problem set 6
... (NB: This exercise requires some non-trivial commutative algebra. In particular, the fact that the coequalizer of two module homomorphisms f, g is the cokernel of the map f − g might come in handy.) Exercise 5 (Ultrafilters as a monad). Recall that an ultrafilter F on a set X is a family of subsets ...
... (NB: This exercise requires some non-trivial commutative algebra. In particular, the fact that the coequalizer of two module homomorphisms f, g is the cokernel of the map f − g might come in handy.) Exercise 5 (Ultrafilters as a monad). Recall that an ultrafilter F on a set X is a family of subsets ...
Probabilistic Skyline Operator over sliding Windows
... (1−P(a14)) smaller than q, the entry (i.e. all elements contained) will be removed from SN,q. On the other hand if the minimum value of Pnew multiplied by (1 − P(a14)) is not smaller than q, then the entry (i.e. all elements contained) remains in SN,q. ...
... (1−P(a14)) smaller than q, the entry (i.e. all elements contained) will be removed from SN,q. On the other hand if the minimum value of Pnew multiplied by (1 − P(a14)) is not smaller than q, then the entry (i.e. all elements contained) remains in SN,q. ...
Workout 4 Solutions
... The digits of a two-digit number “AB” are reversed to form a second two-digit number, and the lesser of the two-digit numbers is subtracted from the greater. What prime number must be a factor of the difference if A = B? Since A = B, then either A > B or A < B. Let’s assume for definiteness that A ...
... The digits of a two-digit number “AB” are reversed to form a second two-digit number, and the lesser of the two-digit numbers is subtracted from the greater. What prime number must be a factor of the difference if A = B? Since A = B, then either A > B or A < B. Let’s assume for definiteness that A ...
Decrease-and
... 3.2. Multiplication a la Russe (Russian peasant method) We can multiply two positive integers using only addition and division by 2. The algorithm is based on the observation that N*M = (N/2) * (M * 2) if N is even, and N*M = ((N-1)/2) * (M*2) + M if N is odd The base case is N = 1: 1*M = M Example: ...
... 3.2. Multiplication a la Russe (Russian peasant method) We can multiply two positive integers using only addition and division by 2. The algorithm is based on the observation that N*M = (N/2) * (M * 2) if N is even, and N*M = ((N-1)/2) * (M*2) + M if N is odd The base case is N = 1: 1*M = M Example: ...
(2)ааf(x) = 2x
... Problem Set October 31st 1. In an AC circuit, the voltage (V) can be found by multiplying the current (C) and impedance (I). Find the impedance, in ohms, if the voltage is 2622i volts and the current is 3 4i amps. 2. Write an equation of a trigonometric function that has an amplitude of 4 ...
... Problem Set October 31st 1. In an AC circuit, the voltage (V) can be found by multiplying the current (C) and impedance (I). Find the impedance, in ohms, if the voltage is 2622i volts and the current is 3 4i amps. 2. Write an equation of a trigonometric function that has an amplitude of 4 ...
Abstracting Planning Problems with Preferences and Soft Goals
... reached. Only those variables that are both a precondition and an effect of some action are included. For each such action, the cost of that action is associated with the variable. The variables are added to the abstract problem in descending order of their minimal associated costs. This abstract st ...
... reached. Only those variables that are both a precondition and an effect of some action are included. For each such action, the cost of that action is associated with the variable. The variables are added to the abstract problem in descending order of their minimal associated costs. This abstract st ...
Standard to Vertex: Using algebraic methods to find exact answers
... 2. Debrief yesterday’s work Q&A(10 minutes) 3. Direct Instruction – Converting to vertex form from standard form. (20 minutes) Solve graphically (Each problem increases in difficulty because the landmarks are not shown in the last problem. 9 x2 4 x 8 ...
... 2. Debrief yesterday’s work Q&A(10 minutes) 3. Direct Instruction – Converting to vertex form from standard form. (20 minutes) Solve graphically (Each problem increases in difficulty because the landmarks are not shown in the last problem. 9 x2 4 x 8 ...
X11 = Space leased at the beginning of month 1 for period of 1 month
... increase. Because as we increase the supply or demand nodes number of decision variables will increase exponentially, plus all the constraints are of equality so we have to introduce artificial variables equal the number of constraints. And due special nature of Transportation Problem, specialized a ...
... increase. Because as we increase the supply or demand nodes number of decision variables will increase exponentially, plus all the constraints are of equality so we have to introduce artificial variables equal the number of constraints. And due special nature of Transportation Problem, specialized a ...
lecture1212
... Since two, say ci and cj (i>j), of the k+1 points must be in the same group (in an opt solution), t 2 opt. Thus, k+1 t 2 opt. For any v in V, by the definition of k+1 , min {|c1v|, |c2v|,…,|ck v|} k+1 . So the algorithm has ratio-2. ...
... Since two, say ci and cj (i>j), of the k+1 points must be in the same group (in an opt solution), t 2 opt. Thus, k+1 t 2 opt. For any v in V, by the definition of k+1 , min {|c1v|, |c2v|,…,|ck v|} k+1 . So the algorithm has ratio-2. ...
Notes for Lecture 11
... Since two, say ci and cj (i>j), of the k+1 points must be in the same group (in an opt solution), t 2 opt. Thus, k+1 t 2 opt. For any v in V, by the definition of k+1 , min {|c1v|, |c2v|,…,|ck v|} k+1 . So the algorithm has ratio-2. ...
... Since two, say ci and cj (i>j), of the k+1 points must be in the same group (in an opt solution), t 2 opt. Thus, k+1 t 2 opt. For any v in V, by the definition of k+1 , min {|c1v|, |c2v|,…,|ck v|} k+1 . So the algorithm has ratio-2. ...
Sangkyum`s slides
... – Branch and bound 1. At each stage, after solving the LP, a fractional variable xj is selected and two subproblems are set up, one where xj=1 and the other where xj=0. (Branch) 2. Solve the LP relaxation of the two subproblems. 3. From each subproblem with a nonintegral solution we branch again to ...
... – Branch and bound 1. At each stage, after solving the LP, a fractional variable xj is selected and two subproblems are set up, one where xj=1 and the other where xj=0. (Branch) 2. Solve the LP relaxation of the two subproblems. 3. From each subproblem with a nonintegral solution we branch again to ...
Linear inverse problems on Erd˝os
... the absence of information, and when there is information available, then two vertices are connected with probability q1 if they are in the same community and with probability q2 if they are in different communities. When G = Kn is the complete graph and X is uniformly distributed, this is the stand ...
... the absence of information, and when there is information available, then two vertices are connected with probability q1 if they are in the same community and with probability q2 if they are in different communities. When G = Kn is the complete graph and X is uniformly distributed, this is the stand ...
18.06 Linear Algebra, Problem set 3 solutions
... The rows of the matrix AB are all linear combinations of the rows of B, and so also satisfy this linear relation. So, column j is the same combination of previous columns of AB, as desired. Since a column is pivot column precisely when it is not a combination of previous columns, this shows that AB ...
... The rows of the matrix AB are all linear combinations of the rows of B, and so also satisfy this linear relation. So, column j is the same combination of previous columns of AB, as desired. Since a column is pivot column precisely when it is not a combination of previous columns, this shows that AB ...
Vertex cover in cubic graphs with large girth
... The length of the smallest cycle (girth of G ) is large C ⊂ V s.t. every edge has at least one endpoint in C ...
... The length of the smallest cycle (girth of G ) is large C ⊂ V s.t. every edge has at least one endpoint in C ...
T(n)
... than half of the entries. Design an O(nlgn) algorithm to decide if the array has a majority element and if so, what the element is. You cannot sort the array, or ask if A[i]>A[j], since comparisons are not supported on the array type. (For example, if the array were of GIF files.) You can, however, ...
... than half of the entries. Design an O(nlgn) algorithm to decide if the array has a majority element and if so, what the element is. You cannot sort the array, or ask if A[i]>A[j], since comparisons are not supported on the array type. (For example, if the array were of GIF files.) You can, however, ...
physics 211 class schedule
... Listed below are the assigned problems and questions for the fall semester. These should be viewed as an ABSOLUTE MINIMUM number of problems to be worked. Usually you will need to work more! Problems are due at the start of class and must be submitted to WebAssign by the authorized date. WebAssign w ...
... Listed below are the assigned problems and questions for the fall semester. These should be viewed as an ABSOLUTE MINIMUM number of problems to be worked. Usually you will need to work more! Problems are due at the start of class and must be submitted to WebAssign by the authorized date. WebAssign w ...
Strategies for Solving Multiplication Problems
... How many people are in the last 8 rows? 8 ⫻ 20 ⫽ 160 That’s 20 people in those last 8 rows. 8 ⫻ 6 ⫽ 48 Now I’ve filled up the last 8 rows with 6 more people in each row. How many people can sit in the auditorium? ...
... How many people are in the last 8 rows? 8 ⫻ 20 ⫽ 160 That’s 20 people in those last 8 rows. 8 ⫻ 6 ⫽ 48 Now I’ve filled up the last 8 rows with 6 more people in each row. How many people can sit in the auditorium? ...
General
... many forms, hence we desire simplification Particularly if we are only interested in the interaction of some of the variables Many problems desire a optimal solution, there are algms (simplex) to find them We may also be interested in asking ...
... many forms, hence we desire simplification Particularly if we are only interested in the interaction of some of the variables Many problems desire a optimal solution, there are algms (simplex) to find them We may also be interested in asking ...
A Constructive Heuristic for the Travelling Tournament Problem
... the home team. The pattern corresponding to Fig. 1, for example, is 1001001. Any team always ends the competition at home, and thus the last element of the pattern array is always 1. This leaves n − 2 0/1 elements to assign. The total number of possible assignments thus equals 2(n−2) . Table 2 indic ...
... the home team. The pattern corresponding to Fig. 1, for example, is 1001001. Any team always ends the competition at home, and thus the last element of the pattern array is always 1. This leaves n − 2 0/1 elements to assign. The total number of possible assignments thus equals 2(n−2) . Table 2 indic ...
Discrete Structures - CSIS121
... traveling salesman wants to visit a number of cities and then return to his starting point. Of course he wants to save time and energy, so he wants to determine the shortest cycle for his trip. We ...
... traveling salesman wants to visit a number of cities and then return to his starting point. Of course he wants to save time and energy, so he wants to determine the shortest cycle for his trip. We ...
Simplification, Optimization and Implication
... many forms, hence we desire simplification Particularly if we are only interested in the interaction of some of the variables Many problems desire a optimal solution, there are algms (simplex) to find them We may also be interested in asking ...
... many forms, hence we desire simplification Particularly if we are only interested in the interaction of some of the variables Many problems desire a optimal solution, there are algms (simplex) to find them We may also be interested in asking ...
06 - The Creativity Process
... Enjoys analyzing potential solutions; and Can get stuck in developing the perfect soluition. ...
... Enjoys analyzing potential solutions; and Can get stuck in developing the perfect soluition. ...