More explanation: The adjacency matrix of a disconnected graph will be block diagonal. Then think about its complement, if two vertices were in different connected component in the …

0 IF it is a simple, connected graph, then for the set of vertices {v: v exists in V}, v is adjacent to every other vertex in V. This type of graph is denoted Kn. For Kn, there will be n vertices and …

This result is immediate by induction once you have established (as lemma) that in every connected graph with at least two vertices there are at least two vertices that can be …

0 Let G be a graph with n vertices. Assume, G is not connected, so there are at least two connecting components. Denote the number of vertices of the components with u und v.

Here's alternative proof that a connected graph with n vertices and n-1 edges must be a tree modified from yours but without having to rely on the first derivation:

Let $G$ be a $k$ -connected graph. Meaning, $G$ has no fewer than $k$ vertices, and for every set of $k-1$ or fewer vertices, if we remove them from $G$, the graph stays connected (Of …

Prove that Strongly Connected Tournament has a Hamiltonian Cycle Ask Question Asked 13 years, 3 months ago Modified 2 years, 6 months ago

Another inductive approach: Try proving it for the case where you have a graph with one node, then show that in a graph with (n + 1) nodes, you can single out some node, remove it from …

What am I missing in here? Connected question: A connected k-regular bipartite graph is 2-connected. Edit: To clarify, my definition of graph allows multiple edges and loops. If a graph …

Assume every connected graph with $k$ vertices and $k-1$ edges is a tree. Let $G$ be a connected graph with $k+1$ vertices that contains at least one cycle (no amount of edges are …