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Cayleys formula states that for a complete graph on nvertices, the number of spanning trees is n^(n-2). For a complete bipartite graph we can use the formula p^q-1 q^p-1. for the number of spanning trees. A generalization of this for any graph is Kirchhoff's theorem or Kirchhoff's matrix tree theorem. This theorem looks at the Laplacian matrix of a graph. ( you may need to look up what that is with some examples). For graphs with a small number of edges and vertices, you can find all the spanning trees and this is often quicker. There are also algorithms such as depth-first and breadth-first for finding spanning trees.

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Q: How do you count spanning trees in a graph?
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Prove that a graph G is connected 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!


What is the least count of graph paper?

The least count of graph paper is determined by the smallest measurable unit on the paper, typically represented by the smallest division or grid square. This value is essential for accurately measuring distances and plotting points on the graph. The least count can vary depending on the type of graph paper, with common values including 1 mm, 0.5 mm, or even 0.1 mm.


How can a graph be used to determine how many solutions an equation has?

Count the number of many times the graph intersects the x-axis. Each crossing point is a root of the equation.


How do you find the perimeter of a rectangle with coordinates on graph paper?

Count the lengths of each side and sum them up (add them all together.)


When creating graph does the x and y axis have to go b the same count?

Yes it does! It has to be equal to connect with each other at the ''o''.

Related questions

How many spanning trees can be drawn with 5 labeled vertices?

No of spanning trees in a complete graph Kn is given by n^(n-2) so for 5 labelled vertices no of spanning trees 125


What is the significance of the cut property in the context of Minimum Spanning Trees (MST)?

In the context of Minimum Spanning Trees (MST), the cut property states that for any cut in a graph, the minimum weight edge that crosses the cut must be part of the Minimum Spanning Tree. This property is significant because it helps in understanding and proving the correctness of algorithms for finding Minimum Spanning Trees.


What is the significance of the cut property of minimum spanning trees (MSTs)?

The cut property of minimum spanning trees (MSTs) states that for any cut in a graph, the minimum weight edge that crosses the cut must be part of the MST. This property is significant because it helps in efficiently finding the minimum spanning tree of a graph by guiding the selection of edges to include in the tree.


What is the cycle property of minimum spanning trees (MSTs) and how does it impact the construction and optimization of MSTs?

The cycle property of minimum spanning trees (MSTs) states that if you have a cycle in a graph and you remove the heaviest edge from that cycle, the resulting graph will still have the same minimum spanning tree. This property impacts the construction and optimization of MSTs by helping to identify and eliminate unnecessary edges, leading to a more efficient and optimal tree structure.


What is Difference between tree and spanning tree?

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.


What is the total number of spanning tree that can be drawn using five labeled vertices?

125 according to Cayley's formula for counting spanning trees. For a complete graph Kn, t(kn) = nn-2 where n is the number of vertices.


How can one find a spanning tree in a given graph?

To find a spanning tree in a given graph, you can use algorithms like Prim's or Kruskal's. These algorithms help identify the minimum weight edges that connect all the vertices in the graph without forming any cycles. The resulting tree will be a spanning tree of the original graph.


What is spanning tree in data structure?

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.


Does every possible minimal spanning tree of a given graph have an identical number of edges?

No, not every possible minimal spanning tree of a given graph has an identical number of edges.


What is the minimum spanning tree of an undirected graph g?

The minimum spanning tree of an undirected graph g is the smallest tree that connects all the vertices in the graph without forming any cycles. It is a subgraph of the original graph that includes all the vertices and has the minimum possible total edge weight.


Prove that 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!


Prove that a graph G is connected 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!