text:
\section{Bounding arities}
\label{bounded}

The expressive power of lifts can be limited in several ways. For example, it is natural to
restrict arities of the newly added relations.  It follows from the above proof that the arities of new relations
in our lifted amalgamation class ${\cal L}$ depend on the size of maximal inclusion minimal cut of the Gaifman graph of the forbidden structure.

In this section we completely characterize the minimal arity of ultrahomogeneous lifts of classes $\mathop{Forb_h}\nolimits({\cal F})$.
This involves a non-trivial Ramsey type statement stated bellow as Lemma \ref{sets}. As a warm up, we first show that generic universal graph for the class $\mathop{Forb_h}\nolimits(C_5)$ can not be
constructed by finite monadic lifts:

Consider, for contradiction, a monadic lift $\relsys{U}'$ which is both a generic relational structure and whose
shadow $\relsys{U}$ is universal for the class $\mathop{Forb_h}\nolimits(C_5)$. Since all extended
relations are monadic, we can view them as finite coloring of vertices.
For $v\in \relsys{U}$ we will denote by $c(v)$ the color of $v$ or, equivalently, set of
all extended relations $\ext{U}{i}$ such that $(v)\in \ext{U}{i}$.

Since graphs in $\mathop{Forb_h}\nolimits(C_5)$ have unbounded chromatic number, we know that
chromatic number of $\relsys{U}$ is infinite. Consider decomposition of
$\relsys{U}$ implied by $c$. Since the range of $c$ is finite, one of the
graphs in this decomposition has infinite chromatic number. Denote this subgraph
by $\relsys{S}$.

In fact is suffices that $\relsys{S}$ is not bipartite. Thus $\relsys{S}$ contains an odd circle. The shortest odd cycle has length $\geq 7$ and thus $\relsys{S}$ contains
 induced
path of length 3 formed by vertices $p_1,p_2,p_3,p_4$.  Additionally there is a vertex $v$ of degree at least 2.  Because the graph is triangle free, the vertices $v_1$ and $v_2$ connected to $v$ are not connected by an edge.

From genericity of $\relsys{U}$ we know that the partial isomorphism mapping
$v_1\to p_1$ and $v_2\to p_4$ can be extended to an automorphism $\varphi$ of $\relsys{U}'$.
The vertex $\varphi(v)$ is connected to $p_1$ and $p_4$ and thus together with
$p_1,p_2,p_4$ contains either triangle or $5$-cycle.  It follows that generic $\relsys{U}'$ cannot be defined by monadic lifts of $\mathop{Forb_h}\nolimits({\cal F})$.

In this section we prove that there is nothing special here about arity 2 and about the pentagon. One can determine the minimal arity of the ultrahomogeneous lifts for general classes $\mathop{Forb_h}\nolimits({\cal F})$. Towards this end we shall need a Ramsey type statement which we formulate after introducing the following:

Let $S$ be finite set with partition $S_1\cup S_2\cup\ldots\cup S_n$. For $v\in S$ we denote by $i(v)$ the index $i$ such that $v\in S_i$. Similarly, for a tuple $\vec{x}=(x_1,x_2,\ldots,x_t)$ of elements of $S$ we denote by $i(A)$ the tuple $(i(x_1), i(x_2),\ldots,i(x_t))$. We make use of the following:
\begin{lem}
\label{sets}
For every $n\geq 2$, $r<n$ and $K$ integers there is relational structure $\relsys{S}=(S,R_{\mathbf{S}})$, with vertices $S=S_1\cup S_2\cup\ldots S_n$ (sets $S_i$ are mutually disjoint) and single relation $R_{\mathbf{S}}$ of arity $2n$ with the following properties:
\begin{enumerate}
  \item Every $(v_1,u_1,v_2,u_2,\ldots v_n,u_n)\in R_{\mathbf{S}}$, $v_1,u_1\in S_1$, $v_2,u_2\in S_2$ satisfies $v_i\neq u_i\in S_i, i=1\ldots n$
  \item For every $\vec{v},\vec{u}\in \relsys{S}$, $\vec{v}\neq\vec{u}$, $\vec{v}$ and $\vec{u}$ has at most $r$ common vertices.
  \item
For every coloring of tuples on $S$ of size $r$ using $K$ colors there is tuple $\vec{v}\in R_{\mathbf{S}}$ such that every two tuples $\vec{x},\vec{x}'$ consisting of vertices of $\vec{v}$ such that $i(\vec{x})=i(\vec{x}')$ have same color.
\end{enumerate}
\end{lem}
\begin{proof}
This statement follows from results obtained by Ne\v set\v ril and R\"odl
\cite{NesetrilRodl}. Although not stated explicitely, this is ``partite version''
of the main result of \cite{NesetrilRodl}. It can be also obtained directly by
means of amalgamation method, see \cite{NCPC,NR2}. In this paper this result plays
auxiliary role only and we omit the proof.
\end{proof}
Given a relational structure $\relsys{S}=(S,R_{\mathbf{S}})$ with relation $R_{\mathbf{S}}$ of arity $2n$ and rooted relational structure $(\relsys{A}, \overrightarrow{R})$ of type $\Delta$ with $\overrightarrow{R}=(r_1,r'_1,r_2,r'_2,\ldots r_n,r'_n)$, we denote by $\relsys{S}*(\relsys{A},\overrightarrow{R})$ the following relational structure $\relsys{B}$ of type $\Delta$:

The vertices of $B$ are equivalence classes of equivalence $\sim$ on $R_{\mathbf{S}} \times A$
generated by the following pairs:

$$(\vec{v}, r_i)\sim (\vec{u}, r_i) \hbox{ iff } \vec{v}_{2i}=\vec{u}_{2i}$$
$$(\vec{v}, r'_i)\sim (\vec{u}, r'_i) \hbox{ iff } \vec{v}_{2i+1}=\vec{u}_{2i+1}$$
$$(\vec{v}, r_i)\sim (\vec{u}, r'_i) \hbox{ iff } \vec{v}_{2i}=\vec{u}_{2i+1}$$

Denote by $[\vec{v}, r_i]$ the equivalence class of $\sim$ containing $(\vec{v}, r_i)$.

We put $\vec{v}\in \rel{B}{j}$ iff $\vec{v}= ([\vec{u},v_1],[\vec{u},v_2],\ldots [\vec{u},v_t])$ for some $\vec{u}\in R_{\mathbf{S}}$ and $(v_1,v_2,\ldots v_t)\in \rel{A}{j}$.

This construction is commonly used in the graph homomorphism context as {\em indicator construction}, see \cite{Hell}. It essentially means replacing every tuple of $R_{\mathbf{S}}$ by disjoint copy of $\relsys{A}$ with roots $\overrightarrow{R}$ identified with vertices of the tuple.

For given vertex $v$ of $\relsys{S}*(\relsys{A},\overrightarrow{R})$ such that $v=[\vec{u}, r_i]$ (or $v=[\vec{u}, r'_i]$) we will call vertex $v'=\vec{u}_{2i}$ (or $v'=\vec{u}_{2i+1}$ respectively) {\em the vertex corresponding to $v$ in $\relsys{S}$}.  Note that this gives correspondence between vertices of $\relsys{S}$ and $\relsys{S}*(\relsys{A},\overrightarrow{R})$ restricted to vertices $[\vec{v}, r_i]$ and $[\vec{v}, r'_i]$.
\begin{figure}
\label{Hconstruct}
\centerline{\includegraphics{pokusc.eps}}
\end{figure}

Finite family finite relational structures is called {\em minimal} iff there
all structures in ${\cal F}$ are cores and there is no homomorphism in between two
structures
in ${\cal F}$.

The following is the main result of this section.

\begin{thm}
\label{aritythm}
Denote by ${\cal F}$ minimal family of finite connected relational structures. There is lift of class $\mathop{Forb_h}\nolimits({\cal F})$ that contains new relations of arity at most $r$ that is amalgamation class iff all minimal cuts of $\relsys{F}\in {\cal F}$ consist of at most $r$ vertices.
\end{thm}
\begin{proof}

Construction of lifted class ${\cal L}$ in proof of Theorem $\ref{mainthm}$ adds
relations of arities corresponding to the sizes of minimal cuts of
$\relsys{F}\in {\cal F}$ so one direction of Theorem \ref{aritythm} follows directly from the proof
of Theorem $\ref{mainthm}$.

In the opposite direction fix class ${\cal F}$, $r\geq 1$ and a relational structure
$\relsys{F}\in {\cal F}$ containing minimal cut $C=\{r_1,r_2,\ldots r_n\}$ of size
$n>r$.  Assume, for contradiction, that there exists lift ${\cal K}$ of class
$\mathop{Forb_h}\nolimits({\cal F})$ such that ${\cal K}$ is an amalgamation class and contains new relations
of arities at most $r$. Denote by $K$ the number of different relational structures on $r$ vertices
appearing in ${\cal K}$.

For the brevity, assume that $\relsys{F}\setminus C$ has only two connected components. Denote by ${\cal P}_1=(\relsys{P}_1,\overrightarrow{R}_1)$ and ${\cal P}_2=(\relsys{P}_2,\overrightarrow{R}_2)$ the pieces generated by $C$ such that $\overrightarrow{R}_1=\overrightarrow{R}_2=(r_1,r_2,\ldots r_n)$. (For 3 and more pieces we can proceed analogously.)

Now we construct relational structure $\relsys{H}$ as follows:
$$H= (P_1\times \{1\})\cup (P_2\times \{2\})$$
and put
$$((v_1,1),\ldots (v_t,1))\in \rel{H}{i}\hbox{ iff }(v_1,\ldots v_t)\in \rel{P_1}{i}$$
$$((v_1,2),\ldots (v_t,2))\in \rel{H}{i}\hbox{ iff }(v_1,\ldots v_t)\in \rel{P_2}{i}$$ with no other tuples.
In other words, $\relsys{H}$ is disjoint union of $\relsys{P}_1$ and $\relsys{P}_2$.  We will consider $\relsys{H}$ rooted by tuple $$\overrightarrow{R}=
((r_1,1),(r_1,2),(r_2,1),(r_2,2),\ldots,(r_n,1),(r_n,2)).$$

Take relational structure $\relsys{S}$ from Lemma \ref{sets} and put $\relsys{D}=\relsys{S}* (\relsys{H},\overrightarrow{R})$. This construction for the two pieces of 5-cycle is shown at Figure \ref{Hconstruct}.
For vertex $v\in S$ denote by $m(v)$ the vertex of $\relsys{D}$ corresponding to $v$ (if it exists) or arbitrary vertex of $\relsys{D}$ otherwise.

Denote by $f$ mapping $[(a,1),\vec{x})]\mapsto a$ for $a\in P_1$ and $[(a,2),\vec{x}]\mapsto a$ for $a\in P_2$. It is easy to check that $f$ is homomorphism $\relsys{D}\to \relsys{F}$.
Additionally for $v\in \relsys{D}$ put $$M(v)=\{\vec{x}; ((a,t),\vec{x})\hbox{ is in equivalence class } v\}.$$
Observe that for vertex $v\in \relsys{D}$ such that $f(v)\in C$, $M(v)$ may contain multiple tuples, while for all other vertices
$M(v)$ contains precisely one tuple.

Assume, to the contrary, that there is homomorphism $\varphi:\relsys{F}\to\relsys{D}$.  By composition we have that $\varphi \circ f$ is homomorphism $\relsys{F}\to\relsys{F}$.  Because $\relsys{F}$ is core, we also know that $\varphi \circ f$ is automorphism of
$\relsys{F}$. It follows that $\varphi$ is injective $\relsys{F}\to\relsys{D}$.
For $v\in \relsys{F}$ denote by $M(v)$ the set $M(\varphi(v''))$ where $v''$ is uniquely defined by $\varphi\circ f(v'')=v$.

It follows that for $v\in F\setminus C$, $M(v)$ consists of single tuple. For tuple $\vec{x}\in \rel{F}{i}$, there is tuple $\varphi(\vec{x})\in \rel{D}{i}$ iff sets $M(v), v\in \varphi(\vec{x})$ have nonempty intersection (i.e. all belongs to single copy of some piece $\relsys{P}_i$) and thus also sets $M(v), v\in \varphi\circ f(\vec{x})$ have nonempty intersection.

As relational systems $\relsys{P}_i\setminus C$ are connected, it follows
that all $M(v), v\in \relsys{P}_i\setminus C$ are equivalent singleton sets.
Denote by $\vec{x}_1$ tuple such that $M(v)=\{\vec{x}_1\}$ for $x\in P_1\setminus C$ and
by $\vec{x}_2$ tuple such that $M(v)=\{\vec{x}_2\}$ for $v\in P_2\setminus C$.
Because copies of pieces in $\relsys{D}$ corresponding to single tuple $\vec{x}\in \cal
X$ are not connected, we have $\vec{x}_1\neq \vec{x}_2$.
 Finally because every vertex in $C$ is connected to $P_1\setminus C$ by some tuple (from minimality of cut $C$), we have $\vec{x}_1\in M(v)$ for every $v\in C$ and
analogously $\vec{x}_2\in M(v)$ for every $v\in C$.  It follows that sets $\vec{x}_1$ and $\vec{x}_2$ overlaps on whole $C$. This $\vec{x}_1$ and $\vec{x}$ overlaps on $r$ or more vertices that is a contradiction with construction of relational system $\relsys{S}$. It follows that there is no homomorphism $\relsys{F}\to \relsys{D}$.

There is also no homomorphism $\relsys{F'}\to \relsys{D}$ for any $\relsys{F'}\in {\cal F}, \relsys{F'}\neq \relsys{F}$ because composing such homomorphism with $f$ would
lead to homomorphism $\relsys{F'}\to \relsys{F}$ that does not exist.  It follows that $\relsys{D}\in \mathop{Forb_h}\nolimits({\cal F})$.

Take generic lift $\relsys{U}\in {\cal K}$.  Every embedding $\Phi:\relsys{D}\to \psi(\relsys{U})$ ($\psi(\relsys{U})$ is shadow of $\relsys{U}$) imply $K$ coloring of $r$-tuples from elements of $D$ (colors are defined by the additional relations of $\relsys{U}$) and thus also
$K$ coloring of $r$-tuples of $\relsys{S}$.  Subsequently, using Lemma \ref{sets}, there is tuple $\vec{v}\in \relsys{S}$, such that $\vec{v}=(u_1,v_1,u_2,v_2,\ldots u_n,v_n)$ and the relations added by lift ${\cal K}$ are equivalent on $\Phi(u_i)$ and $\Phi(v_i), i=1,\ldots, n$. Thus $\relsys{U}$ induce on both sets $\{\Phi(u_1),\Phi(u_2),\ldots \Phi(u_n)\}$ and $\{\Phi(v_1),\Phi(v_2),\ldots \Phi(v_n)\}$ same lift $\relsys{X}$. ($\relsys{X}$ is lift of relational structure induced by $\relsys{F}$ on $C$.)  Subsequently there is partial isomorphism of $\relsys{U}$ mapping $\Phi(u_i)\to \Phi(v_i)$. From genericity of the $\relsys{U}$ this partial isomorphism extends to automorphism $\Psi$ of $\relsys{U}$. From the construction of relational system $\relsys{D}$ this mapping $\Psi$ sends a root of image of piece ${\cal P}_1$ to corresponding roots of image of piece ${\cal P}_2$ and thus shadow of $\relsys{U}$ contains copy of $\relsys{F}\in {\cal F}$, a contradiction.
\end{proof}

\section {Special cases of small arities}
\label{monadic}

By Theorem \ref{aritythm} it follows that only minimal classes of
finite relational structures ${\cal F}$ such that class $\mathop{Forb_h}\nolimits({\cal F})$ has monadic
lift that forms an amalgamation class are precisely classes ${\cal F}$ such that
all minimal vertex cuts of their Gaifman graph have size 1.
Examples forming an amalgamation class include graphs with all its blocks being complete graphs.

Consider even more restricted classes ${\cal F}$ of structures consisting from
(relational) trees only (relational trees are defined bellow). In this case we can claim a much stronger result:
there exists a finite  universal object $\relsys{D}$ which is a retract of an
universal structure $\relsys{U}$.

\subsection{Finite Dualities and Constraint Satisfaction Problems}

\label{Duality}
A Constraint Satisfaction Problem (CSP) is the following decision problem:

\smallskip

\noindent
Instance: A finite structure $\relsys{A}$

\noindent
Question: Does there exists a homomorphism $\relsys{A}\to\relsys{H}$?
\smallskip

We denote by $\mathop{CSP}\nolimits(\relsys{H})$ the class of all finite structures $\relsys{A}$ with $\relsys{A}\to\relsys{H}$.
It is easy to see that class $\mathop{CSP}\nolimits(\relsys{H})$ coincides with a particular instance of lifts and shadow.

A {\em finite duality} (for structures of given type) is any equation
$$\mathop{Forb_h}\nolimits({\cal F})=CSP(\relsys{D})$$
where ${\cal F}$ is s finite set \cite{NesetrilPultr,NTardif,Hell}. $\relsys{D}$ is called {\em dual of ${\cal F}$}. We also write $\relsys{D}_{\cal F}$ for dual of ${\cal F}$ (it is easy to see that $\relsys{D}_{\cal F}$ is up to homomorphism equivalence uniquely determined). The pair $({\cal F},\relsys{D})$ is called {\em dual pair}.  In a sense duality is a simple constraint satisfaction problem: the existence of homomorphism into $\relsys{D}$ (i.e. a $\relsys{D}$-coloring) is equivalently characterized by a finite set of forbidden substructures. Dualities play a role not only in the complexity problems but also in logic, model theory, partial orders and categories. Particularly it follows from \cite{Atserias} and \cite{Rossman} that dualities coincide with those first order definable classes which are homomorphism closed.

Finite dualities for monadic lifts include all classes $CSP(\relsys{H})$. We formulate this as follows:

\begin{prop}
\label{prop21}
For a class ${\cal K}$ of structures the following two statements are equivalent:
\begin{enumerate}
\item ${\cal K}=\mathop{CSP}\nolimits(\relsys{H})$ for finite $\relsys{H}$.
\item There exists a class ${\cal K}'$ of monadic lifts such that
\begin{enumerate}
\item Shadow of ${\cal K}'$ is ${\cal K}$.
\item ${\cal K}'=\mathop{Forb_h}\nolimits({\cal F}')\cap \mathop{Forb_e}\nolimits(\relsys{K}_1)$ where ${\cal F}'$ is a finite set of monadic covering lifts of edges (i.e. every $\relsys{F}\in{\cal F}'$ contains at most one non-unary tuple.) while every vertex belongs to a unary lifted tuple.
\end{enumerate}
\end{enumerate}
\end{prop}

\begin{proof}[Proof (sketch)]
1. obviously imply 2.

In the opposite direction construct $\relsys{H}$ as follows:  Let
$\relsys{H}_0$ be lift with vertex for every consistent combination of new
relations  $\ext{H_0}{i}$ and relations $\rel{H_0}{i}$ empty.  Now construct lift
$\relsys{H}$ on same vertex set as $\relsys{H}_0$ with $\ext{H}{i}=\ext{H_0}{i}$.
Put tuple $\vec{x}\in \rel{H}{i}$ iff the structure induced by $\vec{x}$ is on
$\relsys{H}_0$ with $\vec{x}$ added to $\rel{H}{i}$ is in $\mathop{Forb_h}\nolimits({\cal F}')$.
\end{proof}

In the language of dualities this amounts to saying that the classes $\mathop{CSP}\nolimits(\relsys{H})$ are just classes described by shadow dualities of the simplest kind: forbidden lifts are just vertex colored edges.

A (relation) tree can be defined as follows: The {\em incidence graph} $ig(\relsys{A})$ of relational structure $\relsys{A}$ is the
bipartite graph with parts $A$ and
$$
Block(A) = \{ (i,(a_1, \ldots, a_{\delta_i})) : i \in I,
(a_1, \ldots, a_{\delta_i}) \in \rel{A}{i} \},
$$
and edges $[a, (i,(a_1, \ldots, a_{\delta_i}))]$ such that
$a \in (a_1, \ldots, a_{\delta_i})$. (Here we write
$x \in (x_1, \ldots, x_n)$ when there exists an index $k$
such that $x = x_k$; $Block(A)$ is a multigraph.)  $\relsys{A}$ is called a {\em tree} when
$ig(\relsys{A})$ is a graph tree (see e.g. \cite{MatN}).
The definition of relational trees by the incidence graph $ig(\relsys{A})$ allows us to use graph terminology for relation trees.

Finite dualities are characterized (see \cite{NTardif})
\begin{thm}
For every type $\Delta$ and for every set ${\cal F}$ of relational trees there exists a dual $\Delta$-structure $\relsys{D}_{\cal F}$. Up to homomorphism equivalence there are no other dual pairs.
\end{thm}
\label{dualpairs}
Given ${\cal F}$ various constructions of structure duals are known \cite{NTardif2}.
It follows from this Section that we have a yet another approach to this problem:
\begin{corollary}
\label{nasdual}
Let ${\cal F}$ be a set of finite tree-structures, then there exists finite set of lifted structures ${\cal F}'$ with the following properties:
\begin{itemize}
  \item[$(i)$] $\mathop{Forb_e}\nolimits({\cal F}')$ is amalgamation class (and thus there is universal $\relsys{U}'\in \mathop{Forb_e}\nolimits({\cal F}')$),
  \item[$(ii)$] all lifts in $\mathop{Forb_e}\nolimits({\cal F}')$ are monadic,
  \item[$(iii)$] $\psi(\relsys{U'})=\relsys{U}$ is universal for ${\cal K}$,
  \item[$(iv)$] $\relsys{U}'$ has a finite retract $\extsys{D}_{\cal F}$ and consequently $\psi(\extsys{D}_{\cal F})=\relsys{D}_{\cal F}$ is a dual of ${\cal F}$.
\end{itemize}
\end{corollary}

\begin{proof}[Proof]
Observe that the inclusion minimal cuts of a relational tree
are all having arity 1. Thus for fixed family ${\cal F}$ of finite relational trees
 our Theorem \ref{mainthm} establishes the existence monadic lift
that give an generic structure $\relsys{U}'$ whose shadow is (homomorphism) universal for $\mathop{Forb_h}\nolimits({\cal F})$.

This structure $\relsys{U}'$ is countable.  To get a dual, we find finite
$\relsys{X'}\in {\cal L}$ which is retract of $\relsys{U}'$ and still there is homomorphism $\relsys{Y}\to \relsys{U}'$ iff there is homomorphism $\relsys{Y} \to \relsys{X'}$.

The set ${\cal F}'$ is given by construction in Theorem  $\ref{reprezent}$.
 Observe that every inclusion minimal covering set of every piece of tree is induced by single
tuple and thus the class ${\cal L}$ is defined by forbidden (rooted) homomorphism of structures covered by single tuple.
This means that the generic graph $\relsys{U}'$ has finite retract defined by
all consistent combinations of new relations of its vertices.
\end{proof}

Note that it is also possible to construct $\relsys{D}_{\cal F}$ in a finite way
without using the Fra\"{\i}ss\' e{} limit: for every possible combination of new
relations on single vertex create single vertex of $\relsys{D}_{\cal F}$ and then
keep adding tuples as long as possible so $\relsys{D}_{\cal F}$ is still in ${\cal L}$.

Finally, let us remark that one can prove that $\extsys{U}$ has a finite presentation in the sense of $\cite{HN-Posets}$.

\label{aplikace}

The situation of course changes if we consider more complicated (non-monadic and non-trees) lifted classes $\mathop{Forb_e}\nolimits({\cal F}')$ and in fact it has been proved in \cite{KunN} that any NP language is polynomally equivalent to a class $\psi(\mathop{Forb_e}\nolimits({\cal F}'))$ for a finite set ${\cal F}'$ set of lifted structures. As these structures may be supposed to be connected
(by adding dummy edges) we have the following corollary of Theorem
\ref{univthm}:
\begin{corollary}
\label{NPU}
Every NP problem is polynomally equivalent to membership problem for a class $\mathop{Age}\nolimits(\relsys{U})$ where $\relsys{U}$ is shadow of an ultrahomogeneous structure $\extsys{U}$.
\end{corollary}

Constraint Satisfaction Problems $CSP(\relsys{H})$ for countable templates $\relsys{H}$ were investigated in \cite{BodirskyNesetril} and it has been shown there that the basic results of so called ``algebraic method'' \cite{BufatovJeavonKron} hold also for $\omega_0$-categorical templates \cite{BodirskyNesetril}. Recently, \cite{Bodirsky} proved that $CSP(\relsys{H})$ for general countable template $\relsys{H}$ encodes any problem in NP. It is not known whether $\relsys{H}$ may be chosen $\omega_0$-categorical. Corollary \ref{NPU} complements this: lifts and shadows enable to code any NP language by membership problem for ages of shadows of ultrahomogeneous structures.

\subsection {Forbidden cycles and Urysohn spaces (binary lifts)}

We turn briefly our attention to binary lifts.  This relates some of the earliest results on universal graphs with recently intensively studied
Urysohn spaces:

\label{urysohnsection}
We will consider a finite family ${\cal F}$ consisting of graphs of odd cycles of lengths
$3,5,7,\ldots l$.  As shown by \cite{Mekler} (see also \cite{CherlinShelahShi}) those families have universal
graphs in $\mathop{Forb_e}\nolimits({\cal F})$ and as shown by \cite{CherlinShi} those are the only
classes defined by forbidding a finite set of  cycles.
These classes also form especially easy families of pieces.  In fact each piece
is an undirected path of length at most $l$, where $l$ is the length of longest cycle
in ${\cal F}$ with both ends of the patch being roots.  This allows particularly easy
description of the lifted structure.

We use the following definition which is motivated by metric spaces.
When specialized to graphs, this definition is analogous to (corrected form) of an
$s$-structure \cite{Mekler}. However this approach also gives new easy description (i.e. finite presentation)
of the Urysohn space \cite{metric}.

\begin{defn}
\label{evenoddp}
Pair $(a,b)$ is considered to be {\em even-odd pair} if $a$ is even non-negative integer or $\omega$ and $b$ is odd non-negative integer or $\omega$.

For even-odd pairs $(a,b)$ and $(c,d)$ we say that $(a,b)\leq (c,d)$ iff $a\leq c$ and $b\leq d$.  Consider $a+\omega=\omega$ and $\omega+b=\omega$.  Put: $$(a,b)+(c,d) = (\min(a+c,b+d),\min (a+d,b+c)).$$

For a set $S$, function $d$ from $S$ to even-odd pairs is called {\em even-odd distance function on $S$} if following conditions are satisfied:
\begin{enumerate}
\item $d(x, y) = (0,b)$   if and only if   $x = y$;
\item $d(x, y) = d(y, x)$;
\item $d(x, z) \leq d(x, y) + d(y, z)$.

\end{enumerate}
Finally pair $(S,d)$ where $d$ is even-odd distance function for $S$ is called {\em even-odd metric space}.
\end{defn}

Note that the even-odd metric spaces differ from usual notion of metric space
primarily by the fact that the ordering of values of distance function is not
linear, but forms a 2-dimensional partial order.
Some basic results about metric spaces are valid even in this setting.

Even-odd metric space can form a stronger version of the distance metric on the
graph.  For graph $\relsys{G}$ we can put $d(x,y)=(a,b)$ where $a$ is length of the
shortest walk of even length connecting $x$ and $y$, while $b$
is the length of shortest walk of odd length.

The even-odd distance metric specify length of all possible walks: for a graph
$\relsys{G}$ and a even-odd distance metric $d$ we now have a walk connecting $x$ and
$y$ of length $a$ iff $d(x,y)=(b,c)$ such that $b\leq a$ for $a$ even or $c\leq
a$ for $a$ odd.

It is well known that universal and homogeneous metric space exists for several classes of
metric spaces \cite{d9,Nguyen}.  Analogously we have:
\begin{lem}
\label{Ulema}
There exists a generic even-odd metric space $\relsys{U}$.
\end{lem}
\begin{proof}
 We prove that class ${\cal M}$ of all even-odd metric spaces is an amalgamation class.

To show that ${\cal M}$ has the amalgamation property,
 take a free amalgam $\relsys{D}$ of even-odd metric spaces $\relsys{A}$,
$\relsys{B}$ over $\relsys{C}$. This amalgam is not even-odd metric space, since some distances are not defined.

We can however define a walk from $v_1$ to $v_t$ of length $l$ in $\relsys{D}$ as a sequence of vertices $v_1,v_2,v_3,\ldots v_t$
and distances $d_1,d_2,d_3,\ldots d_{t-1}$ such that $\sum_{i=1}^{t-1} d_i=l$ and $d_i$ is present in
the even-odd pair $d(v_i,v_i+1)$ for $i=1,\ldots,t-1$.

We produce even-odd metric space $\relsys{U}$ on same vertex set as
$\relsys{D}$ where distance of some vertices $a,b\in E$ is even-odd pair
$(l,l')$ such that $l$ is the smallest even value such that there exists walk
joining $a$ and $b$ of length $l$ in $\relsys{D}$. $l'$ is the smallest odd value such
that there exists walk from $a$ to $b$ of length $l'$.

It is easy to see that $\relsys{U}$ is even-odd metric space (every triangular
inequality is supported by the existence of a walk) and other properties of
amalgamation class follows from definition.
\end{proof}

The graphs omitting odd cycles up to length $l$ can be axiomatized by simple
condition on their even-odd distance metric: there are no vertices $x,y$ such
that $d(x,y)=(a,b)$ where $a+b\leq l$. Denote by ${\cal K}_l$ the class of all countable even-odd metric spaces such that there are no vertices $x,y$ such that $d(x,y)=(a,b)$ with $a+b\leq l$.  The existence of universal and homogeneous even-odd metric space $\relsys{U}_l=(U_l,d_l)$ for class ${\cal K}_l$ is simple consequence of Lemma \ref{Ulema}. In fact $\relsys{U}_l$ is a subspace of $\relsys{U}$ induced by all those vertices $v$ of $\relsys{U}$ satisfying $d(v,v)=(0,b)$ and $b>l$.

\begin{thm}
For metric space $\relsys{U}_l=(U_l,d_l)$ denote by $\relsys{G}_l=(U_l,E_l)$ a graph on vertex set $U_l$ where $\{x,y\}\in E_l$ iff $d(x,y)=(a,1)$.

For every choice of odd integer $l\geq 3$, $\relsys{G}_l$ is universal graph for class of all graphs omitting odd cycles of length at most $l$.
\end{thm}
\begin{proof}
Graph $\relsys{G}_l$ is omitting odd cycles up to length $l$ from the fact that any two
vertices $x,y$ on a odd cycle of length $k$ have distance $d(x,y)=(a,b)$ where
$a+b$ is at most $l$.

Now consider any countable graph $\relsys{G}=(V,E)$ omitting odd cycles of length at most
$l$. Construct corresponding even-odd distance metric space $(V,d_\relsys{G})$. By universality argument  $(V,d_\relsys{G})$ is subspace of $\relsys{U}_l$ and thus also $\relsys{G}$ subgraph of $\relsys{G}_l$.
\end{proof}

The explicit construction of the Urysohn space, without using Fra\"{\i}ss\' e{} limit
argument, is described in \cite{metric} and same technique can be carried to
even-odd metric. This is captured by the following definition.

\begin{defn}
\label{defU}
The vertices of ${\cal U}$ are functions $f$ such that:
\begin{enumerate}
\item[(1)] The domain $D_f$ of $f$ is finite (possibly empty) set of functions and $\emptyset$.
\item[(2)] The range of $f$ are even-odd pairs.
\item[(3)] For every $g\in D_f$ and $h\in D_{g}$, we have $h\in D_f$.
\item[(4)] $D_f$ using metric $d_{{\cal U}}$ defined bellow forms a even-odd metric
space.
\item[(5)] $f$ defines an extension of even-odd metric space on vertices $D_f$ by adding a new vertex. This means that $f(\emptyset)=(0,x)$ and for every
$g, h\in D_f$ we have $f(g)+f(h)\leq d_{{\cal U}}(g,h)$ and $f(g)\geq f(h)+d_{\cal U}(g,h)$.

\end{enumerate}
The metric $d_{{\cal U}}(f,g)$ is defined by:
\begin{enumerate}
  \item if $f=g$ then
$d_{{\cal U}}(f,g)=f(\emptyset),$
  \item if $f\in D_g$ then
$d_{{\cal U}}(f,g)=g(f),$
  \item if $g\in D_f$ then
$d_{{\cal U}}(f,g)=f(g),$
  \item if none of above holds then
$d_{{\cal U}}(f,g)=min_{h\in D_f\cap D_g} f(h)+g(h).$

Minimum is taken elementwise on the pairs.
\end{enumerate}
 \end{defn}

\begin{thm}\label{generic}
$({\cal U},d_{\cal U})$ is the generic even-odd metric space.
\end{thm}
This may be seen as yet another incarnation of Kat\v etov functions \cite{Uspen2}, see also \cite{Nguyen}.

\section{Indivisibility results}
We call pair $(A,B)$ a {\em partition of structure} $\relsys{R}$ if $A$ and $B$ are disjoint sets of vertices of $\relsys{R}$ and $A\cup B=R$.  We denote by $\relsys{A}$ the structure induced on $A$ by $\relsys{R}$ and by $\relsys{B}$ the structure induced on $B$ by $\relsys{R}$.

The structure $\relsys{R}$ is {\em weakly indivisible} if for every partion
$(A,B)$ of $R$ for which some finite induced substructure of $\relsys{R}$ does
not have copy in $\relsys{A}$, there exists copy of $\relsys{R}$ in
$\relsys{B}$.

For finite minimal family of finite structures ${\cal F}$, we call structure
$\relsys{A}$ a {\em minimal homomorphic image} of $\relsys{F}\in {\cal F}$ iff
$\relsys{A}$ is homorphic image of $\relsys{F}$ and every proper substructure
of $\relsys{A}$ is in $\mathop{Forb_h}\nolimits({\cal F})$.

The weak indivisibility of homogeneous structures was studied by \cite{Sauer}.
In this section we briefly discuss basic (in)divisibility results on universal
structures of classes $\mathop{Forb_h}\nolimits({\cal F})$.

We say that age of structure $\relsys{A}$
has {\em free vertex amalgamation} if any two finite substructures of $\relsys{A}$ whose intersection is a
singleton have free amalgamation. We make use of the following:

\begin{thm}[\cite{Sauer}]
\label{Sauerthm}
Let $\relsys{H}$ be a countable homogeneous structure with free vertex amalgamation.  Then $\relsys{H}$ is weakly indivisible.
\end{thm}

The construction of universal structures as shadows of ultrahomogeneous makes
this result particularly easy to apply to obtain indivisibility results for
universal structures for classes $\mathop{Forb_h}\nolimits({\cal F})$. This leads to the following partial classification of classes ${\cal F}$ that do admit weakly indivisible structure universal for $\mathop{Forb_h}\nolimits({\cal F})$.
\begin{thm}
Fix minimal finite family of finite structures ${\cal F}$.
\begin{enumerate}
\item The class $\mathop{Forb_h}\nolimits({\cal F})$ contains universal structure that is weakly indivisible if every vertex minimal cut $C$ of every homomorphic image $\relsys{A}$ of $\relsys{F}\in {\cal F}$ is of size at least 2
and additionally the structure induced by $\relsys{A}$ on $C$ is connected and has no cuts of size 1.
\item All universal structures $\relsys{U}$ are
divisible if there is a structure $\relsys{A}$ which is a minimal homomorphic image of $\relsys{F}\in {\cal F}$ such
that $\relsys{A}$ contains cut $C$ of size 1.
\end{enumerate}
\end{thm}

\begin{proof}

To prove $2.$, fix $\relsys{A}$, a minimal homomorphic image of $\relsys{F}\in {\cal F}$
with vertex cut $C$ of size 1.  Denote by ${\cal P}_1,{\cal P}_2,\ldots,{\cal P}_n$
all pieces of $\relsys{A}$ generated by cut $C$.  Fix $\relsys{U}$, universal
structure for $\mathop{Forb_h}\nolimits({\cal F})$. Denote by $U_i$, $i=1...\ldots n$ set of all
vertices $v$ of $\relsys{U}$ such that there is rooted homomorphism of
${\cal P}_i$ to $\relsys{U}$ mapping root of ${\cal P}_i$ to $u$.

Structure induced on $U_i$ on $\relsys{U}$ is not universal as it does not contains a homomorphic image of $\relsys{P}_i$.
 Similarly structure induced on
$U\setminus \bigcup_{i=1}^n U_i$ is not universal since there is no
homomorphic image of $\relsys{P}_1$.  As a result $\relsys{U}$ is divided
to finitely many substructures such that none is universal resulting in
divisibility of $\relsys{U}$.

$1.$ follows from weak indivisibility of class ${\cal L}$.  To apply Theorem
\ref{Sauerthm} we only need to show that class ${\cal L}$ admits free vertex
amalgamation.  This follows directly from the construction of amalgam in proof
of Theorem $\ref{univthm}$: the amalgam constructed is not free in general,
every new tuple added has the property that there is homomorphism from the
structure induced by cut generating the corresponding piece into the vertices
of tuple.   But since we have free amalgam of the shadow and since all cuts of all homomorphic
images do not have cuts of size 1, we have free vertex amalgamation property.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Mack's Attitude May Drive Him to Excel With Raiders
Clowney went No. 1 overall, but some believe Oakland's new linebacker may prove to be the best defensive player from this year's draft
By Doug Williams	• Published at 7:51 am on May 14, 2014
If we're keeping score, defensive end Jadeveon Clowney is ahead of Khalil Mack.
Clowney was the No. 1 pick of last week's NFL draft, going to the Texans. Mack was selected fifth overall by the Raiders.
But Mack, an outside linebacker with great range and a talent for getting to the quarterback, may have something that Clowney doesn't have: a chip on his shoulder. And that may take him to greater heights than Clowney in 2014 and beyond.
49ers 11 hours ago
First-Round Investments Paying Off for 49ers
Warriors Jan 16
Warriors takeaways: What we learned in 134-131 OT loss to Nuggets
Mack's work ethic is highly regarded, and some analysts and scouts believe Mack was the better pro prospect in the draft. Gregg Easterbrook of ESPN.com's Tuesday Morning Quarterback is one who believes Mack is likely to be a better pro than Clowney.
"Clowney, the first defensive player taken, received dozens of college recruiting offers," wrote Easterbrook this week. "Khalil Mack, the second defensive player taken, received just one scholarship offer, from the University of Buffalo. … Clowney has been an athletic celebrity since high school; Mack was an athletic nobody till last fall.
"Celebrities tend to want to lie on the beach sipping mai tais, while nobodies tend to be driven. Don't be surprised if over the next five years, Mack significantly outworks Clowney."
Also, in a poll of six NFL analysts at NFL.com, who were asked to pick the NFL's Offensive and Defensive Rookies of the Year for 2014, Mack received as many votes as Clowney. Gil Brandt, the former Dallas Cowboys talent evaluator, wrote that he believes Mack landed with the perfect team. The Raiders will find ways to make the best use of Mack's talents, he says.
"Raiders defensive coordinator Jason Tarver is a mad scientist with schemes," Brandt wrote. "He will take advantage of Mack's unbelievable pass-rushing talents and design game plans around his new defensive rock. I would be surprised if Mack failed to record double-digit sacks as a rookie."
Mack has been compared by some to Packers outside linebacker Clay Matthews. But Raiders head coach Dennis Allen compares Mack to Von Miller, the Broncos' pass-rushing phenom who has had 35 sacks in 40 games as a pro.
As a collegian, Mack had 28½ sacks in 48 games and 75 tackles for loss.
After being drafted by the Raiders, Mack acknowledged that he believes he was the best defensive player in the draft. And he'd like to prove it.
"I'm at the point now where I'm tired of talking and I want to go out and start proving a little bit. … I cannot wait," he said.

redpajama_set_name: RedPajamaCommonCrawl


Na een mediane follow-up periode van 42 maanden de complete remissiecijfers in de Thalidomide groep was 62%, dit is significant beter dan de 43% uit de controlegroep (p < 0.001), stellen de auteurs. Tegelijkertijd was ook de 5-jaars overleving in de Thalidomidegroep significant hoger: 56% vs. 44% (p = 0.01). Echter: beide groepen hadden een 5-jaars overall overlevingscijfer van ca. 65%. De mediane overlevingstijd na een recidief in de thalidomide groep was 1,1 jaar tegenover 2,7 jaar in de controlegroep (p = 0.001). Diepe adertrombose kwam voor bij 34% van de met thalidomide behandelde patiënten tegenover 18% in de controlegroep (p < 0.001). Een behandeling met laag-moleculair-gewicht heparin verminderde niet het verhoogde risico. Andere bijwerkingen die vaker werden gezien in de thalidomidegroep waren o.a. periferale neuropathie, syncope, maagobstructie, trillingen (angst?) en neutropenia.
Reuters Health Information 2006. © 2006 Reuters Ltd.
NEW YORK (Reuters Health) Mar 08 - Adding thalidomide to high-dose therapy for multiple myeloma makes complete responses more likely and increases event-free survival, but does not improve overall survival and has considerable toxicity, new research shows.
Melphalan-based high-dose therapy has been shown to improve the survival of multiple myeloma. However, it was unclear if adding thalidomide, an agent with activity against advanced and refractory disease, to this therapy could further improve outcomes. To investigate, Dr. Bart Barlogie, from the University of Arkansas for Medical Sciences in Little Rock, and colleagues assessed the outcomes of 668 patients who were randomized to receive two cycles of melphalan-based therapy, supported by hematopoietic-cell transplantation, with or without thalidomide.
The researchers' findings appear in The New England Journal of Medicine for March 9th.
After a median follow-up period of 42 months, the complete response rate in the thalidomide group was 62%, significantly higher than the 43% rate noted in the control group (p < 0.001), the authors state. Similarly, the 5-year event-free survival rate in the thalidomide group was also higher, 56% vs. 44% (p = 0.01).
Due in part to a lower response rate to salvage therapy, the addition of thalidomide did not improve overall survival; both groups had a 5-year overall survival rate of about 65%. The median survival period after relapse in the thalidomide group was 1.1 years compared with a period of 2.7 years in the control group (p = 0.001). Deep vein thrombosis occurred in 34% of thalidomide-treated patients compared with 18% of controls (p < 0.001). Treatment with low-molecular-weight heparin did not eliminate the elevated risk. Other adverse events seen more often in the thalidomide group included peripheral neuropathy, syncope, bowel obstruction, tremor and neutropenia.
"These results indicate that contrary to a widely held belief, a complete response is not a valid surrogate for overall survival in clinical trials," Dr. Michele Cavo and Dr. Michele Baccarani, from the University of Bologna in Italy, comment in a related editorial. "The higher rate of failure to respond to salvage therapy in the thalidomide group needs to be investigated further," they suggest, "especially with respect to the salvage potential of new drugs in patients who had received thalidomide as initial treatment."
N Engl J Med 2006;354:1021-1030,1076-1078.
Reageer op "Thalidomide geeft bij Kahler - Multiple Myeloma meer complete remissies en langere tijd tot recidief optreedt maar geeft weinig tot geen betere overall overleving en geeft aanzienlijk meer bijwerkingen aldus nieuwe fase III studiegegevens. Artikel geplaat"


text:
Ardian acquires a majority stake in Sintetica, a historic swiss pharmaceutical company, to support its internationalization strategy
Mendrisio, May 23rd 2019. Ardian, a world-leading private investment house, announces the signing of a binding agreement for the acquisition of a majority stake in Sintetica SA, a Swiss pharmaceutical company, to support its international expansion in line with the management team's strategy. Under the agreement, current shareholders will reinvest via a minority stake.
Founded in 1921, Sintetica specializes in research, manufacturing and marketing of sterile injectable solutions, mainly in anesthetics, analgesics and adrenergic agonists. The company has manufacturing plants in Mendrisio and Couvet (Switzerland)and employs 280 people. Sintetica has a portfolio of over 50 products which are distributed in more than 45 countries. The company has experienced strong growth in recent years (+31% CAGR in the period 2016-2018) and its turnover in 2018 was approximately CHF 75 million.
Sintetica's unique proposition is characterized by high-quality and innovative products, which are the result of its close collaboration with hospitals and universities. Over the last few years, Sintetica has developed and patented several innovative drugs and ready-to-use formulations in-house, aligned with the latest needs of both patients (e.g. less use of opioid drugs) and hospitals (by supporting, among others, one-day-surgery).
Yann Chareton, Managing Director at Ardian, said: "We are excited to join forces with Sintetica and support the company in its next growth phase by investing in innovation and further improving product quality. Leveraging Ardian's global network and consolidated experience in partnering with entrepreneurs and managers, we intend to support the company in the development of its strategic plan in continuity with the growth path already undertaken."
Luca Bolzani, shareholder and Chairman of Sintetica's Board of Directors, also on behalf of current majority shareholder Daphne, declared: "We are very pleased about the new partnership with Ardian, which will allow Sintetica to further develop its presence in international markets. At the end of a detailed evaluation phase, we have chosen to share our growth project with Ardian, thanks to its credibility and its industrial approach. We are sure that the opportunities arising from the new partnership will be crucial in accelerating our development phase and ensure the necessary continuity to Sintetica's growth. We believe that, thanks to this partnership, the best conditions have been created to further develop the project we started in the early 2000s, which led Sintetica to become one of the most innovative companies in its reference market."
Augusto Mitidieri, CEO of Sintetica, commented: "The company's strategy is based on a global innovation leadership. Quality without compromise and a relentless propensity to innovate are the cornerstones of our strategic plan. People are our most important asset and represent the main reason for our success. We are all pleased to share the new chapter of the company's growth with Ardian, which will contribute to strengthening Sintetica's global expansion."
ABOUT ARDIAN
Ardian is a world-leading private investment house with assets of US$90bn managed or advised in Europe, the Americas and Asia. The company is majority-owned by its employees. It keeps entrepreneurship at its heart and focuses on delivering excellent investment performance to its global investor base.
Through its commitment to shared outcomes for all stakeholders, Ardian's activities fuel individual, corporate and economic growth around the world.
Holding close its core values of excellence, loyalty and entrepreneurship, Ardian maintains a truly global network, with more than 600 employees working from fifteen offices across Europe (Frankfurt, Jersey, London, Luxembourg, Madrid, Milan, Paris and Zurich), the Americas (New York, San Francisco and Santiago) and Asia (Beijing, Singapore, Tokyo and Seoul). It manages funds on behalf of around 880 clients through five pillars of investment expertise: Fund of Funds, Direct Funds, Infrastructure, Real Estate and Private Debt.
www.ardian.com
ABOUT SINTETICA
Founded in 1921 and based in Mendrisio, Sintetica is a Swiss pharmaceutical company specialized in the research, manufacturing and marketing of sterile injectable solutions, including anaesthetic, analgesic and adrenergic drugs, distributed in more than 45 countries. Sintetica operates with an agile distribution model based on two channels: a direct one with hospitals (in Switzerland, Germany, Austria, the United Kingdom and Ireland) and another one based on strategic partnerships through license agreements on a global basis.
With 280 employees and manufacturing plants in Mendrisio and Couvet, the company is globally recognised for its innovative human-centred organisational model, the distinctive go to market philosophy based on a 360 degrees respect and on the high quality of its products.
www.sintetica.com
ARDIAN - Advisor
M&A Advisor: Leopoldo Zambeletti
Commercial Due Diligence: IQVIA – Massimiliano Rubin, Dean Griffiths
Financial Due Diligence: PWC – Emanuela Pettenò, Luca Vergani
Tax: Gitti & Partners – Diego De Francesco, Paolo Ferrandi
Legal: Gianni, Origoni, Grippo, Cappelli & Partners – Gianluca Ghersini, Valentina Dragoni, Raffaella Ceglia
Financing: Gattai, Minoli, Agostinelli & Partners – Lorenzo Vernetti, Silvia Romano, Giorgia Gentilini
Operational Due Diligence: Candesic – Marc Kitten, Floris Wentholt
SINTETICA - Advisor
Financial Due Diligence: PWC – Federico Mussi, Francesco Tieri
M&A Advisor: Rothschild & Co
Legal Advisor: NCTM – Paolo Montironi, Pietro Zanoni, Eleonora Parrocchetti, Alessia Trevisan
Tax: Ludovici Piccone & Partners – Michele Aprile, Loredana Conidi, Daniel Canola
ARDIAN
Viktor Tsvetanov
vtsvetanov@headlandconsultancy.com


Created by vishals@grazitti.com on behalf of 0001fedf9bedbce6cde009f19f51494f5d1d8f4d on Jun 8, 2017 10:44 PM. Last modified by vishals@grazitti.com on Jun 8, 2017 10:44 PM.
Issue Description: You want to push Marketo activities into MS Dynamics.
Issue Resolution: Marketo has the ability to push activities in SFDC, however, the same feature does not exist for MS Dynamics. Instead, you can use Interesting Moments in Marketo Sales Insight to pass activities into Microsoft Dynamics.
Who This Solution Applies To: Customers using Microsoft Dynamics.


text:
Did Steven Mnuchin and Louise Linton use a government plane to watch the eclipse?
GENE BLEVINS/AFP/Getty Images
A watchdog group is attempting to get to the bottom of whether or not Treasury Secretary Steven Mnuchin and his wife, Louise Linton, used a government plane for a trip centered around viewing the total solar eclipse earlier this week.
Citizens for Ethics and Responsibility in Washington (CREW) filed a Freedom of Information Act request to learn "the justification for Secretary Mnuchin's use of a government plane, rather than a commercial flight, for a trip that seems to have been planned around the solar eclipse and to enable the secretary to secure a viewpoint in the path of the eclipse's totality."
Did Steven Mnuchin & Louise Linton use a gov't plane to get a better view of the eclipse? We FOIA'd to find out. https://t.co/uvzWjq1RV7 pic.twitter.com/wFJCYlwcLZ
— Citizens for Ethics (@CREWcrew) August 23, 2017
Mnuchin and Linton officially traveled to Lexington, Kentucky, so Mnuchin could speak alongside Senate Majority Leader Mitch McConnell (R-Ky.) and attend a luncheon. Mnuchin and Linton subsequently boarded the plane to Fort Knox "to tour the bullion reserve at the Army post and view the eclipse."
It is not the only controversy the couple has faced this week. Linton also posted a photo to Instagram of her disembarking the government plane in Kentucky and wrote an ill-advised reply to a commenter who called the getaway "deplorable."


Barska's 8 Megapixel Trail Camera is designed to be your eyes in the woods day or night. Wide, flash range with rapid response time means you won't miss an opportunity for an image. All settings, including camera resolution, video resolution and PIR sensitivity are adjustable for your specific needs.
Barska's Trail Camera allows you to upload images and videos to a computer through the included USB cable to save or share with others. Simply load AA batteries (not included), insert an SD card (not included) select your preferred settings and the trail camera is ready to go. All images are stamped with date and time so you can spot trends and or patterns giving you everything you need to prepare for a great hunt.


The Battle of Graspan. Drawn by J Finnemore, RI, from a sketch by Lester Ralph, our special correspondent with Lord Methuen'. Fold-out illustration published as a Supplement to 'Black and White', January 6, 1900. Australia fought at the Battle of Graspan in the Boer War. Central vertical fold with a couple of staple holes, as issued. Other Boer War images on reverse.


text:
Walgreens Still Selling Cigarettes, But Will Hike Minimum Age For Sales
Filed Under:American Lung Association, Chewing tobacco, Cigarette, cigarettes, Cigars, drug store, smoking, Tobacco, Walgreen, walgreens
DEERFIELD, Ill. (CBSDFW.COM/AP) – Walgreens has decided to raise its minimum age for tobacco sales several weeks after a top federal official chastised the drugstore chain for violating laws restricting access to cigarettes and other tobacco products.
Illinois-based Walgreens Boots Alliance Inc. said Tuesday that it will require customers to be at least 21 years old to purchase tobacco in any of its more than 9,500 stores nationwide. The policy starts September 1.
File photo of a man smoking a cigarette. (credit: Eric Feferberg/AFP/Getty Images)
Former Food and Drug Administration Commissioner Dr. Scott Gottlieb said in March that Walgreens was a top violator of tobacco sales laws among pharmacies that sell those products. Rival CVS Health Corp. stopped selling cigarettes in 2014.
Laws restricting tobacco sales vary nationally.
Most states have set a minimum age of 18, while a dozen have raised that to 21, according to the American Lung Association. Alaska and Alabama set their minimum ages at 19. Many cities and counties also have passed local laws establishing the minimum at 21.
Senate Majority Leader Mitch McConnell said last week that he plans to introduce legislation to raise the minimum age to 21 nationally. The Kentucky Republican called the legislation a top priority.
(© Copyright 2019 CBS Broadcasting Inc. All Rights Reserved. The Associated Press contributed to this report.)


Yesterday morning I finally got around to shaving my beard off. I'd had it for around four years this time and had got very used to it but it really was time for a change. I made some silly joke about my chin getting cold now and then found that when I went out I actually could feel the difference.
I've taken a good critical look at myself in the mirror and, while I still hate the way I look, I do have to say that I prefer being clean-shaven. Every time I put my hand to my chin, which I seem to do quite often, I notice and it feels good.
Today, among all the other tasks – putting a new tax disc on the bike, washing, washing up, possibly a run to the council tip – I hope to be dying my hair. It's all change around here.