Step 1: Understand the statements.
- **(A)** In a directed graph, the sum of lengths of all the adjacency lists is **$|E|$**. This is true because each edge is listed exactly once in the adjacency list of a directed graph.
- **(B)** In an undirected graph, the sum of the lengths of all the adjacency lists is **$2|E|$**. Each edge is counted twice in an undirected graph, once for each endpoint.
- **(C)** For a dense graph, adjacency matrix representation is preferable. This is true because dense graphs have a large number of edges, making an adjacency matrix (which stores the presence of edges between every pair of vertices) more efficient than an adjacency list.
- **(D)** The memory requirement of the adjacency matrix of a graph is dependent on the number of edges. This is false; the memory requirement of an adjacency matrix depends on the number of vertices, not edges.
Step 2: Conclusion.
Thus, the correct answer is **(3) (A), (B), (C) and (D)**.
In C language, mat[i][j] is equivalent to: (where mat[i][j] is a two-dimensional array)
Suppose a minimum spanning tree is to be generated for a graph whose edge weights are given below. Identify the graph which represents a valid minimum spanning tree?
\[\begin{array}{|c|c|}\hline \text{Edges through Vertex points} & \text{Weight of the corresponding Edge} \\ \hline (1,2) & 11 \\ \hline (3,6) & 14 \\ \hline (4,6) & 21 \\ \hline (2,6) & 24 \\ \hline (1,4) & 31 \\ \hline (3,5) & 36 \\ \hline \end{array}\]
Choose the correct answer from the options given below:
Match LIST-I with LIST-II
Choose the correct answer from the options given below:
Consider the following set of processes, assumed to have arrived at time 0 in the order P1, P2, P3, P4, and P5, with the given length of the CPU burst (in milliseconds) and their priority:
\[\begin{array}{|c|c|c|}\hline \text{Process} & \text{Burst Time (ms)} & \text{Priority} \\ \hline \text{P1} & 10 & 3 \\ \hline \text{P2} & 1 & 1 \\ \hline \text{P3} & 4 & 4 \\ \hline \text{P4} & 1 & 2 \\ \hline \text{P5} & 5 & 5 \\ \hline \end{array}\]
Using priority scheduling (where priority 1 denotes the highest priority and priority 5 denotes the lowest priority), find the average waiting time.