How do you trace the path of a Breadth-First Search, such that in the following example:
If searching for key 11
, return the shortest list connecting 1 to 11.
[1, 4, 7, 11]
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Answer
You should have look at http://en.wikipedia.org/wiki/Breadth-first_search first.
Below is a quick implementation, in which I used a list of list to represent the queue of paths.
# graph is in adjacent list representation graph = { '1': ['2', '3', '4'], '2': ['5', '6'], '5': ['9', '10'], '4': ['7', '8'], '7': ['11', '12'] } def bfs(graph, start, end): # maintain a queue of paths queue = [] # push the first path into the queue queue.append([start]) while queue: # get the first path from the queue path = queue.pop(0) # get the last node from the path node = path[-1] # path found if node == end: return path # enumerate all adjacent nodes, construct a # new path and push it into the queue for adjacent in graph.get(node, []): new_path = list(path) new_path.append(adjacent) queue.append(new_path) print bfs(graph, '1', '11')
This prints: ['1', '4', '7', '11']
Another approach would be maintaining a mapping from each node to its parent, and when inspecting the adjacent node, record its parent. When the search is done, simply backtrace according the parent mapping.
graph = { '1': ['2', '3', '4'], '2': ['5', '6'], '5': ['9', '10'], '4': ['7', '8'], '7': ['11', '12'] } def backtrace(parent, start, end): path = [end] while path[-1] != start: path.append(parent[path[-1]]) path.reverse() return path def bfs(graph, start, end): parent = {} queue = [] queue.append(start) while queue: node = queue.pop(0) if node == end: return backtrace(parent, start, end) for adjacent in graph.get(node, []): if node not in queue : parent[adjacent] = node # <<<<< record its parent queue.append(adjacent) print bfs(graph, '1', '11')
The above codes are based on the assumption that there’s no cycles.