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What is the average complexity of exhaustive search when the key is distributed uniformly at random over $N$ keys? | In 1990, Impagliazzo and Levin showed that if there is an efficient average-case algorithm for a distNP-complete problem under the uniform distribution, then there is an average-case algorithm for every problem in NP under any polynomial-time samplable distribution. Applying this theory to natural distributional problems remains an outstanding open question.In 1992, Ben-David et al. showed that if all languages in distNP have good-on-average decision algorithms, they also have good-on-average search algorithms. Further, they show that this conclusion holds under a weaker assumption: if every language in NP is easy on average for decision algorithms with respect to the uniform distribution, then it is also easy on average for search algorithms with respect to the uniform distribution. Thus, cryptographic one-way functions can exist only if there are distNP problems over the uniform distribution that are hard on average for decision algorithms. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the average complexity of exhaustive search when the key is distributed uniformly at random over $N$ keys? | The outline of a formal proof of the O(n log n) expected time complexity follows. Assume that there are no duplicates as duplicates could be handled with linear time pre- and post-processing, or considered cases easier than the analyzed. When the input is a random permutation, the rank of the pivot is uniform random from 0 to n − 1. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select \emph{incorrect} statement. Generic attacks on DES include | The following C code demonstrates a real problem that can arise if #include guards are missing: | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select \emph{incorrect} statement. Generic attacks on DES include | As soon as SPF implementations detect syntax errors in a sender policy they must abort the evaluation with result PERMERROR. Skipping erroneous mechanisms cannot work as expected, therefore include:bad.example and redirect=bad.example also cause a PERMERROR. Another safeguard is the maximum of ten mechanisms querying DNS, i.e. any mechanism except from IP4, IP6, and ALL. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
AES\dots | U+2299 ⊙ CIRCLED DOT OPERATOR | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
AES\dots | Note that one can also overcome the problem with containing dots using the \yahnodots command. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given that $100000000003$ is prime, what is the cardinality of $\mathbf{Z}_{200000000006}^*$? | 103 is a prime number, the largest prime factor of 6 ! + 1 = 721 = 7 ⋅ 103 {\displaystyle 6!+1=721=7\cdot 103} . The previous prime is 101, making them both twin primes. It is the fifth irregular prime, because it divides the numerator of the Bernoulli number The equation 64 3 + 94 3 = 103 3 + 1 {\displaystyle 64^{3}+94^{3}=103^{3}+1} makes 103 part of a "Fermat near miss".There are 103 different connected series-parallel partial orders on exactly six unlabeled elements.103 is conjectured to be the smallest number for which repeatedly reversing the digits of its ternary representation, and adding the number to its reversal, does not eventually reach a ternary palindrome. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given that $100000000003$ is prime, what is the cardinality of $\mathbf{Z}_{200000000006}^*$? | (See OEIS: A060003 for odd Stern numbers) Christian Goldbach conjectured in a letter to Leonhard Euler that every odd integer is of the form p + 2b2 for integer b and prime p. Laurent Hodges believes that Stern became interested in the problem after reading a book of Goldbach's correspondence. At the time, 1 was considered a prime, so 3 was not considered a Stern prime given the representation 1 + 2(12). The rest of the list remains the same under either definition. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. Elliptic Curve Diffie-Hellman is | Diffie–Hellman (RFC 3526) ECDH (RFC 4753) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. Elliptic Curve Diffie-Hellman is | Elliptic-curve Diffie–Hellman (ECDH) is a key agreement protocol that allows two parties, each having an elliptic-curve public–private key pair, to establish a shared secret over an insecure channel. This shared secret may be directly used as a key, or to derive another key. The key, or the derived key, can then be used to encrypt subsequent communications using a symmetric-key cipher. It is a variant of the Diffie–Hellman protocol using elliptic-curve cryptography. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In which attack scenario does the adversary ask for the decryption of selected messages? | In a chosen-plaintext attack the adversary can (possibly adaptively) ask for the ciphertexts of arbitrary plaintext messages. This is formalized by allowing the adversary to interact with an encryption oracle, viewed as a black box. The attacker’s goal is to reveal all or a part of the secret encryption key. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In which attack scenario does the adversary ask for the decryption of selected messages? | In their foundational paper, Goldwasser, Micali, and Rivest lay out a hierarchy of attack models against digital signatures: In a key-only attack, the attacker is only given the public verification key. In a known message attack, the attacker is given valid signatures for a variety of messages known by the attacker but not chosen by the attacker. In an adaptive chosen message attack, the attacker first learns signatures on arbitrary messages of the attacker's choice.They also describe a hierarchy of attack results: A total break results in the recovery of the signing key. A universal forgery attack results in the ability to forge signatures for any message. A selective forgery attack results in a signature on a message of the adversary's choice. An existential forgery merely results in some valid message/signature pair not already known to the adversary.The strongest notion of security, therefore, is security against existential forgery under an adaptive chosen message attack. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
An element of the finite field $\mathrm{GF}(2^8)$ is usually represented by\dots | Let X n {\displaystyle X^{n}} be an n-dimensional vector space over the finite field G F ( q N ) {\displaystyle GF\left({q^{N}}\right)} , where q {\displaystyle q} is a power of a prime and N {\displaystyle N} is a positive integer. Let ( u 1 , u 2 , … , u N ) {\displaystyle \left(u_{1},u_{2},\dots ,u_{N}\right)} , with u i ∈ G F ( q N ) {\displaystyle u_{i}\in GF(q^{N})} , be a base of G F ( q N ) {\displaystyle GF\left({q^{N}}\right)} as a vector space over the field G F ( q ) {\displaystyle GF\left({q}\right)} . Every element x i ∈ G F ( q N ) {\displaystyle x_{i}\in GF\left({q^{N}}\right)} can be represented as x i = a 1 i u 1 + a 2 i u 2 + ⋯ + a N i u N {\displaystyle x_{i}=a_{1i}u_{1}+a_{2i}u_{2}+\dots +a_{Ni}u_{N}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
An element of the finite field $\mathrm{GF}(2^8)$ is usually represented by\dots | Primitive polynomials can be used to represent the elements of a finite field. If α in GF(pm) is a root of a primitive polynomial F(x), then the nonzero elements of GF(pm) are represented as successive powers of α: G F ( p m ) = { 0 , 1 = α 0 , α , α 2 , … , α p m − 2 } . {\displaystyle \mathrm {GF} (p^{m})=\{0,1=\alpha ^{0},\alpha ,\alpha ^{2},\ldots ,\alpha ^{p^{m}-2}\}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider $GF(8)$ defined as $\mathbb{Z}_2[X]/(P(X))$ with $P(x) = X^3 + X + 1$. Compute $X^2 \times (X + 1)$ in $\mathbb{Z}_2[X]/(P(X))$ | G F ( p 2 ) {\displaystyle GF(p^{2})} Let p be a prime such that p ≡ 2 mod 3 and p2 - p + 1 has a sufficiently large prime factor q. Since p2 ≡ 1 mod 3 we see that p generates ( Z / 3 Z ) ∗ {\displaystyle (\mathbb {Z} /3\mathbb {Z} )^{*}} and thus the third cyclotomic polynomial Φ 3 ( x ) = x 2 + x + 1 {\displaystyle \Phi _{3}(x)=x^{2}+x+1} is irreducible over G F ( p ) {\displaystyle GF(p)} . It follows that the roots α {\displaystyle \alpha } and α p {\displaystyle \alpha ^{p}} form an optimal normal basis for G F ( p 2 ) {\displaystyle GF(p^{2})} over G F ( p ) {\displaystyle GF(p)} and G F ( p 2 ) ≅ { x 1 α + x 2 α p: x 1 , x 2 ∈ G F ( p ) } . {\displaystyle GF(p^{2})\cong \{x_{1}\alpha +x_{2}\alpha ^{p}:x_{1},x_{2}\in GF(p)\}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider $GF(8)$ defined as $\mathbb{Z}_2[X]/(P(X))$ with $P(x) = X^3 + X + 1$. Compute $X^2 \times (X + 1)$ in $\mathbb{Z}_2[X]/(P(X))$ | Considering that p ≡ 2 mod 3 we can reduce the exponents modulo 3 to get G F ( p 2 ) ≅ { y 1 α + y 2 α 2: α 2 + α + 1 = 0 , y 1 , y 2 ∈ G F ( p ) } . {\displaystyle GF(p^{2})\cong \{y_{1}\alpha +y_{2}\alpha ^{2}:\alpha ^{2}+\alpha +1=0,y_{1},y_{2}\in GF(p)\}.} The cost of arithmetic operations is now given in the following Lemma labeled Lemma 2.21 in "An overview of the XTR public key system":Lemma Computing xp is done without using multiplication Computing x2 takes two multiplications in G F ( p ) {\displaystyle GF(p)} Computing xy takes three multiplications in G F ( p ) {\displaystyle GF(p)} Computing xz-yzp takes four multiplications in G F ( p ) {\displaystyle GF(p)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be a positive integer. An element $x \in \mathbb{Z}_n$ is \emph{always} invertible when \dots | Indeed, we assume that every nonzero element of the ring Z / n Z {\displaystyle \mathbb {Z} /n\mathbb {Z} } is invertible, so that Z / n Z {\displaystyle \mathbb {Z} /n\mathbb {Z} } must be a field. It implies that n {\displaystyle n} must be prime (cf. Bézout's identity). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be a positive integer. An element $x \in \mathbb{Z}_n$ is \emph{always} invertible when \dots | Let z ∈ N {\displaystyle z\in \mathbb {N} } be an arbitrary natural number. We will show that there exist unique values x , y ∈ N {\displaystyle x,y\in \mathbb {N} } such that z = π ( x , y ) = ( x + y + 1 ) ( x + y ) 2 + y {\displaystyle z=\pi (x,y)={\frac {(x+y+1)(x+y)}{2}}+y} and hence that the function π(x, y) is invertible. It is helpful to define some intermediate values in the calculation: w = x + y {\displaystyle w=x+y\!} t = 1 2 w ( w + 1 ) = w 2 + w 2 {\displaystyle t={\frac {1}{2}}w(w+1)={\frac {w^{2}+w}{2}}} z = t + y {\displaystyle z=t+y\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of these attacks applies to the Diffie-Hellman key exchange when the channel cannot be authenticated? | If Alice and Bob use random number generators whose outputs are not completely random and can be predicted to some extent, then it is much easier to eavesdrop. In the original description, the Diffie–Hellman exchange by itself does not provide authentication of the communicating parties and is thus vulnerable to a man-in-the-middle attack. Mallory (an active attacker executing the man-in-the-middle attack) may establish two distinct key exchanges, one with Alice and the other with Bob, effectively masquerading as Alice to Bob, and vice versa, allowing her to decrypt, then re-encrypt, the messages passed between them. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of these attacks applies to the Diffie-Hellman key exchange when the channel cannot be authenticated? | They will know that all of their private conversations had been intercepted and decoded by someone in the channel. In most cases it will not help them get Mallory's private key, even if she used the same key for both exchanges. A method to authenticate the communicating parties to each other is generally needed to prevent this type of attack. Variants of Diffie–Hellman, such as STS protocol, may be used instead to avoid these types of attacks. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is an acceptable commitment scheme, i.e., one that verifies the hiding and binding property (for a well chosen primitive and suitable $x$ and $r$): | If the commitment C to a value x is computed as C:=Commit(x,open) with open being the randomness used for computing the commitment, then CheckReveal (C,x,open) reduces to simply verifying the equation C=Commit (x,open). Using this notation and some knowledge about mathematical functions and probability theory we formalise different versions of the binding and hiding properties of commitments. The two most important combinations of these properties are perfectly binding and computationally hiding commitment schemes and computationally binding and perfectly hiding commitment schemes. Note that no commitment scheme can be at the same time perfectly binding and perfectly hiding – a computationally unbounded adversary can simply generate Commit(x,open) for every value of x and open until finding a pair that outputs C, and in a perfectly binding scheme this uniquely identifies x. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is an acceptable commitment scheme, i.e., one that verifies the hiding and binding property (for a well chosen primitive and suitable $x$ and $r$): | A better example of a perfectly binding commitment scheme is one where the commitment is the encryption of x under a semantically secure, public-key encryption scheme with perfect completeness, and the decommitment is the string of random bits used to encrypt x. An example of an information-theoretically hiding commitment scheme is the Pedersen commitment scheme, which is computationally binding under the discrete logarithm assumption. Additionally to the scheme above, it uses another generator h of the prime group and a random number r. The commitment is set c = g x h r {\displaystyle c=g^{x}h^{r}} .These constructions are tightly related to and based on the algebraic properties of the underlying groups, and the notion originally seemed to be very much related to the algebra. However, it was shown that basing statistically binding commitment schemes on general unstructured assumption is possible, via the notion of interactive hashing for commitments from general complexity assumptions (specifically and originally, based on any one way permutation) as in. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A 128-bit key ... | The main key supplied from user is of 64 bits. The following operations are performed with it. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A 128-bit key ... | Eight bits are used solely for checking parity, and are thereafter discarded. Hence the effective key length is 56 bits. The key is nominally stored or transmitted as 8 bytes, each with odd parity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a hash function $H$ with $n$ output bits. Tick the \emph{incorrect} assertion. | h {\displaystyle h}: a collision resistant hash function with |q|-bit digests. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a hash function $H$ with $n$ output bits. Tick the \emph{incorrect} assertion. | The number of hash functions, k, must be a positive integer. Putting this constraint aside, for a given m and n, the value of k that minimizes the false positive probability is k = m n ln 2. {\displaystyle k={\frac {m}{n}}\ln 2.} The required number of bits, m, given n (the number of inserted elements) and a desired false positive probability ε (and assuming the optimal value of k is used) can be computed by substituting the optimal value of k in the probability expression above: ε = ( 1 − e − ( m n ln 2 ) n m ) m n ln 2 {\displaystyle \varepsilon =\left(1-e^{-({\frac {m}{n}}\ln 2){\frac {n}{m}}}\right)^{{\frac {m}{n}}\ln 2}} which can be simplified to: ln ε = − m n ( ln 2 ) 2 . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Enigma | A K Peters. Hodges, Andrew (1983). Alan Turing: The Enigma. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Enigma | Mark Saltveit, 1996–. The Enigma. National Puzzlers' League, 1883–. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots | Bob does not want Alice to know which one he receives. Alice generates an RSA key pair, comprising the modulus N {\displaystyle N} , the public exponent e {\displaystyle e} and the private exponent d {\displaystyle d} . She also generates two random values, x 0 , x 1 {\displaystyle x_{0},x_{1}} and sends them to Bob along with her public modulus and exponent. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots | Given block size r, a public/private key pair is generated as follows: Choose large primes p and q such that r | ( p − 1 ) , gcd ( r , ( p − 1 ) / r ) = 1 , {\displaystyle r\vert (p-1),\operatorname {gcd} (r,(p-1)/r)=1,} and gcd ( r , ( q − 1 ) ) = 1 {\displaystyle \operatorname {gcd} (r,(q-1))=1} Set n = p q , ϕ = ( p − 1 ) ( q − 1 ) {\displaystyle n=pq,\phi =(p-1)(q-1)} Choose y ∈ Z n ∗ {\displaystyle y\in \mathbb {Z} _{n}^{*}} such that y ϕ / r ≢ 1 mod n {\displaystyle y^{\phi /r}\not \equiv 1\mod n} .Note: If r is composite, it was pointed out by Fousse et al. in 2011 that the above conditions (i.e., those stated in the original paper) are insufficient to guarantee correct decryption, i.e., to guarantee that D ( E ( m ) ) = m {\displaystyle D(E(m))=m} in all cases (as should be the case). To address this, the authors propose the following check: let r = p 1 p 2 … p k {\displaystyle r=p_{1}p_{2}\dots p_{k}} be the prime factorization of r. Choose y ∈ Z n ∗ {\displaystyle y\in \mathbb {Z} _{n}^{*}} such that for each factor p i {\displaystyle p_{i}} , it is the case that y ϕ / p i ≠ 1 mod n {\displaystyle y^{\phi /p_{i}}\neq 1\mod n} .Set x = y ϕ / r mod n {\displaystyle x=y^{\phi /r}\mod n} The public key is then y , n {\displaystyle y,n} , and the private key is ϕ , x {\displaystyle \phi ,x} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{false} assertion concerning WEP | asserts that if p {\displaystyle p\,\!} is valid then so is q {\displaystyle q\,\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{false} assertion concerning WEP | \operatorname {supp} \psi } Above, c . h . {\displaystyle \operatorname {c.h.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. Which of the following is \emph{not} a group in the general case? | The integers under the multiplication operation, however, do not form a group. This is because, in general, the multiplicative inverse of an integer is not an integer. For example, 4 is an integer, but its multiplicative inverse is 1/4, which is not an integer. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. Which of the following is \emph{not} a group in the general case? | Some non-prime numbers n are such that every group of order n is cyclic. One can show that n = 15 is such a number using the Sylow theorems: Let G be a group of order 15 = 3 · 5 and n3 be the number of Sylow 3-subgroups. Then n3 ∣ {\displaystyle \mid } 5 and n3 ≡ 1 (mod 3). The only value satisfying these constraints is 1; therefore, there is only one subgroup of order 3, and it must be normal (since it has no distinct conjugates). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} statement. | Suppose we are given a Boolean expressions: B 1 = ( v 3 ∨ ¬ v 2 ) ∧ ( ¬ v 1 ∨ ¬ v 3 ) {\displaystyle B_{1}=(v_{3}\lor \neg v_{2})\wedge (\neg v_{1}\lor \neg v_{3})} B 2 = ( v 3 ∨ ¬ v 2 ) ∧ ( ¬ v 1 ∨ ¬ v 3 ) ∧ ( ¬ v 1 ∨ v 2 ) . {\displaystyle B_{2}=(v_{3}\lor \neg v_{2})\wedge (\neg v_{1}\lor \neg v_{3})\wedge (\neg v_{1}\lor v_{2}).} With B 1 {\displaystyle B_{1}} , the algorithm can select v 1 = true {\displaystyle v_{1}={\text{true}}} , so to satisfy the second clause, the algorithm will need to set v 3 = false {\displaystyle v_{3}={\text{false}}} , and resultantly to satisfy the first clause, the algorithm will set v 2 = false {\displaystyle v_{2}={\text{false}}} . If the algorithm tries to satisfy B 2 {\displaystyle B_{2}} in the same way it tried to solve B 1 {\displaystyle B_{1}} , then the third clause will remain unsatisfied. This will cause the algorithm to backtrack and set v 1 = false {\displaystyle v_{1}={\text{false}}} and continue assigning variables further. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} statement. | . In particular, if v i = ∑ j = 1 d i v i j e i j {\displaystyle {\textbf {v}}_{i}=\sum _{j=1}^{d_{i}}v_{ij}{\textbf {e}}_{ij}\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is $\varphi(48)$? | {\displaystyle \varphi _{1}(v)=\!\left({\frac {1}{6}}\right)(1)={\frac {1}{6}}.} By a symmetry argument it can be shown that φ 2 ( v ) = φ 1 ( v ) = 1 6 . {\displaystyle \varphi _{2}(v)=\varphi _{1}(v)={\frac {1}{6}}.} Due to the efficiency axiom, the sum of all the Shapley values is equal to 1, which means that φ 3 ( v ) = 4 6 = 2 3 . {\displaystyle \varphi _{3}(v)={\frac {4}{6}}={\frac {2}{3}}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is $\varphi(48)$? | {\displaystyle \varphi _{1}=p_{\lambda },\quad \varphi _{2}=r^{2}-R^{2},\quad \varphi _{3}={\vec {p}}\cdot {\vec {r}}.} Note the nontrivial Poisson bracket structure of the constraints. In particular, { φ 2 , φ 3 } = 2 r 2 ≠ 0. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the true assertion. | With weak fairness on Tick only a finite number of stuttering steps are permitted between ticks. This temporal logical statement about Tick is called a liveness assertion. In general, a liveness assertion should be machine-closed: it shouldn't constrain the set of reachable states, only the set of possible behaviours.Most specifications do not require assertion of liveness properties. Safety properties suffice both for model checking and guidance in system implementation. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the true assertion. | In computer programming, specifically when using the imperative programming paradigm, an assertion is a predicate (a Boolean-valued function over the state space, usually expressed as a logical proposition using the variables of a program) connected to a point in the program, that always should evaluate to true at that point in code execution. Assertions can help a programmer read the code, help a compiler compile it, or help the program detect its own defects. For the latter, some programs check assertions by actually evaluating the predicate as they run. Then, if it is not in fact true – an assertion failure – the program considers itself to be broken and typically deliberately crashes or throws an assertion failure exception. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. | asserts that if p {\displaystyle p\,\!} is valid then so is q {\displaystyle q\,\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. | (/ indicates line break; some word breaks are uncertain) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Time-Memory Tradeoff Attack ... | As the scope of out-of-order execution increases further into several tens of instructions, the performance benefits of naive speculation decrease. To retain the benefits of aggressive memory dependence speculation while avoiding the costs of mispeculation several predictors have been proposed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Time-Memory Tradeoff Attack ... | The memory-bounded speedup model is the first work to reveal that memory is the performance constraint for high-end computing and presents a quantitative mathematical formulation for the trade-off between memory and computing. It is based on the memory-bounded function,W=G(n), where W is the work and thus also the computation for most applications. M is the memory requirement in terms of capacity, and G is the reuse rate. W=G(M) gives a very simple, but effective, description of the relation between computation and memory requirement. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $f: \mathbb{Z}_{m n} \rightarrow \mathbb{Z}_m \times \mathbb{Z}_n$ be defined by $f (x) = (x \bmod m,x \bmod n)$. Then $f$ is a ring isomorphism between $\mathbb{Z}_{180}$ and: | Let F be a field of characteristic different from 2. Then there is an isomorphism K n M ( F ) / 2 ≅ H e ´ t n ( F , Z / 2 Z ) {\displaystyle K_{n}^{M}(F)/2\cong H_{{\acute {e}}t}^{n}(F,\mathbb {Z} /2\mathbb {Z} )} for all n ≥ 0, where KM denotes the Milnor ring. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $f: \mathbb{Z}_{m n} \rightarrow \mathbb{Z}_m \times \mathbb{Z}_n$ be defined by $f (x) = (x \bmod m,x \bmod n)$. Then $f$ is a ring isomorphism between $\mathbb{Z}_{180}$ and: | But the corresponding ring homomorphism is the inclusion k = k ↪ k {\displaystyle k=k\hookrightarrow k} , which is not an isomorphism and so the restriction f |U is not an isomorphism. Let X be the affine curve x2 + y2 = 1 and let Then f is a rational function on X. It is regular at (0, 1) despite the expression since, as a rational function on X, f can also be written as f ( x , y ) = x 1 + y {\displaystyle f(x,y)={x \over 1+y}} . Let X = A2 − (0, 0). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A Carmichael number $n$ ... | In number theory, a Carmichael number is a composite number n {\displaystyle n} , which in modular arithmetic satisfies the congruence relation: b n ≡ b ( mod n ) {\displaystyle b^{n}\equiv b{\pmod {n}}} for all integers b {\displaystyle b} . The relation may also be expressed in the form: b n − 1 ≡ 1 ( mod n ) {\displaystyle b^{n-1}\equiv 1{\pmod {n}}} .for all integers b {\displaystyle b} which are relatively prime to n {\displaystyle n} . Carmichael numbers are named after American mathematician Robert Carmichael, the term having been introduced by Nicolaas Beeger in 1950 (Øystein Ore had referred to them in 1948 as numbers with the "Fermat property", or "F numbers" for short). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A Carmichael number $n$ ... | Let C ( X ) {\displaystyle C(X)} denote the number of Carmichael numbers less than or equal to X {\displaystyle X} . The distribution of Carmichael numbers by powers of 10 (sequence A055553 in the OEIS): In 1953, Knödel proved the upper bound: C ( X ) < X exp ( − k 1 ( log X log log X ) 1 2 ) {\displaystyle C(X) X 2 7 . {\displaystyle C(X)>X^{\frac {2}{7}}.} In 2005, this bound was further improved by Harman to C ( X ) > X 0.332 {\displaystyle C(X)>X^{0.332}} who subsequently improved the exponent to 0.7039 ⋅ 0.4736 = 0.33336704 > 1 / 3 {\displaystyle 0.7039\cdot 0.4736=0.33336704>1/3} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which symmetric key primitive is used in WPA2 encryption? | A key mixing function to defeat a class of attacks on WEP. A rekeying method to prevent key reuse.TKIP allocated 48 bits to the IV compared to the 24 bits of WEP, so the maximum number is 281,474,976,710,656 (248).In WPA-PSK, each packet was individually encrypted using the IV information, the MAC address, and the pre-shared key as inputs. The RC4 cipher was used to encrypt the packet content with the derived encryption key.Additionally, WPA introduced WPA Enterprise, which provided enhanced security for enterprise-level networks. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which symmetric key primitive is used in WPA2 encryption? | Another, more conservative approach is to employ a cipher designed to prevent related-key attacks altogether, usually by incorporating a strong key schedule. A newer version of Wi-Fi Protected Access, WPA2, uses the AES block cipher instead of RC4, in part for this reason. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. What is the cardinality of $\mathbf{Z}^*_n$? | The cardinality of the set of integers is equal to ℵ0 (aleph-null). This is readily demonstrated by the construction of a bijection, that is, a function that is injective and surjective from Z {\displaystyle \mathbb {Z} } to N = { 0 , 1 , 2 , . . . } | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. What is the cardinality of $\mathbf{Z}^*_n$? | More generally, | Z: n Z | = n {\displaystyle |\mathbb {Z} :n\mathbb {Z} |=n} for any positive integer n. When G is finite, the formula may be written as | G: H | = | G | / | H | {\displaystyle |G:H|=|G|/|H|} , and it implies Lagrange's theorem that | H | {\displaystyle |H|} divides | G | {\displaystyle |G|} . When G is infinite, | G: H | {\displaystyle |G:H|} is a nonzero cardinal number that may be finite or infinite. For example, | Z: 2 Z | = 2 {\displaystyle |\mathbb {Z} :2\mathbb {Z} |=2} , but | R: Z | {\displaystyle |\mathbb {R} :\mathbb {Z} |} is infinite. If N is a normal subgroup of G, then | G: N | {\displaystyle |G:N|} is equal to the order of the quotient group G / N {\displaystyle G/N} , since the underlying set of G / N {\displaystyle G/N} is the set of cosets of N in G. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be any positive integer. Three of the following assertions are equivalent. Tick the remaining one. | Case 3: If n = 3p − 1, then n3 = 27p3 − 27p2 + 9p − 1, which is 1 less than a multiple of 9. For instance, if n = 5 then n3 = 125 = 9×14 − 1. Q.E.D. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be any positive integer. Three of the following assertions are equivalent. Tick the remaining one. | With N = 1, we get the coarsest possible message, which does not give any information. So everything is red on the top left panel. With N = 3, the message is finer. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Birthday attacks \dots | A birthday attack is a type of cryptographic attack that exploits the mathematics behind the birthday problem in probability theory. This attack can be used to abuse communication between two or more parties. The attack depends on the higher likelihood of collisions found between random attack attempts and a fixed degree of permutations (pigeonholes). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Birthday attacks \dots | With 23 individuals, there are 23 × 22/2 = 253 pairs to consider, far more than half the number of days in a year. Real-world applications for the birthday problem include a cryptographic attack called the birthday attack, which uses this probabilistic model to reduce the complexity of finding a collision for a hash function, as well as calculating the approximate risk of a hash collision existing within the hashes of a given size of population. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the number of secret bits in a WEP key? | Once the restrictions were lifted, manufacturers of access points implemented an extended 128-bit WEP protocol using a 104-bit key size (WEP-104). A 64-bit WEP key is usually entered as a string of 10 hexadecimal (base 16) characters (0–9 and A–F). Each character represents 4 bits, 10 digits of 4 bits each gives 40 bits; adding the 24-bit IV produces the complete 64-bit WEP key (4 bits × 10 + 24-bit IV = 64-bit WEP key). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the number of secret bits in a WEP key? | An attacker therefore can assume that all the keys used to encrypt packets share a single WEP key. This fact opened up WEP to a series of attacks which proved devastating. The simplest to understand uses the fact that the 24-bit IV only allows a little under 17 million possibilities. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In a multiplicative cyclic group $G$ of order $m > 1$ with neutral element $e_G$ \ldots | Indeed, let H1 = ⟨A⟩ and H2 = ⟨B⟩ be cyclic subgroups of SL2(Z) generated by A and B accordingly. It is not hard to check that A and B are elements of infinite order in SL2(Z) and that and Consider the standard action of SL2(Z) on R2 by linear transformations. Put and It is not hard to check, using the above explicit descriptions of H1 and H2, that for every nontrivial g ∈ H1 we have g(X2) ⊆ X1 and that for every nontrivial g ∈ H2 we have g(X1) ⊆ X2. Using the alternative form of the ping-pong lemma, for two subgroups, given above, we conclude that H = H1 ∗ H2. Since the groups H1 and H2 are infinite cyclic, it follows that H is a free group of rank two. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In a multiplicative cyclic group $G$ of order $m > 1$ with neutral element $e_G$ \ldots | This follows from elementary group theory, because the exponent of any finite group must divide the order of the group. λ(n) is the exponent of the multiplicative group of integers modulo n while φ(n) is the order of that group. In particular, the two must be equal in the cases where the multiplicative group is cyclic due to the existence of a primitive root, which is the case for odd prime powers. We can thus view Carmichael's theorem as a sharpening of Euler's theorem. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of the following notions means that ``the information must be protected against any malicious modification''? | As shown previously, non-interference policy is too strict for use in most real-world applications. Therefore, several approaches to allow controlled releases of information have been devised. Such approaches are called information declassification. Robust declassification requires that an active attacker may not manipulate the system in order to learn more secrets than what passive attackers already know.Information declassification constructs can be classified in four orthogonal dimensions: What information is released, Who is authorized to access the information, Where the information is released, and When is the information released. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of the following notions means that ``the information must be protected against any malicious modification''? | 20 ICCPR, In the context of limitations on cryptography, restrictions will most often be based on Article 19 (3)(b), i.e., risks for national security and public order. This raises the complex issue of the relation, and distinction, between security of the individual, e.g., from interference with personal electronic communications, and national security. The right to privacy protects against 'arbitrary or unlawful interference' with one's privacy, one's family, one's home and one's correspondence. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Confidentiality means that: | In information security, confidentiality "is the property, that information is not made available or disclosed to unauthorized individuals, entities, or processes." While similar to "privacy," the two words are not interchangeable. Rather, confidentiality is a component of privacy that implements to protect our data from unauthorized viewers. Examples of confidentiality of electronic data being compromised include laptop theft, password theft, or sensitive emails being sent to the incorrect individuals. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Confidentiality means that: | Confidentiality refers to limits on who can get what kind of information. For example, executives concerned about protecting their enterprise's strategic plans from competitors; individuals are concerned about unauthorized access to their financial records. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following acronyms does not designate a mode of operation? | This mode is selected when D7 bit of the Control Word Register is 1. There are three I/O modes: Mode 0 - Simple I/O Mode 1 - Strobed I/O Mode 2 - Strobed Bi-directional I/O | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following acronyms does not designate a mode of operation? | Immediate mode (opcode 0xD4), and register indirect mode (0xD6, 0xD7) are not used. Ey: MOV A,operand Move operand to the accumulator. Immediate mode is not used for this operation (opcode 0xE4), as it duplicates opcode 0x74. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. The brute force attack \dots | Note that one can also overcome the problem with containing dots using the \yahnodots command. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. The brute force attack \dots | . . VERIFY-SELECTION . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
WEP \dots | U+2299 ⊙ CIRCLED DOT OPERATOR | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
WEP \dots | Note that one can also overcome the problem with containing dots using the \yahnodots command. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The DES key schedule\dots | The key schedule of LEA supports 128, 192, and 256-bit keys and outputs 192-bit round keys K i {\displaystyle K_{i}} ( 0 ≤ i < N r {\displaystyle 0\leq i | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The DES key schedule\dots | The very simple key schedule makes IDEA subject to a class of weak keys; some keys containing a large number of 0 bits produce weak encryption. These are of little concern in practice, being sufficiently rare that they are unnecessary to avoid explicitly when generating keys randomly. A simple fix was proposed: XORing each subkey with a 16-bit constant, such as 0x0DAE.Larger classes of weak keys were found in 2002.This is still of negligible probability to be a concern to a randomly chosen key, and some of the problems are fixed by the constant XOR proposed earlier, but the paper is not certain if all of them are. A more comprehensive redesign of the IDEA key schedule may be desirable. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
How many generators do we have in a group of order $13$? | The generators of ≅ ] ≅ , order 128, are {02,1,3} from generators {0,1,2,3}. And ≅ ], order 64, has generators {02,1021,3}. As well, ≅ . Also related = has trionic subgroups: ⅄ = , order 64, and 1=Δ = ≅ ]+, order 32. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
How many generators do we have in a group of order $13$? | The generators are the loops { A 1 , B 1 , … , A n , B n } {\displaystyle \{A_{1},B_{1},\dots ,A_{n},B_{n}\}} (A is simply connected, so it contributes no generators) and there is exactly one relation: A 1 B 1 A 1 − 1 B 1 − 1 A 2 B 2 A 2 − 1 B 2 − 1 ⋯ A n B n A n − 1 B n − 1 = 1. {\displaystyle A_{1}B_{1}A_{1}^{-1}B_{1}^{-1}A_{2}B_{2}A_{2}^{-1}B_{2}^{-1}\cdots A_{n}B_{n}A_{n}^{-1}B_{n}^{-1}=1.} Using generators and relations, this group is denoted ⟨ A 1 , B 1 , … , A n , B n | A 1 B 1 A 1 − 1 B 1 − 1 ⋯ A n B n A n − 1 B n − 1 ⟩ . {\displaystyle \left\langle A_{1},B_{1},\dots ,A_{n},B_{n}\left|A_{1}B_{1}A_{1}^{-1}B_{1}^{-1}\cdots A_{n}B_{n}A_{n}^{-1}B_{n}^{-1}\right.\right\rangle .} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following attacks needs no precomputation. | Since such assurances are not actually available in real practice, sleight of hand in language which implies that they are will generally be misleading. There will always be uncertainty as advances (e.g., in cryptanalytic theory or merely affordable computer capacity) may reduce the effort needed to successfully use some attack method against an algorithm. In addition, actual use of cryptographic algorithms requires their encapsulation in a cryptosystem, and doing so often introduces vulnerabilities which are not due to faults in an algorithm. For example, essentially all algorithms require random choice of keys, and any cryptosystem which does not provide such keys will be subject to attack regardless of any attack resistant qualities of the encryption algorithm(s) used. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following attacks needs no precomputation. | Still users need to deploy the mitigation to eliminate the vulnerability in their systems. In the case of zero-day exploits, nobody has deployed a fix yet. Zero-day attacks are severe threats. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these is \emph{not} a hard computational problem? | Some constraint languages (or non-uniform problems) are known to correspond to problems solvable in polynomial time, and some others are known to express NP-complete problems. However, it is possible that some constraint languages are neither. It is known by Ladner's theorem that if P is not equal to NP, then there exist problems in NP that are neither polynomial-time nor NP-hard. As of 2007, it is not known if such problems can be expressed as constraint satisfaction problems with a fixed constraint language. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these is \emph{not} a hard computational problem? | Somewhat paradoxically, though such a system is not believed to be able to solve all of NP, it can easily solve all NP-complete problems due to self-reducibility. This stems from the fact that if the language L is not NP-hard, the prover is substantially limited in power (as it can no longer decide all NP problems with its oracle). Additionally, the graph nonisomorphism problem (which is a classical problem in IP) is also in compIP, since the only hard operation the prover has to do is isomorphism testing, which it can use the oracle to solve. Quadratic non-residuosity and graph isomorphism are also in compIP. Note, Quadratic non-residuosity (QNR) is likely an easier problem than graph isomorphism as QNR is in UP intersect co-UP. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \textbf{most accurate} answer. For a hash function to be secure (\textbf{but still efficient}) against collision attacks in 2015, the output length should be\dots | Most hash functions are built on an ad hoc basis, where the bits of the message are nicely mixed to produce the hash. Various bitwise operations (e.g. rotations), modular additions and compression functions are used in iterative mode to ensure high complexity and pseudo-randomness of the output. In this way, the security is very hard to prove and the proof is usually not done. Only a few years ago, one of the most popular hash functions, SHA-1, was shown to be less secure than its length suggested: collisions could be found in only 251 tests, rather than the brute-force number of 280. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \textbf{most accurate} answer. For a hash function to be secure (\textbf{but still efficient}) against collision attacks in 2015, the output length should be\dots | Both the block size of the hash function and the output size are completely scalable. The speed can be adjusted by adjusting the number of bitwise operations used by FSB per input bit. The security can be adjusted by adjusting the output size. Bad instances exist and one must take care when choosing the matrix H {\displaystyle H} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tonelli Algorithm is for ... | The Tonelli–Shanks algorithm can (naturally) be used for any process in which square roots modulo a prime are necessary. For example, it can be used for finding points on elliptic curves. It is also useful for the computations in the Rabin cryptosystem and in the sieving step of the quadratic sieve. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tonelli Algorithm is for ... | The Tonelli–Shanks algorithm requires (on average over all possible input (quadratic residues and quadratic nonresidues)) 2 m + 2 k + S ( S − 1 ) 4 + 1 2 S − 1 − 9 {\displaystyle 2m+2k+{\frac {S(S-1)}{4}}+{\frac {1}{2^{S-1}}}-9} modular multiplications, where m {\displaystyle m} is the number of digits in the binary representation of p {\displaystyle p} and k {\displaystyle k} is the number of ones in the binary representation of p {\displaystyle p} . If the required quadratic nonresidue z {\displaystyle z} is to be found by checking if a randomly taken number y {\displaystyle y} is a quadratic nonresidue, it requires (on average) 2 {\displaystyle 2} computations of the Legendre symbol. The average of two computations of the Legendre symbol are explained as follows: y {\displaystyle y} is a quadratic residue with chance p + 1 2 p = 1 + 1 p 2 {\displaystyle {\tfrac {\tfrac {p+1}{2}}{p}}={\tfrac {1+{\tfrac {1}{p}}}{2}}} , which is smaller than 1 {\displaystyle 1} but ≥ 1 2 {\displaystyle \geq {\tfrac {1}{2}}} , so we will on average need to check if a y {\displaystyle y} is a quadratic residue two times. This shows essentially that the Tonelli–Shanks algorithm works very well if the modulus p {\displaystyle p} is random, that is, if S {\displaystyle S} is not particularly large with respect to the number of digits in the binary representation of p {\displaystyle p} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement | Select y ~ ∈ { 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement | - 囗 + お/頁 + selector 4 = 馘 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $(G,+), (H,\times)$ be two groups and $f:G\to H$ be an homomorphism. For $x_1,x_2 \in G$, we have: | Let F be a finitely generated free group, with n generators. Let G1 and G2 be two finitely presented groups. Suppose there exists a surjective homomorphism ϕ: F → G 1 ∗ G 2 {\displaystyle \phi :F\rightarrow G_{1}\ast G_{2}} . Then there exists two subgroups F1 and F2 of F with ϕ ( F 1 ) = G 1 {\displaystyle \phi (F_{1})=G_{1}} and ϕ ( F 2 ) = G 2 {\displaystyle \phi (F_{2})=G_{2}} , such that F = F 1 ∗ F 2 . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $(G,+), (H,\times)$ be two groups and $f:G\to H$ be an homomorphism. For $x_1,x_2 \in G$, we have: | If f , g: G → H {\displaystyle f,g:G\to H} are two group homomorphisms between abelian groups, then their sum f + g {\displaystyle f+g} , defined by ( f + g ) ( x ) = f ( x ) + g ( x ) {\displaystyle (f+g)(x)=f(x)+g(x)} , is again a homomorphism. (This is not true if H {\displaystyle H} is a non-abelian group.) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following terms represents a mode of operation which transforms a block cipher into a stream cipher? | In cryptography, a block cipher mode of operation is an algorithm that uses a block cipher to provide information security such as confidentiality or authenticity. A block cipher by itself is only suitable for the secure cryptographic transformation (encryption or decryption) of one fixed-length group of bits called a block. A mode of operation describes how to repeatedly apply a cipher's single-block operation to securely transform amounts of data larger than a block.Most modes require a unique binary sequence, often called an initialization vector (IV), for each encryption operation. The IV has to be non-repeating and, for some modes, random as well. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following terms represents a mode of operation which transforms a block cipher into a stream cipher? | Stream ciphers represent a different approach to symmetric encryption from block ciphers. Block ciphers operate on large blocks of digits with a fixed, unvarying transformation. This distinction is not always clear-cut: in some modes of operation, a block cipher primitive is used in such a way that it acts effectively as a stream cipher. Stream ciphers typically execute at a higher speed than block ciphers and have lower hardware complexity. However, stream ciphers can be susceptible to security breaches (see stream cipher attacks); for example, when the same starting state (seed) is used twice. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Shannon theorem states that perfect secrecy implies... | Shannon proved, using information theoretic considerations, that the one-time pad has a property he termed perfect secrecy; that is, the ciphertext C gives absolutely no additional information about the plaintext. This is because (intuitively), given a truly uniformly random key that is used only once, a ciphertext can be translated into any plaintext of the same length, and all are equally likely. Thus, the a priori probability of a plaintext message M is the same as the a posteriori probability of a plaintext message M given the corresponding ciphertext. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Shannon theorem states that perfect secrecy implies... | Claude Shannon's development of information theory during World War II provided the next big step in understanding how much information could be reliably communicated through noisy channels. Building on Hartley's foundation, Shannon's noisy channel coding theorem (1948) describes the maximum possible efficiency of error-correcting methods versus levels of noise interference and data corruption. The proof of the theorem shows that a randomly constructed error-correcting code is essentially as good as the best possible code; the theorem is proved through the statistics of such random codes. Shannon's theorem shows how to compute a channel capacity from a statistical description of a channel, and establishes that given a noisy channel with capacity C {\displaystyle C} and information transmitted at a line rate R {\displaystyle R} , then if R < C {\displaystyle R C {\displaystyle R>C} the probability of error at the receiver increases without bound as the rate is increased. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. The Shannon Encryption Model ... | The erroneous-code-word c ∗ {\displaystyle c^{*}} is decoded using a procedure D e c {\displaystyle Dec} , resulting in a decoded-message s ∗ {\displaystyle s^{*}} = D e c ( c ∗ ) {\displaystyle Dec(c^{*})} .The tampering experiment can be used to model several interesting real-world settings, such as data transmitted over a noisy channel, or adversarial tampering of data stored in the memory of a physical device. Having this experimental base, we would like to build special encoding/decoding procedures ( E n c , D e c ) {\displaystyle (Enc,Dec)} , which give us some meaningful guarantees about the results of the above tampering experiment, for large and interesting families F {\displaystyle F} of tampering functions. The following are several possibilities for the type of guarantees that we may hope for. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. The Shannon Encryption Model ... | For this model, errors do not have a known distribution and can be from an adversary, the only constraints being d i s err ≤ t {\displaystyle dis_{\text{err}}\leq t} and that a corrupted word depends only on the input w {\displaystyle w} and not on the secure sketch. It can be shown for this error model that there will never be more than t {\displaystyle t} errors, since this model can account for all complex noise processes, meaning that Shannon's bound can be reached; to do this a random permutation is prepended to the secure sketch that will reduce entropy loss. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. The UMTS improves the security of GSM using | GSM/UMTS various UTRA 5G NR == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement. The UMTS improves the security of GSM using | The Universal Mobile Telecommunications System (UMTS) is one of the new ‘third generation’ 3G mobile cellular communication systems. UMTS builds on the success of the ‘second generation’ GSM system. One of the factors in the success of GSM has been its security features. New services introduced in UMTS require new security features to protect them. In addition, certain real and perceived shortcomings of GSM security need to be addressed in UMTS. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. The extended Euclidean algorithm is typically used to\dots | The standard Euclidean algorithm proceeds by a succession of Euclidean divisions whose quotients are not used. Only the remainders are kept. For the extended algorithm, the successive quotients are used. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n$ be an integer. The extended Euclidean algorithm is typically used to\dots | The extended Euclidean algorithm is the essential tool for computing multiplicative inverses in modular structures, typically the modular integers and the algebraic field extensions. A notable instance of the latter case are the finite fields of non-prime order. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If we need to create a channel that protects confidentiality and we have at our disposal a channel that protects integrity and authenticity, we need to use | Citizens and businesses must have confidence in these networks, especially in terms of preserving their fundamental right to privacy. To achieve this, it was necessary to reform a series of European regulations, especially in the field of telecommunications, but also in terms of cybersecurity and everything that concerns audiovisual media services. The third objective is to enable the market to adapt to changes in its environment. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If we need to create a channel that protects confidentiality and we have at our disposal a channel that protects integrity and authenticity, we need to use | The ISO (International Organization for Standardization) states that confidentiality, integrity, authentication, access control, and non-repudiation should all be considered when creating any secure system. Confidentiality: No unauthorized party can access appropriate messages. Integrity: Messages cannot be changed during transit without being discovered. Authentication: The message needs to be sent by the person/machine who claims to have sent it. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A \textit{Cryptographic Certificate} is the $\ldots$ | A framework for managing digital certificates and encryption keys. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A \textit{Cryptographic Certificate} is the $\ldots$ | The Cryptographic Message Syntax (CMS) is the IETF's standard for cryptographically protected messages. It can be used by cryptographic schemes and protocols to digitally sign, digest, authenticate or encrypt any form of digital data. CMS is based on the syntax of PKCS #7, which in turn is based on the Privacy-Enhanced Mail standard. The newest version of CMS (as of 2009) is specified in RFC 5652 (but see also RFC 5911 for updated ASN.1 modules conforming to ASN.1 2002). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
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