A certain recursive algorithm takes an input list of n elements. Divides the list into yn sub-lists, each with /n elements. Recursively solves each of these yn smaller sub- instances. Then spends an additional 0(n) time to combine the solutions of these sub- instances to obtain the solution of the main instance. As a base case, if the size of the input list is at most a specified positive constant, then the algorithm solves such a small instance directly in 0(1) time. a) Express the recurrence relation that governs T(n), the time complexity of this algorithm. b) Derive the solution to this recurrence relation: T(n) = 0(?). Mention which methods you used to derive your solution.

A certain recursive algorithm takes an input list of n elements. Divides the list into yn sub-lists, each with /n elements. Recursively solves each of these yn smaller sub- instances. Then spends an additional 0(n) time to combine the solutions of these sub- instances to obtain the solution of the main instance. As a base case, if the size of the input list is at most a specified positive constant, then the algorithm solves such a small instance directly in 0(1) time. a) Express the recurrence relation that governs T(n), the time complexity of this algorithm. b) Derive the solution to this recurrence relation: T(n) = 0(?). Mention which methods you used to derive your solution.

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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Suppose a recursive algorithm performs 2 recursive calls. Assume the first recursive call isof size at most 70% the original input size, and the second call is of size at most 25% of theoriginal input size. In addition, the algorithm performs O(n) additional work after makingthese recursive calls. What is the big-Oh run time of this algorithm?
Algorithm A solves an instance of size n by recursively solving eight instances of size n/2, and then combining their solutions in time O(n3). Algorithm B solves an instance of size n by recursively solving twenty instances of size n/3, and then combining their solutions in time O(n2). Algorithm C solves an instance of size n by recursively solving two in- stances of size 2n, and then combining their solutions in time O(n). Which algorithm is preferable, and why?
2. Expand the following recurrence to help you find a closed-form solution, and then use induction to prove your answer is correct. T(n) = T(n-1) + 5 for n> 0; T(0) = 8. 3. Give a recursive algorithm for the sequential search and explain its running time.
Write for the following problem a recursive algorithm whose worst-case timecomplexity is not worse than Θ(n ln n). Given a list of n distinct positiveintegers, partition the list into two sublists, each of size n/2, such that thedifference between the sums of the integers in the two sublists is maximized.You may assume that n is a multiple of 2.
In the divided and conquer algorithm for computing the maxima set of a given set A of n points in 2D, we split the points into left and right halves, recur on each half, and then merge the results. Let MSL and MSR be the return value of these recursive calls. Which of the following statements is true about these sets. Select all that apply. MSR is a maximum set of the right half. Every point in MSL belongs in the maximum set of A. MSL is a maximum set of the left half. Every point in MSR belongs in the maximal set of A.
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Give a recursive algorithm for the sequential search and explain its running time.
A recursive algorithm is designed in such a way that for size n it divides it into 2 subproblems in the ratio 3:2 and it takes linear time to integrate the solution from the 3 subproblem, what will be the recurrence relation of this.
In computer science and mathematics, the Josephus Problem (or Josephus permutation) is a theoretical problem. Following is the problem statement: There are n people standing in a circle waiting to be executed. The counting out begins at some point (rear) in the circle and proceeds around the circle in a fixed direction. In each step, a certain number (k) of people are skipped and the next person is executed. The elimination proceeds around the circle (which is becoming smaller and smaller as the executed people are removed), until only the last person remains, who is given freedom. Given the total number of persons n and a number k which indicates that k-1 persons are skipped and kth person is killed in circle. The task is to choose the place in the initial circle so that you are the last one remaining and so survive. For example, if n = 5 and k = 2, then the safe position is 3. Firstly, the person at position 2 is killed, then person at position 4 is killed, then person at position 1…
Another recursive algorithm is applied to some data A = (a₁, ..., am) where m = 2* (i.e. 2, 4, 8,16 ...) where x is an integer ≥ 1. The running time T is characterised using the following recurrence equations: T(1) = c when the size of A is 1 T(m) = 2T (2) + c otherwise Determine the running time complexity of this algorithm.
Your main task is to write a recursive function sierpinski() that plots a Sierpinski triangle of order n to standard drawing. Think recursively: sierpinski() should draw one filled equilateral triangle (pointed downwards) and then call itself recursively three times (with an appropriate stopping condition). It should draw 1 filled triangle for n = 1; 4 filled triangles for n = 2; and 13 filled triangles for n = 3; and so forth. Sierpinski.java When writing your program, exercise modular design by organizing it into four functions, as specified in the following API: public class Sierpinski { // Height of an equilateral triangle with the specified side length. public static double height(double length) // Draws a filled equilateral triangle with the specified side length // whose bottom vertex is (x, y). public static void filledTriangle(double x, double y, double length) // Draws a Sierpinski triangle of order n, such that the largest filled //…
suppose that n is not 2i for any integer i. How would we change the algorithm so that it handles the case when n is odd? I have two solutions: one that modifies the recursive algorithm directly, and one that combines the iterative algorithm and the recursive algorithm. You only need to do one of the two (as long as it works and does not increase the BigOh of the running time.)