Assume we have a directed graph G; and the graph H is the same graph without the directions (if u is connected to v in G, then v is connected to u in H as well as u to v).

Which of these statements are true?

Question

Assume we have a directed graph G; and the graph H is the same graph without the directions (if u is connected to v in G, then v is connected to u in H as well as u to v).

Which of these statements are true?

BlackTom AI · Accepted Answer

Question restatement: We are given a directed graph G and a graph H that is the same as G but with all edges made undirected (if u is connected to v by a directed edge in G, then there is an undirected edge between u and v in H). We are to determine which of the following statements are true about strong components of G and connected components of H.

Option 1: If u and v are in the same strong component in G, then u and v are in the same component in H as well.
- Why this is true: If u and v lie in the same strong component of G, there exists a directed path from u to v and a directed path from v to u in G. When we convert G to H by ignoring edge directions, those directed paths become undirected paths (since every edge on those paths is still present, just without orientation). Therefore, u and v are connected in H by an undirected path, placing them in the same connected component of H. The underlying idea is that strong connectivity in G guarantees mutual reachability, which naturally implies connectivity in the undirected version H.

Option 2: If C is a strong component in G, then all of the vertices in C are in the same component in H.
- Why this is true: A strong component C in G consists of vertices where every pair has directed paths in both directions within G. If you take any two vertices in C, the directed paths between them remain in G and, when considered in H, become undirected paths because edge directions are ignored in H. Thus, any two vertices in C are connected in H by an undirected path, so all vertices of C lie in the same connected component of H. Since connectivity in H is defined by the existence of a (undirected) path between vertices, the entire strong component C collapses into a single connected component in H.

Additional notes to avoid common misconceptions:
- It might be tempting to think that making edges undirected could somehow split a strong component into multiple parts in H, but the presence of directed paths in G between every pair of vertices in a strong component ensures corresponding undirected paths in H between those same vertices. Consequently, the entire strong component becomes one connected component in H.
- Conversely, a connected component in H does not necessarily correspond to a single strong component in G, since the lack of direction information in H can hide the necessity for two-way reachability that defines a strong component in G. However, this does not affect the validity of the two statements here, which both assert properties that do hold due to the mutual reachability embedded in strong components.

In summary, both statements describe properties that follow from how strong components in a directed graph relate to the connected structure of the underlying undirected graph.

CPSC_V 320 201/202/203 2024W2 Reading Quiz #3 (Graphs)

Assume we have a directed graph G; and the graph H is the same graph without the directions (if u is connected to v in G, then v is connected to u in H as well as u to v). Which of these statements are true?

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