text:
\section{Introduction}
\label{intro}

An improved understanding of the complex processes that drive evolution
has lent support to the idea that reticulate evolutionary events such as
lateral gene transfer or hybridization are more common than originally
thought rendering a phylogenetic tree
(essentially a rooted leaf labelled graph-theoretical tree)
too simplistic a model to fully understand the complex processes
that drive evolution.
Reflecting this, phylogenetic networks have now
joined phylogenetic trees in the center of phylogenetics research.
Influenced by the diversity of questions posed by evolutionary
biologists that can be addressed with a phylogenetic networks, various
alternative definitions of these types of networks have been developed
over the years~\cite{HusonBryant2006}. These include split networks
\cite{BryantMoulton2004,BFSR1995,HHML2004}
as well as ancestral recombination graphs \cite{SongHein2005}, TOM
networks \cite{Willson2006a},
level-$k$ networks\footnote{Note that
these networks were originally introduced in \cite{JanssonSung2004b},
but the definition commonly used now is slightly different with the
main difference being that every vertex of the network with indegree
2 must have outdegree 1 (see e.g.
\cite{IKKS2008} and the references therein).}
with $k$ a non-negative integer that
in a some sense captures how complex the network structure is,
networks for studying the evolution of polyploid organisms
\cite{HuberMoulton2006}, tree-child and tree-sibling
networks \cite{CLRV2008b}, to name just a few.

Apart from split networks which aim
to give an implicit model of evolution and are not the focus of this
note, all other phylogenetic networks mentioned above aim to provide an
explicit model of evolution. Although slightly different in detail, they
are all based on the concept of a leaf-labelled
rooted connected directed acyclic graph (see
the next section for a definition).
For the convenience of the reader, we depict an
example of a phylogenetic network in the form of a
level-$1$ network in Fig.~\ref{FigLevel1}(a).
\begin{figure}[!hgt]
\centering
\begin{tabular}{ccc}
(a) & (b) & (c) \\
\includegraphics[scale=0.45]{Level1Network.eps} &
\includegraphics[scale=0.45]{Level1NetworkTree1.eps} &
\includegraphics[scale=0.45]{Level1NetworkTree2.eps}\\
\end{tabular}
\caption{(a) A level-$1$ phylogenetic network $N$. (b)
and (c) The  phylogenetic trees that form the tree system
$\mathcal T(N)$.
}
\label{FigLevel1}
\end{figure}
Concerning these types of phylogenetic networks, it should be noted that
they are closely related to
{\em galled trees}~\cite{WZZ2001,GEL2003} and that, in addition to constituting
the first layer of the rich hierarchy of level-$k$ networks, they also
form a subclass of the large class of
tree-sibling networks \cite{AVP2008}.

Due to the rich combinatorial structure of phylogenetic networks,
different combinatorial objects have been used to reconstruct them from
biological data. For a set $X$ of taxa (e.g. species or organisms), these
include {\em cluster systems} of $X$,
that is, collections of subsets
of $X$ \cite{BandeltDress1989,HusonRupp2008},
 {\em triplet systems} on $X$, that
is, collections of phylogenetic trees with just three leaves
which are generally called {\em (rooted) triplets}~\cite{JanssonSung2004b,ToHabib2009},
and {\em tree systems}, that is,
collections of phylogenetic trees which all have leaf
set $X$~\cite{Semple2007}.
The underlying rational being that any phylogenetic network
$N$ induces a cluster system $\mathcal C(N)$, a triplet
system $\mathcal R(N)$ and a tree system
$\mathcal T(N)$. Again we defer the precise definitions to
later sections of this note and  remark that for the
level-$1$ network $N$ with leaf set $X=\{a,b\ldots,e\}$
depicted in Fig.~\ref{FigLevel1}(a),
the cluster system $\mathcal C(N) $ consists of $X$,
the five singleton sets of $X$, and the subsets
$\{a,b\}, \{c,d\}, \{b,c,d\}, Y:=\{a,b,c,d\}$, and
the tree system $\mathcal T(N)$ consists of the
phylogentic trees depicted in
Fig.~\ref{FigLevel1}(b) and (c), respectively. Denoting a
phylogenetic tree $t$ on $x,y,z$ such that the root of
$t$ is not the parent vertex of $x$ and $y$ by $z|xy$ (or equivalently
by $xy|z$)
then the triplet system $\mathcal R(N)$ consists
of all triplets of the from $e|xy$ where $x,y\in Y$ distinct,
$x|cd$ with $x\in \{a,b\}$, and
$x|ab$ and $a|bx$ with
$x\in \{c,d\}$.

Although undoubtedly highly relevant for
phylogenetic network reconstruction,
the following fundamental question has however remained
largely unanswered so far (the main exception being the case
when $N$ is in fact a phylogenetic tree in which
case this question is closely related to the well-studied
{\em perfect phylogeny problem} -- see e.g.~\cite{GruenewaldHuber2007} for a recent overview.):
When do the systems
$\mathcal C(N)$, $\mathcal R(N)$, or $\mathcal T(N)$ induced by
a phylogenetic network $N$ {\em encode} $N$, that is, there is no
other phylogenetic network $N'$ for which the corresponding
systems for $N$ and $N'$ coincide?

Complementing the insights for when $N$ is a phylogenetic tree alluded to above,
 answers were recently provided for $\mathcal R(N)$
in case $N$ is a very special type of level-$k$ network, $k\geq 2$,
\cite{IKM2009} and for $\mathcal T(N)$ for the special case that $N$
is a regular network \cite{Willson2009}.
Undoubtedly important first results, there are many types of phylogenetic
networks which are encoded by the tree system
they induce but which are not regular or by the triplet system
they induce but do not belong to  that special  class of
level-$2$ networks. An example for both cases is the level-$1$ network
depicted in Fig.~\ref{FigLevel1}(a). Although one might be tempted to
speculate that all level-$1$ networks enjoy this property, this is not the
case since the level-$1$ networks depicted in Fig.~\ref{FigLevel1}(a) and
Fig.~\ref{Figminimal}(b),
respectively, induce the same tree system and the same triplet system.
The main result of this paper shows that
these observations are not a coincidence.
More precisely, in Theorem~\ref{theorem-montpellier} we
 establish that a level-$1$ network $N$
is encoded by the triplet system $\mathcal R(N)$ (or equivalently
by the tree system $\mathcal T(N)$ or equivalently
the  cluster system $\mathcal S(N)=\mathcal S({\mathcal T}(N))
:=\bigcup_{T\in \mathcal T(N)}\mathcal C(T)$
which arises in the context of the {\em softwired interpretation}
of $N$ \cite{HusonRupp2008} and contains $\mathcal C(N)$)
if and only if, when ignoring directions, $N$ does not
contain a cycle of length 4. Consequently the number
of non-isomorphic (see below) phylogenetic networks $N'$ which all
induce the same tree system (or equivalently
 the same triplet system or the same cluster system $\mathcal S(N)$)
 grows exponentially in the number of cycles of $N$ of length 4.
It is of course highly tempting to speculate that a similar
characterization might hold for higher values of $k$.
However as our examples
show, establishing such a result will require an alternative approach since
 our arguments cannot be extended to level-$2$ networks
and thus to level-$k$ networks with $k\geq 2$.

This note is organized as follows. In the next section, we present
the definition of a level-$1$ network plus surrounding terminology.
In Section~\ref{clusterEncoding}, we
present the definitions of the cluster system $\mathcal C(N)$ and
the tree system $\mathcal T(N)$ induced by a phylogenetic network $N$.
This also completes the definition of the cluster system $ \mathcal S(N)$
given in the introduction.
Subsequent to this, we show that for any
level-$1$ network $N$, the cluster systems $ \mathcal S(N)$
and $\mathcal C(N)$ are
{\em weak hierarchies} (Proposition~\ref{softwired-weak-hierarchy})
which are well-known in cluster analysis.
In addition, we show that this property is not enjoyed by
level-$2$ networks and thus level-$k$ networks with $k\geq 2$.
In Section~\ref{Simple-level-$1$}, we first present the definition of the
triplet system $\mathcal R(N)$ induced by phylogenetic network
$N$. Subsequent to this, we turn our attention to
the special case of encodings of
simple level-$1$ networks. In Section~\ref{alternativeEncoding},
we present our main result (Theorem~\ref{theorem-montpellier}).

To ease the presentation of our results,
in all figures the (unique) root of a
network is the top vertex and all arcs are directed downwards,
away from the root. Furthermore, for any directed graph $G$, we denote the
vertex set of $G$ by $V(G)$ and the set of arcs of $G$ by  $A(G)$.

\section{Basic terminology and results concerning level-$1$ networks}
\label{prelim}

In this section we present the definitions
of a phylogenetic network and of a level-$k$ network, $k\geq 0$.
In addition we also provide  the basic terminology
surrounding these structures.

Suppose $X$ is a finite set.
A {\em  phylogenetic network} $N$ on $X$ is a rooted
directed acyclic graph (DAG) that satisfies the following additional
properties. (i) The set $L(N)$ of {\em leaves}
of $N$, that is vertices with indegee
1 and outdegree 0, is $X$.
(ii) Exactly one vertex of $N$, called the
{\em root}  and denoted by $\rho_N$, has indegree 0 and outdegree 2.
(iii) All vertices of $N$ that are not contained in $L(N)\cup \{\rho_N\}$
are either {\em split vertices}, that is, have indegree 1 and outdegree 2
or {\em reticulation vertices}, that is, have indegree 2 and outdegree 1.
The set of reticulation vertices of $N$ is denoted by $R(N)$.
A phylogenetic network $N$ for which
$R(N)$ is empty is called a {\em (rooted) phylogenetic tree (on $X$)}.
Two phylogenetic networks $N$ and $N'$ which both have
leaf set $X$ are said to be {\em isomorphic} if there exists a bijection
from $V(N)$  to $V(N')$  which is the identity on $X$ and
induces a graph isomorphism between $N$ and $N'$.

To present the definition of a level-$k$ network, we need to
introduce some terminology concerning rooted DAGs first.
Suppose $G$ is a rooted connected DAG with at least 2 vertices.
Then we denote the graph obtained from
$G$ by ignoring the directions on $G$ by $U(G)$. If $H$ is a graph with
at least 2 vertices  then we call
$H$  \emph{biconnected} if $H$ does not contain a vertex
whose removal disconnects it. A {\em biconnected component} of $H$
is a maximal subgraph of $H$ that is biconnected. If $G$ is a phylogenetic
network and $B$ is a rooted sub-DAG such that $U(B)$
is a biconnected component of $U(G)$ then we call $B$ a {\em blob}.

Following~\cite{IKKS2008}, we call a phylogenetic network $N$ a
{\em level-k} network for some non-negative integer $k$
if each blob of $N$ contains at most
$k$ reticulation vertices. Note that some authors
define a level-$1$ network $N$ to be a phylogenetic network
without the above outdegree requirement on the reticulation vertices
of $N$ (see e.g.~\cite{JanssonSung2004b}). Also and sometimes on its own
or in addition to the above, the requirement
that each blob contains at most $k$ reticulation vertices is
sometimes replaced by the requirement that the cycles in $U(N)$ are node
disjoint (see e.g.~\cite{JanssonSung2004b,JanssonSung2006}).
Although in spirit the same definitions, the difference is that a cycle is generally
understood to have at least three vertices which implies that the
network depicted in Fig~\ref{Figminimal}(a) would not be a
level-$1$ network. However
the definition presented in~\cite{IKKS2008} would render
that network a level-$1$ network. Having said that,
the network $N$ depicted in Fig.~\ref{Figminimal}(b) is
a less parsimonious representation of the same biological information
(expressed in  terms of the systems $\mathcal T(N)$,
$\mathcal R(N)$, $\mathcal C(N)$, and $\mathcal S(N)$)
as the level-$1$ network in  Fig.~\ref{Figminimal}(a) in the sense that
the edges in grey are redundant for displaying that information. To avoid
these types of level-$1$ networks which cannot be encoded by any of the
4 systems of interest in this note, we follow~\cite{IKM2009}
and require that every blob in a level-$1$ network
contains at least 4 vertices.
\begin{figure}[!hgt]
\centering
\begin{tabular}{ccc}
(a) & (b)  \\
\includegraphics[scale=0.45]{Level1NetworkNonMinimal.eps} &
\includegraphics[scale=0.45]{Level1NetworkOther.eps} &
\end{tabular}
\caption{
The level-$1$ network $N$ depicted in (a) induces and thus
represents the same triplet system $\mathcal R(N)$, cluster systems
$\mathcal C(N)$ and $\mathcal S(N)$, and tree system $\mathcal T(N)$
as the level-$1$ network $N'$ presented in (b). However,  $N'$ is
a less parsimonious representation of these 4 systems.
}
\label{Figminimal}
\end{figure}

For $k=1,2$, it was shown in~\cite{IKKS2008} (see also~\cite{JanssonSung2006} for the
case $k=1$) that level-$k$ networks can be built up by chaining together
structurally very simple level-$k$ networks called {\em simple level-$k$ networks}.
Defined for general non-negative integers $k$, these atomic building blocks
are precisely those level-$k$ networks that can
be obtained from a {\em level-$k$ generator} by
applying a certain ``leaf hanging'' operation \cite{IKKS2008} to its
``sides''. Such a generator is a biconnected directed acyclic multi-graph
which has a single root, precisely $k$ {\em pseudo-reticulation
vertices} (i.\,e.\,vertices with indegree 2 and outdegree
at most 1) and all other vertices are split vertices where the
root and a split vertex are defined as in the case
of a phylogenetic network. For the convenience of the reader,
we present in Fig~\ref{level-k-generators} the
unique level-$1$ generator and all 4 level-$2$ generators
which originally appeared in slightly different form
in~\cite{IKKS2008}.
\begin{figure}[!hgt]
\centering
\begin{tabular}{cccccc}
\includegraphics[scale=0.5]{FigGenerator1.eps}
& ~ ~ ~
& \includegraphics[scale=0.5]{FigGenerator8a.eps}
& \includegraphics[scale=0.5]{FigGenerator8b.eps}
& \includegraphics[scale=0.5]{FigGenerator8c.eps}
& \includegraphics[scale=0.5]{FigGenerator8d.eps}\\
$\mathcal{G}^1$ & & $\mathcal{G}^2_a$ & $\mathcal{G}^2_b$ &
$\mathcal{G}^2_c$ & $\mathcal{G}^2_d$
\end{tabular}
\caption[level-$1$ and level-$2$ generators]{The unique
level-$1$ generator $\mathcal{G}^1$,
and the four level-2 generators:
$\mathcal{G}^2_a$, $\mathcal{G}^2_b$, $\mathcal{G}^2_c$ and $\mathcal{G}^2_d$.}
\label{level-k-generators}
\end{figure}
Regarding larger values for $k$, it was recently shown
in~\cite{L3Generators}
that there exist $65$ level-3 generators. In addition, it
was shown in~\cite{GBP2009}
that there are $1993$ level-4 generators and that the number
of level-$k$ generators grows exponentially in $k$.
A {\em side} of a generator $G$ is an arc of $G$ or one of its
pseudo-reticulation vertices.

From now on and unless stated otherwise, all phylogenetic
networks have leaf set $X$.

\section{The Systems $\mathcal C(N)$, $\mathcal T(N)$, and
$\mathcal S(N)$ }\label{clusterEncoding}

In this section, we introduce for a phylogenetic
network $N$ the associated systems $\mathcal C(N)$,
$\mathcal T(N)$, and $\mathcal S(N)$ already mentioned in the
introduction. In addition, we prove that
in case $N$ is a level-$1$ network the associated
systems $\mathcal C(N)$ and $\mathcal S(N)$ are weak hierarchies.
We conclude with presenting an example
that shows that level-$k$ networks, $k\geq 2$, do not enjoy this
property in general. We start with some definitions.

Suppose $N$ is a phylogenetic network.
Then we say that a vertex $a\in V(N)$ is {\em below}
a vertex $b\in V(N)$  denoted by $a \preceq_N b$, if there
exists a path $P_{ba}$
(possibly of length 0) from $b$ to $a$.  In this case, we also say
that $b$ is {\em above $a$}. Every vertex $v\in V(N)$
therefore induces a non-empty subset $C(v) =C_N(v)$ of $ X$
which comprises of all leaves of $N$ below $v$ (see e.g.
\cite{SempleSteel2003}).
We collect the subsets $C(v)$ induced by the vertices $v$ of $N$
this way in the set $\mathcal C(N)$, i.e. we put
$\mathcal C(N)=\bigcup_{v\in V(N)}\{C(v)\} $.
For convenience, we refer to any collection $\mathcal C$ of
non-empty subsets of
$X$ as a {\em cluster system (on $X$)} and to the elements of $\mathcal C$
as {\em clusters} of $X$.
It should be noted that in case $N$ is a binary phylogenetic tree,
the cluster system $\mathcal C(N)$ is a {\em hierarchy (on $X$)},
that is, for any two clusters $C_1,C_2\in \mathcal C(N) $
we have that $C_1\cap C_2\in \{\emptyset, C_1,C_2\}$. Hierarchies
are sometimes also called {\em laminar families}, and it is well-known
that the set of clusters $\mathcal C(T)$
induced by a binary phylogenetic tree $T$
uniquely determines that tree (see e.g. \cite{SempleSteel2003}).

In the context of phylogenetic network construction, the concept
of a {\em weak hierarchy (on $X$)} was introduced
in \cite{BandeltDress1989}. These objects are defined as follows.
Suppose $\mathcal C$ is a cluster system on $X$. Then
$\mathcal C$ is called a {\em weak hierarchy (on $X$)} if
\begin{eqnarray}\label{helly}
C_1\cap C_2\cap C_3\in \{C_1\cap C_2, C_2\cap C_3, C_1\cap C_3\}
\end{eqnarray}
holds for any three
elements $C_1,C_2, C_3\in {\mathcal C}$.
Note that the above property is sometimes
also called the {\em weak Helly property} \cite{SempleSteel2003}.
Also note that any hierarchy is in particular a weak hierarchy and that
any subset of a weak hierarchy is again a weak hierarchy. Finally note
that weak hierarchies are
well-known objects in classical hypergraph and abstract
convexity theories \cite{BandeltDress1989}
(see also the reference therein and \cite{BBO04}), and that they
where originally introduced into cluster analysis as {\em medinclus}
in \cite{B88,B89}.

We will establish the main result of this
section (Proposition~\ref{softwired-weak-hierarchy}) by showing
that the cluster system $\mathcal S(N)$ associated to a
level-$1$ network $N$ is a weak hierarchy.
To do this, we first
need to complete the definition of $\mathcal S(N)$
which relies on the definition of the system $\mathcal T(N)$.
We will do this next.

Suppose $N$ is a phylogenetic network. Then  we say that
a phylogenetic tree $T$ is {\em displayed} by $N$ if
the leaf set of $T$ is $X$ and $T$ is a phylogenetic
tree  obtained from $N$ via the following process.
For each reticulation vertex of $N$ delete
one incoming arc and suppress any resulting
degree 2 vertices. In case the root $\rho_N$ of $N$ is rendered
a vertex with out-degree 1 this way, we  identify $\rho_N$ with its unique
child. The set $\mathcal T(N)$ is the collection of all
phylogenetic trees that are displayed by $N$.
To every vertex $v\in V(N)$
a cluster system  $\mathcal S_N(v)$ defined by putting
\begin{align*}
\mathcal S_N(v) =  \{C_T(v): T \in \mathcal T(N)\}
\end{align*}
can be associated. Clearly, $C_N(v) \in \mathcal S_N(v) $ and
$\mathcal S(N)=\bigcup_{v\in V(N)} \mathcal S_N(v)$.

To link clusters of $X$ with level-$1$ networks on $X$,
we say that a cluster $C$ on $X$  is {\em tree-consistent with a
level-$1$ network} $N$ if $C \in \mathcal S(N)$.
More generally, we say that a cluster system $\mathcal C$ is
{\em tree-consistent with a level-$1$ network}
$N$ if $\mathcal C\subseteq \mathcal S(N)$ holds.
Thus, for any level-$1$ network $N$ the cluster system $\mathcal S(N)$
equals the set of all clusters of $X$ that are
tree-consistent with $N$.
Finally, we say that a cluster
system $\mathcal C$ is {\em level-$1$-consistent} if there
exists a level-$1$ network
$N$ such that $\mathcal C$ is tree-consistent with $N$.

We next establish Proposition~\ref{softwired-weak-hierarchy}.
Its proof relies on a characterization of a weak hierarchy $\mathcal H$
on $X$  from \cite[Lemma 1]{BandeltDress1989} in terms of a property of
a certain  $\mathcal H$-closure that can be canonically associated to
$\mathcal H$. More precisely, suppose  $\emptyset\not=Y\subseteq X$
and $\mathcal H$ is a cluster system on $X$. Then the
{\em $\mathcal H$-closure}
$\langle Y\rangle_{\mathcal H}$
of $Y$ is the intersection $\bigcap_{Y\subseteq C,\,\, C\in \mathcal H}C$.
Now a cluster system $\mathcal H$ on $X$ is a weak hierarchy if and
only if for every non-empty subset $A\subseteq X$
there exists elements $a,a'\in A$ such that
$\langle A\rangle_{\mathcal H}=\langle \{a,a'\}\rangle_{\mathcal H}$.
Note that this implies in particular
that the number of elements in a weak hierarchy is at most ${|X|+1}\choose{2}$
\cite{BandeltDress1989}. With regards to this bound it should be
noted that it was recently shown in \cite{KNTX2008}
that the size of a cluster system which is tree-consistent
with a level-$1$ network $N$ is linear in $|X|$.
In view of Proposition~\ref{softwired-weak-hierarchy}, this bound
improves on the previous  bound for this special kind of weak
hierarchy.

\begin{proposition}\label{softwired-weak-hierarchy}
A level-$1$-consistent cluster system
is a weak hierarchy. In particular, the systems  $\mathcal S(N)$
and $\mathcal C(N)$ associated to a level-$1$ network $N$ are
weak hierarchies.
\end{proposition}
\noindent{\em Proof: } Since
every subset of a weak hierarchy is again a weak hierarchy,
it suffices to show that for every level-$1$ network $N$ the
associated cluster system  $\mathcal S(N)$ is a weak hierarchy.
To see this suppose $N$ is a level-$1$ network on $X=\{x_1, \ldots, x_n\}$,
$n\geq 1$.
Consider a graphical representation of $N$ and, starting from the left most
leaf of $N$ in that representation, let $x_1\ldots x_n$ denote
the induced ordering of the leaves of $N$ (note that this might involve
re-labelling some of the elements in $X$). Suppose
$\emptyset \not=A\subseteq X$. Let $i,j\in\{1,\ldots, n\}$ be such that
$x_j\in A$  and every  leaf in
$X$ succeeding $x_j$ in that ordering is not contained in $A$. Similarly,
let $x_i\in A$ be such that every leaf in
$X$ preceding $x_i$ in that ordering is not contained in $A$. We claim
that
$\langle A\rangle_{\mathcal S(N)}=\langle \{x_i,x_j\}\rangle_{\mathcal S(N)}$.
To see this, note that since $N$ is a level-$1$ network, there exists
a subtree $T$ of $N$ such that the leaf set of $T$ is $A$. Note that
$T$ might contain vertices whose indegree and outdegree is one. By
deleting for each reticulation vertex below the root of
$T$ one of its incommming arcs and supressing the resulting degree
2 vertex $T$ can be canonically extended to a subtree $T'$ of some
tree $T''\in\mathcal T(N)$
such that $\{x_i,x_{i+1},\ldots, x_j\} \subseteq L(T')$ and
$L(T')$ is minimal with regards to set inclusion. Note that
$L(T')\in \mathcal S(N)$. But then, by construction,
$\langle A\rangle_{\mathcal S(N)}= L(T')=
\langle \{x_i,x_j\}\rangle_{\mathcal S(N)}$ which proves the claim.
\hfill\hbox{\rule{3pt}{6pt}}\\

We remark in passing that to any cluster system $\mathcal C$ of $X$
a similarity
measure $D_{\mathcal C}:X\times X\to \mathbb R$
can be associated to $\mathcal C$ by putting
$D_{\mathcal C}(a,b)=|\{C\in \mathcal C: a,b\in C\}|$, $a,b\in X$.
Proposition~\ref{softwired-weak-hierarchy}
combined with the main result from \cite{BandeltDress1989}
implies that any tree-consistent cluster system $\mathcal C$
can be uniquely reconstructed from its associated similarity measure
$D_{\mathcal C}$. Using the well-known {\em Farris transform}
(see e.\,g.\,\cite{SempleSteel2003}, and \cite{DHM2007} for a recent
overview) a similarity
measure can be canonically transformed into a distance measure
$D^{\mathcal C}$ on $X$, that is, a map on $X\times X$ into the non-negative reals
that is symmetric, satisfies the triangle inequality, and vanishes on the
main diagonal. The latter measures were recently investigated in
\cite{CJLY2005} from an algorithmical point of view in the context of
representing them in terms of an {\em ultrametric} level-$1$
network. These are generalizations of ultrametric phylogenetic trees
in the sense that every path from the root of the network
to any leaf is of the same length.

We conclude this section with remarking that as the example of
the level-$2$ network $N$ presented in
Fig.~\ref{FigLevel2WeakHierarchy}(a)  shows, the result analogous
to Proposition~\ref{softwired-weak-hierarchy} does not
hold for level-$2$ networks since
$\{\{a,b,c\},\{a,b,d\},$
$\{b,c,d\}\}\subseteq\mathcal S(N)$ but
$\{a,b,c\} \cap \{a,b,d\} \cap \{b,c,d\}=\{b\}$.
\begin{figure}[!hgt]
\centering
\begin{tabular}{cccc}
\includegraphics[scale=0.3]{Level2WeakHierarchy.eps}
& ~~~~~~
&\includegraphics[scale=0.3]{CompatibilityTriplets.eps}
&\includegraphics[scale=0.3]{CompatibilityTree.eps}\\
(a) & & (b) &(c)
\end{tabular}
\caption{(a) A level-2 network $N$ for which $\mathcal S(N)$
is not a weak hierarchy. The phylogenetic network $N$ depicted in (b) does not display the
phylogenetic tree $T$ depicted in (c) but
$\mathcal C(T)$ is tree-consistent with $N$. }
\label{FigLevel2WeakHierarchy}
\end{figure}
Furthermore, we remark that  as the example
of the level-2 network $N$ depicted in
 Fig.~\ref{FigLevel2WeakHierarchy}(b) combined with
the cluster system $\mathcal C(T)$ induced by the phylogenetic tree
$T$ depicted in Fig.~\ref{FigLevel2WeakHierarchy}(c) shows,
a cluster system $\mathcal C(T)$ induced by a phylogenetic tree $T$
can be contained in the cluster system $\mathcal S(N)$ of a level-$2$
network $N$ and $N$ need not display $T$.

\section{Simple level-$1$ Networks}\label{Simple-level-$1$}

In this section we turn our attention
to studying simple level-$1$ networks.
In particular, we establish a fundamental property of these
networks with regards to encodings of level-$1$ networks.
To do this, we require some more definitions. We start with
the definition of the triplet system $\mathcal R(N)$ induced
by a phylogenetic network $N$.

Suppose $N$ is a phylogenetic network. If $Y\subseteq X$
is a subset of $X$ of size 3, then $N$ induces a triplet $t$ on $X$
 by taking
$t$ to be a minimal subtree of $N$ with leaf set $Y$
and suppressing resulting degree two vertices of $t$.
The set of triplets induced on $X$ by $N$ this
way is the triplet system ${\mathcal R}(N)$. Two properties of this
triplet systems should be noted. First, every triplet in ${\mathcal R}(N)$
is consistent with $N$, where a triplet $x|yz$ is called
{\em consistent} with a phylogenetic network $N$ if $x,y,z\in X$
 and there exist two vertices $u,v\in V(N)$  and pairwise internally
vertex-disjoint paths in $N$ from $u$ to $y$, $u$ to $z$, $v$ to
$u$ and $v$ to $x$.
Note that a triplet system ${\mathcal R}$ is called {\em consistent}
with a phylogenetic network $N$ if every triplet in $\mathcal R$
is consistent with $N$. For convenience, we will sometimes
say that a phylogenetic network $N$ is consistent with a
triplet $t$ (or a triplet system ${\mathcal R}$) if $t$ (or ${\mathcal R}$) is  consistent
with $N$. In case ${\mathcal R}$ is consistent with a phylogenetic
network $N$ and $\mathcal R={\mathcal R}(N)$ then  we say that
$\mathcal R$ {\em reflects} $N$. Alternatively, we will say that
$\mathcal R$ is {\em reflected} by $N$. For example,
the triplet set $\mathcal R=\{a|bc, c|ab\}$
is reflected by the three simple level-$1$ networks
$SL_1^i(T)$, $i\in\{1,2,3\}$ on  $\{a,b,c\}$
depicted in Fig.~\ref{simple-level-$1$} which appeared in
slightly different form in \cite{JNS2006}.
\begin{figure}[!hbt]
\begin{center}
\begin{tabular}{ccc}
\includegraphics[scale=0.45]{SimpleLevel1a.eps} &
\includegraphics[scale=0.45]{SimpleLevel1b.eps} &
\includegraphics[scale=0.45]{SimpleLevel1c.eps} \\
$SL_1^1(T)$ & $SL_1^2(T)$ & $SL_1^3(T)$
\end{tabular}
\end{center}
\caption{\label{simple-level-$1$} The three non-isomorphic
simple level-$1$ networks
on $\{a,b,c\}$ that all reflect the triplet
system $\mathcal R=\{a|bc, c|ab\}$.
}
\end{figure}

Second, the triplet system ${\mathcal R}(N)$ is always dense,
where a triplet system ${\mathcal R}$ on $X$ is called {\em dense} if
for any three elements in $a,b,c\in X$ there exits a triplet $t\in {\mathcal R}$
such that $L(t)=\{a,b,c\}$. Arguably unassumingly looking, the
concept of a dense triplet set has proven vital for level-$k$
network reconstruction, $k\geq 1$, from triplet systems. More precisely,
 the only known polynomial time algorithms
for constructing level-$1$ and level-2 networks $N$
consistent with such triplet systems construct $N$, (if it exist)
by essentially building it up recursively from
simple level-$1$ and simple level-$2$ networks
\cite{JanssonSung2006,IKKS2008}. If the assumption
that $\mathcal R$ is dense is dropped however, then it is NP-hard
to decide if there exists a level-$k$ network, $k=1,2$, consistent with
$\mathcal R$ \cite{JanssonSung2006,IKKS2008}.
For larger values of $k$, a polynomial time algorithm
for constructing a level-$k$ network from a dense triple set
was recently presented in \cite{ToHabib2009}.

The next result is rather technical\footnote{
A case analysis based alternative proof of this
result may be found in~\cite{GBP2008}.}
but plays a crucial role in the proof of
our main result (Theorem~\ref{theorem-montpellier}) as it shows that
although all three simple level-$1$ networks depicted
in Fig.~\ref{simple-level-$1$} reflect the same triplet set this property is
lost when adding an additional leaf to a non-cut-arc of each of them.
For a directed graph $G$ these arcs are the elements in $A(G)$
whose removal disconnect $G$.  To establish our result,
we require some more definitions and
notations.

Suppose $N$ is a phylogenetic network and $a,b\in V(N)$
such that $a$ is below $b$. If $c$ is a further vertex in $V(N)$ and
$a \preceq_N b$ and $c\preceq_N b$ holds
then we call $b$ a {\em common ancestor} of $a$ and $c$.
A {\em lowest common ancestor} $lca_N(a,c)$ of $a$ and $c$
is a common ancestor of $a$
and $c$ and no other vertex below $lca_N(a,c)$ is a common ancestor
of $a$ and $c$.
Note that in a level-0 or level-$1$ network $N$, the lowest common
ancestor between any two distinct leaves of $N$ is always unique
whereas this need not be the case for level-$k$ networks with larger $k$.

Now suppose $N$ is
one of the simple level-$1$ networks $SL_1^i(T)$, $i\in \{1,2,3\}$,
on $X=\{a,b,c\}$ depicted in Fig.~\ref{simple-level-$1$}. Let $e=uv\in  A(N) $
be a non-cut arc and suppose that $d\not\in X$. Then we denote
by $N_e\oplus d$ the level-$1$ network obtained from $N$
by adding a new vertex $w$ to $V(N)$ and replacing
$e$ by the arcs $uw$, $wv$, and $wd$. We remark that if the knowledge of $e$ is of no
relevance to the presented argument, then we
will write $N\oplus d$ rather than  $N_e\oplus d$.

\begin{lemma}\label{lemma-montpellier}
Suppose $X=\{a,b,c,d\}$ and $T=\{a|bc, c|ab\}$. Then,
for any two distinct $i,j\in \{1,2,3\}$,
$$
{\mathcal R}(SL_1^i(T)\oplus d)\not={\mathcal R}(SL_1^j(T)\oplus d).
$$
\end{lemma}
\noindent{\em Proof: } Put $N^k:=SL_1^k(T)$, $k\in\{1,2,3\}$, and assume that
there exist distinct $i,j\in\{1,2,3\}$
and non-cut-arcs $e_i\in A(N_i)$
and $e_j\in A(N_j)$ such that
${\mathcal R}(N^i_{e_i}\oplus d)={\mathcal R}(N^j_{e_j}\oplus d)$.
By symmetry, it suffices to consider
the cases $(i,j) \in \{(2,1),(2,3)\}$.
For $k \in \{1,2,3\}$, let $u_k,v_k\in V(N^k)$ such that $e_k=u_kv_k$.
Also for $k \in \{1,2,3\}$, let
 $w_k\not\in V(N^k)$ denote the new vertex in $V(N^k_{e_k}\oplus d)$
such that
by replacing the arc $e_k$ by the
arcs $u_kw_k$, $w_kv_k$, and adding the arc $w_kd$ the new
network $N^k_{e_k}\oplus d$ is obtained from $N^k$.
Note that for all $k\in\{1,2,3\}$, both $N^k$ and $N^{k}\oplus d$
have the same root and the same
reticulation vertex which we denote by $\rho_k$ and $r_k$, respectively.
Furthermore, for all $x,y \in \{a,b,c\}$ we have
$lca_{N^k}(x,y) = lca_{N^k\oplus d}(x,y)$
We distinguish the cases that $u_2=\rho_2$ and
that $u_2\not=\rho_2$.

Suppose first that $u_2=\rho_2$ and put $l=lca_{N^2}(a,b)$.
Then $e_2\in \{\rho_2 r_2, \rho_2 l\}$.
We first establish that $j\not=1$.
Assume for contradiction that $j=1$.
For all $s,t\in V(N^2_{e_2}\oplus d)$ such that $t$ is below $s$
 denote a path from $s$ to $t$ in $N^2_{e_2}\oplus d$ by $P_{st}$.
Observe that $d|ac\in {\mathcal R}(N^2_{e_2}\oplus d)$,
$e_2\in \{\rho_2r_2, \rho_2 l\}$, holds. Indeed,
since $c|ab\in {\mathcal R}(N^2)$ the paths $P_{la}$ and
$P_{lc}$ exist and do not have an internal vertex in common.
Furthermore, $w_2\not=l$ and either the arc $w_2l$
or the arcs $\rho_2 l$ and $\rho_2 w_2$
exist. In both cases the paths $P_{w_2l}$, $P_{\rho_2 l}$ and $P_{\rho_2 w_2}$,
consisting of the arcs $w_2l$, $\rho_2 l$ and $\rho_2 w_2$, respectively,
do not have an internal vertex in common with either
$P_{la}$ or $P_{lc}$. Thus, $d|ac\in {\mathcal R}(N^2_{e_2}\oplus d)$, as required.
By assumption,  $d|ac\in{\mathcal R}(N^1_{e_1}\oplus d)$ follows which is impossible
since $lca_{N^1}(a,c)=\rho_1$ and so $d|ac={\mathcal R}(N^1_e\oplus d)$,
for all non-cut arcs $e\in A(N^1)$.
Thus, $j\not=1$, as required.

If $j=3$ then ${\mathcal R}(N^2_{e_2}\oplus d)={\mathcal R}(N^3_{e_3}\oplus d)$ and
$v_2=r_2$ or  $v_2=l$. If $v_2=r_2$ then
$b|cd\in {\mathcal R}(N^2_{e_2}\oplus d)={\mathcal R}(N^3_{e_3}\oplus d)$ follows.
But then $w_3$ cannot be a vertex on the path in $N^3_{e_3}\oplus d$
from $\rho_3$ to $b$ or on  the path from $\rho_3$ to $r_3$
which avoids $lca_{N^3}(a,c)$
Thus, $u_3=lca_{N^3}(a,b)$ and so
$b|da\in {\mathcal R}(N^3_{e_3}\oplus d)={\mathcal R}(N^2_{e_2}\oplus d)$ which is impossible
as $lca_{N^2\oplus d}(a,d)= \rho_2$ and thus always above
$l$. If
$v_2=l$ then $d|bc,c|bd\in {\mathcal R}(N^2_{e_2}\oplus d)
={\mathcal R}(N^3_{e_3}\oplus d)$
follows which is again impossible since if $w_3$ lies on the path
from $lca_{N^3}(a,c)$ to $r_3$ then
$d|bc\not \in {\mathcal R}(N^3_{e_3}\oplus d)$ and if not then
$u_3=\rho_3$
and so $c|bd\not\in {\mathcal R}(N^3_{e_3}\oplus d)$.  Thus, $j\not=3$.

Now suppose that $u_2\not=\rho_2$. Then $u_2\in \{l,lca_{N^2}(b,c)\}$
Observe that arguments similar
to the previous ones imply that
$a|cd,a|bd,c|ad,c|bd\in {\mathcal R}(N^2_{e_2}\oplus d)$ holds for all
$u\in \{l,lca_{N^2}(b,c) \}$. If $j=1$ then
$u_1\not=\rho_1$ as otherwise  $a|bd$ or $c|bd$
does not belong to ${\mathcal R}(N^1_{e_1}\oplus d)={\mathcal R}(N^2_{e_2}\oplus d)$.
Thus $v_1=r_1$ and
$u_1\in\{lca_{N^1}(b,a),lca_{N^1}(b,c)\}$. If
$u_1=lca_{N^1}(b,a)$ then $a|cd\not\in  {\mathcal R}(N^1_{e_1}\oplus d)$
which is impossible. Swapping the roles of $a$ and $c$ in the
previous argument shows that $u_1=lca_{N^1}(b,c)$
cannot hold either. Thus, $j\not=1$.

If $j=3$ then again since $a|cd,c|ad\in {\mathcal R}(N^1_{e_1}\oplus d)$,
it follows  that $e_3$
must be an arc on the path $P$
from $lca_{N^3}(a,c)$ to
$r_3$. Note that similar arguments as the ones
used above imply that either $d|bc$ or $b|cd$ is contained in
${\mathcal R}(N^2_{e_2}\oplus d)$. But
$b|cd\not\in {\mathcal R}(N^3_{e_3}\oplus d)={\mathcal R}(N^2_{e_2}\oplus d)$
and so $d|bc\in {\mathcal R}(N^2_{e_2}\oplus d)$ must hold.
But this is impossible since then $c|ad,d|bc\in {\mathcal R}(N^2_{e_2}\oplus d)$
but there exists no non-cut-arc $e$ on $P$
such that both triplets are simultaneously contained in
${\mathcal R}(N^3_e\oplus d)$. Thus, $j\not =3$.
\hfill\hbox{\rule{3pt}{6pt}}\\

\section{Encodings of Level-$k$ Networks}\label{alternativeEncoding}
\label{tripletEncoding}
\label{consistency-level-$1$}

In this section, we characterize those level-$1$ networks $N$
that are encoded by the triplet system $\mathcal R(N)$, or equivalently
the tree system $\mathcal T(N)$, or equivalently the
cluster system $\mathcal S(N)$ they induce. In addition, we present an example that
illustrates that our arguments cannot be extended to establish the
corresponding result for level-2 networks and therefore to
level-$k$ networks with $k\geq 3$.

Bearing in mind that there exist triplet system
which can be reflected by more than
one level-$1$ network, we denote the collection of all level-$1$ networks
that reflect a triplet system $\mathcal R$ by
$\mathfrak{L}_1(\mathcal R)$. Clearly, if $\mathcal R$
is reflected by a level-$1$ network $N$ then $N\in \mathfrak{L}_1({\mathcal R}(N))$
and so $|\mathfrak{L}_1({\mathcal R}(N))|\geq 1$. Similarly, we
denote for a tree system $\mathcal T$ the collection
of all level-$1$ networks $N$ for which  $\mathcal T=\mathcal T(N)$ holds by
$\mathfrak{L}_1(\mathcal T)$, and for a cluster system $\mathcal C$ the
collection of all level-$1$ networks $N$ for which
$\mathcal C=\mathcal S(N)$ holds by $\mathfrak{L}_1(\mathcal C)$.
As in the case of triplet systems,
there exist tree systems $\mathcal T$ and cluster systems $\mathcal C$
with $|\mathfrak{L}_1(\mathcal T)|\geq 1$ and
$|\mathfrak{L}_1(\mathcal C)|\geq 1$, respectively.

Clearly, any cluster $C\subseteq X$ induces a triplet system
$\mathcal R(C)$ of triplets on $X$ defined by putting
$$
\mathcal R(C)=\{c_1c_2|x : c_1,c_2\in C \mbox{ and } x\in X-C\}.
$$
Thus, any non-empty cluster system $\mathcal C$ on $X$
induces a triplet system $\mathcal R(\mathcal C)$ defined by
putting
$\mathcal R(\mathcal C):=\bigcup_{C\in\mathcal C} \mathcal R(C)$.
The next result establishes a link between the triplet system induced
by a level-$1$ network $N$ and the triplet system
$\mathcal R(\mathcal S(N))$.

\begin{lemma}\label{new}
Suppose $N$ is a level-$1$ network with at least 3 leaves.
Then
$$
\mathcal R(N)=\bigcup_{T\in \mathcal T(N)}\mathcal R(T)=
\bigcup_{C\in \mathcal S(N)}\mathcal R(C).
$$
\end{lemma}
\noindent{\em Proof: } That $\bigcup_{T\in \mathcal T(N)}\mathcal R(T)=
\bigcup_{C\in \mathcal S(N)}\mathcal R(C)$ holds is trivial.
Also it is straight forward to see that
$\bigcup_{T\in \mathcal T(N)}\mathcal R(T)\subseteq\mathcal R(N)$.
To see the converse set inclusion, suppose that $t\in \mathcal R(N)$.
Let $x_1,x_2,x_3\in X$ such that
$t=x_1x_2|x_3$. Then with $lca(x_1,x_2):=lca_N(x_1,x_2)$
we have $x_3\not\in C_N( lca_N(x_1,x_2))$ and
$lca(x_1,x_2)$ does not equal the root $\rho_N$ of
$N$. Let $P_i$ denote a path from
$\rho_N$  to $x_i$, $i=1,2$ and let $T$ denote
the phylogenetic tree on $X$ obtained from $N$ by modifying
all reticulation vertices $v$ of $N$ in the following way.
If $v\not\in V(P_1)\cup V(P_2)$
then randomly delete one of the incoming arcs of $v$ and suppress the
resulting degree 2 vertex. If this results in the decrease of the outdegree
of the root $\rho_N$ of $N$  then identify $\rho_N$ with is unique child.
If $v\in V(P_i)$, $i=1,2$, then delete that incoming arc of
$v$ that is not an arc of $P_i$ and suppress the resulting degree
2 vertex. Clearly, $T$ is displayed by $N$ and so
$t\in \mathcal \bigcup_{T\in \mathcal T(N)}\mathcal R(T)$.
Thus, $\mathcal R(N)\subseteq \bigcup_{T\in \mathcal T(N)}\mathcal R(T)$
must hold which implies the lemma.
\hfill\hbox{\rule{3pt}{6pt}}\\

Note that as the example of the level-$2$ network depicted in
Fig.~\ref{FigLevel2Counter-changed} shows,
the relationship between the triplet system of a level-$1$ network
$N$ and the triplet system induced by the clusters in $\mathcal S(N)$
does not hold for level-$2$
networks.
\begin{figure}[!hgt]
\centering
\includegraphics[scale=0.3]{Level2Counterexample.eps}
\caption{A level-2 phylogenetic network $N$ with
$c_1c_2|x_1\in \mathcal R(N)$, but $\{c_1,c_2\} \not\in \mathcal S(N)$.
}
\label{FigLevel2Counter-changed}
\end{figure}

To prove the main result of this note (Theorem~\ref{theorem-montpellier})
which we will do next, we require some additional
definitions and notations. Suppose $N$ is a phylogenetic network.
Then we call a subset $\{x,y\}\subseteq  X$ a
{\em cherry} of $N$ if there exists a vertex $v\in V(N)$
such that $vx,vy\in A(N)$. Furthermore, if $N$ is a level-$1$
network and $x\in X$ then
we denote by $N-x$ the level-$1$ network obtained from $N$
by removing $x$ (and its incident arc) and suppressing
the resulting degree 2 vertex. In addition, we say that
$N$ is a {\em strict} level-$1$ network if $N$ is not a
phylogenetic tree. Finally, to a triplet system $\mathcal R$
and some $x\in \bigcup_{t\in \mathcal R}L(t)$, we associate the
triplet set $\mathcal R_x:=\{t\in \mathcal R : x\not\in L(t)\}$.

Armed with these definitions and notations we are now ready to
establish our main result.

\begin{theorem}\label{theorem-montpellier}
Suppose $N$ is a  level-$1$ network with at least 3 leaves.
Then the following statements are equivalent
\begin{enumerate}
\item[(i)] $N$ contains a blob with four vertices.
\item[(ii)]  $|\frak{L}_1(\mathcal R (N))|>1$.
\item[(iii)]  $|\frak{L}_1(\mathcal S (N))|>1$.
\item[(iv)]  $|\frak{L}_1(\mathcal T (N))|>1$.
\end{enumerate}
\end{theorem}
\noindent{\em Proof: }
(i) $\Rightarrow$ (iv):  This is an immediate consequence
of the fact that all simple level-$1$ networks depicted
in Fig.~\ref{simple-level-$1$} induce the same set of phylogenetic trees.

\noindent (iv) $\Rightarrow$ (ii):
Suppose that $N$ is a level-$1$ network such that
$|\frak{L}_1(\mathcal T (N))|>1$. Then there exists a level-$1$ network
$N'$ distinct from $N$ such that $\mathcal T(N)=\mathcal T (N')$.
Combined with Lemma~\ref{new},
$
\mathcal R(N)=\bigcup_{T\in \mathcal T(N)} \mathcal R(T)=
\bigcup_{T\in \mathcal T(N')} \mathcal R(T)=\mathcal R(N')
$
follows and so  $N' \in \frak{L}_1(\mathcal R (N))$.
Thus,  $|\frak{L}_1(\mathcal R (N))|>1$, as required.

\noindent (ii) $\Rightarrow$ (i)
We will show by induction on the number $n$ of leaves of $N$
that if every blob in $N$ contains at least 5 vertices
then $|\frak{L}_1({\mathcal R}(N))|=1$. Suppose $N$ is a level-$1$ network
with $n$ leaves such that every blob of $N$ contains at least 5 vertices.
Note that we may assume that $N$ contains at least one  blob since otherwise
$N$ is a phylogenetic tree and so $|\frak{L}_1({\mathcal R}(N))|=1$
clearly holds.
But then $n\geq  4$. If $n=4$ then, using Lemma~\ref{lemma-montpellier},
it is straightforward to verify that $|\frak{L}_1({\mathcal R}(N))|=1$.
\old{
Suppose first that $N$ has four leaves
$a,b,c,d$, say, and assume without loss of generality
that $T'=\{a|bc, c|ab\} \subseteq {\mathcal R}(N)$.
Then there exist
some $i\in \{1,2,3\}$ such that $N$ equals $SL_1^i(T' )\oplus d$.
But then, by Lemma~\ref{lemma-montpellier},  $|\frak{L}_1({\mathcal R}(N))|=1$
follows.
}

Suppose $n> 4$. Assume for every level-$1$ network $N_0$
with $n_0<n$ leaves that $|\frak{L}_1({\mathcal R}(N_0))|=1$ holds
whenever $N_0$ is a phylogenetic tree or
every blob in $N_0$ contains at least 5 vertices. Suppose for contradiction
that $|\frak{L}_1({\mathcal R}(N))| \geq 2$. Choose some  $N'\in \frak{L}_1({\mathcal R}(N))$
distinct from $N$.
Then $\mathcal R:={\mathcal R}(N)={\mathcal R}(N')$. We distinguish
the cases that $N$ contains a cherry and that it does not.

Suppose first that $N$ contains a cherry $\{x,y\}$.
Without loss of generality, we may assume that this cherry
is as far away from the root of $N$ as possible.
Then since $N$
is a strict level-$1$ network all of whose blobs contain
at least 5 vertices, $N-x$ must enjoy the same property with
regards to its blobs (if $N-x$ still has blobs). But then, by induction
hypothesis, $|\frak{L}_1({\mathcal R}(N-x))|=1$ and so $N-x$
is the unique level-$1$ network that reflects
${\mathcal R}(N-x)=\mathcal R_x$. Since by the choice of $x$, for every leaf $z$ in $N$
distinct from $x$ and $y$, only the triplet $z|xy$ out of the 3 possible
triplets on $\{x,y,z\}$ is contained in $\mathcal R={\mathcal R}(N')$, it follows
that $\{x,y\}$ must also be a cherry in $N'$. But then
$N=N'$ which is impossible.
Thus, $|\frak{L}_1(\mathcal R(N))|=1$ must hold in this case.

Now suppose that $N$ does not contain a cherry. Then there exists
a blob $B$ in $N$ such that all cut-arcs that start with a vertex in $B$
must end in a leaf of $N$. For each such leaf $z$, which we will also
call a leaf of $B$, we denote by $z'$
the vertex of $B$ such that $z'z$ is that cut-arc of $N$. Furthermore,
denote by $p$ the leaf of $B$ such that $p'$ is the reticulation vertex in
$B$. Let $y_1$ and $y_2$ the vertices in $V(N)-V(B)$ such that $y_1'$ and $y_2'$
are  the two parent vertices of
$p'$ in $B$. Note that the root $\rho=\rho_B$ of $B$
could be $y_1'$ or $y_2'$ but not both and that whenever
$y_i'\not=\rho$, $i=1,2$, then $y_i$ is a leaf of $B$ (hence
the abuse of notation). Without
loss of generality, we may assume that the path $P_{\rho {y_1'}}$
from $\rho$ to $y_1'$ in $B$ is at least as long as the path $P_{\rho {y_2'}}$
 from $\rho$ to $y_2'$ in $B$ (where we allow paths
of length zero). Thus, $y_1$ must be a leaf of $B$.
Since $P_{\rho {y_1'}}$ is at least
as long as $P_{\rho {y_2'}}$ and, by assumption on $N$, $B$
contains at least 5 vertices,
there must exist a leaf $y$ of $B$
distinct from $y_1$  such that $y'\in V(P_{\rho {y_1'}})$.
Note that we may assume without loss of generality
that $y'$ is the predecessor of $y_1'$ on that path.
 We distinguish the cases that $|V(B)|>5$ and that $|V(B)|=5$.

Suppose first $|V(B)|>5$.
Since a blob in the level-$1$ network
$N-y_1$ has clearly at least $5$ vertices, we have
$|\frak{L}_1({\mathcal R}(N-y_1))|=1$  by the induction hypothesis. But then
 $N-y_1$ is the unique level-$1$ network that reflects
${\mathcal R}(N-y_1)=\mathcal R_{y_1}$.
 Consequently, since $ \mathcal R={\mathcal R}(N')$ we have  $N'-y_1=N-y_1$.
To see that $N$ equals $N'$ suppose $z$ is a leaf of
$B$ distinct from $y_1,y,p$ (which must exist by assumption on $B$). Then
either $t:=z|yp,p|yz\in \mathcal R$ or $t,y|zp\in \mathcal R$ holds.
We only discuss the case that $t,p|yz\in \mathcal R$ since the case
$t,y|zp\in \mathcal R$
is symmetric.
Let $B^-$ denote the
blob in $N-y_1$ obtained from $B$
by deleting $y_1$ plus its incident arc and suppressing the resulting degree
2 vertex. Since $z$, $y$, and $p$ are leaves of
$B^-$ and the choice of $y_1$ implies that
$y_1y|p, y|y_1p\in {\mathcal R}(N)={\mathcal R}(N')$, it follows that there exists some
blob $B'$ in $N'$ such that
$B^-=B'-y_1$.  Moreover, the suppressed degree 2 vertex of
$V(B')$ is adjacent (in $B'$) with $y'$ and
$p'$, respectively, since
otherwise $y_1|yp\in {\mathcal R}(N')={\mathcal R}(N)$ would hold which contradicts
the choice of $y_1$. Thus  $N=N'$ and so $|\frak{L}_1({\mathcal R}(N))|=1$
must hold in case $|V(B)|>5$.

We conclude with analyzing  the case $|V(B)|=5$.
Then either $\rho=y_2$ and so $B$ has, in addition to
the leaves $y_1,y,p$, precisely one more leaf $z$, or
$\rho\not=y_2$ and the leaves of $B$ are $y_1,y_2,y$ and $p$.
We first consider the case $\rho\not=y_2$.
Consider the level-$1$ network $N-\{y_1,y_1'\}$ obtained from $N$
by removing $y_1$, its parent vertex $y_1'$ and their 3 incident
arcs (plus suppressing resulting degree 2 vertices)
thus effectively turning $B$ into a phylogenetic tree on
the leaves $y,p,y_2$, i.e. the
triplet $t:= y|py_2$. Put $\mathcal R^{t}:=\mathcal R_{y_1}\cup\{t\}$.
Since either $N-\{y_1,y_1'\}$ is a phylogenetic tree or a strict level-$1$
network such that each of its blobs contains at least 5 vertices, the
induction hypothesis implies
$|\frak{L}_1({\mathcal R}(N-\{y_1,y_1'\}))|=1$. Thus,
$N-\{y_1,y_1'\}$ is the unique level-$1$ network that
reflects $\mathcal R^{t}$.
Note that the only way to turn $N-\{y_1,y_1'\}$ into a level-$1$ network that,
in addition to reflecting $\mathcal R^{t}$, is also consistent with $t':=y_2|py\in\mathcal R$
is to replace $t$ by one of the level-$1$ networks $SL_1^j(\{t,t'\})$,
$j\in Y:=\{1,2,3\}$. Denote that element in $Y$ by  $j_{N}$. Since
${\mathcal R}(N)={\mathcal R}(N')$  it follows that the level-$1$ network  obtained
from $N'$ by removing $y_1$, its parent vertex, and their 3 incident
arcs (suppressing resulting degree 2 vertices)
must equal $N-\{y_1,y_1'\}$ with $t$ replaced by one of  $SL_1^j(\{t,t'\})$,
$j\in Y$. Denote that element in $Y$ by  $j_{N'}$.
 Since $\{ y_1|py_2, y_2|py_1, y_2|y_1y, p|y_1y, y|y_1p, y_2|py, t\}
\subseteq \mathcal R={\mathcal R}(N')$ it is easy to check
that $j_N=j_{N'}$ must hold and so
$N$ and $N'$ must be
equal which is again impossible.
Thus, $|\frak{L}_1({\mathcal R}(N))|=1$ must hold in case
$\rho\not=y_2$. Using arguments similar to the previous ones
it is straight-forward $N=N'$ and thus $|\frak{L}_1({\mathcal R}(N))|=1$
must hold in case $\rho=y_2$.

\noindent (iv) $\Rightarrow$ (iii): Suppose that $N$ is a level-$1$ network
with $\frak{L}_1(\mathcal T (N))|>1$. Then
there exists a level-$1$ network
$N'\in \frak{L}_1(\mathcal T (N))$ distinct from $N$ with
$\mathcal T(N)=\mathcal T(N')$. But then
$\mathcal S(N)=\bigcup_{T \in \mathcal T (N)} \mathcal C(T)=
\bigcup_{T \in \mathcal T (N')} \mathcal C(T)=
\mathcal S (N')$ and so $N' \in \frak{L}_1(\mathcal S (N))$. Thus,
$\frak{L}_1(\mathcal S (N))|>1$.

\noindent (iii) $\Rightarrow$ (ii):
Suppose that $N$ is a level-$1$ network with $|\frak{L}_1(\mathcal S (N))|>1$.
Then there exists a level-$1$ network
$N' \in \frak{L}_1(\mathcal S (N))|>1$ distinct from $N$ such that
$\mathcal S (N)=\mathcal S (N')$. But then Lemma~\ref{new}
implies
$
\mathcal R(N)=\bigcup_{C\in \mathcal S(N)} \mathcal R(C)=
\bigcup_{C\in \mathcal S(N')} \mathcal R(C)=\mathcal R(N')
$
and so $N'\in \frak{L}_1(\mathcal R (N))$. Hence,
$\frak{L}_1(\mathcal R(N))|>1$.
\hfill\hbox{\rule{3pt}{6pt}}\\

It should be noted that Theorem~\ref{theorem-montpellier} immediately implies

\begin{corollary}
Let $N$ be a level-$1$ network with at least 3 leaves. The number of
non-isomorphic level-$1$ networks $N'$
that reflect $\mathcal R(N)$ (or equivalently for which
$\mathcal T(N)=\mathcal T(N')$ or equivalently
$\mathcal S(N)=\mathcal S(N')$ holds) is $3^b$, where
$b$ is the number of blobs of $N$ of size four.
\end{corollary}

We remark that the strategy
underlying the proof of Theorem~\ref{theorem-montpellier}
does not immediately extend to level-$k$ networks with $k\geq 2$. The main
reasons for this are that, as already mentioned above, for $k\geq 2$ the
number of distinct level-$k$ generators grows
exponentially in $k$ \cite{GBP2009}.
Also the problem of understanding when two distinct
simple level-$2$ networks reflect the same set of
triplets is far less well understood. For example,
consider the two level-$2$ networks
depicted in Figure~\ref{FigLevel2}. Each one of them is a simple level-$2$
network obtained by hanging leaves of the sides of
the level-$2$ generators $\mathcal G^2_a$ and $\mathcal G^2_b$
depicted in Figure~\ref{level-k-generators}.
As can be quickly
verified, both networks reflect the same triplet set. However adding
additional leaves to both networks by subdividing
the arc one of whose end vertices forms an arc with $x_1$ and the
other forms an arc with $x_2$ and attaching additional leaves
results in two distinct level-$2$ networks that still reflect
the same triplet system.  Regarding the accurate reconstruction of
level-$k$ networks from e.g. triplet data, this results highlights
a serious limitation of level-$2$ networks (and probably
level-$k$ networks in general) as two such network with very different
structure might reflect the same triplet set.

\begin{figure}[!hgt]
\centering
\includegraphics[scale=0.3]{Level2Example.eps}
\caption{Both simple level-$2$ networks reflect the triplet set
$\{a|x_1b,$ $b|x_1a,$ $x_1|ab,$ $a|x_2b,$ $b|x_2a,$ $x_2|ab,$ $x_1|x_2a,$
$a|x_1x_2,$ $x_1|x_2b,$ $b|x_1x_2\}$.
}
\label{FigLevel2}
\end{figure}

We conclude with remarking that
phylogenetic trees on $X$  can also be viewed as
trees together with a bijective labelling map between $X$ and
the leaf set of such trees. Taking this point of view,
phylogenetic trees were generalized in
\cite{HuberMoulton2006}  to {\em MUL-trees}
by allowing two or more leaves of that tree to have the
same label. For example, the tree obtained
from the phylogenetic tree depicted
in Figure~\ref{FigLevel1}(c) by replacing the leaf labelled $a$
by the cherry labelled $\{a,b\}$ is such  a tree. In fact,
this is the MUL-tree induced by
the level-$1$ network  $N$ depicted in Figure~\ref{FigLevel1}(a)
that shows all paths from the root  of $N$ to all leaves of $N$.
For a level-$1$ network $N$ it is easily seen that the MUL-tree $\mathcal M(N)$
induced by $N$ this way is in fact an encoding of $N$ in the sense that
$N$ is the unique level-$1$
network that can give rise to $\mathcal M(N)$.

\bibliographystyle{alpha}

redpajama_set_name: RedPajamaArXiv


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Flyer can be printed at WWW.laslowrollers.com or email below.


There are many articles about Airbnb on this site and the company is often described as being 'the disrupters of the travel industry'. However, despite the articles about Airbnb here, this is not one of them. The major shake-up in the history of the hospitality world was in the first half of the last century.
But in the twentieth century, this began to change. A man named Conrad Hilton started buying up hotels in the 1920s. Later in the century, Kemmon Wilson, another American, was dissatisfied with the accommodations he and his family found when travelling so decided to found a chain of motels which were all identical to each other. Predictability was considered to be a good thing. A couple of years later, a Mr. Marriott entered the hospitality business too – creating cookie-cutter hotels throughout America.
Thanks to these three gentlemen we now have, respectively, The Holiday Inn, the Hilton and Marriott hotel chains. Other similar hoteliers followed in their wake.
The time was ripe. During the middle of the century in the USA interstates were being constructed to facilitate easier travel Automobiles were increasing in numbers and gasoline was cheap. Air travel was becoming accessible to more people. GIs had returned from the war with a taste for travel and a need to enjoy themselves. During this economic boom, the new hotel chains flourished.
That's the time when the traditional hospitality industry had its severest disruption. The population flocked to these new hotel chains because they were predictable; they made the guests feel secure during this new and exciting phase of the tourist.
Fast forward: Airbnb had not yet started when Craigslist became the place to advertise vacation rentals. Shortly after that VRBO started operations. The internet was beginning to make its mark on the hospitality business. But we need to remember that for many, many years homeowners had been renting out their vacation properties, their villas – but the difference now was that they were online.
The next disruption to the traditional hospitality business was also internet based – the OTA; online travel agencies. These are sites such as Expedia, Orbitz, Priceline and so on. They had the advantage of capturing the consumers' email addresses so that these people became 'their' customers. And they find the lowest prices.
Hotels resisted at first but after the events of 9/11, when few people were travelling, hotels began to see the value in using these services.
It's hardly surprising that today's travellers are looking for more than the cookie-cutter, no-surprises experience that the major hotel chains offer. Once you were in a Holiday Inn in, for example, Marachech, your room would be much the same as it would be if you were in Cincinnati, Paris or Miami. Airbnb recognised that today's travellers wanted a more authentic experience.
Times change. People change. Our expectations change. Our needs change.
Today we can travel anywhere (virtually) thanks to the TV, podcasts, YouTube and so on. Today, we want to experience the local culture. We no longer want to be in a Madrid hotel that's the same as the one in Columbus, Ohio.
In the years since Conrad Hilton started buying up hotels with a view to creating a chain, the world has changed. Industries have changed, people have changed. One of the major changes must surely be the internet. One hundred years ago, people would have found it hard to imagine life today.
They would find it unbelievable that we can do our banking from our phones (and that our phones are mobile and fit into our pockets), that we could take part in online auctions, that we can make just about any purchase we wish without going out of the house. We can order our groceries when we're sitting on the beach. We can book accommodation in Cape Town instantly while sitting in a traffic jam in New York.
Airbnb has not been the major disrupter of the hospitality industry that many people, mostly journalists, claim that it has. It's merely a company that offers a service that the public wants.


But things had certainly been cooking with gas when I walked into the meeting room at the University of San Francisco, which the organizers had decked out in white nubby rugs, white pleather couches and those white egg-shaped chairs found only in 1970s decor.
First up was well-known investor and entrepreneur Peter Thiel, poo-poohing Microsoft's prospects of ever making money in search.
By the way, Thiel sold semantic search engine Powerset to Microsoft for upward of $100 million in 2008 to help it, you know, get ahead in search.
Way to insult your money-bearing hosts!
And they say I'm a snarky moderator! Wadhwa is snarktastic!
Wadhwa did not like any of it–not crappy content sites that sully Web search, not the efforts the companies were making to fix things, not the vision the trio had of the future.
And, by the way, Microsoft was not ever going to make money off all the company's efforts.
Way to insult your hosts! I like this event!
Of course, what everyone was interested in was a smackdown between Google and Microsoft, given that the search giant accused the software giant of stealing its results today.
In an excellent, if exhaustive, post by Search Engine Land's Danny Sullivan, Google said Bing was cheating by lifting its search results, which Google said it had proved via a "honeypot" sting operation.
The very presence of the word "honeypot" in any story about search algorithms is superb, in my book, even though this "controversy" is pretty much a he-said-he-said geek-off.
Google's Matt Cutts kept up the cheater pressure at the Bing event, in a short debate with Microsoft's Harry Shum, who was not having any of it.
"It's not like we actually copy anything," he said.
Translation: Actually, we do borrow, just like Facebook's Mark Zuckerberg did to the Winklevii, resulting in a social networking behemoth that will soon take over all search and make this whole debate moot.
Microsoft is rubber, Google is glue. And Facebook, which was not present at the search event, is the real sticky honeypot.


Quag Myer is 11 years old today, born in 2000. Quag is the girl GSP/Britney mix dog of Darlene and Jerry. She is from Broken Bow, NE and is the runt of a 10 dog litter. HAPPY BIRTHDAY QUAG! Don't say it, Yes, I know, but it's my blog.
Anna Pauline Rosentreter is 100 years old today, born in 1911. Anna is the boys' 4th cousin twice removed. Mother - Lucy Powers Jaeger, her mother - Helen Altana Powers, her mother - Lydia Ann Banks, her father - the boys' 4th great-granduncle, David Bradley Banks the 1/2 brother of Samuel Cheney Cotton, via their mother Sally Gold.
Benjamin Dudley Robinson died 131 years ago today in 1880 at the age of 44 in San Antonio Texas. Ben is the husband of the boys' 3rd great-grandaunt, Helen Marshal Cotton, the sister of Hubbard Cotton, the boys' 3rd great-grandfather.
Jeremiah Walker and Esther Tomlin's 287th anniversary is today, married in 1724, he at age 22 and she at 13. This lovely couple are the boys' 8th great-grandparents who begot Benjamin walker who married Sylvia Keith and begot Marshall Walker, who married Hannah Dunbar who begot Cady R. Walker.
Lillie May Stanford died 25 years ago today in 1986 at age 90 in Oklahoma City, OK. Lillie is the boys' 2nd cousin 3 times removed. Her father is Asia Minor Stanford the boys' 1st cousin 4 times removed. Asia's father is the boys' 3rd great-granduncle, Jacob K. Stanford, the brother of David Stanford.


text:
Q: Connecting singnalR through Ocelot gateway in .Net 6 I am trying to connect my signalR to my frontend app through Ocelot gateway but failing to achieve it. The rest of the endpoints are completely accessible through the gateway but when I am trying to connect the signalR through gateway it throws localhost is currently unable to handle this request error.
My Ocelot configuration for signalR hub is
{
  "DownstreamPathTemplate": "/{everything}",
  "DownstreamScheme": "ws",
  "DownstreamHostAndPorts": [
    {
      "Host": "localhost",
      "Port": 5001
    }
  ],
  "UpstreamPathTemplate": "/{everything}",
  "UpstreamHttpMethod": [ "GET", "POST", "PUT", "DELETE", "OPTIONS" ]
}

here is how I register the Ocelot in the program.cs
builder.Configuration.AddJsonFile("ocelot.json", optional: false, reloadOnChange: true);
builder.Services.AddOcelot(builder.Configuration);

app.UseWebSockets();
app.UseOcelot();

here is my signalR hub endpoint
app.UseEndpoints(endpoints =>
    {
        endpoints.MapControllers();
        endpoints.MapHub<NotificationHub>("/notifications");
    });

I just followed the ocelot documentation but could not make it work until now.
here is error ss,

redpajama_set_name: RedPajamaStackExchange


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text:
JOYFULTALK's Jay Crocker
Words by David C. Obenour
Photo by Kyle Cunjak
After two albums of multi-genre inflected trance, Jay Crocker is taking a different direction for his JOYFULTALK pandemic record. Based out of rural Nova Scotia, remoteness is a familiar setting, however isolation lead to its own series of limitations and inspirations. Utilizing samples from frequent collaborators like percussionist Eric Hamelin and the late saxophonist Dan Meichel, Crocker stitched Familiar Science together with added improvisations filling and connecting the concepts. The results land in a unique space between experimentalism, jazz, electronica, funk, post-rock and minimalism.
Off Shelf: One of the first things that hit me listening to Familiar Science was the rhythm section and I wanted to break that apart a little. The album features the work of three different drummers and I wondered if you could talk about that a little – what each musician brought to the recording and how you were able to keep a through line?
Jay Crocker: The initial seeds of the project came through a series of solo drum improvisations, that were sent to me by my longtime friend Eric Hamelin, that I cut up and recombined. Before I left Calgary I was doing a lot of free jazz and free improvisation in the scene there, but I feel like my main collaborator around that time, in that music, was Eric. In winter of 2021 Eric had sent me some solo improvisations he had recorded. I started putting them into the sampler and started to find some really interesting patterns and phrases. I started shaping these patterns and writing melodies for them.
Take it to the Grave features Chris Dadge on drum set. Chris plays an improvised swing over Eric's sample as I improvise on guitar. Chris is another Albertan collaborator and friend that I played with a lot back in Calgary. He runs the Bug Insicion label and improvised music series in Calgary. Chris and I spent a lot of time together discovering new sounds and doing a lot of listening. And was great to get his voice – particular his jazz voice – on this record.
Most of the other tunes are samples of Eric except for Blissed for a minute in which I play drums over a tape loop and Particle Riot which is a drum machine sequence.
OS: There are also a lot of adventurous rhythms that still keep finding that hypnotic head-nodding groove. What is the balance of locking into that, riding it, but knowing when to throw listeners off with something new?
JC: I think if I get bored or disinterested in a certain section of a piece it needs to change or end. Through my compositional process I don't really consider the listener directly but maybe through an evaluation of my own ego in the music. Some things need to be indulged at certain points and other things need to be indulged at other points. Its a fine line between being critical of yourself but also trying to get closer to understanding who you are as an artist.
OS: The bass also is mixed loud and a perfect accompaniment – what role does its part serve in riding out some of these rhythms? How does it ground the music and let it reach further?
JC: The bass is the bridge between rhythm and harmony/melody so it plays kind of the ultimate roll is some music. I played all the electric and synthesizer bass on these recordings and was trying to give it some funk. I was able to ground the samples a little through the bass lines while also giving the music a more recognizable identity.
OS: There are also some really interesting vocal parts thrown in there. What made you add them into a more predominantly instrumental album and how do you think that adds to the songs?
JC: I'm not sure what the inspiration was – maybe some Art Ensemble of Chicago or Sun Ra… John Lurie – but I just felt like a few of those tracks need some wordless vocals. Singing feels good. I do have difficulty hearing my own voice and not wanting to hide in a dark cave but it is an example of understanding what needs to be there regardless of ego or fear or self criticism.
OS: Jazz is one of the genres JOYFULTALK loosely falls under and I wonder how you do and don't identify with that as a genre tag?
JC: I feel like jazz is the music that I still constantly explore and go back to again and again. I love it. It's the form of music that never stops teaching you, loving you, hating you, kissing you, kicking you. It's such a mediation for for me. Sometimes I feel good about it and other times I feel like its so challenging and out of reach. But it's still, for me, about finding my own voice in that music. That's the best part about it… it changes as you change, it changes with you and around you and it is able to reflect the different versions of yourself for better or for worse. For me it's about trying to find the connection between myself and the energy around me and how to interact with that in the moment.
OS: Electronica is probably the other big one, so what parts of that do you find yourself wanting to celebrate or distance yourself from with JOYFULTALK?
JC: I found electronic music through improvisation. As I started moving deeper into improvised music I started moving deeper into electronics simultaneously. I started adding more homebuilt electronics to my improvising palette and eventually it kind of took over. I then started writing scores and pieces for these instruments which led to sequencing and sampling while still adhering to the idea of never being controlled by these instruments and knowing them well enough to be able to perform with them with same fluidity as I can on guitar. I don't think a want to distance myself from it but rather let it just be part of the palette.
OS: This music feels very much alive and ripe for riffing and improvisation in a live setting. How does your approach to recording vary from your live performance? Or is the recording just more a single capture of a live performance?
JC: Familiar Science is very collaged due to my location and the pandemic. I have also developed a very consistent solo practice but this record is a transition into a more live interaction. I think I always had it in the back of my mind that I wanted to be able to perform these pieces with a small group. For the upcoming shows I will be performing the music with drums, guitar/electronics and keys and adding other musicians in different cities.
OS: We're coming off a number of years without live performances. How has the time away changed how you approach playing and performing? Have you re-evaluated anything in the absence?
JC: I live in rural Nova Scotia so out of necessity I had to developed strong solo art making process and before the pandemic I wasn't a lot performing a lot and when I did it would be a in a solo or duo configuration. I think the pandemic has made me miss playing with a group or using a bit more of a "traditional" approach. I think that it is part of why Familiar Science was made the way it was. I was maybe longing to get back to something like a had experienced in the past…
OS: The album also makes use of live samples of the late Dan Meichel on saxophone. Can you talk about how meaningful it was to be able to incorporate that into the recording?
JC: Danny was a great improvisor and a lovely human. I learnt a lot from him and I miss him. I feel like he really had the sound of a true improvisor. I'm not sure I totally understood how good he actually was until now. I was going through old recordings I had made of some large group improv's and there were certain sections where it was mostly his voice poking through so I sampled him and created the music around him. Hagiography is built of a recent sample of Eric on drums with an old sample of Danny playing over top. It's kind of my hologram tribute to him, I think he would've loved it.

redpajama_set_name: RedPajamaCommonCrawl


The online poll tracks a pool of undecided voters before and after the presidential and vice presidential debates. Of the 614 undecided voters surveyed, 456 watched or followed the debate. Statistical margins of error are not applicable to online polls because they are based on samples drawn from opt-in online panels, not on random samples.
Most people polled Friday also thought Biden did better in the debate -- although McCain boasted of Palin's performance at a campaign rally the next day, saying he "almost felt a little sorry" for Biden. McCain also proclaimed "viva la barracuda!" a reference to the Alaska governor's nickname on her high school basketball team. Respondents found 54 percent to 46 percent that Biden posted a better performance.
Palin got poor marks from voters when asked if she looked vice presidential or whether she had a good understanding of the issues -- only 39 percent of the undecided voters thought she looked vice presidential. Nearly two-thirds of those surveyed thought Biden would be a better commander-in-chief.
But Palin did well when people were asked who would do a better job bringing change to Washington. Palin, they found, 58 percent to 42 percent. And roughly two-thirds said they like her, and some 53 percent found her believable. That's a significant number in a debate watched by an estimated 69.9 million viewers -- more than the previous presidential debate.
"I like Palin," said one of the people surveyed in the Ipsos/McClatchy poll, Democrat Greg Brantley, 47, of Irving, Texas.
"She's very outgoing, she's real friendly, she seems to know what she's talking about," said Brantley, who thinks he'll vote for McCain. "She just seems like an average person you could walk down the street and pick up a conversation with her."
Although the most significant findings in the poll are about which direction undecided voters are leaning, there might be something to be learned from Palin's overwhelmingly positive likeability numbers, Young said. It may be that in tough economic times, voters yearn for a candidate who feels their pain.
"We don't we have a 'great communicator' this year, like Ronald Reagan or Bill Clinton," Young said.
Voter Judy Williams of Caliente, Calif., agreed, saying she liked how both vice presidential candidates seemed "sincere and very sympathetic."
Williams also said that Palin exceeded her expectations during the debate and is leaning toward voting for McCain, but she likes Biden so much that if he were on the top of the ticket, she would vote for him for president. She was especially moved when Biden choked up while speaking about the loss of his first wife in a car crash.
"I was very touched by it," she said. "I could definitely sympathize."
But how much does the vice presidential debate matter? Marcee Dellinger, 29, of Yakima, Wash., said she didn't find the vice presidential debate "a persuasive factor" in making a decision. Still undecided, she said she was hoping the next debate between the presidential candidates would give her the information she needs to make up her mind.
"I hope anyway," she said. "I probably lean more toward Republicans, but if I had to vote today I have to say I wouldn't be comfortable 100 percent."


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Victoria McKay, chief operating officer, London Diamond Bourse
Business Big Shots, Hot 100 2015 Gallery
It has been a dramatic year for Hatton Garden's London Diamond Bourse as it was caught up in the fallout from one of Britain's biggest jewellery heists. Fortunately, chief operating offi cer Victoria McKay is the type to take charge, crack on and make the best of a bad situation. She did this admirably offering a pillar of support for affected businesses and individuals looking for someone to turn to. Her achievements for the London Diamond Bourse have been even more signifi cant, resulting in Bonded Warehouse Status. She says: "It took almost two years from start to fi nish and despite initial knockbacks, I was determined I wouldn't take no for an answer — I even ended up getting support from an MP and the London Mayor's Offi ce to make it happen. Subsequently, the scheme we have in place is particularly unique and I understand that no other scheme of its kind exists for a membership organisation in our industry." Since joining LDB, Victoria has been tasked with making its trading fl oor the home of diamonds in the UK, and she remains determined to make London a "hugely successful diamond centre again". She continues: "Membership is growing very fast and this is somewhat down to how we have changed. The Bourse was for a long time considered a closed door organisation and when I started I couldn't believe we weren't showcasing how great the organisation is and how our traditions still have a place in the market." Outside of work, Victoria enjoys spending time with friends and family, great food, dancing and cocktails — hence her mixology-themed shoot. She's also admits to being "a bit of a geek" and is desperately looking forward to the new Star Wars installment due in cinemas later this year. We'll see you at the pictures!


Aiki News is pleased to present this exclusive series on the Omoto religion, One of Japan's foremost "new religions," the Omoto sect has had a turbulent and fascinating history. More Important to practitioners of aikido, its doctrine had a profound influence on Morihei Ueshiba, founder of aikido. Yasuaki Deguchi's memories provide Insight into his grandfather Onisaburo's character.


But perhaps you're thinking – Why?
Well… it's because we think small groups are the heart of Restoration. It's where we connect with God and each other and the Word in a safe and gently accountable way!
If you're new around here, it's a great way to meet people in a contained, finite group. Each trimester only lasts around 10 weeks, and we sign up every trimester so you can always pick a different group next time. Some people stick in the same group for a number of trimesters for continuity, some switch it up every trimester so they get to meet new people. Your choice!
We ask members to commit to being in a small group every trimester (as far as possible) as it's where we develop community and friendships; where we grow the next generation of leaders; where we pray regularly for each other and our partners; where we read the Bible together and work out what it is saying to us individually and corporately.
Looking forward to hearing about your small group adventures of 2015!


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The first delivery
There genuinely is not a lot of crime in Denver, but a single of my employees got carjacked on his first delivery, but kids started working at the marijuana delivery maintenance last week, but he had 3 days of in-store training.
He started delivering orders on Friday! It was an seriously tied up day & I knew that the two of us needed all hands on deck, the first delivery was 6 mi from the marijuana dispensary. It wasn't a delivery in the awful area of town, but the child had to drive through a genuinely seedy neighborhood. The interstate was shut down in that area of city & there are plenty of detours that take you through the city. I sent the child out of the dispensary with $200 worth of marijuana products. He told myself and others that he knew the Denver section well. He did not seem upset about the detours or the side streets, but about 30 minutes after the kids left, he called with an upgrade on the order. The child got carjacked when he left the interstate on a single of the detours. The police were at the scene of the carjacking, but the kid's car was gone. The perpetrators took the car while holding the child hostage with a weapon. He did not bother to take the marijuana out of the car. The Denver police officers did not know where to start with the case. It was the first time they had seen this type of crime in a genuinely long time. I felt terrible for the kid, but thankfully he had full coverage insurance. They will upgrade his stolen car quickly.
Medical Weed Near Me Denver Colorado


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Q: Not getting expected near by place from table I have a place's table, in this, each record has latitude and longitude of that place.
Now I want to find nearby places (within 10 KM) from places table according to given lat-long.
I have 500 + place's records in my database table.
Currently i am using mysql query for the same :
 SELECT FROM provider_venues WHERE IFNULL( ( 111.111 (
 DEGREES(ACOS(COS(RADIANS(26.2743661)) COS(RADIANS(venue_lat))
 COS(RADIANS(73.02130979999993 - venue_long)) +
 SIN(RADIANS(26.2743661)) * SIN(RADIANS(venue_lat))))) ), 0) <= 10

But distance i the am getting from above query is different from the distance i am getting from google API.
I can use google API for this but using google API for every record may not be good approach.
Is there any way to use google API in more optimized way or any other solution to get the expected result faster?


Weight loss shakes seem like an ensured and simple approach to losing weight. Fortunately, most shakes are entirely healthy and safe. The awful news is that you could end up putting on the weight back if you don't have a strong strategy. When you pick the correct shake, adhere to a wholesome weight loss regimen and strategize your lifestyle decisions for after you achieve your goal, you'll have a better chance of succeeding in a long-term.
Weight-loss shakes may respond with diabetes medications, causing your sugar level to plunge hazardously low.
If you suffer from kidney disease or osteoporosis the high protein content in most weight loss shakes could exacerbate your condition. Inquire as to whether there are any brands with more safer protein levels.
Speak with your doctor regarding your plans for diets and consider requesting a referral to a registered dietitian. Tell them about the shakes you've been looking into. They'll measure your weight and height to know your BMI. They could likewise quantify fat stores around your waist and different parts of your body. Cooperate to find out your target weight and to what extent it will take you to achieve it. Make arrangements for follow-up visits.
Measure the suggested measure of shake powder and fluid to combine. Typically, this is one scoop of powder and 8 liquid ounces (237 mL) of milk. Supplant just the dinners recorded in the directions. Most weight loss shakes prescribe no less than one healthy meal, usually dinner.
Don't drink additional shakes to try to limit your hunger. The high protein substance can really make you put on weight if you overdo it. Keep in mind, each gram of protein is 4 calories.
In the event that you have to take your breakfast at 6:00 am and lunch at 12:00 pm, you'll presumably feel hungry by mid-to-late-morning. Oppose the enticement to go after junk food by keeping fiber-rich snacks available. Eat an unpeeled apple or pear, 1 container (237 g) of raspberries, or one medium banana between shakes. Do this again in the middle of lunch and dinner.
Weight loss is completely possible do not let anyone or anything make you think otherwise. However, it takes consistency. Consistently doing every right thing, right.


Holly is one of the fastest growing business communities in Oakland County. Mere minutes from the suburbs, Holly attracts visitors for day trips shopping in our historic downtown to outstanding special events and festivals.
Main Street Holly is your first resource for making your dream of being a business owner come true, with numerous resources available to help you get started. Through our connections with Oakland County and the State of Michigan. Check out our listings of available property listings, and let us help you work with the Village of Holly on any of your permit or regulation needs.
Main Street Holly DDA offers a number of support programs for businesses, including monthly "Pizza Talks" that let you network with other local business owners and learn from experts regarding business, marketing and other resources available to you. Main Street Holly DDA also offers several grant programs that let you grow and improve your business, including incentives for outdoor dining and facade/building improvements.


CLEMSON, SC-Clemson scored an own goal with 8:00 minutes left to go in the second overtime period as the Tigers defeated South Carolina 2-1 in a men's intercollegiate soccer match at Clemson, SC Friday night.
Justin Moore sent a shot, off a free kick, from 23 yards out that bent to the left. A South Carolina defender Josh Alcala deflected the ball into the net and the goal was ruled an own goal to give Clemson a 2-1 victory.
South Carolina's Mike Sambursky scored early in the second half to give South Carolina a 1-0 lead. Sambursky collected the rebound after Ayo Akinssete took a shot and Clemson goalkeeper Phil Marfuggi deflected the shot. Sambursky sent an 18-yard shot into the goal.
Clemson evened the score with :22 seconds left to go in regulartion. Havrid Usry crossed the ball to Jeff Routh. Routh headed the ball in the goal beating the South Carolina goalkeeper Mike Gustavson on the left side near the left post.
"This was a great game," said Clemson Head Coach Trevor Adair. "I feel as if this was a championship game, and this was the first game of the season.
The Tigers are now 34-4-1 in season opening games over the years.
Clemson will play host to Longwood on Tuesday, September 6, in the Tigers' next action.


"Jim Gavin is a smart man and a great manager so he will get every player knowing that a poor or below the standard expected performance and you could sit out the semi-final."
Straight away I think the handicap of eight is too big. I have been very impressed by Kildare all year. Losing Daniel Flynn is a big loss though, he was their star man all year. How will they rejig the forward line will be intrusting? Will they send Cribbin in or another target man?
The fact that Kildare are out of the competition will have to have some effect on the mind of the players, but there is no way Kildare are coming down to Kerry, to be embarrassed after the great season they have had. They won't want to go down with a whimper.
Kerry's back is right up against the wall again, lose and it is all over. They are starting to get a little better I believe; the backline has improved but they still have massive issues upfront and in midfield. Kerry need this win, that is it, all they need is a win and then hope that Galway give them a favour as it is now out of their hands.
Some of the younger players are really starting to get going and if they get through they will have nothing to lose. They really need a good, solid performance and work rate is what we want, I haven't seen Kerry work hard enough this year.
I do expect Kerry to win but the handicap is too big.
The big question in this is, with Galway through already, what way will they approach it? In my mind, Galway will go out to win this game, will they field their strongest team? I don't think so, they play six days later and they know that. Remember Galway have only been beaten once this year and that was by Dublin in a National League Final. There is no way Kevin Walsh wants to get beaten a week before an All-Ireland semi-final no matter who he puts out.
It is very risky to not take this game seriously when you have such a big weekend next Saturday, it's dangerous. Every player will have to go in with the right mindset and that is thinking if I play poorly I could lose my spot next weekend.
Monaghan are fighting for their lives but they seem to be missing a ruthless side. It was missing against Fermanagh, it was missing against Kerry, Kerry were on the racks, Monaghan were on top in so many areas and didn't get the job done. This is a huge game for Monaghan. Excitement and edge will play into this.
I don't think Galway will give near the amount of space to McManus that other sides have given, in many ways this will resemble an Ulster game with the style of play that Galway play but Galway have come through some hard and close contests already this season. Tight battles.
I expect their big players to stand up, they are working so well as a unit now and even though I still have questions about their defence are Monaghan good enough to expose them? For me, no. Monaghan last week only got four points from their team outside of Beggan and McManus. That is a poor return. I don't believe this Galway team want to start losing matches in what will be a huge eight days for them and I expect them to win with a bit to spare.
The game of the weekend, the history of these two is a great read but not since the 1970s has Tyrone had to go up to Donegal in the knock-out stages of the Championship. That is something to look forward to. Donegal have tweaked things for the better this season but they have problems on the injury front. Paddy McBrearty as we know is out and so too is Eoghan Ban Gallagher, these are huge losses, add to that where do they play Michael Murphy and Donegal look to be in a bit of trouble this weekend.
Tyrone did very well last time out against Dublin, I was very impressed with their subs and the impact they had. What Dublin did do very well was to keep Tyrones's leaders quiet. The likes of Mattie Donnelly, Peter Harte & Niall Sludden were all shackled by Dublin. They must step up, if they all step up to where they should be then Tyrone can improve on the Dublin performance.
This is going to be ferocious, intense battle. It is so hard to call, we know how Tyrone are going to play though so we know they will run with the ball and Donegal will sit back and whoever wins the battle in the middle of the field will come out and top and I just think that Tyrone will be the ones to do that and win what I expect to be a very close game.
I know Jim Gavin has spoken to the media and said that this Dublin side will take every match as it comes, that they will take this game as seriously as others and other such soundbites. The reality is Dublin are qualified, this game does not matter at all to them.
Now, I do believe Dublin will start with a strong team and get the job done early and then unload the bench and give a few of these fringe players the opportunity to prove themselves. Jim is a smart man and a great manager so he will get every player knowing that a poor or below the standard expected performance and you could sit out the semi-final.
Roscommon will have a lot of scars from this Super 8's experience. The life was sucked out of them in these two matches. They are a serious side going forward but they leak so many scores at the back, they have a real problem in their defence. Pride is there for Roscommon, they will want to finish the season on a high but Dublin are the wrong team to be going into a game against trying to restore some pride. I can see Dublin push up on their kick-outs and Roscommon will struggle for ball.
Dublin have looked so comfortable for most of the year and this game is only going to go one way. Dublin to win and I fully expect them to cover the handicap.


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Halictus lateronitens is een vliesvleugelig insect uit de familie Halictidae. De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1945 door Cockerell.

lateronitens

redpajama_set_name: RedPajamaWikipedia