A spanning tree is a tree associated with a network. All the nodes of the graph appear on the tree once. A minimum spanning tree is a spanning tree organized so that the total edge weight between nodes is minimized.
yes, but a shortest path tree, not a minimum spanning tree
Minimum cost spanning tree is used for Network designing.(like telephone, electrical, hydraulic, TV cable, computer, road)
with minimum spanning tree algorthim
BPDU
The port will rapidly transition to forwarding.
A spanning tree protocol, or STP, is characteristic to a LAN. It provides a loop-free topology for networks within the system.
A tree is a connected graph in which only 1 path exist between any two vertices of the graph i.e. if the graph has no cycles. A spanning tree of a connected graph G is a tree which includes all the vertices of the graph G.There can be more than one spanning tree for a connected graph G.
Spanning Tree Protocol (STP) does not set up trunk links, it is used to remove logical (Layer 2: Data Link) loops in a network in case there is layer 1 (Physical) redundancy
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!
Proving this is simple. First, you prove that G has a spanning tree, it is connected, which is pretty obvious - a spanning tree itself is already a connected graph on the vertex set V(G), thus G which contains it as a spanning sub graph is obviously also connected. Second, you prove that if G is connected, it has a spanning tree. If G is a tree itself, then it must "contain" a spanning tree. If G is connected and not a tree, then it must have at least one cycle. I don't know if you know this or not, but there is a theorem stating that an edge is a cut-edge if and only if it is on no cycle (a cut-edge is an edge such that if you take it out, the graph becomes disconnected). Thus, you can just keep taking out edges from cycles in G until all that is left are cut-gees. Since you did not take out any cut-edges, the graph is still connected; since all that is left are cut-edges, there are no cycles. A connected graph with no cycles is a tree. Thus, G contains a spanning tree. Therefore, a graph G is connected if and only if it has a spanning tree!
Prims Algorithm is used when the given graph is dense , whereas Kruskals is used when the given is sparse,we consider this because of their time complexities even though both of them perform the same function of finding minimum spanning tree. ismailahmed syed