shortest-path

On the Wikipedia page for Dijkstra's algorithm, they mark visited nodes so they wouldn't be added to the queue again. However, if a node is visited then there can be no distance to that node that is shorter, so doesn't the check alt < dist[v] already account for visited nodes? Am I misunderstanding something about the visited set? for each neighbor v of u: alt := dist[u] + dist_between(u, v); // accumulate shortest dist from source if alt < dist[v] && !visited[v]: dist[v] := alt; // keep the shortest dist from src to v previous[v] := u; insert v into Q; // Add unvisited v into the Q to be

Could you recommend any java library which implements k-shortest algorithm -> searching for alternative ways, not the only shortest one in directed multigraph ? I found only JGraphT but there is actually bug (which i submitted) but it will take a lot of time to fix it i guess, are there any other available implementations ? Except JGraphT i found only small one-man projects :/ OR would be hard to modify Disjktra shortest path alg to show alternative paths ? Thanks Ram Narasimhan 2 Possible Options: Option 1. The class KshortestPath from the MascOpt Package is a good option for a Java

I am looking for an explanation why the AStar / A* algorithm is called AStar. All similar (shortest path problem) algorithms are often named like its developer(s), so what is AStar standing for? There were algorithms called A1 and A2. Later, it was proved that A2 was optimal and in fact also the best algorithm possible, so he gave it the name A* which symbolically includes all possible version numbers. Source: In 1964 Nils Nilsson invented a heuristic based approach to increase the speed of Dijkstra's algorithm. This algorithm was called A1. In 1967 Bertram Raphael made dramatic improvements

given an undirected weighted graph G , and two vertices: start vertex and end vertex what's the most efficient algorithm that finds the shortest path from start to end with ability to turn weight of exactly one edge to zero? EDIT: i know dijkstra algorithm , but as i said , situation is different in this problem: we're allowed to turn one edge to zero, i wanna know how solve this problem efficiently, actually , one way is turn edges weights to zero iteratively! and apply dijkstra algorithmin each step, but , i'm looking for more efficient way thanks You can solve this by using Djikstra's

I am using QuickGraph version 3.6 and I found function SetRootVertex, but no SetTagretVertex. I need this because I am searching short paths in huge graph and this would speed up program a lot. Clases in question are DijkstraShortestPathAlgorithm and AStarShortestPathAlgorithm. I don't think there is a way to this without using events. You could wrap the necessary code in one extension method, making things clear, e.g.: public static class Extensions { class AStarWrapper<TVertex, TEdge> where TEdge : IEdge<TVertex> { private TVertex target; private AStarShortestPathAlgorithm<TVertex, TEdge>

I have a graph with an s and t vertex that I need to find the shortest path between. The graph has a lot of special properties that I would like to capitalize on: The graph is a DAG (directed acyclic graph). I can create a topological sort in O(|V|) time, faster than the traditional O(|V + E|). Within the topological sort, s is the first item in the list and t is the last. I was told that once I have a topological sort of the vertices I could find the shortest path faster than my current standard of Dijkstra's Uniform Cost, but I cannot seem to find the algorithm for it. Pseudo code would be

Suppose there are 3 target nodes in a graph. A vertex-disjoint path means there is not any same node except the end nodes during the path. For any one single node, say node i, how to find all vertex-disjoint paths from node i to the three target nodes? You can solve this problem by reducing it to a max-flow problem in an appropriately-constructed graph. The idea is as follows: Split each node v in the graph into to nodes: v in and v out . For each node v, add an edge of capacity one from v in to v out . Replace each other edge (u, v) in the graph with an edge from u out to v in of capacity 1.