题目
题目

Artificial Intelligence Lecture 5 quiz

简答题

For the graph below, find the path from 1(starting point) to 5 (destination).   For DFS, always prioritize visiting the child with the smaller node index. For example, 1 -> 2 instead of 1-> 4, because 2 < 4.  For BFS, the same node can be visited repeatedly. for example, 1 -> 2, 4 -> 3, 4, 5, 6 -> ... (4 is repeated)   Format: for each algorithm, fill 3 things: 1. the cost of the path found from 1 to 5; 2. the path found by the algorithm from 1 to 5; 3. The node sequence visited (for UCS, this is the "explored").    For example: for BFS, the answer is: 200;1,4,5;1,2,4,3,4,5   Fill in the cost, path, and visited nodes of DFS:

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The prompt asks to fill in three items for DFS: (1) the cost of the path from 1 to 5, (2) the path itself, and (3) the sequence of visited nodes. It also provides an example for BFS and includes an answer placeholder for DFS. First, note the actual graph details (edge weights and connections) are not visible in the provided image or text beyond the one example answer string. Without the graph structure (which edges exist, their costs, and the adjacency order used by DFS), it......Login to view full explanation

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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)?

How does depth-first search complete its search of the search tree?

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()

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