在C#中,如何找到循环依赖链？

Question

当一个部署项目包含第二个部署项目的项目输出,而第二个项目包含第一个项目的输出时,通常会发生此错误.

我有一个检查循环依赖的方法.在输入中,我们有一个包含,例如,一个词典<"A", < "B", "C" >>和 <"B", < "A", "D" >>,这意味着,A取决于B和C,我们有与循环依赖A->B.

但通常情况下我们会遇到更复杂的情况,并且存在一系列依赖关系.如何修改此方法以查找依赖链？例如,我想要一个包含chain的变量A->B->A,而不是类A与class有冲突B.

private void FindDependency(IDictionary<string, IEnumerable<string>> serviceDependence)

Answer 1

在图中查找循环的简单方法是使用递归深度优先图着色算法,其中节点被标记为"访问"或"访问".如果在访问节点时发现它已处于"访问"状态,则表示您有一个循环.标记为"已访问"的节点可以被跳过.例如:

public class DependencyExtensions
{
    enum VisitState
    {
        NotVisited,
        Visiting,
        Visited
    };

    public static TValue ValueOrDefault<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TKey key, TValue defaultValue)
    {
        TValue value;
        if (dictionary.TryGetValue(key, out value))
            return value;
        return defaultValue;
    }

    static void DepthFirstSearch<T>(T node, Func<T, IEnumerable<T>> lookup, List<T> parents, Dictionary<T, VisitState> visited, List<List<T>> cycles)
    {
        var state = visited.ValueOrDefault(node, VisitState.NotVisited);
        if (state == VisitState.Visited)
            return;
        else if (state == VisitState.Visiting)
        {
            // Do not report nodes not included in the cycle.
            cycles.Add(parents.Concat(new[] { node }).SkipWhile(parent => !EqualityComparer<T>.Default.Equals(parent, node)).ToList());
        }
        else
        {
            visited[node] = VisitState.Visiting;
            parents.Add(node);
            foreach (var child in lookup(node))
                DepthFirstSearch(child, lookup, parents, visited, cycles);
            parents.RemoveAt(parents.Count - 1);
            visited[node] = VisitState.Visited;
        }
    }

    public static List<List<T>> FindCycles<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> edges)
    {
        var cycles = new List<List<T>>();
        var visited = new Dictionary<T, VisitState>();
        foreach (var node in nodes)
            DepthFirstSearch(node, edges, new List<T>(), visited, cycles);
        return cycles;
    }

    public static List<List<T>> FindCycles<T, TValueList>(this IDictionary<T, TValueList> listDictionary)
        where TValueList : class, IEnumerable<T>
    {
        return listDictionary.Keys.FindCycles(key => listDictionary.ValueOrDefault(key, null) ?? Enumerable.Empty<T>());
    }
}

然后,您可以使用它:

        var serviceDependence = new Dictionary<string, List<string>>
        {
            { "A", new List<string> { "A" }},
            { "B", new List<string> { "C", "D" }},
            { "D", new List<string> { "E" }},
            { "E", new List<string> { "F", "Q" }},
            { "F", new List<string> { "D" }},
        };
        var cycles = serviceDependence.FindCycles();
        Debug.WriteLine(JsonConvert.SerializeObject(cycles, Formatting.Indented));
        foreach (var cycle in cycles)
        {
            serviceDependence[cycle[cycle.Count - 2]].Remove(cycle[cycle.Count - 1]);
        }
        Debug.Assert(serviceDependence.FindCycles().Count == 0);

更新

您的问题已更新,以请求"最有效的算法"来查找循环依赖项.原始答案中的代码是递归的,因此有可能存在StackOverflowException数千个级别的依赖链.这是一个带有显式堆栈变量的非递归版本:

public static class DependencyExtensions
{
    enum VisitState
    {
        NotVisited,
        Visiting,
        Visited
    };

    public static TValue ValueOrDefault<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TKey key, TValue defaultValue)
    {
        TValue value;
        if (dictionary.TryGetValue(key, out value))
            return value;
        return defaultValue;
    }

    private static void TryPush<T>(T node, Func<T, IEnumerable<T>> lookup, Stack<KeyValuePair<T, IEnumerator<T>>> stack, Dictionary<T, VisitState> visited, List<List<T>> cycles)
    {
        var state = visited.ValueOrDefault(node, VisitState.NotVisited);
        if (state == VisitState.Visited)
            return;
        else if (state == VisitState.Visiting)
        {
            Debug.Assert(stack.Count > 0);
            var list = stack.Select(pair => pair.Key).TakeWhile(parent => !EqualityComparer<T>.Default.Equals(parent, node)).ToList();
            list.Add(node);
            list.Reverse();
            list.Add(node);
            cycles.Add(list);
        }
        else
        {
            visited[node] = VisitState.Visiting;
            stack.Push(new KeyValuePair<T, IEnumerator<T>>(node, lookup(node).GetEnumerator()));
        }
    }

    static List<List<T>> FindCycles<T>(T root, Func<T, IEnumerable<T>> lookup, Dictionary<T, VisitState> visited)
    {
        var stack = new Stack<KeyValuePair<T, IEnumerator<T>>>();
        var cycles = new List<List<T>>();

        TryPush(root, lookup, stack, visited, cycles);
        while (stack.Count > 0)
        {
            var pair = stack.Peek();
            if (!pair.Value.MoveNext())
            {
                stack.Pop();                    
                visited[pair.Key] = VisitState.Visited;
                pair.Value.Dispose();
            }
            else
            {
                TryPush(pair.Value.Current, lookup, stack, visited, cycles);
            }
        }
        return cycles;
    }

    public static List<List<T>> FindCycles<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> edges)
    {
        var cycles = new List<List<T>>();
        var visited = new Dictionary<T, VisitState>();
        foreach (var node in nodes)
            cycles.AddRange(FindCycles(node, edges, visited));
        return cycles;
    }

    public static List<List<T>> FindCycles<T, TValueList>(this IDictionary<T, TValueList> listDictionary)
        where TValueList : class, IEnumerable<T>
    {
        return listDictionary.Keys.FindCycles(key => listDictionary.ValueOrDefault(key, null) ?? Enumerable.Empty<T>());
    }
}

这应该是在相当有效N*log(N) + E,其中N是节点的数量和E是边缘的数目.将Log(N)来自构建visited哈希表,并可以通过使每个节点记住它被淘汰VisitState.这看起来相当合理; 在以下测试工具中,在10000个节点中找到17897个平均长度为4393个周期的时间为125603个总依赖关系,大约为10.2秒:

public class TestClass
{
    public static void TestBig()
    {
        var elapsed = TestBig(10000);
        Debug.WriteLine(elapsed.ToString());
    }

    static string GetName(int i)
    {
        return "ServiceDependence" + i.ToString();
    }

    public static TimeSpan TestBig(int count)
    {
        var serviceDependence = new Dictionary<string, List<string>>();
        for (int iItem = 0; iItem < count; iItem++)
        {
            var name = GetName(iItem);
            // Add several forward references.
            for (int iRef = iItem - 1; iRef > 0; iRef = iRef / 2)
                serviceDependence.Add(name, GetName(iRef));
            // Add some backwards references.
            if (iItem > 0 && (iItem % 5 == 0))
                serviceDependence.Add(name, GetName(iItem + 5));
        }

        // Add one backwards reference that will create some extremely long cycles.
        serviceDependence.Add(GetName(1), GetName(count - 1));

        List<List<string>> cycles;

        var stopwatch = new Stopwatch();
        stopwatch.Start();
        try
        {
            cycles = serviceDependence.FindCycles();
        }
        finally
        {
            stopwatch.Stop();
        }

        var elapsed = stopwatch.Elapsed;

        var averageLength = cycles.Average(l => (double)l.Count);
        var total = serviceDependence.Values.Sum(l => l.Count);

        foreach (var cycle in cycles)
        {
            serviceDependence[cycle[cycle.Count - 2]].Remove(cycle[cycle.Count - 1]);
        }
        Debug.Assert(serviceDependence.FindCycles().Count == 0);

        Console.WriteLine(string.Format("Time to find {0} cycles of average length {1} in {2} nodes with {3} total dependencies: {4}", cycles.Count, averageLength, count, total, elapsed));
        Console.ReadLine();
        System.Environment.Exit(0);

        return elapsed;
    }
}

Answer 2

构建一个包含每个输入的所有直接依赖关系的字典.对于其中的每一个,添加所有唯一的间接依赖项(例如,遍历给定项的每个依赖项,如果父项不存在,则添加它).只要您对字典进行至少一次更改,就重复此操作.如果有一个项目本身就有它的依赖项,那就是一个周期性的依赖:)

当然,这是相对低效的,但它非常简单易懂.如果您正在创建编译器,您可能只需构建所有依赖项的有向图,并搜索其中的路径 - 您可以找到许多用于在有向图中查找路径的现成算法.