Graph Theory: Dirac's Theorem

Summary

This video explains Durac's theorem, which states that a graph with three or more vertices contains a Hamiltonian circuit if the degree of each vertex is at least n/2, where n is the number of vertices. The video demonstrates this by analyzing two graphs. In the first, not all vertices meet the degree condition, so it doesn't contain a Hamiltonian circuit. In the second graph, all vertices meet the condition, and examples of possible Hamiltonian circuits are provided, illustrating how to determine if a graph contains one using Durac's theorem.

Takeaways

• 😀 Durac's theorem applies to graphs with 3 or more vertices.
• 😀 The theorem states that if the degree of every vertex is at least n/2, the graph will contain a Hamiltonian circuit.
• 😀 A Hamiltonian circuit is a path that starts and ends at the same vertex while passing through each vertex exactly once.
• 😀 The degree of a vertex is the number of edges connected to it.
• 😀 To determine if a graph has a Hamiltonian circuit, compare the degree of each vertex with n/2, where n is the total number of vertices.
• 😀 For a graph with 5 vertices, n/2 is 2.5, so each vertex must have a degree greater than 2.5 for the graph to contain a Hamiltonian circuit.
• 😀 In the first graph example, vertex A has a degree of 2, which is less than 2.5, so the graph does not have a Hamiltonian circuit.
• 😀 In the first graph, only one vertex has a degree greater than 2.5, confirming there is no Hamiltonian circuit in this graph.
• 😀 In the second graph example, all vertices have a degree greater than 2.5, so a Hamiltonian circuit must exist.
• 😀 Two examples of Hamiltonian circuits are given: starting from vertex A, traveling through vertices B, E, D, C, and back to A (one circuit), and starting at A, B, C, E, D, and back to A (another circuit).

Q & A

What is Durac's theorem?

Durac's theorem states that if a graph G with n vertices (where n is greater than or equal to 3) has the property that the degree of every vertex is at least n/2, then the graph contains a Hamiltonian circuit.

What is a Hamiltonian circuit?

A Hamiltonian circuit is a path that starts and ends at the same vertex and passes through each vertex of the graph exactly once.

What does the degree of a vertex in a graph represent?

The degree of a vertex represents the number of edges connected to that vertex.

How do we apply Durac's theorem to determine if a graph has a Hamiltonian circuit?

To apply Durac's theorem, calculate n/2 (where n is the number of vertices) and check if the degree of every vertex is at least n/2. If so, the graph contains a Hamiltonian circuit.

In the first graph, does it satisfy Durac's theorem? Why or why not?

No, the first graph does not satisfy Durac's theorem. The degree of vertex A is 2, which is less than n/2 (which is 2.5 for this graph). Therefore, the graph does not contain a Hamiltonian circuit.

What is the significance of the degree of 2.5 in the first example?

In the first example, n is 5, so n/2 equals 2.5. The degree of each vertex must be greater than 2.5 for the graph to contain a Hamiltonian circuit. Since vertex A has a degree of 2, the graph does not satisfy Durac's theorem.

Does the second graph satisfy Durac's theorem?

Yes, the second graph satisfies Durac's theorem. All the vertices have a degree greater than 2.5, which means the graph contains a Hamiltonian circuit.

Can you provide an example of a Hamiltonian circuit in the second graph?

One example of a Hamiltonian circuit in the second graph is starting at vertex A, then traveling to B, E, D, C, and finally returning to A.

Is there more than one possible Hamiltonian circuit in the second graph?

Yes, another example of a Hamiltonian circuit in the second graph could be starting at vertex A, traveling to B, C, E, D, and then returning to A.

Why is the degree of vertex A in the first graph critical in determining the presence of a Hamiltonian circuit?

The degree of vertex A in the first graph is critical because it directly affects whether the graph satisfies Durac's theorem. Since vertex A's degree is less than 2.5, the graph fails to meet the necessary condition for a Hamiltonian circuit.