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Does BFS work with negative edges?

Posted on November 25, 2019 by Author

Table of Contents

  • 1 Does BFS work with negative edges?
  • 2 Is spanning tree a forest?
  • 3 Can an undirected graph have weights?
  • 4 Does BFS visit every node?
  • 5 Is k1 a complete graph?
  • 6 Can I use BFS for shortest path?
  • 7 Why not use BFS to find the cycle of a graph?
  • 8 What is the difference between a BFS and an undirected graph?

Does BFS work with negative edges?

1 Answer. BFS is applicable to unweighted graphs (or graphs where all edges have the same weight). For weighted graphs one can use algorithms such as Dijkstra’s (which is a “modified BFS”), or A* (which is a “modified Dijkstra’s”) . However both Dijkstra’s or A* do not work correctly with negative weights.

Is spanning tree a forest?

Spanning forests For other authors, a spanning forest is a forest that spans all of the vertices, meaning only that each vertex of the graph is a vertex in the forest. For this definition, even a connected graph may have a disconnected spanning forest, such as the forest in which each vertex forms a single-vertex tree.

Why does BFS not work on weighted graph?

BFS will not work on weighted graphs since the path with the fewest edges may not be the shortest if the edges it contains are expensive. However, if all the weights are intergers and they are bounded by a small number, say k, we can still use BFS.

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Does BFS necessarily return the shortest path?

Both BFS and DFS will give the shortest path from A to B if you implemented right.

Can an undirected graph have weights?

An adjacency matrix for an undirected graph is always symmetric. An adjacency matrix can also be used to represent weighted graphs. For example, if M{i,j} = w, then there is an edge from vertex i to vertex j with weight w. The weights of the edges can be stored in nodes of linked lists.

Does BFS visit every node?

A queue (FIFO-First in First Out) data structure is used by BFS. You mark any node in the graph as root and start traversing the data from it. BFS traverses all the nodes in the graph and keeps dropping them as completed. BFS visits an adjacent unvisited node, marks it as done, and inserts it into a queue.

Is Dijkstra a BF?

Conclusion: Dijkstra algorithm is BFS. For their most basic use (i.e., finding the connected components of an undirected graph), depth first search (DFS) and breadth-first search (BFS) are interchangeable.

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Can a forest have cycles?

A disconnected acyclic graph is called a forest. In other words, a disjoint collection of trees is called a forest. The following graph looks like two sub-graphs; but it is a single disconnected graph. There are no cycles in this graph.

Is k1 a complete graph?

A complete digraph is a directed graph in which every pair of distinct vertices is connected by a pair of unique edges (one in each direction). Graph theory itself is typically dated as beginning with Leonhard Euler’s 1736 work on the Seven Bridges of Königsberg….

Complete graph
Notation Kn
Table of graphs and parameters

Can I use BFS for shortest path?

BFS can be used to find shortest path in an unweighted cyclic graph as well. If a graph is unweighted, then BFS can be applied for SP regardless of having loops.

Should I use BFS or DFS?

The only reason to use a BFS would be if you know your (undirected) graph is going to have long paths and small path cover (in other words, deep and narrow). In that case, BFS would require proportionally less memory for its queue than DFS’ stack (both still linear of course).

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Can you use a stack in BFS?

In DFS, think of recursion, you go deeper and deeper, only to go back to the original state, so for that, you can use a stack. With BFS, a stack is useless if you’re visiting all nodes that are exclusively your neighbour.

Why not use BFS to find the cycle of a graph?

That’s why DFS is used to find cycles in directed graphs. BFS provides no such guarantees, so it just doesn’t work. (notwithstanding perfectly good cycle-finding algorithms that include BFS or similar as a sub-procedure)

What is the difference between a BFS and an undirected graph?

BFS provides no such guarantees, so it just doesn’t work. (notwithstanding perfectly good cycle-finding algorithms that include BFS or similar as a sub-procedure) An undirected graph, on the other hand, has a cycle whenever there are two paths between any pair of vertexes, i.e., when it’s not a tree.

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