In C#, how to find chain of circular dependency?
This error usually occurs when one deployment project contains the project outputs of a second deployment project, and the second project contains the outputs of the first project.
I have a method that check circular dependency. In input, we have a dictionary that contains, for example, <"A", < "B", "C" >> and <"B", < "A", "D" >>, this means that A depends on B and C and we have circular dependency with A->B.
But usually we have a more complex situation, with a chain of dependency.
How to modify this method to find a chain of dependency? For example, I want to have a variable that contains chain A->B->A, rather than class A has a conflict with class B.
private void FindDependency(IDictionary<string, IEnumerable<string>> serviceDependence)
Answer
A simple way to find cycles in a graph is to use a recursive depth-first graph-coloring algorithm in which nodes are marked as "visiting" or "visited". If, when visiting a node, you find it is already in the "visiting" state, you have a cycle. Nodes marked as "visited" can be skipped. For instance:
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>());
}
}
Then, you could use it like:
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);
Update
Your question has been updated to request the "most efficient algorithm" for finding cyclic dependencies. The code in the original answer is recursive, so there's a chance of a StackOverflowException for dependency chains thousands of levels deep. Here's a non-recursive version with an explicit stack variable:
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>());
}
}
This should be reasonably efficient at N*log(N) + E where N is the number of nodes and E is the number of edges. The Log(N) comes from building the visited hash table and could be eliminated by making each node remember its VisitState. This seems reasonably performant; in the following test harness, the time to find 17897 cycles of average length 4393 in 10000 nodes with 125603 total dependencies is around 10.2 seconds:
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;
}
}