2 votes

Hamiltonian path in bike-lock graph with $1$ known digit

Partial answer to the cycle version. Case $k=2$, any $n$: The graph has no Hamiltonian cycle. If $n \ge 3$, the graph consists of two $n$-cycles that intersect at a single point $(0,0)$. If $n=2$, the ...
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1 vote
Accepted

A property of directed acyclic graph

Preliminary definition. Let $\mathcal{S}$, $\mathcal{S}'$ be two complementary nonempty sets of indices, i.e., $\mathcal{S}\cup \mathcal{S}'=\left\{1,2,\ldots,p\right\}$ and $\mathcal{S}\cap \mathcal{...
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