41.
Depth-First Search Challenges
Written by Vincent Ngo
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Challenge 1: BFS or DFS
For each of the following two examples, which traversal (depth-first or breadth-first) is better for discovering if a path exists between the two nodes? Explain why.
- Path from A to F.
- Path from A to G.
Challenge 2: Recursive DFS
In this chapter, you went over an iterative implementation of depth-first search. Now write a recursive implementation.
Challenge 3: Detect a cycle
Add a method to Graph to detect if a directed graph has a cycle.
Solutions
Solution to Challenge 1
- Path from A to F: Use depth-first because the path you are looking for is deeper in the graph.
- Path from A to G: Use breadth-first because the path you are looking for is near the root.
Solution to Challenge 2
In the depth-first search chapter, you learned how to implement the algorithm iteratively. Let’s take a look at how you would implement it recursively.
extension Graph where Element: Hashable {
func depthFirstSearch(from start: Vertex<Element>)
-> [Vertex<Element>] {
var visited: [Vertex<Element>] = [] // 1
var pushed: Set<Vertex<Element>> = [] // 2
depthFirstSearch(from: start, // 3
visited: &visited,
pushed: &pushed)
return visited
}
}
func depthFirstSearch(from source: Vertex<Element>,
visited: inout [Vertex<Element>],
pushed: inout Set<Vertex<Element>>) {
pushed.insert(source) // 1
visited.append(source)
let neighbors = edges(from: source)
for edge in neighbors { // 2
if !pushed.contains(edge.destination) {
depthFirstSearch(from: edge.destination, // 3
visited: &visited,
pushed: &pushed)
}
}
}
Solution to Challenge 3
A graph has a cycle when a path of edges and vertices leads back to the same source.
extension Graph where Element: Hashable {
func hasCycle(from source: Vertex<Element>) -> Bool {
var pushed: Set<Vertex<Element>> = [] // 1
return hasCycle(from: source, pushed: &pushed) // 2
}
}
func hasCycle(from source: Vertex<Element>,
pushed: inout Set<Vertex<Element>>) -> Bool {
pushed.insert(source) // 1
let neighbors = edges(from: source) // 2
for edge in neighbors {
if !pushed.contains(edge.destination) &&
hasCycle(from: edge.destination, pushed: &pushed) { // 3
return true
} else if pushed.contains(edge.destination) { // 4
return true
}
}
pushed.remove(source) // 5
return false // 6
}