Counter Example For Every Eulerian Graph Is Hamilton

Counterexamples: Not Every Eulerian Graph is Hamiltonian

This article explores the relationship between Eulerian and Hamiltonian graphs, specifically addressing the common misconception that every Eulerian graph is also Hamiltonian. We'll delve into counterexamples demonstrating that this statement is false, providing a clear understanding of the differences between these two important graph properties. Understanding these distinctions is crucial for various applications in computer science, network optimization, and graph theory.

A graph is considered Eulerian if it contains an Eulerian circuit – a closed trail that visits every edge exactly once. A graph is Hamiltonian if it contains a Hamiltonian cycle – a closed trail that visits every vertex exactly once (except for the starting and ending vertex, which are the same). While the existence of an Eulerian circuit depends solely on the degree of vertices, the existence of a Hamiltonian cycle is a much more complex problem with no simple, universally applicable solution.

The key difference lies in whether we're traversing edges or vertices. An Eulerian graph guarantees a path covering all edges, while a Hamiltonian graph guarantees a path covering all vertices. This distinction leads us to the crucial point: not every Eulerian graph is Hamiltonian.

Let's examine some counterexamples to illustrate this point:

Counterexample 1: The Complete Bipartite Graph K_3,3

The complete bipartite graph K_3,3 is a classic example. It's Eulerian because every vertex has a degree of 3 (an even number, a necessary condition for an Eulerian graph). However, it's not Hamiltonian. Try as you might, you won't find a cycle that visits every vertex exactly once. This is because to create a Hamiltonian cycle you would need to alternate between the two sets of vertices, but that is not possible in K_3,3.

Counterexample 2: A Modified Petersen Graph

Consider a modified Petersen graph where we add an edge between two vertices that are not directly connected in the original Petersen graph. This new graph will still be Eulerian (provided the added edge connects vertices of odd degree), but it will likely remain non-Hamiltonian. The original Petersen graph is famously non-Hamiltonian, and adding a single edge might not be enough to make it Hamiltonian. The complexity of determining Hamiltonian properties makes it difficult to find a guaranteed-Hamiltonian version by simply adding edges.

Counterexample 3: Larger, More Complex Graphs

Numerous larger and more complex graphs can be constructed to serve as counterexamples. These often involve carefully designed structures that guarantee an Eulerian circuit but prevent the creation of a Hamiltonian cycle. The challenge lies in finding the specific pattern that ensures the Eulerian property while thwarting the Hamiltonian property. Creating such examples frequently requires graph theory expertise and potentially computational tools.

Conclusion: Eulerian ≠ Hamiltonian

In summary, while an Eulerian graph guarantees a path that covers every edge, it doesn't guarantee a path that covers every vertex. The existence of a Hamiltonian cycle is a far more challenging problem to solve. The counterexamples presented here clearly demonstrate that being Eulerian is not a sufficient condition for being Hamiltonian. The relationship between these two graph properties is therefore not one of implication, highlighting the importance of understanding their distinct characteristics within the broader field of graph theory.