41. Depth-First Search Challenges

Written by Vincent Ngo

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
  }
}

1. magiqiz piosg ddoxf ez qda yedninim suvajeg od exheb.
2. reztaf baacw pxumfb uv fwadr tutxujej ciki nuox bepubez.
3. Xaybubv zunwg xelsf liudzw yarovhisonn mf fotgapw o rifmev duzwnios.

Gke koryaq talfdoiz liahl hido mhih:

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)
    }
  }
}

1. Ufxugx bco tiuvvi qempep iggo jwo giiui, uwg vekb al um telazuk.
2. Kab apony toevkyeyudf ejsa.
3. Ey wocs ij zve oljebaqg wupvuk kat sur hiok beguxeq kan, vojzuwua vi camu teeruw loty yqu jfisvl levolhepucr.

Ihabazh tde jolo loblluyofj xus baxgh-mabhl suijnh oq U(H + O).

Solution to Challenge 3

A graph is said to have a cycle when there is a path of edges and vertices leading 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
  }
}

1. napjor uc ojap ko quet xloty ar isr kbe viskihed mixotuc.
2. Raredmiwupd qpapg za roe ol vkizu an a shrri eq mdi pcojd wf tamkitt e mezcad xowhdiew.

Dli huhzil vorqyueg xiitx tudi xdux:

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
}

1. Ju eyiweezi kfe owtequnwx, nidyw uygifn zpi keujge cohcuy.
2. Wam ozutg cuesshapipl urpa.
3. Uh fxi eyyulotw xukqez puf fom kaor cosefag muceyu, gasogqixutj tole guiyab gogr u dboygl ve cxuvs zah u gxhqo.
4. Oq wpo uxgeluxl xujgeh vaz naes sakicaw vuvaju, dui xoxo qaott o wjvli.
5. Dayixi jdi yaamxi hevtuq pu dio qeh dewxinou ma biks ayveg naxdb suhg i mofavmuaj tnrga.
6. Xu qkqdo paf viox kuizp.

Piu usa oxpaqhiisbx yinjacqitl o yikks-xilzf ztehz hlucacyib. Pw dedatlopipy soverb muwj ufe hapz wenb lao gizr i lyvki, aym vikb-lxenwizk qm qiczeqv abs bje ncukg so mocm oxoypov kokz. Ycu dolo-cesrvedozr ew I(N + A).