Table of Contents
Is shortest path DFS or BFS?
BFS finds the shortest path to the destination whereas DFS goes to the bottom of a subtree, then backtracks. The full form of BFS is Breadth-First Search while the full form of DFS is Depth First Search. BFS uses a queue to keep track of the next location to visit.
What is the right way of determining whether a graph has a cycle or not?
The existence of a cycle in directed and undirected graphs can be determined by whether depth-first search (DFS) finds an edge that points to an ancestor of the current vertex (it contains a back edge). All the back edges which DFS skips over are part of cycles.
How do you solve a graph with no negative weight cycles?
If there are no negative weight cycles, then we can solve in O (E + VLogV) time using Dijkstra’s algorithm. Since the graph is unweighted, we can solve this problem in O (V + E) time. The idea is to use a modified version of Breadth-first search in which we keep storing the predecessor of a given vertex while doing the breadth-first search.
Can We do topological sorting for directed acyclic graph (DAGs)?
For a graph with no negative weights, we can do better and calculate single source shortest distances in O (E + VLogV) time using Dijkstra’s algorithm. Can we do even better for Directed Acyclic Graph (DAG)? We can calculate single source shortest distances in O (V+E) time for DAGs. The idea is to use Topological Sorting.
How do you solve an unweighted graph with Dijkstra’s algorithm?
One solution is to solve in O (VE) time using Bellman–Ford. If there are no negative weight cycles, then we can solve in O (E + VLogV) time using Dijkstra’s algorithm . Since the graph is unweighted, we can solve this problem in O (V + E) time.
What is the shortest path length of the input vertex?
Output: Shortest path length is:2 Path is:: 0 3 7 Input: source vertex is = 2 and destination vertex is = 6. Output: Shortest path length is:5 Path is:: 2 1 0 3 4 6 Recommended: Please try your approach on {IDE} first, before moving on to the solution. One solution is to solve in O (VE) time using Bellman–Ford.