CS502 Quiz 2 Solution and Discussion

[moaaz]

In strong components algorithm, first of all DFS is run for computing finish times of vertices. CS502

• True Page No.138
• False

This is presented in Figure 8.37. Recall that the component DAG consists of super vertices.

[moaaz]

In strong components algorithm, first of all DFS is run for computing finish times of vertices. CS502

• [d[u], f[u]]⊆[d[v], f[v]]
• [d[u], f[u]]⊇[d[v], f[v]]
• unrelated Page No.129
• Disjoint

The ancestor and descendent relation can be nicely inferred by the parenthesis lemma. u is a descendent of v if and only if [d[u], f[u]]⊆[d[v], f[v]]. u is a ancestor of v if and only if[d[u], f[u]]⊇[d[v], f[v]]. u is unrelated to v if and only if[d[u], f[u]]and[d[v], f[v]]are disjoint. The is shown in Figure 8.26. The width of the rectangle associated with a vertex is equal to the time the vertex was discovered till the time the vertex was completely processed (colored black). Imagine an opening parenthesis ‘(’ at the start of the rectangle and and closing parenthesis ‘)’ at the end of the rectangle. The rectangle (parentheses) for vertex ‘b’ is completely enclosed by the rectangle for ‘a’. Rectangle for ‘c’ is completely enclosed by vertex ‘b’ rectangle.

[moaaz]

For traversing graphs, Breadth-first search can be visualized as a wave front propagating inwards towards root (or source) node. CS502

• True
• False Page No.117

Breadth-first Search:
Here is a more efficient algorithm called the breadth-first search (BFS) Start with s and visit its adjacent nodes. Label them with distance 1. Now consider the neighbors of neighbors of s. These would be at distance 2. Now consider the neighbors of neighbors of neighbors of s. These would be at distance 3. Repeat this until no more unvisited neighbors left to visit. The algorithm can be visualized as a wave front propagating outwards from s visiting the vertices in bands at ever increasing distances from s.

[moaaz]

Digraphs are not used in communication and transportation networks. CS502

• True
• False Page No.135

Strong Components:
We consider an important connectivity problem with digraphs. When diagraphs are used in communication and transportation networks, people want to know that their networks are complete.

[moaaz]

You have an adjacency list for G, what is the time complexity to compute Graph transpose G^T.? CS502 ?

• (V + E) Page No.138
• V E
• V
• E

We will discuss the algorithm without proof. Define GT to be the digraph with the same vertex set at G but in which all edges have been reversed in direction. This is shown in Figure 8.38. Given an adjacency list for G, it is possible to compute G Tin Θ (V+E) time.

[moaaz]

In computing the strongly connected components of a digraph, vertices of the digraph are not partitioned into subsets. CS502

• True
• False Page No.135

We partition the vertices of the digraph into subsets such that the induced sub graph of each subset is strongly connected.

[moaaz]

There are no ________ edges in undirected graph. CS502

• Forward
• Back
• Cross
• Both Forward and Back Page No. 130

For undirected graphs, there is no distinction between forward and back edges.

[moaaz]

We say that two vertices u and v are mutually not reachable if u can reach v and vice versa. CS502

• True Page No. 135
• False

A digraph is strongly connected if for every pair of vertices u, v∈V,u can reach v and vice versa.