What is the maximum number of edges in a directed graph with n nodes?

What is the maximum number of edges in a directed graph with n nodes? Is there any upper bound?

If you have N nodes, there are N - 1 directed edges than can lead from it (going to every other node). Therefore, the maximum number of edges is N * (N - 1).

Directed graph:

Question: What's the maximum number of edges in a directed graph with n vertices?

• Assume there are no self-loops.
• Assume there there is at most one edge from a given start vertex to a given end vertex.

> Each edge is specified by its start vertex and end vertex. There are n > choices for the start vertex. Since there are no self-loops, there are > n-1 choices for the end vertex. Multiplying these together counts all > possible choices.

Answer: n(n−1)

Undirected graph

Question: What's the maximum number of edges in an undirected graph with n vertices?

• Assume there are no self-loops.
• Assume there there is at most one edge from a given start vertex to a given end vertex.

> In an undirected graph, each edge is specified by its two endpoints > and order doesn't matter. The number of edges is therefore the number > of subsets of size 2 chosen from the set of vertices. Since the set of > vertices has size n, the number of such subsets is given by the > binomial coefficient C(n,2) (also known as "n choose 2"). Using the > formula for binomial coefficients, C(n,2) = n(n-1)/2.

Answer: (n*(n-1))/2

In an undirected graph (excluding multigraphs), the answer is n*(n-1)/2. In a directed graph an edge may occur in both directions between two nodes, then the answer is n*(n-1).

In addition to the intuitive explanation Chris Smith has provided, we can consider why this is the case from a different perspective: considering undirected graphs.

To see why in a DIRECTED graph the answer is n*(n-1), consider an undirected graph (which simply means that if there is a link between two nodes (A and B) then you can go in both ways: from A to B and from B to A). The maximum number of edges in an undirected graph is n(n-1)/2 and obviously in a directed graph there are twice as many.

Good, you might ask, but why are there a maximum of n(n-1)/2 edges in an undirected graph? For that, Consider n points (nodes) and ask how many edges can one make from the first point. Obviously, n-1 edges. Now how many edges can one draw from the second point, given that you connected the first point? Since the first and the second point are already connected, there are n-2 edges that can be done. And so on. So the sum of all edges is:

Sum = (n-1)+(n-2)+(n-3)+...+3+2+1 

Since there are (n-1) terms in the Sum, and the average of Sum in such a series is ((n-1)+0)/2 {(last + first)/2}, Sum = n(n-1)/2

If the graph is not a multi graph then it is clearly n * (n - 1), as each node can at most have edges to every other node. If this is a multigraph, then there is no max limit.

Putting it another way:

A complete graph is an undirected graph where each distinct pair of vertices has an unique edge connecting them. This is intuitive in the sense that, you are basically choosing 2 vertices from a collection of n vertices.

nC2 = n!/(n-2)!*2! = n(n-1)/2

This is the maximum number of edges an undirected graph can have. Now, for directed graph, each edge converts into two directed edges. So just multiply the previous result with two. That gives you the result: n(n-1)

In a directed graph having N vertices, each vertex can connect to N-1 other vertices in the graph(Assuming, no self loop). Hence, the total number of edges can be are N(N-1).

In the graph with self loop

max edges= n*n

such as we have 4 nodes(vertex)

4 nodes = 16 edges= 4*4

There can be as many as n(n-1)/2 edges in the graph if not multi-edge is allowed.

And this is achievable if we label the vertices 1,2,...,n and there's an edge from i to j iff i>j.

See here.

The correct answer is n*(n-1)/2. Each edge has been counted twice, hence the division by 2. A complete graph has the maximum number of edges, which is given by n choose 2 = n*(n-1)/2.

Can also be thought of as the number of ways of choosing pairs of nodes n choose 2 = n(n-1)/2. True if only any pair can have only one edge. Multiply by 2 otherwise

Undirected is N^2. Simple - every node has N options of edges (himself included), total of N nodes thus N*N