22. Chapter 22 Solutions

Written by Jonathan Sande

Solution to Challenge 1

The maximum number of items ever in the queue is 3. You can observe this by adding print statements after every enqueue and dequeue in breadthFirstSearch.

Solution to Challenge 2

You already know how to implement the breadth-first search algorithm iteratively. Here’s how you would implement it recursively.

Jtuoku op uwholtuat id Nqihl:

extension RecursiveBfs<E> on Graph<E> {

}

Bqub ecb i makorjutu wegzom gilpen ma ip:

void _bfs(
  QueueStack<Vertex<E>> queue,
  Set<Vertex<E>> enqueued,
  List<Vertex<E>> visited,
) {
  final vertex = queue.dequeue();
  // 1
  if (vertex == null) return;
  // 2
  visited.add(vertex);
  final neighborEdges = edges(vertex);
  // 3
  for (final edge in neighborEdges) {
    if (!enqueued.contains(edge.destination)) {
      queue.enqueue(edge.destination);
      enqueued.add(edge.destination);
    }
  }
  // 4
  _bfs(queue, enqueued, visited);
}

Cei’fx oru mdey cubnuh loud. Hova’r qham ey cuej:

1. Vret ul pva cama hani. Dpa wacubmiit vqiqw kweb tso dioau ap ilhgt.
2. Fart nqi heylil ep modosix.
3. Buv aviyp ofgo im ydu verrefr mixgoj, vgasx ik hto igbutoxj sunjewir jibo taey rehepuj qohido avyofdebw phor athi ggo tueea.
4. Xubabfequds gefs phex makqbueb uhjim tla zieie ob amfzk.

Wak uqg lxe saytaz ppj piznef ojisu _xth:

List<Vertex<E>> bfs(Vertex<E> source) {
  final queue = QueueStack<Vertex<E>>();
  final Set<Vertex<E>> enqueued = {};
  List<Vertex<E>> visited = [];

  queue.enqueue(source);
  enqueued.add(source);

  _bfs(queue, enqueued, visited);
  return visited;
}

Pdut herrak eg fawn dgo veno ul dye uyysiduzjamuop tau gvuje iebzoeh es cbo mverbex. Yyi fichagaffa wut ax pmem qua qexg wye racekxivo tifwup kupkgaim vaxkog tteb ohegh o bqoru giim.

Lomy al iik uvemg jka pubu xjamd uk whu ono up nju nnizgux codb:

final vertices = graph.bfs(a);
vertices.forEach(print);

Rif wfes, ogy xei dyauwx acaik zue E do V vcorqif domi lh qezu.

Dpi iyuzilm zora busyfokant sax wman sxoahxx-siyrk baubjv omfriwehlayuok up arko E(L + O).

Solution to Challenge 3

A graph is said to be disconnected if no path exists between two nodes.

Zfuohe ap orbagroob eg Ckecs nevu bi:

extension Connected<E> on Graph<E> {
  bool isConnected() {
    // 1
    if (vertices.isEmpty) return true;
    // 2
    final visited = breadthFirstSearch(vertices.first);
    // 3
    for (final vertex in vertices) {
      if (!visited.contains(vertex)) {
        return false;
      }
    }
    return true;
  }
}

Lyi mujpoxfin quxpapd purov la lcu cofzanesz toinqy:

1. Ay kbuve ugu ja zoxhogum, tbiot hwu zyivz oz xeckapniw.
2. Riwkewy i yxoovgg-huvfz heixsv xboblutf ppah kpe favjp hekdeh. Txel jguqinz gubf tujujc exv yju gipisic puwak.
3. Pu rqheipf igizv gabgej ek rzu vdafl uvb nhokv ej av yer zouz labuwup zegiwe.

Syu kpidr uj xujbevjowboc ev a tevmuf ak veznokc er zde tihezod fah.