text:
\section{Introduction}

Let $X=X_1\times \cdots\times X_n$ be a product of $n$ finite intervals of integers, and let $f:X\to X$,
\[
x=(x_1,\dots,x_n)\mapsto f(x)=(f_1(x),\dots,f_n(x)).
\]
Such a function $f$ is regarded as a {\em finite dynamical system} with $n$ components (FDS for short). If $X=\{0,1\}^n$ then $f$ is usually called a {\em Boolean network} (BN for short). FDSs have many applications. In particular, since the seminal works of MacCulloch and Pitts, Hopfield, Kauffman and Thomas, they are classical models for the dynamics of neural and gene networks \cite{MP43,H82,GM90,K69,K93,T73,TA90}. They are also used in information theory, for the network coding problem \cite{GR11,GRF16}.

\smallskip
The structure of a FDS $f$ is usually represented by a signed digraph, called {\em interaction graph}. Formally, a {\em signed digraph} is a couple $G=(V,E)$ where $V$ is a set of vertices and $E\subseteq V\times V\times\{+,-\}$ is a set of signed arcs: if $(j,i,s)\in E$ then $G$ has an arc from $j$ to $i$ of sign~$s$. The \BF{interaction graph of $f$} is then defined as the signed digraph  with vertex set $\{1,\dots,n\}$ and with a positive (resp. negative) arc from $j$ to $i$ if there exists $x\in X$ with $x_j<\max(X_j)$ such that
\[
f_i(x_1,\dots,x_j+1,\dots,x_n)-f_i(x_1,\dots,x_j,\dots,x_n)
\]
is positive (resp. negative). If $G$ is the interaction graph of $f$, we say that $f$ is a \BF{FDS  on $G$}.

\smallskip
In molecular biology, the first reliable informations available about a biological system often concern its interaction graph \cite{TK01,N15}. A natural question is then the following: {\em what can be said on the dynamics of $f$ according to its interaction graph only?} Among the many dynamical properties that can be studied, fixed points are of special interest, since they correspond to stable states and often have a strong meaning \cite{R86,TA90,GM90}. For instance, in the context of gene networks, they correspond to stable patterns of gene expression at the basis of particular biological processes \cite{TA90}. As such, they are the property which has been studied the most.

\smallskip
In particular, many works have been stimulated by two general rules, stated in 1981 by the biologist Ren\'e Thomas \cite{T81}, concerning the influence of positive and negative cycles of the interaction graph (cycles are always directed and without repeated vertices, and the sign of a cycle is the product of the signs of its arcs). Roughly, the first (resp. second) rule says that the presence of a positive (resp. negative) cycle is a necessary condition for multiple stable states (resp. permanent periodic behaviors).  Several variations of these rules have been formally stated and proved for several classes of discrete and continuous dynamical systems (see {\em e.g.} \cite{A08,RRT08,RC07,R08,R09,R18,R19} for the discrete case, and \cite{PM95,S98,G98,CD02,S03,S06,KST07,S13,KS19} for the continuous case). In particular, the first rule has many generalizations and strong versions involving local interaction graphs based on the Jacobian matrix (see \cite{R19,KS19} and the references therein). It has also a nice biological interpretation since multiple stable states can account for cell differentiation processes \cite{TK01a}.

\smallskip
In this paper, we are mainly concerned with the basic \BF{discrete versions of Thomas' rules} given by the second and third item of the following theorem. From them we deduce that there is a unique fixed point in the acyclic case, and the first item says something stronger ($f^k$ denotes the $k$th composition of $f$ with itself).

\begin{theorem}\label{thm:thomas}
Let $f$ a FDS with $n$ components on a signed digraph $G$.
\begin{itemize}
\item[{\em (a)}] If $G$ has no cycle, then $f^n$ is constant, and thus $f$ has a unique fixed point \cite{R86,R95}.
\item[{\em (b)}] If $G$ has no negative cycle, then $f$ has at least one fixed point \cite{R10}.
\item[{\em (c)}] If $G$ has no positive cycle, then $f$ has at most one fixed point \cite{RRT08,RC07}.
\end{itemize}
\end{theorem}

The following complementary theorem shows that, in the the Boolean case, stronger conclusions can be obtained under the additional assumption that the interaction graph is {\em strongly connected} ({\em i.e.} for every distinct vertices $i$ and $j$, there is a path from $i$ to $j$).

\begin{theorem}\label{thm:aracena}
Let $f$ be a BN on a strongly connected signed digraph $G$ with at least one arc.
\begin{itemize}
\item[{\em (a)}] If $G$ has no negative cycle, then $f$ has at least two fixed points \cite{A08}.
\item[{\em (b)}] If $G$ has no positive cycle, then $f$ has no fixed point \cite{A08}.
\end{itemize}
\end{theorem}

The above results give conditions on $G$ that force {\em all} the FDSs or BNs on $G$ to have particular properties. In this paper, we take the converse direction: we study conditions on $G$ that allow {\em at least one} FDS on $G$ to have particular properties. Since it is clearly easier to exhibit a particular FDS when we allow the intervals $X_i$ to be large, it is natural to think to bound the size of these intervals. This leads us to focus on {\em degree-bounded systems}, which are usual in the context of gene networks. The formal definition follows.

\smallskip
Let $f$ be an FDS with $n$ components, and let $G$ be its interaction graph. We say that $f$ is \BF{degree-bounded} if, for all $i\in \{1,\dots,n\}$,
\begin{itemize}
\item
$|X_i|=2$ if $d^{\mathrm{out}}_G(i)=0<d^{\mathrm{in}}_G(i)$, and
\item
$|X_i|\leq d^{\mathrm{out}}_G(i)+1$ otherwise.
\end{itemize}
Here, $d^{\mathrm{out}}_G(i)$ is the out-degree of $i$ in $G$, defined as the  number of positive arcs leaving $i$ plus the number of negative arcs leaving $i$. The in-degree $d^{\mathrm{in}}_G(i)$ is defined similarly.

\smallskip
These conditions on the intervals $X_i$ could seem unnatural at first glance. However, Ren\'e Thomas, who introduced them, showed that they are very natural in the context of gene networks. He actually widely popularized the use of degree-bounded systems for modeling gene networks through the so called {\em generalized logical method}; see \cite{TA90,TK01} for a presentation, and see \cite{AMN14} and the references therein for recent applications. The intuition behind the degree-bounded conditions is the following. In continuous modelings of gene networks, the influence of an interaction from $i$ to $j$ of sign $s$ is often described by a sigmoid function, which can be approximated by a step function with a threshold $\theta^s_{ij}$. From this approximation, the system can be discretized in a natural way: the variable $x_i$, which is a non negative real number representing a concentration, is approximated by a discrete variable $\tilde x_i$ defined as the number of out-going interactions $(i,j,s)$ such that $x_i\geq \theta^s_{ij}$. The size of the domain of this discrete variable $\tilde x_i$ is thus at most the number of interactions leaving $i$ plus one (see \cite{TA90,TK01} for details).

\subsection{Nilpotent systems}

Our first result concerns {\bf nilpotent} FDSs, that is, FDSs $f$ such that $f^k$ is constant for some positive integer $k$.

\begin{theorem}\label{thm:nilpotent}
If $G$ is an $n$-vertex connected signed digraph distinct from a signed cycle, then  there exists a degree-bounded system $f$ on $G$ such that $f^{n+1}$ is constant.
\end{theorem}

This shows that a nilpotent degree-bounded system with a linear convergence can be defined on almost all connected signed digraphs. Since such a nilpotent system describes a very simple dynamics $-$ a global convergence in $n+1$ iterations $-$ this shows that {\em the fact that a degree-bounded system describes a complex dynamics cannot be deduced from its interaction graph.}

\smallskip
If $G$ is a signed cycle, then any degree-bounded system $f$ on $G$ is a BN, and we deduce from Theorem~\ref{thm:aracena} that $f$ has either zero or two fixed points, thus $f$ is not nilpotent. Therefore, according to the previous theorem, we have the following characterization:

\begin{corollary}\label{cor:nilpotent}
Let $G$ be a connected signed digraph. There exists a nilpotent degree-bounded system on $G$ if and only if $G$ is not a signed cycle.
\end{corollary}

The existence of nilpotent systems is also the subject of the recent paper \cite{GR16}. It is proved that, for {\em every} $n$-vertex signed digraph $G$,
\begin{itemize}
\item
there exists a FDS $f:X\to X$ on $G$ with $X=\{0,1,2,3\}^n$ such that $f^2$ is constant, and
\item
there exists a FDS $f:X\to X$ on $G$ with $X=\{0,1,2\}^n$ such that $f^{\lfloor \log_2 n\rfloor+2}$ is constant.
\end{itemize}
As shown above, there does not necessarily exist a nilpotent BN on a given signed digraph~$G$, and \cite{GR16} provides several sufficient conditions for the existence of such a BN. This raises the question of a characterization of the signed digraphs on which a nilpotent BN can be defined, in the spirit of the previous corollary, which gives such a characterization in the degree-bounded case. In \cite{GR16} the following question is also raised.

\begin{question}
Is there a constant $c$ such that, for every $n$-vertex signed digraph $G$, if there exists a nilpotent BN on $G$, then there exists a nilpotent BN $f$ on $G$ that converges in at most $cn$ iterations, that is, such that $f^{cn}$ is constant?
\end{question}

Theorem~\ref{thm:nilpotent} and Corollary \ref{cor:nilpotent} clearly give a positive answer in the degree-bounded case.

\subsection{Convergence between systems and weak converses of Thomas' rules}

Our second result generalizes the previous one using the following notion of convergence between systems. Let $f:X\to X$ and $h:Y\to Y$ be two FDSs with $n$ components. We say that \BF{$f$ converges toward $h$} in $k$~steps~if
\begin{itemize}
\item
$f^k(X)\subseteq h(Y)\subseteq Y\subseteq X$, and
\item
$f(x)=h(x)$ for all $x\in Y$.
\end{itemize}
In a signed digraph, a {\em source} is a vertex of in-degree zero, a {\em sink} is a vertex of out-degree zero, and an {\em isolated vertex} is a vertex of in- and out-degree zero.

\begin{theorem}\label{thm:main}
Let $G$ be a connected signed digraph distinct from a signed cycle. Let $H$ be a subgraph of $G$ obtained by removing arcs only. Suppose that every non-isolated source of $H$ is a source of $G$, and that every non-isolated sink of $H$ is a sink of $G$. Then, for every degree-bounded system $h$ on $H$, there exists a degree-bounded system $f$ on $G$ that converges toward $h$ in $k+1$ steps, where $k$ is the number of isolated vertices in $H$.
\end{theorem}

Before discussing some consequences of this theorem, let us first observe that some conditions on the subgraph $H$ are needed to obtain the conclusion. Suppose, for instance, that $H$ is obtained from $G$ by removing an arc from $i$ to $j$ with $d^{\mathrm{in}}_G(i)=d^{\mathrm{out}}_G(i)=d^{\mathrm{in}}_G(j)=1$, and let $h:Y\to Y$ be any degree-bounded system on $H$. Then there is no degree-bounded system $f:X\to X$ on $G$ that converges toward $h$. Indeed, suppose, for a contradiction, that such a $f$ exists. Since $d^{\mathrm{out}}_H(i)< d^{\mathrm{out}}_G(i)=d^{\mathrm{in}}_H(i)=1$, $X_i$ and $Y_i$ are of size two, and thus $X_i=Y_i$. Since $f_j$ only depends on $x_i$ and $X_i=Y_i$, there exists $x,y\in Y$ that only differ in $x_i\neq y_i$ such that $f_j(x)\neq f_j(y)$. Since $f$ converges toward $h$, we deduce that $h_j(x)\neq h_j(y)$ and thus $d^{\mathrm{in}}_H(j)>0$, a contradiction.

\smallskip
A first immediate consequence of Theorem~\ref{thm:main} is Theorem~\ref{thm:nilpotent}. Indeed, consider the subgraph $H$ obtained by removing {\em all} the arcs of $G$, and let $h:Y\to Y$ be any degree-bounded system on $H$. Then $H$ consists in $n$ isolated vertices, and thus $Y$ is a singleton, so that $h:\{\xi\}\to\{\xi\}$ for some $\xi\in \mathbb{Z}^n$. Then, according to the conclusion, there exists a degree-bounded system $f$ on $G$ that converges toward $h$ in $n+1$ steps, thus $f^{n+1}={\mathrm{cst}}=\xi$, and Theorem~\ref{thm:nilpotent} is recovered.

\smallskip
As another consequence, we have the following weak converses of Theorems~\ref{thm:thomas}(b) and \ref{thm:thomas}(c).

\begin{corollary}\label{cor:converse}
Let $G$ be a connected signed digraph.
\begin{itemize}
\item[{\em (a)}]
If $G$ has a negative cycle, then there exists a degree-bounded system on $G$ without fixed~point.
\item[{\em (b)}]
If $G$ has a positive cycle, then there exists a degree-bounded system on $G$ with two fixed~points.
\item[{\em (c)}]
More generally, if $G$ has $k$ vertex-disjoint positive cycles, with $k\geq 1$, then there exists a degree-bounded system on $G$ with $2^k$ fixed points.
\end{itemize}
\end{corollary}

\begin{proof}[{\bf Proof}]
For the first point, suppose that $G$ has a negative cycle $C$, and let $H$ be obtained from $G$ by removing all the arcs of $G$, excepted those of $C$. Let $h$ be any degree-bounded system on $H$. It is easy to see that $h$ has no fixed point. If $G=C$ then $G=H$ and we are done. Otherwise, by Theorem~\ref{thm:main}, there exists a degree-bounded system $f$ on $G$ that converges toward~$h$. Thus $f$ has no fixed point.

\smallskip
The proof of the third point, which implies the second, is similar. Suppose that $G$ has $k$ vertex-disjoint positive cycles, say $C_1,\dots,C_k$. Let $H$ be obtained from $G$ by removing all the arcs of $G$, excepted those of $C_i$, $1\leq i\leq k$. Let $h$ be any degree-bounded system on $H$. It is easy to see that $h$ has $2^k$ fixed points. If $k=1$ and $G=C_1$, then $G=H$ and we are done. Otherwise, by Theorem~\ref{thm:main}, there exists a degree-bounded system $f$ on $G$ that converges toward~$h$. Thus $f$ has $2^k$ fixed points.
\end{proof}

Note that from Thomas' rules and the above weak converses, we obtain the following characterizations: {\em if $G$ is connected, then $G$ has a negative (positive) cycle if and only if there exists a degree-bounded system on $G$ no (two) fixed points.}

\smallskip
Note also that the weak converses of Thomas' rules are false in the Boolean case. For instance, if $G$ consists in two cycles with opposite signs that share exactly one vertex, then it is easy to see that {\em every} BN on $G$ has a unique fixed point. More generally, if $G$ is strongly connected, has a unique negative (positive) cycle, and has at least one positive (negative) cycle, then {\em every} BN on $G$ has at least (at most) one fixed point \cite{R18}.

\smallskip
Another gap between the Boolean case and the degree-bounded case, that shows the relevance of Corollary \ref{cor:converse}(c), is the following. Let $\nu^+$ be the maximum number of vertex-disjoint positive cycles in $G$. According to Corollary \ref{cor:converse}(c), the maximum number of fixed points in a degree-bounded system on $G$ is exponential in $\nu^+$ while, in the Boolean case, it can be linear in $\nu^+$. Indeed, as proved in \cite{ARS17}, for every positive integer $k$ there exists a strongly connected signed digraph $G_k$ with $k$ vertex-disjoint positive cycles such that {\em every} BN on $G_k$ has at most $k+1$ fixed points.

\smallskip
The organization of the paper is the following. A slightly stronger version of Theorem~\ref{thm:nilpotent} is proved in Section \ref{sec:nilpotent}, and the proof of Theorem~\ref{thm:main}, which uses this stronger version, is given in Section~\ref{sec:convergence}.

\section{Nilpotent systems}\label{sec:nilpotent}

Let $G=(V,E)$ be a signed digraph. The set of {\em positive in-neighbors} of $i$ in $G$, denoted $G^+_i$, is the set of $j\in V$ such that $(j,i,+)\in E$. The set of {\em negative in-neighbors} is defined similarly. The set of in-neighbors of $i$ in $G$ is $G_i=G^+_i\cup G^-_i$. These  notations are rather unconventional, but useful in the following for their compactness. Note that $G^+_i$ and $G^-_i$ are not necessarily disjoint: if $j\in G^+_i\cap G^-_i$ then $G$ has both a positive and a negative arc from $j$ to $i$, and we say that $G$ has {\em parallel arcs} from $j$ to $i$. Note also that $d^{\mathrm{in}}_G(i)=|G^+_i|+|G^-_i|$ (thus the in-degree can be greater than the number of in-neighbors and, similarly, the out-degree may be greater than the number of out-neighbors). We denote by $|G|$ be the underlying unsigned digraph of $G$: the vertex set is $V$ and there is an arc from $j$ to $i$ if and only if $j\in G_i$. See Figure~\ref{fig:pseudo-cycle} for an illustration. We say that $G$ is a {\em signed cycle} if $|G|$ is a cycle and $G$ has no parallel arc (thus the signed digraph in Figure~\ref{fig:pseudo-cycle} is not a signed cycle, while its underlying unsigned digraph is a cycle). If $G'=(V',E')$ is another signed digraph, then $G\cup G'=(V\cup V',E\cup E')$. The subgraph of $G$ {\em induced} by $U\subseteq V$, denoted $G[U]$, is the signed digraph with vertex set $U$ and arc set $E\cap U\times U\times \{+,-\}$. We set $G\setminus U=G[V\setminus U]$. We say that $U$ is a {\em strong component} of $G$ is $G[U]$ is strongly connected and $U$ is maximal for this property. A strong component $U$ is {\em initial} if $G$ has no arc from $V\setminus U$ to $U$. A strong component $U$ is {\em trivial} if $G[U]$ is the {\em trivial graph}, {\em i.e.} has one vertex and no arc.

\smallskip
We say that $G$ is {\bf basic} is all the initial strong components of $G$ are trivial. We set
\[
\beta(G)=
\begin{cases}
0&\text{if $G$ is basic},\\
1&\text{otherwise.}
\end{cases}
\]
If $i,j\in V$ then $d_G(j,i)$ denotes the minimum number of arcs in a path from $j$ to $i$. If $j=i$ then $d_G(j,i)=0$ and if there is no path from $j$ to $i$ then $d_G(j,i)=\infty$. If $I\subseteq V$ then $d(I,i)$ is the minimum of $d(j,i)$ for $j\in I$. Let $\mathcal{I}$ be the set of initial strong components of $G$. We set
\[
\lambda(G)=\max_{i\in V}\min_{I\in\mathcal{I}} d_G(I,i)+|I|.
\]

\begin{remark}\label{rem:lambda}
{\em Suppose that $G$ has $n$ vertices. Clearly, for every vertex $i$, there exists $I^*\in\mathcal{I}$ such that $G$ has a path from $I^*$ to $i$. If $P$ is a shortest path from $I^*$ to $i$, then, excepted the initial vertex of $P$, no vertex belong to an initial strong component of $G$. We deduce that the number of arcs in $P$ is at most $n-\sum_{I\in\mathcal{I}}|I|$. As a consequence,
\[
\lambda(G)\leq n-\sum_{I\in\mathcal{I}}|I|+\max_{I\in\mathcal{I}} |I|\leq n.
\]
In particular, if $G$ has at least two initial strong components then $\lambda(G)< n$, and if $G$ is basic and has $k$ sources, then $\lambda(G)\leq n-k+1$. Note also that $\lambda(G)=n$ if $G$ is strongly connected.}
\end{remark}

We are now in position to state the following quantitative version of Theorem~\ref{thm:nilpotent}. We write $f^k={\mathrm{cst}}=\xi$ to mean that $f^k$ is a constant function, always equal to $\xi$.

\begin{figure}
\[
\begin{array}{c}
\begin{tikzpicture}
\node[N] (1) at (150:1){\scriptsize$1$};
\node[N] (2) at (30:1){\scriptsize$2$};
\node[N] (3) at (270:1){\scriptsize$3$};
\path[->,thick]
(3) edge[Green,bend left=25] (1)
(3) edge[red,bend right=25] (1)
(1) edge[Green] (2)
(2) edge[red] (3)
;
\end{tikzpicture}
\\[3mm]
G
\end{array}
\quad\quad
\begin{array}{c}
\begin{tikzpicture}
\node[N] (1) at (150:1){\scriptsize$1$};
\node[N] (2) at (30:1){\scriptsize$2$};
\node[N] (3) at (270:1){\scriptsize$3$};
\path[->,thick]
(1) edge (2)
(2) edge (3)
(3) edge (1)
;
\end{tikzpicture}
\\[3mm]
|G|
\end{array}
\]
{\caption{\label{fig:pseudo-cycle}
Positive (negative) arcs are green (red). This convention is used throughout the paper.
}}
\end{figure}

\begin{theorem}\label{thm:nilpotent2}
Let $G$ be a connected signed digraph distinct from a signed cycle. There exists a degree-bounded system $f:X\to X$ on $G$ such that
\[
f^{\lambda(G)+\beta(G)}={\mathrm{cst}}=\xi,
\]
where $\xi_i=\min(X_i)$ for all sources $i$ of $G$.
\end{theorem}

\begin{proof}[{\bf Proof}]
Suppose that $G$ has vertex set $V=\{1,\dots,n\}$. If $G$ is trivial the theorem is obvious, so assume that $G$ is not trivial. The proof is devised in two cases.

\smallskip
{\bf Case 1: {\boldmath$|G|$\unboldmath} is not a cycle.} Let $\mathcal{I}=\{I_1,\dots,I_k\}$ be the set of initial strong components of $G$. Since $G$ is connected and since $|G|$ is not a cycle, each $I_\ell$ contains at least one vertex $i_\ell$ such that either $i_\ell$ is a source or, for all $j\in G_{i_\ell}$, the out-degree of $j$ in $|G|$ is at least two.  Let $H$ be the subgraph obtained from $G$ by removing all arcs with $i_\ell$ as terminal vertex, $\ell=1,\dots,k$. The initial strong components of $H$ are $\{i_1\},\dots,\{i_k\}$ thus $H$ is basic. Furthermore, for every $\ell=1,\dots,k$ and $j\in G_{i_\ell}$, the out degree of $j$ in $H$ is at least one. Hence,
\begin{equation}\tag{$*$}\label{eq:sink}
\text{\em every sink of $H$ is a sink of $G$}.
\end{equation}
Furthermore,
\[
\lambda(H)\leq\lambda(G).
\]
Indeed, let $i\in V$ and let any $\ell$ such that $G$ has a path $P$ from $I_\ell$ to $i$. Suppose that $P$ is of length  $d_G(I_\ell,i)$. Let $j$ be the first vertex of $P$, and let $P'$ be a path from $i_\ell$ to $j$ of length $d_G(i_\ell,j)$. Since $P'\cup P$ is a path of $H$, $d_H(i_\ell,i)\leq d_G(i_\ell,j)+d_G(I_\ell,i)$, and since $P'$ is contained in $G[I_\ell]$ we have $d_G(i_\ell,j)<|I_\ell|$. Hence $d_H(i_\ell,i)+1\leq d_G(I_\ell,i)+|I_\ell|$. It follows that $\lambda(H)\leq \lambda(G)$.

\smallskip
Let $I=\{i_1,\dots,i_k\}$ and let $X=X_1\times\dots\times X_n$ be defined as follows: for all $i\in V$,
\[
X_i=
\left\{
\begin{array}{lll}
\{0,1,2,3\}&\text{if $i\in G^+_j\cap G^-_j$ for some $j\in I$}&(1)\\
\{0,1,2\}&\text{if $i\in G^+_j\setminus G^-_j$ for some $j\in I$}&(2)\\
\{0,1,2\}&\text{if $i\in G^-_j\setminus G^+_j$ for some $j\in I$}&(3)\\
\{0,1,2\}&\text{if $i\in G^+_j\cap G^-_j$ for some $j\in V\setminus I$}&(4)\\
\{0,1\}&\text{otherwise.}&(5)
\end{array}
\right.
\]
Let $f:X\to X$ be any FDS on $G$. Since $G$ has no isolated vertex, to prove that $f$ is degree-bounded, it is sufficient to prove that $|X_i|\leq\max(2,d^{\mathrm{out}}_G(i)+1)$ for every $i$. This is obvious in cases $(4)$ and $(5)$. In case (1) we have $d^{\mathrm{out}}_G(i)\geq 2$, and if $d^{\mathrm{out}}_G(i)=2$, then $i$ is a sink of $H$ but not a sink of $G$, a contradiction with (\ref{eq:sink}). Hence $d^{\mathrm{out}}_G(i)\geq 3$ and thus $|X_i|=4\leq d^{\mathrm{out}}_G(i)+1$. Similarly, in case (2) and (3) we have $d^{\mathrm{out}}_G(i)\geq 1$, and if $d^{\mathrm{out}}_G(i)=1$, then $i$ is a sink of $H$ but not a sink of $G$, a contradiction with (\ref{eq:sink}). Hence $d^{\mathrm{out}}_G(i)\geq 2$ and thus $|X_i|=3\leq d^{\mathrm{out}}_G(i)+1$. Thus every FDS $f:X\to X$ on $G$ is degree-bounded.

\smallskip
For $1\leq p\leq \lambda(H)$, let $J_p$ be the set of $i\in V$ such that $d_H(I,i)+1=p$. In this way, $J_1=I$, and for all $i\in J_p$ with $p\geq 2$, $G$ has at least on arc from $J_{p-1}$ to $i$. Furthermore, since $\lambda(H)=\max_{i\in V}d_H(I,i)+1$, every vertex $i\in V$ is contained in some $J_p$.

\smallskip
Let $\xi\in X$ be inductively defined as follows:
\begin{itemize}
\item
For all $i\in J_1$:
\[
\xi_i=
\left\{
\begin{array}{ll}
1&\text{if $G^-_i\setminus G^+_i\neq\emptyset$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]
\item
For all $i\in J_p$ with $p\geq 2$:
\[
\xi_i=
\left\{
\begin{array}{ll}
1&\text{if $\xi_j=1$ for all $j\in G^+_i\cap J_{p-1}$}\\
&\quad\text{and $\xi_j=0$ for all $j\in (G^-_i\setminus G^+_i)\cap J_{p-1}$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]
\end{itemize}

\smallskip
Finally, let $f:X\to X$ be defined as follows:
\begin{itemize}
\item
For all $i\in J_1$:
\[
f_i(x)=
\left\{
\begin{array}{ll}
1&\text{if $x_j\geq 2$ for some $j\in G^+_i\setminus G^-_i$}\\
 &\quad\text{or $x_j=2$ for some $j\in G^+_i\cap G^-_i$}\\
 &\quad\quad\text{or $x_j<2$ for some $j\in G^-_i\setminus G^+_i$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]
\item
For all $i\in V\setminus J_1$ with $\xi_i=1$:
\[
f_i(x)=
\left\{
\begin{array}{ll}
1&\text{if $x_j\geq 1$ for some $j\in G^+_i\setminus G^-_i$}\\
&\quad\text{or $x_j=1$ for some $j\in G^+_i\cap G^-_i$}\\
&\quad\quad\text{or $x_j<1$ for some $j\in G^-_i\setminus G^+_i$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]
\item
For all $i\in V\setminus J_1$ with $\xi_i=0$:
\[
f_i(x)=
\left\{
\begin{array}{ll}
1&\text{if $x_j\geq 1$ for all $j\in G^+_i\setminus G^-_i$}\\
&\quad\text{and $x_j=1$ for all $j\in G^+_i\cap G^-_i$}\\
&\quad\quad\text{and $x_j<1$ for all $j\in G^-_i\setminus G^+_i$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]
\end{itemize}

\smallskip
It is straightforward to prove that $G$ is the interaction graph of $f$. Let $\beta=\beta(G)$. Since $\lambda(H)\leq\lambda(G)$, and according to the definition of $\xi$, to conclude the proof it is sufficient to prove that $f^{\lambda(H)+\beta}={\mathrm{cst}}=\xi$. For that we prove the following:
\[\tag{$**$}\label{eq:induction}
\forall 1\leq p\leq\lambda(H),~\forall i\in J_p,~\forall \ell\geq p,\qquad
f^{\ell+\beta}_i={\mathrm{cst}}=\xi_i
\]
We proceed by induction on $p$.
\begin{itemize}
\item
Let $i\in J_1$. If $\beta=0$ then $i$ is a source of $G$, thus $G^-_i\setminus G^+_i=\emptyset$ and we deduce that $f_i={\mathrm{cst}}=0=\xi_i$, thus $f^\ell_i={\mathrm{cst}}=\xi_i$ for all $\ell\geq 1$. Suppose now that $\beta=1$. Let $x\in X$ and $\ell\geq 1$. Since $f(X)\subseteq\{0,1\}^n$ we have $f^\ell(x)\in\{0,1\}^n$. If $\xi_i=1$ there exists $j\in G^-_i\setminus G^+_i$ and since $f^\ell_j(x)<2$ we have $f_i(f^\ell(x))=1=\xi_i$. If $\xi_i=0$ then $G^-_i\setminus G^+_i=\emptyset$, and since $f^\ell_j(x)<2$ for all $j\in G^+_i$, we deduce that $f_i(f^\ell(x))=0=\xi_i$. Thus $f_i^{\ell+1}={\mathrm{cst}}=\xi_i$ in all cases.
\item
Suppose that $i\in J_p$ with $p\geq 2$. Let $x\in X$ and $\ell\geq p$. If $\xi_i=1$ then by the definition of $\xi$ and the induction hypothesis, we have $f^{\ell-1+\beta}_j(x)=\xi_j=1$ for all $j\in G^+_i\cap J_{p-1}$ and $f^{\ell-1+\beta}_j(x)=\xi_j=0$ for all $j\in (G^-_i\setminus G^+_i)\cap J_{p-1}$. Since $(G^+_i\cup G^-_i)\cap J_{p-1}$ is not empty, we deduce from the definition of $f_i$ that $f_i(f^{\ell-1+\beta}(x))=1=\xi_i$. If $\xi_i=0$ then by the definition of $\xi$ and the induction hypothesis, we have $f^{\ell'}_j(x)=\xi_j=0$ for some $j\in G^+_i\cap J_{p-1}$ or $f^{\ell'}_j(x)=\xi_j=1$ for some $j\in (G^+_i\setminus G^-_i)\cap J_{p-1}$, and we deduce from the definition of $f_i$ that $f_i(f^{\ell-1+\beta}(x))=0=\xi_i$. Thus $f_i^{\ell+\beta}={\mathrm{cst}}=\xi_i$ in all cases.
\end{itemize}
This proves (\ref{eq:induction}) and completes the proof of the first case. See Example~\ref{ex1} for an illustration.

\smallskip
{\bf\boldmath Case 2: $|G|$ is a cycle.\unboldmath} Then $\lambda(G)=n$ and $\beta(G)=1$, and since $G$ has no source, we have to prove that there exists a degree-bounded $f$ system on $G$ such that $f^{n+1}={\mathrm{cst}}$. Without loss of generality, suppose that $|G|$ has an arc from $i$ to $i+1$ for all $1\leq i<n$ and an arc from $n$ to $1$. For convenience, identify $n+1$ with $1$ and $0$ with $n$. Let $I$ be the set of vertices $i$ such that $G$ has parallel arcs from $i$ to $i+1$. Since $G$ is not a signed cycle and since $|G|$ is a cycle (as in Figure~\ref{fig:pseudo-cycle}), $I$ is not empty, and without loss of generality we assume that $n\in I$.

\smallskip
Let $X=X_1\times\dots\times X_n$ be defined by
\[
X_i=
\begin{cases}
\{0,1,2\}&\text{if }i\in I\\
\{0,1\}&\text{otherwise.}
\end{cases}
\]
Let $f:X\to X$ be defined by
\[
f_i(x)=
\left\{
\begin{array}{ll}
\max(X_i)&\text{if $x_{i-1}=1$ and $i-1\in G^+_i$}\\
&\quad\text{or $x_{i-1}=0$ and $i-1\in G^-_i\setminus G^+_i$}\\[1mm]
0&\text{otherwise.}
\end{array}
\right.
\]

\smallskip
It is straightforward to show that $f$ is a degree-bounded system on $G$. Let us prove that $f^{n+1}={\mathrm{cst}}$. Since $f_i$ only depends on $x_{i-1}$, we can abusively write $f_i(x)=f_i(x_{i-1})$; this allows us to define $\xi\in X$ by induction in the following way: $\xi_1=0$ and $\xi_i=f_i(\xi_{i-1})$ for all $2\leq i\leq n$. Let us prove, by induction on $i$, that $f^{\ell+1}_i(x)=\xi_i$ for $x\in X$ and $\ell\geq i$. Let $x\in X$. Since $n\in I$, we have $n\in G^+_1\cap G^-_1$ and $f^\ell_n(x)\in\{0,2\}$ for all $\ell\geq 1$, thus $f_1(f^\ell(x))=0=\xi_1$ and the base case is proved. Let $1<i\leq n$ and $\ell\geq i$. By induction hypothesis $f^\ell_{i-1}(x)=\xi_{i-1}$ thus $f_i(f^\ell(x))=f_i(\xi_{i-1})=\xi_i$ and the induction step is proved. Thus $f^{n+1}={\mathrm{cst}}=\xi$.
\end{proof}

\begin{remark}
{\em The bound $\lambda(G)+\beta(G)$ can be reached. For instance, if $G$ satisfies one of the following conditions, then it is straightforward to prove that there is no degree-bounded system $f$ on $G$ such that $f^{\lambda(G)+\beta(G)-1}$ is constant:
\begin{itemize}
\item
$G$ is acyclic,
\item
$G$ is obtained from an acyclic signed digraph by adding a loop on each source,
\item
$G$ consists of two cycles that intersect in a path.
\end{itemize}
It could be interesting to characterize the signed digraphs $G$ for which the bound is reached.}
\end{remark}

\begin{remark}\label{rm:component}
{\em Let $G$ be a signed digraph, and let $G_1,\dots,G_k$ be its connected components. It is easy to see that $\lambda(G)=\max_\ell\lambda(G_\ell)$, and that $\beta(G)=\max_\ell \beta(G_\ell)$. Hence, if $G_\ell$ is not a signed cycle for all $\ell$, by applying the previous theorem on each connected component, we deduce that there exists a degree-bounded system $f:X\to X$ on $G$ such that $f^{\lambda(G)+\beta(G)}={\mathrm{cst}}=\xi$.}
\end{remark}

\begin{example}\label{ex1}
{\em Let $G$ be as follows:
\[
\begin{tikzpicture}
\node[outer sep=1,inner sep=2,circle,draw,thick] (1) at (0,0){\scriptsize$1$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (2) at (2,0){\scriptsize$2$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (3) at (4,0){\scriptsize$3$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (4) at (6,0){\scriptsize$4$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (5) at (8,0){\scriptsize$5$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (6) at (10,0){\scriptsize$6$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (7) at (4,-2){\scriptsize$7$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (8) at (8,-2){\scriptsize$8$};
\path[->,thick]
(1) edge[red,bend left=20] (2)
(2) edge[red,bend left=15] (1)
(2) edge[Green,bend left=45] (1)
(3) edge[Green,bend right=40] (1)
(2) edge[red] (3)
(4) edge[Green,bend right=20] (5)
(5) edge[red,bend right=20] (4)
(3) edge[Green] (7)
(4) edge[red] (7)
(4) edge[Green] (8)
(5) edge[red] (8)
(6) edge[red] (8)
(7) edge[Green, bend right=20] (8)
(7) edge[red, bend right=40] (8)
(8) edge[red, bend right=20] (7)
;
\end{tikzpicture}
\]
The initial strong components are $I_1=\{1,2,3\}$, $I_2=\{4,5\}$ and $I_3=\{6\}$. We have $\lambda(G)=3$, and since $I_1$ and $I_2$ are not trivial, we have $\beta(G)=1$. Then, for the construction described in the proof, we must take $i_1=1$ and $i_3=6$, while $i_2$ should be $4$ or $5$. Let us take $i_2=4$. Then $I=\{1,4,6\}$ and $X$ is as follows:
\[
\begin{array}{l}
X_1=\{0,1\}\\
X_2=\{0,1,2,3\}\\
\end{array}
\qquad
\begin{array}{l}
X_3=\{0,1,2\}\\
X_4=\{0,1\}
\end{array}
\qquad
\begin{array}{l}
X_5=\{0,1,2\}\\
X_6=\{0,1\}\\
\end{array}
\qquad
\begin{array}{l}
X_7=\{0,1,2\}\\
X_8=\{0,1\}\\
\end{array}
\]
Besides, $H$ is obtained by removing the parallel arcs from $2$ to $1$, the arc from $3$ to $1$, and the arc from $5$ to $4$. Thus $\lambda(H)=3$ with $J_1=I=\{1,4,6\}$, $J_2=\{2,5,7,8\}$, and $J_3=\{3\}$. We deduce that
\[
\begin{array}{l}
\xi_1=0\\
\xi_4=1\\
\xi_6=0\\
\end{array}
\qquad
\begin{array}{l}
\xi_2=1\\
\xi_5=1\\
\xi_7=0\\
\xi_8=1
\end{array}
\qquad
\begin{array}{l}
\xi_3=0.
\end{array}
\]
and thus $f$ is defined by
\[
\begin{array}{l}
f_1(x)=
\begin{cases}
1\text{ if $x_2=2$}\\
\text{~~~ or $x_3\geq 2$,}\\
0\text{ otherwise}
\end{cases}
\\[8mm]
f_4(x)=
\begin{cases}
1\text{ if $x_5<2$,}\\
0\text{ otherwise}
\end{cases}
\\[6mm]
f_6(x)=0
\end{array}
~
\begin{array}{l}
f_2(x)=
\begin{cases}
1\text{ if $x_1<1$}\\
0\text{ otherwise}
\end{cases}
\\[5mm]
f_5(x)=
\begin{cases}
1\text{ if $x_4\geq 1$,}\\
0\text{ otherwise}
\end{cases}
\\[5mm]
f_7(x)=
\begin{cases}
1\text{ if $x_3\geq 1$}\\
\text{~~~ and $x_4<1$,}\\
0\text{ otherwise}
\end{cases}
\\[8mm]
f_8(x)=
\begin{cases}
1\text{ if $x_4\geq 1$}\\
\text{~~~ or $x_5<1$}\\
\text{~~~~~~ or $x_6<1$}\\
\text{~~~~~~~~~ or $x_7=1$,}\\
0\text{ otherwise}
\end{cases}
\end{array}
\!\!
\begin{array}{l}
f_3(x)=
\begin{cases}
1\text{ if $x_2<1$,}\\
0\text{ otherwise.}
\end{cases}
\end{array}
\]
It is easy to check that the interaction graph of $f$ is $G$. Furthermore, if $x\in\{0,1\}^n$, then it is easy to see that, for all $\ell\geq 1$,
\[
\begin{array}{l}
f^\ell_1(x)=0=\xi_1\\
f^\ell_4(x)=1=\xi_2\\
f^\ell_6(x)=0=\xi_6\\
\end{array}
\qquad
\begin{array}{l}
f^{\ell+1}_2(x)=1=\xi_2\\
f^{\ell+1}_3(x)=1=\xi_3\\
f^{\ell+1}_7(x)=0=\xi_7\\
f^{\ell+1}_8(x)=1=\xi_8\\
\end{array}
\qquad
\begin{array}{l}
f^{\ell+2}_3(x)=0=\xi_3.
\end{array}
\]
Since $f(x)\in\{0,1\}^n$ for all $x\in X$, we deduce that $f^4(x)=\xi$ for all $x\in X$, as desired.}
\end{example}

\begin{example}
{\em Suppose that $G$ is the $3$-vertex signed digraph as in Figure~\ref{fig:pseudo-cycle}:
\[
\begin{tikzpicture}
\node[outer sep=1,inner sep=2,circle,draw,thick] (1) at (150:1){\scriptsize$1$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (2) at (30:1){\scriptsize$2$};
\node[outer sep=1,inner sep=2,circle,draw,thick] (3) at (270:1){\scriptsize$3$};
\path[->,thick]
(3) edge[Green,bend left=25] (1)
(3) edge[red,bend right=25] (1)
(1) edge[Green] (2)
(2) edge[red] (3)
;
\end{tikzpicture}
\]
Then $|G|$ is a cycle, and we fall in the second case of the proof. We then have $I=\{3\}$, thus $X_1=X_2=\{0,1\}$ and $X_3=\{0,1,2\}$. Therefore, $f$ is defined by
\[
f_1(x)=
\begin{cases}
1\text{ if $x_3=1$}\\
0\text{ otherwise}
\end{cases}
\quad
f_2(x)=
\begin{cases}
1\text{ if $x_1=1$}\\
0\text{ otherwise}
\end{cases}
f_3(x)=
\begin{cases}
2\text{ if $x_2=0$}\\
0\text{ otherwise.}
\end{cases}
\]
It is easy to check that the interaction graph of $f$ is $G$. Furthermore, $f_3(x)\neq 1$ for all $x\in X$. Thus, for all $\ell\geq 2$, we have $f^\ell_1(x)=0$ which implies $f^{\ell+1}_2(x)=0$, which implies $f^{\ell+2}_3(x)=2$. Thus $f^4(x)=(0,0,2)$ for all $x\in X$, as desired.}
\end{example}

\begin{example}
{\em Suppose that $G$ is the $1$-vertex signed digraph as follows:
\[
\begin{tikzpicture}
\node[outer sep=1,inner sep=2,circle,draw,thick] (1) at (0,0){\scriptsize$1$};
\draw[Green,->,thick] (1.120) .. controls (-1,1) and (-1,-1) .. (1.-120);
\draw[red,->,thick] (1.60) .. controls (1,1) and (1,-1) .. (1.-60);
\end{tikzpicture}
\]
Then $|G|$ is a cycle, and we fall in the second case of the proof. We then have $I=\{1\}$, thus $X=X_1=\{0,1,2\}$, and $f=f_1$ is defined by $f(0)=0$, $f(1)=2$ and $f(2)=0$. Thus $f^2(x)=0$ for all $x\in X$, as desired.}
\end{example}

\section{Convergence between systems}\label{sec:convergence}

In this section we prove the following strengthening of Theorem~\ref{thm:main}, which is more suited for a proof by induction. A {\em spanning} subgraph is a subgraph obtained by removing arcs only.

\begin{theorem}\label{thm:main2}
Let $G$ be signed digraph and let $H$ be a spanning subgraph of $G$. Let $I$ be the set of vertices that are isolated in $H$ but not in $G$. Suppose that every source of $H\setminus I$ is a source of $G$, and that every sink of $H\setminus I$ is a sink of $G$. Suppose also that no connected component of $G$ is a signed cycle with only vertices in $I$. Then, for every degree-bounded system $h$ on $H$ there exists a degree-bounded system $f$ on $G$ that converges toward $h$ in at most $|I|+1$ steps.
\end{theorem}

The proof of Theorem~\ref{thm:main2} uses Theorem~\ref{thm:nilpotent2} and the following technical lemma.

\begin{lemma}\label{lem:extension}
Let $G$ be a signed digraph and let $H$ be a spanning subgraph of $G$ such that
\begin{itemize}
\item
$G$ has no arc from a sink of $H$ to a source of~$H$, and
\item
$G$ has no arc from a vertex of $H$ to an isolated vertex of $H$.
\end{itemize}
Let $A$ be the set of sources of $H$ that are not sources of $G$, and let $B$ be the set sinks of $H$ that are not sinks of $G$. Let $h:Y\to Y$ be a degree-bounded system on $H$ and $\xi\in Y$. There exists a degree-bounded system $f:X\to X$ on $G$ such that
\begin{itemize}
\item
$f(X)\subseteq Y$,
\item
$f_i(X)\subseteq h_i(Y)$ for all $i\not\in A$,
\item
$f(x)=h(x)$ for all $x\in Y$ such that $x_i=\xi_i$ for all $i\in B$, and
\item
$f_i(x)=h_i(x)$ for all $x\in Y$ and $i$ such that $G_i\cap B=\emptyset$.
\end{itemize}
\end{lemma}

\begin{proof}[{\bf Poof of Theorem~\ref{thm:main2}, assuming Lemma~\ref{lem:extension}}]
Without loss of generality, suppose that $V=\{1,\dots,n\}$ is the vertex set of $G$ and  that $I=\{1,\dots,m\}$ for some $1\leq m\leq n$. Let $h:Y\to Y$ be a degree-bounded system on $H$. Since $h$ is degree-bounded, for all $i\in I$ there exists $\xi_i\in\mathbb{Z}$ such that $Y_i=\{\xi_i\}$.

\smallskip
Suppose first that $I$ is empty. Then $G$ and $H$ have the same isolated vertices and thus the same sources and sinks. Hence, the conditions of Lemma \ref{lem:extension} are satisfied, and the sets $A$ and $B$ of the statement are both empty. Thus there exists a degree-bounded system $f$ on $G$ that converges toward $h$ in one step, and thus $f$ has the desired properties. Hence, in the following, we assume that $I$ is not empty.

\smallskip
We say that $I$ has the \BF{property $P$} if every connected component $F$ of $G[I]$ satisfies at least one of the following conditions:
\begin{itemize}
\item $F$ is not strongly connected.
\item $F$ has a {\em leaving arc}, that is, $G$ has an arc from a vertex of $F$ to $V\setminus I$.
\item $F$ has no {\em entering arc}, that is, $G$ has no arc from $V\setminus I$ to a vertex of $F$.
\end{itemize}

\begin{figure}
\[
\begin{tikzpicture}
\draw[dotted,rounded corners=5] (-0.5,-0.5) rectangle (4.5,2.5);
\node at (-1.3,1) {$G$};
\node[N] (1) at (1,2){\scriptsize$6$};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\node[N] (5) at (1,0){\scriptsize$4$};
\node[N] (6) at (0,1){\scriptsize$5$};
\path[->,thick]
(1) edge[red] (5)
(5) edge[red] (6)
(5) edge[Green,bend left=30] (6)
(6) edge[Green,bend left=10] (1)
(1) edge[Green, bend left=10] (6)
(4) edge[red,bend left=10] (2)
(2) edge[red] (3)
(2) edge[red,bend left=10] (4)
(3) edge[Green,bend left=10] (4)
(4) edge[Green,bend left=10] (3)
(2) edge[Green] (1)
(4) edge[Green] (1)
(5) edge[red](2)
(5) edge[Green](4)
;
\begin{scope}[shift={(0,-4)}]
\draw[dotted,rounded corners=5] (-0.5,-0.5) rectangle (4.5,2.5);
\node at (-1.3,1) {$H$};

\draw[dotted,rounded corners=5] ({3-0.4},{0-0.4}) rectangle ({4+0.4},{2+0.4});
\node[preaction={fill,white}] at ({3-0.4},1) {\small $I$};
\node at (6.3,1) {$h:Y\to Y$};
\node[N] (1) at (1,2){\scriptsize$6$};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\node[N] (5) at (1,0){\scriptsize$4$};
\node[N] (6) at (0,1){\scriptsize$5$};
\path[->,thick]
(1) edge[red] (5)
(5) edge[red] (6)
(6) edge[Green,bend left=10] (1)
;
\draw[rounded corners=5,->]
(7.8,{1-0.5}) -- (8.6,{1-1.5}) -- (8.6,{-7.1+1.5}) -- (7.8,{-7.1+0.5})
;
\end{scope}
\begin{scope}[shift={(0,-8)}]
\draw[dotted,rounded corners=5] (2.5,-0.5) rectangle (4.5,2.5);
\node at (1.7,1) {$Q$};
\node at (6.3,1) {$g:Z\to Z$};
\node at (6.3,0.25) {[~Theorem~\ref{thm:nilpotent2}~]};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\path[thick,->]
(4) edge[red,bend left=10] (2)
(3) edge[Green,bend left=11] (4)
(4) edge[Green,bend left=11] (3)
;
\draw[rounded corners=5,-]
(7.8,{1-0.5}) -- (8.6,{1-1.5}) -- (8.6,-1)
;\end{scope}
\begin{scope}[shift={(0,-12)}]
\draw[dotted,rounded corners=5] (-0.5,-0.5) rectangle (4.5,2.5);
\node at (-1.3,1) {$\tilde H$};
\node at (6.3,1) {$\tilde h:\tilde Y\to \tilde Y$};
\node[N] (1) at (1,2){\scriptsize$6$};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\node[N] (5) at (1,0){\scriptsize$4$};
\node[N] (6) at (0,1){\scriptsize$5$};
\path[thick,->]
(1) edge[red] (5)
(5) edge[red] (6)
(6) edge[Green,bend left=10] (1)
(4) edge[red,bend left=10] (2)
(3) edge[Green,bend left=11] (4)
(4) edge[Green,bend left=11] (3)
;
\draw[rounded corners=5,->]
(7.8,{1-0.5}) -- (8.6,{1-1.5}) -- (8.6,{-3+1.5}) -- (7.8,{-3+0.5})
;
\node at (10.3,-1) {[~Lemma~\ref{lem:extension}~]};
\end{scope}
\begin{scope}[shift={(0,-16)}]
\draw[dotted,rounded corners=5] (-0.5,-0.5) rectangle (4.5,2.5);
\node at (-1.3,1) {$\tilde G$};
\node at (6.3,1) {$\tilde f:\tilde X\to \tilde X$};
\node[N] (1) at (1,2){\scriptsize$6$};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\node[N] (5) at (1,0){\scriptsize$4$};
\node[N] (6) at (0,1){\scriptsize$5$};
\path[->]
(1) edge[red] (5)
(5) edge[red] (6)
(5) edge[very thick,Green,bend left=30] (6)
(6) edge[Green,bend left=10] (1)
(1) edge[very thick,Green, bend left=10] (6)
(4) edge[red,bend left=10] (2)
(2) edge[very thick,red] (3)
(2) edge[very thick,red,bend left=10] (4)
(3) edge[Green,bend left=10] (4)
(4) edge[Green,bend left=10] (3)
(2) edge[very thick,Green] (1)
(4) edge[very thick,Green] (1)
;
\draw[rounded corners=5,->]
(7.8,{1-0.5}) -- (8.6,{1-1.5}) -- (8.6,{-3+1.5}) -- (7.8,{-3+0.5})
;
\node at (10.3,-1) {[~Lemma~\ref{lem:extension}~]};
\end{scope}
\begin{scope}[shift={(0,-20)}]
\draw[dotted,rounded corners=5] (-0.5,-0.5) rectangle (4.5,2.5);
\node at (-1.3,1) {$G$};
\node at (6.3,1) {$f:X\to X$};
\node[N] (1) at (1,2){\scriptsize$6$};
\node[N] (2) at (3,2){\scriptsize$1$};
\node[N] (3) at (4,1){\scriptsize$2$};
\node[N] (4) at (3,0){\scriptsize$3$};
\node[N] (5) at (1,0){\scriptsize$4$};
\node[N] (6) at (0,1){\scriptsize$5$};
\path[->]
(1) edge[red] (5)
(5) edge[red] (6)
(5) edge[Green,bend left=30] (6)
(6) edge[Green,bend left=10] (1)
(1) edge[Green, bend left=10] (6)
(4) edge[red,bend left=10] (2)
(2) edge[red] (3)
(2) edge[red,bend left=10] (4)
(3) edge[Green,bend left=10] (4)
(4) edge[Green,bend left=10] (3)
(2) edge[Green] (1)
(4) edge[Green] (1)
(5) edge[very thick,red](2)
(5) edge[very thick,Green](4)
;
\end{scope}
\end{tikzpicture}
\]
{\caption{\label{fig:case1}An illustration for the first case in the proof of Theorem~\ref{thm:main2}.}}
\end{figure}

\smallski
\BF{Case 1: $I$ has the property $P$.}  Let $\tilde H$ be the signed digraph obtained from $H$ by adding all the arcs $(i,j,s)$ of $G[I]$ such that $G$ has no arc from $i$ to $V\setminus I$. Let $Q$ be the subgraph of $\tilde H$ induced by $I$ (see Figure~\ref{fig:case1}). Suppose that some connected component of $Q$ is a cycle $C$. Then $G$ has no arc from $C$ to $V\setminus I$, thus $C$ is a connected component of $G[I]$ without leaving arc. Since no connected component of $G$ is a cycle contained in $G[I]$, we deduce that $C$ has at least one entering arc. Thus $C$ is strongly connected, has no leaving arcs and at least one entering arc. This is not possible since $I$ has the property $P$. Thus no connected component of $Q$ is a cycle. Hence, by applying Theorem~\ref{thm:nilpotent2} on each connected component of $Q$, we deduce (cf. Remark~\ref{rm:component}) that there exists a degree-bounded system $g:Z\to Z$ on $Q$ such that
\[
g^{\lambda(Q)+\beta(Q)}={\mathrm{cst}}=(\xi_1,\dots,\xi_m).
\]

\smallskip
Let $\tilde Y=Z_1\times \cdots\times Z_m\times Y_{m+1}\times \cdots\times Y_n$. Since $H$ has no arc from $I$ to $V\setminus I$, if $i\not\in I$ then $h_i$ only depends on variables $x_j$ with $j\not\in I$, and we can define $\tilde h:\tilde Y\to\tilde Y$ as follows, without possible ambiguity:
\[
\tilde h_i(x)=
\begin{cases}
h_i(x_{m+1},\dots,x_n)&\textrm{ if $i\not\in I$}\\
g_i(x_1,\dots,x_m)&\textrm{ if $i\in I$.}
\end{cases}
\]

\smallskip
By construction, $\tilde h$ is a degree-bounded system on $\tilde H$ such that:
\begin{enumerate}
\item[(1)]
$\tilde h_i(\tilde Y)\subseteq h_i(Y)$ for all $i\not\in I$,
\item[(2)]
$\tilde h_i^{\lambda(Q)+\beta(Q)}(\tilde Y)\subseteq \{\xi_i\}=Y_i=h_i(Y)$ for all $i\in I$,
\item[(3)]
$\tilde h(x)=h(x)$ for all $x\in Y$.
\end{enumerate}

\smallskip
Let $\tilde G$ be the signed digraph obtained from $\tilde H$ by adding all the arcs $(i,j,s)$ of $G$ with $j\not\in I$ (see Figure~\ref{fig:case1}). Let $A$ be the set of sources of $\tilde H$ that are not sources of $\tilde G$, and let us prove that $A=\emptyset$. Suppose that $j$ is a source of $\tilde H$. If $j\in I$ then $j$ is obviously a source of $\tilde G$, and if $j\notin I$ then $j$ is a source of $H\setminus I$. Hence, by hypothesis, $j$ is a source of $G$ and thus of $\tilde G$ too. Thus, $A=\emptyset$. This implies that $\tilde G$ has no arc from a sink of $\tilde H$ to a source of $\tilde H$. Furthermore, $\tilde G$ has no arc from a vertex of $\tilde H$ to an isolated vertex $i$ of $\tilde H$ since otherwise $i\not\in I$, thus $i$ is isolated in $H$ but not in $G$, a contradiction. Thus $\tilde G$ and $\tilde H$ satisfy the conditions of Lemma~\ref{lem:extension}. Let $B$ be the set of sinks of $\tilde H$ that are not sinks of $\tilde G$. If $i$ is a sink of $\tilde H$ and $i\not\in I$, then $i$ is a sink of $H\setminus I$ and, by hypothesis, $i$ is sink of $G$ and thus of $\tilde G$ too. Thus $B\subseteq I$. Let $i\in I$ and $j\in B$. Then $G$ has an arc from $j$ to $V\setminus I$, hence, by definition, $Q$ has no arc from $j$ to $i$ and we deduce that $\tilde G$ has no arc from $j$ to $i$. Thus $\tilde G_i\cap B=\emptyset$ for every $i\in I$. We then deduce from Lemma~\ref{lem:extension}, that there exists a degree-bounded system $\tilde f:\tilde X\to\tilde X$ on $\tilde G$ such that
\begin{enumerate}
\item[(4)]
$\tilde f(\tilde X)\subseteq \tilde h(\tilde Y)$,
\item[(5)]
$\tilde f(x)=\tilde h(x)$ for all $x\in\tilde Y$ such that $x_i=\xi_i$ for all $i\in I$, that is, for all $x\in Y$,
\item[(6)]
$\tilde f_i(x)=\tilde h_i(x)$ for all $x\in\tilde Y$ and $i\in I$.
\end{enumerate}

\smallskip
As a consequence, $\tilde f$ has the following four properties:
\begin{enumerate}
\item[(7)]
$\tilde f_i(\tilde X)\subseteq h_i(Y)$ for all $i\not\in I$,
\item[(8)]
$\tilde f^\ell_i(\tilde X)\subseteq \tilde h^\ell_i(\tilde Y)$ for all $i\in I$ and $\ell\geq 1$,
\item[(9)]
$\tilde f^{\lambda(Q)+\beta(Q)}_i(\tilde X)\subseteq h_i(Y)$ for all $i\in I$,
\item[(10)]
$\tilde f(x)=h(x)$ for all $x\in Y$.
\end{enumerate}
Indeed, property (7) results from (1) and (4), and property (10) results from (5) and (3). We now prove (8) by induction on $\ell$. The case $\ell=1$ is given by (4), and if $\ell\geq 2$ and $i\in I$ then
\[
\tilde f^\ell_i(\tilde X)=\tilde f_i(\tilde f^{\ell-1}(\tilde X))\subseteq \tilde f_i(\tilde h^{\ell-1}(\tilde Y))=\tilde h^\ell_i(\tilde Y),
\]
where the inclusion results from the induction hypothesis and the fact that $\tilde f_i$ only depends on variables with indices in $I$ (because $\tilde G$ has no arc from $V\setminus I$ to $I$), and where the last equality results from (6). Finally, (9) results from (8) and (2).

\smallskip
Let $A$ be the set of sources of $\tilde G$ that are not sources of $G$, and let $B$ be the set of sinks of $\tilde G$ that are not sinks of $G$. Since $G$ is obtained from $\tilde G$ by adding only arcs $(i,j,s)$ with $j\in I$ (see Figure~\ref{fig:case1}), we have $A\subseteq I$. Let us prove that $B$ is empty. Let $i$ be a sink of $\tilde G$. If $i\not\in I$ then $i$ is a sink of $H\setminus I$ and thus, by hypothesis, $i$ is a sink of $G$. If $i\in I$ then, by the definition of $\tilde G$, $G$ has no arc from $i$ to $V\setminus I$ thus $i$ is a sink of $G$, and since $i$ is a sink of $\tilde H$, by the definition of $\tilde H$, $G$ has no arc from $i$ to $I$. Thus $i$ is a sink of $G$. This proves that $B=\emptyset$. We deduce that $G$ has no arc from a sink of $\tilde G$ to a source of $\tilde G$. Now, suppose, for a contradiction, that $G$ has an arc from a vertex of $\tilde G$ to an isolated vertex $i$ of $\tilde G$. If $i\not\in I$ then $i$ is isolated in $H\setminus I$ but not in $G$, a contradiction, thus $i\in I$. Hence, $i$ is a trivial connected component of $G[I]$, which is strongly connected, which has no leaving arcs, and which has at least one entering arc, and this contradicts the fact that $I$ has the property $P$. Thus $G$ has no arc from a vertex of $\tilde G$ to an isolated vertex of $\tilde G$. Thus $G$ and $\tilde G$ satisfy the conditions of Lemma~\ref{lem:extension}, with $A\subseteq I$ and $B=\emptyset$. Therefore, there exists a degree-bounded system $f:X\to X$ on $G$ such that
\begin{enumerate}
\item[(11)]
$f(X)\subseteq \tilde X$,
\item[(12)]
$f_i(X)\subseteq \tilde f_i(\tilde X)$ for all $i\not \in A$,
\item[(13)]
$f(x)=\tilde f(x)$ for all $x\in\tilde X$.
\end{enumerate}

\smallskip
As a consequence, $f$ has the following properties:
\begin{itemize}
\item[(14)]
$f_i(X)\subseteq h_i(Y)$ for all $i\not\in I$,
\item[(15)]
$f^{|I|+1}_i(X)\subseteq h_i(Y)$ for all $i\in I$,
\item[(16)]
$f(x)=h(x)$ for all $x\in Y$.
\end{itemize}
Indeed, (14) results from (12), (7) and the fact that $A\subseteq I$. Also, (16) results from (13) and (10). It remains to prove (15). Suppose first that $A=\emptyset$. Then following (12) we have  $f(X)\subseteq \tilde f(\tilde X)$ and we deduce from (13) that $f^\ell(X)\subseteq \tilde f^\ell(\tilde X)$ for all $\ell\geq 1$. Thus by (9) we have
\[
f^{\lambda(Q)+\beta(Q)}_i(X)\subseteq h_i(Y).
\]
Since $\lambda(Q)\leq|I|$ and $\beta(Q)\leq 1$ we obtain (15). Suppose now that $A\neq\emptyset$. Then following (11) and (13) we have $f^{\ell+1}(X)\subseteq \tilde f^\ell(\tilde X)$ for all $\ell\geq 0$. Thus by (9) we have
\[
f^{\lambda(Q)+\beta(Q)+1}_i(X)\subseteq h_i(Y).
\]
If $Q$ is basic then $\beta(Q)=0$ and (15) follows. Otherwise, $Q$ is not basic, and since $A\neq\emptyset$ and $A\subseteq I$, $Q$ has a source. Hence, $Q$ has at least two distinct initial strong components. Thus, as explained in Remark~\ref{rem:lambda}, we have $\lambda(Q)<|I|$ and (15) follows.

\smallskip
We deduce from (14), (15) and (16) that $f$ converges toward $h$ in at most $|I|+1$ steps.

\begin{figure}
\[
\begin{tikzpicture}
\draw[dotted,rounded corners=5] (-0.75,-0.75) rectangle (3.75,2.25);
\node at (-1.55,0.75) {$G$};
\node[N] (4) at (0.0,0.0){\scriptsize$4$};
\node[N] (5) at (1.5,0.0){\scriptsize$5$};
\node[N] (6) at (3.0,0.0){\scriptsize$6$};
\node[N] (1) at (0.0,1.5){\scriptsize$1$};
\node[N] (2) at (1.5,1.5){\scriptsize$2$};
\node[N] (3) at (3.0,1.5){\scriptsize$3$};
\path[->,thick]
(1) edge[Green,bend left=20] (4)
(4) edge[Green,bend left=20] (1)
(2) edge [red] (1)
(4) edge [Green] (5)
(2) edge[Green,bend left=20] (3)
(3) edge[Green,bend left=20] (2)
(5) edge[Green,bend left=20] (6)
(6) edge[red,bend left=20] (5)
;
\begin{scope}[shift={(0,-4)}]
\node at (-1.55,0.75) {$H$};
\node at (5.5,0.75) {$h:Y\to Y$};
\draw[dotted,rounded corners=5] (-0.75,-0.75) rectangle (3.75,2.25);
\draw[dotted,rounded corners=5] (1.05,1.05) rectangle (3.45,1.95);
\node[preaction={fill,white}] at (2.25,1.05) {\scriptsize $\tilde I$};
\draw[dotted,rounded corners=5] (1.05,-0.45) rectangle (3.45,0.45);
\node[preaction={fill,white}] at (2.25,0.45) {\scriptsize $J$};
\draw[dotted,rounded corners=5] (0.9,-0.6) rectangle (3.6,2.1);
\node[preaction={fill,white}] at (0.9,0.75) {\small $I$};
\node[N] (4) at (0.0,0.0){\scriptsize$4$};
\node[N] (5) at (1.5,0.0){\scriptsize$5$};
\node[N] (6) at (3.0,0.0){\scriptsize$6$};
\node[N] (1) at (0.0,1.5){\scriptsize$1$};
\node[N] (2) at (1.5,1.5){\scriptsize$2$};
\node[N] (3) at (3.0,1.5){\scriptsize$3$};
\path[->,thick]
(1) edge[Green,bend left=20] (4)
(4) edge[Green,bend left=20] (1)
;
\draw[rounded corners=5,->]
(7,{0.75-0.5}) -- (7.8,{0.75-1.5}) -- (7.8,{-3.3+1.5}) -- (7,{-3.3+0.5})
;
\node at (10.3,{-2+0.75}) {[~first case of Theorem~\ref{thm:main2}~]};
\end{scope}
\begin{scope}[shift={(0,-8)}]
\node at (-1.55,0.75) {$\tilde G$};
\node at (5.5,0.75) {$\tilde f:\tilde X\to\tilde X$};
\draw[dotted,rounded corners=5] (-0.75,-0.75) rectangle (3.75,2.25);
\node[N] (4) at (0.0,0.0){\scriptsize$4$};
\node[N] (5) at (1.5,0.0){\scriptsize$5$};
\node[N] (6) at (3.0,0.0){\scriptsize$6$};
\node[N] (1) at (0.0,1.5){\scriptsize$1$};
\node[N] (2) at (1.5,1.5){\scriptsize$2$};
\node[N] (3) at (3.0,1.5){\scriptsize$3$};
\path[->,thick]
(1) edge[Green,bend left=20] (4)
(4) edge[Green,bend left=20] (1)
(2) edge [red] (1)
(2) edge[Green,bend left=20] (3)
(3) edge[Green,bend left=20] (2)
;
\draw[rounded corners=5,->]
(7,{0.75-0.5}) -- (7.8,{0.75-1.5}) -- (7.8,{-7.3+1.5}) -- (7,{-7.3+0.5})
;
\end{scope}
\begin{scope}[shift={(0,-12)}]
\draw[dotted,rounded corners=5] (-0.75,-0.75) rectangle (3.75,2.25);
\node at (-1.55,0.75) {$Q$};
\node at (5.5,0.75) {$g:Z\to Z$};
\node at (5.5,0) {[~Theorem~\ref{thm:nilpotent2}~]};
\node[N] (4) at (0.0,0.0){\scriptsize$4$};
\node[N] (5) at (1.5,0.0){\scriptsize$5$};
\node[N] (6) at (3.0,0.0){\scriptsize$6$};
\node[N] (1) at (0.0,1.5){\scriptsize$1$};
\node[N] (2) at (1.5,1.5){\scriptsize$2$};
\node[N] (3) at (3.0,1.5){\scriptsize$3$};
\path[->,thick]
(4) edge [Green] (5)
(5) edge[Green,bend left=20] (6)
(6) edge[red,bend left=20] (5)
;
\draw[rounded corners=5,-]
(7,{0.75-0.5}) -- (7.8,{0.75-1.5}) -- (7.8,-1)
;
\end{scope}
\begin{scope}[shift={(0,-16)}]
\draw[dotted,rounded corners=5] (-0.75,-0.75) rectangle (3.75,2.25);
\node at (-1.55,0.75) {$G$};
\node at (5.5,0.75) {$f:X\to X$};
\node[N] (4) at (0.0,0.0){\scriptsize$4$};
\node[N] (5) at (1.5,0.0){\scriptsize$5$};
\node[N] (6) at (3.0,0.0){\scriptsize$6$};
\node[N] (1) at (0.0,1.5){\scriptsize$1$};
\node[N] (2) at (1.5,1.5){\scriptsize$2$};
\node[N] (3) at (3.0,1.5){\scriptsize$3$};
\path[->,thick]
(1) edge[Green,bend left=20] (4)
(4) edge[Green,bend left=20] (1)
(2) edge [red] (1)
(4) edge [Green] (5)
(2) edge[Green,bend left=20] (3)
(3) edge[Green,bend left=20] (2)
(5) edge[Green,bend left=20] (6)
(6) edge[red,bend left=20] (5)
;
\end{scope}
\end{tikzpicture}
\]
{\caption{\label{fig:case2}An illustration for the second case in the proof of Theorem~\ref{thm:main2}.}}
\end{figure}

\smallski
\BF{Case 2: $I$ has not the property $P$.}
Let $I_1$ be the set of vertices that belongs to a connected component $F$ of $G[I]$ such that $F$ is not strongly connected, or has a leaving arc, or has no entering arcs. Let $J=I\setminus \tilde I$. Since $I$ has not the property $P$, $J$ is not empty. Let $\tilde G$ be the signed digraph obtained from $G$ by removing every arc $(i,j,s)$ with $j\in J$ (see Figure \ref{fig:case2}). Then, $\tilde I$ is the set of vertices isolated in $H$ but not in $\tilde G$. Furthermore, it is straightforward to check that no connected component of $\tilde G$ is a cycle contained in $\tilde G[\tilde I]$ and that every source (resp. sink) of $\tilde G\setminus\tilde I$ is a source (resp. sink) of $\tilde G$. Furthermore, since $\tilde I$ has the property $P$ with respect to $\tilde G$, according to the first case, there exists a degree-bounded system $\tilde f:\tilde X\to\tilde X$ on $\tilde G$ such that
\begin{itemize}
\item[(1)]
$\tilde f_i(\tilde X)\subseteq h_i(Y)$ for all $i\not\in \tilde I$,
\item[(2)]
$\tilde f^{|\tilde I|+1}_i(\tilde X)\subseteq h_i(Y)$ for all $i\in \tilde I$, and thus $\tilde f^{|I|+1}_i(\tilde X)\subseteq h_i(Y)$ for all $i\in \tilde I$,
\item[(3)]
$\tilde f(x)= h(x)$ for all $x\in Y$.
\end{itemize}

\smallskip
Let $\xi\in\mathbb{Z}^n$ be defined by $\xi_i=\max(\tilde X_i)$ for all $i\in \{1,\dots,n\}$. Let $J=I\setminus\tilde I$ and let $Q$ be the spanning subgraph of $G$ that only contains the arcs $(i,j,s)$ of $G$ with $j\in J$. So $\tilde G$ and $Q$ are arc-disjoint and $G=\tilde G\cup Q$ (see Figure~\ref{fig:case2}). Since every connected components $G[J]$ is strongly connected, has no leaving arcs and has an entering arc, we deduce that every connected component of $Q$ is basic. Hence, by applying Theorem~\ref{thm:nilpotent2} on each connected component of $H$ (cf. Remark~\ref{rm:component}) we deduce that there exists a degree-bounded system $g:Z\to Z$ on $Q$ such that
\[
g^{\lambda(Q)}={\mathrm{cst}}=\xi.
\]
and such that $\xi_i=\min(Z_i)$ for all the sources $i$ of $Q$. Since if $i\not\in J$ then $i$ is a source of $Q$ we have $\max(\tilde X_i)=\xi_i=\min(Z_i)$ for all $i\not\in J$. Remark also that $Y_i=\tilde X_i=\{\xi_i\}\subseteq Z_i$ for all $i\in J$.

\smallskip
Let $X$ be defined by:
\[
X=X_1\times\dots\times X_n
\quad\text{with}\quad
X_i=
\begin{cases}
\tilde X_i\cup Z_i&\text{if }i\not\in J\\
Z_i&\text{if }i\in J.
\end{cases}
\]
Let $\pi:X\to Z$ and $\tilde \pi:X\to \tilde X$ be the surjections defined by:
\[
\pi_i(x)=
\begin{cases}
\max(x_i,\xi_i)&\text{if }i\not\in J\\
x_i&\text{if }i\in J.
\end{cases}
\quad\text{and}\quad
\tilde \pi_i(x)=
\begin{cases}
\min(x_i,\xi_i)&\text{if }i\not\in J\\
\xi_i&\text{if }i\in J
\end{cases}
\]
Finally, let $f:X\to X$ be defined by:
\[
f_i(x)=
\begin{cases}
\tilde f_i(\tilde \pi(x))&\textrm{if }i\notin J\\
g_i(\pi(x))&\textrm{if }i\in J.
\end{cases}
\]
Since $\tilde f$ is a degree-bounded system on $\tilde G$ and $g$ is a degree-bounded system on $Q$, it is easy to check that $f$ is a degree-bounded system on $G=\tilde G\cup Q$. $f$ has also the following properties:
\begin{itemize}
\item[(4)]
$f^\ell_i(X)\subseteq\tilde f^\ell_i(\tilde X)$ for all $i\not\in J$ and $\ell\geq 1$.
\item[(5)]
$f^\ell_i(x)=g^\ell_i(\pi(x))$ for all $i\in J$, $x\in X$ and $\ell\geq 1$, so that
$f^{\lambda(Q)}_i={\mathrm{cst}}=\xi_i$ for all $i\in J$,
\item[(6)]
$f(x)=\tilde f(x)$ for all $x\in\tilde X$.
\end{itemize}
We first prove (5) by induction on $\ell$. The case $\ell=1$ results from the definition of $f_i$ for $i\in J$. If $\ell\geq 2$ then, by definition, we have
\[\tag{$*$}\label{eq:h}
f^{\ell}_i(x)=f_i(f^{\ell-1}(x))=g_i(\pi(f^{\ell-1}(x))).
\]
By induction hypothesis, for all $j\in J$ we have $f^{\ell-1}_j(x)=g^{\ell-1}_j(\pi(x))\geq\xi_j$ thus $\pi_j(f^{\ell-1}(x))=g^{\ell-1}_j(\pi(x))$. If $j\not\in J$ then $f^{\ell-1}_j(x)\leq\xi_j$ thus $\pi_j(f^{\ell-1}(x))=\xi_j$. Since $j$ is a source of $Q$ and $\xi$ is a fixed point of~$g$, we have $g_j={\mathrm{cst}}=\xi_j$. Thus, $\pi_j(f^{\ell-1}(x))=g^{\ell-1}_j(\pi(x))$. Consequently, $\pi(f^{\ell-1}(x))=g^{\ell-1}(\pi(x))$ and we deduce from (\ref{eq:h}) that $f^\ell_i(x)=g^\ell_i(\pi(x))$, completing the induction step. We now prove (6). Let $x\in\tilde X$. Then $\tilde \pi(x)=x$ thus $f_i(x)=\tilde f_i(x)$ for all $i\notin J$. Furthermore, $\pi(x)=\xi$ and if $i\in I$ then $\tilde X_i=\{\xi_i\}=\{x_i\}$. Since $\xi$ is a fixed point of $g$ we deduce that $f_i(x)=g_i(\xi)=\xi_i=\tilde f_i(x)$. Hence $f(x)=\tilde f(x)$ and (6) is proved. We finally prove (4) by induction on $\ell$. The case $\ell=1$ is an immediate consequence of the definition of $f_i$ for $i\not\in J$. If $\ell\geq 2$ and $i\not\in J$ then
\[
f^\ell_i(X)=f_i(\tilde f^{\ell-1}(\tilde X))\subseteq f_i(\tilde f^{\ell-1}(\tilde X))=\tilde f^\ell_i(\tilde X),
\]
where the inclusion results from the induction hypothesis and the fact that $f_i$ only depends on variables with indices which are not in $J$ (that is, $\tilde G$ has no arc from $J$ to $V\setminus J$), and where the last equality results from (6).

\smallskip
As a consequence, $f$ has the following three properties:
\begin{itemize}
\item[(7)] $f_i(X)\subseteq h_i(Y)$ for all $i\not\in I$.
\item[(8)] $f^{|I|+1}_i(X)\subseteq h_i(Y)$ for all $i\in I$.
\item[(9)] $f(x)=h(x)$ for all $x\in Y$.
\end{itemize}
Indeed, (7) results from (4) and (1), and (9) results from (6) and (3). It remains to prove (8). From $(4)$ and $(2)$ we have $f^{|I|+1}_i(X)\subseteq h_i(Y)$ for all $i\in\tilde I$, and from (5) we have $f^\ell_i(X)=\{\xi_i\}=h_i(Y)$ for all $i\in J$ and $\ell\geq \lambda(Q)$. We deduce that is sufficient to prove that $\lambda(Q)\leq |I|+1$. Since $J$ corresponds to the set of non-sources of $Q$ and $Q$ is basic, by Remark~\ref{rem:lambda}, we have $\lambda(Q)\leq |J|+1\leq |I|+1$. This completes the proof of (8).

\smallskip
We deduce from (7), (8) and (9) that $f$ converges toward $h$ in at most $|I|+1$ steps.
\end{proof}

We now prove Lemma~\ref{lem:extension}, using the following notations. For all $X\subseteq \mathbb{Z}^n$ and $y\in\mathbb{Z}^n$, $X+y=\{x+y\,|\,x\in X\}$. We denote by $e_i$ the $n$-tuple whose components are all equal to $0$, excepted th $i$th one, which is equal to $1$.

\begin{proof}[{\bf Proof of Lemma~\ref{lem:extension}}]
We proceed by induction on the number of arcs in $G$ that are not in $H$. If no such arcs exist, then $G=H$ and there is nothing to prove. So assume that $G$ has an arc
\[
a=(j,i,s)
\]
that is not in $H$. Let $\tilde G=G\setminus a$ be the spanning subgraph obtained from $G$ by removing $a$. Let $\tilde A$ be the set of sources of $H$ that are not sources of $\tilde G$ and let $\tilde B$ be the set of sinks of $H$ that are no sinks of $\tilde G$. By induction hypothesis, there exists a degree-bounded system $\tilde f:\tilde X\to\tilde X$ on $\tilde G$ such that
\begin{enumerate}
\item[(i)]
$\tilde f(\tilde X)\subseteq Y$,
\item[(ii)]
$\tilde f_k(\tilde X)\subseteq h_k(Y)$ for all $k\not\in \tilde A$.
\item[(iii)]
$\tilde f(x)=h(x)$ for all $x\in Y$ such that $x_k=\xi_k$ for all $k\in \tilde B$.
\item[(iv)]
$\tilde f_k(x)=h_k(x)$ for all $x\in Y$ and vertex $k$ such that $\tilde G_k\cap \tilde B=\emptyset$.
\end{enumerate}
Since $G$ has no arc from a source of $H$ to a sink of $H$, $G$ has no arc from a source of $\tilde G$ to a sink of $\tilde G$. Furthermore, since $G$ has no arc from a vertex of $H$ to an isolated vertex of $H$, $G$ has no arc from a vertex of $\tilde G$ to an isolated vertex of $\tilde G$. This leaves the following three possibilities.

\smallski
\BF{Case 1: $j$ is not a sink of $\tilde G$ and $i$ is not a source of $\tilde G$.}
\begin{itemize}
\ite
If $s=+$ we set $X=\tilde X\cup (\tilde X+e_j)$ and define $f:X\to X$ as follows:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\max(\tilde f_i(\tilde X))&\textrm{if $x_j=\max(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
Hence, $f(X)\subseteq\tilde f(\tilde X)$ and $f(x)=\tilde f(x)$ for all $x\in \tilde X$. Since $i$ is a not a sink of $\tilde G$ we have $A=\tilde A$ and since $j$ is not a source of $\tilde G$, we have $B=\tilde B$. Thus, using the properties (i)-(iv), we deduce that $f$ satisfies the four points of the statement. It remains to prove that $f$ is a degree-bounded system on $G$.

\smallskip
Let $G_f$ be the interaction graph of $f$. Since $f(x)=\tilde f(x)$ for all $x\in \tilde X$, $\tilde G$ is a subgraph of $G_f$. Since it is straightforward to prove that $G_f$ is a subgraph of $G$, to prove that $G_f=G$, it is sufficient to prove that $a$ is an arc of $G_f$. Let $x\in X$ be such that $f_i(x)=\min(\tilde f_i(\tilde X))$ and suppose that $x_j$ is maximal for this property. Since $i$ is not a source of $\tilde G$, we have $\min(\tilde f_i(\tilde X))<\max(\tilde f_i(\tilde X))$ and we deduce from the definition of $f_i$ that $x_j<\max(X_j)$. Consequently, $f_i(x+e_j)>\min(\tilde f_i(\tilde X))$ so $G_f$ has a positive arc from $j$~to~$i$. Thus $G_f=G$, that is, $f$ is an FDS on $G$. Since $\tilde f$ is degree-bounded, it is straightforward to prove that $f$ is degree-bounded too.

\ite
If $s=-$, we proceed similarly with $X=\tilde X\cup (\tilde X+e_j)$ and $f:X\to X$ defined by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\min(\tilde f_i(\tilde X))&\textrm{if $x_j=\max(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
\end{itemize}

\smallski
\BF{Case 2: $j$ is a sink of $\tilde G$ and $i$ is not a source of $\tilde G$.}
Since $j$ is a sink of $\tilde G$ we have $|\tilde X_j|=1$ if $j$ is isolated in $\tilde G$, and $|\tilde X_j|=2$ otherwise. Let
\[
X=
\begin{cases}
\tilde X&\text{if }|\tilde X_j|= 2\\
\tilde X\cup(\tilde X+e_j)&\text{if }|\tilde X_j|= 1\\
\end{cases}
\]
Hence $|X_j|=2$, so either $\xi_j<\max(X_j)$ or $\xi_j>\min(X_j)$, and we consider four cases.
\begin{itemize}
\ite
If $s=+$ and $\xi_j<\max(X_j)$, we define $f:X\to X$ by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\max(\tilde f_i(\tilde X))&\textrm{if $x_j=\max(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
Hence, $f(X)\subseteq\tilde f(\tilde X)$ and $f(x)=\tilde f(x)$ for all $x\in X$ with $x_j=\xi_j$. Since $i$ is a not a source of $\tilde G$ we have $A=\tilde A$, we deduce from (i) and (ii) that $f$ satisfies the first two points of the statement. Since $B=\tilde B\cup\{j\}$ and $j\in G_i\cap B$, we then deduce from (iii) and (iv) that $f$ satisfies the last two points. It remains to prove that $f$ is a degree-bounded system on $G$. We prove that $G$ is the interaction graph of $f$ exactly as in the previous case. Next, since $\tilde f$ is a degree-bounded system on $\tilde G$, and since $|X_j|=2=d^{\mathrm{out}}_G(j)+1$, we deduce that $f$ is degree-bounded. Thus $f$ is indeed a degree-bounded system on $G$.

\ite
If $s=+$ and $\xi_j>\min(X_j)$, we proceed similarly with $f:X\to X$ defined by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\min(\tilde f_i(\tilde X))&\textrm{if $x_j=\min(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.

\end{array}
\]
\ite
If $s=-$ and $\xi_j<\max(X_j)$, we proceed similarly with $f:X\to X$ defined by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\min(\tilde f_i(\tilde X))&\textrm{if $x_j=\max(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.

\end{array}
\]
\ite
If $s=-$ and $\xi_j>\min(X_j)$, we proceed similarly with $f:X\to X$ defined by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\max(\tilde f_i(\tilde X))&\textrm{if $x_j=\min(X_j)$}\\
\tilde f_i(x)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in \tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
\end{itemize}

\smallski
\BF{Case 3: $j$ is not a sink of $\tilde G$ and $i$ is a non-isolated source of $\tilde G$.}
Then $i$ is a source of both $\tilde G$ and $H$, thus $\tilde f_i=\mathrm{cst}=c_i$ and $h_i={\mathrm{cst}}=c'_i$, and following (iii) $c_i=c'_i$. Since $i$ is not isolated in $\tilde G$ we have $|\tilde X_i|\geq 2$ and thus either $c_i<\max(\tilde X_i)$ or $c_i>\min(\tilde X_i)$, and we consider the following four cases. In every cases we set
\[
X=\tilde X\cup (\tilde X+ e_j).
\]

\begin{itemize}
\ite
If $s=+$ and $c_i< \max(\tilde X_i)$, we define $f:X\to X$ by:
\[
\begin{array}{lcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\max(\tilde X_i)&\textrm{if $x_j=\max(X_j)$}\\
c_i&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in\tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]

Hence, $f_i(X)\subseteq \tilde X_i$ and $f_k(X)\subseteq \tilde f_k(\tilde X)$ for all $k\neq i$. Since $A=\tilde A\cup\{i\}$ we deduce from (i) and (ii) that $f$ satisfies the first two points of the statement.  Since $B=\tilde B$ and $h(x)=\tilde f(x)$ for all $x\in \tilde X$, we deduce from (iii) and (iv) that $f$ satisfies the last two points of the statement. It remains to prove that $f$ is a degree-bounded system on $G$.

\smallskip
Let $G_f$ be the interaction graph of $f$. Since $f(x)=\tilde f(x)$ for all $x\in \tilde X$, $\tilde G$ is a subgraph of $G_f$. Since it is straightforward to show that $G_f$ is a subgraph of $G$, to prove that $G_f=G$, it is sufficient to prove that $a$ is an arc of $G_f$. Let $x\in X$ with $x_j=\max(X_j)-1$. Then $f_i(x)=c_i<\max(\tilde X_i)$ and $f_i(x+e_j)=\max(\tilde X_i)$ thus $a$ is indeed an arc of $G_f$. Thus $f$ is a FDS on $G$, and since $\tilde f$ is degree-bounded, it is straightforward to prove that $f$ is degree-bounded too.

\ite
If $s=+$ and $c_i>\min(\tilde X_i)$ we proceed similarly with $f:X\to X$ define by:
\[
\begin{array}{rcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
c_i&\textrm{if $x_j=\max(X_j)$}\\
\min(\tilde X_i)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in\tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
\ite
If $s=-$ and $c_i<\max(\tilde X_i)$, we proceed similarly with $f:X\to X$ defined by:
\[
\begin{array}{rcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
c_i&\textrm{if $x_j=\max(X_j)$}\\
\max(\tilde X_i)&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in\tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
\ite
If $s=-$ and $c_i>\min(X_i)$, we proceed similarly with $f:X\to X$ defined by:
\[
\begin{array}{rcl}
f_i(x)&=&
\left\{
\begin{array}{ll}
\min(\tilde X_i)&\textrm{if $x_j=\max(X_j)$}\\
c_i&\textrm{otherwise}\\
\end{array}
\right.
\\[5mm]
f_k(x)&=&
\left\{
\begin{array}{ll}
\tilde f_k(x-e_j)&\textrm{if $x\not\in\tilde X$}\\
\tilde f_k(x)&\textrm{otherwise}\\
\end{array}
\right.
\quad\forall k\neq i.
\end{array}
\]
\end{itemize}
\end{proof}

\paragraph{Acknowledgment} I wish to thank Emmanuel Bonan and Tony Silva for stimulating discussions.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Rospotrebnadzor voices grievances about quality, safety of McDonald's products
The Russian sanitary watchdog Rospotrebnadzor has grievances about product quality and safety of the McDonald's fast-food restaurant, Rospotrebnadzor chief and Russia's chief sanitary officer Anna Popova told Interfax on Friday.
"We have identified violations which put the product quality and safety of the entire McDonald's chain in doubt," Popova said.
Following an inspection carried out in Veliky Novgorod, eleven administrative-violation protocols were compiled about two McDonalds alone, she said. "These products complied neither with the safety standards, nor energy values. They were extremely remote from the current standards," Popova said.
Read more: RusBurger steals in ahead of Burger King in Crimea
Food in Russia food trade business News

redpajama_set_name: RedPajamaCommonCrawl


People agree that starting a business is risky and although the failure rates might not be as dire as people like to say, 20% fail in their first year and only 50% make it to five years. A decade in, only 30% are still around (Stats courtesy of Fundera.).
A tech startup takes that to a whole new level because of the additional layers of uncertainty that technology imposes. Startup's risks are significant.
Unlike a coffee shop or shoe store which can use relatively accessible market data for planning, most tech startups are operating in newer, untested areas. Some are even pushing the boundaries of medical science or the laws of physics. Startups aren't a single sector so there is no single model geography, market or user-group for a startup.
However, you're an entrepreneur with a vision so you aren't going to stop tackling your world-changing idea just because the odds are stacked against you. But what can you do to improve those odds?
Unfortunately, most articles titled 'startup risks' give you very superficial advice and certainly not enough to turn into an actionable plan. These are similar to growth articles that tell you to 'make more sales'. Duh!
However, that doesn't mean you can't improve your odds and you do that by understanding and managing your risks. What's outlined here is a simple risk management framework that you can apply to your startup to better understand and prepare for the things that threaten your success.
Best of all, it doesn't take long and you already have most of the information you need.
Risk management might sound like the kind of thing that your startup won't need for years, the kind of thing you think about when you have your own in-house lawyer and accounting team. By that stage, you will definitely be thinking about risk management, especially if you plan to IPO. However, if you start managing your startup risks now, you will have a better chance of actually getting to that stage.
Plus, risk management doesn't have to be complicated.
You can start risk management with a simple approach that will scale as you grow.
A simple definition for risk is 'the effect of uncertainty on objectives' (this is the ISO definition). It's how the things we don't understand and manage properly can prevent us from reaching our goal.
A risk assessment is how you can understand what your challenges are.
Risk management is the process to reduce or overcome the risks you identify.
Threat – What is the negative thing that could happen?
Vulnerability – What is in place to prevent this or how prepared are we for this event?
Impact – How badly would this event affect our plans if it happened?
If you understand the above points, that's all you need to know far as definitions or the technicalities of risk management. Now we're going to move on to how to apply this to your business to assess and understand your startup risks.
The simplest way to understand your risk is to ask a series of three questions for each of the main areas of startup risk. These questions allow you to understand the threats to your business, how vulnerable you are and what impact an event might have.
Once you have run through these questions, you will be able to prioritize your risks and start to take preventative action.
To help add some structure, I've grouped similar threat types together to make discussions easier. The categories are shown in bold and the individual threats to think about are also listed.
At this stage, you might think that some of these threats don't apply and you will be tempted to strike these off the list.
Don't. Leave them there for now because you might be surprised what poses a threat once you have thought about it.
You might also need to add to this list, depending upon any specific threat to your startup. Again, you know your business best so feel free to add factors where necessary. The only advice that I would give is to try to avoid having too many specifics at this stage. You can always add more later but we are trying to keep things simple at this stage.
Scoring can get out of hand and become very complicated, very quickly, so it's best to keep it simple. That said, you should have a combination of quantitative (numbers), qualitative (descriptions) and visual (color coding) to help 'score' each element as this helps with calculations, discussions and presentations of the data respectively.
This mix allows you to describe something ('Moderate') in a discussion, apply a value ('3') for scoring and color-code the results ('amber') all within the same system.
This gives you a result between 1 and 100 which you then use to order your risks from highest (worst and most urgent) to lowest. You can read more about risk metrics and this system here.
This is a relatively straightforward assessment as there are only 21 elements so a spreadsheet can work (there's a template for one here). Alternately, you can use software to speed up and simplify the process. Whatever you do, just make sure the system is stable and that everyone is applying the metrics consistently.
Once you have scored each risk, list them in descending order to work out which fire needs putting out first.
Remember, you are in the early stages of a startup so many (all?) of your risks will score pretty highly. You already know that, it just looks worse when it's written down in detail so don't be alarmed if everything comes up red.
Here's the process for understanding and managing your startup risks in full.
Evaluate each element and give it a value for threat, vulnerability and impact to give us an overall score.
Compare the risks to determine which needs to be addressed most urgently.
Again, this doesn't need to be a complicated or time consuming process. Big organizations will spend months on these but you won't have time for that. However, you only need to carve out a couple of hours per month to do this and you will be in much better shape.
And that's all there is to it. Think about your startup and the threats that you face using the categories suggested above. Then, ask yourself the three questions and score the results.
Once you've done that, you will end up with a numeric score for each element so you can prioritize these for action.
As I said, everything is going to run hot in a startup and you will end up with a lot of 'reds,' but now you at least have a better understand what you are dealing with. Most importantly, you might spot some left-field threats that weren't on your radar before.
May the Odds be ever in your Favor….
If you are reading this, then you know running a startup is hard. And turning that into a successful business is even harder.
But we owe it to ourselves, our users and investors (and our families) to give our ideas the best chance of success. The failure rate will still be high – we are doing hard things so that's to be expected – but we can stack the odds in our favor if we really understand what our risks look like. I hope this system you understand and manage your risks.
You can find a lot more detail on risk, risk management and even try the DCDR risk management tool at DCDR.io.
I'd love to know what you think – please leave a comment below.
Here are some examples of the answers that you might find when you ask about the threat from each element. The grid below shows the kinds of statement that would result in a lower or higher results but again, make sure you are thinking about your startup, your market and your challenges when you complete your assessment.
Keep in mind that threats are events that are normally outside the control of the organization. For example, under 'cashflow' the threat is cash flow in your field generally is high or low. So if you are doing biomedicine, your burn rate is going to be very high compared to something wholly browser-based. You might be able to control your cash flow and do things faster and cheaper (maybe that's your secret sauce) but you address that in the next section.
There is a well-defined, validated market for the offering. There is no clear market for the offering.
The market is new and no mature players operate in the space. The market is mature and is dominated by a single entity with significant competitive advantages.
There is a lot of interest in funding ventures in this space There are very few opportunities to raise funds.
The venture has no notable environmental impact The venture imposes a significant burden on the environment due to waste products / resource usage.
There are no specific dependencies required to make the venture operational The venture is wholly dependent on third-party dependencies.
The founding team has the ability and temperament required to make the venture succeed. The founding team has very few applicable skills or experience.
The venture has a deep bench of skilled talent There are no skilled individuals within the venture (e.g. everything is outsourced).
Want this whole article as a PDF? Just send me an email.


If you like to hike or bike and enjoy well maintained trails, think about volunteering. Volunteers are needed to build and maintain trails. And, we can need helpers to lead hikes, "coach" the Sprockidz, work on special events (such as the film fest), map trails, and assist in the office.
If you would like to be on our 'go-to' list of volunteers, please fill out the Volunteer Registration Form when you become a member. We will call or email you when we need your special skills. Also, we send out requests for volunteers through our monthly E-newsletter – the 'Trail Post' so be sure to subscribe to the newsletter.


Salter defeated MacDonald via TKO (injury) at 2:42 of round 1.
Guymon defeated Yoshida via unanimous decision (30-27, 30-27, 30-27).
Beltran defeated Hague via unanimous decision (30-27, 30-26, 29-28).
Hendricks defeated Grant via majority decision (29-27, 29-27, 28-28). Grant had a point deducted due to multiple groin strikes.
Davis defeated Goulet via TKO (punches) at 1:23 of round 2.
Doerksen defeated Lawlor via submission (rear naked choke) at 2:10 of round 2.
Belcher defeated Cote via submission (rear naked choke) at 3:25 of round 2.
Mitrione defeated Slice via TKO (punches) at 4:24 of round 2.
Stephens defeated Stout via split decision (30-27, 29-28, 28-29).
Rua defeated Machida via KO (punches) at 3:35 of round 1 to become the new UFC Light Heavyweight Champion.


I received my February Glossybox the over the weekend. It's always a fun box to open. It's a beauty product subscription service filled with five new products from brands you know, love, and trust.
Again, it came in the cutest box. I really like the boxes themselves since they come with some great designs; you don't want to just throw them out. They are sturdy enough to reuse too.
And inside are five different samples of beauty products. This time, I think each item will be useful or interesting to try.
Illuminating Mask – yes please.
Lip gloss – yes, I have a slight addiction in fact.
Hydrating cleanser – of course I can use that – and this one smells great!
Eye Cream – oh yes, definitely will be putting that on.
Face Primer – from Benefit – only too happy to try this one too.
All great products this time.
You also get this handy card that describes the products, tips, and sometimes contains special offers.
If you'd like to try Glossybox yourself, click here. Enjoy.


BYEI Research Initiative is the research wing of Bangladesh Youth Environmental Initiative, developed to involve youths in research on environmental challenges and opportunities in Bangladesh.
We aim to build the research skills of BYEI affiliates, build awareness of environmental issues through research, and build our voice to engage in environmental debates in Bangladesh. Our research initiative contributes to BYEI broader goals and objectives to disseminate knowledge, raise awareness, and promote positive actions and policies.
International Youth Environment Movements: What motivates us, and what unites us?
Environmental Struggles in Developing Countries: What lessons are there for Bangladesh?
Small-scale Renewable Energy in Bangladesh: Where are we now and where are we going?


What may affect your chances of needing a caesarean section?
This information is about the care and treatment of women who plan or need to have a caesarean section that is set out in NICE clinical guideline 132.
Medical terms are explained in the Glossary.
In September 2012, the information about place of birth was removed from Reducing your chances of needing a caesarean section and Things that do not affect your chances of needing a caesarean section. Place of birth will be addressed in the NICE guideline on intrapartum care (care during labour) that is currently being updated. All other information remains the same.
Yes, if you are planning to have, or need to have, a caesarean section.
extra care you may need if you or your baby develop specific medical conditions in the course of your pregnancy or labour.


This coupon was added about 3 months ago, may expire any time. Verified by HebrewJohn.
This coupon was added about 3 months ago, may expire any time. Verified by Stone.jack.
This coupon was added about 4 months ago, may expire any time. Verified by Tyler23.
No home or vehicle is complete without this Raptor Multi-Tool. This exciting tool features an oxygen tank ring, shears and other tools that can shatter glass and take care of business in an emergency.
Keep your Mako Ti working more efficiently when you have these Replacement Bits. These bits will help you to work through many projects as they offer common flat and Philips head designs.
I met my husband about five years ago. We were both part of the same hiking group. He was actually the leader and it was one of my first weekend backpacking trips where we had to hike into a campsite. I had no idea about the proper type of equipment to bring. My pack was so old and didn't fit correctly. It was also way too full and heavy with a lot of unnecessary items. I was used to running and hiking, and also camping, but I definitely didn't have the modern supplies that most now carry. I didn't even have a camelback! I carried water in several different bottles. I also didn't have trekking poles, which I should have because there were so many steep areas. I just recall Pete helping me so much that weekend. He found me a great stick to use as a pole. He also kept checking back on the trail to make sure I hadn't fallen off the cliff from the weight of my pack. We didn't date right away. But, about a year after that, we saw each other on a different hike, and we've been together since. I love to buy him stuff for our outdoor trips. Leatherman.com has great "survival" tools, which Pete has so many already. However, I saw on the site that we can personalize some of his tools by engraving his name on them. I know he would love the Skeletool because it's an extremely lightweight multi-tool for everyday use.


Kaitlyn Dever arrives for The Hollywood Foreign Press Association (HFPA) And InStyle Celebrate The 2015 Golden Globe Award Season - Arrivals at Fig & Olive Melrose Place on November 20, 2014 in West Hollywood, California.
Kaitlyn Dever opted for simple styling with this loose side-parted 'do at the Golden Globe Award season celebration.
Kaitlyn Dever sported a stark cat eye for a retro touch to her look.


text:
Sans-Souci (uit het Frans: "Zonder zorgen") is een voormalig paleis in het noorden van Haïti. Het is in het begin van de 19e eeuw gebouwd door Henri Christophe, de latere koning Hendrik I. Na een aardbeving in 1842 ligt het in ruïnes. Sinds 1982 behoort het tot het Werelderfgoed, als onderdeel van het Nationaal historisch park Citadel, Sans-Souci, Ramiers.

Het paleis is gebouwd tussen 1810 en 1813. Het ligt in de plaats Milot. Dichtbij ligt de Citadel Laferrière, die ook door Hendrik I gebouwd is.

Zowel de naam als de bouwwijze zijn afgeleid van het paleis Sanssouci in Potsdam. Het paleis werd gekoeld door water uit een bergbeek door buizen onder de vloer te leiden. Het complete terrein neemt 8 hectare in beslag. Hier bevonden zich een drukkerij, weverijen, een hospitaal, scholen, een distilleerderij en militaire barakken.

Nadat hij een beroerte had gehad, pleegde Hendrik I op 8 oktober 1820 in dit paleis zelfmoord. Volgens de overlevering deed hij dit door een zilveren kogel door zijn hoofd te schieten. Hij werd begraven in de Citadel Laferrière.

Tegenwoordig worden de ruïnes door toeristen bezocht. De geschiedenis van Hendrik I in dit paleis wordt getoond in de film Royal Bonbon uit 2002 van Charles Najman.

Bouwwerk in Haïti

redpajama_set_name: RedPajamaWikipedia


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Protect the side that is weakened by .
White can't play at the marked point because of shortage of liberties.
The stones turn the hane at into an atari to which White must react, so that Black can take away the second eye with .
This is punished the same way as her mistake in the follow-up to the correct answer (see above).
Enlarging the eye space at invites Black to play at the vital point . Now and are miai. After , White has to prevent the capture and take a liberty at a upon which Black takes away the other eye at b.
Connecting at in this way makes poor eye shape. Black plays on the vital point of this shape. Even if White continues with at a, Black will be able to almost fill her eye space with a farmer's hat.
Connecting at here is symmetric to Failure 4 and is once again countered by playing on the vital point.


Distance from Hartsfield–Jackson Atlanta International Airport to Las Américas International Airport is 1389 miles / 2235 kilometers / 1207 nautical miles.
Estimated flight time from Hartsfield–Jackson Atlanta International Airport (ATL) to Las Américas International Airport (SDQ) is 3 hours 7 minutes.
There is no time difference between Atlanta and Santo Domingo.
Flight map from Atlanta to Santo Domingo.
Atlanta (ATL) → Santo Domingo (SDQ).


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Endesa maintains its commitment to entrepreneurship and will take part in the new edition of "South Summit 2017"
Published on Friday, 22 September 2017
South Summit 2016
'South Summit' is the leading startup conference in southern Europe, organised by Spain Startup and the IE Business School. For the fifth consecutive year, Endesa will take part in the conference, along with BStartup by Banco Sabadell and Google for Entrepreneurs.
Endesa, as part of its open innovation model, has maintained its commitment to entrepreneurs as a source for providing solutions to meet the technological challenges of the future.
Over 200 international speakers, investment funds, venture capital and business angels have already confirmed their attendance at the conference, which will be held in Madrid from the 4th to the 6th of October in the municipal space of LA N@VE.
The 100 finalist startups have already been selected from among the 3,500 projects submitted from over 70 countries.
Supporting entrepreneurs. This continues to be Endesa's purpose for taking part for the fifth year running in South Summit 2017, the leading startup conference in southern Europe. This year it will be held in Madrid from the 4th to the 6th of October, in the same venue as the last edition: LA N@VE, the centre for inspiration, education and open innovation with which the city of Madrid hopes to transform the economic and social future of the community.
South Summit is organised by Spain Startup and the IE Business School and it will once be the meeting point for the newest startups with the main international investment funds and most innovative corporations, including, Endesa, which, once again is a global partner of the event, which exceeds its own figures every year. Over 12,500 people of 100 different nationalities attended the last edition, 6,700 entrepreneurs, 650 investors, of which, 150 were international and 450 speakers.
Endesa will once again have a stand in the area called market place, where all the other participants will meet and where it will present its initiatives related to entrepreneurship and the company's innovation model; an open model that promotes support for entrepreneurs to develop solutions that will enable the current energy model to be transformed.
This market place is also where the 100 finalist startups will meet, which South Summit has just selected from over 3,500 projects submitted to the Startup Competition, which is the largest number of contestants in all the editions. The one hundred projects have been submitted from participants of over 70 different nationalities, 49% of which are of Spanish origin.
The criteria taken into account to select the finalists includes five fundamental values: the innovation they provide to the market, the economic feasibility, the scalability of their technology, the human team behind the projects and if they are investable.
The countdown for South Summit 2017 has begun and once again the programme will focus on new disruptive technologies, analysing the impact of innovations on the main industries. Among the speakers that have already confirmed their attendance are Kei Shimada, director of innovation and Business Development at Dentsu, a leading company in innovation and Tal Morgenstern, one of the first Snapchat investors. Morgenstern is taking part in this meeting as a partner of the venture capital firm, Lightspeed, focused on investment in Israel, a country with a strong entrepreneurial culture at a global level. Antoinia Norman, a great defender of entrepreneurship as a key factor in the development of the economy and CEO of Branson Centre of Entrepreneurship South Africa will also take part in the conference. As well as Adam Kostyál, Senior Vice President of Listing Services in Europe at Nasdaq; Thierry Petit, Co-Founder & CEO of Showroomprivé; and Gonzalo Salazar-Simpson, a producer of films such as Ocho apellidos vascos or No habrá paz para los malvados.
In addition, the three-day meeting, which will create an environment designed to encourage the emergence of new startups in Spain, the arrival of more entrepreneurs to this country and the creation of an efficient network of investors, will end with the competition award ceremony. Endesa will form part of the jury which will be responsible for choosing the winning startups from different verticals and from the Grand Finale and it will award the first prize to the winner of the competition in the category of Energy & Sustainability.
NEW September, 22 2017
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