CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Two disjoint copies of $K_m$ show that Dirac's bound cannot be lowered to $n/2-1$ for even $n=2m\ge4$

Statement refuted

For even $n\ge4$ , every $n$ -vertex graph with $\delta(G)\ge n/2-1$ is Hamiltonian.

Counterexample

For $m\ge2$ , let $G$ be the disjoint union of two copies of $K_m$ . Then $n=2m$ and $\delta(G)=m-1=n/2-1$ , but $G$ is not Hamiltonian.

Facts & Assumptions

Given: An integer $m\ge2$ and two vertex-disjoint copies of $K_m$ with no edge between them.

[F1]

Every vertex of $K_m$ is adjacent to the other $m-1$ vertices (Empty and complete graphs, complete bipartite graphs, and the convention that $P_n$ and $C_n$ have $n$ vertices, Adjacency, incidence, open and closed neighbourhoods, vertex degree, minimum degree and maximum degree).

[F2]

A Hamiltonian graph has a spanning cycle and is connected (Hamilton paths, Hamilton cycles, Hamiltonian graphs and Hamilton-connected graphs, Connected graphs and connected components defined by the existence of vertex paths).

[L1]

Dirac's theorem uses the stronger threshold $\delta(G)\ge n/2$ (Dirac's theorem: every $n$ -vertex graph with $n\ge3$ and $\delta(G)\ge n/2$ is Hamiltonian).

Verification

technique · direct

1.1

The graph has $n=2m$ vertices. Every vertex has precisely the other $m-1$ vertices in its own copy as neighbours, so $\delta(G)=m-1=n/2-1$ .

givenF1algebra

1.2

The two copies are distinct connected components because no edge joins them. Hence $G$ is disconnected and cannot be Hamiltonian by [F2].

givenF2

1.3

At the endpoint $m=2$ , the construction is two disjoint edges on four vertices, with minimum degree $1=4/2-1$ , so the same failure occurs.

givenalgebra

2.1

Thus lowering the threshold in [L1] by one for even order would make the theorem false.

step 1.1step 1.2step 1.3L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 23 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Applied Combinatorics, Eulerian and Hamiltonian Graphs (standard reference, not scraped)

Two disjoint copies of KmK_mKm​ show that Dirac's bound cannot be lowered to n/2−1n/2-1n/2−1 for even n=2m≥4n=2m\ge4n=2m≥4