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The number of plaintext/ciphertext pairs required for a differential cryptanalysis is\dots | The most powerful form of the attack requires 250 known plaintexts, has a computational complexity of 250, and has a 51% success rate.There have also been attacks proposed against reduced-round versions of the cipher, that is, versions of DES with fewer than 16 rounds. Such analysis gives an insight into how many rounds are needed for safety, and how much of a "security margin" the full version retains. Differential-linear cryptanalysis was proposed by Langford and Hellman in 1994, and combines differential and linear cryptanalysis into a single attack. An enhanced version of the attack can break 9-round DES with 215.8 chosen plaintexts and has a 229.2 time complexity (Biham and others, 2002). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The number of plaintext/ciphertext pairs required for a differential cryptanalysis is\dots | Unlike other forms of cryptanalysis, such as differential and linear cryptanalysis, only one or two known plaintexts are required. The XSL algorithm is tailored to solve the type of equation systems that are produced. Courtois and Pieprzyk estimate that an "optimistic evaluation shows that the XSL attack might be able to break Rijndael 256 bits and Serpent for key lengths 192 and 256 bits." | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given the distribution $P_0$ of a normal coin, i.e. $P_0(0)=P_0(1)=\frac{1}{2}$, and distribution $P_1$ of a biased coin, where $P_1(0)=\frac{1}{3}$ and $P_1(1) = \frac{2}{3}$ , the maximal advantage of a distinguisher using a single sample is\dots | {\displaystyle n\geq {\frac {1}{2\varepsilon ^{2}}}\log \left({\frac {1}{2\delta }}\right).} In order to obtain this lower bound we impose that the total variation distance between two sequences of n {\displaystyle n} samples is at least 1 − 2 δ {\displaystyle 1-2\delta } . This is because the total variation upper bounds the probability of under- or over-estimating the coins' means. Denote P 1 n {\displaystyle P_{1}^{n}} and P 2 n {\displaystyle P_{2}^{n}} the respective joint distributions of the n {\displaystyle n} coin tosses for each coin, then We have ( 1 − 2 δ ) 2 ≤ d T V ( P 1 n , P 2 n ) 2 ≤ 1 − e − D K L ( P 1 n ∥ P 2 n ) = 1 − e − n D K L ( P 1 ∥ P 2 ) = 1 − e − n log ( 1 / ( 1 − 4 ε 2 ) ) 2 {\displaystyle {\begin{aligned}(1-2\delta )^{2}&\leq d_{\mathrm {TV} }\left(P_{1}^{n},P_{2}^{n}\right)^{2}\\&\leq 1-e^{-D_{\mathrm {KL} }(P_{1}^{n}\parallel P_{2}^{n})}\\&=1-e^{-nD_{\mathrm {KL} }(P_{1}\parallel P_{2})}\\&=1-e^{-n{\frac {\log(1/(1-4\varepsilon ^{2}))}{2}}}\end{aligned}}} The result is obtained by rearranging the terms. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given the distribution $P_0$ of a normal coin, i.e. $P_0(0)=P_0(1)=\frac{1}{2}$, and distribution $P_1$ of a biased coin, where $P_1(0)=\frac{1}{3}$ and $P_1(1) = \frac{2}{3}$ , the maximal advantage of a distinguisher using a single sample is\dots | Assume two biased coins, the first with probability p of turning up heads and the second with probability q > p of turning up heads. Intuitively, if both coins are tossed the same number of times, we should expect the first coin turns up fewer heads than the second one. More specifically, for any fixed k, the probability that the first coin produces at least k heads should be less than the probability that the second coin produces at least k heads. However proving such a fact can be difficult with a standard counting argument. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
To how many plaintexts we expect to decrypt a ciphertext in the Rabin cryptosystem when we don't use redundancy? | It has been proven that any algorithm which finds one of the possible plaintexts for every Rabin-encrypted ciphertext can be used to factor the modulus n {\displaystyle n} . Thus, Rabin decryption for random plaintext is at least as hard as the integer factorization problem, something that has not been proven for RSA. It is generally believed that there is no polynomial-time algorithm for factoring, which implies that there is no efficient algorithm for decrypting a random Rabin-encrypted value without the private key ( p , q ) {\displaystyle (p,q)} . The Rabin cryptosystem does not provide indistinguishability against chosen plaintext attacks since the process of encryption is deterministic. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
To how many plaintexts we expect to decrypt a ciphertext in the Rabin cryptosystem when we don't use redundancy? | Decrypting produces three false results in addition to the correct one, so that the correct result must be guessed. This is the major disadvantage of the Rabin cryptosystem and one of the factors which have prevented it from finding widespread practical use. If the plaintext is intended to represent a text message, guessing is not difficult; however, if the plaintext is intended to represent a numerical value, this issue becomes a problem that must be resolved by some kind of disambiguation scheme. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For an interactive proof system, the difference between perfect, statistical and computational zero-knowledge is based on \ldots | A formal definition of zero-knowledge has to use some computational model, the most common one being that of a Turing machine. Let P {\displaystyle P} , V {\displaystyle V} , and S {\displaystyle S} be Turing machines. An interactive proof system with ( P , V ) {\displaystyle (P,V)} for a language L {\displaystyle L} is zero-knowledge if for any probabilistic polynomial time (PPT) verifier V ^ {\displaystyle {\hat {V}}} there exists a PPT simulator S {\displaystyle S} such that ∀ x ∈ L , z ∈ { 0 , 1 } ∗ , View V ^ = S ( x , z ) {\displaystyle \forall x\in L,z\in \{0,1\}^{*},\operatorname {View} _{\hat {V}}\left=S(x,z)} where View V ^ {\displaystyle \operatorname {View} _{\hat {V}}\left} is a record of the interactions between P ( x ) {\displaystyle P(x)} and V ^ ( x , z ) {\displaystyle {\hat {V}}(x,z)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For an interactive proof system, the difference between perfect, statistical and computational zero-knowledge is based on \ldots | This is formalized by showing that every verifier has some simulator that, given only the statement to be proved (and no access to the prover), can produce a transcript that "looks like" an interaction between an honest prover and the verifier in question.The first two of these are properties of more general interactive proof systems. The third is what makes the proof zero-knowledge.Zero-knowledge proofs are not proofs in the mathematical sense of the term because there is some small probability, the soundness error, that a cheating prover will be able to convince the verifier of a false statement. In other words, zero-knowledge proofs are probabilistic "proofs" rather than deterministic proofs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the name of the encryption threat that corresponds to \emph{force the sender to encrypt some messages selected by the adversary}? | Deniable encryption allows the sender of an encrypted message to deny sending that message. This requires a trusted third party. A possible scenario works like this: Bob suspects his wife Alice is engaged in adultery. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the name of the encryption threat that corresponds to \emph{force the sender to encrypt some messages selected by the adversary}? | Computer security Opportunistic encryption Communications security Secure messaging | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $C$ be a perfect cipher with $\ell$-bit blocks. Then, \dots | Let P = P ‖ P ‖ P ‖ P {\displaystyle P=P\|P\|P\|P} be a 128-bit block of plaintext and C = C ‖ C ‖ C ‖ C {\displaystyle C=C\|C\|C\|C} be a 128-bit block of ciphertext, where P {\displaystyle P} and C {\displaystyle C} ( 0 ≤ i < 4 {\displaystyle 0\leq i<4} ) are 32-bit blocks. Let K i = K i ‖ K i ‖ K i ‖ K i ‖ K i ‖ K i {\displaystyle K_{i}=K_{i}\|K_{i}\|K_{i}\|K_{i}\|K_{i}\|K_{i}} ( 0 ≤ i < N r {\displaystyle 0\leq i | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $C$ be a perfect cipher with $\ell$-bit blocks. Then, \dots | A notable property of the Davies–Meyer construction is that even if the underlying block cipher is totally secure, it is possible to compute fixed points for the construction: for any m {\displaystyle m} , one can find a value of h {\displaystyle h} such that E m ( h ) ⊕ h = h {\displaystyle E_{m}(h)\oplus h=h}: one just has to set h = E m − 1 ( 0 ) {\displaystyle h=E_{m}^{-1}(0)} . This is a property that random functions certainly do not have. So far, no practical attack has been based on this property, but one should be aware of this "feature". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The exponent of the group $\mathbb{Z}_9^*$ is | R , C , H , O {\displaystyle \mathbb {R,C,H,O} } stand for the real numbers, complex numbers, quaternions, and octonions. In the symbols such as E6−26 for the exceptional groups, the exponent −26 is the signature of an invariant symmetric bilinear form that is negative definite on the maximal compact subgroup. It is equal to the dimension of the group minus twice the dimension of a maximal compact subgroup. The fundamental group listed in the table below is the fundamental group of the simple group with trivial center. Other simple groups with the same Lie algebra correspond to subgroups of this fundamental group (modulo the action of the outer automorphism group). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The exponent of the group $\mathbb{Z}_9^*$ is | Fix a prime p {\displaystyle p} and suppose the exponents e i {\displaystyle e_{i}} of the cyclic factors of the Sylow p {\displaystyle p} -subgroup are arranged in increasing order: e 1 ≤ e 2 ≤ ⋯ ≤ e n {\displaystyle e_{1}\leq e_{2}\leq \cdots \leq e_{n}} for some n > 0 {\displaystyle n>0} . One needs to find the automorphisms of Z p e 1 ⊕ ⋯ ⊕ Z p e n . {\displaystyle \mathbf {Z} _{p^{e_{1}}}\oplus \cdots \oplus \mathbf {Z} _{p^{e_{n}}}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} statement. When $x\rightarrow+\infty$ \ldots | Set subtraction on the right, and parentheses on the left | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} statement. When $x\rightarrow+\infty$ \ldots | Squeeze { '__'e.1 = ; s.A e.1 = s.A ; = ; }; | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In order to avoid the Bleichenbacher attack in ElGamal signatures, \ldots | A polynomial-time attack, for Ω ( n ) {\displaystyle \Omega (n)} concurrent executions, was shown in 2020 by Benhamouda, Lepoint, Raykova, and Orrù. Schnorr also suggested enhancements for securing blind signatures schemes based on discrete logarithm problem. == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In order to avoid the Bleichenbacher attack in ElGamal signatures, \ldots | Signature algorithms like ElGamal and DSA have parameters which must be set with random information. He shows how one can make use of these parameters to send a message subliminally. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assumption. A language $L$ is in NP if\dots | "LR parsers for natural languages" (PDF). COLING. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assumption. A language $L$ is in NP if\dots | . 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 |
In which of the following groups is the decisional Diffie-Hellman problem (DDH) believed to be hard? | The decisional Diffie–Hellman (DDH) assumption is a computational hardness assumption about a certain problem involving discrete logarithms in cyclic groups. It is used as the basis to prove the security of many cryptographic protocols, most notably the ElGamal and Cramer–Shoup cryptosystems. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In which of the following groups is the decisional Diffie-Hellman problem (DDH) believed to be hard? | The CDH assumption is a weaker assumption than the Decisional Diffie–Hellman assumption (DDH assumption). If computing g a b {\displaystyle g^{ab}} from ( g , g a , g b ) {\displaystyle (g,g^{a},g^{b})} was easy (CDH problem), then one could solve the DDH problem trivially. Many cryptographic schemes that are constructed from the CDH problem rely in fact on the hardness of the DDH problem. The semantic security of the Diffie-Hellman Key Exchange as well as the security of the ElGamal encryption rely on the hardness of the DDH problem. There are concrete constructions of groups where the stronger DDH assumption does not hold but the weaker CDH assumption still seems to be a reasonable hypothesis. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In ElGamal signature scheme and over the random choice of the public parameters in the random oracle model (provided that the DLP is hard), existential forgery is \ldots | When A attempts to forge on a message m, we consider the output of A to be (J, y) where y is the forgery, and J is such that m was the Jth unique query to the random oracle (it may be assumed that A will query m at some point, if A is to be successful with non-negligible probability). (If A outputs an incorrect forgery, we consider the output to be (0, y).) By the forking lemma, the probability (frk) of obtaining two good forgeries y and y' on the same message but with different random oracle outputs (that is, with hJ ≠ h'J) is non-negligible when acc is also non-negligible. This allows us to prove that if the underlying hard problem is indeed hard, then no adversary can forge signatures. This is the essence of the proof given by Pointcheval and Stern for a modified ElGamal signature scheme against an adaptive adversary. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In ElGamal signature scheme and over the random choice of the public parameters in the random oracle model (provided that the DLP is hard), existential forgery is \ldots | This notion is a stronger (more secure) variant of the existential forgery detailed above. Weak existential forgery is the creation (by an adversary) of at least one message/signature pair, ( m , σ ′ ) {\displaystyle \left(m,\sigma '\right)} , given a message and different signature ( m , σ ) {\displaystyle (m,\sigma )} produced by the legitimate signer. Strong existential forgery is essentially the weakest adversarial goal, therefore the strongest schemes are those that are strongly existentially unforgeable. == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following primitives \textit{cannot} be instantiated with a cryptographic hash function? | Basic general information about the cryptographic hash functions: year, designer, references, etc. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following primitives \textit{cannot} be instantiated with a cryptographic hash function? | Selected schemes for the purpose of hashing: SWIFFT. Lattice Based Hash Function (LASH). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In plain ElGamal Encryption scheme \ldots | For the ElGamal encryption we suppose now that Alice is the owner of the XTR public key data ( p , q , T r ( g ) ) {\displaystyle (p,q,Tr(g))} and that she has selected a secret integer k {\displaystyle k} , computed T r ( g k ) {\displaystyle Tr(g^{k})} and published the result. Given Alice's XTR public key data ( p , q , T r ( g ) , T r ( g k ) ) {\displaystyle \left(p,q,Tr(g),Tr(g^{k})\right)} , Bob can encrypt a message M {\displaystyle M} , intended for Alice, using the following XTR version of the ElGamal encryption: Bob selects randomly a b ∈ Z {\displaystyle b\in \mathbb {Z} } with 1 < b < q − 2 {\displaystyle 1 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In plain ElGamal Encryption scheme \ldots | Thus, the scheme serves mainly as a proof of concept, and more efficient provably-secure schemes such as ElGamal have been developed since. Because encryption is performed using a probabilistic algorithm, a given plaintext may produce very different ciphertexts each time it is encrypted. This has significant advantages, as it prevents an adversary from recognizing intercepted messages by comparing them to a dictionary of known ciphertexts. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $X$ be a random variable defined by the visible face showing up when throwing a dice. Its expected value $E(X)$ is: | Let X {\displaystyle X} represent the outcome of a roll of a fair six-sided die. More specifically, X {\displaystyle X} will be the number of pips showing on the top face of the die after the toss. The possible values for X {\displaystyle X} are 1, 2, 3, 4, 5, and 6, all of which are equally likely with a probability of 1/6. The expectation of X {\displaystyle X} is E = 1 ⋅ 1 6 + 2 ⋅ 1 6 + 3 ⋅ 1 6 + 4 ⋅ 1 6 + 5 ⋅ 1 6 + 6 ⋅ 1 6 = 3.5. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $X$ be a random variable defined by the visible face showing up when throwing a dice. Its expected value $E(X)$ is: | The expected value or mean of a random vector X {\displaystyle \mathbf {X} } is a fixed vector E {\displaystyle \operatorname {E} } whose elements are the expected values of the respective random variables. : p.333 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an arbitrary cipher $C$ and a uniformly distributed random permutation $C^*$ on $\{0,1\}^n$. Tick the \textbf{false} assertion. | The idealized abstraction of a (keyed) block cipher is a truly random permutation on the mappings between plaintext and ciphertext. If a distinguishing algorithm exists that achieves significant advantage with less effort than specified by the block cipher's security parameter (this usually means the effort required should be about the same as a brute force search through the cipher's key space), then the cipher is considered broken at least in a certificational sense, even if such a break doesn't immediately lead to a practical security failure.Modern ciphers are expected to have super pseudorandomness. That is, the cipher should be indistinguishable from a randomly chosen permutation on the same message space, even if the adversary has black-box access to the forward and inverse directions of the cipher. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an arbitrary cipher $C$ and a uniformly distributed random permutation $C^*$ on $\{0,1\}^n$. Tick the \textbf{false} assertion. | Attacks that allow distinguishing ciphertext from random data. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textit{incorrect} assertion. Let $P, V$ be an interactive system for a language $L\in \mathcal{NP}$. | {\displaystyle z/y=x.} When the operations are expressed in this way, provability of a given sentence can be encoded as an arithmetic sentence describing termination of an analytic tableau. Provability of consistency can then simply be added as an axiom. The resulting system can be proven consistent by means of a relative consistency argument with respect to ordinary arithmetic. One can further add any true Π 1 0 {\displaystyle \Pi _{1}^{0}} sentence of arithmetic to the theory while still retaining consistency of the theory. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textit{incorrect} assertion. Let $P, V$ be an interactive system for a language $L\in \mathcal{NP}$. | A formal theory using the " ⊥ {\displaystyle \bot } " connective is defined to be consistent, if and only if the false is not among its theorems. In the absence of propositional constants, some substitutes (such as the ones described above) may be used instead to define consistency. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the assertion related to an open problem. | Possible open issues. While there are unclear or open issues, those can be visible. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the assertion related to an open problem. | 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 |
Let $X$ be a plaintext and $Y$ its ciphertext. Which statement is \textbf{not} equivalent to the others? | As an example, take p = 7 {\displaystyle p=7} and q = 11 {\displaystyle q=11} , then n = 77 {\displaystyle n=77} . Take m = 20 {\displaystyle m=20} as our plaintext. The ciphertext is thus c = m 2 mod n = 400 mod 77 = 15 {\displaystyle c=m^{2}{\bmod {n}}=400{\bmod {77}}=15} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $X$ be a plaintext and $Y$ its ciphertext. Which statement is \textbf{not} equivalent to the others? | {\displaystyle {\textrm {ciphertext}}=E_{K3}(D_{K2}(E_{K1}({\textrm {plaintext}}))).} That is, DES encrypt with K 1 {\displaystyle K1} , DES decrypt with K 2 {\displaystyle K2} , then DES encrypt with K 3 {\displaystyle K3} . Decryption is the reverse: plaintext = D K 1 ( E K 2 ( D K 3 ( ciphertext ) ) ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{\emph{incorrect}} assertion. A $\Sigma$-protocol \dots | . ) {\displaystyle \,\vert \psi '\rangle \langle \psi '\vert +(...)} , where the dots denote components of ρ out {\displaystyle \rho _{\operatorname {out} }} resulting from errors not properly corrected by the protocol. It follows that This fidelity is to be compared with the corresponding fidelity obtained when no error-correcting protocol is used, which was shown before to equal 1 − p {\displaystyle {1-p}} . A little algebra then shows that the fidelity after error correction is greater than the one without for p < 1 / 2 {\displaystyle p<1/2} . Note that this is consistent with the working assumption that was made while deriving the protocol (of p {\displaystyle p} being small enough). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{\emph{incorrect}} assertion. A $\Sigma$-protocol \dots | Message with indicator "VFG": XCYBGDSLVWBDJLKWIPEHVYGQZWDTHRQXIKEESQSSPZXARIXEABQIRUCKHGWUEBPF Message with indicator "VFX": YNSCFCCPVIPEMSGIZWFLHESCIYSPVRXMCFQAXVXDVUQILBJUABNLKMKDJMENUNQ Hut 8 would punch these onto banburies and count the repeats for all valid offsets −25 letters to +25 letters. There are two promising positions: XCYBGDSLVWBDJLKWIPEHVYGQZWDTHRQXIKEESQSSPZXARIXEABQIRUCKHGWUEBPF YNSCFCCPVIPEMSGIZWFLHESCIYSPVRXMCFQAXVXDVUQILBJUABNLKMKDJMENUNQ - -- - - - - -- This offset of eight letters shows nine repeats, including two bigrams, in an overlap of 56 letters (16%). The other promising position looks like this: XCYBGDSLVWBDJLKWIPEHVYGQZWDTHRQXIKEESQSSPZXARIXEABQIRUCKHGWUEBPF YNSCFCCPVIPEMSGIZWFLHESCIYSPVRXMCFQAXVXDVUQILBJUABNLKMKDJMENUNQ --- This offset of seven shows just a single trigram in an overlap of 57 letters. Turing's method of accumulating a score of a number of decibans allows the calculation of which of these situations is most likely to represent messages in depth. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The statistical distance between two distributions is \dots | The variation distance of two distributions X {\displaystyle X} and Y {\displaystyle Y} over a finite domain D {\displaystyle D} , (often referred to as statistical difference or statistical distance in cryptography) is defined as Δ ( X , Y ) = 1 2 ∑ α ∈ D | Pr − Pr | {\displaystyle \Delta (X,Y)={\frac {1}{2}}\sum _{\alpha \in D}|\Pr-\Pr|} . We say that two probability ensembles { X k } k ∈ N {\displaystyle \{X_{k}\}_{k\in \mathbb {N} }} and { Y k } k ∈ N {\displaystyle \{Y_{k}\}_{k\in \mathbb {N} }} are statistically close if Δ ( X k , Y k ) {\displaystyle \Delta (X_{k},Y_{k})} is a negligible function in k {\displaystyle k} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The statistical distance between two distributions is \dots | In statistics, probability theory, and information theory, a statistical distance quantifies the distance between two statistical objects, which can be two random variables, or two probability distributions or samples, or the distance can be between an individual sample point and a population or a wider sample of points. A distance between populations can be interpreted as measuring the distance between two probability distributions and hence they are essentially measures of distances between probability measures. Where statistical distance measures relate to the differences between random variables, these may have statistical dependence, and hence these distances are not directly related to measures of distances between probability measures. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{\emph{correct}} assertion. A random oracle $\ldots$ | The truth table of P ↚ Q {\displaystyle P\nleftarrow Q} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{\emph{correct}} assertion. A random oracle $\ldots$ | The truth table of P ↓ Q {\displaystyle P\downarrow Q} is as follows: | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an Sbox $S:\{0,1\}^m \rightarrow \{0,1\}^m$. We have that \ldots | The Cayley transform for general symmetric operators can be adapted to this special case. For every non-negative number a, | a − 1 a + 1 | ≤ 1. {\displaystyle \left|{\frac {a-1}{a+1}}\right|\leq 1.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an Sbox $S:\{0,1\}^m \rightarrow \{0,1\}^m$. We have that \ldots | The third term is ≤ M s ( b − a ) {\displaystyle \leq Ms(b-a)} . Hence, summing the estimates up, we get: | f ( a + t ( b − a ) ) − f ( a ) | ≤ t M | b − a | {\displaystyle |f(a+t(b-a))-f(a)|\leq tM|b-a|} , a contradiction to the maximality of s {\displaystyle s} . Hence, 1 = s ∈ M {\displaystyle 1=s\in M} and that means: | f ( b ) − f ( a ) | ≤ M ( b − a ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. In RSA \ldots | RSA-400 has 400 decimal digits (1,327 bits), and has not been factored so far. RSA-400 = 2014096878945207511726700485783442547915321782072704356103039129009966793396 1419850865094551022604032086955587930913903404388675137661234189428453016032 6191193056768564862615321256630010268346471747836597131398943140685464051631 7519403149294308737302321684840956395183222117468443578509847947119995373645 3607109795994713287610750434646825511120586422993705980787028106033008907158 74500584758146849481 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. In RSA \ldots | Every expression (message send) has a value. Thus ifTrue:ifFalse: can be used: | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the cipher defined by $$\begin{array}{llll} C : & \{0,1\}^{4} & \rightarrow & \{0,1\}^{4} \\ & x & \mapsto & C(x)=x \oplus 0110 \\ \end{array} $$ The value $LP^C(1,1)$ is equal to | For RC5P, analysis was conducted modulo 3. It was observed that the operations in the cipher (rotation and addition, both on 32-bit words) were somewhat biased over congruence classes mod 3. To illustrate the approach, consider left rotation by a single bit: X ⋘ 1 = { 2 X , if X < 2 31 2 X + 1 − 2 32 , if X ≥ 2 31 {\displaystyle X\lll 1=\left\{{\begin{matrix}2X,&{\mbox{if }}X<2^{31}\\2X+1-2^{32},&{\mbox{if }}X\geq 2^{31}\end{matrix}}\right.} Then, because 2 32 ≡ 1 ( mod 3 ) , {\displaystyle 2^{32}\equiv 1{\pmod {3}},\,} it follows that X ⋘ 1 ≡ 2 X ( mod 3 ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the cipher defined by $$\begin{array}{llll} C : & \{0,1\}^{4} & \rightarrow & \{0,1\}^{4} \\ & x & \mapsto & C(x)=x \oplus 0110 \\ \end{array} $$ The value $LP^C(1,1)$ is equal to | Decryption of a ciphertext ( R n + 1 , L n + 1 ) {\displaystyle (R_{n+1},L_{n+1})} is accomplished by computing for i = n , n − 1 , … , 0 {\displaystyle i=n,n-1,\ldots ,0} R i = L i + 1 , {\displaystyle R_{i}=L_{i+1},} L i = R i + 1 ⊕ F ( L i + 1 , K i ) . {\displaystyle L_{i}=R_{i+1}\oplus \operatorname {F} (L_{i+1},K_{i}).} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n=pq$ be a RSA modulus and let $(e,d)$ be a RSA public/private key. Tick the \emph{correct} assertion. | In this example, an RSA modulus purporting to be of the form n = pq is actually of the form n = pqr, for primes p, q, and r. Calculation shows that exactly one extra bit can be hidden in the digitally signed message. The cure for this was found by cryptologists at the Centrum Wiskunde & Informatica in Amsterdam, who developed a Zero-knowledge proof that n is of the form n = pq. This example was motivated in part by The Empty Silo Proposal. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n=pq$ be a RSA modulus and let $(e,d)$ be a RSA public/private key. Tick the \emph{correct} assertion. | This RSA modulus is made public together with the encryption exponent e. N and e form the public key pair (e, N). By making this information public, anyone can encrypt messages to Bob. The decryption exponent d satisfies e d = 1 mod λ ( N ) {\displaystyle ed=1{\bmod {\lambda }}(N)} , where λ ( N ) {\displaystyle \lambda (N)} denotes the Carmichael function, though sometimes φ ( N ) {\displaystyle \varphi (N)} , the Euler’s phi function, is used (note: this is the order of the multiplicative group Z N ∗ {\displaystyle \mathbb {Z} _{N}^{*}} , which is not necessarily a cyclic group). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{true} assertion. A distinguishing attack against a block cipher\dots | Attacks that allow distinguishing ciphertext from random data. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{true} assertion. A distinguishing attack against a block cipher\dots | Before starting off the symmetric encryption process, the input message M ∈ B ∗ {\displaystyle M\in B^{\ast }} is divided into blocks M 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $n >1 $ be a composite integer, the product of two primes. Then, | ( 5 21 ) = 1 but 5 21 − 1 2 ≡ 16 ( mod 21 ) . {\displaystyle {\begin{aligned}\left({\frac {19}{45}}\right)&=1&&{\text{ and }}&19^{\frac {45-1}{2}}&\equiv 1{\pmod {45}}.\\\left({\frac {8}{21}}\right)&=-1&&{\text{ but }}&8^{\frac {21-1}{2}}&\equiv 1{\pmod {21}}.\\\left({\frac {5}{21}}\right)&=1&&{\text{ but }}&5^{\frac {21-1}{2}}&\equiv 16{\pmod {21}}.\end{aligned}}} So if it is unknown whether a number n is prime or composite, we can pick a random number a, calculate the Jacobi symbol (a/n) and compare it with Euler's formula; if they differ modulo n, then n is composite; if they have the same residue modulo n for many different values of a, then n is "probably prime". This is the basis for the probabilistic Solovay–Strassen primality test and refinements such as the Baillie–PSW primality test and the Miller–Rabin primality test. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $n >1 $ be a composite integer, the product of two primes. Then, | Therefore, if (1) concludes that N is composite, it truly is composite. (2) and (3) check if Q has order 2 k {\displaystyle 2^{k}} . Thus, if (2) or (3) conclude N is composite, it is composite. Now, if the algorithm concludes that N is prime, then that means S 1 {\displaystyle S_{1}} satisfies the condition of Theorem 4, and so N is truly prime. There is an algorithm as well for when n is large; however, for this we refer to the aforementioned article. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $C$ be a permutation over $\left\{ 0,1 \right\}^p$. Tick the \emph{incorrect} assertion: | It can be shown that an arbitrary permutation of N objects can be written as a product of transpositions and that the number of transposition in this decomposition is of fixed parity. That is, either a permutation is always decomposed in an even number of transpositions (the permutation is called even and has the parity +1), or a permutation is always decomposed in an odd number of transpositions and then it is an odd permutation with parity −1. Denoting the parity of an arbitrary permutation π by (−1)π, it follows that an antisymmetric wave function satisfies P ^ Ψ ( 1 , 2 , … , N ) ≡ Ψ ( π ( 1 ) , π ( 2 ) , … , π ( N ) ) = ( − 1 ) π Ψ ( 1 , 2 , … , N ) , {\displaystyle {\hat {P}}\Psi {\big (}1,2,\ldots ,N{\big )}\equiv \Psi {\big (}\pi (1),\pi (2),\ldots ,\pi (N){\big )}=(-1)^{\pi }\Psi (1,2,\ldots ,N),} where we associated the linear operator P ^ {\displaystyle {\hat {P}}} with the permutation π. The set of all N! | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $C$ be a permutation over $\left\{ 0,1 \right\}^p$. Tick the \emph{incorrect} assertion: | That is, H n ( p 1 , p 2 , … p n ) = H n ( p i 1 , p i 2 , … , p i n ) {\displaystyle \mathrm {H} _{n}\left(p_{1},p_{2},\ldots p_{n}\right)=\mathrm {H} _{n}\left(p_{i_{1}},p_{i_{2}},\ldots ,p_{i_{n}}\right)} for any permutation { i 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Assume an arbitrary $f:\{0,1\}^p \rightarrow \{0,1\}^q$, where $p$ and $q$ are integers. | The truth table of P ↚ Q {\displaystyle P\nleftarrow Q} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Assume an arbitrary $f:\{0,1\}^p \rightarrow \{0,1\}^q$, where $p$ and $q$ are integers. | Thus f0 is the truth value of ψ. In order to arithmetize ψ we must use the following rules: f i ( a 1 , … , a i ) = { f i + 1 ( a 1 , … , a i , 0 ) ⋅ f i + 1 ( a 1 , … , a i , 1 ) Q i + 1 = ∀ f i + 1 ( a 1 , … , a i , 0 ) ∗ f i + 1 ( a 1 , … , a i , 1 ) Q i + 1 = ∃ {\displaystyle f_{i}(a_{1},\dots ,a_{i})={\begin{cases}f_{i+1}(a_{1},\dots ,a_{i},0)\cdot f_{i+1}(a_{1},\dots ,a_{i},1)&{\mathsf {Q}}_{i+1}=\forall \\f_{i+1}(a_{1},\dots ,a_{i},0)*f_{i+1}(a_{1},\dots ,a_{i},1)&{\mathsf {Q}}_{i+1}=\exists \end{cases}}} where as before we define x ∗ y = 1 − (1 − x)(1 − y). By using the method described in #SAT, we must face a problem that for any fi the degree of the resulting polynomial may double with each quantifier. In order to prevent this, we must introduce a new reduction operator R which will reduce the degrees of the polynomial without changing their behavior on Boolean inputs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In order to have zero-knowledge from $\Sigma$-protocols, we need to add the use of \ldots | . ) {\displaystyle \,\vert \psi '\rangle \langle \psi '\vert +(...)} , where the dots denote components of ρ out {\displaystyle \rho _{\operatorname {out} }} resulting from errors not properly corrected by the protocol. It follows that This fidelity is to be compared with the corresponding fidelity obtained when no error-correcting protocol is used, which was shown before to equal 1 − p {\displaystyle {1-p}} . A little algebra then shows that the fidelity after error correction is greater than the one without for p < 1 / 2 {\displaystyle p<1/2} . Note that this is consistent with the working assumption that was made while deriving the protocol (of p {\displaystyle p} being small enough). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In order to have zero-knowledge from $\Sigma$-protocols, we need to add the use of \ldots | The forward direction is trivial and for the reverse direction suppose Γ {\displaystyle \Gamma } is a total computable functional. To build the truth-table for computing A(n) simply search for a number m such that for all binary strings σ {\displaystyle \sigma } of length m Γ σ ( n ) {\displaystyle \Gamma ^{\sigma }(n)} converges. Such an m must exist by Kőnig's lemma since Γ {\displaystyle \Gamma } must be total on all paths through 2 < ω {\displaystyle 2^{<\omega }} . Given such an m it is a simple matter to find the unique truth-table which gives Γ σ ( n ) {\displaystyle \Gamma ^{\sigma }(n)} when applied to σ {\displaystyle \sigma } . The forward direction fails for weak truth-table reducibility. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A Feistel scheme is used in\dots | The Feistel construction is also used in cryptographic algorithms other than block ciphers. For example, the optimal asymmetric encryption padding (OAEP) scheme uses a simple Feistel network to randomize ciphertexts in certain asymmetric-key encryption schemes. A generalized Feistel algorithm can be used to create strong permutations on small domains of size not a power of two (see format-preserving encryption). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A Feistel scheme is used in\dots | A Feistel network uses a round function, a function which takes two inputs – a data block and a subkey – and returns one output of the same size as the data block. In each round, the round function is run on half of the data to be encrypted, and its output is XORed with the other half of the data. This is repeated a fixed number of times, and the final output is the encrypted data. An important advantage of Feistel networks compared to other cipher designs such as substitution–permutation networks is that the entire operation is guaranteed to be invertible (that is, encrypted data can be decrypted), even if the round function is not itself invertible. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. The SEI of the distribution $P$ of support $G$ \ldots | Let X {\displaystyle \textstyle X} be the support of the distributions of interest. As in the original work of Kearns et al. if X {\displaystyle \textstyle X} is finite it can be assumed without loss of generality that X = { 0 , 1 } n {\displaystyle \textstyle X=\{0,1\}^{n}} where n {\displaystyle \textstyle n} is the number of bits that have to be used in order to represent any y ∈ X {\displaystyle \textstyle y\in X} . We focus in probability distributions over X {\displaystyle \textstyle X} . There are two possible representations of a probability distribution D {\displaystyle \textstyle D} over X {\displaystyle \textstyle X} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. The SEI of the distribution $P$ of support $G$ \ldots | The truth table of P ↚ Q {\displaystyle P\nleftarrow Q} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $X$, $Y$, and $K$ be respectively the plaintext, ciphertext, and key distributions. $H$ denotes the Shannon entropy. Considering that the cipher achieves \emph{perfect secrecy}, tick the \textbf{false} assertion: | The final output is H t | | G t {\displaystyle H_{t}||G_{t}} . The scheme has the rate R H i r o s e = k − n 2 n {\textstyle R_{Hirose}={\frac {k-n}{2n}}} relative to encrypting the message with the cipher. Hirose also provides a proof in the Ideal Cipher Model. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $X$, $Y$, and $K$ be respectively the plaintext, ciphertext, and key distributions. $H$ denotes the Shannon entropy. Considering that the cipher achieves \emph{perfect secrecy}, tick the \textbf{false} assertion: | 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 |
Tick the \emph{correct} assertion. Assume that $C$ is an arbitrary random permutation. | There is a simple randomized polynomial-time algorithm that provides a ( 1 − 1 2 k ) {\displaystyle \textstyle \left(1-{\frac {1}{2^{k}}}\right)} -approximation to MAXEkSAT: independently set each variable to true with probability 1/2, otherwise set it to false. Any given clause c is unsatisfied only if all of its k constituent literals evaluates to false. Because each literal within a clause has a 1⁄2 chance of evaluating to true independently of any of the truth value of any of the other literals, the probability that they are all false is ( 1 2 ) k = 1 2 k {\displaystyle \textstyle ({\frac {1}{2}})^{k}={\frac {1}{2^{k}}}} . Thus, the probability that c is indeed satisfied is 1 − 1 2 k {\displaystyle \textstyle 1-{\frac {1}{2^{k}}}} , so the indicator variable 1 c {\displaystyle \textstyle 1_{c}} (that is 1 if c is true and 0 otherwise) has expectation 1 − 1 2 k {\displaystyle \textstyle 1-{\frac {1}{2^{k}}}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. Assume that $C$ is an arbitrary random permutation. | 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 |
Which class of languages includes some which cannot be proven by a polynomial-size non-interactive proof? | These are the set of languages for which neither membership nor non-membership can be proven in a finite amount of time, and contain all other languages that are not in either RE or co-RE. That is: NRNC = ALL − ( RE ∪ co-RE ) {\displaystyle {\mbox{NRNC}}={\mbox{ALL}}-({\mbox{RE}}\cup {\mbox{co-RE}})} .Not only are these problems undecidable, but neither they nor their complement are recursively enumerable. In January of 2020, a preprint announced a proof that RE was equivalent to the class MIP* (the class where a classical verifier interacts with multiple all-powerful quantum provers who share entanglement); a revised, but not yet fully reviewed, proof was published in Communications of the ACM in November 2021. The proof implies that the Connes embedding problem and Tsirelson's problem are false. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which class of languages includes some which cannot be proven by a polynomial-size non-interactive proof? | A central question of proof complexity is to understand if tautologies admit polynomial-size proofs. Formally, Problem (NP vs. coNP) Does there exist a polynomially bounded propositional proof system? Cook and Reckhow (1979) observed that there exists a polynomially bounded proof system if and only if NP=coNP. Therefore, proving that specific proof systems do not admit polynomial size proofs can be seen as a partial progress towards separating NP and coNP (and thus P and NP). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Graph coloring consist of coloring all vertices \ldots | Every n-vertex complete graph Kn has an exact coloring with n colors, obtained by giving each vertex a distinct color. Every graph with an n-color exact coloring may be obtained as a detachment of a complete graph, a graph obtained from the complete graph by splitting each vertex into an independent set and reconnecting each edge incident to the vertex to exactly one of the members of the corresponding independent set.When k is an odd number, A path or cycle with ( k 2 ) {\displaystyle {\tbinom {k}{2}}} edges has an exact coloring, obtained by forming an exact coloring of the complete graph Kk and then finding an Euler tour of this complete graph. For instance, a path with three edges has a complete 3-coloring. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Graph coloring consist of coloring all vertices \ldots | The equality between the chromatic index and the list chromatic index has been conjectured to hold, even more generally, for arbitrary multigraphs with no self-loops; this conjecture remains open. Many other commonly studied variations of vertex coloring have also been extended to edge colorings. For instance, complete edge coloring is the edge-coloring variant of complete coloring, a proper edge coloring in which each pair of colors must be represented by at least one pair of adjacent edges and in which the goal is to maximize the total number of colors. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What adversarial model does not make sense for a message authentication code (MAC)? | The stochastic model has no real way to fight against this sort of attack and as such is unsuited to dealing with the kind of "intelligent" threats that would be preferable to have defenses against. Therefore, a computationally bounded adversarial model has been proposed as a compromise between the two. This forces one to consider that messages may be perverted in conscious, even malicious ways, but without forcing an algorithm designer to worry about rare cases which likely will never occur. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What adversarial model does not make sense for a message authentication code (MAC)? | Informally, a message authentication code system consists of three algorithms: A key generation algorithm selects a key from the key space uniformly at random. A signing algorithm efficiently returns a tag given the key and the message. A verifying algorithm efficiently verifies the authenticity of the message given the same key and the tag. That is, return accepted when the message and tag are not tampered with or forged, and otherwise return rejected.A secure message authentication code must resist attempts by an adversary to forge tags, for arbitrary, select, or all messages, including under conditions of known- or chosen-message. It should be computationally infeasible to compute a valid tag of the given message without knowledge of the key, even if for the worst case, we assume the adversary knows the tag of any message but the one in question.Formally, a message authentication code (MAC) system is a triple of efficient algorithms (G, S, V) satisfying: G (key-generator) gives the key k on input 1n, where n is the security parameter. S (signing) outputs a tag t on the key k and the input string x. V (verifying) outputs accepted or rejected on inputs: the key k, the string x and the tag t.S and V must satisfy the following: Pr = 1.A MAC is unforgeable if for every efficient adversary A Pr < negl(n),where AS(k, · ) denotes that A has access to the oracle S(k, · ), and Query(AS(k, · ), 1n) denotes the set of the queries on S made by A, which knows n. Clearly we require that any adversary cannot directly query the string x on S, since otherwise a valid tag can be easily obtained by that adversary. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these ciphers does achieve perfect secrecy? | Another way of stating perfect secrecy is that for all messages m 1 , m 2 {\displaystyle m_{1},m_{2}} in message space M, and for all ciphers c in cipher space C, we have Pr k ⇐ K = Pr k ⇐ K {\displaystyle {\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}={\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}} , where Pr {\textstyle \operatorname {Pr} } represents the probabilities, taken over a choice of k {\displaystyle k} in key space K {\displaystyle \mathrm {K} } over the coin tosses of a probabilistic algorithm, E {\displaystyle E} . Perfect secrecy is a strong notion of cryptanalytic difficulty.Conventional symmetric encryption algorithms use complex patterns of substitution and transpositions. For the best of these currently in use, it is not known whether there can be a cryptanalytic procedure that can efficiently reverse (or even partially reverse) these transformations without knowing the key used during encryption. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these ciphers does achieve perfect secrecy? | 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 |
Which of the following problems has not been shown equivalent to the others? | Many mathematical problems have been stated but not yet solved. These problems come from many areas of mathematics, such as theoretical physics, computer science, algebra, analysis, combinatorics, algebraic, differential, discrete and Euclidean geometries, graph theory, group theory, model theory, number theory, set theory, Ramsey theory, dynamical systems, and partial differential equations. Some problems belong to more than one discipline and are studied using techniques from different areas. Prizes are often awarded for the solution to a long-standing problem, and some lists of unsolved problems, such as the Millennium Prize Problems, receive considerable attention. This list is a composite of notable unsolved problems mentioned in previously published lists, including but not limited to lists considered authoritative. Although this list may never be comprehensive, the problems listed here vary widely in both difficulty and importance. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following problems has not been shown equivalent to the others? | ：Miscellaneous 18 problems： | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A proof system is perfect-black-box zero-knowledge if \dots | We define the set of proof systems L EL := { K , KD45 , S4 , S4.2 , S4.3 , S4.3.2 , S4.4 , S5 } {\displaystyle \mathbb {L} _{\textsf {EL}}:=\{{\textsf {K}},{\textsf {KD45}},{\textsf {S4}},{\textsf {S4.2}},{\textsf {S4.3}},{\textsf {S4.3.2}},{\textsf {S4.4}},{\textsf {S5}}\}} . Moreover, for all H ∈ L EL {\displaystyle {\mathcal {H}}\in \mathbb {L} _{\textsf {EL}}} , we define the proof system H C {\displaystyle {\mathcal {H}}^{\textsf {C}}} by adding the following axiom schemes and rules of inference to those of H {\displaystyle {\mathcal {H}}} . For all A ⊆ A G T S {\displaystyle A\subseteq AGTS} , The relative strength of the proof systems for knowledge is as follows: S4 ⊂ S4.2 ⊂ S4.3 ⊂ S4.3.2 ⊂ S4.4 ⊂ S5 . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A proof system is perfect-black-box zero-knowledge if \dots | We define the set of proof systems L EL := { K , KD45 , S4 , S4.2 , S4.3 , S4.3.2 , S4.4 , S5 } {\displaystyle \mathbb {L} _{\textsf {EL}}:=\{{\textsf {K}},{\textsf {KD45}},{\textsf {S4}},{\textsf {S4.2}},{\textsf {S4.3}},{\textsf {S4.3.2}},{\textsf {S4.4}},{\textsf {S5}}\}} . Moreover, for all H ∈ L EL {\displaystyle {\mathcal {H}}\in \mathbb {L} _{\textsf {EL}}} , we define the proof system H C {\displaystyle {\mathcal {H}}^{\textsf {C}}} by adding the following axiom schemes and rules of inference to those of H {\displaystyle {\mathcal {H}}} . For all A ⊆ A G T S {\displaystyle A\subseteq AGTS} , The relative strength of the proof systems for knowledge is as follows: S4 ⊂ S4.2 ⊂ S4.3 ⊂ S4.3.2 ⊂ S4.4 ⊂ S5 . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose that you can prove the security of your symmetric encryption scheme against the following attacks. In which case is your scheme going to be the \textbf{most} secure? | To be strong, an algorithm needs to have a sufficiently long key and be free of known mathematical weaknesses, as exploitation of these effectively reduces the key size. At the beginning of the 21st century, the typical security strength of the strong symmetrical encryption algorithms is 128 bits (slightly lower values still can be strong, but usually there is little technical gain in using smaller key sizes).Demonstrating the resistance of any cryptographic scheme to attack is a complex matter, requiring extensive testing and reviews, preferably in a public forum. Good algorithms and protocols are required, and good system design and implementation is needed as well. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose that you can prove the security of your symmetric encryption scheme against the following attacks. In which case is your scheme going to be the \textbf{most} secure? | Before starting off the symmetric encryption process, the input message M ∈ B ∗ {\displaystyle M\in B^{\ast }} is divided into blocks M 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For a $n$-bit block cipher with $k$-bit key, given a plaintext-ciphertext pair, a key exhaustive search has an average number of trials of \dots | Let P = P ‖ P ‖ P ‖ P {\displaystyle P=P\|P\|P\|P} be a 128-bit block of plaintext and C = C ‖ C ‖ C ‖ C {\displaystyle C=C\|C\|C\|C} be a 128-bit block of ciphertext, where P {\displaystyle P} and C {\displaystyle C} ( 0 ≤ i < 4 {\displaystyle 0\leq i<4} ) are 32-bit blocks. Let K i = K i ‖ K i ‖ K i ‖ K i ‖ K i ‖ K i {\displaystyle K_{i}=K_{i}\|K_{i}\|K_{i}\|K_{i}\|K_{i}\|K_{i}} ( 0 ≤ i < N r {\displaystyle 0\leq i | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For a $n$-bit block cipher with $k$-bit key, given a plaintext-ciphertext pair, a key exhaustive search has an average number of trials of \dots | Suppose that there are 100 million key texts that might plausibly be used, and that on average each has 11 thousand possible starting positions. To an opponent with a massive collection of possible key texts, this leaves possible a brute force search of the order of 2 40 {\displaystyle 2^{40}} , which by computer cryptography standards is a relatively easy target. (See permutation generated running keys above for an approach to this problem). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. For a Vernam cipher... | The Vernam cipher implemented by the Lorenz SZ machines utilizes the Boolean "exclusive or" (XOR) function, symbolised by ⊕ and verbalised as "A or B but not both". This is represented by the following truth table, where x represents "true" and • represents "false". Other names for this function are: exclusive disjunction, not equal (NEQ), and modulo 2 addition (without "carry") and subtraction (without "borrow"). Modulo 2 addition and subtraction are identical. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. For a Vernam cipher... | Verify the ciphertext preamble: If u 1 ≥ P {\displaystyle u_{1}\geq P} or u 2 ≥ P {\displaystyle u_{2}\geq P} or v ≥ P {\displaystyle v\geq P} , then return R e j e c t {\displaystyle Reject} . If u 1 q ≠ 1 r e m P {\displaystyle u_{1}^{q}\neq 1remP} , then return R e j e c t {\displaystyle Reject} . r e j e c t ← 0 {\displaystyle reject\leftarrow 0} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Assume we are in a group $G$ of order $n = p_1^{\alpha_1} p_2^{\alpha_2}$, where $p_1$ and $p_2$ are two distinct primes and $\alpha_1, \alpha_2 \in \mathbb{N}$. The complexity of applying the Pohlig-Hellman algorithm for computing the discrete logarithm in $G$ is \ldots (\emph{choose the most accurate answer}): | In this section, we present the general case of the Pohlig–Hellman algorithm. The core ingredients are the algorithm from the previous section (to compute a logarithm modulo each prime power in the group order) and the Chinese remainder theorem (to combine these to a logarithm in the full group). (Again, we assume the group to be cyclic, with the understanding that a non-cyclic group must be replaced by the subgroup generated by the logarithm's base element.) Input. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Assume we are in a group $G$ of order $n = p_1^{\alpha_1} p_2^{\alpha_2}$, where $p_1$ and $p_2$ are two distinct primes and $\alpha_1, \alpha_2 \in \mathbb{N}$. The complexity of applying the Pohlig-Hellman algorithm for computing the discrete logarithm in $G$ is \ldots (\emph{choose the most accurate answer}): | As an important special case, which is used as a subroutine in the general algorithm (see below), the Pohlig–Hellman algorithm applies to groups whose order is a prime power. The basic idea of this algorithm is to iteratively compute the p {\displaystyle p} -adic digits of the logarithm by repeatedly "shifting out" all but one unknown digit in the exponent, and computing that digit by elementary methods. (Note that for readability, the algorithm is stated for cyclic groups — in general, G {\displaystyle G} must be replaced by the subgroup ⟨ g ⟩ {\displaystyle \langle g\rangle } generated by g {\displaystyle g} , which is always cyclic.) Input. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{\emph{incorrect}} 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 \textbf{\emph{incorrect}} assertion. | "Fl." for flashing, "F." for fixed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} statement. $\Sigma$-protocols \ldots | . ) {\displaystyle \,\vert \psi '\rangle \langle \psi '\vert +(...)} , where the dots denote components of ρ out {\displaystyle \rho _{\operatorname {out} }} resulting from errors not properly corrected by the protocol. It follows that This fidelity is to be compared with the corresponding fidelity obtained when no error-correcting protocol is used, which was shown before to equal 1 − p {\displaystyle {1-p}} . A little algebra then shows that the fidelity after error correction is greater than the one without for p < 1 / 2 {\displaystyle p<1/2} . Note that this is consistent with the working assumption that was made while deriving the protocol (of p {\displaystyle p} being small enough). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} statement. $\Sigma$-protocols \ldots | Protocols which have the above three-move structure (commitment, challenge and response) are called sigma protocols. The Greek letter Σ {\displaystyle \Sigma } visualizes the flow of the protocol. Sigma protocols exist for proving various statements, such as those pertaining to discrete logarithms. Using these proofs, the prover can not only prove the knowledge of the discrete logarithm, but also that the discrete logarithm is of a specific form. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which defense(s) highlight the principle of least privilege in software
security? | Honoring the Principle of least privilege at a granularity provided by the base system such as sandboxing of (to that point successful) attacks to an unprivileged user account helps in reliability of computing services provided by the system. As the chances of restarting such a process are better, and other services on the same machine aren't affected (or at least probably not as much as in the alternative case: i.e. a privileged process gone haywire instead). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which defense(s) highlight the principle of least privilege in software
security? | When applied to users, the terms least user access or least-privileged user account (LUA) are also used, referring to the concept that all user accounts should run with as few privileges as possible, and also launch applications with as few privileges as possible. The principle of (least privilege) is widely recognized as an important design consideration towards enhancing and giving a much needed 'Boost' to the protection of data and functionality from faults (fault tolerance) and malicious behavior. Benefits of the principle include: Intellectual Security. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For which kind of bugs does default LLVM provide sanitizers? | GPL. TinyVM. VMKit of LLVM. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For which kind of bugs does default LLVM provide sanitizers? | Rather the program's behavior is undefined. To make a fuzzer more sensitive to failures other than crashes, sanitizers can be used to inject assertions that crash the program when a failure is detected. There are different sanitizers for different kinds of bugs: to detect memory related errors, such as buffer overflows and use-after-free (using memory debuggers such as AddressSanitizer), to detect race conditions and deadlocks (ThreadSanitizer), to detect undefined behavior (UndefinedBehaviorSanitizer), to detect memory leaks (LeakSanitizer), or to check control-flow integrity (CFISanitizer).Fuzzing can also be used to detect "differential" bugs if a reference implementation is available. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following hold(s) true about update deployment in the secure
development lifecycle? | Secure Design should be a consideration at all points in the development lifecycle (whichever development methodology is chosen). Some pre-built Secure By Design development methodologies exist (e.g. Microsoft Security Development Lifecycle). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following hold(s) true about update deployment in the secure
development lifecycle? | Updates are released on a semiannual basis. The software utilizes Azure Availability Zones for updates. == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
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