Problem Set #1 Quiz Answer
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Graph Search, Shortest Paths, and Data Structures
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Problem Set #1 Quiz Answer
Question 1)
Given an adjacency-list representation of a directed graph, where each vertex maintains an array of its outgoing edges (but *not* its incoming edges), how long does it take, in the worst case, to compute the in-degree of a given vertex? As usual, we use n and m to denote the number of vertices and edges, respectively, of the given graph. Also, let k denote the maximum in-degree of a vertex. (Recall that the in-degree of a vertex is the number of edges that enter it.)
- θ(n)
- θ(m)
- θ(k)
- O
- Cannot determine from the given information
Question 2)
Consider the following problem: given an undirected graph G with n vertices and m edges, and two vertices s and t, does there exist at least one s-t path?
If G is given in its adjacency list representation, then the above problem can be solved in O(m+n) time, using BFS or DFS. (Make sure you see why this is true.)
Suppose instead that G is given in its adjacency *matrix* representation. What running time is required, in the worst case, to solve the computational problem stated above? (Assume that G has no parallel edges.)
- θ(n2)
- θ(n✱m)
- θ(m + n log n)
- θ(m + n)
Question 3)
This problem explores the relationship between two definitions about graph distances. In this problem, we consider only graphs that are undirected and connected. The diameter of a graph is the maximum, over all choices of vertices s and t, of the shortest-path distance between s and t. (Recall the shortest-path distance between s and t is the fewest number of edges in an s-t path.)
Next, for a vertex s, let l(s) denote the maximum, over all vertices t, of the shortest-path distance between s and t. The radius of a graph is the minimum of l(s) over all choices of the vertex s.
Which of the following inequalities always hold (i.e., in every undirected connected graph) for the radius r and the diameter d? [Select all that apply.]
- r ≤ d
- r ≥ d/2
- r ≥ d
- r ≤ d/2
Question 4)
Consider our algorithm for computing a topological ordering that is based on depth-first search (i.e., NOT the “straightforward solution”). Suppose we run this algorithm on a graph G that is NOT directed acyclic. Obviously it won’t compute a topological order (since none exist). Does it compute an ordering that minimizes the number of edges that go backward?
For example, consider the four-node graph with the six directed edges (s, v),(s, w), (v, w), (v,t), (w,t), (t, s). Suppose the vertices are ordered s, v, w, t. Then there is one backwards arc, the (t, s) arc. No ordering of the vertices has zero backwards arcs, and some have more than one.
- Always
- Sometimes yes, sometimes no
- Never
- If and only if the graph is a directed cycle
Question 5)
On adding one extra edge to a directed graph G, the number of strongly connected components…?
- …never decreases by more than 1 (no matter what G is)
- …could remain the same (for some graphs G)
- …will definitely not change (no matter what G is)
- …never decreases (no matter what G is)