Can a Simple Undirected Graph with 11 Vertices and 53 Edges Have an Eulerian Cycle?

In this article, we'll explore whether a simple undirected graph with 11 vertices and 53 edges can have an Eulerian cycle. We'll delve into theorems and reasoning that may help us determine the answer. Let's start by understanding some fundamental theorems in graph theory.

Key Theorems

Theorem 1: If G is a connected graph, G contains an Euler cycle if and only if every vertex has even degree.

Theorem 2: A connected graph G has an Euler path if and only if exactly two vertices have odd degree.

Understanding the Problem

First, we need to analyze the given conditions: the graph has 11 vertices and 53 edges. We'll consider the implications of these conditions on the graph's structure.

Constructing the Graph

Assume that at least one cycle exists. In such a case, the graph must have a structure that allows for a cycle to pass through all vertices at least once. This is a necessary condition for the graph to have an Eulerian cycle.

Edge Parity and Construction

Let's discuss the process of maintaining edge parity while constructing the graph. There are two main methods:

1. Add an edge of even weight: This represents number of edges between two unconnected vertices. This keeps the degree of vertices in parity.

2. Super impose a cycle on the graph with weight n: Edges already in the cycle of the previous graph update their weights such that w w n. Edges that don't exist in the graph are added with their weights set to n.

Note that these methods ensure that the degrees of the vertices remain even, which is a key requirement for an Eulerian cycle.

A Quick Test for Eulerian Cycle

A practical and quick way to determine if a graph can have an Eulerian cycle is to use the following condition:

If the number of edges is even when divided by the number of vertices, then the graph can have an Eulerian cycle. Conversely, if the number of edges is odd, the graph cannot have an Eulerian cycle.

Exploring the Given Numbers

Given the numbers, let’s analyze the particular cases:

1. Complete Graph on 11 Vertices: A complete graph with 11 vertices (denoted as K11) has the maximum number of edges, which is given by the formula (n(n-1))/2. For 11 vertices, the complete graph has (11(11-1))/2 55 edges.

2. Parity Considerations: The graph we are considering has 53 edges, which is less than the maximum 55 edges in a complete K11 graph. Importantly, 53 is an odd number, which hints at a potential non-Eulerian property.

3. Vertex Degrees: With 53 edges, we can determine the parity of the vertex degrees. Recall that for an Eulerian cycle, every vertex must have an even degree. Since the graph can't have more than 55 edges and 53 is odd, the graph will have at least one vertex with an odd degree.

Conclusion

Based on the theorems and parity considerations, we can conclusively state that a simple undirected graph with 11 vertices and 53 edges cannot have an Eulerian cycle. The odd number of edges and the implications of the degree of vertices make this clear.

This problem is likely a homework problem, and the provided hints can be used to determine the answer:

How many edges does a complete graph on 11 vertices have? What can you say about the parity of the vertex degrees of a graph with 11 vertices and 53 edges?