题目
Introduction to Artificial Intelligence (082025-VTK)
单项选择题
How does depth-first search complete its search of the search tree?
选项
A.It selects the deepest node in the tree first
B.It tries nodes from all levels simultaneously.
C.It backtracks and tries other branches within a level before moving to the next level
D.It always starts with the root node
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标准答案
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思路分析
When evaluating how depth-first search (DFS) completes its search, it helps to recall the core behavior of DFS.
Option 1: 'It selects the deepest node in the tree first' implies that DFS always targets the absolute deepest node before exploring other paths, which is not accurate; DFS traverses along a p......Login to view full explanation登录即可查看完整答案
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类似问题
The diagram below represents a problem using a search tree. Which of the following options shows the correct order of node visits to reach Wendy using Depth-First Search (DFS)?
DFS_Graph_4 Context: This question pertains to the use of the Graph Abstract Data Type (ADT) implemented with an Adjacency Map, as studied in our course. Instructions: Begin the traversal at Vertex 'A'. When selecting the next vertex to visit, adhere to alphabetical order. Question: Complete the depth-first search (DFS) for the graph shown below. Guidelines: Initiate the traversal at Vertex 'A', and proceed with the exploration, selecting vertices in alphabetical order where multiple paths are available. def DFS(g, u, discovered): for e in g.incident_edges(u): v = e.opposite(u) if v not in discovered: discovered[v] = e # mark v as discovered via edge e DFS(g, v, discovered)
DFS_Pse_2 This question pertains to the use of the Graph Abstract Data Type (ADT) implemented using an adjacency map, as studied in our course. The algorithms DFS and BFS are used to explore graphs but follow different strategies for traversal. Below is a simplified pseudocode version of a Depth-First Search (DFS) algorithm that uses recursion and a discovereddictionary to track visited vertices: DFS(Graph G, Vertex u, Map discovered): for each edge e incident to u in G: let v be the vertex opposite u on edge e if v is not in discovered: discovered[v] ← e // edge e discovered v DFS(G, v, discovered) In the DFS pseudocode, what does the discovered[v] ← e assignment represent? Graph ADT For reference: class Vertex: def __init__(self, x): self._element = x class Edge: def __init__(self, u, v, x): self._origin = u self._destination = v self._element = x def opposite(self, v): return self._destination if v == self._origin else self._origin class Graph: def __init__(self, directed=False): self._outgoing = {} self._incoming = {} if directed else self._outgoing def incident_edges(self, v): return self._outgoing[v].values()
DFS_Graph_3 Context: This question pertains to the use of the Graph Abstract Data Type (ADT) implemented with an Adjacency Map, as studied in our course. Instructions: Begin the traversal at Vertex 'd'. When selecting the next vertex to visit, adhere to alphabetical order. Question: Complete the depth-first search (DFS) for the graph shown below. Guidelines: Initiate the traversal at Vertex 'd', and proceed with the exploration, selecting vertices in alphabetical order where multiple paths are available. def DFS(g, u, discovered): for e in g.incident_edges(u): v = e.opposite(u) if v not in discovered: discovered[v] = e # mark v as discovered via edge e DFS(g, v, discovered)
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