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Britain could become the first country to sanction the creation of babies with three parents.
The Government has been urged to change the law so women with devastating hereditary diseases have the chance to have healthy children, within three years.
The technique would involve defective DNA in the mother's egg being replaced with material from a donor egg. A child would effectively have two mothers and a father.
dementia, blindness and premature death.
sausage-shaped powerhouses inside cells that turn food into energy.
mitochondrial DNA from the donor.
public for allowing IVF clinics to use the technique.
is born no one can be certain.
Although no country yet sanctions the technique, some babies using it have been born in the US.
The HFEA document will be passed to the Government which will decide whether to try to amend the law.
to the disaster of genetically engineered babies and consumer eugenics'.


text:
China Sanctions 11 Americans Who "Behaved Badly" Over Hong Kong Issue
HomeWorldChina Sanctions 11 Americans Who "Behaved Badly" Over Hong Kong Issue
This Article is From Aug 10, 2020
Washington announced on Friday it was freezing the US assets of Hong Kong Chief Executive Carrie Lam and 10 other senior Chinese officials, in the toughest US action on Hong Kong
WorldAgence France-PresseUpdated: August 10, 2020 3:20 pm IST
US senator Ted Cruz is among those sanctioned by China.
Beijing:
China on Monday sanctioned 11 Americans, including senators Marco Rubio and Ted Cruz, in retaliation for similar US moves prompted by Beijing's crackdown in Hong Kong.
Washington last week accused 11 officials of suppressing "freedom and democratic processes" in Hong Kong, including city leader Carrie Lam, and announced plans to freeze their US assets.
It was the toughest US action yet in response to Beijing's introduction of a sweeping and controversial new national security law for the territory.
Beijing said the measure was a violation of international law and "grossly interferes in China's internal affairs".
"China has decided to impose sanctions on some people that behaved badly on Hong Kong-related issues," foreign ministry spokesman Zhao Lijian said Monday, with Human Rights Watch director Kenneth Roth and National Endowment for Democracy president Carl Gershman also on the list.
Zhao did not give details of what the sanctions would entail.
Republican senators Rubio and Cruz established themselves as two of the most vocal supporters of Hong Kong's democracy movement last year, when the city was convulsed by huge and sometimes violent protests.
Beijing has accused "external forces" of fomenting unrest and responded to the unrest by imposing the security law in late June, sending a political chill through the semi-autonomous finance hub.
Since then, the city's leaders have postponed local elections, citing the coronavirus pandemic.
Authorities have also issued arrest warrants for six exiled pro-democracy activists and launched a crackdown on other activists.
On Monday, Hong Kong media mogul Jimmy Lai, one of the city's most vocal Beijing critics, was arrested under the security law.
- China-US clashes -
The US measures come three months ahead of the November presidential election in which the incumbent Donald Trump, who trails his rival Joe Biden in the polls, is campaigning hard on an increasingly strident anti-Beijing message.
As public disapproval has grown for his handling of the pandemic, Trump has pivoted from his previous focus on striking a trade deal with China to blaming the country for the coronavirus crisis.
Washington and Beijing have sparred on multiple fronts in recent months, and both sides have already imposed sanctions on each other over China's mass internment of mostly Muslim minorities in Xinjiang.
Trump on Thursday ordered Americans to stop doing business within 45 days with popular Chinese apps WeChat and TikTok.
The order claimed TikTok could be used by China to track the locations of federal employees, build dossiers on people for blackmail and conduct corporate espionage.
Zhao said Monday that Washington's Hong Kong-related sanctions "will only make the world more aware of US hegemony, bullying and double standards."
China sanctions AmericansHong KongUS senators
NATO Chief's "Severe Costs" Warning For Russia If Ukraine Attacked

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}

\textbf{Context.}
A map is  a combinatorial object describing the embedding up to homeomorphism of a multigraph on a compact orientable surface (see  \cref{sec:def} for precise definitions).
Map enumeration has been an important research topic for many years now. Tutte first enumerated planar maps and found simple formulas \cite{Tutte}, raising the natural concern of finding bijections explaining those formulas. Such bijections have since been found, mostly thanks to the family of bijections between maps and decorated trees (blossoming trees and mobiles) \cite{Cori1981,theseSch,BDG,AP,BF}.
Some of these bijections have recently been extended to maps on surfaces of higher genus (see for instance \cite{Schaeffer}) and on non orientable surfaces \cite{CD}. These bijections also opened the way to the study of random maps and their scaling limits (see e.g. \cite{CS,Miermont,LG})

Another powerful tool for map enumeration is the KP hierarchy. The Kadomtsev-Petviashvili hierarchy is an infinite set of PDE's on functions with a infinite number of variables $(p_i)_{i\geq 1}$, which arose from mathematical physics (see \cite{Solitons},\cite{Ok} and references therein). The first equation of the hierarchy is
\[{F_{3,1}=\frac{1}{12}F_{1,1,1,1}+F_{2,2}+\frac{1}{2}(F_{1,1})^2},\]
where the indices indicate partial derivatives (e.g. $F_{3,1}=\frac{\partial^2}{\partial p_1\partial p_3}F$).
The link between integrable hierachies and maps was observed in mathematical physics through matrix integrals (see e.g. \cite{LZ} and references therein).
In 2008, Goulden and Jackson \cite{GJ} showed that certain generating functions for maps (with $p_i$ counting vertices of degree $i$) are solutions to the KP hierarchy. It allowed them to derive a very simple recurrence formula
\begin{equation}
\begin{split}
(n+1)T(n,g)=&4n(3n-2)(3n-4)T(n-2,g-1)+4(3n-1)T(n-1,g)\\
&+4\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{g_1+g_2=g\\mathbf{g}_1,g_2\geq 0}}  (3i+2)(3j+2)T(i,g_1)T(j,g_2)+2\mathbb{1}_{n=g=1},
\end{split}
\label{GJ}
\end{equation}
where $T(n,g)$ is the number of rooted cubic maps of genus $g$ with $3n$ edges (equivalently, triangulations of genus $g$ with $3n$ edges). Then, using similar methods, Carrell and Chapuy \cite{CC} proved a recurrence formula on general (i.e. non necessarily cubic nor triangular) maps
\begin{equation}
\begin{split}
(n+1)Q_g(n,f)=&2(2n-1)Q_g(n-1,f)+2(2n-1)Q_g(n-1,f-1)\\
&+(2n-3)(n-1)(2n-1)Q_{g-1}(n-2,f)\\
&+3\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}\sum_{\substack{g_1+g_2=g\\mathbf{g}_1,g_2\geq 0}}(2i+1)(2j+1)Q_{g_1}(i,f_1)Q_{g_2}(j,f_2),
\end{split}
\label{CC}
\end{equation}

where $Q_g(n,f)$ is the number of rooted maps of genus $g$ with $n$ edges and $f$ faces. Taking $f=1$ in \eqref{CC}, one recovers the famous Harer-Zagier recurrence \cite{HZ}.

Finding bijections for formulas arising from the KP hierarchy on maps would allow us to understand maps in greater depth, but for now it is still mainly an open problem. The only special case of \eqref{CC} and \eqref{GJ} for which a bijective proof is known is the case of one-faced maps \cite{CFF}.

\textbf{Contributions of this article.}
In this paper, we present bijective proofs for the planar case (g=0) of Goulden-Jackson and Carrell-Chapuy formulas \eqref{CC} and \eqref{GJ}. Note that contrarily to the one-faced case which is linear, the planar formulas are, as in the general case, quadratic.

We actually prove a more general formula on precubic maps (see \cref{sec:def} for a definition) which implies Goulden-Jackson formula \eqref{GJ}. The Carrell-Chapuy formula \eqref{CC} comes from two separate (but somehow related by their bijective proofs) formulas that were not predicted by the KP hierarchy, one of them being a generalization of the famous R\'emy bijection on trees \cite{Remy} to all planar maps (see formulas \eqref{cut-slide} and \eqref{remy}).

Our bijections rely on a particular exploration of the map and on a "cut and slide" operation. Although non-local, this operation allows us to keep track of the degrees of the vertices (see \cref{sec:deg}). We also take a first step towards the higher genus case by proving the Goulden-Jackson formula for two-faced maps in any genus (see \cref{sec:twofaces}).

\textbf{Comments and related works.}
The bijective study of planar maps is a well understood topic, especially thanks to bijections with tree-like objects. However, as far as we know, our bijection is not equivalent to those bijections (although certain similarities can be observed, such as a search of the dual map, DFS in our case, BFS in the bijections above). A related cut and slide operation appeared in \cite{Bettinelli}, but it is different from ours (see Remark \ref{rem} for more details).

It is a natural question to try to unify bijections in this paper (that apply to planar maps with arbitrary number of faces) and bijections in \cite{CFF} (that apply to unicellular maps wiyth arbitrary genus) in order to prove \eqref{GJ} and \eqref{CC} in full generality. These two approaches seem different at first sight, but they both consist of taking a map with a marked special half edge, and "cutting" at this special half edge in order to obtain one or two maps with less edges, faces, or lower genus. In our case, those special half edges are called \textit{discoveries}, and there are $f-1$ of them in a map with $f$ faces ; in \cite{CFF}, they are called \textit{trisections}, and there are $2g$ of them in a map of genus $g$. Slightly rewriting \eqref{GJ} suggests that in the general case these special half-edges still exist and there would be $2g+f-1$ of them. They would thus be a common generalization of trisections and discoveries. We already found a bijective proof of the precubic recurrence formula for two faced maps (a sketch of proof is included in \cref{sec:twofaces}), but it is already quite complicated and involves several cases. That is why putting it all together seems to be a challenge of its own. In another direction, we can also look at the second KP equation (which is also quadratic), which yields a recurrence formula (very similar to \eqref{precubic}) for planar maps with vertices of degrees $1$ or $4$, and this formula is also proven by our bijection (it is a specialization of the more general formula obtained in \cref{sec:deg}). This suggests that this exploration+cut-and-slide scheme is maybe somehow underlying in the whole KP hierarchy for maps.

\textbf{Structure of the paper.}
In \cref{sec:def}, we will give some definitions on maps and state the main results. The bijections will be described in \cref{sec:bij}, with some proofs postponed to \cref{sec:proof}. \cref{sec:deg} will present refined formulas with control over the degrees of the vertices. In \cref{sec:calcul}, we prove the Carrell-Chapuy formula using formulas \eqref{cut-slide} and \eqref{remy}. In \cref{sec:twofaces}, we give a sketch of proof of the precubic recurrences for two-faced maps in any genus.

\section{Definitions and main results}
\label{sec:def}
\label{sec:results}
\begin{definition}
A map $M$ is the data of a connected multigraph (multiple edges and loops are allowed) $G$ (called the underlying graph) embedded in a compact connected oriented surface $S$, such that $S\setminus G$ is homeomorphic to a collection of disks. The connected components of $S\setminus G$ are called the \textit{faces}. $M$ is defined up to homeomorphism. Equivalently, $M$ is the data of $G$ and a \textbf{rotation system} which describes the clockwise cyclic order of the half-edges around each vertex.
The \textbf{genus} $g$ of $M$ is the genus of $S$ (the number of "handles" in $S$). A \textbf{corner} of $M$ is an angular sector between two consecutive half-edges around a vertex. A \textbf{rooted map} is a map with a distinguished corner. A small arrow is placed in the distinguished corner, thus splitting the corner in two separate corners (left and right of the arrow).
If a rooted map $M$ of genus $g$ has $n$ edges, $v$ vertices and $f$ faces, the Euler formula links those quantities: $v-n+f=2-2g$. $M$ has $2n+1$ corners.

A \textbf{planar map} is a rooted map of genus $0$. It can be drawn on the plane with the root lying on the outer face. A \textbf{precubic map} is a map with vertices of degree $1$ or $3$ only, rooted on a vertex of degree $1$. A \textbf{leaf} is a vertex of degree $1$ that is not the root.
\end{definition}

From now on, all the maps we consider will be rooted, and planar unless stated otherwise.

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\center
\includegraphics[scale=0.5]{map.pdf}
\caption{A planar map}
\label{map}
\end{figure}

In \cref{sec:bij}, we will introduce the concept of discovery, which are special edges obtained by an exploration of the map. An important feature is that a map with $f$ faces has $f-1$ discoveries

\begin{thm}
\label{cs}
There is a bijection between planar maps $M$ with a marked discovery and pairs of planar maps $(M_1,M_2)$ such that $M_1$ has a marked vertex and $M_2$ has a marked leaf. This bijection preserves the total number of edges and faces. It gives the following formula on on planar maps
\begin{equation}
(f-1)Q(n,f)=\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}v_1Q(i,f_1)(2j+1)Q(j,f_2),
\label{cut-slide}
\end{equation}
where $Q(n,f)$ is the number of planar maps with $n$ edges and $f$ faces, and $v_1$ counts the number of vertices in the first map (i.e. $v_1=2+i-f_1$)

This bijection adapts to precubic maps, and the marked vertex of $M_1$ is now a marked leaf. This yields
\begin{equation}
(f-1)\alpha(n,f)=\sum\limits_{i+j=n}\sum\limits_{f_1+f_2=f}\alpha^{(1)}(i,f_1)\alpha^{(1)}(j,f_2),
\label{precubic}
\end{equation}
where $\alpha(n,f)$ counts the number of (planar) precubic maps with $n$ edges and $f$ faces, and $\alpha^{(1)}(n,f)$ counts the number of precubic maps with $n$ edges and $f$ faces and a marked leaf.
\end{thm}

Retracting a leaf into a corner (see \cref{feuille}) is a classical operation that gives a bijection between planar maps $M$ with a marked leaf and planar maps $M'$ with a marked corner, such that $M'$ has as many faces and one edge less than $M$. This operation, along with a more subtle retraction operation on internal nodes, is at the core of the Rémy bijection on plane trees \cite{Remy}. We generalize the operation on internal nodes to all planar maps in the following :
\begin{thm}[Generalized R\'emy bijection]
\label{re}

There is a bijection between planar maps $M$ with a marked node (i.e. a vertex that is not a leaf) and the union of planar maps $M'$ with a marked corner and pairs of planar maps $M_1$ and $M_2$ such that they both have a marked vertex. The total number of faces is preserved, and the total number of edges decreases by one.

Together with the operation of retracting a leaf into a corner, it yields
\begin{equation}
vQ(n,f)=2(2n-1)Q(n-1,f)+\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}v_1Q(i,f_1)v_2Q(j,f_2),
\label{remy}
\end{equation}
where the "$v$-variables" count the number of vertices (i.e $v=2+n-f$, $v_1=2+i-f_1$ and $v_2=2+j-f_2$). This holds for $n>0$.
\end{thm}
Taking $n=3m+2$ and $f=m+2$ in \eqref{precubic}, one recovers
\begin{cor}[Goulden-Jackson \eqref{GJ} planar case]
\label{cor}
\[(n+1)T(n)=4(3n-1)T(n-1)+4\sum_{\substack{i+j=n-2\\i,j\geq 0}}  (3i+2)(3j+2)T(i)T(j),\]
where $T(n)$ counts the number of planar cubic maps with $3n$ edges.
\end{cor}

Combining  \eqref{cut-slide} and \eqref{remy} and doing some manipulations, one recovers
\begin{cor}[Carrell-Chapuy \eqref{CC} planar case]
\begin{equation}
\label{planar}
\begin{split}
\begin{split}
(n+1)Q(n,f)=&2(2n-1)Q(n-1,f)+2(2n-1)Q(n-1,f-1)\\
&+3\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)(2j+1)Q(i,f_1)Q(j,f_2).
\end{split}
\end{split}
\end{equation}
\end{cor}
Formula \eqref{planar} is not straightforwardly derived from \eqref{cut-slide} and \eqref{remy}. The calculations to recover \eqref{planar} from there are displayed in \cref{sec:proof}.
\section{The bijections}
\label{sec:bij}
In this section, we will define the exploration of a planar map and the notion of discoveries that result from it, then we will explain our bijections.

\subsection{The exploration}
\label{explo}
\begin{definition}
The \textbf{exploration} of a planar map (see \cref{exploration}) is defined iteratively in the following way: starting from the root, go along the edges, keeping the edges on the right (progress in clockwise order). When an edge that is at the interface of the current face and a face not yet discovered is found, open this edge into a bud (an outgoing half-edge) and a stem (an ingoing half edge), and continue the process, thus entering the new face. Continue until the root is reached again.

Each edge that has been opened during the process is called a \textbf{discovery}, and the vertex attached to the bud is called a \textbf{discovery vertex}. If there are $f$ faces , there are $f-1$ discoveries (note that several discoveries can share the same discovery vertex).

The exploration is actually equivalent to a DFS of the dual, with a "right first" priority. Thus for each face but the outer face we can define its \textbf{previous face} as the face that is its parent in the covering tree of the dual found by the exploration. The notion of \textbf{previous discovery} can be similarly defined. It also defines an order on the corners (resp. half edges) incident to each vertex, according to the order in which they were visited during the exploration.

Let $e$ be a discovery, incident to faces $f_1$ and $f_2$, such that $f_1$ is the previous face of $f_2$. We say $e$ \textbf{leaves} $f_1$ and \textbf{enters} $f_2$.
\end{definition}

\begin{rem}
The exploration is a dynamic process that modifies the map along the way, but in the end, once the exploration is over and the discoveries have been found, we will deal with the original, unmodified map, with its original edges and faces. It is as if we did the exploration then closed the map back. Alternatively, one can think of an exploration that doesn't open the discoveries but just crosses them.
\end{rem}
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\label{exploration}
\centering

\includegraphics[scale=0.7]{exploration.pdf}
\caption{The exploration of a planar map. The buds are the outgoing arrows, the stems are the ingoing arrows. Left: the original map. Center: the opened map. Right: The original map, with its discoveries and discovery vertices in purple. The red tree describes the partial order among the faces. The corners are labeled in the order they were found during the discovery}
\end{figure}

\begin{lem}
\label{clockwise}
Around each vertex, the order of the corners as defined by the exploration agrees with the clockwise order.
\end{lem}

\begin{proof}
The order between the corners of the map is exactly the same as the order of the corners of the blossoming tree (a blossoming tree is a tree with some buds and stems attached) obtained by opening the discoveries. The exploration of a tree is just a tour of the unique face, and it is clear that the corners of each vertex are in clockwise order.
\end{proof}

We can now relate the bijections and the formulas. In a map with $f$ faces, there are $f-1$ discoveries, so there are $(f-1)Q(n,f)$ maps with $n$ edges, $f$ faces and a marked discovery. A marked leaf can be retracted into a marked corner (see \cref{feuille} left), such that there are $(2i+1)Q(i,f_1)$ maps with $i+1$ edges, $f_1$ faces and a marked leaf. There are $v_2Q(j,f_2)$ maps with $j$ edges, $f_2$ faces and a marked vertex. So \eqref{cut-slide} and \eqref{remy} are indeed consequences of \cref{cs,re}. In a precubic map, one can retract a leaf into a marked side-edge losing two edges (see \cref{feuille} right), so a precubic map with no leaf is equivalent to a cubic map, and we fin

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=0.5]{feuille.pdf}
\caption{Retracting a leaf: in a general map (left), in a precubic map (right)}
\label{feuille}
\end{figure}
\subsection{Cut and slide bijection}

We will describe the bijection of \cref{cs} between maps $M$ with a marked discovery (see \cref{sec:bij} for a precise definition) and pairs of planar maps $(M_1,M_2)$ such that $M_1$ has a marked vertex and $M_2$ has a marked leaf. It will then be straightforward to see that restricting the bijection to precubic maps gives \eqref{precubic}.
\begin{definition}
A discovery is said to be \textbf{disconnecting} if the corner preceding the discovery and the last corner (in the order defined by the exploration) around the discovery vertex lie in the same face.

\label{disco}
\end{definition}
\begin{spli}
Any map with a marked disconnecting discovery can be (bijectively) split into two maps, one with a marked vertex, the other with a marked leaf in the outer face (see \cref{split}).
\end{spli}
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=1.5]{splitting.pdf}
\caption{Splitting a map at a disconnecting discovery. Here we only see what happens locally around the disconnecting discovery. On the left, the discovery and its discovery vertex are in purple, $c$ is the corner preceding the discovery, and $c^*$ is the last corner around the discovery vertex}
\label{split}
\end{figure}

The splitting operation of a disconnecting discovery describes our bijection in the case where the marked discovery is disconnecting, and the reverse bijection in the case where the marked leaf lies in the outer face.

In general, discoveries are not disconnecting. In order to still split the map in two, given a discovery $e$, we will need to find a disconnecting discovery $e'$ that is "canonically" related to $e$. Conversely, the marked leaf in $M_2$ is not always in the outer face, we will need to "propagate" the leaf all the way up to the outer face. The notion of previous discovery will help us with that. This will be the general construction of the bijection that is detailed below.

\begin{lem}
If a discovery leaves the outer face, then it is disconnecting
\label{lemma}
\end{lem}

\begin{proof}
We will show the following stronger result, which directly implies \cref{lemma} : if a vertex has a corner in the outer face, then its last corner lies in the outer face.

In a map $M$, let $v$ be a vertex with one of its corners lying in the outer face. If $v$ is the root vertex, then it is obvious that its last corner lies in the outer face. Else, let $e$ be the first edge around $v$, as defined by the exploration (i.e. the edge we see in the exploration just before the first occurrence of $v$). Let $c$ be one corner around $v$ that lies in the outer face. Suppose it is not the last corner around $v$. Let $c_0$ (resp. $c^*$) be the first (resp. the last) corner around $v$, and let $F_0$ (resp. $F^*$) be the face in which $c_0$ (resp. $c^*$) lies. Suppose $F^*$ is not the outer face. (See \cref{corners})

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=1]{corners.pdf}
\caption{}
\label{corners}
\end{figure}

If a face $F'$ is deeper (in terms of the partial order defined by the exploration) than a face $F$ (and $F\neq F'$), then during the exploration (ignoring all other faces), we first see a part of $F$, then all of $F'$, then the rest of $F$. This would mean that during the exploration of $M$, at the time when we see $c$, $F^*$ hasn't been discovered yet. But that would mean that at the time when we see $e$ (which comes before $c$ in the exploration), $F^*$ hasn't been discovered yet. But then $e$ would be a discovery, and thus $c^*$ would be seen before $c_0$, contradiction
\end{proof}

\begin{bij}
\label{CS1}
The general process is iterative (see \cref{bij} for an example).

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=0.5]{bijection.pdf}
\caption{The bijection (above) and its inverse (below)}
\label{bij}
\end{figure}

\textbf{Cut process:} Start from a map $M$ with a marked discovery $e$, let $v$ be its discovery vertex. If the discovery is disconnecting, then split $M$ at $v$ as described in the splitting operation.

Otherwise, open $e$ into a bud $b_0$ and a stem $s_0$, and consider its previous discovery $e_1$ (in the order defined above). If it is disconnecting, then split it, otherwise open it (into $b_1$ and $s_1$) and consider the previous discovery $e_2$, and so on until a splitting operation is made. Note that a discovery that leaves the outer face is always disconnecting (because of \cref{lemma}), so the algorithm terminates. One ends up with two maps $M_1$ and $M_2'$, such that $M_1$ has a marked vertex and $M_2'$ has a marked leaf $l$ and (possibly) some buds and stems, all lying in the outer face.

\textbf{Slide process:} We will not modify $M_1$. If there are no buds and stems in $M_2'$, we are done. Else, consider $s_0$, and make it a marked leaf. Then consider $l$, and make it a stem. Finally, glue back the buds and stems together canonically: starting from the root of $M_2'$, taking a clockwise tour of the outer face, one encounters a certain number of buds, then the same number of stems. There is only one way to match each bud with each stem such that the map remains planar. Equivalently, if there are $k+1$ buds and $k+1$ stems, match $b_0$ with $s_1$, and so on, until $b_k$ is matched with $l$. We obtain a map $M_2$ with a marked leaf, together with the map $M_1$ with a marked vertex.
\end{bij}
We can now describe the inverse bijection :
\begin{bij}
\label{CS2}
Starting from $M_2$ with a marked leaf $l$ and $M_1$ with a marked vertex, consider $M_2$. $l$ lies in a certain face $F$, and if $F$ is not the outer face, there is a certain discovery $e_0$ that enters $F$. Open it into a bud $b_0$ and a stem $s_0$, then open the previous discovery $e_1$, and repeat the process until a discovery that leaves the outer face has been opened (in that case there is no previous discovery to open). One ends up with a map $M_2'$ with a marked leaf $l$ and possibly some buds and stems, all lying in the outer face. If there are some buds and stems, let $s$ be the stem that was created last in the process. Make $s$ a marked leaf $l^*$, and make $l$ a marked stem $s^*$, then close the map canonically. One now has a map $M_2^*$ with a marked leaf on the outer face and (possibly) a marked edge $e$ (that comes from the closure of $s^*$). This marked edge is actually a discovery in $M_2^*$ (and will be a discovery in the final map). If $e$ doesn't exist, let $l^*=l$, and mark the edge adjacent to $l^*$ (and call it $e$). Now do the inverse of the splitting operation: glue $l^*$ to the root vertex of $M_2^*$ at its first corner, and then glue the root of the resulting map at the last corner of the marked vertex of $M_1$ to obtain a map $M$ with a marked discovery $e$.
\end{bij}
\begin{rem}
This operation restricts to precubic maps, and in this case discovery vertices are always of degree $3$, so that when split, the marked vertex of $M_1$ and the root of $M_2$ are both of degree $1$.
\end{rem}

Those two applications are inverse of each other because of the following property (that will be proved in \cref{sec:proof}) :
\begin{lem}
In the bijection and its inverse, the closure of the buds and stems are discoveries in the resulting map. Moreover, in the bijection, if $e_1$,…$e_k$ are the discoveries created by closing buds and stems, $e_k$ leaves the outer face, and for all $i<k$, $e_{i+1}$ is the previous discovery of $e_i$. In the inverse bijection, if $e_1$,…$e_{k+1}$ are the discoveries created by closing buds and stems, $e_{k+1}$ is disconnecting, and for all $i<k$, $e_{i+1}$ is the previous discovery of $e_i$.
\end{lem}
\begin{rem}
\label{rem}
A similar but different cut and slide operation also appeared in \cite{Bettinelli} (and in some sense also in \cite{slice}). However there are significant differences : the cut path is geodesic (leftmost BFS), and both endpoints of the path need to be specified. Whereas here, the cut path is defined by a leftmost DFS, and only one endpoint of the path (the marked discovery) needs to be specified, the other endpoint (the disconnecting discovery) is uniquely determined. What's more, the bijections in \cite{Bettinelli} imply linear formulas, contrarily to our quadratic formulas.
\end{rem}
\subsection{Generalized R\'emy bijection}
\label{sec:Remy}
We will now describe a generalized R\'emy bijection on planar maps, that also relies on the cut-and-slide operation.

We recall R\'emy's bijection for plane tree that proves the formula $(n+1)Cat(n)=2(2n-1)Cat(n-1)$ where $Cat(n)$ counts the number of rooted plane trees with $n$ edges (see \cref{tree} for an example). Start from a tree $T$ with $n$ edges and a marked vertex $v$. If $v$ a leaf, retract it (as in \cref{feuille}) to obtain a tree $T'$ with $n-1$ edges, with a marked corner. Otherwise, $v$ has a last son $v'$, that is its son that is found last during a clockwise tour of the unique face. We can then contract the edge between $v$ and $v'$, and mark a corner around the merging vertex to remember where to grow the edge back (as in \cref{tree}). The reverse bijection is then straightforward.

\begin{rem}
Leaves are defined as vertices of degree $1$ \textbf{that are not the root vertex}. Thus, if the root vertex has degree $1$, one should apply the second operation to it
\end{rem}
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}

\centering
\includegraphics[scale=0.5]{remy.pdf}
\caption{R\'emy's bijection for plane trees}
\label{tree}
\end{figure}

In a general planar map, let us define precisely the operations of edge contraction and edge growing :

Let $v$ be a non-leaf vertex, it has a last corner $c_2$ (as defined by the exploration), let $e$ be the edge that comes just before $c_2$ in clockwise order around $v$. We call $e$ the last edge around $v$. Let $v'$ be the other end of $e$ (note that it is possible that $v=v'$). $v'$ is called the last son of $v$. We will define the contraction operation only in the case $v\neq v'$.  $e$ is adjacent to two corners of $v$ : $c_2$ and another corner $c_1$. $e$  is also adjacent to two corners of $v'$, $c'$ and $c''$ such that $c'$ appears before $c''$ in the exploration.
$c"$ is the last corner around $v'$

The contraction operation consists in contracting $e$, merging $v$ and $v'$ into a vertex $v^*$, merging $c_1$ and $c'$ into $c^*$ and $c_2$ and $c''$ into $c$. $c$ is the last corner around $v^*$, and $c^*$ is the marked corner.

Conversely, starting with a marked corner $c^*$ adjacent to a vertex $v^*$ whose last corner is $c$, one can grow an edge $e$, splitting $v^*$ into $v$ and $v'$, $c^*$ into $c_1$ and $c'$, and $c$ into $c^*$ and $c_2$. $c_2$ is the last corner around $v$ and $c''$ the last corner around $v$. $v$ is the marked vertex, and $v'$ is its last son. This defines the growing operation.

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=0.7]{growing.pdf}
\caption{The contraction and growing operations}
\label{growing}
\end{figure}

\begin{lem}
Let $M$ be a map with a marked corner $c^*$, and let $M'$ be the map obtained after a growing operation is performed at $c^*$. Let $v$ be the marked vertex and $v'$ its last son. Then $v$ appears before $v'$ in the exploration
\end{lem}
\begin{proof}
The growing operation is local, so both explorations of $M$ and $M'$ look exactly the same before $c^*$ (resp. $c_1$) is found. Let $p$ be the first corner around $v^*$ in $M$, after the growing operation it is adjacent to $v$, and it is reached before any other corner around $v$ or $v'$, thus $v$ appears before $v'$ in the exploration (the special case $p=c^*$ works the same).
\end{proof}

\begin{lem}
Let $v$ be a non-leaf vertex, $e$ its last edge, and $v'$ its last son. If $v'$ is found before $v$ in the exploration (including the case $v=v'$), then $e$ is a discovery.
\end{lem}
\begin{proof}
If $v=v'$, then $e$ is a loop, thus a discovery (since a loop separates a planar map in two, it has to be opened during the exploration.

Otherwise, let $c_1,c_2$ (resp. $c',c''$) be the corners of $v$ (resp. $v'$) that are adjacent to $e$. Note that $c_2$ is the last corner of $v$, and assume $e$ is not a discovery. Thus, it is not opened during the exploration, such that $c''$ comes before $c_2$, and $c_1$ before $c'$ (see \cref{growing}). $c'$ cannot be the first corner of $v'$, otherwise $v$ would be seen before $v'$ in the exploration. Thus, because of \cref{clockwise}, $c'$ comes after $c''$. But that would mean the order among the four corners in the exploration is $c''$, then $c_2$, then $c_1$, then $c'$, which is impossible since $c_2$ is the last corner of $v$.
\end{proof}
Now we are ready to explain the generalized R\'emy bijection :

\begin{bij}
\label{RM}
Take a planar map $M$ and mark a vertex $v$. If it is a leaf, contract it to mark a corner. Else, let $v'$ be its last son, and $e$ be its last edge. If $v'$ is seen after $v$ in the exploration, then contract $e$ and mark the corresponding corner. The inverse operation is the growing operation as described above.This gives the first term of the RHS of \eqref{remy}.
Otherwise, $e$ is a discovery, apply the cut-and-slide operation at $e$. One ends up with two maps $(M_1,M_2)$. $M_1$ has a marked vertex, and $M_2$ has a marked leaf $l$ that is the last son of its neighbor $w$. Contract $l$ to mark $w$. To go backwards, grow $l$ out of the last corner around $w$, then apply the inverse of the cut and slide operation. This gives the second term of the RHS of \eqref{remy}.

\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=0.5]{remy2.pdf}
\caption{The generalized R\'emy bijection. The case similar to trees (above), and the cut-and-slide case (below)}
\label{remy2}
\end{figure}
\end{bij}

\begin{rem}
There is also a generalization of R\'emy's bijection on binary trees to precubic maps, that works exactly the same, we leave it as an exercise to the reader.
\end{rem}

\section{Proofs}

\label{sec:proof}

In this section we will present a proof that \cref{CS1} and \cref{CS2} are inverse to each other, thus implying that the cut and slide operation is a bijection. Then, together with the lemmas presented in \cref{sec:Remy}, it will be enough to prove that the generalized Rémy bijection (\cref{RM}) is indeed a bijection. To do so, we will describe what happens to the different faces of the map during the operation (see \cref{faces}).

Thanks to the exploration, we can consider faces as words on the alphabet of side-edges (each face is described by the list of the side edges that lie inside it, ordered as they appear during the exploration). For any side-edge $e$, we will call $\overline{e}$ its opposite side-edge. By abuse of notation we may also refer to the edge $e$ that is the edge made of the side edges $e$ and $\overline{e}$, but we will stress on the fact that we are referring to an edge.

Fix a planar map $M$. Let $d_0$ be a discovery, $d_1$ its previous discovery (if it exists), and so on, $d_{i+1}$ being the previous discovery of $v_i$, as long as it exists. For all $i$, the discovery $d_i$ discovers the face $F_i$ while leaving face $F_{i+1}$. For all $i$, let $v_i$ be the discovery vertex corresponding to $d_i$, and let us differentiate both side edges of the discovery : $d_i$ sits in $F_{i+1}$ and $\overline{d_i}$ sits in $F_i$. Let $k$ be the smallest $i$ such that $d_i$ is a disconnecting discovery. For all $i$ between $1$ and $k$, we can set $F_i=\overline{d_i}A_id_{i-1}B_i$, that is $A_i$ are the side-edges of $F_i$ encountered before $d_{i-1}$ in the exploration, and $B_i$ are the the side-edges of $F_i$ encountered after $d_{i-1}$ in the exploration. We also set $F_0=\overline{d_0}A_0$ and $F_{k+1}=\overline{d_{k+1}}Cd_kB_{k+1}C'$ where $B_{k+1}$ is the word of side-edges of $F_{k+1}$ that appear between $d_k$ and the last corner around $v_k$.\\
Now let us describe the cut and slide operation when $d_0$ is marked. First, for all $i<k$, all \textbf{edges} $d_i$ are opened into a bud $b_i$ and an stem $s_i$. The \textbf{edge} $d_k$ becomes a marked leaf $s_k$. Then $s_k$ becomes a stem, and $s_0$ becomes the marked leaf. Then $b_i$ is matched with $s_{i+1}$ to create and edge $e_i$. The map has been split into two distinct maps $M_1$ and $M_2$, now let us see what happened to the faces. In $M_1$, there is only one face that has been modified, it is the face of the marked vertex $v^*$, that we call $G^*$. It is straightforward to see that $G^*=\overline{d_{k+1}}CC'$ where the last corner of $v^*$ appears exactly between the end of $C$ and the beginning of $C'$. In $M_2$, the modified faces will be called $G_i$. We have $G_0=\overline{e_0}A_0s_0\overline{s_0}B_1$, $G_i=\overline{e_i}A_ie_{i-1}B_{i+1}$ and $G_k=A_ke_{k-1}B_{k+1}$. $G_k$ is the outer face of the second map, and for all $i$, $G_{i+1}$ is the previous face of $G_i$.\\
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=1]{faces.pdf}
\caption{What happens to the faces during the cut and slide operation. Marked objects are in purple, other discoveries involved are in gold. Blue contour materializes the $A_i$'s and red contour, the $B_i$'s}
\label{faces}
\end{figure}\\
We must now prove that, for all $i<k$, the \textbf{edge} $e_i$ is a discovery that leaves $G_{i+1}$ and discovers $G_i$. We already now that $e_i$ lies in $G_{i+1}$ and $\overline{e_i}$ lies in $G_i$, we just have to prove that the \textbf{edge} $e_i$ is a discovery, i.e $e_i$ is the first side edge $e$ of $G_{i+1}$ such that $\overline{e}$ lies in $G_i$. Assume the contrary. There must be a side edge $f$ in $A_{i+1}$ such that $\overline{f}$ lies in $G_i$. There are two possibilities. Either $\overline{f}$ belongs to $A_i$, but then in the original map, in $F_{i+1}$, $f$ comes before $d_i$ and $\overline{f}$ lies in $F_i$, thus $d_i$ could not be the discovery, contradiction. Or $\overline{f}$ belongs to $B_{i+1}$, but then in the original map, in $F_{i+1}$ we see the side edges $f$, then $d_i$, then $\overline{f}$ in that order. But since we deal with planar maps, an edge which has both sides in the same face is a cut-edge (removing it disconnects the map). But then in that configuration, $d_i$ must be a disconnecting discovery (see \cref{moche}). Indeed, let $M^*$ be the connected component containing $d_i$ that we obtain after removing $e$. Then $d_i$ would be a discovery in the map $M^*$ that leaves the outer face, thus disconnecting (because of \cref{lemma}). This  is a contradiction, because $i<k$. We proved what we wanted. \\
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}
\centering
\includegraphics[scale=0.5]{cutedge.pdf}
\caption{The impossibility of a cut-edge "surrounding" a non disconnecting-discovery. In red, the cut-edge, in purple the discovery}
\label{moche}
\end{figure}\\
Knowing exactly what happens to the faces and that the discoveries are in some sense "preserved" in enough to prove that both the cut and slide operation and its inverse work fine and are actually inverse applications of each other.

\section{Controling the degrees of the vertices}
\label{sec:deg}
Here we present an analogue of \eqref{cut-slide} with control over the degrees of the vertices. If $\textbf{v}=(v_i)_{i\in \mathbb{N}}$ is a sequence of integers, for any $j>0$, we set $\delta_j(\textbf{v})=\textbf{w}$ where $w_j=v_j+1$ and $w_i =v_i$ for $i\neq j$, and $\delta_{-j}(\textbf{v})=\textbf{w'}$ where $w_j'=v_j-1$ and $w_i '=v_i$ for $i\neq j$. Finally, we set $\delta(\textbf{v},j_1,\hdots,j_k)=\delta_{j_1}\circ \hdots\circ \delta_{j_k}(\textbf{v})$.

Let $M(r,f,\textbf{v})$ the number of planar maps with $f$ faces, with root of degree $r$, with $\textbf{v}=(v_i)_{i\in \mathbb{N}}$ s.t. there are $v_i$ vertices of degree $i$ (root included). The cut-and-slide operation only modifies the degrees at the marked leaf and the splitting vertex, so we can immediately derive this more precise formula
\begin{thm}
\begin{align*}
(f-1)M(r,f,\textbf{v})=&\sum_{\substack{j,k\geq 1}}\sum_{\textbf{u}+\textbf{w}=\delta(\textbf{v},1,j,k,-(j+k+1))}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(u_j-\mathbb{1}_{j=r})M(r,f_1,\textbf{u})(w_1-\mathbb{1}_{k=1})M(k,f_2,\textbf{w})\\
&+\sum_{\substack{j+k=r-1}}\sum_{\textbf{u}+\textbf{w}=\delta(\textbf{v},1,-r,k,j)}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}M(j,f_1,\textbf{u})(w_1-\mathbb{1}_{k=1})M(k,f_2,\textbf{w}).
\end{align*}
\end{thm}

\begin{proof}
Notice that, in the cut-slide operation, very few of the vertex degrees are modified : there is the leaf that is created in $M_2$, and the disconnecting discovery vertex that is split in three. Other than that, for all other vertices, although some of their adjacent edges might be split, they are reattached somewhere else so their degree doesn't change.

Let $v$ be the vertex that is split in three : it gives birth to $v_1$, the marked vertex in $M_1$, $v_2$, the root vertex of $M_2$, and $v_3$, that after some possible transformation "becomes" the marked leaf in $M_2$. Say $v$ is of degree $j+k+1$, with $deg(v_1)=j$ and $deg(v_2)=k$.

The $(w_1-\mathbb{1}_{k=1})$ term means that the marked leaf in $M_2$ cannot be the root. Finally, there are two possible cases, depending on whether $v$ is the root of $M$ or not, each implying one term in the RHS.
\end{proof}

\begin{rem}
$\delta_j(\textbf{v})$ means there is one more vertex of degree $j$, and $\delta_{-j}(\textbf{v})$ means there is one less vertex of degree $j$. This notation is somehow complicated but avoids dealing with special cases, e.g. in $\delta(\textbf{v},j,k)$ there is no problem if $j=k$, contrarily to saying something like $v_j'=v_j+1$, $v'_k=v_k+1$ and $v_i'=v_i$ for $i\neq j,k$.
\end{rem}
 Note that this recurrence formula allows us to compute the number of maps with bounded vertex degrees, i.e. it can be specialized to maps with vertex degrees $\leq d$ for some $d$. Taking $d=3$, one recovers \eqref{precubic}.\\

Formula \eqref{remy} also has an analog where the degrees are recorded, but it splits into three different cases, according to whether the marked vertex is a leaf, the root, or another node. If it is a leaf, it is the trivial bijection described in \cref{feuille}. If it is the root vertex, the formula is actually Tutte's formula. We will only include the formula corresponding to the case where the marked vertex is a node. This formula alone suffices to calculate all the terms by induction.
As before, there are extra terms because it depends on whether the modification affects the root or not.

For $p\neq 1$:
\begin{align*}
(v_p-\mathbb{1}_{p=r})M(r,f,\textbf{v})=&\sum_{\substack{j\geq 1}}u_{p+j-2}M(r,f,\textbf{u}=\delta(\textbf{v},-j,-p,j+p-2))\\
&+\sum_{\substack{j,k\geq 0}}\sum_{\substack{\textbf{u}+\textbf{w}=\delta(\textbf{v},-p,p-1,j,k,-(j+k+1))}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(u_j-\mathbb{1}_{j=r})M(r,f_1,\textbf{u})w_{p-1}M(k,f_2,\textbf{w})\\
&+\sum_{k=0}^{r-1} \sum_{\substack{\textbf{u}+\textbf{w}=\delta(\textbf{v},-p,p-1,-r,k,r-k-1)}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}M(r-1-k,f_1,\textbf{u})w_{p-1}M(k,f_2,\textbf{w}).
\end{align*}

\section{The proof of the Carrell-Chapuy formula in the planar case}
\label{sec:calcul}

We are now ready to prove the Carrell-Chapuy formula  in the planar case \eqref{planar}. We will prove it by induction on $n$, only knowing that $Q(0,f)=\mathbb{1}_{f=1}$.\\
Taking the dual of \eqref{cut-slide}, we obtain the following formula, which will be helpful for the proof
\begin{cor}

\begin{equation}
(v-1)Q(n,f)=\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f+1\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)f_2Q(j,f_2)
\label{dual}
\end{equation}
\end{cor}

Starting with \eqref{cut-slide}, and applying \eqref{remy}, then in a second time applying \eqref{cut-slide} backwards, we obtain

\begin{align*}
(f-1)Q(n,f)=&(2n-1)Q(n-1,f-1)+2(2n-1)Q(n-1,f)\\&+2\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)\\&+\sum_{\substack{i+j+k=n-2\\i,j,k\geq 0}}\sum_{\substack{f_1+f_2+f_3=f\\f_1,f_2,f_3\geq 1}}(2i+1)Q(i,f_1)v_2Q(j,f_2)v_3Q(k,f_3)
\\&=(2n-1)Q(n-1,f-1)+2\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)\\&+\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(f_1-1)Q(i,f_1)v_2Q(j,f_2).
\end{align*}
So summing \eqref{remy} to this
\begin{align*}
(n+1)Q(n,f)=&(2n-1)Q(n-1,f-1)+2\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)\\&+\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(i+1)Q(i,f_1)v_2Q(j,f_2).
\end{align*}
Let \[S=\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(i+1)Q(i,f_1)v_2Q(j,f_2).\]
We want to prove
\[S=(2n-1)Q(n-1,f-1)+\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2).\]
 We apply the recursion hypothesis
\begin{align*}
S&=\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2(2i-1)Q(i-1,f_1)+2(2i-1)Q(i-1,f_1-1))v_2Q(j,f_2)\\&+3\sum_{\substack{i+j+k=n-3\\k,l\geq 0}}\sum_{\substack{f_1+f_2+f_3=f\\f_1,f_2,f_3\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)v_3Q(k,f_3)\\&+vQ(n-1,f-1).
\end{align*}
but, according to \eqref{cut-slide}, \[\sum_{\substack{i+j=n-1\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2(2i-1)Q(i-1,f_1-1))v_2Q(j,f_2)=2(f-2)Q(n-1,f-1),\]
and
\begin{align*}
&\sum_{\substack{i+j+k=n-3\\k,l\geq 0}}\sum_{\substack{f_1+f_2+f_3=f\\f_1,f_2,f_3\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)v_3Q(k,f_3)\\&=\sum_{\substack{i+j=n-2\\i,j\geq 0}}\sum_{\substack{f_1+f_2=f\\\sf_1,f_2\geq 1}}(2i+1)Q(i,f_1)(f_2-1)Q(j,f_2).
\end{align*}
So
\begin{align*}
S=&(2(f-2)+v)Q(n-1,f-1)+\sum_{\substack{i+j=n-2}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)(2j+1)Q(j,f_2)\\&+\sum_{\substack{i+j=n-2}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)f_2Q(j,f_2).
\end{align*}
But using \eqref{dual}, we have
\[\sum_{\substack{i+j=n-2}}\sum_{\substack{f_1+f_2=f\\f_1,f_2\geq 1}}(2i+1)Q(i,f_1)f_2Q(j,f_2)=(v-1)Q(n-1,f-1),\]
which finishes the proof.

\begin{rem}
The proof above is not straightforward, and since it uses duality, our method cannot be applied to finding an extended Carrell-Chapuy formula with control over the degrees (if it even exists). It cannot be used either for proving the planar case of the analogue formula on bipartite maps found by Kazarian and Zograf \cite{KZ}.
\end{rem}

\section{A bijection for precubic maps with two faces}
\label{sec:twofaces}
In this section, we will give a sketch the proof of a first step towards uniting higher genus and multiple faces : the case of two-faced precubic maps.

If we restrict \eqref{precubic} to $f=2$, we obtain the following formula
\begin{equation}
\label{deuxfaces}
(2g+1)\alpha_g(n,2)=\alpha^{(3)}_{g-1}(n-6,2)+\sum\limits_{g_1+g_2=g} \sum\limits_{i+j=n}\alpha^{(1)}_{g_1}(i,1)\alpha^{(1)}_{g_2}(j,1)
\end{equation}
where $\alpha_g(n,f)$ counts the number of precubic maps with $n$ edges and $f$ faces, of genus $g$, and $\alpha^{(k)}_g(n,f)$ counts the number of those maps with $k$ marked leaves.

We can define the exploration of a precubic map with two faces in the following way :
\begin{definition}
We will describe an exploration of the map as a canonical labeling of all its corners (see Fig. \ref{twofac}).

As in \cref{explo}, we can define the discovery as the first edge adjacent to both faces that is encountered in a clockwise tour of the root face starting from the root, and the discovery vertex as the vertex that appears right before the discovery in this tour (there is only one discovery since there are 2 faces). Opening the discovery into a bud and a stem creates a blossoming (i.e. with a bud and a stem) unicellular map. In this map, it is possible to label all the corners in their order during the tour of the unique face. Thus, there is a labeling of all the corners of the map, and a discovery vertex (we won't need the discovery itself in what follows).
\end{definition}
The discovery vertex is obviously of degree $3$ (because a leaf is adjacent to only one face).  We can now introduce special vertices and \textit{trisections}.
\begin{definition}
A trisection is a vertex whose corner labels are in counterclockwise order around this vertex. A vertex is said to be \textit{special} if it is a trisection  of if it is the discovery vertex.
\end{definition}
A trisection of the map is exactly a trisection of the unicellular (blossoming) map. In \cite{unicellular,trisections}, it is proven that in a unicellular map of genus $g$, there are $2g$ trisections. What's more, we can easily verify that the discovery vertex is not a trisection thus
\begin{lem}
There are $2g+1$ special vertices in a two-faced precubic map of genus $g$ (see Fig. \ref{twofac} right).
\end{lem}

\begin{figure}[!h]

\captionsetup{width=0.8\textwidth}

\centering
\includegraphics[scale=1]{twoface.pdf}
\caption{A precubic map (left), its exploration (center), and its special points (right, discovery vertex in red, trisections in purple)}
\label{twofac}
\end{figure}

The operation we will consider is fairly simple : take a map of genus $g$ with a marked special vertex, and split it.
There are two possible cases :\begin{enumerate}

\item the map is disconnected into two maps with a marked leaf each, such that the genii, number of edges and faces add up,
\item the resulting map is of genus $g-1$ and has $3$ marked leaves, along with information on how to glue back the leaves together (given $3$ leaves, there are $2$ possible ways of gluing them together).
\end{enumerate}

\begin{rem}
The first case can only appear if the special vertex is the discovery vertex. In the second, case, the special vertex can be either the discovery vertex or a trisection.
\end{rem}
There is a bijection between maps in case (1) and pairs of maps with a marked leaf each, the inverse operation being the same as the case of disconnecting discoveries in the planar case : given two maps with a marked leaf each, it is possible to glue them back together as in \cref{split}, and one obtains a map with a marked discovery vertex.

Case (2) is more complicated : given a map of genus $g-1$ with $3$ marked leaves, that we will call a \textit{tripod}, there are two ways to glue back the leaves. A gluing is said to be \textit{valid} if the resulting map has genus $g$ and the gluing vertex is a special vertex. Tripods can have $0$, $1$ or $2$ valid gluings, and we will provide a classification of tripods with respect to their number of gluings. In the following, we will refer to "marked leaves" as just "leaves" as there is no risk of ambiguity. In order to prove \eqref{deuxfaces}, we will then have to prove there is a bijection between tripods with respectively $0$ and $2$ valid gluings.

\begin{definition}
In a given map, let $v$ be the discovery vertex. It has two corners in the root face, and is thus seen twice in the tour of the root face. We can decompose the root face $F$ as a word on side edges as in \cref{sec:proof}, starting from the root, as $F=TvOv\overline{T}$ (see Fig. \ref{tripod} left).

Take a given tripod $M$, with exactly two leaves in the root face, both in $\overline{T}$. Glue those two leaves to split the root face in two. Let $F'$ be the face thus obtained that is not the root face. We say that $M$ is in the special case if no side edge of $F'$ has its opposite side edge in $T$  (see Fig. \ref{tripod} right).
\end{definition}
\begin{figure}[!h]
\captionsetup{width=0.8\textwidth}

\center
\includegraphics[scale=1]{tripod.pdf}
\caption{The schematic decomposition of the root face (left), and the special case (right, with the leaves in purple)}
\label{tripod}
\end{figure}

\begin{lem} The following classification holds :
\begin{itemize}
\item A tripod with all leaves in the same face or in the special case or with two leaves in $O$ has $1$ valid gluing.
\item A tripod having exactly one leaf in the root face has $2$ valid gluings.
\item A tripod with exactly two leaves in the root face, at least one of which in $T$ or $\overline{T}$ (except for the special case), has no valid gluings.
\end{itemize}
\end{lem}
The proof of this lemma is done by carefully considering the cases implied (we omit it here).

To finish the proof, one needs to find a bijection between tripods with $0$ valid gluings and tripods with $2$ valid gluings. The bijection consists of cleverly "unplugging" the discovery and plugging it somewhere else to "transfer" leaves between the faces. The proof involves several cases and is a bit technical, thus we prefer to omit it. This shows that in average, a tripod has one valid gluing, implying \eqref{deuxfaces}.

\section*{Acknowledgements}
The author wishes to thank Guillaume Chapuy for suggesting the problem and for useful discussions.
\section*{Funding}
This work was supported by ERC-2016-STG 716083 "CombiTop".

redpajama_set_name: RedPajamaArXiv


Fall in love with this brick colonial located in a secluded court location in Cedarcroft. With fresh new design & practical updates, this home looks like the completed project from an HGTV show. So many fantastic changes & in the last year: new central ac, new luxury master bath, new first floor half bath, wall removed between living & dining rooms, kitchen has been opened to the dining room, mother of pearl back splash, new quartz counters, new stainless steel appliances. The added laundry in the reconfigured hall bath is a great convenience. Refinished hardwoods. All rooms freshly painted. Huge yard. Gas heat & cooking. Finished lower level includes a half bath & plush carpet. Detached garage. Click here to learn more.


I had such a great time taking Jordan & Kylie's Lake of the Ozarks engagement pictures ! The first time I got to take their pictures, Kylie had no idea that she was going to get engaged that day! If you haven't yet, you should definitely their INSANE proposal story that we got to be a part of !
Also check out the increadible video of the proposal done by AceLegendary Productions!
Can't wait to shoot their wedding next fall!
Also Check out their save the date trailer! Also done by AceLegendary Productions!


An Oberth-class Federation science vessel.
Federation science vessels was a classification given to starships designated for scientific research by the United Federation of Planets.
↑ Bantam TOS: Star Trek 1 - "Charlie's Law".
Federation science vessel article at Memory Alpha.


At Ashley Industrial Molding, Inc., we provide custom thermoforming of lower volume panels for a range of applications. Five-stage rotary thermoforming equipment provides efficient production of components without the initial capital expenditure required for production molds. Standard material options include ABS, acrylic, and polyethylene, but a wide variety of polymers can be thermoformed, depending on the required application.
Thermoforming is most cost effective at production levels of approximately 300 to 2,000 units per year, depending on size and complexity of the part. If volumes do increase beyond that range, we are very capable of working with customers to ramp up the production process to a more cost effective production method. By offering multiple molding options, we can competitively produce premium quality parts at a range of production volumes.
our thermoforming capability, see the table below, or contact us directly.
Acrylonitrile-Butadiene-Styrene is a low cost thermoplastic material with great impact resistance, easy to machine to close tolerances and thermoforming characteristics. There are flame retardant grades available and the material comes in a wide variety of colored caps and films and textures. Common applications include Cab Roofs, Cosmetic side panels and Covers, as well as interior components.
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This material is available in monolayer UV resistant materials, but bright and light colors can be a challenge. Fading and whitening of bright and light colored TPO material can lead to quality and warranty issues. There are specific cap layers for TPO that can offer the best UV characteristics.


Gaunt Francis wanted an organic looking mat grey stone for a prestigious reception in St Swithins Lane. We narrowed the search down to the available basalts on the market and finally the architect opted for a well-known Italian basalt called Basaltina. The finish was honed/colour resin filled and size 750x750x20mm. A nice size and not particularly common. The order was cut and prepared near Siena, Tuscany. The tiles were cut very nicely and had a clear grain direction and holes regular and slightly elongated as we would expect with the Basaltina Classico.


text:
/**
* PossibleAnswers.js
*
* @description :: TODO: You might write a short summary of how this model works and what it represents here.
* @docs        :: http://sailsjs.org/#!documentation/models
*/

module.exports = {

  attributes: {
    id:
    {
      type: "integer",
      primaryKey: true,
      unique: true,
      autoIncrement: true
    },
    question_id:
    {
      type: "integer",
      required: true,
      model: "questions"
    },
    text:
    {
      type: "string",
      defaultsTo: ""
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  }
};

redpajama_set_name: RedPajamaGithub


text:
Song-By-Song Review of Drake's Scorpion
Drake – Scorpion
Scorpion is Drake's best project since If You're Reading This It's Too Late and his studio best album in 5 years.
Side A: 3.52/5
#1 Survival: 2.75/5
I'm not sure that "Survival" is a good opener for Side A. It's slow and reflective like many Drake tracks, but also menacing in a way, and sets a somber, vengeful tone for the album. Perhaps this was Drake's goal, considering the album is titled "Scorpion" (for it's deceptive, striking tail I suppose). I think "8 Out of 10" would have made a great intro. Especially with the Kanye subliminal disses; Drake striking back like a Scorpion indeed.
#2 Nonstop: 2.75/5
Even at track #2 this song feels out of place. The tempo and vibe is polar opposite to the intro. There's 21 Savage's flow and adlibs blatantly all over this which makes it uncomfortable to listen to with a straight face. But I really like the Mack Daddy Ju "My Head Is Spinnin" sample that creates the chorus. However, not even Drake can save this ill-placed track with bars like "Yeah I'm light-skinned but I'm still a dark nigga".
#3 Elevate: 3.75/5
Elevate saves Side A in a big way, breaking the mediocrity of the first two songs. The intro "Survival" feels bored and the sophomore track "Nonstop" feels random. "Elevate" is vintage Drake as he seamlessly swerves in and out of rapping and harmony.
#4 Emotionless: 4/5
Verse 2 might go down as one of Drake's best verses. In verse 2, Drake addresses Pusha T's "The Story of Adidon" diss: "I wasn't hidin' my kid from the world I was hidin' the world from my kid". He also paints a gorgeous portrait of society in today's social media era: "I know a girl that saves pictures from places she's flown/ To post later and make it look like she still on the go". Me too, Drake. Me too.
#5 God's Plan: 3.5/5
Whose flow is this on the first 8 bars? Let me know so I can go mine their music! I love the rhyme scheme there, and the last 3-4 bars are about 1 syllable short of the others, but still flows like water. Perfect.
#6 I'm Upset: 2/5
Bad song. Contrary to popular belief, this song isn't about Pusha T or their beef. It's about Drake's baby mom and child support. Look at these lyrics:
"Can't go fifty-fifty with no ho/
Every month, I'm supposed to pay her bills/
And get her what she want/
I still got like seven years of doin' what I want/
My dad still got child support from 1991/"
#7 8 Out of 10: 4.5/5
One of the best songs on the album. Vintage 2010 Drake vibes; another banging Boi 1da/Drake collab. 1 of my fav songs on the album and I think it should've been the intro.
#8 Mob Ties: 3.75/5
In this song I hear: Playboi Carti, Ty$, Young Thug, and Future influences. Some say this song references XXXtentacion's death "Louis bags in exchange for body bags". But I won't go there.
A couple facepalm-worthy bars on this song: "Yes I be with Future but I like to reminisce" and of course the infamous "I am not with the RA-RA/ I am a Da-da"
#9 Can't Take A Joke: 4/5
The flow and rhyme scheme on the 2nd verse of this song is bananas. Again, WHO'S FLOW IS THIS?
"I grew up with the Reps Up boys, we crashin' it and splashin' in/
And when I say they crashin' it, I do not mean a accident/
I–I'm still in the studio at 6:45/
And my haters either on they way to work or they arrived/"
#10 Sandra's Rose: 3.25/5
Despite the Premo spot, this isn't one of my standout tracks off Scorpion. The production doesn't really strike me as Premo and Drake's delivery sounds like he's in the clouds – wandering. It doesn't sound like he's rapping with conviction. It's like he's reading off a paper. I had this same issue with the "Do Not Disturb" outro on More Life.
#11 Talk Up ft Jay-Z: 4/5
Again – WHO'S FLOW IS THIS ON VERSE 1? I love the hook, the theme of the track, and Drake's verses. Especially the hook. Who recorded the original reference track? I might need to mine their music!
Sidenote: I wish Hov would go harder and really dot the i's at the end of his verses, but the fact that we're still even getting verses of this caliber from Hov at nearly 50 years old is amazing. He's rapping right in stride with Drake who's 20+ years younger and still in his prime. Salute to Jay-Z, YA KNOW WHAT IM SAYIN??
#12 Is There More?: 4/5
Deeply reflective track; vintage introspective Drake over low-pass filter beats. Drake stunts a bit on this track "Even if you rounded up the numbers and rounded the troops…./ Too bad reality checks don't cover the balances due when it's time to recoup" but also reels it back in asking himself "Is There More To Life?" than living it up. Great closer to Side A.
Scorpion isn't really 1 disc Rap and 1 disc R&B…there's plenty of Rap on disc 2: March 14th, Nice For What, Blue Tint, That's How You Feel, Final Fantasy. These are all on disc 2 and all are rap songs. However, the moods on disc 1 and disc 2 are definitely different, with Side B housing a softer vibe: more R&B sections and more references to women (Jorja, Nicki, Magnolia Shorty, City Girls, Kiki, etc).
Side B: 3.519/5
#1 Peak: 2.75/5
Another questionable intro. I would've preferred "Finesse" here. "Peak" is like a metronome, lulling listeners to sleep right at the beginning of Side B.
#2 Summer Games: 3/5
This seems to be the only dance/pop track on this entire 25-song album. This is very interesting considering More Life was littered with dancehall/pop-aiming singles.
#3 Jaded ft Ty$: 2.75/5
They say "Jaded" is about UK'S new 21 year old songstress, Jorja Smith. I really enjoyed her Lost & Found album that also dropped in June of this year (2018).
On my 1st listen, I couldn't help but notice how deeply sad and DIRECT the music on Lost & Found feels. Some of the songs seem to spiral into a depressive state. I remember hearing "Teenage Fantasy" and every one of her lines hitting me right in the heart. Upon observing "Jaded"'s connection to Jorja Smith I must ask – is Lost & Found about Drake?
Are these "Teenage Fantasy" bars about Drake?
Jorja Smith on "Teenage Fantasy". Are these bars about Drake?
Very, very dope song. I love the added DJ mix at the end of the song, it gives the song a new dimension. Lauryn Hill sample. Murda on the beat. Is this Scorpion's theme song?
#5 Finesse: 3.75/5
Heavy PARTYNEXTDOOR vibes. I feel like "Finesse" and "Peak" are two tracks trying to capture similar vibes, but "Finesse" just does it much, much better. I would've enjoyed this more as the intro to Side B.
#6 Ratchet Happy Birthday: 2.75/5
If anyone was going to make Hip-Hop's 1st unofficial official birthday song since 50 Cent's "In Da Club" and 2Chainz "Birthday Song", it was going to be Drake. However, I don't understand this one right here. This doesn't feel like a birthday anthem – or an anthem at all. I think I heard Drake in autotune, which is a first as far I can remember. Also, this song isn't ratchet whatsoever…ain't no twerk-vibes here.
Only time will tell if "Ratchet Happy Birthday" is played at birthday parties around the world. I enjoy the drums and piano mesh in the production, though.
#7 That's How You Feel: 3.5/5
The Nicki sample in the chorus bounces beautifully with the drums. Nice, moody track. Night-time, windows down, reflecting on a situation-type song.
#8 Blue Tint: 4.25/5
Why is "Blue Tint" smushed in the middle of disc 2? Easily one of the best songs on the album and deserves to be out in the open.
"President doin' us in" is the most political I've ever heard Drake get. Ever. And even then, he's still not even saying anything. Surface-level observation. Guess he just isn't that type of artist.
#9 In My Feelings: 4/5
Probably the most popular song on the album, but seemed like it took a good week for it to gain traction among fans. Drake, as usual, is name-dropping real names on the hook.
This songs features a segue from the City Girls and Magnolia Shorty. "In My Feelings" is the second song to receive a female-sample + chorus/bridge combo on the album (the first being Nicki on "That's How You Feel")
#10 Don't Matter to Me ft Michael Jackson: 3.25/5
Definitely a smooth track but it just doesn't feel right with the MJ feature. I'm sorry. Maybe I'll feel different at a later date, but for now – Meh.
#11 After Dark ft Ty$ & Static Major: 3.75/5
Heavy Ty$ vibes here. Ty$ is in great company among the features on this album: Michael Jackson (RIP), Static Major (RIP), Jay-Z, and Ty$ are the only features. Amazing company.
#12 Final Fantasy: 4/5
Drake GETS OFF on the first verse before the song segways into the R&B transition. WE NEED ANOTHER RAP VERSE!!
#13 March 14th: 3.75/5
Fitting outro and Drake's note-on-wax to his son. Drake breaks into a sorrowful serenade in the tracks last moments, begging for guidance as he transitions "from a boy to a man" with his new son, Adonis. A very vulnerable and honest closing moment on the album. Almost makes you forget that Pusha T delivered audio-Ether in the form of "The Story of Adonis" just a month ago.
★★★1/2: 3.51/5 – Very good; a few bland songs or minor flaws throughout.
Scorpion is Drake's best project in years.
There's a lot on this album that I haven't seen many people unpack. The internet seemed really quick to write this album off. It's pretty dense with bars about Kanye, Push, Diddy, and even newcomer Jorja Smith. This is Drake's best album in YEARS.
Scorpion isn't without it's flaws. There are times where I feel Drake's sacrifices creativity in order to be directly personal – such as naming names of people, events, etc that are enclosed within his inner circle. When this happens, its easy to detach from Drake's songs.
Also, Drake cannot seem to escape corny punchlines and bars (such as "I am not with the RARA, I am a Dada); it's been one of my main criticisms on him his entire career. On "March 14th", Drake cringingly tells his son Adonis "I got this 11 tatted for somebody, now it's yours" (likely a reference to Jorja Smith, whose first project was titled Project 11). Extremely cringeworthy.
I also wish he'd rap with more conviction. Drake's delivery has sounded bored and uninterested after NWTS. This certainly is not the same energy that gave us "All Me" or "Miss Me".
-@TOP5RAPWEBSITE
Sidenote: No pop-dancehall, caribbean/UK accents, or faux grime flows on this album? THANK YOU
Video Review: Drake Scorpion
Drake – Views album review
Kendrick Lamar vs Drake: comparing discographies
Is Drake a Culture Vulture?
Posted in ReviewsTagged city girls drake, Drake, drake adonis, drake eve scorpion, drake jaded jorja smith, drake kanye beef, drake kanye MAGA, drake keke meaning, drake kiki meaning, drake pusha t beef, scorpion drake breakdown, scorpion review drake, scorpion song by song review
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ADDIS ABABA, Ethiopia – Ethiopia's Prime Minister Meles Zenawi, who died of an unknown illness earlier this week, will be buried on Sept. 2, the government spokesman said Thursday.
The government had postponed an emergency session of parliament to swear in a new prime minister as many leaders attended the funeral of a church leader.
Acting Prime Minister Hailemariam Desalegn was to be sworn in Thursday following Zenawi's death on Monday.
"There is no need to rush into it when the nation is grieving," Bereket said. "What all the lawmakers and their constituencies and the nation at large want to do at this time is mourn the great loss and honor the late prime minister. We want to first honor this and handle the prime minister's funeral with due diligence."
"The succession is already completed, and Hailemariam will take over as a prime minister," Bereket said.
Bereket also said the death of Abune Paulos, the head of the Ethiopian Orthodox Church, was another reason for the delay. Hailemariam attended and spoke at the Thursday funeral.
Leaders from around the world are expected to attend Meles's funeral. Meles who ruled Ethiopia for 21 year, died in a Belgium hospital late Monday from an illness that Ethiopian officials have not revealed.
Hailemariam is a relatively young figure on Ethiopia's political scene and it's not clear if old guard leaders will allow him to hold onto the prime minister's seat until 2015 elections.
Ethiopia is a strong U.S. ally on counter-terrorism issues, particularly in Somalia. Meles was hailed for advancing Ethiopia's economic progress but was denounced by human rights groups for a heavy handed approach to political dissent.


Rejoice, everyone! We're bearing some news for you and we're sure you'll be intrigued whether you're living with your family or living solo. Either way, if you're in charge of your household, then what we're about to tell is going to make chores more of an easy task. International appliances company, MABE, has come all the way from Mexico to Egypt with its very first laundry and cooling line in the country.
The Mexican-based company has a 70+ year experience in providing technological solutions for households varying from the production of refrigerators to cooking appliances, and more. The company has had eight production sites in North America. Today, the company manufactures and distributes over nine million products under 16 brands to over 78 countries around the world.
MABE's first line in Egypt will include top load washing machines and laundry center options for a variety of capacities. What makes this line quite attractive is that most of its products offer an ECO function, a technology that allows water saving during each cycle. In addition, the refrigerator line offers Side by Side and Top Mount No Frost configurations featuring top reliability and an energy saving technology. MABE provides up to 20% savings in most of the fridges while ensuring superior food preservation.
A launch event was held in Cairo's prestigious Nile Ritz-Calton hosting the Mexican Ambassador in Egypt, H.E. Mr. José Octavio Tripp. The guests also included MABE's International Vice President, Mr. Ramiro Perez Anzaldo, in addition to Gaballah's top management, Mr. Hesham Fathy and Mr. Gamal Gaballah. The night was hosted by Lebanese superstar, Razan El Maghraby, and Lebanese diva, Nicole Saba, entertained the crowd with a sensational performance.
WE SAID THIS: To know more about MABE, check out their Facebook page.


text:
Q: OpenCV camera stream stopping while in fullscreen mode I want to have two aplications simultaneously run: one that analyzes image from webcam written using OpenCV (the image is acquired through callback function) and an application that goes into fullscreen mode (let's say a 3D game). The problem is that while the fullscreen mode is launched the webcam image stream is stopping - the frames simply don't turn up, the callback function isn't called. This seems to be an issue with OpenCV - to test that a simple application displaying the image form camera has been prepared.
Why the image stream could be blocked by the fullscreen mode? How to bypass this?
Thanks for any hints.

A: Your question does not tell if you have tried to search for the problem in the OpenCV community first, so I post this as a hint in case: http://tech.groups.yahoo.com/group/OpenCV/
Also check out the list of issues, maybe its a known bug:  https://code.ros.org/trac/opencv/report/1

A: I'm not an OpenCV expert so this is closer to a suggestion than an answer - but I've experienced similar on my multi-monitor setup using a number of media players on the second monitor and some fullscreen apps ont he first.
In my limited testing, it comes down to what method is used to render the 3d app - DirectX seems to stop media players, OpenGL doesn't.
So it might not be OpenCV which has a problem - it may be what DirectX does to the hardware during a full-screen game.

A: Actually the behaviour of the OpenCV camera stream is strange. It seems to depend on the native OpenCV window (cvNamedWindow()) that shows the output image form webcam. If the window is on the same screen that went fullscreen the streaming will continue. If the camera window would be placed on another screen, the stream would stop.
Another curious thing is with the screen resolution change. If you change the resolution of the screen and the camera window is not visible (is closed or even minimalized) the image stream would get blocked.
These are just my observations on the topic, maybe it'll be helpful for someone.

redpajama_set_name: RedPajamaStackExchange


Initiating such a curfew would save the airport an estimated $67 million over 10 years, according to the study. The next step for the Jacobs Consultancy, which completed the study, is public outreach about the findings, Gill said.
While Bob Hope Airport's study argues the case to curfew quieter jets made in the 1980s, Van Nuys Regional Airport is compiling its own Part 161 Study to ban older, noisier jets, Gill said. If the FAA does not rule favorably on Bob Hope's study, that would bring the issue of a nighttime curfew to a standstill in Burbank.
"I hope to be able to support the package of Part 161 studies from Burbank and Van Nuys and recognize that the Valley's two neighborhood airports are not appropriate venues for flights between 10 p.m. and 7 a.m.," Rep. Brad Sherman said.
If the FAA does recognize and subsequently rule in favor of the study, then the curfew proposal would be reviewed against the California Environmental Quality Act for any environmental consequences. Gill expects the study is thorough enough that environmental review would not take too long before a final ruling on the curfew.
Editor's Note: We have also been advised that the proposed full Night Time Curfew will also affect one United Airlines flight that departs at 6:50 AM and a FEDEX Flight that arrives before 7:00 AM, as well as the many Ameriflight Cargo flights that arrive and depart at night. Will package deleivery prices be increased or, will next morning deleivery no longer be available to the Burbank community?
The Airport Authority also has authorized and additional $40,000 to the company that is doing the Part 161 Study to answer more questions that the FAA has asked during the Public Comment Period from last year. They have made several trips back to FAA Headquarters in Washington D. D.


text:
+Get Involved
Gov't Backs Down From De-Licencing Debt Collectors Live On Air
We raised concerns about the NSW Government wanting to scrap licencing for debt collectors, and they've backed down.
Shooters Fishers and Farmers Party MLC Robert Borsak celebrated a win today after Minister Victor Dominello completely backed-down from his plan to de-licence debt collectors This came after the Minister's office heard news that Mr Borsak would be speaking on 2GB this morning on the issue.
"This is a win for anybody falling on hard times, brokered by the Shooters Fishers and Farmers Party," Mr Borsak said.
"We don't want a situation in New South Wales where thugs and bikies can knock on people's doors collecting debts.
"Yesterday the NSW Government were saying they would be pushing through their bill to deregulate and de-licence debt collectors through all its stages last night - taking advantage of three less unpaired MPs in the Upper House, but this morning they backed down.
"Labor MLC Sophie Cotsis resigned yesterday to contest the Canterbury by-election, Jan Barham from The Greens was away, and I'm recovering from surgery late last week. The Government wouldn't have needed Crossbench votes to pass legislation, because there are three less votes to oppose them.
"However, they halted the bill halfway through passing it last night, at the last minute.
"This morning I was to appear on 2GB to speak about these irresponsible plans to de-licence debt collectors, but as soon as Dominello's office got wind of the interview they pulled the bill live on air.
"Mike Baird and Troy Grant are afraid of criticism with the growing popularity of our first-time candidate Philip Donato in the Orange by-election who's securing 15.5% of the vote after announcing his candidacy two weeks ago.
"I couldn't believe that they were considering such a plan in the first place. Removing protections from debt collectors would hurt people when they were at their most vulnerable, and would effectively turn New South Wales into a 'Mafia State' All sorts of unsavoury people would be able to knock on people's doors asking for money."
"The Government said that they would remove the existing 'positive licence', instead imposing a 'negative licence' - which means there will be a blacklist of people excluded from the industry.
"Obviously the Baird Government's work experience kid has been at the computer again - first they created the term 'social licence' for banning greyhound racing, now they created the terms 'positive licence' and 'negative licence' to give the false impression they would still be licencing debt collectors.
"I'll be closely watching Dominello's next big plan: to scrap consumer protections for CTP insurance."
The plan to de-licence debt collectors was being created by the Fair Trading Amendment (Commercial Agents) Bill 2016, but they have pulled the bill.
Originally the interview was scheduled for 9:25am, but Minister Dominello's office called 2GB directly to inform the presenters they had changed their mind. The interview finally took place at 9:40am where it was announced live on air that the Government was backing down from their original plans.
Download Robert Borsak's press release here.
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"Set in rural New York state at the turn of the twentieth century, superb new talent James Scott makes his literary debut with The Kept—a propulsive novel reminiscent of the works of Michael Ondaatje, Cormac McCarthy, and Bonnie Jo Campbell, in which a mother and her young son embark on a quest to avenge a terrible and violent tragedy that has shattered their secluded family.
On a cold, snowy day in Upstate New York in 1897, Elspeth Howell is returning home after months away working as a midwife in a distant town. Arriving at the family cabin, she finds her husband and four of her children have been savagely murdered, and one son – Caleb – missing. Before she can search for him, she herself is shot down and grievously injured. As Caleb creeps out from his hiding place in the pantry, he discovers his injured mother and drags her to the barn to begin nursing her back to health. As soon as Elspeth is on her feet, mother and son head off in search of the killers. They follow their trail to a town named Watersbridge, a town from Elspeth's shady past that fills her with dread.
Elspeth holds many secrets close to her heart, more than Caleb could ever imagine, and is positive that her family's murders are retribution for her past wrongs. As Caleb slowly learns more and more about the truth of his mother's sins, he is torn between hate and loyalty but ultimately is focused on gaining justice for his murdered family. Elspeth, too, believes that seeing to the end of the killers is the only way she can atone for her many sins.
The Kept by James Scott is the dark and intense story of innocence lost. Scott's eloquent and masterful writing brings to life the bleak and emotionally heavy atmosphere surrounding Elspeth and Caleb as they travel in search of their family's killers. The serious tones of the novel are complemented by the heavy subject matter and discomfiting scenes that occur during their search. At times hard to read are the stomach-churning moments of violence, so vividly described, that the two witness with regularity.
There is not a cheery moment to be found in this novel. Be prepared for a serious and heart-breaking read. Scott weaves a tale of sin, murder, loss, revenge, and backhanded justice that will hold you in check until the final, gut-wrenching end. Go and get you some, y'all.
The Kept by James Scott: Read it today!


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Montreal Impact: Academy
Montreal Impact vs New York Red Bulls U16/U18 Academies face each other in USSDA League
On a rainy and cold day at Complexe Sportif Claude-Robillard, the Impact and Red Bulls U16/U18 Academies faced off back to back.
By Sofiane Benzaza@sofianebenzaza Sep 30, 2012, 2:48pm EDT
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It was cold, wet and slippery for both teams and between 50 to 75 people attended the games to support the young players of the future for the Montreal Impact.
The Impact U18 team played first at 12:45 EST and lost on a score of 4-1 to the Red Bulls in a game that saw some good and some bad from the Impact.
The U16 teams are facing each other at 2:45 PM EST and the game is ongoing as we speak.
Stay tuned for the final results, scores and quotes from both games, right here at Mount Royal Soccer.com . You can follow the stream and get your updates here.
Impact lose 4-0 to Red Bulls in USSDA U16 game
Impact vs NYRB Academy faceoff: U16 & U18 in USSDA


Unfortunately, individual circumstances are very important when giving advise on these questions and the answers often begin with "It depends…". Let us look at the things you should think about when trying to work out which assets you should include in your CMMS.
What assets should you include in your CMMS?
There are two main considerations that will help you determine which assets to include in your CMMS.
First, your CMMS should be functional: for you, your staff and your guest requestors.
Optimum functionality occurs when your CMMS has enough assets that your users can quickly select the correct asset, without being overwhelmed with detail.
For example, in a school, it would be wise to include in the CMMS an asset "lighting system" in each classroom, in this scenario a user could easily select the lighting system in Classroom A001 and direct a technician to replace the bulb. Conversely, it would be unwise to include every light bulb in every classroom, because a user would find it very difficult to say if it is light number 4 or number 5 in Classroom A001 that has blown. In any case, the technician will attend the classroom and then look for the broken light to fix. Too much detail is meaningless when it comes to the practical aspects of doing maintenance.
Choose a level of asset detail that helps, and does not hinder the people using the CMMS.
Can you extract meaningful data from your CMMS?
If you want to extract meaningful data from your CMMS you must have enough detail in the assets that you include. Without this detail, you will never be able to identify the types of assets on which you are spending too much.
Returning to the school example, the individual classrooms should not be most detailed asset in the CMMS. It is wise to have the lighting system for each classroom included in the CMMS, as well as the air-conditioning and heating systems. With this setup it will be easy to differentiate the maintenance money spent on lighting, air-conditioning and heating systems across the whole school.
In summary, you should make the assets detailed enough that you can examine costs, but not too detailed that a generally intelligent user cannot easily choose the correct asset from a list.
How should I arrange my asset tree?
An asset tree is the system by which your assets are arranged in your CMMS. For example, I have setup a CMMS for my house. In the asset tree for my house, the 'Fridge' is a child asset of the 'kitchen' and the 'kitchen' is, itself, a child asset of the 'My house'.
You should setup your asset tree in such a way that it is easy for the people who do maintenance to find the asset. When thinking about this, consider how you would describe where to find the asset if you were talking on the phone. You would probably say "The fridge in the kitchen has stopped working". This makes it clear you are not talking about the beer fridge in the outside garage. If your facility is going to have more than one asset of the same type it also helps to have an asset naming convention so there is no ambiguity about which asset is being refered to.
In short, you should arrange your asset tree so things can be easily found.
The system described above is fine for items that are stationary, but what about assets like vehicles or welding trolleys that are mobile and found in different locations at different times? You should again ask yourself "how would I describe this over the phone?" Using this approach you will probably find a nice way of collecting together your assets in the tree. For example, your asset tree might group all of your vehicles together as child assets of "Mobile Fleet", and your welding trolley might be grouped under "Mobile workshop tools". Importantly, the users of your system should be able to find the asset they want quickly and logically.
How many assets should I include in my asset tree?
We have done some analysis of the number of assets included in the asset tree of companies who have successfully implemented our CMMS. We defined a successful company as one who are still using their CMMS more than 2 months after first purchasing it and have not stopped using it. The results are shown below. It is very clear that a company with more than 25 Assets per user have more success with their CMMS implementation than one with 25 assets per user or less. Quickly adding 25 or more assets is a great way to make sure your CMMS gets started the right way.


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Home » The Firm » News » Martin F. Lewis to Join Greenhill as a Managing Director
Martin F. Lewis to Join Greenhill as a Managing Director
NEW YORK, February 8, 2007 – Greenhill & Co., Inc. (NYSE: GHL), a leading independent investment bank, announced today that Martin F. Lewis will join the Firm as a Managing Director based in New York. Mr. Lewis, a veteran financial restructuring specialist, will focus on expanding the restructuring advisory practice at Greenhill.
Mr. Lewis, 52, was most recently at Rhone Group and prior to that was a founding member of Miller Buckfire Lewis & Co. and a Managing Director at The Blackstone Group.
Robert F. Greenhill, Chairman and CEO of Greenhill, said "Martin Lewis has had a long and distinguished career advising on a wide range of restructurings and bankruptcy cases. While M&A activity continues to be very strong, Martin's recruitment is evidence of our intention to expand our restructuring advisory capabilities in preparation for when the economic cycle turns."