# Tag Info

8

In addition to the performance problems poncho already mentioned when using RSA signatures without hashing I just want to add on the security warning of poncho: Reordering If you have a message $m>N$ with $N$ being the RSA modulus, then you have to perform at least 2 RSA signatures as $m$ does not longer fit into $Z_N$. Let us assume that it requires ...

7

If you compare DSA with SHA-256 and a 2048 bit group modulus $p$, to RSA with SHA-256, a 2048 bit modulus $n$ and public exponent $e = 65537$, on you will at least perform the following operations: DSA $g^{u_1}y^{u_2}$ - 2*256 squares $\mod p$, up to on average 2*128 multiplications $\mod p$, depending on implementation optimizations. RSA $s^e$ - 16 ...

7

Yes. Any good standard digital signature algorithm will be secure in this setting. Digital signature algorithms are designed to be secure against chosen-message attacks, where the attacker can choose any set of messages and learn the signatures on those messages; the security of the signature scheme means that this doesn't help the attacker at all. This ...

7

For your application: "I need the (underpowered 8-bit) slave to be able to tell if a command issued is really trustable", RSA signature with low public exponent ($e=3$), or Rabin (an analog with $e=2$), is likely the most appropriate, assuming you can't trust the slaves to keep a key secret, which is the only realistic assumption unless that slave uses ...

6

Actually, it does appear to be feasible to construct such a public key, with the caveats that: the public exponent $e$ will be large. While this is legal, most public keys have a small $e$ the modulus may be more vulnerable than usual to a specific (normally nonoptimal) factorization method. However, how vulnerable it is can be controlled. (Note: I'll ...

6

I hope I got your point and try to answer your question. Actually, if I understand you right, then what you call attack can also be seen as an adversary acting in a specific attack model. Therefore I briefly review the security models for digital signature schemes. Goal of an adversary: We start by discussing the goals of an adversary beginning with the ...

6

For many signature schemes, having two signatures using the same randomness for two different hash values allows recovery of the private key. This is used in many security proofs by showing that an adversary that forges a valid signature can be coerced through replaying into producing two signatures of this form. As a consequence, an forger can be twisted ...

6

Why is it common practice to create a hash of the message and sign that instead of signing the message directly? Well, the RSA operation can't handle messages longer than the modulus size. That means that if you have a 2048 bit RSA key, you would be unable to directly sign any messages longer than 256 bytes long (and even that would have problems, ...

6

Well, one reason to hash the data before signing it is because RSA can handle only so much data; we might want to sign messages longer than that. For example, suppose we are using a 2k RSA key; that means that the RSA operation can handle messages up to 2047 bits; or 255 bytes. We often want to sign messages longer than 255 bytes. By hashing the message ...

6

No, signing the hash of the public key cannot introduce a weakness on a secure signature scheme. When we have a signature scheme, we assume that it is secure in an chosen text model, where the attacker has access to the public key, and can ask any text of his choosing to be signed. We can see that any such scheme (such as ECCDSA, or so we believe) cannot ...

5

This sounds like a fair exchange protocol where what is exchanged is a digital signature. Per this paper, these are impossible without trusted third parties. With a trusted third party, they are possible. Indeed people have proposed schemes that do what you describe again relying on a third party in the case of failure.

5

Digital signatures are used to solve this type of problem. That is, a way for $A$ to sign the message for $B$ so that $B$ is highly confident that $A$ signed the message in question. There are lots of signature schemes out there, such as RSA signing, DSA, and others. A MAC is not strictly a digital signature, but has a subset of that functionality and may ...

5

Yes, this looks fine. I assume $A$ and $B_i$ are trusted parties. The protocol as I understand it looks like this: $A$, $B_1$,…,$B_n$ agree on a secret key k. $A$ broadcasts messages ($m_1$,MAC($m_1$,$k$)), … , ($m_j$,MAC($m_j$,$k$) which $B_1$,…,$B_n$ receive and authenticate. I assume $A$ and $B_i$ are trusted parties, so no $B_i$ will itself ...

5

If you search on "timestamp", "timestamping", and "notary" on Crypto.SE and Security.SE, you'll find lots of references. I've collected a number of timestamping services that were mentioned in one of those places; this should provide a number of companies and online services you can check out: http://www.proofofexistence.com/ https://www.btproof.com/ ...

4

First and foremost: it is a bad idea to invent a method to sign or encrypt with RSA (or any crypto). Standards like PKCS#1 or ISO/IEC 9796-2 are here for that purpose, and even these occasionally have more or less subtle flaws. Given comments, I'll assume that the question is about an RSA encryption scheme enciphering message $M$ into $(M||S)^e\bmod N$, and ...

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

Yes, this is possible. Here's one simple scheme: The scheme. We fix an upper bound $B$ for the maximum allowed value of the counter. Let $H$ be a cryptographic hash function and $\text{Sig}(\cdot)$ be an ordinary signature scheme (e.g., a RSA signature). Use $H^k(x)$ to the $k$-fold iteration of $H$ on $x$, e.g., $H^2(x)=H(H(x))$. The signature $S(n)$ ...

4

If he chooses $s$ at random, then the scheme will be stateless but will fail after using the same $s$ twice, which should happen after giving approximately $\:$$\Theta$$\big(\hspace{-0.05 in}$ $2^{H/2}$$\hspace{-0.01 in}\big)\:$ signatures. If he chooses $s$ by applying a PRF to $g(m)$, then the scheme will be deterministic and stateless, but can be ...

4

An RSA signature is a sequence of bytes of the same size of the modulus. If the key uses a 1024-bit modulus $n$, then the signature value is, numerically, an integer in the $1..n-1$ range, and the PKCS#1 standard specifies that this integer should be encoded as a sequence of bytes of the same length as would be needed to encode the modulus, i.e. 128 bytes ...

4

No, the signer is per definition in possession of the secret signing key and thus can always produce signatures for any message of his choice. Consequently, a notion of unforgeability is not meaningful with respect to the signer. For a signature scheme one requires unforgeability for parties who are not in possession of the secret signing key but only the ...

4

No. Cryptography alone cannot solve this problem. Solving this problem requires a combination of technical (e.g., cryptography, systems security) and non-technical (e.g., legal, regulatory, contractual) solutions. Even the technical part is not solely a cryptography question; it as much about systems security.

4

I think you don't quite understand how RSA signatures work (and why they are the size they are). When generating an RSA signature, we follow a two-step process: We take that hash of the message we're signing, and convert (and pad) it into an integer $M$ which is between 0 and $N$ (where $N$ is a large integer that specified by the RSA key) We use the RSA ...

3

As very clearly indicated by the specification, CKM_RSA_X_509 performs "raw" RSA. Raw RSA is simply modular exponentiation. So it performs just the RSASP1 function in the PKCS#1 standards. This means that a user should - at the minimum - also provide a secure padding mechanism. Otherwise the conditions to perform a secure RSA signing operation are not met. ...

3

From these three, ECDSA is faster - it does arithmetic with smaller numbers, and is thus faster. (RSA verification is faster than ECDSA, even though it uses larger numbers, because it computes a exponentiation by a small number.) Still, elliptic curve Schnorr signature should be around 5-10% faster than ECDSA (or even more in a side-channel resistant ...

3

Since you asked in general "Are there any other obvious flaws/issues I'm missing", I would recommend that you would take a critical look at the whole update process flow. I would not consider the quality of the PRNG at your server as an obvious weak point, as long as it is reasonably strong, since on servers you have quite a bit of choice of reasonably good ...

3

Well, lets go through the issues: It seems to be possible to retrieve the (public) key used for creating an ECDSA signature just from the signature alone Nope, not quite. You also need the message being signed. And, with that, it doesn't give you the unique public key; it does allow you to narrow it down to two possibilities (assuming you're using a ...

3

If you implemented it well, and there is no MITM (i.e. use authenticated comm channel), then it is difficult to forge the RSA signatures (meaning you can use them safely). However, in your implementation there are 2 other major factors that you need to consider: One is the use of Javascript as a coding medium: there are many who consider Javascript to be a ...

3

Why the CFS signature is affected Let us review the structure of the CFS signature, which is strongly related to the Niederreiter PKE scheme. In the Niederreiter PKE scheme, a public key is $H \in \mathbb{F}^{n \times k}$, which is a scrambled parity-check matrix of the Goppa codes. A plaintext is a decodable error; for example, we set $S = \{\vec{e} \in ... 3 Short Answer: NO, it is not safe, do NOT do this. Longer Answer: You are true that you can use your RSA keypair for both operations. This approach is used in many applications and scenarios. There are Web Services or Single Sign-On implementations, which enforce you to use the same key pair for both operations. X.509 certificates do not allow you (by ... 3 The probability of such a "collision" occurring randomly, with honestly-generated RSA keys, is extremely low. Mathematically justifying that assertion can be tiresome, but the idea is the following: Let$s$be a RSA signature, i.e. an integer of size$k$bits for some$k$(e.g.$k = 2048$). We consider RSA keys$(n,e)$where$n$is the modulus (of size$k\$ ...

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