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A non-interactive ZK proof is when you play with yourself. Or, more accurately, with an impartial version of yourself. In a normal ZK proof, the prover first issues a bunch of commitments, then the verifier issues challenges that the prover complies with; this proves anything only as long as the verifier is assumed to issue challenges normally without any ...

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The GQ identification scheme is essentially a zero-knowledge proof of a value $x$ such that $x^\mu \equiv J \pmod N$ where $N$ is an RSA modulus and $(\mu,N)$ are system parameters and $J$ is known to the verifier and $x$ only known to the prover. Now your question is not directly concerned with the aforementioned proof where a user shows the possession of ...

5

Let $\#G$ denote the number of elements in the group. In your particular case, $\#G = \varphi{}(n)$ (and even $\#G = n-1$ if $n$ is prime). Let $\xleftarrow{\$}$denote a uniformly random sampling from a finite set of elements. Furthermore,$\mathbb{Z}_m$denotes the set of non-negative integers smaller than$m$and$\stackrel{?}{=}$denotes a equality test ... 5 This is a classical example. Here is the proof system… Bob gives two gloves to Alice so that she is holding one in each hand. Bob can see the gloves at this point, but Bob doesn't tell Alice which is which. Alice then puts both hands behind her back. Next, she either switches the gloves between her hands, or leaves them be, with probability$1/2$each. ... 5 Here's what can happen if you don't do this verification: Suppose Alice, Bob and company generate their public key shares honestly,$h_2, h_3, ..., h_n$Now, Snidely Whiplash (who is also a trustee) is the last to contribute his share, he selects a private key$x_{evil}$and computes$h_{evil} = g^{x_{evil}}$. However, instead of sharing$h_{evil}$as his ... 5 Formally, this is all very complicated, but informally: An interactive proof is a conversation between a prover and a verifier that ends with the verifier either accepting or rejecting. The interactive proof can be zero knowledge, in which case a cheating verifier does not learn anything new by talking to the honest prover. The interactive proof can be a ... 5 Yes. The easiest way is if$K$is an RSA private key, and Bob has the public key. Then, here's how it works; we'll call the ciphertext that Bob has$C$: Bob selects a random number$r$, and computes both$C \cdot r^e \bmod N$and$r^{-1} \bmod N$(where$e$and$N$are the public exponent and the modulus from the public key) Bob sends$C \cdot r^e \bmod ...

4

Alice can prove that the decryption of $C$ is $M$. This can be done using zero knowledge proofs. Simple example: Suppose $C = (x,w)$ is an ElGamal encryption of $M$ under public key $y$, that is $(x,w) = (g^r, y^r M)$ for some $r$. Suppose that the decryption key is $a$, that is, $y=g^a$. Then we know that $M = wx^{-a}$, or $x^a = w/M$. That is, the ...

4

The description of this "kid zero knowledge" example follows the strucure of how interactive proofs that are zero-knowledge usually work: The prover sends a commitment (walks into one of the two sides) The verifier challenges the prover (tosses the coin to decide which side the prover should walk out) The prover gives a response (walks out the side the ...

4

Use the exponential variant of ElGamal, where the plaintext is encoded in the exponent. Elliptic curve ElGamal is fine. In fact, any public key cryptosystem which allows raising ciphertexts to a power such that this operation corresponds homomorphically to multiplication for the plaintext. Your commitments are $c_x = \mathsf{E}(x)$; $c_y = \mathsf{E}(y)$; ...

4

Guillou and Quisquater (link) present a zero-knowledge proof of an RSA signature. Basically, the scheme is as follows: Public knowledge: RSA modulus $n$, public RSA exponent $v$, preimage $X$. Secret knowledge for prover: $A$, such that $A^v = X \mod n$. $$\begin{matrix} \mathcal{P} & & \mathcal{V} \\ r \xleftarrow{\} \mathbb{Z}_n^* ... 4 She can generate a key-pair and include the public key in the book. Having the private counterpart she can at any time proove that she wrote the book by signing an arbitrary statament. 3 Assumption: the normal user can read the message, which is displayed on his screen. Generic attack: the user uses a camera to take a snapshot of the screen when the message is displayed. And voila! What you seek is demonstrated to be impossible. 3 If they don't trust the server they sure shouldn't send any money. The "trusted" third party is used to solve the problem of participants who don't trust each other. So by definition, your problem can only be somewhat mitigated, not solved completely. I'm not sure what you mean by "provably fair". If the server can't prove he cannot cheat, it's not provably ... 3 The problem is one of notation for modular arithmetic, at the point of the question reading if y^2 equals (x * v^e) % n then.. Likely the textbook is about if y^2\equiv x\cdot v^e\pmod n then.. By definition of a\equiv b\pmod n, that holds if and only if n divides a-b (or equivalently: |b-a| is a multiple of n). In the question, 437 ... 3 In very simplified terms, a NI-ZK proof works in 2 stages: First, you run the protocol with a simulator (who is just a verifier, but the random choices are done differently), and then you can give the transcript of the protocol to anyone and convince them that the proof is real. The most important ways to achieve this are: In the random oracle model ... 3 The question is not very clear about exactly what you want to prove and what is publicly known, but here's my answer, based on my best guess at what you mean: Each party should publish (R_1,S_1) and (R_2,S_2). They should also publish (R_3,S_3). Now anyone can verify that (R_3,S_3) is a correctly-formed encryption of the sum of the messages ... 3 Schnorr can be proven zero knowledge when the challenge e is restricted to a small set (typically 0 and 1). Recall that in the Schnorr protocol, the prover knows the logarithm u of y to base g. He chooses a random value r, computes a = g^r and sends a to the verifier. The verifier chooses a random challenge e from some set and sends it ... 3 Short answer: YES. (Though see the note on "hashed" below.) Intro The remote authentication protocols where server does not know the plaintext password are generally known as augmented password authenticated key agreement (PAKE). You can see this wikipedia article for details of PAKE algorithms (augmented and non-augmented). You may find this reference ... 3 Zero-knowledge proofs of knowledge basically allow Alice to convince someone beyond a reasonable doubt that she knows a certain piece of information (i.e., the answer to a certain question), without revealing what exactly that information is. One simple example involves the discrete logarithm problem, which you might be familiar with: given a (large) prime ... 3 More generally, any encryption that is commutative can be used because then:$$(D_k \circ D_K \circ E_k \circ E_K)(m) = m$$I.e. Bob can encrypt the ciphertext E_K(m) with a new key k, then gives that to Alice for decoding with K and finally decodes it himself with k. Stream ciphers are commutative, as is exponentiation modulo n (used in RSA) ... 3 You said in the comments, that H can be a random oracle, but in this case it needs to be a random oracle. Basically your protocol is a Schnorr protocol in disguise, and you throw in the Fiat Shamir heuristic to make it noninteractive. But considering your questions: Well, the basic Schnorr protocol is not zero-knowledge. There exists no simulator for ... 2 So a fundamental property of multi-party anonymous systems is you are only anonymous out of the number of honest participants in the system. If the Stasi control everyone else at the dinner table and know they didn't send the message, then they know you did no matter what protocol you use. In your case with this ring topology, because only your two ... 2 Given the definition of a zero-knowledge proof, it must satisfy three properties: Completeness: if the statement is true, the honest verifier (that is, one following the protocol properly) will be convinced of this fact by an honest prover. Soundness: if the statement is false, no cheating prover can convince the honest verifier that it is true, ... 2 SRP doesn't require very much memory, has a low transmission cost (4 packets), and its CPU usage isn't very high (less than 1 second on a low-end 32-bit CPU, but higher on something like an 8-bit CPU). http://en.wikipedia.org/wiki/Secure_Remote_Password_protocol 2 The usual way to encode long random bitstrings, so that they can be easily memorized and/or entered by humans, is to break them into blocks of (typically) 10 to 12 bits and map each block to an entry in a fixed dictionary of common words. This approach is commonly used for secure passphrase generation, e.g. by Diceware, S/KEY and PGP. Assuming an 11-bit ... 2 Ok, here we are speaking of non-interactive zero-knowledge proof systems for some language L\in NP. We there have a pair \sf (P,V) of probabilistic polynomial time algorithms (called the prover and the verifier) where both have input x\in L and \sf P additionally holds a secret witness w for membership of x in L and wants to convince a ... 2 You can use the techniques in the paper you have linked to show that a list of commitments C_1,\ldots,C_m to the elements in \Sigma are elements in \Psi (the commitment scheme of choice are information-theoretically hiding Pedersen commitments, which are also used in the linked paper) . Basically, this works by the "owner" of the set \Psi publishing ... 2 There are a couple of problems with this zero knowledge proof; it isn't explained very well it gives the verifier more information than it claims to Lets go through them in order: It isn't explained very well The problem is that the terminology it uses in the proof statement differs from the terminology they use in the protocol description; for ... 2 A homomorphic cryptosystem has some operation * on ciphertexts that correspond to some other operation \circ on plaintexts, that is$$\mathcal{D}(c_1 * c_2) = \mathcal{D}(c_1) \circ \mathcal{D}(c_2). Typically, the ciphertexts you get by applying $*$ look like ciphertexts that are produced by the encryption algorithm. For Damgård-Jurik, $*$ is ...

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