Computing the Union Join and Subset Graph of Acyclic Hypergraphs in Subquadratic Time

First, assume that $\mathcal{F}$ contains two distinct sets $S_{i}$ and $S_{j}$ with $S_{i}\subseteq S_{j}$. In that case, we can create a new join tree $T^{\prime}$ as follows. Remove the edge $E_{i}\mathcal{S}$ from $T$ and make $E_{i}$ adjacent to $E_{j}$ instead. Since $S_{i}\subseteq S_{j}$, each element $x\in E_{i}\cap\mathcal{S}$ is also contained in $E_{j}$. Thus, $T^{\prime}$ is a join tree for $H$ and contains the edge $E_{i}E_{j}$.

Next, assume that there is a join tree $T^{\prime}$ for $H$ with the edge $E_{i}E_{j}$. Without loss of generality, let $E_{j}$ be closer to $\mathcal{S}$ in $T^{\prime}$ than $E_{i}$. Recall that $E_{i}\subseteq\mathcal{S}$. Therefore, by properties of join trees, each vertex in $E_{i}$ is also in $E_{j}$. It then directly follows from the construction of $H$ that $S_{i}\subseteq S_{j}$.  $\square$

Recall that we can create a join tree $T$ for $H$ by making each hyperedge $E_{i}$ adjacent to the hyperedge $\mathcal{S}$. To prove Lemma 2, we show that $\mathcal{F}$ contains two distinct sets $S_{i}$ and $S_{j}$ with $S_{i}\subseteq S_{j}$ if and only if $T$ is not a unique join tree for $H$.

First, assume that $\mathcal{F}$ contains two such sets $S_{i}$ and $S_{j}$. In that case, Lemma 1 implies that there is a join tree $T^{\prime}$ for $H$ with the edge $E_{i}E_{j}$. Since $E_{i}E_{j}$ is not an edge in $T$, $T$ is not unique. Next, assume that $T$ is not unique. Then, there is a join tree $T^{\prime}$ and a hyperedge $E_{i}$ such that $E_{i}$ is not adjacent to $\mathcal{S}$ in $T^{\prime}$. Hence, $E_{i}$ is adjacent to some hyperedge $E_{j}$ that is closer to $\mathcal{S}$ in $T^{\prime}$ than $E_{i}$. Since $E_{i}\subseteq\mathcal{S}$, properties of join trees imply that $E_{i}\subseteq E_{j}$. Subsequently, due to Lemma 1, $S_{i}\subseteq S_{j}$.

It follows that a truly subquadratic-time algorithm which determines if an acyclic hypergraph has a unique join tree would imply an equally fast algorithm to solve the Sperner Family problem for any family of sets.  $\square$

Let $G$ be the subset graph for $H$ and $G_{\mathcal{F}}$ be the subset graph for $\mathcal{F}$. Since, by construction of $H$, $E_{i}\subseteq E_{j}$ if and only if $S_{i}\subseteq S_{j}$, $G$ contains the edge $(E_{i},E_{j})$ if and only if $G_{\mathcal{F}}$ contains the edge $(S_{i},S_{j})$. We can therefore construct $G_{\mathcal{F}}$ from $G$ by simply removing the vertex representing $\mathcal{S}$ from $G$ (and its incident edges).

Recall that we can construct $H$ from $\mathcal{F}$ in linear time. Therefore, a truly subquadratic-time algorithm to construct the subset graph of a given acyclic hypergraph would imply an equally fast algorithm to construct a subset graph of a given family of sets.  $\square$

By definition of $G$, properties i and ii are equivalent. It follows from properties of join trees that ii implies v.

We next show that v implies iii. Assume that $E_{i}$ and $E_{j}$ are not adjacent in a join tree $T$. Then there is a path $\langle E_{i}=X_{1},X_{2},\ldots,X_{k}=E_{j}\rangle$ of hyperedges from $E_{i}$ to $E_{j}$ in $T$. For each $p$ with $1\leq p<k$, let $S_{p}=X_{p}\cap X_{p+1}$ be the separator corresponding to the edge $X_{p}X_{p+1}$ of $T$. By properties of join trees, $E_{i}\cap E_{j}\subseteq S_{p}$ for each $S_{p}$. Now assume that each $S_{p}$ contains a vertex $v_{p}\notin E_{i}\cap E_{j}$. Then, $\langle v_{1},v_{2},\ldots,v_{k-1}\rangle$ would form a path in $H$ from $v_{1}\in E_{i}\setminus E_{j}$ to $v_{k-1}\in E_{j}\setminus E_{i}$. That contradicts with property v. Therefore, there is at least one separator $S_{p}$ with $S_{p}\subseteq E_{i}\cap E_{j}$, i. e., there is an edge $X_{p}X_{p+1}$ in $T$ with $E_{i}\cap E_{j}=X_{p}\cap X_{p+1}$.

To show that iii implies ii, consider a join tree $T$ where $E_{i}$ and $E_{j}$ are not adjacent. We can create a join tree $T^{\prime}$ by removing the edge $E_{i}^{\prime}E_{j}^{\prime}$ and adding the edge $E_{i}E_{j}$ instead. Since $E_{i}$ and $E_{j}$ are on different sides of $E_{i}^{\prime}E_{j}^{\prime}$ in $T$, $T^{\prime}$ is also a tree. Additionally, because $E_{i}\cap E_{j}=E_{i}^{\prime}\cap E_{j}^{\prime}$, $T^{\prime}$ is a valid join tree for $H$.

It remains to show that iv is equivalent to iii. We first assume property iii. Let $S=E_{i}\cap E_{j}$ be a separator on the path from $E_{i}$ to $E_{j}$ in some join tree $T$. Since, by properties of join trees, each vertex in $S=E_{i}\cap E_{j}$ is also in $S_{i}$ and $S_{j}$, it follows that $S\subseteq S_{i}$ and $S\subseteq S_{j}$. Now assume property iv. Because $S\subseteq S_{i}\subseteq E_{i}$ and $S\subseteq S_{j}\subseteq E_{j}$, it is also the case that $S\subseteq E_{i}\cap E_{j}$. Since $S$ is on the path from $E_{i}$ to $E_{j}$ in $T$, each vertex that is in both $E_{i}$ and $E_{j}$ also has to be in $S$, i. e., $S\supseteq E_{i}\cap E_{j}$. Therefore, $S=E_{i}\cap E_{j}$.  $\square$

Line 1 runs in $\mathcal{O}(m)$ time, since the nodes of $T$ are the hyperedges of $H$. Recall that $H$ is given as an incidence graph $\mathcal{I}(H)$. Hence, the following are equivalent (with respect to runtime): (i) for each vertex, iterating over all hyperedges containing it; (ii) for each hyperedge, iterating over all vertices it contains; and (iii) iterating over all edges of $\mathcal{I}(H)$. Therefore, line 1, line 1, and line 1 (and subsequently Algorithm 1) run in $\mathcal{O}(N)$ total time.  $\square$

Let $T_{i}$ be the tree after processing $E_{i}$, i. e., $T=T_{1}$ and $T_{m}=T^{\prime}$. Thus, $T_{1}$ is a valid join tree for $H$. Assume, by induction, that $T_{i-1}$ (with $i\geq 2$) is a valid join tree for $H$ too. Recall that, by definition of join trees, the set of hyperedges containing a vertex $v$ form a subtree $T_{v}$ of $T$. The roots of all such $T_{v}$ where $v\in S^{\raisebox{0.3014pt}{$\scriptscriptstyle\uparrow$}}(E_{i})$ are ancestors of $E_{i}$ in $T$ and, thus, form a path. By definition of $j$ (line 1), $E_{j}$ is the lowest of such roots in $T$. It therefore follows that $S^{\raisebox{0.3014pt}{$\scriptscriptstyle\uparrow$}}(E_{i})\subseteq E_{j}$. Subsequently, for each $v\in S^{\raisebox{0.3014pt}{$\scriptscriptstyle\uparrow$}}(E_{i})$, the hyperedges containing $v$ still form a subtree of $T_{i}$ after changing the parent of $E_{i}$ if they did so in $T_{i-1}$. Note that each subtree $T_{u}$ of a vertex $u\notin S^{\raisebox{0.3014pt}{$\scriptscriptstyle\uparrow$}}(E_{i})$ remains unchanged, since it does not contain the edge $E_{i}E_{k}$. Therefore, for each vertex, the hyperedges containing it form a subtree of $T_{i}$ and, thus, $T_{i}$ is a join tree for $H$.  $\square$

Let $E_{k}$ be the lowest common ancestor of $E_{i}$ and $E_{j}$ in $T^{\prime}$. Although $T^{\prime}$ has a potentially different structure than $T$, it is still the case that the parent of a hyperedge in $T^{\prime}$ was an ancestor of it in $T$. Thus, $k\leq i,j$. Note that $P_{ij}$ goes through $E_{k}$ and let $P_{ik}$ and $P_{kj}$ be the respective subpaths of $P_{ij}$. If $P_{ij}$ contains more than two separators $S$ as defined in Lemma 6, at least two of them are either part of $P_{ik}$ or $P_{kj}$. Without loss of generality, let them be on $P_{kj}$ and let $S$ be the lowest such separator. Additionally, let $X$ be the hyperedge directly below $S$, i. e., $S^{\raisebox{0.3014pt}{$\scriptscriptstyle\uparrow$}}(X)=S$. It follows that $X$ is not adjacent to $E_{k}$ in $T^{\prime}$.

Since $S\subseteq S_{i}$, each vertex in $S$ is in all hyperedges on the path from $X$ to $E_{i}$ in $T^{\prime}$, including $E_{k}$. Therefore, $S\subseteq E_{k}$ and $\max\bigl{\{}\,\lambda(v)\bigm{|}v\in S\,\bigr{\}}\leq k$. That is a contradiction, since Algorithm 1 would have made $X$ adjacent to $E_{k}$ or one of its ancestors.  $\square$

Let $E_{i}$ and $E_{j}$ be two hyperedges of $H$. Additionally, let $S_{i}$ and $S_{j}$ be the separators on the path from $E_{i}$ to $E_{j}$ in $T$ (computed in line 2) which are closest to $E_{i}$ and $E_{j}$, respectively. We show the correctness of Algorithm 2 by showing that $E_{i}E_{j}$ is an edge of $G$ if and only if there is a join tree for $H$ with the edge $E_{i}E_{j}$.

First, assume that there is a join tree for $H$ with the edge $E_{i}E_{j}$. Lemma 3 then implies that there is a separator $S\in\mathcal{S}$ such that $S\subseteq S_{i}$, $S\subseteq S_{j}$, and $E_{i}$ and $E_{j}$ are on different sides of $S$ in $T$. Therefore, when processing $S$, the algorithm finds $S_{i}$ and $S_{j}$ (line 2) and consequently adds $E_{i}$ and $E_{j}$ into $\mathbb{E}$ (line 2). Since both hyperedges are on different sides of $S$, Algorithm 2 then also adds the edge $E_{i}E_{j}$ to $G$ (line 2).

We now assume that $E_{i}E_{j}$ is an edge of $G$. Note that Algorithm 2 only adds edges to $G$ in line 2. Thus, there is a separator $S\in\mathcal{S}$ for which the algorithm adds $E_{i}E_{j}$ to $G$. For that $S$, one of $E_{i}$ and $E_{j}$ is in $\mathbb{E}_{1}$ and the other is in $\mathbb{E}_{2}$ (line 2) and, hence, $E_{i}$ and $E_{j}$ are on different sides of $S$ in $T$ (line 2). This implies that $S\subseteq S_{i}$ and $S\subseteq S_{j}$ (line 2 and line 2). Therefore, by Lemma 3, there is a join tree for $H$ with the edge $E_{i}E_{j}$.  $\square$

Creating a join tree for a given acyclic hypergraph $H$ can be implemented in $\mathcal{O}(N)$ time [27]. Modifying that join tree (thereby computing $T$) and computing $\mathcal{S}(H)$ using Algorithm 1 can also be done in $\mathcal{O}(N)$ time (Lemma 4). Thus, line 2 runs in total $\mathcal{O}(N)$ time. Computing the subset graph $G_{\mathcal{S}}$ in line 2 requires $\mathcal{O}\bigl{(}T_{\mathcal{A}}(H)\bigr{)}$ time. Since the hyperedges of $H$ form the vertices of $G$ and since $G$ is created without edges, line 2 runs in $\mathcal{O}(m)$ time.

We show next that a single iteration of the loop starting in line 2 runs in $\mathcal{O}\bigl{(}|\mathbb{E}_{1}|\cdot|\mathbb{E}_{2}|\bigr{)}$ time. That is, the runtime for a single iteration is (asymptotically) equivalent to the number of edges of $G$ created. Note that each iteration creates at least one such edge, namely the edge in $T$ that $S$ represents. Additionally, Lemma 3 and Lemma 6 imply that each edge $E_{i}E_{j}$ is added at most twice to $G$. Therefore, line 2 to line 2 run in $\mathcal{O}\bigl{(}|G|\bigr{)}$ total time.

For a separator $S\in\mathcal{S}$, let $\mathbb{S}$ denote the set of separators $S^{\prime}$ with $S\subseteq S^{\prime}$. Since the subset graph $G_{\mathcal{S}}$ is given, one can compute $\mathbb{S}$ (line 2) in $\mathcal{O}\bigl{(}|\mathbb{S}|\bigr{)}$ time by determining all incoming edges of $S$ in $G_{\mathcal{S}}$. For each $S^{\prime}\in\mathbb{S}$, the algorithm adds, in line 2, exactly one hyperedge into $\mathbb{E}$ plus one additional hyperedge for $S$. Thus, $|\mathbb{E}|=|\mathbb{S}|+1$.

One can determine the hyperedges $E$ and $E^{\prime}$ that form a separator $S^{\prime}$, which one is farther from $S$, and on which side of $S$ they are in $T$ as follows. When creating $S^{\prime}$, add a reference to both hyperedges and include which is the parent and which is the child in $T$. Now assume that each $S^{\prime}$ is also a node of $T$ adjacent to $E$ and $E^{\prime}$. Root $T$ in an arbitrary hyperedge, run a pre-order and post-order on $T$, and let $\mathrm{pre}(x)$ and $\mathrm{post}(x)$ be the indices of a node $x$ in that respective order. For two distinct nodes $x$ and $y$ of $T$ (representing either separators or hyperedges), $x$ is then a descendant of $y$ if and only if $\mathrm{pre}(x)>\mathrm{pre}(y)$ and $\mathrm{post}(x)<\mathrm{post}(y)$. There are four cases when determining which of $E$ and $E^{\prime}$ to add into $\mathbb{E}$: if $S$ and $S^{\prime}$ represent the same edge of $T$, add both hyperedges; if $S^{\prime}$ is a descendant of $S$, add the child-hyperedge; if $S^{\prime}$ is an ancestor of $S$, add the parent-hyperedge; and if $S^{\prime}$ is neither an ancestor nor a descendant of $S$, add the child-hyperedge. Clearly, one side of $S$ contains all its descendants and the other side all remaining hyperedges and separators. That allows us, after a $\mathcal{O}(m)$-time preprocessing, to determine in constant time on which side of $S$ a give a hyperedge is. Therefore, line 2 and line 2 run in $\mathcal{O}\bigl{(}|\mathbb{E}|\bigr{)}$ time.

Line 2 clearly runs in $\mathcal{O}\bigl{(}|\mathbb{E}_{1}|\cdot|\mathbb{E}_{2}|\bigr{)}$ time. Recall that $|\mathbb{S}|+1=|\mathbb{E}|=|\mathbb{E}_{1}|+|\mathbb{E}_{2}|$. Therefore, a single iteration of the loop starting in line 2 also runs in $\mathcal{O}\bigl{(}|\mathbb{E}_{1}|\cdot|\mathbb{E}_{2}|\bigr{)}$ time.  $\square$

We prove the theorem by making a simple linear-time reduction. Consider a family $\mathcal{F}=\{S_{1},S_{2},\ldots,S_{m}\}$ of sets. Create a new vertex $u$ and add it to each set $S_{i}\in\mathcal{F}$. Let $S^{\prime}_{i}$ be the resulting set. The family $\mathcal{F}^{\prime}=\{S^{\prime}_{1},S^{\prime}_{2},\ldots,S^{\prime}_{m}\}$ then forms a hypertree. Clearly, adding $u$ to each set does not change any subset relations. Therefore, $\mathcal{F}$ contains two distinct sets $S_{i}$ and $S_{j}$ with $S_{i}\subseteq S_{j}$ if and only if $\mathcal{F}^{\prime}$ contains two distingt sets $S^{\prime}_{i}$ and $S^{\prime}_{j}$ with $S^{\prime}_{i}\subseteq S^{\prime}_{j}$.  $\square$

Let $H$ be an acyclic hypergraph with a join tree $T$. We first show that the following are equivalent: (i) $H$ has two distinct hyperedges $E_{i}$ and $E_{j}$ with $E_{i}\subseteq E_{j}$, and (ii) $T$ has an edge $EE^{\prime}$ with $E\subseteq E^{\prime}$. Clearly, (ii) implies (i). To show that (i) implies (ii), let $E_{i}$ and $E_{j}$ be two distinct hyperedges of $H$ with $E_{i}\subseteq E_{j}$. It follows from the definition of join trees that $E_{i}\subseteq E_{k}$ for each hyperedge $E_{k}$ on the path from $E_{i}$ to $E_{j}$ in $T$. Therefore, $T$ contains an edge $E_{i}E_{k}$ with $E_{i}\subseteq E_{k}$ and, thus, (i) is equivalent to (ii).

Let $EE^{\prime}$ be an edge of $T$ with $E\subseteq E^{\prime}$, and let $S=E\cap E^{\prime}$ be the separator corresponding to that edge. Note that $|S|=|E|$ in such a case. Hence, it follows that (i) is true if and only if $H$ admits a separator $S=E\cap E^{\prime}$ such that $|S|=|E|$ or $|S|=|E^{\prime}|$.

We can now solve the Sperner Family problem for a given acyclic hypergraph $H$ in linear time as follows. Construct the separator hypergraph for $H$ (see Algorithm 1). For each separator $S=E\cap E^{\prime}$, determine if $|S|=|E|$ or $|S|=|E^{\prime}|$. In that case, return True. Otherwise, if no such separator is found, return False.  $\square$

We first show that $E_{i}\preceq E_{j}$ and $v\in E_{j}$ implies $E_{i}\subseteq E_{j}$. Clearly, $E_{i}\subseteq E_{j}$ if $E_{i}=E_{j}$. Assume now that $E_{i}\nsubseteq E_{j}$, i. e., $E_{i}$ contains a vertex $u\notin E_{j}$. By definition of $v$, $u$ is lower in $M$ than $v$. $E_{i}\preceq E_{j}$ and $v\in E_{j}$ then imply that $E_{i}$, $E_{j}$, $u$, and $v$ form a $\mathrm{\Gamma}$ in $M$. That contradicts with $M$ being $\mathrm{\Gamma}$-free (see Lemma 7). Therefore, $E_{i}\succ E_{j}$ or $v\notin E_{j}$.

Clearly, $v\notin E_{j}$ implies $E_{i}\nsubseteq E_{j}$. Now assume that $E_{i}\succ E_{j}$. Since $E_{i}\neq E_{j}$, there is a lowest vertex $u$ in $M$ which is in one of these hyperedges but not in both. Recall that $M$ is ordered lexicographically. Therefore, $E_{i}\succ E_{j}$ implies that $u\in E_{i}$ ($1$ in $M$) and $u\notin E_{j}$ ($0$ in $M$), i. e., $E_{i}\nsubseteq E_{j}$.  $\square$

We show the correctness of Algorithm 3 by showing that $G$ contains an edge $(E_{i},E_{j})$ if and only if $E_{i}\subseteq E_{j}$. First assume that $G$ contains an edge $(E_{i},E_{j})$. Note that Algorithm 3 only adds $(E_{i},E_{j})$ to $G$ (line 3) if $E_{i}\preceq E_{j}$ and $v\in E_{j}$ (line 3). Therefore, due to Lemma 8, $G$ containing an edge $(E_{i},E_{j})$ implies $E_{i}\subseteq E_{j}$.