コード例 #1
0
        private readonly Collections.Stack<Integer> _path; // Eulerian path; null if no suh path

        #endregion Fields

        #region Constructors

        /// <summary>
        /// Computes an Eulerian path in the specified digraph, if one exists.
        /// </summary>
        /// <param name="g">g the digraph</param>
        public DirectedEulerianPath(Digraph g)
        {
            // find vertex from which to start potential Eulerian path:
            // a vertex v with outdegree(v) > indegree(v) if it exits;
            // otherwise a vertex with outdegree(v) > 0
            var deficit = 0;
            var s = NonIsolatedVertex(g);
            for (var v = 0; v < g.V; v++)
            {
                if (g.Outdegree(v) > g.Indegree(v))
                {
                    deficit += (g.Outdegree(v) - g.Indegree(v));
                    s = v;
                }
            }

            // digraph can't have an Eulerian path
            // (this condition is needed)
            if (deficit > 1) return;

            // special case for digraph with zero edges (has a degenerate Eulerian path)
            if (s == -1) s = 0;

            // create local view of adjacency lists, to iterate one vertex at a time
            var adj = new IEnumerator<Integer>[g.V];
            for (var v = 0; v < g.V; v++)
                adj[v] = g.Adj(v).GetEnumerator();

            // greedily add to cycle, depth-first search style
            var stack = new Collections.Stack<Integer>();
            stack.Push(s);
            _path = new Collections.Stack<Integer>();
            while (!stack.IsEmpty())
            {
                int v = stack.Pop();
                while (adj[v].MoveNext())
                {
                    stack.Push(v);
                    v = adj[v].Current;
                }
                // push vertex with no more available edges to path
                _path.Push(v);
            }

            // check if all edges have been used
            if (_path.Size() != g.E + 1)
                _path = null;

            //assert check(G);
        }
コード例 #2
0
        /// <summary>
        /// Determines whether a digraph has an Eulerian path using necessary
        /// and sufficient conditions (without computing the path itself):
        ///    - indegree(v) = outdegree(v) for every vertex,
        ///      except one vertex v may have outdegree(v) = indegree(v) + 1
        ///      (and one vertex v may have indegree(v) = outdegree(v) + 1)
        ///    - the graph is connected, when viewed as an undirected graph
        ///      (ignoring isolated vertices)
        /// This method is solely for unit testing.
        /// </summary>
        /// <param name="g"></param>
        /// <returns></returns>
        private static bool HasEulerianPath(Digraph g)
        {
            if (g.E == 0)
            {
                return(true);
            }

            // Condition 1: indegree(v) == outdegree(v) for every vertex,
            // except one vertex may have outdegree(v) = indegree(v) + 1
            var deficit = 0;

            for (var v = 0; v < g.V; v++)
            {
                if (g.Outdegree(v) > g.Indegree(v))
                {
                    deficit += (g.Outdegree(v) - g.Indegree(v));
                }
            }
            if (deficit > 1)
            {
                return(false);
            }

            // Condition 2: graph is connected, ignoring isolated vertices
            var h = new Graph(g.V);

            for (var v = 0; v < g.V; v++)
            {
                foreach (int w in g.Adj(v))
                {
                    h.AddEdge(v, w);
                }
            }

            // check that all non-isolated vertices are connected
            var s   = NonIsolatedVertex(g);
            var bfs = new BreadthFirstPaths(h, s);

            for (var v = 0; v < g.V; v++)
            {
                if (h.Degree(v) > 0 && !bfs.HasPathTo(v))
                {
                    return(false);
                }
            }

            return(true);
        }
コード例 #3
0
        private readonly Collections.Stack <Integer> _cycle;  // Eulerian cycle; null if no such cylce

        /// <summary>
        /// Computes an Eulerian cycle in the specified digraph, if one exists.
        /// </summary>
        /// <param name="g">g the digraph</param>
        public DirectedEulerianCycle(Digraph g)
        {
            // must have at least one edge
            if (g.E == 0)
            {
                return;
            }

            // necessary condition: indegree(v) = outdegree(v) for each vertex v
            // (without this check, DFS might return a path instead of a cycle)
            for (var v = 0; v < g.V; v++)
            {
                if (g.Outdegree(v) != g.Indegree(v))
                {
                    return;
                }
            }

            // create local view of adjacency lists, to iterate one vertex at a time
            var adj = new IEnumerator <Integer> [g.V];

            for (var v = 0; v < g.V; v++)
            {
                adj[v] = g.Adj(v).GetEnumerator();
            }

            // initialize stack with any non-isolated vertex
            var s     = NonIsolatedVertex(g);
            var stack = new Collections.Stack <Integer>();

            stack.Push(s);

            // greedily add to putative cycle, depth-first search style
            _cycle = new Collections.Stack <Integer>();
            while (!stack.IsEmpty())
            {
                int v = stack.Pop();
                while (adj[v].MoveNext())
                {
                    stack.Push(v);
                    v = adj[v].Current;
                }
                // add vertex with no more leaving edges to cycle
                _cycle.Push(v);
            }

            // check if all edges have been used
            // (in case there are two or more vertex-disjoint Eulerian cycles)
            if (_cycle.Size() != g.E + 1)
            {
                _cycle = null;
            }

            //assert certifySolution(G);
        }
コード例 #4
0
        private readonly int[] _rank;                        // rank[v] = order where vertex v appers in order

        /// <summary>
        /// Determines whether the digraph <tt>G</tt> has a topological order and, if so,
        /// finds such a topological order.
        /// </summary>
        /// <param name="g">g the digraph</param>
        public TopologicalX(Digraph g)
        {
            // indegrees of remaining vertices
            var indegree = new int[g.V];

            for (var v = 0; v < g.V; v++)
            {
                indegree[v] = g.Indegree(v);
            }

            // initialize
            _rank  = new int[g.V];
            _order = new Collections.Queue <Integer>();
            var count = 0;

            // initialize queue to contain all vertices with indegree = 0
            var queue = new Collections.Queue <Integer>();

            for (var v = 0; v < g.V; v++)
            {
                if (indegree[v] == 0)
                {
                    queue.Enqueue(v);
                }
            }

            for (var j = 0; !queue.IsEmpty(); j++)
            {
                int v = queue.Dequeue();
                _order.Enqueue(v);
                _rank[v] = count++;
                foreach (int w in g.Adj(v))
                {
                    indegree[w]--;
                    if (indegree[w] == 0)
                    {
                        queue.Enqueue(w);
                    }
                }
            }

            // there is a directed cycle in subgraph of vertices with indegree >= 1.
            if (count != g.V)
            {
                _order = null;
            }

            //assert check(G);
        }
コード例 #5
0
        /// <summary>
        /// Determines whether a digraph has an Eulerian cycle using necessary
        /// and sufficient conditions (without computing the cycle itself):
        ///    - at least one edge
        ///    - indegree(v) = outdegree(v) for every vertex
        ///    - the graph is connected, when viewed as an undirected graph
        ///      (ignoring isolated vertices)
        /// </summary>
        /// <param name="g"></param>
        /// <returns></returns>
        private static bool HasEulerianCycle(Digraph g)
        {
            // Condition 0: at least 1 edge
            if (g.E == 0)
            {
                return(false);
            }

            // Condition 1: indegree(v) == outdegree(v) for every vertex
            for (var v = 0; v < g.V; v++)
            {
                if (g.Outdegree(v) != g.Indegree(v))
                {
                    return(false);
                }
            }

            // Condition 2: graph is connected, ignoring isolated vertices
            var h = new Graph(g.V);

            for (var v = 0; v < g.V; v++)
            {
                foreach (int w in g.Adj(v))
                {
                    h.AddEdge(v, w);
                }
            }

            // check that all non-isolated vertices are conneted
            var s   = NonIsolatedVertex(g);
            var bfs = new BreadthFirstPaths(h, s);

            for (var v = 0; v < g.V; v++)
            {
                if (h.Degree(v) > 0 && !bfs.HasPathTo(v))
                {
                    return(false);
                }
            }

            return(true);
        }
コード例 #6
0
 /// <summary>
 /// Initializes a new digraph that is a deep copy of the specified digraph.
 /// </summary>
 /// <param name="g">g the digraph to copy</param>
 public Digraph(Digraph g) : this(g.V)
 {
     E = g.E;
     for (var v = 0; v < V; v++)
     {
         _indegree[v] = g.Indegree(v);
     }
     for (var v = 0; v < g.V; v++)
     {
         // reverse so that adjacency list is in same order as original
         var reverse = new Collections.Stack <Integer>();
         foreach (int w in g._adj[v])
         {
             reverse.Push(w);
         }
         foreach (int w in reverse)
         {
             _adj[v].Add(w);
         }
     }
 }
コード例 #7
0
        private readonly int[] _rank; // rank[v] = order where vertex v appers in order

        #endregion Fields

        #region Constructors

        /// <summary>
        /// Determines whether the digraph <tt>G</tt> has a topological order and, if so,
        /// finds such a topological order.
        /// </summary>
        /// <param name="g">g the digraph</param>
        public TopologicalX(Digraph g)
        {
            // indegrees of remaining vertices
            var indegree = new int[g.V];
            for (var v = 0; v < g.V; v++)
            {
                indegree[v] = g.Indegree(v);
            }

            // initialize
            _rank = new int[g.V];
            _order = new Collections.Queue<Integer>();
            var count = 0;

            // initialize queue to contain all vertices with indegree = 0
            var queue = new Collections.Queue<Integer>();
            for (var v = 0; v < g.V; v++)
                if (indegree[v] == 0) queue.Enqueue(v);

            for (var j = 0; !queue.IsEmpty(); j++)
            {
                int v = queue.Dequeue();
                _order.Enqueue(v);
                _rank[v] = count++;
                foreach (int w in g.Adj(v))
                {
                    indegree[w]--;
                    if (indegree[w] == 0) queue.Enqueue(w);
                }
            }

            // there is a directed cycle in subgraph of vertices with indegree >= 1.
            if (count != g.V)
            {
                _order = null;
            }

            //assert check(G);
        }
コード例 #8
0
        /// <summary>
        /// Determines whether a digraph has an Eulerian path using necessary
        /// and sufficient conditions (without computing the path itself):
        ///    - indegree(v) = outdegree(v) for every vertex,
        ///      except one vertex v may have outdegree(v) = indegree(v) + 1
        ///      (and one vertex v may have indegree(v) = outdegree(v) + 1)
        ///    - the graph is connected, when viewed as an undirected graph
        ///      (ignoring isolated vertices)
        /// This method is solely for unit testing.
        /// </summary>
        /// <param name="g"></param>
        /// <returns></returns>
        private static bool HasEulerianPath(Digraph g)
        {
            if (g.E == 0) return true;

            // Condition 1: indegree(v) == outdegree(v) for every vertex,
            // except one vertex may have outdegree(v) = indegree(v) + 1
            var deficit = 0;
            for (var v = 0; v < g.V; v++)
                if (g.Outdegree(v) > g.Indegree(v))
                    deficit += (g.Outdegree(v) - g.Indegree(v));
            if (deficit > 1) return false;

            // Condition 2: graph is connected, ignoring isolated vertices
            var h = new Graph(g.V);
            for (var v = 0; v < g.V; v++)
                foreach (int w in g.Adj(v))
                    h.AddEdge(v, w);

            // check that all non-isolated vertices are connected
            var s = NonIsolatedVertex(g);
            var bfs = new BreadthFirstPaths(h, s);
            for (var v = 0; v < g.V; v++)
                if (h.Degree(v) > 0 && !bfs.HasPathTo(v))
                    return false;

            return true;
        }
コード例 #9
0
        private readonly Collections.Stack <Integer> _cycle;     // the directed cycle; null if digraph is acyclic

        public DirectedCycleX(Digraph g)
        {
            // indegrees of remaining vertices
            var indegree = new int[g.V];

            for (var v = 0; v < g.V; v++)
            {
                indegree[v] = g.Indegree(v);
            }

            // initialize queue to contain all vertices with indegree = 0
            var queue = new Collections.Queue <Integer>();

            for (var v = 0; v < g.V; v++)
            {
                if (indegree[v] == 0)
                {
                    queue.Enqueue(v);
                }
            }

            for (var j = 0; !queue.IsEmpty(); j++)
            {
                int v = queue.Dequeue();
                foreach (int w in g.Adj(v))
                {
                    indegree[w]--;
                    if (indegree[w] == 0)
                    {
                        queue.Enqueue(w);
                    }
                }
            }

            // there is a directed cycle in subgraph of vertices with indegree >= 1.
            var edgeTo = new int[g.V];
            var root   = -1; // any vertex with indegree >= -1

            for (var v = 0; v < g.V; v++)
            {
                if (indegree[v] == 0)
                {
                    continue;
                }
                root = v;
                foreach (int w in g.Adj(v))
                {
                    if (indegree[w] > 0)
                    {
                        edgeTo[w] = v;
                    }
                }
            }

            if (root != -1)
            {
                // find any vertex on cycle
                var visited = new bool[g.V];
                while (!visited[root])
                {
                    visited[root] = true;
                    root          = edgeTo[root];
                }

                // extract cycle
                _cycle = new Collections.Stack <Integer>();
                var v = root;
                do
                {
                    _cycle.Push(v);
                    v = edgeTo[v];
                } while (v != root);
                _cycle.Push(root);
            }

            //assert check();
        }
コード例 #10
0
        private readonly Collections.Stack <Integer> _path;   // Eulerian path; null if no suh path

        /// <summary>
        /// Computes an Eulerian path in the specified digraph, if one exists.
        /// </summary>
        /// <param name="g">g the digraph</param>
        public DirectedEulerianPath(Digraph g)
        {
            // find vertex from which to start potential Eulerian path:
            // a vertex v with outdegree(v) > indegree(v) if it exits;
            // otherwise a vertex with outdegree(v) > 0
            var deficit = 0;
            var s       = NonIsolatedVertex(g);

            for (var v = 0; v < g.V; v++)
            {
                if (g.Outdegree(v) > g.Indegree(v))
                {
                    deficit += (g.Outdegree(v) - g.Indegree(v));
                    s        = v;
                }
            }

            // digraph can't have an Eulerian path
            // (this condition is needed)
            if (deficit > 1)
            {
                return;
            }

            // special case for digraph with zero edges (has a degenerate Eulerian path)
            if (s == -1)
            {
                s = 0;
            }

            // create local view of adjacency lists, to iterate one vertex at a time
            var adj = new IEnumerator <Integer> [g.V];

            for (var v = 0; v < g.V; v++)
            {
                adj[v] = g.Adj(v).GetEnumerator();
            }

            // greedily add to cycle, depth-first search style
            var stack = new Collections.Stack <Integer>();

            stack.Push(s);
            _path = new Collections.Stack <Integer>();
            while (!stack.IsEmpty())
            {
                int v = stack.Pop();
                while (adj[v].MoveNext())
                {
                    stack.Push(v);
                    v = adj[v].Current;
                }
                // push vertex with no more available edges to path
                _path.Push(v);
            }

            // check if all edges have been used
            if (_path.Size() != g.E + 1)
            {
                _path = null;
            }

            //assert check(G);
        }
コード例 #11
0
ファイル: Digraph.cs プロジェクト: vladdnc/Algorithms-NET
 /// <summary>
 /// Initializes a new digraph that is a deep copy of the specified digraph.
 /// </summary>
 /// <param name="g">g the digraph to copy</param>
 public Digraph(Digraph g)
     : this(g.V)
 {
     E = g.E;
     for (var v = 0; v < V; v++)
         _indegree[v] = g.Indegree(v);
     for (var v = 0; v < g.V; v++)
     {
         // reverse so that adjacency list is in same order as original
         var reverse = new Collections.Stack<Integer>();
         foreach (int w in g._adj[v])
         {
             reverse.Push(w);
         }
         foreach (int w in reverse)
         {
             _adj[v].Add(w);
         }
     }
 }
コード例 #12
0
        private readonly Collections.Stack<Integer> _cycle; // the directed cycle; null if digraph is acyclic

        #endregion Fields

        #region Constructors

        public DirectedCycleX(Digraph g)
        {
            // indegrees of remaining vertices
            var indegree = new int[g.V];
            for (var v = 0; v < g.V; v++)
            {
                indegree[v] = g.Indegree(v);
            }

            // initialize queue to contain all vertices with indegree = 0
            var queue = new Collections.Queue<Integer>();
            for (var v = 0; v < g.V; v++)
                if (indegree[v] == 0) queue.Enqueue(v);

            for (var j = 0; !queue.IsEmpty(); j++)
            {
                int v = queue.Dequeue();
                foreach (int w in g.Adj(v))
                {
                    indegree[w]--;
                    if (indegree[w] == 0) queue.Enqueue(w);
                }
            }

            // there is a directed cycle in subgraph of vertices with indegree >= 1.
            var edgeTo = new int[g.V];
            var root = -1;  // any vertex with indegree >= -1
            for (var v = 0; v < g.V; v++)
            {
                if (indegree[v] == 0) continue;
                root = v;
                foreach (int w in g.Adj(v))
                {
                    if (indegree[w] > 0)
                    {
                        edgeTo[w] = v;
                    }
                }
            }

            if (root != -1)
            {

                // find any vertex on cycle
                var visited = new bool[g.V];
                while (!visited[root])
                {
                    visited[root] = true;
                    root = edgeTo[root];
                }

                // extract cycle
                _cycle = new Collections.Stack<Integer>();
                var v = root;
                do
                {
                    _cycle.Push(v);
                    v = edgeTo[v];
                } while (v != root);
                _cycle.Push(root);
            }

            //assert check();
        }