Question

概览

我试图找出如何使用某种DFS迭代算法遍历有向循环图。这是我目前实施的一个小版本（它没有处理周期）：

class Node(object):

    def __init__(self, name):
        self.name = name

    def start(self):
        print '{}_start'.format(self)

    def middle(self):
        print '{}_middle'.format(self)

    def end(self):
        print '{}_end'.format(self)

    def __str__(self):
        return "{0}".format(self.name)


class NodeRepeat(Node):

    def __init__(self, name, num_repeats=1):
        super(NodeRepeat, self).__init__(name)
        self.num_repeats = num_repeats


def dfs(graph, start):
    """Traverse graph from start node using DFS with reversed childs"""

    visited = {}
    stack = [(start, "")]
    while stack:
        # To convert dfs -> bfs
        # a) rename stack to queue
        # b) pop becomes pop(0)
        node, parent = stack.pop()
        if parent is None:
            if visited[node] < 3:
                node.end()
            visited[node] = 3
        elif node not in visited:
            if visited.get(parent) == 2:
                parent.middle()
            elif visited.get(parent) == 1:
                visited[parent] = 2

            node.start()
            visited[node] = 1

            stack.append((node, None))

            # Maybe you want a different order, if it's so, don't use reversed
            childs = reversed(graph.get(node, []))
            for child in childs:
                if child not in visited:
                    stack.append((child, node))


if __name__ == "__main__":
    Sequence1 = Node('Sequence1')
    MtxPushPop1 = Node('MtxPushPop1')
    Rotate1 = Node('Rotate1')
    Repeat1 = NodeRepeat('Repeat1', num_repeats=2)

    Sequence2 = Node('Sequence2')
    MtxPushPop2 = Node('MtxPushPop2')
    Translate = Node('Translate')
    Rotate2 = Node('Rotate2')
    Rotate3 = Node('Rotate3')
    Scale = Node('Scale')
    Repeat2 = NodeRepeat('Repeat2', num_repeats=3)
    Mesh = Node('Mesh')

    cyclic_graph = {
        Sequence1: [MtxPushPop1, Rotate1],
        MtxPushPop1: [Sequence2],
        Rotate1: [Repeat1],
        Sequence2: [MtxPushPop2, Translate],
        Repeat1: [Sequence1],
        MtxPushPop2: [Rotate2],
        Translate: [Rotate3],
        Rotate2: [Scale],
        Rotate3: [Repeat2],
        Scale: [Mesh],
        Repeat2: [Sequence2]
    }

    dfs(cyclic_graph, Sequence1)

    print '-'*80

    a = Node('a')
    b = Node('b')
    dfs({
        a : [b],
        b : [a]
    }, a)

上面的代码正在测试几个案例，第一个是下图的某种表示：

第二个是最简单的一个图形包含一个＆＃34;无限＆＃34;循环{a->b, b->a}

要求

我们不会存在诸如“无限循环”这样的事情，让我们说出一个无限循环＆＃34;如果找到，将会有一个最大阈值（全局变量）来指示何时停止循环＃34;伪无限循环＆＃34;
所有图形节点都能够创建循环但是会存在一个名为Repeat的特殊节点，您可以在其中指示循环循环的迭代次数
上面发布的mcve是遍历算法的迭代版本，不知道如何处理循环图。理想情况下，解决方案也是迭代的，但如果存在更好的递归解决方案，那就太棒了
我们在这里讨论的数据结构不应被称为＆＃34;有向无环图＆＃34;实际上，因为在这种情况下，每个节点都有其子节点，并且在图形节点连接中没有顺序。
一切都可以连接到编辑器中的任何内容。您将能够执行任何块组合，唯一的限制是执行计数器，如果您进行了无限循环或迭代次数太多，它将会溢出。
算法将保留开始/中间/后节点的方法执行，类似于上面的代码段

问题

任何人都可以提供某种知道如何遍历无限/有限循环的解决方案吗？

参考

如果此时此问题尚不清楚，您可以在此article上详细了解此问题，整个想法将使用遍历算法来实现类似于该文章中所示的类似工具。

这里有一个屏幕截图，展示了这种数据结构的全部功能我想弄清楚如何遍历和运行：

Answer 1

在我开始之前，Run the code on CodeSkulptor!我也希望这些评论能够详细说明我所做的一切。如果您需要更多解释，请查看我对代码下面的递归方法的解释。

# If you don't want global variables, remove the indentation procedures
indent = -1

MAX_THRESHOLD = 10
INF = 1 << 63

def whitespace():
    global indent
    return '|  ' * (indent)

class Node:
    def __init__(self, name, num_repeats=INF):
        self.name = name
        self.num_repeats = num_repeats

    def start(self):
        global indent
        if self.name.find('Sequence') != -1:
            print whitespace()
            indent += 1
        print whitespace() + '%s_start' % self.name

    def middle(self):
        print whitespace() + '%s_middle' % self.name

    def end(self):
        global indent
        print whitespace() + '%s_end' % self.name
        if self.name.find('Sequence') != -1:
            indent -= 1
            print whitespace()

def dfs(graph, start):
    visits = {}
    frontier = [] # The stack that keeps track of nodes to visit

    # Whenever we "visit" a node, increase its visit count
    frontier.append((start, start.num_repeats))
    visits[start] = visits.get(start, 0) + 1

    while frontier:
        # parent_repeat_count usually contains vertex.repeat_count
        # But, it may contain a higher value if a repeat node is its ancestor
        vertex, parent_repeat_count = frontier.pop()

        # Special case which signifies the end
        if parent_repeat_count == -1:
            vertex.end()
            # We're done with this vertex, clear visits so that 
            # if any other node calls us, we're still able to be called
            visits[vertex] = 0
            continue

        # Special case which signifies the middle
        if parent_repeat_count == -2:
            vertex.middle()
            continue  

        # Send the start message
        vertex.start()

        # Add the node's end state to the stack first
        # So that it is executed last
        frontier.append((vertex, -1))

        # No more children, continue
        # Because of the above line, the end method will
        # still be executed
        if vertex not in graph:
            continue

        ## Uncomment the following line if you want to go left to right neighbor
        #### graph[vertex].reverse()

        for i, neighbor in enumerate(graph[vertex]):
            # The repeat count should propagate amongst neighbors
            # That is if the parent had a higher repeat count, use that instead
            repeat_count = max(1, parent_repeat_count)
            if neighbor.num_repeats != INF:
                repeat_count = neighbor.num_repeats

            # We've gone through at least one neighbor node
            # Append this vertex's middle state to the stack
            if i >= 1:
                frontier.append((vertex, -2))

            # If we've not visited the neighbor more times than we have to, visit it
            if visits.get(neighbor, 0) < MAX_THRESHOLD and visits.get(neighbor, 0) < repeat_count:
                frontier.append((neighbor, repeat_count))
                visits[neighbor] = visits.get(neighbor, 0) + 1

def dfs_rec(graph, node, parent_repeat_count=INF, visits={}):
    visits[node] = visits.get(node, 0) + 1

    node.start()

    if node not in graph:
        node.end()
        return

    for i, neighbor in enumerate(graph[node][::-1]):
        repeat_count = max(1, parent_repeat_count)
        if neighbor.num_repeats != INF:
            repeat_count = neighbor.num_repeats

        if i >= 1:
            node.middle()

        if visits.get(neighbor, 0) < MAX_THRESHOLD and visits.get(neighbor, 0) < repeat_count:
            dfs_rec(graph, neighbor, repeat_count, visits)

    node.end()  
    visits[node] = 0

Sequence1 = Node('Sequence1')
MtxPushPop1 = Node('MtxPushPop1')
Rotate1 = Node('Rotate1')
Repeat1 = Node('Repeat1', 2)

Sequence2 = Node('Sequence2')
MtxPushPop2 = Node('MtxPushPop2')
Translate = Node('Translate')
Rotate2 = Node('Rotate2')
Rotate3 = Node('Rotate3')
Scale = Node('Scale')
Repeat2 = Node('Repeat2', 3)
Mesh = Node('Mesh')

cyclic_graph = {
        Sequence1: [MtxPushPop1, Rotate1],
        MtxPushPop1: [Sequence2],
        Rotate1: [Repeat1],
        Sequence2: [MtxPushPop2, Translate],
        Repeat1: [Sequence1],
        MtxPushPop2: [Rotate2],
        Translate: [Rotate3],
        Rotate2: [Scale],
        Rotate3: [Repeat2],
        Scale: [Mesh],
        Repeat2: [Sequence2]
    }

dfs(cyclic_graph, Sequence1)

print '-'*40

dfs_rec(cyclic_graph, Sequence1)

print '-'*40

dfs({Sequence1: [Translate], Translate: [Sequence1]}, Sequence1)

print '-'*40

dfs_rec({Sequence1: [Translate], Translate: [Sequence1]}, Sequence1)

可以找到输入和（格式良好且缩进）输出here。如果您想查看我是如何格式化输出的，请参阅代码，也可以是found on CodeSkulptor。

对，解释。更容易理解但更低效的递归解决方案，我将用来帮助解释，如下：

def dfs_rec(graph, node, parent_repeat_count=INF, visits={}):
    visits[node] = visits.get(node, 0) + 1

    node.start()

    if node not in graph:
        node.end()
        return

    for i, neighbor in enumerate(graph[node][::-1]):
        repeat_count = max(1, parent_repeat_count)
        if neighbor.num_repeats != INF:
            repeat_count = neighbor.num_repeats

        if i >= 1:
            node.middle()

        if visits.get(neighbor, 0) < MAX_THRESHOLD and visits.get(neighbor, 0) < repeat_count:
            dfs_rec(graph, neighbor, repeat_count, visits)

    node.end()  
    visits[node] = 0

我们要做的第一件事是访问节点。我们通过增加字典中节点的访问次数来完成此操作。
然后我们举起节点的start事件。
我们做一个简单的检查，看看节点是否是无子节点（叶子）节点。如果是，我们举起end事件并返回。
现在我们已经确定节点有邻居，我们遍历每个邻居。 旁注：我在递归版本中反转邻居列表（通过使用graph[node][::-1]）来维持邻居遍历的相同顺序（从右到左），如在迭代版本中那样。
1. 对于每个邻居，我们首先计算重复次数。重复计数从祖先节点传播（继承），因此继承的重复计数用于，除非 包含重复计数值。

我们现在已经完成了这个节点;提升end事件并清除其在哈希表中的访问。这样做是为了在其他节点再次调用它时，它不会使可访问性检查失败和/或执行的次数少于所需的次数。

Answer 2

根据comment66244567 - 通过忽略到访问节点的链接并执行广度优先搜索将图形缩减为树，因为这会产生更自然（并且可能更平衡）的树：

protected void Button1_Click(object sender, EventArgs e)
        {
            GridView1.DataSource = null;
           GridView1.DataBind(); 

            connect = new SqlConnection(@"Data Source=LP12;Initial Catalog=Data;Integrated Security=True");
            connect.Open();

            SqlCommand cmd = new SqlCommand();
            cmd.Connection = connect;
            cmd.CommandText = "SELECT DrukSensor, FlowSensor                
+ " FROM SysteemSensorInfo";

            DataSet ds = new DataSet();
            new SqlDataAdapter(cmd).Fill(ds);
            GridView1.DataSource = ds.Tables[0];
            GridView1.DataBind();
            GridView1.AutoGenerateColumns = false;
        }

使用您的图表和def traverse(graph,node,process): seen={node} current_level=[node] while current_level: next_level=[] for node in current_level: process(node) for child in (link for link in graph.get(node,[]) if link not in seen): next_level.append(child) seen.add(child) current_level=next_level，会产生：

def process(node): print node

另一个BFS算法iterative deepening DFS（以速度为代价使用更少的内存）在这种情况下不会赢得任何东西：因为你必须存储对被访问节点的引用，你已经消耗了 O（n）记忆。你不需要产生中间结果（但无论如何你都可以 - 例如In [24]: traverse(cyclic_graph,Sequence1,process) Sequence1 MtxPushPop1 Rotate1 Sequence2 Repeat1 MtxPushPop2 Translate Rotate2 Rotate3 Scale Repeat2 Mesh处理一个级别后的东西。）

如何使用修改后的DFS算法遍历循环有向图

2 个答案: