For each n, find an example of a critical 2-connected graph with a vertex of degree at least n. Definition:
Let G be a critical 2-connected graph; this means that G is 2-connected but no graph G − e for e ∈ E(G) is 2-connected.

Question
For each n, find an example of a critical 2-connected graph with a vertex of degree at least n.

I am pretty new to this, would you give me some examples or hint?

My first thought is a tree, which is minimally connected. Say n = 5, I have a tree T with only 4 edges.Removing any 1 edge, T is disconnected. Is this what the question is looking for? would any tree with order n work for this example? Please help me to understand this question better.
2-connected means the graph is connected, and if you remove any edge it is still connected. In practice this means the graph must contain a connected simple subgraph in which every vertex must have a degree of at least 2. (We'll just consider simple graphs because anything else is just redundant)

If you have a tree, the vertices on the "ends" of the graph only have one edge, so it doesn't work.

The easiest example of a graph that's definitely critically 2-connected is a cycle graph. This is where (if you label your vertices from 1 to k), there's an edge between vertex 1 and vertex 2, vertex 2 and vertex 3, ..., vertex k-1 and vertex k, and vertex k and vertex 1. Now if you remove any edge, say the edge between vertex 1 and vertex 2, then vertex 1 has degree 1 and vertex 2 has degree 1, so they are no longer 2-connected. Good. However you had an extra condition - for each n, you need an example where there is a vertex of degree at least n. For a cycle graph, every vertex has degree 2, so this example only works for n = 1, 2.

Intuitively you need to have one vertex with many edges coming off it, but each of these edges can only connect to a vertex with one other edge. Otherwise if you removed an edge adjacent to the "vertex with degree at least n", it will still be 2-connected and hence you graph wasn't critical. Well the key is to combine this idea with cycle graphs. (All you need are 3-cycles.)

Visualise $G_1$ to be a graph with 3 vertices, where each of these verticies are connected to the other two. You end up with a cycle-graph that looks like a triangle, critically 2-connected. Now make a copy of this graph which shares a vertex with $G_1$. You have a "central vertex" with degree 4, and four verticies around it which all have degree 2. It looks like a bow-tie. This is still critically 2-connected because every edge connects to a vertex with only one other edge. However, we now have a graph for n = 1, 2, 3, and 4.

In the same way, for any n, you:


need to consider any even k greater than it
have k edges radiating out from one vertex
add a vertex to the end of each of these edges
add an edge between every pair of these vertices.


Here's an example of a graph which is critically 2-connected with a vertex of at least degree 8.
2-connected means the graph is connected, and if you remove any edge it is still connected. In practice this means the graph must contain a connected simple subgraph in which every vertex must have a degree of at least 2. (We'll just consider simple graphs because anything else is just redundant)

If you have a tree, the vertices on the "ends" of the graph only have one edge, so it doesn't work.

The easiest example of a graph that's definitely critically 2-connected is a cycle graph. This is where (if you label your vertices from 1 to k), there's an edge between vertex 1 and vertex 2, vertex 2 and vertex 3, ..., vertex k-1 and vertex k, and vertex k and vertex 1. Now if you remove any edge, say the edge between vertex 1 and vertex 2, then vertex 1 has degree 1 and vertex 2 has degree 1, so they are no longer 2-connected. Good. However you had an extra condition - for each n, you need an example where there is a vertex of degree at least n. For a cycle graph, every vertex has degree 2, so this example only works for n = 1, 2.

Intuitively you need to have one vertex with many edges coming off it, but each of these edges can only connect to a vertex with one other edge. Otherwise if you removed an edge adjacent to the "vertex with degree at least n", it will still be 2-connected and hence you graph wasn't critical. Well the key is to combine this idea with cycle graphs. (All you need are 3-cycles.)

Visualise $G_1$ to be a graph with 3 vertices, where each of these verticies are connected to the other two. You end up with a cycle-graph that looks like a triangle, critically 2-connected. Now make a copy of this graph which shares a vertex with $G_1$. You have a "central vertex" with degree 4, and four verticies around it which all have degree 2. It looks like a bow-tie. This is still critically 2-connected because every edge connects to a vertex with only one other edge. However, we now have a graph for n = 1, 2, 3, and 4.

In the same way, for any n, you:


need to consider any even k greater than it
have k edges radiating out from one vertex
add a vertex to the end of each of these edges
add an edge between every pair of these vertices.


Here's an example of a graph which is critically 2-connected with a vertex of at least degree 8.