如何追踪广度优先搜索中的路径？

Question

如何跟踪广度优先搜索的路径，例如在以下示例中：

如果搜索键11，则返回连接的最短列表1 至 11。

[1, 4, 7, 11]

Answer 1

您应该查看http://en.wikipedia.org/wiki/Breadth-first_search 首先。

---

下面是一个快速实现，其中我使用列表列表来表示路径队列。

# 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')

打印：['1', '4', '7', '11']

---

另一种方法是维护从每个节点到其父节点的映射，并在检查相邻节点时记录其父母。搜索完成后，只需根据父映射进行回溯即可。

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')

上面的代码基于没有循环的假设。

Answer 2

非常简单的代码。每次发现节点时，您都​​会不断附加路径。

graph = {
         'A': set(['B', 'C']),
         'B': set(['A', 'D', 'E']),
         'C': set(['A', 'F']),
         'D': set(['B']),
         'E': set(['B', 'F']),
         'F': set(['C', 'E'])
         }
def retunShortestPath(graph, start, end):

    queue = [(start,[start])]
    visited = set()

    while queue:
        vertex, path = queue.pop(0)
        visited.add(vertex)
        for node in graph[vertex]:
            if node == end:
                return path + [end]
            else:
                if node not in visited:
                    visited.add(node)
                    queue.append((node, path + [node]))

Answer 3

我非常喜欢qiao的第一个回答！
这里唯一缺少的是将顶点标记为已访问。

为什么我们需要这样做？
假设有另一个节点 13 从节点 11 连接。现在我们的目标是找到节点 13。
运行一段时间后，队列将如下所示：

[[1, 2, 6], [1, 3, 10], [1, 4, 7], [1, 4, 8], [1, 2, 5, 9], [1, 2, 5, 10]]

请注意，末尾有两条节点号为 10 的路径。
这意味着从节点 10 开始的路径将被检查两次。在这种情况下，它看起来并没有那么糟糕，因为节点号 10 没有任何子节点..但它可能真的很糟糕（即使在这里我们也会无缘无故地检查该节点两次..）
节点号 13 不在这些路径中，因此程序在到达末尾节点号为 10 的第二条路径之前不会返回。我们将重新检查它。.

我们缺少的是设置标记访问过的节点并且不再检查它们..
这是修改后的 qiao 代码：

graph = {
    1: [2, 3, 4],
    2: [5, 6],
    3: [10],
    4: [7, 8],
    5: [9, 10],
    7: [11, 12],
    11: [13]
}


def bfs(graph_to_search, start, end):
    queue = [[start]]
    visited = set()

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            return path
        # We check if the current node is already in the visited nodes set in order not to recheck it
        elif vertex not in visited:
            # enumerate all adjacent nodes, construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex, []):
                new_path = list(path)
                new_path.append(current_neighbour)
                queue.append(new_path)

            # Mark the vertex as visited
            visited.add(vertex)


print bfs(graph, 1, 13)

程序的输出将是：

[1, 4, 7, 11, 13]

没有不必要的重新检查..

Answer 4

我想我应该尝试编写这个代码来娱乐：

graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def bfs(graph, forefront, end):
    # assumes no cycles

    next_forefront = [(node, path + ',' + node) for i, path in forefront if i in graph for node in graph[i]]

    for node,path in next_forefront:
        if node==end:
            return path
    else:
        return bfs(graph,next_forefront,end)

print bfs(graph,[('1','1')],'11')

# >>>
# 1, 4, 7, 11

如果你想要循环，你可以添加这个：

for i, j in for_front: # allow cycles, add this code
    if i in graph:
        del graph[i]

Answer 5

如果图表中包含了循环，这样的效果不是更好吗？

from collections import deque

graph = {
    1: [2, 3, 4],
    2: [5, 6, 3],
    3: [10],
    4: [7, 8],
    5: [9, 10],
    7: [11, 12],
   11: [13]
}


def bfs1(graph_to_search, start, end):
    queue = deque([start])
    visited = {start}
    trace = {}

    while queue:
        # Gets the first path in the queue
        vertex = queue.popleft()
        # Checks if we got to the end
        if vertex == end:
            break

        for neighbour in graph_to_search.get(vertex, []):
            # We check if the current neighbour is already in the visited nodes set in order not to re-add it
            if neighbour not in visited:
            # Mark the vertex as visited
                visited.add(neighbour)
                trace[neighbour] = vertex
                queue.append(neighbour)

path = [end]
while path[-1] != start:
    last_node = path[-1]
    next_node = trace[last_node]
    path.append(next_node)

return path[::-1]

print(bfs1(graph,1, 13))

这样只有新节点才会被访问，而且避免了循环。

Answer 6

我喜欢@Qiao的第一个答案和@Or的补充。为了减少处理量，我想添加 Or 的答案。

在@Or 的回答中，跟踪访问的节点非常棒。我们还可以允许程序比当前更早退出。在 for 循环中的某个时刻，current_neighbour 必须是 end，一旦发生这种情况，就会找到最短路径，程序就可以返回。

我将修改该方法如下，密切注意 for 循环

graph = {
1: [2, 3, 4],
2: [5, 6],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}


    def bfs(graph_to_search, start, end):
        queue = [[start]]
        visited = set()

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            return path
        # We check if the current node is already in the visited nodes set in order not to recheck it
        elif vertex not in visited:
            # enumerate all adjacent nodes, construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex, []):
                new_path = list(path)
                new_path.append(current_neighbour)
                queue.append(new_path)

                #No need to visit other neighbour. Return at once
                if current_neighbour == end
                    return new_path;

            # Mark the vertex as visited
            visited.add(vertex)


print bfs(graph, 1, 13)

输出和其他一切都将是相同的。但是，处理代码所需的时间会更少。这对于较大的图尤其有用。我希望这对将来的人有帮助。

Answer 7

Javascript 版本和搜索第一个/所有路径...

PS，带有 cylces 的图表效果很好。

您可以将其转换为python，它很容易

function search_path(graph, start, end, exhausted=true, method='bfs') {
    // note. Javascript Set is ordered set...
    const queue = [[start, new Set([start])]]
    const visited = new Set()
    const allpaths = []
    const hashPath = (path) => [...path].join(',') // any path hashing method
    while (queue.length) {
        const [node, path] = queue.shift()
        // visited node and its path instant. do not modify it others place
        visited.add(node)
        visited.add(hashPath(path))
        for (let _node of graph.get(node) || []) {
            // the paths already has the node, loops mean nothing though.
            if (path.has(_node))
                continue;
            // now new path has no repeated nodes.
            let newpath = new Set([...path, _node])
            if (_node == end){
                allpaths.push(newpath)
                if(!exhausted) return allpaths; // found and return 
            }
            else {
                if (!visited.has(_node) || // new node till now
                   // note: search all possible including the longest path
                   visited.has(_node) && !visited.has(hashPath(newpath))
                ) { 
                    if(method == 'bfs')
                       queue.push([_node, newpath])
                    else{
                       queue.unshift([_node, newpath])
                    }
                }
            }
        }
    }
    return allpaths
}

输出，如下所示..

[
    [ 'A', 'C' ],
    [ 'A', 'E', 'C'],
    [ 'A', 'E', 'F', 'C' ] // including F in `A -> C`
]