Consider the following pseudo-code: I = 0; J = 0; K = 8; while (I < K – 1) //while-1 { J = J + 1; while (J < K) //while-2 { if (x[I] < x[J]) { temp = x[I]; x[I] = x[J]; x[J] = temp; } } // end of while-2 I = I +1; } // end of while-1 The cyclomatic complexity of the above is ?

Consider the following pseudo-code: I = 0; J = 0; K = 8; while (I < K – 1) //while-1 { J = J + 1; while (J < K) //while-2 { if (x[I] < x[J]) { temp = x[I]; x[I] = x[J]; x[J] = temp; } } // end of while-2 I = I +1; } // end of while-1 The cyclomatic complexity of the above is ?

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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Data structure/ C language / Graph / Dijkstra’s algorithm implement a solution of a very common issue: howto get from one town to another using the shortest route.* design a solution that will let you find the shortest paths betweentwo input points in a graph, representing cities and towns, using Dijkstra’salgorithm. Your program should allow the user to enter the input filecontaining information of roads connecting cities/towns. The programshould then construct a graph based on the information provided from thefile. The user should then be able to enter pairs of cities/towns and thealgorithm should compute the shortest path between the two cities/townsentered.Attached a file containing a list of cities/towns with the following data:Field 1: Vertex ID of the 1st end of the segmentField 2: Vertex ID of the 2nd of the segmentField 3: Name of the townField 4: Distance in KilometerPlease note that all roads are two-ways. Meaning, a record may representboth the roads from feild1 to field2…
Greatest Common Divisor Two integers A and B have the greatest common divisor which is the largest positive integer that divides A and B, evenly, without a remainder. One way to find the Greatest Common Divisor is to consider the prime factors for A = 372 and B = 84 as shown below. 372 223 * 31 84 22*3*7 GCD (372,84) = 2²*3= 12 In a well-documented Python program, hmwk3Q3.py, query the user for two integers. Use a while-loop to find the remainder R when dividing A by B.Replace the value of A with the value of B and replace the value of B with the value of R.Continue this process within the while-loop until the value of B is zero. At that time the value of A will be the GCD. In a comment, provide the response to your program when A= 98,025,733,547 and B = 2,345,109,894,323.
11 - The code segment below has time complexity? for (int i=0; i
Consider the producer-consumer problem where the producer produces items to be consumed by the consumer. A solution of the problem is to be implemented using threads. The main process will run producer and consumer as two different threads. The producer will put the items it generates into a buffer of length 15. The consumer reads the elements of the buffer to get the items. A solution is given below. Each item produced and consumed are also printed to the screen. Run the program several times and find a sample run when the items produced are not correctly consumed. Explain why the program does not always work correctly.
#include<bits/stdc++.h> using namespace std; void bubbleSort(int arr[], int n) { for (int i = 0; i < n - 1; i++) { for (int j = 0; j < n - i - 1; j++) { if (arr[j] > arr[j + 1]) { int temp = arr[j]; arr[j] = arr[j + 1]; arr[j + 1] = temp; } } } } int binarySearch(int arr[], int l, int r, int x, int& comp) { comp++; if (r >= l) { int mid = l + (r - l) / 2; if (arr[mid] == x) { return mid; } if (arr[mid] > x) { return binarySearch(arr, l, mid - 1, x, comp); } return binarySearch(arr, mid + 1, r, x, comp); } return -1; } int main() { int Num[8192]; srand(time(NULL)); for (int i = 0; i < 8192; i++) { Num[i] = rand() % 10001; } clock_t starting_time = clock(); bubbleSort(Num, 8192); clock_t ending_time = clock(); clock_t result =…
(4) {w € [0, 1]* : w does not have 001 as a substring} Testing strings: €, 0, 1, 01, 000, 001, 0001, 0010, 0101, 11000, 110010 The accepted strings are: L U E O 01 000 001 0001 ☐ 0010 0101 11000 110010
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The code box below defines a variable route as a list of directions to navigate a maze. Each instruction is one of the following four basic commands: higher move one step in the positive y direction • lower. move one step in the negative y direction • left: move one step in the negative x direction • right: move one step in the positive x direction Define a function step that takes two arguments, a location (as a tuple of x and y coordinates) and an instruction (higher, lower, left, right) as a string. Given the provided location, it should return the new location (as a tuple of x and y coordinates) when following the specified instruction. If the instruction is invalid, the old location should be returned. Use the function step to determine the final position when starting from the point (0, -4) and following all instructions in the list route. Assign this final position to the variable final_point.
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