bool Graph::searchNeighbors (int nodeId, int neighborId) {Node* current = vertices[nodeId];while (current != nullptr) {if (current->data == neighborId) {return true;current = current->next;}}}return false;void Graph::loadGraph(string edgeListFileName){ifstream file(edgeListFileName);int u, v;if (file.is_open()) {while (file >> u » v) {addNeighbor (u, v);file.close();}}}void Graph::dumpGraph(string adj ListFileName) {ofstream file(adj ListFileName);if (file.is_open() ) {for (size_t i = 0; i < vertices.size(); i++) {file << i << ": ";Node* current = vertices[i];while (current != nullptr) {}file << current->data << " ";current = current->Next;file << endl;}file.close();}} Test 4: Toy graph, adjacency list (0/2)Test failed!

The code you provided seems to be a part of a larger graph implementation, and it is difficult to…

Answered: bool Graph::searchNeighbors (int…

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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Similar questions

How would I convert this adjacency matrix into an adjacency list and then print the adjacency list using Java, I have implemented my own graph data structure and have tried to do this conversion in displayAsList() unsuccessfully. public class Q5Graph {int maxsize;int wmatrix[][];String labels[];int vertexCount; public Q5Graph() { maxsize = 20; wmatrix = new int[maxsize][maxsize]; labels = new String[maxsize];for(int i=0; i < maxsize; i++) for (int j=0; j< maxsize; j++){ wmatrix[i][j] = 0;}vertexCount = 0;} public void addVertex(String label){if (vertexCount == maxsize) { throw new IllegalArgumentException("Graph is already full!"); } else if (hasVertex(label)) { throw new IllegalArgumentException("Vertex '"+label+"' already exists"); } else {labels[vertexCount] = label;vertexCount++;}} public void addEdge(String label1, String label2, int weight){int v1, v2; if(!hasVertex(label1)){throw new IllegalArgumentException("First Vertex '" +…
vector<VertexT> Vertices; map<VertexT, vector<list>> adjList; What is the worst case time complexity for this version of _LookupVertex? bool _LookupVertex(VertexT v) const { if (adjList.count(v) != 0) { return true; } return false; } V is the number of total vertices in the graph, E is the average number of edges per vertex. O(v) O(VlogE) O(logV) O(V^2) O(E+V)
A graph file is provided below. What is the total number of fan-in and fan-out of vertex 4? # Graph example 5 # Number of vertices D # Directed graph 0 1 10 # List of edges 0 3 20 1 3 5 0 2 3 2 1 2 2 4 15 3 4 11
(0)URGENT!!!! Please use this graph class and SOLVE FOR ISOLATEDTRIPLETS QUESTION AND AVERAGE BALANCE RATIO Here is the graph class: public class Graph {private final int V;private int E;private final Bag<Integer>[] adj; /*** Create an empty graph with V vertices.*/@SuppressWarnings("unchecked")public Graph(int V) {if (V < 0) throw new Error("Number of vertices must be nonnegative");this.V = V;this.E = 0;this.adj = new Bag[V];for (int v = 0; v < V; v++) {adj[v] = new Bag<>();}} /*** Return the number of vertices in the graph.*/public int V() { return V; } /*** Return the number of edges in the graph.*/public int E() { return E; } /*** Add the undirected edge v-w to graph.* @throws java.lang.IndexOutOfBoundsException unless both 0 <= v < V and 0 <= w < V*/public void addEdge(int v, int w) {if (v < 0 || v >= V) throw new IndexOutOfBoundsException();if (w < 0 || w >= V) throw new IndexOutOfBoundsException();E++;adj[v].add(w);adj[w].add(v);} /***…
Given the graph class using the adjacency matrix representation,Implement the DFS algorithm (as discussed in the lecture slide) ongraph using the adjacency matrix representation. You have to use thegraph code as a starting point, and add the DFS method in the graphclass similar to the BFS method already present in the Graph class.
"""1.In a directed graph, a strongly connected component is a set of vertices suchthat for any pairs of vertices u and v there exists a path (u-...-v) thatconnects them. A graph is strongly connected if it is a single stronglyconnected component.""" from collections import defaultdict class Graph: """ A directed graph where edges are one-way (a two-way edge can be represented by using two edges). """ def __init__(self,vertex_count): """ Create a new graph with vertex_count vertices. """ self.vertex_count = vertex_count self.graph = defaultdict(list) def add_edge(self,source,target): """ Add an edge going from source to target """ self.graph[source].append(target) def dfs(self): """ Determine if all nodes are reachable from node 0 """ visited = [False] * self.vertex_count self.dfs_util(0,visited) if visited == [True]*self.vertex_count: return True…
Code it. In graph theory, a component, sometimes called a connected component,of an undirected graph is a subgraph in which anytwo vertices are connected to each other by paths.Example: 1 3------------7 | | 2--------4 | | | | output = 2 6--------5''' # Code is Here def dfs(source,visited,adjacency_list): ''' Function that performs DFS ''' visited[source] = True for child in adjacency_list[source]: if not visited[child]: dfs(child,visited,adjacency_list) def count_components(adjacency_list,size): ''' Function that counts the Connected components on bases of DFS. return type : int ''' count = 0 visited = [False]*(size+1) for i in range(1,size+1): if not visited[i]: dfs(i,visited,adjacency_list) count+=1 return count def main(): """ Example application """ node_count,edge_count = map(int, input("Enter the Number of Nodes and Edges…
dict_graph = {} # Read the data.txt file with open('data.txt', 'r') as f: for l in f: city_a, city_b, p_cost = l.split() if city_a not in dict_graph: dict_graph[city_a] = {} dict_graph[city_a][city_b] = int(p_cost) if city_b not in dict_graph: dict_graph[city_b] = {} dict_graph[city_b][city_a] = int(p_cost) # Breadth First Search Method def BreadthFirstSearch(graph, src, dst): q = [(src, [src], 0)] visited = {src} while q: (node, path, cost) = q.pop(0) for temp in graph[node].keys(): if temp == dst: return path + [temp], cost + graph[node][temp] else: if temp not in visited: visited.add(temp) q.append((temp, path + [temp], cost + graph[node][temp])) # Depth First Search Method def DepthFirstSearch(graph, src, dst): stack = [(src, [src], 0)] visited = {src} while stack:…
you get setup to work with graphs.Create a Graph class to store nodes and edges or download a Graph librarysuch as JUNG. Use it to implement Breadth First Search and Depth First SearchFollow the video from class if you need a reference.
Breath First Search Distance Given a directed graph as an adjacency list, you will determine, in this challenge, the breath first search distance of all nodes from a starting vertex, s, assuming that the weight of each edge is 1. In the example graph belove, for s=0, the distances of each vertex from s are 0 1 1 2 3 The output includes the distances for the vertices, 0, 1, 2, 3, 4, in the order of the vertex number. If s=1, then the output will be -1 0 -1 1 2 For the starting vertex 1, the vertex 0 and 2 are unreachable, so the corresponding BFS distances are -1 and -1. Input Format First line in the input contains the starting vertex s Each of the subsequent lines contains two space-separated integers, v and w, that represents an edge from the node v to w. Constraints Node numbers are 0 based, i.e. if there n nodes, the nodes are numnered as 0,1,2,...,n-1. Output Format Print the distace of each vertex from the start vertex in the order of the vertex number. Sample Input 0 0 0 1 0 2…
Think about a set of pairs of real numbers called V intervals on the real line. With a vertex for each interval and edges linking them when they intersect (have any points in common), a collection like this creates an interval graph. Create a software that generates V randomly selected intervals in the unit interval, each of length d, and then builds the accompanying interval graph.Use a BST, please.
Please I need an explanation of the code and how I can implement it graph={}route=[]visited=[] def dfs(start,dest): lmt = input("\nLimit: ") route.append(start) countnode=0 countlimit=0 nodeinlevel=0 while len(route)>0: vertex=route.pop(len(route)-1) if vertex not in visited: visited.append(vertex) route.extend(set(graph[vertex])-set(visited)) countnode+=1 if nodeinlevel==countnode: nodeinlevel=len(route) countnode=0 countlimit+=1 if countlimit==lmt: print ('cut off') break if visited._contains_(dest): return visitedst = input("Starting Point: ")if st not in graph: print(st+"does not exist\n\nStarting Point: "); st = input("Starting Point: ") en = input("\n\nEnding Point: ")if en not in graph: print(en+"does not exist\n\nEnding Point: "); en = input() v = dfs(st, en) print("\nShortest Path: ")…

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