How would we prove that every graph with at least two vertices has two vertices of the same degree?

2) The complete graph with k+1 vertices has degree k, for all vertices, so it is k-regular. Assume the graph G has n vertices. Using the pigeonhole principle, we must pick the n degrees (one for each vertex), from a set of n-1 answers (since we cannot have both 0 and n-1). Hence, one degree must be repeated.

Does there exist a simple graph with two or more vertices such that all the vertices are of different degree?

Thus there is no such graph. A graph with 2 vertices has either 0 or 1 edges, and in either case, the two nodes have the same degree. Assume the theorem holds for k vertices and there are two (or more) of same degree.

When two graphs are having same number of edges vertices and degrees are said to be?

If each vertex of the graph has the same degree k, the graph is called a k-regular graph and the graph itself is said to have degree k. Similarly, a bipartite graph in which every two vertices on the same side of the bipartition as each other have the same degree is called a biregular graph.

Can a graph have two vertices?

These are called simple graphs. It turns out then, that there are only two simple graphs with two vertices. One has an edge and the other doesn’t have any. If we want to allow a graph to have loops or multiple edges we will specifically say so.

How do you know if a simple graph exists?

Note that a simple graph is a graph with no self-loops and parallel edges….Stopping conditions:

1. All the elements remaining are equal to 0 (Simple graph exists).
2. Negative number encounter after subtraction (No simple graph exists).
3. Not enough elements remaining for the subtraction step (No simple graph exists).

Does there exist a simple graph with five vertices of the following degrees?

A simple graph has no parallel edges nor any loops. There are only 5 vertices, so each vertex can only be joined to at most four other vertices, so the maximum degree of any vertex would be 4. Hence, you can’t have a vertex of degree 5.

Is a connected graph and all its vertices are of even degree?

Since all vertices of G have even degree, n≥3. Assume that this is true for n=k, let h be a graph obtained by adding 2 vertex a,b to G and joining a,b to every vertex in G so that every vertex in G still have even degree. Since the statement is true for n=k, G has no bridge, meaning very edge of G lie in some cycle.

Is there a simple graph whose vertices have given degrees?

The answer for c is that there cannot be such a graph – since there are 2 vertices with degree 4, they must be connected to all other vertices. Therefore, the vertex with degree one, is an impossibility.

What do you mean by in degree and out degree of a graph?

The In-Degree of a vertex v written by deg-(v), is the number of edges with v as the terminated vertex. To find the in-degree of a vertex, just count the number of edges ends at the vertex. Out-degree of a vertex. The Out-Degree of a vertex V written by deg+ (v), is the number of edges with v as the initial vertex.

How to find two vertices of the same degree in G?

We can not have a vertex of degree $0$ in G, so the set of vertex degrees is a subset of $S = {1, 2, · · · , n − 1}$. Since the graph Ghas n vertices, by pigeon-hole principle we can find two vertices of the same degree in G.

Can a vertex with degree 0 exist on a graph?

By contradiction – assume that a graph has a vertex v with degree n − 1 . Then v must find exactly n − 1 distinct cities on the other side. Therefore, in that case a vertex with degree 0 can not exist. which amounts to a contradiction – our initial assumption that all the degrees are distinct was false.

Can an edge be infinite in a graph?

An edge may connect a distinct pair of vertices or it may loop back and return to where it started from (without visiting any other vertices). Depending on the cardinality of the set V graphs may be finite or infinite. I will only consider finite graphs.

How many degrees can a graph have with N1 vertices?

But if a degree zero exists then that’s a vertex thats disconnected from the rest of the graph which now has n-1 vertices and the max degree is n-2. So there are more points than the number of possible degrees. So there must be duplication.