One way to avoid Runge's problem is to choose non-equally spaced data points. A particularlygood set to choose are the so-called Chebyshev nodes. To use these, replace the equidistant pointsxk produced by xeqwithxk = COS2k+12n+2for k=0,..., n.(3.4)Generate data points using (3.4) and (3.3), then plot the resulting Lagrange interpolating polynomialand, hence, show that using (3.4) solves the Runge problem.

EN-US 3.2.2.4 Chebyshev nodesOne way to avoid Runge's problem is to choose non-equally…

Answered: One way to avoid Runge's problem is to…

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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For the one-dimensional version of the closest-pair problem, i.e., for the problem of finding two closest numbers among a given set of n real num- bers, design an algorithm that is directly based on the divide-and-conquer technique and determine its efficiency class. Is it a good algorithm for this problem?
PLEASE USE PYTHONGiven a jungle matrix NxM:jungle = [ [1, 0, 0, 0], [1, 1, 0, 1], [0, 1, 0, 0], [1, 1, 1, 1,]]Where 0 means the block is dead end and 1 means the block can be used in the path fromsource to destination.Task:Starting at position (0, 0), the goal is to reach position (N-1, M-1).Your program needs to build and output the solution matrix – a 4x4 matrix with 1’s inpositions used to get from the starting position (0,0) to the ending position (N-1,M-1)with the following constraints:You can only move one space at a timeYou can only in two directions: forward and down.You can only pass thru spaces on the jungle matrix marked ‘1’If you cannot reach the ending position – print a message that you’re trapped in thejungleAlgorithm:If destination is reachedprint the solution matrixElseMark current cell in the solution matrixMove forward horizontally and recursively check if this leads to a solution If there is no solution, move down and recursively check if this leads to a solution If…
Show L = { w | w is an element of {0, 1, 2}* and (the number of 0s in w ≥ the number of 1s in w) and ((the number of 0s + the number of 1s in w) divides the number of 2s in w) } is not context free. You should assume that p is the pumping length and for any string s in L with |s| ≥ p, s can be broken into 5 pieces, s = uvxyz.
You are given k > 2 linked lists, each containing n > k natural numbers sorted in increasing order. All n · k numbers in these lists are distinct. Describe an algorithm that finds the k-th smallest element among all n k numbers. Your algorithm should run in O(k log k) time. Analyze the running time of your algorithm and prove its correctness.
Consider a Skip List with n nodes. Each node k has a random "height" hg, which is the number of linked lists that includes node k. The heights nodes are statistically independent, where P[hx = m] = 2-m, for m 2 1 and P[h = m] = 0 , for m m] 0 such that P[H > m] <= for m = c log, n
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What is the time complexity of functions foo() and bar() respectively in below given code snippets, given that n is the total size of the input and L and R are Vectors def foo(L, R): S = Vector() for x in L: if x in R: S.add_back(x) return S def bar(L, R): R_hash= HashTable() for r in R: R_hash.insert(r, False) S = Vector() for x in L: if R_hash.contains(x): S.add_back(x) return S O(N^2), O(N) O(N), O(N) O(N^2), O(N^2) O(1), O(N)
Use the Transform-and-Conquer algorithm design technique with Instance Simplification variant to design an O(nlogn) algorithm for the problem below. Show the pseudocode. Given a set S of n integers and another integer x, determine whether or not there exist two elements in S whose sum is exactly x.
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