Question

I am trying to understand what is the space complexity of DFS and BFS in a graph. I understand that the space complexity of BFS while using adjacency matrix would be O(v^2) where v is the number of vertices.

By using the adjacency list space complexity would be decreased in average case i.e < v^2. But in the worst case, it would be O(v^2). Even including Queue, it would be O(n^2) (neglecting the lower value)

But, what is the scenario with DFS? Even if we use the adjacency matrix/list. Space Complexity would be O(v^2). But that seems to be a very loose bound, that too without considering stack frames.

Am I correct, regarding the complexities? If, not what are the space complexities of BFS/DFS? and while calculating Space Complexity for DFS, do we consider stack frame or not?

what is the tight bound of space complexity, for BFS and DFS for graph

Answer 1

如伪代码1所示，邻接矩阵或邻接列表的空间消耗不在BFS算法中。邻接矩阵或邻接表是BFS算法的输入，因此不能包含在空间复杂度的计算中。 DFS也是如此。

伪代码1 输入：图Graph和图的起始顶点根

输出：目标状态。父链接将最短路径追溯到根

  procedure BFS(G,start_v):
      let Q be a queue
      label start_v as discovered
      Q.enqueue(start_v)
      while Q is not empty
          v = Q.dequeue()
          if v is the goal:
              return v
         for all edges from v to w in G.adjacentEdges(v) do
             if w is not labeled as discovered:
                 label w as discovered
                 w.parent = v
                 Q.enqueue(w)

BFS的空间复杂度可以表示为O（| V |），其中| V |是一组顶点的基数。对于最坏的情况，您需要将所有顶点都保留在队列中。

DFS的空间复杂度取决于实现方式。下面显示了具有最坏情况空间复杂度O（| E |）的DFS的非递归实现，其中E是边缘集的基数：

procedure DFS-iterative(G,v):
  let S be a stack
  S.push(v)
  while S is not empty
      v = S.pop()
      if v is not labeled as discovered:
         label v as discovered
         for all edges from v to w in G.adjacentEdges(v) do 
              S.push(w）

深度优先搜索已完成，而深度优先搜索未完成。

Space Complexity of DFS and BFS in graph

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