Hamiltonian Path: Existence and How to Decide (Graph-Theoretic Method)

A Hamiltonian path exists in a graph $G=(V,E)$ if and only if there is an ordering of the vertices $v_1,\dots,v_{|V|}$ such that each consecutive pair $\{v_i,v_{i+1}\}\in E$.

To answer your prompt for a specific graph, the general strategy is:

1. check quick necessary conditions (e.g., degree constraints),
2. then either construct a valid Hamiltonian path or prove impossibility by structure-based arguments,
3. if needed, apply a systematic search (backtracking) tailored to the graph.

Because your message references “the following graph” but the actual edges/diagram are not included, I’ll provide a comprehensive, exam-ready decision workflow you can apply directly to that graph as soon as you supply its vertex set and edges.

Key graph-theory terms you’ll use

• Hamiltonian path
• Hamiltonian cycle
• Simple graph
• Degree

Step 1: Do quick necessary checks (fast elimination)

These are not sufficient in general, but they can immediately rule out (a) for many graphs.

1.1 Endpoint degree constraints

A Hamiltonian path has two endpoints (unless $|V|=1$), and internal vertices have degree at least 2 in the path. This yields the following necessary condition:

• If there exists a vertex of degree 0: no Hamiltonian path exists.
• If there exist two vertices of degree 1: a Hamiltonian path might still exist, but it forces those degree-1 vertices to be the two endpoints (so their unique neighbors must fit the path order).
• More generally: in a Hamiltonian path, any vertex with degree $1$ must appear as an endpoint (because it can have only one incident edge in the path).

Use this to quickly narrow which vertices can be endpoints and how forced edges are.

[CalloutBlock] tip If your graph has vertices of very low degree (0 or 1), you often get forced choices early. Those forced choices can then lead either to a full Hamiltonian path (by construction) or a contradiction.

2.2 “Trapping” heuristic (very useful in practice)

When building a Hamiltonian path, avoid creating a vertex $u$ (already adjacent to used vertices) such that the only remaining way to include some unvisited vertex $x$ would require revisiting a used vertex.

A practical method:

• Maintain the set of unvisited vertices.
• After each extension, look for forced degree-1 situations in the remaining induced subgraph on unvisited vertices plus the current endpoint.
• If an unvisited vertex becomes isolated with respect to “future connectivity,” you can backtrack immediately.

[CalloutBlock] warning Don’t rely on heuristics alone for proofs. If you claim “no Hamiltonian path exists,” you must provide a contradiction argument (e.g., forced endpoints leading to impossibility, or partition/connectivity obstruction).

Step 2: Construct a Hamiltonian path (if it exists)

If the graph is small (common in coursework), a successful construction is usually the intended answer for “If so describe it.”

What “describe it” typically means

Provide the vertex order: $v_1 - v_2 - v_3 - \cdots - v_{|V|}$ where each consecutive pair is an edge of $G$.

How to present it cleanly

• Start with a plausible endpoint.
• Show the sequence.
• (Optionally) verify adjacency: list the edges used.

Step 3: Prove nonexistence (if no Hamiltonian path exists)

To say “no Hamiltonian path exists,” you need an argument. Common proof structures include:

3.1 Forced endpoints lead to contradictions

If the graph has degree-1 vertices, they force endpoints. If there are multiple forced endpoints (or forced adjacency choices) that make completing the path impossible, you can conclude nonexistence.

3.2 Cut / separation argument (component barrier)

If you can show that removing a small set of vertices/edges disconnects the remaining graph so that a single simple path cannot traverse all vertices, then no Hamiltonian path exists.

3.3 Parity / bipartite constraints (in some graphs)

For bipartite graphs, any Hamiltonian path alternates between the two parts. Therefore:

• If $|A|$ and $|B|$ sizes differ by more than $1$, then a Hamiltonian path is impossible (but a Hamiltonian cycle has stricter constraints). This is often used when the graph is bipartite and highly unbalanced.

Step 4: If the graph is small, use backtracking (guaranteed method)

Backtracking systematically tries all permutations consistent with edges, but prunes early.

Minimal backtracking outline

• Choose a start vertex candidate.
• Recursively extend the path by trying unused neighbors.
• Prune if:
  • You disconnect the remaining unvisited vertices from the current endpoint structure.
  • Some unvisited vertex cannot have both “required connections” consistent with being internal vs endpoint.

This is essentially an exact algorithm for the decision question.