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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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K4K_4 is Hamiltonian but has no Euler circuit

Statement refuted

Every Hamiltonian finite simple graph has an Euler circuit.

1234

Counterexample

The complete graph K4K_4 has a Hamilton cycle but no Euler circuit.

Facts & Assumptions

Given: The complete graph K4K_4 on vertices 1,2,3,41,2,3,4.

[F2]

A Hamilton cycle visits every vertex exactly once before returning to its start (Hamilton paths, Hamilton cycles, Hamiltonian graphs and Hamilton-connected graphs).

[F4]

A finite simple graph can be regarded as the multigraph with the same vertex and edge sets and the evident endpoint map (Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph).

Verification

technique · direct
1.1

By completeness, 1,2,3,4,11,2,3,4,1 is a cycle through all four vertices, so K4K_4 is Hamiltonian.

F1F2
1.2

Every vertex is adjacent to the other three, so every degree is three. Regard K4K_4 as the corresponding multigraph from [F4]. It is connected, but its degrees are odd; [L1] therefore rules out an Euler circuit.

F1F3F4L1algebra
2.1

Hence Hamiltonicity does not imply existence of an Euler circuit.

step 1.1step 1.2

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: 25 results over 16 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