Q 6. Derive the Time function T(N), best-case complexity and worst-case complexity (showing every single step we discussed in the online class), for the following algorithm. Line Algorithm 1 int Lo = 0; int Hi = data.length -1; int Mid = -1; while (Lo <= Hi) { Mid = Lo + (Hi-Lo)/2; if (data [Mid] = Item) break; if (Item < data [Mid]) Hi = Mid-1; 2 3 4 5 7 8 10 else Lo = Mid+1; 11 } if (data [Mid] == Item) 12 return (Mid); else return (-1); 13 14

Q 6. Derive the Time function T(N), best-case complexity and worst-case complexity (showing every single step we discussed in the online class), for the following algorithm. Line Algorithm 1 int Lo = 0; int Hi = data.length -1; int Mid = -1; while (Lo <= Hi) { Mid = Lo + (Hi-Lo)/2; if (data [Mid] = Item) break; if (Item < data [Mid]) Hi = Mid-1; 2 3 4 5 7 8 10 else Lo = Mid+1; 11 } if (data [Mid] == Item) 12 return (Mid); else return (-1); 13 14

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

Consider the following recursive algorithm: Int active(int list 1 int lowerlndex, iInt upperindex) Int max if (lowerindex == upperindex) return lowerlndex; else max = active(list, lowerlndex + , upperindex); if (list[lowerindex] > = max) return lowerlndex; else return max What is the result of: int x= active([1,5,4,3,2],0,4) Select one: a. 2 b.3
Generate a graph of performance improvement for code below. Tip: Figure 5.17 in the textbook provides an example of a graph depicting performance improvements associated with loop unrolling. code: //normal code int main() { int n, sum=0; n=50; int a[50]; { for(int i=0; i<n; i++) { sum= sum +a[i]; } printf("sum is %d", sum); return 0; } //with unrolling int main() { int n, sum=0; n=50; int a[50] for(int i=0; i<n; i+=10) { scanf("%d", &a[i]); } sum= sum + a[0]; sum= sum + a[1]; sum= sum + a[2]; sum= sum + a[3]; sum= sum + a[4]; sum= sum + a[5]; sum= sum + a[6]; sum= sum + a[7]; sum= sum + a[8]; sum= sum + a[9]; printf("sum is %d", sum); return 0; }
Give a recursive algorithm which takes as input a sequence of numbers and returns the minimum (i.e., smallest) member in the sequence. Your algorithm should not use an iterative loop. FindMin (al, a2,..., an) If (length (alan) = 1), Return (_A_) m: FindMin (al an-1) // The recursive call If (an S m) Else End-if Return (_B__) Return (_C_)
Question 6 What is the time complexity of this recursive function? def foo(x): if x == 0: return 1 else: return foo(x-1) O O O(N) O(N^2) O(log N) 0(1)
What is the time complexity of given loop.
need help in python PYTHON PROGRAMMING Submit a code snippet for a function (in the coding sense) that accepts as an input a list of size n and prints all the integers from 1 up to n. This function must have complexity O(n^2). Hint: Efficiency isn't always the answer. Restraints that have corresponding deductions if not followed are listed below: You must only print each number once (-4/20) The algorithm/code function must have complexity to O(n^2) (-20/20) Update: It needs to be O( n^2 ) hence -20/20 The input must be a list not an integer (-4/20) Example: If the input is a list of size 4, the output is 1 2 3 4
What is the time complexity of the following algorithm? Select one: a. Θ(n) b. Θ(n/2) c. Θ(2n-1) d. Θ(n2)
#1: log (sub)2 (256) = ? #2: What case scenario of an algorithm does Big O notation analyze?
# 1: log(sub)2 (256) = ? #2: What case scenario of an algorithm does Big O notation analyze?
Generate test cases for the following code by using Basic Path Testing. void sort(int a[ ],int N) { int i,j,p,t; } for(i=0;ia[p]) p=j; if(p!=i) {t=a[p]; a[p]=a[i]; a[i]=t; } Draw CFG. How many basic paths for the c) List the basic paths. d) Generate test cases from it.
Q2) Draw program graph for the following code segment then calculate the cyclomatic complexity of the graph. int sea(int[] nums, int item) { int asize = nums.length; int bsize = (int)Math.floor(Math.sqrt(asize)); int pitem = 0; while (nums[Math.min(bsize, asize)-1] = asize) return -1; } while (nums[pitem] < item){ pitem ++; if (pitem == Math.min(bsize, asize)) return -1; if (nums[pitem] == item) return pitem; return -1;