X of taxa involves not only speciations but also recombination, horizontal transfer, or hybridization that cannot be captured by trees. A variety of classes of such networks have been discussed in the literature, including phylogenetic, level-1, tree-child, tree-based, galled tree, regular, or normal networks as models of different types of evolutionary processes. Clusters arise in models of phylogeny as the sets v. The clustering system N conveys key information on N itself. In the special case of rooted phylogenetic trees, T is uniquely determined by its clustering system N and N is a network of type N of type Rooted acyclic graphs appear naturally when the phylogenetic relationship of a set  X of vertices without descendants (the leaves of the network) represents the extant taxa, while the remaining vertices model their ancestors.Networks used to model phylogenetic relationships typically are directed acyclic graphs (DAGs) with a single root, i.e., a unique vertex from which all other vertices can be reached from. As usual in phylogenetics, the subset a and b are more closely related with each other than with c  Dress ; on the T and hierarchies T is uniquely determined by N for a given clustering system T, the Hasse diagram T are isomorphic. For general networks, however, the situation is much more complicated  such that In section“N whenever x and y. These are closely related to weak hierarchies.In “A very prominent role in phylogenetics is played by level-1 networks. Section “X is denoted by A and B overlap if The power set of a given set G is undirected if E is a subset of the set of two-element subsets of V and G is directed if G are of the form degree of a vertex G, denoted by v. If G is directed, we furthermore distinguish the indegree H is a subgraph of G, in symbols induced by some subset We consider graphs P in an undirected (resp. directed) graph G is a subgraph consisting of G is undirected, G is directed, the existence of a undirected path for a subgraph P of a directed graph G that has vertices forward arcbackward arcP are the endpoints of P and all other vertices (in P) are its inner vertices. A path P with vertices induced (in G) if A path K in a directed graph G is a subgraph with vertices undirected cycleK in a directed graph G is a subgraph with K.A directed cycle bipartite if there is a partition of V into subsets W and G connects one vertex in W to one vertex in G is complete bipartite.An undirected graph connected if, for every two vertices u and v. A directed graph is connected if its underlying undirected graph is connected. A connected component of G is a maximal induced subgraph that is connected. A vertex v is a cut vertex in a graph G if G. Similarly, a directed or undirected arc  is a cut arc in G if the graph G.An undirected graph is biconnected if it contains no vertex whose removal disconnects the graph. A block of an undirected or a directed graph is a maximal biconnected subgraph  cycle. Equivalently, a block is non-trivial if it is not a single vertex or a single arc. An arc that is at the same time a  block is a cut arc. For later reference, we state here the following observations that are immediate consequences of the fact that two distinct blocks in a graph share at most one vertex  cycle in B.Any two vertices of a non-trivial block The following well-known result will also be useful throughout:H be a biconnected subgraph of G and P be a path in G that only shares its endpoints with H. Then, the subgraph obtained by adding P to H is again biconnected. contains at least one forward and one backward arc. In a DAG G, a vertex ancestor of v a descendant of u, in symbols G from u to v. We write u and v are comparable and otherwise, incomparable. Moreover, if u is a parent of v, v is a child of u, v that is B a terminal vertex (of B). Note that every terminal vertex v of a non-trivial block B must always have indegree at least 2 since, by Corollary v lies on some undirected cycle in B and, by v in B, its two incident vertices on this cycle must both be in-neighbors. Below we will consider DAGs in which terminal vertices are a type of so-called hybrid vertices.A directed graph r et al. , we callmentclass2pt{minimu, w) in a DAG G is a shortcut if there is a vertex u, w) of N is a shortcut if G has a directed path from u to w avoiding   networks here as a slightly more general class of DAGs than what is customarily considered in most of the literature on the topic.(rooted) network is a directed acyclic graph (N1)root of N, with There is a unique vertex  A network is phylogenetic if (N2)There is no vertex  A vertex with leaf if hybrid vertex if tree vertex if X.A We emphasize that all networks considered here are rooted and thus we always use the term “network” instead of “rooted network.”trees. The set of inner vertices of a network N is strict descendant of x contains v. In contrast to the even more general definition  ofNrooted atv, is the network obtained from the subgraph N[W] induced by the vertices in w if it has indegree 0 and outdegree 1 in N[W] or hybrid vertices of N that have in- and outdegree 1 in N[W].For a vertex N with leaf set X is often called a network onX. Two networks X are graph isomorphic, in symbols graph isomorphism betweenandX, we say that isomorphic in symbols N on X is unique w.r.t. some property (or some set of properties), if A network Many studies into phylogenetic networks require that reticulation events and speciation events are separated, i.e., separated if all its hybrid vertices have outdegree 1.A network is In particular, all leaves have indegree 1 in a separated network (or indegree zero if the network consists of a single vertex).phylogenetic and separated are part of the definition of networks in many publications in the field, see, e.g., Jetten and van Iersel  by Create a new vertex As mentioned above, often only separated networks are considered, stipulating that 1) leaves, i.e., vertices  leaves, u, v) for all v, w) for all Replace arcs  The notation The operation entclass1pt{minimacf. Fig. A.Fig. 3CWe are now in the position to define least-resolved networks:N is least-resolved  such that Repeatedly apply In many applications, phylogenetic networks are considered. However, Apply Now, contraction of an arc The term “ancestor-preserving”—which is rigorously defined below—has been used in particular for mappings between certain network  in N results in a network X and N and Let N be a network on X and  be a shortcut in N. Since  is a shortcut, there is a P in N. Since N is acyclic and u is not a vertex in P since otherwise w has indegree larger than 1 in N. In particular, there is a uw-path P. Since removal of  only decreases the indegree of w and Let N. Clearly, removal of arcs changes nothing about this and thus N. If u, w), then u, w) by the uw-path Now, let X and we have By the latter arguments, Note that deletion of a shortcut from a phylogenetic network does not necessarily result in a phylogenetic network.N is shortcut-free and has no vertex of outdegree 1, then for every vertex v and w are N is phylogenetic.If a network N is shortcut-free, siblings w if and only if Since N be a network on X and N results in a network X or w.In particular, v and Let The proof is rather lengthy and technical and is, therefore, placed to “N be a network on X and X and Let N be a network on X and X or w is the only out-neighbor of u, we do not add any out-neighbors for w. Hence, X.Let X and we have w is the unique out-neighbor of u, one can easily verify that By the latter argument, As an immediate consequence of Lemma N is shortcut-free and does not contain vertices v with Every least-resolved network N be a network and N by applying N and N is phylogenetic, then w is a hybrid vertex and Let N be a network on X. We show first that w is the only vertex in N whose in-neighborhood changes and it has the new vertex w has at least one in-neighbor in N, which becomes an in-neighbor of N and thus w no longer has indegree 0, K comprising the vertices K are in N, then K is a directed cycle in N, a contradiction. If K contains an arc that is not in N, then K must contain the new vertex w and exactly the vertices of K at most once, K that are incident with K are also arcs in N. In particular, there is a N. Together with the fact that N contains a directed cycle, a contradiction. Therefore, Let N creates a network w is the unique out-neighbor of N is recovered from N and The operation N is phylogenetic. Assume first that w is a hybrid vertex and w is replaced by an out-neighbor N is phylogenetic, the latter arguments imply that there is no vertex w is a not hybrid vertex or w is not a hybrid vertex, then w does not change. In addition, w in w is a hybrid vertex and Suppose now that X. The newly created vertex w. The statement “N is recovered from By the latter arguments, The following result shows that the expansion operation does not introduce shortcuts and is an immediate consequence of Lemma N be a network and N by applying N is shortcut-free if and only if Let We remark that an analogue of Corollary N will play a key role in the following. We first establish several technical results that will allow us efficiently reason about the block structure of a network.The blocks of N be a network and u and v are connected by an undirected path P that contains at least 3 vertices and of which all inner vertices w satisfy w.Let u and v, respectively. Let u and v would be u and v, respectively. By construction, P that is the union of the underlying undirected version of P contains at least the three vertices u, v, and w satisfy v and u are There are directed paths u and v are called “up-down-paths” in Bordewich and Semple . Thus, suppose B is a non-trivial block. Suppose, for contradiction, there is a vertex w in P w and all of its parents are contained in a non-trivial block B and w and u. By Observation (1)w and all of its (at least two) parents are contained in B, then clearly (3) mentclass2pt{minimw is contained in a block B but not properly contained, then it must hold k networks:As a consequence, if a hybrid vertex N is level-k if each block B of N contains at most k hybrid vertices distinct from A network N is level-k if each block B of N properly contains at most k hybrid vertices. In Choy et al.  for which As we shall see in Lemma X and a subset closure operator as the map For a clustering system closed if, for all non-emptyA clustering system The following result is well known in the clustering literature.A clustering system For completeness, a proof is provided in “We continue with three simple observations concerning the clusters of networks.N be a network. Then, Let v lies on a directed path from the root x. Furthermore, w lies on a directed path from v. By (N1) and since N is a DAG, there is directed path from x that contains w, and thus By construction, v is a hybrid vertex with unique child w, we have N in Fig. P connecting u or v. However, it requires that u and v have at least one descendant leaf in common, i.e., that We note in passing that the converse of Lemma N be a network and u and v are connected by an undirected path (i)(ii)In particular, P satisfy P is a subgraph of a non-trivial block.Let u and v, respectively, to the leaf x. Let x is contained in both paths. It must hold that u and v would be u and v, respectively, to P satisfies all of the desired properties. Two see that P is a subgraph of a non-trivial block B, observe that, by Lemma u and v are connected by an undirected path P and u and v are connected by two distinct paths that both have at least 3 vertices and that only have the endpoints u and v in common. Hence, u and v lie on a cycle K and thus in a common block B of N. In particular, P is a subgraph of K and thus of B. There are directed paths N be a network and u and v are B of N.Let u and v are u and v are contained in a common non-trivial block B of N. Since B. Thus, neither u nor v is a terminal vertex. In summary, Let w are redundant in the sense that every directed path from w to one of its descendant leaves necessarily passes through the unique child v of w. Thus, we have Clusters of outdegree 1 vertices N be a network. If v is the unique child of w in N, then Let N be a network, B a block in N and H be the set of hybrid vertices h that are properly contained in B and satisfy Let u and v are u and v are connected by an undirected path P which contains a hybrid vertex N. Since B and u and v, Observation h must be properly contained in B by Lemma It suffices to show that mentclass2pt{minimNote that Hasse diagramX as its unique root. The singletons The X. Then, every non-singleton set Let C to some singleton set C and C has a child C has a child C satisfies Let X with corresponding Hasse diagram Let X is the unique cluster in X is the root in Clearly, C is a singleton, and For a given a clustering system C does not overlap any other set. Then, there is no undirected cycle in C is a cut vertex in Let C does not overlap with any other cluster. If K in K cannot intersect both. Now, assume that C is neither a singleton nor X, both C is disconnected and thus C is a cut vertex. In particular, every undirected path connecting a cluster in C and thus the second statement of the lemma follows as an immediate consequence. Suppose that T is isomorphic to the Hasse diagram of its clustering systems Every phylogenetic tree nd Steel . Figure nd Steel ; Baroni nd Steel ; Willsonnd Steel .regular if the map N and (ii)N is shortcut-free.A network N is regular if and only if the following three conditions hold: (a) v to u, i.e., u and v, then neither path consists of a single arc, i.e.,  is not a shortcut. Clearly, conditions (ii) and (c) are equivalent. It therefore suffices to show that condition (i) holds if and only if conditions (a) and (b) are satisfied. Together, (a), (b) and Lemma Proposition 4.1 in Baroni et al.  states tProposition N with For every clustering system Every regular network is least-resolved.N is not least-resolved, i.e., there is a network N by a non-empty sequence of shortcut removals and application of N is shortcut-free and, therefore, the operation that is applied first must be a contraction. Therefore, and since no new vertices are introduced, it must hold N is regular, it holds Suppose, for contradiction, that the regular network N is shortcut-free and satisfies N. Hence, we obtainThe converse of Corollary Not every least-resolved network is regular.Nevertheless, for level-1 networks the terms least-resolved and regular coincide as shown in Corollary A clustering system X is a hierarchy if and only if N is a phylogenetic tree if and only if A clustering system The 1-to-1 correspondence of hierarchies and phylogenetic trees is well known, see, e.g.,  in Proposition Regularity as characterized in Proposition N has the path-cluster-comparability (PCC) property if it satisfies, for all (PCC)u and v are A network Definition N with For every clustering system u and v.We emphasize that (PCC) is a quite restrictive property. For instance, the rooted N be a network satisfying (PCC) and Let u and v are Suppose N be a network. Then, Let vw-path which passes through some child Since Property (PCC) still allows that distinct vertices are associated with the same clusters. It requires, however, that such vertices lie along a common directed path.semi-regular if it is shortcut-free and satisfies (PCC).A network is We introduce the term “semi-regular” because, as we shall see in Theorem N be a semi-regular network and let N is phylogenetic, then u is the unique child of v in this case.Let u be the unique child of v. Thus, it holds w and N being shortcut-free imply w is the unique child of v.Suppose first that N is phylogenetic and assume, for contradiction, that w has a unique child. Therefore, and since N is phylogenetic, there must be a vertex v and N being shortcut-free imply v. Now, suppose, in addition, that As a consequence of Lemmas N be a semi-regular network and let Let N be a semi-regular network, Q(u) are pairwise N. Moreover, if Q(u) is contained in a non-trivial block, or adjacent pairs of vertices in Q(u) form a trivial block.Let Q(u) are pairwise Q(u) forms a directed path in N and there are unique N is acyclic and shortcut-free, this path must be an induced subgraph.By (PCC), u is a hybrid vertex and suppose there is u has a parent in Q(u). Then, there is v and N is shortcut-free, v and Q(u). By similar argument, only P formed by Q(u) have degree 2. Therefore, if one arc in P is contained in an undirected cycle, then all arcs in P are contained in this cycle, in which case Q(u) is contained in a non-trivial block. Otherwise Q(u) consists of a sequence of consecutive cut-arcs. Now, suppose The examples in Fig. A network is regular if and only if it is semi-regular and there is no vertex with outdegree 1.N is regular, it is in particular also semi-regular and thus satisfies (PCC). Furthermore, then u, v satisfying N is semi-regular (and thus shortcut-free) and suppose there is not vertex with outdegree 1. Then, Lemma u, v with N is regular. If N be a network on X. Then, N is regular.Let N is shortcut-free, satisfies (PCC), and has no vertex with outdegree 1  would be a shortcut, contradicting that N is semi-regular, and an arc  would imply A with a vertex in D cannot exist. Since B is a non-trivial block, a and d lie on an undirected cycle K in B. In particular, they are connected by two undirected paths that do not share any inner vertices. Thus, there is an undirected path u. Let P such that A. Since P does not contain u and u and v are N since otherwise u and v are u and v are v is not a terminal vertex since Suppose We note that semi-regularity cannot be omitted in Lemma We continue by showing that whenever (PCC) is satisfied, least-resolved networks are precisely the regular network, To this end, we consider first the implications given by the operations N be a network, N by applying N satisfies (PCC) if and only if Let N and w is the unique child of By Lemma u and v are N also satisfies (PCC), consider u and v are u and v are also u and v are Suppose first N satisfies (PCC). By similar argument as above, it holds, for all u and v are v and v and w are v and w is the unique child of v and w are w in v and v and w are Now, suppose u, w) where Corollary N be a network, N by applying N is semi-regular if and only if Let N be a network satisfying (PCC). The unique regular network N by repeatedly executing one of the following operations (1) and (2) until neither of them is possible: u, w), andremove a shortcut  with apply Let X and that, for all Let mentclass2pt{minimmentclass2pt{minimN satisfies (PCC), i.e., for all u and v are u and v are u and v are also u and v are v with By assumption, entclass1pt{minimaN be a least-resolved network satisfying (PCC). Then, N is uniquely determined by Let N is shortcut-free  and contains no vertex with outdegree 1. By Theorem N is regular. By Corollary As a consequence of Corollary N is regular if and only if N is least-resolved and satisfies (PCC).A network N is separated if all hybrid vertices have outdegree 1. We have already seen above that the Hasse diagrams of the clustering systems cannot produce separated networks with hybrid vertices because u is the only child of v. By Theorem vice versa.Recall that a network entclass1pt{minimaCluster-popping algorithm (Huson and Rupp cluster network (Huson and Rupp The N is a cluster network if (i) it satisfies (PCC), and, for all (b)v is a hybrid vertex and parent of u or vice versa,(c)u is a child of v, then there exists no node w with if (d)v has exactly one child, which is a tree node.every hybrid vertex A network (i’) However, this contradicts the existence of hybrid vertices v, which are required to have exactly one child u by (iv). To see this, observe that We note that the definition of cluster networks usually is expressed using the following condition instead of (PCC): We can rephrase conditions (i)-(iv), and thus the definition of cluster networks, as follows:N is a cluster network if and only if it is semi-regular, separated, and phylogenetic.A network N is a cluster network, i.e., it satisfies conditions (i)-(iv) in Definition N satisfies (PCC) and is separated. Hence, it remains to show that N is shortcut-free and phylogenetic. Suppose, for contradiction, that  is a shortcut in N. Then, there is v is a hybrid vertex with exactly one child, a contradiction. Therefore, u is a tree node, contradicting that N must be shortcut-free and thus semi-regular. Suppose, for contradiction, that N is not phylogenetic. Hence, there is a vertex v with exactly one child u and v must be a hybrid vertex by (ii), a contradiction.Suppose first that N is semi-regular, separated, and phylogenetic. Hence, N satisfies (PCC), i.e., condition (i). Condition (iii) must be satisfied since an arc  with u is the unique child of v. Since N is phylogenetic, v must be a hybrid vertex. Conversely, a hybrid vertex v in a separated network has exactly one child u implying N is separated, it remains to show that the unique child u of a hybrid vertex v is a tree node. Suppose for contradiction that u is a hybrid node. Then, u again has a unique child w. Hence, we have v, w) is an arc in N and, in particular, a shortcut, a contradiction. Therefore, condition (iv) is also satisfied. Conversely, suppose We shall see in Theorem N be a semi-regular phylogenetic network. Then, the multiplicity of each cluster Let N satisfies (PCC) and is shortcut-free. Let u is the unique child of v. Suppose, for contradiction, there is a third vertex w is the unique child of v, a contradiction. If v and u are both the unique child of w, which is not possible since u and v are adjacent in both cases. Recall that a semi-regular network As we shall see in Theorem N and (i)N with cluster C such that sort the (ii)k vertices in C such that sort the (iii)set Let C in N to C in In other words, we map the N and V(N) and X. Writing v is a leaf if and only if Let C in N and k vertices with cluster C in V(N) and X and satisfies v is a leaf if and only if Since N be a semi-regular network. Then, N is the unique semi-regular network whose cluster multiset is Let N and N and X. In the following, we write v, u) must be a shortcut in N, a contradiction. Therefore, X and thus N is the unique semi-regular network whose cluster multiset is Suppose We emphasize that none of the two conditions (PCC) and shortcut-free that define semi-regular networks can be omitted in Theorem N (as any other network for which the property of being phylogenetic has been left out) are not determined by their clustering systems N in Fig. It is an easy task to verify that semi-regular networks N be a cluster network with clustering system C has indegree at least 2 in u and v are adjacent in N.Let N is phylogenetic, separated, and semi-regular, i.e., it satisfies (PCC) and is shortcut-free. By Lemma C has indegree at least 2 in By Theorem u must be the unique child of v. Since N is phylogenetic, this implies that v. Since N is shortcut-free and by Observation C in C in C has indegree at least 2 in Suppose C has at least two distinct parents v, N. Let v has a parent in each of v is a hybrid vertex. Since moreover N is separated, v has a unique child u. By Observation C in N with cluster C. Conversely, suppose N with N is the unique semi-regular separated phylogenetic network with clustering system For every clustering system N be the network obtained from N is a phylogenetic network that satisfies N is semi-regular. In particular, by construction, all hybrid vertices in N have outdegree 1, i.e., N is separated. By Theorem N is a cluster network.By Theorem mentclasspt{minimaN is the unique cluster network with clustering system N and C has multiplicity 2 in C has indegree at least 2 in C has multiplicity 2 in N is the unique cluster network. In particular, by Theorem N is the unique semi-regular separated phylogenetic network with clustering system It remains to show that reticulate networks in , and hence (b) is satisfied trivially. Since N is separated, every hybrid vertex v has exactly one child u. By Obs u is also a hybrid vertex, i.e., there is N is shortcut-free, v and v and N also satisfies condition (a). In summary, every cluster network is reticulate. By Theorem N has the tree-child property if, for every .N be a tree-child network and u and v are Let The converse is not true. Figure normal if it is tree-child and shortcut-free.A network is X comprises not only the leaves in our sense but also the root and all vertices with outdegree 1. Under this assumption, regular networks that are not normal, see Fig. 1 of Willson  and alsoNot every semi-regular network, and in particular not every cluster network, is normal.N with N is unique w.r.t. these properties. In particular, N is the unique cluster network with Let N is phylogenetic, separated, and normal and satisfies N is semi-regular and thus, by Theorem N is unique. Suppose From Proposition N be a binary normal network. Then, N is the unique binary normal network whose cluster set is N is a cluster network.Let The following result appears to be well known, see, e.g., Murakami et al. . An arguEvery phylogenetic level-1 network is tree-child.N be a phylogenetic level-1 network. If v is a hybrid vertex, then, by Lemma v and all its parents. Suppose, for contradiction, there is a non-leaf vertex v whose children are all hybrid vertices. Suppose first that N is phylogenetic, this implies that v is hybrid vertex. Let u be the unique child of v, which is a hybrid vertex by assumption. The hybrid vertices u and v are contained in a common non-trivial block N is level-1. Now, suppose that v and N is level-1. Otherwise, there is at least one N is acyclic, there is N is level-1. Let v with a hybrid child v, u) creates a new tree vertex u as its only child. The modified network is still level-1 but no longer tree-child.Note that “phylogenetic” cannot be omitted in Proposition Next we consider the overlapping clusters in tree-child networks in some more detail:N be a tree-child network and suppose u and v are Let N satisfies (PCC) and thus either u, v are contained in a common non-trivial block B. Set H of hybrid vertices. Since A is non-empty and, by assumption, h, we must have H. Since N is tree-child, there is a leaf x that is reachable from w along a path consisting entirely of tree vertices and thus cannot be a descendant of any hybrid vertex By Corollary N is not necessarily closed.In particular, the case N be a tree-child network with clustering system Let Another class of networks that has received considerable attention in the last decade are tree-based networks  subdividing the arcs of T by introducing vertices with in- and outdegree 1 , and (b) adding arcs  between pairs of vertices, so that N remains acyclic.A network This definition further generalizes the original one for non-binary tree-based networks in Jetten and van Iersel  in the sN on X is tree-based if and only if there is a rooted (not necessarily phylogenetic) spanning tree T with leaf set X, i.e., there are no dummy leaves in T that correspond to inner vertices in N. Clearly, the unique incoming arc  of a tree vertex v must be contained in every rooted spanning tree T of a network N and thus u cannot be a dummy leaf in T. As an immediate consequence, the well-known fact that tree-child network are always tree-based  since the two vertices highlighted in orange correspond to the same cluster but are U of all inner vertices whose children are all hybrid vertices, called omnians in Jetten and van Iersel . Moreover, there are cluster networks and thus phylogenetic separated networks that do not satisfy (PCC).least common ancestor (LCA) of a subset N is an ancestor of all vertices in Y that is Bender et al.  A least N, an LCA does not necessarily exist for a given vertex set. Moreover, an LCA is not unique in general. We write Y. In a (phylogenetic) network N, the root V(N), and thus, a least common ancestor exists for all In general DAGs N be a network and Let y is also an ancestor of all vertices in Z and thus there is a Z, i.e., Consider u only, we write Y exists and is uniquely determined. We leave Y with If u are defined in terms of the fact that no child of u is an ancestor of all vertices in Y. These definitions are equivalent:In Huber and Scholz , least cN be a network and Y if and only if u is an ancestor of all vertices in Y but there is no Y.Let Y if it is ancestor of all vertices in Y and only if-part of the statement follows immediately. Conversely, suppose u is an ancestor of all vertices in Y but there is no u is an ancestor of all vertices in Y but u is a least common ancestor of Y. By definition, We will in particular be concerned here with LCAs of leaves, i.e., the sets v with Suppose N be a network, (i)v with (ii)A.(iii)Let In much of the literature on least common ancestors in DAGs, only pairwise LCAs are considered. Networks with unique pairwise LCAs are of interest because of a close connection with so-called binary clustering systems  if (CL)For every A network N has property (CL). Then, for all Suppose v or one of its descendants. The second statement now follows directly from the definition. If In Lemma N be a network satisfying (CL). Then, (i) Q(u) has a unique w is contained in a directed path from u to v, then Let q is the unique The second statement in Lemma N satisfies (PCC), then it satisfies (CL).If a network N satisfies (PCC). Given a cluster W are pairwise Suppose that Figure N with For every clustering system Since a normal network is tree-child, it satisfies (PCC) by Corollary N satisfies (CL).Every normal network Property (CL), however, does not imply that N is an lca-network if A network From Observation N is an lca-network and Every lca-network satisfies (CL). Moreover, if N be an lca-network on X and Let N satisfies (CL). Hence, we can apply Lemma Since Every lca-network has a closed clustering system.N be an lca-network. We show that u with A. By analogous arguments, A. Since N is an lca-network. Therefore, Let N is a network with a closed clustering system N satisfies (PCC), then it is an lca-network.If A with two distinct vertices u, v are Assume, for contradiction, that there is some set N be a network satisfying (PCC). Then, N is an lca-network if and only if its clustering system Let N be an lca-network satisfying (PCC). By Lemma Let The example in Fig. N is an lca-network if and only if its clustering system A semi-regular network Since every tree-child network satisfies (PCC) we also obtainN is an lca-network if and only if its clustering system A tree-child network Moreover, we haveN with A clustering system N and a unique cluster network N and N with N with Let entclass1pt{minimaentclass1pt{minimaAs in the case of trees, there is a simple connection of the LCA with the closure operator:N be an lca-network with clustering system Let Y. Let u be a vertex in N such that Y, this implies Let entclass1pt{minima) by Eq.  follows Next we show that two sets have the same LCA whenever their LCAs are associated with the same cluster.N be an lca-network on X and let Let Since In this section, we consider an interesting subclass of lca-networks.N on X is a strong lca-network if it is an lca-network and, for every non-empty subset A network Figure We shall see below that strong lca-networks are intimately connected with well-studied types of clustering systems.X isweak hierarchy if a binary if it is pre-binary and, for every C is the unique inclusion-minimal cluster containing x and y.A clustering system Weak hierarchies were introduced in Bandelt and Dress  and subsN be an lca-network. Then, the following conditions are equivalent: (i)For all (ii)Let N is an lca-network, it is in particular pre-binary  in Lemma X is a weak hierarchy if and only if for every non-empty subset . Then, N is a strong lca-network if and only if Let N be a network satisfying (PCC). By Theorem N is an lca-network if and only if N is a strong lca-network precisely if and only if Let Furthermore, we can use the same arguments in the proof of Proposition A clustering system We start by showing that all phylogenetic level-1 networks have the path-cluster-comparability property (PCC).Every phylogenetic level-1 network satisfies (PCC).N is a phylogenetic level-1 network, then it is tree-child by Proposition If N in Fig. We note that “phylogenetic” cannot be dropped in Lemma N be a level-1 network. Then, N is least-resolved if and only if N is regular.Let By Corollary N for a given clustering system We emphasize, however, that there can exist least-resolved networks n be a positive integer. Then, there is no phylogenetic, shortcut-free level-1 network N on n leaves that is not a tree and where Let N be a phylogenetic, shortcut-free level-1 that is not a tree. By Lemma N satisfies (PCC). Since, in addition, N is shortcut-free, N is semi-regular. Since N is not a tree, it must contain a non-trivial block B. By Lemma u and v such that Let As an immediate consequence of Theorem A phylogenetic level-1 network is semi-regular if and only if it is shortcut-free. Furthermore, a level-1 network is regular if and only if it is shortcut-free and has no vertex with outdegree 1.N be a phylogenetic level-1 network and v be a hybrid vertex of N. Then, Let N be a phylogenetic level-1 network and v be a hybrid vertex of N. By Lemma v and all of its (at least two) parents are contained in a common non-trivial block B. Hence, consider first one of the parents v such that v and B and N is level-1, N is phylogenetic, we have N satisfies (PCC), and thus, v and Let v, which together with the arc v, and B and thus equals B. But then B contains two hybrid vertices v and In Case (a), the arc v has a parent B and Lemma V(N) being finite. The latter together with the fact that w of v that is distinct from v. Since v is a hybrid vertex, it has another parent B and satisfies Now, consider Case (b), i.e., w is a parent of v. This together with Lemma w is a parent of v. Finally note that N be a level-1 network. Then, every block B has a unique B is not a single vertex or arc, B and B.Let B follows from Lemma B is acyclic, a u in B does not have an out-neighbor along the cycle, and therefore, u has at least two in-neighbors that are contained in B. Thus, u is a hybrid vertex and, by Lemma u is properly contained in B. By definition of level-1, there is at most one such vertex in B. The statement is trivial for a block that consists only of a single vertex or arc. Uniqueness of the As an immediate consequence, we haveN be a level-1 network and B a block of N. For every Let B . Thus, it is semi-regular, which allows us to use Corollary u of v. Hence, u of v. Moreover, by Corollary w in which case, N is strong lca-network by Lemma Suppose that v has a child N is shortcut-free, u, w, and u and B and u and w and v does have a child v is the unique least common ancestor of Now, suppose N be a shortcut-free level-1 network on X and u, w of v if and only if v is the N.Let u, w of v. Since N is shortcut-free, u and w must be u and w are connected by an undirected path P whose inner vertices x satisfy P and the two arcs  and  form an undirected cycle whose v. Conversely, suppose v is the K in N. Hence, the two vertices v in K are children of v. In particular, v, N is shortcut-free. Therefore, we can apply Lemma Suppose In many applications, vertex- or arc-labeled networks are considered as a scaffold to explain genomic sequence data  one of the Cases (i) or (ii) as specified in Lemma Assume first that x and z as well as y and z, i.e., we have z and v as well. Hence, we have (i) N. For both cases, we have Assume that Case (i) holds for both x and z. In this case, y and z. Hence, neither of the Cases (i) or (ii) as specified in Lemma x and z (resp. y and z), a contradiction. Therefore, Assume now that Case (ii) is satisfied for z and Assume now that In Case (a), v in N, the case N, we must have In Case (b), suppose first, for contradiction, that z and z and c becomes a child of v in In Case (c), In summary, neither of Cases (a), (b), and (c) is possible. Therefore, N be a level-1 network with clustering system Let B, and w with w and w and u are w are located in a common block Let N contains a non-trivial block and thus a hybrid vertex. In particular, therefore, if N is a rooted tree, then Inspection of the proof of Lemma Lemma (Property (L)) A clustering system For later reference, we record an equivalent way of expressing property (L):A clustering system Let N satisfies property (L).The clustering system Let A clustering system If one of the three sets C is not the intersection of two overlapping clusters. In particular, we haveIn order to show that closed clustering systems with property (L) define level-1 networks, we first demonstrate that one can identify the non-trivial blocks directly in a clustering system provided it satisfies (L). We start by introducing subsets of clusters with a given overlap: For a given clustering system T with Let T be phylogenetic tree with T is a tree, we have First, let T such that Assume now that Let D overlap. If D overlap, then (L) implies D. Now, (L) implies Consider a cluster C does not overlap with any cluster in Let D does not overlap with C in any of the three cases. If, on the other hand, D and C also do not overlap. If Alternative (iii) in Lemma Let U(C). To see the “if” direction, note first that D, which is impossible because U(C), and thus also D, contains at least one set overlapping The “only if statement” follows directly from the definition of As a consequence we can use the condition C of a closed clustering system In general, Let X. Since Let Although U(C) does not overlap any cluster Let D cannot overlap since in this case (L) implies D does not overlap U. Let C. To see that C. Clearly, these two supersets overlap with C and So far, If C. Hence, If Lemma Let D satisfies one of the alternatives: (i) Let As an immediate consequence of Corollary Let Furthermore, the sets Let D and D overlap. By (L), Suppose that C in Let C. Hence, D and D and C in C has indegree zero if and only if C has exactly one in-neighbor D and C. Both of these paths must pass through if-direction must also hold. Suppose Lemmas Let C. The statement is trivially true for all C.Let C. It is now easy to verify the graph obtained from (i) a rooted tree whose root has at least two children and (ii) connecting its leaves to an additional vertex is biconnected. Hence, C, and any two vertices lie along a common “undirected” cycle (which necessarily passes through C). By Corollary D in the Hasse diagram (i)D is the unique in-neighbor of (ii)(iii)if Let D is contained in D, i.e., D is its unique in-neighbor.We start with showing Property (i). By Corollary D are adjacent in We continue with showing Property (ii). Since D is the unique in-neighbor of Finally, we show Property (iii). Suppose that Let C, K exists. Since K contains at least one arc of P of K on at least two vertices and where all vertices of P are contained in P are both incident with one cluster in P must be distinct from P that is adjacent to K. Therefore, it suffices to consider the two mutually exclusive cases (a) C overlaps with some cluster C does not overlap any cluster. Furthermore, by Lemma C are contained in C and thus K containing (cf. Eq. ). SupposD is the unique in-neighbor of K, it must be adjacent to another vertex K. Since D is the unique in-neighbor of K in D has at least one child F such that C does not overlap with any cluster in C and C does not overlap with any other cluster. Hence, U as child of U in D. In U does not overlap any set in U is a cut vertex in UD, there is no cycle in By Lemma mentclass2pt{minimX satisfying (L). If Let C does not overlap any cluster. By Lemma C in C. Clearly, We show that the arc We summarize Lemmas X satisfying (L) and with Hasse diagram B is a block of B is a trivial block and, otherwise, in Case (i) a non-trivial one.Let If B is a trivial block of C or the single arc C overlaps with some cluster C and C does not overlap any cluster. Suppose now that B is a non-trivial block of B contains an undirected cycle K on at least 3 clusters. Since K contains at least one cluster C with two in-neighbors K .X satisfying (L). Then, the Hasse diagram Let B of B is a trivial block consisting of a single arc or, if B is a non-trivial block, and thus, by Proposition C is a hybrid vertex. From Corollary C is the only hybrid vertex in By Lemma X that satisfies (L), there is a level-1 phylogenetic network N such that For every closed clustering system N for N is also a level-1 phylogenetic network. By Proposition We summarize Corollary N such that Let N is regular if N can easily be edited into a level-1 network We emphasize, however, that there is no 1-to-1 correspondence between level-1 networks and clustering systems. Recall that a network entclass1pt{minimaN, the regular network N by repeatedly removing shortcuts and contracting arcs  with N in this way.For every level-1 network N be a level-1 network. By Theorem N by repeatedly (1) removing a shortcut and (2) applying u, w) with N does not matter. By Corollary u, w) with N by repeatedly removing shortcuts and contracting arcs  with Let mentclasspt{minimaentclass1pt{minimaentclass1pt{minimaAs a direct consequence of Theorem N such that Let N with v with N is regular and least-resolved.Let N with N is shortcut-free and contains no vertex v with N is phylogenetic. Now, let N be a shortcut-free phylogenetic level-1 network with v with N is a regular network, which is unique by Proposition N is least-resolved. By Corollary entclass1pt{minimaentclass1pt{minimaentclass1pt{minimaMost publications on phylogenetic networks assume that leaves always have indegree 1, see, e.g., Huson et al. .N with v with Let x with N is a phylogenetic network, Corollary N is shortcut-free, Lemma N is level-1 and Lemma N satisfies N has indegree 1 by construction and all vertices with outdegree 1 must be adjacent to leaves.By Corollary N is unique w.r.t. these properties. Let v with v with outdegree 1, we obtain a phylogenetic level-1 network N from N is the unique network with the desired properties. It remains to show that As an immediate consequence, we obtain a characterization of the level-1 networks that are completely determined by the least common ancestor function, and equivalently by their clusters.N be a level-1 network without shortcuts. Then, the following statements are equivalent: (i)(ii)For every Let N is regular, i.e., w. Moreover, assume for contradiction that there leaves If with Eq. , i.e., tFinally, we show that every closed clustering system satisfying (L) is represented by a unique “minimal” separated level-1 network. More precisely, we haveN with N is obtained from the unique regular network Let N with N is also level-1. Since N is semi-regular, it is shortcut-free. Hence, a network with the desired properties exists. To see that N is unique, let By Corollary N has property N a compatible w.r.t.networks if there is N on X such that A frequent task in phylogenetics is the construction of networks based on partial information of putative networks, e.g., subtrees  is a hereditary property, it suffices to show that if X such that N can be constructed in Let The proof  is rather lengthy and technical and is, therefore, placed to Sect. There is a  level-1 network with For every clustering system If Statement (1) is satisfied, then the network computed with Check-L1-Compatibility is a network with B is highly constrained if the unique terminal vertex v of B has only two parents In level-1 networks, the structure a block N be a level-1 network. Then, every non-trivial block is an (undirected) cycle if and only if every hybrid vertex v in N satisfies Let k networks with We note that a similar result does not hold for level-N is a level-1 network and every hybrid vertex v in N satisfies N is outerplanar.If N is a cycle. Therefore, the underlying undirected graph of N does not contain a subdivision of the graph N is outerplanar. By Observation d Harary , N is ougalled trees were introduced as phylogenetic networks in which all cycles are vertex disjoint. Here, we consider a more general version, where cycles are allowed to share a cut vertex and the network is not required to be phylogenetic. More constrained types of networks will be discussed in the subsequent sections.In Gusfield et al. , galled galled tree is a network in which every non-trivial block is an (undirected) cycle.A Every galled tree is level-1.N be a galled tree. Every trivial block B contains at most one hybrid vertex distinct from B and assume, for contradiction, that B properly contains two hybrid vertices h and B is a cycle, the two vertices in B that are adjacent with h must exactly be the two in-neighbors of h. The same holds for h. It is now easy to see that the two path in B that connect h and B are out-neighbors. Hence, these two distinct vertices are B, which contradicts the uniqueness of Let Lemma v.A network is a galled tree if and only if it is level-1 and satisfies B that is distinct from B.In a galled tree, every non-trivial block consists of two internally vertex disjoint paths connecting As we shall see below, this implies that its clustering system satisfies the following property:(N3O)If N with N is level-1 by Lemma B. Clearly, B from v in v in v is a hybrid vertex in B and clearly distinct from i and j were chosen arbitrarily, the paths N has a non-trivial block that is not an undirected cycle. Hence, N is not a galled tree, a contradiction. Suppose there is a galled tree The converse of Lemma N is the clustering system of a shortcut-free galled tree. By Lemma N is level-1, and thus, it is closed and L), and does not contain three pairwise overlapping clusters. By Corollary N is not a galled tree, i.e., it contains a non-trivial block, that is not an undirected cycle. By Observation N is shortcut-free, N must be a galled tree. Suppose first that tisfies NO. Now, spre-pyramidal if there exists a total order X such that, for every Diday ; Bertran(WP)If  Taken together, (L) and (WP) imply (N3O). More precisely, we haveA necessary condition Nebeský  for \docLet Assume, for contradiction, that Pre-pyramidal set systems are also known as “interval hypergraphs.” A characterization in terms of an infinite series of forbidden subhypergraphs has been developed in Tucker ; TrotterLet Starting from Duchet  and (N30).N with Let N is, in particular, a phylogenetic level-1 network (cf. Lemma N satisfies (PCC). By definition, every non-trivial block contains at least 3 vertices. If a block B would contain exactly three vertices, then one easily sees that N contains the shortcut N contains at least 4 vertices. Assume, for contradiction, that N contains a non-trivial block B with at least B refers to an (undirected) cycle in N. Hence, there are two internal vertex disjoint paths in B connecting N is shortcut-free and v and the other path contains vertices B is an undirected cycle, one easily verifies that v and v and v and v and Suppose first that N with v, N. In particular, we have B contains at least five pairwise distinct vertices v, Assume now that there is a shortcut-free phylogenetic galled tree k networks N that are not level-It is worth noting that for paired hierarchies, and in particular also for hierarchies, mentclasspt{minimaThe literature on phylogenetic networks often stipulates that the leaves n et al. . FurtherN is conventional if (i) all leaves have indegree at most 1 and (ii) every hybrid vertex is contained in a unique non-trivial block.A network N is conventional, while We remark that if X satisfying (L). Then, Let B and By Lemma mentclass2pt{minimN is separated, then N is conventional.If N separated. Then, all leaves in N must have indegree at most 1 since they have outdegree 0 and, by definition, hybrid vertices have outdegree 1. By Lemma v that is contained in two non-trivial blocks B and v is hybrid vertex in a separated network, we have v that is contained in two non-trivial blocks cannot exist. Hence, N is conventional. Suppose binary if it is phylogenetic, separated, and all vertices have in- and outdegree at most 2. Equivalently, in a binary network, every tree vertex is either a leaf or has exactly two children, and every hybrid vertex has exactly two parents and one child. As an immediate consequence of the definition, Proposition Recall that a network is Binary level-1 networks are always phylogenetic, separated, conventional, tree-child, galled trees.N be a binary level-1 network. Then, N is a tree.Let N be a binary level-1 network. By definition, N is phylogenetic. If N is a tree, then N is not a tree and thus, N contains hybrid vertices all with outdegree 1 in N. However, Proposition Let Hence, (2-Inc) A clustering system N be a binary network that satisfies (PCC). Then, Let N be a binary network on X with clustering system Let v has at least one but at most two parents in N since N is binary and N with v has at most two parents, at least two of w of v in N. W.l.o.g. assume that v is Assume first that there are (at least) three inclusion-minimal clusters C cannot be a singleton. Since C is not a singleton and N is binary, we can conclude that v has at least one but at most two children in N. Let N with v has at most two children, at least two of w of v in N. Since v is Assume now that there are (at least) three inclusion-maximal clusters X. Then, there is a binary level-1 network N with Let N by applying N is binary. Moreover, since N is a level-1 network. Hence, a binary level-1 network N with N is a binary level-1 network. Conversely, suppose that N is a binary level-1 network on X with N satisfies (PCC) and thus, by Lemma Assume first that Since a phylogenetic level-1 network satisfies (PCC) by Lemma N be a phylogenetic level-1 network. Then, N is a cluster network if and only if it is shortcut-free and separated.Let Observation N be a binary level-1 network. Then, N is a cluster network if and only if it is shortcut-free.Let We finally consider the problem as whether a clustering system w in A given clustering system N be such a network with w in w in N is a galled tree. Hence, by Theorem Assume first that w in n leaves such that each inner vertex has exactly two children and the subgraph induced by the inner vertices is a directed path with the root B, B. Let v be vertex in v as follows: If v. Otherwise, v is a tree vertex. In this case, let B in v that are not contained in some v by a caterpillar n leaves of the caterpillar and the elements in B. We therefore re-attach, for each block B as children of the leaf of the caterpillar that corresponds to B. Note that this construction is well defined since, by Lemma N with Assume now that Using Check-L1-Compatibility and the results in Theorem Determining if a clustering system N whose hybrid vertices w have For all phylogenetic level-1 networks w in N have N is a galled tree by Observation w in If all hybrid vertices N for which there is a binary level-1 network w with B such that N that is obtained from N is still phylogenetic and level-1 and by Lemma The converse of Corollary In this part, we focus on a particular subclass of networks:N is a quasi-binary if B in N.A network We note that, in particular, all binary networks are quasi-binary. Moreover, quasi-binary networks are separated. Therefore and by Corollary Quasi-binary level-1 networks are separated galled trees.N is a phylogenetic quasi-binary level-1 network. In this case, we obtain Property (PCC) as an immediate consequence of Lemma Theorem all networks N for which either precisely one or at least one of the following conditions hold: N is level-k, N does not satisfy (PCC);N is not quasi-binary;N is not shortcut-free;N is not phylogenetic.Then, no network Let N be a phylogenetic quasi-binary level-1 network. Then, N is the unique phylogenetic quasi-binary level-1 network tree whose cluster multiset is Let N and N and V(N) and X. In the following, we write v is a leaf if and only if N and Suppose that both N is acyclic and finite, we can assume w.l.o.g. that v, u) is a shortcut in N. In particular, u is a hybrid vertex and, since by definition the non-trivial blocks in the galled tree N correspond to the undirected cycles, B of N whose unique hybrid vertex is It remains to show that, for all N is quasi-binary, the hybrid vertex u has a unique child c. By the v, u), there is an arc c is the only child of u) such that N is phylogenetic, u), or at least two children. By the choice of , the images of these vertices are adjacent with We continue with showing that z be the parent of the hybrid vertex u that is not v and observe that u must have a parent N is acyclic, it holds that z and v are the only two parents of u in N. Therefore, Let B consists of two paths that have only v and u in common and which do not contain additional hybrid vertices and since v and z are the unique two parents of u, it holds that Now, let Case (a)q is also contained in B. In particular, q is a tree vertex. Since N is a galled tree, B is exactly the undirected cycle that is formed by the shortcut  and the directed path from q as an inner vertex). Hence, q has exactly one child B and, since N is phylogenetic, at least one child that is not in B. Consider an arbitrary such child B. Since all vertices of B lie on a directed path and N is level-1. The vertices N is a level-1 network whose non-trivial blocks correspond to undirected cycles, N contains a non-trivial block q to B and q and N. Since they are q of the tree vertex q is a N correspond to undirected cycles, we have q in N), a contradiction to the construction of mentclass2pt{minimCase (b)N and X and thus N is the unique phylogenetic quasi-binary level-1 network whose cluster multiset is In summary, therefore, N. It utilizes the concepts of cherries and reticulated cherries that have been used extensively in the literature, see, e.g., Bordewich and Semple  are between hierarchies and weak hierarchies (Bertrand and Diatta


Scientific Reports 10.1038/s41598-023-44957-z, published online 25 October 2023Correction to: The original HTML version of this Article contained a display error in Equation 2.This error has now been corrected in the HTML version of the Article; the PDF version was correct from the time of publication.


X is called absolutely -closed if for any homomorphism h[X] is closed in Y. Let X is absolutely X is absolutely X is chain-finite, bounded, group-finite and Clifford + finite. On the other hand, a commutative semigroup X is absolutely X is finite. Also, for a given absolutely X we detect absolutely X.Let  X is complete if and only if X is closed in any metric space containing X as a subspace. A uniform space X is complete if and only if X is closed in any uniform space containing X as a uniform subspace. A topological group G is Raĭkov complete if and only if it is closed in any topological group containing G as a subgroup.In many cases, completeness properties of various objects of General Topology or Topological Algebra can be characterized externally as closedness in ambient objects. For example, a metric space On the other hand, for topological semigroups there are no reasonable notions of (inner) completeness. Nonetheless we can define many completeness properties of semigroups via their closedness in ambient topological semigroups.topological semigroup is a topological space X endowed with a continuous associative binary operation A X is calledclosed if for any isomorphic topological embedding h[X] is closed in Y;injectively-closed if for any injective continuous homomorphism h[X] is closed in Y;absolutely-closed if for any continuous homomorphism h[X] is closed in Y.Let For any topological semigroup we have the implications:X is defined to be  closed if X endowed with the discrete topology has the corresponding closedness property.A semigroup A topological space satisfies the separation axiom zero-dimensional if it has a base of the topology consisting of clopen (= closed-and-open) sets. It is well-known (and easy to see) that every zero-dimensional We will be interested in the  Since H-closed spaces ;complete topological groups ;absolutely maximal topological semigroups ;absolutely H-closed semilattices ;h-complete topological groups ;categorically compact topological groups ;sealed topological groups .Similar complete objects in different classes have been investigated under different names. For instance:mentclass2pt{minimgated in , 45, 51.gated in , 73; \dogated in , 43, 63.gated in , 38, 50.gated in , 68, 71.gated in , 13 provX is calledchain-finite if any infinite set singular if there exists an infinite set AA is a singleton;periodic if for every bounded if there exists n-th power group-finite if every subgroup of X is finite;group-bounded if every subgroup of X is bounded;group-commutative if every subgroup of X is commutative.The following theorem (proved in  is a discrete subspace of Y;topologizable if X is not projectively -discrete if for every homomorphism h[X] is a discrete subspace of Y.Let The study of topologizable and nontopologizable semigroups is a classical topic in Topological Algebra that traces its history back to Markov’s problem  of topolIn Propositions 3.2 and 3.3 of  the follX is X is injectively X is projectively absolutely (Banakh–Bardyla) A semigroup The following two theorems characterizing absolutely X the following conditions are equivalent: X is absolutely X is projectively X is projectively X is finite.For a commutative semigroup X the following conditions are equivalent: X is absolutely X is ideally X is ideally X is chain-finite, bounded, group-finite and Clifford + finite.Let Theorems Let G without elements of order 2 is a subgroup of an absolutely X. If the group G has infinite center, then G is not injectively G is a subgroup of the absolutely X. This example also shows that the equivalences Corollary X letcenter of X, andideal center of X. Every commutative semigroup X has For a semigroup The following theorem proved in , Sect. 5X be a semigroup. X is Z(X) is chain-finite, periodic and nonsingular.If X is Z(X) is group-finite.If X is ideally Z(X) is group-bounded.If (Banakh–Bardyla) Let In  it was pX, its ideal center (Banakh–Bardyla) Let Theorem X absolutely Let The “ideal” version of Problem X has absolutely Let Z-viable semigroups.The “non-ideal” version of Problem Xviable if for any Following Putcha and Weissglass  we call e in a semigroup X is defined to be viable if the setcoideal in X in the sense that X. By X.An idempotent X is viable if and only if each idempotent of X is viable if and only if for every X with By Theorem 3.2 of , a semigFor ideally (absolutely) e be a viable idempotent of a semigroup X and e in X. X is ideally If X is absolutely If (Banakh–Bardyla) Let X is called Z-viable if X is viable. It is clear that each viable semigroup is Z-viable. On the other hand, there exist semigroups which are not Z-viable, see Remark A semigroup A of semigroup X letB of a semigroup X is called bounded if For a subset X is ideally (Banakh–Bardyla) If a semigroup The following theorem gives a partial answer to Problem X is absolutely Z-viable), then the set Z(X) is absolutely If a semigroup For classes X be an absolutely X, one of the following conditions is satisfied: Ce of for every C is countable;Ce of Ce of Then the set X is absolutely Let The cardinal xiom. By , 7.13, tZ(X) of any E(X) endowed with the natural partial order E(X) and also on the partial order of the semilattice reflection X.By Theorem X is called a semilattice congruence if the quotient semigroup semilattice, i.e., a commutative semigroup of idempotents. The intersection X is called the smallest semilattice congruence on X and the quotient semigroup semilattice reflection of X. The smallest semilattice congruence was studied in the monographs  is discrete in Y). Since C is a prime coideal in X, the mapX to the 0-extension Y, endowed with the topology X, the image C is absolutely (resp. projectively) Assume that a semigroup e of a semigroup X is viable if the subsemigroup X. By X.We recall that an idempotent The following lemma was proved in , 2.5.X we have For any semigroup The inclusion E(X): by  or 20]  \documenX we study the order properties of the posets In this section, for any semigroup X be a semigroup and Let X. Then the map Given two viable idempotents In the following four lemmas we prove the statements of Theorem X, the posets For any ideally X, so it suffices to prove that the poset Y is not well-founded, we can find a strictly decreasing sequence Y. For every Y. Consequently, its preimage X.By Proposition X and the complement X. Consider the semigroup S with the topology P. Since X is not ideally It is easy to see that X, the posets For any projectively X onto its semilattice reflection. If the semigroup X is projectively Let X is projectively If a semigroup X onto its semilattice reflection. Consider the set H of all homomorphisms from Let Theorem X, the sets For any absolutely In this section we establish some properties of absolutely X we have implications X is finite;X is absolutely X is projectively X is projectively X is projectively Z(X) is periodic and If X is commutative, then the conditions (1)–(6) are equivalent.For any semigroup The implication X is projectively Z(X) is periodic and group-finite. Then for every Z(X). In both cases, the set To prove that X is projectively Z(X) is periodic. By Theorem To prove that X is commutative, we shall prove that Z(X) is periodic and X is viable and hence Z(X) implies that Now assuming that Z-viable semigroup is absolutely Z(X) is finite and hence If a Z-viability of X yields Z(X) is periodic and hence The In this section we shall prove a sufficient condition of the absolute (Stepp) Every chain-finite semilattice is absolutely X is called E-commutative if A semigroup E-commutative semigroup X is absolutely Each chain-finite group-finite bounded Clifford + finite X is absolutely Y. We should prove that the semigroup h[X] is closed in Y. Replacing Y by h[X] is dense in Y. Since X is bounded, there exists h is a homomorphism, we conclude that Y contains the dense set h[X] and hence coincides with Y. Therefore, To show that X is a chain-finite E-commutative semigroup, the set E(X) is a chain-finite subsemilattice of X. By Theorem E(X) is absolutely h[E(X)] is closed in the Hausdorff topological semigroup Y. The continuity of the map n implies that X is Clifford + finite, the set Y, the set Y and contains the set Since h[X] is not closed in Y, take any point E(X) is chain-finite, we can apply Theorem E(Y). By Theorem L is closed in E(Y). Then its complement E(Y) and its preimage y in Y. Since E(X) is chain-finite, the nonempty subsemilattice X is group-finite, the maximal subgroup y in Y such that Assuming that h[X] is closed in Y. SinceX, we characterize subsets X. We recall that a subset bounded if In this section, given an absolutely The following notion plays a crucial role in our subsequent results.X is defined to be A-centrobounded over a set A semigroup antichain if In the following theorem we endow the set X be an absolutely X;X;X is B-centrobounded over every countable infinite antichain Let X by its 1-extension X contains a two-sided unit 1. By Theorem Z(X) is chain-finite, periodic, nonsingular, and group-finite.Replacing the semigroup n witnesses that X is B-centrobounded over any set Z(X) and the choice of n, we haveThe equivalence It remains to prove the implication There exists a sequence X, for every u and hence the set Z(X) is periodic and group-finite). By the Pigeonhole Principle, for any Z(X) is chain-finite. Therefore, E(Z(X)). Write the infinite set Since the set X onto its semilattice reflection.Let Z(X) imply that the idempotent X and moreover, Let  2.15 in .X is absolutely X, the quotient semigroup X/I is absolutely Since the semigroup I is not empty and the element X/I is well-defined.For convenience, by 0 we denote the element Now we introduce a 0-discrete Hausdorff semigroup topology Q is nonempty, as X contains a unit exactly to omit the easier case Fix any free ultrafilter Q is a subsemilattice of Q. Thus, the semilattice Q contains the smallest element s. Since For any s. ThenThe inclusion Y, we need the following notations. For a real number r by r. For each A implies that Y. On the semigroup To introduce the topology To see that the topology It remains to prove that In each of these cases, So, it remains to consider three cases: (1) Assume that If (2) The case a and b do not belong to the same subgroup (3) If X, is itself Note that Y generated by the base Y. Also, since Let For any The claim is obvious if There exists X is Since m ensures that We claim that for every Y to the set T by the formula:T which satisfies the following conditions:if Let T is Hausdorff and zero-dimensional.The topology V in First we show that the topological space u is an isolated point of Y and T. So, we can take u in 1. If 2. If V is closed in t such that t is disjoint with V. If V. Finally assume that V is clopen.3. If Therefore the topology ab. We must find neighborhoods a, b such that To check the continuity of the semigroup operation in So, it remains to consider three cases:1. (1a) (1b) (1c) 2. 3. Y is a dense subsemigroup of T. Thus, X. The obtained contradiction completes the proof of the implication Observe that the continuity of the binary operation in In this section we shall find some sufficient conditions for centroboundedness, which will be combined with Theorem The following theorem is the main result of this section.X be a semigroup, X, the group X is A-centrobounded if one of the following conditions is satisfied: X is projectively Ce of h[C] is polyfinite;the semigroup X is projectively h[C] is countable;X is absolutely X is absolutely Ce of Let A is countable and hence admits an injective function ath coordinate projection. The definition of the homomorphism h implies that By our assumption, the set X is projectively Ce of C of X is projectively h[C]. By Theorem H. By Theorem h[C] is bounded. Then there exists X is A-centrobounded.1. Assume that h[C] is polyfinite. Then there exist X is A-centrobounded, it suffices to check that for any 2. Assume that the semigroup n and F, there exists a semigroup polynomial For every X is projectively h[C] is countable. By Lemma C in X is projectively h[C]. Being h[C] is zero-closed, see Lemma h[C] is polybounded and by Lemma h[C] is polyfinite. By the preceding statement, the semigroup X is A-centrobounded.3. Assume that the semigroup X is absolutely h[C] of the metric group C is absolutely h[C]. By the absolute h[C], the semigroup h[C] is zero-closed and also h[C] is closed in the zero-dimensional topological group H. Then the metric h[C] is a Polish space.4. Assume that the semigroup h[C] is polybounded. If h[C] is countable, see [h[C] is polybounded. If h[C] is polybounded by Theorem h[C] is polybounded. By Lemma h[C] is polyfinite. By the second statement of this theorem, the semigroup X is A-centrobounded.We claim that ble, see , 6.5. ByX is bounded. For a bounded subset least common multiple of numbers in the set F.5. Assume that X is not A-centrobounded. Writing down the negation of the A-centroboundedness, we obtain sequences (i)(ii)Consider the group n-th coordinate projection. Endow the group C of the absolutely X is absolutely To derive a contradiction, assume that the semigroup Recall that For every m. Let Let For every It follows from X be an absolutely Ce of for every h[C] is polyfinite;the semigroup h[C] is countable;Ce of Then the set X is absolutely Let X. By Theorem S implies that S is Clifford + finite. By Lemma S is absolutely By Theorems X is absolutely If a semigroup Z(X) is chain-finite, group-finite, periodic, and nonsingular. By Lemma X and so is its subset For every Therefore, the commutative semigroup Let X is absolutely X is absolutely Z(X) is periodic and henceGiven an absolutely If X, we shall prove the equivalence of the assertions (1)–(4) of the theorem.Given a class X.The implication The implication


Numerical simulations show that the reflection phase difference Δφuv of the polarization converter is very close to +90° in two separated frequency ranges, so the polarization converter can achieve high-efficiency and dual-band CP-to-LP polarization conversion, it can convert right-handed circular-polarized (RHCP) and left-handed circular-polarized (LHCP) waves into y-polarized and x-polarized waves respectively in the two separated frequency bands of 8.08–9.27 GHz and 13.80–27.11 GHz; moreover, its polarization conversion rate (PCR) is kept larger than 99.7% in the two frequency bands. Finally, to validate the design, a prototype is manufactured and measured, the measured results are in good agreement with the simulated ones.In this paper, to achieve circular-to-linear polarization conversion, a novel polarization converter based on an anisotropic metasurface is proposed. Because the polarization converter is an orthotropic anisotropic structure with a pair of mutually perpendicular symmetric axes  The polarization of electromagnetic (EM) wave refers to the characteristics of the direction and amplitude of the electric field in the EM wave changing with time. For various planar EM waves, the spatial trajectory of the endpoint of the time-varying electric field vector at any spatial-point may be a straight line, a circle or an ellipse, so the polarization of EM wave can be divided into the following three basic types: linear polarization (LP), circular polarization (CP) and elliptical polarization (EM). Different polarized EM waves have different propagation characteristics, a suitable polarization should be chosen for EM waves propagating in different propagation environments, so it is of great significance to effectively modulate the polarization of various electromagnetic waves in real time. At present, the LP and CP waves are both widely used, in order to realize the interconversion of LP and CP waves, many LP-to-CP polarization converter based on various metasurfaces have been proposed, these polarization converters can convert LP wave into CP wave in a certain frequency band through reflection or transmission –16; howey- and x-polarized ones respectively in two separated frequency bands. Moreover, for the CP-to-LP polarization conversion, a detailed theoretical analysis is carried out, in which a new type of CP-to-LP reflection coefficients is proposed, the theoretical analysis shows that magnitudes of the CP-to-LP reflection coefficients of the polarization converter can be completely decided by the reflection phase difference Δφuv under u-polarized and v-polarized incidences; if the reflection phase difference Δφuv can be close to ±90°, the CP-to-LP polarization conversion can be realized.In this work, a novel CP-to-LP polarization converter is proposed based on an anisotropic metasurface, the polarization converter can achieve high-efficiency and dual-band CP-to-LP polarization conversion under both right-handed circular-polarized (RHCP) and left-handed circular-polarized (LHCP) incidences, the RHCP and LHCP incident waves will be converted to u and v along ±45° directions with respect to the vertical y axis. In addition, the geometrical parameters of the unit cell structure are shown in P = 8.00 mm, a = 2.744 mm, b = 2.616 mm, d1 = 0.25mm, w1 = 0.20mm, d2 = 0.20 mm, w2 = 0.20 mm and h1 = 2.50mm and h2 = 2.90 mm; moreover, the two dielectric layers are both chosen as a Polytetrafluoroethylene (PTFE) with a relative permittivity of εr = 1.80 and a loss tangent of tan δ = 0.0008.The proposed dual-band CP-to-LP polarization converter is based on a reflective metasurface, which is a two-dimensional periodic structure, one of its unit cells is shown in u and v axes, so no cross-polarized reflection components will exist when it is under the incidence of u- and v-polarized waves, and its reflection matrix Rlin in the U-V coordinate system can be expressed as follows:The proposed CP-to-LP polarization converter is symmetrical with respect to both When the incident wave of the CP-to-LP polarization converter is assumed as a RHCP or LHCP one, the incident wave can be expressed as:Rlin, the total reflected wave under the RHCP and LHCP incidences can be expressed as:Thus, based on the above reflection matrix u and v are along ±45° directions with respect to the y-axis, as shown in y- and x-polarized unit waves can be expressed as For the symmetric axes ruu and rvv in the reflection matrix Rlin will both be very close to 1.0, but the phases will be different, thus, when we regard the polarization converter as a lossless structure, the following equation is established:φuv = Arg(ruu) − Arg(rvv) ≤ +180°. In this way, the CP-to-LP reflection coefficients in In addition, the CP-to-LP polarization converter can be regarded as a lossless structure because the loss tangent of its two dielectric layers is only 0.0008, the magnitudes of According to φuv is equal to +90°, the magnitude of ry−RHCP and rx−LHCP will be equal to 1.0, the reflected wave under RHCP incidence will be converted to a y-polarized one, and the reflected wave under LHCP incidence will be a x-polarized one; in addition, if Δφuv = −90°, the magnitude of rx−RHCP and ry−LHCP will be 1.0, the reflected waves under RHCP and LHCP incidences will be converted to x-polarized and y-polarized ones, respectively. However, if Δφuv ≠ ±90°, the prefect CP-to-LP polarization conversion can’t be achieved, the reflected wave will be an elliptically polarized one, ry−RHCP(rx−LHCP) and rx−RHCP(ry−LHCP) is a pure imaginary number, which means that the phase difference between them will be just ±90°, thus, the AR of the reflected wave can be expressed as:u- and v-polarized incidences, the simulation result, the phase difference Δφuv between ruu and rvv, is shown in φuv is very close to +90° in the two frequency ranges of 8–9.5 GHz and 13.5–27.5 GHz. Thus, according to the above theoretical prediction, it is known that the polarization converter can achieve CP-to-LP polarization conversion in the two separated frequency ranges, and its reflected waves under RHCP and LHCP incidences will be converted to y-polarized and x-polarized ones, respectively. To show this, according to Δφuv in rx−RHCP| and |ry−LHCP| is kept less than –25 dB and the AR is always larger than 25.0 dB in the two separated frequency bands of 8.08–9.27 GHz and 13.80–27.11 GHz, which shows that the proposed CP-to-LP polarization converter can achieve dual-band CP-to-LP polarization conversion in the two frequency bands. In addition, for To numerically investigate the performance of the proposed CP-to-LP polarization converter, we have carried out numerous numerical simulations using Ansoft HFSS. Firstly, the polarization converter was simulated under Based on the CP-to-LP reflection coefficients and AR shown in Furthermore, to validate the above theoretical prediction, the proposed dual-band CP-to-LP polarization converter has been simulated directly under RHCP and LHCP incidences. The simulation was performed using the commercial software Ansoft HFSS. In the simulation, when the excitation was assumed as a CP plane incident wave, the commonly used Floquet Port could not be available, the excitation must be changed as an incident wave. In this way, in order to analyze the performance of the polarization converter in a certain frequency band, these frequencies must be swept in a discrete manner, thus, the amount of simulation calculation was increased by a little, however, the AR of the reflected wave under the CP plane-wave incidence was directly obtained through far-field report. The simulated results are presented in u- and v-polarized incidences because no ultra-wideband CP antenna exists in common microwave experimental equipment. The measured result, the phase difference Δφuv between ruu and rvv, is shown in φuv is close to +90° in the two frequency ranges of 8–9.5 GHz and 13.5–27.5 GHz, which reasonably agrees with the above simulated results in φuv, the calculated magnitudes of the CP-to-LP reflection coefficients, together with the calculated AR of the reflected wave, are basically the same with the above predicted values, as shown in In order to experimentally validate the performance of the proposed CP-to-LP polarization converter, an experimental sample was fabricated and measured. The grounded dielectric substrate of the experimental sample is shown in u and v, and the reflection phase difference Δφuv under u- and v-polarized incidences is very close to +90° in two separated frequency ranges, the polarization converter can convert RHCP wave into y-polarized one and convert LHCP wave into x-polarized one with a PCR greater than 99.7% in the two separated frequency bands of 8.08–9.27 GHz and 13.80–27.11 GHz. The performance of the CP-to-LP polarization converter has been verified by both simulations and experiment. In summary, the proposed polarization converter can achieve CP-to-LP polarization conversion, moreover, it is not only dual band, but also highly efficient, so it is of great application value to modulate the polarization of various EM waves.In this paper, a high-efficiency and dual-band CP-to-LP polarization converter was proposed based on an anisotropic metasurface. Because the proposed polarization converter is an orthotropic anisotropic structure with a pair of mutually perpendicular symmetric axes S1 Dataset(RAR)Click here for additional data file.


Nature Communications 10.1038/s41467-023-43000-z, published online 11 November 2023Correction to: The original version of this Article contained an error in the Abstract. The correct version states “untapped reservoir” in place of “untapped reservo)ir”. The original version of this Article also contained an error in the Methods section. The correct version states “SNV, Pangenome analysis of SMAG” in place of “SNV, Pangenome analysis of SAMG”. This has been corrected in both the PDF and HTML versions of the Article.


For genotype and haplotype inference, typically, sequencing reads aligned to a reference genome are used. The alignments identify the genomic origin of the reads and help to infer the absence or presence of sequence variants in the genome. Since long sequencing reads often come with high rates of systematic sequencing errors, single nucleotides in the reads are not always correctly aligned to the reference genome, which can thus lead to wrong conclusions about the allele carried by a sequencing read at the variant site. Thus, allele detection is not a trivial task, especially for single-nucleotide polymorphisms and indels.k-merald, improves allele detection accuracy leading to better genotyping performance as compared to the existing WhatsHap implementation using edit-distance-based allele detection, with a decrease of 18% and 24% in error rate for high-coverage Oxford Nanopore and PacBio CLR sequencing reads for sample HG002, respectively. We additionally observed a prominent improvement in genotyping performance for sequencing data with low coverage. For 3To learn the characteristics of sequencing errors, we introduce a method to create an error model in non-variant regions of the genome. This information is later used to distinguish sequencing errors from alternative alleles in variant regions. We show that our method, https://github.com/whatshap/whatshap. Genotyping is a process used for detecting the genotypes of an individual, which further helps in the detection of haplotypes, a task termed as phasing. These processes are widely used in studying the genetic aspects of different diseases and genetic relationships among species. Both genotyping and phasing typically use the alignment between sequencing reads and a reference genome. Thus, prior to genotyping, it is important to determine for each read, whether it carries the reference allele “0” or alternative allele “1” at each of the variant positions it overlaps. Most commonly, short sequencing reads from second-generation sequencing technologies, e.g. Illumina, are used for this purpose because long reads obtained using third-generation sequencing technologies, e.g. Oxford Nanopore technology (ONT) and Pacific BioSciences (PacBio), tend to be more prone to sequencing errors . These k-mer pairs include cases where the two k-mers match, indicating an error-free position, or where they mismatch, indicating a sequencing error. The counts of k-mer pairs are then used to determine the probability for observing each reference-read k-mer pair across the whole genome. We introduce a new approach for global sequence alignment in k-mer space. The read, reference, and alternative sequences in each variant window (excluded during the training phase) are split into k-mers and the strings of k-mers are then aligned. Instead of using a fixed cost value, k-mer mismatches are penalized using the learned error model, i.e. k-mer mismatches that represent common sequencing errors can be allowed in the alignment at a low cost. The sequencing read is then assigned to the allele with the lowest alignment cost. k-merald has been incorporated into the existing WhatsHap implementation and is available as an alternative to the edit-distance-based allele detection.In this study, we propose a new approach, F be the reference sequence excluding all variant windows, where each variant window, v and a flanking region of a fixed number of w base pairs on each side. The training data consists of F and the set of sequencing reads aligned to it, D. Suppose f denotes a k-mer belonging to F, while, d denotes a k-mer belonging to a sequencing read from D. During model training, described in F is traversed from left to right while maintaining, for each f, the count of each mapping d using the mapping positions from the input read alignments. For extracting the reference-read k-mer combinations k-mer AC-GTCT is aligned to the reference k-mer ACTGTCT, the respective * is the nucleotide following ACGTCT in the read sequence. These counts of k-mer combinations k-mer, to obtain a unique matrix M with reference k-mers f shown in columns (j) and read k-mers d represented in rows (i). An entry k-mer k-mer F. Although there are k-mer of length k, many of these combinations are not observed. The F are each given a pseudocount value k-mer combinations can theoretically exist, but have a low probability of occurrence based on our training data. For each reference k-mer f, we define k-mers d aligned to f, i.e. the pair M. The sum of individual counts over all of these pairs is denoted by M, is then converted into a matrix P, storing the probability of observing each possible reference-read k-mer pair k-mer combination As input, we expect a list of candidate variants. In the first step, as shown in learn” module in WhatsHap.Algorithm 1. Model trainingInput:R1: The complete reference sequence, B2: Aligned sequencing reads, V3: Variants for which allele detection is to be performed, Output: M4: 5: whiledo6: ifthen7:  8:   do nothingelse ifthen9:  10:   11:   else12:  13:   fordo14:   R[i]15:    16:    end for17:   end if18:  19:  end while20: return M21: Problem 1V be a set of all variant positions across the reference genome, let v. Determine b at position v.(Allele detection). Let Definition 1b sequence from the window around variant v and let v and v. If x and y, thenb at position v.. Let P, which represents our model of sequencing errors, to define an alignment cost k-mer-based comparisons  haplotagging, the process to label each read with a haplotype of origin, (ii) genotyping, and (iii) phasing.In summary, by design, the algorithm ensures that a read carrying a sequencing error aligns to the reference with a cost lower than to the alternative allele, thus minimizing the risk of a sequencing error being mistaken for a variant allele. The read is assigned the allele resulting in lowest alignment cost. However, equal alignment costs indicate that the algorithm was unable to make an allele detection based on the alignment. In case of multiallelic variants, the alignment is performed using each alternative sequence. Algorithm 2.k-mer alignmentInput:k-mers from the target sequence, 1: List of k-mers from the query sequence, 2: List of 3: Gap penalty, P4: Output: Optimal cost for aligning fordo5: 6:  end for7: fordo8: 9:  end for10: fordo11: fordo12:  13:   14:   15:   16:   end for17:  end for18: return19: k = 7 and w = 25. The error rate for each reference k-mer represents the sum of probabilities of observing each k-mer pair k-mers for ONT, PacBio CLR, and PacBio HiFi, each, reveals that the nature of erroneous k-mers also differs across the sequencing technologies  samples, HG001 (NA12878) and HG002 (NA24385). We used k-mer value of k = 7, variant window w = 25 and k is shown in k-merald for allele detection shows an improved performance in comparison to WhatsHap’s genotyping results based on the conventional edit-distance-based allele detection approach. For 54F1 score values also depict this improvement  and homoploymers  above 200 are more strongly enriched for correct genotypes when using Our training procedure exploits the similarity of a sequenced sample and the reference genome by using variant-free regions for training. In this way, our model can be readily retrained even on a single dataset, which potentially allows it to adapt to subtle differences such as version of the sequencing chemistry and other batch effects. Because the learning procedure is technology agnostic, we anticipate that our method can readily be applied to future long read data types.vbad149_Supplementary_DataClick here for additional data file.