question stringlengths 6 3.53k | text stringlengths 17 2.05k | source stringclasses 1
value |
|---|---|---|
Let $E$ and $F$ be two events. Suppose that they satisfy $p(E|F)=p(E) > 0.$ True or false: Then we must have $p(F|E)=p(F).$ | IF p1 is false, the value of f is given by e2. This conditional expression . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let $E$ and $F$ be two events. Suppose that they satisfy $p(E|F)=p(E) > 0.$ True or false: Then we must have $p(F|E)=p(F).$ | Assuming that P ( A → B ) = P ( B ∣ A ) {\displaystyle P(A\rightarrow B)=P(B\mid A)} holds for a minimally nontrivial set of events and for any P {\displaystyle P} -function leads to a contradiction. Thus P ( A → B ) = P ( B ∣ A ) {\displaystyle P(A\rightarrow B)=P(B\mid A)} can hold for any P {\displaystyle P} -function only for trivial sets of events—that is the triviality result. However, the proof relies on background assumptions that may be challenged. It may be proposed, for instance, that the referent event of an expression like “ A → B {\displaystyle A\rightarrow B} ” is not fixed for a given A {\displaystyle A} and B {\displaystyle B} , but instead changes as the probability function changes. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Determine which of the following conditional statements evaluate to true (more than one answer can be correct): | Let s be the conditional statement if (e) s1 else s2. Then before(e) = before(s), before(s1) = true(e), before(s2) = false(e), and after(s) = after(s1) intersect after(s2). Let s be the while loop statement while (e) s1. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Determine which of the following conditional statements evaluate to true (more than one answer can be correct): | With the following syntax, both expressions are evaluated (with value_if_false evaluated first, then condition, then value_if_false): This alternative syntax provides short-circuit evaluation: | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Determine which of the following compound propositions are satisfiable (more than one answer can be correct): | The problem of determining whether a formula in propositional logic is satisfiable is decidable, and is known as the Boolean satisfiability problem, or SAT. In general, the problem of determining whether a sentence of first-order logic is satisfiable is not decidable. In universal algebra, equational theory, and automated theorem proving, the methods of term rewriting, congruence closure and unification are used to attempt to decide satisfiability. Whether a particular theory is decidable or not depends whether the theory is variable-free and on other conditions. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Determine which of the following compound propositions are satisfiable (more than one answer can be correct): | The satisfiability problem for all the logics introduced is decidable. We list below the computational complexity of the satisfiability problem for each of them. Note that it becomes linear in time if there are only finitely many propositional letters in the language. For n ≥ 2 {\displaystyle n\geq 2} , if we restrict to finite nesting, then the satisfiability problem is NP-complete for all the modal logics considered. If we then further restrict the language to having only finitely many primitive propositions, the complexity goes down to linear in time in all cases. The computational complexity of the model checking problem is in P in all cases. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let P be the statement ∀x(x>-3 -> x>3). Determine for which domain P evaluates to true: | . ≡ P ( t 1 , … , t n ) ∣ A ∧ B ∣ ⊤ ∣ A ∨ B ∣ ⊥ ∣ A ⊃ B ∣ ∀ x . A ∣ ∃ x . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let P be the statement ∀x(x>-3 -> x>3). Determine for which domain P evaluates to true: | x . P ( x ) ) := ( ∀ a ∀ b . P ( a ) ∧ P ( b ) → a = b ) {\displaystyle (!x.P(x)):=(\forall a\forall b.P(a)\land P(b)\rightarrow a=b)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let P(x) is “x is an elephant” and F(x) is “x flies” and the domain consists of all animals. Translate the following statement into English: ∃!x(P(x) ∧ F(x)) | In other words, given a domain of discourse "winged things", p is either found to be a member of this domain or not. Thus there is a relationship W (wingedness) between p (pig) and { T, F }, W(p) evaluates to { T, F } where { T, F } is the set of the boolean values "true" and "false". | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let P(x) is “x is an elephant” and F(x) is “x flies” and the domain consists of all animals. Translate the following statement into English: ∃!x(P(x) ∧ F(x)) | No mammal is a bird. Thus, no dog is a bird.can be notated in the language of monadic predicate calculus as ⇒ ¬ ( ∃ z D ( z ) ∧ B ( z ) ) {\displaystyle \Rightarrow \neg (\exists z\,D(z)\land B(z))} where D {\displaystyle D} , M {\displaystyle M} and B {\displaystyle B} denote the predicates of being, respectively, a dog, a mammal, and a bird. Conversely, monadic predicate calculus is not significantly more expressive than term logic. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let p(x,y) be the statement “x visits y”, where the domain of x consists of all the humans in the world and the domain of y consists of all the places in the world. Use quantifiers to express the following statement: There is a place in the world that has never been visited by humans. | If D is a domain of x and P(x) is a predicate dependent on object variable x, then the universal proposition can be expressed as ∀ x ∈ D P ( x ) . {\displaystyle \forall x\!\in \!D\;P(x).} This notation is known as restricted or relativized or bounded quantification. Equivalently one can write, ∀ x ( x ∈ D → P ( x ) ) . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let p(x,y) be the statement “x visits y”, where the domain of x consists of all the humans in the world and the domain of y consists of all the places in the world. Use quantifiers to express the following statement: There is a place in the world that has never been visited by humans. | This applies to proper names as well as definite descriptions, and functional expressions. Quantifiers, on the other hand, are treated in the usual way as ranging over the domain. This allows for expressions like " ¬ ∃ x ( x = s a n t a ) {\displaystyle \lnot \exists x(x=santa)} " (Santa Claus does not exist) to be true even though they are self-contradictory in classical logic. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following arguments is correct? | Arguments can be either correct or incorrect. An argument is correct if its premises support its conclusion. Deductive arguments have the strongest form of support: if their premises are true then their conclusion must also be true. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following arguments is correct? | Invalid argument, as it is possible that the premises be true and the conclusion false.In the above second to last case (Some men are hawkers ...), the counter-example follows the same logical form as the previous argument, (Premise 1: "Some X are Y." Premise 2: "Some Y are Z." Conclusion: "Some X are Z.") | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
You are given the following collection of premises:
If I go to the museum, it either rains or snows.
I went to the museum on Saturday or I went to the museum on Sunday.
It did not rain and it did not snow on Saturday.
It did not rain on Sunday.
Which conclusions can be drawn from these premises ? (more than one answer can be correct) | An example of this fallacy is as follows: If it is raining, then I have my umbrella. If I have my umbrella, then it is raining.While this may appear to be a reasonable argument, it is not valid because the first statement does not logically guarantee the second statement. The first statement says nothing like "I do not have my umbrella otherwise", which means that having my umbrella on a sunny day would render the first statement true and the second statement false. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
You are given the following collection of premises:
If I go to the museum, it either rains or snows.
I went to the museum on Saturday or I went to the museum on Sunday.
It did not rain and it did not snow on Saturday.
It did not rain on Sunday.
Which conclusions can be drawn from these premises ? (more than one answer can be correct) | (premise) The streets are wet. (premise) Therefore it has rained recently. (conclusion)This argument is logically valid, but quite demonstrably wrong, because its first premise is false — a street cleaner may have passed, the local river could have flooded, etc. A simple logical analysis will not reveal the error in this argument, since that analysis must accept the truth of the argument's premises. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose we have the following function \(f: [0, 2] o [-\pi, \pi] \).
\[f(x) =
egin{cases}
x^2 & ext{ for } 0\leq x < 1\
2-(x-2)^2 & ext{ for } 1 \leq x \leq 2
\end{cases}
\] | Consider the function f ( x ) = { e − 1 x if x > 0 , 0 if x ≤ 0 , {\displaystyle f(x)={\begin{cases}e^{-{\frac {1}{x}}}&{\text{if }}x>0,\\0&{\text{if }}x\leq 0,\end{cases}}} defined for every real number x. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Suppose we have the following function \(f: [0, 2] o [-\pi, \pi] \).
\[f(x) =
egin{cases}
x^2 & ext{ for } 0\leq x < 1\
2-(x-2)^2 & ext{ for } 1 \leq x \leq 2
\end{cases}
\] | A function such as f ( x 1 , x 2 , . . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following functions \( f :\mathbb{Z} imes \mathbb{Z} o \mathbb{Z} \) are surjective? | For any set X, the identity function idX on X is surjective. The function f: Z → {0, 1} defined by f(n) = n mod 2 (that is, even integers are mapped to 0 and odd integers to 1) is surjective. The function f: R → R defined by f(x) = 2x + 1 is surjective (and even bijective), because for every real number y, we have an x such that f(x) = y: such an appropriate x is (y − 1)/2. The function f: R → R defined by f(x) = x3 − 3x is surjective, because the pre-image of any real number y is the solution set of the cubic polynomial equation x3 − 3x − y = 0, and every cubic polynomial with real coefficients has at least one real root. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following functions \( f :\mathbb{Z} imes \mathbb{Z} o \mathbb{Z} \) are surjective? | In mathematics, a surjective function (also known as surjection, or onto function ) is a function f such that every element y can be mapped from some element x such that f(x) = y. In other words, every element of the function's codomain is the image of at least one element of its domain. It is not required that x be unique; the function f may map one or more elements of X to the same element of Y. The term surjective and the related terms injective and bijective were introduced by Nicolas Bourbaki, a group of mainly French 20th-century mathematicians who, under this pseudonym, wrote a series of books presenting an exposition of modern advanced mathematics, beginning in 1935. The French word sur means over or above, and relates to the fact that the image of the domain of a surjective function completely covers the function's codomain. Any function induces a surjection by restricting its codomain to the image of its domain. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \(A = \{a, b, c, d, ..., z\}\) be the set of lower cased English letters.
Let \(S = \{a, b, ab, cd, ae, xy, ord, ...\}\) be the set of all strings using \(A\) as an alphabet.
Given \(s\in S\), \(N(s)\) is the number of vowels in \(s\).
For example,\(N(algrzqi) = 2\), \(N(bebebe) = 3\).
We say \((s, t)\) belongs to relation \(R\) if \(N(s) \leq N(t)\).
Which of the following statements are true (more than one answer can be correct) ? | Thus, one could equally well start with the alphabet in five letters that is S = { a , b , c , d , e } {\displaystyle S=\{a,b,c,d,e\}} . In this example, the set of all words or strings W ( S ) {\displaystyle W(S)} will include strings such as aebecede and abdc, and so on, of arbitrary finite length, with the letters arranged in every possible order. In the next step, one imposes a set of equivalence relations. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \(A = \{a, b, c, d, ..., z\}\) be the set of lower cased English letters.
Let \(S = \{a, b, ab, cd, ae, xy, ord, ...\}\) be the set of all strings using \(A\) as an alphabet.
Given \(s\in S\), \(N(s)\) is the number of vowels in \(s\).
For example,\(N(algrzqi) = 2\), \(N(bebebe) = 3\).
We say \((s, t)\) belongs to relation \(R\) if \(N(s) \leq N(t)\).
Which of the following statements are true (more than one answer can be correct) ? | The first order objects of S2S are finite binary strings. The second order objects are arbitrary sets (or unary predicates) of finite binary strings. S2S has functions s→s0 and s→s1 on strings, and predicate s∈S (equivalently, S(s)) meaning string s belongs to set S. Some properties and conventions: By default, lowercase letters refer to first order objects, and uppercase to second order objects. The inclusion of sets makes S2S second order, with "monadic" indicating absence of k-ary predicate variables for k>1. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A friend asked you to prove the following statement true or false: if a and b are rational numbers, a^b must be irrational. Examining the case where a is 1 and b is 2, what kind of proof are you using ? | If it is rational, our statement is proved. If it is irrational, ( 2 2 ) 2 = 2 {\displaystyle ({\sqrt {2}}^{\sqrt {2}})^{\sqrt {2}}=2} proves our statement. Dov Jarden Jerusalem In a bit more detail: Recall that 2 {\displaystyle {\sqrt {2}}} is irrational, and 2 is rational. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
A friend asked you to prove the following statement true or false: if a and b are rational numbers, a^b must be irrational. Examining the case where a is 1 and b is 2, what kind of proof are you using ? | Dov Jarden gave a simple non-constructive proof that there exist two irrational numbers a and b, such that ab is rational:Consider √2√2; if this is rational, then take a = b = √2. Otherwise, take a to be the irrational number √2√2 and b = √2. Then ab = (√2√2)√2 = √2√2·√2 = √22 = 2, which is rational. Although the above argument does not decide between the two cases, the Gelfond–Schneider theorem shows that √2√2 is transcendental, hence irrational. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let us define the relation R = {(1, 1)} on the set A = {1, 2, 3}. Which of the following properties does R satisfy ? (multiple answers) | Formally, a relation R over a set X can be seen as a set of ordered pairs (x, y) of members of X. The relation R holds between x and y if (x, y) is a member of R. For example, the relation "is less than" on the natural numbers is an infinite set Rless of pairs of natural numbers that contains both (1,3) and (3,4), but neither (3,1) nor (4,4). The relation "is a nontrivial divisor of" on the set of one-digit natural numbers is sufficiently small to be shown here: Rdiv = { (2,4), (2,6), (2,8), (3,6), (3,9), (4,8) }; for example 2 is a nontrivial divisor of 8, but not vice versa, hence (2,8) ∈ Rdiv, but (8,2) ∉ Rdiv. If R is a relation that holds for x and y one often writes xRy. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let us define the relation R = {(1, 1)} on the set A = {1, 2, 3}. Which of the following properties does R satisfy ? (multiple answers) | The binary relation R on the set {1, 2, 3, 4} is defined so that aRb holds if and only if a divides b evenly, with no remainder. For example, 2R4 holds because 2 divides 4 without leaving a remainder, but 3R4 does not hold because when 3 divides 4, there is a remainder of 1. The following set is the set of pairs for which the relation R holds. {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}.The corresponding representation as a logical matrix is ( 1 1 1 1 0 1 0 1 0 0 1 0 0 0 0 1 ) , {\displaystyle {\begin{pmatrix}1&1&1&1\\0&1&0&1\\0&0&1&0\\0&0&0&1\end{pmatrix}},} which includes a diagonal of ones, since each number divides itself. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If A is an uncountable set and B is an uncountable set, A − B cannot be : | If an uncountable set X is a subset of set Y, then Y is uncountable. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If A is an uncountable set and B is an uncountable set, A − B cannot be : | As implied above, the subcountability property cannot be adopted for all sets, given the theory proves P ω {\displaystyle {\mathcal {P}}_{\omega }} to be a set. The theory has many of the nice numerical existence properties and is e.g. consistent with Church's thesis principle as well as with ω → ω {\displaystyle \omega \to \omega } being subcountable. It also has the disjunctive property. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following sets can you use Cantor’s Diagonalization Argument to prove it is uncountable (multiple answers) ? | In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers. : 20– Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began. The diagonal argument was not Cantor's first proof of the uncountability of the real numbers, which appeared in 1874. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following sets can you use Cantor’s Diagonalization Argument to prove it is uncountable (multiple answers) ? | The best known example of an uncountable set is the set R of all real numbers; Cantor's diagonal argument shows that this set is uncountable. The diagonalization proof technique can also be used to show that several other sets are uncountable, such as the set of all infinite sequences of natural numbers and the set of all subsets of the set of natural numbers. The cardinality of R is often called the cardinality of the continuum, and denoted by c {\displaystyle {\mathfrak {c}}} , or 2 ℵ 0 {\displaystyle 2^{\aleph _{0}}} , or ℶ 1 {\displaystyle \beth _{1}} (beth-one). The Cantor set is an uncountable subset of R. The Cantor set is a fractal and has Hausdorff dimension greater than zero but less than one (R has dimension one). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
You need to quickly find if a person's name is in a list: that contains both integers and strings such as:
list := ["Adam Smith", "Kurt Gödel", 499, 999.95, "Bertrand Arthur William Russell", 19.99, ...]
What strategy can you use? | Since two different people cannot have the same number, it is possible to prove or disprove the presence of a participant simply by looking up his or her number. However it is possible that only the names of attendees were registered. If the name of a person is not found in the list, we may safely conclude that that person was not present; but if it is, we cannot conclude definitely without further inquiries, due to the possibility of homonyms (for example, two people named John Smith). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
You need to quickly find if a person's name is in a list: that contains both integers and strings such as:
list := ["Adam Smith", "Kurt Gödel", 499, 999.95, "Bertrand Arthur William Russell", 19.99, ...]
What strategy can you use? | This can be compiled into an executable which matches and outputs strings of integers. For example, given the input: abc123z. !&*2gj6 the program will print: Saw an integer: 123 Saw an integer: 2 Saw an integer: 6 | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let S(x) be the statement “x has been in a lake” and L(x) be the statement “x lives in Lausanne” and the domain of x consists of all the humans in the world.
The sentence : “there exists exactly one human that lives in Lausanne and that has never been in a lake” corresponds to the statement (multiple choices possible): | (∃x)(M(x) & W(x)) The statement O may be interpreted as "There exists at least one x that is both a man and non-white." (∃x)(M(x) & ~W(x)) The statement Y may be interpreted as "There exists at least one x that is both a man and white and there exists at least one x that is both a man and non-white." (∃x)(M(x) & W(x)) & (∃x)(M(x) & ~W(x)) The statement U may be interpreted as "One of two things, either whatever x may be, if x is a man, then x is white or whatever x may be, if x is a man, then x is non-white." (x)(M(x) → W(x)) w (x)(M(x) → ~W(x)) | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let S(x) be the statement “x has been in a lake” and L(x) be the statement “x lives in Lausanne” and the domain of x consists of all the humans in the world.
The sentence : “there exists exactly one human that lives in Lausanne and that has never been in a lake” corresponds to the statement (multiple choices possible): | a = "Socrates is a man." b = "All men are mortal." c = "Socrates is mortal." | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \( f : A
ightarrow B \) be a function from A to B such that \(f (a) = |a| \).
f is a bijection if: | That is, f is bijective if, for any y ∈ Y , {\displaystyle y\in Y,} the preimage f − 1 ( y ) {\displaystyle f^{-1}(y)} contains exactly one element. The function f is bijective if and only if it admits an inverse function, that is, a function g: Y → X {\displaystyle g\colon Y\to X} such that g ∘ f = id X {\displaystyle g\circ f=\operatorname {id} _{X}} and f ∘ g = id Y . {\displaystyle f\circ g=\operatorname {id} _{Y}.} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \( f : A
ightarrow B \) be a function from A to B such that \(f (a) = |a| \).
f is a bijection if: | A function is bijective if it is both injective and surjective. A bijective function is also called a bijection or a one-to-one correspondence. A function is bijective if and only if every possible image is mapped to by exactly one argument. This equivalent condition is formally expressed as follows: The function f: X → Y {\displaystyle f\colon X\to Y} is bijective, if for all y ∈ Y {\displaystyle y\in Y} , there is a unique x ∈ X {\displaystyle x\in X} such that f ( x ) = y . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \( P(n) \) be a proposition for a positive integer \( n \) (positive integers do not include 0).
You have managed to prove that \( orall k > 2, \left[ P(k-2) \wedge P(k-1) \wedge P(k)
ight]
ightarrow P(k+1) \).
You would like to prove that \( P(n) \) is true for all positive integers. What is left for you to do ? | This proof, due to Euler, uses induction to prove the theorem for all integers a ≥ 0. The base step, that 0p ≡ 0 (mod p), is trivial. Next, we must show that if the theorem is true for a = k, then it is also true for a = k + 1. For this inductive step, we need the following lemma. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Let \( P(n) \) be a proposition for a positive integer \( n \) (positive integers do not include 0).
You have managed to prove that \( orall k > 2, \left[ P(k-2) \wedge P(k-1) \wedge P(k)
ight]
ightarrow P(k+1) \).
You would like to prove that \( P(n) \) is true for all positive integers. What is left for you to do ? | {\displaystyle 0<\left|x-{\frac {\,p\,}{q}}\right|<{\frac {1}{\;q^{n}\,}}~.} If the claim is true, then the desired conclusion follows. Let p and q be any integers with q > 1 . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is equivalent to \((10001)_2\) ? (Multiple answers can be correct) | Since: 1000009 = 1000 2 + 3 2 = 972 2 + 235 2 {\displaystyle \ 1000009=1000^{2}+3^{2}=972^{2}+235^{2}} we have from the formula above: Thus, 1000009 = ⋅ {\displaystyle 1000009=\left\cdot \left\,} = ( 2 2 + 17 2 ) ⋅ ( 7 2 + 58 2 ) {\displaystyle =\left(2^{2}+17^{2}\right)\cdot \left(7^{2}+58^{2}\right)\,} = ( 4 + 289 ) ⋅ ( 49 + 3364 ) {\displaystyle =(4+289)\cdot (49+3364)\,} = 293 ⋅ 3413 {\displaystyle =293\cdot 3413\,} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following is equivalent to \((10001)_2\) ? (Multiple answers can be correct) | Thus A = 1 {\displaystyle A=1} and since F 2 = 1 {\displaystyle F_{2}=1} the claim is proven. We now take some k ≥ 2 {\displaystyle k\geq 2} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which sets are countable (Multiple answers can be correct) : | A set is hereditarily countable if and only if it is countable, and every element of its transitive closure is countable. If the axiom of countable choice holds, then a set is hereditarily countable if and only if its transitive closure is countable. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which sets are countable (Multiple answers can be correct) : | In mathematics, a set is countable if either it is finite or it can be made in one to one correspondence with the set of natural numbers. Equivalently, a set is countable if there exists an injective function from it into the natural numbers; this means that each element in the set may be associated to a unique natural number, or that the elements of the set can be counted one at a time, although the counting may never finish due to an infinite number of elements. In more technical terms, assuming the axiom of countable choice, a set is countable if its cardinality (the number of elements of the set) is not greater than that of the natural numbers. A countable set that is not finite is said to be countably infinite. The concept is attributed to Georg Cantor, who proved the existence of uncountable sets, that is, sets that are not countable; for example the set of the real numbers. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the value of \(f(4)\) where \(f\) is defined as \(f(0) = f(1) = 1\) and \(f(n) = 2f(n - 1) + 3f(n - 2)\) for integers \(n \geq 2\)? | . . , 1 } } | {\displaystyle F_{n+2}=F_{n}+F_{n-1}+...+|\{\{1,1,...,1,2\}\}|+|\{\{1,1,...,1\}\}|} ... where the last two terms have the value F 1 = 1 {\displaystyle F_{1}=1} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the value of \(f(4)\) where \(f\) is defined as \(f(0) = f(1) = 1\) and \(f(n) = 2f(n - 1) + 3f(n - 2)\) for integers \(n \geq 2\)? | . N = I 1 ( f 1 + f 2 + f 3 ) + . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following are true regarding the lengths of integers in some base \(b\) (i.e., the number of digits base \(b\)) in different bases, given \(N = (FFFF)_{16}\)? | In base ten, a sixteen-bit integer is certainly adequate as it allows up to 32767. However, this example cheats, in that the value of n is not itself limited to a single digit. This has the consequence that the method will fail for n > 3200 or so. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Which of the following are true regarding the lengths of integers in some base \(b\) (i.e., the number of digits base \(b\)) in different bases, given \(N = (FFFF)_{16}\)? | In base 13 and higher bases (such as hexadecimal), 11 is represented as B, where ten is A. In duodecimal, 11 is sometimes represented as E or ↋, and ten as T, X, or ↊. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In a lottery, a bucket of 10 numbered red balls and a bucket of 5 numbered green balls are used. Three red balls and two green balls are drawn (without replacement).
What is the probability to win the lottery? (The order in which balls are drawn does not matter). | As an example, consider a bucket containing 30 balls. The balls are either red, black or white. Ten of the balls are red, and the remaining 20 are either black or white, with all combinations of black and white being equally likely. In option X, drawing a red ball wins a person $100, and in option Y, drawing a black ball wins them $100. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
In a lottery, a bucket of 10 numbered red balls and a bucket of 5 numbered green balls are used. Three red balls and two green balls are drawn (without replacement).
What is the probability to win the lottery? (The order in which balls are drawn does not matter). | The chances of winning a lottery jackpot can vary widely depending on the lottery design, and are determined by several factors, including the count of possible numbers, the count of winning numbers drawn, whether or not order is significant, and whether drawn numbers are returned for the possibility of further drawing. In a simple 6-from-49 lotto, a player chooses six numbers from 1 to 49 (no duplicates are allowed). If all six numbers on the player's ticket match those produced in the official drawing (regardless of the order in which the numbers are drawn), then the player is a jackpot winner. For such a lottery, the chance of being a jackpot winner is 1 in 13,983,816.In bonusball lotteries where the bonus ball is compulsory, the odds are often even lower. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in our "Consensus-Based Total-Order Broadcast" algorithm, if the set of messages decided on by consensus is not sorted deterministically at all? | Out-of-order delivery occurs when sequenced packets arrive out of order. This may happen due to different paths taken by the packets or from packets being dropped and resent. HOL blocking can significantly increase packet reordering.Reliably broadcasting messages across a lossy network among a large number of peers is a difficult problem. While atomic broadcast algorithms solve the single point of failure problem of centralized servers, those algorithms introduce a head-of-line blocking problem. The Bimodal Multicast algorithm, a randomized algorithm that uses a gossip protocol, avoids head-of-line blocking by allowing some messages to be received out-of-order. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in our "Consensus-Based Total-Order Broadcast" algorithm, if the set of messages decided on by consensus is not sorted deterministically at all? | Saks' research in computational complexity theory, combinatorics, and graph theory has contributed to the study of lower bounds in order theory, randomized computation, and space–time tradeoff. In 1984, Saks and Jeff Kahn showed that there exist a tight information-theoretical lower bound for sorting under partially ordered information up to a multiplicative constant.In the first super-linear lower bound for the noisy broadcast problem was proved. In a noisy broadcast model, n + 1 {\displaystyle n+1} processors P 0 , P 1 , … , P n {\displaystyle P_{0},P_{1},\ldots ,P_{n}} are assigned a local input bit x i {\displaystyle x_{i}} . Each processor may perform a noisy broadcast to all other processors where the received bits may be independently flipped with a fixed probability. The problem is for processor P 0 {\displaystyle P_{0}} to determine f ( x 1 , x 2 , … , x n ) {\displaystyle f(x_{1},x_{2},\ldots ,x_{n})} for some function f {\displaystyle f} . Saks et al. showed that an existing protocol by Gallager was indeed optimal by a reduction from a generalized noisy decision tree and produced a Ω ( n log ( n ) ) {\displaystyle \Omega (n\log(n))} lower bound on the depth of the tree that learns the input.In 2003, P. Beame, Saks, X. Sun, and E. Vee published the first time–space lower bound trade-off for randomized computation of decision problems was proved. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, then process i has not failed, | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, then process i has not failed, | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we implement TRB with an eventually perfect failure detector ◇P, under the assumption that at least one process can crash? | Before arguing that the Chandra–Toueg consensus algorithm satisfies the three properties above, recall that this algorithm requires n = 2*f + 1 processes, where at most f of which are faulty. Furthermore, note that this algorithm assumes the existence of eventually strong failure detector (which are accessible and can be used to detect the crash of a node). An eventually strong failure detector is one that never identifies some specific non-faulty (or correct) process as having failed, after some initial period of confusion, and, at the same time, eventually identifies all faulty processes as failed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we implement TRB with an eventually perfect failure detector ◇P, under the assumption that at least one process can crash? | All failure detectors in the system will eventually suspect the p because of the infinite loop created by failure detectors. This example also shows that a weak completeness failure detector can also suspect all crashes eventually. The inspection of crashed programs does not depend on completeness. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Show that P is the weakest failure detector for Group Membership. | The following is an example abstracted from the Department of Computer Science at Yale University. It functions by boosting the completeness of a failure detector. From the example above, if p crashes, then the weak-detector will eventually suspect it. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Show that P is the weakest failure detector for Group Membership. | One of the key insights in non-adaptive group testing is that significant gains can be made by eliminating the requirement that the group-testing procedure be certain to succeed (the "combinatorial" problem), but rather permit it to have some low but non-zero probability of mis-labelling each item (the "probabilistic" problem). It is known that as the number of defective items approaches the total number of items, exact combinatorial solutions require significantly more tests than probabilistic solutions — even probabilistic solutions permitting only an asymptotically small probability of error.In this vein, Chan et al. (2011) introduced COMP, a probabilistic algorithm that requires no more than t = e d ( 1 + δ ) ln ( n ) {\displaystyle t=ed(1+\delta )\ln(n)} tests to find up to d {\displaystyle d} defectives in n {\displaystyle n} items with a probability of error no more than n − δ {\displaystyle n^{-\delta }} . This is within a constant factor of the t = O ( d log 2 n ) {\displaystyle t=O(d\log _{2}n)} lower bound.Chan et al. (2011) also provided a generalisation of COMP to a simple noisy model, and similarly produced an explicit performance bound, which was again only a constant (dependent on the likelihood of a failed test) above the corresponding lower bound. In general, the number of tests required in the Bernoulli noise case is a constant factor larger than in the noiseless case.Aldridge, Baldassini and Johnson (2014) produced an extension of the COMP algorithm that added additional post-processing steps. They showed that the performance of this new algorithm, called DD, strictly exceeds that of COMP, and that DD is 'essentially optimal' in scenarios where d 2 ≥ n {\displaystyle d^{2}\geq n} , by comparing it to a hypothetical algorithm that defines a reasonable optimum. The performance of this hypothetical algorithm suggests that there is room for improvement when d 2 < n {\displaystyle d^{2} | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Use Total Order Broadcast to implement an Asset Transfer sequential object. | Out-of-order delivery occurs when sequenced packets arrive out of order. This may happen due to different paths taken by the packets or from packets being dropped and resent. HOL blocking can significantly increase packet reordering.Reliably broadcasting messages across a lossy network among a large number of peers is a difficult problem. While atomic broadcast algorithms solve the single point of failure problem of centralized servers, those algorithms introduce a head-of-line blocking problem. The Bimodal Multicast algorithm, a randomized algorithm that uses a gossip protocol, avoids head-of-line blocking by allowing some messages to be received out-of-order. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Use Total Order Broadcast to implement an Asset Transfer sequential object. | Commitment ordering (or commit ordering) is a general serializability technique that achieves distributed serializability (and global serializability in particular) effectively on a large scale, without concurrency control information distribution (e.g., local precedence relations, locks, timestamps, or tickets), and thus without performance penalties that are typical to other serializability techniques (Raz 1992). The most common distributed concurrency control technique is strong strict two-phase locking (SS2PL, also named rigorousness), which is also a common centralized concurrency control technique. SS2PL provides both the serializability, strictness, and commitment ordering properties. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If a process j≠i fails, then process i has failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If a process j≠i fails, then process i has failed | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, then process i has not failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, then process i has not failed | Generally, processes are considered to be the same if no context, that is other processes running in parallel, can detect a difference. Unfortunately, making this intuition precise is subtle and mostly yields unwieldy characterisations of equality (which in most cases must also be undecidable, as a consequence of the halting problem). | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we devise a broadcast algorithm that does not ensure the causal delivery property but only (in) its non-uniform variant:
No correct process pi delivers a message m2 unless pi has already delivered every message m1 such that m1 → m2? | Out-of-order delivery occurs when sequenced packets arrive out of order. This may happen due to different paths taken by the packets or from packets being dropped and resent. HOL blocking can significantly increase packet reordering.Reliably broadcasting messages across a lossy network among a large number of peers is a difficult problem. While atomic broadcast algorithms solve the single point of failure problem of centralized servers, those algorithms introduce a head-of-line blocking problem. The Bimodal Multicast algorithm, a randomized algorithm that uses a gossip protocol, avoids head-of-line blocking by allowing some messages to be received out-of-order. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we devise a broadcast algorithm that does not ensure the causal delivery property but only (in) its non-uniform variant:
No correct process pi delivers a message m2 unless pi has already delivered every message m1 such that m1 → m2? | The P.I.P.S. environment does not support true signalling. Basic signal support is emulated using threads. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Devise an algorithm that, without consensus, implements a weaker specification of NBAC by replacing the termination property with weak termination.
Weak termination: Let p be a distinguished process, known to all other processes. If p does not crash then all correct processes eventually decide. Your algorithm may use a perfect failure detector. | Older implementations will fail if there are any updates broadcast over the memory bus. This is called weak LL/SC by researchers, as it breaks many theoretical LL/SC algorithms. Weakness is relative, and some weak implementations can be used for some algorithms. LL/SC is more difficult to emulate than CAS. Additionally, stopping running code between paired LL/SC instructions, such as when single-stepping through code, can prevent forward progress, making debugging tricky.Nevertheless, LL/SC is equivalent to CAS in the sense that either primitive can be implemented in terms of the other, in O(1) and in a wait-free manner. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Devise an algorithm that, without consensus, implements a weaker specification of NBAC by replacing the termination property with weak termination.
Weak termination: Let p be a distinguished process, known to all other processes. If p does not crash then all correct processes eventually decide. Your algorithm may use a perfect failure detector. | The following are correctness arguments to satisfy the algorithm of changing a failure detector W to a failure detector S. The failure detector W is weak in completeness, and the failure detector S is strong in completeness. They are both weak in accuracy. It transforms weak completeness into strong completeness. It preserves the perpetual accuracy. It preserves the eventual accuracy.If all arguments above are satisfied, the reduction of a weak failure detector W to a strong failure detector S will agree with the algorithm within the distributed computing system. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, then process i has failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, then process i has failed | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, nothing can be said about process i | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If all processes j≠i fail, nothing can be said about process i | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Prove that √2 is irrational. | A proof that the square root of 2 is irrational can be expressed this way: formulas(assumptions). 1*x = x. % identity x*y = y*x. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Prove that √2 is irrational. | The classic proof that the square root of 2 is irrational is a refutation by contradiction. Indeed, we set out to prove the negation ¬ ∃ a, b ∈ N {\displaystyle \mathbb {N} } . a/b = √2 by assuming that there exist natural numbers a and b whose ratio is the square root of two, and derive a contradiction. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The "Consensus-Based Total-Order Broadcast" algorithm transforms a consensus abstraction (together with a reliable broadcast abstraction) into a total-order broadcast abstraction. Describe a transformation between these two primitives in the other direction, that is, implement a (uniform) consensus abstraction from a (uniform) total-order broadcast abstraction. | In fault-tolerant distributed computing, an atomic broadcast or total order broadcast is a broadcast where all correct processes in a system of multiple processes receive the same set of messages in the same order; that is, the same sequence of messages. The broadcast is termed "atomic" because it either eventually completes correctly at all participants, or all participants abort without side effects. Atomic broadcasts are an important distributed computing primitive. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
The "Consensus-Based Total-Order Broadcast" algorithm transforms a consensus abstraction (together with a reliable broadcast abstraction) into a total-order broadcast abstraction. Describe a transformation between these two primitives in the other direction, that is, implement a (uniform) consensus abstraction from a (uniform) total-order broadcast abstraction. | Out-of-order delivery occurs when sequenced packets arrive out of order. This may happen due to different paths taken by the packets or from packets being dropped and resent. HOL blocking can significantly increase packet reordering.Reliably broadcasting messages across a lossy network among a large number of peers is a difficult problem. While atomic broadcast algorithms solve the single point of failure problem of centralized servers, those algorithms introduce a head-of-line blocking problem. The Bimodal Multicast algorithm, a randomized algorithm that uses a gossip protocol, avoids head-of-line blocking by allowing some messages to be received out-of-order. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we devise a Best-effort Broadcast algorithm that satisfies the causal delivery property, without being a causal broadcast algorithm, i.e., without satisfying the agreement property of a reliable broadcast? | Designing an algorithm for atomic broadcasts is relatively easy if it can be assumed that computers will not fail. For example, if there are no failures, atomic broadcast can be achieved simply by having all participants communicate with one "leader" which determines the order of the messages, with the other participants following the leader. However, real computers are faulty; they fail and recover from failure at unpredictable, possibly inopportune, times. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Can we devise a Best-effort Broadcast algorithm that satisfies the causal delivery property, without being a causal broadcast algorithm, i.e., without satisfying the agreement property of a reliable broadcast? | Lower bounds have been computed for many of the data streaming problems that have been studied. By far, the most common technique for computing these lower bounds has been using communication complexity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Is the decision rule of the FloodSet algorithm so critical? In other words, is there any alternative decision rule we can have? If so, name one. | Different flooding algorithms can be applied for different problems, and run with different time complexities. For example, the flood fill algorithm is a simple but relatively robust algorithm that works for intricate geometries and can determine which part of the (target) area that is connected to a given (source) node in a multi-dimensional array, and is trivially generalized to arbitrary graph structures. If there instead are several source nodes, there are no obstructions in the geometry represented in the multi-dimensional array, and one wishes to segment the area based on which of the source nodes the target nodes are closest to, while the flood fill algorithm can still be used, the jump flooding algorithm is potentially much faster as it has a lower time complexity. Unlike the flood fill algorithm, however, the jump flooding algorithm cannot trivially be generalized to unstructured graphs. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Is the decision rule of the FloodSet algorithm so critical? In other words, is there any alternative decision rule we can have? If so, name one. | A general criticism of non-probabilistic decision rules, discussed in detail at decision theory: alternatives to probability theory, is that optimal decision rules (formally, admissible decision rules) can always be derived by probabilistic methods, with a suitable utility function and prior distribution (this is the statement of the complete class theorems), and thus that non-probabilistic methods such as info-gap are unnecessary and do not yield new or better decision rules. A more general criticism of decision making under uncertainty is the impact of outsized, unexpected events, ones that are not captured by the model. This is discussed particularly in black swan theory, and info-gap, used in isolation, is vulnerable to this, as are a fortiori all decision theories that use a fixed universe of possibilities, notably probabilistic ones. Sniedovich raises two points to info-gap decision theory, one substantive, one scholarly: 1. the info-gap uncertainty model is flawed and oversold One should consider the range of possibilities, not its subsets. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Explain why any fail-noisy consensus algorithm (one that uses an eventually perfect failure detector ◇P) actually solves uniform consensus (and not only the non-uniform variant). | Before arguing that the Chandra–Toueg consensus algorithm satisfies the three properties above, recall that this algorithm requires n = 2*f + 1 processes, where at most f of which are faulty. Furthermore, note that this algorithm assumes the existence of eventually strong failure detector (which are accessible and can be used to detect the crash of a node). An eventually strong failure detector is one that never identifies some specific non-faulty (or correct) process as having failed, after some initial period of confusion, and, at the same time, eventually identifies all faulty processes as failed. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Explain why any fail-noisy consensus algorithm (one that uses an eventually perfect failure detector ◇P) actually solves uniform consensus (and not only the non-uniform variant). | An ε-approximate consensus halving can be computed by an algorithm based on Tucker's lemma, which is the discrete version of Borsuk-Ulam theorem. An adaptation of this algorithm shows that the problem is in the complexity class PPA. This holds even for arbitrary bounded and non-atomic valuations. However, the run-time of this algorithm may be exponential in the problem parameters. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Prove that x + |x - 7| ≥ 7 | + x 5 5 ! − x 7 7 ! | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Prove that x + |x - 7| ≥ 7 | + x 5 5 ! − x 7 7 ! + ⋯ . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the communication complexity of the FloodSet algorithm in number of messages? | Lower bounds have been computed for many of the data streaming problems that have been studied. By far, the most common technique for computing these lower bounds has been using communication complexity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the communication complexity of the FloodSet algorithm in number of messages? | There are ⌈ log ( n − 1 ) ⌉ {\displaystyle \lceil \log(n-1)\rceil } phases in total. Each winner sends in the order of 2 k {\displaystyle 2^{k}} messages in each phase. So, the messages complexity is O ( n log n ) {\displaystyle O(n\log n)} . | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If a process j≠i fails, then process i has not failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If a process j≠i fails, then process i has not failed | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the communication complexity of the FloodSet algorithm in number of bits? | Lower bounds have been computed for many of the data streaming problems that have been studied. By far, the most common technique for computing these lower bounds has been using communication complexity. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What is the communication complexity of the FloodSet algorithm in number of bits? | While Alice and Bob can always succeed by having Bob send his whole n {\displaystyle n} -bit string to Alice (who then computes the function f {\displaystyle f} ), the idea here is to find clever ways of calculating f {\displaystyle f} with fewer than n {\displaystyle n} bits of communication. Note that, unlike in computational complexity theory, communication complexity is not concerned with the amount of computation performed by Alice or Bob, or the size of the memory used, as we generally assume nothing about the computational power of either Alice or Bob. This abstract problem with two parties (called two-party communication complexity), and its general form with more than two parties, is relevant in many contexts. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If no process j≠i fails, then process i has failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If no process j≠i fails, then process i has failed | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in the uniform reliable broadcast algorithm if the completeness property of the failure detector is violated? | The following are correctness arguments to satisfy the algorithm of changing a failure detector W to a failure detector S. The failure detector W is weak in completeness, and the failure detector S is strong in completeness. They are both weak in accuracy. It transforms weak completeness into strong completeness. It preserves the perpetual accuracy. It preserves the eventual accuracy.If all arguments above are satisfied, the reduction of a weak failure detector W to a strong failure detector S will agree with the algorithm within the distributed computing system. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in the uniform reliable broadcast algorithm if the completeness property of the failure detector is violated? | The protocol provides the following guarantees: Strong Completeness: Full completeness is guaranteed (e.g. the crash-failure of any node in the group is eventually detected by all live nodes). Detection Time: The expected value of detection time (from node failure to detection) is T ′ ˙ 1 1 − e − q f {\displaystyle T'{\dot {}}{\frac {1}{1-e^{-q_{f}}}}} , where T ′ {\displaystyle T'} is the length of the protocol period, and q f {\displaystyle q_{f}} is the fraction of non-faulty nodes in the group. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Design an algorithm that implements consensus using multiple TRB instances. | Weighted voting for replicated data Consensus in the presence of partial synchrony | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Design an algorithm that implements consensus using multiple TRB instances. | An ε-approximate consensus halving can be computed by an algorithm based on Tucker's lemma, which is the discrete version of Borsuk-Ulam theorem. An adaptation of this algorithm shows that the problem is in the complexity class PPA. This holds even for arbitrary bounded and non-atomic valuations. However, the run-time of this algorithm may be exponential in the problem parameters. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, then process i has failed | The algorithm assumes that: the system is synchronous. processes may fail at any time, including during execution of the algorithm. a process fails by stopping and returns from failure by restarting. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
If process i fails, then eventually all processes j≠i fail
Is the following true? If some process j≠i does not fail, then process i has failed | there is a failure detector which detects failed processes. message delivery between processes is reliable. each process knows its own process id and address, and that of every other process. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in the uniform reliable broadcast algorithm if the accuracy property of the failure detector is violated? | The following are correctness arguments to satisfy the algorithm of changing a failure detector W to a failure detector S. The failure detector W is weak in completeness, and the failure detector S is strong in completeness. They are both weak in accuracy. It transforms weak completeness into strong completeness. It preserves the perpetual accuracy. It preserves the eventual accuracy.If all arguments above are satisfied, the reduction of a weak failure detector W to a strong failure detector S will agree with the algorithm within the distributed computing system. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
What happens in the uniform reliable broadcast algorithm if the accuracy property of the failure detector is violated? | Designing an algorithm for atomic broadcasts is relatively easy if it can be assumed that computers will not fail. For example, if there are no failures, atomic broadcast can be achieved simply by having all participants communicate with one "leader" which determines the order of the messages, with the other participants following the leader. However, real computers are faulty; they fail and recover from failure at unpredictable, possibly inopportune, times. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Implement a uniform reliable broadcast algorithm without using any failure detector, i.e., using only BestEffort-Broadcast(BEB). | Designing an algorithm for atomic broadcasts is relatively easy if it can be assumed that computers will not fail. For example, if there are no failures, atomic broadcast can be achieved simply by having all participants communicate with one "leader" which determines the order of the messages, with the other participants following the leader. However, real computers are faulty; they fail and recover from failure at unpredictable, possibly inopportune, times. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Implement a uniform reliable broadcast algorithm without using any failure detector, i.e., using only BestEffort-Broadcast(BEB). | The ESBT-broadcast (Edge-disjoint Spanning Binomial Tree) algorithm is a pipelined broadcast algorithm with optimal runtime for clusters with hypercube network topology. The algorithm embeds d {\displaystyle d} edge-disjoint binomial trees in the hypercube, such that each neighbor of processing element 0 {\displaystyle 0} is the root of a spanning binomial tree on 2 d − 1 {\displaystyle 2^{d}-1} nodes. To broadcast a message, the source node splits its message into k {\displaystyle k} chunks of equal size and cyclically sends them to the roots of the binomial trees. Upon receiving a chunk, the binomial trees broadcast it. | https://www.kaggle.com/datasets/conjuring92/wiki-stem-corpus |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.