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

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$K_{3,3}$ has a Hamilton cycle

Example

Let the two parts of $K_{3,3}$ be $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$ . Then

a_1,b_1,a_2,b_2,a_3,b_3,a_1

is a Hamilton cycle.

Facts & Assumptions

Given: The displayed bipartition of $K_{3,3}$ .

[F1]

Every vertex in $A$ is adjacent to every vertex in $B$ , and there are no edges within a part (Empty and complete graphs, complete bipartite graphs, and the convention that $P_n$ and $C_n$ have $n$ vertices).

[F2]

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

Verification

technique · direct

1.1

Each consecutive pair in the displayed cyclic list has one endpoint in $A$ and one in $B$ , so all six required edges occur by [F1].

2.1

Before returning to $a_1$ , the list contains each of the six vertices exactly once. It is therefore a Hamilton cycle by [F2].

step 1.1F2∎

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: 11 results over 7 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)

K3,3K_{3,3}K3,3​ has a Hamilton cycle