Abstract
Given a connected hypergraph with vertex set V, a convexity space on is a subset of the powerset of V that contains ∅, V, and the singletons; furthermore, is closed under intersection and every set in is connected in . The members of are called convex sets. The convex hull of a subset X of V is the smallest convex set containing X. By a cluster of we mean any nonempty subset of V in which every two vertices are separated by no convex set. We say that a convexity space on is decomposable if it satisfies the following three axioms: (i) the maximal clusters of form an acyclic hypergraph, (ii) every maximal cluster of is a convex set, and (iii) for every nonempty vertex set X, a vertex does not belong to the convex hull of X if and only if it is separated from X by a convex cluster. We prove that a decomposable convexity space on is fully specified by the maximal clusters of in that (1) there is a closed formula which expresses the convex hull of a set in terms of certain convex clusters of and (2) is a convex geometry if and only if the subspaces of induced by maximal clusters of are all convex geometries. Finally, we prove the decomposability of some known convexities in graphs and hypergraphs taken from the literature (such as “monophonic” and “canonical” convexities in hypergraphs and “all-paths” convexity in graphs).
1. Introduction
A (finite) convexity space [1] over a finite nonempty set is a subset of the powerset of that contains and and is closed under intersection. The members of are called convex sets. The convex hull of a subset of , denoted by , is the smallest convex set containing . It is well known that (i) , (ii) if then , and (iii) . A convexity space over is a point-convexity space [2] if, for every element of , the singleton is a convex set. Abstract convexity theory has been applied to graphs [3–6] and hypergraphs [4, 7]. A convexity space on a connected hypergraph is a point-convexity space such that every convex set is connected. If is a convex set, the subspace of induced by is the convexity space on the (connected) subhypergraph induced by .
Most convexities in graphs and hypergraphs have been stated in terms of “feasible” paths [8]; accordingly, a vertex set is convex if contains all vertices on every feasible path joining two vertices in . For example, geodetic convexity, monophonic convexity, and all-paths convexity are obtained using shortest paths, chordless paths, or all paths, respectively.
We note that in graphs monophonic convexity (-convexity, for short) and all-paths convexity (ap-convexity, for short) enjoy similar properties. For example, Duchet [5] proved that “a vertex does not belong to the -convex hull of a set if and only if is separated from by a clique,” and known results on ap-convexity [3, 9] imply that “a vertex does not belong to the ap-convex hull of a set if and only if is separated from by a cut-vertex.” This similarity also applies to the algorithms to compute -convex hulls [7, 9–11] and ap-convex hulls [9].
In this paper, we show that -convexity and ap-convexity belong to a wide class of convexity spaces on hypergraphs, which we call “decomposable” and define by means of three axioms. To this end, we need to introduce the notion of a “cluster.”
Let be a convexity space on a connected hypergraph with vertex set . A cluster of is a nonempty subset of in which every two vertices are separated by no convex set. The cluster hypergraph of is the set of maximal clusters of . A convexity space on is decomposable if it satisfies the following three axioms.
Axiom 1.The cluster hypergraph of is acyclic.
Axiom 2. Every edge of the cluster hypergraph of is a convex set.
Axiom 3. For every nonempty vertex set , a vertex does not belong to the convex hull of if and only if it is separated from by a convex cluster.
Our main result is that if is a decomposable convexity space on , then enjoys the following two properties.(D1) there is a closed formula which expresses the convex hull of a set in terms of certain convex clusters of .(D2) is a convex geometry if and only if the subspaces of induced by maximal clusters of are all convex geometries.
The rest of the paper is organized as follows. In Section 2 we recall more-or-less standard definitions and some preliminary results on acyclic hypergraphs, which will be used in the sequel. In Section 3 we state general properties of the cluster hypergraph. In Section 4 we prove some properties of convexity spaces that satisfy Axioms 1 and 2. In Section 5 we prove that decomposable convexity spaces enjoy properties (D1) and (D2). In Section 6, we give three examples of decomposable convexity spaces taken from the literature. Finally, Section 7 contains a closing note.
2. Preliminaries
Let be a finite set. A hypergraph [12] on is a nonempty set of subsets of whose union recovers . The elements of are the vertices of and the sets in are the (hyper) edges of . If , then we call the empty hypergraph. A nonempty hypergraph is trivial if it has exactly one edge. A hypergraph is simple if no edge is a subset of another edge. A partial hypergraph of hypergraph is a nonempty subset of . A cover of is a simple hypergraph on such that each edge of is contained in some edge of .
A vertex is a leaf if it belongs to exactly one edge.
Two vertices are adjacent if they belong together to some edge. A vertex is adjacent to a vertex set if contains a vertex such that and are adjacent.
A clique is a nonempty vertex set in which every two vertices are adjacent. A partial edge is a nonempty vertex set included in some edge. A hypergraph is conformal if every clique is a partial edge. The clique hypergraph of a hypergraph is the hypergraph whose edges are precisely the maximal cliques of .
A (vertex) path is a sequence of distinct vertices , such that if then and are adjacent for ; the path is said to join and from to and to pass through any set such that . A hypergraph is connected if every two vertices are joined by a path. The connected components of a hypergraph are its maximal partial hypergraphs that are connected.
Let be a nonempty subset of ; the subhypergraph of induced by is the hypergraph on , denoted by , whose edges are the nonempty intersections of with edges of . A subset of is connected if is connected.
Let be a nonempty proper subset of ; by we denote the subhypergraph of induced by . If is a connected component of , the neighborhood of , denoted by , is the set of vertices in that are adjacent to the vertex set of .
A vertex is separated from a set by if and is disjoint from the vertex set of the connected component of containing .
In the next two subsections we deal with “separators” and “acyclic hypergraphs.”
2.1. Separators
Let be a connected hypergraph with vertex set . A subset of is a separator of if is not connected. If is a separator of , then is said to separate every two vertices that are in distinct connected components of . Note that separates two vertices and if and only if and every path joining and passes through . A separator separates a subset of if separates two distinct vertices in . A set is a minimal separator for a vertex pair if and are separated by and are separated by no proper subset of . A set is a minimal vertex separator (MVS) [13] of if there exists a vertex pair for which is a minimal separator.
2.2. Acyclic Hypergraphs
A hypergraph is acyclic if there exists a running-intersection ordering of its edges, that is, if then, for each there exists such that Several equivalent definitions of acyclicity exist [14]. A test for acyclicity is given by the following algorithm which reduces a hypergraph to the empty hypergraph if and only if is acyclic [14].
Graham Reduction. Repeatedly apply the following two operations until neither can be longer applied: (vertex deletion): remove a vertex if it is a leaf. (edge deletion): remove an edge if it is contained in another edge.
A more efficient algorithm to test acyclicity was given in [15]. The following two facts state well known properties of MVSs of an acyclic hypergraph [14, 16] and they will be used in the sequel.
Fact 1. Let be an acyclic, simple, and connected hypergraph. The MVSs of are exactly the partial edges of that are removed during the Graham reduction of .
Fact 2. Let be an acyclic, simple, and connected hypergraph. If is an MVS of , then there exist two edges and of such that and, for every and , is the only minimal separator for .
For our purposes, we need a modification of Graham reduction when there is a set of sacred vertices that may not be deleted by vertex deletion. Let denote the result of applying Graham reduction to a hypergraph with the vertex deletion rule modified to disallow the removal of any vertex in , the application of rules proceeding until such time as no more applications are possible [16]. We call the Graham reduction of with sacred set . It is well known [16] that is defined uniquely, that is, if a deletion rule is applicable to a vertex or edge, applying a rule elsewhere does not make the first inapplicable; so, the reduction procedure proceeds to do everything that is ever possible, independent of the order of reductions chosen. In other words, the Graham reduction process enjoys the so-called Church-Rosser property [16], which makes it a pruning process according to the terminology used in [17]. It is also well known [16] that if is an acyclic, simple and connected hypergraph, then(i) is an acyclic, simple, and connected hypergraph; (ii)the edges of are the maximal edges of an induced subhypergraph of whose vertex set contains ;(iii) is a subset of the vertex set of ;(iv)the MVSs of are the MVSs of plus the partial edges removed during the Graham reduction process.
The following lemma contains one more property of which will be used later on.
Lemma 1 (see [18]). Let be an acyclic hypergraph on , a nonempty proper subset of , and a vertex in . The following holds:(i)a vertex of belongs to the vertex set of if and only if is separated from by no partial edge of ; (ii)if is not a vertex of , then there exists an edge of that separates from in ; (iii)if is an edge of , then is the union of with the neighborhoods of the connected components of whose vertex sets are not disjoint from .
Figure 1 illustrates part (iii) of Lemma 1 for the two cases in which and .
Corollary 2. Let be an acyclic hypergraph on , a nonempty proper subset of , and a vertex in . If A is an edge of , is not a vertex of , and is a partial edge of , then A is contained in every partial edge of that separates from in .
Proof. First of all, observe that the connected component of that contains has an empty intersection with . Let be a partial edge of that separates from in , and let be the vertex set of a connected component of such that . Then must contain the neighborhood of for, otherwise, and each vertex in would be in the same connected component of . Moreover, for, otherwise, and some vertex in would be in the same connected component of . From part (iii) of Lemma 1, it follows that .
3. The Cluster Hypergraph
Let be a convexity space on a connected hypergraph . Recall from the Introduction that a cluster of is a nonempty vertex set that is separated by no convex set of , and the cluster hypergraph of is the hypergraph, henceforth denoted by , whose edges are precisely the maximal clusters of . In this section, we shall state some general properties of the cluster hypergraph.
Theorem 3. Let be a connected hypergraph and a convexity space on . The cluster hypergraph of is a cover of the clique hypergraph of and is a conformal hypergraph.
Proof. Since every clique of is a cluster, every maximal clique of is a partial edge of , that is, is a cover of the clique hypergraph of . Moreover, since two vertices and of are adjacent (in ) if and only if is a cluster of , one has that every clique of is a cluster of and, hence, a partial edge of .
By Theorem 3, is a cover of so that every separator of is also a separator of . We now prove that the converse also holds for convex separators of .
Lemma 4. Let be a connected hypergraph, a convexity space on , the cluster hypergraph of , and a convex set of . Two vertices that are separated by in are also separated by in .
Proof. Let and be two vertices that are separated by in . Since is a convex set, the vertex pair is not a cluster of and, hence, and are not adjacent in . Suppose, by contradiction, that and are not separated by in . Then, there exists in a path , , such that for all , . Since each vertex pair is a partial edge of , the set is a cluster of and, since is a convex set, and cannot be separated by in so that in there exists a path from to such that no vertex on belongs to . Combining the paths , we obtain a path in from to and no vertex on belongs to , which contradicts the hypothesis that and are separated by in .
The next lemma is useful for the sequel.
Lemma 5. Let be a connected hypergraph, a convexity space on , the cluster hypergraph of , and a convex set. If the vertex sets of a connected component of and of a connected component of are not disjoint, then they coincide.
Proof. Let be the vertex set of a connected component of and the vertex set of a connected component of such that . By Theorem 3, is a cover of and, since , one has . Suppose, by contradiction, that , and let and . Since neither nor are in , , and are separated by in . By Lemma 4, and are separated by in , which contradicts the fact that and are both in .
The next result relates the MVSs of the cluster hypergraph of with “minimal vertex convex separators” of , which we define as follows. A convex set of is a minimal convex separator of for a vertex pair if and are separated by and are separated by no convex proper subset of . A set is a minimal vertex convex separator (MVCS) of if there exists a vertex pair for which is a minimal convex separator.
Theorem 6. Let be a connected hypergraph, a convexity space on , and the cluster hypergraph of . Every MVCS of is the convex hull of an MVS of .
Proof. Let be an MVCS of , and let be a vertex pair for which is a minimal convex separator of . Since is a convex set, separates and in by Lemma 4; therefore, must contain a minimal separator of for . Let be such an MVS of . Since , one has and, hence, neither nor belongs to . Therefore, since separates and in and since , also the set separates and in and, since is a cover of , the set separates and in . Finally, since and is a minimal convex separator of for , one has .
Example 7. Consider the (hyper)graph with vertex set that is shown in Figure 2, and the convexity space that only contains , , the five singletons, and the set . The MVCSs of are the set (which is a minimal convex separator for the vertex pairs , , and ) and the set (which is a minimal convex separator for the vertex pair ). The cluster hypergraph of is . The MVSs of are the set (which is a minimal separator for the vertex pairs , , and ) and the set (which is a minimal separator for the vertex pair ). The MVCS of is the convex hull of the MVS of , and the MVCS of is the convex hull of the MVS of .
4. Axioms 1 and 2
In this section we state some properties of convexity spaces that satisfy Axioms 1 and 2. Recall that a convexity space on satisfies Axiom 1 if the cluster hypergraph of is acyclic, and satisfies Axiom 2 if every maximal cluster of is a convex set. First of all, observe that the following condition is equivalent to Axiom 2.
Fact 3. Let be a connected hypergraph. A convexity space on satisfies Axiom 2 if and only if the convex hull of every cluster is a cluster.
Proof of (“only if”). Let be any cluster of and a maximal cluster containing . By Axiom 2, is a convex set. Since , one has and, hence, is a cluster.
Proof of (“if”). Let be any maximal cluster of . We need to prove that . Of course, . By hypothesis, is a cluster. Let be a maximal cluster of that contains . Then, one has . Since is a maximal cluster, one has so that .
The following are two properties of convexity spaces that satisfy both Axioms 1 and 2.
Theorem 8. Let be a connected hypergraph, a convexity space on and the cluster hypergraph of . If satisfies Axioms 1 and 2, then every MVS of is a convex set.
Proof. Since satisfies Axiom 1, is an acyclic hypergraph. Let be any MVS of . By Fact 2, for some two edges and of . Since satisfies Axiom 2, and are convex sets of so that is a convex set of .
Corollary 9. Let be a connected hypergraph, a convexity space on , and the cluster hypergraph of . If satisfies Axioms 1 and 2, then the MVCSs of are exactly the MVSs of .
Proof. We first prove that (i) every MVCS of is an MVS of and, then, that (ii) every MVS of is an MVCS of .
Proof of (i). Let be an MVCS of . By Theorem 6, there exists an MVS of such that . By Theorem 8, so that .
Proof of (ii). Let be an MVS of . Since is acyclic, by Fact 2 there exists a vertex pair for which is the only minimal separator in . Therefore, and are not adjacent in so that is not a cluster of . Let be a minimal convex separator of for . By Lemma 4, separates and in . Since is the only minimal vertex separator for in , one has that . On the other hand, since is convex by Theorem 8 and since is a cover of , is also a convex separator of for . Finally, since is a minimal convex separator of for , one has .
Example 10. Consider again the (hyper)graph of Figure 2 and the convexity space on that only contains , , the singletons and the four sets , , and . The MVCSs of are and and the cluster hypergraph is . Since is an acyclic hypergraph and its edges are all convex sets, satisfies Axioms 1 and 2. Note that the MVCSs of are exactly the MVSs of .
5. Decomposable Convexity Spaces
Recall that a convexity space on is decomposable if satisfies Axioms 1 and 2 and, for every subset of , the convex hull of equals the set of vertices that are separated from by no convex cluster of .
Example 10 (continued). The convexity space is not decomposable since the convex hull of the set is the whole vertex set but the vertex is separated from by the convex cluster .
In the next two subsections, we shall prove the properties (D1 and D2) mentioned in the introduction.
5.1. Property (D1)
Given any convexity space on that satisfies Axioms 1 and 2, we first provide a closed formula for the set of vertices that are separated from a set by no convex cluster of . It will follow that if is decomposable, then the same formula is an expression of the convex hull of .
Lemma 11. Let be a hypergraph on , a convexity space on that satisfies Axioms 1 and 2, the cluster hypergraph of , and a nonempty subset of . The set of vertices that are separated from by no convex cluster of is given by the union of the convex hulls of edges of .
Proof. Let . We now prove that if and only if is separated from by some convex cluster of .
Proof of (“if”). Let be a vertex of that is separated from by a convex cluster of . Let such a convex cluster. Since is a convex set, separates from in by Lemma 5. Moreover, since is a cluster of , is a partial edge of . In order to prove that , we first show that is not a vertex of and, then, for every edge of .
Since in the vertex is separated from by the partial edge of , is not a vertex of by part (i) of Lemma 1. Consider now any edge of . If then, since is a subset of the vertex set of , one has that . Assume that and suppose, by contradiction, that . By Corollary 2, one has that and, hence, . To sum up, by assumption, and since separates from , which is in contradiction with the inclusion .
Proof of (“only if”). Assume that . We now prove that there exists a convex cluster that separates from in . Recall that the vertex set of is a subset of and is a subset of for every edge of . Therefore, since , is not a vertex of and for every edge of . Since is not a vertex of , by part (ii) of Lemma 1 there exists an edge of that separates from in and, since and , one has that and, hence, separates from in . Since is a cover of , is separated from by the convex cluster in .
At this point, we are in a position to prove property (D1) of decomposable convexity spaces.
Theorem 12. Let be a hypergraph on , a decomposable convexity space on , the cluster hypergraph of , and a subset of . The convex hull of is given by the union of the convex hulls of the edges of , that is,
Proof. By Lemma 11 and Axiom 3.
Theorem 12 also leads to the following characterization of convex sets.
Theorem 13. Let be a connected hypergraph on , a decomposable convexity space on , and the cluster hypergraph of . A nonempty subset of is convex if and only if (a) equals the vertex set of , and(b)every edge of is a convex set.
Proof of (“if”). By condition (a), equals the vertex set of and, by condition (b), for every edge of . By (2), one has and, hence, is a convex set.
Proof of (“only if”). First of all, recall that for any subset of , one has that is a subset of the vertex set of . Moreover, for every edge of GR(). Therefore, one has Assume that is a convex set. Then so that, by (2), one has that and for every edge of , which are exactly conditions (a) and (b).
5.2. Property (D2)
We begin with a few basic definitions and preliminary results. A vertex in a convex set is an extreme point of if is a convex set. A convexity space is a convex geometry if it enjoys the Minkowsky-Krein-Milman property: every convex set is the convex hull of the set of its extreme points.
In order to prove property (D2), we make use of the following two characterizations of convex geometries.
Lemma 14 (e.g., see [17]). A convexity space is a convex geometry if and only if, for every set in , , there exists such that belongs to .
The next characterization of convex geometries is based on the notion of a “descending path,” which is defined as follows. Let and be two convex sets with . A descending path [17] from to in the lattice of is an ordering of the elements of such that, for each , , the set is convex.
Lemma 15 (see [4]). A convexity space is a convex geometry if and only if, for every set in , , there exists a descending path from to in the lattice of .
Corollary 16. Let be a convex geometry and let and be two convex sets. If is a subset of , , then there exists a descending path from to in the lattice of the induced subspace .
Proof. Since is a convex set, by Lemma 15, there exists a descending path , . Thus, each set is convex, . Of course, each vertex in must appear in . Let , , be the positions in of the vertices in . Since both and are convex sets, the set is convex, ; moreover, , , so that . Therefore, is a descending path from to in the lattice of .
At this point, we are in a position to prove property (D2) of decomposable convexity spaces.
Theorem 17. Let be a connected hypergraph, a decomposable convexity space on , and the cluster hypergraph of . The convexity space is a convex geometry if and only if the subspaces of induced by edges of are all convex geometries.
Proof. Proof of (“only if”). Let be any edge of . Since is decomposable, is a convex set by Axiom 2. Let be the subspace of induced by . Consider any convex set in , . By Corollary 16, there exists a descending path from to in the lattice of , which by Lemma 15 proves that is a convex geometry.
Proof of (“if”). Let be any convex set with . By Lemma 14, we need to prove that there exists a vertex such that is convex. By Theorem 13, equals the vertex set of and every edge of is a convex set. Distinguish two cases depending on whether or not is a partial hypergraph of (i.e., every edge of is an edge of ).
Case 1. is a partial hypergraph of . Since , all the edges of that are not edges of have been deleted during the Graham reduction of with sacred set . Let be the edge of that has been deleted; then, its residual part, say , was found to be contained in an edge of . Then, is an MVS of (see the properties of the Graham reduction). Since is decomposable, is a convex set by Axiom 2, and is a convex set by Theorem 8.
Case 2. is not a partial hypergraph of . Then there exists an edge of that is strictly contained in an edge of . Let be an edge of that contains . Since is decomposable, is a convex set by Axiom 2, and is a convex set by condition (b) of Theorem 13.
In both cases, by hypothesis, is a convex geometry. Therefore, since is a proper subset of , by Lemma 14 there exists such that the set is convex. We will show that is a convex set, which by Lemma 14 proves that is a convex geometry. Since , by the Church-Rosser property of the Graham reduction procedure, we can assume that is the last vertex that is deleted during the Graham reduction of with sacred set . Therefore, in Case 1 the hypergraph can be viewed as being obtained from by adding the edge , and in Case 2 the hypergraph can be viewed as being obtained from by deleting the edge and adding the edge . Since is a convex set, by condition (a) of Theorem 13 the vertex set of equals , so that the vertex set of equals , which proves that satisfies condition (a) of Theorem 13. Morever, an edge of is either or an edge of and, since is a convex set and every edge of is a convex set by condition (b) of Theorem 13, one has that every edge of is a convex set, which proves that satisfies condition (b) of Theorem 13. So, by Theorem 13, is convex.
It is worth observing that, by Theorem 13, every descending path in a decomposable convex geometry on is a Graham elimination ordering of vertices of the cluster hypergraph (or, equivalently, is a perfect elimination ordering of the chordal graph on the vertex set of where two vertices are adjacent if and only if they form a cluster of ).
6. Examples
In this section we prove that -convexity spaces [4, 5, 7] and -convexity spaces [7] in hypergraphs are decomposable, and that ap-convexity spaces [3, 9] in graphs are decomposable.
6.1. -Convexity in Hypergraphs
Let be a connected hypergraph with vertex set . A path in is chordless if, for every two nonconsecutive vertices and on , and are not adjacent. A subset of is -convex if either or contains all vertices on every chordless path joining two vertices in [4, 5]. Equivalently, a subset of is -convex if either or or the neighborhood of every connected component of is a clique of [5, 7]. Note that every clique of is an -convex set. By we denote the -convex hull of . Efficient algorithms for computing -convex hulls in graphs were given in [9–11]. A similar algorithm for computing -convex hulls in hypergraphs was given in [7]; it makes use of the acyclic hypergraph whose edges are the maximal sets that are not separable by cliques of [19]. This hypergraph is called the prime hypergraph of [7]. Let denote the prime hypergraph of . We can summarize the algorithm in [7] into the following formula: where
Fact 4. If is an edge of , then and, hence, . It follows that every edge of is an -convex set.
Fact 5. If is cluster of (i.e., is a partial edge of ), then and, hence, .
In order to prove the decomposability of -convexity, we need the following lemma.
Lemma 18. Let be a connected hypergraph, and let be the m-convexity space on . The cluster hypergraph of equals the prime hypergraph of .
Proof. It is sufficient to prove that two vertices of are separated by an -convex set of if and only if they are separated by a clique of .
Proof of (if). The statement follows from the fact that every clique of is an -convex set of .
Proof of (only if). We need to prove that if two vertices and of are separated by an -convex set of , then they are separated by a clique of . Let be an -convex set separating and . Thus, and belong to two distinct connected components of . Let be the connected component of containing , and let be the neighborhood of . Then, is also a connected component of and the vertex belongs to another connected component of . Therefore, and are separated by and, since is an -convex set of and is the neighborhood of a connected component of , is a clique of , which proves the statement.
Theorem 19. Let be a connected hypergraph. The -convexity space on is decomposable.
Proof. Let be the -convexity space on . By Lemma 18, the cluster hypergraph of is acyclic so that, by Fact 4, every-edge of the cluster hypergraph of is an -convex set. Therefore, satisfies Axioms 1 and 2. Moreover, by Lemmas 11 and 18 and by Fact 5, the right-hand side of (5) equals the set of vertices that are separated from by no -convex cluster of . Finally, by the equality in (5), also satisfies Axiom 3, which proves the statement.
6.2. -Convexity in Hypergraphs
Let be a connected hypergraph with vertex set . A subset of is -convex if either or or the neighborhood of every connected component of is a partial edge of [7]. Note that every partial edge of is a -convex set. By we denote the -convex hull of .
An efficient algorithm for computing -convex hulls in hypergraphs was given in [12]; it makes use of the acyclic hypergraph whose edges are the maximal sets that are not separable by partial edges of [20]. The hypergraph is called the compact hypergraph of [7, 18, 20]. Let denote the compact hypergraph of . We can summarize the algorithm in [7] into the following formula: where
Fact 6. If is an edge of , then and, hence, . It follows that every edge of is a -convex set.
Fact 7. If is cluster of (i.e., is a partial edge of ), then and, hence, .
In order to prove the decomposability of -convexity, we need the following lemma whose proof is analogous to the proof of Lemma 18.
Lemma 20. Let be a connected hypergraph and let be the -convexity space on . The cluster hypergraph of equals the compact hypergraph of .
Theorem 21. Let be a connected hypergraph. The -convexity space on is decomposable.
Proof. Let be the -convexity space on .By Lemma 20, the cluster hypergraph of is acyclic so that, by Fact 6, every-edge of the cluster hypergraph of is a -convex set. Therefore, satisfies Axioms 1 and 2. Moreover, by Lemmas 11 and 20 and by Fact 7, the right-hand side of (7) equals the set of vertices that are separated from by no -convex cluster of . Finally, by the equality in (7), also satisfies Axiom 3, which proves the statement.
6.3. ap-Convexity in Graphs
Let be a connected graph with vertex set . A subset of is ap-convex if either or contains all vertices on every path joining two vertices in [3, 9]. By we denote the ap-convex hull of .
Let be the block hypergraph of ; that is, is the acyclic hypergraph whose edges are the maximal sets that are not separable by cut vertices of . From results in [3, 9], it follows that the ap-convex hull of a subset of is given by the following formula: where is the edge of that contains .
At this point, it is a mere exercise to prove that the cluster hypergraph of equals the block hypergraph of and that the ap-convexity space on is decomposable.
7. Closing Note
Given a convexity space on a connected hypergraph , we introduced the notions of a cluster of and of the cluster hypergraph of . Using them, we define decomposability of by means of three axioms and proved that decomposability implies the existence of a closed formula (see (2) above) which expresses the convex hull of a vertex set in terms of the convex hulls of edges of , where is the cluster hypergraph of . Moreover, we proved that a decomposable convexity space is a convex geometry if and only if the subspaces of induced by the edges of are all convex geometries. Finally, we showed that both -convexity and -convexity in hypergraphs are decomposable, and that ap-convexity in graphs is decomposable. An open problem is the search for other decomposable known hypergraph convexities. We conjecture that not only ap-convexity is decomposable in hypergraphs too, but also every convexity that is coarser than -convexity (such as “simple-path convexity” [4]) is decomposable. On the other hand, it is easy to see that geodetic convexity (-convexity) is not decomposable in the general case as is proven by the following example. However, since -convexity and -convexity are equivalent in distance-hereditary graphs [4], one has that -convexity is decomposable in distance-hereditary graphs.
Example 22. Consider the (hyper)graph shown in Figure 3 and the -convexity space on . Recall from the introduction that a vertex set is -convex if contains all vertices on every shortest path joining two vertices in .
It is easy to see that the sets and are the only minimal vertex -convex separators of . Moreover, the cluster hypergraph of is , , , . Since is not an acyclic hypergraph, the -convexity space on is not decomposable.
