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- Develop a test client for Graph that reads a graph from the input stream named as command-line argument and then prints it, relying on toString().Create a Graph test client that reads a graph from the command-line parameter input stream and outputs it using function toString() { [native code] } ()1- CREATE JAVA CODE TO BUILD A GRAPH ( import from file )? 2-PRINT THE TOPOLOGICAL ORDER CRILE 3- GIVEN A SOURCE NODE PRINT THE CONTENTS USING BREADTH FIRST AND DEPTH traversal ? 4- CHECK THE GRAPH IS CONNECTED OR NOT 5- PRINT THE ADJACENCY MATRIX FOR THE GRAPH
- A client can add instances of DirectedEdge class to a List<Edge>, providedthat class Edge extends DirectedEdge. Group of answer choices True FalseCreate a Graph test client that receives a graph from the command-line argument input stream and prints it using function toString() { [native code] } ().Create an implementation of LinkedList provided. For each implementation create a tester to verify the implementation of thatdata structure performs as expected Your task is to: Create a client ( a class with a main ) ‘StagBusClient’ which builds a bus route by performing the following operations on your linked list: Create (insert) 4 stations List the stations Check if a station is in the list (print result) Check for a station that exists, and one that doesn’t Remove a station List the stations Add a station before another station. List the stations Add a station after another station. Print the stations LinkedList.java package linkedList; public interface LinkedList { public Boolean isItemInList(String thisItem); // true if it is, false if not public Boolean addItem(String thisItem); // true if added, false if it was already there, or an error public Integer itemCount(); public void listItems(); public Boolean deleteItem(String thisItem); // true if deleted, false if not…
- THIS IS MY CODE HELP ME ACHIEVE POINTS OUTLINED BELOW : #include <stdio.h>#include <stdlib.h>#include <string.h>#include <float.h>#include "graph.h"#include "dijkstra.h" #define INFINITY DBL_MAX /* find shortest paths between source node id and all other nodes in graph. *//* upon success, returns an array containing a table of shortest paths. *//* return NULL if *graph is uninitialised or an error occurs. *//* each entry of the table array should be a Path *//* structure containing the path information for the shortest path between *//* the source node and every node in the graph. If no path exists to a *//* particular desination node, then next should be set to -1 and weight *//* to DBL_MAX in the Path structure for this node */Path *dijkstra(Graph *graph, int id, int *pnEntries){ int n; int i, j; int* nv = get_vertices(graph, &n); int *S = malloc(n * sizeof(int)); double *D = malloc(n * sizeof(double)); int *R = malloc(n…83. Reconfiguration of the stations in transparent bridges are unnecessary, if the bridge has added or deleted from a. Packages b. Frames c. Segments d. SystemRefer to image to answer this; Show the current state of the runtime stack at the call pointed to below –include static and dynamic links.
- Create the following graph. Implement the following functions. 1. addEdge() that takes two vertices as two parameters and creates an Edge between them. 2. nonAdjacentVertices() that takes a single vertex as parameter, and returns an array of vertices that are NOT adjacent to that vertex. 3. Write a function addUnknownVertices() that takes a vertex as a parameters, and creates an Edge with the vertices that are NOT adjacent to it. 4. Write a method searchVertex() which takes an array of Vertices as parameters. The first member of the array will be the starting vertex and the last member will be the Vertex you want to reach. For example searchVertex(Vertex 0, Vertex 4, Vertex 3) will output “Vertex cannot be reached” since you can reach from 0 to 4, but cannot from 4 to 3. Again, searchVertex(Vertex 0, Vertex 1, Vertex 2, Node 3) will output “Vertex can be reached” since you can reach from 0 to 1, 1 to 2, 2 to 3.What exactly is Direct Mapping, and how exactly does it function?Given a linked list of N nodes such that it may contain a loop. A loop here means that the last node of the link list is connected to the node at position X(1-based index). If the link list does not have any loop, X=0. Remove the loop from the linked list, if it is present, i.e. unlink the last node which is forming the loop. Example 1: Input: N = 3 value[] = {1,3,4} X = 2 Output: 1 Explanation: The link li. st looks like 1 -> 3 -> 4 ^ | |____|