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You are given an i.i.d source with symbols taking value in the alphabet $\mathcal{A}=\{a,b,c,d\}$ and probabilities $\{1/8,1/8,1/4,1/2\}$. Consider making blocks of length $n$ and constructing a Huffman code that assigns a binary codeword to each block of $n$ symbols. Choose the correct statement regarding the average codeword length per source symbol. | In this example, the weighted average codeword length is 2.25 bits per symbol, only slightly larger than the calculated entropy of 2.205 bits per symbol. So not only is this code optimal in the sense that no other feasible code performs better, but it is very close to the theoretical limit established by Shannon. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
You are given an i.i.d source with symbols taking value in the alphabet $\mathcal{A}=\{a,b,c,d\}$ and probabilities $\{1/8,1/8,1/4,1/2\}$. Consider making blocks of length $n$ and constructing a Huffman code that assigns a binary codeword to each block of $n$ symbols. Choose the correct statement regarding the average codeword length per source symbol. | This leaves us with a single node and our algorithm is complete. The code lengths for the different characters this time are 1 bit for A and 3 bits for all other characters. This results in the lengths of 1 bit for A and per 3 bits for B, C, D and E, giving an average length of We see that the Huffman code has outperformed both types of Shannon–Fano code, which had expected lengths of 2.62 and 2.28. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A bag contains the letters of LETSPLAY. Someone picks at random 4 letters from the bag without revealing the outcome to you. Subsequently you pick one letter at random among the remaining 4 letters. What is the entropy (in bits) of the random variable that models your choice? Check the correct answer. | In information theory, the entropy of a random variable is the average level of "information", "surprise", or "uncertainty" inherent to the variable's possible outcomes. Given a discrete random variable X {\displaystyle X} , which takes values in the alphabet X {\displaystyle {\mathcal {X}}} and is distributed according to p: X → {\displaystyle p\colon {\mathcal {X}}\to }: where Σ {\displaystyle \Sigma } denotes the sum over the variable's possible values. The choice of base for log {\displaystyle \log } , the logarithm, varies for different applications. Base 2 gives the unit of bits (or "shannons"), while base e gives "natural units" nat, and base 10 gives units of "dits", "bans", or "hartleys". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A bag contains the letters of LETSPLAY. Someone picks at random 4 letters from the bag without revealing the outcome to you. Subsequently you pick one letter at random among the remaining 4 letters. What is the entropy (in bits) of the random variable that models your choice? Check the correct answer. | Symmetrically, Player B presumes Player A expects the random numbers 2-100 and chooses a1, so B chooses b2. As a result, the players had a final result of (1, a2, b2), with a payoff of 1 for both - the lowest possible payoff (total or individual). Now suppose that it is common knowledge that the random number is 1 - that is, both players are also aware that the other player knows the random number is 1, in addition to knowing this themselves. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following congruence classes has a multiplicative inverse? | In mathematics, particularly in the area of arithmetic, a modular multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m. In the standard notation of modular arithmetic this congruence is written as a x ≡ 1 ( mod m ) , {\displaystyle ax\equiv 1{\pmod {m}},} which is the shorthand way of writing the statement that m divides (evenly) the quantity ax − 1, or, put another way, the remainder after dividing ax by the integer m is 1. If a does have an inverse modulo m, then there are an infinite number of solutions of this congruence, which form a congruence class with respect to this modulus. Furthermore, any integer that is congruent to a (i.e., in a's congruence class) has any element of x's congruence class as a modular multiplicative inverse. Using the notation of w ¯ {\displaystyle {\overline {w}}} to indicate the congruence class containing w, this can be expressed by saying that the modulo multiplicative inverse of the congruence class a ¯ {\displaystyle {\overline {a}}} is the congruence class x ¯ {\displaystyle {\overline {x}}} such that: a ¯ ⋅ m x ¯ = 1 ¯ , {\displaystyle {\overline {a}}\cdot _{m}{\overline {x}}={\overline {1}},} where the symbol ⋅ m {\displaystyle \cdot _{m}} denotes the multiplication of equivalence classes modulo m. Written in this way, the analogy with the usual concept of a multiplicative inverse in the set of rational or real numbers is clearly represented, replacing the numbers by congruence classes and altering the binary operation appropriately. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following congruence classes has a multiplicative inverse? | A reduced residue system modulo 10 could be {1, 3, 7, 9}. The product of any two congruence classes represented by these numbers is again one of these four congruence classes. This implies that these four congruence classes form a group, in this case the cyclic group of order four, having either 3 or 7 as a (multiplicative) generator. The represented congruence classes form the group of units of the ring Z / 10 Z {\displaystyle \mathbb {Z} /10\mathbb {Z} } . These congruence classes are precisely the ones which have modular multiplicative inverses. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Is $(\mathbb{Z} / 8\mathbb{Z}^*, \cdot)$ isomorphic to $(\mathbb{Z} / k\mathbb{Z}, +)$ for some $k$? | Finally, use that O {\displaystyle O} is a Dedekind domain and therefore each ideal can be written as a product of prime ideals. In other words, the map ( ⋅ ) {\displaystyle (\cdot )} is a K × {\displaystyle K^{\times }} -equivariant group homomorphism. As a consequence, the map above induces a surjective homomorphism { ϕ: C K , fin → Cl K α K × ↦ ( α ) K × {\displaystyle {\begin{cases}\phi :C_{K,{\text{fin}}}\to \operatorname {Cl} _{K}\\\alpha K^{\times }\mapsto (\alpha )K^{\times }\end{cases}}} To prove the second isomorphism, it has to be shown that ker ( ϕ ) = O ^ × K × . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Is $(\mathbb{Z} / 8\mathbb{Z}^*, \cdot)$ isomorphic to $(\mathbb{Z} / k\mathbb{Z}, +)$ for some $k$? | There is a natural ring homomorphism K 0 ( X ) → H 2 ∗ ( X , Q ) , {\displaystyle K^{0}(X)\to H^{2*}(X,\mathbb {Q} ),} the Chern character, such that K 0 ( X ) ⊗ Q → H 2 ∗ ( X , Q ) {\displaystyle K^{0}(X)\otimes \mathbb {Q} \to H^{2*}(X,\mathbb {Q} )} is an isomorphism. The equivalent of the Steenrod operations in K-theory are the Adams operations. They can be used to define characteristic classes in topological K-theory. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a $(k+1,k)$ block code that to a binary sequence $x_1,\dots,x_k$ associates the codeword $x_1,\dots,x_k,x_{k+1}$, where $x_{k+1}= x_1+\ldots+x_k$ mod $2$. This code can detect all the errors of odd weight. | We can improve this situation. If we use the generator polynomial g ( x ) = p ( x ) ( 1 + x ) {\displaystyle g(x)=p(x)(1+x)} , where p {\displaystyle p} is a primitive polynomial of degree r − 1 {\displaystyle r-1} , then the maximal total block length is 2 r − 1 − 1 {\displaystyle 2^{r-1}-1} , and the code is able to detect single, double, triple and any odd number of errors. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a $(k+1,k)$ block code that to a binary sequence $x_1,\dots,x_k$ associates the codeword $x_1,\dots,x_k,x_{k+1}$, where $x_{k+1}= x_1+\ldots+x_k$ mod $2$. This code can detect all the errors of odd weight. | An exceptional block design is the Steiner system S(5,8,24) whose automorphism group is the sporadic simple Mathieu group M 24 {\displaystyle M_{24}} . The codewords of the extended binary Golay code have a length of 24 bits and have weights 0, 8, 12, 16, or 24. This code can correct up to three errors. So every 24-bit word with weight 5 can be corrected to a codeword with weight 8. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A generator matrix $G$ of binary $(6,3)$ linear code maps the information vectors $m_1 = (1,0,1)$ and $m_2=(1,1,1)$ into the codewords $c_1 = (1,1,0,0,0,1)$ and $c_2=(1,0,0,0,1,0)$ respectively. Which of the following is true? | If G is a matrix, it generates the codewords of a linear code C by w = s G {\displaystyle w=sG} where w is a codeword of the linear code C, and s is any input vector. Both w and s are assumed to be row vectors. A generator matrix for a linear q {\displaystyle _{q}} -code has format k × n {\displaystyle k\times n} , where n is the length of a codeword, k is the number of information bits (the dimension of C as a vector subspace), d is the minimum distance of the code, and q is size of the finite field, that is, the number of symbols in the alphabet (thus, q = 2 indicates a binary code, etc.). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A generator matrix $G$ of binary $(6,3)$ linear code maps the information vectors $m_1 = (1,0,1)$ and $m_2=(1,1,1)$ into the codewords $c_1 = (1,1,0,0,0,1)$ and $c_2=(1,0,0,0,1,0)$ respectively. Which of the following is true? | ExampleFrom the above matrix we have 2k = 24 = 16 codewords. Let a → {\displaystyle {\vec {a}}} be a row vector of binary data bits, a → = , a i ∈ { 0 , 1 } {\displaystyle {\vec {a}}=,\quad a_{i}\in \{0,1\}} . The codeword x → {\displaystyle {\vec {x}}} for any of the 16 possible data vectors a → {\displaystyle {\vec {a}}} is given by the standard matrix product x → = a → G {\displaystyle {\vec {x}}={\vec {a}}G} where the summing operation is done modulo-2. For example, let a → = {\displaystyle {\vec {a}}=} . Using the generator matrix G {\displaystyle G} from above, we have (after applying modulo 2, to the sum), x → = a → G = ( 1 0 1 1 ) ( 1 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 0 0 1 1 1 1 ) = ( 1 0 1 1 2 3 2 ) = ( 1 0 1 1 0 1 0 ) {\displaystyle {\vec {x}}={\vec {a}}G={\begin{pmatrix}1&0&1&1\end{pmatrix}}{\begin{pmatrix}1&0&0&0&1&1&0\\0&1&0&0&1&0&1\\0&0&1&0&0&1&1\\0&0&0&1&1&1&1\\\end{pmatrix}}={\begin{pmatrix}1&0&1&1&2&3&2\end{pmatrix}}={\begin{pmatrix}1&0&1&1&0&1&0\end{pmatrix}}} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Bob designs a uniquely decodable code $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with following codeword lengths. egin{center}egin{tabular}{ |c|c|c|c|c|c|c| } \hline& $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline$|\Gamma(\cdot)|$ & 1 & 1 & 2 & 2 & 3 & 3 \ \hline\end{tabular}\end{center} Which of the following is true? | Let each source symbol from the alphabet S = { s 1 , s 2 , … , s n } {\displaystyle S=\{\,s_{1},s_{2},\ldots ,s_{n}\,\}} be encoded into a uniquely decodable code over an alphabet of size r {\displaystyle r} with codeword lengths ℓ 1 , ℓ 2 , … , ℓ n . {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}.} Then ∑ i = 1 n r − ℓ i ⩽ 1. {\displaystyle \sum _{i=1}^{n}r^{-\ell _{i}}\leqslant 1.} Conversely, for a given set of natural numbers ℓ 1 , ℓ 2 , … , ℓ n {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r {\displaystyle r} with those codeword lengths. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Bob designs a uniquely decodable code $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with following codeword lengths. egin{center}egin{tabular}{ |c|c|c|c|c|c|c| } \hline& $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline$|\Gamma(\cdot)|$ & 1 & 1 & 2 & 2 & 3 & 3 \ \hline\end{tabular}\end{center} Which of the following is true? | Let C {\displaystyle {\mathcal {C}}} be a ( n , k , d ) q {\displaystyle (n,k,d)_{q}} error-correcting code; in other words, C {\displaystyle {\mathcal {C}}} is a code of length n {\displaystyle n} , dimension k {\displaystyle k} and minimum distance d {\displaystyle d} over an alphabet Σ {\displaystyle \Sigma } of size q {\displaystyle q} . The list-decoding problem can now be formulated as follows: Input: Received word x ∈ Σ n {\displaystyle x\in \Sigma ^{n}} , error bound e {\displaystyle e} Output: A list of all codewords x 1 , x 2 , … , x m ∈ C {\displaystyle x_{1},x_{2},\ldots ,x_{m}\in {\mathcal {C}}} whose hamming distance from x {\displaystyle x} is at most e {\displaystyle e} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $H(S_n) = H(S_{n-1})$. | In general, if P(n) is the probability of throwing at least n sixes with 6n dice, then: P ( n ) = 1 − ∑ x = 0 n − 1 ( 6 n x ) ( 1 6 ) x ( 5 6 ) 6 n − x . {\displaystyle P(n)=1-\sum _{x=0}^{n-1}{\binom {6n}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{6n-x}\,.} As n grows, P(n) decreases monotonically towards an asymptotic limit of 1/2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $H(S_n) = H(S_{n-1})$. | Here legal means that the coefficients are non-negative and sum to six, so that each die has six sides and every face has at least one spot. (That is, the generating function of each die must be a polynomial p(x) with positive coefficients, and with p(0) = 0 and p(1) = 6.) Only one such partition exists: x ( x + 1 ) ( x 2 + x + 1 ) = x + 2 x 2 + 2 x 3 + x 4 {\displaystyle x\;(x+1)\;(x^{2}+x+1)=x+2x^{2}+2x^{3}+x^{4}} and x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x + x 3 + x 4 + x 5 + x 6 + x 8 {\displaystyle x\;(x+1)\;(x^{2}+x+1)\;(x^{2}-x+1)^{2}=x+x^{3}+x^{4}+x^{5}+x^{6}+x^{8}} This gives us the distribution of spots on the faces of a pair of Sicherman dice as being {1,2,2,3,3,4} and {1,3,4,5,6,8}, as above. This technique can be extended for dice with an arbitrary number of sides. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $H(S_1,\ldots,S_n) = \sum_{i=1}^n H(S_i|S_1\ldots S_{i-1})$. | In general, if P(n) is the probability of throwing at least n sixes with 6n dice, then: P ( n ) = 1 − ∑ x = 0 n − 1 ( 6 n x ) ( 1 6 ) x ( 5 6 ) 6 n − x . {\displaystyle P(n)=1-\sum _{x=0}^{n-1}{\binom {6n}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{6n-x}\,.} As n grows, P(n) decreases monotonically towards an asymptotic limit of 1/2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $H(S_1,\ldots,S_n) = \sum_{i=1}^n H(S_i|S_1\ldots S_{i-1})$. | Here legal means that the coefficients are non-negative and sum to six, so that each die has six sides and every face has at least one spot. (That is, the generating function of each die must be a polynomial p(x) with positive coefficients, and with p(0) = 0 and p(1) = 6.) Only one such partition exists: x ( x + 1 ) ( x 2 + x + 1 ) = x + 2 x 2 + 2 x 3 + x 4 {\displaystyle x\;(x+1)\;(x^{2}+x+1)=x+2x^{2}+2x^{3}+x^{4}} and x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x + x 3 + x 4 + x 5 + x 6 + x 8 {\displaystyle x\;(x+1)\;(x^{2}+x+1)\;(x^{2}-x+1)^{2}=x+x^{3}+x^{4}+x^{5}+x^{6}+x^{8}} This gives us the distribution of spots on the faces of a pair of Sicherman dice as being {1,2,2,3,3,4} and {1,3,4,5,6,8}, as above. This technique can be extended for dice with an arbitrary number of sides. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, the source is stationary. | In general, if P(n) is the probability of throwing at least n sixes with 6n dice, then: P ( n ) = 1 − ∑ x = 0 n − 1 ( 6 n x ) ( 1 6 ) x ( 5 6 ) 6 n − x . {\displaystyle P(n)=1-\sum _{x=0}^{n-1}{\binom {6n}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{6n-x}\,.} As n grows, P(n) decreases monotonically towards an asymptotic limit of 1/2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, the source is stationary. | Here legal means that the coefficients are non-negative and sum to six, so that each die has six sides and every face has at least one spot. (That is, the generating function of each die must be a polynomial p(x) with positive coefficients, and with p(0) = 0 and p(1) = 6.) Only one such partition exists: x ( x + 1 ) ( x 2 + x + 1 ) = x + 2 x 2 + 2 x 3 + x 4 {\displaystyle x\;(x+1)\;(x^{2}+x+1)=x+2x^{2}+2x^{3}+x^{4}} and x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x + x 3 + x 4 + x 5 + x 6 + x 8 {\displaystyle x\;(x+1)\;(x^{2}+x+1)\;(x^{2}-x+1)^{2}=x+x^{3}+x^{4}+x^{5}+x^{6}+x^{8}} This gives us the distribution of spots on the faces of a pair of Sicherman dice as being {1,2,2,3,3,4} and {1,3,4,5,6,8}, as above. This technique can be extended for dice with an arbitrary number of sides. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $\lim_{n o\infty}H(S_n) = \log_2(6)$. | In general, if P(n) is the probability of throwing at least n sixes with 6n dice, then: P ( n ) = 1 − ∑ x = 0 n − 1 ( 6 n x ) ( 1 6 ) x ( 5 6 ) 6 n − x . {\displaystyle P(n)=1-\sum _{x=0}^{n-1}{\binom {6n}{x}}\left({\frac {1}{6}}\right)^{x}\left({\frac {5}{6}}\right)^{6n-x}\,.} As n grows, P(n) decreases monotonically towards an asymptotic limit of 1/2. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following loaded dice with $6$ faces: $P(S_1=6)=5/6$ and $P(S_1 = x)=1/30$ for $x\in\{1,2,3,4,5\}$. Suppose we throw it indefinitely. Hence, we have a source $S=S_1 S_2 S_3\ldots$. Then, $\lim_{n o\infty}H(S_n) = \log_2(6)$. | Here legal means that the coefficients are non-negative and sum to six, so that each die has six sides and every face has at least one spot. (That is, the generating function of each die must be a polynomial p(x) with positive coefficients, and with p(0) = 0 and p(1) = 6.) Only one such partition exists: x ( x + 1 ) ( x 2 + x + 1 ) = x + 2 x 2 + 2 x 3 + x 4 {\displaystyle x\;(x+1)\;(x^{2}+x+1)=x+2x^{2}+2x^{3}+x^{4}} and x ( x + 1 ) ( x 2 + x + 1 ) ( x 2 − x + 1 ) 2 = x + x 3 + x 4 + x 5 + x 6 + x 8 {\displaystyle x\;(x+1)\;(x^{2}+x+1)\;(x^{2}-x+1)^{2}=x+x^{3}+x^{4}+x^{5}+x^{6}+x^{8}} This gives us the distribution of spots on the faces of a pair of Sicherman dice as being {1,2,2,3,3,4} and {1,3,4,5,6,8}, as above. This technique can be extended for dice with an arbitrary number of sides. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the group $(\mathbb{Z} / 23 \mathbb{Z}^*, \cdot)$. Find how many elements of the group are generators of the group. (Hint: $5$ is a generator of the group.) | the group elements, g, in G are functions of the parameters: and all parameters set to zero returns the identity element of the group: Group elements are often matrices which act on vectors, or transformations acting on functions. The generators of the group are the partial derivatives of the group elements with respect to the group parameters with the result evaluated when the parameter is set to zero: In the language of manifolds, the generators are the elements of the tangent space to G at the identity. The generators are also known as infinitesimal group elements or as the elements of the Lie algebra of G. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the group $(\mathbb{Z} / 23 \mathbb{Z}^*, \cdot)$. Find how many elements of the group are generators of the group. (Hint: $5$ is a generator of the group.) | (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 |
In RSA, we set $p = 7, q = 11, e = 13$. The public key is $(m, e) = (77, 13)$. The ciphertext we receive is $c = 14$. What is the message that was sent? (Hint: You may solve faster using Chinese remainder theorem.). | The public key in the RSA system is a tuple of integers ( N , e ) {\displaystyle (N,e)} , where N is the product of two primes p and q. The secret key is given by an integer d satisfying e d ≡ 1 ( mod ( p − 1 ) ( q − 1 ) ) {\displaystyle ed\equiv 1{\pmod {(p-1)(q-1)}}} ; equivalently, the secret key may be given by d p ≡ d ( mod p − 1 ) {\displaystyle d_{p}\equiv d{\pmod {p-1}}} and d q ≡ d ( mod q − 1 ) {\displaystyle d_{q}\equiv d{\pmod {q-1}}} if the Chinese remainder theorem is used to improve the speed of decryption, see CRT-RSA. Encryption of a message M produces the ciphertext C ≡ M e ( mod N ) {\displaystyle C\equiv M^{e}{\pmod {N}}} , which can be decrypted using d {\displaystyle d} by computing C d ≡ M ( mod N ) {\displaystyle C^{d}\equiv M{\pmod {N}}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In RSA, we set $p = 7, q = 11, e = 13$. The public key is $(m, e) = (77, 13)$. The ciphertext we receive is $c = 14$. What is the message that was sent? (Hint: You may solve faster using Chinese remainder theorem.). | Setting up an RSA system involves choosing large prime numbers p and q, computing n = pq and k = φ(n), and finding two numbers e and d such that ed ≡ 1 (mod k). The numbers n and e (the "encryption key") are released to the public, and d (the "decryption key") is kept private. A message, represented by an integer m, where 0 < m < n, is encrypted by computing S = me (mod n). It is decrypted by computing t = Sd (mod n). Euler's Theorem can be used to show that if 0 < t < n, then t = m. The security of an RSA system would be compromised if the number n could be efficiently factored or if φ(n) could be efficiently computed without factoring n. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an RSA encryption where the public key is published as $(m, e) = (35, 11)$. Which one of the following numbers is a valid decoding exponent? | 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 |
Consider an RSA encryption where the public key is published as $(m, e) = (35, 11)$. Which one of the following numbers is a valid decoding exponent? | If the public exponent is small and the plaintext m {\displaystyle m} is very short, then the RSA function may be easy to invert, which makes certain attacks possible. Padding schemes ensure that messages have full lengths, but additionally choosing the public exponent e = 2 16 + 1 {\displaystyle e=2^{16}+1} is recommended. When this value is used, signature verification requires 17 multiplications, as opposed to about 25 when a random e {\displaystyle e} of similar size is used. Unlike low private exponent (see Wiener's attack), attacks that apply when a small e {\displaystyle e} is used are far from a total break, which would recover the secret key d. The most powerful attacks on low public exponent RSA are based on the following theorem, which is due to Don Coppersmith. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $K = (K_1, K_2,..., K_n)$, where each $K_i$ is $0$ or $1$ with probability $1/2$. Let $K'=(K'_1, K'_2, ..., K'_n)$ such that, for each $i$, $K'_i \in {0, 1}$ and $K'_{i} = \sum_{j = 1}^i K_j ext{mod} 8.$ True or false: Using $K'$ as the key one can achieve perfect secrecy if the message is $n$ bits. | Another way of stating perfect secrecy is that for all messages m 1 , m 2 {\displaystyle m_{1},m_{2}} in message space M, and for all ciphers c in cipher space C, we have Pr k ⇐ K = Pr k ⇐ K {\displaystyle {\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}={\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}} , where Pr {\textstyle \operatorname {Pr} } represents the probabilities, taken over a choice of k {\displaystyle k} in key space K {\displaystyle \mathrm {K} } over the coin tosses of a probabilistic algorithm, E {\displaystyle E} . Perfect secrecy is a strong notion of cryptanalytic difficulty.Conventional symmetric encryption algorithms use complex patterns of substitution and transpositions. For the best of these currently in use, it is not known whether there can be a cryptanalytic procedure that can efficiently reverse (or even partially reverse) these transformations without knowing the key used during encryption. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $K = (K_1, K_2,..., K_n)$, where each $K_i$ is $0$ or $1$ with probability $1/2$. Let $K'=(K'_1, K'_2, ..., K'_n)$ such that, for each $i$, $K'_i \in {0, 1}$ and $K'_{i} = \sum_{j = 1}^i K_j ext{mod} 8.$ True or false: Using $K'$ as the key one can achieve perfect secrecy if the message is $n$ bits. | The piling-up lemma allows the cryptanalyst to determine the probability that the equality: X 1 ⊕ X 2 ⊕ ⋯ ⊕ X n = 0 {\displaystyle X_{1}\oplus X_{2}\oplus \cdots \oplus X_{n}=0} holds, where the X's are binary variables (that is, bits: either 0 or 1). Let P(A) denote "the probability that A is true". If it equals one, A is certain to happen, and if it equals zero, A cannot happen. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a message $T$ and a key $K$ chosen independently from $T$. True or false: If there exists a perfectly secret encryption scheme using $K$, then $H(T) \leq H(K)$. | In a symmetric-key system, Bob knows Alice's encryption key. Once the message is encrypted, Alice can safely transmit it to Bob (assuming no one else knows the key). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a message $T$ and a key $K$ chosen independently from $T$. True or false: If there exists a perfectly secret encryption scheme using $K$, then $H(T) \leq H(K)$. | A cryptosystem is considered secure in terms of indistinguishability if no adversary, given an encryption of a message randomly chosen from a two-element message space determined by the adversary, can identify the message choice with probability significantly better than that of random guessing (1⁄2). If any adversary can succeed in distinguishing the chosen ciphertext with a probability significantly greater than 1⁄2, then this adversary is considered to have an "advantage" in distinguishing the ciphertext, and the scheme is not considered secure in terms of indistinguishability. This definition encompasses the notion that in a secure scheme, the adversary should learn no information from seeing a ciphertext. Therefore, the adversary should be able to do no better than if it guessed randomly. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(n,k)$ linear code with minimum distance $d_{\min} = 4$. Let $\mathcal{C}'$ be the code obtained by adding a parity-check bit $x_{n+1}=x_1 \oplus x_2 \oplus \cdots \oplus x_n$ at the end of each codeword of $\mathcal{C}$. Let $d_{\min}'$ be the minimum distance of $\mathcal{C}'$. Which of the following is true? | {\displaystyle \min _{c\in C,\ c\neq c_{0}}d(c,c_{0})=\min _{c\in C,\ c\neq c_{0}}d(c-c_{0},0)=\min _{c\in C,\ c\neq 0}d(c,0)=d.} In other words, in order to find out the minimum distance between the codewords of a linear code, one would only need to look at the non-zero codewords. The non-zero codeword with the smallest weight has then the minimum distance to the zero codeword, and hence determines the minimum distance of the code. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(n,k)$ linear code with minimum distance $d_{\min} = 4$. Let $\mathcal{C}'$ be the code obtained by adding a parity-check bit $x_{n+1}=x_1 \oplus x_2 \oplus \cdots \oplus x_n$ at the end of each codeword of $\mathcal{C}$. Let $d_{\min}'$ be the minimum distance of $\mathcal{C}'$. Which of the following is true? | Let B {\displaystyle {\mathcal {B}}} be a binary code consisting of M {\displaystyle M} codewords of length n {\displaystyle {\mathit {n}}} and minimum distance d min {\displaystyle {d_{\min }}} , such that c ∈ B {\displaystyle {\mathit {c}}\in {\mathcal {B}}} implies that c ¯ ∈ B {\displaystyle {\mathit {\bar {c}}}\in {\mathcal {B}}} . For w > 0 {\displaystyle {\mathit {w}}>0} , consider the constant-weight subcode B w = { u ∈ B: w H ( u ) = w } {\displaystyle {\mathcal {B_{\mathit {w}}}}=\{u\in {\mathcal {B}}:{\mathit {w_{H}}}(u)={\mathit {w}}\}} , where w H ( ⋅ ) {\displaystyle {w_{H}(\cdot )}} denotes Hamming weight. Choose w > 0 {\displaystyle {\mathit {w}}>0} such that n ≥ 2 w + ⌈ d min / 2 ⌉ {\displaystyle {\mathit {n}}\geq {\mathit {2w}}+\lceil {\mathit {d_{\min }}}/2\rceil } , and consider a DNA code, C w {\displaystyle {\mathcal {C}}_{w}} , with the following choice for its even and odd components: E = { a b ¯: a , b ∈ B w } {\displaystyle {\mathcal {E}}=\left\{a{\bar {b}}:a,b\in {\mathcal {B}}_{w}\right\}} , O = { a b R C: a , b ∈ B , a < l e x b } {\displaystyle {\mathcal {O}}=\left\{ab^{RC}:a,b\in {\mathcal {B}},a<_{lex}b\right\}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a $(n,k)$ Reed-Solomon code on $\mathbb{F}_q$. Let $\mathcal{C}'$ be the $(2n,k)$ code such that each codeword of $\mathcal{C}'$ is a codeword of $\mathcal{C}$ repeated twice, i.e., if $(x_1,\dots,x_n) \in\mathcal{C}$, then $(x_1,\dots,x_n,x_1,\dots,x_n)\in\mathcal{C'}$. What is the minimum distance of $\mathcal{C}'$? | In order to prove a lower bound for the distance of a code C ∗ {\displaystyle C^{*}} we prove that the Hamming distance of an arbitrary but distinct pair of codewords has a lower bound. So let Δ ( c 1 , c 2 ) {\displaystyle \Delta (c^{1},c^{2})} be the Hamming distance of two codewords c 1 {\displaystyle c^{1}} and c 2 {\displaystyle c^{2}} . For any given m 1 ≠ m 2 ∈ ( F q k ) K , {\displaystyle m_{1}\neq m_{2}\in \left(\mathbb {F} _{q^{k}}\right)^{K},} we want a lower bound for Δ ( C ∗ ( m 1 ) , C ∗ ( m 2 ) ) . {\displaystyle \Delta (C^{*}(m_{1}),C^{*}(m_{2})).} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a $(n,k)$ Reed-Solomon code on $\mathbb{F}_q$. Let $\mathcal{C}'$ be the $(2n,k)$ code such that each codeword of $\mathcal{C}'$ is a codeword of $\mathcal{C}$ repeated twice, i.e., if $(x_1,\dots,x_n) \in\mathcal{C}$, then $(x_1,\dots,x_n,x_1,\dots,x_n)\in\mathcal{C'}$. What is the minimum distance of $\mathcal{C}'$? | Furthermore, there are two polynomials that do agree in k − 1 {\displaystyle k-1} points but are not equal, and thus, the distance of the Reed–Solomon code is exactly d = n − k + 1 {\displaystyle d=n-k+1} . Then the relative distance is δ = d / n = 1 − k / n + 1 / n = 1 − R + 1 / n ∼ 1 − R {\displaystyle \delta =d/n=1-k/n+1/n=1-R+1/n\sim 1-R} , where R = k / n {\displaystyle R=k/n} is the rate. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(6,3)$ linear code containing the codewords $\mathbf{x}_1 = 011011$, $\mathbf{x}_2 = 101101$ and $\mathbf{x}_3 = 111000$. True or false: The minimum distance of the code is $d_{\min} = 3$. | {\displaystyle \min _{c\in C,\ c\neq c_{0}}d(c,c_{0})=\min _{c\in C,\ c\neq c_{0}}d(c-c_{0},0)=\min _{c\in C,\ c\neq 0}d(c,0)=d.} In other words, in order to find out the minimum distance between the codewords of a linear code, one would only need to look at the non-zero codewords. The non-zero codeword with the smallest weight has then the minimum distance to the zero codeword, and hence determines the minimum distance of the code. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(6,3)$ linear code containing the codewords $\mathbf{x}_1 = 011011$, $\mathbf{x}_2 = 101101$ and $\mathbf{x}_3 = 111000$. True or false: The minimum distance of the code is $d_{\min} = 3$. | A code whose minimum distance is at least 3, have a check matrix all of whose columns are distinct and non zero. If a check matrix for a binary code has m {\displaystyle m} rows, then each column is an m {\displaystyle m} -bit binary number. There are 2 m − 1 {\displaystyle 2^{m}-1} possible columns. Therefore, if a check matrix of a binary code with d m i n {\displaystyle d_{min}} at least 3 has m {\displaystyle m} rows, then it can only have 2 m − 1 {\displaystyle 2^{m}-1} columns, not more than that. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(6,3)$ linear code containing the codewords $\mathbf{x}_1 = 011011$, $\mathbf{x}_2 = 101101$ and $\mathbf{x}_3 = 111000$. True or false: A generator matrix for the code is egin{equation*} G = egin{pmatrix} 1 &0 &0 &0 &1 &1 \ 0 &1 &0 &0 &0 &1 \ 0 &0 &1 &0 &1 &1 \end{pmatrix} \end{equation*} | From this, the generator matrix G can be obtained as {\displaystyle {\begin{bmatrix}I_{k}|P\end{bmatrix}}} (noting that in the special case of this being a binary code P = − P {\displaystyle P=-P} ), or specifically: G = ( 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 ) . {\displaystyle \mathbf {G} ={\begin{pmatrix}1&0&0&1&0&1\\0&1&0&1&1&1\\0&0&1&1&1&0\\\end{pmatrix}}.} Finally, by multiplying all eight possible 3-bit strings by G, all eight valid codewords are obtained. For example, the codeword for the bit-string '101' is obtained by: ( 1 0 1 ) ⊙ ( 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 0 ) = ( 1 0 1 0 1 1 ) {\displaystyle {\begin{pmatrix}1&0&1\\\end{pmatrix}}\odot {\begin{pmatrix}1&0&0&1&0&1\\0&1&0&1&1&1\\0&0&1&1&1&0\\\end{pmatrix}}={\begin{pmatrix}1&0&1&0&1&1\\\end{pmatrix}}} ,where ⊙ {\displaystyle \odot } is symbol of mod 2 multiplication. As a check, the row space of G is orthogonal to H such that G ⊙ H T = 0 {\displaystyle G\odot H^{T}=0} The bit-string '101' is found in as the first 3 bits of the codeword '101011'. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(6,3)$ linear code containing the codewords $\mathbf{x}_1 = 011011$, $\mathbf{x}_2 = 101101$ and $\mathbf{x}_3 = 111000$. True or false: A generator matrix for the code is egin{equation*} G = egin{pmatrix} 1 &0 &0 &0 &1 &1 \ 0 &1 &0 &0 &0 &1 \ 0 &0 &1 &0 &1 &1 \end{pmatrix} \end{equation*} | If G is a matrix, it generates the codewords of a linear code C by w = s G {\displaystyle w=sG} where w is a codeword of the linear code C, and s is any input vector. Both w and s are assumed to be row vectors. A generator matrix for a linear q {\displaystyle _{q}} -code has format k × n {\displaystyle k\times n} , where n is the length of a codeword, k is the number of information bits (the dimension of C as a vector subspace), d is the minimum distance of the code, and q is size of the finite field, that is, the number of symbols in the alphabet (thus, q = 2 indicates a binary code, etc.). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be the $(6,3)$ linear code on $\mathbb{F}_3$ whose parity-check matrix is egin{equation*} H = egin{pmatrix} 2 &0 &1 &1 &1 &0 \ 1 &2 &0 &0 &1 &1 \ 0 &0 &0 &1 &1 &1 \end{pmatrix}. \end{equation*} True or false: The sequence $\mathbf{y} = 111000$ is a codeword of $\mathcal{C}$. | This is a (6, 3) linear code, with n = 6 and k = 3. Again ignoring lines going out of the picture, the parity-check matrix representing this graph fragment is H = ( 1 1 1 1 0 0 0 0 1 1 0 1 1 0 0 1 1 0 ) . {\displaystyle \mathbf {H} ={\begin{pmatrix}1&1&1&1&0&0\\0&0&1&1&0&1\\1&0&0&1&1&0\\\end{pmatrix}}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be the $(6,3)$ linear code on $\mathbb{F}_3$ whose parity-check matrix is egin{equation*} H = egin{pmatrix} 2 &0 &1 &1 &1 &0 \ 1 &2 &0 &0 &1 &1 \ 0 &0 &0 &1 &1 &1 \end{pmatrix}. \end{equation*} True or false: The sequence $\mathbf{y} = 111000$ is a codeword of $\mathcal{C}$. | In coding theory, a parity-check matrix of a linear block code C is a matrix which describes the linear relations that the components of a codeword must satisfy. It can be used to decide whether a particular vector is a codeword and is also used in decoding algorithms. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(5,2)$ linear code with generator matrix egin{equation*} G = egin{pmatrix} 1 &0 &1 &0 &1 \ 0 &1 &0 &1 &1 \end{pmatrix} \end{equation*} and consider a minimum-distance decoder obtained by choosing the coset leaders of the standard array of $\mathcal{C}$ so that the error probability is minimized under a binary symmetric channel with bit-flip probability $\epsilon < rac{1}{2}$. True or false: The word $00101$ is certainly not one of the coset leaders. | As a result, the half-the minimum distance acts as a combinatorial barrier beyond which unambiguous error-correction is impossible, if we only insist on unique decoding. However, received words such as y {\displaystyle y} considered above occur only in the worst-case and if one looks at the way Hamming balls are packed in high-dimensional space, even for error patterns e {\displaystyle e} beyond half-the minimum distance, there is only a single codeword c {\displaystyle c} within Hamming distance e {\displaystyle e} from the received word. This claim has been shown to hold with high probability for a random code picked from a natural ensemble and more so for the case of Reed–Solomon codes which is well studied and quite ubiquitous in the real world applications. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal{C}$ be a binary $(5,2)$ linear code with generator matrix egin{equation*} G = egin{pmatrix} 1 &0 &1 &0 &1 \ 0 &1 &0 &1 &1 \end{pmatrix} \end{equation*} and consider a minimum-distance decoder obtained by choosing the coset leaders of the standard array of $\mathcal{C}$ so that the error probability is minimized under a binary symmetric channel with bit-flip probability $\epsilon < rac{1}{2}$. True or false: The word $00101$ is certainly not one of the coset leaders. | An example of a standard array for the 2-dimensional code C = {00000, 01101, 10110, 11011} in the 5-dimensional space V (with 32 vectors) is as follows: The decoding procedure is to find the received word in the table and then add to it the coset leader of the row it is in. Since in binary arithmetic adding is the same operation as subtracting, this always results in an element of C. In the event that the transmission errors occurred in precisely the non-zero positions of the coset leader the result will be the right codeword. In this example, if a single error occurs, the method will always correct it, since all possible coset leaders with a single one appear in the array. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $b$ be the maximum number of linearly independent columns of a parity check matrix $H$ of a linear code. True or false: Then, the minimum distance of the code is $b+1$. | A code whose minimum distance is at least 3, have a check matrix all of whose columns are distinct and non zero. If a check matrix for a binary code has m {\displaystyle m} rows, then each column is an m {\displaystyle m} -bit binary number. There are 2 m − 1 {\displaystyle 2^{m}-1} possible columns. Therefore, if a check matrix of a binary code with d m i n {\displaystyle d_{min}} at least 3 has m {\displaystyle m} rows, then it can only have 2 m − 1 {\displaystyle 2^{m}-1} columns, not more than that. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $b$ be the maximum number of linearly independent columns of a parity check matrix $H$ of a linear code. True or false: Then, the minimum distance of the code is $b+1$. | Hence the rate of Hamming codes is R = k / n = 1 − r / (2r − 1), which is the highest possible for codes with minimum distance of three (i.e., the minimal number of bit changes needed to go from any code word to any other code word is three) and block length 2r − 1. The parity-check matrix of a Hamming code is constructed by listing all columns of length r that are non-zero, which means that the dual code of the Hamming code is the shortened Hadamard code, also known as a Simplex code. The parity-check matrix has the property that any two columns are pairwise linearly independent. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following mysterious binary encoding:egin{center} egin{tabular}{c|c} symbol & encoding \ \hline $a$ & $??0$\ $b$ & $??0$\ $c$ & $??0$\ $d$ & $??0$ \end{tabular} \end{center} where with '$?$' we mean that we do not know which bit is assigned as the first two symbols of the encoding of any of the source symbols $a,b,c,d$. What can you infer on this encoding assuming that the code-words are all different? | For the example mentioned above, the encoding becomes: (1,1,2), ('B','A','C','D') This means that the first symbol B is of length 1, then the A of length 2, and remains of 3. Since the symbols are sorted by bit-length, we can efficiently reconstruct the codebook. A pseudo code describing the reconstruction is introduced on the next section. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following mysterious binary encoding:egin{center} egin{tabular}{c|c} symbol & encoding \ \hline $a$ & $??0$\ $b$ & $??0$\ $c$ & $??0$\ $d$ & $??0$ \end{tabular} \end{center} where with '$?$' we mean that we do not know which bit is assigned as the first two symbols of the encoding of any of the source symbols $a,b,c,d$. What can you infer on this encoding assuming that the code-words are all different? | ASCII reserves the first 32 codes (numbers 0–31 decimal) for control characters known as the "C0 set": codes originally intended not to represent printable information, but rather to control devices (such as printers) that make use of ASCII, or to provide meta-information about data streams such as those stored on magnetic tape. They include common characters like the newline and the tab character. In 8-bit character sets such as Latin-1 and the other ISO 8859 sets, the first 32 characters of the "upper half" (128 to 159) are also control codes, known as the "C1 set". They are rarely used directly; when they turn up in documents which are ostensibly in an ISO 8859 encoding, their code positions generally refer instead to the characters at that position in a proprietary, system-specific encoding, such as Windows-1252 or Mac OS Roman, that use the codes to instead provide additional graphic characters. Unicode defines additional control characters, including bi-directional text direction override characters (used to explicitly mark right-to-left writing inside left-to-right writing and the other way around) and variation selectors to select alternate forms of CJK ideographs, emoji and other characters. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose that you possess a $D$-ary encoding $\Gamma$ for the source $S$ that does not satisfy Kraft's Inequality. Specifically, in this problem, we assume that our encoding satisfies $\sum_{i=1}^n D^{-l_i} = k+1 $ with $k>0$. What can you infer on the average code-word length $L(S,\Gamma)$? | For Shannon's method, the word lengths satisfy l i = ⌈ − log 2 p i ⌉ ≤ − log 2 p i + 1. {\displaystyle l_{i}=\lceil -\log _{2}p_{i}\rceil \leq -\log _{2}p_{i}+1.} Hence the expected word length satisfies Here, H ( X ) = − ∑ i = 1 n p i log 2 p i {\displaystyle H(X)=-\textstyle \sum _{i=1}^{n}p_{i}\log _{2}p_{i}} is the entropy, and Shannon's source coding theorem says that any code must have an average length of at least H ( X ) {\displaystyle H(X)} . Hence we see that the Shannon–Fano code is always within one bit of the optimal expected word length. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose that you possess a $D$-ary encoding $\Gamma$ for the source $S$ that does not satisfy Kraft's Inequality. Specifically, in this problem, we assume that our encoding satisfies $\sum_{i=1}^n D^{-l_i} = k+1 $ with $k>0$. What can you infer on the average code-word length $L(S,\Gamma)$? | This is an immediate consequence of the Kraft-McMillan inequality. Kraft's inequality states that given a sequence of strings { x i } i = 1 n {\displaystyle \{x_{i}\}_{i=1}^{n}} there exists a prefix code with codewords { σ i } i = 1 n {\displaystyle \{\sigma _{i}\}_{i=1}^{n}} where ∀ i , | σ i | = k i {\displaystyle \forall i,|\sigma _{i}|=k_{i}} if and only if: ∑ i = 1 n s − k i ≤ 1 {\displaystyle \sum _{i=1}^{n}s^{-k_{i}}\leq 1} where s {\displaystyle s} is the size of the alphabet S {\displaystyle S} . Without loss of generality, let's suppose we may order the k i {\displaystyle k_{i}} such that: k 1 ≤ k 2 ≤ . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=2$ there exists a uniquely-decodable code with the same lengths of $\Gamma$. | Let each source symbol from the alphabet S = { s 1 , s 2 , … , s n } {\displaystyle S=\{\,s_{1},s_{2},\ldots ,s_{n}\,\}} be encoded into a uniquely decodable code over an alphabet of size r {\displaystyle r} with codeword lengths ℓ 1 , ℓ 2 , … , ℓ n . {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}.} Then ∑ i = 1 n r − ℓ i ⩽ 1. {\displaystyle \sum _{i=1}^{n}r^{-\ell _{i}}\leqslant 1.} Conversely, for a given set of natural numbers ℓ 1 , ℓ 2 , … , ℓ n {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r {\displaystyle r} with those codeword lengths. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=2$ there exists a uniquely-decodable code with the same lengths of $\Gamma$. | The source coding theorem for symbol codes places an upper and a lower bound on the minimal possible expected length of codewords as a function of the entropy of the input word (which is viewed as a random variable) and of the size of the target alphabet. Note that, for data that exhibits more dependencies (whose source is not an i.i.d. random variable), the Kolmogorov complexity, which quantifies the minimal description length of an object, is more suitable to describe the limits of data compression. Shannon entropy takes into account only frequency regularities while Kolmogorov complexity takes into account all algorithmic regularities, so in general the latter is smaller. On the other hand, if an object is generated by a random process in such a way that it has only frequency regularities, entropy is close to complexity with high probability (Shen et al. 2017). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=4$ then $\Gamma$ is necessarily uniquely-decodable. | Let each source symbol from the alphabet S = { s 1 , s 2 , … , s n } {\displaystyle S=\{\,s_{1},s_{2},\ldots ,s_{n}\,\}} be encoded into a uniquely decodable code over an alphabet of size r {\displaystyle r} with codeword lengths ℓ 1 , ℓ 2 , … , ℓ n . {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}.} Then ∑ i = 1 n r − ℓ i ⩽ 1. {\displaystyle \sum _{i=1}^{n}r^{-\ell _{i}}\leqslant 1.} Conversely, for a given set of natural numbers ℓ 1 , ℓ 2 , … , ℓ n {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r {\displaystyle r} with those codeword lengths. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=4$ then $\Gamma$ is necessarily uniquely-decodable. | Let C {\displaystyle {\mathcal {C}}} be a ( n , k , d ) q {\displaystyle (n,k,d)_{q}} error-correcting code; in other words, C {\displaystyle {\mathcal {C}}} is a code of length n {\displaystyle n} , dimension k {\displaystyle k} and minimum distance d {\displaystyle d} over an alphabet Σ {\displaystyle \Sigma } of size q {\displaystyle q} . The list-decoding problem can now be formulated as follows: Input: Received word x ∈ Σ n {\displaystyle x\in \Sigma ^{n}} , error bound e {\displaystyle e} Output: A list of all codewords x 1 , x 2 , … , x m ∈ C {\displaystyle x_{1},x_{2},\ldots ,x_{m}\in {\mathcal {C}}} whose hamming distance from x {\displaystyle x} is at most e {\displaystyle e} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=4$ then $\Gamma$ is necessarily prefix-free. | Let each source symbol from the alphabet S = { s 1 , s 2 , … , s n } {\displaystyle S=\{\,s_{1},s_{2},\ldots ,s_{n}\,\}} be encoded into a uniquely decodable code over an alphabet of size r {\displaystyle r} with codeword lengths ℓ 1 , ℓ 2 , … , ℓ n . {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}.} Then ∑ i = 1 n r − ℓ i ⩽ 1. {\displaystyle \sum _{i=1}^{n}r^{-\ell _{i}}\leqslant 1.} Conversely, for a given set of natural numbers ℓ 1 , ℓ 2 , … , ℓ n {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r {\displaystyle r} with those codeword lengths. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=4$ then $\Gamma$ is necessarily prefix-free. | There is a bijection γ from the Lyndon words in an ordered alphabet to a basis of the free Lie algebra on this alphabet defined as follows: If a word w has length 1 then γ ( w ) = w {\displaystyle \gamma (w)=w} (considered as a generator of the free Lie algebra). If w has length at least 2, then write w = u v {\displaystyle w=uv} for Lyndon words u, v with v as long as possible (the "standard factorization"). Then γ ( w ) = {\displaystyle \gamma (w)=} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=3$ then $\Gamma$ is extbf{not} uniquely-decodable | Let each source symbol from the alphabet S = { s 1 , s 2 , … , s n } {\displaystyle S=\{\,s_{1},s_{2},\ldots ,s_{n}\,\}} be encoded into a uniquely decodable code over an alphabet of size r {\displaystyle r} with codeword lengths ℓ 1 , ℓ 2 , … , ℓ n . {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}.} Then ∑ i = 1 n r − ℓ i ⩽ 1. {\displaystyle \sum _{i=1}^{n}r^{-\ell _{i}}\leqslant 1.} Conversely, for a given set of natural numbers ℓ 1 , ℓ 2 , … , ℓ n {\displaystyle \ell _{1},\ell _{2},\ldots ,\ell _{n}} satisfying the above inequality, there exists a uniquely decodable code over an alphabet of size r {\displaystyle r} with those codeword lengths. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider a source $S$ with some distribution $P_S$ over the alphabet $\mathcal{A} = \{a, b, c, d, e, f\}$. Consider the following encoding $\Gamma$ over a code alphabet $\mathcal{D}$ of size $D$ with the following codeword lengths: egin{center} egin{tabular}{ |c|c|c|c|c|c|c| } \hline & $a$ & $b$ & $c$ & $d$ & $e$ & $f$ \ \hline $l(\Gamma(\cdot))$ & 1 & 1 & 1 & 2 & 2 & 4 \ \hline \end{tabular} \end{center} True or false: If $D=3$ then $\Gamma$ is extbf{not} uniquely-decodable | For the example mentioned above, the encoding becomes: (1,1,2), ('B','A','C','D') This means that the first symbol B is of length 1, then the A of length 2, and remains of 3. Since the symbols are sorted by bit-length, we can efficiently reconstruct the codebook. A pseudo code describing the reconstruction is introduced on the next section. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following sequence of random variables $S_1,\ldots,S_n,\ldots$ Assume that the limit $H^\star(\mathcal{S})=k$ exists and is finite. Suppose that there exists $\hat{n}>0$ such that for all $i\geq \hat{n}$ one has that the marginal distributions of $S_{i+1}$ and $S_i$ satisfy $p_{S_{i+1}}=p_{S_i}$. Denote with $\mathcal{Y}_{\hat{n}}$ the alphabet of the source $S_{\hat{n}}$. True or false: Can one use this information to infer that the following holds: $|\mathcal{Y}_{\hat{n}}| \geq 2^k $? | Infinitely divisible distributions appear in a broad generalization of the central limit theorem: the limit as n → +∞ of the sum Sn = Xn1 + … + Xnn of independent uniformly asymptotically negligible (u.a.n.) random variables within a triangular array X 11 X 21 X 22 X 31 X 32 X 33 ⋮ ⋮ ⋮ ⋱ {\displaystyle {\begin{array}{cccc}X_{11}\\X_{21}&X_{22}\\X_{31}&X_{32}&X_{33}\\\vdots &\vdots &\vdots &\ddots \end{array}}} approaches — in the weak sense — an infinitely divisible distribution. The uniformly asymptotically negligible (u.a.n.) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the following sequence of random variables $S_1,\ldots,S_n,\ldots$ Assume that the limit $H^\star(\mathcal{S})=k$ exists and is finite. Suppose that there exists $\hat{n}>0$ such that for all $i\geq \hat{n}$ one has that the marginal distributions of $S_{i+1}$ and $S_i$ satisfy $p_{S_{i+1}}=p_{S_i}$. Denote with $\mathcal{Y}_{\hat{n}}$ the alphabet of the source $S_{\hat{n}}$. True or false: Can one use this information to infer that the following holds: $|\mathcal{Y}_{\hat{n}}| \geq 2^k $? | In order that the probability distribution of a random variable X {\displaystyle X} be uniquely defined by its moments α k = E X k {\displaystyle \alpha _{k}=EX^{k}} it is sufficient, for example, that Carleman's condition be satisfied: A similar result even holds for moments of random vectors. The problem of moments seeks characterizations of sequences μ n ′: n = 1 , 2 , 3 , … {\displaystyle {{\mu _{n}}':n=1,2,3,\dots }} that are sequences of moments of some function f, all moments α k ( n ) {\displaystyle \alpha _{k}(n)} of which are finite, and for each integer k ≥ 1 {\displaystyle k\geq 1} let where α k {\displaystyle \alpha _{k}} is finite. Then there is a sequence μ n ′ {\displaystyle {\mu _{n}}'} that weakly converges to a distribution function μ {\displaystyle \mu } having α k {\displaystyle \alpha _{k}} as its moments. If the moments determine μ {\displaystyle \mu } uniquely, then the sequence μ n ′ {\displaystyle {\mu _{n}}'} weakly converges to μ {\displaystyle \mu } . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_{0},S_{1},S_{2},\dots$ be an infinite sequence produced by a source $\mathcal{S}$. All $S_{n}$ take values in $\{0,1\}$, and $S_{n+1}$ depends only on $S_n$, that is, $p_{S_{n+1} | S_0, \dots, S_n}(s_{n+1} | s_0, \dots, s_n) = p_{S_{n+1} | S_n}(s_{n+1} | s_n)$. The probability $p_{S_{n+1}|S_{n}}$ is schematically represented in the graph below: egin{center} ikzset{ state/.style = { draw, circle, minimum size = 20pt, font = ontsize{12}{12}\selectfont, } } egin{tikzpicture}[> = latex]
ode[state] (a) {$0$};
ode[state, right of = a] (b) {$1$}; \path[->] (a) edge[bend left, above] node {$1/2$} (b); \path[->] (a) edge[loop left] node {$ rac{1}{2}$} (); \path[->] (b) edge[bend left, below] node {$1/4$} (a); \path[->] (b) edge[loop right] node {$ rac{3}{4}$} (); \end{tikzpicture} \end{center} For instance, the edge from $0$ to $1$ means that $p_{S_{n+1}|S_{n}}(1|0) = rac{1}{2}$. We also have that $p_{S_0}(0)=1$. True or false: For every $n\geq 0$, $H(S_n|S_0,\ldots,S_{n-1})
eq H(S_n|S_{n-1}) $. | The probability of obtaining any one particular random graph with m edges is p m ( 1 − p ) N − m {\displaystyle p^{m}(1-p)^{N-m}} with the notation N = ( n 2 ) {\displaystyle N={\tbinom {n}{2}}} .A closely related model, the Erdős–Rényi model denoted G(n,M), assigns equal probability to all graphs with exactly M edges. With 0 ≤ M ≤ N, G(n,M) has ( N M ) {\displaystyle {\tbinom {N}{M}}} elements and every element occurs with probability 1 / ( N M ) {\displaystyle 1/{\tbinom {N}{M}}} . The latter model can be viewed as a snapshot at a particular time (M) of the random graph process G ~ n {\displaystyle {\tilde {G}}_{n}} , which is a stochastic process that starts with n vertices and no edges, and at each step adds one new edge chosen uniformly from the set of missing edges. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_{0},S_{1},S_{2},\dots$ be an infinite sequence produced by a source $\mathcal{S}$. All $S_{n}$ take values in $\{0,1\}$, and $S_{n+1}$ depends only on $S_n$, that is, $p_{S_{n+1} | S_0, \dots, S_n}(s_{n+1} | s_0, \dots, s_n) = p_{S_{n+1} | S_n}(s_{n+1} | s_n)$. The probability $p_{S_{n+1}|S_{n}}$ is schematically represented in the graph below: egin{center} ikzset{ state/.style = { draw, circle, minimum size = 20pt, font = ontsize{12}{12}\selectfont, } } egin{tikzpicture}[> = latex]
ode[state] (a) {$0$};
ode[state, right of = a] (b) {$1$}; \path[->] (a) edge[bend left, above] node {$1/2$} (b); \path[->] (a) edge[loop left] node {$ rac{1}{2}$} (); \path[->] (b) edge[bend left, below] node {$1/4$} (a); \path[->] (b) edge[loop right] node {$ rac{3}{4}$} (); \end{tikzpicture} \end{center} For instance, the edge from $0$ to $1$ means that $p_{S_{n+1}|S_{n}}(1|0) = rac{1}{2}$. We also have that $p_{S_0}(0)=1$. True or false: For every $n\geq 0$, $H(S_n|S_0,\ldots,S_{n-1})
eq H(S_n|S_{n-1}) $. | Let S i {\displaystyle S_{i}\,} denote the event that the source sequence X 1 n ( i ) {\displaystyle X_{1}^{n}(i)} was generated at the source, so that P ( S i ) = P ( X 1 n ( i ) ) . {\displaystyle P(S_{i})=P(X_{1}^{n}(i))\,.} Then the probability of error can be broken down as P ( E ) = ∑ i P ( E ∣ S i ) P ( S i ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_{0},S_{1},S_{2},\dots$ be an infinite sequence produced by a source $\mathcal{S}$. All $S_{n}$ take values in $\{0,1\}$, and $S_{n+1}$ depends only on $S_n$, that is, $p_{S_{n+1} | S_0, \dots, S_n}(s_{n+1} | s_0, \dots, s_n) = p_{S_{n+1} | S_n}(s_{n+1} | s_n)$. The probability $p_{S_{n+1}|S_{n}}$ is schematically represented in the graph below: egin{center} ikzset{ state/.style = { draw, circle, minimum size = 20pt, font = ontsize{12}{12}\selectfont, } } egin{tikzpicture}[> = latex]
ode[state] (a) {$0$};
ode[state, right of = a] (b) {$1$}; \path[->] (a) edge[bend left, above] node {$1/2$} (b); \path[->] (a) edge[loop left] node {$ rac{1}{2}$} (); \path[->] (b) edge[bend left, below] node {$1/4$} (a); \path[->] (b) edge[loop right] node {$ rac{3}{4}$} (); \end{tikzpicture} \end{center} For instance, the edge from $0$ to $1$ means that $p_{S_{n+1}|S_{n}}(1|0) = rac{1}{2}$. We also have that $p_{S_0}(0)=1$. True or false: $H(\mathcal{S})= h(1/3)$, where $h$ is the binary entropy. | Let β i ( t ) = P ( Y t + 1 = y t + 1 , … , Y T = y T ∣ X t = i , θ ) {\displaystyle \beta _{i}(t)=P(Y_{t+1}=y_{t+1},\ldots ,Y_{T}=y_{T}\mid X_{t}=i,\theta )} that is the probability of the ending partial sequence y t + 1 , … , y T {\displaystyle y_{t+1},\ldots ,y_{T}} given starting state i {\displaystyle i} at time t {\displaystyle t} . We calculate β i ( t ) {\displaystyle \beta _{i}(t)} as, β i ( T ) = 1 , {\displaystyle \beta _{i}(T)=1,} β i ( t ) = ∑ j = 1 N β j ( t + 1 ) a i j b j ( y t + 1 ) . {\displaystyle \beta _{i}(t)=\sum _{j=1}^{N}\beta _{j}(t+1)a_{ij}b_{j}(y_{t+1}).} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_{0},S_{1},S_{2},\dots$ be an infinite sequence produced by a source $\mathcal{S}$. All $S_{n}$ take values in $\{0,1\}$, and $S_{n+1}$ depends only on $S_n$, that is, $p_{S_{n+1} | S_0, \dots, S_n}(s_{n+1} | s_0, \dots, s_n) = p_{S_{n+1} | S_n}(s_{n+1} | s_n)$. The probability $p_{S_{n+1}|S_{n}}$ is schematically represented in the graph below: egin{center} ikzset{ state/.style = { draw, circle, minimum size = 20pt, font = ontsize{12}{12}\selectfont, } } egin{tikzpicture}[> = latex]
ode[state] (a) {$0$};
ode[state, right of = a] (b) {$1$}; \path[->] (a) edge[bend left, above] node {$1/2$} (b); \path[->] (a) edge[loop left] node {$ rac{1}{2}$} (); \path[->] (b) edge[bend left, below] node {$1/4$} (a); \path[->] (b) edge[loop right] node {$ rac{3}{4}$} (); \end{tikzpicture} \end{center} For instance, the edge from $0$ to $1$ means that $p_{S_{n+1}|S_{n}}(1|0) = rac{1}{2}$. We also have that $p_{S_0}(0)=1$. True or false: $H(\mathcal{S})= h(1/3)$, where $h$ is the binary entropy. | Its formula is obtained by first calculating the dominant eigenvalue λ {\displaystyle \lambda } and corresponding eigenvector ψ {\displaystyle \psi } of the adjacency matrix, i.e. the largest λ ∈ R {\displaystyle \lambda \in \mathbb {R} } with corresponding ψ ∈ R n {\displaystyle \psi \in \mathbb {R} ^{n}} such that ψ A = λ ψ {\displaystyle \psi A=\lambda \psi } . Then stochastic matrix and stationary probability distribution are given by S i j = A i j λ ψ j ψ i {\displaystyle S_{ij}={\frac {A_{ij}}{\lambda }}{\frac {\psi _{j}}{\psi _{i}}}} for which every possible path of length l {\displaystyle l} from the i {\displaystyle i} -th to j {\displaystyle j} -th vertex has probability 1 λ l ψ j ψ i {\displaystyle {\frac {1}{\lambda ^{l}}}{\frac {\psi _{j}}{\psi _{i}}}} .Its entropy rate is log ( λ ) {\displaystyle \log(\lambda )} and the stationary probability distribution ρ {\displaystyle \rho } is ρ i = ψ i 2 ‖ ψ ‖ 2 2 {\displaystyle \rho _{i}={\frac {\psi _{i}^{2}}{\|\psi \|_{2}^{2}}}} .In contrast to GRW, the MERW transition probabilities generally depend on the structure of the entire graph (are nonlocal). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A colleague challenges you to create a $(n-1,k,d_{min})$ code $\mathcal C'$ from a $(n,k,d_{min})$ code $\mathcal C$ as follows: given a generator matrix $G$ that generates $\mathcal C$, drop one column from $G$. Then, generate the new code with this truncated $k imes (n-1)$ generator matrix. The catch is that your colleague only gives you a set $\mathcal S=\{\vec s_1,\vec s_2, \vec s_3\}$ of $3$ columns of $G$ that you are allowed to drop, where $\vec s_1$ is the all-zeros vector, $\vec s_2$ is the all-ones vector, and $\vec s_3$ is a canonical basis vector. From the length of the columns $s_i$ you can infer $k$. You do not know $n$, neither do you know anything about the $n-3$ columns of $G$ that are not in $\mathcal S$. However, your colleague tells you that $G$ is in systematic form, i.e., $G=[I ~~ P]$ for some unknown $P$, and that all of the elements in $\mathcal S$ are columns of $P$. Which of the following options in $\mathcal S$ would you choose as the column of $G$ to drop? | Let G be a generator matrix of C. We can always suppose that the first row of G is of the form r = (1, ..., 1, 0, ..., 0) with weight d. G = {\displaystyle G={\begin{bmatrix}1&\dots &1&0&\dots &0\\\ast &\ast &\ast &&G'&\\\end{bmatrix}}} The matrix G ′ {\displaystyle G'} generates a code C ′ {\displaystyle C'} , which is called the residual code of C . {\displaystyle C.} C ′ {\displaystyle C'} obviously has dimension k ′ = k − 1 {\displaystyle k'=k-1} and length n ′ = N ( k , d ) − d . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A colleague challenges you to create a $(n-1,k,d_{min})$ code $\mathcal C'$ from a $(n,k,d_{min})$ code $\mathcal C$ as follows: given a generator matrix $G$ that generates $\mathcal C$, drop one column from $G$. Then, generate the new code with this truncated $k imes (n-1)$ generator matrix. The catch is that your colleague only gives you a set $\mathcal S=\{\vec s_1,\vec s_2, \vec s_3\}$ of $3$ columns of $G$ that you are allowed to drop, where $\vec s_1$ is the all-zeros vector, $\vec s_2$ is the all-ones vector, and $\vec s_3$ is a canonical basis vector. From the length of the columns $s_i$ you can infer $k$. You do not know $n$, neither do you know anything about the $n-3$ columns of $G$ that are not in $\mathcal S$. However, your colleague tells you that $G$ is in systematic form, i.e., $G=[I ~~ P]$ for some unknown $P$, and that all of the elements in $\mathcal S$ are columns of $P$. Which of the following options in $\mathcal S$ would you choose as the column of $G$ to drop? | The RM(2,3) code is generated by the set: { v 0 , v 1 , v 2 , v 3 , v 1 ∧ v 2 , v 1 ∧ v 3 , v 2 ∧ v 3 } {\displaystyle \{v_{0},v_{1},v_{2},v_{3},v_{1}\wedge v_{2},v_{1}\wedge v_{3},v_{2}\wedge v_{3}\}} or more explicitly by the rows of the matrix: ( 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 ) {\displaystyle {\begin{pmatrix}1&1&1&1&1&1&1&1\\1&0&1&0&1&0&1&0\\1&1&0&0&1&1&0&0\\1&1&1&1&0&0&0&0\\1&0&0&0&1&0&0&0\\1&0&1&0&0&0&0&0\\1&1&0&0&0&0&0&0\\\end{pmatrix}}} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A binary prefix-free code $\Gamma$ is made of four codewords. The first three codewords have codeword lengths $\ell_1 = 2$, $\ell_2 = 3$ and $\ell_3 = 3$. What is the minimum possible length for the fourth codeword? | Given A set of symbols and their weights (usually proportional to probabilities). Find A prefix-free binary code (a set of codewords) with minimum expected codeword length (equivalently, a tree with minimum weighted path length from the root). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A binary prefix-free code $\Gamma$ is made of four codewords. The first three codewords have codeword lengths $\ell_1 = 2$, $\ell_2 = 3$ and $\ell_3 = 3$. What is the minimum possible length for the fourth codeword? | To maintain the prefix-free property, B's codeword may not start 00, so the lexicographically first available word of length 3 is 010. Continuing like this, we get the following code: Alternatively, we can use the cumulative probability method. Note that although the codewords under the two methods are different, the word lengths are the same. We have lengths of 2 bits for A, and 3 bits for B, C, D and E, giving an average length of which is within one bit of the entropy. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n \geq 2$ be a positive integer, and $M$ a uniformly distributed binary message of length $2n$. Let $P_K(M)$ denote the one-time pad encryption of $M$ with key $K$. Let $K_1$ be a uniformly distributed binary key length $n$. Let $K_2$ be the complement of $K_1$. Let $K_3$ be the reverse of $K_1$. Let $K_i||K_j$ denote the concatenation of the two keys. True or false: Encryption with the key $K_4 = ( K_1|| K_1 ) $, $P_{K_4}(M)$ provides perfect secrecy. | Shannon proved, using information theoretic considerations, that the one-time pad has a property he termed perfect secrecy; that is, the ciphertext C gives absolutely no additional information about the plaintext. This is because (intuitively), given a truly uniformly random key that is used only once, a ciphertext can be translated into any plaintext of the same length, and all are equally likely. Thus, the a priori probability of a plaintext message M is the same as the a posteriori probability of a plaintext message M given the corresponding ciphertext. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n \geq 2$ be a positive integer, and $M$ a uniformly distributed binary message of length $2n$. Let $P_K(M)$ denote the one-time pad encryption of $M$ with key $K$. Let $K_1$ be a uniformly distributed binary key length $n$. Let $K_2$ be the complement of $K_1$. Let $K_3$ be the reverse of $K_1$. Let $K_i||K_j$ denote the concatenation of the two keys. True or false: Encryption with the key $K_4 = ( K_1|| K_1 ) $, $P_{K_4}(M)$ provides perfect secrecy. | The encryption exponent e and λ ( N ) {\displaystyle \lambda (N)} also must be relatively prime so that there is a modular inverse. The factorization of N and the private key d are kept secret, so that only Bob can decrypt the message. We denote the private key pair as (d, N). The encryption of the message M is given by C ≡ M e mod N {\displaystyle C\equiv M^{e}{\bmod {N}}} and the decryption of cipher text C {\displaystyle C} is given by C d ≡ ( M e ) d ≡ M e d ≡ M mod N {\displaystyle C^{d}\equiv (M^{e})^{d}\equiv M^{ed}\equiv M{\bmod {N}}} (using Fermat's little theorem). Using the Euclidean algorithm, one can efficiently recover the secret key d if one knows the factorization of N. By having the secret key d, one can efficiently factor the modulus of N. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n \geq 2$ be a positive integer, and $M$ a uniformly distributed binary message of length $2n$. Let $P_K(M)$ denote the one-time pad encryption of $M$ with key $K$. Let $K_1$ be a uniformly distributed binary key length $n$. Let $K_2$ be the complement of $K_1$. Let $K_3$ be the reverse of $K_1$. Let $K_i||K_j$ denote the concatenation of the two keys. True or false: Encryption with the key $K_6 = ( K_1|| K_3 ) $, $P_{K_6}(M)$ provides perfect secrecy. | Shannon proved, using information theoretic considerations, that the one-time pad has a property he termed perfect secrecy; that is, the ciphertext C gives absolutely no additional information about the plaintext. This is because (intuitively), given a truly uniformly random key that is used only once, a ciphertext can be translated into any plaintext of the same length, and all are equally likely. Thus, the a priori probability of a plaintext message M is the same as the a posteriori probability of a plaintext message M given the corresponding ciphertext. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $n \geq 2$ be a positive integer, and $M$ a uniformly distributed binary message of length $2n$. Let $P_K(M)$ denote the one-time pad encryption of $M$ with key $K$. Let $K_1$ be a uniformly distributed binary key length $n$. Let $K_2$ be the complement of $K_1$. Let $K_3$ be the reverse of $K_1$. Let $K_i||K_j$ denote the concatenation of the two keys. True or false: Encryption with the key $K_6 = ( K_1|| K_3 ) $, $P_{K_6}(M)$ provides perfect secrecy. | Another way of stating perfect secrecy is that for all messages m 1 , m 2 {\displaystyle m_{1},m_{2}} in message space M, and for all ciphers c in cipher space C, we have Pr k ⇐ K = Pr k ⇐ K {\displaystyle {\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}={\underset {k\Leftarrow \mathrm {K} }{\operatorname {Pr} }}} , where Pr {\textstyle \operatorname {Pr} } represents the probabilities, taken over a choice of k {\displaystyle k} in key space K {\displaystyle \mathrm {K} } over the coin tosses of a probabilistic algorithm, E {\displaystyle E} . Perfect secrecy is a strong notion of cryptanalytic difficulty.Conventional symmetric encryption algorithms use complex patterns of substitution and transpositions. For the best of these currently in use, it is not known whether there can be a cryptanalytic procedure that can efficiently reverse (or even partially reverse) these transformations without knowing the key used during encryption. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (7, 223)$ are valid encoding/decoding exponents. | Thus any d satisfying d⋅e ≡ 1 (mod φ(n)) also satisfies d⋅e ≡ 1 (mod λ(n)). However, computing d modulo φ(n) will sometimes yield a result that is larger than necessary (i.e. d > λ(n)). Most of the implementations of RSA will accept exponents generated using either method (if they use the private exponent d at all, rather than using the optimized decryption method based on the Chinese remainder theorem described below), but some standards such as FIPS 186-4 (Section B.3.1) may require that d < λ(n). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (7, 223)$ are valid encoding/decoding exponents. | 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 |
Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (319, 23)$ are valid encoding/decoding exponents. | Thus any d satisfying d⋅e ≡ 1 (mod φ(n)) also satisfies d⋅e ≡ 1 (mod λ(n)). However, computing d modulo φ(n) will sometimes yield a result that is larger than necessary (i.e. d > λ(n)). Most of the implementations of RSA will accept exponents generated using either method (if they use the private exponent d at all, rather than using the optimized decryption method based on the Chinese remainder theorem described below), but some standards such as FIPS 186-4 (Section B.3.1) may require that d < λ(n). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider an RSA encryption where the $(p, q)$ are determined as $(53, 61)$. True or false: $(e,d) = (319, 23)$ are valid encoding/decoding exponents. | 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 |
Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: $(G,*)$ is a group if and only if $b=d.$ | Nevertheless, it is a monomorphism in this category. This follows from the implication q ∘ h = 0 ⇒ h = 0, which we will now prove. If h: G → Q, where G is some divisible group, and q ∘ h = 0, then h(x) ∈ Z, ∀ x ∈ G. Now fix some x ∈ G. Without loss of generality, we may assume that h(x) ≥ 0 (otherwise, choose −x instead). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: $(G,*)$ is a group if and only if $b=d.$ | The general theory of abstract algebra allows an "addition" operation to be any associative and commutative operation on a set. Basic algebraic structures with such an addition operation include commutative monoids and abelian groups. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: If $b
ot=d,$ then $(G,*)$ cannot be a group. | Nevertheless, it is a monomorphism in this category. This follows from the implication q ∘ h = 0 ⇒ h = 0, which we will now prove. If h: G → Q, where G is some divisible group, and q ∘ h = 0, then h(x) ∈ Z, ∀ x ∈ G. Now fix some x ∈ G. Without loss of generality, we may assume that h(x) ≥ 0 (otherwise, choose −x instead). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G$ be a set and $*$ a commutative operation on pairs of elements from $G.$ Suppose there exists an element $e\in G$ such that $a*e=e*a=a$ for all $a \in G.$ Also, suppose there exist elements $b,c,d \in G$ such that $b*c=d*c.$. True or false: If $b
ot=d,$ then $(G,*)$ cannot be a group. | The general theory of abstract algebra allows an "addition" operation to be any associative and commutative operation on a set. Basic algebraic structures with such an addition operation include commutative monoids and abelian groups. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G_1, G_2, G_3$, be valid generator matrices of dimensions $\mathbb F^{k_i imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $ ext{rank}(G_i) = k_i$. True or false: Assuming $k_1 = k_2 + k_3$, the matrix $egin{pmatrix} G_1 &\vline &egin{matrix} G_2 &\vline &0\ \cline{1-3} 0 &\vline &G_3 \end{matrix} \end{pmatrix}$ is also a valid generator matrix. | Its generators are the matrices g k = ( 1 + 2 ( 2 + 2 ) α e i k π 4 ( 2 + 2 ) α e − i k π 4 1 + 2 ) , {\displaystyle g_{k}={\begin{pmatrix}1+{\sqrt {2}}&(2+{\sqrt {2}})\alpha e^{\tfrac {ik\pi }{4}}\\(2+{\sqrt {2}})\alpha e^{-{\tfrac {ik\pi }{4}}}&1+{\sqrt {2}}\end{pmatrix}},} where α = 2 − 1 {\displaystyle \alpha ={\sqrt {{\sqrt {2}}-1}}} and k = 0 , … , 3 {\displaystyle k=0,\ldots ,3} , along with their inverses. The generators satisfy the relation g 0 g 1 − 1 g 2 g 3 − 1 g 0 − 1 g 1 g 2 − 1 g 3 = 1. {\displaystyle g_{0}g_{1}^{-1}g_{2}g_{3}^{-1}g_{0}^{-1}g_{1}g_{2}^{-1}g_{3}=1.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G_1, G_2, G_3$, be valid generator matrices of dimensions $\mathbb F^{k_i imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $ ext{rank}(G_i) = k_i$. True or false: Assuming $k_1 = k_2 + k_3$, the matrix $egin{pmatrix} G_1 &\vline &egin{matrix} G_2 &\vline &0\ \cline{1-3} 0 &\vline &G_3 \end{matrix} \end{pmatrix}$ is also a valid generator matrix. | {\displaystyle {\begin{pmatrix}a&0\\0&{\frac {1}{a}}\\\end{pmatrix}},a\in \mathbb {R} ^{\times };\quad {\begin{pmatrix}1&x\\0&1\\\end{pmatrix}},x\in \mathbb {R} ;\quad S={\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}}.} One calculates the claim for these generators and obtains the claim for all g ∈ S L 2 ( R ) {\displaystyle g\in \mathrm {SL} _{2}(\mathbb {R} )} . Q.E.D. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G$, be a valid generator matrix of dimensions $\mathbb F^{k imes n}$. Recall that ``valid'' means that $k \leq n$ and $ ext{rank}(G) = k$. Let $D_1 \in \mathbb F^{k imes k}$ and $D_2 \in \mathbb F^{n imes n}$ be diagonal matrices with non-zero diagonal elements. True or false: $D_1 \cdot G \cdot D_2$ is also a valid generator matrix. | Its generators are the matrices g k = ( 1 + 2 ( 2 + 2 ) α e i k π 4 ( 2 + 2 ) α e − i k π 4 1 + 2 ) , {\displaystyle g_{k}={\begin{pmatrix}1+{\sqrt {2}}&(2+{\sqrt {2}})\alpha e^{\tfrac {ik\pi }{4}}\\(2+{\sqrt {2}})\alpha e^{-{\tfrac {ik\pi }{4}}}&1+{\sqrt {2}}\end{pmatrix}},} where α = 2 − 1 {\displaystyle \alpha ={\sqrt {{\sqrt {2}}-1}}} and k = 0 , … , 3 {\displaystyle k=0,\ldots ,3} , along with their inverses. The generators satisfy the relation g 0 g 1 − 1 g 2 g 3 − 1 g 0 − 1 g 1 g 2 − 1 g 3 = 1. {\displaystyle g_{0}g_{1}^{-1}g_{2}g_{3}^{-1}g_{0}^{-1}g_{1}g_{2}^{-1}g_{3}=1.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G$, be a valid generator matrix of dimensions $\mathbb F^{k imes n}$. Recall that ``valid'' means that $k \leq n$ and $ ext{rank}(G) = k$. Let $D_1 \in \mathbb F^{k imes k}$ and $D_2 \in \mathbb F^{n imes n}$ be diagonal matrices with non-zero diagonal elements. True or false: $D_1 \cdot G \cdot D_2$ is also a valid generator matrix. | {\displaystyle {\begin{pmatrix}a&0\\0&{\frac {1}{a}}\\\end{pmatrix}},a\in \mathbb {R} ^{\times };\quad {\begin{pmatrix}1&x\\0&1\\\end{pmatrix}},x\in \mathbb {R} ;\quad S={\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}}.} One calculates the claim for these generators and obtains the claim for all g ∈ S L 2 ( R ) {\displaystyle g\in \mathrm {SL} _{2}(\mathbb {R} )} . Q.E.D. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G_1, G_2$, be valid generator matrices of dimensions $\mathbb F^{k_i imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $ ext{rank}(G_i) = k_i$. True or false: Assuming $k_1 = k_2$ and $n_1=n_2$, the matrix $G_{7} + G_{8}$ is also a valid generator matrix. | Its generators are the matrices g k = ( 1 + 2 ( 2 + 2 ) α e i k π 4 ( 2 + 2 ) α e − i k π 4 1 + 2 ) , {\displaystyle g_{k}={\begin{pmatrix}1+{\sqrt {2}}&(2+{\sqrt {2}})\alpha e^{\tfrac {ik\pi }{4}}\\(2+{\sqrt {2}})\alpha e^{-{\tfrac {ik\pi }{4}}}&1+{\sqrt {2}}\end{pmatrix}},} where α = 2 − 1 {\displaystyle \alpha ={\sqrt {{\sqrt {2}}-1}}} and k = 0 , … , 3 {\displaystyle k=0,\ldots ,3} , along with their inverses. The generators satisfy the relation g 0 g 1 − 1 g 2 g 3 − 1 g 0 − 1 g 1 g 2 − 1 g 3 = 1. {\displaystyle g_{0}g_{1}^{-1}g_{2}g_{3}^{-1}g_{0}^{-1}g_{1}g_{2}^{-1}g_{3}=1.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $G_1, G_2$, be valid generator matrices of dimensions $\mathbb F^{k_i imes n_i}$, all over the same field $\mathbb F$. Recall that ``valid'' means that for all $i$, $k_i \leq n_i$ and $ ext{rank}(G_i) = k_i$. True or false: Assuming $k_1 = k_2$ and $n_1=n_2$, the matrix $G_{7} + G_{8}$ is also a valid generator matrix. | {\displaystyle {\begin{pmatrix}a&0\\0&{\frac {1}{a}}\\\end{pmatrix}},a\in \mathbb {R} ^{\times };\quad {\begin{pmatrix}1&x\\0&1\\\end{pmatrix}},x\in \mathbb {R} ;\quad S={\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}}.} One calculates the claim for these generators and obtains the claim for all g ∈ S L 2 ( R ) {\displaystyle g\in \mathrm {SL} _{2}(\mathbb {R} )} . Q.E.D. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal C_1$ be a linear code over $\mathbb F_3^n$, and let $\mathcal C_2$ be a linear code over $\mathbb F_2^n$. True or false: $\mathcal C_1 \cap \mathcal C_2$ is necessarily a linear code over $\mathbb F_3^n$. | Let C ⊆ F 2 n {\displaystyle C\subseteq F_{2}^{n}} be a linear code such that C ⊥ {\displaystyle C^{\perp }} has distance greater than ℓ + 1 {\displaystyle \ell +1} . Then C {\displaystyle C} is an ℓ {\displaystyle \ell } -wise independent source. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathcal C_1$ be a linear code over $\mathbb F_3^n$, and let $\mathcal C_2$ be a linear code over $\mathbb F_2^n$. True or false: $\mathcal C_1 \cap \mathcal C_2$ is necessarily a linear code over $\mathbb F_3^n$. | The construction thus gives the linear map: (cf. Lemma 1) C k G → H 2 k ( M ; C ) , f ↦ . {\displaystyle \mathbb {C} _{k}^{G}\to H^{2k}(M;\mathbb {C} ),\,f\mapsto \left.} In fact, one can check that the map thus obtained: C G → H ∗ ( M ; C ) {\displaystyle \mathbb {C} ^{G}\to H^{*}(M;\mathbb {C} )} is an algebra homomorphism. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $$G= egin{pmatrix} 1 &1 &1 &0 &1 &0\ 0 &1 &1 &1 &0 &0\ 0 &1 &1 &0 &0 &0\ 0 &1 &1 &1 &0 &1 \end{pmatrix}$$ be the generator matrix of a $(6,4)$ linear code $\mathcal C$ over $\mathbb F_2$. True or false: $G$ admits a systematic form (i.e., it can be put into systematic form via elementary row operations). | Let G be a generator matrix of C. We can always suppose that the first row of G is of the form r = (1, ..., 1, 0, ..., 0) with weight d. G = {\displaystyle G={\begin{bmatrix}1&\dots &1&0&\dots &0\\\ast &\ast &\ast &&G'&\\\end{bmatrix}}} The matrix G ′ {\displaystyle G'} generates a code C ′ {\displaystyle C'} , which is called the residual code of C . {\displaystyle C.} C ′ {\displaystyle C'} obviously has dimension k ′ = k − 1 {\displaystyle k'=k-1} and length n ′ = N ( k , d ) − d . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $$G= egin{pmatrix} 1 &1 &1 &0 &1 &0\ 0 &1 &1 &1 &0 &0\ 0 &1 &1 &0 &0 &0\ 0 &1 &1 &1 &0 &1 \end{pmatrix}$$ be the generator matrix of a $(6,4)$ linear code $\mathcal C$ over $\mathbb F_2$. True or false: $G$ admits a systematic form (i.e., it can be put into systematic form via elementary row operations). | Using the systematic construction for Hamming codes from above, the matrix A is apparent and the systematic form of G is written as G = ( 1 0 0 0 0 1 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 0 0 0 1 1 1 1 0 ) 4 , 8 . {\displaystyle \mathbf {G} =\left({\begin{array}{cccc|cccc}1&0&0&0&0&1&1&1\\0&1&0&0&1&0&1&1\\0&0&1&0&1&1&0&1\\0&0&0&1&1&1&1&0\end{array}}\right)_{4,8}.} The non-systematic form of G can be row reduced (using elementary row operations) to match this matrix. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $$G= egin{pmatrix} 1 &1 &1 &0 &1 &0\ 0 &1 &1 &1 &0 &0\ 0 &1 &1 &0 &0 &0\ 0 &1 &1 &1 &0 &1 \end{pmatrix}$$ be the generator matrix of a $(6,4)$ linear code $\mathcal C$ over $\mathbb F_2$. True or false: If one substitutes the last row of $G$ by $(1,0,0,1,1,1)$, the thereby obtained matrix generates the same code $\mathcal C$. | Let G be a generator matrix of C. We can always suppose that the first row of G is of the form r = (1, ..., 1, 0, ..., 0) with weight d. G = {\displaystyle G={\begin{bmatrix}1&\dots &1&0&\dots &0\\\ast &\ast &\ast &&G'&\\\end{bmatrix}}} The matrix G ′ {\displaystyle G'} generates a code C ′ {\displaystyle C'} , which is called the residual code of C . {\displaystyle C.} C ′ {\displaystyle C'} obviously has dimension k ′ = k − 1 {\displaystyle k'=k-1} and length n ′ = N ( k , d ) − d . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $$G= egin{pmatrix} 1 &1 &1 &0 &1 &0\ 0 &1 &1 &1 &0 &0\ 0 &1 &1 &0 &0 &0\ 0 &1 &1 &1 &0 &1 \end{pmatrix}$$ be the generator matrix of a $(6,4)$ linear code $\mathcal C$ over $\mathbb F_2$. True or false: If one substitutes the last row of $G$ by $(1,0,0,1,1,1)$, the thereby obtained matrix generates the same code $\mathcal C$. | Let C ( x ) = c = ( c 0 , … , c 2 n − 1 ) {\displaystyle C(x)=c=(c_{0},\dots ,c_{2^{n}-1})} be the codeword in C {\displaystyle C} corresponding to message x {\displaystyle x} . Let G = ( ↑ ↑ ↑ g 0 g 1 … g 2 n − 1 ↓ ↓ ↓ ) {\displaystyle G={\begin{pmatrix}\uparrow &\uparrow &&\uparrow \\g_{0}&g_{1}&\dots &g_{2^{n}-1}\\\downarrow &\downarrow &&\downarrow \end{pmatrix}}} be the generator matrix of C {\displaystyle C} . By definition, c i = x ⋅ g i {\displaystyle c_{i}=x\cdot g_{i}} . From this, c i + c j = x ⋅ g i + x ⋅ g j = x ⋅ ( g i + g j ) {\displaystyle c_{i}+c_{j}=x\cdot g_{i}+x\cdot g_{j}=x\cdot (g_{i}+g_{j})} . By the construction of G {\displaystyle G} , g i + g j = g i + j {\displaystyle g_{i}+g_{j}=g_{i+j}} . Therefore, by substitution, c i + c j = x ⋅ g i + j = c i + j {\displaystyle c_{i}+c_{j}=x\cdot g_{i+j}=c_{i+j}} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathbb F$ be a field of cardinality $q$ and let $0<k<n\leq q$ be unspecified integers. As seen in the lecture, we generate a $(n,k,d_{min})$ Reed-Solomon code with the following mapping: $$\mathbb F^k
ightarrow \mathbb F^n ~~,~~ \vec u \mapsto \vec c =(P_{\vec u}(a_1),P_{\vec u}(a_2),\ldots,P_{\vec u}(a_n))$$ for $a_i \in \mathbb F$ all distinct and $P$ a polynomial of degree $k-1$ with coefficient vector $\vec u\in\mathbb F^k$. Now, we construct a $(n,k',d'_{min})$ code $\mathcal C'$ similarly to the above one by assigning $a_1\leftarrow a_2$ while leaving $n,P$ and $a_2,\ldots,a_n$ unchanged. As before, the code is generated by evaluating $P_{\vec u}(a_2,a_2,a_3,\dots,a_n)$ over all possible coefficients vectors $\vec u \in \mathbb F^k$. This is by definition not an RS code, however it is still a well-defined linear block code. True or false: We know for certain that $d'_{min}=d_{min}-1$. | Consider a ( n , k ) {\displaystyle (n,k)} Reed–Solomon code over the finite field F = G F ( q ) {\displaystyle \mathbb {F} =GF(q)} with evaluation set ( α 1 , α 2 , … , α n ) {\displaystyle (\alpha _{1},\alpha _{2},\ldots ,\alpha _{n})} and a positive integer r {\displaystyle r} , the Guruswami-Sudan List Decoder accepts a vector β = ( β 1 , β 2 , … , β n ) {\displaystyle \beta =(\beta _{1},\beta _{2},\ldots ,\beta _{n})} ∈ {\displaystyle \in } F n {\displaystyle \mathbb {F} ^{n}} as input, and outputs a list of polynomials of degree ≤ k {\displaystyle \leq k} which are in 1 to 1 correspondence with codewords. The idea is to add more restrictions on the bi-variate polynomial Q ( x , y ) {\displaystyle Q(x,y)} which results in the increment of constraints along with the number of roots. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $\mathbb F$ be a field of cardinality $q$ and let $0<k<n\leq q$ be unspecified integers. As seen in the lecture, we generate a $(n,k,d_{min})$ Reed-Solomon code with the following mapping: $$\mathbb F^k
ightarrow \mathbb F^n ~~,~~ \vec u \mapsto \vec c =(P_{\vec u}(a_1),P_{\vec u}(a_2),\ldots,P_{\vec u}(a_n))$$ for $a_i \in \mathbb F$ all distinct and $P$ a polynomial of degree $k-1$ with coefficient vector $\vec u\in\mathbb F^k$. Now, we construct a $(n,k',d'_{min})$ code $\mathcal C'$ similarly to the above one by assigning $a_1\leftarrow a_2$ while leaving $n,P$ and $a_2,\ldots,a_n$ unchanged. As before, the code is generated by evaluating $P_{\vec u}(a_2,a_2,a_3,\dots,a_n)$ over all possible coefficients vectors $\vec u \in \mathbb F^k$. This is by definition not an RS code, however it is still a well-defined linear block code. True or false: We know for certain that $d'_{min}=d_{min}-1$. | Using low-degree polynomials over a finite field F {\displaystyle \mathbb {F} } of size q {\displaystyle q} , it is possible to extend the definition of Reed–Muller codes to alphabets of size q {\displaystyle q} . Let m {\displaystyle m} and d {\displaystyle d} be positive integers, where m {\displaystyle m} should be thought of as larger than d {\displaystyle d} . To encode a message x ∈ F k {\textstyle x\in \mathbb {F} ^{k}} of width k = ( m + d m ) {\displaystyle k=\textstyle {\binom {m+d}{m}}} , the message is again interpreted as an m {\displaystyle m} -variate polynomial p x {\displaystyle p_{x}} of total degree at most d {\displaystyle d} and with coefficient from F {\displaystyle \mathbb {F} } . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the source $S_1, S_2, \dots$ such that $S_1$ is uniformly distributed on $\mathbb{Z}/10\mathbb{Z}^*$, and for every $n\geq 1$, $S_{n+1}$ is distributed uniformly on $\mathbb{Z}/(S_n+1)\mathbb{Z}^*$. Let $H(\mathcal{S}) = \lim_{n o\infty} H(S_n)$. True or false: $H(\mathcal{S}) = 0$. | A sequence (s1, s2, s3, ...) of real numbers is said to be well-distributed on if for any subinterval of we have lim n → ∞ | { s k + 1 , … , s k + n } ∩ | n = d − c b − a {\displaystyle \lim _{n\to \infty }{\left|\{\,s_{k+1},\dots ,s_{k+n}\,\}\cap \right| \over n}={d-c \over b-a}} uniformly in k. Clearly every well-distributed sequence is uniformly distributed, but the converse does not hold. The definition of well-distributed modulo 1 is analogous. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Consider the source $S_1, S_2, \dots$ such that $S_1$ is uniformly distributed on $\mathbb{Z}/10\mathbb{Z}^*$, and for every $n\geq 1$, $S_{n+1}$ is distributed uniformly on $\mathbb{Z}/(S_n+1)\mathbb{Z}^*$. Let $H(\mathcal{S}) = \lim_{n o\infty} H(S_n)$. True or false: $H(\mathcal{S}) = 0$. | The multiplication theorem for the Hurwitz zeta function ζ ( s , a ) = ∑ n = 0 ∞ ( n + a ) − s {\displaystyle \zeta (s,a)=\sum _{n=0}^{\infty }(n+a)^{-s}} gives a distribution relation ∑ p = 0 q − 1 ζ ( s , a + p / q ) = q s ζ ( s , q a ) . {\displaystyle \sum _{p=0}^{q-1}\zeta (s,a+p/q)=q^{s}\,\zeta (s,qa)\ .} Hence for given s, the map t ↦ ζ ( s , { t } ) {\displaystyle t\mapsto \zeta (s,\{t\})} is a distribution on Q/Z. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_1$ be a random variable taking values in $\{a,b\}$ with probability $p_{S_1}(a) = rac{1}{4}$ and $p_{S_1}(b) = rac{3}{4}$. Let $S_2$ be a random variable, independent of $S_1$, taking values in $\{c,d\}$ with probability $p_{S_2}(c) = q$ and $p_{S_2}(d) = 1-q$, for some $q\in[0,1]$. Let $\Gamma_H$ be the binary Huffman code for the sequence $S = S_1 S_2$, and let $L(S,\Gamma_H)$ be the average codeword-length of $\Gamma_H$. True or false: $1\leq L(S, \Gamma_H) \leq 2$ for all $q\in[0,1]$. | Let Σ1, Σ2 denote two finite alphabets and let Σ∗1 and Σ∗2 denote the set of all finite words from those alphabets (respectively). Suppose that X is a random variable taking values in Σ1 and let f be a uniquely decodable code from Σ∗1 to Σ∗2 where |Σ2| = a. Let S denote the random variable given by the length of codeword f (X). If f is optimal in the sense that it has the minimal expected word length for X, then (Shannon 1948): H ( X ) log 2 a ≤ E < H ( X ) log 2 a + 1 {\displaystyle {\frac {H(X)}{\log _{2}a}}\leq \mathbb {E} <{\frac {H(X)}{\log _{2}a}}+1} Where E {\displaystyle \mathbb {E} } denotes the expected value operator. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S_1$ be a random variable taking values in $\{a,b\}$ with probability $p_{S_1}(a) = rac{1}{4}$ and $p_{S_1}(b) = rac{3}{4}$. Let $S_2$ be a random variable, independent of $S_1$, taking values in $\{c,d\}$ with probability $p_{S_2}(c) = q$ and $p_{S_2}(d) = 1-q$, for some $q\in[0,1]$. Let $\Gamma_H$ be the binary Huffman code for the sequence $S = S_1 S_2$, and let $L(S,\Gamma_H)$ be the average codeword-length of $\Gamma_H$. True or false: $1\leq L(S, \Gamma_H) \leq 2$ for all $q\in[0,1]$. | The average code length is L C ( X ) = ∑ x ∈ X p ( x ) L ( x ) = ∑ x ∈ X p ( x ) ( ⌈ log 2 1 p ( x ) ⌉ + 1 ) {\displaystyle LC(X)=\sum _{x\in X}p(x)L(x)=\sum _{x\in X}p(x)\left(\left\lceil \log _{2}{\frac {1}{p(x)}}\right\rceil +1\right)} . Thus for H(X), the entropy of the random variable X, H ( X ) + 1 ≤ L C ( X ) < H ( X ) + 2 {\displaystyle H(X)+1\leq LC(X) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S$ be a random variable taking values in $\{a,b,c,d,e\}$ with the following probabilities. $$egin{array}{|c|c|c|c|c|c|} \hline & a & b & c & d & e \ \hline p_S(\cdot) & 1/3 & 1/3 & 1/9 & 1/9 & 1/9\ \hline \end{array}$$ Let $\Gamma_D$ be the $D$-ary Huffman code for $S$. Let $L(S,\Gamma_D)$ be the average codeword-length of $\Gamma_D$, and let $H_D(S)$ be the $D$-ary entropy of $S$. True or false: If $D=3$, then $L(S,\Gamma_D) = H_D(S)$. | The average code length is L C ( X ) = ∑ x ∈ X p ( x ) L ( x ) = ∑ x ∈ X p ( x ) ( ⌈ log 2 1 p ( x ) ⌉ + 1 ) {\displaystyle LC(X)=\sum _{x\in X}p(x)L(x)=\sum _{x\in X}p(x)\left(\left\lceil \log _{2}{\frac {1}{p(x)}}\right\rceil +1\right)} . Thus for H(X), the entropy of the random variable X, H ( X ) + 1 ≤ L C ( X ) < H ( X ) + 2 {\displaystyle H(X)+1\leq LC(X) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $S$ be a random variable taking values in $\{a,b,c,d,e\}$ with the following probabilities. $$egin{array}{|c|c|c|c|c|c|} \hline & a & b & c & d & e \ \hline p_S(\cdot) & 1/3 & 1/3 & 1/9 & 1/9 & 1/9\ \hline \end{array}$$ Let $\Gamma_D$ be the $D$-ary Huffman code for $S$. Let $L(S,\Gamma_D)$ be the average codeword-length of $\Gamma_D$, and let $H_D(S)$ be the $D$-ary entropy of $S$. True or false: If $D=3$, then $L(S,\Gamma_D) = H_D(S)$. | We give an example of the result of Huffman coding for a code with five characters and given weights. We will not verify that it minimizes L over all codes, but we will compute L and compare it to the Shannon entropy H of the given set of weights; the result is nearly optimal. For any code that is biunique, meaning that the code is uniquely decodeable, the sum of the probability budgets across all symbols is always less than or equal to one. In this example, the sum is strictly equal to one; as a result, the code is termed a complete code. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
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