Prove that a graph of order n with at least ((n-1)(n-2))/(2)+1 edges must be connected. Give an

Prove that a graph of order n with at least...

Key Concepts

Disconnected Graph Example Construction

The construction of a disconnected graph with one fewer edge than the connectivity threshold involves taking a complete graph on (n-1) vertices and adding an isolated vertex. This graph has exactly ((n-1)(n-2))/2 edges, and its disconnected nature (due to the isolated vertex) serves as a sharp example showing that the edge threshold is best possible. This example illustrates how a small decrease in the number of edges can lead to a qualitatively different property (disconnectedness) in a graph.

Threshold Number of Edges for Connectivity

The threshold number of edges for guaranteeing connectivity in a graph of order n is significant because it identifies the point at which the graph can no longer be disconnected without violating the edge count. In this context, the threshold is given by ((n-1)(n-2))/2 + 1, which arises from considering the maximum number of edges a disconnected graph can have. This result is useful in extremal graph theory and has implications for understanding network structures.

Graph Connectivity

Graph connectivity is a fundamental concept in graph theory that refers to whether there is a path between every pair of vertices in a graph. A connected graph is one where such a path exists for any two vertices, implying that the graph is 'in one piece'. Proving connectivity by edge count is common because a certain minimum number of edges often guarantees that the graph is connected.

Maximum Edge Count in a Disconnected Graph

One key aspect of the problem is understanding the maximum number of edges a disconnected graph of a given order can possess. This is typically achieved by partitioning the graph into two components, where one component is as large as possible while the other is as small as possible, such as isolating a single vertex. The largest number of edges then comes from creating a complete graph on the (n-1) vertices, which has ((n-1)(n-2))/2 edges.

Transcript

00:01 So for this question, we're trying to show that a simple graph with n vertices must be connected if it has greater than, oh, sorry, greater than n minus 1 times n minus 2 over 2 edges.

00:31 So one way to do this is to show that if this simple graph, so our logic is if the graph is not, connected, then the max number of edges is n minus 1 times n minus 2 over 2.

00:58 That's one way to solve this problem, right? because if the max number of edges in a non -connected graph is this number, then that means that if you have more than this number of edges, then you must be connected.

01:11 So that's the, essentially the overview of how we're going to solve this problem.

01:16 So without loss of generality, suppose that the graph is not connected.

01:21 Then let's say one component has k vertices, where k is just a number from like one to n minus one.

01:27 So k vertices in one component, that means that there are n minus k vertices split among at least one other component, right? we don't know if they're all in a separate component or if they're like all separate at least one other component.

01:52 So if we show that this graph can't have more than this number of edges, we're good.

02:03 So we want to maximize, maximize the number of edges in this graph, in this graph, in the graph above.

02:14 And the way to do that is by assuming, right, as we said before, the maximum number of edges in a simple graph occurs when the graph is complete and so let's assume that this is a complete component.

02:30 So if we maximize that, we get that this we have k choose two vertices in this complete component.

02:36 And then to maximize the number of edges in this, we assume that they're all in one component and that it's a complete component as well.

02:48 So the number of met.

02:50 So we've maximized the number of edges in this graph by assuming that both of these components are complete and then that gives us k squared minus n k plus n squared minus two n over two that's this is the maximum number of edges in the in this in this graph and so we want to maximize this expression right um so we realize if we look at uh it's a this is a parabola shaping upward right if we find its vertex its vertex it's vertex it's vertex a occurs by calculus at n over 2.

03:30 We just take the derivative.

03:32 So we will get 2k minus n is equal to 0.

03:36 So n is equal to k over 2.

03:39 No, k is equal to n over 2, sorry.

03:42 And so our vertex is here.

03:44 That means that its maximum will be at the bounds, right? it'll be greatest, farthest away from n over 2.

03:51 So we know that the bound, the domain, right, for k is 1 to n minus 1.

04:02 Luckily, that's the metric about n over 2...