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The Advanced Encryption Standard (AES) is based on arithmetics on\dots | The Advanced Encryption Standard (AES), also known by its original name Rijndael (Dutch pronunciation: ), is a specification for the encryption of electronic data established by the U.S. National Institute of Standards and Technology (NIST) in 2001.AES is a variant of the Rijndael block cipher developed by two Belgian cryptographers, Joan Daemen and Vincent Rijmen, who submitted a proposal to NIST during the AES selection process. Rijndael is a family of ciphers with different key and block sizes. For AES, NIST selected three members of the Rijndael family, each with a block size of 128 bits, but three different key lengths: 128, 192 and 256 bits. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Advanced Encryption Standard (AES) is based on arithmetics on\dots | The estimated strength of the encryption corresponds to about 60–80 bits for common symmetrical ciphers. Cryptographically, this effective key length is quite low, but appropriate in that the protocol was not designed as a secure transport protocol but rather as a fast and efficient obfuscation method. AES was proposed as the encryption method, but not adopted because it consumed too much CPU time. The required Diffie–Hellman keys to achieve a security equal to AES would have been much bigger or require elliptic curve cryptography, making the handshake more expensive in terms of used CPU time. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In \emph{all} finite Abelian groups $(G,+)$ \ldots | Consider a finite abelian group G = ⊕i Cdi , where the d1 | d2 | ... | dr are invariant factors. Then D ( G ) ≥ M ( G ) = 1 − r + ∑ i d i . {\displaystyle D(G)\geq M(G)=1-r+\sum _{i}{d_{i}}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{incorrect} assertion. In \emph{all} finite Abelian groups $(G,+)$ \ldots | The finite group G {\displaystyle G} is abelian if and only if p ( G ) = 1 {\displaystyle p(G)=1} . One has p ( G ) = k ( G ) # G {\displaystyle p(G)={\frac {k(G)}{\#G}}} where k ( G ) {\displaystyle k(G)} is the number of conjugacy classes of G {\displaystyle G} .If G {\displaystyle G} is not abelian then p ( G ) ≤ 5 / 8 {\displaystyle p(G)\leq 5/8} (this result is sometimes called the 5/8 theorem) and this upper bound is sharp: there are infinitely many finite groups G {\displaystyle G} such that p ( G ) = 5 / 8 {\displaystyle p(G)=5/8} , the smallest one being the dihedral group of order 8. There is no uniform lower bound on p ( G ) {\displaystyle p(G)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The needed number of DES operations to attack DES is about ... | 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 needed number of DES operations to attack DES is about ... | An attack can be plausibly carried out. A wounded primitive has an attack taking between 280 and around 2100 operations. An attack is not possible right now, but future improvements are likely to make it possible. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which algorithm can be typically used in order to generate a prime number? | In computational number theory, a variety of algorithms make it possible to generate prime numbers efficiently. These are used in various applications, for example hashing, public-key cryptography, and search of prime factors in large numbers. For relatively small numbers, it is possible to just apply trial division to each successive odd number. Prime sieves are almost always faster. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which algorithm can be typically used in order to generate a prime number? | Agrawal and Biswas use a more sophisticated technique, which divides P n {\displaystyle {\mathcal {P}}_{n}} by a random monic polynomial of small degree. Prime numbers are used in a number of applications such as hash table sizing, pseudorandom number generators and in key generation for cryptography. Therefore, finding very large prime numbers (on the order of (at least) 10 350 ≈ 2 1024 {\displaystyle 10^{350}\approx 2^{1024}} ) becomes very important and efficient primality testing algorithms are required. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots | Given block size r, a public/private key pair is generated as follows: Choose large primes p and q such that r | ( p − 1 ) , gcd ( r , ( p − 1 ) / r ) = 1 , {\displaystyle r\vert (p-1),\operatorname {gcd} (r,(p-1)/r)=1,} and gcd ( r , ( q − 1 ) ) = 1 {\displaystyle \operatorname {gcd} (r,(q-1))=1} Set n = p q , ϕ = ( p − 1 ) ( q − 1 ) {\displaystyle n=pq,\phi =(p-1)(q-1)} Choose y ∈ Z n ∗ {\displaystyle y\in \mathbb {Z} _{n}^{*}} such that y ϕ / r ≢ 1 mod n {\displaystyle y^{\phi /r}\not \equiv 1\mod n} .Note: If r is composite, it was pointed out by Fousse et al. in 2011 that the above conditions (i.e., those stated in the original paper) are insufficient to guarantee correct decryption, i.e., to guarantee that D ( E ( m ) ) = m {\displaystyle D(E(m))=m} in all cases (as should be the case). To address this, the authors propose the following check: let r = p 1 p 2 … p k {\displaystyle r=p_{1}p_{2}\dots p_{k}} be the prime factorization of r. Choose y ∈ Z n ∗ {\displaystyle y\in \mathbb {Z} _{n}^{*}} such that for each factor p i {\displaystyle p_{i}} , it is the case that y ϕ / p i ≠ 1 mod n {\displaystyle y^{\phi /p_{i}}\neq 1\mod n} .Set x = y ϕ / r mod n {\displaystyle x=y^{\phi /r}\mod n} The public key is then y , n {\displaystyle y,n} , and the private key is ϕ , x {\displaystyle \phi ,x} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion. In RSA with public key $(e,N)$ and private key $(d,N)$ \ldots | Bob does not want Alice to know which one he receives. Alice generates an RSA key pair, comprising the modulus N {\displaystyle N} , the public exponent e {\displaystyle e} and the private exponent d {\displaystyle d} . She also generates two random values, x 0 , x 1 {\displaystyle x_{0},x_{1}} and sends them to Bob along with her public modulus and exponent. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion concerning WPA2 | A tool to automatize WEP cracking and logging of WPA handshakes. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion concerning WPA2 | WS-ReliableMessaging WS-Reliability WS-RM Policy Assertion | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement regarding Kerckhoffs' principle. | 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} statement regarding Kerckhoffs' principle. | Suppose we are given a Boolean expressions: B 1 = ( v 3 ∨ ¬ v 2 ) ∧ ( ¬ v 1 ∨ ¬ v 3 ) {\displaystyle B_{1}=(v_{3}\lor \neg v_{2})\wedge (\neg v_{1}\lor \neg v_{3})} B 2 = ( v 3 ∨ ¬ v 2 ) ∧ ( ¬ v 1 ∨ ¬ v 3 ) ∧ ( ¬ v 1 ∨ v 2 ) . {\displaystyle B_{2}=(v_{3}\lor \neg v_{2})\wedge (\neg v_{1}\lor \neg v_{3})\wedge (\neg v_{1}\lor v_{2}).} With B 1 {\displaystyle B_{1}} , the algorithm can select v 1 = true {\displaystyle v_{1}={\text{true}}} , so to satisfy the second clause, the algorithm will need to set v 3 = false {\displaystyle v_{3}={\text{false}}} , and resultantly to satisfy the first clause, the algorithm will set v 2 = false {\displaystyle v_{2}={\text{false}}} . If the algorithm tries to satisfy B 2 {\displaystyle B_{2}} in the same way it tried to solve B 1 {\displaystyle B_{1}} , then the third clause will remain unsatisfied. This will cause the algorithm to backtrack and set v 1 = false {\displaystyle v_{1}={\text{false}}} and continue assigning variables further. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $H$ be a hash function based on the Merkle-Damg{\aa}rd construction. The Merkle-Damg{\aa}rd theorem says that \dots | Benaloh and de Mare define a one-way hash function as a family of functions h ℓ: X ℓ × Y ℓ → Z ℓ {\displaystyle h_{\ell }:X_{\ell }\times Y_{\ell }\to Z_{\ell }} which satisfy the following three properties: For all ℓ ∈ Z , x ∈ X ℓ , y ∈ Y ℓ {\displaystyle \ell \in \mathbb {Z} ,x\in X_{\ell },y\in Y_{\ell }} , one can compute h ℓ ( x , y ) {\displaystyle h_{\ell }(x,y)} in time poly ( ℓ , | x | , | y | ) {\displaystyle {\text{poly}}(\ell ,|x|,|y|)} . (Here the "poly" symbol refers to an unspecified, but fixed, polynomial.) No probabilistic polynomial-time algorithm will, for sufficiently large ℓ {\displaystyle \ell } , map the inputs ℓ ∈ Z , ( x , y ) ∈ X ℓ × Y ℓ , y ′ ∈ Y ℓ {\displaystyle \ell \in \mathbb {Z} ,(x,y)\in X_{\ell }\times Y_{\ell },y'\in Y_{\ell }} , find a value x ′ ∈ X ℓ {\displaystyle x'\in X_{\ell }} such that h ℓ ( x , y ) = h ℓ ( x ′ , y ′ ) {\displaystyle h_{\ell }(x,y)=h_{\ell }(x',y')} with more than negligible probability. For all ℓ ∈ Z , x ∈ X ℓ , y 1 , y 2 ∈ Y ℓ {\displaystyle \ell \in \mathbb {Z} ,x\in X_{\ell },y_{1},y_{2}\in Y_{\ell }} , one has h ( h ( x , y 1 ) , y 2 ) = h ( h ( x , y 2 ) , y 1 ) {\displaystyle h(h(x,y_{1}),y_{2})=h(h(x,y_{2}),y_{1})} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $H$ be a hash function based on the Merkle-Damg{\aa}rd construction. The Merkle-Damg{\aa}rd theorem says that \dots | Let h: S × X → { 0 , 1 } m {\textstyle h\colon {\mathcal {S}}\times {\mathcal {X}}\rightarrow \{0,\,1\}^{m}} be a 2-universal hash function. If m ≤ H ∞ ( X ) − 2 log ( 1 ε ) {\textstyle m\leq H_{\infty }(X)-2\log \left({\frac {1}{\varepsilon }}\right)} then for S uniform over S {\displaystyle {\mathcal {S}}} and independent of X, we have: δ ≤ ε . {\textstyle \delta \left\leq \varepsilon .} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a blockcipher $\mathsf{B}:\mathcal{K}\times\{0,1\}^n \rightarrow \{0,1\}^n$ with $|\mathcal{K}|=N$ for which we choose a secret key $K\in\mathcal{K}$ uniformly at random. Tick the \textit{correct} assertion. | Let k {\displaystyle k} be a positive integer and let P = { 0 , 1 } k {\displaystyle P=\{0,1\}^{k}} be the set of messages. Let f: Y → Z {\displaystyle f:\,Y\rightarrow Z} be a one-way function. For 1 ≤ i ≤ k {\displaystyle 1\leq i\leq k} and j ∈ { 0 , 1 } {\displaystyle j\in \{0,1\}} the signer chooses y i , j ∈ Y {\displaystyle y_{i,j}\in Y} randomly and computes z i , j = f ( y i , j ) {\displaystyle z_{i,j}=f(y_{i,j})} . The private key, K {\displaystyle K} , consists of 2 k {\displaystyle 2k} values y i , j {\displaystyle y_{i,j}} . The public key consists of the 2 k {\displaystyle 2k} values z i , j {\displaystyle z_{i,j}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a blockcipher $\mathsf{B}:\mathcal{K}\times\{0,1\}^n \rightarrow \{0,1\}^n$ with $|\mathcal{K}|=N$ for which we choose a secret key $K\in\mathcal{K}$ uniformly at random. Tick the \textit{correct} assertion. | Even if k-1 players collude to generate the number r*, as long as the kth player truthfully generates a random r ′ {\displaystyle r'} , the sum r = r ∗ + r ′ {\displaystyle r=r*+r'} is still uniformly random in {0, 51}. Measured in terms of the number of single-agent encryptions, the algorithm in is optimal when no collisions occur, in the sense that any protocol that is fair to every player must perform at least as many encryption operations. At minimum, every agent must encrypt every card that is actually used. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The $n^2$ problem ... | For n ∈ { 1 , 2 , . . . } | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The $n^2$ problem ... | . . , n M ) x 2 ∗ ( n 1 , . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A passive adversary\dots | (Each of these q interactions is a query.) The attacker guesses how the coin landed. He wins if his guess is correct.The attacker, which we can model as an algorithm, is called an adversary. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A passive adversary\dots | Unlike traditional cryptographic applications, such as encryption or signature, one must assume that the adversary in an MPC protocol is one of the players engaged in the system (or controlling internal parties). That corrupted party or parties may collude in order to breach the security of the protocol. Let n {\displaystyle n} be the number of parties in the protocol and t {\displaystyle t} the number of parties who can be adversarial. The protocols and solutions for the case of t < n / 2 {\displaystyle t | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{true} statement regarding RSA Cryptosystem. | 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} statement regarding RSA Cryptosystem. | In the random oracle model, if RSA is ( t ′ , ϵ ′ ) {\displaystyle (t',\epsilon ')} -secure, then the full domain hash RSA signature scheme is ( t , ϵ ) {\displaystyle (t,\epsilon )} -secure where, t = t ′ − ( q hash + q sig + 1 ) ⋅ O ( k 3 ) ϵ = ( 1 + 1 q sig ) q sig + 1 ⋅ q sig ⋅ ϵ ′ {\displaystyle {\begin{aligned}t&=t'-(q_{\text{hash}}+q_{\text{sig}}+1)\cdot {\mathcal {O}}\left(k^{3}\right)\\\epsilon &=\left(1+{\frac {1}{q_{\text{sig}}}}\right)^{q_{\text{sig}}+1}\cdot q_{\text{sig}}\cdot \epsilon '\end{aligned}}} .For large q sig {\displaystyle q_{\text{sig}}} this reduces to ϵ ∼ exp ( 1 ) ⋅ q sig ⋅ ϵ ′ {\displaystyle \epsilon \sim \exp(1)\cdot q_{\text{sig}}\cdot \epsilon '} . This means that if there exists an algorithm that can forge a new FDH signature that runs in time t, computes at most q hash {\displaystyle q_{\text{hash}}} hashes, asks for at most q sig {\displaystyle q_{\text{sig}}} signatures and succeeds with probability ϵ {\displaystyle \epsilon } , then there must also exist an algorithm that breaks RSA with probability ϵ ′ {\displaystyle \epsilon '} in time t ′ {\displaystyle t'} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Vernam cipher\dots | (2014). An Introduction to Mathematical Cryptography (2nd ed.). doi:10.1007/978-1-4939-1711-2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Vernam cipher\dots | Goldreich, Oded (2001 and 2004). Foundations of Cryptography. Cambridge University Press. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which protocol does \emph{not} use RC4. | RFC 7366: "Encrypt-then-MAC for Transport Layer Security (TLS) and Datagram Transport Layer Security (DTLS)". RFC 7465: "Prohibiting RC4 Cipher Suites". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which protocol does \emph{not} use RC4. | WEP TKIP (default algorithm for WPA, but can be configured to use AES-CCMP instead of RC4) BitTorrent protocol encryption Microsoft Office XP (insecure implementation since nonce remains unchanged when documents get modified) Microsoft Point-to-Point Encryption Transport Layer Security / Secure Sockets Layer (was optional and then the use of RC4 was prohibited in RFC 7465) Secure Shell (optionally) Remote Desktop Protocol (optionally) Kerberos (optionally) SASL Mechanism Digest-MD5 (optionally, historic, obsoleted in RFC 6331) Gpcode.AK, an early June 2008 computer virus for Microsoft Windows, which takes documents hostage for ransom by obscuring them with RC4 and RSA-1024 encryption PDF Skype (in modified form)Where a protocol is marked with "(optionally)", RC4 is one of multiple ciphers the system can be configured to use. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is a mode of operation for blockciphers that requires a non-repeating IV (a nonce) to be secure? | Some cryptographic primitives require the IV only to be non-repeating, and the required randomness is derived internally. In this case, the IV is commonly called a nonce (a number used only once), and the primitives (e.g. CBC) are considered stateful rather than randomized. This is because an IV need not be explicitly forwarded to a recipient but may be derived from a common state updated at both sender and receiver side. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is a mode of operation for blockciphers that requires a non-repeating IV (a nonce) to be secure? | Furthermore, it does not suffer from the short-cycle problem that can affect OFB.If the IV/nonce is random, then they can be combined with the counter using any invertible operation (concatenation, addition, or XOR) to produce the actual unique counter block for encryption. In case of a non-random nonce (such as a packet counter), the nonce and counter should be concatenated (e.g., storing the nonce in the upper 64 bits and the counter in the lower 64 bits of a 128-bit counter block). Simply adding or XORing the nonce and counter into a single value would break the security under a chosen-plaintext attack in many cases, since the attacker may be able to manipulate the entire IV–counter pair to cause a collision. Once an attacker controls the IV–counter pair and plaintext, XOR of the ciphertext with the known plaintext would yield a value that, when XORed with the ciphertext of the other block sharing the same IV–counter pair, would decrypt that block.Note that the nonce in this diagram is equivalent to the initialization vector (IV) in the other diagrams. However, if the offset/location information is corrupt, it will be impossible to partially recover such data due to the dependence on byte offset. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
How many generators are there in $\mathbb{Z}_n$? | (Z/pZ)n is generated by n elements, and n is the least possible number of generators. In particular, the set {e1, ..., en} , where ei has a 1 in the ith component and 0 elsewhere, is a minimal generating set. Every elementary abelian group has a fairly simple finite presentation. ( Z / p Z ) n ≅ ⟨ e 1 , … , e n ∣ e i p = 1 , e i e j = e j e i ⟩ {\displaystyle (\mathbb {Z} /p\mathbb {Z} )^{n}\cong \langle e_{1},\ldots ,e_{n}\mid e_{i}^{p}=1,\ e_{i}e_{j}=e_{j}e_{i}\rangle } | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
How many generators are there in $\mathbb{Z}_n$? | The basic generator set G 0 {\displaystyle {\mathcal {G}}_{0}} has n − k {\displaystyle n-k} Pauli sequences of finite support: G 0 = { G i ∈ F ( Π Z + ): 1 ≤ i ≤ n − k } . {\displaystyle {\mathcal {G}}_{0}=\left\{\mathbf {G} _{i}\in F(\Pi ^{\mathbb {Z} ^{+}}):1\leq i\leq n-k\right\}.} The constraint length ν {\displaystyle \nu } of the code is the maximum degree of the generators in G 0 {\displaystyle {\mathcal {G}}_{0}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Using salt for UNIX passwords \ldots | Earlier versions of Unix used a password file /etc/passwd to store the hashes of salted passwords (passwords prefixed with two-character random salts). In these older versions of Unix, the salt was also stored in the passwd file (as cleartext) together with the hash of the salted password. The password file was publicly readable for all users of the system. This was necessary so that user-privileged software tools could find user names and other information. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Using salt for UNIX passwords \ldots | The security of passwords is therefore protected only by the one-way functions (enciphering or hashing) used for the purpose. Early Unix implementations limited passwords to eight characters and used a 12-bit salt, which allowed for 4,096 possible salt values. This was an appropriate balance for 1970s computational and storage costs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
One can find a collision in a hash function $h\colon \{0,1\}^* \rightarrow \{0,1\}^n$ with expected time complexity\dots | In this approach, we base the security of hash function on some hard mathematical problem and we prove that finding collisions of the hash functions is as hard as breaking the underlying problem. This gives a somewhat stronger notion of security than just relying on complex mixing of bits as in the classical approach. A cryptographic hash function has provable security against collision attacks if finding collisions is provably polynomial-time reducible from problem P which is supposed to be unsolvable in polynomial time. The function is then called provably secure, or just provable. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
One can find a collision in a hash function $h\colon \{0,1\}^* \rightarrow \{0,1\}^n$ with expected time complexity\dots | Since m {\displaystyle m} is a constant, hashing requires time O ( n log n log log n ) {\displaystyle O(n\log n\log \log n)} . They proved that the hash function family is collision resistant by showing that if there is a polynomial-time algorithm that succeeds with non-negligible probability in finding b ≠ b ′ ∈ D m {\displaystyle b\neq b'\in D^{m}} such that h ( b ) = h ( b ′ ) {\displaystyle h(b)=h(b')} , for a randomly chosen hash function h ∈ R m {\displaystyle h\in R^{m}} , then a certain problem called the “shortest vector problem” is solvable in polynomial time for every ideal of the ring Z / ⟨ f ⟩ {\displaystyle \mathbb {Z} /\langle f\rangle } . Based on the work of Lyubashevsky and Micciancio in 2006, Micciancio and Regev defined the following algorithm of hash functions based on ideal lattices: Parameters: Integers q , n , m , d {\displaystyle q,n,m,d} with n ∣ m {\displaystyle n\mid m} , and vector f ∈ Z n {\displaystyle \in \mathbb {Z} ^{n}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement regarding the Enigma machine. | Tentatively making these assumptions, the following partial decrypted message is obtained. heVeTCSWPeYVaWHaVSReQMthaYVaOeaWHRtatePFaMVaWHKVSTYhtZetheKeetPeJVSZaYPaRRGaReM WQhMGhMtQaReWGPSReHMtQaRaKeaTtMJTPRGaVaKaeTRaWHatthattMZeTWAWSQWtSWatTVaPMRtRSJ GSTVReaYVeatCVMUeMWaRGMeWtMJMGCSMWtSJOMeQtheVeQeVetQSVSTWHKPaGARCStRWeaVSWeeBtV eZMtFSJtheKaGAaWHaPSWYSWeWeaVtheStheVtheRGaPeRQeVeeBGeeHMWYPFhaVHaWHYPSRRFQMtha PPtheaCCeaVaWGeSJKTVWMRheHYSPHtheQeMYhtSJtheMWReGtQaROeVFVeZaVAaKPeaWHtaAMWYaPP thMWYRMWtSGSWRMHeVatMSWMGSTPHhaVHPFKPaZeNTCMteVJSVhMRSCMWMSWVeRCeGtMWYMt Using these initial guesses, Eve can spot patterns that confirm her choices, such as "that". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement regarding the Enigma machine. | On the other hand, if A paid, she announces ¬ ( 1 ⊕ 0 ) = 0 {\displaystyle \lnot (1\oplus 0)=0} . The three public announcements combined reveal the answer to their question. One simply computes the XOR of the three bits announced. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
One of the following ciphers is a \emph{block} cipher. Which one? | In cryptography, a block cipher is a deterministic algorithm that operates on fixed-length groups of bits, called blocks. Block ciphers are the elementary building blocks of many cryptographic protocols. They are ubiquitous in the storage and exchange of data, where such data is secured and authenticated via encryption. A block cipher uses blocks as an unvarying transformation. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
One of the following ciphers is a \emph{block} cipher. Which one? | A block cipher is one of the most basic primitives in cryptography, and frequently used for data encryption. However, by itself, it can only be used to encode a data block of a predefined size, called the block size. For example, a single invocation of the AES algorithm transforms a 128-bit plaintext block into a ciphertext block of 128 bits in size. The key, which is given as one input to the cipher, defines the mapping between plaintext and ciphertext. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Compute $\phi(90)$. | There are several formulae for computing φ(n). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Compute $\phi(90)$. | })=\phi (m)^{i!}} . Since N is finite, for i great enough, ϕ ( m ) i ! | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $p$ and $q$ be two prime numbers and $n=pq$. Let $K_p=(e,n)$ and $K_s=(d,n)$ be the RSA public and private keys respectively. Recall that the encryption of a message $m$ is $c=m^e \bmod{n}$ and the decryption is $m=c^d \bmod{n}$. Which assertion is \emph{always true}? | On the other hand, if n = pq is the product of two distinct prime numbers, then φ(n) = (p − 1)(q − 1). In this case, finding f from n and e is as difficult as computing φ(n) (this has not been proven, but no algorithm is known for computing f without knowing φ(n)). Knowing only n, the computation of φ(n) has essentially the same difficulty as the factorization of n, since φ(n) = (p − 1)(q − 1), and conversely, the factors p and q are the (integer) solutions of the equation x2 – (n − φ(n) + 1) x + n = 0. The basic idea of RSA cryptosystem is thus: if a message x is encrypted as y = xe (mod n), using public values of n and e, then, with the current knowledge, it cannot be decrypted without finding the (secret) factors p and q of n. Fermat's little theorem is also related to the Carmichael function and Carmichael's theorem, as well as to Lagrange's theorem in group theory. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $p$ and $q$ be two prime numbers and $n=pq$. Let $K_p=(e,n)$ and $K_s=(d,n)$ be the RSA public and private keys respectively. Recall that the encryption of a message $m$ is $c=m^e \bmod{n}$ and the decryption is $m=c^d \bmod{n}$. Which assertion is \emph{always true}? | 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 |
Select the non-associative operation. | Note that this operation is in general not associative. The main difficulty is how to choose the map r e d {\displaystyle red} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the non-associative operation. | A partial operation is associative if x ∗ ( y ∗ z ) = ( x ∗ y ) ∗ z {\displaystyle x*(y*z)=(x*y)*z} for every x, y, z in X for which one of the members of the equality is defined; the equality means that the other member of the equality must also be defined. Examples of non-total associative operations are multiplication of matrices of arbitrary size, and function composition. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
MD5 is | MD5 is a hash function producing a 128-bit hash value. Apache Avro uses a 128-bit random number as synchronization marker for efficient splitting of data files. == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
MD5 is | The MD5 message-digest algorithm is a widely used hash function producing a 128-bit hash value. MD5 was designed by Ronald Rivest in 1991 to replace an earlier hash function MD4, and was specified in 1992 as RFC 1321. MD5 can be used as a checksum to verify data integrity against unintentional corruption. Historically it was widely used as a cryptographic hash function; however it has been found to suffer from extensive vulnerabilities. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database, and may be preferred due to lower computational requirements than more recent Secure Hash Algorithms. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $N=3\cdot5\cdot13$. Then a quadratic residue in $\mathbb{Z}_N^*$ has | Take p = 97 {\displaystyle p=97} . A possible quadratic non-residue for 97 is 13, since 13 97 − 1 2 ≡ − 1 ( mod 97 ) {\displaystyle 13^{\frac {97-1}{2}}\equiv -1{\pmod {97}}} . so we let x = 13 97 − 1 4 = 22 ( mod 97 ) {\displaystyle x=13^{\frac {97-1}{4}}=22{\pmod {97}}} . The Euclidean algorithm applied to 97 and 22 yields: The first two remainders smaller than the square root of 97 are 9 and 4; and indeed we have 97 = 9 2 + 4 2 {\displaystyle 97=9^{2}+4^{2}} , as expected. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $N=3\cdot5\cdot13$. Then a quadratic residue in $\mathbb{Z}_N^*$ has | The first thing to notice when working within the ring Z of integers is that if the prime number q is ≡ 3 (mod 4) then a residue r is a quadratic residue (mod q) if and only if it is a biquadratic residue (mod q). Indeed, the first supplement of quadratic reciprocity states that −1 is a quadratic nonresidue (mod q), so that for any integer x, one of x and −x is a quadratic residue and the other one is a nonresidue. Thus, if r ≡ a2 (mod q) is a quadratic residue, then if a ≡ b2 is a residue, r ≡ a2 ≡ b4 (mod q) is a biquadratic residue, and if a is a nonresidue, −a is a residue, −a ≡ b2, and again, r ≡ (−a)2 ≡ b4 (mod q) is a biquadratic residue.Therefore, the only interesting case is when the modulus p ≡ 1 (mod 4). Gauss proved that if p ≡ 1 (mod 4) then the nonzero residue classes (mod p) can be divided into four sets, each containing (p−1)/4 numbers. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion regarding WEP and WPA2. | A tool to automatize WEP cracking and logging of WPA handshakes. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion regarding WEP and WPA2. | WEP WPA (TKIP in hardware) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Plain RSA (with an $\ell$-bit modulus) \dots | The proof of the correctness of RSA is based on Fermat's little theorem, stating that ap − 1 ≡ 1 (mod p) for any integer a and prime p, not dividing a.We want to show that for every integer m when p and q are distinct prime numbers and e and d are positive integers satisfying ed ≡ 1 (mod λ(pq)). Since λ(pq) = lcm(p − 1, q − 1) is, by construction, divisible by both p − 1 and q − 1, we can write for some nonnegative integers h and k.To check whether two numbers, such as med and m, are congruent mod pq, it suffices (and in fact is equivalent) to check that they are congruent mod p and mod q separately.To show med ≡ m (mod p), we consider two cases: If m ≡ 0 (mod p), m is a multiple of p. Thus med is a multiple of p. So med ≡ 0 ≡ m (mod p). If m ≢ {\displaystyle \not \equiv } 0 (mod p), m e d = m e d − 1 m = m h ( p − 1 ) m = ( m p − 1 ) h m ≡ 1 h m ≡ m ( mod p ) , {\displaystyle m^{ed}=m^{ed-1}m=m^{h(p-1)}m=(m^{p-1})^{h}m\equiv 1^{h}m\equiv m{\pmod {p}},} where we used Fermat's little theorem to replace mp−1 mod p with 1.The verification that med ≡ m (mod q) proceeds in a completely analogous way: If m ≡ 0 (mod q), med is a multiple of q. So med ≡ 0 ≡ m (mod q). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Plain RSA (with an $\ell$-bit modulus) \dots | 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 |
Pick the \emph{correct} statement. | Make a decisive statement about each. If the subject agrees – says, 'That's right', or 'That describes me all right', or similar – leave it immediately. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Pick the \emph{correct} statement. | . . VERIFY-SELECTION . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following elements belong to $\mathbf{Z}_{35}^*$? | ( z − α 4 ) ! ( β 1 − z ) ! ( β 2 − z ) ! | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following elements belong to $\mathbf{Z}_{35}^*$? | The elements of Z {\displaystyle \mathbb {Z} } ×15 are the congruence classes {1, 2, 4, 7, 8, 11, 13, 14}; there are φ(15) = 8 of them. x x, x2, x3, ... (mod 15) 1: 1 2: 2, 4, 8, 1 4: 4, 1 7: 7, 4, 13, 1 8: 8, 4, 2, 1 11: 11, 1 13: 13, 4, 7, 1 14: 14, 1 Since there is no number whose order is 8, there are no primitive roots modulo 15. Indeed, λ(15) = 4, where λ is the Carmichael function. (sequence A002322 in the OEIS) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
When using the plain ElGamal cryptosystem over the group $\mathbb{Z}_p^*$ with a fixed $\ell$-bit prime $p$ and a fixed generator $g\in\mathbb{Z}_p^*$, the \textit{tightest} complexity of generating a new key-pair is\ldots | Regev proposed a public-key cryptosystem based on the hardness of the LWE problem. The cryptosystem as well as the proof of security and correctness are completely classical. The system is characterized by m , q {\displaystyle m,q} and a probability distribution χ {\displaystyle \chi } on T {\displaystyle \mathbb {T} } . The setting of the parameters used in proofs of correctness and security is q ≥ 2 {\displaystyle q\geq 2} , usually a prime number between n 2 {\displaystyle n^{2}} and 2 n 2 {\displaystyle 2n^{2}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
When using the plain ElGamal cryptosystem over the group $\mathbb{Z}_p^*$ with a fixed $\ell$-bit prime $p$ and a fixed generator $g\in\mathbb{Z}_p^*$, the \textit{tightest} complexity of generating a new key-pair is\ldots | Shahram Khazaei, Simon Fischer, and Willi Meier give a cryptanalysis of the ASG allowing various tradeoffs between time complexity and the amount of output needed to mount the attack, e.g. with asymptotic complexity O ( L 2 .2 2 L / 3 ) {\displaystyle O(L^{2}.2^{2L/3})} and O ( 2 2 L / 3 ) {\displaystyle O(2^{2L/3})} bits, where L {\displaystyle L} is the size of the shortest of the three LFSRs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
We represent $GF(2^8)$ as $\mathbb{Z}_2[X]/P(X)$ where $P(X) = X^8 + X^4+X^3+X+1$. Then, $(X^7+X^6)\times (X + 1)=$\dots | Let X n {\displaystyle X^{n}} be an n-dimensional vector space over the finite field G F ( q N ) {\displaystyle GF\left({q^{N}}\right)} , where q {\displaystyle q} is a power of a prime and N {\displaystyle N} is a positive integer. Let ( u 1 , u 2 , … , u N ) {\displaystyle \left(u_{1},u_{2},\dots ,u_{N}\right)} , with u i ∈ G F ( q N ) {\displaystyle u_{i}\in GF(q^{N})} , be a base of G F ( q N ) {\displaystyle GF\left({q^{N}}\right)} as a vector space over the field G F ( q ) {\displaystyle GF\left({q}\right)} . Every element x i ∈ G F ( q N ) {\displaystyle x_{i}\in GF\left({q^{N}}\right)} can be represented as x i = a 1 i u 1 + a 2 i u 2 + ⋯ + a N i u N {\displaystyle x_{i}=a_{1i}u_{1}+a_{2i}u_{2}+\dots +a_{Ni}u_{N}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
We represent $GF(2^8)$ as $\mathbb{Z}_2[X]/P(X)$ where $P(X) = X^8 + X^4+X^3+X+1$. Then, $(X^7+X^6)\times (X + 1)=$\dots | Such a g − a {\textstyle g-a} has a nontrivial factor in common with f ( x ) {\textstyle f(x)} , which can be computed via the gcd. As p {\textstyle p} is small, we can cycle through all possible a {\textstyle a} . For the case of large primes, which are necessarily odd, one can exploit the fact that a random nonzero element of F p ∗ {\textstyle \mathbb {F} _{p}^{*}} is a square with probability 1 / 2 {\textstyle 1/2} , and that the map x → x p − 1 2 {\textstyle x\to x^{\frac {p-1}{2}}} maps the set of non-zero squares to 1 {\textstyle 1} , and the set of non-squares to − 1 {\textstyle -1} . Thus, if we take a random element g ∈ Fix p ( F q / f ( x ) ) {\textstyle g\in {\text{Fix}}_{p}(\mathbb {F} _{q}/f(x))} , then with good probability g p − 1 2 − 1 {\textstyle g^{\frac {p-1}{2}}-1} will have a non-trivial factor in common with f ( x ) {\textstyle f(x)} .For further details one can consult. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textit{wrong} assertion. | "I made an error in this transmission. Transmission will continue with the last word correctly sent." | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textit{wrong} assertion. | (/ indicates line break; some word breaks are uncertain) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement: hash functions can be used to construct | Select y ~ ∈ { 1 , . . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Select the \emph{incorrect} statement: hash functions can be used to construct | Selected schemes for the purpose of hashing: SWIFFT. Lattice Based Hash Function (LASH). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{false} assertion about Diffie and Hellman. | Diffie–Hellman (RFC 3526) ECDH (RFC 4753) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{false} assertion about Diffie and Hellman. | "Timing Attacks on Implementations of Diffie-Hellman, RSA, DSS, and Other Systems". Advances in Cryptology – CRYPTO '96. pp. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A hash function $h$ is collision-resistant if\dots | h {\displaystyle h}: a collision resistant hash function with |q|-bit digests. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A hash function $h$ is collision-resistant if\dots | Collision resistance prevents an attacker from creating two distinct documents with the same hash. A function meeting these criteria may still have undesirable properties. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. Bluetooth 2.0 Pairing is secure when ... | Current mobile software generally must allow a connection using a temporary state initiated by the user in order to be 'paired' with another device to copy content. There seem to have been, in the past, available reports of phones being Bluesnarfed without pairing being explicitly allowed. After the disclosure of this vulnerability, vendors of mobile phone patched their Bluetooth implementations and, at the time of writing, no current phone models are known to be vulnerable to this attack. Any device with its Bluetooth connection turned on and set to "discoverable" (able to be found by other Bluetooth devices in range) may be susceptible to Bluejacking and possibly to Bluesnarfing if there is a vulnerability in the vendor's software. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. Bluetooth 2.0 Pairing is secure when ... | Pairing mechanisms changed significantly with the introduction of Secure Simple Pairing in Bluetooth v2.1. The following summarizes the pairing mechanisms: Legacy pairing: This is the only method available in Bluetooth v2.0 and before. Each device must enter a PIN code; pairing is only successful if both devices enter the same PIN code. Any 16-byte UTF-8 string may be used as a PIN code; however, not all devices may be capable of entering all possible PIN codes. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Bluetooth pairing v2.0 is based on\dots | updated Bluetooth stack that opens up even more tethered connectivity options | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Bluetooth pairing v2.0 is based on\dots | Pairing mechanisms changed significantly with the introduction of Secure Simple Pairing in Bluetooth v2.1. The following summarizes the pairing mechanisms: Legacy pairing: This is the only method available in Bluetooth v2.0 and before. Each device must enter a PIN code; pairing is only successful if both devices enter the same PIN code. Any 16-byte UTF-8 string may be used as a PIN code; however, not all devices may be capable of entering all possible PIN codes. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Kerckhoffs principle says that | Kerckhoffs's principle (also called Kerckhoffs's desideratum, assumption, axiom, doctrine or law) of cryptography was stated by Dutch-born cryptographer Auguste Kerckhoffs in the 19th century. The principle holds that a cryptosystem should be secure, even if everything about the system, except the key, is public knowledge. This concept is widely embraced by cryptographers, in contrast to security through obscurity, which is not. Kerckhoffs's principle was phrased by American mathematician Claude Shannon as "the enemy knows the system", i.e., "one ought to design systems under the assumption that the enemy will immediately gain full familiarity with them". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Kerckhoffs principle says that | A generalization some make from Kerckhoffs's principle is: "The fewer and simpler the secrets that one must keep to ensure system security, the easier it is to maintain system security." Bruce Schneier ties it in with a belief that all security systems must be designed to fail as gracefully as possible: principle applies beyond codes and ciphers to security systems in general: every secret creates a potential failure point. Secrecy, in other words, is a prime cause of brittleness—and therefore something likely to make a system prone to catastrophic collapse. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A MAC forgery is\dots | The problem is that the MAC is only authenticating a string of bytes, while Alice and Bob need to authenticate the way the message was constructed as well. If not, then it may be possible for an attacker to substitute a message with a valid MAC but a different meaning. Systems can manage this problem by adding metadata such as a protocol number or by formatting messages with an explicit structure, such as XML. == References == | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A MAC forgery is\dots | If the check fails, a short poem is displayed, reading "Your karma check for today: There once was a user that whined, his existing OS was so blind, he'd do better to pirate an OS that ran great, but found his hardware declined. Please don't steal Mac OS! Really, that's way uncool. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion. | "Fl." for flashing, "F." for fixed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{incorrect} assertion. | If so, attempt to position cursor at that line. If it exists, begin interpretation there; if not, report an error. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The encryption in GSM is done by ... | In 2008 it was reported that a GSM phone's encryption key can be obtained using $1,000 worth of computer hardware and 30 minutes of cryptanalysis performed on signals encrypted using A5/1. However, GSM also supports an export weakened variant of A5/1 called A5/2. This weaker encryption cypher can be cracked in real-time. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The encryption in GSM is done by ... | There are flaws in the implementation of the GSM encryption algorithm that allow passive interception. The equipment needed is available to government agencies or can be built from freely available parts.In December 2011, German researcher Karsten Nohl revealed that it was possible to hack into mobile phone voice and text messages on many networks with free decryption software available on the Internet. He blamed the mobile phone companies for relying on outdated encryption techniques in the 2G system, and said that the problem could be fixed very easily. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. GSM anonymity is broken by ... | The GSM industry has identified a number of potential fraud attacks on mobile operators that can be delivered via abuse of SMS messaging services. The most serious threat is SMS Spoofing, which occurs when a fraudster manipulates address information in order to impersonate a user that has roamed onto a foreign network and is submitting messages to the home network. Frequently, these messages are addressed to destinations outside the home network—with the home SMSC essentially being "hijacked" to send messages into other networks.The only sure way of detecting and blocking spoofed messages is to screen incoming mobile-originated messages to verify that the sender is a valid subscriber and that the message is coming from a valid and correct location. This can be implemented by adding an intelligent routing function to the network that can query originating subscriber details from the home location register (HLR) before the message is submitted for delivery. This kind of intelligent routing function is beyond the capabilities of legacy messaging infrastructure. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \textbf{false} statement. GSM anonymity is broken by ... | See id. at 1257. Claim 20 is directed to a problem unique to text-message telecommunication between a mobile device and a computer. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{true} assertion related to the ElGamal signature scheme. | 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 \emph{true} assertion related to the ElGamal signature scheme. | One can verify that a signature ( r , s ) {\displaystyle (r,s)} is a valid signature for a message m {\displaystyle m} as follows: Verify that 0 < r < p {\displaystyle 0 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Kerckhoff's principle is not followed when security requires that \dots | First, a 19th century rule known as Kerckhoffs's principle, later formulated as Shannon's maxim, teaches that "the enemy knows the system" and the secrecy of a cryptosystem algorithm does not provide any advantage. Second, secret methods are not open to public peer review and cryptanalysis, so potential mistakes and insecurities can go unnoticed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The Kerckhoff's principle is not followed when security requires that \dots | A generalization some make from Kerckhoffs's principle is: "The fewer and simpler the secrets that one must keep to ensure system security, the easier it is to maintain system security." Bruce Schneier ties it in with a belief that all security systems must be designed to fail as gracefully as possible: principle applies beyond codes and ciphers to security systems in general: every secret creates a potential failure point. Secrecy, in other words, is a prime cause of brittleness—and therefore something likely to make a system prone to catastrophic collapse. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion about $\mathbf{Z}_{pq}^*$, where $p$ and $q$ are distinct prime numbers \ldots | Let π ( x ) {\displaystyle \pi (x)} , the prime-counting function, denote the number of primes less than or equal to x {\displaystyle x} . If q {\displaystyle q} is a positive integer and a {\displaystyle a} is coprime to q {\displaystyle q} , we let π ( x ; q , a ) {\displaystyle \pi (x;q,a)} denote the number of primes less than or equal to x {\displaystyle x} which are equal to a {\displaystyle a} modulo q {\displaystyle q} . Dirichlet's theorem on primes in arithmetic progressions then tells us that π ( x ; q , a ) ≈ π ( x ) φ ( q ) {\displaystyle \pi (x;q,a)\approx {\frac {\pi (x)}{\varphi (q)}}} where φ {\displaystyle \varphi } is Euler's totient function. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Tick the \emph{correct} assertion about $\mathbf{Z}_{pq}^*$, where $p$ and $q$ are distinct prime numbers \ldots | This yields a contradiction, as 1 = | 1 | ∗ ≤ | a | ∗ | p | ∗ k + | b | ∗ | q | ∗ k < | a | ∗ + | b | ∗ 2 ≤ 1. {\displaystyle 1=|1|_{*}\leq |a|_{*}|p|_{*}^{k}+|b|_{*}|q|_{*}^{k}<{\frac {|a|_{*}+|b|_{*}}{2}}\leq 1.} This means that there exists a unique prime p such that | p | ∗ < 1 {\displaystyle |p|_{*}<1} and that for all other prime q, one has | q | ∗ = 1 {\displaystyle |q|_{*}=1} (from the hypothesis of this second case). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the inverse of 28 in $\mathbf{Z}_{77}$? | With increasing dimension, expressions for the inverse of A get complicated. For n = 4, the Cayley–Hamilton method leads to an expression that is still tractable: A − 1 = 1 det ( A ) ( 1 6 I − 1 2 A + A 2 tr A − A 3 ) . {\displaystyle \mathbf {A} ^{-1}={\frac {1}{\det(\mathbf {A} )}}\left({\frac {1}{6}}\left\mathbf {I} -{\frac {1}{2}}\mathbf {A} \left+\mathbf {A} ^{2}\operatorname {tr} \mathbf {A} -\mathbf {A} ^{3}\right).} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the inverse of 28 in $\mathbf{Z}_{77}$? | Then for the inverse g ( z ) {\displaystyle g(z)} (satisfying f ( g ( z ) ) ≡ z {\displaystyle f(g(z))\equiv z} ), we have g ( z ) = ∑ n = 1 ∞ z n n ! = ∑ n = 1 ∞ 1 n z n , {\displaystyle {\begin{aligned}g(z)&=\sum _{n=1}^{\infty }\left{\frac {z^{n}}{n! | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the time complexity to perfom a left-to-right multiplication of two $\ell$-bit integers? | A fundamental question of longstanding theoretical interest is to prove lower bounds on the complexity and exact operation counts of fast Fourier transforms, and many open problems remain. It is not rigorously proved whether DFTs truly require Ω ( N log N ) {\textstyle \Omega (N\log N)} (i.e., order N log N {\displaystyle N\log N} or greater) operations, even for the simple case of power of two sizes, although no algorithms with lower complexity are known. In particular, the count of arithmetic operations is usually the focus of such questions, although actual performance on modern-day computers is determined by many other factors such as cache or CPU pipeline optimization. Following work by Shmuel Winograd (1978), a tight Θ(N) lower bound is known for the number of real multiplications required by an FFT. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the time complexity to perfom a left-to-right multiplication of two $\ell$-bit integers? | There is a trivial lower bound of Ω(n) for multiplying two n-bit numbers on a single processor; no matching algorithm (on conventional machines, that is on Turing equivalent machines) nor any sharper lower bound is known. Multiplication lies outside of AC0 for any prime p, meaning there is no family of constant-depth, polynomial (or even subexponential) size circuits using AND, OR, NOT, and MODp gates that can compute a product. This follows from a constant-depth reduction of MODq to multiplication. Lower bounds for multiplication are also known for some classes of branching programs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $X$ and $K$ be two independent random variables in a group $G$ with $n$ elements and where $K$ is uniformly distributed over $G$. Let $Y = X+K$. Then, for any $y\in G$, $\Pr[Y=y]=$\dots | Consider a group G {\displaystyle G} and subgroups G 1 , G 2 , … , G n {\displaystyle G_{1},G_{2},\dots ,G_{n}} of G {\displaystyle G} . Let G I {\displaystyle G_{I}} denote ⋂ i ∈ I G i {\displaystyle \bigcap _{i\in I}G_{i}} for I ⊆ { 1 , … , n } {\displaystyle I\subseteq \{1,\dots ,n\}} ; this is also a subgroup of G {\displaystyle G} . It is possible to construct a probability distribution for n {\displaystyle n} random variables X 1 , … , X n {\displaystyle X_{1},\dots ,X_{n}} such that H ( X I ) = log | G | | G I | {\displaystyle H(X_{I})=\log {\frac {|G|}{|G_{I}|}}} . (The construction essentially takes an element a {\displaystyle a} of G {\displaystyle G} uniformly at random and lets X i {\displaystyle X_{i}} be the corresponding coset a G i {\displaystyle aG_{i}} ). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $X$ and $K$ be two independent random variables in a group $G$ with $n$ elements and where $K$ is uniformly distributed over $G$. Let $Y = X+K$. Then, for any $y\in G$, $\Pr[Y=y]=$\dots | If the X1, X2, ..., XN are independent, but not identically distributed random variables, where G X i {\displaystyle G_{X_{i}}} denotes the probability generating function of X i {\displaystyle X_{i}} , then G S N ( z ) = ∑ i ≥ 1 f i ∏ k = 1 i G X i ( z ) . {\displaystyle G_{S_{N}}(z)=\sum _{i\geq 1}f_{i}\prod _{k=1}^{i}G_{X_{i}}(z).} For identically distributed Xi this simplifies to the identity stated before. The general case is sometimes useful to obtain a decomposition of SN by means of generating functions. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of these primitives can be surely considered insecure today? | Cryptographic primitives, on their own, are quite limited. They cannot be considered, properly, to be a cryptographic system. For instance, a bare encryption algorithm will provide no authentication mechanism, nor any explicit message integrity checking. Only when combined in security protocols, can more than one security requirement be addressed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of these primitives can be surely considered insecure today? | A computer programmer rarely invents a new programming language while writing a new program; instead, they will use one of the already established programming languages to program in. Cryptographic primitives are one of the building blocks of every crypto system, e.g., TLS, SSL, SSH, etc. Crypto system designers, not being in a position to definitively prove their security, must take the primitives they use as secure. Choosing the best primitive available for use in a protocol usually provides the best available security. However, compositional weaknesses are possible in any crypto system and it is the responsibility of the designer(s) to avoid them. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If an adversary mounts a chosen plaintext attack, it means he can\dots | Attacks that lead to disclosure of the key or plaintext. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If an adversary mounts a chosen plaintext attack, it means he can\dots | In a chosen-plaintext attack the adversary can (possibly adaptively) ask for the ciphertexts of arbitrary plaintext messages. This is formalized by allowing the adversary to interact with an encryption oracle, viewed as a black box. The attacker’s goal is to reveal all or a part of the secret encryption key. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
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