When an edge is removed from a tree then the resulting graph has two connected components?

Removing an edge from any connected graph creates a graph with at most two connected components. The original graph doesn’t need to be a tree; it just needs to be connected.

How many edges must be removed from the graph to get a spanning tree?

This has 15 edges, and the spanning tree requires 1 edge arriving at all vertices. So we can see that 2 vertices aren’t connected to the degree 6 vertex, hence we require 6 edges plus these 2 connecting edges. So we have 8 edges total. so 15−8=7, so 7 edges need be removed.

What is a tree edge?

Tree Edge: It is an edge that is present in the tree obtained after performing DFS on the graph. Back Edge: It is an edge (u, v) such that v is an ancestor of node u but not part of the DFS Traversal of the tree. Edge from 5 to 4 is a back edge.

How many edges must be removed from a connected graph with n vertices and m edges to produce a spanning tree explain in terms of M and N and give an example?

In fact, a graph may have more than one spanning tree, as a rule for producing a spanning tree with n vertices and m edges is to remove (m – n + 1 ) edges. For example, suppose we are given the following undirected graph containing three edges and three vertices.

What is spanning tree in graph?

A spanning tree of a connected graph G can also be defined as a maximal set of edges of G that contains no cycle, or as a minimal set of edges that connect all vertices.

How many maximum edges will be removed from the complete graph?

If the graph is a complete graph, then the spanning tree can be constructed by removing maximum (e-n+1) edges.

Would the Minimum Spanning Tree change due to the change in weights?

The answer is yes. The simplest proof is that, if G has n vertices, then any spanning tree of G has n − 1 edges. Therefore incrementing each edge weight by 1 increases the cost of every spanning tree by a constant, n − 1.