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Consider the exhaustive search of a uniformly distributed key in a set of size $N$. Think of the possible strategies and their complexities. Which of the following is \textbf{not} possible (We assume that memory access is constant.) | 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 |
Consider the exhaustive search of a uniformly distributed key in a set of size $N$. Think of the possible strategies and their complexities. Which of the following is \textbf{not} possible (We assume that memory access is constant.) | Gasarch, Glenn, and Kruskal have performed a comparison of different computational methods for large subsets of { 1 , … n } {\displaystyle \{1,\dots n\}} with no arithmetic progression. Using these methods they found the exact size of the largest such set for n ≤ 187 {\displaystyle n\leq 187} . Their results include several new bounds for different values of n {\displaystyle n} , found by branch-and-bound algorithms that use linear programming and problem-specific heuristics to bound the size that can be achieved in any branch of the search tree. One heuristic that they found to be particularly effective was the thirds method, in which two shifted copies of a Salem–Spencer set for n {\displaystyle n} are placed in the first and last thirds of a set for 3 n {\displaystyle 3n} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} 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{true} assertion. | {\textsf {T}}\rangle } . The reasoning is as follows. The object booleanWitness has the member T that is assigned the type Boolean as its value. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Pohlig-Hellman algorithm can be used to \dots | Donald Knuth, The Art of Computer Programming, Vol 2: Seminumerical Algorithms, section 3.4.1. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Pohlig-Hellman algorithm can be used to \dots | "Section 2.3.4: The Bellman-Ford-Moore algorithm". Digraphs: Theory, Algorithms and Applications (First ed.). Springer. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. In a zero-knowledge interactive proof of knowledge, \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 |
Tick the \textbf{true} assertion. In a zero-knowledge interactive proof of knowledge, \ldots | As the program of the prover does not necessarily spit out the knowledge itself (as is the case for zero-knowledge proofs) a machine with a different program, called the knowledge extractor is introduced to capture this idea. We are mostly interested in what can be proven by polynomial time bounded machines. In this case the set of knowledge elements is limited to a set of witnesses of some language in NP. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $X,Y$ be two random variables over the same probability space. Then, | Finally, the two random variables X and Y are equal if they are equal as functions on their measurable space: X ( ω ) = Y ( ω ) for all ω . {\displaystyle X(\omega )=Y(\omega )\qquad {\hbox{for all }}\omega .} This notion is typically the least useful in probability theory because in practice and in theory, the underlying measure space of the experiment is rarely explicitly characterized or even characterizable. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. Let $X,Y$ be two random variables over the same probability space. Then, | Given two random variables X , Y {\displaystyle X,Y} with values ( 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} 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{false} assertion. | {\textsf {T}}\rangle } . The reasoning is as follows. The object booleanWitness has the member T that is assigned the type Boolean as its value. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. The advantage of a distinguisher of two distributions $P_0$ and $P_1$ | The truth table of P ↚ Q {\displaystyle P\nleftarrow Q} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. The advantage of a distinguisher of two distributions $P_0$ and $P_1$ | Using P ( ¬ B | A ) = 1 − P ( B | A ) {\displaystyle P(\neg B\vert A)=1-P(B\vert A)} twice, one may use Bayes' theorem to also express P ( ¬ B | ¬ A ) {\displaystyle P(\neg B\vert \neg A)} in terms of P ( A | B ) {\displaystyle P(A\vert B)} and without negations: P ( ¬ B | ¬ A ) = 1 − ( 1 − P ( A | B ) ) P ( B ) P ( ¬ A ) {\displaystyle P(\neg B\vert \neg A)=1-\left(1-P(A\vert B)\right){\frac {P(B)}{P(\neg A)}}} ,when P ( ¬ A ) = 1 − P ( A ) ≠ 0 {\displaystyle P(\neg A)=1-P(A)\neq 0} . From this we can read off the inference P ( A | B ) = 1 ⟹ P ( ¬ B | ¬ A ) = 1 {\displaystyle P(A\vert B)=1\implies P(\neg B\vert \neg A)=1} .In words: If certainly B {\displaystyle B} implies A {\displaystyle A} , we infer that certainly ¬ A {\displaystyle \neg A} implies ¬ B {\displaystyle \neg B} . Where P ( B ) ≠ 0 {\displaystyle P(B)\neq 0} , the two implications being certain are equivalent statements. In the probability formulas, the conditional probability P ( A | B ) {\displaystyle P(A\vert B)} generalizes the logical implication B ⟹ A {\displaystyle B\implies A} , where now beyond assigning true or false, we assign probability values to statements. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The number of plaintext/ciphertext pairs required for a linear cryptanalysis is\dots | But this may not be enough assurance; a linear cryptanalysis attack against DES requires 243 known plaintexts (with their corresponding ciphertexts) and approximately 243 DES operations. This is a considerable improvement over brute force attacks. Public-key algorithms are based on the computational difficulty of various problems. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The number of plaintext/ciphertext pairs required for a linear cryptanalysis is\dots | A generalization of LC—multiple linear cryptanalysis—was suggested in 1994 (Kaliski and Robshaw), and was further refined by Biryukov and others. (2004); their analysis suggests that multiple linear approximations could be used to reduce the data requirements of the attack by at least a factor of 4 (that is, 241 instead of 243). A similar reduction in data complexity can be obtained in a chosen-plaintext variant of linear cryptanalysis (Knudsen and Mathiassen, 2000). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. For a cipher $C$, decorrelation theory says that \ldots | In cryptography, decorrelation theory is a system developed by Serge Vaudenay in 1998 for designing block ciphers to be provably secure against differential cryptanalysis, linear cryptanalysis, and even undiscovered cryptanalytic attacks meeting certain broad criteria. Ciphers designed using these principles include COCONUT98 and the AES candidate DFC, both of which have been shown to be vulnerable to some forms of cryptanalysis not covered by the theory. According to Vaudenay, the decorrelation theory has four tasks: 1) the definition of a measurement for the decorrelation, which usually relies on a matrix norm; 2) the construction of simple primitive or "decorrelation module" with a quite good decorrelation; 3) the construction of cryptographic algorithms with decorrelation modules so that the primitive can be inherited by the algorithm; and, 4) proving that the decorrelation provides security against attacks. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. For a cipher $C$, decorrelation theory says that \ldots | We say that assumption A {\displaystyle A} is stronger than assumption B {\displaystyle B} when A {\displaystyle A} implies B {\displaystyle B} (and the converse is false or not known). In other words, even if assumption A {\displaystyle A} were false, assumption B {\displaystyle B} may still be true, and cryptographic protocols based on assumption B {\displaystyle B} may still be safe to use. Thus when devising cryptographic protocols, one hopes to be able to prove security using the weakest possible assumptions. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given a function $f:\left\{ 0,1 \right\}^p \rightarrow \left\{ 0,1 \right\}^q$, given $a\in\left\{ 0,1 \right\}^p$ and $b \in \left\{ 0,1 \right\}^q$, we define $DP^{f}(a,b) = \Pr_{X}[f(X \oplus a) = f(X) \oplus b]$. We have that $\ldots$ | f →: P ( X ) → P ( Y ) {\displaystyle f^{\rightarrow }:{\mathcal {P}}(X)\to {\mathcal {P}}(Y)} with f → ( A ) = { f ( a ) | a ∈ A } {\displaystyle f^{\rightarrow }(A)=\{f(a)\;|\;a\in A\}} f ←: P ( Y ) → P ( X ) {\displaystyle f^{\leftarrow }:{\mathcal {P}}(Y)\to {\mathcal {P}}(X)} with f ← ( B ) = { a ∈ X | f ( a ) ∈ B } {\displaystyle f^{\leftarrow }(B)=\{a\in X\;|\;f(a)\in B\}} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Given a function $f:\left\{ 0,1 \right\}^p \rightarrow \left\{ 0,1 \right\}^q$, given $a\in\left\{ 0,1 \right\}^p$ and $b \in \left\{ 0,1 \right\}^q$, we define $DP^{f}(a,b) = \Pr_{X}[f(X \oplus a) = f(X) \oplus b]$. We have that $\ldots$ | Let q = p + ε. Taking a = nq in (1), we obtain: Pr ( 1 n ∑ X i ≥ q ) ≤ inf t > 0 E e t n q = inf t > 0 ( E e t q ) n . {\displaystyle \Pr \left({\frac {1}{n}}\sum X_{i}\geq q\right)\leq \inf _{t>0}{\frac {E\left}{e^{tnq}}}=\inf _{t>0}\left({\frac {E\left}{e^{tq}}}\right)^{n}.} Now, knowing that Pr(Xi = 1) = p, Pr(Xi = 0) = 1 − p, we have ( E e t q ) n = ( p e t + ( 1 − p ) e t q ) n = ( p e ( 1 − q ) t + ( 1 − p ) e − q t ) n . {\displaystyle \left({\frac {\operatorname {E} \left}{e^{tq}}}\right)^{n}=\left({\frac {pe^{t}+(1-p)}{e^{tq}}}\right)^{n}=\left(pe^{(1-q)t}+(1-p)e^{-qt}\right)^{n}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In linear cryptanalysis,\dots | Forking lemma Leftover hash lemma Piling-up lemma (linear cryptanalysis) Yao's XOR lemma | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In linear cryptanalysis,\dots | A generalization of LC—multiple linear cryptanalysis—was suggested in 1994 (Kaliski and Robshaw), and was further refined by Biryukov and others. (2004); their analysis suggests that multiple linear approximations could be used to reduce the data requirements of the attack by at least a factor of 4 (that is, 241 instead of 243). A similar reduction in data complexity can be obtained in a chosen-plaintext variant of linear cryptanalysis (Knudsen and Mathiassen, 2000). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The worst case complexity of an exaustive search (with memory) against DES is\dots | When top-down parser tries to parse an ambiguous input with respect to an ambiguous CFG, it may need exponential number of steps (with respect to the length of the input) to try all alternatives of the CFG in order to produce all possible parse trees, which eventually would require exponential memory space. The problem of exponential time complexity in top-down parsers constructed as sets of mutually recursive functions has been solved by Norvig in 1991. His technique is similar to the use of dynamic programming and state-sets in Earley's algorithm (1970), and tables in the CYK algorithm of Cocke, Younger and Kasami. The key idea is to store results of applying a parser p at position j in a memorable and to reuse results whenever the same situation arises. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The worst case complexity of an exaustive search (with memory) against DES is\dots | In the geometric view of the online binary search tree problem, an access sequence x 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Who invented linear cryptanalysis? | A linear cryptanalysis is a form of cryptanalysis based on finding affine approximations to the action of a cipher. Linear cryptanalysis is one of the two most widely used attacks on block ciphers; the other being differential cryptanalysis.The discovery is attributed to Mitsuru Matsui, who first applied the technique to the FEAL cipher (Matsui and Yamagishi, 1992). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Who invented linear cryptanalysis? | In cryptography, linear cryptanalysis is a general form of cryptanalysis based on finding affine approximations to the action of a cipher. Attacks have been developed for block ciphers and stream ciphers. Linear cryptanalysis is one of the two most widely used attacks on block ciphers; the other being differential cryptanalysis. The discovery is attributed to Mitsuru Matsui, who first applied the technique to the FEAL cipher (Matsui and Yamagishi, 1992). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For a blockcipher $B:\{0,1\}^k\times \{0,1\}^n \rightarrow \{0,1\}^n$ that has decorrelation $Dec^q_{\| \cdot \|_{\infty}}(B,C^*)=d$ (from a perfect cipher $C^*$), the best advantage of \textit{any} distinguisher that makes $q$ queries is \ldots | Suppose a code of 2 n R {\displaystyle 2^{nR}} codewords. Let W be drawn uniformly over this set as an index. Let X n {\displaystyle X^{n}} and Y n {\displaystyle Y^{n}} be the transmitted codewords and received codewords, respectively. n R = H ( W ) = H ( W | Y n ) + I ( W ; Y n ) {\displaystyle nR=H(W)=H(W|Y^{n})+I(W;Y^{n})} using identities involving entropy and mutual information ≤ H ( W | Y n ) + I ( X n ( W ) ; Y n ) {\displaystyle \leq H(W|Y^{n})+I(X^{n}(W);Y^{n})} since X is a function of W ≤ 1 + P e ( n ) n R + I ( X n ( W ) ; Y n ) {\displaystyle \leq 1+P_{e}^{(n)}nR+I(X^{n}(W);Y^{n})} by the use of Fano's Inequality ≤ 1 + P e ( n ) n R + n C {\displaystyle \leq 1+P_{e}^{(n)}nR+nC} by the fact that capacity is maximized mutual information.The result of these steps is that P e ( n ) ≥ 1 − 1 n R − C R {\displaystyle P_{e}^{(n)}\geq 1-{\frac {1}{nR}}-{\frac {C}{R}}} . As the block length n {\displaystyle n} goes to infinity, we obtain P e ( n ) {\displaystyle P_{e}^{(n)}} is bounded away from 0 if R is greater than C - we can get arbitrarily low rates of error only if R is less than C. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For a blockcipher $B:\{0,1\}^k\times \{0,1\}^n \rightarrow \{0,1\}^n$ that has decorrelation $Dec^q_{\| \cdot \|_{\infty}}(B,C^*)=d$ (from a perfect cipher $C^*$), the best advantage of \textit{any} distinguisher that makes $q$ queries is \ldots | by only comparing the 8 bits of each intermediate value, the authors was able to orchestrate a MITM attack on the cipher, despite there being 20 rounds between the two subciphers. Using partial-matching increased the amount of false positives, but nothing that noticeably increased the complexity of the attack. == Notes == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
I want to send a value to Bob without him knowing which value I sent and such that I cannot change my mind later when I reveal it in clear. I should use \dots | Alice keeps a private. Bob chooses an element b from B and sends wb to Alice. Bob keeps b private. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
I want to send a value to Bob without him knowing which value I sent and such that I cannot change my mind later when I reveal it in clear. I should use \dots | Alice (the sender) wishes to send a value x to Bob (the receiver). The communication channel between Alice and Bob is imperfect, and can introduce errors. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} 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{false} assertion. | {\textsf {T}}\rangle } . The reasoning is as follows. The object booleanWitness has the member T that is assigned the type Boolean as its value. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. $x\in \mathbf{Z}_{n}$ is invertible iff \ldots | We then have X X − 1 = = = I n {\displaystyle \mathbf {X} \mathbf {X} ^{-1}=\left=\left=\mathbf {I} _{n}} , where δ i j {\displaystyle \delta _{i}^{j}} is the Kronecker delta. We also have X − 1 X = = = = I n {\displaystyle \mathbf {X} ^{-1}\mathbf {X} =\left=\left=\left=\mathbf {I} _{n}} , as required. If the vectors x i {\displaystyle \mathbf {x} _{i}} are not linearly independent, then ( x 1 ∧ x 2 ∧ ⋯ ∧ x n ) = 0 {\displaystyle (\mathbf {x} _{1}\wedge \mathbf {x} _{2}\wedge \cdots \wedge \mathbf {x} _{n})=0} and the matrix X {\displaystyle \mathbf {X} } is not invertible (has no inverse). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. $x\in \mathbf{Z}_{n}$ is invertible iff \ldots | {\displaystyle \mathbf {B} ={\begin{pmatrix}-1&{\tfrac {3}{2}}\\1&-1\end{pmatrix}}.} The matrix B {\displaystyle \mathbf {B} } is invertible. To check this, one can compute that det B = − 1 2 {\textstyle \det \mathbf {B} =-{\frac {1}{2}}} , which is non-zero. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following circuits does not change an input difference. | The voltage difference in the low state, where the voltages on the wires are exchanged, is 0 V − V S = − V S {\displaystyle 0\,\mathrm {V} -V_{S}=-V_{S}} . The difference between high and low logic levels is therefore V S − ( − V S ) = 2 V S {\displaystyle V_{S}-(-V_{S})=2V_{S}\,} . This is twice the difference of the single-ended system. If the voltage noise on one wire is uncorrelated to the noise on the other one, it takes twice as much noise to cause an error with the differential system as with the single-ended system. In other words, differential signalling doubles the noise immunity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following circuits does not change an input difference. | Examples are the classic transistor emitter-coupled Schmitt trigger, the op-amp inverting Schmitt trigger, etc. Modified input voltage (parallel feedback): when the input voltage crosses the threshold in some direction the circuit changes its input voltage in the same direction (now it adds a part of its output voltage directly to the input voltage). Thus the output augments the input voltage and does not affect the threshold. These circuits can be implemented by a single-ended non-inverting amplifier with 'parallel positive feedback' where the input and the output sources are connected through resistors to the input. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these attacks is not a side channel attack? | In computer security, a side-channel attack is any attack based on extra information that can be gathered because of the fundamental way a computer protocol or algorithm is implemented, rather than flaws in the design of the protocol or algorithm itself (e.g. flaws found in a cryptanalysis of a cryptographic algorithm) or minor, but potentially devastating, mistakes or oversights in the implementation. (Cryptanalysis also includes searching for side-channel attacks.) Timing information, power consumption, electromagnetic leaks, and sound are examples of extra information which could be exploited to facilitate side-channel attacks. Some side-channel attacks require technical knowledge of the internal operation of the system, although others such as differential power analysis are effective as black-box attacks. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which one of these attacks is not a side channel attack? | Side-channel attacks do not attack the cipher as a black box, and thus are not related to cipher security as defined in the classical context, but are important in practice. They attack implementations of the cipher on hardware or software systems that inadvertently leak data. There are several such known attacks on various implementations of AES. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. The maximum advantage of an \textbf{adaptive} distinguisher limited to $q$ queries between two random functions $F$ and $F^*$ is always\dots | Sensitivity conjecture for Boolean functions (Hao Huang, 2019) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. The maximum advantage of an \textbf{adaptive} distinguisher limited to $q$ queries between two random functions $F$ and $F^*$ is always\dots | For a Boolean function f: { 0 , 1 } n → { 0 , 1 } {\displaystyle f:\{0,1\}^{n}\to \{0,1\}} , the sensitivity of f {\displaystyle f} is defined to be the maximum sensitivity of f {\displaystyle f} over all x {\displaystyle x} , where the sensitivity of f {\displaystyle f} at x {\displaystyle x} is the number of single-bit changes in x {\displaystyle x} that change the value of f ( x ) {\displaystyle f(x)} . Sensitivity is related to the notion of total influence from the analysis of Boolean functions, which is equal to average sensitivity over all x {\displaystyle x} . The sensitivity conjecture is the conjecture that sensitivity is polynomially related to query complexity; that is, there exists exponent c , c ′ {\displaystyle c,c'} such that, for all f {\displaystyle f} , D ( f ) = O ( s ( f ) c ) {\displaystyle D(f)=O(s(f)^{c})} and s ( f ) = O ( D ( f ) c ′ ) {\displaystyle s(f)=O(D(f)^{c'})} . One can show through a simple argument that s ( f ) ≤ D ( f ) {\displaystyle s(f)\leq D(f)} , so the conjecture is specifically concerned about finding a lower bound for sensitivity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. A distinguisher can \ldots | 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{false} assertion. A distinguisher can \ldots | asserts that if p {\displaystyle p\,\!} is true then so is q {\displaystyle q\,\!} , the inference p ⊢ q {\displaystyle p\vdash q\,\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider any block cipher $C$ and a uniformly distributed random permutation $C^*$ on $\{0,1\}^\ell$. Then, for any $n \ge 1$ we always have\dots | For this case we prove a weaker version of the theorem: One-way permutation → pseudorandom generator A one-way permutation is a one-way function that is also a permutation of the input bits. A pseudorandom generator can be constructed from one-way permutation ƒ as follows: Gl: {0,1}l→{0,1}l+1 = ƒ(x).B(x), where B is hard-core predicate of ƒ and "." is a concatenation operator. Note, that by the theorem proven above, it is only needed to show the existence of a generator that adds just one pseudorandom bit. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider any block cipher $C$ and a uniformly distributed random permutation $C^*$ on $\{0,1\}^\ell$. Then, for any $n \ge 1$ we always have\dots | 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 |
Let $X$, $Y$, and $K$ be respectively the plaintext, ciphertext, and key distributions. $H$ denotes the Shannon entropy. The consequence of perfect secrecy is \dots | 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 |
Let $X$, $Y$, and $K$ be respectively the plaintext, ciphertext, and key distributions. $H$ denotes the Shannon entropy. The consequence of perfect secrecy is \dots | 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 |
Tick the \textbf{true} assertion. A first preimage attack on a hash function H is \ldots | For a hash function for which L is the number of bits in the message digest, finding a message that corresponds to a given message digest can always be done using a brute force search in approximately 2L evaluations. This is called a preimage attack and may or may not be practical depending on L and the particular computing environment. However, a collision, consisting of finding two different messages that produce the same message digest, requires on average only about 1.2 × 2L/2 evaluations using a birthday attack. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. A first preimage attack on a hash function H is \ldots | Hence, the Merkle–Damgård hash construction reduces the problem of finding a proper hash function to finding a proper compression function. A second preimage attack (given a message m 1 {\displaystyle m_{1}} an attacker finds another message m 2 {\displaystyle m_{2}} to satisfy hash ( m 1 ) = hash ( m 2 ) {\displaystyle \operatorname {hash} (m_{1})=\operatorname {hash} (m_{2})} can be done according to Kelsey and Schneier for a 2 k {\displaystyle 2^{k}} -message-block message in time k × 2 n / 2 + 1 + 2 n − k + 1 {\displaystyle k\times 2^{n/2+1}+2^{n-k+1}} . Note that the complexity of this attack reaches a minimum of 2 3 n / 4 + 2 {\displaystyle 2^{3n/4+2}} for long messages when k = 2 n / 4 {\displaystyle k=2^{n/4}} and approaches 2 n {\displaystyle 2^{n}} when messages are short. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In an interactive proof system for a language $L$, having $\beta$-soundness means that\dots | Every interactive proof system defines a formal language of strings L {\displaystyle L} . Soundness of the proof system refers to the property that no prover can make the verifier accept for the wrong statement y ∉ L {\displaystyle y\not \in L} except with some small probability. The upper bound of this probability is referred to as the soundness error of a proof system. More formally, for every prover ( P ~ ) {\displaystyle ({\tilde {\mathcal {P}}})} , and every y ∉ L {\displaystyle y\not \in L}: Pr < ϵ . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In an interactive proof system for a language $L$, having $\beta$-soundness means that\dots | Most proofs of soundness are trivial. For example, in an axiomatic system, proof of soundness amounts to verifying the validity of the axioms and that the rules of inference preserve validity (or the weaker property, truth). If the system allows Hilbert-style deduction, it requires only verifying the validity of the axioms and one rule of inference, namely modus ponens. (and sometimes substitution) Soundness properties come in two main varieties: weak and strong soundness, of which the former is a restricted form of the latter. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A proof system is computational-zero-knowledge if \dots | As the program of the prover does not necessarily spit out the knowledge itself (as is the case for zero-knowledge proofs) a machine with a different program, called the knowledge extractor is introduced to capture this idea. We are mostly interested in what can be proven by polynomial time bounded machines. In this case the set of knowledge elements is limited to a set of witnesses of some language in NP. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A proof system is computational-zero-knowledge if \dots | 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 |
Tick the \textbf{false} assertion. Assume that $C$ is a 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 \textbf{false} assertion. Assume that $C$ is a random permutation. | If x {\displaystyle \mathbf {x} } is a random permutation of the integers 1 , 2 , … , n {\displaystyle 1,2,\ldots ,n} and A n ≡ a s n ( x ) {\displaystyle A_{n}\equiv {\rm {as}}_{n}(\mathbf {x} )} , then it is possible to show that E = 2 n 3 + 1 6 and Var = 8 n 45 − 13 180 . {\displaystyle E={\frac {2n}{3}}+{\frac {1}{6}}\qquad {\text{and}}\qquad \operatorname {Var} ={\frac {8n}{45}}-{\frac {13}{180}}.} Moreover, as n → ∞ {\displaystyle n\rightarrow \infty } , the random variable A n {\displaystyle A_{n}} , appropriately centered and scaled, converges to a standard normal distribution. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Standard encryption threats do not include: | The encryption features are prohibited. The digital signature features (from SES) are prohibited. The "patched data" features (from PKPatchMaker) are prohibited. Archives may not span multiple volumes or be segmented. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Standard encryption threats do not include: | Recent incidents of terrorism have led to further calls for restrictions on encryption. Even though, in the interest of public safety, there are many proposals to interfere with the free deployment of strong encryption, these proposals do not hold up against close scientific scrutiny. These proposals side-step a more fundamental point, related to what is at stake for users. More advanced security measures seem necessary for governments, considering the existing threat landscape for users of digital communications and computing.While many governments consider that encryption techniques could present a hurdle in the investigation of crime and the protection of national security, certain countries, such as Germany or the Netherlands have taken a strong position against restrictions on encryption on the Internet. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In Linear Cryptanalysis, the corresponding mask circuit of \ldots | Output: Decrypted message M ∈ B ∗ ∪ R e j e c t {\displaystyle M\in B^{\ast }\cup {Reject}} . Decrypt the ciphertext: If L ( ψ ) < 3 L B ( P ) + 16 {\displaystyle L(\psi )<3L_{B}(P)+16} , then return R e j e c t {\displaystyle Reject} . Compute: note that 0 ≤ u 1 , u 2 , v < 256 l {\displaystyle 0\leq u_{1},u_{2},v<256^{l}} , where l = L B ( P ) {\displaystyle l=L_{B}(P)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In Linear Cryptanalysis, the corresponding mask circuit of \ldots | 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{true} assertion. In a zero-knowledge interactive proof for $L$, \ldots | "Formalization of Forcing in Isabelle/ZF". Archive of Formal Proofs. arXiv:2001.09715. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. In a zero-knowledge interactive proof for $L$, \ldots | We explain how the four inference rules are used in proofs, using the proof of ¬ p ≡ p ≡ ⊥ {\textstyle \lnot p\equiv p\equiv \bot } . The logic symbols ⊤ {\textstyle \top } and ⊥ {\textstyle \bot } indicate "true" and "false," respectively, and ¬ {\textstyle \lnot } indicates "not." The theorem numbers refer to theorems of A Logical Approach to Discrete Math. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the Squared Euclidean Imbalance? | The imbalance in feature i {\displaystyle i} is: c i = ∑ j = 1 m X j {\displaystyle c_{i}=\sum _{j=1}^{m}{X_{j}}} . Since the X j {\displaystyle X_{j}} are independent random variables, by the Chernoff bound, for every a > 0 {\displaystyle a>0}: P r o b ≤ 2 exp ( − a 2 / 2 m ) {\displaystyle Prob\left\leq 2\exp(-a^{2}/2m)} Select: a = 4 m ln n {\displaystyle a={\sqrt {4m\ln n}}} and get: P r o b ≤ 2 n 2 {\displaystyle Prob\left\leq {\frac {2}{n^{2}}}} By the union bound, P r o b ≤ 2 n {\displaystyle Prob\left\leq {\frac {2}{n}}} . == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the Squared Euclidean Imbalance? | The squared Euclidean distance is a measure of how "close" two sets are. In particular, if two sets are compact, then their squared Euclidean distance is zero if and only if they are equal. Thus, we may quantify how close to convexity Q {\displaystyle Q} is by upper-bounding d 2 ( C o n v ( Q ) , Q ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. A cipher with a good decorrelation of order 2 protects against \ldots | 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 |
Tick the \textbf{false} assertion. A cipher with a good decorrelation of order 2 protects against \ldots | 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 |
For any function $f:\{0,1\}^p\rightarrow \{0,1\}^q$ and for any $a\in\{0,1\}^p$, we have\ldots | f →: P ( X ) → P ( Y ) {\displaystyle f^{\rightarrow }:{\mathcal {P}}(X)\to {\mathcal {P}}(Y)} with f → ( A ) = { f ( a ) | a ∈ A } {\displaystyle f^{\rightarrow }(A)=\{f(a)\;|\;a\in A\}} f ←: P ( Y ) → P ( X ) {\displaystyle f^{\leftarrow }:{\mathcal {P}}(Y)\to {\mathcal {P}}(X)} with f ← ( B ) = { a ∈ X | f ( a ) ∈ B } {\displaystyle f^{\leftarrow }(B)=\{a\in X\;|\;f(a)\in B\}} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For any function $f:\{0,1\}^p\rightarrow \{0,1\}^q$ and for any $a\in\{0,1\}^p$, we have\ldots | We show that f() ⊆ . Indeed, for every x ∈ W we have x = f(x) and since w is the least upper bound of W, x ≤ f(w). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion regarding plain Rabin, i.e., Rabin without any 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 |
Tick the \textbf{incorrect} assertion regarding plain Rabin, i.e., Rabin without any redundancy. | Patterns may assert that previous text or subsequent text contains a pattern without consuming matched text (zero-width assertion). For example, /\w+(?=\t)/ matches a word followed by a tab, without including the tab itself. Look-behind assertions cannot be of uncertain length though (unlike Perl) each branch can be a different fixed length. \K can be used in a pattern to reset the start of the current whole match. This provides a flexible alternative approach to look-behind assertions because the discarded part of the match (the part that precedes \K) need not be fixed in length. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. A cipher $C$ perfectly decorrelated at order 2 implies\dots | 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 |
Tick the \emph{incorrect} assertion. A cipher $C$ perfectly decorrelated at order 2 implies\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{false} assertion. A distinguisher \ldots | 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{false} assertion. A distinguisher \ldots | asserts that if p {\displaystyle p\,\!} is true then so is q {\displaystyle q\,\!} , the inference p ⊢ q {\displaystyle p\vdash q\,\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion regarding the security of the Diffie-Hellman key exchange over a subgroup $\langle g \rangle \subset \mathbb{Z}_p^*$. | G F ( p t ) {\displaystyle GF\left(p^{t}\right)} Let now ⟨ γ ⟩ {\displaystyle \langle \gamma \rangle } be a multiplicative group of order ω {\displaystyle \omega } . The security of the Diffie–Hellman protocol in ⟨ γ ⟩ {\displaystyle \langle \gamma \rangle } relies on the Diffie–Hellman (DH) problem of computing γ x y given γ , γ x and γ y {\displaystyle \gamma ^{xy}{\text{ given }}\gamma ,\gamma ^{x}{\text{ and }}\gamma ^{y}} . We write D H ( γ x , γ y ) = γ x y {\displaystyle DH(\gamma ^{x},\ \gamma ^{y})=\gamma ^{xy}} . There are two other problems related to the DH problem. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion regarding the security of the Diffie-Hellman key exchange over a subgroup $\langle g \rangle \subset \mathbb{Z}_p^*$. | The group G satisfies the requisite condition for secure communication as long as there is no efficient algorithm for determining gab given g, ga, and gb. For example, the elliptic curve Diffie–Hellman protocol is a variant that represents an element of G as a point on an elliptic curve instead of as an integer modulo n. Variants using hyperelliptic curves have also been proposed. The supersingular isogeny key exchange is a Diffie–Hellman variant that was designed to be secure against quantum computers, but it was broken in July 2022. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In Differential Cryptanalysis, the corresponding differential circuit of \ldots | Higher-order differential cryptanalysis Truncated differential cryptanalysis Impossible differential cryptanalysis Boomerang attack | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} assertion. In Differential Cryptanalysis, the corresponding differential circuit of \ldots | 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 \emph{correct} assertion. Linear cryptanalysis \ldots | In order for the whole system to work, one has to postulate that: ∀ m ∈ M , I D ∈ { 0 , 1 } ∗: D e c r y p t ( E x t r a c t ( P , K m , I D ) , P , E n c r y p t ( P , m , I D ) ) = m {\displaystyle \forall m\in {\mathcal {M}},ID\in \left\{0,1\right\}^{*}:\mathrm {Decrypt} \left(\mathrm {Extract} \left({\mathcal {P}},K_{m},ID\right),{\mathcal {P}},\mathrm {Encrypt} \left({\mathcal {P}},m,ID\right)\right)=m} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. Linear cryptanalysis \ldots | 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 |
Let $p>2$ be a prime. Then \dots | Let p be a prime number. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $p>2$ be a prime. Then \dots | If p is an odd prime, then every prime q that divides 2p − 1 must be 1 plus a multiple of 2p. This holds even when 2p − 1 is prime. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which assertion has not been proven? | It is based on the premise that there is no proof for a certain claim. From this premise, the conclusion is drawn that this claim must therefore be false. For example, "Nobody has ever proved to me there's a God, so I know there is no God". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which assertion has not been proven? | It is based on the premise that there is no proof for a certain claim. From this premise, the conclusion is drawn that this claim must therefore be false. For example, "Nobody has ever proved to me there's a God, so I know there is no God". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In hypothesis testing \ldots | asserts that if p {\displaystyle p\,\!} is true then so is q {\displaystyle q\,\!} , the inference p ⊢ q {\displaystyle p\vdash q\,\!} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In hypothesis testing \ldots | The method proposed in Section 8.7 of Duda, Hart & Stork (2001) suggests successively guessing grammar rules (productions) and testing them against positive and negative observations. The rule set is expanded so as to be able to generate each positive example, but if a given rule set also generates a negative example, it must be discarded. This particular approach can be characterized as "hypothesis testing" and bears some similarity to Mitchel's version space algorithm. The Duda, Hart & Stork (2001) text provide a simple example which nicely illustrates the process, but the feasibility of such an unguided trial-and-error approach for more substantial problems is dubious. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the cipher defined using the key $K\in \{0,1\}^{64} $ by $$\begin{array}{llll} C : & \{0,1\}^{64} & \rightarrow & \{0,1\}^{64} \\ & x & \mapsto & C(x)=x \oplus K \\ \end{array} $$ Let $x=1\dots 11$, the value $\mathsf{LP}^{C_K}(x,x)$ 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 using the key $K\in \{0,1\}^{64} $ by $$\begin{array}{llll} C : & \{0,1\}^{64} & \rightarrow & \{0,1\}^{64} \\ & x & \mapsto & C(x)=x \oplus K \\ \end{array} $$ Let $x=1\dots 11$, the value $\mathsf{LP}^{C_K}(x,x)$ 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 |
Tick the \textbf{incorrect} assertion. Let $H:\left\{ 0,1 \right\}^*\rightarrow\left\{ 0,1 \right\}^n$ be a hash function. | If this hash value equals the hash value of the pattern, it performs a full comparison at that position. In order for this to work well, the hash function should be selected randomly from a family of hash functions that are unlikely to produce many false positives, that is, positions of the text which have the same hash value as the pattern but do not actually match the pattern. These positions contribute to the running time of the algorithm unnecessarily, without producing a match. Additionally, the hash function used should be a rolling hash, a hash function whose value can be quickly updated from each position of the text to the next. Recomputing the hash function from scratch at each position would be too slow. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion. Let $H:\left\{ 0,1 \right\}^*\rightarrow\left\{ 0,1 \right\}^n$ be a hash function. | , h ( k , y ) } {\displaystyle \left\{h(0,x),...,h(k,x)\right\}=\left\{h(0,y),...,h(k,y)\right\}} are identical. This makes a collision twice as likely as the hoped-for 1 / | T | 2 {\displaystyle 1/|T|^{2}} . There are additionally a significant number of mostly-overlapping hash sets; if h 2 ( y ) = h 2 ( x ) {\displaystyle h_{2}(y)=h_{2}(x)} and h 1 ( y ) = h 1 ( x ) ± h 2 ( x ) {\displaystyle h_{1}(y)=h_{1}(x)\pm h_{2}(x)} , then h ( i , y ) = h ( i ± 1 , x ) {\displaystyle h(i,y)=h(i\pm 1,x)} , and comparing additional hash values (expanding the range of i {\displaystyle i} ) is of no help. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In which group is the discrete logarithm problem believed to be hard? | There exist groups for which computing discrete logarithms is apparently difficult. In some cases (e.g. large prime order subgroups of groups (Zp)×) there is not only no efficient algorithm known for the worst case, but the average-case complexity can be shown to be about as hard as the worst case using random self-reducibility.At the same time, the inverse problem of discrete exponentiation is not difficult (it can be computed efficiently using exponentiation by squaring, for example). This asymmetry is analogous to the one between integer factorization and integer multiplication. Both asymmetries (and other possibly one-way functions) have been exploited in the construction of cryptographic systems. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In which group is the discrete logarithm problem believed to be hard? | The discrete logarithm problem is considered to be computationally intractable. That is, no efficient classical algorithm is known for computing discrete logarithms in general. A general algorithm for computing logb a in finite groups G is to raise b to larger and larger powers k until the desired a is found. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider two distributions $P_0,P_1$ with the same supports and a distinguisher $\mathcal{A}$ that makes $q$ queries. Tick the \textit{incorrect} assertion. | Consider two uniform distributions, with the support of p = {\displaystyle p=} enclosed within q = {\displaystyle q=} ( C ≤ A < B ≤ D {\displaystyle C\leq A | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider two distributions $P_0,P_1$ with the same supports and a distinguisher $\mathcal{A}$ that makes $q$ queries. Tick the \textit{incorrect} assertion. | Consider two probability distributions P {\displaystyle P} and Q {\displaystyle Q} . Usually, P {\displaystyle P} represents the data, the observations, or a measured probability distribution. Distribution Q {\displaystyle Q} represents instead a theory, a model, a description or an approximation of P {\displaystyle P} . The Kullback–Leibler divergence is then interpreted as the average difference of the number of bits required for encoding samples of P {\displaystyle P} using a code optimized for Q {\displaystyle Q} rather than one optimized for P {\displaystyle P} . Note that the roles of P {\displaystyle P} and Q {\displaystyle Q} can be reversed in some situations where that is easier to compute, such as with the expectation–maximization algorithm and evidence lower bound computations. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the complexity of prime number generation for a prime of length $\ell$? | To derive a certificate from this theorem, we first encode Mx, My, A, B, and q, then recursively encode the proof of primality for q < n, continuing until we reach a known prime. This certificate has size O((log n)2) and can be verified in O((log n)4) time. Moreover, the algorithm that generates these certificates can be shown to be expected polynomial time for all but a small fraction of primes, and this fraction exponentially decreases with the size of the primes. Consequently, it's well-suited to generating certified large random primes, an application that is important in cryptography applications such as generating provably valid RSA keys. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the complexity of prime number generation for a prime of length $\ell$? | Let π ( x ) {\displaystyle \pi (x)} be the number of primes smaller than x ∃ c 1 , c 2 > 0: π ( x + x ) − π ( x ) ≥ c 2 x log c 1 x {\displaystyle \exists c_{1},c_{2}>0:\pi (x+{\sqrt {x}})-\pi (x)\geq {\frac {c_{2}{\sqrt {x}}}{\log ^{c_{1}}x}}} for sufficiently large x. If one accepts this conjecture then the Goldwasser–Kilian algorithm terminates in expected polynomial time for every input. Also, if our N is of length k, then the algorithm creates a certificate of size O ( k 2 ) {\displaystyle O(k^{2})} that can be verified in O ( k 4 ) {\displaystyle O(k^{4})} .Now consider another conjecture, which will give us a bound on the total time of the algorithm. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In ElGamal signature scheme, if we avoid checking that $0 \leq r < p$ then \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, if we avoid checking that $0 \leq r < p$ then \ldots | The ElGamal signature scheme is a digital signature scheme based on the algebraic properties of modular exponentiation, together with the discrete logarithm problem. The algorithm uses a key pair consisting of a public key and a private key. The private key is used to generate a digital signature for a message, and such a signature can be verified by using the signer's corresponding public key. The digital signature provides message authentication (the receiver can verify the origin of the message), integrity (the receiver can verify that the message has not been modified since it was signed) and non-repudiation (the sender cannot falsely claim that they have not signed the message). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. MAC is \ldots | DEBUG gdb DDT, a PDP-10 debugger from DEC used as a command shell for the MIT Incompatible Timesharing System Firebug/Chromebug, a JavaScript shell and debugging environment as a Firefox plugin | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} assertion. MAC is \ldots | COBOL On-Line debugger. APL. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For $K$ a field, $a,b\in K$ with $4a^3+27b^2 \neq 0$, $E_{a,b}(K)$ is | Let K be any field of characteristic not 2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
For $K$ a field, $a,b\in K$ with $4a^3+27b^2 \neq 0$, $E_{a,b}(K)$ is | {\displaystyle N_{B/A}(xB)=N_{L/K}(x)A.} Let L / K {\displaystyle L/K} be a Galois extension of number fields with rings of integers O K ⊂ O L {\displaystyle {\mathcal {O}}_{K}\subset {\mathcal {O}}_{L}} . Then the preceding applies with A = O K , B = O L {\displaystyle A={\mathcal {O}}_{K},B={\mathcal {O}}_{L}} , and for any b ∈ I O L {\displaystyle {\mathfrak {b}}\in {\mathcal {I}}_{{\mathcal {O}}_{L}}} we have N O L / O K ( b ) = K ∩ ∏ σ ∈ Gal ( L / K ) σ ( b ) , {\displaystyle N_{{\mathcal {O}}_{L}/{\mathcal {O}}_{K}}({\mathfrak {b}})=K\cap \prod _{\sigma \in \operatorname {Gal} (L/K)}\sigma ({\mathfrak {b}}),} which is an element of I O K {\displaystyle {\mathcal {I}}_{{\mathcal {O}}_{K}}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
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