Assume a graph G is simple (ie. no self loop or parallel edges). Let v be any vertex in the graph. Let boolean[] marked be initalized to all false. Consider: boolean dfs(Graph G, int v) { } marked[v]= true; for (int w G.adj(v)) { if (w == u) continue; if (marked[w]) return true; if (dfs (G, v, w)) return true; } return false; If the call dfs(v) returns true, then: a. The graph has cycles b. The graph is bipartite c. The graph is connected Not enough information b a

Assume a graph G is simple (ie. no self loop or parallel edges). Let v be any vertex in the graph. Let boolean[] marked be initalized to all false. Consider: boolean dfs(Graph G, int v) { } marked[v]= true; for (int w G.adj(v)) { if (w == u) continue; if (marked[w]) return true; if (dfs (G, v, w)) return true; } return false; If the call dfs(v) returns true, then: a. The graph has cycles b. The graph is bipartite c. The graph is connected Not enough information b a

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Concept explainers

Polynomial Time

The problems for which the deterministic polynomial-time algorithm exists belong to the complexity class (p) and it is central in the field of computational complexity theory. According to Cobham’s thesis, polynomial time is a synonym for “feasible”, “fast”, and “efficient”.

Question

Assume a graph G is simple (i.e., no self-loop or parallel edges).

Let v be any vertex in the graph.
Let boolean[] marked be initialized to all false.
Consider:

```java
boolean dfs(Graph G, int v) {
marked[v] = true;
for (int w : G.adj(v)) {
if (w == u) continue;
if (marked[w]) return true;
if (dfs(G, w)) return true;
}
return false;
}
```

If the call dfs(v) returns true, then:
a. The graph has cycles
b. The graph is bipartite
c. The graph is connected

- ○ Not enough information
- ○ b
- ○ c
- ○ a

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