3. Draw the recursion tree when n = 12, where n represents the length of the array, for the following recursive method: int sumSquares (int[] array, int first, int last) ( if (first == last) return array [first] * array[first]; int mid (first + last) / 2; return sumSquares (array, first, mid) + sumSquares (array, mid + 1, last); } • Determine a formula that counts the numbers of nodes in the recursion tree. What is the Big-Ⓒ for execution time? • Determine a formula that expresses the height of the tree. What is the Big-Ⓒ for memory? • Write an iterative solution for this same problem and compare its efficiency with this recursive solution.

3. Draw the recursion tree when n = 12, where n represents the length of the array, for the following recursive method: int sumSquares (int[] array, int first, int last) ( if (first == last) return array [first] * array[first]; int mid (first + last) / 2; return sumSquares (array, first, mid) + sumSquares (array, mid + 1, last); } • Determine a formula that counts the numbers of nodes in the recursion tree. What is the Big-Ⓒ for execution time? • Determine a formula that expresses the height of the tree. What is the Big-Ⓒ for memory? • Write an iterative solution for this same problem and compare its efficiency with this recursive 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

See similar textbooks

Similar questions

9. Ackermann's Function Ackermann's function is a recursive mathematical algorithm that can be used to test how well a computer performs recursion. Write a method ackermann (m, n), which solves Ackermann's function. Use the following logic in your method: If m = 0 then return n + 1 If n = 0 then return ackermann (m Otherwise, return ackermann(m 1, 1) 1, ackermann (m, n - 1))
Write recursive methods for the following operations: • removeDup(String s): takes string s and returns another string r constructed from s after removing all duplicates. For example, when s = "pineapples", r will be "pineals". • intersection(String s1, String s2): takes two duplicate-free strings and returns the string consisting of all letters that appear in both s1 and s2. • union(String s1, String s2): takes two duplicate-free strings and returns the string consisting of all letters that appear in either s1 or s2. • difference(String s1, String s2): takes two duplicate-free strings and returns the string consisting of all letters that appear only in s1. For example, when s1 = "aples" and s2 = "pears", intersection(s1, s2) will return "apes", union(s1, s2) will "aplesr", and difference(s1, s2) "l". Calling method removeDup() is not allowed in intersection(s1, s2) , union(s1, s2) or difference(s1, s2) method. Your program should check strings are duplicate free before they are…
please answer if the question is true or false In the Recursive Binary Search shown below, there are two base cases: Line 10: the search is exhausted without a finding a value that matches the key Line 16: the search finds a value that matches the key 1 public class RecursiveBinarySearch { 2 public static int binarySearch(int[] list, int key) { 3 int low = 0; 4 int high = list.length - 1; 5 return binarySearch(list, key, low, high); 6 } 7 8 private static int binarySearch(int[] list, int key, 9 int low, int high) { 10 if (low > high) // The list has been exhausted without a match 11 return -low - 1; 12 13 int mid = (low + high) / 2; 14 if (key < list[mid]) 15 return binarySearch(list, key, low, mid - 1); 16 else if (key == list[mid]) 17 return mid; 18 else 19 return binarySearch(list, key, mid + 1, high); 20 } 21 22 public static void main(String[]…
Which is the base case of the following recursion function: def mult3(n): if n == 1: return 3 else: return mult3(n-1) + 3 else n == 1 mult3(n) return mult3(n-1) + 3
1. Let product(n,m) be a recursive addition-subtraction method for multiplying two positive integers. Recursive cases for m = 1 and m < 1 make this method. The return value should be n plus a recursive product() call with n and m - 1. Test a Java method.
MazeSolver attempts to recursively traverse a Maze. The goal is to get from the * given starting position to the bottom right, following a path of 1’s. Arbitrary * constants are used to represent locations in the maze that have been TRIED * and that are part of the solution PATH.
For function addOdd(n) write the missing recursive call. This function should return the sum of all postive odd numbers less than or equal to n. Examples: addOdd(1) -> 1addOdd(2) -> 1addOdd(3) -> 4addOdd(7) -> 16 public int addOdd(int n) { if (n <= 0) { return 0; } if (n % 2 != 0) { // Odd value return <<Missing a Recursive call>> } else { // Even value return addOdd(n - 1); }}
Complete the following recursive function that returns the sum of all the numbers in a list that are positive and even numbers. Replace the bold text with the correct expression. #lang racket (define (positiveEvenNums 1st) (if (null? 1st) (if (THE CAR OF THE LIST IS POSITIVE AND EVEN) (+ (car 1st) (positiveEvenNums (cdr 1st))) (positiveEvenNums (cdr 1st))))) (positiveEvenNums '(-3 8 4 3 -2 0 5))
Pascal's triangle is a useful recursive definition that tells us the coefficients in the expansion of the polynomial (x + a)^n. Each element in the triangle has a coordinate, given by the row it is on and its position in the row (which you could call a column). Every number in Pascals triangle is defined as the sum of the item above it and the item above it and to the left. If there is a position that does not have an entry, we treat it as if we had a 0 there. *picture of the pascals triangle* Given the following recursive function signature, write the recursive function that takes a row and a column and finds the value at that position in the triangle. Assume that the triangle starts at row 0 and column 0. Examples: pascal(2, 1) -> 2, pascal(1, 2) -> 0 public int pascal(int row, int column) { }
part 1. draw the recursive call tree for the following function when its called with a(7) int a (int n) { if (n % 4== 0) { return n+1;} else { return a(n-1) + a (n*4); } } part 2. what is the returning value?
In the Recursive Binary Search shown below, there are two base cases: Line 10: the search is exhausted without a finding a value that matches the key Line 16: the search finds a value that matches the key 1 public class RecursiveBinarySearch { 2 public static int binarySearch(int[] list, int key) { 3 int low = 0; 4 int high = list.length - 1; 5 return binarySearch(list, key, low, high); 6 } 7 8 private static int binarySearch(int[] list, int key, 9 int low, int high) { 10 if (low > high) // The list has been exhausted without a match 11 return -low - 1; 12 13 int mid = (low + high) / 2; 14 if (key < list[mid]) 15 return binarySearch(list, key, low, mid - 1); 16 else if (key == list[mid]) 17 return mid; 18 else 19 return binarySearch(list, key, mid + 1, high); 20 } 21 22 public static void main(String[] args) { 23 int[] list = {3, 5, 7, 8, 12, 17,…
Answer must be solved Recursively neighbor(n, prev) Takes a positive integer n and a default argument prev and returns the number of digits in n that have the same digit to its left or to its right. Floor division (//) and modulo (%) operators are useful here. Forbidden methods, operators, and libraries: • You are not allowed to convert n to other data type such str() or list() • You are not allowed to use pow, the ** operator, or import any function (such as math.log) • You are not allowed to declare global variables • You are not allowed to use for/while loops in any form Preconditions n: int -> positive integer prev: int -> positive integer sequence of items Returns: int Examples: >>> neighbor(777) 3 >>> neighbor(469704) 0 >>> neighbor(411477) # 411477 4 >>> neighbor(7951452114587444436999) # 7951452114587444436999 9 >>> neighbor(22222222222222233) 17