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arxiv: 1812.08809 · v1 · pith:2VE5SNV2new · submitted 2018-12-20 · 💻 cs.DM · math.CO

Arc-disjoint strong spanning subdigraphs in compositions and products of digraphs

classification 💻 cs.DM math.CO
keywords digraphstrongdecompositiongooddotsverticesdigraphsproduct
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A digraph $D=(V,A)$ has a good decomposition if $A$ has two disjoint sets $A_1$ and $A_2$ such that both $(V,A_1)$ and $(V,A_2)$ are strong. Let $T$ be a digraph with $t$ vertices $u_1,\dots , u_t$ and let $H_1,\dots H_t$ be digraphs such that $H_i$ has vertices $u_{i,j_i},\ 1\le j_i\le n_i.$ Then the composition $Q=T[H_1,\dots , H_t]$ is a digraph with vertex set $\{u_{i,j_i}\mid 1\le i\le t, 1\le j_i\le n_i\}$ and arc set $$A(Q)=\cup^t_{i=1}A(H_i)\cup \{u_{ij_i}u_{pq_p}\mid u_iu_p\in A(T), 1\le j_i\le n_i, 1\le q_p\le n_p\}.$$ For digraph compositions $Q=T[H_1,\dots H_t]$, we obtain sufficient conditions for $Q$ to have a good decomposition and a characterization of $Q$ with a good decomposition when $T$ is a strong semicomplete digraph and each $H_i$ is an arbitrary digraph with at least two vertices. For digraph products, we prove the following: (a) if $k\geq 2$ is an integer and $G$ is a strong digraph which has a collection of arc-disjoint cycles covering all vertices, then the Cartesian product digraph $G^{\square k}$ (the $k$th powers with respect to Cartesian product) has a good decomposition; (b) for any strong digraphs $G, H$, the strong product $G\boxtimes H$ has a good decomposition.

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