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10

If both $G_1$ and $G_2$ have prime order $r$, then this means that there are generators $g_1$ and $g_2$; thus, for every $u_1 \in G_1$, there is an integer $x_1$ modulo $r$ such that $u_1 = g_1^{x_1}$. Therefore, every pairing value $e(u_1, u_2)$ is equal to $e(g_1^{x_1},g_2^{x_2}) = e(g_1, g_2)^{x_1x_2}$ by bilinearity. It follows that $e(g_1,g_2)$ is a ...


10

The cornerstone of the argument is the following: If the cycle attack works, then you can factor $n$ (see details below). The attacker can choose $e$. I.e., when trying to factor $n$, the attacker is not constrained to use the specific $e$ which you selected for your public key; he can invent his own $e$, since he will do all the computations himself. ...


10

$g^x \cdot g^y \;\;\; = \;\;\; (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x$ of them] $\ldots \cdot g\cdot g\cdot g) \: \cdot \: (\hspace{.02 in}g\cdot g\cdot g\cdot \ldots$ [$\hspace{.03 in}y$ of them] $\ldots \cdot g\cdot g\cdot g)$ $= \;\;\; g\cdot g\cdot g\cdot \ldots$ [$\hspace{.02 in}x\hspace{-0.05 in}+\hspace{-0.05 in}y$ of them] ...


9

It depends. If the order $m$ of $g$'s group is known and $a$ has an inverse modulo $m$ (which is the case if and only if $a$ is coprime to $m$), then it is easy: Calculate the inverse $b:=a^{-1}\bmod m$ (for instance, using the Euclidean algorithm), and compute the power $(g^a)^b$. By Lagrange's theorem, this equals $g$. However, there are cases for which ...


8

As far as I understand, the HSP is a hard problem such that: some types of HSP (namely those operating in an abelian group) can (theoretically) be solved efficiently on a quantum computer (assuming one can be built); many types of public key cryptosystems can be reduced to the HSP: if you can solve the HSP you can break the key. In particular, integer ...


8

A safe prime is a prime number $p$ for which $(p-1)/2$ is also prime. The order of an element $g$ of the group $\mathbf{Z}^*_p$ (the integers modulo $p$, excluding 0) is the smallest integer $n$ such that $g^n\equiv 1\pmod{p}$; this is always a factor of $p-1$. The orders of the subgroups of the group generated by $g$ are the factors of the order of $g$; ...


7

The question makes a number of statements that are incorrect. It is not correct that a fixed point is guaranteed to exist. It is not correct that if you hold the plaintext constant and vary the key, then a fixed point is guaranteed to exist. Moreover, the existence of fixed points has only an extremely tenuous connection to security. Assume $E$ is a ...


7

Basically, every time you choose a group where the required hard problem is not hard, then you will run into a problem. Even if we have a problem instance that is of size that is considered secure in the setting of asymmetric cryptography. Lets for instance implement a discrete logarithm style cryptosystem in the group $Z_n$ with addition and let $g$ be a ...


6

Well, the reason that a specific cryptographical object needs to work in a specific subgroup probably has to do with the details of that object, and the cryptographical properties it needs from the subgroup. One obvious possibility is that they need to avoid leaking information via the Jacobi symbol; that is an easily computed function that maps values in ...


6

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 ...


6

If $\mathcal{G}$ is of order $N$ (who doesn't look like a prime number btw) and $g$ is a generator of $\mathcal{G}$ then $g$ has order N. Since $(g^{nq})^p=1$ and $\forall 1\le k <p, (g^{nq})^k\neq 1$ (g has order $N$ and $knq<N$) then $g^{nq}$ generates a subgroup of $\mathcal{G}$ of order $p$ and there's only one such subgroup : $\mathcal{G_p}$. The ...


6

DrLecter gave a good answer, I just wanted to include another well-known example. The Pohlig-Hellman algorithm can be used to compute discrete logs in groups whose order is a smooth integer. If two parties executing a textbook Diffie-Hellman key exchange use as their modulus a prime $p$ such that $p-1$ has only small factors (is 'smooth') an eavesdropping ...


6

Note that you do not have an efficiently computable homomorphism from $G_1$ to $G_2$, but in Type-2 you have an efficiently computable homomorphism $\psi: G_2 \rightarrow G_1$ and in Type-3 you do not have one. But what I don't understand is what is the use of the homomorphism in cryptography? Well, if you have a tuple $(aP',bP',cP')\in G_2^3$ with ...


5

With addition and $\mathbb{Z}_n$, each party chooses a secret $x$ and sends $xg \pmod n$ over the wire, for an agreed upon generator $g$. Division by $g$ modulo $n$ is easily computable, and reveals $x$. In other words, a prerequisite for DH to be secure is that the equivalent to discrete logarithm is hard in the chosen group. With $\mathbb{Z}_n$ and ...


5

This is due to the Extended Euclidean algorithm, which allows us to compute inverses modulo any number. If the modulus is prime, things are even more easier to explain. For prime $p$, we know that $g^{p-1} \equiv 1 \pmod{p}$. Therefore, $y = g^{p-2} \equiv 1/g \pmod {p}$. Therefore, $(xg).y \equiv x \pmod{p}$, revealing the secret key. If modulus is not ...


5

There are some known groups in which computational Diffie-Hellman assumption is equivalent to discrete logarithm problem. Besides, It has been shown that the equivalence holds "when a small amount of extra information depending on the group order is provided". Furthermore, those extra informations has been computed for certain elliptic curve groups used in ...


5

Ok, I will start with a cryptographic bilinear map. Cryptographic Bilinear Map A cryptographic bilinear map $e: G_1\times G_2 \rightarrow G_T$ as the name says is a map that is linear in both components, i.e., it holds that for all $g\in G_1$ and $h\in G_2$ and all $a,b\in Z_p$ (where $p$ is the order of all groups) we have that $e(g^a,h^b)=e(g,h)^{ab}$. ...


4

A composite order group is like having a 2-dimensional vector space, because of the Chinese Remainder Theorem. More concretely in the context of a bilinear map, if $g$ is a generator with order $N=pq$, then $g_p = g^q$ generates an order-$p$ subgroup, and $g_q = g^p$ generates an order-$q$, and $e(g_p, g_q) = 1$. They cancel each other out, and so you can ...


4

The answer comes from Euler's Theorem. Note: math below is done modulo $N$ unless otherwise specified and draws heavily from group theory. That theorem says that any element of a group (say $m$) raised to the order of the group, in this case $\phi(N)$ is congruent to $1$ (i.e., $m^{\phi(N)}\equiv 1\bmod{N}$). Furthermore, this holds for multiples of ...


4

Almost all cryptographic algorithms which use groups actually work in subgroups generated by a conventional element; even if the group as a whole is non-abelian, the subgroup is cyclic, thus abelian. The Anshel-Anshel-Goldfeld protocol tries to use non-commutativity itself, and relies on "how much non-abelian" the group is. All asymmetric cryptographic ...


4

Every element $g$ in a group $G$ generates a subgroup of $G$ of order $r$, where $r$ is the smallest (non-zero) integer such that $g^r = 1$. Moreover, if $g^s = 1$ for some positive value $s$, then $s$ is a multiple of $r$. Finally, $r$ necessarily divides the order of $G$ (i.e. the number of elements in $G$). Therefore, if your group order is $N = ab$ for ...


4

Your first two paragraphs made a series of statements; these statements were less than perfectly accurate, and D.W. attempted to address those. You then went on and asked What I don't understand is how a key that is longer than a block size provides any extra security. From what I understand this would suggest the existence of many fixed points, ...


4

Not at all. It's very trivial: $$(g^{ab})^{b^{-1}}=g^a$$


4

As for the question "how difficult would it be to solve a random instance of a discrete log problem modulo an RSA modulus", well, it turns out that we can give a fairly solid answer; which is essentially "about as difficult as factoring the modulus". Here's is a demonstration that the discrete log problem is not drastically easier than factoring the ...


4

The problem you are referring to seems to be the Decisional Linear Assumption (DLIN), which states that given $(u,v,u^a,v^b)\in \mathbb{G}^4$, it is hard to distinguish a couple $(h,h^{a+b}) \in \mathbb{G}^2$ from a totally random couple $(h,h') \in \mathbb{G}^2$. There is also the Computational Linear Assumption (CLIN), which states that it is hard to ...


4

Firstly, $|\mathbb Z_n|=n$, whereas $|\mathbb Z_n^*|=\varphi(n)<n$. So, by the pigeon-hole principal there cannot be a mathematically invertible function $f:\mathbb Z_n\to\mathbb Z_n^*$. So, lets relax our idea of what 'invertible' means a bit. How about ensuring every element of $\mathbb Z_n^*$ has a preimage? Yep, we can do that. To use a couple of ...


4

The number of points on the curve $|E({\mathbb F}_p)|$ is defined as $|E({\mathbb F}_p)|=p+1-t$ where $t$ is the so called trace of Frobenius. Using Hasse's theorem one can bound $t$ as $|t| \leq 2\sqrt p$, which gives you an estimation for the number of points for $E({\mathbb F}_p)$. Now you could use a naive algorithm and simply run through all elements ...


4

Originally we had restrictions like this in place as the alternative was emails about bugs in ciphers like RSA resulting from people not understanding the implications of modular arithmetic (if you are going to push right to the boundary you really have to understand what's going on). Having said that, I actually think we'd outgrown this by the time ElGamal ...


4

I'll add something to the previous answer. The first way to construct multilinear maps is pretty recent and was introduced by Sanjam Garg, Craig Gentry and Shai Halevi. What we want is given groups $G_1,\ldots,G_n$ and $G_T$ a map: $$e:G_1\times\cdots\times G_n\to G_T$$ that satisfies the linearity property in DrLecter's answer. It's worth nothing here, ...


4

One way to do this, if you're working with a multiplicative group $Z^*_p$, is to pick a prime $p$ so that $p-1$ has a large prime factor $q$; once you have this, then to generate a generator of order $q$, you pick a random value $h$, compute $g = h^{(p-1)/q}$, and if that is not 1, then $g$ is a generator of your group. Obvious questions: How do you find ...



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