CMPUT 650: Learning To Make Decisions
... A binary relation R from a set A to a set B is a subset of the Cartesian Product AB, that is, a set of ordered pairs (a,b) with aA and b B A binary relation R on a set A is a subset of the Cartesian product AA, that is, ordered pairs of the form (a,b) where a A and b A Instead of (x,y) R A ...
... A binary relation R from a set A to a set B is a subset of the Cartesian Product AB, that is, a set of ordered pairs (a,b) with aA and b B A binary relation R on a set A is a subset of the Cartesian product AA, that is, ordered pairs of the form (a,b) where a A and b A Instead of (x,y) R A ...
MARTIN`S CONJECTURE, ARITHMETIC EQUIVALENCE, AND
... to the larger class of functions f so that f ≡m g for some uniformly Turing invariant g. Steel has conjectured that this is true of all Turing invariant functions. Conjecture 1.4 (Steel [26]). If f : 2ω → 2ω is Turing invariant, then there exists a uniformly Turing invariant g which is defined a.e., ...
... to the larger class of functions f so that f ≡m g for some uniformly Turing invariant g. Steel has conjectured that this is true of all Turing invariant functions. Conjecture 1.4 (Steel [26]). If f : 2ω → 2ω is Turing invariant, then there exists a uniformly Turing invariant g which is defined a.e., ...
Nick%27s Mathematica..
... What is the 1000th digit to the right of the decimal point in the decimal representation of (1 + ...
... What is the 1000th digit to the right of the decimal point in the decimal representation of (1 + ...
Computer Science Foundation Exam
... You must do ALL of them. Each counts for 15% of the Discrete Structures exam grade. Show the steps of your work carefully. Problems will be graded based on the completeness of the solution steps and not graded based on the answer alone. Credit cannot be given when your results are unreadable. Questi ...
... You must do ALL of them. Each counts for 15% of the Discrete Structures exam grade. Show the steps of your work carefully. Problems will be graded based on the completeness of the solution steps and not graded based on the answer alone. Credit cannot be given when your results are unreadable. Questi ...
Equivalent Fractions
... Equivalent fractions are different ways to represent the same number, the same point on the number line. Initially, students find equivalence in fractions less than 1 whole (e.g., 1 half = 2 fourths). They then express whole numbers as fractions, using number bonds and number lines, to show how many ...
... Equivalent fractions are different ways to represent the same number, the same point on the number line. Initially, students find equivalence in fractions less than 1 whole (e.g., 1 half = 2 fourths). They then express whole numbers as fractions, using number bonds and number lines, to show how many ...
CHAPTER 5: EQUIVALENCE RELATIONS AND EQUIVALENCE
... Example 2: For x, y ∈ R define x y to mean that x − y ∈ Z. Prove that is an equivalence relation on R. Proof: To see that is reflexive, let x ∈ R. Then x − x = 0 and 0 ∈ Z, so x x. To see that is symmetric, let a, b ∈ R. Suppose a b. Then a − b ∈ Z – say a − b = m, where m ∈ Z. Then b − a ...
... Example 2: For x, y ∈ R define x y to mean that x − y ∈ Z. Prove that is an equivalence relation on R. Proof: To see that is reflexive, let x ∈ R. Then x − x = 0 and 0 ∈ Z, so x x. To see that is symmetric, let a, b ∈ R. Suppose a b. Then a − b ∈ Z – say a − b = m, where m ∈ Z. Then b − a ...
Nelson`s Strong Negation, Safe Beliefs and the - CEUR
... Intuitively, we can explain safe beliefs as follows: Our formula or program represents our knowledge. We want its semantics to be a complete theory. For some atoms a in the language, either a or ¬a can be infered directly from the information in the program. If this is not the case, we will have to ...
... Intuitively, we can explain safe beliefs as follows: Our formula or program represents our knowledge. We want its semantics to be a complete theory. For some atoms a in the language, either a or ¬a can be infered directly from the information in the program. If this is not the case, we will have to ...
Complexity Results for Checking Equivalence of Stratified Logic
... Recent research in nonmonotonic logic programming under the answer-set semantics focuses on different notions of program equivalence. However, previous results do not address the important classes of stratified programs and its subclass of acyclic (i.e., recursion-free) programs, although they are re ...
... Recent research in nonmonotonic logic programming under the answer-set semantics focuses on different notions of program equivalence. However, previous results do not address the important classes of stratified programs and its subclass of acyclic (i.e., recursion-free) programs, although they are re ...
Modal logic and the approximation induction principle
... from the richest characterizations, which correspond to the canonical process equivalences, there are also finitary versions (denoted with a superscript ∗ ), which allow only conjunctions over a finite set. Intermediate equivalences based on formulas with arbitrary conjunctions but of finite depth a ...
... from the richest characterizations, which correspond to the canonical process equivalences, there are also finitary versions (denoted with a superscript ∗ ), which allow only conjunctions over a finite set. Intermediate equivalences based on formulas with arbitrary conjunctions but of finite depth a ...
Numerical experiments on the condition number of the interpolation
... big, or to use alternative strategies that (at extra cost) bypass the interpolation matrix [25,24]. To avoid contamination of our results by roundoff errors, all computations of condition number in this paper, with the exception of those in Fig. 12, were performed in variable precision arithmetic in ...
... big, or to use alternative strategies that (at extra cost) bypass the interpolation matrix [25,24]. To avoid contamination of our results by roundoff errors, all computations of condition number in this paper, with the exception of those in Fig. 12, were performed in variable precision arithmetic in ...
Permutation and Probability
... The order in which the people are being chosen does not matter because the positions for which they are being chosen are the same. They are all going to be members of the student council, with the same duties. (Combination) However, if Mr. Fant’s class was choosing 4 out of 7 students to be presiden ...
... The order in which the people are being chosen does not matter because the positions for which they are being chosen are the same. They are all going to be members of the student council, with the same duties. (Combination) However, if Mr. Fant’s class was choosing 4 out of 7 students to be presiden ...
Permutations and determinants Math 130 Linear Algebra
... n matrix, which we’ll do in a moment, it will be Permutations. A permutation of the set defined as a sum/difference of n! terms, each term {1, 2, 3, . . . , n} is a list of its n elements where being a product of n elements, one element chosen each element appears exactly once in the list. For out o ...
... n matrix, which we’ll do in a moment, it will be Permutations. A permutation of the set defined as a sum/difference of n! terms, each term {1, 2, 3, . . . , n} is a list of its n elements where being a product of n elements, one element chosen each element appears exactly once in the list. For out o ...
Equivalence relations and Counting
... clique every possible connection is present, while between the cliques there are no connections at all. ...
... clique every possible connection is present, while between the cliques there are no connections at all. ...
MATH 521–01 Problem Set #1 solutions 1. Prove that for every
... Proof. We shall prove specifically that any N > 1 such that N ≡ 1 (mod (k + 1))! satisfies this rule; we shall do this by induction on k. In the base case k = 1, we are asserting that whenever N ≡ 1 (mod 2) and N > 1, it is the case that N + 1 is composite. This is trivial because if N ≡ 1 (mod 2) t ...
... Proof. We shall prove specifically that any N > 1 such that N ≡ 1 (mod (k + 1))! satisfies this rule; we shall do this by induction on k. In the base case k = 1, we are asserting that whenever N ≡ 1 (mod 2) and N > 1, it is the case that N + 1 is composite. This is trivial because if N ≡ 1 (mod 2) t ...
A STUDY OF EULERIAN NUMBERS FOR PERMUTATIONS IN THE
... ascent number. Therefore, the identities for Dn,k presented here correspond to those in [1] that are obtained by replacing k with n − k − 1. In order to study these numbers, we make use of an operator on permutations in [n], which was introduced in [9]. The operator σ is defined by adding one to all ...
... ascent number. Therefore, the identities for Dn,k presented here correspond to those in [1] that are obtained by replacing k with n − k − 1. In order to study these numbers, we make use of an operator on permutations in [n], which was introduced in [9]. The operator σ is defined by adding one to all ...
Document
... • Example: What is the minimum number of students required in a discrete mathematics class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D and F? Solution: The minimum number of students needed to ensure that at least 6 students receive the sam ...
... • Example: What is the minimum number of students required in a discrete mathematics class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D and F? Solution: The minimum number of students needed to ensure that at least 6 students receive the sam ...
Yet another triangle for the Genocchi numbers
... The study of Genocchi numbers, it is claimed, goes back to Euler. In what sense that is true is somewhat unclear, but during the last two or three decades the Genocchi numbers have been studied by Dumont and some collaborators [2–7]. In recent years there has been a flurry of activity in this field, ...
... The study of Genocchi numbers, it is claimed, goes back to Euler. In what sense that is true is somewhat unclear, but during the last two or three decades the Genocchi numbers have been studied by Dumont and some collaborators [2–7]. In recent years there has been a flurry of activity in this field, ...
Full text
... and copy out the visible letters, which are (serially, row by row) ITTIAOHTSOLOC. We then rotate the grid counterclockwise through 90° and again copy out the visible letters, which are IOLESIHIMLTAIM. Two more rotations gives us UMKAFGHGSYSODand NILHFLHNAIMFRE. Running these four groups together and ...
... and copy out the visible letters, which are (serially, row by row) ITTIAOHTSOLOC. We then rotate the grid counterclockwise through 90° and again copy out the visible letters, which are IOLESIHIMLTAIM. Two more rotations gives us UMKAFGHGSYSODand NILHFLHNAIMFRE. Running these four groups together and ...
CHAP03 Sets, Functions and Relations
... If R and S are relations on the set X then the sum of R and S is the relation R + S defined on X by: x(R+S)y if xRy or xSy. As sets, this is simply the union: S + T = S ∪ T. Example 10: The relation “spouse of” means “husband or wife of”. If H = “husband of” and W = “wife of” then H + W is the relat ...
... If R and S are relations on the set X then the sum of R and S is the relation R + S defined on X by: x(R+S)y if xRy or xSy. As sets, this is simply the union: S + T = S ∪ T. Example 10: The relation “spouse of” means “husband or wife of”. If H = “husband of” and W = “wife of” then H + W is the relat ...
Equivalence Relations
... With poker chips, some are red, some are white and some are blue. You can group them into three subsets – the red ones, the white ones, and the blue ones. You can say two red chips are alike in having the same color, and a red and a blue chip are not alike because they have different colors. Wit ...
... With poker chips, some are red, some are white and some are blue. You can group them into three subsets – the red ones, the white ones, and the blue ones. You can say two red chips are alike in having the same color, and a red and a blue chip are not alike because they have different colors. Wit ...