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7

Simplified SSLv3/TLS from this book Note, $R_{(Alice|Bob)}$ is a random nonce chosen by Alice or Bob respectively, and $\{S\}_{Bob}$ is encryption with Bob's public key. pre-master secret As stated in one of the answer you link to, "The point of a premaster secret is to provide greater consistency between TLS cipher suites." In the figure above, the ...


5

You could view it as a form of One-Time-Pad encryption. In this view one of the two "messages" is the key and the other is the ciphertext (it does not matter which you think of as which as their roles are symmetric). It may, however, be more natural to think of as a simple secret sharing scheme. Roughly speaking a secret sharing scheme is a scheme where a ...


5

Shamir Secret Sharing (SSS) is based on constructing a polynomial of degree $t-1$, whose independent term is the secret $S$. Each share is actually a point of the polynomial. The security of SSS is based on the fact that, when one wants to interpolate a polynomial of degree $t-1$, one needs at least $t$ points of the polynomial. It can be seen graphically ...


4

If you perform the distribution digitally (using networks) then you have a problem. Unless you use another one time pad you lose the perfect confidentiality as the distribution itself won't deliver perfect security. But using another one time pad is pointless: you would lose exactly as many key bits as you are distributing, while you are only protecting the ...


4

Actually, you can do Shamir Secret Sharing over any finite field $GF(p^k)$, for any prime $p$ and any integer $k$. If $k=1$, you have the $GF(p)$ field you mentioned; however it works on extension fields as well. We often pick $p=2$ and $k$ a multiple of 8; this makes everything nice even number of bytes (at the cost of doing our calculations in $GF(2^k)$). ...


4

Take a linear polynomial: $y=mx+b$. If I tell you that the point $(1,5)$ is on the line, can you tell me $m$ and $b$? No, because in fact there are infinitely many lines that pass through the point $(1,5)$. It takes 2 points to uniquely identify a line. In general it takes $t$ points to uniquely identify a degree $t-1$ polynomial. Furthermore, given $t-1$ ...


4

Is the Kurihara algorithm really what it purports to be (dramatically faster but equally secure replacement for Shamir Secret Sharing)? The algorithm being referred to is in this paper, and I believe that the speed benefits are at best marginal, if not nonexistent. As for the speed benefits being marginal, well, normally we use secret sharing as a part ...


4

Shamir's $(t,n)$ secret sharing scheme involves picking a random polynomial $p$ (over a finite field) of degree $t-1$, such that $p(0) = s$ is the secret value to be shared (this is easy to do, since $p(0)$ is just the constant term of the polynomial), and then evaluating the polynomial at $n$ distinct non-zero points $x_1, \dotsc, x_n$ to construct $n$ ...


3

This is because $t$ shares uniquely defines the polynomial of degree $t-1$. $t-1$ shares still leaves $k$ possible and equally likely polynomials, for $k$ the size of the field, so the secret is information theoretically hidden. Think of a degree 1 polynomial, essentially a line. If you know just one point on the line, you cannot say anything about the ...


3

There is a one-to-one correspondence between the value of the second share and the value of the secret: each of the $2^8$ possible values of the second share will give a (distinct) value of the secret among the $2^8$ possible values for it, and vice versa. Hence, guessing the second share is exactly the same thing as guessing the secret.


3

There are no security advantages to evaluating the polynomial at random places instead of sequential. The information theoretic security proof of Shamir secret sharing does not depend on the evaluation points being chosen in any specific manner.


3

To answer your question, the cryptographic complexity will reduce by $2^{k-n}$ where $k$ is the key length and $n$ is the number of bits known. A slightly better scheme that doesn't use a cryptographic key sharing scheme can be as follows: Generate three random numbers $B_1$, $B_2$, $B_3$ with bit length equal to the size of your key $K_p$. Generate each ...


3

Just use Shamir's Secret sharing: wiki link It's designed by Adi Shamir, who is the "S" in RSA. It's fairly simple to use and there are no known weaknesses in it. While it doesn't split the data, splitting the key may work just as well (unless you absolutely must split the data.)


2

If you have $t-1$ shares of an $(t,n)$ system, you have a chance at learning of the roots of the system... as long as some of your shares have a 0 for the $y$ coordinate. You can't learn any other roots. Demonstration that the possession of a share with a $y$ coordinate of 0 gives you knowledge of a root (and forgive me if this is too obvious): A share ...


2

When a TLS/SSL session starts(after the hellos and cipher decisions) the server gives the client it's cert. The key in the cert could perform different actions depending on the key-agreement algorithm decided on by the client and server. Let's say they agree on RSA key agreement. This means the cert contains the server public RSA key and the server has a ...


2

In secret sharing, one person typically wants to share a secret. Only that one person knows that secret, so they are the dealer. In MPC, which VIFF implements protocols for, you actually have multiple parties, each with a secret input(s). So, each of these parties is the dealer for their individual secret input(s). In this sense, there is no central dealer ...


1

Since Bob already knows $K_{AB}$ he can run $D_{K_{AB}}(E_{K_{AB}}(E_{K_{AC}}(K)))$. Which will leave him with $E_{K_{AC}}(K)$. And to decrypt that any further (in order to get $K$) he would need either Alice or Charlie's help.


1

There is no such thing as an incorrect secret. There is just some secret, that must be uniquely defined by the distribution phase.


1

A Flaw If you try working through your example concretely then you should see at least one issue with decoding. Consider values mod $8$: $ A = { 3 } \\ B = {7} \\ secret = {5 } $ Using your equations we have: $ Q = 3 + 7 + 5 = 7 \\ U = 3 - 7 + 5 = 1 $ And for decoding: $ B = (U - Q) / 2 \\ B = (1 - 7) / 2 \\ B = -6 / 2 ({\text mod} 8) $ But what ...


1

I'm not sure about your specific system; this only addresses "can a key derived from the password make a good shared secret?" The most common password hashing functions are actually designed for exactly this purpose -- deriving a cryptographic key from a (weak) password. That's actually what PBKDF2 stands for: "password-based key derivation function #2" ...


1

Reform the problem. Instead of each participant picking their givee (which they give to), have them select a giver (which they receive from). Each participant randomly generates a number (appropriately large) and anonymously submits it (e.g., via the tor network) to the site. This number represents them as giver. After all participants have entered, the ...


1

Just to be specific we are talking about information theoretically secure secret sharing, e.g. Shamir sharing. Lets say a secret $secret$ is shared in $n$ shares $S = \{s_1, \ldots, s_n\}$. Assume further party $P_i$ is given share $s_i$. If I understand your question correctly, you are asking if we give $P_i$ some share $r \ne s_i$ is there anyway for $P_i$ ...


1

Can you use threshold encryption and a mixnet? It might not be the fastest thing in the world but it uses well-understood components. Setup Every player generates an ElGamal keypair and proves knowledge of their secret key. The joint public key is the product of all public keys. (If you're worried about reset attacks, look up "Pedersen threshold key ...



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