SSD Set System, Graph Decomposition and Hamiltonian Cycle

Abstract

In this paper, we first study what we call Superset-Subset-Disjoint (SSD) set system. Based on properties of SSD set system, we derive the following (I) to (IV): (I) For a nonnegative integer $k$ and a graph $G=(V,E)$ with $|V|\ge2$, let $X_1,X_2,\dots,X_q\subsetneq V$ denote all maximal proper subsets of $V$ that induce $k$-edge-connected subgraphs. Then at least one of (a) and (b) holds: (a) $\{X_1,X_2,\dots,X_q\}$ is a partition of $V$; and (b) $V\setminus X_1, V\setminus X_2,\dots,V\setminus X_q$ are pairwise disjoint. (II) For $k=1$ and a strongly-connected digraph $G$, whether $V$ is in (a) and/or (b) can be decided in $O(n+m)$ time and we can generate all such $X_1,X_2,\dots,X_q$ in $O(n+m+|X_1|+|X_2|+\dots+|X_q|)$ time, where $n=|V|$ and $m=|E|$. (III) For a digraph $G$, we can enumerate in linear delay all vertex subsets of $V$ that induce strongly-connected subgraphs. (IV) A digraph is Hamiltonian if there is a spanning subgraph that is strongly-connected and in the case (a).

Figures (4)

• Figure 1: Strongly-connected system is not SD; $Y$ is a MinRS of $S$ and $S'$ is a PSS of $S$, where we see that $Y\not\subseteq S'$ and $Y\cap S'\ne\emptyset$.
• Figure 2: System classes and examples
• Figure 3: A counterexample of the converse of Lemma \ref{['lem:str_RS_in_DT']}(iii). (a) A strongly-connected graph $G$. (b) The dominator tree ${\mathit{DT}}_s$ of $G_s$. Although every vertex of $V \setminus A = \{s, v\}$ is reachable from $s$ in (b), $G - A$ is not strongly-connected.
• Figure 4: An example where a leaf of the dominator tree does not belong to a MinRS. The graph $G$ is strongly-connected and $\textup{{{Min}\xspace{RS}\xspace}\xspace}_{{\mathcal{S}}_1}(V) = \{ \{s\}, \{v_2, v_3, v_4\}, \{v_5, v_6\} \}$. Although vertex $v_8$ is a leaf of ${\mathit{DT}}_s$, it does not belong to any MinRS of $V$.

Theorems & Definitions (32)

• Definition 1
• Theorem 1
• Theorem 2
• Theorem 3
• Theorem 4
• Theorem 5
• Lemma 1
• Lemma 2
• Lemma 3
• Lemma 4