CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02

How statement and proof provenance work

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 $n = 2 m \geq 4$

Statement refuted

For even $n \geq 4$ , every $n$ -vertex graph with $δ (G) \geq n / 2 - 1$ is Hamiltonian.

Counterexample

For $m \geq 2$ , let $G$ be the disjoint union of two copies of $K_{m}$ . Then $n = 2 m$ and $δ (G) = m - 1 = n / 2 - 1$ , but $G$ is not Hamiltonian.

Facts & Assumptions

Given: An integer $m \geq 2$ 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 $δ (G) \geq n / 2$ (Dirac's theorem: every $n$ -vertex graph with $n \geq 3$ and $δ (G) \geq n / 2$ is Hamiltonian).

Verification

technique · direct

1.1

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

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 · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

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