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- DFS is run on the following graph. F is the starting vertex. F A B E Which vertices are in the stack before the first iteration of the while loop? Given vertices B, C are in the stack after the first iteration of the while loop: Which vertices are in the stack after the second iteration of the while loop? Which vertices are in visitedSet after the second iteration of the while loop? Ex: A, B, C, D, E, F (commas between values)Data Structures Weighted Graph Applications Demonstration Look at Figure 29.23 which illustrates a weighted graph with 6 verticies and 8 weighted edges. Simply provide: Minimal Spanning Tree as an illustration or a textual list of edges (in our standard vertex order). Single-Source Shortest Path route from vertex 0 to the other 5 (described as one path/route for each). You may draw the two solutions and attach the illustration or describe them in text (a list of edges for the one and the vertex to vertex path the other). Be sure the final trees or path lists are clearly visible in your solution.Question 11 Complete the function body below. def max_neighbours(g): Input : adjacency matrix of graph (g) Out put: vertex with maximum number of neighbours in g # your code here #
- 30. In the implementation for a breadth-first search we studied, a queue was used. The code below replaces the queue with a stack. List the pre-order enumeration that the vertices in the graph below are visited using this modified method, starting from vertex 0. } 3 /** stack-based search */ static void modSearch (Graph G, int start) { Stack S new AStack (G.n()); S.push(start); } 6 G.setMark (start, VISITED); while (S.length) > 0) { } modSearch: intv S.pop(); PreVisit (G, v); for (int w = G.first (v); w < G.n(); w = G.next(v, w)) if (G.getMark (w) == UNVISITED) { G.setMark (w, VISITED); S.push(w);What prerequisites must be met in order to determine whether a linked list T is empty, (i) simple singly linked list, (ii) headed singly linked list, (iii) simple circularlylinked list or (iv) headed circularly linked list?Explain STACK as ADT. List out the application of Stack Define Graph. List its type with example. What is Queue. List the few application of Queue. Explain ADT operation for Array implementation of Queue Explain the following operations in a single linked list. Insert an element Delete an element
- WRITE A PROGRAM IN C++ BFS traverses the graph in a concentric fashion, visiting all vertices that are near to a starting vertex, then all unvisited vertices two edges away from it, and so on, until all vertices in the same linked component as the initial vertex are visited. BFS utilises queue in place of a stack. Input graph as Adjacency Matrix Constraints NA Output Format space separated nodes, which are rechable Sample Input 0 0 1 1 1 0 0 0 1 0 0 0 0 0 0 1 0 Sample Output 0 2 3 4Hippity hoppity, abolish loopity def frog_collision_time(frog1, frog2): A frog hopping along on the infinite two-dimensional lattice grid of integers is represented as a 4-tuple of the form (sx, sy, dx, dy) where (sx, sy) is its starting position at time zero, and (dx, dy) is its constant direction vector for each hop. Time advances in discrete integer steps 0, 1, 2, 3, ... so that each frog makes one hop at every tick of the clock. At time t, the position of that frog is given by the formula (sx+t*dx, sy+t*dy) that can be nimbly evaluated for any t.Given two frogs frog1 and frog2 that are guaranteed to initially stand on different squares, return the time when both frogs hop into the same square. If these two frogs never simultaneously arrive at the same square, return None.This function should not contain any loops whatsoever. The result should be calculated using conditional statements and integer arithmetic. Perhaps the best way to get cracking is to first solve a simpler version…Hippity hoppity, abolish loopity def frog_collision_time(frog1, frog2): A frog hopping along on the infinite two-dimensional lattice grid of integers is represented as a 4- tuple of the form (sx, sy, dx, dy) where (sx, sy) is its starting position at time zero, and (dx, dy) is its constant direction vector for each hop. Time advances in discrete integer steps 0, 1, 2, 3, ... so that each frog makes one hop at every tick of the clock. At time t, the position of that frog is given by the formula (sx+t*dx, sy+t*dy) that can be nimbly evaluated for any t. Given two frogs frog1 and frog2 that are guaranteed to initially stand on different squares, return the time when both frogs hop into the same square. If these two frogs never simultaneously arrive at the same square, return None. This function should not contain any loops whatsoever. The result should be calculated using conditional statements and integer arithmetic. Perhaps the best way to get cracking is to first solve a simpler…
- To apply the binary search algorithm, the data items should be represented as: а. a tree b. a list implemented as a linked-list a list implemented as an array C. d. an ordered list implemented as an array Given a graph with M edges, what is the total degree of all the vertices? a. b. M? с. М/2 d. 2MQuestion 3 Construct an adjacency list for the graph given below. Fill your answers in the given order (a1 through a14) >[a1] - [a2] A Ta3] Ta4] [a5] > [a6] [a7] [a8] [a9] D. [a10] [a11] [a13] > Ta14]Part 2) Adjacency-matrix graph representation: Java implementation Adjacency-list graph representation: Java implementation Create two classes Graph, one for each representation. Add variable members, constructors, addEdge, searchEdge (given two nodes return true if adjacent or false otherwise), and Iterable (iterates through adjacent nodes).