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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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P3P_3 has a Hamilton path but no Hamilton cycle

Statement refuted

Every finite simple graph with a Hamilton path has a Hamilton cycle.

Counterexample

The path graph P3P_3 has vertices 0,1,20,1,2 and edges 01,1201,12.

Facts & Assumptions

Given: The path graph P3P_3.

[F2]

A Hamilton path contains every vertex exactly once (Hamilton paths, Hamilton cycles, Hamiltonian graphs and Hamilton-connected graphs).

Verification

technique · direct
1.1

The path 0,1,20,1,2 uses both edges and contains every vertex exactly once, so it is a Hamilton path.

F1F2
1.2

Deleting the middle vertex 11 leaves two isolated vertices, hence two components. This violates [L1] for the singleton set S={1}S=\{1\}, so P3P_3 has no Hamilton cycle.

F1L1algebra
2.1

Therefore a Hamilton path need not extend to a Hamilton cycle.

step 1.1step 1.2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources