图G的二分性判定、极大路径及欧拉回路相关技术问询

Let’s break down each of your questions one by one:

1. Is G non-bipartite due to the odd cycle b-h-d? #

Absolutely correct. Bipartite graphs are defined as graphs that can be 2-colored such that no adjacent vertices share the same color. A core theorem here states a graph is bipartite if and only if it contains no odd-length cycles. The cycle b-h-d is a triangle (length 3, an odd number), which directly violates this condition. Your reasoning is perfect.

2. Is e-c-a-b-h-d-f-g a maximal path? #

Yes, this is a valid maximal path. A maximal path is one that cannot be extended by adding any adjacent vertices not already in the path. Here’s why:

• The endpoints are e and g. e connects only to c and b (both already in the path), and g connects only to f and d (also in the path).
• Every vertex in the graph is included in this path, so there are no unused vertices to add. This fits the definition of a maximal path exactly.

3. Is c-e-b-a-c-h-b-d-f-g-d-h-f-c an Euler circuit? #

Yes, this is a valid Euler circuit. For a circuit to qualify as an Euler circuit, two rules must hold:

1. Every edge in the graph is traversed exactly once.
2. The circuit starts and ends at the same vertex.

Let’s verify:

• Edge coverage: Listing all edges in the circuit shows each edge appears exactly once: c-e, e-b, b-a, a-c, c-h, h-b, b-d, d-f, f-g, g-d, d-h, h-f, f-c. All edges in the graph are accounted for with no duplicates.
• Closed circuit: The path starts and ends at c, satisfying the closed circuit requirement.
• Prerequisite check: All vertices in G have even degrees (2 or 4), which is a necessary condition for an Euler circuit to exist—this circuit meets all the criteria.

内容的提问来源于stack exchange，提问作者Zainab Husain