denote a valid ending ...

4

You got some notation wrong. There is no algorithm like "AES-GCM-SHA-256". AES is a block cipher, i.e. a pseudorandom permutation of 128-bit blocks. It itself only allows encryption for messages of size 128 bits (= 16 bytes), with a limited security guarantee. When you mean "encrypt the data using AES", you actually mean "use AES with some mode of ...

4

As noted in archie's answer to your earlier question, the EAX paper first defines a generic encrypt-then-MAC composition method called EAX2, with separate keys for the encryption and MAC components, and proves its security (Appendix C). It then defines EAX as EAX2 instantiated with CTR mode and OMAC, and with the same key used for both components, and then ...

4

No, the scheme described in the question does not provide integrity. A forgery is possible when the message's size is allowed to vary (which is presumably the case since some padding is used), and the adversary can choose some segment of the message with knowledge of the message before that segment. That is, a message $M=M_b||M_c||M_d$ with the beginning ...

4

An OCB like mode seems impossible with stream-ciphers. It's coupled tightly to the concept of a keyed permutation i.e. a (tweakable) block-cipher. Many authenticated encryption actually combine two distinct primitives. It's just that the specification and API only expose the combination. Essentially these xor a key-stream into the message to encrypt it ...

4

In comparison against CBC mode and HMAC, GCM mode is quite commonly better alternative. But, I'll go to detail where it neccessarily is not. Just like Richie Frame, I also do not agree that CBC + HMAC is always the best comparison target. I've added few other details. Hope you find them useful. Against CBC and HMAC I'll discuss downsides first. The ...

4

The first 32 bytes of XSalsa20 output are used as key for the one-time-mac Poly1305. Poly 1305 needs a new 32 byte key for each message, using part of the key-stream is a natural way to obtain those. Requiring those empty bytes makes implementing the API easier. The implementer only needs to call XSalsa20 on the zero padded input buffer once, receiving both ...

4

The simple way to build authenticated encryption using a Feistel Network is to build a Feistel based block cipher, then use one of the many modes of operation that turn a block cipher into an authenticated encryption scheme (eg CCM,OCB,GCM). For a good survey on the subject of modes-of-operation I would recommend this paper by Rogaway. It does not cover the ...

3

The answer to this question follows directly from the answers to Should we MAC-then-encrypt or encrypt-then-MAC? and the comment thread here. In short: Your scheme is computationally secure (IND-CCA2 and INT-CTXT) assuming that HMAC is a computationally secure privacy-preserving MAC; but your scheme is wildly impractical, as fgrieu explains, so it is not ...

3

The #1 thing you can do is: don't derive your keys as a function of a password/passphrase. That's a security breach just waiting to happen. Using something like scrypt mitigates the risk somewhat, but by no means does it eliminate the risk. This is likely to be the weakest link in your cryptographic scheme. Instead, use a truly random value as your ...

3

Does it negatively affect security to calculate a hash value of the ciphertext before MAC calculation? Like exchanging step 2. with this: HMAC-SHA256(SHA256(ciphertext)). Technically, yes, but not significantly. In order to attack the scheme you propose, the attacker would have to be able to do at least one of two things: (1) Find an attack on the ...

3

Sounds like you might be after deterministic authenticated encryption. Check out SIV mode. This mode doesn't use an IV at all (though it does generate one internally, outside of the attacker's control). Being deterministic, it leaks equality of (header, plaintext) pairs. The basic idea is to put the (header, plaintext) pair through a PRF, such as HMAC, and ...

3

The main pitfall is possibly thinking that it provides authentication. The result is still a public key scheme. It won't help to include the IV in the associated data, since that is already covered by the authenticated part of AEAD. Including the RSA-encrypted symmetric key in the associated data would help by making the resulting hybrid encryption CCA2 ...

3

When we transmit information across an insecure channel, we wish for our data to be secure. So, what does this mean? To discuss these we'll use the standard cryptographic situation of Alice and Bob. Alice wants to send something (the plaintext) across an insecure channel (what this means will be discussed) to Bob. This channel will be listened to by Eve ...

2

I'm thinking there's a third potential solution. Each time you close a log file, you could append the name of the next new log file, timestamp it, then sign the log file. When it is time to create a new log file, you would read the previous log file, validate the signature, validate the time stamp, read the new log file name, and create it. You'd ...

2

Edit: You've clarified in the comments that confidentiality of the logfile's contents is important. Given an AEAD function $C = E_k(iv, plaintext, aad)$, a safe construct is $$C = E_k(iv, contents, filename).$$ There is no need to include either the key or the IV in the additional authenticated data, and I would recommend against doing so. It is ...

2

In entity authentication, the parties do not necessarily exchange a session key, but authenticate to each other by providing some proof related to their identites. For instance, if one party signs a challenge given by its partner, this party proves its identity. Implicit key authentication merely guarantees that the partner (with whom you are talking right ...

2

There aren't any known attacks on the PRFness of HMAC-SHA256 better than brute force. (So you can truncate that MAC to length L where $\:\:\frac1{2^L}+\epsilon\:\:$ is an acceptable risk of forgery.) To reduce the impact of a forgery without making the ciphertext any longer, one should use a format-preserving encryption (FPE) scheme that is secure against ...

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