Alphabeta MathHow 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.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
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 have vertices and edges
Thus consists of two triangles sharing only . It is Eulerian but not Hamiltonian.
Facts & Assumptions
Given: The explicitly displayed finite simple graph .
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).
If a graph has a Hamilton cycle, deleting a nonempty proper set leaves at most components (If has a Hamilton cycle, then deleting any nonempty proper vertex set leaves at most connected components).
A triangle is the cycle graph (Empty and complete graphs, complete bipartite graphs, and the convention that and have vertices).
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
Regard as the corresponding multigraph from [F2]. It is connected, , and each of has degree two. Hence every degree is even, and [L1] gives an Euler circuit; explicitly, uses every edge once.
Deleting leaves the two disjoint edges and , hence two connected components. Since , [L2] shows that cannot have a Hamilton cycle.
Thus is Eulerian and non-Hamiltonian, refuting the proposed implication.
Depends on
- Euler's theorem and Hierholzer's construction: a connected finite undirected multigraph has an Euler circuit if and only if every degree is even
- If $G$ has a Hamilton cycle, then deleting any nonempty proper vertex set $S$ leaves at most $|S|$ connected components
- Empty and complete graphs, complete bipartite graphs, and the convention that $P_n$ and $C_n$ have $n$ vertices
- Multigraphs, loops and directed graphs as variants distinct from the default finite simple graph
Used by
Nothing in the library uses this result yet.
Dependency tree · next 3 levels
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Sources
- Applied Combinatorics, Eulerian and Hamiltonian Graphs (standard reference, not scraped)