Detect cycle in directed graph leetcode

first, before moving on to the solution. Solution Approach: Depth First Traversal can be used to detect cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is a For a disconnected graph, we get the DFS forest as output. To detect cycle, we can check for cycle in individual trees by checking back edges. In the In this post, a different solution is discussed. The solution is from WHITE : Vertex is not processed yet. Initially, all vertices are WHITE. GRAY: Vertex is being processed (DFS for this vertex has started, but not finished which means that all descendants (in DFS tree) of this vertex are not processed yet (or this vertex is in the function call stack) BLACK : Vertex and all its descendants are processed. While doing DFS, if an edge is encountered from current vertex to a GRAY vertex, then this edge is back edge and hence there is a cycle. Algorithm: • Create a recursive function that takes the edge and color array (this can be also kept as a global variable) • Mark the current node as GREY. • Traverse all the adjacent nodes and if any node is marked GREY then return true as a loop is bound to exist. • If any adjacent vertex is WHITE then call the recursive function for that node. If the function returns true. Return true. • If no adjacent node is grey or has not returned true then mark the current Node as BLACK and return false. Implementation: Output Graph contains cycle Complexity Analysis: • Time complexity: O(V...

Approach: The problem can be solved based on the following idea: To find cycle in a directed graph we can use the only if there is a back edge [i.e., a node points to one of its ancestors] present in the graph. To detect a back edge, we need to keep track of the nodes visited till now and the nodes that are in the current recursion stack [i.e., the current path that we are visiting]. If during recursion, we reach a node that is already in the recursion stack, there is a cycle present in the graph. Note: If the graph is disconnected then get the DFS forest and check for a cycle in individual trees by checking back edges. Follow the below steps to Implement the idea: • Create a recursive dfs function that has the following parameters – current vertex, visited array, and recursion stack. • Mark the current node as visited and also mark the index in the recursion stack. • Iterate a loop for all the vertices and for each vertex, call the recursive function if it is not yet visited (This step is done to make sure that if there is a forest of graphs, we are checking each forest): • In each recursion call, Find all the adjacent vertices of the current vertex which are not visited: • If an adjacent vertex is already marked in the recursion stack then return true. • Otherwise, call the recursive function for that adjacent vertex. • While returning from the recursion call, unmark the current node from the recursion stack, to represent that the current node is no longer a part of the ...

1 CS Level Up Series Introduction 2 Dynamic Arrays ... 26 more parts... 3 Linked Lists 4 Stacks 5 Queues 6 Hash Tables 7 Binary Search Trees 8 Binary Heaps 9 Priority Queues 10 Graphs 11 Tries 12 Binary & Bit Manipulation 13 Common Sorting Algorithms 14 Searching Algorithms 15 Permutations, Combinations, & Subsets 16 NP-Complete & Fibonacci Heap 17 Detecting Graph Cycles With Depth-First Search 18 Finding Shortest Paths In Graphs (using Dijkstra's & BFS) 19 Topological Sorting of Directed Acyclic Graphs (DAGs) 20 Finding Articulation Points & Bridges in Undirected Graphs 21 Finding Strongly Connected Components in Directed Graphs using Tarjan's Algorithm 22 Checking If An Undirected Graph Is Bipartite 23 Extending An Iterator 24 Union-find (Disjoint-set) 25 Minimum Spanning Tree (Kruskal's Algorithm) 26 Sliding Window Technique 27 String Searching Using Rabin-Karp 28 Fenwick Tree (Binary Indexed Tree) 29 Dynamic Programming 30 Traveling Salesman Problem • • • • • Deadlock Detection in the book: Elements of Programming Interviews To get the most out of this article, it is helpful if you already know what a graph is, and ideally what depth-first search is. Some takeaways: • A graph cycle is when there is a "loop" or circular reference. This can be a series of edges that connect back to an origin vertex. • If a graph has a cycle it is a cyclic graph. • To determine if a graph has a cycle, we can traverse the graph and look for a back edge. • A back edge is one that connects a...

Let A be the adjacency matrix for the graph G = (V,E). A(i,j) = 1 if the nodes i and j are connected with an edge, A(i,j) = 0 otherwise. My objective is the one of understanding whether G is acyclic or not. A cycle is defined in the following way: • i and j are connected: A(i,j) = 1 • j and k are connected: A(j,k) = 1 • k and i are connected: A(k,i) = 1 I have implemented a solution which navigates the matrix as follows: • Start from an edge (i,j) • Select the set O of edges which are outgoing from j, i.e., all the 1s in the j-th row of A • Navigate O in a DFS fashion • If one of the paths generated from this navigation leads to the node i, then a cycle is detected Obviously this solution is very slow, since I have to evaluate all the paths in the matrix. If A is very big, the required overhead is very huge. I was wondering whether there is a way of navigating the adjacency matrix so as to find cycles without using an expensive algorithm such as DFS. I would like to implement my solution in MATLAB. Thanks in advance, Eleanore. I came across this question when answering this walks of length k from i to j in G. The key thing here is that a walk is allowed to repeat vertices. So even if a diagonal entries of A^k is positive, each walk the entry is counting may contain repeated vertices, and so wouldn't count as a cycle. Counterexample: A path of length 4 would contain a 4-cycle according to Danil's answer (not to mention that the answer would imply P=NP because it would solve...

1 CS Level Up Series Introduction 2 Dynamic Arrays ... 26 more parts... 3 Linked Lists 4 Stacks 5 Queues 6 Hash Tables 7 Binary Search Trees 8 Binary Heaps 9 Priority Queues 10 Graphs 11 Tries 12 Binary & Bit Manipulation 13 Common Sorting Algorithms 14 Searching Algorithms 15 Permutations, Combinations, & Subsets 16 NP-Complete & Fibonacci Heap 17 Detecting Graph Cycles With Depth-First Search 18 Finding Shortest Paths In Graphs (using Dijkstra's & BFS) 19 Topological Sorting of Directed Acyclic Graphs (DAGs) 20 Finding Articulation Points & Bridges in Undirected Graphs 21 Finding Strongly Connected Components in Directed Graphs using Tarjan's Algorithm 22 Checking If An Undirected Graph Is Bipartite 23 Extending An Iterator 24 Union-find (Disjoint-set) 25 Minimum Spanning Tree (Kruskal's Algorithm) 26 Sliding Window Technique 27 String Searching Using Rabin-Karp 28 Fenwick Tree (Binary Indexed Tree) 29 Dynamic Programming 30 Traveling Salesman Problem • • • • • Deadlock Detection in the book: Elements of Programming Interviews To get the most out of this article, it is helpful if you already know what a graph is, and ideally what depth-first search is. Some takeaways: • A graph cycle is when there is a "loop" or circular reference. This can be a series of edges that connect back to an origin vertex. • If a graph has a cycle it is a cyclic graph. • To determine if a graph has a cycle, we can traverse the graph and look for a back edge. • A back edge is one that connects a...

Let A be the adjacency matrix for the graph G = (V,E). A(i,j) = 1 if the nodes i and j are connected with an edge, A(i,j) = 0 otherwise. My objective is the one of understanding whether G is acyclic or not. A cycle is defined in the following way: • i and j are connected: A(i,j) = 1 • j and k are connected: A(j,k) = 1 • k and i are connected: A(k,i) = 1 I have implemented a solution which navigates the matrix as follows: • Start from an edge (i,j) • Select the set O of edges which are outgoing from j, i.e., all the 1s in the j-th row of A • Navigate O in a DFS fashion • If one of the paths generated from this navigation leads to the node i, then a cycle is detected Obviously this solution is very slow, since I have to evaluate all the paths in the matrix. If A is very big, the required overhead is very huge. I was wondering whether there is a way of navigating the adjacency matrix so as to find cycles without using an expensive algorithm such as DFS. I would like to implement my solution in MATLAB. Thanks in advance, Eleanore. I came across this question when answering this walks of length k from i to j in G. The key thing here is that a walk is allowed to repeat vertices. So even if a diagonal entries of A^k is positive, each walk the entry is counting may contain repeated vertices, and so wouldn't count as a cycle. Counterexample: A path of length 4 would contain a 4-cycle according to Danil's answer (not to mention that the answer would imply P=NP because it would solve...

Approach: The problem can be solved based on the following idea: To find cycle in a directed graph we can use the only if there is a back edge [i.e., a node points to one of its ancestors] present in the graph. To detect a back edge, we need to keep track of the nodes visited till now and the nodes that are in the current recursion stack [i.e., the current path that we are visiting]. If during recursion, we reach a node that is already in the recursion stack, there is a cycle present in the graph. Note: If the graph is disconnected then get the DFS forest and check for a cycle in individual trees by checking back edges. Follow the below steps to Implement the idea: • Create a recursive dfs function that has the following parameters – current vertex, visited array, and recursion stack. • Mark the current node as visited and also mark the index in the recursion stack. • Iterate a loop for all the vertices and for each vertex, call the recursive function if it is not yet visited (This step is done to make sure that if there is a forest of graphs, we are checking each forest): • In each recursion call, Find all the adjacent vertices of the current vertex which are not visited: • If an adjacent vertex is already marked in the recursion stack then return true. • Otherwise, call the recursive function for that adjacent vertex. • While returning from the recursion call, unmark the current node from the recursion stack, to represent that the current node is no longer a part of the ...

first, before moving on to the solution. Solution Approach: Depth First Traversal can be used to detect cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is a For a disconnected graph, we get the DFS forest as output. To detect cycle, we can check for cycle in individual trees by checking back edges. In the In this post, a different solution is discussed. The solution is from WHITE : Vertex is not processed yet. Initially, all vertices are WHITE. GRAY: Vertex is being processed (DFS for this vertex has started, but not finished which means that all descendants (in DFS tree) of this vertex are not processed yet (or this vertex is in the function call stack) BLACK : Vertex and all its descendants are processed. While doing DFS, if an edge is encountered from current vertex to a GRAY vertex, then this edge is back edge and hence there is a cycle. Algorithm: • Create a recursive function that takes the edge and color array (this can be also kept as a global variable) • Mark the current node as GREY. • Traverse all the adjacent nodes and if any node is marked GREY then return true as a loop is bound to exist. • If any adjacent vertex is WHITE then call the recursive function for that node. If the function returns true. Return true. • If no adjacent node is grey or has not returned true then mark the current Node as BLACK and return false. Implementation: Output Graph contains cycle Complexity Analysis: • Time complexity: O(V...

Approach: The problem can be solved based on the following idea: To find cycle in a directed graph we can use the only if there is a back edge [i.e., a node points to one of its ancestors] present in the graph. To detect a back edge, we need to keep track of the nodes visited till now and the nodes that are in the current recursion stack [i.e., the current path that we are visiting]. If during recursion, we reach a node that is already in the recursion stack, there is a cycle present in the graph. Note: If the graph is disconnected then get the DFS forest and check for a cycle in individual trees by checking back edges. Follow the below steps to Implement the idea: • Create a recursive dfs function that has the following parameters – current vertex, visited array, and recursion stack. • Mark the current node as visited and also mark the index in the recursion stack. • Iterate a loop for all the vertices and for each vertex, call the recursive function if it is not yet visited (This step is done to make sure that if there is a forest of graphs, we are checking each forest): • In each recursion call, Find all the adjacent vertices of the current vertex which are not visited: • If an adjacent vertex is already marked in the recursion stack then return true. • Otherwise, call the recursive function for that adjacent vertex. • While returning from the recursion call, unmark the current node from the recursion stack, to represent that the current node is no longer a part of the ...

first, before moving on to the solution. Solution Approach: Depth First Traversal can be used to detect cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is a For a disconnected graph, we get the DFS forest as output. To detect cycle, we can check for cycle in individual trees by checking back edges. In the In this post, a different solution is discussed. The solution is from WHITE : Vertex is not processed yet. Initially, all vertices are WHITE. GRAY: Vertex is being processed (DFS for this vertex has started, but not finished which means that all descendants (in DFS tree) of this vertex are not processed yet (or this vertex is in the function call stack) BLACK : Vertex and all its descendants are processed. While doing DFS, if an edge is encountered from current vertex to a GRAY vertex, then this edge is back edge and hence there is a cycle. Algorithm: • Create a recursive function that takes the edge and color array (this can be also kept as a global variable) • Mark the current node as GREY. • Traverse all the adjacent nodes and if any node is marked GREY then return true as a loop is bound to exist. • If any adjacent vertex is WHITE then call the recursive function for that node. If the function returns true. Return true. • If no adjacent node is grey or has not returned true then mark the current Node as BLACK and return false. Implementation: Output Graph contains cycle Complexity Analysis: • Time complexity: O(V...