Question6 Tick all statements about graph representations that are correct.Select one or more alternatives: Given an adjacency-matrix representation of a graph, printing all the edges of the graph can always be achieved in [math] time, where [math] is the number of edges. For a complete graph, the storage cost for the array-of-edges representation is always [math], where [math] is the number of vertices. It is always the case that if a node [math] belongs to a clique of 5 nodes, then there are at least 5 elements in the adjacency list for [math] given an adjacency-list representation of the graph. Deleting all the edges incident on vertex [math] in a graph is achieved in [math] time in the array-of-edges representation, where [math] is the number of edges. ResetMaximum marks: 2 Flag question undefined

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BlackTom AI · Accepted Answer

Question6 asks which statements about graph representations are correct, with multiple answers possible. 
Option 1: 'Given an adjacency-matrix representation of a graph, printing all the edges of the graph can always be achieved in O(E) time, where E is the number of edges.' In an adjacency-matrix representation, edges are captured in a V x V matrix (V = number of vertices). To print all edges, you typically scan the entire matrix and output each existing edge, which costs Θ(V^2) time, not Θ(E) in general. Since E can be as large as Θ(V^2) but can be much smaller in sparse graphs, the statement claiming O(E) time universally is false. 
Option 2: 'For a complete graph, the storage cost for the array-of-edges representation is always O(V^2), where V is the number of vertices.' A complete graph has E = V(V−1)/2 edges. An array-of-edges representation stores one entry per edge, so storage grows as Θ(E) = Θ(V^2). Since E scales with V^2, this is correctly described as O(V^2). 
Option 3: 'It is always the case that if a node v belongs to a clique of 5 nodes, then there are at least 5 elements in the adjacency list for v given an adjacency-list representation of the graph.' In a clique of size 5, each node is connected to the other 4 nodes within the clique, so the degree of v is at least 4 due to those connections. The adjacency list for v would contain at least these 4 neighbors, not necessarily 5. If there are no other connections, it would be exactly 4. Therefore claiming at least 5 elements is incorrect. 
Option 4: 'Deleting all the edges incident on vertex v in a graph is achieved in O(E) time in the array-of-edges representation, where E is the number of edges.' In an array-of-edges (edge list) representation, removing all edges incident to a given vertex typically requires scanning the entire edge list to filter out edges touching v, which is a linear pass over E elements, i.e., O(E) time. Although one could attempt more sophisticated data structures, the standard operation in a simple edge list yields O(E). Hence this statement is true. 
Overall, option-by-option reasoning shows that Option 2 and Option 4 are correct, while Options 1 and 3 are not correct.

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