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

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Two triangles sharing one vertex form an Eulerian graph with no Hamilton cycle

Statement refuted

Every finite simple graph with an Euler circuit has a Hamilton cycle.

Counterexample

Let $G$ have vertices $x, a, b, c, d$ and edges

$x a, a b, b x, x c, c d, d x .$

Thus $G$ consists of two triangles sharing only $x$ . It is Eulerian but not Hamiltonian.

Facts & Assumptions

Given: The explicitly displayed finite simple graph $G$ .

[L1]

A connected finite multigraph has an Euler circuit exactly when every degree is even (Euler's theorem and Hierholzer's construction: a connected finite undirected multigraph has an Euler circuit if and only if every degree is even).

[L2]

If a graph has a Hamilton cycle, deleting a nonempty proper set $S$ leaves at most $∣ S ∣$ components (If $G$ has a Hamilton cycle, then deleting any nonempty proper vertex set $S$ leaves at most $∣ S ∣$ connected components).

[F1]

A triangle is the cycle graph $C_{3}$ (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[F2]

A finite simple graph can be regarded as the multigraph with the same vertex and edge sets and with each edge assigned its two endpoints (Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph).

Verification

technique · direct

1.1

Regard $G$ as the corresponding multigraph from [F2]. It is connected, $\deg (x) = 4$ , and each of $a, b, c, d$ has degree two. Hence every degree is even, and [L1] gives an Euler circuit; explicitly, $x, a, b, x, c, d, x$ uses every edge once.

givenF1F2L1algebra

1.2

Deleting $S = {x}$ leaves the two disjoint edges $a b$ and $c d$ , hence two connected components. Since $2 > ∣ S ∣ = 1$ , [L2] shows that $G$ cannot have a Hamilton cycle.

givenL2algebra

2.1

Thus $G$ is Eulerian and non-Hamiltonian, refuting the proposed implication.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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)