图表需要具有的最小边数是多少？

Question

Suppose we can construct a graph with N number of vertices, what's the minimum number of edges that graph needs to have so that it can't have an articulation point? In other words, it needs to be biconnected.

A complete graph obviously doesn't have any articulation point, but we can still remove some of its edges and it may still not have any. So it seems it can have lesser number of edges than the complete graph. With N vertices, there are a number of ways in which we can construct graph. So this minimum number should satisfy any of those graphs.

Clarification as the title is confusing for users - What is the smallest m (as a function of n) such that every n-vertex graph having at least m edges is necessarily biconnected?

Answer 1

Finding the smallest number of edges to add to make a graph 2-connected (if that is the question) is in the unweighted case solvable in polynomial time; there's an algorithm in Eswaran&Tarjan: Augmentation Problems , where also some general bounds are given.

Answer 2

EDIT: I'm interpreting your question to be: "What is the minimum number of edges that guarantees that a graph is biconnected?" Ie what is the smallest m (as a function of n) such that every n-vertex graph having at least m edges is necessarily biconnected? This meaning seems to be clear from the body of your question, but your title suggests a different question, namely "What is the minimum number of edges with which it is possible to construct a biconnected graph on n vertices?" (That question has an easy answer: It's a single n-edge cycle. This is tight, since every n-vertex graph with n-1 or fewer edges is a forest, and no forest is biconnected.)

EDIT: Fixed erroneous claim below, thanks commenter user1990169

Well, here's an easy lower bound: A leaf in a connected component of 3 or more vertices is always the neighbour of an articulation point, so if you take a complete graph on n-1 vertices (which has (n-1)(n-2)/2 edges) and add a single leaf (vertex plus edge to that vertex), you get a non-biconnected graph. So f(n) >= (n-1)(n-2)/2 + 2.

EDIT: Upper bound taken directly from David Eisenstat's comment

The proof of the corresponding upper bound is that, given a graph with an articulation vertex, its removal disconnects a vertices from b vertices, where a + b = n - 1 and a ≥ 1 and b ≥ 1. The number of edges in this new graph is at most a ( a - 1)/2 + b ( b - 1)/2, so the original total is at most a ( a - 1)/2 + b ( b - 1)/2 + a + b ≤ ( n - 1) ( n - 2)/2 + 1 by optimizing a single-variable function.