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10

Yes, of course there is a benefit to signing unencrypted emails. The article you cite is solely about the combination of signature and encryption; it doesn't directly say anything about signing unencrypted emails. There is an important concern raised by the article which does apply to unencrypted emails, but that's because that concern applies equally ...


8

The article you linked to predates the S/MIME 3.2 spec. If your client is sending S/MIME 3.2 messages, it should support header protection. Refer to RFC 5751 Section 3.1: In order to protect outer, non-content-related message header fields (for instance, the "Subject", "To", "From", and "Cc" fields), the sending client MAY wrap a full MIME message ...


6

OpenPGP as defined by RFC 4880 knows two different encodings. Binary encoding Obviously, there is no reasonable limitation to an (ASCII) character subset in binary encoding. Radix 64 Radix 64 is also often entitled ASCII armored. In the end, it is a base64 encoding with a checksum. The content may consist of [a-zA-Y0-0+/=]. ASCII-armored OpenPGP ...


6

It mainly depends on how the algorithm was selected. If it was selected by a public competition like for AES, then it is likely to be secure. If it was forced in by the NSA such as Dual-EC random number generator, then you may have some doubts. Other questions you may want to ask yourself are: Is this an "original" algorithm or was the problem that it ...


5

SafeCurves lists some ways to compare the security of elliptic curves. Their security criteria are split to "ECDLP security" and "ECC security". Failing the former basically means "there is no way to use this curve securely in general" while the latter "it is difficult to implement this curve securely". None of the (few) BouncyCastle-supported curves that ...


5

Yes! (restrictions apply). ISO/IEC 9796-2 (scheme 1, SHA-1 hash, option 1 also known as implicit hash identifier, alternative signature production function) is a fully standard signature scheme, based on RSA, widely used in the Smart Card industry for public key certificates and message authentication, that adds only 22 bytes of signature overhead (if the ...


5

Towards the security of the signature scheme, no precaution against timing attack is necessary when verifying an asymmetric signature. That's because there is no secret involved, thus no information leak to fear. However it can happen that the message, or the signature itself, is intended to be secret; a leak by timing dependency (during computation of the ...


4

The benefit to signing a non-encrypted email is that any recipient can verify that it was indeed you who wrote that non-encrypted email, unless your key was compromised (or the signing protocol has an exploit).


4

$q$ does not divide $s^e-h(m)$, but $p$ does, so since the gcd must divide both $s^e-h(m)$ and $n$ it's $p$. To be even more explicit, we know that $p$ divides both $s^e-h(m)$ and $n$. The only larger divisor of $n$ that is also divisible by $p$ is $n$ itself, but if $n$ would divide $s^e-h(m)$, then $q$ would also divide $s^e-h(m)$, which we already assumed ...


4

I think you have some misunderstanding here. Finding collisions when knowing the trapdoor is a required feature, but leaking the trapoor when knowing collisions is an undesirable "feature" (which some constructions suffer from). A chameleon hash function (aka trapdoor commitment) allows you given the trapdoor to find pairs $(m,r)$ and $(m',r')$ with $m\neq ...


4

Yes, you can, but you would need access to raw or textbook RSA encryption and you would have to implement the PKCS#1 v1.5 or PSS padding primitives yourself. Beware that PKCS#1 v1.5 compatible padding is different for encryption signature generation. If you only have PKCS#1 v1.5 encryption or OAEP encryption available then the encryption routine will ...


4

Well, if the hash function is weak, then the attacker might be able to take a valid signature for a signed message, and find a second message for which the signature for this first would also validate for the second. For example, if Alice signs the message "I like chocolate", what Bob might do is find a second message "Alice owes Bob $13,106,107.57", and ...


4

The main benefit of adding randomness in RSA signature padding is that it simplifies and strengthens security arguments. At least that's claimed by PKCS#1v2, paragraph above 8.1.1 (emphasis mine) RSASSA-PSS is different from other RSA-based signature schemes in that it is probabilistic rather than deterministic, incorporating a randomly generated salt ...


3

The problem about Man-in-the-Middle attack on Diffie-Hellman is that both sides are not confident about other side's public key (g^a and g^b). If they were sure that they have correct public key of their's friend Man-in-the-Middle attack wouldn't be possible, because MITM attack is based on the forgery of public keys by adversary! If for instance Bob and ...


3

You can prove that a document was signed after a certain date by including data that was not known to anyone before that date, such as stock market data. You cannot prove that a document was signed before a certain date by purely cryptographic means. Information doesn't go stale, so when you show a signature, it could have been signed at any time. You can ...


3

This sounds like "fair exchange," the subject of many good research papers. In general you need a third party to give any security guarantees, but "optimistic fair exchange" involves the third party only when one of the parties tries to cheat (i.e., when both play honestly there is no involvement from the third party). Incidentally, Diffie-Hellman is most ...


3

In general, no. There are: $$ {2^{64} \choose 2^n} = \frac{2^{64}!}{2^{n}!(2^{64}-2^n)!} $$ possible ways of selecting $2^n$ distinct 64-bit vectors. This is a huge number; using Stirling's approximation of factorials, when $2^{n}$ is substantially smaller than $2^{64}$ (i.e. when $n$ is smaller than $55$ or so), this number of combinations is approximately ...


3

If you want $N$ serial numbers, your serial numbers will have to use $n$ bits for uniqueness, where $n = \log_2 N$. So if you have 100 bits to use for the serial, you could use 20 to get about a million serials and have 80 bits to use for a cryptographic MAC or signature. Now there are two approaches, the symmetric and the asymmetric. In the symmetric ...


3

If you look at exact security, the height matters. The reason is that it defines the number of OTS key pairs and hence the possible number of one time signatures per MSS key pair. To forge a MSS signature, it is enough to generate a forgery for 1 out of $2^h$ OTS signatures. Hence you get a reduction in the bit security of $h$ bits.


3

No, not really. Elliptic curve signatures are the smallest you'll find in common use. An $n$-bit elliptic curve produces $2n$-bit ECDSA signatures. The smallest standard curves that offer some security are 160-bit, and those are not really recommended (e.g. NIST recommends 224+ bits). That would give you 40 byte signatures. Lower than 64, but not 32. So 40 ...


3

The only way you could do this if if you could affect the padding schemes appropriately. Mathematically, textbook RSA encryption with the private key is the same as textbook RSA signature generation. Nobody should use textbook RSA, however. In practice, padding schemes are used and they differ between the two operations. So unless you can turn off padding ...


3

The initial idea of Fiat and Shamir was to eliminate the interaction in public coin protocols (note that public coin means that the random choices of the verifier are made public) and was used to convert three move public coin identification scheme into conceptually simple signature schemes (it has later been proven by Pointcheval and Stern that under the ...


3

According to this answer, RSA with the "usual" "padding scheme, described in PKCS#1 as the 'old-style, v1.5' padding," can be made to satisfy that; one would need to specify NULL or omission and require that the public exponent's prime factors are all easily findable and sufficiently bigger than the 4th root of the modulus.


3

It is logically impossible to transfer a private key. The key will continue to be a signature key, but it will cease to be "private" the minute it is transferred. A signature key that isn't private isn't a private key. If you want the document to be signed by the user (in any semantically coherent sense), this operation has to take place on a device ...


3

To start with, it's certainly not a bad idea to avoid SHA-1 when other algorithms exist, which do not have the SHA-1 weaknesses to anyone's knowledge. The security of SHA-1 depends on how you're using it. The vulnerability is what's known as a collision vulnerability: an attacker has the ability to create two input strings with the same SHA-1 hash with less ...


3

If we signed a secret message $m$ by publishing its signature $σ$ computed as $m^d\bmod N$, at least two very bad things would happen: The message would not be so secret anymore That's because anyone knows the public key $(N,e)$, and thus from $σ$ can compute $σ^e\bmod N$, which is $m\bmod N$. This reveals a lot of information about $m$, which goes ...


3

An attack is described in Section 4.1.6 of the SEC1 document. Regarding xagawa's answer: The attack you describe is different from that described in Section 4.5 of the Blake-Wilson--Menezes paper. Specifically, their attack: (a) does not require knowledge of the secret ephemeral key $k$, and (b) changes the reference point $P$, which is not allowed in ...


3

First to explain you, why you get 512-bit outputs from a 256-bit curve: The output is basically a point (x-coordinate is enough) and a message-dependant value, with the x-coordinate being expressed as integer. You can verify the signature by checking for a specific relationship between the point and the message-dependant value and the public key point. In ...


3

In this notation, $f^k(x)$ means "apply $f$ $k$ times in succession". For example, $f^3(x)$ is defined to be $f(f(f(x)))$. Because of this definition $f^a(f^b(x)) = f^{a+b}(x)$ holds trivially (even though we known nothing else about $f$), as the the left side means "do $f$ $b$ times, and then do it $a$ times", while the right means "do $f$ $a+b$ times". ...


3

My answer is not original, as I am simply summing up information from different questions already solved in this site. Nevertheless, I thought it could be interesting to collect everything in one answer. The first thing you have to know is to differentiate between a digital signature and a message authentication code (MAC). In this case, HMAC is a MAC and ...



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