Is a graph with N-1 edges connected?

Is a graph with N-1 edges connected?

Proof: We know that the minimum number of edges required to make a graph of n vertices connected is (n-1) edges. We can observe that removal of one edge from the graph G will make it disconnected. Thus a connected graph of n vertices and (n-1) edges cannot have a circuit. Hence a graph G is a tree.

Does MST have N-1 edges if the graph has n edges?

If you remove the minimum spanning tree (MST) of n−1 edges (a simple connected graph has one) from a graph with n edges (or more), you still have one edge. Between the vertices of this edge, a path should be in the MST, forming a cycle in the original graph.

Why does a complete graph have n n-1 )/ 2 edges?

But this method counts every edge twice, because every edge going out from one vertex is an edge going into another vertex. Hence, you have to divide your result by 2. This leaves you with n(n−1)/2.

Does a tree always have N-1 edges?

Every vertex that is added to the tree contributes one edge to the tree. Thus, the number of edges required to add (n+1)th node = 1. Thus the total number of edges will be (n – 1) + 1 = n -1+1 = n = (n +1) – 1. Thus, P(n+1) is true.

How can you prove an acyclic graph has n-1 edges?

The spanning graph for this is an acyclic copy of this where all the vertices are present, and all the edges are a subset of the graph with the condition that each connection is distinct. Apparently the MST should have n-1 nodes. How can this be proven? Every acyclic graph can be represented as a tree, if all the nodes are connected.

How can we prove that$ G$ is acyclic?

Since $G$ is acyclic, there is at least one vertex $v$ with degree $1$. The graph induced by deleting this node is acyclic with $n$ vertices and $n-1$ edges: it is a connected tree by assumption. Linking $v$ back to the tree does not create a cycle, it follows that $G$ is a connected tree.

Are there V-1 edges in an undirected graph?

For Undirected Graph – It will be a spanning tree (read about spanning tree) where all the nodes are connected with no cycles and adding one more edge will form a cycle. In the spanning tree, there are V-1 edges.

Which is the root of an undirected graph?

BTW – for undirected graphs, words like “root”, “parent” and “child” don’t mean much, at least not formally. Any vertex in any undirected tree can be considered the root, and which vertex you happen to call the root decides for all edges which vertex is the parent and which is the child.