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The problems: The Discrete Logarithm problem: Given $y$, find $x$ so that $g^x = y$. The Computational Diffie-Hellman problem: Given $y_1 = g^{x_1}$ and $y_2 = g^{x_2}$ (but not $x_1$ and $x_2$), find $y = g^{x_1·x_2}$. The Decisional Diffie-Hellman problem: Given $y_1, y_2, y_3$, decide if they are of the form $y_1 = g^{x_1}$, $y_2 = g^{x_2}$ and ...

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I have asked a similar question to Arjen Lenstra a few years ago: I was investigating three 2048-bit primes of low Hamming weight: $p_1 = 2^{2048} - 2^{1056} + 2^{736} - 2^{320} + 2^{128} + 1$ $p_2 = 2^{2048} - 2^{1376} + 2^{992} + 2^{896} + 2^{640} - 1$ $p_3 = 2^{2048} - 2^{2016} + 2^{1984} - 2^{1856} - 2^{1824} + 2^{1792} - 2^{1760} + 2^{1696} + 2^{1664} ... 8 Discrete logarithms in$\mathbb{F}_{p}$share the same asymptotic complexity as integer factorization for general numbers:$L_p[1/3,1.923]$for general integers,$L_p[1/3,1.587]$for special integers. Discrete logarithms in$\mathbb{F}_{p^n}$have the same asymptotic complexity as factoring special integers, i.e.$L_{p^n}[1/3, 1.587]$, via the Function ... 8 The generic discrete logarithm problem is this: Given a group$(G, ·)$with generator$g$and$y \in G$, find$x \in \mathbb N$such that$y = g^x$. The "classic" discrete logarithm problem (the one used in "classic" DSA and ECDSA) is this with some subgroup of the (multiplicative) group of a prime field, i.e.$(\mathbb Z/p \mathbb Z)^*$: Given a ... 6 Short answer: Yes. The discrete logarithm can be attacked in a multitude of ways: Baby-step giant-step (BSGS), Pollard's Rho, Pohlig-Hellman, and the several variants of Index Calculus, the best of which currently is the Number Field Sieve. Let$n$be the order of the generator of our field$\mathbb{F}_p$; it is$n = p-1$. We are trying to find$x$given ... 6 256-bit discrete logarithms on a prime field are definitely not of the order of magnitude used in cryptographic applications. Secure sizes for this problem are in the thousands of bits, very much like integer factorization. To break that example discrete logarithm, you probably want to use Index Calculus, more specifically the Linear Sieve. Resorting to the ... 5 Oh, and while you did not specifically ask about this, there is another point I believe that is important to highlight; DH and SRP are different protocols, and have different requirements on the generator they use. In particular, taking a generator that is designed to be used securely within DH can void the security properties of SRP. Here's what's going ... 5 Well, yes, that is generally good advice about DH. Here is some background on this: support you were given a value$g^x \bmod p$, and you were also told that$1 \le x \le A$for some value$A$. If so, then there are several known attacks (such as Big Step/Little Step and Pollard's Rho) that can recover$x$in about$\sqrt A$steps. If we have as our ... 5 For$p = 2q+1$, one can note that elements of$\mathbb{G}_q$are exactly the non-zero quadratic residues modulo$p$: Since$p$is prime,$\mathbb{Z}_p$is a field. Hence, the polynomial$X^q-1$, being of degree$q$, cannot have more than$q$roots in$\mathbb{Z}_p$. So$\mathbb{G}_q$contains all the$q$values of order$1$or$q$. If$x$is a non-zero ... 5 Antoine Joux very kindly sent me the following on the topic: People worry that [logarithms over fields with composite exponent] might be easier, this is why they use prime exponent. For some factorization of the exponent, viewing the finite field as a tower of extensions$(p^{n_1})^{n_2}$indeed makes things easier. See "The function field sieve ... 5 For ElGamal to be secure, the 'discrete log problem' (which is, given$g$and$g^x$, find$x$) must be intractable. You give a generic way to attack the discrete log problem for a group with$n$elements with something like$n$steps (I say about because your approach isn't the simplest version of this type of attack; the simplest does take$n$steps); ... 5 No, it is not possible to compute$\lambda$easily. Specifically, if you have a black box that, given a random instance$c$,$c^\lambda \bmod n^2$, was able to recover$\lambda$with nontrivial probability, you can use that to factor$n$with nontrivial probability. Hence, if we believe the factorization problem is hard, we must also believe that this ... 4 For the second case, mapping numbers from$\mathbb{Z}_q$to$\mathbb{G}_q$and back when:$p=aq+1$with an$a$such that, e.g., |p|=1024 and |q|=160 It appears an efficient subgroup encoding/decoding scheme does not exist. Although it has not been proven that one cannot exist, notable cryptographers have conjectured it in the literature. For example, ... 4 Probably the easiest solution for the case a=2 is to map$m\in\{1\ldots q\}$to$(m/p)m$where$(m/p)$is the Legendre symbol. The inverse can be obtained by mapping a quadratic residue$x\in Z/(pZ)^*$either to x or -x depending on which of the two residue classes contains an integer in$\{1\ldots q\}$. This is of course a well know solution, but I can't ... 4 There is a reduction from DL to RSA if the DL oracle accepts composite modulus. For prime modulus, a reduction is not known. I copied the following from this wikipedia page with minor edits. Let$n = pq$be RSA modulus. Generate random number$a$co-prime to$n$and random number$x < n$but very close to$n$. Compute$b = a^x \text{ mod } n$but ... 4 The discrete logarithm problem can be attacked with either a specific or a generic algorithm. A specific algorithm is one that tries to exploit structural weaknesses of the specific group in which discrete logarithm is used; e.g. Index Calculus when we are talking about exponentiation modulo a big prime. Generic algorithms only use the group law and thus ... 4 To complete @Samuel's answer, there are a few shortcuts that can be used when n is composite; however, they only contribute small constant factors, hence they do not change the asymptotic behavior: If n can be divided by r, then one can first solve the discrete logarithm in the subfield GF(2r). In a sieve-based algorithm, this can provide up to half the ... 4 Ok, lets try again from scratch. First, some group theory definitions: A group is a set of elements and an operation$\times$which satisfies a specific set of properties (closure, associativity, an identity element, inverses) A subgroup is a subset of elements of a group which, along with the group operation from the main group, satisfies those same set ... 4 It means that if you have an oracle access to CDH, then you can solve DDH, but we do not know if there exists a reduction the other way round. Technically, suppose$\mathsf{CDH}$is an oracle that finds$g^{xy}$, given$(g^x,g^y)$, then for a DDH instance, say$(a,b,c)$, you can feed$\mathsf{CDH}$with$(a,b)$and output$1$if output of$\mathsf{CDH}$... 4 I don't know the general answer, however, it appears that Baby-step Giant-step is able to give you the solution in$O(\sqrt{in})$time (where$n$is the size of$G$); this is$O(\sqrt{i})$times longer than it takes the same algorithm to solve a single discrete-log problem. The first observation is that if you know the group order$n$and a group generator ... 4 I'm not sure how to answer the question about "trust" in the assumption. I suppose that is a matter of personal belief more than science. We don't know the truthfulness about any cryptographic assumptions, although obviously some have received more scrutiny than others. I'd say that the KEA assumptions have received relatively little attention, and so should ... 4 Here's the deal. The discrete log problem is feasible in the special case where the exponent is known to come from a small range of possibilities (e.g., the exponent is not too large). Suppose we are given$y=g^m$, and we want to find$m$. Suppose moreover we know that$m$is small:$0 \le m < 2^{30}$, say. Then it turns out it is easy to recover$m$, ... 3 In addition to Jalaj's correct answer, I also want to dispute one of your statements: "From my understanding, since the Discrete Log(DL) Problem is considered hard, then so is CDH" Actually, that's not the case. Now, if you can solve the DL problem, the CDH problem is easy (and so the CDH problem is no harder than the DL problem). However, there's no ... 3 It sounds like you are thinking about this the wrong way. Verifying a commitment is very fast, if you choose the commitment scheme properly. In particular, I recommend that you use a hash-function-based commitment scheme: C(i) = Hash(x(i) || open(i)). Then verifying an opened commitment requires just one hash evaluation, which is very fast. Based on my ... 3 The flaw in your reasoning lies in your assumption that finding the discrete log of$f$is any easier than finding the discrete log of$g$. It just isn't. That assumption is not correct. And if you try to recurse and apply your procedure to$f$, you'll recurse for a very long time, and your algorithm will take exponentially long. Let me summarize your ... 2 According to this: To summarize: solving the discrete logarithm problem for a composite modulus is exactly as hard as factoring and solving it modulo primes. So, given your question "Would the ability to efficiently find Discrete Logs have any impact on the security of RSA?" the answer would be yes. Furthermore, if you can solve DLP for composite ... 2 You need to refine your definition of discrete logarithm to get a precise answer, as the discrete logarithm problem can be defined for any group, Being able to compute the discrete logarithm on the group of points of a degenerate elliptic curves defined over the ring$Z_n$also yields the factorization of$n$(see Silverman's xedni calculus). 2 When we say an algorithm is 'exponential' or 'polynomial', when mean that the running time is bounded by an exponential or polynomial function of the size of the input in bits. That is, the input of the function is not$p$, but$\log p$. And so,$O(\sqrt p ) = O( e^{1/2 \log p})$is bounded by an exponential in$\log p$. Also note that the Wiki article is ... 2 On the third case, I have a comment. The third oracle may help the adversary using Cheon's algorithm for the DL problem. Let$q$be a prime order of the subgroup$\mathbb{G}$of$(\mathbb{Z}/p\mathbb{Z})^{\times}$. In the third case, the adversary has an oracle$a \mapsto a^k$for any$a$. Hence, it can obtain$g^{k^i}$from$g^{k^{i-1}}\$ and so on. When ...

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In practise the view would be that no, it does not get any easier. Indeed many popular deployed schemes depend on it. For example the Trusted Authority in the Boneh-Franklin IBE scheme has a master secret s and issues private keys to users in the form s.ID_i, where ID_i is a point on an elliptic curve, and ID_i is related to the identity of the i-th user. It ...

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