Give an example of a connected, weighted, undirected graph such that neither the BFS-tree nor the DFS-tree is an MST, regardless of how the adjacency lists are ordered and at which vertex BFS or DFS are started. Argue that your graph satisfies the above requirements.

Adjacency-lists data structure. The standard graph representation for graphs that arenot dense is called the adjacency-lists data structure, where we keep track of all thevertices adjacent to each vertex on a linked list that is associated with that vertex. Wemaintain an array of lists so that, given a vertex, we can immediately access its list. Toimplement lists, we use our Bag ADT from Section 1.3 with a linked-list implementation, so that we can add new edges in constant time and iterate through adjacent vertices in constant time per adjacent vertex. The Graph implementation on page 526 is basedon this approach, and the figure on the facing page depicts the data structures built bythis code for tinyG.txt. To add an edge connecting v and w, we add w to v’s adjacencylist and v to w’s adjacency list. Thus, each edge appears twice in the data structure. ThisGraph implementation achieves the following performance characteristics:■ Space usage proportional to V + E■ Constant time to add…

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Give an example of a connected, weighted, undirected graph such that no matter at which vertex you start a BFS (breadth first search) the resulting BFS-tree is not a MST (minimum spanning tree), regardless of how the adjacency lists are ordered. Justify correctness of your example.

We have an issue if G is a connected, undirected graph. An edge that connects adjacent edges in a graph is known as a bridge. Every bridge must be an edge in order for the depth-first search tree of G to exist. Give one or two examples to support your point.

Write Algorithm to computes SCDS of any vertex set R in a DP graph with essential dominating pair (u, v) with d(u, v) >9.

Let R be the reduction from 3-SAT to VERTEX-COVER as discussed in class.

Prove that a minimum spanning tree of a graph component must include a shortest edge. It is possible, in Dijkstra’s algorithm, that an edge removed from the priority queue is not useful: it takes us to a previously visited node. Some of these extraneous edges can be avoided by not placing an edge in the priority queue if the destination has already been visited. Is it still possible to encounter an edge to a previously visited node?

Justify whether true or false:  For any undirected connected graph G, there exists at least two nodes such that removing thenodes won’t disconnect the graph.

An undirected, connected graph with least weight edges is a subgraph that contains all vertices. There may be several spanning trees for a given graph. In a spanning tree, the total weight is the sum of the weights of all of the edges. The number of edges in an MST is (vertices - 1).