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} -functi... | 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 automa... | 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 ... | 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 ... | 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 tru... | 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 ... | 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... | 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 ... | (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 th... | 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... | 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 un... | 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... | 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 a... | 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... | 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, lowerca... | 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... | 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 ... | 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 wh... | 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 \omeg... | 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-t... | 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 ... | 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 pe... | 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)) ... | 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... | 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 ... | 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 ... | 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... | 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 ba... | 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... | 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 ... | 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 or... | 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 ... | 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" ... | 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 ... | 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 t... | 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 h... | 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 ... | 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 ... | 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... | 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 ... | 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 pr... | 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 (... | 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 com... | 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 (... | 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 ... | 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 participan... | 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) nod... | 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 s... | 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 ... | 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 ... | 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 comp... | 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 pr... | 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'{... | 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 ... | 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 pr... | 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 participan... | 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 participan... | 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... | 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.