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{"url":"https:\/\/gamedev.stackexchange.com\/questions\/124782\/unity-wwise-no-audio-in-my-game-build","text":"# Unity Wwise. No audio in my game build\n\nI'm building a game in Unity using Wwise as the audio engine. I'm on a Mac and whenever I build the game, there is no audio. Apparently this is because the sound banks are not in the correct folder. They should be in Assets\/StreamingAssets\/Audio\/GeneratedSoundBanks\/ followed by a sub-folder with the name of whatever platform you are using. I should have a Mac folder located here however there isn't one. Instead my sound banks just sit in the GeneratedSoundBanks folder which works when i test the game in unity, however there is no audio when I build the game. Apparently this has something to do with environment variables or a python path but I have no idea how to fix or edit these as I am new to coding as well as Wwise and Unity. If I build the game for web player I get an error CS0246: The type of namespace name 'AKRESULT' could not be found. Web GL says it doesn't work for the unity wise integration package I have either (and going to an older one just messed my whole projects audio up completely). If anyone could give me any help at all, it would be very much appreciated. Thank You.\n\nI'm using: Unity 5.3.5.1f, Wwise v2016.1 (32-bit), WwiseUnityIntergration_v2016.1_Mac.unitypackage.\n\nYou need to copy the GeneratedSoundbanks folder in your Wwise project, and paste it into the Assets\/StreamingAssets\/Audio\/ directory of your Unity project before running the build process in Unity.","date":"2020-07-13 12:22:58","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.22865357995033264, \"perplexity\": 1366.0326460383717}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-29\/segments\/1593657143365.88\/warc\/CC-MAIN-20200713100145-20200713130145-00266.warc.gz\"}"} | null | null |
\section{Introduction}
A $G$\emph{-flow} is a continuous action of a group $G$ on a compact Hausdorff space. Given two $G$-flows $\,^{\!\,_{\!\,_{\!\,_{\bullet}}}}\!: G \times X \to X$ and $\,^{\!\,_{\!\,_{\!\,_{\circ}}}}\!: G \times Y \to Y$, a \emph{quotient map} from $\,^{\!\,_{\!\,_{\!\,_{\bullet}}}}$ to $\,^{\!\,_{\!\,_{\!\,_{\circ}}}}$ is a continuous surjection $\pi: X \to Y$ that preserves the action of $G$, in the sense that $\pi(g \,^{\!\,_{\!\,_{\!\,_{\bullet}}}} x) = g \,^{\!\,_{\!\,_{\!\,_{\circ}}}} \pi(x)$ for every $g$ and $x$.
\begin{maintheorem1}
Let $G$ be a countable discrete group, and consider the trivial flow on the space $(G \times \omega)^*$ induced by the natural action of $G$ on $G \times \omega$, namely $g \,^{\!\,_{\!\,_{\!\,_{\bullet}}}} (h,n) = (gh,n)$.
\begin{enumerate}
\item Every $G$-flow of weight $\leq\! \aleph_1$ is a quotient of this flow.
\item Consequently, the Continuum Hypothesis implies this flow is universal for $G$-flows of weight $\leq\! \mathfrak{c}$.
\end{enumerate}
\end{maintheorem1}
\noindent Here $(G \times \omega)^* = \beta(G \times \omega) \setminus (G \times \omega)$ is the Stone-\v{C}ech remainder of the discrete space $G \times \omega$. The \emph{weight} of a flow means the weight of its underlying topological space $X$, that is, the smallest cardinality of a basis for $X$.
What is called the ``Main Theorem'' above is in fact just a corollary to a stronger result, which we call the ``Main Lemma,'' but its statement is more technical and will be postponed until later. Roughly, the main lemma characterizes the weight-$\aleph_1$ quotients of any trivial $S$-flow on $\omega^*$ for any countable discrete semigroup $S$. This improves on some former work of the author in \cite{Brian}, which gives a topological characterization of the weight-$\aleph_1$ quotients of a particular $(\mathbb{N},+)$-flow on $\omega^*$ (the one generated by the shift map).
The assumption of the Continuum Hypothesis (henceforth abbreviated \ensuremath{\mathsf{CH}}\xspace) in part (2) of this theorem is necessary, in the sense that the conclusion does not follow from \ensuremath{\mathsf{ZFC}}\xspace alone. Even more is true: \ensuremath{\mathsf{ZFC}}\xspace does not prove the existence of universal weight-$\mathfrak{c}$ flows for any group, as the following observation shows.
\begin{theorem}\label{thm:noflows}
It is consistent that no (semi)group $G$ admits a universal $G$-flow of weight $\leq\! \mathfrak{c}$.
\end{theorem}
\begin{proof}
Observe that for every (semi)group $G$ and every compact Hausdorff space $X$, there is at least one $G$-flow on $X$, namely the trivial flow $(g,x) \mapsto x$.
It follows from this observation that if $\,^{\!\,_{\!\,_{\!\,_{\bullet}}}}\!: G \times Y \to Y$ is a universal $G$-flow of weight $\leq\! \mathfrak{c}$, then every compact Hausdorff space $X$ of weight $\leq\! \mathfrak{c}$ is a continuous image of $Y$.
In Section 6 of \cite{Dow&Hart}, Dow and Hart show that it is consistent that no such space $Y$ exists.
\end{proof}
\vspace{2mm}
If $G = (\mathbb{Z},+)$, then each $G$-flow on $\omega^*$ is generated by an autohomeomorphism of $\omega^*$ and, conversely, every autohomeomorphism of $\omega^*$ generates a $G$-flow. Thus, via Stone duality, results concerning $(\mathbb{Z},+)$-flows on $\omega^*$ can be translated into results concerning automorphisms of the Boolean algebra $\mathcal{P}(\w)/\mathrm{fin}$.
Given two automorphisms $\varphi,\psi: \mathcal{P}(\w)/\mathrm{fin} \to \mathcal{P}(\w)/\mathrm{fin}$, we say that $\varphi$ \emph{embeds} in $\psi$ if there is a self-embedding $e: \mathcal{P}(\w)/\mathrm{fin} \to \mathcal{P}(\w)/\mathrm{fin}$ such that $e \circ \varphi = \psi \circ e$. Equivalently, $\varphi$ embeds in $\psi$ if there is a subalgebra $\AA$ of $\mathcal{P}(\w)/\mathrm{fin}$ such that $(\mathcal{P}(\w)/\mathrm{fin},\varphi)$ is isomorphic to $(\AA,\psi \!\restriction\! \AA)$.
\begin{center}
\begin{tikzpicture}
\node at (5,0) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (8.5,0) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (8.5,2) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (5,2) {$\mathcal{P}(\w)/\mathrm{fin}$};
\draw[->] (5.95,2) -- (7.55,2); \node at (6.75,2.25) {$\psi$};
\draw[->] (5.95,0) -- (7.55,0); \node at (6.75,-.25) {$\varphi$};
\draw[right hook->] (8.5,.35) -- (8.5,1.68); \node at (8.7,1) {\small $e$};
\draw[right hook->] (5,.35) -- (5,1.68); \node at (4.8,1) {\small $e$};
\node at (11.5,1.25) {\Large $\varphi \hookrightarrow \psi$};
\end{tikzpicture}
\end{center}
\begin{maintheorem2}
Assuming \ensuremath{\mathsf{CH}}\xspace, there is a trivial automorphism $\t$ of the Boolean algebra $\mathcal{P}(\w)/\mathrm{fin}$ such that every other automorphism embeds in $\t$.
\end{maintheorem2}
By applying Stone duality, this result follows directly from a special case of the Main Theorem, namely the case $G = (\mathbb{Z},+)$. The automorphism $\t$ mentioned in the theorem is the one generated by the map $(z,n) \mapsto (z+1,n)$ on $\mathbb{Z} \times \omega$.
In fact, we will go a bit further and investigate precisely which automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$ are universal in the sense of this dualized Main Theorem. The investigation is carried out in \ensuremath{\mathsf{ZFC}}\xspace+\ensuremath{\mathsf{CH}}\xspace, because those are the axioms needed for applyling the Main Lemma. We will classify precisely which trivial automorphisms are universal under \ensuremath{\mathsf{CH}}\xspace, and show that there are many nontrivial universal automorphisms as well.
\vspace{2mm}
The paper is organized as follows.
Section~\ref{sec:theorem} begins by establishing the terminology necessary for stating the Main Lemma, and ends by stating it and deriving the Main Theorem from it. We will include a brief outline of the proof, and some hints as to where the difficulties lie, but the proof of this result is put off until the end of the paper.
Section~\ref{sec:dynamics} looks at the automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$ in light of the Main Lemma, and initiates an investigation of the ``embeds in'' relation on the set of automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$, focusing on universal automorphisms. Most of the theorems in Section~\ref{sec:dynamics} have relatively simple proofs that use the Main Lemma, Stone duality, and basic ideas and techniques from topological dynamics. Most of the results are established in \ensuremath{\mathsf{ZFC}}\xspace+\ensuremath{\mathsf{CH}}\xspace so that the Main Lemma can be applied; but some observations using \ensuremath{\mathsf{OCA+MA}}\xspace are made in order to establish the independence of some of the results proved using \ensuremath{\mathsf{CH}}\xspace. Section~\ref{sec:dynamics} includes several open questions.
At last, Section~\ref{sec:proof} gives a proof of the Main Lemma. A significant portion of this section is devoted to introducing and developing the topological tools needed to prove the result. These tools are used in the proof of this result and nowhere else in the paper, and this is part of the reason for postponing the proof until the end of the paper. The other part is the author's opinion that the material in Sections \ref{sec:dynamics} is simply more exciting.
\section{The main lemma}\label{sec:theorem}
\subsection{Semigroup actions and Stone-\v{C}ech remainders}
An \emph{action} of a semigroup $S$ on a set $X$ is a function $\varphi: S \times X \to X$ with the property that $\varphi(q,\varphi(p,x)) = \varphi(pq,x)$ for all $p,q \in S$.
We will almost always write $\varphi_p(x)$ for $\varphi(p,x)$. Using this notation, an action of $S$ on $X$ is a function $\varphi: S \times X \to X$ such that $\varphi_p \circ \varphi_q = \varphi_{pq}$ for all $p,q \in S$. An action is called \emph{separately finite-to-one} if each of the functions $\varphi_p$ is finite-to-one.
All the semigroups we consider in this paper are discrete. Thus an action $\varphi$ of a semigroup $S$ on a topological space $X$ is called \emph{continuous} if each of the functions $\varphi_p$ is continuous. (Because $S$ is discrete, this is equivalent to the requirement that $\varphi$ be continuous on $S \times X$.)
Thus an $S$-flow is simply a collection $\set{\varphi_p}{p \in S}$ of continuous functions on a compact Hausdorff space $X$, with the property that $\varphi_p \circ \varphi_q = \varphi_{pq}$ for all $p,q \in S$. In what follows, we will often consider an $S$-flow to be a $S$-indexed collection of functions on $X$, rather than a single function on $S \times X$.
If $D$ is a countable discrete space, then $\b D$ and $D^*$ are, respectively, the Stone-\v{C}ech compactification and the Stone-\v{C}ech remainder of $D$. In what follows we will usually have $D = \omega$, $D = S$, or $D = S \times \omega$ for some countable discrete semigroup $S$.
Every function $f: D \to D$ extends (uniquely) to a continuous function $\b f: \b D \to \b D$, called the \emph{Stone extension} of $f$, defined by
$$\b f(\mathcal{U}) = \set{A \subseteq D}{f^{-1}(A) \in \mathcal{U}}$$
for all $\mathcal{U} \in \b D$..
If $f: D \to D$ is finite-to-one, then $\b f$ maps $D^*$ to $D^*$ and we denote $\b f \!\restriction\! D^*$ by $f^*$.
We say that $f^*$ is \emph{induced} by $f$, and any function on $D^*$ that is obtained in this way is called \emph{trivial}.
Suppose $\varphi: S \times D \to D$ is a separately finite-to-one action of a semigroup $S$ on a countable discrete set $D$. For each $p \in S$ the function $\varphi_p: D \to D$ induces a trivial map $(\varphi_p)^*: D^* \to D^*$.
Taken together, these functions define an action of $S$ on $D^*$.
Formally, we define a function $\varphi^*: S \times D^* \to D^*$ by setting
$$(\varphi^*)_p = (\varphi_p)^*$$
for all $p \in S$ (and naturally, we write $\varphi^*_p$ instead of $(\varphi_p)^*$ or $(\varphi^*)_p$). One may easily check that $\varphi^*$ is an action of $S$ on $D^*$, and, because each of the functions $\varphi_p^*$ is continuous, this makes $\varphi^*$ an $S$-flow. Any $S$-flow on $D^*$ arising in this way is called \emph{trivial}, and we say that the action $\varphi^*$ on $D^*$ is \emph{induced} by the action $\varphi$ on $D$.
\subsection{Metrizable reflections.}
Recall that $H_\k$ denotes the set of all sets hereditarily smaller than $\k$. In what follows, the letter $H$ will always denote a set of the form $H_\k$ for some sufficiently large value of $\k$. The precise value of $\k$ does not matter very much for our purposes, but the reader who wishes to be economical is welcome to verify that $\k = \aleph_2$ is sufficient, because every structure we consider can be coded inside $H_{\aleph_2}$.
To be more precise, let us described a way of coding a weight-$\aleph_1$ flow in $H_{\aleph_2}$. Every topological space of weight $\aleph_1$ is (up to homeomorphism) a closed subspace of $[0,1]^{\aleph_1}$. A ``code'' for a closed subspace of $[0,1]^{\aleph_1}$ could be an $\omega_1$-length list of all the basic open neighborhoods in $[0,1]^{\aleph_1}$ that intersect $X$. A continuous function on $X$ could be coded as a directed graph, with these basic open neighborhoods as its vertices, the arrows in the graph indicating the relation $f(U \cap X) \cap V \neq \emptyset$.
We will look at countable elementary submodels of the structure $(H,\in)$. Roughly, the idea is that if $(M,\in) \preceq (H,\in)$ then only a countable portion of the ``code'' for some weight-$\aleph_1$ flow is captured by $M$. Decoding this flattened fragment of the code leads to a metrizable flow, which we call a metrizable reflection of the original. The process is not unlike file compression: an elementary submodel acts as a data compressor on our topological flows, turning a file with $\aleph_1$ bits of information into a zip file with $\aleph_0$ bits that still carries essentially the same message.
In what follows, we will work at a topological level and avoid the tedious business of encoding and decoding spaces. On the one hand, this ``pointed'' approach forces us to work with larger objects than necessary, which obscures the fact that our theorems are naturally set in $H_{\aleph_2}$; on the other hand, we will see that points and sequences of points feature prominently in our proofs, and eliminating them would obscure the ideas behind the proofs.
If $(M,\in)$ is a countable elementary submodel of $(H,\in)$, then $M \cap \omega_1$ is a countable ordinal number. We denote this ordinal by $\delta^M$.
The points of $[0,1]^{\aleph_1}$ are functions $x: \omega_1 \to [0,1]$, and we define the projection maps
$$\pi_\a(x) = x(\a) \qquad \text{ and } \qquad \Pi_\delta(x) = x \!\restriction\! \delta$$
for $\a,\delta \in \omega_1$.
If for each $\a < \delta$ we have a function $x_\a: Z \to [0,1]$, then $\Delta_{\a < \delta}x_\a$ denotes the unique function $Z \to [0,1]^\delta$ with $\pi_\a \circ \Delta_{\a < \delta}x_\a = x_\a$ for all $\a < \omega_1$.
\begin{definition}$\ $
\begin{itemize}
\item Let $X$ be a closed subspace of $[0,1]^{\omega_1}$, and let $M$ be a countable elementary submodel of $H$. The \emph{reflection of $X$ in $M$} is the space
$$X^M = \Pi_{\delta^M}[X].$$
Any space of this form is called a \emph{metrizable reflection} of $X$.
\item If $f: X \to X$ is continuous, then the reflection of $f$ in $M$, denoted $f^M$, is the continuous function $f^M: X^M \to X^M$ defined by the equation
$$\Pi_{\delta^M} \circ f = f^M \circ \Pi_{\delta^M},$$
In other words, $f^M$ is the unique continuous function making the following diagram commute:
\begin{center}
\begin{tikzpicture}
\draw [fill=white,white] (-1,0) circle (5pt); \node at (-1,0) {$X^M$};
\draw [fill=white,white] (2,0) circle (5pt); \node at (2,0) {$X^M$};
\draw [fill=white,white] (2,2) circle (5pt); \node at (2,2) {$X$};
\draw [fill=white,white] (-1,2) circle (5pt); \node at (-1,2) {$X$};
\draw[->] (-.6,2) -- (1.6,2); \node at (.5,2.3) {$f$};
\draw[->,dashed] (-.6,0) -- (1.6,0); \node at (.5,-.3) {$f^M$};
\draw[->] (2,1.68) -- (2,.3); \node at (2.45,1) {\small $\Pi_{\delta^M}$};
\draw[->] (-1,1.68) -- (-1,.3); \node at (-1.4,1) {\small $\Pi_{\delta^M}$};
\end{tikzpicture}
\end{center}
Any function of this form is called a \emph{metrizable reflection} of $f$.
\item Similarly, if $\psi$ is an $S$-flow on $X$, then the \emph{reflection of $\psi$ in $M$} is denoted $\psi^M$ and is defined by putting $(\psi^M)_p = (\psi_p)^M$ for all $p \in S$ (and, of course, this map will be denoted simply by $\psi^M_p$). Any flow of this form is called a \emph{metrizable reflection} of $\psi$.
\end{itemize}
\end{definition}
This definition is essentially due to Bandlow; see \cite{Bandlow}. It is not immediately clear that the function $f^M$ is well-defined, but this turns out to be a consequence of elementarity, articulated in the following lemma.
\begin{lemma}
Let $X$ be a subspace of $[0,1]^{\omega_1}$ and let $f: X \to X$ be continuous. If $M$ is an elementary submodel of $H$, then
$$\Pi_{\delta^M}(x) = \Pi_{\delta^M}(y) \qquad \text{implies} \qquad \Pi_{\delta^M} \circ f(x) = \Pi_{\delta^M} \circ f(y)$$
for all $x,y \in X$. In other words, the function $f^M: X^M \to X^M$ is well-defined.
\end{lemma}
This lemma was first proved by Noble and Ulmer in \cite{N&U} and later rediscovered by Shchepin in \cite{Shchepin} (neither source phrases it quite this way, but their proofs show this nonetheless).
Notice that if $\psi$ is a flow, then any of its metrizable reflections $\psi^M$ is a quotient of $\psi$, with $\Pi_{\delta^M}$ being the natural quotient mapping.
\subsection{Sequences that almost match a flow.}
A sequence $\seq{x_n}{n \in D}$ of points in a space $X$ is called \emph{tail-dense} if for any finite $F \subseteq D$, the set $\set{x_n}{n \in D \setminus F}$ is dense in $X$.
If $X$ is a metric space and $x,y \in X$, we write $x \approx_\varepsilon y$ to mean $\mathrm{dist}(x,y) < \varepsilon$.
\begin{definition}\label{def:philike}
Let $X$ be a compact metric space and $D$ a countable set.
\begin{itemize}
\item Let $f: X \to X$ be continuous, and let $p: D \to D$ be any function.
\begin{itemize}
\item[$\circ$] A $D$-indexed sequence $\seq{x_n}{n \in D}$ of points in $X$ is said to be \emph{$p$-like up to $\varepsilon$ with respect to $f$}
if
$$f(x_n) \approx_\varepsilon x_{p(n)} \qquad \text{for all but finitely many } n \in D.$$
When $f$ is clear from context, we say simply that the sequence is $p$-like up to $\varepsilon$.
\item[$\circ$] If $\seq{x_n}{n \in D}$ is $p$-like up to $\varepsilon$ for every $\varepsilon > 0$, then we say that $\seq{x_n}{n \in D}$ is \emph{$p$-like (with respect to $f$)}.
\end{itemize}
\item Let $\psi: S \times X \to X$ be an $S$-flow, and let $\varphi: S \times D \to D$ be an action of $S$ on $D$.
\begin{itemize}
\item[$\circ$] A $D$-indexed sequence $\seq{x_n}{n \in D}$ of points in $X$ is said to be \emph{$\varphi$-like up to $\varepsilon$ with respect to $\psi$} if for every $p \in S$, it is $\varphi_p$-like up to $\varepsilon$ with respect to $\psi_p$. When $\psi$ is clear from context, we say simply that the sequence is $p$-like up to $\varepsilon$.
\item[$\circ$] If $\seq{x_n}{n \in D}$ is $\varphi$-like up to $\varepsilon$ for every $\varepsilon > 0$, then we say that $\seq{x_n}{n \in D}$ is $\varphi$-like (with respect to $f$).
\end{itemize}
\end{itemize}
\end{definition}
In other words, $\seq{x_n}{n \in D}$ is $\varphi$-like up to $\varepsilon$ if, on a tail of the sequence, the action of $\psi_p$ on the $x_n$ ``approximately'' (up to an error of $\varepsilon$) matches the action of $\varphi_p$ on their indices. The first part of the definition is illustrated below.
\vspace{2mm}\begin{center}
\begin{tikzpicture}[style=thick, xscale=.67,yscale=1]
\draw (-5,-.6) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,-.6);
\draw (-5,.1) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,.1);
\draw (-5,.8) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,.8);
\draw (-5,1.5) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,1.5);
\draw (-5,2.2) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,2.2);
\draw (-5,-1.3) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (-5,-1.3);
\draw[->] (-4.9,2.1) .. controls (-4.7,1.9) and (-4.7,1.8) .. (-4.9,1.6);
\draw[->] (-4.9,1.4) .. controls (-4.7,1.2) and (-4.7,1.1) .. (-4.9,.9);
\draw[->] (-4.9,.7) .. controls (-4.7,.5) and (-4.7,.4) .. (-4.9,.2);
\draw[->] (-4.9,0) .. controls (-4.7,-.2) and (-4.7,-.3) .. (-4.9,-.5);
\draw[->] (-4.9,-.7) .. controls (-4.7,-.9) and (-4.7,-1) .. (-4.9,-1.2);
\draw (-5,-1.7) node {\small $\vdots$} -- (-5,-1.7);
\draw (-5.3,2.18) node {\scriptsize $1$} -- (-5.3,2.18);
\draw (-5.3,1.48) node {\scriptsize $2$} -- (-5.3,1.48);
\draw (-5.3,.78) node {\scriptsize $3$} -- (-5.3,.78);
\draw (-5.3,.08) node {\scriptsize $4$} -- (-5.3,.08);
\draw (-5.3,-.62) node {\scriptsize $5$} -- (-5.3,-.62);
\draw (-5.3,-1.32) node {\scriptsize $6$} -- (-5.3,-1.32);
\node at (-7.5,1) {\small $D = \mathbb{N}$};
\node at (-7.5,.3) {\footnotesize $s(n) = n+1$};
\draw (-1.5,-2.1) -- (-1.5,2.5) -- (9,2.5) -- (9,-2.1) -- (-1.5,-2.1);
\node at (8,-1.6) {$X$};
\node at (1.5,1.8) {an $s$-like sequence};
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (.4,-1) {};
\node at (.25,-.7) {\small $x_1$};
\draw[->] (.47,-1) .. controls (.9,-.9) and (1,-.8) .. (1.23,-.47);
\node at (1.1,-1) {\scriptsize $f$};
\node[circle, black!30, draw, fill=black!30, inner sep=0pt, minimum width=3pt] at (1.28,-.4) {}
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (2,-.2) {};
\node at (1.85,.1) {\small $x_2$};
\draw[->] (2,-.2) .. controls (2.8,0) and (3.2,0) .. (4,-.2);
\node at (3,.2) {\scriptsize $f$};
\node[circle, black!30, draw, fill=black!30, inner sep=0pt, minimum width=3pt] at (4.11,-.24) {}
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (4.4,-.45) {};
\node at (4.27,-.72) {\small $x_3$};
\draw[->] (4.4,-.45) .. controls (5.2,-.67) and (5.6,-.67) .. (6.4,-.45);
\node at (5.4,-.85) {\scriptsize $f$};
\node[circle, black!30, draw, fill=black!30, inner sep=0pt, minimum width=3pt] at (6.52,-.43) {}
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (6.8,-.4) {};
\node at (7.05,-.7) {\small $x_4$};
\draw[->] (6.8,-.4) .. controls (7.25,.1) and (7.25,.6) .. (6.8,1.1);
\node at (7.4,.35) {\scriptsize $f$};
\node[circle, black!30, draw, fill=black!30, inner sep=0pt, minimum width=3pt] at (6.75,1.18) {}
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (6.82,1.3) {};
\node at (7,1.55) {\small $x_5$};
\draw[->] (6.82,1.3) .. controls (6.25,1.6) and (5.6,1.6) .. (5,1.25);
\node at (5.9,1.73) {\scriptsize $f$};
\node[circle, black!30, draw, fill=black!30, inner sep=0pt, minimum width=3pt] at (4.9,1.19) {}
\node[circle, draw, fill=black, inner sep=0pt, minimum width=4pt] at (5,1.11) {};
\node at (5.5,1.05) {\small $x_6$};
\node at (4,1.02) {$\dots$};
\end{tikzpicture}\end{center}\vspace{2mm}
As the picture indicates, for the successor map $s(n) = n+1$, an $s$-like sequence is simply a sequence $\seq{x_n}{n \in \mathbb{N}}$ of points in $X$ such that $\lim_{n \to \infty} \mathrm{dist}(f(x_n),x_{n+1}) = 0$. Such sequences are sometimes called ``asymptotic pseudo-orbits'' in the topological dynamics literature.
\subsection{The main lemma and the main theorem.}
We are finally in a position to state the full version of the main result of this paper:
\begin{theorem}[the Main Lemma]\label{thm:main}
Let $S$ be a countable discrete semigroup, let $X$ be a compact Hausdorff space of weight $\leq\!\aleph_1$, and let $D$ be a countable set. Let $\varphi$ be a separately finite-to-one action of $S$ on $D$, and let $\psi: S \times X \to X$ be an $S$-flow. The following are equivalent:
\begin{enumerate}
\item $\psi$ is a quotient of $\varphi^*$.
\item Every metrizable quotient of $\psi$ is a quotient of $\varphi^*$.
\item Some metrizable reflection of $\psi$ is a quotient of $\varphi^*$.
\item Every metrizable quotient of $\psi$ contains a tail-dense $\varphi$-like sequence.
\item Some metrizable reflection of $\psi$ contains a tail-dense $\varphi$-like sequence.
\end{enumerate}
\end{theorem}
As we will see in the following two sections, if we are given a flow $\psi$ and an action $\varphi: D \to D$, it is often very easy to decide when $(4)$ holds. Thus this theorem enables us to decide, with relative ease, when a weight-$\aleph_1$ flow is the quotient of a trivial flow on $\omega^*$.
\begin{theorem}[the Main Theorem]\label{thm:main1}
Let $G$ be a countable discrete group, and consider the trivial flow $\varphi^*$ on the space $(G \times \omega)^*$ induced by the natural action $\varphi$ of $G$ on $G \times \omega$, namely $\varphi_g(h,n) = (gh,n)$.
\begin{enumerate}
\item Every $G$-flow of weight $\leq\! \aleph_1$ is a quotient of $\varphi^*$.
\item Assuming \ensuremath{\mathsf{CH}}\xspace, $\varphi^*$ is a universal $G$-flow of weight $\leq\! \mathfrak{c}$.
\end{enumerate}
\end{theorem}
\begin{proof}
By Theorem~\ref{thm:main}, it suffices to show that every metrizable $G$-flow contains a tail-dense $\varphi$-like sequence of points.
Suppose $\psi: G \times X \to X$ is a metrizable $G$-flow. Let $\seq{d_n}{n \in \omega}$ be a tail-dense sequence of points in $X$. Define a sequence $\seq{x_{(g,n)}}{(g,n) \in G \times \omega}$ of points in $X$ by putting
$$x_{(g,n)} = \psi_g(d_n)$$
for all $g \in G$ and $n \in \omega$. This sequence is obviously $\varphi$-like. Note that, if $e$ is the identity element of $G$, $x_{(e,n)} = d_n$ for all $n$; thus, by our choice of the $d_n$, this sequence is tail-dense.
\end{proof}
We will see in the next section that the proof of this result extends to the semigroup $(\mathbb{N},+)$ as well. For the time being, non-discrete groups and uncountable groups both seem to be out of reach.
\begin{question}
Is it consistent to have a universal $(\mathbb{R},+)$-flow of weight $\mathfrak{c}$? Is the natural flow on $(\mathbb{R} \times \omega)^*$ an example under \ensuremath{\mathsf{CH}}\xspace?
\end{question}
\subsection{A few comments on the proof}
By forgetting about groups and actions (i.e., applying the ``forgetful functor'' mapping the category of $G$-flows to the category of compact Hausdorff spaces, as in the proof of Theorem~\ref{thm:noflows}), the Main Theorem reduces to Parovi\v{c}enko's Theorem \cite{Parovicenko}, a classic result of set-theoretic topology:
\begin{itemize}
\item Every compact Hausdorff space of weight $\leq\! \aleph_1$ is a continuous image of $\omega^* = \b\omega \setminus \omega$.
\end{itemize}
One may view Theorem~\ref{thm:main} as the ``dynamic version'' of Parovi\v{c}enko's theorem, its natural analogue in the category of $G$-flows, which identifies (under \ensuremath{\mathsf{CH}}\xspace) a natural universal object for the category. It is natural, then, that the proof of our main theorem shares some features with a proof of Parovi\v{c}enko's theorem.
Of the various proofs of Parovi\v{c}enko's theorem, ours is closest in spirit to that of B{\l}aszczyk and Szyma\'nski in \cite{B&S}. Their proof begins by writing a given compact Hausdorff space $X$ as a length-$\omega_1$ inverse limit of compact metrizable spaces: $X = \varprojlim \seq{X_\a}{\a < \omega_1}$. They then construct a coherent transfinite sequence of continuous surjections $\pi_\a: \omega^* \to X_\a$, and define $\pi: \omega^* \to X$ to be the inverse limit of this sequence. The $\pi_\a$ are constructed recursively, using a variant of the following lifting lemma at successor stages:
\begin{lemma}\label{lem:easylift}
Let $Z$ and $Y$ be compact metric spaces, and let $f: Z \to Y$ be a continuous surjection. If $\pi_Y: \omega^* \to Y$ is a continuous surjection, then it can be lifted to a continuous surjection $\pi_Z: \omega^* \to Z$ such that $\pi_Y = f \circ \pi_Z$.
\end{lemma}
\vspace{-1mm}
\begin{center}
\begin{tikzpicture}
\draw [fill=white,white] (-1,0) circle (5pt); \node at (-1,0) {$Y$};
\draw [fill=white,white] (2,0) circle (5pt); \node at (2,0) {$Z$};
\draw [fill=white,white] (2,2) circle (5pt); \node at (2,2) {$\ \omega^*$};
\draw[->] (1.65,0) -- (-.65,0); \node at (.5,-.25) {$f$};
\draw[->] (1.75,1.8) -- (-.7,.18); \node at (.5,1.3) {$\pi_Y$};
\draw[->,dashed] (2,1.68) -- (2,.3); \node at (2.3,1) {$\pi_Z$};
\end{tikzpicture}
\end{center}
In proving our Main Lemma, the first part of B{\l}aszczyk and Szyma\'nski's proof goes through: every $G$-flow of weight $\aleph_1$ is a length-$\omega_1$ inverse limit of metrizable $G$-flows. However, we run into trouble with the analogue of Lemma~\ref{lem:easylift}: the analogous lemma for flows is false (see Example 3.4 in \cite{Brian}).
The key to getting around this problem is to build our inverse limit system using metrizable reflections. Specifically, suppose $\psi$ is a weight-$\aleph_1$ flow. If $\seq{M_\a}{\a < \omega_1}$ is a continuous increasing chain of countable elementary submodels of $H$, then it gives rise to a sequence $\seq{\psi^{M_\a}}{\a < \omega_1}$ of metrizable flows, and it is clear that
$\psi = \textstyle \varprojlim_{\a < \omega_1}\psi^{M_\a}$.
Furthermore, the elementarity between the models gives this inverse limit system a strong degree of coherence, and ultimately is the key that unlocks a workable substitute for Lemma~\ref{lem:easylift}.
This technique is by now old news. It was first used by Dow and Hart in \cite{Dow&Hart} to prove a ``connected version'' of Parovi\v{c}enko's theorem:
\begin{itemize}
\item Every continuum of weight $\leq\! \aleph_1$ is a continuous image of $\mathbb H^*$, the Stone-\v{C}ech remainder of the half-line $\mathbb H = [0,\infty)$.
\end{itemize}
Their techniques were first adapted to a dynamical setting by the author in \cite{Brian}, where it was proved that a weight $\leq\!\aleph_1$ dynamical system $(X,f)$ is a quotient of $(\omega^*,\sigma)$ if and only if it is weakly incompressible. (Here $\sigma$ denotes the shift map, and $(X,f)$ is weakly incompressible if $f(\closure{U}) \not\subseteq U$ for every open $U$ with $\emptyset \neq U \neq X$.)
We should mentioned that the results in \cite{Brian} follow almost immediately from the Main Lemma (this is proved in the next section), and that the Dow-Hart theorem follows from the results in \cite{Brian} (a short, easy proof is given in Section 5 of \cite{Brian}). Thus the Main Lemma is stronger than the theorems in \cite{Brian} and \cite{Dow&Hart}, but the main idea of the proof is the same.
Several technical difficulties arise when applying this idea to arbitrary flows. The main achievement of the proof here is simply to overcome these difficulties and show that the technique can be applied in this very general setting. But this achievement is primarily technical and therefore not as exciting as some of the applications that motivated it. We will set these technicalities aside for now and instead turn to the applications: finding universal dynamical systems of various kinds, and classifying the universal automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$.
\section{Autohomeomorphisms of $\omega^*$ (and automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$)}\label{sec:dynamics}
\subsection{Dynamical systems and Stone duality}
If $S = (\mathbb{N},+)$, then every continuous $f: X \to X$ generates an $S$-flow, namely $(n,x) \mapsto f^n(x)$. Conversely, every $S$-flow on $X$ is generated in this way. Thus $(\mathbb{N},+)$-flows are really the same thing as continuous self-maps. Similarly, $(\mathbb{Z},+)$-flows are really the same thing as self-homeomorphisms.
\begin{definition}
A \emph{dynamical system} is a continuous function from a compact Hausdorff space to itself.
\end{definition}
The terminology we have employed for flows carries over naturally to dynamical systems, sometimes becoming simpler by virtue of this translation:
\begin{itemize}
\item Given two dynamical systems $f: X \to X$ and $g: Y \to Y$, $f$ is a \emph{quotient} of $g$ if there is a continuous surjection $\pi: Y \to X$ such that $\pi \circ g = f \circ \pi$. This is denoted $g \twoheadrightarrow f$.
\end{itemize}
This is more of a lemma than a definition: our claim is that $g \twoheadrightarrow f$ if and only if the corresponding $(\mathbb{N},+)$-flow generated by $f$ is a quotient of the corresponding $(\mathbb{N},+)$-flow generated by $g$. This is trivial to verify: it amounts to checking that if $\pi \circ g = f \circ \pi$ then $\pi \circ g^n = f^n \circ \pi$ for all $n$.
\begin{itemize}
\item Let $p: D \to D$ and let $\phi$ represent the $(\mathbb{N},+)$-action on $D$ generated by $p$, namely $\phi_n = p^n$. If $f: X \to X$ is a metrizable dynamical system, then a sequence $\seq{x_n}{n \in D}$ of points in $X$ is \emph{$p$-like} if and only if it is $\phi$-like with respect to the flow generated by $f$.
\end{itemize}
Again, this is more of a lemma than a definition. It is easy to verify using the uniform continuity of $f$, and we omit the proof.
We are mainly interested in a specific kind of dynamical system: homeomorphisms from $\omega^*$ to itself. Via Stone duality, results about autohomeomorphisms of $\omega^*$ translate into results about automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$. In this section and the next we investigate some consequences of Theorem~\ref{thm:main} for autohomeomorphisms of $\omega^*$ and automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$. We assume a basic familiarity with Stone duality, and refer the reader to \cite{johnstone} for more details.
A function $p: D \to D$ is a \emph{mod-finite permutation} of $D$ if there are cofinite subsets $A,B \subseteq D$ such that $p$ restricts to a bijection $A \to B$. In this case, $p$ induces an automorphism of $\mathcal{P}(D)/\mathrm{fin} \cong \mathcal{P}(\w)/\mathrm{fin}$, which we denote $p^\uparrow$. Equivalenty, $p^\uparrow$ is the map defined by setting $p^\uparrow([A]) = [p(A)]$ for every $A \subseteq D$. This function is called the \emph{lifting} of $p$, and any automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ obtained in this way is called \emph{trivial}. Trivial automorphisms are dual to trivial autohomeomorphisms, in the sense that
\begin{itemize}
\item[$\circ$] An automorphism $\varphi$ of $\mathcal{P}(\w)/\mathrm{fin}$ is trivial if and only if the dual autohomeomorphism $\varphi^{\mathrm{St}}$ is trivial.
\item[$\circ$] Moreover, if $p$ is a mod-finite permutation of $\omega$ then $(p^\uparrow)^{\mathrm{St}} = p^*$ and $(p^*)^{\mathrm{St}} = p^\uparrow$.
\end{itemize}
Given two automorphisms $\gamma,\varphi$ of $\mathcal{P}(\w)/\mathrm{fin}$, $\varphi$ embeds in $\gamma$ if there is a subalgebra $\AA$ of $\mathcal{P}(\w)/\mathrm{fin}$ and an isomorphism $e: \mathcal{P}(\w)/\mathrm{fin} \to \AA$ such that $e \circ \varphi = \gamma \circ e$. This is denoted $\varphi \hookrightarrow \gamma$. Embeddings of automorphisms are dual to quotients of autohomeomorphisms, in the sense that
\begin{itemize}
\item[$\circ$] If $f$ and $g$ are autohomeomorphisms of $\omega^*$, then $g \twoheadrightarrow f$ if and only if $f^{\mathrm{St}} \hookrightarrow g^{\mathrm{St}}$. Conversely, if $\varphi$ and $\gamma$ are automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$, then $\varphi \hookrightarrow \gamma$ if and only if $\gamma^{\mathrm{St}} \twoheadrightarrow \varphi^{\mathrm{St}}$.
\item[$\circ$] Moreover, if $q$ is a quotient mapping from $g$ to $f$ then $q^{\mathrm{St}}$ is an embedding from $g^{\mathrm{St}}$ to $f^{\mathrm{St}}$, and if $e$ is an embedding from $\varphi$ to $\gamma$ then $e^{\mathrm{St}}$ is a quotient mapping from $\gamma^{\mathrm{St}}$ to $\varphi^{\mathrm{St}}$.
\end{itemize}
\vspace{1mm}
\begin{center}
\begin{tikzpicture}
\node at (0,0) {$\omega^*$};
\node at (2,0) {$\omega^*$};
\node at (2,2) {$\omega^*$};
\node at (0,2) {$\omega^*$};
\draw[->] (.35,2) -- (1.65,2); \node at (1,2.25) {$g$};
\draw[->] (.35,0) -- (1.65,0); \node at (1,-.25) {$f$};
\draw[->>] (2,1.68) -- (2,.3); \node at (2.2,1) {\small $q$};
\draw[->>] (0,1.68) -- (0,.3); \node at (-.2,1) {\small $q$};
\node at (1,-1.25) {\Large $g \twoheadrightarrow f$};
\node at (6,0) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (9,0) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (9,2) {$\mathcal{P}(\w)/\mathrm{fin}$};
\node at (6,2) {$\mathcal{P}(\w)/\mathrm{fin}$};
\draw[->] (8.05,2) -- (6.95,2); \node at (7.5,2.25) {$\gamma$};
\draw[->] (8.05,0) -- (6.95,0); \node at (7.5,-.25) {$\varphi$};
\draw[right hook->] (9,.35) -- (9,1.68); \node at (9.2,1) {\small $e$};
\draw[right hook->] (6,.35) -- (6,1.68); \node at (5.8,1) {\small $e$};
\node at (7.5,-1.25) {\Large $\varphi \hookrightarrow \gamma$};
\draw[double,double equal sign distance,-implies] (3.6,1) -- (4.5,1);
\draw[double,double equal sign distance,-implies] (4.4,1) -- (3.5,1);
\end{tikzpicture}
\end{center}
\subsection{Universal dynamical systems and automorphisms}
An automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ is \emph{universal} if every other automorphism embeds in it. We will call an automorphism universal with respect to some property $P$ if it has property $P$, and every other automorphism with property $P$ embeds in it.
Let $t$ be the permutation of $\omega \times \mathbb{Z}$ given by $t(n,z) = (n,z+1)$.
\vspace{2mm}\begin{center}
\begin{tikzpicture}[style=thin]
\draw (-3,-2) node {\large$t$};
\draw (-1.2,-1.5) node {$\dots$};
\draw[fill=black] (0,-1.5) circle (1.5pt);
\draw[fill=black] (1,-1.5) circle (1.5pt);
\draw[fill=black] (2,-1.5) circle (1.5pt);
\draw[fill=black] (3,-1.5) circle (1.5pt);
\draw[fill=black] (4,-1.5) circle (1.5pt);
\draw (5.2,-1.5) node {$\dots$};
\draw[->] (-.5,-1.5) -- (-.15,-1.5);
\draw[->] (.15,-1.5) -- (.85,-1.5);
\draw[->] (1.15,-1.5) -- (1.85,-1.5);
\draw[->] (2.15,-1.5) -- (2.85,-1.5);
\draw[->] (3.15,-1.5) -- (3.85,-1.5);
\draw[->] (4.15,-1.5) -- (4.5,-1.5);
\draw (-1.2,-2.2) node {$\dots$};
\draw[fill=black] (0,-2.2) circle (1.5pt);
\draw[fill=black] (1,-2.2) circle (1.5pt);
\draw[fill=black] (2,-2.2) circle (1.5pt);
\draw[fill=black] (3,-2.2) circle (1.5pt);
\draw[fill=black] (4,-2.2) circle (1.5pt);
\draw (5.2,-2.2) node {$\dots$};
\draw[->] (-.5,-2.2) -- (-.15,-2.2);
\draw[->] (.15,-2.2) -- (.85,-2.2);
\draw[->] (1.15,-2.2) -- (1.85,-2.2);
\draw[->] (2.15,-2.2) -- (2.85,-2.2);
\draw[->] (3.15,-2.2) -- (3.85,-2.2);
\draw[->] (4.15,-2.2) -- (4.5,-2.2);
\draw (-1.2,-2.9) node {$\dots$};
\draw[fill=black] (0,-2.9) circle (1.5pt);
\draw[fill=black] (1,-2.9) circle (1.5pt);
\draw[fill=black] (2,-2.9) circle (1.5pt);
\draw[fill=black] (3,-2.9) circle (1.5pt);
\draw[fill=black] (4,-2.9) circle (1.5pt);
\draw (5.2,-2.9) node {$\dots$};
\draw[->] (-.5,-2.9) -- (-.15,-2.9);
\draw[->] (.15,-2.9) -- (.85,-2.9);
\draw[->] (1.15,-2.9) -- (1.85,-2.9);
\draw[->] (2.15,-2.9) -- (2.85,-2.9);
\draw[->] (3.15,-2.9) -- (3.85,-2.9);
\draw[->] (4.15,-2.9) -- (4.5,-2.9);
\draw (.3,-3.5) node {$\vdots$};
\draw (2,-3.5) node {$\vdots$};
\draw (3.7,-3.5) node {$\vdots$};
\end{tikzpicture}\end{center}\vspace{2mm}
\begin{theoremdual}\label{thm:main2}
Assuming \ensuremath{\mathsf{CH}}\xspace,
\begin{enumerate}
\item $t^*$ is universal for bijective dynamical systems of weight $\leq\!\mathfrak{c}$.
\item $t^\uparrow$ is a universal automorphism.
\end{enumerate}
\end{theoremdual}
\begin{proof}
$(1)$ follows directly from Theorem~\ref{thm:main1} (Main Theorem 1) by setting $G = (\mathbb{Z},+)$.
$(2)$ follows from $(1)$ via Stone duality.
\end{proof}
Part $(1)$ of this theorem can be strengthened as follows:
\begin{theorem}\label{thm:surjections}
Assuming \ensuremath{\mathsf{CH}}\xspace, $t^*$ is universal for surjective dynamical systems of weight $\leq\!\mathfrak{c}$.
\end{theorem}
\begin{proof}
By Theorem~\ref{thm:main}, it suffices to prove that if $X$ is a compact metric space and $f: X \to X$ a continuous surjection, then $X$ contains a tail-dense $t$-like sequence of points.
Suppose $f: X \to X$ is a continuous surjection, with $X$ metrizable.
Let $\seq{d_n}{n \in \omega}$ be a tail-dense sequence of points in $X$. For all $n \in \mathbb{N}$ and $z \geq 0$, define $x_{n,z} = f^z(d_n)$. For negative $z$, where $f^z(d_n)$ may consist of many points, we define $x_{n,z}$ by recursion. For fixed $n$, use backwards recursion on $z$ to choose a sequence $x_{n,-1},x_{n,-2},x_{n,-3},\dots$ of points in $X$ such that $f(x_{n,z}) = x_{n,z+1}$ for all $z < 0$. The sequence $\seq{x_{n,z}}{n \in \omega, z \in \mathbb{Z}}$ defined in this way is clearly $t$-like. Because each $d_n$ appears in the sequence, it is tail-dense.
\end{proof}
If we wish for something universal for all dynamical systems, not just the surjective ones, then $t^*$ will not work, because every quotient of a surjective function is surjective.
Let $u$ be the map on $\omega \times \omega$ given by $(m,n) \mapsto (m,n+1)$.
\vspace{2mm}
\begin{center}
\begin{tikzpicture}[style=thin]
\draw (-2,-5.5) node {\large$u$};
\draw[fill=black] (0,-5) circle (1.5pt);
\draw[fill=black] (1,-5) circle (1.5pt);
\draw[fill=black] (2,-5) circle (1.5pt);
\draw[fill=black] (3,-5) circle (1.5pt);
\draw[fill=black] (4,-5) circle (1.5pt);
\draw (5.2,-5) node {$\dots$};
\draw[->] (.15,-5) -- (.85,-5);
\draw[->] (1.15,-5) -- (1.85,-5);
\draw[->] (2.15,-5) -- (2.85,-5);
\draw[->] (3.15,-5) -- (3.85,-5);
\draw[->] (4.15,-5) -- (4.5,-5);
\draw[fill=black] (0,-5.7) circle (1.5pt);
\draw[fill=black] (1,-5.7) circle (1.5pt);
\draw[fill=black] (2,-5.7) circle (1.5pt);
\draw[fill=black] (3,-5.7) circle (1.5pt);
\draw[fill=black] (4,-5.7) circle (1.5pt);
\draw (5.2,-5.7) node {$\dots$};
\draw[->] (.15,-5.7) -- (.85,-5.7);
\draw[->] (1.15,-5.7) -- (1.85,-5.7);
\draw[->] (2.15,-5.7) -- (2.85,-5.7);
\draw[->] (3.15,-5.7) -- (3.85,-5.7);
\draw[->] (4.15,-5.7) -- (4.5,-5.7);
\draw[fill=black] (0,-6.4) circle (1.5pt);
\draw[fill=black] (1,-6.4) circle (1.5pt);
\draw[fill=black] (2,-6.4) circle (1.5pt);
\draw[fill=black] (3,-6.4) circle (1.5pt);
\draw[fill=black] (4,-6.4) circle (1.5pt);
\draw[->] (.15,-6.4) -- (.85,-6.4);
\draw[->] (1.15,-6.4) -- (1.85,-6.4);
\draw[->] (2.15,-6.4) -- (2.85,-6.4);
\draw[->] (3.15,-6.4) -- (3.85,-6.4);
\draw[->] (4.15,-6.4) -- (4.5,-6.4);
\draw (5.2,-6.4) node {$\dots$};
\draw (.3,-7) node {$\vdots$};
\draw (2,-7) node {$\vdots$};
\draw (3.7,-7) node {$\vdots$};
\end{tikzpicture}\end{center
\begin{theorem}
Assuming \ensuremath{\mathsf{CH}}\xspace, $u^*$ is universal for dynamical systems of weight $\leq\!\mathfrak{c}$.
\end{theorem}
\begin{proof}
This follows directly from Theorem~\ref{thm:main1} by setting $G = (\mathbb{N},+)$ (see the comments following the proof, where it is pointed out that the theorem holds for $(\mathbb{N},+)$, despite its not being a group).
\end{proof}
Turning back to automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$, it is natural to ask whether $t^\uparrow$ is the only one that is universal. In some sense of the word ``only'' the answer is no, as the following theorem shows.
\begin{theorem}\label{thm:universals}
Assuming \ensuremath{\mathsf{CH}}\xspace, there are $2^\mathfrak{c}$ universal automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$.
\end{theorem}
\begin{proof}
By Theorem 1.2.6 in \cite{JvM}, \ensuremath{\mathsf{CH}}\xspace implies that $\omega^*$ is homeomorphic to $(\omega \times \mathbb{Z} \times 2^\mathfrak{c})^*$, where $2^\mathfrak{c}$ denotes the weight-$\mathfrak{c}$ Cantor cube. We may view $2^\mathfrak{c}$ as a topological group. For each $g \in 2^\mathfrak{c}$, the map $\varphi_g$ defined by
$$\varphi_g(n,z,x) = (n,z+1,gx)$$
induces an autohomeomorphism $\varphi_g^*$ of $(\omega \times \mathbb{Z} \times 2^\mathfrak{c})^*$, and it is easy to see that distinct $g$ give rise to distinct autohomeomorphisms. For each $g \in 2^\mathfrak{c}$, the map $q_g: \omega \times \mathbb{Z} \times 2^\mathfrak{c} \to \omega \times \mathbb{Z}$ defined by $q_g(n,z,x) = (n,z)$ induces a continuous surjection $q_g^*: (\omega \times \mathbb{Z} \times 2^\mathfrak{c})^* \to (\omega \times \mathbb{Z})^*$. For each $g \in 2^\mathfrak{c}$, we have $q_g \circ \varphi_g = t \circ q_g$, and it follows that $q_g^* \circ \varphi_g^* = t^* \circ q_g^*$. Thus $\varphi_g^* \twoheadrightarrow t^*$ for all $g \in 2^\mathfrak{c}$. But the $\twoheadrightarrow$ relation is clearly transitive, so by Theorem~\ref{thm:main2}, each $\varphi_g^*$ is universal for bijective dynamical systems of weight $\leq\!\mathfrak{c}$.
By Stone duality, $\varphi_g^\uparrow$ is a universal automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ for all $g \in 2^\mathfrak{c}$.
\end{proof}
Despite this result, one still wonders whether any two universal automorphisms are ``really'' different. Let us make this notion precise:
\begin{definition}$\ $
\begin{itemize}
\item Two autohomeomorphisms $f$ and $g$ of a space $X$ are \emph{isomorphic} if there is a third autohomeomorphism $h$ such that $h \circ f = g \circ h$.
\item Dually, two automorphisms $\varphi$ and $\gamma$ of $\mathcal{P}(\w)/\mathrm{fin}$ are \emph{isomorphic} if there is a third automorphism $\eta$ such that $\eta \circ \gamma = \varphi \circ \eta$.
\end{itemize}
\end{definition}
\emph{A priori}, the isomorphism class of an automorphism might be as large as $2^\mathfrak{c}$. Thus it is not clear from Theorem~\ref{thm:universals} whether there are any non-isomorphic universal automorphisms. Later in this section, we will characterize the universal trivial automorphisms precisely (under \ensuremath{\mathsf{CH}}\xspace). Then we will see that even among the trivial automorphisms, there are non-isomorphic universal maps.
Of course, there are many automorphisms that are not universal. The identity map is the most obvious example: it is anti-universal, in the sense that no other automorphism embeds in it. After characterizing the universal trivial automorphisms (under \ensuremath{\mathsf{CH}}\xspace), we will see that ``most'' of them (in the sense of Baire category) are not universal. On our way to this classification, we must analyze two other critical mod-finite permutations of $\omega$.
\subsection{Universal automorphisms with additional properties}
In this section we analyze two other mod-finite permutations of $\omega$, and show that their liftings are universal for automorphisms with certain properties.
Let $r$ denote the permutation of $\omega$ consisting of infinitely many disjoint finite cycles, one cycle of size $n!$ for each $n \in \mathbb{N}$. Let $s$ denote the successor map $s(n) = n+1$. The maps $s^*$ and $s^\uparrow$ are both called the \emph{shift map}.
\vspace{2mm}\begin{center}
\begin{tikzpicture}[style=thin]
\draw (-3,2) node {\large$r$};
\draw[fill=black] (0,2) circle (1.5pt);
\draw[fill=black] (1.5,2.3) circle (1.5pt);
\draw[fill=black] (1.5,1.7) circle (1.5pt);
\draw[fill=black] (3,2) circle (1.5pt);
\draw[fill=black] (3.3,2.5) circle (1.5pt);
\draw[fill=black] (3.3,1.5) circle (1.5pt);
\draw[fill=black] (3.9,2.5) circle (1.5pt);
\draw[fill=black] (3.9,1.5) circle (1.5pt);
\draw[fill=black] (4.2,2) circle (1.5pt);
\draw (5.5,2) node {$\dots$};
\draw[->] (.08,2.05) .. controls (.35,2.22) and (.35,1.78) .. (.08,1.95);
\draw[->] (1.6,2.22) .. controls (1.7,2.1) and (1.7,1.9) .. (1.6,1.78);
\draw[->] (1.4,1.78) .. controls (1.3,1.9) and (1.3,2.1) .. (1.4,2.22);
\draw[->] (3,2.1) .. controls (3,2.2) and (3.15,2.4) .. (3.22,2.45);
\draw[->] (3.38,2.56) .. controls (3.5,2.64) and (3.7,2.64) .. (3.83,2.55);
\draw[->] (4,2.46) .. controls (4.05,2.45) and (4.18,2.3) .. (4.2,2.1);
\draw[->] (3.2,1.54) .. controls (3.15,1.55) and (3.02,1.7) .. (3,1.9);
\draw[->] (3.82,1.44) .. controls (3.7,1.36) and (3.5,1.36) .. (3.37,1.45);
\draw[->] (4.2,1.9) .. controls (4.2,1.8) and (4.05,1.6) .. (3.98,1.55);
\draw (-3,.3) node {\large$s$};
\draw[fill=black] (0,.3) circle (1.5pt);
\draw[fill=black] (1,.3) circle (1.5pt);
\draw[fill=black] (2,.3) circle (1.5pt);
\draw[fill=black] (3,.3) circle (1.5pt);
\draw[fill=black] (4,.3) circle (1.5pt);
\draw (5.2,.3) node {$\dots$};
\draw[->] (.15,.3) -- (.85,.3);
\draw[->] (1.15,.3) -- (1.85,.3);
\draw[->] (2.15,.3) -- (2.85,.3);
\draw[->] (3.15,.3) -- (3.85,.3);
\draw (4.15,.3) -- (4.5,.3);
\end{tikzpicture}\end{center}\vspace{2mm}
We adopt the convention that ``$A \subset B$'' means $A$ is a strict subset of $B$.
\begin{definition} Let $X$ be a zero-dimensional compact Hausdorff space, and let $f: X \to X$ be a dynamical system.
\begin{itemize}
\item $f$ is \emph{chain transitive} if for every clopen $U \subseteq X$ with $\emptyset \neq U \neq X$, $f(U) \not\subseteq U$.
\item $f$ is \emph{chain recurrent} if for every clopen $U \subseteq X$, $f(U) \not\subset U$.
\end{itemize}
Dually, if $\varphi$ is an automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ then
\begin{itemize}
\item $\varphi$ is \emph{chain transitive} if for every $a \in \mathcal{P}(\w)/\mathrm{fin}$ with $[\emptyset] \neq a \neq [\omega]$, $\varphi(a) \not\leq a$.
\item $\varphi$ is \emph{chain recurrent} if for every $a \in \mathcal{P}(\w)/\mathrm{fin}$, $\varphi(a) \not< a$.
\end{itemize}
\end{definition}
For now, we have stated the definitions of chain transitive/recurrent dynamical systems for zero-dimensional spaces only. Soon we will extend the definition to all dynamical systems, but first let us state the theorem to be proved concerning $s$ and $r$.
\begin{theorem}\label{thm:dynamics}
Assuming \ensuremath{\mathsf{CH}}\xspace,
\begin{enumerate}
\item $s^*$ is universal for chain transitive dynamical systems of weight $\leq\!\mathfrak{c}$.
\item $r^*$ is universal for chain recurrent dynamical systems of weight $\leq\!\mathfrak{c}$.
\end{enumerate}
By Stone duality, it follows that
\begin{enumerate}
\item $s^\uparrow$ is universal for chain transitive automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$.
\item $r^\uparrow$ is universal for chain recurrent automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$.
\end{enumerate}
\end{theorem}
Theorem~\ref{thm:dynamics}$(1)$ is the main result of \cite{Brian}. The (straightforward) proof in this section therefore shows that the results in \cite{Brian} are a special case of Theorem~\ref{thm:main}.
We will prove Theorem~\ref{thm:dynamics} through a sequence of lemmas.
\begin{lemma}\label{lem:ct1}
$\ $
\begin{enumerate}
\item $s^*$ is chain transitive.
\item $r^*$ is chain recurrent. Moreover, if $\tilde r\!$ is any permutation of $\omega$ consisting only of finite cycles, then $\tilde r^*\!$ is chain recurrent.
\end{enumerate}
\end{lemma}
\begin{proof}
For $(2)$, it is easy to see that $\tilde r(A) \not\subset A$ for any $A \subseteq \omega$. It follows that $\tilde r^*(A^*) \not\subset A^*$ for any $A \subseteq \omega$, so that $\tilde r^*\!$ is chain recurrent. $(1)$ is proved similarly; it also appears as Lemma 5.3 in \cite{WRB}.
\end{proof}
\begin{lemma}\label{lem:ct2}
Both chain transitivity and chain recurrence are preserved by taking quotients.
\end{lemma}
\begin{proof}
This is proved in chapter 4 of \cite{akin}. The result is stated there for metrizable dynamical systems only, but the proof does not use this.
\end{proof}
Observe that these two lemmas suffice already for the ``only if'' direction of Theorem~\ref{thm:dynamics}. For the ``if'' direction, we will need alternative characterizations of chain transitivity and recurrence. Let us expand our ``$\approx_\varepsilon$'' notation for metric spaces to arbitrary spaces as follows. If $X$ is a topological space and $\O$ is an open cover of $X$, we write $x \approx_\O y$ to mean that $x,y \in U$ for some $U \in \O$. Notice that for a metric space, ``$\approx_\varepsilon$'' coincides with ``$\approx_\O$'' when $\O$ is the open cover consisting of all open sets of diameter $\leq \! \varepsilon$.
\begin{definition}
Let $f: X \to X$ be a dynamical system.
If $\O$ is an open cover of $X$, then an $\O$\textit{-chain from $a$ to $b$} is a sequence $\langle x_i : i \leq n \rangle$ of points in $X$ with $x_0 = a$, $x_n = b$, and $n \geq 1$, such that $f(x_i) \approx_{\O} x_{i+1}$ for every $i < n$. If $X$ is metrizable, an \emph{$\varepsilon$-chain} is defined to be an $\O$-chain where $\O$ is the open cover consisting of all open sets of diameter $\leq \! \varepsilon$
\end{definition}
Roughly, an $\O$-chain is an orbit that is computed with small errors at each step, the size of the errors being restricted by $\O$. The following lemma is well-known, at least for metrizable dynamical systems, and a proof can be found in chapter 4 of \cite{akin} (the proofs given there are for metrizable systems, but the reader can check that metrizability is never actually used; see also Lemma 5.2 in \cite{WRB}).
\begin{lemma}\label{lem:chains}
Let $f: X \to X$ be a zero-dimensional dynamical system.
\begin{enumerate}
\item $f$ is {chain transitive} if and only if for every $a,b \in X$ and every open cover $\O$ of $X$, there is an $\O$-chain from $a$ to $b$.
\item $f$ is {chain recurrent} if and only if for every $a \in X$ and every open cover $\O$ of $X$, there is an $\O$-chain from $a$ to $a$.
\end{enumerate}
\end{lemma}
This lemma explains the origin of the terms ``chain transitive'' and ``chain recurrent''. We also take parts $(1)$ and $(2)$ of this lemma as the definition of chain transitivity and chain recurrence, respectively, in the case that $X$ is not zero-dimensional.
\begin{lemma}\label{lem:chainz}
Let $X$ be a metric space and $f: X \to X$ a dynamical system.
\begin{enumerate}
\item $f$ is chain transitive if and only if $X$ contains a tail-dense $s$-like sequence.
\item $f$ is chain recurrent if and only if $X$ contains a tail-dense $r$-like sequence.
\end{enumerate}
\end{lemma}
\begin{proof}
For $(1)$, the forward direction is proved by Bowen in \cite{bowen}. The idea is to fix a tail-dense sequence $\seq{d_n}{n \in \omega}$ of points in $X$, and then to expand this sequence by connecting $d_n$ to $d_{n+1}$ with a $\nicefrac{1}{n}$-chain.
For the converse direction, suppose $(X,f)$ is a metrizable dynamical system and that $\seq{x_n}{n \in \omega}$ is a tail-dense $s$-like sequence of points in $X$. Let $\varepsilon > 0$, and fix $N \in \mathbb{N}$ such that $f(x_n) \approx_\varepsilon x_{n+1}$ for all $n \geq N$. Let $a,b \in X$. Because every tail of $\seq{x_n}{n \in \omega}$ is dense, there is some $m \geq N$ and some $n \geq m$ such that $f(a) \approx_\varepsilon x_m$ and $b \approx_\varepsilon x_n$. Then
$$\<a,x_m,x_{m+1},\dots,x_{n-2},x_{n-1},b\>$$
is an $\varepsilon$-chain from $a$ to $b$.
The proof of $(2)$ is similar, but slightly more invovled. For the forward direction, let $f: X \to X$ be a chain recurrent dynamical system, with $X$ metrizable, and let $\seq{d_k}{k \in \omega}$ be a tail-dense sequence of points in $X$. For every $k \in \mathbb{N}$, fix $n_k \in \mathbb{N}$ such that there is a $\nicefrac{1}{k}$-chain
$$\<d_k,x^k_1,x^k_2,\dots,x^k_{n_k-2},x^k_{n_k-1},d_k\>$$
of length $n_k+1$ from $d_k$ to itself (and set $x^k_0 = d_k$ for convenience). Furthermore, suppose that the function $k \mapsto n_k$ is strictly increasing. This assumption sacrifices no generality, because we can always increase the length of our chain by repeating it if necessary:
$$\<d_k,x^k_1,x^k_2,\dots,x^k_{n_k-2},x^k_{n_k-1},d_k,x^k_1,x^k_2,\dots,x^k_{n_k-2},x^k_{n_k-1},d_k\>.$$
For convenience, we take $D = \bigcup_{n \in \mathbb{N}}\{n\} \times n!$ to be the domain of $r$, with $r(n,m) = (n,m+1)$ and with the addition understood modulo $n!$.
For $n \in \mathbb{N}$ and $m < n!$, define
\[
x_{n,m} =
\begin{cases}
\text{any point} & \text{ if } n < n_1, \\
x^k_{m\,(\text{mod }n_k)} & \text{ if }n_k \leq n < n_{k+1} \text{ and } 0 \leq m < n!.
\end{cases}
\]
In other words, we define our $r$-like sequence by mapping the cycle of length $n!$ onto the $\nicefrac{1}{k}$-chain from $d_k$ to itself whenever $n_k \leq n < n_{k+1}$.
Notice that if $n_k \leq n < n_{k+1}$, then $n_k$ divides $n!$, so that
$$\<x_{n,0},x_{n,1},x_{n,2},\dots,x_{n,n!-2},x_{n,n!-1},x_{n,0}\>$$
is a $\nicefrac{1}{k}$-chain from $d_k$ to itself; it is just the $\nicefrac{1}{k}$-chain we began with, repeated $\nicefrac{n!}{n_k}$ times.
If $\varepsilon > 0$, fix $k$ such that $\nicefrac{1}{k} < \varepsilon$; then $f(x_{n,m}) \approx_\varepsilon x_{n,m+1}$ for all $n \geq n_k$. Thus the sequence $\seq{x_{n,m}}{n \in \mathbb{N},m < n!}$ is $r$-like. It is clear that the sequence is also tail-dense, because it contains all the $d_k$.
For the converse direction, suppose $f: X \to X$ is a dynamical system and $\seq{x_{n,m}}{n \in \mathbb{N}, m < n!}$ is a tail-dense $r$-like sequence of points in $X$. Let $\varepsilon > 0$ and let $a \in X$; we must find an $\varepsilon$-chain from $a$ to itself. Let $\delta > 0$ be small enough that
\begin{itemize}
\item if $x \approx_{\delta} y \approx_{\delta} z$, then $x \approx_{\varepsilon} z$, and
\item if $x \approx_{\delta} y$ and $f(y) \approx_{\delta} z$, then $f(x) \approx_{\varepsilon} z$.
\end{itemize}
Fix $N \in \mathbb{N}$ such that $f(x_{n,m}) \approx_\delta x_{n,m+1}$ for every $n \geq N$, where the addition is understood modulo $n!$. Because every our sequence is tail-dense, there is some $n \geq N$ and $m < n!$ such that $a \approx_{\delta} x_{n,m}$. It follows from our choice of $\delta$ that
$$\<a,x_{n,m+1},x_{n,m+1}, \dots , x_{n,m+n!-1}, a\>$$
is an $\varepsilon$-chain from $a$ to $a$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:dynamics}]
For $(1)$, the ``only if'' part follows from Lemmas \ref{lem:ct1}$(1)$ and \ref{lem:ct2}$(1)$. For the ``if'' part, Lemma~\ref{lem:chainz}$(1)$ shows that every metrizable chain transitive dynamical system contains a tail-dense $s$-like sequence. If $(X,f)$ is chain transitive then so is each of its metrizable quotients by Lemma~\ref{lem:ct1}$(1)$, so the theorem follows from Theorem~\ref{thm:main}. $(2)$ is proved the same way.
\end{proof}
One may easily check that the proof of Theorem~\ref{thm:dynamics} goes through, with only minor modifications, for $s^{-1}$ instead of $s$. Thus $s^\uparrow$ and $(s^\uparrow)^{-1}$ both are universal chain transitive automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$ (under \ensuremath{\mathsf{CH}}\xspace). No other trivial maps are chain transitive, up to re-indexing the domain of $s$ or $s^{-1}$ (this is Lemma 5.5 in \cite{WRB}).
Let us also observe that the proof of Theorem~\ref{thm:dynamics} still works if $r$ is replaced with any permutation $\tilde r\!$ such that
\begin{itemize}
\item $\tilde r\!$ decomposes into a union of finite cycles, and
\item for each $k \in \mathbb{N}$ all but finitely many of periods of these cycles are divisible by $k$.
\end{itemize}
\subsection{Every trivial automorphism embeds in its inverse}
In this subsection, we will break briefly from classifying the universal trivial automorphisms to observe an interesting consequence of Theorem~\ref{thm:dynamics}$(2)$.
\begin{corollary}\label{cor:inverses}
Assuming \ensuremath{\mathsf{CH}}\xspace, $(s^*)^{-1}$ is a quotient of $s^*$.
\end{corollary}
\begin{proof}[Proof]
It is enough to show that $(s^*)^{-1}$ is chain transitive. But we already know that $s^*$ is chain transitive, and in general an invertible map is chain transitive if and only if its inverse is (any two points can still be connected by a chain -- the chains just run in the opposite direction).
\end{proof}
This corollary was first observed in Section 5 of \cite{Brian}, where it was also proved that the assumption of \ensuremath{\mathsf{CH}}\xspace cannot be dropped: under \ensuremath{\mathsf{OCA+MA}}\xspace, the shift map is not a quotient of its inverse (see Theorem 5.7 in \cite{Brian}).
Corollary~\ref{cor:inverses} constitutes some small progress on what seems to be a difficult open question:
\begin{question}\label{q:shiftmap}
Is it consistent that the shift map $s^\uparrow$ is isomorphic to its inverse? Does it follow from \ensuremath{\mathsf{CH}}\xspace?
\end{question}
For more partial progress on this question, see \cite{geschke}. In this subsection, we will see that Corollary~\ref{cor:inverses} extends to every trivial map under \ensuremath{\mathsf{CH}}\xspace, and that the answer to Question~\ref{q:shiftmap} may tell us something about all of them.
\begin{theorem}\label{thm:inverses}
Assuming \ensuremath{\mathsf{CH}}\xspace,
\begin{enumerate}
\item every trivial autohomeomorphism of $\omega^*$ is a quotient of its inverse.
\item every trivial automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ embeds in its inverse.
\end{enumerate}
Furthermore, if \ensuremath{\mathsf{CH}}\xspace holds and the answer to Question~\ref{q:shiftmap} is positive, then
\begin{enumerate}
\item every trivial autohomeomorphism of $\omega^*$ is isomorphic to its inverse.
\item every trivial automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ is isomorphic to its inverse.
\end{enumerate}
\end{theorem}
Before proving this theorem, we will require a bit more terminology.
\begin{definition}
Let $p: \omega \to \omega$ be a mod-finite permutation of $\omega$. $A \subseteq \omega$ is \emph{fixed by $p$} if for all $n$ in the domain of $p$, $p(n) \in A$ if and only if $n \in A$.
\begin{itemize}
\item $A \subseteq \omega$ is a \emph{$\mathbb{Z}$-orbit} if $A$ is fixed by $p$ and if $A = \set{a_z}{z \in \mathbb{Z}}$, where
\begin{itemize}
\item[$\circ$] $a_{n+1} = p(a_n)$ for all $n \in \mathbb{N}$,
\item[$\circ$] all the $a_n$ are distinct.
\end{itemize}
\item $A \subseteq \omega$ is an \emph{$\mathbb{N}$-orbit} if $A$ is fixed by $p$ and if $A = \set{a_n}{n \in \mathbb{N}}$, where
\begin{itemize}
\item[$\circ$] $a_{n+1} = p(a_n)$ for all $n \in \mathbb{N}$,
\item[$\circ$] all the $a_n$ are distinct.
\end{itemize}
\item $A \subseteq \omega$ is a \emph{backwards $\mathbb{N}$-orbit} if it is an $\mathbb{N}$-orbit for the mod-finite permutation $p^{-1}$.
\item $A \subseteq \omega$ is an \emph{$n$-cycle} if $|A| = n$ and $p$ cyclically permutes the elements of $A$. $A$ is called a \emph{finite cycle} if it is an $n$-cycle for some $n \in \mathbb{N}$, and in this case $n$ is called the \emph{period} of the cycle.
\end{itemize}
\end{definition}
\begin{lemma}
For every mod-finite permutation $p$ of $\omega$, there is a mod-finite permutation $q$ of $\omega$ with $p^* = q^*$ and $p^\uparrow = q^\uparrow$, such that $q$ induces a partition of $\omega$ into the four types of orbits described above. Furthermore, only finitely many members of this partition are either $\mathbb{N}$-orbits or backwards $\mathbb{N}$-orbits, and only one of these two types is included at all.
\end{lemma}
To prove the first assertion of this lemma, it suffices to construct a mod-finite permutation $q$ such that $p$ and $q$ differ in only finitely many places and the domain of $q$ can be partitioned into the four types of orbits described above. This is an easy exercise, and we omit the proof. For the second assertion, it suffices to notice that an $\mathbb{N}$-orbit and a backwards $\mathbb{N}$-orbit can be combined into a $\mathbb{Z}$-orbit (by extending the domain of $p$ by one point). In light of this lemma, we may and do assume from now on that every mod-finite permutation induces a partition of $\omega$ into the four types of orbits as described above.
\begin{proof}[Proof of Theorem~\ref{thm:inverses}]
Let $p$ be a mod-finite permutation of $\omega$. Assume that $p$ has $n$ $\mathbb{N}$-orbits and no backwards $\mathbb{N}$-orbits (the proof is nearly identical if we assume it has $n$ backwards $\mathbb{N}$-orbits and no $\mathbb{N}$-orbits). Let $A_0,A_1,\dots,A_{n-1}$ denote the $\mathbb{N}$-orbits of $p$, and let $B = \omega \setminus \bigcup_{i < n}A_i$.
Observe that $p \!\restriction\! B$ is a bijection, and is naturally isomorphic to its inverse (each $\mathbb{Z}$-orbit and each $m$-cycle remains a $\mathbb{Z}$-orbit or an $m$-cycle -- it just runs in the opposite direction under $(p \!\restriction\! B)^{-1}$). Thus $p^* \!\restriction\! B^*$ is isomorphic to $(p^*)^{-1} \!\restriction\! B^*$. By Corollary~\ref{cor:inverses}, $p^* \!\restriction\! A_i^*$ is a quotient of $(p^*)^{-1} \!\restriction\! A_i^*$ for each $i < n$. If the answer to Question~\ref{q:shiftmap} is positive, then $p^* \!\restriction\! A_i^*$ is isomorphic to $(p^*)^{-1} \!\restriction\! A_i^*$ for each $i < n$.
Pasting these quotient mappings (or isomorphisms) together proves part $(1)$ of the theorem, and part $(2)$ follows via Stone duality.
\end{proof}
It is an open question whether any of the results of this section can be extended to non-trivial automorphisms:
\begin{question}
Assuming \ensuremath{\mathsf{CH}}\xspace, does every automorphism of $\mathcal{P}(\w)/\mathrm{fin}$ embed in its inverse (even the nontrivial ones)? Is every automorphism isomorphic to its inverse?
\end{question}
\subsection{Classifying the trivial universal maps}
We mentioned already at the end of Section 3.3 that the proof of Theorem~\ref{thm:dynamics} shows something a little stronger than the statement of the theorem. We will use this stronger statement in this subsection:
\begin{lemma}\label{lem:cycles}
Assuming \ensuremath{\mathsf{CH}}\xspace,
\begin{enumerate}
\item Both $s^*$ and $(s^*)^{-1}$ are universal for chain transitive dynamical systems of weight $\leq\!\mathfrak{c}$.
\item Let $\tilde r\!$ be a permutation of $\omega$ consisting only of finite cycles, and suppose that for every $k \in \mathbb{N}$, all but finitely many of the finite cycles of $\tilde r\!$ have period divisible by $k$. Then
$\tilde r^*$ is universal for chain recurrent dynamical systems of weight $\leq\!\mathfrak{c}$.
\end{enumerate}
\end{lemma}
\begin{definition}
If $p$ is a mod-finite permutation of $\omega$, let us say that $p$ is \emph{pan-divisible} if for every $k \in \mathbb{N}$, all but finitely many of the finite cycles of $p$ have period divisible by $k$.
\end{definition}
For example, $r$, $s$, and $t$ are all pan-divisible: $r$ is pan-divisible because any fixed $k$ divides $n!$ for large enough $n$, and $s$ and $t$ are pan-divisible vacuously, because they have no finite cycles.
\begin{theorem}\label{thm:orbitz}
Assume \ensuremath{\mathsf{CH}}\xspace, and let $p$ be a mod-finite permutation of $\omega$.
The following are equivalent:
\begin{enumerate}
\item $p$ has infinitely many $\mathbb{Z}$-orbits and is pan-divisible.
\item $p^*$ is a universal for surjective dynamical systems of weight $\leq\mathfrak{c}$.
\item $p^\uparrow$ is a universal automorphism of $\mathcal{P}(\w)/\mathrm{fin}$.
\end{enumerate}
\end{theorem}
Observe that one may endow the set of mod-finite permutations of $\omega$ with a natural topology that makes it into a Polish space. As a consequence of this theorem, the set of mod-finite permutations of $\omega$ that induce universal automorphisms is meager. In this sense, ``most'' mod-finite permutations do not give rise to universal automorphisms.
\begin{proof}[Proof of Theorem~\ref{thm:orbitz}]
By applying Stone duality, it is easy to see that $(2) \Rightarrow (3)$. We will prove the theorem by showing $(1) \Rightarrow (2)$ and $(3) \Rightarrow (1)$.
To prove $(1) \Rightarrow (2)$, suppose $p$ is a mod-finite permutation of $\omega$ that has infinitely many $\mathbb{Z}$-orbits and is pan-divisible. Let $A$ denote the union of the $\mathbb{Z}$-orbits of $p$, let $C$ denote the union of the finite cycles of $p$, and let $B = \omega \setminus (A \cup C)$ denote the union of the $\mathbb{N}$-orbits or the backwards $\mathbb{N}$-orbits. If $B \cup C$ is finite, then $p^* = t^*$ and the result follows from Theorem~\ref{thm:surjections}. So let us suppose $B \cup C$ is infinite.
Let $f: X \to X$ be a surjective dynamical system of weight $\leq\!\mathfrak{c}$; we must show $p^* \twoheadrightarrow f$. Roughly, the idea is to use the universality of $p^* \!\restriction\! A^*$ to find a quotient mapping $p^* \!\restriction\! A^* \to f$, and then to extend this mapping to all of $\omega^*$ in a ``harmless'' way.
We will use the following well-known fact about dynamical systems:
\begin{lemma}
There is a closed subspace $Y \subseteq X$ such that $f$ maps $Y$ into itself, and the dynamical system $f \!\restriction\! Y: Y \to Y$ is chain transitive.
\end{lemma}
\begin{proof}[Proof of the lemma:]
A closed subset $Y$ of $X$ is called \emph{minimal} if $Y \neq \emptyset$, $f$ maps $Y$ into itself, and for every $y \in Y$ the orbit $\set{f^z(y)}{z \in \mathbb{Z}}$ of $y$ is dense in $Y$. It is well-known that every dynamical system contains minimal closed sets (hint: apply Zorn's lemma to the poset of all nonempty, closed, $f$-invariant subsets of $X$). Any such set clearly suffices.
\end{proof}
By Theorem~\ref{thm:surjections}, $p^* \!\restriction\! A^*$ is universal for surjective dynamical systems of weight $\leq\!\mathfrak{c}$. Let $q_A: A^* \to X$ be a quotient mapping from $p^* \!\restriction\! A^*$ to $f$.
By Theorem~\ref{thm:dynamics}, $p^* \!\restriction\! B^*$ is universal for chain transitive dynamical systems of weight $\leq\!\mathfrak{c}$. Let $q_B: B^* \to Y$ be a quotient mapping from $p^* \!\restriction\! B^*$ to $f \!\restriction\! Y$.
By Lemma~\ref{lem:cycles}, $p^* \!\restriction\! C^*$ is universal for chain recurrent dynamical systems of weight $\leq\!\mathfrak{c}$. Let $q_C: C^* \to Y$ be a quotient mapping from $p^* \!\restriction\! C^*$ to $f \!\restriction\! Y$.
Pasting these three maps together, we obtain a map $q = q_A \cup q_B \cup q_C$ defined on all of $\omega^*$, and it is clear that $q$ is a quotient mapping from $p^*$ to $f$. This completes the proof of $(1) \Rightarrow (2)$.
To show $(3) \Rightarrow (1)$, we must show two things:
\begin{enumerate}[(a)]
\item if $p^\uparrow$ is universal, then $p$ has infinitely many $\mathbb{Z}$-orbits, and
\item if $p^\uparrow$ is universal, then $p$ is pan-divisible.
\end{enumerate}
For (a), it is easiest to prove the contrapositive of the Stone dual. That is, we will show that if $p$ has only finitely many $\mathbb{Z}$-orbits then $p^*$ is not a universal autohomeomorphism of $\omega^*$.
If $p$ has only finitely many $\mathbb{Z}$-orbits, then (by removing one point in each of the $\mathbb{Z}$ orbits from the domain of $p$) each of them may be decomposed into an $\mathbb{N}$-orbit and a backwards $\mathbb{N}$-orbit. Thus $\omega$ can be decomposed into three sets, $A$, $B$, and $C$, such that $A$ consists of finitely many $\mathbb{N}$-orbits, $B$ consists of finitely many backwards $\mathbb{N}$-orbits, and $C$ consists of finite cycles. So $p^* \!\restriction\! A^*$ is a union of finitely many copies of $s^*$, and it follows from Lemma~\ref{lem:ct1}$(1)$ that $p^* \!\restriction\! A^*$ is chain recurrent. Similarly, $p^* \!\restriction\! B^*$ is chain recurrent. Lastly, $p^* \!\restriction\! C^*$ is chain recurrent as well by Lemma~\ref{lem:ct1}. It follows that $p^*$ is chain recurrent. Moreover, every quotient of $p^*$ is chain recurrent by Lemma~\ref{lem:ct2}. But some autohomeomorphisms of $\omega^*$ are not chain recurrent (for example, the map $t^*$), so this shows $p^*$ is not universal.
For $(2)$, we will prove the contrapositive: if $p$ is not pan-divisible, then $p^\uparrow$ is not universal.
Supposing $p$ is not pan-divisible, there is some $n \in \mathbb{N}$ and an infinite $A \subseteq \omega$ such that $p \!\restriction\! A$ is an infinite union of finite cycles, none of which have period divisible by $n$. Let $c_n$ denote the permutation of $\omega$ that is a disjoint union of infinitely many $n$-cycles. We claim $c_n^\uparrow \not\hookrightarrow p^\uparrow$.
Aiming for a contradiction, suppose $c_n^\uparrow \hookrightarrow p^\uparrow$ and let $\mathbb{B}$ be a subalgebra of $\mathcal{P}(\w)/\mathrm{fin}$ such that $c_n^\uparrow$ is isomorphic to $p^\uparrow \!\restriction\! \mathbb{B}$. Every member of $\mathcal{P}(\w)/\mathrm{fin}$ is periodic under $c_n^\uparrow$ with a period dividing $n$. In other words, $(c_n^\uparrow)^n(a) = a$ for every $a \in \mathcal{P}(\w)/\mathrm{fin}$, and it follows that $(p^\uparrow)^n(b) = b$ for every $b \in \mathbb{B}$. On the other hand, using our assumption about $p$ and $A$, it is easy to check that if $a < [A]$ then $(p^\uparrow)^n(a) \neq a$. From this it follows that for all $b \in \mathbb{B}$, either $b = [A]$ or $b \leq [\omega \setminus A]$. But then $\mathbb{B}$ is not a subalgebra of $\mathcal{P}(\w)/\mathrm{fin}$, and this is the desired contradiction.
\end{proof}
In light of Theorem~\ref{thm:orbitz}, it is now easy to show that not all universal automorphisms are isomorphic. This shows, in particular, that two automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$ may embed in each other without being isomorphic.
\begin{definition}
If $p$ and $q$ are mod-finite permutations of $\omega$, let $p \vee q$ denote the mod-finite permutation of $\omega \times 2$ that acts like $p$ on $\omega \times \{0\}$ and like $q$ on $\omega \times \{1\}$. By reindexing, $p \vee q$ is considered a mod-finite permutation of $\omega$.
\end{definition}
\begin{example}
By Theorem~\ref{thm:orbitz}, if \ensuremath{\mathsf{CH}}\xspace holds then both $t^*$ and $(t \vee r)^*$ are universal autohomeomorphisms of $\omega^*$. However, these autohomeomorphisms are not isomorphic. To see this, consider the following property of an autohomeomorphism $h$:
\begin{itemize}
\item[$(\dagger)$] $h$ is chain recurrent, but its restriction to a clopen subset never yields a chain transitive dynamical system.
\end{itemize}
It is fairly easy to check that
\begin{itemize}
\item with $(t \vee r)^\uparrow$, there is an invariant, clopen subset of $\omega^*$ with property $(\dagger)$, namely the part that is a copy of $r^*$.
\item with $t^\uparrow$, no invariant, clopen subset of $\omega^*$ has property $(\dagger)$.
\end{itemize}
In both cases we leave the details of checking this to the reader. It follows that $(t \vee r)^\uparrow$ and $t^\uparrow$ are not isomorphic.
\end{example}
\subsection{Universal automorphisms under \ensuremath{\mathsf{OCA+MA}}\xspace}
We will end this section with a few observations concerning the structure of the $\hookrightarrow$ and $\twoheadrightarrow$ relations under $\ensuremath{\mathsf{OCA+MA}}\xspace$. This will establish the independence of some of the results proved in the earlier parts of this section.
We observed already in subsection 3.4 that the results there fail under \ensuremath{\mathsf{OCA+MA}}\xspace, which implies that the shift map and its inverse are not quotients of each other. Working backwards, we will next show that the results of subsection 3.3 are independent, and after that we will consider subsection 3.2, and whether there might be universal automorphisms under \ensuremath{\mathsf{OCA+MA}}\xspace.
We do not need to apply either $\ensuremath{\mathsf{OCA}}\xspace$ or $\ensuremath{\mathsf{MA}}\xspace$ directly for any of the results in this section. Instead, we may content ourselves with applying a consequence of $\ensuremath{\mathsf{OCA+MA}}\xspace$, a general version of which was proved by Farah in \cite{farah}. We quote the result without proof:
\begin{theorem}\label{thm:farah}
Assuming \ensuremath{\mathsf{OCA+MA}}\xspace, if $F: \omega^* \to \omega^*$ is continuous, then there is some $A \subseteq \omega$ such that the image of $F \!\restriction\! (\omega \setminus A)^*$ is nowhere dense, and $F \!\restriction\! A^*$ is induced by a finite-to-one function $A \to \omega$.
\end{theorem}
Note that a special case of this theorem is that under \ensuremath{\mathsf{OCA+MA}}\xspace, all automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$ are trivial (this consequence of \ensuremath{\mathsf{OCA+MA}}\xspace was known before Theorem~\ref{thm:farah} and is due to Velickovic \cite{velickovic}; the same result under \ensuremath{\mathsf{PFA}}\xspace is due to Shelah-Stepr\={a}ns \cite{S&S}; consistency was first proved by Shelah \cite{shelah}).
\begin{definition}
Let $p$ be a mod-finite permutation of $\omega$.
\begin{itemize}
\item The \emph{cyclic part} of $p$ is the union of its finite cycles. If $C$ is the cyclic part of $p$, then $p^* \!\restriction\! C^*$ is called the cyclic part of $p^*$.
\item If the cyclic part of $p$ is finite, then $p$ and $p^*$ are called \emph{acyclic}.
\end{itemize}
\end{definition}
\begin{theorem}\label{lem:farah2}
Assuming \ensuremath{\mathsf{OCA+MA}}\xspace, $t^*$ has no cyclilc maps as quotients, and no cyclic map has $s^*$ or $(s^*)^{-1}$ as a quotient. Consequently, $t^*$ is not universal, and $r^*$ is not universal for chain transitive autohomeomorphisms.
\end{theorem}
\begin{proof}
We will prove first that $t^*$ has no cyclic maps as quotients. Suppose $c$ is a cyclic mod-finite permutation of $\omega$, and suppose that $Q: (\omega \times \mathbb{Z})^* \rightarrow \omega^*$ is a quotient mapping. By Theorem~\ref{thm:farah}, there is some $A \subseteq \omega \times \mathbb{Z}$ such that $Q \!\restriction\! A^*$ is induced by a finite-to-one function $q: A \to \omega$, and the image of $Q \!\restriction\! (\omega \times \mathbb{Z} \setminus A)^*$ is nowhere dense.
Because the image of $Q \!\restriction\! (\omega \times \mathbb{Z} \setminus A)^*$ is nowhere dense, $A$ must be infinite and $q(A)$ must be co-finite: otherwise $Q$ could not be a surjection.
Because $q(A)$ is co-finite, we may assume (by removing finitely many points from $A$, if necessary) that $q(A)$ is an infinite union of disjoint cycles, say $q(A) = \bigcup_{n \in \omega}C_n$.
For each $n \in \omega$, fix $a_n \in A$ with $q(a_n) \in C_n$. The set $q^{-1}(C_n)$ is finite, so $t^k(a_n) \notin C_n$ for all sufficiently large $k$. On the other hand, $c^k(q(a_n)) \in C_n$ for all $k$. Thus there is for each $n$ some $k_n \in \mathbb{N}$ such that $q(t^{k_n}(a_n)) \in C_n$ but $q(t^{k_n+1}(a_n)) \notin C_n$. Let $b_n = t^{k_n}(a_n)$ for each $n$, and let $B = \set{b_n}{n \in \mathbb{N}}$.
$B$ is an infinite subset of $A$, and by design, we have $q(t(b)) \neq c(q(b))$ for all $b \in B$. It follows that if $\mathcal{U} \in B^*$ then $q^*(t^*(\mathcal{U})) \neq c^*(q^*(\mathcal{U}))$. This contradicts the supposition that $Q$ is a quotient mapping and establishes $t^* \not\twoheadrightarrow c^*$.
Next we will prove that no cyclic map has $s^*$ as a quotient. Suppose $c$ is a cyclic mod-finite permutation of $\omega$ as before, and suppose that $Q$ is a quotient mapping from $c^*$ to $s^*$. By Theorem~\ref{thm:farah}, there is some $A \subseteq \omega$ such that $Q \!\restriction\! A^*$ is induced by a finite-to-one function $q: A \to \omega$, and the image of $Q \!\restriction\! (\omega \setminus A)^*$ is nowhere dense.
As before, $A$ must be infinite and $q(A)$ must be co-finite, since otherwise $Q$ could not be a surjection.
Thus for all but finitely many $n \in \mathbb{N}$, we may find some $a_n \in A$ such that $q(a_n) = n$. Because $c$ is cyclic, there is some $k > 0$ such that $c^{k}(a_n) = a_n$, although
$$q(c^{k}(a_n)) = q(a_n) = n \neq n+k = q(a_n)+k = s^{k}(q(a_n)).$$
Thus for each $n$, there is some $k_n \geq 0$ such that $q(c^{k_n}(a_n)) = s^{k_n}(q(b))$ but $q(c^{k_n+1}(a_n)) \neq s^{k_n+1}(q(b))$ (for example, we could take $k_n$ to be the least $k$ satisfying the above inequality, minus one). Let $b_n = c^{k_n}(a_n)$, and observe that $q(c(b_n)) \neq s(q(b_n))$. Let $B = \set{b_n}{n \in \mathbb{N}}$.
$B$ is an infinite subset of $A$, and we have $q(c(b)) \neq s(q(b))$ for all $b \in B$. It follows that if $\mathcal{U} \in B^*$ then $q^*(c^*(\mathcal{U})) \neq s^*(q^*(\mathcal{U}))$. This contradicts the supposition that $Q$ is a quotient mapping and establishes $c^* \not\twoheadrightarrow s^*$.
\end{proof}
\begin{theorem}
Assuming \ensuremath{\mathsf{OCA+MA}}\xspace,
\begin{enumerate}
\item there is no universal chain transitive automorphism.
\item there is no universal chain recurrent automorphism.
\end{enumerate}
\end{theorem}
\begin{proof}
For $(1)$, it was proved as Lemma 5.5 in \cite{WRB} that $s^\uparrow$ and its inverse are the only chain transitive trivial automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$. Thus, under \ensuremath{\mathsf{OCA+MA}}\xspace, there are precisely two chain train transitive automorphisms. However, we claim that neither one is a quotient of the other, so neither one is universal. That \ensuremath{\mathsf{OCA+MA}}\xspace implies $(s^\uparrow)^{-1}$ is not a quotient of $s^\uparrow$ was proved (via Theorem~\ref{thm:farah}) as Theorem 5.7 in \cite{Brian}. The proof given there is easily adapted to show that $s^\uparrow$ is not a quotient of $(s^\uparrow)^{-1}$.
To prove $(2)$, we will need a definition and a few claims.
If $p$ is a mod-finite permutation of $\omega$ and has finitely many $\mathbb{Z}$-orbits, then the \emph{index} of $p$, denoted $\delta(p)$, is the number of $\mathbb{N}$-orbits of $p$, plus the number of backwards $\mathbb{N}$-orbits, plus twice the number of $\mathbb{Z}$-orbits. If $p$ has infinitely many $\mathbb{Z}$-orbits, then we set $\delta(p) = \infty$.
\begin{claim}
If $p$ and $q$ are mod-finite permutations of $\omega$ and $p^* \twoheadrightarrow q^*$, then $\delta(q) \leq \delta(p)$.
\end{claim}
\begin{proof}[Proof of claim:]
By removing one point from the domain of $p$ (or $q$), each $\mathbb{Z}$-orbit can be decomposed into an $\mathbb{N}$-orbit and a backwards $\mathbb{N}$-orbit. Thus we may assume that $p$ and $q$ have no $\mathbb{Z}$-orbits, and that $\delta(p)$ (or $\delta(q)$) is equal to the number of $\mathbb{N}$-orbits and backwards $\mathbb{N}$-orbits in $p$ (or in $q$, respectively).
Suppose $\delta(q) > \delta(p)$ and suppose $Q$ is a quotient mapping from $p^*$ to $q^*$.
Let $A$ be an $\mathbb{N}$-orbit or a backwards $\mathbb{N}$-orbit in $q$. Fix $B \subseteq \omega$ such that $B^* = Q^{-1}(A^*)$. Notice that we cannot have $p(B) \setminus B$ infinite (because $Q$ is a quotient mapping and $q^*(B^*) = A^*$). Thus, by modifying $B$ on a finite set if necessary, we may assume $B$ is a union of orbits in $p$. By Lemma~\ref{lem:farah2}, $B$ is not a union of finite cycles. Thus $B$ contains either an $\mathbb{N}$-orbit or a backwards $\mathbb{N}$-orbit of $p$. This establishes an injection from the set of $\mathbb{N}$-orbits and backwards $\mathbb{N}$-orbits of $q$ to the set of $\mathbb{N}$-orbits and backwards $\mathbb{N}$-orbits of $p$, which proves the claim.
\end{proof}
\begin{claim}
If $p$ is a mod-finite permutation of $\omega$, then $p^*$ is chain recurrent if and only if $\delta(p) \neq \infty$.
\end{claim}
\begin{proof}[Proof of claim:]
Let $C$ denote the union of all the finite cycles of $p$, and let $A = \omega \setminus C$ denote the union of its infinite orbits.
If $\delta(p) \neq \infty$, then $p^* \!\restriction\! A^*$ is a finite union of copies of $s^*$ and $(s^*)^{-1}$. Each of these is chain transitive, and $p^* \!\restriction\! C^*$ is chain recurrent by Lemma~\ref{lem:ct1}. Thus $p^*$ is chain recurrent.
On the other hand, if $\delta(p) = \infty$ then $p$ contains infinitely many $\mathbb{Z}$-orbits, and it is easy to check that this implies $p^*$ fails to be chain recurrent.
\end{proof}
From these two claims it follows that there is no universal chain recurrent automorphism. In fact a little more is true: there is no ``jointly universal'' finite family, which means that given any finitely many chain recurrent automorphisms, there is a chain recurrent automorphism (any one with higher index) that is a quotient of none of them.
\end{proof}
Let us now turn to the question of whether $\ensuremath{\mathsf{OCA+MA}}\xspace$ admits a universal automorphism of $\mathcal{P}(\w)/\mathrm{fin}$. We have seen already that $t^\uparrow$ is no longer universal under \ensuremath{\mathsf{OCA+MA}}\xspace. We will prove:
\begin{theorem}\label{thm:ocamauniv}
Assuming \ensuremath{\mathsf{OCA+MA}}\xspace,
\begin{enumerate}
\item an automorphism embeds in $t^\uparrow$ if and only if it is acyclic.
\item an automorphism embeds in $r^\uparrow$ if and only if it is cyclic.
\item every automorphism embeds in either $t^\uparrow$ or $(t \vee r)^\uparrow$.
\end{enumerate}
\end{theorem}
Part $(3)$ of this theorem asserts there is a jointly universal pair of automorphisms under \ensuremath{\mathsf{OCA+MA}}\xspace. We do not know whether this pair can be trimmed down to a single universal automorphism:
\begin{question}\label{q:univ}
$(\ensuremath{\mathsf{OCA+MA}}\xspace)$ Is $(t \vee r)^\uparrow$ a universal automorphism?
\end{question}
If $f: X \to X$ and $g: Y \to Y$ are dynamical systems, then a \emph{subquotient mapping} from $g$ to $f$ is a continuous (but not necessarily surjective) function $q: Y \to X$ such that $q \circ g = f \circ q$.
Let $z$ denote the permutation of $\mathbb{Z}$ mapping $n$ to $n+1$.
\vspace{2mm}\begin{center}
\begin{tikzpicture}[style=thin]
\draw (-3,-1.7) node {};
\draw (-3,-1.3) node {};
\draw (-3,-1.5) node {\large$z$};
\draw (-1.2,-1.5) node {$\dots$};
\draw[fill=black] (0,-1.5) circle (1.5pt);
\draw[fill=black] (1,-1.5) circle (1.5pt);
\draw[fill=black] (2,-1.5) circle (1.5pt);
\draw[fill=black] (3,-1.5) circle (1.5pt);
\draw[fill=black] (4,-1.5) circle (1.5pt);
\draw (5.2,-1.5) node {$\dots$};
\draw[->] (-.5,-1.5) -- (-.15,-1.5);
\draw[->] (.15,-1.5) -- (.85,-1.5);
\draw[->] (1.15,-1.5) -- (1.85,-1.5);
\draw[->] (2.15,-1.5) -- (2.85,-1.5);
\draw[->] (3.15,-1.5) -- (3.85,-1.5);
\draw[->] (4.15,-1.5) -- (4.5,-1.5);
\end{tikzpicture}\end{center}\vspace{2mm}
\begin{lemma}\label{lem:subquotient}
$(\ensuremath{\mathsf{ZFC}}\xspace)$ For every autohomeomorphism $\varphi$ of $\omega^*$, there is a subquotient mapping from $\b z$ to $\varphi$. Consequently, there is a subquotient mapping from $s^*$ to $\varphi$ and there is a subquotient mapping from $(s^*)^{-1}$ to $\varphi$.
\end{lemma}
\begin{proof}
Let $\varphi$ be any autohomeomorphism of $\omega^*$, and let $\mathcal{U}_0$ be any element of $\omega^*$. Then
$$\mathcal{U} \mapsto \U\mbox{-}\!\lim_{n \in \Z} \varphi^n(\mathcal{U}_0)$$
is a subquotient mapping from $\b z$ to $\varphi$. It is clear that $\b z \!\restriction\! \mathbb{N}^*$ is isomorphic to $s^*$ and that $\b z \!\restriction\! (\mathbb{Z} \setminus \mathbb{N})^*$ is isomorphic to $(s^*)^{-1}$, so the ``consequently'' part of the lemma follows.
\end{proof}
\begin{lemma}\label{lem:z}
$(\ensuremath{\mathsf{ZFC}}\xspace)$ Both $s^*$ and $(s^*)^{-1}$ are quotients of $z^*$.
\end{lemma}
\begin{proof}
To find a quotient mapping from $z^*$ to $s^*$, let $q_+$ be an isomorphism from $z^* \!\restriction\! \mathbb{N}^*$ to $s^*$ and let $q_-$ be a subquotient mapping from $z^* \!\restriction\! (\mathbb{Z} \setminus \mathbb{N})^*$ to $s^*$. Pasting these together, $q_+ \cup q_-$ is a quotient mapping from $z^*$ to $s^*$. A quotient mapping from $z^*$ to $(s^*)^{-1}$ can be found in exactly the same way.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:ocamauniv}]
To prove $(1)$, let $p$ be an acyclic mod-finite permutation of $\omega$. First suppose $p$ has infinitely many $\mathbb{Z}$-orbits and $k$ other orbits (either $\mathbb{N}$-orbits or backwards $\mathbb{N}$-orbits). Let $B$ denote the union of these $k$ orbits. Let $A \subseteq \omega$ denote the union of some (any) $k$ orbits of $t$. By Lemma~\ref{lem:z}, there is a quotient mapping from $t^* \!\restriction\! A^*$ to $p^* \!\restriction\! B^*$. Furthermore, $t^* \!\restriction\! (\omega \setminus A)^*$ and $p^* \!\restriction\! (\omega \setminus B)^*$ are isomorphic, since both $t \!\restriction\! (\omega \setminus A)$ and $p \!\restriction\! (\omega \setminus B)$ consist of infinitely many $\mathbb{Z}$-orbits. By pasting together a quotient mapping $t^* \!\restriction\! A^* \to p^* \!\restriction\! B^*$ and an isomorphism $t^* \!\restriction\! (\omega \setminus A)^* \to p^* \!\restriction\! (\omega \setminus B)^*$, we obtain a quotient mapping $t^* \to p^*$.
Next suppose $p$ has only finitely many $\mathbb{Z}$ orbits. Let $k$ denote the total number of orbits of $p$. Let $A$ denote the union of any $k$ of the orbits of $t$, and let $B = \omega \times \mathbb{Z} \setminus A$. Using Lemma~\ref{lem:z}, it is clear that there is a quotient mapping from $t^* \!\restriction\! A^*$ to $p^*$. $B$ consists of infinitely many $\mathbb{Z}$-orbits, and by collapsing them all onto $\mathbb{Z}$ in the natural way we obtain a quotient mapping from $t^* \!\restriction\! B^*$ to $\b z$. Composing this with a subquotient mapping from $\b z$ to $p^*$, we get a subquotient mapping from $t^* \!\restriction\! B^*$ to $p^*$. Pasting this together with a quotient mapping from $t^* \!\restriction\! A^*$ to $p^*$ gives the desired quotient mapping from $t^*$ to $p^*$.
To prove $(2)$, let $c$ be a cyclic permutation of $\omega$. It is easy to find a finite-to-one map $q: \bigcup_{n \in \mathbb{N}}\{n\} \times n! \rightarrow \omega$ such that $q \circ r = c \circ q$. Then $q^*$ is a quotient mapping from $r^*$ to $c^*$.
For $(3)$, let $p$ be a mod-finite permutation of $\omega$. If $p$ is acyclic, then $p^*$ is a quotient of $t^*$ by $(1)$. Suppose $p$ is not acyclic. Let $C$ denote the union of the finite cycles of $p$, and let $A = \omega \setminus C$. If $A$ is infinite, then $p^* \!\restriction\! C^*$ is a quotient of $r^*$ and $p^* \!\restriction\! A^*$ is a quotient of $t^*$, so pasting quotient mappings together gives a quotient mapping from $(t \vee r)^*$ to $p^*$. If $A$ is finite, then $p^*$ is a quotient of $r^*$ and there is a subquotient mapping from $t^*$ to $p^*$ (there is a natural quotient mapping from $t^*$ to $\b z$, and we may compose this with a subquotient mapping from $\b z$ to $p^*$). Again, pasting these mappings together gives a quotient mapping from $(t \vee r)^*$ to $p^*$.
\end{proof}
Note that this argument shows in \ensuremath{\mathsf{ZFC}}\xspace that every trivial automorphism embeds in either $t^\uparrow$ or $(t \vee r)^\uparrow$. \ensuremath{\mathsf{OCA+MA}}\xspace is only used to eliminate the possibility of nontrivial automorphisms.
\begin{question}
$(\ensuremath{\mathsf{ZFC}}\xspace)$ Does every automorphism embed in a trivial one?
\end{question}
If the answer to this question is positive, then, in \ensuremath{\mathsf{ZFC}}\xspace, we have a jointly universal pair of automorphisms of $\mathcal{P}(\w)/\mathrm{fin}$.
\begin{question}\label{q:subquotient}
$(\ensuremath{\mathsf{ZFC}}\xspace)$ Is there a subquotient mapping from $r^*$ to $s^*$?
\end{question}
If the answer to Question~\ref{q:subquotient} is positive, then so is the answer to Question~\ref{q:univ}. By the proof of Theorem~\ref{thm:ocamauniv}$(3)$, $(t \vee r)^\uparrow$ fails to be universal only if there is no quotient mapping from $(t \vee r)^*$ to some acyclic permutation $p^*$. If there is a subquotient mapping from $r^*$ to $s^*$, then this cannot happen (find a quotient mapping from $t^*$ onto $p^*$, and paste this together with a subquotient mapping from $r^*$ to $s^*$ to $p^*$).
If the answer to both of these questions is positive, then it would imply (in \ensuremath{\mathsf{ZFC}}\xspace) that $(t \vee r)^\uparrow$ is a universal automorphism of $\mathcal{P}(\w)/\mathrm{fin}$.
\section{Proof of the main lemma}\label{sec:proof}
This section contains the proof of Theorem~\ref{thm:main}. Before beginning the proof, we will need several definitions and lemmas concerning ultrafilter limits.
\subsection{Limits along an ultrafilter, part I.}
Suppose $X$ is a compact Hausdorff space and $f: D \to X$ is a function. Then there is a unique continuous function $\b f: \b\omega \to X$ that extends $f$, the \emph{Stone extension} of $f$. For a $D$-indexed sequence $\seq{x_n}{n \in D}$ of points in $X$ and $\mathcal{U} \in \b D$, we will write $\U\mbox{-}\!\lim_{n \in D} x_n$ for the image of $\mathcal{U}$ under the Stone extension of the function $n \mapsto x_n$. Equivalently,
$$x = \U\mbox{-}\!\lim_{n \in D} x_n \ \ \Leftrightarrow \ \ \set{n \in D}{x_n \in V} \in \mathcal{U} \text{ for every neighborhood } V \ni x.$$
Thus every ultrafilter $\mathcal{U} \in D^*$ gives rise to an operator $\U\mbox{-}\!\lim_{n \in D}$ on $D$-indexed sequences in compact Hausdorff spaces, which picks out a single limit point of the sequence.
Furthermore, these operators commute with continuous functions:
\begin{lemma}\label{lem:ulims1}
Let $X$ be a compact Hausdorff space, and let $\seq{x_n}{n \in D}$ be a $D$-indexed sequence of points in $X$ for some countable set $D$.
\begin{enumerate}
\item If $f: X \to X$ is continuous and $\mathcal{U} \in D^*$, then
$$\textstyle f \!\left( \U\mbox{-}\!\lim_{n \in D} x_n \right) = \U\mbox{-}\!\lim_{n \in D} f(x_n).$$
\item If $g: D \to D$ is a finite-to-one function and $\mathcal{U} \in D^*$, then
$$\textstyle \U\mbox{-}\!\lim_{n \in D} x_{g(n)} = g^*(\U)\mbox{-}\!\lim_{n \in D} x_n.$$
\end{enumerate}
\end{lemma}
\noindent A proof can be found in Chapter 3 of \cite{H&S}. The same is true for the following lemma.
\begin{lemma}\label{lem:ulims2}
Let $X$ be a compact Hausdorff space, and let $\seq{x_n}{n \in D}$ be a $D$-indexed sequence of points in $X$ for some countable set $D$.
\begin{enumerate}
\item The map $\mathcal{U} \mapsto \U\mbox{-}\!\lim_{n \in D} x_n$ is a continuous function $D^* \to X$.
\item $x \in X$ is in the image of this function if and only if every neighborhood of $x$ contains infinitely many of the $x_n$. In particular, the map $\mathcal{U} \mapsto \U\mbox{-}\!\lim_{n \in D} x_n$ is a surjection if and only if every open subset of $X$ contains infinitely many of the $x_n$.
\end{enumerate}
\end{lemma}
Lemma~\ref{lem:ulims2} tells us how $\mathcal{U}$-limits can be used to define continuous surjections with domain $D^*$. To prove the main lemma, we need to develop a generalization of Lemma~\ref{lem:ulims2} that tells us when $f(\U\mbox{-}\!\lim_{n \in D} y_n) = \U\mbox{-}\!\lim_{n \in D} z_n$, even if $f$ is not defined on any of the $y_n$. In the proof of the main lemma, we will have functions $\psi_p$ defined on a subspace $X$ of $[0,1]^{\omega_1}$, but with none of the $\psi_p$ defined on $[0,1]^{\omega_1} \setminus X$. In that setting, we will need to know how $\psi_p(\U\mbox{-}\!\lim_{n \in D} y_n)$ behaves, even if some (or all) of the $y_n$ are in $[0,1]^{\omega_1} \setminus X$.
\begin{definition}
Let $Y$ be a topological space and let $D$ be a countable set.
Two $D$-indexed sequences $\seq{x_n}{n \in D}$ and $\seq{y_n}{n \in D}$ of points in $Y$ are \emph{tail-similar} if, for every open cover $\O$ of $Y$, $x_n \approx_\O y_n$ for all but finitely many $n \in D$.
\end{definition}
\begin{lemma}\label{lem:ulims3}
Let $Y$ be a compact Hausdorff space, let $D$ be a countable set, and let $\seq{x_n}{n \in D}$ and $\seq{y_n}{n \in D}$ be $D$-indexed sequences of points in $Y$.
\begin{enumerate}
\item $x = \U\mbox{-}\!\lim_{n \in D} x_n$ if and only if $\set{n}{x \approx_\O x_n} \in \mathcal{U}$ for every open cover $\O$ of $Y$.
\item $\U\mbox{-}\!\lim_{n \in D} x_n = \U\mbox{-}\!\lim_{n \in D} y_n$ if and only if $\set{n}{x_n \approx_\O y_n} \in \mathcal{U}$ for every open cover $\O$ of $Y$.
\item $\U\mbox{-}\!\lim_{n \in D} x_n = \U\mbox{-}\!\lim_{n \in D} y_n$ for all $\mathcal{U} \in D^*$ if and only if the sequences $\seq{x_n}{n \in D}$ and $\seq{y_n}{n \in D}$ are tail-similar.
\end{enumerate}
\end{lemma}
\begin{proof}
For $(1)$, the ``if'' part follows easily from the definition of a $\mathcal{U}$-limit. For the ``only if'' part, suppose $\set{n \in D}{x \approx_\O x_n} \notin \mathcal{U}$ for some open cover $\O$. Let $V$ be any member of $\O$ containing $x$. Then $\set{n \in D}{x_n \in V} \notin \mathcal{U}$, which yields $\set{n \in D}{x_n \in Y \setminus V} \in \mathcal{U}$. By the definition of the $\mathcal{U}$-limit, this implies $\U\mbox{-}\!\lim_{n \in D} x_n \in Y \setminus V$, so that $\U\mbox{-}\!\lim_{n \in D} x_n \neq x$, completing the proof of $(1)$.
$(2)$ follows easily from $(1)$ and $(3)$ follows easily from $(2)$.
\end{proof}
\begin{definition}
Let $X$ be a closed subset of a compact Hausdorff space $Y$, and $f: X \to X$.
If $\O$ is an open cover of $Y$ and $y,z \in Y$, then we write $y \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z$
to mean that there is some $x \in X$ with $x \approx_\O y$ and $f(x) \approx_\O z$.
\end{definition}
\noindent The symbol $y \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z$ is read ``$f$ maps $y$ to $z$ modulo $\O$" and can be thought of as asserting that the expression ``$f(y)=z$'', though it may formally be mere nonsense, is approximately correct.
\vspace{2mm}\begin{center}
\begin{tikzpicture}[style=thick, xscale=1,yscale=.8]
\draw (-1.5,-2) -- (-1.5,3) -- (9,3) -- (9,-2) -- (-1.5,-2);
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\draw[dotted,fill=black!5] (6.5,.8) circle (.75);
\draw[line width=.75mm] (0,-1) .. controls (3,.5) and (5,1) .. (8,1);
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\draw (1.43,.87) node {\small $y$} -- (1.43,.87);
\draw (6.7,.23) node [circle, draw, fill=black, inner sep=0pt, minimum width=3pt] {} -- (6.7,.23);
\draw (6.51,.32) node {\small $z$} -- (6.51,.32);
\draw (3.5,1.4) node {$f$} -- (3.5,1.4);
\draw[->] (2.3,.2) .. controls (3.3,1.25) and (4.7,1.45) .. (6,1.05);
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\draw (6.5,1.2) node {\scriptsize $f(x)$} -- (6.5,1.2);
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\draw (-.5,2.2) node {$Y$} -- (-.5,2.2);
\draw (4.6,-.7) node {\Large $y \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z$} -- (4.6,-.7);
\end{tikzpicture}\end{center}\vspace{2mm}
\begin{lemma}\label{lem:ulims4}
Let $X$ be a closed subspace of a compact Hausdorff space $Y$, and let $D$ be a countable set. Let $f: X \to X$ be continuous and let $\seq{y_n}{n \in D}$ and $\seq{z_n}{n \in D}$ be $D$-indexed sequences of points in $Y$. The following are equivalent:
\begin{enumerate}
\item For every open cover $\O$ of $Y$, $y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n$ for all but finitely many $n \in D$.
\item For every $\mathcal{U} \in D^*$, $\U\mbox{-}\!\lim_{n \in D} y_n$ and $\U\mbox{-}\!\lim_{n \in D} z_n$ are both in $X$, and $$\textstyle f(\U\mbox{-}\!\lim_{n \in D} y_n) = \U\mbox{-}\!\lim_{n \in D} z_n.$$
\end{enumerate}
\end{lemma}
\begin{proof}
To prove $(1)$ implies $(2)$, first note that if $(1)$ holds, then the definition of a $\mathcal{U}$-limit implies (using the fact that $\mathcal{U}$ is non-principal) that both $\U\mbox{-}\!\lim_{n \in D} y_n$ and $\U\mbox{-}\!\lim_{n \in D} z_n$ are in $X$. Let $y = \U\mbox{-}\!\lim_{n \in D} y_n$ and $z = f(y)$, and let $V$ be an open neighborhood of $z$ (in $Y$). We need to show that $\set{n}{z_n \in V} \in \mathcal{U}$; then, because $V$ is arbitrary, we will know $\U\mbox{-}\!\lim_{n \in D} z_n = z$ by definition.
To show $\set{n}{z_n \in V} \in \mathcal{U}$, we make use of Lemma 3.1 from \cite{Brian}, a special case of which states:
\begin{itemize}
\item There is an open cover $\O$ of $Y$ such that if $y',z' \in Y$ with $y' \approx_\O y$ and $y' \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z'$, then $z' \in V$.
\end{itemize}
(The lemma in \cite{Brian} is stated only for $Y = [0,1]^\delta$, but one can easily check that the proof does not depend on this.) Heuristically, this lemma just expresses the continuity of $f$, which requires that points near $y$ map to points near $z$.
Let $\O$ be an open cover of $Y$ as described above. By the definition of a $\mathcal{U}$-limit, we have
$\set{n \in D}{y_n \approx_\O y} \in \mathcal{U}.$
Because $y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n$ for all but finitely many $n \in D$, and because $\mathcal{U}$ is a non-principal ultrafilter on $D$,
$$\set{n \in D}{y_n \approx_\O y \text{ and } y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n} \in \mathcal{U}.$$
By our choice of $\O$,
$$\set{n \in D}{z_n \in V} \supseteq \set{n \in D}{y_n \approx_\O y \text{ and } y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n}$$
so that $\set{n}{z_n \in V} \in \mathcal{U}$ as required.
For the converse, suppose that $\U\mbox{-}\!\lim_{n \in D} y_n$ and $\U\mbox{-}\!\lim_{n \in D} z_n$ are both in $X$ for every $\mathcal{U} \in D^*$, but $(1)$ does not hold. We will show that this implies $\textstyle f(\U\mbox{-}\!\lim_{n \in D} y_n) \neq \U\mbox{-}\!\lim_{n \in D} z_n$ for some $\mathcal{U} \in D^*$.
If $(1)$ fails, then there is some open cover $\O$ of $Y$ and some infinite $A \subseteq D$ such that $\neg (y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n)$ for all $n \in A$. Fix $\mathcal{U} \in A^*$ and observe that, by the definition of a $\mathcal{U}$-limit,
$$B = \set{n \in D}{\textstyle y_n \approx_\O \U\mbox{-}\!\lim_{n \in D} y_n} \in \mathcal{U}.$$
If $n \in B$ and $z_n \approx_\O f(\U\mbox{-}\!\lim_{n \in D} y_n)$, then $y_n \xmapsto{\ \!_{\,\!_{f}}\,}_{\!_{\O}} z_n$. Thus, for $n \in B \cap A$, we have $z_n \not\approx_\O f(\U\mbox{-}\!\lim_{n \in D} y_n)$. But $B \cap A \in \mathcal{U}$, so this implies $\U\mbox{-}\!\lim_{n \in D} z_n \neq f(\U\mbox{-}\!\lim_{n \in D} y_n)$ as desired.
\end{proof}
\subsection{$\varphi$-like sequences revisited.}
In this subsection we extend the notion of $\varphi$-like sequences to a broader context, where $X$ is no longer required to be metrizable, and where the sequence of points may lie in some space $Y \supseteq X$.
\begin{definition}\label{def:philike}
Let $X$ be a closed subset of a compact Hausdorff space $Y$, and let $D$ be a countable set.
\begin{itemize}
\item Let $f: X \to X$ be continuous, and let $p: D \to D$ be any function.
\begin{itemize}
\item[$\circ$] A $D$-indexed sequence $\seq{y_n}{n \in D}$ of points in $Y$ is said to be \emph{$p$-like for $\O$ with respect to $f$}, where $\O$ is an open cover of $Y$, if
$$y_n \xmapsto{\ \!_{\,\!_{\psi_p}}\,}_{\!_{\O}} y_{\varphi(n)} \qquad \text{for all but finitely many }n \in D.$$
When $f$ is clear from context, we say simply that the sequence is $p$-like for $\O$.
\item[$\circ$] If $\seq{y_n}{n \in D}$ is $p$-like for $\O$ for every open cover $\O$ of $Y$, then we say that $\seq{y_n}{n \in D}$ is \emph{$p$-lilke (with respect to $f$)}.
\end{itemize}
\item Let $\psi: S \times X \to X$ be an $S$-flow, and let $\varphi: S \times D \to D$ be an action of $S$ on $D$.
\begin{itemize}
\item[$\circ$] A $D$-indexed sequence $\seq{y_n}{n \in D}$ of points in $Y$ is said to be \emph{$\varphi$-like for $\O$ with respect to $\psi$}, where $\O$ is an open cover of $Y$, if for every $p \in S$ it is $\varphi_p$-like for $\O$ with respect to $\psi_p$.
When $\psi$ is clear from context, we say simply that the sequence is $\varphi$-like for $\O$.
\item[$\circ$] If $\seq{y_n}{n \in D}$ is $\varphi$-like for $\O$ for every open cover $\O$ of $Y$, then we say that $\seq{y_n}{n \in D}$ is {$\varphi$-lilke (with respect to $f$)}.
\end{itemize}
\end{itemize}
\end{definition}
Observe that this new definition contains the old one as a special case, namely when $X=Y$ and $X$ is metrizable.
\vspace{1mm}
\begin{center}
\begin{tikzpicture}[style=thick, xscale=.67,yscale=1]
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\draw (-5,-1.7) node {\small $\vdots$} -- (-5,-1.7);
\draw (-5.3,2.18) node {\scriptsize $1$} -- (-5.3,2.18);
\draw (-5.3,1.48) node {\scriptsize $2$} -- (-5.3,1.48);
\draw (-5.3,.78) node {\scriptsize $3$} -- (-5.3,.78);
\draw (-5.3,.08) node {\scriptsize $4$} -- (-5.3,.08);
\draw (-5.3,-.62) node {\scriptsize $5$} -- (-5.3,-.62);
\draw (-5.3,-1.32) node {\scriptsize $6$} -- (-5.3,-1.32);
\node at (-7.5,1) {\small $D = \mathbb{N}$};
\node at (-7.5,.3) {\footnotesize $s(n) = n+1$};
\draw (-1.5,-2.1) -- (-1.5,2.5) -- (9,2.5) -- (9,-2.1) -- (-1.5,-2.1);
\node at (8,-1.6) {$Y$};
\node at (1.5,1.8) {an $s$-like sequence};
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\draw (-.25,-1) node {$X$} -- (-.25,-1);
\draw[->] (1.46,-.35) .. controls (1.77,-.5) and (2.07,-.5) .. (2.17,-.1);
\draw (1.92,-.6) node {\scriptsize $f$} -- (1.92,-.6);
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\node at (1,.51) {\scriptsize $x_1$};
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\node at (2.2,.61) {\scriptsize $x_2$};
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\node at (6.7,1.23) {\scriptsize $x_6$};
\node at (7.5,1.25) {\small $\dots$};
\end{tikzpicture}\end{center}\vspace{1mm}
\begin{lemma}\label{lem:psilike}
Let $X$ be a closed subspace of some compact Hausdorff space $Y$, let $\psi$ be an $S$-flow on $X$, and let $\seq{x_n}{n \in D}$ be a $D$-indexed sequence of points in $Y$ for some countable set $D$. If $\varphi$ is an action of $S$ on $D$, then
\begin{enumerate}
\item[$(\dagger)$] The map $\pi: D^* \to Y$ defined by $\pi(\mathcal{U}) = \U\mbox{-}\!\lim_{n \in D} x_n$ is a quotient mapping from $\varphi^*$ to $\psi$ if and only if
\begin{enumerate}
\item $\pi[D^*] = X$, and
\item $\seq{x_n}{n \in D}$ is $\varphi$-like.
\end{enumerate}
\end{enumerate}
\end{lemma}
\begin{proof}
We will prove the ``if'' direction of $(\dagger)$ first. Using $(a)$ and Lemma~\ref{lem:ulims2}, $\pi$ is a continuous surjection from $D^*$ onto $X$. We need to show that $\pi$ preserves the action of $S$, in the sense that $\pi \circ \varphi^*_p = \psi_p \circ \pi$ for all $p \in S$. This is a direct application of Lemma~\ref{lem:ulims4}:
\begin{align*}
\psi_p \circ \pi(\mathcal{U}) & = \psi_p(\pi(\mathcal{U})) = \psi_p(\textstyle \U\mbox{-}\!\lim_{n \in D} x_n) \\
& = \U\mbox{-}\!\lim_{n \in D} x_{\varphi_p(n)} = \varphi^*_g(\U)\mbox{-}\!\lim_{n \in D} x_n \\
& = \pi(\varphi^*_p(\mathcal{U})) = \pi \circ \varphi^*_p(\mathcal{U})
\end{align*}
(Lemma~\ref{lem:ulims4} was applied to get the third equality, and Lemma~\ref{lem:ulims1}(2) to get the fourth).
For the ``only if'' direction, suppose $\pi$ is a quotient mapping and let us prove that (b) holds (note that (a) holds by the definition of a quotient mapping). Aiming for a contradiction, suppose $\seq{x_n}{n \in D}$ is not $\varphi$-like. Fix $p \in S$, an open cover $\O$ of $Y$, and an infinite $A \subseteq D$ such that, for every $n \in A$, it is false that $x_n \xmapsto{\ \!_{\,\!_{\psi_p}}\,}_{\!_{\O}} x_{\varphi(n)}$; that is, if $n \in A$ and $x_n \approx_{\O} x \in X$, then $\psi_p(x) \not\approx_{\O} x_{\varphi(n)}$.
Let $\mathcal{U} \in D^*$ with $A \in \mathcal{U}$, and let $x = \U\mbox{-}\!\lim_{n \in D} x_n$. Fix $O \in \O$ with $x \in O$, set $B = \set{n \in A}{x_n \in O}$, and observe that $B \in \mathcal{U}$. Now $x_n \approx_\O x$ for all $n \in B$, and by our choice of $\O$ this means $\psi_p(x) \not\approx_\O x_{\varphi(n)}$ for all $n \in B$. In particular, if $\psi_g(x) \in V \in \O$ then $\set{n \in B}{x_{\varphi_g(n)} \in V} \subseteq \mathbb{N} \setminus B$. Thus $\set{n \in B}{x_{\varphi_p(n)} \in V} \notin \mathcal{U}$ and $\U\mbox{-}\!\lim_{n \in D} x_{\varphi_p(n)} \neq \psi_p(x)$. Hence
$$\pi(\varphi^*p(\mathcal{U})) = \U\mbox{-}\!\lim_{n \in D} x_{\varphi_p(n)} \neq \psi_p(x) = \psi_p(\pi(\mathcal{U}))$$
so that $\pi$ is not a quotient mapping.
\end{proof}
\subsection{The proof}
We are finally ready to prove the Main Lemma. We reproduce the statement of the lemma here for convenience:
\begin{theorem34}
Let $S$ be a countable discrete semigroup, let $X$ be a compact Hausdorff space of weight $\leq\!\aleph_1$, and let $D$ be a countable set. Let $\varphi$ be a separately finite-to-one action of $S$ on $D$, and let $\psi: S \times X \to X$ be an $S$-flow. The following are equivalent:
\begin{enumerate}
\item $\psi$ is a quotient of $\varphi^*$.
\item Every metrizable quotient of $\psi$ is a quotient of $\varphi^*$.
\item Some metrizable reflection of $\psi$ is a quotient of $\varphi^*$.
\item Every metrizable quotient of $\psi$ contains a tail-dense $\varphi$-like sequence.
\item Some metrizable reflection of $\psi$ contains a tail-dense $\varphi$-like sequence.
\end{enumerate}
\end{theorem34}
\begin{proof}
The structure of the proof is as follows. The implications $(3) \Rightarrow (2)$ and $(5) \Rightarrow (4)$ are trivial. We will first prove that $(1) \Rightarrow (3)$, and then prove a lemma showing that $(2) \Leftrightarrow (4)$ and $(3) \Leftrightarrow (5)$. Finally, we will show $(4) \Rightarrow (1)$ to finish the proof. Showing $(4) \Rightarrow (1)$ is the longest and most difficult part of the proof.
To show that $(1)$ implies $(3)$, it is enough to show that every metrizable reflection of $\psi$ is a quotient of $\psi$, because the composition of two quotient mappings is again a quotient mapping. But this is obvious: if $M$ is a countable elementary submodel of $H$, so that $\psi^M$ is a metrizable reflection of $\psi$, then the natural projection $\Pi_{\delta^M}: X \to X^M$ is a quotient mapping.
The following lemma immediately implies that $(2) \Leftrightarrow (4)$ and $(3) \Leftrightarrow (5)$.
\begin{lemma}
Suppose $Z$ is a compact metric space and $\mu: G \times Z \to Z$ is a $G$-flow. Let $\varphi: G \times D \to D$ be an action of $G$ on a countable set $D$. Then $\mu$ is a quotient of $\varphi^*$ if and only if $Z$ contains a tail-dense $\varphi$-like sequence.
\end{lemma}
\begin{proof}
In order to prove our lemma, we will need the following folklore result:
\vspace{2mm}
\noindent \emph{Fact: } If $F: D^* \to Z$ is a continuous function, then there is a $D$-indexed sequence $\seq{z_n}{n \in D}$ of points in $Z$ such that $F(\mathcal{U}) = \U\mbox{-}\!\lim_{n \in D} z_n$ for all $\mathcal{U} \in D^*$.
\vspace{2mm}
\begin{proof}[Proof of fact:] This fact is well-known in the case $Z = [0,1]$ (indeed, some authors take this as the defining property of $D^*$). One way to prove this fact in general is to apply Urysohn's theorem, which allows us to view $Z$ (up to homeomorphism) as a closed subspace of $[0,1]^\omega$. For each $m \in \omega$, the map $\pi_m \circ F$ is a continuous function $D^* \to [0,1]$, so the special case of the fact allows us to find a $D$-indexed sequence $\seq{s_n^m}{n \in D}$ of points in $Z$ such that $\pi_m \circ F(\mathcal{U}) = \U\mbox{-}\!\lim_{n \in D} s_n^m$ for all $\mathcal{U} \in D^*$. Taking $s_n = \Delta_{n \in \omega}s^m_n$ now gives us a $D$-indexed sequence $\seq{s_n}{n \in D}$ of points in $[0,1]^\omega$ (but not necessarily in $Z$) such that $F(\mathcal{U}) = \U\mbox{-}\!\lim_{n \in D} s_n$ for all $\mathcal{U} \in D^*$. Finally, we may get $z_n$ from $s_n$ by using the metrizability of $[0,1]^\omega$. Define $z_n$ to be any point of $Z$ such that $\mathrm{dist}(z_n,s_n) = \mathrm{dist}(Z,s_n)$ (such a point exists because $Z$ is closed). One easily shows that the sequences $\seq{z_n}{n \in D}$ and $\seq{s_n}{n \in D}$ are tail-similar, so that Lemma~\ref{lem:ulims3}(3) completes the proof.
\end{proof}
Returning to the proof of the lemma, suppose $\pi: D^* \to Z$ is a quotient mapping from $\varphi^*$ to $\mu$. Using the above fact, there is a $D$-indexed sequence $\seq{z_n}{n \in D}$ of points in $Z$ such that $\pi(\mathcal{U}) = \U\mbox{-}\!\lim_{n \in D} z_n$ for all $\mathcal{U} \in D^*$. By Lemma~\ref{lem:psilike}, the sequence $\seq{z_n}{n \in D}$ is $\varphi$-like, which proves the forward direction. Conversely, if a sequence $\seq{z_n}{n \in D}$ of points in $Z$ is $\varphi$-like, then Lemma~\ref{lem:psilike} states that the function $\mathcal{U} \mapsto \U\mbox{-}\!\lim_{n \in D} z_n$ is a quotient mapping from $\varphi^*$ to $\mu$.
\end{proof}
\vspace{3mm}
It remains to show that $(4)$ implies $(1)$. Let $M$ be a countable elementary submodel of $H$ with $X, \psi \in M$, so that $\psi^M$ is a metrizable reflection of $\psi$, and suppose that $X^M$ contains a $\varphi$-like sequence of points. By Lemma~\ref{lem:psilike}, it suffices to construct a $\varphi$-like sequence of points in $[0,1]^{\omega_1}$; this is what we shall do.
We will construct a map $q_\xi: D \to [0,1]$ for every $\xi < \omega_1$ via a length-$\omega_1$ transfinite recursion. In the end, the diagonal mapping $Q = \Delta_{\xi < \omega_1}q_\xi$ will define a $\varphi$-like sequence $\seq{Q(n)}{n \in \omega}$ in $[0,1]^{\omega_1}$.
Before beginning this construction, we will need a little more terminology. Let us say that $\O$ is a \emph{nice} open cover for $[0,1]^{\omega_1}$ if it is an open cover consisting of finitely many sets of the form
$$\pi_{\a_1}^{-1}[I_1] \cap \pi_{\a_2}^{-1}[I_2] \cap \dots \cap \pi_{\a_k}^{-1}[I_k]$$
where each $\a_j$ is an ordinal $<\! \omega_1$ and each $I_j$ is an open interval in $[0,1]$ with rational endpoints.
Observe that the nice open covers suffice to describe the topology of $[0,1]^{\omega_1}$ (in the sense that every open cover of $[0,1]^{\omega_1}$ is refined by a nice open cover). This proves the following simple but useful observation:
\begin{observation1} \label{ob:nice}
If a sequence $\seq{x_n}{n \in D}$ of points in $[0,1]^{\omega_1}$ is $\varphi$-like for every nice open cover, then it is $\varphi$-like.
\end{observation1}
Also observe that every nice open cover $\O$ is (uniformly) definable from a finite list of ordinals. For example, the open set above is definable from $\a_1,\a_2,\dots,\a_k$ (we do not need to mention the intervals $I_1,I_2,\dots,I_k$ in a definition of this set, because the intervals have rational endpoints, and rational numbers are definable without parameters), and $\O$ consists of finitely many open sets like this one. We say that a nice open cover is defined using ordinals $<\!\delta$ if each ordinal $\a_i$ appearing as above in the definition of some $U \in \O$ is $<\!\delta$.
Finally, observe that whether a sequence is $\varphi$-like for some nice open cover $\O$ depends only on the projection of that sequence onto the ordinals used to define $\O$. Specifically, we have:
\begin{observation2}\label{ob:bdd}
Let $\delta \leq \omega_1$ and let $\O$ be a nice open cover of $[0,1]^{\omega_1}$ defined using only ordinals $<\!\delta$. If $\seq{x_n}{n \in D}$ and $\seq{y_n}{n \in D}$ are $D$-indexed sequences in $[0,1]^{\omega_1}$ such that
$$\Pi_{\delta}(x_n) = \Pi_{\delta}(y_n) \qquad \text{for all } n \in D,$$
then either both sequences are $\varphi$-like for $\O$ or neither is.
\end{observation2}
Abusing our terminology slightly, if $\delta < \omega_1$ then a sequence $\seq{x_n}{n \in D}$ of points in $[0,1]^\delta$ will be called $\varphi$-like for some nice open cover $\O$ of $[0,1]^{\omega_1}$ provided that $\O$ is defined using ordinals $<\!\delta$, and that any sequence $\seq{y_n}{n \in D}$ of points in $[0,1]^{\omega_1}$ with
$$\Pi_{\delta'}(x_n) = \Pi_{\delta'}(y_n) \qquad \text{for all } n \in D$$
is $\varphi$-like for $\O$. Similarly, a sequence of points in $[0,1]^\delta$ is called $\varphi$-like if it is $\varphi$-like for every nice open cover of $[0,1]^{\omega_1}$ defined using ordinals $<\!\delta$.
For the remainder of the proof, it will be convenient to identify $D$ with $\omega \setminus \{0\} = \mathbb{N}$. This sacrifices no generality as, until now, $D$ was an arbitrary countable set.
With these things in place, let us turn to our recursive construction. Using the L\"{o}wenheim-Skolem theorem, fix a sequence $\seq{M_\a}{\a < \omega_1}$ of countable elementary submodels of $H$ such that
\begin{enumerate}
\item $M_0 = M$.
\item for each $\a$, $\seq{M_\b}{\b \leq \a} \in M_{\a+1}$.
\item for limit $\a$, $M_\a = \bigcup_{\b < \a}M_\b$.
\end{enumerate}
For convenience, we will write $\delta_\a$ instead of $\delta^{M_\a}$ for each $\a < \omega_1$.
As stated above, we will construct a map $q_\xi: D \to [0,1]$ for every $\xi < \omega_1$ via a length-$\omega_1$ transfinite recursion. Step $\a$ of the recursion will be used to construct simultaneously all the maps $q_\xi$ with $\xi \in \delta_\a \setminus \bigcup_{\b < \a}\delta_\b$. The construction will ensure that at the end of stage $\a$, the map $\Delta_{\xi < \delta_\a}q_\xi$ defines a $\varphi$-like sequence of points in $[0,1]^{\delta_\a}$.
For the base step of the recursion, we use our assumption that the metrizable reflection $\psi^M$ of $\psi$ contains a $\varphi$-like sequence. Let $\seq{r_n}{n \in D}$ be a $\varphi$-like sequence of points in $X^{M_0}$, and recall that $X^{M_0} \subseteq [0,1]^{\delta_0}$. Taking $q_\xi(n) = \pi_\xi(r_n)$, the map $\Delta_{\xi < \delta_0}q_\xi$ defines a $\varphi$-like sequence of points in $[0,1]^{\delta_0}$ as desired.
Without loss of generlity, we may (and do) assume that each of the real numbers $q_\xi(n)$, for $\xi < \delta_0$ and $n \in D$, is rational. To justify this assumption, use the fact that $[0,1]^{\delta_0}$ is metrizable to choose for each $r_n$ a point $r_n' \in M_0$ with rational coordinates such that $\mathrm{dist}(r_n,r_n') < \nicefrac{1}{n}$. These two sequences are clearly similar. Hence $\seq{r'_n}{n \in D}$ is $\varphi$-like, and replacing $\seq{r_n}{n \in D}$ with $\seq{r_n'}{n \in D}$ in the definition of the $q_\xi$ makes each number of the form $q_\xi(n)$ rational.
[Note: The reason for making the $q_\xi(n)$ all rational is that we will need $q_\xi(n) \in M_0$ in order to make use of the elementarity assumption $M_0 \preceq M_1$ at the next stage of the recursion. Making each $q_\xi(n)$ rational is just a convenient way to accomplish this.]
This completes the base step of the recursion. Observe that the construction of the $q_\xi$, $\xi < \delta_0$, can be carried out in $M_1$, because $M_0 \in M_1$.
For later stages of the recursion, we assume that the following three inductive hypotheses hold at the end of stage $\a$: letting $y^\a_n = \Delta_{\xi < \delta_\a}q_\xi(n)$ for convenience, then
\begin{itemize}
\item [(H1)] the sequence $\seq{y^\a_n}{n \in \omega}$ is $\varphi$-like in $[0,1]^{\delta_\a}$.
\item [(H2)] the sequence $\seq{y^\a_n}{n \in \omega}$ is a member of $M_{\a+1}$.
\item [(H3)] for each $\xi < \a$, $q_\xi(n)$ is a rational number.
\end{itemize}
Notice that these hypotheses hold at the end of stage $\a=0$ described above.
For limit $\a$ there is nothing to do: clause $(3)$ in our choice of the $M_\a$ guarantees that $\delta_\a = \bigcup_{\b < \a}\delta_\b$ for all limit ordinals $\a$, so at stage $\a$ the maps $q_\xi$ are already defined for every $\xi < \delta_\a$. The hypotheses $(\text{H}2)$ and $(\text{H}3)$ clearly hold at $\a$ if they hold for every $\b < \a$. For $(\text{H}1)$, note that every nice open cover of $[0,1]^{\omega_1}$ is defined from only finitely many ordinals. Thus, because $\delta_\a = \bigcup_{\b < \a}\delta_\b$, any nice open cover of $[0,1]^{\omega_1}$ defined from ordinals $<\!\delta_\a$ is already defined from ordinals $<\!\delta_\b$ for some $\b < \a$. $(\text{H}1)$ at $\a$ now follows from Observation 2 and the fact that $(\text{H}1)$ holds at $\b$ for every $\b < \a$.
For the successor stages of the recursion, fix $\a < \omega_1$ and suppose the functions $q_\xi$ have already been constructed for every $\xi < \delta_\a$. For each $n$, let $y^\a_n = \Delta_{\xi < \delta_\a}q_\xi(n)$.
We will show how to obtain $q_\xi$ for $\delta_\a \leq \xi < \delta_{\a+1}$.
There are only countable many nice open covers of $[0,1]^{\omega_1}$ defined using ordinals $<\!\delta_{\a+1}$. Also, any finitely many of these covers have a common refinement that is also a nice open cover of $[0,1]^{\omega_1}$ defined using ordinals $<\!\delta_{\a+1}$. Thus we may find a countable sequence $\seq{\O_m}{m < \omega}$ of nice open covers of $[0,1]^{\omega_1}$ defined using ordinals $<\!\delta_{\a+1}$ such that
\begin{enumerate}
\item $\O_m \in M_{\a+1}$ for every $m \in \omega$,
\item $\O_m$ refines $\O_\ell$ whenever $\ell \leq m$, and
\item if $\O$ is any basic open cover of $[0,1]^{\delta_{\a+1}}$, then some $\O_m$ refines $\O$.
\end{enumerate}
Note that this part of the construction occurs ``outside'' $M_{\a+1}$, because we are using the fact that $\delta_{\a+1}$ is countable.
Fix $m \in \omega$ and consider $\O_m$. The set of ordinals used in the definition of $\O_m$ is finite and may be split into two parts: those ordinals that are below $\delta_\a$, and those that are in the interval $[\delta_\a,\delta_{\a+1})$. Let us call these two finite sets of ordinals $F_m^\downarrow$ and $F_m^\uparrow$, respectively.
The ordinals in $F_m^\uparrow$ are not in $M_\a$, but for each ordinal $\xi \in F_m^\uparrow$ we may use elementarity to find an ``avatar'' ordinal $\zeta \in M_{\a}$ that acts like $\xi$, at least with respect to some prescribed first-order formula. The idea behind defining the $q_\xi$ for $\xi \geq \delta_\a$ is to find a sequence of increasingly faithful avatars, and then to define $q_\xi$ by diagonalizing across the avatar functions $q_\zeta$.
More precisely, let $F_m^\uparrow = \set{\xi_i}{i \leq \ell_m}$. For any first-order formula
$$\Phi(\xi_1,\xi_2,\dots,\xi_{\ell_m},a)$$
where $a$ is a parameter from $M_\a$, we may apply elementarity to find a set $E = \set{\zeta_i}{i \leq \ell_m} \subseteq M_\a$ such that
$$M_{\a+1} \models \Phi(\xi_1,\xi_2,\dots,\xi_{\ell_m},a) \qquad \Leftrightarrow \qquad M_\a \models \Phi(\zeta_1,\zeta_2,\dots,\zeta_{\ell_m},a)$$
Formulas containing more and more information about $X$ and $\psi$ will yield increasingly faithful avatars of the ordinals in $F_m^\uparrow$.
Let us enumerate $S = \{p_0,p_1,p_2,p_3,\dots\}$. For every $m < \omega$ it is possible to write down in the language of first-order logic a (very long) formula $\Phi^m(x_1,x_2,\dots,x_{\ell_m},X,\psi)$ that does all of the following:
\begin{enumerate}
\item $\Phi^m$ declares that each $x_i$ is a countable ordinal.
\item $\Phi^m$ defines a nice open cover in terms of $F^0_m$ and $x_1,x_2,\dots,x_{\ell_m}$. This is done in the natural way, so that the open cover defined by setting $x_1=\xi_1,x_2=\xi_2,\dots,x_{\ell_m}=\xi_{\ell_m}$ is $\O_m$.
\item $\Phi^m$ declares that certain properties hold of the nice open cover it defines and how that open cover interacts with $X$ and with the maps $\psi_{p_0},\psi_{p_1},\dots,\psi_{p_m}$. This is again done in the natural way, so that $\Phi^m(\xi_1,\xi_2,\dots,\xi_{\ell_m},X,\psi)$ declares the complete list of the following combinatorial properties of $\O_m$:
\begin{enumerate}
\item for all $\mathcal J \subseteq \O_m$, $\Phi^m$ asserts either that $\bigcap \mathcal J \cap X = \emptyset$ or that $\bigcap \mathcal J \cap X \neq \emptyset$,
\item if $\mathcal J \subseteq \O_m$, $\bigcap \mathcal J \cap X \neq \emptyset$, $U \in \O_m$, and $i \leq m$, then $\Phi^m$ asserts either that $\psi_{p_i}[\bigcup \mathcal J \cap X] \cap U = \emptyset$ or that $\psi_{p_i}[\bigcup \mathcal J \cap X] \cap U \neq \emptyset$.
\end{enumerate}
\end{enumerate}
Put simply, the formula $\Phi^m$ records the definition of the open cover $\O_m$ and its behavior with respect to the maps $\psi_{p_0},\psi_{p_1},\dots,\psi_{p_m}$.
If $E = \seq{\zeta_i}{i \leq \ell_m}$ is a finite sequence of ordinals $<\!\delta_{\a+1}$ such that $\Phi^m(\zeta_1,\zeta_2,\dots,\zeta_{\ell_m},X,\psi)$ holds, let us write $\O^m(E)$ for the basic open cover defined by $\Phi^m$. For example, $\O^m(F_m^\uparrow) = \O_m$ for all $k$.
Given two points $x,x'$ in $[0,1]^{\omega_1}$, the information contained in $(2)$ is enough to determine precisely which elements of $\O^m(E)$ contain each of $x$ and $x'$. Once that is known, the information in $(3)$ is enough to determine whether or not $x \xmapsto{\ \!_{\,\!_{\psi_p}}\,}_{\!_{\O^m(E)}} x'$ for any $p = p_0,p_1,\dots,p_m$. More formally, we have:
\begin{observation3}\label{obs:itworks,bitches!}
Suppose $F = \<\zeta_1,\dots,\zeta_{\ell_m}\>$ is a finite sequence of ordinals $<\!\delta_\a$, that $E = \<\xi_1,\dots,\xi_{\ell_m}\>$ is a finite sequence of ordinals $<\!\delta_{\a+1}$, and that
$$M_\a \models \Phi^m(\zeta_1,\zeta_2,\dots,\zeta_{\ell_m},X,\psi) \ \ \ \text{and} \ \ \ M_{\a+1} \models \Phi^m(\xi_1,\xi_2,\dots,\xi_{\ell_m},X,\psi).$$
Suppose further that $x,x' \in [0,1]^{\delta_\a}$ and $y,y' \in [0,1]^{\delta_{\a+1}}$, and that
$$\pi_{\zeta_i}(x) = \pi_{\xi_i}(y) \qquad \text{and} \qquad \pi_{\zeta_i}(x') = \pi_{\xi_i}(y')$$
for all $i \leq \ell$. Then
$$x \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m(F)}} x' \qquad \text{implies} \qquad y \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m(E)}} y'$$
for all $j \leq m$.
\end{observation3}
By the inductive hypothesis $(\text{H}3)$, $\Phi^m(F^\uparrow_m)$ is expressible in $M_{\a+1}$. Furthermore, our choice of $\Phi^m$ and $F^\uparrow_m$ ensures $M_{\a+1} \models \Phi^m(F^\uparrow_m)$. Thus
$$M_{\a+1} \models \exists x_1,x_2,\dots,x_{\ell_m}\Phi^m(x_1,x_2,\dots,x_{\ell_m},X,\psi).$$
By elementarity, $M_{\a} \models \exists x_1,x_2,\dots,x_{\ell_m}\Phi^m(x_1,x_2,\dots,x_{\ell_m},X,\psi)$, whence there is a finite sequence $E^m = \seq{\zeta_i}{i \leq \ell_m}$ of ordinals in $M_\a$ such that
$$M_\a \models \Phi^m(\zeta_1,\zeta_2,\dots,\zeta_{\ell_m},X,\psi).$$
If $\xi \in F^\uparrow_m$, then we will denote by $\zeta^m_\xi$ the corresponding member of $E^m$.
For each $m$, let $k(m)$ be the least natural number with the property that for all $n \geq k(m)$ and all $j \leq m$,
$$y^\a_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m(E^m)}} y^\a_{\varphi_{p_j}(n)}.$$
This $k(m)$ exists by our inductive hypothesis $(\text{H}1)$, because $\O^m(E^m)$ is a nice open cover of $[0,1]^{\omega_1}$ defined with ordinals $<\!\delta_\a$. If $m < m'$, then $\O_{m'}$ refines $\O_m$ and there are more functions $\psi_{p_j}$ to consider; so $k(m) \leq k(m')$. In other words, the function $m \mapsto k(m)$ is non-decreasing.
We are now in a position to define the maps $q_\xi$ for $\delta_\a \leq \xi < \delta_{\a+1}$:
$$
q_\xi(n) = \begin{cases}
0 & \textrm{ if $k(m) < n \leq k(m+1)$ and $\xi \notin F_m^\uparrow$,} \\
q_{\zeta^m_\xi}(i) & \textrm{ if $k(m) < n \leq k(m+1)$ and $\xi \in F_m^\uparrow$.}
\end{cases}
$$
Roughly, this says that $q_\xi$ assumes the behavior of its avatar function $q_{\zeta_\xi^m}$ on the interval between $k(m)$ and $k(m+1)$, provided some suitable avatar has already been found. As $m$ increases, $\zeta^m_\xi$ becomes a better and better avatar, because the formula $\Phi^m$ includes more and more information about the topology of $[0,1]^{\omega_1}$ and $X$ and about the flow $\psi$.
With the $q_\xi$ thus defined, we need to check that our inductive hypotheses $(\text{H}1)$-$(\text{H}3)$ remain true at the next stage of the recursion. As before, $(\text{H}2)$ and $(\text{H}3)$ are easy to check. Indeed, $(\text{H}3)$ follows trivially from the definition of the $q_\xi$ for $\delta_\a \leq \xi < \delta_{\a+1}$. For $(\text{H}2)$, note that because $\seq{M_\b}{\b \leq \a+1} \in M_{\a+2}$, the above construction of the $q_\xi$, $\delta_\a \leq \xi < \delta_{\a+1}$, can be carried out in $M_{\a+2}$. Thus the result of this construction, namely the sequence $\seq{\Delta_{\b < \delta_{\a+1}}q_\b(n)}{ n \in \omega}$, is a member of $M_{\a+2}$ as well.
For $(\text{H}1)$, let us write $y^{\a+1}_n = \Delta_{\xi < \delta_{\a+1}}(n)$ for each $n$. Let $\O_m$ be one of the nice open covers considered in our construction; we wish to show that $\seq{y^{\a+1}_n}{n \in D}$ is $\varphi$-like with respect to $\O_m$. Because $m$ is arbitrary, this suffices to show that $(\text{H}1)$ holds at $\a+1$.
Fix $n$ with $k(m) < n \leq k(m+1)$. Then
$$y^{\a}_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m(E^m)}} y^{\a}_{\varphi_{p_j}(n)}$$
for each $j \leq m$ by the choice of $k(m)$.
From this, from Observation 3, and from our choice of $E^m$, we deduce
$$y^{\a+1}_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m(F^\uparrow_m)}} y^{\a+1}_{\varphi_{p_j}(n)}$$
or, equivalently,
$$y^{\a+1}_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m}} y^{\a+1}_{\varphi_{p_j}(n)}$$
for each $j \leq m$. If $m' > m$ and $k(m') < n \leq k(m'+1)$, then similarly
$$y^{\a+1}_n \mapstoooef y^{\a+1}_{\varphi_{p_j}(n)}$$
for each $j \leq m$. Because $\O^{m'}$ refines $\O^m$, this implies
$$y^{\a+1}_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m}} y^{\a+1}_{\varphi_{p_j}(n)}$$
for each $j \leq m$ again. Thus this relation holds for all but finitely many $n$, namely for all $n > k(m)$. In other words, we have shown that, for every $m$.
\begin{equation}
\text{if } j \leq m \text{ then} \ y^{\a+1}_n \xmapsto{\ \!_{\,\!_{\psi_{p_j}}}\,}_{\!_{\O^m}} y^{\a+1}_{\varphi_{p_j}(n)} \ \text{for all } n > k(m).
\tag{$*$}
\end{equation}
That $(*)$ holds for every $m$ implies that $\seq{y^{\a+1}_n}{n \in D}$ is $\varphi$-like in $[0,1]^{\delta_{\a+1}}$. This proves that $(\text{H}1)$ holds at $\a+1$ and completes the successor step of our recursion.
We claim that the map $Q = \Delta_{\a < \omega_1}q_\a$ is as required; i.e., the sequence $\seq{Q(n)}{n < \omega}$ is a $\varphi$-like sequence in $[0,1]^{\omega_1}$. The argument is essentially the same as it was for the preservation of $(\text{H}1)$ at limit stages of the recursion. If $\O$ is a nice open cover of $[0,1]^{\omega_1}$, then $\O$ is defined by finitely many ordinals. Thus there is some $\a < \omega_1$ such that $\O$ is defined from ordinals $<\!\a$. Because $(\text{H}1)$ is true at $\a$, Observation 2 implies that $\seq{Q(n)}{n < \omega}$ is a $\varphi$-like sequence for $\O$. Observation 1 tells us that it suffices to consider only the nice open covers, so $\seq{Q(n)}{n < \omega}$ is $\varphi$-like. An application of Lemma~\ref{lem:psilike} completes the proof.
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,149 |
Frognal is a small area of Hampstead, North West London in the London Borough of Camden. Frognal is reinforced as the name of a minor road, which goes uphill from Finchley Road and at its upper end is in the west of Hampstead village.
History
The first reference to Frognal is as a tenement in the 15th century, probably on the site of the later Frognal House (now 99 Frognal). By the mid-eighteenth century it was a significant settlement, sought after by eminent lawyers, and infill development continued through the nineteenth and twentieth centuries.
The organist at St Andrew's Presbyterian Church was the father of composer John Tavener.
Architecture
Frognal has a diverse architecture, with many architecturally notable buildings. The central area, lacking large council estates, has undergone less change than some other parts of Hampstead. University College School, an independent day school founded in 1830, relocated to Frognal (the road) in 1907. Frognall Grove, Grade II listed, (1871–72) was a large house inherited by the architect George Edmund Street, who made additions to it. It was later subdivided into four semi-detached houses.
Notable residents
Gerald Abraham, musicologist, lived at 106 Frognal from the 1940s onwards.
Sir Walter Besant, the author, lived at 106 Frognal, and died at 18 Frognal Gardens in 1901.
Webster Booth, the tenor and Anne Ziegler, the soprano, lived at Frognal Cottage, 102 Frognal, from 1949 to 1952.
Dennis Brain (1921–1957), the horn player, lived at 37 Frognal.
Kathleen Ferrier (1912–1953), the contralto, lived at 2 Frognal Mansions, 97 Frognal, from 1942.
Charles Forte, restaurant and hotel owner (Trust House Forte) lived at Greenaway Gardens in Frognal.
The Labour Party leader Hugh Gaitskell lived at 16 Frognal Gardens and ran a salon of influence in the 1940s, and as Chancellor of the Exchequer in 1950.
General Charles de Gaulle lived from 1942 to 1944 in 99 Frognal.
Kate Greenaway (1846–1901), the illustrator, lived in a house in Frognal designed for her by Richard Norman Shaw in 1885.
Tamara Karsavina, the ballerina, lived at 108 Frognal in the 1950s.
E. V. Knox (1881–1971), the editor of Punch, lived at 110 Frognal from 1945.
William Page, historian and general editor of the Victoria County History, lived at Frognal Cottage (now 102 Frognal) from 1906 until 1922.
Sir Bernard Spilsbury (1877–1947), the pathologist, died at 20 Frognal.
Anton Walbrook, the actor, died at 69 Frognal in 1967.
Alistair Sim, the actor, lived at 8 Frognal Gardens.
Nearest places
Hampstead village
West Hampstead
Swiss Cottage, Hampstead
Belsize Park, Hampstead
Fortune Green, West Hampstead
Overlapping (in many definitions):
South Hampstead
Finchley Road (a linear area)
Rail and London Underground stations
Finchley Road & Frognal
Hampstead
Finchley Road
See also
One Oak, Frognal
Shepherd's Well, Frognal Way
Sun House, Frognal
University College School
Basil Feldman, Baron Feldman of Frognal
Susan Garden, Baroness Garden of Frognal
References
Further reading
Areas of London
Districts of the London Borough of Camden
Streets in the London Borough of Camden
Places formerly in Middlesex | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,537 |
Produced by Suzanne Shell, Diane Monico, and The Online
Distributed Proofreading Team at http://www.pgdp.net
[Illustration: (cover)]
[Illustration: (frontispiece)]
"SOME SAY"
NEIGHBOURS IN CYRUS
BY
LAURA E. RICHARDS
Author of "Captain January," "Melody," "Queen Hildegarde,"
"Five-Minute Stories," "When I Was Your Age,"
"Narcissa," "Marie," "Nautilus," etc.
TWELFTH THOUSAND
[Illustration]
BOSTON
DANA ESTES & COMPANY
PUBLISHERS
_Copyright, 1896_,
BY ESTES & LAURIAT
_All rights reserved_
Colonial Press:
C. H. Simonds & Co., Boston, Mass., U.S.A.
Electrotyped by Geo. C. Scott & Sons
"SOME SAY"
TO MY
Dear Sister,
FLORENCE HOWE HALL,
THIS VOLUME
IS AFFECTIONATELY
DEDICATED
* * * * *
"SOME SAY."
Part I.
"And some say, she expects to get him married to Rose Ellen before the
year's out!"
"I want to know if she does!"
"Her sister married a minister, and her father was a deacon, so mebbe
she thinks she's got a master-key to the Kingdom. But I don't feel so
sure of her gettin' this minister for Rose Ellen. Some say he's so
wropped up in his garden truck that he don't know a gal from a
gooseberry bush. He! he!"
The shrill cackle was answered by a slow, unctuous chuckle, as of a
fat and wheezy person; then a door was closed, and silence fell.
The minister looked up apprehensively; his fair face was flushed, and
his mild, blue eyes looked troubled. He gazed at the broad back of his
landlady, as she stood dusting, with minute care, the china ornaments
on the mantelpiece; but her back gave no sign. He coughed once or
twice; he said, "Mrs. Mellen!" tentatively, first low, then in his
ordinary voice, but there was no reply. Was Mrs. Mellen deaf? he had
not noticed it before. He pondered distressfully for a few moments;
then dropped his eyes, and the book swallowed him again. Yet the sting
remained, for when presently the figure at the mantelpiece turned
round, he looked up hastily, and flushed again as he met his hostess'
gaze, calm and untroubled as a summer pool.
"There, sir!" said Mrs. Mellen, cheerfully. "I guess that's done to
suit. Is there anything more I can do for you before I go?"
The minister's mind hovered between two perplexities; a glance at the
book before him decided their relative importance.
"Have you ever noticed, Mrs. Mellen, whether woodcocks are more apt to
fly on moonshiny nights, as White assures us?"
"Woodbox?" said Mrs. Mellen. "Why, yes, sir, it's handy by; and when
there's no moon, the lantern always hangs in the porch. But I'll see
that Si Jones keeps it full up, after this."
Decidedly, the good woman was deaf, and she had not heard. Could those
harpies be right? If any such idea as they suggested were actually in
his hostess' mind, he must go away, for his work must not be
interfered with, and he must not encourage hopes,--the minister
blushed again, and glanced around to see if any one could see him.
But he was so comfortable here, and Miss Mellen was so intelligent, so
helpful; and this seemed the ideal spot on which to compile his New
England "Selborne."
He sighed, and thought of the woodcock again. Why should the bird
prefer a moonshiny night? Was it likely that the creature had any
appreciation of the beauties of nature? Shakespeare uses the woodcock
as a simile of folly, to express a person without brains. Ha!
The door opened, and Rose Ellen came in, her eyes shining with
pleasure, her hands full of gold and green.
"I've found the 'Squarrosa,' Mr. Lindsay!" she announced. "See, this
is it, surely!"
The minister rose, and inspected the flowers delightedly. "This is it,
surely!" he repeated. "Stem stout, hairy above; leaves large, oblong,
or the lower spatulate-oval, and tapering into a marginal petiole,
serrate veiny; heads numerous; seeds obtuse or acute; disk-flowers, 16
x 24. This is, indeed, a treasure, for Gray calls it 'rare in New
England.' I congratulate you, Miss Mellen."
"Late, sir?" said Mrs. Mellen, calmly. "Oh, no, 'tisn't hardly five
o'clock yet. Still, 'tis time for me to be thinkin' of gettin'
supper."
"Don't you want I should make some biscuit for supper, mother?" asked
Rose Ellen, coming out of her rapt contemplation of the goldenrod that
Gray condescended to call rare, he to whom all things were common.
Her mother made no answer.
"Don't you want I should make a pan of biscuit?" Rose Ellen repeated.
Still there was no reply, and the girl turned to look at her mother in
some alarm.
"Why, mother, what is the matter? why don't you answer me?"
"Your mother's deafness," the minister put in, hurriedly, "seems
suddenly increased: probably a cold,--"
"Was you speakin' to me, Rose Ellen?" said Mrs. Mellen.
"Why, yes!" said the girl, in distress.
"Why, mother, how did you get this cold? you seemed all right when I
went out."
"Gettin' old!" cried Mrs. Mellen. "'Tis nothin' of the sort, Rose
Ellen! I've took a cold, I shouldn't wonder. I went out without my
shawl just for a minute. I expect 'twas careless, but there! life is
too short to be thinkin' all the time about the flesh, 'specially when
there's as much of it as I have. I've ben expectin' I should grow hard
of hearin', though, these two years past. The Bowlers do, you know,
Rose Ellen, 'long about middle life. There was your Uncle Lihu. I can
hear him snort now, sittin' in his chair, like a pig for all the
world, and with no idea he was makin' a sound."
"But it's come on so sudden!" cried Rose Ellen, in distress.
"That's Bowler!" said her mother. "Bowler for all the world! They take
things suddin, whether it's hoarsin' up, or breakin' out, or what it
is. There! you've heard me tell how my Aunt Phoebe 'Lizabeth come out
with spots all over her face, when she was standin' up to be married.
Chicken-pox it was, and they never knew where she got it; but my
grand'ther said 'twas pure Bowler, wherever it come from."
She gazed placidly at her daughter's troubled face; then, patting her
with her broad hand, pushed her gently out of the room before her.
"Mr. Lindsay's heard enough of my bein' hard of hearin', I expect,"
she said, cheerfully, as they passed into the kitchen.
"Don't you fret, Rose Ellen! You won't have to get a fog-horn yet
awhile. I don't know but it would be a good plan for you to mix up a
mess o' biscuit, if you felt to: Mr. Lindsay likes your biscuit real
well, I heard him say so."
"That's what I was going to do," said Rose Ellen, still depressed. "I
wish't you'd see the doctor, mother. I don't believe but he could help
your hearing, if you take it before it's got settled on you."
"Well, I won't, certain!" said Mrs. Mellen. "The idea, strong and well
as I be! Bowler blood's comin' out, that's all; and the only wonder is
it hasn't come out before."
All that day, and the next, the minister did not seem like himself. He
was no more absent-minded than usual, perhaps,--that could hardly be.
But he was grave and troubled, and the usual happy laugh did not come
when Rose Ellen checked him gently as he was about to put pepper into
his tea. Several times he seemed about to speak: his eye dwelt
anxiously on the cream-jug, in which he seemed to be seeking
inspiration; but each time his heart failed him, and he relapsed with
a sigh into his melancholy reverie.
Rose Ellen was silent, too, and the burden of the talk fell on her
mother. At supper on the second day, midway between the ham and the
griddle-cakes, Mrs. Mellen announced:
"Rose Ellen, I expect you'd better go down to Tupham to-morrow, and
stay a spell with your grandm'ther. She seems to be right poorly, and
I expect it'd be a comfort to her to have you with her. I guess you'd
better get ready to-night, and Calvin Parks can take you up as he goes
along."
Rose Ellen and the minister both looked up with a start, and both
flushed, and both opened wide eyes of astonishment.
"Why, mother!" said the girl. "I can't go away and leave you now, with
this cold on you."
Her mother did not hear her, so Rose Ellen repeated the words in a
clear, high-pitched voice, with a note of anxiety which brought a
momentary shade to Mrs. Mellen's smooth brow. The next moment,
however, the brow cleared again.
"I guess you'd better go!" she said again. "It'd be a pity if Mr.
Lindsay and I couldn't get along for a month or six weeks; and I wrote
mother yesterday that you would be up along to-morrow, so she'll be
looking for you. I don't like to have mother disappointed of a thing
at her age, it gives her the palpitations."
"You--wrote--that I was coming!" repeated Rose Ellen. "And you never
told me you was writing, mother? I--I should have liked to have known
before you wrote."
"Coat?" said Mrs. Mellen. "Oh, your coat'll do well enough, Rose
Ellen. Why, you've only just had it bound new, and new buttons put on.
I should take my figured muslin, if I was you, and have Miss Turner
look at it and see how you could do it over: she has good ideas,
sometimes, and it'd be a little different from what the girls here was
doin', maybe. Anyway, I'd take it, and your light sack, too. 'Twon't
do no harm to have 'em gone over a little."
Rose Ellen looked ready to cry, but she kept the tears back
resolutely.
"I--don't--want to leave you, with this deafness coming on!" she
shouted, her usually soft voice ringing like a bugle across the
tea-table.
"There! there! don't you grow foolish," her mother replied, with
absolute calm.
"Why, I can hear ye as well as ever, when you raise your voice a mite,
like that. I should admire to know why you should stay at home on my
account. I suppose I know my way about the house, if I be losin' my
hearing just a dite. It isn't going to spoil my cooking, that I can
see; and I guess Mr. Lindsay won't make no opposition to your going,
for any difference it'll make to him."
Mr. Lindsay, thus appealed to, stammered, and blushed up to his eyes,
and stammered again; but finally managed to say, with more or less
distinctness, that of course whatever was agreeable to Mrs. and Miss
Mellen was agreeable to him, and that he begged not to be considered
in any way in the formation of their plans.
"That's just what I was thinking!" said his hostess. "A man don't want
no botheration of plans. So that's settled, Rose Ellen."
Rose Ellen knew it was settled. She was a girl of character and
resolution, but she had never resisted her mother's will, nor had any
one else, so far as she knew. She cried a good deal over her packing,
and dropped a tear on her silk waist, the pride of her heart, and was
surprised to find that she did not care. "There's no one there to care
whether I look nice or not!" she said aloud; and then blushed
furiously, and looked around the room, fearfully, to be sure that she
was alone.
Early next morning the crack of a whip was heard, and Calvin Parks's
voice, shouting cheerfully for his passenger. The minister, razor in
hand, peeped between his shutters, and saw Rose Ellen come from the
house, wiping her eyes, and looking back, with anxious eyes. A wave of
feeling swept through him, and he felt, for the moment, that he hated
Mrs. Mellen. He had never hated any one before in his innocent life;
while he was pondering on this new and awful sensation, the pale,
pretty face had sunk back in the depths of the old red-lined stage,
the whip cracked, and Calvin drove away with his prey.
Mrs. Mellen came out on the steps, and looked after the stage. Then,
with a movement singularly swift for so stout a person, she made a few
paces down the walk, and, turning, looked up at the windows of the
houses on either side of her own. In both houses a figure was leaning
from a window, thrown half out over the sill, in an attitude of eager
inquiry. At sight of Mrs. Mellen they dodged back, and only a slight
waving of curtains betrayed their presence. The good woman folded her
arms deliberately, and stood for five minutes, absorbed in the distant
landscape; then she turned, and went slowly back to the house.
"There!" she said, as she closed the door behind her. "That'll keep
'em occupied for one while!" and there was infinite content in her
tone.
Mr. Lindsay, coming in to breakfast, found his hostess beaming behind
the teakettle, placid and cheerful as usual. He still hated her, and
found difficulty in replying with alacrity to her remarks on the
beauty of the morning.
"I expect you and me'll have a right cozy time together!" she
announced. "You no need to put yourself out to talk to me, 'cause I
reelly don't seem to be hearing very good; and I won't talk to you,
save and except when you feel inclined. I know an elder does love to
have a quiet house about him. My sister married a minister, and my
father was a deacon himself, so I'm accustomed to the ways of the
ministry."
Mr. Lindsay stirred his tea, gloomily. The words recalled to his mind
those which had so disturbed him a day or two ago, just when all this
queer business of the deafness had come on. He remembered the spiteful
tones of the two neighbours, and recalled how the words had hissed in
his ears. He had thought of going away himself, lest he should
encourage false hopes in the breast of his gentle young friend--or her
mother; surely Rose Ellen,--as he said the name to himself, he felt
his ears growing pink, and knew that he had not said the name before,
even to himself; straightway said it again, to prove the absurdity of
something, he was not sure what, and felt his throat dry and hot. Now
Rose Ellen herself was gone, and for an indefinite time. She had not
gone willingly, of that he was sure; but it was equally evident that
her mother had no such thoughts as those two harridans had suggested.
He glanced up furtively, to meet a broad, beaming glance, and the
question whether he felt feverish any.
"You seem to flush up easy!" said Mrs. Mellen. "I should be careful,
if I was you, Mr. Lindsay, and not go messing round ponds and such at
this season of the year. It's just this time we commonly look for
sickness rising in the air."
Mr. Lindsay stirred his tea again, and sighed. His mind seemed
singularly distracted; and that, too, when the most precious moments
of the year were passing. He must put all other matters out of his
head, and think only of his great work.
Had the Blackburnian Warbler been seen in this neighbourhood, as he
had been told? He could hardly believe in such good fortune. The shy,
mistrustful bird, hunting the thickest foliage of the tallest forest
trees,--how should his landlady's daughter have seen it when she was
seeking for ferns? yet her description had been exactly that of the
books: "Upper parts nearly uniform black, with a whitish scapular
stripe and a large white patch in the middle of the wing coverts; an
oblong patch--" but she had not been positive about the head. No, but
she _was_ positive as to the bright orange-red on chin, throat, and
forepart of the breast, and the three white tail-feathers. Ah! why was
she gone? why was she not here to show him the way, as she promised,
to the place where she had seen the rare visitor? He might possibly
have found the nest, that rare nest which Samuels never saw, which
only Audubon had described: "composed externally of different
textures, and lined with silky fibres and thin, delicate strips of
bark, over which lies a thick bed of feathers and horsehair."
It should be found in a small fork of a tree, should it? five or six
feet from the ground, near a brook? well, he might still search, the
next time he went out; meanwhile, there were the ferns to analyze, and
that curious moss to determine, if might be. "But mosses are almost
hopeless!" he said aloud, with an appealing glance across the table,
where he was wont to look for sympathy and encouragement.
"Soap dish?" said Mrs. Mellen, with alacrity.
"Well, I don't wonder you ask, Mr. Lindsay. Why, I found it full of
frogs' eggs this very morning, and I hove 'em away and scalt it out.
It's drying in the sun this minute, and I'll bring it right up to your
room directly."
She beamed on him, and left the room. Mr. Lindsay groaned; looked
about him for help, but found none, and retired, groaning, to his
study.
Part II.
The minister had had a delightful but exhausting afternoon. He had
gone to look for the nest of a marsh-hen, which he had some reason to
think might be in a certain swamp, about five miles from the village.
He did not find the nest, but he found plenty of other things: his
pockets bulged with mosses and roots, his hat was wound with a curious
vine that might possibly be Clematis Verticillaris, and both hands
were filled with specimens of every conceivable kind. Incidentally the
mosquitoes and black flies had found him: his face was purple, and,
like that of the lady at the Brick Lane Branch tea-party, "swellin'
wisibly;" and blood was trickling down his well-shaped nose from a
bramble-scratch. He had fallen down once or twice in the bog, with
results to his clothes; and altogether he presented a singular figure
to the view of his parishioners as he strode hastily through the
street. Heads were thrust out of windows, staid eyes rolled in horror,
but the minister saw nothing. He was tired, and absorbed in his new
possessions. It was good to sit down in his study, and spread his
treasures out on the broad table, and gloat over them. A clump of damp
moss rested quietly on his new sermon, "The Slough of Despond," but he
took no note. He was looking for a place to put this curious little
lizard in, and after anxious thought selected the gilt celluloid box,
lined with pink satin, which the Mission Circle had given him on
Christmas for his collars and cuffs. He felt, vaguely, that it was not
the right place for the lizard, but there seemed to be nothing else in
reach,--except the flitter-work pen-box, and Rose Ellen had made that
for him. Ah! if Rose Ellen were here now, how much she could help him!
it was so much easier for two to analyze than one. He at the
microscope, and Rose Ellen corroborating, correcting from the
textbook,--it was a perfect arrangement.
The minister sighed heavily. Mrs. Mellen brought in his tea, for it
was Wednesday evening, and he preferred an early cup of tea, and a
modest supper after the meeting. Food distracted his mind, he was apt
to say, from thought, a statement which his landlady treated with
indulgent contempt, as she had never known him to remark the
difference between "riz" bread and the soda article.
She set the cup down before him, and he promptly dipped a fern root
into it; then started back with a cry of dismay.
"Well indeed, sir!" said Mrs. Mellen, "I should think so, truly! What
did you do that for, and spoil your tea?"
"The--tea--a--that is, it is of no consequence about the tea!" said
Mr. Lindsay, hastily.
"I fear I have injured the root. I thought it was water. Dear! dear!
Miss Mellen was in the habit of bringing me a glass of water when I
brought plants home."
Mrs. Mellen said nothing, but brought the water, and a fresh cup of
tea; but Mr. Lindsay had fallen into the depths of the moss, and took
no notice of either.
She left the room, but presently returned, knitting in hand, and
stood, unnoticed, in the doorway, glancing from time to time at the
minister. He certainly was "a sight to behold," as she said to
herself. She may have thought other things beside, but her face gave
no sign. Presently the bell began to ring for Wednesday evening
meeting. Mrs. Mellen glanced again at the minister, but he heard
nothing. The botany was open before him, and he was muttering strange
words that sounded like witch-talk.
"Stamens six, hypogenous! anthers introrse! capsule cartilaginous,
loculicidally three-valved, scurfy-leaved epiphytic!" What did it all
mean? A slow flush crept over the woman's broad, placid face; her
eyelids quivered, her eye roamed restlessly about the room. She
shifted her weight from one foot to the other, and breathed heavily,
as if in distress; and still her eyes came back to the slender figure
in the great chair, bent in absorbed interest over the table.
Ding! dong! ding! the notes came dropping through the air, clear and
resonant. Even a deaf person might hear them, perhaps. Mrs. Mellen was
evidently struggling with herself. Once she opened her lips as if to
speak; once she stepped forward with outstretched hand, as if to shake
the man into wakefulness and attention; but she did not speak, and her
hand dropped again; and presently the bell stopped, and Sophronia
Mellen went away to her sitting-room, hanging her head.
Half an hour later there was knocking at the door, and the sound of
many voices, anxious voices, pitched high and loud, on account of Mrs.
Mellen's deafness.
"How's Mr. Lindsay? When was he took sick? Have ye had the doctor?"
"Do you think it's ketchin', Mis' Mellen? Think of all the young
children in this parish, if anythin' should get the rounds! My! it's
awful!"
"How does he look? Some say he was pupple in the face when they see
him coming home through the street. Most everybody did see him, and he
was a sight! Apoplexy, most likely!"
"Has he ever had fits, think? he don't look fitty, but you never can
tell."
"Have ye sent for his folks? You'd feel better to, I sh'd think, if
he's taken; some say he has a mother rollin' in wealth, down Brunswick
way."
"Well, some say he ain't nothin' of the sort. Christiana Bean saw an
aunt of his once, and she hadn't flesh enough on her to bait a
mouse-trap with, Christiana said so."
"Does he know you, Mis' Mellen? it's awful to see folks out of their
heads; I don't know how any one kin bear to see it; you'd better let
me come in and spell you a bit; you look clean tuckered out with the
fright you've had."
Mrs. Mellen stood and looked quietly at the crowd of "members" that
surged and cackled about her.
"I could hear better if one'd speak at a time!" she said, mildly.
"Did you want to see Elder Lindsay? it--it must be gettin' near
meetin'-time, isn't it?"
"Meetin'-time! meetin's over, and Mr. Lindsay never come nigh. Do you
mean to say he ain't sick? do you mean to say--"
"What _do_ you mean to say, Mis' Mellen?"
Mrs. Mellen held the door in her hand, and still gazed quietly at the
excited throng. At length,--
"Whatever's the matter with Mr. Lindsay," she said, in clear incisive
tones, "I ain't going to let in no lunatic asylum to drive him clean
out of his mind. Deacon Strong and Deacon Todd, if you'll step this
way, I presume Mr. Lindsay'll be pleased to see you. And if the rest
of you 'ud go home quiet, mebbe it might seem more consistent. There
has been a meetin', you say? the Baptists will be just about comin'
out now."
An hour later, the two deacons were taking their leave of Mr. Lindsay.
They stood, hat in hand, and were looking at the young man with
pitying eyes. They were elderly men, of kind disposition.
"Well, Mr. Lindsay," Mr. Todd was saying; "I guess we've said about
all there is to say. Of course 'twas a pity, and such things make
talk; but 'twon't occur again, I dare say. Some say--"
"It _may_ occur again," cried the young minister. He was sitting with
his head in his hand, and despair in his face.
"It may occur again! I seem to have no mind, no memory! I am unfit to
be a minister of the Christian Church. My brethren, what shall I do?"
The elder men exchanged glances: then Deacon Strong stepped forward
and laid his hands on the young man's shoulder, for he loved him.
"Mr. Lindsay," he said, kindly, "so far as I can see, there's only one
thing the matter with you; you want a wife!"
"A wife!" repeated Charles Lindsay. His tone suggested that he had
never heard of the article.
"A wife!" the deacon said again, with emphasis; and his fellow deacon
nodded assent.
"A sensible, clever young woman, who will help you in parish matters,
and be a comfort to you in every way,--a--hem! yes, in every way." The
deacon reddened through his beard, and glanced at Deacon Todd; but the
latter was a kind man, and knew Mrs. Strong, and gazed out of the
window.
"And--and tell you when it was time for meeting. I don't know as you'd
have to look more'n a hundred miles for the very young woman that
would make the right kind of helpmeet for you, but you know best about
that. Anyway, Mr. Lindsay, it is not good for man to be alone, we have
Scripture for that: and it's quite evident that it's particularly bad
for you to be alone, with your--a--your love of nature" (the deacon
caught sight of the lizard, peering disconsolately out of the gilt
celluloid box, and brought his remarks to a hasty conclusion). "And so
we'll be going, Mr. Lindsay, and don't you fret about to-night's
meeting, for we'll make it all right."
Mr. Lindsay bowed them out, with vague thanks, and muttered
expressions of regret. He hardly heard their adieux; the words that
were saying themselves over and over in his head were,--
"You want a wife!"
Did he want a wife? Was that what was the matter with him? Was that
why he went about all day and every day, these last weeks, feeling as
if half of him were asleep? He had always been a strong advocate of
the celibacy of the clergy, as far as his own case went. Nothing, he
had always assured himself, should ever come between him and his work.
A wife would be a perpetual distraction: she would want money, and
amusement, and a thousand things that he never thought about; and she
would interfere with his sermons, and with his collections, and--and
altogether, he would never marry. But now,--
And what was it that happened only the other day, here in the village?
A man and his wife had been quarrelling, to the scandal of the whole
congregation. They were an elderly couple, and when it came to
smashing crockery and emptying pails of water over each other, the
minister felt it his duty to interfere. So he called on the wife,
intending to reason with her first alone, and then, when she was
softened and convinced, to call in the husband and reconcile them, and
perhaps pray with them, since both were "members." But before he had
spoken a dozen of his well-arranged and logical sentences, he was
interrupted by loud and tearful outcries.
The lady never thought it would come to this, no, never! Some thought
she had enough to bear without this, but she knew how to submit to
the will of Providence, and no one should say she struv nor hollered.
She knew what was due to a minister, even if he was only just in
pants; she only hoped Mr. Lindsay wouldn't see fit to say anything to
her husband. Take Reuben Meecher when he was roused, and tigers was
tame by him: and if he should know that his wife was spoke to so, by
them as wasn't born or thought of when they was married, and nobody
couldn't say but they had lived respectable for forty years, and now
to come to this! The lady was well used to ministers, and some of the
most aged in the country, and she knew what was due to them; but for
her part, she thought 'twas well for ministers, as well as others, to
speak of what they'd had exper'ence in, and then there would be no
feeling!
The visit was not a success, nor did it cheer the minister to hear the
old couple chuckling to each other as he went sadly away, and to feel
that they were laughing at him.
But he was very humble, and he laid the spiteful words to heart.
What did he know? What had he to say to his people, when it came to
the real, terrible things of life? What had he had in his whole life,
save kindness and a sheltered home, and then study, and a little
divinity, and a little science? He sat and gazed at the image of
himself in his mind's glass, and found it a gibbering phantom, with
emptiness where should be eyes, and dry dust where should be living
waters.
As he sat thus sadly pondering, the sound of voices struck upon his
ear. The window was open, and now that his mind was awake, there was
no question of his hearing, when the two next-door neighbours leaned
out of their back windows, across Mrs. Mellen's back yard. He had
grown to loathe the sound of those two voices, the shrill cackling
one, and the fat chuckle that was even more hateful. What were they
saying now?
"You don't tell me she wants to git him for herself? speak jest a dite
louder! She can't hear ye, and he's so muddled up he never heard the
bell for meetin', some say; but there's others think he'd ben
drinkin', and Deacon Strong and Deacon Todd jest leagued together with
Sophrony Mellen to hide it. He was black in the face when he came
home, and reelin' in his walk, for I see him with these eyes."
Charles Lindsay started as if stung by a venomous snake. He put out
his hand to the window, but now the sharp voice broke in, anxious to
have its turn.
"Well, I shouldn't be a mite surprised if 'twas so, Mis' Bean, and
you've had experience, I'm sure, in such matters, after what you
suffered with Mr. Bean. But what I was sayin', some do say Phrony
Mellen's bound to have the minister for herself, and that's why she
sent Rose Ellen off, traipsin' way down to Tupham, when her grandma'am
don't need her no more'n a toad needs a tail."
"I want to know if they say that!" replied Mrs. Bean. "But you know,
some say Rose Ellen's got a beau down to Tupham, and that's why she
went off without askin' leave or license, and her ma deef and all. I
see her go myself, and she went off early in the mornin', and if ever
I see a person what you may call slink away secret, like she'd done
somethin' to be 'shamed of, 'twas that girl. _She_ knew what she was
goin' for, well enough. Rose Ellen ain't no fool, for all she's as
smooth as baked custard. Now you mark my words, Mis' Peake,--"
At this moment, the back door opened with a loud clang. Mrs. Mellen
stood on the doorstep, and her eyes were very bright. She said
nothing, but gazed calmly up and down the yard, as if considering the
beauty of the night. Then, after a few minutes, she turned and
scrutinized her neighbours' windows. Nothing was to be seen, only a
white muslin curtain waved gently in the moonlight: nothing was to be
heard, only a faint rustle, probably of the same curtain.
"It's an elegant night!" said Mrs. Mellen, aloud. "I thought I heard
voices, but my hearin' does play me such tricks, these days."
Her calm, sensible voice fell like balm on the distracted ears of the
minister. He was soothed, he knew not why. The horrors that those
harpies suggested,--could there be truth in them? Rose Ellen with
a--his mind refused to frame the detestable word! Was there anything
true in the world? Was it all scandal and hatefulness and untruth?
He rose and paced his study in anguish of mind, but his ears were
still awake,--he thought he never should regain the joy of losing
himself,--and now another sound came to them, the sound of wheels. Why
did his heart stop, and then beat violently? What was there in the
sound of wheels? It was the late stage, of course, and Calvin Parks
was driving fast, as usual, to get to his home, five miles away,
before ten o'clock at night. But that stage came from Tupham, and
Tupham meant Rose Ellen. Rose Ellen, who was as smooth as baked
custard, and who had a--the wheels were slacking; the steady beat of
the horses' feet stopped; the stage had paused at the Widow Mellen's
door.
"Here we be!" said Calvin Parks. "Take my hand, Rosy! so, thar she
goes! Hope ye'll find yer ma right smart! Give her my respects and
tell her,--wal, I swan!"
For the door flew open, and out ran the minister, torn and stained and
covered with dust, and caught Rose Ellen by both hands and drew her
almost forcibly into the house.
"Mother!" cried the girl. "How is she? I--I got so scared, not hearing
from her, I couldn't stay another day, Mr. Lindsay!"
"Oh,--your mother?" said Mr. Lindsay, incoherently. "She--a--she seems
to be in excellent health, except for her deafness. It is I who am
ill, Rose Ellen: very ill, and wanting you more than I could bear!"
"Wanting me?" faltered Rose Ellen, with lips wide, with blue eyes
brimming over. "You, Mr. Lindsay, wanting me?"
"Yes, Rose Ellen!" cried the minister. They were still standing in the
passage, and he was still holding her hands, and it was quite absurd,
only neither of them seemed to realize it.
"I have always wanted you, but I have only just found it out. I cannot
live at all without you: I have been only half alive since you went
away. I want you for my own, for always."
"Oh, you can have me!" cried Rose Ellen, and the blue eyes brimmed
over altogether with happy shining tears. "Oh, I was yours all the
time, only I didn't know you--I didn't know--"
She faltered, and then hurried on. "It--it wasn't only that I was
scared about mother, Mr. Lindsay. I couldn't stay away from--oh, some
said--some said you were going to be married, and I couldn't bear it,
no, I couldn't!"
But when Charles Lindsay heard that, he drew Rose Ellen by both hands
into the study, and shut the door. And only the lizard knew what
happened next.
* * * * *
It was a month later.
There had been a wedding, the prettiest wedding that the village had
ever seen. The whole world seemed turned to roses, and the sweetest
rose of all, Rose Ellen Lindsay, had gone away on her husband's arm,
and Deacon Strong and Deacon Todd were shaking hands very hard, and
blowing peals of joy with their pocket-handkerchiefs. Mrs. Mellen had
preserved her usual calm aspect at the wedding, and looked young
enough to be her own daughter, "some said," in her gray silk and white
straw bonnet. But when it was all over, the wedding party gone, and
the neighbours scattered to their homes again, Sophronia Mellen did a
strange thing. She went round deliberately, and opened every window of
her house. The house stood quite apart, with only the two houses close
beside it on either hand, and no others till you came quite into the
street itself. She opened every window to its utmost. Then she took a
tin pan, and a pair of tongs, and leaned out of the front parlour
window, and screamed three times, at the top of her lungs, beating
meanwhile with all her might upon the pan. Then she went to the next
window, and screamed and banged again, and so on all over the house.
There were twenty windows in her house, and by the time she had gone
the round, she was crimson and breathless. Nevertheless, she managed
to put her last breath into a shriek of such astounding volume that
the windows fairly rang. One last defiant clang of the tongs on the
tin pan and then she sat down quietly by the back parlour window, and
settled herself well behind the curtain, and prepared to enjoy herself
thoroughly. "They shall have their fill this time!" she murmured to
herself; "and I shall get all the good of it."
For some minutes there was dead silence: the event had been too awful
to be treated lightly. At length a rustling was heard, and very
cautiously a sharp nose, generously touched with colour, was protruded
from the window of the left-hand house.
"Mis' Bean," said the owner of the nose. "Be you there?"
"Well, I should say I was!" was the reply; and Mrs. Bean's fat curls
shook nervously out of her window.
"Maria Peake, what do you s'pose this means? Ain't it awful? Why, I've
got palpitations to that degree,--don't s'pose there's a robber in
the house, do ye? with all them weddin' presents about, 'twould be a
dreadful thing! 'Tain't likely he would spare her life, and she tryin'
to give the alarm like that! Most likely she's layin' dead this
minute, and welterin' in her--"
"Sssssssh!" hissed Mrs. Peake, in a deadly whisper. "Melissa Bean, you
won't let a person hear herself think. 'Tain't no robber, I tell ye!
She's gone out of her mind, Phrony Mellen has, as sure as you're a
breathin' woman!"
"You don't tell me she has!" Mrs. Bean leaned further out, her eyes
distended with awful curiosity, her fat lips dropping apart. She was
not a pleasant object, the hidden observer thought; but she was no
worse than the skinny cabbage-stalk which now stretched itself far out
from the opposite window.
"I tell ye," Mrs. Peake hissed, still in that serpent-whisper, the
most penetrating sound that ever broke stillness, "She's as crazy as a
clo'esline in a gale o' wind. Some say she's wore an onsettled eye for
six weeks past, and she glared at me yesterday, when I run in to borry
an egg, same as if I was one wild animal and she was another. Ssssh!
'Tis Bowler, I tell ye! They go that way, jest as often as they git a
chance! I call it an awful jedgment on Elder Lindsay, bein' married
into that family. Some say his mother besought him on her bended
knees, but he was clean infatooated. I declare to you, Mis' Bean, I'm
terrified most to death, to think of you and me alone here, so near to
a ravin' lunatic. I don't think nothin' of robbers, alongside o'
madness. She might creep in while you're standin' there,--your house
is more handy by than mine, 'count of there bein' no fence, and--"
"Yah! bah! ha! ha! ha! hurrah!" sounded in sharp, clear tones from
Mrs. Mellen's window. Two ghastly faces, white with actual terror,
gazed at each other for an instant, then disappeared; and immediately
after was heard a sound of bolts being driven home, and of heavy
furniture being dragged about.
But Mrs. Mellen sat and fanned herself, being somewhat heated, and
gazed calmly at the beauty of the prospect.
"I've enjoyed myself real well!" she said. "I couldn't free my mind,
not while Rosy and Mr. Lindsay was round; I've had a real good time."
She fanned herself placidly, and then added, addressing the universe
in general, with an air of ineffable good will:
"I shouldn't wonder if my hearin' improved, too, kind o' suddin, same
as it came on. That's Bowler, too! It's real convenient, bein' a
Bowler!"
* * * * *
NEIGHBOURS IN CYRUS
NEIGHBOURS IN CYRUS.
"Hi-Hi!" said Miss Peace, looking out of the window. "It is really
raining. Isn't that providential, now?"
"Anne Peace, you are enough to provoke a saint!" replied a peevish
voice from the furthest corner of the room. "You and your providences
are more than I can stand. What do you mean this time, I _should_ like
to know? the picnic set for to-day, and every soul in the village
lottin' on goin', 'xcept those who _would_ like best to go and can't.
I've been longin' for these two years to go to a picnic and it's never
ben so's I could. And now, jest when I _could_ ha' gone, this
affliction must needs come to me. And then to have you rejoicin'
'cause it rains!"
The speaker paused for breath, and Miss Peace answered mildly: "I'm
real sorry for you, Delia, you know I am; and if the' was any way of
getting you to the grove,--but what I was thinking of, you know I
couldn't finish Jenny Miller's dress last night, do what I could; and
seeing it raining now, thinks I, they'll have to put off the picnic
till to-morrow or next day, and then Jennie can go as nice as the
rest. She does need a new dress, more than most of the girls who has
them. And she's so sweet and pretty, it's a privilege to do for her.
That's all I was thinking, Delia."
Mrs. Delia Means sniffed audibly, then she groaned.
"Your leg hurting you?" cried Miss Peace, with ready sympathy.
"Well, I guess you'd think so," was the reply. "If _you_ had red-hot
needles run into your leg. Not that it's any matter to anybody."
"Hi-hi," said Miss Peace, cheerily. "It's time the bandages was
changed, Delia. You rest easy just a minute, and I'll run and fetch
the liniment and give you a rub before I put on the new ones."
Mrs. Means remaining alone, it is proper to introduce her to the
reader. She and Miss Peace were the rival seamstresses of Cyrus
Village; that is, they would have been rivals, if Mrs. Means had had
her way; but rivalry was impossible where Anne Peace was one of the
parties. She had always maintained stoutly that Delia Means needed
work a sight more than she did, having a family, and her husband so
weakly and likely to go off with consumption 'most any time. Many and
many a customer had Anne turned from her door, with her pleasant
smile, and "I don't hardly know as I could, though I should be pleased
to accommodate you; but I presume likely Mis' Means could do it for
you. She doos real nice work, and I don't know as she's so much drove
just now as I am."
Delia Case had been a schoolmate of Anne Peace's. She was a pretty
girl, with a lively sense of her own importance and a chronic taste
for a grievance. She had married well, as every one thought, but in
these days her husband had lost his health and Delia was obliged to
put her shoulder to the wheel. She sewed well, but there was a sigh
every time her needle went into the cloth, and a groan when it came
out.
"A husband and four children, and have to sew for a living!"--this was
the burden of her song; and it had become familiar to her neighbours
since David Means had begun to "fail up," as they say in Cyrus.
Anne Peace had always been the faithful friend of "Delia Dumps." (It
was Uncle Asy Green who had given her the name which stuck to her
through thick and thin--Uncle Asy believed in giving people their due,
and thought "Anne made a dreffle fool of herself, foolin' round with
that woman at all.") Anne had been her faithful friend, and never
allowed people to make fun of her if she were present.
A week before my story opens, when Mrs. Means fell down and broke her
leg, just as she was passing Miss Peace's house, the latter lady
declared it to be a special privilege.
"I can take care of her," she explained to the doctor, when he
expressed regret at being obliged to forbid the sufferer's being moved
for some weeks, "just as well as not and better. David isn't fit to
have the care of her, and--well, doctor, I can say to you, who know it
as well as I do, that Delia mightn't be the best person for David to
have round him just now, when he needs cheering up. Then, too, I can
do her sewing along with my own, as easy as think; work's slack now,
and there's nothing I'm specially drove with. I've been wishing right
along that I could do something to help, now that David is so poorly.
I'm kin to David, you know, so take it by and large, doctor, it doos
seem like a privilege, doesn't it?"
The doctor growled. He was not fond of Mrs. Means.
"If you can get her moved out of Grumble Street and into Thanksgiving
Alley," he said, "it'll be a privilege for this village; but you can't
do it, Anne. However, there's no use talking to you, you incorrigible
optimist. You're the worst case I ever saw, Anne Peace, and I haven't
the smallest hope of curing you. Put the liniment on her leg as I
told you, and I'll call in the morning. Good day!"
"My goodness me, what was he saying to you?" Mrs. Means asked as Anne
went back into the bedroom. "You've got something that you'll never
get well of? Well, Anne Peace, that does seem the cap sheaf on the
hull. Heart complaint, I s'pose it is; and what would become of me, if
you was to be struck down, as you might be any minute of time, and me
helpless here, and a husband and four children at home and he failin'
up. You did look dretful gashly round the mouth yisterday, I noticed
it at the time, but of course I didn't speak of it. Why, here I should
lay, and might starve to death, and you cold on the floor, for all the
help I should get." Mrs. Means shed tears, and Anne Peace answered
with as near an approach to asperity as her soft voice could command.
"Don't talk foolishness, Delia. I'm not cold yet, nor likely to be.
Here, let me 'tend to your leg; it's time I was getting dinner on this
minute."
It continued to rain on the picnic day; no uncertain showers, to keep
up a chill and fever of fear and hope among the young people, but a
good, honest downpour, which everybody past twenty must recognize as
being just the thing the country needed. Jenny Miller came in, smiling
all over, though she professed herself "real sorry for them as was
disappointed." "Tudie Peaslee sat down and cried, when she saw 'twas
rainin'," she said, as she prepared to give her dress the final
trying-on. "There, Miss Peace. I did try to feel for her, but I just
couldn't, seems though. Oh, ain't that handsome? that little puff is
too cute for anything! I do think you've been smart, Miss Peace. Not
that you ever was anything else."
"You've a real easy figure to fit, Jenny," Miss Peace replied,
modestly. "I guess that's half the smartness of it. It doos set good,
though, I'm free to think. The styles is real pretty this summer,
anyhow. Don't that set good, Delia?"
She turned to Mrs. Means, who was lying on the sofa (they call it a
l'unge in Cyrus), watching the trying-on with keenly critical eyes.
"Ye-es," she said. "The back sets good enough, but 'pears to me
there's a wrinkle about the neck that I shouldn't like to see in any
work of mine. I've always ben too particklar, though; it's time thrown
away, but I can't bear to send a thing out 'cept jest as it should
be."
"It _don't_ wrinkle, Mis' Means!" cried Jenny, indignantly. "Not a
mite. I was turning round to look at the back of the skirt, and that
pulled it; there ain't a sign of a wrinkle, Miss Peace, so don't you
think there is."
Mrs. Means sniffed, and said something about the change in young
folks' manners since she was a girl. "If I'd ha' spoke so to my
elders--I won't say betters, for folks ain't thought much of when they
have to sew for a livin', with a husband and four children to keer
for--I guess I should ha' found it out in pretty quick time."
"Hi-hi!" said Miss Peace, soothingly. "There, Delia, Jenny didn't mean
anything. Jenny, I guess I'll have to take you into the bedroom, so's
I can pull this skirt out a little further. This room doos get so
cluttered with all my things round." She hustled Jenny, swelling like
an angry partridge, into the next room, and closed the door carefully.
"You don't want to anger Mis' Means, dear," she said gently, taking
the pins out of her mouth for freer speech. "She may be jest a scrap
pudgicky now and again, but she's seen trouble, you know, and she doos
feel it hard to be laid up, and so many looking to her at home. Turn
round, dear, jest a dite--there!"
"I can't help it, Miss Peace," said Jenny. "There's no reason why Mis'
Means should speak up and say the neck wrinkled, when anybody can see
it sets like a duck's foot in the mud. I don't mind what she says to
me, but I ain't goin' to see you put upon, nor yet other folks ain't.
I should like to know! and that wrapper she cut for Tudie Peaslee set
so bad, you'd think she'd fitted it on the pump in the back yard,
Mis' Peaslee said so herself."
"Hi-hi!" cried Anne Peace, softly, with an apprehensive glance toward
the door; "don't speak so loud, Jennie. Tudie ain't so easy a form to
fit as you, not near. And you say she was real put about, do ye, at
the picnic being put off?"
"She was so!" Jenny assented, seeing that the subject was to be
changed. "She'd got her basket all packed last night, she made so sure
'twas goin' to be fine to-day. Chicken sandwiches, she had, and baked
a whole pan of sponge-drops, jest because some one--you know who--is
fond of 'em." Miss Peace nodded sagely, with her mouth full of pins,
and would have smiled if she could; "and now they've put it off till
Saturday, 'cause the minister can't go before then, and every livin'
thing will be spoiled."
"Dear, dear!" cried Miss Anne, her kind face clouding over; "that does
seem too bad, don't it? all those nice things! and Tudie makes the
best sponge-cakes I ever eat, pretty nigh."
Jenny smiled, and stretched her hand toward a basket she had brought.
"They won't really be wasted, Miss Peace," she said. "Tudie thought
you liked 'em, and I've got some of 'em here for you, this very
minute. You was to eat 'em for your own supper, Tudie told me to tell
you so."
"Well, I do declare, if that isn't thoughtful!" exclaimed Miss Peace,
looking much gratified. "Tudie is a sweet girl, I must say. Delia is
real fond of cake, and she's been longing for some, but it doos seem
as if I couldn't find time to make it, these days."
"I should think not!" cried Jenny (who was something of a pepper-pot,
it must be confessed), "I should think not, when you have her to take
care of, and her work and yours to do, and all. And, Miss
Peace,--Tudie meant the sponge-drops for _you_, every one. She told me
so."
"Yes, dear, to be sure she did, and that's why I feel so pleased, just
as much as if I had eaten them. But bread _is_ better for me,
and--why! if she hasn't sent a whole dozen. One, two, three--yes, a
dozen, and one over, sure as I stand here. Now, that I call generous.
And, I'll tell you what, dearie! Don't say a word, for I wouldn't for
worlds have Tudie feel to think I was slighting her, or didn't
appreciate her kindness; but--well, I _have_ wanted to send some
little thing round to that little girl of Josiah Pincher's, that has
the measles, and I do suppose she'd be pleased to death with some of
these sponge-drops. Hush! don't say a word, Jenny! it would be a real
privilege to me, now it would. And you know it isn't that I don't
think the world of Tudie, and you, too; now, don't you?"
Jenny protested, half-laughing, and half-crying; for Tudie Peaslee had
declared herself ready to bet that Miss Peace would not eat a single
one of the sponge-drops, and Jenny had vowed she should. But would she
or would she not, before ten minutes were over she had promised to
leave the sponge-drops at the Pinchers' door as she went by, for
little Geneva. There was no resisting Miss Peace, Tudie was right; but
suddenly a bright idea struck Jenny, just as she was putting on her
hat and preparing to depart. Seizing one of the sponge-drops, she
broke off a bit, and fairly popped it into Miss Peace's mouth, as the
good lady was going to speak. "It's broke, now," she cried, in high
glee, "it's broke in two, and you can't give it to nobody. Set right
down, Miss Peace, and let me feed you, same as I do my canary bird."
She pushed the little dressmaker into a chair, and the bits followed
each other in such quick succession that Miss Peace could make no
protest beyond a smothered, "Oh, don't ye, dear; now don't! that's
enough!--my stars, Jenny, what do you think my mouth's made of?"
(Crunch!) "There, dear, there! It is real good--oh, dear! not so fast.
I _shall_ choke! Tell Tudie--no, dearie, not another morsel!"
(Crunch.) "Well, Jenny Miller, I didn't think you would act so, now I
didn't."
The sponge-cake was eaten, and Jenny, with a triumphant kiss on the
little rosy, withered-apple cheek, popped her head in at the parlour
door to cry, "Good day, Mis' Means!" and flew laughing away with her
victory and her cakes.
"Well, Anne Peace," was Mrs. Means's greeting, as her hostess came
back, looking flushed and guilty, and wiping her lips on her apron,
"how you can stand havin' that Miller girl round here passes me. She'd
be the death of me, I know that; but it's lucky other folks ain't so
feelin' as I am, I always say. Of all the forward, up-standin' tykes
ever I see--but there! it ain't to be supposed anybody cares whether
I'm sassed or whether I ain't."
Saturday was bright and fair, and Anne Peace stood at the window with
a beaming smile, watching the girls troop by on their way to the
picnic. She had moved Mrs. Means's sofa out of the corner, so that she
could see, too, and there was a face at each window. Miss Peace was a
little plump, partridge-like woman, with lovely waving brown hair, and
twinkling brown eyes. She had never been a beauty, but people always
liked to look at her, and the young people declared she grew prettier
every year. Mrs. Means was tall and weedy, with a figure that used to
be called willowy, and was now admitted to be lank; her once fair
complexion had faded into sallowness, and her light hair had been
frizzed till there was little left of it. Her eyebrows had gone up,
and the corners of her mouth had gone down, so that her general effect
was depressing in the extreme.
"There go Tudie and Jenny!" cried Miss Peace, in delight. "If they
ain't a pretty pair, then I never saw one, that's all. Jenny's dress
doos set pretty, if I do say it; and after all, it's her in it that
makes it look so well. There comes the minister, Delia. Now I'm glad
the roses are out so early. He doos so love roses, Mr. Goodnow does.
And the honeysuckle is really a sight. Why, this is the first time you
have fairly seen the garden, Delia, since you came. Isn't it looking
pretty?"
"I never did see how you could have your garden right close 't onto
the street that way, Anne," was the reply. "Everybody 't comes by
stoppin' and starin', and pokin' their noses through the fence. Look
at them boys, now! why, if they ain't smellin' at the roses, the
boldfaced brats. Knock at the winder, Anne, and tell 'em to git out.
Shoo! be off with you!" She shook her fist at the window, but,
fortunately, could not reach it.
"Hi-hi!" said Anne Peace. "You don't mean that, Delia. What's roses
for but to smell? I do count it a privilege, to have folks take
pleasure in my garden." She threw up the window, and nodded pleasantly
to the children. "Take a rose, sonny, if you like 'em," she said.
"Take two or three, there's enough for all. Whose little boys are
you?" she added, as the children, in wondering delight, timidly broke
off a blossom or two. "Mis' Green's, over to the Corners! Now I want
to know! have you grown so 't I didn't know you? and how's your
mother? Jest wait half a minute, and I'll send her a little posy.
There's some other things besides roses, perhaps she'd like to have a
few of."
She darted out, and filled the boys' hands with pinks and mignonette,
<DW29>s and geraniums.
It was not a large garden, this of Anne Peace's, but every inch of
space was made the most of. The little square and oblong beds lay
close to the fence, and from tulip-time to the coming of frost they
were ablaze with flowers. Nothing was allowed to straggle, or to take
up more than its share of room. The roses were tied firmly to their
neat green stakes; the crown-imperials nodded over a spot of ground
barely large enough to hold their magnificence; while the phlox and
sweet-william actually had to fight for their standing-room.
It was a pleasant sight, at all odd times of the day, to see Miss
Peace bending over her flowers, snipping off dead leaves, pruning, and
tending, all with loving care.
Many flower-lovers are shy of plucking their favourites, and I recall
one rose-fancier, whose gifts, like those of the Greeks, were dreaded
by his neighbours, as the petals were always ready to drop before he
could make up his mind to cut one of the precious blossoms; but this
was not the case with Anne Peace. Dozens of shallow baskets hung in
her neat back entry, and they were filled and sent, filled and sent,
all summer long, till one would have thought they might almost find
their way about alone. It is a positive fact that her baskets were
always brought back, "a thing imagination boggles at;" but perhaps
this was because the neighbours liked them better full than empty.
"Makin' flowers so cheap," Mrs. Means would say, "seems to take the
wuth of 'em away, to my mind; but I'm too feelin', I know that well
enough. Anne, she's kind o' callous, and she don't think of things
that make me squinch, seem's though."
Weeks passed on, the broken leg was healed, and Mrs. Means departed to
her own house. "I s'pose you'll miss me, Anne," she said, at parting,
"I shall you; and you have ben good to me, if 't _has_ ben kind o'
dull here, so few comin' and goin'." (Miss Peace's was generally the
favourite resort of all the young people of the village, and half the
old ones, but the "neighbouring" had dropped off, since Mrs. Means
had been there.) "Good-by, Anne, and thank you for all you've done. I
feel to be glad I've been company for you, livin' alone as you do,
with no husband nor nothin' belongin' to you."
"Good-by, Delia," replied Anne Peace, cheerfully. "Don't you fret
about me. I'm used to being alone, you know; and it's been a
privilege, I'm sure, to do what I could for you, so long as we've been
acquainted. My love to David, and don't forget to give him the syrup I
put in the bottom of your trunk for him."
"'Twon't do him any good!" cried Mrs. Means, as the wagon drove away,
turning her head to shout back at her hostess. "He's bound to die,
David is. He'll never see another spring, I tell him, and then I shall
be left a widder, with four children and--"
"Oh, gerlang! gerlang, _up_!" shouted Calvin Parks, the stage-driver,
whose stock of patience was small; the horse started, and Mrs. Means's
wails died away in the distance.
In this instance the predictions of the doleful lady seemed likely to
be verified; for David Means continued to "fail up." Always a slight
man, he was now mere skin and bone, and his cheerful smile grew
pathetic to see. He was a distant cousin of Anne Peace's, and had
something of her placid disposition; a mild, serene man, bearing his
troubles in silence, finding his happiness in the children whom he
loved almost passionately. He had married Delia Case because she was
pretty, and because she wanted to marry him; had never known, and
would never know, that he might have had a very different kind of
wife. Perhaps Anne Peace hardly knew herself that David had been the
romance of her life, so quickly had the thought been put away, so
earnestly had she hoped for his happiness; but she admitted frankly
that she "set by him," and she was devoted to his children.
"Can nothing be done?" she asked the good doctor one day, as they came
away together from David's house, leaving Delia shaking her head from
the doorsteps. "Can nothing be done, doctor? it doos seem as if I
couldn't bear to see David fade away so, and not try anything to stop
it."
Doctor Brown shook his head thoughtfully. "I doubt if there's much
chance for him, Anne," he said kindly. "David is a good fellow, and if
I saw any way--it might be possible, if he could be got off to Florida
before cold weather comes on--there is a chance; but I don't suppose
it could be managed. He has no means, poor fellow, save what he
carries in his name."
"Florida?" said Anne Peace, thoughtfully; and then she straightway
forgot the doctor's existence, and hurried off along the street, with
head bent and eyes which saw nothing they rested on.
Reaching her home, where all the flowers smiled a bright welcome,
unnoticed for once, her first action was to take out of a drawer a
little blue book, full of figures, which she studied with ardour. Then
she took a clean sheet of paper, and wrote certain words at the top of
it; then she got out her best bonnet.
Something very serious was on hand when Miss Peace put on her best
bonnet. She had only had it four years, and regarded it still as a
sacred object, to be taken out on Sundays and reverently looked at,
then put back in its box, and thought about while she tied the strings
of the ten-year-old velvet structure, which was quite as good as new.
Two weddings had seen the best bonnet in its grandeur, and three
funerals; but no bells, either solemn or joyous, summoned her to-day,
as she gravely placed the precious bonnet on her head, and surveyed
her image with awestruck approval in the small mirror over the
mantelpiece.
"It's _dreadful_ handsome!" said Miss Peace, softly. "It's too
handsome for me, a great sight, but I want to look my best now, if
ever I did."
It was at Judge Ransom's door that she rang first; a timid, apologetic
ring, as if she knew in advance how busy the judge would be, and how
wrong it was of her to intrude on his precious time. But the judge
himself opened the door, and was not at all busy, but delighted to
have a chance to chat with his old friend, whom he had not seen for a
month of Sundays. He made her come in, and put her in the biggest
armchair (which swallowed her up so that hardly more than the bonnet
was visible), and drew a footstool before her little feet, which
dangled helplessly above it; then he took his seat opposite, in
another big chair, and said it was a fine day, and then waited, seeing
that she had something of importance to say.
Miss Peace's breath came short and quick, and she fingered her
reticule nervously. She had not thought it would be quite so dreadful
as this. "Judge," she said--and paused, frightened at the sound of her
voice, which seemed to echo in a ghostly manner through the big room.
"Well, Miss Peace!" said the judge, kindly. "Well, Anne, what is it?
How can I serve you? Speak up, like a good girl. Make believe we are
back in the little red schoolhouse again, and you are prompting me in
my arithmetic lesson."
Anne Peace laughed and . "You're real kind, judge," she said.
"I wanted--'twas only a little matter"--she stopped to clear her
throat, feeling the painful red creep up her cheeks, and over her
brow, and into her very eyes, it seemed; then she thought of David,
and straightway she found courage, and lifted her eyes and spoke out
bravely. "David Means, you know, judge; he is failing right along, and
it doos seem as if he couldn't last the winter. But Doctor Brown
thinks that if he should go to Florida, it might be so 't he could be
spared. So--David hasn't means himself, of course, what with his poor
health and his large family, and some thought that if we could raise a
subscription right here, among the folks that has always known David,
it might be so 't he could go. What do you think, judge?"
The judge nodded his head, thoughtfully.
"I don't see why it couldn't be done, Miss Peace," he said, kindly.
"David is a good fellow, and has friends wherever he is known; I
should think it might very well be done, if the right person takes it
up."
"I--I've had no great experience," faltered Anne Peace, looking down,
"but I'm kin to David, you know, and as he has no one nearer living, I
took it upon myself to carry round a paper and see what I could raise.
I came to you first, judge, as you've always been a good friend to
David. I've got twenty-five dollars already--"
"I thought you said you came to me first," said the judge, holding out
his hand for the paper. "What's this? A friend, twenty-five dollars?"
"Yes," said Anne Peace, breathlessly. "They--they didn't wish their
name mentioned--"
"Oh, they didn't, didn't they?" muttered the judge, looking at her
over his spectacles. Such a helpless look met his--the look of
hopeless innocence trying to deceive and knowing that it was not
succeeding--that a sudden dimness came into his own eyes, and he was
fain to take off his spectacles and wipe them, just as if he had been
looking through them. And through the mist he seemed to see--not Miss
Anne Peace, in her best bonnet and her cashmere shawl, but another
Anne Peace, a little, brown-eyed, slender maiden, sitting on a brown
bench, looking on with rapture while David Means ate her luncheon.
It was the judge's turn to clear his throat.
"Well, Anne," he said, keeping his eyes on the paper, "this--this
unknown friend has set a good example, and I don't see that I can do
less than follow it. You may put my name down for twenty-five, too."
"Oh, judge," cried Miss Peace, with shining eyes. "You are too good. I
didn't expect, I'm sure--well, you _are_ kind!"
"Not at all! not at all!" said the judge, gruffly (and indeed,
twenty-five dollars was not so much to him as it was to "them," who
had made the first contribution).
"You know I owe David Means something, for licking him when he--"
"Oh, don't, Dan'el--judge, I should say," cried Anne Peace, in
confusion. "Don't you be raking up old times. I'm sure I thank you a
thousand times, and so will Delia, when she--"
"No, she won't," said the judge. "Tell the truth, Anne Peace! Delia
will say I might have given fifty and never missed it. There! I won't
distress you, my dear. Good day, and all good luck to you!" and so
ended Miss Peace's first call.
With such a beginning, there was no doubt of the success of the
subscription. Generally, in Cyrus, people waited to see what Judge
Ransom and Lawyer Peters gave to any charity, before making their own
contribution. "Jedge Ransom has put down five dollars, has he? Well
he's wuth so much, and I'm wuth so much. Guess fifty cents will be
about the right figger for me:" this is the course of reasoning in
Cyrus. But with an unknown friend starting off with twenty-five
dollars and Judge Ransom following suit, it became apparent to every
one that David Means must go to Florida, whatever happened. The dollar
and five-dollar subscriptions poured in rapidly, till, one happy day,
Anne Peace stood in her little room and counted the full amount out on
the table, and then sat down (it was not her habit to kneel, and she
would have thought it too familiar, if not actually popish) and
thanked God as she had never found it necessary to thank Him for any
of the good things of her own life.
So David Means went to Florida, and his wife and two children went
with him. This had been no part of the original plan, but at the bare
idea of his going without her, Mrs. Means had raised a shrill cry of
protest. "What? David go down there, and she and the children stay
perishing at home? she guessed not. If Florida was good for David, it
was good for her, too, and she laid up ever sence spring, as she might
say, and with no more outing than a woodchuck in January. Besides, who
was to take care of David, she'd like to know? Mis' Porter's folks,
who had a place there? She'd like to know if she was to be beholden to
Jane Porter's folks for taking care of her lawful husband, and like
enough laying him out, for she wasn't one to blind herself, nor yet to
set herself against the will of Providence." Doctor Brown stormed and
fumed, but Anne Peace begged him to be quiet, and "presumed likely"
she could raise enough to cover the expenses for Delia and the two
older children. 'Twas right and proper, of course, that his wife
should go with him, and David wouldn't have any pleasure in the trip
if he hadn't little Janey and Willy along. He did set so by those
children, it was a privilege to see them together; he was always one
to make of children, David was.
She did raise the extra money, this sweet saint, but she ate no meat
for a month, finding it better for her health. Joey and Georgie Means,
however, never wanted for their bit of steak at noon, and grew fat and
rosy under Miss Peace's kindly roof.
It was a pathetic sight when the sick man took leave of the little
group of friends and neighbours that gathered on the platform at the
station to bid him farewell. He had lost courage, poor David; perhaps
he had not very much to start with, and things had gone hard with him
for a long time. He knew he should never see these faces again, this
homely, friendly place. He gazed about with wistful eyes, noting every
spot in the bare little station. He had known it all by heart, ever
since he was a child, for his father had been station-master. He could
have built the whole thing over, with his eyes shut, he thought, and
now he should never see it again. Yet he was glad to go, in a way,
glad to think, at least, that he should die warm, as his wife
expressed it, and that his tired eyes were going to look on green and
blossoming things, instead of the cold, white beauty which meant
winter to him.
He had scarcely ever left Cyrus for more than a day or two; he had a
vague idea that it was not creditable to go to the other world, and be
able to give so little account of this one. Now, at least, he should
be able to look his seafaring grandfather and his roving uncle in the
face, if so be he should happen to meet them "over yender."
He stood on the platform with his youngest child clasped close in his
arms. This was the hardest part of all, to leave the children. His
wife and the two older children had already taken their places in the
car, and the good-natured conductor stood with his watch in his hand,
willing to give David every second he conscientiously could. He came
from East Cyrus himself, and was a family man.
Anne Peace stood close by, holding fast the hand of little Joey.
Strange sounds were in her ears, which she did not recognize as the
beating of her own heart; she kept looking over her shoulder, to see
what was coming. Her eyes never left David's face, but they were
hopeful, even cheerful eyes. She thought he would come back much
better, perhaps quite well. Doctor had said there was a chance, and
she did hear great things of Florida.
And now the conductor put up his watch and hardened his heart. "Come,
David, better step inside now. All aboard!"
"Good-by, David!" cried Doctor Brown, waving a friendly hand.
"Good-by, David!" cried Anne Peace, lifting little Joey in her arms,
though he was far too heavy for her.
"Look at father, Joey dear, throw a kiss to father; good-by, good-by,
David!" The train moved out of the station, but David Means, his eyes
fixed on the faces of his children, had forgotten to look at Anne
Peace.
Winter came, and a bitter winter it was. No one in Cyrus could
remember such steady cold, since the great winter of sixty years ago,
when the doctor's grandfather was frozen to death, driving across the
plains to visit a poor woman. The horse went straight to the place,
his head being turned that way and his understanding being good; but
when the farmer came out with his lantern, there sat the old doctor
stiff and dead in his sleigh. Those were the days when people, even
doctors, had not learned how to wrap up, and would drive about all
winter with high, stiff hats and one buffalo robe, not tucked in, as
we have them nowadays, but dropping down at their feet. There was
small chance of our Doctor Brown's freezing to death, in his
well-lined sleigh, with his fur cap pulled down over his nose and his
fur coat buttoned up to his chin and the great robes tucked round him
in a scientific manner. Still, for all that, it was a bitter winter,
and a good many people in Cyrus and elsewhere, who had no fur coats,
went cold by day and lay cold by night, as one good lady pathetically
expressed it. There was little snow, and what there was fell in
wonderful crystals, fairy studies in geometry, which delighted the
eyes of Joey and Georgie Means, as they trotted to school, with Miss
Peace's "nuby" over one little head and her shawl over the other.
Every morning the sun rose in a clear sky, shining like steel; every
evening the same sky glowed with wonderful tints of amethyst and
tender rose, fading gradually, till all was blue once more, and the
stars had it all their own way, throbbing with fierce, cold light.
It was a great winter for Joey and Georgie! They never thought of its
being too cold, for every morning their toes were toasted over the
fire before schooltime, as if they had been muffins, and they were
sent off nice and hot, with a baked potato in each pocket, in case
their hands should be cold through the two pairs of thick mittens
which Aunt Peace had provided.
Then, when they came home, dinner was waiting, such a dinner as they
were not in the habit of having; a little mutton pie, or a smoking
Irish stew, with all the dumplings and gravy they wanted (and they
wanted a great deal), and then pancakes, tossed before their very
eyes, with a spoonful of jam in the middle of each, or blanc-mange
made in the shape of a cow, which tasted quite different from any
other blanc-mange that ever was. Also, they had the freedom of the
corn-popper, and might roast apples every evening till bedtime.
Doctor Brown shook his head occasionally, and told Anne Peace she
would unfit those children for anything else in life than eating good
things; but it was very likely that was jealousy, he added, for
certainly his medicines had never given the children these rosy cheeks
and sparkling eyes.
And when bedtime came, and the two little brown heads were nestled
down in the pillows of the big four-poster in the warm room, Anne
Peace would humbly give thanks that they had been well and happy
through another day, and then creep off to the cold, little room which
she had chosen this winter, "because it was more handy." Often, when
awakened in the middle of the night by the sharp, cracking frost
noises, which tell of intensest cold, she would creep in to feel of
the children, and make sure that they were as warm as two little
dormice, which they always were. I do not know how many times she took
a blanket or comforter off her own bed to add to their store; but I do
know that she would not let Jenny Miller go into her room to see. She
almost rejoiced in the excessive cold, saying to herself with
exultation, "Fifteen below! well, there! and I s'pose it's like summer
in Florida, this minute of time!" And then she fancied David sitting
under an orange-tree, fanning himself, and smiled, and went meekly to
work to break the ice in her water-butt.
Every week letters came from David Means to his children, telling them
of the beauty all around him and wishing they were there. He said
little of his health, but always assured them that Janey and Willy
were real smart, and sent his love to Anne Peace and his remembrance
to all friends at home.
The letters were short, and each time they grew a little shorter, till
by and by it was only a postal card, written in a faint and trembling
hand, but saying that the weather was fine, and father was so glad to
get their little letter, and he would write more next time, but was
very busy just now. When she read one of these, Anne Peace would go
away into her little cold room for a while, and then would come back
smiling and say that now they must write a real _good_ letter to
father, and tell him how well they were doing at school.
At last came a week when there was no postal card; another week, and
there came a letter edged with black and written in Mrs. Means's hand.
The children were at school when it came, and Jenny Miller, coming in
by chance to bring a pot of head-cheese of her mother's making, found
Miss Peace crouching in the corner of the sofa, weeping quietly, with
the letter lying on her lap.
"Why, Miss Peace," cried Jenny, frightened at the sight of tears in
those steadfast eyes, "What is the matter? Do tell me, dear! Why,
you're real cold in here. I do believe the fire has gone out. You've
had bad news, Miss Peace, have you? Do tell me, that's a dear soul,
and don't cry."
"Yes," said Anne Peace. "The fire is out, Jenny, and David is dead."
She held out the letter, saying something about
"privilege--think--rest;" but Jenny Miller was already on her knees,
putting kindlings into the stove at a reckless rate. Then, when the
fire was crackling merrily, she ran to fetch a shawl and wrapped it
round the poor trembling shoulders, and chafed the cold hands in her
own warm, young fingers. But soon Miss Peace grew uneasy; she was not
used to being "done for," having only the habit of doing for others.
She pointed eagerly to the letter. "Read it, Jenny," she said,
anxiously. "I--I am all right, dear. It's come rather sudden, that's
all, and those poor little children--but read the letter." The words
died away, and Jenny, sitting down beside her, took the paper and
read.
It began "Friend Anne," and went on to say that the writer's poor
husband died yesterday, and she was left, as she always knew she
should be, a widow with four children. It did seem to her as if he
might have been let die to home, instead of being carted all the way
down there and then have to send the remains back. She had to promise
him she would send them back, though it did seem a pity with the
beautiful "semetary" they had there, and full of Northern folks as it
would hold and the undertaker a perfect gentleman, if she ever saw
one. But the widow hoped she knew her duty, and she would not wish to
be thought wanting in anything.
Now she supposed they would want to know how David passed away, though
she had no "strenth" to write, not having had her clothes off for days
or, you might say, weeks, nor slep' one consektive hour the last ten
nights. Well, he had seemed to gain a little when they first came, but
it wasn't no real gain, for he lost it all again and more too. The
pounds just fell off from that man, it seemed as if you could see them
go. The last month he fairly pined away, and she thought right to let
the folks at home know that he was called to depart, but he wouldn't
hear to it. "He said, Delia, he said, if you want me to die easy, he
said, don't let on to no one at home but what I'm doing all right." So
she set by and held her peace, though it went against her conscience.
Last Monday he couldn't leave his bed, and she said, "David, she said,
you never will leave it till you're carried," and he said, p'raps
'twas so, but yet he wouldn't allow it, for fear of scaring the
children. So that night he sat up in bed and his arms went out and he
said "Home!" just that word, two or three times over, and dropped
back and was gone. There she was, a widow with four small children,
and what she should do she didn't know.
Away there in a strange land as you might say, if it _was_ all one
country, it did seem as if them as sent them might have thought of
that and let them stay at home among their own folks. Not but what
there was elegant folks there. Everybody hed been as kind as could be;
one lady who was in "morning" herself had lent her a bonnet to wear to
the funeral (for she wasn't one to send the remains off without
anything being said over them); it was a real handsome bonnet, and she
had taken a pattern of it, to have one made for herself. The lady was
from New York way, and real stylish.
Mrs. Means intended to stay on a spell, as the money was not all gone,
and her strength needed setting up, after all she had been through.
Mr. Tombs, the undertaker, said he never saw any one bear afflicktion
so; she told him she was used to it. He was a perfect gentleman, and a
widower himself, so he could feel for her. Miss Peace might be
thankful that she was never called on to bear afflicktion, with no one
but herself to look out for; not but what 'twas lonesome for her, and
Mrs. Means supposed she'd be glad enough to keep Georgie and Joey on
a spell longer for company. Tell them they are poor orphans now, with
no father to earn their bread. The writer wished her husband's remains
to be buried in his father's lot, as she had no money to buy one. Miss
Peace might see if any one felt to put up a moniment for David; he
hadn't an enemy in the world, and he never begredged a dollar when he
had it to give, for anything there was going. If he had thought a
little more about her, and less about everybody's cat and dog, she
might have something now to put bread in her children's mouths, let
alone her own. Not that she had any appetite, a flea wouldn't fatten
on what she ate. Lawyer Peters was his mother's third cousin if she
was living. He spent more on those girls of his than would clothe the
writer and her children for a year.
The remains went by the same boat with this letter, so Miss Peace
would know when to expect them. Mrs. Means looked to her to see that
David had a decent funeral; a handsome one she couldn't expect, folks
in Cyrus were close enough about all that didn't go on their own
backs, though she shouldn't wish it said.
So now there was no more, from Miss Peace's unfortunate friend, "the
Widow Means."
After reading this precious epistle, Jenny Miller found herself,
perhaps for the first time in her life, with nothing to say. She
could only sit and press her friend's hand, and thrill, as a girl
will, at the touch of a sorrow which she only now began dimly to
guess. It was Miss Peace who broke the silence, speaking in her usual
quiet tone.
"Thank you, Jenny, dear! I'm sure it was a privilege, having you come
in just now. David Means was kin to me, you know, and I always set by
him a great deal; and then the poor little children!" she faltered
again for an instant, but steadied her voice and went on: "You'd
better go home now, dear, for the fire is going beautiful, and I don't
need anything. I--I shall have to see to things for the funeral, you
know. And don't forget to thank your mother for the cheese. It looks
real good, and Georgie doos like it the best of anything for
breakfast. I guess I'll get on my bonnet, and go to see Abel Mound,
the sexton."
But here Jenny found her voice, and protested. Miss Peace should not
have anything at all to do with all that. 'Twasn't fitting she should,
as the nearest kin poor Mr. Means had in Cyrus. Her father would see
to it all, Jenny knew he would, and Doctor Brown would help him. She
would go herself and speak to the doctor this minute. Miss Peace would
have to be here to tell the children when they came home from school,
poor little things! and that was all she should do about it.
Anne Peace hesitated; and then Jenny had an inspiration, or, as she
put it in telling Tudie Peaslee afterwards, "a voice spoke to her."
"Miss Peace," she said timidly, "I--I don't suppose you would feel to
pick those flowers you were going to send over to Tupham for the
Sunday-school festival? I know they kind o' lot on the flowers you
send, 'cause they're always so fresh, and you do them up so pretty.
But if you don't feel to do it, I can send them word, or ask some one
else"--
"The idea!" cried Anne Peace, brightening up. "I forgot the flowers,
Jenny, I did so! I should be pleased to pick them, and I'll do it this
minute. There--there isn't anything I should like so well. And I do
thank you, dear, and if you really think your father wouldn't mind
seeing--I am sure it is a privilege to have such neighbours, I always
say. There couldn't anybody be more blessed in neighbours than I have
always been."
In ten minutes Miss Peace was at work in her garden, cutting,
trimming, tying up posies, and finding balm for her inward wound in
the touch of the rose-leaves, and in the smell of mignonette, David's
favourite flower. No one in Cyrus had such mignonette as Miss Peace,
and people thought she had some special receipt for making it grow and
blossom luxuriantly; but she always said no, it was only because she
set by it. Folks could most always grow the things they set most
store by, she thought.
So the Sunday-school festival at Tupham Corner was a perfect blaze of
flowers, and the minister in his speech made allusion to generous
friends in other parishes, who sent of their wealth to swell our
rejoicings, and of their garden produce to gladden our eyes; but while
the eyes of Tupham were being gladdened, Anne Peace was brushing
Joey's and Georgie's hair, and tying black ribbons under their little
chins, smiling at them through her tears, and bidding them be brave
for dear father's sake, who was gone to the best home now, and would
never be sick any more, or tired, or--or sad.
It was a quiet funeral: almost a cheerful one, the neighbours said, as
they saw the little room filled with bright flowers (they all seemed
to smell of mignonette, there was so much of it hidden among the
roses), and the serene face of the chief mourner, who stood at the
head of the coffin, with a child in either hand. It was an unusual
thing, people felt. Generally, at Cyrus funerals, the mourners stayed
up-stairs, leaving the neighbours to gather round the coffin in the
flower-scented room below; but it did not seem strange in Anne Peace,
somehow, and, after the first glance, no one could fancy any one else
standing there. The old minister, who had christened both David and
Anne on the same day, said a few gentle, cheering words, and the choir
sang "Lead, kindly Light;" then the procession went its quiet way to
the churchyard, and all was over.
Jenny Miller and the doctor followed Miss Peace home from the
churchyard, but made no attempt to speak to her. She seemed
unconscious of any one save the children, to whom she was talking in
low, cheerful tones. The doctor caught the words "rest," "home,"
"happiness;" and as she passed into the house he heard her say
distinctly: "Blessed privilege! My children now, my own! my own!"
"So they are!" said Doctor Brown, taking off his glasses to clear
them. "So they are, and so they will remain. I don't imagine Delia
will ever come back, do you, Jenny?"
"No," said Jenny, "I don't. She'll marry the undertaker before the
year is out."
And she did.
THE END.
* * * * *
Transcriber's Notes
Original spelling and punctuation have been preserved except for the
joining of common contractions.
Page 8: Added closing quotes: (you seemed all right when I went out.")
Page 9: Phoebe had oe ligature in original book. (you've heard me tell
how my Aunt Phoebe 'Lizabeth)
Page 56: Removed extra quotation mark before I: ("You are too good. I
didn't expect, I'm sure--well, you are kind!")
End of Project Gutenberg's "Some Say", by Laura Elizabeth Howe Richards
*** | {
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} | 6,181 |
ZYTARUK: Dialing for doctors, lining up for labs
Clearly, our medical system needs a serious booster shot of efficiency, in both arms.
Tom Zytaruk
Jan. 27, 2016 4:00 p.m.
'It remains abundantly clear that
So let it be written…
I am fortunate to have a good family doctor. He is competent, caring, and when I have an appointment, he takes time to listen to me and consider the best course of action. I don't get shunted out of his office like I'm just another ear tag in the cattle line.
In other words, I have no complaint with him. But securing an appointment with my doctor can be downright stressful, as the competition to see him is fierce.
It starts the second the clinic opens, early in the morning. I call the clinic, get a machine or busy signal, hit redial and repeat until I get through, to hopefully secure a coveted appointment for that day. Often within minutes he's booked up.
I know other patients have to be performing the same ritual. It's kind of like dialing up a morning radio show, hoping you get through for a chance to win the big prize.
But let's say you get lucky and score an appointment. Maybe then you're sent to a medical lab for a test.
Negotiating this can also be stressful. I think back upon arriving at LifeLabs on a Saturday morning, before the office opened at 7 o'clock. It was still dark outside. When I stepped out of the elevator, 13 people were already lined up in the hallway in front of me.
A bunch of people arrived right behind me, and that snaking line soon became more like an anaconda than a garter snake. The place should be called LineLabs.
Frustrating as it was for the patients in queue, I can only imagine what it must be like for the lab employees who arrive at work in the morning, coffee cup in hand, to a huge lineup of anxious clients. If I came to work every day with 30 or 40 people waiting to speak with me before I even got to my desk, I think I'd go mad. Well, madder than I already am, anyway.
Something's gotta give here.
The last time a doctor made a house call on my behalf was when I was a teenager living in North Delta. I am now 50 years old.
Conversely, just a few years ago when we were last in Guatemala, my wife's native country, I got sick — no big deal, just a flu — and they called a doctor. Bam! He was at the house, medical bag in hand, within the hour. And the service was exceptional.
Guatemala City's population, I might add, is roughly 3,354,000 whereas the population of Metro Vancouver is 2,513,000.
The Central American city has a much higher population density than we do here and yet doctors there gladly make house calls within the hour, while trying to see a doctor up here — at their clinic, mind you — can be quite the adventure.
Who's the Third World country, eh? Makes you wonder.
Again, I consider myself fortunate to actually have a family doctor, scheduling aggravations or no. In 2013 B.C.'s provincial government launched a $132-million initiative called A GP for Me, to address the shortage of doctors here.
By the end of that year, roughly 200,000 British Columbians did not have a family doctor and there was no significant change in 2014 either. The numbers aren't yet in for 2015.
Some headway has been made in White Rock and South Surrey, where the program has found doctors for 12,000 residents who didn't have one. In Surrey and North Delta, 15,000 residents were actively looking for a family doctor in 2013-14 but it's hoped the next evaluation will reveal some improvement. As it now stands, while 14 new doctors have since set up practice here, Surrey is still growing by roughly 1,200 new residents per month, and 27 per cent of our local doctors are expected to retire in the next nine years.
It is encouraging they are trying to fix this but it remains abundantly clear that, whether you have a family doctor or not, the availability of prompt medical service has not kept up with population growth. Unless this improves, long lineups and fevered phone calling for appointments will continue to be the norm.
That said, still I wonder why we can't do what's done in Guatemala, with its higher population density.
Is there a quick fix?
Don't hold your breath too long. You might need to see a doctor for that, and you could be in for a long wait…
So let it be done.
Tom Zytaruk is a staff writer with the Now. Email him at tom.zytaruk@thenownewspaper.com
B.C. VIEWS: This is your province on weed
LETTER: Shut down oil production until after U.S. election | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,669 |
This report forecasts US engineering plastic resin demand in pounds to 2019. Total demand is segmented by type of resin in terms of nylon, acrylonitrile-butadiene-styrene, polycarbonate, thermoplastic polyester, and other resins. Total demand is also segmented by market as follows: motor vehicles, electrical and electronic, medical and consumer, construction, and other markets.
To illustrate historical trends, total demand is provided in an annual series from 2004 to 2014; the various segments are reported at five-year intervals for 2009 and 2014. A Market Environment section provides pertinent background on historical market size and trends, key economic indicators, regulatory and environmental factors, and trends in trade. A Segmentation and Forecasts section defines resins and markets, discusses market drivers and constraints, identifies substitute products when applicable, and assesses the impact of key drivers and constraints on each resin and market segment over the forecast period. | {
"redpajama_set_name": "RedPajamaC4"
} | 4,246 |
Q: Como cambiar el idioma a la barra de herramientas del ReportViewer Winforms Resulta que tengo una aplicación de Windows Forms y necesito que el idioma de la barra de herramientas del Report Viewer me salga en español y no en Inglés
Gracias
Saludos
MI código es el siguiente donde armo el informe:
CultureInfo.CurrentUICulture = new CultureInfo("es-ES", false);
string report = "DALISF.rdlc";
reportViewer1.LocalReport.ReportEmbeddedResource = report;
reportViewer1.LocalReport.ReportPath = Path.Combine(@"..\FORMATOS\", report);
ReportDataSource rds1 = new ReportDataSource("Familias", Agregar);
reportViewer1.LocalReport.DataSources.Clear();
reportViewer1.LocalReport.DataSources.Add(rds1);
///<summary>
/// Vista de impresión de la página
///</summary>
reportViewer1.SetDisplayMode(DisplayMode.PrintLayout);
// Esta linea que tenemos abajo sirve para rellenar toda la página en donde se muestra el informe
reportViewer1.Dock = DockStyle.Fill;
///<summary>
/// Todo el ancho de la página
///</summary>
reportViewer1.ZoomMode = ZoomMode.PageWidth;
//reportViewer1.PrintDialog();
// Añadimos los parámetros que van a tener los reportes en cuestión (tenemos que ponerle los mismos nombres que tienen en el informe)
mAnadirParametros("CodigoDesdeFamilia", DesdeCodigo);
mAnadirParametros("CodigoHastaFamilia", HastaCodigo);
mAnadirParametros("NombreDesdeFamilia", DesdeNombreFam);
mAnadirParametros("NombreHastaFamilia", HastaNombreFam);
mAnadirParametros("Infopie", "Para cualquier duda o consulta contacte con CTI Canarias. Tlf: 922 365 002 - info@cticanarias.com");
reportViewer1.LocalReport.SetParameters(reportv); //Añadimos los parámetros al reporte.
reportViewer1.RefreshReport(); // El informe se refresca cada vez que lo carga.
reportViewer1.RenderingComplete += new RenderingCompleteEventHandler(RenderingCompleteEventHandler);
A: Ya lo arreglé usando la interfaz IReportViewerMessages mediante la cúal implementamos la misma y tenemos diversas acciones para cambiar al idioma que deseemos la barra de herramientas del report viewer. Por aquí les dejo la documentación por si a alguien más le sirve:
https://docs.microsoft.com/es-es/previous-versions/ms255036(v=vs.140)
A: Deberias cambiar la cultura pero de toda la aplicacion, para eso en el Main() de Program.cs deberias agregar la linea
CultureInfo ci = new CultureInfo("es-ES");
Thread.CurrentThread.CurrentCulture = ci;
Thread.CurrentThread.CurrentUICulture = ci;
De esta forma defines la cultura de frma global para que afecte tambien a los controles
A: Para que todos aquellos que tengan el mismo problema puedan solucionarlo, voy al poner el snippet con el código necesario para que funciona, en el idioma en que deseen.
Para ello vamos a una clase determinada en donde cargamos la interfaz. Para cargarla ponemos lo siguiente:
public class CCustomMessageClass : IReportViewerMessages
{
public string BackButtonToolTip
{
get { return ("Volver al informe primario"); }
}
public string BackMenuItemText
{
get { return ("Add your custom text here."); }
}
public string ChangeCredentialsText
{
get { return ("Add your custom text here."); }
}
public string CurrentPageTextBoxToolTip
{
get { return ("Página actual"); }
}
public string DocumentMapButtonToolTip
{
get { return ("Add your custom text here."); }
}
public string DocumentMapMenuItemText
{
get { return ("Add your custom text here."); }
}
public string ExportButtonToolTip
{
get { return ("Exportar"); }
}
public string ExportMenuItemText
{
get { return ("Exportar"); }
}
public string FalseValueText
{
get { return ("Add your custom text here."); }
}
public string FindButtonText
{
get { return ("Buscar"); }
}
public string FindButtonToolTip
{
get { return ("Buscar"); }
}
public string FindNextButtonText
{
get { return ("Buscar siguiente"); }
}
public string FindNextButtonToolTip
{
get { return ("Buscar siguiente"); }
}
public string FirstPageButtonToolTip
{
get { return ("Primera página"); }
}
public string LastPageButtonToolTip
{
get { return ("Última página"); }
}
public string NextPageButtonToolTip
{
get { return ("Página siguiente"); }
}
public string NoMoreMatches
{
get { return ("No se encontraron más entradas coincidentes"); }
}
public string NullCheckBoxText
{
get { return ("Add your custom text here."); }
}
public string NullCheckBoxToolTip
{
get { return ("Add your custom text here."); }
}
public string NullValueText
{
get { return ("Add your custom text here."); }
}
public string PageOf
{
get { return ("de"); }
}
public string PageSetupButtonToolTip
{
get { return ("Configurar página"); }
}
public string PageSetupMenuItemText
{
get { return ("Configurar página"); }
}
public string ParameterAreaButtonToolTip
{
get { return ("Add your custom text here."); }
}
public string PasswordPrompt
{
get { return ("Add your custom text here."); }
}
public string PreviousPageButtonToolTip
{
get { return ("Página anterior"); }
}
public string PrintButtonToolTip
{
get { return ("Imprimir"); }
}
public string PrintLayoutButtonToolTip
{
get { return ("Diseño de impresión"); }
}
public string PrintLayoutMenuItemText
{
get { return ("Diseño de impresión"); }
}
public string PrintMenuItemText
{
get { return ("Imprimir"); }
}
public string ProgressText
{
get { return ("Cargando"); }
}
public string RefreshButtonToolTip
{
get { return ("Actualizar"); }
}
public string RefreshMenuItemText
{
get { return ("Actualizar"); }
}
public string SearchTextBoxToolTip
{
get { return ("Buscar texto en informe"); }
}
public string SelectAValue
{
get { return ("Selecciona un valor"); }
}
public string SelectAll
{
get { return ("Selecciona todo"); }
}
public string StopButtonToolTip
{
get { return ("Detener representación"); }
}
public string StopMenuItemText
{
get { return ("Detener"); }
}
public string TextNotFound
{
get { return ("No se han encontrado resultados que coincidan con el criterio de búsqueda."); }
}
public string TotalPagesToolTip
{
get { return ("Total de páginas"); }
}
public string TrueValueText
{
get { return ("Add your custom text here."); }
}
public string UserNamePrompt
{
get { return ("Add your custom text here."); }
}
public string ViewReportButtonText
{
get { return ("Ver informe"); }
}
public string ViewReportButtonToolTip
{
get { return ("Ver informe"); }
}
public string ZoomControlToolTip
{
get { return ("Zoom"); }
}
public string ZoomMenuItemText
{
get { return ("Add your custom text here."); }
}
public string ZoomToPageWidth
{
get { return ("Ancho de página"); }
}
public string ZoomToWholePage
{
get { return ("Toda la página"); }
}
}
Y luego a la hora de cargar el reporte con parámetros y demás le añadimos la interfaz que acabamos de añadir de la siguiente manera:
1º Implementamos una instancia de la interfaz recién creada:
CCustomMessageClass myMessageClass = new CCustomMessageClass();
2º Para utilizarla lo hacemos de la siguiente manera (una vez que se añadan los parámetros al informe):
reportViewer1.Messages = myMessageClass;
reportViewer1.RefreshReport();
| {
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} | 3,668 |
Royal Oak è una città degli Stati Uniti d'America, nella Contea di Oakland, nello Stato del Michigan. È un sobborgo settentrionale di Detroit.
Altri progetti
Collegamenti esterni | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 3,942 |
San Giorgio di Piano es un municipio situado en el territorio de la Provincia de Bolonia, en Emilia-Romaña, (Italia).
Demografía
Enlaces externos
Sitio web oficial de San Giorgio di Piano
Localidades de la provincia de Bolonia
San Giorgio di Piano | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 6,113 |
{"url":"https:\/\/gamedev.stackexchange.com\/questions\/165646\/creating-rotation-matrix-in-a-left-handed-coordinate-system-from-a-direction-vec","text":"# creating rotation matrix in a left handed coordinate system from a direction vector\n\nI have been looking at creating rotation matrices from a direction vector in the OpenCV coordinate space and there is one thing that has me slightly confused. Here is an OpenCV rotation matrix, that I got from an opencv function (let us call this matrix r):\n\narray([[-0.4136457 , -0.19724711, 0.88881427],\n[-0.57926765, 0.810177 , -0.08978985],\n[-0.70238609, -0.55200255, -0.44938511]])\n\n\nNow, starting from the z-axes unit vector, I am trying to recreate this vector. So, I have the following code:\n\ndef basis(v):\nv = v \/ np.linalg.norm(v)\nif v[0] > 0.9:\nb1 = np.asarray([0.0, 1.0, 0.0])\nelse:\nb1 = np.asarray([1.0, 0.0, 0.0])\n\nb1 -= v * np.dot(b1, v)\nb1 *= np.reciprocal(np.linalg.norm(b1))\nb2 = np.cross(v, b1)\nreturn b1, b2, v\n\n\nI can call this function as:\n\nx, y, z = basis(r[:, 2])\n\n\nThen I compute the rotation matrix as:\n\navg = np.asarray([[x[0], y[0], z[0]],\n[x[1], y[1], z[1]],\n[x[2], y[2], z[2]]])\n\n\nNow running this code returns:\n\narray([[ 0.4582676 , 0. , 0.88881427],\n[ 0.17414826, -0.98061724, -0.08978985],\n[ 0.8715866 , 0.19593324, -0.44938511]])\n\n\nSo, the signs along the x and y-axes are flipped.\n\nNow, in my basis function, if I change the line b1 = np.asarray([1.0, 0.0, 0.0]) to b1 = np.asarray([-1.0, 0.0, 0.0]). It returns with the correct sign like:\n\narray([[-0.4582676 , 0. , 0.88881427],\n[-0.17414826, 0.98061724, -0.08978985],\n[-0.8715866 , -0.19593324, -0.44938511]])\n\n\nI am guessing this has something to do with the handedness as the opencv origin is at the top left corner and the y axes is increasing in the downward direction rather than upward but the thing that has me confused is why does chaging the sign of the x coordinate of the unit vector makes a difference? I was expecting to have to change the other condition i.e. b1 = np.asarray([0.0, 1.0, 0.0]) to have the sign flip in the y coordinate.\n\n\u2022 This follows directly from the left-hand rule, doesn't it? If you hold your left hand so that your fingers curl along the angle from v to b1, your thumb points in the direction of their cross product b2. If you negate b1, then you have to turn your hand around to do the same trick, pointing your thumb in the opposite direction. Starting from a negated x gives you a negated y. Have I misunderstood what you're asking? \u2013\u00a0DMGregory Nov 26 '18 at 13:25","date":"2019-01-21 11:34:11","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5000855326652527, \"perplexity\": 1196.382951363942}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-04\/segments\/1547583792338.50\/warc\/CC-MAIN-20190121111139-20190121133139-00373.warc.gz\"}"} | null | null |
\section{Introduction}
There is strong theoretical evidence that quantum chromodynamics (QCD) at high-energy (or small Bjorken-$x$) leads to a non-linear regime where gluon recombination or unitarity effects become important \cite{sg,mv}, resulting in a saturation of parton densities in hadrons and nuclei.
The quest for experimental evidence of the possible signature of gluon saturation phenomenon has been the program of various past or existing experiments from HERA and RHIC to the LHC, and future experiments such as an Electron-Ion Collider (EIC) \cite{eic} and the LHeC \cite{lhec}. Nevertheless, experimental evidence that can unarguably point towards gluon saturation phenomenon, has been elusive so far. This is because the experiments currently at our disposal are limited in their kinematic coverage, and often other approaches provide alternative descriptions of the same sets of data.
An effective field theory approach that describes the high-energy limit of QCD is the colour glass condensate (CGC), see the review \cite{Gelis:2010nm}. In this formalism, the standard quantum evolution equations (with large logarithms of $1/x$ resummed), lead to a situation in which the occupancy of the slow modes in the hadron is so high than they can be treated classically, with the fast modes considered as sources.
The corresponding renormalisation group equations, known in the limit of scattering of a dilute probe on a dense hadron, are the so-called Jalilian-Marian-Iancu-McLerran-Weigert-Leonidov-Kovner (JIMWLK) hierarchy of equations \cite{jimwlk} or, in the large $N_c$ limit, the Balitsky-Kovchegov (BK) equation \cite{bk}, presently known to next-to-leading accuracy \cite{Balitsky:2008zza,Kovner:2013ona}.
One of the most crucial tests of the CGC (or saturation) approach has been its success in the description of the highly precise combined data of the proton structure at HERA \cite{rcbk,ip-sat-a,b-cgc-a} alongside data from exclusive diffractive processes in electron-proton collisions, such as exclusive vector meson production and deeply virtual Compton scattering (DVCS) \cite{ip-sat-a,b-cgc-a}. Nevertheless, the standard DGLAP-type approaches - without inclusion of any saturation effect - give an equally good description of the same data. While the CGC description can be considered more economical due to the use of a significantly smaller number of fitting parameters, it is limited to small-$x$ data and restricted to the gluon sector. On the other hand, in addition to DIS and diffractive processes \cite{rcbk,ip-sat-a,b-cgc-a,ip-sat0, ip-sat1, watt-bcgc,ip-g2}, within the CGC framework it is also possible to simultaneously describe other high-energy hadronic interactions in a regime not currently accessible to approaches that rely on collinear factorisation. For example, in proton-proton \cite{pp-LR} and nuclear collisions \cite{jav-d,aa-LR,hic-ap, raju-glasma,all-pa} several observables have been successfully addressed: single inclusive hadron \cite{pa-raju,jav-pa,pa-R,pa-t,pa-jam,pa-ana} and prompt photon \cite{pa-R,jr-p} production, and semi-inclusive photon-hadron \cite{jr-p,p-h} and dihadron \cite{di-all} productions. For a recent review, see Ref.\,\cite{Albacete:2014fwa} and references therein.
Exclusive diffractive vector meson production provides a rich testing ground of many QCD novel properties \cite{ip-sat-a,b-cgc-a,ip-sat0, ip-sat1, watt-bcgc,ip-g2,diff3,ann2,diff4,stan,diff2,ggg,exc}. In particular, by measuring the squared momentum transfer $t$, one can study the transverse spatial distribution of the gluons in the hadron wave function that cannot be probed in inclusive DIS. In this respect, new experimental measurements are under way. The LHCb and ALICE collaborations have recently released new data on $J/\psi$ photoproduction with photon-proton center-of-mass energies up to about 1.3 TeV \cite{lhcb,lhcbn,TheALICE:2014dwa}, the highest energy ever measured so far in this kind of reaction.
Alongside this, the H1 Collaboration also recently reported some new data for $J/\psi$ with improved precision \cite{h1-2013}. On the other hand, the recently released high-precision combined HERA data \cite{Aaron:2009aa,Abramowicz:1900rp} that were not available at the time of previous studies of diffractive processes \cite{eic,lhec,ip-sat0, ip-sat1, watt-bcgc,watt-2}, provide extra important constraints on saturation models \cite{ip-sat-a,b-cgc-a}.
In this work, we analyse these data on exclusive photoproduction of vector mesons off the proton and provide predictions for the kinematics accessible in future experiments. We show that the freedom to choose the charm mass in the range consistent with global analysis of inclusive observables, results in sizable uncertainties for the total cross-section of elastic photoproduction of vector mesons. Nevertheless, we show that, even with these uncertainties, the recent LHC data \cite{lhcbn,TheALICE:2014dwa} seem to favour the saturation picture. We systematically study elastic diffractive production of different vector mesons $J/\psi$, $\psi(2s)$, $\phi$ and $\rho$ off protons and investigate which vector meson production is more sensitive to saturation physics and what measurement can potentially be a better probe of the signal. In particular, we study $\psi(2s)$ diffractive production by constructing the $\psi(2s)$ forward wave function via a fit to the leptonic decay, and we provide various predictions for diffractive $\psi(2s)$ production as well as the ratio of $\psi(2s)/J/\psi$ at HERA and the LHC.
Furthermore, we find that the corresponding $t$-distributions of differential cross-section may unambiguously discriminate among saturation and non-saturation models. This is due to the emergence of a pronounced dip (or multiple dips) in the $t$-distribution of diffractive photoproduction of vector mesons at relatively large $|t|$ (but within reach of future experiments \cite{eic,lhec}) which is directly related to saturation physics. In this way, we go beyond existing recent works on $J/\psi$ and $\psi(2s)$ production both in the dipole model \cite{Ducati:2013tva,Goncalves:2014wna} and in pQCD \cite{n-pqcd,Jones:2013eda,Guzey:2014axa,Guzey:2013qza}, and of lighter mesons in the dipole model \cite{Santos:2014vwa}.
This paper is organised as follows. In section II, we introduce the formulation of the colour dipole approach for calculating exclusive diffractive processes. In section III, we introduce the IP-Sat and b-CGC dipole models. In section IV, we present a detailed numerical analysis and our main results. In subsection A we first show our results and predictions for the total diffractive cross-section of different vector mesons, while in subsection B we discuss the origin of the dips in the $t$-distribution of diffractive photoproduction of vector mesons and provide predictions for future experiments. We summarise our main results in section V.
\section{ Exclusive diffractive processes in the colour-dipole formalism}
In the colour dipole formalism, the underlying mechanism for diffractive production of different vector mesons and for inclusive DIS is similar. Namely, one must calculate the probability of finding a colour dipole of transverse size $r$ with impact parameter $b$ in the wave function of a (real or virtual) photon or of a vector meson. Similar to the case of the inclusive DIS process, the scattering amplitude for the exclusive diffractive process $\gamma^*+p\to V+p$, with a final state vector meson $V=J/\psi, \psi(2s), \phi,\rho$ (or a real photon $V=\gamma$ in DVCS), can be written in terms of a convolution of the dipole amplitude $\mathcal{N}$ and the overlap of the wave functions of the photon and the exclusive final state particle (see \cite{ip-sat1,ip-sat-a, b-cgc-a} and the references therein),
\begin{equation} \label{am-i}
\mathcal{A}^{\gamma^* p\rightarrow Vp}_{T,L} = \mathrm{2i}\,\int\!\mathrm{d}^2\vec{r}\int\!\mathrm{d}^2\vec{b}\int_0^1\!\mathrm{d}{z}\;(\Psi_{V}^{*}\Psi)_{T,L}(r,z,m_f,M_V;Q^2)\;\mathrm{e}^{-\mathrm{i}[\vec{b}-(1-z)\vec{r}]\cdot\vec{\Delta}}\mathcal{N}\left(x,r,b\right),
\end{equation}
with $\vec{\Delta}^2=-t$ and $t$ being the squared momentum transfer. In this equation, $\mathcal{N}$ is the imaginary part of the forward $q\bar{q}$ dipole-proton scattering amplitude with transverse dipole size $r$ and impact parameter $b$. The parameter $z$ is the fraction of the light cone momentum of the virtual photon carried by the quark and $m_f$ denotes the mass of the quark with flavour $f$.
The above expression can be understood in light front time as distinct chronological subprocesses, namely the $\gamma^\star$ first fluctuates into a quark-antiquark pair (the so-called colour $q\bar{q}$-dipole) which then interacts with the target. Finally the $q\bar{q}$ pair recombines to form the final state vector meson. In \eq{am-i} summations over the quark helicities and over the quark flavour $f=u,d,s, c$ are implicit. The phase factor $\exp\left(i(1-z)\vec{r}\cdot\vec{\Delta}\right)$ in the above equation is due to the non-forward wave-function contribution \cite{bbb}. In \eq{am-i}, the $\Psi_{V}^{*}\Psi$ is the forward overlap wave function of photon and vector meson (see below).
The differential cross-section of the exclusive diffractive processes can then be written in terms of the scattering amplitude as \cite{ip-sat1,watt-bcgc,b-cgc-a},
\begin{equation}
\frac{\mathrm{d}\sigma^{\gamma^* p\rightarrow Vp}_{T,L}}{\mathrm{d} t} = \frac{1}{16\pi}\left\lvert\mathcal{A}^{\gamma^* p\rightarrow Vp}_{T,L}\right\rvert^2\;(1+\beta^2) R_g^{2},
\label{vm}
\end{equation}
with
\begin{eqnarray} \label{eq:beta}
\beta &=& \tan\left(\frac{\pi\delta}{2}\right), \nonumber\\
R_g(\delta) &=& \frac{2^{2\delta+3}}{\sqrt{\pi}}\frac{\Gamma(\delta+5/2)}{\Gamma(\delta+4)}, \nonumber\\
\delta &\equiv& \frac{\partial\ln\left(\mathcal{A}_{T,L}^{\gamma^* p\rightarrow Vp}\right)}{\partial\ln(1/x)}, \
\end{eqnarray}
where the factor $(1+\beta^2)$ takes into account the missing real part of amplitude (notice that the amplitude in \eq{am-i} is purely imaginary), with $\beta$ being the ratio of real to imaginary parts of the scattering amplitude \cite{ip-sat1,ip-sat-a,b-cgc-a}. The factor $R_g$ incorporates the skewness effect, coming from the fact that the gluons attached to the $q\bar{q}$ can carry different light-front fractions $x,x^{\prime}$ of the proton \cite{ske,mrt,ske-n}.
The skewedness factor given in \eq{eq:beta} was obtained at NLO level, in the limit that $x^{\prime}\ll x\ll 1$ and at small $t$ assuming that the diagonal gluon density of target has a power-law form \cite{ske}. Note that there are uncertainties with respect to the actual incorporation of the skewness correction at small $x$ in dipole models\footnote{In the IP-Sat model \cite{ip-sat-a}, the skewness effect can be simply incorporated by multiplying the gluon distribution $xg(x,\mu^2)$ by a factor $R_g(\gamma)$ with $\gamma \equiv \frac{\partial\ln\left[xg(x,\mu^2)\right]}{\partial\ln(1/x)}$. This is consistent with the prescription given in Eqs.\,(\ref{vm}, \ref{eq:beta}) \cite{b-cgc-a}.} \cite{Rezaeian:2013eia}. However, these uncertainties will not affect our main results and conclusions.
The forward photon wave functions at leading order is well known in QCD, see e.g. Refs.~\cite{Dosch,Lepage}. The normalized photon wave function for the longitudinal photon polarization ($\lambda = 0$) and the transverse photon polarisations ($\lambda = \pm 1$) are given by~\cite{beta1},
\begin{eqnarray}
\Psi_{h\bar{h},\lambda=0}(r,z,Q) &=& e_f \sqrt{4\pi\alpha_{\mathrm{em}}}\, \sqrt{N_c}\,
\delta_{h,-\bar h} \, 2Qz(1-z)\, \frac{K_0(\epsilon r)}{2\pi}, \\
\Psi_{h\bar{h},\lambda=\pm 1}(r,z,Q) &=&
\pm e_f \sqrt{4\pi\alpha_{\mathrm{em}}} \, \sqrt{2N_c}\,
\left\{
\mathrm{i}e^{\pm \mathrm{i}\theta_r}[
z\delta_{h,\pm}\delta_{\bar h,\mp} -
(1-z)\delta_{h,\mp}\delta_{\bar h,\pm}] \partial_r \, + \,
m_f \delta_{h,\pm}\delta_{\bar h,\pm}
\right\}\, \frac{K_0(\epsilon r)}{2\pi},
\label{tspinphot}
\end{eqnarray}
where $N_c$ is the number of colours, the subscripts $h$ and $\bar h$ denote the helicities of the quark and the antiquark respectively and $\theta_r$ is the azimuthal angle between the vector $\vec{r}$ and the $x$-axis in the transverse plane. We have used a notation $\epsilon^2 \equiv z(1-z)Q^2+m_f^2$ where the subscript $f$ denotes the flavour.
Following Refs.\,\cite{ip-sat0,ip-sat1,ip-sat-a,b-cgc-a,beta,FF}, we assume that the forward vector meson wave functions $\Psi_V$ are effectively dominated by the $q\bar{q}$ Fock component and have the same spin and polarization structure as in the case of the photon:
\begin{eqnarray}
\Psi^V_{h\bar{h},\lambda=\pm 1}(r,z) &=&
\pm\sqrt{2N_c}\, \frac{1}{z(1-z)}
\left\{
\mathrm{i}e^{\pm \mathrm{i}\theta_r}[
z\delta_{h,\pm}\delta_{\bar h,\mp} -
(1-z)\delta_{h,\mp}\delta_{\bar h,\pm}] \partial_r \, + \,
m_f \delta_{h,\pm}\delta_{\bar h,\pm}
\right\}\, \phi_T(r,z), \\
\Psi^V_{h\bar{h},\lambda=0}(r,z) &=& \sqrt{N_c}
\delta_{h,-\bar h}
\left[ M_V\,+ \, \delta \, \frac{m_f^2 - \nabla_r^2}{M_Vz(1-z)}\,
\right] \phi_L(r,z),\
\end{eqnarray}
where $\nabla_r^2 \equiv (1/r)\partial_r + \partial_r^2$, $M_V$ is the meson mass and the effective charge is defined $\hat{e}_f=2/3$, $1/3$, or $1/\sqrt{2}$, for $J/\psi$ (and $\psi(2s)$), $\phi$ or $\rho$ mesons respectively\footnote{See \cite{Santos:2014vwa} for a study of the impact of different forms of the wave function on $\rho$ production.}. The longitudinally polarised vector meson wave function is slightly more complicated than in the case of the photon since the coupling of the quarks to the meson is non-local \cite{beta}. For the scalar parts of the wave functions $\phi_{T,L}(r,z)$, we employ the boosted Gaussian wave-functions with the Brodsky-Huang-Lepage prescription \cite{stan1}. The boosted Gaussian wave-functions were found to provide a very good description of exclusive diffractive HERA data \cite{ip-sat1,ip-sat-a,b-cgc-a}. For the ground state vector meson ($1s$) and its first excited state $2s$, the scalar function $\phi_{T,L}(r,z)$, has the following general form \cite{beta,FF},
\begin{eqnarray}
\phi_{T,L}^{1s}(r,z) &=& \mathcal{N}_{T,L} z(1-z)
\exp\left(-\frac{m_f^2 \mathcal{R}_{1s}^2}{8z(1-z)} - \frac{2z(1-z)r^2}{\mathcal{R}_{1s}^2} + \frac{m_f^2\mathcal{R}_{1s}^2}{2}\right), \\
\phi_{T,L}^{2s}(r,z) &=& \mathcal{N}_{T,L} z(1-z)
\exp\left(-\frac{m_f^2 \mathcal{R}_{2s}^2}{8z(1-z)} - \frac{2z(1-z)r^2}{\mathcal{R}_{2s}^2} + \frac{m_f^2\mathcal{R}_{2s}^2}{2}\right) \nonumber\\
&\times&\Big[1+\alpha_{2s}\left(2+\frac{m_f^2 \mathcal{R}_{2s}^2}{4z(1-z)} - \frac{4z(1-z)r^2}{\mathcal{R}_{2s}^2} - m_f^2\mathcal{R}_{2s}^2\right)\Big],\
\end{eqnarray}
where the parameter $\alpha_{2s}$ controls the position of the node of the radial wave function of the $V(2s)$. The boosted Gaussian wave function\footnote{ Note that the Coulomb term \cite{beta} has been ignored in the wave function here because adding it introduces another parameter to the wave function (plus an unknown running coupling) and a singular behaviour at the origin, However, this should not be important at high energy for large dipole sizes, and its contribution should be either negligible or simply absorbed into the remaining parameters of the wave function. On the phenomenology side, there is no strong evidence of Coulomb contribution even at lower energy at HERA, and indeed a good fit of vector meson wave function to leptonic decay can be found even without it as done in \cite{b-cgc-a} and here, see table \ref{t-1}.} has several advantages over other commonly used models, namely it is more self-consistent, it is fully boost invariant and it has the proper short-distance limit at $m_f\to 0$.
The normalisation and orthogonality conditions allow the missing higher order Fock component of the wave functions to be effectively absorbed into the overall normalisation factor,
\begin{eqnarray}
&& \frac{N_c}{2\pi}\int_0^1\!\frac{\mathrm{d}{z}}{z^2(1-z)^2}\int\!\mathrm{d}^2\vec{r}\;
\left\{m_f^2(\phi_T^{1s(2s)})^2+\left[z^2+(1-z)^2\right]
\left(\partial_r\phi_T^{1s(2s)}\right)^2\right\} =1,\\
&& \frac{N_c}{2\pi} \int_0^1\!
\mathrm{d}{z}\,
\int\!\mathrm{d}^2\vec{r}\;
\left[
M_V\phi_L^{1s(2s)}+
\delta\,
\frac{m_f^2-\nabla_r^2}{M_V z(1-z)}\,\phi_L^{1s(2s)}\right]^2 =1, \\
&&\frac{N_c}{2\pi}\int_0^1\!\frac{\mathrm{d}{z}}{z^2(1-z)^2}\int\!\mathrm{d}^2\vec{r}\;
\left\{m_f^2 \phi_T^{1s}\phi_T^{2s} +\left[z^2+(1-z)^2\right]
\partial_r\phi_T^{1s} \partial_r\phi_T^{2s}\right\} =0.\
\end{eqnarray}
Another important input is the leptonic decay width of the vector meson which is given by
\begin{equation}
\Gamma_{V\to e^+e^-} = \frac{4\pi\alpha_{\rm em}^2f_V^2}{3M_V},
\end{equation}
where the decay widths are given by \cite{ip-sat0,ip-sat1},
\begin{gather}
f_{V,T} = \hat{e}_f\, \left.\frac{N_c}{2\pi M_V}
\int_0^1\!\frac{\mathrm{d}{z}}{z^2(1-z)^2}
\left\{m_f^2-\left[z^2+(1-z)^2\right]\nabla_r^2\right\}\phi_T(r,z)\right\rvert_{r=0},\\
%
\label{eq:nnz_fvl}
f_{V,L} = \hat{e}_f\, \left.\frac{N_c}{\pi}
\int_0^1\!
\mathrm{d}{z}\,
\left[
M_V +\delta\, \frac{m_f^2-\nabla_r^2}{M_Vz(1-z)}\right]
\phi_L(r,z)\right\rvert_{r=0}.
\end{gather}
In the above, consistent with underlying dynamics of the vector meson production in the colour-dipole factorisation, we assumed that the leptonic decay $V\to \gamma^{\star}\to e^+e^-$ can be also described by a factorized from in which the vector meson contributes mainly through its properties at the origin.
The vector meson wave function in the boosted Gaussian model, has only 3 (4 for $2s$ state) parameters, namely $\mathcal{N}_{T,L}, \mathcal{R}$ and $\alpha_{2s}$ which are determined from normalisation, the orthogonality conditions and a fit to the experimental leptonic decay width. For the case of $1s$ ground state vector meson production we have $\alpha_{2s}=0$. Unfortunately we do not have experimental data for leptonic decay width for longitudinal and transverse polarisations component separately. Therefore, we assume that the measured experimental value is the average between those for longitudinal and transverse polarisations. Note that the parameters of the wave function cannot be uniquely extracted from the conditions indicated above; namely, several sets of solutions exist. In order to put more constrain on the parameters of the wave function, it is natural to assume that $\mathcal{N}_{T}\approx \mathcal{N}_{L}$ (or $\mathcal{N}_{T}=\mathcal{N}_{L}$). This is because in the boosted Gaussian wave function there is only one radius parameter which should dynamically give the correct normalisation for both longitudinal and the transverse polarisations component up to a prefactor that mimics the missing higher order Fock components.
The parameters for $J/\psi$, $\psi(2s)$ and $\rho$ determined from the above conditions are given in table \ref{t-1}. In this table we also compare the value of $\Gamma_{e^+e^-}$ obtained from our fit with the experimental result $\Gamma^{exp}_{e^+e^-}$. Note that in order to estimate the possible theoretical uncertainties associated with the condition $\mathcal{N}_{T}=\mathcal{N}_{L}$, in table \ref{t-1}, we also give a parameter set extracted by relaxing this condition. The preferred values of $\mathcal{N}_{T}$ and $\mathcal{N}_{L}$ are similar as we expect. It is also shown in table \ref{t-1} that a different value for the charm and light quark masses mainly affects the normalisation of the wave function. In \fig{f-phi}, we show the scalar part of the light-cone wave function of $J/\psi$ and $\psi(2s)$ using the parameter set corresponding to $m_c=1.4$ GeV. The position of node in $\psi(2s)$ wave function changes with the value of $z$.
Note that in Ref.\,\cite{ip-sat1} it was assumed that $f_{V,T}= f_{V,L}$ while running $\mathcal{N}_{T}$ and $\mathcal{N}_{L}$ freely in a fit. We do not impose this condition here, although the values of $f_{V,T}$ and $ f_{V,L}$ obtained in our scheme become rather similar. In our approach, for the case of $J/\psi$ and $\rho$, we obtained $\Gamma_{e^+e^-}=5.54$ KeV and $7.02$ KeV while in the approach of Ref.\,\cite{b-cgc-a} for the same quark masses we have $\Gamma_{e^+e^-}=6.79$ and $9.52$ KeV respectively, compared to the experimental value of $\Gamma^{exp}_{e^+e^-}=5.55\pm 0.14$ for $J/\psi$ and $\Gamma^{exp}_{e^+e^-}=7.04\pm 0.06$ for $\rho$ \cite{pdg-2012}. We checked that with the new parameter sets given in table \ref{t-1}, the description of the diffractive $J/\psi$ production at HERA and the LHC will be similar compared to the one with the old vector meson wave function parameter sets.
\begin{figure}[t]
\includegraphics[width=0.5\textwidth,clip]{plot-psi.eps}
\caption{The scalar part of the light-cone wave function of $J/\psi$ and $\psi(2s)$ with $m_c=1.4$ GeV for two different values of $z$. }
\label{f-phi}
\end{figure}
\begin{table}
\centering
\begin{tabular}{c|c|c|c|c|c||c|c|c}
\hline\hline
Meson & $m_f/GeV$ & $\mathcal{N}_L$ & $\mathcal{N}_T$ & $\mathcal{R}^2$/$\text{GeV}^2$ & $\alpha_{2s}$ & $M_V$/GeV & $\Gamma^{exp}_{e^+e^-}$/KeV &$\Gamma_{e^+e^-}$/KeV \\ \hline
$J/\psi$ & $m_c=1.4$& $0.57$ & $0.57$ & $2.45$ & $0$ & $3.097$ &$5.55\pm 0.14$ &$5.54$ \\ \hline
$J/\psi$ & $m_c=1.27$& $0.592$ & $0.596$ & $2.45$ & $0$ & $3.097$ &$5.55\pm 0.14$ &$5.46$ \\ \hline
$\psi(2s)$ &$m_c=1.4$& 0.67 &0.67 & 3.72& -0.61& 3.686& $2.37\pm 0.04$& 2.39\\ \hline
$\psi(2s)$ &$m_c=1.27$& 0.69 &0.70 & 3.72& -0.61& 3.686& $2.37\pm 0.04$& 2.35\\ \hline
$\rho$ & $m_{u,d,s}=0.01$& 0.894 & 1.004& 13.3& 0&0.775 & $7.04\pm 0.06$ & 7.06\\ \hline
$\rho$ & $m_{u,d,s}=0.14$ & 0.852& 0.908& 13.3& 0&0.775 & $7.04\pm 0.06$ & 7.02\\ \hline
\end{tabular}
\caption{Parameters of the boosted Gaussian vector meson wave functions for $J/\psi$, $\psi(2s)$ and $\rho$ obtained for two different values of quark masses. }
\label{t-1}
\end{table}
\section{Impact-parameter dependent dipole models}
The common ingredient for the total (and reduced) cross-sections (i.e. for the proton structure functions in DIS) and for exclusive diffractive vector meson production \eq{am-i}, is the universal $q\bar{q}$ dipole-proton forward scattering amplitude. Although the impact-parameter dependence of the dipole amplitude is less important for inclusive processes, it is crucial for describing exclusive diffractive ones. Note that the impact-parameter profile of the dipole amplitude entails intrinsically non-perturbative physics, which is beyond the QCD weak-coupling approach to small-$x$ physics \cite{al,bk-c,ana}.
Therefore, the impact-parameter dependence of the dipole amplitude, unfortunately, can only be treated phenomenologically at this time.
Supported by experimental data, it is generally assumed a Gaussian profile for gluons where the width of the profile, as only free parameter, is fixed via a fit to diffractive data at HERA. We use two well-known impact-parameter dependent saturation models, the so-called IP-Sat \cite{ip-sat-a,ip-sat0} and b-CGC \cite{watt-bcgc, b-cgc-a} models which both have been very successful in phenomenological applications from HERA to RHIC and the LHC.
In the IP-Sat model \cite{ip-sat0}, the proton-dipole forward scattering amplitude is given by
\begin{eqnarray}
\mathcal{N}\left(x,r,b\right) &=&1-\exp\left(-\frac{\pi^{2}r^{2}}{2N_{c}}\alpha_{s}\left(\mu^{2}\right)xg\left(x,\mu^{2}\right)T_{G}(b)\right)\,, \label{ip-sat} \\
T_{G}(b)&=& \frac{1}{2\pi B_G}\exp\left(-b^2/2B_G\right) \label{ip-b}, \
\end{eqnarray}
where $T_G(b)$ is the gluon impact-parameter profile and $xg\left(x,\mu^{2}\right)$ is the gluon density, evolved with dipole transverse size $r$ up to the scale $\mu^{2}=4/r^{2}+\mu_{0}^{2}$ with LO DGLAP gluon evolution (neglecting its coupling to quarks) with
initial gluon distribution at the scale $\mu_0^2$
\begin{eqnarray}
xg\left(x,\mu_{0}^{2}\right) &=&A_{g}\,x^{-\lambda_{g}}(1-x)^{5.6} \label{g}.\
\end{eqnarray}
We take the corresponding one-loop running-coupling value of $\alpha_s$ for four flavours, with $\Lambda_{\text{QCD}}=0.156$ GeV fixed by the experimentally measured value of $\alpha_s$ at the $Z^0$ mass. The contribution from bottom quarks is neglected. The IP-Sat dipole amplitude can be derived at the classical level in the CGC~\cite{mv}. Through eikonalization it explicitly maintains unitarity while matching smoothly the high $Q^2$ perturbative QCD limit via DGLAP evolution. The eikonalization of the gluon distribution in the IP-Sat model represents a resummation of higher twist contributions which become important at small $x$. The first term of the expansion of the exponential in \eq{ip-sat} corresponds to the leading-order pQCD expansion for the dipole amplitude in the colour-transparency region, as opposed to the saturation case, and it is here called the 1-Pomeron model.
In the b-CGC dipole model \cite{watt-bcgc}, the colour dipole-proton forward scattering amplitude is given by
\begin{eqnarray} \label{CA5}
\mathcal{N}\left( x, r, b\right)\,\,=\,\, \left\{\begin{array}{l}\,\,\,N_0\,\left( \frac{r Q_s}{2}\right)^{2\gamma_{eff}}\ \ {\rm for } \ \ r Q_s\,\leq\,2\,,\\ \\
1\,\,-\,\,\exp\left( -\mathcal{A} \ln^2\left( \mathcal{B} r Q_s\right)\Rb\ \ {\rm for} \ \ \ rQ_s\,>\,2\,,\end{array}
\right.
\end{eqnarray}
where the effective anomalous dimension $\gamma_{eff}$ and the saturation scale $Q_s$ of the proton explicitly depend on the impact parameter and are defined as
\begin{eqnarray} \label{g-eff}
\gamma_{eff}&=&\gamma_s\,\,+\,\,\frac{1}{\kappa \lambda Y}\ln\left(\frac{2}{r Q_s}\right), \nonumber\\
Q_s\equiv Q_{s}(x,b)&=&\left(\frac{x_0}{x}\right)^{\frac{\lambda}{2}}\,\exp\left\{- \frac{b^2}{4\gamma_s B_{CGC}}\right\} \text{GeV}, \
\end{eqnarray}
where $Y=\ln(1/x)$ and $\kappa= \chi''(\gamma_s)/\chi'(\gamma_s)$, with $\chi$ being the LO BFKL
characteristic function. The parameters $\mathcal{A}$ and
$\mathcal{B}$ in \eq{CA5} are determined uniquely from the matching of the dipole amplitude and
its logarithmic derivatives at $r Q_s=2$. The b-CGC model is constructed by smoothly interpolating between two analytically known limiting cases \cite{IIM}, namely the solution of the BFKL equation in the vicinity of the saturation line for small dipole sizes, and the solution of the BK equation deep inside the saturation region for large dipole sizes \cite{LT,lt3}.
Although both the b-CGC and the IP-Sat models include saturation effects and depend on impact-parameter,
the underlying dynamics of two models is quite different, namely saturation in the b-CGC and the IP-Sat models is probed through the increase of the gluon density (in the dilute regimes) driven by BFKL and DGLAP evolutions, respectively. For detailed comparisons of two saturation models, see Ref.\,\cite{b-cgc-a}.
The parameters of the dipole amplitudes in the IP-Sat ($\mu_0, A_g, \lambda_g$) and b-CGC ($N_0$, $\gamma_s, x_0, \lambda$) models were determined via a fit to the recent combined HERA data for the reduced cross-sections \cite{Aaron:2009aa,Abramowicz:1900rp} in the range $Q^2\in [0.75, 650]\, \text{GeV}^2$ and $x\le 0.01$. The widths of the impact-parameter profiles, $B_{G}$ and $B_{CGC}$ in the IP-Sat and b-CGC models respectively, were iteratively fixed to give a good description of the $t$-dependence of exclusive diffractive $J/\psi$ production at HERA (at small-$t$ where data lie), while at the same time this consistently fixes the normalisation of the inclusive reduced cross-section without further adjustment and give an excellent description of all other diffractive data (for different vector mesons and DVCS production) at small $x$ \cite{ip-sat-a,b-cgc-a}. The values of parameters of the models can be found in Refs.\, \cite{ip-sat-a,b-cgc-a}. Note that in both the IP-Sat and b-CGC models, the fit to the recent combined HERA data at $x\le 0.01$ becomes stable for $Q^2 \ge Q^2_{min} =0.75\, \text{GeV}^2$: one observes a steady increase in $\chi^2$ with decreasing values of $Q^2_{min}$ \cite{ip-sat-a,b-cgc-a}. Therefore, our photoproduction results at $Q\approx 0$ may be considered as a test of the model beyond the kinematics where it was fitted. But the generic features of our results there, are expected not to be affected by this extrapolation.
For vector meson production, the dipole amplitude in \eq{am-i} is evaluated at $x=x_{Bj}\left(1+M^2_V/Q^2\right)$, where $M_V$ denotes the mass of the vector meson\footnote{At $Q^2=0$, we have $x=M^2_V/(W_{\gamma p}^2-M^2_N)$ where $M_N$ denotes the nucleon mass and $W_{\gamma p}$ is the center-of-mass energy of the photon-proton system.} and $x_{Bj}$ is Bjorken-$x$.
We stress again that in the master equations (\ref{am-i}), (\ref{vm}), (\ref{eq:beta}), the small-$x$ dynamics encoded in the dipole amplitude $\mathcal{N}\left( x, r, b\right)$, including its impact-parameter dependence, is the same for different vector mesons $J/\psi, \psi(2s), \phi,\rho$ and for DVCS, while the overlap wave functions between the photon and the vector mesons $\Psi_{V}^{*}\Psi$, control the typical transverse dipole size which contributes at a given kinematics.
\section{Main numerical results and predictions}
\subsection{Total cross-section of exclusive diffractive production }
\begin{figure}[t]
\includegraphics[width=0.55\textwidth,clip]{plot-jpsi-w-photo-oldf.eps}
\caption{Total $J/\psi$ cross-section as a function of $W_{\gamma p}$, compared to results from the b-CGC and IP-Sat models with parameters of the models determined via a fit to the recent combined data from HERA \cite{ip-sat-a,b-cgc-a} and the old $F_2$ structure function \cite{watt-bcgc} (dashed-dotted line, labeled b-CGC 2008).
The data are from fixed target experiments \cite{fix}, the H1, ZEUS \cite{h1-2013,Chekanov:2002xi,Chekanov:2004mw,Aktas:2005xu}, LHCb \cite{lhcbn} and ALICE (preliminary data) \cite{TheALICE:2014dwa} Collaborations. We also show the LHeC pseudo-data obtained from a simulation \cite{lhec}.}
\label{f-vw1-ad}
\end{figure}
\begin{figure}[t]
\includegraphics[width=0.6\textwidth,clip]{plot-jpsi-w-photof.eps}
\caption{Total $J/\psi$ cross-section as a function of $W_{\gamma p}$, compared to results from the CGC/Saturation (orange band) calculated from the b-CGC and IP-Sat models \cite{ip-sat-a,b-cgc-a}. The CGC band includes the uncertainties associated with our freedom to choose the charm mass within the range $m_c= 1.2 \div 1.4$ GeV. The results of pQCD fits at LO and NLO \cite{pqcd-j} are taken from \cite{h1-2013}. The experimental data are the same as in \fig{f-vw1-ad}. }
\label{f-vw1}
\end{figure}
We first focus on the total cross-section of elastic diffractive production of various vector mesons. Here and thereafter, for the total cross-section we perform the integral over $|t|\in[0,1]\,\text{GeV}^2$ (unless it is explicitly given). The advantage of the $J/\psi$ over other vector mesons is that because of its large mass, the calculation both for the cross-section and the overlap wave function are under better theoretical control and can be treated perturbatively.
In \fig{f-vw1-ad}, we compare the results for the total $J/\psi$ cross-section as a function of center-of-mass energy of the photon-proton system $W_{\gamma p}$, obtained using the IP-Sat and b-CGC dipole models with a fixed charm mass $m_c=1.27$ GeV. Both models with parameters extracted via a fit to the recent combined HERA data \cite{ip-sat-a,b-cgc-a}, give consistent results with the LHCb data \cite{lhcb}. However, the b-CGC model with the parameters extracted via a fit to the old data (the $F_2$ structure function) \cite{watt-bcgc}, underestimates the recent LHCb data. The results obtained from the b-CGC and IP-Sat models are slightly different at very high energies due to the fact the power-law behaviour of the saturation scale in these two models is different \cite{b-cgc-a}.
\begin{figure}[t]
\includegraphics[width=0.49\textwidth,clip]{plot-jpsi-w-photo-massf.eps}
\includegraphics[width=0.49\textwidth,clip]{plot-jpsi-w-photo-pomf.eps}
\caption{Left: Total $J/\psi$ cross-section as a function of $W_{\gamma p}$, compared to results from the IP-Sat model with different charm mass $m_c$. Right: Total $J/\psi$ cross-section as a function of $W_{\gamma p}$, compared to the results from the IP-Sat (saturation) 1-Pomeron models with different charm mass $m_c$. The experimental data are the same as in \fig{f-vw1-ad}. }
\label{f-vw2}
\end{figure}
\begin{figure}[th]
\includegraphics[width=0.47\textwidth,clip]{plot-psi-w-pho2.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-psi-w-pho3.eps}
\caption{Left: Total $\psi(2s)$ diffractive photoproduction cross-section as a function of $W_{\gamma p}$, compared to results from the IP-Sat and b-CGC models with different charm mass $m_c$. Right: Similar to the left panel, the results of the CGC/saturation (orange band) and 1-Pomeron models are compared. The experimental data are from the H1 collaboration \cite{h1-psi} for quasi-elastic ($Z>0.95$) photoproduction of $\psi(2s)$ while all theory curves are for elastic diffractive production with elasticity $Z=1$. }
\label{f-psi1}
\end{figure}
In \fig{f-vw1}, we compare the results obtained from saturation models and from a pQCD approach at LO and NLO \cite{pqcd-j} with all available data from fixed target experiments to the recent ones from the H1, ZEUS, LHCb and ALICE collaborations\footnote{For the purpose of illustrating the precision that could be achieved in future experiments, both in \fig{f-vw1-ad} and in \fig{f-vw1} we also show the LHeC pseudo-data obtained from a simulation \cite{lhec} based on a power-law extrapolation of HERA data.} \cite{lhcbn,TheALICE:2014dwa,h1-2013,fix,Chekanov:2002xi,Chekanov:2004mw,Aktas:2005xu}. The band labeled "CGC" includes the saturation results obtained from the IP-Sat and b-CGC models with the parameters of models constrained by the recent combined HERA data. Note that the LHCb data points in \fig{f-vw1} were not used for fixing the model parameters, and therefore our CGC results in \fig{f-vw1} at high energy can be considered as predictions. Also note that diffractive $J/\psi$ production is sensitive to the charm quark mass at low $Q^2$. This is because the scale in the integrand of the cross-section is set by the charm quark mass for low virtualities $Q^2<m_c^2$. The CGC band in \fig{f-vw1} also includes the uncertainties associated with choosing the charm mass within the range $m_c=1.2 \div 1.4$ GeV extracted from a global analysis of existing data at small-x $x<0.01$ \cite{ip-sat-a,b-cgc-a}. In \fig{f-vw1}, we compare with the LHCb updated data released in 2014 \cite{lhcbn} which are significantly more precise compared to earlier measurements \cite{lhcb} (see also \fig{f-vw2} right panel). It is seen that the ALICE \cite{TheALICE:2014dwa} and LHCb \cite{lhcbn} data are in good agreement with the CGC predictions while there seem to be some tensions between the experimental data and the pQCD results (labeled MNRT LO and NLO) at high $W_{\gamma p}$. It was recently shown that including the LHCb data in the pQCD fit, allows a better constraint on the low-$x$ gluon distribution \cite{n-pqcd}.
\begin{figure}[t]
\includegraphics[width=0.47\textwidth,clip]{plot-psi-w-photo.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-psi-q2-22.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-psi-w-fixedQ2.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-psi-t.eps}
\caption{ Top Left: The ratio of the cross-section for $\psi(2s)$ and $J/\psi$ ($R=\psi(2s)/J/\psi$) for diffractive photoproduction as a function of $W_{\gamma p}$. Top right: The ratio $R$ for diffractive production as a function of $Q^2$ at a fixed $W_{\gamma p}=95$ GeV. Bottom left: The ratio $R$ for diffractive production as a function of $W_{\gamma p}$ at a fixed $Q^2=10$ GeV. Bottom right: The ratio $R$ for diffractive production as a function of $|t|$ at a fixed $Q^2=10$ GeV and $W_{\gamma p}=120$ GeV. In all panels, the theoretical curves are the results from the IP-Sat and b-CGC models with different parameter sets corresponding to different charm masses. The experimental data are from the H1 Collaboration \cite{h1-ratioi}.}
\label{f-ratio}
\end{figure}
In \fig{f-vw2}, we show the charm-mass dependence of the total $J/\psi$ cross-section as a function of $W_{\gamma p}$.
Within the saturation models, a lower charm mass about $m_c\approx 1.27 $ GeV is preferred. However, in the non-saturation version of the IP-Sat model (1-Pomeron), a larger charm mass about $m_c\approx 1.4 $ GeV provides a better agreement with experimental data (see the right panel of that figure). In \fig{f-vw2} right panel, we also show ALICE preliminary data \cite{TheALICE:2014dwa}, the LHCb updated data (labeled LHCb 2014) \cite{lhcbn} and earlier LHCb data \cite{lhcb} (labeled LHCb 2013). It is seen that the combined ALICE and LHCb updated 2014 data are more in favour of the saturation than of the 1-Pomeron model results at high $W_{\gamma p}$.
Nevertheless, in order to clearly discriminate among models one should first more accurately determine the charm mass. This can be done by precise measurements of the charm structure function $F^{c}_2$ or a reduced cross-section for charm production in a wider range of kinematics, including at small virtualities, than those currently available at HERA (restricted to $Q^2\ge 2.5\,\text{GeV}^2$ and $x \ge 3\times 10^{-5}$ \cite{Abramowicz:1900rp}). Such measurements can in principle be done in the projected LHeC \cite{lhec}.
\begin{figure}[t]
\includegraphics[width=0.47\textwidth,clip]{plot-rho-w-photo.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-rho-w-mass.eps}
\caption{Left: Total diffractive $\rho$ cross-section as a function of $W_{\gamma p}$ at different virtualities $Q^2=0, 2.4,3.3,6, 13.5\,\text{GeV}^2$.
The CGC band (orange band) includes the results obtained from the b-CGC and IP-Sat models and also the uncertainties associated with our freedom to choose the light quark mass within the range $m_{u,d,s}= 0.01 \div 0.14$ GeV. Right: Total diffractive $\rho$ cross-section compared to results from the CGC/saturation and 1-Pomeron model with two different masses $m_{u,d,s}=0.01$ and $0.14$ GeV.
The experimental data are from \cite{h1-rho,zeus-rho}.}
\label{f-rho1}
\end{figure}
In \fig{f-psi1}, we show the total cross-section of elastic diffractive photoproduction of $\psi(2s)$ as a function of $W_{\gamma p}$ obtained from the IP-Sat and b-CGC saturation models with different charm masses corresponding to different parameter sets of the dipole amplitude. Note that the experimental data \cite{h1-psi} are for quasi-elastic ($Z>0.95$) photoproduction of $\psi(2s)$ while all theory curves are for elastic diffractive production with elasticity $Z=1$. The elasticity is defined as $Z=E_{\psi(2s)}/E_{\gamma}\approx (W^2-M_Y^2)/(W^2-m_p^2)$ where $M_Y$ is the effective mass of the hadrons produced in the dissociation of the proton.
In the right panel, we compare the results obtained from the 1-Pomeron and the saturation models. It is seen that within theoretical uncertainties associated with charm mass, the 1-Pomeron and the saturation models give rather similar results in the range of energy shown in \fig{f-psi1}. This is mainly due to the fact that the $\psi(2s)$ is heavier than $J/\psi$, therefore effective dipole sizes $r\sim 1/\epsilon$ which contribute to the total cross-section are smaller for $\psi(2s)$ than for $J/\psi$. Note that although the scalar part of the $\psi(2s)$ wave function extends to large dipole sizes (see \fig{f-phi}), due to the existence of the node, there is large cancellation between dipole sizes above and below the node position. As a result, the total cross-section of $\psi(2s)$ is suppressed compared to $J/\psi$ production, see Figs.\,\ref{f-psi1},\ref{f-ratio}.
In \fig{f-ratio}, we show the ratio of the cross-section for $\psi(2s)$ and $J/\psi$ for diffractive production $R=\psi(2s)/J/\psi$ as functions of $W_{\gamma p}$ at $Q=0$ (top left panel), $Q^2$ at a fixed $W_{\gamma p}=95$ GeV (top right panel), $W_{\gamma p}$ at a fixed $Q^2=10$ GeV$^2$ (bottom left panel) and $|t|$ at a fixed $Q^2=10$ GeV$^2$ and $W_{\gamma p}=120$ GeV (bottom left panel). It is seen that at a fixed high virtualities, the ratio $R$ has little dependence to $|t|$ and $W_{\gamma p}$ (bottom panel), while the ratio $R$ increases with virtualities at a fixed $W_{\gamma p}$ (top right panel).
It is also seen in \fig{f-ratio} (top left panel) that the photoproduction ratio $R (Q=0)$ increases with $W_{\gamma p}$ and becomes sensitive to different saturation models. Therefore, precise measurements of the ratio of diffractive photoproduction of $\psi(2s)$ and $J/\psi$ at HERA and the LHC can provide valuable extra constraint on the saturation models.
In \fig{f-rho1}, we show total diffractive $\rho$ meson cross-section as a function of $W_{\gamma p}$ at different virtualities $Q^2=0, 2.4,3.3,6, 13.5\,\text{GeV}^2$, compared to results obtained from the b-CGC and the IP-Sat models. In the case of photoproduction, similar to experimental measurement, we perform the integral over $t\in [0,0.5]\,\text{GeV}^2$. The orange band labeled CGC includes results from both the IP-Sat and b-CGC models with uncertainties associated to our freedom to choose different light-quark masses within a range $m_{u,d,s}= 0.01 \div 0.14$ GeV. We also compare the CGC/saturation results with those obtained from the 1-Pomeron model with two different light quark masses $m_{u,d,s}= 0.01$ and $0.14$ GeV. It is seen that 1-Pomeron results are significantly different from the saturation models, and HERA data can already rule out the 1-Pomeron model with light quark masses. Notice that increasing the light quark masses to $m_{u,d,s}\approx 0.35 \div 0.4 $ GeV (not shown in \fig{f-rho1}), significantly reduces the cross-section in the 1-Pomeron model and brings it closer to the saturation results with $m_{u,d,s}= 0.01 \div 0.14$ GeV. However, a dipole model with such a large light-quark masses does not provide a good description of the structure functions at very low virtualities \cite{ip-sat-a,b-cgc-a}.
This may indicate the existence of large non-linear effects for the diffractive photoproduction of the $\rho$ meson. Note that, as we already pointed out, the effective dipole size which contributes to the cross-section is proportional to the inverse of the meson mass at $Q=0$. Therefore the total diffractive cross-section of lighter vector meson such as the $\rho$ meson should be a better probe of saturation physics (see also below).
\subsection{$t$-distribution of the diffractive production off protons and the origin of dips}
In \fig{f-vt0}, left panel, we compare the saturation and non-saturation models results for the $t$-distribution of the exclusive photoproduction of $J/\psi$ at $Q\approx 0$ with available data from HERA. It can be observed that at low $|t|$ where currently experimental data are available, one cannot discriminate between the saturation and non-saturation (1-Pomeron) models and all three models: IP-Sat, b-CGC and 1-Pomeron, provide a good description. However, at large $|t|$ the models give drastically different results, namely both the IP-Sat and b-CGC saturation models produce a dip while the 1-Pomeron model does not. In \fig{f-vt0}, right panel, we show the charm-mass dependence of the $t$-distribution of exclusive $J/\psi$ photoproduction. The appearance and position of the dip are only slightly affected by the choice of charm mass. Therefore, in this respect, theoretical uncertainties due to the charm mass are less important for the $t$-distribution than for the total cross-section.
\begin{figure}[t]
\includegraphics[width=0.49\textwidth,clip]{plot-jpsi-t-lhcbf.eps}
\includegraphics[width=0.49\textwidth,clip]{plot-jpsi-t-mcf.eps}
\caption{Left: Differential vector meson cross-sections for $J/\psi$, as a function of $|t|$ within the IP-Sat, b-CGC and 1-Pomeron models with a fixed $m_c=1.27$ GeV at HERA. Right: Results obtained from the IP-Sat and 1-Pomeron models are compared for two values of the charm mass $m_c=1.27, 1.4$ GeV. The experimental data are from the H1 Collaboration \cite{h1-2013,Aktas:2005xu}. }
\label{f-vt0}
\end{figure}
\begin{figure}[h]
\includegraphics[width=0.49\textwidth,clip]{plot-jpsi-t-lhcb2.eps}
\caption{Differential $J/\psi$ cross-section, as a function of $|t|$ within the IP-Sat (saturation) and IP-Sat (1-Pomeron) models with a fixed $m_c=1.27 $ GeV at LHC/LHeC energies $W_{\gamma p}=1, 5$ TeV and $Q^2=0, 10\,\text{GeV}^2$. }
\label{f-vt1}
\end{figure}
In \fig{f-vt1}, we show our predictions for the $t$-distribution of exclusive $J/\psi$ photoproduction at LHC/LHeC energies $W_{\gamma p}=1, 5$ TeV at two virtualities $Q^2=0, 10\,\text{GeV}^2$ obtained from the IP-Sat (saturation) and the 1-Pomeron models. In the saturation model, the dip shifts to smaller values of $|t|$ for smaller $Q$ and for higher $W_{\gamma p}$. Note that saturation effects are expected to become more important at low virtualities and high energies.
In \fig{f-vt2}, we compare the results obtained from the IP-Sat and b-CGC models with those from the 1-Pomeron model, for the $t$-distribution of the elastic photoproduction of vector mesons\footnote{In the case of $\phi$ meson, we use boosted Gaussian wavefunction with parameters given in Ref.\cite{ip-sat1}. For other vector mesons, we use parameters for the wavefunction given in table \ref{t-1}.} $J/\psi$, $\psi(2s)$, $\phi$ and $\rho$ off the proton at an energy accessible at the LHC/LHeC, $W_{\gamma p}=1$ TeV, for $Q=0$. Drastic different patterns for the diffractive $t$-distribution also emerge between saturation and non-saturation models for lighter vector meson production such as $\rho$ and $\phi$, with the appearance of multiple dips. Note that the prospects at the LHeC \cite{lhec} indicate that access to values of $|t|$ around 2 GeV$^2$, required to observe the dips for $J/\psi$, is challenging. On the other hand, the accuracy that can be expected at lower $|t|$ should allow to observe the bending of the distributions. And lower values of $|t|$ for lighter vector mesons should be clearly accessible, probably even at the EIC \cite{eic} but for smaller $W_{\gamma p}$.
\begin{figure}[t]
\includegraphics[width=0.42\textwidth,clip]{plot-jpsi-t-lhcb1.eps}
\includegraphics[width=0.42\textwidth,clip]{plot-psi-t-lhc.eps}
\includegraphics[width=0.42\textwidth,clip]{plot-phi-t-lhcb.eps}
\includegraphics[width=0.42\textwidth,clip]{plot-rho-t-lhcb.eps}
\caption{ Differential diffractive vector meson photoproduction cross-sections for $J/\psi, \psi(2s), \phi, \rho$, as a function of $|t|$ within the IP-Sat (saturation), b-CGC and 1-Pomeron models at a fixed $W_{\gamma p}=1$ TeV and $Q=0$. The thickness of points includes the uncertainties associated with our freedom to choose different values for the charm quark mass within the range $m_c \approx 1.2\div 1.4$ GeV (corresponding to different dipole parameter sets) and $m_{u,d,s}\approx 0.01$ GeV. }
\label{f-vt2}
\end{figure}
\begin{figure}[t]
\includegraphics[width=0.47\textwidth,clip]{plot-t-dipolef.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-dipolef.eps}
\caption{ Left: The $t$-distribution of dipole amplitude defined in \eq{d-f} with different values of cutoff on dipole transverse-size $\Lambda_r$ in the IP-Sat (saturation) and IP-Sat (1-Pomeron) models at a fixed $x=10^{-6}$. Right: The dipole amplitude in different models as a function of dipole size $r$ at $x=10^{-6}$ for two values of impact parameter $b=0, 3\,\text{GeV}^{-1}$. }
\label{f-sd}
\end{figure}
The emergence of a single or multiple dips in the $t$-distribution of the vector mesons in the saturation models is directly related to the saturation (unitarity) features of the dipole scattering amplitude $\mathcal{N}$ at large dipole sizes. In order to see more clearly this effect, let us define a $t$-distribution of the dipole amplitude in the following way:
\begin{equation} \label{d-f}
\frac{d\sigma^{\text{dipole}}}{dt}=2\pi\Big{|}\int_{0}^{\Lambda_r} rdr\, \int \mathrm{d}^2\vec{b}\, \mathrm{e}^{-\mathrm{i}\vec{b}\cdot\vec{\Delta}} \mathcal{N}\left(x,r,b\right)\Big{|}^2,
\end{equation}
where $\Lambda_r$ is an upper bound on the dipole size. The above expression is in fact very similar to Eqs.\,(\ref{am-i}), (\ref{vm}), see also Ref.\,\cite{ip-sat0}. Note that in \eq{am-i}, the overlap of photon and vector meson wave functions gives the probability for finding a colour dipole of transverse size $r$ in the vector meson wave function and it naturally gives rise to an implicit dynamical cutoff $\Lambda_r$ which varies with kinematics and the mass of the vector meson.
The cutoff $\Lambda_r$ is larger at lower virtualities and for lighter vector mesons. On the other hand, quantum evolution leads to unitarity constrains on the amplitude at lower dipole sizes with decreasing values of $x$ or increasing energies.
Thus, by varying the cutoff $\Lambda_r$, one probes different regimes of the dipole from colour transparency to the saturation regime.
In the 1-Pomeron model, since the impact-parameter profile of the dipole amplitude is a Gaussian for all values of $r$, its Fourier transform becomes exponential for all values of $t$ irrespective of the value of the cut-off. For low $\Lambda_r$, the integrand in \eq{d-f} is in the colour transparency regime (or the 1-Pomeron limit of the IP-Sat model), and the $b$-dependence of the amplitude is Gaussian and consequently its Fourier transform is exponential for all values of $t$.
However, in a case with a large cutoff $\Lambda_r$, the typical dipole size which contributes to the integral is within the unitarity or black-disc limit, see e.g. \cite{Frankfurt:2011cs}, with $\mathcal{N}\to 1$ (see \fig{f-sd} right panel). Then, the Fourier transform of the dipole amplitude
leads to a dip or multi-dips, as seen in \fig{f-sd} (left panel).
The saturation effect becomes more important at smaller Bjorken-$x$ or larger $W_{\gamma p}$, and lower virtualities $Q$ where the the contribution of large dipole sizes becomes more important, leading to a large effective $\Lambda_r$ and consequently to the dip-type structure.
For lighter vector mesons, the overlap extends to larger dipole sizes resulting in a dip structure as seen in \fig{f-sd}. The full calculation computed from \eq{vm} and shown in \fig{f-vt2}, indeed supports the fact that lighter vector mesons (which naturally have a larger $\Lambda_r$) develop multiple dips within the same kinematic region in which the heavier vector meson has a single dip (with a correspondingly smaller $\Lambda_r$), consistent with the expectation in the saturation picture shown in \fig{f-sd} (left panel). The exact position of dips and whether the t-distribution has multiple or a single minimum depend on the value of dynamical cutoff $\Lambda_r$ (via the kinematics and the mass of vector mesons) and the impact-parameter profile of the saturation scale. In the case of $\psi(2s)$ vector meson, although the scalar part of the $\psi(2s)$ wave function extends to large dipole sizes, due to the node effect, there is large cancellation between dipole sizes above and below the node position. As a result, the total cross-section of $\psi(2s)$ is suppressed compared to $J/\psi$ production as seen in \fig{f-ratio} and the dip in the $t$-distribution moves slightly to higher $|t|$ compared to diffractive $J/\psi$ production. We recall that $\psi(2s)$ is slightly heavier than $J/\psi$ and consequently the dip (for a heavier vector meson) moves toward higher $|t|$ compared to $J/\psi$ production.
Admittedly, the impact parameter dependence in saturation models lies in the domain of non-perturbative physics as commented previously and is, at present, put by hand and adjusted to data. A Gaussian profile is usually considered, but one could also try another profile whose Fourier transform leads to dips in the diffractive distribution. Therefore, the presence of dips cannot be considered, per se, as a signal of saturation.
But it is important to note that the main difference between a dipole model with linear and non-linear evolution (incorporating saturation effects through some specific model as those employed in this work) is that the former does not lead to the black-disc limit and, therefore, the dips do not systematically shift toward lower $|t|$ by increasing $W_{\gamma p}$, $1/x$, and $r$ or $1/Q$, while the latter does.
Non-linear evolution evolves any realistic profile in $b$, like a Gaussian or Woods-Saxon distribution, and makes it closer to a step-like function in the $b$-space by allowing an increase in the periphery of the hadron (the dilute region) while limiting the growth in the denser center, see Fig. \ref{f-w6} for illustration. This leads to the appearance of dips with non-linear evolution even if the dips were not present at the initial condition at low energies or for large $x$ (e.g. a Gaussian profile), or to the receding of dips towards lower values of $|t|$ even if they were already present in the initial condition (e.g. with a Woods-Saxon type profile). In \fig{f-w6}, we show the evolution of the effective impact-parameter profile of the dipole amplitude defined as $T^{eff}(b)=\mathcal{N}(x,r,b)/\sigma^{dipole}(x,r)$ with $x$ and $r$ in different models. It is clearly seen that in the saturation models, by increasing $1/x$ or $r$, the effective impact-parameter profile $T^{eff}(b)$ naturally evolves towards a step-like function with a dynamical median extended to a larger $b$ while, in contrast, in the 1-Pomeron dipole model $T^{eff}(b)$ does not change with $r$ and $x$. Note that the typical impact-parameter of collisions is approximately related to the inverse of $|t|$, namely $|t|\propto 1/b$, see Eqs.\,(\ref{am-i}), (\ref{vm}) or \eq{d-f}. Now, at small $|t|$, the typical collisions are mostly peripheral and the system is in the dilute regime with a Gaussian profile. Therefore, saturation effects become less relevant, and there will be no dip in the $t$-distribution. On the other hand, at large $|t|$ the typical collisions are central, and interactions probe the high-density region of the target proton. Then saturation effects become important and distort the impact-parameter profile leading to diffractive dips.
The position of the dip in the $t$-distribution is presently rather model dependent. This is mainly due to the fact that the appearance of dips probes the dipole scattering amplitude in the saturation regime, where current available data at small $x$ do not constrain sufficiently the dipole models \cite{b-cgc-a,Albacete:2014fwa}. The exact position of the dip can only be numerically computed and depends on the effective dipole transverse size probed by the system (via a convolution between vector meson overlap wavefunction and the dipole amplitude) and impact-parameter profile of the saturation scale. Nevertheless, it is qualitatively expected that the dip becomes stronger or moves to lower $|t|$ for the case that the saturation or unitarity effects probed by the system at a given kinematics and impact parameter become more important. The saturation scale in the IP-Sat and b-CGC models is approximately similar at the HERA kinematics for the typical impact-parameter probed in the total $\gamma^\star p$ cross-section of about $b\approx 2\div 3\,\text{GeV}^{-1}$ \cite{b-cgc-a}. However, at very small $x$ and large $|t|$ the effective impact-parameter profile of the dipole amplitude in these two saturation models is different. This is shown in \fig{f-w6} where it is seen that in the b-CGC model because of non-trivial correlations between $x$ and $b$, the effective impact-parameter profile of dipole tends to flatten sooner with lowering $x$ and/or increasing dipole transverse size $r$ compared to the IP-Sat model. Therefore, the black disk limit is probed slightly faster in the b-CGC model than in the IP-Sat model, and consequently the dip (or dips) appears at lower $|t|$ in the b-CGC model compared to the IP-Sat model. This general expectation is, remarkably, in accordance with the results obtained from full computation for different vector mesons shown in \fig{f-vt2}. We also numerically verified that changing kinematics ($W_{\gamma p}$ and $Q$) does not alter this feature.
\begin{figure}[t]
\includegraphics[width=0.47\textwidth,clip]{plot-dipole-bf.eps}
\includegraphics[width=0.47\textwidth,clip]{plot-dipole-bxf.eps}
\caption{Effective impact-parameter profile of the dipole amplitude defined as $T^{eff}(b)=\mathcal{N}(x,r,b)/\sigma^{dipole}(x,r)$ in different models as a function of impact-parameter $b$, for a fixed value of $x$ and two different values of dipole transverse size $r$ (left panel), and for a fixed $r$ and two different values of $x$ (right panel). }
\label{f-w6}
\end{figure}
\section{conclusion}
In this paper, we investigated the exclusive production of vector mesons in high-energy collisions. We extended previous studies \cite{ip-sat0,ip-sat1} by using saturation models fitted to the most recent inclusive DIS data and including in our analysis the recent experimental diffractive data from the LHCb \cite{lhcb,lhcbn} and ALICE \cite{TheALICE:2014dwa} collaborations, and the combined HERA analysis \cite{Aaron:2009aa,Abramowicz:1900rp}. We showed that the recent LHC data on diffractive $J/\psi$ photoproduction are in good agreement with the saturation/CGC predictions while there are some tensions between recent LHCb and ALICE data with the 1-Pomeron model and pQCD results, see Figs.\,\ref{f-vw1},\,\ref{f-vw2} (right panel). This can be considered as the first hint of saturation effects at work in diffractive photoproduction of vector mesons off proton at the LHC.
We provided predictions for the total cross-section of diffractive photoproduction of $J/\psi, \psi(2s)$ and $\rho$ within the gluon saturation/CGC picture at the LHC and future colliders. To single out the non-linear effects due to saturation, we also compared with those results obtained in the 1-Pomeron model. We also provided predictions for the ratio of diffractive production of $\psi(2s)$ to $J/\psi$, namely $R=\psi(2s)/J/\psi$ at HERA and the LHC.
We showed that while at high virtualities $R$ has little $|t|$ and $W_{\gamma p}$ dependence, it moderately increases with virtuality $Q$ at a fixed $W_{\gamma p}$. We also found that the photoproduction ratio $R (Q=0)$ increases with $W_{\gamma p}$ and becomes sensitive to different saturation models.
We showed that the $t$-differential cross-section of exclusive production of vector mesons in high-energy collisions offers a unique opportunity to probe the saturation regime. We quantified some theoretical uncertainties and showed that the appearance of a dip or dips in the diffractive $t$-distribution of the different vector mesons ($J/\psi, \psi(2s)$, $\phi,\rho$) is a robust prediction of the saturation picture. In non-saturation models, dips are either absent or expected to lie at larger $|t|$ and not to shift towards smaller $|t|$ with increasing energy. The position of the dip is presently rather model dependent, see \fig{f-vt0}. This is mainly due to the fact that the appearance of dips probes the dipole scattering amplitude at large dipole sizes in the saturation regime, where current available data at small $x$ do not constrain sufficiently the dipole models \cite{b-cgc-a,Albacete:2014fwa}.
On the positive side, future experimental data, in particular at the LHeC, on the energy and $t$ diffractive distributions of different vector meson production off protons, will provide valuable constraints on saturation models and allow us to unravel the relevance of non-linear effects in the accessible kinematic region.
We recall that the $t$-distribution of all vector mesons, as well as DVCS, at HERA, can be correctly reproduced by fixing the impact-parameter profile of the colour dipole amplitude at small $|t|$, despite the fact that the vector meson and DVCS wave functions are very different \cite{ip-sat-a,b-cgc-a}. This strongly hints at universality of the extracted impact-parameter distribution of gluons in the periphery of the proton. On the other hand, at large $|t|$ where we do not have currently experimental data, one can probe the transverse spatial distribution of gluons in the center of the proton where the black-disc limit could be at work. Therefore, the $t$-distribution of diffractive vector mesons would provide the most important information on the relevance of saturation dynamics. Besides, the impact parameter distribution of gluons in protons and nuclei (a natural extension of our work that can be explored in electron-nucleus colliders \cite{eic,lhec}, see also \cite{Frankfurt:2011cs,Caldwell:2010zza,ip-g1,tu}) is a crucial ingredient for a detailed characterisation of the initial conditions in heavy ion collisions. Note that the effects of fluctuations and correlations on the proton are not incorporated into our formulation. This is an important issue that certainly deserves separate study.
Note that diffractive vector meson production off a nucleus is quite different from a proton target. The diffractive interaction with the nuclear target can either be elastic (coherent) or inelastic (incoherent) - in the latter case the nucleus subsequently radiates a photon or breaks up into colour neutral fragments, while in the former, the nucleus stays intact. The coherent cross section is obtained by averaging the amplitude before squaring it, $|\langle\mathcal{A}\rangle_ N |^2$, and the incoherent one is the variance of the amplitude with respect to the initial nucleon configurations N of the nucleus $\langle |\mathcal{A}|^2\rangle_ N -|\langle \mathcal{A}\rangle_ N |^2$ which according to the Good-Walker picture measures the fluctuations or lumpiness of the gluon density inside the nucleus. In the case of a nucleus, the diffractive production rate is controlled by two different scales of $1/R_p$ and $1/R_A$ with $R_p$ and $R_A$ being the proton and nucleus size. At momentum scales corresponding to the nucleon size $|t|\sim 1/R_p^2$ the diffractive cross section is almost purely incoherent. The $t$-distribution in coherent diffractive production off nucleus gives rise to a dip-type structure for both saturation and non-saturation models, while in the case of incoherent production at small $|t|$, both saturation and non-saturation models do not lead to dips \cite{ip-g1,tu}. This is in drastic contrast to the diffractive production off proton where only saturation models lead to dip-type structure in the $t$-distribution at values of $|t|$ that can be experimentally accessible. Therefore, diffractive production off nucleus is a sensible probe of unitarity effects at the nuclear level while being less sensitive to the unitarity limit and saturation effects inside the proton.
Finally, note that diffractive dips in $t$-distribution were also observed in elastic hadronic reactions \cite{dip-el,dip-el-2}. However, it remains to be understood whether the origin of the dips in elastic hadronic reactions and diffractive DIS is the same. In contrast to diffractive DIS, the differential cross-section of elastic proton-proton collisions is not currently computable in the weak coupling regime due to the absence of a large scale, and some phenomenological models are often employed (for a review see Ref.\,\cite{rev-el}). Nevertheless, in both cases multiple parton interactions or multiple Pomeron exchanges seem to play an important role in the appearance of dips in the $t$-distribution, see e.g. \cite{all-el,Brogueira:2011jb}.
\begin{acknowledgments}
We thank Jorge Dias de Deus, Edmond Iancu, Paul Laycock, Genya Levin, Magno Machado and Jan Nemcik for useful discussions. The work of NA is supported by the European Research Council grant HotLHC ERC-2011-StG-279579; by the People Programme (Marie
Curie Actions) of the European Union Seventh Framework Programme FP7/2007-2013/
under REA grant agreement n318921; by Ministerio de Ciencia e Innovaci\'on of Spain under project FPA2011-22776; by Xunta de Galicia (Conseller\'{\i}a de Educaci\'on and Conseller\'\i a de Innovaci\'on e Industria - Programa Incite); and by the Spanish Consolider-Ingenio 2010 Programme CPAN and FEDER.
The work of AHR is supported in part by Fondecyt grant 1110781.
\end{acknowledgments}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,238 |
using UnityEngine;
using System.Collections;
public class RainbowTrail : MonoBehaviour
{
public float time = 2.0f;
public int rate = 10;
private Vector3[] arv3;
private int head;
private int tail= 0;
private float sliceTime;
private RaycastHit hit;
public Player player;
public GameObject explosion;
public FinishManager fm;
void Start (){
sliceTime = 1.0f / rate;
arv3 = new Vector3[(Mathf.RoundToInt(time * rate) + 1)];
for (int i= 0; i < arv3.Length; i++)
arv3[i] = player.transform.position;
head = arv3.Length-1;
StartCoroutine(CollectData());
}
IEnumerator CollectData (){
while (true) {
if (player != null && player.transform.position != arv3[head]) {
head = (head + 1) % arv3.Length;
tail = (tail + 1) % arv3.Length;
arv3[head] = player.transform.position;
}
yield return new WaitForSeconds(sliceTime);
}
}
void Update (){
if (Hit ()) {
Destroy(hit.collider.gameObject);
Instantiate(explosion, hit.transform.position, Quaternion.identity);
int playerNum;
if (hit.collider.gameObject.Equals(player.gameObject)) {
if(player.getPlayerNum() == 1 ) {
playerNum = 2;
} else {
playerNum = 1;
}
} else {
playerNum = player.getPlayerNum();
}
fm.finish(playerNum);
}
}
bool Hit (){
int i= head;
int j= (head - 1);
if (j < 0) j = arv3.Length - 1;
while (j != head) {
if (Physics.Linecast(arv3[i], arv3[j], out hit))
return true;
i = i - 1;
if (i < 0) i = arv3.Length - 1;
j = j - 1;
if (j < 0) j = arv3.Length - 1;
}
return false;
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,035 |
{"url":"https:\/\/solvedlib.com\/what-is-time-to-impact-drop-4451lb-weight-with,93088","text":"# What is time to impact? Drop a 4451lb weight with a flat surface area of 107.2535...\n\n###### Question:\n\nWhat is time to impact? Drop a 4451lb weight with a flat surface area of 107.2535 square feet from 1 mile high and object falls flat at constant acceleration with no aerodynamic drag until terminal velocity. Coefficient of drag = C(d)= 1.28, Gravitation constant = g =32.2 ft\/sec squared\n\n#### Similar Solved Questions\n\n##### 7. Promote communication about patient preferences and values to plan care related to treatment for kidney...\n7. 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[nch g Ani 439 363 3i6 Nong...\n##### Arrhuopologlses ate Most Inicrested In t2era37Mutple ChokePaleozoicNeozokcCenozolProterozoiecnasar\nArrhuopologlses ate Most Inicrested In t2 era 37 Mutple Choke Paleozoic Neozokc Cenozol Proterozoie cnasar...\n##### Risk-utility balancing refers to the fact that: Select one: O a. manufacturers must disclose all risks...\nRisk-utility balancing refers to the fact that: Select one: O a. manufacturers must disclose all risks to using their product before putting it on the market O b. all products can be made completely safe if manufacturers are willing to spend enough money C. some products are inherently dangerous and...\n##### Borax Solubility: Investigating the Relationship between Thermodynamics and Equilibrium21Look Up the Kye values for the follwing compounds: Iead (II) sulfate, aluminum hydrox- Ide silver chlonde, silver bromide; and copper sulfide_ Which one most soluble? least soluble? Explain_A student is curious about the value for NaCl: The student boks up tha value in the Tcant appandix of his textbook but = find value for NaCl Why not?Draw the structure fr the tetraborate ion1G iotuc Oil 31 ;1 47 TS 0;3e e\nBorax Solubility: Investigating the Relationship between Thermodynamics and Equilibrium 21 Look Up the Kye values for the follwing compounds: Iead (II) sulfate, aluminum hydrox- Ide silver chlonde, silver bromide; and copper sulfide_ Which one most soluble? least soluble? Explain_ A student is curio...\n##### 18. Identify which of the following statements is true. A) With limited exceptions, a loss can...\n18. Identify which of the following statements is true. A) With limited exceptions, a loss can be recognized by a liquidating corporation when it makes a liquidating distribution of property that has declined in value. B) When computing the corporate-level gain on a liquidating distribution, the FMV...\n##### Part A If the system is released from rest, determine the speed of the 22 kg...\nPart A If the system is released from rest, determine the speed of the 22 kg cylinder A after A has moved downward a distance of 3 m. The differential pulley has a mass of 17 kg with a radius of gyration about its center of mass of ko = 110 mm (Figure 1) Express your answer with the appropriate unit...\n##### Problem 1. The function solution to the diflerential equation y\" 2N\/ + Aly = 0. Find the second linearly independent solution as follows_Lct y u(sJe^ be solution to the differential equation Determine u(r). Hint: plug it into the equation then solve the resulting equation and find general solution for u(z)\nProblem 1. The function solution to the diflerential equation y\" 2N\/ + Aly = 0. Find the second linearly independent solution as follows_ Lct y u(sJe^ be solution to the differential equation Determine u(r). Hint: plug it into the equation then solve the resulting equation and find general sol...\n##### What is the percentage difference? Total initial momentum = 7kg.m\/s Total final momentum = 9 kg.m\/s\nWhat is the percentage difference? Total initial momentum = 7 kg.m\/s Total final momentum = 9 kg.m\/s...\n##### The U.S. Census Bureau collects data on the ages of marriedpeople. Suppose that eight married couples are randomly selectedand have the ages given in the following table. Determinethe 90%90% confidence interval for the true meandifference between the ages of married males and marriedfemales.Let d=(age of husband)?(age of wife)d=(age of husband)?(age of wife).Assume that the ages are normally distributed for the populationsof both husbands and wives in the U.S.Husband68686363232363636666474736363\nThe U.S. Census Bureau collects data on the ages of married people. Suppose that eight married couples are randomly selected and have the ages given in the following table. Determine the 90%90% confidence interval for the true mean difference between the ages of married males and married females. Le...\n##### ( pts) windox being built und the top is A semicircle and thc bottom is rectangle. If thcre 12 melers (riming matcrials. whal must the dimension; Of thc - wndo be I0 Iet in the most. light?\n( pts) windox being built und the top is A semicircle and thc bottom is rectangle. If thcre 12 melers (riming matcrials. whal must the dimension; Of thc - wndo be I0 Iet in the most. light?...\n##### 7. An object moving along a straight line path has a differentiable position function y = s(t); s(t) measures the object's position relative to the origin at time t. It is known that at time t = 9 seconds, the object's position is s(9) = 4 feet (i.e. 4 feet to the right of the origin): Furthermore, the object's instantaneous velocity at t = 9 is -1.2 feet per second, and its acceleration at the same instant is 0.08 feet per second per second:Use local linearity to estimate the pos\n7. An object moving along a straight line path has a differentiable position function y = s(t); s(t) measures the object's position relative to the origin at time t. It is known that at time t = 9 seconds, the object's position is s(9) = 4 feet (i.e. 4 feet to the right of the origin): Fur...\nFSA is a privately held firm. As an analyst trying to determine the value of FSA\u2019s common stock and bonds, you have estimated the market value of the firm\u2019s assets to be $1 million and the standard deviation of the asset return to be .3. The debt of FSA, which consists of zero-coupon ban... 5 answers ##### Assignment Submission For this assignment; you submit answers by question parts The - Assignment Scoring number of subr Your last submission is used for your score{ points GHCoINg12 3.4018ML Let f(x) = x2 + X and g(x) = 3x2 1. Find the function. f : 9 (f . 9)(x)Find the domain. (Enter your answer using interval notation )Nede HeneRaad ItMaater ItWam @ Assignment Submission For this assignment; you submit answers by question parts The - Assignment Scoring number of subr Your last submission is used for your score { points GHCoINg12 3.4018ML Let f(x) = x2 + X and g(x) = 3x2 1. Find the function. f : 9 (f . 9)(x) Find the domain. (Enter your answer ... 5 answers ##### Use the extended Euclidean Algorithm (using matrices) to find integers and such that 204x 75v = 1 (show all steps)204,b = 75,0 =204 Use the extended Euclidean Algorithm (using matrices) to find integers and such that 204x 75v = 1 (show all steps) 204,b = 75,0 = 204... 5 answers ##### -\/3 POINTSSPRECALC7 6.6.016.MY NOTESASK YOUR TEACHERSolve triangle ABC: (If an answer does not exist, enter DNE: Round your answers to one decimal place:) c = 17Need Help?LecaneelThbtoln Muter-\/1 POINTSSPRECALC7 6.6.022.Mi:MY NOTESASK YOUR TEACHERFind the indicated side x. (Use either the Law of Sines or the Law of Cosines, as appropriate. Assume b = 10 and \u00e2\u201a\u00ac = 19. Round your answer to one decimal place:)Need Help?Rand FHaeberItHalk la uitular -\/3 POINTS SPRECALC7 6.6.016. MY NOTES ASK YOUR TEACHER Solve triangle ABC: (If an answer does not exist, enter DNE: Round your answers to one decimal place:) c = 17 Need Help? Lecaneel Thbtoln Muter -\/1 POINTS SPRECALC7 6.6.022.Mi: MY NOTES ASK YOUR TEACHER Find the indicated side x. (Use either th... 1 answer ##### Describe several factors that affect the quality of dietary protein. How does eating too much, or... Describe several factors that affect the quality of dietary protein. How does eating too much, or too little protein affect health?... 1 answer ##### Rotenone is a potent inhibitor of Complex I. Atpenin is a potent inhibitor of Complex II.... Rotenone is a potent inhibitor of Complex I. Atpenin is a potent inhibitor of Complex II. Which do you expect to inhibit the proton motive force-why? (5 pts)... 5 answers ##### Find the solution of the Bernoulli'$ ODE, dy+y= dx yz Xy > 0\nFind the solution of the Bernoulli' \\$ ODE, dy+y= dx yz Xy > 0...\n##### Question 143 ptsA 120-V rms voltage at 1.0 kHz is applied to a 2.0-mH inductor; a 4.0-HF capacitor and a resistor: If the rms value of the current in this circuit is 0.60 A_ what is the value of the resistor at resonance?300 Q120 Q420 9240 Q200 Q\nQuestion 14 3 pts A 120-V rms voltage at 1.0 kHz is applied to a 2.0-mH inductor; a 4.0-HF capacitor and a resistor: If the rms value of the current in this circuit is 0.60 A_ what is the value of the resistor at resonance? 300 Q 120 Q 420 9 240 Q 200 Q...\n##### QUESTION 6 Which is the correct formula for computing the overhead rate? A. estimated use of...\nQUESTION 6 Which is the correct formula for computing the overhead rate? A. estimated use of the cost driver for production\/estimated overhead for the activity B. estimated overhead for the product\/estimated use of the cost driver for the activity C. estimated use of the cost driver for production\/e...\n##### Anda 7Let denote the usual matrix multiplication It To 01 fo 0 0 0 1 0 0 1 0 0 thenosia nonora dalanteggio max: Jo.00p Contrassegna comanda(a) det(A? B? ) (D) det(A? B2) (c) 'det(A? B2) = 2 (C) det(A? . B2)PrecedenteSuccesslm\nanda 7 Let denote the usual matrix multiplication It To 01 fo 0 0 0 1 0 0 1 0 0 then osia non ora dala nteggio max: Jo.00 p Contrassegna comanda (a) det(A? B? ) (D) det(A? B2) (c) 'det(A? B2) = 2 (C) det(A? . B2) Precedente Successlm...\n##### Two charges, Q1= 2.60 HC, and Q2 = 6.10 pC are located Qz at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure: 2 1 What is the magnityde of the electric field at point P; located at (5.50 cm, 0) , due to Q1 alone? 5.50x106 N\/C You are correct. Previous Tries What is the X-component of the total electric field at P? 30.9E6 NJC By the principle of linear superposition, the total electric field at \"q Position P is the vector sum of the electric field contribution from charge and Q\nTwo charges, Q1= 2.60 HC, and Q2 = 6.10 pC are located Qz at points (0,-3.50 cm ) and (0,+3.50 cm), as shown in the figure: 2 1 What is the magnityde of the electric field at point P; located at (5.50 cm, 0) , due to Q1 alone? 5.50x106 N\/C You are correct. Previous Tries What is the X-component of t...\n##### Calculate the volume in milliliters of a 1.48 mol\/L calcium bromide solution that contains 100 mmol...\nCalculate the volume in milliliters of a 1.48 mol\/L calcium bromide solution that contains 100 mmol of calcium bromide (CaBry). Round your answer to 3 significant digits....\n##### The nurse in the outpatient orthopedic clinic is reviewing the medical records of several patients being...\nThe nurse in the outpatient orthopedic clinic is reviewing the medical records of several patients being seen in the clinic today. In the most recent note by the provider, the nurse reads the following: \u201c67-year-old, moderately obese woman examined today for increasing back and left shoulder p...","date":"2022-12-01 00:53:07","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5568481683731079, \"perplexity\": 3963.139008531692}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-49\/segments\/1669446710777.20\/warc\/CC-MAIN-20221130225142-20221201015142-00729.warc.gz\"}"} | null | null |
'use strict';
const assert = require('assert').strict;
const log4js = require('log4js');
const httpLogger = log4js.getLogger('http');
const settings = require('../../utils/Settings');
const hooks = require('../../../static/js/pluginfw/hooks');
const readOnlyManager = require('../../db/ReadOnlyManager');
hooks.deprecationNotices.authFailure = 'use the authnFailure and authzFailure hooks instead';
// Promisified wrapper around hooks.aCallFirst.
const aCallFirst = (hookName, context, pred = null) => new Promise((resolve, reject) => {
hooks.aCallFirst(hookName, context, (err, r) => err != null ? reject(err) : resolve(r), pred);
});
const aCallFirst0 =
async (hookName, context, pred = null) => (await aCallFirst(hookName, context, pred))[0];
exports.normalizeAuthzLevel = (level) => {
if (!level) return false;
switch (level) {
case true:
return 'create';
case 'readOnly':
case 'modify':
case 'create':
return level;
default:
httpLogger.warn(`Unknown authorization level '${level}', denying access`);
}
return false;
};
exports.userCanModify = (padId, req) => {
if (readOnlyManager.isReadOnlyId(padId)) return false;
if (!settings.requireAuthentication) return true;
const {session: {user} = {}} = req;
if (!user || user.readOnly) return false;
assert(user.padAuthorizations); // This is populated even if !settings.requireAuthorization.
const level = exports.normalizeAuthzLevel(user.padAuthorizations[padId]);
return level && level !== 'readOnly';
};
// Exported so that tests can set this to 0 to avoid unnecessary test slowness.
exports.authnFailureDelayMs = 1000;
const checkAccess = async (req, res, next) => {
const requireAdmin = req.path.toLowerCase().startsWith('/admin');
// ///////////////////////////////////////////////////////////////////////////////////////////////
// Step 1: Check the preAuthorize hook for early permit/deny (permit is only allowed for non-admin
// pages). If any plugin explicitly grants or denies access, skip the remaining steps. Plugins can
// use the preAuthzFailure hook to override the default 403 error.
// ///////////////////////////////////////////////////////////////////////////////////////////////
let results;
let skip = false;
const preAuthorizeNext = (...args) => { skip = true; next(...args); };
try {
results = await aCallFirst('preAuthorize', {req, res, next: preAuthorizeNext},
// This predicate will cause aCallFirst to call the hook functions one at a time until one
// of them returns a non-empty list, with an exception: If the request is for an /admin
// page, truthy entries are filtered out before checking to see whether the list is empty.
// This prevents plugin authors from accidentally granting admin privileges to the general
// public.
(r) => (skip || (r != null && r.filter((x) => (!requireAdmin || !x)).length > 0)));
} catch (err) {
httpLogger.error(`Error in preAuthorize hook: ${err.stack || err.toString()}`);
if (!skip) res.status(500).send('Internal Server Error');
return;
}
if (skip) return;
if (requireAdmin) {
// Filter out all 'true' entries to prevent plugin authors from accidentally granting admin
// privileges to the general public.
results = results.filter((x) => !x);
}
if (results.length > 0) {
// Access was explicitly granted or denied. If any value is false then access is denied.
if (results.every((x) => x)) return next();
if (await aCallFirst0('preAuthzFailure', {req, res})) return;
// No plugin handled the pre-authentication authorization failure.
return res.status(403).send('Forbidden');
}
// This helper is used in steps 2 and 4 below, so it may be called twice per access: once before
// authentication is checked and once after (if settings.requireAuthorization is true).
const authorize = async () => {
const grant = async (level) => {
level = exports.normalizeAuthzLevel(level);
if (!level) return false;
const user = req.session.user;
if (user == null) return true; // This will happen if authentication is not required.
const encodedPadId = (req.path.match(/^\/p\/([^/]*)/) || [])[1];
if (encodedPadId == null) return true;
let padId = decodeURIComponent(encodedPadId);
if (readOnlyManager.isReadOnlyId(padId)) {
// pad is read-only, first get the real pad ID
padId = await readOnlyManager.getPadId(padId);
if (padId == null) return false;
}
// The user was granted access to a pad. Remember the authorization level in the user's
// settings so that SecurityManager can approve or deny specific actions.
if (user.padAuthorizations == null) user.padAuthorizations = {};
user.padAuthorizations[padId] = level;
return true;
};
const isAuthenticated = req.session && req.session.user;
if (isAuthenticated && req.session.user.is_admin) return await grant('create');
const requireAuthn = requireAdmin || settings.requireAuthentication;
if (!requireAuthn) return await grant('create');
if (!isAuthenticated) return await grant(false);
if (requireAdmin && !req.session.user.is_admin) return await grant(false);
if (!settings.requireAuthorization) return await grant('create');
return await grant(await aCallFirst0('authorize', {req, res, next, resource: req.path}));
};
// ///////////////////////////////////////////////////////////////////////////////////////////////
// Step 2: Try to just access the thing. If access fails (perhaps authentication has not yet
// completed, or maybe different credentials are required), go to the next step.
// ///////////////////////////////////////////////////////////////////////////////////////////////
if (await authorize()) return next();
// ///////////////////////////////////////////////////////////////////////////////////////////////
// Step 3: Authenticate the user. (Or, if already logged in, reauthenticate with different
// credentials if supported by the authn scheme.) If authentication fails, give the user a 401
// error to request new credentials. Otherwise, go to the next step. Plugins can use the
// authnFailure hook to override the default error handling behavior (e.g., to redirect to a login
// page).
// ///////////////////////////////////////////////////////////////////////////////////////////////
if (settings.users == null) settings.users = {};
const ctx = {req, res, users: settings.users, next};
// If the HTTP basic auth header is present, extract the username and password so it can be given
// to authn plugins.
const httpBasicAuth = req.headers.authorization && req.headers.authorization.startsWith('Basic ');
if (httpBasicAuth) {
const userpass =
Buffer.from(req.headers.authorization.split(' ')[1], 'base64').toString().split(':');
ctx.username = userpass.shift();
// Prevent prototype pollution vulnerabilities in plugins. This also silences a prototype
// pollution warning below (when setting settings.users[ctx.username]) that isn't actually a
// problem unless the attacker can also set Object.prototype.password.
if (ctx.username === '__proto__') ctx.username = null;
ctx.password = userpass.join(':');
}
if (!(await aCallFirst0('authenticate', ctx))) {
// Fall back to HTTP basic auth.
const {[ctx.username]: {password} = {}} = settings.users;
if (!httpBasicAuth || !ctx.username || password == null || password !== ctx.password) {
httpLogger.info(`Failed authentication from IP ${req.ip}`);
if (await aCallFirst0('authnFailure', {req, res})) return;
if (await aCallFirst0('authFailure', {req, res, next})) return;
// No plugin handled the authentication failure. Fall back to basic authentication.
res.header('WWW-Authenticate', 'Basic realm="Protected Area"');
// Delay the error response for 1s to slow down brute force attacks.
await new Promise((resolve) => setTimeout(resolve, exports.authnFailureDelayMs));
res.status(401).send('Authentication Required');
return;
}
settings.users[ctx.username].username = ctx.username;
// Make a shallow copy so that the password property can be deleted (to prevent it from
// appearing in logs or in the database) without breaking future authentication attempts.
req.session.user = {...settings.users[ctx.username]};
delete req.session.user.password;
}
if (req.session.user == null) {
httpLogger.error('authenticate hook failed to add user settings to session');
return res.status(500).send('Internal Server Error');
}
const {username = '<no username>'} = req.session.user;
httpLogger.info(`Successful authentication from IP ${req.ip} for user ${username}`);
// ///////////////////////////////////////////////////////////////////////////////////////////////
// Step 4: Try to access the thing again. If this fails, give the user a 403 error. Plugins can
// use the authzFailure hook to override the default error handling behavior (e.g., to redirect to
// a login page).
// ///////////////////////////////////////////////////////////////////////////////////////////////
if (await authorize()) return next();
if (await aCallFirst0('authzFailure', {req, res})) return;
if (await aCallFirst0('authFailure', {req, res, next})) return;
// No plugin handled the authorization failure.
res.status(403).send('Forbidden');
};
/**
* Express middleware to authenticate the user and check authorization. Must be installed after the
* express-session middleware.
*/
exports.checkAccess = (req, res, next) => {
checkAccess(req, res, next).catch((err) => next(err || new Error(err)));
};
| {
"redpajama_set_name": "RedPajamaGithub"
} | 8,773 |
Q: Что означает "опыт коммерческой разработки" на С++ Какие нужны знания что бы утверждать, что есть опыт коммерческой разработки на С++. Просьба, не писать необдуманных ответов.
A: это значит только одно - приложение, которое разрабатывалось, прямо или косвенно приносило деньги. То есть:
*
*приложение продавалось Вами непосредственно.
*приложение продавалось фирмой, которая платила Вам зарплату.
*приложение было "бесплатным" сервером, а для доступа к нему, людям нужно было покупать отдельное (не зависящее от Вас) приложение, например, клиент на адроиде.
Отсюда вывод. Вам не знания нужны, а просто список приложений. То есть, ответ должен быть такой: "работал N лет в фирме ABC, которая делала продукт ZZZ (можно посмотреть на сайте www.previouscompany.com). Я в этом продукте делал функциональность YYY".
A: В общем-то элементарно - опыт коммерческой разработки С++ подразумевает работу С++-программистом, причем вне зависимости от типа - будь то работа в офисе, удаленная работа, фриланс или какие-то другие формы сотрудничества, в том числе и без трудового договора или договора найма (то есть важен именно опыт таковой работы. Хотя, весьма возможно, что опыт фрилансерской деятельности может по разным причинам цениться ниже. Однако же его тоже можно отнести к пресловутому "опыту коммерческой разработки"). Разумеется, написание лабораторных работ по 500 рублей за штуку сюда не входит, и это еще один критерий - имеет значение скорее всего сам факт разработки сколько-нибудь существенного ПО, а следовательно и понимания принципов разработки и способность кандидата подтвердить свои навыки. Возможно, что собеседующего (ведь речь о вакансиях, не так ли?) устроит и факт деятельного участия в каком-нибудь более-менее существенном опенсорсном проекте (хотя не уверен - для некоторых собеседующих, а уж тем более рекрутеров даже факт наличия богатого прошлого на Гитхабе или регулярные коммиты в ядро Linux мало о чем скажет (несколько утрирую, но, думаю, мысль понятна)). В общем, в основном "опыт коммерческой разработки" - это (помимо опыта работы и записи в трудовой книжке) то, что отличает студента без опыта от хотя бы джуниора, и позволяет понять, что соискатель занимался чем-то более серьезным, нежели курсовые проекты, формошлёпство и велосипедостроение | {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,353 |
The Political Activist
'In a year when so much felt out of our control, getting involved civically and electing the politicians you want to represent you is something that you can control.' – Noah Goldmann, an SU student who worked on a Senate campaign last year
As Told To Maggie Hicks
Illustration by Samantha Currier
Noah Goldmann is an SU student who worked on a Senate campaign last year.
I think a lot of people only realize the impact of politics on their lives when there is a crisis.
Once COVID hit, the stakes felt higher. I mean, the stakes were higher. A lot of people were very suddenly realizing how great an impact their elected representatives can have on their day-to-day lives, and that was true for me, too.
In 2016, my mom and I went and knocked on doors in Philadelphia with Hillary Clinton … but we all know how that turned out. After that election, I was like, "Oh my God, I could've done so much more." I promised myself that the next time an election rolled around, I would do everything I could to make the difference that I wanted to see in the world.
Last summer, I went to Montana to work on Steve Bullock's Senate campaign. When I went, in July, it had the second lowest COVID rate in the country. So, I came from New York, where no one was doing anything, to Montana where everyone was doing most things.
A lot of our campaigning was in person. We never opened our office to volunteers, but we would meet them outside and we would have events outside. We had a lot of first-time volunteers – in a year when so much felt out of our control, getting involved civically and electing the politicians you want to represent you is something that you can control.
"I felt so grateful to be able to do something where I could actually be in the world and be relating to other people."
Noah Goldmann
By October and November, we were regularly working 80 hours a week. We spent every moment of the day trying to elect the politicians we wanted to elect. But in the midst of all of that, I felt so grateful to be able to do something where I could actually be in the world and be relating to other people. I wanted to wake up the day after the election knowing that I did everything I could do.
Throughout the months I was there, the COVID rates slowly ticked up and up, and by November, they were some of the highest in the country. It was kind of funky because Steve Bullock was the current governor of Montana when I was there, so he was trying to balance running his race for Senate with controlling the COVID outbreak in the state. It's a pretty fine line to walk between opening up and shutting down, especially as a politician who wants to get re-elected.
There's always a part of me that thinks that I could've worked harder or made a few extra phone calls or put in an hour more a couple days before the election. But moving to Montana, a state that I had never previously visited, and working so many hours and making thousands and thousands of phone calls and recruiting hundreds of volunteers and holding probably 100 events with our candidates – it was what I could do.
I am lucky enough to find myself in a spot where I can fight to make the world a better place, and lots of people need to just use their time to get by. So, I'm going to fight for myself, but also for them.
This as-told-to interview is part of COVID in the Community, a series created by students in the Reporting classes at the Newhouse School in Spring 2021. COVID in the Community documents the experiences of Syracuse area residents living through this extraordinary time.
Maggie Hicks
is a contributor for The NewsHouse.
Co Authors :
Samantha Currier
is a graphic designer for The NewsHouse. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 2,673 |
Q: Creating Environment variables in Jenkins I am trying to create a custom Environment variables in Jenkins by doing this
List of key-value pairs
Name: Build_Date
Value: Date()
The value part is not resolving to a date - any ideas?
Here is where I am trying to config the above
Thanks
A: I confirm that you can use the EnvInject plugin with a Groovy script:
Here is the Groovy script:
// Generate a global BUILD_ID_LONG variable with date and time
// =======================================
TimeZone.setDefault(TimeZone.getTimeZone('UTC'))
def now = new Date()
def map = [BUILD_ID_LONG: now.format("yyyyMMdd_HHmm")]
return map
Next, you can use the ${BUILD_ID_LONG} variable in your build steps.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,927 |
Q: Problema de troca de servidor Eu tinha um site com um sistema em um servidor, precisei migrar para outro e agora aparece erro em quase todas as funções PHP do sistema. Eu já dei as permissões nas pastas, já que esta hospedado em linux, porém não consigo mesmo assim.
Quando tento criar uma conta nova, por exemplo ele da os seguintes erros:
Warning: fsockopen() [function.fsockopen]: unable to connect to
autors.com.br:587 (Connection refused) in
/home/storage/1/f2/68/autors/public_html/smtp.class.php on line 134
erro 222 Warning: fgets(): supplied argument is not a valid stream
resource in /home/storage/1/f2/68/autors/public_html/smtp.class.php on
line 122
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,802 |
package com.gemstone.gemfire.distributed.internal;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.PrintStream;
import java.util.List;
import java.util.Set;
import com.gemstone.gemfire.GemFireIOException;
import com.gemstone.gemfire.distributed.internal.membership.InternalDistributedMember;
import com.gemstone.gemfire.distributed.internal.membership.MembershipManager;
import com.gemstone.gemfire.internal.LogWriterImpl;
import com.gemstone.gemfire.internal.PureLogWriter;
import com.gemstone.gemfire.internal.i18n.LocalizedStrings;
/**
* A log intended for recording product-use.
*
* This class wraps a {@link LogWriterImpl} which it uses to record messages.
* It wipes the log when it gets too large. The size of the log file is limited to 5mb by
* default and can be adjusted with the system property <b>max_view_log_size</b>,
* though the size is not allowed to be less than 1mb.
*
* @author Bruce Schuchardt
* @since 2013
*/
public final class ProductUseLog implements MembershipListener {
protected static long MAX_PRODUCT_USE_FILE_SIZE = Long.getLong("max_view_log_size", 5000000);
private final int logLevel;
private final File productUseLogFile;
private PureLogWriter logWriter;
private InternalDistributedSystem system;
static {
// don't allow malicious setting of the file size to something too small
if (MAX_PRODUCT_USE_FILE_SIZE < 1000000) {
MAX_PRODUCT_USE_FILE_SIZE = 1000000;
}
}
public ProductUseLog(File productUseLogFile) {
this.productUseLogFile = productUseLogFile;
this.logLevel = LogWriterImpl.INFO_LEVEL;
createLogWriter();
}
/** adds the log as a membership listener to the given system and logs the view when members join */
public void monitorUse(InternalDistributedSystem system) {
this.system = system;
DM dmgr = system.getDistributionManager();
dmgr.addMembershipListener(this);
MembershipManager mmgr = dmgr.getMembershipManager();
if (mmgr != null) {
log("Log opened with new distributed system connection. " + system.getDM().getMembershipManager().getView());
} else { // membership manager not initialized?
log("Log opened with new distributed system connection. Membership view not yet available in this VM.");
}
}
public synchronized void log(String logMessage) {
if (!this.logWriter.isClosed()) {
if (this.productUseLogFile.length() + logMessage.length() + 100 > MAX_PRODUCT_USE_FILE_SIZE) {
clearLog();
}
this.logWriter.info(logMessage);
}
}
/**
* Closes the log. It may be reopened with reopen(). This does not remove
* the log from any distributed systems it is monitoring.
*/
public synchronized void close() {
if (!this.logWriter.isClosed()) {
this.logWriter.close();
}
}
/**
* returns true if the log has been closed
*/
public synchronized boolean isClosed() {
return this.logWriter.isClosed();
}
/** reopens a closed log */
public synchronized void reopen() {
if (this.logWriter.isClosed()) {
createLogWriter();
}
}
private synchronized void clearLog() {
this.logWriter.close();
this.productUseLogFile.delete();
createLogWriter();
}
private synchronized void createLogWriter() {
FileOutputStream fos;
try {
fos = new FileOutputStream(productUseLogFile, true);
} catch (FileNotFoundException ex) {
String s = LocalizedStrings.InternalDistributedSystem_COULD_NOT_OPEN_LOG_FILE_0.toLocalizedString(productUseLogFile);
throw new GemFireIOException(s, ex);
}
PrintStream out = new PrintStream(fos);
this.logWriter = new PureLogWriter(this.logLevel, out);
}
@Override
public void memberJoined(InternalDistributedMember id) {
log("A new member joined: " + id + ". " + system.getDM().getMembershipManager().getView());
}
@Override
public void memberDeparted(InternalDistributedMember id, boolean crashed) {
}
@Override
public void memberSuspect(InternalDistributedMember id,
InternalDistributedMember whoSuspected) {
}
@Override
public void quorumLost(Set<InternalDistributedMember> failures,
List<InternalDistributedMember> remaining) {
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,902 |
Fra i suoi lavori si annoverano:
Il monumento alla Resistenza in Mantova nel 1965, al quale collaborò anche Aldo Bergonzoni
Il monumento ai Martiri del Libano in Beirut nel 1958 con Giuseppe Persichetti
Il monumento al Partigiano della città di Parma
Cappella della Vergine di Zahle, Libano
Monumento a Riad Al Solh in Beirut, Libano, con Giuseppe Persichetti
Sistemazione delle ceneri di Luigi Pirandello per conto del Comune di Agrigento
È stato vincitore di numerosi concorsi nazionali di progettazione, tra cui:
Secondo premio nel Concorso Nazionale per il Monumento ai Picciotti di Palermo, per un progetto presentato assieme al padre ed a Renato Guttuso nel 1965
Secondo premio per il Monumento ai Partigiani di Udine (mai realizzato)
Attivo in tutta Italia nell'ambito dell'edilizia residenziale privata e pubblica, realizzò tra l'altro il quartiere Q2 a Latina, la variante autostradale di Terracina lungo la SS. 7 - Appia e le Torri di Vigne Nuove alla Bufalotta (Roma).
Note | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,657 |
{"url":"http:\/\/m-phi.blogspot.com\/2013\/05\/nominalism-vs-syntax.html","text":"## Friday, 24 May 2013\n\n### Nominalism vs. Syntax\n\nIt is difficult to maintain, consistently, the following two claims:\n(i) There are no abstracta.\n(ii) There are syntactical entities (and they behave as our standard accounts say they do).\nConsider, for example, how one defines a language $L$. Beginning with two building blocks, $A$ and $B$, we say that $\\{A,B\\}$ is the alphabet. It's usually implicit, but sometimes needs to be stated, that $A \\neq B$. (One has to state this in a formalized theory of syntax.)\n\nFor the many of the usual purposes of syntactical theory, it does not matter what these building blocks $A$ and $B$ are. They could be two eggs. They could be the numbers 7 and $\\aleph_{57}$. They could be the letter types \"a\" and \"$\\aleph$\". Or they could be two of my guitars. Or they could be two tokens of the letters \"a\" and \"$\\aleph$''. It does not matter. And the fact that it doesn't matter plays an important role in G\u00f6del's incompleteness results, where the leading ideas involve the structural interplay between the properties of numbers, sequences, syntactical entities and finitary computations (plus, times, exponentation, and so on).\n\nLet our alphabet $\\Sigma = \\{A,B\\}$. Next we consider the set $\\Sigma^{\\ast}$ of finite sequences drawn from $\\Sigma$. Finite sequences drawn from an alphabet are usually called,\n\u2022 strings\n\u2022 words\n\u2022 expressions\nThese are the syntactical entities that one is discussing, quantifying over, referring to, etc. The crucial point is that these are sequences from the alphabet. In particular, $\\Sigma^{\\ast}$ is closed under sequence concatenation. So,\nif $\\alpha, \\beta \\in \\Sigma^{\\ast}$, then $\\alpha ^{\\frown} \\beta \\in \\Sigma^{\\ast}$.\nAnd:\n$|\\Sigma^{\\ast}| = \\aleph_0$.\nThis means that there are $\\aleph_0$-many syntactical entities. The terms $a_0, a_2, \\dots, a_n$ occurring in a sequence $\\alpha = (a_0, a_1, \\dots, a_n)$ may well be concreta. But the sequence $\\alpha$ itself is a (possibly\u00a0mixed) abstractum. More exactly, a sequence is usually understood as a function:\n$\\alpha : \\{0,1,\\dots,n\\} \\to \\Sigma$.\nThis is not mandated. What is mandated is that sequences are individuated in a certain way:\n$\\alpha = \\beta$ if and only if $\\alpha$ and $\\beta$ have the same terms, in the same order.\nSo, e.g,, if $(a_0, a_1, \\dots, a_n) = (b_0, b_1, \\dots, b_k)$, then $n = k$, and $a_0 = b_0$,\u00a0$a_1 = b_1$,\u00a0and so on.\n\nNormally, one goes on to define certain special subsets $X, Y, \\dots$ of $\\Sigma^{\\ast}$. Perhaps these are the formulas, or terms, and whatnot. Usually, the definitions satisfy certain computational constraints: e.g., perhaps an inductive definition. So, $X$ might be, e.g., a recursive set or a recursively enumerable set. But for this discussion here, these subsets don't matter. They're subsets, and we discussing the enclosing set, of all strings from the alphabet.\n\nReturn to (i) and (ii). Suppose (i) is true. So, there are no abstracta. Hence, there are, a fortiori, no mixed abstracta; and therefore, there are no sequences; and, therefore, there are no strings; and therefore no syntactical entities, except a very, very small number of tokens, which are not closed under concatenation. Hence, (ii) is false.\n\nOne might suggest that these claims (i) and (ii) are \"really\" consistent under some reinterpretation $I$. But what exactly is this $I$? How is $I$ defined? Is it a secret?\n\nI think that the optimal nominalistic responses to the inconsistency of (i) and (ii) are:\n\u2022 either to accept the inconsistency and thus simply accept that (ii) (i.e., syntax) is false (see Quine & Goodman 1947, \"Steps Toward a Constructive Nominalism\"),\n\u2022 or to reinterpret (ii,) to make it \"true under a reinterpretation\", so that \"syntactical entity\" refers perhaps to possibilia (i.e., possible concrete tokens: see Burgess & Rosen 1997, A Subject with No Object, for some discussion of this) or perhaps to some kind of physical entity (such as perhaps spacetime regions), assuming there are sufficiently many.\nI'm not optimistic about either kind of approach. The first approach is an error theory, and is too damaging to science. An error theory for morality is one thing; an error theory for science is another! The second, \"hermeneutic\", approach invokes possibilia and this raises similar sceptical and metaphysical worries as abstracta do. (See the final chapter of Shapiro 1997, Philosophy of Mathematics: Structure and Ontology.) It also raises the question of what grounds one might give for the reinterpretation. A classic discussion of some of these topics is Burgess 1983, \"Why I am not a Nominalist\".\n\nSo far as I can tell, the more recent \"weaseling\" approach to nominalism---which I think is extremely interesting---proposed by Melia 2000 (\"Weaseling Away the Indispensability Argument\", Mind) and endorsed and developed recently by Yablo 2012 (\"Explanation, Extrapolation and Existence\", Mind) doesn't seem to apply in the syntactic case. But I'm not sure.\n\n#### 1 comment:\n\n1. I'm skeptical of your reasoning.\n\nWhen we study a formal language, or try to define it, then we are doing mathematics and what we are defining is a mathematical idealization of language. So, of course, the entities in this idealization are mathematical entities and count as abstracta.\n\nThe question of the existence of abstracta, including those you discuss, still seems entirely distinct from the question of the existence of syntactic entities - the actual syntactic entities that are used, rather than their mathematical idealizations.","date":"2018-02-19 23:53:35","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7639774680137634, \"perplexity\": 1020.7976253517711}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-09\/segments\/1518891812855.47\/warc\/CC-MAIN-20180219231024-20180220011024-00007.warc.gz\"}"} | null | null |
The return of Sales Bootcamp helps Hoosiers break into tech
TechPoint's Sales Bootcamp has done it again. The popular career prep program has helped a new class of recent graduates and career changers find their place in the challenging and lucrative business of tech sales. Having gone on hiatus in early 2020 due to the COVID pandemic, Sales Bootcamp made its return last month with […]
By Joe Perin March 24, 2022
Indy tech community comes together to further social good
This summer, 12 Indianapolis technology groups joined forces for the Third Annual Indy Tech Gives, a social fundraising campaign to raise donations for central Indiana charities. Participants compete as individuals and teams, raising funds for the benefiting charity selected by their company. This approach to corporate social responsibility unites employees around a common goal and […]
By Jenna Jameson August 19, 2021
OneCause wins Pandemic Pivot of the Year Mira Award for small to medium sized businesses
OneCause won the award for Pandemic Pivot of the Year for small- to mid-sized businesses during TechPoint's 22nd annual Mira Awards gala honoring the best of tech in Indiana. On March 13th, 2020, OneCause went from supporting 200 plus in-person events a week, with 2,100 total on the books for spring … to zero. The […]
By Joshua Hall April 22, 2021
Meet the Pandemic Pivot of the Year nominees (small-medium businesses) for the 22nd annual Mira Awards
The Pandemic Pivot of the Year (Small-Medium Businesses) award recognizes tech employers who responded in remarkable ways to the COVID-19 pandemic. As vaccines roll out and we collectively navigate towards this "new normal" everyone keeps talking about, we want to stop and take a moment to reflect on how far we've come. Striving and surviving […]
By TechPoint Staff April 20, 2021
Meet the Tech Product of the Year nominees for the 22nd annual Mira Awards
Recognizing ground-breaking and disruptive new tech products coming from innovative organizations in Indiana In the past year, Indy tech gurus of all shapes and sizes have combined their talents and resources to solve everyday problems people face—especially in light of the ongoing COVID-19 pandemic. In a time of mass disruption, our 2021 nominees set out […]
Meet the Innovation of the Year nominees for the 22nd annual Mira Awards
Honoring scientific achievements, R&D efforts, and other trailblazing discoveries that have the potential to change industries The term "innovator" refers to a person or entity that introduces new methods, ideas, or products into the world. True innovation doesn't go with the flow or by the book; it challenges the status quo and pushes boundaries in […]
Philanthropy During A Pandemic: Kicking-off the Year-end Giving Season
Year-end giving during a pandemic takes on a whole new meaning in this virtual world. GivingTuesday has always been one of the largest single charitable giving days of the year. In a year of great need and uncertainty, 2020 was like no other. The philanthropic community and its supporters rallied for a record breaking day, […]
By Steve Johns December 8, 2020
Supporting the Virtual Fundraising Pivot
The current pandemic has had wide reaching effects on all types of businesses, including nonprofit organizations. Despite the real threat posed by COVID-19, the need to support these important missions and philanthropic organizations remains unchanged. Now more than ever communities are relying on local nonprofits, yet many are experiencing new challenges to fund their causes. […]
By Steve Johns May 14, 2020 | {
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} | 2,128 |
\section{Introduction}
Radical-ion pairs have been at the focus of chemical,
biological and physical research for the last few decades, mostly
because of their relevance to photosynthesis \cite{boxer_photo}.
It is a chain of electron-transfer reactions \cite{boxer_rev_1}
that converts the photon energy absorbed by pigment molecules in
the photosynthetic reaction center to energy capable of driving
further chemical reactions of biological importance. The
radical-ion pairs formed in this intricate photochemical machinery
further complicate the picture, since the spin interactions of
their unpaired electrons make the whole process
magnetic-sensitive. The importance of the magnetic sensitivity of
photosynthetic reactions \cite{hab1,schulten_weller,boxer_rev_2}
was realized early on and intensely studied theoretically as well
as experimentally. The timing of photosynthetic reactions
apparently rests on the complicated interplay between
non-radiative electron-transfer rates, spin-state mixing rates and
spin-selective recombination rates, to name a few of the
participating processes. It is therefore not surprising that
theories and experimental techniques from all of the
aforementioned disciplines have been summoned to address this
problem.
Parallel to these developments, it was early on realized
\cite{schulten1,schulten2} that the magnetic sensitivity of radical-ion pair (RIP)
recombination reactions might be at the basis of the biochemical
magnetic compass used by several species for navigation in earth's
magnetic field \cite{w1}. After the development of extensive theoretical
models \cite{hore1,ritz1,weaver,solovyov} and several supporting experiments \cite{mouritsen,ritz2,maeda},
the so-called RIP magnetoreception mechanism is now established as forming the
basis of the biochemical compass, at least of migratory avian
species.
The magnetic sensitivity of RIP reactions \cite{brock} can be
easily understood: the RIP is formed from a photoexcited
donor-acceptor dyad molecule (DA) in the singlet spin state.
Magnetic interactions with the external magnetic field and
hyperfine interactions with the molecule's nuclear spins induce
the so-called singlet-triplet (S-T) mixing. Since the
recombination of the RIP is spin-selective, i.e. the singlet
$^{\rm S}$(D$^{+}$A$^{-}$) RIP can recombine to the neutral DA
molecule, whereas the triplet $^{\rm T}$(D$^{+}$A$^{-}$)
recombines, if at all, to a triplet metastable excited state
$^{\rm T}$DA, it is seen that the magnetic field can directly
influence the end-result of the photoexcitation-recombination
cycle.
What has gone unnoticed so far, however, is the fact that the
spin-selective charge recombination process is fundamentally a
quantum measurement \cite{kom_prl}. The tunneling of the RIP into
a vibrational excited state of the DA molecule is a scattering
process the outcome of which depends on the RIP'a spin state. In
other words, the intermolecular recombination dynamics
continuously interrogate the radical-ion pair's spin state. This
interrogation can be readily described by quantum measurement
theory \cite{braginsky} and leads to purely quantum phenomena such
as the quantum Zeno effect \cite{misra}. The theoretical
description of magnetic-sensitive RIP recombination processes was
so far based on phenomenological classical reaction theory
\cite{steiner} which masked the quantum effects that are dominant
in the regime of low-magnetic fields. This is the reason that
classical theories were rather successful, i.e. most measurements
were performed at high magnetic fields, where classical reaction
theories are valid. Several puzzling experimental observations at
low fields were however left unexplained. As will be shown in the
following, the quantum Zeno effect crucially affects the relevant
time constants of the RIP recombination process, and naturally
leads to an understanding of several experimental findings that
could not be reconciled with classical RIP reaction theories. In
this article we will present the detailed quantum theory of the
RIP recombination process and elucidate the fundamentally
different interpretation of the low-magnetic-field effects that
emerges. We will outline the connection of the full quantum theory
to be presented with the classical reaction theories. We will also
provide an explanation of experimental findings that have
challenged the classical theory, in particular measurements
regarding the effect of deuteration on the magnetic-sensitivity of
RIP recombination yields.
\section{The quantum description of RIP recombination}
In Figure 1 we diagrammatically describe the evolution of the RIP
recombination. The RIP is at time $t=0$ created in the singlet
state $^{\rm S}$(D$^{+}$A$^{-}$), which coherently mixes with the
triplet $^{\rm T}$(D$^{+}$A$^{-}$) under the unitary action of the
magnetic hamiltonian ${\cal H}$. At the same time, the charge
recombination is performing a quantum measurement of the singlet-
and triplet-state projection operators $Q_{S}$ and $Q_{T}$, with
the measurement rate being $k_{S}$ and $k_{T}$, respectively. At
some point the measurement is over. If the measurement result is
$q_{S}=1$ (obviously meaning that $q_{T}=0$), the RIP can tunnel
to an excited vibrational state of the neutral DA molecule, $^{\rm
S}$(DA)$^{*}$, and from there decay to the DA ground state at a
rate $\gamma_{S}$. In the event that the measurement produces the
result $q_{S}=0$ (obviously meaning that $q_{T}=1$), the RIP
similarly ends up at a metastable triplet state of the DA
molecule, $^{\rm T}$DA, and this happens at a rate $\gamma_{T}$.
In this work we will not deal with the decay rates $\gamma_{S}$
and $\gamma_{T}$, since they not are relevant for the magnetic
dynamics.
\begin{figure}
\centering
\includegraphics[width=8 cm]{quantum_recombination_scheme.eps}
\caption{Quantum measurement scheme of the radical-ion pair charge recombination process. The coherent evolution of the RIP induced by
the magnetic Hamiltonian is interrupted by the measurements performed by the charge-recombination process. At some random time the measurement
is completed and a definite measurement result is available, so that the recombination can proceed accordingly.}
\label{scheme}
\end{figure}
Quantum measurement theory \cite{braginsky} describes the effect
of a continuous measurement of a system observable $q$ performed
at a rate $k$ on the system's time evolution by the Liouville
equation: \begin{equation} \frac{d\rho}{dt}=-i[{\cal H},\rho]-k[q,[q,\rho]]
\end{equation} where $\rho$ is the quantum system's density matrix and the
first term describes the unitary evolution due to a Hamiltonian
interaction ${\cal H}$. In the case of the RIP recombination
process, the measured system observables are $Q_{S}$ and $Q_{T}$.
Therefore, the RIP's density matrix evolution equation will be
\begin{equation} \frac{d\rho}{dt}=-i[{\cal
H},\rho]-k_{S}[Q_{S},[Q_{S},\rho]]-k_{T}[Q_{T},[Q_{T},\rho]]\label{ev1}
\end{equation} Measurements performed on a coherently evolving quantum
system always cause decoherence. That is, \eqref{ev1} describes
the evolution of an initially pure state of the RIP to a mixed
state, during which the coherence induced by the Hamiltonian
${\cal H}$ is also dissipated. This is due to the
measurement-induced back-action. The RIP magnetic Hamiltonian to
be considered is comprised of the Zeeman interaction of the two
unpaired electrons with the external magnetic field and by the
hyperfine interactions with the molecule's nuclear magnetic
moments. We will consider for simplicity only two nuclei, with
nuclear spins $I_{1}$ and $I_{2}$, and different hyperfine
couplings (in frequency units) $A_{1}$ and $A_{2}$ to each of the
electrons. We will consider isotropic hyperfine interactions, as
we are not interested in describing directional effects
\cite{hore3} in this work. The magnetic Hamiltonian thus reads
\begin{equation} {\cal H}=h_{1}+h_{2} \end{equation} where \begin{equation} h_{j}=\omega
s_{jz}+a_{j}\mathbf{I}_{j}\cdot\mathbf{s}_{j}\label{ham12} \end{equation} is
the interaction Hamiltonian for each electron and $j=1,2$. The
Larmor frequency is \begin{equation} \omega=g_{s}\mu_{B}B=2.8 {\rm {{MHz}\over
{G}}}B[{\rm G}] \end{equation} and throughout we are using units such that
$\hbar=1$. For earth's field of $\sim$ 0.5 G, $\omega=0.7~{\rm
{\mu s^{-1}}}$. In the following, $\omega$ will be the unit of
frequency, i.e. we will normalize all rates by $\omega$. The
normalization is performed by $\omega$ and not by the hyperfine
frequency scale $a_{1}$ (or $a_2$) for the following reason. If
the mechanism is to serve for the measurement of the magnetic
field $\omega$, with precision $\delta\omega$, then from
Heisenberg's time-energy uncertainty relation follows that in a
measurement time $\tau$, the achievable precision $\delta\omega$
is such that $\tau\delta\omega\sim 1$. The precision
$\delta\omega$ is understood to be the width of the measured
distribution of frequencies $\omega$. Since the measurement time
$\tau$ is determined by the decay rate $\lambda$ of the RIP S-T
mixing, $\tau\sim 1/\lambda$, the relative width of the measured
distribution is $\delta\omega/\omega\sim \lambda/\omega$.
Therefore Heisenberg's uncertainty poses the requirement that
$\lambda/\omega<1$, if the mechanism is to be sensitive to the
magnetic field itself. If however, the signal-to-noise of the
actual detection process is $s$, than the measurement sensitivity
is enhanced to $\lambda/s\omega$. For example, if the
magnetic-field change is detected by the change in the recombined
molecules yield and the latter can be measured at the level of
1\%, then $\delta\omega/\omega\sim 0.01$ if $\lambda/\omega\sim
1$. It is obvious that the condition $\lambda/\omega<1$ can be
more easily satisfied at high fields, since the classical
recombination rates $k$ (and hence the decay rates $\lambda$) are
usually much larger than 1 ${\rm \mu s^{-1}}$. In the following,
we are going to show how the quantum Zeno effect allows the
existence of long S-T lifetimes, so that the condition
$\lambda/\omega\sim 1$ is satisfied even if $k\gg 1~{\rm \mu
s^{-1}}$.
We will now derive some basic properties of \eqref{ev1}. We denote
by $S=\rm Tr\{\rho Q_{S}\}$ and $T=\rm Tr\{\rho Q_{T}\}$ the probability
to find the RIP in the singlet and triplet spin state,
respectively. Due to the fact that $Q_{S}$ and $Q_{T}$ are
orthogonal projection operators, we have $Q_{S}^{2}=Q_{S}$,
$Q_{T}^{2}=Q_{T}$, $Q_{S}Q_{T}=0$ and $Q_{S}+Q_{T}=1$. Therefore,
using the identity $\rm Tr\{[A,B]\}=0$, it follows from \eqref{ev1}
that
\begin{align}
\frac{dS}{d\tau}=-i\rm Tr\{[Q_{S},{\cal H}]\rho\}\\
\frac{dT}{d\tau}=-i\rm Tr\{[Q_{T},{\cal H}]\rho\}
\end{align}
and \begin{equation} \rm Tr\{\rho\}=S+T=1 \end{equation} Does the fact that $\rm Tr\{\rho\}=1$
mean that the RIP never recombines? Not at all. Every time the
RIP's spin state approaches the point $S=1$ (or $T=1$), there is a
non-zero probability of singlet- or triplet channel charge
recombination. If the RIP recombines, it ceases to exist in the
Hilbert space defined by the eigenstates of ${\cal H}$, and
together with is ceases the meaning of the density matrix $\rho$
and its normalization. This is formally described by a quantum
stochastic process \cite{gisin, wiseman} taking place in an
enlarged Hilbert space. A brief analysis has appeared in
\cite{kom_prl} and a more detailed presentation of the RIP's spin
state time evolution will be presented in a forthcoming
manuscript. Such a description requires the complete solution of
the density matric equation \eqref{ev1}. In the following we will
present some general arguments that do not necessitate such a
solution.
\section{Eigenvalue spectrum of the Liouville equation}
Since the nuclear spin multiplicity of the problem we are
considering is $(2I_{1}+1)(2I_{2}+1)$, and the electron spin
multiplicity is 4, the density matrix is a square $N$-dimensional
matrix, where $N=4(2I_{1}+1)(2I_{2}+1)$. Equation \eqref{ev1} thus
represents a set of $N^2$ coupled first-order differential
equations. We can rewrite them in the form $dR/dt=AR$, where
$R=(\rho_{11}~\rho_{12}~...~\rho_{NN})^{T}$ is a column vector of
dimension $N^2$ containing all density matrix elements and $A$ is
a $N^2$-dimensional square matrix. The eigenvalues of $A$ are
playing a central role in the considerations of this work, since
they are governing the time evolution of the RIP's spin state.
These eigenvalues are found by numerically diagonalizing the
matrix $A$, and in general are complex numbers of the form
$-\lambda+i\Omega$, where $\lambda\geq 0$ is termed the decay rate
and $\Omega$ the mixing frequency of this particular eigen-mode.
There are $n=N^2$ such eigenmodes. For example, if $I_{1}=1/2$ and
$I_{2}=1$, then $n=576$. It is easily seen that $n$ grows
exponentially with the number of participating spins, and the
numerical diagonalization quickly becomes a formidable problem.
This is the reason we limit the calculations to the case of just
two nuclear spins.
\begin{figure}
\centering
\includegraphics[width=8 cm]{eigenvalues1_jpg.eps}
\caption{The decay rates and mixing frequencies of \eqref{ev1} are shown as a function of $k_{S}/\omega$, for $k_{T}=0.2k_{S}$,
$I_{1}=I_{2}=1/2$, and $a_{2}=2a_{1}=3\omega$.}
\label{fig:eig}
\end{figure}
A result of such a diagonalization is shown in Figure 2 for
typical values of the problem's parameters. We assume that
$k_{T}<k_{S}$, so that the recombination rate scale is determined
by $k_{S}$. We will distinguish two physically distinct regimes:
(i) $k_{S}/\omega\ll 1$, and (ii) $k_{S}/\omega\gg 1$. We term the
first regime classical, since it is in this case that classical
reaction theories are valid, whereas the second is termed quantum,
because it is in this parameter regime that the quantum Zeno
effect is at play. Before examining the magnetic sensitivity in
these two regimes, we note the following. Since the time evolution
of the density matrix is described by $n$ eigenmodes, every
density matrix element can be written as \begin{equation}
\rho_{ij}(t)=\sum_{l=1}^{n}{c_{ij}^{(l)}e^{(-\lambda_{l}+i\Omega_{l})t}},
\end{equation} where the coefficients $c_{ij}^{(l)}$ are determined from the
initial conditions. Therefore \begin{equation} \langle
Q_{S}(t)\rangle=\sum_{ij}{\rho_{ij}(t)q_{ji}}=\sum_{l=1}^{n}{e^{(-\lambda_{l}+i\Omega_{l})t}}A_{l}\label{QSt}
\end{equation} where $\rho_{ij}=\langle i|\rho|j\rangle$ and $|i\rangle$
with $i=1,...,n$ are the basis states, and \begin{equation}
A_{l}=\sum_{ij}{c_{ij}^{(l)}q_{ji}} \end{equation} with $q_{ji}=\langle
j|Q_{S}|i\rangle$. The reality of $\langle Q_{S}(t)\rangle$ is
based on the appearance of positive and negative mixing
frequencies, as shown in Figure 2B.
\subsection{Classical Regime}
We here study the magnetic sensitivity of RIP reactions when
$k_{S}/\omega<1$, i.e. in the regime of small recombination rates
and/or large magnetic fields. This is termed the classical regime.
We first note that in this regime the decay rates $\lambda$ scale
proportionally to $k_{S}$, $\lambda\sim k_{S}$. Secondly, the
mixing frequencies are constant, i.e. independent of the
recombination rate $k_{S}$. Hence all $n$ terms contribute to
$\langle Q_{S}(t)\rangle$, i.e. $n_{\rm slow}=n$ in \eqref{QStsl}.
In this case, the magnetic sensitivity of $\langle
Q_{S}(t)\rangle$ stems from the dependence of $q_{ji}$ on
$\omega$, i.e. a change in $\omega$ will result in a change of
$A_{l}$. As can be seen in Figure 3B, the matrix elements $q_{ij}$
and hence $A_{l}$ are varying strongly for fields $\omega/a>1$
until the very high field regime $\omega/a\gg 1$, where the
magnetic sensitivity drops to zero. In this high-field regime, the
${\rm T}_{\pm}$ branches of the energy eigenvalues have separated
away and there is small mixing only between ${\rm T}_{0}$ and S,
as can be seen in Figure 3A.
\begin{figure}
\centering
\includegraphics[width=8 cm]{energies_and_qij_jpg.eps}
\caption{(A) Energy eigenvalues of the magnetic Hamiltonian for nuclear spins $I_{1}=I_{2}=1/2$ and hyperfine couplings
$a_{1}=a$ and $a_{2}=2a$.(B) The corresponding matrix elements $q_{ij}$.
In this case the $Q_{S}$-matrix is 16-dimensional, so there are
256 matrix elements $q_{ij}$ of $Q_{S}$ in the basis formed by
the eigenvectors of ${\cal H}$. It has been checked that the
eigenvalues of the $Q_{S}$-matrix in this representation are 0
and 1. (C) Zoom in at low magnetic fields, in order to better display the change of $q_{ij}$ in this regime.}
\label{energies}
\end{figure}
The above considerations form the classical interpretation of
magnetic field effects on RIP recombination reactions, that have
already been analyzed in detail \cite{hore1,ritz1} (and references
therein).
\subsection{Quantum Regime}
The situation changes dramatically as the ratio $k_{S}/\omega$
increases. Indeed, it is seen from Figure 3C that for low magnetic
fields, $\omega/a\ll 1$, a 10\% change in $\omega$ produces a mere
2\% change in $q_{ij}$. According to classical theory, this is
further suppressed by the ratio $\omega/k_{S}$, resulting in a
negligible change of $\langle Q_{S}(t)\rangle$. However,
significant recombination yield changes are indeed observed
experimentally.
\begin{figure}
\centering
\includegraphics[width=8 cm]{eigenvalues2_jpg.eps}
\caption{(A) Decay rates only the lower branch is shown) as a function of $\omega$ for $k_{S}=15$, $k_{T}=0.2k_{S}$, nuclear spins
$I_{1}=I_{2}=1/2$ and hyperfine couplings
$a_{1}=1.5$ and $a_{2}=3$. The dashed red line is the $\lambda=\omega$ line. The magnetic sensitivity derives from the changing number
of decay rates crossing this line.}
\label{eig2}
\end{figure}
The reason is that in this regime quantum effects become dominant,
fundamentally altering the nature of the magnetic sensitivity of
the recombination process. As seen in Figure 2A, it is in this
regime that the decay rates split into two branches, the fast
(upper branch) and slow (lower branch) decay rates. The latter are
a manifestation of the quantum Zeno effect \cite{misra}, the
physical significance of has been already described elsewhere
\cite{kom_bj}. We will here elaborate on the new interpretation of
magnetic sensitivity of RIP reactions in this quantum regime. Let
us suppose that some of the decay rates $\lambda_{l}$ are slow,
i.e. $\lambda_{l}\ll\Omega_{l}$, and some other are fast, i.e.
$\lambda_{l}\gg\Omega_{l}$. Let us further suppose that there are
$n_{\rm slow}$ of the former and hence $n_{\rm fast}=n-n_{\rm
slow}$ of the latter. Thus $n_{\rm fast}$ terms in \eqref{QSt}
will quickly decay away, leaving the $n_{\rm slow}$ terms to
determine the long-time behavior of $\langle Q_{S}(t)\rangle$:
\begin{equation} \langle Q_{S}(t)\rangle_{t\gg1/\lambda_{\rm
fast}}=\sum_{l=1}^{n_{\rm
slow}}{e^{(-\lambda_{l}+i\Omega_{l})t}}A_{l}\label{QStsl} \end{equation} It
thus follows that in this regime it is mainly the change of the
number $n_{\rm slow}$ with the magnetic field $\omega$ that is
responsible for the magnetic sensitivity of $\langle
Q_{S}(t)\rangle$. This number is a strongly changing function of
$\omega$ as shown in Figure 4.
\section{Phenomenological Liouville Equations}
The phenomenological density matrix evolution equations used so
far are \begin{equation} \frac{d\rho}{dt}=-i[{\cal
H},\rho]-k_{S}(Q_{S}\rho+\rho Q_{S})-k_{T}(Q_{T}\rho+\rho
Q_{T})\label{classL} \end{equation} The decay rates of the eigenvalue
spectrum of \eqref{classL} are shown in Figure 5. It is seen that
all the decay rates scale in proportion to $k_{S}$, i.e. the
quantum Zeno effect has disappeared. This is expected, because
equation \eqref{classL} by design forces the normalization of the
density matrix, $\rm Tr\{\rho\}$, to an exponential decay: \begin{equation}
\frac{d\rm Tr\{\rho\}}{dt}=-k_{S}\langle Q_{S}\rangle-k_{T}\langle
Q_{T}\rangle\label{trace}\end{equation} There is an analogous situation in
atomic physics experiments: transit time broadening \cite{budker}.
When an atomic coherence (for example a ground state spin
coherence) of an atomic vapor is probed by a laser, the atoms stay
in the interaction volume (defined by the laser beam) for a
limited time $\tau_{tr}$, determined by the atomic velocity
distribution. The measured spin coherence lifetime will thus be
limited by $\tau_{tr}$, which will add to the spin-resonance width
by $1/\tau_{tr}$. This is the transit-time broadening, which would
be described by an equation similar to \eqref{trace}: \begin{equation}
\frac{d\rm Tr\{\rho\}}{dt}=-\frac{\rm Tr\{\rho\}}{\tau_{tr}}\end{equation}
\begin{figure}
\centering
\includegraphics[width=8 cm]{classical_decay_rates_jpg.eps}
\caption{Decay rates of classical reaction theory Liouville equation \eqref{classL}. The calculation was done for $k_{T}=0.2k_{S}$,
$I_{1}=I_{2}=1/2$, $a_{1}=1.5\omega$ and $a_{2}=3\omega$.}
\label{classical}
\end{figure}
This equation phenomenologically describes the disappearance of
atoms with a time constant $\tau_{\rm tr}$. The situation with
radical-ion pairs, however, is fundamentally different, since in
this case there is a selective quantum measurement continuously
going on. An attempt to artificially introduce the crucial term
$Q_{S}\rho Q_{S}$ existing in \eqref{ev1} and missing from
\eqref{classL} was made in \cite{hab2} based on phenomenological
considerations, without any further consequences on the
theoretical description of RIP recombination dynamics. All
theoretical treatments, summarized in \cite{steiner}, have
thereafter used the phenomenological density matrix equation
\eqref{classL}.
\section{Explanation of Experimental Observations Inconsistent with Classical Reaction Theory}
We will now apply the interpretation of the low-field magnetic
sensitivity of RIP recombination reaction previously put forward
to address experimental observations that seem paradoxical within
the framework of classical reaction theory. Unusually long time
constants that have been observed in RIP reactions several times
have been explained elsewhere \cite{kom_prl}. Recently observed
\cite{maeda} magnetic sensitivity of RIP reaction rates which is
in stark contrast with classical reaction theory has been shown
\cite{kom_nat} to be fully consistent with the quantum theory.
Several experimental observations relevant to avian
magnetoreception based on magnetic-sensitive RIP reactions have
also been accounted for \cite{kom_bj}. Here we focus on some other
puzzling observations, namely the effect of the RIP's deuteration
on the magnetic sensitivity of the RIP recombination.
\subsection{Magnetic Sensitivity and Deuteration}
The effects of deuteration on the magnetic sensitivity of RIP
recombination reactions have early on \cite{blank} signalled the
inadequacies of classical reaction theories, as has been noted in
\cite{boxer_photo}. As stated in \cite{blank}, "The identical
triplet quantum yields in H and D samples does not support the
generally accepted idea that hyperfine interactions are
responsible for spin rephasing in the P870$^{+}$I$^{-}$ radical
pair". Indeed, it is rather straightforward to derive, based on
the classical spin-state mixing description, the result of a
change of hyperfine couplings. The triplet recombination yield is
crudely expected \cite{schulten_wolynes} to scale as $a/k_{S}$,
where $a$ is the dominant hyperfine coupling and $k_{S}$ the
singlet recombination rate. Any change in $a$ should thus lead to
a proportional change in the yield, contrary to the 1979
observations.
It is rather straightforward to explain this apparent
contradiction by use of the eigenvalue spectrum of the quantum
evolution equation \eqref{ev1}. Indeed, if we calculate the
eigenvalue spectrum for a molecule with $I_{1}=I_{2}=1/2$ and
hyperfine couplings $a_{1}=1.5\omega$ and $a_{2}=2.8a_{1}$, we
find that in the quantum regime ($k_{S}/\omega\gg 1$) the number
of decay rates $\lambda$ such that $\lambda/\omega<1$ are 58.2\%
of the total (256). If we repeat the calculation for $I_{1}=1/2$,
$I_{2}=1$, $a_{1}=1.5\omega$ and $a_{2}=0.86a_{1}$, i.e. we
replaced the second nucleus, which before was assumed to be a
hydrogen, with a deuterium, we find that the previous percentage
changes to 62.5\% (the total number of eigenvalues is now 576)
i.e. the relative yield change is only 7\%. However, the change in
the hyperfine coupling is about 300\%, clearly exemplifying the
disparity of classical expectations with observed data.
\begin{figure}
\centering
\includegraphics[width=8 cm]{Py_DMA_jpg.eps}
\caption{Magnetic sensitivity of the reaction yield of the Py-DMA radical-ion pair. What is calculated is the
drop in the singlet-recombination yield. The magnetic Hamiltonian contains two nuclear spins, with $I_{1}$ and $I_2$ being either 1/2 or 1, dependent
on which of the four RIPs is considered. The hyperfine couplings are taken from \cite{timmel_sc}. The singlet and triplet recombination rates used
were $k_{s}=50~{\rm \mu s}^{-1}$ and $k_{T}=0.2k_{S}$, respectively. Similar results where obtained for $k_{S}=10,20~{\rm \mu s}^{-1}$.
(A) The decay rates of Py-${\rm h}_{10}^{-}$/DMA-${\rm h}_{11}^{+}$ as a function of magnetic field. Similar plots are obtained for the other
radical-ion pairs. The magnetic sensitivity is found by counting the decay rates that are smaller than $\omega$.
(B) Estimated singlet-yield drop from the count of the slow decay rates for all four Py-DMA RIPs. It is seen that the protonated pyrene
pairs show no magnetic response.}
\label{pydma}
\end{figure}
We will now focus on more recent data regarding the effect
deuteration has on the magnetic sensitivity \cite{timmel_sc,hore2}
of the the pyrene (Py)-dimethylaniline (DMA) RIP recombination
dynamics. What has been observed is that the low-field-effect is
most pronounced when only one of the radicals exhibits strong
hyperfine couplings, with the second radical being deuterated and
hence having substantially reduced hyperfine couplings (the
magnetic moment of deuterium is about one third of the proton's).
In particular, in the case of protonated pyrene RIPs Py-${\rm
h}_{10}^{-}$/DMA-${\rm h}_{11}^{+}$ and Py-${\rm
h}_{10}^{-}$/DMA-${\rm d}_{11}^{+}$, there was almost no
magnetic-sensitivity of the exciplex fluorescence. On the
contrary, the recombination dynamics of deuterated pyrene RIPs
Py-${\rm d}_{10}^{-}$/DMA-${\rm h}_{11}^{+}$ and Py-${\rm
d}_{10}^{-}$/DMA-${\rm d}_{11}^{+}$ showed a clear low-field
sensitivity. In Figure 6 we show the calculated behavior of the
magnetic sensitivity. For each of the four RIPs we calculate the
eigenvalue spectrum and count the decay rates $\lambda$ that are
smaller than $\omega$. This spectrum is shown for one of the RIPs
in Figure 6A. Half the percentage of those decay rates is a good
estimate \cite{kom_bj} for the decrease in the singlet
recombination yield. It is seen that the relative yield change with
magnetic field displayed in Figure 6B is in perfect agreement with
observations. The absolute values of the yield drop
are not accurately predicted due to the simplicity of the two-spin
model used (they also depend on the particular value of $k_{S}$).
In summary, we have analyzed the physical interpretation of low
magnetic field effects that is bestowed upon the radical-ion pair
recombination reactions by their full quantum-mechanical
description. The explanatory power of this theory has been applied
in understanding experimental observations on the magnetic
sensitivity of RIP reactions that were conflicting classical
reaction theories. It has been shown that the eigenvalue spectrum
of the density matrix equation describing the time evolution of
the RIP's spin state accounts for the observations regarding RIP
deuteration. In particular, the slow decay rates imposed by the
quantum Zeno effect on the eigenvalue spectrum, and their
dependence on the problem's physical parameters, are shown to
contain rich dynamical information. Finally, it is rather crucial
to address the effect of low magnetic fields on photosynthetic
reactions under this new perspective.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,701 |
La è una stazione ferroviaria della città di Izumisano, nella prefettura di Osaka in Giappone. La stazione è gestita dalla JR West e serve la linea Hanwa ed è il punto d'origine della linea Kansai Aeroporto, sebbene tutti i treni proseguano comunque sulla Hanwa fino a Osaka e oltre.
Servizi
JR West
■ Linea Hanwa
■ Linea Kansai Aeroporto
Espresso limitato Kuroshio
Espresso limitato Haruka
Struttura
La stazione è costituita da due marciapiedi a isola con 4 binari in superficie.
Stazioni adiacenti
Altri progetti
Collegamenti esterni
Hineno
Hineno | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,119 |
## Contents
1. We Crash and Burn a Party
2. I Have a Word with Chaos
3. We Win a Box Full of Nothing
4. I Consult the Pigeon of War
5. A Dance with Death
6. Amos Plays with Action Figures
7. I Get Strangled by an Old Friend
8. My Sister, the Flowerpot
9. Zia Breaks Up a Lava Fight
10. 'Take Your Daughter to Work Day' Goes Horribly Wrong
11. Don't Worry, Be Hapi
12. Bulls with Freaking Laser Beams
13. A Friendly Game of Hide-and-Seek (with Bonus Points for Painful Death!)
14. Fun with Split Personalities
15. I Become a Purple Chimpanzee
16. Sadie Rides Shotgun (Worst. Idea. Ever.)
17. Brooklyn House Goes to War
18. Death Boy to the Rescue
19. Welcome to the Fun House of Evil
20. I Take a Chair
21. The Gods Are Sorted; My Feelings Are Not
22. The Last Waltz (for Now)
Glossary
Egyptian Gods and Goddesses Mentioned in _The Serpent's Shadow_
### _Books by Rick Riordan_
_The Percy Jackson series_
PERCY JACKSON AND THE LIGHTNING THIEF*
PERCY JACKSON AND THE SEA OF MONSTERS*
PERCY JACKSON AND THE TITAN'S CURSE*
PERCY JACKSON AND THE BATTLE OF THE LABYRINTH
PERCY JACKSON AND THE LAST OLYMPIAN
THE DEMIGOD FILES
CAMP HALF-BLOOD CONFIDENTIAL
PERCY JACKSON AND THE GREEK GODS
PERCY JACKSON AND THE GREEK HEROES
_The Heroes of Olympus series_
THE LOST HERO*
THE SON OF NEPTUNE*
THE MARK OF ATHENA
THE HOUSE OF HADES
THE BLOOD OF OLYMPUS
THE DEMIGOD DIARIES
_The Kane Chronicles series_
THE RED PYRAMID*
THE THRONE OF FIRE
THE SERPENT'S SHADOW
_The Percy Jackson and Kane Chronicles Adventures_
DEMIGODS AND MAGICIANS: THE SON OF SOBEK,
THE STAFF OF SERAPIS & THE CROWN OF PTOLEMY
_The Magnus Chase series_
MAGNUS CHASE AND THE SWORD OF SUMMER
MAGNUS CHASE AND THE HAMMER OF THOR
HOTEL VALHALLA: GUIDE TO THE NORSE WORLDS
_The Trials of Apollo series_
THE HIDDEN ORACLE
THE DARK PROPHECY
_* Also available as a graphic novel_
_To three great editors who shaped my writing career:_
_Kate Miciak, Jennifer Besser and Stephanie Lurie – the magicians who have brought my words to life_
## WARNING
_This is a transcript of an audio recording. Twice before, Carter and Sadie Kane have sent me such recordings, which I transcribed as_ The Red Pyramid _and_ The Throne of Fire. _While I'm honoured by the Kanes' continued trust, I must advise you that this third account is their most troubling yet. The tape arrived at my home in a charred box perforated with claw and teeth marks that my local zoologist could not identify. Had it not been for the protective hieroglyphs on the exterior, I doubt the box would have survived its journey. Read on and you will understand why._
SADIE
## 1. We Crash and Burn a Party
SADIE KANE HERE.
If you're listening to this, congratulations! You survived Doomsday.
I'd like to apologize straight away for any inconvenience the end of the world may have caused you. The earthquakes, rebellions, riots, tornadoes, floods, tsunamis and of course the giant snake who swallowed the sun – I'm afraid most of that was our fault. Carter and I decided we should at least explain how it happened.
This will probably be our last recording. By the time you've heard our story, the reason for that will be obvious.
Our problems started in Dallas, when the fire-breathing sheep destroyed the King Tut exhibit.
That night the Texas magicians were hosting a party in the sculpture garden across the street from the Dallas Museum of Art. The men wore tuxedos and cowboy boots. The women wore evening dresses and hairdos like explosions of candy floss.
(Carter says it's called cotton candy in America. I don't care. I was raised in London, so you'll just have to keep up and learn the proper way of saying things.)
A band played old-timey country music on the pavilion. Strings of fairy lights glimmered in the trees. Magicians did occasionally pop out of secret doors in the sculptures or summon sparks of fire to burn away pesky mosquitoes, but otherwise it seemed like quite a normal party.
The leader of the Fifty-first Nome, JD Grissom, was chatting with his guests and enjoying a plate of beef tacos when we pulled him away for an emergency meeting. I felt bad about that, but there wasn't much choice, considering the danger he was in.
'An attack?' He frowned. 'The Tut exhibit has been open for a month now. If Apophis was going to strike, wouldn't he have done it already?'
JD was tall and stout, with a rugged, weathered face, feathery red hair and hands as rough as bark. He looked about forty, but it's hard to tell with magicians. He might have been four hundred. He wore a black suit with a skinny bolo tie and a large silver Lone Star belt buckle, like a Wild West marshal.
'Let's talk on the way,' Carter said. He started leading us towards the opposite side of the garden.
I must admit my brother acted remarkably confident.
He was still a monumental dork, of course. His hair had a chunk missing on the left side where his griffin had given him a 'love bite', and you could tell from the nicks on his face that he hadn't quite mastered the art of shaving. But since his fifteenth birthday he'd shot up in height and put on muscle from hours of combat training. He looked poised and mature in his black linen clothes, especially with that _khopesh_ sword at his side. I could almost imagine him as a leader of men without laughing hysterically.
[Why are you glaring at me, Carter? That was quite a generous description.]
Carter manoeuvred round the buffet table, grabbing a handful of tortilla chips. 'Apophis has a pattern,' he told JD. 'The other attacks all happened on the night of the new moon, when darkness is greatest. Believe me, he'll hit your museum tonight. And he'll hit it hard.'
JD Grissom squeezed round a cluster of magicians drinking champagne. 'These other attacks...' he said. 'You mean Chicago and Mexico City?'
'And Toronto,' Carter said. 'And... a few others.'
I knew he didn't want to say more. The attacks we'd witnessed over the summer had left us both with nightmares.
True, full-out Armageddon hadn't come yet. It had been six months since the Chaos snake Apophis had escaped from his Underworld prison, but he still hadn't launched a large-scale invasion of the mortal world as we'd expected. For some reason, the serpent was biding his time, settling for smaller attacks on nomes that seemed secure and happy.
_Like this one_ , I thought.
As we passed the pavilion, the band finished their song. A pretty blonde woman with a fiddle waved her bow at JD.
'Come on, sweetie!' she called. 'We need you on steel guitar!'
He forced a smile. 'Soon, hon. I'll be back.'
We walked on. JD turned to us. 'My wife, Anne.'
'Is she also a magician?' I asked.
He nodded, his expression turning dark. 'These attacks. Why are you so sure Apophis will strike _here_?'
Carter's mouth was full of tortilla chips, so his response was, 'Mhm-hmm.'
'He's after a certain artefact,' I translated. 'He's already destroyed five copies of it. The last one in existence happens to be in your Tut exhibit.'
'Which artefact?' JD asked.
I hesitated. Before coming to Dallas, we'd cast all sorts of shielding spells and loaded up on protective amulets to prevent magical eavesdropping, but I was still nervous about speaking our plans aloud.
'Better we show you.' I stepped round a fountain, where two young magicians were tracing glowing _I Love You_ messages on the paving stones with their wands. 'We've brought our own crack team to help. They're waiting at the museum. If you'll let us examine the artefact, possibly take it with us for safekeeping –'
'Take it _with_ you?' JD scowled. 'The exhibit is heavily guarded. I have my best magicians surrounding it night and day. You think you can do better at Brooklyn House?'
We stopped at the edge of the garden. Across the street, a two-storey-tall King Tut banner hung from the side of the museum.
Carter took out his mobile phone. He showed JD Grissom an image on the screen – a burnt-out mansion that had once been the headquarters for the One Hundredth Nome in Toronto.
'I'm sure your guards are good,' Carter said. 'But we'd rather not make your nome a target for Apophis. In the other attacks like this one... the serpent's minions didn't leave any survivors.'
JD stared at the phone's screen, then glanced back at his wife, Anne, who was fiddling her way through a two-step.
'Fine,' JD said. 'I hope your team is top-notch.'
'They're amazing,' I promised. 'Come on, we'll introduce you.'
Our crack squad of magicians was busy raiding the gift shop.
Felix had summoned three penguins, which were waddling around wearing paper King Tut masks. Our baboon friend, Khufu, sat on top of a bookshelf reading _The History of the Pharaohs_ , which would've been quite impressive except he was holding the book upside down. Walt – oh, dear Walt, _why_? – had opened the jewellery cabinet and was examining charm bracelets and necklaces as if they might be magical. Alyssa levitated clay pots with her earth elemental magic, juggling twenty or thirty at a time in a figure of eight.
Carter cleared his throat.
Walt froze, his hands full of gold jewellery. Khufu scrambled down the bookshelf, knocking off most of the books. Alyssa's pottery crashed to the floor. Felix tried to shoo his penguins behind the till. (He does have rather strong feelings about the usefulness of penguins. I'm afraid I can't explain it.)
JD Grissom drummed his fingers against his Lone Star belt buckle. 'This is your amazing team?'
'Yes!' I tried for a winning smile. 'Sorry about the mess. I'll just, _um_...'
I pulled my wand from my belt and spoke a word of power: ' _Hi-nehm!_ '
I'd got better at such spells. Most of the time, I could now channel power from my patron goddess Isis without passing out. And I hadn't exploded once.
The hieroglyph for _Join together_ glowed briefly in the air:
Broken bits of pottery flew back together and mended themselves. Books returned to the shelf. The King Tut masks flew off the penguins, revealing them to be – gasp – penguins.
Our friends looked rather embarrassed.
'Sorry,' Walt mumbled, putting the jewellery back in the case. 'We got bored.'
I couldn't stay mad at Walt. He was tall and athletic, built like a basketball player, in tracksuit trousers and sleeveless tee that showed off his sculpted arms. His skin was the colour of hot cocoa, his face every bit as regal and handsome as the statues of his pharaoh ancestors.
Did I fancy him? Well, it's complicated. More on that later.
JD Grissom looked over our team.
'Nice to meet you all.' He managed to contain his enthusiasm. 'Come with me.'
The museum's main foyer was a vast white room with empty café tables, a stage and a ceiling high enough for a pet giraffe. On one side, stairs led up to a balcony with a row of offices. On the other side, glass walls looked out at the night-time skyline of Dallas.
JD pointed up at the balcony, where two men in black linen robes were patrolling. 'You see? Guards are everywhere.'
The men had their staffs and wands ready. They glanced down at us, and I noticed their eyes were glowing. Hieroglyphs were painted on their cheekbones like war paint.
Alyssa whispered to me: 'What's up with their eyes?'
'Surveillance magic,' I guessed. 'The symbols allow the guards to see into the Duat.'
Alyssa bit her lip. Since her patron was the earth god Geb, she liked solid things, such as stone and clay. She didn't like heights or deep water. She _definitely_ didn't like the idea of the Duat – the magical realm that coexisted with ours.
Once, when I'd described the Duat as an ocean under our feet with layers and layers of magical dimensions going down forever, I thought Alyssa was going to get seasick.
Ten-year-old Felix, on the other hand, had no such qualms. 'Cool!' he said. 'I want glowing eyes.'
He traced his finger across his cheeks, leaving shiny purple blobs in the shape of Antarctica.
Alyssa laughed. 'Can you see into the Duat now?'
'No,' he admitted. 'But I can see my penguins much better.'
'We should hurry,' Carter reminded us. 'Apophis usually strikes when the moon is at the top of its transit. Which is –'
' _Agh!_ ' Khufu held up all ten fingers. Leave it to a baboon to have perfect astronomical sense.
'In ten minutes,' I said. 'Just brilliant.'
We approached the entrance of the King Tut exhibit, which was rather hard to miss because of the giant golden sign that read KING TUT EXHIBIT. Two magicians stood guard with full-grown leopards on leashes.
Carter looked at JD in astonishment. 'How did you get complete access to the museum?'
The Texan shrugged. 'My wife, Anne, is president of the board. Now, which artefact did you want to see?'
'I studied your exhibit maps,' Carter said. 'Come on. I'll show you.'
The leopards seemed quite interested in Felix's penguins, but the guards held them back and let us pass.
Inside, the exhibit was extensive, but I doubt you care about the details. A labyrinth of rooms with sarcophagi, statues, furniture, bits of gold jewellery – blah, blah, blah. I would have passed it all by. I've seen enough Egyptian collections to last several lifetimes, thank you very much.
Besides, everywhere I looked, I saw reminders of bad experiences.
We passed cases of _shabti_ figurines, no doubt enchanted to come to life when called upon. I'd killed my share of those. We passed statues of glowering monsters and gods whom I'd fought in person – the vulture Nekhbet, who'd once possessed my gran (long story); the crocodile Sobek, who'd tried to kill my cat (longer story); and the lion goddess Sekhmet, whom we'd once vanquished with hot sauce (don't even ask).
Most upsetting of all: a small alabaster statue of our friend Bes, the dwarf god. The carving was aeons old, but I recognized that pug nose, the bushy sideburns, the potbelly and the endearingly ugly face that looked as if it had been hit repeatedly with a frying pan. We'd only known Bes for a few days, but he'd literally sacrificed his soul to help us. Now, each time I saw him I was reminded of a debt I could never repay.
I must have lingered at his statue longer than I realized. The rest of the group had passed me and were turning into the next room, about twenty metres ahead, when a voice next to me said, 'Psst!'
I looked around. I thought the statue of Bes might have spoken. Then the voice called again: 'Hey, doll. Listen up. Not much time.'
In the middle of the wall, eye-level with me, a man's face bulged from the white textured paint as if trying to break through. He had a beak of a nose, cruel thin lips and a high forehead. Though he was the same colour as the wall, he seemed very much alive. His blank eyes managed to convey a look of impatience.
'You won't save the scroll, doll,' he warned. 'Even if you did, you'd never understand it. You need my help.'
I'd experienced many strange things since I'd begun practising magic, so I wasn't particularly startled. Still, I knew better than to trust any old white-spackled apparition who spoke to me, especially one who called me _doll_. He reminded me of a character from those silly Mafia movies the boys at Brooklyn House liked to watch in their spare time – someone's Uncle Vinnie, perhaps.
'Who are you?' I demanded.
The man snorted. 'Like you don't know. Like there's _anybody_ who doesn't know. You've got two days until they put me down. You want to defeat Apophis, you'd better pull some strings and get me out of here.'
'I have no idea what you're talking about,' I said.
The man didn't sound like Set the god of evil, or the serpent Apophis, or any of the other villains I'd dealt with before, but one could never be sure. There was this thing called _magic_ , after all.
The man jutted out his chin. 'Okay, I get it. You want a show of faith. You'll never save the scroll, but go for the golden box. That'll give you a clue about what you need, if you're smart enough to understand it. Day after tomorrow at sunset, doll. Then my offer expires, 'cause that's when _I_ get permanently –'
He choked. His eyes widened. He strained as if a noose were tightening round his neck. He slowly melted back into the wall.
'Sadie?' Walt called from the end of the corridor. 'You okay?'
I looked over. 'Did you see that?'
'See what?' he asked.
_Of course not_ , I thought. What fun would it be if other people saw my vision of Uncle Vinnie? Then I couldn't wonder if I were going stark raving mad.
'Nothing,' I said, and I ran to catch up.
The entrance to the next room was flanked by two giant obsidian sphinxes with the bodies of lions and the heads of rams. Carter says that particular type of sphinx is called a _criosphinx_. [Thanks, Carter. We were all dying to know that bit of useless information.]
' _Agh!_ ' Khufu warned, holding up five fingers.
'Five minutes left,' Carter translated.
'Give me a moment,' JD said. 'This room has the heaviest protective spells. I'll need to modify them to let you through.'
' _Uh_ ,' I said nervously, 'but the spells will still keep out enemies, like giant Chaos snakes, I hope?'
JD gave me an exasperated look, which I tend to get a lot.
'I _do_ know a thing or two about protective magic,' he promised. 'Trust me.' He raised his wand and began to chant.
Carter pulled me aside. 'You okay?'
I must have looked shaken from my encounter with Uncle Vinnie. 'I'm fine,' I said. 'Saw something back there. Probably just one of Apophis's tricks, but...'
My eyes drifted to the other end of the corridor. Walt was staring at a golden throne in a glass case. He leaned forward with one hand on the glass as if he might be sick.
'Hold that thought,' I told Carter.
I moved to Walt's side. Light from the exhibit bathed his face, turning his features reddish brown like the hills of Egypt.
'What's wrong?' I asked.
'Tutankhamen died in that chair,' he said.
I read the display card. It didn't say anything about Tut dying in the chair, but Walt sounded very sure. Perhaps he could sense the family curse. King Tut was Walt's great-times-a-billion-uncle, and the same genetic poison that killed Tut at nineteen was now coursing through Walt's bloodstream, getting stronger the more he practised magic. Yet Walt refused to slow down. Looking at the throne of his ancestor, he must have felt as if he were reading his own obituary.
'We'll find a cure,' I promised. 'As soon as we deal with Apophis...'
He looked at me, and my voice faltered. We both knew our chances of defeating Apophis were slim. Even if we succeeded, there was no guarantee Walt would live long enough to enjoy the victory. Today was one of Walt's _good_ days, and still I could see the pain in his eyes.
'Guys,' Carter called. 'We're ready.'
The room beyond the criosphinxes was a 'greatest hits' collection from the Egyptian afterlife. A life-sized wooden Anubis stared down from his pedestal. On top of a replica of the scales of justice sat a golden baboon, which Khufu immediately started flirting with. There were masks of pharaohs, maps of the Underworld and loads of canopic jars that had once been filled with mummy organs.
Carter passed all that by. He gathered us round a long papyrus scroll in a glass case on the back wall.
'This is what you're after?' JD frowned. 'The Book of Overcoming Apophis? You do realize that even the best spells against Apophis aren't very effective.'
Carter reached in his pocket and produced a bit of burnt papyrus. 'This is all we could salvage from Toronto. It was another copy of the same scroll.'
JD took the papyrus scrap. It was no bigger than a postcard and too charred to let us make out more than a few hieroglyphs.
' "Overcoming Apophis..." ' he read. 'But this is one of the most common magic scrolls. Hundreds of copies have survived from ancient times.'
'No.' I fought the urge to look over my shoulder, in case any giant serpents were listening in. 'Apophis is after only one particular version, written by this chap.'
I tapped the information plaque next to the display. ' "Attributed to Prince Khaemwaset," ' I read, ' "better known as Setne." '
JD scowled. 'That's an evil name... one of most villainous magicians who ever lived.'
'So we've heard,' I said, 'and Apophis is destroying only _Setne's_ version of the scroll. As far as we can tell, only six copies existed. Apophis has already burned five. This is the last one.'
JD studied the burnt papyrus scrap doubtfully. 'If Apophis has truly risen from the Duat with all his power, why would he care about a few scrolls? No spell could possibly stop him. Why hasn't he already destroyed the world?'
We'd been asking ourselves the same question for months.
'Apophis is afraid of this scroll,' I said, hoping I was right. 'Something in it must hold the secret to defeating him. He wants to make sure all copies are destroyed before he invades the world.'
'Sadie, we need to hurry,' Carter said. 'The attack could come any minute.'
I stepped closer to the scroll. It was roughly two metres long and a half-metre tall, with dense lines of hieroglyphs and colourful illustrations. I'd seen loads of scrolls like this describing ways to defeat Chaos, with chants designed to keep the serpent Apophis from devouring the sun god Ra on his nightly journey through the Duat. Ancient Egyptians had been quite obsessed with this subject. Cheery bunch, those Egyptians.
I could read the hieroglyphs – one of my many amazing talents – but the scroll was a lot to take in. At first glance, nothing struck me as particularly helpful. There were the usual descriptions of the River of Night, down which Ra's sun boat travelled. Been there, thanks. There were tips on how to handle the various demons of the Duat. Met them. Killed them. Got the T-shirt.
'Sadie?' Carter asked. 'Anything?'
'Don't know yet,' I grumbled. 'Give me a moment.'
I found it annoying that my bookish brother was the combat magician, while _I_ was expected to be the great reader of magic. I barely had the patience for magazines, much less musty scrolls.
_You'd never understand it_ , the face in the wall had warned. _You need my help._
'We'll have to take it with us,' I decided. 'I'm sure I can figure it out with a little more –'
The building shook. Khufu shrieked and leaped into the arms of the golden baboon. Felix's penguins waddled around frantically.
'That sounded like –' JD Grissom blanched. 'An explosion outside. The party!'
'It's a diversion,' Carter warned. 'Apophis is trying to draw our defences away from the scroll.'
'They're attacking my friends,' JD said in a strangled voice. 'My wife.'
'Go!' I said. I glared at my brother. 'We can handle the scroll. JD's _wife_ is in danger!'
JD clasped my hands. 'Take the scroll. Good luck.'
He ran from the room.
I turned back to the display. 'Walt, can you open the case? We need to get this out of here as fast –'
Evil laughter filled the room. A dry, heavy voice, deep as a nuclear blast, echoed all around us: ' _I don't think so, Sadie Kane._ '
My skin felt as if it were turning to brittle papyrus. I remembered that voice. I remembered how it felt being so close to Chaos, as if my blood were turning to fire, and the strands of my DNA were unravelling.
' _I think I'll destroy you with the guardians of Ma'at_ ,' Apophis said. ' _Yes, that will be amusing._ '
At the entrance to the room, the two obsidian criosphinxes turned. They blocked the exit, standing shoulder to shoulder. Flames curled from their nostrils.
In the voice of Apophis, they spoke in unison: 'No one leaves this place alive. Goodbye, Sadie Kane.'
SADIE
## 2. I Have a Word with Chaos
WOULD YOU BE SURPRISED TO LEARN that things went badly from there?
I didn't think so.
Our first casualties were Felix's penguins. The criosphinxes blew fire at the unfortunate birds, and they melted into puddles of water.
'No!' Felix cried.
The room rumbled, much stronger this time.
Khufu screamed and jumped on Carter's head, knocking him to the floor. Under different circumstances that would've been funny, but I realized Khufu had just saved my brother's life.
Where Carter had been standing, the floor dissolved, marble tiles crumbling as if broken apart by an invisible jackhammer. The area of disruption snaked across the room, destroying everything in its path, sucking artefacts into the ground and chewing them to bits. Yes... _snaked_ was the right word. The destruction slithered exactly like a serpent, heading straight for the back wall and the Book of Overcoming Apophis.
'Scroll!' I shouted.
No one seemed to hear. Carter was still on the floor, trying to prise Khufu off his head. Felix knelt in shock at the puddles of his penguins, while Walt and Alyssa tried to pull him away from the fiery criosphinxes.
I slipped my wand from my belt and shouted the first word of power that came to mind: ' _Drowah!_ '
Golden hieroglyphs – the command for _Boundary_ – blazed in the air. A wall of light flashed between the display case and the advancing line of destruction:
I'd often used this spell to separate quarrelling initiates or to protect the snack cupboard from late-night nom-nom raids, but I'd never tried it for something so important.
As soon as the invisible jackhammer reached my shield, the spell began to fall apart. The disturbance spread up the wall of light, shaking it to pieces. I tried to concentrate, but a much more powerful force – Chaos itself – was working against me, invading my mind and scattering my magic.
In a panic, I realized I couldn't let go. I was locked in a battle I couldn't win. Apophis was shredding my thoughts as easily as he'd shredded the floor.
Walt knocked the wand out of my hands.
Darkness washed over me. I slumped into Walt's arms. When my vision cleared, my hands were burnt and steaming. I was too shocked to feel the pain. The Book of Overcoming Apophis was gone. Nothing remained except a pile of rubble and a massive hole in the wall, as if a tank had smashed through.
Despair threatened to close up my throat, but my friends gathered round me. Walt held me steady. Carter drew his sword. Khufu showed his fangs and barked at the criosphinxes. Alyssa wrapped her arms round Felix, who was sobbing into her sleeve. He had quickly lost his courage when his penguins were taken away.
'So that's it?' I shouted at the criosphinxes. 'Burn up the scroll and run away as usual? Are you so afraid to show yourself in person?'
More laughter rolled through the room. The criosphinxes stood unmoving in the doorway, but figurines and jewellery rattled in the display cases. With a painful creaking sound, the golden baboon statue that Khufu had been chatting up suddenly turned its head.
' _But I am everywhere._ ' The serpent spoke through the statue's mouth. 'I can destroy anything you value... and anyone you value.'
Khufu wailed in outrage. He launched himself at the baboon and knocked it over. It melted into a steaming pool of gold.
A different statue came to life – a gilded wooden pharaoh with a hunting spear. Its eyes turned the colour of blood. Its carved mouth twisted into a smile. 'Your magic is weak, Sadie Kane. Human civilization has grown old and rotten. I will swallow the sun god and plunge your world into darkness. The Sea of Chaos will consume you all.'
As if the energy were too much for it to contain, the pharaoh statue burst. Its pedestal disintegrated, and another line of evil jackhammer magic snaked across the room, churning up the floor tiles. It headed for a display against the east wall – a small golden cabinet.
_Save it_ , said a voice inside me – possibly my subconscious, or possibly the voice of Isis, my patron goddess. We'd shared thoughts so many times it was hard to be sure.
I remembered what the face in the wall had told me... _Go for the golden box. That'll give a clue about what you need._
'The box!' I yelped. 'Stop him!'
My friends stared at me. From somewhere outside, another explosion shook the building. Chunks of plaster rained from the ceiling.
'Are these children the best you could send against me?' Apophis spoke from an ivory _shabti_ in the nearest case – a miniature sailor on a toy boat. 'Walt Stone... you are the luckiest. Even if you survive tonight, your sickness will kill you before my great victory. You won't have to watch your world destroyed.'
Walt staggered. Suddenly I was supporting him. My burnt hands hurt so badly, I had to fight down a surge of nausea.
The line of destruction trundled across the floor, still heading for the golden cabinet. Alyssa thrust out her staff and barked a command.
For a moment, the floor stabilized, smoothing into a solid sheet of grey stone. Then new cracks appeared, and the force of Chaos blasted its way through.
'Brave Alyssa,' the serpent said, 'the earth you love will dissolve into Chaos. You will have no place to stand!'
Alyssa's staff burst into flames. She screamed and threw it aside.
'Stop it!' Felix yelled. He smashed the glass case with his staff and demolished the miniature sailor along with a dozen other _shabti_.
Apophis's voice simply moved to a jade amulet of Isis on a nearby mannequin. 'Ah, little Felix, I find you amusing. Perhaps I'll keep you as a pet, like those ridiculous birds you love. I wonder how long you'll last before your sanity crumbles.'
Felix threw his wand and knocked over the mannequin.
The crumbling trail of Chaos was now halfway to the golden cabinet.
'He's after that box!' I managed to say. 'Save the box!'
Granted, it wasn't the most inspiring call to battle, but Carter seemed to understand. He jumped in front of the advancing Chaos, stabbing his sword into the floor. His blade cut through the marble tile like ice cream. A blue line of magic extended to either side – Carter's own version of a force field. The line of disruption slammed against the barrier and stalled.
' _Poor Carter Kane._ ' The serpent's voice was all around us now – jumping from artefact to artefact, each one bursting from the power of Chaos. ' _Your leadership is doomed. Everything you tried to build will crumble. You will lose the ones you love the most._ '
Carter's blue defensive line began to flicker. If I didn't help him quickly...
'Apophis!' I yelled. 'Why wait to destroy me? Do it now, you overgrown rat snake!'
A hiss echoed through the room. Perhaps I should mention that one of my many talents is making people angry. Apparently it worked on snakes, too.
The floor settled. Carter released his shielding spell and almost collapsed. Khufu, bless his baboon wits, leaped to the golden cabinet, picked it up and bounded off with it.
When Apophis spoke again, his voice hardened with anger. ' _Very well, Sadie Kane. It's time to die._ '
The two ram-headed sphinxes stirred, their mouths glowing with flames. Then they lunged straight at me.
Fortunately one of them slipped in a puddle of penguin water and skidded off to the left. The other would've ripped my throat out had it not been tackled by a timely camel.
Yes, an actual full-sized camel. If you find that confusing, just think how the criosphinx must have felt.
Where did the camel come from, you ask? I may have mentioned Walt's collection of amulets. Two of them summoned disgusting camels. I'd met them before, so I was less than excited when a ton of dromedary flesh flew across my line of sight, ploughed into the sphinx and collapsed on top of it. The sphinx growled in outrage as it tried to free itself. The camel grunted and farted.
'Hindenburg,' I said. Only one camel could possibly fart that badly. 'Walt, why in the world –?'
'Sorry!' he yelled. 'Wrong amulet!'
The technique worked, at any rate. The camel wasn't much of a fighter, but it was quite heavy and clumsy. The criosphinx snarled and clawed at the floor, trying unsuccessfully to push the camel off, but Hindenburg just splayed his legs, made alarmed honking sounds and kept farting.
I moved to Walt's side and tried to get my bearings.
The room was quite literally in chaos. Tendrils of red lightning arced between exhibits. The floor was crumbling. The walls cracked. Artefacts were coming to life and attacking my friends.
Carter fended off the other criosphinx, stabbing it with his _khopesh_ , but the monster parried his strikes with its horns and breathed fire.
Felix was surrounded by a tornado of canopic jars that pummelled him from every direction as he swatted them with his staff. An army of tiny _shabti_ had surrounded Alyssa, who was chanting desperately, using her earth magic to keep the room in one piece. The statue of Anubis chased Khufu around the room, smashing things with its fists as our brave baboon cradled the golden cabinet.
All around us, the power of Chaos grew. I felt it in my ears like a coming storm. The presence of Apophis was shaking apart the entire museum.
How could I help all my friends at once, protect that gold cabinet _and_ keep the museum from collapsing on top of us?
'Sadie,' Walt prompted. 'What's the plan?'
The first criosphinx finally pushed Hindenburg off its back. It turned and blew fire at the camel, which let loose one final fart and shrank back into a harmless gold amulet. Then the criosphinx turned towards me. It did not look pleased.
'Walt,' I said, 'guard me.'
'Sure.' He eyed the criosphinx uncertainly. 'While you do what?'
_Good question_ , I thought.
'We have to protect that cabinet,' I said. 'It's some sort of clue. We have to restore Ma'at, or this building will implode and we'll all die.'
'How do we restore Ma'at?'
Instead of answering, I concentrated. I lowered my vision into the Duat.
It's hard to describe what it's like to experience the world on many levels at once – it's a bit like looking through 3D glasses and seeing hazy colourful auras around things, except the auras don't always match the objects, and the images are constantly shifting. Magicians have to be careful when they look into the Duat. Best-case scenario, you'll get mildly nauseous. Worst-case scenario, your brain will explode.
In the Duat, the room was filled with the writhing coils of a giant red snake – the magic of Apophis slowly expanding and encircling my friends. I almost lost my concentration along with my dinner.
_Isis_ , I called. _A little help?_
The goddess's strength surged through me. I stretched out my senses and saw my brother battling the criosphinx. Standing in Carter's place was the warrior god Horus, his sword blazing with light.
Swirling around Felix, the canopic jars were the hearts of evil spirits – shadowy figures that clawed and snapped at our young friend, though Felix had a surprisingly powerful aura in the Duat. His vivid purple glow seemed to keep the spirits at bay.
Alyssa was surrounded by a dust storm in the shape of a giant man. As she chanted, Geb the earth god lifted his arms and held up the ceiling. The _shabti_ army surrounding her blazed like a wildfire.
Khufu looked no different in the Duat, but as he leaped around the room evading the Anubis statue, the golden cabinet he was carrying flapped open. Inside was pure darkness – as if it were full of octopus ink.
I wasn't sure what that meant, but then I looked at Walt and gasped.
In the Duat, he was shrouded in flickering grey linen – mummy cloth. His flesh was transparent. His bones were luminous, as if he were a living X-ray.
_His curse_ , I thought. _He's marked for death._
Even worse: the criosphinx facing him was the centre of the Chaos storm. Tendrils of red lightning arced from its body. Its ram face changed into the head of Apophis, with yellow serpentine eyes and dripping fangs.
It lunged at Walt, but, before it could strike, Walt threw an amulet. Golden chains exploded in the monster's face, wrapping round its snout. The criosphinx stumbled and thrashed like a dog in a muzzle.
'Sadie, it's all right.' Walt's voice sounded deeper and more confident, as if he were older in the Duat. 'Speak your spell. Hurry.'
The criosphinx flexed its jaws. The gold chains groaned. The other criosphinx had backed Carter against a wall. Felix was on his knees, his purple aura failing in a swirl of dark spirits. Alyssa was losing her battle against the crumbling room as chunks of the ceiling fell around her. The Anubis statue grabbed Khufu's tail and held him upside down while the baboon howled and wrapped his arms round the gold cabinet.
Now or never: I had to restore order.
I channelled the power of Isis, drawing so deeply on my own magic reserves that I could feel my soul start to burn. I forced myself to focus, and I spoke the most powerful of all divine words: 'Ma'at.'
The hieroglyph burned in front of me – small and bright like a miniature sun:
'Good!' Walt said. 'Keep at it!' Somehow he'd managed to pull in the chains and grab the sphinx's snout. While the creature bore down on him with all its force, Walt's strange grey aura was spreading across the monster's body like an infection. The criosphinx hissed and writhed. I caught a whiff of decay like the air from a tomb – so strong that I almost lost my concentration.
'Sadie,' Walt urged, 'maintain the spell!'
I focused on the hieroglyph. I channelled all my energy into that symbol for order and creation. The word shone brighter. The coils of the serpent burned away like fog in sunlight. The two criosphinxes crumbled to dust. The canopic jars fell and shattered. The Anubis statue dropped Khufu on his head. The army of _shabti_ froze around Alyssa, and her earth magic spread through the room, sealing cracks and shoring up walls.
I felt Apophis retreating deeper into the Duat, hissing in anger.
Then I promptly collapsed.
'I told you she could do it,' said a kindly voice.
My mother's voice... but of course that was impossible. She was dead, which meant I spoke with her only occasionally, and only in the Underworld.
My vision returned, hazy and dim. Two women hovered over me. One was my mum – her blonde hair clipped back, her deep blue eyes sparkling with pride. She was transparent, as ghosts tend to be, but her voice was warm and very much alive. 'It isn't the end yet, Sadie. You must carry on.'
Next to her stood Isis in her white silky gown, her wings of rainbow light flickering behind her. Her hair was glossy black, woven with strands of diamonds. Her face was as beautiful as my mum's, but more queenly, less warm.
Don't misunderstand. I knew from sharing Isis's thoughts that she cared for me in her own way, but gods are not human. They have trouble thinking of us as more than useful tools or cute pets. To gods, a human lifespan doesn't seem much longer than that of the average gerbil.
'I would not have believed it,' Isis said. 'The last magician to summon Ma'at was Hatshepsut herself, and even she could only do it while wearing a fake beard.'
I had no idea what that meant. I decided I didn't want to know.
I tried to move but couldn't. I felt as if I were floating at the bottom of a bathtub, suspended in warm water, the two women's faces rippling at me from just above the surface.
'Sadie, listen carefully,' my mother said. 'Don't blame yourself for the deaths. When you make your plan, your father will object. You must convince him. Tell him it's the only way to save the souls of the dead. Tell him...' Her expression turned grim. 'Tell him it's the only way he'll see _me_ again. You _must_ succeed, my sweet.'
I wanted to ask what she meant, but I couldn't seem to speak.
Isis touched my forehead. Her fingers were as cold as snow. 'We must not tax her any further. Farewell for now, Sadie. The time rapidly approaches when we must join together again. You are strong. Even stronger than your mother. Together we will rule the world.'
'You mean, _Together we will defeat Apophis_ ,' my mother corrected.
'Of course,' Isis said. 'That's what I meant.'
Their faces blurred together. They spoke in a single voice: 'I love you.'
A blizzard swept across my eyes. My surroundings changed, and I was standing in a dark graveyard with Anubis. Not the musty old jackal-headed god as he appeared in Egyptian tomb art, but Anubis as I usually saw him – a teenaged boy with warm brown eyes, tousled black hair and a face that was ridiculously, annoyingly gorgeous. I mean, _please_ – being a god, he had an unfair advantage. He could look like anything he wanted. Why did he always have to appear in _this_ form that twisted my insides to pretzels?
'Wonderful,' I managed to say. 'If you're here, I must be dead.'
Anubis smiled. 'Not dead, though you came close. That was a risky move.'
A burning sensation started in my face and worked its way down my neck. I wasn't sure if it was embarrassment, anger or delight at seeing him.
'Where have you been?' I demanded. 'Six months, not a word.'
His smile melted. 'They wouldn't let me see you.'
'Who wouldn't let you?'
'There are rules,' he said. 'Even now they're watching, but you're close enough to death that I can manage a few moments. I need to tell you: you have the right idea. Look at what _isn't_ there. It's the only way you might survive.'
'Right,' I grumbled. 'Thanks for not speaking in riddles.'
The warm sensation reached my heart. It began to beat, and suddenly I realized I'd been _without_ a heartbeat since I'd passed out. That probably wasn't good.
'Sadie, there's something else.' Anubis's voice became watery. His image began to fade. 'I need to tell you –'
'Tell me in person,' I said. 'None of this "death vision" nonsense.'
'I can't. They won't let me.'
'You still sound like a little boy. You're a god, aren't you? You can do what you like.'
Anger smouldered in his eyes. Then, to my surprise, he laughed. 'I'd forgotten how irritating you are. I'll try to visit... _briefly_. We have something to discuss.' He reached out and brushed the side of my face. 'You're waking now. Goodbye, Sadie.'
'Don't leave.' I grasped his hand and held it against my cheek.
The warmth spread throughout my body. Anubis faded away.
My eyes flew open. 'Don't leave!'
My burnt hands were bandaged, and I was gripping a hairy baboon paw. Khufu looked down at me, rather confused. _'Agh?_ '
Oh, fab. I was flirting with a monkey.
I sat up groggily. Carter and our friends gathered round me. The room hadn't collapsed, but the entire King Tut exhibit was in ruins. I had a feeling we would not be invited to join the Friends of the Dallas Museum any time soon.
'Wh-what happened?' I stammered. 'How long –?'
'You were dead for two minutes,' Carter said, his voice shaky. 'I mean, _no heartbeat_ , Sadie. I thought... I was afraid...'
He choked up. Poor boy. He really would have been lost without me.
[Ouch, Carter! Don't pinch.]
'You summoned Ma'at,' Alyssa said in amazement. 'That's like... impossible.'
I suppose it was rather impressive. Using divine words to create an object like an animal or a chair or a sword – that's hard enough. Summoning an element like fire or water is even trickier. But summoning a concept, like Order – that's just not done. At the moment, however, I was in too much pain to appreciate my own amazingness. I felt as if I'd just summoned an anvil and dropped it on my head.
'Lucky try,' I said. 'What about the golden cabinet?'
' _Agh!_ ' Khufu gestured proudly to the gilded box, which sat nearby, safe and sound.
'Good baboon,' I said. 'Extra Cheerios for you tonight.'
Walt frowned. 'But the Book of Overcoming Apophis was destroyed. How will a cabinet help us? You said it was some kind of clue... ?'
I found it hard to look at Walt without feeling guilty. My heart had been torn between him and Anubis for months now, and it just wasn't fair of Anubis to pop into my dreams, looking all hot and immortal, when poor Walt was risking his life to protect me and getting weaker by the day. I remembered how he had looked in the Duat, in his ghostly grey mummy linen...
No. I couldn't think about that. I forced myself to concentrate on the golden cabinet.
_Look at what isn't there_ , Anubis had said. Stupid gods and their stupid riddles.
The face in the wall – Uncle Vinnie – had told me the box would give us a hint about how to defeat Apophis, _if_ I was smart enough to understand it.
'I'm not sure what it means yet,' I admitted. 'If the Texans let us take it back to Brooklyn House...'
A horrible realization settled over me. There were no more sounds of explosions outside. Just eerie silence.
'The Texans!' I yelped. 'What's happened to them?'
Felix and Alyssa bolted for the exit. Carter and Walt helped me to my feet, and we ran after them.
The guards had all disappeared from their stations. We reached the museum foyer, and I saw columns of white smoke outside the glass walls, rising from the sculpture garden.
'No,' I murmured. 'No, no.'
We tore across the street. The well-kept lawn was now a crater as big as an Olympic pool. The bottom was littered with melted metal sculptures and chunks of stone. Tunnels that had once led into the Fifty-first Nome's headquarters had collapsed like a giant anthill some bully had stepped on. Around the rim of the crater were bits of smoking evening wear, smashed plates of tacos, broken champagne glasses and the shattered staffs of magicians.
_Don't blame yourself for the deaths_ , my mother had said.
I moved in a daze to the remains of the patio. Half the concrete slab had cracked and slid into the crater. A charred fiddle lay in the mud next to a gleaming bit of silver.
Carter stood next to me. 'We – we should search,' he said. 'There might be survivors.'
I swallowed back a sob. I wasn't sure how, but I sensed the truth with absolute certainty. 'There aren't any.'
The Texas magicians had welcomed us and supported us. JD Grissom had shaken my hand and wished me luck before running off to save his wife. But we'd seen the work of Apophis in other nomes. Carter had warned JD: _The serpent's minions don't leave any survivors._
I knelt down and picked up the gleaming piece of silver – a half-melted Lone Star belt buckle.
'They're dead,' I said. 'All of them.'
CARTER
## 3. We Win a Box Full of Nothing
ON THAT HAPPY NOTE, Sadie hands me the microphone. [Thanks a lot, sis.]
I wish I could tell you that Sadie was wrong about the Fifty-first Nome. I'd love to say we found all the Texas magicians safe and sound. We didn't. We found nothing except the remnants of a battle: burnt ivory wands, a few shattered _shabti_ , scraps of smouldering linen and papyrus. Just like in the attacks on Toronto, Chicago and Mexico City, the magicians had simply vanished. They'd been vaporized, devoured or destroyed in some equally horrible way.
At the edge of the crater, one hieroglyph burned in the grass: _Isfet_ , the symbol for Chaos. I had a feeling Apophis had left it there as a calling card.
We were all in shock, but we didn't have time to mourn our comrades. The mortal authorities would be arriving soon to check out the scene. We had to repair the damage as best we could and remove all traces of magic.
There wasn't much we could do about the crater. The locals would just have to assume there'd been a gas explosion. (We tended to cause a lot of those.)
We tried to fix the museum and restore the King Tut collection, but it wasn't as easy as cleaning up the gift shop. Magic can only go so far. So if you go to a King Tut exhibit someday and notice cracks or burn marks on the artefacts, or maybe a statue with its head glued on backwards – well, sorry. That was probably our fault.
As police blocked the streets and cordoned off the blast zone, our team gathered on the museum roof. In better times we might have used an artefact to open a portal to take us back home, but over the last few months, as Apophis had got stronger, portals had become too risky to use.
Instead I whistled for our ride. Freak the griffin glided over from the top of the nearby Fairmont Hotel.
It's not easy finding a place to stash a griffin, especially when he's pulling a boat. You can't just parallel-park something like that and put a few coins in the meter. Besides, Freak tends to get nervous around strangers and swallow them, so I'd settled him on top of the Fairmont with a crate of frozen turkeys to keep him occupied. They have to be frozen. Otherwise he eats them too fast and gets hiccups.
[Sadie is telling me to hurry up with the story. She says you don't care about the feeding habits of griffins. Well, excuse me.]
Anyway, Freak came in for a landing on the museum roof. He was a beautiful monster, if you like psychotic falcon-headed lions. His fur was the colour of rust and, as he flew, his giant hummingbird wings sounded like a cross between chain saws and kazoos.
'FREEEEEK!' Freak cawed.
'Yeah, buddy,' I agreed. 'Let's get out of here.'
The boat trailing behind him was an Ancient Egyptian model – shaped like a big canoe made from bundles of papyrus reeds, enchanted by Walt so that it stayed airborne no matter how much weight it carried.
The first time we'd flown Air Freak, we'd strung the boat underneath Freak's belly, which hadn't been very stable. And you couldn't simply ride on his back, because those high-powered wings would chop you to shreds. So the sleigh-boat was our new solution. It worked great, except when Felix yelled down at the mortals, 'Ho, ho, ho, Merry Christmas!'
Of course, most mortals can't see magic clearly, so I'm not sure what they _thought_ they saw as we passed overhead. No doubt it caused many of them to adjust their medication.
We soared into the night sky – the six of us and a small cabinet. I still didn't understand Sadie's interest in the golden box, but I trusted her enough to believe it was important.
I glanced down at the wreckage of the sculpture garden. The smoking crater looked like a ragged mouth, screaming. Fire trucks and police cars had surrounded it with a perimeter of red and white lights. I wondered how many magicians had died in that explosion.
Freak picked up speed. My eyes stung, but it wasn't from the wind. I turned so my friends couldn't see.
_Your leadership is doomed._
Apophis would say anything to throw us into confusion and make us doubt our cause. Still, his words hit me hard.
I didn't like being a leader. I always had to appear confident for the sake of the others, even when I wasn't.
I missed having my dad to rely on. I missed Uncle Amos, who'd gone off to Cairo to run the House of Life. As for Sadie, my bossy sister, she always supported me, but she'd made it clear she didn't want to be an authority figure. Officially, _I_ was in charge of Brooklyn House. Officially, I called the shots. In my mind, that meant if we made mistakes, like getting an entire nome wiped off the face of the earth, then the fault was mine.
Okay, Sadie would never actually blame me for something like that, but that's how I felt.
_Everything you tried to build will crumble..._
It seemed incredible that not even a year had passed since Sadie and I had first arrived at Brooklyn House, completely clueless about our heritage and our powers. Now we were running the place – training an army of young magicians to fight Apophis using the path of the gods, a kind of magic that hadn't been practised in thousands of years. We'd made so much progress – but, judging from how our fight against Apophis had gone tonight, our efforts hadn't been enough.
_You will lose the ones you love the most..._
I'd already lost so many people. My mom had died when I was seven. My dad had sacrificed himself to become the host of Osiris last year. Over the summer, many of our allies had fallen to Apophis, or had been ambushed and 'disappeared' thanks to the rebel magicians who didn't accept my Uncle Amos as the new Chief Lector.
Who else could I lose... Sadie?
No, I'm not being sarcastic. Even though we'd grown up separately for most of our lives – me travelling around with Dad, Sadie living in London with Gran and Gramps – she was still my sister. We'd grown close over the last year. As annoying as she was, I needed her.
Wow, that's depressing.
[And there's the punch in the arm I was expecting. Ow.]
Or maybe Apophis meant someone else, like Zia Rashid...
Our boat rose above the glittering suburbs of Dallas. With a defiant squawk, Freak pulled us into the Duat. Fog swallowed the boat. The temperature dropped to freezing. I felt a familiar tingle in my stomach, as if we were plunging from the top of a roller coaster. Ghostly voices whispered in the mist.
Just when I started to think we were lost, my dizziness passed. The fog cleared. We were back on the East Coast, sailing over New York Harbor towards the night-time lights of the Brooklyn waterfront and home.
The headquarters of the Twenty-first Nome perched on the shoreline near the Williamsburg Bridge. Regular mortals wouldn't see anything but a huge dilapidated warehouse in the middle of an industrial yard, but to magicians Brooklyn House was as obvious as a lighthouse – a five-storey mansion of limestone blocks and steel-framed glass rising from the top of the warehouse, glowing with yellow and green lights.
Freak landed on the roof, where the cat goddess Bast was waiting for us.
'My kittens are alive!' She took my arms and looked me over for wounds, then did the same to Sadie. She tutted disapprovingly as she examined Sadie's bandaged hands.
Bast's luminous feline eyes were a little unsettling. Her long black hair was tied back in a plait, and her acrobatic bodysuit changed patterns as she moved – by turns tiger stripes, leopard spots or calico. As much as I loved and trusted her, she made me a little nervous when she did her 'mother cat' inspections. She kept knives up her sleeves – deadly iron blades that could slip into her hands with the flick of her wrists – and I was always afraid she might make a mistake, pat me on the cheek and end up decapitating me. At least she didn't try to pick us up by the scruffs of our necks or give us a bath.
'What happened?' she asked. 'Everyone is safe?'
Sadie took a shaky breath. 'Well...'
We told her about the destruction of the Texas nome.
Bast growled deep in her throat. Her hair poofed out, but the plait held it down so her scalp looked like a heated pan of Jiffy-Pop popcorn. 'I should've been there,' she said. 'I could have helped!'
'You couldn't,' I said. 'The museum was too well protected.'
Gods are almost never able to enter magicians' territory in their physical forms. Magicians have spent millennia developing enchanted wards to keep them out. We'd had enough trouble reworking the wards on Brooklyn House to give Bast access without opening ourselves up to attacks by less friendly gods.
Taking Bast to the Dallas Museum would've been like trying to get a bazooka through airport security – if not totally impossible then at least pretty darn slow and difficult. Besides, Bast was our last line of defence for Brooklyn House. We needed her to protect our home base and our initiates. Twice before, our enemies had almost destroyed the mansion. We didn't want there to be a third time.
Bast's bodysuit turned pure black, as it tended to do when she was moody. 'Still, I'd never forgive myself if you...' She glanced at our tired, frightened crew. 'Well, at least you're back safe. What's the next step?'
Walt stumbled. Alyssa and Felix caught him.
'I'm fine,' he insisted, though he clearly wasn't. 'Carter, I can get everyone together if you want. A meeting on the terrace?'
He looked like he was about to pass out. Walt would never admit it, but our main healer, Jaz, had told me that his level of pain was almost unbearable all the time now. He was only able to stay on his feet because she kept tattooing pain-relief hieroglyphs on his chest and giving him potions. In spite of that, I'd asked him to come to Dallas with us – another decision that weighed on my heart.
The rest of our crew needed sleep, too. Felix's eyes were puffy from crying. Alyssa looked like she was going into shock.
If we met now, I wouldn't know what to say. I had no plan. I couldn't stand in front of the whole nome without breaking down. Not after having caused so many deaths in Dallas.
I glanced at Sadie. We came to a silent agreement.
'We'll meet tomorrow,' I told the others. 'You guys get some sleep. What happened with the Texans...' My voice caught. 'Look, I know how you feel. I feel the same way. But it wasn't your fault.'
I'm not sure they bought it. Felix wiped a tear from his cheek. Alyssa put her arm round him and led him towards the stairwell. Walt gave Sadie a glance I couldn't interpret – maybe wistfulness or regret – then followed Alyssa downstairs.
' _Agh?_ ' Khufu patted the golden cabinet.
'Yeah,' I said. 'Could you take it to the library?'
That was the most secure room in the mansion. I didn't want to take any chances after all we'd sacrificed to save the box. Khufu waddled away with it.
Freak was so tired he didn't even make it to his covered roost. He just curled up where he was and started snoring, still attached to the boat. Travelling through the Duat takes a lot out of him.
I undid his harness and scratched his feathery head. 'Thanks, buddy. Dream of big fat turkeys.'
He cooed in his sleep.
I turned to Sadie and Bast. 'We need to talk.'
It was almost midnight, but the Great Room was still buzzing with activity. Julian, Paul and a few of the other guys were crashed on the couches, watching the sports channel. The ankle-biters (our three youngest trainees) were colouring pictures on the floor. Potato-chip bags and soda cans littered the coffee table. Shoes were tossed randomly across the snakeskin rug. In the middle of the room, the two-storey-tall statue of Thoth, the ibis-headed god of knowledge, loomed over our initiates with his scroll and quill. Somebody had put one of Amos's old pork-pie hats on the statue's head, so he looked like a bookie taking bets on the football game. One of the ankle-biters had coloured the god's obsidian toes pink and purple with crayons. We're big on respect here at Brooklyn House.
As Sadie and I came down the stairs, the guys on the couch got to their feet.
'How did it go?' Julian asked. 'Walt just came through, but he wouldn't say –'
'Our team is safe,' I said. 'The Fifty-first Nome... not so lucky.'
Julian winced. He knew better than to ask for details in front of the little kids. 'Did you find anything helpful?'
'We're not sure yet,' I admitted.
I wanted to leave it at that, but our youngest ankle-biter, Shelby, toddled over to show me her crayon masterpiece. 'I kill a snake,' she announced. 'Kill, kill, kill. Bad snake!'
She'd drawn a serpent with a bunch of knives sticking out of its back and Xs in its eyes. If Shelby had made that picture at school, it probably would've earned her a trip to the guidance counsellor, but here even the littlest ones understood something serious was happening.
She gave me a toothy grin, shaking her crayon like a spear. I stepped back. Shelby might've been a kindergartner, but she was already an excellent magician. Her crayons sometimes morphed into weapons, and the things she drew tended to peel off the page – like the red, white and blue unicorn she had summoned for the Fourth of July.
'Awesome picture, Shelby.' I felt like my heart was being wrapped tight in mummy linen. Like all the littlest kids, Shelby was here with her parents' consent. The parents understood that the fate of the world was at stake. They knew Brooklyn House was the best and safest place for Shelby to master her powers. Still, what kind of childhood was this for her, channelling magic that would destroy most adults, learning about monsters that would give anybody nightmares?
Julian ruffled Shelby's hair. 'Come on, sweetie. Draw me another picture, okay?'
Shelby said, 'Kill?'
Julian steered her away. Sadie, Bast and I headed to the library.
The heavy oaken doors opened to a staircase that descended into a huge cylindrical room like a well. Painted on the domed ceiling was Nut, the sky goddess, with silver constellations glittering on her dark blue body. The floor was a mosaic of her husband, Geb, the earth god, his body covered with rivers, hills and deserts.
Even though it was late, our self-appointed librarian, Cleo, still had her four _shabti_ statues at work. The clay men rushed around dusting shelves, rearranging scrolls and sorting books in the honeycombed compartments along the walls. Cleo herself sat at the worktable, jotting notes on a papyrus scroll while she talked to Khufu, who squatted on the table in front of her, patting our new antique cabinet and grunting in Baboon, like: _Hey, Cleo, wanna buy a gold box?_
Cleo wasn't much in the bravery department, but she had an incredible memory. She could speak six languages, including English, her native Portuguese (she was Brazilian), Ancient Egyptian and a few words of Baboon. She'd taken it upon herself to create a master index to all our scrolls and had been gathering more scrolls from all over the world to help us find information on Apophis. It was Cleo who'd found the connection between the serpent's recent attacks and the scrolls written by the legendary magician Setne.
She was a great help, though sometimes she got exasperated when she had to make room in _her_ library for our school texts, Internet stations, large artefacts and Bast's back issues of _Cat Fancy_ magazine.
When Cleo saw us coming down the stairs, she jumped to her feet. 'You're alive!'
'Don't sound so surprised,' Sadie muttered.
Cleo chewed her lip. 'Sorry, I just... I'm glad. Khufu came in alone, so I was worried. He was trying to tell me something about this gold box, but it's empty. Did you find the Book of Overcoming Apophis?'
'The scroll burned,' I said. 'We couldn't save it.'
Cleo looked like she might scream. 'But that was the last copy! How could Apophis destroy something so valuable?'
I wanted to remind Cleo that Apophis was out to destroy the entire world, but I knew she didn't like to think about that. It made her sick with fear.
Getting outraged about the scroll was more manageable for her. The idea that anybody could destroy a book of any kind made Cleo want to punch Apophis in the face.
One of the _shabti_ jumped onto the table. He tried to stick a scanner label on the golden cabinet, but Cleo shooed the clay man away.
'All of you, back to your places!' She clapped her hands, and the four _shabti_ returned to their pedestals. They reverted to solid clay, though one was still wearing rubber gloves and holding a feather duster, which looked a little odd.
Cleo leaned in and studied the gold box. 'There's nothing inside. Why did you bring it?'
'That's what Sadie, Bast and I need to discuss,' I said. 'If you don't mind, Cleo.'
'I don't mind.' Cleo kept examining the cabinet. Then she realized we were all staring at her. 'Oh... you mean privately. Of course.'
She looked a little upset about getting kicked out, but she took Khufu's hand. 'Come on, _babuinozinho_. We'll get you a snack.'
' _Agh!_ ' Khufu said happily. He adored Cleo, possibly because of her name. For reasons none of us quite understood, Khufu loved things that ended in - _o_ , like avocados, Oreos and armadillos.
Once Cleo and Khufu were gone, Sadie, Bast and I gathered round our new acquisition.
The cabinet was shaped like a miniature school locker. The exterior was gold, but it must've been a thin layer of foil covering wood, because the whole thing wasn't very heavy. The sides and top were engraved with hieroglyphs and pictures of the pharaoh and his wife. The front was fitted with latched double doors, which opened to reveal... well, not much of anything. There was a tiny pedestal marked by gold footprints, as if an Ancient Egyptian Barbie doll had once stood there.
Sadie studied the hieroglyphs along the sides of the box. 'It's all about Tut and his queen, wishing them a happy afterlife, blah, blah. There's a picture of him hunting ducks. Honestly? That was his idea of paradise?'
'I like ducks,' Bast said.
I moved the little doors back and forth on their hinges. 'Somehow I don't think the ducks are important. Whatever was inside here, it's gone now. Maybe grave robbers took it, or –'
Bast chuckled. 'Grave robbers took it. Sure.'
I frowned at her. 'What's so funny?'
She grinned at me, then Sadie, before apparently realizing we didn't get the joke. 'Oh... I see. You actually don't know what this is. I suppose that makes sense. Not many have survived.'
'Not many what?' I asked.
'Shadow boxes.'
Sadie wrinkled her nose. 'Isn't that a sort of school project? Did one for English once. Deadly boring.'
'I wouldn't know about school projects,' Bast said haughtily. 'That sounds suspiciously like _work_. But this is an _actual_ shadow box – a box to hold a shadow.'
Bast didn't sound like she was kidding, but it's hard to tell with cats.
'It's in there right now,' she insisted. 'Can't you see it? A little shadowy bit of Tut. Hello, shadow Tut!' She wriggled her fingers at the empty box. 'That's why I laughed when you said grave robbers might have stolen it. Ha! That would be a trick.'
I tried to wrap my mind around this idea. 'But... I've heard Dad lecture on, like, every possible Egyptian artefact. I never once heard him mention a shadow box.'
'As I told you,' Bast said, 'not many have survived. Usually the shadow box was buried far away from the rest of the soul. Tut was quite silly to have it placed in his tomb. Perhaps one of the priests put it there against his orders, out of spite.'
I was totally lost now. To my surprise, Sadie was nodding enthusiastically.
'That must've been what Anubis meant,' she said. ' _Pay attention to what's not there._ When I looked into the Duat, I saw darkness inside the box. And Uncle Vinnie said it was a clue to defeating Apophis.'
I made a 'Time out' T with my hands. 'Back up. Sadie, where did you see Anubis? And since when do we have an uncle named Vinnie?'
She looked a little embarrassed, but she described her encounter with the face in the wall, then the visions she'd had of our mom and Isis and her godly almost-boyfriend Anubis. I knew my sister's attention wandered a lot, but even _I_ was impressed by how many mystical side trips she'd managed, just walking through a museum.
'The face in the wall could've been a trick,' I said.
'Possibly... but I don't think so. The face said we would need his help, and we had only two days until something happened to him. He told me this box would show us what we needed. Anubis hinted I was on the right track, saving this cabinet. And Mum...' Sadie faltered. 'Mum said this was the only way we'd ever see her again. Something is happening to the spirits of the dead.'
Suddenly I felt like I was back in the Duat, wrapped in freezing fog. I stared at the box, but I still didn't see anything. 'How do shadows tie in to Apophis and spirits of the dead?'
I looked at Bast. She dug her fingernails into the table, using it like a scratching post, the way she does when she's tense. We go through a lot of tables.
'Bast?' Sadie asked gently.
'Apophis and shadows,' Bast mused. 'I'd never considered...' She shook her head. 'These are really questions you should ask Thoth. He's much more knowledgeable than I.'
A memory surfaced. My dad had given a lecture at a university somewhere... Munich, maybe? The students had asked him about the Egyptian concept of the soul, which had multiple parts, and my dad mentioned something about shadows.
_Like one hand with five fingers_ , he'd said. _One soul with five parts._
I held up my own fingers, trying to remember. 'Five parts of the soul... what are they?'
Bast stayed silent. She looked pretty uncomfortable.
'Carter?' Sadie asked. 'What does that have to do –?'
'Just humour me,' I said. 'The first part is the _ba_ , right? Our personality.'
'Chicken form,' Sadie said.
Trust Sadie to nickname part of your soul after poultry, but I knew what she meant. The _ba_ could leave the body when we dreamed, or it could come back to the earth as a ghost after we died. When it did, it appeared as a large glowing bird with a human head.
'Yeah,' I said. 'Chicken form. Then there's the _ka_ , the life force that leaves the body when it dies. Then there's the _ib_ , the heart –'
'The record of good and bad deeds,' Sadie agreed. 'That's the bit they weigh on the scales of justice in the afterlife.'
'And fourth...' I hesitated.
'The _ren_ ,' Sadie supplied. 'Your secret name.'
I was too embarrassed to look at her. Last spring she'd saved my life by speaking my secret name, which had basically given her access to my most private thoughts and darkest emotions. Since then she'd been pretty cool about it, but still... that's not the kind of leverage you want to give your little sister.
The _ren_ was also the part of the soul that our friend Bes had given up for us in our gambling match six months ago with the moon god Khonsu. Now Bes was a hollow shell of a god, sitting in a wheelchair in the Underworld's divine nursing home.
'Right,' I said. 'But the fifth part...' I looked at Bast. 'It's the shadow, isn't it?'
Sadie frowned. 'The shadow? How can a shadow be part of your soul? It's just a silhouette, isn't it? A trick of the light.'
Bast held her hand over the table. Her fingers cast a vague shadow over the wood. 'You can never be free of your shadow – your _sheut_. All living beings have them.'
'So do rocks, pencils and shoes,' Sadie said. 'Does that mean _they_ have souls?'
'You know better,' Bast chided. 'Living beings are different from rocks... well, _most_ are, anyway. The _sheut_ is not just a physical shadow. It's a magical projection – the silhouette of the soul.'
'So this box...' I said. 'When you say it holds King Tut's shadow –'
'I mean it holds one fifth of his soul,' Bast confirmed. 'It houses the pharaoh's _sheut_ so it will not be lost in the afterlife.'
My brain felt like it was about to explode. I knew this stuff about shadows must be important, but I didn't see how. It was like I'd been handed a puzzle piece, but it was for the wrong puzzle.
We'd failed to save the _right_ piece – an irreplaceable scroll that might've helped us beat Apophis – and we'd failed to save an entire nome full of friendly magicians. All we had to show from our trip was an empty cabinet decorated with pictures of ducks. I wanted to knock King Tut's shadow box across the room.
'Lost shadows,' I muttered. 'This sounds like that _Peter Pan_ story.'
Bast's eyes glowed like paper lanterns. 'What do you think _inspired_ the story of Peter Pan's lost shadow? There have been folktales about shadows for centuries, Carter – all handed down since the days of Egypt.'
'But how does that _help_ us?' I demanded. 'The Book of Overcoming Apophis would've helped us. Now it's gone!'
Okay, I sounded angry. I _was_ angry.
Remembering my dad's lectures made me want to be a kid again, travelling the world with him. We'd been through some weird stuff together, but I'd always felt safe and protected. He'd always known what to do. Now all I had left from those days was my suitcase, gathering dust in my closet upstairs.
It wasn't fair. But I knew what my dad would say about that: _Fair means everyone gets what they need. And the only way to get what you need is to make that happen yourself._
Great, Dad. I'm facing an impossible enemy, and what I _need_ in order to defeat him just got destroyed.
Sadie must've read my expression. 'Carter, we'll figure it out,' she promised. 'Bast, you were about to say something earlier about Apophis and shadows.'
'No, I wasn't,' Bast murmured.
'Why are you so nervous about this?' I asked. 'Do gods _have_ shadows? Does Apophis? If so, how do they work?'
Bast gouged some hieroglyphs in the table with her fingernails. I was pretty sure the message read: DANGER _._
'Honestly, children... this is a question for Thoth. Yes, gods have shadows. Of course we do. But – but it's not something we're supposed to talk about.'
I'd rarely seen Bast look so agitated. I wasn't sure why. This was a goddess who'd fought Apophis face to face, claw to fang, in a magical prison for thousands of years. Why was she scared of shadows?
'Bast,' I said, 'if we can't figure out a better solution, we'll have to go with Plan B.'
The goddess winced. Sadie stared dejectedly at the table. Plan B was something only Sadie, Bast, Walt and I had discussed. Our other initiates didn't know about it. We hadn't even told our Uncle Amos. It was _that_ scary.
'I – I would hate that,' Bast said. 'But, Carter, I really don't know the answers. And if you start asking about shadows you'll be delving into very dangerous –'
There was a knock on the library doors. Cleo and Khufu appeared at the top of the stairs.
'Sorry to disturb,' Cleo said. 'Carter, Khufu just came down from your room. He seems anxious to talk with you.'
' _Agh!_ ' Khufu insisted.
Bast translated from baboon-speak. 'He says there's a call for you on the scrying bowl, Carter. A _private_ call.'
As if I weren't stressed enough already. Only one person would be sending me a scrying vision, and if she were contacting me so late at night it had to be bad news.
'Meeting adjourned,' I told the others. 'See you in the morning.'
CARTER
## 4. I Consult the Pigeon of War
I WAS IN LOVE WITH A BIRDBATH.
Most guys checked their phone for texts, or obsessed over what girls were saying about them online. Me, I couldn't stay away from the scrying bowl.
It was just a bronze saucer on a stone pedestal, sitting on the balcony outside my bedroom. But whenever I was in my room I found myself stealing glances at it, resisting the urge to rush outside and check for a glimpse of Zia.
The weird thing was – I couldn't even call her my girlfriend. What do you call somebody when you fall in love with her replica _shabti_ , then rescue the real person only to find she doesn't share your feelings? And Sadie thinks _her_ relationships are complicated.
Over the past six months, since Zia had gone to help my uncle at the First Nome, the bowl had been our only contact. I'd spent so many hours staring into it, talking with Zia, I could hardly remember what she looked like without enchanted oil rippling across her face.
By the time I reached the balcony, I was out of breath. From the surface of the oil, Zia stared up at me. Her arms were crossed; her eyes so angry, they looked like they might ignite. (The first scrying bowl Walt had made actually _did_ ignite, but that's another story.)
'Carter,' she said, 'I'm going to strangle you.'
She was beautiful when she threatened to kill me. Over the summer she'd let her hair grow out so that it swept over her shoulders in a glossy black wave. She wasn't the _shabti_ I'd first fallen for, but her face still had a sculpted beauty – delicate nose, full red lips, dazzling amber eyes. Her skin glowed like terracotta warm from the kiln.
'You heard about Dallas,' I guessed. 'Zia, I'm sorry –'
'Carter, _everyone_ has heard about Dallas. Other nomes have been sending Amos _ba_ messengers for the past hour, demanding answers. Magicians as far away as Cuba felt ripples in the Duat. Some claimed you blew up half of Texas. Some said the entire Fifty-first Nome was destroyed. Some said – some said you were dead.'
The concern in her voice lifted my spirits a little, but it also made me feel guiltier.
'I wanted to tell you in advance,' I said. 'But by the time we realized Apophis's target was Dallas we had to move immediately.'
I told her what had happened at the King Tut exhibit, including our mistakes and casualties.
I tried to read Zia's expression. Even after so many months, it was hard to guess what she was thinking. Just _seeing_ her tended to short-circuit my brain. Half the time I could barely remember how to speak in complete sentences.
Finally she muttered something in Arabic – probably a curse.
'I'm glad you survived – but the Fifty-first destroyed... ?' She shook her head in disbelief. 'I knew Anne Grissom. She taught me healing magic when I was young.'
I remembered the pretty blonde lady who had played with the band, and the ruined fiddle at the edge of the explosion.
'They were good people,' I said.
'Some of our last allies,' Zia said. 'The rebels are already blaming you for their deaths. If any more nomes desert Amos...'
She didn't have to finish that thought. Last spring, the worst villains in the House of Life had formed a hit squad to destroy Brooklyn House. We'd defeated them. Amos had even given them amnesty when he became the new Chief Lector. But some refused to follow him. The rebels were still out there – gathering strength, turning other magicians against us. As if we needed more enemies.
'They're blaming me?' I asked. 'Did they contact you?'
'Worse. They broadcasted a message to you.'
The oil rippled. I saw a different face – Sarah Jacobi, leader of the rebels. She had milky skin, spiky black hair and dark, permanently startled eyes lined with too much kohl. In her pure white robes she looked like a Halloween ghoul.
She stood in a room lined with marble columns. Behind her glowered half a dozen magicians – Jacobi's elite killers. I recognized the blue robes and shaven head of Kwai, who'd been exiled from the North Korean nome for murdering a fellow magician. Next to him stood Petrovich, a scar-faced Ukrainian who'd once worked as an assassin for our old enemy Vlad Menshikov.
The others I couldn't identify, but I doubted that any of them was as bad as Sarah Jacobi herself. Until Menshikov had released her, she'd been exiled in Antarctica for causing an Indian Ocean tsunami that killed more than a quarter of a million people.
'Carter Kane!' she shouted.
Because this was a broadcast, I knew it was just a magical recording, but her voice made me jump.
'The House of Life demands your surrender,' she said. 'Your crimes are unforgivable. You must pay with your life.'
My stomach barely had time to drop before a series of violent images flashed across the oil. I saw the Rosetta Stone exploding in the British Museum – the incident that had unleashed Set and killed my father last Christmas. How had Jacobi got a visual of that? I saw the fight at Brooklyn House last spring, when Sadie and I had arrived in Ra's sun boat to drive out Jacobi's hit squad. The images she showed made it look like _we_ were the aggressors – a bunch of hooligans with godly powers beating up on poor Jacobi and her friends.
'You released Set and his brethren,' Jacobi narrated. 'You broke the most sacred rule of magic and cooperated with the gods. In doing so, you unbalanced Ma'at, causing the rise of Apophis.'
'That's a lie!' I said. 'Apophis was rising anyway!'
Then I remembered I was yelling at a video.
The scenes kept shifting. I saw a high-rise building on fire in the Shibuya district of Tokyo, headquarters of the 234th Nome. A flying demon with the head of a samurai sword crashed through a window and carried off a screaming magician.
I saw the home of the old Chief Lector, Michel Desjardins – a beautiful Paris townhouse on the Rue des Pyramides – now in ruins. The roof had collapsed. The windows were broken. Ripped scrolls and soggy books littered the dead garden, and the hieroglyph for Chaos smouldered on the front door like a cattle brand.
'All this you have caused,' Jacobi said. 'You have given the Chief Lector's mantle to a servant of evil. You have corrupted young magicians by teaching the path of the gods. You've weakened the House of Life and left us at the mercy of Apophis. We will not stand for this. Any who follow you will be punished.'
The vision changed to Sphinx House in London, headquarters for the British nome. Sadie and I had visited there over the summer and managed to make peace with them after hours of negotiations. I saw Kwai storming through the library, smashing statues of the gods and raking books off the shelves. A dozen British magicians stood in chains before their conqueror, Sarah Jacobi, who held a gleaming black knife. The leader of the nome, a harmless old guy named Sir Leicester, was forced to his knees. Sarah Jacobi raised her knife. The blade fell, and the scene shifted.
Jacobi's ghoulish face stared up at me from the surface of the oil. Her eyes were as dark as the sockets of a skull.
'The Kanes are a plague,' she said. 'You must be destroyed. Surrender yourself and your family for execution. We will spare your other followers as long as they renounce the path of the gods. I do not seek the office of Chief Lector, but I must take it for the good of Egypt. When the Kanes are dead, we will be strong and united again. We will undo the damage you've caused and send the gods and Apophis back to the Duat. Justice comes swiftly, Carter Kane. This will be your only warning.'
Sarah Jacobi's image dissolved in the oil, and I was alone again with Zia's reflection.
'Yeah,' I said shakily. 'For a mass murderer, she's pretty convincing.'
Zia nodded. 'Jacobi has already turned or defeated most of our allies in Europe and Asia. A lot of the recent attacks – against Paris, Tokyo, Madrid – those were Jacobi's work, but she's blaming them on Apophis – or Brooklyn House.'
'That's ridiculous.'
'You and I know that,' she agreed. 'But the magicians are scared. Jacobi is telling them that if the Kanes are destroyed Apophis will go back to the Duat and things will return to normal. They _want_ to believe it. She's telling them that following you is a death sentence. After the destruction of Dallas –'
'I get it,' I snapped.
It wasn't fair for me to get mad at Zia, but I felt so helpless. Everything we did seemed to turn out wrong. I imagined Apophis laughing in the Underworld. Maybe that's why he hadn't attacked the House of Life in full force yet. He was having too much fun watching us tear each other apart.
'Why didn't Jacobi direct her message at Amos?' I asked. 'He's the Chief Lector.'
Zia glanced away as if checking on something. I couldn't see much of her surroundings, but she didn't seem to be in her dorm room at the First Nome or in the Hall of Ages. 'Like Jacobi said, they consider Amos a servant of evil. They won't talk to him.'
'Because he was possessed by Set,' I guessed. 'That wasn't _his_ fault. He's been healed. He's fine.'
Zia winced.
'What?' I asked. 'He _is_ fine, isn't he?'
'Carter, it's – it's complicated. Look, the main problem is Jacobi. She's taken over Menshikov's old base in St Petersburg. It's almost as much of a fortress as the First Nome. We don't know what she's up to or how many magicians she has. We don't know when she'll strike or where. But she's going to attack soon.'
_Justice comes swiftly. This will be your only warning._
Something told me Jacobi wouldn't attack Brooklyn House again, not after she'd been humiliated last time. But if she wanted to take over the House of Life and destroy the Kanes what else could her target be?
I locked eyes with Zia, and I realized what she was thinking.
'No,' I said. 'They'd never attack the First Nome. That would be suicide. It's survived for five thousand years.'
'Carter... we're weaker than you realize. We were never fully staffed. Now many of our best magicians have disappeared, possibly gone over to the other side. We've got some old men and a few scared children left, plus Amos and me.' She spread her arms in exasperation. 'And half the time _I'm_ stuck here –'
'Wait,' I said. 'Where are you?'
Somewhere to Zia's left, a man's voice warbled, 'Hell-ooooo!'
Zia sighed. 'Great. He's up from his nap.'
An old man stuck his face in the scrying bowl. He grinned, showing exactly two teeth. His bald wrinkly head made him look like a geriatric baby. 'Zebras are here!'
He opened his mouth and tried to suck the oil out of the bowl, making the whole scene ripple.
'My lord, no!' Zia pulled him back. 'You can't drink the enchanted oil. We've talked about this. Here, have a cookie.'
'Cookies!' he squealed. 'Wheee!' The old man danced off with a tasty treat in his hands.
Zia's senile grandfather? Nope. That was Ra, god of the sun, first divine pharaoh of Egypt and arch-enemy of Apophis. Last spring we'd gone on a quest to find him and revive him from his twilight sleep, trusting he would rise in all his glory and fight the Chaos snake for us.
Instead, Ra woke up senile and demented. He was excellent at gumming biscuits, drooling and singing nonsense songs. Fighting Apophis? Not so much.
'You're babysitting _again_?' I asked.
Zia shrugged. 'It's after sunrise here. Horus and Isis watch him most nights on the sun boat. But during the day... well, Ra gets upset if I don't come to visit, and none of the other gods want to watch him. Honestly, Carter...' She lowered her voice. 'I'm afraid of what they'd do if I left Ra alone with them. They're getting tired of him.'
'Wheee!' Ra said in the background.
My heart sank. Yet another thing to feel guilty about: I'd saddled Zia with nanny duty for a sun god. Stuck in the throne room of the gods every day, helping Amos run the First Nome every night, Zia barely had time to sleep, much less go on a date – even if I could get up the courage to ask her.
Of course, that wouldn't matter if Apophis destroyed the world, or if Sarah Jacobi and her magical killers got to me. For a moment I wondered if Jacobi was right – if the world _had_ gone sideways because of the Kane family, and if it would be better off without us.
I felt so helpless I briefly considered calling on the power of Horus. I could've used some of the war god's courage and confidence. But I suspected that joining my thoughts with Horus's wouldn't be a good idea. My emotions were jumbled enough without another voice in my head, egging me on.
'I know that expression,' Zia chided. 'You can't blame yourself, Carter. If it weren't for you and Sadie, Apophis would have already destroyed the world. There's still hope.'
_Plan B_ , I thought. Unless we could figure out this mystery about shadows and how they could be used to fight Apophis, we'd be stuck with Plan B, which meant certain death for Sadie and me even if it worked. But I wasn't going tell Zia that. She didn't need any more depressing news.
'You're right,' I said. 'We'll figure out something.'
'I'll be back at the First Nome tonight. Call me then, okay? We should talk about –'
Something rumbled behind her, like a stone slab grinding across the floor.
'Sobek's here,' she whispered. 'I hate that guy. Talk later.'
'Wait, Zia,' I said. 'Talk about what?'
But the oil turned dark, and Zia was gone.
I needed to sleep. Instead, I paced my room.
The dorm rooms at Brooklyn House were amazing – comfortable beds, HD TVs, high-speed wireless Internet and magically restocking mini-fridges. An army of enchanted brooms, mops and dusters kept everything tidy. The closets were always full of clean, perfectly fitting clothes.
Still, my room felt like a cage. Maybe that's because I had a baboon for a roommate. Khufu wasn't here much (usually downstairs with Cleo or letting the ankle-biters groom his fur), but there was a baboon-shaped depression on his bed, a box of Cheerios on the nightstand and a tyre swing installed in the corner of the room. Sadie had done that last part as a joke, but Khufu loved it so much I couldn't take it down. The thing was I'd got used to him being around. Now that he spent most of his time with the kindergartners, I missed him. He'd grown on me in an endearing, annoying way, kind of like my sister.
[Yeah, Sadie. You saw that one coming.]
Screensaver pictures floated across my laptop monitor. There was my dad at a dig site in Egypt, looking relaxed and in charge in his khaki fatigues, his sleeves rolled up on his dark muscular arms as he showed off the broken stone head of some pharaoh's statue. Dad's bald scalp and goatee made him look slightly devilish when he smiled.
Another picture showed Uncle Amos onstage at a jazz club, playing his saxophone. He wore round dark glasses, a blue porkpie hat and a matching silk suit, impeccably tailored as always. His cornrows were braided with sapphires. I'd never actually seen Amos play onstage, but I liked this photo because he looked so energetic and happy – not like he did these days, with the weight of leadership on his shoulders. Unfortunately the photo also reminded me of Anne Grissom, the Texas magician with her fiddle, having so much fun earlier this evening just before she died.
The screensaver changed. I saw my mom bouncing me on her knee when I was a baby. I had this ridiculous 'fro back then, which Sadie always teases me about. In the photo, I'm wearing a blue sleepsuit stained with puréed yams. I'm holding my mom's thumbs, looking startled as she bounces me up and down, like I'm thinking, _Get me off this ride!_ My mom is as beautiful as always, even in an old T-shirt and jeans, her hair tied back in a bandanna. She smiles down at me like I'm the most wonderful thing in her life.
That photo hurt to look at, but I kept looking at it.
I remembered what Sadie had told me – that something was affecting the spirits of the dead, and we might not see our mom again unless we figured it out.
I took a deep breath. My dad, my uncle, my mom – all of them powerful magicians. All had sacrificed so much to restore the House of Life.
They were older, wiser and stronger than me. They'd had decades to practise magic. Sadie and I had had nine months. Yet we needed to do something no magician had ever managed – defeat Apophis himself.
I went to my closet and took down my old travelling case. It was just a black leather carry-on bag, like a million others you might see in an airport. For years I'd lugged it around the world as I travelled with my dad. He'd trained me to live with only the possessions I could carry.
I opened the suitcase. It was empty now except for one thing: a statuette of a coiled serpent carved in red granite, engraved with hieroglyphs. The name – _Apophis_ – was crossed out and overwritten with powerful binding spells, but still this statuette was the most dangerous object in the whole house – a representation of the enemy.
Sadie, Walt and I had made this thing in secret (over Bast's strong objections). We'd only trusted Walt because we needed his charm-making skills. Not even Amos would have approved such a dangerous experiment. One mistake, one miscast spell, and this statue could turn from a weapon against Apophis into a gateway allowing him free access to Brooklyn House. But we'd had to take the risk. Unless we found some other means of defeating the serpent, Sadie and I would have to use this statue for Plan B.
'Foolish idea,' said a voice from the balcony.
A pigeon was perched on the railing. There was something very un-pigeonlike about its stare. It looked fearless, almost dangerous; and I recognized that voice, which was more manly and warlike than you'd normally expect from a member of the dove family.
'Horus?' I asked.
The pigeon bobbed its head. 'May I come in?'
I knew he wasn't just asking out of courtesy. The house was heavily enchanted to keep out unwanted pests like rodents, termites and Egyptian gods.
'I give you permission to enter,' I said formally. 'Horus, in the form of a... _uh_... pigeon.'
'Thank you.' The pigeon hopped off the railing and waddled inside.
'Why?' I asked.
Horus ruffled his feathers. 'Well, I looked for a falcon, but they're a little scarce in New York. I wanted something with wings, so a pigeon seemed the best choice. They've adapted well to cities, aren't scared of people. They're noble birds, don't you think?'
'Noble,' I agreed. 'That's the first word that comes to mind when I think of pigeons.'
'Indeed,' Horus said.
Apparently sarcasm didn't exist in Ancient Egypt, because Horus never seemed to get it. He fluttered onto my bed and pecked at a few Cheerios left over from Khufu's lunch.
'Hey,' I warned, 'if you poop on my blankets –'
'Please. War gods do not poop on blankets. Well, except for that one time –'
'Forget I said anything.'
Horus hopped to the edge of my suitcase. He peered down at the statuette of Apophis. 'Dangerous,' he said. 'Much too dangerous, Carter.'
I hadn't told him about Plan B, but I wasn't surprised that he knew. Horus and I had shared minds too many times. The better I got at channelling his powers, the better we understood each other. The downside of godly magic was that I couldn't always shut off that connection.
'It's our emergency back-up,' I said. 'We're trying to find another way.'
'By looking for that scroll,' he recalled. 'The last copy of which burned up tonight in Dallas.'
I resisted the urge to spike the pigeon. 'Yes. But Sadie found this shadow box. She thinks it's some sort of clue. You wouldn't know anything about using shadows against Apophis, would you?'
The pigeon turned its head sideways. 'Not really. My understanding of magic is fairly straightforward. Hit enemies with a sword until they're dead. If they rise again, hit them again. Repeat as necessary. It worked against Set.'
'After how many years of fighting?'
The pigeon glared at me. 'What's your point?'
I decided to avoid an argument. Horus was a war god. He loved to fight, but it had taken him years to defeat Set, the god of evil. And Set was small stuff next to Apophis – the primordial force of Chaos. Whacking Apophis with a sword wasn't going to work.
I thought about something Bast had said earlier, in the library.
'Would Thoth know more about shadows?' I asked.
'Probably,' Horus grumbled. 'Thoth isn't good for much except studying his musty old scrolls.' He regarded the serpent figurine. 'Funny... I just remembered something. Back in the old days, the Egyptians used the same word for _statue_ and _shadow_ , because they're both smaller copies of an object. They were both called a _sheut_.'
'What are trying to tell me?'
The pigeon ruffled its feathers. 'Nothing. It just occurred to me, looking at that statue while you were talking about shadows.'
An icy feeling spread between my shoulder blades.
Shadows... statues.
Last spring Sadie and I had watched as the old Chief Lector Desjardins cast an execration spell on Apophis. Even against minor demons, execration spells were dangerous. You're supposed to destroy a small statue of the target and, in doing so, utterly destroy the target itself, erasing it from the world. Make a mistake, and things start exploding – including the magician who cast it.
Down in the Underworld, Desjardins had used a makeshift figurine against Apophis. The Chief Lector had died casting the execration and had only managed to push Apophis a little deeper into the Duat.
Sadie and I hoped that with a more powerful magic statue both of us working together might be able to execrate Apophis completely, or at least throw him so deep into the Duat that he'd never return.
That was Plan B. But we knew such a powerful spell would tap so much energy it would cost us our lives. Unless we found another way.
Statues as shadows, shadows as statues.
Plan C began forming in my mind – an idea so crazy I didn't want to put it into words.
'Horus,' I said carefully, 'does Apophis have a shadow?'
The pigeon blinked its red eyes. 'What a question! Why would you... ?' He glanced down at the red statue. 'Oh... _Oh._ That's clever, actually. Certifiably insane; but clever. You think Setne's version of the Book of Overcoming Apophis, the one Apophis was so anxious to destroy... you think it contained a secret spell for –'
'I don't know,' I said. 'It's worth asking Thoth. Maybe he knows something.'
'Maybe,' Horus said grudgingly. 'But I still think a frontal assault is the way to go.'
'Of course you do.'
The pigeon bobbed its head. 'We are strong enough, you and I. We should combine forces, Carter. Let me share your form as I once did. We could lead the armies of gods and men and defeat the serpent. Together, we'll rule the world.'
The idea might have been more tempting if I hadn't been looking at a plump bird with Cheerio dust on its plumage. Letting the pigeon rule the world sounded like a bad idea.
'I'll get back to you on that,' I said. 'First, I should talk to Thoth.'
'Bah.' Horus flapped his wings. 'He's still in Memphis, at that ridiculous sports stadium of his. But if you plan on seeing him I wouldn't wait too long.'
'Why not?'
'That's what I came to tell you,' Horus said. 'Matters are getting complicated among the gods. Apophis is dividing us, attacking us one by one, just as he's doing with you magicians. Thoth was the first to suffer.'
'Suffer... how?'
The pigeon puffed up. A wisp of smoke curled from its beak. 'Oh dear. My host is self-destructing. It can't hold my spirit for much longer. Just hurry, Carter. I'm having trouble keeping the gods together, and that old man Ra isn't helping our morale. If you and I don't lead our armies soon, we may not have any armies left to lead.'
'But –'
The pigeon hiccuped another wisp of smoke. 'Gotta go. Good luck.'
Horus flew out of the window, leaving me alone with the statuette of Apophis and a few grey feathers.
I slept like a mummy. That was the good part. The bad part was that Bast let me sleep until the afternoon.
'Why didn't you wake me up?' I demanded. 'I've got things to do!'
Bast spread her hands. 'Sadie insisted. You had a rough night last night. She said you needed your rest. Besides, I'm a cat. I respect the sanctity of sleep.'
I was still mad, but part of me knew Sadie was right. I'd expended a lot of magical energy the previous night and had gone to sleep really late. Maybe – just maybe – Sadie had my best interests at heart.
[I just caught her making faces at me, so maybe not.]
I showered and dressed. By the time the other kids got back from school, I was feeling almost human again.
Yes, I said _school_ , as in normal old school. We'd spent last spring tutoring all the initiates at Brooklyn House, but with the start of the fall semester Bast had decided that the kids could use a dose of regular mortal life. Now they went to a nearby academy in Brooklyn during the day and learned magic in the afternoons and on weekends.
I was the only one who stayed behind. I'd _always_ been home-schooled. The idea of dealing with lockers, schedules, textbooks and cafeteria food on top of running the Twenty-first Nome was just too much for me.
You'd think the other kids would have complained, especially Sadie. But, in fact, attending school was working out okay for them. The girls were happy to have more friends (and less dorky boys to flirt with, they claimed). The guys could play sports with actual teams rather than one-on-one with Khufu using Egyptian statues for hoops. As for Bast, she was happy to have a quiet house so she could stretch out on the floor and snooze in the sunlight.
At any rate, by the time the others got home, I'd done a lot of thinking about my conversations with Zia and Horus. The plan I'd formulated last night still seemed crazy, but I decided that it might be our best shot. After briefing Sadie and Bast, who (disturbingly) agreed with me, we decided it was time to tell the rest of our friends.
We gathered for dinner on the main terrace. It's a nice place to eat, with invisible barriers that keep out the wind, and a great view of the East River and Manhattan. The food magically appeared, and it was always tasty. Still, I dreaded eating on the terrace. For nine months we'd had all our important meetings there. I'd come to associate sit-down dinners with disasters.
We filled up our plates from the buffet as our guardian albino crocodile, Philip of Macedonia, splashed happily in his swimming pool. Eating next to a twenty-foot-long crocodile took some getting used to, but Philip was well trained. He only ate bacon, stray waterfowl and the occasional invading monster.
Bast sat at the head of the table with a can of Purina Fancy Feast. Sadie and I sat together at the opposite end. Khufu was off babysitting the ankle-biters, and some of our newer recruits were inside doing their homework or catching up on spell crafting, but most of our main people were present – a dozen senior initiates.
Considering how badly last night had turned out, everyone seemed in strangely good spirits. I was kind of glad they didn't yet know about Sarah Jacobi's video death threat. Julian kept bouncing in his chair and grinning for no particular reason. Cleo and Jaz were whispering together and giggling. Even Felix seemed to have recovered from his shock in Dallas. He was sculpting tiny _shabti_ penguins out of his mashed potatoes and bringing them to life.
Only Walt looked glum. The big guy had nothing on his dinner plate except three carrots and a wedge of Jell-O. (Khufu insisted Jell-O had major healing properties.) Judging from the tightness round Walt's eyes and the stiffness of his movements, I guessed his pain was even worse than last night.
I turned to Sadie. 'What's going on? Everybody seems... distracted.'
She stared at me. 'I keep forgetting you don't go to school. Carter, it's the first dance tonight! Three other schools will be there. We _can_ hurry up the meeting, can't we?'
'You're kidding,' I said. 'I'm thinking about plans for Doomsday, and you're worried about being late to a dance?'
'I've mentioned it to you a dozen times,' she insisted. 'Besides, we need something to boost our spirits. Now, tell everyone your plan. Some of us still have to decide what to wear.'
I wanted to argue, but the others were looking at me expectantly.
I cleared my throat. 'Okay. I know there's a dance, but –'
'At seven,' Jaz said. 'You _are_ coming, right?'
She smiled at me. Was she... flirting?
[Sadie just called me dense. Hey, I had other things on my mind.]
' _Uh_... s-so anyway,' I stammered. 'We need to talk about what happened in Dallas, and what happens next.'
That killed the mood. The smiles faded. My friends listened as I reviewed our mission to the Fifty-first Nome, the destruction of the Book of Overcoming Apophis and the retrieval of the shadow box. I told them about Sarah Jacobi's demand for my surrender and the turmoil among the gods that Horus had mentioned.
Sadie stepped in. She explained her weird encounter with the face in the wall, two gods and our ghost mother. She shared her gut feeling that our best chance to defeat Apophis had something to do with shadows.
Cleo raised her hand. 'So... the rebel magicians have a death warrant out for you. The gods can't help us. Apophis could arise at any time, and the last scroll that might've helped us to defeat him has been destroyed. But we shouldn't worry, because we have an empty box and a vague hunch about shadows.'
'Why, Cleo,' Bast said with admiration, 'you have a catty side!'
I pressed my hands against the surface of the table. It would've taken very little effort to summon the strength of Horus and smash it to kindling. But I doubted that would help my reputation as a calm, collected leader.
'This is more than a vague hunch,' I said. 'Look, you've all learned about execration spells, right?'
Our crocodile, Philip, grunted. He slapped the pool with his tail and made it rain on our dinner. Magical creatures are a little sensitive about the word _execration_.
Julian dabbed the water off his grilled cheese sandwich. 'Dude, you can't execrate Apophis. He's _massive_. Desjardins tried it and got killed.'
'I know,' I said. 'With a standard execration, you destroy a statue that represents the enemy. But what if you could crank up the spell by destroying a more powerful representation – something more connected to Apophis?'
Walt sat forward, suddenly interested. 'His shadow?'
Felix was so startled he dropped his spoon, crushing one of his mashed-potato penguins. 'Wait, what?'
'I got the idea from Horus,' I said. 'He told me statues were called shadows in ancient times.'
'But that was just, like, symbolic,' Alyssa said. 'Wasn't it?'
Bast set down her empty Fancy Feast can. She still looked nervous about the whole topic of shadows, but when I'd explained to her that it was either this or Sadie and me dying she'd agreed to support us.
'Maybe not,' the cat goddess said. 'I'm no expert on execration, mind you. Nasty business. But it's possible that a statue used for execration was originally meant to represent the target's shadow, which is an important part of the soul.'
'So,' Sadie said, 'we could cast an execration spell on Apophis, but instead of destroying a statue we could destroy his actual shadow. Brilliant, eh?'
'That's nuts,' Julian said. 'How do you destroy a shadow?'
Walt shooed a mashed-potato penguin away from his Jell-O. 'It's not nuts. Sympathetic magic is all about using a small copy to manipulate the actual target. It's possible the whole tradition of making little statues to represent people and gods – maybe at one time those statues actually _contained_ the target's _sheut_. There are lots of stories about the souls of the gods inhabiting statues. If a shadow was trapped in a statue, you might be able to destroy it.'
'Could you make a statue like that?' Alyssa asked. 'Something that could bind the shadow of... of Apophis himself?'
'Maybe.' Walt glanced at me. Most of the folks at the table didn't know we'd already made a statue of Apophis that might work for that purpose. 'Even if I could, we'd have to find the shadow. Then we'd need some pretty advanced magic to capture it and destroy it.'
'Find a shadow?' Felix smiled nervously, like he hoped we were joking. 'Wouldn't it be right _under_ him? And how do you capture it? Step on it? Shine a light on it?'
'It'll be more complicated than that,' I said. 'This ancient magician Setne, the guy who wrote his own version of the Book of Overcoming Apophis, I think he must have created a spell to catch and destroy shadows. That's why Apophis was so anxious to burn the evidence. That's his secret weakness.'
'But the scroll is gone,' Cleo said.
'There's still someone we can ask,' Walt said. 'Thoth. If anyone knows the answers, he will.'
The tension around the table seemed to ease. At least we'd given our initiates something to hope for, even if it were a long shot. I was grateful we had Walt on our side. His charm-making ability might be our only hope of binding a shadow to the statue, and his vote of confidence carried weight with the other kids.
'We need to visit Thoth right away,' I said. 'Tonight.'
'Yes,' Sadie agreed. 'Right after the dance.'
I glared at her. 'You aren't serious.'
'Oh, yes, brother dear.' She smiled mischievously, and for a second I was afraid she might invoke my secret name and force me to obey. 'We're attending the dance tonight. And you're coming with us.'
SADIE
## 5. A Dance with Death
CHEERS, CARTER. At least you have the sense to hand me the microphone for _important_ things.
Honestly, he drones on and on about his plans for the Apocalypse, but he makes no plans at all for the school dance. My brother's priorities are severely skewed.
I don't think I was being selfish wanting to go to the dance. Of _course_ we had serious business to deal with. That's exactly why I insisted on partying first. Our initiates needed a morale boost. They needed a chance to be normal kids, to have friends and lives outside Brooklyn House – something worth fighting for. Even armies in the field fight better when they take breaks for entertainment. I'm sure some general somewhere has said that.
By sunset, I was ready to lead my troops into battle. I'd picked out quite a nice black strapless dress and put black lowlights in my blonde hair, with just a touch of dark make-up for that risen-from-the-grave look. I wore simple flats for dancing (despite what Carter says, I do not wear combat boots all the time; just ninety per cent of the time), the silver _tyet_ amulet from my mother's jewellery box and the pendant Walt had given me for my last birthday with the Egyptian symbol of eternity, _shen_.
Walt had an identical amulet among his own collection of talismans, which provided us with a magic line of communication and even the ability to summon the other person to our side in emergencies.
Unfortunately, the _shen_ amulets didn't mean we were dating exclusively. Or even dating at all. If Walt had _asked_ me, I think I would've been fine with it. Walt was so kind and gorgeous – perfect, really, in his own way. Perhaps if he'd asserted himself a bit more I would've fallen for him and been able to let go of that _other_ boy, the godly one.
But Walt was dying. He had this silly idea that it would be unfair to me if we started a relationship under those circumstances. As if that would stop me. So we were stuck in this maddening limbo – flirting, talking for hours, a few times even sharing a kiss when we let our guard down – but eventually Walt would always pull away and shut me out.
Why couldn't things be simple?
I bring this up because I literally ran into Walt as I was coming down the stairs.
'Oh!' I said. Then I noticed he was still wearing his old muscle shirt, jeans and no shoes. 'You're not ready yet?'
'I'm not going,' he announced.
My mouth fell open. 'What? Why?'
'Sadie... you and Carter will need me when you visit Thoth. If I'm going to make it, I have to rest.'
'But...' I forced myself to stop. It wasn't right for me to pressure him. I didn't need magic to see that he really was in great pain.
Centuries of magical healing knowledge at our disposal, yet nothing we tried seemed to help Walt. I ask you: what's the point of being a magician if you can't wave your wand and make the people you care about feel better?
'Right,' I said. 'I – I was just hoping...'
Anything I said would've sounded bratty. I wanted to dance with him. Gods of Egypt, I'd _dressed up_ for him. The mortal boys at school were all right, I suppose, but they seemed quite shallow compared to Walt (or, yes, fine – compared to Anubis). As for the other boys of Brooklyn House – dancing with them would have made me feel a bit odd, like I was dancing with my cousins.
'I could stay,' I offered, but I suppose I didn't sound very convincing.
Walt managed a faint smile. 'No, go, Sadie. Really. I'm sure I'll be feeling better when you get back. Have a good time.'
He brushed past me and climbed the steps.
I took several deep breaths. Part of me did want to stay and look after him. Going without him didn't seem right.
Then I glanced down into the Great Room. The older kids were joking and talking, ready to leave. If _I_ didn't go, they might feel obliged to stay, too.
Something like wet cement settled in my stomach. All the joy and excitement suddenly went out of the evening for me. For months I'd been struggling to adjust to life in New York after so many years in London. I'd been forced to balance life as a young magician with the challenges of being an ordinary schoolgirl. Now, just when this dance had seemed to offer me a chance to combine both worlds and have a lovely night out, my hopes were dashed. I'd still have to go and pretend to have fun. But I'd only be doing it out of duty, to make the others feel better.
I wondered if this was what being a grown-up felt like. Horrible.
The only thing that cheered me up was Carter. He emerged from his room dressed like a junior professor, in a coat and tie, button-down shirt and trousers. Poor boy – of course he'd never been to a dance any more than he'd been to school. He had no clue whatsoever.
'You look... wonderful.' I tried to keep a straight face. 'You do realize it's not a funeral?'
'Shut up,' he grumbled. 'Let's get this over with.'
The school the kids and I attended was Brooklyn Academy for the Gifted. Everyone called it BAG. We had no end of jokes about this. The students were Baggies. The glamour girls with nose jobs and Botox lips were Plastic Bags. Our alumni were Old Bags. And, naturally, our headmistress, Mrs Laird, was the Bag Lady.
Despite the name, the school was quite nice. All the students were gifted in some sort of art, music or drama. Our schedules were flexible, with lots of independent study time, which worked perfectly for us magicians. We could pop off to battle monsters as needed; and, as magicians, it wasn't difficult for us to pass ourselves off as gifted. Alyssa used her earth magic to make sculptures. Walt specialized in jewellery. Cleo was an amazing writer, since she could retell stories that had been forgotten since the days of Ancient Egypt. As for me, I needed no magic. I was a natural at drama.
[Stop laughing, Carter.]
You might not expect this in the middle of Brooklyn, but our campus was like a park, with acres of green lawns, well-tended trees and hedges, even a small lake with ducks and swans.
The dance was held in the pavilion in front of the administration building. A band played in the gazebo. Lights were strung in the trees. Teacher chaperones walked the perimeter on 'bush patrol', making sure none of the older students sneaked off into the shrubbery.
I tried not to think about it, but the music and crowd reminded me of Dallas the night before – a very different sort of party, which had ended badly. I remembered JD Grissom clasping my hand, wishing me luck before he ran off to save his wife.
Horrible guilt welled inside me. I forced it down. It wouldn't do the Grissoms any good for me to start crying in the middle of the dance. It certainly wouldn't help my friends enjoy themselves.
As our group dispersed into the crowd, I turned to Carter, who was fiddling with his tie.
'Right,' I said. 'You need to dance.'
Carter looked at me in horror. 'What?'
I called over one of my mortal friends, a lovely girl named Lacy. She was a year younger than I, so she looked up to me greatly. (I know, it's hard not to.) She had cute blonde pigtails, a mouthful of braces and was possibly the only person at the dance _more_ nervous than my brother. She'd seen pictures of Carter before, however, and seemed to find him _hot_. I didn't hold that against her. In most ways, she had excellent taste.
'Lacy – Carter,' I introduced them.
'You look like your pictures!' Lacy grinned. The bands of her braces were alternating pink and white to match her dress.
Carter said, ' _Uh_ –'
'He doesn't know how to dance,' I told Lacy. 'I'd be ever so grateful if you'd teach him.'
'Sure!' she squealed. She grabbed my brother's hand and swept him away.
I started to feel better. Perhaps I could have fun tonight, after all.
Then I turned and found myself face to face with one of my _not-_ so-favourite mortals – Drew Tanaka, head of the popular-girl clique, with her supermodel goon squad in tow.
'Sadie!' Drew threw her arm round me. Her perfume was a mixture of roses and tear gas. 'So glad you're here, sweetie. If I'd known you were coming, you could've ridden in the limo with us!'
Her friends made sympathetic 'Aww' sounds and grinned to show they were not at all sincere. They were dressed more or less the same, in the latest silky designer bits their parents had no doubt commissioned for them during the last Fashion Week. Drew was the tallest and most glamorous (I use the word as an insult) with awful pink eyeliner and frizzy black curls that were apparently Drew's own personal crusade to bring back the 1980s perm. She wore a pendant – a glittering platinum and diamond _D_ – possibly her initial, or her grade average.
I gave her a tight smile. 'A limo, wow. Thanks for that. But, between you, your friends and your egos, I doubt there would've been extra room.'
Drew pouted. 'That's not nice, hon! Where is Walt? Is the poor baby still sick?'
Behind her, some of the girls coughed into their fists, mimicking Walt.
I wanted to pull my staff from the Duat and turn them all into worms for the ducks. I was pretty sure I could manage that, and I doubted anyone would miss them, but I kept my temper.
Lacy had warned me about Drew the first day of school. Apparently the two of them had gone to some summer camp together – blah, blah, I didn't really listen to the details – and Drew had been just as much of a tyrant there.
That did not, however, mean she could be a tyrant with _me_.
'Walt's at home,' I said. 'I _did_ tell him you'd be here. Funny, that didn't seem to motivate him much.'
'What a shame,' Drew sighed. 'You know, maybe he's not really sick. He might just be allergic to you, hon. That does happen. I should go to his place with some chicken soup or something. Where does he live?'
She smiled sweetly. I didn't know if she actually fancied Walt or if she just pretended because she hated me. Either way, the idea of turning her into an earthworm was becoming more appealing.
Before I could do anything rash, a familiar voice behind me said, 'Hello, Sadie.'
The other girls let out a collective gasp. My pulse quickened from 'slow walk' to 'fifty-metre dash'. I turned and found that – yes, indeed – the god Anubis had crashed our dance.
He had the nerve to look amazing, as usual. He's _so_ annoying that way. He wore skinny black trousers with black leather boots and a biker's jacket over an Arcade Fire T-shirt. His dark hair was naturally dishevelled as if he'd just woken up, and I fought the urge to run my fingers through it. His brown eyes glittered with amusement. Either he was happy to see me or he enjoyed seeing me flustered.
'Oh... my... god,' Drew whimpered. 'Who...'
Anubis ignored her (bless him for that) and held out his elbow for me – a sweet old-fashioned gesture. 'May I have this dance?'
'I suppose,' I said, as noncommittally as I could.
I looped my arm through his, and we left the Plastic Bags behind us, all of them muttering, 'Oh my god! Oh my god!'
_No, actually_ , I wanted to say. _He's_ my _amazingly hot boy god. Find your own._
The uneven paving stones made for a dangerous dance floor. All around us, kids were tripping over each other. Anubis didn't help matters, as all the girls turned and gawked at him as he led me through the crowd.
I was glad Anubis had my arm. My emotions were so jumbled I felt dizzy. I was ridiculously happy that he was here. I felt crushingly guilty that poor Walt was at home alone while I strolled arm in arm with Anubis. But I was relieved that Walt and Anubis weren't both here together. That would've been _beyond_ awkward. The relief made me feel guiltier, and so on. Gods of Egypt, I was a mess.
As we reached the middle of the dance floor, the band suddenly switched from a dance number to a love ballad.
'Was that your doing?' I asked Anubis.
He smiled, which wasn't much of an answer. He put one hand on my hip and clasped my other hand, like a proper gentleman. We swayed together.
I'd heard of dancing on air, but it took me a few steps to realize we were actually levitating – a few millimetres off the ground, not enough for anyone to notice, just enough for us to glide across the stones while others stumbled.
A few metres away, Carter looked quite awkward as Lacy showed him how to slow-dance. [Really, Carter, it isn't quantum physics.]
I gazed up at Anubis's warm brown eyes and his exquisite lips. He'd kissed me once – for my birthday, last spring – and I'd never quite got over it. You'd think a god of death would have cold lips, but that wasn't the case at all.
I tried to clear my head. I knew Anubis must be here for some reason, but it was awfully hard to focus.
'I thought... _Um_ ,' I gulped, and barely managed not to drool on myself.
_Oh, brilliant, Sadie_ , I thought. _Let's try for a complete sentence, now, shall we?_
'I thought you could only appear in places of death,' I said.
Anubis laughed gently. 'This _is_ a place of death, Sadie. The Battle of Brooklyn Heights, 1776. Hundreds of American and British troops died right where we're dancing.'
'How romantic,' I muttered. 'So we're dancing on their graves?'
Anubis shook his head. 'Most never received proper burials. That's why I decided to visit you here. These ghosts could use a night of entertainment, just like your initiates.'
Suddenly, spirits were twirling all around us – luminous apparitions in eighteenth-century clothes. Some wore the red uniforms of British regulars. Others had ragtag militia outfits. They pirouetted with lady ghosts in plain farm dresses or fancy silk. A few of the posh women had piles of curly hair that would have made even Drew jealous. The ghosts seemed to be dancing to a different song. I strained my ears and could faintly hear violins and a cello.
None of the regular mortals seemed to notice the spectral invasion. Even my friends from Brooklyn House were oblivious. I watched as a ghostly couple waltzed straight through Carter and Lacy. As Anubis and I danced, Brooklyn Academy seemed to fade and the ghosts became more real.
One soldier had a musket wound in his chest. A British officer had a tomahawk sticking out of his powdered wig. We danced between worlds, waltzing side by side with smiling, gruesomely slaughtered phantoms. Anubis certainly knew how to show a girl a good time.
'You're doing it again,' I said. 'Taking me out of phase, or whatever you call it.'
'A little,' he admitted. 'We need privacy to talk. I promised you I'd visit in person –'
'And you did.'
'– but it's going to cause trouble. This may be the last time I can see you. There's been grumbling about our situation.'
I narrowed my eyes. Was the god of the dead blushing?
'Our situation,' I repeated.
'Us.'
The word set my ears buzzing. I tried to keep my voice even. 'As far as I'm aware, there _is_ no official "us". Why would this be the last time we can talk?'
He was definitely blushing now. 'Please, just listen. There's so much I need to tell you. Your brother has the right idea. The shadow of Apophis is your best hope, but only one person can teach you the magic you need. Thoth may guide you somewhat, but I doubt he'll reveal the secret spells. It's too dangerous.'
'Hold on, hold on.' I was still reeling from the comment about _us_. And the idea that this might be the last time I saw Anubis... That sent my brain cells into panic mode, thousands of tiny Sadies running around in my skull, screaming and waving their arms.
I tried to focus. 'You mean Apophis _does_ have a shadow? It could be used to execrate –'
'Please don't use that word.' Anubis grimaced. 'But, yes, all intelligent entities have souls, so all of them have shadows, even Apophis. I know this much, being the guide of the dead. I have to make souls my business. Could his shadow be used against him? In theory, yes. But there are many dangers.'
'Naturally.'
Anubis twirled me through a pair of colonial ghosts. Other students watched us, whispering as we danced, but their voices sounded distant and distorted, as if they were on the far side of a waterfall.
Anubis studied me with a sort of tender regret. 'Sadie, I wouldn't set you on this path if there were another way. I don't want you to die.'
'I can agree with that,' I said.
'Even _talking_ about this sort of magic is forbidden,' he warned. 'But you need to know what you're dealing with. The _sheut_ is the least understood part of the soul. It's... how to explain... a soul of last resort, an afterimage of the person's life force. You've heard that the souls of the wicked are destroyed in the Hall of Judgement –'
'When Ammit devours their hearts,' I said.
'Yes.' Anubis lowered his voice. 'We say that this completely destroys the soul. But that's not true. The shadow lingers. Occasionally, not often, Osiris has decided to, ah, _review_ a judgement. If someone was found guilty, but new evidence comes to light, there must be a way to retrieve a soul from oblivion.'
I tried to grasp that. My thoughts felt suspended in midair like my feet, not able to connect with anything solid. 'So... you're saying the shadow could be used to, _um_ , _reboot_ a soul? Like a computer's back-up drive?'
Anubis looked at me strangely.
'Ugh, I'm sorry.' I sighed. 'I've been spending too much time with my geeky brother. He speaks like a computer.'
'No, no,' Anubis said. 'It's actually a good analogy. I'd just never thought of it that way. Yes, the soul isn't completely destroyed until the shadow is destroyed, so in extreme cases, with the right magic, it's possible to reboot the soul using the _sheut_. Conversely, if you were to destroy a god's shadow, or even Apophis's shadow as part of an ex– _um_ , the sort of spell you mentioned –'
'The _sheut_ would be infinitely more powerful than a regular statue,' I guessed. 'We could destroy him, possibly without destroying ourselves.'
Anubis glanced around us nervously. 'Yes, but you can see why this sort of magic is secret. The gods would never want such knowledge in the hands of a mortal magician. This is why we always hide our shadows. If a magician were able to capture a god's _sheut_ and use it to threaten us –'
'Right.' My mouth felt dry. 'But I'm on your side. I'd only use the spell on Apophis. Surely Thoth will understand that.'
'Perhaps.' Anubis didn't sound convinced. 'Start with Thoth, at least. Hopefully he'll see the need to assist you. I fear, though, you may still need better guidance – more _dangerous_ guidance.'
I gulped. 'You said only one person could teach us the magic. Who?'
'The only magician crazy enough to ever research such a spell. His trial is tomorrow at sunset. You'll have to visit your father before then.'
'Wait. What?'
Wind blew through the pavilion. Anubis's hand tightened on mine.
'We have to hurry,' he said. 'There's more I need to tell you. Something is happening with the spirits of dead. They're being... Look, there!'
He pointed to a pair of nearby spectres. The woman danced barefoot in a simple white linen dress. The man wore breeches and a frock coat like a colonial farmer, but his neck was canted at a funny angle, as if he'd been hanged. Black mist coiled round the man's legs like ivy. Another three waltz steps, and he was completely engulfed. The murky tendrils pulled him into the ground, and he disappeared. The woman in white kept dancing by herself, apparently unaware that her partner had been consumed by evil fingers of smog.
'What – what was _that_?' I asked.
'We don't know,' Anubis said. 'As Apophis grows stronger, it's happening more frequently. Souls of the dead are disappearing, being drawn further down into the Duat. We don't know where they're going.'
I almost stumbled. 'My mother. Is she all right?'
Anubis gave me a pained look, and I knew the answer. Mum had warned me – we might never see her again unless we discovered a way to defeat Apophis. She'd sent me that message urging me to find the serpent's shadow. It _had_ to be connected to her dilemma somehow.
'She's missing,' I guessed. My heart pounded against my ribs. 'It's got something to do with this business about shadows, hasn't it?'
'Sadie, I wish I knew. Your father is – he's trying his best to find her, but –'
The wind interrupted him.
Have you ever stuck your hand out of a moving car and felt the air push against you? It was a bit like that, but ten times more powerful. A wedge of force pushed Anubis and me apart. I staggered backwards, my feet no longer levitating.
'Sadie...' Anubis reached out, but the wind pushed him further away.
'Stop that!' said a squeaky voice between us. 'No public displays of affection on _my_ watch!'
The air took on human form. At first it was just a faint silhouette. Then it became more solid and colourful. Before me stood a man in an old-fashioned aviator's outfit – leather helmet, goggles, scarf and a bomber's jacket, like photos I'd seen of the Royal Air Force pilots during World War II. He wasn't flesh and blood, though. His form swirled and shifted. I realized he was put together from blown rubbish: specks of dirt, scraps of paper, bits of dandelion fuzz, dried leaves – all churning about, but held together in such a tight collage by the wind that from a distance he might have passed for a normal mortal.
He wagged his finger at Anubis. 'This is the final insult, boy!' His voice hissed like air from a balloon. 'You have been warned _numerous_ times.'
'Hold on!' I said. 'Who are you? And Anubis is hardly a boy. He's five thousand years old.'
'Exactly,' the aviator snapped. 'A mere child. And I didn't give you permission to speak, girl!'
The aviator exploded. The blast was so powerful my ears popped and I fell on my bum. Around me, the other mortals – my friends, teachers and all the students – simply collapsed. Anubis and the ghosts seemed unaffected. The aviator formed again, glaring down at me.
I struggled to my feet and tried to summon my staff from the Duat. No such luck.
'What have you done?' I demanded.
'Sadie, it's all right,' Anubis said. 'Your friends are only unconscious. Shu just lowered the air pressure.'
'Shoe?' I demanded. 'Shoe who?'
Anubis pressed his fingers to his temples. 'Sadie... this is Shu, my great-grandfather.'
Then it struck me: Shu was one of those ridiculous godly names I'd heard before. I tried to place it. 'Ah. The god of... flip-flops. No, wait. Leaky balloons. No –'
'Air!' Shu hissed. 'God of the air!'
His body dissolved into a tornado of debris. When he formed again, he was in Ancient Egyptian costume – bare-chested with a white loincloth and a giant ostrich feather sprouting from his braided headband.
He changed back into RAF clothes.
'Stick with the pilot's outfit,' I said. 'The ostrich feather really doesn't work for you.'
Shu made an unfriendly whooshing sound. 'I'd _prefer_ to be invisible, thank you very much. But you mortals have polluted the air so badly it's getting harder and harder. It's _dreadful_ what you've done, the last few millennia! Haven't you people heard of "Spare the Air" days? Carpooling? Hybrid engines? And don't get me started on cows. Did you know that every cow belches and farts over a hundred gallons of methane a day? There are one and a half billion cows in the world. Do you have _any_ idea what that does to my respiratory system?'
' _Uh_...'
From his jacket pocket, Shu produced an inhaler and puffed on it. 'Shocking!'
I raised an eyebrow at Anubis, who looked mortally embarrassed (or perhaps immortally embarrassed).
'Shu,' he said, 'we were just talking. If you'll let us finish –'
'Oh, _talking_!' Shu bellowed, no doubt releasing his own share of methane. 'While holding hands and dancing, and other degenerate behaviour. Don't play innocent, boy. I've been a chaperone before, you know. I kept your grandparents apart for aeons.'
Suddenly I remembered the story of Nut and Geb, the sky and earth. Ra had commanded Nut's father, Shu, to keep the two lovers apart so they would never have children who might some day usurp Ra's throne. That strategy hadn't worked, but apparently Shu was still trying.
The air god waved his hand in disgust at the unconscious mortals, some of whom were just starting to groan and stir. 'And now, Anubis, I find you in this den of iniquity, this morass of questionable behaviour, this... this –'
'School?' I suggested.
'Yes!' Shu nodded so vigorously, his head disintegrated into a cloud of leaves. 'You heard the decree of the gods, boy. You've become _entirely_ too close to this mortal. You are hereby banned from further contact!'
'What?' I shouted. 'That's ridiculous! Who decreed this?'
Shu made a sound like a deflating balloon. Either he was laughing or giving me a windy raspberry. 'The entire council, girl! Led by Lord Horus and Lady Isis!'
I felt as if I were dissolving into scraps of rubbish myself.
Isis and Horus? I couldn't believe it. Stabbed in the back by my two supposed friends. Isis and I were going to have words about this.
I turned to Anubis, hoping he'd tell me it was a lie.
He raised his hands miserably. 'Sadie, I was trying to tell you. Gods are not allowed to become directly... _um_ , _involved_ with mortals. That's only possible when a god inhabits a human form, and... and, as you know, I've never worked that way.'
I gritted my teeth. I wanted to argue that Anubis had quite a _nice_ form, but he'd told me often that he could only manifest in dreams or in places of death. Unlike other gods, he'd never taken a human host.
It was so _unfair_. We hadn't even dated properly. One kiss six months ago, and Anubis was grounded from seeing me forever?
'You can't be serious.' I'm not sure who made me angrier – the fussy air god chaperone or Anubis himself. 'You're not really going to let them rule you like this?'
'He has no choice!' Shu cried. The effort made him cough so badly that his chest exploded into dandelion fluff. He took another blast from his inhaler. 'Brooklyn ozone levels – deplorable! Now, off with you, Anubis. No more contact with this mortal. It is _not_ proper. And as for you, girl, stay away from him! You have more important things to do.'
'Oh, yes?' I said. 'And what about you, Mr Trash Tornado? We're preparing for war, and the most important thing you can do is keep people from waltzing?'
The air pressure rose suddenly. Blood roared in my head.
'See here, girl,' Shu growled. 'I've already helped you more than you deserve. I heeded that Russian boy's prayer. I brought him here all the way from St Petersburg to speak with you. So, shoo!'
The wind blasted me backwards. The ghosts blew away like smoke. The unconscious mortals began to stir, shielding their faces from the debris.
'Russian boy?' I shouted over the gale. 'What on earth are you talking about?'
Shu disbanded into rubbish and swirled around Anubis, lifting him off his feet.
'Sadie!' Anubis tried to fight his way towards me, but the storm was too strong. 'Shu, at least let me tell her about Walt! She has a right to know!'
I could barely hear him above the wind. 'Did you say, _Walt_?' I shouted. 'What about him?'
Anubis said something I couldn't make out. Then the flurry of debris completely obscured him.
When the wind died, both gods were gone. I stood alone on the dance floor, surrounded by dozens of kids and adults who were starting to wake up.
I was about to run to Carter to make sure he was all right. [Yes, Carter, honestly I was.]
Then, at the edge of the pavilion, a young man stepped into the light.
He wore a grey military outfit with a wool coat too heavy for the warm September night. His enormous ears seemed to be the only things holding up his oversized hat. A rifle was slung across his shoulder. He couldn't have been more than seventeen and, though he was definitely not from any of the schools at the dance, he looked vaguely familiar.
_St Petersburg_ , Shu had said.
Yes. I'd met this boy briefly last spring. Carter and I had been running from the Hermitage Museum. This boy had tried to stop us. He'd been disguised as a guard, but revealed himself as a magician from the Russian nome – one of the servants of the evil Vlad Menshikov.
I grabbed my staff from the Duat – successfully this time.
The boy raised his hands in surrender.
' _Nyet!_ ' he pleaded. Then, in halting English, he said: 'Sadie Kane. We... need... to talk.'
SADIE
## 6. Amos Plays with Action Figures
HIS NAME WAS LEONID, and we agreed not to kill each other.
We sat on the steps of the gazebo and talked while the students and teachers struggled to wake up around us.
Leonid's English was not good. My Russian was nonexistent, but I understood enough of his story to be alarmed. He'd escaped the Russian Nome and somehow convinced Shu to whisk him here to find me. Leonid remembered me from our invasion of the Hermitage. Apparently I'd made a strong impression on the young man. No surprise. I am rather memorable.
[Oh, stop laughing, Carter.]
Using words, hand gestures and sound effects, Leonid tried to explain what had happened in St Petersburg since the death of Vlad Menshikov. I couldn't follow it all, but this much I understood: _Kwai, Jacobi, Apophis, First Nome, many deaths, soon, very soon._
Teachers began corralling students and calling parents. Apparently they feared the mass blackout might have been caused by bad punch or hazardous gas (Drew's perfume, perhaps) and they'd decided to evacuate the area. I suspected we'd have police and paramedics on the scene shortly. I wanted to be gone before then.
I dragged Leonid over to meet my brother, who was stumbling around, rubbing his eyes.
'What happened?' Carter asked. He scowled at Leonid. 'Who –?'
I gave him the one-minute version: Anubis's visit, Shu's intervention, the Russian's appearance. 'Leonid has information about an impending attack on the First Nome,' I said. 'The rebels will be after him.'
Carter scratched his head. 'You want to hide him at Brooklyn House?'
'No,' I said. 'I've got to take him to Amos straight away.'
Leonid choked. 'Amos? He turn into Set – eat face?'
'Amos will _not_ eat your face,' I assured him. 'Jacobi's been telling you stories.'
Leonid still looked uneasy. 'Amos not become Set?'
How to explain without making it sound worse? I didn't know the correct Russian for: _He was possessed by Set but it wasn't his fault, and he's much better now._
'No Set,' I said. 'Good Amos.'
Carter studied the Russian. He looked at me with concern. 'Sadie, what if this is a trap? You _trust_ this guy?'
'Oh, I can handle Leonid. He doesn't want me to morph him into a banana slug, do you, Leonid?'
' _Nyet,_ ' Leonid said solemnly. 'No banana slug.'
'There, you see?'
'What about visiting Thoth?' Carter asked. 'That can't wait.'
I saw the worry in his eyes. I imagined he was thinking the same thing I was: our mum was in trouble. The spirits of the dead were disappearing, and it had something to do with the shadow of Apophis. We had to find the connection.
'You visit Thoth,' I said. 'Take Walt. And, _uh_ , keep an eye on him, all right? Anubis wanted to tell me something about him, but there wasn't time. And in Dallas when I looked at Walt in the Duat...'
I couldn't make myself finish. Just thinking about Walt wrapped in mummy linen brought tears to my eyes.
Fortunately, Carter seemed to get the general idea. 'I'll keep him safe,' he promised. 'How will you get to Egypt?'
I pondered that. Leonid had apparently flown here via Shu Airways, but I doubted that fussy aviator god would be willing to help me, and I didn't want to ask.
'We'll risk a portal,' I said. 'I know they've been a bit wonky, but it's just one quick jump. What could go wrong?'
'You could materialize inside a wall,' Carter said. 'Or wind up scattered through the Duat in a million pieces.'
'Why, Carter, you care! But, really, we'll be fine. And we haven't got much choice.'
I gave him a quick hug – I know, horribly sentimental, but I wanted to show solidarity. Then, before I could change my mind, I took Leonid's hand and raced across campus.
My head was still spinning from my talk with Anubis. How dare Isis and Horus keep us apart when we weren't even together! And what had Anubis wanted to tell me about Walt? Perhaps he'd wanted to end our ill-fated relationship and give his blessing for me to date Walt. (Lame.) Or perhaps he wanted to declare his undying love and fight Walt for my affections. (Highly unlikely, nor would I appreciate being fought over like a basketball.) Or perhaps – most probable – he'd wanted to break some bad news.
Anubis had visited Walt on several occasions that I knew of. They'd both been rather tight-lipped about what was discussed, but since Anubis was the guide of the dead I assumed he'd been preparing Walt for death. Anubis might have wanted to warn me that the time was nigh – as if I needed another reminder.
Anubis: off-limits. Walt: at death's door. If I lost both of the guys I liked, well... there wasn't much point in saving the world.
All right, that was a _slight_ exaggeration. But only slight.
On top of that, my mum was in trouble and Sarah Jacobi's rebels were planning some horrible attack on my uncle's headquarters.
Why, then, did I feel so... _hopeful_?
An idea started to tug at me – a tiny glimmer of possibility. It wasn't just the prospect that we might find a way to defeat the serpent. Anubis's words kept playing in my mind: _The shadow lingers. There must be a way to retrieve a soul from oblivion._
If a shadow could be used to bring back a mortal soul that had been destroyed, could it do the same for a god?
I was so lost in thought that I barely noticed when we reached the fine-arts building. Leonid stopped me.
'This for portal?' He pointed to a block of carved limestone in the courtyard.
'Yes,' I said. 'Thanks.'
Long story short: when I started at BAG, I reckoned it would be good to have an Egyptian relic close by for emergencies. So I did the logical thing: I borrowed a chunk of limestone frieze from the nearby Brooklyn Museum. Honestly, the museum had enough rocks. I didn't think they'd miss this one.
I'd left a facsimile in its place and asked Alyssa to present the actual Egyptian frieze to her art teacher as her class project – an attempt to simulate an ancient art form. The teacher had been duly impressed. He'd installed 'Alyssa's' artwork in the courtyard outside his classroom. The carving showed mourners at a funeral, which I thought appropriate for a school setting.
It wasn't a powerful or important piece of art, but all relics of Ancient Egypt have some amount of power, like magical batteries. With the right training, a magician can use them to jump-start spells that would otherwise be impossible, such as opening portals.
I'd got rather good at this particular magic. Leonid watched my back as I began to chant.
Most magicians wait for 'auspicious moments' to open gates. They spend years memorizing a timetable of important anniversaries like the time of day each god was born, the alignment of the stars and whatnot. I suppose I should have worried about such things, but I didn't. Given the thousands of years of Egyptian history, there were so _many_ auspicious moments that I simply chanted until I hit one. Of course, I had to hope my portal didn't open during an _inauspicious_ moment. That could have caused all sorts of nasty side effects – but what's life without taking a few risks?
[Carter is shaking his head and muttering. I have no idea why.]
The air rippled in front of us. A circular doorway appeared – a swirling vortex of golden sand – and Leonid and I jumped through.
I'd like to say my spell worked perfectly and we ended up in the First Nome. Sadly, I was a bit off the mark.
The portal spat us out roughly a hundred metres above Cairo. I found myself free-falling through the cool night air towards the city lights below.
I didn't panic. I could have cast any number of spells to get out of this situation. I could have even assumed the form of a kite (the bird of prey, not the kind with a string), although that wasn't my favourite way to travel. Before I could decide on a plan of action, Leonid grabbed my hand.
The direction of the wind changed. Suddenly we were gliding over the city in a controlled descent. We set down softly in the desert just outside the city limits near a cluster of ruins that I knew from experience hid an entrance to the First Nome.
I looked at Leonid in amazement. 'You summoned the power of Shu!'
'Shu,' he said grimly. 'Yes. Necessary. I do... forbidden.'
I smiled with delight. 'You clever boy! You learned the path of the gods on your own? I knew there was a reason I didn't turn you into a banana slug.'
Leonid's eyes widened. 'No banana slug! Please!'
'It was a compliment, silly,' I said. 'Forbidden is good! Sadie likes forbidden! Now, come on. You need to meet my uncle.'
No doubt Carter would describe the underground city in excruciating detail, with exact measurements of each room, boring history on every statue and hieroglyph, and background notes on the construction of the magical headquarters of the House of Life.
I will spare you that pain.
It's big. It's full of magic. It's underground.
There. Sorted.
At the bottom of the entry tunnel, we crossed a stone bridge over a chasm, where I was challenged by a _ba_. The glowing bird spirit (with the head of a famous Egyptian I probably should've known) asked me a question: _What colour are the eyes of Anubis?_
Brown. _Duh._ I suppose he was trying to trick me with an easy question.
The _ba_ let us pass into the city proper. I hadn't visited in six months, and I was distressed to see how few magicians were about. The First Nome had never been crowded. Egyptian magic had withered over the centuries as fewer and fewer young initiates learned the arts. But now most shops in the central cavern were closed. At the market stalls, no one was haggling over the price of _ankh_ s or scorpion venom. A bored-looking amulet salesman perked up as we approached, then slumped as we passed by.
Our footsteps echoed in the silent tunnels. We crossed one of the subterranean rivers, then wound our way through the library quarter and the Chamber of Birds.
[Carter says I should tell you why it's called that. It's a cave full of all sorts of birds. Again – _duh_. (Carter, why are you banging your head against the table?)]
I brought my Russian friend down a long corridor, past a sealed tunnel that had once led up to the Great Sphinx of Giza, and finally to the bronze doors of the Hall of Ages. It was my uncle's hall now, so I strolled right in.
Impressive place? Certainly. If you filled it with water, the hall would've been large enough for a pod of whales. Running down the middle, a long blue carpet glittered like the River Nile. Along either side marched rows of columns, and between them shimmered curtains of light displaying scenes from Egypt's past – all sorts of horrible, wonderful, heart-wrenching events.
I tried to avoid looking at them. I knew from experience that those images could be dangerously absorbing. Once I'd made the mistake of touching the lights, and the experience had almost turned my brain into oatmeal.
The first section of light was gold – the Age of the Gods. Further along, the Old Kingdom glowed silver, then the Middle Kingdom in coppery brown, and so on.
Several times as we walked, I had to pull Leonid back from scenes that caught his eye. Honestly, I wasn't much better.
I got teary-eyed when I saw a vision of Bes entertaining the other gods by doing cartwheels in a loincloth. (I cried because I missed seeing him so full of life, I mean, though the sight of Bes in a loincloth _is_ enough to make anyone's eyes burn.)
We passed the bronze curtain of light for the New Kingdom. I stopped abruptly. In the shifting mirage, a thin man in priestly robes held a wand and a knife over a black bull. The man muttered as if blessing the animal. I couldn't tell much about the scene, but I recognized the man's face – a beaky nose, high forehead, thin lips that twisted in a wicked smile as he ran the knife along the poor animal's throat.
'That's him,' I muttered.
I walked towards the curtain of light.
' _Nyet._ ' Leonid grabbed my arm. 'You tell me the lights are bad, stay away.'
'You – you're right,' I said. 'But that's Uncle Vinnie.'
I was positive it was the same face that had appeared in the wall at the Dallas Museum, but how could that be? The scene I was looking at must have happened thousands of years ago.
'Not Vinnie,' Leonid said. 'Khaemwaset.'
'Sorry?' I wasn't sure if I'd heard him correctly, or even in what language he'd spoken. 'Is that a name?'
'He is...' Leonid slipped into Russian, then sighed in exasperation. 'Too difficult to explain. Let us see Amos, who will not eat my face.'
I forced myself to look away from the image. 'Good idea. Let's keep going.'
At the end of the hall, the curtains of red light for the Modern Age changed to dark purple. Supposedly this marked the beginning of a new age, though none of us knew exactly what sort of era it would be. If Apophis destroyed the world, I guessed it would be the Age of Extremely Short Lives.
I'd expected to see Amos sitting at the foot of the pharaoh's throne. That was the traditional place for the Chief Lector, symbolizing his role as the pharaoh's main advisor. Of course, the pharaohs rarely needed advising these days, as they'd all been dead for several thousand years.
The dais was empty.
That stumped me. I'd never considered where the Chief Lector hung out when he wasn't on display. Did he have a dressing room, possibly with his name and a little star on the door?
'There.' Leonid pointed.
Once again, my clever Russian friend was right. On the back wall, behind the throne, a faint line of light shone along the floor – the bottom edge of a door.
'A creepy secret entrance,' I said. 'Well done, Leonid.'
On the other side, we found a sort of war room. Amos and a young woman in camouflage clothes stood at opposite ends of a large table inlaid with a full-colour world map. The table's surface was crowded with tiny figurines – painted ships, monsters, magicians, cars and markers with hieroglyphs.
Amos and the camouflage girl were so engrossed in their work, moving figurines across the map, they didn't notice us at first.
Amos wore traditional linen robes. With his barrel-shaped figure, they made him look a bit like Friar Tuck, except with darker skin and cooler hair. His braided locks were decorated with gold beads. His round glasses flashed as he studied the map. Draped round his shoulders was the leopard-skin cape of the Chief Lector.
As for the young woman... oh, gods of Egypt. It was _Zia._
I'd never seen her in modern clothes before. She wore camouflage cargo pants, hiking boots and an olive-coloured tank top that flattered her coppery skin. Her black hair was longer than I remembered. She looked so much more grown-up and gorgeous than she'd been six months ago I was glad Carter hadn't come along. He would've had difficulty picking up his jaw from the floor.
[Yes, you would have, Carter. She looked quite stunning, in a Commando Girl sort of way.]
Amos moved one of the figurines across the map. 'Here,' he told Zia.
'All right,' she said. 'But that leaves Paris undefended.'
I cleared my throat. 'Are we interrupting?'
Amos turned and broke into a grin. 'Sadie!'
He crushed me in a hug, then rubbed my head affectionately.
'Ow,' I said.
He chuckled. 'I'm sorry. It's just so good to see you.' He glanced at Leonid. 'And this is –'
Zia cursed. She wedged herself between Amos and Leonid. 'He's one of the Russians! Why is _he_ here?'
'Calm down,' I told her. 'He's a friend.'
I explained about Leonid's appearance at the dance. Leonid tried to help, but he kept slipping into Russian.
'Wait,' Amos said. 'Let's make this easier.'
He touched Leonid's forehead. ' _Med-wah._ '
In the air above us, the hieroglyph for _Speak_ burned red:
'There,' Amos said. 'That should help.'
Leonid's eyebrows shot up. 'You speak Russian?'
Amos smiled. 'Actually for the next few minutes, we'll all be speaking Ancient Egyptian, but it will sound to each of us like our native tongue.'
'Brilliant,' I said. 'Leonid, you'd best make the most of your time.'
Leonid took off his army cap and fidgeted with the brim. 'Sarah Jacobi and her lieutenant, Kwai... they mean to attack you.'
'We know that,' Amos said dryly.
'No, you don't understand!' Leonid's voice trembled with fear. 'They are evil! They are working with Apophis!'
Perhaps it was a coincidence, but when he said that name several figurines on the world map sparked and melted. My heart felt much the same way.
'Hold on,' I said. 'Leonid, how do you know this?'
His ears turned pink. 'After the death of Menshikov, Jacobi and Kwai came to our nome. We gave them refuge. Soon Jacobi took over, but my comrades did not object. They, ah, hate the Kanes very much.' He looked at me guiltily. 'After you broke into our headquarters last spring... well, the other Russians blame you for Menshikov's death and the rise of Apophis. They blame you for everything.'
'Quite used to that,' I said. 'You didn't feel the same?'
He pinched his oversized cap. 'I saw your power. You defeated the _tjesu-heru_ monster. You could have destroyed me, but you didn't. You did not seem evil.'
'Thanks for that.'
'After we met, I became curious. I began reading old scrolls, learning to channel the power of the god Shu. I have always been a good air elementalist.'
Amos grunted. 'That took courage, Leonid. Exploring the path of the gods on your own in the middle of the Russian Nome? You were brave.'
'I was foolhardy.' Leonid's forehead was damp with sweat. 'Jacobi has killed magicians for lesser crimes. One of my friends, an old man named Mikhail, he once made the mistake of saying all Kanes might not be bad. Jacobi arrested him for treason. She gave him to Kwai, who does magic with – with lightning... terrible things. I heard Mikhail screaming in the dungeon for three nights before he died.'
Amos and Zia exchanged grave looks. I had a feeling this wasn't the first time they'd heard about Kwai's torture methods.
'I'm so sorry,' Amos said. 'But how can you be sure Jacobi and Kwai are working for Apophis?'
The young Russian glanced at me for reassurance.
'You can trust Amos,' I promised. 'He'll protect you.'
Leonid chewed his lip. 'Yesterday I was in one of the chambers deep under the Hermitage, a place I thought was secret. I was studying a scroll to summon Shu – very forbidden magic. I heard Jacobi and Kwai approaching, so I hid. I overheard the two of them speaking, but their voices were... splintered. I don't know how to explain.'
'They were possessed?' Zia asked.
'Worse,' Leonid said. 'They were each channelling dozens of voices. It was like a war council. I heard many monsters and demons. And presiding over the meeting was one voice, deeper and more powerful than the rest. I'd never heard anything like it, as if darkness could speak.'
'Apophis,' Amos said.
Leonid had gone very pale. 'Please understand, most magicians in St Petersburg, they are not evil. They are only scared and desperate to survive. Jacobi has convinced them she will save them. She has misled them with lies. She says the Kanes are demons. But she and Kwai... _they_ are the monsters. They are no longer human. They have set up a camp at Abu Simbel. From there, they will lead the rebels against the First Nome.'
Amos turned to his map. He traced his finger south along the River Nile to a small lake. 'I sense nothing at Abu Simbel. If they are there, they've managed to hide themselves completely from my magic.'
'They are there,' Leonid promised.
Zia scowled. 'Under our very noses, within easy striking distance. We should've killed the rebels at Brooklyn House when we had the chance.'
Amos shook his head. 'We are servants of Ma'at – order and justice. We don't kill our enemies for things they might do in the future.'
'And now our enemies will kill us,' Zia said.
On the table map, two more figurines sparked and melted in Spain. A miniature ship broke into pieces off the coast of Japan.
Amos grimaced. 'More losses.'
He chose a cobra figurine from Korea and pushed it towards the shipwreck. He swept away the melted magicians from Spain.
'What _is_ that map?' I asked.
Zia moved a hieroglyph token from Germany to France. 'Iskandar's war map. As I once told you, he was an expert at statuary magic.'
I remembered. The old Chief Lector had been so good he'd made a replica of Zia herself... but I decided not to bring that up.
'Those tokens stand for actual forces,' I guessed.
'Yes,' Amos said. 'The map shows us our enemy's movements, at least most of them. It also allows us to send our forces by magic to where they are needed.'
'And, _uh_ , how are we doing?'
His expression told me all I needed to know.
'We are spread too thin,' Amos said. 'Jacobi's followers strike wherever we are weakest. Apophis sends his demons to terrorize our allies. The attacks seem coordinated.'
'Because they are,' Leonid said. 'Kwai and Jacobi are under the serpent's control.'
I shook my head in disbelief. 'How could Kwai and Jacobi be so stupid? Don't they understand Apophis is going to destroy the world?'
'Chaos is seductive,' Amos said. 'No doubt Apophis has made them promises of power. He whispers in their ears, convincing them they are too important to be destroyed. They believe they can make a new world better than the old, and the change is worth any price – even mass annihilation.'
I couldn't grasp how anyone could be so deluded, but Amos spoke as if he understood. Of course, Amos had been through this. He'd been possessed by Set, god of evil and Chaos. Compared to Apophis, Set was a minor nuisance, but he'd still been able to turn my uncle – one of the most powerful magicians in the world – into a helpless puppet. If Carter and I hadn't defeated Set and forced him to return to the Duat... well, the consequences wouldn't have been pretty.
Zia picked up a falcon figurine. She moved it towards Abu Simbel, but the little statue began to steam. She was forced to drop it.
'They've put up powerful wards,' she said. 'We won't be able to eavesdrop.'
'They will attack in three days,' Leonid said. 'At the same time, Apophis will rise – at dawn on the autumn equinox.'
' _Another_ equinox?' I grumbled. 'Didn't that _last_ bit of nastiness happen on one of those? You Egyptians have an unhealthy obsession with equinoxes.'
Amos gave me a stern look. 'Sadie, as I'm sure you're aware, the equinox is a time of great magic significance, when day and night are equal. Besides, the autumn equinox marks the last day before darkness overtakes the light. It is the anniversary of Ra's retreat into the heavens. I feared that Apophis might make his move at that time. It's a most inauspicious day.'
'Inauspicious?' I frowned. 'But inauspicious is bad. Why would they... oh.'
I realized for the forces of Chaos, our bad days must've been their good days. That meant they probably had a lot of good days.
Amos leaned on his staff. His hair seemed to be turning grey before my eyes. I remembered Michel Desjardins, the last Chief Lector, and how quickly he had aged. I couldn't bear the idea of that happening to Amos.
'We don't have the strength to defeat our enemies,' he said. 'I will have to use other means.'
'Amos, no,' Zia said. 'Please.'
I wasn't sure what they were talking about. Zia sounded horrified, and anything that scared her I didn't want to know about.
'Actually,' I said, 'Carter and I have a plan.'
I told them about our idea of using Apophis's own shadow against him. Perhaps saying this in front of Leonid was reckless, but he had risked his life to warn us about Sarah Jacobi's plans. He had trusted me. The least I could do was return the favour.
When I finished explaining, Amos gazed at his map. 'I've never heard of such magic. Even if it's possible –'
'It _is_ ,' I insisted. 'Why else would Apophis delay his Doomsday attack so he could track down and destroy every scroll by this fellow Setne? Apophis is afraid we'll figure out the spell and stop him.'
Zia crossed her arms. 'But you can't. You just said all copies were destroyed.'
'We'll ask Thoth for help,' I said. 'Carter's on his way there now. And in the meantime... I have an errand to run. I may be able to test our theory about shadows.'
'How?' Amos asked.
I told him what I had in mind.
He looked as if he wanted to object, but he must've seen the defiance in my eyes. We're related, after all. He knows how stubborn Kanes can be when they set their minds to something.
'Very well,' he said. 'First you must eat and rest. You can leave at dawn. Zia, I want you to go with her.'
Zia looked startled. 'Me? But I might... I mean, is it wise?'
Again I got the feeling I'd missed an important conversation. What had Amos and Zia been discussing?
'You'll be fine,' Amos assured her. 'Sadie will need your help. And I will arrange for someone else to watch Ra during the day.'
She looked quite nervous, which wasn't like her. Zia and I had had our differences in the past, but she'd never been short of confidence. Now I almost felt worried for her.
'Cheer up,' I told her. 'It'll be a laugh. Quick trip to the Netherworld, fiery lake of doom. What could go wrong?'
CARTER
## 7. I Get Strangled by an Old Friend
SO, YEAH.
Sadie goes off on a side adventure with some guy, leaving me to do the boring work of figuring out how to save the world. Why does this sound familiar? Oh, right. That's the way Sadie always is. If it's time to move forward, you can count on her to veer sideways on some ADHD tangent of her own.
[Why are you thanking me, Sadie? That wasn't a compliment.]
After the Brooklyn Academy dance, I was pretty miffed. Bad enough being forced to slow-dance with Sadie's friend Lacy. But passing out on the dance floor, waking up with Lacy snoring in my armpit and then finding out I'd missed visits from two gods – that was just embarrassing.
After Sadie and the Russian guy left, I got our crew back to Brooklyn House. Walt was confused to see us so soon. I pulled him and Bast aside for a quick conference on the terrace. I explained what Sadie had told me about Shu, Anubis and the Russian dude Leonid.
'I'll take Freak to Memphis,' I said. 'Be back as soon as I talk to Thoth.'
'I'm going with you,' Walt said.
Sadie had told me to take him along, of course, but, looking at him now, I had second thoughts. Walt's cheeks were sunken. His eyes were glassy. I was alarmed by how much worse he looked since just yesterday. I know this is horrible, but I couldn't help thinking about Egyptian burial practices – how they'd pack a body with embalming salts to slowly dry it up from the inside. Walt looked like he'd been started on that process.
'Look, man,' I said, 'Sadie asked me to keep you safe. She's worried about you. So am I.'
He clenched his jaw. 'If you plan on using a shadow for your spell, you'll have to capture it with that figurine. You'll need a _sau_ , and I'm the best you've got.'
Unfortunately, Walt was right. Neither Sadie nor I had the skill to capture a shadow, if that were even possible. Only Walt had that kind of charm-making talent.
'All right,' I muttered. 'Just... keep your head down. I don't want my sister going nuclear on me.'
Bast poked Walt's arm, the way a cat might nudge a bug to see if it were still alive. She sniffed his hair.
'Your aura is weak,' Bast said, 'but you should be all right to travel. Try not to exert yourself. No magic unless absolutely necessary.'
Walt rolled his eyes. 'Yes, Mother.'
Bast seemed to like that.
'I'll watch the other kittens,' she promised. ' _Er_ , I mean initiates. You two be careful. I don't have much love for Thoth, and I don't want you caught up in his problems.'
'What problems?' I asked.
'You'll see. Just come back to me. All this guard duty is cutting into my nap schedule!'
She shooed us towards Freak's stable and headed back downstairs, muttering something about catnip.
We hitched up the boat. Freak squawked and buzzed his wings, anxious to go. He looked like he'd had a good rest. Besides, he knew that a journey meant more frozen turkeys for him.
Soon we were flying over the East River.
Our ride through the Duat seemed bumpier than usual, like airplane turbulence, except with ghostly wailing and heavy fog. I was glad I'd had a light dinner. My stomach churned.
The boat shuddered as Freak brought us out of the Duat. Below us spread a different night-time landscape – the lights of Memphis, Tennessee, curving along the banks of the Mississippi River.
On the shoreline rose a glassy black pyramid – an abandoned sports arena that Thoth had appropriated for his home. Bursts of multicoloured light peppered the air, reflections rippling across the pyramid. At first I thought Thoth was hosting a fireworks exhibition. Then I realized his pyramid was under attack.
Clambering up the sides was a gruesome assortment of demons – humanoid figures with chicken feet or paws or insect legs. Some had fur. Some had scales or shells like tortoises. Instead of heads, many had weapons or tools sprouting from their necks – hammers, swords, axes, chain saws, even a few screwdrivers.
At least a hundred demons were climbing towards the top, digging their claws into the seams of the glass. Some tried to smash their way through, but, wherever they struck, the pyramid flickered with blue light, repelling their attacks. Winged demons swirled through the air, screeching and diving at a small group of defenders.
Thoth stood at the peak. He looked like a scruffy college lab assistant in a white medical coat, jeans and T-shirt, a day-old beard and wild Einstein hair – which doesn't sound very intimidating, but you should see him in combat. He threw glowing hieroglyphs like grenades, causing iridescent explosions all around him. Meanwhile his assistants, a troop of baboons and long-beaked birds called ibises, engaged the enemy. The baboons slammed basketballs into the demons, sending them toppling back down the pyramid. The ibises ran between the monsters' legs, jabbing their beaks in the most sensitive places they could find.
As we flew closer, I lowered my vision into the Duat. The scene there was even scarier. The demons were connected by red coils of energy that formed one massive translucent serpent. The monster encircled the entire pyramid. At the top, Thoth shone in his ancient form – a giant, white-kilted man with the head of an ibis, hurling bolts of energy at his enemies.
Walt whistled. 'How can the mortals not notice a battle like that?'
I wasn't sure, but I remembered some of the recent disaster news. Huge storms had been causing floods all along the Mississippi River, including here in Memphis. Hundreds of people had been displaced. Magicians might be able to see what was really happening, but any regular mortals still in the city probably thought this was just a major thunderstorm.
'I'll help Thoth,' I said. 'You stay in the boat.'
'No,' Walt said. 'Bast said I should use magic only in an emergency. This qualifies.'
I knew Sadie would kill me if I let Walt get hurt. On the other hand, Walt's tone told me he wasn't going to back down. He can be almost as stubborn as my sister when he wants to be.
'Fine,' I said. 'Hold on.'
A year ago, if I'd faced a fight like this, I would have curled into a ball and tried to hide. Even our battle at the Red Pyramid last Christmas seemed minor compared to dive-bombing an army of demons with no back-up except one sick guy and a slightly dysfunctional griffin.
But a lot had happened in the past year. Now this was just another bad day in the life of the Kane family.
Freak came screaming down out of the night sky and banked hard to the right, shooting across the side of the pyramid. He gulped down smaller demons and shredded the larger ones with his buzz-saw wings. Some that survived got run over by our boat.
As Freak began to climb again, Walt and I jumped out, scrambling for footing on the glassy slope. Walt threw an amulet. In a flash of light, a golden sphinx appeared, with a lion's body and the head of woman. After our experience at the Dallas Museum, I didn't much care for sphinxes, but thankfully this one was on our side.
Walt jumped on its back and rode into battle. The sphinx snarled and pounced on a reptilian demon, tearing it to pieces. Other monsters scattered. I couldn't blame them. A massive gold lion would have been scary enough, but the growling woman's head made it even more horrifying, with merciless emerald eyes, a shining Egyptian crown and a fanged mouth with way too much lipstick.
As for me, I summoned my _khopesh_ from the Duat. I called on the power of Horus, and the glowing blue avatar of the war god formed around me. Soon I was encased in a twenty-foot-tall hawk-headed apparition.
I stepped forward. The avatar mirrored my movements. I swiped my sword at the nearest demons, and the avatar's massive glowing blade ploughed them down like bowling pins. Two of the monsters actually had bowling pins for heads, so I guess that was appropriate.
The baboons and ibises were slowly making headway against the surge of demons. Freak flew round the pyramid, snapping up winged demons or smacking them out of the air with his boat.
Thoth kept flinging hieroglyphic grenades.
'Bloated!' he cried. The corresponding hieroglyph flew through the air, bursting against a demon's chest in a spray of light. Instantly, the demon swelled like a water balloon and rolled screaming down the pyramid.
'Flat!' Thoth blasted another demon, who collapsed and shrivelled into a monster-shaped doormat.
'Intestinal problems!' Thoth yelled. The poor demon who got zapped with that one turned green and doubled over.
I waded through monsters, tossing them aside and slicing them to dust. Everything was going great until a winged demon did a kamikaze dive into my chest. I toppled backwards, slamming against the pyramid with such force that I lost my concentration. My magical armour dissolved. I would've skidded all the way down the pyramid if the demon hadn't grabbed my throat and held me in place.
'Carter Kane,' he hissed. 'You are stupidly persistent.'
I recognized that face – like an anatomy-class cadaver with muscle and sinew but no skin. His lidless eyes glowed red. His fangs were bared in a murderous grin.
'You,' I grunted.
'Yes,' the demon chuckled, his claws tightening round my neck. 'Me.'
Face of Horror – Set's lieutenant from the Red Pyramid, and the secret mouthpiece of Apophis. We'd killed him in the shadow of Washington Monument, but I guess that didn't mean anything. Now he was back, and, judging by his rasping voice and glowing red eyes, he was still possessed by my least favourite snake.
I didn't remember his being able to fly, but now leathery bat wings sprouted from his shoulders. He straddled me with his chicken legs, his hands digging into my windpipe. His breath smelled like fermented juice and skunk spray.
'I could have killed you many times,' the demon said, 'but you interest me, Carter.'
I tried to fight him off. My arms had turned to lead. I could barely hold my sword.
Around us, the sounds of battle became muted. Freak flew overhead, but his wings beat so sluggishly I could actually see them. A hieroglyph exploded in slow motion like dye in water. Apophis was dragging me deeper into the Duat.
'I can feel your turmoil,' said the demon. 'Why do you fight this hopeless battle? Don't you realize what will happen?'
Images raced through my mind.
I saw a landscape of shifting hills and fiery geysers. Winged demons turned in the sulphurous sky. Spirits of the dead skittered across the hills, wailing in desperation and clawing for handholds. They were all being pulled in the same direction – towards a blot of darkness on the horizon. Whatever it was, its gravity was as powerful as a black hole. It sucked in the spirits, bending the hills and plumes of fire towards it. Even the demons in the air struggled.
Huddled in the shelter of a cliff, the glowing white form of a woman tried to anchor herself against the dark current. I wanted to cry. The woman was my mother. Other ghosts flew past her, wailing helplessly. My mom tried to reach out, but she couldn't save them.
The scene shifted. I saw the Egyptian desert at the edge of Cairo under a blazing sun. Suddenly the sands erupted. A giant red serpent rose from the Underworld. He lunged at the sky and somehow, impossibly, swallowed the sun in a single gulp. The world darkened. Frost spread across the dunes. Cracks appeared in the ground. The landscape crumbled. Whole neighbourhoods of Cairo sank into chasms. A red ocean of Chaos swelled up from the Nile, dissolving the city and desert, washing away pyramids that had stood for millennia. Soon there was nothing left but a boiling sea under a starless black sky.
'No gods can save you, Carter.' Apophis sounded almost sympathetic. 'This fate has been decreed since the beginning of time. Yield to me, and I will spare you and those you love. You will ride the Sea of Chaos. You will be master of your own destiny.'
I saw an island floating across the boiling ocean – a small patch of green earth like an oasis. My family and I could be together on that island. We could survive. We could have anything we wanted just by imagining it. Death would mean nothing.
'All I ask is a token of goodwill,' Apophis urged. 'Give me Ra. I know you hate him. He represents everything that is wrong with your mortal world. He has grown senile, rotten, weak and useless. Surrender him to me. I will spare you. Think on this, Carter Kane. Have the gods promised you anything as fair?'
The visions faded. Face of Horror grinned down at me, but suddenly his features contorted in pain. A fiery hieroglyph burned across his forehead – the symbol for _desiccate_ – and the demon crumbled to dust.
I gasped for breath. My throat felt like it was packed with hot coals.
Thoth stood over me, looking grim and tired. His eyes swirled with kaleidoscopic colours, like portals to another world.
'Carter Kane.' He offered me a hand and helped me up.
All the other demons were gone. Walt stood at the peak of the pyramid with the baboons and ibises, who were climbing over the golden sphinx lady like she was a merry-go-round animal. Freak hovered nearby, looking full and happy from eating so many demons.
'You shouldn't have come,' Thoth chided. He brushed demon dust off his T-shirt, which had a flaming heart logo and the words HOUSE OF BLUES. 'It was much too dangerous, especially for Walt.'
'You're welcome,' I croaked. 'It looked like you needed help.'
'The demons?' Thoth waved dismissively. 'They'll be back just before sunrise. They've been attacking every six hours for the past week. Quite annoying.'
'Every six hours?' I tried to imagine that. If Thoth had been fighting off an army like that several times a day for a week... I didn't see how even a god could have that much power.
'Where are the other gods?' I asked. 'Shouldn't they be helping you?'
Thoth wrinkled his nose as if he smelled a demon with intestinal problems. 'Perhaps you and Walt should come inside. Now that you're here, we have a lot to talk about.'
I'll say this for Thoth. He knew how to decorate a pyramid.
The former arena's basketball court was still there, no doubt so his baboons could play. (Baboons love basketball.) The JumboTron still hung from the ceiling, flashing a series of hieroglyphs that announced things like: GO TEAM! DEFENCE! and THOTH 25 – DEMONS 0 in Ancient Egyptian.
The stadium seating had been replaced with a series of tiered balconies. Some were lined with computer stations, like mission control for a rocket launch. Others had chemistry tables cluttered with beakers, Bunsen burners, vials of smoking goo, jars of pickled organs and stranger things. The nosebleed section was devoted to scroll cubbies – a library easily as big as the one in the First Nome. And behind the left backboard rose a three-storey-tall whiteboard covered in computations and hieroglyphs.
Hanging from the girders, instead of championship banners and retired numbers, were black tapestries embroidered with gold incantations.
Courtside was Thoth's living area – a free-standing gourmet kitchen, a plush collection of couches and easy chairs, piles of books, buckets of Lego and Tinker Toys, a dozen flat-screen TVs showing different news programmes and documentaries, and a small forest of electric guitars and amplifiers – everything a scatterbrained god needed to be able to do twenty things at once.
Thoth's baboons took Freak into the locker room to groom him and let him rest. I think they were worried he might eat the ibises, since they did look a bit like turkeys.
Thoth turned to Walt and me, looking us over critically. 'You need rest. Then I'll fix you some dinner.'
'We don't have time,' I said. 'We have to –'
'Carter Kane,' Thoth scolded. 'You've just battled Apophis, got the Horus knocked out of you, been dragged through the Duat and half strangled. You're no good to anyone until you get some sleep.'
I wanted to protest, but Thoth pressed his hand to my forehead. Weariness washed over me.
'Rest,' Thoth insisted.
I collapsed on the nearest couch.
I'm not sure how long I slept, but Walt got up first. When I woke, he and Thoth were deep in conversation.
'No,' Thoth said. 'It's never been done. And I'm afraid you don't have time....' He faltered when he noticed me sitting up. 'Ah. Good, Carter. You're awake.'
'What did I miss?'
'Nothing,' he said, a little too cheerily. 'Come and eat.'
His kitchen counter was laden with fresh-cut brisket, sausage, ribs and cornbread, plus an industrial-sized dispenser of ice tea. Thoth had once told me that barbecue was a form of magic, and I guess he was right. The smell of food made me temporarily forget my troubles.
I scarfed down a brisket sandwich and drank two glasses of tea. Walt nibbled on a rib, but he didn't seem to have much of an appetite.
Meanwhile Thoth picked up a Gibson guitar. He struck a power chord that shook the arena floor. He'd got better since I'd last heard him. The chord actually sounded like a chord, not like a mountain goat being tortured.
I gestured around with a piece of cornbread. 'This place is looking good.'
Thoth chuckled. 'Better than my last headquarters, eh?'
The first time Sadie and I had crossed paths with the god of knowledge, he'd been holed up at a local university campus. He had tested our worth by sending us on a quest to trash Elvis Presley's house (long story), but hopefully we were past the testing phase now. I preferred hanging out courtside eating barbecue.
Then I thought about the visions Face of Horror had shown me – my mother in danger, a darkness swallowing the souls of the dead, the world dissolving in a sea of Chaos – except for one small island floating across the waves. The memory kind of killed my appetite.
'So...' I pushed my plate away. 'Tell me about the demon attacks. And what were you saying to Walt?'
Walt stared at his half-eaten pork rib.
Thoth strummed a minor chord. 'Where to start... ? The attacks began seven days ago. I'm cut off from the other gods. They haven't come to my rescue, I imagine, because they're having similar problems. Divide and conquer – Apophis understands that basic military principle. Even if my brethren _could_ help me... well, they have other priorities. Ra was recently brought back, as you may recall.'
Thoth gave me a hard look, like I was an equation he couldn't balance. 'The sun god must be guarded on his nightly journey. That takes a lot of godpower.'
My shoulders sagged. I didn't need one more thing to feel guilty about. I also didn't think it was fair of Thoth to act so critical of me. Thoth had been on our side, more or less, about bringing back the sun god. Maybe seven days of demon attacks had started to change his mind.
'Can't you just leave?' I asked.
Thoth shook his head. 'Perhaps you can't see so deeply into the Duat, but the power of Apophis has completely encircled this pyramid. I am quite stuck.'
I gazed up at the arena's ceiling, which suddenly seemed much lower. 'Which means... we're stuck too?'
Thoth waved aside the question. ' _You_ should be able to pass back through. The serpent's net is designed to catch a god. You and Walt aren't large or important enough to be caught.'
I wondered if that were true, or if Apophis was allowing me to come and go – to have the choice of surrendering Ra.
_You interest me, Carter_ , Apophis had said. _Yield to me, and I will spare you._
I took a deep breath. 'But, Thoth, if you're on your own... I mean, how much longer can you last?'
The god brushed at his lab coat, which was covered with scribbles in a dozen languages. The word _time_ fluttered off his sleeve. Thoth caught it, and suddenly he was checking a gold pocket watch.
'Let's see. Judging from the weakening of the pyramid's defences and the rate at which my power is being expended, I'd say I could withstand nine more attacks, or just over two days, which would take us to dawn on the equinox. Ha! That can't be a coincidence.'
'And then?' Walt asked.
'Then my pyramid will be breached. My minions will be killed. I'm guessing Doomsday will happen all over, in fact. The fall equinox would be a sensible time for Apophis to rise. He'll probably cast me into the abyss, or possibly scatter my essence across the universe in a billion pieces. Hmm... the physics of a god's death.' His pocket watch turned into a pen. He scribbled something on the neck of his guitar. 'That would make an excellent research paper.'
'Thoth,' Walt prompted. 'Tell Carter what you told me, about why you're being targeted.'
'I thought that was obvious,' Thoth said. 'Apophis wants to distract me from helping you. That _is_ why you've come, isn't it? To find out about the serpent's shadow?'
For a moment I was too stunned to speak. 'How did you know?'
' _Please._ ' Thoth played a Jimi Hendrix riff, then set down his guitar. 'I _am_ the god of knowledge. I knew sooner or later you'd come to the conclusion that your only hope of victory was a shadow execration.'
'A shadow execration,' I repeated. 'That's an actual spell with an actual name? It could work?'
'In theory.'
'And you didn't volunteer this information – _why?_ '
Thoth snorted. 'Knowledge of any value can't be given. It must be sought and earned. You're a teacher now, Carter. You should know this.'
I wasn't sure whether to strangle him or hug him. 'So, I'm seeking the knowledge. I'm earning the knowledge. How do I defeat Apophis?'
'I'm so glad you asked!' Thoth beamed at me with his multicoloured eyes. 'Unfortunately, I can't tell you.'
I glanced at Walt. 'Do you want to kill him, or should I?'
'Now, now,' Thoth said. 'I can guide you a little. But you'll have to connect the freckles, as they say.'
'Dots,' I said.
'Yes,' he said. 'You're on the right track. The _sheut_ could be used to destroy a god, or even Apophis himself. And, yes, like all sentient beings, Apophis has a shadow, though he keeps that part of his soul well hidden and well guarded.'
'So where is it?' I asked. 'How do we use it?'
Thoth spread his hands. 'The second question I can't answer. The first question I'm not allowed to answer.'
Walt shoved his plate aside. 'I've been trying to get it out of him, Carter. For a god of knowledge, he isn't very helpful.'
'Come on, Thoth,' I said. 'Can't we do a quest for you or something? Couldn't we blow up Elvis's house again?'
'Tempting,' the god said. 'But you must understand, giving a mortal the location of an immortal's shadow – even Apophis's – would be a grave crime. The other gods already think I'm a sell-out. Over the centuries, I've divulged too many secrets to mankind. I taught you the art of writing. I taught you magic and founded the House of Life.'
'Which is why magicians still honour you,' I said. 'So help us one more time.'
'And give humans knowledge that could be used to destroy the gods?' Thoth sighed. 'Can you understand why my brethren might object to such a thing?'
I clenched my fists. I thought about my mother's spirit huddling beneath a cliff, fighting to stay put. The dark force _had_ to be Apophis's shadow. Apophis had shown me that vision to make me despair. As his power grew, his shadow grew stronger, too. It was pulling in the spirits of the dead, consuming them.
I could guess the shadow was somewhere in the Duat, but that didn't help. It was like saying _somewhere in the Pacific Ocean_. The Duat was huge.
I glared at Thoth. 'Your other option is not to help us and let Apophis destroy the world.'
'Point taken,' he admitted, 'which is why I'm still talking to you. There _is_ a way you could find the shadow's location. Long ago, when I was young and naive, I wrote a book – a field study, of sorts – called the Book of Thoth.'
'Catchy name,' Walt muttered.
' _I_ thought so!' Thoth said. 'At any rate, it described every form and disguise each god can take, their most secret hiding places – all sorts of embarrassing details.'
'Including how to find their shadows?' I asked.
'No comment. At any rate, I never meant for humans to read the book, but it was stolen in ancient times by a crafty magician.'
'Where is it now?' I asked. Then I held up my hands. 'Wait... let me guess. You can't tell us.'
'Honestly, I don't know,' Thoth said. 'This crafty magician hid the book. Fortunately he died before he could take full advantage of it, but he _did_ use its knowledge to formulate a number of spells, including the shadow execration. He wrote down his thoughts in a special variation of the Book of Overcoming Apophis.'
'Setne,' I said. 'That's the magician you're talking about.'
'Indeed. His spell was only theoretical, of course. Even _I_ never had that knowledge. And, as you know, all copies of his scroll have now been destroyed.'
'So it's hopeless,' I said. 'Dead end.'
'Oh, no,' Thoth said. 'You could ask Setne himself. He wrote the spell. He hid the Book of Thoth that, ahem, may or may not describe the shadow's location. If he were so inclined, he could help you.'
'But hasn't Setne been dead for thousands of years?'
Thoth grinned. 'Yes. And that's only the first problem.'
Thoth told us about Setne, who'd apparently been pretty famous in Ancient Egypt – like Robin Hood, Merlin and Attila the Hun rolled into one. The more I heard, the less I wanted to meet him.
'He was a pathological liar,' Thoth said. 'A scoundrel, a traitor, a thief and a brilliant magician. He prided himself on stealing books of knowledge, including mine. He battled monsters, adventured in the Duat, conquered gods and broke into sacred tombs. He created curses that couldn't be lifted and unearthed secrets that should have stayed buried. He was quite the evil genius.'
Walt tugged at his amulets. 'Sounds like you admire him.'
The god gave him a sidelong grin. 'Well, I appreciate the pursuit of knowledge, but I couldn't endorse Setne's methods. He'd stop at nothing to possess the secrets of the universe. He wanted to be a god, you see – not the _eye_ of a god. A full-fledged immortal.'
'Which is impossible,' I guessed.
'Hard, not impossible,' Thoth said. 'Imhotep, the first mortal magician – he was made a god after his death.' Thoth turned towards his computers. 'That reminds me, I haven't seen Imhotep in millennia. I wonder what he's up to. Perhaps I should Google him –'
'Thoth,' Walt said, 'concentrate.'
'Right. So, Setne. He created this spell for destroying any being – even a god. I could never endorse such knowledge falling into the hands of a mortal, but, hypothetically speaking, if you needed the spell to defeat Apophis, you might be able to convince _Setne_ to teach you the enchantment and lead you to the shadow of Apophis.'
'Except Setne's dead,' I said. 'We keep coming back to that.'
Walt sat up. 'Unless... you're suggesting we find his spirit in the Underworld. But, if Setne was so evil, wouldn't Osiris have condemned him in the Hall of Judgement? Ammit would've eaten his heart, and he would have ceased to exist.'
'Normally, yes,' Thoth said. 'But Setne is a special case. He's quite... persuasive. Even before the court of the Underworld, he was able to, ah, manipulate the legal system. Many times, Osiris sentenced him to oblivion, but Setne always managed to evade punishment. He got a lighter sentence, or he made a plea bargain, or he simply escaped. He's managed to survive – as a spirit, at least – all these aeons.'
Thoth turned his swirling eyes towards me. 'But recently, Carter Kane, your father became Osiris. He's been cracking down on rebellious ghosts, trying to restore Ma'at to the Underworld. The next time the sun sets, approximately fourteen hours from now, Setne is scheduled for a new trial. He will come before your father. And this time –'
'My dad won't let him go.' I felt like the demon's hands were closing round my throat again.
My father was fair but stern. He didn't take excuses from anyone. All the years we'd travelled together, I could never even get away with leaving my shirt untucked. If Setne was as bad as Thoth said, my father would show him no mercy. He'd toss this guy's heart to Ammit the Devourer like it was a doggie biscuit.
Walt's eyes shone with excitement. He looked more animated than I'd seen him in a long time. 'We can plead with your dad,' he said. 'We can get Setne's trial delayed, or ask for a reduced sentence in exchange for Setne's help. The laws of the Underworld allow that.'
I frowned. 'How do you know so much about dead people's court?'
I regretted saying that immediately. I realized that he'd probably been preparing himself to face that courtroom. Maybe that's what he'd been discussing with Thoth earlier.
_I'm afraid you don't have much time_ , Thoth had said.
'Sorry, man,' I said.
'It's okay,' Walt said. 'But we have to try. If we can convince your dad to spare Setne –'
Thoth laughed. 'That would be amusing, wouldn't it? If Setne got off yet again, because his evil ways were the only thing that might save the world?'
'Hilarious,' I said. The brisket sandwich wasn't sitting well in my stomach. 'So you're suggesting we go to my father's court and try to save the ghost of an evil psychotic magician. Then we ask this ghost to lead us to Apophis's shadow and teach us how to destroy it, while trusting that he won't escape, kill us or betray us to the enemy.'
Thoth nodded enthusiastically. 'You'd have to be crazy! I certainly hope you are.'
I took a deep breath. 'I guess I'm crazy.'
'Excellent!' Thoth cheered. 'One more thing, Carter. To make this work, you'll need Walt's help, but he's running out of time. His only chance –'
'It's fine,' Walt snapped. 'I'll tell him myself.'
Before I could ask what he meant, the overtime buzzer blared from the arena's speakers.
'It's almost dawn,' Thoth said. 'You two had better leave, before the demons return. Good luck. And, by all means, give Setne my regards – if you live that long, of course.'
CARTER
## 8. My Sister, the Flowerpot
THE RIDE BACK WASN'T FUN.
Walt and I held on to the boat while our teeth chattered and our eyes jiggled. The magic fog had turned the colour of blood. Ghostly voices whispered angrily, like they'd decided to riot and loot the ethereal world.
Sooner than I expected, Freak pushed his way out of the Duat. We found ourselves over the New Jersey dockyards, our boat trailing steam as Freak bobbed wearily through the air. In the distance, the Manhattan skyline gleamed gold in the sunrise.
Walt and I hadn't spoken during the trip. The Duat tends to put a damper on conversation. Now he regarded me sheepishly.
'I should explain some things,' he said.
I can't pretend I wasn't curious. As his sickness had progressed, Walt had got more and more secretive. I wondered what he'd been talking about with Thoth.
But it wasn't my business. After Sadie learned my secret name last spring and got a free tour of my innermost thoughts, I'd become sensitive about respecting people's privacy.
'Look, Walt, it's your personal life,' I said. 'If you don't want to tell –'
'But it's not just personal. You need to know what's going on. I – I won't be around much longer.'
I gazed down at the harbour, the Statue of Liberty passing below us. For months I'd known Walt was dying. It never got easier to accept. I remembered what Apophis had said at the Dallas Museum: Walt wouldn't live long enough to see the end of the world.
'Are you sure?' I asked. 'Isn't there some way –?'
'Anubis is sure,' he said. 'I've got until sunset tomorrow, at the very latest.'
I didn't want to hear another impossible deadline. By sunset tonight, we had to save the ghost of an evil magician. By sunset tomorrow, Walt would die. And the sunrise after that, if we were really lucky, we could look forward to Doomsday.
I never liked being thwarted. Whenever I felt like something was impossible, I usually tried even harder out of sheer stubbornness.
But at this point I felt like Apophis was having a good laugh at my expense.
_Oh, you're not a quitter?_ he seemed to be asking. _How about now? What if we give you a few more impossible tasks? Are you a quitter now?_
Anger made a small hard knot in my gut. I kicked the side of the boat and nearly broke my foot.
Walt blinked. 'Carter, it's –'
' _Don't_ say it's all right!' I snapped. 'It's _not_ all right.'
I wasn't mad at him. I was mad at the unfairness of his stupid curse and the fact that I kept failing people who depended on me. My parents had died to give Sadie and me a chance to save the world, which we were close to botching. In Dallas, dozens of good magicians had died because they'd tried to help me. Now we were about to lose Walt.
Sure, he was important to Sadie. But I relied on him just as much. Walt was my unofficial lieutenant at Brooklyn House. The other kids listened to him. He was a calming presence in every crisis, the deciding vote in every debate. I could trust him with any secret – and even with making the execration statue of Apophis, which I couldn't tell my uncle about. If Walt died...
'I won't let it happen,' I said. 'I refuse.'
Wild thoughts ran through my mind: maybe Anubis was lying to Walt about his imminent death, trying to push Walt away from Sadie. (Okay, unlikely. Sadie wasn't that much of a prize.)
[Yeah, Sadie, I really said that. Just checking to see if you were still paying attention.]
Maybe Walt could beat the odds. People survived cancer miraculously. Why not ancient curses? Maybe we could put him in suspended animation like Iskandar had done for Zia, until we found an antidote. Sure, his family had been searching for a cure unsuccessfully for centuries. Jaz, our best healer, had tried everything with no luck. But maybe we'd overlooked something.
'Carter,' Walt said. 'Will you let me finish? We've got to make plans.'
'How can you be so calm?' I demanded.
Walt fingered his _shen_ necklace, the twin of the one he'd given Sadie. 'I've known about my curse for years. I won't let it stop me from doing what I need to. One way or another, I'm going to help you beat Apophis.'
'How?' I said. 'You just told me –'
'Anubis has an idea,' Walt said. 'He's been helping me make sense of my powers.'
'You mean...' I glanced at Walt's hands. Several times I'd seen him turn objects to ashes simply by touching them, the way he'd done to that criosphinx in Dallas. The power didn't come from any of his magic items. None of us understood it and, as Walt's disease progressed, he seemed less and less able to control it, which made me think twice about giving the guy a high five.
Walt flexed his fingers. 'Anubis thinks he understands why I have that ability. And there's more. He thinks there might be a way to extend my life.'
That was such good news that I let out a shaky laugh. 'Why didn't you say so? He can cure you?'
'No,' Walt said. 'Not a cure. And it's risky. It's never been done before.'
'That's what you were talking to Thoth about.'
Walt nodded. 'Even if Anubis's plan works, there could be... side effects. You might not like it.' He lowered his voice. 'Sadie might not like it.'
Unfortunately, I had a vivid imagination. I envisioned Walt turning into some sort of undead creature – a withered mummy, a ghostly _ba_ or a disfigured demon. In Egyptian magic, side effects could be pretty extreme.
I tried not to let my emotions show. 'We want you to live. Don't worry about Sadie.'
I could tell from Walt's eyes that he worried about Sadie a lot. Seriously, what did he _see_ in my sister?
[Stop hitting me, Sadie. I'm just being honest.]
Walt flexed his fingers. Maybe it was my imagination, but I thought I detected wisps of grey steam curling from his hands, as if just talking about his strange power had made it turn active.
'I won't make the decision yet,' Walt said. 'Not until I'm on my last breath. I want to talk to Sadie first, explain to her...'
He rested his hand on the side of the boat. That was a mistake. The woven reeds turned grey under his touch.
'Walt, stop!' I yelped.
He jerked his hand away, but it was too late. The boat crumbled to ashes.
We lunged for the ropes. Thankfully they did not crumble – maybe because Walt was paying more attention now. Freak squawked as the boat disappeared, and suddenly Walt and I were dangling under the griffin's belly, holding on to the ropes for dear life and bonking into each other as we flew above the skyscrapers of Manhattan.
'Walt!' I yelled over the wind. 'You _really_ need to get a handle on that power!'
'Sorry!' he shouted back.
My arms were aching, but somehow we made it to Brooklyn House without plummeting to our deaths. Freak set us down on the roof, where Bast was waiting, her mouth agape.
'Why are you swinging from ropes?' she demanded.
'Because it's so fun,' I growled. 'What's the news?'
Behind the chimneys, a frail voice warbled: 'Ha-lllooooo!'
The ancient sun god Ra popped out. He gave us a toothless grin and hobbled around the roof, muttering, 'Weasels, weasels. Cookie, cookie, cookie!' He reached into the folds of his loincloth and tossed cookie crumbs in the air like confetti – and, yes, it was just as disgusting as it sounds.
Bast tensed her arms, and her knives shot into her hands. Probably just an involuntary reflex, but she looked tempted to use those blades on someone – anyone. She reluctantly slipped the blades back into her sleeves.
'The news?' she said. 'I'm on babysitting duty, thanks to your Uncle Amos, who asked me for a favour. And Sadie's _shabti_ is waiting for you downstairs. Shall we?'
Explaining Sadie and her _shabti_ would take a whole separate recording.
My sister had no talent for crafting magical statues. That didn't stop her from trying. She'd got this harebrained idea that she could create the perfect _shabti_ to be her avatar, speak with her voice and do all her chores like a remote-controlled robot. All her previous attempts had exploded or gone haywire, terrorizing Khufu and the initiates. Last week she'd created a magical Thermos with googly eyes that levitated around the room, yelling, 'Exterminate! Exterminate!' until it smacked me in the head.
Sadie's latest _shabti_ was Sadie Junior – a gardener's nightmare.
Not being much of an artist, Sadie had fashioned a vaguely human figure out of red ceramic flowerpots, held together by magic, string and duct tape. The face was an upside-down pot with a smiley face drawn in black marker.
'About time.' The pot creature was waiting in my room when Walt and I came in. Its mouth didn't move, but Sadie's voice echoed from inside the face pot as if she were trapped within the _shabti_. That thought made me happy.
'Stop smiling!' she ordered. 'I can see you, Carter. Oh... and, _uh_ , hullo, Walt.'
The pot monster made squeaky grinding noises as it stood up straight. One clunky arm rose and tried to fix Sadie's nonexistent hair. Leave it to Sadie to be self-conscious around boys, even when she's made out of pots and duct tape.
We traded stories. Sadie told us about the impending attack on the First Nome that was supposed to go down at sunrise on the equinox and the alliance between Sarah Jacobi's forces and Apophis. Wonderful news. Just great.
In return, I told Sadie about our visit with Thoth. I shared the visions Apophis had shown me about our mother's precarious situation in the Duat (which made the pot monster shudder) and the end of the world (which didn't seem to surprise her at all). I didn't tell Sadie about Apophis's offer to spare me if I gave up Ra. I didn't feel comfortable announcing that with Ra just outside the door, singing songs about cookies. But I told her about the evil ghost Setne, whose trial would start at sunset in the Hall of Judgement.
'Uncle Vinnie,' Sadie said.
'Pardon?' I asked.
'The face that spoke to me at the Dallas Museum,' she said. 'It was obviously Setne himself. He warned me that we would need his help to understand the shadow execration spell. He said we'd have to "pull some strings" and free him before sunset tonight. He meant the trial. We'll have to convince Dad to free him.'
'I did mention that Thoth said he's a murderous psychopath, right?'
The pot monster made a clucking sound. 'Carter, it'll be fine. Befriending psychopaths is one of our specialities.'
She turned her flowerpot head towards Walt. 'You'll be coming along, I hope?'
Her tone had a hint of reproach, like she was still upset that Walt hadn't attended the school-dance/mass-blackout party.
'I'll be there,' he promised. 'I'm fine.'
He shot me a warning look, but I wasn't going to contradict him. Whatever he and Anubis were plotting, I could wait for him to explain it to Sadie. Jumping in the middle of the whole Sadie-Walt-Anubis drama sounded about as much fun as diving into a food processor.
'Right,' Sadie said. 'We'll meet you two at the Hall of Judgement before sunset tonight. That should give us time to finish up.'
'Finish up?' I asked. 'And who is _us_?'
It's hard to read expressions on a smiley-face pot, but Sadie's hesitation told me enough. 'You aren't in the First Nome any more,' I guessed. 'What are you doing?'
'A small errand,' Sadie said. 'I'm off to see Bes.'
I frowned. Sadie went to see Bes in his nursing home almost every week, which was fine and all, but why now? ' _Uh_ , you do understand we're in a hurry.'
'It's necessary,' she insisted. 'I've got an idea that might help us with our shadow project. Don't fret. Zia's with me.'
'Zia?' It was my turn to feel self-conscious. If I were a flowerpot, I would've checked my hair. 'That's why Bast is watching Ra today? Why exactly are you and Zia –?'
'Stop worrying,' Sadie chided. 'I'll take good care of her. And no, Carter, she hasn't been talking about you. I have no idea how she feels about you.'
' _What?_ ' I wanted to punch Sadie Junior in her ceramic face. 'I didn't say anything like that!'
'Now, now,' she chided. 'I don't think Zia cares what you wear. It's not a date. Just please brush your teeth for once.'
'I'm going to kill you,' I said.
'Love you too, brother dear. Ta!'
The pottery creature crumbled into pieces, leaving a mound of shards and a red clay face smiling up at me.
Walt and I joined Bast outside my room. We leaned on the rail overlooking the Great Room while Ra skipped back and forth on the balcony, singing nursery songs in Ancient Egyptian.
Down below, our initiates were getting ready for the school day. Julian had a breakfast sausage sticking out of his mouth as he rummaged through his backpack. Felix and Sean were arguing over who stole whose maths textbook. Little Shelby was chasing the other ankle-biters with a fistful of crayons that shot rainbow-coloured sparks.
I'd never had a big family, but living at Brooklyn House I felt like I had a dozen brothers and sisters. Despite the craziness, I enjoyed it... which made my next decision even harder.
I told Bast about our plan to visit the Hall of Judgement.
'I don't like it,' she said.
Walt managed a laugh. 'Is there a plan you'd like better?'
She tilted her head. 'Now that you mention it, no. I don't like plans. I'm a cat. Still, if half the things I've heard about Setne are true –'
'I know,' I said. 'But it's our only shot.'
She wrinkled her nose. 'You don't want me to come along? You're sure? Maybe I could get Nut or Shu to watch Ra –'
'No,' I said. 'Amos is going to need help at the First Nome. He doesn't have the numbers to fend off an attack from both the rebel magicians and Apophis.'
Bast nodded. 'I can't enter the First Nome, but I can patrol outside. If Apophis shows himself, I will engage him in battle.'
'He'll be at full strength,' Walt warned. 'He's getting stronger by the hour.'
She lifted her chin defiantly. 'I've fought him before, Walt Stone. I know him better than anyone. Besides, I owe it to Carter's family. And to Lord Ra.'
'Kitty!' Ra appeared behind us, patted Bast on the head and skipped away. 'Meow, meow, meow!'
Watching him prance around, I wanted to scream and throw things. We'd risked everything to revive the old sun god, hoping we'd get a divine pharaoh who could stand toe to toe with Apophis. Instead we got a wrinkly, bald troll in a loincloth.
_Give me Ra_ , Apophis had urged. _I know you hate him._
I tried to put it out of my mind, but I couldn't quite shake that image of an island in the Sea of Chaos – a personal paradise where the people I loved would be safe. I knew it was a lie. Apophis would never deliver on that promise. But I could understand how Sarah Jacobi and Kwai might be tempted.
Besides, Apophis knew how to strike a nerve. I _did_ resent Ra for being so weak. Horus agreed with me.
_We don't need the old fool._ The war god's voice spoke inside my head. _I'm not saying you should give him to Apophis, but he is useless. We should put him aside and take the throne of the gods for ourselves._
He made it sound so tempting – such an obvious solution.
But, no. If Apophis wanted me to give up Ra, then Ra must be valuable in some way. The sun god still had a role to play. I just had to figure out what it was.
'Carter?' Bast frowned. 'I know you're worried about me, but your parents saved me from the abyss for a reason. Your mother foresaw that I would make a difference in the final battle. I will fight Apophis to the death if necessary. He won't get past me.'
I wavered. Bast had already helped us so much. She had almost been destroyed fighting the crocodile god Sobek. She'd enlisted her friend Bes to help us and then seen him reduced to an empty shell. She'd helped us restore her old master, Ra, to the world and now she was stuck babysitting him. I didn't want to ask her to face Apophis again, but she was right. She knew the enemy better than anyone – except maybe Ra, when he was in his right mind.
'All right,' I said. 'But Amos will need more help than you can give, Bast. He'll need magicians.'
Walt frowned. 'Who? After the disaster in Dallas, we don't have many friends left. We could contact São Paulo and Vancouver – they're still with us – but they won't be able to spare many people. They'll be worried about protecting their own nomes.'
I shook my head. 'Amos needs magicians who know the path of the gods. He needs _us_. All of us.'
Walt digested that silently. 'You mean, abandon Brooklyn House.'
Below us, the ankle-biters shrieked with joy as Shelby tried to tag them with her sparking crayons. Khufu sat on the fireplace mantel eating Cheerios, watching ten-year-old Tucker bounce a basketball off the statue of Thoth. Jaz was putting a bandage on Alyssa's forehead. (Probably she'd been attacked by Sadie's rogue Thermos, which was still on the loose.) In the middle of all this, Cleo was sitting on the sofa, engrossed in a book.
Brooklyn House was the first real home some of them had ever known. We'd promised to keep them safe and teach them to use their powers. Now I was about to send them unprepared into the most dangerous battle of all time.
'Carter,' Bast said, 'they're not ready.'
'They _have_ to be,' I said. 'If the First Nome falls, it's all over. Apophis will attack us in Egypt, at the source of our power. We have to stand together with the Chief Lector.'
'One last battle.' Walt gazed sadly at the Great Room, maybe wondering whether or not he'd die before that battle happened. 'Should we break the news to others?'
'Not yet,' I said. 'The rebel magicians' attack on the First Nome won't happen until tomorrow. Let the kids have one last day at school. Bast, when they come home this afternoon, I want you to lead them to Egypt. Use Freak, use whatever magic you have to. If all goes well in the Underworld, Sadie and I will join you before the attack.'
'If all goes well,' Bast said dryly. 'Yes, that happens a lot.'
She glanced over at the sun god, who was trying to eat the doorknob to Sadie's room. 'What about Ra?' she asked. 'If Apophis is going to attack in two days...'
'Ra has to keep making his nightly journey,' I said. 'That's part of Ma'at. We can't mess with it. But on the morning of the equinox he'll need to be in Egypt. He'll have to face Apophis.'
'Like _that_?' Bast gestured towards the old god. 'In his loincloth?'
'I know,' I admitted. 'It sounds crazy. But Apophis still thinks Ra is a threat. Maybe facing Apophis in battle will remind Ra who he is. He might rise to the challenge and become... what he used to be.'
Walt and Bast didn't answer. I could tell from their expressions that they didn't buy it. Neither did I. Ra was gumming Sadie's doorknob with intent to kill, but I didn't think he'd be much good against the Lord of Chaos.
Still, it felt good to have a plan of action. That was much better than standing around, dwelling on the hopelessness of our situation.
'Use today to organize,' I told Bast. 'Gather up the most valuable scrolls, amulets, weapons – anything we can use to help the First Nome. Let Amos know you're coming. Walt and I will head to the Underworld and meet Sadie. We'll rendezvous with you in Cairo.'
Bast pursed her lips. 'All right, Carter. But be careful of Setne. However bad you think he is? He's ten times worse.'
'Hey, we defeated the god of evil,' I reminded her.
Bast shook her head. 'Set is a god. He doesn't change. Even with a god of Chaos, you can pretty much predict how he'll act. Setne, on the other hand... he has both power _and_ human unpredictability. Don't trust him. Swear to me.'
'That's easy,' I said. 'I promise.'
Walt folded his arms. 'So how are we going to get to the Underworld? Portals are unreliable. We're leaving Freak here, and the boat is destroyed –'
'I have another boat in mind,' I said, trying to believe it was a good idea. 'I'm going to summon an old friend.'
SADIE
## 9. Zia Breaks Up a Lava Fight
I'D BECOME QUITE AN EXPERT at visiting the godly nursing home – which was a sad statement on my life.
The first time Carter and I found our way there, we had travelled the River of Night, plunged down a fiery waterfall and almost died in a lake of lava. Since then, I'd discovered I could simply call on Isis to transport me, as she could open doorways to many locations in the Duat. Honestly, though, dealing with Isis was almost as annoying as swimming through fire.
After my _shabti_ conversation with Carter, I joined Zia on a limestone cliff overlooking the Nile. It was already midday in Egypt. Getting over portal-lag had taken me longer than I'd expected. After changing into more sensible clothes, I'd had a quick lunch and one more strategy talk with Amos deep in the Hall of Ages. Then Zia and I had climbed back to the surface. Now we stood at a ruined shrine to Isis on the river just south of Cairo. It was a good place to summon the goddess, but we didn't have much time.
Zia still wore her combat outfit – camouflage cargo pants and an olive vest top. Her staff was slung over her back, and her wand hung at her belt. She rummaged through her pack, checking her supplies one last time.
'What did Carter say?' she asked.
[That's right, brother dear. I stepped out of earshot before I contacted you, so Zia didn't hear any of those teasing comments. Honestly, I'm not _that_ mean.]
I told her what we'd discussed, but I couldn't bring myself to share how my mum's spirit was in danger. I'd known about the problem in general terms since I'd spoken with Anubis, of course, but the knowledge that our mother's ghost was huddled under a cliff somewhere in the Duat, resisting the pull of the serpent's shadow – well, that bit of information had lodged in my chest like a bullet. If I tried to touch it, I feared it would go straight to my heart and kill me.
I explained about my villainous ghost friend Uncle Vinnie and how we intended to solicit his help.
Zia looked appalled. 'Setne? As in _the_ Setne? Does Carter realize –?'
'Yep.'
'And Thoth suggested this?'
'Yep.'
'And you're actually going along with it?'
'Yep.'
She gazed down the Nile. Perhaps she was thinking of her home village, which had stood on the banks of this river until it was destroyed by the forces of Apophis. Perhaps she was imagining her entire homeland crumbling into the Sea of Chaos.
I expected her to tell me that our plan was insane. I thought she might abandon me and go back to the First Nome.
But I suppose she had got used to the Kane family – poor girl. She must've known by now that _all_ our plans were insane.
'Fine,' she said. 'How do we reach this... nursing home of the gods?'
'Just a mo.' I closed my eyes and concentrated.
_Yoo-hoo, Isis?_ I thought. _Anyone home?_
_Sadie_ , the goddess answered immediately.
In my mind she appeared as a regal woman with dark braided hair. Her dress was gossamer white. Her prismatic wings shimmered like sunlight rippling through clear water.
I wanted to smack her.
_Well, well_ , I said. _If it isn't my good friend who decides whom I can and can't date._
She had the nerve to look surprised. _Are you speaking of Anubis?_
_Right, first try!_ I should've left it at that since I needed Isis's help. But seeing her floating there all shiny and queenly made me angrier than ever. _Where do you get the nerve, eh? Going behind my back, lobbying to keep Anubis away from me. How is that your business?_
Surprisingly, Isis kept her temper. _Sadie, there are things you don't understand. There are rules._
_Rules?_ I demanded. _The world is about to end, and you're worried about which boys are socially acceptable for me?_
Isis steepled her fingers. _The two issues are more connected than you realize. The traditions of Ma'at must be followed, or Chaos wins. Immortals and mortals can only interact in specific, limited ways. Besides, you cannot afford to be distracted. I'm doing you a favour._
_A favour!_ I said. _If you want to do me a_ real _favour, we need passage to the Fourth House of the Night – the House of Rest, Sunny Acres or whatever you want to call it. After that, you can stay out of my private life!_
Perhaps that was rude of me, but Isis had crossed the line. Besides, why should I act properly with a goddess who had previously rented space in my head? Isis should have known me better!
The goddess sighed. _Sadie, proximity to the gods is dangerous. It must be regulated with utmost care. You know this. Your uncle is still tainted from his experience with Set. Even your friend Zia is struggling._
_What do you mean?_ I asked.
_If you join with me, you'll understand_ , Isis promised. _Your mind will be clear. It's past time we united again and combined our strength._
There it was: the sales pitch. Every time I called on Isis, she tried to persuade me to meld with her as we'd done before – mortal and god inhabiting one body, acting with a single will. Each time, I said no.
_So_ , I ventured, _proximity to the gods is dangerous, but you'reanxious to join forces with me again. I'm glad you're looking out for my safety._
Isis narrowed her eyes. _Our situation is different, Sadie. You need my strength._
Certainly it was tempting. Having the full power of a goddess at my command was quite a rush. As the Eye of Isis, I would feel confident, unstoppable, completely without fear. One could get addicted to such power – and that was the problem.
Isis could be a good friend, but her agenda wasn't always best for the mortal world – or for Sadie Kane.
She was driven by her loyalty to her son Horus. She'd do anything to see him on the throne of the gods. She was ambitious, vengeful, power-hungry and envious of anyone who might have more magic than she did.
She claimed my mind would be clearer if I let her in. What she really meant was that I'd start seeing things her way. It would be harder to separate my thoughts from hers. I might even come to believe she was right by keeping Anubis and me apart. (Horrifying idea.)
Unfortunately, Isis had a point about joining forces. Sooner or later we'd have to. There was no other way I'd have the power to challenge Apophis.
But now wasn't the time. I wanted to remain Sadie Kane as long as possible – just my own wonderful self without any godly hitchhiker.
_Soon_ , I told Isis. _I have things to do first. I need to be sure my decisions are my own. Now, about that doorway to the House of Rest..._
Isis was quite good at looking hurt and disapproving at the same time, which must have made her an impossible mother. I almost felt sorry for Horus.
_Sadie Kane_ , she said, _you are my favourite mortal, my chosen magician. And still you do not trust me._
I didn't bother to contradict her. Isis knew how I felt.
The goddess spread her arms in resignation. _Very well. But the path of the gods is the only answer. For all the Kanes, and for that one._ She nodded in Zia's direction. _You will need to advise her, Sadie. She must learn the path quickly._
_What do you mean?_ I asked again. I really wished she would stop talking in riddles. Gods are so annoying that way.
Zia was a much more experienced magician than I was. I didn't know how I could advise her. Besides, Zia was a fire elementalist. She tolerated us Kanes, but she had never shown the slightest interest in the path of the gods.
_Good luck_ , Isis said. _I will await your call._
The image of the goddess rippled and vanished. When I opened my eyes, a square of darkness the size of a doorway hovered in the air.
'Sadie?' Zia asked. 'You were silent for so long I was getting worried.'
'No need.' I tried for a smile. 'Isis just likes to talk. Next stop, the Fourth House of the Night.'
I'll be honest. I never quite understood the difference between the swirling sand portals that magicians can summon with artefacts and the doors of darkness that gods are able to conjure. Perhaps the gods use a more advanced wireless network. Perhaps they simply have better aim.
Whatever the reason, Isis's portal worked much more reliably than the one I'd created to get to Cairo. It deposited us right in the lobby of Sunny Acres.
As soon as we stepped through, Zia scanned our surroundings and frowned. 'Where is everyone?'
Good question. We'd arrived at the correct godly nursing home – the same potted plants, the same massive lobby with windows looking out on the Lake of Fire, the same rows of limestone columns plastered with tacky posters of smiling seniors and mottos like: _These Are Your Golden Centuries!_
But the nurses' station was unattended. IV poles were clustered in one corner like they were having a conference. The sofas were empty. The coffee tables were littered with half-played games of draughts and senet. Ugh, I _hate_ senet.
I stared at an empty wheelchair, wondering where its occupant had gone, when suddenly the chair burst into flames, collapsing in a pile of charred leather and half-melted steel.
I stumbled backwards. Behind me, Zia held a ball of white-hot fire in her hand. Her eyes were as wild as a cornered animal's.
'Are you mad?' I yelled. 'What are you –?'
She lobbed her second fireball at the nurses' station. A vase full of daisies exploded in a shower of flaming petals and pottery shards.
'Zia!'
She didn't seem to hear me. She summoned another fireball and took aim at the sofas.
I should have run for cover. I wasn't prepared to die saving badly upholstered furniture. Instead, I lunged at her and grabbed her wrist. 'Zia, stop it!'
She glared at me with flames in her eyes – and I mean that quite literally. Her irises had become discs of orange fire.
This was terrifying, of course, but I stood my ground. Over the past year I'd got rather used to surprises – what with my cat being a goddess, my brother turning into a falcon and Felix producing penguins in the fireplace several times a week.
'Zia,' I said firmly. 'We can't burn down the nursing home. What's got into you?'
A look of confusion passed over her face. She stopped struggling. Her eyes returned to normal.
She stared at the melted wheelchair, then the smouldering remains of the bouquet on the carpet. 'Did I –?'
'Decide those daisies needed to die?' I finished. 'Yes, you did.'
She extinguished her fireball, which was lucky, as it was starting to bake my face. 'I'm sorry,' she muttered. 'I – I thought I had it under control...'
'Under control?' I let go of her hand. 'You mean to say you've been throwing _a lot_ of random fireballs lately?'
She still looked bewildered, her gaze drifting around the lobby. 'N-no... maybe. I've been having blackouts. I come to, and I don't remember what I've done.'
'Like just now?'
She nodded. 'Amos said... at first he thought it might be a side effect of my time in that tomb.'
Ah, the tomb. For months, Zia had been trapped in a watery sarcophagus while her _shabti_ ran about impersonating her. The Chief Lector Iskandar had thought this would protect the real Zia – from Set? From Apophis? We still weren't sure. At any rate, it didn't strike me as the most brilliant idea for a supposedly wise two-thousand-year-old magician to have come up with. During her slumbers, Zia had endured horrible nightmares about her village burning and Apophis destroying the world. I suppose that might lead to some nasty posttraumatic stress.
'You said Amos thought that _at first_ ,' I noted. 'There's more to the story, then?'
Zia gazed at the melted wheelchair. The light from outside turned her hair the colour of rusted iron.
'He was here,' she murmured. 'He was here for aeons, trapped.'
I took a moment to process that. 'You mean Ra.'
'He was miserable and alone,' she said. 'He had been forced to abdicate his throne. He left the mortal world and lost the will to live.'
I stamped out a smouldering daisy on the carpet. 'I don't know, Zia. He looked quite happy when we woke him up, singing and grinning and so on.'
'No.' Zia walked towards the windows, as if drawn by the lovely view of brimstone. 'His mind is still sleeping. I've spent time with him, Sadie. I've watched his expressions while he naps. I've heard him whimpering and mumbling. That old body is a cage, a prison. The true Ra is trapped inside.'
She was starting to worry me now. Fireballs I could deal with. Incoherent rambling – not so much.
'I suppose it makes sense you'd have sympathy with Ra,' I ventured. 'You're a fire elementalist. He's a fiery sort of god. You were trapped in that tomb. Ra was trapped in a nursing home. Perhaps that's what caused your blackout just now. This place reminded you of your own imprisonment.'
That's right – Sadie Kane, junior psychologist. And why not? I'd spent enough time diagnosing my crazed mates Liz and Emma back in London.
Zia stared out at the burning lake. I had the feeling that my attempt at therapy might not have been so therapeutic.
'Amos tried to help me,' she said. 'He knows what I'm going through. He cast a spell on me to focus my mind, but...' She shook her head. 'It's been getting worse. This is the first day in weeks that I _haven't_ taken care of Ra and, the more time I spend with him, the fuzzier my thoughts get. When I summon fire now, I have trouble controlling it. Even simple spells I've done for years – I channel too much power. If that happens during a blackout...'
I understood why she sounded frightened. Magicians have to be careful with spells. If we channel too much power, we might inadvertently exhaust our reserves. Then the spell would tap directly into the magician's life force – with unpleasant consequences.
_You will need to advise her_ , Isis had told me. _She must learn the path quickly._
An uncomfortable thought began to form. I remembered Ra's delight when he had first met Zia, the way he'd tried to give her his last remaining scarab beetle. He'd babbled on and on about zebras... possibly meaning Zia. And now Zia was starting to empathize with the old god, even trying to burn down the nursing home where he'd been trapped for so long.
That couldn't be good. But how could I advise her when I had no idea what was happening?
Isis's warnings rattled around in my head: the path of the gods was the answer for all the Kanes. Zia was struggling. Amos was still tainted by his time with Set.
'Zia...' I hesitated. 'You said Amos knows what you're going through. Is that why he asked Bast to watch Ra today? To give you time away from the sun god?'
'I – I suppose.'
I tried to steady my breathing. Then I asked the harder question: 'In the war room, Amos said he might have to use other means to fight his enemies. He hasn't... _um_ , he hasn't been having trouble with Set?'
Zia wouldn't meet my eyes. 'Sadie, I promised him –'
'Oh, gods of Egypt! He's _calling_ on Set? Trying to channel his power, after all Set did to him? Please, no.'
She didn't answer, which was an answer in itself.
'He'll be overwhelmed!' I cried. 'If the rebel magicians find out that the Chief Lector is meddling with the god of evil, just as they suspected –'
'Set isn't just the god of evil,' Zia reminded me. 'He is Ra's lieutenant. He defended the sun god against Apophis.'
'You think that makes it all better?' I shook my head in disbelief. 'And now Amos thinks you're having trouble with Ra? Does he think Ra is trying to...' I pointed to Zia's head.
'Sadie, please...' Her voice trailed off in misery.
I suppose it wasn't fair for me to press her. She seemed even more confused than I was.
Still, I hated the idea of Zia being disorientated so close to our final battle – blacking out, throwing random fireballs, losing control of her own power. Even worse was the possibility that Amos had some sort of link with Set – that he might actually have _chosen_ to let that horrible god back into his head.
The thought tied my gut into _tyets_ – Isis knots.
I imagined my old enemy Michel Desjardins scowling. Ne voyez-vous pas _, Sadie Kane? This is what comes from the path of the gods. This is why the magic was forbidden._
I kicked the melted remains of the wheelchair. One bent wheel squeaked and wobbled.
'We'll have to table that conversation,' I decided. 'We're running out of time. Now... where have all the old folks gone?'
Zia pointed out the window. 'There,' she said calmly. 'They're having a beach day.'
We made our way down to the black sand beach by the Lake of Fire. It wouldn't have been my top vacation spot, but elderly gods were lounging on deck chairs under brightly coloured umbrellas. Others snored on beach towels or sat in their wheelchairs and stared at the boiling vista.
One shrivelled bird-headed goddess in a one-piece swimming costume was building a sand pyramid. Two old men – I assumed they were fire gods – stood waist-deep in the blazing surf, laughing and splashing lava in each other's faces.
Tawaret the caretaker beamed when she saw us.
'Sadie!' she called. 'You're early this week! And you've brought a friend.'
Normally, I wouldn't have stood still as an upright grinning female hippo charged towards me for a hug, but I'd got used to Tawaret.
She'd traded her high heels for flip-flops. Otherwise she was dressed in her usual white nurse's uniform. Her mascara and lipstick were tastefully done, for a hippo, and her luxuriant black hair was pinned under a nurse's cap. Her ill-fitting blouse opened over an enormous belly – possibly a sign of permanent pregnancy, as she was the goddess of childbirth, or possibly a sign of eating too many cupcakes. I'd never been entirely sure.
She embraced me without crushing me, which I greatly appreciated. Her lilac perfume reminded me of my gran, and the tinge of sulphur on her clothes reminded me of Gramps.
'Tawaret,' I said, 'this is Zia Rashid.'
Tawaret's smile faded. 'Oh... Oh, I see.'
I'd never seen the hippo goddess so uneasy. Did she somehow know that Zia had melted her wheelchair and torched her daisies?
As the silence got awkward, Tawaret recovered her smile. 'Sorry, yes. Hello, Zia. It's just that you look... well, never mind! Are you a friend of Bes's too?'
' _Uh_ , not really,' Zia admitted. 'I mean, I suppose, but –'
'We're here on a mission,' I said. 'Things in the upper world have gone a bit wonky.'
I told Tawaret about the rebel magicians, Apophis's plans for attack and our mad scheme to find the serpent's shadow and stomp it to death.
Tawaret mashed her hippoish hands together. 'Oh dear. Doomsday tomorrow? Bingo night was supposed to be Friday. My poor darlings will be so disappointed...'
She glanced down the beach at her senile charges, some of whom were drooling in their sleep or eating black sand or trying to talk to the lava.
Tawaret sighed. 'I suppose it would be kinder not to tell them. They've been here for aeons, forgotten by the mortal world. Now they have to perish along with everyone else. They don't deserve such a fate.'
I wanted to remind her that _no one_ deserved such a fate – not my friends, not my family and certainly not a brilliant young woman named Sadie Kane, who had her whole life ahead of her. But Tawaret was so kind-hearted I didn't want to sound selfish. She didn't seem concerned for herself at all, just the fading gods she cared for.
'We're not giving up yet,' I promised.
'But this plan of yours!' Tawaret shuddered, causing a tsunami of jiggling hippo flesh. 'It's impossible!'
'Like reviving the sun god?' I asked.
She conceded that with a shrug. 'Very well, dear. I'll admit you've done the impossible before. Nevertheless...' She glanced at Zia, as if my friend's presence still made her nervous. 'Well, I'm sure you know what you're doing. How can I help?'
'May we see Bes?' I asked.
'Of course... but I'm afraid he hasn't changed.'
She led us down the beach. The past few months I'd visited Bes at least once a week, so I knew many of the elderly gods by sight. I spotted Heket the frog goddess perched on top of a beach umbrella as if it were a lily pad. Her tongue shot out to catch something from the air. Did they have flies in the Duat?
Further on, I saw the goose god Gengen-Wer, whose name – I kid you not – meant the Great Honker. The first time Tawaret told me that, I almost spewed tea. His Supreme Honkiness was waddling along the beach, squawking at the other gods and startling them out of their sleep.
Yet every time I visited, the crowd changed. Some gods disappeared. Others popped up – gods of cities that no longer existed; gods who had only been worshipped for a few centuries before being replaced by others; gods so old they'd forgotten their own names. Most civilizations left behind pottery shards or monuments or literature. Egypt was so old it had left behind a landfill's worth of deities.
Halfway down the beach, we passed the two old codgers who'd been playing in the lava. Now they were wrestling waist-deep in the lake. One pummelled the other with an _ankh_ and warbled, 'It's _my_ pudding! _My_ pudding!'
'Oh dear,' Tawaret said. 'Fire-embracer and Hot Foot are at it again.'
I choked back a laugh. 'Hot Foot? What sort of godly name is that?'
Tawaret studied the fiery surf, as if looking for a way to navigate through it without getting incinerated. 'They're gods from the Hall of Judgement, dear. Poor things. There used to be forty-two of them, each in charge of judging a different crime. Even in the old days, we could never keep them all straight. Now...' She shrugged. 'They're quite forgotten, sadly. Fire-embracer, the one with the _ankh_ – he used to be the god of robberies. I'm afraid it made him paranoid. He always thinks Hot Foot has stolen his pudding. I'll have to break up the fight.'
'Let me,' Zia said.
Tawaret stiffened. 'You, my... dear?'
I got the feeling she was going to say something other than _dear_.
'The fire won't bother me,' Zia assured her. 'You two go ahead.'
I wasn't sure how Zia could be so confident. Perhaps she simply preferred swimming in flames to seeing Bes in his present state. If so, I couldn't blame her. The experience was unsettling.
Whatever the case, Zia strode towards the surf and waded straight in like a flame-retardant _Baywatch_ lifeguard.
Tawaret and I continued along the beach. We reached the dock where Ra's sun boat had anchored the first time Carter and I had visited this place.
Bes sat at the end of the pier in a comfy leather chair, which Tawaret must have brought down especially for him. He wore a fresh red-and-blue Hawaiian shirt and khaki shorts. His face was thinner than it had been last spring, but otherwise he looked unchanged – the same scraggly nest of black hair, the same bristly mane that passed for a beard, the same lovably grotesque face that reminded me of a pug dog's.
But Bes's soul was gone. He stared vacantly at the lake, not reacting at all when I knelt next to him and gripped his furry hand.
I remembered the first time he'd saved my life – picking me up in a limo full of rubbish, driving me to Waterloo Bridge, then scaring away two gods who had been chasing me. He had jumped out of the car wearing nothing but a pair of Speedos and screamed, 'Boo!'
Yes, he'd been a true friend.
'Dear Bes,' I said, 'we're going to try to help you.'
I told him everything that had happened since my last visit. I knew he couldn't hear me. Since his secret name had been stolen, his mind simply wasn't there. But talking to him made me feel better.
Tawaret sniffled. I knew she had loved Bes forever, though Bes hadn't always returned her feelings. He couldn't have had a better caretaker.
'Oh, Sadie...' The hippo goddess wiped away a tear. 'If you truly could help him, I – I'd do anything. But how is it possible?'
'Shadows,' I said. 'This bloke Setne... he found a way to use shadows for an execration spell. If the _sheut_ is a back-up copy of the soul, and if Setne's magic could be used in reverse.'
Tawaret's eyes widened. 'You believe you could use Bes's shadow to bring him back?'
'Yes.' I knew it sounded mad, but I _had_ to believe. Saying it aloud to Tawaret, who cared about Bes even more than I did... well, I simply couldn't let her down. Besides, if we could do this for Bes, then who knew? Perhaps we could use the same magic to get the sun god Ra back in fighting shape. First things first, however. I intended to keep my promise to the dwarf god.
'Here's the tricky bit,' I said. 'I'm hoping you can help me locate Bes's shadow. I don't know much about gods and their _sheuts_ and whatnot. I understand that you often hide them?'
Tawaret shifted nervously, her feet creaking on the pier boards. ' _Um_ , yes...'
'I'm hoping they're a bit like secret names,' I forged on. 'Since I can't ask Bes where he keeps his shadow, I thought I'd ask the person who was closest to him. I thought you'd have the best chance of knowing.'
Seeing a hippo blush is quite odd. It almost made Tawaret look delicate – in a massive sort of way.
'I – I saw his shadow once,' she admitted. 'During one of our best moments together. We were sitting on the temple wall in Saïs.'
'Sorry?'
'A city in the Nile Delta,' Tawaret explained. 'The home of a friend of ours – the hunting goddess Neith. She liked to invite Bes and me on her hunting excursions. We would, ah, flush her prey for her.'
I imagined Tawaret and Bes, two gods with super-ugly powers, ploughing through the marshes hand in hand, yelling, 'Boo!' to scare up bevies of quail. I decided to keep that image to myself.
'At any rate,' Tawaret continued, 'one night after dinner, Bes and I were sitting alone on the walls of Neith's temple, watching the moon rise over the Nile.'
She gazed at the dwarf god with such adoring eyes that I couldn't help but imagine myself on that temple wall, sharing a romantic evening with Anubis... no, Walt... no... Gah! My life was horrid.
I sighed unhappily. 'Go on, please.'
'We talked about nothing in particular,' Tawaret remembered. 'We held hands. That was all. But I felt so close to him. Just for a moment, I looked at the mud-brick wall next to us, and I saw Bes's shadow in the torchlight. Normally gods don't keep their shadows so close. He must've trusted me a great deal to reveal it. I asked him about it, and he laughed. He said, "This is a good place for my shadow. I think I'll leave it here. That way it can always be happy, even when I'm not." '
The story was so sweet and sad I could hardly bear it.
Down the shore, the old god Fire-embracer shrieked something about pudding. Zia was standing in the surf, trying to keep the two gods apart as they splashed her with lava from both sides. Amazingly, it didn't seem to bother her.
I turned to Tawaret. 'That night in Saïs – how long ago was it?'
'A few thousand years.'
My heart sank. 'Any chance the shadow would still be there?'
She shrugged helplessly. 'Saïs was destroyed centuries ago. The temple is gone. Farmers pulled down the ancient buildings and used the mud bricks for fertilizer. Most of the land has reverted to marshes.'
Blast. I'd never been a fan of Egyptian ruins. From time to time, I'd been tempted to pull down a few temples myself. But just this once I wished the ruins had survived. I wanted to cuff those farmers.
'Then there's no hope?' I asked.
'Oh, there's always hope,' Tawaret said. 'You could search the area, calling on Bes's shadow. You're his friend. It might appear to you if it's still there. And if Neith is still in the area she might be able to help. That is if she doesn't hunt you instead...'
I decided not dwell on that possibility. I had enough problems. 'We'll have to try. If we can find the shadow and puzzle out the proper spell –'
'But, Sadie,' the goddess said, 'you have so little time. You have to stop Apophis! How can you help Bes, too?'
I looked at the dwarf god. Then I bent down and kissed his bumpy forehead. 'I made a promise,' I said. 'Besides, we'll need him if we're going to win.'
Did I really believe that? I knew Bes couldn't scare Apophis away simply by yelling, 'Boo!' no matter how ghastly he looked in his Speedos. In the sort of battle we were facing, I wasn't sure one more god would even make a difference. And I was even less sure that this reverse shadow idea could work on Ra. But I had to try with Bes. If the world ended the day after tomorrow, I would _not_ go to my death without first knowing I'd done everything I could to save my friend.
Of all the goddesses I'd met, Tawaret was the most likely to understand my motives.
She put her hands protectively on Bes's shoulders. 'In that case, Sadie Kane, I wish you luck – for Bes, and for all of us.'
I left her on the dock, standing behind Bes as if the two gods were enjoying a romantic sunset together.
On the beach, I rejoined Zia, who was brushing ashes out of her hair. Except for a few burn holes in her trousers, she looked perfectly fine.
She gestured at Fire-embracer and Hot Foot, who were once again playing nicely in the lava. 'They're not so bad,' Zia said. 'They just needed some attention.'
'Like pets,' I said. 'Or my brother.'
Zia actually smiled. 'Did you find the information you need?'
'I think so,' I said. 'But first we need to get to the Hall of Judgement. It's almost time for Setne's trial.'
'How do we get there?' Zia asked. 'Another doorway?'
I stared across the Lake of Fire, pondering that problem. I remembered the Hall of Judgement being on an island somewhere on this lake, but Duat geography is a bit dodgy. For all I knew, the hall was on a totally different level of the Duat, or the lake was six billion miles wide. I didn't fancy the idea of walking round the shore through unknown territory or taking a swim. And I certainly didn't feel like arguing with Isis again.
Then I saw something across the fiery waves – the silhouette of a familiar steamboat approaching, twin smokestacks trailing luminous gold smoke and a paddlewheel churning through the lava.
My brother – bless his heart – was absolutely mad.
'Problem solved,' I told Zia. 'Carter will give us a ride.'
SADIE
## 10. 'Take Your Daughter to Work Day' Goes Horribly Wrong
AS THEY APPROACHED THE DOCK, Carter and Walt waved at us from the bow of the _Egyptian Queen_. Next to them stood the captain, Bloodstained Blade, who looked quite dashing in his riverboat pilot's uniform, except for the fact that his head was a blood-speckled double-sided axe.
'That's a demon,' Zia said nervously.
'Yes,' I agreed.
'Is it safe?'
I raised an eyebrow at her.
'Of course not,' she muttered. 'I'm travelling with the Kanes.'
The crew of glowing orbs zipped around the boat, pulling lines and lowering the gangplank.
Carter looked tired. He wore jeans and a rumpled shirt with specks of barbecue sauce on it. His hair was wet and flat on one side as if he'd fallen asleep in the shower.
Walt looked much better – well, really, there was no contest. He wore his usual sleeveless shirt and tracksuit trousers, and managed a smile for me even though his posture made it obvious he was in pain. The _shen_ charm on my necklace seemed to heat up, or perhaps that was just my body temperature rising.
Zia and I climbed the gangplank. Bloodstained Blade bowed, which was quite unnerving, as his head could've sliced a watermelon in half.
'Welcome aboard, Lady Kane.' His voice was a metallic hum from the edge of his frontal blade. 'I am at your service.'
'Thanks ever so,' I said. 'Carter, may I speak with you?'
I grabbed his ear and pulled him towards the deckhouse.
'Ow!' he complained as I dragged him along. I suppose doing that in front of Zia wasn't nice, but I thought I might as well give her pointers on how best to handle my brother.
Walt and Zia followed us into the main dining room. As usual, the mahogany table was laden with platters of fresh food. The chandelier illuminated colourful wall murals of Egyptian gods, the gilded columns and ornately moulded ceiling.
I let go of Carter's ear and snarled, 'Have you lost your mind?'
'Ow!' he yelled again. 'What is your problem?'
'My problem,' I said, lowering my voice, 'is that you summoned this boat again and its demon captain, who Bast warned would slit our throats if he ever got the opportunity!'
'He's under a magic binding,' Carter argued. 'He was _fine_ last time.'
'Last time _Bast_ was with us,' I reminded him. 'And if you think I trust a demon named Bloodstained Blade further than I can –'
'Guys,' Walt interrupted.
Bloodstained Blade entered the dining room, dipping his axe head under the doorframe. 'Lord and Lady Kane, the journey is short from here. We will arrive at the Hall of Judgement in approximately twenty minutes.'
'Thanks, BSB,' Carter said as he rubbed his ear. 'We'll join you on deck soon.'
'Very good,' said the demon. 'What are your orders when we arrive?'
I tensed, hoping Carter had thought ahead. Bast had warned us that demons needed very clear instructions to stay under control.
'You'll wait for us while we visit the Hall of Judgement,' Carter announced. 'When we return, you'll take us where we wish to go.'
'As you say.' Bloodstained Blade's tone had a hint of disappointment – or was that my imagination?
After he left, Zia frowned. 'Carter, in this case I agree with Sadie. How can you trust that creature? Where did you get this ship?'
'It belonged to our parents,' Carter said.
He and I shared a look, silently agreeing that was enough said. Our mum and dad had sailed this riverboat up the Thames to Cleopatra's Needle the night Mum had died releasing Bast from the Abyss. Afterwards, Dad had sat in this very room, grieving, with only the cat goddess and the demon captain for company.
Bloodstained Blade had accepted us as his new masters. He'd followed our orders before, but that was little comfort. I didn't trust him. I didn't like being on this ship.
On the other hand, we needed to get to the Hall of Judgement. I was hungry and thirsty, and I supposed I could endure a twenty-minute voyage if it meant enjoying a chilled Ribena and a plate of tandoori chicken with naan.
The four of us sat round the table. We ate while we compared stories. All in all, it was quite possibly the most awkward double date in history. We had no shortage of dire emergencies to talk about, but the tension in the room was as thick as Cairo smog.
Carter hadn't seen Zia in person for months. I could tell he was trying not to stare. Zia was clearly uncomfortable sitting so close to him. She kept leaning away, which no doubt hurt his feelings. Perhaps she was just worried about having another fireball-throwing episode. As for me, I was elated to be next to Walt, but at the same time I was desperately worried about him. I couldn't forget how he'd looked wrapped in glowing mummy linen, and I wondered what Anubis had wanted to tell me about Walt's situation. Walt tried to hide it, but he was obviously in great pain. His hands trembled as he picked up his peanut-butter sandwich.
Carter told me about the pending evacuation of Brooklyn House, which Bast was overseeing. My heart nearly broke when I thought of little Shelby, wonderful silly Felix, shy Cleo and all the rest going off to defend the First Nome against an impossible attack, but I knew Carter was right. There was no other choice.
Carter kept hesitating, as if waiting for Walt to contribute information. Walt stayed silent. Clearly he was holding something back. Somehow or other, I'd have to get Walt alone and grill him for details.
In return, I told Carter about our visit to the House of Rest. I shared my suspicions that Amos might be calling on Set for extra power. Zia didn't contradict me, and the news didn't sit well with my brother. After several minutes of swearing and pacing the room, he finally calmed down enough to say, 'We can't let that happen. He'll be destroyed.'
'I know,' I said. 'But we can best help him by moving forward.'
I didn't mention Zia's blackout in the nursing home. In Carter's present state of mind, I thought that might be too much for him. But I did tell him what Tawaret had said about the possible location of Bes's shadow.
'The ruins of Saïs...' He frowned. 'I think Dad mentioned that place. He said there wasn't much left. But even if we could find the shadow we don't have time. We've got to stop Apophis.'
'I made a promise,' I insisted. 'Besides, we _need_ Bes. Think of it as a trial run. Saving his shadow will give us a chance to practise this sort of magic before we try it on Apophis – _um_ , in reverse, of course. It might even give us a way to revive Ra.'
'But –'
'She's got a point,' Walt interrupted.
I'm not sure who was more surprised – Carter, or me.
'Even if we get Setne's help,' Walt said, 'trapping a shadow in a statue is going to be difficult. I'd feel better if we could try it on a friendly target first. I could show you how it's done while – while I still have time.'
'Walt,' I said, 'please, don't talk like that.'
'When you face Apophis,' he continued, 'you'll have only one chance to get the spell right. It would be better to have some practice.'
_When you face Apophis_. He said it so calmly, but his meaning was clear: he wouldn't be around when that happened.
Carter nudged his half-eaten pizza. 'I just... I don't see how we can do it all in time. I know this is a personal mission for you, Sadie, but –'
'She has to,' Zia said gently. 'Carter, you once went off on personal mission in the middle of a crisis, didn't you? That worked out.' She put her hand on Carter's. 'Sometimes you have to follow your heart.'
Carter looked like he was trying to swallow a golf ball. Before he could say anything, the ship's bell sounded.
In the corner of the dining room, a loudspeaker crackled with Bloodstained Blade's voice: 'My lords and ladies, we have reached the Hall of Judgement.'
The black temple looked just as I remembered. We made our way up the steps from the dock and passed between rows of obsidian columns that marched into the gloom. Sinister-looking scenes of Underworld life glittered on the floor and in friezes circling the pillars – black designs on black stone. Despite the reed torches that burned every few metres, the air was so hazy with volcanic ash I couldn't see far in front of us.
As we moved deeper into the temple, voices whispered around us. Out of the corner of my eye, I saw groups of spirits drifting across the pavilion – ghostly shapes camouflaged in the smoky air. Some moved aimlessly – crying softly or tearing at their clothes in despair. Others carried armfuls of papyrus scrolls. These ghosts looked more solid and purposeful, as if they were waiting for something.
'Petitioners,' Walt said. 'They've brought their case files, hoping for an audience with Osiris. He was gone so long... there must be a real backlog of cases.'
Walt's step seemed lighter. His eyes looked more alert, his body less weighed down by pain. He was so close to death I'd feared this trip to the Underworld might be hard for him, but if anything he seemed more at ease than the rest of us.
'How do you know?' I asked.
Walt hesitated. 'I'm not sure. It just seems... correct.'
'And the ghosts without scrolls?'
'Refugees,' he said. 'They're hoping this place will protect them.'
I didn't ask what from. I remembered the ghost at the Brooklyn Academy dance who'd been engulfed in black tendrils and dragged underground. I thought about the vision Carter had described – our mother huddled beneath a cliff somewhere in the Duat, resisting the pull of a dark force in the distance.
'We need to hurry.' I started to forge ahead, but Zia grabbed my arm.
'There,' she said. 'Look.'
The smoke parted. Twenty metres ahead stood a massive set of obsidian doors. In front of them, an animal the size of a greyhound sat on its haunches – an oversized jackal with thick black fur, fluffy pointed ears and a face somewhere between a fox and a wolf. Its moon-coloured eyes glittered in the darkness.
It snarled at us, but I wasn't put off. I may be biased, but I think jackals are cute and cuddly, even if they _were_ known for digging up graves in Ancient Egypt.
'It's just Anubis,' I said hopefully. 'This is where we met him last time.'
'That's not Anubis,' Walt warned.
'Of course it is,' I told him. 'Watch.'
'Sadie, don't,' Carter said, but I walked towards the guardian.
'Hullo, Anubis,' I called. 'It's just me, Sadie.'
The cute fuzzy jackal bared his fangs. His mouth began to froth. His adorable yellow eyes sent an unmistakable message: _One more step and I'll chew your head off._
I froze. 'Right... that's not Anubis, unless he's having a really bad day.'
'This is where we met him before,' Carter said. 'Why isn't he here?'
'It's one of his minions,' Walt ventured. 'Anubis must be... elsewhere.'
Again, he sounded awfully sure, and I felt a strange pang of jealousy. Walt and Anubis seemed to have spent more time talking with each other than with me. Walt was suddenly an expert on all things deathly. Meanwhile, I couldn't even be _near_ Anubis without invoking the wrath of his chaperone – Shu, the god of hot air. It wasn't fair!
Zia moved next to me, gripping her staff. 'So, what now? Do we have to defeat it to pass?'
I imagined her lobbing some of her daisy-destroying fireballs. That's all we needed – a yelping, flaming jackal running through my father's courtyard.
'No,' Walt said, stepping forward. 'It's just a gatekeeper. It needs to know our business.'
'Walt,' Carter said, 'if you're wrong...'
Walt raised his hands and slowly approached the jackal. 'I am Walt Stone,' he said. 'This is Carter and Sadie Kane. And this is Zia...'
'Rashid,' Zia supplied.
'We have business at the Hall of Judgement,' Walt said.
The jackal snarled, but it sounded more inquisitive, not so _chew-your-head-off_ hostile.
'We have testimony to offer,' Walt continued. 'Information relevant to the trial of Setne.'
'Walt,' Carter whispered, 'when did you become a junior lawyer?'
I shushed him. Walt's plan seemed to be working. The jackal tilted its head as if listening, then rose and padded away into the darkness. The obsidian double doors swung open silently.
'Well done, Walt,' I said. 'How did you... ?'
He faced me, and my heart did a somersault. Just for a moment I thought he looked like... No. Obviously my mixed-up emotions were playing with my mind. ' _Um_ , how did you know what to say?'
Walt shrugged. 'I took a guess.'
Just as quickly as they'd opened, the doors began to close.
'Hurry!' Carter warned. We sprinted into the courtroom of the dead.
At the start of the autumn term – my first experience in an American school – our teacher had asked us to write down our parents' contact information and what they did for a living, in case they could help with career day. I had never heard of career day. Once I understood what it was, I couldn't stop giggling.
_Could your dad come talk about his work?_ I imagined the headmistress asking.
_Possibly, Mrs Laird..._ I'd say. _Except he's dead, you see. Well, not completely dead. He's more of a resurrected god. He judges mortal spirits and feeds the hearts of the wicked to his pet monster. Oh, and he has blue skin. I'm sure he'd make quite an impression on career day for all those students aspiring to grow up and become Ancient Egyptian deities._
The Hall of Judgement had changed since my last visit. The room tended to mirror the thoughts of Osiris, so it often looked like a ghostly replica of my family's old apartment in Los Angeles, from the happier times when we all lived together.
Now, possibly because Dad was on duty, the place was fully Egyptian. The circular chamber was lined with stone pillars carved in lotus flower designs. Braziers of magic fire washed the walls in green and blue light. In the centre of the room stood the scales of justice, two large golden saucers balanced from an iron T.
Kneeling before the scales was the ghost of a man in a pinstriped suit, nervously reciting from a scroll. I understood why he was tense. On either side of him stood a large reptilian demon with green skin, a cobra head and a wicked-looking pole arm poised over the ghost's head.
Dad sat at the far end of the room on a golden dais, with a blue-skinned Egyptian attendant at his side. Seeing my father in the Duat was always disorientating, because he appeared to be two people at once. On one level, he looked like he had in life – a handsome, muscular man with chocolate-brown skin, a bald scalp and a neatly trimmed goatee. He wore an elegant silk suit and a dark travelling coat, like a businessman about to board a private jet.
On a deeper level of reality, however, he appeared as Osiris, god of the dead. He was dressed as a pharaoh in sandals, an embroidered linen kilt and rows of gold and coral neckbands on his bare chest. His skin was the colour of a summer sky. Across his lap lay a crook and flail – the symbols of Egyptian kingship.
As strange as it was seeing my father with blue skin and a skirt, I was so happy to be near him again I quite forgot about the court proceedings.
'Dad!' I ran towards him.
[Carter says I was foolish, but Dad _was_ the king of the court, wasn't he? Why shouldn't I be allowed to run up to say hello?]
I was halfway across when the snake demons crossed their pole arms and blocked my path.
'It's all right,' Dad said, looking a bit startled. 'Let her through.'
I flew into his arms, knocking the crook and flail out of his lap.
He hugged me warmly, chuckling with affection. For a moment I felt like a little girl again, safe in his embrace. Then he held me at arm's length, and I could see how weary he was. He had bags under his eyes. His face was gaunt. Even the powerful blue aura of Osiris, which normally surrounded him like the corona of a star, flickered weakly.
'Sadie, my love,' he said in a strained voice. 'Why have you come? I'm _working_.'
I tried not to feel hurt. 'But, Dad, this is important!'
Carter, Walt and Zia approached the dais. My father's expression turned grim.
'I see,' he said. 'First let me finish this trial. Children, stand here on my right. And, please, don't interrupt.'
My dad's attendant stamped his foot. 'My lord, this is most irregular!'
He was an odd-looking fellow – an elderly blue Egyptian man with a huge scroll in his arms. Too solid to be a ghost, too blue to be human, he was almost as decrepit as Ra, wearing nothing but a loincloth, sandals and an ill-fitting wig. I suppose that glossy black wedge of fake hair was meant to look manly in an Ancient Egyptian sort of way, but, along with the kohl eyeliner and the rouge on his cheeks, the old boy looked like a grotesque Cleopatra impersonator.
The roll of papyrus he held was simply enormous. Years ago, I'd gone to synagogue with my friend Liz, and the Torah they kept there was _tiny_ in comparison.
'It's all right, Disturber,' my father told him. 'We may continue now.'
'But, my lord –' The old man (was his name really Disturber?) became so agitated he lost control of his scroll. The bottom dropped out and unravelled, bouncing down the steps like a papyrus carpet.
'Oh, bother, bother, bother!' Disturber struggled to reel in his document.
My father suppressed a smile. He turned back to the ghost in the pinstriped suit, who was still kneeling at the scales. 'My apologies, Robert Windham. You may finish your testimony.'
The ghost bowed and scrapped. 'Y-yes, Lord Osiris.'
He referred to his notes and began rattling off a list of crimes he wasn't guilty of – murder, theft and selling cattle under false pretences.
I turned to Walt and whispered, 'He's a modern chap, isn't he? What's he doing in Osiris's court?'
I was a bit troubled to find that Walt once again had an answer.
'The afterlife looks different to every soul,' he said, 'depending on what they believe. For that guy, Egypt must've made a strong impression. Maybe he read the stories when he was young.'
'And if someone doesn't believe in _any_ afterlife?' I asked.
Walt gave me a sad look. 'Then that's what they experience.'
On the other side of the dais, the blue god Disturber hissed at us to be quiet. Why is it when adults try to silence kids they always make more noise than the noise they're trying to stop?
The ghost of Robert Windham seemed to be winding down his testimony. 'I haven't given false witness against my neighbours. _Um_ , sorry, I can't read this last line –'
'Fish!' Disturber yelped crossly. 'Have you stolen any fish from the holy lakes?'
'I lived in Kansas,' the ghost said. 'So... no.'
My father rose from his throne. 'Very well. Let his heart be weighed.'
One of the snake demons produced a linen parcel the size of a child's fist.
Next to me, Carter inhaled sharply. 'His _heart_ is in there?'
'Shh!' Disturber said so loudly his wig almost fell off. 'Bring forth the Destroyer of Souls!'
On the far wall of the chamber, a doggy door burst open. Ammit ran into the room in great excitement. The poor dear wasn't very coordinated. His miniature lion chest and forearms were sleek and agile, but his back half was a stubby and much-less-agile hippo bum. He kept sliding sideways, swerving into pillars and knocking over braziers. Each time he crashed, he shook his lion's mane and crocodile snout and yipped happily.
[Carter is scolding me, as always. He says Ammit is female. I'll admit I can't prove it either way, but I've always thought of Ammit as a boy monster. He's much too hyper to be otherwise, and the way he marks his territory... but never mind.]
'There's my baby!' I cried, quite carried away. 'There's my Poochiekins!'
Ammit ran at me and leaped into my arms, nuzzling me with his rough snout.
'My lord Osiris!' Disturber lost the bottom of his scroll again, which unravelled around his legs. 'This is an outrage!'
'Sadie,' Dad said firmly, 'please do not refer to the Devourer of Souls as Poochiekins.'
'Sorry,' I muttered, and let Ammit down.
One of the snake demons set Robert Windham's heart on the scales of justice. I'd seen many pictures of Anubis performing this duty, and I wished he were here now. Anubis would've been _much_ more interesting to watch than some snake demon.
On the opposite scale, the Feather of Truth appeared. (Don't get me started on the Feather of Truth.)
The scales wavered. The two saucers stopped, just about even. The pinstriped ghost sobbed with relief. Ammit whimpered disappointedly.
'Most impressive,' my father said. 'Robert Windham, you have been found sufficiently virtuous, despite the fact you were an investment banker.'
'Red Cross donations, baby!' the ghost yelled.
'Yes, well,' Dad said dryly, 'you may proceed to the afterlife.'
A door opened to the left of the dais. The snake demons hauled Robert Windham to his feet.
'Thank you!' he yelled, as the demons escorted him out. 'And if you need any financial advice, Lord Osiris, I still believe in the long term viability of the market –'
The door shut behind him.
Disturber sniffed indignantly. 'Horrible man.'
My father shrugged. 'A modern soul who appreciated the ancient ways of Egypt. He couldn't have been all bad.' Dad turned to us. 'Children, this is Disturber, one of my advisors and gods of judgement.'
'Sorry?' I pretended not to have heard. 'Did you say he's _disturbed_?'
'Disturber is my name!' the god shouted angrily. 'I judge those who are guilty of losing their temper!'
'Yes.' Despite my father's weariness, his eyes sparkled with amusement. 'That was Disturber's traditional duty, although now that he's my last minister he helps me with all my cases. There used to be forty-two judgement gods for different crimes, you see, but –'
'Like Hot Foot and Fire-embracer,' Zia said.
Disturber gasped. 'How do you know of them?'
'We saw them,' Zia said. 'In the Fourth House of the Night.'
'You – saw –' Disturber almost dropped his scroll altogether. 'Lord Osiris, we must save them immediately! My brethren –'
'We will discuss it,' Dad promised. 'First, I want to hear what brings my children to the Duat.'
We took turns explaining: the rebel magicians and their secret alliance with Apophis, their impending attack on the First Nome and our hope to find a new sort of execration spell that might stop Apophis for good.
Some of our news surprised and troubled our father – like the fact that many magicians had fled the First Nome, leaving it so poorly defended that we'd sent our initiates from Brooklyn House to help and that Amos was flirting with the powers of Set.
'No,' Dad said. 'No, he can't! These magicians who've abandoned him – inexcusable! The House of Life must rally to the Chief Lector.' He began to rise. 'I should go to my brother –'
'My lord,' Disturber said, 'you are not a magician any more. You are Osiris.'
Dad grimaced, but he eased back into his throne. 'Yes. Yes, of course. Please, children, continue.'
Some of our news Dad already knew. His shoulders slumped when we mentioned the spirits of the dead who were disappearing and the vision of our mum lost somewhere in the deep Duat, fighting against the pull of a dark force that Carter and I were certain was the shadow of Apophis.
'I have searched for your mother everywhere,' Dad said despondently. 'This force that is taking the spirits – whether it's the serpent's shadow or something else – I cannot stop it. I can't even _find_ it. Your mother...'
His expression turned brittle as ice. I understood what he was feeling. For years he had lived with guilt because he couldn't prevent our mum's death. Now she was in danger again and, even though he was the lord of the dead, he felt helpless to save her.
'We can find her,' I promised. 'All of this is connected, Dad. We have a plan.'
Carter and I explained about the _sheut_ and how it might be used for a king-sized execration spell.
My father sat forward. His eyes narrowed. 'Anubis _told_ you this? He revealed the nature of the _sheut_ to a mortal?'
His blue aura flickered dangerously. I'd never been scared of my dad, but I'll admit I took a step back. 'Well... it wasn't just Anubis.'
'Thoth helped,' Carter said. 'And some of it we guessed –'
'Thoth!' my father spat. 'This is dangerous knowledge, children. Much too dangerous. I won't have you –'
'Dad!' I shouted. I think I surprised him, but my patience had finally snapped. I'd had quite enough of gods telling me what I _shouldn't_ or _couldn't_ do. 'Apophis's shadow is what's drawing the souls of the dead. It has to be! It's feeding on them, getting stronger as Apophis prepares to rise.'
I hadn't really processed that idea before, but as I spoke the words they felt like the truth – horrifying, but the truth.
'We've got to find the shadow and capture it,' I insisted. 'Then we can use it to banish the serpent. It's our only chance – unless you want us to use a _standard_ execration. We've got the statue ready to go for that, don't we, Carter?'
Carter patted his backpack. 'The spell will kill us,' he said. 'And it probably won't work. But if that's our only option...'
Zia looked horrified. 'Carter, you didn't tell me! You made a statue of – of _him_? You'd sacrifice yourself to –'
'No,' our father said. The anger drained out of him. He slumped forward and put his face in his hands. 'No, you're right, Sadie. A small chance is better than none. I just couldn't bear it if you...' He sat up and took a breath, trying to regain his composure. 'How can I help? I assume you came here for a reason, but you're asking for magic I don't possess.'
'Yes, well,' I said, 'that's the tricky part.'
Before I could say more, the sound of a gong reverberated through the chamber. The main doors began to open.
'My lord,' Disturber said, 'the next trial begins.'
'Not now!' my father snapped. 'Can't it be delayed?'
'No, my lord.' The blue god lowered his voice. 'This is _his_ trial. You know...'
'Oh, by the twelve gates of the night,' Dad cursed. 'Children, this trial is very serious.'
'Yes,' I said. 'Actually, that's what –'
'We'll talk afterwards,' Dad cut me off. 'And please, whatever you do, don't speak to the accused or make eye contact with him. This spirit is particularly –'
The gong sounded again. A troop of demons marched in, surrounding the accused. I didn't have to ask who he was.
Setne had arrived.
The guards were intimidating enough – six red-skinned warriors with guillotine blades for heads.
Even without them, I could tell Setne was dangerous from all the magical precautions. Glowing hieroglyphs spiralled around him like the rings of Saturn – a collection of anti-magic symbols like: _Suppress_ , _Dampen_ , _Stay_ , _Shut up_ , _Powerless_ and _Don't even think about it._
Setne's wrists were bound together with pink strips of cloth. Two more pink bands were tied round his waist. One was fastened round his neck, and two more connected his ankles so he shuffled as he walked. To the casual observer, the pink ribbons might've looked like the Hello Kitty incarceration play set, but I knew from personal experience that they were some of the most powerful magic bonds in the world.
'The Seven Ribbons of Hathor,' Walt whispered. 'I wish I could make some of those.'
'I've got some,' Zia murmured. 'But the recharge time is _really_ long. Mine won't be ready until December.'
Walt looked at her in awe.
The guillotine demons fanned out on either side of the accused.
Setne himself didn't look like trouble, certainly not someone worthy of so much security. He was quite small – not _Bes_ small, mind you, but still a diminutive man. His arms and legs were scrawny. His chest was a xylophone of ribs. Yet he stuck out his chin and smiled confidently as if he owned the world – which isn't easy when one is wearing only a loincloth and some pink ribbons.
Without a doubt, his face was the same one I'd seen in the wall at the Dallas Museum and again in the Hall of Ages. He'd been the priest who sacrificed that bull in the shimmering vision from the New Kingdom.
He had the same hawkish nose, heavy-lidded eyes and thin cruel lips. Most priests from ancient times were bald, but Setne's hair was dark and thick, slicked back with oil like a 1950s tough boy. If I'd seen him in Piccadilly Circus (with more clothes on, hopefully) I would've steered clear, assuming he was handing out advertisements or trying to sell scalped tickets to a West End show. Sleazy and annoying? Yes. Dangerous? Not really.
The guillotine demons pushed him to his knees. Setne seemed to find that amusing. His eyes flickered over the room, registering each one of us. I tried not to make eye contact, but it was difficult. Setne recognized me and winked. Somehow I knew that he could read my jumbled emotions quite well and that he found them funny.
He inclined his head towards the throne. 'Lord Osiris, all this fuss for me? You shouldn't have.'
My father didn't answer. With a grim expression, he gestured at Disturber, who shuffled through his scroll until he found the proper spot.
'Setne, also known as Prince Khaemwaset –'
'Oh, wow...' Setne grinned at me, and I fought the urge to smile back. 'Haven't heard _that_ name in a while. That's ancient history, right there!'
Disturber huffed. 'You stand accused of heinous crimes! You have blasphemed against the gods four thousand and ninety-two times.'
'Ninety-one,' Setne corrected. 'That crack about Lord Horus – that was just a misunderstanding.' He winked at Carter. 'Am I right, pal?'
How in the world did he know about Carter and Horus?
Disturber shuffled his scroll. 'You have used magic for evil purposes, including twenty-three murders –'
'Self-defence!' Setne tried to spread his hands, but the ribbons restrained him.
'– including one incident where you were _paid_ to kill with magic,' Disturber said.
Setne shrugged. 'That was self-defence for my employer.'
'You plotted against three separate pharaohs,' Disturber continued. 'You tried to overthrow the House of Life on six occasions. Most grievous of all, you robbed the tombs of the dead to steal books of magic.'
Setne laughed easily. He glanced at me as if to say, _Can you believe this guy?_
'Look, Disturber,' he said, 'that _is_ your name, right? A handsome, intelligent judgement god like you – you've got to be overworked and underappreciated. I feel for you, I really do. You've got better things to do than dig up my old history. Besides, all these charges – I answered them already in my previous trials.'
'Oh.' Disturber looked confused. He adjusted his wig self-consciously and turned to my dad. 'Should we let him go, then, my lord?'
'No, Disturber.' Dad sat forward. 'The prisoner is using Divine Words to influence your mind, warping the most sacred magic of Ma'at. Even in his bindings, he is dangerous.'
Setne examined his fingernails. 'Lord Osiris, I'm flattered, but, honestly, these charges –'
'Silence!' Dad thrust his hand towards the prisoner. The swirling hieroglyphs glowed brighter around him. The Ribbons of Hathor tightened.
Setne began to choke. His smug expression melted, replaced by absolute hatred. I could feel his anger. He wanted to kill my father, kill us all.
'Dad!' I said. 'Please, don't!'
My father frowned at me, clearly unhappy with the interruption. He snapped his fingers, and Setne's bonds eased. The ghost magician coughed and retched.
'Khaemwaset, son of Ramesses,' my father said calmly, 'you have been sentenced to oblivion more than once. The first time you managed to plead for a reduced sentence, volunteering to serve the pharaoh with your magic –'
'Yes,' Setne croaked. He tried to recover his poise, but his smile was twisted with pain. 'I'm skilled labour, my lord. It would be a crime to destroy me.'
'Yet you escaped en route,' my father said. 'You killed your guards and spent the next three hundred years sowing Chaos across Egypt.'
Setne shrugged. 'It wasn't _that_ bad. Just a bit of fun.'
'You were captured and sentenced again,' my father continued, 'three more times. In each instance, you connived your way to freedom. And since the gods have been absent from the world you've run amok, doing as you pleased, committing crimes and terrorizing mortals.'
'My lord, that's unfair,' Setne protested. 'First of all, I _missed_ you gods. Honestly, it was a dull few millennia without you. As for these so-called crimes, well, some people might say the French Revolution was a first-class party! I know _I_ enjoyed myself. And Archduke Ferdinand? A total bore. If you knew him, you would've assassinated him too.'
'Enough!' Dad said. 'You are done. I am the host of Osiris now. I will not tolerate the existence of a villain like you, even as a spirit. This time you are out of tricks.'
Ammit yipped excitedly. The guillotine guards chopped their blades up and down as if they were clapping. Disturber cried, 'Hear, hear!'
As for Setne... he threw back his head and laughed.
My father looked stunned, then outraged. He raised his hand to tighten the Ribbons of Hathor, but Setne said, 'Wait, my lord. Here's the thing. I'm _not_ out of tricks. Ask your children over there. Ask their friends. Those kids need my help.'
'No more lies,' my father growled. 'Your heart shall be weighed, _again_ , and Ammit will devour –'
'Dad!' I shrieked. 'He's right! We _do_ need him.'
My father turned towards me. I could practically see the grief and rage roiling inside him. He'd lost his wife again. He was powerless to assist his brother. A battle for the end of the world was about to begin, and his children were on the front line. Dad _needed_ to serve justice on this ghost magician. He needed to feel that he could do something right.
'Dad, please, listen,' I said. 'I know it's dangerous. I know you'll hate this. But we came here because of Setne. What we told you earlier about our plan – Setne's got the knowledge we need.'
'Sadie's right,' Carter said. 'Please, Dad. You asked how you could help. Give us custody of Setne. He's the key to defeating Apophis.'
At the sound of that name, a cold wind blew through the courtroom. The braziers sputtered. Ammit whimpered and put his paws over his snout. Even the guillotine demons shuffled nervously.
'No,' Dad said. 'Absolutely not. Setne is influencing you with his magic. He is a servant of Chaos.'
'My lord,' Setne said, his tone suddenly soft and respectful, 'I'm a lot of things, but a servant of the snake? No. I don't want the world destroyed. There's nothing in that for me. Listen to the girl. Let her tell you her plan.'
The words worked their way into my mind. I realized Setne was using magic, commanding me to speak. I steeled myself against the urge. Sadly, Setne was ordering me to do something I loved – talk. It all came spilling out: how we'd tried to save the Book of Overcoming Apophis in Dallas, how Setne had spoken with me there, how we'd found the shadow box and struck on the idea of using the _sheut_. I explained my hopes to revive Bes and destroy Apophis.
'It's impossible,' Dad said. 'Even if it wasn't, Setne can't be trusted. I would never release him, especially not to my children. He'd kill you at the first opportunity!'
'Dad,' Carter said, 'we're not children any more. We can do this.'
The agony in my father's face was hard to bear. I forced back my tears and approached the throne.
'Dad, I know you love us.' I gripped his hand. 'I know you want to protect us, but you risked everything to give us a chance at saving the world. Now it's time we did that. This is the only way.'
'She's right.' Setne managed to sound regretful, as if he were sorry he might get a reprieve. 'Also, my lord, it's the only way to save the spirits of the dead before the shadow of Apophis destroys them all – including your wife.'
My father's face turned from sky blue to deep indigo. He gripped the throne like he wanted to tear off the armrests.
I thought Setne had gone too far.
Then my father's hands relaxed. The anger in his eyes changed to desperation and hunger.
'Guards,' he said, 'give the prisoner the Feather of Truth. He will hold it while he explains himself. If he lies, he will perish in flames.'
One of the guillotine demons plucked the feather from the scales of justice. Setne looked unconcerned as the glowing plume was placed in his hands.
'Right!' he began. 'So your kids are correct. I did create a shadow execration spell. In theory, it could be used to destroy a god – or even Apophis. I never tried. Unfortunately, it can only be cast by a living magician. I died before I could test it. Not that I wanted to kill any gods, my lord. I was just thinking I'd use it to blackmail them into doing my bidding.'
'Blackmail... the gods,' Dad growled.
Setne smiled guiltily. 'This was back in my misguided youth. Anyway, I recorded the formula in several copies of the Book of Overcoming Apophis.'
Walt grunted. 'Which have all been destroyed.'
'Okay,' Setne said, 'but my original notes would still be in the margins of the Book of Thoth that I... that I stole. See? Being honest. I guarantee you even Apophis hasn't found that book. I hid it too well. I can show you where it is. The book will explain how to find the shadow of Apophis, how to capture it and how to cast the execration.'
'Can't you just tell us how?' Carter asked.
Setne pouted. 'Young master, I'd _love_ to. But I don't have the whole book memorized. And it's been millennia since I wrote that spell. If I told you one wrong word in the incantation, well... we wouldn't want any mistakes. But I can lead you to the book. Once we get it –'
' _We?_ ' Zia asked. 'Why can't you just give us directions to the book? Why do you need to come along?'
The ghost grinned. 'Because, doll, I'm the only one who can retrieve it. Traps, curses... you know. Besides, you'll need my help deciphering the notes. The spell is complicated! But don't worry. All you gotta do is keep these Ribbons of Hathor on me. It's Zia, right? You've got experience using them.'
'How did you know –?'
'If I cause you any trouble,' Setne continued, 'you can tie me up good like a Harvest Day present. But I won't try to escape – at least not until I lead you to the Book of Thoth and then get you safely to the shadow of Apophis. Nobody knows the deepest levels of the Duat like I do. I'm your best hope for a guide.'
The Feather of Truth didn't react. Setne didn't go up in flames, so I guessed he wasn't lying.
'Four of us,' Carter said. 'One of him.'
'Except he killed his guards last time,' Walt pointed out.
'So we'll be more careful,' Carter said. 'All of us together should be able to keep him under control.'
Setne winced. 'Oh, except... see, Sadie's got her little side task, doesn't she? She's gotta find the shadow of Bes. And, actually, it's a good idea.'
I blinked. 'It is?'
'Absolutely, doll,' Setne said. 'We don't have much time. More specifically, your friend Walt there doesn't have much time.'
I wanted to kill the ghost, except he was already dead. I suddenly hated that smug smile.
I gritted my teeth. 'Go on.'
'Walt Stone – sorry, pal, but you won't survive long enough to get the Book of Thoth, travel to the shadow of Apophis and use the spell. There just isn't time left on your clock. But getting Bes's shadow – that won't take as long. It'll be a good test of the magic. If it works, great! If it doesn't... well, we've only lost one dwarf god.'
I wanted to stomp his face, but he gestured for patience.
'What I'm thinking,' he said, 'is we split up. Carter and Zia, you two go with me to get the Book of Thoth. Meanwhile, Sadie takes Walt to the ruins of Saïs to find the dwarf's shadow. I'll give you some notes on how to capture it, but the spell is just theory. In practice, you'll need Walt's charm-making skill to pull it off. He'll have to improvise if anything goes wrong. If Walt succeeds, then Sadie will know how to capture a shadow. If Walt dies afterwards – and I'm sorry, but casting a spell like that will probably do him in – then Sadie can rendezvous with us in the Duat, and we'll hunt down the snake's shadow. Everybody wins!'
I wasn't sure whether to weep or scream. I only managed to keep my calm because I sensed that Setne would find any reaction extremely funny.
He faced my father. 'What do you say, Lord Osiris? It's a chance to get your wife back, defeat Apophis, restore Bes's soul, save the world! All I ask is that when I come back, the court take my good deeds into consideration when you sentence me. How fair is that, huh?'
The chamber was silent except for the crackling fires in the braziers.
Finally Disturber seemed to shake himself out of a trance. 'My lord... what is your ruling?'
Dad looked at me. I could tell he hated this plan. But Setne had tempted him with the one thing he couldn't pass up: a chance to save our mum. The vile ghost had promised me one last day alone with Walt, which I wanted more than anything, and a chance to save Bes, which was a close second. He'd put Carter and Zia together and promised them a chance to save the world.
He'd put hooks in all of us and reeled us in like fish from a sacred lake. But, despite the fact that I knew we were being played, I couldn't find a reason to say no.
'We have to, Dad,' I said.
He lowered his head. 'Yes, we do. May Ma'at protect us all.'
'Oh, we'll have fun!' Setne said cheerfully. 'Shall we get going? Doomsday isn't gonna wait!'
CARTER
## 11. Don't Worry, Be Hapi
TYPICAL.
Sadie and Walt go off looking for a friendly shadow, while Zia and I escort a psychotic murderous ghost to his heavily trapped stash of forbidden magic. Gee, who got the better end of that deal?
The _Egyptian Queen_ burst out of the Underworld and into the Nile like a breaching whale. Its paddlewheel churned through the blue water. Its smokestacks billowed golden smoke into the desert air. After the gloom of the Duat, the sunlight was blinding. Once my eyes adjusted, I saw we were chugging downriver, heading north, so we must have surfaced somewhere to the south of Memphis.
On either side, marshy green riverbanks columned with palm trees stretched into the humid haze. A few houses dotted the landscape. A battered pick-up truck rumbled down the riverfront road. A sailboat glided by on our port side. No one paid us any attention.
I wasn't sure exactly where we were. It could've been anywhere along the Nile. But judging from the position of the sun it was already late morning. We'd eaten and slept in my father's realm, figuring we wouldn't be able to close our eyes once we had custody of Setne. It hadn't felt like much of a rest, but obviously we'd spent more time down under than I realized. The day was slipping by. Tomorrow at dawn, the rebels would attack the First Nome, and Apophis would rise.
Zia stood next to me at the bow. She'd showered and changed into a spare set of combat clothes – a camo tank top, olive cargo pants tucked into her boots. Maybe that doesn't sound glamorous, but in the morning sunlight she was beautiful. Best of all, she was here in person – not a reflection in the scrying bowl, not a _shabti_. When the wind changed direction, I caught the scent of her lemon shampoo. Our forearms touched as we leaned against the rail, but she didn't seem to mind. Her skin was feverishly warm.
'What are you thinking?' I asked.
She had trouble focusing on me. Up close, the flecks of green and black in her amber eyes were sort of hypnotizing. 'I was thinking about Ra,' she said. 'Wondering who's taking care of him today.'
'I'm sure he's fine.'
But I felt a little disappointed. Personally, I was thinking about the moment when Zia had taken my hand in the dining room last night: _Sometimes you have to follow your heart._ This might be our last day on earth. If it was, I should really tell Zia how I felt about her. I mean, I thought she knew, but I didn't _know_ that she knew, so... Oh, man. Headache.
I started to say, 'Zia –'
Setne materialized next to us. 'All better!'
In the daylight, he looked almost like flesh and blood, but when he turned in a circle, showing off his new clothes, his face and hands flickered holographically. I'd given him permission to put on something besides the loincloth. In fact, I'd insisted. But I hadn't expected an outfit so mind-boggling.
Maybe he was trying to live up to Sadie's nickname for him: Uncle Vinnie. He wore a black suit jacket with padded shoulders, a red T-shirt, a crisp pair of jeans and blindingly white running shoes. Round his neck was a heavy gold chain of interlocking _ankh_ s. On each pinky he wore a ring the size of a jawbreaker, with the symbol of power – _was_ – set in diamonds. His hair was combed back with even more grease. His eyes were lined with kohl. He looked like the Ancient Egyptian Mafia.
Then I noticed something missing from his ensemble. He didn't seem to be wearing the Ribbons of Hathor.
I'll admit: I panicked. I yelled the command word Zia had taught me: ' _Tas!_ '
The symbol for _Bind_ flared in Setne's face:
The Ribbons of Hathor reappeared round his neck, wrists, ankles, chest and waist. They expanded aggressively, cocooning Setne in a pink tornado until he was wrapped tight as a mummy, with nothing showing but his eyes.
'Mm!' he protested.
I took a deep breath. Then I snapped my fingers. The bindings shrank back to their normal size.
'What was _that_ for?' Setne demanded.
'I didn't see the ribbons.'
'You didn't...' Setne laughed. 'Carter, Carter, Carter. Come on, pal. That's just an illusion – a cosmetic change. I can't _really_ get out of these things.'
He held out his wrists. The ribbons vanished then reappeared. 'See? I'm just concealing them, 'cause pink doesn't go with my outfit.'
Zia snorted. 'Nothing goes with that outfit.'
Setne shot her an irritated look. 'No need to get personal, doll. Just relax, okay? You saw what happens – one word from you and I'm tied up good. No problems.'
His tone sounded so reasonable. Setne was no problem. Setne would cooperate. I could just relax.
In the back of my mind, the voice of Horus said, _Careful._
I raised my mental guard. Suddenly I was aware of hieroglyphs floating in the air around me – half-visible wisps of smoke. I willed them to disappear, and they fizzled like gnats in a bug zapper. 'Stop it with the magic words, Setne. I'll relax when our business is done and you're back in my dad's custody. Now, where are we going?'
A moment of surprise passed over Setne's face. He hid it with a smile. 'Sure, no problem. Glad to see that _path-of-the-gods_ magic is working out for you. How you doing in there, Horus?'
Zia snarled impatiently. 'Just answer the question, you maggot, before I burn that smile off your face.'
She thrust out her hand. Flames wreathed her fingers.
'Zia, whoa,' I said.
I'd seen her get angry before, but the _burn-your-smile-off_ tactic seemed a little harsh even for her.
Setne didn't seem concerned. From his jacket, he pulled a strange white comb – were those human finger bones? – and brushed his greasy hair.
'Poor Zia,' he said. 'The old man is getting to you, isn't he? Having any trouble with, ah, temperature control yet? I've seen a few people in your situation spontaneously combust. Not pretty.'
His words obviously rattled Zia. Her eyes seethed with loathing, but she closed her fist and extinguished the flames. 'You vile, despicable –'
'Take it easy, doll,' Setne said. 'I'm just expressing concern. As for where we're going – south of Cairo, the ruins of Memphis.'
I wondered what he'd meant about Zia. I decided this wasn't the time to ask. I didn't want Zia's flaming fingers in _my_ face.
I tried to recall what I knew about Memphis. I remembered it was one of the old capitals of Egypt, but it had been destroyed centuries ago. Most of the ruins were buried under modern Cairo. Some were scattered in the desert to the south. My dad had probably taken me to excavation sites in that area once or twice, but I didn't have any clear recollection. After a few years, all the dig sites sort of blended together.
'Where exactly?' I demanded. 'Memphis was a big place.'
Setne wiggled his eyebrows. 'You got that right. Man, the times I used to have in Gamblers' Alley... but forget it. The less you know, pal, the better. We don't want our snaky Chaos friend gleaning information from your mind, do we? Speaking of which, it's a miracle he hasn't already seen your plans and sent some nasty monster to stop you. You seriously need to work on your mental defences. Reading your mind is _way_ too easy. As for your girlfriend here...'
He leaned towards me with a grin. 'Would you like to know what _she's_ thinking?'
Zia understood the Ribbons of Hathor better than I did. Instantly, the band round Setne's neck tightened and became a lovely pink collar with a leash. Setne gagged and clawed at his throat. Zia grabbed the other end of the lead.
'Setne, you are I are going to the wheelhouse,' she announced. 'You will give the captain _exact_ information about where we're going, or you'll never breathe again. Understood?'
She didn't wait for a response. He couldn't have given one anyway. She dragged him across the deck and up the stairs like a very bad dog.
As soon as they'd disappeared into the pilot's house, someone next to me chuckled. 'Remind me not to get on _her_ bad side.'
Horus's instincts kicked in. Before I knew what was happening, I'd summoned my _khopesh_ from the Duat and was resting the curved edge against my visitor's throat.
'Really?' said the god of Chaos. 'This is how you greet an old friend?'
Set leaned casually against the rail in a black three-piece suit and a matching pork-pie hat. The outfit was striking against his blood-red skin. The last time I'd seen him, he'd been bald. Now he had braided cornrows decorated with rubies. His black eyes glittered behind small round glasses. With a chill, I realized he was impersonating Amos.
'Stop that.' I pressed my blade against his throat. 'Stop mocking my uncle!'
Set looked offended. 'Mocking? My dear boy, imitation is the sincerest form of flattery! Now, please, can we talk like civilized semi-divine beings?'
With one finger he pushed the _khopesh_ away from his neck. I lowered my blade. Now that I was over my initial shock, I had to admit I was curious about what he wanted.
'Why are you here?' I demanded.
'Oh, pick a reason. The world ends tomorrow. Perhaps I wanted to say goodbye.' He grinned and waved. 'Bye! Or perhaps I wanted to explain. Or give you a warning.'
I glanced towards the wheelhouse. I couldn't see Zia. No alarm bells were ringing. No one else seemed to have noticed that the god of evil had just materialized on our boat.
Set followed my gaze. 'How about that Setne, huh? I love that guy.'
'You would,' I muttered. 'Was he named after you?'
'Nah. _Setne_ is just his nickname. His real name is Khaemwaset, so you can see why he likes Setne better. I hope he doesn't kill you right away. He's a lot of fun... until he kills you.'
'Is that what you wanted to explain?'
Set adjusted his glasses. 'No, no. It's the thing with Amos. You've got the wrong idea.'
'You mean that you possessed him and tried to destroy him?' I asked. 'That you almost shattered his mind? And that now you want to do it again?'
'The first two – true. The last one – no. Amos called _me_ , kid. You gotta understand, I could never have invaded his mind in the first place if he didn't share some of my qualities. He _understands_ me.'
I clenched my sword. 'I understand you, too. You're evil.'
Set laughed. 'You figure that out all by yourself? The god of evil is evil? Sure I am, but not _pure_ evil. Not _pure_ Chaos, either. After I spent some time in Amos's head, he understood. I'm like that improvisational jazz he loves – chaos within order. That's our connection. And I'm still a god, Carter. I'm... what do you call it? The _loyal opposition_.'
'Loyal. Yeah, right.'
Set gave me a sly smile. 'Okay, I want to rule the world. Destroy anyone who gets in my way? Of course. But that snake Apophis – he takes things too far. He wants to pull the whole of creation down into a big soupy primordial mess. Where's the fun in that? If it comes down to Ra or Apophis, I fight on Ra's side. That's why Amos and I have a deal. He's learning the path of Set. I'm going to help him.'
My arms trembled. I wanted to cut Set's head off, but I wasn't sure I had the strength. I also wasn't sure it would hurt him. I knew from Horus that gods tended to laugh off simple injuries like decapitation.
'You expect me to believe you'll cooperate with Amos?' I asked. 'Without trying to overpower him?'
'Sure, I'll _try_. But you should have more faith in your uncle. He's stronger than you think. Who do you think sent me here to explain?'
An electric charge went through my body. I wanted to believe Amos had everything under control, but this was _Set_ talking. He did remind me a lot of the ghost magician Setne – and that wasn't a good thing.
'You've done your explaining,' I said. 'Now you can leave.'
Set shrugged. 'Okay, but it does seem like there was one more thing...' He tapped his chin. 'Oh, right. The warning.'
'The warning?' I repeated.
'Because usually when Horus and I fight it would be _me_ who was responsible for what's about to kill you. But this time it's not. I thought you should know. Apophis is _so_ copying my moves, but like I said...' He took off his pork-pie hat and bowed, the rubies glittering in his cornrows. 'Imitation is flattery.'
'What are you –?'
The riverboat lurched and groaned as if we'd hit a sandbar. Up in the wheelhouse, the alarm bell _ding_ ed. The glowing crew orbs zipped around the deck in a panic.
'What's happening?' I grabbed the rail.
'Oh, that'd be the giant hippo,' Set said casually. 'Good luck!'
He disappeared in a cloud of red smoke as a monstrous shape rose from the Nile.
You might not think a hippo could inspire terror. Screaming 'Hippo!' doesn't have the same impact as screaming 'Shark!'. But I'm telling you – as the _Egyptian Queen_ careened to one side, its paddlewheel lifting completely out of the water, and I saw that monster emerge from the deep, I nearly discovered the hieroglyphs for _accident in my pants_.
The creature was easily as big as our riverboat. Its skin glistened purple and grey. As it rose near the bow, it fixed its eyes on me with unmistakable malice and opened a maw the size of an airplane hangar. Its bottom peglike teeth were taller than me. Looking down the creature's throat, I felt like I was seeing a bright pink tunnel straight to the Underworld. The monster could have eaten me right there, along with the front half of the boat. I would have been too paralysed to react.
Instead, the hippo bellowed. Imagine someone revving a dirt bike, then blowing a trumpet. Now imagine those sounds amplified twenty times, coming at you in a blast of breath that smells of rotten fish and pond scum. That's what a giant hippo's war cry is like.
Somewhere behind me, Zia yelled, 'Hippo!' Which I thought was a little late.
She stumbled towards me over the rocking deck, the tip of her staff on fire. Our ghostly pal Setne floated behind her, grinning with delight.
'There it is!' Setne shook his diamond pinky rings. 'Told ya Apophis would send a monster to kill you.'
'You're so smart!' I shouted. 'Now, how do we stop it?'
'BRRRAAHHHHH!' The hippo shoved its face against the _Egyptian Queen_. I tumbled backwards and slammed against the deckhouse.
Out of the corner of my eye I saw Zia blast a column of fire at the creature's face. The flames went straight up its left nostril, which just made the hippo mad. It snorted smoke and bashed the ship harder, catapulting Zia into the river.
'No!' I staggered to my feet. I tried to summon the avatar of Horus, but my head was throbbing. My focus was shot.
'Want some advice?' Setne wafted next to me, unaffected by the rocking of the ship. 'I could give you a spell to use.'
His evil smile didn't exactly fill me with confidence.
'Just stay put!' I pointed at his hands and yelled, ' _Tas!_ '
The Ribbons of Hathor tied his wrists together.
'Oh, come on!' he complained. 'How am I supposed to comb my hair like this?'
The hippo peered at me over the rail – its eye like a greasy black dinner plate. Up in the wheelhouse, Bloodstained Blade rang the alarm bell and shouted at the crew, 'Hard to port! Hard to port!'
Somewhere over the side, I heard Zia choking and splashing, which at least meant she was alive, but I had to keep the hippo away from her and give the _Egyptian Queen_ time to disengage. I grabbed my sword, charged up the tilting deck and leaped straight onto the monster's head.
My first discovery: hippos are slippery. I scrambled for a handhold – not easy while wielding a sword – and almost slid off the other side of the hippo's head before I hooked my free arm round its ear.
The hippo roared and shook me like a dangly earring. I caught a glimpse of a fishing boat sailing calmly by as if nothing were wrong. The crew orbs of the _Egyptian Queen_ zipped around a large crack in the stern. Just for a moment, I saw Zia floundering in the water, about twenty yards downstream. Then her head went under. I summoned all my strength and drove my sword into the hippo's ear.
' _BRRRAAHHHHH!_ ' The monster thrashed its head. I lost my grip and went sailing across the river like a three-point shot.
I would've hit the water hard, but at the last second I changed into a falcon.
I know... that sounds crazy. _Oh, by the way, I just happened to change into a falcon._ But it was fairly easy magic for me, since the falcon was Horus's sacred animal. Suddenly, instead of falling, I was soaring over the Nile. My vision was so sharp I could see field mice in the marshes. I could see Zia struggling in the water, as well as every bristle on the hippo's massive snout.
I dove at the monster's eye, raking it with my claws. Unfortunately it was heavily lidded and covered with some kind of membrane. The hippo blinked and bellowed in annoyance, but I could tell that I hadn't done any real damage.
The monster snapped at me. I was much too fast. I flew to the ship and perched on the wheelhouse roof, trying to catch my breath. The _Egyptian Queen_ had managed to turn. It was slowly putting distance between itself and the monster, but the hull had taken serious damage. Smoke billowed from the cracks in the stern. We were listing to starboard, and Bloodstained Blade kept ringing his alarm bell, which was really annoying.
Zia was working to stay afloat, but she'd drifted further downstream from the hippo and didn't seem to be in immediate danger. She tried to summon fire – which isn't easy to do when you're floundering in a river.
The hippo lumbered back and forth, apparently looking for the pesky bird that had poked it in the eye. The monster's ear was still bleeding, though my sword was no longer there – maybe at the bottom of the river somewhere. Finally the hippo turned its attention to the ship.
Setne materialized next to me. His arms were still tied, but he looked like he was enjoying himself. 'You ready for that advice now, pal? I can't cast the spell myself 'cause I'm dead and all, but I can tell you what to say.'
The hippo charged. It was less than fifty yards away, closing fast. If it hit the ship at that speed, the _Egyptian Queen_ would break into kindling.
Time seemed to slow down. I tried to gather my focus. Emotions are bad for magic, and I was completely panicked, but I knew I'd only get one shot at this. I spread my wings and flew straight at the hippo. Halfway there, I transformed back into a human, dropped like a stone and summoned the avatar of Horus.
If it hadn't worked, I would've ended my life as an insignificant grease spot on the chest of a charging hippo.
Thankfully, the blue aura flickered around me. I landed in the river encased in the glowing body of a twenty-foot-tall hawk-headed warrior. Compared to the hippo, I was still tiny, but I got its attention when I drove my fist into its snout.
That worked really well for about two seconds. The monster forgot all about the ship. I sidestepped and made it turn towards me, but I was way too slow. Wading through the river in avatar form was about as easy as running through a room full of bouncy balls.
The monster lunged. It twisted its head and clamped its mouth round my waist. I staggered, trying to break free, but its jaws were like a vice grip. Its teeth sank into the magical shielding. I didn't have my sword. All I could do was pummel its head with my glowing blue fists, but my power was fading rapidly.
'Carter!' Zia screamed.
I had maybe ten seconds to live. Then the avatar would collapse, and I'd be swallowed or bitten in half.
'Setne!' I yelled. 'What's that spell?'
'Oh, _now_ you want the spell,' Setne called from the ship. 'Repeat after me: _Hapi, u-ha ey pwah_.'
I didn't know what that meant. Setne might've been tricking me into self-destructing or transforming into a chunk of Swiss cheese. But I was out of options. I shouted: ' _Hapi, u-ha ey pwah!_ '
Blue hieroglyphs – brighter than I'd ever summoned – blazed above the hippo's head:
Seeing them written out, I suddenly understood their meaning: _Hapi, arise and attack._ But what did _that_ mean?
At least they distracted the hippo. It let go of me and snapped at the hieroglyphs. My avatar failed. I plunged into the water, my magic exhausted, my defences gone – just tiny little Carter Kane in the shadow of a sixteen-ton hippo.
The monster swallowed the hieroglyphs and snorted. It shook its head as if it had just gulped down a chili pepper.
_Great_ , I thought. _Setne's awesome magic summoned an appetizer for the devil hippo._
Then, from the boat, Setne yelled, 'Wait for it! Three, two, one...'
The Nile boiled around me. A huge mass of brown seaweed erupted beneath me and lifted me skywards. Instinctively I held on, slowly realizing that the seaweed wasn't seaweed. It was _hair_ on top of a colossal head. The giant man rose from the Nile, higher and higher, until the hippo looked almost cute in comparison. I couldn't tell much about the giant from the top of his head, but his skin was darker blue than my father's. He had shaggy brown hair full of river muck. His belly was hugely swollen, and he seemed to be wearing nothing but a loincloth made of fish scales.
' _BRRRAAHHHHH!_ ' The hippo lunged, but the blue giant grabbed its bottom teeth and stopped it cold. The force of the impact nearly shook me off his head.
'Yay!' the blue giant bellowed. 'Hippo toss! I love this game!' He swung his arms in a golf-swing motion and launched the monster out of the water.
Few things are stranger than watching a giant hippo fly. It careened wildly, kicking its stubby legs as it sailed over the marshlands. Finally it crashed into a limestone cliff in the distance, causing a minor avalanche. Boulders collapsed on top of the hippo. When the dust settled, there was no sign of the monster. Cars kept driving down the river road. Fishing boats went about their business, as if blue giants fighting hippos was nothing remarkable on this stretch of the Nile.
'Fun!' the blue giant cheered. 'Now, who summoned me?'
'Up here!' I yelled.
The giant froze. He carefully patted his scalp until he found me. Then he picked me up with two fingers, waded over to the riverbank and gently set me down.
He pointed to Zia, who was struggling to reach the shore, and the _Egyptian Queen_ , which was drifting downstream, listing and smoking from the stern. 'Are those friends of yours?'
'Yes,' I said. 'Could you help them?'
The giant grinned. 'Be right back!'
A few minutes later the _Egyptian Queen_ was safely moored. Zia sat next to me on the shore, wringing Nile water out of her hair.
Setne hovered next to us, looking quite smug, even though his arms were still tied. 'So maybe _next time_ you'll trust me, Carter Kane!' He nodded at the giant, who loomed over us, still grinning like he was _really_ excited to be here. 'May I present my old friend Hapi!'
The blue giant waved at us. 'Hi!'
His eyes were completely dilated. His teeth were brilliant white. A mass of stringy brown hair fell round his shoulders, and his skin rippled in different shades of watery blue. His belly was much too big for his body. It sagged over his fish-scale skirt like he was either pregnant or had swallowed a blimp. He was, without a doubt, the tallest, fattest, bluest, most cheerful hippie giant I had ever met.
I tried to place his name, but I couldn't.
'Hapi?' I asked.
'Why, yes, I am happy!' Hapi beamed. 'I'm always happy because I'm Hapi! Are you happy?'
I glanced at Setne, who seemed to find this terribly amusing.
'Hapi is the god of the Nile,' the ghost explained. 'Along with his other duties, Hapi is the provider of bountiful harvests and all good things, and so he is always –'
'Happy,' I guessed.
Zia frowned up at the giant. 'Does he have to be so big?'
The god laughed. Immediately he shrank down to human size, though the crazy cheerful look on his face was still pretty unnerving.
'So!' Hapi rubbed his hands with anticipation. 'Anything else I can do for you kids? It's been centuries since anybody summoned me. Since they built that stupid Aswan Dam, the Nile doesn't flood every year like it used to. Nobody depends on me any more. I could _kill_ those mortals!'
He said this with a smile, as if he'd suggested bringing the mortals some home-baked cookies.
I did some quick thinking. It's not often a god offers to do you favours – even if that god is psychotically over-caffeinated. 'Actually, yes,' I said. 'See, Setne suggested I summon you to deal with the hippo, but –'
'Oh, Setne!' Hapi chuckled and pushed the ghost playfully. 'I _hate_ this guy. Absolutely despise him! He's the only magician who ever learned my secret name. Ha!'
Setne shrugged. 'It was nothing, really. And, I gotta say, you came in handy many times back in the old days.'
'Ha, ha!' Hapi's smile became painfully wide. 'I'd love to rip off your arms and legs, Setne. That would be amazing!'
Setne's expression remained calm, but he drifted a little further away from the smiling god.
' _Um_ , anyway,' I said. 'We're on a quest. We need to find this magic book to defeat Apophis. Setne was leading us to the ruins of Memphis, but now our boat is busted. Do you think –?'
'Oh!' Hapi clapped excitedly. 'The world is going to end tomorrow. I forgot!'
Zia and I exchanged looks.
'Right...' I said. 'So, if Setne told you exactly where we were going, could you take us there? And, _um_ , if he won't tell you, then you could rip his limbs off. That would be fine.'
'Yay!' Hapi cried.
Setne gave me a murderous look. 'Yeah, sure. We're going to the _serapeum_ – the temple of the Apis Bull.'
Hapi smacked his knee. 'I should have figured! Brilliant place to hide something. That's pretty far inland, but, sure, I can send you there if you want. And, just so you know, Apophis has demons scouring the riverbanks. You'd never get to Memphis without my help. You'd get torn into a million pieces!'
He seemed genuinely pleased to share that news.
Zia cleared her throat. 'Okay, then. We'd love your help.'
I turned towards the _Egyptian Queen_ , where Bloodstained Blade stood at the railing, awaiting further orders. 'Captain,' I called, 'wait here and continue repairing the ship. We'll –'
'Oh, the ship can go too!' Hapi interrupted. 'That's no problem.'
I frowned. I wasn't sure how the river god was going to move the ship, especially since he'd told us Memphis was inland, but I decided not to ask.
'Belay that order,' I called to the captain. 'The ship is coming with us. Once we reach Memphis, you'll continue repairs and await further orders.'
The captain hesitated. Then he bowed his axe-blade head. 'I obey, my lord.'
'Great!' Hapi said.
He held out his palm, which contained two slimy black orbs like fish eggs. 'Swallow these. One each.'
Zia wrinkled her nose. 'What are they?'
'They'll take you where you want to go!' the god promised. 'They're Hapi pills.'
I blinked. 'What now?'
The ghost Setne cleared his throat. He looked like he was trying not to laugh. 'Yeah, you know. Hapi invented them. So that's what they're called.'
'Just eat them!' Hapi said. 'You'll see.'
Reluctantly, Zia and I took the pills. They tasted even worse than they looked. Instantly, I felt dizzy. The world shimmered like water.
'It was nice to meet you!' Hapi cried, his voice turning murky and distant. 'You do realize you're walking into a trap, don't you? Okay! Good luck!'
With that, my vision went blue and my body melted into liquid.
CARTER
## 12. Bulls with Freaking Laser Beams
BEING LIQUIDATED IS NOT FUN. I will never be able to walk by another LIQUIDATION SALE sign without getting seasick and feeling like my bones are turning to tapioca.
I know I'm going to sound like a public service announcement here, but for all you kids at home: if somebody offers you Hapi pills, just say, 'No!'
I felt myself seeping inland through the mud, travelling at incredible speed. When I hit the hot sand, I evaporated, rising above the ground as a cloud of moisture, pushed west by the winds into the desert. I couldn't exactly see, but I could feel the movement and the heat. My molecules agitated as the sun dispersed me.
Suddenly the temperature dropped again. I sensed cool stone around me – a cave or an underground room, maybe. I coalesced into moisture, splashed to the floor as a puddle, then rose and solidified into Carter Kane once more.
For my next trick, I buckled to my knees and lost my breakfast.
Zia stood near me, hugging her stomach. We seemed to be in the entry tunnel of a tomb. Below us, stone steps led into the darkness. A few feet above, desert sunlight blazed.
'That was _horrible_ ,' Zia gasped.
I could only nod. Now I understood the science lesson my dad had once taught me in home-schooling – matter has three forms: solid, liquid and gas. In the last few minutes I'd been all three. And I didn't like it.
Setne materialized just outside the doorway, smiling down at us. 'So, did I come through again, or what?'
I didn't remember loosening his bonds, but his arms were now free. That would've worried me more if I hadn't felt so sick.
Zia and I were still wet and muddy from our swim in the Nile, but Setne looked immaculate – jeans and T-shirt freshly pressed, Elvis hair perfect, not even a spot on his white running shoes. That disgusted me so much I staggered into the sunlight and threw up on him. Unfortunately, my stomach was mostly empty and he was a ghost, so nothing much happened.
'Hey, pal!' Setne adjusted his golden _ankh_ necklace and straightened his jacket. 'Some respect, all right? I did you a favour.'
'A favour?' I gulped back the horrible taste in my mouth. 'Don't – _ever_ –'
'Never Hapi again,' Zia finished for me. 'Never.'
'Aw, c'mon!' Setne spread his hands. 'That was a smooth trip! Look, even your ship made it.'
I squinted. Mostly we were surrounded by flat, rocky desert, like the surface of Mars, but beached on a nearby sand dune was a slightly broken riverboat – the _Egyptian Queen_. The stern wasn't on fire any more, but the ship looked like it had taken more damage in transit. A section of railing was broken. One of the smokestacks was leaning dangerously. For some reason, a huge slimy tarp of fish scales was hanging off the pilot's house like a snagged parachute.
Zia muttered, 'Oh, gods of Egypt – please don't let that be Hapi's loincloth.'
Bloodstained Blade stood at the bow, facing our direction. He had no expression, being an axe head, but from the way his arms were crossed I could tell he was not a Hapi camper.
'Can you fix the ship?' I called to him.
'Yes, my lord,' he hummed. 'Given a few hours. Sadly, we seem to be stuck in the middle of a desert.'
'We'll worry about that later,' I said. 'Get the ship repaired. Wait here for us to return. You'll receive more instructions at that time.'
'As you say.' Bloodstained Blade turned and started humming at the glowing orbs in a language I didn't understand. The crew stirred into a flurry of activity.
Setne smiled. 'See? Everything's good!'
'Except we're running out of time.' I looked at the sun. I figured it was one or two in the afternoon, and we still had a lot to do before Doomsday tomorrow morning. 'Where does that tunnel go? What's a _serapeum_? And why did Hapi say it was a trap?'
'So many questions,' Setne said. 'Come on, you'll see. You're gonna love this place!'
I did not love this place.
The steps down led to a wide hall chiselled from golden bedrock. The barrelled ceiling was so low I could touch it without stretching my arms. I could tell that archaeologists had been here, from the bare electric bulbs that cast shadows across the arches. Metal beams braced the walls, but the cracks in the ceiling didn't help me feel safe. I'd never been comfortable in enclosed spaces.
Every thirty feet or so, square alcoves opened up on either side of the main hall. Each niche held a massive freestanding stone sarcophagus.
After passing the fourth such coffin, I stopped. 'Those things are way too big for humans. What's in there?'
'Bull,' Setne said.
'Excuse me?'
Setne's laugh echoed through the hall. I figured that if there were any sleeping monsters in this place they were awake now.
'These are the burial chambers for the Apis Bull.' Setne gestured around him proudly. 'I built all this, you know, back when I was Prince Khaemwaset.'
Zia ran her hand along the white stone lid of the sarcophagus. 'The Apis Bull. My ancestors thought it was an incarnation of Osiris in the mortal world.'
' _Thought?_ ' Setne snorted. 'It _was_ his incarnation, doll. At least some of the time – like on festival days and whatnot. We took our Apis Bull seriously back then.'
He patted the coffin like he was showing off a used car. 'This bad boy here? He had the perfect life. All the food he could eat. Got a harem of cows, burnt offerings, a special gold cloth for his back – all the perks. Only had to show himself in public a few times a year for big festivals. When he turned twenty-five, he got slaughtered in a big ceremony, mummified like a king and put down here. Then a new bull took his place. Nice gig, huh?'
'Killed at twenty-five,' I said. 'Sounds awesome.'
I wondered how many mummified bulls were down that hallway. I didn't want to find out. I liked being right here, where I could still see the exit and the sunlight outside. 'So why is this place called a – what was it?'
' _Serapeum,_ ' Zia answered. Her face was illuminated with golden light – probably just the electrical bulbs reflecting off the stone, but it seemed like she was glowing. 'Iskandar, my old teacher – he told me about this place. The Apis Bull was a vessel for Osiris. In later times, the names were merged: Osiris-Apis. Then the Greeks shorted it to Serapis.'
Setne sneered. 'Stupid Greeks. Moving in on our territory. Taking over our gods. I'm telling you, I got no love for those guys. But, yeah, that's how it happened. This place became known as a _serapeum_ – a house for dead bull gods. Me, I wanted to call it the Khaemwaset Memorial of Pure Awesomeness, but my dad wouldn't go for it.'
'Your dad?' I asked.
Setne waved aside the question. 'Anyway, I hid the Book of Thoth down here before I died because I knew no one would ever disturb it. You'd have to be frothing-at-the-mouth crazy to mess with the sacred tomb of the Apis Bull.'
'Great.' I felt like I was turning back into liquid.
Zia frowned at the ghost. 'Don't tell me – you hid the book in one of these sarcophagi with a mummified bull, and the bull will come to life if we disturb it?'
Setne winked at her. 'Oh, I did better than that, doll. Archaeologists have discovered _this_ part of the complex.' He gestured at the electric lights and metal support beams. 'But I'm gonna take you on a _behind-the-scenes_ tour.'
The catacombs seemed to go on forever. Hallways split off in different directions, all of them lined with sarcophagi for holy cows. After descending a long slope, we ducked through a secret passage behind an illusionary wall.
On the other side, there were no electric lights. No steel beams braced the cracked ceiling. Zia summoned fire at the tip of her staff and burned away a canopy of cobwebs. Our footprints were the only marks on the dusty floor.
'Are we close?' I asked.
Setne chuckled. 'It's just getting good.'
He led us further into the maze. Every so often, he stopped to deactivate traps with a command or a touch. Sometimes he made me do it – supposedly because he couldn't cast certain spells, being dead – though I got the feeling he thought it would be incredibly funny if I failed and died.
'How come you can touch some things but not other things?' I asked. 'You seem to have a real selective ability.'
Setne shrugged. 'I don't make the rules of the spirit world, pal. We can touch money and jewellery. Picking up trash and messing with poison spikes, no. We get to leave that dirty work to the living.'
Whenever the traps were disabled, hidden hieroglyphs glowed and vanished. Sometimes we had to jump over pits that opened in the floor, or swerve when arrows shot from the ceiling. Paintings of gods and pharaohs peeled off the walls, formed into ghostly guardians and faded. The whole time, Setne kept a running commentary.
'That curse would've made your feet rot off,' he explained. 'This one over here? That summons a plague of fleas. And this one – oh, man. This is one of my favourites. It turns you into a dwarf! I hate those short little guys.'
I frowned. Setne was shorter than me, but I decided to let it go.
'Yes, indeed,' he continued. 'You're lucky to have me along, pal. Right now, you'd be a flea-bitten dwarf with no feet. And you haven't even seen the worst of it! Right this way.'
I wasn't sure how Setne remembered so many details about this place from so long ago, but he was obviously proud of these catacombs. He must have relished designing horrible traps to kill intruders.
We turned down another corridor. The floor sloped again. The ceiling got so low I had to stoop. I tried to stay calm, but I was having trouble breathing. All I could think about were those tons of stone over my head, ready to collapse at any moment.
Zia took my hand. The tunnel was so narrow we were walking single file, but I glanced back at her.
'You okay?' I asked.
She mouthed the words: _Watch him._
I nodded. Whatever trap Hapi had warned us about, I had a feeling we hadn't seen it yet, even though we were surrounded by traps. We were alone with a murderous ghost, deep underground in his home territory. I didn't have my _khopesh_ any more. For some reason I hadn't been able to summon it from the Duat. And I couldn't use my warrior avatar in such a tiny tunnel. If Setne turned on us, my options would be limited.
Finally the corridor widened. We reached a dead end – a solid wall flanked by two statues of my dad... I mean, Osiris.
Setne turned. 'Okay, here's the score, you guys. I'm gonna have to cast a disenchantment to open this wall. The spell takes a few minutes. I don't want you freaking out halfway through and wrapping me in pink ribbons, or things could get ugly. Half-finished magic right here, and this whole tunnel could collapse on top of us.'
I managed to avoid screaming like a little girl – but only barely.
Zia cranked the fire on her staff to white-hot. 'Careful, Setne. I know what a proper disenchantment sounds like. If I suspect you're casting anything else, I'll blast you into ectoplasmic dust.'
'Relax, doll.' Setne cracked his knuckles. His diamond pinky rings flashed in the firelight. 'You gotta keep that scarab under control, or you're gonna turn _yourself_ into ashes.'
I frowned. 'Scarab?'
Setne glanced back and forth between us and laughed. 'You mean she hasn't told you? And you haven't figured it out? You _kids_ today! I _love_ the ignorance!'
He turned towards the wall and began to chant. Zia's fire ebbed to a cooler red flame. I gave her a questioning look.
She hesitated – then touched the base of her throat. She hadn't been wearing a necklace before. I was sure of that. But, when she touched her throat, an amulet blinked into existence – a glittering golden scarab on a gold chain. She must have hidden it with a glamour – a magical illusion like Setne had done with the Ribbons of Hathor.
The scarab looked metal, but I realized I'd seen it before, and I'd seen it _alive_. Back when Ra had imprisoned Apophis in the Underworld, he'd given up part of his soul – his incarnation as Khepri, scarab of the morning sun – to keep his enemy confined. He'd buried Apophis under a landslide of living beetles.
By the time Sadie and I had found that prison last spring, millions of scarabs had been reduced to desiccated shells. When Apophis broke free, only one golden beetle survived: the last remnant of Khepri's power.
Ra had tried to swallow that scarab. (Yes, disgusting. I know.) When that didn't work... he'd offered it to Zia.
I didn't remember Zia taking the scarab, but somehow I knew that amulet was the same bug.
'Zia –'
She shook her head adamantly. 'Later.'
She gestured at Setne, who was in the middle of his spell.
Okay, probably not a good time to talk. I didn't want the tunnel coming down on us. But my mind was reeling.
_You haven't figured it out?_ Setne had taunted me.
I knew Ra was fascinated with Zia. She was his favourite babysitter. Setne mentioned that Zia was having temperature control problems. _The old man is getting to you_ , he'd said. And Ra had given Zia that scarab – literally a piece of his soul – as if she were his high priestess... or maybe someone even more important.
The tunnel rumbled. The dead-end wall dissolved into dust, revealing a chamber beyond.
Setne glanced back at us with a smile. 'Showtime, kids.'
We followed him into a circular room that reminded me of the library at Brooklyn House. The floor was a sparkling mosaic of pastures and rivers. On the walls, painted priests were adorning painted cows with flowers and feathery headdresses for some kind of festival, while Ancient Egyptians waved palm fronds and shook bronze noisemakers called _sistrums_. The domed ceiling depicted Osiris on his throne, passing judgement over a bull. For an absurd moment, I wondered if Ammit devoured the hearts of wicked cows and if he liked the beefy taste.
In the middle of the chamber, on a coffin-shaped pedestal, stood a life-sized statue of the Apis Bull. It was made of dark stone – basalt, maybe – but painted so skillfully it looked alive. Its eyes seemed to follow me. Its hide glistened black except for a small white diamond on the front of its chest, and over its back was a gold blanket cut and embroidered to resemble hawk's wings. Between its horns sat a frisbee of gold – a sun disc crown. Beneath that, sticking out of the bull's forehead like a curly unicorn horn, was a rearing cobra.
A year ago I would've said, 'Freaky, but at least it's just a statue.' Now, I'd had lots of experience with Egyptian statues coming to life and trying to stomp the _ankh_ out of me.
Setne didn't seem worried. He strolled right up to the stone bull and patted its leg. 'The Shrine of Apis! I built this chamber just for my chosen priests and me. Now all we have to do is wait.'
'Wait for what?' Zia asked. Being a smart girl, she was hanging back by the entrance with me.
Setne checked his non-existent watch. 'It won't be long. Just a timer, sort of. Come on in! Make yourself comfortable.'
I edged my way inside. I waited for the doorway to solidify behind me, but it stayed open. 'You sure the book is still here?'
'Oh, yeah.' Setne walked round the statue, checking the base. 'I just need to remember which of these panels on the dais is going to pop open. I wanted to make this entire room out of gold, you know? That would've been much cooler. But Dad cut back on my funding.'
'Your dad.' Zia stepped next to me and slipped her hand into mine, which I didn't mind. The golden scarab necklace glinted round her neck. 'You mean Ramesses the Great?'
Setne's mouth twisted in a cruel sneer. 'Yeah, that's how his PR department branded him. Me, I liked to call him Ramesses II, or Ramesses Number Two.'
'Ramesses?' I said. 'Your dad is _the_ Ramesses?'
I suppose I hadn't processed how Setne fitted into Egyptian history. Looking at this scrawny little guy with his greasy hair, his shoulder-padded jacket and his ridiculous bling, I couldn't believe he was related to a ruler so famous. Even worse, it made him related to _me_ , since our mom's side of the family traced its magic heritage from Ramesses the Great.
[Sadie says she can see the family resemblance between Setne and me. (Shut up, Sadie.)]
I guess Setne didn't like my looking surprised. He stuck his beaky nose in the air. 'You should know what it's like, Carter Kane – growing up in the shadow of a famous dad. Always trying to live up to his legend. Look at you, son of the great Dr Julius Kane. You finally make a name for yourself as a big-shot magician, what does your dad do? He goes and becomes a god.'
Setne laughed coldly. I'd never felt any resentment towards my father before; I'd always thought it was cool being Dr Kane's son. But Setne's words rolled over me, and anger started to build in my chest.
_He's playing with you_ , said the voice of Horus.
I knew Horus was right, but that didn't make me feel better.
'Where's the book, Setne?' I asked. 'Enough delays.'
'Don't warp your wand, pal. It won't be much longer.' He gazed at the picture of Osiris on the ceiling. 'There he is! The blue dude himself. I'm telling you, Carter, you and I are a lot alike. I can't go anywhere in Egypt without seeing my dad's face, either. Abu Simbel? There's Papa Ramesses glaring down at me – four copies of him, each sixty feet tall. It's like a nightmare. Half the temples in Egypt? He commissioned them and put up statues of himself. Is it any wonder I wanted to be the world's _biggest_ magician?' He puffed up his scrawny chest. 'And I made it, too. What I don't understand, Carter Kane, is why you haven't taken the pharaoh's throne yet. You've got Horus on your side, itching for power. You should merge with the god, become the pharaoh of the world and, ah...' He patted the Apis statue. 'Take the bull by the horns.'
_He's right_ , Horus said. _This human has wisdom._
_Make up your mind_ , I complained.
'Carter, don't listen to him,' Zia said. 'Setne, whatever you're up to – stop. Now.'
'What _I'm_ up to? Look, doll –'
'Don't call me that!' Zia said.
'Hey, I'm on your side,' Setne promised. 'The book's right here in the dais. As soon as the bull moves –'
'The bull _moves_?' I asked.
Setne narrowed his eyes. 'Didn't I mention that? I got the idea from this holiday we used to have in the old days, the Festival of Sed. Awesome fun! You ever been to that Running of the Bulls in, what is it, Spain?'
'Pamplona,' I said. Another wave of resentment got the best of me. My dad had taken me to Pamplona once, but he hadn't let me go out in the street while the bulls were running through town. He'd said it was too dangerous – as if his secret life as a magician weren't _way_ more dangerous than that.
'Right, Pamplona,' Setne agreed. 'Well, you know where that tradition started? Egypt. The pharaoh would do this ritual race with the Apis Bull to renew his kingly power, prove his strength, get blessed by the gods – all that junk. In later times, it became just a charade, no real danger. But at the beginning it was the real thing. Life and death.'
On the word _death_ , the bull statue moved. He bent his legs stiffly. Then he lowered his head and glared at me, snorting out a cloud of dust.
'Setne!' I reached for my sword, but of course it wasn't there. 'Make that thing stop, or I'll wrap you in ribbons so fast –'
'Oh, I wouldn't do that,' Setne warned. 'See, I'm the only one who can pick up the book without getting zapped by about sixteen different curses.'
Between the bull's horns, its golden sun disc flashed. On its forehead, the cobra writhed to life, hissing and spitting gobs of fire.
Zia drew her wand. Was it my imagination, or was her scarab necklace starting to steam? 'Call off that creature, Setne. Or I swear –'
'I can't, doll. Sorry.' He grinned at us from behind the bull's dais. He didn't look very sorry. 'This is part of the security system, see? If you want the book, you've got to distract the bull and get it out of here, while I open the dais and grab the Book of Thoth. I have complete faith in you.'
The bull pawed his pedestal and leaped off. Zia pulled me back into the hallway.
'That's it!' Setne shouted. 'Just like the Sed Festival. Prove you're worthy of the pharaoh's throne, kid. Run or die!'
The bull charged.
A sword would've been really nice. I would've settled for a matador's cape and a spear. Or an assault rifle. Instead, Zia and I ran back through the catacombs and quickly realized that we were lost. Letting Setne lead us into the maze had been a stupid idea. I should've dropped breadcrumbs or marked the walls with hieroglyphs or something.
I hoped the tunnels would be too narrow for the Apis Bull. No such luck. I heard rock walls rumbling behind us as the bull shouldered his way through. There was another sound I liked even less – a deep hum followed by an explosion. I didn't know what that was, but it was good incentive to run faster.
We must have passed through a dozen halls. Each had twenty or thirty sarcophagi. I couldn't believe how many Apises had been mummified down here – centuries' worth of bull. Behind us, our monstrous stone friend bellowed as he smashed his way through the tunnels.
I glanced back once and was sorry I did. The bull was closing fast, the cobra on his forehead spewing fire.
'This way!' Zia cried.
She pulled me down a side corridor. At the far end, what looked like daylight spilled from an open doorway. We sprinted towards it.
I was hoping for an exit. Instead we stumbled into another circular chamber. There was no bull statue in the middle, but spaced round the circumference were four giant stone sarcophagi. The walls were painted with pictures of bovine paradise – cows being fed, cows frolicking in meadows, cows being worshipped by silly little humans. The daylight streamed from a shaft in the domed ceiling, twenty feet above. A beam of sunshine sliced through the dusty air and hit the middle of the floor like a spotlight, but there was no way we could use the shaft to escape. Even if I turned into a falcon, the opening was too narrow and I wasn't about to leave Zia alone.
'Dead end,' she said.
' _HRUUUFF!_ ' The Apis Bull loomed in the doorway, blocking our exit. His hood ornament cobra hissed.
We backed into the room until we stood in the warm sunlight. It seemed cruel to die here, stuck under thousands of tons of rock but able to see the sun.
The bull pawed the floor. He took a step forward then hesitated, as if the sunlight bothered him.
'Maybe I can talk to him,' I said. 'He's connected to Osiris, right?'
Zia looked at me like I was crazy – which I was – but I didn't have any better ideas.
She readied her wand and staff. 'I'll cover you.'
I stepped towards the monster and showed my empty hands. 'Nice bull. I'm Carter Kane. Osiris is my dad, sort of. How about we call a truce and –'
The cobra spewed fire in my face.
It would've turned me into an extra-crispy Carter, but Zia shouted a command. As I stumbled backwards, her staff absorbed the blast, sucking in the flames like a vacuum cleaner. She sliced the air with her wand, and a shimmering red wall of fire erupted around the Apis Bull. Unfortunately, the bull just stood there and glared at us, completely unharmed.
Zia cursed. 'We seem to be at an impasse with the fire magic.'
The bull lowered its horns.
My war-god instincts took control. 'Take cover!'
Zia dived one way. I dived the other. The bull's sun disc glowed and hummed, then shot a golden beam of heat right where we'd been standing. I barely made it behind a sarcophagus. My clothes were steaming. The bottoms of my shoes were melted. Where the beam had hit, the floor was blackened and bubbling, as if the rock had reached boiling point.
'Cows with laser beams?' I protested. 'That's _completely_ unfair!'
'Carter!' Zia called from across the room. 'You okay?'
'We'll have to split up!' I shouted back. 'I'll distract it. You get out of here!'
'What? No!'
The bull turned towards the sound of her voice. I had to move fast.
My avatar wouldn't be much good in an enclosed space like this, but I needed the war god's strength and speed. I summoned the power of Horus. Blue light flickered around me. My skin felt as thick as steel, my muscles as powerful as hydraulic pistons. I rose to my feet, smashed my fists into the sarcophagus and reduced it to a pile of stone and mummy dust. I picked up a chunk of the lid – a three-hundred-pound stone shield – and charged at the bull.
We smashed into each other. Somehow I held my ground, but it took every bit of my magical strength. The bull bellowed and pushed. The cobra spat flames that rolled over the top of my shield.
'Zia, get out of here!' I shouted.
'I'm not leaving you!'
'You've got to! I can't –'
The hairs on my arms stood up even before I heard the humming sound. My slab of stone disintegrated in a flash of gold and I flew backwards, crashing into another sarcophagus.
My vision blurred. I heard Zia shout. When my eyes could focus again, I saw her standing in the middle of the room, wrapped in sunlight, chanting a spell I didn't recognize. She'd got the bull's attention, which had probably saved my life. But, before I could cry out, the bull aimed his sun disc and shot a super-heated laser beam straight at Zia.
'No!' I screamed.
The light blinded me. The heat sucked all the oxygen out of my lungs. There was no way Zia could have survived that hit.
But when the golden light faded Zia was still there. Around her burned a massive shield shaped like... like a scarab shell. Her eyes glowed with orange fire. Flames swirled around her. She looked at the bull and spoke a deep rasping voice that definitely wasn't hers: 'I am Khepri, the rising sun. I will not be denied.'
Only later did I realize that she'd spoken in Ancient Egyptian.
She thrust out her hand. A miniature comet shot towards the Apis Bull and the monster burst into flames, turning and stomping, suddenly panicked. His legs crumbled. He collapsed and broke into a smoking pile of charred rubble.
The room was suddenly quiet. I was afraid to move. Zia was still wreathed in fire, and it seemed to be getting hotter – burning yellow, then white. She stood as if in a trance. The golden scarab round her neck was definitely smoking now.
'Zia!' My head throbbed, but I managed to rise.
She turned towards me and hefted another fireball.
'Zia, no!' I said. 'It's me. Carter.'
She hesitated. 'Carter... ?' Her expression turned to confusion, then fear. The orange flames faded in her eyes, and she collapsed in the pool of sunlight.
I ran to her. I tried to gather her in my arms, but her skin was too hot to touch. The golden scarab had left a nasty burn on her throat.
'Water,' I muttered to myself. 'I need water.'
I'd never been good at divine words, but I shouted: ' _Maw!_ '
The symbol blazed above us:
Several cubic gallons of water materialized in midair and crashed down on us. Zia's face steamed. She coughed and spluttered, but she didn't wake. Her fever still felt dangerously high.
'I'll get you out of here,' I promised, lifting her in my arms.
I didn't need the strength of Horus. I had so much adrenalin coursing through my body I didn't feel any of my own injuries. I ran right by Setne when he passed me in the hall.
'Hey, pal!' He turned and jogged along next to me, waving a thick papyrus scroll. 'Good job! I got the Book of Thoth!'
'You almost killed Zia!' I snapped. 'Get us out of here – NOW!'
'Okay, okay,' Setne said. 'Calm down.'
'I'm taking you back to my dad's courtroom,' I growled. 'I'm going to _personally_ stuff you down Ammit's mouth, like a branch into a wood chipper.'
'Whoa, big man.' Setne led me up a sloping passage back to the electrical lighting of the excavated tunnels. 'How about we get you out of here first, huh? Remember, you still need me to decipher this book and find the serpent's shadow. Then we'll see about the wood chipper, okay?'
'She can't die,' I insisted.
'Right, I got that.' Setne led me through more tunnels, picking up speed. Zia seemed to weigh nothing. My headache had disappeared. Finally we burst into the sunlight and ran for the _Egyptian Queen_.
I'll admit I wasn't thinking straight.
When we got back on board, Bloodstained Blade reported on the ship's repairs, but I barely heard him. I ploughed right past him and carried Zia inside to the nearest cabin. I set her on the bed and rummaged through my pack for medical supplies – a water bottle, some magic salve Jaz had given me, a few written charms. I was no _rekhet_ like Jaz. My healing powers consisted mostly of bandages and aspirin, but I began to work.
'Come on,' I mumbled. 'Come on, Zia. You're going to be fine.'
She was so warm that her drenched clothes had almost dried. Her eyes were rolled back in her head. She started muttering, and I could've sworn she said, 'Dung balls. Time to roll the dung balls.'
It might've been funny – except for the fact that she was dying.
'That's Khepri talking,' Setne explained. 'He's the divine dung beetle, rolling the sun across the sky.'
I didn't want to process that – the idea that the girl I liked had been possessed by a dung beetle and was now having dreams about pushing a giant sphere of flaming poo across the sky.
But there was no question: Zia had used the path of the gods. She'd called on Ra – or at least one of his incarnations, Khepri.
Ra had chosen her, the way Horus had chosen me.
Suddenly it made sense that Apophis had destroyed Zia's village when she was young, and that the old Chief Lector Iskandar had gone to such lengths to train her and then hide her in a magical sleep. If she held the secret to reawakening the sun god...
I dabbed some ointment on her throat. I pressed a cold washcloth to her forehead, but it didn't seem to help.
I turned to Setne. 'Heal her!'
'Oh, _um_...' He winced. 'See, healing magic isn't really my thing. But at least you've got the Book of Thoth! If she dies, it wasn't for nothing –'
'If she dies,' I warned, 'I will... I will...' I couldn't think of a torture painful enough.
'I see you need some time,' Setne said. 'No problem. How about I go tell your captain where we're heading? We should get back to the Duat, back onto the River of Night as soon as possible. Do I have your permission to give him orders?'
'Fine,' I snapped. 'Just get out of my sight.'
I don't know how much time passed. Zia's fever seemed to subside. She started breathing more easily and slipped into a gentler sleep. I kissed her forehead and stayed by her side, holding her hand.
I was dimly aware of the ship's moving. We dropped into a momentary free-fall, then hit water with a shudder and a loud splash. I felt a river rolling under the hull once again and, from the tingling in my gut, I guessed we were back in the Duat.
The door creaked open behind me, but I kept my eyes on Zia.
I waited for Setne to say something – probably to brag about how well he'd done navigating us back to the River of Night – but he stayed silent.
'Well?' I asked.
The sound of splintering wood made me jump.
Setne wasn't at the door. Instead, Bloodstained Blade loomed over me, his axe head having just split the doorframe. His fists were clenched.
He spoke in an angry, cold hum: 'Lord Kane, it's time to die.'
SADIE
## 13. A Friendly Game of Hide-and-Seek (with Bonus Points for Painful Death!)
I SEE. LEAVE OFF WITH THE AXE-MURDERING DEMON. Trying to make my part of the story seem boring, eh? Carter, you are _such_ an attention hog.
Well, as you were cruising down the Nile in a lavishly appointed riverboat, Walt and I were travelling in a bit less style.
From the realm of the dead, I ventured another conversation with Isis to negotiate a doorway into the Nile Delta. Isis must have been cross with me (I can't imagine why) because she deposited Walt and me waist-deep in a swamp, our feet completely stuck in the mud.
'Thanks!' I yelled at the sky.
I tried to move but couldn't. Clouds of mosquitoes gathered around us. The river was alive with bubbling and splashing noises, which made me think of pointy-toothed tiger fish and the water elementals Carter had once described to me.
'Any ideas?' I asked Walt.
Now that he was back in the mortal world, he seemed to have lost his vitality. He looked... I suppose the phrase would be _hollowed out_. His clothes fitted more loosely. The whites of his eyes were tinted an unhealthy yellow. His shoulders hunched, as if the amulets round his neck weighed him down. Seeing him like this made me want to cry – which is not something I do easily.
'Yeah,' he said, digging through his bag. 'I have just the thing.'
He brought out a _shabti_ – a white wax figurine of a crocodile.
'Oh, you didn't,' I said. 'You wonderfully naughty boy.'
Walt smiled. For a moment he almost looked like his old self. 'Everyone was abandoning Brooklyn House. I figured it wasn't right to leave him behind.'
He tossed the figurine in the river and spoke a command word. Philip of Macedonia erupted from the water.
Being surprised by a giant crocodile in the Nile is something you usually want to avoid, but Philip was a welcome sight. He smiled at me with his massive croc teeth, his pink eyes gleaming and his white scaly back floating just above the surface.
Walt and I grabbed hold. In no time, Philip had pulled us free of the muck. Soon we were perched on his back, making our way upriver. I rode in front, straddling Philip's shoulders. Walt sat behind at Philip's midsection. Philip was such a roomy crocodile that this left considerable space between Walt and me – possibly more than I would've preferred. Nevertheless we had a lovely ride, except for being drenched, caked in mud and swarmed by mosquitoes.
The landscape was a maze of waterways, grassy islands, reed beds and muddy shoals. It was impossible to tell where the river ended and the land began. Occasionally in the distance we saw ploughed fields or the rooftops of small villages, but mostly we had the river to ourselves. We saw several crocodiles, but they all steered clear of us. They would have been quite insane to bother Philip.
Like Carter and Zia, we'd got a late start leaving the Underworld. I was alarmed at how far the sun had already climbed in the sky. The heat turned the air into a soupy haze. My shirt and trousers were soaked through. I wished I'd brought a change of clothes, though it wouldn't have made much difference, as my pack was damp, too. Also, with Walt around, there was no place to change.
After a while, I got bored with watching the Delta. I turned and sat cross-legged, facing Walt. 'If we had some wood, we could start a campfire on Philip's back.'
Walt laughed. 'I don't think he'd like that. Plus, I'm not sure we want to send up smoke signals.'
'You think we're being watched?'
His expression turned serious. 'If I were Apophis, or even Sarah Jacobi...'
He didn't need to finish that thought. Any number of villains wanted us dead. Of _course_ they'd be looking for us.
Walt rummaged through his collection of necklaces. I didn't notice the gentle curves of his mouth at all, or how his shirt clung to his chest in the humid air. No – all business, that's me.
He chose an amulet shaped like an ibis – Thoth's sacred animal. Walt whispered to it and threw it into the air. The charm expanded into a beautiful white bird with a long curved beak and black-tipped wings. It circled above us, buffeting my face with wind, then flew off slowly and gracefully over the wetlands. It reminded me of a stork from those old cartoons – the birds who bring babies in bundles. For some ridiculous reason, that thought made me blush.
'You're sending it to scout ahead?' I guessed.
Walt nodded. 'It'll look for the ruins of Saïs. Hopefully they're close by.'
_Unless Isis sent us to the wrong end of the Delta_ , I thought.
Isis didn't reply, which was proof enough she was miffed.
We glided upstream on Crocodile Cruise Line. Normally I wouldn't have felt uncomfortable having so much face time with Walt, but there was so much to say, and no good way to say it. Tomorrow morning, one way or another, our long fight against Apophis would be over.
Of course I was worried about _all_ of us. I'd left Carter with the sociopathic ghost of Uncle Vinnie. I hadn't even got up the courage to tell him that Zia occasionally became a fireball-lobbing maniac. I worried about Amos and his struggle with Set. I worried about our young initiates, virtually alone at the First Nome and no doubt terrified. I felt heartbroken for my father, who sat on his Underworld throne grieving for our mother – yet again – and of course I feared for my mother's spirit, on the verge of destruction somewhere in the Duat.
More than anything, I was concerned about Walt. The rest of us had _some_ chance of surviving, however slim. Even if we prevailed, Walt was doomed. According to Setne, Walt might not even survive our trip to Saïs.
I didn't need anyone to tell me that. All I had to do was lower my vision into the Duat. A grey sickly aura swirled around Walt, growing weaker and weaker. How long, I wondered, before he turned into the mummified vision I'd seen in Dallas?
Then again, there was the _other_ vision I'd seen at the Hall of Judgement. After talking to the jackal guardian, Walt had turned to me, and just for a moment I thought he was...
'Anubis wanted to be there,' Walt interrupted my thoughts. 'I mean, in the Hall of Judgement – he wanted to be there for you, if that's what you were wondering about.'
I scowled. 'I was wondering about _you_ , Walt Stone. You're running out of time, and we haven't had a proper talk about it.'
Even saying _that_ much was difficult.
Walt trailed his feet in the water. He'd set his shoes to dry on Philip's tail. Boys' feet are not something I find attractive, especially when they've just been removed from mucky trainers. However, Walt's feet were quite nice. His toes were almost the same colour as the swirling silt in the Nile.
[Carter is complaining about my comments on Walt's feet. Well, _pardon me_. It was easier to focus on his toes than on the sad look on his face!]
'Tonight at the latest,' he said. 'But, Sadie, it's okay.'
Anger swelled inside me, taking me quite by surprise.
'Stop it!' I snapped. 'It's not anywhere _close_ to okay! Oh, yes, you've told me how grateful you are to have known me, and learned magic at Brooklyn House, and helped with the fight against Apophis. All very noble. But it's not –' My voice broke. 'It's not okay.'
I pounded my fist on Philip's scaly back, which wasn't fair to the crocodile. Yelling at Walt wasn't fair either. But I was tired of tragedy. I wasn't _designed_ for all this loss and sacrifice and horrible sadness. I wanted to throw my arms round Walt, but there was a wall between us – this knowledge that he was doomed. My feelings for him were so mixed up I didn't know whether I was driven by simple attraction or guilt or (dare I say it) love – or stubborn determination not to lose someone else I cared about.
'Sadie...' Walt gazed across the marshes. He looked quite helpless, and I suppose I couldn't blame him. I was being rather impossible. 'If I die for something I believe in... that's okay with me. But death doesn't have to be the end. I've been talking with Anubis, and –'
'Gods of Egypt, not _that_ again!' I said. ' _Please_ don't talk about him. I know exactly what he's been telling you.'
Walt looked startled. 'You do? And... you don't like the idea?'
'Of course not!' I yelled.
Walt looked absolutely crestfallen.
'Oh, come off it!' I said. 'I know Anubis is the guide for the dead. He's been preparing you for the afterlife. He's told you that it'll be right. You'll die a noble death, get a speedy trial and go straight into Ancient Egyptian Paradise. _Wonderful!_ You'll be a ghost like my poor mother. Perhaps it's not the end of the world for _you_. If it makes you feel better about your fate, then fine. But I don't want to hear about it. I don't need another... another person I can't be with.'
My face was burning. It was bad enough that my mother was a spirit. I could never properly hug her again, never go shopping with her, never get advice about _girl_ sorts of things. Bad enough that I'd been cut off from Anubis – that horribly frustrating gorgeous god who'd wrapped my heart into knots. Deep down, I'd always known a relationship with him was impossible given our age difference – five thousand years or so – but having the other gods decree him off-limits just rubbed salt in the wound.
Now to think of Walt as a spirit, out of reach as well – that was simply too much.
I looked up at him, afraid my bratty behaviour would have made him feel even worse.
To my surprise, he broke into a smile. Then he laughed.
'What?' I demanded.
He doubled over, still laughing, which I found quite inconsiderate.
'You find this funny?' I shouted. 'Walt Stone!'
'No...' He hugged his sides. 'No, it's just... You don't understand. It's not like that.'
'Well, then, what _is_ it like?'
He got control of himself. He seemed to be collecting his thoughts when his white ibis dived out of the sky. It landed on Philip's head, flapped its wings and cawed.
Walt's smile melted. 'We're here. The ruins of Saïs.'
Philip carried us ashore. We put on our shoes and waded across the marshy ground. In front of us stretched a forest of palm trees, hazy in the afternoon light. Herons flew overhead. Orange-and-black bees hovered over the papyrus plants.
One bee landed on Walt's arm. Several more circled his head.
Walt looked more perplexed than worried. 'The goddess who's supposed to live around here, Neith... didn't she have something to do with bees?'
'No idea,' I admitted. For some reason, I felt the urge to speak quietly.
[Yes, Carter. It _was_ a first for me. Thanks for asking.]
I peered through the palm forest. In the distance, I thought I saw a clearing with a few clumps of mud brick sticking above the grass like rotten teeth.
I pointed them out to Walt. 'The remains of a temple?'
Walt must have felt the same instinct for stealth that I did. He crouched in the grass, trying to lower his profile. Then he glanced back nervously at Philip of Macedonia. 'Maybe we shouldn't have a three-thousand-pound crocodile trampling through the woods with us.'
'Agreed,' I said.
He whispered a command word. Philip shrank back to a small wax statuette. Walt pocketed our croc, and we began sneaking towards the ruins.
The closer we got, the more bees filled the air. When we arrived at the clearing, we found an entire colony swarming like a living carpet over a cluster of crumbling mud-brick walls.
Next to them, sitting on a weathered block of stone, a woman leaned on a bow, sketching in the dirt with an arrow.
She was beautiful in a severe way – thin and pale with high cheekbones, sunken eyes and arched eyebrows, like a supermodel walking the line between glamorous and malnourished. Her hair was glossy black, braided on either side with flint arrowheads. Her haughty expression seemed to say: _I'm much too cool to even look at you._
There was nothing glamorous about her clothes, however. She was dressed for the hunt in desert-coloured fatigues – beige, brown and ochre. Several knives hung from her belt. A quiver was strapped to her back, and her bow looked like quite a serious weapon – polished wood carved with hieroglyphs of power.
Most disturbing of all, she seemed to be waiting for us.
'You're noisy,' she complained. 'I could've killed you a dozen times already.'
I glanced at Walt, then back at the huntress. ' _Um_... thanks? For not killing us, I mean.'
The woman snorted. 'Don't thank me. You'll have to do better than that if you want to survive.'
I didn't like the sound of that, but, generally speaking, I don't ask heavily armed women to elaborate on such statements.
Walt pointed to the symbol the huntress was drawing in the dirt – an oval with four pointy bits like legs.
'You're Neith,' Walt guessed. 'That's your symbol – the shield with crossed arrows.'
The goddess raised her eyebrows. 'Think much? Of _course_ I'm Neith. And, yes, that's my symbol.'
'It looks like a bug,' I said.
'It's not a bug!' Neith glowered. Behind her, the bees became agitated, crawling over the mud bricks.
'You're right,' I decided. 'Not a bug.'
Walt wagged his finger as if he'd just had a thought. 'The bees... I remember now. That was one name for your temple – the House of the Bee.'
'Bees are tireless hunters,' Neith said. 'Fearless warriors. I like bees.'
' _Uh_ , who doesn't?' I offered. 'Charming little... buzzers. But, you see, we're here on a mission.'
I began to explain about Bes and his shadow.
Neith cut me off with a wave of her arrow. 'I know why you're here. The others told me.'
I moistened my lips. 'The others?'
'Russian magicians,' she said. 'They were terrible prey. After that, a few demons came by. They weren't much better. They all wanted to kill you.'
I moved a step closer to Walt. 'I see. And so you –'
'Destroyed them, of course,' Neith said.
Walt made a sound somewhere between a grunt and a whimper. 'Destroyed them because... they were evil?' he said hopefully. 'You knew the demons and those magicians were working for Apophis, right? It's a conspiracy.'
'Of course it's a conspiracy,' Neith said. 'They're _all_ in on it – the mortals, the magicians, the demons, the tax collectors. But I'm on to them. Anyone who invades my territory pays.' She gave me a hard smile. 'I take trophies.'
From under the collar of her army jacket, she dug out a necklace. I winced, expecting to see some grisly bits of... well, I don't even want to say. Instead, the cord was strung with ragged squares of cloth – denim, linen, silk.
'Pockets,' Neith confided, a wicked gleam in her eyes.
Walt's hands went instinctively to the sides of his workout pants. 'You, _um_... took their _pockets_?'
'Do you think me cruel?' Neith asked. 'Oh, yes, I collect the pockets of my enemies.'
'Horrifying,' I said. 'I didn't know demons had pockets.'
'Oh, yes.' Neith glanced in either direction, apparently to be sure no one was eavesdropping. 'You just have to know where to look.'
'Right...' I said. 'So, anyway, we've come to find Bes's shadow.'
'Yes,' the goddess said.
'And I understand you're a friend of Bes and Tawaret's.'
'That's true. I like them. They're ugly. I don't think they're in the conspiracy.'
'Oh, definitely not! So could you, perhaps, show us where Bes's shadow is?'
'I could. It dwells in my realm – in the shadows of ancient times.'
'In the... what now?'
I was _so_ sorry I asked.
Neith nocked her arrow and shot it towards the sky. As it sailed upwards, the air rippled. A shockwave spread across the landscape, and I felt momentarily dizzy.
When I blinked, I found that the afternoon sky had turned a more brilliant blue, striped with orange clouds. The air was crisp and clean. Flocks of geese flew overhead. The palm trees were taller; the grass was greener –
[Yes, Carter, I know it sounds silly. But the grass really _was_ greener on the other side.]
Where the mud-brick ruins had been, a proud temple now stood. Walt, Neith and I were just outside the walls, which rose ten metres and gleamed brilliant white in the sun. The whole complex must have been at least a kilometre square. Halfway down the left wall, a gate glittered with gold filigree. A road lined with stone sphinxes led to the river, where sailboats were docked.
Disorientating? Yes. But I'd had a similar experience once before, when I'd touched the curtains of light in the Hall of Ages.
'We're in the past?' I guessed.
'A shadow of it,' Neith said. 'A memory. This is my refuge. It may be your burial ground, unless you survive the hunt.'
I tensed. 'You mean... you hunt _us_? But we're not your enemy! Bes is your friend. You should be helping us!'
'Sadie's right,' Walt said. ' _Apophis_ is your enemy. He's going to destroy the world tomorrow morning.'
Neith snorted. 'The end of the world? I've seen– _that_ coming for aeons. You soft mortals have ignored the warning signs, but I'm prepared. I've got an underground bunker stockpiled with food, clean water and enough weapons and ammunition to hold off a zombie army.'
Walt knitted his eyebrows. 'A zombie army?'
'You never know!' Neith snapped. 'The point is I'll survive the apocalypse. I can live off the land!' She jabbed a finger at me. 'Did you know the palm tree has six different edible parts?'
' _Um_ –'
'And I'll never be bored,' Neith continued, 'since I'm also the goddess of weaving. I have enough twine for a millennium of macramé!'
I had no reply, as I wasn't sure what macramé was.
Walt raised his hands. 'Neith, that's great, but Apophis is rising tomorrow. He'll swallow the sun, plunge the world into darkness and let the whole earth crumble back into the Sea of Chaos.'
'I'll be safe in my bunker,' Neith insisted. 'If you can prove to me that you're friend and not foe, maybe I'll help you with Bes. Then you can join me in the bunker. I'll teach you survival skills. We'll eat rations and weave new clothes from the pockets of our enemies!'
Walt and I exchanged looks. The goddess was a nutter. Unfortunately, we needed her help.
'So you want to hunt us,' I said. 'And we're supposed to survive –'
'Until sunset,' she said. 'Evade me that long, and you can live in my bunker.'
'I've got a counter-offer,' I said quickly. 'No bunker. If we win, you help us find Bes's shadow, but you'll also fight on our side against Apophis. If you're really a war goddess and a huntress and all that, you should enjoy a good battle.'
Neith grinned. 'Done! I'll even give you a five-minute head start. But I should warn you: I never lose. When I kill you, I'll take your pockets!'
'You drive a hard bargain,' I said. 'But fine.'
Walt elbowed me. ' _Um_ , Sadie –'
I shot him a warning look. As I saw it, there was no way we could escape this hunt, but I _did_ have an idea that might keep us alive.
'We've begun!' Neith cried. 'You can go anywhere in my territory, which is basically the entire delta. It doesn't matter. I'll find you.'
Walt said, 'But –'
'Four minutes, now,' Neith said.
We did the only sensible thing. We turned and ran.
'What is macramé?' I yelled as we barrelled through the rushes.
'A kind of weaving,' Walt said. 'Why are we talking about this?'
'Dunno,' I admitted. 'Just cur–'
The world turned upside down – or, rather, I did. I found myself hanging in a scratchy twine net with my feet in the air.
' _That's_ macramé,' Walt said.
'Lovely. Get me down!'
He pulled a knife from his pack – practical boy – and managed to free me, but I reckoned we'd lost most of our head start. The sun was lower on the horizon, but how long would we have to survive – thirty minutes? An hour?
Walt rifled through his pack and briefly considered the white wax crocodile. 'Philip, maybe?'
'No,' I said. 'We can't fight Neith head-on. We have to avoid her. We can split –'
'Tiger. Boat. Sphinx. Camels. No invisibility,' Walt muttered, examining his amulets. 'Why don't I have an amulet for invisibility?'
I shuddered. The last time I'd tried invisibility, it hadn't gone very well. 'Walt, she's a hunting goddess. We probably couldn't fool her with any sort of concealment spell, even if you had one.'
'Then what?' he asked.
I put my finger on Walt's chest and tapped the one amulet he wasn't considering – a necklace that was the twin to mine.
'The _shen_ amulets?' He blinked. 'But how can those help?'
'We split up and buy time,' I said. 'We can share thoughts through the amulets, yes?'
'Well... yes.'
'And they can teleport us to each other's side, right?'
Walt frowned. 'I – I designed them for that, but –'
'If we split up,' I said, 'Neith will have to choose one of us to track. We get as far apart as possible. If she finds me first, you teleport me out of danger with the amulet. Or vice versa. Then we split up again, and we keep at it.'
'That's brilliant,' Walt admitted. 'If the amulets will work quickly enough. And if we can keep the mental connection. And if Neith doesn't kill one of us before we can call for help. And –'
I put my finger to his lips. 'Let's just leave it at "That's brilliant".'
He nodded, then gave me a hasty kiss. 'Good luck.'
The silly boy shouldn't do things like that when I need to stay focused. He dashed off to the north and, after a dazed moment, I ran south.
Squishy combat boots are not the best for sneaking around.
I considered wading into the river, thinking perhaps the water would obscure my trail, but I didn't want to go for a swim without knowing what was under the surface – crocs, snakes, evil spirits. Carter once told me that most Ancient Egyptians couldn't swim, which had seemed ridiculous to me at the time. How could people living next to a river not swim? Now I understood. No one in his right mind would want to take a dip in that water.
[Carter says a swim in the Thames or the East River would be almost as bad for your health. All right, fair point. (Now shut up, brother dear, and let me get back to the brilliant Sadie-saves-the-day part.)]
I ran along the banks, crashing through reeds, jumping straight over a sunning crocodile. I didn't bother to check if it was chasing me. I had bigger predators to worry about.
I'm not sure how long I ran. It seemed like miles. As the riverbank widened, I veered inland, trying to stay under the cover of the palm trees. I heard no signs of pursuit, but I had a constant itch in the middle of my shoulder blades where I expected an arrow.
I stumbled through a clearing where some Ancient Egyptians in loincloths were cooking over an open fire next to a small thatched hut. Perhaps the Egyptians were just shadows from the past, but they looked real enough. They seemed quite startled to see a blonde girl in combat clothes stumble into their encampment. Then they saw my staff and wand and immediately grovelled, putting their heads to the dirt and mumbling something about _Per Ankh_ – the House of Life.
' _Um_ , yes,' I said. ' _Per Ankh_ official business. Carry on. Bye.'
Off I raced. I wondered if I would appear on a temple wall painting some day – a blonde Egyptian girl with purple highlights running sideways through the palm trees, screaming, 'Yikes!' in hieroglyphics as Neith chased after me. The thought of some poor archaeologist trying to figure that out almost lifted my spirits.
I reached the edge of the palm forest and stumbled to a stop. Before me, ploughed fields spread into the distance. Nowhere to run or hide.
I turned back.
_THUNK!_
An arrow hit the nearest palm tree with such force that dates rained down on my head.
_Walt_ , I thought desperately, _now, please._
Twenty metres away, Neith rose from the grass. She had smeared river mud on her face. Palm fronds stuck from her hair like bunny ears.
'I've hunted feral pigs with more skill than you,' she complained. 'I've hunted _papyrus_ plants with more skill!'
_Now, Walt_ , I thought. _Dear, dear Walt. Now._
Neith shook her head in disgust. She nocked an arrow. I felt a tugging sensation in my stomach, as if I were in a car and the driver suddenly slammed on the brakes.
I found myself sitting in a tree next to Walt, on the lowest bough of a large sycamore.
'It worked,' he said.
Wonderful Walt!
I kissed him properly – or as properly as possible given our situation. There was a sweet smell about him I hadn't noticed before, as if he'd been eating lotus blossoms. I imagined that old school rhyme: 'Walt and Sadie / sitting in a tree / K-I-S-S-I-N-G.' Fortunately, anyone who might tease me was still five thousand years in the future.
Walt took a deep breath. 'Is that a thank-you?'
'You look better,' I noticed. His eyes weren't as yellow. He seemed to be moving with less pain. This should have delighted me, but instead it made me worried. 'That lotus smell... did you drink something?'
'I'm okay.' He looked away from me. 'We'd better split up and try again.'
That didn't make me any less worried, but he was right. We had no time to chat. We both jumped to the ground and headed off in opposite directions.
The sun was almost touching the horizon. I began to feel hopeful. Surely we wouldn't have to hold out much longer.
I almost stumbled into another macramé net, but fortunately I was on the lookout for Neith's arts and crafts projects. I sidestepped the trap, pushed through a stand of papyrus plants and found myself back at Neith's temple.
The golden gates stood open. The wide avenue of sphinxes led straight into the complex. No guards... no priests. Maybe Neith had killed them all and collected their pockets, or perhaps they were all down in the bunker, preparing for a zombie invasion.
Hmm. I reckoned that the last place Neith might look for me was in her home base. Besides, Tawaret had seen Bes's shadow up on those ramparts. If I could find the shadow without Neith's help, all the better.
I ran for the gates, keeping a suspicious eye on the sphinxes. None of them came alive. Inside the massive courtyard were two freestanding obelisks tipped with gold. Between them glowered a statue of Neith in Ancient Egyptian garb. Shields and arrows had been piled round her feet like spoils of war.
I scanned the surrounding walls. Several stairways led up to the ramparts. The setting sun cast plenty of long shadows, but I didn't see any obvious dwarf silhouettes. Tawaret had suggested I call to the shadow. I was about to try when I heard Walt's voice in my mind: _Sadie!_
It's awfully hard to concentrate when someone's life depends on you.
I grasped the _shen_ amulet and muttered, 'Come on. Come on.'
I pictured Walt standing next to me, preferably without an arrow in him. I blinked – and there he was. He almost knocked me down with a hug.
'She – she would've killed me,' Walt gasped. 'But she wanted to talk first. She said she liked our trick. She was proud to slay us and take our pockets.'
'Super,' I said. 'Split up again?'
Walt glanced over my shoulder. 'Sadie, look.'
He pointed to the north-west corner of the walls, where a tower jutted from the ramparts. As the sky turned red, shadows slowly melted from the side of the tower, but one shadow remained – the silhouette of a stout little man with frizzy hair.
I'm afraid we forgot our plan. Together, we ran to the steps and climbed the wall. In no time, we were standing on the parapets, staring at the shadow of Bes.
I realized we must have been in the exact spot where Tawaret and Bes had held hands on the night Tawaret had described. Bes had told the truth – he'd left his shadow here so it could be happy, even when he wasn't.
'Oh, Bes...' My heart felt like it was shrinking into a wax _shabti_. 'Walt, how do we capture it?'
A voice behind us said, 'You don't.'
We turned. A few metres away, Neith stood on the ramparts. Two arrows were nocked in her bow. At this range, I imagined she'd have no trouble hitting us both at once.
'A good try,' she admitted. 'But I always win the hunt.'
SADIE
## 14. Fun with Split Personalities
AN EXCELLENT TIME TO CALL ON ISIS?
Perhaps. But even if Isis had answered I doubted I could summon any magic faster than Neith could shoot. And on the off-chance I actually defeated the huntress I had the feeling Neith would consider it cheating if I used another goddess's power against her. She'd probably decide I was part of the Russian/zombie/tax-collector conspiracy.
As mad as Neith was, we needed her help. She'd be much more useful shooting arrows at Apophis than sitting in her bunker making jackets out of our pockets and knotted twine.
My mind raced. How to win over a hunter? I didn't know much about hunters, except for old Major McNeil, Gramps's friend from the pensioners' home, who used to tell stories constantly about... Ah.
'It's a shame, really,' I blurted out.
Neith hesitated, as I'd hoped she would.
'What is?' she asked.
'Six edible parts of a palm tree.' I laughed. 'It's seven actually.'
Neith frowned. 'Impossible!'
'Oh, yes?' I raised my eyebrows. 'Have _you_ ever lived off the land in Covent Garden? Have _you_ ever trekked through the wilds of Camden Lock and lived to tell about it?'
Neith's bow dipped ever so slightly. 'I do not know those places.'
'I thought not!' I said triumphantly. 'Oh, the stories we could've shared, Neith. The tips for survival. Once I went for a whole week on nothing but stale biscuits and the juice of the Ribena.'
'Is that a plant?' Neith asked.
'With every nutrient you need for survival,' I said. 'If you know where to buy – I mean harvest it.'
I lifted my wand, hoping she would see this as a dramatic move, not a threat. 'Why once, in my bunker at Charing Cross Station, I stalked the deadly prey known as Jelly Babies.'
Neith's eyes widened. 'They are dangerous?'
'Horrible,' I agreed. 'Oh, they seem small alone, but they always appear in great numbers. Sticky, fattening – quite deadly. There I was, alone with only two quid and a Tube pass, beset by Jelly Babies, when... Ah, but never mind. When the Jelly Babies come for you... you will find out on your own.'
She lowered her bow. 'Tell me. I must know how to hunt Jelly Babies.'
I looked at Walt gravely. 'How many months have I trained you, Walt?'
'Seven,' he said. 'Almost eight.'
'And have I ever deemed you worthy of hunting Jelly Babies with me?'
' _Uh_... no.'
'There you have it!' I knelt and began tracing on the rampart floor with my wand. 'Even Walt is not ready for such knowledge. I could draw for you here a picture of the dreaded Jelly Baby, or even – gods forbid! – the McVitie's Hobnob. But that knowledge might destroy a lesser hunter.'
'I am the goddess of hunting!' Neith inched closer, staring in awe at the glowing markings – apparently not realizing I was making protective hieroglyphs. 'I must know.'
'Well...' I glanced at the horizon. 'First, you must understand the importance of timing.'
'Yes!' Neith said eagerly. 'Tell me of this.'
'For instance...' I tapped the hieroglyphs and activated my spell. 'It's sunset. We're still alive. We win.'
Neith's expression hardened. 'Trickery!'
She lunged at me, but the protective glyphs flared, pushing back the goddess. She raised her bow and shot her arrows.
What happened next was surprising on many levels. First, the arrows must have been heavily enchanted, because they sailed right through my defences. Second, Walt lunged forward with impossible speed. Faster than I could scream (which I did), Walt snatched the arrows out of the air. They crumbled to grey dust, scattering in the wind.
Neith stepped back in horror. 'It's _you_. This is unfair!'
'We won,' Walt said. 'Honour your agreement.'
A look passed between them that I didn't quite understand – some sort of contest of wills.
Neith hissed through clenched teeth. 'Very well. You may go. When Apophis rises, I will fight at your side. But I will not forget how you trespassed on my territory, child of Set. And you –'
She glared at me. 'I lay this hunter's curse upon you: some day you will be tricked by _your_ prey as I have been tricked today. May you be set upon by a pack of wild Jelly Babies!'
With that terrifying threat, Neith dissolved into a pile of twine.
'Child of Set?' I narrowed my eyes at Walt. 'What exactly –?'
'Look out!' he warned. All around us, the temple began to crumble. The air rippled as the magic shockwave contracted, transforming the landscape back to present-day Egypt.
We barely made it to the base of the stairs. The last walls of the temple were reduced to a pile of worn mud bricks, but the shadow of Bes was still visible against them, slowly fading as the sun went down.
'We need to hurry,' Walt said.
'Yes, but how do we capture it?'
Behind us, someone cleared his throat.
Anubis leaned against a nearby palm tree, his expression grim. 'I'm sorry to intrude. But, Walt... it's time.'
Anubis was sporting the formal Egyptian look. He wore a golden neck collar, a black kilt, sandals and pretty much nothing else. As I've mentioned before, not many boys could pull off this look, especially with kohl eyeliner, but Anubis managed.
Suddenly his expression turned to alarm. He sprinted towards us. For a moment I had an absurd vision of myself on the cover of one of Gran's old romance novels, where the damsel wilts into the arms of one half-dressed beefy guy while another stands by, casting her longing looks. Oh, the horrible choices a girl must make! I wished I'd had a moment to clean up. I was still covered in dried river muck, twine and grass, like I'd been tarred and feathered.
Then Anubis pushed past me and gripped Walt's shoulders. Well... that was unexpected.
I quickly realized, however, that he'd stopped Walt from collapsing. Walt's face was beaded with sweat. His head drooped, and his knees gave out as if someone had cut the last string holding him together. Anubis lowered him gently to the ground.
'Walt, stay with me,' Anubis urged. 'We have business to finish.'
'Business to finish?' I cried. I'm not sure what came over me, but I felt as if I'd just been Photoshopped out of my own book cover. And if there was one thing I wasn't used to it was being ignored. 'Anubis, what are you _doing_ here? What is going on with you two? _And what business?_ '
Anubis frowned at me, as if he'd forgot my presence. That didn't do much to help my mood. 'Sadie –'
'I tried to tell her,' Walt groaned. Anubis helped him sit up, though Walt still looked awful.
'I see,' Anubis said. 'Couldn't get a word in edgewise, I guess?'
Walt managed a weak smile. 'You should've seen her talking to Neith about Jelly Babies. She was like... I don't know, a verbal freight train. The goddess never stood a chance.'
'Yes, I saw,' Anubis said. 'It was endearing, in an annoying sort of way.'
'I beg your pardon?' I wasn't sure which of them to slap first.
'And when she turns red like that,' Anubis added, as if I were some interesting specimen.
'Cute,' Walt agreed.
'So have you decided?' Anubis asked him. 'This is our last chance.'
'Yes. I can't leave her.'
Anubis nodded and squeezed his shoulder. 'Neither can I. But the shadow first?'
Walt coughed, his face contorting in pain. 'Yes. Before it's too late.'
I can't pretend I was thinking clearly, but one thing was obvious: these two had been talking behind my back _much_ more than I'd realized. What on earth had they been telling each other about me? Forget Apophis swallowing the sun – _this_ was my ultimate nightmare.
How could they _both_ not leave me? Hearing that from a dying boy and a god of death sounded quite ominous. They'd formed some sort of conspiracy....
Oh, lord. I was beginning to think like Neith. Soon I'd be huddled in an underground bunker eating army rations and cackling as I sewed together the pockets of all the boys who'd dumped me.
With difficulty, Anubis helped Walt over to the shadow of Bes, now rapidly disappearing in the twilight.
'Can you do it?' Anubis asked.
Walt murmured something I couldn't make out. His hands were shaking, but he pulled a block of wax from his bag and began kneading it into a _shabti_. 'Setne tried to make it sound so complicated, but I see now. It's simple. No wonder the gods wanted this knowledge kept out of mortal hands.'
'Excuse me,' I interrupted.
They both looked at me.
'Hi, I'm Sadie Kane,' I said. 'I don't mean to barge in on your chummy conversation, but what in _blazes_ are you doing?'
'Capturing Bes's shadow,' Anubis told me.
'But...' I couldn't seem to make words come out. So much for being a verbal freight train. I'd become a verbal train wreck. 'But if that's the business you were talking about, then what was all that about _deciding_ , and _leaving me_ , and –'
'Sadie,' Walt said, 'we're going to lose the shadow if I don't act now. You need to watch the spell, so you can do this with the shadow of the serpent.'
'You are _not_ going to die, Walt Stone. I forbid it.'
'It's a simple incantation,' he continued, quite ignoring my plea. 'A regular summons, with the words _shadow of Bes_ substituted for _Bes_. After the shadow is absorbed, you'll need a binding spell to anchor it. Then –'
'Walt, stop it!'
He was shivering so badly his teeth chattered. How could he think about giving me a magic lesson now?
'– then for the execration,' he said, 'you'll need to be in front of Apophis. The ritual is exactly the same as normal. Setne lied about that part – there's nothing special about his enchantment. The only hard part is finding the shadow. For Bes, just reverse the spell. You should be able to cast it from a distance, since it's a beneficial spell. The shadow will _want_ to help you. Send out the _sheut_ to find Bes and it should... should bring him back.'
'But –'
'Sadie.' Anubis put his arms round me. His brown eyes were full of compassion. 'Don't make him talk more than he has to. He needs his strength for this spell.'
Walt began to chant. He raised the lump of wax, which now resembled a miniature Bes, and pressed it against the shadow on the wall.
I sobbed. 'But he'll die!'
Anubis held me. He smelled of temple incense – copal and amber and other ancient fragrances.
'He was born under the shadow of death,' Anubis said. 'That's why we understand each other. He would've collapsed long before now, but Jaz gave him one last potion to hold off the pain – to give him a final burst of energy in an emergency.'
I remembered the sweet smell of lotus on Walt's breath. 'He took it just now. When we were running from Neith.'
Anubis nodded. 'It's worn off. He'll only have enough energy to finish this spell.'
'No!' I meant to scream and hit him, but I'm afraid I rather melted and wept instead. Anubis sheltered me in his arms and I snivelled like a little girl.
I have no excuse. I simply couldn't stand the thought of losing Walt, even to bring back Bes. Just once, couldn't I succeed at something without a massive sacrifice?
'You have to watch,' Anubis told me. 'Learn the spell. It's the only way to save Bes. And you'll need the same enchantment to capture the serpent's shadow.'
'I don't care!' I cried, but I did watch.
As Walt chanted, the figurine absorbed the shadow of Bes like a sponge soaking up liquid. The wax turned as black as kohl.
'Don't worry,' Anubis said gently. 'Death won't be the end for him.'
I pounded on his chest without much force. 'I don't want to hear that! You shouldn't even be here. Didn't the gods put a restraining order on you?'
'I'm not supposed to be near you,' Anubis agreed, 'because I have no mortal form.'
'How, then? There's no graveyard. This isn't _your_ temple.'
'No,' Anubis admitted. He nodded at Walt. 'Look.'
Walt finished his spell. He spoke a single command word: ' _Hi-nehm._ '
The hieroglyph for _Join together_ blazed silver against the dark wax:
It was the same command I'd used to repair the gift shop in Dallas, the same command Uncle Amos had used last Christmas when he had demonstrated how to put a broken saucer back together. And, with horrible certainty, I knew it would be the last spell Walt ever cast.
He slumped forward. I ran to his side. I cradled his head in my arms. His breathing was ragged.
'Worked,' he muttered. 'Now... send the shadow to Bes. You'll have to –'
'Walt, please,' I said. 'We can get you to the First Nome. Their healers might be able to –'
'No, Sadie...' He pressed the figurine into my hands. 'Hurry.'
I tried to concentrate. It was almost impossible, but I managed to reverse the wording of an execration. I channelled power into the figurine and imagined Bes as he once was. I urged the shadow to find its master, to reawaken his soul. Instead of erasing Bes from the world, I tried to draw him back into the picture, this time with permanent ink.
The wax statue turned to smoke and disappeared.
'Did – did it work?' I asked.
Walt didn't answer. His eyes were closed. He lay perfectly still.
'Oh, please... no.' I hugged his forehead, which was rapidly cooling. 'Anubis, do something!'
No answer. I turned, and Anubis was gone.
'Anubis!' I screamed so loudly it echoed off the distant cliffs. I set Walt down as gently as I could. I stood and turned in a full circle, my fists clenched. 'That's it?' I shouted at the empty air. 'You take his soul and leave? I _hate_ you!'
Suddenly Walt gasped and opened his eyes.
I sobbed with relief.
'Walt!' I knelt next to him.
'The gate,' he said urgently.
I didn't know what he meant. Perhaps he'd had some sort of near-death vision? His voice sounded clearer, free of pain, but still weak. 'Sadie, hurry. You know the spell now. It will work on... on the serpent's shadow.'
'Walt, what happened?' I brushed the tears from my face. 'What gate?'
He pointed feebly. A few metres away, a door of darkness hovered in the air. 'The whole quest was a trap,' he said. 'Setne... I see his plan now. Your brother needs your help.'
'But what about you? Come with me!'
He shook his head. 'I'm still too weak. I will do my best to summon reinforcements for you in the Duat – you'll need them – but I can barely move. I'll meet you at sunrise, at the First Nome, if – if you're sure you don't hate me.'
'Hate you?' I was completely baffled. 'Why on earth would I hate you?'
He smiled sadly – a smile that wasn't quite like him.
'Look,' he said.
It took me a moment to understand his meaning. A cold feeling washed over me. How had Walt survived? Where was Anubis? And what had they been conspiring about?
Neith had called Walt a child of Set, but he wasn't. Set's only child was Anubis.
_I tried to tell her_ , Walt had said.
_He was born under the shadow of death_ , Anubis had told me. _That's why we understand each other._
I didn't want to, but I lowered my vision into the Duat. Where Walt lay, I saw a different person, like a superimposed image... a young man lying weak and pale, in a gold neckband and black Egyptian kilt, with familiar brown eyes and a sad smile. Deeper still, I saw the glowing grey radiance of a god – the jackal-headed form of Anubis.
'Oh... no, no.' I got up and stumbled away from him. From _them_. Too many puzzle pieces fell together at once. My head was spinning. Walt's ability to turn things to dust... it was the path of Anubis. He'd been channelling the god's power for months. Their friendship, their discussions, the _other_ way Anubis hinted at for saving Walt...
'What have you done?' I stared at him in horror. I wasn't even sure what to call him.
'Sadie, it's me,' Walt said. 'Still me.'
In the Duat, Anubis spoke in unison: 'Still me.'
'No!' My legs trembled. I felt betrayed and cheated. I felt as if the world were already crumbling into the Sea of Chaos.
'I can explain,' he said in two voices. 'But Carter needs your help. Please, Sadie –'
'Stop it!' I'm not proud of how I acted, but I turned and fled, leaping straight through the doorway of darkness. At the moment I didn't even care where it led, as long as it was away from that deathless creature I had thought I loved.
CARTER
## 15. I Become a Purple Chimpanzee
JELLY BABIES? SERIOUSLY?
I hadn't heard that part. My sister never ceases to amaze me [and no, Sadie, that's not a compliment, either].
Anyway, while Sadie was having her supernatural guy drama, I was confronting an axe-murdering riverboat captain who apparently wanted to change his name to Even-More-Bloodstained Blade.
'Back down,' I told the demon. 'That's an order.'
Bloodstained Blade made a humming sound that might've been laughter. He swung his head to the left – kind of an Elvis Presley dance move – and smashed a hole in the wall. Then he faced me again, splinters all over his shoulders.
'I have other orders,' he hummed. 'Orders to kill!'
He charged like a bull. After the mess we'd just been through in the _serapeum_ , a bull was the last thing I wanted to deal with.
I thrust out my fist. ' _Ha-wi!_ '
The hieroglyph for _Strike_ glowed between us:
A blue fist of energy slammed into Bloodstained Blade, pushing him out the door and straight through the wall of the opposite stateroom. A hit like that would have knocked out a human, but I could hear BSB digging out from the rubble, humming angrily.
I tried to think. It would've been nice to keep smashing him with that hieroglyph over and over, but magic doesn't work that way. Once spoken, a divine word can't be used again for several minutes, sometimes even hours.
Besides, divine words are top-of-the-line magic. Some magicians spend years mastering a single hieroglyph. I'd learned the hard way that saying too many will burn through your energy really fast, and I didn't have much to spare.
First problem: keep the demon away from Zia. She was still half conscious and totally defenceless. I summoned as much magic as I could and said: ' _N'dah!_ ' _– Protect._
Blue light shimmered around her. I had a horrible flashback to when I found Zia in her watery tomb last spring. If she woke up encased in blue energy and thought she was imprisoned again...
'Oh, Zia,' I said, 'I didn't mean –'
'KILL!' Bloodstained Blade rose from the wreckage of the opposite room. A feather pillow was impaled on his head, raining goose fluff all over his uniform.
I dashed into the hall and headed for the stairs, glancing back to be sure the captain was following me and not going after Zia. Lucky me – he was right on my tail.
I reached the deck and yelled, 'Setne!'
The ghost was nowhere to be seen. The crew lights were going crazy, buzzing around frantically, bonking into walls, looping round the smokestacks, lowering and raising the gangplank for no apparent reason. I guess without Bloodstained Blade to give them directions they were lost.
The riverboat careened down the River of Night, weaving drunkenly in the current. We slipped between two jagged rocks that would have pulverized the hull, then dropped over a cataract with a jaw-rattling _thunk_. I glanced up at the wheelhouse and saw no one steering. It was a miracle that we hadn't crashed already. I had to get the boat under control.
I ran for the stairs.
When I was halfway there, Bloodstained Blade appeared out of nowhere. He sliced his head across my gut, ripping open my shirt. If I'd had a larger belly – no, I don't want to think about it. I stumbled backwards, pressing my hand against my navel. He'd only grazed the skin, but the sight of blood on my fingers made me feel faint.
_Some warrior_ , I scolded myself.
Fortunately, Bloodstained Blade had embedded his axe head in the wall. He was still trying to tug it free, grumbling, 'New orders: _Kill Carter Kane. Take him to the Land of Demons. Make sure it's a one-way trip._ '
The Land of Demons?
I bolted up the stairs and into the wheelhouse.
All around the boat, the river churned into whitewater rapids. A pillar of stone loomed out of the fog and scraped against our starboard side, ripping off part of the railing. We twisted sideways and picked up speed. Somewhere ahead of us, I heard the roar of millions of tons of water cascading into oblivion. We were rushing towards a waterfall.
I looked around desperately for the shore. It was hard to see through the thick fog and gloomy grey light of the Duat, but a hundred yards or so off the bow I thought I saw fires burning, and a dark line that might've been a beach.
The Land of Demons sounded bad, but not as bad as dropping off a waterfall and getting smashed to pieces. I ripped the cord off the alarm bell and lashed the pilot's wheel in place, pointing us towards the shore.
'Kill Kane!'
The captain's well-polished boot slammed me in the ribs and sent me straight through the port window. Glass shattered, raking my back and legs. I bounced off a hot smokestack and landed hard on the deck.
My vision blurred. The cut across my stomach stung. My legs felt like they'd been used for a tiger's chew toy and, judging from the hot pain in my side, I may have broken some ribs in the fall.
All in all, not my best combat experience.
_Hello?_ Horus spoke in my mind. _Any intention of calling for_ _help, or are you happy to die on your own?_
_Yeah_ , I snapped back at him. _The sarcasm is real helpful._
Truthfully, I didn't think I had enough energy left to summon my avatar, even with Horus's help. My fight with the Apis Bull had nearly tapped me out, and that was before I got chased by an axe demon and kicked out of a window.
I could hear Bloodstained Blade stomping his way back down the stairs. I tried to rise and almost blacked out from the pain.
_A weapon_ , I told Horus. _I need a weapon._
I reached into the Duat and pulled out an ostrich feather.
'Really?' I yelled.
Horus didn't answer.
Meanwhile the crew lights zipped around in a panic as the boat barrelled towards the shore. The beach was easier to see now – black sand littered with bones and plumes of volcanic gas shooting from fiery crevices. Oh, good. Just the sort of place I wanted to crash-land.
I dropped the ostrich feather and reached into the Duat again.
This time I pulled out a pair of familiar weapons – the crook and flail, symbols of the pharaoh. The crook was a gold-and-red shepherd's rod with a curved end. The flail was a pole arm with three wicked-looking spiked chains. I'd seen lots of similar weapons. Every pharaoh had a set. But _these_ looked disturbingly like the original pair – the weapons of the sun god that I'd found last spring buried in Zia's tomb.
'What are these doing here?' I demanded. 'These should be with Ra.'
Horus remained silent. I got the feeling he was as surprised as I was.
Bloodstained Blade stormed round the side of the wheelhouse. His uniform was ripped and covered in feathers. His blades had some new nicks, and he'd got the emergency bell wrapped round his left boot so it clanged as he walked. But he still looked better than me.
'Enough,' he hummed. 'I have served the Kanes too long!'
Towards the bow of the ship, I heard the _crank, crank, crank_ of the gangplank lowering. I glanced over and saw Setne strolling calmly across as the river churned beneath him. He stopped at the edge of the plank and waited as the boat raced towards the black-sand beach. He was preparing to jump to safety. And tucked under his arm was a large papyrus scroll – the Book of Thoth.
'Setne!' I screamed.
He turned and waved, smiling pleasantly. 'It'll be fine, Carter! I'll be right back!'
' _Tas!_ ' I yelled.
Instantly the Ribbons of Hathor encased him, scroll and all, and Setne pitched overboard into the water.
I hadn't planned on that, but I didn't have time to worry about it. Bloodstained Blade charged, his left foot going _clump, BONG!, clump, BONG!_ I rolled sideways as his axe head cut the floor, but he recovered more quickly than I could. My ribs felt like they'd been dipped in acid. My arm was too weak to lift Ra's flail. I raised the crook for defence, but I had no idea how to use it.
Bloodstained Blade loomed over me, humming with evil glee. I knew I couldn't evade another attack. I was about to become two separate halves of Carter Kane.
'We are done!' he bellowed.
Suddenly, he erupted in a column of fire. His body vaporized. His metal axe head dropped, impaling itself in the deck between my feet.
I blinked, wondering if this was some sort of demon trick, but Bloodstained Blade was truly and completely gone. Beside the axe head, all that remained were his polished boots, a slightly melted alarm bell and some charred goose feathers floating in the air.
A few feet away, Zia leaned against the wheelhouse. Her right hand was wrapped in flames.
'Yes,' Zia muttered to the smoking axe-blade. 'We're done.'
She extinguished her fire, then stumbled over and embraced me. I was so relieved I could almost ignore the searing pain in my side.
'You're okay,' I said, which sounded dumb under the circumstances, but she rewarded me with a smile.
'Fine,' she said. 'Had a moment of panic. Woke up with blue energy all around me, but –'
I happened to glance behind her, and my stomach turned inside out.
'Hold on!' I yelled.
The _Egyptian Queen_ rammed into the shore at full speed.
I now understand the whole thing about wearing seat belts.
Hanging on did absolutely no good. The boat ran aground with such force that Zia and I shot into the air like human cannonballs. The hull cracked apart behind us with an almighty _ka-blam!_ The landscape hurtled towards my face. I had half a second to contemplate whether I would die by smacking into the ground or falling into a flaming crevice. Then, from above me, Zia grabbed my arm and hoisted me skywards.
I caught a glimpse of her, grim-faced and determined, holding on to me with one hand and hanging from the talons of a giant vulture with the other. Her amulet. I hadn't thought about it in months, but Zia had a vulture amulet. She'd somehow managed to activate it, because she's just awesome that way.
Unfortunately, the vulture wasn't strong enough to hold two people aloft. It could only slow our fall so, instead of being smashed flat, Zia and I rolled hard against the black sandy soil, tumbling over each other right to the edge of a fiery crevice.
My chest felt like it had been stomped flat. Every muscle in my body ached, and I had double vision. But, to my amazement, the sun god's crook and flail were clasped tightly in my right hand. I hadn't even realized I still had them.
Zia must've been in better shape than me (of course, I'd seen roadkill in better shape than me). She found the strength to drag me away from the fissure and down towards the beach.
'Ouch,' I said.
'Lie still.' Zia spoke a command word, and her vulture shrank back into a charm. She rummaged through her backpack, brought out a small ceramic jar and began rubbing blue paste on the cuts, burns and bruises that covered my upper body. The pain in my side eased immediately. The wounds disappeared. Zia's hands were smooth and warm. The magical unguent smelled like blossoming honeysuckle. It wasn't the worst experience I'd had all day.
She scooped another dollop of salve and looked at the long cut across my stomach. ' _Um_... you should do this part.'
She scraped the salve onto my fingers and let me apply it. The gash mended. I sat up slowly and took care of the glass cuts on my legs. Inside my chest, I swear I could feel my ribs mending. I took a deep breath and was relieved to find it didn't hurt.
'Thank you,' I said. 'What is that stuff?'
'Nefertem's Balm,' she said.
'It's a bomb?'
Her laughter made me feel almost as good as the salve. ' _Healing_ balm, Carter. It's made of blue lotus flower, coriander, mandrake, ground malachite and a few other special ingredients. Very rare, and this is my only jar. So don't get injured any more.'
'Yes, ma'am.'
I was pleased that my head had stopped spinning. My double vision was returning to normal.
The _Egyptian Queen_ wasn't in such good shape. The remains of the hull were scattered across the beach – boards and railings, ropes and glass, mixed with the bones that had already been there. The wheelhouse had imploded. Fire curled from the broken windows. The fallen smokestacks bubbled golden smoke into the river.
As we watched, the stern cracked off and slid underwater, dragging the glowing orbs of light with it. Maybe the magical crew was bound to the boat. Maybe they weren't even alive. But I still felt sorry for them as they disappeared under the murky surface.
'We won't be going back that way,' I said.
'No,' Zia agreed. 'Where are we? What happened to Setne?'
_Setne._ I'd almost forgotten about that ghostly scumbag. I would've been fine with his sinking to the bottom of the river, except that he'd taken the Book of Thoth.
I scanned the beach. To my surprise, I spotted a slightly battered pink mummy about twenty yards down the shore, squirming and struggling through the flotsam, apparently trying to inchworm his way to freedom.
I pointed him out to Zia. 'We could leave him like that, but he's got the Book of Thoth.'
She gave me one of those cruel smiles that made me glad she wasn't my enemy. 'No hurry. He won't get far. How about a picnic?'
'I like the way you think.'
We spread out our supplies and tried to clean up as best we could. I busted out some bottled water and protein bars – yeah, look at me, the Boy Scout.
We ate and drank and watched our gift-wrapped pink ghost try to crawl away.
'How did we get here, exactly?' Zia asked. Her golden scarab still glittered at her throat. 'I remember the _serapeum_ , the Apis Bull, the room with the sunlight. After that, it's fuzzy.'
I described what had happened as best I could – her magic scarab shield, her suddenly awesome powers from Khepri, the way she'd fried the Apis Bull and almost combusted herself. I explained how I'd got her back to the ship and how Bloodstained Blade had turned psycho.
Zia winced. 'You granted Setne permission to give Bloodstained Blade orders?'
'Yeah. Maybe not my best idea.'
'And he brought us here – to the Land of Demons, the most dangerous part of the Duat.'
I'd heard of the Land of Demons, but I didn't know much about it. At the moment, I didn't want to learn. I'd already escaped death so many times today that I just wanted to sit here, rest and talk with Zia – and maybe enjoy watching Setne struggle to get somewhere in his cocoon.
'You, _uh_ , feeling okay?' I asked Zia. 'I mean, about the stuff with the sun god...'
She gazed across the pitted landscape of black sand, bones and fire. Not many people can look good in the light of super-heated volcanic gas plumes. Zia managed.
'Carter, I wanted to tell you, but I didn't understand what was happening to me. I was frightened.'
'It's okay,' I said. 'I was the Eye of Horus. I understand.'
Zia pursed her lips. 'Ra is different, though. He's much older, much more dangerous to channel. And he's trapped in that old husk of a body. He can't restart his cycle of rebirth.'
'That's why he needs you,' I guessed. 'He woke up talking about _zebras_ – you. He offered you that scarab when he first met you. He wants you to be his host.'
A crevice spewed fire. The reflection in Zia's eyes reminded me of how she'd looked when she merged with Khepri – her pupils filled with orange flames.
'When I was entombed in that... that sarcophagus,' Zia said, 'I almost lost my mind, Carter. I still have nightmares. And when I tap into Ra's power I have the same sense of panic. He feels imprisoned, helpless. Reaching out to him is like... it's like trying to save somebody who's drowning. They grab on to you and take you down with them.' Zia shook her head. 'Maybe that doesn't make sense. But his power tries to escape through me, and I can barely control it. Every time I black out, it gets worse.'
'Every time?' I said. 'Then you've blacked out before?'
She explained what had happened in the House of Rest when she'd tried to destroy the nursing home with her fireballs. Just a minor little detail Sadie forgot to tell me.
'Ra is too powerful,' she said. 'I'm too weak to control him. In the catacombs with the Apis Bull, I might've killed you.'
'But you didn't,' I said. 'You saved my life – _again_. I know it's hard, but you can control the power. Ra needs to break out of his prison. The whole shadow magic idea that Sadie wants to try with Bes? I get the feeling that won't work with Ra. The sun god needs _rebirth_. You understand what that's like. I think that's why he gave you Khepri, the rising sun.' I pointed to her scarab amulet. 'You're the key to bringing him back.'
Zia took a bite of her protein bar. 'This tastes like styrofoam.'
'Yeah,' I admitted. 'Not as good as Macho Nachos. I still owe you that date at the mall food court.'
She laughed weakly. 'I wish we could do that right now.'
'Usually girls aren't so eager to go out with me. _Um_... not that I've ever asked –'
She leaned over and kissed me.
I'd imagined this many times, but I was so unprepared I didn't act very cool about it. I dropped my protein bar and breathed in her cinnamon fragrance. When she pulled away, I was gaping like a fish. I said something like, 'Hum-uh-huh.'
'You are kind, Carter,' she said. 'And funny. And, despite the fact you were just pushed out of a window and hurled from an explosion, you're even handsome. You've also been very patient with me. But I'm afraid. I've never been able to hold on to anyone I cared about – my parents, Iskandar... If I'm too weak to control the power of Ra and I end up hurting you –'
'No,' I said immediately. 'No, you won't, Zia. Ra didn't choose you because you're weak. He chose you because you're strong. And, _um_...' I looked down at the crook and flail lying at my side. 'These just sort of appeared... I think they showed up for a reason. You should take them.'
I tried to hand them over, but she curled my fingers round them.
'Keep them,' she said. 'You're right: they didn't appear by accident, but they appeared in _your_ hands. They may be Ra's, but Horus must be pharaoh.'
The weapons seemed to heat up, or maybe that was because Zia was holding my hands. The idea of using the crook and flail made me nervous. I'd lost my _khopesh_ – the sword used by the pharaoh's guards – and gained the weapons of the pharaoh himself. Not just any pharaoh, either... I was holding the implements of Ra, the first king of the gods.
Me, Carter Kane, a home-schooled fifteen-year-old who was still learning how to shave and could barely dress himself for a school dance – somehow I'd been deemed worthy of the most powerful magic weapons in creation.
'How can you be sure?' I asked. 'How could these be for me?'
Zia smiled. 'Maybe I'm getting better at understanding Ra. He needs Horus's support. I need you.'
I tried to think of what to say, and whether I had the nerve to ask for another kiss. I'd never pictured my first date being on a bone-littered riverbank in the Land of Demons, but at that moment there was no place I'd rather be.
Then I heard a _bonk_ – the sound of someone's head hitting a thick piece of wood. Setne let out a muffled curse. He'd managed to inchworm himself right into a broken section of keel. Dazed and off-balance, he rolled into the water and started to sink.
'We'd better fish him out,' I said.
'Yes,' Zia agreed. 'We don't want the Book of Thoth to get damaged.'
We hauled Setne onto the beach. Zia carefully dispelled just the ribbons round his chest so she could pull the Book of Thoth out from under his arm. Thankfully, the papyrus scroll appeared intact.
Setne said, 'Mmm-hmmpfh!'
'Sorry, not interested,' I said. 'We've got the book, so we'll be leaving you now. I don't feel like being stabbed in the back any more or listening to your lies.'
Setne rolled his eyes. He shook his head vigorously, mumbling what was probably a very good explanation of why he'd been within his rights to turn my demon servant against me.
Zia opened the scroll and studied the writing. After a few lines, she began to frown. 'Carter, this is... really dangerous stuff. I'm only skimming, but I see descriptions of the gods' secret palaces, spells to make them reveal their true names, information on how to recognize all the gods no matter what form they try to take...'
She looked up fearfully. 'With knowledge like this, Setne could have caused _a lot_ of damage. The only good thing... as far as I can tell, most of these spells can only be used by a living magician. A ghost wouldn't be able to cast them.'
'Maybe that's why he kept us alive this long,' I said. 'He needed our help to get the book. Then he planned on tricking us into casting the spells he wanted.'
Setne mumbled in protest.
'Can we find Apophis's shadow without him?' I asked Zia.
'Mm-mm!' Setne said, but I ignored him.
Zia studied a few more lines. 'Apophis... the _sheut_ of Apophis. Yes, here it is. It lies in the Land of Demons. So we're in the right place. But this map...' She showed me part of the scroll, which was so dense with hieroglyphs and pictures I couldn't even tell it _was_ a map. 'I have no idea how to read it. The Land of Demons is huge. From what I've read, it's constantly shifting, breaking apart and reforming. And it's full of demons.'
'Imagine that.' I tried to swallow the bitter taste from my mouth. 'So we'll be as out of place here as demons are in the mortal world. We won't be able to go anywhere unseen, and everything that meets us will want to kill us.'
'Yes,' Zia agreed. 'And we're running out of time.'
She was right. I didn't know exactly what time it was in the mortal world, but we had descended into the Duat in the late afternoon. By now, the sun might have gone down. Walt wasn't supposed to survive past sunset. For all I knew, he might be dying right now, and my poor sister... No. It was too painful to think about.
But at dawn tomorrow Apophis would rise. The rebel magicians would attack the First Nome. We didn't have the luxury to roam around a hostile land, fighting everything in our path until we'd found what we were looking for.
I glared down at Setne. 'I'm guessing you can guide us to the shadow.'
He nodded.
I turned to Zia. 'If he does or says anything you don't like, incinerate him.'
'With pleasure.'
I commanded the ribbons to release just his mouth.
'Holy Horus, pal!' he complained. 'Why did you tie me up?'
'Well, let's see... maybe because you tried to get me _killed_?'
'Aw, that?' Setne sighed. 'Look, pal, if you're going to overreact every time I try to kill you –'
' _Overreact?_ ' Zia summoned a white-hot fireball into her hand.
'Okay, okay!' Setne said. 'Look, that demon captain was going to turn on you anyway. I just helped things along. And I did it for a reason! We needed to get here, to the Land of Demons, right? Your captain would never have agreed to set that course unless he thought he could kill you. This is his homeland! Demons don't _ever_ bring mortals here unless they're for snacks.'
I had to remember Setne was a master liar. Whatever he told me was complete and utter Apis-quality bull. I steeled my willpower against his words, but it was still difficult not to find them reasonable.
'So you were going to let Bloodstained Blade kill me,' I said, 'but it was for a good cause.'
'Aw, I knew you could take him,' Setne said.
Zia held up the scroll. 'And that's why you were running away with the Book of Thoth?'
'Running? I was going to scout ahead! I wanted to find the shadow so I could lead you there! But that's not important. If you let me go, I can still bring you to the shadow of Apophis, and I can get you there unseen.'
'How?' Zia asked.
Setne sniffed indignantly. 'I've been practising magic since your ancestors were in diapers, doll. And while it's true I can't do all the mortal spells I'd like...' He glanced wistfully at the Book of Thoth. 'I _have_ picked up some tricks only ghosts can do. Untie me and I'll show you.'
I looked at Zia. I could tell we were thinking the same thing: terrible idea, but we didn't have a better one.
'I can't believe we're seriously considering this,' she grumbled.
Setne grinned. 'Hey, you're being smart. This is your best shot. Besides, I _want_ you to succeed! Like I said, I don't want Apophis destroying _me_. You won't regret it.'
'I'm pretty sure I will.' I snapped my fingers, and the Ribbons of Hathor unravelled.
Setne's brilliant plan? He turned us into demons.
Well, okay... it was actually just a glamour, so we _looked_ like demons, but it was the best illusion magic I'd ever seen.
Zia took one look at me and started to giggle. I couldn't see my own face, but she told me I now had a massive bottle opener for a head. I _did_ notice that my skin was fuchsia, and I had hairy bowed legs like a chimpanzee.
I didn't blame Zia for laughing, but she didn't look much better. She was now a big muscular girl demon with bright green skin, a zebra-hide dress and the head of a piranha.
'Perfect,' Setne said. 'You'll blend right in.'
'What about you?' I asked.
He spread his hands. He was still wearing his jeans, white sneakers and black jacket. His diamond pinky rings and gold _ankh_ chain flashed in the volcanic firelight. The only difference was that his red T-shirt now read: GO, DEMONS!
'You can't improve on perfection, pal. This outfit works anywhere. The demons won't even bat an eye – assuming they have eyes. Now, come on!'
He drifted inland, not waiting to see if we would follow.
Every once in a while, Setne checked the Book of Thoth for directions. He explained that the shadow would be impossible to find in this moving landscape without consulting the book, which served as a combination compass, tourist's guide and Farmer's Almanac timetable.
He promised us it would be a short journey, but it seemed pretty long to me. Any more time in Demon Land and I'm not sure I would have come out sane. The landscape was like an optical illusion. We spotted a vast mountain range in the distance, then walked fifty feet and discovered the mountains were so tiny we could jump over them. I stepped into a small puddle and suddenly found myself drowning in a flooded sinkhole fifty feet wide. Huge Egyptian temples crumbled and rearranged themselves as if some invisible giant were playing with blocks. Limestone cliffs erupted out of nowhere, already carved with monumental statues of grotesque monsters. The stone faces turned and watched us as we passed.
Then there were the demons. I'd seen lots of them under Camelback Mountain, where Set built his red pyramid, but here in their native environment they were even larger and more horrible. Some looked like torture victims, with gaping wounds and twisted limbs. Others had insect wings or multiple arms or tentacles made from darkness. As for their heads, pretty much every zoo animal and Swiss Army knife attachment was well represented.
The demons roamed in hordes across the dark landscape. Some built fortresses. Others tore them down. We saw at least a dozen large-scale battles. Winged demons circled through the smoky air, occasionally snatching up unsuspecting smaller monsters and carrying them off.
But none of them bothered us.
As we stumbled along, I became more and more aware of the presence of Chaos. A cold churning started in my gut, spreading through my limbs like my blood cells were turning to ice. I'd felt this before at the prison of Apophis, when Chaos sickness had almost killed me, but this place seemed even more poisonous.
After a while, I realized everything in the Land of Demons was being pulled in the direction in which we were travelling. The whole landscape was bending and crumbling, the fabric of matter unweaving. I knew the same force was pulling at the molecules of my body.
Zia and I should have died. But, as bad as the cold and the nausea were, I sensed that they should have been worse. Something was protecting us, an invisible layer of warmth keeping the Chaos at bay.
_It is her_ , said the voice of Horus, with grudging respect. _Ra sustains us._
I looked at Zia. She still appeared to be a piranha-headed green she-demon, but the air around her shimmered like vapour off a hot road.
Setne kept glancing back. Each time, he seemed surprised to find us still alive. But he shrugged and kept going.
The demons became fewer and further between. The landscape got even more twisted. Rock formations, sand dunes, dead trees, even pillars of fire all leaned towards the horizon.
We came to a cratered field, peppered with what looked like huge black lotus blossoms. They rose up quickly, spread their petals and burst. Only when we got closer did I realize they were knots of shadowy tendrils, like Sadie had described at the Brooklyn Academy dance. Each time one burst, it spat out a spirit that had been dragged from the upper world. These ghosts, no more than pale bits of mist, clawed desperately for something to anchor them, but they were quickly dispersed and sucked away in the same direction in which we were travelling.
Zia frowned at Setne. 'You're not affected?'
The ghost magician turned. For once his expression was grim. His colour was paler, his clothes and jewellery bleached out. 'Let's just keep moving, huh? I hate this place.'
I froze. Ahead of us stood a cliff I recognized – the same one I'd seen in the vision Apophis had shown me. Except now there were no spirits huddled in its shelter.
'My mother was there,' I said.
Zia seemed to understand. She took my hand. 'It might be a different cliff. The landscape is always changing.'
Somehow I knew it was the same place. I had the feeling Apophis had left it intact just to taunt me.
Setne twisted his pinky rings. 'The serpent's shadow feeds on spirits, pal. None of them last long. If your mom was here –'
'She was strong,' I insisted. 'A magician, like you. If you can fight it, she could too.'
Setne hesitated. Then he shrugged. 'Sure, pal. We're close now. Better keep going.'
Soon I heard a roar in the distance. The horizon glowed red. We seemed to be moving faster, as if we'd stepped on an automated walkway.
Then we came over the crest of a hill, and I saw our destination.
'There you go,' Setne said. 'The Sea of Chaos.'
Before us spread an ocean of mist, fire or water – it was impossible to tell which. Greyish-red matter churned, boiling and smoking, surging just like my stomach. It stretched as far as I could see – and something told me it had no end.
The ocean's edge wasn't so much a beach as a reverse waterfall. Solid ground poured into the sea and disappeared. A house-sized boulder trundled over the hill to our right, slid down the beach and dissolved in the surf. Chunks of solid ground, trees, buildings and statues constantly flew over our heads and sailed into the ocean, vaporizing as they touched the waves. Even the demons weren't immune. A few winged ones strayed over the beach, realized too late that they'd flown too close and disappeared screaming into the swirling misty soup.
It was pulling us, too. Instead of walking forward, I was instinctively backpedalling now, just to stay in one place. If we got any closer, I was afraid I wouldn't be able to stop.
Only one thing gave me hope. A few hundred yards to the north, jutting into the waves, was a single solid strip of land like a jetty. At the far end rose a white obelisk like the Washington Monument. The spire glowed with light. I had a feeling it was ancient – even older than the gods. As beautiful as the obelisk was, I couldn't help thinking of Cleopatra's Needle on the banks of the River Thames, where my mother had died.
'We can't go down there,' I said.
Setne laughed. 'The Sea of Chaos? That's where we all came from, pal. Haven't you heard how Egypt was formed?'
'It rose from this sea,' Zia said, almost in a trance. 'Ma'at appeared from Chaos – the first land, creation from destruction.'
'Yep,' Setne said. 'The two great forces of the universe. And there they are.'
'That obelisk is... the first land?' I asked.
'Dunno,' Setne said. 'I wasn't there. But it's the _symbol_ of Ma'at, for sure. Everything else, that's Apophis's power, always chewing away at creation, always eating and destroying. You tell me, which force is more powerful?'
I tried to swallow. 'Where is Apophis's shadow?'
Setne chuckled. 'Oh, it's here. But to see it, to catch it, you'll have to cast the spell from out there – at the edge of the jetty.'
'We'll never make it,' Zia said. 'One false step –'
'Sure,' Setne agreed cheerfully. 'It'll be fun!'
CARTER
## 16. Sadie Rides Shotgun (Worst. Idea. Ever.)
HERE'S SOME FREE ADVICE: don't walk towards Chaos.
With every step, I felt like I was being dragged into a black hole. Trees, boulders and demons flew past us and were sucked into the ocean, while lightning flickered through the red-grey mist. Under our feet, chunks of the ground kept cracking and sliding into the tide.
I grasped the crook and flail in one hand and held Zia's hand with the other. Setne whistled and floated along beside us. He tried to act cool, but, from the way his colours were fading and his greased hair pointed towards the ocean like a comet's tail, I figured he was having a tough time holding his ground.
Once I lost my balance. I almost tumbled into the surf, but Zia pulled me back. A few steps later, a fish-headed demon flew out of nowhere and slammed into me. He grabbed my leg, trying desperately to avoid getting sucked in. Before I could decide whether or not to help him, he lost his grip and disappeared into the sea.
The most horrible thing about the journey? Part of me was tempted to give up and let Chaos draw me in. Why keep struggling? Why not end the pain and the worry? So what if Carter Kane dissolved into trillions of molecules?
I knew those thoughts weren't really mine. The voice of Apophis was whispering in my head, tempting me as it had before. I concentrated on the glowing white obelisk – our lighthouse in the storm of Chaos. I didn't know if that spire was really the first part of creation, or how that myth sat with the Big Bang or with God creating the world in seven days or whatever else people might believe. Maybe the obelisk was just a manifestation of something larger – something my mind couldn't comprehend. Whatever the case, I knew the obelisk stood for Ma'at, and I had to focus on it. Otherwise I was lost.
We reached the base of the jetty. The rocky path felt reassuringly solid under my feet, but the pull of Chaos was strong on either side. As we inched forward, I remembered photos I'd seen of construction workers building skyscrapers back in the old days, fearlessly walking across girders six hundred feet in the air with no safety harnesses.
I felt like that now, except I wasn't fearless. The winds buffeted me. The jetty was ten feet wide, but I still felt like I was going to lose my balance and pitch into the waves. I tried not to look down. The stuff of Chaos churned and crashed against the rocks. It smelled like ozone, car exhaust and formaldehyde mixed together. The fumes alone were almost enough to make me pass out.
'Just a little further,' Setne said.
His form flickered unevenly. Zia's green demon disguise blinked in and out. I held up my arm and saw my glamour shimmering in the wind, threatening to collapse. I didn't mind losing the shocking-purple bottle-opening chimp look, but I hoped the wind would tear away only the illusion, not my actual skin.
Finally, we reached the obelisk. It was carved with tiny hieroglyphs, thousands of them, white on white, so they were almost impossible to read. I spotted the names of gods, enchantments to invoke Ma'at, and some Divine Words so powerful they almost blinded me. Around us, the Sea of Chaos heaved. Each time the wind blew, a glowing shield in the shape of a scarab flickered around Zia – the magical carapace of Khepri, sheltering us all. I suspected it was the only thing keeping us from instant death.
'What now?' I asked.
'Read the spell,' Setne said. 'You'll see.'
Zia handed me the scroll. I tried to find the right lines, but I couldn't see straight. The glyphs blurred together. I should have anticipated this problem. Even when I _wasn't_ standing next to the Sea of Chaos, I'd never been good at incantations. I wished Sadie were there.
[Yes, Sadie. I actually said that. Don't gasp so loud.]
'I – I can't read it,' I admitted.
'Let me help.' Zia traced her finger down the scroll. When she found the hieroglyphs she wanted, she frowned.
'This is a simple summoning spell.' She glared at Setne. 'You said the magic was complicated. You said we'd need your help. How could you lie while holding the Feather of Truth?'
'I didn't lie!' Setne protested. 'The magic _is_ complicated for me. I'm a ghost! Some spells – like summoning spells – I can't cast at all. And you _did_ need my help to find the shadow. You needed the Book of Thoth for that, and you needed me to interpret it. Otherwise, you'd still be shipwrecked at the river.'
I hated to admit it, but I said, 'He's got a point.'
'Sure I do,' Setne said. 'Now that you're here, the rest isn't so bad. Just force the shadow to show itself, and then I – _er_ – you can capture it.'
Zia and I exchanged a nervous look. I imagined she felt the same way I did. Standing at the edge of creation, facing an endless Sea of Chaos, the _last_ thing I wanted to do was cast a spell that would summon part of Apophis's soul. It was like shooting off a flare gun, signalling, _Hey, big nasty shadow! Here we are! Come and kill us!_
I didn't see that we had much choice, though.
Zia did the honours. It was an easy invocation, the kind a magician might use to summon a _shabti_ or an enchanted dust mop or pretty much any minor creature from the Duat.
When Zia finished, a tremor spread in all directions, as if she'd dropped a massive stone into the Sea of Chaos. The disturbance rippled up the beach and over the hills.
' _Um_... what was that?' I asked.
'Distress signal,' Setne said. 'I'm guessing the shadow just called on the forces of Chaos to protect it.'
'Wonderful,' I said. 'We'd better hurry, then. Where's the –? Oh...'
The _sheut_ of Apophis was so large it took me a moment to understand what I was looking at. The white obelisk seemed to cast a shadow across the sea, but, as the shadow darkened, I realized that it wasn't the silhouette of the obelisk. Rather, the shadow writhed across the surface of the water like the body of a giant snake. The shadow grew until the head of the serpent almost reached the horizon. It lashed across the sea, darting its tongue and biting at nothing.
My hands shook. My insides felt like I'd just chugged a big glass of Chaos water. The serpent's shadow was so massive, radiating so much power, that I didn't see how we could possibly capture it. What had I been thinking?
Only one thing kept me from total panic.
The serpent wasn't completely free. Its tail seemed to be anchored to the obelisk, as if someone had driven a spike to keep it from escaping.
For a disturbing moment, I felt the serpent's thoughts. I saw things from Apophis's point of view. It was trapped by the white obelisk – seething and in pain. It hated the world of mortals and gods, which pinned it down and constricted its freedom. Apophis despised creation the way I might despise a rusty nail driven through my foot, keeping me from walking.
All Apophis wanted was to snuff out the obelisk's blinding light. He wanted to annihilate the earth, so he could go back to the darkness and swim forever in the unrestricted expanses of Chaos. It took all of my willpower not to feel sorry for the poor little world-destroying, sun-devouring serpent.
'Well,' I said hoarsely. 'We found the shadow. Now what do we do with it?'
Setne chuckled. 'Oh, I can take it from here. You guys did great. _Tas!_ '
If I hadn't been so distracted, I might have seen what was coming, but I didn't. My demon glamour suddenly turned into solid bands of mummy linen, covering my mouth first, then wrapping round my body with blinding speed. I toppled and fell, completely encased except for my eyes. Zia hit the rocks next to me, also cocooned. I tried to breathe, but it was like inhaling through a pillow.
Setne leaned over Zia. He carefully extracted the Book of Thoth from beneath her bindings and tucked it under his arm. Then he smiled down at me.
'Oh, Carter, Carter.' He shook his head as if he were mildly disappointed. 'I like you, pal. I really do. But you are _way_ too trusting. After that business on the riverboat, you _still_ gave me permission to cast a glamour spell on you? Come on! Changing a glamour into a straitjacket is _sooo_ easy.'
'Mmm!' I grunted.
'What's that?' Setne cupped his ear. 'Hard to talk when you're all bound up, isn't it? Look, it's nothing personal. I couldn't cast that invocation spell myself, or I would have done it ages ago. I needed you two! Well... one of you, anyway. I figured I'd be able to kill either you or your girlfriend along the way, make the other one easier for me to handle. I never thought _both_ of you would survive this far. Impressive!'
I wriggled and almost toppled into the water. For some reason, Setne pulled me back to safety.
'Now, now,' he chided. 'No point killing yourself, pal. Your plan isn't ruined. I'm just going to alter it. I'll trap the shadow. That part I can do myself! But, instead of casting the execration, I'll blackmail Apophis, see? He'll destroy only what I _let_ him destroy. Then he retreats back into Chaos, or his shadow gets stomped and the big snake goes bye-bye.'
'Mmm!' I protested, but it was getting harder to breathe.
'Yeah, yeah.' Setne sighed. 'This is the part where you say, "You're mad, Setne! You'll never get away with it!" But the thing is I will. I've been getting away with impossible stuff for thousands of years. I'm sure the snake and I can come to a deal. Oh, I'll let him kill Ra and the rest of the gods. Big deal. I'll let him destroy the House of Life. I'll _definitely_ let him tear down Egypt and every cursed statue of my dad, Ramesses. I want that blowhard erased from existence! But the whole mortal world? Don't worry about it, pal. I'll spare most of it. I've gotta have someplace to rule, don't I?'
Zia's eyes flared orange. Her bonds started to smoke, but they held her fast. Her fire receded, and she slumped against the rocks.
Setne laughed. 'Nice try, doll. You guys sit tight. If you make it through the big shake-up, I'll come back and get you. Maybe you can be my jesters or something. You two crack me up! But in the meantime I'm afraid we're done here. No miracle's gonna drop from the sky and save you.'
A rectangle of darkness appeared in the air just above the ghost's head. Sadie dropped out of it.
I'll say this for my sister: she has great timing and she's quick on the draw. She crashed into the ghost and sent him sprawling. Then she noticed us wrapped up like presents, quickly realized what was going on and turned towards Setne.
' _Tas!_ ' she yelled.
'Noooo!' Setne was wrapped in pink ribbons until he looked like a forkful of spaghetti.
Sadie stood and stepped back from Setne. Her eyes were puffy like she'd been crying. Her clothes were covered in dried mud and leaves.
Walt wasn't with her. My heart sank. I was almost glad my mouth was covered, because I wouldn't have known what to say.
Sadie took in the scene – the Sea of Chaos, the serpent's writhing shadow, the white obelisk. I could tell she felt the pull of Chaos. She braced her feet, leaning away from the sea like the anchorperson in a tug-of-war. I knew her well enough to tell she was steeling herself, pushing her emotions back inside and forcing her sorrow down.
'Hullo, brother dear,' she said in a shaky voice. 'Need some help?'
She managed to dispel the glamour on us. She looked surprised to find me holding Ra's crook and flail. 'How in the world –?'
Zia briefly explained what we'd been up to – from the fight with the giant hippo through to Setne's most recent betrayals.
'All that,' Sadie marvelled, 'and you had to drag my brother along too? You poor girl. But how can we even survive here? The Chaos power...' She focused on Zia's scarab pendant. 'Oh. I really am thick. No wonder Tawaret looked at you strangely. You're channelling the power of Ra.'
'Ra chose me,' Zia said. 'I didn't want this.'
Sadie got very quiet – which wasn't like her.
'Sis,' I said, as gently as possible, 'what happened to Walt?'
Her eyes were so full of pain that I wanted to apologize for even asking. I hadn't seen her look like that since... well, since our mom died, when Sadie was little.
'He's not coming,' she said. 'He's... gone.'
'Sadie, I'm so sorry,' I said. 'Are you –?'
'I'm fine!' she snapped.
Translation: _I'm most definitely not fine, but if you ask again I'll stuff wax in your mouth._
'We have to hurry,' she continued, trying to modulate her voice. 'I know how to capture the shadow. Just give me the figurine.'
I had a moment of panic. Did I still _have_ the statue of Apophis that Walt had made? Coming all the way here and forgetting it would've been a major bonehead move.
Fortunately, it was still at the bottom of my pack.
I handed it to Sadie, who stared at the careful red carving of the coiled serpent, the hieroglyphs of binding around the name _Apophis_. I imagined she was thinking of Walt and all the effort he'd put into making it.
She knelt at the edge of the jetty, where the obelisk's base met the shadow.
'Sadie,' I said.
She froze. 'Yeah?'
My mouth felt like it was full of glue. I wanted to tell her to forget the whole thing.
Seeing her at the obelisk, with that massive shadow coiling towards the horizon... I just knew something would go wrong. The shadow would attack. The spell would backfire somehow.
Sadie reminded me so much of our mom. I couldn't shake the impression that we were repeating history. Our parents had tried to restrain Apophis once before, at Cleopatra's Needle, and our mom had died. I'd spent years watching my dad deal with his guilt. If I stood by now while Sadie got hurt...
Zia took my hand. Her fingers were trembling, but I was grateful for her presence. 'This will work,' she promised.
Sadie blew a strand of hair from her face. 'Listen to your girlfriend, Carter. And stop distracting me.'
She sounded exasperated, but there was no irritation in her eyes. Sadie understood my concerns as clearly as she knew my secret name. She was just as scared as I was, but in her own annoying way she was trying to reassure me.
'May I continue?' she asked.
'Good luck,' I managed.
Sadie nodded.
She touched the figurine to the shadow and began to chant.
I was afraid the waves of Chaos might dissolve the figurine or, worse, pull Sadie in. Instead, the serpent's shadow began to thrash. Slowly it shrank, writhing and snapping its mouth as if it were being hit with a cattle prod. The figurine absorbed the darkness. Soon the shadow was completely gone, and the statue was midnight black. Sadie spoke a simple binding spell on the figurine: ' _Hi-nehm._ '
A long hiss escaped from the sea – almost like a sigh of relief – and the sound echoed across the hills. The churning waves turned a lighter shade of red, as if some murky sediment had been dredged away. The pull of Chaos seemed to lessen just slightly.
Sadie stood. 'Right. We're ready.'
I stared at my sister. Sometimes she teased me that she'd eventually catch up to me in age and be my older sibling. Looking at her now, with that determined glint in her eyes and the confidence in her voice, I could almost believe her. 'That was amazing,' I said. 'How did you know the spell?'
She scowled. Of course, the answer was obvious: she'd watched Walt do the same spell on Bes's shadow... before whatever happened to Walt.
'The execration will be easy,' she said. 'We have to be facing Apophis, but otherwise it's the same spell we've been practising.'
Zia prodded Setne with her foot. 'That's another thing this maggot lied about. What should we do with him? We'll have to get the Book of Thoth out of those bindings, obviously, but after that should we shove him into the drink?'
'MMM!' Setne protested.
Sadie and I exchanged looks. We silently agreed that we couldn't dissolve Setne – even as horrible as he was. Maybe we'd seen too many awful things over the past few days, and we didn't need to see any more. Or maybe we knew that Osiris had to be the one to decide Setne's punishment, since we had promised to bring the ghost back to the Hall of Judgement.
Maybe, standing next to the obelisk of Ma'at, surrounded by the Sea of Chaos, we both realized that restraining ourselves from vengeance is what made us different from Apophis. Rules had their place. They kept us from unravelling.
'Drag him along,' Sadie said. 'He's a ghost. Can't be _that_ heavy.'
I grabbed his feet, and we made our way back down the jetty. Setne's head bonked against the rocks, but that didn't concern me. It took all my concentration to keep putting one foot in front of the other. Moving away from the Sea of Chaos was even harder than moving towards it.
By the time we reached the beach, I was exhausted. My clothes were drenched in sweat. We trudged across the sand and finally crested the hill.
'Oh...' I uttered some words that were _definitely_ not divine.
In the cratered field below us, demons had gathered – hundreds of them, all marching in our direction. As Setne had guessed, the shadow _had_ sent a distress signal to the forces of Apophis, and the call had been answered. We were trapped between the Sea of Chaos and a hostile army.
At this point, I was starting to wonder, _Why me?_
All I wanted was to infiltrate the most dangerous part of the Duat, steal the shadow of the primordial Lord of Chaos and save the world. Was that too much to ask?
The demons were maybe two football fields away, closing rapidly. I estimated that there were at least three or four hundred of them, and more kept pouring onto the field. Several dozen winged monsters were even closer, spiralling lower and lower overhead. Against this army, we had two Kanes, Zia and a gift-wrapped ghost. I didn't like those odds.
'Sadie, can you make a gate to the surface?' I asked.
She closed her eyes and concentrated. She shook her head. 'No signal from Isis. Possibly we're too close to the Sea of Chaos.'
That was a scary thought. I tried to summon the avatar of Horus. Nothing happened. I guess I should have known it would be hard to channel his powers down here, especially after I had asked him for a weapon back on the ship and the best he could do was an ostrich feather.
'Zia?' I said. 'Your powers from Khepri are still working. Can you get us out of here?'
She clutched her scarab amulet. 'I don't think so. All Khepri's energy is being spent shielding us from Chaos. He can't do any more.'
I considered running back to the white obelisk. Maybe we could use it to open a portal. But I quickly dismissed the idea. The demons would be on us before we ever got there.
'We're not going to get out of this,' I decided. 'Can we cast the execration on Apophis right now?'
Zia and Sadie spoke in unison: 'No.'
I knew they were right. We had to stand face to face with Apophis for the spell to work. But I couldn't believe we'd come all this way, just to be stopped now.
'At least we can go out fighting.' I unhooked the crook and flail from my belt.
Sadie and Zia readied their staffs and wands.
Then, at the other end of the field, a wave of confusion spread through the demons' ranks. They slowly began turning away from us, running in different directions. Behind the demon army, fireballs lit the sky. Plumes of smoke rose from newly opened craters in the ground. A battle seemed to be breaking out at the wrong end of the field.
'Who are they fighting?' I asked. 'Each other?'
'No.' Zia pointed, a smile spreading across her face. 'Look.'
It was hard to see through the hazy air, but a wedge of combatants was slowly forcing its way through the back ranks of the demons. Their numbers were smaller – maybe a hundred or so – but the demons gave way to them. Those that didn't were cut down, trampled or blown up like fireworks.
'It's the gods!' Sadie said.
'That's impossible,' I said. 'The gods wouldn't march into the Duat to rescue us!'
'Not the big gods, no.' She grinned at me. 'But the old forgotten ones from the House of Rest would! Anubis _said_ he was calling for reinforcements.'
'Anubis?' I was really confused now. When had she seen Anubis?
'There!' Sadie shouted. 'Oh –!'
She seemed to forget how to speak. She just waved her finger towards our new friends. The battle lines opened momentarily. A sleek black car barrelled into combat. The driver had to be a maniac. He ploughed down demons, going out of his way to hit them. He jumped over fiery crevices and spun in circles, flashing his lights and honking his horn. Then he came straight at us, until the front ranks of demons started to scatter. Only a few brave winged demons had the nerve to chase him.
As the car got closer, I could see it was a Mercedes limo. It climbed the hill, trailed by bat demons, and screeched to a stop in a cloud of red dust. The driver's door opened, and a small hairy man in blue Speedos stepped out.
I had never been so happy to see someone so ugly.
Bes, in all his horrible warty glory, climbed onto the roof of his car. He turned to face the bat demons. His eyes bulged. His mouth opened impossibly wide. His hair stood out like porcupine quills, and he yelled, 'BOO!'
The winged demons screamed and disintegrated.
'Bes!' Sadie ran towards him.
The dwarf god broke into a grin. He slid down to the hood, so he was almost Sadie's height when she hugged him.
'There's my girl!' he said. 'And, Carter, get your sorry hide over here!'
He hugged me, too. I didn't even mind him rubbing his knuckles on my head.
'And, Zia Rashid!' Bes cried generously. 'I got a hug for you too –'
'I'm good,' Zia said, stepping back. 'Thanks.'
Bes bellowed with laughter. 'You're right. Time for warm and fuzzy later. We gotta get you guys out of here!'
'The – the shadow spell?' Sadie stammered. 'It actually worked?'
'Of course it worked, you crazy kid!' Bes thumped his hairy chest and suddenly he was wearing a chauffeur's uniform. 'Now, get in the car!'
I turned to grab Setne... and my heart nearly stopped. 'Oh, holy Horus...' The magician was gone. I scanned the terrain in every direction, hoping he'd just inchwormed away. There was no sign of him.
Zia blasted fire at the spot where he'd been lying. Apparently, the ghost hadn't merely become invisible, because there was no scream.
'Setne was right there!' Zia protested. 'Tied up in the Ribbons of Hathor! How could he just disappear?'
Bes frowned. 'Setne, eh? I hate that weasel. Have you got the serpent's shadow?'
'Yeah,' I said, 'but Setne has the Book of Thoth.'
'Can you cast the execration without it?' Bes asked.
Sadie and I exchanged looks.
'Yes,' we both said.
'Then we'll have to worry about Setne later,' Bes said. 'We don't have much time!'
I guess if you have to travel through the Land of Demons, a limo is the way to go. Unfortunately, Bes's new sedan was no cleaner than the one we'd left at the bottom of the Mediterranean last spring. I wondered if he pre-ordered them already littered with old Chinese-food containers, stomped-on magazines and dirty laundry.
Sadie rode shotgun. Zia and I climbed in the back. Bes slammed the accelerator and played a game of hit-the-demon.
'Five points if you can hit that bloke with the cleaver head!' Sadie screamed.
_Boom!_ Cleaver-head went flying over the hood.
Sadie applauded. 'Ten points if you can hit those two dragonfly things at once.'
_Boom, boom!_ Two very large bugs hit the windshield.
Sadie and Bes laughed like crazy. Me, I was too busy yelling, 'Crevice! Look out! Flaming geyser! Go left!'
Call me practical. I wanted to live. I grabbed Zia's hand and tried to hang on.
As we approached the heart of the battle, I could see the gods pushing back the demons. It looked like the entire Sunny Acres Godly Retirement Community had unleashed their geriatric wrath on the forces of darkness. Tawaret the hippo goddess was in the lead, wearing her nurse's outfit and high heels, swinging a flaming torch in one hand and a hypodermic needle in the other. She bonked one demon on the head, then injected another in the rump, causing him to pass out immediately.
Two old guys in loincloths were hobbling around, throwing fireballs into the sky and incinerating flying demons. One of the old dudes kept screaming, 'My pudding!' for no apparent reason.
Heket the frog goddess leaped around the battlefield, knocking out monsters with her tongue. She seemed to have a special fondness for the demons with insect heads. A few yards away, the senile cat goddess Mekhit was smashing demons with her walker, yelling, 'Meow!' and hissing.
'Should we help them?' Zia asked.
Bes chuckled. 'They don't need help. This is the most fun they've had in centuries. They have a purpose again! They're going to cover our retreat while I get _you_ to the river.'
'But we don't have a ship any more!' I protested.
Bes raised a furry eyebrow. 'You sure about that?' He slowed the Mercedes and rolled down the window. 'Hey, sweetie! You okay here?'
Tawaret turned and gave him a huge hippo smile. 'We're fine, honeycakes! Good luck!'
'I'll be back!' he promised. He blew her a kiss, and I thought Tawaret was going to faint from happiness.
The Mercedes peeled out.
'Honeycakes?' I asked.
'Hey, kid,' Bes growled, 'do I criticize _your_ relationships?'
I didn't have the guts to look at Zia, but she squeezed my hand. Sadie stayed quiet. Maybe she was thinking about Walt.
The Mercedes leaped one last flaming chasm and slammed to a stop on the beach of bones.
I pointed to the wreckage of the _Egyptian Queen_. 'See? No boat.'
'Oh, yeah?' Bes asked. 'Then what's that?'
Upriver, light blazed in the darkness.
Zia inhaled sharply. 'Ra,' she said. 'The sun boat approaches.'
As the light got closer, I saw she was right. The gold-and-white sail gleamed. Glowing orbs flitted around the deck of a boat. The crocodile-headed god Sobek stood at the bow, knocking aside random river monsters with a big pole. And sitting in a fiery throne in the middle of the sun barque was the old god Ra.
'Halllloooooo!' he yelled across the water. 'We have cooooookies!'
Sadie kissed Bes on the cheek. 'You're brilliant!'
'Hey, now,' the dwarf mumbled. 'You're gonna make Tawaret jealous. It just so happened the timing was right. If we'd missed the sun boat, we'd have been out of luck.'
That thought made me shudder.
For millennia, Ra had followed this cycle – sailing into the Duat at sunset, travelling along the River of Night until he emerged into the mortal world again at sunrise. But it was a one-way trip, and the boat kept to a tight schedule. As Ra passed through the various Houses of the Night, their gates closed until the next evening, making it easy for mortal travellers like us to get stranded. Sadie and I had experienced that once before, and it hadn't been fun.
As the sun boat drifted towards the shore, Bes gave us a lopsided grin. 'Ready, kids? I got a feeling things up in the mortal world aren't going to be pretty.'
That was the first unsurprising thing I'd heard all day.
The glowing lights extended the boat's gangplank, and we climbed aboard for what might be the last sunrise in history.
SADIE
## 17. Brooklyn House Goes to War
I WAS SORRY TO LEAVE THE LAND OF DEMONS.
[Yes, Carter, I'm quite serious.]
After all, I'd had a rather successful visit there. I'd saved Zia and my brother from that horrid ghost Setne. I'd captured the serpent's shadow. I'd witnessed the Charge of the Old Folks' Brigade in all its glory and, most of all, I'd been reunited with Bes. Why wouldn't I have fond memories of the place? I might even take a beach holiday there some day, rent a cabana on the Sea of Chaos. Why not?
The flurry of activity also distracted me from less pleasant thoughts. But once we'd arrived at the riverbank and I'd had a few moments to breathe I started thinking about how I'd learned the spell to rescue Bes's shadow. My elation turned to despair.
Walt – oh, Walt. What had he done?
I remembered how lifeless and cold he'd been, cradled in my arms amid the mud-brick ruins. Then suddenly he had opened his eyes and gasped.
_Look_ , he'd said to me.
On the surface, I'd seen Walt as I'd always known him. But in the Duat... the boy god Anubis shimmered, his ghost-grey aura sustaining Walt's life.
_Still me_ , they had said in unison. Their double voice had made my skin tingle.
_I'll meet you at sunrise_ , they had promised, _at the First Nome, if you're sure you don't hate me._
Did I hate him? Or was it _them_? Gods of Egypt, I wasn't even sure what to call him any more! I certainly didn't know how I felt or if I wanted to see him again.
I tried to put those thoughts aside. We still needed to defeat Apophis. Even with his captured shadow, there was no guarantee we would succeed in casting the spell. I doubted Apophis would stand idly by while we tried to obliterate him from the universe. And it was entirely possible that the execration would require more magic than Carter and I had, combined. If we burned up, my dilemma with Walt would hardly be a problem.
Nevertheless, I couldn't stop thinking about him/them – the way their warm brown eyes merged together so perfectly and how natural Anubis's smile looked on Walt's face.
Argh! This was _not_ helpful.
We climbed aboard the sun barque – Carter, Zia, Bes and me. I was relieved beyond words that my favourite dwarf would be accompanying us to our final battle. I needed a reliably ugly god in my life right now.
At the bow, our old enemy Sobek regarded me with a crocodile smile, which I suppose was the only kind of smile he had. 'So... the little Kane children have returned.'
'So,' I snapped, 'the crocodile god wants his teeth kicked in.'
Sobek threw back his scaly green head and laughed. 'Well said, girl! You have iron in your bones.'
I suppose that was meant as a compliment. I chose to sneer at him and turn away.
Sobek only respected strength. In our first encounter, he had drowned Carter in the Rio Grande and smacked me across the Texas–Mexico border. We hadn't got much chummier since. From what I'd heard, he had only agreed to join our side because Horus and Isis had threatened him with extreme bodily harm. That didn't say much about his loyalty.
The glowing crew orbs fluttered around me, humming in my mind – little happy greetings of: _Sadie. Sadie. Sadie._ Once upon a time, they had _also_ wanted to kill me, but, since I'd awakened their old master Ra, they'd become quite friendly.
'Yes, hullo, boys,' I muttered. 'Lovely to see you. Excuse me.'
I followed Carter and Zia to the fiery throne. Ra gave us a toothless grin. He was still as old and wrinkly as ever, but something seemed different about his eyes. Before, his gaze had always slid over me as if I were part of the scenery. Now, he actually focused on my face.
He held out a plate of macaroons and chocolate biscuits, which were a bit melted from the heat of his throne. 'Cookies? Wheee!'
' _Uh_ , thanks.' Carter took a macaroon.
Naturally, I opted for the chocolate. I hadn't eaten a proper meal since we'd left our father's court.
Ra set down the platter and wobbled to his feet. Bes tried to help, but Ra waved him off. He tottered towards Zia.
'Zia,' he warbled happily, as if singing a nursery rhyme. 'Zia, Zia, Zia.'
With a jolt, I realized it was the first time I'd heard him use her actual name.
He reached out to touch her scarab amulet. Zia backed away nervously. She glanced at Carter for reassurance.
'It's okay,' Carter promised.
She took a deep breath. She unclasped her necklace and pressed it into the old man's hands. A warm glow expanded from the scarab, enveloping both Zia and Ra in a brilliant golden light.
'Good, good,' Ra said. 'Good...'
I expected the old god to get better. Instead, he began to crumble.
It was one of the most alarming things I'd seen in a very alarming day. First his ears fell off and melted to dust. Then his skin started turning to sand.
'What's happening?' I cried. 'Shouldn't we do something?'
Carter's eyes widened with horror. His mouth opened, but no words came out.
Ra's smiling face dissolved. His arms and legs cracked apart like a desiccated sand sculpture. His particles scattered across the River of Night.
Bes grunted. 'That was fast.' He didn't seem particularly shocked. 'Usually it takes longer.'
I stared at him. 'You've seen this _before_?'
Bes gave me a crooked grin. 'Hey, I took my turns working on the sun barque in the old days. We've _all_ seen Ra go through his cycle. But it's been a long, long time. Look.'
He pointed at Zia.
The scarab had disappeared from her hands, but golden light still radiated around her like a full-body halo. She turned towards me with a brilliant smile. I'd never seen her so at ease, so pleased.
'I see now.' Her voice was much richer, a chorus of tones descending in octaves through the Duat. 'It's all about balance, isn't it? My thoughts and his. Or is it mine and hers... ?'
She laughed like a child on her first bike ride. 'Rebirth, at last! You were right, Sadie and Carter! After so many aeons in the darkness, I am finally reborn through Zia's compassion. I'd forgotten what it is like to be young and powerful.'
Carter stepped back. I couldn't blame him. The memory of Walt and Anubis merging was still fresh in my mind, so I had a sense what Carter was feeling; it was more than a little creepy hearing Zia describe herself in the third person.
I lowered my vision deeper into the Duat. In Zia's place stood a tall man in leather and bronze armour. In some ways, he still looked like Ra. He was still bald. His face was still wrinkled and weathered with age, and he had the same kindly smile (only with teeth). Now, though, his posture was straight. His body rippled with muscles. His skin glowed like molten gold. He was the world's buffest, most golden grandpa.
Bes knelt. 'My lord Ra.'
'Ah, my small friend.' Ra ruffled the dwarf god's hair. 'Rise! It's good to see you.'
At the bow, Sobek came to attention, holding his long iron staff like a rifle. 'Lord Ra! I knew you would return.'
Ra chuckled. 'Sobek, you old reptile. You would snap me up for dinner if you thought you could get away with it. Horus and Isis kept you in line?'
Sobek cleared his throat. 'As you say, my king.' He shrugged. 'I can't help my nature.'
'No matter,' Ra said. 'We'll need your strength soon enough. Are we approaching sunrise?'
'Yes, my king.' Sobek pointed ahead of us.
I saw light at the end of the tunnel – literally. As we neared the end of the Duat, the River of Night widened. The exit gates stood about a kilometre ahead, flanked by statues of the sun god. Past that, daylight glowed. The river turned to clouds and poured into the morning sky.
'Very good,' Ra said. 'Steer us to Giza, Lord Sobek.'
'Yes, my king.' The croc god thrust his iron staff into the water, poling us along like a gondolier.
Carter still hadn't moved. The poor boy stared at the sun god with a mixture of fascination and shock.
'Carter Kane,' Ra said with affection, 'I know this is difficult for you, but Zia cares for you greatly. Nothing about her feelings has changed.'
I coughed. 'Ah... request? Please don't kiss him.'
Ra laughed. His image rippled, and I saw Zia in front of me again.
'It's all right, Sadie,' she promised. 'Now would not be the time.'
Carter turned awkwardly. ' _Um_... I'll just... be over there.' He bumped into the mast, then staggered towards the stern of the boat.
Zia knitted her brow in concern. 'Sadie, go take care of him, will you? We'll be reaching the mortal world soon. I must stay vigilant.'
For once, I didn't argue. I went to check on my brother.
He was sitting by the tiller in crash position, his head between his knees.
'All right?' I asked. Stupid question, I know.
'She's an old man,' he muttered. 'The girl I like is a buff old man with a voice deeper than mine. I kissed her on the beach, and now...'
I sat next to him. The glowing orbs fluttered around us in excitement as the ship approached the daylight.
'Kissed her, eh?' I said. 'Details, please.'
I thought he might feel better if I could get him talking. I'm not sure if it worked, but at least it got his head out from between his knees. He told me about his journey with Zia through the _serapeum_ and the destruction of the _Egyptian Queen_.
Ra – I mean Zia – stood at the bow between Sobek and Bes, very carefully _not_ looking back at us.
'So you told her it was all right,' I summed up. 'You encouraged her to help Ra. And now you're having second thoughts.'
'Do you blame me?' he asked.
'We've both hosted gods ourselves,' I said. 'It doesn't have to be permanent. And she's still Zia. Besides, we're heading into battle. If we don't survive, do you want to spend your last few hours pushing her away?'
He studied my expression. 'What happened to Walt?'
Ah... _touché_. At times, it seemed that Carter knew _my_ secret name as well as I knew his.
'I... I don't know exactly. He's alive, but only because –'
'He's hosting Anubis,' Carter finished.
'You knew?'
He shook his head. 'Not until I saw that look on your face. But it makes sense. Walt has a knack for... whatever it is. That grey obliteration touch. Death magic.'
I couldn't answer. I'd come back here to comfort Carter and reassure him that everything would be all right. Now, somehow, he'd managed to turn the tables.
He put his hand briefly on my knee. 'This could work, sis. Anubis can keep Walt alive. Walt could live a normal life.'
'You call that _normal_?'
'Anubis has never had a human host. This is his chance to have an actual body, to be flesh and blood.'
I shivered. 'Carter, it isn't like Zia's situation. _She_ can separate at any time.'
'So let me get this straight,' Carter said. 'The two guys you liked – one who was dying and one who was off-limits because he's a god – are now one guy, who isn't dying and isn't off-limits. And you're complaining.'
'Don't make me sound ridiculous!' I shouted. 'I'm not ridiculous!'
The three gods looked back at me. All right. Fine. I _did_ sound ridiculous.
'Look,' Carter said, 'let's agree to freak out about this later, okay? Assuming we don't die.'
I took a shaky breath. 'Deal.'
I helped my brother up. Together we joined the gods at the bow as the sun boat emerged from the Duat. The River of Night disappeared behind us, and we sailed across the clouds.
The Egyptian landscape spread out red and gold and green in the dawn. To the west, sandstorms swirled across the desert. To the east, the Nile snaked its way through Cairo. Directly below us, at the edge of the city, three pyramids rose on the plains of Giza.
Sobek struck his staff against the bow of the ship. He shouted like a herald: 'At last, Ra has truly returned! Let his people rejoice! Let his throngs of worshippers assemble!'
Perhaps Sobek said that as a formality, or to suck up to Ra, or possibly just to make the old sun god feel worse. Whatever the case, nobody down below was assembling. Definitely nobody was rejoicing.
I'd seen this vista many times, but something was wrong. Fires burned across the city. The streets seemed strangely deserted. There were no tourists, no humans at all around the pyramids. I'd never seen Giza so empty.
'Where is everyone?' I asked.
Sobek hissed in disgust. 'I should have known. The weak humans are in hiding or scared away because of the unrest in Egypt. Apophis has planned this well. His chosen battleground will be clear of mortal annoyances.'
I shivered. I'd heard about the troubles in Egypt lately, along with all the strange natural disasters, but I hadn't thought of it as part of Apophis's plan.
If this was his chosen battleground...
I focused more closely on the plains of Giza. Peering into the Duat, I realized the area wasn't empty after all. Encircling the base of the Great Pyramid was an enormous serpent formed from a swirling tornado of red sand and darkness. His eyes were burning points of light. His fangs were forks of lightning. Wherever he touched, the desert boiled, and the pyramid itself shook with a horrible resonance. One of the oldest structures in human history was about to crumble.
Even from high above, I could feel the presence of Apophis. He radiated panic and fear so strongly I could sense the mortals across Cairo cowering in their homes, afraid to go out. The whole land of Egypt was holding its breath.
As we watched, Apophis reared his massive cobra head. He struck at the desert floor, biting a house-sized crater in the sand. Then he recoiled as if he'd been stung, and hissed with anger. At first, I couldn't tell what he was fighting. I called on Isis's bird-of-prey sight and spotted a small lithe figure in a leopard-skin leotard, knives flashing in both hands as she leaped with inhuman agility and speed, striking at the serpent and evading his bite. All by herself, Bast was holding Apophis at bay.
My mouth tasted like old pennies. 'She's alone. Where are the others?'
'They await the pharaoh's orders,' Ra said. 'Chaos has left them divided and confused. They will not march to battle without a leader.'
'Then lead them!' I demanded.
The sun god turned. His form shimmered, and for a moment I saw Zia in front of me instead. I wondered if she would blast me to cinders. I had a feeling that would be quite easy for her now.
'I will face my old enemy,' she said calmly, still with Ra's voice. 'I won't let my loyal cat fight alone. Sobek, Bes – attend me.'
'Yes, my king,' Sobek said.
Bes cracked his knuckles. His chauffeur's outfit vanished, replaced by only his Dwarf Pride Speedos. 'Chaos... get ready to meet Ugly.'
'Wait,' Carter said. 'What about us? We've got the serpent's shadow.'
The ship was descending rapidly now, coming in for a landing just south of the pyramids.
'First things first, Carter.' Zia pointed to the Great Sphinx, which stood about three hundred metres from the pyramids. 'You and Sadie must help your uncle.'
Between the Sphinx's paws, a trail of smoke rose from a tunnel entrance. My heart missed a beat. Zia had once told us how that tunnel was sealed to keep archaeologists from finding their way into the First Nome. Obviously, the tunnel had been forced open.
'The First Nome is about to fall,' Zia said. Her form shifted again, and it was the sun god standing before me. I really wished he/she/they would make up their mind.
'I will hold off Apophis as long as I can,' Ra said. 'But if you don't help your uncle and your friends immediately there will be no one left to save. The House of Life will crumble.'
I thought about poor Amos and our young initiates, surrounded by a mob of rebel magicians. We couldn't let them be slaughtered.
'She's right,' I said. ' _Er_ , _he's_ right. Whichever.'
Carter nodded reluctantly. 'You'll need these, Lord Ra.'
He offered the sun god the crook and flail, but Ra shook his head. Or Zia shook her head. Gods of Egypt, this is confusing!
'When I told you the gods waited for their pharaoh,' Ra said, 'I meant you, Carter Kane, the Eye of Horus. I am here to fight my old enemy, not to assume the throne. That is your destiny. Unite the House of Life, rally the gods in my name. Never fear, I will hold Apophis until you come.'
Carter stared at the crook and flail in his hands. He looked every bit as terrified as he had when Ra had crumbled to sand.
I couldn't blame him. Carter had just been ordered to assume the throne of creation and lead an army of magicians and gods into battle. A year ago, even six months ago, the idea of my brother being given that kind of responsibility would've horrified me as well.
Strangely, I didn't mind it now. Thinking of Carter as the pharaoh was actually comforting. I'm sure I'll regret saying this, and I'm sure Carter will never let me forget it, but the truth was I'd been relying on my brother ever since we'd moved to Brooklyn House. I'd come to depend on his strength. I trusted him to make the right decisions, even when he didn't trust himself. When I had learned his secret name, I'd seen one very clear trait woven into his character: leadership.
'You're ready,' I told him.
'Indeed,' Ra agreed.
Carter looked up, a bit stunned, but I suppose he could tell I wasn't teasing him – not this time.
Bes punched him in the shoulder. 'Course you're ready, kid. Now, stop wasting time and go save your uncle!'
Looking at Bes, I tried not to get teary-eyed. I'd already lost him once.
As for Ra, he seemed so confident, but still he was confined to the form of Zia Rashid. She was a strong magician, yes, but she was new to this hosting business. If she wavered even slightly or overextended herself...
'Good luck, then.' Carter swallowed. 'I hope...'
He faltered. I realized the poor boy was trying to say goodbye to his girlfriend, possibly for the last time, and he couldn't even kiss her without kissing the sun god.
Carter began to change shape. His clothes, his pack, even the crook and flail melted into plumage. His form shrank until he was a brown-and-white falcon. Then he spread his wings and dived off the side of the boat.
'Oh, I hate this part,' I muttered.
I called on Isis and invited her in: _Now. It's time to act as one._
Immediately her magic flowed into me. It felt as if someone had switched on enough hydroelectric generators to light up a nation and channelled all that power straight into me. I turned into a kite (the bird) and soared into the air.
For once, I had no problem turning back to human. Carter and I rendezvoused at the feet of the Great Sphinx and studied the newly blasted tunnel entrance. The rebels hadn't been too subtle. Stone blocks the size of cars had been reduced to rubble. The surrounding sand had blackened and melted to glass. Either Sarah Jacobi's crew had used a _ha-di_ spell or several sticks of dynamite.
'This tunnel...' I said. 'Doesn't the other end open just across from the Hall of Ages?'
Carter nodded grimly. He pulled out the crook and flail, which were now glowing with ghostly white fire. He plunged into the darkness. I summoned my staff and wand and followed him inside.
As we descended, we saw evidence of battle. Explosions had scorched the walls and steps. One portion of the ceiling had buckled. Carter was able to clear a path with the strength of Horus, but, as soon as we were through, the tunnel collapsed behind us. We wouldn't be exiting that way.
Below us, I heard the sounds of combat – divine words being cast; fire, water and earth magic clashing. A lion roared. Metal clanged on metal.
A few metres further and we found the first casualty. A young man in a tattered grey military uniform was propped against the wall, holding his stomach and wheezing painfully.
'Leonid!' I cried.
My Russian friend was pale and bloody. I put my hand on his forehead. His skin was cold.
'Below,' he gasped. 'Too many. I try –'
'Stay here,' I said, which I realized was silly, since he could hardly move. 'We'll be back with help.'
He nodded bravely, but I looked at Carter and knew we were thinking the same thing. Leonid might not last that long. His uniform coat was soaked with blood. He kept his hand over his gut, but he'd clearly been savaged – either by claws or knives or some equally horrible magic.
I cast a _Slow_ spell on Leonid, which would at least steady his breathing and stem the flow of blood, but it wouldn't help much. The poor boy had risked his life to escape St Petersburg. He'd come all the way to Brooklyn to warn me about the impending attack. Now he'd tried to defend the First Nome against his former masters, and they'd cut him down and walked right over him, leaving him to suffer a lingering death.
'We _will_ be back,' I promised again.
Carter and I stumbled on.
We reached the bottom of the steps and were instantly thrown into battle. A _shabti_ lion leaped at my face.
Isis reacted faster than I could have. She gave me a single word to speak: ' _Fah!_ '
And the hieroglyph for _Release_ shimmered in the air:
The lion shrank to a wax statuette and bounced harmlessly off my chest.
All around us, the corridor was in mayhem. In either direction our initiates were locked in combat with enemy magicians. Directly in front of us, a dozen rebels had formed a wedge blocking the doors to the Hall of Ages, and our friends seemed to be trying to get past them.
For a moment, that seemed backwards to me. Shouldn't our side be defending the doors? Then I realized what must have happened. The attack on the sealed tunnel had surprised our allies. They'd rushed to help Amos, but by the time they'd got to the doors the enemies were already inside. Now this lot was keeping our reinforcements from reaching Amos, while our uncle was inside the hall, possibly alone, facing Sarah Jacobi and her elite hit squad.
My pulse raced. I charged into battle, flinging spells from Isis's incredibly diverse menu. It felt good to be a goddess again, I must admit, but I had to keep careful track of my energy. If I let Isis have free reign, she would destroy our enemies in seconds, but she would also burn me up in the process. I had to temper her inclination to rend the puny mortals to pieces.
I threw my wand like a boomerang and hit a large, bearded magician who was yelling in Russian as he fought sword to sword against Julian.
The Russian disappeared in a golden flash. Where he'd been standing, a hamster squeaked in alarm and scurried away. Julian grinned at me. His sword blade was smoking and the turn-ups of his trousers were on fire, but otherwise he looked all right.
'About time!' he said.
Another magician charged him, and we had no further time to chat.
Carter waded forward, swinging his flail and crook as if he had trained with them all his life. An enemy magician summoned a rhino – which I thought quite rude, considering the tight space we were in. Carter lashed it with his flail, and each spiked chain became a rope of fire. The rhino crumbled, cut into three pieces, and melted into a pile of wax.
Our other friends weren't doing too badly, either. Felix used an ice spell that I'd never seen before – encasing his enemies in big fluffy snowmen, complete with carrot noses and pipes. His army of penguins waddled around him, pecking at enemy magicians and stealing their wands.
Alyssa was fighting with another earth elementalist, but this Russian woman was clearly outmatched. She'd probably never faced the power of Geb before. Each time the Russian summoned a stone creature or tried to throw boulders, her attacks dissolved into rubble. Alyssa snapped her fingers, and the floor turned to quicksand under her opponent's feet. The Russian sank up to her shoulders, quite stuck.
At the north end of the corridor, Jaz crouched next to Cleo, tending her arm, which had been turned into a sunflower. Cleo had got off better than her opponent, though. At her feet lay a human-sized volume of the novel _David Copperfield_ , which I had a feeling had once been an enemy magician.
(Carter tells me David Copperfield _is_ a magician. He finds this funny for some reason. Just ignore him. I do.)
Even our ankle-biters had got into the act. Young Shelby had scattered her crayons down the hallway to trip the enemy. Now she was wielding her wand like a tennis racket, running between the legs of adult magicians, swatting them on the bottom and yelling, 'Die, die, die!'
Aren't children adorable?
She swatted a large metal warrior, a _shabti_ no doubt, and he transformed into a rainbow-coloured pot-bellied pig. If we lived through the day, I had a bad feeling Shelby would want to keep it.
Some of the First Nome residents were helping us, but depressingly few. A handful of tottering old magicians and desperate merchants threw talismans and deflected spells.
Slowly but surely, we waded towards the doors, where the main wedge of enemies seemed to be focused on a single attacker.
When I realized who it was, I was tempted to turn _myself_ into a hamster and scamper away, squeaking.
Walt had arrived. He ripped through the enemy line with his bare hands – throwing one rebel magician down the hallway with inhuman strength, touching another and instantly encasing the man in mummy linen. He grabbed the staff of a third rebel, and it crumbled to dust. Finally he swept his hand towards the remaining enemies, and they shrank to the size of dolls. Canopic jars – the sort used to bury a mummy's internal organs – sprang up around each of the tiny magicians, sealing them in with lids shaped like animal heads. The poor magicians yelled desperately, banging on the clay containers and wobbling about like a line of very unhappy bowling pins.
Walt turned to our friends. 'Is everyone all right?'
He looked like normal old Walt – tall and muscular with a confident face, soft brown eyes and strong hands. But his clothes had changed. He wore jeans, a dark Dead Weather T-shirt and a black leather jacket – Anubis's outfit, sized up to fit Walt's physique. All I had to do was lower my vision into the Duat, just a bit, and I saw Anubis standing there in all his usual annoying gorgeousness. Both of them – occupying the same space.
'Get ready,' Walt told our troops. 'They've sealed the doors, but I can –'
Then he noticed me, and his voice faltered.
'Sadie,' he said. 'I –'
'Something about opening the doors?' I demanded.
He nodded mutely.
'Amos is in there?' I asked. 'Fighting Kwai and Jacobi and who knows what else?'
He nodded again.
'Then stop staring at me and _open the doors_ , you annoying boy!'
I was talking to both of them. It felt quite natural. And it felt good to let my anger out. I'd deal with those two – that one – whatever he was – later. Right now, my uncle needed me.
Walt/Anubis had the nerve to smile.
He put his hand on the doors. Grey ash spread across the surface. The bronze crumbled to dust.
'After you,' he told me, and we charged into the Hall of Ages.
SADIE
## 18. Death Boy to the Rescue
THE GOOD NEWS: Amos wasn't entirely alone.
The bad news: his back-up was the god of evil.
As we poured into the Hall of Ages, our rescue attempt sputtered to a stop. We hadn't expected to see a deadly aerial ballet with lightning and knives. The normal floating hieroglyphs that filled the room were gone. The holographic curtains on either side of the hall flickered weakly. Some had collapsed altogether.
As I'd suspected, an assault team of enemy magicians had locked themselves in here with Amos, but it looked like they were regretting their choice.
Hovering midair in the centre of the hall, Amos was cloaked in the strangest avatar I'd ever seen. A vaguely human form swirled around him – part sandstorm, part fire, rather like the giant Apophis we'd seen upstairs, except a lot happier. The giant red warrior laughed as he fought, spinning a ten-metre black iron staff with careless force. Suspended in his chest, Amos copied the giant's moves, his face beaded with sweat. I couldn't tell if Amos was directing Set or trying to restrain him. Possibly both.
Enemy magicians flew circles round him. Kwai was easy to spot, with his bald head and blue robes, darting through the air like one of those martial arts monks who could defy gravity. He shot bolts of red lightning at the Set avatar, but they didn't seem to have much effect.
With her spiky black hair and flowing white robes, Sarah Jacobi looked like the Schizophrenic Witch of the West, especially as she was surfing about on a storm cloud like a flying carpet. She held two black knives like barbershop razors, which she threw over and over in a horrific juggling act, launching them into the Set avatar, then catching them as they returned to her hands. I'd seen knives like that before – _netjeri_ blades, made from meteoric iron. They were mostly used in funeral ceremonies, but they seemed to work quite well as weapons. With every strike, they disrupted the avatar's sandy flesh a little more, slowly wearing it down. As I watched her throw her knives, anger clenched inside me like a fist. Some instinct told me that Jacobi had stuck my Russian friend Leonid with those knives before leaving him to die.
The other rebels weren't quite as successful with their attacks, but they were certainly persistent. Some blasted Set with gusts of wind or water. Others launched _shabti_ creatures, like giant scorpions and griffins. One fat bloke was pelting Amos with bits of cheese. Honestly, I'm not sure I would have chosen a Cheese Master for my elite hit squad, but perhaps Sarah Jacobi got peckish during her battles.
Set seemed to be enjoying himself. The giant red warrior slammed his iron staff into Kwai's chest and sent him spiralling through the air. He kicked another magician into the holographic curtains of the Roman Age, and the poor man collapsed with smoke coming out of his ears, his mind probably overloaded with visions of toga parties.
Set thrust his free hand towards the Cheese Master. The fat magician was swallowed in a sandstorm and began to scream, but, just as quickly, Set retracted his hand. The storm died. The magician dropped to the floor like a rag doll, unconscious but still alive.
'Bah!' the red warrior bellowed. 'Come on, Amos, let me have _some_ fun. I only wanted to strip the flesh from his bones!'
Amos's face was tight with concentration. Clearly he was doing his best to control the god, but Set had many other enemies to play with.
'Pull!' The red god shot lightning at a stone sphinx and blasted it to dust. He laughed insanely and swatted his staff at Sarah Jacobi. 'This is fun, little magicians! Don't you have any more tricks?'
I'm not sure how long we stood in the doorway, watching the battle. Probably not more than a few seconds, but it seemed like an eternity.
Finally Jaz choked back a sob. 'Amos... he's possessed again.'
'No,' I insisted. 'No, this is different! He's in control.'
Our initiates gazed at me with disbelief. I understood their panic. I remembered better than anyone how Set had nearly broken my uncle's sanity. It was hard to comprehend that Amos would ever willingly channel the red god's power. Yet he was doing the impossible. He was winning.
Still, even the Chief Lector couldn't channel that much power for long.
'Look at him!' I pleaded. 'We have to help him! Amos isn't possessed. He's controlling Set!'
Walt frowned. 'Sadie, that – that's impossible. Set can't be controlled.'
Carter raised his crook and flail. 'Obviously he _can_ be, because Amos is doing it. Now, are we going to war, or what?'
We charged forward, but we'd hesitated too long. Sarah Jacobi had noticed our presence. She yelled down at her followers: 'Now!'
She may have been evil, but she was not a fool. Their assault on Amos thus far had simply been to distract him and weaken him. On her cue, the real attack began. Kwai blasted lightning at Amos's face just as the other magicians drew out magic ropes and threw them at the Set avatar.
The red warrior staggered as the ropes tightened all at once, lashing round his legs and arms. Sarah Jacobi sheathed her knives and produced a long black lariat. Sailing her storm cloud above the avatar, she deftly lassoed his head and pulled the noose tight.
Set roared with outrage, but the avatar began to shrink. Before we could even close the distance, Amos was kneeling on the floor of the Hall of Ages, surrounded by only the thinnest of glowing red shields. Magical ropes now bound him tight. Sarah Jacobi stood behind him, holding the black lasso like a leash. One of her _netjeri_ blades was pressed against Amos's neck.
'Stop!' she commanded us. 'This ends _now_.'
My friends hesitated. The rebel magicians turned and faced us warily.
Isis spoke in my mind: _Regrettable, but we must let him die. He hosts Set, our old enemy._
_That's my uncle!_ I replied.
_He has been corrupted_ , Isis said. _He is already gone._
'No!' I yelled. Our connection wavered. You can't share the mind of a god and have a disagreement. To be the Eye, you must act in perfect unison.
Carter seemed to be having similar trouble with Horus. He summoned the hawk-warrior avatar, but almost immediately it dissipated and dropped Carter to the floor.
'Come on, Horus!' he growled. 'We _have_ to help.'
Sarah Jacobi's laugh sounded like metal scraped through sand.
'Do you see?' She pulled tight on the noose round Amos's neck. ' _This_ is what comes from the path of the gods! Confusion. Chaos. _Set_ himself in the Hall of Ages! Even you misguided fools cannot deny this is wrong!'
Amos clawed at his throat. He growled in outrage, but it was Set's voice that spoke. 'I try to do something nice and _this_ is my thanks? You should have let me kill them, Amos!'
I stepped forward, careful to make no sudden movements. 'Jacobi, you don't understand. Amos is channelling Set's power, but he's in control. He could have killed you, but he didn't. Set was a lieutenant of Ra. He's a useful ally, properly managed.'
Set snorted. 'Useful, yes! I don't know about the _properly managed_ business. Let me go, puny magicians, so I can crush you!'
I glared at my uncle. 'Set! Not helping!'
Amos's expression changed from anger to concern. 'Sadie!' he said with his own voice. 'Go: fight Apophis. Leave me here!'
'No,' I said. 'You're the Chief Lector. We'll fight for the House of Life.'
I didn't look behind me, but I hoped that my friends would agree. Otherwise my last stand would be very, very short.
Jacobi sneered. 'Your uncle is a servant of Set! You and your brother are sentenced to death. The rest of you, lay down your weapons. As your new Chief Lector, I will give you amnesty. Then we will battle Apophis together.'
'You're in _league_ with Apophis!' I yelled.
Jacobi's face turned stony cold. 'Treason.'
She thrust out her staff. ' _Ha-di._ '
I raised my wand, but Isis wasn't helping me this time. I was just Sadie Kane, and my defences were slow. The explosion ripped through my weak shields and threw me backwards into a curtain of light. Images from the Age of the Gods crackled around me – the founding of the world, the crowning of Osiris, the battle between Set and Horus – like having sixty different movies downloaded into my brain while being electrocuted. The light shattered, and I lay on the floor, dazed and drained.
'Sadie!' Carter charged towards me, but Kwai blasted him with a bolt of red lightning. Carter fell to his knees. I didn't even have the strength to cry out.
Jaz ran towards him. Little Shelby yelled, 'Stop it! Stop it!' Our other initiates seemed stunned, unable to move.
'Give up,' Jacobi said. I realized she was speaking with words of power, just like the ghost Setne had done. She was using magic to paralyse my friends. 'The Kanes have brought you nothing but trouble. It's time this ended.'
She lifted her _netjeri_ blade from Amos's throat. Quick as light, she threw it at me. As the blade flew, my mind seemed to speed up. In that millisecond, I understood that Sarah Jacobi wouldn't miss. My end would be as painful as poor Leonid's, who was bleeding to death alone in the outer tunnel. Yet I could do nothing to defend myself.
A shadow crossed in front of me. A bare hand snatched the blade out of the air. The meteoric iron turned grey and crumbled.
Jacobi's eyes widened. She hastily drew her second knife.
'Who are you?' she demanded.
'Walt Stone,' he said, 'blood of the pharaohs. And Anubis, god of the dead.'
He stepped in front of me, shielding me from my enemies. Maybe my vision was double because I'd cracked my head, but I saw the two of them with equal clarity – both handsome and powerful, both quite angry.
'We speak with one voice,' Walt said. 'Especially on this matter. _No one_ harms Sadie Kane.'
He thrust out his hand. The floor split open at Sarah Jacobi's feet, and souls of the dead sprang up like weeds – skeletal hands, glowing faces, fanged shadows and winged _ba_ with their claws extended. They swarmed Sarah Jacobi, wrapping her in ghostly linen, and dragged her screaming into the chasm. The floor closed behind her, leaving no trace that she had ever existed.
The black noose slackened round Amos's neck, and the voice of Set laughed with delight. 'That's my boy!'
'Shut up, Father,' Anubis said.
In the Duat, Anubis looked as he always had, with his tousled dark hair and lovely brown eyes, but I'd never seen him filled with such rage. I realized that anyone who dared to hurt me would suffer his full wrath, and Walt wasn't going to hold him back.
Jaz helped Carter to his feet. His shirt was burnt, but he looked all right. I suppose a blast of lightning wasn't the worst thing that had happened to him lately.
'Magicians!' Carter managed to stand tall and confident, addressing both our initiates and the rebels. 'We're wasting time. Apophis is above, about to destroy the world. A few brave gods are holding him back for _our_ sakes, for the sake of Egypt and the world of mortals, but they can't do it alone. Jacobi and Kwai led you astray. Unbind the Chief Lector. We _have_ to work together.'
Kwai snarled. Red electricity arced between his fingers. 'Never. We do not bow to gods.'
I managed to rise.
'Listen to my brother,' I said. 'You don't trust the gods? They are already helping us. Meanwhile, Apophis wants us to fight one another. Why do you think your attack was timed for this morning, at the same moment Apophis is rising? Kwai and Jacobi have sold you out. The enemy is right in front of you!'
Even the rebel magicians now turned to stare at Kwai. The remaining ropes fell away from Amos.
Kwai sneered. 'You're too late.'
His voice hummed with power. His robes turned from blue to blood red. His eyes glowed, his pupils turning to reptilian slits. 'Even now, my master destroys the old gods, sweeping away the foundations of your world. He will swallow the sun. All of you will die.'
Amos got to his feet. Red sand swirled around him, but I had no doubt who was in charge now. His white robes shimmered with power. The leopard-skin cape of the Chief Lector gleamed on his shoulders. He held out his staff and multicoloured hieroglyphs filled the air.
'House of Life,' he said. 'To war!'
Kwai did not give up easily.
I suppose that's what happens when the Serpent of Chaos is invading your thoughts and filling you with unlimited rage and magic.
Kwai sent a chain of red lightning across the room, knocking over most of the other magicians, including his own followers. Isis must have protected me, because the electricity rippled over me with no effect. Amos didn't seem bothered in his swirling red tornado. Walt stumbled, but only briefly. Even Carter in his weakened state managed to turn aside the lightning with his pharaoh's crook.
The others weren't as lucky. Jaz collapsed. Then Julian. Then Felix and his squad of penguins. All our initiates and the rebels they'd been fighting crumpled unconscious to the floor. So much for a massive offensive.
I summoned the power of Isis. I began to cast a binding charm, but Kwai wasn't done with his tricks. He raised his hands and created his own sandstorm. Dozens of whirl winds spun through the hall, thickening and forming into creatures of sand – sphinxes, crocodiles, wolves and lions. They attacked in every direction, even pouncing on our defenceless friends.
'Sadie!' Amos warned. 'Protect them!'
I quickly changed spells – casting hasty shields over our unconscious initiates. Amos blasted the monsters one after the other, but they just kept re-forming.
Carter summoned his avatar. He charged at Kwai, but the red magician blasted him backwards with a new surge of lightning. My poor brother slammed into a stone column, which collapsed on top of him. I could only hope his avatar had taken the brunt of the impact.
Walt released a dozen magical creatures at once – his sphinx, his camels, his ibis, even Philip of Macedonia. They charged at the sand creatures, trying to keep them away from the fallen magicians.
Then Walt turned to face Kwai.
'Anubis,' Kwai hissed. 'You should have stayed in your funeral parlour, boy god. You are outmatched.'
By way of answer, Walt spread his hands. On either side of him, the floor cracked open. Two massive jackals leaped from the crevices, their fangs bared. Walt's form shimmered. Suddenly he was dressed in Egyptian battle armour, a _was_ staff twirling in his hands like a deadly fan blade.
Kwai roared. He blasted the jackals with waves of sand. He hurled lightning and words of power at Walt, but Walt deflected them with his staff, reducing Kwai's attacks to grey ashes.
The jackals harried Kwai from either side, sinking their teeth into his legs, while Walt stepped in and swung his staff like a golf club. He hit Kwai so hard that I imagined it echoed all the way through the Duat. The magician fell. His sand creatures vanished.
Walt called off his jackals. Amos lowered his staff. Carter rose from the rubble, looking dizzy but unharmed. We gathered round the fallen magician.
Kwai should have been dead. A line of blood trickled from his mouth. His eyes were glassy. But, as I studied his face, he took a sharp breath and laughed weakly.
'Idiots,' he rasped. ' _Sahei._ '
A blood-red hieroglyph burned against his chest:
His robes erupted in flames. Before our eyes, he dissolved into sand and a wave of cold – the power of Chaos – rippled through the Hall of Ages. Columns shook. Chunks of stone fell from the ceiling. A slab the size of an oven crashed into the steps of the dais, almost crushing the pharaoh's throne.
_'Bring down,'_ I said, realizing what the hieroglyph meant. Even Isis seemed terrified by the invocation. ' _Sahei_ is _Bring down_.'
Amos swore in Ancient Egyptian – something about donkeys trampling Kwai's ghost. 'He used up his life force to cast this curse. The hall is already weakened. We'll have to leave before we're buried alive.'
I glanced around us at the fallen magicians. Some of our initiates were starting to stir, but there was no way we could get them all to safety in time.
'We have to stop it!' I insisted. 'We have four gods present! Can't we save the hall?'
Amos furrowed his brow. 'The power of Set will not help me in this. He can only destroy, not restore.'
Another column toppled. It broke across the floor, barely missing one of the unconscious rebels.
Walt – who looked quite good in armour, by the way – shook his head. 'This is beyond Anubis. I'm sorry.'
The floor rumbled. We had only seconds to live. Then we would be just another bunch of entombed Egyptians.
'Carter?' I asked.
He regarded me helplessly. He was still weak, and I realized his battle magic wouldn't be much good in this situation.
I sighed. 'So it comes down to me, as always. Fine. You three shield the others as best you can. If this doesn't work, get out quickly.'
'If _what_ doesn't work?' Amos said as more chunks of ceiling rained down around us. 'Sadie, what are you planning?'
'Just a word, dear uncle.' I raised my staff and called on the power of Isis.
She immediately understood what I needed. Together, we tried to find calm in the Chaos. I focused on the most peaceful, well-ordered moments of my life – and there weren't many. I remembered my sixth birthday party in Los Angeles with Carter, my dad and mum – the last clear memory I had of all of us together as a family. I imagined listening to music in my room at Brooklyn House while Khufu ate Cheerios on my dresser. I imagined sitting on the terrace with my friends, having a restful breakfast as Philip of Macedonia splashed in his pool. I remembered Sunday afternoons at Gran and Gramps's flat – Muffin on my lap, Gramps's rugby game on the telly and Gran's horrible biscuits and weak tea on the table. Good times, those were.
Most important, I faced down my own chaos. I accepted my jumbled emotions about whether I belonged in London or New York, whether I was a magician or a schoolgirl. I was Sadie Kane, and if I survived today I could freaking well balance it all. And, yes, I accepted Walt and Anubis... I gave up my anger and dismay. I imagined both of them with me and, if that was peculiar, well then it fitted right in with the rest of my life. I made peace with the idea. Walt was alive. Anubis was flesh and blood. I stilled my restlessness and let go of my doubts.
' _Ma'at_ ,' I said.
I felt as if I'd struck a tuning fork against the foundation of the earth. Deep harmony resonated outwards through every level of the Duat.
The Hall of Ages stilled. Columns rose and repaired themselves. The cracks in the ceiling and floor sealed. Holographic curtains of light blazed again along either side of the hall, and hieroglyphs once more filled the air.
I collapsed into Walt's arms. Through my fuzzy vision, I saw him smiling down at me. Anubis, too. I could see them both, and I realized I didn't have to pick.
'Sadie, you did it,' he said. 'You're so amazing.'
'Uh-huh,' I muttered. 'Goodnight.'
They tell me I was only out a few seconds, but it felt like centuries. When I came to, the other magicians were back on their feet. Amos smiled down at me. 'Up you come, my girl.'
He helped me to my feet. Carter hugged me quite enthusiastically, almost as if he appreciated me properly for once.
'It's not over,' Carter warned. 'We have to get to the surface. Are you ready?'
I nodded, though neither of us was in good shape. We'd used up too much energy in the fight for the Hall of Ages. Even with the gods' help, we were in no condition to face Apophis. But we had little choice.
'Carter,' Amos said formally, gesturing to the empty throne. 'You are blood of the pharaohs, Eye of Horus. You carry the crook and flail, bestowed by Ra. The kingship is yours. Will you lead us, gods and mortals, against the enemy?'
Carter stood straight. I could see the doubt and fear in him, but possibly that was just because I knew him. I'd spoken his secret name. On the outside, he looked confident, strong, adult – even kingly.
[Yes, I said that. Don't get a big head, brother dear. You're still a huge dork.]
'I'll lead you,' Carter said. 'But the throne will have to wait. Right now, Ra needs us. We have to get to the surface. Can you show us the quickest way?'
Amos nodded. 'And the rest of you?'
The other magicians shouted assent – even the former rebels.
'We aren't many,' Walt observed. 'What are your orders, Carter?'
'First we get reinforcements,' he said. 'It's time I summoned the gods to war.'
CARTER
## 19. Welcome to the Fun House of Evil
SADIE SAYS I LOOKED CONFIDENT?
Good one.
Actually, being offered kingship of the universe (or supreme command over gods and magicians or whatever) pretty much had me shaking in my shoes.
I was grateful that it had happened as we headed into combat, so I didn't have time to think about it too much or freak out.
_Go with it_ , Horus said. _Use my courage._
For once I was glad to let him take the lead. Otherwise when we reached the surface and I saw how bad things were I would've run back inside, screaming like a kindergartner.
(Sadie says that's not fair. Our kindergartners weren't screaming. They were more anxious for combat than I was.)
Anyway, our little band of magicians popped out of a secret tunnel halfway up Khafre's pyramid and stared down at the end of the world.
To say Apophis was huge would be like saying the _Titanic_ took on a little water. Since we'd been underground, the serpent had grown. Now he coiled under the desert for miles, wrapping round the pyramids and tunnelling under the outskirts of Cairo, lifting entire neighbourhoods like old carpeting.
Only the serpent's head was above ground, but it rose almost as tall as the pyramids. It was formed of sandstorm and lightning, like Sadie described, and when it fanned out its cobra's crest it displayed a blazing hieroglyph no magician would ever write – _Isfet_ , the sign of Chaos:
The four gods battling Apophis looked tiny in comparison. Sobek straddled the serpent's back, chomping down again and again with his powerful crocodile jaws and smashing away with his staff. His attacks connected, but they didn't seem to bother Apophis.
Bes danced around in his Speedos, swinging a wooden club and yelling, 'Boo!' so loudly the people in Cairo were probably cowering under the beds. But the giant Chaos snake did not look terrified.
Our cat friend Bast wasn't having much luck either. She leaped onto the serpent's head and slashed wildly with her knives, then jumped away before Apophis could shake her off, but the serpent only seemed interested in one target.
Standing in desert between the Great Pyramid and the Sphinx, Zia was surrounded in brilliant golden light. It was hard to look directly at her, but she was shooting fireballs like a Roman candle – each one exploding against the serpent's body and disrupting his form. The serpent retaliated, biting chunks out of the desert, but he couldn't seem to find Zia. Her location shifted like a mirage – always several feet away from wherever Apophis struck.
Still, she couldn't keep this up forever. Looking into the Duat, I could see the four gods' auras weakening, and Apophis kept getting larger and stronger.
'What do we do?' Jaz asked nervously.
'Wait for my signal,' I said.
'Which is what?' Sadie asked.
'I don't know yet. I'll be back.'
I closed my eyes and sent my _ba_ into the heavens. Suddenly I stood in the throne room of the gods. Stone columns soared overhead. Braziers of magical fire stretched into the distance, their light reflecting on the polished marble floor. In the centre of the room, Ra's sun boat rested on its dais. His throne of fire sat empty.
I seemed to be alone – until I called out.
'Come to me.' Horus and I spoke in unison. 'Fulfil your oath of loyalty.'
Trails of glowing smoke drifted into the room like slow-motion comets. Lights blazed to life, swirling between the columns. All around me, the gods materialized.
A swarm of scorpions scuttled across the floor and merged to form the goddess Serqet, who glared at me distrustfully from beneath her scorpion-shaped crown. Babi the baboon god climbed down from the nearest column and bared his fangs. Nekhbet the vulture goddess perched on the prow of the sun boat. Shu the wind god blew in as a dust devil, then took the appearance of a World War II pilot, his body created entirely from dust, leaves and scraps of paper.
There were dozens more: the moon god Khonsu in his silver suit; the sky goddess Nut, her galactic blue skin glimmering with stars; Hapi the hippie with his green fish-scale skirt and his crazy smile; and a severe-looking woman in camouflage hunting clothes, a bow at her side, grease paint on her face and two ridiculous palm fronds sticking out of her hair – Neith, I assumed.
I'd hoped for more friendly faces, but I knew Osiris couldn't leave the Underworld. Thoth was still stuck in his pyramid. And many other gods – probably the ones most likely to help me – were also under siege from the forces of Chaos. We'd have to make do.
I faced the assembled gods and hoped my legs weren't shaking too badly. I still felt like Carter Kane, but I knew that when they looked at me they were seeing Horus the Avenger.
I brandished the crook and flail. 'These are the symbols of the pharaoh, given to me by Ra himself. He has named me your leader. Even now, he is facing Apophis. We must join the battle. Follow me and do your duty.'
Serqet hissed. 'We only follow the strong. Are you strong?'
I moved with lightning speed. I lashed the flail across the goddess, cutting her into a flaming pile of baked scorpions.
A few live critters scuttled out of the wreckage. They moved to a safe distance and began to re-form, until the goddess was whole again, cowering behind a brazier of blue flames.
The vulture goddess Nekhbet cackled. 'He is strong.'
'Then come,' I said.
My _ba_ returned to earth. I opened my eyes.
Above Khafre's pyramid, storm clouds gathered. With a clap of thunder they parted, and the gods charged into battle – some riding war chariots, some in floating warships, some on the backs of giant falcons. The baboon god Babi landed on top of the Great Pyramid. He pounded his chest and howled.
I turned to Sadie. 'How's that for a signal?'
We clambered down the pyramid to join the fight.
First tip on fighting a giant Chaos serpent: don't.
Even with a squadron of gods and magicians at your back, it's not a battle you're likely to win. I got clued in to this as we charged closer and the world seemed to fracture. I realized Apophis wasn't just coiling in and out of the desert, wrapping himself round the pyramids. He was coiling in and out of the Duat, splintering reality into different layers. Trying to find him was like running through a fun house full of mirrors, each mirror leading to another fun house filled with more mirrors.
Our friends began to split up. All around us, gods and magicians became isolated, some sinking deeper into the Duat than others. We fought a single enemy, but we were each fighting only a fragment of his power.
At the base of the pyramid, snaky coils encircled Walt. He tried to force his way out, blasting the serpent with grey light that turned his scales to ashes, but the serpent just regenerated, closing tighter and tighter round Walt. A few hundred feet away, Julian had summoned a full Horus avatar, a giant green hawk-headed warrior with a _khopesh_ in either hand. He sliced away at the serpent's tail – or at least one version of it – while the tail lashed around and tried to impale him. Deeper in the Duat, the goddess Serqet stood in nearly the same place. She had turned herself into a giant black scorpion and was confronting another image of the serpent's tail, parrying it with her stinger in a bizarre sword fight. Even Amos had been waylaid. He faced the wrong direction (or so it looked to me) and sliced his staff through the empty air, shouting command words at nothing.
I hoped that we were weakening Apophis by forcing him to deal with so many of us at once, but I couldn't see any sign of the serpent's power decreasing.
'He's dividing us!' Sadie shouted. Even standing right next to me, she seemed to be speaking from the other side of a roaring wind tunnel.
'Grab hold!' I held out the pharaoh's crook. 'We have to stay together!'
She took the other end of the crook, and we forged ahead.
The closer we got to the serpent's head, the harder it was to move. I felt like we were running through layers of clear syrup, each thicker and more resistant than the last. I looked around us and realized most of our allies had fallen away. Some I couldn't even see because of the Chaos distortion.
Ahead of us, a bright light shimmered as if through fifty feet of water.
'We have to get to Ra,' I said. 'Concentrate on him!'
What I was really thinking: _I have to save Zia._ But I was pretty sure Sadie knew that without my spelling it out.
I could hear Zia's voice, summoning waves of fire against her enemy. She couldn't be much further – maybe twenty feet in mortal distance? Through the Duat it might have been a thousand miles.
'Almost there!' I said.
_You're too late, little ones_ , the voice of Apophis hummed in my ears. _Ra will be my breakfast today._
A snake coil as big as a subway car slammed into the sand at our feet, almost crushing us. The scales rippled with Chaos energy, making me want to double over with nausea. Without Horus shielding me, I'm pretty sure I would have been vaporized just standing so close to it. I swung my flail. Three lines of fire cut through the snake's hide, blasting it to shreds of red and grey fog.
'Okay?' I asked Sadie.
She looked pale, but nodded. We trudged on.
A few of the most powerful gods still fought around us. Babi the baboon was riding one version of the serpent's head, pounding his massive fists between Apophis's eyes, but the serpent seemed only mildly annoyed. The hunter goddess Neith hid behind a pile of stone blocks, sniping at another snakehead with her arrows. She was pretty easy to spot because of the palm fronds in her hair, and she kept yelling something about a Jelly Baby conspiracy. Further on, another serpent's mouth sank its fangs into Nekhbet the vulture goddess, who shrieked in pain and exploded into a pile of black feathers.
'We're running out of gods!' Sadie cried.
Finally we reached the middle of the Chaos storm. Walls of red and grey smoke swirled around us, but the roar died in the centre as if we'd stepped into the eye of a hurricane. Above us rose the true head of the serpent – or at least the manifestation that held most of his power.
How did I know this? His skin looked more solid, glistening with golden red scales. His mouth was a pink cavern with fangs. His eyes glowed, and his cobra's hood spread so wide it blocked a quarter of the sky.
Before him stood Ra, a shining apparition too bright to look at directly. If I glanced from the corner of my eye, however, I could see Zia at the centre of the light. She now wore the clothes of an Egyptian princess – a silky dress of white and gold, a golden necklace and armbands. Even her staff and wand were gilded. Her image danced in the hot vapour, causing the serpent to misjudge her location every time he struck.
Zia shot tracers of red flame towards Apophis – blinding his eyes and burning away patches of his skin – but the damage seemed to heal almost instantly. He was growing stronger and larger. Zia wasn't so fortunate. If I concentrated, I could sense her life force, her _ka_ , growing weaker. The luminous glow at the centre of her chest was becoming smaller and more concentrated, like a flame reduced to a pilot light.
Meanwhile, our feline friend Bast was doing her best to distract her old enemy. Over and over she jumped on the serpent's back, slashing with her knives and mewling in anger, but Apophis just shook her off, throwing her into the storm.
Sadie scanned the area with alarm. 'Where's Bes?'
The dwarf god had disappeared. I was beginning to fear the worst when a small grumpy voice near the edge of the storm called, 'Some help, maybe?'
I hadn't paid much attention to the ruins around us. The plains of Giza were littered with big stone blocks, trenches and old building foundations from previous excavations. Under a nearby car-sized wedge of limestone, the dwarf god's head was sticking out.
'Bes!' Sadie cried as we ran to his side. 'Are you all right?'
He glared up at us. 'Do I look all right, kid? I have a ten-ton block of limestone on my chest. Snake-breath over there knocked me flat and dropped this thing on top of me. Most blatant act of dwarf cruelty ever!'
'Can you move it?' I asked.
He gave me a look almost as ugly as his _Boo!_ face. 'Gee, Carter, I didn't think of that. It's so comfortable under here. _Of course_ I can't move it, you dolt! Blocks of stone don't scare easily. Help a dwarf out, huh?'
'Stand back,' I told Sadie.
I summoned the strength of Horus. Blue light encased my hand, and I karate-chopped the stone. It cracked right down the middle, falling on either side of the dwarf god.
It would've been more impressive if I hadn't yelped like a puppy and cradled my fingers. Apparently I needed to work on the karate trick more, because my hand felt like it was boiling in oil. I was pretty sure I'd broken some bones in there.
'All right?' Sadie asked.
'Yeah,' I lied.
Bes climbed to his feet. 'Thanks, kid. Now it's time for some snake-bashing.'
We ran to help Zia, which turned out to be a bad idea. She glanced over and saw us – and, just for a moment, she was distracted.
'Carter, thank the gods!' She spoke in two-part harmony – partly her, partly the deep commanding voice of Ra, which was a little hard to take. Call me close-minded, but hearing my girlfriend talk like a five-thousand-year-old male god was not on my top ten list of Things I Find Attractive. Still, I was so glad to see her I almost didn't care.
She lobbed another fireball down the throat of Apophis. 'You're just in time. Our snaky friend is getting stro–'
'Look out!' Sadie screamed.
This time, Apophis wasn't fazed by the fire. He struck immediately – and he didn't miss. His mouth hit like a wrecking ball.
When Apophis rose again, Zia was gone. There was a crater in the sand where she'd been standing, and a human-sized lump illuminated the snake's gullet from the inside, glowing as it travelled down his throat.
Sadie tells me that I went a little insane. Honestly, I don't remember. The next thing I can recall, my voice was raw from screaming and I was staggering away from Apophis, my magic almost exhausted, my broken hand throbbing, my crook and flail smoking with red-grey ooze – the blood of Chaos.
Apophis had three gashes in his neck that weren't closing. Otherwise, he looked fine. It's hard to tell if a snake has an expression, but I was pretty sure he was gloating.
'As it was foretold!' He spoke aloud, and the earth shook. Cracks spread across the desert as if it had suddenly become thin ice. The sky turned black, lit only by stars and streaks of red lightning. The temperature began to drop. 'You cannot cheat destiny, Carter Kane! I have swallowed Ra. Now the end of the world is at hand!'
Sadie fell to her knees and sobbed. Despair swept over me, worse than the cold. I felt Horus's power fail, and I was just Carter Kane again. All around us, in different levels of the Duat, gods and magicians stopped battling as terror spread through their ranks.
With catlike agility, Bast landed next to me, breathing hard. Her hair was puffed out so much, it looked like a sea urchin covered with sand. Her bodysuit was ripped and torn. She had a nasty bruise on the left side of her jaw. Her knives were steaming and pitted with corrosion from the serpent's poison.
'No,' she said firmly. 'No, no, no. What's our plan?'
'Plan?' I tried to make sense of her question. Zia was gone. We'd failed. The ancient prophecy had come true, and I would die knowing that I was a complete and utter loser. I looked at Sadie, but she seemed just as shell-shocked.
'Wake up, kid!' Bes waddled up to me and kicked me in the kneecap, which was as high as he could reach.
'Ow!' I protested.
'You're the leader now,' he growled. 'So you'd _better_ have a plan. I didn't come back to life to get killed again!'
Apophis hissed. The ground continued to crack, shaking the foundations of the pyramids. The air was so cold that my breath turned to mist.
'Too late, poor children.' The serpent's red eyes stared down at me. 'Ma'at has been dying for centuries. Your world was only a temporary speck in the Sea of Chaos. All that you built meant nothing. _I_ am your past and your future! Bow to me now, Carter Kane, and perhaps I will spare you and your sister. I will enjoy having survivors to witness my triumph. Is that not preferable to death?'
My limbs felt heavy. Somewhere inside, I was a scared little boy who wanted to live. I'd lost my parents. I'd been asked to fight a war that was _way_ too big for me. Why should I keep going when it was hopeless? And if I could save Sadie...
Then I focused on the serpent's throat. The glow of the swallowed sun god sank lower and lower into Apophis's gullet. Zia had given her life to protect us.
_Never fear_ , she'd said. _I will hold Apophis until you come._
Anger cleared my thoughts. Apophis was trying to sway me, the way he'd corrupted Vlad Menshikov, Kwai, Sarah Jacobi and even Set, the god of evil himself. Apophis was the master of eroding reason and order, of destroying everything that was good and admirable. He was selfish, and he wanted me to be selfish as well.
I remembered the white obelisk rising from the Sea of Chaos. It had stood for thousands of years, against all odds. It represented courage and civilization, making the right choice instead of the easy choice. If I failed today, that obelisk would finally crumble. Everything humans had built since the first pyramids of Egypt would be for nothing.
'Sadie,' I said, 'you have the shadow?'
She got to her feet, her shocked expression turning to rage. 'I thought you'd never ask.'
From her bag she produced the granite figurine, now midnight black with the shadow of Apophis.
The serpent recoiled, hissing. I thought I detected fear in his eyes.
'Don't be foolish,' Apophis snarled. 'That ridiculous spell will not work – not now, when I am triumphant! Besides, you are too weak. You would never survive the attempt.'
Like all effective threats, it had the ring of truth. My magic reserves were nearly tapped out. Sadie's couldn't be much better. Even if the gods helped, we would likely burn ourselves up casting an execration.
'Ready?' Sadie asked me, her tone defiant.
'Attempt it,' Apophis warned, 'and I will raise your souls from Chaos again and again, just so I can kill you slowly. I will do the same for your father and mother. You will know an eternity of pain.'
I felt like I'd swallowed one of Ra's fireballs. My fists clenched round the crook and flail, despite the throbbing pain in my hand. The power of Horus surged back into me – and once again we were in absolute agreement. I was his Eye. I _was_ the Avenger.
'Mistake,' I told the serpent. 'You should _never_ threaten my family.'
I threw the crook and flail. They smashed into Apophis's face and erupted in a column of fire like a nuclear blast.
The serpent howled in pain, engulfed in flames and smoke, but I suspected I'd only bought us a few seconds.
'Sadie,' I said, 'are you ready?'
She nodded and offered me the figurine. Together, we held it and prepared for what might be the last spell of our lives. There was no need to consult a scroll. We'd been practising for this execration for months. We both knew the words by heart. The only question was whether the shadow would make the difference. Once we started, there would be no stopping. And, whether we failed or succeeded, we would probably burn up.
'Bes and Bast,' I said, 'can you two keep Apophis away from us?'
Bast smiled and hefted her knives. 'Protect my kittens? You don't even need to ask.' She glanced at Bes. 'And, in case we die, I'm sorry about all the times I toyed with your emotions. You deserved better.'
Bes snorted. 'That's okay. I finally came to my senses and found the right girl. Besides, you're a cat. It's your nature to think you're the centre of the universe.'
She stared at him blankly. 'But I _am_ the centre of the universe.'
Bes laughed. 'Good luck, kids. Time to bring on the ugly.'
'DEATH!' Apophis screamed, emerging from the column of fire with his eyes blazing.
Bast and Bes – the two greatest friends and protectors we'd ever had – charged to meet Apophis.
Sadie and I began the spell.
CARTER
## 20. I Take a Chair
LIKE I SAID, I'M NOT GOOD WITH INCANTATIONS.
Doing one right requires unbroken concentration, correct pronunciation and perfect timing. Otherwise you're liable to destroy yourself and everyone within ten feet, or turn yourself into some form of marsupial.
Trying to cast a spell with someone else – that's doubly hard.
Sure, Sadie and I had studied the words, but it's not like we could actually _do_ the execration in advance. With a spell like that, you only get one shot.
As we began, I was aware of Bast and Bes battling the serpent, and our other allies locked in combat at different levels of the Duat. The temperature kept dropping. Crevices widened in the ground. Red lightning spread across the sky like cracks in a black dome.
It was hard to keep my teeth from chattering. I concentrated on the stone figurine of Apophis. As we chanted, the statue began to smoke.
I tried not to think about the last time I'd heard this incantation. Michel Desjardins had died casting it, and he had faced only a partial manifestation of the serpent, not Apophis at his full power after triumphantly devouring Ra.
_Focus_ , Horus told me.
Easy for him to say. The noise, cold and explosions around us made it almost impossible – like trying to count backwards from a hundred while people scream random numbers in your ears.
Bast was thrown over our heads and landed against a stone block. Bes roared in anger. He slammed his club into the snake's neck so hard that Apophis's eyes rattled in his head.
Apophis snapped at Bes, who grabbed one fang and hung on for dear life as the serpent raised his head and shook his mouth, trying to dislodge the dwarf god.
Sadie and I continued to chant. The serpent's shadow steamed as the figurine heated up. Gold and blue light swirled around us as Isis and Horus did their best to shield us. Sweat stung my eyes. Despite the frosty air, I began to feel feverish.
When we came to the most important part of the spell – the naming of the enemy – I finally began to sense the true nature of the serpent's shadow. Funny how that works: sometimes you don't really understand something until you destroy it. The _sheut_ was more than just a copy or a reflection, more than a 'back-up disk' for the soul.
A person's shadow stood for his legacy, his impact on the world. Some people cast hardly any shadow at all. Some cast long, deep shadows that endured for centuries. I thought about what the ghost Setne had said – how he and I had each grown up in the shadow of a famous father. I realized now that he hadn't just meant it as a figure of speech. My dad cast a powerful shadow that still affected me and the whole world.
If a person cast no shadow at all, he couldn't be alive. His existence became meaningless. Execrating Apophis by destroying his shadow would cut his connection to the mortal world completely. He'd never be able to rise again. I finally understood why he'd been so anxious to burn Setne's scrolls, and why he was afraid of this spell.
We reached the last lines. Apophis dislodged Bes from his fang, and the dwarf sailed into the side of the Great Pyramid.
The serpent turned towards us as we spoke the final words: 'We exile you beyond the void. You are no more.'
'NO!' Apophis roared.
The statue flared, dissolving in our hands. The shadow disappeared in a puff of vapour, and an explosive wave of darkness knocked us off our feet.
The serpent's legacy on Earth shattered – the wars, murders, turmoil and anarchy Apophis had caused since ancient times finally lost power, no longer casting their shadow across our future. Souls of the dead were expelled from the blast – thousands of ghosts that had been trapped and crushed within the shadow of Chaos. A voice whispered in my mind: _Carter_ , and I sobbed with relief. I couldn't see her, but I knew that our mother was free. Her spirit was returning to its place in the Duat.
'Short-sighted mortals!' Apophis writhed and began to shrink. 'You haven't just killed me. You've exiled the gods!'
The Duat collapsed, layer upon layer, until the plains of Giza were one reality again. Our magician friends stood in a daze around us. The gods, however, were nowhere to be seen.
The serpent hissed, his scales falling away in smoking pieces. 'Ma'at and Chaos are linked, you fools! You cannot push me away without pushing away the gods. As for Ra, he shall die within me, slowly digested –'
He was cut short (literally) when his head exploded. Yes, it was just as gross as it sounds. Flaming bits of reptile flew everywhere. A ball of fire rolled up from the serpent's neck. The body of Apophis crumbled into sand and steaming goo, and Zia Rashid stepped out of the wreckage.
Her dress was in tatters. Her golden staff had cracked like a wishbone, but she was alive.
I ran towards her. She stumbled and collapsed against me, completely exhausted.
Then someone else rose from the smoking ruins of Apophis.
Ra shimmered like a mirage, towering over us as a muscular old man with golden skin, kingly robes and the pharaoh's crown. He stepped forward and daylight returned to the sky. The temperature warmed. The cracks in the ground sealed themselves.
The sun god smiled down at me. 'Well done, Carter and Sadie. Now, I must withdraw as the other gods have done, but I owe you my life.'
'Withdraw?' My voice didn't sound like mine. It was deeper, more gravelly – but it wasn't Horus's voice either. The war god seemed to be gone from my mind. 'You mean... forever?'
Ra chuckled. 'When you're as old as I am, you learn to be careful with that word _forever_. I thought I was leaving forever the first time I abdicated. For a while, at least, I must retreat into the sky. My old enemy Apophis was not wrong. When Chaos is pushed away, the gods of order, Ma'at, must also distance themselves. Such is the balance of the universe.'
'Then... you should take these.' Again I offered him the crook and flail.
Ra shook his head. 'Keep them for me. You are the rightful pharaoh. And take care of my favoured one...' He nodded at Zia. 'She will recover, but she will need support.'
Light blazed around the sun god. When it faded, he was gone. Two dozen weary magicians stood round a smoking, serpent-shaped mark in the desert as the sun rose over the pyramids of Giza.
Sadie rested her hand on my arm. 'Brother dear?'
'Yeah?'
'That was a bit too close.'
For once, I had no argument with my sister.
The rest of the day was a blur. I remember helping Zia to the healing rooms of the First Nome. My own broken hand took only minutes to fix, but I stayed with Zia until Jaz told me I needed to go. She and the other healers had dozens of wounded magicians to treat – including the Russian kid Leonid, who, amazingly, was expected to pull through – and while Jaz thought I was very sweet I was very much in the way.
I wandered through the main cavern and was shocked to see it full of people. Portals around the world had started working again. Magicians were flooding in to help with clean-up and pledge their support to the Chief Lector. Everybody loves to show up at the party once all the hard work is done.
I tried not to feel bitter about it. I knew that many of the other nomes had been fighting their own battles. Apophis had done his best to divide and conquer us. Still, it left a bad taste in my mouth. Many people stared in awe at Ra's crook and flail, which still hung from my belt. A few people congratulated me and called me a hero. I kept walking.
As I past the staff vendor's cart, someone said, ' _Psssst!_ '
I glanced towards the nearest alley. The ghost Setne was leaning against the wall. I was so startled I thought I must be hallucinating. He couldn't possibly be here, still in his horrible jacket and jewellery and jeans, his Elvis hair perfectly combed, the Book of Thoth tucked under his arm.
'You did good, pal,' he called. 'Not the way I would've handled it, but not bad.'
Finally I unfroze. ' _Tas!_ '
Setne just grinned. 'Yeah, we're done playing that game. But don't worry, pal. I'll see you around.'
He disappeared in a puff of smoke.
I'm not sure how long I stood there before Sadie found me.
'All right?' she asked.
I told her what I'd seen. She winced, but didn't look very surprised. 'I suppose we'll have to deal with that git sooner or later, but for now you'd better come with me. Amos has called a general assembly in the Hall of Ages.' She slipped her arm through mine. 'And try to smile, brother dear. I know it's hard. But you're a role model now, as horrifying as I find that.'
I did my best, though it was difficult to put Setne out of my mind.
We passed several of our friends helping with the restoration. Alyssa and a squad of earth elementalists were reinforcing walls and ceilings, trying to make sure the caverns didn't collapse on us.
Julian was sitting on the steps of the Scrying House, chatting up a few girls from the Scandinavian nome. 'Yeah, you know,' he was telling them, 'Apophis saw me coming with my big combat avatar, and he pretty much knew it was over.'
Sadie rolled her eyes and pulled me along.
Little Shelby and the other ankle-biters ran up to us, grinning and breathless. They'd helped themselves to some charms from one of the unmanned shopping kiosks, so they looked like they'd just come back from Egyptian Mardi Gras.
'I killed a snake!' Shelby told us. 'A big snake!'
'Really?' I asked. 'All by yourself?'
'Yes!' Shelby assured me. 'Kill, kill, kill!' She stomped her feet, and sparks flew from her shoes. Then she ran off, chasing her friends.
'That girl has a future,' Sadie said. 'Reminds me of myself when I was young.'
I shuddered. What a disturbing thought.
Gongs began ringing throughout the tunnels, summoning everyone to the Hall of Ages. By the time we got there, the hall was absolutely jammed with magicians – some in robes, some in modern clothes, some in pyjamas like they'd teleported straight from bed. On either side of the carpet, holographic curtains of light shimmered between the columns just as they had before.
Felix ran up to us, all smiles, with a herd of penguins behind him. (Herd? Flock? Gaggle? Oh, whatever.)
'Check it out!' he said happily. 'I learned this one during the battle!'
He spoke a command word. At first I thought it was _shish kebab_ , but later he told me it was: ' _Se-kebeb!_ ' _– Make cold._
Hieroglyphs appeared on the floor in frosty white:
The chill spread until a twenty-foot-wide section of the floor was coated in thick white ice. The penguins waddled across it, flapping their wings. One unfortunate magician stepped back and slipped so badly his staff went flying.
Felix pumped his fist. 'Yes! I found my path. I'm supposed to follow the god of ice!'
I scratched my head. 'There's a god of ice? Egypt is a desert. Who's the ice god?'
'I have no idea!' Felix beamed. He slid across the ice and went running off with his penguins.
We made our way down the hall. Magicians were trading stories, mingling and checking in with old friends. Hieroglyphs floated through the air, brighter and thicker than I'd ever seen, like a rainbow alphabet soup.
Finally the crowd noticed Sadie and me. A hush spread through the room. All eyes turned towards us. The magicians parted, clearing the way to the throne.
Most of the magicians smiled as we walked past. A few whispered thanks and congratulations. Even the former rebel magicians seemed genuinely pleased to see us. But I did catch a few angry looks. No matter that we'd defeated Apophis; some of our fellow magicians would always doubt us. Some would never stop hating us. The Kane family still needed to watch our backs.
Sadie scanned the crowd anxiously. I realized she was looking for Walt. I'd been so focused on Zia that I hadn't thought about how worried Sadie must be. Walt had disappeared after the battle, along with the rest of the gods. He didn't seem to be here now.
'I'm sure he's fine,' I told her.
'Shh.' Sadie smiled at me, but her eyes said: _If you embarrass me in front of all these people, I will strangle you._
Amos waited for us at the steps of the throne. He'd changed into a crimson suit that went surprisingly well with his leopard-skin cape. His hair was braided with garnets, and his glasses were tinted red. The colour of Chaos? I got the feeling he was playing up his connection to Set – which all the other magicians had definitely heard about by now.
For the first time in history, our Chief Lector had the god of evil, strength and Chaos on speed dial. That might make people trust him less, but magicians were like the gods – they respected strength. I doubted Amos would have much trouble enforcing his rule any more.
He smiled as we approached. 'Carter and Sadie, on behalf of the House of Life, I thank you. You have restored Ma'at! Apophis has been execrated, and Ra has once again risen into the heavens, but this time in triumph. Well done!'
The hall erupted in cheering and applause. Dozens of magicians raised their staffs and sent up miniature firework displays.
Amos embraced us. Then he stepped aside and gestured me towards the throne. I hoped that Horus might give me some words of encouragement, but I couldn't feel his presence at all.
I tried to control my breathing. That chair had been empty for thousands of years. How could I be sure it would even hold my weight? If the throne of the pharaohs broke under my royal butt, that would be a great omen.
Sadie nudged me. 'Go on, then. Don't be stupid.'
I climbed the steps and eased myself onto the throne. The old chair creaked, but it held me.
I gazed out over the crowd of magicians.
Horus wasn't there for me. But somehow that was okay. I glanced over at the shimmering curtains of light – the New Age, glowing purple – and I had a feeling it was going to be an age of good things, after all.
My muscles began to relax. I felt like I'd stepped out of the war god's shadow, just as I'd stepped out of my father's. I found the words.
'I accept the throne.' I held up the crook and flail. 'Ra has given me authority to lead the gods and magicians in times of crisis, and I'll do my best. Apophis has been banished, but the Sea of Chaos is always there. I've seen it with my own eyes. Its forces will always try to erode Ma'at. We can't think that all our enemies are gone.'
The crowd stirred nervously.
'But for now,' I added, 'we are at peace. We can rebuild and expand the House of Life. If war comes again, I'll be here as the Eye of Horus and as pharaoh. But as Carter Kane...'
I rose and placed the crook and flail on the throne. I stepped down from the dais. 'As Carter Kane, I'm a kid who has a lot of catching up to do. I've got my own nome to run at Brooklyn House. And I've got to graduate from high school. So I'm going to leave day-to-day operations where they should be – in the hands of the Chief Lector, Amos Kane, steward of the pharaoh.'
Amos bowed to me, which felt a little strange. The crowd applauded wildly. I wasn't sure if they approved of me, or if they were just relieved that a kid wasn't going to be giving them daily orders from the throne. Either way, I was okay with it.
Amos embraced Sadie and me again.
'I'm proud of you both,' he said. 'We'll speak soon, but right now, come...' He gestured to the side of the dais, where a door of darkness had opened in the air. 'Your parents would like to see you.'
Sadie looked at me nervously. 'Uh-oh.'
I nodded. Strange how I went instantly from the pharaoh of the universe to a kid worried about getting grounded. As much as I wanted to see my parents, I'd broken an important promise to my father... I'd lost track of a dangerous prisoner.
The Hall of Judgement had turned into Party Central. Ammit the Devourer ran around the scales of justice, yapping excitedly with a birthday hat on his crocodile head. The guillotine-headed demons lounged on their pole arms, holding glasses of what looked like champagne. I didn't know how they could drink with those guillotine heads, but I didn't want to find out. Even the blue judgement god Disturber seemed to be in a good mood. His Cleopatra wig was sideways on his head. His long scroll had unravelled halfway across the room, but he was laughing and talking with the other judgement gods who had been rescued from the House of Rest. Fire-embracer and Hot Foot kept dropping cinders on his papyrus, but Disturber didn't seem to notice or care.
At the far end of the room, Dad sat on his throne, holding hands with our ghostly mom. To the left of the dais, spirits from the Underworld played in a jazz ensemble. I was pretty sure I recognized Miles Davis, John Coltrane and a few of my dad's other favourites. Being the god of the Underworld has its perks.
Dad beckoned us forward. He didn't look mad, which was a good sign. We made our way through the crowd of happy demons and judgement gods. Ammit yapped at Sadie and purred as she scratched under his chin.
'Children.' Dad held out his arms.
It felt strange being called children. I didn't feel like a child any more. Children weren't asked to fight Chaos serpents. They didn't lead armies to stop the end of the world.
Sadie and I both hugged our dad. I couldn't hug Mom, of course, since she was a ghost, but I was happy enough to see her safe. Except for the glowing aura around her, she looked just like she did when she was alive – dressed in jeans and her _ankh_ T-shirt, her blonde hair gathered back in a bandanna. If I didn't look directly at her, I could have almost mistaken her for Sadie.
'Mom, you survived,' I said. 'How –?'
'All thanks to you two.' Mom's eyes sparkled. 'I held on as long as I could, but the shadow was too powerful. I was consumed, along with so many other spirits. If you hadn't destroyed the _sheut_ when you did and released us, I would've been... Well, it doesn't matter now. You've done the impossible. We are so proud.'
'Yes,' Dad agreed, squeezing my shoulder. 'Everything we've worked for, everything we've hoped for – you have accomplished. You've exceeded my highest expectations.'
I hesitated. Was it possible he didn't know about Setne?
'Dad,' I said, ' _um_... we didn't succeed at _everything_. We lost your prisoner. I still don't understand how he escaped. He was tied up and –'
Dad raised his hand to stop me. 'I heard. We may never know how Setne escaped exactly, but you can't blame yourselves.'
'We can't?' Sadie asked.
'Setne has evaded capture for aeons,' Dad said. 'He's outwitted gods, magicians, mortals and demons. When I let you take him, I suspected he would find a way to escape. I just hoped you could control him long enough to get his help. And you did.'
'He got us to the shadow,' I admitted. 'But he also stole the Book of Thoth.'
Sadie bit her lip. 'Dangerous stuff, that book. Setne may not be able to cast all the spells himself, being a ghost, but he could still cause all sorts of mischief.'
'We will find him again,' Dad promised. 'But for now let's celebrate your victory.'
Our mom reached out and brushed her ghostly hand through Sadie's hair. 'May I borrow you a moment, my dear? I have something I'd like to discuss with you.'
I wasn't sure what that was about, but Sadie followed our mom towards the jazz band. I hadn't noticed before, but two of the ghostly musicians looked very familiar and rather out of place. A big red-headed man in Western clothes sat at a steel guitar, grinning and tapping his boots as he traded solos with Miles Davis. Next to him, a pretty blonde woman played the fiddle, leaning down from time to time to kiss the red-headed man on the forehead. JD Grissom and his wife, Anne, from the Dallas Museum, had finally found a party that didn't have to end. I'd never heard steel guitar and fiddle with a jazz band before, but somehow they made it work. I suppose Amos was right: music and magic both needed a little chaos within the order.
As Mom and Sadie talked, Sadie's eyes widened. Her expression turned serious. Then she smiled shyly and blushed, which wasn't like Sadie at all.
'Carter,' my dad said, 'you did well in the Hall of Ages. You will make a good leader. A wise leader.'
I wasn't sure how he knew about my speech, but a lump formed in my throat. My dad doesn't hand out compliments lightly. Being with him again, I remembered how much easier life had been, travelling with him. He'd always known what to do. I could always count on his calming presence. Until that Christmas Eve in London when he had disappeared, I hadn't appreciated just how much I had relied on him.
'I know it's been hard,' Dad said, 'but you will lead the Kane family into the future. You have truly stepped out of my shadow.'
'Not completely,' I said. 'I wouldn't want that. As dads go, you're pretty, _um_ , shadowy.'
He laughed. 'I'll be here if you need me. Never doubt that. But, as Ra said, the gods will have a harder time contacting the mortal world, now that Apophis has been execrated. As Chaos retreats, so must Ma'at. Nevertheless, I don't think you'll _need_ much help. You've succeeded on your own strength. Now _you_ are the one casting the long shadow. The House of Life will remember you for ages to come.'
He hugged me once more, and it was easy to forget that he was the god of the dead. He just seemed like my dad – warm and alive and strong.
Sadie came over, looking a little shaken.
'What?' I asked.
She giggled for no apparent reason, then got serious again. 'Nothing.'
Mom drifted next to her. 'Off you go, you two. Brooklyn House is waiting.'
Another door of darkness appeared by the throne. Sadie and I stepped through. For once I wasn't worried about what waited on the other side. I knew we were going home.
Life got back to normal with surprising speed.
I'll let Sadie tell you about the events at Brooklyn House and her own drama. I'll fast-forward to the interesting stuff.
[Ouch! I thought we agreed: no pinching!]
Two weeks after the battle with Apophis, Zia and I were sitting in the food court at the Mall of America in Bloomington, Minnesota.
Why there? I'd heard the Mall of America was the biggest in the country, and I figured we'd start big. It was an easy trip through the Duat. Freak was happy to sit on the roof and eat frozen turkeys while Zia and I explored the mall.
[That's right, Sadie. For our first real date, I picked up Zia in a boat pulled by a deranged griffin. So what? Like _your_ dates aren't weird?]
Anyway, when we got to the food court, Zia's jaw dropped. 'Gods of Egypt...'
The restaurant choices were pretty overwhelming. Since we couldn't decide, we got a little of everything: Chinese, Mexican (the Macho Nachos), pizza and ice cream – the four basic food groups. We grabbed a table overlooking the amusement park at the centre of the mall.
A lot of other kids were hanging out in the food court. Many of them stared at us. Well... not at _me_. They were mostly looking at Zia and no doubt wondering what a girl like her was doing with a guy like me.
She'd healed up nicely since the battle. She wore a simple sleeveless dress of beige linen and black sandals – no make-up, no jewellery except for her gold scarab necklace. She looked way more glamorous and mature than the other girls in the mall.
Her long black hair was tied back in a ponytail, except for a little strand that curled behind her right ear. She'd always had luminous amber eyes and warm coffee-and-milk skin, but, since hosting Ra, she seemed to glow even more. I could feel her warmth from across the table.
She smiled at me over her bowl of chow mein. 'So, this is what typical American teenagers do?'
'Well... sort of,' I said. 'Though I don't think either of us will ever pass for _typical_.'
'I hope not.'
I had trouble thinking straight when I looked at her. If she'd asked me to jump over the railing, I probably would've done it.
Zia twirled her fork through her noodles. 'Carter, we haven't talked much about... you know, my being the Eye of Ra. I can guess how strange that was for you.'
See? Just your typical teenage conversation in the mall.
'Hey, I understand,' I said. 'It wasn't strange.'
She raised an eyebrow.
'Okay, it was strange,' I admitted. 'But Ra needed your help. You were amazing. Have you, _uh_ , talked to him since... ?'
She shook her head. 'He's retreated from the world, just like he said. I doubt I'll be the Eye of Ra again – unless we face another Doomsday.'
'So, with our luck, not for a few more weeks, you mean.'
Zia laughed. I loved her laugh. I loved that little curl of hair behind her ear.
[Sadie says I'm being ridiculous. Like she's one to talk.]
'I had a meeting with your Uncle Amos,' Zia said. 'He has lots of help at the First Nome now. He thought it would be good for me to spend some time away, try to live a more... typical life.'
My heart tripped and stumbled straight into my ribs. 'You mean, like, leave Egypt?'
Zia nodded. 'Your sister suggested I stay at Brooklyn House, attend American school. She says... how did she put it? _Americans are an odd bunch, but they grow on you._ '
Zia scooted round the table and took my hand. I sensed about twenty jealous guys glaring at me from the other tables of the food court.
'Would you mind if I stayed in Brooklyn House? I could help teach the initiates. But if that would make you uncomfortable –'
'No!' I said much too loudly. 'I mean, no, I don't mind. Yes, I'd like that. A lot. Quite a bit. Totally fine.'
Zia smiled. The temperature in the food court seemed to go up another ten degrees. 'So that's a yes?'
'Yes. I mean, unless it would make _you_ uncomfortable. I wouldn't want to make things awkward or –'
'Carter?' she said gently. 'Shut up.'
She leaned over and kissed me.
I did as she commanded, no magic necessary. I shut up.
SADIE
## 21. The Gods Are Sorted; My Feelings Are Not
AH, MY THREE FAVOURITE WORDS: _Carter, shut up._
Zia really has come a long way since we first met. I think there's hope for her, even if she does fancy my brother.
At any rate, Carter has wisely left the last bit of the story for me to tell.
After the battle with Apophis, I felt horrible on many levels. Physically, I was knackered. Magically, I'd used up every last bit of energy. I was afraid I might have permanently damaged myself, as I had a smouldering feeling behind my sternum that was either my exhausted magic reservoir or very bad heartburn.
Emotionally, I wasn't much better. I had watched Carter embrace Zia when she emerged from the steaming goo of the serpent, which was all very well, but it only reminded me of my own turmoil.
Where was Walt? (I'd decided to call him that, or I would drive myself crazy figuring out his identity.) He had been standing nearby just after the battle. Now he was gone.
Had he left with the other gods? I was already worried about Bes and Bast. It wasn't like them to disappear without saying goodbye. And I wasn't keen on what Ra had said about the gods leaving the earth for a while.
_You cannot push me away without pushing away the gods_ , Apophis had warned.
The serpent might have mentioned that _before_ we execrated him. I had just made my peace with the whole Walt/Anubis idea – or _mostly_ , at any rate – and now Walt had vanished. If he'd been declared off-limits again, I was going to crawl into a sarcophagus and never come out.
While Carter was with Zia in the infirmary, I wandered the corridors of the First Nome, but found no sign of Walt. I tried to contact him with the _shen_ amulet. No answer. I even tried to contact Isis for advice, but the goddess had gone silent. I didn't like that.
So, yes, I was quite distracted in the Hall of Ages during Carter's little acceptance speech: _I'd like to thank all the little people for making me pharaoh, et cetera, et cetera._
I was glad to visit the Underworld and be reunited with my mum and dad. At least _they_ weren't off-limits. But I was quite disappointed not to find Walt there. Even if he wasn't allowed in the mortal world, shouldn't he be in the Hall of Judgement, taking over the duties of Anubis?
That's when my mother pulled me aside. (Not literally, of course. Being a ghost, she couldn't pull me anywhere.) We stood to the left of the dais where the dead musicians played lively music. JD Grissom and his wife, Anne, smiled at me. They seemed happy, and I was glad for that, but I still had trouble seeing them without feeling guilty.
My mum tugged at her necklace – a ghostly replica of my own _tyet_ amulet. 'Sadie... we've never got to talk much, you and I.'
Bit of an understatement, since she died when I was six. I understood what she meant, though. Even after our reunion last spring, she and I had never really chatted. Visiting her in the Duat was rather hard, and ghosts don't have e-mail or Skype or mobile phones. Even if they had had a proper Internet connection, 'friending' my dead mother on Facebook would have felt rather odd.
I didn't say any of that. I just nodded.
'You've grown strong, Sadie,' Mum said. 'You've had to be brave for so long it must be hard for you to let your defences down. You're afraid to lose any more people you care about.'
I felt lightheaded, as if I were turning into a ghost, too. Had I become see-through, like my mother? I wanted to argue and protest and joke. I didn't want to hear my mother's commentary, especially when it was so accurate.
At the same time, I was so mixed up inside about Walt, so worried about what had happened to him, I wanted to break down and cry on my mother's shoulder. I wanted her to hug me and tell me it was all right. Unfortunately, one can't cry on the shoulder of a ghost.
'I know,' Mum said sadly, as if reading my thoughts. 'I wasn't there for you when you were small. And your father... well, he had to leave you with Gran and Gramps. They tried to provide you with a normal life, but you're so much _more_ than normal, aren't you? And now here you are, a young woman...' She sighed. 'I've missed so much of your life I don't know if you'll want my advice now. But for what it's worth: trust your feelings. I can't promise that you'll never get hurt again, but I can promise you the risk is worth it.'
I studied her face, unchanged since the day she had died: her wispy blonde hair, her blue eyes, the rather mischievous curve of her eyebrows. Many times, I'd been told that I looked like her. Now I could see it clearly. As I'd got older, it was quite striking how much our faces looked alike. Put some purple highlights in her hair and Mum would've made an excellent Sadie stunt double.
'You're talking about Walt,' I said at last. 'This is a heart-to-heart chat about _boys_?'
Mum winced. 'Yes, well... I'm afraid I'm rubbish at this. But I had to try. When I was a girl, Gran wasn't much of a resource for me. I never felt I could talk to her.'
'I should think not.' I tried to imagine talking about guys with my grandmother while Gramps yelled at the telly and called for more tea and burnt biscuits.
'I think,' I ventured, 'that mothers normally warn _against_ following one's heart, getting involved with the wrong sort of boy, getting a bad reputation. That sort of thing.'
'Ah.' Mum nodded contritely. 'Well, you see, I can't do that. I suppose I'm not worried about you doing the wrong thing, Sadie. I _am_ worried that you might be afraid to trust someone – even the right someone. It's _your_ heart, of course. Not mine. But I'd say Walt is more nervous than you are. Don't be too hard on him.'
'Hard on _him_?' I almost laughed. 'I don't even know where he is! And he's hosting a god who – who –'
'Whom you also like,' Mum supplied. 'And that's confusing, yes. But they are really one person, now. Anubis has so much in common with Walt. Neither has ever had a real life to look forward to. Now, together, they do.'
'You mean...' The horrible burning sensation behind my sternum began to ease, ever so slightly. 'You mean I _will_ see him again? He's not exiled or whatever nonsense the gods are going on about?'
'You will see him,' my mother affirmed. 'Because they are one, inhabiting a single mortal body, they may walk the earth, as the Ancient Egyptian god-kings did. Walt and Anubis are both good young men. They are both nervous, and quite awkward in the mortal world, and scared about how people will treat them. And they both feel the same way about you.'
I was probably blushing terribly. Carter stared at me from the top of the dais, no doubt wondering what was wrong. I didn't trust myself to meet his eyes. He was a bit too good at reading my expression.
'It's so _hard_ ,' I complained.
Mother laughed softly. 'Yes, it is. But if it's any consolation... dealing with _any_ man means dealing with multiple personalities.'
I glanced up at my father, who was flickering back and forth between Dr Julius Kane and Osiris, the Smurf-blue god of the Underworld.
'I take your point,' I said. 'But where _is_ Anubis? I mean Walt. Ugh! There I go again.'
'You will see him soon,' Mum promised. 'I wanted you to be prepared.'
My mind said: _This is too confusing, too unfair. I can't handle a relationship like this._
But my heart said: _Shut up! Yes, I can!_
'Thanks, Mum,' I said, no doubt failing miserably to look calm and collected. 'This business with the gods pulling away. Does that mean we won't see you and Dad as much?'
'Probably,' she admitted. 'But you know what to do. Keep teaching the path of the gods. Bring the House of Life back to its former glory. You and Carter and Amos will make Egyptian magic stronger than ever. And that's good... because your challenges are not over.'
'Setne?' I guessed.
'Yes, him,' Mum said. 'But there are other challenges as well. I haven't completely lost the gift of prophecy, even in death. I see murky visions of other gods and rival magic.'
That _really_ didn't sound good.
'What do you mean?' I asked. 'What _other gods_?'
'I don't know, Sadie. But Egypt has always faced challenges from outside – magicians from elsewhere, even gods from elsewhere. Just be vigilant.'
'Lovely,' I muttered. 'I preferred talking about boys.'
Mother laughed. 'Once you return to the mortal world, there will be one more portal. Look for it tonight. Some old friends of yours would like a word.'
I had a feeling I knew whom she meant.
She touched a ghostly pendant round her neck – the _tyet_ symbol of Isis.
'If you need me,' Mum said, 'use your necklace. It will call to me, just as the _shen_ necklace calls to Walt.'
'That would've been handy to know sooner.'
'Our connection wasn't strong enough before. Now... I think it is.' She kissed my forehead, though it felt like only a faint cool breeze. 'I'm proud of you, Sadie. You have your whole life ahead of you. Make the most of it!'
That night at Brooklyn House, a swirling sand portal opened on the terrace, just as my mother had promised.
'That's for us,' I said, getting up from the dinner table. 'Come on, brother dear.'
On the other side of the portal, we found ourselves at the beach by the Lake of Fire. Bast was waiting, tossing a ball of yarn from hand to hand. Her pure black bodysuit matched her hair. Her feline eyes danced in the red light of the waves.
'They're waiting for you.' She pointed up the steps to the House of Rest. 'We'll talk when you come back down.'
I didn't need to ask why she wasn't coming. I heard the melancholy in her voice. She and Tawaret had never got along because of Bes. Obviously, Bast wanted to give the hippo goddess some space. But, also, I wondered if my old friend was starting to realize that she'd let a good man get away.
I kissed her on the cheek. Then Carter and I climbed the stairs.
Inside the nursing home, the atmosphere was festive. Fresh flowers decorated the nurses' station. Heket the frog goddess walked upside down along the ceiling, hanging party streamers, while a group of elderly dog-headed gods danced and sang the hokey-pokey – a very slow version, but still impressive. _You put your walker in / you put your IV out_ – and so forth. The ancient lion-headed goddess Mekhit was slow-dancing with a tall male god. She purred loudly with her head on his shoulder.
'Carter, look,' I said. 'Is that –?'
'Onuris!' Tawaret answered, trotting over in her nurse's outfit. 'Mekhit's husband! Isn't it wonderful? We were sure he'd faded ages ago, but when Bes called the old gods to war Onuris came tottering out of a supply closet. Many others appeared, too. They were finally needed, you see! The war gave them a reason to exist.'
The hippo goddess crushed us in an enthusiastic hug. 'Oh, my dears! Just look how happy everyone is! You've given them new life.'
'I don't see as many as before,' Carter noticed.
'Some went back to the heavens,' Tawaret said. 'Or off to their old temples and palaces. And, of course, your dear father, Osiris, took the judgement gods back to his throne room.'
Seeing the old gods so happy warmed my heart, but I still felt a twinge of worry. 'Will they stay this way? I mean, they won't fade again?'
Tawaret spread her stubby hands. 'I suppose that depends on you mortals. If you remember them and make them feel important, they should be fine. But come, you'll want to see Bes!'
He sat in his usual chair, staring blankly out of the window at the Lake of Fire. The scene was so familiar I feared he'd lost his _ren_ again.
'Is he all right?' I cried, running up to him. 'What's wrong with him?'
Bes turned, looking startled. 'Besides being ugly? Nothing, kid. I was just thinking – sorry.'
He rose (as much as a dwarf can rise) and hugged us both.
'Glad you kids could make it,' Bes said. 'You know Tawaret and I are going to build a home on the lakeside. I've got used to this view. She'll keep working at the House of Rest. I'll be a house dwarf for a while. Who knows? Maybe I'll get some little dwarf hippo babies to look after!'
'Oh, Bes!' Tawaret blushed fiercely and batted her hippo eyelids.
The dwarf god chuckled. 'Yeah, life is good. But if you kids need me just holler. I've always had more luck coming to the mortal world than most gods.'
Carter scowled fretfully. 'Do you think we'll need you a lot? I mean, of course we want to see you! I just wondered –'
Bes grunted. 'Hey, I'm an ugly dwarf. I've got a sweet car, an excellent wardrobe and amazing powers. Why _wouldn't_ you need me?'
'Good point,' Carter agreed.
'But, _uh_ , don't call _too_ often,' Bes said. 'After all, my honeycakes and I got a few millennia of quality time to catch up on.'
He took Tawaret's hand, and for once I didn't find the name of this place – Sunny Acres – quite so depressing.
'Thank you for everything, Bes,' I said.
'Are you kidding?' he said. 'You gave me my life back, and I don't just mean my shadow.'
I got the distinct feeling the two gods wanted some time by themselves, so we said our goodbyes and headed down the steps to the lake.
The white sand portal was still swirling. Bast stood next to it, engrossed in her ball of yarn. She laced it between her fingers to make a rectangle like a cat's cradle. (No, I didn't mean that as a pun, but it _did_ seem appropriate.)
'Having fun?' I asked.
'Thought you'd want to see this.' She held up the cat's cradle. A video image flickered across its surface like on a computer screen.
I saw the Hall of the Gods with its soaring columns and polished floors, its braziers burning with a hundred multicoloured fires. On the central dais, the sun boat had been replaced with a golden throne. Horus sat there in his human form – a bald muscular teen in full battle armour. He held a crook and flail across his lap, and his eyes gleamed – one silver, one gold. At his right stood Isis, smiling proudly, her rainbow wings shimmering. On his left stood Set, the red-skinned Chaos god with his iron staff. He looked quite amused, as if he had all sorts of wicked things planned for later. The other gods knelt as Horus addressed them. I scanned the crowd for Anubis – with or without Walt – but, again, I didn't see him.
I couldn't hear the words, but I reckoned it was a similar speech to the one Carter had delivered to the House of Life.
'He's doing the same thing I did,' Carter protested. 'I bet he even stole my speech. That copycat!'
Bast clucked disapprovingly. 'No need to call names, Carter. Cats are not copiers. We are all unique. But, yes, what you do as pharaoh in the mortal world will often be mirrored in the world of the gods. Horus and you, after all, rule the forces of Egypt.'
'That,' I said, 'is a truly scary thought.'
Carter swatted me lightly on the arm. 'I just can't believe that Horus left without even a goodbye. It's as if he tossed me aside as soon as he was done using me and then forgot about me.'
'Oh, no,' Bast said. 'Gods wouldn't do that. He simply had to leave.'
But I wondered. Gods were rather selfish creatures, even those who weren't cats. Isis hadn't given me a proper goodbye or thank-you either.
'Bast, you're coming with us, aren't you?' I pleaded. 'I mean, this silly exile can't apply to you! We need our nap instructor at Brooklyn House.'
Bast wadded up her ball of yarn and tossed it down the steps. Her expression was quite sad for a feline. 'Oh, my kittens. If I could, I would pick you up by the scruffs of your necks and carry you forever. But you've grown. Your claws are sharp, your eyesight is keen, and cats must make their own way in the world. I must say farewell for now, though I'm sure we'll meet again.'
I wanted to protest that I hadn't grown up and I didn't even have claws.
[Carter disagrees, but what does he know?]
But part of me knew Bast was right. We'd been lucky to have her with us for so long. Now we had to be adult cats – _er_ , humans.
'Oh, Muffin...' I hugged her fiercely and could feel her purring.
She ruffled my hair. Then she rubbed Carter's ears, which was quite funny.
'Go on, now,' she said. 'Before I start to mewl. Besides...' She fixed her eyes on the ball of yarn, which had rolled to the bottom of the steps. She crouched and tensed her shoulders. 'I have some hunting to do.'
'We'll miss you, Bast,' I said, trying not to cry. 'Good hunting.'
'Yarn,' she said absently, creeping down the steps. 'Dangerous prey, yarn...'
Carter and I stepped through the portal. This time it deposited us onto the roof of Brooklyn House.
We had one more surprise. Standing by Freak's roost, Walt was waiting. He smiled when he saw me, and my legs felt wobbly.
'I'll, _um_ , be inside,' Carter said.
Walt walked over, and I tried to remember how to breathe.
SADIE
## 22. The Last Waltz (for Now)
HE'D CHANGED HIS LOOK AGAIN.
His amulets were gone except for one – the _shen_ that matched mine. He wore a black muscle shirt, black jeans, a black leather duster and black combat boots – a sort of mix of Anubis's and Walt's styles, but it made him look like someone entirely different and new. Yet his eyes were quite familiar – warm, dark brown and lovely. When he smiled, my heart fluttered as it always had.
'So,' I said, 'is this another goodbye? I've had quite enough goodbyes today.'
'Actually,' Walt said, 'it's more of a hello. My name's Walt Stone, from Seattle. I'd like to join the party.'
He held out his hand, still smiling slyly. He was repeating exactly what he'd said the first time we met, when he arrived at Brooklyn House last spring.
Instead of taking his hand, I punched him in the chest.
'Ow,' he complained. But I doubt that I'd hurt him. He had quite a solid chest.
'You think you can just merge with a god and _surprise_ me like that?' I demanded. ' _Oh, by the way, I'm actually two minds in one body._ I don't appreciate being taken off guard.'
'I did try to tell you,' he said. 'Several times. Anubis did, too. We kept getting interrupted. Mostly by you talking a lot.'
'No excuse.' I folded my arms and scowled as best I could. 'My mum seems to think I should go easy on you because this is all very new to you. But I'm still cross. It's confusing enough, you know, liking someone, without their morphing into a _god_ whom I also like.'
'So you do like me.'
'Stop trying to distract me! Are you truly asking to stay here?'
Walt nodded. He was very close now. He smelled good, like vanilla candles. I tried to remember if that was Walt's scent or Anubis's. Honestly, I couldn't recall.
'I've still got a lot to learn,' he said. 'I don't need to stick with charm-making any more. I can do more intensive magic – the path of Anubis. No one's ever done that before.'
'Discovering new magical ways to annoy me?'
He tilted his head. 'I could do amazing tricks with mummy linen. For instance, if someone talks too much, I could summon a gag –'
'Don't you dare!'
He took my hand. I gave him a defiant scowl, but I didn't take back my hand.
'I'm still Walt,' he said. 'I'm still mortal. Anubis can stay in this world as long as I'm his host. I'm hoping to live a good long life. Neither of us ever thought that was possible. So I'm not going anywhere, unless you want me to leave.'
My eyes probably answered for me: _No, please. Not ever._ But I couldn't very well give him the satisfaction of my saying that out loud, could I? Boys can get so full of themselves.
'Well,' I grumbled, 'I suppose I could tolerate it.'
'I owe you a dance.' Walt put his other hand on my waist – a traditional pose, very old-fashioned, as Anubis had done when we waltzed at the Brooklyn Academy. My gran would've approved.
'May I?' he asked.
'Here?' I said. 'Won't your chaperone Shu interrupt?'
'Like I said, I'm mortal now. He'll let us dance, though I'm sure he's keeping an eye on us to make sure we behave.'
'To make sure _you_ behave,' I snipped. 'I'm a proper young lady.'
Walt laughed. I supposed it was funny. _Proper_ wasn't the first word normally used to describe me.
I pounded his chest again, though I'll admit not very hard. I put my hand on his shoulder.
'I'll have you remember,' I warned, 'that my father is your employer in the Underworld. You'd best mind your manners.'
'Yes, ma'am,' Walt said. He leaned down and kissed me. All my anger melted into my shoes.
We started to dance. There was no music, no ghostly dancers, no floating on air – nothing magic about it. Freak watched us curiously, no doubt wondering how this activity was going to produce turkeys to feed the griffin. The old tar roof creaked under our feet. I was still quite tired from our long battle, and I hadn't cleaned up properly. No doubt I looked horrid. I wanted to melt into Walt's arms, which is basically what I did.
'So you'll let me stick around?' he asked, his breath warm on my scalp. 'Let me experience a typical teenage life?'
'I suppose.' I looked up at him. It took no effort at all to slip my vision into the Duat and see Anubis there, just under the surface. But it really wasn't necessary. This was a new boy in front of me, and he was everything I liked. 'Not that I'm an expert myself, but there is one rule I insist on.'
'Yes?'
'If anyone asks you if you're taken,' I said, 'the answer is _yes_.'
'I think I can live with that,' he promised.
'Good,' I said. 'Because you don't want to see me be cross.'
'Too late.'
'Shut up and dance, Walt.'
We did – with the music of a psychotic griffin screaming behind us, and the sirens and horns of Brooklyn wailing below. It was quite romantic.
So there you have it.
We've returned to Brooklyn House. The various catastrophes plaguing the world have lessened – at least somewhat – and we are dealing with an influx of new initiates as the school year gets properly under way.
It should be obvious now why this may be our last recording. We're going to be so busy training and attending school and living our lives that I doubt we'll have time or reason to send out any more audio pleas for help.
We'll put this tape in a secure box and send it along to the chap who's been transcribing our adventures. Carter seems to think the postal service will do, but I think I'll give it to Khufu to carry through the Duat. What could possibly go wrong?
As for us, don't think our lives will be all fun and games. Amos couldn't leave a mob of teens unsupervised and, as we don't have Bast any more, Amos has sent a few adult magicians to Brooklyn House as teachers (read: chaperones). But we all know who's really in charge – _me_. Oh, yes, and perhaps Carter a little bit.
We're not done with trouble, either. I'm still worried about that murderous ghost Setne, who's on the loose in the world with his devious mind, horrible fashion sense and the Book of Thoth. I'm also puzzling over my mother's comments about rival magic and other gods. No idea what that means, but it doesn't sound good.
In the meantime, there are still hotspots of evil magic and demon activity all over the world that we have to take care of. We've even got reports of unexplainable magic as close as Long Island. Probably have to check that out.
But for now I plan on enjoying my life, annoying my brother as much as possible and making Walt into a proper boyfriend while keeping the other girls away from him – most likely with a flamethrower. My work is never done.
As for you lot out there, listening to this recording – we're never too busy for new initiates. If you have the blood of the pharaohs, what are you waiting for? Don't let your magic go to waste. Brooklyn House is open for business.
## Glossary
Commands used by Carter, Sadie and others
| | **_Drowah_** 'Boundary'
---|---|---
| | **_Fah_** 'Release'
| | **_Ha-di_** 'Destroy'
| | **_Hapi, u-ha ey pwah_** 'Hapi, arise and attack'
| | **_Ha-tep_** 'Be at peace'
| | **_Ha-wi_** 'Strike'
| | **_Hi-nehm_** 'Join together'
| | **_Isfet_** 'Chaos'
| | **_Ma'at_** 'Restore order'
| | **_Maw_** 'Water'
| | **_Med-wah_** 'Speak'
| | **_N'dah_** 'Protect'
| | **_Sa-hei_** 'Bring down'
| | **_Se-kebeb_** 'Make cold'
| | **_Tas_** 'Bind'
##### OTHER EGYPTIAN TERMS
**_Ankh_** a hieroglyphic symbol for 'life'
**_Ba_** one of the five parts of the soul: the personality
**Barque** the pharaoh's boat
**Canopic jar** vessel used to store a mummy's organs
**Criosphinx** a creature with a body of a lion and head of a ram
**Duat** magical realm that coexists with our world
**Hieroglyphics** the writing system of Ancient Egypt, which used symbols or pictures to denote objects, concepts or sounds
**_Ib_** one of the five parts of the soul: the heart
**Isfet** the symbol for total Chaos
**_Ka_** one of the five parts of the soul: the life force
**_Khopesh_** a sword with a hook-shaped blade
**Ma'at** order of the universe
**_Netjeri_** a knife made from meteoric iron for the opening of the mouth in a ceremony
**_Per Ankh_** the House of Life
**Pharaoh** a ruler of Ancient Egypt
**_Rekhet_** healer
**_Ren_** one of the five parts of the soul: the secret name; identity
**Sarcophagus** a stone coffin, often decorated with sculpture and inscriptions
**_Sau_** a charm maker
**Scarab** beetle
**_Shabti_** a magical figurine made out of clay
**_Shen_** eternal; eternity
**_Sheut_** one of the five parts of the soul: the shadow; can also mean 'statue'
**_Sistrum_** bronze noisemaker
**_Tjesu heru_** a snake with two heads – one on its tail – and dragon legs
**_Tyet_** the symbol of Isis
**_Was_** power; staff
## Egyptian Gods and Goddesses Mentioned in _The Serpent's Shadow_
**Apophis** the god of Chaos
**Anubis** the god of funerals and death
**Babi** the baboon god
**Bast** the cat goddess
**Bes** the dwarf god
**Disturber** a god of judgement who works for Osiris
**Geb** the earth god
**Gengen-Wer** the goose god
**Hapi** the god of the Nile
**Heket** the frog goddess
**Horus** the war god, son of Isis and Osiris
**Isis** the goddess of magic, wife of her brother Osiris and mother of Horus
**Khepri** the scarab god, Ra's aspect in the morning
**Khonsu** the moon god
**Mekhit** minor lion goddess, married to Onuris
**Neith** the hunting goddess
**Nekhbet** the vulture goddess
**Nut** the sky goddess
**Osiris** the god of the Underworld, husband of Isis and father of Horus
**Ra** the sun god, the god of order; also known as Amun-Ra
**Sekhmet** the lion goddess
**Serqet** the scorpion goddess
**Set** the god of evil
**Shu** the air god, great-grandfather of Anubis
**Sobek** the crocodile god
**Tawaret** the hippo goddess
**Thoth** the god of knowledge
## **Your story starts here...**
Do you **love books** and **discovering new stories?**
Then **puffin.co.uk** is the place for you. . .
• Thrilling adventures, fantastic fiction and laugh-out-loud fun
• Brilliant videos featuring your favourite authors and characters
• Exciting competitions, news, activities, the Puffin blog and SO MUCH more. . .
**puffin.co.uk**
## Do you love listening to stories?
## Want to know what happens behind the scenes in a recording studio?
Hear funny sound effects, exclusive author interviews and the best books read by famous authors and actors on the Puffin Podcast at www.puffin.co.uk
#ListenWithPuffin
## **It all started with a scarecrow...**
**Puffin is over seventy years old.** Sounds ancient, doesn't it? But Puffin has never been so lively. We're always on the lookout for the next big idea, which is how it began all those years ago.
Penguin Books was a big idea from the mind of a man called Allen Lane, who in 1935 invented the quality paperback and changed the world. **And from great Penguins, great Puffins grew, changing the face of children's books forever.**
The first four Puffin Picture Books were hatched in 1940 and the first Puffin story book featured a man with broomstick arms called Worzel Gummidge. In 1967 Kaye Webb, Puffin Editor, started the Puffin Club, promising to **'make children into readers'**. She kept that promise and over 200,000 children became devoted Puffineers through their quarterly instalments of _Puffin Post_.
Many years from now, we hope you'll look back and remember Puffin with a smile. **No matter what your age or what you're into, there's a Puffin for everyone.** The possibilities are endless, but one thing is for sure: whether it's a picture book or a paperback, a sticker book or a hardback, **if it's got that little Puffin on it – it's bound to be good.**
**www.puffin.co.uk**
##### PUFFIN BOOKS
UK | USA | Canada | Ireland | Australia
India | New Zealand | South Africa
Puffin Books is part of the Penguin Random House group of companies whose addresses can be found at global.penguinrandomhouse.com.
www.penguin.co.uk
www.puffin.co.uk
www.ladybird.co.uk
First published in the USA by Disney • Hyperion Books, an imprint of Disney Book Group, 2012
Published simultaneously in Great Britain in Puffin Books 2012
Published in this edition 2013
Text copyright © Rick Riordan, 2012
Hieroglyph art by Michelle Gengaro-Kokmen
All rights reserved
The moral right of the author and illustrator has been asserted
Cover illustration © Jacey, 2017.
ISBN: 978-0-141-96754-7
# Contents
1. Cover
2. Title Page
3. Contents
4. Dedication
5. 1. We Crash and Burn a Party
6. 2. I Have a Word with Chaos
7. 3. We Win a Box Full of Nothing
8. 4. I Consult the Pigeon of War
9. 5. A Dance with Death
10. 6. Amos Plays with Action Figures
11. 7. I Get Strangled by an Old Friend
12. 8. My Sister, the Flowerpot
13. 9. Zia Breaks Up a Lava Fight
14. 10. 'Take Your Daughter to Work Day' Goes Horribly Wrong
15. 11. Don't Worry, Be Hapi
16. 12. Bulls with Freaking Laser Beams
17. 13. A Friendly Game of Hide-and-Seek (with Bonus Points for Painful Death!)
18. 14. Fun with Split Personalities
19. 15. I Become a Purple Chimpanzee
20. 16. Sadie Rides Shotgun (Worst. Idea. Ever.)
21. 17. Brooklyn House Goes to War
22. 18. Death Boy to the Rescue
23. 19. Welcome to the Fun House of Evil
24. 20. I Take a Chair
25. 21. The Gods Are Sorted; My Feelings Are Not
26. 22. The Last Waltz (for Now)
27. Glossary
28. Egyptian Gods and Goddesses Mentioned in _The Serpent 's Shadow_
29. Puffin Web Fun
30. Puffin Audio
31. The Story of Puffin
32. Copyright Page
1. Cover
2. Contents
3. Begin Reading
| {
"redpajama_set_name": "RedPajamaBook"
} | 1,145 |
Q: ODI 12C - Business Key auto increment Using ODI 12C, I Have a dimension with a combination of 2 Primary Keys. I Just want to perform a business key with auto-increment .
NOTE : - I Don't need a SCD behavior
A: I did this :
1 - Create sequence in Database Schema ( with no cache )
2 - Call it in the the needed mapping :
<%=odiRef.getObjectName("L", "SEQ_NAME", "D")%>.nextval
3 - Set the mapping to be :
Active for inserts only, uncheck NOT NULL condition, EXCUTE ON HINT :TARGET
4 - PUT CKM TO : CKM SQL
5 - Run
A: You can create a sequence in your project or in the global objects. There are three types of it:
*
*Standard sequence, where the last value will be stored in the repository and ODI will increment it each time it retrieves a value.
*Specific sequence, same thing as above except that you choice in which table in which schema the value is stored
*Native sequence, which just use an underlying database sequence. This is the most popular choice as it will be a lot better for performance.
You can then call it using :<SEQUENCE_NAME>_NEXTVAL. For a native sequence it will get a new value for each row. For a standard or specific sequence it will give a new value for each row processed by the agent. So if you do row-by-row operation, it will give a new value for each row, but if you do batch operation, it will use the same value everywhere.
You can also call it using #<SEQUENCE_NAME>_NEXTVAL but this one will be substitued only once, before pushing the SQL on the database so all rows will have the same value.
If you use a native sequence on an Oracle database, you will have to set the Execute On Hint to Target so the sequence call is in the outermost select.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 8,426 |
BlackJack---Ruby
================
Rudimentary implementation of BlackJack in Ruby
To run the game, type the following into the terminal
'ruby blackjackplay.rb'
The other files are to be read but not run directly.
They are classes:
BlackJackGame.rb
This class basically contains all information about the game itself.
Card.rb
This class contains information about a single card.
For example, an instance of a card would have information
about the card's suit, value, and number/symbol
Deck.rb
This class was made to generate a deck for the game.
Hand.rb
This is a class that contains all information about a players hand.
DealersHand.rb
This class inherits from the Hand.rb class and contains information
regarding the dealer's hand.
Player.rb
This class contains all information about the player and what actions
and states they can be in as well as their money and their bets.
Dealer.rb
This class inherits from Player.rb and contains information about
the dealer. This is a special case because they aren't exactly affected
much in terms of money since they represent the House/Casino so they
themselves don't really gain or lost anything.
Enjoy! :)
| {
"redpajama_set_name": "RedPajamaGithub"
} | 1,318 |
@class WeiboModel;
@class WeiboView;
@interface WeiboCell : UITableViewCell {
InterfaceImageView *_userImage; //用户头像视图
UILabel *_nickLabel; //昵称
UILabel *_repostCountLabel; //转发数
UILabel *_commentLabel; //回复数
UILabel *_sourceLabel; //发布来源
UILabel *_createLabel; //发布时间
}
@property (nonatomic, retain) WeiboModel *weiboModel; //微博数据模型对象
@property (nonatomic, retain) WeiboView *weiboView; //微博视图
@end
| {
"redpajama_set_name": "RedPajamaGithub"
} | 9,152 |
Q: Not getting Google Places Photos from place id. Invalid Api key? On debugging PhotoMetaDataResult result parameter zzUX shows Places_API_KEY_INVALID. Debugging Image . Although I have "Google Places API Key for Android" in Manifests file.
public class RestaurantList extends AppCompatActivity implements GoogleApiClient.ConnectionCallbacks, GoogleApiClient.OnConnectionFailedListener{
String placeId;
GoogleApiClient mGoogleApiClient;
protected void onCreate(Bundle savedInstanceState) {
mGoogleApiClient = new GoogleApiClient.Builder(this).addApi(Places.GEO_DATA_API)
.addApi(Places.PLACE_DETECTION_API)
.addConnectionCallbacks(this)
.addOnConnectionFailedListener(this).build();
placePhotosTask();
}
abstract class PhotoTask extends AsyncTask<String, Void, PhotoTask.AttributedPhoto> {
private int mHeight;
private int mWidth;
public PhotoTask(int width, int height) {
mHeight = height;
mWidth = width;
}
/**
* Loads the first photo for a place id from the Geo Data API.
* The place id must be the first (and only) parameter.
*/
@Override
protected AttributedPhoto doInBackground(String... params) {
if (params.length != 1) {
return null;
}
final String placeId = params[0];
AttributedPhoto attributedPhoto = null;
PlacePhotoMetadataResult result = Places.GeoDataApi
.getPlacePhotos(mGoogleApiClient, placeId).await();
if (result.getStatus().isSuccess()) {
PlacePhotoMetadataBuffer photoMetadataBuffer = result.getPhotoMetadata();
if (photoMetadataBuffer.getCount() > 0 && !isCancelled()) {
// Get the first bitmap and its attributions.
PlacePhotoMetadata photo = photoMetadataBuffer.get(0);
CharSequence attribution = photo.getAttributions();
// Load a scaled bitmap for this photo.
Bitmap image = photo.getScaledPhoto(mGoogleApiClient, mWidth, mHeight).await()
.getBitmap();
attributedPhoto = new AttributedPhoto(attribution, image);
}
// Release the PlacePhotoMetadataBuffer.
photoMetadataBuffer.release();
}
return attributedPhoto;
}
/**
* Holder for an image and its attribution.
*/
class AttributedPhoto {
public final CharSequence attribution;
public final Bitmap bitmap;
public AttributedPhoto(CharSequence attribution, Bitmap bitmap) {
this.attribution = attribution;
this.bitmap = bitmap;
}
}
}
private void placePhotosTask() {
final String placeId = "ChIJrTLr-GyuEmsRBfy61i59si0"; // Australian Cruise Group
// Create a new AsyncTask that displays the bitmap and attribution once loaded.
new PhotoTask(mImageView.getWidth(), mImageView.getHeight()) {
@Override
protected void onPreExecute() {
// Display a temporary image to show while bitmap is loading.
//mImageView.setImageResource(R.drawable.empty_photo);
}
@Override
protected void onPostExecute(AttributedPhoto attributedPhoto) {
if (attributedPhoto != null) {
// Photo has been loaded, display it.
mImageView.setImageBitmap(attributedPhoto.bitmap);
// Display the attribution as HTML content if set.
if (attributedPhoto.attribution == null) {
mText.setVisibility(View.GONE);
} else {
mText.setVisibility(View.VISIBLE);
mText.setText(Html.fromHtml(attributedPhoto.attribution.toString()));
}
}
}
}.execute(placeId);
}
protected void onDestroy() {
photoMetadataBuffer.release();
mGoogleApiClient.disconnect();
super.onDestroy();
}
@Override
protected void onStart() {
super.onStart();
mGoogleApiClient.connect();
}
}
On debugging PhotoMetaDataResult result parameter zzUX shows Places_API_KEY_INVALID. Although I have "Google Places API Key for Android" in Manifests file. Thanks in advance.
Debugging image
A: You did not include your manifest, so I can only guess but from the google api docs it seems they have changed the meta data from
<meta-data
android:name="com.google.android.geo.API_KEY"
android:value="@string/google_maps_key" />
to
<meta-data
android:name="com.google.android.maps.v2.API_KEY"
android:value="@string/google_maps_key" />
try changing that value in your manifest and see if that helps!
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,686 |
Jiji people also known as Bajiji (Wajiji in Swahili) are a Bantu ethnic and linguistic group based in Kigoma Region, Tanzania.
If tribes are classified by language and not by race, Bajiji (Jiji people) are part of Baha (Ha people) since their language is the same. Traditionally they were organized into a separate kingdom, Bujiji (Swahili Ujiji, same as the Swahili town of Ujiji near Kigoma), and formed part of Buha (Uha, Ha territory) with other kingdoms: Heru, Bushingo (Ushingo), Ruguru (Luguru), Muhambwe and Buyungu, all of them in Kigoma Region, Tanzania.
In the 18th and 19th century there was a sultanate that involved the Jiji and Manyema groups under Mwene Mbonwean Sultanate of Ujiji.
References
Ethnic groups in Tanzania
Indigenous peoples of East Africa | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 1,402 |
The clear blue skies and prairie fields of southern Manitoba are the backdrop for the natural attraction of the R.M. of Pipestone.
Situated in Manitoba's southwest corner, the R.M .of Pipestone is a great place to visit and an even better place to make your home. Enjoy all that each community has to offer. Play a round of golf, take in a movie, relax at your campsite, or walk down the self-guided nature trail.
Don't miss out on the slo-pitch tournament at Pipestone Fun Days, the mud bog competition at Sinclair Redneck Days or the tarp auction at Cromer's Stick Horse Rodeo. Summer, winter, spring or fall, the residents of the R.M. of Pipestone welcome you and invite you back for another visit. | {
"redpajama_set_name": "RedPajamaC4"
} | 2,335 |
KiraSweeet . TheNinjaGirl. TheNinjaGirl. AngelicaaRose.
AKANEXX0cplhotdesire90VINCEHALlSummerWells .GiovanniUnoRickyRickycplhotdesire90ZarinaMa .EllieSweettNewAmoreMikeVicentMayssaRosse .GiovanniUnoAnetteTurnerXAlishaHallSexB0y4WildDream .Knopo4kkaLovelyGirlMbickmusclesbig1sophie4love .SharonredhairCamTestJessHugeHolesSprayZarinaMa .DarkCountessesyourLOVEspell69domwolfzMiaXLust .jolievaleryamiewoodsAshlynnGrayscerazad .brookekatyAlishaHallSofiiSexyTatoobulgaroboy .Mariano4kaBlack11VSwhite11bulgaroboytheLegendaryCOCK .latinosexardientOliviaFrostAngelicaaRoseSamantaWurtzbach .
CandyCoatsMirandaNelWendyMilnDarkCountesses .ShyLiliXCockreleifcplhotdesire90RedHeadMATUREX .ChaBlissAnetteTurnerXCamTestJessAntonellaLynx .AKANEXX0AppleBelleSwitchMoniqFrenchboymat .brookekatyAKANEXX0divakitty2TerryLady .SexB0y4WildDreamAlishaHallAnabellaSecretAnitchkaMaya .1sophie4loveBlondNextDoorXCantiBlackKarinaKross .MsKarrmadivakitty2SallyMatherskinkysonya .AntonellaLynxOliviaFrostNaughtySlimBruceANDean .RedHeadMATUREXSamantaWurtzbachBlondNextDoorXJackMarcos .TheNinjaGirlCamTestJessAKANEXX0crazyhotbody38 . | {
"redpajama_set_name": "RedPajamaC4"
} | 3,399 |
\section{Introduction} \label{sec:intro}
Long gamma-ray bursts (GRBs) are bright bursts of gamma-rays followed by extremely luminous multi-wavelength afterglow, from the X-rays to the radio wavelengths. They have been shown to be associated with the collapse of massive stars \citep{hjorth2003very, stanek2003spectroscopic}. GRBs have been observed across the cosmic history, from $z \sim 0.01$ to $z \sim 8.2$ \citep{tanvir2009gamma, salvaterra2009grb, fynbo2000hubble}.
These attributes make them a viable probe for tracing the star-formation history of the universe, especially at $z > 2$ where other probes are scarce.
However, the exact relation between GRB rates and cosmic star formation rate (SFR) is still an unsolved problem \citep{greiner2015gamma, schulze2015optically, perley2016swift_1, perley2016swift_2}. Various observations of $z < 1.5$ GRB hosts have raised questions on whether GRBs can be used as unbiased tracers of star formation \citep{boissier2013method, perley2013population, vergani2015long, schulze2015optically, perley2016swift_2}. Particularly, GRB hosts at $z < 1$ show a strong bias towards faint, low-mass ($\mathrm{M_{*} < 10^{10}~M_{\odot}}$), star-forming galaxies and lower metallicities (below solar metallicity) compared to other
star-formation tracers, even after taking into account GRBs with high line-of-sight dust obscuration \citep{graham2013metal, perley2013population, kelly2014host, vergani2015long, japelj2016long, perley2016swift_2}. However, this bias appears to subside at $z > 2$ \citep{greiner2015gamma} since the mean metallicity of typical star-forming galaxies is below solar. A significant amount of star formation at these redshifts happens in dusty massive starbursts (submm-bright; see \cite{casey2014dusty} for a review). Thus, high-mass, (relatively) metal-rich, dusty galaxies with high star formation rates may form a significant fraction of the GRB host population at $z > 2$ \citep{perley2013population, greiner2016probing, perley2016swift_2}. To understand whether GRBs truly trace star formation at $z > 2$, it is important to measure the total SFR (i.e. dust-obscured + dust-unobscured).
Radio observations provide a probe of recent total star formation rate. In star-forming galaxies, the radio luminosity at frequencies below a few $\times$ 10 GHz is dominated by the synchrotron emission from relativistic electrons, previously accelerated by supernova remnants, propagating in the interstellar magnetic field \citep{condon1992radio}. The relativistic electrons probably have lifetimes $\leq$ 100 Myr, thus this component traces recent ($ <$ 100 Myr) star formation.
There are about 100 GRB host observations at radio frequencies down to limits between $3 - 500$ $\mu$Jy (see \cite{greiner2016probing} for details). So far, there have been 19 cm-wave observations of GRB hosts at $z > 2$, out of which two were detections: GRB 080207A and GRB 090404 \citep{greiner2016probing, perley2013population, perley2015connecting, perley2016late}. However, none of these high-z GRBs have high-resolution, high-SNR afterglow spectra.
GRBs with high-resolution afterglow spectra can be excellent test cases for examining the biases in GRB host population at high-z since a measure of the host metallicity may be derived from these spectra to help characterize the galaxy population traced by GRBs at $z > 2$. The availability of a high-resolution rest-frame UV spectrum of the GRB afterglow implies that the rest-frame UV is largely unobscured (${A_{UV} \la}$ 2-3 mag). The radio observations of these GRB hosts may be used to find out whether this lack of obscuration is simply due to a clear line-of-sight or due to an overall lack of dust obscuration in the host galaxy. Dusty sightlines do not necessarily imply dusty host galaxies. This needs to be tested, especially in light of past cm-wave observations of \cite{hatsukade2012constraints} and \cite{perley2013population}, where the deep upper limits on the radio flux from the galaxy hosts of so-called `dark GRBs' (i.e. UV-dark afterglow due to high line-of-sight extinction) imply that the dark GRBs do not always occur in galaxies enshrouded by dust or in galaxies exhibiting extreme star formation rates (few $\times$ 100 $-$ 1000 $\mathrm{M_{\odot} yr^{-1}}$).
New radio-based SFR constraints are particularly needed for massive ($\mathrm{M_{*} \gtrsim 10^{10} M_{\odot}}$) GRB hosts at $z > 2$ since the massive star-forming galaxies at high-z are likely to be significantly dusty \citep{casey2014dusty, shapley2011physical}. One of our objectives is therefore to understand whether massive GRB hosts at $z > 2$ share this characteristic of typical massive star-forming galaxies at $z > 2$.
This pre-selection of $z > 2$ GRB hosts based on high-resolution afterglow spectra is also useful to inform the total SFR of the GRB hosts in the CGM-GRB sample \citep{gatkine2019cgm}, particularly for the massive GRB hosts which are likely to have a substantial dust-obscured star formation component. The high-resolution spectra quantitatively trace the kinematics of the circumgalactic and interstellar media of the host. The total star formation (obscured $+$ unobscured) is a major driver of galactic outflows that feed the circumgalactic medium (CGM). Therefore, constraining the total SFR is necessary for studying the CGM-galaxy connection.
In this paper, we report deep, late-time radio observations of four $z > 2$ GRB hosts with existing high-resolution afterglow spectra. The sample includes GRB 080810 which is the highest-redshift GRB host yet ($z = 3.35$) with deep radio observations. These results were obtained using Karl Jansky Very Large Array (VLA) in C-band ($4-8$ GHz). Section \ref{sec:data} describes the target selection, VLA observations, and analysis. In section \ref{sec:Total_SFR_estimate}, we derive the constraints on the radio-based SFRs and discuss the obscured fraction of the SFR in each GRB host individually. The implications of these results for dust obscuration in GRB hosts are discussed in Section \ref{sec: discussion} and the key conclusions are summarized in Section \ref{sec:summary}.
\section{Sample and observations} \label{sec:data}
\subsection{Sample Selection}
\label{subsec:sample}
The CGM-GRB sample is a sample of 27 $z > 2$ GRBs with high-resolution (resolving power $R$ $>$ 6000) and high signal-to-noise ratio (median SNR $\sim$ 10) afterglow spectra \citep{gatkine2019cgm}. None of these GRBs have previously reported late-time radio observations. A subset of these objects is selected by imposing various criteria. Only GRBs that occurred at least six years ago are considered to ensure that the radio flux contribution from the afterglow is minimal \citep{perley2015connecting}. From the remaining 17, only GRB hosts with existing $\mathrm{M_{\star}}$ measurements and $\mathrm{M_{\star} > 10^{9.5}~M_{\odot}}$ are selected since their UV-based SFR is expected to be most affected by dust obscuration. This resulted in a set of four GRB hosts: GRB 021004, GRB 080310, GRB 080810, and GRB 121024A. Further, the VLA observations of GRB 080810 reported here (at $z = 3.35$) make it the the highest-redshift GRB with a late-time radio observation of the host. Table \ref{tab:GRB_list} summarizes the sample and its key properties.
\begin{deluxetable*}{ccccccccccc}[t]
\tablecaption{Summary of the VLA observations \label{tab:GRB_list}}
\tablehead{
\colhead{GRB\tablenotemark{a}} &
\colhead{$z$} &
\colhead{R.A.} &
\colhead{Dec.} &
\colhead{Date} &
\colhead{\makecell{$t_{int}$\\ (min)}} & \colhead{\makecell{Total $t_{int}$\\ (min)}} &
\colhead{\makecell{3$\sigma$ Limit\\ ($\mu$Jy)}} & \colhead{\makecell{Beam size\\ ($''$)}} &
\colhead{\makecell{Flux/\\ bandpass}} &
\colhead{\makecell{Complex\\ gain}}}
\startdata
021004 & 2.323 & 00:26:54.68 & +18:55:41.6 & \makecell{2018 Dec 16\\ 2018 Dec 18} & \makecell{120.5\\ 150} & 270.5 & 3.0 & 3.7 $\times$ 4.5 & 3C48 & \makecell{J0010+1724}\\
080310 & 2.427 & 14:40:13.80 & $-$00:10:30.7 & \makecell{2018 Dec 04\\ 2018 Dec 11\\ 2018 Dec 15\\ 2018 Dec 18\\ 2018 Dec 24 } & \makecell{90\\ 66\\ 90\\ 90\\ 66} & 402 & 6 & 3.2 $\times$ 4.0 & 3C286 & J1445+0958\\
080810 & 3.35 & 23:47:10.49 & $+$00:19:11.5 & \makecell{2018 Dec 09\\ 2018 Dec 22\\ 2019 Jan 05\\ 2019 Jan 10} & \makecell{71\\ 135\\ 71\\ 66} & 343 & 3.8 & 3.7 $\times$ 4.9 & 3C48 & J2323-0317\\
121024A & 2.298 & 04:41:53.30 & $-$12:17:26.6 & \makecell{2018 Dec 17} & \makecell{123} & 123 & 18 & 3.9 $\times$ 5.6 & 3C138 & J0437-1844\\
\enddata
\tablenotetext{a}{All the observations were performed in the C-band ($4-8$ GHz) in C array configuration of the VLA}
\end{deluxetable*}
\subsection{VLA Observations} \label{sec:methods}
We performed the radio observations using the fully upgraded Karl G. Jansky Very Large array (VLA) using C-band receivers spanning 4 $-$ 8 GHz and with a central frequency of 6 GHz. We used 3-bit samplers to utilize the entire 4096 MHz bandwidth of the C band to maximize the continuum sensitivity. The dual polarization setup was used. The observations were conducted in the C array configuration during the months of December 2018 to January 2019 (program VLA 18B-312, PI: Gatkine). The integration time for each GRB host is listed in Table \ref{tab:GRB_list} (typical $\sim$ 4.5 hours). A nearby complex gain (amplitude and phase) calibrator was observed every 30 $-$ 40 minutes during any scheduling block and a standard flux calibrator was observed every hour. The 3-$\sigma$ rms and the synthesized beam size for each source are listed in Table \ref{tab:GRB_list}.
The data reduction was carried out using the Common Astronomy Software Applications package (CASA) version 5.5.0. The standard {\fontfamily{qcr}\selectfont CASA} pipeline was used to flag and calibrate the observations. Imaging and deconvolution was performed using the {\fontfamily{qcr}\selectfont tclean} function in {\fontfamily{qcr}\selectfont CASA}. Natural weighting was employed while cleaning the measurement sets to maximize the continuum sensitivity. In the case of GRB 121024A, additional flagging was performed to clip the outlier visibilities and channels heavily affected with radio frequency interference. Further, self-calibration was performed to clean the image around a bright source at a separation of 6$^\prime$, a robust weighting was employed, and a multi-term multi-frequency synthesis (mtmfs, with 2 terms) deconvolver was used to account for spectral index gradient in the much brighter contaminating source.
The synthesized beam size for C-configuration observations is significantly coarser (beam size $\sim$ 4\arcsec) than the angular extent of the galaxy (1 kpc translates to $\sim$ 0.1\arcsec\ at $z \sim 2.5$). Therefore, the host galaxies are unresolved and can be treated as point sources here. The 1$\sigma$ flux-density level was derived by sampling a blank region spanning $\sim$100 $\times$ synthesized beam area around the target.
The maps for GRB 021004 and GRB 080810 have rms sensitivities close to that predicted by the VLA noise calculator. However, GRB 121024A and GRB 080310 had particularly bright sources near the half-power response of the primary beam. At this location in the primary beam, the amplitude response is variable owing to antenna pointing errors, which result in amplitude gain errors in the visibilities that are a function of field position in addition to antenna, frequency, and time. Standard self-calibration does not work well if there are position-dependent errors; antenna pointing errors limited the dynamic range of the maps for GRB 080310 and especially GRB 121024A, and consequently our sensitivity for these objects.
\begin{figure*}
\centering
\includegraphics[width=0.9\textwidth]{Figure1.pdf}
\figcaption{\label{fig:VLA_postage}
Contour maps of the radio flux density in $30\arcsec\ \times 30\arcsec$ fields centered on the four GRBs of our sample. The location of the GRB and 2\arcsec\ error circle are marked as red crosses and black circles, respectively. The synthesized beam is shown in the bottom left corner. The contours are marked as -12, -6, -3, -1.5, 1.5, 3, 6, 12 $\times \sigma$ with negative values marked as dotted contours. None of the GRB hosts are detected at the 3$\sigma$ level.}
\end{figure*}
\section{Radio- and UV-based SFR}\label{sec:Total_SFR_estimate}
\subsection{Radio-based SFR}\label{sec:radio_SFR}
\begin{figure}
\centering
\includegraphics[width=0.45\textwidth]{Figure2.pdf}
\figcaption{\label{fig:Flux_vs_z}
Curves showing the radio flux density averaged over $4-8$ GHz for various star formation rates ($\mathrm{M_{\odot}/yr}$) over a redshift range $z \sim 0 - 4$ using a spectral index of $\alpha = 0.7$. The 3$\sigma$ upper limits of various GRBs are shown with downward triangles. The horizontal dotted lines are drawn to guide the eye.}
\end{figure}
As described in Section \ref{sec:intro}, the radio continuum at frequencies below a few $\times$ 10 GHz traces the total (i.e. dust-obscured $+$ dust-unobscured) star formation activity in the last 100 Myr \citep{condon1992radio}. The radio-far-IR relation for star-forming galaxies which quantifies the radio-SFR relation is shown to hold true at intermediate and high redshifts \citep{sargent2010vla}. On the other hand, the UV/optical light (including the emission lines) primarily probes the portion of the SFR that is not significantly obscured by dust (i.e. dust-unobscured SFR) even with dust attenuation included in the modeling (see the example of GRB 100621A in \citealt{stanway2014radio}). Thus, a significant discrepancy between the UV-based and radio-based SFR measures would imply the presence of substantial dust obscuration within the galaxy.
\noindent In the discussion below, we use the following naming system:\\
\newline $\mathrm{SFR_{total}}$: Radio-based total SFR
\newline $\mathrm{SFR_{unobscured}}$: UV-based unobscured SFR,
\newline $\mathrm{SFR_{obscured}}$: the portion of SFR that is obscured due to the dust (= $\mathrm{SFR_{total} - SFR_{unobscured}}$).\\
Here, we observe the GRB hosts in C-band ($4-8$ GHz) at $z \sim 2-3.5$, thus we are sensitive to $\nu_{rest} = 25 \pm 10$ GHz. The rest-frame radio luminosity is produced by three mechanisms: non-thermal synchrotron emission ($\epsilon_{1}$), free-free emission ($\epsilon_{2}$), and thermal emission from dust ($\epsilon_{3}$), as shown in \cite{yun2002radio}. The thermal dust component is insignificant ($<$ 1\%) at the frequencies of interest. The radio-SFR relation for star-forming galaxies \citep{yun2002radio} is thus given by:
\begin{mathletters}
\begin{eqnarray} \label{eqn:M_CGM}
S(\nu_{obs}) & = & \Big( \epsilon_{1} + \epsilon_{2} + \epsilon_{3} \Big) \times \frac{(1+z)SFR}{D_{L}^{2}}
\end{eqnarray}
\end{mathletters}
\noindent where,
\noindent
$\epsilon_{1} = 25f_{nth}\nu_{0}^{-\alpha}$\\
$\epsilon_{2} = 0.71\nu_{0}^{-0.1}$ \\
$\epsilon_{3} = 1.3 \times 10^{-6} \frac{\nu_{0}^{3}[1 - e^{-(\nu_{0}/2000)^{\beta}}]}{e^{0.048\nu_{0}/T_{d} -1}}$. \\
Here, the symbols $\epsilon_{1}$, $\epsilon_{2}$, and $\epsilon_{3}$ represent
the contributions from non-thermal synchrotron, free-free, and dust thermal emission respectively. $D_{L}$ is luminosity distance in Mpc, SFR is star formation rate in $\mathrm{M_{\odot} yr^{-1}}$, $\nu_{0}$ is rest-frame frequency in GHz, $f_{nth}$ is the scaling factor, $\alpha$ is the synchrotron spectral index, $T_{d}$ is the dust temperature in K, and $\beta$ is the dust emissivity. For the typical values of $T_{d}$ ($\sim$ 60 K) and $\beta$ (1.35), the dust emission is insignificant for $\nu_{\rm rest} \sim 25$ GHz. hence, we neglect this term. The non-thermal synchrotron emission is the most dominant contributor in the given frequency range. Since we do not have a robust measurement of the actual spectral index for any of our objects, we assume a canonical average value of $\alpha = -0.7$. Past literature has used values ranging from $-0.6$ to $-0.75$ \citep{hatsukade2012constraints, perley2013population, perley2015connecting, stanway2014radio, greiner2016probing}. This range of $\alpha$ affects the radio luminosity by 25\%.
This equation assumes a Salpeter initial mass function (IMF). Due to various assumptions in the calibration of radio-based SFRs, it is subject to a systematic uncertainty of about a factor of $\sim$2 \citep{yun2002radio, bell2003estimating, murphy2011calibrating}.
Figure \ref{fig:Flux_vs_z} shows the observed flux densities averaged over $4-8$ GHz for various star formation rates as a function of redshift and the respective $3\sigma$ upper limits of our targets. The UV- and radio-derived SFRs for our four targets are summarized in Table \ref{tab:GRB_SFR} along with the stellar masses and ratios of radio-based (total) and UV-based (dust-unobscured) SFRs.
\subsection{Late-time afterglow emission}
The GRBs have long-lived radio afterglows. Therefore, any estimates of SFR using the radio emission can only be made after the afterglow has faded considerably to ensure minimal/no contamination due to the afterglow. We compiled the past early-time radio observations of the afterglows of our target GRBs available in the literature and extrapolated the afterglow decay using a canonical long GRB radio light curve model (forward shock model) with a $t^{-1}$ decay \citep{chandra2012radio} as follows:
\begin{equation}
f(t)=\begin{cases}
F_{m}t_{m}^{-1/2}t^{1/2}, & \text{if $t<t_{m}$}.\\
F_{m}t^{-1}, & \text{if $t > t_{m}$ }.
\end{cases}
\end{equation}
Here, $F_{m}$ is the peak flux density at a given frequency and $t_{m}$ is the time of the peak in that frequency. For this extrapolation, we used a conservative approach. We use the latest flux density measurement in C-band (if available) as the peak flux density. If it is not available (eg: GRB 121024A), we extrapolate the flux density using the standard GRB radio afterglow model described in \cite{chandra2012radio}. The typical values of $t_{m}$ range between rest-frame 3 and 6 days at a rest-frame frequency of $\sim$ 25 GHz (which we probe since our targets are at $z \sim 2-3.5$). We translate this $t_{m}$ to the observer frame for each GRB and plot the radio afterglow evolution in Figure \ref{fig:GRB_Radio_evolution}. The three lines show the decay with $t_{m}$ = 3, 4.5, and 6 days (in the rest frame). No early-time radio observations are available for GRB 080310. The conservative approach used here gives the upper limit of radio flux density due to the afterglow and further shows that the late-time radio fluxes for our observations are dominated by the host galaxy and are not likely to be contaminated by the afterglow.
\begin{figure*}
\centering
\includegraphics[width=1\textwidth]{Figure3.pdf}
\figcaption{\label{fig:GRB_Radio_evolution}
Radio evolution of the afterglows of GRB 021004, GRB 080810, and GRB 121024A, extrapolated using the canonical afterglow evolution model described in Section \ref{sec:radio_SFR}. }
\end{figure*}
\subsection{SFR in each GRB host}
We summarize the UV-derived and radio-derived SFRs for the four GRBs in the following subsections.
Using the VLA observations, we obtain an estimate of the total SFR ($\mathrm{SFR_{total}}$), independent of assumptions on the dust extinction (in the line of sight or otherwise).
\newline We also compare the observed ratio $\mathrm{SFR_{total}/SFR_{unobscured}}$ for our GRB hosts with the same ratio for star-forming galaxies with a similar stellar mass at a redshift range $z \sim 2 - 2.5$, as derived from the CANDELS survey \citep{whitaker2017constant} and summarize this in Figure \ref{fig:SFR_vs_Mstar}.
\subsubsection{GRB 021004}
GRB 021004 is one of the best studied GRBs from the gamma-rays to radio wavelengths. The optical afterglow was detected 3.2 minutes after the prompt high-energy emission and was followed up extensively \citep{fynbo2005afterglow}. The extremely blue host galaxy of GRB 021004 was identified and studied through late-time imaging in the rest-frame UV and optical bands. HST ACS imaging in the F606W band revealed that the host galaxy has a very compact core with a half-light radius of only 0.4 kpc (at $z$ = 2.323). The impact parameter of the afterglow position is only 0.015\arcsec, corresponding to a distance of 119 pc, which is one of the smallest for long GRBs \citep{fynbo2005afterglow, blanchard2016offset}. The line-of-sight extinction $A_{V}$ is 0.20 $\pm$ 0.02 mag.\ (using the SMC extinction law) as derived after 1 week of afterglow decay \citep{fynbo2005afterglow}. The Ly$\alpha$-derived neutral hydrogen column density ($\mathrm{N_{HI}}$) along the line of sight is modest ($\sim$ 10$^{19}$ cm$^{-2}$; \citealt{prochaska2008survey}).
\cite{castro2010grb} derived the host SFR of 40 $\mathrm{M_{\odot} yr^{-1}}$ (without any dust correction) by attributing all of the H$\alpha$ emission to star formation. Given the small $A_{V}$, the dust correction was assumed to be minimal from the afterglow SED. On the other hand, \cite{jakobsson2005ly+} have estimated a lower limit of SFR as 10.6 $\mathrm{M_{\odot} yr^{-1}}$ by converting the Ly$\alpha$ flux to SFR \citep{kennicutt1998star} and assuming a 100\% Ly$\alpha$ escape fraction.
We derive a 3$\sigma$ upper limit on the C-band flux density of 3.0 $\mu$Jy, corresponding to a radio SFR limit of 85 $\mathrm{M_{\odot} yr^{-1}}$ at $z \sim 2.323$. This result is consistent with the low $A_{V}$ derived from the optical-NIR SED and therefore suggests that the host galaxy as a whole is not significantly affected by dust. This observation identifies a galaxy that is able to sustain a SFR of $\sim$ 40 $\mathrm{M_{\odot} yr^{-1}}$ at $z \sim 2.3$ without significant dust obscuration. Using the non-extinction-corrected H$\alpha$ emission, we get $\mathrm{SFR_{unobscured}}$ = 40 $\mathrm{M_{\odot} yr^{-1}}$, so the ratio $\mathrm{SFR_{total}/SFR_{unobscured}}$ is $< 2.1$ for this $\mathrm{M_{*} > 10^{10} M_{\odot}}$ galaxy. In contrast, the corresponding ratio derived for the main sequence of star-forming galaxies at $z \sim 2.5$ from \cite{whitaker2017constant} is $\sim 6$.
Given the small impact parameter of the afterglow ($119$ pc) from the centroid of the galaxy, the apparent lack of significant dust extinction along the line of sight to the GRB, and in the host galaxy as a whole from the radio observations, is puzzling.
\begin{figure}
\centering
\includegraphics[width=0.45\textwidth]{Figure4.pdf}
\figcaption{\label{fig:SFR_vs_Mstar}
The SFR $-$ $M_{\star}$ relation decomposed into total (black star), obscured (red circle), and unobscured components (blue triangle) of the star formation rate for the galaxies in the CANDELS survey at $z \sim 2-2.5$ \citep{whitaker2017constant}. The gray
band corresponds to the typical 0.3 dex width of the observed relation. Individual GRBs in our sample are shown in various colors with their UV-derived SFR (tracing the dust-unobscured SFR) and the radio-derived SFR (tracing the total SFR).}
\end{figure}
\subsubsection{GRB 080310}
The afterglow of GRB 080310 was detected 1.5 minutes after the prompt high-energy emission and was followed up extensively \citep[see][for a full discussion]{littlejohns2012origin}. The redshift of this GRB is 2.427 \citep{prochaska2008grb, vreeswijk2008grb}. \cite{perley2008grb} estimated a low line-of-sight extinction $A_{V}$ of 0.10 $\pm$ 0.05 mag.\ using an SMC-like extinction law (at an average time of $t_0$ $+$ 1750 s). The line-of-sight $\mathrm{N_{HI}}$ is modest ($\sim$ 10$^{18.8}$ cm$^{-2}$).
The late-time host galaxy imaging using the Low Resolution Imaging Spectrometer (LRIS) on the Keck-I telescope yielded a non-detection with a g-band limiting magnitude of 27.0 \citep{perley2009host}. We estimate a SFR upper limit of 4.5 $\mathrm{M_{\odot} yr^{-1}}$ using the UV luminosity-SFR relation for GRB host galaxies described in \cite{savaglio2009galaxy}. \cite{perley2016swift_2} estimated log($\mathrm{M_{*}/M_{\odot}}$) = 9.8 $\pm$ 0.1 using {\em Spitzer} 3.6 $\mu m$ imaging. However, we caution the reader of the possibility that the {\em Spitzer} 3.6 $\mu m$ flux is contaminated by the diffraction spike from a nearby star despite careful modeling and subtraction of the spike \citep{perley2016swift_2}.
The VLA observations constrain the SFR to less than 180 $\mathrm{M_{\odot} yr^{-1}}$ (3-$\sigma$ upper limit). However, this limit is not sufficiently deep to constrain the dust obscuration in the host galaxy of GRB 080310.
\subsubsection{GRB 080810}
This is the highest-redshift GRB in our sample at $z = 3.35$. The afterglow of GRB 080810 was detected 80 seconds after the prompt emission by the X-ray telescope (XRT; \cite{burrows2005swift}) and UV-optical telescope (UVOT; \cite{roming2005swift}) on board the Neil Gehrels Swift Observatory \citep{gehrels2004swift}. \cite{prochaska2008grb} obtained the optical spectra of the afterglow using the Keck HIRES spectrograph starting 37 minutes after the trigger and derived a redshift of 3.35. The Ly$\alpha$-derived line-of-sight $\mathrm{N_{HI}}$ is small ($\sim$ 10$^{17.5}$ cm$^{-2}$). We refer the readers to \cite{page2009multiwavelength} for a discussion of the extensive multi-wavelength follow-up of this GRB.
Extensive late-time ground-based photometry and spectroscopy of the host galaxy of GRB 080810 revealed an extended structure with a bright compact region \citep[see][for more details]{wiseman2017gas}. Further, a strong detection of redshifted Ly$\alpha$ emission at a redshift of 3.36 confirmed the association of the GRB and the detected host galaxy \citep{wiseman2017gas}. They estimate a modest host extinction of $A_{V} \sim$ 0.4 mag.\ from SED fitting. \cite{greiner2015gamma} convert the extinction-corrected UV luminosity to SFR (using the $\mathrm{L_{UV}-SFR}$ relation in \cite{duncan2014mass} and $A_{1600} \sim$ 1.3 mag.) to obtain SFR $\sim$ 100 $\mathrm{M_{\odot} yr^{-1}}$, which is further corroborated by SED fitting \citep{wiseman2017gas}. The uncorrected SFR is $\sim$ 30 $\mathrm{M_{\odot} yr^{-1}}$. The stellar mass, derived from the {\em Spitzer} 3.6 $\mu m$ photometry, is log($\mathrm{M_{*}/M_{\odot}}$) = 10.2 $\pm$ 0.1 \citep{perley2016swift}.
Here we report the first ever deep late-time radio observation of a GRB with a spectroscopic redshift $z > 3.1$. We derive a 3-$\sigma$ upper limit on the C-band flux density of 3.8 $\mu$Jy, corresponding to a radio-based SFR upper limit of 235 $\mathrm{M_{\odot}~yr^{-1}}$ at $z \sim 3.35$. The dust-corrected SFR from the UV SED is therefore consistent with the total SFR limit derived from the radio observations. This further implies that the modest $A_{V}$ estimated from the UV SED fitting reasonably takes into account the dust correction.
Using the uncorrected UV SFR, we derive a ratio $\mathrm{SFR_{total}/SFR_{unobscured}}$ $< 7$ for this $\mathrm{M_{*} > 10^{10}~M_{\odot}}$ galaxy. This is consistent with the corresponding ratio derived for the main sequence of star-forming galaxies at $z \sim 2-2.5$ from \cite{whitaker2017constant}, which again gives $\mathrm{SFR_{total}/SFR_{unobscured}} \sim 6$. However, if the dust-corrected UV SFR ($\sim 102$ $\mathrm{M_{\odot} yr^{-1}}$) is used instead, we get $\mathrm{SFR_{radio}/SFR_{UV, corr}} < 2.3$. Here, we extrapolate the non-evolution of this ratio from $z \sim 2.5$ to 3.3 for the star-forming galaxies on the main sequence at a given $\mathrm{M_{*}}$, as presented in \cite{whitaker2017constant}.
\subsubsection{GRB 121024A}
The afterglow of GRB 121024A was followed up 93 seconds after the prompt emission by the X-ray telescope (XRT; \cite{burrows2005swift}) on board the Neil Gehrels Swift Observatory \citep{gehrels2004swift}.
\cite{tanvir2012grb} obtained the optical/NIR spectra of the afterglow using the X-shooter spectrograph on the Very Large Telescope (VLT) and determined a redshift of 2.298. The line-of-sight $\mathrm{N_{HI}}$ of 10$^{21.5}$ cm$^{-2}$ indicates that this is a damped Ly$\alpha$ system. We refer the readers to \cite{friis2015warm} for a detailed summary of the extensive multi-wavelength follow-up of this GRB.
Various emission lines including H$\alpha$, H$\beta$, [O II] $\lambda\lambda$3727, 3729 doublet, [N II] $\lambda$6583, and [O III] $\lambda\lambda$4959, 5007 were detected in the X-shooter NIR spectrum of the afterglow. Extensive optical and NIR photometry of the host galaxy was obtained using VLT/HAWK-I, NOT, and GTC \citep[see][for details]{friis2015warm}. The stellar population synthesis modelling of the host yielded a modest extinction $A_{V}$ of 0.15 $\pm$ 0.15 mag.\ and $\mathrm{log(M_{*}/M_{\odot})}$ = 9.9$^{+0.2}_{-0.3}$.
\cite{friis2015warm} estimate the SFR from the extinction-corrected H$\alpha$ and [O II] fluxes as 42 $\pm$ 11 and 53 $\pm$ 15 $\mathrm{M_{\odot}~yr^{-1}}$ using conversion factors from \cite{kennicutt1998star}. However, note that the extinction correction to the SFR is small ($\sim 15\%$). They further corroborate this SFR by stellar population synthesis modelling.
The 3-$\sigma$ upper limit on the C-band flux density of GRB 121024A is 18 $\mu$Jy. The relatively higher background is due to a bright source at 6$^\prime$ angular separation. Using the VLA observations, we obtain a 3-$\sigma$ upper limit of the total SFR as 500 $\mathrm{M_{\odot} yr^{-1}}$. However, this limit is not sufficiently deep to constrain the dust obscuration in the host galaxy of GRB 121024A. The limiting $\mathrm{SFR_{total}/SFR_{unobscured}} < 12.5$ is consistent with the corresponding expected ratio ($\sim 5$) from \cite{whitaker2017constant} for a star-forming galaxy of this stellar mass on the main sequence at $z \sim 2-2.5$.
\begin{deluxetable*}{ccccccc}[t]
\tablecaption{Summary of GRB host properties \label{tab:GRB_SFR}}
\tablehead{
\colhead{GRB\tablenotemark{a}} &
\colhead{$z$} &
\colhead{\makecell{$\mathrm{log(N_{HI})}$\tablenotemark{a} \\ (cm$^2$) }} &
\colhead{\makecell{M$_*$\\ ($\mathrm{M_{\odot}}$)}} &
\colhead{\makecell{SFR$\mathrm{_{UV}}$\\ ($\mathrm{M_{\odot}~yr^{-1}}$)}} &
\colhead{\makecell{SFR$\mathrm{_{Radio}}$\tablenotemark{b}\\ ($\mathrm{M_{\odot}~yr^{-1}}$)}} &
\colhead{$\mathrm{\frac{SFR_{total}}{SFR_{UV}}}$} }
\startdata
021004 & 2.323 & 19.00 $\pm$ 0.2\tablenotemark{c} & 10.2 $\pm$ 0.18\tablenotemark{g} & 40 $\pm$ 10 & $<$ 85 & $<$ 2.1 \\
080310 & 2.427 & 18.80 $\pm$ 0.1\tablenotemark{d} & 9.78 $\pm$ 0.2\tablenotemark{h} & $<$ 5 & $<$ 180 & $-$ \\
080810 & 3.35 & 17.5 $\pm$ 0.15\tablenotemark{e} & 10.24$\pm$ 0.1\tablenotemark{h}& 102 $\pm$ 48 & $<$ 235 & $<$ 2.3 \\
121024A & 2.298 & 21.5 $\pm$ 0.1\tablenotemark{f} & 9.9$^{+0.2}_{-0.3}$\tablenotemark{f} & 40 $\pm$ 4 & $<$ 500 & $<$ 12.5\\
\enddata
\vspace{3ex}
$^{a}${Ly$\alpha$-derived $\mathrm{N_{HI}}$}
$^{b}${3$\sigma$ upper limit}, $^{c}${\cite{prochaska2008survey}}, $^{d}${\cite{fox2008high}}, $^{e}${\cite{page2009multiwavelength}}, $^{f}${\cite{friis2015warm}}, $^{g}${\cite{savaglio2009galaxy}}, $^{h}${\cite{perley2016swift_2}}
\end{deluxetable*}
\section{Discussion} \label{sec: discussion}
\noindent
Our observations have targeted massive ($\mathrm{M_{*} > 10^{9.5}~M_{\odot}}$) high-z GRBs ($z \sim 2 - 3.5$) with high-resolution and high SNR rest-frame UV afterglow spectra (i.e.\ a rest-frame UV-bright afterglow). The results presented here, in concert with previous observations for the so-called `dark' GRBs \citep[rest-frame UV/optically dark afterglows;][]{perley2013population, perley2015connecting, greiner2016probing}, show that the fraction of dust-obscured star formation ($\mathrm{SFR_{obscured}/SFR_{total}}$) in most of the GRB hosts (even at $z > 2$) is less than 90\% \citep[with a few exceptions such as GRB 090404;][]{perley2013population}. There is no evidence for extreme dust obscuration in the star-forming regions of GRB host galaxies, within the uncertainties of the radio flux$-$SFR relation. This result is summarized in Figures \ref{fig:SFR_vs_Mstar} and \ref{fig:SFR_vs_z_lit}.
The two best-constrained
cases of GRB 021004 ($z = 2.323$) and GRB 080810 ($z = 3.35$) show that the radio-based total SFR $\lesssim$ $2$ $\times$ UV-based unobscured SFR. Further, the ratio of total-to-unobscured SFRs for these GRB hosts is significantly smaller than the corresponding ratio for the star-forming galaxies on the main sequence \citep{whitaker2017constant}.
Particularly, GRB 021004 provides a striking example of lack of significant dust obscuration in the central region of a star-forming galaxy at $z > 2$, given that the separation of the GRB from the galaxy centroid is only 119 pc \citep{chen2012near}. The sightline extinction, derived from the afterglow is also small ($\mathrm{A_{V}}$ = $0.2 \pm 0.02$ mag.). Two possible scenarios can explain these results: a)
the GRB occurred in a locally dusty cloud but globally, the host galaxy lacks significant amount of dust. The low sightline extinction would then imply that the burst
occurred along a clear sightline within its star-forming cloud. b) the GRB occurred in a star-forming region which has cleared the dust from past star formation and the overall galaxy also lacks significant amount of dust. The GRB sightline would then be a representative sightline. Similarly, for GRB 080810 ($z = 3.35$), the deep radio limit suggests that the overall star formation activity in this GRB host is not heavily obscured by dust, and possibly even less obscuration than the star-forming main sequence at $z = 3.35$ \citep{speagle2014highly, whitaker2017constant}.
The limits in our observations are 3 times deeper than the previous limits on the SFR at $z > 2$ \citep{perley2015connecting}, and thus provide tighter constraints on whether GRB hosts at these redshifts are more likely to be dusty starburst galaxies or not.
The results from our limited sample suggest that the GRBs with UV-bright afterglows (i.e. optically thin sighltines in UV) at $z \sim 2-3.5$ are likely to be star-forming galaxies with SFRs moderately higher ($< 5 \times$) than the star-forming main sequence \citep{speagle2014highly}, but without significant dust obscuration in their star-forming regions.
However, it is possible that this result only applies to the GRBs with UV-bright afterglows due to our selection criteria. At the same time, the dust extinction along a sightline may not necessarily represent the dust obscuration on a galaxy scale, for optically thin as well as optically thick sightlines (in UV).
More radio observations of GRB hosts at $z > 2$ with a depth at least $2 \times SFR_{UV}$ are necessary to confirm this hypothesis. This is required for GRBs with UV-bright afterglows as well as with UV/optically dark afterglows to rule out any selection bias based on the line-of-sight extinction.
\begin{figure}
\centering
\includegraphics[width=0.5\textwidth]{Figure5.pdf}
\figcaption{\label{fig:SFR_vs_z_lit}
The comparison of the radio-derived SFR (tracing the total SFR) and UV-derived SFR (tracing the dust-unobscured SFR) as a function of redshift for the four GRBs presented here (in the foreground), and GRBs in the literature in the background. P13: \cite{perley2013population} and one data point (GRB 060814) from \cite{greiner2016probing}.}
\end{figure}
\section{Summary} \label{sec:summary}
\noindent
If the GRBs are unbiased tracers of star formation at high redshifts ($z > 2$), then we should expect that a large fraction of GRB hosts are highly dust-obscured starbursting galaxies, since these are well known to be major contributors to the cosmic star formation at high redshifts.
The goal of our study was to investigate the galaxy-scale dust obscuration in the GRB hosts with optically thin sightlines in the UV.
We conducted deep radio observations of a
subset of four massive ($\mathrm{M_{*} > 10^{9.5}}$ $\mathrm{M_{\odot}}$) GRB hosts at $z > 2$ for which high signal-to-noise (typical SNR $\sim$ 10) and high-resolution ($\Delta v$ $<$ 50 km s$^{-1}$) rest-frame UV spectra of the afterglow are available. The selected targets are GRB 021004, GRB 080310, GRB 080810, and GRB 121024A. We measured the total SFR (= obscured $+$ unobscured SFR) of the hosts using VLA C-band observations and compared them against the unobscured component of the SFR, measured from UV-optical photometry. The depth of the radio observations in this study has allowed us to put tight constraints on the ratio of the total-to-unobscured SFRs ($\mathrm{SFR_{total}/SFR_{unobscured}}$).
We find that the radio-based star formation rates are in general not substantially higher than those obtained from the optical/UV measurements. Thus, the fraction of total star formation that is obscured by dust ($\mathrm{SFR_{obscured}/SFR_{total}}$) in most of the GRB hosts, even at $z > 2$, is less than 90\%. Particularly, for the two best-constrained objects, GRB 021004 ($z = 2.323$) and GRB 080810 ($z = 3.35$), we find that the upper limit of the radio-based `total SFR' is less than twice the UV-based `unobscured SFR' of the GRB hosts (thus, $\mathrm{SFR_{obscured}/SFR_{total}} < 50\%$). Our results suggest that the
dust obscuration in the star-forming regions of these galaxies is small, and sometimes (e.g.\ for GRB 021004 and GRB 080810) even smaller than the dust obscuration seen in typical main-sequence star-forming galaxies at these redshifts.
The present upper limits on the radio-based SFRs prevent us from determining where the GRB host population lies with respect to the main sequence of star-forming galaxies at $z > 2$. Deeper radio observations to a depth of $2 \times \mathrm{SFR_{UV}}$ are required to answer this question.
P.G. was supported by NASA Earth and Space Science Fellowship (ASTRO18F-0085) for this research. The authors are grateful to Drs.\ S. Bradley Cenko and Daniel Perley for their useful comments in the early stages of this paper. We thank Dr.\ Ashley Zauderer and Nicholas Ferraro for helpful discussions about data analysis. The National Radio Astronomy Observatory is a facility of the National Science Foundation operated under cooperative agreement by Associated Universities, Inc.
\vspace{2mm}
\facilities{VLA}
\software{{\fontfamily{qcr}\selectfont CASA} \citep{mcmullin2007casa}, {\fontfamily{astropy}\selectfont astropy} \citep{robitaille2013astropy} }
\bibliographystyle{apj}
\section{}
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is a 1969 Japanese cel-animated action-comedy musical film produced by Tōei Animation (then Tōei Dōga) and the second film to be directed by Kimio Yabuki. The screenplay and lyrics, written by Hisashi Inōe and Morihisa Yamamoto, is based on the European literary fairy tale of the same name by Charles Perrault, expanded with elements of Alexandre Dumas-esque swashbuckling adventure and cartoon animal slapstick, with many other anthropomorphic animals (kemono in Japanese) in addition to the title character. The Tōei version of the character himself is named Pero, after Perrault. It is the 15th film that Tōei Dōga produced.
The film was released straight to television in the United States by AIP-TV.
The film is particularly notable for giving Toei Animation its mascot and logo and for its roll call of top key animators of the time: Yasuo Ōtsuka, Reiko Okuyama, Sadao Kikuchi, Yōichi Kotabe, Akemi Ōta, Hayao Miyazaki and Akira Daikubara, supervised by animation director Yasuji Mori and given a relatively free rein and adequate support to create virtuosic and distinctive sequences, making it a key example of the Japanese model of division of labour in animation by which animators are assigned by scene rather than character. Most famous of these sequences is a chase across castle parapets animated in alternating cuts by Ōtsuka and Miyazaki which would serve as the model for similar sequences in such later films as Miyazaki's feature-directing début The Castle of Cagliostro and The Cat Returns. Miyazaki is also the manga artist of a promotional comic book adaptation of the film originally serialised in the Sunday Chūnichi Shimbun during 1969, in which it is credited to Tōei Dōga as a whole, and republished in 1984 in a book about the making of the film. The film was re-released 9 years later in the 1978 Summer Toei Manga Matsuri on July 22.
Since becoming Toei Animation's mascot, Pero's face can be seen on the company's primary logo at the beginning or ending to some of Toei's other animated features, both from Japan and some of their outsourced work for other companies. In 2016, a new 3D on-screen logo featuring Pero was revealed in celebration of the company's 60th anniversary. The 3D on-screen logo without the 60th Anniversary wordmark and the company's motto was used since 2019. The former on-screen logo is currently used as a print logo.
Cast
Plot
The enigmatic Puss, Pero, is declared an outlaw by his feline home village for saving mice, an act that defies the nature of cats and is therefore illegal. After Pero escapes to avoid punishment, the Feline King dispatches three bumbling assassins to find and capture Pero, warning them that they face execution should they fail. Pero begins his journey, (all the while dodging his would-be captors throughout the adventure) soon meeting young Pierre, a poor, neglected miller's son who is ousted from his home. The two quickly become good friends and set off together across the countryside. They eventually arrive at a bustling kingdom where a ceremony has begun in which to select a suitable prince who shall wed the lonely, innocent Princess Rosa. Pero sees potential in Pierre as the perfect candidate and hurries into the castle to begin his plan, much to Pierre's opposition. Misfortune soon enshrouds the kingdom as Lucifer, an ogre sorcerer, appears displaying his awesome magical abilities with promises of power and riches if Rosa becomes his bride. Despite the King's excited willingness, Rosa sternly declines Lucifer's offer which enrages him with disappointment. He threatens the King with a terrifying demonstration of the darkness that will befall his country if Rosa is not surrendered to him in three days time. Pero, now dumbfounded, witnesses this shocking event and what was once a simple mission of persuasion has now become more than he ever bargained for.
Release
The film was released in Japan theatrically on March 18, 1969.
The AIP English dub version was first released on VHS twice, first by Vestron Video in 1982 (under licensed from then-owner Orion Pictures) and by Media Home Entertainment in 1985, then by Hi-Tops Video. In 2006, Discotek Media has released a DVD version of the film containing the original Japanese version with English subtitles, the AIP English dub, and a music and effects track in Region 1 NTSC format in the United States, under the title The Wonderful World of Puss 'n Boots. On December 2, 2020, Toei released the film on Blu-ray in Japan.
Reception
The film placed 58th in a list of the 150 best animated films and series of all time compiled by Tokyo's Laputa Animation Festival from an international survey of animation staff and critics in 2003.
A 1998 re-release of the film earned a distribution income of () at the Japanese box office.
Manga
Hayao Miyazaki created a 12-chapter manga series as a promotional tie-in for the film. The series was printed in colour and ran in the Sunday edition of Tokyo Shimbun, from January to March 1969. The series was released in pocket book form by Tokuma Shoten in February 1984.
Sequels
The 1969 Puss 'n Boots was followed by two sequels. The series itself is known collectively as .
The Three Musketeers in Boots
The second film, the misleadingly-titled (1972), actually departs from the Dumasian Europe of the first for a Western setting and was released on VHS in the early 1980s in the United States by MPI Home Video as Ringo Rides West and in the United Kingdom by Mountain Video as Ringo Goes West, with Pero renamed to Ringo for its English dub. It is also marketed by Tōei as Return of Pero and popularly known today as The Three Musketeers in Boots, despite the film featuring no musketeers, as the setting is the American Old West.
Puss 'n Boots Travels Around the World
The third, (1976), was licensed by Turner Program Services and given a dub where Pero is renamed Pussty and was directed by Peter Fernandez. The dub was released on video in the US by RCA/Columbia Pictures Home Video.
The film was a major success in the Soviet Union, drawing 42.4million admissions at the Soviet box office. This was equivalent to approximately million Rbls ().
The video game Nagagutsu o Haita Neko: Sekai Isshū 80 Nichi Dai Bōken is based on this third film and was also released, in a heavily revised version, in the United States under the title Puss 'n Boots: Pero's Great Adventure, where it is better known than the film itself. The game was released for the Famicom in Japan and was later released outside of Japan for the Nintendo Entertainment System. The game and its plot, based on the third film, was used as a plot in one of the episodes of the second season of Captain N: The Game Master, entitled "Once Upon A Time Machine", which have re-designs of Pero and the two villains of the film, Count Gourmon (Gruemon in the game's instruction manual) and Dr. Garigari. Dr. Garigari had actually appeared a decade before in 1965 as the antagonist to Toei's animated TV series Hustle Punch and was recycled to be used in this film.
Dream short film
In 2018, Toei produced a 6 minute short film entitled Dream, where a young boy dreams about whales as well as Pero who also appears in the dream. This version of Pero has small black eyes, but that was to match the animation style of the short.
References
Sources
External links
Puss 'n Boots at Tōei Animation's corporate website
Nagagutsu o Haita Neko at Tōei Animation's Japanese website
1969 films
1969 anime films
1972 films
1972 anime films
1976 films
1976 anime films
1960s action films
1970s action films
1969 comedy films
1972 comedy films
1960s fantasy films
1970s fantasy films
1960s musical films
1970s musical films
Animated films about cats
Anime and manga based on fairy tales
Japanese film series
Films based on Puss in Boots
Toei Animation films
Toei Company films
Animated comedy films
Animated musical films
Japanese comedy films
Japanese musical films
Discotek Media
Japanese animated fantasy films
Japanese fantasy adventure films
1976 comedy films
1960s children's animated films
Films directed by Kimio Yabuki | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 2,029 |
José Echegaray y Eizaguirre (* 19. April 1832 in Madrid; † 4., 12., 14. oder 16. September 1916 ebenda) war ein spanischer Dramatiker, Politiker und Literaturnobelpreisträger. Als Neuromantiker trug er wesentlich zur Weiterentwicklung des spanischen Dramas bei; seine Stücke wurden in ganz Europa aufgeführt.
Leben
Echegaray wurde in Madrid geboren, wohin er nach der in Murcia verbrachten Kindheit zurückkehrte, um an der Escuela de Caminos (eigentlich Escuela Técnica Superior de Ingenieros de Caminos, Canales y Puertos) Ingenieurwissenschaften zu studieren. Nach dem Abschluss 1853 arbeitete er zunächst einige Monate als Ingenieur und arbeitete dann bis 1868 als Dozent an der Escuela de Caminos. 1857 heiratete er Ana Perfecta Estrada, mit der er eine Tochter hatte. 1866 wurde er in die Academia de Ciencias Físicas, Exactas y Naturales (Akademie für Naturwissenschaften) aufgenommen, der er in späteren Jahren auch als Präsident vorstand. Sein Interesse für Wirtschaftswissenschaft erwachte, als er an der Escuela de Caminos den Wirtschaftswissenschaftler Gabriel Rodríguez kennenlernte.
An der Revolution von 1868, die Spanien zu einer konstitutionellen Monarchie machte, war Echegaray nicht beteiligt, obwohl er mit deren zentraler Figur, Juan Prim, befreundet und im Ateneo von Madrid als Redner bekannt war. Er nahm als Abgeordneter an Parlamentsdebatten und verfassunggebenden Versammlungen teil und wurde zum Minister für wirtschaftliche Entwicklung und Finanzen ernannt. In dieser Eigenschaft gründete er die Bank von Spanien und führte Papiergeld ein. Nach ersten Erfolgen als Dramatiker (El libro talonario von 1874) legte er seine Ämter nieder, nahm jedoch weiterhin am politischen Leben teil. 1905 wurde er erneut Finanzminister.
1904 wurde Echegaray zusammen mit Frédéric Mistral der Nobelpreis für Literatur verliehen, sehr zum Missfallen der jüngeren Schriftsteller seines Landes (siehe Generation von 98), die in ihm einen Vertreter literarischer Traditionen sahen, die es zu überwinden galt.
Werk
Echegaray verfasste über 70 Dramen in Vers und Prosa. Immer wieder änderte er seinen literarischen Stil, blieb aber stets der romantischen Schule Spaniens treu. Als einer der letzten Vertreter dieser Richtung ließ er sich durchaus auch von Schriftstellern anderer europäischer Länder beeinflussen, wie etwa Henrik Ibsen, Alexandre Dumas und Hermann Sudermann. Er veröffentlichte oft unter den Pseudonymen Jorge Hayaseca oder Jorge Hayaseca y Eizaguirre. Mit seinem literarischen Schaffen wurde er einer der Wegbereiter für Jacinto Benavente.
Werkauswahl
La esposa del vengado (Drama, 1874, deutsch Die Frau des Rächers, Ü:1883)
O locura, o santidad (Drama, 1877, deutsch Wahnsinn oder Heiligkeit, Ü: 1889)
En el seno de la muerte (Drama, 1879, deutsch Im Schoße des Todes, Ü: 1882)
La muerte en los labios (Drama, 1880, deutsch Der Tod auf den Lippen, Ü: 1970)
El gran Galeoto (Drama, 1881 dt. Galeotto, 1901, 1974 unter dem Titel Der große Kuppler)
Vida alegre y muerte triste (Drama, 1885 deutsch. Lustiges Leben – trauriger Tod, Ü: 1892)
De mala raze (Drama, 1886 deutsch Adelina (auch: Von schlechter Rasse), Ü: 1970)
Mariana (Drama, 1892 deutsch Mariana Ü: o. J. 1896–1900)
Mancha que limpia (Drama, 1895 deutsch Mathilde oder Der Flecken, der reinigt, Ü: 1898)
Herencias malas (Drama, 1902 deutsch Schlechte Erbschaften, Ü: 1904)
En el puño de la espada (Drama, 1875)
En el pilar y en la cruz (Drama, 1878)
Conflicto entre dos deberes (1882)
El hijo de Don Juan (Drama, 1892)
Weblinks
Literatur von und über José Echegaray im Katalog des Ibero-Amerikanischen Instituts in Berlin
Literatur von und über José Echegaray im Katalog der Bibliothek des Instituto Cervantes in Deutschland
Nobelpreisträger für Literatur
Autor
Literatur (19. Jahrhundert)
Literatur (Spanisch)
Literatur (Spanien)
Drama
Ritter des Ordens vom Goldenen Vlies
Mitglied der Junta para Ampliación de Estudios
Person als Namensgeber für einen Merkurkrater
Person (Baskenland)
Spanier
Geboren 1832
Gestorben 1916
Mann | {
"redpajama_set_name": "RedPajamaWikipedia"
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Q: PHP Class Instance into MySQL database I'm working on a website in which I would like to use PHP and MySQL.
Mostly I insert hundreds of variables into my MySQL database but I am getting tired of this.
I have searched the internet for a solution but without any succes.
My website uses Object Oriented PHP and I would like to insert the whole instance into the database.
Example:
class User
{
private $userID;
private $username;
// etc
public function __construct($userID, $username)
{
$this->userID = $userID;
$this->username = $username
}
// Other functions go here
}
I would like to know if I could save a whole instance ($user = new User($x, $y)) into my MySQL database. If possible, what field type do I need?
Thanks in advance.
A: What you're looking for is probably an ORM (object relational mapper), which allows you to map directly your objects to tables in the database.
Several PHP ORMs are available, for instance :
*
*Doctrine, perhaps the most widely used one,
*Propel
*Redbean which has a bit of a different approach, but I like that one quite a lot.
You should also be aware of the fact that using such a library might impact performance quite a bit, but with caching and other solutions this can be mitigated.
A: You should serialize the object somehow and store it in a text field. You can use either PHP's native serialize()/unserialize() or, if they are simple value-objects, you could use json_encode()/json_decode(). See this thread for more information/compare of the two
A: You can't save an instance into database. Because instance is occupied memory where using the compiler it saves states of the class. However Manu Letroll has given a good solution you should use orm to make work easier.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 4,420 |
{"url":"https:\/\/kops.uni-konstanz.de\/entities\/publication\/e9a2de5c-e99f-4c1a-a043-1650023212f6","text":"## Heterogeneous Social Preferences and the Dynamics of Free Riding in Public Good Experiments\n\n2008\nG\u00e4chter, Simon\n##### Series\nResearch paper series \/ Thurgauer Wirtschaftsinstitut; 27\n##### Publication type\nWorking Paper\/Technical Report\n##### Abstract\nWe provide a direct test of the role of social preferences and beliefs in voluntary cooperation and its decline. We elicit individuals cooperation preference in one experiment and use them as well as subjects elicited beliefs to make predictions about contributions to a public good played repeatedly. We find substantial heterogeneity in people s preferences. With simulation methods based on this data, we show that the decline of cooperation is driven by the fact that most people have a preference to contribute less than others. Belief formation and virtual learning do not contribute to the decline of cooperation.\n330 Economics\n##### Keywords\nPublic goods experiments,social preferences,conditional cooperation,free riding\n##### Cite This\nISO 690FISCHBACHER, Urs, Simon G\u00c4CHTER, 2008. Heterogeneous Social Preferences and the Dynamics of Free Riding in Public Good Experiments\nBibTex\n@techreport{Fischbacher2008Heter-11874,\nyear={2008},\nseries={Research paper series \/ Thurgauer Wirtschaftsinstitut},\ntitle={Heterogeneous Social Preferences and the Dynamics of Free Riding in Public Good Experiments},\nnumber={27},\nauthor={Fischbacher, Urs and G\u00e4chter, Simon}\n}\n\nRDF\n<rdf:RDF\nxmlns:dcterms=\"http:\/\/purl.org\/dc\/terms\/\"\nxmlns:dc=\"http:\/\/purl.org\/dc\/elements\/1.1\/\"\nxmlns:rdf=\"http:\/\/www.w3.org\/1999\/02\/22-rdf-syntax-ns#\"\nxmlns:bibo=\"http:\/\/purl.org\/ontology\/bibo\/\"\nxmlns:dspace=\"http:\/\/digital-repositories.org\/ontologies\/dspace\/0.1.0#\"\nxmlns:foaf=\"http:\/\/xmlns.com\/foaf\/0.1\/\"\nxmlns:void=\"http:\/\/rdfs.org\/ns\/void#\"\nxmlns:xsd=\"http:\/\/www.w3.org\/2001\/XMLSchema#\" >\n<dcterms:abstract xml:lang=\"eng\">We provide a direct test of the role of social preferences and beliefs in voluntary cooperation and its decline. We elicit individuals cooperation preference in one experiment and use them as well as subjects elicited beliefs to make predictions about contributions to a public good played repeatedly. We find substantial heterogeneity in people s preferences. With simulation methods based on this data, we show that the decline of cooperation is driven by the fact that most people have a preference to contribute less than others. Belief formation and virtual learning do not contribute to the decline of cooperation.<\/dcterms:abstract>\n<dcterms:title>Heterogeneous Social Preferences and the Dynamics of Free Riding in Public Good Experiments<\/dcterms:title>\n<dc:contributor>G\u00e4chter, Simon<\/dc:contributor>\n<dcterms:available rdf:datatype=\"http:\/\/www.w3.org\/2001\/XMLSchema#dateTime\">2011-03-25T09:40:49Z<\/dcterms:available>\n<dcterms:isPartOf rdf:resource=\"https:\/\/kops.uni-konstanz.de\/server\/rdf\/resource\/123456789\/46\"\/>\n<dspace:isPartOfCollection rdf:resource=\"https:\/\/kops.uni-konstanz.de\/server\/rdf\/resource\/123456789\/46\"\/>\n<dc:creator>Fischbacher, Urs<\/dc:creator>\n<dspace:hasBitstream rdf:resource=\"https:\/\/kops.uni-konstanz.de\/bitstream\/123456789\/11874\/1\/49_08_05_TWI_RPS_Fischerbacher_Gaechter.pdf\"\/>\n<foaf:homepage rdf:resource=\"http:\/\/localhost:8080\/\"\/>\n<dc:language>eng<\/dc:language>\n<dc:format>application\/pdf<\/dc:format>","date":"2023-03-31 14:21:31","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.36385881900787354, \"perplexity\": 4667.955790333823}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-14\/segments\/1679296949642.35\/warc\/CC-MAIN-20230331113819-20230331143819-00593.warc.gz\"}"} | null | null |
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\title[Self-organisation in cellular automata with coalescent particles: Qualitative and quantitative approaches]{Self-organisation in cellular automata with coalescent particles: Qualitative and quantitative approaches}
\author{Benjamin Hellouin de Menibus}
\address{Andrés Bello University and Centro de Modelamiento Matemático (Universidad de Chile), Santiago. }
\email{benjamin.hellouin@gmail.com}
\urladdr{http://mat-unab.cl/~hellouin/}
\author{Mathieu Sablik}
\address{Aix Marseille Université, CNRS, Centrale Marseille, I2M UMR 7373\\
13453, Marseille, France}
\email{sablik@univ-amu.fr}
\urladdr{http://www.i2m.univ-amu.fr/~sablik/}
\keywords{Cellular automata, Particules, limit measures, Brownian motion}
\date{}
\begin{document}
\maketitle
\begin{abstract}
This article introduces new tools to study self-organisation in a family of simple cellular automata which contain some particle-like objects with good collision properties (coalescence) in their time evolution. We draw an initial configuration at random according to some initial $\s$-ergodic measure $\mu$, and use the limit measure to descrbe the asymptotic behaviour of the automata.
We first take a qualitative approach, i.e. we obtain information on the limit measure(s). We prove that only particles moving in one particular direction can persist asymptotically. This provides some previously unknown information on the limit measures of various deterministic and probabilistic cellular automata: $3$ and $4$-cyclic cellular automata (introduced in~\cite{Fisch}), one-sided captive cellular automata (introduced in~\cite{Theyssier-2004}), N. Fatès' candidate to solve the density classification problem~\cite{Fates}, self stabilization process toward a discrete line~\cite{Regnault-Remilia-2015}\dots
In a second time we restrict our study to to a subclass, the gliders cellular automata. For this class we show quantitative results, consisting in the asymptotic law of some parameters: the entry times (generalising~\cite{EntryTimes}), the density of particles and the rate of convergence to the limit measure.
\end{abstract}
\section{Introduction}
A cellular automaton is a complex system defined by a local rule which acts synchronously and uniformly on the configuration space. These simple models exhibit a wide variety of dynamical behaviors and even in the one-dimensional case (the focus of this article) they are not well understood. Formally, given a finite alphabet $\A$, a configuration is an element of the set $\az$ which is compact for the product topology and a \define{cellular automata} $F:\az\to\az$ is defined by a local function $f:\A^{[-r,r]}\to \A$, for some \emph{radius} $r>0$, which acts synchronously and uniformly on every cell of the configuration:
\[F(x)_i=f(x_{[i-r,i+r]})\textrm{ for all }x\in\az\textrm{ and }i\in\Z.\]
Equivalently, cellular automata can be defined as continuous fonctions that commute with the \emph{shift map} $\sigma$ defined by $\sigma(x)_i=x_{i+1}$ for all $x\in\az$ and $i\in\Z$.
Even though cellular automata have been introduced by J. Von Neumann~\cite{VonNeumann}, the impulsion for their systematic study was given by the work of S. Wolfram~\cite{Wolfram84}. He instigated a systematic study of \emph{elementary} cellular automata, which are the cellular automata defined on the alphabet $\{0,1\}$ with radius $1$ (there are $2^{2^3}=256$ such cellular automata and each of them is associated a number $\# n$). In particular he proposed a classification according to the observation of the space-time diagrams produced by the time evolution of cellular automata starting from a random configuration.
One of these classes corresponds to a particular form of self-organisation: from a random configuration, after a short transitional regime, regions consisting in a simple repeated pattern emerge and grow in size, while the boundaries between them persist under the action of the cellular automaton and can be followed from an instant to the next. Therefore their movement (time evolution) can be defined inductively, and in this case we call these boundaries \define{particles}. In the simplest case, these particles evolve at constant speed and are annihilated when colliding with other particles; however, they can sometimes exhibit a periodic behavior or even perform a random walk, and the collisions may give birth to new particles following some more or less complicated rules.
This type of behavior was first observed empirically in elementary cellular automata \#18, \#122, \#126, \#146, and \#182 \cite{Grassb, Grassa}, then \#54, \#62, \#184 \cite{Boccara}, etc. The interest for these automata stems from their dynamics that appeared neither too simple nor too chaotic, giving hope to better understand their underlying structure. In Figure~\ref{fig:particles}, we show the space-time diagrams of a sample of such automata iterated on a (uniform) random configuration.
Roughly speaking, studying particles in cellular automata requires two steps:
\begin{itemize}
\itemsep0em
\item identifying and describing the particles, usually as finite words;
\item describing the particle dynamics and understanding its effect on the properties of the CA.
\end{itemize}
Historically, this study was often performed on individual or small groups of similar-looking CA, and the first step was done in a case-by-case manner. See for example \cite{Fisch, Fisch-2} for the 3-state cyclic automaton, \cite{BelitskyFerrari-1995, Belitsky-2005} for Rule $\#184$ and other automata with similar dynamics, \cite{Grassa, Eloranta-1992} for Rule $\#18$\dots Other works such as \cite{Eloranta} skip the first step and study particle dynamics in an abstract manner instead, deducing dynamical properties of automata by making assumptions on the dynamics of their particles (althought examples of such particles are provided).
The first general formalism of particles in cellular automata was introduced by M. Pivato: homogeneous regions are correspond to words from a subshift $\Sigma$ and particles are defects in a configuration of $\Sigma$. He developed some invariants to characterise the persistence of a defect~\cite{Pivato-2007-algebra, Pivato-2007-spectral} and he described the different possible dynamics of propagation of a defect~\cite{Pivato-2007-defect}.\sk
In the present work we focus on the second step first. More precisely, we are interested in how the existence of some particle set with good dynamical properties affects the typical asymptotic behavior. Then we apply this general framework to a variety of examples, finding the sets of particles by Pivato's methods or otherwise, to explain the self-organisation that is observed experimentally.
Let us define more formally what we mean by typical asymptotic behavior. Starting from a $\s$-invariant measure $\mu\in\Ms(\az)$ (i.e. $\mu(\s^{-1}(U))=\mu(U)$ for all borelian set $U$), we consider the iteration of a cellular automata $F$ on this measure:
\[
\begin{array}{rcccl}
\F:&\Ms(\az)&\longrightarrow&\Ms(\az)&\\
&\mu&\longmapsto &\F\mu&\textrm{ where }\F\mu(U)=\mu(F^{-1}(U))\textrm{ for all borelian }U.
\end{array}
\]
We then study the asymptotic properties of the sequence $(\F^t\mu)_{t\in\N}$, and particularly the set of cluster points called the \define{limit measure(s)}. Sometimes our information on the $\mu$-limit set, which is the union of the supports of all limit measures. Equivalently, it is the set of configurations containing only patterns whose probability to appear in the space-time diagram does not tend to zero as time tends to infinity. This set introduced in~\cite{Kurka-Maass-2000} corresponds to the configurations which are observed asymptotically when the automaton is iterated on a random configuration.
When studying typical asymptotic behaviour in this sense, it is unreasonable to except a general result since a wide variety of limit measures can be reached in the general case~\cite{Hellouin-Sablik-2013} and any nontrivial property of the $\mu$-limit set is undecidable \cite{Delacourt-2011}. That is why we consider restricted cases for the dynamics of the particles. To determine the $\mu$-limit set in some cases, P. K\r{u}rka suggests an approach based on particle weight function which assigns weights to certain words~\cite{Kurka-2003}. However, this method does not cover any case when a defect can remain in the $\mu$-limit set. Hence we aim at a more general approach, in terms of particle dynamics as well as initial measures.
One of our main motivations for this study is the class of captive cellular automata, where the local rule cannot make a colour appear if it is not already present in the neighbourhood. These automata were introduced by G. Theyssier in \cite{Theyssier-2004} for their algebraic properties, but he also noticed an interesting phenomenon: when drawing a captive cellular automaton at random (fixed alphabet and neighbourhood), most captive automata exhibited the type of behavior we described above. Any kind of general result regarding self-organisation of captive cellular automata remains a challenging open problem.
\sk
This article is divided into two main sections, corresponding to improved versions of results previously published in conferences~\cite{Hellouin-Sablik-2011,Hellouin-Sablik-2012}. In Section~\ref{sec:particles}, we present a qualitative result generalised from~\cite{Hellouin-Sablik-2011} with an improved formalism, shorter proofs and a new application to probabilist cellular automata (Section~\ref{sec:probabilist}). Then, in Section~\ref{sec:brown}, we refine our approach on a subclass to obtain some quantitative results. Sections~\ref{sec:walks} to \ref{sec:brownian} were published in \cite{Hellouin-Sablik-2012}; we correct some inaccuracies in the proofs and extend the study to other parameters.
\paragraph{Qualitative approach}\ In Section~\ref{sec:particles}, we prove a qualitative result:. For an initial $\s$-ergodic measure $\mu$, assuming particles have good collision properties (coalescence), only particles moving in one particular direction can persist aymptotically. We introduce our own formalism of particle system in Section~\ref{sec:particlesdef} so as to be able to describe the dynamics of the particles, and Section~\ref{sec:selforg} is dedicated to the proof itself. Section~\ref{sec:defects} present a simplified version of Pivato's formalism which is by far the simplest way to find such a particle system in most examples.
We spend Section~\ref{sec:particlesexamples} on various examples of automata where this result can be applied:
\begin{description}
\item[Section~\ref{section.184}] we characterize the $\mu$-limit set of the ``traffic'' automaton (rule $\#184$), a simple case that may clarify the formalism. The results were known for initial Bernoulli measures~\cite{BelitskyFerrari-1995, Belitsky-2005} but our method applies for every $\s$-ergodic measure;
\item[Section~\ref{section.cyclic}] we consider the family of $n$-cyclic cellular automata introduced in~\cite{Fisch-2,Fisch}. Using our method, we go further in the study of these simple automata: in particular, for $n=3$ or $4$, we show that the limit measure is unique and is a convex combination of Dirac measures supported by uniform configurations.
\item[Section~\ref{section.captive}] we characterize the $\mu$-limit set of all one-sided captive cellular automata. This is a first step to the study of asymptotic behaviour of captive cellular automata.
\item[Section~\ref{section.randomwalk}] last, we apply our formalism to a cellular automaton where the particules do not have a linear speed but instead perform random walks by drawing randomness from the initial measure.
\end{description}
However, our results are not general enough to apply to defects of a sofic subshift that can have a particle-like behavior, such as in Rule $\#18$ (see the bottom right picture in Figure~\ref{fig:particles} and \cite{Eloranta-1992}), or to more complicated particle systems such as those observed in general captive cellular automata.
Finally, in Section~\ref{sec:probabilist}, we generalize our method to probabilistic cellular automata. As an application, we partially describe limit measures of the probabilistic cellular automata proposed by N. Fatès in \cite{Fates} as a candidate to solve the density classification problem. Another application proposed in Section~\ref{section.Regnault-Remilia-line} present a generalization to the infinite line of a self stabilization process toward a discrete line proposed in~\cite{Regnault-Remilia-2015}.
\paragraph{Quantitative approach}\ In Section~\ref{sec:brown}, we improve the previous qualitative results with a quantitative approach, considering the time evolution of some parameters when the particle dynamics are very simple. This research direction was inspired by \cite{EntryTimes}, where the authors consider the waiting time before a particle crosses the central column (called entry time). Using the same approach as in \cite{BelitskyFerrari-1995, Kurka-Maass-2000}, we show that the behavior of these automata can be described by a random walk process (Section~\ref{sec:walks}), and we approximate this process by a Brownian motion using scale invariance (Section~\ref{sec:brownian}). Thanks to this tool, we answer negatively a conjecture proposed in~\cite{EntryTimes} by determining the correct asymptotic law for the entry time of a particle in the central column (Section~\ref{sec:entry}). We then use the same approach on various natural parameters such as the density of particles at time $t$ (Section~\ref{sec:density}) or the rate of convergence to the limit measure (Section~\ref{sec:rate}). This generalises some known results on initial Bernoulli measures from \cite{EntryTimes} and \cite{Belitsky-2005}, in particular relaxing the conditions on the initial measures. In Section~\ref{sec:extensions}, we exhibit various examples with similar dynamics on which these results apply.
In all the article, space-time diagrams were produced using the Sage mathematical software \cite{Sage} and follow the convention $\square = 0, \blacksquare = 1, \textcolor{red}{\blacksquare} = 2, \textcolor{blue}{\blacksquare} = 3 $.
\begin{figure}[!ht]
\begin{scriptsize}
\begin{center}
\begin{tabular}{p{0.5cm}p{0.5cm}|p{5cm}p{5cm}p{5cm}}
&\multirow{4}{*}{\begin{sideways}{ Cases with quantitative results (Section~\ref{sec:brown})}\end{sideways} } &(-1,1)-gliders CA (Sec.~\ref{sec:walks})&Rule 184 or traffic rule (Sec.~\ref{section.184}) & 3-state cyclic CA (Sec.~\ref{section.cyclic})\\
\multirow{7}{*}{\begin{sideways}{\normalsize Cases with qualitative result (Section~\ref{sec:particles})}\end{sideways} } &&\includegraphics[width=5cm]{Factgr.png}&\includegraphics[width=5cm]{184.png}&\includegraphics[width=5cm]{cycle3.png}\\
&&(0,1)-gliders CA (Sec.~\ref{sec:walks})&One-sided captive CA (Sec.~\ref{section.captive})&One-sided captive CA (Sec.~\ref{section.captive})\\
&&\includegraphics[width=5cm, trim = 0 0 0 225,clip]{gliders.png}&\includegraphics[width=5cm]{captif.png}&\includegraphics[width=5cm]{onesided.png}\\
\cline{2-5}
&& 4-state cyclic CA (Sec.~\ref{section.cyclic})&5-state cyclic CA (Sec.~\ref{section.cyclic})&Random walk CA (Sec.~\ref{section.randomwalk})\\
&&\includegraphics[width=5cm]{cycle4.png}&\includegraphics[width=5cm]{cycle5.png}&\includegraphics[width=5cm]{WalkShort}\\
\cline{2-5}
&\multirow{2}{*}{\begin{sideways}{ Prob. CA (Sec.~\ref{sec:probabilist})}\end{sideways} }&Fatès'candidate for density (Sec.~\ref{section.FatesDensity}) &Self-stabilization of the line (Sec.~\ref{section.Regnault-Remilia-line})&Generic one-sided captive PCA\\
&&\includegraphics[width=5cm]{fates.png}&\includegraphics[width=5cm]{line.png}&\includegraphics[width=5cm]{PCA.png}\\
\hline
\hline
&&Generic captive CA&Generic captive CA&Generic captive CA\\
\multirow{2}{*}{\begin{sideways}{\normalsize Unknown cases}\end{sideways} } &&\includegraphics[width=5cm, trim = 0 0 0 225,clip]{captif1.png}&
\includegraphics[width=5cm, trim = 0 0 0 225,clip]{captif2.png}&
\includegraphics[width=5cm]{captif3}
\\
&&Generic captive CA&Generic captive CA&Rule 18\\
&&
\includegraphics[width=5cm]{captif4.png}
&
\includegraphics[width=5cm]{captif5.png}
&
\includegraphics[width=5cm]{18.png}\\
\end{tabular}
\end{center}
\end{scriptsize}
\caption{Space-time diagrams of some cellular automata with particles, starting from
a configuration drawn uniformly at random. }\label{fig:particles}
\end{figure}
\section{Particle-based organisation: qualitative results}\label{sec:particles}
In this section, we take a qualitative approach to self-organisation: that is, we assume some properties on the
dynamics of the particles of some cellular automata and try to deduce properties of the $\mu$-limit measures set of the cellular automaton, with no regard to how fast this organisation takes place.
\subsection{Particles}\label{sec:particlesdef}
\subsubsection{Definition of symbolic system}
Given a finite alphabet $\A$, a \define{word} is a finite sequence of elements of $\A$. Denote $\A^{\ast}=\bigcup_n\A^n$ the set of all words where $\A^0$ is the empty word $\varepsilon$. An infinite sequence indexed by $\Z$ is called a \define{configuration} and the set of configuration $\az$ is a compact set. For a word $u\in\A^{\ast}$ the \define{cylinder} $[u]$ is the set of configuration where $u$ appears at the position $0$.
On $\az$ we define the \define{shift} map $\s(x)_i=x_{i+1}$ for all $x\in\az$ and $i\in\Z$. A \define{subshift} is a closed $\s$-invariant subset of $\az$. Equivalently a subshift can be define with a set of forbidden patterns $\mathcal{F}$ as the set of configuration where no pattern of $\mathcal{F}$ appear. If $\mathcal{F}$ is finite, we call it a \define{subshift of finite type} or SFT.
Given two finite alphabets $\A$ and $\B$, a \define{morphism} is a continuous function $\pi:\az\to\B^{\Z}$ which commutes with the shift (i.e. $\s(\pi(x))=\pi(\s(x))$ for all $x\in\az$). Equivalently a morphism can be defined by a local map $f:\A^{\mathcal{N}}\to\B$ where $\mathcal{N}\subset\Z$ is a finite set called the neighborhood such that
\[\pi(x)_i=f(x_{[i-r,i+r]})\textrm{ for all $x\in\az$ and $i\in\Z$.}\]
The \define{radius} is the minimal $r\in\N$ such that $\mathcal{N}\subset[-r,r]$. A \define{cellular automaton} is a morphism $\az\to\az$, that is, the input and the output are defined on the same alphabet. In particular a cellular automaton can be iterated and it makes sense to study its dynamics.
\subsubsection{Particle system}
\begin{definition}[Particle system]~
Let $F:\az\to\az$ be a cellular automaton. A \define{particle system} for $F$ is a tuple $(\P, \pi, \phi)$, where:
\begin{itemize}
\itemsep0em
\item $\P$ is a finite set whose elements are called \define{particles};
\nomentry{$\P$}{Set of particles in a particle system}
\item $\pi : \az \mapsto (\P\cup\{0\})^\Z$ is a morphism;
\nomentry{$\pi$}{Morphism}
\item $\phi : \az\times \Z \mapsto 2^\Z$ where $2^\Z$ denotes the subsets of $\Z$ is a function called \define{update function},
such that the update function satisfies the following properties for all $x\in\az$ and $k\in\Z$,
denoting $\p_{\P, \pi}(x) = \{k\in\Z\ :\ \pi(x)_k\in\P\}$,
and omitting $\P$ and $\pi$ when the particle system is fixed by the context:
\nomentry{$\p(x)$}{Set of coordinates where $x$ contains a particle}
\begin{description}
\item[Locality] There is a constant $r>0$ (its \define{radius}) such that $\phi(x,k)\subset[k-r,k+r]$.
The particles cannot ``jump'' arbitrarily far. By constant we mean it does not depend on $x$ and $k$.
\item[Surjectivity] $\p(F(x))=\phi(x,\Z)$.
A particle at time $t+1$ cannot appear from nowhere; it must be the image of some particle at time $t$.
\item[Particle control]$\forall k\in\Z, \begin{array}{l}k\in\p(x) \Rightarrow \forall k'\in\phi(x,k), k'\in \p(F(x));\\
k\notin\p(x) \Rightarrow \phi(x,k) = \emptyset.\end{array}$
If a particle is sent somewhere, it remains a particle; conversely, a particle cannot come from a non-particle.
\item[Disjunction] $k < k' \Rightarrow \phi(x,k) = \phi(x,k') \text{ or } \max\phi(x,k) < \min\phi(x,k')$.
Two different particles crossing is considered an interaction, in which case their common image is the resulting set of particles.
This assumption excludes half-progression half-interaction cases where two particles share a part of their image.
\end{description}
\end{itemize}
Intuitively, the update function associate to each particle at time $t$ (given as a coordinate in a configuration)
its set of images at time $t+1$ under the action of the cellular automaton.
This image can be one particle if the particle simply persists, but also $\emptyset$ if it disappears or many particles.
Particles that interact share the same image.
\nomentry{$\phi$}{Update function in a particle system}
\end{definition}
The four conditions ensure that the update function accurately describe the time evolution of the particles. Notice that since the morphism and update function are defined locally, the conditions can be checked in an
automatic manner by simple enumeration of patterns up to a certain length.\sk
In the context of a fixed particle system for $F$, we use shorthands for the composition of the update function, defined inductively:
\[\phi^t(x,k) = \bigcup_{k'\in\phi(x,k)}\phi^{t-1}(F(x),k')\mbox{ and }\phi^{-1}\circ\phi(x,k) = \{k'\in\Z\ |\
\phi(x,k') = \phi(x,k)\}.\] If $\phi(x,k)$ is a singleton, we use ``$\phi(x,k)$'' instead of ``the only member
of $\phi(x,k)$'' as an abuse of notation.
\subsubsection{Coalescence}
We postpone the discussion on how to actually find a particle system in a given cellular automata to
Section~\ref{sec:defects}. We now look for assumptions on the dynamics of the particles that let us
deduce that some particles disappear asymptotically. Simulations suggest that this is the case when the
particles are forced to collide, and that these collisions are destructive in the sense that the total number of
particles decreases; thus we introduce the notion of coalescence.
\begin{definition}[Coalescence]
Let $F:\az\to\az$ be a cellular automaton, and $(\P,\pi,\phi)$ a particle system for $F$.
This particle system is \define{coalescent} if, for every $x\in\az$ and $k\in\p(x)$, the particle has
one of two possible behaviors:
\begin{description}
\itemsep0em
\item[Progression] $|\phi(x,k)| = |\phi^{-1}(\phi(x,k))| = 1$, and $\pi(x)_k =
\pi(F(x))_{\phi(x,k)}$
(the particle persists and its type does not change), or
\nomentry{$\prog(x)$}{Set of coordinates where $x$ contains a progressing particle}
\item[Destructive interaction] $|\phi(x,k)| < |\phi^{-1}(\phi(x,k))|$
(particles collide and strictly fewer particles are created).
\nomentry{$\inter(x)$}{Set of coordinates where $x$ contains a interacting particle}
\end{description}
\end{definition}
Progressing and interacting particles of a configuration $x\in\az$ are denoted $\prog_{\P,\pi,\phi}(x)$ and $\inter_{\P,\pi,\phi}(x)$, respectively, and
$\P,\pi$ and $\phi$ are omitted when the particle system is clear from the context.
$k\in\prog_{\P,\pi,\phi}(x)$ is the case when we use ``$\phi(x,k)$'' to mean ``the only member of the singleton $\phi(x,k)$''.\sk
Notice that, regardless of coalescence, we have because of locality $|\phi(x,k)| + |\phi^{-1}(\phi(x,k))|\leq 2r+2$, where $r$ is the radius of the update function.
See Figure~\ref{fig:visual} for a visual proof.
\begin{figure}[ht]
\begin{tikzpicture}
\draw (0,0) node {$x$};
\draw (0,1) node {$F(x)$};
\draw[black!30] (1,0) -- (13,0) (1,1) -- (13,1);
\draw[very thick] (3,-.1) -- (3,.1) (10,-.1) -- (10,.1) (3,0) -- (10,0);
\draw[very thick] (5,.9) -- (5,1.1) (8,.9) -- (8,1.1) (5,1) -- (8,1);
\draw (6.5,-.5) node {$\phi^{-1}(\phi(x,k))$};
\draw (6.5,1.5) node {$\phi(x,k)$};
\draw[->] (3.1,.1) -- (7.9,.9);
\draw (4.2,.5) node {$\leq r$};
\draw (8.8,.5) node {$\leq r$};
\draw[->] (9.9,.1) -- (5.1,.9);
\end{tikzpicture}
\caption{Visual proof that $|\phi(x,k)| + |\phi^{-1}(\phi(x,k))|\leq 2r+2$.}
\label{fig:visual}
\end{figure}
Even though the main result makes no reference to the speed of a particle, introducing this notion lets us
state a corollary that is easier to use as well as corresponding more clearly to the intuition.
\begin{definition}[Speed]
Let $F$ be a cellular automata and $(\P,\pi,\phi)$ be a particle system for $F$.
A particle $p\in\P$ has \define{speed $\bm{v\in\Z}$} if for any configuration $x\in\az$ and $k\in\Z$ such that $\pi(x)_k=p$, we have one of
the following:
\begin{description}
\itemsep0em
\item[Eventual interaction] $\exists t, \phi^t(x,k) \in\inter(F^t(x))$;
\item[Progression at speed v] $\forall t, \phi^t(x,k)\in\prog(F^t(x))$ and $\phi^t(x,k)-k\underset{t\to\infty}\sim
vt$.
\end{description}
\end{definition}
\subsection{Probability measures and $\mu$-limit sets}
The $\mu$-limit set was introduced in~\cite{Kurka-Maass-2000} to describe the asymptotic behaviour corresponding to empirical observations. It consists in the patterns whose probability to appear does not tend to 0 when the initial point is chosen at random. To define it formally, let us introduce some notations.
Denote $\Ms(\az)$ the set of $\s$-invariant probability measure on $\az$ (i.e. $\mu$ such that $\mu(\s^{-1}(U))=\mu(U)$ for all borelian $U$). A measure is ergodic if every $\sigma$-invariant borelian set has measure $0$ or $1$, and we denote $\Merg(\az)$ the set of egordic probability measures. \cite{Walters} gives a good introduction to ergodic probability measures.
\paragraph{Examples}~The \define{Bernoulli measure} $\lambda_{(p_a)_{a\in\A}}$ associated with a sequence $(p_a)_{a\in\A}$ of elements of $[0,1]$ whose sum is $1$ is defined by $\lambda_{(p_a)_{a\in\A}}([u])=p_{u_0}p_{u_1}\dots p_{u_{|u|-1}}$ for all $u\in\A^{\ast}$. If all elements of $(p_a)_{a\in\A}$ have the same value $\frac{1}{|A|}$ we call it the uniform Bernoulli measure denoted $\lambda$. For any finite word $u\in\A^{\ast}$, we also define $\meas{u}$ the unique $\s$-invariant probability measure supported by the $\s$-periodic configuration $^{\omega}u^{\omega}$.\sk
Given a cellular automaton $F:\az\to\az$ and an initial measure $\mu\in\Ms(\az)$, we define the measure $\F\mu$ by $\F\mu(U)=\mu(F^{-1}(U))$ for any borelian $U$. Since $F$ commutes with $\s$, one has $\F\mu\in\Ms(\az)$. Moreover if $\mu\in\Merg(\az)$, then $\F\mu\in\Merg(\az)$ as well. This allows to define the following action:
\[
\begin{array}{rccc}
\F:&\Ms(\az)&\longrightarrow&\Ms(\az)\\
&\mu&\longmapsto &\F\mu.
\end{array}
\]
We consider the set of cluster points of the sequence $(\F^t\mu)_{t\in\N}$ called the \define{limit measures set} and denoted by $\V(F,\mu)$. The closure of the union of the supports of these measures is called the \define{$\mu$-limit set} and it is denoted by $\Lambda_{\mu}(F)$. Equivalently, it can be defined as the subshift
\[\Lambda_{\mu}(F)=\left\{x\in\az : u\in\A^{\ast} \textrm{ does not appear in $x$ if and only if }\lim_{t\to\infty}\F^t\mu([u])=0\right\}.\]
\subsection{A particle-based self-organisation result}\label{sec:selforg}
Define the \define{frequency} with which the pattern $u$ appears in the configuration $x$ as
\[\freq(u,x)=\limsup_{n\to\infty}\frac{\card\{i\in[-n,n]:x_{[i,i+|u|-1]}=u\}}{2n+1}.\]
Similarly we define $\freq(S,x)$ where $S$ is a set of patterns.
We introduce the following notations for all the subsequent proofs. For $n\in\N$, let $\Ball_n$ be the set $[-n,n]\subset \Z$.
Let $F:\az\to\az$ be a cellular automaton. In the context of a fixed particle system $(\P,\pi,\phi)$, we
introduce the \define{densities of particles} in a configuration $x\in\az$:
\[\mbox{for }p\in\P,\ \D_p(x) = \freq(p,\pi(x))\quad\mbox{and}\quad \D(x) = \freq(\P,\pi(x));\]
\[\D_\prog(x) = \limsup_{t\to\infty}\frac 1{2t+1}|\prog(x)\cap\Ball_t|\quad\mbox{ and similarly for
}\D_\inter(x),\]
the last two definitions applying only if the particle system is coalescent.
For $\mu\in\Merg(\az)$, the $\limsup$ can be replaced by a simple limit in the definition of frequency for $\mu$-almost all configurations, this is the Birkhoff's ergodic theorem.
This implies for example that $\D(x) = \sum_{p\in\P}\D_p(x)$ for $\mu$-almost all $x$.
First of all, the following proposition clarifies how controlling the frequency of interactions gives us information about the evolution of the density of the different kinds of particles.
\begin{proposition}[Evolution of densities]\label{prop:technical}
Let $F : \az\to\az$ be a cellular automaton, $\mu\in\Merg(\az)$, and $(\P,\pi,\phi)$ a coalescent particle system for $F$, we denote $r$ the radius of the update function $\phi$. Then, for $\mu$-almost all
$x\in\az$:
\begin{enumerate}
\item $\D(F(x))\leq \D(x) - \frac{1}{r+1}\D_\inter(x)$;
\item $\forall p\in\P, \D_p(F(x))\leq \D_p(x) + \D_\inter(x)$.
\end{enumerate}
\end{proposition}
\begin{proof} (i) By surjectivity of the update function, we have $\p(F(x))=\bigcup_{k\in\p(x)}\phi(x,k)$.
Furthermore, by locality, \begin{align*}\forall x\in\az,\ \forall n\in\N,\
\p(F(x))\cap \Ball_n \subseteq& \bigcup_{k\in\p(x)\cap\Ball_{n+r}}\phi(x,k)\\
&\subseteq\bigcup_{k\in\prog(x)\cap\Ball_{n+r}}\phi(x,k)\quad \sqcup \bigcup_{k\in\inter(x)\cap\Ball_{n+r}}\phi(x,k).
\end{align*}
The second line being obtained by coalescence: since $\p(x) = \prog(x)\sqcup \inter(x)$,
particles in $F(x)$ are either images of progressing particles or of interacting particles.
By disjunction:
\begin{align*}
\forall x\in\az,\ \left|\bigcup_{k\in\prog(x)\cap\Ball_{n+r}}\phi(x,k)\right| &= |\prog(x)\cap\Ball_{n+r}|\\
\mbox{and }\quad\forall x\in\az,\ \left|\bigcup_{k\in\inter(x)\cap\Ball_{n+r}}\phi(x,k)\right| &\leq
\frac{r}{r+1}\left|\phi^{-1}\left(\bigcup_{k\in\inter(x)\cap\Ball_{n+r}}\phi(x,k)\right)\right| \\
&\leq \frac{r}{r+1}\left|\inter(x)\cap\Ball_{n+2r}\right|.
\end{align*}
This first equality is because progressing particles are ``one-to-one''. The ratio $\frac r{r+1}$ is due to the condition of coalescence plus the remark that $|\phi(x,k)| +
|\phi^{-1}(\phi(x,k))|\leq 2r+2$. The last inequality is by locality.
\[
\forall x\in\az,\ |\p(F(x))\cap \Ball_n| \leq |\prog(x)\cap\Ball_{n+r}| + \frac{r}{r+1}\left|\inter(x)\cap\Ball_{n+2r}\right|.\]
Then, passing to the limit:
\[\forall_\mu x\in\az,\ \D(F(x))\leq \D_\prog(x)+\frac{r}{r+1}\D_\inter(x) =
\D(x)-\frac1{r+1}\D_\inter(x).\]
(ii) Similarly, for any particle $p\in\P$, one has for all $x\in\az$ and $n\in\N$:
\[\{k\in\Ball_n\ |\ \pi(F(x))_k=p\} \subseteq
\bigcup_{k\in\p(x)\cap\Ball_{n+r}}\phi(x,k)\quad\mbox{(locality)}.\]
For $k\in\prog(x)$, if $\pi(F(x))_{\phi(x,k)} = p$, then by definition of coalescence $\pi(x)_k = p$.
For $\mu$-almost all $x$, using $\p(x) = \prog(x)\sqcup \inter(x)$, we conclude that $\D_p(F(x))\leq \D_p(x)+\D_{inter}(x)$
by passing to the limit.
\end{proof}
We state our main result. A simple version (Corollary~\ref{cor:MainResult}) states that in a coalescent particle
system with a $\s$-ergodic initial measure, if all particles can be assigned a speed, then only particles with one fixed
speed can remain asymptotically. The more general result is designed to handle more difficult cases such as particles
performing random walks, as we can see on the last example of Section~\ref{sec:particlesexamples}.
\begin{definition}[Clashing]
Let $F:\az\to\az$ be a cellular automaton, $(\P,\pi,\phi)$ a coalescent particle system for $F$, and $\P_1$ and $\P_2$ two subsets
of $\P$. We say that $\P_1$ \define{clashes with} $\P_2$ $\mu$-almost surely if, for every $n\in\N^\ast$ and $\mu$-almost
all $x\in\az$,
\[\pi(x)_0\in\P_1 \mbox{ and }\pi(x)_n\in\P_2 \Longrightarrow \exists t\in\N,\phi^t(x,0)\in\inter(F^t(x))\mbox{ or }\phi^t(x,n)\in\inter(F^t(x))\]
\end{definition}
The intuition behind clashing particles in the following: if two clashing particles are present, then they end up interacting (almost surely) and thus decreasing
the global frequency of particles. This is why they cannot both persist asymptotically. Note that clashing is oriented left to right: particles with speed +1
clash with particles of speed -1, but the converse is not true.
\begin{theorem}[Main qualitative result]\label{prop:MainResult}
Let $F:\az\to\az$ be a cellular automaton, $\mu$ an initial $\s$-ergodic measure and $(\P,\pi,\phi)$ a coalescent
particle system for $F$ where $\P$ can be partitioned into sets $\P_1\dots \P_n$ such that, for every $i<j$, $\P_i$
clashes with $\P_j$ $\mu$-almost surely.
Then all particles appearing in the $\mu$-limit set belong to the same subset, i.e. there exists a $i$ such that
\[\forall p\in\P,\ p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow p\in\P_i.\]
If furthermore there exists a $j$ such that $\P_j$ clashes with itself $\mu$-almost surely, then this set of particles
does not appear in the $\mu$-limit set, i.e.
\[\forall p\in\P,\ p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow p\notin\P_j.\]
\end{theorem}
\begin{corollary}[Version with speedy particles]\label{cor:MainResult}
Let $F:\az\to\az$ be a cellular automaton, $\mu$ an initial $\s$-ergodic measure and $(\P,\pi,\phi)$ a coalescent
particle system for $F$.
If each particle $p\in\P$ has speed $v_p\in\R$,
then there is a speed $v\in\R$ such that:
\[\forall p\in\P, p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow v_p=v.\]
\end{corollary}
\begin{proof}[Proof of Theorem~\ref{prop:MainResult}]
Let $i=1,\, j=2$ for clarity and assume there are two particles $p_1\in\P_1, p_2\in\P_2$ appearing in
$\L(\pi(\Lambda_\mu(F)))$. By definition, this means that $\pi_\ast\F^t\mu([p_i]) \underset{t\to\infty}\nrightarrow 0$
for $i\in\{1,2\}$.\sk
For all $x\in\az$, $(\D(F^t(x)))_{t\in\N}$ is a decreasing sequence of positive reals. For $t\in\N$, by
Birkhoff's ergodic theorem applied to $\pi_\ast\F^t\mu$, we have for $\mu$-almost all~$x$
$\D(F^t(x)) = \pi_\ast\F^t\mu([\P])$. Therefore, for $\mu$-almost all $x$, $(\D(F^t(x)))_{t\in\N} =
(\pi_\ast\F^t\mu([\P]))_{t\in\N}$ and therefore $\pi_\ast\F^t\mu([\P])\underset{t\to\infty}\longrightarrow d_\infty \geq 0$.
For $x\in\az$, denote $\D_{\P_i}(x) = \freq(\P_i, \pi(x))$. By the first point of Proposition~\ref{prop:technical},
we can see that for $\mu$-almost all $x\in\az$,
\[\sum_{t\in\N}\D_\inter(F^t(x)) \leq (r+1)\left(\sum_{t\in\N}\D(F^{t+1}(x))-\D(F^t(x))\right)\leq (r+1)(\D(x)-d_\infty) < +\infty.\]
Again by Birkhoff's theorem, $(\D_{\P_i}(F^t(x)))_{t\in\N} = (\pi_\ast\F^t\mu([\P_i]))_{t\in\N}$ for $\mu$-almost
all $x$. By the second point of Proposition~\ref{prop:technical},
\[\mbox{For }i\in\{1,2\},\ \sup_{n\in\N}|\D_{\P_i}(F^{t+n}(x))-\D_{\P_i}(F^t(x))|\leq \sum_{n=0}^\infty \D_\inter(F^{t+n}(x))\to 0.\]
Thus $\pi_\ast\F^t\mu([\P_i])$ is a Cauchy sequence and admits a limit $d_i\neq 0$.\sk
Since clashing particles are present with positive frequency, they generate interactions that decrease the global density of particles.
We will reach a contradiction with the fact that the global density tends to a limit.\sk
Fix $\varepsilon < \frac{d_1\cdot d_2}{2r+5}$ and $T$ large enough such that for $t\geq T,\
\pi_\ast\F^t\mu([\P])-d_\infty<\varepsilon$
and $|\pi_\ast\F^t\mu([\P_i])-d_i| < \varepsilon$ for $i\in\{1,2\}$.
By Birkhoff's ergodic theorem applied on $\pi_\ast\F^T\mu$, we have:
\[\frac 1K\sum_{k=0}^K\pi_\ast\F^T\mu\left([p_1]_0\cap[p_2]_k\right) \underset{K\to\infty}\longrightarrow
\pi_\ast\F^T\mu([p_1])\cdot\pi_\ast\F^T\mu([p_2]),\]
and $\pi_\ast\F^T\mu([p_1])\cdot\pi_\ast\F^T\mu([p_2]) \geq d_1\cdot d_2-(d_1+d_2-\varepsilon)\varepsilon\geq d_1\cdot d_2-2\varepsilon$.
In particular, one can find a $k$ such that $\pi_\ast\F^T\mu([p_1]_0\cap[p_2]_k)>d_1\cdot d_2-3\varepsilon$.
By Birkhoff's theorem, this means that words of $V_k = p_1(\P\cup\{0\})^{k-1}p_2\subset(\P\cup\{0\})^\ast$
have frequency at least $d_1\cdot d_2-3\varepsilon$ in $F^T(x)$, for $\mu$-almost all $x\in\az$.\sk
Since $\P_1$ and $\P_2$ clash $\mu$-almost surely, any occurrence of $V_k$ yields an interaction:
\begin{align*}
\forall_\mu x\in\az,\ \D(F^T(x))-d_\infty &\geq \frac 1{r+1} \sum_{t=T}^\infty \D_\inter(F^t(x))&\mbox{Proposition~\ref{prop:technical}(i)}\\
&\geq \frac 1{2r+2}\freq(V_k,F^T(x))&\mbox{$\P_1$ and $\P_2$ clash $\mu$-almost surely}\\
&\geq \frac 1{2r+2}(d_1\cdot d_2-3\varepsilon)>\varepsilon,
\end{align*}
which is a contradiction with the definition of $\varepsilon$.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{cor:MainResult}]
Consider the set of speeds $\{v_p\ :\ p\in\P\}$ and order it as $v_1>v_2> \dots>v_n$. Now partition the set of particles
into $(\P_{v_i})_{0\leq i\leq n}$ where $\P_{v_i}$ is the set of particles with speed $v_i$, and apply the theorem.
We only need to show that for any $i<j$, $\P_{v_i}$ clashes with $\P_{v_j}$ $\mu$-almost surely. Let
$p_i\in\P_{v_i}$ and $p_j\in\P_{v_j}$, and $x\in\az$ such that $\pi(x)_0=p_i$ and $\pi(x)_n=p_j$ for some $n\in\N^\ast$. If
both particles satisfy the second property in the definition of speed (Progression at speed $v$), then for some $t$
large enough we have $\phi^t(x,0)>\phi^t(x,n)$, which is forbidden by coalescence since two particles in progression
cannot cross.
Thus at some time $t$ we have either $\phi^t(x,0)\in\inter(F^t(x))$ or $\phi^t(x,n)\in\inter(F^t(x))$.
\end{proof}
\subsection{Defects}\label{sec:defects}
Before giving a series of examples where this result can be used to describe the typical asymptotic behavior of a cellular automaton, we
present the formalism introduced by Pivato in~\cite{Pivato-2007-algebra,Pivato-2007-spectral} that defines particles as defects with
respect to a $F$-invariant subshift $\Sigma$. Indeed, this formalism gives us an easier way to find the particle systems
in our examples.
Intuitively, the $F$-invariant subshift describes the homogeneous regions that persist under the action of $F$ in the space-time diagram, and defects are the borders between these regions. This allows us to define $\P$ and $\pi$ in a way that corresponds to the intuition, even though it gives no information on the dynamics of the particles (the update function $\phi$).
\subsubsection{General definitions}
For a cellular automaton $F$, consider $\Sigma$ a $F$-invariant subshift. The \define{defect field} of $x\in\az$ with
respect to $\Sigma$ is defined as:
\[\mathcal{F}^{\Sigma}_x:\begin{array}{ccl}
\Z&\to&\N\cup\{\infty\}\\
k&\mapsto&\displaystyle \max \left\{n\in \N : x_{k+[-\lfloor\frac n2\rfloor, \lceil\frac
n2\rceil]}\in \L_n(\Sigma)\right\}
\end{array}
,\]
where the result is possibly $0$ or $\infty$ if the set is empty or infinite. Intuitively, this function returns the size of the largest word admissible for $\Sigma$ centred on the cell $k$. A \define{defect} in a configuration $x$ relative to
$\Sigma$ is a local minimum of $\mathcal{F}^{\Sigma}_x$. Then the interval $[k,l]$ between two defects forms a
homogeneous region in the sense that $x_{[k+1,l]}\in \mathcal L(\Sigma)$.\sk
However, it is not true that we can always make a correspondence between defects and a finite set of words (forbidden
patterns), so as to obtain a finite set of particles and a morphism. This is the case only when the set of
forbidden patterns is finite, that is, when $\Sigma$ is a SFT. In this case, a defect corresponds to the centre of the occurrence of a forbidden word.
This is a limitation of our result.\sk
The examples given in Figure~\ref{fig:particles} suggest that defects can usually be classified using one of these approaches:
\vspace{-.4\baselineskip}
\begin{itemize}
\itemsep0em
\item Regions correspond to different subshifts and defects behave according to their surrounding regions
(\define{interfaces} - e.g. cyclic automaton);
\item Regions correspond to the same periodic subshift and defects correspond to a ``phase change''
(\define{dislocations} - e.g. rule 184 automaton).
\end{itemize}
\subsubsection{Interfaces}
Let $\Sigma$ be a SFT and assume $\Sigma$ can be decomposed as a disjoint union $\Sigma_1 \sqcup \dots \sqcup \Sigma_n$
of $F$-invariant $\s$-transitive SFTs (the \define{domains}). The intuition is that between two defects, each region
belongs to the language of only one of the domains, and we can classify defects according to which domain the
regions surrounding them on the left and the right correspond to. Since each domain is $F$-invariant, this
classification is conserved under the action of $F$ for non-interacting defects.
Formally, since the different domains $(\Sigma_k)_{k\in[1,n]}$ are disjoint SFTs, there is a length $\alpha>0$
such that $(\L_{\alpha}(\Sigma_k))_{k\in[1,n]}$ are disjoint where $\L_{\alpha}(\Sigma_k)$ is the set of length $\alpha$ of $\Sigma_k$ (if two subshifts share arbitrarily long words, they share a configuration by closure).
In particular, if $u\in\L_{\alpha}(\Sigma)$, then there is a unique $k$
such that $u\in\L(\Sigma_k)$: we say that $u$ \define{belongs to the domain $k$}. Thus, for a given configuration, we
can assign a choice of a domain to each homogeneous region between two consecutive defects, and this choice is
unique if this region is larger than $\alpha$.
\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}
\foreach \x/\colour/\k/\f in
{-1/black/1/7,0/black/1/5,1/black/1/3,2/black/1/1,3/white/0/2,4/white/0/4,5/white/0/5,6/white/0/3,7/white/0/1,8/gray/2/2
,9/gray/2/4,10/gray/2/5,11/gray/2/3, 12/gray/2/1, 13/white/0/2, 14/white/0/4, 15/white/0/6}
{
\filldraw[draw = black] [fill=\colour!80] (0.7*\x, 0) rectangle (0.7*\x+0.7,0.7);
\draw (0.7*\x+0.35, 0.95) node {\small \k};
\draw (0.7*\x+0.35, -0.25) node {\small \f};
}
\draw [draw = red,thick] (2.1,-0.4) -- (2.1,1);
\draw [draw = red, thick] (5.6,-0.4) -- (5.6,1);
\draw [draw = red, thick] (9.1,-0.4) -- (9.1,1);
\draw[red] (2.1, 1.4) node {$1-0$};
\draw[red] (5.8, 1.4) node {$0-2$};
\draw[red] (9.3, 1.4) node {$2-0$};
\draw (-1.4, -0.25) node {$\mathcal F_a(x)$};
\draw (-1.8, 0.35) node {$x$};
\draw (-1.4, 1) node {domain};
\draw (-1.1, 0.35) node {\dots};
\draw (11.6, 0.35) node {\dots};
\draw [pattern = north east lines] (1.4,0) rectangle (2.1,0.7);
\draw [pattern = north east lines] (4.9,0) rectangle (5.6,0.7);
\draw [pattern = north east lines] (8.4,0) rectangle (9.1,0.7);
\end{tikzpicture}
\caption{Interfaces between monochromatic domains, marked by slanted patterns. Red lines show the
visual intuition of a domain change.}\label{fig:defect}
\end{center}
\end{figure}
We call these defects \define{interface defects} and we can classify them according to the domain of the surrounding
regions. Let $\P = \{p_{ij}\ :\ (i,j)\in[1,n]^2\}$ be the set of particles.
Define the morphism $\pi : \az \to (\P\cup\{0\})^\Z$ of order $\max(r,2\alpha)$, where $r$ is the radius of $\Sigma$, in the following way.
For $x\in\az$ and $k\in\Z$:
\begin{itemize}
\itemsep0em
\item if $x_{k+[-\lfloor\frac r2\rfloor, \lceil\frac r2\rceil]}\in\L(\Sigma)$, then $\pi(x)_k=0$;
\item else, let $\left\{\begin{array}{l}
u_1 = x_{[k-m,k]}$ where $m = \max\{n\leq \alpha\ :\ x_{[k-n,k]}\in\L(\Sigma)\}\\
u_2 = x_{[k+1,k+m]}$ where $m = \max\{n\leq \alpha\ :\ x_{[k+1,k+m]}\in\L(\Sigma)\}\\
d_i \mbox{ a domain to which }u_i\mbox{ belongs }(i\in\{1,2\})
\end{array}\right.$
and put $\pi(x)_k=p_{d_1d_2}$.
\end{itemize}
Notice that the domain choice (choice of $d_i$) is unique when domains are larger than $\alpha$ cells; otherwise, the choice between the possible $d_i$ is arbitrary, or fixed beforehand.
\subsubsection{Dislocations}
Contrary to interface defects that mark a change between languages of different SFT, \emph{dislocation defects} mark a ``change of phase'' inside a single SFT.
Let $\Sigma$ be a $\s$-transitive SFT of order $r>1$. We say that $\Sigma$ is \define{$P$-periodic} if there exists a
partition $V_1,\dots,V_P$ of $\L_{r-1}(\Sigma)$ such that
\[a_1\cdots a_r\in\L_r(\Sigma)\quad\Leftrightarrow \quad \exists i\in\Z /P\Z,\
a_1\cdots a_{r-1}\in V_i\mbox{ and }a_2 \cdots a_r\in V_{i+1}.\]
The \define{period} of $\Sigma$ is the maximal $P\in\N$ such that $\Sigma$ is $P$-periodic. For example, the orbit of a
finite word $u\in\A^\ast$, defined as $\{\s^k(\mathstrut^\infty u^\infty) : k\in\Z\}$ is a periodic SFT of period less than $|u|$.
We thus associate to each $x\in\Sigma$ its \define{phase} $\varphi(x) \in\Z/P\Z$ such that $x_{[0,r-2]}\in
V_{\varphi(x)}$. Obviously, $\varphi(\s^k(x)) = \varphi(x)+k\mod p$. For $x\in\az$, we say that an homogeneous region
$[a,b]$ (i.e. a region such that $x_{[a,b]}\in\Sigma$) is \define{in phase $k$} if $\exists y\in\Sigma, \varphi(y)=k,
x_{[a,b]}=y_{[a,b]}$. If $b-a>r-2$, the phase of a region is unique and means $x_{[a,a+r-2]}\in V_{k+a\mod p}$.
\begin{figure}[!ht]
\begin{center}
\begin{tikzpicture}
\foreach \x/\colour/\k/\f in {-1/black/1/7,
0/white/1/5,1/black/1/3,2/white/1/1,3/white/0/2,4/black/0/4,5/white/0/6,6/black/0/5,7/white/0/3,8/black/0/1,9/black/1/2,
10/white/1/4,11/black/1/5,12/white/1/3,13/black/1/1, 14/black/0/2, 15/white/0/4, 16/black/0/6}
{
\filldraw[draw = black] [fill=\colour!80] (0.7*\x, 0) rectangle (0.7*\x+0.7,0.7);
\draw (0.7*\x+0.35, 0.95) node {\small \k};
\draw (0.7*\x+0.35, -0.25) node {\small \f};
}
\draw [draw = red,thick] (2.1,-0.4) -- (2.1,1.1);
\draw[red] (2.1, 1.4) node {$1-0$};
\draw [draw = red, thick] (6.3,-0.4) -- (6.3,1.1);
\draw[red] (6.3, 1.4) node {$0-1$};
\draw [draw = red, thick] (9.8,-0.4) -- (9.8,1.1);
\draw[red] (9.8, 1.4) node {$0-1$};
\draw (-1.4, -0.25) node {$\mathcal{F}_a(x)$};
\draw (-1.6, .35) node {$x$};
\draw (-1.4, 1) node {phase};
\draw (-1.1, 0.35) node {\dots};
\draw (12.3, 0.35) node {\dots};
\draw [pattern = north east lines] (1.4,0) rectangle (2.1,0.7);
\draw [pattern = north east lines] (5.6,0) rectangle (6.3,0.7);
\draw [pattern = north east lines] (9.1,0) rectangle (9.8,0.7);
\end{tikzpicture}
\end{center}
\caption{Dislocations in the checkerboard subshift ($P=2$), marked by slanted patterns. Red lines show the visual
intuition of a change of phase, with the surrounding local phases.}
\label{fig:dislo}
\end{figure}
As we can see in Figure~\ref{fig:dislo}, the
finite word corresponding to a defect (here $00$ or $11$) does not depend only on the
phase of the surrounding region but also on the position of the defect. More precisely, since $\varphi(\s(x)) =
\varphi(x)+1$, a defect in position $j$ with a region in phase $f$ to its left and a defect in position $0$ with a
region in phase $f+j\mod P$ to its left ``observe'' the same finite word to their left.
Therefore, we define for each defect its \define{local phases} $\varphi([i,j])+ j\mod P$ (left) and
$\varphi([j,k])+j\mod P$ (right), where $j$ is the position of the defect and $[i,j]$ and $[j,k]$ are the surrounding
homogeneous regions.
Now we classify the defects according to the local phase of the surrounding regions. Let $\P = \{p_{ij}\ :\
(i,j)\in\Z/P\Z^2\}$ be the set of particles. Since defects correspond to the centre of occurrences of forbidden words
and the phase of a region can be locally distinguished, the morphism $\pi : \az \to (\P\cup\{0\})$ of order $2r-2$
is defined exactly as in the interface case. The choice of local phase is unique if the region is larger than $r-1$ cells.\sk
In the general case, those two formalisms can be mixed by fixing a decomposition $\Sigma = \bigsqcup_{i\in\A}\Sigma_i$
where some of the $\Sigma_i$ have nonzero periods. We can classify defects according to the domains and local phase of
the surrounding regions in a similar manner. Except for arbitrary choices for small regions,
obtaining the set of particles and the morphism from the SFT decomposition can be done in an entirely automatic way.
\subsection{Examples}\label{sec:particlesexamples}
\subsubsection{Rule 184}\label{section.184}
We consider the rule $\#184$ or ``traffic'' automaton $F_{184}:\{0,1\}^{\Z}\to\{0,1\}^{\Z}$ defined by the following local rule:
$f_{184}(x_{-1}x_0x_1)=1$ if and only if $x_0x_1=11$ or $x_{-1}x_0=10$.
This time evolution of this automaton can be seen as a road where the symbol $1$ represent vehicles and the symbol $0$ an empty space. The vehicles go forward if the cell just before them is empty and stay put otherwise. In this context, the rule $\#184$ has been very well studied, especially in the case of initial Bernoulli measures \cite{BelitskyFerrari-1995, Belitsky-2005}. We use this example mostly as a simple case to better understand the formalism, althought our method has the advantage to hold for more general probability measures.
\begin{proposition}
Let $F_{184}$ be the traffic automaton and $\mu \in \Merg$. Then:
\begin{align*}
\mu([00])>\mu([11])&\Rightarrow 11\notin \Lambda_\mu(F_{184});\\
\mu([00])<\mu([11])&\Rightarrow 00\notin \Lambda_\mu(F_{184});\\
\mu([00])=\mu([11])&\Rightarrow \Lambda_\mu(F_{184}) = \{\mathstrut^\infty 01^\infty, \mathstrut^\infty 10^\infty\}.
\end{align*}
\end{proposition}
\begin{figure}[!ht]
\begin{center}
\includegraphics[width = 7cm]{Factr.png}$\begin{array}{c}\to\vspace{3cm} \end{array}$\includegraphics[width =
7cm]{Factgr.png}\vspace{-1.2cm}
\end{center}
\caption{Particle system for the traffic automaton.}
\label{fig:Factors}
\end{figure}
\begin{proof}
We consider the checkerboard SFT $\Sigma = \{\per{(01)},\per{(10)}\}$, which is $2$-periodic and $F_{184}$-invariant. Using the
dislocation formalism, we define the phases $\varphi(\per{(01)}) = 0$ and
$\varphi(\per{(10)}) = 1$, obtaining a set of particles defined by their local phases $\{p_{01}, p_{10}\}$.
The corresponding morphism of order $r=2$ is defined by the local rule:
\[\begin{array}{ccc}
00&\to& p_{01}\\
11&\to& p_{10}\\
\mbox{otherwise}&\to& 0
\end{array}.
\]
Indeed, consider $x\in\az$ with a defect $x_{01} = 00$. The phase of the $0$ in position $0$ is $0$ and the phase of the $0$ in position $1$ is $1$,
so this corresponds to a particle $p_{01}$. Changing the position of the defect would not change the particle since the local phase would be
modified accordingly.
The update function is defined in the intuitive manner: with $p_{01}$ evolving at speed $-1$ and $p_{10}$ at speed $+1$
and both particles being sent to $\emptyset$ in case of collision.
\[\forall x\in\az,\forall k\in\Z, \phi(x,k)=\left\{\begin{array}{cl}
\{k+1\}&\mbox{if }\pi(x)_k = p_{10}\mbox{ and }\pi(x)_{k+1}\neq p_{01} \mbox{ and }\pi(x)_{k+2}\neq p_{01}\\
\{k-1\}&\mbox{if }\pi(x)_k = p_{01}\mbox{ and }\pi(x)_{k-1}\neq p_{10} \mbox{ and }\pi(x)_{k-2}\neq p_{10}\\
\emptyset&\text{otherwise (and in particular if }\pi(x)_k=0)\end{array}\right.\]
We now check that the particle system satisfies all necessary conditions. To do that, one should verify that the update
function is defined properly, that is, show that
for all $x\in\az$ and $k\in\Z$ we have: \[\pi(F(x))_{k+1}=p_{10}\Leftrightarrow \pi(x)_k = p_{10}\mbox{ and }\pi(x)_{k+1}\neq p_{01}
\mbox{ and }\pi(x)_{k+2}\neq p_{01},\]
and similarly for $p_{01}$. This is tedious due to the high number of cases but can be easily automated by enumerating all patterns of length $4$.
The different conditions follow from this claim:
\begin{description}
\itemsep0em
\item[Locality] Obvious by definition of $\phi$.
\item[Surjectivity] The $(\Rightarrow)$ direction of the claim implies that $\pi(F(x))_{k+1}=p_{10}
\Rightarrow \pi(x)_k=p_{10}$ and $\{k+1\} = \phi(x,k)$. The other cases are similar.
\item[Particle control] The first condition is simply the $(\Leftarrow)$ direction of the claim. The second condition
is by definition of $\phi$.
\item[Disjunction] For $k<k'$, to have $\phi(x,k)>\phi(x,k')$, the only way would be to have $\pi(x)_{k'}=p_{01}$,
$\pi(x)_k=p_{10}$ and $k'=k+1$. In that case, by definition, $\phi(x,k)=\phi(x,k')=\emptyset$.
\item[Coalescence and speeds] Obvious by definition of $\phi$.
\end{description}
Therefore we can apply Corollary~\ref{cor:MainResult} and only one type of particle remains in
$\Lambda_\mu(F_{184})$.\sk
Furthermore, since the collisions are of the form $p_{01} + p_{10} \to \emptyset$, it is clear that for all $x\in\az$,
$\D_{p_{01}}(F_{184}(x)) - \D_{p_{01}}(x)=\D_{p_{10}}(F_{184}(x)) - \D_{p_{10}}(x)$. Therefore, which particle remains
is decided according to whether $\mu([00])>\mu([11])$ or the opposite, both particles disappearing in case of equality.
\end{proof}
\subsubsection{$n$-state cyclic automaton}\label{section.cyclic}
The $n$-state cyclic automaton $C_n$ is a particular captive cellular automaton defined on the alphabet $\A = \Z/n\Z$ by
the local rule \[c_n(x_{i-1}, x_i, x_{i+1}) = \left\{\begin{array}{ll}x_i+1&\mbox{if }x_{i-1} = x_i+1 \mbox{ or
}x_{i+1}= x_i+1;\\x_i&\mbox{otherwise.}\end{array}\right.\]
See Figure~\ref{fig:particles} for an example of space-time diagram.\sk
This automaton was introduced by~\cite{Fisch}. In this paper, the author shows that for all Bernoulli measure $\mu$, the
set $[i]_0$ (for $i\in\A$) is a $\mu$-attractor iff $n\geq5$. Simulations starting from a random configuration suggest
the following: for $n=3$ or $4$, monochromatic regions keep increasing in size; for $n\geq 5$, we observe the
convergence to a fixed point where small regions are delimited by vertical lines. We use the main result
to explain this observation.
\begin{proposition}
Define:
\begin{align*}
u_+ = \{ab\in\A^2 : (b-a)\mod n = +1\};\\
u_- = \{ab\in\A^2 : (b-a)\mod n = -1\};\\
u_0 = \{ab\in\A^2 : (b-a)\mod n \neq \pm1\}.
\end{align*}
Then, for any measure $\mu\in\Merg((\Z/nZ)^\Z)$, only one of those three sets may intersect the language of
$\Lambda_\mu(C_n)$.\sk
If furthermore $\mu$ is a Bernoulli measure, then the persisting set can only be $u_0$.
\end{proposition}
\begin{proof}
We consider the interface defects relatively to the decomposition $\Sigma = \bigsqcup_{i\in\A}\Sigma_i$, where $\Sigma_i
= \{\mathstrut^{\infty}i^{\infty}\}$. $\Sigma$ is a $C_n$-invariant SFT of order $r=2$, and defects are exactly
transitions between
colours. Thus we define $\P = \{p_{ab}\ :\ ab\in\A^2\}$. One cell is enough to distinguish between
domains ($\alpha=1$) and we obtain a morphism $\pi$ of order $2$ defined by the local rule:
\[ \begin{array}{ccc}
\A^2&\to&\P\cup\{0\}\\
a\cdot a&\mapsto& 0\\
a\cdot b&\mapsto& p_{ab}
\end{array}\quad\quad\mbox{for all }a,b\in\A.
\]
Simulations suggest that $p_{ab}$ evolves at constant speed $+1$ if $ab\in u_+$, $-1$ if $ab\in u_-$ and
$0$ if $ab\in u_0$. Particles progress at their assigned speed unless they meet another particle, in which case they collide
and disappear. We group together the particles of same speed, writing $p_+ = \{p_{ab}\ :\ ab\in u_+ \}$ and $p_-$ and
$p_0$ similarly. Formally, for $x\in\az$ and $k\in\Z$ the update function is defined as:
\[ \phi(x,k)=\left\{\begin{array}{cl}
\{k+1\}&\mbox{if }\pi(x)_k \in p_+\mbox{ and }\pi(x)_{k+1}\notin p_0\cup p_- \mbox{ and }\pi(x)_{k+2}\notin p_-;\\
\{k-1\}&\mbox{if }\pi(x)_k \in p_-\mbox{ and }\pi(x)_{k-1}\notin p_0\cup p_+ \mbox{ and }\pi(x)_{k-2}\notin p_+;\\
\{k\}&\mbox{if }\pi(x)_k \in p_0\mbox{ and }\pi(x)_{k+1}\notin p_- \mbox{ and }\pi(x)_{k-1}\notin p_+;\\
\emptyset&\text{otherwise (and in particular if }\pi(x)_k=0).\end{array}\right.\]
As previously, checking that this particle system satisfies all conditions necessary to apply
Corollary~\ref{cor:MainResult} is tedious but can be automated, since it consists mostly in checking that the update function
actually describes the dynamics of the particles on all words of length $6$. Since $[p_+] = \pi([u_+])$ and so on, we obtain the result.
\textbf{If $\mu$ is a Bernoulli measure:} Consider the ``mirror'' application $\gamma((a_k)_{k\in\Z}) = (a_{-k})_{k\in\Z}$.
$\gamma$ is continuous, and thus measurable. We have
$\mu(\gamma([u])) = \mu([u^{-1}]) = \mu([u])$, where $(u_1\cdots u_n)^{-1} = u_n\cdots u_1$. But $\pi(x)_k\in
p_+\Leftrightarrow \pi(\gamma(x))_{-k}\in p_-$, and conversely; since $F\circ\gamma =
\gamma\circ F$, all measures $\F^t\mu$ are $\gamma$-invariant, and thus no particle in $p_+$ or $p_-$ can persist in
$\L(\pi(\Lambda_{\mu}(F)))$ (since otherwise, the symmetrical particle would persist too).
\end{proof}
For small values of $n$ or particular initial measures, this proposition can be refined in the following manner:
\begin{description}
\item[$\bm{n=3}$] $p_0$ is empty. Given the combinatorics of collisions, where a particle in $p_+$ can only
disappear by colliding with a particle in $p_-$, we see that particles in $p_+$ persist if and only if
$\pi_\ast\mu([p_+])>\pi_\ast\mu([p_-])$, and symmetrically. In the equality case (in particular, for any Bernoulli
measure), no defect can persist in the $\mu$-limit set, which means that $\Lambda_{\mu}(F)$ is a set of
monochromatic configurations.
\item[$\bm{n=4}$] If $\mu$ is a Bernoulli measure, the result of~\cite{Fisch} shows that $[i]_0$ cannot be a
$\mu$-attractor for any $i$. In other words, for $\mu$-almost all $x$, $F^t(x)$ does not converge, which means that particles in
$p_+$ or $p_-$ cross the central column infinitely often (even though their probability to appear tends to $0$).
This could not happen if particles in $p_0$ were persisting in $\pi(\Lambda_{\mu}(F))$, and thus $\Lambda_{\mu}(F)$ is a set of
monochromatic configurations.
\item[$\bm{n\geq5}$] If $\mu$ is a nondegenerate Bernoulli measure, the result of~\cite{Fisch} shows that $[i]_0$ is a
$\mu$-attractor for all $i$. This means that some particles in $p_0$ persist in $\pi(\Lambda_{\mu}(F))$, and any
configuration of $\Lambda_{\mu}(F)$ contains only homogeneous regions separated by vertical lines.
\end{description}
For $n=3$ or $4$, since $\Lambda_{\mu}(F)$ is a set of
monochromatic configurations we deduce that the sequence $(F^n\mu)_{n\in\N}$ converges to a convex combination of Dirac measures. However this method does not give any insight as to the coefficient of each component. As shown in~\cite{these-benjamin}, if $\mu$ is a Bernoulli measure then \[C_{3\ast}^t\mu \underset{t\to\infty}{\longrightarrow}\mu([2])\meas{0}+\mu([0])\meas{1}+\mu([1])\meas{2}.\]
The problem is open for the 4-cyclic cellular automaton.
\subsubsection{One-sided captive cellular automata}\label{section.captive}
We consider the family of captive cellular automata $F:\az\to\az$ of neighbourhood $\{0,1\}$, which means that the local
rule $f:\A^{\{0,1\}}\to\A$ satisfies $f(a_0a_1)\in\{a_0,a_1\}$. See Figure~\ref{fig:particles} for an example of
space-time diagram.
\begin{proposition}
Let $F$ be a one-sided captive automaton and $\mu\in\Merg(\az)$.
Define:
\begin{align*}
u_+ &= \{ab\in\A^2\ :\ a\neq b,\ f(a,b) = a\}\\
u_- &= \{ab\in\A^2\ :\ a\neq b,\ f(a,b) = b\}
\end{align*}
Then either $u_+\notin\L(\Lambda_\mu(F))$ or $u_-\notin\L(\Lambda_\mu(F))$.\sk
If moreover, for all $a,b\in\mathcal{A}$, the local rule satisfies $f(ab)=f(ba)$ and $\mu$ is a Bernoulli measure,
then $\Lambda_{\mu}(F)\subseteq \{\mathstrut^\infty a^\infty\ :\ a\in\A\}$ (no particle remains).
\end{proposition}
\begin{proof}
We consider the interface defects relative to the decomposition $\Sigma = \bigsqcup_{i\in\A}\Sigma_i$ where $\Sigma_i =
\{\mathstrut^{\infty}i^{\infty}\}$ and obtain the same particles $\P$ and morphism $\pi$ as the $n$-state cyclic
automata. $p_{ab}$ evolve at speed $-1$ if $f(a,b)=b$ and $0$ if $f(a,b)=a$, and we define $p_{-1}$ and $p_0$
accordingly.
The update function is defined as follows:
\[\forall x\in\az,\ \forall k\in\Z,\ \phi(x,k)=
\left\{\begin{array}{cl}
\{k\}&\text{if } \pi(x)_k\in p_0 \mbox{ and } \pi(x)_{k+1}\notin p_{-1}\\
\{k-1\}&\text{if }\pi(x)_k\in p_{-1} \mbox{ and } \pi(x)_{k-1}\notin p_0 \\
\emptyset&\text{otherwise}\end{array}\right.\]
As in the two previous examples, we can check that the update function describes the particles dynamics on all words of length 3,
and deduce the properties of locality, growth, surjectivity, coalescence and speed from there. We then apply the main result.
\textbf{If $\mu$ is a Bernoulli measure:} Then $\mu$ is invariant under the mirror application
$\gamma$ and $F\circ\gamma = \gamma\circ F$ by hypothesis. As in the previous example, we conclude that no particle can persist in $\Lambda_\mu(F)$.
\end{proof}
\subsubsection{An automaton performing random walks}\label{section.randomwalk}
Let $F$ be defined on the alphabet $\A = (\Z/2\Z)^2$ on the neighbourhood $\{-2,\dots, 2\}$ by the local rule $f$
defined as follows:
\[f : (a_{-2},b_{-2}),\dots,(a_2,b_2) \mapsto (a_{-2}+a_2, c)\mbox{ where } c = \begin{array}{l}1 \mbox{ if }(a_{-1},b_{-1}) =
(0,1)\mbox{ or }(a_0,b_0) = (1,1);\\0\mbox{ otherwise.}\end{array}\]
Intuitively, the first layer performs addition mod $2$ at distance 2, while the ones on the second layer behave as particles, moving
right if the first layer contains a 1 and not moving if it contains a 0. Two colliding particles simply merge.
\begin{figure}[!ht]
\begin{center}
\includegraphics[width = \textwidth]{Walks}
\end{center}
\caption{Automaton performing random walks iterated on the uniform measure. $\blacksquare$ is a particle, while the
second layer is represented by $\square$ (0) or $\textcolor{gray}\blacksquare$ (1).}
\end{figure}
\begin{proposition}
Let $\nu\in\Merg((\Z/2\Z)^\Z)$ and $\mu = \lambda\times\nu$, where $\lambda$ is the uniform measure on $(\Z/2\Z)^\Z$. Then $\F^t\mu\underset{t\to\infty}\longrightarrow\lambda\times\meas{0}$.
\end{proposition}
\begin{proof}
Pivato's formalism is not necessary here. Consider the set of particles $\P = \{1\}$ and the morphism $\pi$ that is the projection on the second layer.
The update function is defined as:
\[\forall x\in\az,\ \forall k\in\Z,\ \phi(x,k) = \left\{\begin{array}{cc}
\{k+1\}&\mbox{if }x_k = (1,1);\\
\{k\}&\mbox{if }x_k = (0,1);\\
\emptyset&\mbox{otherwise.}
\end{array}\right.
\]
Intuitively, each particle performs a random walk with independent steps and no bias. Thus
Corollary~\ref{cor:MainResult} is not sufficient to conclude, and we need to use the general result of
Theorem~\ref{prop:MainResult} by proving that $\{1\}$ clashes with itself.\sk
Let $k\in\N$. We prove that, when $x$ is chosen according to $\mu_k$ the conditional measure of $\mu$ relative to the
event $\pi(x)_0=\pi(x)_k=1$, $\phi^t(x,k)-\phi^t(x,0)$ performs an unbiased and independent random walk with a ``death condition'' on 0
(particle collision).
Writing $(a_n^t,b_n^t) = F^t(x)_n$, we have $a_0^t = \sum_{n=0}^t\binom nta_{-2t+4n}^0\mod 2$ by
straightforward induction.
Consider the evolution of $\phi^t(x,k)-\phi^t(x,0)$ at each step:
\begin{align*}
\delta_t(x) &= (\phi^{t+1}(x,k)-\phi^t(x,k))-(\phi^{t+1}(x,0)-\phi^{t}(x,0))\\
&= a^t_{\phi^t(x,k)} - a^t_{\phi^t(x,0)}\\
&= \left(\sum_{n=0}^t\binom nt a_{\phi^t(x,k)-2t+4n}^0\mod 2\right)-\left( \sum_{n=0}^t\binom nt
a_{\phi^t(x,0)-2t+4n}^0\mod 2\right).
\end{align*}
Notice that:
\begin{itemize}
\itemsep0em
\item $a^0_{\phi^t(x,k)+2t}$ has coefficient $1$ in the left-hand term and $0$ in the right-hand term;
\item $a^0_{\phi^t(x,0)-2t}$ has coefficient $0$ in the left-hand term and $1$ in the right-hand term.
\end{itemize}
Because the particle cannot move by more than one cell per step,
these variables did not appear in the expression of any previous $\delta_{t'}(x), t'<t$. Because the initial measure
is uniform, all variables $a_n^0$ are chosen independently and fairly between 0 and 1. Since both terms are sums of variables taking values
in $\Z/2\Z$, this is enough to show that the terms are independent of each other, independent from all previous
$\delta_{t'}(x), t'<t$, and are fairly distributed between 0 and 1. Therefore $\delta_t(x)$ takes value 0 with probability
$\frac 12$, -1 with probability $\frac 14$ and +1 with probability $\frac 14$, independently from all previous
$\delta_{t'}$.\sk
Therefore $\phi^t(x,k)-\phi^t(x,0)$ performs an unbiased and independent random walk, which implies that
$\mu_k(\{x\ :\ \forall t, \phi^t(x,k)>\phi^t(x,0)\}) = 0$ (standard result in
one-dimensional random walks). Since particles cannot cross, they almost surely end up being in interaction, and therefore $\{1\}$ clashes with itself $\mu$-almost surely. Applying the theorem,
we find that no particle can remain in $\Lambda_\mu(F)$.
More precisely, if we write $\pi_i$ the morphism projecting on the $i$-th coordinate, $\pi_{2\ast}\F^t\mu\to\meas{0}$.
Since the addition mod 2 automaton is surjective, it leaves the uniform measure invariant.
Therefore $\pi_{1\ast}\F^t\mu = \lambda$, and we conclude that $\F^t\mu \to \lambda\times\meas{0}$.
\end{proof}
\subsection{Probabilistic cellular automata}\label{sec:probabilist}
\subsubsection{Adaptation of our formalism for probabilistic cellular automata}
This approach can be adapted to non-deterministic cellular automata, and in particular probabilistic
cellular automata. We use here a generalised version of the standard definition.
\begin{definition}
Let $\A$ be a finite alphabet and $\Nb\subset\Z$. We define the application that applies a bi-infinite sequence of
local rules to a configuration componentwise:
\[\Phi_\Nb:\begin{array}{rl}
(\A^{\A^{\Nb}})^\Z\times\az &\to \az\\
((f_i)_{i\in\Z},(x_i)_{i\in\Z}) &\mapsto (f_i((x_{i+r})_{r\in\Nb})_{i\in\Z}.
\end{array}
\]
\end{definition}
\begin{definition}[Generalised probabilistic cellular automaton]
A \define{generalised probabilistic cellular automaton} $\tilde F$ on the alphabet $\A$ with neighbourhood $\Nb$ is
defined by a measure on bi-infinite sequence of local rules $\nu\in \Ms(\aarz)$.
For a configuration $x\in\az$, $\tilde F : \az\to\Msaz$ is then defined as:
\[\mbox{For any borelian }U,\ \tilde F(x)(U)=\int_{\aarz} 1_U(\Phi_\Nb(f,x))\ud\nu(f).\]
\end{definition}
A deterministic cellular automaton $F$ defined by a local rule $f$ corresponds in this formalism to a Dirac $\nu =
\meas{f}$ (in which case the image measure is a Dirac on the image configuration),
and usual probabilistic cellular automata correspond to the case where $\nu$ is a Bernoulli measure; in other
words, the local rule that applies at each coordinate is drawn independently among a finite set of local rules $\A^\Nb\to\A$.
\begin{definition}[Action on the space of measures]
A generalised probabilistic cellular automaton defined by a measure $\nu\in \Ms(\aarz)$ extends
naturally to an action $\tilde\F:\Msaz\to\Msaz$ by defining \[\tilde\F\mu(U) =
\int_{\az}\int_{\aarz} 1_U(\Phi_\Nb(f,x))\ud\nu(f)\ud\mu(x).\]
The $\mu$-limit measures set of $\tilde F$, $\V(\tilde F, \mu)$, is the set of adherence values of the sequence $(\tilde\F^t\mu)_{t\in\N}$,
and the $\mu$-limit set can be defined as \[\Lambda_\mu(\tilde F) =
\overline{\bigcup_{\eta\in\V(\tilde F, \mu)}\supp\ \eta}.\]
\end{definition}
The definitions of a particle system extend directly, except that the update function also depends on the
choice of the local rules as well as on the configuration. Therefore we write $\phi(x,n,(f_i))$ instead of $\phi(x,n)$,
where $x\in\az, n\in\Z$ and $(f_i)\in\aarz$, and the composition notation is simplified as follows (inductively):
\[\phi^t\left(x,n,(f^k)_{0\leq k< t}\right) = \bigcup_{m\in\phi(x,n,f^1)}
\phi^{t-1}\left(\Phi_\Nb(f^{t-1}, x), m, f^{t-1}\right),\]
where each $f^t\in\aarz$ is a bi-infinite sequence of local rules.
A particle system is said to be coalescent $\nu$-almost surely if the coalescence conditions hold for all $x\in\az$ and $\nu$-almost every
$f\in\aarz$, and a particle $p\in\P$ has speed $v$ $\nu^\infty$-almost surely if the speed conditions hold for $\nu^\infty$-almost every sequence
$(f^t)_{t\in\N}$, where $\nu^\infty$ is the product measure (i.e. each $f^t$ is drawn independently according to $\nu$).
The clashing conditions are extended similarly. \sk
\begin{theorem}[Qualitative result for probabilistic automata]\label{prop:MainResultProb}
Let $\tilde F:\az\to\Msaz$ be a probabilistic cellular automaton defined by $\nu\in\Ms(\aarz)$, $\mu$ an initial
$\s$-ergodic measure and $(\P,\pi,\phi)$ a $\nu^\infty$-almost surely coalescent particle system for $\tilde F$ where
$\P$ can be partitioned into sets $\P_1\dots \P_n$ such that, for any $i<j$, $\P_i$ clashes with $\P_j$
$\mu,\nu^\infty$-almost surely.
Then all particles appearing in the $\mu$-limit set belong to the same subset, i.e. there exists a $i$ such that
\[\forall p\in\P,\ p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow p\in\P_i.\]
If furthermore there exists a $j$ such that $\P_j$ clashes with itself $\mu,\nu^\infty$-almost surely, then this set of
particles does not appear in
the $\mu$-limit set, i.e. \[\forall p\in\P,\ p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow p\notin\P_j.\]
\end{theorem}
\begin{corollary}[Main result with speedy particles - probabilistic automata]
Let $\tilde F:\az\to\az$ be a probabilistic cellular automaton defined by $\nu\in\Ms(\aarz)$, $\mu$ an initial
$\s$-ergodic measure and $(\P,\pi,\phi)$ a $\nu^\infty$-almost surely coalescent particle system for $\tilde F$.
Assume that each particle $p\in\P$ has speed $v_p\in\R$ $\nu^\infty$-almost
surely, then there is a speed $v\in\R$ such that:
\[\forall p\in\P,\ p\in\L(\pi(\Lambda_\mu(F)))\Rightarrow v_p=v.\]
\end{corollary}
The proof of these statements are exactly the same as the proofs of Theorem~\ref{prop:MainResult} and
Corollary~\ref{cor:MainResult}, except that every statement in the proof holds $\nu^\infty$-almost surely.
This Theorem and Corollary can be applied for different probabilistic cellular automata, for exemple when we mix two one sided captive CA (see Figure~\ref{fig:PCACaptive}). We are going to detail two exemples from the literature and obtain new information about its limit measures (Section~\ref{section.FatesDensity} and~\ref{section.Regnault-Remilia-line}).
\begin{figure}[h!]
\begin{center}
\includegraphics[width=9cm]{PCA.png}
\end{center}
\caption{Exemple of probabilistic cellular automata where the update of each cell is chosen between two one sided captive CA.}
\label{fig:PCACaptive}
\end{figure}
\subsubsection{Example: Fatès' density classifying candidate}\label{section.FatesDensity}
For any real $p\in[0,1]$, consider the probabilistic automaton $\tilde F$ on the alphabet $\{0,1\}$ defined on the neighbourhood $\Nb = \{-1,0,1\}$ by
local rules drawn independently between the traffic rule (rule \#184 defined in Section~\ref{section.184}) with probability $p$ and the majority rule (rule \#232 where $F(x)_i=1 $ if and only if $x_{-1}+x_0+x_1\geq 2$) with probability $1-p$. This corresponds to the case where $\nu$ is a Bernoulli measure.
This automaton was introduced by Fatès in \cite{Fates} as a candidate to solve the density classification problem. Even
though the following result does not answer this question, it is new to our knowledge.
\begin{figure}[!ht]
\begin{center}
\begin{tabular}{c|c|c}
\includegraphics[width = .3\textwidth]{trafmaj1}&
\includegraphics[width = .3\textwidth]{trafmaj2}&
\includegraphics[width = .3\textwidth]{trafmaj3}\\
$p = \frac 14$&$p = \frac 12$&$p = \frac 34$
\end{tabular}
\end{center}
\caption{Dynamics of the traffic-majority automaton iterated on the initial measure $\Ber(\frac 35, \frac 25)$.
Density classification is more efficient with $p$ close to 1.}
\end{figure}
\begin{proposition}
Let $\mu\in\Merg(\az)$ and $p$ be a real in $[0,1]$. Then \[\Lambda_\mu(\tilde{F}) \subset \{\per{0}, \per{1}, \per{(01)}, \per{(10)}\}.\]
As a consequence, any limit measure of $(\tilde{F}^t_\ast\mu)_{t\in\N}$ is a convex combination of $\meas{0}, \meas{1}$ and $\meas{01}$.
\end{proposition}
\begin{proof}
The cases $p=0,1$ correspond to deterministic automata and can be treated easily.\sk
The visual intuition suggests to consider interface defects according to the decomposition $\Sigma_0\sqcup\Sigma_1\sqcup\Sigma_2$, where
$\Sigma_0 = \{\per{0}\}$, $\Sigma_1 = \{\per{1}\}$ (monochromatic subshifts) and $\Sigma_2 = \{ \per{(01)}, \per{(10)}\}$ (checkerboard subshift),
since those SFTs are invariant under the action of both rules. The set of particles would be $\P = \{p_{i,j} : i\neq j\in\{0,1,2\}\}$.
However, as Figure~\ref{fig:trafmaj} shows, the particle $p_{10}$ can ``explode'' and give birth to two particles $p_{12}$ and
$p_{20}$, contradicting the condition of coalescence. To solve this problem, we tweak the particle system by replacing
each particle $p_{10}$ by one particle $p_{12}$ and one particle $p_{20}$.
\begin{center}
\captionsetup{type=figure}
\begin{tabular}{cl}
\hspace{-5.8cm}
\includegraphics[width=\textwidth, trim = 0 -25 0 0]{trafmaj.png}&
\hspace{-12.5cm}
\begin{tikzpicture}
\draw (1.4, -.8) node {\footnotesize{``explosion''}};
\draw[->, thick] (.8,-1) -- (0,-1.65);
\draw (-.4,-2.2) node {\footnotesize{$p_{12}$}};
\draw[->, thick] (-.4,-2.1) -- (-.15,-1.95);
\draw (.2,-2.2) node {\footnotesize{$p_{20}$}};
\draw[->, thick] (.2,-2.1) -- (-0.05,-1.95);
\draw (2.2,0.2) node {\footnotesize{$p_{02}+p_{20} = \emptyset$}};
\draw[->, thick] (2.2,0) -- (2.65,-.5);
\draw (3.3,.8) node {\footnotesize{$p_{02}+p_{21} = p_{01}$}};
\draw[->, thick] (3.5,.6) -- (4.35, .15);
\draw (5.3,-2.1) node {\footnotesize{$p_{01}$}};
\end{tikzpicture}
\end{tabular}
\caption{Fatès' traffic-majority probabilistic automaton, with $p=\frac 34$.}
\label{fig:trafmaj}
\end{center}
The corresponding morphism $\pi$ is defined on the neighbourhood $\{0,\dots,3\}$ by the local rule:
\[\begin{array}{cccccc}
0011&\mapsto&p_{01}&
\quad\_110&\mapsto&p_{12}\\
0010&\mapsto&p_{02}&
\quad \_100&\mapsto&p_{20}\\
1011&\mapsto&p_{21}&
\quad \mbox{otherwise}&\mapsto&0
\end{array}
\]
where the wildcards $\_$ can take both values.
In the absence of interactions, the update function $\phi(x,k,f)$ can be defined in the following manner.
Regardless of the rule that is applied, $p_{01},p_{02}$ and $p_{21}$ move at a constant speed $0$, $+1$ and $-1$ respectively. A particle $p_{12}$ move at speed $-1$ if rule $\#184$ is applied at its position and at speed $+1$ otherwise (independent random walk with bias $1-2p$), except if a particle $p_{20}$ prevents its movement to the right, in which case it does not move. The particle $p_{20}$ behaves symmetrically. Furthermore all interactions are of the form $p_{ij}+p_{ji}\to\emptyset$ or $p_{ij}+p_{jk}\to p_{ik}$ (when $(i,j,k) \neq (1,2,0)$, by the last remark). A formal definition of the update function would be tedious, but it is entirely described by these remarks. The various conditions of locality, disjunction, particle control, surjectivity and coalescence are proved similarly to the previous examples.\sk
Assume $p\geq \frac12$. We show that no particle can remain asymptotically by applying the main result on the
sets $(\P_i)_{0\leq i\leq 4}$: $\{p_{02}\}$, $\{p_{20}\}$, $\{p_{01}\}$, $\{p_{12}\}$ and $\{p_{21}\}$. We need only to
show the clashes relative to the second and fourth sets since all other clashes are consequences of the speed of these
particles.
Let $k\in\N$ and $x$ be such that $\pi(x)_0 = p_{02}$ and $\pi(x)_k \in \{p_{12}, p_{20}\}$. Since $p_{02}$ progresses at speed 1,
the distance $\phi^t(x,k)-\phi^t(x,0)$ cannot increase, and it decreases by at least one with probability $p$ (respectively $1-p$).
It is clear that the particles end up in interaction $\nu^\infty$-almost
surely. Showing that $p_{12}$ and $p_{20}$ clash with $p_{21}$ is symmetric.
Let $x$ be such that $\pi(x)_0 = p_{20}$ and $\pi(x)_k = p_{01}$. As long as there are no interactions, the
distance $\phi^t(x,k)-\phi^t(x,0) = -\phi^t(x,0)$ performs an independent random walk of bias $2p-1$, where a
increasing step is sometimes replaced by a constant step. Such a random walk reaches $0$ $\nu^\infty$-almost surely,
which shows that the particles end up in interaction. The clashes between
$p_{01}$ and $p_{12}$, and between $\{p_{20}\}$ and $\{p_{12}\}$, are proved in a similar manner.
The same proof holds for $p\leq \frac12$ by exchanging the roles of $p_{20}$ and $p_{12}$.\sk
Applying Theorem~\ref{prop:MainResultProb}, we conclude that only one particle $p_{ij}$ can remain in the $\mu$-limit set.
However, if we consider $V_k = \{ x\in\Lambda_\mu(F) : \pi(x)_k = p_{ij}\}$, we notice that configurations in $V_k$ are
of the form $y\cdot z$, where $y\in\A^{]-\infty, k]}$ is admissible for $\Sigma_i$ and $z\in\A^{[k+1, +\infty[}$ is
admissible for $\Sigma_j$; in particular, they contain only one particle, and the $(V_k)_{k\in\Z}$ are disjoint. By
$\s$-invariance, for any measure $\eta\in\V(\tilde F,\mu)$, $\eta(V_k)$ is
independent from $k$ and $\eta\left(\bigcup_kV_k\right) = \sum_k\eta(V_k)\leq 1$. Consequently, $\eta(V_k)=0$, which means
$V_k \notin\supp(\eta)$,
and we conclude that no particle remain in the $\mu$-limit set. In other words, $\Lambda_\mu(F) \subset
\Sigma_0\cup\Sigma_1\cup\Sigma_2$.
\end{proof}
\subsubsection{Example: Approximation of a line}\label{section.Regnault-Remilia-line}
In~\cite{Regnault-Remilia-2015}, the authors introduce a random process that, starting from a finite word on $\{0,1\}$, organises bits through local flips to obtain asymptotically a discrete line whose slope corresponds to the frequency of symbols $1$ in the initial finite word. Different local rules are used to obtain every rational slope in this way. If the flips are performed in parallel instead, we obtain a probabilistic cellular automata for every slope $\alpha\in\mathbb{Q}\cap[0,1]$. We consider its action on any initial $\s$-ergodic measure satisfying $\mu([1])=\alpha$. Theorem~\ref{prop:MainResultProb} shows that the sequence of measures converges towards the measure supported by a periodic configuration representing a discrete line of slope $\alpha$. To simplify the presentation, we consider here that $\alpha=\frac{1}{2}$; the method can be easily generalised to other slopes.\bigskip
Define the following local rules:
\begin{itemize}
\item $f_0$ is the identity;
\item $f_{-1}(x_{-2}, x_{-1},x_0,x_1)=\begin{cases}x_0&\text{ if }x_{-2}x_{-1}x_0x_1=0101\textrm{ or }1010, \\
x_{-1}&\text{ otherwise};\end{cases} $
\item $f_1(x_{-1},x_0,x_1,x_2)=\begin{cases}x_{0}&\text{ if }x_{-1}x_0x_1x_2=0101\textrm{ or }1010,\\x_1&\text{ otherwise}.\end{cases} $
\end{itemize}
Let $\tilde F_{\textrm{line}}$ be a probabilistic cellular automaton (represented in figure~\ref{fig:line}) defined by a $\s$-ergodic measure $\nu\in\mathcal{M}_{\s}(\{f_0,f_1,f_{-1}\}^{\Z})$ whose support is the subshift of finite type defined by the set of forbidden patterns \[\{f_0f_{-1},\ f_1f_1,\ f_1f_0,\ f_{-1}f_{-1},\ f_{-1}f_{1}\}.\] To put it more simply, any time the local rules in two consecutive cells are $f_1$ and $f_{-1}$ (which happens with positive probability), the probabilistic CA permutes these two letters, except if they are at the center of a four-letter words $1010$ or $0101$. In any other situation, it acts as the identity.
We consider the interface defects with regards to the checkerboard SFT $\Sigma = \{\per{(01)},\per{(10)}\}$. The defects $11$ and $00$ form a $\nu^\infty$-almost surely coalescent particle system for $\tilde F_{\textrm{line}}$. Thus if the initial measure $\mu$ is $\s$-ergodic such that $\mu([1])=\mu([0])$, the sequence $\tilde F_{\textrm{line}}^n\mu$ converges toward the $\s$-invariant measure supported by the $\s$-periodic orbit $\per{(01)}$ which correspond to the discrete line of slope $\frac{1}{2}$.
\begin{figure}[h!]
\begin{center}
\includegraphics[width=9cm]{line.png}
\end{center}
\caption{Exemple of space time diagram of $\tilde F_{\textrm{line}}$ where $\nu$ is chosen to the Markov measure which maximizes the entropy of the subshift of finite type where the forbidden patterns are $\{f_0f_{-1},\ f_1f_1,\ f_1f_0,\ f_{-1}f_{-1},\ f_{-1}f_{1}\}$.}
\label{fig:line}
\end{figure}
\newcommand\Mix{\mathcal Mix}
\newcommand\G{G_\ast}
\section{Particle-based organisation: quantitative results}\label{sec:brown}
For some cellular automata with simple defect dynamics, the previous results can be refined with a quantitative approach: that is, to determine the asymptotic distribution of random variables related to the particles. In \cite{EntryTimes}, P. K\r urka, E. Formenti and A. Dennunzio considered $T_n(x)$, the entry time after time $n$ on the initial configuration $x$, which is the waiting time before a particle appears in a given position after time $n$. They restricted their study to a gliders automaton, which is a cellular automaton on 3 states: a background state and two particles evolving at speeds 0 and -1 that annihilate on contact. Thus, we have one entry time for each type of particle ($T_n^+(x)$ and $T_n^-(x)$). When the initial configuration is drawn according to the Bernoulli measure of parameters $(\frac 12, 0, \frac 12)$, which means that each cell contains, independently, a particle of each type with probability $\frac 12$, they proved that: \[\forall \alpha\in\R^+,~\mu\left(\frac {T_n^-(x)}n
\leq \alpha\right)\underset{n\to\infty}{\longrightarrow}\frac 2\pi \arctan\sqrt \alpha.\]
They also called to develop formal tools in order to be able to handle more complex automata, starting with the $(-1,1)$ symmetric case.
In Section~\ref{sec:entry}, we extend this result to allow arbitrary values for the particle speeds $v_-$ and $v_+$, and relax the conditions on the initial measure to some $\alpha$-mixing conditions. Then, when $v_-< 0$ and $v_+\geq 0$, we have:
\[\forall x\in\R^+,~\mu\left(\frac{T_n^-(x)}n\leq \alpha\right) \underset{n\to\infty}{\longrightarrow} \frac
2\pi\arctan\left(\sqrt{\frac{-v_-\alpha}{v_+-v_-+v_+\alpha}}\right),\] and symmetrically if we exchange $+$ and $-$. The proof relies on the fact that the behavior of gliders automata can be characterised by some random walk process; this idea was introduced by V. Belitsky and P. Ferrari in \cite{BelitskyFerrari-1995} and was already used in \cite{Kurka-Maass-2000} and \cite{EntryTimes}. In our case, a particle appearing in a position corresponds to a minimum between two concurrent random walks. The new tool here is that under $\alpha$-mixing conditions, we rescale this process and approximate it with a Brownian motion. Thus we obtain the explicit asymptotic distribution of entry times.
This method, consisting in associating a random walk to each gliders automata and studying this random walk using scale invariance, is not limited to this particular conjecture concerning entry times. Indeed, we see in the next two sections that it can be used to study the asymptotic behavior of two other, arguably more natural, parameters: the particle density at time $t$ and the rate of convergence to the limit measure. However, we obtain only an upper bound instead of an explicit asymptotic distribution. There is no doubt this method can be adapted to other parameters in a similar way.
Furthermore, these results can be extended to other automata with similar behavior, such as those in Figure~\ref{fig:particles}, by factorising them onto a gliders automaton. This point is discussed in Section~\ref{sec:extensions}. This method is more difficult to generalize when there is birth of particle, even in simple case as the $4$-cyclic cellular automaton.
\subsection{Gliders automata and random walks}\label{sec:walks}
In this section we give the definition and the first properties of the class of cellular automata studied.
\begin{definition}[Gliders automata]
Let $v_-<v_+ \in \Z$. The \define{$(v_-,v_+)$-gliders automaton} (or GA) $G$ is the cellular automaton of neighbourhood
$[-v_+, -v_-]$ defined on the alphabet $\A = \{-1, 0, +1\}$ by the local rule:
\[f(x_{-v_+}\dots x_{-v_-}) = \left\{\begin{array}{cl}+1&\tx{if }x_{-v_+}=+1\tx{ and }\forall N\leq -v_-,\sum_{n=-v_++1}^{N}
x_n\geq 0\\-1&\tx{if }x_{-v_-}=-1\tx{ and }\forall N\geq -v_+,\sum_{n=N}^{-v_--1} x_n\leq
0\\0&\tx{otherwise.}\end{array}\right.\]
\nomentry{$(v_-,v_+)$-GA}{Gliders automaton with particle speeds $v_-,v_+$}
\end{definition}
In all the following, $\A = \{-1,0,+1\}$ and the diagrams are represented with the convention $\square = 0, \blacksquare
= +1, \textcolor{red}{\blacksquare} = -1$.
\begin{figure}[!ht]
\begin{center}
\includegraphics[width = \textwidth, trim = 0 0 0 300, clip]{gliders.png}
\caption{Space-time diagram of the $(-1,0)$-gliders automaton on a random initial configuration.}
\end{center}
\end{figure}
Our results apply on automata with simple defects dynamics, namely, automata admitting a particle system with $\P =
\{\pm1\}$ and whose update function corresponds to a gliders automaton. We first prove our results for gliders automata
before generalising them in Section~\ref{sec:extensions}. Let us introduce some tools that turn the study of the dynamics of
a gliders automaton into the study of some random walk.
\begin{definition}[Random walk associated with a configuration]\label{def:walk}
Let $x\in\{-1,0,1\}^\Z$. Define the partial sums $S_x$ by:
\[S_x(0) = 0\quad \mbox{and}\quad \forall k\in\Z,\ S_x(k+1)-S_x(k) = x_k.\]
We extend $S_x$ to $\R$ by putting $S_x(t) = (\lceil t\rceil-t)S_x(\lfloor t\rfloor) + (t-\lfloor t\rfloor)S_x(\lceil t\rceil)$
for $t\in\R$. We also introduce the rescaled process $S_x^k:t\mapsto\frac{S_x(kt)}{\sqrt k}$.
\nomentry{$S_x(t)$}{Random walk associated with the configuration x}
\nomentry{$S^k_x(t)$}{Rescaled process corresponding to $S_x$}
\end{definition}
This random walk is simpler to study than the space-time diagram of the gliders automaton, and actually contains the
same amount of information, as shown by the following technical lemmas.
\begin{definition}
Let $f:\R\to\R$ and $U\subset\R$. We define $\argmin{U} f$ by:
\[\forall t\in U,\ t= \argmin{[t_0,t_1]} f \Longleftrightarrow \forall t'\in U\backslash\{t\}, f(t) < f(t').\]
In other words, $t$ realises the strict minimum of $f$ on $U$; this point is not always defined.
\end{definition}
\begin{lemma}\label{lem:MarAlea}
Let $G$ be the $(v_-,v_+)$-gliders automaton. For all $j\in\Z$ and $n\geq1$,
\begin{align*}
j = \argmin{[j,\ j+n]}S_{G(x)} \Longleftrightarrow j-v_+ = \argmin{[j-v_+,\ j+n-v_-]}{S_x},\\
j = \argmin{[j-n,\ j]} S_{G(x)} \Longleftrightarrow j-v_- = \argmin{[j-n-v_+,\ j-v_-]} S_x.
\end{align*}
\end{lemma}
\begin{proof}
We prove those equivalences by induction on $n$. At each step, we prove only the first
equivalence, the other one being symmetric.
\begin{description}
\item[Base case.]
\begin{align*}
S_{G(x)}(j) < S_{G(x)}(j+1) &\Leftrightarrow G(x)_j = +1\\
&\Leftrightarrow x_{j-v_+} = +1 \tx{ and } \forall N\leq -v_-,\sum_{t=-v_++1}^N x_{j+t}\geq
0\\
&\Leftrightarrow S_{x}(j-v_+) < \underset{[j+1-v_+,\ j+1-v_-]}{\min} S_x.
\end{align*}
\item[Induction.] Assume both equivalences hold for some $n\geq 1$.
Suppose $j = \argmin{[j,\ j+n+1]}S_{G(x)}$. In particular $j = \argmin{[j,\ j+n]}S_{G(x)}$, and by
induction hypothesis $j-v_+ = \argmin{[j-v_+,\ j+n-v_-]}{S_x}$.
We distinguish two cases:
\begin{itemize}
\item if $S_x(j+n-v_-+1) > S_x(j-v_+)$, then $j-v_+ = \argmin{[j-v_+,\ j+n-v_-+1]}{S_x}$ and we conclude;
\item otherwise, this means that $S_x(j+n-v_-+1) = S_x(j-v_+)$ (the walk can decrease by at most one at each step),
and thus \[j+n-v_-+1 = \argmin{[j-v_++1,\ j+n-v_-+1]}{S_x}.\]
By induction hypothesis, \[j+n+1 = \argmin{[j+1,\ j+n+1]}{S_{G(x)}},\]
and in particular $S_{G(x)}(j+n+1)<S_{G(x)}(j+1)$. Therefore $S_{G(x)}(j+n+1)\leq S_{G(x)}(j)$, a
contradiction with the first assumption.
\end{itemize}
The converse is proved in a similar manner.
\end{description}
\end{proof}
\begin{lemma}\label{lem:Min} Let $G$ be the $(v_-,v_+)$-gliders automaton. For all $j\in\Z$ and $k\geq 0,$
\begin{align*}G^t(x)_j= -1&\Longleftrightarrow j-v_-t+1 = \argmin{[j-v_+t,\ j-v_-t+1]}S_x\\
G^t(x)_j= +1&\Longleftrightarrow j-v_+t = \argmin{[j-v_+t,\ j-v_-t+1]} S_x.
\end{align*}
This is illustrated in Figure~\ref{fig:LemMin}.
\end{lemma}
\begin{figure}[!ht]
\begin{center}
\begin{tabular}{cc}
\includegraphics{Random.png}&\vspace{-4.155cm}\\
\begin{tikzpicture}
\foreach \x/\y/\z in {-2/0/0, -1/0/0, 0/0/1, 1/1/1, 2/1/1, 3/1/1, 4/1/1, 5/1/2, 6/2/2, 7/2/2, 8/2/2, 9/2/2, 10/2/2,
11/2/1, 12/1/1, 13/1/1, 14/1/1, 15/1/2, 16/2/2, 17/2/2, 18/2/3, 19/3/3, 20/3/3, 21/3/2, 22/2/2, 23/2/2, 24/2/2, 25/2/2,
26/2/2, 27/2/2, 28/2/1, 29/1/1, 30/1/1, 31/1/1, 32/1/1, 33/1/0, 34/0/0, 35/0/0, 36/0/0}
{
\draw[very thick] (-3.25+.21*\x,-1.55+.4*\y) -- (-3.25+.21*\x+.23,-1.55+.4*\z);
}
\draw (-3.7,-0.75) node {$S_x$};
\draw (-3.35,-1.8) node {\small $j-k+1$};
\draw (3.35,-1.8) node {\small $j+k$};
\draw[<->, thick] (-4, 0) -- (5,0);
\draw (-4.2,0) node {$a$};
\draw[->] (-3.5, 0) -- (-3.5,3.25);
\draw (-3.7,1.625) node {$k$};
\draw (0,-0.1) -- (0,0.1);
\draw (0, -0.3) node {\small $j$};
\filldraw[blue] (-0.13,3.2) rectangle (0.07, 3.4);
\draw[blue, thick] (-0.12,3.21) -- (-3.2,0.13) -- (-3.2, -1.55);
\draw[blue, thick] (0.06,3.21) -- (3.13,0.13) -- (3.13,-1.55) ;
\draw[red, thick] (-3.21, -1.55) -- (3.14,-1.55);
\draw (-0.5,3.7) node {\small $G^k(x)_j$};
\end{tikzpicture}&
\end{tabular}
\end{center}
\caption{Illustration of Lemma \ref{lem:Min}. A strict minimum is reached on $j-k+1$.}
\label{fig:LemMin}
\end{figure}
\begin{proof}
By induction on $t$, proving only the first equivalence at each step:
\begin{description}
\item[Base case $(t=0)$] By definition of $S_x$, $S_x(j+1)<S_x(j) \Leftrightarrow x_j=-1$.
\item[Induction] Assume that both equivalences hold for a given itme $t$. By applying the induction
hypothesis on $G(x)$,
$G^{t+1}(x)_j= -1 \Leftrightarrow j-v_-t+1 = \argmin{[j-v_+t,\ j-v_-t+1]} S_{G(x)}$ and we
conclude by applying Lemma \ref{lem:MarAlea}.
\end{description}
\end{proof}
\subsection{Entry times}\label{sec:entry}
The main result of Section~\ref{sec:particles} implies that, for any $\sigma$-ergodic initial measure $\mu$,
$\Lambda_G(\mu)$ contains at most one kind of particle, which one depending on whether $\mu([+1]) > \mu([-1])$ or the
opposite.
When $\mu([+1]) = \mu([-1])$, $\Lambda_G(\mu)$ only contains the particleless configuration $\per{0}$. In other words,
$G_\ast^t\mu \to \meas{0}$, which means that the probability of
seeing a particle in any fixed finite window tends to 0 as $t\to\infty$.
\begin{definition}[Entry times]
Let $v_-< 0 \leq v_+\in\Z$, $G$ the $(v_-,v_+)$-GA and $x \in\{-1,0,1\}^\Z$. We define:
\[T_n^-(x) = \min\{k\in\N\ :\ \exists i\in[0,|v_-|-1],\ G^{k+n}(x)_i=-1\},\]
\nomentry{$T_n^+, T_n^-$}{Waiting time after time $n$ before a particle crosses the central column}
with $T_n^-(x) = \infty$ if this set is empty. This is the \define{entry time} of
$x$ into the set $\{b\in\{-1,0,1\}^\Z\ :\ \exists i\in[0,|v_-|-1], b_i=-1\}$ after time $n$ at position 0. We define $T_n^+(x)$
in a symmetrical manner.
\end{definition}
The size of the considered window is such that any particle ``passing through'' the column 0 appears in this
window exactly once (See Figure~\ref{fig:entry}). Of course entry times for particles of speed 0 make no sense. From now on, we only consider $T^-$ for simplicity, all the results being valid for $T^+$.
\begin{figure}[h!]
\begin{center}
\includegraphics[scale = 0.8]{Entry3.png}
\hspace{-9.31cm}
\begin{tikzpicture}
\draw[blue,thick] (0,0) -- (0,4.2);
\draw[blue,thick] (0.42,0) -- (0.42,4.2);
\draw[blue,thick] (0,1) rectangle (0.42,1.14);
\draw[blue,thick] (0,1) -- (0.42,1.14) (0.42,1) -- (0,1.14);
\filldraw[blue!15!white] (0.02,1.16) rectangle (0.40,3.09);
\draw[<->, thick] (-0.2,1.14) -- (-0.2,3.1);
\draw (-0.7, 2.12) node {\small $T_n^-(x)$};
\draw[<->,thick] (-0.2,1) -- (-0.2,0);
\draw (-0.5, 0.5) node {\small $n$};
\draw (-4,-0.01) -- (5,-0.01) (-4,0.13) -- (5,0.13);
\draw (-4.25,0.2) node {\small $x$};
\end{tikzpicture}
\end{center}
\caption{An entry time for the (-3,1)-gliders automaton.}\label{fig:entry}
\end{figure}
As a consequence of Birkhoff's ergodic theorem, when $\mu([-1]) > \mu([+1])$, $-1$ particles persist $\mu$-almost surely and their
density converges to a positive number. Therefore:
\begin{itemize}
\itemsep0em
\item $\mu(T^+_n(x)=\infty) \underset{n\to\infty}{\longrightarrow} 1$;
\item $\forall \alpha>0,\mu\left(\frac {T_n^-(x)}n \leq \alpha\right)\underset{n\to\infty}{\longrightarrow} 1$,
\end{itemize}
and symmetrically. This is why we only consider the case $\mu([-1]) = \mu([+1])$. K{\r u}rka and al. proved the
following result:
\begin{theorem}[\cite{EntryTimes}]
For the $(-1,0)$-GA (``Asymmetric gliders'') with an initial measure $\mu = \Ber\left(\frac
12,0,\frac 12\right)$:\[\forall \alpha>0,~\mu\left(\frac {T_n^-(x)}n \leq
\alpha\right)\underset{n\to\infty}{\longrightarrow}\frac 2\pi \arctan\sqrt \alpha.\]
\end{theorem}
In the same article, they conjectured that this result could be extended to any initial Bernoulli measure of parameters
$(p,1-2p,p)$ for $0\leq p\leq \frac12$ by replacing the right-hand term by $\frac 2\pi \arctan\sqrt{2p\alpha}$. We will
prove
that this conjecture is actually incorrect.\sk
To state our result, we introduce two particular subclasses of $\Msaz$. We recall the definition of the $\alpha$-mixing coefficients of a measure $\mu\in\Msaz$:
\[\alpha_\mu(n) = \sup \{|\mu(A\cap B) - \mu(A)\mu(B)|: A\in \mathcal \Ba_{]-\infty,0]}, B\in \mathcal \Ba_{[n,+\infty[}\}.\]
Define:
\begin{itemize}
\itemsep0em
\item $\Ber_=$ the set of Bernoulli measures on $\{-1, 0, +1\}^\Z$ and parameters $(p,1-2p,p)$ for some
$0<p\leq \frac 12$;
\item $\Mix$ the set of measures $\mu\in\Ms(\{-1, 0, +1\}^\Z)$ satisfying:
\begin{itemize}
\itemsep0em
\item $\int_{\az} x_0\ \ud\mu(x) = 0$;
\item $\sum_{k=0}^\infty \int_{\az} x_0\cdot x_k\ \ud\mu(x)$ converges absolutely to a real $\s_\mu^2>0$ (asymptotic variance);
\item $\exists \varepsilon>0,\sum_{n\geq 0}\alpha_\mu (n)^{\frac 14 - \varepsilon} < \infty$.
\end{itemize}
\end{itemize}
In particular, $\Ber_= \subset \Mix$.
\begin{theorem}[Quantitative result for entry time]\label{thm:Entry}
For any $(v_-,v_+)$-GA with $v_-< 0$ and $v_+\geq 0$ and any initial measure $\mu\in\Mix$,
\[\forall \alpha>0,~\mu\left(\frac{T_n^-(x)}n\leq \alpha\right) \underset{n\to\infty}{\longrightarrow} \frac
2\pi\arctan\left(\sqrt{\frac{-v_-\alpha}{v_+-v_-+v_+\alpha}}\right).\]
\end{theorem}
Notice that this limit is independent from $\mu$ (as long as $\mu\in\Mix$), disproving the conjecture when
$\mu\in\Ber_=$.
\subsection{Brownian motion and proof of the main result}\label{sec:brownian}
The third hypothesis for $\Mix$ is chosen so that the large-scale behavior of the partial sums $S_x(t)$ can be
approximated by a Brownian motion. This invariance principle is the core of our proofs.
The first and second conditions ensures that the Brownian motion obtained this way have no bias and nonzero variance,
respectively.
\begin{definition}[Brownian motion]
A \define{Brownian motion} (or \define{Wiener process}) $B$ of mean $0$ and variance $\sigma^2$ is a continuous
time stochastic process taking values in $\R$ such that:
\begin{itemize}
\renewcommand{\labelitemi}{\labelitemii}
\itemsep0em
\item $B(0) = 0$,
\item $t\mapsto B(t)$ is almost surely continuous,
\item $B(t_2) - B(t_1)$ follow the normal law of mean 0 and variance $(t_2 - t_1)\s^2$;
\item For $t_1<t_2\leq t'_1<t'_2$, increments $B(t_2) - B(t_1)$ and $B(t'_2) - B(t'_1)$ are independent.
\end{itemize}
\end{definition}
\nomentry{$B(t)$}{Brownian motion value at time $t$}
See \cite{Morters-Peres} for a general introduction to Brownian motion.
\begin{proposition}[Rescaling property]
Let $B$ be a Brownian motion. Then, for any $k>0$, $t\mapsto \frac 1{\sqrt k}B(kt)$ is a Brownian motion with same mean and variance.
\end{proposition}
We now state some invariance principles, which consists in appoximating rescaled random walks by Brownian motion.
We use a strong version, which guarantees an almost sure convergence by considering a copy of the process in a richer probability space.
\begin{theorem}[\cite{Shao-Lu2}, Corollary 9.3.1]\label{thm:Invariance}
Let $X = (X_i)_{i\in\N}$ be a family of random variables taking values in $\{-1,0,1\}$. We denote
$\alpha_X(n)$ its $\alpha$-mixing coefficients defined as:
\[\alpha_X(n) = \sup \{|P(A\cap B) - P(A)P(B)|: t\in\N, A\in X_{[0,t]}, B\in X_{[t+n,+\infty[}\},\]
where $X_{[a,b]}$ is the sigma-algebra generated by $(X_a,\dots, X_b)$.\sk
Assume that:
\begin{enumerate}
\item $\forall i, \mathbb E(X_i)=0$;
\item $\frac 1t\mathbb E\left(\sum_{i,j=1}^{\lfloor t\rfloor} X_i\cdot X_j\right)$ converges absolutely to some positive real $\s^2$;
\item $\exists \varepsilon>0, \sum_{n=1}^\infty \alpha_X(n)^{\frac 14+\varepsilon}$.
\end{enumerate}
Then we can define two processes $X' = (X'_i)_{i\in\N}$ and $B$ on a richer probability space $(\Omega, \mathbb P)$ such that:
\begin{enumerate}
\item $X$ and $X'$ have the same distribution;
\item $B$ is a Brownian motion of mean 0, variance $\s^2$;
\item for any $\varepsilon>0$,
\[\left|\sum_{i=1}^{\lfloor t\rfloor} X_i - B(t)\right| = O\left(t^{\frac 14 + \varepsilon}\right)\quad\mathbb P\mbox{-almost surely}.\]
\end{enumerate}
\end{theorem}
\begin{corollary}\label{thm:Brown}
Let $\mu \in \Mix$.
For any fixed constants $q<r\in\mathbb R$, we can define a process $X' = (X'_i)_{i\in\Z}$
and a family of processes $(t\mapsto B_n(t))_{n\in\N}$ on a richer probability space $(\Omega, \mathbb P)$ such that:
\begin{enumerate}
\item $X'$ has distribution $\mu$;
\item every $B_n$ is a Brownian motion of mean 0 and variance $\s_\mu^2>0$;
\item for any $\varepsilon>0$, denoting $S_{X'}$ the piecewise linear function defined by $S_{X'}(0) = 0$ and $S_{X'}(k+1)-S_{X'}(k) = X'_k$ for all $k\in\Z$,
\[\forall n\in\N, \sup_{t\in[q,r]}\left|\frac{S_{X'}(nt)}{\sqrt{n}} - B_n(t)\right| = O\left(n^{-\frac 14 + \varepsilon}\right)\quad\mathbb P\mbox{-almost surely}.\]
\end{enumerate}
\end{corollary}
\begin{proof}
We apply Theorem~\ref{thm:Invariance} on $(x_i)_{i\in\N}$, where $x$ has distribution $\mu$. Because $\mu$ is $\s$-invariant,
this is a stationary process.
The first and third conditions are satisfied by definition of $\Mix$. For the second condition,
\[\frac 1n\mathbb E(S_X(n)^2) = \frac 1n\sum_{0\leq i,j\leq n}\mathbb E(x_i\cdot x_j) \underset{n\to\infty}\longrightarrow\sigma_\mu^2\]
by stationarity. We obtain two processes $X^1 = (X^1_i)_{i\in\N}$ and $B^1$ on a richer probability space $(\Omega, \mathbb P)$ such that
$X^1$ has the same distribution as $x$, $B^1$ is a Brownian motion of mean 0, variance $\s_\mu^2$, and:
\[\forall \varepsilon>0, \left|\sum_{i=1}^{\lfloor t\rfloor} X^1_i - B^1(t)\right| = \underset{t\to+\infty}O\left(t^{\frac 14 + \varepsilon}\right)\quad\mathbb P\mbox{-almost surely}.\]
Since the variables $X^1_i$ take value in $\{-1,0,1\}$, we have for any $t$ $\left|\sum_{i=1}^{\lfloor t\rfloor} X^1_i - S_{X^1}(t)\right|<1$
(a staircase and piecewise linear function having the same values on $\N$). Therefore:
\[\forall \varepsilon>0, \left|S_{X^1}(t) - B^1(t)\right| = \underset{t\to+\infty}O\left(t^{\frac 14 + \varepsilon}\right)\quad\mathbb P\mbox{-almost surely}.\]
\[\forall \varepsilon>0, \forall n\in\N,\ \frac 1{\sqrt n}\left|S_{X^1}(tn)-B^1(tn)\right|=
\underset{n\to\infty}{O}\left(n^{-\frac 14+\epsilon}\right)\cdot\underset{t\to\infty}{O}\left(|t|^{\frac 14+\epsilon}\right) \quad\mathbb P\mbox{-almost surely}. \]
For any $r\in\R_+^2$, taking the sup for $t\in[0,r]$, we obtain:
\[\forall \varepsilon>0, \forall n\in\N,\ \sup_{t\in[0,r]}\left|\frac {S_{X^1}(tn)}{\sqrt n}-\frac {B^1(tn)}{\sqrt n}\right|=\underset{n\to\infty}{O}\left(n^{-\frac14+\epsilon}
\right)\quad\mathbb P\mbox{-almost surely}.\]
By rescaling property $B^1_n : t \mapsto \frac {B^1(tn)}{\sqrt n}$ is a Brownian motion of same mean and variance as $B^1$.\sk
To extend the result to negative values, we apply the theorem again to $(x_{-i-1})_{i\in\N}$, obtaining a process $X^2$ and a Brownian motion $B^2$
satisfying the same asymptotic bound on $t\to -\infty$. Joining both parts, we can see that the process $X' = \dots X^2_{-2}, X^2_{-1}, X^1_{0}, X^1_{1}\dots$ have distribution $\mu$
and $B_n : t\mapsto B_n^1(t)$ if $t\geq 0$, $B_n^2(t)$ if $t<0$ is a Brownian motion.
\end{proof}
For a survey of invariance principles under different assumptions, see \cite{Merlevede-Rio}.\sk
Using this last result, we prove the main theorem.
\begin{proof}[Proof of Theorem~\ref{thm:Entry}]
For any $x\in \{-1,0,1\}^\Z$, Lemma \ref{lem:Min} applied on the column 0 gives:
\begin{align*}
T_n^-(x)& =\min\left\{k\in\N\ |\ \exists j\in [0,-v_-[~, S_x(-v_-(n+k)+j+1) <\min_{[-v_+(n+k)+j,\ -v_-(n+k)+j]}S_x\right\}\\
& =\min\left\{k\in\N\ |\ \exists j\in [0,-v_-[~, S_x(-v_-(n+k)+j+1) < \min_{[-v_+(n+k)+j,\ -v_-n]}S_x\right\}
\end{align*}
Note that if this condition is reached on $k\in\N$, since $S_x$ is piecewise linear, it is attained for $t$ as soon as
$t>k-1$ and reciprocally. Thus:
\[T_n^-(x)=\inf\left\{t\geq 0\ |\ \exists j\in [0,-v_-[~, S_x(-v_-(n+t)+j+2) < \min_{[-v_+(n+t)+j+1,-v_-n]}S_x\right\}\]
Replacing $j$ by $0$ in this expression adds to the infimum a value comprised between $0$ and $\frac {-v_--1}{-v_-}$ (remember $v_-<0$).
Since the infimum is necessarily an integer, we compensate by taking the integer part:
\begin{align*}
T_n^-(x)&=\left\lfloor\inf\left\{t\geq 0\ |\ S_x(-v_-(n+t)+2) < \min_{[-v_+(n+t)+1,-v_-n]}S_x\right\}\right\rfloor\\
&=\left\lfloor\inf\left\{t\geq 0\ |\ S_x^n\left(-v_-\left(1+\frac tn\right)+\frac2n\right) < \min_{[-v_+(1+\frac
tn)+\frac1n,-v_-]}S_x^n\right\}\right\rfloor\\
&=\left\lfloor n\cdot\inf\left\{t\geq 0\ |\ S_x^n\left(-v_-(1+t)+\frac 2n\right) <
\min_{[-v_+(1+t)+\frac1n,-v_-]}S_x^n\right\}\right\rfloor
\end{align*}
Dividing by $n$, since $S_x^n$ is $\sqrt n$-Lipschitz and $t-\frac 1n \leq \frac{\lfloor nt\rfloor}n \leq
t$ for all $t,n\in\R\times\N$:
\begin{align*}
\mu\left(\min_{[-v_-,-v_-(1+\alpha)]}S_x^n +\frac4{\sqrt n}< \min_{[-v_+(1+\alpha),-v_-]}S_x^n\right) \leq
\mu\left(\frac{T_n^-(x)}n \leq \alpha\right) \\ \mu\left(\frac{T_n^-(x)}n \leq \alpha\right) \leq
\mu\left(\min_{[-v_-,-v_-(1+\alpha)]}S_x^n -\frac3{\sqrt n}< \min_{[-v_+(1+\alpha),-v_-]}S_x^n\right)\tag{1}\label{eq1}
\end{align*}
Using Corollary~\ref{thm:Brown}, we build a process $X'$ and a family of processes $(B_n)_{n\in\N}$ on a richer probability space $(\Omega, \mathbb P)$ such that
$X'$ is distributed according to $\mu$ and the $B_n$ are Brownian motions.
\[\forall n\in\N, \sup_{[-v_+(1+\alpha),-v_-(1+\alpha)]}\left|\frac{S_{X'}(nt)}{\sqrt{n}} - B_n(t)\right| =
O\left(n^{-\frac 14 + \varepsilon}\right)\quad\mathbb P\mbox{-almost surely}.\]
By symmetry, $B_n^l(t) = B_n(-v_--t)-B_n(-v_-)$ and $B_n^r(t) =
B_n(-v_-+t)-B_n(-v_-)$ are two independent Brownian motions on $[0,\ v_--v_+(1+\alpha)]$ and $[0,\ -v_-\alpha]$, respectively. Consequently, for any
$\varepsilon >0$ and $n$ large enough:
\begin{align*}\mu\left(\min_{[-v_-,-v_-(1+\alpha)]}S_x^n -\varepsilon< \min_{[-v_+(1+\alpha),-v_-]}S_x^n\right)&=
\mathbb P\left(\min_{[-v_-,-v_-(1+\alpha)]}S_{X'}^n -\varepsilon< \min_{[-v_+(1+\alpha),-v_-]}S_{X'}^n\right)\\
&\leq \mathbb P\left(\min_{[-v_-,-v_-(1+\alpha)]}B_n-2\varepsilon<\min_{[-v_+(1+\alpha),-v_-]}B_n\right)\\
&\leq \mathbb P\left(\min_{[0,-v_-\alpha]}B_n^l-2\varepsilon<\min_{[0,-v_-+v_+(1+\alpha)]}B_n^r\right)\tag{2}\label{eq2}
\end{align*}
and a symmetrical lower bound for the first term of (\ref{eq1}). We evaluate this last term.\sk
For any Brownian motion $B$ and $b>0$, we have by rescaling $\displaystyle\mathbb P\left(\min_{[0,b]} B\geq m\right) =
\mathbb P\left(\min_{[0,1]} B\geq \frac m{\sqrt b}\right)$. Furthermore,
since $B_n^l$ and $B_n^r$ are independent, so are $\displaystyle\min_{[0,1]} B^l_x$ and $\displaystyle\min_{[0,1]}
B^r_x$. Denote $\mu_m$ the law of the minimum of a Brownian motion on $[0,1]$, which is defined by the density function:
\[\begin{array}{rcll}
\R&\to&\R&\\
t&\mapsto &\frac{e^{-t^2}}2&\mbox{ if }t\leq 0,\\
&&0 &\mbox{ otherwise.}
\end{array}\quad\mbox{(see \cite{Morters-Peres}).}
\]
This means that for any $y,z>0$:
\begin{align*}
\mathbb P\left(\min_{[0,y]}B^l_n< \min_{[0,z]} B^r_n\right) &=\int_{-\infty}^0\int_{-\infty}^01_{\{\sqrt y\cdot m_1\leq \sqrt
z\cdot m_2\}}d\mu_m(m_2)d\mu_m(m_1)\\
&\underset{(i)}=\frac 4{2\pi}\int_{-\infty}^0\int_{-\infty}^{\frac {\sqrt{z}\cdot m_2}{\sqrt y}}
e^{\frac{-m_1^2}2}e^{\frac {-m_2^2}2} dm_1dm_2\\
&\underset{(ii)}=\frac 2{\pi}\int_\pi^{\pi+\arctan\left(\sqrt {\frac yz}\right)}\int_0^{+\infty} re^{\frac{-r^2}2}
drd\theta\\
&=\frac 2{\pi}\arctan\left(\sqrt {\frac yz}\right)\label{eq3}\tag{3}
\end{align*}
(i) by using the law of the minimum of a Brownian motion, (ii) by passing in polar variables. For $\varepsilon>0$, a
similar calculation gives:
\begin{align*}\left|\mathbb P\left(\min_{[0,y]}B^l_n-2\varepsilon< \min_{[0,z]}B^r_n\right)-\mathbb P\left(\min_{[0,y]}B^l_n<
\min_{[0,z]}B^r_n\right)\right| &\leq \frac 4{2\pi}\int_{-\infty}^0\int_{\frac {\sqrt z\cdot m_2}{\sqrt y}}^{\frac
{\sqrt z\cdot m_2+2\varepsilon}{\sqrt y}} e^{\frac{-m_1^2}2}e^{\frac {-m_2^2}2} dm_1dm_2\\
&\leq\frac {8\varepsilon}{2\pi\sqrt y}\int_{-\infty}^0 e^{\frac{-ym_2^2}{2z}}e^{\frac {-m_2^2}2} dm_2\\
&\underset{\varepsilon \to 0}{\longrightarrow}0\label{eq4}\tag{4}
\end{align*}
To sum up, the right-hand term in (\ref{eq2}) converges to $\frac 2\pi\arctan\left(\sqrt{\frac{-v_-\alpha}{v_+-v_-+v_+\alpha}}\right)$
as $\varepsilon\to0$. The first term in (\ref{eq1}) can be bounded from below by the same method. Since $\varepsilon$ can be taken as
small as possible by taking $n$ large enough, the theorem follows.
\end{proof}
\subsection{Particle density}\label{sec:density}
\begin{definition}[Particle density in a configuration]
The \define{$-1$ particle density} in $x\in \{-1,0,1\}^\Z$ is defined as $d_-(x) = \freq(-1,x)$.
$d_+(x)$ is defined in a symmetrical manner.
\end{definition}
In all the following, any result on $d_-$ also holds for $d_+$ by symmetry.
\begin{theorem}[Decrease rate of the particle density]\label{thm:density}
Let $G$ be a $(v_-,v_+)$-GA with initial measure $\mu\in\Mix$. Then:
\[\forall_\mu x\in\{-1,0,1\}^\Z,\ \forall \varepsilon > 0,\ d_-(G^t(x)) = O \left(t^{-\frac 14+\varepsilon}\right)\]
If furthermore $\mu\in\Ber_=$:
\[\forall_\mu x\in\{-1,0,1\}^\Z,\ d_-(G^t(x)) \sim t^{-\frac 12}\]
\end{theorem}
\begin{proof}
When $\mu\in \Mix$, it is in particular $\s$-ergodic, and so are its images $\G^t\mu$. By Birkhoff's
ergodic theorem, one has $d_-(G^t(x)) = \G^t\mu([-1]) = \mu(G^t(x)_0=-1)$ for $\mu$-almost all $x\in\{-1,0,1\}^\Z$.\sk
We first prove the theorem when $G$ is the $(-1,0)$-gliders automaton. By Lemma \ref{lem:Min},
\[\mu(G^t(x)_0= -1)=\mu\left(S_x(t+1) < \min_{[0,t]} S_x\right).\]
\paragraph{Equivalent ($\mu\in\Ber_=$):}
By symmetry, \[\mu\left(S_x(t+1) < \min_{[0,t]} S_x\right) = \mu\left(S_x(0) < \min_{[1,t+1]} S_x\right),\] which is the probability that the random walk starting from 0 remains strictly positive during $t$ steps, also known as
its probability of survival. According to \cite{Redner}, when the random walk is symmetric and the steps are independent, we have the equivalent
$\mu(G^t(x)_0=-1) \sim \frac1{\sqrt t}$.
\paragraph{Upper bound:}
\[\mu\left(S_x(t+1) < \min_{[0,t]} S_x\right) \leq \mu\left(S_x^{t+1}(1) = \min_{[0,1]} S^{t+1}_x\right).\]
Using Corollary~\ref{thm:Brown}, we have:
\begin{align*}
\mu\left(S_x^{t+1}(1)=\min_{[0,1]} S^{t+1}_x\right) &= \mathbb P\left(S_{X'}^{t+1}(1)=\min_{[0,1]} S^{t+1}_{X'}\right)\\
&\leq \mathbb P\left(B_{t+1}(1)\leq\min_{[0,1]}B_{t+1}+C_{t+1}\right)\\
&\leq\mathbb P\left(B_{t+1}(0)\leq\min_{[0,1]}B_{t+1}+C_{t+1}\right),
\end{align*}
where $\displaystyle C_{t+1} = \sup_{[0,1]}\left|S_{X'}^{t+1}-B_{t+1}\right| = O\left(t^{-\frac 14+\varepsilon}\right)$ $\mathbb P$-almost surely,
and where the third line is obtained by symmetry of the Brownian motion.
Furthermore $\displaystyle\mathbb P\left(\min_{[0,1]}B_{t+1}>-C_{t+1}\right) = \int_{-C_{t+1}}^0 e^{-x^2/2} dx \leq C_{t+1} = O \left(t^{-\frac
14+\varepsilon}\right)$.
\paragraph{General case (any $v_- < v_+$):} Let $G'$ be the $(v_-,v_+)$-GA. Then
\[G' = \s^{v_+}\circ G^{v_+-v_-}.\]
To conclude, it is enough to see that the particle density is $\s$-invariant and decreasing under the action of $G$.
\end{proof}
\subsection{Rate of convergence}\label{sec:rate}
In this section, we estimate the rate of convergence to the limit measure. For that we fix a distance on the space $\Msaz$ of $\s$-invariant measures, which induces the weak$^\ast$ topology:
\[\dm(\mu,\nu)=\sum_{n\in\N}\frac{1}{2^n}\max_{u\in\A^n}\left|\mu([u])-\nu([u])\right|.\]
\begin{theorem}[Rate of convergence to the limit measure]\label{thm:rate}
Let $G$ be the $(v_-,v_+)$-GA with initial measure $\mu\in\Mix$. Then:
\[ \forall \varepsilon>0,\ \dm(G^t_\ast\mu,\meas{0})=O\left(t^{-1/4+\varepsilon}\right)\]
If furthermore $\mu\in\Ber_=$: \[\dm(G^t_\ast\mu,\meas{0})=\Omega\left(t^{-1/2}\right)\]
\end{theorem}
\begin{proof}
We first prove the theorem when $G$ is the $(-1,0)$-gliders automaton. By defining $0^\ell\in\A^\ell$ the word
containing only zeroes, the distance can be rewritten:
\[\forall t\in\N, \dm(G_\ast^t\mu,\meas{0}) = \sum_{\ell=1}^\infty\frac
1{2^\ell}G^t_\ast\mu\left(\az\backslash[0^\ell]\right).\]
\textbf{Lower bound when $\mu\in\Ber_=$:} $\dm(G_\ast^t\mu,\meas{0}) >
G^t_\ast\mu\left(\az\backslash[0]\right)$. We conclude with Theorem \ref{thm:density}.\
\textbf{Upper bound:} We give an upper bound for $G^t_\ast\mu(\az\backslash[0^\ell]) = \mu(\exists 0\leq d\leq \ell, G^t(x)_d=\pm1)$ for $\ell\in\N$
and $t\in\N$. By Lemma \ref{lem:Min},
\[\forall d\in\Z,\ G^t(x)_d = +1 \Leftrightarrow S_x(d)<\min_{[d+1,d+t]}S_x.\]
Therefore:
\begin{align*}
G^t_\ast\mu\left(\bigcap_{d=0}^\ell [+1]_d\right)&\leq \mu\left(\min_{[0,\ell]} S_x < \min_{[\ell+1,t]}S_x\right)\\
&\leq \mu\left(\min_{[0,t]}S_x \geq -\ell\right)\\
&\leq \mu\left(\min_{[0,1]}S^t_x \geq -\frac \ell{\sqrt t}\right)
\end{align*}
By Corollary~\ref{thm:Brown}, using the same notations as in the previous proofs:
\begin{align*}
G^t_\ast\mu\left(\exists 0\leq d\leq \ell, x_d = +1\right) &\leq \mathbb P\left(\min_{[0,1]}S^t_{X'} \geq -\frac \ell{\sqrt t}\right)\\
&\leq \mathbb P\left(\min_{[0,1]}B_t \geq -\frac \ell{\sqrt t} - C_t\right) \quad\quad\textrm{ where }C_t = O\left(t^{-\frac
14+\varepsilon}\right)\\
&= O\left(t^{-\frac 14+\varepsilon}\right)
\end{align*}
for any $\varepsilon>0$, following the same calculations as in Section~\ref{sec:density}. The case of $-1$ particles is symmetrical, and we conclude.
\paragraph{General case:} Apply the same method as in the previous section, considering that $\dm$ and all
considered measures are $\s$-invariant and that any CA is Lipschitz w.r.t $\dm$.
\end{proof}
\subsection{Extension to other cellular automata}\label{sec:extensions}
\begin{definition}Let $F_1, F_2$ be two CAs on $\az$ and $\B^{\Z}$, respectively. We say that $F_1$
\define{factorises onto} $F_2$ if there exists a \define{factor} $\pi: \az\to \mathcal \B^{\Z}$ such that
$\pi\circ F_1 = F_2 \circ\pi$.\end{definition}
In other words, $F_1$ admits a particle system ($\P, \pi, \phi)$ with $\B =\P\cup\{0\}$ and where $\phi$
is a cellular automaton on $\bz$.
In this section, we extend the Theorems~\ref{thm:Entry} and \ref{thm:density} to automata that factorise
onto a gliders automaton, and discuss conditions for the extension of Theorem~\ref{thm:rate}. In Section
\ref{sec:defects}, we exhibited a general method to find such a factor using experimental intuition when such a factor
is not obvious.
In other words, using the formalism from Section \ref{sec:particles}, we extend the theorems to automata that admit a
particle system $(\P,\pi,\phi)$, where $\P = \{-1,+1\}$ and $\phi$ acts as a gliders automaton.\sk
In order to extend the theorem to such CAs, starting from an initial measure $\mu$, we must first ensure that
$\pi_\ast\mu\in\Mix$. We show that the third condition in the definition of $\Mix$ is invariant under morphism.
\begin{proposition}
Let $\pi: \az \to \bz$ be a morphism, $\mu\in\Msaz$ and $k>0$ any real such that $\sum_{n\geq 0}\alpha_\mu (n)^k <
\infty$. Then, $\sum_{n\geq 0}\alpha_{\pi_\ast\mu} (n)^k < \infty$.
\end{proposition}
\begin{proof}
We keep the notations from the definition of $\alpha_\mu(n)$. $\pi$ is defined by a local rule with neighbourhood
$\Nb \subset [-r,r]$ for some $r>0$. Then, $\pi^{-1}
\Ba_{-\infty}^0 \subset \Ba_{-\infty}^r$ and $\pi^{-1}\Ba_n^\infty \subset \Ba_{n-r}^{+\infty}$. By
$\s$-invariance, we have for all $n$ $\alpha_{\pi_\ast\mu}(n) < \alpha_\mu(n-2r)$, and the result follows.
\end{proof}
Hence, if $\mu\in\Mix$, we only have to prove that $\pi_\ast\mu$ weighs evenly the sets of particles $-1$ and
$+1$, and that the corresponding asymptotic variance is not zero. Under these assumptions, we can extend some of the previous
results with the forbidden patterns playing the role of the particles.
\begin{corollary}
Let $F:\az\to\az$ be a CA and $\mu\in\Msaz$. Suppose that $F$ factorises onto a $(v_-,v_+)$-GA via a factor
$\pi$ such that $\pi_\ast\mu\in\Mix$.
Then Theorem~\ref{thm:Entry} and the first point of Theorem~\ref{thm:density} hold if we replace ``$x_k = \pm1$'' by
``$\pi(x)_k=\pm1$''.
\end{corollary}
\paragraph{Examples:}
\begin{figure}[!ht]
\begin{center}
\includegraphics[width = 0.47\textwidth, trim = 0 0 0 50, clip]{cycle3cut.png}
\hspace{0.5cm}
\includegraphics[width = 0.47\textwidth, trim = 0 0 0 50, clip]{captifcut.png}
\hspace{0.5cm}
\includegraphics[width = 0.47\textwidth, trim = 0 0 0 50, clip]{Prod.png}
\end{center}
\caption{The 3-state cyclic CA, a one-sided captive CA and the product CA.}
\end{figure}
\begin{description}
\item[Traffic automaton] Let $\A = \{0,1\}$ and $F_{184}$ be the elementary CA corresponding to rule $\#184$.
$F_{184}$ factorises on the $(-1,+1)$-gliders automaton, using the factor found in
Section~\ref{sec:particlesexamples}:
\[\begin{array}{ccc}
00 &\mapsto& +1\\
11 &\mapsto& -1\\
\mbox{otherwise} &\to& 0
\end{array}
\]
This factor is represented in Figure~\ref{fig:Factors}. If $\mu$ is a measure such
that $\pi_\ast\mu\in\Mix$, then Theorem~\ref{thm:Entry} and the first point of Theorem~\ref{thm:density} hold.
For example, this is true for the 2-step Markov measure defined by the
matrix $\left(\begin{matrix}p&1-p\\1-p&p\end{matrix}\right)$ and the eigenvector $\left(\begin{matrix}1/2\\
1/2\end{matrix}\right)$ with $p>0$. A particular case is the Bernoulli measure of parameters $(\frac 12,\frac 12)$.
Theorem~\ref{thm:rate} can also be extended by considering $\dm(F_{184\ast}^t\mu, \meas{01})$, since this distance can be
bounded knowing the density of particles.
\item[3-state cyclic automaton] Let $\A = \Z/3\Z$ and $C_3$ be the 3-state cyclic automaton. We
consider the factor $\pi$ defined in Section~\ref{sec:particlesexamples}:
\[\begin{array}{cccl}
ab &\mapsto& +1 &\mbox{ if } a=b+1\mod 3\\
ab &\mapsto& -1 &\mbox{ if } a=b-1\mod 3\\
ab &\mapsto& 0 &\mbox{ if } a=b
\end{array}
\]
If $\mu$ is such that $\pi_\ast\mu\in\Mix$, then Theorem~\ref{thm:density} applies. This is true in particular when
$\mu$ is any 2-step Markov measure defined by a matrix $(p_{ij})_{1\leq i,j\leq 3}$ satisfying $p_{01}+p_{12}+p_{20} =
p_{10}+p_{21}+p_{02}$, all of these values being nonzero, with $(\mu_i)_{1\leq i\leq 3}$ its only eigenvector. This
includes any nondegenerate Bernoulli measure. However, even when the limit measure is known (e.g. starting from the uniform measure),
Theorem~\ref{thm:rate} does not apply directly.
\item[One-sided captive automata] Let $F$ be any one-sided captive cellular automaton defined by a local rule $f$.
As explained in Section \ref{sec:particlesexamples}, $F$ factorises onto the $(-1,0)$-gliders automaton with a factor defined by:
\[\begin{array}{cccl}
ab &\mapsto& +1 &\mbox{ if } a\neq b, f(a,b)=a\\
ab &\mapsto& -1 &\mbox{ if } a\neq b, f(a,b)=b\\
ab &\mapsto& 0 &\mbox{ if } a=b
\end{array}
\]
For an initial measure $\mu$, if
$\pi_\ast\mu\in\Mix$, then Theorem~\ref{thm:Entry} and the first point of Theorem~\ref{thm:density} apply.\sk
Notice that this class of automata contains the identity ($\forall a,b\in\A, f(a,b) = b$) and the shift $\s$ ($\forall
a,b\in\A, f(a,b) = a$). However, since we have in each case $\pi^{-1}(+1) = \emptyset$ or $\pi^{-1}(-1) = \emptyset$, it
is impossible to find an initial measure that weighs evenly each kind of particle, and so $\pi_\ast\mu$ cannot belong in
$\Mix$.
The limit measure, however, depends on the exact rule, and Theorem~\ref{thm:rate} does not apply directly.
\end{description}
\paragraph{Counter-example:}
\begin{description}
\item [Product automaton] Let $\A = \Z/2\Z$ and $F_{128}$ be the CA of neighbourhood $\{-1,0,1\}$ defined by the local
rule
$f(x_{-1},x_0,x_{1}) = x_{-1}\cdot x_0\cdot x_{1}$. Using the formalism from Section \ref{sec:defects}, we can see that
$F_{128}$ factorises onto the $(-1,1)$-GA by the factor
\[\pi:\left\{\begin{array}{ccc} 01&\to&+1\\
10&\to&-1\\
\textrm{otherwise}&\to&0
\end{array}\right.
\]
If $\mu$ is any Bernoulli measure, then $\pi_\ast\mu$ satisfies all conditions of $\Mix$ except that $\s_\mu = 0$;
indeed, we
can check that for $\pi_\ast\mu$-almost all configurations, the particles $+1$ and $-1$ alternate. Hence, only one particle
can cross any given column after time 0, and therefore
$\forall \alpha>0,\ \mu\left(\frac{T_n^-(x)}n\leq \alpha\right)\underset{n\to\infty}{\longrightarrow} 0$. Furthermore,
any particle survives up to time $t$ only if it is the border of a initial cluster of black cells larger than $2t$ cells,
which happens with a probability $\mu([1])^{2t}$ exponentially decreasing in $t$.
\end{description}
Even though we showed that the asymptotic distributions of entry times are known for some class of cellular automata and a large class of measures, this covers only very specific dynamics. It is not known how these results extend for more than 2 particles and/or other kind of particle interaction. In particular, there is no obvious stochastic process
characterising the behavior of such automata that would play the role of $S_x$ in our proofs.
\subsection*{Acknowledgements}
This work was partially supported by the ANR project QuasiCool (ANR-12-JS02-011-01) and the ANR project Valet (ANR-13-JS01-0010).
\newpage
\newcommand{\etalchar}[1]{$^{#1}$}
| {
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Carry on Reilly
PAUL REILLY isn't ready to finish his Huddersfield Giants career just yet.
The 31-year-old heads to Hull for Saturday's big play-off clash (7.00) knowing defeat would bring the curtain down on his 11 years with his hometown club.
The former England international will leave the Galpharm for Wakefield Trinity Wildcats at the end of the season.
And before then he is determined to play as many more games as possible in the claret and gold.
"As far as I'm concerned, I don't intend finishing with the Giants this weekend!" announced Reilly, who is confident of making a full recovery from a dead leg suffered during last Saturday's 25-24 defeat at Hull KR.
"If we're to reach the Grand Final, we've still got three games to play, which means I could still play for the Giants four more times.
"That's what I've set my heart on doing, and that's something everyone at the club knows we are capable of achieving.
"I know it's going to be tough, but we have shown in the past that when it comes to one-off knockout games we can do it.
"Not many people gave us a chance of beating Leeds in the Challenge Cup semi-final last season, but we did it.
"And we know we have the ability to go to Hull and win. It's another tough one, but we'll be ready for the challenge." | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 1,347 |
\section{Introduction}
\label{sec:intro}
Through superradiant scattering, energy can be extracted from rapidly rotating black holes (see \cite{Brito:2015oca} for a review). In global anti-de Sitter space (AdS), the reflecting boundary causes these black holes to be unstable to the superradiant instability \cite{Hawking:1999dp,Reall:2002bh,Cardoso:2004hs,Kunduri:2006qa,Cardoso:2006wa,Murata:2008xr,Kodama:2009rq,Dias:2011at,Dias:2013sdc,Cardoso:2013pza}, whose ultimate endpoint remains an open problem \cite{Dias:2011ss,Dias:2015rxy,Niehoff:2015oga,Chesler:2018txn}. Because small black holes typically have high angular frequency, the ultimate configuration of low-energy states in AdS likewise remains unknown.
Early work on this problem involved studying perturbations of the Kerr-AdS and Myers-Perry-AdS black holes
(the higher-dimensional analog of Kerr-AdS \cite{Myers:1986un,Hawking:1998kw,Gibbons:2004uw,Gibbons:2004js}, see \cite{Emparan:2008eg} for a review),
where quasi-normal spectra were obtained in \cite{Dias:2013sdc,Cardoso:2013pza}.
For specific modes, the onset of the superradiant instability occurs at a frequency where $\Im(\omega)=0$, but $\Re(\omega)\neq0$. This suggests the existence of a time-periodic black hole with a single helical Killing vector that branches from these onsets. Such black holes, called \emph{black resonators}, which can be viewed as black holes with a gravitational hair, were later constructed in \cite{Dias:2015rxy}, where it was found that they have higher entropy (horizon area) than the corresponding Kerr-AdS black hole with the same mass and angular momentum.
It is therefore entropically permissible for Kerr-AdS black holes to evolve towards black resonators. However, though black resonators are stable to the mode that generated them, they are still rapidly rotating and hence remain unstable to other, typically higher, superradiant modes \cite{Dias:2011ss,Dias:2015rxy,Niehoff:2015oga,Chesler:2018txn,Green:2015kur}. It appears, therefore, that the instability leads to a cascade with higher and higher modes growing in time. If there is indeed an unceasing energy cascade towards higher modes, there will eventually be a significant amount of energy placed in sub-Planckian length scales, which can be viewed as a violation of the weak cosmic censorship conjecture \cite{Penrose:1969pc}.
To date, there is only a single study of time-evolution involving the rotational superradiance of AdS \cite{Chesler:2018txn}. In Kerr-AdS, there is typically one unstable mode that dominates the dynamics at early times. Evolution then proceeds towards a black resonator until the instabilities of the black resonator itself begin to take over and drive the continuing evolution. The evolution in \cite{Chesler:2018txn} was not continued further due to numerical limitations.
Because of the lack of symmetries and the long time-scales involved, the study of the superradiant instability is a significant numerical challenge. It is therefore fortunate that a simplification in a more limited setting has been found. By moving to five dimensions and allowing both angular momenta to be equal, black resonators with a cohomogeneity-1 ansatz (i.e. the metric functions depend only on a single variable, and the solution can be obtained by solving ODEs) were constructed in \cite{Ishii:2018oms}. The scalar, electromagnetic, and gravitational quasinormal modes of these black resonators were studied shortly thereafter in \cite{Ishii:2020muv}.\footnote{For the superradiant instability of an electromagnetic perturbation isolated from these modes, a cohomogeneity-1 photonic black resonator was also constructed in \cite{Ishii:2019wfs}.} As anticipated, black resonators are unstable to higher modes.
These higher-mode instabilities of black resonators also have their individual onsets, from which new black resonators having multiple frequencies could be generated.
We will refer such solutions as multi black resonators.
Here, we set out to construct such multi black resonator solutions and to study their relationship with black resonators and Myers-Perry-AdS black holes. Ideally, we would study multi black resonators that are generated by gravitational perturbations of black resonators. However, these perturbations break too many symmetries of the original black resonator solution. We therefore focus on multi black resonators that are generated by scalar fields. Ordinarily, a scalar field would also break most of these symmetries, but we will rely on a multiplet scalar constructed using Wigner D-matrices, from which the cohomogeneity-1 structure can be preserved. A scalar doublet version of such a system was previously studied in \cite{Dias:2011at,Choptuik:2017cyd}\footnote{See also \cite{Stotyn:2013yka,Stotyn:2013spa} for a study covering different number of dimensions.}. The higher scalar multiplets we introduce can coexist with oscillations of the metric, but do not contribute additional extra oscillating frequencies to the metric.
Even in this limited setting, the full space of solutions is intricate. There are Myers-Perry-AdS black holes, black holes with scalar hair, black resonators, and now hairy black resonators, which are a kind of multi black resonator. In addition, there are boson stars, geons, and now graviboson stars, which are all horizonless solutions that serve as the zero-size limit of hairy Myers-Perry-AdS, black resonators, and hairy black resonators, respectively. All of these solutions compete thermodynamically when they share the same energy and angular momenta. We will compute the full phase diagram of this system. Perhaps surprisingly, we find that hairy black resonators are never dominant in such a phase diagram.
Another advantage of the multiplet scalar model is that different nonlinear solutions generated by different mode-instabilities can be consistently compared with one another, while maintaining a cohomogeneity-1 ansatz. This will allow us to show that black holes generated from higher scalar mode instabilities have higher entropy and occupy a larger region of phase space than those from lower modes. Similar conclusions reached by previous work were only argued by extrapolating perturbative calculations, and here we are able to perform full nonlinear calculations and compute the actual phase boundaries.
This paper is structured as follows. In the next section, we review some basic properties of isometries of $S^3$ and Wigner D-matrices that we use in the construction of our ansatz. Then in section \ref{sec:scalar_coh1}, we describe details of our ansatz. Sections \ref{sec:geonMPAdS} and \ref{sec:BR} review the Myers-Perry-AdS solution, geons, and black resonators in this ansatz, which were studied in \cite{Ishii:2018oms,Ishii:2020muv}. Section \ref{sec:HMPAdS} discusses hairy black holes and boson stars which are higher multiplet versions of those in \cite{Dias:2011at}. Then, in section \ref{sec:HBR}, we present the entirely new hairy black resonator and graviboson star solutions. The entire phase diagram of all solutions then pieced together in section \ref{sec:phase}, and then we compare the results to a higher wavenumber in section \ref{sec:phasej5}. We then finish with some concluding remarks in section \ref{sec:conclude}. The appendix contains technical details.
\section{Hypersphere isometries and Wigner D-matrices}
Our metric ansatz will contain deformations of an $S^3$ whose perturbations can naturally be written in terms of Wigner D-matrices $D^j_{mk}(\theta,\phi,\chi)$. Let us therefore begin with a review of the Wigner D-matrices, which we will later use for designing a cohomogeneity-1 scalar field ansatz.
(See Refs.~\cite{Sakurai:2011zz,Hu:1974hh,Murata:2007gv,Kimura:2007cr,Murata:2008yx,Murata:2008xr,Ishii:2020muv}
for an introduction to the Wigner D-matrix and its applications in gravitational perturbation theory.)
We focus on the $SO(4)\simeq SU(2)_L \times SU(2)_R$ isometry of $S^3$, whose metric can be written as
\begin{equation}
\mathrm{d}\Omega_3^2=\frac{1}{4}(\sigma_1^2+\sigma_2^2+\sigma_3^2)\ ,
\end{equation}
where $\sigma_i \ (i=1,2,3)$ are 1-forms defined by
\begin{equation}
\begin{split}
\sigma_1 &= -\sin\chi \mathrm{d}\theta + \cos\chi\sin\theta \mathrm{d}\phi\ ,\\
\sigma_2 &= \cos\chi \mathrm{d}\theta + \sin\chi\sin\theta \mathrm{d}\phi\ ,\\
\sigma_3 &= \mathrm{d}\chi + \cos\theta \mathrm{d}\phi \ .
\end{split}
\label{inv1form}
\end{equation}
These satisfy the $SU(2)$ Maurer-Cartan equation $\mathrm{d}\sigma_i = (1/2) \epsilon_{ijk} \sigma_j \wedge \sigma_k$.
The coordinate ranges are $0\leq \theta < \pi $, $0\leq \phi <2\pi$, and $0\leq \chi <4\pi$, and have a twisted periodicity $(\theta,\phi,\chi) \simeq (\theta,\phi+2\pi,\chi+2\pi) \simeq (\theta,\phi,\chi+4\pi)$.
The Killing vectors generating $SU(2)_L$ and $SU(2)_R$, denoted by $\xi_i$ and $\bar{\xi}_i$, respectively, are given by
\begin{equation}
\begin{split}
\xi_1 &= \cos\phi\partial_\theta +
\frac{\sin\phi}{\sin\theta}\partial_\chi -
\cot\theta\sin\phi\partial_\phi\ ,\\
\xi_2 &= -\sin\phi\partial_\theta +
\frac{\cos\phi}{\sin\theta}\partial_\chi -
\cot\theta\cos\phi\partial_\phi\ ,\\
\xi_3 &= \partial_\phi\ ,
\end{split}
\label{lkv}
\end{equation}
and
\begin{equation}
\begin{split}
\bar{\xi}_1 &= -\sin \chi \partial_\theta + \frac{\cos \chi}{\sin \theta} \partial_\phi - \cot \theta \cos \chi \partial_\chi \ , \\
\bar{\xi}_2 &= \cos \chi \partial_\theta + \frac{\sin \chi}{\sin \theta} \partial_\phi - \cot \theta \sin \chi \partial_\chi \ , \\
\bar{\xi}_3 &= \partial_\chi \ .
\end{split}
\label{rkv}
\end{equation}
Note that $\bar{\xi}_i$ are the dual vectors of $\sigma_i$:
$(\sigma_i)_\alpha (\bar{\xi}_j)^\alpha=\delta_{ij} \ (\alpha=\theta,\phi,\chi)$.
Using language from quantum mechanics, we can define the ``angular momentum'' operators
\begin{equation}
L_i= i\xi_i \ ,\quad R_i= i \bar{\xi}_i \ ,
\label{SU2LR}
\end{equation}
which satisfy the commutation relations $[L_i,L_j]=i \epsilon_{ijk}L_k$ and $[R_i,R_j]=-i \epsilon_{ijk}R_k$.
These operators are Hermitian under the inner product on the $S^3$,
\begin{equation}
(f,g)\equiv
\frac{1}{8}\int^\pi_0\mathrm{d}\theta\int^{2\pi}_0\mathrm{d}\chi \int_0^{4\pi}\mathrm{d}\chi \, \sin\theta f^\ast(\theta,\phi,\chi)g(\theta,\phi,\chi)\ .
\label{innerprod}
\end{equation}
Under the $SU(2)_L$ and $SU(2)_R$, the 1-forms introduced in Eq.~(\ref{inv1form}) transform as
\begin{equation}
L_i \sigma_j=0\ ,\quad R_i \sigma_j = -i\epsilon_{ijk}\sigma_k\ ,
\label{LRsigma}
\end{equation}
where the operations of $L_i$ and $R_i$ are defined by Lie derivatives.
From the first equation of \eqref{LRsigma}, one can see that $\sigma_i$ are invariant under $SU(2)_L$. For this reason, they are called $SU(2)$-invariant 1-forms.
The second equation means that $R_i$ generate the three-dimensional rotation of the ``vector'' $(\sigma_1,\sigma_2,\sigma_3)$.
In particular, $R_3$ generates $U(1)_R \subset SU(2)_R$, which corresponds to rotation in the $(\sigma_1,\sigma_2)$-plane.
The invariance of $\mathrm{d}\Omega_3^2$ under $SU(2)_L\times SU(2)_R$ can be easily checked by using Eq.~(\ref{LRsigma}).
The generators $L_i$ and $R_i$ share the same Casimir operator: $L^2\equiv L_1^2+L_2^2+L_3^2=R_1^2+R_2^2+R_3^2$, and the set of commutative operators is given by ($L^2$, $L_z$, $R_z$).
The Wigner D-matrix $D^j_{mk}(\theta,\phi,\chi)$ is defined to be the eigenfunction of these operators:
\begin{equation}
L^2 D^j_{mk} = j(j+1) D^j_{mk}\ ,\quad
L_z D^j_{mk} = m D^j_{mk}\ ,\quad
R_z D^j_{mk} = k D^j_{mk}\ ,
\label{Deigen}
\end{equation}
where the ranges of the quantum numbers $(j,m,k)$ are
\begin{equation}
\begin{split}
j&= 0, \, 1/2, \, 1,\, 3/2, \ldots \ , \\
m&= -j, \, -j+1,\ldots, \, j \ , \\
k&= -j, \, -j+1, \ldots, \, j \ .
\end{split}
\label{jmkrange}
\end{equation}
The Wigner D-matrices are orthogonal under the inner product \eqref{innerprod},
\begin{equation}
(D^{j'}_{m'k'},D^j_{mk})=\frac{2\pi^2}{2j+1}\delta_{mm'}\delta_{kk'}\delta_{jj'}\ .
\label{Dorth}
\end{equation}
We can also define the ladder operators $L_{\pm} = L_x \pm i L_y$ and $R_{\pm} = R_y \pm i R_x$, which shift the ``orbital angular momenta'' of $D^j_{mk}$ as
\begin{equation}
\begin{split}
L_+ D^j_{mk} &= \varepsilon_{m+1} D^j_{(m+1)k}\ , \quad
L_- D^j_{mk} = \varepsilon_{m} D^j_{(m-1)k}\ , \\
R_+ D^j_{mk} &= \epsilon_{k+1} D^j_{m(k+1)}\ , \quad \;
R_- D^j_{mk} = \epsilon_{k} D^j_{m(k-1)}\ ,
\end{split}
\label{Dladder}
\end{equation}
where $\varepsilon_m=\sqrt{(j+m)(j-m+1)}$ and $\epsilon_k=\sqrt{(j+k)(j-k+1)}$.
The Wigner D-matrices satisfy a convenient formula for summation,
\begin{equation}
\sum_{m=-j}^j (D^j_{mk'})^\ast D^j_{mk}=\delta_{k'k}\ .
\label{DmDm}
\end{equation}
This can be proved easily using the ladder operators. Using Eqs.~(\ref{Deigen}) and (\ref{Dladder}), we find $L_i (\sum_{m=-j}^j (D^j_{mk'})^\ast D^j_{mk}) = 0$.
Hence, the left hand side of \eqref{DmDm} is a constant.
Integrating this equation over the $S^3$ as Eq.~(\ref{innerprod}) and using Eq.~(\ref{Dorth}), we find that the constant is $\delta_{kk'}$.
For later convenience, we introduce a $(2j+1)$-component vector $\vec{D}_k$ by
\begin{equation}
\vec{D}_k=
\begin{pmatrix}
D^j_{m=j, k} \\
D^j_{m=j-1, k} \\
\vdots\\
D^j_{m=-j, k}
\end{pmatrix}
\ .
\end{equation}
Although $\vec{D}_k$ also depends on the index $j$, we suppress it for notational simplicity because we will generally keep $j$ fixed once the content of the scalar field is specified.
In this notation, Eq.~(\ref{DmDm}) is simply written as
\begin{equation}
\vec{D}_{k'}^\ast \cdot \vec{D}_k = \delta_{k'k}\ .
\label{orthD}
\end{equation}
\section{Cohomogeneity-1 spacetime with rotating scalar field}
\label{sec:scalar_coh1}
We now describe the ansatz for resonating cohomogeneity-1 spacetimes. We will show that the energy-momentum tensor of the matter field we introduce is consistent with the symmetries of the metric and therefore that the equations of motion reduce to a consistent set of ordinary differential equations.
We consider the following five-dimensional Einstein-scalar system with a negative cosmological constant:
\begin{equation}
S=\frac{1}{16\pi G_5}\int \mathrm{d}^5x\sqrt{-g}\left(R+\frac{12}{L^2}-\partial^\mu \vec{\Pi}^\ast \cdot \partial_\mu \vec{\Pi}\right)\ ,
\label{Estnscalar}
\end{equation}
where $\vec{\Pi}$ denotes a $(2j+1)$-component complex scalar multiplet, $G_5$ is the five-dimensional Newton constant, and $L$ is the AdS radius.
Hereafter, we set $L=1$.
For the metric, we take the cohomogeneity-1 ansatz \cite{Ishii:2018oms}
\begin{multline}
\mathrm{d}s^2=-(1+r^2)f(r)\mathrm{d}\tau^2 + \frac{\mathrm{d}r^2}{(1+r^2)g(r)}\\
+\frac{r^2}{4}
\{
\alpha(r)\sigma_1^2+\frac{1}{\alpha(r)}\sigma_2^2+\beta(r)(\sigma_3+2h(r)\mathrm{d}\tau)^2
\}\;,
\label{SU2metric}
\end{multline}
where $\sigma_i$ were defined in \eqref{inv1form}.
For the scalar multiplet, we take
\begin{equation}
\vec{\Pi}(\tau,r,\theta,\phi,\chi)= \sum_{k\in K} \Phi_k(r) \vec{D}_k(\theta,\phi,\chi)\ ,
\label{PiSU2}
\end{equation}
where $\Phi_k(r)$ are real scalar fields, and $K$ is defined by
\begin{equation}
\begin{split}
&K=\{j,j-2,j-4,\cdots,-j\}\quad (j:\textrm{integer}), \\
&K=\{j,j-2,j-4,\cdots,-j+1\}\quad (j:\textrm{half integer})\ .
\end{split}
\label{Kdef}
\end{equation}
We first comment on the metric ansatz \eqref{SU2metric} before later addressing the scalar. This metric ansatz preserves $SU(2)_L$ but breaks $SU(2)_R$. If $\alpha(r)=1$, then a $U(1)_R \subset SU(2)_R$ symmetry generated by $R_3$ is restored. To see this, we can use the fact that
\begin{equation}
\sigma_1^2+\sigma_2^2=\mathrm{d}\theta^2+\sin^2\theta \mathrm{d}\phi^2=\mathrm{d}\Omega_2^2\;,
\end{equation}
which is independent of $\chi$.
In the metric \eqref{SU2metric}, we also assume invariance under two discrete transformations $P_1$ and $P_2$ defined by
\begin{equation}
P_1(\tau,\chi,\phi)=(-\tau,-\chi,-\phi)\ ,\quad
P_2(\tau,\chi,\phi)=(\tau,\chi+\pi,\phi)\ .
\label{P1P20}
\end{equation}
The 1-forms $(\mathrm{d}\tau,\sigma_1,\sigma_2,\sigma_3)$ are transformed by $P_1$ and $P_2$ as
\begin{equation}
\begin{split}
&P_1(\mathrm{d}\tau,\sigma_1,\sigma_2,\sigma_3)=(-\mathrm{d}\tau,-\sigma_1,\sigma_2,-\sigma_3), \\
&P_2(\mathrm{d}\tau,\sigma_1,\sigma_2,\sigma_3)=(\mathrm{d}\tau,-\sigma_1,-\sigma_2,\sigma_3)\ .
\end{split}
\label{P1P2}
\end{equation}
Because of the invariance under $P_1$ and $P_2$, cross terms such as $\sigma_1\sigma_2$ do not appear in Eq.\eqref{SU2metric}.
By examining boundary conditions, it turns out that the metric ansatz \eqref{SU2metric} is taken to be in a frame where asymptotic infinity is rotating. We will search for black holes with a Killing horizon generated by $\partial_\tau$. This condition in turn enforces that $f(r_h)=g(r_h)=0$. For black hole solutions with $\alpha(r)\neq 1$ we must also satisfy $h(r_h)= 0$ (see appendix~\ref{app:bc}).\footnote{If $\alpha(r) \neq 1$, the $U(1)_R$ isometry $\chi \to \chi + \mathrm{const.}$ is broken, and therefore there is no continuous shift of $h(r)$ that does not change the metric.}
Meanwhile, $h(r)$ approaches a constant value $\Omega$ at infinity $r\to\infty$, and the asymptotic form of the metric becomes
\begin{equation}
ds^2\simeq -r^2\mathrm{d}\tau^2 +\frac{\mathrm{d}r^2}{r^2}+\frac{r^2}{4}\{\sigma_1^2+\sigma_2^2+(\sigma_3+2\Omega\mathrm{d}\tau)^2\}\;,
\end{equation}
from which we see that the boundary metric is $R^{(\tau)}\times S^3$, but with rotation in the $\sigma_3$ directions.
This is the rotating frame at infinity.
The meaning of $\Omega$ becomes clear by moving to the non-rotating frame at infinity, which is the natural frame for interpreting conserved charges and other quantities of the black hole.
We can switch to the non-rotating frame by applying the following coordinate transformation:
\begin{equation}
\mathrm{d}t=\mathrm{d}\tau\ ,\quad \mathrm{d}\psi=\mathrm{d}\chi+2\Omega \mathrm{d}\tau\ .
\label{tpsidef}
\end{equation}
In the new frame, the horizon generator is written as\footnote{We use a canonically normalized angular coordinate $\psi/2\in [0,2\pi)$.}
\begin{equation}
\frac{\partial}{\partial \tau}=
\frac{\partial}{\partial t} + \Omega \frac{\partial}{\partial (\psi/2)}\ .
\end{equation}
Therefore, $\Omega$ corresponds to the angular velocity of the horizon.
When $\Omega>1$, the norm of $\partial_\tau$, $g_{\tau\tau}$, becomes positive at infinity.
This implies that there is no global time-like Killing vector in the domain of outer communications, and therefore the spacetime is non-stationary for $\Omega>1$.
If $\alpha\neq1$, the components of the new metric transformed from Eq.~\eqref{SU2metric} become explicitly time dependent. Thus our metric can be said to describe time periodic solutions, even though the metric ansatz is cohomogeneity-1.\footnote{See also Ref.~\cite{Garbiso:2020dys} for a five-dimensional cohomogeneity-1 geometry with periodic time dependence in asymptotically Poincar\'{e} AdS space with a $S^1$ direction.} In our metric ansatz, we can therefore distinguish between time-periodic solutions (with $\alpha\neq1$) and solutions that are stationary (with $\alpha=1$).
Now we comment on the scalar field \eqref{PiSU2}. This ansatz is precisely the ``double stepping'' ansatz ($k$ decreases by 2 in the sum) introduced in the perturbative analysis of black resonators and geons~\cite{Ishii:2020muv}. Indeed, the modes in $K$ decouple from those of its complement $K^c=\{j,j-1,\cdots,-j\}\setminus K$. This fact ultimately stems from the discrete isometry $P_2$ of the metric \eqref{SU2metric}. Under $P_2$, the Klein-Gordon equation for $\vec{\Pi}$ can be decomposed into even and odd parts. More specifically, because $\vec{D}_k\propto e^{-ik\chi}$, the Wigner D-matrices with $k\in K$ and $k\in K^c$ acquire different phase factors of $\pm 1$.
The Klein-Gordon equations for $k\in K$ and $k\in K^c$ are therefore decoupled.
In this paper, we will consider only $k\in K$.
Note also that we have defined our ansatz with fixed $j$ and $2j+1$ multiplet. However, a solution with a particular $j$ is also a solution to the theory with larger multiplets than $2j+1$, simply by setting the extra components of the multiplet to zero. Solutions with different $j$ can therefore be consistently compared with one another.
Finally, we show that the scalar field ansatz \eqref{PiSU2} is consistent with the metric \eqref{SU2metric}, i.e.~the Einstein and Klein-Gordon equations reduce to a consistent set of ODEs.
The Einstein equations from Eq.~(\ref{Estnscalar}) are given by $G_{\mu\nu}-6g_{\mu\nu}=T_{\mu\nu}$, where
the energy-momentum tensor is
\begin{equation}
T_{\mu\nu}=\mathcal{T}_{(\mu\nu)}-\frac{1}{2}g_{\mu\nu} \mathcal{T}\ ,\quad
\mathcal{T}_{\mu\nu}=\partial_{\mu} \vec{\Pi}^\ast\cdot \partial_{\nu} \vec{\Pi} \ ,\quad
\mathcal{T}=g^{\mu\nu}\mathcal{T}_{\mu\nu}\ .
\label{Tdef}
\end{equation}
To derive the explicit expression of the energy-momentum tensor for Eq.~(\ref{PiSU2}), it is convenient to
introduce 1-forms $\sigma_\pm$ defined by
\begin{equation}
\sigma_{\pm}=\frac{1}{2}(\sigma_1\mp i \sigma_2)=\frac{1}{2}e^{\mp i\chi}(\mp i \mathrm{d}\theta+\sin\theta \mathrm{d}\phi)
\end{equation}
and use the basis $e^a=\{\mathrm{d}\tau,\mathrm{d}r,\sigma_+,\sigma_-,\sigma_3\}$ $(a=\tau,r,+,-,3)$.
Their dual vectors are given by $e_a=\{\partial_\tau,\partial_r,e_+,e_-,\partial_\chi\}$ where
\begin{equation}
e_\pm= \bar{\xi}_1\pm i\bar{\xi}_2
=\pm R_{\mp}
=e^{\pm i \chi}(\pm i \partial_\theta + \frac{1}{\sin\theta}\partial_\phi - \cot\theta \partial_\chi)\ .
\end{equation}
One can check that $e^a_\mu e_b^\mu=\delta_b^a$.
In this basis, the derivatives of the scalar field can be evaluated by using Eqs.~(\ref{Deigen}) and (\ref{Dladder}) as
\begin{equation}
\begin{split}
&\partial_\tau \vec{\Pi} =0\ ,\quad
\partial_r \vec{\Pi} =\sum_{k\in K}\Phi'_k \vec{D}_k\ ,\quad
\partial_+ \vec{\Pi} =\sum_{k\in K}\epsilon_k \Phi_k \vec{D}_{k-1}\ ,\\
&\partial_- \vec{\Pi} =-\sum_{k\in K}\epsilon_{k+1} \Phi_k \vec{D}_{k+1}\ ,\quad
\partial_3 \vec{\Pi} =-i\sum_{k\in K}k \Phi_k \vec{D}_k\ ,
\end{split}
\label{Pider}
\end{equation}
where $\partial_a \equiv e_a^\mu\partial_\mu$.
Some components of $\mathcal{T}_{ab}$
vanish because of the double stepping coupling~(\ref{Kdef}). For example, we find
\begin{equation}
\mathcal{T}_{r+}
= \sum_{k,k'\in K}\epsilon_k \Phi'_{k'} \Phi_k \vec{D}_{k'}^\ast \cdot \vec{D}_{k-1}
=\sum_{k,k'\in K}\epsilon_k \Phi'_{k'} \Phi_k \delta_{k',k-1}=0\ ,
\end{equation}
where in the second equality we used Eq.~(\ref{orthD}), and the last one follows from the fact that $k'$ and $k-1$ cannot be equal because of the double stepping of $k$~(\ref{Kdef}).
To evaluate the non-vanishing components of $\mathcal{T}_{ab}$, we also use the orthogonality of the Wigner D-matrices (\ref{orthD}).
The upshot is that the energy momentum tensor is given by
\begin{multline}
\mathcal{T}_{(ab)} e^a e^b
= \sum_{k\in K}\bigg[
\Phi'_{k}{}^2 \mathrm{d}r^2
-\epsilon_{k-1}\epsilon_k \Phi_{k-2}\Phi_k (\sigma_+^2+\sigma_-^2)\\
+(\epsilon_{k}^2+\epsilon_{k+1}^2) \Phi_{k}^2 \sigma_+ \sigma_-
+k^2 \Phi_{k}^2 \sigma_3^2
\bigg]\ .
\label{Tab}
\end{multline}
This result, invariant under $SU(2)_L$, is consistent with the spacetime~(\ref{SU2metric}), and the Einstein and Klein-Gordon equations reduce to a consistent set of coupled ODEs.
The explicit form of the equations of motion is summarized in appendix~\ref{tech}, where technical details in solving the equations are also explained.
In our ansatz of the scalar field~(\ref{PiSU2}),
the conserved current of the complex scalar field $J_\mu$ is given by
\begin{equation}
J_\mu \mathrm{d}x^\mu \equiv \textrm{Im}[\,\vec{\Pi}\cdot \partial_\mu \vec{\Pi}^\ast\,] \mathrm{d}x^\mu
=\sum_{k\in K}k^2\Phi_k^2 \sigma_3\ .
\end{equation}
We have $J_\psi \neq 0$ if and only if $\Phi_k\neq 0$ for $k\neq 0$.
This indicates that there is a rotating flow of the scalar field and
it carries angular momentum when the scalar field has non-trivial $\psi$-dependence.
\section{Geons and superradiant instability of Myers-Perry-AdS}
\label{sec:geonMPAdS}
\begin{figure}
\begin{center}
\includegraphics[scale=0.5]{Egeon.pdf}
\end{center}
\caption{Energy of geons $E_\textrm{geon}$ as a function of the angular momentum $J$.
}
\label{Egeon}
\end{figure}
If $\alpha(r)=1$ in Eq.~(\ref{SU2metric}), there is an exact solution describing a rotating black hole with both angular momenta set equal. It is part of the Myers-Perry-AdS family of solutions, which we will abbreviate MPAdS. In our ansatz, the metric functions are
\begin{equation}
\begin{split}
&g(r)=1-\frac{2\mu (1-a^2)}{r^2(1+r^2)} +\frac{2a^2\mu}{r^4(1+r^2)}\ ,\quad
\beta(r)=1+\frac{2 a^2\mu}{r^4}\ ,\\
&h(r)=\Omega-\frac{2\mu a}{r^4+2 a^2\mu}\ ,\quad
f(r)=\frac{g(r)}{\beta(r)}\ ,\quad
\alpha(r)=1\ .
\end{split}
\label{MPAdSFunctions}
\end{equation}
The event horizon $r=r_h$ is located at the largest root of $g(r_h)=0$, and the isometry group of this solution is $R^{(\tau)} \times SU(2)_L \times U(1)_R$.
The solution is parametrised by $\mu$ and $a$, with $\mu=0$ corresponding to pure AdS.
As written, $\Omega$ is merely a gauge parameter that allows us to move between rotating and non-rotating frames at infinity.
For consistency of notation and convenience, we set
\begin{equation}
\Omega=\frac{2\mu a}{r_h^4+2 a^2\mu}
\end{equation}
so that $\Omega$ is the angular velocity of the horizon. For this choice, we have $h(r_h)=0$ and, thus, $\partial_\tau$ becomes the horizon generator. MPAdS solutions are bounded by extremality, which occurs at
\begin{equation}
\Omega_\mathrm{extr}=\frac{\sqrt{1+2r_h^2}}{\sqrt 2 r_h}\,.
\end{equation}
Before discussing perturbations of MPAdS, let us briefly describe a family of solutions called geons. Geons are horizonless, nonlinear extensions of gravitational normal modes of pure AdS. Within our ansatz, these normal modes are given by a perturbation of the form $\alpha=1+\delta\alpha(r)$ about AdS, with $\Omega$ appearing as an eigenvalue. Geons therefore carry angular momentum, with $\Omega$ as an angular frequency. Fig.~\ref{Egeon} is the energy of geons $E_\textrm{geon}$ as a function of the angular momentum $J$. Later in this paper, we will primarily use the difference of the energy $E$ from $E_\textrm{geon}$ in figures for better visibility. Near vacuum AdS, the geons have frequency parameter $\Omega\simeq 3/2$, and hence the energy scales as $E_\textrm{geon}\simeq (3/2)J$.
\begin{figure}
\centering
\subfigure
{\includegraphics[scale=0.5]{Onsets_rh_Omega_ver2.pdf}\label{onset_Om}
}
\subfigure
{\includegraphics[scale=0.5]{Onsets.pdf}\label{onset_EJ}
}
\caption{
(Left) Onset of the superradiant instability of MPAdS against gravitational perturbation (red) and scalar field perturbations with $j=k=9/2,\,5$ (blue) in the $(\Omega,r_h)$-plane. Above the red and blue curves,
MPAdS is unstable to gravitational and scalar perturbations, respectively.
The extreme limit of MPAdS is shown by the black curve.
(Right) The same data as the left panel are shown in the $(E-E_\textrm{geon},J)$-plane, though for the scalar field perturbations only the onset of the mode with $j=k=9/2$ is shown for visibility.
}
\label{onset}
\end{figure}
Now let us return to perturbations of MPAdS black holes. When the MPAdS black hole has sufficiently high angular frequency, it is unstable to superradiance against both gravitational and scalar field perturbations.
Among the unstable superradiant modes is one that breaks only the $U(1)_R$ isometry of the metric.
Its onset mode of the gravitational perturbation can be found in our ansatz by perturbing the metric function as $\alpha(r)=1+\delta\alpha(r)$ about MPAdS and linearising the equation of motion~(\ref{EOMa}) in $\delta\alpha(r)$.
The onset mode of the scalar field perturbation is just given by the probe scalar field satisfying the Klein-Gordon equation~(\ref{EOMa}).
In the MPAdS background, modes with different $k$ decouple in the Klein-Gordon equation. We will focus on modes with $k=j$, as these are the most dominant.
In Fig.~\ref{onset}, we show results for gravitational and scalar field onsets with $j=k=9/2$ and $j=k=5$. In the left figure, we show the onset of the instabilities in the $(\Omega,r_h)$- and $(E-E_\textrm{geon},J)$-planes. The scalar and gravitational onset curves intersect at green dots. For reference, we also show where MPAdS is extremal, and where $\Omega=1$. Recall that MPAdS is unstable for $\Omega>1$, but not necessarily to the $j=k=9/2$ and $j=k=5$ modes.
In the right figure, we show the same results in the $(E-E_\textrm{geon},J)$-plane,
but only show the result for the scalar for $j=k=9/2$ for visibility.
The onset of the instability for gravitational and scalar field perturbations are shown by red and blue curves, respectively.
These onset curves intersect at green dots.
The extreme MPAdS is also shown by the black curve.
Onset curves of scalar field perturbations terminate at black dots on the extreme MPAdS.
In the insets in the right panel, we zoom in on the regions around the green and black dots for visibility.
Note that, in the left inset, we take $E-2.75J$ as the vertical axis for visibility.
\section{Black resonators}
\label{sec:BR}
As mentioned in the previous section, MPAdS is unstable to superradiance against gravitational perturbations.
A new family of cohomogeneity-1 solutions with $\alpha(r)\neq 1$ branches off from the onset of the instability \cite{Ishii:2018oms}. These onsets were given by the red curves labelled ``gravitational perturbations'' in Fig.~\ref{onset}. Because $\alpha\neq1$ for this new family, these black holes are time-periodic as seen in the non-rotating frame at infinity and are known as black resonators. These black resonators have $R \times SU(2)_L$ isometries. Like MPAdS black holes, these black resonators are a two-parameter family. The horizonless limit of black resonators are geons \cite{Dias:2011ss,Horowitz:2014hja,Martinon:2017uyo,Fodor:2017spc}, which we have already described in the previous section and in Fig.~\ref{Egeon}. Like black resonators, geons also have $R \times SU(2)_L$ isometries and are time-periodic in the non-rotating frame at infinity.
\begin{figure}
\begin{center}
\includegraphics[scale=0.6]{BR_entropy_pm3d.pdf}
\end{center}
\caption{
Entropy of black resonators as a function of $(E-E_\textrm{geon},J)$.
The upper solid curve indicates the onset of the gravitational superradiant instability of the MPAdS, where the black resonators branch off. The extreme MPAdS is shown by a black curve. In the upper-right region below the onset curve, we simply do not have numerical data.
}
\label{Sbr}
\end{figure}
In Fig.~\ref{Sbr}, the entropy of black resonators $S$ is shown by the colour map as a function of $(E-E_\textrm{geon},J)$.
The extreme MPAdS is shown by a black curve. Only MPAdS exists in the upper side of the black curve.
The black resonators branch off from the onset of the superradiant instability shown by the solid curve on the upper edge of the plotted region. In the upper-right white region below the onset curve, we do not have numerical data for black resonators.
For a large $J$, the MPAdS at the onset is very close to the extremality, and numerical construction of the black resonator becomes difficult.
In the limit of geons, $E-E_\textrm{geon} \to 0$, the entropy also approaches zero.
In Ref.~\cite{Ishii:2018oms}, it has been found that the angular velocity of the black resonator and geon always satisfied $\Omega>1$.
Therefore, the black resonator and geon are superradiant and non-stationary.
\section{Hairy Myers-Perry-AdS and boson stars}
\label{sec:HMPAdS}
Thus far, we have discussed MPAdS, black resonators, and geons, which are all solutions that satisfy $\vec{\Pi}=0$. Now we turn to solutions with $\vec{\Pi}\neq0$, beginning with those with $\alpha=1$.
A special case of our ansatz~(\ref{PiSU2}) is given by the scalar field with ``single-$k$'':
\begin{equation}
\vec{\Pi}(\tau,r,\theta,\phi,\chi) = \Phi_k(r)\vec{D}_{k}(\theta,\phi,\chi) \quad (\textrm{no summation})\ .
\label{Pisinglek}
\end{equation}
In this case, the matter stress tensor Eq.~(\ref{Tab}) reduces to
\begin{equation}
\mathcal{T}_{(ab)} e^a e^b
= \Phi'_{k}{}^2 \mathrm{d}r^2 +\frac{1}{4}(\epsilon_{k}^2+\epsilon_{k+1}^2) \Phi_{k}^2 (\sigma_1^2+\sigma_2^2)
+k^2 \Phi_{k}^2 \sigma_3^2\ ,
\label{Tab2}
\end{equation}
where we returned to $\sigma_{1,2}$ from $\sigma_\pm$.
In this expression, the coefficients of $\sigma_1$ and $\sigma_2$ coincide, and therefore it has the invariance under $U(1)_R \in SU(2)_R$ generated by the angular momentum operator $R_3$.
The metric (\ref{SU2metric}) therefore also has this isometry, which implies that $\alpha(r)=1$ in the single-$k$ case.
\begin{figure}
\centering
\subfigure[Energy of boson star]
{\includegraphics[scale=0.45]{boson_star_EvsJ_ver3.pdf}\label{Ebs}
}
\subfigure[Entropy of hairy MPAdS]
{\includegraphics[scale=0.55]{S_Omega_HMPAdS.pdf}\label{SHMPAdS}
}
\subfigure[Domain of hairy MPAdS]
{\includegraphics[scale=0.5]{HMPAdS_region.pdf}\label{HMPAdSdomain}
}
\caption{
(a) Energy of boson stars as a function of the angular momentum $J$.
The existence of multiple turning points are shown in the insets for $j=1/2$.
(b) Entropy of the hairy MPAdS for $k=j=9/2$, parametrized by $J$ and $E-E_\textrm{geon}$.
The solid curve on the left edge indicates the onset of the scalar field superradiant instability of the MPAdS, from which the hairy MPAdS branches off.
(c) Domain of existence of the hairy MPAdS. In orange and red regions, the hairy MPAdS exists. In the red region, the entropy and other physical quantities become multi-valued because of the turning points.
}
\end{figure}
Superradiant instabilities can be induced by scalar fields as well as gravitational fields. These onset curves were shown earlier for the scalar field with $j=k=9/2$ and $j=k=5$ by the blue curves in Fig.~\ref{onset}. For the scalar field, we refer to the solutions branching off from the onset of the superradiant instability as the \textit{hairy MPAdS black holes}, as they contain scalar hair. Because black resonators can be interpreted as black holes with gravitational hair, hairy black holes are the scalar field counterparts to black resonators.
A hairy MPAdS black hole becomes a boson star in the horizonless limit. Boson stars can also be described as nonlinear scalar normal modes of pure AdS. Boson stars are therefore the scalar field counterparts to geons. We obtain the perturbative solution of the boson star near pure AdS in appendix~\ref{largej}, where we also discuss the large-$j$ limit for the perturbative solution.
In the following, we focus on the scalar field with $k=j$, which is the most relevant mode for the superradiant instability for a given $j$.
We note that the hairy MPAdS black hole with $j=1/2$ was constructed in Ref.~\cite{Dias:2011at}.\footnote{
The ansatz~(\ref{Pisinglek}) for $j=1/2$ was first considered in Ref.~\cite{Hartmann:2010pm} for constructing rotating boson stars in asymptotically flat spacetime.}
Our treatment (\ref{Pisinglek}) gives a generalization to $j\geq 1$.
In Fig.~\ref{Ebs}, the energy of boson stars are shown as a function of the angular momentum $J$.
The curves correspond to $j=1/2,1,3/2,\cdots,5$ from left to right.
The difference of the energy from that of the gravitational geon (see Fig.~\ref{Egeon}) is used in the vertical axis.
These results near small $E$ and $J$ agree with perturbation theory about pure AdS. More specifically, the scalar field (\ref{Pisinglek}) with $k=j$ has the lowest normal mode at $\Omega=1+2/j$, from which the boson star branches off.\footnote{The normal mode frequency for the scalar field with the quantum numbers $(j,k)$ is given by $\Omega = (2+j+n)/k$ where $n$ is the radial overtone number. The lowest mode has $n=0$. One can also see that $k=j$ gives the lowest $|\Omega|$.}
For small $J$, the energy of the boson star is given by $E\simeq(1+2/j) J$.
Comparing this with that of the gravitational geon $E_\textrm{geon}\simeq (3/2)J$, one finds that $E-E_\textrm{geon}$ in small $J$ is $\mathcal{O}(J^2)$ for $j=4$ and negative for $j \ge 9/2$.
As $J$ is increased, there are turning points in the energy of boson stars, indicating the change of stability of the boson stars.
It is common for solutions past the turning points to be unstable \cite{poincare1885,Sorkin:1981jc,Sorkin:1982ut,Arcioni:2004ww}.
In the insets of the figure, we zoom in on the curve for $j=1/2$
(here $E-4.5J$ is used in the vertical axis for visibility).
The curve is folded multiple times.
The entropy of the hairy MPAdS for $j=9/2$ is shown in Fig.~\ref{SHMPAdS}.
The angular velocity of the horizon is also shown by the colour map.
For the result of $j=5$, see section \ref{sec:phasej5}.
The hairy MPAdS branches off from the onset of the scalar field superradiant instability denoted by the solid curve on the left edge of the plot region.
Turning points of the energy also exist for hairy MPAdS as shown by the green curve,
and as a result the entropy and other physical quantities of the hairy MPAdS become multi-valued.
We see $\Omega>1$, and this indicates that the hairy MPAdS and boson star are superradiant and non-stationary.
Fig.~\ref{HMPAdSdomain} shows the domain of existence of the hairy MPAdS in the ($E-E_\textrm{geon},J$)-plane. In orange and red regions, the hairy MPAdS exists.
In particular, in the red region, there are multiple solutions for a fixed $(E,J)$, and physical quantities are multi-valued.
Our numerical calculation indicates that the curve of turning points terminates at the black dot: the intersecting point between the extreme MPAdS and the onset of scalar field superradiant instability of MPAdS.
A part of the turning points is shown by the dashed blue curve.
We drew this part by interpolation of our numerical data.
\section{Hairy black resonator and graviboson star}
\label{sec:HBR}
Finally, we consider the most general case in our ansatz (\ref{SU2metric}) and (\ref{PiSU2}) that has both $\alpha(r) \neq 1$ and $\vec{\Pi}(r) \neq 0$.
Ref.~\cite{Ishii:2020muv} has located the onset of the superradiant instability of black resonators for scalar fields.
At the onset of the instability, there is a $\tau$-independent perturbation of the scalar field in the form of Eq.~(\ref{PiSU2}). This is expected to lead to a new family of black resonator solutions with a nontrivial scalar hair.
Similarly, the hairy MPAdS black hole is expected to be unstable against gravitational perturbations with $\alpha(r)\neq 1$ (see appendix~\ref{InstHMPAdS}), and a new family of black resonator solutions is expected to branch from the onset of this instability.
We can therefore have scalar hair on black resonators, and gravitational ``resonator''-type excitations on hairy MPAdS black holes. It turns out both of these excitations are part of the same family of solutions, which we call \textit{hairy black resonators}. Hairy black resonators have their own family of horizonless solutions, which we refer to as \textit{graviboson stars}, as they resemble a combination of a geon and a boson star.
\begin{figure}
\centering
\subfigure[Domain of hairy black resonator]
{\includegraphics[scale=0.5]{HBR_triangle2.pdf}\label{trig}
}
\subfigure[Entropy]
{\includegraphics[scale=0.5]{entropy_3sols_ver2.pdf}\label{S3sol}
}
\subfigure[Cross section at $J=0.8$]
{\includegraphics[scale=0.6]{entropy_3sols_Jslice.pdf}\label{S3sol_Jfix}
}
\caption{
(a) Domain of existence of the hairy black resonator. It exists inside the ``triangle'' surrounded by purple curves.
The equal-entropy-curve between the hairy MPAdS and black resonator is plotted by the orange curve.
The angular velocity is shown by the color map.
(b) Entropies of the black resonator (blue), hairy MPAdS (orange), and hairy black resonator (green) for $j=9/2$.
The entropy of the hairy black resonator is never the largest.
(c) Cross section of (a) at $J=0.8$.
}
\end{figure}
Fig.~\ref{trig} shows the domain of existence of the hairy black resonator.
The hairy black resonator exists in the coloured ``triangular'' domain surrounded by the purple curves.
The colour map corresponds to the angular velocity of the horizon.
The top and bottom-left edges of the triangle correspond to the onset of instability of the hairy MPAdS and black resonator, respectively.
The hairy MPAdS is unstable in the upper side of the top purple curve.
The black resonator is unstable in the lower side of the bottom-left purple curve.
The bottom-right edge is the horizonless limit of the hairy black resonator: graviboson star.
We also show the equal-entropy-curve between the hairy MPAdS and black resonator by the orange curve.
The angular velocity of the hairy black resonator always satisfies $\Omega>1$. Therefore, the hairy black resonator is also superradiant and non-stationary.
We can now compare the entropy of the hairy black resonators to that of black resonators and hairy black holes. Fig.~\ref{S3sol} gives a summary of the entropies of the black resonator (blue), hairy MPAdS (orange), and hairy black resonator (green) for $j=9/2$, which is the smallest $j$ for which black resonators can be unstable. Fig.~\ref{S3sol_Jfix} corresponds to its slice at $J=0.8$.
We find that the entropy of the hairy black resonator is never the largest among the available solutions. Instead, the most entropic solution is either a hairy MPAdS black hole or a black resonator (or MPAdS, but only in regions where none of the other solutions exist).
The entropy changes continuously, but the field configurations are discontinuous across this transition.
The hairy black resonator and graviboson star can be interpreted as a simple example of a multi-oscillating solution.
Recall that the Wigner D-matrix depends on $\chi$ as $D_k(\theta,\phi,\chi) \propto e^{-ik\chi}$.
Then, in the the non-rotating frame at infinity~(\ref{tpsidef}),
the scalar field of the hairy black resonator
can be written as
\begin{equation}
\vec{\Pi}(t,r,\theta,\phi,\psi)
= \sum_{k\in K} e^{2ik\Omega t}\Phi_k (r) \vec{D}_k(\theta,\phi,\psi)\ .
\end{equation}
This has the eigenfrequencies $\omega=2k\Omega$ $(k\in K)$. Since the solution has periodic time dependence on several frequencies, it is multi-oscillating.
In Ref.~\cite{Choptuik:2019zji}, multi-oscillating boson stars with non-commensurate frequencies have been constructed by solving partial differential equations. In this paper, resonating solutions are obtained by solving ordinary differential equations although the frequencies are commensurate.
One important difference in our solutions from those of \cite{Choptuik:2019zji} is the presence of a horizon. In order for solutions to remain steady-state (i.e. independent of $\tau$), fields cannot pass through the horizon. This restricts the frequency of the fields to be multiple of the angular frequency of the horizon, and hence any multi-oscillating solutions must have commensurate frequencies. We expect non-commensurate multi-oscillating geons and boson stars to exist within this theory \eqref{Estnscalar}, but they would neither fall within our ansatz nor be the horizonless limit to a black hole.
\section{Phase diagram}
\label{sec:phase}
Finally, we put all the solutions together in a phase diagram in Fig.~\ref{phasediagram}. We take $j=9/2$ in an ensemble with fixed-$(E,J)$.
We use $E-E_\textrm{geon}$ as the vertical axis for visibility.
\begin{itemize}
\item The extreme MPAdS is located on the black curve. Regular MPAdS black holes exist above this curve, while the MPAdS develops a naked singularity below it.
\item The red curve and line are for the gravitational black resonators and geons.
The curve on the top corresponds to the onset of the gravitational superradiant instability of the MPAdS.
The black resonators branch off from this curve to the bottom.
The horizontal red line in the bottom ($E=E_\textrm{geon}$) expresses the family of the gravitational geons.
The black resonators lie between the red curve and line.
\item The blue curves are associated with the hairy MPAdSs and boson stars.
The upper-left part of the blue curve, from $J=0$ to the black dot at $J=2.285$, is the onset of the scalar field superradiant instability on MPAdS black holes for $k=j=9/2$.
The onset coincides with the extreme MPAdS at the black dot, where the onset terminates.
The family of hairy MPAdS black holes branches off from this curve.
\begin{figure}
\centering
\subfigure[Phase diagram]
{\includegraphics[scale=0.5]{phase_diagram_multiplet_scalar.pdf}\label{phasediagram}
}
\subfigure[Black hole having maximal entropy]
{\includegraphics[scale=0.5]{phase_bdry.pdf}\label{phasebdry}
}
\caption{
(a) Phase diagram of asymptotically AdS solutions in Einstein-multiplet complex scalar fields system for $j=9/2$.
(b) Black hole having maximal entropy among the MPAdS, hairy MPAdS, black resonator, and hairy black resonator for $j=9/2$.
}
\end{figure}
The other blue curve corresponds to the family of boson stars.
There is a maximum in $(E,J)$ at the top right, which is a turning point for the curve for the boson star.
A collection of turning points for hairy MPAdSs, denoted by a blue dashed curve, extends from the top-right tip to the left as $r_h$ is increased.
It appears to extend toward the black dot.
The hairy MPAdSs exist in the region that is apparently enclosed by the blue curves for the onset and the boson star before the turning point, and the blue dashed curve.
In the upper-right region enclosed by the upper boson star onset curve and the dashed curve, physical quantities of hairy MPAdSs become multi-valued.
\item The purple curves denote the boundary of the existing region for the hairy black resonators and graviboson stars.
The bottom-right side of the distorted purple triangle curve is the locations of the family of graviboson stars.
The bottom-left side is the onset of the scalar field superradiant instability of the black resonator \cite{Ishii:2020muv}. The top side is the onset of the gravitational instability of hairy MPAdS.
The hairy black resonators exist in the region enclosed by the purple curves.
\item The orange curve gives the location where the entropies of the black resonators and hairy MPAdSs become equal.
Across the transition, the entropy is continuous, but the field configurations are discontinuous.
The hairy MPAdS has the higher entropy to the right of this curve, while the other side is dominated by the gravitational black resonators.
\end{itemize}
Fig.~\ref{phasebdry} is the phase diagram of the black hole solutions with the maximum entropy in the $(E,J)$-plane. The hairy black resonators never have the largest entropy and hence do not appear in this figure.
\section{Phase diagram for higher $j$}
\label{sec:phasej5}
We can also obtain solutions for the multiplet complex scalar with $j>9/2$. Here, we consider $j=5$. Then, we need to consider $11$-component complex scalar fields at least. Note that by setting one of the scalar multiplet components to zero, the $j=9/2$ solution is also a solution for the theory with $j=5$. So though we have defined a scalar field ansatz for particular $j$'s, the solutions with different $j$ can be consistently compared to one another, so long as we choose the scalar field to have a sufficiently large multiplet.
In Ref.~\cite{Ishii:2020muv}, it was shown that, for an integer $j$, the scalar field is decomposed
into even and odd parity modes under the parity transformation
$P_1$ defined in Eq.(\ref{P1P20}). The even and odd parity modes satisfy $\Phi_{-k}= \Phi_k$ and $\Phi_{-k}=-\Phi_k$ $(k\in K)$, respectively.
In this section, we only consider the even parity mode.
In Fig.~\ref{SHMPAdS5}, we compare the entropies of the hairy MPAdS for $j=9/2$ and $j=5$.\footnote{
In the case of $\alpha(r)=1$, the equations of motion are identical for
$\Phi_{j}(r)=\cos \lambda\, \phi(r)$, $\Phi_{-j}(r)=\sin \lambda \, \phi(r)$
and $\Phi_k=0$ $(|k|\neq j)$ for any value of $\lambda$.
The even and odd parity modes correspond to $\lambda=\pm \pi/2$, and both modes give the same equations of motion.
}
We find that the solution for $j=5$ has higher entropy than that for $j=9/2$ at least in the region labeled as ``hairy MPAdS''
in Fig.~\ref{phasebdry}.
This result suggests that, in a theory with a $(2j'+1)$-component complex scalar field,
a hairy MPAdS with $j<j'$ evolves into that with $j=j'$ by the superradiant instability in the region of a small angular momentum
if we assume $SU(2)_L$ spacetime symmetry.
(Note that the hairy MPAdS with $j=j'$ should be further unstable to $SU(2)_L$-breaking perturbations~\cite{Green:2015kur}.)
Fig.~\ref{phasediagram5} is the phase diagram of solutions with $j=5$.
For the explanation of each curve, see section~\ref{sec:phase}.
The diagram is qualitatively similar to that for $j=9/2$.
Fig.~\ref{phasebdry2j10} is the phase diagram of the black hole solutions with the maximum entropy for $j=5$.
The region in which the hairy MPAdS with $j=5$ entropically dominates is bigger compared to the case of $j=9/2$.
This indicates that, the larger the quantum number $j$ is, the wider the region covered by the hairy MPAdS will be in Fig.~\ref{phasebdry}.
If we extrapolate to arbitrarily large $j$, the black resonator would never dominate the phase diagram in a theory with an infinite number of complex scalar fields.
\begin{figure}
\centering
\subfigure[Entropy for $j=9/2$ and $j=5$]
{\includegraphics[scale=0.65]{entropy_HMPAdS_2sol.pdf}\label{SHMPAdS5}
}
\subfigure[Phase diagram for $j=5$]
{\includegraphics[scale=0.5]{phase_diagram_multiplet_scalar_2j_10.pdf}\label{phasediagram5}
}
\subfigure[Black hole having maximal entropy]
{\includegraphics[scale=0.5]{phase_bdry_2j10.pdf}\label{phasebdry2j10}
}
\caption{
(a) Entropy of the hairy MPAdS for $j=9/2$ (orange) and $j=5$ (green).
(b) Phase diagram of asymptotically AdS solutions in Einstein-multiple complex scalar fields system for $j=5$.
(c) Black hole having maximal entropy among the MPAdS, hairy MPAdS, black resonator, and hairy black resonator for $j=5$.}
\end{figure}
\section{Conclusion}
\label{sec:conclude}
To summarise our results, we have studied asymptotically global AdS solutions of Einstein gravity coupled to a $(2j+1)$ complex scalar multiplet within a cohomogeneity-1 ansatz. The following solutions are available within our ansatz: Myers-Perry-AdS black holes, black resonators (black holes with gravitational hair), black holes with scalar hair, and hairy black resonators (black holes with both gravitational and scalar hair). The latter three of these branch from various superradiant instabilities and have zero horizon limits that are geons, boson stars, and graviboson stars, respectively. The phase diagram of all of these solutions was shown in Figs.~\ref{phasediagram} and \ref{phasediagram5} for $j=9/2$ and $j=5$, respectively.
The entropy of the hairy black resonator is never the largest among the three available solutions as shown in Fig.~\ref{phasebdry} for $j=9/2$.
This seems natural in the view of the perturbative stability. Inside the triangular region enclosed by purple curves in Fig.~\ref{phasediagram}, both of the black resonator and hairy MPAdS is stable against corresponding perturbations.
Thus, the black resonator and hairy MPAdS would not evolve into the hairy black resonator. This is consistent with the fact that the hairy black resonator is entropically subdominant.
Finally, we were able to compare both $j=9/2$ and $j=5$ solutions, and we find that the $j=5$ hairy MPAdS solutions are dominant and cover a larger portion of phase space than those of $j=9/2$. It is natural to expect that the trend continues to higher $j$.
We can make this claim stronger by the following argument.
The phase boundary between hairy MPAdS and black resonators always lies between two points: (1) The intersection between gravitational and scalar onsets of MPAdS, and (2) where the boson stars intersect geons.
Results in Ref.~\cite{Ishii:2020muv} suggest that the point (1) is located at a higher angular momentum for a higher $j$.
Also, by explicitly constructing boson stars for $j\leq 11/2$, we found that the same applies to the point (2) at least for $j=9/2,5,11/2$.
The study of time evolution of this system is an interesting future direction.
If we assume $SU(2)_L$-symmetry in the spacetime,
a time-dependent ansatz for this system would give a $1+1$ dimensional evolution system.
Curiously, though black resonators are unstable to superradiant scalar perturbations and hairy MPAdS are unstable to superradiant gravitational perturbations, the solutions that branch from these instabilities (namely, the hairy black resonators) are never entropically dominant. Therefore, these unstable black resonators or hairy MPAdS cannot evolve to hairy black resonators. Instead, unstable black resonators will likely evolve towards hairy MPAdS, removing its gravitational hair. Similarly, unstable hairy MPAdS will evolve towards black resonators, shedding its scalar hair.
Hairy black resonators themselves can (entropically) evolve to either black resonators or hairy MPAdS,
most likely to whichever is dominant. In all of these case, scalars with $k<j$ will be suppressed, and either the gravitational or $k=j$ scalar field instability will survive.
It would be especially interesting to study the time evolution of a system with large $j$. As we have mentioned, a large $j$ multiplet contains smaller $j$ solutions within it, so the full system contains a tower of lower wavenumbers, many of which are unstable modes in black resonators or MPAdS. These modes have different growth rates, with the largest wavenumber typically being the slowest. However, the hairy black hole with the largest wavenumber is likely the most dominant entropically. A time evolution would therefore tell us how these competing instabilities interact with each other, and how a cascade to higher wavenumbers proceeds.
Though the growth rates of high modes are extremely small and present a significant numerical challenge, such a calculation seems more feasible in this 1+1 setting than the full 3+1 setting of Kerr-AdS. Furthermore, the high-wavenumbers are associated with angular directions rather than the radial direction. As the 1+1 equations do not directly see the angular gradients, the numerical resolution can be kept relatively low, allowing faster time evolution due to the larger Courant number.
A time-dependent ansatz for this system would give a $1+1$ dimensional evolution system. In the discrete isometries~(\ref{P1P2}), we can only assume $P_2$-invariance for the time-dependent spacetime, and the equations of motion require a non-trivial cross term like $\gamma(t,r)\sigma_1\sigma_2$ in the metric for consistency.
Despite the fact that we can construct solutions with several values of $j$, this comes at a cost. In particular, we would like to remind the reader that for each value of $j$ we need $2j+1$ complex scalar fields to make our co-homogeneity one ansatz work. Furthermore, the phases of each of these scalars must be fine tuned so that the overall dependence in the angles do cancel. Perhaps more importantly, all these scalars are minimally coupled to gravity.\footnote{One could potentially add a mass term, and much of our discussion would still go through.} One might ask whether such scalars are easy to come by in consistent reductions from some higher dimensional supergravity theory such as type IIB and the answer appears to be no. To our knowledge the largest known conjectured truncation of type IIB supergravity\footnote{This has actually never been shown in full generality, partially because of the self dual condition imposed
on the Ramond-Ramond F5 form flux, even though interesting progress has been recently made in \cite{Ciceri:2014wya}.} arises when considering compactifications of the form AdS$_5\times S^5$, with the lower dimensional theory being five-dimensional $\mathcal{N}=8$ gauged supergravity comprising a total 42 scalars, 15 gauge fields and 12 form fields. However, these scalars appear to be non-minimally coupled to gravity, and to have very complicated potentials (see for instance \cite{Ciceri:2014wya}), thus giving very little hope that our model will find a precise holographic realisation.
\acknowledgments
We would like to thank Oscar~Dias for useful conversations.
The work of T.~I.~was supported in part by JSPS KAKENHI Grant Number JP18H01214 and JP19K03871.
The work of K.~M.~was supported in part by JSPS KAKENHI Grant Number JP18H01214 and JP20K03976.
BW acknowledges support from ERC Advanced Grant GravBHs-692951 and MEC grant FPA2016-76005-C2-2-P. J.~E.~S. is supported in part by STFC grants PHY-1504541 and ST/P000681/1. J.~E.~S. also acknowledges partial support from a J. Robert Oppenheimer Visiting Professorship.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 381 |
( na numeração romana) foi um ano comum do século IV do Calendário Juliano, da Era de Cristo, a sua letra dominical foi B (52 semanas), teve início a um sábado e terminou também a um sábado.
Acontecimentos
Eusébio torna-se bispo de Cesareia Marítima (data aproximada).
Nascimentos
Cirilo de Jerusalém, teólogo (data eventual, aproximada)
Mortes | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 5,552 |
\section{\bf #1}}}
\newcommand{\SUBSECTION}[1]{\bigskip{\large\subsection{\bf #1}}}
\newcommand{\SUBSUBSECTION}[1]{\bigskip{\large\subsubsection{\bf #1}}}
\begin{titlepage}
\begin{flushright}
{UGVA-DPNC 1995/6-166}
\\
{DESY 95-100}
\\
June 1995
\end{flushright}
\begin{center}
\vspace*{2cm}
{\large \bf
BHAGENE3, A MONTE CARLO EVENT GENERATOR
FOR LEPTON PAIR PRODUCTION AND
WIDE ANGLE BHABHA SCATTERING IN $e^{+} e^{-}$
COLLISIONS NEAR THE Z-PEAK
}
\vspace*{1.5cm}
\end{center}
\begin{center}
{\bf J.H.Field \footnote{ E-mail JFIELD@CERNVM.CERN.CH}}
\end{center}
\begin{center}
{
D\'{e}partement de Physique Nucl\'{e}aire et Corpusculaire
Universit\'{e} de Gen\`{e}ve . 24, quai Ernest-Ansermet
CH-1211 Gen\`{e}ve 4.
}
\end{center}
\begin{center}
{\bf T.Riemann \footnote{ E-mail riemann@ifh.de}}
\end{center}
\begin{center}
{Deutsches Elektronen-Synchrotron DESY
\\
Institut f\"ur Hochenergiephysik, Platanenallee 6, D--15738 Zeuthen, Germany
}
\end{center}
\vspace*{2cm}
\begin{abstract}
A new Monte Carlo event generator for wide angle Bhabha scattering
and muon pair production in $e^+e^-$ collisions is described. The
program includes complete one-loop electroweak corrections, and QED
radiative corrections. The O($\alpha$) QED correction uses the exact
matrix element. Higher order QED corrections are included in an
improved soft photon approximation with exponentiation of initial
state radiation. Events are generated in the full phase space of the
final state including explicit mass effects in the region of collinear
mass singularities. The program is intended for centre of mass energies
around and above the Z peak and for Bhabha scattering at angles greater than
$10^{\circ}$ .
\end{abstract}
\vspace*{1cm}
\end{titlepage}
\begin{center}
{\bf PROGRAM SUMMARY}
\end{center}
\par \underline{\it Title of the program: } BHAGENE3
\par \underline{\it Computer: } IBM 3090
\par \underline{\it Operating system:} VM/CMS
\par \underline{\it Programming language used:} FORTRAN 77
\par \underline{\it High speed storage required:} 678k words
\par \underline{\it No. of bits in a word:} 32
\par \underline{\it Peripherals used:} Line printer
\par \underline{\it Number of cards in combined program and test deck:}
about 6900
\par \underline{\it Keywords:} Radiative corrections, Monte Carlo simulation,
Wide angle Bhabha scattering, Muon pair production, Photons,
Quantum electrodynamics, Electroweak theory .
\par \underline{\it Nature of physical problem:} Calculation of all 1-loop
electroweak corrections, and of QED corrections up to O($\alpha^3$) for wide
angle Bhabha scattering and muon pair production in the vicinity of the Z peak.
\par \underline{\it Method of solution:} The 1-loop electroweak virtual
corrections are calculated analytically. The O($\alpha$) QED radiative
corrections use the
exact matrix element. The higher order QED corrections are given by an improved
soft photon approximation. Event configurations containing real photons are
first generated according to an approximate model, and are then reweighed
according to the theoretical distributions. The weight throwing technique is
used to produce unit weight events in the full final state phase space.
\par \underline{\it Restrictions on the complexity of the problem: } Some
terms in $m_l$/$E$ ($l=e,\mu$) are neglected, so that fermion threshold effects
are not properly taken into account. The approximation used for O($\alpha^2$)
and higher order real photon radiation may be unreliable in events with
multiple
hard photon radiation. For cross section accuracy at the \% level it is not
recommended to use the program for Bhabha scattering angles less than
$10^{\circ}$.
\par \underline{\it Typical running times:} For Bhabha events: initialisation
about 140 sec, generation 709 unit weight events per second. For muon pairs
the initialisation time is shorter by a factor of about 3 and the event
generation rate is roughly doubled. The time unit is an IBM3090 CPU second.
\par \underline{\it Unusual features of the program:} Extensive use is made of
one and two dimensional look-up tables for fast and flexible generation of
Monte-Carlo variables. The space required for these tables means that the fast
memory requirements may be greater than in other comparable programs. These
look-up tables as well as average event weights are created in the
initialisation
phase of the program which is, in consequence, relatively time-consuming.
\par \underline{\it References:}
\begin{itemize}
\item[[1]] J.H.Field, Phys. Lett. B323 (1994) 423
\item[[2]] M.B\"{o}hm, A.Denner and W.Hollik,
Nucl. Phys. B304 (1988) 687.
\item[[3]] F.A.Berends, R.Kleiss and W.Hollik,
Nucl. Phys. B304 (1988) 712.
\item[[4]] D.Bardin, W.Hollik and T.Riemann,
Z. Phys. C Particles and Fields 49 (1991) 485.
\end{itemize}
\SECTION{\bf{Introduction}}
Data on the leptonic decays of the Z taken since 1990 at LEP and SLC
have made possible many precise tests of the Standard Model (SM) of
weak and electromagnetic interactions~\cite{x1}. These electroweak
analyses require Monte Carlo event generators, incorporating both the
SM predictions and the numerically important pure QED radiative
corrections, in order to properly take into account detector acceptance,
efficiency and resolution effects, for arbitrary experimental cuts.
For fermion pair production (excluding $e^{+}e^{-}$), the most widely
used generator is KORALZ~\cite{x2}. This incorporates one-loop
electroweak corrections using the SM, multiple bremsstrahlung photons in
the initial state and a single bremsstrahlung photon in the final state
as well as a complete simulation of $\tau$ decays, including polarisation
effects and QED radiative corrections associated with the $\tau$ decay
products.
For wide angle Bhabha scattering two generators have been written prior
to that, BHAGENE3, described in the present paper. The first, BABAMC
{}~\cite{x3,x4} includes one-loop SM electroweak corrections and QED
corrections
(real and virtual) to O($\alpha$). The second, BHAGEN~\cite{x5}
includes higher order QED corrections (by exponentiation) in both initial
and final states, but is limited to `quasi-elastic' configurations where
the photons are soft, or almost collinear with the radiating lepton.
This limitation is a severe disadvantage when comparing with actual
experimental data, since in general very loose cuts (typically none at
all on the lepton-photon angles) are made in order to minimise the size
of the QED radiative corrections. More recently a new Monte Carlo
generator
UNIBAB~\cite{x6} has been written, which is a general purpose program
in many ways comparable to BHAGENE3.
Electroweak corrections are included and arbitrary numbers of initial
and final state photons generated using a LLA structure function
formalism. Unlike BHAGENE3 however, O($\alpha$) initial/final
interference effects are not included in the current version.
Bhabha generators also exist ~\cite{x7,x8} for the small angle region of
lepton scattering angle $\theta_{l}$ (dominated by the $t$-channel photon
exchange diagram) typical of the angular acceptance of the luminosity
monitors of the LEP/SLC experiments.
The generator described in this paper, BHAGENE3, is adapted to the
description of wide angle Bhabha scattering ($10^{\circ} < \theta_{l}
< 170^{\circ}$), and of $\mu$-pair production, in the region of the Z
peak. $\tau$-decays are not incorporated. Electroweak corrections in the
SM, including the most important two-loop effects, are implemented as
described in Ref.\cite{x9}. Multiple hard photon generation, ( $n_{\gamma}
\leq 2(3)$ for initial (final) states) is also included. No restrictions
are necessary on the kinematical configuration of the generated events.
However, the approximate nature of the hard photon generation algorithm
should be borne in mind when multiple, very hard photon configurations
are considered~\cite{x10}. A brief description of the program has been
given previously, and comparisons of cross-sections and charge
asymmetries made~\cite{x10} with the analytical or semi-
analytical programs:
ZFITTER~\cite{x11}, ALIBABA~\cite{x12} and TOPAZ0~\cite{x13}.
The approach adopted for the QED radiative corrections is to treat the
O($\alpha$) correction exactly (in particular the relevant fermion mass
terms are systematically included throughout, so that collinear photon
radiation is correctly generated) as in BABAMC~\cite{x3,x4} and
MUSTRAAL~\cite{x14}. Higher order real photon radiation is treated, not
by using exact O($\alpha^{2}$) amplitudes~\cite{x15,x16,x17}, but by an
improved soft photon approximation. This may be justified~\cite{x10} by
the relatively small size of the O($\alpha^{2}$),O($\alpha^{3}$), ...
hard photon corrections as compared to that at O($\alpha$). As in
Ref.[3,4,14] the final state phase space is divided into two regions.
A cut ($E^{\gamma}_{0}$) is applied on the energy of each photon. For
$ E^{\gamma} < E^{\gamma}_{0}$ the photon energy and angular
distributions are integrated over, so as to yield a Virtual, Soft (VS)
corrected cross-section. The corresponding generated events have a
`Born topology'; the outgoing leptons being exactly back-to-back with
energy: $ E_{l} = E_{beam} = E $. The generation of such V,S events
is described in Section 3 below. For $ E^{\gamma} > E^{\gamma}_{0} $ a
dilepton event with 1,2,3 hard photons is generated over the full
available phase-space. Denoting the number of initial/final state
photons by $n^{I}_{\gamma}/n^{F}_{\gamma}$, the different possibilities
are:
\[ n^{I}_{\gamma}/n^{F}_{\gamma} = 1/0, 0/1, 2/0, 1/1, 0/2, 0/3 \]
This `hard photon' generation is described in Section 4 below.
A complete description of the program, its structure and parameters,
may be found below in Section 5. In this Introduction only a brief
account is given of the most important techniques employed (see
also Ref.[10]). The first step in the execution of the program is an
initialisation phase where all electroweak parameters are calculated
within the SM from standard input parameters such as the fine
structure constant, $\alpha$, the Fermi constant G, and the masses of
the Z ($M_{Z}$), the top quark ($M_{t}$) and the Higgs boson ($M_{H}$).
Parameters needed for the QED radiative corrections, taking into account
the lepton flavour (e,$\mu$) selected and the beam energy E are also
calculated at this stage. Next the probabilities for VS, `Initial
State hard' (IS) and `Final State hard' (FS) events are calculated.
For the VS events this is done by angular integration of the
differential cross-section, including weak and electromagnetic
radiative corrections. For IS, FS events approximate factorisable
cross-sections are used that may be readily integrated analytically
and/or numerically, to give the corresponding probabilities. Also,
during the initialisation phase, a number of one or two dimensional
Look Up Tables (LUT) for the total photon energy, photon energy
splitting fractions, and lepton scattering angle, are produced.
For each variable the Integrated Probability Distribution (IPD)
is calculated by analytical and/or numerical integration.
The
IPD is then inverted by linear interpolation (see Appendix C) to
yield the LUT. The average weights $\overline{W}_{I}$, $\overline{W}_{F}$
of the IS, IF events are also found, during initialisation, by
re-weighting events according to the `exact' hard cross-section,
to be described below in Section 4.
The event generation phase of the program execution is now entered.
A
VS , IS or FS event is first chosen. For IS, FS events $n^{I}_{\gamma}$
or $n^{F}_{\gamma}$ are then chosen according to a Poisson distribution,
and the appropriate sub-generator is entered. The extensive use of LUT
rather than a Weight Rejection Procedure (WRP) as in Ref.[4,13] gives
a very fast and efficient event generation algorithm. Arbitrary user
defined cuts may be applied during the event generation phase.
Four-vectors
are written out for each unit weight event which is chosen by
using a WRP. In the last stage of execution the
`exact' cross section corresponding to the user-supplied cuts is is
calculated and printed out as well as the user-defined histograms
or other distributions that are up-dated during the generation phase.
Finally in this Introduction a few remarks on the limitations of the
program and some recommended restrictions on its use. Only cross-sections
summed over final and averaged over initial states of the
lepton and photon polarisations are calculated. Only leptonic (not
quark anti-quark) final states may be generated. As in Ref.[3,4,14]
not all terms $ \approx m_{l} / E $ are included, so that the
near-threshold region of lepton pair production is not properly
described. As an approximate model is used to generate hard photons
at O($\alpha^{2}$) and higher, based on an improved soft photon
approximation, cross-sections for configurations with multiple
hard photons should be regarded with caution. Some comparisons
with exact O($\alpha^{2}$) calculations are given in Ref.[10].
The precision of the LUT's used for the lepton scattering angle
requires that $10^{\circ} < \theta_{l} < 170^{\circ}$ in Bhabha
scattering, if a cross-section accuracy of $\approx 1\%$ is required.
\SECTION{\bf{Weak Virtual Corrections}}
In the initialisation, the $W$ mass is determined iteratively from $\alpha,
G_{\mu}, M_Z$ in subroutine SETCON:
\begin{eqnarray}
M_W &=& M_Z \sqrt{1-\sqrt{1-\frac{4\pi\alpha}
{\sqrt{2}G_{\mu}M_Z^2(1-\Delta r)}}}
\label{mwiter}
\end{eqnarray}
where $\Delta r$ is calculated in subroutine SEARC1.
The improved Born approximation for the Bhabha differential cross section
consists of the
sum of contributions with $t$ channel and $s$ channel exchange of the
photon and $Z$ boson and their interferences:
\begin{eqnarray}
\frac{d \sigma^{WC}}{dc}
=
\frac{\pi \alpha^2}{2 s} \sum_{A} \sum_{a} T_a(A)
& A=\gamma, \gamma Z, Z,
& a= s,st,t
\label{v1}
\end{eqnarray}
Here $s$ and $t$ are the usual Mandelstam variables and $c = \cos \theta_l $
where $\theta_l$ is the lepton scattering angle.
In Bhabha scattering the final state helicities are not measurable, while
the beams may be longitudinally polarized.
We introduce the notation:
\begin{eqnarray}
\lambda_1 &=& 1-\lambda_+ \lambda_-
\\
\lambda_2 &=& \lambda_+ - \lambda_-
\\
\lambda_3 &=& 1+\lambda_+ \lambda_-
\label{lamda}
\end{eqnarray}
where $\lambda_{+(-)}$ is the degree of longitudinal polarisation
of the positron (electron) beam.
The s channel cross section contributions are:
\begin{eqnarray}
T_s(\gamma)
&=&
\left| F_A(s)\right|^2
\left[ \lambda_1 \left( 1+c^2 \right) \right]
\\
T_s(\gamma Z)
&=&
2 \Re e
\Bigl\{
\rho(s) \chi(s)
F_A^*(s)
\Bigl[
[ \lambda_1 v_{ee}(s) + \lambda_2 v_e(s) ] (1+c^2)
\nonumber \\
&&
+~ [(\lambda_1+\lambda_2) v_e(s) ] 2c
\Bigr] \Bigr\}
\\
T_s(Z)
&=&
\left|\rho(s) \chi(s) \right|^2
\Biggl\{
\Bigl[
\lambda_1 \left( 1 + 2 |v_e(s)|^2 + |v_{ee}(s)|^2 \right)
\nonumber \\
& &
+~2 \lambda_2 \Re e \left( v_e(s) [1+v_{ee}(s)^*]\right) \Bigr] (1+c^2)
\nonumber \\
&&
+~2 \Re e
\left[
\lambda_1\left(|v_e(s)|^2 + v_{ee}(s)\right)+\lambda_2 v_e(s)
\left( 1+v_{ee}^*(s) \right)
\right]
2c
\Biggr\}
\label{v2}
\end{eqnarray}
Here, we use the following abbreviations:
\begin{eqnarray}
\chi(s) &=& \kappa \frac{s}{s-M_Z^2 + is\Gamma_Z / M_Z}
\label{chiz}
\\
\kappa &=& \frac{G_{\mu}}{\sqrt{2}} \, \frac{M_Z^2}{8\pi \alpha}
\\
v_e(s) &=& 1 - 4 s_W^2 \kappa_e(s)
\\
v_{ee}(s) &=& -1 + 2 v_e(s) + 16 s_W^4 \kappa_{ee}(s)
\label{v3}
\end{eqnarray}
In our expression~(\ref{v1}) for $d\sigma^{WC}/dc$
the axial couplings are equal to unity.
The effective
couplings $v_e(s), v_{ee}(s)$ and the weak form factor $\rho(s)$ are complex
valued.
They contain virtual weak
corrections which are called from the electroweak
library {\tt BHASHA} which is part of the program package described
here.
It has been derived from the electroweak library {\tt DIZET}~\cite{x18}
which is used in the package {\tt ZFITTER}~\cite{x11} and is
intended for the description of $s$ channel fermion pair production in the
region of the Z resonance.
The complete theoretical description of the renormalisation procedure adopted
in the unitary gauge and related topics may be found in~\cite{x19},
the formulae for the $Z$ width in~\cite{x20} and $s$ channel scattering
in~\cite{x11}, and those for Bhabha scattering in~\cite{x9}.
The complex valued
$s$ channel corrections from the fermionic vacuum polarisation are contained
in $F_A(s)$,
\begin{eqnarray}
F_A(s)
&=& \frac{\alpha(-|s|)}{\alpha}
\label{fas}
\end{eqnarray}
and are explained below.
We now describe the t channel contributions:
\begin{eqnarray}
T_t(\gamma)
&=&
F_A(t)^2
\left[
2 \lambda_1
\frac{(1+c)^2}{(1-c)^2}
+ 8 \lambda_3
\frac{1}{(1-c)^2}
\right]
\\
T_t(\gamma Z)
&=&
2 \rho(t) \chi(t) F_A(t)
\nonumber \\
&& \times \left\{
2 \left[ \lambda_1 \left( 1+v_{ee}(t) \right) - \lambda_2 v_e(t) \right]
\frac{(1+c)^2}{(1-c)^2}
- 8 \lambda_3 \left(1-v_{ee}(t)\right)
\frac{1}{(1-c)^2}
\right\}
\nonumber \\
\\
T_t(Z)
&=&
\left[\rho(t) \chi(t) \right]^2
\Biggl\{
2\Biggl[
\lambda_1 \left( 1+4v_e(t)^2+2v_{ee}(t) +v_{ee}(t)^2\right)
\nonumber \\
& &
+4\lambda_2 v_e(t) \left(1+v_{ee}(t)\right) \Biggr]
\frac{(1+c)^2}{(1-c)^2}
+~8\lambda_3\left(1-v_{ee}(t) \right)
\frac{1}{(1-c)^2}
\Biggr\}
\label{v5}
\end{eqnarray}
The following additional abbreviations are used:
\begin{eqnarray}
\chi(t) &=& \kappa \frac{t}{t-M_Z^2}
\\
t&=& -\frac{s}{2}\left(1-c\right)
\label{v6}
\end{eqnarray}
The form factors $F_A(t), v_e(t), v_{ee}(t), \rho(t)$ are real valued.
For wide angle Bhabha scattering, the values of $|t|$ may become substantially
smaller than $s$.
Nevertheless, the form factors do not vary much since they depend only
logarithmically on the scale.
There is one exception to this.
Bremsstrahlung diagrams with initial state radiation
and $t$ channel photon exchange may yield substantial cross section
contributions.
There, the effective value $|t'|$ of $|t|$
may become extremely small and the different value of the running QED coupling
has to be taken into account properly:
\begin{eqnarray}
F_A(t)
&=& \frac{\alpha(|t|)}{\alpha}
\label{fat}
\end{eqnarray}
For this reason the running alpha correction is included not only in
$d\sigma^{WC}/dc$ but also in the cross sections for final states with
hard photons (see below).
Finally, we describe in this section the contributions from the $\gamma
Z$ interference:
\begin{eqnarray}
T_{st}(\gamma)
&=&
- 2 \Re e \left[ F_A^*(s) F_A(t) \right] \lambda_1
\frac{(1+c)^2}{(1-c)}
\\
T_{st}(\gamma Z)
&=&
- 2 \Re e
\left\{
\chi(t) \rho(t) F_A^*(s)
\left[ \lambda_1 \left( 1+v_{ee}(t) \right) - 2 \lambda_2 v_e(t) \right]
+ \left( t \leftrightarrow s \right)
\right\}
\frac{(1+c)^2}{(1-c)}
\nonumber \\
\\
T_{st}(Z)
&=&
-2 \Re e \left\{ \chi(s) \rho(s) \chi(t) \rho(t) \left[
\lambda_1 \left( [1+v_{ee}(s)] [1+v_{ee}(t)]
\right. \right. \right.
\nonumber \\
& &
\left. \left. \left.
+ 4 v_e(s) v_e(t) \right)
- \lambda_2 \left( v_e(s) [1+v_{ee}(t)] + v_e(t) [1+v_{ee}(s)] \right)
\right] \right\}
\frac{(1+c)^2}{(1-c)}
\nonumber \\
\label{v7}
\end{eqnarray}
The running of the QED coupling~$\alpha(Q^2)$
is taken into account as follows:
\begin{eqnarray}
\alpha(Q^2)
&=&
\frac{\alpha}
{1-\Delta \alpha}
\\
\Delta \alpha &=&
\Delta \alpha_l + \Delta \alpha_{udcsb} + \Delta \alpha_t
\label{dalf}
\end{eqnarray}
where we use the convention that in the $s$ channel it is $Q^2=-s$, and in the
$t$ channel $Q^2=|t|$.
The $\Delta \alpha$ is calculated in function XFOTF1.
For leptons:
\begin{eqnarray}
\Delta \alpha_l &=& \sum_{f=e,\mu,\tau}
Q_f^2 N_f
\Delta F_f(Q^2)
\label{vac0}
\\
\Delta F_{f}(Q^2)
&=&
\frac{\alpha}{\pi}
\left\{-\frac{5}{9}+\frac{4}{3}\frac{m_f^2}{Q^2}+\frac{1}{3}
\beta_f
\left(1-\frac{2m_f^2}{Q^2}\right)
\left[
\ln \left| \frac{\beta_f+1}{\beta_f-1} \right| - i \pi \theta(-Q^2-4m_f^2)
\right] \right\}
\label{vac1}
\nonumber \\
\\
\beta_f
&=&
\sqrt{1+\frac{4m_f^2}{Q^2}}
\label{vac2}
\end{eqnarray}
and their colour factor $N_f$ and charge $Q_f$ are unity.
In the weak library, the $\Delta F_f$ is calculated by function XI3,
$\Delta F_f = 2$ XI3.
The contribution from the light quarks has been parametrised in two different
ways.
Either it is calculated with~(\ref{vac0})
(with flag setting NPAR(2)=2); in this case,
we use effective quark masses~\cite{x21}:
$m_u=m_d=0.041, m_s=0.15, m_c=1.5, m_b=4.5$ GeV.
The other, preferred approach (with NPAR(2)=3) uses a parametrisation of the
hadronic vacuum polarisation~\cite{x19}, which is contained in function XADRQQ.
The $t$ quark corrections are:
\begin{eqnarray}
\Delta \alpha_t &=&
Q_t^2 N_t
\Delta F_t(Q^2) + \Delta \alpha
^{\rm{2loop},\alpha\alpha_s}
\label{vact}
\end{eqnarray}
where the latter term contains higher order corrections and is calculated from
functions ALQCDS, ALQCD:
\begin{eqnarray}
\Delta \alpha
^{\rm{2loop},\alpha\alpha_s}
&=&
\frac{\alpha\alpha_s}{3\pi^2}Q_t^2\frac{m_t^2}{Q^2}
{\cal R}e
\Biggl\{
\Pi_t^{VF}(Q^2) + \frac{45}{4}
\Biggr\}
\label{daqcd}
\end{eqnarray}
$\Pi_t^{VF}(Q^2)$ is a two loop self energy
function~\cite{x22,x23}.
Setting the corresponding flag NPAR(3) to zero (default value) results in the
neglect of this very small correction. For NPAR(3)=1,2 approximate, exact
calculations according to Eqn.(2.29) are made.
The first one is not a really good approximation, the second is
very time consuming.
For the function $\Delta F_f(Q^2)$ one may derive the following two
approximations.
For light fermions
with $m_f^2 << |Q^2|$, the following approximate formula is valid:
\begin{eqnarray}
\Delta F_f(Q^2)
\rightarrow
\frac{\alpha}{3\pi}
\left[ \ln\frac{|Q^2|}{m_f^2}-\frac{5}{3}-i\pi\theta(-Q^2)\right]
\label{runalf}
\end{eqnarray}
While, heavy fermions with $m_f^2 >> |Q^2|$
practically decouple:
\begin{eqnarray}
\Delta F_f(Q^2)
\rightarrow
\frac{\alpha}{3\pi}
\left( \frac{4}{15}\frac{-Q^2}{m_f^2}\right)
\label{runalt}
\end{eqnarray}
At LEP~1, the effective QED coupling may be treated as a constant in the $s$
channel~\cite{x24}:
\begin{eqnarray}
F_A(M_Z^2) \approx \frac{137.036}{128.87}
\label{v4}
\end{eqnarray}
The form factors $\kappa_{e}, \kappa_{ee}, \rho$ are calculated
in subroutine
ROKAP($\ldots$, s, t, u,$\ldots$, QE, QF, $\ldots$, XFF,$\ldots$).
XFF is a vector of 4 functions\footnote{
The third of these four functions, $v_f$, is for Bhabha scattering equal to the
second one, $v_e$.
}
which depend on $s$,$t$,$u$ and the charges of the
initial and final state fermions (here minus one):
\begin{eqnarray}
u = -t -s
\label{u}
\end{eqnarray}
and
\begin{eqnarray}
\rho(s) &=& \mbox{XFF}(1; u, -s, t; -1,-1)
\nonumber \\
v_e(s) &=& \mbox{XFF}(2; u, -s, t; -1,-1)
\nonumber \\
v_{ee}(s)&=& \mbox{XFF}(4; u, -s, t; -1,-1)
\\
\rho(t) &=& \mbox{XFF}(1; s, -t, u; -1,-1)
\nonumber \\
v_e(t) &=& \mbox{XFF}(2; s, -t, u; -1,-1)
\nonumber \\
v_{ee}(t)&=& \mbox{XFF}(4; s, -t, u; -1,-1)
\label{v8}
\end{eqnarray}
These corrections are switched on and off with flag NPAR(1);
the order $\alpha^2m_t^4$ contributions to them with NPAR(6),
the $\alpha\alpha_s m_t^2$ corrections with flag NPAR(3), and
the $ZZ$ and $WW$ box terms with NPAR(4).
The latter are negligibly small at LEP~1\footnote{
This statement depends to a certain extent on the calculational scheme
chosen.}.
They introduce in the weak
corrections to the $s$ channel a dependence on the scattering angle.
In the $t$ channel, correspondingly,
the weak corrections will depend not only on the scattering
angle, but
also on s.
The total Z width~\cite{x20} is used in~(\ref{chiz}) and
is calculated in subroutine ZWRATE:
\begin{eqnarray}
\Gamma_Z
&=&
\frac{G_{\mu}M_Z^3}{12\pi\sqrt{2}} \sum_f N_C(f) R_f^{QCD}
\sqrt{1-\frac{4m_f^2}{M_Z^2}}
\left(1+\frac{3}{4}\frac{\alpha}{\pi}Q_f^2\right)
\rho_f^Z
\nonumber \\
&&\times~
\left\{
\left(1+\frac{2m_f^2}{M_Z^2}\right)
\left[1-4 \kappa_f^Z s_W^2 |Q_f| +8 \left(\kappa_f^Z\right)^2 s_W^4 Q_f^2
\right]
-\frac{3m_f^2}{M_Z^2}
\right\}
\label{gammaz}
\end{eqnarray}
The form factors $\rho_f^Z, \kappa_f^Z$ are not identical with the process
dependent form factors for the scattering process although the leading terms of
the latter, if calculated at the $Z$ peak, yield a good approximation of the
first ones.
The overall factor $N_C(f)$ is equal to 1 for leptons and 3 for quarks and
$R_f^{QCD}$ describes final state QCD corrections in case of quarks:
$R_f^{QCD} = 1+\alpha_s/\pi + \ldots$.
The numerical value has to be chosen in accordance with the
scheme and order of the QCD corrections in mind and is set by the user
with the parameters XPAR(10) for the $b$ quark and XPAR(9) for the other, light
quarks.
More details on the $Z$ decay rate may be found in~\cite{x11,x25} and
references therein.
\SECTION{\bf{Virtual and Soft Photonic corrections}}
Three distinct contributions to the virtual and soft radiative
correction may be distinguished:
1) Purely weak virtual corrections
as described in the preceeding Section. These include all self-energy and
vertex corrections involving W, Z and H (Higgs) bosons, as well as ZZ
and WW box diagrams.
2) Electromagnetic virtual corrections: vertex corrections involving
only photons, $\gamma \gamma$ and $\gamma$Z box diagrams and the effect
of both leptonic and hadronic Vacuum Polarisation Insertions (VPI) in all
off-shell photon propagators.
3) Corrections due to real soft photons.
For 1) BHAGENE3 uses the `Dubna-Zeuthen' (DZ) renormalisation scheme
described in Refs.\cite{x19} and [9]. The weak virtual corrections modify the
vector and
axial-vector coupling constants appearing in the $s$, $t$ channel Z-exchange
diagrams according to Eqns. 3.15-3.18 of Ref.[9]. For 2), the O($\alpha$)
vertex and $\gamma \gamma$ and $\gamma$Z box contributions are taken from
Ref.[3], and the VPI contribution from Ref.[27]. Vertex corrections to
O($\alpha^{2}$) and the corresponding exponentiated soft photon
correction, 3), (the integral over the angles and energies of all photons
with total energy below a fixed cut-off: $ y = E^{\gamma}_{tot}/E < y_{0}
$) are given by Ref.[28,29]. The differential cross-section, corrected
for weak loop effects 1), and also the VPI part of 2), is denoted by
$ d^{WC}\sigma/dc $ where $c = \cos \theta_{l}$ , and $\theta_{l}$ is the
lepton scattering angle. Including the soft real photon, the O($\alpha$)
virtual, and the leading log O($\alpha^{2}$) virtual corrections as well
as the $\gamma \gamma$, $\gamma$Z box diagrams, gives the following
expression for the Virtual,Soft differential cross-section:
\begin{eqnarray}
\frac{d\sigma^{VS}}{dc} & = & C_{V}^{i}\left\{\frac{d\sigma^{WC}}{dc}
+\frac{d\sigma^{EC}}{dc}\left[\exp(\beta_{e}\ln\frac{1}{y_{0}})-1
\right]\right\} \nonumber
\\ & & \mbox{} +\frac{d\sigma^{EC}}{dc}\left[
C_{V}^{f}+(\beta_{int}+\beta_{f})\ln\frac{1}{y_{0}}\right]+
\sum_{i=1}^{N}\frac{d\sigma_{i}^{EC}}{dc}\delta_{i}^{EBOX}
\end{eqnarray}
where
\begin{eqnarray*}
\beta_{e} & = & \frac{2\alpha}{\pi}\left[\ln\frac{s}{m_{e}^{2}}-1
\right]\\
\beta_{f} & = & \frac{2\alpha}{\pi}\left[\ln\frac{s}{m_{f}^{2}}-1
\right]\\
\beta_{int} & = & \frac{4\alpha}{\pi}\ln\frac{t}{u}\\
C_{V}^{i} & = & 1+\frac{\alpha}{\pi}\left[\frac{3}{2}\ln(\frac{s}
{m_{e}^{2}})+\frac{\pi^{2}}{3}-2\right]+\frac{9}{8}(\frac{\alpha}{\pi}
)^{2}\ln^{2}(\frac{s}{m_{e}^{2}})-\frac{\pi^{2}}{12}\beta_{e}^{2}
\\
C_{V}^{f} & = & 1+\frac{\alpha}{\pi}\left[\frac{3}{2}\ln(\frac{s}
{m_{f}^{2}})+\frac{\pi^{2}}{3}-2\right]\\
m_{f} & = & \mbox{mass of final state leptons} \end{eqnarray*}
The differential cross section d$\sigma^{EC}$/dc is corrected for the
`running' of $\alpha$ in the s and t-channel photon exchange diagrams by
the replacement:
\begin{equation}
\alpha \rightarrow \alpha (u) = \frac{\alpha}{1+\Pi^{\gamma} (u)}
\end{equation}
where $\alpha$ is the on-shell QED fine structure constant
($\alpha^{-1}$ = 137.036...) and $\Pi^{\gamma}(u)=-\Delta \alpha$ is the photon
proper self energy function, parameterised according to Ref.[27].
The most important virtual weak corrections are also included in
d$\sigma^{EC}$/dc via the replacements ~\cite{x9}
\begin{equation}
G_{\mu} \rightarrow \rho_{e}^{DZ} G_{\mu}, \mbox{\hspace{1cm}} s_{W}^{2}
\rightarrow \kappa_{e}^{DZ}(s) s_{W}^{2}
\end{equation}
where $G_{\mu}$ is the Fermi constant, determined from the muon
lifetime, and $s_{W} = \sin \theta_{W}$ and $\theta_{W}$ is the weak
mixing angle in the on-shell scheme:
\begin{equation}
s_{W}^{2} \equiv 1 - M_{W}^{2}/M_{Z}^{2} \end{equation}
The replacements (3.3) modify the tree level axial vector and vector
coupling constants:
\begin{eqnarray}
g_{A} & = & \frac{1}{4} \left[\frac{\sqrt{2}G_{\mu}M_{Z}^{2}}{\pi \alpha}
\right]\\
g_{V} & = & g_{A}\left[1-4s_{W}^{2}\right] \end{eqnarray}
Since however d$\sigma^{EC}$/dc is multiplied, in Eqn.(3.1) by the
electromagnetic correction of $O(\alpha)$, the effect of the weak
corrections in these terms is quite negligible ( $ \ll 0.1\% $ ) and
so d$\sigma^{EC}$/dc may be considered to be simply
\underline{E}electromagnetically \underline{C}orrected. The label i
in the last term in Eqn.(3.1) denotes the different partial
cross-sections and interference terms (see for example Refs.[3,10,30])
resulting from the two (four) Feynman diagrams that contribute for
$ f \neq e $ ( $ f = e $ ). $\delta_{i}^{EBOX}$ are the corrections
due to $\gamma \gamma $ and $\gamma$ Z box diagrams ~\cite{x3}.
In (3.1) initial state radiation is exponentiated whereas final state
radiation and initial/final interference effects are calculated only to
O($\alpha$). This simplification is justified by the following argument.
The large logarithmic terms at O($\alpha^{2}$), O($\alpha^{3}$),...
associated with initial state radiation do not cancel in the
cross-section when the VS and `hard' contributions, separated by the
arbitrary cut $y_{0}$ on the scaled photon energy, are added, and give
a sizeable (several $\%$) correction. On the other hand, for final state
radiation, because of the KLN theorem ~\cite{x31} the leading logarithms
cancel exactly in the fully integrated cross-section, and cancel almost
completely when only loose cuts are applied. For the level of accuracy
aimed for in BHAGENE3 ( $\approx 0.5 \%$ fractional error in the cross-
section) it is then sufficient to consider, in cross-section calculations
(both $VS$ and `hard'), final state radiation and initial/final
interference effects \footnote{ See Ref.[32] for a discussion of the
effect of initial/final interference in the
charge asymmetry for fermion pair production ($f \neq e$).} effects to
O($\alpha$).
The differential cross-section (3.1) is integrated over the angular range
$ c_{MIN} < c < c_{MAX} $ :
\begin{equation}
\sigma^{VS} = \int_{c_{MIN}}^{c_{MAX}}\frac{d\sigma^{VS}}{dc}dc
\end{equation}
The probability that a $VS$ type event is generated is proportional to
$\sigma^{VS}$. The lepton scattering angle $\theta_{l}$ is generated by
first calculating an integrated probability distribution in the form of
a histogram with content $P_{i}$ in bin $i$ where:
\begin{equation}
P_{i} =\left[ \int_{c_{MIN}}^{c_{i}}\frac{d\sigma^{VS}}{dc}dc\right]/
\sigma^{VS} \end{equation}
Inverting this distribution by linear interpolation yields a LUT for
$c$ with bin index $j$ :
\begin{equation} c_{j} = f(P_{j}) \end{equation}
$c$ is then generated using:
\begin{equation} c_{j} = f(Rn) \end{equation}
In Eqn(3.10) and subsequently $Rn$ is a random number uniformly
distributed
in the interval $ 0 \leq Rn \leq 1 $. $c$ is calculated by linear
interpolation in the LUT (3.9) using the bins $k, k+1$, adjacent to
$P = Rn$ ($ P_{k} \leq Rn \leq P_{k+1}$).
A similar technique (see also Appendix
C) is used throughout the program for the generation of one or two
dimensional distributions. The advantages are:
(i) Complete generality.
(ii) 100\% efficiency. Multiple trials, as in the weight rejection
technique are unnecessary.
(iii) Fast generation. The simple operations needed to perform
linear interpolation in a LUT use, in general, much less computing
time than the calculation of a weight.
The main disadvantage is a certain overhead in computing time during the
initialisation phase, when the LUT's are created. This becomes
progressively less important as large samples of events are generated.
The number of bins used in the LUT depends on the steepness of the
function to be generated and the desired accuracy. Because of the sharp
peaking of the angular distribution near $c$=1 in Bhabha scattering, due
to $t$-channel photon exchange, a finer binning of the integrated
probability distribution (3.8) is used in this region. 200 bins are
assigned to each of the angular regions:
\[ c_{MIN} < c < 0.8c_{MAX} \] and
\[ 0.8c_{MAX} < c < c_{MAX} \]
In the LUT (3.9), 500 uniformly spaced bins are used.
\SECTION{\bf{Hard Photon Corrections}}
\SUBSECTION{
{Generation of Hard Photon Events}}
The `exact' hard photon cross-section is calculated by first generating
events according to simple factorised differential cross-sections,
valid for collinear initial state or final state radiation, and then
reweighting each event according to the procedure described, for example,
in the first of Refs.[4]. `Exact' is written in quotes because the hard
cross section formula used is not the result of an exact, fixed order,
matrix element calculation, but rather an improved soft approximation
comparable to that of Ref.[33]. From now on this distinction will be
understood and the term `exact' used without quotation marks. The exact
cross-section formula (see Appendix B) is derived by exponentiating
the initial state radiation terms of the O($\alpha$) hard cross-section
{}~\cite{x4} in such a way that consistency with Eqn.(3.1) is obtained in
the soft photon limit:
\begin{equation} y = 2 \sum_{i} E_{i}^{\gamma} / \sqrt{s} \ll 1
\end{equation}
The use of such a `soft' approximation for the generation of `hard'
photons, may be justified, in the region of the Z peak, by the small
width of the Z relative to its mass : $\Gamma_{Z}/M_{Z} \approx 0.03$
which strongly damps out hard initial state radiation. Typically $y_{0}$
in Eqn.(3.1) is chosen to be 0.005, corresponding, at the Z peak, to
$ \sum E_{i}^{\gamma} = 225$ $MeV$, so that the majority of `hard'
photons are indeed soft at the scale of $ \Gamma_{Z} = 2.7$ $GeV$.
Exponentiation is not applied to the final state radiation terms, since
the KLN Theorem~\cite{x31} guarantees the smallness of O($\alpha^{2}$),
O($\alpha^{3}$),... corrections in this case, as already discussed in
Section 2 above. Similarly, in accordance with Refs.[11-13,32] the
initial/final interference is also treated only to O($\alpha$).
The approximate differential cross-sections $d \sigma_{A}^{I}$ ,
$d \sigma_{A}^{F}$ for initial, final state radiation respectively,
are:
\begin{equation}
d \sigma_{A}^{I} = \frac{\alpha sy}{16 \pi^{3} \kappa_{+} \kappa_{-}}
\frac{d \sigma_{0} (s',t^{\star})}{d(-t^{\star})}
\left[(1+\delta_{V}^{i}) y^{\beta_{e}}-y + \frac{y^{2}}{2}\right]
d \Omega_{+} d \Omega_{\gamma} dy
\end{equation}
\begin{equation}
d \sigma_{A}^{F} = \frac{\alpha sy}{16 \pi^{3} \kappa'_{+} \kappa'_{-}}
\frac{d \sigma_{0} (s,t)}{d(-t)}
d \Omega_{+} d \Omega_{\gamma} dy
\end{equation}
The notation closely follows that of Ref.[3,4].
4-vectors are defined according to:
\[ e^{+}(p_{+})e^{-}(p_{-}) \rightarrow l^{+}(q_{+})l^{-}(q_{-})
\gamma(k) \]
and then:
\begin{eqnarray}
s = (p_{+}+p_{-})^{2} \mbox{\hspace{1.0cm}} t = (p_{+}-q_{+})^{2}
\mbox{\hspace{1.0cm}} & & u = (p_{+}-q_{-})^{2} \nonumber \\
s' = (q_{+}+q_{-})^{2} \mbox{\hspace{1.0cm}} t' = (p_{-}-q_{-})^{2}
\mbox{\hspace{1.0cm}} & & u' = (p_{-}-q_{+})^{2} \\
\kappa_{\pm} = p_{\pm} \cdot k \mbox{\hspace{1.5cm}}
\kappa'_{\pm} = q_{\pm} \cdot k \mbox{\hspace{1.5cm}} & & \nonumber
\end{eqnarray}
\begin{eqnarray}
d\Omega_{+} = d(\cos \theta_{+}) d \phi_{+} \mbox{\hspace{1.0cm}}
d\Omega_{\gamma} = d(\cos \theta_{\gamma})d \phi_{\gamma}
\mbox{\hspace{1.0cm}} & &
\end{eqnarray}
$\theta_{+}$, $\phi_{+}$ are the polar and azimuthal angles of the
$\l^{+}$ relative to the incoming $e^{+}$ direction, while
$\theta_{\gamma}$, $\phi_{\gamma}$ are the polar and azimuthal
angles of the photon relative to the incoming $e^{+}$ ( $\l^{+}$)
directions for initial (final) state radiation. Also in Eqns.(4.2,4.3):
\begin{equation} t^{\star} = \frac{-s'}{2}(1-\cos \theta^{\star})
\end{equation}
\begin{equation}
\delta_{V}^{i} = \frac{\alpha}{\pi}\left[\frac{3}{2}\ln(\frac{s}
{m_{e}^{2}})+\frac{\pi^{2}}{3}-2\right] \end{equation}
where $\theta^{\star}$ is the $l^{+}$ scattering angle relative to the
incoming $e^{+}$ direction in the Outgoing Dilepton Rest Frame (ODLR).
$d \sigma_{0}$ is the Born level (uncorrected) differential cross-
section for $ e^{+}e^{-} \rightarrow l^{+}l^{-} $. The probabilities
to generate `initial state' (I) or `final state' (F) events, according
to the approximate models are proportional to the cross-sections
$\sigma_{A}^{I}$, $\sigma_{A}^{F}$ given by integrating Eqns.(4.2,4.3)
respectively, over the angles and energies of the outgoing leptons and
photon:
\begin{equation} \sigma_{A}^{I} = \frac{\alpha}{4 \pi} \ln\left(\frac
{s}{m_{e}^{2}}\right) \int_{y_{MIN}}^{y_{MAX}}\frac{dy}{y} \int_{c_{MIN}
^{\star}}^{c_{MAX}^{\star}} \frac{d \sigma_{0}( s',t^{\star})}{d(-t^{
\star})} \left[ (1+\delta_{V}^{i})y^{\beta_{e}}-y+\frac{y^{2}}{2} \right]
d c^{\star} \end{equation}
where
\begin{eqnarray}
s' = s(1-y) \mbox{\hspace{1.0cm}} t^{\star} = -\frac{s}{2}(1-y)
(1-c^{\star})\mbox{\hspace{1.0cm}} & & c^{\star} = \cos \theta^{\star}
\nonumber
\end{eqnarray}
\begin{equation} \sigma_{A}^{F} = \frac{\alpha}{4 \pi} \ln\left(\frac
{s}{m_{f}^{2}}\right) \ln\left(\frac{y_{MAX}}{y_{MIN}}\right)
\int_{c_{MIN}^{+}}^{c_{MAX}^{+}} \frac{d \sigma_{0}(s,t)}{d(-t)}
d c^{+} \end{equation}
where
\[ c^{+} = \cos \theta_{+} \]
The angular integrals over the lepton scattering angle may be performed
analytically~\cite{x4,x34}. The results are given below in Appendix A.
The expression for the exact differential cross-section
$d \sigma^{EXACT}$ is very lengthy and may be found in Appendix B.
The approximate cross-section $d \sigma_{A}^{I}$, and the exact cross-
section $d \sigma^{EXACT}$ contain, because they are exponentiated,
contributions from 2,3,.. hard photons, even though they are functions
of the angular variables of a single `photon'. In fact the angular
variables of all, except one, of the photons are already integrated
out. On performing the angular integration for this `photon'
\footnote{Note that the energy of this `photon' is in fact the summed
energy of all real photons.} a distribution differential in only the
total photon energy (the derivative of Eqn.(3.1) with respect to
$y_{0}$, with the replacement $y_{0} \rightarrow y$) is obtained.
The ansatz employed is to use the same average weight for all pure
initial state or all pure final state topologies, and, in the case
of one initial and one final state photon in the same event, to
use the harmonic mean of the initial and final state weights.
The calculation of the average weights $\overline{W}'_{I}$,
[$\overline{W}'_{f}$] for events generated according to Eqn.(4.2),
[(4.3)] proceeds as follows. At the end of the initialisation phase
of the program $N(1,0)$,and $N(0,1)$ events are generated with relative
probabilities proportional to $\sigma_{A}^{I}$ and $\sigma_{A}^{F}$
according to Eqns.(4.2) and (4.3). The details of the event generation
algorithm used in each case are given below in Sections 4.3.1 and 4.4.1.
The average weights are then given by the expressions:
\begin{equation}
\overline{W}'_{I} = \frac{1}{N(1,0)} \sum_{i=1}^{N(1,0)}
\frac{d \sigma^{EXACT}(\alpha_{k}^{i})}{d \sigma_{A}^{I}(\alpha_{k}^{i})
+ d \sigma_{A}^{F}(\alpha_k^i)} \end{equation}
\begin{equation}
\overline{W}'_{F} = \frac{1}{N(0,1)} \sum_{j=1}^{N(0,1)}
\frac{d \sigma^{EXACT}(\alpha_{k}^{j})}{d \sigma_{A}^{I}(\alpha_{k}^{j})
+ d \sigma_{A}^{F}(\alpha_k^j)} \end{equation}
where $\alpha_{k}^{i}$, ($\alpha_{k}^{j}$) are the $k$ kinematical
variables necessary to completely specify the kinematical configuration
of the $i$th I event ($j$th F event). In the subsequent event generation
phase weights are assigned to multiphoton events according to Table 1.
The relative probabilities of radiating 1,2,3 photons are given by a
Poisson distribution function~\cite{x35}:
\begin{eqnarray}
P_{I}^{(\geq 2 \gamma)}/ P_{I}^{(\geq 1 \gamma)} & = & 1-e^{-r_{e}} \\
P_{F}^{(\geq 2 \gamma)}/ P_{F}^{(\geq 1 \gamma)} & = & 1-e^{-r_{f}} \\
P_{F}^{(\geq 3 \gamma)}/ P_{F}^{(\geq 1 \gamma)} & = & 1-(1+r_{f})
e^{-r_{f}} \end{eqnarray}
where
\begin{equation}
r_{e} = \beta_{e} \ln \left( \frac{1}{y_{0}} \right), \mbox{\hspace{1cm}}
r_{f} = \beta_{f} \ln \left( \frac{1}{y_{0}} \right)
\end{equation}
These probabilities are used to construct the `a priori' event generation
probabilities $P(n_{\gamma}^{I},n_{\gamma}^{F})$ presented in Table 2.
The following definitions are used in this Table:
\begin{eqnarray}
P_{VS} & = & \sigma^{VS}/ \sigma^{A}_{TOT} \\
P_{I} & = & \sigma_{A}^{I}/ \sigma^{A}_{TOT} \\
P_{F} & = & \sigma_{A}^{F}/ \sigma^{A}_{TOT} \end{eqnarray}
where \[ \sigma_{TOT}^{A} = \sigma^{VS} + \sigma_{A}^{I} + \sigma_{A}^
{F} \]
and
\begin{equation}
\rho_{IF} = (S_{I}+S_{F})/(S_{I}+S_{F}+\sqrt{S_{I}S_{F}})\end{equation}
where
\begin{equation}
S_{I} = P_{I}(1-e^{-r_{e}}), \mbox{\hspace{1cm}}
S_{F} = P_{F}(1-e^{-r_{f}})
\end{equation}
In Table 2 $P(2,0)$, $P(0,3)$ are actually the probabilities for
$ \geq 2$, ($ \geq 3$) initial, (final) state photons, so so that initial
states with 2,3,.. photons and final states with 3,4,.. photons are
assigned exactly 2 and 3 photons respectively. $P(1,1)$ is derived from
$P(2,0)$ and $P(0,2)$ using a factorisation ansatz. The probabilities
in Table 2 sum to unity.
For the case of single photon I or F events the weight $W$ is calculated
as:
\begin{equation}
W =
\frac{d \sigma^{EXACT}(\alpha_{k})}{d \sigma_{A}^{I}(\alpha_{k})
+ d \sigma_{A}^{F}(\alpha_k)} \end{equation}
The generation of the kinematical variables $\alpha_{k}$ necessary to
completely define the event configuration is described below.
After checking for consistency with the imposed kinematical cuts
(events failing the cuts are assigned weight zero) events with unit
weight are written out by requiring that:
\begin{equation} W/W_{MAX} > Rn \end{equation}
$W_{MAX}$ is chosen as small as possible (typically $W_{MAX} = 2$) in
order to maximise the efficiency of event generation.
\SUBSECTION{
{Cross-Section Calculation}}
The calculation of the cross-section printed out after completion of
the event generation loop is now described. Suppose that $N$ event
generation trials are made in the loop with $N(n_{\gamma}^{I},
n_{\gamma}^{F})$ trials in each different final state channel
chosen according to the probabilities in Table 2.
Event weights are defined as follows :
\begin{eqnarray}
W_I^i & = &
\frac{d \sigma^{EXACT}(\alpha_{k}^i)}{d \sigma_{A}^{I}(\alpha_{k}^i)
+ d \sigma_{A}^{F}(\alpha_k^i)}, \mbox{\hspace{1cm}} 1 < i < N(1,0) \\
W_F^j & = &
\frac{d \sigma^{EXACT}(\alpha_{k}^j)}{d \sigma_{A}^{I}(\alpha_{k}^j)
+ d \sigma_{A}^{F}(\alpha_k^j)}, \mbox{\hspace{1cm}} 1 < j < N(0,1) \\
W_{VS}^l & = & 1,
\mbox{\hspace{4cm}} 1 < l < N(0,0) \end{eqnarray}
where the $\alpha_k^{i,j}$ are defined after Eqns.(4.10,4.11).
The cross-section $\sigma^{CUT}$ corresponding to the imposed
kinematical cuts (trial events that fail the cuts are assigned weight
zero) is given by:
\begin{equation}
\sigma^{CUT} = \overline{W}(\sigma^{V,S}+\sigma_A^I+\sigma_A^F)
\end{equation}
where
\begin{eqnarray}
\overline{W} & = & [(N(1,0)+N(2,0)\overline{W}_I+(N(0,2)+N(0,2)+N(0,3))
\overline{W}_F \nonumber \\
& & +N(1,1)\sqrt{\overline{W}_I \overline{W}_F}+
N(0,0) \overline{W}_{VS}]/N
\end{eqnarray}
\begin{eqnarray}
\overline{W}_I & = & \frac{1}{N(1,0)} \sum_{i=1}^{N(1,0)} W_I^i \\
\overline{W}_F & = & \frac{1}{N(0,1)} \sum_{j=1}^{N(0,1)} W_F^j \\
\overline{W}_{VS} & = & \frac{1}{N(0,0)} \sum_{l=1}^{N(0,0)}
W_{VS}^l \end{eqnarray}
The weights $\overline{W}_I$, $\overline{W}_F$ defined by (3.26), (3.27)
are identical to $\overline{W}'_{I}$, $\overline{W}'_f$ defined by
(4.10,4.11). Since however, the former are more precise than the
latter, due to the typically smaller statistical errors of the
Monte-Carlo integration in the main generation loop, they are preferred
for the calculation of $\sigma^{CUT}$. The error on $\sigma^{CUT}$,
$\Delta \sigma^{CUT}$ due to the statistical error of the Monte-Carlo
integration is~\cite{x36}:
\begin{equation}
\frac{\Delta \sigma^{CUT}}{\sigma^{CUT}} = [SW2-(SW)^2/N_{IF}]^{\frac
{1}{2}}/SW \end{equation}
where
\begin{eqnarray*}
SW2 & = & \sum_{i=1}^{N(1,0)}(W_I^i)^2+\sum_{j=1}^{N(0,1)}(W_F^j)^2 \\
SW & = & N(1,0)\overline{W}_I+N(0,1)\overline{W}_F \\
N_{IF} & = & N(1,0)+N(0,1) \end{eqnarray*}
\SUBSECTION{
{Initial State Radiation Events}}
\SUBSUBSECTION{\tt{One Initial State Photon}}
The final state 4-vectors of events with a single initial state photon
are generated by the subroutine ZINIGB. These events are first generated
according to the approximate differential cross-section (4.2) and then
re-weighted according to the exact differential cross-section using
Eqn.(4.23).
During the initialisation phase a LUT for the scaled photon energy
$y = E_{\gamma}/E$, with 200 uniformly spaced bins from $y_{MIN}$ to
$y_{MAX}$ is created, as described in Section 2 above. This is done by
integrating Eqn.(4.2) first over $\cos\theta^\star$, then over $y$. The
angular integral is done analytically (see Appendix A), and the y
integral by numerical integration \footnote{Using the CERN
Library program DGAUSS}. At the same time a two dimensional LUT
(200 bins in $y$, 500 in $\cos\theta^{\star}$) is created. Details of the
creation and use of such a two dimensional LUT may be found in Appendix
C. The subroutines used to create these one dimensional (1-D) and two
dimensional (2-D) LUT are SETYD, SETCTD respectively.
The kinematical variables $\alpha_{k}$ used to completely specify
the configuration are, in order of their generation:
\[ y,\mbox{\hspace{0.5cm}} \cos\theta_{\gamma},\mbox{\hspace{0.5cm}}
\phi_+^{\star},
\mbox{\hspace{0.5cm}}\cos\theta^{\star},\mbox{\hspace{0.5cm}}
\phi_{\gamma} \]
The scaled photon energy y is chosen from the 1-D LUT. The angle
$\theta_{\gamma}$ between the incoming $e^{+}$ and the photon is chosen
according to the distribution:
\begin{equation}
\frac{dn}{dc_{\gamma}} \simeq \frac{1}{1-\beta_{IN}^{2}c_{\gamma}^{2}}
\end{equation}
where
\[ c_{\gamma} = \cos \theta_{\gamma},\mbox{\hspace{1.5cm}}
\beta_{IN} = \sqrt{1-\frac{4m_e^2}{s}} \]
In this case the integrated probability distribution is calculated
analytically and inverted to give:
\begin{equation}
c_{\gamma} = (a-1)/[\beta_{IN}(a+1)] \end{equation}
where
\[ a = [(1-\beta_{IN})/(1+\beta_{IN})]\exp [2Rn\ln \frac{1+\beta_{IN}}
{1-\beta_{IN}}] \]
The azimuthal angle $\phi_+^{\star}$ between the planes P1 (defined by
photon and the incoming $e^{+}$), and P2 (defined by the outgoing $l^+$
and the incoming $e^+$), in the outgoing dilepton rest frame (ODLR),
is generated uniformly. Next the 2-D LUT is used, together with the
previously generated value of y to give $\cos\theta^{\star}$. Finally
the angle $\phi_{\gamma}$, giving the orientation of the plane defined
by the photon and the incoming $e^+$ in the incoming $e^+,e^-$ (LAB)
frame, is generated uniformly. To obtain the 4-vectors in the LAB
system of the outgoing $l^+,l^-,\gamma$ a Lorentz transformation
between the ODLR and the LAB frames is required. The corresponding
boost direction is parallel to the photon direction. Denoting by
$\alpha_B$ the angle between the photon and the incoming $e^+$ in
the ODLR frame, the transformation gives:
\begin{eqnarray}
\sin \alpha_B & = & \frac{ \sin \theta_{\gamma}}{1-\beta^{\star}
\cos\theta_{\gamma}} \\
\cos \alpha_B & = & \frac{ \cos \theta_{\gamma}-\beta^{\star}}
{1-\beta^{\star}
\cos\theta_{\gamma}} \end{eqnarray}
where
\begin{equation}
\beta^{\star} = y/(2-y) \end{equation}
The 3-momentum components of the $l^+$ in the ODLR frame:
\begin{eqnarray}
q_+^{(1)\star} & = & q^{\star}s^{\star} \sin \phi_+^{\star} \\
q_+^{(2)\star} & = & q^{\star}s^{\star} \cos \phi_+^{\star} \\
q_+^{(3)\star} & = & q^{\star}c^{\star} \end{eqnarray}
where
\[ s^{\star} = \sin \theta^{\star},\mbox{\hspace{1.0cm}}
c^{\star} = \cos \theta^{\star},\mbox{\hspace{1.0cm}}
q^{\star} = \sqrt{\frac{s(1-y)}{4}-m_l^2} \]
are transformed into the LAB frame using the Lorentz transformation
defined by Eqns.(4.34-4.36). Conservation of 3-momentum in the LAB
frame (the 3-momenta of the $l^+$ and photon being known) then determines
the 3-momentum of the $l^-$. Energy conservation of the complete event is
now checked. In the case of a deviation of more than $\pm 0.5\%$ the
event is rejected and a new one is generated. A similar check is made
for all other hard photon event topologies described in the following
Sections. If the event is accepted, the 3-vectors of all particles
are rotated about the incoming $e^+$ axis, so that the azimuthal
angle of the photon becomes $\phi_{\gamma}$.
\SUBSUBSECTION{\tt{Two Initial State Photons}}
Events with two initial state photons are generated by subroutine
ZINI2G according to an improved soft photon approximation
{}~\cite{x3,x14,x37}. The differential cross-section is:
\begin{eqnarray}
d\sigma^I_{2\gamma} & = & C_{n} d\sigma_{0}(s')
\frac{p_{+} \cdot p_{-}}{(p_{+} \cdot k_{1})
(p_{-} \cdot k_{1})}\frac{p_{+} \cdot p_{-}}{(p_{+} \cdot k_{2})(
p_{-} \cdot k_{2})}\nonumber \\
& & \mbox{} \times (1-\frac{k_{1}}{E}+\frac{k_{1}^{2}}{2E^{2}})
(1-\frac{k_{2}}{E}+\frac{k_{2}^{2}}{2E^{2}}) \frac{d^{3}k_{1}}{k_{1}}
\frac{d^{3}k_{2}}{k_{2}} \nonumber \\
& & \nonumber \\
& = & 4C_{n}\frac{d\sigma_{0}(s') d(y_{1}) d(y_{2}) dy_{1}
d\Omega_{1} dy_{2}d\Omega_{2}}{(1-\beta_{IN}^{2}\cos^{2}\Theta_
{\gamma 1})(1-\beta_{IN}^{2}\cos^{2}\Theta_{\gamma 2})}
\end{eqnarray}
where
\[
s' = M_{l^+l^-}^2,\mbox{\hspace{0.5cm}} E =\sqrt{s}/2,\mbox{\hspace{0.5cm}}
y_i = 2 k_i/\sqrt{s}, \mbox{\hspace{0.5cm}} d(y)=(1-y+
\frac{y^{2}}{2})/y
\]
Here $d\sigma_0(s')$ is the Born differential cross-section and $k_1$,
$k_2$;$d\Omega_1$,$d\Omega_2$ are, respectively, the energies and solid
angle elements of the radiated photons. $\theta_{\gamma 1}$,
$\theta_{\gamma 2}$ are the angles between the photons and the incoming
$e^+$. The factors $d(y_i)$ are the Gribov-Lipatov~\cite{x38} kernels.
In Eqn.(4.40) recoil effects are neglected, whereas in the event
generation algorithm, the change in direction of the radiator due to
the recoil from the first radiated photon (in the case that both photons
are radiated from the same incoming $e^{\pm}$) is allowed for. The photon
angles relative to the radiator are still however generated according to
Eqn.(4.40), so that effects due to the virtuality of the first
recoiling $e^{\pm}$ are not taken into account. Also symmetrisation
effects due to different possible time ordering in the radiation of
two photons from the same incoming line are neglected.
The kinematical variables $\alpha_k$ used to define the configuration of
an event with two initial state photons are, in order of generation:
\[ y,\mbox{\hspace{0.5cm}} \cos\theta_{\gamma 1},\mbox{\hspace{0.5cm}}
\cos\theta_{\gamma 2},\mbox{\hspace{0.5cm}} u,
\mbox{\hspace{0.5cm}} \phi_{\gamma 12},\mbox{\hspace{0.5cm}}
\phi_+^{\star}, \mbox{\hspace{0.5cm}} \cos \theta^{\star},
\mbox{\hspace{0.5cm}} \phi_{\gamma 1} \]
Here $y$ is the scaled total photon energy:
\begin{equation}
2y_{MIN} \leq y = 2(k_1+k_2)/\sqrt{s} \leq 2y_{MAX}
\end{equation}
$u$ is defined as $\ln(y_1/y_2)$ where $y_1$ and $y_2$ are the scaled
energies of the photons:
\begin{equation}
y_{MIN} \leq y_i = 2k_i/\sqrt{s} \leq y_{MAX}
\end{equation}
and $y_1 > y_2$. The angles $\theta_{\gamma 1}$, $\theta_{\gamma 2}$
are generated according to the same distribution (Eqn.(4.32)) as the
angle $\theta_{\gamma}$ for the case of single initial state photon
events.
$\phi_{\gamma 12}$ is the angle between the planes
defined by the incoming beam direction and each of the photons.
The angle $\phi_+^{\star}$ is that between the plane formed by the
summed momentum vector of the two photons and the beam direction,
and the plane formed by the $l^+$ and the beam direction, all in the ODLR
frame. The other variables are defined in the same way as for single
initial state radiation events.
The variables $y$ and $u$ are chosen using LUT generated at
initialisation.
The joint distribution in $y$ and $u$ is derived from that for $y_1$ and
$y_2$ (Eqn.(4.40)):
\begin{equation}
\frac{d^2n}{dy_1dy_2} \simeq \int_{c_{MIN}^{\star}}^{c_{MAX}^{\star}}
d\sigma_0(\tilde{s}',c^{\star})dc^{\star}d(y_1)d(y_2)
\end{equation}
where:
\begin{equation}\tilde{s}' = s(1-y_1-y_2+y_1y_2) \simeq s'\end{equation}
In general the outgoing dilepton effective mass $\sqrt{s'}$ depends also
on the angles of the radiated photons \footnote{The exact expression, in
the present case, is given after Eqn.(4.57) below}.
The expression (4.44) holds for
back-to-back photons, the configuration corresponding to the maximum
value of $y$ in (4.40), and therefore spanning the full possible phase
space. For collinear photons, where $y_{MAX} = 1$, the $y_1y_2$ term in
(4.44) is absent. Separate LUT are thus generated for events with
(approximately) back-to-back or collinear photons. If $\cos \theta_{\gamma 1}$,
$\cos \theta_{\gamma 2}$ have opposite signs (back-to-back configuration)
then Eqn.4.44 is used to define $\tilde{s}'$ in Eqn.4.44 and $y_{MAX}
\simeq 2$. If $\cos \theta_{\gamma 1}$ , $\cos \theta_{\gamma 2}$
have the same sign (collinear configuration), the $y_1y_2$ term in
Eqn 4.44 is omitted and $y_{MAX} \simeq 1$.
By the change of variables:
\begin{eqnarray}
y_1 & = & \frac{y}{2}(1+R) \\
y_2 & = & \frac{y}{2}(1-R) \\
R & = & (e^u-1)/(e^u+1) \\
W & = & \ln(1/y) \end{eqnarray}
(4.43) may be re-written as:
\begin{equation}
\frac{d^2n}{dWdu} \simeq \int_{c_{MIN}^{\star}}^{c_{MAX}^{\star}}
d\sigma_0(\tilde{s}',c^{\star})dc^{\star}d_1d_2
\end{equation}
where:
\[ d_i = 1-y_i+\frac{y_i^2}{2},\mbox{\hspace{0.5cm}}
y_1 = e^{u-W}/(e^u+1),
\mbox{\hspace{0.5cm}} y_2 = e^{-W}/(e^u+1) \]
1-D LUTs for $y=e^{-W}$ are created at initialisation by subroutines
SETYDJ,SETYDL for back-to-back, collinear photons respectively
using Eqn.(4.49). The integral over $c^{\star}$ is performed
analytically, and those over $W$ and $u$ numerically, using, respectively
Simpson's rule and Gaussian integration. At the same time 2-D LUTs in
the variables $W$ and $u$ are created by the subroutines SETRI, SETRL
for back-to-back, collinear photons respectively.
The integration
limits on $W = \ln(1/y)$ are derived from Eqn.(4.41).
Overall energy
conservation leads to the following limits for the variable $u$ :
\begin{eqnarray}
u_{MAX} & = & \ln [(y+y_1^{MAX}-y_2^{MIN})/(y-y_1^{MAX}+y_2^{MIN})]
\\
u_{MIN} & = & -u_{MAX} \end{eqnarray}
where:
\begin{eqnarray*}
y_1^{MAX} & = & \min [ y_{MAX},y-y_{MIN} ] \\
y_2^{MIN} & = & \max [ y_{MIN},y-y_{MAX} ] \end{eqnarray*}
Because the factorised formula (4.49) used to generate the
photon energies does not take into account the angles of the radiated
photons, overall overall energy conservation of the event is checked,
after generation of the variables $\alpha_k$ and before construction
of 4-vectors. In the LAB system, the 3-momentum of the outgoing
dilepton is equal to the total 3-momentum of the two photons. Energy
conservation then requires that:
\begin{eqnarray}
\sqrt{s} & = & k_1+k_2 +[(\vec{k_1}+\vec{k_2})^2+M_{l^+l^-}^2]^\frac
{1}{2} \\
& > & k_1+k_2 +|\vec{k_1}+\vec{k_2}| \end{eqnarray}
The inequality (4.53) follows from the condition that the dilepton
effective mass $M_{l^+l^-}$ is $ > 0 $. (4.53) may also be written as:
\begin{eqnarray}
2 > & &y_1 + y_2 + [ y_1^2 +y_2^2
+ 2y_1y_2(\sin \theta_{\gamma 1}\sin \theta_{\gamma 2}
\sin \phi_{\gamma 12}+\cos \theta_{\gamma 1} \theta_{\gamma 2}]^\frac{1}
{2} \end{eqnarray}
If the R.H.S. of (4.54) is found to be $ > 1.99 $ the current event is
rejected and a new one generated.
In constructing the 4-vectors of the photons, the recoil of the
radiating $e^{\pm}$ is allowed for. The recoil angle, $\alpha_R$ of
the $e^{\pm}$, after photon 1 has been radiated, is:
\begin{equation}
\alpha_R = \arctan \left( \frac{y_1 \sin \theta_{\gamma 1}}
{1-y_1| \cos \theta_{\gamma 1}|} \right) \end{equation}
In the case of radiation of the two photons from the same incoming
line, given by the condition:
\[ \cos \theta_{\gamma 1} \cos \theta_{\gamma 2} > 0 \]
the angle $\theta_{\gamma 2}$ is chosen relative to the direction
of the recoiling $e^{\pm}$, rather than the beam direction.
The remainder of the event generation follows closely the procedure
for the case of one initial state photon, described above. The lepton
scattering angle is chosen using the same 2-D LUT as in the one photon
case. The boost between th LAB and the ODLR frames is now along the
direction of the vector sum ($ \vec{K}$) of the 3-momenta of the two
photons. The angle $\theta_{\gamma}$ in Eqns.(4.34,4.35) is replaced
by $\theta_{K}$, the angle between the incoming $e^+$ direction and
$\vec{K}$. The Lorentz transformation parameter $\beta^{\star}$ of
Eqn.(4.36) is replaced by the expression:
\begin{equation}
\beta^{\star} = \frac{2| \vec{K} |}{\sqrt{s}(2-y)}
\end{equation}
and the lepton momentum in the ODLR frame used in Eqns.(4.37-4.39)
is now given by:
\begin{equation}
q^{\star} = \sqrt{\frac{s'}{4}-m_l^2}
\end{equation}
where
\[ s' = \frac{s}{4}(2-y)^2-(\vec{K})^2 \]
Finally, since events were generated using the approximate value of $s'$,
$\tilde{s}'$ given by Eqns.(4.43) and (4.44) (with the appropriate
modification for collinear photons) the events are re-weighted using a
WRP, to take into account the exact value of $s'$ as given above
after Eqn.(4.57). The weight used is:
\[ d \sigma_{0}(s',c^{*}) / d \sigma_{0}(\tilde{s}',c^{*}) \]
\SUBSECTION{
{Final State Radiation Events}}
\SUBSUBSECTION{\tt{One Final State Photon}}
Events with a single final state photon are generated by the subroutine
ZFINGB, according to the differential cross-section Eqn.(4.3). They
are subsequently re-weighted according to the exact differential
cross-section using Eqn.(4.24)
The kinematical variables $\alpha_{k}$ used to define the event
configuration are, in order of their generation:
\[ y,\mbox{\hspace{0.5cm}} \cos\theta_{\gamma}^{\star},
\mbox{\hspace{0.5cm}} \cos \theta_{\pm},
\mbox{\hspace{0.5cm}}
\phi_{\gamma}',
\mbox{\hspace{0.5cm}}
\phi_{\pm} \]
Following (4.8) the scaled photon energy, $y$, is given by:
\begin{equation}
y = y_{MIN} \exp[ Rn \ln \left( \frac{y_{MAX}}{y_{MIN}} \right)]
\end{equation}
The angle, $\theta_{\gamma}^{\star}$, between the $l^+$ and the photon in
the ODLR frame, is chosen according to the distribution:
\begin{equation}
\frac{dn}{dc_{\gamma}^{\star}} \simeq \frac{1}{1-\beta_{FI}^2
(c_{\gamma}^{\star})^2} \end{equation}
where
\[ c_{\gamma}^{\star} = \cos \theta_{\gamma}^{\star},
\mbox{\hspace{1.5cm}}
\beta_{FI} = \sqrt{1-\frac{4m_l^2}{s}} \]
The scaled energies of the $l^+,l^-$ in the LAB frame, $y_+,y_-$, are
obtained by a Lorentz transformation:
\begin{equation}
y_{\pm} = \frac{2}{\sqrt{s}} \gamma^{\star}
[q^{(0)\star} \pm \beta^{\star} q^{\star}
c_{\gamma}^{\star}]
\end{equation}
where
\begin{eqnarray*}
q^{(0)\star} = \frac{\sqrt{s(1-y)}}{2},
\mbox{\hspace{1.5cm}}& &
q^{\star} = \sqrt{(q^{(0)\star})^2-m_l^2} \\
\beta^{\star} = y/(2-y),
\mbox{\hspace{1.5cm}}& &
\gamma^{\star} = (1-\frac{y}{2})/\sqrt{1-y}
\end{eqnarray*}
The LAB scattering angles $\theta_{\pm}$ of the $l^{\pm}$ are found
using the same 2-D LUT in $y$ and $\cos \theta^{\star}$ as used for initial
state radiation. Here $y$ is set to zero so that $s' \rightarrow s$
and $\theta^{\star} \rightarrow \theta_+$. Following Ref[14], the angle
$\theta_+$ is assigned to the $l^+$, or $\pi-\theta_+$ to the $l^-$,
according to whether, or not :
\begin{equation} P_s > Rn \end{equation}
where
\[ P_s = y_+^2/(y_+^2+y_-^2) \]
This has the effect that, if the photon is radiated from the $l^{\pm}$
then the scattering angle is most likely assigned to the other lepton
$l^{\mp}$, whose direction is unaffected by recoil effects. In the case
that the $\theta_+$ is assigned to the $l^+$, the 3-axis is chosen along
the $l^+$ direction and the 3-momentum components of the $l^+$,$\gamma$
are given, in the LAB system, by the expressions:
\begin{eqnarray}
q_+^{(1)} & = & 0 \nonumber \\
q_+^{(2)} & = & 0 \\
q_+^{(3)} & = & \frac{\sqrt{s}}{2} y_+ \beta_+ \nonumber \end{eqnarray}
\begin{eqnarray}
k^{(1)} & = & \frac{\sqrt{s}}{2} y s_{\gamma}' \cos \phi_{\gamma}'
\nonumber \\
k^{(2)} & = & \frac{\sqrt{s}}{2} y s_{\gamma}' \sin \phi_{\gamma}' \\
k^{(3)} & = & \frac{\sqrt{s}}{2} y c_{\gamma}'
\nonumber \end{eqnarray}
where
\[ c_{\gamma}' = 1-2[1-(1-y)/y_+]/y,\mbox{\hspace{1.0cm}}
s_{\gamma}' = \sqrt{1-(c_{\gamma}')^2},\mbox{\hspace{1.0cm}}
\beta_+ = \sqrt{1-\frac{4m_l^2}{s y_+^2}} \]
Here $\phi_{\gamma}'$ is the azimuthal angle of the photon about the
$l^+$ direction. If $\pi-\theta_+$ is assigned to the $l^-$, the
3-axis is chosen along the $l^-$ direction, and the 3-momentum
components of the $l^-$, $\gamma$ are given in the LAB system as:
\begin{eqnarray}
q_-^{(1)} & = & 0 \nonumber \\
q_-^{(2)} & = & 0 \\
q_-^{(3)} & = & \frac{\sqrt{s}}{2} y_- \beta_- \nonumber \end{eqnarray}
\begin{eqnarray}
k^{(1)} & = & \frac{\sqrt{s}}{2} y s_{\gamma}''\cos \phi_{\gamma}'
\nonumber \\
k^{(2)} & = & \frac{\sqrt{s}}{2} y s_{\gamma}''\sin \phi_{\gamma}' \\
k^{(3)} & = & \frac{\sqrt{s}}{2} y c_{\gamma}''
\nonumber \end{eqnarray}
where
\[ c_{\gamma}'' = 1-2[1-(1-y)/y_-]/y,\mbox{\hspace{1.0cm}}
s_{\gamma}'' = \sqrt{1-(c_{\gamma}'')^2},\mbox{\hspace{1.0cm}}
\beta_- = \sqrt{1-\frac{4m_l^2}{s y_-^2}} \]
The momentum components in the standard LAB system with 3-axis along
the incoming $e^+$ direction, are found by rotating the above 3-vectors
by $\theta_+$ (or $\pi- \theta_+$) about the 1-axis, as appropriate.
The azimuthal angle $\phi_+$,($\phi_-$) of the $l^+$, ($l^-$) about the
3-axis is then generated uniformly. Finally the 3-momentum of the
remaining lepton is given by momentum conservation in the LAB frame.
\SUBSUBSECTION{\tt{Two Final State Photons}}
Events with two final state photons are generated by the subroutine
ZFIN2G according to the factorised differential cross-section
{}~\cite{x3,x14,x37}:
\begin{eqnarray}
d\sigma^F_{2\gamma} & = & C_{n} d\sigma_{0}(s)
\frac{q_{+} \cdot q_{-}}{(q_{+} \cdot k_{1})
(q_{-} \cdot k_{1})}\frac{q_{+} \cdot q_{-}}{(q_{+} \cdot k_{2})(
q_{-} \cdot k_{2})}\nonumber \\
& & \mbox{} \times (1-\frac{k_{1}}{E}+\frac{k_{1}^{2}}{2E^{2}})
(1-\frac{k_{2}}{E}+\frac{k_{2}^{2}}{2E^{2}}) \frac{d^{3}k_{1}}{k_{1}}
\frac{d^{3}k_{2}}{k_{2}} \nonumber \\
& & \nonumber \\
& = & 4C_{n}\frac{d\sigma_{0}(s) d(y_{1}) d(y_{2}) dy_{1}
d\Omega_{1} dy_{2}d\Omega_{2}}{(1-\beta_{FI}^{2}\cos^{2}\Theta_
{\gamma 1}')(1-\beta_{FI}^{2}\cos^{2}\Theta_{\gamma 2}')}
\end{eqnarray}
The variables in (4.66) are defined after Eqns.(4.40) and (4.59), with
the exception of $\theta_{\gamma 1}'$, $\theta_{\gamma 2}'$, which are
the angles between the photons 1,2 and the $l^+$ in the LAB system. To
take into account recoil effects however, when $\cos \theta_{\gamma 2}'
< 0$ the photon 2 is assigned the angle $\pi-\theta_{\gamma 2}'$
relative to the $l^-$ direction rather than $\theta_{\gamma 2}'$
relative to the $l^+$ direction.
The kinematical variables $\alpha_{k}$ in the order of generation are:
\[ y,\mbox{\hspace{0.5cm}} cos\theta_{\gamma 1}',\mbox{\hspace{0.5cm}}
cos\theta_{\gamma 2}',\mbox{\hspace{0.5cm}} u,
\mbox{\hspace{0.5cm}}cos\theta_{\pm},\mbox{\hspace{0.5cm}}
\phi_{\gamma 12}', \mbox{\hspace{0.5cm}} \phi_{\gamma T}',
\mbox{\hspace{0.5cm}} \phi_{\pm} \]
Here $\theta_{\pm}$ are
the scattering angles of the $l^{\pm}$ relative to the incoming $e^+$
in the LAB frame. The azimuthal angle $\phi_{\gamma12}'$ is chosen to
be that between the planes defined by : $(\vec{k_1}, \vec{q_+})$;
$(\vec{k_2}, \vec{q_+})$ for $\cos \theta_{\gamma 2}' > 0$ and between
the planes: $(\vec{k_1}, \vec{q_-})$; $(\vec{k_2}, \vec{q_-})$ for
$\cos \theta_{\gamma 2}' < 0$. $\phi_{\gamma T}'$ is the azimuthal
angle about the $l^+$ direction of the total momentum vector
$\vec{K_2}$ of the two photons. $\phi_{\pm}$ are the azimuthal angles
of the $l^{\pm}$ relative to the incoming $e^+$. All three angles
are generated uniformly in the interval 0 to $2 \pi$ radians. The
polar angles $\theta_{\gamma 1}'$, $\theta_{\gamma 2}'$ are generated
according to Eqn.(4.32) with the replacements $\theta_{\gamma}
\rightarrow \theta_{\gamma}'$, $\beta_{IN} \rightarrow \beta_{FI}$.
The variables $y$ and $u$ ($ y = y_1+y_2$, $u =\ln (y_1/y_2)$) are chosen
from, respectively, 1-D and 2-D LUT's generated at initialisation.
The distribution (4.49) is here replaced by:
\begin{equation}
\frac{d^2n}{dWdu} \simeq d_1d_2
\end{equation}
where $W = \ln(1/y)$. The subroutines that generate the 1-D LUT for
$y$ and the 2-D LUT for $W$ are SETYDK, SETRK respectively. The angle
$\theta_+$ is generated as described above for single final state
photon events.
The construction of the 3-vectors of the final state particles is
now described. Unlike for the case of single final state photon
radiation it is here more convenient to work entirely in the LAB
frame. Choosing the 3-axis along the $l^+$ direction, with the
2-axis in the plane containing $\vec{q_+}$ and $\vec{k_1}$ the
momentum components of photon 1 are:
\begin{eqnarray}
k^{(1)}_1 & = & 0
\nonumber \\
k^{(2)}_1 & = & \frac{\sqrt{s}}{2} y_1 s_{\gamma 1}' \\
k^{(3)}_1 & = & \frac{\sqrt{s}}{2} y_1 c_{\gamma 1}'
\nonumber \end{eqnarray}
where
\[ s_{\gamma 1}' = \sin \theta_{\gamma 1}',\mbox{\hspace{1.0cm}}
c_{\gamma 1}' = \cos \theta_{\gamma 1}' \]
In the case that
$c_{\gamma 2}' > 0$ (photon 2 radiated from the $l^+$) the momentum
components of photon 2 are given by:
\begin{eqnarray}
k^{(1)}_2& = &\frac{\sqrt{s}}{2}y_2 s_{\gamma 2}'\cos \phi_{\gamma 12}'
\nonumber \\
k^{(2)}_2& = &\frac{\sqrt{s}}{2}y_2 s_{\gamma 2}'\sin \phi_{\gamma 12}'
\\
k^{(3)}_2& = &\frac{\sqrt{s}}{2}y_2 c_{\gamma 2}'
\nonumber \end{eqnarray}
For $c_{\gamma 2}' < 0$ (photon 2 radiated from the $l^-$) the direction
of the 3-axis in Eqn.(4.69) is chosen opposite to the direction of the
$l^-$. In this case the momentum components are still constructed
according to Eqn.(4.69), but they are subsequently rotated about the
1-axis through the acollinearity angle $\alpha_{A}$ between the $l^+$,
$l^-$ directions. This angle is calculated from the equations:
\begin{eqnarray}
\sin \alpha_A & = & y_1 s_{\gamma 1}' / y_- \nonumber \\
y_- & = & 2-y_1-(1-y_1)/[1-0.5(1-c_{\gamma 1}') y_1] \\
\cos \alpha_A & = & \sqrt{1-\sin^2 \alpha_A} \nonumber
\end{eqnarray}
By this procedure the strong peaking of the radiated photons along the
$l^{\pm}$ directions is properly accounted for, even in the case of
recoils generated by the first radiated photon \footnote{ In Eqn.(4.69)
the angle $\theta_{\gamma 2}'$ is that between the photon and the
radiating (virtual) $l^-$, rather than that between the photon and the
outgoing $l^-$ as in Eqn.(4.66). This difference is of little importance
for soft or nearly collinear photons. For hard non-collinear photons
however the rate will be over-estimated as compared to Eqn.(4.66). In
the case that both photons are radiated from the $l^+$ the ansatz used
tends, on the contrary to underestimate the rate of such photons as
compared to Eqn.(4.66)}. Before proceeding to construct the 3-vectors
of the outgoing leptons energy conservation is checked using Eqn.(3.51).
If the R.H.S. is $ > 0.99 \sqrt{s}$ the event is rejected and a new one
is generated.
The scaled energies $y_{\pm}$ of the $l^{\pm}$ are next calculated:
\begin{eqnarray}
y_+ & = & 2[1-y(1-\frac{y}{4})-\frac{(\vec{K_2})^2}{s}]
/(2-y+\frac{2K_2^{(3)}}{\sqrt{s}}) \\
y_- & = & 2-y-y_+ \end{eqnarray}
As in Eqns.(4.68,4.69) the 3-axis is along the $l^+$ direction.
The angle $\theta_+$ is assigned to $l^+$ or $\pi-\theta_+$ to the $l^-$,
as described above for single photon final state radiation. The
azimuthal angle of the total photon momentum $\vec{K_2}$ is then
generated uniformly by assigning the angle $\phi_{\gamma T}'$. The
3-axis is now rotated into the direction of the incoming $e^+$ and the
angle $\phi_+$ (or $\phi_-$) is assigned. Finally the 3-momentum
of the remaining lepton is found using momentum conservation in
the LAB system.
\SUBSUBSECTION{\tt{Three Final State Photons}}
Events with three final state photons are generated, according to the
obvious generalisation of Eqn.(4.66), by the subroutine ZFIN3G.
The event configuration is defined by the following kinematical
variables:
\[ y,\mbox{\hspace{0.5cm}} \cos\theta_{\gamma 1}',\mbox{\hspace{0.5cm}}
\cos\theta_{\gamma 2}',\mbox{\hspace{0.5cm}} \cos\theta_{\gamma 3}',
\mbox{\hspace{0.5cm}} u',\mbox{\hspace{0.5cm}} u,
\mbox{\hspace{0.5cm}} \cos\theta_{\pm},\mbox{\hspace{0.5cm}}
\phi_{12}', \mbox{\hspace{0.5cm}} \phi_{\gamma3}',
\mbox{\hspace{0.5cm}} \phi_{\gamma T}',
\mbox{\hspace{0.5cm}} \phi_{\pm} \]
$\theta_{\gamma 3}'$ is similarly defined to $\theta_{\gamma 1}'$,
$\theta_{\gamma 2}'$ but refers to the third (least energetic) photon.
The variable $u'$ is given by:
\begin{equation} u' = \ln\left(\frac{y_1+y_2}{y_3}\right)
\end{equation}
The definition of $\phi_{\gamma 3}'$ is similar to that given above
for the case of 2 final state photons except that the total photon
momentum vector $\vec{K_3}$ corresponds to three photons. The angle
$\phi_{\gamma 3}'$ is that between the plane defined by the total
momentum of the first two (most energetic) photons $\vec{K_2}$
and the $l^+$ (or $l^-$), and that defined by the direction of
photon 3 and $l^+$ (or $l^-$). $l^+$ ($l^-$) is taken for
$\cos\theta_{\gamma 3}' > 0 ( < 0 )$.
All other variables have been defined
previously.
Following the soft photon limit of the generalisation of Eqn.(4.66) the
photon energies are first generated according to the distribution:
\begin{equation}
\frac{d^3n}{dy_1dy_2dy_3} \simeq \frac{1}{y_1y_2y_3} \end{equation}
By a change of variables:
\[ W=\ln[1/(y_1+y_2+y_3)],\mbox{\hspace{1.0cm}}
u = \ln(y_1/y_2),\mbox{\hspace{1.0cm}}
u' = \ln[(y_1+y_2)/y_3] \]
(4.74) simplifies to
\begin{equation}
\frac{d^3n}{dWdudu'} \simeq constant \end{equation}
A 1-D LUT for the variable $W$, and a 2-D LUT for $u'$ (for a given $W$)
are created at initialisation by the subroutines SETYD3, SETRK3
respectively. The limits on the variable $u$ in the nested integration:
\[ I = \int dW \left\{\int du'\left [ \int du \right] \right\} \]
are
\begin{eqnarray}
u_{MIN} = \ln \left( \frac{y'-y_1^{MAX}}{y_1^{MAX}}\right) \\
u_{MAX} = \ln \left( \frac{y'-y_1^{MIN}}{y_1^{MIN}}\right)
\end{eqnarray}
where
\begin{eqnarray*}
y^{MIN}_1 = \max[y'-y_{MAX},y_{MIN}] \\
y^{MAX}_1 = \min[y'-y_{MIN},y_{MAX}]
\end{eqnarray*}
and
\[ y'=y_1+y_2=y e^{u'}/(1+e^{u'}) \]
The limits on the $u'$ integration are given by Eqns.(4.50,4.51) with
the
replacements $u \rightarrow u'$ and
\begin{eqnarray*}
y^{MIN}_1 \rightarrow \max[2y_{MIN},y-y_{MAX}] \\
y^{MAX}_1 \rightarrow \min[y_{MAX},y-2y_{MIN}]
\end{eqnarray*}
The integral over $u$ is done analytically, that over $u'$ by Gaussian
integration, and that over $W$ by Simpson's rule. During the event
generation phase the value of $W$ is chosen from the 1-D LUT, The value
of $u'$ is then chosen using the 2-D LUT. Finally, $u$ is given by:
\begin{equation} u = u_{MIN}+Rn(u_{MIN}-u_{MAX}) \end{equation}
where $u_{MIN}$, $u_{MAX}$ are given by Eqns.(4.76,4.77).
$y_1$, $y_2$, $y_3$ are then derived from $W$,$u$, $u'$ via the
equations:
\begin{eqnarray}
y & = & e^W \nonumber \\
y' & = & ye^{u'}/(1+e^{u'}) \nonumber \\
y_1& = & y'e^{u}/(1+e^{u}) \\
y_2& = & y'-y \nonumber \\
y_3& = & y -y' \nonumber
\end{eqnarray}
The photon spectra are then modified by weight rejection to take into
account the hard photon corrections given by the Gribov-Lipatov~\cite{x38}
kernels.
Defining the weight function:
\begin{equation} {\cal W} = d(y_1)d(y_2)d(y_3)y_1y_2y_3 \end{equation}
where $d(y)$ is defined after Eqn.(4.40), the event is rejected if:
\[ {\cal W} < Rn \]
The construction of the momentum vectors of the final state particles
follows closely that described in the previous section for the case
of two photons. The 3-vectors of the two most energetic photons are
given by Eqns(4.68,4.69) above, and the angle $\theta_{\gamma 2}'$
is chosen relative to the $l^+$ direction (opposite of the $l^-$
direction ) according to whether $\cos \theta_{\gamma 2}'$ is $> 0$ or
($ < 0$). Energy conservation is then checked using Eqn.(4.54). To take
into account recoil effects in the radiation of the third photon
the acollinearity angle between the $l^+$ and $l^-$ after the first
two photons have been radiated is calculated, and $\theta_{\gamma 3}'$
is chosen relative to the $l^+$ direction (opposite of the $l^-$
direction, for $\cos \theta_{\gamma 3}' > 0$ ($ < 0$). That is,
the procedure used above for the second photon is iterated. The
azimuthal angle $\phi_{\gamma 3}'$ is chosen to be that between
the planes defined by ($\vec{K_2},\vec{q_+}$) ; ($\vec{k_3},\vec{q_+}$)
for $\cos \theta_{\gamma 3}' > 0$, and that between
($\vec{K_2},\vec{q_-}$) ; ($\vec{k_3},\vec{q_-}$)
for $\cos \theta_{\gamma 3}' < 0$. Since the relative directions of
all three photons are now fixed energy conservation is again checked
using the generalisation of Eqn.(4.53):
\begin{equation}
\sqrt{s} > k_1+k_2 +k_3+|\vec{k_1}+\vec{k_2}+\vec{k_3}|
\end{equation}
If the RHS of Eqn(4.81) is $ > 0.99\sqrt{s}$, the event is rejected.
Eqns.(4.71,4.72)
with the replacement $\vec{K_2} \rightarrow \vec{K_3}$ is
now used to calculate $y_+$, $y_-$. The scattering angle $\theta_+$
is assigned to the $l^+$, or $\pi-\theta_+$ to the $l^-$, according to
Eqns.(4.61). The angles $\phi_{\gamma T}'$ and $\phi_+$ or
$\phi_-$ are assigned as described above for the two photon case.
Finally the 3-momentum of the remaining lepton is calculated from
overall momentum conservation.
\SUBSUBSECTION{\tt{Events with One Initial State and One Final State
Photon}}
Events with one initial state and one final state photon are
generated by the subroutine ZINF2G according to the differential
cross-section (c.f Eqns.(4.40,4.66)):
\begin{eqnarray}
d\sigma^IF_{2\gamma} & = & C_{n} d\sigma_{0}(s'')
\frac{p_{+} \cdot p_{-}}{(p_{+} \cdot k_{1})
(p_{-} \cdot k_{1})}\frac{q_{+} \cdot q_{-}}{(q_{+} \cdot k_{2})(
q_{-} \cdot k_{2})}\nonumber \\
& & \mbox{} \times (1-\frac{k_{1}}{E}+\frac{k_{1}^{2}}{2E^{2}})
(1-\frac{k_{2}}{E}+\frac{k_{2}^{2}}{2E^{2}}) \frac{d^{3}k_{1}}{k_{1}}
\frac{d^{3}k_{2}}{k_{2}} \nonumber \\
& & \nonumber \\
& = & 4C_{n}\frac{d\sigma_{0}(s'') d(y_{1}) d(y_{2}) dy_{1}
d\Omega_{1} dy_{2}d\Omega_{2}}{(1-\beta_{IN}^{2}\cos^{2}\Theta_
{\gamma 1})(1-\beta_{FI}^{2}\cos^{2}\Theta_{\gamma 2}')}
\end{eqnarray}
where
\[ s'' = s(1-y_1) \]
and photons 1,(2) are radiated in the initial,(final) state.
The kinematical variables used, in this case, to define the event
configuration are:
\[ y,\mbox{\hspace{0.5cm}} cos\theta_{\gamma 1} ,\mbox{\hspace{0.5cm}}
cos\theta_{\gamma 2}',\mbox{\hspace{0.5cm}} u,
\mbox{\hspace{0.5cm}}cos\theta_+'',\mbox{\hspace{0.5cm}}
\phi_+'', \mbox{\hspace{0.5cm}} \phi_{\gamma 2}'',
\mbox{\hspace{0.5cm}} \phi_{\gamma 1} \]
Here $\theta_{\gamma 2}''$ is the angle between photon 2 and the $l^+$
in the rest frame of $l^+l^-\gamma 2$ (ODLGR frame). In this frame the
1-axis is chosen perpendicular to the plane defined by the incoming
$e^+$ and the direction of photon 1. $\theta_+''$, $\phi_+''$ are
the polar and azimuthal angles of the $l^+$ relative to the
incoming $e^+$ direction (allowing, if necessary, for the recoil
generated by photon 1) in the ODLGR frame. The other variables
have been previously defined.
The variables $y$, $u$ are generated using 1-D, 2-D LUT created at
initialisation by the subroutines SETYDI, SETR respectively.
The procedure is the same as that described in Section 4.3.2 above.
The 2-D LUT is created according to Eqn.(4.49) with the replacement:
$\tilde{s}' \rightarrow s''$. The configuration of $l^+l^- \gamma 2$
is generated as described in Section 4.4.1, for single final state
radiation, but in the ODLGR frame rather than the LAB frame. Recoil
effects from the initial state photon are accounted for by rotation by
the angle $\alpha_B$ given by Eqns.(4.34-4.35) with the replacement
$y \rightarrow y_1$ in the formula (4.36) for $\beta^{\star}$. Since
the scaled photon energies $y_1$, $y_2$ are defined in the LAB frame,
the energy of photon 2 must first be calculated, by Lorentz
transformation, in the ODLGR frame, from its angles in this frame
(specified by $\theta_{\gamma 2}''$, $\phi_{\gamma 2}''$,
$\theta_+''$, $\phi_+''$ ) and its LAB energy $\frac{\sqrt{s}}{2} y_2$.
For simplicity, in this case, $\theta_+''$ is always assigned to the
$l^+$, rather than $\theta_+''$ to the $l^+$ or $\pi-\theta_+''$ to the
$l^-$ according to Eqn(4.61) \footnote{The procedures described in
Ref.[14] for assigning the scattering angle in the case of single
initial or final state photons are not directly applicable in this case}.
Finally the 4-vectors of $l^+$, $l^-$ and photon 2 are tranformed back into
the lab. frame and the event is rotated about the incoming $e^+$ direction
so that photon 1 has azimuthal angle $\phi_{\gamma 1}$.
\SECTION{\bf{Program Structure and Performance}}
\SUBSECTION{General Organisation. How to use the Program}
The program has a very short main program containing definitions of the most
important input parameters, which are stored in the labelled common block ICOM.
These variables are described in Table 3. The execution of the program has
three distinct phases:
\begin{itemize}
\item Initialisation
\item Generation of a single unit weight event
\item Termination
\end{itemize}
Each of these phases is entered via a call to subroutine BHAGENE3 in the main
program:
\\[.5cm]
CALL BHAGENE3(MODE,CTP1,CTP2,CTM1,CTM2,CTAC,EP0,EM0)
\\[.5cm]
MODE is set to $-1,0,1$ for the initialisation, generation and termination
phases respectively. The other parameters of BHAGENE3 define the kinematical
cuts to be applied to the generated events:
\\[.5cm]
\hspace*{2cm} CTP1 $=$ minimum value of $\cos \theta_{l^+}$
\\
\hspace*{2cm} CTP2 $=$ maximum value of $\cos \theta_{l^+}$
\\
\hspace*{2cm} CTM1 $=$ minimum value of $\cos \theta_{l^-}$
\\
\hspace*{2cm} CTM2 $=$ maximum value of $\cos \theta_{l^-}$
\\
\hspace*{2cm} CTAC $=$ maximum value of $\cos \phi_{col}$
\\
\hspace*{2cm} EP0 $=$ minimum energy of $l^+$ ($GeV$)
\\
\hspace*{2cm} EM0 $=$ minimum energy of $l^-$ ($GeV$)
\\
\\[.5cm]
All these cuts are applied in the laboratory (incoming $e^+,e^-$ centre of
mass) system. The angle $\phi_{col}$ is the collinearity angle between the
$l^+$ and the $l^-$ (CATC $=$ -1 for a back-to-back configuration).
In the calls of BHAGENE3 with MODE $=$ 0,1 only this parameter need be
specified.
A typical main program to generate 5000 unit weight events might be:
\\[.5cm]
\\
PROGRAM BHAMAIN
\\
IMPLICIT REAL*8 (A-H,O-Z))
\\
COMMON/ICOM/OIMZ,OIMT,OIMH,OMAS,IOCHA,IOEXP,OW,OCTC1,OCTC2,IOXI
\\
OIMZ=91.18D0~~~~~~~~~~~~~~~~~~! {\sf Z Mass (GeV)}
\\
OIMT=150.0D0~~~~~~~~~~~~~~~~~~! {\sf Top quark mass (GeV)}
\\
OIMH=100.0D0~~~~~~~~~~~~~~~~~~! {\sf Higgs boson mass (GeV)}
\\
OMAS=0.12D0~~~~~~~~~~~~~~~~~~~! {\sf Alphas }
\\
IOCHA=1~~~~~~~~~~~~~~~~~~~~~~~~~! {\sf 0 for Muon pairs, 1 for
electron pairs }
\\
IOEXP=1~~~~~~~~~~~~~~~~~~~~~~~~~! {\sf 0 for O(Alpha), 1 for
exponentiation }
\\
OW=91.00D0~~~~~~~~~~~~~~~~~~~~~! {\sf Collision energy (GeV) }
\\
OCTC1=-0.8D0~~~~~~~~~~~~~~~~~~! {\sf Lower cos(theta) limit in ODLR frame }
\\
OCTC2=0.8D0~~~~~~~~~~~~~~~~~~~! {\sf Upper cos(theta) limit in ODLR frame }
\\
IOEXI=713883717~~~~~~~~~~~~~~~! {\sf Intial random number }
\\
\\
\hspace*{3cm} ! {\sc KINEMATICAL CUTS FOR GENERATED EVENTS }
\\
\\
CTP1=-0.7D0~~~~~~~~~~~~~~~~~~~~! {\sf Lower cos(theta) for l+ in the
lab frame}
\\
CTP2=0.7D0~~~~~~~~~~~~~~~~~~~~~! {\sf Upper cos(theta) for l+ in the
lab frame}
\\
CTM1=-0.7D0~~~~~~~~~~~~~~~~~~~~! {\sf Lower cos(theta) for l- in the lab
frame}
\\
CTM2=0.7D0~~~~~~~~~~~~~~~~~~~~~! {\sf Upper cos(theta) for l- in the lab
frame}
\\
CTAC=-0.9D0~~~~~~~~~~~~~~~~~~~~! {\sf Upper cosine of collinearity angle }
\\
EP0=2.0D0~~~~~~~~~~~~~~~~~~~~~~~~! {\sf Minimum energy of l+ (GeV) }
\\
EM0=2.0D0~~~~~~~~~~~~~~~~~~~~~~~! {\sf Minimum energy of l- (GeV) }
\\
\\
\hspace*{3cm} ! {\sc INITIALISATION PHASE }
\\
\\
CALL BHAGENE3(-1,CTP1,CTP2,CTM1,CTM2,CTAC,EP0,EM0)
\\
\\
\hspace*{3cm} ! {\sc GENERATION LOOP }
\\
\\
DO J=1,5000
\\
CALL BHAGENE3(0)
\\
ENDDO
\\
\\
\hspace*{3cm} ! {\sc TERMINATION PHASE }
\\
\\
CALL BHAGENE3(1)
\\
STOP
\\
END
\\
\\
Other initialisation parameters of interest to users are defined in
BHAGENE3 itself. A list of the most important of these can be found in Table 4.
\SUBSECTION{Initialisation Phase}
A flow chart of the initialisation phase of the program is shown in Fig 1.
The functionality of the different subprograms shown there has been described
above. The main physical quantities calculated are also indicated. In order to
estimate the average weights of events with $\geq$ two hard photons the
initialisation phase actually includes the generation of $4\times 10^{4}$
events,
which is relatively time consuming. Users should be aware of this.
\SUBSECTION{Generation Phase}
As shown in the flow chart in Fig 2. each of the different event topologies is
generated by a different subprogram. The calculations performed by the five
different
hard photon subgenerators : ZINIGB, ZFINGB, ZFIN2G, ZFIN3G, and ZINF2G
are described above
in Section 4. The 4-vectors of the generated events are written in the
labelled
common
block C4VEC :
\\[.5cm]
\hspace{2cm} COMMON/C4VEC/PPV(4),PMV(4),
QPV(4),QMV(4),GAM1V(4),GAM2V(4),GAM3V(4),\newline
WEIGHT,NPHOT,ISG
\\[.5cm]
The 4-vectors (defined as: $p_x,p_y,p_z,E$ ) are in the order:
($e^+,e^-,l^+,l^-,
\gamma_1,\gamma_2,\gamma_3 $ ). WEIGHT is the event weight, NPHOT the number
of
filled
photon 4-vectors and ISG a flag indicating the subgenerator that produced
the event
(see Table 2.). The photons are ordered in energy, the first photon being the
most
energetic.
\SUBSECTION{Termination Phase}
A flow chart of the termination phase is shown in Fig 3. The exact
cross-section
($\sigma^{CUT}$) and its error ($\Delta\sigma^{CUT}$) are printed out,
together
with
the input parameters (including those calculated in the initialisation phase).
Other
cross-sections used to calculate the event generation probabilities
$P(n_{\gamma}^I,n_{\gamma}^F)$ are also printed out.
A sample output is shown in Fig 4.
\SUBSECTION{Program Performance}
As mentioned above, after a relatively lengthy initialisation procedure, during
which all look-up tables are created and average weights are calculated for
multiphoton events, the event generation procedure is itself fast.
Typical times for the initialisation phase are 140, 48 IBM 3090 CPU seconds for
$e^+e^-$, $\mu^+\mu^-$ pair generation, respectively. Average times to generate
a single unit weight $e^+e^-$, ($\mu^+\mu^-$) event are $1.41\times 10^{-3}$,
($0.64\times 10^{-3}$) IBM 3090 CPU seconds. The combined weight distribution
for initial and final state radiation $e^+e^-$ events for parameters and cuts
as
in the sample main program given above, (whose output is shown in Fig. 4) is
presented in Fig. 5. The weight distribution can be seen to be well centered
around 1.0, resulting in an efficient generation of unit weight events by the
weight throwing procedure ~\cite{x4}. The fractions of events with weights
$>$ 2.0, (3.0) is $\simeq 6 \times 10^{-3}$, ($ 2 \times 10^{-4}$)
respectively. The maximum weight may then be chosen to be 2.0, allowing
efficient generation of event samples with cross-sections known at, or below
the \% level~\cite{x10}.
\pagebreak
{\bf \large {Appendix A}}
Following Ref.[39], the Born differential cross-section for Bhabha
scattering, including both $s$ and $t$ channel Z exchange may be written
as:
\[
\frac{d\sigma_{0}}{dt} = \frac{2 \pi \alpha^2}{s^2}[B_0+B_2(\frac{t}{s})
^2+B_3(1+\frac{t}{s})^2] \hspace{2cm} (A1) \]
where
\begin{eqnarray*}
t & = & -\frac{s}{2}(1-\cos\theta_+) \\
B_0 & = & [\frac{s}{t}+c_- \chi_t(t)]^2 \\
B_2 & = & |1+c_- \chi_s(s)|^2
\end{eqnarray*}
\[ B_3 = \frac{1}{2}\{|1+\frac{s}{t}+a_+[\chi_t(t)+\chi_s(s)]|^2+
(a_+ \rightarrow a_-) \} \]
\begin{eqnarray*}
c_- = g_V^2-g_A^2 & , & a_{\pm} = (g_V \pm g_A)^2 \\
\chi_t(t) = \frac{s}{t-M_Z^2} & , & \chi_s(s) = \frac{s}{s-M_Z^2
+is(\Gamma_Z/M_Z)}
\end{eqnarray*}
$\theta_+$ is the scattering angle between the incoming and outgoing
$e^{\pm}$. $M_Z$, $\Gamma_Z$ are the mass and width of the Z. The Z width
is neglected in the t-channel exchange amplitude. The vector and
axial-vector coupling constants $g_A$, $g_V$ are given, in the Standard
Model by Eqns. 3.3-3.6. The cross-section for $\mu$-pair production is
recovered on setting $s/t$ to zero in $B_0$, $B_3$.
Analytical integration of $A$1 over $t$ yields the result ~\cite{x34}
( see also the first of Ref.[4]):
\[
\int_{t_{MIN}}^{t_{MAX}} d \sigma_0 = \frac{2 \pi \alpha^2}{s^2}
[S(t_{MIN})-S(t_{MAX})] \hspace{2cm} (A2) \]
The function $S(t)$ is the sum of the following 5 terms derived from
Eqn. $A$1 :
\begin{eqnarray*}
\int B_0 dt &=& s^2 [-\frac{1}{t} + \frac{2c_-}{M_Z^2} \ln
\frac{t-M_Z^2}{t} - \frac{c_-^2}{t-M_Z^2}] \\
& & \\
\int B_3 dt &=& \frac{1}{2}\{ s^2[ -\frac{2}{t} + \frac{2(a_+ +a_-)}
{M_Z^2} \ln \frac{t-M_Z^2}{t}-\frac{(a_+^2+a_-^2)}{t-M_Z^2}] \\
& & +2s[(b_++b_-) \ln t + (a_+b_+ +a_-b_-) \ln (t-M_Z^2)] + (b'_+
+ b'_-)t \} \\
& & \\
\int \frac{2 B_3 t}{s} dt &= & s\{2 \ln t + 2 (a_+ + a_-) \ln (t-M_Z^2)+
(a_+^2 + a_-^2)[\ln(t-M_Z^2)- \frac{M_Z^2}{t-M_Z^2}]\} \\
& & + 2\{(b_+ + b_-)t +(a_+ b_+ + a_-b_-)[t+M_Z^2\ln(t-{M_Z^2}]\}
+(b'_+ + b'_-) \frac{t^2}{2s} \\
& & \\
\int B_2 \frac{t^2}{s^2} dt &= &\frac{B_2 t^3}{3 s^2} \\
& & \\
\int B_3 \frac{t^2}{s^2} dt &=& \frac{1}{2}\{[2t+2(a_+ + a_-)(t+M_Z^2\ln(t-M_Z^
2))+(a_+^2 +a_-^2)(t+2M_Z^2 \ln (t-M_Z^2) \\
& & -\frac{M_Z^4}{t-M_Z^2})] + \frac{2}{s}[(b_+ + b_-)
\frac{t^2}{2} + (a_+ b_+ + a_- b_-)
(\frac{t^2}{2}+t M_Z^2+M_Z^4\ln(t-M_Z^2))] \\
& & +(b'_+ +b'_-)\frac{t^3}{3s^2}\}
\end{eqnarray*}
where
\begin{eqnarray*}
b_{\pm} &=& 1 + a_{\pm} Re\chi_s \\
b'_{\pm} &=& 1 + a_{\pm}2 Re\chi_s + a_{\pm}^2|\chi_s|^2
\end{eqnarray*}
{\bf \large {Appendix B}}
The exact hard photon cross-section, with exponentiated initial state
radiation, is derived from formulae given in the second of Refs.[4] :
\[ \frac{d\sigma^{EXACT}}{d\Omega_+ d\Omega_{\gamma}dy} =
\frac{\alpha^3 y}{16\pi^2 s} X(\alpha_k) \hspace{2cm} (B1) \]
The function $X$($\alpha_k$) of the kinematical variables $\alpha_k$
specifying the event configuration, (defined in Sections 4.3.1 to
4.4.4), has contributions from $s$-channel exchanges ($ss$), $t$-channel
exchanges ($tt$) and $s-t$ interference ($st$). In each of these cases
separate
contributions from initial state radiation (INI), final state
radiation (FIN), and initial/final interference (INT) may be
distinguished. Thus :
\[ X=\sum_{i,j} X_i^j \hspace{.5cm} i=ss,tt,st;\hspace{.5cm}
j=INI,FIN,INT \hspace{1cm} (B2) \]
where
\begin{eqnarray*}
X_{ss}^{INI} & = & \frac{1}{s'\kappa_+\kappa_-}\left[B_1(c_-,s')
[f(s)t^2+t'^{2}]+B_5(s')[f(s)u^2+u'^{2}]\right] \\
& & -m_e^2f(s) \left[\frac{B_{ss}(s',t,u)}{\kappa_-^2} +
\frac{B_{ss}(s',t',u')}{\kappa_+^2}\right] \\
& & \\
X_{ss}^{FIN} & = & \frac{1}{s\kappa'_+\kappa'_-}\left[B_1(c_-,s)
[t^2+t'^{2}]+B_5(s)[u^2+u'^{2}]\right] \\
& & -m_l^2 \left[\frac{B_{ss}(s,t,u')}{\kappa_{-}^{'2}} +
\frac{B_{ss}(s,t',u)}{\kappa_{+}^{'2}}\right] \\
& & \\
X_{ss}^{INT} & = & \frac{1}{ss'}\left[\frac{u}{\kappa_+\kappa'_-}
+\frac{u'}{\kappa_-\kappa'_+}-\frac{t}{\kappa_+ \kappa'_+}
-\frac{t'}{\kappa_- \kappa'_-} \right] \\
& & \times \left[B_4(s,s')(t^2+t'^2)+B_6(s,s')(u^2+u'^2) \right] \\
& & +\frac{\vec{p_+}\cdot(\vec{q_+}\times \vec{q_-})(s-s')}
{2 \sqrt{s} s' \kappa_+ \kappa_- \kappa'_+ \kappa'_-}
B_7(s,s')(u^2-u'^2) \\
& & \\
X_{tt}^{INI} & = & \frac{s}{\kappa_+\kappa_-}\left[ B_6(t,t')
[\frac{f(s)u^2+u'^2}{tt'}]+B_4(t,t')[\frac{f(s)s^2+s'^2}{tt'}]\right] \\
& & -m_e^2f(s) \left[\frac{B_{tt}(s',t,u)}{\kappa_-^2} +
\frac{B_{tt}(s',t',u')}{\kappa_+^2}\right] \\
& & \\
X_{tt}^{FIN} & = & \frac{s'}{\kappa'_+\kappa'_-}\left[ B_6(t,t')
[\frac{u^2+u'^2}{tt'}]+B_4(t,t')[\frac{s^2+s'^2}{tt'}]\right]\\
& & -m_l^2 \left[\frac{B_{tt}(s,t,u')}{\kappa_-^{'2}} +
\frac{B_{tt}(s,t',u)}{\kappa_+^{'2}}\right] \\
& & \\
X_{tt}^{INT} & = & \frac{s^2+s'^2}{tt'} \left[-\frac{t}{\kappa_+
\kappa'_+}B_1(c_-,t')-\frac{t'}{\kappa_- \kappa'_-}B_(c_-,t)\right] \\
& & + \frac{u^2+u'^2}{tt'} \left[-\frac{t}{\kappa_+ \kappa'_+}
B_5(t')-\frac{t'}{\kappa_- \kappa'_-}B_5(t)\right] \\
& & + \left(\frac{u}{\kappa_+ \kappa'_-}+
\frac{u'}{\kappa_- \kappa'_+}\right)
\left[ B_6(t,t')
[\frac{u^2+u'^2}{tt'}]+B_4(t,t')[\frac{s^2+s'^2}{tt'}]\right]
\\ & & \\
X_{st}^{INI} & = & \frac{f(s)u^2+u'^2}{s'\kappa_+ \kappa_-}\left[
-\left(\frac{u'+t'}{t'}\right)B_6(s',t')-\left(\frac{u+t}{t}\right)
B_6(s',t)\right] \\
& & -m_e^2f(s)\left[\frac{B_{st}(s',t,u)}{\kappa_-^2}
+\frac{B_{st}(s',t',u')}{\kappa_+^2}\right] \\
& & \\
X_{st}^{FIN} & = & \frac{u^2+u'^2}{s\kappa'_+ \kappa'_-}\left[
-\left(\frac{u'+t}{t}\right)B_6(s,t)-\left(\frac{u+t'}{t'}\right)
B_6(s,t')\right] \\
& & -m_l^2f(s)\left[\frac{B_{st}(s,t,u')}{\kappa_-^{'2}}
+\frac{B_{st}(s,t',u)}{\kappa_+^{'2}}\right] \\
& & \\
\end{eqnarray*}
\begin{eqnarray}
X_{st}^{INT} & = & \left[\frac{u'}{\kappa_- \kappa'_+}+
\frac{u'+s'}{\kappa_+\kappa'_+}\right]\left(\frac{u^2+u'^2}{s't'}\right)
B_6(s',t') \nonumber \\
& & +\left[\frac{u}{\kappa_+ \kappa'_-}+
\frac{u+s'}{\kappa_- \kappa'_-}\right]\left(\frac{u^2+u'^2}{s't}\right)
B_6(s',t) \nonumber \\
& & +\left[\frac{u'}{\kappa'_+ \kappa_-}+
\frac{u'+s}{\kappa'_- \kappa_-}\right]\left(\frac{u^2+u'^2}{st}\right)
B_6(s,t) \nonumber \\
& & +\left[\frac{u}{\kappa_+ \kappa'_-}+
\frac{u+s}{\kappa'_+\kappa_+}\right]\left(\frac{u^2+u'^2}{st'}\right)
B_6(s,t') \nonumber \\
& & +\frac{\sqrt{s}[\vec{p_+}\cdot(\vec{q_+}\times\vec{q_-})]
(u^2-u'^2)}{2\kappa_+\kappa_-\kappa'_+\kappa'_-}
[-\left(\frac{t-t'}{tt'}\right)B_7(t,t')
+2\frac{B_7(s,t)}{st(\kappa_-\kappa'_+\kappa'_-)} \nonumber \\
& & +2\frac{B_7(s,t')}{st'(\kappa_+\kappa'_+\kappa'_-)}
-2\frac{B_7(s',t)}{s't(\kappa_+\kappa_-\kappa'_-)}
-2\frac{B_7(s',t')}{s't'(\kappa_+\kappa_-\kappa'_+)}] \nonumber
\end{eqnarray}
\begin{eqnarray*}
B_1(c,s) & = & 1+[\frac{2c}{s}(s-M_Z^2)+c^2]|\chi_s(s)|^2 \\
& & \\
B_4(s,s') & = & 1+c_-[(1-\frac{M_Z^2}{s})|\chi_s(s)|^2
+(1-\frac{M_Z^2}{s'})|\chi_s(s')|^2] \\
& & +c_-^2[(1-\frac{M_Z^2}{s})(1-\frac{M_Z^2}{s'})
+\frac{M_Z^2 \Gamma_Z^2}{ss'}]|\chi_s(s)|^2|\chi_s(s')|^2 \\
& & \\
B_5(s) & = & 1+[2(g_V^2+g_A^2)(1-\frac{M_Z^2}{s})
+(g_V^4+g_A^4+6g_V^2g_A^2)]|\chi_s(s)|^2
\end{eqnarray*}
\begin{eqnarray*}
B_6(s,s') & = & 1+(g_V^2+g_A^2)\left[(1-\frac{M_Z^2}{s})|\chi_s(s)|^2+
(1+\frac{M_Z^2}{s'})|\chi_s(s')|^2 \right] \\
& = & +(g_V^4+g_A^4+6g_V^2g_A^2)\left[(1-\frac{M_Z^2}{s})
(1-\frac{M_Z^2}{s'})+\frac{M_Z^2 \Gamma_Z^2}{ss'}\right]
|\chi_s(s)|^2 |\chi_s(s')|^2
\\ & & \\
B_7(s,s') & = & -4M_Z \Gamma_Z \left\{ g_A g_V [\frac{|\chi_s(s)|^2}{s}-
\frac{|\chi_s(s')|^2}{s'} ]
\right.
\\
& &
\left.
+2 g_V g_A(g_V^2+g_A^2)(\frac{1}{s}-\frac{1}{s'})
|\chi_s(s)|^2 |\chi_s(s')|^2
\right\}
\\ & & \\
B_{ss} (s,t,u) & = & \left[ B_{1}(a_{+}^{2},s)+B_{1}(a_{-}^{2},s) \right]
(\frac{u}{s})^2+2B_{1}(c_{-},s)(\frac{t}{s})^2
\\ & & \\
B_{tt}(s,t,u) & = & \left[ B_1(a_+^2,t)+B_1(a_-^2,t) \right]
(\frac{u}{t})^2+2B_1(c_-,t)(\frac{s}{t})^2
\\ & & \\
B_{st}(s,t,u) & = & 4 B_6(s,t) \frac{u^2}{st}
\end{eqnarray*}
The constants $c_-$, $a_{\pm}$ are defined in Appendix A and $g_V$,
$g_A$ in Eqns 2.3-2.6. Note that the functions $\chi_s$, $\chi_t$ are
assigned according to the arguments of $B_1$, $B_4$ ... . For example
$B_4(s,t')$ is given by the replacement $\chi_s(s') \rightarrow
\chi_t(t')$ in the expresssion above for $B_4(s,s')$.
Exponentiation of the initial state radiation is included via the
function:
\[ f(s) = 2 C_V^i \exp \beta_e \ln (y) - 1 \hspace{2cm} (B3) \]
where $\beta_e$, $C_V^i$ are defined after Eqn 2.1. This ensures that
the $y \rightarrow 0$ limit of Eqn. $B$1 is identical to the derivative
of Eqn 2.1 with respect to $y_0$, in which the replacement
$y_0 \rightarrow y$ is made. That is `soft' and `hard' photons are
treated in a consistent way.
{\bf \large {Appendix C}}
Two dimensional look up tables are generated according to a
generalisation of the procedure described in the text (Eqns 3.7-3.10)
for a one dimensional look up table. Consider for example the case of
the Born differential cross-section $d\sigma_0/dc$ for a range of
different CM energies : $ s_{min} < s < s_{max} $. Let $k$ be the bin
index for $s$. The equations 2.7, 2.8 generalise to:
\[ \sigma_0^k = \int_{c_{min}}^{c_{max}} \frac{d\sigma_0^k}{dc}dc
\hspace{2cm} (C1) \]
where
\[ d \sigma_0^k \equiv d \sigma_0 (s_k) \]
\[ P_i^k =\frac{1}{\sigma_0^k}\int_{c_{min}}^{c_i} \frac{d \sigma_0^k}
{dc}dc \hspace{2cm} (C2) \]
For each value of $k$ the distribution $P_i^k$ is inverted by linear
interpolation to yield a look up table with bin index $j$ :
\[ c_i^j = f^k(P_i^k) \hspace{3cm} (C3) \]
To generate the angular distribution for a given value of $s$, the
adjacent bins in $s$ of index $k$, $k+1$ are first located :
\[ s_k < s < s_{k+1}\]
The closest bins in the look up tables of index $k$, $k+1$ to the
random number $Rn$ are the found:
\[P_i^k < Rn < P_{i+1}^k \]
\[P_{i'}^{k+1} < Rn < P_{i'+1}^{k+1} \]
With:
\[ \delta_i = (Rn-P_i^k)/(P_{i+1}^k-P_i^k) \]
\[ \delta_{i'}=(Rn-P_{i'}^{k+1})/(P_{i'+1}^{k+1}-P_{i'}^{k+1}) \]
Two values of $c$ are now calculated according to :
\[ c^k = f^k(P_i^k)+[f^k(P_{i+1}^k)-f^k(P_i^k)]\delta_i \]
\[ c^{k+1} = f^{k+1}(P_{i'}^k)+[f^{k+1}(P_{i'+1}^{k+1})-f^{k+1}(P_{i'}^{k+1})]
\delta_{i'} \]
with
\[ \Delta_k = (s-s_k)/(s_{k+1}-s_k) \]
the generated value of $c$ is, finally, given by the equation :
\[ c = c^k + (c^{k+1}-c^k) \Delta_k \hspace{2cm} (C4) \]
\pagebreak
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,825 |
August 24, 2021 August 24, 2021 ColinOnCars
Colin-on-Cars – Supporting leopard and toad conservation
The most gracious of all the wild cat family, the leopard, is being helped by the Ford Wildlife Foundation (FWF) that is lending a supportive hand to the Cape Leopard Trust (CLT) and the Endangered Wildlife Trust (EWT).
The two Ford Rangers, which are provided to the Trusts for two years, enable the conservation teams to confidently complete their work which often involves travelling long distances and accessing remote areas.
The CLT and EWT teams consist of passionate and enthusiastic members that have dedicated themselves to protecting threatened and endangered animal species. They have joined forces in a unique project that sees them working together to ensure the continued survival of the cape leopards and western leopard toads. By working together, the two organisations hope to improve awareness of two species that are important indicators of ecosystem health.
"We really appreciate the support from the Ford Wildlife Foundation," says Dr Kathryn Williams, Research and Conservation Director, Cape Leopard Trust. "The Ford Ranger allows us to improve our knowledge on the ecology, presence, and threats to both species in the Cape's Overberg region."
The project, affectionately named a Tale of Two Leopards, aims to investigate the presence of both the leopard and the leopard toad – an endangered amphibian named for its striking resemblance to a leopard – in order to ensure their survival, secure their habitat and prey base, and promote their coexistence with people.
In addition to the Tale of Two Leopards projects, the Cape Leopard Trust is currently undertaking its largest ever camera survey, the first comprehensive such study of the Overberg. The camera survey seeks to improve knowledge of leopards in the Western Cape through a dedicated online app or data portal, which allows landowners, conservation NGO's and local communities to become citizen scientists by reporting leopard sightings.
Local community members can submit verifiable records of data, such as camera-trap photos, digital images of leopard signs such as droppings, scratch marks on trees or feeding sites and direct leopard observations to app.capeleopard.org.za. All submissions are kept anonymous.
"Broad-scale research is essential to conserve species that occupy large home ranges such as the leopard. It is extremely challenging for scientists to collect large quantities of data at this scale, but conservation stewards can help. Keep your eyes peeled for signs of this elusive cat and contribute your data to help us conserve leopards in the Western Cape," says Dr Williams.
For more than 30 years, Ford Motor Company of Southern Africa (FMCSA) has actively been involved in the conservation of wildlife and ecosystems in South Africa and sub-Saharan Africa. The Ford Wildlife Foundation (FWF), which was established in 2014, continues FMCSA's long-standing support of conservation projects in Southern Africa through the provision of 4×4 Ranger Double Cabs to partner organisations. During the two-year loan period, the vehicles are monitored and serviced by Ford's extensive dealer network to ensure optimum performance and efficiency.
For more information on the Ford Wildlife Foundation, visit the website: https://www.ford.co.za/about-ford/wildlife-foundation/
Previous Colin-on-Cars: Road Review – Hyundai Kona 2.0 Nu Executive
Next Colin-on-Cars – Hot action expected in East London | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,014 |
Der Ruppberg ist ein hoher Berg im Thüringer Wald. Er liegt in den Stadtgebieten von Steinbach-Hallenberg und Zella-Mehlis im thüringischen Landkreis Schmalkalden-Meiningen.
Der Berg, dessen waldloser Gipfel im Ostsüdosten des Stadtgebiets von Steinbach-Hallenberg liegt, gilt als Hausberg der im Südosten befindlichen Stadt Zella-Mehlis. Am Gipfel tritt das Porphyrgestein, aus dem der Berg besteht, deutlich zutage.
Geographische Lage
Der Ruppberg erhebt sich im mittleren Thüringer Wald im Naturpark Thüringer Wald. Er liegt zwischen der Kernstadt von Steinbach-Hallenberg im Westnordwesten, dem Dorf Oberschönau im Nordwesten, der Stadt Oberhof im Ostnordosten, der Stadt Zella-Mehlis im Südosten, die der Berg mit vulkanähnlicher Gestalt überragt, und Bermbach, einem Ortsteil der Stadt Steinbach-Hallenberg im Westen.
Nördlich des Ruppbergs liegt der Kanzlersgrund mit dem Haselbach (Hasel) genannten Oberlauf der Schwarza. Jenseits davon erhebt sich der dem Hauptkamm des Gebirges vorgelagerte Hohe Stein (). Über den Hauptkamm mit dortigem Finsterbachkopf () verläuft der Rennsteig. Nordwestlich schließt sich der Große Hermannsberg () an, östlich liegt der Gebrannte Stein (). Nach Südosten fällt die Landschaft in das Tal der Lichtenau mit Zella-Mehlis ab.
Geschichte
Bereits die Kelten errichteten wohl eine Wallanlage auf dem Ruppberg. Wallgrabenreste sind im Norden und Süden unterhalb der beiden Gipfelplateaus noch im Gelände zu erahnen. Später, vermutlich vom Ende des 10. Jahrhunderts bis um 1200, soll sich hier eine Burg (das Schloss Ruprechtsburg) befunden haben, die den Herren von Nordeck gehörte. Es ist anzunehmen, dass die Herren von Nordeck Besitzer einer dynastischen Kleinherrschaft waren. Der letzte Vertreter dieses Geschlechts (Gebhard von Nordeck) verstarb 1115 oder 1120. Angeblich noch im 18. Jahrhundert sollen Mauerreste sichtbar gewesen und Kleinfunde gemacht wurden sein. Im Jahre 1619 wurde das Amt Hallenberg (bis zu diesem Zeitpunkt zugehörig zum Kurfürstentum Sachsen) gegen den hessischen Anteil der Zent Benshausen ausgewechselt (Benshäuser Tauschvertrag). Infolgedessen befindet sich der Ruppberg bis heute auf Steinbach-Hallenberger Flurgemarkung.
Die erste Schutzhütte wurde 1898 von der Ruppberggemeinde errichtet. Die heutige Hütte stammt aus dem Jahr 1958.
Beschreibung
Gipfel
Ein Wanderweg erreicht den Ruppberg bei der Hütte des Ruppbergvereins, nördlich davon befindet sich ein kleiner Nebenfelsen mit einem Fahnenmast, auf dem immer dann die Fahne gehisst wird, wenn die Hütte geöffnet ist. Im Süden erhebt sich der Hauptgipfel mit einem Gipfelstein, auf dem sichtbare markante Punkte eingetragen sind.
Ausblick
Zu den Sichtzielen mit Blick vom Ruppberggipfel gehören:
Wanderrouten
Der kürzeste Aufstieg führt vom an der Straße Zella-Mehlis–Oberschönau (L 2691) liegenden Ruppbergparkplatz zum Gipfel; der letzte Wegteil ist recht steil aber mit Stufen und Geländern ausgestattet (Aufstieg ca. 30 Minuten). In der warmen Jahreszeit verkehrt ein Bus vom Mehliser Markt zum Parkplatz.
Vom Mehliser Markt über das Friedrich-Ludwig-Jahn-Denkmal, die Wiese am Buchenbrunnen, Braukopfhütte (ca. 2 Stunden)
Vom Mehliser Markt durch das Ruppertstal, am Grillplatz zum Waldhaus abzweigen, über die Wiese zum Ruppbergparkplatz (ca. 2 Stunden)
Vom Sportplatz Zella-Mehlis über den Stachelsrain, die Lämmerröder, Dammwiese, Heinrichsbacher Stein, Parkplatz (ca. 2 Stunden, 30 Minuten)
Von Benshausen auf dem Rhön-Rennsteig-Wanderweg (ca. 1 Stunde, 30 Minuten)
Vom Rondell (Schnittpunkt B 247 und Rennsteig) bei Oberhof über das Schützenbergmoor, vorbei am Veilchenbrunnen, Ruppbergparkplatz (ca. 2 Stunden, 20 Minuten)
Von Steinbach-Hallenberg über das Knüllfeld, Rupprasen auf den Gipfel (ca. 2 Stunden, 30 Minuten)
Weitere Bilder
Weblinks
Ruppberg (Homepage des Ruppbergvereins), auf ruppberg.de
360-Grad-Panorama vom Ruppberggipfel (Beschriftung von Sichtzielen zuschaltbar), auf panorama-photo.net
Einzelnachweise
Berg im Naturpark Thüringer Wald
Berg im Landkreis Schmalkalden-Meiningen
Geographie (Zella-Mehlis) | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 9,030 |
DIR=$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)
# Make sure perf is available.
if [ ! -x "$(command -v perf)" ]
then
echo "Cannot find perf, please make sure it's installed."
exit 1
fi
# Install cargo-flamegraph
installed_flamegraph=0
if [ ! -x "$(command -v cargo-flamegraph)" ]; then
echo "cargo-flamegraph not installed; installing ..."
cargo install flamegraph
installed_flamegraph=1
fi
# Create flamegraph
cargo flamegraph --bin=alacritty -- $@
# Unintall cargo-flamegraph if it has been installed with this script
if [ $installed_flamegraph == 1 ]; then
read -p "Would you like to uninstall cargo-flamegraph? [Y/n] " -n 1 -r
echo
if [[ "$REPLY" =~ ^[^Nn]*$ ]]; then
cargo uninstall flamegraph
fi
fi
| {
"redpajama_set_name": "RedPajamaGithub"
} | 252 |
// Copyright (c) Microsoft Corporation. All rights reserved.
// Licensed under the MIT License. See License.txt in the project root for license information.
namespace ODataValidator.Rule
{
#region Namespace.
using System;
using System.ComponentModel.Composition;
using System.Linq;
using System.Net;
using Newtonsoft.Json.Linq;
using ODataValidator.Rule.Helper;
using ODataValidator.RuleEngine;
using ODataValidator.RuleEngine.Common;
#endregion
/// <summary>
/// Class of service implemenation feature to verify .
/// </summary>
[Export(typeof(ExtensionRule))]
public class ServiceImpl_SystemQueryOptionSkipToken : ServiceImplExtensionRule
{
/// <summary>
/// Gets the service implementation feature name
/// </summary>
public override string Name
{
get
{
return "ServiceImpl_SystemQueryOptionSkipToken";
}
}
/// <summary>
/// Gets the service implementation feature description
/// </summary>
public override string Description
{
get
{
return this.CategoryInfo.CategoryFullName + ",$skiptoken";
}
}
/// <summary>
/// Gets the service implementation feature specification in OData document
/// </summary>
public override string V4SpecificationSection
{
get
{
return "";
}
}
/// <summary>
/// Gets the service implementation feature level.
/// </summary>
public override RequirementLevel RequirementLevel
{
get
{
return RequirementLevel.Must;
}
}
/// <summary>
/// Gets the service implementation category.
/// </summary>
public override ServiceImplCategory CategoryInfo
{
get
{
var parent = new ServiceImplCategory(ServiceImplCategoryName.RequestingData);
return new ServiceImplCategory(ServiceImplCategoryName.SystemQueryOption, parent);
}
}
/// <summary>
/// Verifies the service implementation feature.
/// </summary>
/// <param name="context">The Interop service context</param>
/// <param name="info">out parameter to return violation information when rule does not pass</param>
/// <returns>true if the service implementation feature passes; false otherwise</returns>
public override bool? Verify(ServiceContext context, out ExtensionRuleViolationInfo info)
{
if (context == null)
{
throw new ArgumentNullException("context");
}
bool? passed = null;
info = null;
var svcStatus = ServiceStatus.GetInstance();
var pagingLimitEntitySetUrls = MetadataHelper.GetPagingLimitEntitySetURLs();
if (null == pagingLimitEntitySetUrls || !pagingLimitEntitySetUrls.Any())
{
return passed;
}
var pagingLimitEntitySetUrl = pagingLimitEntitySetUrls.First();
string url = svcStatus.RootURL.TrimEnd('/') + "/" + pagingLimitEntitySetUrl;
int actualNum = JsonParserHelper.GetEntitiesCountFromFeed(url);
var resp = WebHelper.Get(new Uri(url), Constants.V4AcceptHeaderJsonFullMetadata, RuleEngineSetting.Instance().DefaultMaximumPayloadSize, context.RequestHeaders);
var detail = new ExtensionRuleResultDetail("ServiceImpl_SystemQueryOptionSkipToken", url, HttpMethod.Get, string.Empty);
info = new ExtensionRuleViolationInfo(new Uri(url), string.Empty, detail);
if (null != resp && HttpStatusCode.OK == resp.StatusCode)
{
var jObj = JObject.Parse(resp.ResponsePayload);
var jArr = jObj.GetValue("value") as JArray;
passed = jArr.Count < actualNum;
}
else
{
passed = false;
}
return passed;
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,238 |
Q: How to override Drawer in Material UI In my component I need to override css parameter "overflow-y". This parameter is described in class .MuiDrawer-paper.
Usually to override css is piece of cake via makeStyles. But in this component has two divs. Parent container and daughter div. And when I set overrided class like:
const useStyles = makeStyles((theme) => ({
paper: {
overflowY: 'unset',
},
)};
...
className={classes.paper}
Parent div gets this class and it does not have any sense. Because I need to override daughter class.
I tried to do some thing like this:
className={{ paper: classes.paper }}
But in this case class wan't picked... What should I do?
A: The correct way to override material ui classes is to make use of classes prop on Drawer component instead of className.
Read more about overriding classes
const useStyles = makeStyles((theme) => ({
paper: {
overflowY: 'unset',
},
)};
...
<Drawer
classes={{
paper: classes.paper,
}}
anchor="left"
open={open}
/>
A: The top-voted answer in this thread contains legacy code (makeStyles). I was able to override CSS on a child element of a MUI component via the styled API (migration guide):
const StyledMUIComponent = styled(MUIComponentName)({
"& .child-class-to-target": {
overflowY: 'unset'
}
})
A: I have 2 options.
*
*using !important
const useStyles = makeStyles((theme) => ({
paper: {
overflowY: 'unset !important',
},
)};
*using styles property.
<Drawer style={{overflowY: 'unset'}} />
I prefer to use styles property.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 388 |
{"url":"https:\/\/en.universaldenker.org\/formulas\/833","text":"# Formula Acoustic Doppler Effect Observer frequency \u00a0\u00a0 Emitter frequency \u00a0\u00a0 Emitter speed \u00a0\u00a0 Observer speed\n\n## Observer frequency\n\nUnit\nFrequency perceived by an observer (who may hear a loud ambulance, for example).\n\n## Emitter frequency\n\nUnit\nFrequency emitted for example by the siren of an ambulance.\n\n## Speed of sound\n\nUnit\nSpeed of sound at which sound waves propagate. In air, the speed of sound is: $$c ~\\approx~ 340 \\, \\frac{\\text m}{\\text s}$$ at 20\u00b0C.\n\n## Emitter speed\n\nUnit\nSpeed at which the emitter (for example an ambulance) moves relative to the observer.\n\n\"$$c ~-~ v_{\\text s}$$\" is used when the emitter moves towards the observer. \"$$c ~+~ v_{\\text s}$$\" when the emitter is moving away from the observer. If the emitter is stationary, then $$v_{\\text s} = 0$$.\n\n## Observer speed\n\nUnit\nSpeed at which the observer moves relative to the emitter.\n\nUse \"$$c ~+~ v$$\" when the observer moves towards the emitter. \"$$c ~-~ v$$\" when the observer moves away from the emitter. If the observer is standing still, use $$v = 0$$.","date":"2022-01-24 13:29:15","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.917445182800293, \"perplexity\": 2597.499833726818}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2022-05\/segments\/1642320304570.90\/warc\/CC-MAIN-20220124124654-20220124154654-00718.warc.gz\"}"} | null | null |
Битката при прохода Касерин е сражение, проведено през Втората световна война, като част от Тунизийската кампания.
Всъщност тя е серия от битки, проведени около прохода Касерин, клисура с широчина 2 мили (3 км) в масива Гранд Дорсал на Атласките планини в Западен централен Тунис. Силите на Оста, участвали в сблъсъка, са основно от германско-италианската танкова армия (преструктурираната германска Танкова армия "Африка"), водена от фелдмаршал Ервин Ромел и Пета танкова армия, водена от генерал Ханс Юрген фон Арним. Силите на съюзниците са основно от 2-ри корпус на американската армия, командвана от генерал-майор Лойд Фредендал, който е част от британската Първа армия, командвана от генерал-лейтенант Кенет Артър Ноуъл Андерсън.
Това е първият значителен сблъсък между американските и германските сили през Втората световна война. Неопитните американски войници, водени по неадекватен начин от своите командири, претърпяват тежки загуби и са отблъснати на 50 мили (80 км) от техните първоначални позиции западно от прохода Фаид. Битката е описана като "когато аматьорите се срещат за пръв път с професионалистите". Впоследствие американската армия въвежда бързи промени от организацията на ниво единици до замяна на комадващите офицери. При следващата битка, в някои случаи само няколко седмици по-късно, те се представят значително по-добре.
Обстановка в Северна Африка преди битката
Американските и британските сили слизат на няколко точки по брега на Френско Мароко и Алжир на 8 ноември, 1942 г. по време на Операция Факел. Това се случва само дни след като генерал Бърнард Монтгомъри постига успех по време на втората битка за Ел Аламейн. Разбирайки опасността от война на два фронта, германски и италиански войски са прехвърлени от Сицилия, за да окупират Тунис – един от лесно защитимите райони в Северна Африка и само на една нощ разстояние с кораб от базите в Сицилия.
Дори след десанта, проведен по време на операция "Факел" от съюзниците, има малко организирана защита в западната пустиня. Още по-важно е, че никакви усилия не са положени от съюзническите морски и въздушни сили да прекъснат потока от войници и материали в Тунис, това става по-късно след като значителни сили са били прехвърлени. В допълнение съюзниците се придвижват много бавно, за да установят контакт с германците, докато се опитват да преговарят с местните командири на Вишиска Франция. Няколко опита са направени да се отцепи Тунис през ноември и декември 1942 г. преди германските войски да бъдат подсилени, но слабата координация и лесният за защита терен позволява на малък брой германски и италиански части разположени там да ги осуетят.
На 23 януари 1943 г. британската Осма армия, водена от Монтгомъри, превзема Триполи, лишавайки по този начин Ромел от основната му база за продоволствие. Ромел е бил подготвен за такова развитие, и планира да блокира южния подстъп към Тунис от Триполи като окупира широката линия от защитни укрепления, известна като линията Марет, конструирана от французите, за да отблъскват италианските атаки от Либия. Укрепленията се простират от Атласките планини на запад до залива Сидра на изток, позволявайки дори на малък брой германски/италиански части да задържат настъплението на съюзниците.
Фаид
Пречка на този план е фактът, че съюзническите войски вече са пресекли Атласките планини и са установили предна база във Фаид, на склоновете на източния ръкав на планините. Това ги поставя в отлична позиция да отрежат Ромел от силите разположени на север и да прекъснат снабдяването му от Тунис. Естествено Оста не може да позволи това да се случи.
Елементи от Пета танкова армия на фон Арним достигат позициите на съюзниците на източния склон на Атласките планини на 30 януари. Германската 21-ва танкова дивизия се среща с френските защитници на Фаид и ги преодолява без особени затруднения. Три опита са направени да се спре тяхното настъпление от американската 1-ва бронетанкова дивизия, но всеки от опитите се изправя срещу класическия блицкриг – всеки път, когато получават нареждане да оформят защитна позиция, американците намират тези позиции вече превзети и са атакувани от германски войници като претърпяват тежки загуби. След 3 дни американският 2-ри корпус е принуден да отстъпи.
По-голямата част от Тунис пада в германски ръце, а проходите към равнините са блокирани. Съюзниците все още държат вътрешността на планината, но това слабо засяга Ромел, тъй като всички изходи на изток са блокирани. През следващите две седмици Ромел и останалите командири, разположени на север, обсъждат какво да предприемат. Имайки предвид по-късните им действия, това закъснение им коства много.
В крайна сметка Ромел решава, че може да подобри проблемите със снабдяването си и допълнително да отслаби американската заплаха в неговия фланг, като атакува по посока на две американски бази, разположени на запад от западния ръкав на планините в Алжир. Въпреки че има слаб интерес да задържи вътрешно планинските долини, едно бързо нападение ще му осигури запасите и ще наруши допълнително действията на американците.
На 24 февруари 21-ва танкова дивизия отново започва да се придвижва на запад, атакувайки Сиди Боу Зид, намиращ се на около 10 мили (16 км) от Фаид във вътрешността на Атласките планини. Битката се води през целия ден, но поради слабото използване на бронирана техника от страна на американците те губят битката. В края на деня бойното поле е владение на Танковата армия. Контраатаката на следващия ден е отбита без затруднение и на 16 февруари германците се насочват към Сбейтла.
Поради липсата на лесно защитим терен американските сили отстъпват, за да оформят нови защитни линии на по-лесно защитимия проход Касерин в западния ръкав на планината. Дотогава американските сили губят 2546 войници, 103 танка, 280 превозни средства, 18 полеви артилерийски оръдия, 3 противотанкови оръдия и една цяла противовъздушна батарея.
Военни действия
На 19 февруари Ромел започва нападението. Следващия ден той лично води атаката с формираната 10-а танкова дивизия, отпусната му от 5-а танкова армия, разположена на север, докато 21-ва танкова дивизия, също част от 5-а танкова армия, продължава да атакува на север през прохода Сбиба.
В рамките на няколко минути защитните линии на американците са пробити. Техните леки оръдия и танкове нямат никакъв шанс срещу по-тежката германска техника, а и имат малък или никакъв опит в боя с бронирана техника. Германските танкове Панцер IV и Тигър отблъскват всички атаки с лекота. Танковете M3 Лий и M3 Стюарт, с които се сблъскват, им отстъпват по огнева мощ и екипажите им са по-неопитни. В същото време американските командири се свързват с тяхното командване за разрешение да организират контраатака или артилерийски бараж, често получавайки разрешение, след като тези линии са вече подминати от настъпващия противник. Още веднъж 1-ва бронетанкова дивизия получава заповеди да заеме безполезни позиции, и до втория ден на офанзивата два от техните три бойни командни поста са прегазени, докато третият в повечето случаи не е на активна служба.
След като нахлуват в прохода германските сили се разделят на две групи, всяка напредваща по един от двата пътя, водещи на северозапад извън прохода. Ромел остава с основната група на 10-а танкова дивизия, придвижваща се по северния от двата пътя, водещи към Тала, докато смесена италианско-германска група поема по южния път по посока на Хаидра. Остатъкът от бойно командване Б на 1-ва бронетанкова дивизия се придвижва 20 мили (30 км), за да се изправи срещу южната група на 20 февруари, но не успява да спре настъплението и през следващия ден.
Моралът на американските части започва да пада и до вечерта много части отстъпват, изоставяйки екипировката си на бойното поле. Проходът е напълно отворен и е изглеждало, че временните складове за муниции в Тебеса са били изключително близо. Все пак отчаяната съпротива на няколко изолирани групи, останали в тила, сериозно забавят настъплението на германците и на втория ден операциите по прочистване все още са продължавали, докато бронетанковият авангард продължавал настъплението по пътищата.
До нощта на 21 февруари 10-а танкова дивизия е в близост до малкото градче Тала, разполагащо с два пътя към Тебеса. Ако градът паднел и германската дивизия решала да се предвижи по южния от двата пътя, американската 9-а пехотна дивизия, разположена на север, е щяла да бъде лишена от припаси, а команден център Б на 1-ва бронетанкова дивизия е щяла да бъде в капан между 10-а танкова дивизия и техните поддържащи части придвижващи се на север по втория път. Същата нощ малки групи от британските, френските, и американските части освободени от защитната линия на север са изпратени постепенно на защитната линия при Тала. Цялата артилерия на американската 9-а пехотна дивизия, 48 оръдия, които са започнали да се придвижват на 17 февруари от техните позиции на запад, се разполагат през нощта. Когато битката започва отново на следващия ден, защитата е значително по-силна и фронтовата линия се държи основно от британска пехота с изключително силна поддръжка от американската артилерия.
Прекалено разтегнатите линии и недостатъчните продоволствия карат Ромел да прекрати офанзивата. Притеснявайки се, че приближаващата британска 8-а Армия може да пробие линията Марет ако не я подсили, той спира нападението и започва да отстъпва на изток. На 23 февруари американците провеждат масивно въздушно нападение срещу прохода, което ускорява германското отстъпление и до вечерта на 25 февруари проходът е превзет.
Други действия
Атаката срещу Сбиба е спряна на 19 февруари от елементи на британската 1-ва пехотна бригада, 2-ри батальон на Колдстрийм гардс.
Последици от битката
След битката и двете страни изучават резултатите. Като цяло Ромел гледа с презрение на американската техника и бойни умения и не ги смята за заплаха. Но все пак хвали няколко американски части като 2-ри батальон на 13-и бронетанков полк от 1-ва бронетанкова дивизия, с командир Орландо Уард. Той описва защитата на Сбеитла, проведена от тази бойна част, като "добре замислена и изпълнена". Известно време след битката германските части използват голяма част от пленените американски превозни средства.
Съюзниците разглеждат сериозно резултатите от битката и незабавно генерал Дуайт Айзенхауер започва реструктуриране на съюзническото командване като създава ново главно командване 18-а Армия под командването на генерал сър Харълд Александър. С това цели да затегне контрола над корпусите и армиите на трите съюзнически нации, участващи в конфликта, и да подобри тяхната координация (имало е значителни спънки по време на действията в предходните месеци).
Най-важното за американските сили е, че командващият 2-ри корпус Лойд Фредендал е освободен от командване и е изпратен на небойно назначение за остатъка от войната. Айзенхауер потвърждава чрез генерал-майор Омар Брадли, че и други освен подчинените на Фредендал нямат вяра в него като техен командир. Командващият на британската Първа армия генерал-лейтенант Кенет Артър Ноуъл Андерсън също е смятал Фредендал за некомпетентен. На 6 март генерал-майор Джордж Патън е назначен за командир на 2-ри корпус с изричната задача да подобри боеспособността. Брадли е назначен като помощник командир на 2-ри корпус, а впоследствие поема командването му. Още няколко офицера са преместени или повишени. Бригаден генерал Стафорд Лирой Ъруин, командващ артилерията на 9-а дивизия при Касерин, се превръща в успешен дивизионен командир. На командирите са дадени по-широки правомощия за вземане на моментни решения без да се допитват до главното командване и са били насърчавани да изграждат командните постове в близост до фронтовата линия. За разлика от Фредендал, който изгражда голяма и силно укрепена щабквартира на голямо разстояние от фронта и рядко е посещавал фронтовата линия. Освен това той е имал навика да разпръсква частите под негово командване, така че получените по този начин изолирани позиции са били лесни за обкръжение и унищожение.
Положени са усилия да се подобри концентрираната артилерийска и въздушна поддръжка, извиквана по необходимост, която преди това е била трудна за координиране. Докато практиката на повикване за артилерийски обстрел срещу дадена позиция на американската артилерия довежда до голямо подобрение, проблемите с координирането на въздушна поддръжка не са удволетворени задоволително до битката за Нормандия (повече от година по-късно).
Вместо да се назначават различни задачи на елементите от всяка дивизия, както е правил Фредендал, единиците се придържат една към друга. 2-ри Корпус незабавно започва да използва дивизиите си като едно цяло, вместо да ги разпръсква на части със силно отделени мисии. До времето на пристигането си в Сицилия, техните сили значително нарастват.
В литературата и киното
Поредицата от книги "Brotherhood of War" на У. Е. Б. Грифин започва с американски офицер, пленен по време на битката при прохода Касерин.
Филмът от 1970 г. "Patton" започва с образа на генерал Омар Брадли, разглеждащ последиците от битката при прохода Касерин.
Филмът от 1980 г. "The Big Red One" разглежда битката при прохода Касерин като първото голямо сражение на отряд.
Във филма "Спасяването на редник Райън", сержант Майкъл Хорват отбелязва, че той и капитан Милър са се били в битката при прохода Касерин.
Вижте също
Немски Африкански корпус
Ервин Ромел
Ханс Юрген фон Арним
Бележки
Използвана литература
Външни препратки
Статия в сайта Leaders & Battles Database
Битките при прохода Касерин (електронна книга, разпространявана от Американската армия)
Касерин
История на Тунис
Битки на Средиземноморския театър
Битки на Германия през Втората световна война
Битки на Италия през Втората световна война
Битки на САЩ през Втората световна война
Битки на Великобритания през Втората световна война
Битки на Франция през Втората световна война | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,656 |
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ask patches [ ; saturates cell
if is-DEM < 800[
set cell-storage cell-storage + fill-rate
]
]
ask patches [
if any? neighbors4 with [ any? turtles-here ] [
set cell-storage cell-storage + fill-rate
]
]
ask patches [
if cell-storage > 0 [
if cell-storage > 5 [
set cell-storage 5
if not any? turtles-here [
sprout 1 [
set color blue
set size 10
set shape "circle"
]
]
]
set pcolor cell-storage + 82
]
]
end
Currently trying to figure out how to let the flood disappear after/or within this the tick, so that it can reoccur in the next one. I´m not aiming to reset the tick counter, it should reach 200.
Tried resetting the ticks, but only to manage resetting everything.
Any ideas ?
Thank you very much in adavance
Cheers
A: You can simply use the display primitive to update the view without waiting for the tick counter to advance.
I am not familiar with your whole model but here's a conceptual example that may help:
to water_rise
...
end
to go
repeat 100 [
water-rise
display
]
tick
end
This code would execute your water-rise procedure 100 times, update the view with new patch colors after each execution, and only increase the tick counter by 1 after the repeat loop is done.
Here is the link to NetLogo Dictionary entry about the display primitive:
http://ccl.northwestern.edu/netlogo/docs/dictionary.html#display
Note: due to its design, NetLogoWeb does not support the display primitive. So, you need to use the desktop version.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 3,459 |
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"redpajama_set_name": "RedPajamaC4"
} | 4,794 |
Carriera
Ha recitato in diversi musical a Los Angeles, Broadway e a Londra, tra cui Hair (Off Broadway, 1967; Los Angeles, 1968), Jesus Christ Superstar (Broadway, 1971), Tricks (Broadway, 1973), Raisin (Washington, 1973), Let My People Come (Off Broadway, 1974), Timbuktu! (Broadway, 1974), [[Ain't Misbehavin' (musical)|Ain't Misbehavin''']] (Londra, 1979), La piccola bottega degli orrori (Londra, 1983), King (Londra, 1990), Children of Eden (Londra, 1991), 70, Girls, 70 (Londra, 1991), Once on this Island (Londra, 1994), Carousel (Londra, 1995), Il re ed io (Londra, 1996), The Goodbye Girl (Londra, 1997), La bella e la bestia (Londra, 1998) e Follies (Londra, 2002).
Attualmente si è ritirata dalle scene e insegna recitazione nella nativa Riverside.
Filmografia
Cinema
Supergirl - La ragazza d'acciaio (Supergirl), regia di Jeannot Szwarc (1984)
Televisione
The Tomorrow People - serie TV, 3 episodi (1992)
Painted Lady, regia di Julian Jarrold - film TV (1997)
Bugs - Le spie senza volto - serie TV, 3 episodi (1998-1999)
A Christmas Carol,'' regia di Catherine Morshead - film TV (2000)
Note
Collegamenti esterni
Cantanti da musical | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 4,147 |
With a complex health care system, developing strong relationships and advocating on behalf of our members is more important than ever before.
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The Council on Public Affairs and Communications (CPAC) helps to facilitate government relations on behalf of Doctors of BC to meet the goals of our strategic framework and to promote and support physician leadership across the province.
In this way, we help ensure our doctors are always on the front line, delivering quality health care and being natural advocates for system reform, patient rights, and professional satisfaction. | {
"redpajama_set_name": "RedPajamaC4"
} | 6,101 |
\section{Introduction: the plurality of dynamics}\label{introduction}
What are the dynamical equations of physics? A first try:
\begin{quote}
Before the twentieth century, they were the equations of classical mechanics: Hamilton's equations, say,
\begin{equation}
\dot{q}^i= \pbp{H}{p_i}\,\,\,\,\,\,\dot{p}_i=-\pbp{H}{q^i}.
\end{equation}
Now we know that they are the equations of quantum mechanics:\footnote{For the purposes of this article I assume that the Schr\"{o}dinger equation is exact in quantum mechanics, and that `wavefunction collapse', whatever its significance, is at any rate not a \emph{dynamical} process. This clearly excludes the GRW theory \cite{grw,pearle,bassighirardireview} and similar dynamical-collapse theories (see \citeN{alberttimechance} for extensive consideration of their statistical mechanics); it includes the Everett interpretation (\citeNP{everett,wallacebook}) and hidden-variable theories like Bohmian mechanics (\citeNP{bohm,bohmhiley,durrincushing}); in \citeN{wallaceorthodoxy} I argue that it includes orthodox quantum mechanics, provided `orthodoxy' is understood via physicists' actual practice.} that is, Schr\"{o}dinger's equation,
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}} \ket{\psi}= -\frac{i}{\hbar}\op{H}\ket{\psi}.
\end{equation}
\end{quote}
This answer is misleading, bordering on a category error. Hamilton's equations, and Schr\"{o}dinger's, are not concrete systems of equations but frameworks in which equations may be stated. For any choice of phase space or Hilbert space, and any choice of (classical or quantum) Hamiltonian on that space, we get a set of dynamical equations. In classical mechanics, different choices give us:
\begin{itemize}
\item The Newtonian equations for point particles (better: rigid spheres) interacting under gravity, generalisable to interactions under other potential-based forces;
\item The simple harmonic oscillator equations that describe springs and other vibrating systems;
\item Euler's equations, describing the rotations of rigid bodies;
\item Euler's (other) equations, describing fluid flow in the regime where viscosity can be neglected.
\item (With a little care as to how the Hamiltonian dynamical framework handles them), the field equations of classical electromagnetism and general relativity.
\end{itemize}
In quantum mechanics, we also have
\begin{itemize}
\item The quantum version of the Newtonian equations, applicable to (e.g.) nonrelativistic point particles interacting under some potential;
\item The quantum version of the harmonic oscillator;
\item The quantum field theories of solid-state physics, describing such varied systems as superconductors and vibrating crystals;
\item The quantum field theories of particle physics, in particular the Standard Model, generally thought by physicists to underly pretty much all phenomena in which gravity can be neglected and most in which it cannot.
\end{itemize}
All of these equations are widely used in contemporary physics, and all have been thoroughly confirmed empirically in applications to the systems to which they apply. And this is already something of a puzzle: how is it that physics doesn't just get by with one such system, the \emph{right} such system? Even if that system is far too complicated to solve in practice, that in itself does not guarantee us the existence of other, simpler, equations, applicable in domains where the Right System cannot be applied? Any systematic understanding of physics has to grapple with the plurality of different dynamical equations it uses. Indeed, it must grapple with the plurality of \emph{frameworks}, for the classical equations, too, remain in constant use in contemporary physics. The problem of `the quantum-classical transition' is poorly understood if it is seen as a transition from one dynamical system (quantum mechanics) to another; it is, rather, a problem of understanding the many concrete classical dynamical equations we apply in a wide variety of different situations.
But even this account greatly understates the plurality of dynamics. Some more examples still:
\begin{itemize}
\item The equations of radioactive, or atomic, decay.
\item The Navier-Stokes equation, describing fluid flow when viscocity \emph{cannot} be neglected.
\item The Langevin equation, a stochastic differential equation describing the behaviour of a large body in a bath of smaller ones, and the related Fokker-Planck equation, describing the evolution of a probability distribution under the Langevin equation.
\item The Vlasov equation, describing the evolution of the particles in a plasma --- or the stars in a cluster or galaxy --- in the so-called ``collisionless'' regime.
\item The Boltzmann equation, describing the evolution of the atoms or molecules in a dilute (but not `collisionless') gas.
\item The Balescu-Lenard equation, describing the evolution of plasmas outside the collisionless regime.
\item The master equations of chemical dynamics, describing the change in chemical (or nuclear) composition of a fluid when particles of one species can react to form particles of another.
\end{itemize}
It is (as we shall see) a delicate matter which of these equations is classical and which quantum. But at any rate, none of them fit into the dynamical framework of either Hamiltonian classical mechanics or unitary quantum mechanics. In fact, for the most part they have two features that are foreign to both:
\begin{enumerate}
\item They are \emph{probabilistic}, either through being stochastic or, more commonly, by describing the evolution of probability distributions in a way that cannot be reduced to deterministic evolution of individual states. (In the quantum case, they describe the evolution of mixed states in a way that cannot be reduced to deterministic --- much less unitary --- evolution of pure states.)
\item They are \emph{irreversible}: in a sense to be clarified shortly, they build in a clear direction of time, whereas both Hamilton's equations and the Schr\"{o}dinger equation describe dynamics which are in an important sense time-reversal-invariant.
\end{enumerate}
And again, these equations --- and others like them --- are very widely used throughout physics, and very thoroughly tested in their respective domains.
One can imagine a world in which physics is simply a patchwork of overlapping domains each described by its own domain-specific set of equations, with few or no connections between them. (Indeed, Nancy Cartwright~\citeyear{Cartwright1983,cartwrightdappled} makes the case that the actual world is like this.) But the consensus in physics is that in the \emph{actual} world, these various dynamical systems are connected: equation A and equation B describe the same system at different levels of detail, and equation A turns out to be derivable from equation B (perhaps under certain assumptions). Indeed, it is widely held in physics that \emph{pretty much every} dynamical equation can ultimately, in principle, be derived from the Standard Model, albeit sometimes through very long chains of inference via lots of intermediate steps.
I don't think it does great violence to physics usage to say that (non-equilibrium) statistical mechanics just \emph{is} the process of carrying out these constructions, especially (but not only) in the cases where the derived equations are probabilistic and/or irreversible. But in any case, in this paper I want to pursue a characterisation of the \emph{conceptual foundations} of (non-equilibrium) statistical mechanics as concerned with understanding how such constructions can be carried out and what additional assumptions they make. The main conceptual puzzles then concern precisely (1) the probabilistic and (2) the irreversible features of (much) emergent higher-level dynamics; they can be dramatised as a contradiction between two apparent truisms:
\begin{enumerate}
\item Derivation of probabilistic, irreversible higher-level physics from non-probabilistic, time-reversal-invariant lower-level physics is \emph{impossible}: probabilities cannot be derived from non-probabilistic inputs without some explicitly probabilistic assumptions, and more importantly, as a matter of logic time-reversal invariant low-level dynamics cannot allow us to derive time-reversal-noninvariant results.
\item Derivation of probabilistic, irreversible higher-level physics from non-probabilistic, time-reversal-invariant lower-level physics is \emph{routine}: in pretty much all of the examples I give, and more besides, apparent derivations can be found in the textbooks. And these derivations have novel predictive success, both qualitatively (in many cases, such as the Boltzmann, Vlasov and Balescu-Lenard equations, higher-level equations in physics are derived from the lower level and then tested, rather than established phenomenologically and only then derived) and quantitatively (statistical mechanics provides calculational methods to work out the coefficients and parameters in higher-level equations from lower-level inputs, and does so very successfully).
\end{enumerate}
I want to suggest that we can make progress on resolving this apparent paradox through three related strategies:
\begin{enumerate}
\item Focussing on the derivation of quantitative equations of motion, rather than on the qualitative problem of how irreversibility could \emph{possibly} emerge from underlying microphysics. The advantage of the former strategy is that we actually have concrete `derivations' of irreversible, probabilistic microphysics all over modern statistical mechanics; identifying the actual assumptions made in those derivations, and in particular focussing in on those assumptions which have a time-reversal-noninvariant character, is \emph{prima facie} a much more tractable task than trying to work out in the abstract what those assumptions might be.
\item Looking at examples from (relatively) contemporary physics. Insofar as philosophy of statistical mechanics has considered the quantitative features of non-equilibrium statistical mechanics, it has \emph{generally} focussed on Boltzmann's original derivation of the eponymous equation, and of later attempts to improve the mathematical rigor of that equation whilst keeping its general conceptual structure. But contemporary statistical mechanics tends to treat the Boltzmann equation in a different (more explicitly probabilistic) way, as well as incorporating a wide range of additional examples.
\item Engaging fully with quantum as well as classical mechanics. It is common to see, in foundational discussions, the claim that the relevant issues go over from classical to quantum \emph{mutatis mutandis}, so that we can get away with considering only classical physics. \citeN[p.12]{sklarstatmech} is typical:
\begin{quote}
the particular conceptual problems on which we focus --- the origin and rationale of probability distribution assumptions over initial states, the justification of irreversible kinetic equations, and so on --- appear, for the most part, in similar guise in the development of both the classical and quantum versions of the theory. The hope is that by exploring these issues in the technically simpler classical case, insights will be gained that will carry over to the understanding of the corrected version of the theory.
\end{quote}
I will argue that this position, while defensible as regards irreversibility, fails radically for probability --- something that could perhaps have been predicted when we recall that quantum theory is itself an inherently probabilistic theory.
\end{enumerate}
In sections \ref{bbgky}--\ref{coulddobetter} I look at modern non-equilibrium statistical mechanics through two detailed classes of example: dilute gases (including the famous Boltzmann equation, where I contrast a modern, probabilistic analysis (section \ref{boltzmann-modern}) with Boltzmann's own approach (section \ref{boltzmann-contrast}) and linear systems (in particular Brownian motion in an oscillator environment, which gives rise to the stochastic Langevin equation and the related Fokker-Planck equation). In both cases, the name of the game is to derive closed-form equations from some coarse-grained description of the system. And two common features of these derivations stand out: explicit and irreducible appeal to the Liouville form of classical mechanics, and a concrete, quantitative condition for irreversibility expressed in terms of an initial-state condition that constrains the residual (\mbox{i.\,e.\,}, non-coarse-grained) features.
In sections \ref{mixedstates}--\ref{conclusion} I consider the quantum case, first in general terms (section \ref{mixedstates}), where the analogy between quantum mixed states
and classical probability distributions turns out to be superficial, and then in the concrete context of the quantum versions of the dilute gas and of Brownian motion --- where ``quantum version'' does not mean ``some new system, analogous to the old'', but ``the same old system, analysed correctly once we remember that quantum mechanics is the correct underlying dynamics''. I conclude that while the problem of irreversibility looks pretty similar classically and quantum-mechanically, the problem of probability is radically transformed: statistical-mechanical probabilities reduce entirely to quantum-mechanical ones.
The physics discussed in this paper is standard and I do not give original references. I have followed \citeN{balescubook} (for dilute gases), \citeN{zehtime} and Zwanzig~\citeyear{Zwanzig1960,Zwanzig1966} for the general linear-systems formalism, \citeN{zwanzigbook} for the Langevin and Fokker-Planck equations, \citeN{pazzurekreview} for the perturbative derivation of the Fokker-Planck equation, \citeN{peres} for the phase-space version of quantum mechanics, and \citeN{liboffkinetic} for the quantum version of the BBGKY hierarchy.
\section{The BBGKY hierarchy and the Vlasov equation}\label{bbgky}
Consider $N$ indistinguishable classical point particles (for very large $N$: $10^6$ at least, perhaps larger), interacting via the Hamiltonian
\begin{eqnarray}
H_N&=&\sum_{1\leq i \leq N}\frac{\vctr{p}_i \cdot \vctr{p}_i}{2m} + \sum_{1\leq i < j \leq N}V(\vctr{q}^i-\vctr{q}^j)
\nonumber
\\
&\equiv & H_1(\vctr{p}_i,\vctr{q}^i) + \sum_{1\leq i < j \leq N}V(\vctr{q}^i-\vctr{q}^j).
\end{eqnarray}
With the right choices of the interaction term $V$, this might describe the particles in a `classical' gas or plasma, or the stars in a galaxy or cluster.
A standard move in contemporary statistical mechanics is to consider the evolution of a \emph{probability distribution} $\rho$ over phase space under the Hamiltonian dynamics defined by this Hamiltonian. (Thus probability is introduced explicitly and by hand; I offer no justification for this at this stage, beyond the fact that it is in fact routinely done!) It is well known that any such distribution evolves by the Liouville equation,
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}}\rho=\pb{H_N}{\rho}=\sum_{1\leq i \leq N}\left( \pbp{H_N}{\vctr{q}^i} \cdot \pbp{\rho}{\vctr{p}_i}- \pbp{H_N}{\vctr{q}^i} \cdot \pbp{\rho}{\vctr{p}_i}\right).
\end{equation}
If we assume the distribution $\rho$ is symmetric under particle interchange (or, equivalently, if we just work with the symmetrised version of $\rho$, which is empirically equivalent), and if we write $x^i$, schematically, for the six coordinates $\vctr{q}^i,\vctr{p}_i$, then we can define, for each $M\leq N$, the $M$-particle marginal probability as
\begin{equation}
\rho_M = \prod_{M<i\leq N}\int \dr{x}^i \rho(x^1,\ldots x^N).
\end{equation}
$\rho_m$ represents the probability that a randomly-selected $M$-tuple of particles will be found in a given region of their $M$-particle phase space.
We can, iteratively, define $m$-particle correlation functions: the 2-particle correlation function is
\begin{equation}
c_2(x^1,x^2)=\rho_2(x^1,x^2)-\rho_1(x^1)\rho_1(x^2),
\end{equation}
the three-particle correlation function is
\begin{eqnarray}
c_3(x^1,x^2,x^3)&=&\rho_3(x^1,x^2,x^3)-\rho_1(x^1)\rho_1(x^2)\rho_1(x^3) \nonumber \\
& - & \rho_2(x^1,x^2)c_2(x^3) - \rho_2(x^1,x^3)c_2(x^2) - \rho_2(x^2,x^3)c_2(x^1)
\end{eqnarray}
and so forth.
The \emph{BBGKY hierarchy} (named for Bogoliubov, Born, Green, Kirkwood and Yvon) is a rewriting of the Liouville equation in terms of the $n$-particle marginals. For each $M$, an equation of motion can be written down that is \emph{almost} a closed first-order differential equation for $\rho_M$, but which has a term in $\rho_{M+1}$:
\begin{eqnarray}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_M}&=&\pb{H_M}{\rho_M} \nonumber \\
&+ &(N\!-\!M)\sum_{1\leq i\leq M}\int \dr{x}^{M+1}\vctr{\nabla}V(\vctr{q}^i-\vctr{q}^{M+1})\cdot \pbp{\rho_{M+1}}{\vctr{p}_i}
\end{eqnarray}
So the first equation in the hierarchy can be used to determine the evolution of $\rho_1$, but only given the value of $\rho_2$; the latter can in turn be determined from the \emph{second} equation, and so forth. For instance, the first equation in the hierarchy,
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho}_1 = \pb{H_1}{\rho_1} + (N\!-\!1)\int \dr{x^2} \vctr{\nabla}V(\vctr{q}^1-\vctr{q}^2)\cdot \pbp{\rho_2}{\vctr{p}_2},
\end{equation}
is the equation for free-particle motion plus a correction term linear in the two-particle marginal.
The last equation in the hierarchy is just the $N$-particle Liouville equation, and the full system of equations does not take us beyond that equation. The value of the hierarchy is that it allows us to define various approximations. If we can justify assuming that the $M+1$-particle marginals are negligible, we can truncate the hierarchy at the $M$th equation and obtain a \emph{closed} system of equations for $\rho_M$. The grounds for that assumption will have to be assessed on a case-by-case basis, but crucially, any such assumption does not \emph{in itself} contain any statement of time asymmetry.
For instance, the \emph{collisionless approximation} takes $c_2\simeq 0$, \mbox{i.\,e.\,} $\rho_2(x^1,x^2)\simeq \rho_1(x^1)\rho_1(x^2)$ (and also $N-1 \simeq N$) and closes the first equation in the hierarchy to give a self-contained equation in $\rho_1$: the \emph{Vlasov equation},
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_1}(\vctr{q},\vctr{p},t) = \pb{H_1}{\rho_1}(\vctr{q},\vctr{p},t) +N \rho_1(\vctr{q},\vctr{p},t)\int\dr{\vctr{p}'} \dr{\vctr{q}'}\rho_1(\vctr{q}',\vctr{p}',t)\vctr{\nabla}V(\vctr{q}-\vctr{q}'),
\end{equation}
widely used in plasma physics and galactic dynamics.
This approximation is justified by (a) assuming the initial multiparticle correlations are negligible; (b) making additional assumptions about the initial state and the dynamics which jointly entail that negligible multiparticle correlations remain negligible. Under these assumptions the equation, which on its face describes a one-particle \emph{probability distribution}, can also be taken to describe the \emph{actual distribution} of particles when averaged over regions large compared to (total system volume / N).
I make no claim that the assumptions used to derive the Vlasov equation have been rigorously demonstrated to entail it. But they don't appear to differ from the general run of the mill in mainstream theoretical physics in their level of rigor. And, since they do not distinguish a direction of time, nor does the Vlasov equation: it is probabilistic, but time-reversal-invariant. (It is, however, nonetheless highly non-trivial, despite its superficial resemblance to the one-particle version of the Poisson equation: in particular, the second term in the equation is non-linear in $\rho_1$.) Irreversibility will require a more complicated equation, to which I now turn.
\section{The Boltzmann equation: a modern approach}\label{boltzmann-modern}
Let's relax the collisionless assumption, but only slightly: in the \emph{dilute-gas assumption}, we assume that \emph{three}-particle correlations are negligible. We also assume a two-particle interaction potential $V$ that decreases rapidly with distance, so that the nonlinear term in the Vlasov equation is negligible. The first two equations in the BBGKY hierarchy are now a closed set of equations in $\rho_1$ and $c_2$, which after some manipulation (see \citeN[ch.7]{balescubook} for the details) yields the following (schematically expressed) \emph{integro-differential equation} for $\rho_1$:
\begin{equation} \label{protoboltzmann}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_1}(t) = \{H_1,\rho_1\} + \int_0^t \dr{\tau} K(\tau)( \rho_1 \otimes \rho_1)(t-\tau)+ \Lambda(t) \cdot c_2(0)
\end{equation}
Here I write $(\rho_1\otimes \rho_1) (x^1,x^2,t)\equiv \rho_1(x^1,t)\rho_1(x^2,t)$, and suppress the dependence of $\rho_1$ and $c_2$ on anything except time. $\Lambda(t)$ and $K(t)$ are time-dependent linear operators whose precise form will not be needed.
This equation holds only for $t>0$, and might appear to describe explicitly time-reversal-noninvariant dynamics. It does not: it has been derived from the dilute-gas approximation without further assumptions and the latter assumption is time-reversal (and time-translation) invariant. The time reverse of the equation holds for $t<0$, and the whole system distinguishes no preferred direction or origin of time.
It can now be demonstrated --- and again, this is a \emph{mathematical} result, the rigor of which can be questioned but which involves no additional physical assumptions --- that $K(\tau) \rho_1 \otimes \rho_1(t-\tau)$ decreases very rapidly with time, so that (for times short compared to the recurrence time $T$ of the system) we can approximate the third term in (\ref{protoboltzmann}) by
\begin{equation}
\int_0^t \dr{\tau} K(\tau) (\rho_1 \otimes \rho_1)(t-\tau) \simeq \left(\int_0^T \dr{\tau}K(\tau)\right)(\rho_1\otimes \rho_1)(t).
\end{equation}
And in fact the expression on the right hand side is equal to the well-known \emph{Boltzmann collision term} $\kappa[\sigma,\rho_1(t)]$, which depends on $\rho_1(t)$ (quadratically) and on the scattering cross-section $\sigma(\vctr{p}\vctr{p}'\rightarrow \vctr{k}\vctr{k'})$ for two-particle scattering under the interaction potential $V$. (The latter can be calculated by standard methods of scattering theory.)
The equation (\ref{protoboltzmann}) can now be approximated as
\begin{equation} \label{protoboltzmann2}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_1}(t) = \{H_1,\rho_1(t)\} + \kappa[\sigma,\rho_1(t)] + \Lambda(t) \cdot c_2(0).
\end{equation}
This differs from the full \emph{Boltzmann equation},
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_1}(t) = \{H_1,\rho_1(t)\} + \kappa[\sigma,\rho_1(t)] ,
\end{equation}
only by the final term $\Lambda(t)\cdot c_2(0)$, which is a dependency of the rate of change of $\rho_1$ at time $t$ on the \emph{initial} two-particle correlation function.
In textbooks (see, \mbox{e.\,g.\,}, \citeN[ch.7]{balescubook}, one can find heuristic arguments to the effect that this term can be expected to be negligible (perhaps after some short `transient' period): indeed, Balescu quotes Prigogine and co-workers as referring to the $\Lambda(t)\cdot c_2(0)$ term as the \emph{destruction term}. But this assumption needs to be treated cautiously: the Boltzmann equation is time-reversal noninvariant, and so the assumption that this term vanishes must in some way build in assumptions that break the time symmetries of the problem. And indeed, the heuristic arguments would fail if a precisely arranged pattern of delicate correlations were present at time 0: they rely on certain assumptions of genericity about those correlations.
Following \citeN{wallacelogic}, let's call an initial two-particle correlation function (and, by extension, an initial probability distribution satisfying our other assumptions) \emph{forward compatible} if $\Lambda(t)\cdot c_2(0)$ is negligible for at least $0<t \ll T$. Then we have established that systems whose initial state is forward compatible will obey the Boltzmann equation.
How are we to think about this assumption? I can see three options:
\begin{enumerate}
\item At one extreme, we could simply posit that a system's state is forward compatible, and derive from this that it obeys the Boltzmann equation. This is dangerously close to circularity: to say that a state is forward compatible is close to saying just \emph{that} it obeys the Boltzmann equation. The formal machinery developed here allows us to sidestep that circularity: to say that a correlation function satisfies $\Lambda(t)\cdot c_2(0)\simeq 0$ is equivalent to saying that its forward time evolution satisfies the Boltzmann equation only given significant, non-trivial mathematical work. But we are still left with little clarity as to what this obscure mathematical expression means, and in particular, we fail to connect to the heuristic arguments that this term can in some sense be `expected' to vanish for `reasonable' choices of $c_2(0)$.
\item At the other extreme, we could exploit the linearity of $\Lambda(t)$ to observe that if $c_2(0)$ \emph{vanishes}, the troublesome term is removed entirely. So this state, at least, is certainly forward compatible: positing that the initial state is uncorrelated thus suffices to guarantee that the Boltzmann equation holds for $0<t \ll T$. However, it is much stronger than is required (we expect that a great many other choices of correlation are also forward compatible) and somewhat physically implausible (a totally uncorrelated dilute gas will swiftly build up some correlations, and indeed we can write an explicit equation for them; if they really were negligible at all times, the Boltzmann equation would reduce to the Vlasov equation). And again, this assumption fails to engage with the heuristic arguments for the vanishing of $\Lambda(t)\cdot c_2(0)$ for `reasonable' initial states.
\item An intermediate strategy (suggested in \citeN{wallacelogic}) is to assume \emph{Simplicity}: the assumption that the initial state's correlations can be described in a reasonably simple mathematical form. (We need to explicitly bar descriptions that involve starting with a simple sate, evolving it forward, and then time-reversing it!) We have heuristic, but extremely strong, grounds to assume that Simple states are forward compatible.
\end{enumerate}
The second and third of these assumptions, counter-intuitively, are time-reversal invariant: Simple states time reverse to simple states, uncorrelated states to uncorrelated ones. But they are not time \emph{translation} invariant: their forward \emph{and backward} time evolutes violate the condition. Imposing either condition guarantees (or, for the simplicity assumption, is heuristically likely to guarantee) that the Boltzmann equation holds for $0<t\ll T$, and that its time reverse holds for $0>t>-T$. So such conditions can only be imposed at the beginning of a system's existence, on pain of experimental disconfirmation.
Superficially, it is easy to make the argument that simplicity, at least, can naturally be expected from a physical process that creates a system: such processes cannot plausibly be expected to generate delicate patterns of correlations. (The same might be said for the straightforward assumption of forward compatibility.) But again, this is not innocent: the time reversal of a system's final state, just before its destruction, is certainly not forward compatible, and so our argument builds in time-directed assumptions. Assuming that we seek a dynamical explanation for irreversibility (and are not content, for instance, to appeal to unanalysable notions of agency or causation) then a familiar regress beckons, and we are led to assume some condition analogous to forward compatibility or simplicity, applied to the very early Universe. (I discuss this issue in more detail in \citeN{wallacelogic}.)
In any case, for the purposes of \emph{this} paper, what matters is that we have identified the time-asymmetric assumption being made in (a modern derivation of) the Boltzmann equation: it is an assumption about the initial condition of the system, phrased probabilistically (it is a constraint on the two-particle correlation function, which is inherently probabilistic) and with no implications as to the bulk distribution of particles in the system (insofar as these are coded by the one-particle marginal, given that all higher-order correlations are small). It will be instructive to compare this to the situation in Boltzmann's own derivation of the Boltzmann equation.
\section{The Boltzmann equation: contrast with the historical approach}\label{boltzmann-contrast}
The equation derived by Boltzmann himself\footnote{Here I follow Brown \emph{et al}'s~\citeyear{brownboltzmann}'s account of Boltzmann's work; see that paper for original references.} has the same functional form as what I have called the Boltzmann equation (albeit Boltzmann confined his attention to the case where $\rho_1$ is spatially constant, so that the free-particle term $\pb{H_1}{\rho_1}$ vanishes). But its interpretation is quite different. To Boltzmann, $\rho_1$ was not a probability distribution: it was a smoothed version of the \emph{actual} distribution function of particles over 1-particle phase space, so that $\rho_1(\vctr{q},\vctr{p})\delta V$ is the actual fraction of particles in a region $\delta V$ around $(\vctr{q},\vctr{p})$. (The smoothing is necessary because, with a finite number of point particles, $\rho_1$ would otherwise just be a sum of delta functions.)
Boltzmann makes a number of simplifying assumptions about the dynamics (notably: that collisions are hard-sphere events; that three-body collisions can be neglected; that long-distance interactions can be neglected) which are time-reversal invariant and broadly equivalent to the time-reversal-invariant assumptions made in the modern derivation. He then assumes the famous \emph{Stosszahlansatz} (SZA): the assumption that the subpopulation of particles which are about to undergo a collision has the same distribution of momenta as the population as a whole. Given this assumption, \emph{imposed over a finite time interval $[0,\tau]$}, he then deduces that over that time interval the Boltzmann equation holds. The assumption is thus explicitly time-reversal-noninvariant: assuming that $\rho_1$ is not stationary, if it holds over an interval then its time reverse cannot hold over that interval. (If it did, then both the Boltzmann equation and its time reverse would hold over that interval, contradicting Boltzmann's famous $H$-theorem, which tells us that the function $H[\rho]$ monotonically increases during the evolution of any non-stationary distribution under the Boltzmann equation.)
From a formal perspective, the probabilistic derivation in section \ref{boltzmann-modern} has significant advantages over the historical approach. (Perhaps unsurprisingly, given that it has in fact largely supplanted the historical approach in theoretical physics.) In particular:
\begin{enumerate}
\item The SZA is a condition that must hold \emph{throughout an interval of time} if that system is to obey the Boltzmann equation over that interval of time. But whether a condition holds at time $t>0$ is dynamically determined by the system's state at time $0$, and Boltzmann's derivation provides little insight as to what constraint on the state at time 0 suffices to impose the SZA at later times. Lanford's celebrated proof of the Boltzmann equation (Lanford~\citeyear{lanford1,lanford2,lanford2}; carefully discussed by \citeN{valentelanford} and \citeN{uffinkvalentelanford}) demonstrates that imposing (a generalisation of) the SZA at time 0 suffices to guarantee that it holds over some $[0,\tau]$ --- but $\tau$ is extremely short, only a fifth of a mean free collision time.
By contrast, the probabilistic approach provides an explicit condition --- $\Lambda(t)\cdot c_2(0)\simeq 0$ for $0<t<T$ --- for when the Boltzmann equation holds, and allows us to state in closed form at least one correlation function --- $c_2(0)=0$ --- that guarantees that the condition will continue to hold. (To be fair, Lanford's results are at a much higher level of rigor than the probabilistic derivation, so a reader's assessment of this point will depend on their degree of tolerance of the mathematical practices of mainstream theoretical physics.)
\item The framework of the probabilistic approach readily generalises. On slightly different assumptions about the dynamics, for instance, it leads to Landau's kinetic equation (governing weak collision processes) or to the Lenard-Balescu equation (governing collisional plasmas). Boltzmann's original approach does not seem to have offered a comparably effective framework for the construction of other statistical-mechanical equations.
\item Most significantly, and as I will discuss further in section \ref{boltzmann-quantum}, the probabilistic approach readily transfers to quantum mechanics, where Boltzmann's original approach appears to fail completely.
\end{enumerate}
Against this, Boltzmann's original approach might be thought to have a major \emph{conceptual advantage} over the probabilistic approach, precisely because it eschews problematic notions of probability that seem to have no natural place in classical mechanics given that the latter features a deterministic dynamics. (Harvey Brown presses the point in his contribution to this volume.) To be blunt (the critic might say) then if progress in theoretical physics has moved from Boltzmann's original, relative-frequency, conception of the Boltzmann equation to a conception that relies on a mysterious concept of objective probability, then so much the worse for progress.
In my view, the ultimate resolution of this problem is quantum-mechanical: when we consider quantum statistical mechanics, these probabilities receive a natural interpretation. But even setting quantum theory aside, it is not obvious that the criticism is well-founded. For one thing, probabilistic notions are frequently appealed to in discussions of the Boltzmann equation, even where the latter is understood \emph{a la} Boltzmann: see Brown \emph{et al} \citeyear{brownboltzmann} as regards Boltzmann's own derivation, and \citeN{uffinkvalentelanford} in the context of Lanford's more rigorous analysis. So it is not after all clear that some notion of probability is not required --- in which case, why object to incorporating it into the equations themselves?
More importantly, conceptual analyses have to answer to the actual shape of theoretical physics, at least insofar as the latter is empirically successful. I have already noted that several generalisations of the Boltzmann equation can be derived within the probabilistic framework but (so far as I know) have not been demonstrated within Boltzmann's conception of the equation.
Now, these equations, whatever their derivation, are expressions of the one-particle marginal and can be reinterpreted without empirical consequence as equations in the actual relative frequency of particles' phase-space locations. But classical statistical mechanics \emph{also} offers examples where the predictions of the theory are themselves probabilistic in nature. In the next section, I present one such example; it will also serve as a demonstration that the analysis of irreversibility in section \ref{boltzmann-modern} is more general than just the Boltzmann equation.
\section{Mori-Zwanzig projection and the Fokker-Planck equation}\label{fokkerplanck}
Let's now consider a different Hamiltonian:
\begin{equation}
H(Q,P,q,p)= \frac{P^2}{2M}+V(Q) +\sum_{1\leq i \leq N}\omega_i(q^{i2}+p_i^2)+ \sum_{1\leq i \leq N}\lambda_i Qq^i
\end{equation}
(where functional dependence on $q,p$ schematically depicts dependence on all of the $q^i,p_i$).
This describes a single particle in one dimension (with position $Q$ and momentum $P$) (the `system') interacting with a bath of harmonic oscillators (the `environment'); it is one common model for Brownian motion. Again we introduce probabilities explicitly via a probability density $\rho$ over the phase space for the $N+1$ particles, and take Liouville's equation as the basic dynamical equation for this system. That distribution can be decomposed as
\begin{equation}
\rho(Q,P,q,p)=\rho_S(Q,P)\rho_E(q,p) + C(Q,P,q,p)
\end{equation}
where $\rho_S$ and $\rho_E$ are, respectively, the marginal probability distributions over system and environment, and $C$ is the correlation function between the two.
Heuristically we might hope to find:
\begin{itemize}
\item that there is an autonomous dynamics for the single particle (assuming that it is massive compared to the bath particles and that there are very many of the latter, so that their influence is a kind of background noise);
\item that the environment marginal $\rho_E$ is pretty much constant during the system's evolution, provided that it starts off invariant under the self-Hamiltonian
\begin{equation}
H_E = \sum_{1\leq i \leq N}\omega_i(q^{i2}+p_i^2)
\end{equation}
of the bath.
\end{itemize}
With that in mind, we define the following projection map on the space of distributions:
\begin{equation}
J \rho (Q,P,q,p) = \rho_S(Q,P) \mc{E}(q,p).
\end{equation}
Here $\rho_S$ is the system marginal distribution of $\rho$, and $\mc{E}$ is some \emph{fixed} distribution for the environment, satisfying
\begin{equation}
\pb{H_E}{\mc{E}}=0.
\end{equation}
(Typically we take $\mc{E}$ to be the canonical distribution for some given temperature.)
Then we can write $\rho$ itself as
\begin{equation}
\rho = J \rho + (1-J) \rho \equiv \rho_r + \rho_i.
\end{equation}
This is actually a special case of a general process --- the \emph{Mori-Zwanzig projection} --- for constructing autonomous dynamical equations in statistical mechanics. The components $\rho_r = J \rho$ and $\rho_i =(1-J)\rho$ are called, respectively, the \emph{relevant} and \emph{irrelevant} parts of $\rho$, and the hope is that an autonomous dynamics can be found for $\rho_r$ alone --- given certain assumptions about $\rho_i$. The decomposition of the two-particle marginal in the Boltzmann distribution into a product of one-particle marginals and a residual correlation term has a similar structure -- but in that case, the projection that defines the decomposition is nonlinear).
Without any approximation, we can obtain an integro-differential equation for $\rho_r$ somewhat similar in form to equation (\ref{protoboltzmann}), obtained in our analysis of the Boltzmann equation:
\begin{eqnarray}
\ensuremath{\frac{\dr{}}{\dr{t}}}\rho_r(t)&=&J L_H \rho_r(t) \nonumber
\\
&+ &J \int_0^t \dr{\tau} \e{(1-J)\tau L_H}(1-J)L_H \rho_r(\tau) \nonumber
\\
&+& J L_H \e{(1-J)t L_H} \rho_i(0),
\end{eqnarray}
where
\begin{equation}
L_H \rho \equiv \pb{H}{\rho}.
\end{equation}
This equation is exact, and so displays no irreversibility. However, if $J$ has been chosen appropriately then we would hope to find:
\begin{itemize}
\item That the kernel in the second term falls off sufficiently rapidly with $\tau$ that it can be approximated as
\begin{eqnarray}
\int_0^t \dr{\tau} \e{(1-J)\tau L_H}(1-J)L_H \rho_r(\tau) & \simeq & \left( \int_0^T \dr{\tau} \e{(1-J)\tau L_H}(1-J)L_H\right)\rho_r(t) \nonumber \\
&\equiv& K \rho(t)
\end{eqnarray}
where $T$ is some very long time of order the recurrence time.
\item That the third term can plausibly be expected to vanish given some `reasonable' constraints on $\rho_i(0)$.
\end{itemize}
Given the first assumption, and defining
\begin{equation}
\Lambda(t)=J L_H \e{(1-J)t L_H},
\end{equation}
we are left with the equation
\begin{equation}
\ensuremath{\frac{\dr{}}{\dr{t}}}\rho_r(t)=J L_H \rho_r(t)+ K \rho_r(t) + \Lambda(t) \rho_i(0)
\end{equation}
which is very similar in form to the proto-Boltzmann equation (\ref{protoboltzmann2}). No time-asymmetric assumption has been made to get this far; we now have to make an initial-time assumption about $\rho_i(0)$ to guarantee that $\rho_i(0)$ is forward compatible, \mbox{i.\,e.\,} that $\Lambda(t)\rho_i(0)\simeq 0$ for $0<t<T$. As with the Boltzmann case, we can guarantee this by taking $\rho_i(0)=0$; as with the Boltzmann case, this is usually overkill.
Returning to the specific case of the oscillator bath, we can calculate the second term approximately by working to second order in perturbation theory. Given a forward-compatible initial state (which, in this case, will typically require at least that $\rho_E(0)=\mc{E}$), we get the equation
\begin{equation}\label{FP}
\ensuremath{\frac{\dr{}}{\dr{t}}} \rho_S(t) = \{\tilde{H}_S,\rho_S(t)\}\nonumber + \eta \pb{X}{P\rho_S(t)}+ \alpha \pb{X}{\pb{X}{\rho_S(t)}} + f \pb{X}{\pb{P}{\rho}}
\end{equation}
where
\begin{equation}
\tilde{H}_S = P^2/2M + V(Q)+ \Delta V(Q)
\end{equation}
and $\xi, \eta$, $f$ and $\Delta Q$ are (somewhat complicated) functions of the $\lambda_i$ coefficients and the frequencies $\omega_i$ of the oscillator bath. For reasonable assumptions, $f$ is usually negligible; equation (\ref{FP}) is then the \emph{Fokker-Planck equation}, and we can get insight into it by recognising that it is the equation for the probability distribution of a particle evolving under the \emph{Langevin equation}, the stochastic differential equation
\begin{equation}
\dot{Q(t)}=P(t)/M; \,\,\,\, \dot{P}(t)=- \eta P(t) + \xi(t)
\end{equation}
where $\xi(t)$ is a random variable satisfying
\begin{equation}
\langle \xi(t)\rangle =0; \langle \xi(t_1)\xi(t_2)\delta t^2\rangle =2 \alpha \delta t.
\end{equation}
(For details of both, see \citeN[chs.1-2]{zwanzigbook}.)
These equations --- which are highly effective at describing the physics of Brownian motion --- make explicitly probabilistic predictions: for instance, that the root-mean-square value of the particle's distance from its starting place after time $t$ is (after an initial transient phase) equal to $2\alpha t/\eta^2$. Unlike the case of Brownian motion, the equations cannot be reinterpreted as non-probabilistic equations concerning the relative frequency of particles in a single system.
Incidentally, since the underlying Hamiltonian of this system is quadratic, the appeal to second-order perturbation theory is dispensible: with a bit of care, we can solve it \emph{exactly} and confirm the approximate validity of equation (\ref{FP}). See \citeN{pazzurekreview} for details.
\section{Classical statistical mechanics: could do better?}\label{coulddobetter}
I began with the observation that non-equilibrium statistical mechanics is concerned with establishing the relations between dynamical equations at different levels of description, especially in the cases where the higher-level equations are probabilistic and/or irreversible; I noted that it is mysterious how probability or irreversibility can be derived from an underlying dynamics which has neither feature, but that paying attention to the details of such derivations might be illuminating.
We have now seen two such derivations in detail (conceptual detail at any rate; the mathematics was left schematic), and I have argued that
\begin{enumerate}
\item The probabilistic features of the equations of statistical mechanics are not readily removable: even in the case of the Boltzmann equation, which can be reinterpreted as a non-probabilistic equation, its derivation is better understood and more readily generalised when understood probabilistically, and in the case of the Fokker-Planck and Langevin equations, the equations cannot even be reinterpreted non-probabilistically.
\item The irreversibility of at least a wide class of classical statistical-mechanical equations (specifically, those which are either derived from the BBGKY hierarchy via the same approximation scheme we used for the Boltzmann equation, or derived by the Mori-Zwanzig method from a linear projection) can be tracked to a cleanly-stated assumption about the initial microstate --- an assumption, however, which is again stated probabilistically.
\item No clue has been gleaned about the origins of the probabilities used in these equations: in each case, our starting point was the Louville equation, applied to a probability distribution placed by \emph{fiat} on phase space.
\end{enumerate}
So we are in the unsatisfactory position of having acquired considerable evidence as to the importance and ineliminability of probabilities in non-equilibrium statistical mechanics, without gaining any insight at all into the origins of these probabilities. In the rest of the paper, I will demonstrate that this dilemma is resolved, or at least radically transformed, when we move from classical to quantum mechanics.
\section{Mixed states and probability distributions in quantum statistical mechanics}\label{mixedstates}
At first sight, there is a straightforward translation scheme between classical and quantum that maps the previous section's results directly across to quantum theory. To phase-space points correspond Hilbert-space rays. To Hamilton's equation corresponds Schr\"{o}dinger's. To probability distributions on phase space correspond density operators. To Liouville's equation corresponds its quantum counterpart,
\begin{equation} \label{quantum-liouville}
\dot{\rho}=L_H \rho \ensuremath{=_{_{df}}} \frac{i}{\hbar}\comm{\rho}{H}.
\end{equation}
Appearances, however, can be deceptive.
Von Neumann originally \emph{introduced} the density operator to represent ignorance of which quantum state a system is prepared in: the idea is that if the system is prepared in state $\ket{\psi_i}$ with probability $p_i$, then for any measured observable $\op{X}$ its expectation value is
\begin{equation}
\langle \op{X}\rangle=\sum_i p_i \matel{\psi_i}{X}{\psi_i}=\textsf{Tr}\left[\op{X}\left(\sum_i p_i \proj{\psi_i}{\psi_i}\right)\right],
\end{equation}
so that if we define
\begin{equation} \label{density-operator}
\rho=\sum_i p_i \proj{\psi_i}{\psi_i},
\end{equation}
the formula $\langle\op{X}\rangle=\textsf{Tr}(\op{X}\rho)$ neatly summarises our empirical predictions. But it was recognised from the start that (\ref{density-operator}) cannot be inverted to recover the $p_i$ and $\ket{\psi_i}$: many assignments of probabilities to pure states give rise to the same $\rho$. By contrast, for any two distinct probability distributions over phase space there is (trivially) some measurement whose expectation value is different on the two distributions.
(It is true that if we add the extra information that the $\ket{\psi_i}$ are orthogonal, and if we assume that $\rho$ is non-degenerate, then the $p_i$ and the $\ket{\psi_i}$ can after all be recovered. But (i) there is no particular problem in making sense of probability distributions over non-orthogonal states; (ii) the density operators used in statistical mechanics --- particularly the microcanonical and canonical distributions --- are massively degenerate.)
That's a puzzling disanalogy, to be sure, but hardly fatal. In these post-positivist days, surely we can cope with a little underdetermination? But it gets worse. For recall: a joint state of system $A$ and system $B$ is generically \emph{entangled}: it cannot be written as a product of pure states, and indeed there is no pure state of system $A$ that correctly predicts the results of measurements made on system $A$ alone. Indeed, the correct mathematical object to represent the state of system $A$ alone, when it is entangled with another system, is again a density operator $\rho$, obtained by carrying out the partial trace over the degrees of freedom of system $B$.
(But \emph{this} density operator cannot be understood as a probabilistic mixture of pure states, on pain of failing to reproduce in-principle-measureable Bell-type results.)
The corollary is that we can only represent a quantum system (or, indeed, our information about a quantum system, if that's your preferred way of understanding probabilities here) by a probability measure over pure states if with probability 1 the system is not entangled with its surroundings. And in the case of the macroscopically large systems studied in statistical mechanics, this is a wildly implausible assumption: even if for some (already wildly implausible) reason we were confident that at time 0 there was no entanglement between system and environment, some will develop extremely rapidly. (But note, conversely, that simply assuming (or idealising) that the system is currently dynamically isolated from its surroundings in no way rules out the possibility that it is entangled with them.)
If we want to place a probability measure over states of a system, then, in general the only option will be to place it over the system's possible \emph{mixed} states. Mathematically, that's straightforward enough: if a system has probability $p_i$ of having mixed state $\rho_i$, then the expected value of a measurement of $\op{X}$ is
\begin{equation}
\langle \op{X}\rangle = \sum_i p_i \textsf{Tr}(\op{X}\rho_i)=\textsf{Tr}\left[\op{X}\left(\sum_i p_i \rho_i\right)\right].
\end{equation}
But is clear that we get the \emph{exact same} predictions by simply assigning the system the single mixed state $\sum_i p_i \rho_i$. In other words, probability distributions \emph{over} mixed states are indistinguishable from \emph{individual} mixed states.
Note how different this is from the classical case. Even formally (never mind conceptually), the move from phase-space dynamics to distributional dynamics is an extension of the mathematical framework of classical physics. But in quantum mechanics, the move from pure states to density operators is forced upon us by considerations of entanglement quite independently of any explicit probabilistic assumption, and so there is nothing formally novel about their introduction in statistical mechanics.
Put another way, \emph{any mathematical claim} in quantum statistical mechanics can be interpreted as a claim about the actual (perhaps mixed) state of the system. Nothing formally requires us to interpret it as any kind of probability distribution over quantum states.
Now, perhaps this would be unimpressive if the foundations of quantum statistical mechanics made major \emph{conceptual} use of probabilistic ideas in deriving their results: if, for instance, the arguments given for the mathematical form of quantum statistical equilibrium relied on probabilistic ideas. But a casual perusal of the (fairly minimal) literature on quantum statistical mechanics serves to disabuse one of this notion. From Tolman's classic text~\citeN{tolman} to modern textbook discussions, the norm is to start with classical statistical mechanics and then move from a probability distribution over phase-space points to a probability distribution over \emph{eigenstates of energy drawn from a particular basis}. I don't really know any way to make sense of this, but \emph{certainly} it can't be made sense of as a probability distribution over possible states of the system given that (a) there is absolutely no reason, given the massive degeneracy that typically characterises the Hamiltonian of macroscopic systems, to prefer one energy basis over another; (b) there is no a priori reason, even ignoring entanglement, to assume with probability 1 that the system is in an eigenstate of energy; (iii) we shouldn't ignore entanglement anyway, so assuming the system is in \emph{any} pure state is unmotivated.\footnote{Very occasionally (the only place I know is Binney \emph{et al}~\citeyear{binney}) it is observed that a probability distribution uniform over all pure states with respect to the natural(unitary-invariant) measure on states gives the same density operator as a distribution uniform over a particular orthonormal basis, and that this is the justification for the latter; even here, though, the problem of mixed states is left unanswered.}
This is not to say that there is not high-quality, rigorous work on the foundations of quantum statistical mechanics, or even on the form of the equilibrium distributions (for recent examples see Goldstein \emph{et al}~\citeyear{goldsteinequilibrium}, Malabarba \emph{et al}~\citeyear{malabarba}, and \citeN{shortequilibrium}). But insofar as these arguments avoid the above fallacies, they do so because they try to establish (or at any rate can be interpreted as trying to establish) that the actual mixed state of a system at equilibrium is the microcanonical or canonical ensemble, not that the correct probability distribution over such states reproduces that ensemble.
I conclude that there is no justification, and no need, for interpreting the mixed states used routinely in quantum statistical mechanics as any kind of probability distribution over quantum states, rather than as the (mixed) quantum states of individual systems. And if that is the case, then --- since in the real world ``classical'' systems are just quantum systems that we can get away with approximating as classical --- it becomes at least very tempting to interpret the `probability' distributions of classical mechanics simply as classical limits of individual quantum states, without any need at all for a distinctively statistical-mechanical conception of probability.
In fact --- at least for the non-equilibrium systems we have considered so far --- this move is not just tempting, but compulsory. As we will see, it is simply not viable to interpret these systems as probability distributions over localised quasi-classical systems, once quantum mechanics is taken into account.
\section{The quantum mechanics of dilute gases}\label{dilutegassection}
The dilute gas of hard spheres is the paradigm example in classical statistical mechanics, both in considering the approach \emph{to} equilibrium (the Boltzmann equation assumes this system) and in analysing equilibrium itself. But what happens when quantum mechanics is used to analyse a system of this kind?
Let us begin with some parameters. The typical gas at standard temperature and pressure has a molecular mass of around $10^{-26}$ or $10^{-27}$ kg, a molecular scale of around $10^{-10}$ m, a density of about 1 $\mathrm{kg}\,\mathrm{m}^{-3}$, and typical molecular velocities of around $10^2$ or $10^3$ $\mathrm{m}\,\mathrm{s}^{-1}$; for definiteness, let's assume our gas is confined in a 1$\mathrm{m}^3$ box, and that it consists of $10^{27}$ hard spheres of mass $10^{-27}$ kg and cross-sectional area $10^{-20} \mathrm{m}^2$ moving at a mean speed of $10^{3}$ $\mathrm{m}\,\mathrm{s}^{-1}$.
If classical \emph{micro}dynamics is to be a good approximation for this system, presumably this means that the particles in the gas must be, and remain, localised. So let's start the system off in some (probably unknown) state in which each molecule is a localised wavepacket with a reasonably definite position and momentum. We can now use approximately classical means to consider how those wavepackets evolve prior to their first collisions.
Elementary methods tell us that for a gas with these parameters, the mean free path --- the mean distance travelled by a given particle before it collides with another --- is $10^{-7}$m. The typical particle will cover this distance in $10^{-10}$ seconds.
How wide will its wavepacket be at that time? Let its original width be $\Delta q(0)$, so that its initial spread $\Delta p$ in momentum is at least $\sim \hbar/\Delta q(0)$. Roughly speaking, its additional spread in position as a result of this spread in momentum after time $t$ will be $t \times \Delta p/m$, so that its total spread after time $t$ is given by
\begin{equation}
\Delta q(t)=\Delta q(0)+t \Delta p/m = \Delta q(0)+\frac{\hbar t}{m} \frac{1}{\Delta q(0)}
\end{equation}
Elementary calculus then tells us that this is minimised for $\Delta q(0)=\sqrt{\hbar t/m}$, so that
\begin{equation}
\Delta q(t)\geq 2 \sqrt{\hbar t/m}.
\end{equation}
For the parameters governing our ideal gas, this means that the minimum size of the packet after collision is approximately $3\times 10^{-8}$ m.\footnote{Don't be fooled into thinking that `decoherence' will somehow preserve localisation to a greater degree than this. These are microscopic particles: they don't decohere, or rather: their decoherence is caused by their collision with other particles, and we're explicitly considering the between-collision phase.}
But this is thirty times the diameter of the molecule! To a rough approximation, the first set of scattering events in our dilute gas will be well modelled by \emph{plane-wave} scattering off a hard-sphere scattering surface. The resultant joint state of two scattering particles will be a sum of two terms: one (much the larger one) corresponding to the state of the two particles in the absence of their interaction, and one an entangled superposition of the particles after scattering, with significant amplitude for any direction of scattering. Each particle's individual mixed state will be highly delocalised.
This is fairly obviously nothing like the classical microdynamics, in which each pair of particles deterministically scatters into a specific post-collision state determined entirely by the pre-collision state. A quantum-mechanical dilute gas, even if it is initially highly localised, rapidly evolves into a massively entangled and massively delocalised mess. So the classical microdynamics that underpins the derivation of Boltzmann's equation (either in the `modern' form I presented in section \ref{boltzmann-modern}, or in Boltzmann's original derivation) is \emph{wildly false} for real, physical gases.
Let's be clear about what we have learned here. The point is not that there are quantum regimes where the classical Boltzmann equation breaks down. There are such regimes, of course: the statistical mechanics literature is replete with discussions of ``quantum'' gases, but in general this refers to situations where the intermolecular interaction has some complex quantum form, or (more usually) where the gas is sufficiently dense that the effects of Bose-Einstein or Fermi-Dirac statistics come into play, or where (as in the photon or phonon gas) particle number is not conserved. But I am not considering that situation; nor am I considering the situations where the \emph{quantum} Boltzmann equation is appropriate (as discussed by Brown, this volume). There is abundant empirical evidence that the particle distribution in ordinary, nonrelativistic, dilute gases is governed by the \emph{classical} Boltzmann equation --- and yet, we have seen that in those gases, the microdynamics
assumed in the derivations of that equation is simply incorrect.
Why, then, do these derivations work? To address this question, it will be useful to adopt a representation of quantum-mechanical systems as functions on phase space (as distinct from the usual representations on configuration or momentum space); doing so will also cast light more generally on the relation between quantum states and classical probability distributions.
\section{The quantum/classical transition on phase space}\label{wignersection}
In the position representation of a quantum state, information about the probability of a given result on a \emph{position} measurement can be readily read off the quantum state via the mod-squared amplitude rule, whereas information about results of \emph{momentum} measurements are encoded rather inaccessibly in the phase variation of the wavefunction. In the momentum representation, the reverse is true. A phase space representation can be seen as enabling us to more readily access both lots of information.
An alternative way of understanding quantum mechanics on phase space is more direct: if the position representation is optimised for position measurements, and the momentum representation for momentum measurements, how can we represent the state in a way optimised for phase-space measurements? One immediate objection is that the uncertainty principle forbids joint measurements of position and momentum; in fact, though, it forbids only \emph{sharp} joint measurements. The by-now well-established formalism of ``positive operator valued measurements'' (POVMs) allows for a wide variety of unsharp measurements, including phase-space measurements.
Indeed, we can straightforwardly write down such a family of phase space measurements: choose any wave-packet state $\ket{\Omega}$, reasonably well localized in position and momentum but otherwise arbitrary (Gaussians are good choices) and let $\ket{\vctr{q},\vctr{p}}$ denote the state obtained by translating the original state first by $\vctr{q}$ in position and then by $\vctr{p}$ in momentum. (Here $\vctr{q}$ and $\vctr{p}$ are vectors representing the $M$ position and $M$ momentum coordinates of our system; if it consists of $N$ particles, for instance, $M=3N$.) Then the (continuously infinite) family of operators $(2 \pi)^{-M}\proj{\vctr{q},\vctr{p}}{\vctr{q},\vctr{p}}$ is a POVM. (See appendix for proof; the equivalent result for Gaussian choices of the wavepacket is standard, but I am not aware of a proof in quite this form.)
Since these wavepacket states are not orthogonal, they do not define a \emph{sharp} measurement; however, if we partition phase space into disjoint cells $C_i$ which are large compared to the spread of the wave-packet state (and thus, at a minimum, large compared to $\hbar^M$), the operators
\begin{equation}
\op{\Pi}_i=\int_{C_i}\proj{\vctr{q},\vctr{p}}{\vctr{q},\vctr{p}}
\end{equation}
will approximately satisfy
\begin{equation}
\op{\Pi}_i\op{\Pi}_j \simeq \delta_{ij}\op{\Pi}_i.
\end{equation}
The \emph{Husimi function}, defined for quantum state $\rho$ by
\begin{equation}
H_\rho(\vctr{q},\vctr{p})=(2\pi)^{-M}\bra{\vctr{q},\vctr{p}}\rho\ket{\vctr{q},\vctr{p}},
\end{equation}
can thus be consistently interpreted as a phase-space probability distribution provided we do not probe it on too-small length scales. In fact, the Husimi function can be inverted to recover $\ket{\psi}$, so in principle we could use it as our phase-space representation; however, for our purposes it is inconvenient, and a better choice is the \emph{Wigner function},
\begin{equation}
W_\rho(\vctr{q},\vctr{p})=\frac{1}{(\hbar \pi)^M}\int \dr{\vctr{x}}\e{2i\vctr{p}\cdot \vctr{x}/\hbar}\bra{\vctr{q}\!-\!\vctr{x}}\rho\ket{\vctr{q}\!+\!\vctr{x}}.
\end{equation}
The mathematical form of the Wigner function is not transparently connected with anything physical, but in fact, if we use the Wigner function (which is real, though not necessarily positive) as a probability measure, it will approximately give the correct phase-space measurement probabilities if averaged over regions large compared to $\hbar^M$. (Indeed, the Husimi function can be obtained from the Wigner function merely by smearing it in position and momentum with the respective mod-squared wavefunctions of the wavepacket state; one advantage of the Wigner function is that it abstracts away from the need to specify a particular choice of wavepacket state.)
The main virtue of the Wigner function is that its dynamics can be conveniently compared to the classical case. Recall that classical probability distributions evolve by Liouville's equation:
\begin{equation}
\dot{\rho}=\pb{H}{\rho}.
\end{equation}
Transforming the Schr\"{o}dinger equation to the Wigner representation tells us that the Wigner function satisfies
\begin{equation}
\dot{W}=\mb{H}{W}\ensuremath{=_{_{df}}}\frac{2i}{\hbar}\sin \left(\frac{\hbar}{2i}\pb{\cdot}{\cdot}\right)\cdot (H,W).
\end{equation}
Here $\mb{\cdot}{\cdot}$, the \emph{Moyal bracket}, is best understood via its series expansion: assuming $H$ has the standard kinetic-energy-plus-potential-energy form, we can expand it as
\begin{equation}
\dot{W}=\mb{H}{W}=\pb{H}{W}+\frac{\hbar^2}{24}\frac{\partial^3V}{\partial \vctr{q}^3}
\frac{\partial^3W}{\partial \vctr{p}^3}+ O(\hbar^4).
\end{equation}
So the dynamics of the Wigner function is the dynamics of the phase-space probability distribution, together with correction terms in successively higher powers of $\hbar$, which suggests that \emph{ceteris paribus}\footnote{The full story is more complicated, as stressed by \citeN{pazzurekreview}; see also my discussion in \citeN[ch.3]{wallacebook}.} the corrections become negligible for macroscopically large systems --- and are exactly zero for free particles in any case.
Let's pause and consider these technical results from a conceptual viewpoint. The natural question one ends up asking first, when confronted with something like the Wigner function, is: why can't we just suppose that this \emph{is} a classical probability distribution, and thus resolve the paradoxes of quantum theory? And the usual answer given is: the Wigner function is generally\footnote{Specifically, it's nonnegative for pure states only if those states are Gaussian; cf \citeN{hudson1974}.} not nonnegative, which probability distributions have to be. But this is not the real problem (if it were, we could shift to the Husimi distribution, which \emph{is} reliably nonnegative). The real problem is that there is no underlying microdynamics on phase-space points such that the Wigner function's (or the Husimi function's) dynamics can be seen as a probabilistic dynamics for that microdynamics.
Put another way, suppose we were simply given the Liouville equation. Our interpretation of that equation as the evolution equation for a probability distribution rests on the fact that we can get back the Liouville equation as the dynamics for probability distributions induced by Hamilton's equations. (That the latter are deterministic is not relevant; similarly, given the Fokker-Planck equation we can justify reading it probabilistically by observing that it is the probabilistic dynamics induced by the stochastic Langevin equation.)\footnote{A technical aside: at least formally, any linear map of a vector space to itself that preserves the $L^1$ norm and which preserves the subset of vectors with nonnegative coefficients in a given basis can be interpreted as generated from a possibly-stochastic dynamics with respect to this basis. So what is doing the technical work here is the fact that neither the Wigner function nor the Husimi function have dynamical equations which can be extended to the space of all functions on phase space without violating the basis-preservation rule, \mbox{i.\,e.\,} without mapping some positive functions to negative ones. In the case of the Wigner function, the dynamical rule fails to preserve positivity even on the subspace of functions that correspond to quantum states, but this is not a requirement.}
The point is this: the Wigner function is not a function on phase space in the \emph{original} understanding of phase space. It is not a function on a space which can be \emph{physically interpreted} as the space of possible positions and momenta of point particles, for there are no such particles. It does, however, illustrate that in certain regimes (those where the higher-order terms in its evolution can be neglected, and in which the system is not probed on length-scales where the uncertainty principle comes into play), there is an approximate isomorphism between the dynamics of the quantum \emph{state} and that of the classical \emph{probability distribution}. The classical limit of quantum mechanics, to quote \citeN{ballentine}, is classical \emph{statistical} mechanics.
To strengthen the point, consider the Wigner (or Husimi) functions of pure states. They certainly do not correspond to delta functions on phase space, but to extended distributions on phase space. Even quantum wave-packet states are taken to Gaussian packets, not to delta functions: to probability distributions (or quantum generalisations thereof), not to individual microstates.
Now: in certain contexts, those wavepackets avoid spreading out and evolve so as to mirror the evolution of phase-space \emph{points} under classical microdynamics. In \emph{these} contexts, it is perhaps reasonable to think of classical \emph{microdynamics} as emerging as a limiting regime of quantum physics. But the contexts in which this occurs are relatively narrow (corresponding to isolated systems with large masses and non-chaotic dynamics; cf \citeN{zurekpazchaos} and \citeN[ch.3]{wallacebook}) and do not include many of the typical contexts in which ``classical'' statistical mechanics is applied, as we saw in the previous section.
\section{The classical Boltzmann equation from a quantum perspective}\label{boltzmann-quantum}
In the limiting case of a non-interacting dilute gas of atoms, Hamiltonian microdynamics is wildly false --- yet Liouville dynamics is exactly correct. This suggests a route to understanding the classical Boltzmann equation in a quantum universe: if Liouvillian dynamics (applied, of course, to Wigner functions) is \emph{approximately correct} even for a dilute interacting gas, then the modern derivation of the Boltzmann equation via the BBGKY hierarchy will still go through. And it is heuristically plausible that Liouvillian dynamics are approximately correct for such systems: after all, the particles in the gas spend most of their time moving freely, and in the free-particle regime Liouvillian and quantum dynamics coincide.
In fact, we can easily go beyond heuristic plausibility. The BBGKY hierarchy, and the various marginals, can be defined for the Moyal-bracket dynamics and the N-particle Wigner function \emph{exactly} as for the Poisson-bracket dynamics and the classical phase-space distribution: the functional form is identical in both cases. As such, we can replicate the derivation of the Boltzmann equation by replacing Poisson with Moyal brackets throughout. Making the same assumptions and approximations at each step, we again obtain equation (\ref{protoboltzmann}):
\[
\ensuremath{\frac{\dr{}}{\dr{t}}}{\rho_1}(t) = \{H_1,\rho_1(t)\} + \kappa[\sigma,\rho_1(t)] + \Lambda(t) \cdot c_2(0),
\]
which reduces to the Boltzmann equation if the forward-compatibility condition $\Lambda(t) \cdot c_2(0)=0$ is satisfied.
\emph{Formally} speaking, this differs from the classical Boltzmann equation only in that $\sigma$ is the quantum-mechanical, and not the classical, cross-section. But in fact this is exactly what we need to explain the applicability of the Boltzmann equation: in actual usage, the cross-section used is invariably that calculated by quantum mechanics if quantum and classical predictions differ. (In the physics of nuclear or chemical reactions, for instance, we use a generalisation of the Boltzmann equation to handle multiple particle species, and include quantum-derived cross sections for reactions that turn particles of one species into particles of another.)
Conceptually, this equation is completely different, though: $\rho_1$, and $c_2$, are features not of a probability distribution over an unknown classical microstate, but of a single quantum state. There is not even any particular requirement for that state to be a pure state (albeit that the `overkill' condition, $c_2=0$, does require the true $N$-particle state to be mixed). In particular, there is nothing probabilistic about the forward-compatibility requirement: it is a constraint on the actual quantum state.
Or, better: there \emph{is} something probabilistic about $\rho$, and about the forward-compatibility requirement, but only in the sense that there is something probabilistic about the quantum state itself (however that probabilistic nature is to be understood). The probabilities in the Boltzmann equation are not a new species of probability: they are Born-rule quantum probabilities, in the classical limit where the Liouville equation (though not Hamilton's equations) are approximately valid.
At this point, let us revisit section \ref{boltzmann-contrast}'s comparision between a BBGKY derivation of the Boltzmann equation and Boltzmann's own, probability-free, derivation. There, I suggested that the BBGKY derivation has the major virtues of (1) providing a single and relatively tractable initial-time condition for the Boltzmann equation to hold at later times; (2) generalising much more readily to other statistical-mechanical contexts, and (3) most importantly, applying to quantum mechanics; against this, I conceded that in the classical context it has a greater reliance on a somewhat mysterious notion of statistical-mechanical probability (albeit that notion is clearly necessary in classical statistical mechanics, as the Langevin equation also demonstrates).
From a quantum perspective, this disadvantage is entirely negated and the force of (3) becomes clear. In a quantum universe, the statistical-mechanical probabilities of the (classical) Boltzmann equation are nothing over and above quantum probability, something which (however mysterious it might be) we are already committed to. And in a quantum universe, the BBGKY derivation remains largely unchanged, while Boltzmann's original derivation relies on assumptions about the microdynamics of the system in question that are clearly, demonstrably, false.
\section{The Fokker-Planck equation from a quantum perspective}\label{fokkerplanck-quantum}
The Boltzmann equation is in practice used simply to make predictions about the relative frequency of the particle distribution, so that its probabilistic nature is somewhat obscured; I introduced the Langevin and Fokker-Planck equations in section \ref{fokkerplanck} to demonstrate that often the predictions (and not just the methods of derivation) of classical statistical mechanics are probabilistic: often the equations that emerge in non-equilibrium statistical mechanics are stochastic. In fact --- at least in this particular example --- those probabilities also turn out to be quantum-mechanical probabilities, so that (for instance) Brownian motion, in our universe, needs to be understood as a quantum-mechanically random process, not simply as a consequence of our ignorance of the classical microstate or something similar.
To elaborate: in section \ref{fokkerplanck} I derived the Fokker-Planck equation starting from the Liouville equation appropriate to a Hamiltonian that couples one large, with many small, harmonic oscillators, as a special case of the Mori-Zwanzig method of projection. The latter translates across to quantum mechanics \emph{mutatis mutandis}, simply be reinterpreting $\rho$ as a density operator and redefining
\begin{equation}
L_H \rho = - \frac{i}{\hbar}[H,\rho].
\end{equation}
In the particular case of the coupled oscillators, since the system is quadratic its phase-space dynamics are \emph{identical} whether understood classically or quantum-mechanically, so that equation (\ref{FP}) can be taken directly over to the quantum context simply by replacing Poisson brackets with commutators and probability distributions with mixed states (and taking a little care with non-commuting operators):
\begin{eqnarray}
\ensuremath{\frac{\dr{}}{\dr{t}}} \rho_S(t) & = & -\frac{i}{\hbar} [\tilde{H}_S,\rho_S(t)]\nonumber + \gamma \frac{i}{\hbar}[X,(P\rho_S(t)+ \rho_S(t)P)] \nonumber \\ & & - D [X,[X,\rho_S(t)]] - \frac{1}{\hbar}f[X,[P,\rho_S(t)]].
\end{eqnarray}
I have absorbed some factors of $\hbar$ into the coefficients, following the conventions of \citeN{pazzurekreview}.
This is an equation for the reduced state $\rho_S(t)$ of the large oscillator, derived by separating the combined state $\rho(t)$ of large and small oscillators into
\begin{equation}
\rho(t) = \rho_S(t)\otimes \mc{E} + \rho_I(t),
\end{equation}
for some fixed state $\mc{E}$ of the small oscillators, invariant under their self-Hamiltonian.
It is valid (to second order in perturbation theory, and after some short `transient' period for the coefficients to settle down to their long-term values) whenever $\rho_I(0)$ satisfies the forward compatibility conditions, and in particular whenever $\rho_I(0)=0$, \mbox{i.\,e.\,} when the system and environment are initially uncorrelated. Quantum-mechanical features of the equation show themselves only when it is applied to highly non-classical \emph{states}, such as coherent superpositions of wavepackets in different regions: in that context, it is the standard \emph{decoherence master equation} of environment-induced decoherence, and rapidly decoheres that superposition. But given as input a wavepacket, or an incoherent mixture of wavepackets, then it exactly reproduces the classical probability dynamics.
Classically speaking (and assuming that the $f$ coefficient is negligible) we have already noted that those dynamics can be interpreted as the probability-distribution evolution defined by the stochastic Langevin equation. But in the classical case the probabilities arise entirely because the input environment state $\epsilon$ is a probability distribution: the apparently-stochastic behaviour of the system is a consequence of an explicitly probabilistic assumption about the environment. (Normally we assume $\mc{E}$ is the canonical distribution of the oscillators.)
This is not the case in quantum mechanics. Firstly, even if we do choose a canonical state
\begin{equation}
\mc{E}_C= \frac{\e{-\beta H_E}}{\textsf{Tr} \e{-\beta H_E}}
\end{equation}
for the environment, it is (as I argued in section \ref{density-operator}) debatable at best whether this state should be interpreted probabilistically. But suppose it is so interpreted: then we should actually regard $\mc{E}_C$ as a probabilistic mixture of energy eigenstates, and standard statistical arguments tell us that this probability distribution is dominated by eigenstates $\ket{\{n_i\}}$ (where $n_i$ is the excitation number of the $i$th oscillator) satisfying
\begin{equation}
\sum_i ( n_i + 1/2) \omega_i \simeq \langle H_E\rangle_{\mc{E}_C}.
\end{equation}
Each such eigenstate is a perfectly valid choice for the environment state $\mc{E}$, and if we explicitly check the formulae for the coefficients in \citeN{pazzurekreview}, we find that for the overwhelming majority of such states, the coefficients have pretty much the same value as they would if calculated with the mixed state $\mc{E}_C$.
But the initial state $\proj{\psi}{\psi}\otimes \proj{\{n_i\}}{\{n_i\}}$ is \emph{pure}, and (whatever the right interpretation of quantum-mechanical \emph{mixed} states) contains no purely statistical-mechanical probability assumptions. So the probabilities in the Fokker-Planck equation derived from that initial state are purely quantum-mechanical. And if we hold on (implausibly) to the statistical interpretation of $\mc{E}_C$, we find that its statistical-mechanical probabilities make virtually no contribution to the probabilities in \emph{its} Fokker-Planck equation. This is radically unlike the classical case, in which perfect knowledge of the environment's microstate suffices (in principle) to eliminate the probabilities entirely from the large oscillator's dynamics.
As a final observation: it is fairly easy to show that the equilibrium state of a Fokker-Planck oscillator is a thermal state at the same temperature as its bath. Since this result holds even when the environment's initial state is a pure state as above, at least in \emph{this} case the equilibrium state again has to be interpreted purely quantum-mechanically, with no trace of statistical-mechanical probabilities.
\section{Conclusion}\label{conclusion}
Classical non-equilibrium statistical mechanics, in its contemporary formulations and across at least the wide class of applications considered in sections \ref{bbgky}--\ref{fokkerplanck}, adds to Hamiltonian classical mechanics two additional components: probability (via a move from Hamiltonian to Liouvillian dynamics) and irreversibility.
Irreversibility is introduced by some kind of boundary condition; that much could have been deduced \emph{a priori} given that it is not present in the underlying dynamics (Hamiltonian or Liouvillian), but the common form of that condition in BBGKY-derived or Mori-Zwanzig-derived statistical mechanics differs from the two most-commonly-discussed such conditions in the philosophical literature:
\begin{itemize}
\item Unlike the commonly-discussed \emph{past hypothesis} that the initial state of the system (most commonly, the entire Universe) is low in entropy, the forward-compatibility conditions are conditions on the \emph{microstate}, invisible at the coarse-grained level. (Low entropy, insofar as that is a condition on the \emph{coarse-grained} state, is neither necessary nor sufficient for the Boltzmann or Fokker-Planck equations to hold.)
\item Forward compatibility is closer to the \emph{Stosszahlansatz} of Boltzmann's own approach to non-equilibrium statistical mechanics, which is also a condition on the microstructure of the system, invisible at the coarse-grained level. But forward compatibility is (i) a condition on the Liouvillian, not the Hamiltonian, state (and thus explicitly probabilistic); (ii) mathematically formulated as a condition on the initial state, not as a requirement that must hold throughout the evolution period; (iii) demonstrably satisfied by at least one class of initial states, namely those for which the `irrelevant' part of the state vanishes.
\end{itemize}
Probability might be the most mysterious part of classical statistical mechanics as long as it is understood in isolation from quantum theory --- hence the temptation of Brown and others to eliminate it. But as with other cases in the philosophy of physics --- the enormous success of quantum theory in the face of the measurement problem, and of mainstream (non-algebraic) quantum field theory in the face of renormalisation --- it will not do simply to deny the coherence of a probability-based statistical mechanics given its empirical success.
But in any case, this mystery is largely dissolved by quantum mechanics. Though it is not often explicitly stated, the default assumption in foundations of statistical mechanics is that quantum and classical statistical mechanics are related as in Figure 1: mixed-state quantum mechanics is obtained from pure-state quantum mechanics via explicit addition of (mysterious) probabilities, just as Liouvillian classical mechanics is obtained from Hamiltonian classical mechanics, and the quantum/classical transition is from pure-state quantum mechanics to Hamiltonian classical mechanics.
\begin{figure}[H]\label{figure1}\caption{Quantum and classical statistical mechanics: standard view}
\begin{center}
\begin{tikzcd}[column sep = 8.0em, row sep = 5.0em]
CM \arrow{r}{+ \mbox{ probability}} & CSM \\
QM \arrow{u}[description]{\mbox{classical limit}}\arrow{r}{+ \mbox{ probability}} & QSM \\
\end{tikzcd}
\end{center}
\end{figure}
We have seen that this is not viable:
\begin{itemize}
\item The analogy between pure/mixed states and Hamiltonian/Liouvillian states is superficial: mixed states are already required as part of the formalism of quantum mechanics by subsystem considerations (section \ref{mixedstates}) and the addition of mysterious additional probabilities does not therefore generalise the dynamics in the quantum case as it does in the classical. Any equation of quantum mechanics in which probability distributions over (pure or mixed) states can always be reinterpreted as an equation about individual mixed states, so there is no inference from the empirical success of a given part of quantum statistical mechanics to the need for statistical-mechanical probabilities.
\item Study of the quantum-classical transition shows that quantum states (pure or mixed) correspond to phase-space distributions, not to phase-space points, and so argues for an understanding of that transition as a transition from quantum mechanics to Liouvillian, not Hamiltonian, classical mechanics.
\item Elementary considerations of wave-packet spreading, applied to dilute-gas systems with realistic parameters, reveals that Hamiltonian mechanics completely fails for such systems: inter-particle collisions are more like plane-wave interactions than classical hard-sphere scattering, and lead rapidly to a highly entangled state. So interpretation of the Boltzmann equation as a one-particle-marginal probability distribution (or relative-frequency description of) the trajectories of well-localised, deterministically evolving point particles is unjustified in our quantum universe. On the other hand, the derivation of that equation from Liouvillian dynamics via the BBGKY hierarchy remains valid even when the Liouvillian dynamics are understood as an approximate description of Schr\"{o}dinger dynamics, and indeed can be replicated exactly, \emph{mutatis mutandis}, in the Wigner-function formalism.
\item In a stochastic dynamics like the Brownian motion described by the Langevin equation, once we consider it quantum-mechanically we find that the stochasticity is itself quantum-mechanical, and persists even if the system and its environment begins in a pure state.
\end{itemize}
There is, in short, abundant evidence that `statistical-mechanical' probabilities are just quantum-mechanical probabilities. Considerations of the structure of quantum theory, of phase-space descriptions of quantum mechanics, and of specific models, all tell us that Liouvillian states need to be understood as quantum states in the regime where the Liouville equation approximates the Schr\"{o}dinger equation, and not as application of a new notion of probability to Hamiltonian dynamics. Hamiltonian dynamics is an approximation to Liouvillian dynamics in certain regimes, not its underpinning.\footnote{Of course, this is entirely compatible with the use of Hamiltonian dynamics as a formal method to analyse Liouvillian dynamics.} The correct relation between classical and quantum statistical mechanics is given by Figure 2.
\begin{figure}[H]\label{figure2}\caption{Quantum and classical statistical mechanics: improved view}
\begin{center}
\begin{tikzcd}[column sep = 4.0em,row sep=5.0em]
CM \arrow[leftarrow]{r}{\mbox{large-mass}}[swap]{\mbox{non-chaotic regime}} & CSM \\
\mbox{pure-state } QM \arrow[phantom]{r}{\subseteq} & \mbox{mixed-state } QM \arrow{u}[description]{\mbox{decoherent regime}}\\
\end{tikzcd}
\end{center}
\end{figure}
So in quantum statistical mechanics, one of the two puzzles of non-equilibrium statistical mechanics --- probability --- is resolved, or at any rate reduced to the problem of understanding quantum probabilities. The other puzzle --- the status of the assumptions underpinning irreversibility --- translates over to the quantum context \emph{mutatis mutandis}; a fuller understanding of them must await another day.
\section*{Acknowledgements}
Thanks to David Albert, Katherine Brading, Shelly Goldstein, Owen Maroney, Tim Maudlin, Wayne Myrvold, Simon Saunders, Chris Timpson, and especially Harvey Brown, for useful discussions.
\section*{Appendix}
Here I prove that if $\ket{\Omega}$ is an arbitrary state in NRQM with $M$ degrees of freedom, and that
\begin{equation} \ket{\vctr{q},\vctr{p}}\ensuremath{=_{_{df}}} \op{T}_\vctr{p}\op{T}_\vctr{q}\ket{\Omega}\end{equation}
is the state obtained by translating $\ket{\Omega}$ first by $\vctr{q}$ in position and then by $\vctr{p}$ in momentum, then
\begin{equation}
\op{\Pi}\ensuremath{=_{_{df}}}\frac{1}{(2\pi)^M}\int \dr{\vctr{q}}\dr{\vctr{p}}\proj{\vctr{q},\vctr{p}}{\vctr{q},\vctr{p}}=\op{\mathsf{1}}.
\end{equation}
From this it follows that the continous family of operators $(2 \pi)^{-M}\proj{\vctr{q},\vctr{p}}{\vctr{q},\vctr{p}}$ is a POVM, or more rigorously that if phase space is decomposed into countably many disjoint Liouville-measureable sets $C_i$, then the family of operators
\begin{equation}
\op{\Pi}_i=\int_{C_i}\proj{\vctr{q},\vctr{p}}{\vctr{q},\vctr{p}}
\end{equation}
is a POVM. In this section, $\hbar=1$.
To prove this, let $\ket{\psi}$ and $\ket{\varphi}$ be arbitrary Hilbert-space vectors.
Inserting two complete sets of position eigenstates, we obtain
\begin{equation} \label{app1}
\matel{\psi}{\Pi}{\varphi}=(2\pi)^{-M}\!\!
\int \dr{\vctr{q}}\dr{\vctr{p}}\dr{\vctr{x}}\dr{\vctr{y}}
\bk{\psi}{\vctr{x}}\bk{\vctr{x}}{\vctr{q},\vctr{p}}\bk{\vctr{q},\vctr{p}}{\vctr{y}}
\bk{\vctr{y}}{\varphi}.
\end{equation}
Now,
\begin{equation}
\bk{\vctr{x}}{\vctr{q},\vctr{p}}=
\bra{\vctr{x}}\!\op{T}_\vctr{p}\op{T}_\vctr{q}\!\ket{\Omega}=\e{i\vctr{x}\cdot \vctr{p}}\bra{\vctr{x}}\!\op{T}_\vctr{q}\!\ket{\Omega}
=\e{i\vctr{x}\cdot \vctr{p}}\bk{\vctr{x}\!+\!\vctr{q}}{\Omega}.
\end{equation}
So we can rewrite (\ref{app1}) as
\begin{equation}
\matel{\psi}{\Pi}{\varphi}=(2\pi)^{-M}\!\!
\int \dr{\vctr{q}}\dr{\vctr{p}}\dr{\vctr{x}}\dr{\vctr{y}}
\bk{\psi}{\vctr{x}}\bk{\vctr{x}\!+\!\vctr{q}}{\Omega}\bk{\Omega}{\vctr{y}\!+\!\vctr{q}}
\bk{\vctr{y}}{\varphi}
\e{i(\vctr{x}-\vctr{y})\cdot \vctr{p}}.
\end{equation}
The integral over $\vctr{p}$ can now be evaluated, and gives a delta function factor,
\begin{equation}
\int \dr{\vctr{p}}\e{i(\vctr{x}-\vctr{y})\cdot \vctr{p}}=(2\pi)^M \delta(\vctr{x}-\vctr{y}),
\end{equation}
which we can in turn use to perform the integration over \vctr{y}, obtaining
\begin{equation}
\matel{\psi}{\Pi}{\varphi}=
\int \dr{\vctr{q}}\dr{\vctr{x}}
\bk{\psi}{\vctr{x}}\bk{\vctr{x}\!+\!\vctr{q}}{\Omega}\bk{\Omega}{\vctr{x}\!+\!\vctr{q}}
\bk{\vctr{x}}{\varphi}
\end{equation}
After a change of variables $\vctr{z}=\vctr{x}+\vctr{q}$, this is just
\begin{equation}
\matel{\psi}{\Pi}{\varphi}=
\left(\int \dr{\vctr{x}}\bk{\psi}{\vctr{x}}\bk{\vctr{x}}{\varphi}\right)
\left(\int \dr{\vctr{z}}\bk{\vctr{z}}{\Omega}\bk{\Omega}{\vctr{z}}\right)
=\bk{\psi}{\varphi}\bk{\Omega}{\Omega}=\bk{\psi}{\varphi}.
\end{equation}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 7,874 |
\section{Introduction and main results}
There exists a large literature on stochastic comparisons between order statistics arising from possibly heterogeneous populations, see Balakrishnan and Zhao (2013) for a review. In reliability theory, order statistics play a prominent role as the lifetime of a $k$-out-of-$n$ system can be represented by the $n-k+1$th order statistic of the $n$ component lifetimes. Because of the complexity of the distribution of order statistics arising from heterogeneous populations, stochastic comparisons with those from a homogeneous population are helpful, and can provide bounds on tail probabilities and hazard rates. Of particular interest is the following result of Bon and P\v{a}lt\v{a}nea (2006). Let $X_1,\ldots, X_n$ be independent exponential random variables with rates $\lambda_1,\ldots, \lambda_n$ respectively. Let $Y_1,\ldots, Y_n$ be another set of independent exponential random variables with a common rate $\gamma$. Then the $k$th order statistic of the heterogeneous sample is larger than that of the homogeneous sample in the usual stochastic order, that is, $Y_{k:n}\leq_{\rm st} X_{k:n}$, if and only if
\begin{equation}
\label{paltanea}
\gamma\geq \left({n \choose k}^{-1} s_k(\lambda_1,\ldots, \lambda_n)\right)^{1/k},
\end{equation}
where $s_k$ denotes the $k$th elementary symmetric function. For the sample maximum, Khaledi and Kochar (2000) showed that (\ref{paltanea}) with $k=n$ implies that $X_{n:n}$ is larger than $Y_{n:n}$ according to both the hazard rate order and the dispersive order. In the context of failsafe systems, P\v{a}lt\v{a}nea (2008) showed that (\ref{paltanea}) with $k=2$ is equivalent to $Y_{2:n}\leq_{\rm hr} X_{2:n}$. In the $k=2$ case, Zhao et al. (2009) obtained related results for the likelihood ratio ordering.
In this paper we aim to extend such results to general $k$. Specifically, we have
\begin{theorem}
\label{thm1}
For each $1\leq k\leq n$, (\ref{paltanea}) is equivalent to
$Y_{k:n}\leq_{\rm hr} X_{k:n}$, and also equivalent to $Y_{k:n}\leq_{\rm disp} X_{k:n}$.
\end{theorem}
The statement concerning $\leq_{\rm hr}$ in Theorem~\ref{thm1} confirms a conjecture of P\v{a}lt\v{a}nea (2008). Theorem~\ref{thm1} is established by considering the star order between $X_{k:n}$ and $Y_{k:n}$. For random variables $X$ and $Y$ supported on $(0, \infty)$ with distribution functions $F$ and $G$ respectively, we say $X$ is smaller than $Y$ in the star order, denoted by $X\leq_* Y$ (or $F\leq_* G$), if $G^{-1}F(x)/x$ is increasing in $x>0$, where $G^{-1}$ denotes the right continuous inverse of $G$. Equivalently, $X\leq_* Y$ if and only if, for each $c>0$, $F(cx)$ crosses $G(x)$ at most once, and from below, as $x$ increases on $(0, \infty)$. For definitions and properties of various stochastic orders including $\leq_*$, see Shaked and Shanthikumar (2007).
Kochar and Xu (2009) showed $Y_{n:n}\leq_* X_{n:n}$, a key result that allowed Xu and Balakrishnan (2012) to obtain stochastic comparison results for the ranges of heterogeneous exponential samples; see Genest et al. (2009) for related work. Our Theorem~\ref{thm2} confirms the conjecture of Xu and Balakrishnan (2012) (see also Balakrishnan and Zhao (2013), Open Problem 5). A special case of our Theorem~\ref{thm2} was obtained by Kochar and Xu (2011) for multiple-outlier models.
\begin{theorem}
\label{thm2}
For each $1\leq k\leq n$, we have $Y_{k:n}\leq_* X_{k:n}$.
\end{theorem}
The star ordering is a strong variability ordering. It implies the Lorenz ordering $\leq_{\rm L}$, which in turn implies the ordering of the coefficients of variation. Our Theorem~\ref{thm2} is a strengthening of the main result of Da et al. (2014) who showed $Y_{k:n}\leq_{\rm L} X_{k:n}$.
Theorems~\ref{thm1} and \ref{thm2} can be extended from order statistics to general spacings.
\begin{corollary}
\label{coro1}
For $1\leq m< k \leq n$ we have $Y_{k:n} - Y_{m:n} \leq_* X_{k:n} - X_{m:n}$.
\end{corollary}
\begin{proposition}
\label{prop1}
For $1\leq m < k\leq n$ we have $Y_{k:n} - Y_{m:n} \leq_{\rm order} X_{k:n} - X_{m:n}$ if and only if
\begin{equation}
\label{char}
{n-m\choose k-m} \gamma^{k-m} \geq \sum_{\mathbf{r}} s_{k-m}^{[\mathbf{r}]}(\lambda) \prod_{j=1}^m \frac{ \lambda_{r_j}}{\Lambda -\sum_{i=1}^{j-1} \lambda_{r_i}}
\end{equation}
where $\mathbf{r} = (r_1, \ldots, r_m)$, the outer sum is over all permutations of $\{1, \ldots, n\}$ using $m$ at a time, $\lambda = (\lambda_1, \ldots, \lambda_n)$, $\Lambda=\sum_{i=1}^n \lambda_i$, $s_{k-m}^{[\mathbf{r}]}(\lambda) = s_{k-m}(\lambda\backslash\{\lambda_{r_1}, \ldots,\lambda_{r_m}\})$,
and $\leq_{\rm order}$ is any of $\leq_{\rm st}, \leq_{\rm hr}$ or $\leq_{\rm disp}$.
\end{proposition}
In the special case of $m=1$ and $k=n$, which corresponds to comparing the sample ranges, the condition (\ref{char}) reduces to
$$\gamma \geq \left(\frac{\prod_{i=1}^n \lambda_i}{\Lambda / n} \right)^{1/(n-1)},$$
and we recover Theorem~4.1 of Xu and Balakrishnan (2012). In the case of ordinary spacings, that is, $k=m+1$, Proposition~\ref{prop1} can also be derived using the log-concavity arguments of Yu (2009). As noted by P\v{a}lt\v{a}nea (2011), however, it is difficult to implement such an argument in general.
In Section~2 we prove Theorem~\ref{thm2}, from which Theorem~\ref{thm1}, Corollary~\ref{coro1} and Proposition~\ref{prop1} are deduced. We refer to P\v{a}lt\v{a}nea (2008) for numerical illustrations and Da et al. (2014) for potential applications of these results. It would be interesting to see if results similar to Theorem~\ref{thm1} and Proposition~\ref{prop1} can be obtained for the likelihood ratio order, for which Theorem~\ref{thm2}, based on the star order, is not helpful.
\section{Derivation of main results}
We need the following closure property of the star order with respect to mixtures.
\begin{lemma}[Lemma~3.1 of Xu and Balakrishnan (2012)]
\label{lem1}
Let $F$ be a distribution function with density $f$ supported on $(0, \infty)$ such that $f(e^x)$ is log concave in $x\in\mathbb{R}$. Let $G_1,\ldots, G_n$ be distribution functions with densities $g_1,\ldots, g_n$ supported on $(0,\infty)$ such that $F\leq_* G_i$ for each $i=1,\ldots, n$. Then $F\leq_* \sum_{i=1}^n p_i G_i$ for $p_i>0$ such that $\sum_{i=1}^n p_i =1$.
\end{lemma}
We also need the following representation of order statistics from heterogeneous exponential samples.
\begin{lemma}[P\v{a}lt\v{a}nea (2011), Theorem~4.1]
\label{lem2}
Let $S=\{X_1, \ldots, X_n\}$ be a set of $n>1$ independent exponential random variables with rates $\lambda_1,\ldots, \lambda_n$ respectively. Denote $S^{[i]}=S \backslash \{X_i\}$ and let $X_{j:(n-1)}^{[i]}$ be the $j$th order statistic from $S^{[i]}$ with distribution function $F_{j:(n-1)}^{[i]}$ for $j\leq n-1$. Then for $k\leq n-1$ the $(k+1)$th order statistic $X_{(k+1):n}$ from $S$ is the sum of two independent random variables: $X_{1:n}$, which has an exponential distribution with rate $\Lambda = \sum_{j=1}^n \lambda_j$, and a mixture of order statistics with distribution function $\Lambda^{-1} \sum_{i=1}^n \lambda_i F_{k:(n-1)}^{[i]}$.
\end{lemma}
Finally, we need some stochastic comparison results concerning convolutions of gamma variables. For independent random variables $Z_1,\ldots, Z_n\sim {\rm gamma}(\alpha)$ let $F_\theta$ denote the distribution function of the weighted sum $\sum_{i=1}^n \theta_i Z_i$ where $\theta=(\theta_1,\ldots, \theta_n)$ is a vector of positive weights. We use $\prec_w$ to denote weak sub-majorization (Marshall et al. (2009)). The following Lemma is a consequence of Theorem~1 of Yu (2011).
\begin{lemma}
\label{lem.yu}
For $\alpha >0$, if $\log \eta \prec_w \log \theta$ then $F_\eta\leq_{\rm st} F_\theta$.
\end{lemma}
Lemma~\ref{lem3} asserts a unique crossing between the distribution functions of two weighted sums of iid gamma variables when the weights form a special configuration. Lemma~\ref{lem3} is an important step in Yu's (2016) investigation of the unique crossing conjecture of Diaconis and Perlman (1990); it is also a key tool in deriving our main results (we only need the $\alpha=1$ case).
\begin{lemma}[Theorem~1 of Yu (2016)]
\label{lem3}
Suppose $\alpha \geq 1$. Suppose $0<\theta_1\leq \cdots \leq \theta_n$ and $\eta_1\leq \cdots \leq \eta_n$ and (a) there exists $2\leq k\leq n$ such that $\theta_i<\eta_i$ for $i< k$ and $\theta_i > \eta_i$ for $i\geq k$;
(b) $\prod_{i=1}^n \eta_i > \prod_{i=1}^n \theta_i$. Then there exists $x_0\in (0, \infty)$ such that $F_\eta(x) < F_\theta(x)$ for $x\in (0, x_0)$ and the inequality is reversed for $x> x_0$.
\end{lemma}
\begin{proof}[Proof of Theorem~\ref{thm2}]
Let us use induction. The $k=1$ case is trivial. Suppose the claim holds for a certain $k\geq 1$ and all $n\geq k$. We shall prove that it holds for $k+1$ and all $n\geq k+1$. Assume $\lambda_i,\, i=1,\ldots,n,$ are not all equal, and let $F_{k:n}^{(\tau)}$ denote the distribution function of the $k$th order statistic from a sample of $n$ iid exponential variables with rate $\tau$. In the notation of Lemma~\ref{lem2}, the induction hypothesis yields $F_{k:(n-1)}^{(\tau)} \leq_* F_{k:(n-1)}^{[i]}$ for $\tau>0,\ n\geq k+1,$ and $i=1,\ldots, n$. As noted by Xu and Balakrishnan (2012), $F_{k:(n-1)}^{(\tau)}$ has a density $f$ such that $f(e^x)$ is log-concave. By Lemma~\ref{lem1}, we have
\begin{equation}
\label{star}
F_{k:(n-1)}^{(\tau)} \leq_* \sum_{i=1}^n \frac{\lambda_i}{\Lambda} F_{k:(n-1)}^{[i]}.
\end{equation}
As noted by Da et al. (2014), when $x\downarrow 0$ we have
$$F_{k:(n-1)}^{[i]}(x) = s_k^{[i]}(\mathbf{\lambda}) x^k + o(x^k),\quad F_{k:(n-1)}^{(\tau)} = {n-1 \choose k} \tau^k x^k + o(x^k),$$
where $\mathbf{\lambda} = (\lambda_1, \ldots, \lambda_n)$ and $s_k^{[i]}(\mathbf{\lambda}) = s_k(\mathbf{\lambda}\backslash \{\lambda_i\})$. Thus, if ${n-1 \choose k} \tau^k \geq \sum_{i=1}^n \lambda_i s_k^{[i]}(\mathbf{\lambda})/\Lambda $, or equivalently $\tau\geq \tau^*$ with $\tau^* = (n s_{k+1}(\mathbf{\lambda})/\Lambda /{n\choose k+1})^{1/k}$,
then
\begin{equation}
\label{st}
F_{k:(n-1)}^{(\tau)} \leq_{\rm st} \sum_{i=1}^n \frac{\lambda_i}{\Lambda} F_{k:(n-1)}^{[i]}.
\end{equation}
Analogous considerations as $x\to\infty$ (P\v{a}lt\v{a}nea, 2011) reveal that if $\tau \leq \tau_*\equiv \sum_{i=1}^{n-k} \lambda_{(i)}/(n-k)$, where $\lambda_{(1)}, \ldots, \lambda_{(n)}$ are $\lambda$ arranged in increasing order, then (\ref{st}) holds with the direction of $\leq_{\rm st}$ reversed. For $\tau\in (\tau_*, \tau^*)$, star ordering implies that there exists $x_0>0$ such that
\begin{equation}
\label{one.cross}
F_{k:(n-1)}^{(\tau)}(x)\leq \sum_{i=1}^n \frac{\lambda_i}{\Lambda} F_{k:(n-1)}^{[i]}(x),\quad x\in (0, x_0),
\end{equation}
and the inequality is reversed if $x\in (x_0, \infty)$. One can show strict inequality, that is, the crossing point is unique. Indeed, if there exists another $\tilde{x}\in (0, x_0)$, for example, such that equality holds in (\ref{one.cross}) at $x=\tilde{x}$, then a slight increase in $\tau$ would produce at least two crossings, near $x_0$ and $\tilde{x}$, respectively (equality cannot hold for all $x\in [\tilde{x}, x_0]$ unless the two distributions are identical, because these distribution functions can be written as linear combinations of exponential functions and are therefore analytic).
In view of Lemma~\ref{lem2}, we can convolve both sides of (\ref{star}) with an exponential with rate $\Lambda$ and obtain that ${\rm expo}(\Lambda) * F_{k:(n-1)}^{(\tau)}$ crosses $F_{(k+1):n}$ (the distribution function of $X_{(k+1):n}$) exactly once, and from below, for $\tau\in (\tau_*, \tau^*)$. That there is at most one crossing follows from variation diminishing properties of TP2 kernels (Karlin 1968). Upon close inspection there is exactly one crossing at a unique point. In particular, because $\Lambda > (n-k) \tau> (n-k) \tau_*$, convolving with ${\rm expo}(\Lambda)$ cannot reverse the sign of $F_{k:(n-1)}^{(\tau)}(x) - \Lambda^{-1} \sum_{i=1}^n \lambda_i F_{k:(n-1)}^{[i]}(x)$ as $x\to\infty$. We then recognize the crossing point, denoted by $x_*$, as a decreasing, continuous function of $\tau$, because $F_{k:(n-1)}^{(\tau)}$ stochastically decreases in $\tau$. Furthermore, the above analysis at the end points $\tau_*$ and $\tau^*$ shows that $x_*(\tau)\uparrow \infty$ as $\tau\downarrow \tau_*$ and $x_*(\tau)\downarrow 0$ as $\tau\uparrow \tau^*$.
The distribution of $Y_{(k+1):n}$ is the convolution
$$F_{(k+1):n}^{(\gamma)}= {\rm expo}(n\gamma) * {\rm expo}((n-1)\gamma) *\cdots * {\rm expo}((n-k)\gamma) .$$
Suppose $F_{(k+1): n}^{(\gamma)}$ crosses $F_{(k+1):n}$ at some $x^*>0$. Then we can choose $\tau\in (\tau_*, \tau^*)$ such that $x_*(\tau) = x^*$, that is, $ {\rm expo}(\Lambda) * F_{k:(n-1)}^{(\tau)}$ crosses $F_{(k+1):n}$ at exactly $x^*$, from below. If $\gamma \geq \Lambda/n$ then Maclaurin's inequality yields $\gamma\geq \tau^*> \tau$, which implies that
\begin{equation}
\label{no.cross}
F_{(k+1):n}^{(\gamma)} \leq_{\rm st} {\rm expo}(\Lambda) * F_{k:(n-1)}^{(\tau)},
\end{equation}
contradicting the existence of $x^*$. If $\gamma\leq \tau$ then the inequality (\ref{no.cross}) is reversed, and there is again no crossing. Thus we must have $\gamma\in (\tau, \Lambda/n)$. If $\prod_{i=0}^k ((n-i)\gamma) \geq \Lambda \prod_{i=1}^k ((n-i) \tau)$ then one can show (the $\log$ applies element-wise)
$$-\log(n\gamma, (n-1)\gamma,\ldots, (n-k)\gamma)\prec_w -\log(\Lambda, (n-1)\tau, \ldots, (n-k)\tau),
$$
which again implies (\ref{no.cross}) by Lemma~\ref{lem.yu}. For the remaining case, $\prod_{i=0}^k ((n-i)\gamma) < \Lambda \prod_{i=1}^k ((n-i) \tau)$, the conditions of Lemma~\ref{lem3} are satisfied. It follows that $F_{(k+1):n}^{(\gamma)}$ crosses ${\rm expo}(\Lambda) * F_{k:(n-1)}^{(\tau)}$ exactly once, from below, at the same point $x^*$. We deduce that $F_{(k+1):n}^{(\gamma)}$ crosses $F_{(k+1):n}$ exactly once, from below, at $x^*$. Since $\gamma>0$ is arbitrary, we have $Y_{(k+1):n}\leq_* X_{(k+1):n}$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm1}]
Theorem~\ref{thm1} can be deduced from Theorem~\ref{thm2} as discussed by Xu and Balakrishnan (2012). Specifically, given $Y_{k:n}\leq_* X_{k:n}$, we have the equivalence
$$Y_{k:n}\leq_{\rm st} X_{k:n}\Longleftrightarrow Y_{k:n}\leq_{\rm disp} X_{k:n}.$$
Since $Y_{k:n}$ has increasing failure rate, dispersive ordering implies hazard rate ordering.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{coro1}]
Extending Lemma~\ref{lem2} we can write the distribution function of $X_{k:n}-X_{m:n}$ as
\begin{equation}
\label{spacing}
\sum_{\mathbf{r}} \prod_{j=1}^m \frac{\lambda_{r_j}}{\Lambda -\sum_{i=1}^{j-1} \lambda_{r_i}} F_{(k-m):(n-m)}^{[\mathbf{r}]},
\end{equation}
where the notation is the same as in the statement of Proposition~\ref{prop1}, and $F_{(k-m):(n-m)}^{[\mathbf{r}]}$ denotes the distribution function of the $(k-m)$th order statistics of $\{X_1, \ldots, X_n\}\backslash\{X_{r_1}, \ldots, X_{r_m}\}$. The distribution of $Y_{k:n} - Y_{m:n}$ is simply $F_{(k-m):(n-m)}^{(\gamma)}$. The claim therefore follows from Theorem~\ref{thm2} and Lemma~\ref{lem1}.
\end{proof}
\begin{proof}[Proof of Proposition~\ref{prop1}]
Given the star ordering, we can establish the characterization for $\leq_{\rm st}$ by examining the distribution function (\ref{spacing}) near $x=0$ (we did this for $m=1$ in the proof of Theorem~\ref{thm2}). Characterizations for $\leq_{\rm hr}$ and $\leq_{\rm disp}$ follow as in the proof of Theorem~\ref{thm1}.
\end{proof}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,216 |
Q: Android ContextMenu for View (Not Activity) I am using the following method to add a ContextMenu to a custom view i have built but i want to know how to react to the press of that contextmenu.
This is not an Activity so i cannot do this:
@override
public boolean onOptionsItemSelected(MenuItem item) {
Here is the code
private View.OnCreateContextMenuListener vC = new View.OnCreateContextMenuListener() {
@Override
public void onCreateContextMenu(ContextMenu arg0, View arg1,
ContextMenuInfo arg2) {
// TODO Auto-generated method stub
arg0.add(0, 0, 0, "Call");
arg0.add(0, 1, 0, "Map");
arg0.add(0, 2, 0, "Market");
}
};
Update:
Here is a very simplified verion of my class.
public final class NewView extends View {
public NewView(Context context, AttributeSet attrs) {
super(context, attrs);
cntxt = context;
this.setLongClickable(true);
this.setOnLongClickListener(vLong);
this.setOnCreateContextMenuListener(vC);
}
private View.OnLongClickListener vLong = new View.OnLongClickListener() {
public boolean onLongClick(View view) {
showContextMenu();
return true;
}
};
private View.OnCreateContextMenuListener vC = new View.OnCreateContextMenuListener() {
@Override
public void onCreateContextMenu(ContextMenu arg0, View arg1,
ContextMenuInfo arg2) {
// TODO Auto-generated method stub
arg0.add(0, 0, 0, "Call");
arg0.add(0, 1, 0, "Map");
arg0.add(0, 2, 0, "Market");
}
};
}
A: Noone seem to answer exhaustively TS' question (and mine eventually), thus let me post a solution to this. Given the code above you can attach custom MenuItem's click handlers:
private View.OnCreateContextMenuListener vC = new View.OnCreateContextMenuListener() {
@Override
public void onCreateContextMenu(ContextMenu menu, View v,
ContextMenuInfo menuInfo) {
menu.add(0, 0, 0, "Call")
.setOnMenuItemClickListener(mMenuItemClickListener);
menu.add(0, 1, 0, "Map")
.setOnMenuItemClickListener(mMenuItemClickListener);
menu.add(0, 2, 0, "Market")
.setOnMenuItemClickListener(mMenuItemClickListener);
}
};
private OnMenuItemClickListener mMenuItemClickListener = new OnMenuItemClickListener() {
@Override
public boolean onMenuItemClick(MenuItem item) {
switch (item.getItemId()) {
case 0:
// do "Call"
return true;
case 1:
// do "Map"
return true;
case 2:
// do "Market"
return true;
}
return false;
}
};
};
A: Use item.getItemId() and create switch and cases based on the number returned by getItemId()
Something like this.
@override
public boolean onOptionsItemSelected(MenuItem item) {
switch(item.getItemId())
{
case 1:
Log.i("FIRST ITEM: ", "CALL");
break;
case 2:
Log.i("2nd ITEM: ", "MAP");
break;
case 3:
Log.i("3rd ITEM: ", "Market");
break;
default:
}
}
I hope this is what you meant by reacting on menu items selection. :)
A: This does seem to be an odd inconsistency in Android - a View can create a context menu, but the handling of said menu can only happen in completely different code?
I'm also solving this with setOnMenuItemClickListener(). The documentation suggests this is reasonable, but it requires effort to set it up if you are using a MenuInflater.
public void onCreateContextMenu(final ContextMenu menu, final View v,
final ContextMenuInfo menuInfo) {
super.onCreateContextMenu(menu, v, menuInfo);
final MenuInflater menuInflater = this.menuInflater;
menuInflater.inflate(R.menu.context_date, menu);
final int length = menu.size();
for (int index = 0; index < length; index++) {
final MenuItem menuItem = menu.getItem(index);
menuItem.setOnMenuItemClickListener(this);
}
}
A: You could assign a variable to your View from your activity after you inflate the main layout (in `onCreate).
myView = (View) findViewById(R.id.my_view);
Next, do registerForContextMenu(myView);
Finally, you can override onCreateContextMenu where you can add what happens when the context menu appears.
A: In the view :
Let your view implements ContextMenuInfo
public class MyView extends View implements ContextMenuInfo
Then, override getContextMenuInfo to return your view.
@Override
protected ContextMenuInfo getContextMenuInfo() {
return this ;
}
In you activity :
Register your view for ContextMenu.
registerForContextMenu(myView);
Then access the view in your activity onContextItemSelected.
public boolean onContextItemSelected(MenuItem item) {
switch (item.getItemId()) {
...
case R.id.action_ctx_wordview_read_tts:
MyView myView = (MyView) item.getMenuInfo();
...
}
}
Note :
When I add an onClickListener to the MyView, the contextMenu stoped working. I had to add the following code to solve the problem.
tvWord.setOnLongClickListener(new View.OnLongClickListener() {
@Override
public boolean onLongClick(View v) {
return false;
}
});
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,070 |
Q: How to handle interrupt signal when reading from stdin in bash I'm playing around with bash read functionality. I like what I have so far as a simple layer on top of my current shell. The read -e does tab-complete and previous commands, and sending EOF with ctrl+d gets me back to my original shell. Here's my reference:
Bash (or other shell): wrap all commands with function/script
I'd like some help handling SIGINT, ctrl+c. In a normal shell, if you start typing and hit ^C halfway, it immediately ends the line. For this simple example, after ^C, I still have to hit return before it's registered.
How do I keep the nice things that readline does, but still handle SIGINT correctly? Ideally, it would send a continue statement to the while read loop, or somehow send a \n to STDIN where my read is waiting.
Example code:
#!/bin/bash
# Emulate bash shell
gtg=1
function handleCtrl-C {
# What do I do here?
gtg=0
return
}
trap handleCtrl-C INT
while read -e -p "> " line
do
if [[ $gtg == 1 ]] ; then
eval "$line"
fi
gtg=1
done
A: I think I came up with something finally I liked. See SIGINT to cancel read in bash script? for that answer.
A: Reading man 7 signal tells that some system calls have a restartable flag set as a result will return back to the command
For some system calls, if a signal is caught while the call is
executing and the call is prematurely terminated, the call is
auto-matically restarted. Any handler installed with signal(3) will
have the SA_RESTART flag set, meaning that any restartable system call
will not return on receipt of a signal. The affected system calls
include read(2), write(2), sendto(2), recvfrom(2),sendmsg(2), and
recvmsg(2) on a communications channel or a low speed device and
during a ioctl(2) or wait(2). However, calls that have already
committed are not restarted, but instead return a partial success (for
example, a short read count). These semantics could be changed with
siginterrupt(3).
You can try printing the value input to line and verify that the read is resumed after CtrlC return until new line is hit. Type in something like "exit", followed by Ctrl-C and then "exit" the output comes out as "exitexit". Make the following change and run for the above test case
echo ">$line<"
if [ $gtg == 1 ] ; then
You'll the output as
You can verify this with a C program as well.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,026 |
Q: How to use OAuth 2 for authentication in Node?
I am very confused about how to authenticate a user with OAuth 2. But I am comfortable when using JWT. Please can you make me understand with an example using OAuth 2?
*
*When user gets registered.
*User Login.
I'll be very thankful to you.
A: I would suggest to first get familiar with OAuth2 workflow and then you could use something like this.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,334 |
package com.stratio.mojo.unix.core;
/*
* The MIT License
*
* Copyright 2009 The Codehaus.
*
* Permission is hereby granted, free of charge, to any person obtaining a copy of
* this software and associated documentation files (the "Software"), to deal in
* the Software without restriction, including without limitation the rights to
* use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies
* of the Software, and to permit persons to whom the Software is furnished to do
* so, subject to the following conditions:
*
* The above copyright notice and this permission notice shall be included in all
* copies or substantial portions of the Software.
*
* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
* SOFTWARE.
*/
import fj.data.*;
import static fj.data.List.*;
import com.stratio.mojo.unix.*;
import static com.stratio.mojo.unix.UnixFsObject.*;
import com.stratio.mojo.unix.util.*;
import com.stratio.mojo.unix.util.line.*;
import static com.stratio.mojo.unix.util.line.LineStreamUtil.*;
import org.joda.time.*;
/**
* @author <a href="mailto:trygvis@inamo.no">Trygve Laugstøl</a>
*/
public class AssemblyOperationUtil
{
// private static final List<Replacer> filters = nil();
//
// public static RegularFile fromFileObject( RelativePath toFile, FileObject fromFile, FileAttributes attributes )
// throws FileSystemException
// {
// FileContent content = fromFile.getContent();
//
// return regularFile( toFile, new LocalDateTime( content.getLastModifiedTime() ), content.getSize(), attributes,
// filters );
// }
//
// public static Directory dirFromFileObject( RelativePath toFile, FileObject fromFile, FileAttributes attributes )
// throws FileSystemException
// {
// if ( !fromFile.getType().equals( FileType.FOLDER ) )
// {
// throw new FileSystemException( "Not a directory: " + fromFile.getName().getPath() + ", was: " +
// fromFile.getType() );
// }
//
// FileContent content = fromFile.getContent();
//
// return UnixFsObject.directory( toFile, new LocalDateTime( content.getLastModifiedTime() ), attributes );
// }
public static void streamIncludesAndExcludes( LineStreamWriter streamWriter, List<String> includes,
List<String> excludes )
{
if ( !includes.isEmpty() )
{
streamWriter.add( " Includes: " ).addAllLines( prefix( includes, " " ) );
}
else
{
streamWriter.add( " No includes set" );
}
if ( !excludes.isEmpty() )
{
streamWriter.add( " Excludes: " ).addAllLines( prefix( excludes, " " ) );
}
else
{
streamWriter.add( " No excludes set" );
}
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 2,470 |
Q: Selenium Webdriver - Python - leboncoin (classified) - pb to select a dropdown list & box I continue my path on the learning curve of Python/Selenium.
Selenium Webdriver - Python - leboncoin - pb to select a button with an accent
After successfully connected to the site, and selecting the button, I would like to populate the fields to submit a classified ad.
At this stage, there are 2 fields I am trying to select, a dropdown list and a box.
the Html is the below
Dropdown list
<div class="selectWrapper single">
<select id="category" class="select" name="category">
<option value="0">«Choisissez une catégorie»</option>
<option id="cat71" value="71" style="background-color:#E6E6E6" disabled="">-- EMPLOI --</option>
<option id="cat33" value="33">Offres d'emploi</option>
<option id="cat1" value="1" style="background-color:#E6E6E6" disabled="">-- VEHICULES --</option>
<option id="cat2" value="2">Voitures</option>
<option id="cat3" value="3">Motos</option>
<option id="cat4" value="4">Caravaning</option>
<option id="cat5" value="5">Utilitaires</option>
<option id="cat6" value="6">Equipement Auto</option>
<option id="cat44" value="44">Equipement Moto</option>
<option id="cat50" value="50">Equipement Caravaning</option>
<option id="cat7" value="7">Nautisme</option>
<option id="cat51" value="51">Equipement Nautisme</option>
<option id="cat8" value="8" style="background-color:#E6E6E6" disabled="">-- IMMOBILIER --</option>
<option id="cat9" value="9">Ventes immobilières</option>
<option id="cat10" value="10">Locations</option>
<option id="cat11" value="11">Colocations</option>
<option id="cat13" value="13">Bureaux & Commerces</option>
<option id="cat66" value="66" style="background-color:#E6E6E6" disabled="">-- VACANCES --</option>
<option id="cat12" value="12">Locations & Gîtes</option>
<option id="cat67" value="67">Chambres d'hôtes</option>
<option id="cat68" value="68">Campings</option>
<option id="cat69" value="69">Hôtels</option>
<option id="cat70" value="70">Hébergements insolites</option>
<option id="cat14" value="14" style="background-color:#E6E6E6" disabled="">-- MULTIMEDIA --</option>
<option id="cat15" value="15">Informatique</option>
<option id="cat43" value="43">Consoles & Jeux vidéo</option>
<option id="cat16" value="16">Image & Son</option>
<option id="cat17" value="17">Téléphonie</option>
<option id="cat18" value="18" style="background-color:#E6E6E6" disabled="">-- MAISON --</option>
<option id="cat19" value="19">Ameublement</option>
<option id="cat20" value="20">Electroménager</option>
<option id="cat45" value="45">Arts de la table</option>
<option id="cat39" value="39">Décoration</option>
<option id="cat46" value="46">Linge de maison</option>
<option id="cat21" value="21">Bricolage</option>
<option id="cat52" value="52">Jardinage</option>
<option id="cat22" value="22">Vêtements</option>
<option id="cat53" value="53">Chaussures</option>
<option id="cat47" value="47">Accessoires & Bagagerie</option>
<option id="cat42" value="42">Montres & Bijoux</option>
<option id="cat23" value="23">Equipement bébé</option>
<option id="cat54" value="54">Vêtements bébé</option>
<option id="cat24" value="24" style="background-color:#E6E6E6" disabled="">-- LOISIRS --</option>
<option id="cat25" value="25">DVD / Films</option>
<option id="cat26" value="26">CD / Musique</option>
<option id="cat27" value="27">Livres</option>
<option id="cat28" value="28">Animaux</option>
<option id="cat55" value="55">Vélos</option>
<option id="cat29" value="29">Sports & Hobbies</option>
<option id="cat30" value="30">Instruments de musique</option>
<option id="cat40" value="40">Collection</option>
<option id="cat41" value="41">Jeux & Jouets</option>
<option id="cat48" value="48">Vins & Gastronomie</option>
<option id="cat56" value="56" style="background-color:#E6E6E6" disabled="">-- MATERIEL PROFESSIONNEL --</option>
<option id="cat57" value="57">Matériel Agricole</option>
<option id="cat58" value="58">Transport - Manutention</option>
<option id="cat59" value="59">BTP - Chantier Gros-oeuvre</option>
<option id="cat60" value="60">Outillage - Matériaux 2nd-oeuvre</option>
<option id="cat32" value="32">Équipements Industriels</option>
<option id="cat61" value="61">Restauration - Hôtellerie</option>
<option id="cat62" value="62">Fournitures de Bureau</option>
<option id="cat63" value="63">Commerces & Marchés</option>
<option id="cat64" value="64">Matériel Médical</option>
<option id="cat31" value="31" style="background-color:#E6E6E6" disabled="">-- SERVICES --</option>
<option id="cat34" value="34">Prestations de services</option>
<option id="cat35" value="35">Billetterie</option>
<option id="cat49" value="49">Evénements</option>
<option id="cat36" value="36">Cours particuliers</option>
<option id="cat65" value="65">Covoiturage</option>
<option id="cat37" value="37" style="background-color:#E6E6E6" disabled="">-- -- --</option>
<option id="cat38" value="38">Autres</option>
</select>
</div>
<span class="label-error full " data-for="category"></span>
</div>
Box
<div class="inputWrapper">
<input id="subject" class="nude" name="subject" maxlength="50" value="" type="text">
<span class="unit"></span>
</div>
As there are no accent, It thought I could directly use their tag and the code is the below
Python code:
Select(driver.find_element_by_id("category")).select_by_visible_text("Locations")
driver.find_element_by_id("subject").send_keys("text to insert")
For both, I have an error.
Error message
Traceback (most recent call last):
File "connectionwebdriver2", line 31, in test_connectionwebdriver2
driver.find_element_by_xpath("//*[@id='cat10']").click()
File "/Users/olivierhoen/scrap/lib/python3.6/site-packages/selenium/webdriver/remote/webdriver.py", line 385, in find_element_by_xpath
return self.find_element(by=By.XPATH, value=xpath)
File "/Users/olivierhoen/scrap/lib/python3.6/site-packages/selenium/webdriver/remote/webdriver.py", line 955, in find_element
'value': value})['value']
File "/Users/olivierhoen/scrap/lib/python3.6/site-packages/selenium/webdriver/remote/webdriver.py", line 312, in execute
self.error_handler.check_response(response)
File "/Users/olivierhoen/scrap/lib/python3.6/site-packages/selenium/webdriver/remote/errorhandler.py", line 242, in check_response
raise exception_class(message, screen, stacktrace)
selenium.common.exceptions.NoSuchElementException: Message: An element could not be located on the page using the given search parameters.
Though I have read different posts and selenium documentation, I am a bit stuck
Thanks in advance for your support
Best,
Full Python code (I have set up specific credential)
# -*- coding: utf-8 -*-
from selenium import webdriver
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.keys import Keys
from selenium.webdriver.support.ui import Select
from selenium.webdriver.support.ui import WebDriverWait
from selenium.common.exceptions import NoSuchElementException
from selenium.common.exceptions import NoAlertPresentException
import unittest, time, re
class Connectionwebdriver2(unittest.TestCase):
def setUp(self):
self.driver = webdriver.Safari()
self.driver.maximize_window()
self.base_urldr = "https://compteperso.leboncoin.fr/account/index.html"
self.verificationErrors = []
self.accept_next_alert = True
self.driver.implicitly_wait(10)
def test_connectionwebdriver2(self):
driver = self.driver
wait = WebDriverWait(driver,10)
driver.get(self.base_urldr)
driver.find_element_by_name("st_username").clear()
driver.find_element_by_name("st_username").send_keys("thecoingood@gmail.com")
driver.find_element_by_name("st_passwd").clear()
driver.find_element_by_name("st_passwd").send_keys("thecoingood1")
driver.find_element_by_id("connect_button").click()
#driver.get("https://www2.leboncoin.fr/ai?ca=15_s")
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//li[@class='deposer']"))).click()
driver.find_element_by_xpath("//*[@id='cat10']").click()
driver.find_element_by_xpath("//*[@id='subject']").send_keys("test")
time.sleep(10)
A: This should work for you :)
from selenium import webdriver
driver = webdriver.Firefox()
driver.get("https://www.leboncoin.fr/ai/form/0")
driver.find_element_by_xpath("//*[@id='cat10']").click()
driver.find_element_by_xpath("//*[@id='subject']").send_keys("test")
and to find the xpath of an element do this:
Gif
Edit:
Just did some modifications to your code to test it, and it works perfectly in firefox.
driver = webdriver.Firefox()
driver.maximize_window()
base_urldr = "https://compteperso.leboncoin.fr/account/index.html"
verificationErrors = []
accept_next_alert = True
driver.implicitly_wait(10)
wait = WebDriverWait(driver,10)
driver.get(base_urldr)
driver.find_element_by_name("st_username").clear()
driver.find_element_by_name("st_username").send_keys("thecoingood@gmail.com")
driver.find_element_by_name("st_passwd").clear()
driver.find_element_by_name("st_passwd").send_keys("thecoingood1")
driver.find_element_by_id("connect_button").click()
#driver.get("https://www2.leboncoin.fr/ai?ca=15_s")
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//li[@class='deposer']"))).click()
driver.find_element_by_xpath("//*[@id='cat10']").click()
driver.find_element_by_xpath("//*[@id='subject']").send_keys("test")
time.sleep(10)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,505 |
{"url":"https:\/\/codegolf.stackexchange.com\/questions\/135102\/formic-functions-ant-queen-of-the-hill-contest\/135156","text":"# Formic Functions - Ant Queen of the Hill Contest\n\nNew tournaments whenever needed. New players and new updates very welcome.\n\nNot actual game footage.\n\nEach player starts with one ant - a queen, who collects food. Each piece of food can be held or used to produce a worker. Workers also collect food to be brought back to the queen.\n\n16 players compete in one arena. The winner is the queen holding the most food after she has taken 30,000 turns. The catch is that the ants can only communicate by changing the colors of the arena squares, which may also be changed by rival ants...\n\n# Watching the game\n\nThis is a JavaScript competition, which means you can watch the game play out live in your browser by clicking the link below.\n\nMany thanks to Helka Homba for the original Stack Snippet King of the Hill contests, Red vs. Blue - Pixel Team Battlebots, and Block Building Bot Flocks, which provided the idea of a web browser hosted KotH and heavily informed the code for this one.\n\nHuge thanks also for all the feedback and testing from the wonderful people in the Sandbox and in Chat.\n\n(Click the image to see the full leaderboard and joint places explanation - only a few players are showing here to save space.)\n\nThis leaderboard is based on the players as they were on Sunday 2nd September 2018.\n\n# Screenshots\n\nSome images of how the arena looks towards the end of a game. Click images to view full size.\n\nTo get an idea of what is happening in the arena and how all these patterns form, you can run the game and hover the mouse over the arena to zoom in and see the ants at work. Also see the fascinating explanations in the answers.\n\n# The arena\n\nThe arena is a toroidal (edge wrapping) grid of square cells. It has width 2500 and height 1000. All cells start as color 1.\n\nInitially exactly 0.1% of cells will contain food. The 2500 pieces of food will be scattered uniformly randomly. No new food will be introduced during the game.\n\nThe queens will be placed randomly on empty cells, with no guarantee that they will not be adjacent to each other (although this is very unlikely).\n\n# Ant abilities\n\n\u2022 Sight: Each ant sees the 9 cells in its 3 by 3 neighbourhood. It has no knowledge of any other ants outside this neighbourhood. It sees the contents of each of the 9 cells (other ants and food), and also each cell's color.\n\u2022 No memory: Each ant makes its decisions based on what it sees - it does not remember what it did in the previous turn and has no way of storing state other than in the colors of the arena cells.\n\u2022 No orientation: An ant does not know where it is or which way it faces - it has no concept of North. The 3 by 3 neighbourhood will be presented to it at a randomly rotated orientation that changes each turn so it cannot even walk in a straight line unless it has colors to guide it. (Making the same move every turn will result in a random walk rather than a straight line.)\n\u2022 Moving, color marking and producing workers: See Output below.\n\u2022 Immortality: These are highland ants that cannot die. You can confuse rival ants by changing the colors around them, or constrain them from moving by surrounding them with 8 ants of your own, but they cannot be harmed apart from this.\n\u2022 Carrying food: A worker can carry up to 1 piece of food. A queen can carry an arbitrary amount of food.\n\u2022 Transferal of food: If a worker is adjacent to a queen (in any of the 8 directions), food will be automatically transferred in one of the following ways:\n\u2022 A laden worker adjacent to its own queen will transfer its food to its queen.\n\u2022 An unladen worker adjacent to an enemy queen will steal 1 piece of food, if present.\n\nA worker cannot steal from a worker, and a queen cannot steal from a queen. Also a worker cannot take food from its own queen, and a queen cannot steal from an enemy worker.\n\nNote that ants take turns sequentially and food transferral occurs at the end of each ant's individual turn and does not take up a turn. It happens regardless of whether a worker moves next to a queen or a queen moves next to a worker, and still happens if both ants involved stand still for their move.\n\n# Coding\n\n### Provide a function body\n\nEach ant is controlled by an ant function. Each turn the player's ant function is called separately for each ant (not just once per player, but once for the queen and once for each worker that player controls). Each turn, the ant function will receive its input and return a move for that particular ant.\n\nPost an answer containing a code block showing the body of a JavaScript function, and it will be automatically included in the controller (just refresh the controller page). The name of the player forms the title of the answer, in the form # PlayerName (which will be truncated to a maximum of 40 characters in the controller tables).\n\n### No state, no time, no random\n\nA function must not access global variables and must not store state between turns. It may use built in functions that do not involve storing state. For example, the use of Math.abs() is fine, but Date.getTime() must not be used.\n\nAn ant function may only use a pseudo random number generator that it supplies itself, that does not store state. For example, it may use the colors\/food\/ants visible as the seed each turn. Math.random() is explicitly forbidden, since like nearly all pseudorandom number generators, it stores state in order to progress to the next number in sequence.\n\nA simple random strategy is still possible due to the random orientation of the input - an ant that always chooses the same direction will perform a random walk rather than a straight line path. See the example answers for simple ways of using this randomness and avoiding this randomness.\n\nAn ant function is permitted to contain further functions within its body. See the existing answers for examples of how this can be useful.\n\n### Console.log\n\nYou can log to the console during testing a new challenger player, but once posted as an answer here the player will have no access to console.log. Attempting to use it will result in an error and disqualification until edited. This should help to keep leaderboard tournaments fast, while still allowing debugging code pasted into the new challenger text area.\n\n# Input and output\n\n### Input\n\nThe orientation of the input will be chosen at random for each ant and for each turn. The input will be rotated by 0, 90, 180 or 270 degrees, but will never be reflected.\n\nCells are numbered in English reading order:\n\n0 1 2\n3 4 5\n6 7 8\n\n\nThe ant function will receive an array called view, containing an object for each of the 9 visible cells. Each object will have the following:\n\ncolor: a number from 1 to 8\nfood: 0 or 1\nant: null if there is no ant on that cell, or otherwise an ant object\n\n\nIf a cell contains an ant, the ant object will have the following:\n\nfood: 0 or more (maximum 1 for a worker)\ntype: 1 to 4 for a worker, or 5 for a queen\nfriend: true or false\n\n\nThe ant can determine its own details by looking at the ant in the central cell, view[4].ant. For example, view[4].ant.type is 5 for a queen, or a number from 1 to 4 for a worker (indicating its type).\n\n### Output\n\nOutput is returned as an object representing the action to take. This can have any of the following:\n\ncell: a number from 0 to 8 (mandatory)\ncolor: a number from 1 to 8 (optional)\ntype: a number from 1 to 4 (optional)\n\n\nIf color and type are omitted or zero, then cell indicates the cell to move to.\n\nIf color is non-zero, the indicated cell is set to that color.\n\nIf type is non-zero, a worker ant of that type is created on the indicated cell. Only a queen can create a new worker, and only if she has food, as this costs one piece of food per worker.\n\nExample outputs:\n\n{cell:0}: move to cell 0\n{cell:4}: move to cell 4 (that is, do nothing, as 4 is the central cell)\n{cell:4, color:8}: set own cell to color 8\n{cell:6, type:1}: create a type 1 worker on cell 6\n{cell:6, color:1}: set cell 6 to color 1\n{cell:6, color:0}: equivalent to just {cell:6} - move rather than set color\n{cell:6, type:0}: equivalent to just {cell:6} - move rather than create worker\n{cell:6, color:0, type:0}: move to cell 6 - color 0 and type 0 are ignored\n\n\nInvalid outputs:\n\n{cell:9}: cell must be from 0 to 8\n{cell:0, color:9}: color must be from 1 to 8\n{cell:0, type:5}: type must be from 1 to 4 (cannot create a new queen)\n{cell:4, type:1}: cannot create a worker on a non-empty cell\n{cell:0, color:1, type:1}: cannot set color and create worker in the same turn\n\n\nAn ant moving onto a cell containing food will automatically pick up the piece of food.\n\n# Worker type\n\nEach worker has a type, a number from 1 to 4. This has no meaning to the controller, and is for the player to do with as they wish. A queen could produce all her workers as type 1, and give them all the same behaviour, or she could produce several types of workers with different behaviours, perhaps type 1 as foragers and type 2 as guards.\n\nThe worker type number is assigned by you when a worker is created, and cannot be changed thereafter. Use it however you see fit.\n\n# Turn order\n\nAnts take turns in a set order. At the start of a game the queens are assigned a random order which does not change for the rest of the game. When a queen creates a worker, that worker is inserted into the turn order at the position before its queen. This means that all other ants belonging to all players will move exactly once before the new worker takes its first turn.\n\n# Limit on number of players\n\nObviously an unlimited number of players cannot fit into the arena. Since there are now more than 16 answers, each game will feature a randomly chosen 16 of them. The average performance over many games will give a leaderboard featuring all the players, without ever having more than 16 in a single game.\n\n# Time limit per turn\n\nEach time the ant function is called, it should return within 15 milliseconds. Since the time limit may be exceeded due to fluctuations outside the ant function's control, an average will be calculated. If at any point the average is above 15 milliseconds and the total time taken by that particular ant function across all calls so far is more than 10 seconds, the relevant player will be disqualified.\n\n# Disqualification\n\nThis means the player will not be eligible to win and their ant function will not be called again during that game. They will also not be included in any further games. If a player is disqualified on the tournament machine during a leaderboard game then it will be excluded from all future leaderboard games until edited.\n\nA player will be disqualified for any of the following for any of its ants (queen or worker):\n\n\u2022 Exceeding the time limit as described (averaged over 10 seconds).\n\u2022 Returning an invalid move as described under Output.\n\u2022 The cell to move to contains an ant.\n\u2022 The cell to move to contains food and the ant is already a laden worker.\n\u2022 The cell to produce a worker on is not empty (contains food or an ant).\n\u2022 A worker is trying to produce a worker.\n\nIt may seem harsh to disqualify for invalid moves, rather than simply interpreting this as no move. However, I believe that enforcing correct implementations will lead to more interesting strategies over time. This is not intended to be an additional challenge, so a clear reason will be displayed when a player is disqualified, with the specific input and output alongside to aid in fixing the code.\n\nYou may provide multiple answers, provided that they do not team up against the others. Provided each answer is working solely towards its own victory, you are permitted to tailor your strategy to take advantage of weaknesses in specific other strategies, including changing the color of the cells to confuse or manipulate them. Bear in mind that as more answers come in, the likelihood of meeting any particular player in a given game will diminish.\n\nYou may also edit your answers whenever you choose. It is up to you whether you post a new answer or edit an existing one. Provided the game is not flooded with many near-identical variations, there should be no problem.\n\n# Scoring\n\nAt the end of each game, a player's score is the number of other players who have less food carried by their queen. Food carried by workers is not counted. This score is added to the leaderboard, which is displayed in order of average score per game.\n\nJoint places indicate that the order of players is not yet consistent between 6 subsets of the games played so far. The list of games is split into 6 subsets because this is the minimum number that will give a probability of less than 5% that a given pair of players will be assigned distinct places in the wrong order.\n\n# Chat\n\nTo keep the comments section clear here, please use the dedicated chat room for any questions and discussion. Comments on this post are likely to be cleared after a while, whereas messages in the chat room will be kept permanently.\n\nJust to let you know, I'll be more inclined to upvote answers that include a clear and interesting explanation of how the code works.\n\n\u2022 @DestructibleLemon for the sake of anyone reading through these comments, I've answered that in the chat room \u2013\u00a0trichoplax Jul 23 '17 at 4:58\n\u2022 \u2013\u00a0Draco18s no longer trusts SE Jul 27 '17 at 3:02\n\u2022 Hey, I made a thing! You might find it interesting since it's inspired by this challenge and includes a Formic Functions test implementation. \u2013\u00a0Dave Sep 9 '17 at 15:54\n\u2022 @Dave Your controller is blazingly fast :) - but let me mention that its scoring seems to differ from the original in cases where queens are tied for food at the end of a game. The score should be the number of other participants whose queens hold (strictly) less food. E.g., if three players have 0 food at the end, they should all score zero for this game, not three. \u2013\u00a0GNiklasch Jan 20 '18 at 14:52\n\u2022 @GNiklasch thanks; fixed. Also I see that your ant dominates the game now. Impressive! \u2013\u00a0Dave Jan 21 '18 at 12:11\n\n# Straighter\n\nvar i, j\nvar orthogonals = [1, 3, 7, 5] \/\/ These are the non-diagonal cells\n\/\/ Color own cell if white\nif (view[4].color != 6) {\nreturn {cell:4, color:6}\n}\nvar specified = null;\n\/\/ Otherwise move to a white cell opposite a colored cell\nfor (i=0; i<4; i++) {\nj = (i+2) % 4\nif (view[orthogonals[i]].color !== 6 &&\nview[orthogonals[j]].color == 6 && !view[orthogonals[i]].ant) {\nspecified = {cell:orthogonals[i]};\n} else if (view[orthogonals[i]].food) {\nreturn {cell:orthogonals[i]}\n}\n}\nif(specified) { return specified; }\n\/\/ Otherwise move to one of the vertical or horizontal cells if not occupied\nfor (i=1; i<9; i+=2) {\nif (!view[i].ant) {\nreturn {cell:i}\n}\n}\n\n\/\/ Otherwise move to one of the diagonal cells if not occupied\nfor (i=0; i<9; i+=2) {\nif (!view[i].ant) {\nreturn {cell:i};\n}\n}\n\n\/\/ Otherwise don't move at all\nreturn {cell:4};\n\n\nSince, while testing trail eraser, I noticed that it sometimes nears the top of the leaderboard when it doesn't find any trails, I'm posting an ant using the same movement but no workers. Basicaly a repost of smart straight line from meta, except ignores other colors and doesn;t make workers (thereby avoiding disqualification)\n\n\u2022 Note that this does have some synergy with trail-eraser in that they use the same color, and trail eraser ignores green incursions, but they aren't written to collaborate together \u2013\u00a0pppery Jul 24 '17 at 1:41\n\n# Antdom Walking Artist\n\nThis is a fancy queen-only random walk entry with a slight twist-- the queen ant wanders around coloring cells if she feels that there aren't enough non-colored cells nearby. If she sees food, she'll instantly grab it. Hopefully she could gather some food, unless other ants steal them away.\n\nA lot of worker ants would stop by and be appalled at the \"marvelous\" art. This means that other worker ants may eventually forget to return to their own queen ants.\n\nWorkers can be spawn occasionally. They walk along horizontal or vertical lines if they haven't found any food yet, and will walk on diagonal lines once they found some food. Unfortunately, there is little coordination between the queen and the workers, and the workers will forever roam free.\n\n## How it works\n\n1. If this is a Queen ant (type 5):\n\na. If there is food in either of the 8 neighboring cells, fetch food.\n\nb. Otherwise, if there are strictly less than 4 colored cells in the 8 neighboring cells, color the lowest-indexed white cell with the color code next_color = (8 - #colored).\n\nc. Otherwise, when time is just right, (i.e. if seed = 666 mod 2333 and me['food'] = 2 mod 3), spawn a worker of random type.\n\nd. Otherwise, walk to the lowest-indexed neighboring cell that contains no ant.\n\n2. If this is a Worker ant (type 1-4):\n\na. If the worker is unladen (does not carry food) and if there is food in either of the 8 neighboring cells, fetch food.\n\nb. Otherwise, if the worker is walking on a white cell, it gets colored by Math.abs(next_color - 2) + 2.\n\nc. Otherwise, attempt to walk along a straight line in the vertical or horizontal direction. If this is not possible, walk to the lowest-indexed neighboring cell that contains no ant (and no food if the worker is laden).\n\n## Code\n\nvar action = { cell: 4, color: 0, type: 0 },\ncolors = view.map((v) => v.color),\nfoods = view.map((v) => v.food === 1),\nants = view.map((v) => v.ant),\nme = ants[4];\n\nvar seed = colors.reduce((s, v) => s * 8 + (v - 1));\n\n\/\/ Search the neighboring cells for food and walk on to them.\nvar next_food = foods.indexOf(true);\n\n\/\/ As the last measure, find a move that avoids other ants and food.\nvar get_direction = (view) => {\nvar direction = 4, complete = false;\nfor (var i = 0; !complete && i < 4; ++i) {\nif (view[i]['ant'] === null && view[i]['food'] === 0) {\ncomplete = true;\ndirection = i;\n}\n}\nfor (var i = 5; !complete && i < 9; ++i) {\nif (view[i]['ant'] === null && view[i]['food'] === 0) {\ncomplete = true;\ndirection = i;\n}\n}\nreturn direction;\n};\n\n\/\/ Attempt to walk along a diagonal.\nvar get_diagonal_opposite = (view, food) => {\nvar direction = null;\nif (view[0]['color'] === 1 && view[8]['color'] > 1 && view[0]['ant'] === null && view[0]['food'] <= food) {\ndirection = 0;\n} else if (view[8]['color'] === 1 && view[0]['color'] > 1 && view[8]['ant'] === null && view[8]['food'] <= food) {\ndirection = 8;\n} else if (view[2]['color'] === 1 && view[6]['color'] > 1 && view[2]['ant'] === null && view[2]['food'] <= food) {\ndirection = 2;\n} else if (view[6]['color'] === 1 && view[2]['color'] > 1 && view[6]['ant'] === null && view[6]['food'] <= food) {\ndirection = 6;\n}\nreturn direction;\n};\n\n\/\/ Attempt to walk along a horizontal or vertical line.\nvar get_lateral_opposite = (view, food) => {\nvar direction = null;\nif (view[1]['color'] === 1 && view[7]['color'] > 1 && view[1]['ant'] === null && view[1]['food'] <= food) {\ndirection = 1;\n} else if (view[7]['color'] === 1 && view[1]['color'] > 1 && view[7]['ant'] === null && view[7]['food'] <= food) {\ndirection = 7;\n} else if (view[3]['color'] === 1 && view[5]['color'] > 1 && view[3]['ant'] === null && view[3]['food'] <= food) {\ndirection = 3;\n} else if (view[5]['color'] === 1 && view[3]['color'] > 1 && view[5]['ant'] === null && view[5]['food'] <= food) {\ndirection = 5;\n}\nreturn direction;\n};\n\n\/\/ A random color. Starts from 8. Reduce 1 if any neighboring cells are colored.\nvar next_color = 8 - colors.reduce((s, v, i) => s + ((i !== 4 && v > 1) ? 1 : 0), 0);\nif (me['type'] === 5) {\n\/\/ Queen Ant\nif (next_food !== -1) {\n\/\/ Moves onto food.\naction['cell'] = next_food;\n} else if (next_color > 4) {\n\/\/ If not enough colored cells, go around and color.\nvar target = colors.indexOf(1);\naction['cell'] = target;\naction['color'] = next_color;\n} else if (seed % 2333 === 666 && me['food'] % 3 === 2) {\n\/\/ If queen has (3k + 2) food pellets, randomly spawn worker ants.\nvar type = seed % 4 + 1;\nif (ants[1] === null) {\naction['type'] = type;\naction['cell'] = 1;\n} else if (ants[0] === null) {\naction['type'] = type;\naction['cell'] = 0;\n}\n} else {\naction['cell'] = get_direction(view);\n}\n} else if (me['type'] <= 4) {\n\/\/ Gatherers\nif (me['food'] === 0) {\n\/\/ 1. Collect food if one is nearby.\n\/\/ 2. Color current cell.\n\/\/ 3. Move in a vertical\/horizontal line.\nif (next_food !== -1) {\naction['cell'] = next_food;\n} else if (colors[4] < 2) {\naction['cell'] = 4;\naction['color'] = Math.abs(next_color - 2) + 2;\n} else {\n\/\/ Do not color everything. That will expose oneself.\naction['cell'] = get_lateral_opposite(view, 1) || get_direction(view);\n}\n} else {\n\/\/ 1. Color current cell.\n\/\/ 2. Move diagonally.\nif (colors[4] < 2) {\n\/\/ Color current cell.\naction['cell'] = 4;\naction['color'] = Math.abs(next_color - 2) + 2;\n} else {\naction['cell'] = get_diagonal_opposite(view, 0) || get_direction(view);\n}\n}\n}\n\nif (action['color'] === 0) {\ndelete action['color'];\n}\nif (action['type'] === 0) {\ndelete action['type'];\n}\n\nreturn action;\n\n\u2022 Edit 1: Spawning free-roaming workers with a small probability. \u2013\u00a0Frenzy Li Jul 23 '17 at 13:47\n\u2022 This ones does not appear to behave in a deterministic manner. That is, all alone by itself on the same seed will produce different results. I believe this is against the rules, per no use of Math.random(). I don't know what it is that is doing it, but it's making it very difficult for me to test a certain behavior when this ant does not behave deterministically. \u2013\u00a0Draco18s no longer trusts SE Jul 25 '17 at 19:17\n\u2022 Hmm, maybe it is. And just that the minor changes I've been making have resulted in a different number of ants which messes with things. Nevermind! \u2013\u00a0Draco18s no longer trusts SE Jul 25 '17 at 20:29","date":"2020-08-10 18:21:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.2885514497756958, \"perplexity\": 2274.5325897326925}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-34\/segments\/1596439737168.97\/warc\/CC-MAIN-20200810175614-20200810205614-00281.warc.gz\"}"} | null | null |
<?php
/**
* Created by PhpStorm.
* User: Bharatwaaj
* Date: 07-06-2017
* Time: 15:06
*/
include 'firebaseConnectivity.php';
?>
<script type="text/javascript">
// Counting Notification count from firebase
var no_of_notifications = 0, firstChildrenvalue = 0;
database.ref('users').once('value', function (snapshot) {
firstChildrenvalue = snapshot.numChildren();
});
database.ref('users').on('child_added', function (snapshot) {
if (firstChildrenvalue >= no_of_notifications) {
document.getElementById("no_of_newjoin").innerHTML = 0;
document.getElementById("no_of_notifications").innerHTML = 0;
document.getElementById("notify_no").innerHTML = 0;
}
else{
document.getElementById("no_of_newjoin").innerHTML = no_of_notifications-firstChildrenvalue;
document.getElementById("no_of_notifications").innerHTML = no_of_notifications-firstChildrenvalue;
document.getElementById("notify_no").innerHTML = no_of_notifications-firstChildrenvalue;
}
no_of_notifications++;
});
</script>
<header class="main-header">
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<a href="dashboard.php" class="logo">
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<!-- logo for regular state and mobile devices -->
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</a>
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| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,090 |
How Land O'Lakes' on-site clinic personalizes employee benefits
The company's clinic sculpted its 'wellness journey' into a holistic approach to employee well-being, its director of benefits told attendees at #SHRM19.
Katie Clarey @kclarey21
LAS VEGAS — As Land O'Lakes scouted a location for its new headquarters that opened one year ago, it sought a place that could house talent-luring amenities. The facility it chose is now home to 1,800 employees, a daycare, a fitness center and an on-site health clinic.
The company's amenity-driven property hunt came along a larger "wellness journey" that's been progressing for several decades, Land O'Lakes Director of Benefits Emily Maher told attendees at the Society for Human Resource Management 2019 Annual Conference. In particular, Land O'Lakes' on-site clinic — the main subject of Maher's talk — sculpted its wellness journey into a holistic approach to employee well-being.
Financial foundation: The key to the on-site clinic
As Land O'Lakes established the on-site clinic at its new Arden Hills, Minnesota, headquarters, it focused on making benefits about its employees and making them personal, Maher said. A medical assistant, nurse practitioner, physical therapist and personal trainer staff the 850-square-foot clinic, which features two care suites, a physical therapy room and a lab.
Before Maher and her benefits team could make the clinic a reality, they needed executive buy-in, which included a discussion about money. "You must have the financial basis to put in an on-site clinic," she said.
To make a convincing case that the clinic would produce a worthwhile return on investment, those proposing the amenity had to show that employees would save time and money by seeking medical care at their workplace rather than a facility in the community. "If our employees can save money, it saves them money and it saves us money," Maher said.
Smaller medical bills aren't the only result of an on-site clinic that appeals to executives. "I will tell you that although it's important for us to demonstrate the ROI, there are other facets of the clinic that are important," Maher said.
The advantage of on-site data
Using the insurance claims and pharmacy data collected by the on-site clinic through its medical provider partner, Land O'Lakes identified its top health risks: diabetes, depression and obesity.
From there, it could introduce targeted solutions and initiatives through the staff at the on-site clinic, paired with others throughout the workplace. The onsite clinic gave the company the reach to impact at least 1,800 employees, Maher said.
Once it identified depression as a top risk, for example, Land O'Lakes introduced a suicide awareness program, an employee assistance program (EAP) and another emotional wellness program called Learn to Live. "The statistics on using EAPs are incredibly low in most organizations," Maher said, noting that most are happy to see an employee usage rate of 4%.
More employees at the Land O'Lakes headquarters are using the program because the staff at the on-site clinic refer employees to the EAP, whether during a conversation at an appointment, a chat at the Land O'Lakes cafeteria or a session during an employee resources meeting, Maher said. This surge in usage is owed to the personal relationships employees develop with the clinic staff, Maher said: "It's things like this that drive home the need for integrated opportunity and personalized experience."
The data also revealed gaps in employees' care. Employees who hadn't sought medical care in years popped in to see the medical assistant or nurse practitioner during their lunch breaks, which gave the on-site clinic staff — who had access to their care history — the opportunity to address long-ignored or unnoticed problems, Maher said. "So many opportunities are available to us by leveraging the data we have," Maher said. "It has shored up our wellness plan."
Follow Katie Clarey on Twitter
Filed Under: Corporate News | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,638 |
Q: log4net not logging in worker threads with MinimalLock model I have a rater "interesting" problem, where log4net doesn't write the log messages if they come from inside a worker thread in ASP.NET MVC. This seems to only be an issue when I've added <lockingModel type="log4net.Appender.FileAppender+MinimalLock" /> to my log4net configuration and when the website is running on my server (IIS 8.0). On my own computer log4net logs the messages just fine - even with the MinimalLock configuration and if I remove this 'MinimalLock' it will log messages in worker threads on the server as well.
Here is an example:
public class MvcApplication : System.Web.HttpApplication
{
private static readonly log4net.ILog log = log4net.LogManager.GetLogger(
System.Reflection.MethodBase.GetCurrentMethod().DeclaringType);
protected void Application_Start()
{
// This is logged just fine - with or without MinimalLock,
// both local and on server
log.Info("Logging ouside worker!");
// This is only logged without MinimalLock configured
// if the site is hosted on my server runnning IIS 8.0.
var thread = new Thread(() => log.Info("Logging inside worker!"));
thread.Start();
}
}
and here is my log4net configuration:
<log4net>
<root>
<level value="ALL" />
<appender-ref ref="LogFileAppender" />
</root>
<appender name="LogFileAppender" type="log4net.Appender.RollingFileAppender">
<file value="App_Data\log.log" />
<appendToFile value="true" />
<rollingStyle value="Size" />
<maxSizeRollBackups value="10" />
<maximumFileSize value="10MB" />
<staticLogFileName value="true" />
<lockingModel type="log4net.Appender.FileAppender+MinimalLock" />
<layout type="log4net.Layout.PatternLayout">
<conversionPattern value="%date [%thread] %level %logger - %message%newline" />
</layout>
</appender>
</log4net>
Without the MinimalLock lockingModel configured, I'm unable to read the log file on the server, so removing it isn't really an option. I'm also unable to avoid logging inside of worker threads, as I need to log potential errors inside of the workers spawned by Quartz.net.
I suspect it is an issue of permissions on the server, but I have very limited access to the server, as I'm not the administrator (I basically only have FTP access - they won't even grant me external access to the MSSQL DB).
Here's the question
If my suspicion is correct, what changes should I request from my server administrator in order to fix the problem? Which permissions should be granted to which user, in order for MinimalLock to work properly in my hosted environment?
If it is not an issue with permissions, what could it be instead?
Any help is much appreciated. Thanks in advance.
Update:
After enabling the debug log for log4net, I found the following errors when I run the example code above (full path omitted):
log4net:ERROR [RollingFileAppender] Unable to acquire lock on file App_Data\log.log. Access to the path 'App_Data\log.log' is denied.
log4net:ERROR [RollingFileAppender] Unable to acquire lock on file App_Data\log.log. Access to the path 'App_Data\log.log' is denied.
log4net:ERROR [RollingFileAppender] Unable to acquire lock on file App_Data\log.log. Access to the path 'App_Data\log.log' is denied.
log4net: ConfigureAndWatchHandler: Changed [web.config]
log4net: ConfigureAndWatchHandler: Changed [web.config]
log4net: Hierarchy: Shutdown called on Hierarchy [log4net-default-repository]
log4net:ERROR [RollingFileAppender] Unable to acquire lock on file App_Data\log.log. Access to the path 'App_Data\log.log' is denied.
log4net:ERROR [RollingFileAppender] Could not close writer [log4net.Util.CountingQuietTextWriter]
log4net.Appender.FileAppender+LockingStream+LockStateException: The file is not currently locked
at log4net.Appender.FileAppender.LockingStream.AssertLocked()
at log4net.Appender.FileAppender.LockingStream.Flush()
at System.IO.StreamWriter.Flush(Boolean flushStream, Boolean flushEncoder)
at System.IO.StreamWriter.Dispose(Boolean disposing)
at System.IO.StreamWriter.Close()
at log4net.Util.QuietTextWriter.Close()
at log4net.Appender.TextWriterAppender.CloseWriter()
These only occur on the server.
A: I managed to solve the issue.
In my hosting environment, the worker threads are associated with a different Windows identity than their parent. The identity used by the worker threads only had read permission on my App_Data/Logs folder. As my access to the server is limited, I created a .cshtml file to programmatically add write and modify permissions to the identity of the worker threads.
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,140 |
{"url":"http:\/\/gmatclub.com\/forum\/gmat-problem-solving-ps-140\/index-650.html?sk=r&sd=a","text":"GMAT Problem Solving (PS) - Page 14\nCheck GMAT Club App Tracker for the Latest School Decision Releases http:\/\/gmatclub.com\/AppTrack\n\n It is currently 04 Dec 2016, 17:28\n\n# Events & Promotions\n\n###### Events & Promotions in June\nOpen Detailed Calendar\n\n# GMAT Problem Solving (PS)\n\n new topic Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous \u00a0 \u00a01\u00a0\u00a0...\u00a0\u00a012\u00a0\u00a0\u00a013\u00a0\u00a0\u00a014\u00a0\u00a0\u00a015\u00a0\u00a0\u00a016\u00a0\u00a0...\u00a0\u00a0271\u00a0 \u00a0 Next Search for:\nTopics Author Replies \u00a0 Views Last post\n\nAnnouncements\n\n171\n 150 Hardest and easiest questions for PS\nBunuel\n\n7\n\n29941\n\n11 Sep 2016, 02:25\n\n879\n GMAT PS Question Directory by Topic & Difficulty\nbb\n\n0\n\n325684\n\n22 Feb 2012, 10:27\n\nTopics\n\n3\n A student took five papers in an examination, where the full marks\nexcelingmat\n\n1\n\n1247\n\n12 Oct 2015, 10:23\n\n The volume of a certain substance is always directly proportional to\nBunuel\n\n1\n\n327\n\n20 Dec 2015, 05:45\n\n A plane was originally flying at an altitude of x feet when it ascende\nBunuel\n\n1\n\n343\n\n23 Nov 2015, 19:03\n\n The arithmetic mean and standard deviation of a certain\nMaithiliGokarn\n\n1\n\n3369\n\n15 Mar 2012, 07:35\n\n2\n What is the largest integer k such that 10! is divisible by\narindamsur\n\n1\n\n1947\n\n11 Jun 2014, 01:07\n\n Data Sufficiency\nagdimple333\n\n1\n\n1072\n\n04 May 2011, 10:56\n\n1\n Sequence, Exponents\nMUTHUTECH08\n\n1\n\n571\n\n03 Sep 2015, 07:22\n\n The rhombus (AFCE) is inscribed in a rectange\nkuttingchai\n\n1\n\n545\n\n12 Aug 2015, 08:48\n\n Inequality comparison\nskbjunior\n\n1\n\n1035\n\n24 Apr 2011, 22:57\n\n In a certain diving competition, 5 judges score each dive on a scale\nBunuel\n\n1\n\n392\n\n16 Nov 2015, 23:10\n\n Exactly 36% of the numbers in set S are even multiples of 3\nenigma123\n\n1\n\n1088\n\n11 Feb 2012, 15:44\n\n3\n The function f is defined for all positive integers n by\nMSoS\n\n1\n\n3376\n\n21 Feb 2012, 05:48\n\n RTD \"Kiss\" Problem\nGmatSlayer112\n\n1\n\n2300\n\n13 Jan 2012, 05:05\n\n3\n Overlapping Sets PS\ngmatpapa\n\n1\n\n1427\n\n08 Apr 2011, 08:50\n\n2\n The figure above represents a frame; the shaded regions represent the\nBunuel\n\n1\n\n495\n\n29 Nov 2015, 10:15\n\n2\n Cindy paddles her kayak upstream at m kilometers per hour, and then\nBunuel\n\n1\n\n396\n\n13 Dec 2015, 12:29\n\n2\n In the below diagram, points A and B lie on the circle with center O,\nBunuel\n\n1\n\n665\n\n31 Aug 2015, 16:00\n\n1\n How many positive integers less than 20 can be expressed as\nenigma123\n\n1\n\n1259\n\n11 Feb 2012, 17:07\n\n9\n The number of digits in the number (4^11)(5^25) = ?\namithyarli\n\n1\n\n653\n\n18 Sep 2015, 05:33\n\n1\n If it takes a tub 3 minutes to drain 5\/7 of its content, how much more\naiming4mba\n\n1\n\n1227\n\n22 Jul 2010, 22:31\n\n How many times they interesect ?\nAlchemist1320\n\n1\n\n1238\n\n25 Jun 2011, 00:15\n\n If x and y are integers such that x > y, which of the following CANNOT\nBunuel\n\n1\n\n399\n\n20 Dec 2015, 04:51\n\n If n + xy = n and x is not equal to 0, which of the follow\namgelcer\n\n1\n\n1293\n\n19 Nov 2013, 02:32\n\n Each night before he goes to bed, Jordan likes to pick out\niwillcrackgmat\n\n1\n\n3816\n\n13 Mar 2012, 19:31\n\n If x = y + y^2 , and y is a negative integer, when y decreases in valu\nBunuel\n\n1\n\n361\n\n10 Jan 2016, 09:11\n\n A can do a job in 12 hrs B can do a job in 6 hours. 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If there are n differen\ndflorez\n\n1\n\n1709\n\n30 Dec 2015, 07:58\n\n Diagonal of a cube\nmelguy\n\n1\n\n2210\n\n29 May 2011, 18:59\n\n2\n What least number should be subtracted from 2590 so that the\nsunny3011\n\n1\n\n2941\n\n08 Jun 2014, 03:19\n\n1\n A parking garage rents parking spaces for $10 per week or$30 per\nBunuel\n\n1\n\n942\n\n29 Nov 2015, 10:14\n\n If a = 105 and a^3 = 21 \u00d7 25 \u00d7 45 \u00d7 b, what is the value of b?\nBunuel\n\n1\n\n398\n\n20 Dec 2015, 04:58\n\n if the ratio of 4 to 5 1\/2 is equal to the ratio of y to 2 3\/8 then y\nRSOHAL\n\n1\n\n777\n\n22 Dec 2015, 13:02\n\n A sporting goods store sold 64 Frisbees in one week, some for $3 and Bunuel 1 382 19 Jan 2016, 11:46 How many 4 digit nos that do not contain the digits 3 or 6 a abhi47 1 10903 02 Feb 2012, 14:06 Every year Taylor goes to the same carnival, and he attempts to shoot Bunuel 1 256 06 Apr 2016, 07:48 A comittee of three people is to be chosen from four married andih 1 2415 09 Apr 2012, 13:08 1 A track shown above, there is a rectangle in the centre and 2 sides a MathRevolution 1 300 24 Feb 2016, 21:01 In the figure above, what is the distance from point P to point Q? reto 1 777 12 May 2015, 20:53 1 In the figure above, ED = 1, CD = 2, and AE =6*3^(1\/2). What is the Bunuel 1 364 27 Dec 2015, 21:33 1 If 4^(2x + 2) = 16^(3x \u2212 1), what is the value of x ? Bunuel 1 375 20 Dec 2015, 05:00 Don't really know how to solve this problem Olenkap82 1 1153 25 Aug 2011, 09:40 From the sale of sleeping bags, a retailer made a gross enigma123 1 3647 10 Mar 2012, 13:41 1 At a certain restaurant, the price of a sandwich is$ 4.00 more than\nBunuel\n\n1\n\n369\n\n06 Jan 2016, 11:00\n\n What is the total surface area in square meters of a rectangular solid\nBunuel\n\n1\n\n237\n\n10 Jan 2016, 07:42\n\n PS on MGMAT Test\nsumeeet4u\n\n1\n\n1000\n\n29 Apr 2011, 06:14\n\n1\n In a certain sculpture, coils of wire are arranged in rows. \u00a0 Tags:\u00a0Sequences\nx13069\n\n1\n\n1248\n\n02 Jan 2010, 15:01\n\n A number x is multiplied with itself and then added to the product of\nwhitehalo\n\n1\n\n237\n\n27 Feb 2016, 23:52\n\n Dividing by 3\u20448 and then multiplying by 5\u20446 is the same as dividing by\nBunuel\n\n1\n\n269\n\n10 Jan 2016, 07:52\n\n Algebra\nagdimple333\n\n1\n\n1323\n\n01 May 2011, 14:56\n\n2\n Nathan took out a student loan for 1200\\$ at 10 percent annual interest\nRSOHAL\n\n1\n\n758\n\n22 Dec 2015, 20:27\n\n new topic Question banks Downloads My Bookmarks Reviews Important topics Go to page Previous \u00a0 \u00a01\u00a0\u00a0...\u00a0\u00a012\u00a0\u00a0\u00a013\u00a0\u00a0\u00a014\u00a0\u00a0\u00a015\u00a0\u00a0\u00a016\u00a0\u00a0...\u00a0\u00a0271\u00a0 \u00a0 Next Search for:\n Who is online In total there are 5 users online :: 0 registered, 0 hidden and 5 guests (based on users active over the past 15 minutes) Users browsing this forum: No registered users and 5 guests Statistics Total posts 1564692 | Total topics 189616 | Active members 479464 | Our newest member ushamirra\n\n Powered by phpBB \u00a9 phpBB Group and phpBB SEO Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.","date":"2016-12-05 01:28:20","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.33790990710258484, \"perplexity\": 2979.8345867460553}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-50\/segments\/1480698541517.94\/warc\/CC-MAIN-20161202170901-00245-ip-10-31-129-80.ec2.internal.warc.gz\"}"} | null | null |
using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.ComponentModel.DataAnnotations;
namespace Vilicus.Dal.Models
{
public enum Status
{
Submitted = 1, // Database friendly id
Retracted,
Rejected,
Accepted,
[Description("In Work")]
InWork,
Resolved
}
public enum Urgency
{
Passive = 1,
Low,
Normal,
High,
Critical
}
public class Tag
{
public int Id { get; set; }
public string Name { get; set; }
public virtual ICollection<TicketTagMapping> TicketTagMapping { get; set; }
}
public class TicketTagMapping
{
public int Id { get; set; }
public int TicketId { get; set; }
public int TagId { get; set; }
public virtual Ticket Ticket { get; set; }
public virtual Tag Tag { get; set; }
}
public class Ticket
{
[Required]
public int Id { get; set; }
[Required]
public string Subject { get; set; }
[Required]
public string Description { get; set; }
[Required]
public DateTime TicketTime { get; set; }
[Required]
public Urgency UrgencyId { get; set; }
[Required]
public Status StatusId { get; set; }
[Required]
public int UserId { get; set; }
public virtual User User { get; set; }
public virtual ICollection<TicketTagMapping> TicketTagMapping { get; set; }
public virtual ICollection<Message> Messages { get; set; }
}
public class Message
{
public int Id { get; set; }
public string Text { get; set; }
public DateTime MessageTime { get; set; }
public int UserId { get; set; }
public int TicketId { get; set; }
public virtual User User { get; set; }
public virtual Ticket Ticket { get; set; }
}
public class User
{
public int Id { get; set; }
public string FirstName { get; set; }
public string LastName { get; set; }
public virtual Client Employer { get; set; }
public virtual ICollection<Ticket> Tickets { get; set; }
public virtual ICollection<Message> Messages { get; set; }
}
public class Client
{
public int Id { get; set; }
public string Name { get; set; }
public virtual ICollection<User> Employees { get; set; }
}
}
| {
"redpajama_set_name": "RedPajamaGithub"
} | 7,888 |
Q: html5 template-element AngularJS Im experimenting with template elements and Angular.
<template id="loginTemplate">
<div id="login" class="menuTop menu top" data-ng-controller="loginCtrl">
<form class="form-inline text-center" id="loginForm" data-ng-submit="login()">
<input type="email" class="input-big" placeholder="Email" data-ng-model="loginEmail">
<input type="password" class="input-mini" placeholder="Password" data-ng-model="loginPass">
<input type="text" class="input-small" placeholder="Display name" data-ng-model="loginName" data-ng-show="register">
<button type="submit" class="btn-small">Sign in</button>
<button type="button" class="btn-small" data-ng-hide="register" data-ng-click="register=true">Register</button>
</form>
</div>
</template>
JS:
(function init(){
"use strict";
var loginTemplate = document.querySelector("#loginTemplate").content;
document.body.appendChild(loginTemplate.cloneNode(true));
return;
})();
This works great in rendering my Angular controller data on init.
However when I want to conditionally load the next template it doesn't work.
<template id="menuTemplate">
<div id="menu" class="menuTop menu top" data-ng-controller="menuCtrl">
{{ time.now }}
</div>
</template>
JS:
socket.on("logon", function ( data ){
//this fails
$scope.user.name = data.userName;
var menuTemplate = document.querySelector("#menuTemplate").content;
document.body.removeChild(document.getElementById("login"));
document.body.appendChild(menuTemplate.cloneNode(true));
}
This will show me the {{ variables }}.
Loading this menuTemplate first makes it work, but loading it conditionally doesn't.
Maybe Angular doesn't support template-elements, as they have said support will be available in 2.0, but initial load works so maybe someone has it working elsewhere?
Maybe the .clonenode breaks the angular link when performed after load? I dont know.
edit:Have a fiddle but have been unable to make it work as it does in chrome canary.
http://jsfiddle.net/CSK8n/
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 9,426 |
{"url":"https:\/\/en.academic.ru\/dic.nsf\/enwiki\/1666731\/Ostwald_ripening","text":"# Ostwald ripening\n\n\ufeff\nOstwald ripening\nBasic schematic of the Ostwald ripening process\n\nOstwald ripening is an observed phenomenon in solid solutions or liquid sols which describes the change of an inhomogeneous structure over time. In other words, over time, small crystals or sol particles dissolve, and redeposit onto larger crystals or sol particles.[1]\n\nDissolution of small crystals or sol particles and the redeposition of the dissolved species on the surfaces of larger crystals or sol particles was first described by Wilhelm Ostwald in 1896.[2] Ostwald ripening is generally found in water-in-oil emulsions, while flocculation is found in oil-in-water emulsions [3]\n\n## Mechanism\n\nThis thermodynamically-driven spontaneous process occurs because larger particles are more energetically favored than smaller particles.[4] This stems from the fact that molecules on the surface of a particle are energetically less stable than the ones in the interior. Consider a cubic crystal of atoms: all the atoms inside are bonded to 6 neighbors and are quite stable, but atoms on the surface are only bonded to 5 neighbors or less, which makes these surface atoms less stable. Large particles are more energetically favorable since, continuing with our example, more atoms are bonded to 6 neighbors and fewer atoms are at the unfavorable surface. As the system tries to lower its overall energy, molecules on the surface of a small particle (energetically unfavorable, with only 3 or 4 or 5 bonded neighbors) will tend to detach from the particle, as per the Kelvin equation, and diffuse into the solution. When all small particles do this, it increases the concentration of free atoms in solution. When the free atoms in solution are supersaturated, the free atoms have a tendency to condense on the surface of larger particles.[4]. Therefore, all smaller particles shrink, while larger particles grow, and overall the average size will increase. After an infinite amount of time, the entire population of particles will have become one, huge, spherical particle to minimize the total surface area.\n\nIn 1961, Lifshitz and Slyozov [5] performed a mathematical investigation of Ostwald ripening in the case where diffusion of material is the slowest process. The derivation first states how a single particle grows in a solution, and this equation describes where the boundary is between small, shrinking particles and large, growing particles. Through a lengthy and abstract mathematical derivation, they conclude that the average radius of the particles <R>, grows as follows:\n\n$\\langle R \\rangle ^3 - \\langle R \\rangle _0 ^3 = \\frac {8 \\gamma c_{\\infty}v^2D} {9R_g T} t$\n\nwhere\n\n$\\langle R \\rangle$ = average radius of all the particles\n\u03b3 = particle surface tension or surface energy\n$c_{\\infty}$ = solubility of the particle material\nv = molar volume of the particle material\nD = diffusion coefficient of the particle material\nRg = ideal gas constant\nT = absolute temperature\n\nand\n\nt = time\n\nTwo things should be noted about this growth law: (a) The quantity $\\langle R \\rangle ^3$ is different from $\\langle R^3 \\rangle$, and only the latter one can be used to calculate average volume, and (b) the statement that $\\langle R \\rangle$ goes as t1 \/ 3 relies on $\\langle R \\rangle _0$ being zero; but because nucleation is a separate process from growth, this places $\\langle R \\rangle _0$ outside the bounds of validity of the equation. In contexts where the actual value of $\\langle R \\rangle _0$ is irrelevant, an approach that respects the meanings of all terms is to take the time derivative of the equation to eliminate $\\langle R \\rangle _0$ and t. Another such approach is to change the $\\langle R \\rangle _0$ to $\\langle R \\rangle _i$ with the initial time i having a positive value.\n\nAlso contained in the Lifshitz and Slyozov derivation is an equation for the size distribution function f(R,t) of particles. For convenience, the radius of particles is divided by the average radius to form a new variable, $\\rho = R \/ \\langle R \\rangle$:\n\n$f(R,t) = \\frac {4}{9} \\left(\\frac {3}{3+\\rho}\\right)^\\frac {7}{3} \\left(\\frac {1.5} {1.5 - \\rho}\\right)^\\frac {11}{3} \\exp \\left(- \\frac {1.5}{1.5 - \\rho}\\right) \\rho < 1.5$\n\nIronically, at the same time as Lifshitz and Slyozov published their findings, Carl Wagner performed his own mathematical investigation of Ostwald ripening [6], examining both systems where diffusion was slow and also where attachment and detachment at the particle surface was slow. Although his calculations and approach were different, Wagner made exactly the same conclusions as Lifshitz and Slyozov for slow-diffusion systems. This duplicate derivation went unnoticed for years because the two scientific papers were published on opposite sides of the Iron Curtain in 1961.[citation needed] It was not until 1975 that Kahlweit addressed the fact that the theories were identical [7] and combined them into the Lifshitz-Slyozov-Wagner or LSW Theory of Ostwald ripening. Many experiments and simulations have proven LSW theory to be robust and accurate. Even some systems that undergo spinodal decomposition have been shown to quantitatively obey LSW theory after initial stages of growth [8].\n\nFor the curious, Wagner derived that when attachment and detachment of molecules is slower than diffusion, then the growth rate becomes\n\n$\\langle R \\rangle ^2 = \\frac {64 \\gamma c_{\\infty} v^2 k_s} {81 R_g T} t$\n\nwhere ks is the reaction rate constant of attachment with units of length per time. Since the average radius is usually something that can be measured in experiments, it is fairly easy to tell if a system is obeying the slow-diffusion equation or the slow-attachment equation. If, of course, the experimental data obeys neither equation, then it is likely that another mechanism is taking place and Ostwald ripening is not occurring.\n\nAlthough LSW theory and Ostwald ripening were intended for solids ripening in a fluid, Ostwald ripening is also observed in liquid-liquid systems. For example, in an oil-in-water emulsion polymerization,[3] In this case, Ostwald ripening causes the diffusion of monomers (i.e. individual molecules or atoms) from smaller droplets to larger droplets due to greater solubility of the single monomer molecules in the larger monomer droplets. The rate of this diffusion process is linked to the solubility of the monomer in the continuous (water) phase of the emulsion. This can lead to the destabilization of emulsions (for example, by creaming and sedimentation).[9]\n\n## Specific examples\n\nOil droplets in pastis mixed with water grow by Ostwald ripening.\n\nAn everyday example of Ostwald ripening is the re-crystallization of water within ice cream which gives old ice cream a gritty, crunchy texture. Larger ice crystals grow at the expense of smaller ones within the ice cream, thereby creating a coarser texture.[10]\n\nAnother gastronomical example is in the ouzo effect, where the droplets in the cloudy microemulsion grow by Ostwald ripening.\n\nIn geology, it is the textural coarsening, aging or growth of phenocrysts and crystals in solid rock which is below the solidus temperature. It is often ascribed as a process in the formation of orthoclase megacrysts,[11] as an alternative to the physical processes governing crystal growth from nucleation and growth rate thermochemical limitations.\n\nIn chemistry, the term refers to the growth of larger crystals from those of smaller size which have a higher solubility than the larger ones. In the process, many small crystals formed initially slowly disappear, except for a few that grow larger, at the expense of the small crystals. The smaller crystals act as fuel for the growth of bigger crystals. Limiting Ostwald ripening is fundamental in modern technology for the solution synthesis of quantum dots.[12] Ostwald ripening is also the key process in the digestion of precipitates, an important step in gravimetric analysis. The digested precipitate is generally purer, and easier to wash and filter.\n\nOstwald ripening can also occur in emulsion systems, with molecules diffusing from small droplets to large ones through the continuous phase. When a miniemulsion is desired, an extremely hydrophobic compound is added to stop this process from taking place.\n\n## References\n\n1. ^ IUPAC, Compendium of Chemical Terminology, 2nd ed. (the \"Gold Book\") (1997). Online corrected version: \u00a0(2006\u2013) \"Ostwald ripening\".\n2. ^ W. Ostwald. 1896. Lehrbuch der Allgemeinen Chemie, vol. 2, part 1. Leipzig, Germany.\n3. ^ a b Hubbard, Arthur T. (2004). Encyclopedia of Surface and Colloid Science. CRC Press. p.\u00a04230. ISBN\u00a00824707591. Retrieved 2007-11-13.\n4. ^ a b Ratke, Lorenz; Voorhees, Peter W. (2002). Growth and Coarsening: Ostwald Ripening in Material Processing. Springer. pp.\u00a0117\u2013118. ISBN\u00a03540425632. Retrieved 2007-11-15.\n5. ^ I.M. Lifshitz, V.V. Slyozov (1961). \"The Kinetics of Precipitation from Supersaturated Solid Solutions\". Journal of Physics and Chemistry of Solids 19 (1\u20132): 35\u201350. Bibcode 1961JPCS...19...35L. doi:10.1016\/0022-3697(61)90054-3.\n6. ^ C. Wagner (1961). \"THEORIE DER ALTERUNG VON NIEDERSCHLAGEN DURCH UMLOSEN (OSTWALD-REIFUNG)\". Zeitschrift fur Elektrochemie 65 (7): 581\u2013591.\n7. ^ M. Kahlweit (1975). \"Ostwald Ripening of Precipitates\". Advances in Colloid and Interface Science 5 (1): 1\u201335. doi:10.1016\/0001-8686(75)85001-9.\n8. ^ N. Vladimirova, A. Malagoli, R. Mauri (1998). \"Diffusion-driven phase separation of deeply quenched mixtures\". Physical Review E 58 (6): 7691\u20137699. Bibcode 1998PhRvE..58.7691V. doi:10.1103\/PhysRevE.58.7691.\n9. ^ Branen, Alfred Larry (2002). Food Additives. CRC Press. p.\u00a0724. ISBN\u00a00824793439. Retrieved 2007-11-15.\n10. ^ Clark, Chris (2004). The Science of Ice Cream. Royal Society of Chemistry. pp.\u00a078\u201379. ISBN\u00a00854046291. Retrieved 2007-11-13.\n11. ^ Mock, A. (2003). \"Using Quantitative Textural Analysis to Understand the Emplacement of Shallow-Level Rhyolitic Laccoliths\u2014a Case Study from the Halle Volcanic Complex, Germany\". Journal of Petrology 44 (5): Pp. 833\u2013849. doi:10.1093\/petrology\/44.5.833. Retrieved 2007-11-14.\n12. ^ Vengrenovich, R.D.; Gudyma, Yu. V.; Yarema, S. V. (December 2001). \"Ostwald ripening of quantum-dot nanostructures\". Semiconductors 35 (12): pp.1378\u20131382. Bibcode 2001Semic..35.1378V. doi:10.1134\/1.1427975. Retrieved 2007-11-14.\n\nWikimedia Foundation. 2010.\n\n### Look at other dictionaries:\n\n\u2022 Ostwald \u2014 may refer to: Friedrich Wilhelm Ostwald, the physico chemist (Nobel Prize of Chemistry, 1909) The Ostwald s rule related to the polymorphism: the least stable phase is first formed The Ostwald Process, a synthesis method for making nitric acid\u2026 \u2026 \u00a0 Wikipedia\n\n\u2022 Ostwald\u2013Freundlich equation \u2014 See also: Surface tension#Thermodynamics The Gibbs\u2013Thomson effect (also called the Gibbs\u2013Kelvin effect or Kelvin effect) relates surface curvature to vapor pressure and chemical potential and is a consequence of surface tension. It is named after \u2026 \u00a0 Wikipedia\n\n\u2022 Sintering \u2014 is a method used to create objects from powders. It is based on atomic diffusion. Diffusion occurs in any material above absolute zero but it occurs much faster at higher temperatures. In most sintering processes the powdered material is held in\u2026 \u2026 \u00a0 Wikipedia\n\n\u2022 Ouzo effect \u2014 The ouzo effect during the preparation of absinthe. The ouzo effect (also louche effect and spontaneous emulsification) is a phenomenon observed when water is added to ouzo and other anise flavored liqueurs and spirits, such as pastis, raki, arak \u2026 \u00a0 Wikipedia\n\n\u2022 Flocculation \u2014 Flocculent redirects here. For the galaxy type, see Flocculent Spiral Galaxy. Flocculation, in the field of chemistry, is a process wherein colloids come out of suspension in the form of floc or flakes by the addition of a clarifying agent. The\u2026 \u2026 \u00a0 Wikipedia\n\n\u2022 Gibbs-Thomson effect \u2014 The Gibbs Thomson effect (also called the Gibbs Kelvin effect or Kelvin effect) relates surface curvature to vapor pressure and chemical potential. It is named after Josiah Willard Gibbs and William Thomson, 1st Baron Kelvin. (It is not to be\u2026 \u2026 \u00a0 Wikipedia\n\n\u2022 \u041e\u0441\u0442\u0432\u0430\u043b\u044c\u0434\u043e\u0432\u0441\u043a\u043e\u0435 \u0441\u043e\u0437\u0440\u0435\u0432\u0430\u043d\u0438\u0435 \u2014 \u041f\u0435\u0440\u0435\u043a\u043e\u043d\u0434\u0435\u043d\u0441\u0430\u0446\u0438\u044f (\u0432 \u0438\u043d\u043e\u0441\u0442\u0440\u0430\u043d\u043d\u043e\u0439 \u043b\u0438\u0442\u0435\u0440\u0430\u0442\u0443\u0440\u0435: \u043a\u043e\u0430\u043b\u0435\u0441\u0446\u0435\u043d\u0446\u0438\u044f coalescence, \u041e\u0441\u0442\u0432\u0430\u043b\u044c\u0434\u043e\u0432\u0441\u043a\u043e\u0435 \u0441\u043e\u0437\u0440\u0435\u0432\u0430\u043d\u0438\u0435 Ostwald ripening[1]) \u043f\u0440\u043e\u0446\u0435\u0441\u0441 \u043a\u043e\u043d\u0434\u0435\u043d\u0441\u0430\u0446\u0438\u0438 \u043f\u0435\u0440\u0435\u0441\u044b\u0449\u0435\u043d\u043d\u043e\u0439 \u0444\u0430\u0437\u044b \u0432\u0435\u0449\u0435\u0441\u0442\u0432\u0430 \u043d\u0430 \u043f\u043e\u0437\u0434\u043d\u0438\u0445 \u0432\u0440\u0435\u043c\u0435\u043d\u0430\u0445 \u0440\u0430\u0437\u0432\u0438\u0442\u0438\u044f, \u043a\u043e\u0433\u0434\u0430 \u0440\u043e\u0441\u0442 \u043a\u0440\u0443\u043f\u043d\u044b\u0445 \u0437\u0451\u0440\u0435\u043d \u043d\u043e\u0432\u043e\u0439 \u0444\u0430\u0437\u044b (\u043d\u0430\u043f\u0440\u0438\u043c\u0435\u0440, \u043a\u0430\u043f\u0435\u043b\u044c \u2026 \u00a0 \u0412\u0438\u043a\u0438\u043f\u0435\u0434\u0438\u044f\n\n\u2022 Cathedral-Peak-Granodiorit \u2014 Der Matterhorn Peak besteht aus Cathedral Peak Granodiorit Der Cathedral Peak Granodiorit wurde nach seiner Typlokalit\u00e4t, dem im US amerikanischen Yosemite Nationalpark gelegenen Cathedral Peak benannt. Das Granodioritmassiv ist ein Bestandteil\u2026 \u2026 \u00a0 Deutsch Wikipedia\n\n\u2022 Precipitation (chemistry) \u2014 This article is about the chemical phenomemon. For other uses, see Precipitation. Precipitate redirects here. For the album by Interpol, see Precipitate EP. Chemical Precipitation Precipitation is the formation of a solid in a solution or inside\u2026 \u2026 \u00a0 Wikipedia\n\n\u2022 Polymer-bonded explosive \u2014 A polymer bonded explosive, also called PBX or plastic bonded explosive, is an explosive material in which explosive powder is bound together in a matrix using small quantities (typically 5\u201310% by weight) of a synthetic polymer ( plastic ). Note\u2026 \u2026 \u00a0 Wikipedia","date":"2020-02-21 07:09:26","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 15, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8114835023880005, \"perplexity\": 4538.251100463851}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2020-10\/segments\/1581875145443.63\/warc\/CC-MAIN-20200221045555-20200221075555-00090.warc.gz\"}"} | null | null |
Health officers
The Medical Heritage Library
JSTOR Early Journal Content
JSTOR Early Journal Content, The Journal of Infectious Diseases
Augustus C. Long Health Sciences Library
macnutt, j. scott
macnutt, j. scott (joseph scott), b. 1885
London School of Hygiene & Tropical Medicine Library & Archives Service
A manual for health officers, [electronic resource]
1915 by MacNutt, J. Scott (Joseph Scott), b. 1885; London School of Hygiene and Tropical Medicine; London School of Hygiene and Tropical Medicine
Contains bibliographies
Topics: Health officers, Public health
A manual for health officers
1915 by MacNutt, J. Scott (Joseph Scott), b. 1885
Source: http://books.google.com/books?id=HXUUppzYkKYC&oe=UTF-8
1915 by MacNutt, J. Scott
Topics: Hygiene, cbk
Vol 11: Is Typhoid Fever a "Rural" Disease?
Sep 1, 1912 by MacNutt, J. Scott
"Is Typhoid Fever a "Rural" Disease?" is an article from The Journal of Infectious Diseases, Volume 11 . View more articles from The Journal of Infectious Diseases . View this article on JSTOR . View this article's JSTOR metadata . You may also retrieve all of this items metadata in JSON at the following URL: https://archive.org/metadata/jstor-30046864
Source: http://www.jstor.org/stable/10.2307/30046864
Vol 7: On the Mills-Reincke Phenomenon and Hazen's Theorem concerning the Decrease in Mortality from Diseases Other than Typhoid Fever following the Purification of Public Water-Supplies
Aug 24, 1910 by Macnutt, J. Scott
"On the Mills-Reincke Phenomenon and Hazen's Theorem concerning the Decrease in Mortality from Diseases Other than Typhoid Fever following the Purification of Public Water-Supplies" is an article from The Journal of Infectious Diseases, Volume 7 . View more articles from The Journal of Infectious Diseases . View this article on JSTOR . View this article's JSTOR metadata . You may also retrieve all of this items metadata in JSON at the following URL:... | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,480 |
#ifndef FMOD_EXAMPLES_COMMON_H
#define FMOD_EXAMPLES_COMMON_H
#include "common_platform.h"
#include "fmod.h"
#include <stddef.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdarg.h>
#include <stdio.h>
#include <assert.h>
#define NUM_COLUMNS 50
#define NUM_ROWS 25
#ifndef Common_Sin
#define Common_Sin sin
#endif
#ifndef Common_snprintf
#define Common_snprintf snprintf
#endif
#ifndef Common_vsnprintf
#define Common_vsnprintf vsnprintf
#endif
enum Common_Button
{
BTN_ACTION1,
BTN_ACTION2,
BTN_ACTION3,
BTN_ACTION4,
BTN_LEFT,
BTN_RIGHT,
BTN_UP,
BTN_DOWN,
BTN_MORE,
BTN_QUIT
};
/* Cross platform functions (common) */
void Common_Fatal(const char *format, ...);
void Common_Draw(const char *format, ...);
void ERRCHECK(FMOD_RESULT result);
/* Functions with platform specific implementation (common_platform) */
void Common_Init(void **extraDriverData);
void Common_Close();
void Common_Update();
void Common_Sleep(unsigned int ms);
void Common_Exit(int returnCode);
void Common_DrawText(const char *text);
void Common_LoadFileMemory(const char *name, void **buff, int *length);
void Common_UnloadFileMemory(void *buff);
void Common_Format(char *buffer, int bufferSize, const char *formatString...);
bool Common_BtnPress(Common_Button btn);
bool Common_BtnDown(Common_Button btn);
const char *Common_BtnStr(Common_Button btn);
const char *Common_MediaPath(const char *fileName);
const char *Common_WritePath(const char *fileName);
void Common_Mutex_Create(Common_Mutex *mutex);
void Common_Mutex_Destroy(Common_Mutex *mutex);
void Common_Mutex_Enter(Common_Mutex *mutex);
void Common_Mutex_Leave(Common_Mutex *mutex);
#endif
| {
"redpajama_set_name": "RedPajamaGithub"
} | 6,887 |
the medium is the messiah
Apocalyptabuse, or How to Survive "The End"
By S. Brent Plate | May 6, 2020
Call this fantasized thinking apocalyptabuse: the demoralizing mythic-psychic warfare that deprives people of hope, makes us fear that The End is near, and thereby cuts off our aspirations of any earthly life to come.
It's the End and Nothing Feels Fine
By Kelly J. Baker | April 1, 2020
I don't use the word "apocalypse" or "apocalyptic" lightly. I'm a scholar of bad endings. And the pandemic that we face right now feels like it could be a very bad ending.
Transit of Shadow: On Eclipses
By Ed Simon | August 21, 2017
Totally predictable, yet totally magical.
KtB Wants You…To Write About the Eclipse
By Brook Wilensky-Lanford | August 3, 2017
The total solar eclipse passing over the continental U.S. this August 21st is the first to do so in 99 years, but American eclipse stories go back much farther than that. It was a "Dark Day" in the spring of 1780 that apparently convinced Shaker leader Mother Ann Lee to "open" her revelation to new…
Goodbye, Tim LaHaye
By Kelly J. Baker | July 28, 2016
When I first encountered Tim LaHaye and Jerry Jenkins' Left Behind (1995), I was sitting on the a sage-colored corduroy coach at my mom's house. Left Behind rested on a nearby table. It was 1998. I was in my first year of college and already determined to take courses on world religions. My determination proved…
Zombies and Guns
By Kelly J. Baker | October 27, 2015
When did a movie monster become a reason for purchasing weapons? An excerpt from The Zombies Are Coming.
Look For The Signs
By Kelly J. Baker | May 1, 2015
The Messengers is another supernatural mystery from a network, the CW, known for vampires, zombies, demons, angels, and other monsters that go bump in the night. I watched the first two episodes with the hope that maybe this show would mimic Supernatural, my long-time favorite that emphasizes moral ambiguity, the peril of good intentions, and…
Kingsman: The One-Percent Apocalypse
By Kelly J. Baker | February 26, 2015
I went to watch Kingsman: The Secret Service the day after Valentine's Day. My husband and I were eager to see a movie, any movie really, that didn't involve talking animals. The choices were limited for those us who refuse to engage Fifty Shades of Grey, so we settled upon Kingsman because it was the…
Breaking: God is Not a Wizard
By Mary Valle | October 29, 2014
Pope Francis recently said that, regarding the Big Bang, there's no conflict behind that concept and faith in God. He went on to say "when we read about creation in Genesis, we run the risk of imagining God was a magician, with a magic wand able to do everything. But that is not so."…
Apocalypse, Later
By Brook Wilensky-Lanford | May 21, 2014
We don't always celebrate the anniversaries of non-events, but here at KtB we thought we'd call your attention to the fact that we have now survived Harold Camping's non-end-of-the-world, predicted for May 21, 2011, for exactly three years. Congratulations! The infamously inaccurate Family Radio pastor's apocalypse was fascinating for Camping's utter conviction in its reality,…
It's All Fun and Games Until Odin Gets Eaten by a Wolf
By Eric Scott | February 21, 2014
You may have seen the links going around. Certainly I have, though I guess I'm an obvious target. "Vikings of the World Unite: The Apocalypse is Upon Us." "Viking apocalypse: End of the world predicted to happen on Saturday (but don't cancel your weekend plans yet.)" Word on the street is that Ragnarök is scheduled…
Occupy, from the Grave
By Atchu Novad | September 18, 2013
A few weeks ago, Atchu was announced dead. disturbing to even mention someone's passing, so afraid we are of the Reaper. like Obi-Wan Kenobi, Obama or any other who has crossed the river of death, i am dead as fuck. Mofos-who-don't-get-metaphors-or-art:: alex is fine:: his chubby body lingers on. Atchu on the other hand, committed…
Only the End of the World Again
By Eric Scott | December 21, 2012
I don't think there's much point in reiterating the genesis of this post; if you're reading Killing the Buddha, I'm guessing that you're exposed to the same media universe as the rest of us, and you're aware that a significant portion of the population believes a significant portion of the population believes the world is…
The End of Macha
By Dylan Harris | December 20, 2012
With the end of the world approaching, I will be snuggling up somewhere beneath a mosquito net, floating down a river through the mostly still-wild and mysterious Bolivian Amazon. I look forward to staring up at the stars, trying to piece together exactly what the Mayans had in mind (or didn't—they may have just gotten…
Apocalypse Week…Again
By Brook Wilensky-Lanford | December 18, 2012
Here at Killing the Buddha, we've been all over the apocalypse for a long time. The apocalypse is kind of like Christmas: it threatens to happen just about every year, or then maybe it does, or doesn't, and anyway life goes on. As the apparently Mormon-inspired sci-fi show Battlestar Galactica likes to say "All of…
The GOP attack on Dems Jerusalem platform isn't about the Jews…
By Jeff Sharlet | September 6, 2012
It's about Christians. In case you missed it, the Democratic Party dropped its longstanding insistence that Jerusalem is and will be the capital of Israel, despite the fact that most of the world doesn't recognize it as such. Why not? Oh, the fear of setting off a never-ending, apocalyptic war. But for some of us,…
By Nathaniel Page | November 22, 2011
An exercise in recreational destruction.
A Kinder, Gentler Apocalypse
Remember all the world-ending-on-May-21 hype earlier this year? Apparently God was playing more of his "I'm gonna pretend to high-five you, then pull my hand away at the last minute and say 'Psych!'" games with all of us. Harold Camping of Family Radio was admittedly "flabbergasted" when it didn't happen as planned. But lo! It has… | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 8,579 |
Slowloris is een denial of service-aanvalstool oorspronkelijk gemaakt door Robert "RSnake" Hansen (in Perl) waarmee een enkele machine de webserver van een andere machine onbereikbaar kan maken met erg weinig bandbreedte, dit in tegenstelling tot veel andere denial of service-aanvallen.
Slowloris probeert zo veel mogelijk (losse) HTTP-verbindingen met het doelwit open te houden gedurende de aanval. Dit wordt bereikt door verbindingen met de doelwebserver te openen en met regelmaat een gedeeltelijk HTTP-verzoek te verzenden. Betrokken servers houden hierdoor deze verbindingen open, en bereiken hierdoor op een gegeven moment hun limiet op lopende verbindingen. Dit resulteert in het niet langer beschikbaar zijn van deze webserver wanneer een legitieme gebruiker wil verbinden; de webserver is immers bezig met de vele verbindingen van de aanvaller.
Kwetsbare webservers
Deze aanval kan - volgens de auteur van de aanval - werken op de volgende webservers:
Apache 1.x en 2.x
dhttpd
Websense "blokpagina's" (niet bevestigd)
Trapeze Wireless Web Portal (niet bevestigd)
Verizon's MI424-WR FIOS kabelmodem (niet bevestigd)
Verizon's Motorola settopbox (poort 8082 en vereist autorisatie - niet bevestigd)
BeeWare WAF (niet bevestigd)
Deny All WAF (oudere versies waren kwetsbaar, nieuwe versies bevatten een patch om niet langer kwetsbaar te zijn voor deze aanval)
Flask
Referenties
Denial-of-service | {
"redpajama_set_name": "RedPajamaWikipedia"
} | 7,148 |
Tag Archives: indiana jones
"The Curse of Oak Island" is must see TV!
Graphic from: http://kittyinaz.files.wordpress.com/2013/05/jolly-roger-4.jpg
A curse. Pirates. A treasure. Booby traps.
It has all the trappings of the next "Indiana Jones" or "National Treasure" movie. The main difference? This is for real.
Oak Island, Nova Scotia. Three boys discover a pit ("the Money Pit") on and begin to dig. As they get deeper, strange artifacts begin to pop up. Flagstones. Wooden platforms. Small metal artifacts. And then, at around 27 metres (that's more than 80 feet), they find a stone cipher that when translated says, "Forty feet below, two million pounds are buried." The boys dig a little more, and then leave for the night. When they return, the shaft is underwater. They conclude a booby trap has been triggered by their digging, which flooded the shaft to protect the treasure from plunderers (oakislandtreasure).
In modern times, Dan Blankenship and his associates dig a shaft parallel to the Money Pit (called 10X), fortifying it with steel. He takes some video in which he insists he can see a body, a treasure chest, and other buried items. 10X eventually floods as well. Due to disagreements over land ownership, digging on the island ceases until brothers Rick and Marty Lagina buy a controlling stake in the island's tourism company and are able to resume excavations.
I saw the first episode of History's "The Curse of Oak Island" yesterday; I haven't been that excited since seeing "In Search of Noah's Ark" when I was a kid. The premiere episode explains the history of the Money Pit and 10X and documents the Lagina brothers' excavation of the pits as they search for the fabled treasure. Viewers get to see the Blankenship video and meet Dan Blankenship (now 80 and just as obsessed as ever) and his son who are active members of the Laginas' team. The first thing they do is send a camera into the Money Pit, but the footage comes back inconclusive and the files mysteriously disappear from the computer midway through the viewing. Perhaps this is part of the curse, they wonder.
Next, the team drills into the hole while one of the brothers searches the fill. They find bits of blue transfer ceramic, but not much else. Lastly, air is pumped into the shaft in an effort to remove the water. They jerry-rig a sediment holding tank and use a shovel to start mucking about, but turn up only a single metal artifact. Later, the team takes a boat ride to see the island from the water. They hypothesize the presence of five box-drains, used to draw sea water into the shafts and plan to dive at a later date to confirm or debunk their existence.
In "The Curse of Oak Island," the Laginas and the Blankenships document their real-life adventure as they search for pirate treasure. The curse promises that seven will die before the secret of the Money Pit is revealed, which only serves to bolster the excitement which makes this an hour of tv worth watching. What I like about "The Curse of Oak Island" is that, unlike other salvaging shows, the Laginas do things legally. They let us know the credentials of the team members, as well as the permits needed and the legalities and cost of the dig and that the task they have undertaken is dangerous, with the implicit message not to try this at home.
"The Curse of Oak Island" is not only exciting television, it's also responsible television. And that's good archaeology.
Did you see "The Curse of Oak Island"? What did you think?
This entry was posted in Uncategorized and tagged 10X, archaeology, blankenship, booby trap, curse, dig, history, indiana jones, lagina, money pit, national treasure, nova scotia, oak island, pirate, the curse of oak island, treasure on January 27, 2014 by eliseabram.
(Former) Archaeologist's Lament
Zero Hour, another adventure/thriller television series with archaeological roots was cancelled last week after only three episodes. The pilot episode saw Anthony Edwards (of ER fame) as the editor of a sceptics magazine whose wife is kidnapped after she purchases a historic clock. When he takes the clock apart, Edwards finds a diamond upon which a map has been etched. He follows the map to a buried German submarine in the arctic, where he is pursued by a man who was somehow genetically engineered by Nazi scientists.
Though all of this sounds spectacularly interesting as a series concept, the idea was poorly executed as it suffered from less than believable dialogue and unusual casting. In spite of this, though Zero Hour had potential, it was more than likely doomed by its affiliation with archaeology.
Movies with archaeological ties generally do well at the box office. Consider Stargate, Indiana Jones,Tomb Raider, The Mummy, and National Treasure. The same cannot be said for television shows of the same genre which are few and far between. Two of these are Veritas: the Quest, and the British Bonekickers. In Veritas, a team of people search the globe for artifacts that piece together Earth's great mystery, though what that may be is not revealed in the show's short run. All that is known is that it somehow involves the group's leader, and his son and deceased wife. In Bonekickers, archaeologists participate in episodic digs, some with ties to popular legends or high profile historical eras. The only other archaeology-types around are those on Bones, and that's more forensic anthropology than archaeology per se. I dreamed of seeing the Primeval cast hunker down to an archaeological dig when they found modern artifacts in a dinosaurian era, or modern people digging up the remains of the Terra Nova settlement (though that apparently took place in a different timeline than ours) but, alas, that was never to come to pass.
Many play fast and loose with the term "archaeology", such as in "Antique Archaeology", the shop ran by the American Pickers, for example. Even worse is the Savage Family Diggers/ American Diggers franchise which sees ex-wrestler Rick Savage knock on people's doors asking to dig on their properties for a percentage of the profit. While they may "save" artifacts from being destroyed or remaining forever buried and decomposing, they are, in effect, destroying archaeological sites. And while I readily acknowledge that laws in The States differ from those in Canada, the fact that they do they painstaking research to find the sites then do nothing to save the subtleties of the sites' historic occupation, does little to elevate them from pot-hunting status. Yet they have been awarded their own series of shows which creates the illusion that what they are doing is lucrative and not at all deplete of morals.
Archaeology as a discipline is in danger of extinction. Even when I practiced it, the threat of satellite imagery and ground penetrating radar to document sites threatened to render those of us who saw it as a noble pursuit, obsolete. In his article entitled "Archaeology Is Not a Strong Brand", Martin Rundkvist takes the profusion of available archaeology-named domains to indicate that the word no longer packs significant punch. He avers that the "little regional bits of the past and archaeological practice" have rendered the word, and the discipline by default, unexciting. I maintain the reason for this could be the dearth of local archaeological projects in North America (certainly in central Ontario). I got out of the discipline because, though I loved it dearly and could imagine doing nothing else with my life, I could not make a living at it. I began my career making enough money to live comfortably, had the position remained opened twelve months of the year. Each year I returned to the field being offered fewer and fewer dollars per hour until I was earning little more than minimum wage 6 months a year (if I were lucky) and UIC was breathing down my back to get re-trained in order to dump my hard-earned degree and get a year-round office job. I chose, instead, to go back to school and complete teacher training. It took some time, but I have come to terms with perpetuating archaeology through my writing. I always fancied returning to the discipline in retirement, but I doubt I will be able to tote buckets of wet dirt at that advanced age. No, I must remain content with fanaticizing about fantastical archaeology, rather than practicing actual archaeology, barring my winning the lottery, that is.
Rundkvist, Martin. Archaeology Is Not a Strong Brand. Aardvarchaeology. 2 Mar 2013. < http://scienceblogs.com/aardvarchaeology/2013/03/02/archaeology-is-not-a-strong-brand/>. 12 Mar 13.
This entry was posted in Uncategorized and tagged american diggers, archaeology, bonekickers, indiana jones, national treasure, ontario, primeval, savage family diggers, stargate, the mummy, tomb raider, veritas, zero hour on March 12, 2013 by eliseabram. | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 7,672 |
of the gardens. And, there's Nelly waiting to greet us!
never stops Nelly-kitty from being super friendly, or from joining us on our tour!
Like preferring the Front of the greenhouse to the back. It's so pretty!
Especially since it looks out Over the gardens!
times now. And I'm still impressed Every time!
curly little Blooms... but I must have one!
This is another Must have... Just look at those leaves!
hope I live long enough to see our Arbor this over-grown!
Roses as far as the Eye can see!
Every color is present and Accounted for!
And this is a Favorite of mine!
This one too - Just look at all these roses!
roof in one other place and that's Carmel-by-the-sea. My favorite!
Was this always the Rose garden?
This is some fabulously colored Corner, huh?
This grass looks like Fireworks!
I wonder how many Times a day she makes these rounds?
Any way you look at it... This is a gorgeous Garden!
Especially looking at it from here!
What a perfect Blossom. See the little Bee there in the middle, on the left?
trees and I love every one of them!
We need a Greenhouse at home!
I can't imagine a lovelier way to spend this Saturday morning!
How beautiful! Wow...I'd love to spend the afternoon there! | {
"redpajama_set_name": "RedPajamaC4"
} | 5,272 |
{"url":"https:\/\/tvtropes.org\/pmwiki\/article_history.php?article=Main.CensorshipTropes","text":"# History Main \/ CensorshipTropes\n\nGo To\n\n* {{Bowdlerise}}: Has some elements in common with:\n** EditedForSyndication: Sometimes used to remove content, but mostly used to cut scenes for more commercials or to cut out scenes that are either too long or are superfluous to the plot [[note]](if it's the rerun of a sketch comedy show, like ''Series\/SaturdayNightLive'', ''SCTV'', ''Fridays'', or ''Series\/{{MADtv}}'', then sketches are usually edited to fit a 30-minute or 60-minute time frame, for having too many technical errors, [[ClumsyCopyrightCensorship to get rid of copyrighted material]], or because the sketch was too controversial)[[\/note]].\n** TooSoon: Bowdlerization due to [[FunnyAneurysmMoment Funny Aneurysm]] or HarsherInHindsight moments of the \"life imitates art\" variety.\n** ClumsyCopyrightCensorship: Removing or covering up product placement and\/or copyrighted music or footage from reruns, reairings, and when put on different media (i.e. DVD and online streaming).\n** MissingEpisode and BannedEpisode (though censorship isn't the only reason a TV show episode is pulled. Other reasons include: TroubledProduction issues, writers' or actors' strikes, budget cuts, CreatorBacklash, CreatorBreakdown, or the usual 'the show got canceled early and a lot of the planned episodes were never finished').\n*** BannedInChina: Also has elements in common with MissingEpisode and BannedEpisode, as some international versions of American TV shows have been banned due to ValuesDissonance.\n\nto:\n\n* BannedEpisode: When an episode of a TV series (or other installment of a media franchise) is prohibited from being shown again.\n* BannedInChina: Media that has been banned from an entire country, usually due to the national government's censorship laws.\n* {{Bowdlerise}}: Has some elements in common with:\n** EditedForSyndication: Sometimes used to remove content, but mostly used to cut scenes for\nGeneral toning-down of content considered obscene into something more commercials or to cut out scenes that are either too long or are superfluous to the plot [[note]](if it's the rerun of a sketch comedy show, like ''Series\/SaturdayNightLive'', ''SCTV'', ''Fridays'', or ''Series\/{{MADtv}}'', then sketches are usually edited to fit a 30-minute or 60-minute time frame, for having too many technical errors, [[ClumsyCopyrightCensorship to get rid of copyrighted material]], or because the sketch was too controversial)[[\/note]].\npalatable.\n** TooSoon: Bowdlerization due to [[FunnyAneurysmMoment Funny Aneurysm]] or HarsherInHindsight moments of the \"life imitates art\" variety.\n** ClumsyCopyrightCensorship: Removing or covering up product placement and\/or copyrighted music or footage from reruns, reairings, and when put on different media (i.e. DVD and online streaming).\n** MissingEpisode and BannedEpisode (though censorship isn't the only reason a TV show episode is pulled. Other reasons include: TroubledProduction issues, writers' or actors' strikes, budget cuts, CreatorBacklash, CreatorBreakdown, or the usual 'the show got canceled early and a lot of the planned episodes were never finished').\n*** BannedInChina: Also has elements in common with MissingEpisode and BannedEpisode, as some international versions of American TV shows have been banned due to ValuesDissonance.\n\"LifeImitatesArt\" variety.\n\n* ContentWarnings: Mostly played straight; mostly PlayedForLaughs, depending on context\n\nto:\n\n* ClumsyCopyrightCensorship: Removing or covering up ProductPlacement and\/or {{UsefulNotes\/copyright}}ed music or footage from reruns, reairings, and when put on different media (i.e. DVD and online streaming).\n* ContentWarnings: Mostly played straight; mostly sometimes PlayedForLaughs, depending on contextcontext.\n\n** GettingCrapPastTheRadar: DefyingTheCensors on an underhanded level. Leads to CrossesTheLineTwice or RefugeInAudacity if done with explicit content or if done in excess.\n\n** GettingCrapPastTheRadar: DefyingTheCensors on an underhanded level. Leads to CrossesTheLineTwice or RefugeInAudacity if done with explicit content or if done in excess.\n* EditedForSyndication: Sometimes used to remove content, but mostly used to cut scenes for more commercials or to cut out scenes that are either too long or are superfluous to the plot [[note]](if it's the rerun of a sketch comedy show, like ''Series\/SaturdayNightLive'', ''SCTV'', ''Fridays'', or ''Series\/{{MADtv}}'', then sketches are usually edited to fit a 30-minute or 60-minute time frame, for having too many technical errors, [[ClumsyCopyrightCensorship to get rid of copyrighted material]], or because the sketch was too controversial)[[\/note]].\n\n* PoliticallyCorrectHistory\n\nto:\n\n* PoliticallyCorrectHistory** PoliticallyCorrectHistory: Distortion of historical truth by removing or downplaying its more [[ValuesDissonance unpleasant aspects]].\n\nCompare TheseTropesShouldWatchTheirLanguage.\n\nto:\n\n* AbridgedForChildren\n\n* DroppedAbridgedOnIt\n\n* PreAprovedSermon\n\nto:\n\n* PreAprovedSermonPreApprovedSermon\n\n* PreaPprovedSermon\n\nto:\n\n* PreaPprovedSermonPreAprovedSermon\n\n* PreaPprovedSermon\n\n* Creator\/NintendoTreehouse\n\nto:\n\n* Creator\/NintendoTreehouse\n\nto:\n\n* Creator\/NintendoTreehouse\n\n* MusicIsPolitics\n\n* ConvenientlyInterruptedDocument\n\n* GosKino\n\nto:\n\n* GosKinoUsefulNotes\/GosKino\n\n** OrwellianRetcon\n\n** OrwellianRetcon\n\n->''\"'''WARNING:''' The following program is a realish documentary, and may contain language which is vulgar, offensive, or grammatically awkward. Such language has been censored whenever possible, but for the sake of higher ratings, we may have let a few gratuitous and especially titillating instances slide.\"''\\\\\n--The opening of the \"[[SomethingCompletelyDifferent American Dicks]]\" episode of ''WesternAnimation\/{{Duckman}}''\n\n->''\"'''Chapter 12'''\\\\\nThe German censors -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- [[TakeThat Idiots]] -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --\"''\\\\\n--Heinrich Heine, ''Ideas: The Book of Le Grand''\n\nInteresting tropes about [[spoiler:[[CensoredForComedy blank]]]]ing out things not meant for corrupting sensitive eyes or ears. This index is not about the [[Administrivia.TheContentPolicyAndThe5PCircuit Content Policy]] of Wiki\/TVTropes itself.\n\nto:\n\n->''\"'''WARNING:''' The following program is a realish documentary, and may contain language which is vulgar, offensive, or grammatically awkward. Such language has been censored whenever possible, but for the sake of higher ratings, we may have let a few gratuitous and especially titillating instances slide.\"''\\\\\n--The opening of the\n\"''\n-->-- ''WesternAnimation\/{{Duckman}}'',\n\"[[SomethingCompletelyDifferent American Dicks]]\" episode of ''WesternAnimation\/{{Duckman}}''\n\n->''\"'''Chapter 12'''\\\\\nThe German censors -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- [[TakeThat Idiots]] -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --\"''\\\\\n--Heinrich Heine, ''Ideas: The Book of Le Grand''\n\nopening.\n\nInteresting tropes about [[spoiler:[[CensoredForComedy blank]]]]ing [[spoiler:blank]][[CensoredForComedy ing]] out things not meant for corrupting sensitive eyes or ears. This index is not about the [[Administrivia.TheContentPolicyAndThe5PCircuit Content Policy]] of Wiki\/TVTropes itself.\n\n* {{Gag Censor}}: any visual censorship technique PlayedForLaughs\n\nto:\n\n* {{Gag Censor}}: GagCensor: any visual censorship technique PlayedForLaughs\nPlayedForLaughs.\n\nShowing 15 edit(s) of 147","date":"2018-07-21 01:49:01","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9391177296638489, \"perplexity\": 9476.07675717954}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-30\/segments\/1531676592150.47\/warc\/CC-MAIN-20180721012433-20180721032433-00111.warc.gz\"}"} | null | null |
{"url":"https:\/\/www.enotes.com\/homework-help\/lim-x-gt-oo-e-x-x-4-evaluate-limit-using-782891","text":"# `lim_(x->oo) e^x\/x^4` Evaluate the limit, using L\u2019H\u00f4pital\u2019s Rule if necessary.\n\nkseddy123 | Certified Educator\n\nGiven to solve,\n\n`lim_(x->oo) e^x\/(x^4)`\n\nas `x->oo` then the ` e^x\/(x^4) =oo\/oo` form\n\nso upon applying the L 'Hopital rule we get the solution as follows,\n\nas for the general equation it is as follows\n\n`lim_(x->a) f(x)\/g(x) is = 0\/0` or `(+-oo)\/(+-oo)` then by using the L'Hopital Rule we get\u00a0 the solution with the\u00a0 below form.\n\n`lim_(x->a) (f'(x))\/(g'(x))`\n\nso , now evaluating\n\n`lim_(x->oo) (e^x)\/(x^4)`\n\n= `lim_(x->oo) ((e^x)')\/((x^4)')`\n\n= `lim_(x->oo) ((e^x))\/((4x^3))`\n\nagain `((e^x))\/((4x^3))` is of the form `oo\/oo` so , we can apply again L'Hopital Rule .\n\n=`lim_(x->oo) ((e^x)')\/((4x^3)')`\n\n=`lim_(x->oo) ((e^x))\/(((4*3)x^2))`\n\n=`lim_(x->oo) ((e^x))\/((12x^2))`\n\nagain\u00a0`((e^x))\/((12x^2))` is of the form `oo\/oo ` so , we can apply again L'Hopital Rule .\n\n=`lim_(x->oo) ((e^x)')\/((12x^2)')`\n\n=`lim_(x->oo) ((e^x))\/(((12*2)x))`\n\n= `lim_(x->oo) ((e^x))\/(((24)x))`\n\nagain\u00a0`((e^x))\/(((24)x))` is of the form `oo\/oo` so , we can apply again L'Hopital Rule .\n\n= `lim_(x->oo) ((e^x)')\/(((24)x)')`\n\n=`lim_(x->oo) ((e^x))\/(24)`\n\non plugging the value`x= oo` , we get\n\n=`((e^(oo)))\/(24)`\n\n=`oo`","date":"2017-10-18 11:30:16","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9577139019966125, \"perplexity\": 2686.3033184237843}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2017-43\/segments\/1508187822930.13\/warc\/CC-MAIN-20171018104813-20171018124813-00098.warc.gz\"}"} | null | null |
Q: Visual Studio update - 'could not load type' error editing tt file I updated to VS2019 v16.10 and since then, I get an error every time I type into my tt file (reversePoco).
From reading what little I can find, it may be related to an extension but I see no way to identify which. I didn't even have LanguageService installed, so I installed it but no change.
Has anyone else encountered this and found a solution?
Thanks.
A: I had the same issue. I solved on v. 16.11.13 disabling the extension named "Devart T4 Editor", which is apparently shipped with Visual Studio (Menu Extensions > Manage Extensions > Installed > Devart T4 Editor > Disable).
You will loss the syntax highlighting on the .TT files, but you can install other working extension like T4Language (I tried this one) or t4editor.
(Source)
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,700 |
Penn Center: A Little-known Haven of the Civil Rights Movement
by Carrie Tatro May 1, 2018
Penn Center, St. Helena Island, South Carolina, has served as a bastion of peace and a place of refuge since post-Civil War reconstruction through the civil rights movement of the 1960s. Today, it functions as an important cultural resource center. Screen shot from YouTube/HowStuffWorks
Nestled off the beaten path in the heart of the Gullah/Geechee Sea Islands of South Carolina's lowcountry, is an American treasure waiting to be discovered by lucky vacationers and passersby. Situated beneath ancient moss-laden live oaks and tucked between glimmering salt marshes and the Atlantic coast, Penn Center, located for 156 years on sleepy St. Helena Island in Beaufort County, may be the most important African-American historical landmark and educational-cultural center you've probably never heard of.
Penn and Reconstruction
Established six months before President Abraham Lincoln issued the Emancipation Proclamation and three years before the 13th amendment legally abolished slavery, the Penn School was founded in 1862 by Pennsylvania Quaker and Unitarian missionaries as a main component of the Port Royal Experiment, an early example of Reconstruction in progress, even as the Civil War raged into the spring of 1865. Financed from donations raised by abolitionists, Penn (named after Quaker activist William Penn) was the first school founded in a Confederate state for the sole purpose of educating ex-slaves.
What began in the living room of the abandoned Oaks Plantation before the first school house was built, eventually grew to become a 50-acre campus (the land was donated by freedman and future businessman, Hastings Gantt) with 19 now-historic buildings, including the York W. Bailey Museum, which showcases an archive of rare photographs of African-Americans as well as scarce artifacts related to Gullah/Geechee history and culture.
In 1862 when the U.S. Navy seized the Port Royal Sound from Confederate troops, wealthy plantation owners fled St. Helena and the surrounding Sea Islands, reluctantly abandoning their prized crop of world-renowned Sea Island cotton and liberating between 10,000 and 32,000 slaves, who suddenly found themselves free and autonomous. Northern abolitionists and humanitarians saw the need to educate the freed slaves in hopes that their efforts would become a model for helping formerly enslaved people gain citizenship.
The first classes were taught by white abolitionists Laura Towne and Ellen Murray and briefly by Charlotte Forten, who was the first northern African-American teacher at Penn. The earliest curriculum followed the New England, euro-centric model of "socialized" education that included reading, writing, arithmetic, geography, history and music.
In the early 1900s Rossa B. Cooley and Grace House, two other northern white women, revised the curriculum to follow Booker T. Washington's Hampton-Tuskegee model of industrial education. (Cooley, a photographer, documented the school in more than 3,000 photographs that now reside in the Penn School collection). Under the tutelage of Cooley and House, classical studies like algebra and Latin were eliminated and courses such as masonry, carpentry and the domestic arts were added. Although up to the end of World War II, the state of South Carolina required that African-Americans be educated only through the seventh grade, Penn provided schooling through the twelfth grade and offered adult education classes as well.
By the late 1940s, the population of St. Helena had dwindled significantly, as native islanders, young people in particular, moved from isolated Beaufort County to states in the North, or sought better employment opportunities in the larger cities of the South. In response, the Board of Trustees at Penn redefined the purpose of the school, and launched the Penn Community Service Center in 1948. As such, Penn Center became one of the few places in the Jim Crow South where interracial activist groups could convene overnight in integrated facilities without the threat of violence or legal consequences.
During the 1950s and '60s, under the direction of devout Quakers, Courtney and Elizabeth Siceloff, the Penn Center became a major, though somewhat secret, facilitator for civil rights activism and social justice, not just for South Carolina, but for the entire nation. In the 2014 book, "Penn Center: A History Preserved," authors Orville Burton and Wilbur Cross tell us that the Siceloffs listened to the islander's concerns and broke away from the condescending notion that the black community needed to be "taught" citizenship to become "civilized" and "Americanized." And that they came to understand "the Christian commitment and theological worldview of the southern African Americans before Martin Luther King, Jr. brought it to the attention of the world."
During the 1960s, Penn Center hosted numerous interracial human rights conferences with groups including the NAACP, the World Peace Foundation, the SRC, SCCHR, CORE, the Southern Christian Leadership Conference (SCLC), Student Non-Violent Coordinating Committee (SNCC) and the Peace Corps, to name a few. Because of its isolated location, these integrated groups could stealthily meet to organize and strategize in a clandestine setting, safely under the radar of local authorities, the public and the press.
Martin Luther King, Jr. (right) with famed nonviolence advocate Ira Sandperl, singer Joan Baez and civil rights leader Jesse Jackson (partially hidden) at the Penn Center in 1966.
Screen shot from YouTube/HowStuffWorks.
It was former Atlanta mayor Andrew Young, in his role as leader of the SCLC, who introduced Martin Luther King, Jr. to the serenity and security of the coastal backwater that was Penn Center. King and his lieutenants, other luminaries of the civil rights movement and countless unnamed activists met with the SCLC at Penn five times between 1964 and 1967. It became a bastion of peace and a place of refuge where King could unwind, breathe freely and express himself openly, saying things in front of groups at Penn that he couldn't say on the national stage. Folk singer Joan Baez, who attended a retreat in 1966, recalled in "Penn Center: A History Preserved," King saying that "he couldn't take the pressure anymore, that he just wanted to go back...and preach in his little church, and he was tired of being a leader."
King composed many of his speeches at Penn, including his "I Have a Dream" speech, which he wrote while staying in the Hastings Gantt cottage where he often retreated. At Penn, King was able to voice publicly his unpopular anti-Vietnam stance and his concerns for the 40 million Americans living in poverty which led to his strongly held belief that there was something intrinsically wrong with capitalism. It was at the Penn Center that King explained what threatened the Beloved Community, the "three basic evils in America: the evil of racism, the evil of excessive materialism, the evil of militarism," which he called the "inseparable triplets" that any movement would necessarily have to address in order to elicit change.
According to Burton and Cross, even though Penn Center was remote, it still faced opposition from some white people regarding King's visits.
Walter Mack, a future executive director of Penn, told the authors how the community worked together to keep King out of the public eye: "The record of him coming was kept secret, even from the local sheriff ... They wouldn't tell anybody. You never knew who would want to hurt Dr. King."
The Hastings Gantt cottage at Penn Center, where MLK wrote his iconic "I Have a Dream" speech.
Screen shot from YouTube/HowStuffWorks
Joseph McDomick, retired Magistrate of St. Helena Island, said, "Even when he came here, it had to be kept secret ... We couldn't notify any law enforcement people because we didn't know who would be in that little group that would be after doing him in."
In December 1967 King held his fifth and final meeting with the SCLC at Penn. They discussed the Poor People's Campaign, and King told the gathering, "I don't know if I'll see all of you before April, but I send you forth."
Four months later, on April 4, 1968, he was assassinated at the Lorraine Motel in Memphis, Tennessee, the day after he'd told striking sanitation workers, "We've got to give ourselves to this struggle until the end."
Late lowcountry author Pat Conroy, who attended high school in Beaufort, South Carolina and wrote lovingly and extensively about the Sea Islands in books like "The Water Is Wide" and "The Prince of Tides," and who'd met Martin Luther King, Jr., Julian Bond and other civil rights leaders at Penn, said in a 2010 speech on its campus, "I watched my whole country change because of meetings that had taken place at Penn Center."
In a Nov. 11, 2016 article published in The Hill, current executive director, Dr. Rodell Lawrence, wrote, "Most Americans came from somewhere else to this continent and Penn Center provides us with a direct link to the African origins of slaves that occupied America's southeastern seaboard. It is a window to a place in which many African Americans emerged from bondage, and set out on a new journey as free men and women. It is a place and a time to celebrate. Penn Center vividly embodies the American ideal of "liberty and justice for all" and in every sense is a true historic national monument."
To that end, President Barack Obama, by executive order, made a swath of Beaufort County, South Carolina, including the Penn Center, a National Historical Monument to the Reconstruction Era one week before he left office in January 2017.
Now That's Sobering
Unlike the more isolated Penn Center, which never faced racial violence or backlash from hate groups during Jim Crow segregation, Koinonia, an integrated collective farm and "social gospel" community founded in 1942 by two Baptist ministers in southwest Georgia, faced ongoing violence and harassment from the Ku Klux Klan. Both violence and a local boycott of the farm's products prompted Koinonia to sell its produce through a mail order business that continues to this day.
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How the Civil Rights Movement Worked | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,924 |
(function () {
// TODO: figure out how to handle dynamic configuration of LRSes
var session = null,
versions = TinCan.versions(),
i,
version,
doSendStatementSyncTest,
doGetStatementSyncTest,
doVoidStatementSyncTest,
doStateSyncTest,
doStateSyncContentTypeJSONTest,
doActivityProfileSyncTest,
doActivityProfileSyncContentTypeJSONTest,
NATIVE_CORS = false;
if (typeof XMLHttpRequest !== "undefined" && typeof (new XMLHttpRequest()).withCredentials !== "undefined") {
NATIVE_CORS = true;
}
QUnit.module(
"TinCan-sync No LRS",
{
setup: function () {
session = new TinCan ();
},
teardown: function () {
session = null;
}
}
);
test(
"tincan.sendStatement (prepared, sync)",
function () {
var preparedStmt,
sendResult,
//
// Cloud is lowercasing the mbox value so just use a lowercase one
// and make sure it is unique to prevent merging that was previously
// possible, but leaving the commented version as a marker for something
// that ought to be tested against a 1.0.0 spec
//
//actorMbox = "mailto:TinCanJS-test-TinCan+" + Date.now() + "@tincanapi.com";
actorMbox = "mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com";
preparedStmt = session.prepareStatement(
{
actor: {
mbox: actorMbox
},
verb: {
id: "http://adlnet.gov/expapi/verbs/attempted"
},
target: {
id: "http://tincanapi.com/TinCanJS/Test/TinCan_sendStatement/prepared-sync"
}
}
);
sendResult = session.sendStatement(preparedStmt);
ok(sendResult.hasOwnProperty("statement"), "sendResult has property: statement");
deepEqual(sendResult.statement, preparedStmt, "sendResult property value: statement");
ok(sendResult.hasOwnProperty("results"), "sendResult has property: results");
ok(sendResult.results.length === 0, "sendResult results property: length");
}
);
QUnit.module(
"TinCan-sync Single LRS",
{
setup: function () {
session = {};
for (i = 0; i < versions.length; i += 1) {
v = versions[i];
if (TinCanTestCfg.recordStores[v]) {
session[v] = new TinCan (
{
recordStores: [
TinCanTestCfg.recordStores[v]
]
}
);
}
}
},
teardown: function () {
session = null;
}
}
);
doSendStatementSyncTest = function (v) {
test(
"tincan.sendStatement (prepared, sync): " + v,
function () {
var preparedStmt,
sendResult,
//
// Cloud is lowercasing the mbox value so just use a lowercase one
// and make sure it is unique to prevent merging that was previously
// possible, but leaving the commented version as a marker for something
// that ought to be tested against a 1.0.0 spec
//
//actorMbox = "mailto:TinCanJS-test-TinCan+" + Date.now() + "@tincanapi.com";
actorMbox = "mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com";
preparedStmt = session[v].prepareStatement(
{
actor: {
mbox: actorMbox
},
verb: {
id: "http://adlnet.gov/expapi/verbs/attempted"
},
target: {
id: "http://tincanapi.com/TinCanJS/Test/TinCan_sendStatement/prepared-sync/" + v
}
}
);
sendResult = session[v].sendStatement(preparedStmt);
ok(sendResult.hasOwnProperty("statement"), "sendResult has property: statement (" + v + ")");
deepEqual(sendResult.statement, preparedStmt, "sendResult property value: statement (" + v + ")");
ok(sendResult.hasOwnProperty("results"), "sendResult has property: results (" + v + ")");
ok(sendResult.results.length === 1, "sendResult results property: length (" + v + ")");
ok(sendResult.results[0].hasOwnProperty("err"), "sendResult result 0 has property: err (" + v + ")");
ok(sendResult.results[0].hasOwnProperty("xhr"), "sendResult result 0 has property: xhr (" + v + ")");
deepEqual(sendResult.results[0].err, null, "sendResult result 0 property value: err (" + v + ")");
}
);
};
doGetStatementSyncTest = function (v) {
test(
"tincan.getStatement (sync): " + v,
function () {
var sendResult,
getResult,
//
// Cloud is lowercasing the mbox value so just use a lowercase one
// and make sure it is unique to prevent merging that was previously
// possible, but leaving the commented version as a marker for something
// that ought to be tested against a 1.0.0 spec
//
//actorMbox = "mailto:TinCanJS-test-TinCan+" + Date.now() + "@tincanapi.com";
actorMbox = "mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com";
sendResult = session[v].sendStatement(
{
actor: {
mbox: actorMbox
},
verb: {
id: "http://adlnet.gov/expapi/verbs/attempted"
},
target: {
id: "http://tincanapi.com/TinCanJS/Test/TinCan_getStatement/sync/" + v
}
}
);
getResult = session[v].getStatement(sendResult.statement.id);
ok(getResult.hasOwnProperty("statement"), "getResult has property: statement (" + v + ")");
ok(getResult.hasOwnProperty("err"), "getResult has property: err (" + v + ")");
ok(getResult.hasOwnProperty("xhr"), "getResult has property: xhr (" + v + ")");
deepEqual(getResult.err, null, "getResult property value: err (" + v + ")");
TinCanTest.assertHttpRequestType(getResult.xhr, "getResult property value is: xhr (" + v + ")");
// clear the "stored" and "authority" properties since we couldn't have known them ahead of time
getResult.statement.stored = null;
getResult.statement.authority = null;
if (v === "0.9") {
sendResult.statement.inProgress = false;
}
if (v !== "0.9" && v !== "0.95") {
//
// in 1.0.0 the version should be 1.0.0, in 1.0.1 it was supposed to be
// returned as 1.0.1 but the spec still said 1.0.0, in the future it is
// expected that if the LRS supports a particular version that is what
// it returns
//
sendResult.statement.version = "1.0.0";
}
deepEqual(getResult.statement, sendResult.statement, "getResult property value: statement (" + v + ")");
}
);
};
doVoidStatementSyncTest = function (v) {
test(
"tincan.getVoidedStatement (sync): " + v,
function () {
var sendResult,
getResult,
//
// Cloud is lowercasing the mbox value so just use a lowercase one
// and make sure it is unique to prevent merging that was previously
// possible, but leaving the commented version as a marker for something
// that ought to be tested against a 1.0.0 spec
//
//actorMbox = "mailto:TinCanJS-test-TinCan+" + Date.now() + "@tincanapi.com";
actorMbox = "mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com",
actor = new TinCan.Agent(
{ mbox: actorMbox }
),
compareVoidSt;
// need to create a new statement that we can then void
sendResult = session[v].sendStatement(
{
actor: actor,
verb: {
id: "http://adlnet.gov/expapi/verbs/attempted"
},
target: {
id: "http://tincanapi.com/TinCanJS/Test/TinCan_getVoidedStatement/sync/" + v
}
}
);
voidResult = session[v].voidStatement(
sendResult.statement,
null,
{
actor: actor
}
);
compareVoidSt = new TinCan.Statement(
{
actor: actor,
verb: new TinCan.Verb(
{
id: "http://adlnet.gov/expapi/verbs/voided"
}
),
target: new TinCan.StatementRef(
{
id: sendResult.statement.id
}
)
},
{
doStamp: false
}
);
ok(voidResult.hasOwnProperty("statement"), "voidResult has property: statement (" + v + ")");
compareVoidSt.id = voidResult.statement.id;
compareVoidSt.timestamp = voidResult.statement.timestamp;
deepEqual(voidResult.statement, compareVoidSt, "voidResult property value: statement (" + v + ")");
ok(voidResult.hasOwnProperty("results"), "voidResult has property: results (" + v + ")");
ok(voidResult.results.length === 1, "voidResult results property: length (" + v + ")");
ok(voidResult.results[0].hasOwnProperty("err"), "voidResult result 0 has property: err (" + v + ")");
ok(voidResult.results[0].hasOwnProperty("xhr"), "voidResult result 0 has property: xhr (" + v + ")");
deepEqual(voidResult.results[0].err, null, "voidResult result 0 property value: err (" + v + ")");
TinCanTest.assertHttpRequestType(voidResult.results[0].xhr, "voidResult result 0 property value is: xhr (" + v + ")");
getVoidedResult = session[v].getVoidedStatement(sendResult.statement.id);
ok(getVoidedResult.hasOwnProperty("statement"), "getResult has property: statement (" + v + ")");
ok(getVoidedResult.hasOwnProperty("err"), "getResult has property: err (" + v + ")");
ok(getVoidedResult.hasOwnProperty("xhr"), "getResult has property: xhr (" + v + ")");
deepEqual(getVoidedResult.err, null, "getResult property value: err (" + v + ")");
TinCanTest.assertHttpRequestType(getVoidedResult.xhr, "getResult property value is: xhr (" + v + ")");
// clear the "stored" and "authority" properties since we couldn't have known them ahead of time
getVoidedResult.statement.stored = null;
getVoidedResult.statement.authority = null;
if (v === "0.9") {
sendResult.statement.inProgress = false;
}
if (v === "0.9" || v === "0.95") {
sendResult.statement.voided = true;
}
if (v !== "0.9" && v !== "0.95") {
//
// in 1.0.0 the version should be 1.0.0, in 1.0.1 it was supposed to be
// returned as 1.0.1 but the spec still said 1.0.0, in the future it is
// expected that if the LRS supports a particular version that is what
// it returns
//
sendResult.statement.version = "1.0.0";
}
deepEqual(getVoidedResult.statement, sendResult.statement, "getVoidedResult property value: statement (" + v + ")");
}
);
};
doStateSyncTest = function (v) {
test(
"tincan state (sync): " + v,
function () {
var setResult,
key = "setState (sync)",
val = "TinCanJS",
mbox ="mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com",
options = {
agent: new TinCan.Agent(
{
mbox: mbox
}
),
activity: new TinCan.Activity(
{
id: "http://tincanapi.com/TinCanJS/Test/TinCan_setState/sync/" + v
}
)
};
setResult = session[v].setState(key, val, options);
if (! NATIVE_CORS) {
deepEqual(
setResult,
{
xhr: null,
err: new Error("Unsupported content type for IE Mode request")
},
"setResult when using XDomainRequest"
);
return;
}
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
ok(setResult.err === null, "setResult.err is null");
getResult = session[v].getState(key, options);
ok(getResult.hasOwnProperty("state"), "getResult has property: state (" + v + ")");
ok(getResult.state instanceof TinCan.State, "getResult state property is TinCan.State (" + v + ")");
deepEqual(getResult.state.contents, val, "getResult state property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.state.contentType), "application/octet-stream", "getResult state property contentType (" + v + ")");
//
// reset the state to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.state.etag;
setResult = session[v].setState(key, val + 1, options);
delete options.lastSHA1;
deleteResult = session[v].deleteState(key, options);
}
);
};
doStateSyncContentTypeJSONTest = function (v) {
test(
"tincan state (sync, JSON content type): " + v,
function () {
var setResult,
key = "setState (sync, json content)",
val = {
testObj: {
key1: "val1"
},
testBool: true,
testNum: 1
},
mbox ="mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com",
options = {
contentType: "application/json",
agent: new TinCan.Agent(
{
mbox: mbox
}
),
activity: new TinCan.Activity(
{
id: "http://tincanapi.com/TinCanJS/Test/TinCan_setState/syncContentType/" + v
}
)
};
setResult = session[v].setState(key, val, options);
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
getResult = session[v].getState(key, options);
ok(getResult.hasOwnProperty("state"), "getResult has property: state (" + v + ")");
ok(getResult.state instanceof TinCan.State, "getResult state property is TinCan.State (" + v + ")");
deepEqual(getResult.state.contents, val, "getResult state property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.state.contentType), "application/json", "getResult state property contentType (" + v + ")");
//
// reset the state to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.state.etag;
setResult = session[v].setState(key, val + 1, options);
delete options.lastSHA1;
deleteResult = session[v].deleteState(key, options);
}
);
};
doActivityProfileSyncTest = function (v) {
test(
"tincan activityProfile (sync): " + v,
function () {
var setResult,
key = "activityProfile (sync)",
val = "TinCanJS",
options = {
activity: new TinCan.Activity(
{
id: "http://tincanapi.com/TinCanJS/Test/TinCan_setActivityProfile/sync/" + v
}
)
};
setResult = session[v].setActivityProfile(key, val, options);
if (! NATIVE_CORS) {
deepEqual(
setResult,
{
xhr: null,
err: new Error("Unsupported content type for IE Mode request")
},
"setResult when using XDomainRequest"
);
return;
}
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
getResult = session[v].getActivityProfile(key, options);
ok(getResult.hasOwnProperty("profile"), "getResult has property: profile (" + v + ")");
ok(getResult.profile instanceof TinCan.ActivityProfile, "getResult profile property is TinCan.ActivityProfile (" + v + ")");
deepEqual(getResult.profile.contents, val, "getResult profile property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.profile.contentType), "application/octet-stream", "getResult profile property contentType (" + v + ")");
setResult = session[v].setActivityProfile(key, val + 1, options);
//
// reset the profile to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.profile.etag;
setResult = session[v].setActivityProfile(key, val + 2, options);
delete options.lastSHA1;
deleteResult = session[v].deleteActivityProfile(key, options);
}
);
};
doActivityProfileSyncContentTypeJSONTest = function (v) {
test(
"tincan activityProfile (sync, JSON content type): " + v,
function () {
var setResult,
key = "activityProfile (sync)",
val = {
testObj: {
key1: "val1"
},
testBool: true,
testNum: 1
},
options = {
activity: new TinCan.Activity(
{
id: "http://tincanapi.com/TinCanJS/Test/TinCan_setActivityProfile/syncJSON/" + v
}
),
contentType: "application/json"
};
setResult = session[v].setActivityProfile(key, val, options);
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
getResult = session[v].getActivityProfile(key, options);
ok(getResult.hasOwnProperty("profile"), "getResult has property: profile (" + v + ")");
ok(getResult.profile instanceof TinCan.ActivityProfile, "getResult profile property is TinCan.ActivityProfile (" + v + ")");
deepEqual(getResult.profile.contents, val, "getResult profile property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.profile.contentType), "application/json", "getResult profile property contentType (" + v + ")");
setResult = session[v].setActivityProfile(key, val + 1, options);
//
// reset the profile to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.profile.etag;
setResult = session[v].setActivityProfile(key, val + 2, options);
delete options.lastSHA1;
deleteResult = session[v].deleteActivityProfile(key, options);
}
);
};
doAgentProfileSyncTest = function (v) {
test(
"tincan agentProfile (sync): " + v,
function () {
var setResult,
key = "agentProfile (sync)",
val = "TinCanJS",
mbox ="mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com",
options = {
agent: new TinCan.Agent(
{
mbox: mbox
}
)
};
setResult = session[v].setAgentProfile(key, val, options);
if (! NATIVE_CORS) {
deepEqual(
setResult,
{
xhr: null,
err: new Error("Unsupported content type for IE Mode request")
},
"setResult when using XDomainRequest"
);
return;
}
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
getResult = session[v].getAgentProfile(key, options);
ok(getResult.hasOwnProperty("profile"), "getResult has property: profile (" + v + ")");
ok(getResult.profile instanceof TinCan.AgentProfile, "getResult profile property is TinCan.AgentProfile (" + v + ")");
deepEqual(getResult.profile.contents, val, "getResult profile property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.profile.contentType), "application/octet-stream", "getResult profile property contentType (" + v + ")");
setResult = session[v].setAgentProfile(key, val + 1, options);
//
// reset the profile to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.profile.etag;
setResult = session[v].setAgentProfile(key, val + 2, options);
delete options.lastSHA1;
deleteResult = session[v].deleteAgentProfile(key, options);
}
);
};
doAgentProfileSyncContentTypeJSONTest = function (v) {
test(
"tincan agentProfile (sync, JSON content type): " + v,
function () {
var setResult,
key = "agentProfile (sync)",
val = {
testObj: {
key1: "val1"
},
testBool: true,
testNum: 1
},
mbox ="mailto:tincanjs-test-tincan+" + Date.now() + "@tincanapi.com",
options = {
agent: new TinCan.Agent(
{
mbox: mbox
}
),
contentType: "application/json"
};
setResult = session[v].setAgentProfile(key, val, options);
ok(setResult.hasOwnProperty("err"), "setResult has property: err (" + v + ")");
ok(setResult.hasOwnProperty("xhr"), "setResult has property: xhr (" + v + ")");
getResult = session[v].getAgentProfile(key, options);
ok(getResult.hasOwnProperty("profile"), "getResult has property: profile (" + v + ")");
ok(getResult.profile instanceof TinCan.AgentProfile, "getResult profile property is TinCan.AgentProfile (" + v + ")");
deepEqual(getResult.profile.contents, val, "getResult profile property contents (" + v + ")");
deepEqual(TinCan.Utils.getContentTypeFromHeader(getResult.profile.contentType), "application/json", "getResult profile property contentType (" + v + ")");
setResult = session[v].setAgentProfile(key, val + 1, options);
//
// reset the profile to make sure we test the concurrency handling
//
options.lastSHA1 = getResult.profile.etag;
setResult = session[v].setAgentProfile(key, val + 2, options);
delete options.lastSHA1;
deleteResult = session[v].deleteAgentProfile(key, options);
}
);
};
for (i = 0; i < versions.length; i += 1) {
version = versions[i];
if (TinCanTestCfg.recordStores[version]) {
doSendStatementSyncTest(version);
doGetStatementSyncTest(version);
doVoidStatementSyncTest(version);
doStateSyncTest(version);
doStateSyncContentTypeJSONTest(version);
doActivityProfileSyncTest(version);
doActivityProfileSyncContentTypeJSONTest(version);
doAgentProfileSyncTest(version);
doAgentProfileSyncContentTypeJSONTest(version);
}
}
}());
| {
"redpajama_set_name": "RedPajamaGithub"
} | 251 |
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Counting citizen science, and making citizen scientists count
Physics professor tackles science-laureate-like role at local science centre.
Jennifer Pascoe - 09 March 2018
Sivakoff was just named the inaugural science fellow at the TELUS World of Science Edmonton (TWoSE), where he will focus on delivering public programming as a way to give people an opportunity to get involved in science projects in the local community.
Gregory Sivakoff believes that to continue moving society forward, scientists need to engage the community to contribute to the cause.
"Our entire civilization is strongly based on the advances of sciences," said Sivakoff, associate professor in the Department of Physics best known for his work exploring black holes. "It's very easy to forget how much work has gone and continues to go into advancing science to develop our future. So to me, it's important to promote science as a way to ensure our civilization continues to make the scientific advances that will benefit the next generation."
Sivakoff was just named the inaugural science fellow at the TELUS World of Science Edmonton (TWoSE). Through this role over the coming year, he will focus on delivering public programming in astronomy and beyond as a way to give people an opportunity to get involved in science projects in the local community and around the world, developing an interest in space beyond a pop culture phenomenon or casual interest.
"Citizen scientists are increasingly involved in astronomical research, and the universe is really a rich environment for discovery. It's the largest stage we have. And there are organizations that enable people with relatively modest equipment--sometimes even just their eyes--to become part of the scientific research scene."
Sivakoff counts the opportunity he had to get involved in research at a young age as key to holding his interest and inspiring his own journey through academia, including a prestigious Fulbright scholarship.
"I've always appreciated the ability to get involved with research earlier or differently than you might normally expect."
As for the initial seed of inspiration that spurred Sivakoff on to taking the chance with research, he credits his home environment as creating fertile ground for a scientific mind, something he sees being made possible by organizations like TWoSE.
"Science has always been a part of my family. I always enjoyed science and math in school, and my dad and I had a conversation when I was about to go to to college. He said, 'You like science, you like math. Space is big. You ought to be able to find your place in it.'"
The senior Sivakoff certainly nailed that prediction. A self-described "acronym fiend"--(in)famous in the field of astronomy for coming up with project names as tantalizing as MAVERICS and JACPOT--Sivakoff is part of roughly a dozen national and international radio telescope research collaborations.
Sivakoff will be sharing his expertise as a guest in the forthcoming Black Holes 101 massive open online course, launching later this year out of the University of Alberta's Faculty of Science, free to anyone with an internet connection anywhere in the world.
"Black holes are dominant in popular culture. At the same time, they're just a great vehicle for talking about general aspects of astronomy and physics and getting people further enmeshed in science."
Faculty of Science Quick Links
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Main UAlberta Switchboard: 780-492-3111
Contact Us: Faculty of Science Directory | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 4,528 |
AI in HeathCare Artificial Intelligence News/Geeks
Eyes are one of the most important organs of our body. In our population, there are more than 39 million people as being truly blind and246 million have low vision. What if there will be something that will tell whether you will go blind or not?
Dr. Ramasamy Kim the head of retina services at an eye hospital in southern India and his team has created an amazing machine learning algorithm that helps in diagnosing eye problem caused by diabetes and tells whether you will go blind or not?
Over the past few years, Kim and his team at the Aravind eye hospital have studied and worked over thousands of images of the retina. In total the team has examined about 15,000 images.
All these images were collected across the country, showing the interior surface of the eyeball, known as the fundu. With this huge dataset, the researchers identified the very first condition that leads to diabetes that can lead to blindness.
After finding this important spot, the team starts grading each image, marking abnormal spots, lesions, and indications of bleeding. They have created a complete database of retinal images from all over the world, compiled by Google and its health-tech subsidiary, Verily.
The data then put into the machine learning algorithm created by the team with a set of instructions built into a computer program to identify eye complications arising from diabetes.
Kim's machine learning points toward several yellow spots that look like pinpricks of light emerging from the reddish-brown orb. In diabetes, the tiny blood vessels of the retina heavily get damage.
That's what causes these tiny blood vessels to leak blood and cholesterol. In many cases, the patient does not even feel this. There is ultimately no symptoms until the patience retinal tissue begins to swell.
After a few days, the patients start noticing dullness in vision and eyesight begins to fail. At this point, it gets completely difficult to treat and in some cases, irreversible.
Kim says, it is remarkable that a machine can be trained to identify even the earliest stages of the diabetic retinopathy and grade it with as much accuracy as a human doctor.
"It's been very exciting to watch this take shape," says Dr. Renu Rajan, a retinal surgeon involved in the project.
Recommended: Autoblow World's first AI sex toy
"When a machine is trained to be capable of identifying abnormal patterns in this way, it saves a doctor so much time in diagnosis; time that could be better spent helping the patient manage the condition and in aftercare."
"Big data" and artificial intelligence has come in for stinging criticism in recent years, over issues of privacy, fairness, accountability, and transparency."
Kim and his team have been involved in a pilot project to test the algorithm from the starting of 2016.
The machine-learning algorithm created by the team is at its early stage and the results it generates is being checked against a manual grading process.
"If there is a big difference between the results generated from the AI software and the manual diagnosis by an ophthalmologist, a senior retina specialist will make the final decision on the grading," says Kim.
"These discrepancies are analyzed, helping to improve and fine-tune the data that will be fed into it."
Kim hopes that in the near future India's remotest villages will mostly get benefited by AI-powered machines set up like vending booths, the villagers will be capable of taking photographs of people's inner eye and offering a digital diagnosis.
"Think of it as a screening and referral tool, that could tell you with a great deal of precision whether you needed to see a specialist or not and if so, how urgently you needed to see one," says Dr. H Parida, a retina specialist."
Google and Verily have been working with well-established eye care centers in order to build the raw data that their algorithm needs.
"By partnering with well-known institutions like Aravind eye hospital and Shankara Nethralaya, we can continue our research and pilot studies in implementing AI-powered screening technology, and then extend it to clinical practice," says Dr. Lily Peng, a Google product manager.
← This New Artificial Intelligence System Can Spot Fake Online Reviews
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Researchers Created An AI That Can Edit The Facial Expressions Of Actors In A Video | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 0 |
Q: Why parallel For loop in C# takes more time than simple For loop? I have a CFD (Computational fluid dynamics ) in c# which takes too much time to calculate results. For improving my code, I started to learn TPL and using parallel codes. For the loops that order is not important I can use TPL and for the loops with order the PLINQ is the only way.
Am I correct?
As the first step, I changed For loops to Parallel.For and interestingly found out that run time increased !
Sample of my code:
for (int i = 0; i < nx; i++)
{
for (int j = 0; j < ny; j++)
{
if (!Cells[i, j, 0].IsVirtual)
{
// calculate x velocity
// calculate y velocity
}
}
}
With parallel tasks:
Parallel.for (0,nx, i =>
{
for (int j = 0; j < ny; j++)
{
if (!Cells[i, j, 0].IsVirtual)
{
// calculate x velocity
// calculate y velocity
}
}
});
How can I speed up my code?
Each my outputs takes 10 min which is very long time and I ned at least 5000 outputs.
A: For small loops the overhead of managing the threads is probably impacting the overall execution time. You would probably see different results if each iteration took longer to execute.
A: Over large datasets (e.g. at least 500,000 cells) you might be hitting problems with cache invalidation because you're iterating through memory inefficiently.
You might see a performance increase if you change it to this (regardless of whether or not you use TPL) (note how I flipped the iteration from i,j to j,i):
for (int j = 0; j < ny; j++)
{
for (int i = 0; i < nx; i++)
{
if (!Cells[i, j, 0].IsVirtual)
{
// calculate x velocity
// calculate y velocity
}
}
}
See here for an explanation: Why does the order of the loops affect performance when iterating over a 2D array?
| {
"redpajama_set_name": "RedPajamaStackExchange"
} | 5,206 |
The deal would have given Sinclair access to more than 70 percent of U.S. TV-owning households.
Tribune Media has officially terminated its $3.9 billion agreement with Sinclair Broadcast Group, the company announced this morning.
This comes after roughly 15 months of Sinclair revising its local station divestiture plan in an effort to appease the federal government's long-standing concerns. But Tribune finally took matters into its own hands and walked away.
The Sinclair-Tribune deal was struck in May 2017 and would have given Sinclair—already the largest owner of local stations in the US —ownership of 233 local TV stations across the nation, including 42 Tribune stations, and access to more than 70 percent of U.S. TV-owning households.
Tribune will now most likely sell its stations to another local broadcast group, with Sinclair and its conservative-skewing news operation very likely to try and make a different deal to acquire different local TV stations. | {
"redpajama_set_name": "RedPajamaC4"
} | 274 |
\section{Introduction}
\label{intro}
Let $\mathcal{A}$ be the class of normalized analytic functions defined on the open unit disk $\mathbb{D}:=\{z\in \mathbb{C}:|z|<1\}$, given by
\begin{equation}
f(z) = z+\sum_{n=2}^{\infty}a_nz^n\label{form}
\end{equation}
and let $\mathcal{S}$ be the subclass of $\mathcal{A}$ consisting of univalent functions.
Let $\mathcal{P}$ be the class of analytic functions defined on $\mathbb{D}$ with positive real part, given by $p(z)=1+\sum_{n=1}^{\infty}p_n z^n$. Let $h$ and $g$ be two analytic functions. We say $h_1$ is subordinate to $h_2$, denoted by $h_1\prec h_2$, if there exists a Schwarz function $w$ with $w(0)=0$ and $|w(z)|\leq |z|$ such that $h_1(z)=h_2(w(z))$.
Bieberbach conjecture \cite{goodman vol1}[Page no. ] contributed significantly to the growth of geometric function theory and the rise of new coefficient problems, we refer \cite{goodman vol1} for more information. Since then, more number of subclasses of $\mathcal{S}$ involving $\mathcal{S}^{*}$ and $\mathcal{C}$ of starlike and convex functions respectively have been introduced. In 1992, Ma and Minda \cite {ma-minda} introduced the following two classes:
\begin{equation}
\mathcal{S}^{*}(\varphi)=\bigg\{f\in \mathcal {A}:\dfrac{zf'(z)}{f(z)}\prec \varphi(z) \bigg\}\label{mindaclass}
\end{equation}
and
\begin{equation}
\mathcal{C}(\varphi)=\bigg\{f\in \mathcal {A}:1+\dfrac{zf''(z)}{f'(z)}\prec \varphi(z) \bigg\},\label{mindaclassc}
\end{equation}
which unifies various subclasses of $\mathcal{S}^{*}$ and $\mathcal{C}$ respectively.
Here $\varphi$ is an analytic univalent function satisfying the conditions $\RE\varphi(z)>0$, $\varphi(\mathbb{D})$ symmetric about the real axis and starlike with respect to $\varphi(0)=1$ with $\varphi'(0)>0$.
The concept of Hankel determinants was introduced in \cite {pomi} and is a topic of interest for many authors even today. The definition of the $q^{th}$ Hankel determinants $H_{q,n}(f)$ of analytic functions $f \in \mathcal{A}$ of the form (\ref{form}) under the assumption that $a_1:=1$ is as follows:
\begin{equation}
H_{q,n}(f) =\begin{vmatrix}
a_n&a_{n+1}& \ldots &a_{n+q-1}\\
a_{n+1}&a_{n+2}&\ldots &a_{n+q}\\
\vdots& \vdots &\ddots &\vdots\\
a_{n+q-1}&a_{n+q}&\ldots &a_{n+2q-2}\label{5hqn}
\end{vmatrix}
\end{equation}
for $n,q\in \mathbb{N}$.
The third order Hankel determinant expression, which is denoted by $H_{3,1}(f)$, is derived from (\ref{5hqn}) by taking $q=3$ and $n=1$, defined as:
\begin{equation}
H_{3,1}(f) =2 a_2a_3a_4-a_3^3-a_4^2-a_2^2a_5+a_3a_5.\label{1h3}
\end{equation}
Over the past, several authors established sharp bound of second-order Hankel determinants, see \cite{alarif,krishna bezilevic}.
However, it is more complicated to compute the bound of the third-order Hankel determinant, which is obtained by taking $q=3$ in (\ref{5hqn}), and there are few articles concerning sharp bounds. Zaprawa \cite{zap} estimated that
\begin{equation*}
|H_{3,1}(f)|\leq \begin{cases}
1,& f\in \mathcal{S}^{*}\\
49/540,& f\in \mathcal{C}.
\end{cases}
\end{equation*}
Later, for the class $\mathcal{S}^{*}$, Kwon et al.\cite{sharpstarlike} obtained $|H_{3,1}(f)|\leq 8/9$, which is recently best improved to $4/9$ by Kowalczyk et al. \cite{4/9}.
Lecko et al.\cite{lecko 1/2 bound} has obtained $|H_{3,1}(f)|\leq 1/9$, which is sharp for functions belonging to the class $\mathcal{S}^{*}(1/2)$. For more on Hankel determinants, we refer \cite{kowal,rath,lecko 1/2 bound,4/9}.
By specializing the function $\varphi$ in $\mathcal{S}^*(\varphi)$ and $\mathcal{C}(\varphi)$, we obtain some well-known subclasses of $\mathcal{S}^*$ and $\mathcal{C}$, which are illustrated below:
\begin{enumerate}
\item Janowski~\cite{1janowski} introduced, $\mathcal{S}^*((1+Az)/(1+Bz))=:\mathcal{S}^*[A,B]$ and $\mathcal{C}((1+Az)/(1+Bz))=:\mathcal{C}[A,B]$ for $-1\leq B<A\leq 1$.
\item On taking $\varphi(z)=\sqrt{1+z}$, Sok\'{o}\l \ and Stankiewicz \cite{stan} introduced the class, $\mathcal{S}^{*}_ L:=\mathcal{S}(\sqrt{1+z})$ and $\mathcal{C}_ L:=\mathcal{C}(\sqrt{1+z})$
\end{enumerate}
Similarly, the classes $\mathcal{S}_{e}^{*}$ and $\mathcal{C}_{e}$ were introduced and studied by Mendiratta et al.\cite{mendi} by considering $\varphi(z)=e^z$ in (\ref{mindaclass}) and (\ref{mindaclassc}) respectively, defined as:
\begin{equation*}
\mathcal{S}_{e}^{*}=\bigg\{f\in \mathcal {A}:\dfrac{zf'(z)}{f(z)}\prec e^{z}\bigg\}\quad \text{and}\quad \mathcal{C}_{e}=\bigg\{f\in \mathcal {A}:1+\dfrac{zf''(z)}{f'(z)}\prec e^{z}\bigg\}.
\end{equation*}
Several radius problems \cite{mendi} and implications of differential subordination \cite{adibastarlikenessexponential} have been established for various subclasses associated with the exponential function.
Zaprawa \cite{zap2019} has improved the bounds of the third Hankel determinants related to the classes $\mathcal {S}^{*}_{e}$ and $\mathcal{C}_{e}$, which is given b
\begin{equation}\label{improved}
|H_{3,1}(f)|\leq \begin{cases}
0.385,& f\in \mathcal{S}^{*}_{e}\\
0.021,& f\in \mathcal{C}_{e}.
\end{cases}
\end{equation}
However, the results were not sharp.
In the present investigation, we obtain the sharp bound of the result given in \eqref{improved}.
We estimate sharp third hankel determinant, bounds of the sixth and seventh coefficients and fourth Hankel determinant for functions in $\mathcal {S}^{*}_{e}$ and $\mathcal{C}_{e}$, respectively in the subsequent sections.
\section{Hankel Determinants for $\mathcal{S}^{*}_{e}$}
\subsection{Preliminaries}
Let $f\in \mathcal{S}_{e}^{*},$ then there exists a Schwarz function $w(z)$ such that
\begin{equation}\label{3 formulaai}
\dfrac{zf'(z)}{f(z)}=e^{w(z)}.
\end{equation}
Suppose that $p(z)=1+p_1z+p_2z^2+\cdots\in \mathcal{P}$ and consider $w(z)=(p(z)-1)/(p(z)+1)$. Further, by substituting the expansions of $w(z)$, $p(z)$ and $f(z)$ in equation (\ref{3 formulaai}) and then comparing the coefficients, we obtain the expressions of $a_i (i=2,3,...,7)$ in terms of $p_j (j=1,2,...,5)$, given as follows:
\begin{equation}\label{5 ea2}
a_2=\dfrac{1}{2}p_1,\quad a_3=\dfrac{1}{16}\bigg(4p_2+p_1^2\bigg), \quad a_4=\dfrac{1}{288}\bigg(-p_1^3 +12 p_1 p_2 + 48 p_3\bigg),
\end{equation}
\begin{equation}\label{5 ea5}
a_5=\dfrac{1}{1152}\bigg( p_1^4 -12 p_1^2 p_2 +24 p_1 p_3 + 144 p_4\bigg),
\end{equation}
\begin{equation}\label{5 ea6}
a_6=\frac{1}{57600}\bigg(-17 p_1^5 + 220 p_1^3 p_2 - 480 p_1 p_2^2 - 480 p_1^2 p_3 - 480 p_2 p_3 + 720 p_1 p_4 + 5760p_5\bigg),
\end{equation}
and
\begin{align}
a_7&=\frac{1}{8294400}\bigg(881 p_1^6 - 13260 p_1^4 p_2 + 48240 p_1^2 p_2^2 - 14400 p_2^3 + 29040 p_1^3 p_3-106560 p_1 p_2 p_3\nonumber\\
&\quad\quad \quad \quad \quad\quad - 57600 p_3^2 - 56160 p_1^2 p_4 - 86400 p_2 p_4 + 69120 p_1 p_5 \bigg).\label{5 ea7}
\end{align}
The formula for $p_i$ $(i=2,3,4)$, which is included in the Lemma \ref{pformula} below, plays a vital role in establishing the sharp bound for Hankel determinants and forms the foundation for our main results. See the recent survey article~\cite{surveycaratheodory} for more information on the coefficients of the class $\mathcal{P}$.
\begin{lemma}\cite{rj,lemma1}\label{pformula}
Let $p\in \mathcal {P}$ has the form $1+\sum_{n=1}^{\infty}p_n z^n.$ Then
\begin{equation}
p_2=\frac{p_1^2+\gamma (4-p_1^2)}{2},\label{b2}
\end{equation}
\begin{equation}
p_3=\frac{p_1^3+2p_1(4-p_1^2)\gamma -p_1(4-p_1^2) {\gamma}^2+2(4-p_1^2)(1-|\gamma|^2)\eta}{4}, \label{b3}
\end{equation}
and \begin{equation}
p_4=\frac{p_1^4+(4-p_1^2)\gamma (p_1^2({\gamma}^2-3\gamma+3)+4\gamma)-4(4-p_1^2)(1-|\gamma|^2)(p_1(\gamma-1)\eta+\bar{\gamma}{\eta}^2-(1-|\eta|^2)\rho)}{8}, \label{b4}
\end{equation}
for some $\gamma$, $\eta$ and $\rho$ such that $|\gamma|\leq 1$, $|\eta|\leq 1$ and $|\rho|\leq 1.$
\end{lemma}
\subsection{Sharp Third Hankel Determinant for $\mathcal{S}^{*}_{e}$}
The following theorem presents the sharp bound for $|H_{3,1}(f)|$ for functions belonging to the class $\mathcal{S}^{*}_{e}$.
\begin{theorem}\label{5 sharph31}
Let $f\in \mathcal {S}^{*}_{e}.$ Then
\begin{equation}
|H_{3,1}(f)|\leq 1/9.\label{5 9.5}
\end{equation}
This result is sharp.
\end{theorem}
\begin{proof}
Since the class $\mathcal {P}$ is invariant under rotation, the value of $p_1$ belongs to the interval [0,2]. Let $p:=p_1$ and then substitute the values of $a_i(i=2,3,4,5)$ in equation (\ref{1h3}) from equations (\ref{5 ea2}) and (\ref{5 ea5}). We get
\begin{align*}
H_{3,1}(f)&=\dfrac{1}{331776}\bigg(-211 p^6 + 420 p^4 p_2 - 1872 p^2 p_2^2 - 5184 p_2^3 + 2544 p^3 p_3+10944 p p_2 p_3\\
&\quad \quad\quad \quad \quad \quad - 9216 p_3^2 - 7776 p^2 p_4 + 10368 p_2 p_4\bigg).
\end{align*}
After simplifying the calculations through (\ref{b2})-(\ref{b4}), we obtain
$$H_{3,1}(f)=\dfrac{1}{331776}\bigg(\beta_1(p,\gamma)+\beta_2(p,\gamma)\eta+\beta_3(p,\gamma){\eta}^2+\phi(p,\gamma,\eta)\rho\bigg),$$
for $\gamma,\eta,\rho\in \mathbb {D}.$ Here
\begin{align*}
\beta_1(p,\gamma):&=-13p^6-36{\gamma}^2p^2(4-p^2)^2-360{\gamma}^3p^2(4-p^2)^2+72{\gamma}^4p^2(4-p^2)^2+78{\gamma}p^4(4-p^2)\\
&\quad+120p^4{\gamma}^2(4-p^2)-324p^4{\gamma}^3(4-p^2)-1296{\gamma}^2p^2(4-p^2),\\
\beta_2(p,\gamma):&=24(1-|\gamma|^2)(4-p^2)(17p^3+54{\gamma}p^3+30p\gamma(4-p^2)-12p{\gamma}^2(4-p^2)),\\
\beta_3(p,\gamma):&=144(1-|\gamma|^2)(4-p^2)(-16(4-p^2)-2|\gamma|^2(4-p^2)+9p^2\bar{\gamma}),\\
\phi(p,\gamma,\eta):&=1296(1-|\gamma|^2)(4-p^2)(1-|\eta|^2)(2(4-p^2)\gamma-p^2).
\end{align*}
By choosing $x=|\gamma|$, $y=|\eta|$ and utilizing the fact that $|\rho|\leq 1,$ the above expression reduces to the following:
\begin{align*}
|H_{3,1}(f)|\leq \dfrac{1}{331776}\bigg(|\beta_1(p,\gamma)|+|\beta_2(p,\gamma)|y+|\beta_3(p,\gamma)|y^2+|\phi(p,\gamma,\eta)|\bigg)\leq M(p,x,y),
\end{align*}
where
\begin{equation}
M(p,x,y)=\dfrac{1}{331776}\bigg(m_1(p,x)+m_2(p,x)y+m_3(p,x)y^2+m_4(p,x)(1-y^2)\bigg),\label{5 new}
\end{equation}
with
\begin{align*}
m_1(p,x):&=13p^6+36x^2p^2(4-p^2)^2+360x^3p^2(4-p^2)^2+72x^4p^2(4-p^2)^2+78xp^4(4-p^2)\\
&\quad +120p^4x^2(4-p^2)+324p^4x^3(4-p^2)+1296x^2p^2(4-p^2),\\
m_2(p,x):&=24(1-x^2)(4-p^2)(17p^3+54xp^3+30px(4-p^2)+12px^2(4-p^2)),\\
m_3(p,x):&=144(1-x^2)(4-p^2)(16(4-p^2)+2x^2(4-p^2)+9p^2x),\\
m_4(p,x):&=1296(1-x^2)(4-p^2)(2x(4-p^2)+p^2).
\end{align*}
In the closed cuboid $U:[0,2]\times [0,1]\times [0,1]$, we now maximise $M(p,x,y)$, by locating the maximum values in the interior of the six faces, on the twelve edges, and in the interior of $U$.
\begin{enumerate}
\item We start by taking into account every internal point of $U$. Assume that $(p,x,y)\in (0,2)\times (0,1)\times (0,1)$. We calculate $\partial{M}/\partial y$
to identify the points of maxima in the interior of $U$. We get
\begin{align*}
\dfrac{\partial M}{\partial y}&=\dfrac{(4 - p^2)(1 - x^2)}{13824} \bigg(24 p x (5 + 2 x) + p^3 (17 + 24 x - 12 x^2)+96 (8 - 9 x + x^2) y \\
&\quad \quad\quad \quad\quad\quad\quad\quad\quad- 12 p^2 (25 - 27 x + 2 x^2) y\bigg).
\end{align*}
Now $\dfrac{\partial M}{\partial y}=0$ gives
\begin{equation*}
y=y_0:=\dfrac{p(17 p^2 + 120 x + 24 p^2 x + 48 x^2 - 12 p^2 x^2)}{12 (-64 + 25 p^2 + 72 x - 27 p^2 x - 8 x^2 + 2 p^2 x^2)}.
\end{equation*}
The existence of critical points requires that $y_0$ belong to $(0,1)$, which is only possible when
\begin{align}
300p^2+864x+24p^2x^2&>17p^3+120px+24p^3x+48px^2-12p^3x^2\nonumber\\
&\quad+768+864x+24p^2x^2.\label{5 h1}
\end{align}
Now, we find the solution satisfying the inequality (\ref{5 h1})
for the existence of critical points using the hit and trial method. If we assume $p$ tends to 0, then there does not exist any $x\in (0,1)$ satisfying the equation (\ref{5 h1}). But, when $p$ tends to 2, the equation (\ref{5 h1}) holds for all $x<37/54.$ We also observe that there does not exist any $p\in (0,2)$ when $x\in (37/54,1)$.
Similarly, if we assume $x$ tends to 0, then for all $p>1.68218$, the equation (\ref{5 h1}) holds. After calculations, we observe that there does not exist any $x\in (0,1)$ when $p\in (0,1.68218).$
Thus, the domain for the solution of the equation is $(1.68218,2)\times (0,37/54).$ Now, we examine that $\frac{\partial M}{\partial y}|_{y=y_0}\neq 0$ in $(1.68218,2)\times (0,37/54).$
So, we conclude that the function $M$ has no critical point in $(0,2)\times (0,1)\times (0,1).$
\item The interior of each of the cuboid $U$'s six faces is now being considered.\\
On $p=0, M(p,x,y)$ turns into
\begin{equation}
s_1(x,y):=\dfrac{(1-x^2)(8y^2 + x^2y^2+9x(1-y^2))}{72},\quad x,y\in (0,1).\label{5 9.4}
\end{equation}
Since
\begin{equation*}
\dfrac{\partial s_1}{\partial y}=\dfrac{(1 - x^2)(x-1)(x-8)y}{36}\neq 0,\quad x,y\in (0,1),
\end{equation*}
indicates that $s_1$ has no critical points in $(0,1)\times(0,1)$. \\
\noindent On $p=2, M(p,x,y)$ reduces to
\begin{equation}
M(2,x,y):=\dfrac{13}{5184},\quad x,y\in (0,1).\label{5 9.3}
\end{equation}
On $x=0, M(p,x,y)$ becomes
\begin{align}
s_2(p,y):&=\dfrac{13p^6 + (4-p^2)(408p^3y+2304y^2(4-p^2)+1296p^2(1-y^2)}{331776}
\label{5 9.1}
\end{align}
with $p\in (0,2)$ and $y\in (0,1).$ To determine the points of maxima, we solve $\partial s_2/\partial p=0$ and $\partial s_2/\partial y=0$. After solving $\partial s_2/\partial y=0,$ we get
\begin{equation}
y=\dfrac{17p^3}{12(25p^2-64)}(=:y_p).\label{5 y}
\end{equation}
In order to have $y_p\in (0,1)$ for the given range of $y$, $p_0:=p>\approx 1.68218$ is required. Based on calculations, $\partial s_2/\partial p=0$ gives
\begin{equation}
1728 p - 864 p^3 + 13 p^5 + 816 p^2 y - 340 p^4 y - 7872 p y^2 +
2400 p^3 y^2=0.\label{5 9}
\end{equation}
After substituting equation (\ref{5 y}) into equation (\ref{5 9}), we have
\begin{equation}
21233664 p - 27205632 p^3 + 11472192 p^5 - 1613016 p^7 + 2700 p^9=0.\label{5 40}
\end{equation}
A numerical calculation suggests that $p\approx 1.35596\in (0,2)$ is the solution of (\ref{5 40}). So, we conclude that $s_2$ does not have any critical point in $(0,2)\times(0,1)$.\\
\noindent On $x=1, M(p,x,y)$ reforms into
\begin{equation}
s_3(p,y):=M(p,1,y)=\dfrac{12672 p^2 - 2952 p^4 - 41 p^6}{331776}, \quad p\in (0,2).\label{5 9.2}
\end{equation}
While computing $\partial s_3/\partial p=0$, $p_0:=p\approx 1.43461$ comes out to be the critical point. Undergoing simple calculations, $s_3$ achieves its maximum value $\approx 0.0398426$ at $p_0.$\\
\noindent On $y=0, M(p,x,y)$ can be viewed as
\begin{align*}
s_4(p,x):&=\dfrac{1}{331776}\bigg(41472 x (1-x^2) + 576 p^2 (9 - 36 x + x^2 + 46 x^3 + 2 x^4)\\
&\quad \quad\quad\quad\quad\quad -24 p^4 (54 - 121 x - 8 x^2 + 174 x^3 + 24 x^4)\\
&\quad \quad\quad\quad\quad\quad +p^6 (13 - 78 x - 84 x^2 + 36 x^3 + 72 x^4)\bigg).
\end{align*}
After undergoing further calculations such as,
\begin{align*}
\dfrac{\partial s_4}{\partial x}&=\dfrac{1}{331776}\bigg(-82944 x^2 + 41472 (1-x^2) + 576 p^2 (-36 + 2 x + 138 x^2 + 8 x^3)\\
&\quad\quad \quad\quad \quad \quad-
24 p^4 (-121 - 16 x + 522 x^2 + 96 x^3) + p^6 (-78 - 168 x \\
&\quad\quad \quad\quad \quad \quad+ 108 x^2 + 288 x^3)\bigg)
\end{align*}
and \begin{align*}
\dfrac{\partial s_4}{\partial p}&=\dfrac{1}{331776}\bigg(6 p^5 (13 - 78 x - 84 x^2 + 36 x^3 + 72 x^4)-96 p^3 (54 - 121 x- 8 x^2 \\
&\quad\quad \quad\quad\quad + 174 x^3+ 24 x^4)+1152 p (9 - 36 x + x^2 + 46 x^3 + 2 x^4) \bigg),
\end{align*}
we observe that no solution in $(0,2)\times (0,1)$ exists of the system of equations $\partial s_4/\partial x=0$ and $\partial s_4/\partial p=0$.\\
On $y=1, M(p,x,y)$ reduces to
\begin{align*}
s_5(p,x):&=\dfrac{1}{331776}\bigg(2304 p x (5 + 2 x - 5 x^2 - 2 x^3) - 4608 (-8 + 7 x^2 + x^4)\\
&\quad\quad \quad\quad\quad+576 p^2 (-32 + 9 x + 38 x^2 + x^3 + 6 x^4)-24 p^5 (17 + 24 x \\
&\quad\quad \quad\quad\quad - 29 x^2- 24 x^3 + 12 x^4)+96 p^3 (17 - 6 x - 41 x^2+ 6 x^3\\
&\quad\quad \quad\quad\quad + 24 x^4)-24 p^4 (-96 + 41 x + 130 x^2+ 12 x^3 + 36 x^4)\\
&\quad\quad \quad\quad\quad +p^6 (13 - 78 x - 84 x^2 + 36 x^3+ 72 x^4)\bigg).
\end{align*}
\noindent The system of equations $\partial s_5/\partial x=0$ and $\partial s_5/\partial p=0$ also do not have any solution in $(0,2)\times (0,1).$\\
\item We next examine the maxima attained by $M(p,x,y)$ on the edges of the cuboid $U$. From equation (\ref{5 9.1}), we have $M(p,0,0)=r_1(p):=(5184p^2-1296p^4+13p^6)/331776.$ It is easy to observe that $r_1'(p)=0$ whenever $p=\delta_0:=0$ and $p=\delta_1:=1.4367\in [0,2]$ as its points of minima and maxima respectively.
Hence,
\begin{equation*}
M(p,0,0)\leq 0.0159535, \quad p\in [0,2].
\end{equation*}
Now considering the equation (\ref{5 9.1}) at $y=1,$ we get $M(p,0,1)=r_2(p):=(36864 - 18432 p^2 + 1632 p^3 + 2304 p^4 - 408 p^5 + 13 p^6)/331776.$ It is easy to observe that $r_2'(p)<0$ in $[0,2]$ and hence $p=0$ serves as the point of maxima. So,
\begin{equation*}
M(p,0,1)\leq \dfrac{1}{9}, \quad p\in [0,2].
\end{equation*}
Through computations, equation (\ref{5 9.1}) shows that $M(0,0,y)$ attains its maxima at $y=1.$ This implies that
\begin{equation*}
M(0,0,y)\leq \dfrac{1}{9}, \quad y\in [0,1].
\end{equation*}
Since, the equation (\ref{5 9.2}) does not involve $x$, we have $M(p,1,1)=M(p,1,0)=r_3(p):=(12672p^2 - 2952 p^4-41 p^6)/331776.$ Now, $r_3'(p)=4224 p - 1968 p^3 - 41 p^5=0$ when $p=\delta_2:=0$ and $p=\delta_3:=1.43461$ in the interval $[0,2]$ with $\delta_2$ and $\delta_3$ as points of minima and maxima respectively. Hence
\begin{equation*}
M(p,1,1)=M(p,1,0)\leq 0.0398426,\quad p\in [0,2].
\end{equation*}
After considering $p=0$ in (\ref{5 9.2}), we get, $M(0,1,y)=0.$ The equation (\ref{5 9.3}) has no variables. So, on the edges,
the maximum value of $M(p,x,y)$ is
\begin{equation*}
M(2,1,y)=M(2,0,y)=M(2,x,0)=M(2,x,1)=\dfrac{13}{5184},\quad x,y\in [0,1].
\end{equation*}
Using equation (\ref{5 9.4}), we obtain $M(0,x,1)=r_4(x):=(8 - 7 x^2 - x^4)/72.$ Upon calculations, we see that $r_4(x)$ is a decreasing function in $[0,1]$ and attains its maxima at $x=0.$ Hence
\begin{equation*}
M(0,x,1)\leq \dfrac{1}{9},\quad x\in [0,1].
\end{equation*}
Again utilizing the equation (\ref{5 9.4}), we get $M(0,x,0)=r_5(x):=x(1-x^2)/8.$ On further calculations, we get $r_5'(x)=0$ for $x=\delta_4:=1/\sqrt{3}.$ Also, $r_5(x)$ is an increases in $[0,\delta_4)$ and decreases in $(\delta_4,1].$ So, it reaches its maximum value at $\delta_4.$ Thus
\begin{equation*}
M(0,x,0)\leq 0.0481125,\quad x\in [0,1].
\end{equation*}
\end{enumerate}
Given all the cases, the inequality (\ref{5 9.5}) holds.\\
Let the function
$f_1(z)\in \mathcal{S}^{*}_{e}$, be defined as
\begin{equation*}
f_1(z)=z\exp\bigg(\int_{0}^{z}\dfrac{e^{t^3}-1}{t}dt\bigg)=z+\dfrac{z^4}{3}+\dfrac{5z^7}{36}+\cdots,\label{5 extremal}
\end{equation*}
with $f(0)=0$ and $f'(0)=1$, acts as an extremal function for the bound of $|H_{3,1}(f)|$ for $a_2=a_3=a_5=0$ and $a_4=1/3$.
\end{proof}
\subsection{Fourth Hankel Determinant for $\mathcal{S}^{*}_{e}$}
To continue, we state the next lemma, which will be utilized frequently.
\begin{lemma}\cite{bellv,shelly}\label{2 pomi lemma}
Let $p=1+\sum_{n=1}^{\infty}p_nz^n\in \mathcal{P}.$ Then
\begin{equation*}
|p_n|\leq 2, \quad n\geq 1,\label{2 caratheodory1}
\end{equation*}
\begin{equation*}
|p_{n+k}-\nu p_n p_k|\leq \begin{cases}
2, & 0\leq \nu\leq 1;\\
2|2\nu-1|,& otherwise,
\end{cases}\label{2 caratheodory2}
\end{equation*}
and \begin{equation*}
|p_1^3-\nu p_3|\leq
\begin{cases}2|\nu-4|,& \nu\leq 4/3;\\ \\
2\nu\sqrt{\dfrac{\nu}{\nu-1}},& 4/3<\nu.
\end{cases}\label{2 caratheodory3}
\end{equation*}
\end{lemma}
We derive the expression of the fourth Hankel determinant when $q=4$ and $n=1$ are put into equation (\ref{5hqn}) as follows :
\begin{equation}\label{5 h41}
H_{4,1}(f)=a_7H_{3,1}(f)-a_6T_1+a_5T_2-a_4T_3,
\end{equation}
where
\begin{equation}\label{5t1}
T_1:=a_6(a_3-a_2^2)+a_3(a_2a_5-a_3a_4)-a_4(a_5-a_2a_4),
\end{equation}
\begin{equation}\label{5t2}
T_2:=a_3(a_3a_5-a_4^2)-a_5(a_5-a_2a_4)+a_6(a_4-a_2a_3),
\end{equation}
and
\begin{equation}\label{5t3}
T_3:=a_4(a_3a_5-a_4^2)-a_5(a_2a_5-a_3a_4)+a_6(a_4-a_2a_3).\end{equation}
\noindent Now, using Lemma \ref{2 pomi lemma}, we first determine the bounds of $T_1$, $T _2$, and $T_3$.\\ By substituting the values of $a_i$'s $(i=2,3,...,6)$ in (\ref{5t1}) using (\ref{5 ea2})-(\ref{5 ea6}), we obtain
\begin{align*}
5529600T_1&=581 p_1^7+ 5040 p_1^4 p_3 + 25920 p_1^2 p_2 p_3- 7068 p_1^5 p_2 + 11040 p_1^3 p_4-115200 p_3 p_4 \\
& \quad +7920 p_1^3 p_2^2- 69120 p_2^2 p_3 +74880 p_1 p_2 p_4-25920 p_1 p_2^3 + 57600 p_1 p_3^2 \\
&\quad+ 138240 p_2 p_5-103680 p_1^2 p_5
\end{align*}
or
\begin{align}
5529600|T_1|&\leq | p_1^4(581 p_1^3+ 5040 p_3)| + |p_1^2 p_2(25920 p_3- 7068 p_1^3) |+|p_4 ( 11040 p_1^3 -115200 p_3 )|\nonumber \\
& \quad+| p_2^2(7920 p_1^3 - 69120 p_3)| + | p_1 p_2( 74880 p_4-25920 p_2^2)|+ |57600 p_1 p_3^2|\nonumber\\
&\quad + | p_5(138240 p_2- 103680 p_1^2)|.\label{5 et1}
\end{align}
Using Lemma \ref{2 pomi lemma} and the triangle inequality, we arrive at
\begin{align}
|p_1^4(581 p_1^3+ 5040 p_3)|\leq 235648, \quad |p_1^2 p_2(25920 p_3- 7068 p_1^3) |\leq 4976640\sqrt{\frac{15}{1571}},\label{5 et11}
\end{align}
\begin{align}
|p_4(11040 p_1^3 -115200 p_3) | \leq 1843200\sqrt{\frac{15}{217}}, \quad | p_2^2(7920 p_1^3 - 69120 p_3)|\leq 442368\sqrt{\frac{30}{17}}, \label{5 et12}
\end{align}
\begin{align}
|p_1 p_2 (74880 p_4-25920 p_2^2)|\leq 599040, \quad |57600 p_1 p_3^2|\leq 460800, \label{5 et13}
\end{align}
and\begin{equation} |p_5(138240 p_2 - 103680 p_1^2)| \leq 552960.\label{5 et14}\end{equation}
By considering equation (\ref{5 et1}) in view of (\ref{5 et11})-(\ref{5 et14}),
\begin{align*}
|T_1|&\leq \frac{1848448 + 4976640 \sqrt{\frac{15}{1571}} + 1843200 \sqrt{\frac{15}{217}} + 442368 \sqrt{\frac{30}{17}} }{5529600} \nonumber\\
&\approx 0.616137
\end{align*}
\noindent Now, we calculate the bound of $T_2$ in the similar way by substituting the values of $a_i$'s $(i=2,3,...,6)$ in (\ref{5t2}) from equations (\ref{5 ea2})-(\ref{5 ea6}), as follows:
\begin{align}
22118400T_2&= 235 p_1^8+ 8712 p_1^5 p_3 + 37440 p_1^3 p_2 p_3- 1156 p_1^6 p_2 -63360 p_1 p_2^2 p_3-14640 p_1^4 p_2^2\nonumber\\
&\quad + 161280 p_1 p_3 p_4 - 8400 p_1^4 p_4 + 368640 p_3 p_5- 76800 p_1^3 p_5- 8640 p_1^2 p_2^3 \nonumber \\
&\quad+172800 p_2^2 p_4 - 345600 p_4^2 - 40320 p_1^2 p_3^2- 184320 p_2 p_3^2+178560 p_1^2 p_2 p_4\nonumber\\
&\quad-184320 p_1 p_2 p_5\nonumber
\end{align}
or
\begin{align}
22118400|T_2|&\leq |p_1^5(235 p_1^3+ 8712 p_3)| + |p_1^3 p_2(37440 p_3- 1156 p_1^3)|+|- p_1 p_2^2(63360 p_3+14640 p_1^3)|\nonumber\\
&\quad + |p_1p_4(161280p_3 - 8400 p_1^3) | + |p_5(368640 p_3 - 76800 p_1^3)| +|-8640 p_1^2 p_2^3|\nonumber \\
&\quad+|p_4(172800 p_2^2 - 345600 p_4)|+|-p_3^2(184320 p_2 + 40320 p_1^2) |\nonumber\\
&\quad+|p_1 p_2(178560 p_1 p_4-184320 p_5) |. \label{5 et2}
\end{align}
Lemma \ref{2 pomi lemma} and the triangle inequality lead us to
\begin{align}
|p_1^5(235 p_1^3+ 8712 p_3)|\leq 617728,\quad |p_1^3 p_2(37440 p_3- 1156 p_1^3)|\leq 14376960 \sqrt{\frac{65}{9071}},\label{5 et21}
\end{align}
\begin{align}
| p_1 p_2^2(63360 p_3+14640 p_1^3)|\leq 1950720,\quad |p_1p_4(161280 p_3 - 8400 p_1^3)|\leq 737280 \sqrt{\frac{42}{13}}, \label{5 et22}
\end{align}
\begin{align}
|p_5(368640 p_3- 76800 p_1^3)|\leq 2949120 \sqrt{\frac{6}{19}}, \quad |8640 p_1^2 p_2^3|\leq 276480,\label{5 et23}
\end{align}
\begin{align}
|p_4(172800 p_2^2 - 345600 p_4)|\leq 1382400,\quad |p_3^2(184320 p_2+40320 p_1^2 ) |\leq 2119680, \label{5 et24}
\end{align}
and
\begin{align}
|p_1 p_2(178560 p_1 p_4-184320 p_5)|\leq 1474560.\label{5 et25}
\end{align}
By substituting the values from equations (\ref{5 et21})-(\ref{5 et25}) in equation (\ref{5 et2}),
\begin{align*}
|T_2|&\leq \frac{7821568 + 14376960 \sqrt{\frac{65}{9071}} + 2949120 \sqrt{\frac{6}{19}} + 737280 \sqrt{\frac{42}{13}}}{22118400} \nonumber\\
&\approx 0.543487
\end{align*}
Next, we determine the bound of $T_3$, by replacing the values of $a_i$'s $(i=2,3,...,6)$ from equations (\ref{5 ea2})-(\ref{5 ea6}) in (\ref{5t3}), as follows:
\begin{align*}
597196800T_3&=6120 p_1^8+ 143424 p_1^5 p_3 - 425 p_1^9-9000 p_1^6 p_3+ 9000 p_1^7 p_2+ 172800 p_1^4 p_2 p_3\\
&\quad + 302400 p_1^3 p_3^2- 2764800 p_3^3+1036800 p_1^3 p_2 p_4+6220800 p_2 p_3 p_4-17280 p_1^4 p_2^2\\
&\quad+ 9953280 p_3 p_5- 2073600 p_1^3 p_5+ 967680 p_1^3 p_2 p_3-64512 p_1^6 p_2-1036800 p_1 p_2 p_3^2 \\
&\quad- 32400 p1^5 p_2^2-777600 p_1^2 p_2^2 p_3 + 1244160 p_1 p_3 p_4-259200 p_1^4 p_4-97200 p_1^5 p_4\\
&\quad + 1555200 p_1 p_2^2 p_4 - 4665600 p_1 p_4^2- 414720 p_1 p_2^2 p_3- 4976640 p_1 p_2 p_5-829440 p_2 p_3^2\\
&\quad- 829440 p_1^2 p_3^2+414720 p_1^2 p_2^3- 622080 p_1^2 p_2 p_4-172800 p_1^3 p_2^3
\end{align*}
or
\begin{align}
597196800|T_3|&\leq |p_1^5(6120 p_1^3+143424 p_3)|+|p_1^6( 425 p_1^3+9000 p_3)|+ |p_1^4 p_2(9000 p_1^3+ 172800p_3)|\nonumber\\
&\quad + | p_3^2(302400 p_1^3- 2764800 p_3)|+|p_2 p_4(1036800 p_1^3 +6220800 p_3)|+|17280 p_1^4 p_2^2|\nonumber\\
&\quad+ |p_5(9953280 p_3-2073600 p_1^3)|+ |p_1^3 p_2 (967680 p_3-64512 p_1^3)|+|1036800 p_1 p_2 p_3^2|\nonumber\\
&\quad +|p_1^2 p_2^2(32400 p1^3+777600 p_3)|+ |p_1p_4(1244160 p_3-259200 p_1^3)|+|97200 p_1^5 p_4 |\nonumber \\
&\quad + |p_1p_4(1555200 p_2^2 - 4665600 p_4)|+|p_1 p_2( 414720 p_2 p_3+ 4976640 p_5)|\nonumber\\
&\quad+|p_3^2(829440 p_2 + 829440 p_1^2)|+|p_1^2 p_2(414720p_2^2- 622080 p_4)| +|172800 p_1^3 p_2^3|.\label{5 et3}
\end{align}
By applying Lemma \ref{2 pomi lemma} and the triangle inequality, \begin{align}
|p_1^5(6120 p_1^3+143424 p_3)|\leq 10745856, \quad |p_1^6( 425 p_1^3+9000 p_3)|\leq 1369600, \label{5 et31}
\end{align}
\begin{align}
|p_1^4 p_2(9000 p_1^3+ 172800p_3)|\leq 13363200,\quad | p_3^2(302400 p_1^3- 2764800 p_3)|\leq 58982400 \sqrt{\frac{3}{19}},\label{5 et32}
\end{align}
\begin{align}
|p_2 p_4(1036800 p_1^3 +6220800 p_3)|\leq 82944000, \quad |p_5(9953280 p_3-2073600 p_1^3)|\leq 79626240 \sqrt{\frac{6}{19}}, \label{5 et33}
\end{align}
\begin{align}
|p_1^3 p_2 (967680 p_3-64512 p_1^3)|\leq 2211840 \sqrt{210},\quad |p_1^2 p_2^2(32400 p1^3+777600 p_3)|\leq 29030400,\label{5 et34}
\end{align}
\begin{align}
|p_1p_4(1244160 p_3-259200 p_1^3)|\leq 19906560 \sqrt{\frac{6}{19}}, \quad |p_1p_4(1555200 p_2^2 - 4665600 p_4)|\leq 37324800,\label{5 et35}
\end{align}
\begin{align}
|p_1 p_2( 414720 p_2 p_3+ 4976640 p_5)|\leq 46448640, \quad |p_3^2(829440 p_2 + 829440 p_1^2)|\leq 19906560,\label{5 et36}
\end{align}
\begin{align}
|17280 p_1^4 p_2^2|+|1036800 p_1 p_2 p_3^2|+|97200 p_1^5 p_4|+|172800 p_1^3 p_2^3|\leq 34974720,\label{5 et37}
\end{align}
and \begin{align}
|p_1^2 p_2(414720p_2^2- 622080 p_4)|\leq 9953280. \label{5 et38}
\end{align}
So, by using equations (\ref{5 et31})-(\ref{5 et38}) in equation (\ref{5 et3}), we get
\begin{align*}
|T_3|&\leq \frac{286061056 + 58982400 \sqrt{\frac{3}{19}} + 99532800 \sqrt{\frac{6}{19}} + 2211840 \sqrt{210}}{597196800}\nonumber\\
&\approx 0.665582
\end{align*}
\begin{remark}\label{5st}
On the basis of the above calculations, the bounds of $T_1$, $T_2$ and $T_3$ are $0.616137$, $0.543487$ and $0.665582$ respectively.
\end{remark}
Now, to proceed ahead, we seek the bounds of initial coefficients $a_i$ for $i=2,3,4,5$. These bounds have been derived in \cite{zap2019}, presented below in the form of the following remark.
\begin{remark}\label{5 zapa5s}
For $f\in \mathcal{S}^{*}_{e},$ $|a_2|\leq 1,$ $|a_3|\leq 3/4$, $|a_4|\leq 17/36$ and $|a_5|\leq 25/72$. Here the first three bounds are sharp.
\end{remark}
It is significantly more difficult to find coefficient boundaries for $n>5$. To solve this problem, we use the Lemma \ref{2 pomi lemma} to establish the bounds for the sixth and seventh coefficient for functions of the class $\mathcal{S}^{*}_{e}$ in the following Lemma.
\begin{lemma}\label{5 a6a7lemma}
Let $f\in \mathcal{S}_{e}^{*}.$ Then $|a_6|\leq 587/1800\approx 0.326111$ and $|a_7|\leq 1397/4320\approx 0.32338$.
\end{lemma}
\begin{proof}
By suitably rearranging the terms given in equation (\ref{5 ea6}), we have
\begin{equation*}
57600a_6= 220 p_1^3 p_2 - 480 p_1^2 p_3 - 480 p_1 p_2^2 + 720 p_1 p_4-17 p_1^5- 480 p_2 p_3+ 5760p_5.
\end{equation*}
Using triangle inequality, it can be viewed as
\begin{align}
57600|a_6|&\leq |p_1^2(220 p_1 p_2-480 p_3) |+ |p_1( 720 p_4 - 480p_2^2)|+|-17 p_1^5|\nonumber\\
&\quad +|5760p_5- 480 p_2 p_3|.\label{5 a6}
\end{align}
Using Lemma \ref{2 pomi lemma}, we arrive at the following inequalities:
\begin{equation}
|p_1^2(220 p_1 p_2-480 p_3) |\leq 3840,\quad |p_1( 720 p_4 - 480p_2^2)|\leq 2880, \quad |17 p_1^5|\leq 544,\label{5 a61} \end{equation}
and
\begin{equation}
|5760p_5- 480 p_2 p_3|\leq 11520. \label{5 a62}
\end{equation}
By using equation (\ref{5 a61}) and (\ref{5 a62})
in equation (\ref{5 a6}), we obtain
\begin{equation*}
|a_6|\leq \frac{587}{1800}\approx 0.326111.
\end{equation*}
Similarly, considering equation (\ref{5 ea7}), we have
\begin{align*}
8294400a_7&=881 p_1^6 - 13260 p_1^4 p_2 + 48240 p_1^2 p_2^2 - 14400 p_2^3 + 29040 p_1^3 p_3- 56160 p_1^2 p_4\nonumber\\
&\quad +69120 p_1 p_5-106560 p_1 p_2 p_3 - 57600 p_3^2 - 86400 p_2 p_4.
\end{align*}
Through the triangle inequality, it can also be seen as
\begin{align}
8294400|a_7|&\leq|p_1^4(881 p_1^2 - 13260 p_2)| + |p_2^2 (48240 p_1^2 - 14400 p_2)|+| p_1 (69120 p_5-106560p_2 p_3)|\nonumber\\
&\quad+ | p_1^2(29040 p_1 p_3- 56160 p_4)|+|- 57600 p_3^2|+| - 86400 p_2 p_4 |.\label{5a7}
\end{align}
Lemma \ref{2 pomi lemma} takes us to the following:
\begin{align}
|p_1^4(881 p_1^2 - 13260 p_2)|\leq 424320,\quad |p_2^2(48240 p_1^2 - 14400 p_2)|\leq 656640, \label{5 ea71}
\end{align}
\begin{align}\label{5 ea72}
|p_1(69120 p_5-106560 p_2 p_3)|\leq 576000, \quad |p_1^2(29040 p_1 p_3- 56160 p_4)|\leq 449280,
\end{align}
and \begin{align}\label{5 ea73}
| 57600 p_3^2 |+| 86400 p_2 p_4 |\leq 576000.
\end{align}
By substituting the values from equations (\ref{5 ea71})-(\ref{5 ea73}) in (\ref{5a7}), we have
\begin{equation*}
|a_7|\leq
\frac{1397}{4320}\approx 0.32338.
\end{equation*}
\end{proof}
\begin{theorem}\label{5 h41bound}
Let $f\in \mathcal{S}^{*}_{e}.$ Then
\begin{equation*}
|H_{4,1}(f)|\leq 0.29059.\end{equation*}
\end{theorem}
The proof of the above theorem follows by substituting the values obtained from Theorem \ref{5 sharph31},
Remark \ref{5st}, Remark \ref{5 zapa5s} and Lemma \ref{5 a6a7lemma} in the equation (\ref{5 h41}), therefore, it is skipped here.
\section{Hankel Determinants for $\mathcal{C}_{e}$}
\subsection{Preliminaries}
When $f\in \mathcal{C}_{e}$, we replace the L.H.S of equation (\ref{3 formulaai}) by $1+zf''(z)/f'(z)$ and arrive at the following equation
\begin{equation*}\label{3 formulaaic}
1+\dfrac{zf''(z)}{f'(z)}=e^{w(z)}.
\end{equation*}
Proceeding on the similar lines as done for the class $\mathcal{S}^{*}_{e}$,
we obtain $a_i (i=2,3,...,7)$ in terms of $p_j (j=1,2,...,5)$, then compare the corresponding coefficients as follows:
\begin{equation}\label{5 ea2c}
a_2=\dfrac{1}{4}p_1,\quad a_3=\dfrac{1}{48}\bigg(p_1^2+4p_2\bigg), \quad a_4=\dfrac{1}{1152}\bigg(-p_1^3 +12 p_1 p_2 + 48 p_3\bigg),
\end{equation}
\begin{equation}\label{5 ea5c}
a_5=\dfrac{1}{5760}\bigg( p_1^4 -12 p_1^2 p_2 +24 p_1 p_3 + 144 p_4\bigg),
\end{equation}
\begin{align}
a_6&=\frac{1}{345600}\bigg(-17 p_1^5 + 220 p_1^3 p_2 - 480 p_1 p_2^2 - 480 p_1^2 p_3 - 480 p_2 p_3+720 p_1 p_4+ 5760 p_5\bigg),\label{5 ea6c}
\end{align}
and
\begin{align}
a_7&=\frac{1}{58060800}\bigg(881 p_1^6 - 13260 p_1^4 p_2 + 48240 p_1^2 p_2^2 - 14400 p_2^3 +29040 p_1^3 p_3 - 106560 p_1 p_2 p_3 \nonumber\\\label{5 ea7c}
&\quad\quad\quad\quad\quad\quad - 57600 p_3^2 - 56160 p_1^2 p_4- 86400 p_2 p_4 + 69120 p_1 p_5\bigg).
\end{align}
\subsection{Sharp Third Hankel Determinant for $\mathcal{C}_{e}$}
In this section, we establish the best bound of $|H_{3,1}(f)|$ for functions that belong to the class $\mathcal{C}_{e}$.
\begin{theorem}\label{5 ceh31}
Let $f\in \mathcal {C}_{e}.$ Then
\begin{equation}
|H_{3,1}(f)|\leq \dfrac{1}{144}.\label{5 sharph31c}
\end{equation}
This bound is sharp.
\end{theorem}
\begin{proof}
We follow the same steps which were used to prove Theorem \ref{5 sharph31}. The values of $a _i's(i=2,3,4,5)$ from equations (\ref{5 ea2c}) and (\ref{5 ea5c}) are substituted into equation (\ref{1h3}). Thus
\begin{align*}
H_{3,1}(f)&=\dfrac{1}{6635520}\bigg(-173 p^6 + 552 p^4 p_2 - 1872 p^2 p_2^2 - 3840 p_2^3 + 2208 p^3 p_3 + 8064 p p_2 p_3\\
&\quad\quad \quad \quad \quad \quad-11520 p_3^2 - 6912 p^2 p_4 + 13824 p_2 p_4\bigg).
\end{align*}
Using (\ref{b2})-(\ref{b4}) for simplification, we arrive at
$$H_{3,1}(f)=\dfrac{1}{6635520}\bigg(\alpha_1(p,\gamma)+\alpha_2(p,\gamma)\eta+\alpha_3(p,\gamma){\eta}^2+\psi(p,\gamma,\eta)\rho\bigg),$$
where $\gamma,\eta,\rho\in \mathbb {D},$
\begin{align*} \alpha_1(p,\gamma):&=-5p^6-180{\gamma}^2p^2(4-p^2)^2+1536{\gamma}^3(4-p^2)^2-240{\gamma}^3p^2(4-p^2)^2+144{\gamma}^4p^2(4-p^2)^2\\
&\quad+12{\gamma}p^4(4-p^2)-120p^4{\gamma}^2(4-p^2),\\
\alpha_2(p,\gamma):&=(1-|\gamma|^2)(4-p^2)(240p^3-288p{\gamma}(4-p^2)-576p\gamma^2(4-p^2)),\\
\alpha_3(p,\gamma):&=(1-|\gamma|^2)(4-p^2)(-2880(4-p^2)-576|\gamma|^2(4-p^2)),\\
\psi(p,\gamma,\eta):&=3456\gamma(1-|\gamma|^2)(4-p^2)^2(1-|\eta|^2).
\end{align*}
Since $|\rho|\leq 1$, also for the simplicity of the calculations, assume $x=|\gamma|$ and $y=|\eta|$,
\begin{align*}
|H_{3,1}(f)|\leq \dfrac{1}{6635520}\bigg(|\alpha_1(p,\gamma)|+|\alpha_2(p,\gamma)|y+|\alpha_3(p,\gamma)|y^2+|\psi(p,\gamma,\eta)|\bigg)\leq N(p,x,y),
\end{align*}
where
\begin{equation}
N(p,x,y)=\dfrac{1}{6635520}\bigg(n_1(p,x)+n_2(p,x)y+n_3(p,x)y^2+n_4(p,x)(1-y^2)\bigg),\label{3s new}
\end{equation}
with
\begin{align*}
n_1(p,x):&=5p^6+180x^2p^2(4-p^2)^2+1536x^3(4-p^2)^2+240x^3p^2(4-p^2)^2+144x^4p^2(4-p^2)^2\\
&\quad +12xp^4(4-p^2)+120p^4x^2(4-p^2),\\
n_2(p,x):&=(1-x^2)(4-p^2)(240p^3+288px(4-p^2)+576px^2(4-p^2)),\\
n_3(p,x):&=(1-x^2)(4-p^2)(2880(4-p^2)+576x^2(4-p^2)),\\
n_4(p,x):&=3456x(1-x^2)(4-p^2)^2.
\end{align*}
We must maximise $N(p,x,y)$ in the closed cuboid $V:[0,2]\times [0,1]\times [0,1]$. By identifying the maximum values on the twelve edges, the interior of $V$, and the interiors of the six faces, we can prove this.
\begin{enumerate}
\item We start by taking into account, every interior point of $V$. Assume that $(p,x,y)\in (0,2)\times (0,1)\times (0,1).$ We partially differentiate equation (\ref{3s new}) with respect to $y$ to locate the points of maxima in the interior of $V$. We obtain
\begin{align*}
\dfrac{\partial N}{\partial y}&=\dfrac{(1 - x^2)(4 - p^2) }{138240} \bigg(24 p x (1 + 2 x)-p^3 (-5 + 6 x + 12 x^2)+ 96 (5-6x+x^2)y \\
& \quad \quad \quad\quad\quad\quad \quad\quad\quad -24 p^2 (5 - 6 x + x^2) y\bigg).
\end{align*}
Now $\dfrac{\partial N}{\partial y}=0$ gives
\begin{equation*}
y=y_1:=\dfrac{5 p^3 + 6 p x (4-p^2)(1+2x)}{24 (4- p^2) (6 x - x^2-5)}.
\end{equation*}
Since $y_1$ must be a member of $(0,1)$ for critical points to exist, this is only possible if
\begin{equation}
24 (20 + (p-24) x + (4 + 2 p - p^2) x^2)+p^3 (5 - 6 x - 12 x^2)<24p^2(5-6x).\label{5c h1}
\end{equation}
Now, we find the solutions satisfying the inequality (\ref{5c h1})
for the existence of critical points using the hit and trial method. If we assume $p$ tends to 0 and 2, then no such $x\in (0,1)$ exists satisfying equation (\ref{5c h1}). Similarly, if we take $x$ tending to $0$ and $1$, then there does not exist any $p\in (0,2)$ satisfying equation (\ref{5c h1}). Therefore, we conclude that the function $N$ has no critical point in $(0,2)\times (0,1)\times (0,1).$
\item Now, we study the interior of each of the six faces of the cuboid $V$.\\
When $p=0, N(p,x,y)$ becomes
\begin{equation}
c_1(x,y):=\dfrac{y^2(15 - 12 x^2 - 3 x^4) + 18 x (1 - y^2) - 2 x^3 (5-9 y^2)}{2160},\quad x,y\in (0,1).\label{5c 9.4}
\end{equation}
Since
\begin{equation*}
\dfrac{\partial c_1}{\partial y}=\dfrac{y(1-x)^2(x+1)(5-x)}{360}\neq 0,\quad x,y\in (0,1),
\end{equation*}
we note that, in $(0,1)\times(0,1)$, $c_1$ does not have any critical point. \\
\noindent When $p=2, N(p,x,y)$ settles into
\begin{equation}
N(2,x,y):=\dfrac{1}{20736},\quad x,y\in (0,1).\label{5c 9.3}
\end{equation}
\noindent When $x=0, N(p,x,y)$ turns into
\begin{equation}
c_2(p,y):=\dfrac{(p^3 + 96 y - 24 p^2 y)^2}{1327104},\quad p\in (0,2)\quad \text{and}\quad y\in (0,1). \label{5c 9.1}
\end{equation}
We solve $\partial c_2/\partial p=0$ and $\partial c_2/\partial y=0$ to locate the points of maxima. On solving $\partial c_2/\partial y=0,$ we obtain
\begin{equation*}
y=-\dfrac{p^3}{24(4-p^2)}(=:y_p)
\end{equation*}
Upon calculations, we observe that such $y_p$ does not belong to $(0,1)$. Consequently, no such critical point of $c_2$ exists in $(0,2)\times(0,1)$.\\
When $x=1, N(p,x,y)$ becomes
\begin{equation}
N(p,1,y)=c_3(p,y):=\dfrac{24576 - 3264 p^2 - 2448 p^4 + 437 p^6}{6635520}, \quad p\in (0,2).\label{5c 9.2}
\end{equation}
And while computing $\partial c_3/\partial p=0$, we notice that $c_3$ has no critical point in $(0,2).$\\
\noindent When $y=0, N(p,x,y)$ reduces to
\begin{align*}
c_4(p,x):&=\dfrac{1}{6635520}\bigg(6144 x (9 - 5 x^2) + 192 p^2 x (-144 + 15 x + 100 x^2 + 12 x^3)\\
&\quad \quad\quad \quad\quad \quad -48 p^4 x (-73 + 20 x + 80 x^2 + 24 x^3)\\
&\quad \quad\quad \quad \quad\quad+p^6 (5 - 12 x + 60 x^2 + 240 x^3 + 144 x^4)\bigg).
\end{align*}
Calculations lead to,
\begin{align*}
\dfrac{\partial c_4}{\partial x}&=\dfrac{1}{6635520}\bigg(-61440 x^2 - 6144 (-9 + 5 x^2) + 192 p^2 x (15 + 200 x + 36 x^2)\\
&\quad \quad\quad \quad \quad \quad -48 p^4 x (20 + 160 x + 72 x^2) +
192 p^2 (-144 + 15 x + 100 x^2 \\
&\quad \quad\quad \quad \quad \quad+ 12 x^3)-48 p^4 (-73 + 20 x + 80 x^2 + 24 x^3) +
p^6 (-12 + 120 x \\
&\quad \quad\quad \quad \quad \quad + 720 x^2 + 576 x^3) \bigg)
\end{align*}
and \begin{align*}
\dfrac{\partial c_4}{\partial p}&=\dfrac{1}{6635520}\bigg(384 p x (-144 + 15 x + 100 x^2 + 12 x^3)-192 p^3 x (-73 + 20 x \\
&\quad\quad\quad\quad\quad \quad + 80 x^2+ 24 x^3)+6 p^5 (5 - 12 x + 60 x^2 + 240 x^3 + 144 x^4)\bigg).
\end{align*}
No solution exist for the system of equations, $\partial c_4/\partial x=0$ and $\partial c_4/\partial p=0$, according to a numerical calculation, in $(0,2)\times (0,1).$\\
\noindent When $y=1$, $N(p,x,y)$ reduces to
\begin{align*}
c_5(p,x):&=\frac{1}{6635520}\bigg(5 p^6 +(4-p^2)( 12 p^4 x + 120 p^4 x^2 + 180 p^2 (4 - p^2) x^2 \\
& \quad\quad\quad\quad\quad\quad+1536 (4-p^2) x^3 + 240 p^2 (4-p^2) x^3 + 144 p^2 (4-p^2) x^4 \\
&\quad\quad\quad\quad\quad\quad+ 3456 (4-p^2)x (1 - x^2)+48 (1- x^2) (p^3 (5-6x \\
&\quad\quad\quad\quad\quad\quad- 12 x^2)+24 p x (1 + 2 x)))\bigg).
\end{align*}
The two equations $\partial c_5/\partial x=0$ and $\partial c_5/\partial p=0$ also do not assume any solution in $(0,2)\times (0,1).$
\item Next, we check the maximum values of $N(p,x,y)$ obtained on the edges of the cuboid $V$. From equation (\ref{5c 9.1}), we have $N(p,0,0)=t_1(p):=p^6/1327104.$ It is easy to observe that $t_1'(p)=0$ for $p=0$
in the interval $[0,2]$. The maximum value of $t_1(p)$ is $0.$
Now the equation (\ref{5c 9.1}) reduces to $N(p,0,1)=t_2(p):=(96 - 24 p^2 + p^3)^2/1327104$ at $y=1$. Since, $t_2'(p)<0$ in $[0,2]$, hence $p=0$ is the point of maxima. Thus
\begin{equation*}
N(p,0,1)\leq \dfrac{1}{144}, \quad p\in [0,2].
\end{equation*}
Through computations, equation (\ref{5c 9.1}) shows that $N(0,0,y)$ attains its maxima at $y=1.$ Hence
\begin{equation*}
N(0,0,y)\leq \dfrac{1}{144}, \quad y\in [0,1].
\end{equation*}
Since, the equation (\ref{5c 9.2}) is free from $x$, we have $N(p,1,1)=N(p,1,0)=t_3(p):=(24576 - 3264 p^2 - 2448 p^4 + 437 p^6)/6635520.$ Now, we observe that $t_3'(p)<0$ in $[0,2]$, consequently, $t_3(p)$ attains its maximum at $p=0$. Hence
\begin{equation*}
N(p,1,1)=N(p,1,0)\leq 0.0037037,\quad p\in [0,2].
\end{equation*}
On substituting $p=0$ in equation (\ref{5c 9.2}), we get, $N(0,1,y)=1/270.$ The equation (\ref{5c 9.3}) does not contain any variable such as $p$, $x$ and $y$. Therefore, the maxima of $N(p,x,y)$ on the edges
is given by
\begin{equation*}
N(2,1,y)=N(2,0,y)=N(2,x,0)=N(2,x,1)=\dfrac{1}{20736},\quad x,y\in [0,1].
\end{equation*}
Using equation (\ref{5c 9.4}), we obtain $N(0,x,1)=t_4(x):=(15 - 12 x^2 + 8 x^3 - 3 x^4)/2160.$ Upon calculations, we see that $t_4$ is a decreasing function in $[0,1]$ and its maximum value is achieved at $x=0.$ Hence
\begin{equation*}
N(0,x,1)\leq \dfrac{1}{144},\quad x\in [0,1].
\end{equation*}
On again using equation (\ref{5c 9.4}), we get $N(0,x,0)=t_5(x):=x(9-5x^2)/1080.$ On further calculations, we get $t_5'(x)=0$ for $x=\beta_0:=\sqrt{3/5}.$ Also, $t_5(x)$ increases in $[0,\beta_0)$ and decreases in $(\beta_0,1].$ So, $\beta_0$ is the point of maxima. Thus
\begin{equation*}
N(0,x,0)\leq 0.00430331,\quad x\in [0,1].
\end{equation*}
\end{enumerate}
Because of all the cases discussed above, the inequality (\ref{5 sharph31c}) holds.\\
The function $f_2(z)\in \mathcal{C}_{e}$, defined as
\begin{equation*}
f_2(z)=\int_{0}^{z}\bigg(\exp\bigg(\int_{0}^{y}\dfrac{e^{t^3}-1}{t}dt\bigg)\bigg)dy=z+\dfrac{z^4}{12}+\dfrac{5z^7}{252}+\cdots,
\end{equation*}
with $f(0)=f'(0)-1=0$, plays the role of an extremal function for the bounds of $|H_{3,1}(f)|$ having values $a_3=a_5=0$ and $a_4=1/12.$
\end{proof}
\subsection{Fourth Hankel Determinant for $ \mathcal{C}_{e}$}
By selecting $q=4$ and $n=1$ in the equation (\ref{5hqn}), the expression of $|H_{4,1}(f)|$ can be obtained for functions in the class $\mathcal{C}_{e}$, which is given as follows:
\begin{equation}\label{5 ch41}
H_{4,1}(f)=a_7H_{3,1}(f)-a_6U_1+a_5U_2-a_4U_3.
\end{equation}
Here
\begin{equation}\label{5cu1form}
U_1:=a_6(a_3-a_2^2)+a_3(a_2a_5-a_3a_4)-a_4(a_5-a_2a_4),
\end{equation}
\begin{equation}\label{5cu2form}
U_2:=a_3(a_3a_5-a_4^2)-a_5(a_5-a_2a_4)+a_6(a_4-a_2a_3),
\end{equation}
and
\begin{equation}\label{5cu3form}
U_3:=a_4(a_3a_5-a_4^2)-a_5(a_2a_5-a_3a_4)+a_6(a_4-a_2a_3).
\end{equation}
\noindent We start by determining the bounds for $U_1$, $U_2$, and $U_3$.\\
By substituting the values of $a_i$'s $(i=2,3,...,6)$ in (\ref{5cu1form}) from equations (\ref{5 ea2c})-(\ref{5 ea6c}), we obtain
\begin{align*}
132710400U_1&=487 p_1^7 - 6304 p_1^5 p_2 + 11440 p_1^3 p_2^2 - 24960 p_1 p_2^3 +5280 p_1^4 p_3+ 34560 p_1 p_3^2 \\
& \quad + 19200 p_1^2 p_2 p_3-53760 p_2^2 p_3 + 57600 p_1 p_2 p_4 - 138240 p_3 p_4\\
&\quad + 184320 p_2 p_5-92160 p_1^2 p_5+ 8640 p_1^3 p_4,
\end{align*}
can also be viewed as the following due to the triangle inequality,
\begin{align}
132710400|U_1|&\leq |p_1^5(487 p_1^2 - 6304 p_2)| + |p_1 p_2^2(11440 p_1^2 - 24960 p_2)| + |p_1 p_3(5280 p_1^3+ 34560p_3)|\nonumber \\
& \quad+| p_2 p_3 (19200 p_1^2-53760 p_2) | + | p_4(57600 p_1 p_2 - 138240 p_3)|\nonumber\\
&\quad + |p_5(184320 p_2 -92160 p_1^2)|+ |8640 p_1^3 p_4|.\label{5 ecu1}
\end{align}
Using Lemma \ref{2 pomi lemma}, we arrive at
\begin{align}
|p_1^5(487 p_1^2 - 6304 p_2)| \leq 40320, \quad |p_1 p_2^2(11440 p_1^2 - 24960 p_2)|\leq 399360,\label{5 ecu11}
\end{align}
\begin{align}
|p_1 p_3(5280 p_1^3+ 34560p_3)| \leq 445440, \quad | p_2 p_3( 19200 p_1^2-53760 p_2) | \leq 430080, \label{5 ecu12}
\end{align}
\begin{align}
| p_4(57600 p_1 p_2 -138240 p_3)|\leq 55296, \quad |p_5(184320 p_2-92160 p_1^2)|\leq 73728, \label{5 ecu13}
\end{align}
and\begin{equation} |8640 p_1^3 p_4| \leq 138240.\label{5 ecu14}\end{equation}
By substituting the values obtained from equations (\ref{5 ecu11})-(\ref{5 ecu14}) in (\ref{5 ecu1}),
\begin{align*}
|U_1|&\leq
\frac{4121}{345600}\approx 0.0119242
\end{align*}
We replace the values of $a_i$'s $(i=2,3,...,6)$ from equations (\ref{5 ea2c})-(\ref{5 ea7c}) in equation (\ref{5cu2form}) and proceed on the same lines to obtain the bound of $U_2$
\begin{align*}
1592524800U_2&= 463 p_1^8 - 2732 p_1^6 p_2 - 23472 p_1^4 p_2^2 - 14400 p_1^2 p_2^3 + 14592 p_1^5 p_3 - 108288 p_1^2 p_3^2 \\
&\quad +92928 p_1^3 p_2 p_3- 138240 p_1 p_2^2 p_3 + 373248 p_1^2 p_2 p_4- 25344 p_1^4 p_4 +276480 p_2^2 p_4 \\
&\quad - 995328 p_4^2+1105920 p_3 p_5- 276480 p_1 p_2 p_5+ 221184 p_1 p_3 p_4\nonumber\\
&\quad - 161280 p_1^3 p_5- 322560 p_2 p_3^2,
\end{align*}
by implementing the triangle inequality,
\begin{align}
1592524800|U_2|&\leq |p_1^6( 463 p_1^2 - 2732p_2)| + | p_1^2 p_2^2(- 23472 p_1^2 - 14400 p_2)|+| p_1^2 p_3(14592 p_1^3 - 108288 p_3)|\nonumber\\
&\quad + |p_1p_2 p_3(92928 p_1^2 - 138240p_2)| + |p_1^2p_4(373248 p_2- 25344 p_1^2) |\nonumber \\
&\quad+| p_4(276480 p_2^2 - 995328 p_4)|+|p_5(1105920 p_3- 276480 p_1 p_2) |\nonumber\\
&\quad+|221184 p_1 p_3 p_4|+| - 161280 p_1^3 p_5|+|- 322560 p_2 p_3^2|. \label{5 ecu2}
\end{align}
By applying Lemma \ref{2 pomi lemma}, we have
\begin{align}
| p_1^6(463 p_1^2 - 2732 p_2)|\leq 349696,\quad |p_1^2 p_2^2( 23472 p_1^2+14400 p_2)|\leq 1963008,\label{5 ecu21}
\end{align}
\begin{align}
|p_1^2 p_3(14592 p_1^3 - 108288 p_1^2 p_3^2|\leq 866304\sqrt{\frac{282}{61}},\quad |p_1p_2 p_3(92928 p_1^2- 138240p_2)|\leq 2211840, \label{5 ecu22}
\end{align}
\begin{align}
|p_1^2p_4(373248 p_2- 25344 p_1^2)|\leq 5971968, \quad | p_4(276480 p_2^2 - 995328 p_4)|\leq 3981312,\label{5 ecu23}
\end{align}
\begin{align}
| p_5(1105920 p_3- 276480 p_1 p_2)|\leq 4423680, \label{5 ecu24}
\end{align}
and
\begin{align}
|221184 p_1 p_3 p_4|+|161280 p_1^3 p_5|+|322560 p_2 p_3^2|\leq 6045696.\label{5 ecu25}
\end{align}
By substituting the values from equations (\ref{5 ecu21})-(\ref{5 ecu25}) in (\ref{5 ecu2}),
\begin{align*}
|U_2|&\leq \frac{24947200 + 866304 \sqrt{\frac{282}{61}}}{1592524800}\approx 0.0168348
\end{align*}
Again, substitute the values of $a_i$'s $(i=2,3,...,6)$ from equations (\ref{5 ea2c})-(\ref{5 ea7c}) in (\ref{5cu3form}) and proceed to calculate the bound of $U_3$ in the same manner.
\begin{align*}
38220595200U_3&=11424 p_1^8- 128256 p_1^6 p_2 + 10812 p_1^7 p_2 - 503 p_1^9 + 69120 p_1^4 p_2^2 + 552960 p_1^2 p_2^3\\
&\quad - 42192 p_1^5 p_2^2- 181440 p_1^3 p_2^3+206208 p_1^4 p_2 p_3-11664 p_1^6 p_3+ 1889280 p_1^3 p_2 p_3\\
&\quad -1658880 p_1 p_2^2 p_3 - 2211840 p_1^2 p_3^2-2211840 p_2 p_3^2 + 283392 p_1^3 p_3^2- 967680 p_1 p_2 p_3^2 \\
&\quad + 3317760 p_1 p_3 p_4 - 483840 p_1^4 p_4 + 1271808 p_1^3 p_2 p_4 - 117504 p_1^5 p_4 \\
&\quad + 1658880 p_1 p_2^2 p_4- 5971968 p_1 p_4^2+ 6635520 p_2 p_3 p_4 - 331776 p_1^2 p_3 p_4 \\
&\quad + 26542080 p_3 p_5-6635520 p_1 p_2 p_5+ 244224 p_1^5 p_3- 794880 p_1^2 p_2^2 p_3 \\
&\quad- 2764800 p_3^3 - 829440 p_1^2 p_2 p_4- 3870720 p_1^3 p_5,
\end{align*}
can be visualized as the following with the help of the triangle inequality,
\begin{align}
38220595200|U_3|&\leq | p_1^6(11424 p_1^2- 128256 p_2)|+| p_1^7(10812 p_2 - 503 p_1^2) |+ | p_1^2 p_2^2(69120 p_1^2 + 552960p_2)|\nonumber\\
&\quad + | p_1^3 p_2^2(- 42192 p_1^2- 181440 p_2)|+|p_1^4 p_3(206208p_2-11664 p_1^2 )|\nonumber\\
&\quad+ |p_1 p_2p_3(1889280 p_1^2-1658880p_2)|+|p_3^2(-2211840 p_1^2 -2211840 p_2)| \nonumber\\
&\quad +|p_1 p_3^2(283392 p_1^2- 967680 p_2)|+|p_1 p_4(3317760p_3 - 483840 p_1^3) |\nonumber \\
&\quad +|p_1^3 p_4(1271808p_2-117504 p_1^2 )|+|p_1p_4(1658880p_2^2- 5971968 p_4)|\nonumber\\
&\quad+|p_3 p_4 (6635520 p_2 - 331776 p_1^2) |+ |p_5(26542080 p_3-6635520 p_1 p_2)|\nonumber\\
&\quad+| 244224 p_1^5 p_3- 794880 p_1^2 p_2^2 p_3- 2764800 p_3^3 - 829440 p_1^2p_2 p_4- 3870720 p_1^3 p_5|.\label{5 ecu3}
\end{align}
By applying Lemma \ref{2 pomi lemma}, we get
\begin{align}
| p_1^6(11424 p_1^2- 128256 p_2)|\leq 16416768, \quad | p_1^7(10812 p_2 - 503 p_1^2) |\leq 2767872 \label{5 ecu31}
\end{align}
\begin{align}
|p_1^2 p_2^2( 69120 p_1^2 + 552960p_2)|\leq 22118400,\quad |p_1^3 p_2^2(- 42192 p_1^2- 181440p_2)|\leq 17012736 ,\label{5 ecu32}
\end{align}
\begin{align}
|p_1^4 p_3(206208 p_2-11664 p_1^2)|\leq 13197312, \quad |p_1 p_2p_3(1889280 p_1^2-1658880 p_2)|\leq 33914880, \label{5 ecu33}
\end{align}
\begin{align}
|p_3^2(2211840 p_1^2+2211840 p_2)|\leq 5308416,\quad |p_1 p_3^2(283392 p_1^2- 967680p_2)|\leq 15482880,\label{5 ecu34}
\end{align}
\begin{align}
|p_1 p_4(3317760p_3 - 483840 p_1^3)|\leq 106168320 \sqrt{\frac{3}{41}}, \quad |p_1^3 p_4(1271808p_2 -11750 p_1^2 )|\leq 40697856,\label{5 ecu35}
\end{align}
\begin{align}
|p_1p_4(1658880p_2^2- 5971968p_4)|\leq 47775744, \quad |p_3 p_4 (6635520 p_2 - 331776 p_1^2)|\leq 53084160,\label{5 ecu36}
\end{align}
\begin{align}
|p_5(26542080 p_3-6635520 p_1 p_2)|\leq 106168320,\label{5 ecu37}
\end{align}
and \begin{align}
| 244224 p_1^5 p_3|+|794880 p_1^2 p_2^2 p_3|+|2764800 p_3^3 |+|829440 p_1^2 p_2p_4|+| 3870720 p_1^3 p_5|\leq 138387456. \label{5 ecu38}
\end{align}
So, by using equations (\ref{5 ecu31})-(\ref{5 ecu38}) in (\ref{5 ecu3}), we get
\begin{align*}
|U_3|&\leq \frac{560108544 + 106168320 \sqrt{\frac{3}{41}}}{38220595200}\approx0.015406
\end{align*}
\begin{remark}\label{5cu}
The bounds of $U_1$, $U_2$ and $U_3$, based on the above calculations, are $0.0119242$, $0.0168348$, and $0.015406$ respectively.
\end{remark}
The bounds of $a_i$ $(i=2,3,4,5)$ for functions in the class $\mathcal{C}_{e}$ are obtained in \cite{zap2019}, presented below in the following remark:
\begin{remark}\label{zapavaluesc}
For $f\in \mathcal{C}_{e},$ $|a_2|\leq 1/2,$ $|a_3|\leq 1/4$, $|a_4|\leq 17/144$ and $|a_5|\leq 5/72$. The first three bounds are sharp.
\end{remark}
Next, we calculate the bounds of the sixth and seventh coefficient of functions belonging to the class $\mathcal{C}_{e}$ to establish our main result along the lines of Lemma \ref{5 a6a7lemma}.
\begin{lemma}\label{5 ca6a7lemma}
Let $f\in \mathcal{C}_{e}.$ Then $|a_6|\leq 587/10800\approx 0.0543519$ and $|a_7|\leq 0.0343723$.
\end{lemma}
\begin{proof}
A suitable rearrangement of the terms given in equation (\ref{5 ea6c}) provides us
\begin{equation*}
345600a_6= 5760 p_5- 480 p_2 p_3 +720 p_1 p_4- 480 p_1 p_2^2-17 p_1^5 + 220 p_1^3 p_2- 480 p_1^2 p_3. \\
\end{equation*}
Further, through the triangle inequality, it can be viewed as
\begin{align}
345600|a_6|&\leq |5760 p_5- 480 p_2 p_3|+|p_1(720p_4- 480 p_2^2) |+| -17 p_1^5| \nonumber\\
&\quad +|p_1^2( 220 p_1 p_2- 480 p_3)|.\label{5 ca6}
\end{align}
Using Lemma \ref{2 pomi lemma}, we arrive at
\begin{equation}
|5760 p_5- 480 p_2 p_3|\leq 11520,\quad |p_1(720p_4- 480 p_2^2) |\leq 2880,\label{5 ca61} \end{equation}
\begin{equation}
|p_1^2( 220 p_1 p_2- 480 p_3)|\leq 3840,\quad \text{and}\quad | 17 p_1^5|\leq 544. \label{5 ca62}
\end{equation}
By using equation (\ref{5 ca61}) and (\ref{5 ca62}) in (\ref{5 ca6}), we have
\begin{equation*}
|a_6|\leq \frac{587}{10800}\approx 0.0543519.
\end{equation*}
Similarly, considering equation (\ref{5 ea7c}), we have
\begin{align*}
58060800a_7&=881 p_1^6 - 13260 p_1^4 p_2 + 48240 p_1^2 p_2^2 -106560 p_1 p_2 p_3 + 29040 p_1^3 p_3- 57600 p_3^2 \nonumber\\
&\quad + 69120 p_1 p_5- 56160 p_1^2 p_4 - 86400 p_2 p_4 - 14400 p_2^3.
\end{align*}
It can also be seen as with the aid of the triangle inequality,
\begin{align}
58060800|a_7|&\leq | p_1^4 (881 p_1^2 - 13260p_2)| + | p_1 p_2(48240 p_1 p_2-106560 p_3)| + |p_3(29040 p_1^3 - 57600 p_3)| \nonumber\\
&\quad +| p_1(69120 p_5- 56160 p_1 p_4)|+|- p_2(86400 p_4+14400 p_2^2)|.\label{5ca7}
\end{align}
Lemma \ref{2 pomi lemma} takes us at
\begin{align}
| p_1^4(881 p_1^2 - 13260 p_2)|\leq 424320,\quad |p_1 p_2(48240 p_1 p_2-106560 p_3)| \leq 852480, \label{5 eca71}
\end{align}
\begin{align}\label{5 eca72}
|p_3(29040 p_1^3 - 57600 p_3)|\leq 921600\sqrt{\frac{15}{119}}, \quad |p_1(69120p_5- 56160 p_1 p_4 )|\leq 276480,
\end{align}
and \begin{align}\label{5 eca73}
| p_2(86400 p_4 + 14400 p_2^2)|\leq 460800.
\end{align}
By substituting the values from equations (\ref{5 eca71})-(\ref{5 eca73}) in (\ref{5ca7}), we have
\begin{equation*}
|a_7|\leq \frac{2014080 + 921600 \sqrt{\frac{15}{119}}}{58060800}\approx 0.0403246.
\end{equation*}
\end{proof}
We obtain the following result by omitting the proof as it directly follows from Theorem \ref{5 ceh31},
Remark \ref{5cu}, Remark \ref{zapavaluesc}, Lemma \ref{5 ca6a7lemma} and equation (\ref{5 ch41}).
\begin{theorem}\label{5 ch41bound}
Let $f\in \mathcal{C}_{e}.$ Then
\begin{equation*}
|H_{4,1}(f)|\leq 0.00101775.\end{equation*}
\end{theorem}
\subsection*{Acknowledgment}
Neha is thankful to the Department of Applied Mathematics, Delhi Technological University, New Delhi-110042 for providing Research Fellowship.
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 1,512 |
FreeBASIC é um compilador BASIC distribuído como software livre através da licença (GPL). O compilador foi desenvolvido para ser sintaticamente similar ao QuickBASIC, sem deixar de suportar novos recursos como ponteiros, tipos de dados não sinalizados, inline-assembly, um pré-processador, namespaces, métodos em tipos de dados definidos pelo usuário (TYPE's), entre outros.
Ele compila para DOS, Microsoft Windows e Linux, e está sendo portado para outras plataformas.
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Ferramentas de programação para Linux | {
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Phnom Penh Post - CHRC to form legal assistance team
CHRC to form legal assistance team
Niem Chheng | Publication date 07 August 2019 | 10:37 ICT
CHRC President Keo Remy. Post Pix
Wed, 7 August 2019
Niem Chheng
The Cambodian Human Rights Committee (CHRC) recently posted on its Facebook page that it would form a legal aid group "to defend those whose rights have been violated, especially the poor".
Ministry of Justice and CHRC spokesman Chin Malin on Tuesday said the team would add to the existing legal aid groups available in Cambodia.
"This additional mechanism is to be implemented so that people would have more options when it comes to accessing legal services.
"We already have similar mechanisms provided by the Ministry of Justice, the Bar Association of the Kingdom of Cambodia [BAKC] as well as the ministries of Interior, Defence, Women Affairs.
"There are also private mechanisms such as that of [Prime Minister] Hun Sen's legal aid group," Malin said.
He said the committee's legal aid group would be responsible for three main tasks, including providing legal consultancy to people, facilitating in conflict resolution outside the judicial system and defending the rights of defendants at court trials or hearings.
Commenting on the matter, the Office of the UN High Commissioner for Human Rights (OHCHR) country representative Simon Walker stressed that everyone had the right to legal assistance, "so any step to provide legal assistance is welcome".
"OHCHR supports streamlined and coordinated legal aid and, in this context, has been working with the Ministry of Justice on the development of the legal aid policy.
"OHCHR encourages the swift adoption of this policy with adequate resourcing to support it," he told The Post via email on Tuesday.
Similarly, Yung Phanit, the deputy head of the BAKC's department of legal defence for the poor, said he welcomed CHRC's initiative.
He said the move would reduce the requests for lawyers which have continuously increased at the BAKC.
"If their targets are poor people, their work would reduce the burden that the BAKC is facing," Phanit said.
Contact author: Niem Chheng
Gov't aims to provide legal aid for journalists
Workshop educates on legal aid
Government legal aid team to meet
Prime Minister Hun Sen announces free legal aid for garment workers
PM orders lawyers for poor women | {
"redpajama_set_name": "RedPajamaCommonCrawl"
} | 5,875 |
{"url":"https:\/\/academic.oup.com\/cercor\/article\/18\/9\/2086\/354095\/The-Dynamical-Response-Properties-of-Neocortical","text":"## Abstract\n\nCortical neurons are often classified by current\u2013frequency relationship. Such a static description is inadequate to interpret neuronal responses to time-varying stimuli. Theoretical studies suggested that single-cell dynamical response properties are necessary to interpret ensemble responses to fast input transients. Further, it was shown that input-noise linearizes and boosts the response bandwidth, and that the interplay between the barrage of noisy synaptic currents and the spike-initiation mechanisms determine the dynamical properties of the firing rate. To test these model predictions, we estimated the linear response properties of layer 5 pyramidal cells by injecting a superposition of a small-amplitude sinusoidal wave and a background noise. We characterized the evoked firing probability across many stimulation trials and a range of oscillation frequencies (1\u20131000 Hz), quantifying response amplitude and phase-shift while changing noise statistics. We found that neurons track unexpectedly fast transients, as their response amplitude has no attenuation up to 200 Hz. This cut-off frequency is higher than the limits set by passive membrane properties (\u223c50 Hz) and average firing rate (\u223c20 Hz) and is not affected by the rate of change of the input. Finally, above 200 Hz, the response amplitude decays as a power-law with an exponent that is independent of voltage fluctuations induced by the background noise.\n\n## Introduction\n\nThe response of a single neuron to a changing input is limited by the neuron's maximal spike frequency. Inputs which vary faster can only be encoded in the collective activity of a population. This can be observed in cortical rhythms when individual cells fire irregularly and at much lower spiking rate than the population rhythm revealed through local field potentials (Buzsaki and Draguhn 2004). Individual cells tend to fire more often at the peak of the oscillation but cannot emit a spike for every cycle. However, whereas 1 cell is in the refractory period another 1 may fire during the next cycle, so that the population can globally sustain fast rhythms. It is therefore of central importance to investigate how neurons respond to time-varying inputs and to identify the impact of synaptic background noise (Par\u00e9 et al. 1998; Shadlen and Newsome 1998; Steriade 2001).\n\nPrevious theoretical studies (Knight 1972a; Gerstner 2000) extensively addressed these issues in models of spiking neurons. They emphasized the role of background noise in simplifying the neuronal response dynamics and allowing arbitrarily fast time-varying inputs to be encoded undistorted. Brunel et al. (2001) confirmed these theoretical findings for a more realistic mathematical description of synaptic background noise and quantitatively linked the temporal correlations of the background inputs (i.e., the synaptic filtering) to the response dynamics. However, by a more accurate description of the spike-initiation mechanisms in nonlinear integrate-and-fire neurons and conductance-based models, it was predicted that the linear response of a neuron is always dominated by a low-pass behavior, whose cut-off frequency is independent of the background noise as well as the rate of change of the input (Fourcaud-Trocm\u00e9 et al. 2003; Fourcaud-Trocm\u00e9 and Brunel 2005; Naundorf et al. 2005).\n\nBy investigating how the instantaneous firing rate is modulated by a noisy input with a small sinusoidal component, we experimentally estimated the linear response properties of layer 5 pyramidal cells of the rat somatosensory cortex, over a wide frequency range of input oscillations (i.e., 1\u20131000 Hz). We evaluated the extent of response linearity, tested the ability of cells to track temporally varying inputs, and investigated the impact of background noise. In the limit of small input amplitude, this allows one to predict the spiking activity of a population of weakly interacting neurons, on the basis of the single-cell responses to elementary sinusoidally modulated currents. This also allows to study how neurons take part in collective rhythms, inferring the preferred global frequency in recurrent networks (Fuhrmann et al. 2002; Brunel and Wang 2003; Wang 2003; Geisler et al. 2005) where each cell responds to a correlated foreground rhythm (i.e., the signal) while experiencing a distinct synaptic background activity.\n\nAlthough the response properties of cortical neurons to stationary fluctuating inputs have been previously characterized (Chance et al. 2002; Rauch et al. 2003; Giugliano et al. 2004; Higgs et al. 2006; La Camera et al. 2006; Arsiero et al. 2007), this is the 1st time that the response of cortical neurons to temporally modulated inputs is investigated over a wide range of input frequencies and through analysis of the background noise.\n\n## Materials and Methods\n\n### Experimental Preparation and Recordings\n\nTissue preparation was as described in Rauch et al. (2003). Briefly, neocortical slices (sagittal, 300 \u03bcm thick) were prepared from 14- to 52-days-old Wistar rats. Large layer 5 (L5), regular-spiking pyramidal cells (McCormick et al. 1985) of the somatosensory cortex with a thick apical dendrite were visualized by differential interference contrast microscopy. Some neurons were filled with biocytin and stained (Hsu et al. 1981), to check that the entire neuronal apical dendrite was indeed in the plane of the slice, which was always the case. Whole-cell patch-clamp recordings were made at 32 \u00b0C from the soma (10\u201320 M\u03a9 access resistance) with extracellular solution containing (in mM): 125 NaCl, 25 NaHCO3, 2.5 KCl, 1.25 NaH2PO4, 2 CaCl2, 1 MgCl2, 25 glucose, bubbled with 95% O2, 5% CO2, perfused at a minimal rate of 1 mL\/min. Electrode resistance and capacitance were 6.97 \u00b1 0.18 M\u03a9 and 23.73 \u00b1 1.11 pF, respectively, when filled with an intracellular solution containing (in mM): 115 K-gluconate, 20 KCl, 10 4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid (HEPES), 4 adenosine triphosphate-Mg, 0.3 Na2-guanosine triphosphate, 10 Na2-phosphocreatine, pH adjusted to 7.3 with KOH. All the chemicals were from Sigma or Merck (Switzerland). Other pipette solutions were reported not to alter significantly the response properties of the cells under very similar experimental conditions (Rauch et al. 2003). A BVC-700A bridge amplifier (Dagan Corporation, MN) was used in current-clamp mode and bridge balance and capacitance neutralization were routinely applied. Hyperpolarizing current steps and linear swept sine waves (ZAP) were injected to obtain estimates of the passive properties of patched neurons, such as the total membrane capacitance Cm and apparent input resistance Rin (Iansek and Redman 1973), as well as the membrane impedance amplitude profile (Hutcheon et al. 1996). Signals were low-pass filtered at 2.5 kHz, sampled at 5\u201315 kHz, and captured on the computer.\n\nFinally, care was taken to ensure that the neuronal response was consistent and reproducible throughout the whole recording session (see Fig. 1b). The total whole-cell resistance Rin and the resting membrane voltage Em were continuously monitored (during T2 and T1, respectively, located as in Fig. 1). Data collection began after these observables attained stable values and the experiment was stopped in case of any drift.\n\nFigure 1.\n\nIn vivo\u2013like stimulation protocol and the stability of in vitro recording conditions. In vivo irregular background synaptic inputs were emulated in vitro by injection of noisy currents under current-clamp. Specifically, gaussian currents characterized by mean I0, standard deviation s and correlation time \u03c4, were injected into the soma of layer 5 pyramidal cells. A deterministic sinusoidally oscillating waveform of amplitude I1 and modulation frequency f was then superimposed to the background noise (a\u2014lower trace), and the stimulation trials were interleaved by a recovery interval Trec. The initial segments of each stimulus (i.e., lasting T1, T2, and T3) were used to monitor the stability of the recording conditions on a trial-by-trial basis. Panel b shows a typical experimental session, plotting over time the whole-cell resistance Rin (estimated during T2, b\u2014upper panel), the resting membrane potential Em (averaged during T1, b\u2014middle panel), as well as the reproducibility of the cell discharge rate rfix, evaluated in response to a stationary noise, characterized by fixed statistics (s, \u03c4)fix (during T3). Continuous lines in (b) represent average values of each observable across the whole experiment, whereas the gray shading in (b\u2014lower panel) indicates a confidence level of approximately 68%, which describes the variance allowed for the data. The middle panel shows a layer V pyramidal cell of the somatosensory cortex of the rat stained with Biocytin.\n\nFigure 1.\n\nIn vivo\u2013like stimulation protocol and the stability of in vitro recording conditions. In vivo irregular background synaptic inputs were emulated in vitro by injection of noisy currents under current-clamp. Specifically, gaussian currents characterized by mean I0, standard deviation s and correlation time \u03c4, were injected into the soma of layer 5 pyramidal cells. A deterministic sinusoidally oscillating waveform of amplitude I1 and modulation frequency f was then superimposed to the background noise (a\u2014lower trace), and the stimulation trials were interleaved by a recovery interval Trec. The initial segments of each stimulus (i.e., lasting T1, T2, and T3) were used to monitor the stability of the recording conditions on a trial-by-trial basis. Panel b shows a typical experimental session, plotting over time the whole-cell resistance Rin (estimated during T2, b\u2014upper panel), the resting membrane potential Em (averaged during T1, b\u2014middle panel), as well as the reproducibility of the cell discharge rate rfix, evaluated in response to a stationary noise, characterized by fixed statistics (s, \u03c4)fix (during T3). Continuous lines in (b) represent average values of each observable across the whole experiment, whereas the gray shading in (b\u2014lower panel) indicates a confidence level of approximately 68%, which describes the variance allowed for the data. The middle panel shows a layer V pyramidal cell of the somatosensory cortex of the rat stained with Biocytin.\n\nThe results reported here represent data from L5 pyramidal cells (n = 67) of the somatosensory cortex. The average resting membrane potential was Em = \u221266 \u00b1 4.4 mV, the apparent input resistance (Rin) was 45 \u00b1 2.6 M\u03a9, the membrane time-constant (\u03c4m) was 18.32 \u00b1 0.8 ms. The total capacitance Cm was estimated as 448 \u00b1 19 pF. Liquid junction potentials were left uncorrected.\n\n### Injection of Sinusoidal Noisy Currents\n\nTo probe the response dynamics of pyramidal cells under in vivo\u2013like conditions, independent realizations of a noisy current were computer-synthesized and injected somatically in current-clamp configuration (see Fig. 1a). Each experiment consisted of the repeated injection of current stimuli I(t), lasting T = 10\u201330 s each, interleaved by a recovery Trec of 30 s. A deterministic sinusoidally oscillating current with frequency f was superimposed to the noisy current component and injected (Fig. 2c,d), so that\n\n(1)\nInoise(t)was generated as a realization of an Ornstein\u2013Uhlenbeck stochastic process with zero-mean and variance s2 (Rauch et al. 2003), and independently synthesized for each repetition by iterating the equation\n(2)\nwhere \u03bet represents a random variable from a normal distribution (Press et al. 1992), and it was updated at every time step dt (i.e. 5\u201315 kHz). Inoise(t) is then an exponentially filtered white-noise and it aims at mimicking in vitro the barrage of a large numbers of balanced background excitatory and inhibitory synaptic inputs at the soma (Destexhe et al. 2001, 2003; Rauch et al. 2003; Arsiero et al. 2007). Inoise(t) is characterized by a steady-state Gaussian amplitude-distribution with zero-mean and variance s2, and by a steady-state autocorrelation function exponentially decaying with time constant \u03c4. The value of \u03c4 corresponds to the decay time-constants of individual synaptic currents and it was varied in the range 5\u2013100 ms, thereby referring to fast (AMPA- and GABAA-mediated) as well as slow (NMDA- and GABAB-mediated) synaptic currents (Tuckwell 1988; Rauch et al. 2003). The choice of s2 was aimed at mimicking the membrane voltage fluctuations observed in cortical recordings in vivo, which are around 3\u20135 mV (Par\u00e9 et al. 1998), and it is also effectively representative of nonzero cross-correlations of background inputs (Rudolph and Destexhe 2004).\n\nFigure 2.\n\nAnalyzing the discharge response to the oscillatory input signal over a background of irregular synaptic inputs. Irregular spike trains were evoked in the same neuron by sinusoidally modulated noisy current injections. The time of occurrence of each action potential (a, b) was referred to its peak and represented by a thick vertical mark. Lower panels show the spike raster-plots collected for different input modulation frequencies, f = 10 Hz and f = 250 Hz. The instantaneous firing rate r(t) (c, d\u2014upper panels) reveals a sinusoidal modulation in time. This was estimated by the peristimulus time histograms (PSTHs) (bars) across repeated trials and successive input cycles, and quantified by the best-fit sinusoid with frequency f (black thick line). For the sake of comparison, the sinusoidal component of I(t) (c, d\u2014lower panels) was plotted in red and superimposed to the actual injected waveform. Although the mean firing rate r0 remains constant, its modulation r1 and phase-shift \u03a6 depend on the input frequency f.\n\nFigure 2.\n\nAnalyzing the discharge response to the oscillatory input signal over a background of irregular synaptic inputs. Irregular spike trains were evoked in the same neuron by sinusoidally modulated noisy current injections. The time of occurrence of each action potential (a, b) was referred to its peak and represented by a thick vertical mark. Lower panels show the spike raster-plots collected for different input modulation frequencies, f = 10 Hz and f = 250 Hz. The instantaneous firing rate r(t) (c, d\u2014upper panels) reveals a sinusoidal modulation in time. This was estimated by the peristimulus time histograms (PSTHs) (bars) across repeated trials and successive input cycles, and quantified by the best-fit sinusoid with frequency f (black thick line). For the sake of comparison, the sinusoidal component of I(t) (c, d\u2014lower panels) was plotted in red and superimposed to the actual injected waveform. Although the mean firing rate r0 remains constant, its modulation r1 and phase-shift \u03a6 depend on the input frequency f.\n\nThe number of repetitions for the same set of stimulation parameters (I0, I1, s, \u03c4, f) was 5\u201320, approximately ensuring an accuracy of at least 10% on the estimate of the instantaneous firing rate, with a confidence of 68% (see Rauch et al. 2003). Waveforms were injected in a random order to minimize the effect of slow drifts in the recording conditions. Although the explored range for f was 1\u20131000 Hz, the effect of distinct values for \u03c4 and for (I0, s) was also investigated (as in Figs 5 and 6). Stimulations by a single sinusoid at the time were preferred to probing simultaneously the entire frequency-domain, with the aim of shortening each stimulation epoch in favor of the stability of the recordings (Fig. 1) and of the signal-to-noise ratio.\n\n### Injection of Noisy Broadband Waveforms\n\nWe also injected periodic broadband waveforms instead of sinusoids, under background noise Inoise(t). In analogy to equation (1), the stimulation current is defined as\n\n(3)\n\nSimilar signals were preferred to a superposition of many sinusoids as they let us to compare our results with those of Mainen and Sejnowski (1995), who did not consider any background component in their stimulation protocol. A set of waveforms iT(t) of duration T = 100 ms was generated once and for all by iterating equation (2) offline, using \u03c4 = 1 ms and s = I1. Thus, each iT(t) was a segment of a frozen colored noise, with zero-mean and significant spectral energy content approximately up to \u03c4\u22121 = 1 kHz. We could generate distinct waveforms by choosing different initialization seed \u03be0 in equation (2). In order to allow a repeated stimulation by iT(t) and efficient data collection, i(t) was constructed by \u201cgluing\u201d together hundreds of identical and nonoverlapping replicas of iT(t). \u03be0 was selected to minimize the absolute difference |iT(0) \u2212 iT(T)| and thus reducing discontinuities at the boundaries between 2 successive replicas.\n\n### Data Analysis\n\nThe membrane voltage was recorded in response to each noisy (independent) periodic stimulus realization (see Fig. 2c,d, lower panels). Raw traces were offline processed in Matlab (The Mathworks, Natick, MA) to extract individual spike times {tk}, k = 1,2,3,\u2026, after discarding an initial transient where spike-frequency adaptation and other voltage-dependent currents might not be at \u201cregime\u201d (i.e., 1\u20133 s out of T\u2014see Fig. 1). Most of the data analysis was devoted to quantitatively estimating the response rate r(t) evoked by the periodic noisy current stimulation I(t).\n\nThe peristimulus time histogram (PSTH) of the spike times was constructed over all repetitions by aligning the evoked spike trains according to successive cycles of the same stimulus I(t), for the sake of direct comparison with the analysis performed by Fourcaud-Trocm\u00e9 et al. (2003). The bin size was chosen as one-thirtieth of the input period 1\/f, so that the stimulus duration T corresponds to the same a priori statistical accuracy on the estimate of r(t), irrespectively of f. A sinusoid of frequency f was then fit to the PSTH by the Levenberg\u2013Marquardt algorithm, in the least-squares sense (Press et al. 1992), obtaining estimates of the instantaneous firing rate amplitude r1(f) and the phase \u03a6(f) and their confidence intervals.\n\nThe analysis of the neuronal response to broadband waveforms injections (eq. 3) was performed by means of PSTH over 0.5-ms-wide bins, and evoked spike trains were aligned according to the corresponding successive cycles of iT(t). The spikes collected during an initial transient of each stimulation trial were discarded. By taking an average-window moving across successive stimulation cycles, the stationarity of the mean number of spikes emitted in each cycle of duration T was monitored as a strict necessary condition for further data analysis and phenomenological model identification. This procedure allowed us to detect and remove the effect of brief transient fluctuations in the input resistance.\n\n### Phenomenological Model\n\nAlong the lines of phenomenological \u201ccascade\u201d predictive models of neural response properties (French 1976; Victor and Shapley 1979c; Carandini et al. 1996; Kim and Rieke 2001; Powers et al. 2005; Slee et al. 2005), and in closer analogy to classic Fourier System Identification (Brogan 1991), we considered an input\u2013output relationship based on linear ordinary differential equations (i.e., a linear filter, eq. 5), similarly to Powers et al. (2005). Unlike that approach, we focused on the transformation of the input signal component (i.e., sinusoids or iT(t)) into firing rates r(t). Thus, the identification of these transformations depended on the statistics of the background noise (i.e. I0, s, and \u03c4). Instead of the time-domain, the linear filtering was operatively specified and identified in the frequency-domain (eq. 6). This allowed us to consider a reduced number of free parameters.\n\nIn detail, the input is 1st fed into a threshold-linear element H(x) (see Fig. 7a):\n\n(4)\nwhere iT(t) is the input signal measured in nA. Then H(iT(t)) is transformed into y(t) according to the following equation,\n(5)\nwhere n > m (Brogan 1991). The filter model alone as employed in Figures 5 and 6, can simply be obtained by setting H(x) = x in equation (5). Under periodic regimes, equation (5) is equivalent to the product where and are the (discrete) Fourier transforms of y(t) and H(iT(t)), respectively, and can be written as\n(6)\nwhere and G0 is a real number that represents the low-frequency gain. {zi} and {\u03c0i} are the roots of the polynomials with coefficients {bi} and {ai} and act as the lower or upper cut-off frequencies of elementary high-pass or low-pass filters, respectively, arranged in cascade and have the physical meaning of the inverse of intrinsic time-constants. The filter input\u2013output gain and phase-shift across input modulation frequencies f are fully specified by G0 and by the number of distinct {zi} and {\u03c0i} (i.e., m and n) and their values. For instance, equation (6) accounts for the high-frequency (f \u2192 +\u221e) power-law \u2220f\u2212\u03b1 observed in our experiments, with \u03b1 = nm, strictly integer. Identical input\u2013output relationships are commonly employed to describe electrical filters, composed of linear resistors, capacitors and inductors (Horowitz and Hill 1989). Finally, a constant propagation delay \u0394t was further included, together with an output offset, so that\n(7)\nwhere the phase of \u2220 was indicated by \u2220 and expressed in degrees.\n\nIn summary, equations (5\u20137) describe a linear transformation preceded by a static, or no-memory, threshold-linear stage (eq. 4). The cascade ordering \u201cnonlinear\u2013linear\u201d was preferred to \u201clinear\u2013nonlinear\u201d for slightly better fit performances. All the parameters (i.e., P1, P2, G0, {zi},{\u03c0i}, \u0394t) were adjusted to minimize the discrepancies between actual data and model predictions, employing Simulated Annealing techniques (Press et al. 1992). The chosen cost\u2013function to minimize was represented by the \u03c72 that quantified the mean quadratic discrepancy between actual data and model prediction, weighted by the confidence interval (Press et al. 1992). Large deviations are therefore weighted on the basis of the confidence on these data estimates. For the identification of the full cascade model in the time-domain, \u03c72 was complemented by 1st-derivative mean discrepancies.\n\n### Statistics\n\nNinety-five percent confidence accuracy intervals on the nonlinear least-square parameter estimates were determined for r0, r1(f), and \u03a6(f) by the Levenberg\u2013Marquardt fit algorithm, providing error bars in the plots of Figures 5 and 6 as in Fourcaud-Trocm\u00e9 et al. (2003). For Figure 1b (lower panel) and Figure 6b, the gray shading represents the asymmetric 68% confidence accuracy interval (i.e., corresponding to 1 standard deviation) for the mean firing rate rfix, as in Rauch et al. (2003).\n\nIn the case of identification of the phenomenological filter models, the \u03c72-test was used to evaluate the quality of the fits (Press et al. 1992), implicitly taking into account the number of free parameters.\n\nKendall's Tau nonparametric (rank-order) test (Press et al. 1992) was finally employed to assess correlations among spike-shape features and stimulation parameters, providing a measure c of correlation together with its significance level P, which represents the probability of obtaining the same value for c from statistically independent samples (i.e., false positive).\n\n## Results\n\n### The Linear Response to Time-Varying Noisy Inputs\n\nDue to irregular spontaneous activity and the high degree of convergence, cortical neurons receive a continuous barrage of excitatory and inhibitory potentials in the intact brain. At the same time cortical cells participate in a variety of oscillations, whose frequency spans several orders of magnitude (e.g., 0.05\u2013500 Hz) during distinct behavioral states (Buzsaki and Draguhn 2004). What is the impact of the background activity on neuronal responsiveness and on collective oscillations? We approached these issues by studying the linear response properties of single neurons characterizing their instantaneous discharge rate r(t) in response to a noisy background current with a small sinusoidal component, hereafter referred to as the \u201csignal.\u201d This allowed us not only to investigate how cortical neurons participate in an oscillatory regime, but especially how cells track temporally varying inputs under distinct background conditions (Fig. 1). We systematically varied the input oscillation frequency f, its amplitude I1 and offset I0, as well as the statistics (s, \u03c4) of the background noise (eqs. 1 and 2). Because no correlation between the shape of the action potentials and these stimulation parameters was found, we restricted our analysis to the timing of each spike. However, very small correlations c exist between (I0, I1, s) and the maximal upstroke velocity and spike duration (|c| \u2264 0.1; P < 10\u22123), but they are consequence of nonideal bridge-balancing. Weak correlations c were instead found between the rat postnatal day and the spike upstroke velocity (c = 0.21; P < 10\u221212), downstroke velocity (c = \u22120.23; P < 10\u221214), and spike duration (c = \u22120.27; P < 10\u221219), as observed by many investigators.\n\nThe firing rate r(t) was estimated from the peristimulus time histograms (PSTHs) of the spike times over hundreds of cycles of the input current and over several stimulation trials. It was interpreted as the instantaneous discharge probability or, equivalently, as the firing rate of a cortical population composed of independent neurons.\n\nIn the limit of small-signal input amplitude I1, r(t) could be well approximated by a sine wave oscillating at the same frequency f as the input current (Fig. 2c,d, upper panels):\n\n(8)\nr(t) is fully described in terms of mean firing rate r0, modulation amplitude r1(f), and phase-shift \u03a6(f) relative to the input current, as in linear dynamical transformations. At the beginning of each experiment, the stimulation parameters were selected in a way that r0 was in the range 10\u201320 Hz, the membrane voltage fluctuations induced by the noise were 1\u20135 mV, and the discharge modulation amplitude r1 was 0 < r1 < r0. Figure 2 reports typical spike responses evoked by input modulations at f = 10 Hz and at f = 250 Hz, recorded in the same cell. Individual firing times across successive input cycles and trial repetitions showed high variability (Fig. 2a,b, lower panels), as a consequence of the noise component uncorrelated with the sinusoidal signal oscillations.\n\nScaling the input amplitude I1 in the range 20\u2013200 pA while keeping I0, s, and f fixed resulted in a linear scaling of the output amplitude r1 (n = 3, not shown). However, for large input modulation depth (i.e. I1 > 0.3 I0), the amplitudes of output superimposed sinusoidal oscillations characterized by multiple frequencies of f (i.e., higher harmonic components) increased (n = 3), revealing the presence of input\u2013output distortions as the limit of small input amplitude was exceeded. Thus, in most of the experiments we employed I1 smaller than 30% of I0, to fulfill the validity of the linear approximation where higher harmonics in the output could be neglected. Although similar values of I1 are not infinitesimal with respect to I0, this choice was confirmed to be reasonable by studying and predicting the neuronal discharge in response to more complex inputs signals across a wide range of firing rates, as discussed in the experiments of Figure 7.\n\nConsistent with the hypothesis of linearity, no significant difference between the sum of the responses to individual sinusoids and the response to the sum of multiple sinusoids injected simultaneously was observed (Movshon et al. 1978; Victor 1979; Carandini et al. 1996) (n = 20, not shown).\n\n### Cortical Neurons Track Fast Inputs\n\nOur experimental characterization aimed at identifying the linear neuronal response properties and at studying the way background noise affects them (Sakai 1992; Chichilnisky 2001; Fourcaud-Trocm\u00e9 et al. 2003; Naundorf et al. 2005; Apfaltrer et al. 2006). In the framework of classic Fourier decomposition of any input signal to a neuron, r1(f) and \u03a6(f) give quantitative information on how the neuronal encoding differentially attenuates and delays each frequency component f of the input, in the limit of small-signal amplitude.\n\nFigure 3 summarizes population data and reports the unexpectedly wide bandwidth of the output temporal modulation depth r1\/r0 and output phase-shift \u03a6. Although r0 was unaffected by f, r1 decreased significantly only for f > 100\u2013200 Hz, regardless of the intensity and temporal correlations of the background noise. The profile of r1(f) across frequencies did not match the membrane impedance, which was dominated by voltage-dependent resonances in the low-frequency range (i.e., 5\u201310 Hz\u2014previously related to h-currents and M-currents) and by a low-pass behavior at high frequencies (not shown) with strong attenuation above 50 Hz (Gutfreund et al. 1995; Hutcheon et al. 1996).\n\nFigure 3.\n\nModulation depth (r1\/r0) and phase-shift \u03a6 of the response to a noisy oscillatory input. The instantaneous firing rate r(t) evoked by small sinusoidal currents over a noisy background revealed sinusoidal oscillations with amplitude r1 and phase-shift \u03a6, around a mean r0 (quantified as in Fig. 2c,d). Surprisingly, pyramidal neurons can relay fast input modulations, up to several hundred cycles per second. The high-frequency response behavior matches a power-law relationship (i.e., r1f\u03b1) with a linear phase-shift (i.e., \u03a6 \u223c f). These plots were obtained for 67 cells, averaging across available repetitions and distinguishing between offset-currents I0 above (suprathreshold regime) and below (subthreshold regime) the DC rheobase of the corresponding cell (as in Fig. 5). Data points corresponding to distinct input modulation frequencies were pooled together in nonoverlapping bins with size 0.1\u201310 Hz (low frequencies) and 100\u2013200 Hz (high frequencies). Error bars represent the SE across the data points available (32 \u00b1 25) for each bin. Markers shape and color identify the suprathreshold or weak-noise regime (black) and the subthreshold or strong-noise regime (red), characterized by distinct values for I0 and s2, adapted to yield a similar mean rate r0 \u223c 20 Hz (i.e., 19.7 \u00b1 1.5 Hz).\n\nFigure 3.\n\nModulation depth (r1\/r0) and phase-shift \u03a6 of the response to a noisy oscillatory input. The instantaneous firing rate r(t) evoked by small sinusoidal currents over a noisy background revealed sinusoidal oscillations with amplitude r1 and phase-shift \u03a6, around a mean r0 (quantified as in Fig. 2c,d). Surprisingly, pyramidal neurons can relay fast input modulations, up to several hundred cycles per second. The high-frequency response behavior matches a power-law relationship (i.e., r1f\u03b1) with a linear phase-shift (i.e., \u03a6 \u223c f). These plots were obtained for 67 cells, averaging across available repetitions and distinguishing between offset-currents I0 above (suprathreshold regime) and below (subthreshold regime) the DC rheobase of the corresponding cell (as in Fig. 5). Data points corresponding to distinct input modulation frequencies were pooled together in nonoverlapping bins with size 0.1\u201310 Hz (low frequencies) and 100\u2013200 Hz (high frequencies). Error bars represent the SE across the data points available (32 \u00b1 25) for each bin. Markers shape and color identify the suprathreshold or weak-noise regime (black) and the subthreshold or strong-noise regime (red), characterized by distinct values for I0 and s2, adapted to yield a similar mean rate r0 \u223c 20 Hz (i.e., 19.7 \u00b1 1.5 Hz).\n\nAbove 200 Hz the output modulation depth decayed as a negative power-law, which appears as a straight (dashed) line in the double-logarithmic plot of Figure 3. The power-law exponent estimated by linear regression through the population data of Figure 3 was close to 2 (\u03b1 = \u22121.80) and it matched the value obtained by averaging the exponents estimated in single experiments (\u03b1 = \u22121.81 \u00b1 0.31, n = 6\u2014see Fig. 4a). A similar qualitative dependence, induced by system linearization, was anticipated by theoretical studies (Gerstner 2000; Knight 1972a) and could be replicated quantitatively in the case of integer power-law exponents through canonical phase oscillator models (Naundorf et al. 2005), nonlinear integrate-and-fire models (Fourcaud-Trocm\u00e9 et al. 2003), and conductance-based neuronal modeling (Fourcaud-Trocm\u00e9 et al. 2003). Integer values of \u03b1 also relate to the number of best-fit free parameters of the phenomenological band-pass filters used in Figures 5\u20137 (see the Methods\u2014eq. 6), introduced to fit the experimental data as discussed in the following sections.\n\nFigure 4.\n\nThe high-frequency dynamical response properties of a typical cortical neuron, plotted in linear scale. The modulation amplitude (a) r1(f), elicited by noisy oscillatory inputs, shows a power-law behavior (see also Fig. 3) captured by 1\/f\u03b1, with \u03b1 \u223c 2, whereas the phase \u03a6 of the response (b) decreases linearly with increasing frequencies f (i.e., \u03a6 \u2192 \u2212360\u00b0\u00b7f\u00b7\u0394t). Stimulation parameters (I0, I1, s) = (400, 150, 500) pA and \u03c4 = 5 ms.\n\nFigure 4.\n\nThe high-frequency dynamical response properties of a typical cortical neuron, plotted in linear scale. The modulation amplitude (a) r1(f), elicited by noisy oscillatory inputs, shows a power-law behavior (see also Fig. 3) captured by 1\/f\u03b1, with \u03b1 \u223c 2, whereas the phase \u03a6 of the response (b) decreases linearly with increasing frequencies f (i.e., \u03a6 \u2192 \u2212360\u00b0\u00b7f\u00b7\u0394t). Stimulation parameters (I0, I1, s) = (400, 150, 500) pA and \u03c4 = 5 ms.\n\nFigure 5.\n\nThe intensity of background fluctuations affects the dynamical response of cortical neurons. The impact of the noise variance s2 was examined across a wide range of input frequencies f, in 4 distinct cells (ad), under the same conditions of Figure 3. Strong background noise smoothes r1(f) at intermediate frequencies, as in a programmable equalizer. Linear instead of logarithmic scale has been employed here for the y-axis. Each subpanel (top to bottom) reports r1(f) and \u03a6(f), identifying the suprathreshold or weak-noise regime (\u201csupra\u201d\u2014black markers) and the subthreshold or strong-noise regime (\u201csub\u201d\u2014red markers) by different marker shapes and colors. Each regime is characterized by distinct values for I0 and s2, adapted to yield a similar mean rate r0 \u223c 20 Hz. Experimental data points (markers) have been plotted together with the best-fit predictions from a phenomenological filter model (continuous traces). For these cells, band-pass 2nd-order filters (i.e., n = 2\u2014eq. 6) were found to describe the experimental data with high significance (see Supplemental Table S1). Error bars represent the 95% confidence intervals, obtained by the Levenberg\u2013Marquardt fit algorithm. High-frequency error bars were large because of the poor signal-to-noise ration as well as for the ambiguity of the (periodic) estimates of \u03a6(f). Although I1 = 50 pA and \u03c4 = 5 ms were fixed for all cells and both regimes, the remaining stimulation parameters were: (suprathreshold) (I0, s)a = (500, 50), (I0, s)b = (400, 20), (I0, s)c = (250, 25) and (I0, s)d = (350, 50) pA; (subthreshold) (I0, s)a = (300, 400), (I0, s)b = (150, 325), (I0, s)c = (100, 250) and (I0, s)d = (100, 450) pA.\n\nFigure 5.\n\nThe intensity of background fluctuations affects the dynamical response of cortical neurons. The impact of the noise variance s2 was examined across a wide range of input frequencies f, in 4 distinct cells (ad), under the same conditions of Figure 3. Strong background noise smoothes r1(f) at intermediate frequencies, as in a programmable equalizer. Linear instead of logarithmic scale has been employed here for the y-axis. Each subpanel (top to bottom) reports r1(f) and \u03a6(f), identifying the suprathreshold or weak-noise regime (\u201csupra\u201d\u2014black markers) and the subthreshold or strong-noise regime (\u201csub\u201d\u2014red markers) by different marker shapes and colors. Each regime is characterized by distinct values for I0 and s2, adapted to yield a similar mean rate r0 \u223c 20 Hz. Experimental data points (markers) have been plotted together with the best-fit predictions from a phenomenological filter model (continuous traces). For these cells, band-pass 2nd-order filters (i.e., n = 2\u2014eq. 6) were found to describe the experimental data with high significance (see Supplemental Table S1). Error bars represent the 95% confidence intervals, obtained by the Levenberg\u2013Marquardt fit algorithm. High-frequency error bars were large because of the poor signal-to-noise ration as well as for the ambiguity of the (periodic) estimates of \u03a6(f). Although I1 = 50 pA and \u03c4 = 5 ms were fixed for all cells and both regimes, the remaining stimulation parameters were: (suprathreshold) (I0, s)a = (500, 50), (I0, s)b = (400, 20), (I0, s)c = (250, 25) and (I0, s)d = (350, 50) pA; (subthreshold) (I0, s)a = (300, 400), (I0, s)b = (150, 325), (I0, s)c = (100, 250) and (I0, s)d = (100, 450) pA.\n\nFigure 6.\n\nThe timescale of background fluctuations affects the dynamical response of cortical neurons. The effect of the timescale of fluctuations (i.e., correlation time \u03c4) was examined across a wide range of input frequencies f, in 4 cells (ad). At high input frequencies pyramidal neurons are insensitive to the noise-color, in the sense that they do not speed up or slow down their fastest reaction time, for \u201cwhite\u201d or \u201ccolored\u201d background noise. Linear instead of logarithmic scale has been employed here for the y-axis. The panels (top to bottom) report r1(f) and \u03a6(f), with different marker shapes and colors referring to 2 stimulation regimes, indicated as \u03c4slow (red markers) and \u03c4fast (black markers). Although \u03c4fast was fixed to 5 ms and \u03c4slow was (ad) 45\u201350 ms, in (d) the range 5\u2013100 ms could be explored. As in Figure 5, experimental data points (markers) have been plotted together with the best-fit predictions from a phenomenological filter model (continuous and dashed traces). For these cells, band-pass third-order filters (i.e., n = 3\u2014eq. 6) were found to describe the data with high significance (see Supplemental Table S2). Error bars represent the 95% confidence intervals obtained by the Levenberg\u2013Marquardt fit algorithm. High-frequency error bars were large because of the poor signal-to-noise ration as well as for the ambiguity of the (periodic) estimates of \u03a6(f). Stimulation parameters were: (I0, I1, s)a = (250, 50, 100), (I0, I1, s)b = (300, 50, 100), (I0, I1, s)c = (300, 50, 100), and (I0, I1, s)d = (300, 50, 75) pA.\n\nFigure 6.\n\nThe timescale of background fluctuations affects the dynamical response of cortical neurons. The effect of the timescale of fluctuations (i.e., correlation time \u03c4) was examined across a wide range of input frequencies f, in 4 cells (ad). At high input frequencies pyramidal neurons are insensitive to the noise-color, in the sense that they do not speed up or slow down their fastest reaction time, for \u201cwhite\u201d or \u201ccolored\u201d background noise. Linear instead of logarithmic scale has been employed here for the y-axis. The panels (top to bottom) report r1(f) and \u03a6(f), with different marker shapes and colors referring to 2 stimulation regimes, indicated as \u03c4slow (red markers) and \u03c4fast (black markers). Although \u03c4fast was fixed to 5 ms and \u03c4slow was (ad) 45\u201350 ms, in (d) the range 5\u2013100 ms could be explored. As in Figure 5, experimental data points (markers) have been plotted together with the best-fit predictions from a phenomenological filter model (continuous and dashed traces). For these cells, band-pass third-order filters (i.e., n = 3\u2014eq. 6) were found to describe the data with high significance (see Supplemental Table S2). Error bars represent the 95% confidence intervals obtained by the Levenberg\u2013Marquardt fit algorithm. High-frequency error bars were large because of the poor signal-to-noise ration as well as for the ambiguity of the (periodic) estimates of \u03a6(f). Stimulation parameters were: (I0, I1, s)a = (250, 50, 100), (I0, I1, s)b = (300, 50, 100), (I0, I1, s)c = (300, 50, 100), and (I0, I1, s)d = (300, 50, 75) pA.\n\nFigure 7.\n\nPrediction of the discharge response to a broadband input signal over a background noise. We challenged the significance of the linear response properties, searching for best-fit parameters of a phenomenological cascade model to predict the instantaneous firing rate in response to a broadband input iT(t) (b\u2014upper panel). Such a model, sketched in (a), has the structure of a classic Hammerstein model (Sakai 1992), where a static, or no-memory, threshold-linear element is followed by a linear system, as for the band-pass filters of Figures 5 and 6 (see the Methods). In (b), only the broadband current signal is shown (top), together with the corresponding spiking pattern elicited across different cycles and repetitions (middle). In the lower panel, the best-fit output r(t) of the model (red dots) was compared with the instantaneous firing probability (continuous blue line) obtained as a PSTH with a 68% confidence interval (gray shaded area), estimated over the corresponding raster plot (middle). The cascade model captures the input\u2013output response properties of cortical neurons to fast inputs with acceptable accuracy (see Supplemental Table S3).\n\nFigure 7.\n\nPrediction of the discharge response to a broadband input signal over a background noise. We challenged the significance of the linear response properties, searching for best-fit parameters of a phenomenological cascade model to predict the instantaneous firing rate in response to a broadband input iT(t) (b\u2014upper panel). Such a model, sketched in (a), has the structure of a classic Hammerstein model (Sakai 1992), where a static, or no-memory, threshold-linear element is followed by a linear system, as for the band-pass filters of Figures 5 and 6 (see the Methods). In (b), only the broadband current signal is shown (top), together with the corresponding spiking pattern elicited across different cycles and repetitions (middle). In the lower panel, the best-fit output r(t) of the model (red dots) was compared with the instantaneous firing probability (continuous blue line) obtained as a PSTH with a 68% confidence interval (gray shaded area), estimated over the corresponding raster plot (middle). The cascade model captures the input\u2013output response properties of cortical neurons to fast inputs with acceptable accuracy (see Supplemental Table S3).\n\nEven though the inspection of Figure 3 seems to indicate that the points at highest frequencies can be fitted by 1\/f, Figure 4a supports the conclusion that 1\/f2 is a more precise characterization. Nevertheless, numerical simulations showed that the high-frequency asymptotic behavior might be reached at frequencies which are much higher than the cut-off frequency (Fourcaud-Trocm\u00e9 et al. 2003), so that assessing the precise value of \u03b1 might not be conclusive on the basis of our observations.\n\nAs opposed to typical linear systems, the phase-shift at high frequencies did not saturate but decreased linearly with f (see Fig. 4b). This is reminiscent of the presence of a constant time delay \u0394t between input and output. This delay was in the range 0.3\u20131.1 ms, sometimes much larger than the \u201cthreshold-to-peak voltage\u201d lag $\u03c4sp$ during a spike. $\u03c4sp$ quantifies the rising phase of each action potential, upon conventional definition of \u201cthreshold\u201d as the membrane voltage corresponding to a rate of change of 10 mV\/ms, and it was in the range of 0.3\u20130.5 ms. As expected from the previous report (Fourcaud-Trocm\u00e9 et al. 2003), \u0394tmodel was always equal to $\u03c4sp$ in single-compartmental computer simulations (not shown). However, the mismatch between $\u03c4sp$ and \u0394t observed in some cells might be explained in terms of relevant additional axo-somatic and somato-axonic propagation latencies of about 0.2 ms each. This was measured directly by Palmer and Stuart (2006), who reported that cortical cells initiate action potentials at the distal end of the initial axon segment (see also Shu et al. 2006).\n\n### The Background Noise Affects the Neuronal Dynamical Response at Intermediate Frequencies\n\nIn the absence of background fluctuations, a neuron discharges only when its input current surpasses a certain threshold (i.e., the rheobase current). When the input current is noisy and fluctuations are induced in the membrane voltage, the neuron can be brought to spiking even when its average input is below the threshold (i.e., \u201csubthreshold\u201d). Thus, the mean firing rate of the neuron r0 is determined by both the mean current I0 and the standard deviation s of the noise. At the beginning of each experiment, I0 and s were tuned to obtain the same mean firing rate r0, chosen in the range 10\u201320 Hz. This allowed us to evoke 2 different discharge regimes, reflected in the degree of the irregular firing: the suprathreshold or weak-noise regime and the subthreshold or strong-noise regime. In the weak-noise regime, the background input fluctuation amplitude s was set to 20\u201350 pA and its mean I0 was chosen above rheobase. Conversely, in the strong-noise regime, I0 was set below rheobase, and s was increased until r0 matched the value obtained in the suprathreshold regime.\n\nFigure 5 summarizes the results of these experiments, reporting the responses of 4 typical cells (see also Fig. 3). It shows that the intensity s of the background noise, mimicking presynaptic firing as well as presynaptic background cross-correlations (Rudolph and Destexhe 2004), differentially affects the neuronal response. This occurs especially at intermediate frequencies, flattening the response profile, and smoothening resonances as predicted in theoretical studies (Knight 1972a; Brunel et al. 2001; Fourcaud-Trocm\u00e9 et al. 2003; Richardson et al. 2003). The modulation of the neuronal discharge does not appear significantly attenuated at frequencies lower than 100\u2013300 Hz in both regimes (see also Fig. 2), as for Figure 3 but plotted in linear instead of logarithmic scale for the vertical axis. At low input frequencies f (1\u201320 Hz), an increase in r1 and a phase-advance were always observed (see Figs 5 and 6). These effects are apparent when analyzing single-cell responses rather than population averages (compare Figs 3 and 5).\n\nIn general, uniform and dense sampling of the frequency axis was not practicable, given the limited time window for stability and reproducibility of the neuronal response in typical recordings (see the Methods). This resulted in privileging high frequencies in some experiments (e.g., see Fig. 4) while neglecting intermediate frequencies in others, and vice versa (e.g., Fig. 6). This prompted us to test a posteriori whether data points collected simultaneously on the response magnitude r1 and phase \u03a6 were consistent with the hypothesis of linearity, while providing meaningful interpolations between samples (see Fig. 5d). In fact, the mutual relationship between r1 and phase \u03a6 cannot be arbitrary in a linear system. Therefore, a filter model (eqs. 6 and 7, see the Methods) was routinely employed to fit the data from each experiment. This model captured the neuronal response to the input signal component and its best-fit attenuation and phase-shift were plotted in Figures 5 and 6 as thick continuous lines. As in electrical filters made of linear resistors, capacitors and inductors (Horowitz and Hill 1989), the number and location of the model intrinsic time-constants account for integer power-law behavior and for low frequencies resonances and phase-advance, while matching the profiles of r1 and \u03a6 simultaneously. Changing the background noise level (black and red colors in Figs 3 and 5) resulted only in a shift in the best-fit values of the intrinsic time-constants of the model and required no modification of their number. This shift was smaller for faster time-constants (i.e., less than \u00b1 30%, for time constants below \u223c3 ms\u2014see Supplemental Table S1), indicating that the high-frequency response of the neuron was generally unaffected by the noise intensity.\n\n### Background Temporal Correlations Do Not Speed up Neuronal Reaction Times\n\nThe timescale of background fluctuations (i.e., the \u201ccolor\u201d of the noise) was systematically varied in our experiments (Fig. 6). This is set by the correlation time \u03c4 of the noise (eq. 2) that mimics the decay time-constant of synaptic currents. In previous theoretical studies, the dependence of \u03a6 on \u03c4 was emphasized (Brunel et al. 2001), suggesting that synaptic noise might have an impact on the reaction times to fast inputs transients reducing the response phase-lag to zero and removing amplitude attenuations (Knight 1972a; Gerstner 2000). Here, we explored the effect of changing the values of \u03c4 in the range 5\u2013100 ms, thereby mimicking the contribution of fast (AMPA- and GABAA-mediated), slow (NMDA- and GABAB-mediated) synaptic currents. Both \u03a6 and r1 showed sensitivity to \u03c4 for intermediate frequencies, but not in the high-frequency regime, as plotted in Figure 6 for 4 typical cells. This is consistent with the results of the simulations of a conductance-based model neuron (not shown), and with the predictions of Fourcaud-Trocm\u00e9 et al. (2003).\n\nAs discussed in the previous section and shown in Figure 5, r1(f) and $\u03a6(f)$ could be simultaneously fit by the frequency response of a linear filter model. A change of the noise time-constant \u03c4 shifted the best-fit parameters, but required no modification of their number. The shift was smaller for faster time-constants (i.e., less than \u00b1 20%, for time constants generally below \u223c3 ms\u2014see Supplemental Table S2).\n\n### Significance of the Linear Response Properties to Predict Neuronal Responses\n\nThe good accuracy of the linear filter to fit the experimental data (Figs 5 and 6, continuous lines) prompted us to test up to which extend linear properties dominate the input\u2013output response in pyramidal neurons. In fact, ideal linear systems process each Fourier-component of their input independently and distortion-free, so that the frequency-domain response of the system is sufficient to predict the corresponding output.\n\nWe investigated the response r(t) to a broadband signal iT(t), instead of sinusoids (see the Methods). With the aim of approaching the conditions of the periodic regime studied in the previous sections, iT(t) was cyclically repeated with a period of T = 100 ms. With the additional background noise, these experiments generalize and extend those of Mainen and Sejnowski (1995), who looked at fast stimulus transients and neuronal response reliability. Furthermore, our approach allows one to study the response of a cortical population, where neurons experience uncorrelated background activity, weakly interacting with each other and receiving the same input signal. In Figure 7b, a sample waveform of the broadband input was plotted, together with the raster-plots of the spikes evoked across hundreds of cycles and repetitions. In analogy to the analysis shown in Figure 2, the peristimulus time histograms (PSTHs) computed from the raster-plot was used to estimate the instantaneous firing rate r(t).\n\nAlthough instantaneous input amplitudes were not small compared with I0, the phenomenological filter employed in Figures 5 and 6 could predict the time-varying neuronal response with satisfying accuracy over a wide range of output firing rates (Fig. 7), tracking fast input transients. However, to account for large negative input amplitudes that occasionally occur, a minimal current-threshold was needed in cascade to the linear filter (eqs. 4, 6, and 7). Without it, the correct dynamical range of the response could not be replicated and the fitting procedure led to low prediction performances. The order \u201cnonlinear\u2013linear\u201d, sketched in Figure 7a, was preferred to the \u201clinear\u2013nonlinear\u201d (Sakai 1992) as it systematically led to slightly superior fit performances, as well as to a possible interpretation as the neuronal rheobase.\n\n## Discussion\n\nIn the present work we studied the basic questions of how neurons encode time-varying inputs into spike trains, how efficiently they achieve it and what the impact of the background noise is. This is of central importance to understand network activities like network-driven persistent oscillatory regimes, which depend on the single-cell dynamical response properties and on recurrent connectivity.\n\nPrevious studies used deterministic oscillating inputs in invertebrate (Knight 1972b; French et al. 2001) and vertebrate neurons in hippocampus and entorhinal cortex (Schreiber et al. 2004), in thalamocortical neurons (Smith et al. 2000), in spinal interneurons and motoneurons (Baldissera et al. 1984), in the vestibular system (du Lac and Lisberger 1995; Ris et al. 2001), in the auditory (Liu et al. 2006), and visual systems (Victor and Shapley 1979a, 1979b; Sakai 1992; Carandini et al. 1996; Nowak et al. 1997), with emphasis on spike timing and reliability (Fellous et al. 2001; Schaette et al. 2005) and synchronization (Gutkin et al. 2005). Our results extend those studies in 2 ways: 1) by examining the contribution of background fluctuations and 2) by systematically exploring the dynamical response properties up to the high-frequency range (1 kHz).\n\nBy the interpretation of the instantaneous firing rate as a population activity, our analysis suggests that cortical ensembles are extremely efficient in tracking transients that are much faster than the membrane time-constant (\u223c20 ms\u2014see the Methods) and the average interspike interval (\u223c1\/r0 \u2245 50 ms) of individual cells. This finding was anticipated by many theoretical studies and it correlates with the previous observations that single cortical neurons (Mainen and Sejnowski 1995) and hypoglossal motoneurons (Powers et al. 2005) may have phase-locked firing responses to fast-varying current inputs, as well as with the study of Bair and Koch (1996), who observed large cut-off frequencies in the power spectra of the responses of middle temporal cortical neurons to in vivo random visual stimulation. However, our results extend the previous studies to the case of high-frequency phase-locking of the population firing rates, under noisy background. Although this is not unexpected (Knight 1972a), our findings disprove that the noise and its temporal correlations make a neuronal population respond instantaneously to an input (Knight 1972a; Gerstner 2000; Brunel et al. 2001; Silberberg et al. 2004). In fact, both noise intensity s and correlation time \u03c4 modulate the neuronal response only at low and intermediate input frequencies and do not affect the low-pass filtering profile of the response, in agreement with Fourcaud-Trocm\u00e9 et al. (2003) and with Naundorf et al. (2005).\n\nThe location of the observed cut-off frequency was higher than the predictions from single-compartmental conductance based model neurons (Fourcaud-Trocm\u00e9 et al. 2003). In those studies, the cut-off was of the order of r0 and increased with the increasing sharpness of the action potentials. Similarly, Naundorf et al. (2005) observed an increase in the neuronal response at input frequencies much higher than r0 (i.e. up to 200 Hz) for increasing action potential onset speed, while studying a phase-oscillator point neuron model. We propose that the effective spike sharpness could be higher than what was previously modeled at the soma. We speculate that a multicompartmental description that incorporates the details of axonic spike initiation (McCormick et al. 2007; Shu et al. 2007) might quantitatively support our experimental observations.\n\nWe observed a phase-advance at low input frequencies that was previously interpreted mechanistically on the basis of ion currents responsible for spike-frequency adaptation (Fleidervish et al. 1996; Ahmed et al. 1998; Fuhrmann et al. 2002; Compte et al. 2003; Paninski et al. 2003), as well as of resonances of the membrane impedance (Brunel et al. 2003; Richardson et al. 2003). These hypotheses are consistent with the input frequency range \u223c1\u201310 Hz (i.e. (100 ms)\u22121\u2013(1000 ms)\u22121) of the phase-advance and with its sensitivity to the levels of the background noise we observed in our experiments.\n\nThe use of current-clamp was a meaningful choice for an immediate comparison to the analytical and numerical studies of Fourcaud-Trocm\u00e9 et al. (2003), Geisler et al. (2005) and many others. A more realistic somatic conductance\u2013injection is expected to change quantitatively but not qualitatively our conclusions (see also Apfaltrer et al. 2006). Even when excitatory and inhibitory fluctuating conductances significantly alter the effective membrane time-constant \u03c4m of the neuron (Destexhe et al. 2003), their additional temporal modulation will not affect further \u03c4m, in the limit of small amplitude considered here. Previous theoretical studies directly showed that the location of the cut-off frequency as well as of the resonances due to subthreshold resonances (Richardson et al. 2003) shift with distinct conductance-states of the neuron, but pointed out that the power-law exponent \u03b1 and the sensitivity to the background noise remain unaffected (Fourcaud-Trocm\u00e9 et al. 2003; Geisler et al. 2005). Nevertheless, in order to carefully extend the discussion of Rauch et al. (2003) (see also La Camera et al. 2004; Richardson and Gerstner 2005) towards a mapping between the dynamical response properties induced by current-driven stimuli to those induced by conductance-driven stimuli, our results will require to be reevaluated under dynamic-clamp recordings (Robinson 1994; Destexhe et al. 2001).\n\nThe response amplitude r1(f) decays as a power-law in the high-frequency range and the exponent \u03b1 of the power-law 1\/f\u03b1 was approximately 2. This is in contrast to what is predicted for the Wang-Buzsaki model (Wang and Buzsaki 1996) and for the exponential integrate-and-fire neuron (Fourcaud-Trocm\u00e9 et al. 2003), but it is consistent with a polynomial VI dependence of the spike-initiating mechanisms (not shown). This steeper power-law is unlikely a measurement artifact. The glass pipette used to inject sinusoidal input currents has indeed low-pass filter properties in \u201ccascade\u201d to the neuron. However, these filtering properties occur mainly between input\u2013output voltages due to parallel parasitic capacitances. Input\u2013output currents are unlikely to be prefiltered due to inductive electrical effects and viscosity in the movement of charge carriers in the pipette solutions, as these are negligible phenomena in the frequency range we investigated.\n\nFinally, the local slope (i.e., gain) of the static fI curve affects neuronal responses regardless of the input modulation frequencies (Fourcaud-Trocm\u00e9 et al. 2003). Previously reported gain-modulations induced by background noise (Chance et al. 2002; Higgs et al. 2006) are qualitatively distinct than the effects shown in Figures 5 and 6, as they act by scaling the firing rate output of the neurons across all the input frequency-bands.\n\nOur experimental results then suggest that the action potential is a major evolutionary breakthrough, not only for making possible long-distance propagation of signals, but more importantly because it represents a powerful large-bandwidth digital intercellular communication channel, through population coding. In fact, our work shows that population coding with spikes has no significant attenuation in the range 0\u2013200 Hz, while it compensates the heavy drawbacks of the analog intracellular membrane properties, which filter out input frequencies faster than \u223c50 Hz.\n\n### Relations to Reverse-Correlation Methods\n\nNeural coding and the dynamical characterization of the input\u2013output transformation operated by neurons, have been previously addressed by using methods of stimulus reconstruction (Bialek et al. 1991; Rieke et al. 1995) or reverse correlation (de Boer and Kuyper 1968; Gerstner and Kistler 2002). The last identifies the typical input current preceding a spike. Such a procedure estimates the 1st-order Wiener kernel and thus the linear component of the system (Kroller 1992) even though underlying nonlinearities might be present \u201cin cascade\u201d (Chichilnisky 2001). The reverse correlation kernel is proportional to the impulse response of the linear response of the neuron and it characterizes the \u201cmeaning\u201d of each spike (Kroller 1992). The frequency-domain characterization that we considered so far directly relates to such an impulse response upon Fourier transform, although here we focused only on the encoding of the input signal (and not of the overall waveform) into the output. We thus generalized the previous experimental investigations to include the effect of background noise.\n\nConsistently with the weak impact of background noise at high input frequencies that we reported, one might expect that similar cut-off frequencies (i.e. \u223c100\u2013200 Hz) were quantitatively observed by previous investigators, although they might have not included any background noise. For instance, the \u201cstimulus kernel\u201d identified by Powers et al. (2005) in motoneurons by injecting stationary \u201cwhite\u201d-noise inputs, appears to be dominated by a single decay time-constant in the order of 5\u201310 ms, indeed matching the 100\u2013200 Hz cut-off frequencies of our data. Similarly, the low-noise phase-advance properties we observed (e.g., Fig. 2a) and the filter model intrinsic (high-pass) time-constants identified in our experiments, quantitatively correlate with the \u201cfeed-back\u201d kernel computed by the same authors after selecting short and long interspike-intervals to unveil the effect of spike-frequency adaptation.\n\nFinally, with the aim of further exploring the relationship of our approach with the previous ones, we directly computed the spike-triggered average (STA) of the input current preceding a spike, in 3 experiments where a broadband signal iT(t) was injected. In Figure 8, we compare the Fourier transform of the STA to the frequency response of the best-fit linear filter model optimized to match the instantaneous firing rate (as in Fig. 7). As expected interpreting the STA as the 1st-order kernel reveals striking similarities between the 2 approaches, especially for frequencies higher than 200 Hz.\n\nFigure 8.\n\nComparison between the 1st-order kernels computed by reverse-correlation techniques and the best-fit frequency response of the linear filter model of Figure 7. The modulation amplitude r1(f) (dashed line, identifying eq. 6) was compared with the fast Fourier transform (FFT) of the STA (markers) of the input current preceding a spike. The last was evaluated correlating the signal component iT(t) with the timing of each action potential, in 3 experiments. As expected from interpreting the STA as the 1st-order kernel, striking similarities are apparent.\n\nFigure 8.\n\nComparison between the 1st-order kernels computed by reverse-correlation techniques and the best-fit frequency response of the linear filter model of Figure 7. The modulation amplitude r1(f) (dashed line, identifying eq. 6) was compared with the fast Fourier transform (FFT) of the STA (markers) of the input current preceding a spike. The last was evaluated correlating the signal component iT(t) with the timing of each action potential, in 3 experiments. As expected from interpreting the STA as the 1st-order kernel, striking similarities are apparent.\n\n### The Phenomenological Filter Model\n\nLinear response properties are relevant to predict the response to complex noisy waveforms, even though the hypothesis of small input amplitude was not strictly respected by iT(t). As discussed by Carandini et al. (1996), our experiments support the idea that in vivo membrane potential fluctuations linearize the response to stimulus-related input components (Masuda et al. 2005). Consistently, the neuronal response r(t) could not be captured by employing a static nonlinearity alone (not shown), even though for stationary noisy stimuli a similar description is appropriate (Rauch et al. 2003; Giugliano et al. 2004; La Camera et al. 2006; Arsiero et al. 2007). The additional cascade threshold-linear element simply relates to the presence of a minimal input threshold. It is interesting to note that the piecewise-linear profile of such nonlinearity reflects the minor role played by distortions and harmonics in our experiments.\n\nSummarizing, a simple \u201ccascade\u201d model could quantitatively capture the time course of the instantaneous discharge rate (see also Shelley et al. 2002; Gutkin et al. 2005; Schaette et al. 2005), although it neglected the precise firing times. On the other hand, these can be captured by spiking neuron models, as in Jolivet et al. (2006) and Paninski (2006), identifying the parameters of an exponential (or quadratic) integrate-and-fire including spike-frequency adaptation as in Brette and Gerstner (2005).\n\n### Cortical Rhythms\n\nSlow inputs produced a phase-advance of the output response whereas fast inputs a phase-lag, relative to the input modulation (Fuhrmann et al. 2002). This has been proposed to have important consequences for emerging population dynamics in recurrent networks, as the signals propagation between pre- and postsynaptic spikes does not only depend on the synaptic delays but also on the (oscillation frequency-dependent) delay introduced by the postsynaptic neuron itself. The fact that spike timing depends on f is particularly relevant for the emergence of population rhythms including fast ripples (Buzsaki et al. 1992; Csicsvari et al. 1999; Grenier et al. 2003; Buzsaki and Draguhn 2004; Buzsaki et al. 2004). In fact, the spikes of a presynaptic neuron, which is engaged in network-driven oscillations, generate periodic synaptic currents. Then the postsynaptic neuron experiences the periodic maxima of these currents after a synaptic delay, and responds to such a current signal reaching the maximum of its firing rate with an additional delay \u03a6(f) and attenuation r1\/I1. If both presynaptic and postsynaptic neurons are participating in the same global rhythm, the overall delay between the pre- and postsynaptic spikes must be consistent with the period of the global oscillation and no strong attenuation should occur at that frequency, as shown in computer simulations by Fuhrmann et al. (2002), Brunel and Wang (2003), and Geisler et al. (2005). Therefore, not every oscillation frequency f is compatible with a given recurrent network architecture, synaptic coupling and firing regime.\n\nWe showed that the phase-shift and response amplitude of L5 pyramidal cells depends on the background fluctuations (Figs 5 and 6). This suggests that the frequency of emerging rhythms can be modulated by a background network embedding those neurons, as the phase of single-cell response is affected. More general, any network activity that relies on the timing of recurrent spikes is governed not only by the synaptic dynamics but is also controlled by the response properties (\u03a6(f), r1) of single cells.\n\n## Funding\n\nSwiss National Science Foundation grant (No. 31-61335.00) to H.-R.L.; the Silva Casa foundation, the National Institute of Mental Health grant (2R01MH-62349), and the Swartz Foundation to C.G. and X.-J.W.; and the Human Frontier Science Program (LT00561\/2001-B) to M.G.\n\n## Supplementary Material\n\nSupplementary material can be found at: http:\/\/www.cercor.oxfordjournals.org\/\n\nWe are grateful to Drs L.F. Abbott, W. Gerstner, and M.J.E. Richardson for helpful discussions and to Drs A. Amarasingham, M. Arsiero, T. Berger, N. Brunel, A. Destexhe, G. Fuhrmann, E. Vasilaki for comments on an earlier version of the manuscript. 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\section{Introduction}
For a fixed integer $k \in \N$ and a function (or sequence) $f \colon [n] \to \R$, a \emph{length-$k$ monotone subsequence of $f$} is a tuple of $k$ indices, $(i_1, \dots, i_k) \in [n]^k$, such that $i_1 < \dots < i_k$ and $f(i_1) < \dots < f(i_k)$. More generally, for a permutation $\pi \colon [k] \to [k]$, a \emph{$\pi$-pattern of $f$} is given by a tuple of $k$ indices $i_1 < \dots < i_k$ such that $f(i_{j_1}) < f(i_{j_2})$ whenever $j_1, j_2 \in [k]$ satisfy $\pi(j_1) < \pi(j_2)$. A sequence $f$ is $\pi$-free if there are no subsequences of $f$ with order pattern $\pi$. Recently, Newman, Rabinovich, Rajendraprasad, and Sohler~\cite{NRRS17} initiated the study of property testing for forbidden order patterns in a sequence. Their paper was the first to analyze algorithms for finding $\pi$-patterns in sublinear time (for various classes of the permutation $\pi$); additional algorithms and lower bounds for several classes of permutations have later been obtained by Ben-Eliezer and Canonne~\cite{BC18}.
Of particular interest of $\pi$-freeness testing is the case where $\pi = (12\dots k)$, i.e., $\pi$ is a monotone permutation. In this case, avoiding length-$k$ monotone subsequence may be equivalently rephrased as being decomposable into $k-1$ monotone non-increasing subsequences. Specifically, a function $f \colon [n] \to \R$ is $(12\dots k)$-free if and only if $[n]$ can be partitioned into $k-1$ disjoint sets $A_1, \ldots, A_{k-1}$ such that, for each $i\in[k-1]$, the restriction $f|_{A_i}$ is non-increasing. When interested in algorithms for testing $(12\dots k)$-freeness that have a \emph{one-sided error},\footnote{An algorithm for testing property $\calP$ is said to have \emph{one-sided error} if the algorithm always outputs ``yes'' if $f \in \calP$, i.e., has perfect completeness.} the algorithmic task becomes the following:
\begin{quote}\itshape
For $k \in \N$ and $\eps > 0$, design a randomized algorithm that, given query access to a function $f \colon [n] \to \R$ guaranteed to be $\eps$-far from being $(12\dots k)$-free,\footnote{A function $f \colon [n] \to \R$ is \emph{$\eps$-far} from $\pi$-free if any $\pi$-free function $g \colon [n] \to \R$ satisfies $\Pr_{\bi \sim [n]}[f(\bi) \neq g(\bi)]\geq \eps$.} outputs a length-$k$ monotone subsequence of $f$ with probability at least $9/10$.
\end{quote}
The task above is a natural generalization of monotonicity testing of a function $f \colon [n] \to \R$ with algorithms that make a one-sided error,
a question which dates back to the early works in property testing, and has received significant attention since in various settings (see, e.g.,~\cite{DGLRRS99,GGLRS00,FLNRRS02,AMW13,BRY14b,Bel18,PRV18, BE19}, and the recent textbook \cite{G17}). For the problem of testing monotonicity, Erg\"{u}n, Kannan, Kumar, Rubinfeld, and Viswanathan \cite{EKKRV00} were the first to give a non-adaptive algorithm which tests monotonicity of functions $f\colon [n] \to \R$ with one-sided error making $O(\log(n) / \eps)$ queries. (Recall that an algorithm is \emph{non-adaptive} if its queries do not depend on the answers to previous queries, or, equivalently, if all queries to the function can be made in parallel.) Furthermore, they showed that $\Omega(\log n)$ queries are necessary for non-adaptive algorithms. Subsequently, Fischer \cite{F04} showed that $\Omega(\log n)$ queries are necessary even for adaptive algorithms. Generalizing from monotonicity testing (when $k=2$),
Newman et al.\ gave in~\cite{NRRS17} the first sublinear-time algorithm for $(12\dots k)$-freeness testing, whose query complexity is $(\log(n)/\eps)^{O(k^2)}$. Their algorithm is non-adaptive and has one-sided error; as such, it outputs a length-$k$ monotone subsequence with probability at least $9/10$ assuming the function $f$ is $\eps$-far from $(12\dots k)$-free. However, other than the aforementioned lower bound of $\Omega(\log n)$ which follows from the case $k=2$, no lower bounds were known for larger~$k$.
The main contribution of this work is to settle the dependence on $n$ in the query complexity of testing for $(12\dots k)$-freeness with non-adaptive algorithms making one-sided error. Equivalently, we settle the complexity of non-adaptively finding a length-$k$ monotone subsequence under the promise that the function $f \colon [n] \to \R$ is $\eps$-far from $(12\dots k)$-free.
\begin{theorem}\label{thm:intro-ub}
Let $k \in \N$ be a fixed parameter. For any $\eps > 0$, there exists an algorithm that, given query access to a function $f \colon [n] \to \R$ which is $\eps$-far from $(12\dots k)$-free, outputs a length-$k$ monotone subsequence of $f$ with probability at least $9/10$. The algorithm is non-adaptive and makes $(\log n)^{\lfloor \log_2 k \rfloor} \cdot \poly(1/\eps)$ queries to $f$.
\end{theorem}
Our algorithm thus significantly improves on the $(\log(n)/\eps)^{O(k^2)}$-query non-adaptive algorithm of \cite{NRRS17}. Furthermore, its dependence on $n$ is optimal; indeed, in the next theorem we prove a matching lower bound for all fixed $k \in \N$.
\begin{theorem}
\label{thm:intro-lb}
Let $k \in \N$ be a fixed parameter. There exists a constant $\eps_0 > 0$ such that any non-adaptive algorithm which, given query access to a function $f\colon [n] \to \R$ that is $\eps_0$-far from $(12\dots k)$-free, outputs a length-$k$ monotone subsequence with probability $9/10$, must make $\Omega((\log n)^{\lfloor \log_2 k \rfloor})$ queries. Moreover, one can take $\eps_0 = 1/(4k)$.
\end{theorem}
\noindent We further note that the lower bound holds even for the more restricted case where $f \colon [n] \to [n]$ is a permutation.
\subsection{Related work}
Testing monotonicity of a function over a partially ordered set $\mathcal{X}$ is a well-studied and fruitful question, with works spanning the past two decades.
Particular cases include when $\calX$ is the line $[n]$ \cite{EKKRV00,F04,Bel18,PRV18,BE19}, the Boolean hypercube $\{0,1\}^d$~\cite{DGLRRS99,BBM12,BCGM12,CS13,CST14,CDST15,KMS15,BB15,CS16,CWX17, CS18}, and the hypergrid $[n]^d$~\cite{BRY14a,CS14, BCS18}.
We refer the reader to~\cite[Chapter~4]{G17} for more on monotonicity testing, or for an overview of the field of property testing (as introduced in~\cite{RS96,GGR98}) in general.
This paper is concerned with the related line of work on finding order patterns in sequences and permutations. For the exact case, Guillemot and Marx \cite{GM14} showed that an order pattern $\pi$ of length $k$ can be found in a sequence $f$ of length $n$ in time $2^{O(k^2 \log k)} n$; in particular, the problem of finding order patterns is fixed-parameter tractable (in the parameter $k$). Fox \cite{Fox13} later improved the running time to $2^{O(k^2)} n$. A very recent work of Kozma \cite{Koz19} provides the state-of-the-art for the case where $k = \Omega(\log n)$.
In the sublinear regime, the most relevant works are the aforementioned papers of Newman et al.~\cite{NRRS17} and Ben-Eliezer and Canonne~\cite{BC18}. In particular, \cite{NRRS17} shows an interesting dichotomy for testing $\pi$-freeness: when $\pi$ is monotone, the non-adaptive query complexity
is polylogarithmic in $n$ for fixed $k$ and $\eps$, whereas for non-monotone $\pi$, the query complexity is $\Omega(\sqrt{n})$.
Two related questions are that of estimating the \emph{distance to monotonicity} and the length of the \emph{longest increasing subsequence} (LIS),
which have also received significant attention from both the sublinear algorithms perspective~\cite{PRR06, ACCL07, SS17}, as well as the streaming perspective~\cite{GJKK07,GG10,SS13,EJ15,NSaks15}.
In particular, Saks and Seshadhri gave in~\cite{SS17} a randomized algorithm which, on input $f \colon [n] \to \R$, makes $\poly(\log n, 1/\delta)$ queries and outputs $\hat{m}$ approximating up to additive error $\delta n$ the length of the longest increasing subsequence of $f$. This paper also studies monotone subsequences of the input function, albeit from a different (and incomparable) end of the problem. Loosely speaking, in \cite{SS17} the main object of interest is a very \emph{long} monotone subsequence (of length linear in $n$), and the task at hand is to get an estimate for its total length, whereas in our setting, there are $\Omega(n)$ disjoint copies of \emph{short} monotone subsequences (of length $k$, which is a constant parameter), and these short subsequences may not necessarily combine to give one long monotone subsequence.
\subsection{Our techniques: Upper bound}
We now give a detailed overview of the techniques underlying our upper bound, Theorem~\ref{thm:intro-ub}, and provide some intuition behind the algorithms and notions we introduce. The starting point of our discussion will be the algorithm of Newman et al.~\cite{NRRS17}, which we re-interpret in terms of the language used throughout this paper; this will set up some of the main ideas behind our structural result (stated in Section~\ref{sec:structural}), which will be crucial in the analysis of the algorithm.
For simplicity, let $\eps > 0$ be a small constant and let $k \in \N$ be fixed. Consider a function $f \colon [n] \to \R$ which is $\eps$-far from $(12\dots k)$-free. This implies that there is a set $T \subseteq [n]^k$ of $\eps n/k$ \emph{disjoint} $(12\dots k)$-patterns. Specifically, the set $T$ is comprised of $k$-tuples $(i_1,\dots, i_k) \in [n]$ where $i_1 < \dots < i_k$ and $f(i_1) < \dots < f(i_k)$ and each $i \in [n]$ appears in at most one $k$-tuple in $T$.\footnote{To see why such $T$ exists, take $T$ to be a maximal set of disjoint $(12\dots k)$-patterns. Suppose $|T| < \eps n/k$ and consider the function $g$ given by greedily eliminating all $(12\dots k)$-patterns in $f$, and note that $g$ is $(12\dots k)$-free and differs on $f$ in less than $\eps n$ indices.}
A key observation made in \cite{NRRS17} is that if, for some $c \in [k-1]$, $(i_1, \dots, i_c, i_{c+1}, \dots, i_k)$ and $(j_1, \dots, j_c, j_{c+1}, \dots, j_k)$ are two $k$-tuples in $T$ which satisfy $i_c < j_{c+1}$ and $f(i_c) < f(j_{c+1})$, then their combination
\[
(i_1,\dots, i_c, j_{c+1}, \dots, j_k)
\]
is itself a length-$k$ monotone subsequence of $f$. Therefore, in order to design efficient sampling algorithms, one should analyze to what extent parts of different $(12\dots k)$-tuples from $T$ may be combined to form length-$k$ monotone subsequences of $f$.
Towards this goal, assign to each $k$-tuple $(i_1, \dots, i_k)$ in $T$ a \emph{distance profile} $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1, \dots, d_{k-1}) \in [\eta]^{k-1}$, where $\eta = O(\log n)$.\footnote{We remark that the notion of a distance profile is solely used for the introduction and for explaining \cite{NRRS17}, and thus does not explicitly appear in subsequent sections.} This distance profile is a $(k-1)$-tuple of non-negative integers satisfying
\[ 2^{d_j} \leq i_{j+1} - i_j < 2^{d_j + 1} \qquad\qquad j \in [k-1]\,; \]
and let $\textsf{gap}(i_1,\dots, i_k) = c \in [k-1]$ be the smallest integer where $d_c \geq d_{j}$ for all $j \in [k-1]$ (i.e., $d_c$ denotes an (approximately) maximum length between two adjacent indices in the $k$-tuple). Suppose, furthermore, that for a particular $c \in [k-1]$, the subset $T_c \subseteq T$ of $k$-tuples whose gap is at $c$ satisfies $|T_c| \geq \eps n / k^2$ (such a $c \in [k-1]$ must exist since the $T_c$'s partition $T$). If $(i_1, \dots, i_k) \in T_c$ and $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1, \dots, d_k)$, then the probability that a uniformly random element $\bell$ of $[n]$ ``falls'' into that gap is
\begin{align}
\Prx_{\bell \sim [n]}\left[ i_{c} \leq \bell \leq i_{c+1}\right] &\geq \frac{2^{d_c}}{n}. \label{eq:prob}
\end{align}
Whenever this occurs for a particular $k$-tuple $(i_1, \dots, i_k)$ and $\bell \in [n]$, we say that $\bell$ \emph{cuts} the tuple $(i_1,\dots, i_k)$. Note that the indices $i_{c+1}, \dots, i_{k}$ are contained within the interval $[\bell, \bell + k \cdot 2^{d_c + 1}]$ and the indices $i_1, \dots, i_c$ are contained within the interval $[\bell - k \cdot 2^{d_c + 1}, \bell]$. As a result, if we denote by $\delta_d(\bell) \in [0,1]$, for each $d \in [\eta]$, the \emph{density} of $k$-tuples from $T_c$ lying inside $[\bell - k \cdot 2^{d + 1}, \bell + k \cdot 2^{d + 1}]$ (i.e., the fraction of this interval comprised of elements of $T_c$), we have
\begin{align}
\Ex_{\bell \sim [n]} \left[ \sum_{d \in [\eta]} \delta_{d}(\bell) \right] &= \sum_{d \in [\eta]} \sum_{\substack{(i_1,\dots, i_k) \in T_c \\ \textsf{dist-prof}(i_1,\dots, i_k)_c =d}} \Prx_{\bell \sim [n]}\left[ i_c \leq \bell \leq i_{c+1}\right] \cdot \frac{1}{2 \cdot k \cdot 2^{d+1}} \gsim \frac{|T_c|}{n} \gsim \eps. \label{eq:expect}
\end{align}
For any $\ell$ achieving the above inequality, since $\eta = O(\log n)$, there exists some $d^* \in [\eta]$ such that $\delta_{d^*}(\ell) \gsim \eps / \log n$. Consider now the set of $k$-tuples $T_{c, d^*}(\ell)\subseteq T_c$ contributing to $\delta_{d^*}(\ell)$, i.e., those $k$-tuples in $T_c$ which are cut by $\ell$ and lie in $[\ell - k \cdot 2^{d^*+1}, \ell + k \cdot 2^{d^*+1}]$. Denote $r_{\mathrm{med}} = \textsf{median}\{ f(i_c) : (i_1, \dots, i_k) \in T_{c, d^*}(\ell)\}$, and let
\begin{align*}
T_{L} &= \left\{ (i_1,\dots, i_c) : (i_1, \dots, i_k) \in T_{c, d^*}(\ell) \text{ and } f(i_c) \leq r_{\mathrm{med}} \right\}, \qquad \text{and}\\
T_{R} &= \left\{ (i_{c+1},\dots, i_k) : (i_1, \dots, i_k) \in T_{c, d^*}(\ell) \text{ and } f(i_c) \geq r_{\mathrm{med}} \right\},
\end{align*}
where we note that $T_L$ and $T_R$ both have size at least $|T_{c, d^*}(\ell)| / 2$. If the algorithm finds a $c$-tuple in $T_{L}$ and a $(k-c)$-tuple in $T_R$, by the observation made in \cite{NRRS17} that was mentioned above, the algorithm could combine the tuples to form a length-$k$ monotone subsequence of $f$. At a high level, one may then recursively apply these considerations on $[\ell - k \cdot 2^{d^*+1}, \ell]$ with $T_L$ and $[\ell, \ell + k \cdot 2^{d^*+1}]$ with $T_R$. A natural algorithm then mimics the above reasoning algorithmically, i.e., samples a parameter $\bell \sim [n]$, and tries to find the unknown parameter $d^* \in [\eta]$ in order to recurse on both the left and right sides; once the tuples have length $1$, the algorithm samples within the interval to find an element of $T_L$ or $T_R$. This is, in essence, what the algorithm from~\cite{NRRS17} does, and this approach leads to a query complexity of $(\log n)^{O(k^2)}$.
In particular, suppose that at each (recursive) iteration, the parameter $c$, corresponding to the gap of tuples in $T$, always equals $1$. Note that this occurs when all $(12\dots k)$-patterns $(i_1,\dots, i_k)$ in $T$ have $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1,\dots, d_{k-1})$ with
\begin{align}
d_1 \geq d_2 \geq \dots \geq d_{k-1}. \label{eq:decreasing}
\end{align}
Then, if $k$ is at $k_0$, a recursive call leads to a set $T_L$ containing $1$-tuples, and $T_{R}$ containing $(k_0 - 1)$-tuples. This only decreases the length of the subsequences needed to be found by $1$ (so there will be $k-1$ recursive calls), while the algorithm pays for guessing the correct value of $d^*$ out of $\Omega(\log n)$ choices, which may decrease the density of monotone $k_0$-subsequences within the interval of the recursive call by a factor as big as $\Omega(\log n)$.\footnote{Initially, the density of $T$ within $[n]$ is $\eps$, and the density of $T_L$ or $T_R$ in $[\ell - k\cdot 2^{d^*+1}, \ell]$ and $[\ell, \ell + k \cdot 2^{d^*+1}]$ is $\eps / \log n$.} As a result, the density of the length-$k_0$ monotone subsequence in the relevant interval could be as low as $\eps / (\log n)^{k_0}$, which means that $(\log n)^{\Omega(k_0)}$ samples will be needed for the $k_0$-th round according to the above analysis, giving a total of $(\log n)^{\Omega(k^2)}$ samples
(as opposed to $O((\log n)^{\lfloor \log_2 k\rfloor})$, which is the correct number, as we prove).\smallskip
In order to overcome the above difficulty, we consider a particular choice of a family $T$ of length-$k$ monotone subsequences given by the ``greedy'' procedure (see Figure~\ref{fig:greedy}). Loosely speaking, this procedure begins with $T = \emptyset$ and iterates through each index $i_1 \in [n] \setminus T$.
Each time, if $(i_1)$ can be extended to a length-$k$ monotone subsequence (otherwise it continues to the next available index), the procedure sets $i_2$ to be the first index, after $i_1$ and not already in $T$, such that $(i_1, i_2)$ can be extended to a length-$k$ monotone subsequence; then, it finds an index $i_3$ which is the next first index after $i_2$ and not in $T$ such that $(i_1, i_2, i_3)$ can be extended; and so on, until it has obtained a length-$k$ monotone subsequence starting at $i_1$. It then adds the subsequence as a tuple to $T$, and repeats. This procedure eventually outputs a set $T$ of disjoint, length-$k$ monotone subsequences of $f$ which has size at least $\eps n/ k^2$, and satisfies another crucial ``interleaving'' property (see Lemma~\ref{lem:rematching}):
\begin{quote}
($\star$) If $(i_1, \dots, i_k)$ and $(j_1, \dots, j_k)$ are $k$-patterns from $T$ and $c \in [k-1]$ satisfy $j_1 < i_1$, $j_c < i_c$, and $i_{c+1} < j_{c+1}$, then $f(i_{c+1}) < f(j_{c+1})$.
\end{quote}
Moreover, a slight variant of~\eqref{eq:prob} guarantees that for any $(i_1, \dots, i_k) \in T_c$ with $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1, \dots, d_{k-1})$,
\begin{align*}
\Prx_{\bell \sim [n]}\left[ i_c + 2^{d_c}/3 \leq \bell \leq i_{c+1} - 2^{d_c} / 3\right] \gsim \frac{2^{d_c}}{n}.
\end{align*}
Whenever the above event occurs, we say $\bell \sim [n]$ \emph{cuts} $(i_1,\dots, i_k)$ at $c$ \emph{with slack}, and note that $i_1, \dots, i_c$ lie in $[\bell - k \cdot 2^{d_c+1}, \bell]$ and $i_{c+1}, \dots, i_k$ in $[\bell, \bell + k \cdot 2^{d_c+1}]$. We denote, similarly to the above, $\delta_{d}(\bell) \in [0, 1]$ to be the density of $k$-tuples from $T_c$ which are cut with slack by $\bell$, and conclude~\eqref{eq:expect}. We then utilize ($\star$) to make the following claim: suppose two $k$-tuples $(i_1, \dots, i_k), (j_1, \dots, j_k) \in T_c$ satisfy $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1, \dots, d_{k-1})$, and $\textsf{dist-prof}(j_1,\dots, j_k) = (d_1', \dots, d_{k-1}')$, where $d_{c} \leq d_{c}' - a\log k$, for some constant $a$ which is not too small. If $(i_1, \dots, i_k)$ and $(j_1, \dots, j_k)$ are cut at $c$ with slack, this means that $\ell$ lies roughly in the middle of $i_{c}$ and $i_{c+1}$ and of $j_c$ and $j_{c+1}$, and since the distance between $i_c$ and $i_{c+1}$ is much smaller than that between $j_c$ and $j_{c+1}$, the index $j_1$ will come before $i_1$, the index $j_c$ will come before $i_{c}$, but the index $i_{c+1}$ will come before $j_{c+1}$. By ($\star$), $f(i_{c+1}) < f(j_{c+1})$ (cf. Lemma~\ref{lem:At:properties}). In other words, the value, under the function $f$, of $(c+1)$-th indices from tuples in $T_{c, d}(\ell)$ increases as $d$ increases.
As a result, if $\ell \in [n]$ satisfies $\sum_{d \in [\eta]} \delta_{d}(\ell) \gsim \eps$, and $\delta_{d}(\ell) \ll \eps$ for all $d \in [\eta]$, that is, if the summands in~\eqref{eq:expect} are \emph{spread out}, an algorithm could find a length-$k$ monotone subsequence by sampling, for many values of $d \in [\eta]$, indices which appear as the $(c+1)$-th index of tuples in $T_{c, d}(\ell)$. We call such values of $\ell$ the starts of \emph{growing suffixes} (as illustrated in Figure~\ref{fig:growing-suffix}). In Section~\ref{sec:case1}, we describe an algorithm that makes $\tilde{O}(\log n / \eps)$ queries and finds, with high probability, a length-$k$ monotone subsequence if there are many such growing suffixes (see Lemma~\ref{lem:case1}). The algorithm works by randomly sampling $\bell \sim [n]$ and hoping that $\bell$ is the start of a growing suffix; if it is, the algorithm samples enough indices from the segments $[\ell + 2^{d}, \ell + 2^{d+1}]$ to find a $(c+1)$-th index of some tuple in $T_{c, d}(\ell)$, which gives a length-$k$ monotone subsequence.
The other case corresponds to the scenario where $\ell \in [s]$ satisfies $\sum_{d \in [\eta]} \delta_d(\ell) \gsim \eps$, but the summands are \emph{concentrated} on few values of $d \in [\eta]$. In this case, we may consider a value of $d^* \in [\eta]$ which has $\delta_{d^*}(\ell) \gsim \eps$, and then look at the intervals $[\ell - k \cdot 2^{d^*+1}, \ell]$ and $[\ell, \ell + k \cdot 2^{d^* + 1}]$. We can still define $T_{L}$ and $T_R$, both of which have size at least $|T_{c, d^*}| / 2$ and have the property that any $c$-tuple from $T_L$ can be combined with any $(k-c)$-tuple from $T_R$. Additionally, since $\delta_{d^*}(\ell) \gsim \eps$, we crucially do \emph{not} suffer a loss in the density of $T_L$ and $T_R$ in their corresponding intervals~--~a key improvement over the $\Omega(\log n)$ loss in density incurred by the original approach we first discussed. We refer to these intervals as \emph{splittable intervals} (cf. Figure~\ref{fig:splittable}), and observe that they lead to a natural recursive application of these insights to the intervals $[\ell - k \cdot 2^{d^*+1}, \ell]$ and $[\ell, \ell + k \cdot 2^{d^*+1}]$. The main structural result, given in Theorem~\ref{thm:tree}, does exactly this, and encodes the outcomes of the splittable intervals in an object we term a \emph{$k$-tree descriptor} (see Section~\ref{ssec:tree:descriptors}) whenever there are not too many growing suffixes. Intuitively, a $k$-tree descriptor consists of a rooted binary tree $G$ on $k$ leaves, as well as some additional information, which corresponds to a function $f \colon [n] \to \R$ without many growing suffixes. Each internal node $v$ in $G$ corresponds to a recursive application of the above insights, i.e., $v$ has $k_0$ leaves in its subtree, a parameter $c_v \in [k_0 - 1]$ encoding the gap of sufficiently many $k_0$-tuples, and a collection of disjoint intervals of the form $[\ell - k \cdot 2^{d^*}, \ell + k \cdot 2^{d^*}]$ where $\ell$ cuts $(12\dots k_0)$-patterns with slack at $c_v$ and satisfies~\eqref{eq:expect}; the left child of $v$ has $c$ leaves and contains the $(12\dots c)$-patterns in $T_L$ and intervals $[\ell - k \cdot 2^{d^*}, \ell]$; the right child of $v$ has $k_0 -c$ leaves and contains the $(12\dots (k_0 - c))$-patterns in $T_R$ and intervals $[\ell, \ell + k\cdot 2^{d^*}]$ (see Figure~\ref{fig:tree-descriptor}).
The algorithm for this case is more involved than the previous, and leads to the $O((\log n)^{\lfloor \log_2 k \rfloor})$-query complexity stated in Theorem~\ref{thm:intro-ub}. The algorithm proceeds in $r_0 = 1 + \lfloor \log_2 k \rfloor$ rounds, maintaining a set $\bA \subseteq [n]$, initially empty:
\begin{itemize}
\item \textit{Round 1}: For each $i \in [n]$, include $i$ in $\bA$ independently with probability $\Theta(1/(\eps n))$.
\item \textit{Round $r$, $2 \leq r \leq r_0$}: For each $i \in \bA$ from the previous round, and each $j = 1, \dots, O(\log n)$, consider the interval $B_{i,j} = [i - 2^j, i + 2^j]$. For each $i' \in B_{i,j}$, include $i'$ in $\bA$ independently with probability $\Theta(1/(\eps 2^j))$.
\end{itemize}
At the end of all rounds, the algorithm queries $f$ at all indices in $\bA$, and outputs a $(12\dots k)$-pattern from $\bA$, if one exists.
Recall the case considered in the sketch of the algorithm of~\cite{NRRS17}, when the function $f$ has all $(12\dots k)$-patterns $(i_1, \dots, i_{k})$ in $T$ satisfying $\textsf{dist-prof}(i_1, \dots, i_k) = (d_1, \dots, d_{k-1})$ with $d_1 \ge d_2 \ge \ldots \ge d_{k-1}$. In this case, the $k$-tree descriptor $G$ consists of a rooted binary tree of depth $k$. The root has a left child which is a leaf (corresponding to $1$-tuples of first indices of some tuples in $T$, stored in $T_L$) and a right child (corresponding to suffixes of length $(k-1)$ of some tuples in $T$, stored in $T_R$) is an internal node. The root node corresponds to one application of the structural result, and the right child corresponds to a $(k-1)$-tree descriptor for the tuples in $T_R$. Loosely speaking, as $d_2 \ge \ldots \ge d_{k-1}$ the same reasoning repeats $k-1$ times, and leads to a path of length $k-1$ down the right children of the tree, the right child of the $(k-1)$-th internal node corresponding to a $1$-tuple (i.e., a leaf).\footnote{This is somewhat inaccurate, as in each step, after forming $T_L$ and $T_R$, we apply the greedy algorithm again and obtain new sets $T_L'$ and $T_R'$, which may violate the assumption $d_1 \ge d_2 \ge \ldots \ge d_k$. We ignore this detail at the moment to simplify the explanation.}
To gain some intuition, we analyze how the algorithm behaves on these instances. Suppose that in round 1, the algorithm samples an element $i \in [n]$ which is the $k$-th index of a $1$-tuple stored in the right-most leaf of $G$. In particular, this index belongs to the set $T_R$ of the $(k-1)$-th internal node, as a second index of a cut $(12)$-pattern in the $(k-1)$-th recursive call of the structural result. Similarly, $i$ also belongs to that set $T_R$ of the $(k-2)$-th internal node, as a part the third index of a cut $(123)$-pattern in the $(k-2)$-th recursive call. We may continue with all these inclusions to the root, i.e., $i$ is the $k$-th element of some $(12\dots k)$-pattern in $T$, which is cut in the first call to the structural result. Round 2 of the algorithm will consider the $k-1$ intervals $B_{i, d_{k-1}'}, B_{i, d_{k-2}'}, \dots, B_{i, d_{1}'}$, where $d_j'= d_j + \Theta(\log k)$, since it iterates through all $O(\log n)$ intervals of geometrically increasing lengths.\footnote{Note that the intervals $B_{i, d_{j}'}$ and $B_{i,d_{j+1}'}$ may be the same, for instance when $d_j = d_{j+1}$.} One can check that for each $j \in [k-1]$, the interval $B_{i, d_{j}'}$ contains $[\ell_j - k \cdot 2^{d_j}, \ell_j]$, where $\ell_j$ is some index which cut the $(k-j+1)$-tuple $(i_j,\dots, i_k)$ with slack in the $j$-th recursive call of the structural result. Recall that the set $T_L$ of $1$-tuples has density $\Omega(\eps)$ inside $[\ell_j - k \cdot 2^{d_j}, \ell_j]$ and may be combined with any $(k-j)$-tuple from $T_R$. Following this argument, in the \emph{second} round of the algorithm, $\bA$ will include some index of $T_L$ (for each $j \in [k-1]$), and these indices combine to form a $(12\dots k)$-pattern~--~that is, with high probability, after two rounds, the algorithm succeeds in finding a monotone subsequence of length $k$.
Generalizing the above intuition for all possible distance profiles necessitates the use of $1 + \lfloor \log_2 k \rfloor$ rounds, and requires extra care. At a high level, consider an arbitrary $k$-tree descriptor $G$ for $\Omega(\eps n)$ many $(12\dots k)$-patterns in $f$. Denote the root $u$, and consider the unique leaf $w$ of $G$ where the root-to-$w$ path $(u_1, \dots, u_h)$ with $u_1 = u$ and $u_h = w$, satisfies that at each internal node $u_{l}$, the next node $u_{l+1}$ is the child with larger number of leaves in its subtree.\footnote{Ties are broken by picking the left child.} We call such a leaf a \emph{primary index} of $G$. The crucial property of the primary index is that the root-to-leaf path of $w$, $(u_1, u_2, \dots u_h)$, is such that the siblings of the nodes on this path\footnote{For example, if $(u_1, \dots, u_h)$ is the root-to-$w$ path where $u_1$ is the root and $u_h = w$, the sibling nodes along the path are given by $u_{2}', u_{3}', \dots, u_{h}'$, where $u_{l}'$ is the sibling of $u_l$. Namely, if the $l$-th node on the root-to-$w$ path is a left child of the $(l-1)$th node, then $u_{l}'$ is the right child of the $(l-1)$-th node. Analogously, if the $l$-th node is a right child of the $(l-1)$-th node, then $u_l'$ is the left child of the $(l-1)$-th node.} have strictly fewer than $k / 2$ leaves in their subtrees.
The relevant event in the first round of the algorithm is that of sampling an index $i \in [n]$ which belongs to a $1$-tuple of the primary index $w$ of $G$. This occurs with probability at least $1 - 1/(100 k)$, since we sample each element of $[n]$ with probability $\Theta(1/(\eps n))$ while there are at least $\Omega(\eps n)$ many $(12\dots k)$-patterns. Now, roughly speaking, letting $(u_1, \dots, u_h)$ be the root-to-$w$ path in $G$, and $(u_2', \dots, u_h')$ be the sibling nodes, the subtrees of $G$ rooted at $u_2', \dots, u_h'$ will be tree descriptors for the function $f$ restricted to $B_{i, j}$'s and within these interval, the density of tuples is at least $\Omega(\eps)$. As a result, the second round of the algorithm, recursively handles each subtree rooted at $u_2', \dots, u_h'$ with one fewer round. Since the subtrees have strictly fewer than $k/2$ leaves, $\lfloor \log_2 k \rfloor - 1$ rounds are enough for an inductive argument. Moreover, since the total number of nodes is at most $2k$ and each recursive call succeeds with probability at least $1-1/(100 k)$, by a union bound we may assume that all recursive calls succeed.
Unrolling the recursion, the query complexity $\Theta((\log n)^{\lfloor \log_2 k \rfloor})$ can be explained with a simple combinatorial game. We start with a rooted binary tree $G$ on $k$ leaves. In one round, whenever $G$ is not simply a leaf, we pick the leaf $w$ which is the primary index of $G$, and replace $G$ with a collection of subtrees obtained by cutting out the root-to-$w$ path in $G$. These rounds ``pay'' a factor of $\Theta(\log n)$, since the algorithm must find intervals on which the collection of subtrees form tree descriptors of $f$ (restricted to these intervals). In the subsequent rounds, we recurse on each subtree simultaneously, picking the leaf of the primary index in each, and so on. After $\lfloor \log_2 k \rfloor$ many rounds, the trees are merely leaves, and the algorithm does not need to pay the factor $\Theta(\log n)$ to find good intervals, as it may simply sample from these intervals.
The execution of the above high-level plan is done in Section~\ref{sec:case2}, where Lemma~\ref{lem:case2} is the main inductive lemma containing the analysis of the main algorithm (shown in Figure~\ref{fig:sample-splittable} and Figure~\ref{fig:sample-splittable-2}).
\subsection{Our techniques: Lower bound}
In order to highlight the main ideas behind the proof of Theorem~\ref{thm:intro-lb} (the lower bound on the query complexity), we first cover the simpler case of $k=2$. This case corresponds to a lower bound of $\Omega(\log n)$ on the number of queries needed for non-adaptive and one-sided algorithms for monotonicity testing. Such a lower bound is known, even for adaptive algorithms with two-sided error \cite{EKKRV00, F04}. We rederive and present the well-known non-adaptive one-sided lower bound in our language; after that, we generalize it to the significantly more involved case $k > 2$. For the purpose of this introduction, we give an overview assuming that both $n$ and $k$ are powers of $2$; as described in Section \ref{sec:lowerbounds}, a simple ``padding'' argument generalizes the result to all $n$ and $k$.
For any $n \in \N$ which is a power of $2$ and $t \in [n]$, consider the \emph{binary representation} $B_n(t) = (b^t_1, b^t_2, \ldots, b^t_{\log_2 n}) \in \{0,1\}^{\log_2 n}$ of $t$, where $t = b^t_1 \cdot 2^{0} + b^t_2 \cdot 2^{1} + \dots + b^t_{\log_2 n} \cdot 2^{\log_2 n - 1}$.
For $i \in [\log_2 n]$, the \emph{bit-flip operator}, $F_i \colon [n] \to [n]$, takes an input $t \in [n]$ with binary representation $B_n(t)$ and outputs the number $F_i(t) = t' \in [n]$ with binary representation obtained by flipping the $i$-th bit of $B_n(t)$. Finally, for any two distinct elements $x,y \in [n]$, let $M(x,y) \in [\log_2 n]$ be the index of the most significant bit in which they differ, i.e., the largest $i$ where $b_i^x \neq b_i^y$.
As usual for lower bounds on randomized algorithms, we rely on Yao's minimax principle \cite{Y77}. In particular, our lower bounds proceed by defining, for each $n$ and $k$ (which are powers of $2$), a distribution $\calD_{n,k}$ supported on functions $f \colon [n] \to \R$ which are all $\eps$-far from $(12\dots k)$-free. We show that any deterministic and non-adaptive algorithm which makes fewer than $q$ queries, where $q = c_k (\log_2 n)^{\log_2 k}$ and $c_k > 0$ depends only on $k$, fails to find a $(12\ldots k)$-pattern in a random $\boldf \sim \calD_{n,k}$, with probability at least $1/10$. Note that any deterministic, non-adaptive algorithm which makes fewer than $q$ queries is equivalently specified by a set $Q \subseteq [n]$ with $|Q| < q$. Thus, the task of the lower bound is to design a distribution $\calD_{n,k}$ supported on functions $f \colon [n] \to \R$, each of which is $\eps$-far from $(12\dots k)$-free, such that for any $Q \subseteq [n]$ with $|Q| < c_k (\log_2 n)^{\log_2 k}$ the following holds
\begin{align}
\Prx_{\boldf \sim \calD_{n,k}}\left[ \exists i_1, \dots, i_k \in Q: i_1 < \dots < i_k \text{ and } \boldf(i_1) < \dots < \boldf(i_k) \right] \leq \frac{9}{10}. \label{eq:yao}
\end{align}
\medskip\noindent\textsc{Lower bound for $k=2$ (monotonicity).}
The case of $k=2$ relies on the following idea: for any $i \in [\log_2 n]$, one can construct a function (in fact, a permutation) $f_i \colon [n] \to [n]$ which is $1/2$-far from $(12)$-free, and furthermore, all pairs of distinct elements $(x, y) \in [n]^2$ where $x < y$ and $f_i(x) < f_i(y)$ satisfy $M(x, y) = i$.
One can construct such a function $f_i \colon [n] \to [n]$, for any $i \in[\log_2 n]$ in the following way.
First, let $f^{\downarrow} \colon [n] \to [n]$ be the decreasing permutation, $f^{\downarrow}(x) = n+1-x$ for any $x \in [n]$.
Now take $f_i$ to be $f^{\downarrow} \circ F_i$, where $\circ$ denotes function composition, that is, $f_i(x) = f^{\downarrow}(F_i(x))$ for any $x \in [n]$.
Finally, set $\calD_{n,2}$ to be the uniform distribution over the functions $f_1, f_2, \ldots, f_{\log n}$ (see Figure~\ref{fig:mon-construct}).
\begin{figure}
\centering
\begin{picture}(340, 200)
\put(0,0){\includegraphics[width=0.7\linewidth]{mon-construct.pdf}}
\put(45, 60){$2^{i}$}
\put(19, 108){$2^{i-1}$}
\end{picture}
\caption{Example of a function $f_i$ lying in the support of $\calD_{n,2}$. One may view the domain as being divided into intervals of length $2^{i}$ (displayed as intervals lying between dotted red lines) and the permutation $f^{\downarrow}$ flipped across adjacent intervals of length $2^{i-1}$ inside a segment of length $2^i$ (displayed as intervals lying between dotted blue lines). Note that all $(12)$-patterns in $f_i$ above have the $i$th bit flipped.}
\label{fig:mon-construct}
\end{figure}
Towards proving \eqref{eq:yao} for the distribution $\calD_{n, 2}$, we introduce the notion of \emph{binary profiles} captured by a set of queries. For any fixed $Q \subseteq [n]$, the binary profiles captured in $Q$ are given by the set
\[
\textsf{bin-prof}(Q) = \{ i \in[\log_2 n] \ :\ \text{there exist $x,y \in Q$ such that } M(x,y)=i\}.
\]
Since all $(12)$-patterns $(x, y)$ of $f_i$ have $M(x, y) = i$, the probability over $\boldf \sim \calD_{n,2}$ that an algorithm whose set of queries is $Q$, finds a $(12)$-pattern in $\boldf$ is at most $|\textsf{bin-prof}(Q)| / \log_2 n$.
We show that for any set $Q \subseteq [n]$, $|\textsf{bin-prof}(Q)| \leq |Q| - 1$. This completes \eqref{eq:yao}, and proves the lower bound of $\frac{9}{10}\log_2 n$ for $k=2$.
The proof that $|\textsf{bin-prof}(Q)| \leq |Q|-1$ for any set $Q \subset [n]$, i.e., the number of \emph{captured profiles} is bounded by the number of queries, follows by induction on $|Q|$.
The base case $|Q| \leq 2$ is trivial. When $|Q| > 2$,
let $i_\text{max} = \max \textsf{bin-prof}(Q)$. Consider the partition of $Q$ into $Q_0$ and $Q_1$, where
\[
Q_0 = \{x \in Q: b_{i_\text{max}}^x = 0\} \qquad \text{ and }\qquad Q_1 = \{ y \in Q : b_{i_{\text{max}}}^y = 1\}.
\]
Since $\textsf{bin-prof}(Q) = \textsf{bin-prof}(Q_0) \cup \textsf{bin-prof}(Q_1) \cup \{i_\text{max}\}$, we conclude that
\[
|\textsf{bin-prof}(Q)| \leq |\textsf{bin-prof}(Q_0)| + |\textsf{bin-prof}(Q_1)| + 1 \leq |Q_0| - 1 + |Q_1| - 1 + 1 = |Q|-1,
\]
where the second inequality follows from the inductive hypothesis.
\medskip\noindent\textsc{Generalization to $k>2$: Proof of Theorem \ref{thm:intro-lb}.}
We now provide a detailed sketch of the proof of Theorem~\ref{thm:intro-lb}. The main objects and notions used are defined, while leaving technical details to Section~\ref{sec:lowerbounds}. Let $k = 2^h$ for $h \in\N$; the case $h=1$ corresponds to the previous discussion.
We first define the distributions $\calD_{n,k}$ supported on permutations $f \colon [n] \to [n]$ which are $\Omega(1/k)$-far from $(12\dots k)$-free. Recall that the function $f_i$ in the case $k=2$ was constructed by ``flipping'' bit $i$ in the representation of $f^{\downarrow}$, that is, $f_i = f^{\downarrow} \circ F_i$. Generalizing this construction, for any $i_1 < i_2 < \dots < i_h \in [\log_2 n]$ we let $f_{i_1, \ldots, i_{h}} \colon [n] \to [n]$ denote the result of flipping bits $i_1, i_2, \ldots, i_h$ in the representation of $f^{\downarrow}$:
\[ f_{i_1, \ldots, i_{h}} \eqdef f^{\downarrow} \circ F_{i_h} \circ \ldots \circ F_{i_1}.
\]
It can be shown that $f_{i_1, \ldots, i_h}$ is $(1/k)$-far from $(12\dots k)$-free (see Figure~\ref{fig:construct-recurse}).
We take $\calD_{n,k}$ as the uniform distribution over all functions of the form $f_{i_1,\ldots, i_{h}}$, where $i_1 < \dots < i_h \in [\log_2 n]$.
\begin{figure}
\centering
\begin{picture}(370, 200)
\put(0,0){\includegraphics[width=0.8\linewidth]{construct-recurse.pdf}}
\put(65, 52){$2^{i_h}$}
\put(19, 90){$2^{i_h-1}$}
\put(195, 150){$f_{i_1, \dots, i_{h-1}}$}
\end{picture}
\caption{Example of a function $f_{i_1,\dots, i_h}$ lying in the support of $\calD_{n,k}$, where $k = 2^h$. Similarly to the case of $k=2$ (shown in Figure~\ref{fig:mon-construct}), the domain is divided into intervals of length $2^{i_h}$ (shown between red dotted lines), and functions are flipped across adjacent intervals of length $2^{i_h-1}$ within an interval of length $2^{i_h}$ (shown between blue dotted lines). Inside each grey region is a recursive application of the construction, $f_{i_1, \dots, i_{h-1}}$, after shifting the range.}
\label{fig:construct-recurse}
\end{figure}
Towards the proof of \eqref{eq:yao} for the distribution $\calD_{n,k}$, we
generalize the notion of a binary profile.
Consider any $k$-tuple of indices $(x_1, \dots, x_k) \in [n]^{k}$ satisfying $x_1 < \dots < x_k$.
We say that $(x_1, x_2, \ldots, x_k)$ has \emph{$h$-profile of type $(i_1, \ldots, i_h)$} if,
\[
\text{for every $j \in [k-1]$}, \qquad M(x_{j}, x_{j+1}) = i_{M(j-1, j)}.
\]
For instance, when $h=3$ (i.e., $k=8$) the tuple $(x_1, \dots, x_k)$ has $h$-profile of type $(i_1, i_2, i_3)$ if the sequence $(M(x_j, x_{j+1}))_{j = 1}^{7}$ is $(i_1, i_2, i_1, i_3, i_1, i_2, i_1)$. See Figure~\ref{fig:profiles} for a visual demonstration of a $3$-profile.\footnote{Unlike the case $k=2$, not all tuples $(x_1, \ldots, x_k)$ with $x_1 < \ldots < x_k$ have an $h$-profile. For what follows we will only be interested in tuples that \emph{do} have a profile.}
\begin{figure}
\centering
\begin{picture}(600, 200)
\put(0,0){\includegraphics[width=\linewidth]{profiles.pdf}}
\put(120, 95){$2^{i_3}$}
\put(405, 42){$2^{i_2}$}
\put(415, 70){$2^{i_1}$}
\put(240, 155){$x_1$}
\put(258, 155){$x_2$}
\put(295, 155){$x_3$}
\put(312, 155){$x_4$}
\put(375, 155){$x_5$}
\put(395, 155){$x_6$}
\put(430, 155){$x_7$}
\put(448, 155){$x_8$}
\end{picture}
\caption{Example of a function $f_{i_1, i_2, i_3}$ in the support of $\calD_{n, 8}$, with a $k$-tuple $(x_1, \dots, x_8)$ whose $h$-profile has type $(i_1, i_2, i_3)$.}
\label{fig:profiles}
\end{figure}
It can be shown that for any $i_1 < \dots < i_h \in [\log_2 n]$, the function $f = f_{i_1, \ldots, i_{h}}$ has the following property. If $x_1 < \dots < x_k \in [n]$ satisfy $f(x_1) < \dots < f(x_k)$, i.e., the $k$-tuple $(x_1, \dots, x_k)$ is a $(12\dots k)$-pattern of $f_{i_1, \dots, i_k}$, then $(x_1, \dots, x_k)$ has an $h$-profile of type $(i_1, \dots, i_h)$.
We thus proceed similarly to the case $k=2$. For any $Q \subseteq [n]$, we define the set of all $h$-profiles captured by $Q$ as follows
\[
\textsf{bin-prof}_h(Q) = \left\{(i_1, \ldots, i_{h}) :\ \begin{array}{l} \text{there exist $x_1, \dots,x_k \in Q$ where } x_1 < \dots < x_k \\
\text{and $(x_1, \dots, x_k)$ has $h$-profile of type $(i_1, \ldots, i_{h})$} \end{array} \right\}.
\]
The proof that $|\textsf{bin-prof}_h(Q)| \leq |Q| - 1$ for any $Q \subseteq [n]$ follows by induction on $h$. The base case $h=1$ was covered in the discussion on $k=2$. For $h > 1$, we define subsets
\[
\emptyset = B_{\log_2 n+1} \subseteq B_{\log_2 n} \subseteq \ldots \subseteq B_{1} = Q,
\]
where, given $B_{i+1}$, the set $B_i \supseteq B_{i+1}$ is an arbitrary maximal subset of $Q$ containing $B_{i+1}$, so that no two elements $x \neq y \in B_i$ satisfy $M(x,y) < i$. Additionally, for each $j \in [\log_2 n]$ we let
\[
N_j = \left\{ (i_2, \dots, i_h) : 1 \leq j < i_2 \dots < i_h \leq \log_2 n \text{ and } (j, i_2, \dots, i_h) \in \textsf{bin-prof}_h(Q) \right\}.
\]
The key observation is that $N_j \subseteq \textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})$. To see this, note first that any $(j, i_2, \ldots, i_{h}) \in \textsf{bin-prof}_h(Q)$ also satisfies $(j, i_2, \ldots, i_{h}) \in \textsf{bin-prof}_h(B_j)$. Indeed, suppose that a tuple $(x_1, \ldots, x_k)$ with $x_1 < \dots < x_k \in Q$ has $h$-profile $(j, i_2, \ldots, i_{h})$. By the maximality of $B_j$, we know that for every $1 \leq \ell \leq k$ there exists $y_\ell \in B_j$ such that either $x_{\ell} = y_{\ell}$ or $M(x_{\ell}, y_{\ell}) < j$. This implies that $\{y_1, \ldots, y_{k}\} \subseteq B_j$ has $h$-profile $(j, i_2, \ldots, i_{h})$.
Now, suppose that $y_1, \dots, y_k \in B_j$ satisfies $y_1< \ldots< y_{k}$ and has $h$-profile of type $(j, i_2, \ldots, i_{h})$ in $B_j$. For any $1 \leq t \leq k/2$ we have $M(y_{2t-1}, y_{2t}) = j$. Therefore, at most one of $y_{2t-1}, y_{2t}$ is in $B_{j+1}$, and hence, for any such $t$ there exists $z_t \in \{y_{2t-1}, y_{2t}\} \setminus B_{j+1} \subseteq B_{j} \setminus B_{j+1}$. It follows that $(z_1, \ldots, z_{k/2}) \in B_j \setminus B_{j+1}$ has $(h-1)$-profile $(i_2, \ldots, i_{h})$. This concludes the proof that $N_j \subseteq \textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})$.
We now use the last observation to prove that $|\textsf{bin-prof}_h(Q)| \leq |Q|-1$. Note that
$$
\textsf{bin-prof}_h(Q) = \bigcup_{j=1}^{\log_2 n} \{(j, i_2, \ldots, i_h) : (i_2, \ldots, i_h) \in N_j \}
\qquad \text{and} \qquad
Q = \bigcup_{j=1}^{\log_2 n} (B_j \setminus B_{j+1}),
$$
where both unions are disjoint unions. By the induction assumption, $|N_j| \leq |\textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})| < |B_j \setminus B_{j+1}|$ for any $j$ if $N_j$ is non-empty; If $N_j$ is empty, then $|N_j| \leq |\textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})| \le |B_j \setminus B_{j+1}|$ trivially holds. Hence
\[
|Q| = \sum_{j=1}^{\log_2 n} |B_j \setminus B_{j+1}| >\sum_{j=1}^{\log_2 n} |N_j| = |\textsf{bin-prof}_h(Q)|,
\]
where the strict inequality follows because if $\textsf{bin-prof}_h(Q)$ is non-empty then $N_j$ is non-empty for some $j$. This completes the proof.
\subsection{Organization}
We start by introducing the notation that we shall use throughout the paper in Section~\ref{sec:preliminaries}. In Section~\ref{sec:structural} we prove our main structural result, and formally define the notions that underlie it: namely, Theorem~\ref{thm:tree}, along with the definitions of growing suffixes and representation by tree descriptors (Definitions~\ref{def:growing-suffixes} and~\ref{def:tree-rep}). Section~\ref{sec:algorithm} then leverages this dichotomy to describe and analyze our testing algorithm, thus establishing the upper bound of Theorem~\ref{thm:intro-ub} (see Theorem~\ref{thm:ub} for a formal statement). Finally, we complement this algorithm with a matching lower bound in Section~\ref{sec:lowerbounds}, where we prove Theorem~\ref{thm:intro-lb}.
While Section~\ref{sec:algorithm} crucially relies on Section~\ref{sec:structural}, these two sections are independent of~Section~\ref{sec:lowerbounds}, which is mostly self-contained.
\subsection{Notation and Preliminaries}\label{sec:preliminaries}
We write $a \lsim b$ if there exists a universal positive constant $C > 0$ such that $a \leq C b$, and $a \asymp b$ if we have both $a \lsim b$ and $b \lsim a$. At times, we write $\poly(k)$ to stand for $O(k^{C})$, where $C > 0$ is a large enough universal constant. Unless otherwise stated, all logarithms will be in base 2. We frequently denote $\calI$ as a collection of disjoint intervals, $I_1, \dots, I_s$, and then write $\calS(\calI)$ for the set of all sub-intervals which lie within some interval in $\calI$. For two collections of disjoint intervals $\calI_0$ and $\calI_1$, we say that $\calI_1$ is a \emph{refinement} of $\calI_0$ if every interval in $\calI_1$ is contained within an interval in $\calI_0$. (We remark that it is not the case that intervals in $\calI_1$ must form a partition of intervals in $\calI_0$.) For a particular set $A \subseteq [n]$ and an interval $I \subseteq [n]$, we define the \emph{density} of $A$ in $I$ as the ratio $|A \cap I| / |I|$. Given a set $S$, we write $\bx \sim S$ to indicate that $\bx$ is a random variable given by a sample drawn uniformly at random from $S$, and $\calP(S)$ for the power set of $S$. Given a sequence $f$ of length $n$, we shall interchangeably use the notions \emph{$(12\dots k)$-copy}, \emph{$(12 \dots k)$-pattern}, and \emph{length-$k$ increasing subsequence}, to refer to a tuple $(x_1, \ldots, x_k) \in [n]^k$ such that $x_1 < \ldots < x_k$ and $f(x_1) < \ldots < f(x_k)$.
\section{Structural Result}\label{sec:structural}
\subsection{Rematching procedure}
Let $f \colon [n] \to \R$ be a function which is $\eps$-far from $(12\dots k)$-free.
Let $T$ be a set of $k$-tuples representing monotone subsequences of length $k$ within $f$, i.e.,
\[
T \subseteq \left\{ (i_1, \dots, i_k) \in [n]^k : i_1 < \dots < i_k \text{ and } f(i_1) < \dots < f(i_k) \right\},
\]
and for such $T$ let $E(T)$ be the set of indices of subsequences in $T$, so
\[
E(T) = \bigcup_{(i_1, \ldots, i_k) \in T} \{i_1, \ldots, i_k\}.
\]
\begin{observation} \label{obs:disjoint-families}
If $f \colon [n] \to \R$ is $\eps$-far from $(12 \dots k)$-free, then there exists a set $T \subseteq [n]^k$ of disjoint length-$k$ increasing subsequences of $f$ such that $|T| \ge \eps n / k$.
\end{observation}
To see why the observation holds, take $T$ to be a maximal disjoint set of such $k$-tuples. Then we can obtain a $(12 \dots k)$-free sequence from $f$ by changing only the entries of $E(T)$ (e.g.\ for every $i \in E(T)$ define $f(i) = f(j)$ where $j$ is the largest $[n] \setminus E(T)$ which is smaller than $i$. If there is no $j \in [n] \setminus E(T)$ where $j < i$, let $f(i) = \max_{\ell \in [n]} f(\ell)$). Since $f$ is $\eps$-far from being $(1 2 \dots k)$-free, we have $|E(T)| \ge \eps n$, thus $|T| \ge \eps n / k$.
In this section, we show that from a function $f \colon [n] \to \R$ which is $\eps$-far from $(12\dots k)$-free and a set $T_0$ of disjoint, length-$k$ monotone subsequences of $f$, a greedy rematching algorithm finds a set $T$ of disjoint, length-$k$ monotone subsequences of $f$ where $E(T) \subseteq E(T_0)$ with some additional structure, which will later be exploited in the structural lemma and the algorithm. The greedy rematching algorithm, $\texttt{GreedyDisjointTuples}$, is specified in Figure~\ref{fig:greedy}; \red{for convenience, in view of its later use in the algorithm, we phrase it in terms of an arbitrary parameter $k_0$, not necessarily the (fixed) parameter $k$ itself.}
\begin{lemma}\label{lem:rematching}
Let $k_0 \in \N$, $f \colon [n] \to \R$, and let $T_0 \subseteq [n]^{k_0}$ be a set of disjoint monotone subsequences of $f$ of length $k_0$. Then there exists a set $T \subseteq [n]^{k_0}$ of disjoint $k_0$-tuples with $E(T) \subseteq E(T_0)$ such that the following holds.
\begin{enumerate}
\item\label{en:cond-1}
The set $T$ holds disjoint monotone subsequences of length $k_0$.
\item\label{en:cond-2}
The size of $T$ satisfies $|T| \geq |T_0| / k_0$.
\item\label{en:cond-3}
For any two $(i_1, \dots, i_{k_0}), (j_1, \dots, j_{k_0}) \in T$ and any $\ell \in [k_0 - 1]$, if $i_1 < j_1$, $i_{\ell} < j_{\ell}$ and $i_{\ell+1} > j_{\ell+1}$ then $f(i_{\ell+1}) > f(j_{\ell+1})$.
\end{enumerate}
\end{lemma}
\begin{figure}[ht!]
\begin{framed}
\centering
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{GreedyDisjointTuples}\hspace{0.05cm}(f, k_0, T_0)$
\vspace{0.3cm}
\noindent {\bf Input:} A function $f \colon [n] \to \R$, integer $k_0\in\N$, and a set $T_0$ of disjoint monotone subsequences of $f$ of length $k_0$.\\
{\bf Output:} a set $T \subseteq [n]^{k_0}$ of disjoint monotone subsequences of $f$ of length $k_0$.
\begin{enumerate}
\item Let $T = \emptyset$ and $i$ be the minimum element in $E(T_0)$. Repeat the following.
\begin{itemize}
\item[i.]
Let $i_1 \gets i$. If there exists $j_2, \ldots, j_{k_0} \in E(T_0) \setminus E(T)$ such that $(i_1, j_2, \dots, j_{k_0})$ is an increasing subsequence of $f$, pick $i_2, \ldots, i_{k_0} \in E(T_0) \setminus E(T)$ recursively as follows: for $\ell = 2, \ldots, k_0$, let $i_{\ell}$ be the smallest element in $E(T_0) \setminus E(T)$ for which there exist $j_{\ell+1}, \dots, j_{k_0} \in E(T_0) \setminus E(T)$ such that $(i_1, \dots, i_{\ell}, j_{\ell+1}, \dots, j_{k_0})$ is an increasing subsequence of $f$.
\item[ii.]
If $(i_1, \dots, i_{k_0})$ is a monotone subsequence found by (i), set $T \gets T \cup \{ (i_1, \dots, i_{k_0})\}$.
\item[iii.]
Let $i$ be the next element of $E(T_0)\setminus E(T)$, if such an element exists; otherwise, proceed to \ref{itm:alg-rematch-end}.
\end{itemize}
\item \label{itm:alg-rematch-end}
Output $T$.
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{GreedyDisjointTuples}$ subroutine.} \label{fig:greedy}
\end{figure}
\begin{proofof}{Lemma~\ref{lem:rematching}}
We show that the subroutine $\texttt{GreedyDisjointTuples}(f, k_0, T_0)$, described in Figure~\ref{fig:greedy}, finds a set $T$ with $E(T) \subseteq E(T_0)$ satisfying properties \ref{en:cond-1}, \ref{en:cond-2}, and \ref{en:cond-3}. Property \ref{en:cond-1} is clear from the description of $\texttt{GreedyDisjointTuples}(f,k_0, T_0)$. For \ref{en:cond-2}, suppose $|T| < |T_0|/k_0$, then, there exists a tuple $(i_1, \dots, i_{k_0}) \in T_0$ with $\{ i_1, \dots, i_{k_0}\} \cap E(T) = \emptyset$. Since $\texttt{GreedyDisjointTuples}(f, k_0, T_0)$ increases the size of $T$ throughout the execution, $\{ i_1, \dots, i_{k_0} \} \cap T = \emptyset$ at every point in the execution of the algorithm. This is a contradiction; when $i = i_1$, a monotone subsequence disjoint from $T$ would have been found, and $i_1$ included in $T$. Finally, for \ref{en:cond-3}, consider the iteration when $i = i_1$, and note that at this moment, $T \cap \{ i_1, \dots, i_{k_0}, j_1, \dots, j_{k_0}\} = \emptyset$. Suppose that $i_{\ell} < j_{\ell}$, $j_{\ell+1} < i_{\ell+1}$; if $f(j_{\ell+1}) \geq f(i_{\ell+1})$, then $(i_1, \dots, i_{\ell}, j_{\ell+1}, \dots, j_{k_0})$ is an increasing subsequence in $E(T_0) \setminus E(T)$, which means that $j_{\ell+1}$ would have been preferred over $i_{\ell+1}$, a contradiction.
\end{proofof}
\begin{definition}[$c$-gap]
Let $(i_1, \dots, i_{k_0})$ be a monotone subsequence of $f$ and let $c \in [k_0-1]$. We say that $(i_1, \dots, i_{k_0})$ is a \emph{$c$-gap subsequence} if $c$ is the smallest integer such that $i_{c+1} - i_c \geq i_{b+1} - i_b$ for all $b \in [{k_0}-1]$.
\end{definition}
Note that for a set $T$ of disjoint length-$k_0$ monotone subsequences of $f$, we may partition the $k_0$-tuples of $T$ into $(T_1, \dots, T_{k_0-1})$ where for each $c \in [k_0-1]$, $T_c$ holds the $c$-gap monotone subsequences of $T$. As these sets form a partition of $T$, the following lemma is immediate from Lemma~\ref{lem:rematching}.
\begin{lemma}\label{lem:big-split}
Let $f \colon [n] \to \R$, and let $T_0$ be a set of disjoint length-$k_0$ monotone subsequences of $f$. Then there exist $c \in [k_0-1]$ and a family $T \subseteq [n]^{k_0}$ of disjoint monotone subsequences of $f$, with $E(T) \subseteq E(T_0)$ such that the following holds.
\begin{enumerate}
\item
The subsequences in $T$ are all $c$-gap subsequences.
\item
$|T| \ge |T_0| / k_0^2$.
\item\label{itm:property-greedy}
For any two $(i_1, \dots, i_{k_0}), (j_1, \dots, j_{k_0}) \in T$ and any $\ell \in [k_0 - 1]$, if $i_1 < j_1$, $i_{\ell} < j_{\ell}$ and $i_{\ell+1} > j_{\ell+1}$ then $f(i_{\ell+1}) > f(j_{\ell+1})$.
\end{enumerate}
\end{lemma}
\subsection{Growing suffixes and splittable intervals}
We now proceed to set up notation and prepare for the main structural theorem for sequences $f \colon [n] \to \R$ which are $\eps$-far from $(12\dots k)$-free. In order to simplify the presentation of the subsequent discussion, consider fixed $k \in \N$ and $\eps \in (0, 1)$, as well as a fixed sequence $f \colon [n] \to \R$ which is $\eps$-far from $(12\dots k)$-free. We will, at times, suppress polynomial factors in $k$ by writing $\poly(k)$ to refer to a large enough polynomial in $k$, whose degree is a large enough universal constant.
By Observation~\ref{obs:disjoint-families} and Lemma~\ref{lem:big-split}, there exists an integer $c \in [k-1]$ and a set $T$ of disjoint monotone subsequences of $f$ which have a $c$-gap, satisfying $|T| \geq \eps n / \poly(k)$ and property \ref{itm:property-greedy} from Lemma~\ref{lem:big-split}. For the rest of the subsection, we consider a fixed setting of such $c \in [k-1]$ and set $T$.
We will show (in Theorem~\ref{thm:main-structure}) that one of the following two possibilities holds.
Either there is a large set of what we call \emph{growing suffixes} (see Definition~\ref{def:growing-suffixes} for a formal definition), or there are disjoint intervals which we call \emph{splittable} (see Definition~\ref{def:splittable} for a formal definition). Intuitively, a growing suffix will be given by the suffix $(a, n]$ and will have the property that by dividing $(a, n]$ into $\Theta(\log_2(n - a))$ segments of geometrically increasing lengths, there are many monotone subsequences $(i_1, \dots, i_k)$ of $f$ lying inside $(a, n]$ where each $i_t$ belongs to a different segment. In the other case, an interval $[a,b]$ is called splittable if it can be divided into three sub-intervals of roughly equal size,
which we refer to as the left, middle, and right intervals, with the following property: the left interval contains a large set $T_{L}$ of $(12\dots c)$-patterns, the right interval contains a large set $T_R$ of $(12\dots (k-c))$-patterns, and combining any $(12\dots c)$-pattern in $T_L$ with any $(12\dots (k-c))$-pattern in $T_R$ yields a $(12\dots k)$-pattern.
For each index $a \in [n]$, let $\eta_a = \lceil \log_2(n-a) \rceil$. Let $S_1(a), \dots, S_{\eta_a}(a) \subseteq [n]$ be disjoint intervals given by $S_t(a) = [a + 2^{t-1}, a + 2^t) \cap [n]$. The collection of intervals $S(a) = (S_t(a) : t \in [\eta_a])$ partitions the suffix $(a, n]$ into intervals of geometrically increasing lengths (except possibly the last interval, which may be shorter), and we refer to the collection $S(a)$ as the \emph{growing suffix} at $a$.
\begin{definition}\label{def:growing-suffixes}
Let $\alpha, \beta \in [0,1]$.
We say that an index $a \in [n]$ starts an \emph{$(\alpha, \beta)$-growing suffix} if, when considering the collection of intervals $S(a) = \{ S_t(a) : t \in [\eta_a]\}$, for each $t \in [\eta_a]$ there is a subset $D_t(a) \subseteq S_t(a)$ of indices
such that the following properties hold.
\begin{enumerate}
\item
We have $|D_t(a)|/|S_t(a)| \le \alpha$ for all $t \in [\eta_a]$, and
$\sum_{t=1}^{\eta_a} |D_t(a)|/|S_t(a)| \geq \beta$.
\label{en:grow-cond-3}
\item
For every $t, t' \in [\eta_a]$ where $t < t'$, if $b \in D_t(a)$ and $b' \in D_{t'}(a)$, then $f(b) < f(b')$.
\label{en:grow-cond-2}
\end{enumerate}
\end{definition}
Intuitively, our parameter regime will correspond to the case when $\alpha$ is much smaller than $\beta$, specifically, $\alpha \leq \beta / \poly(k)$, for a sufficiently large-degree polynomial in $k$. If $a \in [n]$ starts an $(\alpha, \beta)$-growing suffix with these parameters, then the $\eta_a$ segments, $S_1(a), \dots, S_{\eta_a}(a)$, contain many monotone subsequences of length $k$ which are algorithmically easy to find (given access to the start $a$).
Indeed, by (\ref{en:grow-cond-2}), it suffices to find a $k$-tuple $(i_1, \ldots, i_k)$ such that $i_1 \in D_{t_1}, \ldots, i_k \in D_{t_k}$, for some $t_1, \ldots, t_k \in [\eta_a]$ with $t_1 < \ldots < t_k$ (see Figure~\ref{fig:growing-suffix}).
By (\ref{en:grow-cond-3}), the sum of densities is at least $\beta$, yet each density is less than $\alpha \le \beta / \poly(k)$. In other words, the densities of $D_{1}(a), \dots, D_{\eta_a}(a)$ within $S_{1}(a), \dots, S_{\eta_a}(a)$, respectively, must be spread out, which implies, intuitively, that there are many ways to pick suitable $i_1, \ldots, i_k$.
\begin{figure}
\begin{picture}(300, 150)
\put(0,0){\includegraphics[width=\linewidth]{growing-suffix.pdf}}
\put(162, 5){$a$}
\end{picture}
\caption{Depiction of an $(\alpha, \beta)$-growing suffix at index $a \in [n]$ (see Definition~\ref{def:growing-suffixes}). The labeled segments $S_t(a)$ are shown, as well as the subsets $D_t(a)$. Notice that for all $j$, all the elements in $D_t(a)$ lie below those in $D_{t+1}(a)$. In Section~\ref{sec:case1}, we show that if an algorithm knows that $a$ starts an $(\alpha, \beta)$-growing suffix, for $\alpha \leq \beta / \poly(k)$, then sampling $\poly(k) / \beta$ many random indices from each $S_t(a)$ finds a monotone pattern with probability at least $0.9$.}
\label{fig:growing-suffix}
\end{figure}
\begin{definition}\label{def:splittable}
Let $\alpha,\beta \in (0,1]$ and $c \in [k_0-1]$. Let $I \subseteq [n]$ be an interval, let $T \subseteq I^{k_0}$ be a set of disjoint, length-$k_0$ monotone subsequences of $f$ lying in $I$, and define
\begin{align*}
T^{(L)} &= \{ (i_1, \dots, i_c) \in I^c : (i_1, \dots, i_c) \text{ is a prefix of a $k_0$-tuple in $T$}\}, \text{ and }\\
T^{(R)} &= \{ (j_1, \dots, j_{k_0-c}) \in I^{k_0-c} : (j_1, \dots, j_{k_0-c}) \text{ is a suffix of a $k_0$-tuple in $T$}\}.
\end{align*}
We say that the pair $(I, T)$ is \emph{$(c, \alpha,\beta)$-splittable} if $|T|/|I| \geq \beta$; $f(i_c) < f(j_1)$ for every $(i_1, \dots, i_c) \in T^{(L)}$ and $(j_1, \dots, j_{k_0-c}) \in T^{(R)}$; and there is a partition of $I$ into three adjacent intervals $L, M, R \subseteq I$ (that appear in this order, from left to right) of size at least $\alpha |I|$, satisfying $T^{(L)} \subseteq L^c$ and $T^{(R)} \subseteq R^{k_0-c}$.
A collection of disjoint interval-tuple pairs $(I_1, T_1), \dots, (I_s, T_s)$ is called a \emph{$(c, \alpha,\beta)$-splittable collection of $T$} if each $(I_j, T_j)$ is $(c, \alpha, \beta)$-splittable and the sets $(T_j : j \in [s])$ partition $T$.
\end{definition}
\begin{figure}
\begin{picture}(300, 120)
\put(0,0){\includegraphics[width=\linewidth]{splittable.pdf}}
\put(235, 10){$I$}
\put(180, 34){$L$}
\put(285, 34){$R$}
\put(235, 34){$M$}
\put(180, 60){$T^{(L)}$}
\put(285, 105){$T^{(R)}$}
\end{picture}
\caption{Depiction of a $(c, \alpha,\beta)$-splittable interval, as defined in Definition~\ref{def:splittable}. The interval $I$ is divided into three adjacent intervals, $L, M$, and $R$, and the disjoint monotone sequences are divided so that $T^{(L)}$ contains the indices $(i_1, \dots, i_c)$ and $T^{(R)}$ contains the indices $(i_{c+1}, \dots, i_k)$. Furthermore, we have that every $(i_1, \dots, i_c) \in T^{(L)}$ and $(j_{c+1}, \dots, j_k) \in T^{(R)}$ have $f(i_c) < f(j_{c+1})$, so that any monotone pattern of length $c$ in $E(T^{(L)})$ may be combined with any monotone pattern of length $k-c$ in $E(T^{(R)})$ to obtain a monotone pattern of length $k$ within $I$.}
\label{fig:splittable}
\end{figure}
We now state the main theorem of this section, whose proof will be given in Section~\ref{sec:proof1}.
\begin{theorem}\label{thm:main-structure}
Let $k, k_0 \in \N$ be positive integers satisfying $1\leq k_0\leq k$, and let $\delta\in (0,1)$ and let $C > 0$. Let $f \colon [n] \to \R$ be a function and let $T_0 \subseteq [n]^{k_0}$ be a set of $\delta n$ disjoint monotone subsequences of $f$ of length $k_0$.
Then there exists an $\alpha \geq \Omega(\delta/k^5)$ such that at least one of the following conditions holds.
\begin{enumerate}
\item \label{en:suffix}
Either there exists a set $H \subseteq [n]$, of indices that start an $(\alpha, C k \alpha)$-growing suffix, satisfying $\alpha |H| \geq \delta n / \poly(k, \log(1/\delta))$; or
\item \label{en:split}
There exists an integer $c$ with $1 \le c < k_0$, a set $T$, with $E(T) \subseteq E(T_0)$, of disjoint length-$k_0$ monotone subsequences, and a $(c, 1/(6k),\alpha)$-splittable collection of $T$, of disjoint interval-tuple pairs $(I_1, T_1), \dots, (I_s, T_s)$, such that
\[
\alpha \sum_{h=1}^s |I_h| \geq \frac{|T_0|}{\poly(k, \log(1/\delta))}.
\]
\end{enumerate}
\end{theorem}
We remark that the above theorem is stated with respect to the two parameters, $k_0$ and $k$, for ease of applicability. In particular, in the next section, we will apply Theorem~\ref{thm:main-structure} multiple times, and it will be convenient to have $k$ be fixed and $k_0$ be a varying parameter. In that sense, even though the monotone subsequences in question have length $k_0$, the relevant parameters that Theorem~\ref{thm:main-structure} lower bounds only depend on $k$.
Consider the following scenario: $f \colon [n] \to \R$ is a sequence which is $\eps$-far from $(12\dots k)$-free, so by Observation~\ref{obs:disjoint-families}, there exists a set $T_0$ of disjoint, length-$k$ monotone subsequences of $f$ of size at least $\eps n / k$. Suppose that upon applying Theorem~\ref{thm:main-structure} with $k_0 = k$ and $\delta = \eps / k$, (\ref{en:split}) holds. Then, there exists a $(c, 1/(6k),\alpha)$-splittable collection of a large subset of disjoint, length-$k$ monotone subsequences $T$ into disjoint interval-tuple pairs $(I_1, T_1), \dots, (I_s, T_s)$. For each $h \in [s]$, the pair $(I_h, T_h)$ is $(c, 1/(6k),\alpha)$-splittable, so let $I_h = L_h \cup M_h \cup R_h$ be the left, middle, and right intervals of $I_h$; furthermore, let $T_{h}^{(L)}$ be the $(12\dots c)$-patterns in $L_h$ which appear as prefixes of $T_h$, and $T_h^{(R)}$ be the $(12\dots (k-c))$-patterns in $R_h$ which appear as suffixes of $T_h$ in $R_h$. Thus, the restricted function $f_{|L_h} \colon L_h \to \R$ contains $|T_h|$ disjoint $(12\dots c)$-patterns, and $f_{|R_h} \colon R_h \to \R$ contains $|T_h|$ disjoint $(12\dots (k-c))$-patterns. This naturally leads to a recursive application of Theorem~\ref{thm:main-structure} to the function $f_{|L_h}$ with $k_0 = c$, and to the function $f_{|R_h}$ with $k_0 = k-c$, for all $h \in [s]$.
\subsection{Tree descriptors}\label{ssec:tree:descriptors}
We now introduce the notion of \emph{tree descriptors}, which will summarize information about a function $f$ after applying Theorem~\ref{thm:main-structure} recursively. Then, we state the main structural result for functions that are $\eps$-far from $(12\dots k)$-free. The goal is to say that every function which is $\eps$-far from $(12\dots k)$-free either has many growing suffixes, or there exists a tree descriptor which describes the behavior of many disjoint, length-$k$ monotone subsequences in the function. The following two definitions make up the notion of a tree descriptor representing a function. Figure~\ref{fig:tree-descriptor} shows an example of Definitions~\ref{def:tree-descriptor} and~\ref{def:tree-rep}.
\begin{definition}\label{def:tree-descriptor}
Let $k_0 \in \N$ and $\delta \in (0,1)$.
A \emph{$(k_0, \delta)$-weighted-tree} is a pair $(G, \varrho)$, where
\begin{itemize}
\item
$G = (V, E, w)$ is a rooted binary tree with edges labeled by a function $w \colon E \to \{0,1\}$. Every non-leaf node has two outgoing edges, $e_0, e_1$ with $w(e_0) = 0$ and $w(e_1) = 1$. The set of leaves $V_{\ell} \subseteq V$ satisfies $|V_{\ell}| = k_0$, and $\leq_{G}$ is the total order defined on the leaves by the values of $w$ on a root-to-leaf path.\footnote{Specifically, for $l_1, l_2 \in V_{\ell}$ at depths $d_1$ and $d_2$, with root to leaf paths $(r, u^{(1)}, \dots, u^{(d_1-1)}, l_1)$ and $(r, v^{(1)}, \dots, v^{(d_2-1)}, l_2)$, then $l_1 \leq_{G} l_2$ if and only if $(w(r, u^{(1)}), w(u^{(1)}, u^{(2)}), \dots, w(u^{(d_1-1)}, l_1)) \leq (w(r, v^{(1)}), w(v^{(1)}, v^{(2)}), \dots, w(v^{(d_2-1)}, l_2))$ in the natural partial order on $\{0,1\}^*$.}
\item
$\varrho \colon V \to [\lceil \log(1 / \delta) \rceil]$ is a function that assigns a positive integer to each node of $G$.
\end{itemize}
\end{definition}
In the next definition, we show how we use weighted trees to represent a function $f$ and a set of disjoint, length-$k_0$ monotone subsequences.
\begin{definition}\label{def:tree-rep}
Let $k, k_0 \in \N$ be such that $1 \leq k_0 \leq k$, let $\alpha \in (0,1)$, let $I \subseteq \N$ be an interval, and let $f \colon I \to \R$ be a function. Let $T \subseteq I^{k_0}$ be a set of disjoint monotone subsequences of $f$.
A triple $(G, \varrho, \mathsf{I})$ is called a $(k, k_0, \delta)$-\emph{tree descriptor}\footnote{We shall sometimes refer to this as a $k_0$-tree descriptor, in particular when $k, \delta$ are not crucial to the discussion.} of $(f,T,I)$, if $(G, \varrho)$ is a $(k_0, \delta)$-weighted tree, $\mathsf{I}$ is a function $\mathsf{I} \colon V \rightarrow \calP(\calI)$ (where $V = V(G)$), and the following recursive definition holds.
\begin{enumerate}
\item If $k_0 = 1$ (so $T \subseteq I$),
\begin{itemize}
\item
The graph $G= (V, E, w)$ is the rooted tree with one node, $r$, and no edges.
\item
The function $\varrho \colon V \to [\lceil \log(1 / \delta) \rceil]$ (simply mapping one node) satisfies $ 2^{-\varrho(r)} \le |T|/|I| \le 2^{-\varrho(r) + 1}$.
\item
The map $\mathsf{I} \colon V \to \calS(I)$ is given by $\mathsf{I}(r) = \{ \{ t \} : t \in T\}$.
\end{itemize}
\item If $k_0 > 1$,
\begin{itemize}
\item
The graph $G = (V, E, w)$ is a rooted binary tree with $k_0$ leaves. We refer to the root by $r$, the left child of the root (namely, the child incident with the edge given $0$ by $w$) by $v_{\sf left}$, and the right child of the root (the child incident with the edge given $1$) by $v_{\sf right}$. Let $c$ be the number of leaves in the subtree of $v_{\sf left}$, so $v_{\sf right}$ has $k_0 - c$ leaves in its subtree.
\item
Write $\mathsf{I}(r) = \{I_1, \dots, I_s\}$. Then $I_1, \dots, I_s$ are disjoint sub-intervals of $I$, and, setting $T_i = (I_i)^{k_0} \cap T$, the pairs $(I_1, T_1), \dots, (I_s, T_s)$ form a $(c, 1/(6k), 2^{-\varrho(r)})$-splittable collection of $T$, and
\[
2^{-\varrho(r)} \sum_{h = 1}^s |I_h| \ge \frac{|T|}{\poly(k, \log(1/\delta))^k}.
\]
\item
For each $h\in [s]$ there exists a partition $(L_h, M_h, R_h)$ of $I_h$ that satisfies Definition~\ref{def:splittable}, such that the sets $T_h^{(L)}$, of prefixes of length $c$ of subsequences in $T_h$, and $T_h^{(R)}$, of suffixes of length $k_0 - c$ of subsequences in $T_h$, satisfy $T_h^{(L)} \subseteq (L_h)^c$ and $T_h^{(R)} \subseteq (R_h)^{k_0 - c}$. Moreover, the following holds.
The tuple $(G_{\sf left}, \varrho_{\sf left}, \mathsf{I}_{h, \, \sf left})$ is a $(k, c, \delta)$-tree descriptor of $f$, $T_h^{(L)}$, and $L_h$, where $G_{\sf left}$ is the subtree rooted at $v_{\sf left}$, $\varrho_{\sf left}$ is the restriction of $\varrho$ to the subtree $G_{\sf left}$, and $\mathsf{I}_{h,\, \sf left}$ is defined by $\mathsf{I}_{h,\, \sf left}(v) \eqdef \{J \in \mathsf{I}(v) \colon J \subseteq L_h\}$ for all $v \in G_{\sf left}$.
Analogously, the tuple $(G_{\sf right}, \varrho_{\sf right}, \mathsf{I}_{h,\, \sf right})$ is a $(k,k_0 - c, \delta)$-tree descriptor of $f$, $T_h^{(R)}$, and $R_h$, where $G_{\sf right}, \varrho_{\sf right}, \mathsf{I}_{h,\, \sf right}$ are defined analogously.
\end{itemize}
\end{enumerate}
\end{definition}
We remark that it is \emph{not} the case that for every function $f \colon I \to \R$ defined on an interval $I$, and for every $T \subseteq I^{k_0}$ which is a set of disjoint, length-$k_0$ monotone subsequences of $f$, there must exist a $k_0$-tree descriptor which represents $(f, T,I)$.
The goal will be to apply Theorem~\ref{thm:main-structure} recursively whenever we are in (\ref{en:split}), and to find a sufficiently large set $T$ of disjoint length-$k$ monotone subsequences, as well as a $k$-tree descriptor which represents $(f, T,I)$.
\begin{figure}
\begin{picture}(300, 280)
\put(0,0){\includegraphics[width=\linewidth]{tree-descriptor.pdf}}
\put(154, 187){$1$}
\put(209, 187){$2$}
\put(262, 187){$3$}
\put(317, 187){$4$}
\put(236, 282){\textcolor{white}{$r$}}
\put(180, 235){$v_0$}
\put(288, 235){\textcolor{white}{$v_1$}}
\put(60, 39){$i_1$}
\put(78, 39){$j_2$}
\put(138, 39){$l_3$}
\put(168, 39){$h_4$}
\end{picture}
\caption{Depiction of a tree descriptor $(G, \varrho, \mathsf{I})$ representing $(f, T, I)$, as defined in Definitions~\ref{def:tree-descriptor} and~\ref{def:tree-rep}. The graph $G$ displayed above is a rooted tree with four leaves, which are ordered and labeled left-to-right. The root node $r$, filled in black, has its corresponding intervals from $\mathsf{I}(r)$ shown below the sequence as three black intervals. Each of the black intervals in $\mathsf{I}(r)$ is a $(2, \alpha,\beta)$-splittable interval, for $\alpha \approx 1/3$ and $\beta \geq 1/6$. Then, the root has the left child $v_0$, filled in red, and the right child $v_1$, filled in blue. The red intervals are those belonging to $\mathsf{I}(v_0)$, and the blue intervals are those belonging to $\mathsf{I}(v_1)$. Each black interval in $\mathsf{I}(r)$ has a left part, which contains intervals in $\mathsf{I}(v_0)$, and a right part, which contains intervals in $\mathsf{I}(v_1)$. The red and blue intervals in $\mathsf{I}(v_0)$ and $\mathsf{I}(v_1)$ are also $(1, \alpha,\beta)$-splittable, and the left part of the red intervals contains indices which will form the 1 in the monotone pattern of length $4$, and the right part of the red intervals contains indices which will form the 2. Likewise, the left part of blue intervals will contain the indices corresponding to 3, and the right part of the blue intervals will contain indices corresponding to 4. The regions where the indices from $T$ lie are shown above the sequence, where the indices 1--4 of some monotone pattern in $T$ lie in regions which are progressively darker. In order to see how a monotone subsequence may be sampled given that $(G, \ell, \mathsf{I})$ is a tree descriptor for $(f, T, I)$ with sufficiently large $T$, consider indices $i_1$ and $j_2$ that belong to some subsequences from $T$, and lie in different shaded regions of the same red interval, within a black interval; and furthermore, $l_3$ and $h_4$ belong to some subsequence from $T$, and lie in different shaded regions of the same blue interval, within the same black interval as $i_1$ and $j_2$; then, the subsequence $(i_1, j_2, l_3, h_4)$ is a monotone subsequence even though $(i_1, j_2, l_3, h_4) \notin T$.
}
\label{fig:tree-descriptor}
\end{figure}
\subsection{The structural dichotomy theorem}
We are now in a position to state the main structural theorem of far-from-$(12\dots k)$-free sequences, which guarantees that every far-from-$(12\dots k)$-free sequence either has many growing suffixes, or can be represented by a tree descriptor. The algorithm for finding a $(12\dots k)$-pattern will proceed by considering the two cases independently. The first case, when a sequence has many growing suffixes, is easy for algorithms; we will give a straight-forward sampling algorithm making roughly $O_k(\log n/\eps)$ queries. The second case, when a sequence is represented by a tree descriptor is the ``hard'' case for the algorithm.
\begin{theorem}[Main structural result]
\label{thm:tree}
Let $k \in \N$, $\eps > 0$, and let $f\colon [n] \to \R$ be a function which is $\eps$-far from $(12 \dots k)$-free. Then one of the following holds, where $C > 0$ is a large constant.
\begin{itemize}
\item
There exists a parameter $\alpha \ge \eps / \poly(k, \log(1/\eps))^k$, and a set $H \subseteq [n]$ of indices which start an $(\alpha, C k \alpha)$-growing suffix, with
\[
\alpha |H| \geq \frac{\eps n}{\poly(k, \log(1/\eps))^{k}},
\]
\item
or there exists a set $T \subseteq [n]^k$ of disjoint monotone subsequences of $f$ satisfying
\[
|T| \geq \frac{\eps n}{\poly(k, \log(1/\eps))^{k^2}}
\]
and a $(k, k, \beta)$-tree descriptor $(G, \varrho, \mathsf{I})$ which represents $(f, T, [n])$, where $\beta \ge \eps / \poly(k, \log(1/\eps))^{k^2}$.
\end{itemize}
\end{theorem}
\begin{proof}
We shall prove the following claim, by induction, for all $k_0 \in [k]$. Here $C > 0$ is a large constant, and $C' > 0$ is a large enough constant such that $\alpha \ge \delta / (C' k^{5})$ in the statement of Theorem~\ref{thm:main-structure}, applied with the constant $C$.
\begin{description}
\item[Claim.]
Let $K = C' k^5$ and let $P(\cdot, \cdot)$ be the function from the statement of Theorem~\ref{thm:main-structure}; so $P(x,y) = \poly(x, \log y)$, and we may assume that $P$ is increasing in both variables.
Let $A(\cdot, \cdot)$ and $B(\cdot, \cdot)$ be increasing functions, such that
\begin{align} \label{eqn:ABC}
\begin{split}
& A(k_0, 1/\delta) \ge 12k \lceil \log(K^{k_0}/\delta) \rceil \cdot P( k, 1/\delta) \cdot A(k_0-1, K/\delta) \\
& A(1, 1/\delta) = 1/\delta \\
& B(k_0, 1/\delta) \ge 2 \cdot P(k, K/\delta) \cdot \left(2k \lceil \log(K B(k_0-1, K/\delta)/\delta) \rceil\right)^{2k_0} \cdot B(k_0-1, K/\delta) \\
& B(1, 1/\delta) = 1/\delta \\
\end{split}
\end{align}
Note that there exists such $A(\cdot, \cdot)$ and $B(\cdot, \cdot)$ with $A(k, 1/\delta) = (\poly(k, \log(1/\delta)))^k$ and $B(k, 1/\delta) = (\poly(k, \log(1/\delta)))^{k^2}$.
Let $I \subseteq \N$ be an interval, let $g$ be a sequence $g \colon I \to \R$, let $T_0 \subseteq I^{k_0}$ be a set of disjoint length-$k_0$ monotone subsequences, and define $\delta \eqdef |T_0|/|I|$. Then
\begin{enumerate}
\item \label{case:H}
Either there exists $\alpha \geq \delta / K^{k_0}$, which is an integer power of $1/2$, along with a set $H \subseteq I$ of $(\alpha, C k \alpha)$-growing suffix start points such that
\[
\alpha |H| \geq \frac{\delta|I|}{A(k_0, 1/\delta)},
\]
\item \label{case:split}
Or there exists a set $T \subseteq I^{k_0}$ of disjoint $k_0$-tuples satisfying $E(T) \subseteq E(T_0)$ and
\[
|T| \geq \frac{|T_0|}{B(k_0, 1/\delta)}
\]
and a ($k, k_0, \alpha$)-tree descriptor $(G, \varrho, \mathsf{I})$ for $(g, T, I)$, where $\alpha \ge \delta / B(k_0, 1/\delta)$.
\end{enumerate}
\end{description}
Note that since $f$ is $\eps$-far from $(12 \dots k)$-free, there is a set $T_0 \subseteq [n]^k$ of at least $\eps n / k$ disjoint length-$k$ monotone subsequences. By applying the above claim for $k_0 = k$, $T_0$, $[n]$ and $f$, the theorem follows. Thus, it remains to prove the claim; we proceed by induction.
\begin{description}
\item[if $k_0=1$:]
Note that here $T_0$ is a subset of $I$.
We define the ($k,1, \delta$)-tree descriptor $(G, \varrho, \mathsf{I})$ which represents $f, T = T_0, I$ in the natural way:
\begin{itemize}
\item
$G= (V, E)$ is a rooted tree with one node: $V = \{r\}$ and $E = \emptyset$.
\item
$\varrho\colon V\to\N$ is given by $\varrho(r) = \lceil \log(1/\delta) \rceil$, so $2^{-\varrho(r)} \leq |I\cap T|/|I| \leq 2^{-\varrho(r)+1}$.
\item
$\mathsf{I} \colon V \to \calS(I)$ is given by $\mathsf{I}(r) = \{ \{ t \} : t \in T\}$.
\end{itemize}
\item[if $2\leq k_0\leq k$:]
By Theorem~\ref{thm:main-structure}, there exists $\alpha \ge \delta/ K$ such that one of~\eqref{en:suffix} and~\eqref{en:split}, from the statement of the theorem, holds.
\begin{itemize}
\item
If~\eqref{en:suffix} holds, there is a set $H \subseteq I$ of $(\alpha, C k \alpha)$-growing suffix start points with
\[
\alpha |H| \geq \frac{\delta |I|}{P(k, 1/\delta)};
\]
note that we may assume that $\alpha$ is an integer power of $1/2$.\footnote{to be precise and to ensure that we can take $\alpha$ to be an integer power of $2$, it might be better to apply Theorem~\ref{thm:main-structure} with constant $2C$, to allow for some slack; this does not change the argument.}
\item
Otherwise,~\eqref{en:split} holds, and we are given an integer $c\in[k_0-1]$, a set $T$ of disjoint length-$k_0$ monotone subsequences, with $E(T) \subseteq E(T_0)$, and a $(c, 1/(6k),\alpha)$-splittable collection of $T$ into disjoint interval-tuple pairs $(I_1, T_1), \dots, (I_s, T_s)$, such that
\[
\alpha \sum_{h=1}^s |I_h|
\geq \frac{|T_0|}{P(k, 1/\delta)}
= \frac{\delta |I|}{P(k, 1/\delta)}.
\]
Recall that by definition of splittability, $|T_h| / |I_h| \ge \alpha$ for every $h \in [s]$.
\end{itemize}
If \eqref{en:suffix} holds, we are done; so we assume that \eqref{en:split} holds.
For each $h \in [s]$, since $(I_h, T_h)$ is a $(c, 1/(6k), \alpha)$-splittable pair, there exists a partition $(L_h, M_h, R_h)$ that satisfies the conditions stated in Definition~\ref{def:splittable}. Let $T_h^{(L)}$ be the collection of prefixes of length $c$ of subsequences in $T_h$, and let $T_h^{(R)}$ be the collection of suffixes of length $k_0 - c$ of subsequences in $T_h$.
We apply the induction hypothesis to each of the pairs $(L_h, T_h^{(L)})$ and $(R_h, T_h^{(R)})$.
We consider two cases for each $h \in [s]$.
\begin{enumerate}
\item
\eqref{case:H} holds for either $(L_h, T_h^{(L)})$ or $(R_h, T_h^{(R)})$. This means that there exists $\beta_h$, which is an integer power of $1/2$, and which satisfies $\beta_h \ge \alpha / K^{\max\{c, k_0 - c\}} \ge \alpha / K^{k_0 - 1} \ge \delta / K^{k_0}$, and a set $H_h \subseteq I_h$ of start points of $(\beta_h, Ck \beta_h)$-growing subsequences, such that (using $|R_h|, |L_h| \ge |I_h| / (6k)$)
\[
\beta_h |H_h| \ge
\frac{\alpha |I_h|}{6k \cdot A(k_0-1, 1/\alpha)}
\]
\item
Otherwise, \eqref{case:split} holds for both $(L_h, T_h^{(L)})$ and $(R_h, T_h^{(R)})$. Setting $\beta = \alpha / B(k_0-1, 1/\alpha)$, this means that there exists a $(k, c, \beta)$-tree descriptor $(G_h^{(L)}, \varrho_h^{(L)}, \mathsf{I}_h^{(L)})$, for $(g, \calL_h, L_h)$ where $\calL_h \subseteq (L_h)^c$ is a set of length-$c$ monotone subsequences, such that $E(\calL_h) \subseteq E(T_h^{(L)})$ and
\begin{equation} \label{eqn:calL-h}
|\calL_h| \ge
\frac{|T_h^{(L)}|}{B(k_0-1, 1/\alpha)},
\end{equation}
and, similarly, there exists a $(k, k_0 - c, \beta)$-tree descriptor $(G_h^{(R)}, \varrho_h^{(R)}, \mathsf{I}_h^{(R)})$ for $(g, \calR_h, L_h)$, where $\calR_h \subseteq (R_h)^{k_0-c}$ is a set of length-$(k_0 - c)$ monotone subsequences, such that $E(\calR_h) \subseteq E(T_h^{(R)})$ and
\begin{equation} \label{eqn:calR-h}
|\calR_h| \ge
\frac{|T_h^{(R)}|}{B(k_0-1, 1/\alpha)}.
\end{equation}
For convenience, we shall assume that $|\calL_h| = |\calR_h|$, by possibly removing some elements of the largest of the two (and reflecting this in the corresponding tree descriptor).
\end{enumerate}
Suppose first that
\[
\sum_{h \colon \, \text{first case holds for $h$}} |I_h| \ge \frac{1}{2} \cdot \sum_{h = 1}^s |I_h|.
\]
Since each $\beta_h$ is an integer power of $1/2$, there are at most $\lceil \log(K^{k_0} / \delta) \rceil$ possible values for $\beta_h$. Hence, there exists some $\beta$ (with $\beta \ge \delta / K^{k_0}$) such that the collection $S$, of indices $h \in [s]$ for which the first case holds for $h$ and $\beta_h = \beta$, satisfies
\[
\sum_{h \in S} |I_h| \ge \frac{1}{2 \lceil \log(K^{k_0}/\delta) \rceil} \cdot \sum_{h = 1}^s |I_h|.
\]
Let $H = \bigcup_{h \in S} H_h$. Then $H$ is a set of start points of $(\beta, Ck \beta)$-growing suffixes, with
\begin{align*}
\beta |H|
&\ge \frac{\alpha}{6k \cdot A(k_0-1, 1/\alpha)} \cdot \sum_{h \in S} |I_h|
\ge \frac{\alpha}{12 k \lceil \log(K^{k_0}/\delta) \rceil \cdot A(k_0-1, 1/\alpha)} \cdot \sum_{h = 1}^s |I_h| \\
&\ge \frac{\delta |I|}{12 k \lceil \log(K^{k_0}/\delta) \rceil \cdot P(k, 1/\delta) \cdot A(k_0-1, 1/\alpha) }
\ge \frac{\delta|I|}{A(k_0, 1/\delta)},
\end{align*}
where the last inequality follows from \eqref{eqn:ABC}.
This proves the claim in this case.
Next, we may assume that
\[
\sum_{h \colon \, \text{second case holds for $h$}} |I_h| \ge \frac{1}{2} \cdot \sum_{h = 1}^s |I_h|.
\]
Note that the number of quadruples $(G_h^{(L)}, \varrho_h^{(L)}, G_h^{(R)}, \varrho_h^{(R)})$ (whose elements are as above) is at most $(2c)^{2c} (2(k_0-c))^{2(k_0-c)} (\lceil \log(1/\beta) \rceil)^{2k_0} \le (2k \lceil \log(1/\beta) \rceil)^{2k_0}$, since the number of trees on $l$ vertices is at most $l^l$, and we have at most $\lceil \log(1/\beta) \rceil$ possible weights to assign to each of the vertices. It follows that there exists such a quadruple $(G_L^\ast, \varrho_L^\ast, G_R^\ast, \varrho_R^\ast)$ such that if $S$ is the set of indices $h$ that were assigned this quadruple, then
\begin{align} \label{eqn:intervals-dense}
\begin{split}
\alpha \cdot \sum_{h \in S} |I_h|
&\ge \frac{\alpha}{(2k\lceil \log(1/\beta) \rceil)^{2k_0}} \cdot \sum_{\text{second case holds for $h$}}|I_h| \\
&\ge \frac{\alpha}{2\cdot(2k\lceil \log(1/\beta) \rceil)^{2k_0}} \cdot \sum_{h = 1}^s|I_h|
\ge \frac{|T_0|}{2 \cdot P(k, 1/\delta) \cdot (2 k \lceil \log(1/\beta) \rceil)^{2k_0}}.
\end{split}
\end{align}
We form a set $\calT_h$ of monotone length-$k_0$ subsequences by matching elements from $\calL_h$ with elements from $\calR_h$ for each $h \in S$; that they can be matched follows from the assumption that $|\calL_h| = |\calR_h|$, and that these form monotone subsequences follows from the assumptions on $\calL_h, \calR_h$. Set $\calT := \cup_{h \in S} \calT_h$.
Note that $(I_h, \calT_h)$ is $(k_0, c, \beta)$-splittable by \eqref{eqn:calL-h} and \eqref{eqn:calR-h} (using $\beta = \alpha / B(k_0-1, 1/\alpha)$).
Let $(G, \varrho)$ be the $(k, k_0, \beta)$-weighted-tree obtained by taking a root $r$, with weight $\varrho(r) = \lceil \log(1 / \beta) \rceil$, adding the tree $(G_L^\ast, \varrho^\ast)$ as a subtree to its left (i.e., the root of this tree is joined to $r$ by an edge with value $0$) and adding the tree $(G_R^\ast, \varrho^\ast)$ as a subtree to its right.
Now, we form a $(G, \varrho, \mathsf{I})$-tree descriptor by setting
\[
\mathsf{I}(v) =
\left\{
\begin{array}{ll}
\{I_h : h \in S\} & v = r \\
\bigcup_{h \in S} \mathsf{I}_h^{(L)}(v) & v \in G_L^{\ast} \\
\bigcup_{h \in S} \mathsf{I}_h^{(R)}(v) & v \in G_R^{\ast}.
\end{array}
\right.
\]
We claim that $(G, \varrho, \mathsf{I})$ is a $(k, k_0, \beta)$-tree descriptor for $(g, \calT, I)$.
Indeed, $((I_h, \calT_h))_{h \in S}$ is a $(c, 1/(6k), 2^{-\varrho(r)})$-splittable collection of $\calT$, and, by \eqref{eqn:intervals-dense} and because $|T_0| \ge |\calT|$
\begin{align*}
2^{-\varrho(r)} \sum_{h \in S} |I_h|
& \ge
\frac{\alpha}{2} \cdot \sum_{h \in S} |I_h| \ge
\frac{|\calT|}{4 \cdot P(k, 1/\delta) \cdot (2k\lceil \log(1/\beta) \rceil)^{2k_0}} =
\frac{|\calT|}{\poly(k, \log(1/\delta))^k}.
\end{align*}
The remaining requirements in the recursive defnition of a tree descriptor (see Definition~\ref{def:tree-rep}) follow as $(G_L^{\ast}, \varrho^{\ast}, \mathsf{I}_h^{(L)})$ is a $(k, c, \beta)$-tree descriptor for $(g, \calL_h, L_h)$ and $(G_L^{\ast}, \varrho^{\ast}, \mathsf{I}_h^{(R)})$ is a $(k, k_0-c, \beta)$-tree descriptor for $(g, \calR_h, R_h)$ for every $h \in S$. Since $\beta = \alpha / B(k_0-1, 1/\alpha) \ge \delta / B(k_0, 1/\delta)$, it follows that $(G, \varrho, \mathsf{I})$ is a $(k, k_0, \delta / B(k_0, 1/\delta))$-tree descriptor for $(g, \calT, I)$.
It remains to lower-bound the size of $\calT$. Using \eqref{eqn:calR-h} and \eqref{eqn:intervals-dense}, we have
\begin{align*}
|\calT|
& = \sum_{h \in S} |\calR_h|
\ge \frac{1}{B(k_0-1, 1/\alpha)}\cdot \sum_{h \in S} |T_h|
\ge \frac{\alpha}{B(k_0-1, 1/\alpha)} \cdot \sum_{h \in S} |I_h| \\
& \ge \frac{|T_0|}{2 \cdot P(k, 1/\delta) \cdot (2k \lceil \log(1/\beta) \rceil)^{2k_0} \cdot B(k_0-1, 1/\alpha)} \ge \frac{|T_0|}{B(k_0, 1/\delta)}.
\end{align*}
This completes the proof of the inductive claim in this case.
\qedhere
\end{description}
\end{proof}
\subsection{Proof of Theorem~\ref{thm:main-structure}}\label{sec:proof1}
We now prove Theorem~\ref{thm:main-structure}. For the rest of this section, let $k, k_0 \in \N$, with $1 \leq k_0 \leq k$, be fixed, and let $f \colon [n] \to \R$ be a fixed function. Let $T_0$ be a set of $\delta n$ disjoint monotone subsequences of $f$ of length $k_0$. We apply Lemma~\ref{lem:big-split} to the set $T_0$; this specifies an integer $c \in [k_0-1]$ and a subset $T$ of at least $\delta n / k^2$ disjoint monotone subsequences of length $k_0$ satisfying the conclusion of Lemma~\ref{lem:big-split}.
\begin{definition}
Let $(i_1, \dots, i_{k_0}) \in [n]^{k_0}$ be a monotone subsequence with a $c$-gap. We say that $(i_1, \dots, i_{k_0})$ is \emph{at scale $t$} if $2^{t} \leq i_{c+1} - i_c \leq 2^{t+1}$, where $t \in \{0, \dots, \lfloor \log n \rfloor\}$.
\end{definition}
\begin{definition}
Let $(i_1, \dots, i_{k_0}) \in [n]^{k_0}$ be a monotone subsequence with a $c$-gap. For $\gamma \in (0,1)$, we say that $\ell \in [n]$ \emph{$\gamma$-cuts $(i_1, \dots, i_{k_{0}})$ at $c$ with slack} if
\begin{equation} \label{eq:slack}
i_c + \gamma(i_{c+1} - i_c) \leq \ell \leq i_{c+1} - \gamma(i_{c+1} - i_c).
\end{equation}
\end{definition}
We hereafter consider the parameter setting of $\gamma \eqdef 1/3$. For $\ell \in [n]$, $t \in \{0, \dots, \lfloor \log n \rfloor\}$, and any subset $U \subset T$ of disjoint $(12\dots k_0)$-patterns in $f$ let
\begin{align}
A_t(\ell, U) &= \{ (i_1, \dots, i_{k_0}) \in U : (i_1, \dots, i_{k_0}) \text{ is at scale $t$ and is $\gamma$-cut at $c$ with slack by $\ell$} \}. \label{eq:cut-sets}
\end{align}
We note that for each $(i_1, \dots, i_{k_0}) \in A_t(\ell, U)$, the index $i_{c+1}$ is in $[\ell, \ell + 2^{t+1}]$, and since $A_t(\ell, U)$ is made of disjoint monotone sequences, $|A_t(\ell, U)| \leq 2^{t+1}$.
\begin{lemma}
\label{lem:At:properties}
For every $\ell \in [n]$, $t \in \{0, \dots, \lfloor \log n \rfloor \}$, and $U \subset T$,
\begin{itemize}
\item
Every $(i_1, \dots, i_{k_0}) \in A_t(\ell, U)$ satisfies
\[
\ell - (k-1)2^{t+1} \leq i_1, \dots, i_c \le \ell - \gamma 2^t \qquad \qquad
\ell + \gamma 2^t \le i_{c+1}, \dots, i_{k_0} \leq \ell + (k-1) 2^{t+1}.
\]
\item
Let $t_1 \geq t_2 + 1 + \log(1/\gamma) + \log(c+1)$, $(i_1, \dots, i_{k_0}) \in A_{t_1}(\ell, U)$ and $(j_1, \dots, j_{k_0}) \in A_{t_2}(\ell, U)$. Then $f(j_{c+1}) < f(i_{c+1})$.
\end{itemize}
\end{lemma}
\begin{proof}
Fix any $\ell \in [n]$, $t\in \{0, \dots, \lfloor \log n \rfloor\}$ and $U \subset T$. To establish the first bullet, consider any $(i_1, \dots, i_{k_0}) \in A_t(\ell, U)$.
By definition of a $c$-gap sequence, we have
\[
i_1 \ge i_{c+1} - c (i_{c+1} - i_c) \ge \ell - (k-1) 2^{t+1},
\]
using $i_{c+1} - i_c \le 2^{t+1}$ and $i_{c+1} \ge \ell$. By \eqref{eq:slack}, we have $i_c \le \ell - \gamma 2^t$ (using $i_{c+1} - i_c \ge 2^t$).
The first inequality follows as $i_1 < \dots < i_c$. The inequality for $i_{c+1}, \dots, i_{k_0}$ follows similarly.
For the second bullet, let $(i_1, \dots, i_{k_0}) \in A_{t_1}(\ell, U)$ and $(j_1, \dots, j_{k_0}) \in A_{t_2}(\ell, U)$ and suppose that $2^{t_1} \geq 2^{t_2+1}\cdot (c+1)/\gamma$.
We have $i_c \le \ell - \gamma 2^{t_1}$ and $j_c \ge \ell - 2^{t_2+1}$ (using \eqref{eq:slack} and \eqref{eq:cut-sets}), from which it follows that $j_c > i_c$.
Similarly, $i_1 < i_c \le \ell - \gamma 2^{t_1}$ and $j_1 \ge \ell - (c - 1) 2^{t_2 + 1}$, implying that $j_1 > i_1$, and $i_{c+1} \ge \ell + \gamma 2^{t_1}$ and $j_{c+1} \le \ell + 2^{t_2+1}$, which implies that $i_{c+1} > j_{c+1}$. The inequality $f(j_{c+1}) < f(i_{c+1})$ follows from the assumption that $T$ satisfies \eqref{itm:property-greedy} from Lemma~\ref{lem:big-split}.
\end{proof}
The proof of Theorem~\ref{thm:main-structure} will follow by considering a random $\bell \sim [n]$ and the sets $A_1(\bell, T), \dots, A_{\lfloor\log n\rfloor}(\bell, T)$. By looking at how the sizes of the sets $A_1(\bell, T), \dots, A_{\log n-1}(\bell, T)$ vary, we will be able to say that $\bell$ is the start of a growing suffix, or identify a splittable interval. Towards this goal, we first establish a simple lemma; here $v(\ell, U)$ is defined to be $\sum_{t=0}^{\lfloor \log n \rfloor} |A_t(\ell, U)| / 2^t$.
\begin{lemma}\label{lem:expect}
Let $U \subset T$ be any subset and $\bell \sim [n]$ be sampled uniformly at random. Then
\[
\Ex_{\bell \sim [n]} v(\ell, U) \geq \frac{|U|}{3n}.
\]
\end{lemma}
\begin{proof}
Fix a sequence $i = (i_1, \dots, i_{k_0}) \in U$, and let $t(i)\in \{0, \dots, \lfloor\log n\rfloor\}$ be its scale.
Then, the probability (over a uniformly random $\bell$ in $[n]$) that $i$ belongs to $A_{t(i)}(\bell, U)$ is lower bounded as
\[
\Prx_{\bell\sim[n]}[ i\in A_{t(i)}(\bell, U) ] \geq \frac{(1-2\gamma)2^{t(i)}}{n} = \frac{2^{t(i)}}{3n}.
\]
Therefore,
$
\sum_{t=0}^{\log n-1} \sum_{i\in U \colon t(i) = t} \Pr_{\bell\sim[n]}[ i\in A_{t}(\bell, U) ] /2^t \geq |U|/(3n)
$, or, equivalently, since $\Pr_{\bell\sim[n]}[ i\in A_{t}(\bell, U) ] = 0$ for $t\neq t(i)$,
\[
\Ex_{\bell \sim [n]}\left[ \sum_{t=0}^{\log n-1} \frac{ |A_{t}(\bell, U)| }{2^t} \right]
= \Ex_{\bell \sim [n]}\left[ \sum_{t=0}^{\log n-1} \sum_{i\in U} \frac{ \mathbbm{1}\{ i \in A_{t}(\bell, U)\} }{2^t} \right] \geq \frac{|U|}{3n},
\]
establishing the lemma.
\end{proof}
We next establish an auxiliary lemma that we will use in order to find growing suffixes.
\begin{lemma}\label{lem:move-back}
Let $\ell \in [n]$ and $U \subset T$ be such that every $t \in \{0, \dots, \lfloor \log n \rfloor\}$ satisfies $|A_t(\ell, U)| / 2^t \leq \beta$. Then, if $\ell' \in [n]$ is any index satisfying
\begin{align}
\max \{ i_c : (i_1, \dots, i_{k_0}) \in A_t(\ell, U), t \in \{0, \dots, \lfloor \log n \rfloor\} \leq \ell' \leq \ell, \label{eq:bound-on-ell}
\end{align}
then $\ell'$ is the start of an $(4\beta, v(\ell, U) / (12 \log k))$-growing suffix.
\end{lemma}
\begin{proof}
Let $\Delta = 1 + \log(1/\gamma) + \log(c+1) $, and notice that $3 \le \Delta \le 3\log k$. Then, there exists a set $\calT \subseteq \{0,\dots, \lfloor \log n \rfloor\}$ such that
\begin{enumerate}
\item
All distinct $t, t' \in \calT$ satisfy $|t - t' | \geq \Delta$; and,
\item
$\sum_{t \in \calT} \frac{|A_t(\ell, U)|}{2^t} \geq \frac{1}{\Delta + 1} \sum_{t=0}^{\log n-1} \frac{|A_t(\ell, U)|}{2^t} = \frac{v(\ell, U)}{\Delta + 1}$.
\end{enumerate}
(Such a set exists by an averaging argument.) Now, consider the sets
\[
D_t(\ell)=
\begin{cases}
\{ i_{c+1} : (i_1,\dots,i_{k_0})\in A_t(\ell, U) \} &\text{ if } t\in\calT\\
\emptyset &\text{ if } t\in\{0,\dots,\lfloor \log n \rfloor\}\setminus \calT .
\end{cases}
\]
Considering any $\ell' \in [n]$ satisfying (\ref{eq:bound-on-ell}), we have the following for all $t \in \{0, \dots, \lfloor \log n \rfloor\}$ with $D_{t}(\ell) \neq \emptyset$: $\ell - 2^{t+1} \leq \ell' \leq \ell$; $\min D_t(\ell) \geq \ell + 2^t /3$; and $\max D_t(\ell) \leq \ell' + 2^{t+1}$. Therefore, $D_t(\ell) \subset S_{t-1}(\ell') \cup S_{t}(\ell') \cup S_{t+1}(\ell')$. (Recall that $S_t(a) = [a + 2^{t-1}, a + 2^t)$.)
For each $t \in \calT$, let $n(t) \in \{ t-1, t, t+1\}$ satisfying $|D_t(\ell) \cap S_{n(t)}(\ell') | \geq |D_t(\ell)|/ 3$, and notice that all $n(t) \in \{0, \dots, \lfloor \log n \rfloor\}$ are distinct since $\Delta \geq 3$.
The first condition in Definition~\ref{def:growing-suffixes} holds as the densities of $D_t(\ell) \cap S_{n(t)}(\ell')$ in the corresponding intervals $S_{n(t)}(\ell')$ are upper bounded by $|D_t(\ell)| / |S_{n(t)}(\ell')| \le |A_t(\ell, U)|/2^{t-2} \leq 4\beta$, and the sum of these densities satisfies
\[
\sum_{t \in \calT} \frac{|D_t(\ell)\cap S_{n(t)}(\ell')|}{|S_{n(t)}(\ell')|} \geq \sum_{t \in \calT} \frac{|D_t(\ell)|}{3 \cdot 2^t} = \sum_{t \in \calT} \frac{|A_t(\ell, U)|}{3 \cdot 2^t} \geq \frac{v(\ell, U)}{3(\Delta + 1)},
\]
which is at least $v(\ell, U) / (12 \log k)$.
The second condition in Definition~\ref{def:growing-suffixes} holds, because for any choice of $b \in D_t(\ell), b' \in D_{t'}(\ell)$ with $t < t'$, we have $t' \ge t + \Delta$ (by the choice of $\calT$), and hence $f(b) < f(b')$ by the second item of Lemma~\ref{lem:At:properties}.
\end{proof}
\begin{lemma}\label{lem:easy-growing}
For every $\eta > 0$, there exists a subset $U \subset T$ such that every $(i_1, \dots, i_{k_0}) \in U$ has $i_{c}$ as the start of an $(1, \eta)$-growing suffix, and every $\ell \in [n]$ satisfies $v(\ell, T \setminus U) \leq 12\eta \log(k)$.
\end{lemma}
\begin{proof}
Define sets $U_j$, elements $\ell_j$, and $k_0$-tuples $(i_{j,1}, \ldots, i_{j, k_0})$ recursively as follows. Set $U_0 := \emptyset$, and given a set $U_{j-1}$, if $v(\ell, T \setminus U_{j-1}) \le 12 \eta \log k$ for every $\ell \in [n]$, stop; otherwise, let $\ell_{j} \in [n]$ be such that $v(\ell_j, T \setminus U_j) > 12 \eta \log k$ and define $U_j = U_{j-1} \cup \{(i_{j,1}, \ldots, i_{j,k_0})\}$, where
\[
i_{j,c} = \max\{i_c : (i_1, \ldots, i_{k_0}) \in T \setminus U_j \text{ and $(i_1, \ldots, i_{k_0})$ is $\gamma$-cut by $\ell_j$}\}.
\]
Let $j^*$ be the maximum $j$ for which $U_j$ was defined, and set $U := U_{j^*}$.
Every $k_0$-tuple in $U$ is of the form $(i_{j,1}, \ldots, i_{j, k_0})$ for some $j \le j^*$. By Lemma~\ref{lem:move-back}, applied with $\ell = \ell_j$, $U = T \setminus U_{j-1}$, $i_{j,c}$, it follows that $i_{j,c}$ is the start of an $(1, \eta)$-growing suffix, for every $j$ for which $U_j$ was defined. Lemma~\ref{lem:easy-growing} follows.
\end{proof}
We let $C > 0$ be a large enough constant. Let $U \subset T$ be the set obtained from Lemma~\ref{lem:easy-growing} with $\eta = Ck$, and suppose that $|U| \geq |T|/2$. Then, we may let $\alpha = 1$ and $H = \{ i_c : (i_1, \dots, i_{k_0}) \in U\}$. Notice that every index in $H$ is the start of an $(\alpha, Ck\alpha)$-growing suffix, and since $|H| \geq |T|/2$, we obtain the first item in Theorem~\ref{thm:main-structure}. Suppose then, that $|U| < |T|/2$, and consider the set $V = T\setminus U$. By definition of $V$, we now have $v(\ell, V) \leq 12 Ck \log k$ for every $\ell \in [n]$. Let $b_0$ be the largest integer which satisfies $2^{b_0} \leq 12 C k\log k$ and $b_1$ be the smallest integer which satisfies $2^{-b_1} \leq \delta/(12 k^2)$, so $2^{b_0} \lsim 2^{b_1} \asymp k^2/\delta$. For $-b_0 \leq j \leq b_1$, consider the pairwise-disjoint sets
\begin{align} \label{eq:def-B}
B_j &= \left\{ \ell \in [n] : 2^{-j} \leq v(\ell, V) \leq 2^{-j + 1}\right\},
\end{align}
and note that by Lemma~\ref{lem:expect}, since $|V| \geq |T| / 2 \ge \delta n / 2k^2$,
\[
\frac{1}{n} \sum_{j =-b_0}^{b_1} |B_j| \cdot 2^{-j+1}
\geq \frac{1}{n} \sum_{\ell \in [n]} v(\ell, V) \geq \frac{\delta}{6k^2}.
\]
Thus, denoting
\[
\mu \eqdef \frac{\delta}{6 k^2(b_1 + b_0 +1)} \asymp \frac{\delta}{k^2\log(k/\delta)},
\]
there is an integer $-b_0 \leq j^\ast \leq b_1$ that satisfies
\begin{align}
|B_{j^\ast}| \cdot 2^{-j^\ast} &\geq \mu n. \label{eq:B-lb}
\end{align}
\newcommand{\texttt{GreedyDisjointIntervals}}{\texttt{GreedyDisjointIntervals}}
\begin{lemma}
There exists a deterministic algorithm, $\emph{\texttt{GreedyDisjointIntervals}}(f, B, j)$, which takes three inputs: a function $f \colon [n] \to \R$, a set $B \subseteq [n]$ of integers, and an integer $j \in [-b_0, b_1]$, and outputs a collection $\calI$ of interval-tuple pairs
or a subset $H \subseteq B$. An execution of the algorithm $\emph{\texttt{GreedyDisjointIntervals}}(f, B_{j^\ast}, j^\ast)$ where $\mu$, $B_{j^\ast}$ and $j^\ast$ are defined in~\eqref{eq:B-lb}, satisfies one of the following two conditions, where $C > 0$ is a large constant.
\begin{itemize}
\item
The algorithm returns a set $H \subseteq B$ of indices that start a $(4 \cdot 2^{-j^\ast} /(Ck\log k), 2^{-j^\ast}/(12\log k))$-growing suffix, and $|H| \geq 2^{j^\ast-1} \mu n$; or
\item
The algorithm returns a $(c, 1/(6k),2^{-j^\ast}/(8Ck^2\log k))$-splittable collection $(I_1, T_1), \dots, (I_s, T_s)$, where $\sum_{h = 1}^{s} |I_h| \geq 2^{j^\ast-2} \mu n$.
\end{itemize}
\end{lemma}
\begin{figure}[ht!]
\begin{framed}
\centering
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{GreedyDisjointIntervals}\vspace{0.3cm}(f, B, j)$
\noindent {\bf Input:} A function $f \colon [n] \to \R$, a set $B \subseteq [n]$ and an integer $j$, such that every $\ell \in B$ satisfies $2^{-j} \leq v(\ell, V) \leq 2^{-j+1}$.\\
{\bf Output:} a set of disjoint intervals-tuple pairs $(I_1, T_1), \dots, (I_s, T_s)$ or a subset $H \subseteq B$.
\begin{enumerate}
\item
Let $\calI$ be a collection of interval-tuple pairs, which is initially empty.
\item
Consider the map $q \colon B \to \{ 0, \dots, \lfloor \log n \rfloor\} \cup \{ \bot \}$ defined by
\begin{align*}
q(\ell) = \left\{
\begin{array}{lc}
\bot & \forall t \in \{0, \dots, \lfloor \log n \rfloor \}, \frac{|A_t(\ell,V)|}{2^{t}} < \frac{2^{-j}}{ Ck\log k} \\
\max\left\{ t : \frac{|A_t(\ell, V)|}{2^t} \geq \frac{2^{-j}}{Ck \log k} \right\} & \text{otherwise}
\end{array}
\right. .
\end{align*}
\item\label{step:return:H} Let $H = \{ \ell \in B : q(\ell) = \bot\}$, and \Return $H$ if $|H| \geq |B| / 2$.
\item Otherwise, let $D \gets B \setminus H$ and repeat the following until $D = \emptyset$:
\begin{itemize}
\item\label{step:qt}
Pick any $\ell \in D$ where $q(\ell) = \max_{\ell' \in D} q(\ell')$, and let $t = q(\ell)$.
\item\label{step:IT}
Let $I \gets [\ell - k 2^{t+1}, \ell + k 2^{t+1}] \cap [n]$ and $T' \gets A_t(\ell, V)$.
\item
Obtain $T''$ from $T'$ as follows: find a value $\nu$ such that at least $|T'|/2$ of tuples $(i_1,\dots, i_{k_0})\in T'$ satisfy $f(i_c) \leq \nu$, and at least $|T'|/2$ of tuples $(i_1,\dots, i_{k_0})\in T'$ satisfy $f(i_{c+1}) > \nu$ ($\nu$ could be taken to be the median of the multiset $\{f(i_c) : (i_1, \dots, i_{k_0}) \in T'\}$). Recombine these prefixes and suffixes (matching them in one-to-one correspondence) to obtain a set of disjoint $k_0$-tuples $T''$ of size $|T''|\geq |T'|/2$.
\item\label{step:update:C}
Append $(I, T'')$ to $\calI$, and let $D \gets D \setminus [\ell - 2 \cdot k 2^{t+1}, \ell + 2 \cdot k 2^{t+1}]$.
\end{itemize}
\item\label{step:return:I} \Return $\calI$.
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{GreedyDisjointIntervals}$ subroutine.} \label{fig:greedy-intervals}
\end{figure}
\begin{proof}
It is clear that the algorithm always terminates, and outputs either a collection $\calI$ of interval-tuple pairs or a subset $H \subseteq B$.
Suppose that the input of the algorithm, $(f, B_{j^\ast}, j^\ast)$, satisfies~\eqref{eq:B-lb}, and consider the two possible types of outputs.\medskip
If the algorithm returns a set $H\subseteq B_{j^\ast}$ (in step~\ref{step:return:H}), then we have $|H| \geq \frac{|B|}{2} \geq \frac{1}{2}\cdot 2^{j^\ast}\mu n$ (the second inequality by~\eqref{eq:B-lb}). (To see why the elements of $H$ start $(4 \cdot 2^{-j^\ast} /(Ck\log k), 2^{-j^\ast}/(12\log k))$-growing suffixes (Definition~\ref{def:growing-suffixes}), notice that we may apply Lemma~\ref{lem:move-back} with $\ell' = \ell$ and $\beta = 2^{-j^\ast} / (Ck \log k)$.)
If, instead, the algorithm returns a collection $\calI = ((I_h, T_h) : h \in [s])$ in step~\ref{step:return:I}, we have that, by construction, each $T_h$ is obtained from a set $T'_h=A_{t}(\ell, V)$ for some $\ell$ with $q(\ell)\neq \bot$. Consequently, for all $h \in[s]$ we have
\begin{equation} \label{eq:density-T-h}
\frac{|T_h|}{|I_h|} \geq \frac{|T'_h|}{2|I_h|}
\geq \frac{|A_{q(\ell)}(\ell, V)|}{4k \cdot 2^{q(\ell)+1}}
\geq \frac{1}{8k} \cdot \frac{2^{-j^\ast}}{Ck\log k}.
\end{equation}
(from the definition of $q(\ell)$). To argue that $\sum_{h=1}^s |I_h|$ is large, observe that, since we did not output the set $H$, we must have had $|D|>|B_{j^\ast}|/2$. Since, when adding $(I_h, T_h)$ (corresponding to some $\ell_h$) to $\calI$ we remove at most $4k 2^{q(\ell)+1}=2|I_h|$ elements from $D$, in order to obtain an empty set $D$ and reach step~\ref{step:return:I} we must have $\sum_{h=1}^s |I_h| \geq |B_{j^\ast}|/4$, which is at least $2^{j^\ast}\mu n/4$ by~\eqref{eq:B-lb}. Moreover, the sets $I_h$ are disjoint: this is because of our choice of maximal $q(\ell)$ in step~\ref{step:IT}, which ensures that after removing $[\ell - 2k 2^{q(\ell)+1}, \ell + 2k 2^{q(\ell)+1}]$ in step~\ref{step:update:C} there cannot remain any $\ell'\in D$ with $[\ell' - k 2^{q(\ell')+1}, \ell' + k 2^{q(\ell')+1}]\cap I_h \neq \emptyset$.
Thus, it remains to prove that $\calI$ is a $(c,1/(6k),2^{-j^\ast}/(8Ck^2\log k))$-splittable collection. To do so, consider any $(I_h, T_h)\in\calI$.
The first condition in Definition~\ref{def:splittable} of splittable pairs, namely that $|T_h| / |I_h| \ge 2^{-j^\ast}/(8Ck^2 \log k)$ holds due to \eqref{eq:density-T-h}.
Recalling step~\ref{step:IT}, we have $I_h = [\ell - k 2^{t+1}, \ell + k 2^{t+1}]$ for some $\ell$, where $t = q(\ell)$, and $T_h$ obtained from $T_h' = A_{t}(\ell, V)$. Set
\[
L_h\eqdef [\ell - k 2^{t+1}, \ell - \gamma 2^{t}], \quad M_h\eqdef (\ell - \gamma 2^{t}, \ell + \gamma 2^{t}),\quad R_h\eqdef [\ell + \gamma 2^{t}, \ell + k 2^{t+1}].
\]
This is a partition of $I_h$ into three adjacent intervals whose size is at least $|I_h| / (6k)$ (recall that $\gamma = 1/3$). Moreover, for every $(i_1,\dots,i_{k_0})\in T_h'$, the $c$-prefix $(i_1,\dots,i_c)$ is in $(L_h)^c$ while the $(k_0-c)$-suffix $(i_{c+1},\dots,i_{k_0})$ is in $(R_h)^{k_0-c}$, by the first item of Lemma~\ref{lem:At:properties}. Since $T_h$ is obtained from a subset of these very prefixes and suffices, the conclusion holds for $T_h$ as well. Moreover, our construction of $T_h$ from $T'_h$ guarantees that the last requirement in Definition~\ref{def:splittable} holds: for every prefix $(i_1, \dots, i_c)$ of a tuple in $T_h$ and suffix $(j_1, \dots, j_{k_0-c})$ of a tuple in $T_h$, we have $f(i_c) < f(j_1)$. This shows that $(I_h,T_h)$ is $(c,1/(6k),2^{-j^\ast}/(8Ck^2 \log k))$-splittable, and overall that $\calI$ is a $(c,1/(6k),2^{-j^\ast}/(8Ck^2\log k))$-splittable collection as claimed.
\end{proof}
Theorem~\ref{thm:main-structure} follows by executing $\texttt{GreedyDisjointIntervals}(f, B_{j^\ast}, j^\ast)$.
If the algorithm outputs a set $H \subseteq B_{j^\ast}$, set $\alpha = 4 \cdot 2^{-j^\ast} / (Ck\log k)$, so we have identified a subset $H$ of $(\alpha, C' \alpha k)$-growing suffixes (where $C' = C / 48$) satisfying $\alpha |H| \geq \delta n / \poly(k, \log(1/\delta)) = |T_0| / \poly(k, \log(1/\delta))$ (using the definition of $\mu$ before \eqref{eq:B-lb}).
Otherwise, set $\alpha = 2^{-j^\ast} / (8 Ck^2\log k)$, and the algorithm outputs a $(c, 1/(6k),\alpha)$-splittable collection $\{(I_1, T_1), \dots, (I_s, T_s)\}$ of the set $T' := \cup_{h \in [s]} T_h$. Clearly, $E(T') \subseteq E(T)$, and moreover, $\alpha \sum_{h = 1}^s |I_h| \ge \delta n / \poly(k, \log(1/\delta)) = |T_0| / \poly(k, \log(1/\delta))$.
In fact, $2^{-j^*} = \Omega(\delta / k^2)$ and so $\alpha \ge \Omega(\delta / (k^4 \log k))$.
\section{The Algorithm}\label{sec:algorithm}
\subsection{High-level plan}
\newcommand{\texttt{Sampler}}{\texttt{Sampler}}
We now present the algorithm for finding monotone subsequences of length $k$.
\begin{theorem}\label{thm:ub}
Consider any fixed value of $k \in \N$. There exists a non-adaptive and randomized algorithm, $\emph{\texttt{Sampler}}_k(f, \eps)$, which takes two inputs: query access to a function $f \colon [n] \to \R$ and a parameter $\eps > 0$. If $f$ is $\eps$-far from $(12\dots k)$-free, then $\emph{\texttt{Sampler}}_k(f, \eps)$ finds a $(12\dots k)$-pattern with probability at least $9/10$. The query complexity of $\emph{\texttt{Sampler}}_k(f, \eps)$ is at most
\[ \frac{1}{\eps} \left(\frac{\log n}{\eps}\right)^{\lfloor \log_2 k \rfloor} \cdot \poly(\log(1/\eps))\,. \]
\end{theorem}
The particular dependence on $k$ and $\log(1/\eps)$ obtained from Theorem~\ref{thm:ub} is on the order of $(k \log(1/\eps))^{O(k^2)}$.
The algorithm is divided into two cases, corresponding to the two outcomes from an application of Theorem~\ref{thm:tree}. Suppose $f \colon [n] \to \R$ is a function which is $\eps$-far from being $(12\dots k)$-free. By Theorem~\ref{thm:tree} one of the followin holds, where $C > 0$ is a large constant.
\begin{description}
\item[Case 1:]
there exist $\alpha \ge \eps / \polylog(1/\eps)$ and a set $H \subseteq [n]$ of $(\alpha, Ck\alpha)$-growing suffixes where $\alpha |H| \geq \eps n / \polylog(1/\eps)$, or
\item[Case 2:]
there exist a set $T \subseteq [n]^k$ of disjoint, length-$k$ monotone sequences, that satisfies $|T| \geq \eps n / (\polylog(1/\eps))$, and a $k$-tree descriptor $(G, \varrho, \mathsf{I})$ which represents $(f, T, [n])$.
\end{description}
Theorem~\ref{thm:ub} follows from analyzing the two cases independently, and designing an algorithm for each.
\newcommand{\texttt{Sample-Suffix}}{\texttt{Sample-Suffix}}
\begin{lemma}[Case 1]\label{lem:case1}
Consider any fixed value of $k \in \N$, and let $C > 0$ be a large enough constant. There exists a non-adaptive and randomized algorithm, $\emph{\texttt{Sample-Suffix}}_k(f, \eps)$ which takes two inputs: query access to a function $f \colon [n] \to \R$ and a parameter $\eps > 0$. Suppose there exist $\alpha \in (0,1)$ and a set $H \subseteq [n]$ of $(\alpha, Ck \alpha)$-growing suffixes satisfying $\alpha |H| \geq \eps n / \polylog(1/\eps)$,\footnote{Here we think of $k$ as fixed, so $\polylog(1/\eps)$ is allowed to depend on $k$. In this lemma, the expression stands for $(k \log (1/\eps))^k$.} then $\emph{\texttt{Sample-Suffix}}_k(f, \eps)$ finds a length-$k$ monotone subsequence of $f$ with probability at least $9/10$.
The query complexity of $\emph{\texttt{Sample-Suffix}}_k(f, \eps)$ is at most
\[
\frac{\log n}{\eps} \cdot \polylog(1/\eps).
\]
\end{lemma}
Lemma~\ref{lem:case1} above, which corresponds to the first case of Theorem~\ref{thm:tree}, is proved in Section~\ref{sec:case1}.
\newcommand{\texttt{Sample-Splittable}}{\texttt{Sample-Splittable}}
\begin{lemma}[Case 2]\label{lem:case2}
Consider any fixed value of $k \in \N$. There exists a non-adaptive, randomized algorithm, $\emph{\texttt{Sample-Splittable}}_k(f, \eps)$ which takes two inputs: query access to a sequence $f \colon [n] \to \R$ and a parameter $\eps > 0$. Suppose there exists a set $T \subseteq [n]^k$ of disjoint, length-$k$ monotone subsequences of $f$ where $|T| \geq \eps n / \polylog(1/\eps)$,\footnote{in this case the $\polylog(1/\eps)$ term stands for $(k \log(1/\eps))^{O(k^2)}$} as well as a $(k, k, \alpha)$-tree descriptor $(G, \varrho, \mathsf{I})$ that represents $(f, T, [n])$, where $\alpha \ge \eps / \polylog(1/\eps)$, then $\emph{\texttt{Sample-Splittable}}_k(f, \eps)$ finds a length-$k$ monotone subsequence of $f$ with probability at least $9/10$. The query complexity of $\emph{\texttt{Sample-Splittable}}_k(f, \eps)$ is at most
\[ \frac{1}{\eps} \left(\frac{\log n}{\eps} \right)^{\lfloor \log_2 k \rfloor} \cdot \polylog(1/\eps). \]
\end{lemma}
\begin{proof}[Proof of Theorem~\ref{thm:ub} assuming Lemmas~\ref{lem:case1} and~\ref{lem:case2}]
The algorithm $\texttt{Sampler}_k(f, \eps)$ executes both $\texttt{Sample-Suffix}_k(f, \eps)$ and $\texttt{Sample-Splittable}_k(f, \eps)$; if either algorithm finds a length-$k$ monotone subsequence of $f$, output such a subsequence. We note that by Theorem~\ref{thm:tree}, either case 1, or case 2 holds. If case 1 holds, then by Lemma~\ref{lem:case1}, $\texttt{Sample-Suffix}(f, \eps)$ outputs a length-$k$ monotone subsequence with probability at least $9/10$, and if case 2 holds, then by Lemma~\ref{lem:case2}, $\texttt{Sample-Splittable}_k(f, \eps)$ outputs a length-$k$ monotone subsequence with probability at least $9/10$. Thus, regardless of which case holds, a length-$k$ monotone subsequence will be found with probability at least $9/10$. The query complexity then follows from the maximum of the two query complexities.
\end{proof}
\subsection{Proof of Lemma~\ref{lem:case1}: an algorithm for growing suffixes}\label{sec:case1}
We now prove Lemma~\ref{lem:case1}. Let $C > 0$ be a large constant, and let $k \in \N$ be fixed. Let $\eps > 0$ and $f \colon [n] \to \R$ be a function which is $\eps$-far from $(12\dots k)$-free. Furthermore, as per the assumption of case 1 of the algorithm, we assume that there exists a parameter $\alpha \in (0, 1)$ as well as a set $H \subseteq [n]$ of $(\alpha, Ck\alpha)$-growing suffixes, where $\alpha |H| \geq \eps n / \polylog(1/\eps)$.
\newcommand{\texttt{Growing-Suffix}}{\texttt{Growing-Suffix}}
\begin{figure}[ht!]
\begin{framed}
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{Growing-Suffix}\hspace{0.05cm}(f, \alpha_0, a)$
\vspace{0.3cm}
\noindent {\bf Input:} Query access to a function $f \colon [n] \to \R$, a parameter $\alpha_0 \in (0, 1)$, and an index $a \in [n]$.\\
{\bf Output:} a subset of $k$ indices $i_1 < \dots < i_k$ where $f(i_1) < \dots <f(i_k)$, or \fail.
\begin{enumerate}
\item
Let $\eta_a = \lceil \log(n-a) \rceil$ and consider the sets $S_j(a) = (a + \ell_{j-1}, a+\ell_j] \cap [n]$ for all $j \in [\eta_a]$ and $\ell_j = 2^j$.
\item
For each $j \in [\eta_a]$, let $\bA_j \subseteq S_j(a)$ be obtained by sampling uniformly at random $T \eqdef 1 / \alpha_0$ times from $S_j(a)$.
\item
For each $j \in [\eta_a]$ and each $b \in \bA_j$, query $f(b)$ .
\item
If there exist indices $i_1, \dots, i_k \in \bA_1 \cup \dots \cup \bA_{\eta_i}$ satisfying $i_1 < \dots < i_k$ and $f(i_1) < \dots < f(i_k)$, \Return such indices $i_1, \dots, i_k$. Otherwise, \Return \fail.
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{Growing-Suffix}$ subroutine.} \label{alg:growing-suffix}
\label{fig:growing-suffix-alg}
\end{figure}
The algorithm, which underlies the result of Lemma~\ref{lem:case1}, proceeds by sampling uniformly at random an index $\ba \sim [n]$, and running a sub-routine which we call $\texttt{Growing-Suffix}$, with $\ba$ as input. The sub-routine is designed so that if $\ba$ is the start of an $(\alpha, Ck\alpha)$-growing suffix then the algorithm will find a length-$k$ monotone subsequence of $f$ with probability at least $99/100$. The sub-routine, $\texttt{Growing-Suffix}$, is presented in Figure~\ref{fig:growing-suffix-alg}.
\begin{lemma}\label{lem:growing}
Let $f \colon [n] \to \R$ be a function, let $\alpha, \alpha_0, \beta \in (0, 1)$ be parameters satisfying $\beta \ge Ck \alpha$ and $\alpha_0 \le \alpha$, and suppose that $a \in [n]$ starts a $(\alpha, \beta)$-growing suffix in $f$. Then $\emph{\texttt{Growing-Suffix}}(f, \alpha_0, a)$ finds a length-$k$ monotone subsequence of $f$ with probability at least $99/100$.
\end{lemma}
\begin{proof}
Recall, from Definition~\ref{def:growing-suffixes}, that if $a \in [n]$ is the start of a $(\alpha, \beta)$-growing suffix of $f$ then there exist a collection of sets, $D_{1}(a), \dots, D_{\eta_a}(a)$ and parameters $\delta_1(a), \dots, \delta_{\eta_a}(a) \in (0, \alpha]$, where every $j \in [\eta_a]$ has
\[
D_j(a) \subseteq S_j(a), \qquad
|D_j(a)| = \delta_j(a) \cdot |S_j(a)|,\qquad
\text{and} \qquad \sum_{j=1}^{\eta_a} \delta_j(a) \geq \beta.
\]
Further, if, for some $j_1, \dots, j_k \in [\eta_i]$, we have $j_1 < \dots < j_k$ and for all $\ell \in [k]$, $\bA_{j_{\ell}} \cap D_{j_{\ell}}(a) \neq \emptyset$, then the union $D_{j_1}(a) \cup \ldots \cup D_{j_k}(a)$ contains a length-$k$ monotone subsequence. In view of this, for each $j \in [\eta_a]$, consider the indicator random variable
\[ \bE_j \eqdef \ind\{ \bA_{j} \cap D_j(a) \neq \emptyset \}, \]
and observe that by the foregoing discussion $\texttt{Growing-Suffix}(f, \alpha_0, a)$ samples a length-$k$ monotone subsequence of $f$ whenever $\sum_{j=1}^{\eta_a} \bE_j \geq k$.
We note that the $\bE_j$'s are independent, and that
\[
\Prx[\bE_j = 1] = 1 - \left( 1 - \delta_j(a)\right)^T \geq \min\left\{ \frac{T \cdot \delta_j(a)}{10}, \frac{1}{10} \right\}.
\]
Let $J \subseteq [\eta_a]$ be the set of indices satisfying $T \cdot \delta_j(a) \geq 1$ (recall that $T = 1 / \alpha_0$). Then, if $|J| \geq C k$ we have
\[
\Ex\!\left[ \sum_{j=1}^{\eta_a} \bE_j \right] \geq \frac{C k}{10},
\]
since every variable $j \in J$ contributes at least $1/10$.
On the other hand, if $|J| \leq C k / 2$, then, since $\delta_j(a) \leq \alpha$ for every $j$, we have $\sum_{j \in [\eta_a] \setminus J} \delta_j(a) \geq \beta - |J| \cdot \alpha \geq \beta / 2$ (using $\beta \ge Ck \alpha$) so that
\[
\Ex\!\left[ \sum_{j=1}^{\eta_a} \bE_j\right] \geq \Ex\!\left[ \sum_{j\in [\eta_a] \setminus J} \bE_j \right] \geq \frac{T}{10} \cdot \frac{\beta}{2} \geq \frac{Ck}{20}.
\]
In either case, $\Ex[\sum_{j\in[\eta_a]} \bE_j] \geq C k / 20$, and since the events $\bE_i$ are independent, via a Chernoff bound we obtain that $\sum_j \bE_j$ is larger than $k$ with probability at least $99/100$.
\end{proof}
\begin{figure}[ht!]
\begin{framed}
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{Sample-Suffix}_k\hspace{0.05cm}(f, \eps)$
\vspace{.3cm}
\noindent {\bf Input:} Query access to a function $f \colon [n] \to \R$, and a parameter $\eps \in (0, 1)$.\\
{\bf Output:} a subset of $k$ indices $i_1 < \dots < i_k$ where $f(i_1) < \dots <f(i_k)$, or \fail.
\begin{enumerate}
\item Repeat the following for all $j = 1, \dots, O(\log(1/\eps))$, letting $\alpha_j = 2^{-j}$:
\begin{itemize}
\item For $t_j = \alpha_j \cdot \polylog(1/\eps) /\eps$ iterations, sample $\ba \sim [n]$ uniformly at random and run $\texttt{Growing-Suffix}(f, \alpha_j, \ba)$, and if it returns a length-$k$ monotone subsequence of $f$, \Return that subsequence.
\end{itemize}
\item If the algorithm has not already output a monotone subsequence, \Return \fail.
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{Sample-Suffix}$ subroutine.} \label{fig:sample-suffix}
\end{figure}
\noindent With this in hand, we can now establish~Lemma~\ref{lem:case1}.
\begin{proof}[Proof of Lemma~\ref{lem:case1}]
First, note that the query complexity of $\texttt{Sample-Suffix}_k(f, \eps)$ is
\[
\sum_{j=1}^{O(\log(1/\eps))} t_j \cdot O( \log n / \alpha_j) = \frac{\log n \cdot \polylog(1/\eps)}{\eps}.
\]
Consider the iteration of $j$ where $\alpha_j \leq \alpha \leq 2 \alpha_j$ (note that since $\alpha \ge \eps / \polylog(1 / \eps)$, there exists such $j$). Then, since $|H| \ge \eps / (\alpha \cdot \polylog(1/\eps))$, we have that $t_j \ge Cn / |H|$ (for a sufficiently large constant $C$). Thus, with probability at least $99/100$, some iteration satisfies $\ba \in H$. When this occurs, $\texttt{Growing-Suffix}(f, \alpha_j, \ba)$ will output a length-$k$ monotone subsequence with probability at least $99/100$, by Lemma~\ref{lem:growing}, and thus by a union bound we obtain the desired result.
\end{proof}
\subsection{Proof of Lemma~\ref{lem:case2}: an algorithm for splittable intervals}\label{sec:case2}
\newcommand{\texttt{Sample-Helper}}{\texttt{Sample-Helper}}
We now prove Lemma~\ref{lem:case2}. We consider a fixed setting of $k \in \N$ and $\eps > 0$, and let $f \colon [n] \to \R$ be any sequence which is $\eps$-far from being $(12\dots k)$-free. Furthermore, as per case 2 of the algorithm, we assume that there exists a set $T \subseteq [n]^k$ of disjoint length-$k$ monotone subsequences of $f$ where
\[
|T| \geq \frac{\eps n}{\polylog(1/\eps)},
\]
and $(G, \varrho, \mathsf{I})$ is a $(k, k, \alpha)$-tree descriptor which represents $(f, T, [n])$, where $\alpha \ge \eps / \polylog(1/\eps)$. In what follows, we describe a sub-routine, $\texttt{Sample-Splittable}_k(f, \eps)$ in terms of two parameters $\rho, q \in \R$. The parameter $\rho > 0$ is set to be sufficiently large and independent of $n$, satisfying
\begin{equation}\label{eq:case2:setting:rho}
\rho \geq \frac{\eps}{\polylog(1/\eps)}.
\end{equation}
One property which we will want to satisfy is that if we take a random subset of $[n]$ by including each element independently with probability $1/(\rho n)$, we will include an element belonging to $E(T)$ with probability at least $1 - 1/(C k)$, for a large constant $C > 0$.
The parameter $q$ will be an upper bound on the query complexity of the algorithm, which we set to a high enough value satisfying:
\[ q = O\left(\frac{1}{\rho} \left( \frac{\log n}{\rho}\right)^{\lfloor \log_2 k \rfloor}\right) \leq \frac{1}{\eps} \cdot \left( \frac{\log n}{\eps} \right)^{\lfloor \log_2 k \rfloor} \cdot \polylog(1/\eps). \]
\begin{figure}[ht!]
\begin{framed}
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{Sample-Splittable}_k\hspace{0.05cm}(f, \eps)$
\vspace{.3cm}
\noindent {\bf Input:} Query access to a sequence $f \colon [n] \to \R$, and a parameter $\eps \in (0, 1)$.\\
{\bf Output:} a subset of $k$ indices $i_1 < \dots < i_k$ where $f(i_1) < \dots <f(i_k)$, or \fail.
\begin{enumerate}
\item Let $r = \lfloor \log_2 k \rfloor$ and run $\texttt{Sample-Helper}(r,[n],\rho)$, to obtain a set $\bA \subseteq [n]$.
\item If $|\bA| > q$, \Return \fail; otherwise, for each $a \in \bA$, query $f(a)$. If there exists a monotone sequence of $f$ of length $k$, then \Return that subsequence. If not, \Return \fail.
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{Sample-Splittable}$ subroutine.} \label{fig:sample-splittable}
\end{figure}
\begin{figure}[ht!]
\begin{framed}
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{Sample-Helper}\hspace{0.05cm}(r,I, \rho)$
\vspace{.3cm}
\noindent {\bf Input:} An integer $r \in \N$, an interval $I \subseteq [n]$, and a parameter $\rho \in (0, 1)$.\\
{\bf Output:} a subset of $A \subseteq I$.
\begin{enumerate}
\item Let $\bA_0 = \emptyset$. For every index $a \in I$, let $\bA_0 \gets \bA_0 \cup \{ a \}$ with probability $1 / (\rho |I|)$.
\item If $r = 0$, \Return $\bA_0$.
\item If $r > 0$, proceed with the following:
\begin{itemize}
\item
For every index $a \in \bA_0$, consider the $O(\log n)$ intervals given by $B_{a, j} = [a - \ell_j, a + \ell_j]$, for $j = 1, \dots,O(\log n)$ and $\ell_j = 2^j$, and let $\bR_{a, j} \gets \texttt{Sample-Helper}(r-1, B_{a, j}, \rho)$.
\item
Let $\bA$ be the set
\[
\bA \gets \bigcup_{\substack{a \in \bA_0, \, j = O(\log n)}} \bR_{a,j}.
\]
\item \Return the set $(\bA_0 \cup \bA) \cap I$.
\end{itemize}
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{Sample-Helper}$ subroutine.} \label{fig:sample-splittable-2}
\end{figure}
The descriptions of the main algorithm $\texttt{Sample-Splittable}_k$ and the sub-routine $\texttt{Sample-Helper}$, are given in Figure~\ref{fig:sample-splittable} and Figure~\ref{fig:sample-splittable-2}. Note that, for any $r \in \N$, if we let $\calD_r$ be the distribution of $|\bA|$, where $\bA$ is the output of a call to $\texttt{Sample-Helper}(r, [n], \rho)$. Then, we have that $\calD_0 = \Bin(n, \rho)$, and for $r > 0$, $\calD_r$ is stochastically dominated by the random variable
\[ \sum_{i=1}^{\by_0} \sum_{j=1}^{O(\log n)} \bx_{r-1}^{(i, j)}, \]
where $\by_0 \sim \Bin(n, 1/(\rho n))$ and $\bx_{r-1}^{(i,j)} \sim \calD_{r-1}$ for all $i \in \N$ and $j \in [O(\log n)]$ are all mutually independent. As a result, for $r \geq 1$,
\[ \Ex\left[ |\bA| \right] \leq \frac{1}{\rho} \cdot \log n \cdot \Ex_{\bx \sim \calD_{r-1}}[\bx], \]
and since $\Ex_{\bx \sim\calD_0}[\bx] = 1/\rho$, we have:
\[ \Ex\left[ |\bA| \right] \leq \frac{1}{\rho} \left( \frac{\log n}{\rho} \right)^{r}. \]
We may then apply Markov's inequality to conclude that $|\bA| \leq q$ with probability at least $99/100$. As a result, we focus on proving that the probability that the set $\bA$ contains a monotone subsequence of $f$ of length $k$ is at least $99/100$. This would imply the desired result by taking a union bound.
In addition to the above, we define another algorithm, $\texttt{Sample-Helper}^*$, in Figure~\ref{fig:sample-help}, which will be a \emph{helper} sub-routine. We emphasize that $\texttt{Sample-Helper}^*$ is not executed in the algorithm itself, but will be useful in order to analyze $\texttt{Sample-Helper}$.
\begin{figure}[H]
\begin{framed}
\begin{minipage}{.98\textwidth}
\noindent Subroutine $\texttt{Sample-Helper}^*\hspace{0.05cm}(r,I, \rho, \calI)$
\vspace{.3cm}
\noindent {\bf Input:} An integer $r \in \N$, an interval $I \subseteq [n]$, a parameter $\rho \in (0, 1)$, and a collection of disjoint intervals $\calI$ of $[n]$.\\
{\bf Output:} two subsets $\bA, \bA_0 \subseteq I$.
\begin{enumerate}
\item\label{sample:helper*:step:1} Let $\bA_0 = \emptyset$. For every index $a \in I$ which lies inside an interval in $\calI$, let $\bA_0 \gets \bA_0 \cup \{ a \}$ with probability $1 / (\rho |I|)$.
\item\label{sample:helper*:step:2} If $r = 0$, \Return $\bA_0$.
\item\label{sample:helper*:step:3} If $r > 0$, proceed with the following:
\begin{itemize}
\item
For every index $a \in \bA_0$, consider the $O(\log n)$ intervals given by $B_{a, j} = [a - \ell_j, a + \ell_j]$, for $j = 1, \dots O(\log n)$, and $\ell_j = 2^j$, and let $(\bR_{a, j}, \bR_{a, j, 0}) \gets \texttt{Sample-Helper}^*(r-1, B_{a, j}, \rho, \calI)$.
\item
Let $\bA$ to be the set
\[
\bA \gets \bigcup_{\substack{a \in \bA_0, \,\, j = O(\log n)}} \bR_{a,j}.
\]
\item
\Return the set $(\bA \cap I, \bA_0 \cap I)$.
\end{itemize}
\end{enumerate}
\end{minipage}
\end{framed}\vspace{-0.2cm}
\caption{Description of the $\texttt{Sample-Helper}^*$ subroutine.} \label{fig:sample-help}
\end{figure}
Before proceeding, we require a ``coupling lemma.'' Its main purpose is to prove the intuitive fact that if $\calI_0, \calI_1$ are collections of disjoint intervals, and the latter is a refinement of the former (namely, each intervals in $\calI_1$ is contained in an interval of $\calI_0$), then $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_0)$ is more likely to find a length-$k$ monotone subsequence than $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_1)$ does.
\begin{lemma}\label{lem:coupling}
Let $r \in \N$ be an integer, $f \colon [n] \to \R$ a function, $\rho \in (0,1)$ a parameter, and $\calI_0$ and $\calI_1$ collections of disjoint intervals in $[n]$, such that each interval in $\calI_1$ lies inside an interval from $\calI_0$.
Denote by $(\bA^{(i)}, \bA_0^{(i)})$ the random pair of sets given by the output of $\emph{\texttt{Sample-Helper}}^*(r, [n], \rho, \calI_i)$, for $i = 0,1$.
Lastly, let $\emph{\ensuremath{\mathcal{E}}} \colon \calP([n]) \times \calP([n]) \to \{0,1\}$ be any \emph{monotone} function; that is, it satisfies $\emph{\ensuremath{\mathcal{E}}}(S_1, S_2) \leq \emph{\ensuremath{\mathcal{E}}}(S_1', S_2')$ for any $S_1 \subseteq S_1' \subseteq [n]$ and $S_2 \subseteq S_2' \subseteq [n]$. Then,
\[ \Prx[ \emph{\ensuremath{\mathcal{E}}}(\bA^{(0)}, \bA^{(0)}_0) = 1] \geq \Prx[ \emph{\ensuremath{\mathcal{E}}}(\bA^{(1)}, \bA^{(1)}_0) = 1]. \]
\end{lemma}
\begin{proof}
Consider an execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_0)$ which outputs a pair $(\bA^{(0)}, \bA_0^{(0)})$. Let $\bA^{(1)}$ and $\bA^{(1)}$ be the subsets of $\bA^{(0)}$ and $\bA^{(0)}$, respectively, obtained by running a parallel execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_1)$, which follows the execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_0)$, but whenever an element which is not in an interval of $\calI_1$ is considered, it is simply ignored (i.e., it is not included in $\bA^{(0)}$ or in $\bA_0^{(0)}$ and no recursive calls based on such elements are made). It is easy to see that this coupling yields a pair $(\bA^{(1)}, \bA_0^{(1)})$ with the same distribution as that given by running $\texttt{Sample-Helper}^*(r, [n], \rho, \calI_1)$.
As $\emph{\ensuremath{\mathcal{E}}}(\cdot, \cdot)$ is increasing, if $\emph{\ensuremath{\mathcal{E}}}(\bA^{(0)}, \bA^{(0)}_0)$ holds then so does $\emph{\ensuremath{\mathcal{E}}}(\bA^{(1)}, \bA^{(1)}_0)$. The lemma follows.
\end{proof}
The following corollary is a direct consequence of Lemma~\ref{lem:coupling}. Specifically, we use the facts that $\texttt{Sample-Splittable}_k(f, \eps)$ calls $\texttt{Sample-Helper}(\lfloor \log_2 k \rfloor, [n], \rho)$, which is equivalent to calling $\texttt{Sample-Helper}(\lfloor \log_2 k \rfloor, [n], \rho, \{ [n] \})$, and that finding a $(12\dots k)$-pattern in $\calI$ is a monotone event.
\begin{corollary}\label{cor:helper}
Let $\calI$ be any collection of disjoint intervals in $[n]$. Suppose $(\bA, \bA_0)$ is the random pair of sets given by the output of $\emph{\texttt{Sample-Helper}}^*(\lfloor \log_2 k \rfloor, n, \rho, \calI)$, then,
\begin{align*}
&\Prx[\emph{\texttt{Sample-Splittable}}_k(f, \eps) \text{ finds a $(12\dots k)$-pattern of $f$}] \geq \\
&\qquad\qquad\qquad\qquad\qquad\qquad\Prx[\bA \text{ contains a $(12\dots k)$-pattern in $f_{|\calI}$} ].
\end{align*}
\end{corollary}
\begin{definition}
Let $k_0 \in \N$ be a positive integer, and let $(G, \varrho)$ be a $k_0$-tree descriptor (for this definition we do not care about the third component of the descriptor, $\mathsf{I}$). We say that $p \in [k_0]$ is the \emph{primary index} of $(G, \varrho)$ if the leaf with rank $p$ under $\leq_{G}$ is the unique leaf whose root-to-leaf path $(u_1, \dots, u_d)$ satisfies the following: for each $d' \in [d-1]$, denoting the left and right children of $u_{d'}$ by $v_l$ and $v_r$, respectively, $u_{d' + 1}$ is $v_l$ if the number of leaves in the subtree rooted at $v_l$ is at least the number of leaves in the subtree rooted at $v_r$, and otherwise, $u_{d' + 1}$ is $v_r$.
\end{definition}
From Corollary~\ref{cor:helper}, we note that Lemma~\ref{lem:case2} follows from the following lemma.
\begin{lemma} \label{lem:case2-alg}
Let $k, k_0, n \in \N$ satisfy $1 \leq k_0 \leq k$, let $C$ be a large enough constant, and let $\alpha, \rho \in (0,1)$ be such that $\rho \ge C \alpha$ and $\alpha \ge \rho / \polylog(1/\rho)$.
Let $f \colon [n] \to \R$ be a function, let $\calI$ be a collection of disjoint intervals in $[n]$, for each $I \in \calI$ let $T_I \subseteq I^{k_0}$ be a set of disjoint, length-$k_0$ monotone subsequence of $f$, and suppose that
\[
\sum_{I \in \calI} |T_I| \ge \alpha n / 4.
\]
Suppose that $(G, \varrho)$ is a $(k, k_0, \alpha)$-weighted-tree such that for every $I \in \calI$ there exists a function $\mathsf{I}_I :V(G) \to \calS(I)$, such that $(G, \varrho, \mathsf{I}_I)$ is a tree descriptor that represents $(f, T_I, I)$.
Given any $r \in \N$ satisfying $\lfloor \log_2 k_0 \rfloor \leq r$, let $(\bA, \bA_0)$ be the pair of sets output by the sub-routine $\emph{\texttt{Sample-Helper}}^*(r, [n], \rho, \calI)$.
With probability at least $1 - k_0 / (100 k)$, there exist indices $i_1, \dots, i_{k_0} \in [n]$ with the following properties.
\begin{enumerate}
\item
$(i_1, \ldots, i_{k_0})$ is a length-$k_0$ monotone subsequence of $f$.
\item
There is an interval $I \in \calI$ such that $i_1, \ldots, i_{k_0} \in I \cap E(T_I)$.
\item
$i_1, \ldots, i_{k_0} \in \bA$ and $i_p \in \bA_0$, where $p$ is the primary index of $(G, \varrho)$.
\end{enumerate}
\end{lemma}
\begin{proof}
The proof proceeds by induction on $k_0$. Consider the base case, when $k_0 = 1$. In this case, $\lfloor \log_2 k_0 \rfloor = 0$, so for any $r \geq 0$, $\texttt{Sample-Helper}^*(r, [n],\rho, \calI)$ runs step~\ref{sample:helper*:step:1}. As a result, $\texttt{Sample-Helper}^*(r, [n], \rho,\calI)$ samples each element inside an interval of $\calI$ independently with probability $1/(\rho n)$. In order to satisfy the requirements of the lemma in this case, we need $\bA_0$ to contain an element of $\cup_{I \in \calI} T_I$. By the assumption on the size of this union, and because each of the elements of the union lives inside some interval from $\calI$, such an element will exist with sufficiently high probability via a Chernoff bound.
For the inductive step, assume that Lemma~\ref{lem:case2-alg} is fulfilled whenever $k_0 < K$, for $K \in \N$ satisfying $1 < K \leq k$, and we will prove, assuming this inductive hypothesis, that Lemma~\ref{lem:case2-alg} holds for $k_0 = K$. So consider a setting $k_0 = K$. Let $\calI$, $(G, \varrho)$ and $\mathsf{I}_I$ be as in the statement of the lemma. Denote the root of $(G, \varrho)$ by $v_{\sf root}$, and its left and right children by $v_{\sf left}$ and $v_{\sf right}$. Let $c$ be the number of leaves in the subtree $(G_{\sf left}, \varrho_{\sf left})$ rooted at $v_{\sf left}$, so $k_0 - c$ is the number of leaves in the subtree $(G_{\sf right}, \varrho_{\sf right})$ rooted at $v_{\sf right}$. We shall assume that $c \ge k_0 - c$; the other case follows by an analogous argument.
For each $I \in \calI$, the collection of pairs $(J, T_{I,J})$, where $J \in \mathsf{I}_I(v_{\sf root})$ and $T_{J} = T_I \cap J^{k_0}$ is the restriction of $T_I$ to $J$, is a $(c, 1/(6k), \alpha)$-splittable collection of $I$. Let $\calJ$ be the collection of all such intervals $J$ (note that they are pairwise disjoint and that $\calJ$ is a refinement of $\calI$).
Let $(L_J, M_J, R_J)$ be the partition of $J$ into left, middle and right intervals, respectively, and let $T_J^{(L)}$ and $T_J^{(R)}$ be sets of $c$-prefixes and $(k_0 - c)$-suffixes of $k_0$-tuples from $T_{I, J}$, as given by Definition~\ref{def:splittable}. Set
\[
\calL = \{L_J : J \in \calJ\}, \qquad \calR = \{R_J : J \in \calJ\}, \qquad T^{(L)} = \bigcup_{J \in \calJ} T_J^{(L)}, \qquad T^{(R)} = \bigcup_{J \in \calJ} T_J^{(R)}.
\]
Note that $(G_{\sf left}, \varrho_{\sf left}, \mathsf{I}_{J, {\sf left}})$ is a $(k, c, \alpha)$-tree descriptor for $(f, T_{J}, J)$, with appropriate $\mathsf{I}_{J, {\sf left}}$. Similarly, $(G_{\sf right}, \varrho_{\sf right}, \mathsf{I}_{J, {\sf right}})$ is a $(k, k_0 - c, \alpha)$-tree descriptor for $(f, T_{J}, J)$, with appropriate $\mathsf{I}_{J, {\sf right}}$.
\newcommand{ {\sf a}}{ {\sf a}}
We consider an execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI)$ which outputs a random pair of sets $(\bA, \bA_0)$. Let $\bA^{(L)}$ and $\bA_0^{(L)}$ be the subsets of $\bA$ and $\bA_0$, respectively, obtained by running a parallel execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calL)$, where, as in the proof of Lemma~\ref{lem:coupling}, we follow the execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI)$, but whenever an element which is not in $\calL$ is considered, we ignore it. As stated above, this coupling yields a pair $(\bA^{(L)}, \bA_0^{(L)})$ with the distribution given by running $\texttt{Sample-Helper}^*(r, [n], \rho, \calL)$.
For $a \in \bA_0^{(L)}$, and any $j \in [O(\log n)]$, let $(\bA^{(a,j)}, \bA_0^{(a,j)})$ be the output of the recursive call (inside the execution of $\texttt{Sample-Helper}^*(r, [n], \rho, \calI)$) of $\texttt{Sample-Helper}^*(r-1, B_{a, j}, \rho, \calR)$.
We define the collection:
\[
\calS = \left\{ (S_0, S) : \begin{array}{l}
S_0 \subseteq S \subseteq E(T^{(L)}) \\
\text{there exist $i_1, \dots, i_c \in S$ forming a $(12\dots c)$-pattern such that $i_p \in S_0$} \\
\text{there exist $J \in \calJ$ such that $i_1, \dots, i_c \in L_J$}
\end{array} \right\}.
\]
For each $(S_0, S) \in \calS$, we let $ {\sf a}(S_0, S) \in E(T^{(L)})$ be some $i_{p} \in S$ such that there exist $c-1$ indices $i_1, \dots, i_{p-1}, i_{p+1}, i_{c}$, such that $(i_1, \ldots, i_c)$ forms a $(12\dots c)$-pattern in $S$, and $i_1, \ldots, i_p \in L_J$ for some $J \in \calJ$. Let $\texttt{seg}(S_0, S)$ be this interval $J$, and let $\texttt{len}(S_0, S) \in [O(\log n)]$ be the smallest $j$ for which $R_J \subseteq B_{a,j}$, where $a = {\sf a}(S_0, S)$.
Let $\bE_L$ be the event that
\[
\left(\bA^{(L)} \cap E(T^{(L)}), \bA_0^{(L)} \cap E(T^{(L)}) \right) \in \calS,
\]
and let $\bE_L(S_0, S)$ be the event that
\[
\bA^{(L)} \cap E(T^{(L)}) = S_0 \qquad \qquad \bA_0^{(L)} \cap E(T^{(L)}) = S,
\]
so $\bE_L = \cup_{(S_0, S) \in \calS} \bE_L(S_0, S)$, and the events $\bE_L(S_0, S)$ are pairwise disjoint.
By the induction hypothesis, applied with the family $\{L_J : J \in \calJ\}$ and the corresponding sets $T_J^{(L)}$ (using $\sum_{J \in \calJ} |T_J^{(L)}| = \sum_{J \in \calJ} |T_J| \ge \alpha n / 4$), we have
\[
\Pr[\bE_L] \ge 1 - c/(100k).
\]
Let $\bE_R(a, j)$ be the event that $a \in \bA_0$, and in the recursive run of $\texttt{Sample-Helper}^*(r-1, B_{a, j}, \rho, \calR)$ inside $\texttt{Sample-Helper}^*(r, [n], \rho, \calI)$, there exist indices $i_1', \ldots, i_{k_0-c}'$ such that
\begin{itemize}
\item
$(i_1', \ldots, i_{k_0-c}')$ form a length $(k_0-c)$-monotone subsequence.
\item
$i_1', \ldots, i_{k_0-c}' \in E(T_J^{(R)})$, where $J$ is the interval in $\calJ$ with $i \in J$.
\item
$i_1', \ldots, i_{k_0-c}' \in \bA^{(a,j)}$ and $i_q' \in \bA_0^{(a,j)}$, where $q$ is the primary index of $(G_{\sf right}, \varrho_{\sf right})$.
\end{itemize}
Let $\bF_R(a, j)$ be the event that in a run of $\texttt{Sample-Helper}^*(r-1, B_{a, j}, \rho, \calR)$, there exist $i_1', \ldots, i_{k_0-c}'$ as above.
Fix some $(S_0, S) \in \calS$, and let $a = {\sf a}(S_0, S)$, $J = \texttt{seg}(S_0, S)$ and $j = \texttt{len}(S_0, S)$. We claim that
\[
\Pr[\bE_R(a, j) \,\,|\,\, \bE_L(S_0, S)] = \Pr[\bF_R(a, j)].
\]
Indeed, by conditioning on $\bE_L(S_0, S)$ we know that $a \in \bA_0$, so there will be a recursive run of $\texttt{Sample-Helper}^*(r-1, B_{a, j}, \rho, \calR)$, and moreover the event $\bE_L(S_0, S)$ will have no influence on the outcomes of this run.
Note that $|T_J^{(R)}| \ge \alpha |R_J| \ge \alpha |B_{a, J}| / 4$.
By the induction hypothesis, applied with the interval $B_{a, J}$ in place of $[n]$, the family $\{R_J\}$ and the corresponding set $T_J^{(R)}$, and the tree $(G_{\sf right}, \varrho_{\sf right})$, we find that $\Pr[\bF_R(a, j)] \ge 1 - (k_0-c)/(100k)$.
We note that if both $\bE_L(S_0, S)$ and $\bE_R(a, j)$ hold, then there are indices $i_1, \ldots, i_c, i_1', \ldots, i_{k_0 - c}'$ such that
\begin{itemize}
\item
$(i_1, \ldots, i_c)$ is a length-$c$ monotone subsequence in $E(T_J^{(L)})$, and $(i_1', \ldots, i_{k_0 - c}')$ is a length-$c$ monotone subsequence in $E(T_J^{(R)})$. In particular, $(i_1, \ldots, i_c, i_1', \ldots, i_{k_0 - c}')$ is a length-$k_0$ monotone subsequence that lies in $E(T_j)$.
\item
$i_1, \ldots, i_c, i_1', \ldots, i_{k_0 - c}' \in \bA$ and $i_p \in \bA_0$ (recall that $p$ is the primary index of both $G$ and $G_{\sf left}$).
\end{itemize}
I.e.\ if these two events hold, then the requirements ot the lemma are satisfied.
It follows that the requirements ot the lemma are satisfied with at least the following probability, using the fact that the events $\bE_L(S_0, S)$ are disjoint.
\begin{align*}
& \sum_{(S_0, S) \in \calS} \Pr[\bE_R( {\sf a}(S_0,S), \texttt{len}(S_0,S)) \text{ and } \bE_L(S_0, S)] \\
= & \sum_{(S_0, S) \in \calS} \Pr[\bE_R( {\sf a}(S_0,S), \texttt{len}(S_0,S)) \,\, | \,\, \bE_L(S_0, S)] \times \Pr[\bE_L(S_0, S)] \\
\ge & \sum_{(S_0, S) \in \calS} \Pr[\bF_R( {\sf a}(S_0,S), \texttt{len}(S_0,S))] \times \Pr[\bE_L(S_0, S) ]\\
\ge & \left(1 - \frac{k_0-c}{100k}\right) \cdot \sum_{(S_0, S) \in \calS} \Pr[\bE_L(S_0, S)] \\
\ge & \left(1 - \frac{k_0-c}{100k}\right) \cdot \Pr[\bE_L] \\
\ge & \left(1 - \frac{k_0-c}{100k}\right) \cdot \left(1 - \frac{c}{100k}\right) \ge 1 - \frac{k_0}{100k}.
\end{align*}
This completes the proof of Lemma~\ref{lem:case2-alg}.
\end{proof}
\section{Lower Bounds}\label{sec:lowerbounds}
In this section, we prove our lower bound for non-adaptive testing of $(12\dots k)$-freeness with one-sided error, Theorem~\ref{thm:intro-lb}. Below we give a precise quantitative version of our lower bound statement for the case where $k$ and $n$ are both a power of $2$, from which one can derive the general case, as we shall explain soon.
\begin{theorem}
\label{thm:lower_bound}
Let $k \leq n \in \N$ be powers of $2$ and let $0 < p < 1$. There exists a constant $\eps_0 > 0$ such that any non-adaptive algorithm which, given query access to a function $f\colon [n] \to \R$ that is $\eps_0$-far from $(12\dots k)$-free, outputs a length-$k$ monotone subsequence with probability at least $p$, must make at least $ p \binom{\log_2 n}{ \log_2 k }$ queries. Moreover, one can take $\eps_0 = 1/k$.
\end{theorem}
As is usual for arguments of this type, to prove Theorem~\ref{thm:lower_bound} we follow Yao's minimax principle~\cite{Y77}. We construct a distribution $\calD_{n,k}$ over sequences that are $(1/k)$-far from $(12\ldots k)$-free, such that any deterministic algorithm, that makes fewer than $p \binom{\log_2 n}{ \log_2 k }$ queries, fails to find a $(12\ldots k)$-copy in a sequence drawn from this distribution, with probability larger than $1-p$. Here, a deterministic non-adaptive algorithm that makes $q$ queries amounts to deterministically picking a $q$-element subset $Q$ of $[n]$ in advance (without seeing any values in the sequence), and querying all elements of $Q$.
\paragraph{Handling general $k$ and $n$.}
We first explain how to prove our general lower bound, Theorem~\ref{thm:intro-lb}, using the lower bound distribution $\calD_{n,k}$ of the case where $n$ and $k$ are powers of $2$, given in Theorem~\ref{thm:lower_bound}, as a black box. The reduction relies on standard ``padding'' techniques.
Given integers $k, n$ with $k \le n$, write $k = 2^h + t$ for $h,t \in \N$ with $t < 2^h$, and let $k' = 2^h$. Let $n'$ be the largest power of $2$ which is not larger than $nk' / k$, and note that $n' \geq n/4$ and $k' \leq n'$.
We construct our lower bound distribution $\calD_{n,k}$ as follows. Given any $f' \colon [n'] \to \R$ in $\calD_{n', k'}$, we partition the set $\{n'+1, n'+2, \ldots, n\}$ into $t$ consecutive intervals $I_1, \ldots, I_t$, each of size at least $n' / k'$, and extend $f'$ to a sequence $f \colon [n] \to \R$ satisfying the following conditions.
\begin{itemize}
\item $f(x) = f'(x)$ for any $x \in [n']$.
\item $f$ is decreasing within any $I_i$, that is, $f(x) > f(y)$ for $x < y \in I_i$.
\item $f(x) < f(y)$ for any $x \in [n']$ and $y \in I_1$, and for any $x \in I_i$ and $y \in I_j$ where $i < j$.
\end{itemize}
Clearly, we can construct such a sequence $f$ from any given sequence $f'$. Moreover, it is possible to make sure that the values $f(x)$ with $x \in [n]$ are distinct, and thus by relabeling $f$ can be taken to be a permutation.
Furthermore, any $(12\dots k')$-copy in $f'$ can be extended to a $(12\dots k)$-copy in $f$ by appending exactly one arbitrary element from each $I_i$ to it, for a total of $t = k - k'$ additional elements.
Building on the fact that $f'$ is $(1/k')$-far from $(12\dots k')$-free and that $n' \geq n/4$ and $n-n' \ge n(k-k')/k$, we conclude that $f$ is $(1/4k')$-far from $(12\dots k)$-free.
Form a distribution $\calD_{n,k}$ by picking a random $\boldf$ according to the distribution $\sim \calD_{n',k'}$ and extending it to a sequence $f'$ as above.
\noindent The rest of this section is devoted to the proof of Theorem~\ref{thm:lower_bound}.
\subsection{Basic binary profiles and monotonicity testing}
\label{subsec:lower_bound_monotonicity}
In a sense, the proof of our lower bound, Theorem~\ref{thm:lower_bound}, is a (substantial) generalization of the non-adaptive lower bound for testing monotonicity. In order to introduce the machinery required for the proof, we present, in this subsection, a simple proof of the classical $\Omega(\log n)$ non-adaptive one-sided lower bound for monotonicity testing \cite{EKKRV00} using basic versions of the tools we shall use for the full proof. Then, in Subsection~\ref{subsec:lower_bound_full_proof} we proceed to present our tools in their full generality, and provide the proof of Theorem~\ref{thm:lower_bound}.
Intuitively, one way to explain why monotonicity testing requires $\Omega(\log n)$ queries relies on the following reasoning. There exist $\Omega(\log n)$ different distance ``profiles'' our queries should capture; and it can be shown that in general, a small set of queries cannot capture many types of different profiles all at once. At a high level, our new lower bound is an extension of this argument, which uses a more general type of profiles. We start, then, with a formal definition of the basic profiles required for the case of monotonicity testing. Below we restate the required definitions related to the binary representation of numbers in $[n]$.
\begin{definition}[Binary representation]
For any $n \in \N$ which is a power of $2$ and $t \in [n]$, the \emph{binary representation} $B_n(t)$ of $t$ is the unique tuple $(b^t_1, b^t_2, \ldots, b^t_{\log_2 n}) \in \{0,1\}^{\log_2 n}$ satisfying $t = b^t_1 \cdot 2^{0} + b^t_2 \cdot 2^{1} + \dots + b^t_{\log_2 n} \cdot 2^{\log_2 n - 1}$.
For $i \in [\log_2 n]$, the \emph{bit-flip operator}, $F_i \colon [n] \to [n]$, is defined as follows. Given $t \in [n]$ with $B_n(t) = (b^t_1, \ldots, b^t_{\log n})$, we set $F_n(t) = t'$ where $t' \in [n]$ is the unique integer satisfying $B_n(t') = (b^t_1, \ldots, b^{t}_{i-1}, 1-b^t_i, b^t_{i+1}, \ldots, b^t_{\log n})$.
Finally, for any two distinct elements $x,y \in [n]$, let $M(x,y) \in [\log_2 n]$ denote the index of the most significant bit in which they differ, i.e., the largest $i$ with $b_i^x \neq b_i^y$.
\end{definition}
\noindent Note that the bit-flip operator $F_i$ is a permutation on $[n]$.
\paragraph{The construction.}
We start by providing our lower bound construction $\calD_{n,2}$, supported on sequences that are far from $(12)$-free.
Let $f^{\downarrow} \colon [n] \to [n]$ denote the (unique) decreasing permutation on $[n]$, i.e., the function $f^{\downarrow}(x) = n+1-x$ for any $x \in [n]$.
For any $i \in [\log n]$, define $f_i \colon [n] \to [n]$ to be the composition of $f^{\downarrow}$ with the bit-flip operator $F_i$, that is, $f_i(x) = f^{\downarrow}(F_i(x))$ for any $x \in [n]$. Note that $f_i$ is a permutation, as a composition of permutations. See Figure~\ref{fig:mon-construct} for a visualization of the construction. Finally, define $\calD_{n,2}$ as the uniform distribution over the sequences $f_1, f_2, \ldots, f_{\log n}$.
The next lemma characterizes the set of all $(1,2)$-patterns in $f_i$.
\begin{lemma}
\label{lem:monotone_copies}
Let $i \in [\log n]$. A pair $x < y \in [n]$ forms a $(1,2)$-copy in $f_i$ if and only if $M(x,y) = i$.
\end{lemma}
\begin{proof}
Let $x < y \in [n]$. If $M(x,y) > i$, then $F_i(x) < F_i(y)$ holds and so $f_i(x) = f^{\downarrow}(F_i(x)) > f^{\downarrow}(F_i(y)) = f_i(y)$, implying that $(x,y)$ is not a $(1,2)$-copy. If $M(x,y) < i$ then $x$ and $y$ share the bit in index $i$ of the binary representation, and thus flipping it either adds $2^{i-1}$ to both $x$ and $y$ or decreases $2^{i-1}$ from both of them. In both cases, $F_i(x) < F_i(y)$, and like the previous case we get $f_i(x) > f_i(y)$.
Finally, if $M(x,y)=i$ then one can write $x = z + 0 \cdot 2^{i-1} + x'$ and $y = z + 1 \cdot 2^{i-1} + y'$, where $z$ corresponds to the $\log n - i$ most significant bits in the binary representation (which are the same in $x$ and $y$), and $x', y' < 2^{i-1}$ correspond to the $i-1$ least significant bits. Therefore, $F_i(x) = z + 1 \cdot 2^{i-1} + x' > z + 0 \cdot 2^{i-1} + y' = F_i(y)$ and thus $f_i(x) = f^{\downarrow}(F_i(x)) < f^{\downarrow}(F_i(y)) = f_i(y)$, as desired.
\end{proof}
We conclude that each of the sequences $f_i$ is $(1/2)$-far from $(12)$-free.
\begin{lemma}
\label{lem:monotonicity_construction}
For any $i \in [\log n]$, the sequence $f_i$ contains a collection $\calC$ of $n/2$ disjoint $(1,2)$-copies.
\end{lemma}
\begin{proof}
For any $x \in [n]$ whose binary representation $B_n(x) = (b^x_1, \ldots, b^x_{\log n})$ satisfies $b^x_i = 0$, we have $M(x, F_i(x)) = i$. By Lemma~\ref{lem:monotone_copies}, $(x, F_i(x))$ is thus a $(1,2)$-copy. Picking
$$
\calC = \{ (x, F_i(x)) \ : \ x \in [n],\ b^x_i = 0 \},
$$
and noting that the pairs in $\calC$ are disjoint, the proof follows.
\end{proof}
\paragraph{Binary Profiles.}
We now formally define our notion of \emph{binary profiles}, and describe why they are useful for proving lower bounds for problems of this type.
\begin{definition}[Binary profiles captured]
Let $n \in \N$ be a power of $2$ and let $Q \subseteq [n]$.
The set of \emph{binary profiles} captured by $Q$ is defined as
$$
\textsf{bin-prof}(Q) = \{i \in [\log n] \ : \ \text{there exist $x,y \in Q$ satisfying $M(x,y)=i$} \}.
$$
\end{definition}
The next lemma asserts that the number of binary profiles that set captures does not exceed (or even match) the size of the set.
\begin{lemma}
\label{lem:binary_dist_prof}
Let $Q \subseteq [n]$ be a subset of size $q > 0$. Then $|\textsf{bin-prof}(Q)| \leq q-1$.
\end{lemma}
\begin{proof}
We proceed by induction on $q$. For $q \leq 2$, the statement clearly holds. Otherwise, let $i_{\text{max}} = \max \textsf{bin-prof}(Q)$ be the maximum index of a bit in which two elements $x,y \in Q$ differ.
For $j=0,1$, define
$$
Q_j = \{x \in Q \ :\ \text{the binary representation of $x$ is $B_n(x) = (b^x_1, \ldots, b^x_{\log n})$, and $b^x_{i_{\text{max}}} = j$}\}.
$$
\noindent
Clearly, for any $x \in Q_0$ and $y \in Q_1$, we have $M(x,y) = i_{\text{max}}$. We can therefore write $\textsf{bin-prof}(Q)$ as
$$
\textsf{bin-prof}(Q) = \textsf{bin-prof}(Q_0) \cup \textsf{bin-prof}(Q_1) \cup \{i_{\text{max}}\},
$$
from which we conclude that
$$
|\textsf{bin-prof}(Q)| \leq |\textsf{bin-prof}(Q_0)| + |\textsf{bin-prof}(Q_1)| + 1 \leq |Q_0| - 1 + |Q_1| - 1 + 1 = |Q| - 1,
$$
where the second inequality follows from the induction hypothesis.
\end{proof}
\paragraph{Proof for the case $k=2$ using binary profiles.}
After collecting all the ingredients required to prove the case $k=2$ of Theorem~\ref{thm:lower_bound}, we now conclude the proof.
Fix $0 < p < 1$, let $n$ be a power of two, and consider the distribution $\calD_{n,2}$ defined above, supported on sequences that are $(1/2)$-far from $(12)$-free (see Lemma~\ref{lem:monotonicity_construction}).
Let $Q \subseteq [n]$ be any subset of size at most $p \log n$. It suffices to show that, for $\boldf \sim \calD_{n,2}$, the probability that $Q$ contains a $(12)$-copy in $\boldf$ is less than $p$. By Lemma~\ref{lem:monotone_copies}, $Q$ contains a $(12)$-copy with respect to $f_i$ if and only if $i \in \textsf{bin-prof}(Q)$. Thus, the above probability is equal to $|\textsf{bin-prof}(Q)| / \log n$, which, by Lemma~\ref{lem:binary_dist_prof}, is at most $(|Q|-1) / \log n < p$, as desired.
\subsection{Hierarchical binary profiles and the lower bound}
\label{subsec:lower_bound_full_proof}
To prove Theorem~\ref{thm:lower_bound} in its full generality, we significantly extend the proof presented in Subsection~\ref{subsec:lower_bound_monotonicity} for the case $k=2$, relying on a generalized hierarchical (and more involved) notion of a binary profile.
Let $n > k \geq 2$ be powers of $2$, and write $k = 2^h$ (so $h \in \N$).
We show that there exist $\binom{\log_2 n}{h} = \binom{\log_2 n}{\log_2 k}$ different types of \emph{binary $h$-profiles} (see Definition~\ref{def:h-profile}) with the following properties. First, a subset $Q \subseteq [n]$ can capture at most $|Q|-1$ such profiles (Lemma~\ref{lem:lower_bound_general_construction} below, generalizing Lemma~\ref{lem:monotonicity_construction}); and second, for each such profile there exists a sequence (in fact, a permutation) that is $(1/k)$-far from $(12\ldots k)$-free, such that any set of queries $Q$ that finds $(12\ldots k)$-pattern with respect to this sequence must capture the given profile (Lemma~\ref{lem:hierarchical_copies} below, generalizing Lemma~\ref{lem:monotone_copies}).
\paragraph{Hierarchical binary profiles.}
While the proof for the case $k=2$ relied on a rather basic variant of a binary profile, our lower bound for general $k$ requires a more sophisticated, hierarchical type of profile, described below.
\begin{definition}[binary $h$-profiles] \label{def:h-profile}
Let $(x_1, \dots, x_k) \in [n]^{k}$ be a $k$-tuple of indices satisfying $x_1 < \dots < x_k$. For an $h$-tuple $(i_1, \dots, i_h) \in [\log_2 n]^h$ satisfying $i_1 < \dots < i_k$,
we say that $(x_1, \ldots, x_k)$ has \emph{$h$-profile of type $(i_1,\ldots, i_{h})$} if,
\[
M(x_j, x_{j+1}) = i_{M(j-1, j)} \qquad \text{for every $j \in [k-1]$}.
\]
\end{definition}
For example, when $h=3$ (and $k=8$), a tuple $(x_1, \ldots, x_8) \in [n]^8$ with $x_1 < \ldots < x_8$ has binary $3$-profile of type $(i_1, i_2, i_3)$ if the sequence $(M(x_j, x_{j+1}))_{j=1}^7$ is $(i_1, i_2, i_1, i_3, i_1, i_2, i_1)$.
See Figure~\ref{fig:profiles} for a visual depiction of such a binary $3$-profile.
Similarly to the case $k=2$, given a set of queries $Q \subseteq [n]$, we shall be interested in the collection of $h$-profiles captured by $Q$.
\begin{definition}[Binary $h$-profiles captured]
Let $n \geq k \geq 2$ be powers of $2$ where $k = 2^h$.
For any $Q \subseteq [n]$, we denote the set of all $h$-profiles captured by $Q$ by
\[
\textsf{bin-prof}_h(Q) = \left\{(i_1, \ldots, i_{h}) :\
\begin{array}{l} \text{there exist $x_1, \dots,x_k \in Q$ where } x_1 < \dots < x_k \\
\text{and $(x_1, \dots, x_k)$ has $h$-profile of type $(i_1, \ldots, i_{h})$}
\end{array}
\right\}.
\]
\end{definition}
The next lemma is one of the main ingredients of our proof, generalizing Lemma~\ref{lem:binary_dist_prof}. It shows that a set $Q$ of queries cannot capture $|Q|$ or more different $h$-profiles.
\begin{lemma}
\label{lem:tree_prof_lower_bound}
Let $h, n \in \N$ where $n \geq 2^h$ is a power of $2$. For any $\emptyset \neq Q \subseteq [n]$, we have $|\textsf{bin-prof}_h(Q)| \leq |Q|-1$.
\end{lemma}
\begin{proof}
We proceed by induction on $h$. The case $h=1$ was settled in Lemma~\ref{lem:binary_dist_prof}. Suppose now that $h > 1$, and define
$$\emptyset = B_{\log n + 1} \subseteq B_{\log n} \subseteq \ldots \subseteq B_{1} = Q$$
as follows. Set $B_{\log n + 1} = \emptyset$, and given $B_{i+1}$, define the set $B_i \supseteq B_{i+1}$ as an arbitrary maximal subset of $Q$ containing $B_{i+1}$ which does not have two elements with $M(x,y) < i$.
Additionally, for each $j \in [\log_2 n]$, define
\[ N_j = \left\{ (i_2, \dots, i_h) : 1 \leq j < i_2 \dots < i_h \leq \log_2 n \text{ and } (j, i_2, \dots, i_h) \in \textsf{bin-prof}_h(Q) \right\}. \]
\begin{claim}
Let $j < i_2 < \ldots < i_h \in [\log n]$, and suppose that $(j, i_2, \ldots, i_h) \in \textsf{bin-prof}_h(Q)$. Then $(j, i_2, \ldots, i_h) \in \textsf{bin-prof}_h(B_j)$.
\end{claim}
\begin{proof}
Suppose that a tuple $(x_1, \ldots, x_k)$ with $x_1 < \dots < x_k \in Q$ has $h$-profile $(j, i_2, \ldots, i_{h})$. By the maximality of $B_j$, we know that for every $1 \leq \ell \leq k$ there exists $y_\ell \in B_j$ such that either $x_{\ell} = y_{\ell}$ or $M(x_{\ell}, y_{\ell}) < j$. Indeed, if this were not the case, then $B'_j \eqdef B_j \cup \{x_{\ell}\}$ would be a set that strictly contains $B_{j}$ and does contain two elements $x \neq y$ with $M(x,y) = j$, a contradiction to the maximality of $B_{j}$.
By definition of a profile, we conclude that $\{y_1, \ldots, y_{k}\} \subseteq B_j$ has $h$-profile $(j, i_2, \ldots, i_{h})$.
\end{proof}
\begin{claim}
For any $j \in [\log n]$, we have $N_j \subseteq \textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})$.
\end{claim}
\begin{proof}
Suppose that $(i_2, \ldots, i_h) \in N_j$, then $(j, i_2, \ldots, i_h) \in \textsf{bin-prof}_h(Q)$. By the previous lemma, we know that $(j, i_2, \ldots, i_h) \in \textsf{bin-prof}_h(B_j)$.
Therefore, there exists a tuple $(y_1, \dots, y_k)$ where $y_1 < \ldots< y_{k} \in B_j$, that has $h$-profile of type $(j, i_2, \ldots, i_{h})$.
For any $t \in k/2$, it holds that $M(y_{2t-1}, y_{2t}) = j$. Therefore, at most one of $y_{2t-1}, y_{2t}$ is in $B_{j+1}$, and hence, for any such $t$ there exists $z_t \in \{y_{2t-1}, y_{2t}\} \setminus B_{j+1} \subseteq B_{j} \setminus B_{j+1}$. Consider the tuple $(z_1, \ldots, z_{k/2})$, whose elements are contained in $B_j \setminus B_{j+1}$. It follows from our choice of $z_t$ that $M(z_t, z_{t+1}) = M(y_{2t}, y_{2t+2})$ for any $t \in [k/2]$, from which we conclude that $(z_1, \ldots, z_{k/2})$
has $(h-1)$-profile $(i_2, \ldots, i_{h})$. In other words, $(i_2, \ldots, i_h) \in \textsf{bin-prof}_h(B_j \setminus B_{j+1})$, as desired.
\end{proof}
We are now ready to finish the proof of Lemma~\ref{lem:tree_prof_lower_bound}. Observe that $\textsf{bin-prof}_h(Q)$ and $Q$ can be written as the following disjoint unions:
\[
\textsf{bin-prof}_h(Q) = \bigcup_{j=1}^{\log_2 n} \{(j, i_2, \ldots, i_h) : (i_2, \ldots, i_h) \in N_j \}
\qquad \text{and} \qquad
Q = \bigcup_{j=1}^{\log_2 n} (B_j \setminus B_{j+1}).
\]
It follows from the last claim and the induction assumption that
\begin{equation}
\label{eqn:N_j}
|N_j| \leq |\textsf{bin-prof}_{h-1}(B_j \setminus B_{j+1})| \le |B_j \setminus B_{j+1}|,
\end{equation}
where for $j$ with $N_j \neq \emptyset$ there is a strict inequality.
Now,
if $N_j$ is empty for all $j$ then, trivially, $|\textsf{bin-prof}_h(Q)|= 0 \leq |Q|-1$. Otherwise, there exists some non-empty $N_j$, for which \eqref{eqn:N_j} yields a strict inequality, and we get
\[
|Q| = \sum_{j=1}^{\log_2 n} |B_j \setminus B_{j+1}| >\sum_{j=1}^{\log_2 n} |N_j| = |\textsf{bin-prof}_h(Q)|,
\]
establishing the proof of the Lemma~\ref{lem:tree_prof_lower_bound}.
\end{proof}
\paragraph{The construction.}
For any $i_1 < i_2 < \ldots < i_h \in [\log n]$, we define $f_{i_1, \ldots, i_h} \colon [n] \to [n]$ as
\[
f_{i_1, \ldots, i_{h}} \eqdef f^{\downarrow} \circ F_{i_h} \circ \ldots \circ F_{i_1},
\]
where, as before, $\circ$ denotes function composition. In other words, for any $x \in [n]$ we have $f_{i_1, \ldots, i_h}(x) = f^{\downarrow}(F_{i_{h}}(F_{i_{h-1}}(\dots(F_{i_1}(x)\dots))))$. Note that $f_{i_1, \ldots, i_h}$ is indeed a permutation, as a composition of permutations.
(See Figure~\ref{fig:construct-recurse}, which visually describes the construction of $f_{i_1, \ldots, i_h}$ recursively, as a composition of $F_{i_h}$ with $f_{i_1, \ldots, i_{h-1}}$.)
We take $\calD_{n,k}$ to be the uniform distribution over all sequences of the form $f_{i_1, \ldots, i_h}$ with $i_1 < i_2 < \ldots < i_k$. The size of the support of $\calD_{n,k}$ is $\binom{\log_2 n}{h} = \binom{\log_2 n}{\log_2 k}$.
\paragraph{Structural properties of the construction.}
Recall that our lower bound distribution $\calD_{n,k}$ is supported on the family of permutations $f_{i_1, \ldots, i_{h}}$, where $i_1 < \ldots < i_h \in [\log n]$, described above.
We now turn to show that these $f_{i_1, \ldots, i_h}$ satisfy two desirable properties. First, to capture a $(12\dots k)$-copy in $(f_{i_1, \ldots, i_h})$, our set of queries $Q$ must satisfy $(i_1, \ldots, i_h) \in \textsf{bin-prof}_h(Q)$ (Lemma~\ref{lem:hierarchical_copies}). And second, each such $f_{i_1, \ldots, i_k}$ is $(1/k)$-far from $(12\dots k)$-free (Lemma~\ref{lem:lower_bound_general_construction}).
\begin{lemma}
\label{lem:hierarchical_copies}
Let $(x_1, \ldots, x_k) \in [n]^k$ be a $k$-tuple where $x_1 < \ldots < x_k$, and let $f = f_{i_1, \ldots, i_{h}}$ be defined as above. Then $f(x_1) < f(x_2) < \ldots < f(x_k)$ (i.e., $(x_1, \ldots, x_k)$ is a $(12\dots k)$-copy with respect to $f_{i_1, \ldots, i_h}$) if and only if $(x_1, \ldots, x_k)$ has binary $h$-profile of type $(i_1, i_2 \ldots, i_{h})$. Furthermore, $f_{i_1, \ldots, i_{h}}$ does not contain increasing subsequences of length $k+1$ or more.
\end{lemma}
\begin{proof}
The proof is by induction on $h$, with the base case $h=1$ covered by Lemma~\ref{lem:monotone_copies}; in particular, it follows from Lemma~\ref{lem:monotone_copies} that $f_{i}$ has no increasing subsequence of length $3$, since there exist no $x < y < z \in [n]$ with $M(x,y) = M(y,z) = i$.
For the inductive step, we need the following claim, which generalizes Lemma~\ref{lem:monotone_copies}.
\begin{claim}
\label{claim:lower_bigger_general}
A pair $x < y \in [n]$ satisfies $f_{i_1, \ldots, i_h}(x) < f_{i_1, \ldots, i_h}(y)$ if and only if $M(x,y) \in \{i_1, \ldots, i_h\}$.
\end{claim}
\begin{proof}
Let $F_{i_1, \ldots, i_h} = F_{i_h} \circ \ldots \circ F_{i_1}$.
Since $f_{i_1, \ldots, i_h} = f^{\downarrow} \circ F_{i_1, \ldots, i_h}$, it suffices to show that $F_{i_1, \ldots, i_h}(x) > F_{i_1, \ldots, i_h}(y)$ if any only if $M(x, y) \in \{i_1, \ldots, i_h\}$. To do so, we prove the following two statements.
\begin{itemize}
\item For any $x < y \in [n]$, $F_i(x) > F_i(y)$ if and only if $M(x,y) = i$.
\item For any $x < y \in [n]$, $M(F_i(x), F_i(y)) = M(x,y)$.
\end{itemize}
Indeed, using these two statements, the proof easily follows by induction: the value of $M(x,y)$ never changes regardless of which bit-flips we simultaneously apply to $x$ and $y$. Now, applying any of the bit-flips $F_i$ to $x$ and $y$, where $i \neq M(x,y)$, does not change the relative order between them, while applying $F_{M(x,y)}$ does change their relative order. This means that a change of relative order occurs if and only if $M(x,y) \in \{i_1, \ldots, i_h\}$, which settles the claim.
The proof of the first statement was essentially given, word for word, in the proof of Lemma~\ref{lem:monotone_copies}. The second statement follows
by a simple case analysis of the cases where $i$ is bigger than, equal to, or smaller than $M(x,y)$, showing that in any of these cases, $M(F_i(x), F_i(y)) = M(x,y)$.
\end{proof}
Suppose now that $(x_1, \ldots, x_k) \in [n]^k$ is a tuple with $x_1 < \ldots < x_k$ and a binary $h$-profile of type $(i_1, \ldots, i_h)$ is a $(12 \dots k)$-copy in $f_{i_1, \ldots, i_h}$. By definition of a binary $h$-profile, we have that $M(x_j, x_{j+1}) \in \{i_1, \ldots, i_h\}$ for any $j \in [k-1]$, which, by the claim, implies that $f_{i_1, \ldots, i_h}(x_j) < f_{i_1, \ldots, i_h}(x_{j+1})$. It thus follows that $(x_1, \ldots, x_k)$ is a $(12 \dots k)$-copy in $f_{i_1, \ldots, i_h}$, as desired.
Conversely, suppose that a tuple $(x_1, \ldots, x_k) \in [n]^k$ with $x_1 < \ldots < x_k$ is a $(12 \dots k)$-copy in $f_{i_1, \ldots, i_k}$.
We need to show that $(x_1, \ldots, x_{k})$ has binary $h$-profile of type $(i_1, \ldots, i_h)$, that is, $M(x_j, x_{j+1}) = i_{M(j-1, j)}$ for every $j \in [k-1]$.
Define $r = \argmax_j \{M(x_j, x_{j+1})\}$, and note that $r$ is unique; otherwise, we would have $x < y < z \in [n]$ so that $M(x,y) = M(y,z)$, a contradiction.
\begin{claim}
$M(x_r, x_{r+1}) = i_h$.
\end{claim}
\begin{proof}
By Claim~\ref{claim:lower_bigger_general}, we know that $M(x_r, x_{r+1}) \in \{i_1, \ldots, i_h\}$. Suppose to the contrary that $M(x_r, x_{r+1}) \leq i_{h-1}$. Then, $M(x_j, x_{j+1}) \leq M(x_r, x_{r+1}) \leq i_{h-1}$ for every $j \in [k-1]$, and by Claim~\ref{claim:lower_bigger_general}, for any $j \in [k-1]$ we have $f_{i_1, \ldots, i_{h-1}}(x_j) \leq f_{i_1, \ldots, i_{h-1}}(x_{j+1})$, that is, $(x_1, x_2, \ldots, x_k)$ is a $(12 \dots k)$-copy in $f_{i_1, \ldots, i_{h-1}}$. This contradicts the last part of the inductive hypothesis.
\end{proof}
\begin{claim}
$r = k/2$.
\end{claim}
\begin{proof}
Without loss of generality, suppose to the contrary that $r > k/2$ (the case where $r< k/2$ is symmetric). As the tuple $(x_1, \ldots, x_r)$ is an increasing subsequence for $f_{i_1, \ldots, i_h}$, we have $M(x_j, x_{j+1}) \in \{i_1, \ldots, i_h\}$ for any $j \in [r-1]$. By the maximality and uniqueness of $r$, $M(x_j, x_{j+1}) < i_h$ for any $j \in [r-1]$. Thus, it follows from Claim~\ref{claim:lower_bigger_general} that $(x_1, \ldots, x_r)$ is a $(12\dots r)$-copy in $f_{i_1, \ldots, i_{h-1}}$, contradicting the last part of the inductive hypothesis.
\end{proof}
It thus follows from the two claims that $M(x_{k/2}, x_{k/2+1}) = i_h$. Since $M(x_j, x_{j+1}) \in \{i_1, \ldots, i_{h-1}\}$ for any $j \in [k-1] \setminus \{k/2\}$, we conclude, again from Claim~\ref{claim:lower_bigger_general}, that $(x_1, \ldots, x_{k/2})$ and $(x_{k/2+1}, \ldots, x_k)$ both induce length-$(k/2)$ increasing subsequences in $f_{i_1, \ldots, i_{h-1}}$. By the inductive hypothesis, they both have binary $(h-1)$-profile $(i_1, \ldots, i_{h-1})$. Combined with the last two claims, we conclude that $(x_1, \ldots, x_k)$ has binary $h$-profile $(i_1, \ldots, i_h)$, as desired.
It remains to verify that $f_{i_1, \ldots, i_h}$ does not contain an increasing subsequence of length $k+1$. If, to the contrary, it does contain one, induced on some tuple $(x_1, \ldots, x_{k+1}) \in [n]^{k+1}$ where $x_1 < \ldots < x_{k+1}$, then, applying the last two claims to the length-$k$ two tuples $(x_1, \ldots, x_k)$ and $(x_2, \ldots, x_{k+1})$, we conclude that $M(x_{k/2}, x_{k/2+1}) = M(x_{k/2+1}, x_{k/2 +2}) = i_h$. However, as discussed above, there cannot exist $x<y<z \in [n]$ with $M(x,y) = M(y,z)$ -- a contradiction.
\end{proof}
It remains to prove that each $f_{i_1, \ldots, i_h}$ is indeed $(1/k)$-far from $(12\dots k)$-free. After we spent quite some effort to characterize \emph{all} $(12 \dots k)$-copies in $f_{i_1, \ldots, i_h}$, this upcoming task is much simpler.
\begin{lemma}
\label{lem:lower_bound_general_construction}
Let $n \geq k \geq 2$ be powers of two and write $k = 2^h $.
The sequence $f_{i_1, \ldots, i_h} \colon [n] \to [n]$, defined above, contains $n / k$ disjoint $(12 \dots k)$-copies.
\end{lemma}
\begin{proof}
Fix $i_1 < \ldots < i_h$ as in the statement of the lemma. We say that $x,y \in [n]$ with binary representations $B_n(x) = (b^x_1, \ldots, b^x_{\log n})$ and $B_n(y) = (b^y_1, \ldots, b^y_{\log n})$ are \emph{$(i_1, \ldots, i_h)$-equivalent} if $b^x_i = b^y_i$ for any $i \in [\log n] \setminus \{i_1, \ldots, i_{h}\}$. Clearly, this is an equivalence relation, partitioning $[n]$ into $n/k$ equivalence classes, each of size exactly $k = 2^h$. Moreover, it is straightforward to verify that the elements $x_1 < x_2 < \ldots < x_k$ of any equivalence class satisfy $M(x_j, x_{j+1}) \in \{i_1, \ldots, i_h\}$ for any $j \in [k-1]$, and thus, by Claim~\ref{claim:lower_bigger_general}, $(x_1, \ldots, x_k)$ constitutes a $(12 \dots k)$-copy in $f_{i_1, \ldots, i_k}$.
\end{proof}
\paragraph{Proof of Theorem~\ref{thm:lower_bound}.}
It now remains to connect all the dots for the proof of Theorem~\ref{thm:lower_bound}.
\begin{proof}
Fix $0 < p < 1$, let $n \geq k$ be powers of $2$, and write $k = 2^h$.
As before, we follow Yao's minimax principle \cite{Y77}, letting $\calD_{n,k}$ be the uniform distribution over all $\binom{\log_2 n}{h} = \binom{\log_2 n}{\log_2 k}$ sequences (in fact permutations) $f_{i_1, \ldots, i_h} \colon [n] \to [n]$, where $i_1 < \ldots < i_h \in [\log n]$. Recall that, by Lemma~\ref{lem:lower_bound_general_construction}, this distribution is supported on sequences that are $(1/k)$-far from $(12\dots k)$-free.
It suffices to show that, for $\boldf \sim \calD_{n,k}$, the probability for any subset $Q \subseteq [n]$ of size at most $p \binom{\log_2 n}{h}$ to capture a $(12 \dots k)$-copy in $\boldf$ is less than $p$.
Indeed, by Lemma~\ref{lem:hierarchical_copies}, $Q$ captures a copy in $f_{i_1, \ldots, i_h}$ if and only if $(i_1, \ldots, i_h) \in \textsf{bin-prof}_h(Q)$, so the success probability for any given $Q$ is exactly $|\textsf{bin-prof}_h(Q)| / \binom{\log_2 n}{h} < |Q| / \binom{\log_2 n}{h} \leq p$ for any $Q \subseteq [n]$ with $|Q| \leq p \binom{\log_2 n}{h}$, where the first inequality follows from Lemma~\ref{lem:tree_prof_lower_bound}. The proof of Theorem~\ref{thm:lower_bound} follows.
\end{proof}
\begin{flushleft}
\bibliographystyle{alpha}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 8,417 |
ALLENTOWN, Pa. (AP) — A Reading man and the Allentown Roman Catholic Diocese have settled his lawsuit over allegations a priest groped him in 1989, when he was a 12-year-old altar boy.
The Morning Call of Allentown said Friday the diocese confirmed the settlement but had no other comment.
Attorney William Rush says the agreement also addresses the man's medical treatment needs.
Rush's client had alleged Rev. Richard Ford, who's now dead, molested him in the rectory of Holy Guardian Angels Church in Reading.
An investigative grand jury reported in August that about 300 Catholic priests had sexually abused more than 1,000 children over decades in six Pennsylvania dioceses, including Allentown. | {
"redpajama_set_name": "RedPajamaC4"
} | 8,098 |
\section{Introduction}
Mathematical practices are changing due to the availability of
mathematical knowledge on the Web. This paper deals with the question
whether mathematicians have special needs or preferences when
accessing this knowledge and if yes, what are those? In particular, we
focus on how mathematicians think of search on the Web: what are their
cognitive categories, what kinds of searches do they distinguish, and
which attributes do they associate with tools for math access?
The usability study~\cite{EuDML:UsabilityStudy:2010} conducted
interviews with mathematicians and essentially stated that
mathematicians didn't know how to use the offerings of mathematical
search interfaces. To get a better understanding we wanted to dig
deeper. In~\cite{Zhao:MIR:2008} {\name{Zhao}} concentrates on
user-centric and math-aware requirements for math search. The former
are based on mathematicians' specific information needs and search
behaviors, the latter are the needs for structured indizes by the
system. In contrast, we focus on eliciting attributions of existing
math search interfaces by mathematicians versus non-mathematicians. We
hope to learn what exactly sets mathematicians apart, since from this
knowledge we can deduce implications for future mathematical designs.
We decided on using repertory grid interviews as main
methodology to elicit evaluation schemes with respect to selected
math search interfaces (``{\bf{\mathUI}}'') and to understand how mathematicians
classify those {\mathUI}s. The main advantage of the method is its
semi-empirical nature. On the one hand, it allows to get deep insights
into the topic at hand through deconstruction and intense discussion of
each subject's idiosyncratic set of constructs and their resp. mapping
to the set of {\mathUI}s. On the other hand, the grids produced in
such \rgi sessions can be analyzed with a General Procrustes Analysis
to obtain statistically significant correlations between the elicited
constructs or the chosen {\mathUI}s. We used the \textsf{Idiogrid}~\cite{Grice:Idiogrid:2002} and the \textsf{OpenRepGrid}~\cite{OpenRepGrid} software for this.
Information search is not a single act, but a process with many strategies and options: {\myCitation{In fact, we move fluidly between models of ask, browse,
filter, and search without noting the shift. We scan feeds, ask
questions, browse answers, and search
again.}}~\cite[p.7]{Morville:SearchPatterns:2010}. Therefore, we can consider the term ``search'' as an umbrella term for (at least) the following approaches:
\begin{description}
\item[Finding] = already knowing what one is looking for
(\cite{NavarroEtAl:CognStrategiesSearching:1999,Shneiderman:ClarifyingSearch:1997}
call it ``fact-finding'')
\item[Browsing] = getting an overview over a topic or an idea of a
concept (\cite{NavarroEtAl:CognStrategiesSearching:1999} calls it
``exploration of availability'')
\item[Surfing] = surrendering to the links, drifting from one to
another (see~\cite{WiseEtAl:SearchingVsSurfing:2009})
\item[Solving/Information Gathering] = creating a search plan, i.e., specifying a sequence
of actions that achieves the solution of a problem
(see~\cite[65ff.]{RussellNorvig:aiama95},~\cite{Kellar:InformationSeekingTasks:2007})
\item[Asking] = posing a question to find an answer
(see~\cite{Taylor:ProcessOfAskingQuestions:1962})
\end{description}
Our question here is, what search approach is used with which
assessment attributes for what kind of math search tool? The answer
could enable us to design specifically for more math search approaches
by learning from the used ones.
We start out in Section~\ref{sec:study} with a description of the \rgi
study. In Section~\ref{sec:findings} we present the elicited interview
data and note the patterns that emerge from this data. The patterns
state interesting, prototypical attributions of mathematicians, which
separate the data gathered from the group of mathematicians from the
one of non-mathematicians. To demo the utility of such patterns, we
apply them in a discussion of an interesting, confusing evaluation of
two specific {\mathUI}s in Section~\ref{sec:example}. We conclude in
Section~\ref{sec:conclusion} by hinting at general design implications
for mathematical (search) interfaces based on the found set of
patterns.
\section{The Study}\label{sec:study}
The aim of our study was to find out what distinguishes mathematicians from
non-mathematicians when using a web interface for searching relevant content, here math
content. From the outset it was clear that observational methods wouldn't work as the
working context of a mathematician is typically neither restrained to certain locations
nor time slots. Surveys (or structured interviews) were out of question as the answers
require a deep insight of subjects into their own math search behavior, which
cannot be assumed in general. Unstructured interviews could have been made use of to get
such deep insights, but we would either have to do
too many to be able to soundly interpret them or too few to draw general
conclusions. Finally,
the option of semi-structured interviews as methodology was discarded, since it became
clear in the first pilot study trials that mathematicians tend to describe ``truths'' and
``falsities''. In particular, they try to scrutinize the interview or interviewer and
manipulate the outcome towards what they think is the correct answer. Thus, the
interviewer has to trade her observational stance with a continuously sparring stance,
which hinders the process of gaining deep insights.
In the end, we opted for the methodology of repertory grid interviews,
as they allow a semi-empirical analysis, and interviewees understand
quickly that they are not asked to decide on rights or wrongs. The
\textbf{Repertory Grid Interview (\rgi)
Technique}~\cite{HassenzahlWessler:RepGridTechnology00,Jankowicz:2003,Kelly:BriefIntroductionToPCT}
explores personal constructs, i.e., how persons perceive and
understand the world around them. It has been used as a usability/user
experience method to research users' personal constructs when
interacting with software artifacts
(see~\cite{HeideckerHassenzahl_RGTFuerAttraktivitaet_2007,HertzumEtAl:PersonalUsabilityConstructs2012,TanHunter:RGTinIT2002}
for examples). \rgi has the advantage that it can deliver valuable
insights into the perception of users even with relative low numbers
of study subjects (seeo~\cite{Kohlhase:HumanSpreadsheetInteraction:2013} for more details).
\begin{table}[ht]\centering\footnotesize\vspace{-2.em}
\begin{tabular}{|p{0.18\columnwidth}|p{0.58\columnwidth}|p{0.22\columnwidth}|}\hline
\tabhead{Element Name}&\tabhead{Short Description}&\tabhead{URL}\\\hline
\elemZBMathNew & {\small{ ``an {\textit{abstracting and reviewing service}} in pure and applied mathematics'' }} & {\scriptsize{\url{zbMath.org}}}\\\hline
\elemZBMathOld & {\small{ the former interface of \elemZBMathNew }} & {\scriptsize{not available}}\\\hline
\elemMathSciNet & {\small{ ``searchable {\textit{database of reviews, abstracts and bibliographic information}} for much of the mathematical sciences literature'' }} & {\scriptsize{\url{ams.org/mathscinet}}}\\\hline
\elemGoogleScholar & {\small{ ``search of {\textit{scholarly literature}} across many disciplines and sources'' }} & {\scriptsize{\url{scholar.google.com}}}\\\hline
\elemGoogle & {\small{ ``Search the {\textit{world's information}}, including webpages, images, videos and more'' }} & {\scriptsize{\url{google.com}}}\\\hline
\elemOffice & {\small{ the personal {\textit{office}} as math search interface}} & --- \\\hline
\elemTIB & {\small{ The {\textit{online catalogue}} of the Uni Hannover Library }} & {\scriptsize{\url{tib.uni-hannover.de}}}\\\hline
\elemVifamath & {\small{ ``The Virtual Library of Mathematics'' - a {\textit{meta online catalogue}}}} & {\scriptsize{\url{vifamath.de}}}\\\hline
\elemLibrary & {\small{ a {\textit{physical library}} known by the subject }} & --- \\\hline
\elemArxiv & {\small{ ``{\textit{Open e-print archive}} with over [\dots] 10000 [articles] in mathematics'' }} & {\scriptsize{\url{arxiv.org}}}\\\hline
\elemResearchGate & {\small{ ``a {\textit{network}} dedicated to science and research'' }} & {\scriptsize{\url{researchgate.net}}}\\\hline
\elemMathoverflow & {\small{ ``a {\textit{question and answer site}} for professional mathematicians'' }} & {\scriptsize{\url{mathoverflow.net}}}\\\hline
\elemColleagues & {\small{ personal {\textit{colleagues}} as math search interface }} & --- \\\hline
\elemMSCMap & {\small{ ``accessing math via {\textit{interactive maps}}'' based on an MSC metric }} & {\scriptsize{\url{map.mathweb.org}}}\\\hline
\elemCatchup & {\small{ an interface for {\textit{catching up}} with the newest articles in math }} & {\scriptsize{\url{arxiv.org/catchup}}}\\\hline
\elemFormulaSearch & {\small{ ``allows to search for {\textit{mathematical formulae}} in documents indexed in zbMath'' }} & {\scriptsize{\url{zbmath.org/formulae}}}\\\hline
\elemBibliography & {\small{ a {\textit{bibliography} as math search interface}}} & \\\hline
\end{tabular}\vspace{0.5em}\\
\caption{The \rgi Elements in the Study}\label{tab:elements}\vspace{-3.0em}
\end{table}
\subsection{The \rgi Elements}
As we want to cover a broad range of different types of math search
interfaces we opted for a set of 17 {\mathUI}s as {\bf{\rgi elements}}
-- ranging from standard {\mathUI}s like ``Zentralblatt Mathematik
(zbMath)'' or ``MathSciNet'' via social media platforms like
``mathoverflow'' to scientific prototypes like the ``MSC map''
interface (MSC = Math Subject Classification, see~\cite{MSC-SKOS}). To
avoid being limited to digital {\mathUI}s, we included traditional
search situations like asking colleagues or personal office spaces as
well. Table~\ref{tab:elements} summarizes the 17 elements used in the
{\rgi}s and gives short descriptions -- the ones from their websites
where available -- and their web addresses if applicable. Note that
wikis (e.g., ``Wikipedia'' or ``PlanetMath'') were excluded as the
tension between searching for articles versus encyclopedia entries was
perceived problematic in the pilot study, so we opted for the
former. As we were only interested in the search behavior of
mathematicians we disregarded mathematical software whose main task is
computation or verification.
\subsection{The \rgi Set-Up}
At the beginning of each interview the interviewer introduced the
interviewee to all {\mathUI}s based on print-outs. Both the home page
with its search facilities and the search result pages were
discussed. The front page of each print-out presented the homepage
initialized with the phrase ``Cauchy sequence'' in the search box if
applicable. The back page displayed the search result wrt to this
query. For {\mathUI}s with special features extra pages were
attached. For {\elemFormulaSearch} the {\LaTeX} query corresponding to
$?a_{?n}\in\mathtt{N}$ was used.
An \rgi interview iterates the following process until the
interviewee's individual construct space seems to be exhausted:
\begin{compactenum}[\em i.]
\item The interviewee randomly chooses three \rgi elements.
\item He declares which two of the three elements seem more similar.
\item He determines the aspect under which these two are more similar
and the aspect under which the third one is different. Those aspects
are the ``{\bf{poles}}'' of an interviewee-dependent evaluation
dimension, the so-called ``{\bf{construct}}''.
\end{compactenum}
To get a sense of what the users consider important properties of
{\mathUI}s, we extended this set-up by encouraging most interviewees
to judge the ``fitness'' of each {\mathUI} for mathematical search.
As is typical with {\rgi}s, the interviews were very
intense. Therefore, the findings are not only based on the
actual data elicited in the {\rgi} but also on the deep
discussions taking place during each interview.
\subsection{The {\rgi} Data}
We conducted interviews with 24 people, all of which were interested
in accessing math on the web. Out of these, 18 had a degree in
mathematics. For the final analysis we decided to use 22 {\rgi}s:
interviews with a group of 11 professional mathematicians working in a
scientific environment (``\Math''), a group of 5 content experts for
mathematical information (``\zbMath''), and a group of 6
non-mathematicians (``\noMath''). Only 3 of the
participants were female.
Each interview took between 1.75 and 3 hours, in which an average of 4
constructs were elicited. The \Math group created 50 constructs,
{\zbMath} reported 28 constructs and \noMath 29 constructs. The
rating scale for these 107 elicited constructs was a 7-point Likert
scale.
\section{Findings}\label{sec:findings}
As already mentioned, the {\rgi} method is semi-empirical. This means
that there will be a quantitative and a qualitative analysis of the
data gathered. Due to space limitations we will focus on presenting and interpreting the most interesting, statistically significant quantitative results in form of dendrograms and qualitative results in form of patterns. Note that here, the theory emerges from the data, thus, it provides us
with patterns but not with proofs.
With the {\bf{Generalized Procrustes Analysis (GPA)}} method
(see~\cite{Gower:GeneralizedProcrustesAnalysis_1975})
3-dim\-ensional data matrices can be analyzed with a multivariate
statistical technique. In particular, in our RGI we can compare the
individual (dim 1) natural lan\-guage con\-structs (dim 2) rated on
our fixed set of {\mathUI}s (dim 3). We conducted a GPA
with {\textsf{Idiogrid}} for each data set and refer
to~\cite{Koh:FramingsOfInformation:2013} for a de\-tailed description
of an analoguous GPA procedure. To provide a shared set of (virtual)
standard constructs on which the individual ratings of the \rgi
elements of each interviewee can be compared, the GPA method produces ``{\bf{abstract constructs}}'' of the form ``Con\_i -
ConOpo\_i'' with poles ``Con\_i'' and ``ConOpo\_i''.
\begin{figure}[h]\vspace*{-2em}
\includegraphics[width=\columnwidth]{pics_final/distinction_inMath_infoMath_noMath_2}\vspace*{-1em}
\caption{Dendogram of the Abstract Construct Clusters
(wrt. Euclidean distance and Ward clustering) of \Math, \zbMath
and \noMath: we can clearly discern a ``common'' cluster, which is
equally shared by all three, a strong \Math cluster and a fairly
strong \noMath cluster.}\vspace{-1.0em}
\label{fig:distinction}\vspace{-1.0em}
\end{figure}
Based on a pre-study we suspected a distinction of the interviewee
group not only into mathematicians and non-mathematicians, but into
research mathematicians, mathematics practitioners and
non-mathematicians. Therefore we compared the
element evaluations of the \Math, \zbMath, and \noMath group. We subjected
the union of the group-specific sets of abstract constructs to a
cluster analysis run by \textsf{OpenRepGrid} resulting in the
dendrogram in Fig.~\ref{fig:distinction}. Recall that
\textbf{dendrograms} are a visual representation of correlation
data. Two constructs in Fig.~\ref{fig:distinction} are closely
correlated, if their scores on the \rgi elements are similar. The
distance to the next upper level of two constructs/groups of
constructs {\emph{indicates}} this relative closeness. Please note
that we left out the scale in the dendrograms, as we are not
interested in the absolute numbers, only in their relative
groupings. This also means, that we won't use arguments in our
discussion of findings based on this scale. Nevertheless, we can for
example, conclude from Fig.~\ref{fig:distinction} that \con{6}{\Math}
and \con{7}{\Math} are the most correlated constructs. For the
conversion of \textsf{Idiogrid} data to \textsf{OpenRepGrid} data we
developed the according software.
The interview data seen in Fig.~\ref{fig:distinction} indeed suggest a difference
between how people in the \Math, \zbMath and the \noMath group think about {\mathUI}s.
The \zbMath interviewees' point of view lies between the one of \Math and \noMath
subjects. In particular, there are \zbMath abstract constructs in every cluster and there
is no cluster dominated by the \zbMath abstract constructs. As this user group dilutes
possible similarities or dissimilarities wrt the user group in focus -- the professional
mathematicians -- we further on only analyzed the \Math and \noMath data in depth. From here
on we will call \Math members ``mathematicians'' and \noMath members
``non-mathematicians''.
\begin{figure}\vspace*{-2em}
\includegraphics[width=\columnwidth]{pics_final/elementCluster_2}\vspace*{-1em}
\caption{Cluster Dendrograms of {\mathUI} Elements for \Math and
\noMath}\label{fig:elementCluster}\vspace*{-1em}
\end{figure}
Fig.~\ref{fig:elementCluster} gives a visualization of the element clusters of group
\Math resp. \noMath as dendrograms. The difference between the clusters is evident; we
will elaborate the interpretations in the next paragraphs.
There are three main element clusters for \Math in
Fig.~\ref{fig:elementCluster}. Clearly, one of these contains the {\mathUI} elements
whose main purpose it is to find mathematical content (``{\bf{math search cluster}}'').
In the math search cluster both, {\elemFormulaSearch} and {\elemMSCMap}, are innovative
mathematical services, nevertheless they are identified as being most similar to the
standard {\mathUI}s {\elemZBMathNew}, {\elemMathSciNet} and {\elemZBMathOld}. This shows
that \myPatterns[pat:Familiarity]{Mathematicians do not assess {\mathUI}s based on
familiarity.}
Another cluster includes all {\mathUI} elements that provide a
personal touch in the search process (``{\bf{personal search
cluster}}''). Here, the term ``personal'' labels the
interactive adaptation and customization of the search or search
results in a process driven by human interactions. In the interviews it became
quite clear, that anything involving human beings or communities was
highly distinctive and predominantly highly appreciated. So
\myPatterns[pat:Community]{Mathematicians trust human and community resources.}
Note that we don't mean a naive trust here, but a trust given the
sensible precautions. Even though the element clusters of the \noMath
interviewees also include a personal search cluster (see
Fig.~\ref{fig:elementCluster}), the elements {\elemBibliography} and
{\elemResearchGate} are missing and replaced by {\elemLibrary}. The
\Math participants explicitly commented that they don't have
confidence in the librarians' expertise in math. Interestingly,
mathematicians showed a lot of skepticism wrt {\elemResearchGate} but
not because they could not rely on the links the {\elemResearchGate}
members would provide, but rather because they mistrusted
{\elemResearchGate}'s competence in judging the relevance of links.
An indication of this is also given by the well-known observation that
mathematicians like anecdotes about fellow mathematicians like no
other community of practice.
The third cluster groups the remaining elements. Noticeably
{\elemGoogle} and {\elemGoogleScholar}, which mathematicians nowadays
use heavily for mathematical searches, are in this
cluster. Nevertheless, these elements
are not specific to math search, therefore we label this cluster as
the ``{\bf{general search cluster}}''.
According to {\name{Zhao}}'s usability study in~\cite{Zhao:MIR:2008},
mathematicians use {\myCitation{three main approaches: general keyword
search, browsing math-specific resources and personal contact.}}
This can also be seen in our three clusters for the \Math
group.
For the \noMath element clusters we only want to point out that the
clusters are indeed very different from the ones in the \Math
dendrogram. For example, for mathematicians the {\mathUI}s
\elemMathSciNet, \elemZBMathNew and \elemZBMathOld correlate the
highest, whereas for non-mathematicians each of them correlates more
with a different element than with each other. The only similarities
seem to be the obvious correlation between \elemGoogle and
\elemGoogleScholar, and the same very high correlation distance
between the personal search cluster and the others.
\begin{figure}[h
\includegraphics[width=\columnwidth]{pics_final/constructCluster_math_2}\vspace*{-1em}
\caption{Cluster Dendrogram of Construct Clusters in \Math. The first
two levels of the dendrogram were contracted for a more readable
image. Moreover, the numbers in parentheses in each construct
encode the individual interviewee issuing
it.}\label{fig:constructClustersMath}\vspace*{-2em}
\end{figure}
For a more precise qualitative analysis consider the dendrogram in
Fig.~\ref{fig:constructClustersMath}. First we decided on fitting
catego\-ries/sub\-ca\-te\-go\-ries for each cluster. We looked, for instance, at the first
main cluster and decided on the category ``fit for math''. Then we elaborated on its four
subclusters, e.g., for the fourth cluster we selected ``preconditions for search'' as a
subcategory. Note that there are blue-colored abstract constructs ``Con\_i -
ConOpo\_i'' among the constructs. We can interpret them now as characteristic constructs
of the corresponding major subcluster, so we associate each abstract construct with its
subcategory. Out of convenience, we call them by their explicit pole name together with
the corresponding data set, thus we say for example, ``\con{4}{\Math} `means' math
specificity''.
According to \name{Kuhltau et al.}
in~\cite{KuhltauEtAl:InformationSeekingRevisited:2008,Kuhltau:SeekingMeaning:2004} the
information search process can be described by a six-phase framework consisting of
{\emph{initiation}} (prompting a search), {\emph{selection}} (identifying information
needs), {\emph{exploration}} (pondering available tools and thus search strategies),
{\emph{formulation}} (formalizing search queries), {\emph{collection}} (gathering
information and goal-oriented cherry picking in search results), and {\emph{search
closure}} (giving up on the search). In our study we are not interested in the entire
search process, but in the interactions with the user interface, so we focus on the
iterative acts of selection, exploration, formulation and collection. In these phases a
user seeking information translates a search intension into a query or series of queries
optimizing for the relevance of the final collection of search results. Interestingly, the
four phases are mirrored in the construct clusters of the \Math group (see
Fig.~\ref{fig:constructClustersMath} on the left).
\begin{figure}\center\vspace{-2.5em}
\includegraphics[width=0.8\columnwidth]{pics_final/structureCoefficients_inMath}\vspace*{-1em}
\caption{Abstract Construct Ranking in \Math via Structure Coefficients}\label{fig:structureCoefficientsMath}\vspace*{-2.0em}
\end{figure}
To obtain a {\bf{ranking for the abstract constructs}} consider the
structure coefficients of the abstract constructs (wrt their ratings
on the three main principal components $PC_i$) for each interviewee
group in Fig.~\ref{fig:structureCoefficientsMath}
and~\ref{fig:structureCoefficientsNoMath}. The Euclidean {\bf{length}} of the
resp. 3-dimensional construct vector indicates its construct's
relevance. To distinguish between two abstract constructs that are in
the same subcluster in Fig.~\ref{fig:constructClustersMath}, we
compare their structure coefficients. For any distinctive deviation we
take a closer look in the biplots for the resp. PC-dimension and
elaborate on its meaning.
It is obvious that the rankings of the \Math group are distinct from
the ones of the \noMath group. What mathematicians care about the most
is the relevance of the search result with respect to their search
intension. So they seek interfaces and databases that allow them to
formulate precisely that in accordance with the respective search
philosophy they want to apply ({\con{6}{\Math}}, {\con{7}{\Math}},
{\con{5}{\Math}}, {\con{1}{\Math}}). As this describes a search
process that enables the user to find exactly what he is looking for,
we have \myPatterns[pat:Finding]{Finding is the primary mathematical
search task.}
Note that the math search cluster of the \Math group in Fig.~\ref{fig:elementCluster}
also has a clear focus on ``finding''. For the \noMath math search cluster this is much
less clear, e.g., the {\elemVifamath} \mathUI, which concentrates on collecting mathematical
information (from legacy math articles to images of mathematicians), but not on its
findability, thus mimicking a physical math library without noticeable presence of other
people. The interviewer observed that interviewees aligned the distinct kinds of search
like finding, browsing or solving with the clusters, but that the evaluation of search
activities was different for mathematicians and non-mathematicians. The former had a clear
preference for finding, followed by browsing and solving/asking, and even a hint of
rejection for surfing. In contrast, the \noMath participants indicated a preference for
browsing and surfing, followed by solving/asking and finally finding. Note that the
position of ``finding'' in this ranking may be well due to the fact, that only one
participant in the \noMath group worked in a scientific environment.
It is conspicuous that even though there was an obvious \mathUI
cluster with respect to ``people'' (the personal search cluster) for
the \Math group in Fig.~\ref{fig:constructClustersMath}, there is no
appreciation of ``socialness'' in their ranked list of constructs in
Fig.~\ref{fig:structureCoefficientsMath}. In particular,
mathematicians distinguish certain {\mathUI}s, i.e., the tools, as socially driven, but as
professionals they do not appreciate ``socialness'' as a value per
se in their evaluation schemes. In the theory of ``Communities of Practice
({\bf{CoP}})''~\cite{SituatedLearning}, {\bf{practices}} are not only typical
customs shared within a community, but they are tools that define the
community. Whereas in other CoPs
social interaction is a tool for achieving {\emph{social}} bindings,
in the mathematical CoP, social interaction is a tool for doing
mathematics, i.e., it is a mathematical practice. Therefore, we note
that \myPatterns[pat:SocialInteractionTool]{Mathematicians appreciate
social interaction as a {\emph{mathematical}} tool. In particular,
it is a mathematical practice to collaborate and exchange feedback.}
In this sense, we confirm {\name{Brown}}'s dictum
in~\cite{Brown:InformationSeekingScientists:1999} that mathematicians
may rely more heavily on their social network than other disciplines.
\begin{figure}\center\vspace{-2.0em}
\includegraphics[width=0.8\columnwidth]{pics_final/structureCoefficients_noMath}\vspace*{-1em}
\caption{Abstract Construct Ranking in \noMath via Structure
Coefficients}\label{fig:structureCoefficientsNoMath}\vspace*{-2.0em}
\end{figure}
Let us recall from Fig.~\ref{fig:distinction}, that some {\mathUI}
elements were in a subcluster shared by all three user groups. That
is, with respect to these constructs the \Math, \zbMath as well as
\noMath interviewees agreed on the evaluation of the given
{\mathUI}s. In particular, {\mathUI} scores correlate on
\con{3}{\zbMath} (``usability''),
{\con{1}{\noMath}} (``input design''),
\con{5}{\zbMath} (``simple design''),
{\con{6}{\noMath}} (``usability and interactivity''), and {\con{2}{\Math}} (``supportiveness:result''), {\con{1}{\Math}} (``adequacy: search philosophy'').
We note the different flavor of the non-\Math constructs versus the \Math
constructs. Where the former aim for design aspects, the latter are only concerned with
fitness of the {\mathUI} for achieving the search intension. It becomes even clearer if we
consider the phrasing ``usability'' in the non-\Math group and ``supportiveness'' in the
\Math group: Whereas usability is a neutral measure for all kinds of qualities while using
an object, supportiveness is a task-oriented requirement in the use-flow of a human
person. The media-theoretic difference is that the first doesn't tell us anything about
whether the user adopts a {\mathUI} as a mere tool or as a medium (in the sense of
{\name{McLuhan} as {\myCitation{any extension of the human body [\ldots] as a side-effect
of a technology}}~\cite[564]{McLuhan:ExtensionsOfMan:2003}}, i.e., a technology that
empowers its users): \myPatterns[pat:Medium]{Mathematicians aim at adopting a search tool
as a medium.} One consequence is that once they have adopted it as a medium, they won't
easily change to other media. Not surprisingly, this shared construct cluster also
supports a long-standing belief that \myPatterns[pat:Function]{Mathematicians appreciate
function over form.}
Even though the {\mathUI} elements' scores were highly correlated in
the shared cluster, their respective conceptualization can still
disagree. To understand the conceptualization, we look at the
meanings of the distinct constructs and the location of an element wrt
constructs. For instance, for mathematicians {\elemGoogleScholar}
enables a top-down approach (as search philosophy =\con{1}{\Math}) by
using a very general technique of ranking the search results (high
supportiveness for presenting search results =\con{2}{\Math}), but
offers a very textual input design (=\con{1}{\noMath}) and a
medium-rated effectiveness (as part of usability = \con{6}{\noMath})
for non-mathematicians. Here, note that the evaluation by the
mathematicians concerns the outcome, whereas the non-mathematicians
rather assess it by the input. This argument can be made more
generally, as the resp. construct clusters for the \Math
resp. \noMath groups favor the result resp. the input. Interestingly
one cluster category of the \Math group didn't make it into the
consensus grid. In particular, the category ``supportiveness of
input'' has no representative among the abstract constructs of
\Math. We conclude \myPatterns[pat:Outcome]{Mathematicians care
more for the outcome than the input.} This also means that
mathematicians seem to be willing to trade input hardships (like more
complex interfaces) for output satisfaction (i.e., having perfect
{\bf{precision}} -- all found results fit the search intension-- and
{\bf{recall}} -- all fitting results were found).
In Fig.~\ref{fig:distinction} we observe that the constructs
{\con{5}{\Math}}, {\con{6}{\Math}}, {\con{7}{\Math}}, and
{\con{8}{\Math}} are part of an abstract construct cluster only
containing \Math constructs; they are the enhanced (yellow colored)
constructs in Fig.~\ref{fig:structureCoefficientsMath}. Here, the
math interfaces scored similarly according to the attributes
``relevance of results: driven by user'', ``relevance of results:
driven by data'', ``precision of input'', and ``preconditions of
search''. Thus, we can interpret that a mathematical search interface
that empowers the user by enabling him to fine-tune the search query
is considered to strongly improve the relevance of the result. This
interpretation is supported by Pattern~\myPattern{pat:Outcome}, thus
we note that \myPatterns[pat:Empowerment]{Mathematicians want to be
empowered in the search process.}
Moreover, mathematicians obviously realize that this precision comes at a cost: the
underlying data have to be structured enough. Therefore, if the data do not allow such a
fine-tuning right away, they are willing to iteratively refine their query themselves. A
direct consequence seems to be that mathematicians want as much support in formulating a
search query as they can get.
Whereas non-mathematicians will agree that Pattern~\myPattern{pat:Outcome} is different
from their own approach, wrt the above consequence their attitude might be different as
the pattern describes the disregard of input facilities by mathematicians and the latter
the total investment of time and energy towards satisfying the search
intension. We already
observed that this is a cluster of elements marked by mathematicians only. That is, this
kind of evaluation scheme didn't occur to non-mathematicians, thus it isn't a dominant
one.
For Pattern~\myPattern{pat:Finding} we argued with the abstract construct ranking within
Fig.~\ref{fig:structureCoefficientsMath}. Interestingly, three of the four first ranked
items in that list belong to the uniquely mathematical cluster. The fourth one
(\con{8}{\Math}, ``preconditions for search'') occurred in the \Math group only, that is,
it discriminates between mathematicians' and non-mathematicians' search behavior ever
more. As mathematicians take the preconditions for search into account in the exploration
phase of an information search process, they value their anticipation of the search
outcome. This has two consequences: \myPatterns[pat:Transparency]{Mathematicians base
their information search process on transparency of the search result.}
Additionally, if they put a lot of thought into the exploration phase, they expect to be
rewarded by a good search result. So we hold \myPatterns[pat:Expectations]{Mathematicians
expect to find meaningful information in the search result.} In the interviews, it was
striking how much awe {\elemGoogle} evoked. Pattern~\myPattern{pat:Expectations} solves
this riddle: Considering the low amount of work to be invested in the exploration phase,
the expectations towards the search results are really low. Therefore, the relevance of
{\elemGoogle} searches amazes mathematicians tremendously.
\section{Understanding the Mathematical Perspective on {\mathUI}s: an
Example}\label{sec:example}
To see the utility of the elicited patterns, we will now discuss the {\mathUI}s \elemMathSciNet and \elemZBMathNew under a mathematical perspective, which is informed by our elicited patterns.
\begin{wrapfigure}{r}{7.2cm}\vspace*{-2em}
\includegraphics[width=0.6\columnwidth]{pics_final/MSN}\vspace*{-1em}
\caption{\mathUI of {\elemMathSciNet}}\label{fig:MSN}\vspace*{-2.5em}
\end{wrapfigure}
Let us start with {\elemMathSciNet} as seen in Fig.~\ref{fig:MSN} and {\elemZBMathNew}
shown in Fig.~\ref{fig:zbMath}. From above we know that mathematicians don't discern
between {\elemMathSciNet} and {\elemZBMathNew}. This immediately rai\-ses the question why
this might be the case. Evidently both layouts use a lot of vacuity to focus the users' attention
and use bright colors sparingly. But we know because of Pattern~\myPattern{pat:Function},
that the form is not important to mathematicians, so the reason for their alignment cannot
stem from these observations.
Unfortunately, at first glance the similarity of the start
page already ends here: {\elemZBMathNew} provides a simple search, i.e., a one-step
search, {\elemMathSciNet} a multi-dimensional, structured search.
Moreover, {\elemZBMathNew} offers innovative extra services
like mathematical software search and formula search,
{\elemMathSciNet} an extra citation service. The former offers inline
search fields to specify the search. The latter provides social media
connections. If we look at the search result page of each, we will find that there are as many differences.
Now let us take a closer look, for example, at the difference between
{\elemZBMathNew}'s simple search and {\elemMathSciNet}'s structured
search. We know because of Pattern~\myPattern{pat:Outcome}, that
mathematicians value the outcome higher than the input. Therefore, as
{\elemZBMathNew} offers not only the functionality of
{\elemMathSciNet}s multi-dimensional search via inline search fields
in the simple search but also choicewise a link to a structured
search, the functionality seems to be the same for mathematicians. The
input inefficiencies can be neglected, the potential outcome is the
same.
\begin{wrapfigure}{l}{7.2cm}\vspace*{-2.0em}
\includegraphics[width=0.6\columnwidth]{pics_final/zbMath}\vspace*{-1em}
\caption{\mathUI of {\elemZBMathNew}}\label{fig:zbMath}\vspace*{-2.0em}
\end{wrapfigure}
What about the clear differences in functionality in these
{\mathUI}s? Note that the social media links weren't recognized once
in the interviews with the mathematicians, which also fits
Pattern~\myPattern{pat:SocialInteractionTool}, stating that they
appreciate the communities' practices, but not the links to a
community themselves. Then, it seems rather evident that
{\elemZBMathNew} offers {\emph{more}} functionality, as
{\elemMathSciNet}'s extra functionality consists only of the citations
index. So shouldn't Pattern~\myPattern{pat:Empowerment} kick in and
lead to a distinctive perception of both systems?
We can counter-argue with two patterns. On the one hand,
Pattern~\myPattern{pat:Finding} tells us that finding is the major
kind of search a mathematician is conducting. The additional services
{\elemZBMathNew} provides the user with are essentially no services
that support finding, they rather support browsing. This is clear for
the mathematical software search. The facetted search with its
abilities to refine a search in the process also supports browsing
behavior explicitly. In contrast, the formula search feature was
designed for finding, but in the interviews, mathematicians indicated
that they simply don't believe in the finding capability of the
software (unfair as it is). In~\cite{Zhao:MIR:2008}, interestingly, a
similar phenomenon was observed. The underlying reason for this
disbelief could lie in Pattern~\myPattern{pat:Medium}, namely that
they have adopted {\elemZBMathNew} as a medium, and that uses string
search. Therefore, their conceptualization of this service doesn't fit
yet and is a challenge to change.
On the other hand, {\elemZBMathNew} is rather new. The older version didn't have as many
relevant extra features as this new one. Thus, Pattern~\myPattern{pat:Medium} strikes
again. Quite a few interviewees reported that they use {\elemMathSciNet} and even when
they became aware that {\elemZBMathNew} has more to offer now, they didn't mention any
intention to change over.\medskip
We can summarize that the patterns help us understand the perception of mathematicians
much better. This new-found understanding in turn triggers new design challenges and
ultimately better, more math-oriented designs.
\section{Conclusion}\label{sec:conclusion}
We have presented an \rgi study that was concerned with mathematical
search interfaces. To be able to understand the idiosyncracies of
mathematicians, we interviewed mathematicians as well as
non-mathematicians, with a focus on the former. From the quantitative
data and its qualitative interpretation several patterns emerged:
\begin{description}
\item [P~\myPattern{pat:Familiarity}] {\myCitation{Mathematicians
do not assess {\mathUI}s based on familiarity.}}
\item [P~\myPattern{pat:Community}] {\myCitation{Mathematicians trust human and community resources.}}
\item [P~\myPattern{pat:Finding}] {\myCitation{Finding is the
primary mathematical search task.}}
\item [P~\myPattern{pat:SocialInteractionTool}]
{\myCitation{Mathematicians appreciate social interaction as a
mathematical tool. In particular, it is a mathematical
practice to collaborate and exchange feedback.}}
\item [P~\myPattern{pat:Medium}] {\myCitation{Mathematicians aim
at adopting a search tool as a medium.}}
\item [P~\myPattern{pat:Function}] {\myCitation{Mathematicians
appreciate function over form.}}
\item [P~\myPattern{pat:Outcome}] {\myCitation{Mathematicians
care more for the outcome than the input.}}
\item [P~\myPattern{pat:Empowerment}] {\myCitation{Mathematicians
want to be empowered in the search process.}}
\item [P~\myPattern{pat:Transparency}]
{\myCitation{Mathematicians base their information search process
on transparency of the search result.}}
\item [P~\myPattern{pat:Expectations}]{\myCitation{Mathematicians
expect to find meaningful information in the search result.}}
\end{description}
With these patterns many design issues for {\mathUI}s can be understood and elaborated on much deeper now.
For instance, {\name{Libbrecht}} posed
in\cite{Libbrecht:TooPreciseTopicQueries:2013} the question whether
(mathematical) search queries may become too precise (so that the
search result becomes too small). But this question does only make
sense for browsing queries not for finding queries. The
Pattern~\myPattern{pat:Empowerment} suggests that the solutions should
be finetuned to the distinct kind of searches. If that is not
possible, the default case should be ``finding'' because of
Pattern~\myPattern{pat:Finding}.
Pattern~\myPattern{pat:Medium} indicates that a change from one tool
to another is not easily done by mathematicians. In particular, a
change of media will only occur if the innovation is disruptive, a
mere incremental innovation won't suffice. Therefore, phrasing a major
change (like the one from {\elemZBMathOld} to {\elemZBMathNew}) as a mere
update won't convince mathematicians to switch, and because of
Pattern~\myPattern{pat:Outcome}, neither will an announcement of
change that essentially points to the new Google-like layout of the
homepage.
Moreover, our data suggest that the search approach ``finding'' is
used by mathematicians predominantly when interacting with elements
from the math search cluster, ``browsing'' when interacting with
{\mathUI}s in the general search cluster and ``solving/asking'' when
using elements in the personal search cluster. Thus, we can look for
the properties of the resp. cluster to extend {\mathUI}s by more
search approaches. Note that \elemGoogle is best-known for its
browsing qualities, only for specific kinds of queries it is now also
successful in finding. Under this aspect \elemGoogle is also often
used by mathematicians.
Our future work is concerned with general design implications based on
the foundational work conducted in this paper. For example, one simple
consequence concerns the development process of math user interface
development: Specify the user group of your math service beforehand
and appreciate the credo of ``participatory design'' that strongly
admonishes developers to acknowledge the fact that {\emph{``You are
not the user!''}}. In our study, e.g., the mathematics
practitioners turned out to be different from the professional
mathematicians. Another consequence might be that we should make
mathematicians more effective by supporting their
{\emph{interventions}} in formulating a search query.
Finally, we like to note that the \rgi methodology -- even though strenuous at times -- seems to be a worthy methodology for use
with mathematicians.
\subsubsection*{Acknowledgement}
I thank all my interviewees for their motivation and patience with the
\rgi method. Moreover, I appreciated the supportive work environment
at zbMath, especially discussions with Wolfram Sperber. This work has
been funded by the Leibniz association under grant SAW-2012-FIZ. The final publication is available at {\url{http://link.springer.com}}.
\renewcommand\baselinestretch{.97} \printbibliography
\end{document}
| {
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options = select.children('option');
active.text(
select.children('option[selected]').val() ?
select.children('option[selected]').val() :
options.eq(0).val()
);
options.each(function(){
var optionOuter = $('<li></li>',{
class : 'animated_item'
}),
optionInner = $('<a></a>',{
text : $(this).val(),
href : '#'
}),
tpl = optionOuter.append(optionInner);
list.append(tpl);
});
list.on("click", "a", function(event){
event.preventDefault();
var v = $(this).text();
active.text(v);
select.val(v);
if(Core.TRANSITIONSUPPORTED){
$(this).closest('.dropdown').add(active).removeClass("active");
}else{
$(this).closest('.dropdown').add(active).removeClass("active visible");
}
});
Core.mainAnimation.prepareDropdown.call(list);
});
},
/**
** @return jQuery
**/
tabs : function(options){
return this.each(function(){
var $container = $(this),
tabs = {
init : function(){
$container.addClass('initialized');
this.nav = $container.children('.tabs_nav').length ? $container.children('.tabs_nav') : $container.children('.ts_nav');
this.subContainer = $container.children('.tab_containers_wrap').length ? $container.children('.tab_containers_wrap') : $container.children('.ts_containers_wrap');
this.tab = this.subContainer.children('.tab_container').length ? this.subContainer.children('.tab_container') : this.subContainer.children('.ts_container');
this.detectManyTabs(this.nav);
this.startState();
var self = this;
$(window).afterResize(function(){
self.responsive.bind(self)();
}, 300);
this.nav.on('click', 'a:not(.all)', { tabs : this }, this.openSubContainer);
},
detectManyTabs : function(collection){
if(collection.hasClass('tabs_nav') && collection.children('li').length >= 5 && $container.hasClass('type_2')){
$container.addClass('many_tabs');
}
},
startState : function(){
var i = this.nav.children('.active').index();
if(i < 0){
i = 0;
this.nav.children().eq(i).addClass('active');
}
var active = this.tab.eq(i);
active.siblings().addClass('invisible');
this.showTab(active);
},
openSubContainer : function(event){
var tabs = event.data.tabs,
tab = $($(this).attr('href'));
$(this).parent().addClass('active').siblings().removeClass('active');
tabs.showTab(tab);
event.preventDefault();
},
showTab : function(element){
var height = element.outerHeight();
element.removeClass('invisible').siblings().addClass('invisible');
this.subContainer.css('height', height);
},
responsive : function(){
var height = this.tab.not('.invisible').outerHeight();
this.subContainer.css('height', height);
}
}
tabs.init();
});
},
/**
** Call function after window resize and delay
** @param fn - function that will be called
** @param delay - Delay, after which function will be called
** @param namepsace - namespace for event
**/
afterResize : function(fn, delay, namespace){
var ns = namespace || "";
return this.each(function(){
$(this).on('resize' + ns, function(){
setTimeout(function(){
fn();
}, delay);
});
});
},
/**
** Accordion plugin
** @param toggle - set to true, if need to be toggle
** @return jQuery
**/
accordion : function(toggle){
return this.each(function(){
var $this = $(this),
active = $this.children('dt.active').length ? $this.children('dt.active') : $this.children('dt:first').addClass('active');
$this.children('dt').not(active).next().hide();
$this.on('click', 'dt', function(){
if(!toggle){
$(this).addClass('active')
.siblings('dt')
.removeClass('active')
.end()
.next('dd')
.stop()
.slideDown(300)
.siblings('dd')
.stop()
.slideUp(300);
}
else{
$(this).toggleClass('active').next().stop().slideToggle();
}
});
});
}
});
},
generateBackToTopButton : function(){
$('<button></button>', {
id : "back_to_top",
class : "def_icon_btn middle_btn theme_button animated transparent"
}).appendTo($('body'));
},
/**
** Generate stylized button instead default browser's button with type="file"
**/
attachButton: function(){
$('input[type="file"]').each(function(){
var wrap = $('<div></div>'),
btn = $('<button></button>', {
class: 'button_dark_grey middle_btn attach_file_btn',
type: 'button',
text: 'Browse'
});
wrap.append(btn);
$(this).before(wrap).hide();
});
$('.attach_file_btn').on('click', function(){
$(this).parent().next().trigger('click');
});
},
/**
** Generate rel attribute from valid data-rel attribute
**/
fancyboxValidationFix : function(){
var fb = $('[class*="fancybox_item"]');
if(!fb.length) return;
fb.each(function(){
$(this).attr('rel', $(this).data('rel'));
});
},
/**
** Handling animation when page has been scrolled
**/
animatedContent : function(){
$("[data-animation]").each(function() {
var self = $(this);
if($(window).width() > 767) {
self.appear(function() {
var delay = (self.attr("data-animation-delay") ? self.attr("data-animation-delay") : 1);
if(delay > 1) self.css("animation-delay", delay + "ms");
self.removeClass('transparent').addClass("visible " + self.attr("data-animation"));
}, {accX: 0, accY: -250});
}
else {
self.removeClass("transparent").addClass("visible");
}
});
},
/**
** handling main animation of theme
**/
mainAnimation : {
container : $('.dropdown'),
init : function(){
this.prepareEachDropdown();
this.bindEvents();
},
prepareDropdown : function(){
var self = Core.mainAnimation;
var $this = $(this),
hasDropdown = $this.find('.dropdown').length ? 'children' : 'find';
$this.data("fn", hasDropdown);
var items = $this[$this.data('fn')]('.animated_item'),
len = items.length;
$this.data("len", len);
self.defaultState($this);
if(Core.TRANSITIONSUPPORTED){
items.children('a').on(Core.TRANSITIONEND, function(event){
event.stopPropagation();
});
items.eq(0).on(Core.TRANSITIONEND, function(event){
if($this.hasClass("active") || event.originalEvent.propertyName !== "transform" || !event.target.classList.contains("animated_item")) return false;
self.defineNewState($this);
$this.removeClass("visible");
});
items.eq($this.data("len") - 1).on(Core.TRANSITIONEND, function(){
if(!$this.hasClass("active")) return false;
self.defineNewState($this,true);
});
}
},
prepareEachDropdown : function(){
var self = this;
self.container.each(self.prepareDropdown);
},
bindEvents : function(){
$('body').on('click.dropdown','[class*="open_"]',function(event){
$(this).add($(this).next()).toggleClass('active');
if(Core.TRANSITIONSUPPORTED){
$(this).next().addClass("visible");
}
else{
$(this).next().toggleClass('visible');
}
});
},
defineNewState : function(container, reverse){
container.addClass(container.data('fn'));
if(reverse){
var len = container.data("len"),
items = container[container.data('fn')]('.animated_item');
for(var i = len,j = 0; i >= 0, j < len; i--, j++){
items.eq(j).attr('style','transition-delay:' + i / 10 + 's');
}
}
else{
this.defaultState(container);
}
},
defaultState : function(container){
var items = container[container.data('fn')]('.animated_item');
items.each(function(i){
$(this).attr('style', 'transition-delay:' + (i+1) / 10 + 's');
});
}
},
/**
** store small events
**/
events: {
closeButton : function(){
$('body').on('click.close_button', '.close',function(){
$(this).parent().animate({
opacity : 0
},function(){
/* shopping cart addition*/
var $this = $(this),
ISSC = $this.closest('.shopping_cart').length,
collection = $this.parent().index() != 0 && ISSC ?
$this.add($this.parent()) : $this;
/* end shopping cart addition*/
collection.slideUp(function(){
if(!ISSC) return;
var parent = $(this).closest('.shopping_cart'),len;
$(this).remove();
len = parent.find('.animated_item').length;
parent.data("len", len);
Core.mainAnimation.defineNewState(parent,true);
});
});
});
},
/**
** product raring
**/
productRating : function(){
$('.rating').each(function(){
var i = $(this).children('.active:last').index();
$(this).on('mouseenter mouseleave','li',function(){
$(this).add($(this).siblings()).removeClass('active');
$(this).add($(this).prevAll()).addClass('active');
}).on('mouseleave',function(){
$(this).children('li').removeClass('active').eq(i).addClass('active').prevAll().addClass('active');
});
});
},
backToTop : function(offset){
var w = $(window),
b = $('#back_to_top');
w.on("scroll", function(){
if(w.scrollTop() > offset && !b.hasClass('visible')){
b.removeClass('transparent').addClass('bounceInRight visible');
}
else if(w.scrollTop() <= offset && b.hasClass('visible')){
b.removeClass('bounceInRight').addClass('bounceOutRight');
}
});
b.on('click', function(){
$('html,body').animate({
scrollTop : 0
},400,'swing');
}).on(Core.ANIMATIONEND, function(){
if(b.hasClass('bounceOutRight')){
b.removeClass('visible bounceOutRight').addClass('transparent');
}
});
},
socialFeeds : function(){
$('.social_feeds').find('[class*="open_"]').on('click', function(){
$(this).parent().siblings().children('.dropdown.active').removeClass(Core.TRANSITIONSUPPORTED ? 'active' : 'active visible');
});
},
qty : function(){
$('body').on('click.qty','.qty [data-direction]', function(){
var $this = $(this),
d = $this.data('direction'),
input = $this.siblings('input'),
c = +input.val();
if(c == 1 && d == "minus") return;
input.val(d == "minus" ? --c : ++c);
}).on('keydown.qty','.qty input', function(event){
var kC = event.keyCode;
if( !((kC >= 48 && kC <= 57) || (kC >= 96 && kC <= 105) || kC == 8) ) event.preventDefault();
});
},
productPreview : function(){
var cContainer = $('[data-output]'),
output = $(cContainer.data('output'));
cContainer.find('[data-large-image]:first').addClass('active');
cContainer.on('click','[data-large-image]', function(){
var $this = $(this);
if($this.hasClass('active')) return;
$this.addClass('active').parent().siblings().children('img').removeClass('active');
var src = $(this).data('large-image');
output.children('img').stop().animate({
opacity : 0
},
function(){
$(this).attr('src',src).stop().animate({
opacity : 1
});
});
});
},
openModalWindow : function(){
$('body').on('click.modal', '[data-modal-url]', function(event){
$.arcticmodal({
url : $(this).data('modal-url')
});
event.preventDefault();
});
},
/**
** Object that handles and sends user data
**/
contactForm : {
init : function(){
$('body').on('submit.contactform', '.contactform', { cf : this }, this.checkValidation);
},
checkValidation : function( event ){
var $this = $(this),
cf = event.data.cf,
errorMessage = "";
event.preventDefault();
errorMessage += cf.emptyFields($this.find('input, textarea'));
errorMessage += cf.minLength($this.find('textarea'),20);
if(errorMessage){
cf.showMessage(errorMessage, $this, 'error');
return false;
}
if(Core.XHRLEVEL2){
var fd = new FormData();
$.each($this.find("input[type='file']"), function(i, tag) {
$.each($(tag)[0].files, function(i, file) {
fd.append(tag.name, file);
});
});
var params = $this.serializeArray();
$.each(params, function (i, val) {
fd.append(val.name, val.value);
});
cf.ajaxSend(fd, $this);
}
else cf.ajaxSend($this.serialize(), $this);
},
emptyFields : function(collection){
var message = "";
var sortedCollection = $.map(collection, function(el){
if(el.hasAttribute('required')) return el;
});
sortedCollection.forEach(function(el){
if(!$(el).val()){
message = "All required fields must be filled!\n";
}
});
return message;
},
minLength : function(el, len){
if(!el[0].hasAttribute('required')) return '';
var message = "";
if(el.val().length < len) message = el[0].title + " must contain at least "+len+" characters!\n";
return message;
},
ajaxSend : function(data, container){
var cf = this;
$.ajax({
url: 'php/contact-send.php',
type: 'POST',
data: data,
cache: false,
contentType: false,
processData: false,
success : function(data){
cf.showMessage(data, container, data.indexOf('success') != -1 ? 'success' : 'error');
if(data.indexOf('success') != -1){
container.find('input, textarea').val('');
}
},
error : function(data){
cf.showMessage(data, container, 'error');
}
});
},
createFeedBackTemplate : function(message, type){
return $('<div></div>', {
text : message,
class : 'alert_box_' + type
}).hide();
},
showMessage : function(message, container, type){
var form = container;
if(form.hasClass('showed_message')) return false;
var alert = this.createFeedBackTemplate(message, type);
form.append(alert).addClass('showed_message');
alert.slideDown(300).delay(3000).slideUp(300, function(){
$(this).remove();
form.removeClass('showed_message');
});
}
},
/**
** Object that handles and sends user data
**/
subscribe : {
init : function(){
$('body').on('submit.subscribe', '.subscribe', { cf : this }, this.checkValidation);
},
checkValidation : function( event ){
var $this = $(this),
cf = event.data.cf;
event.preventDefault();
if(!$this.find('input').val()){
cf.showMessage("Enter your email address!", $this, 'error');
return false;
}
cf.ajaxSend($this.serialize(), $this);
},
ajaxSend : function(data, container){
var cf = this;
$.ajax({
url: 'php/subscribe.php',
type: 'POST',
data: data,
success : function(data){
cf.showMessage(data, container, data.indexOf('success') != -1 ? 'success' : 'error');
if(data.indexOf('success') != -1){
container.find('input').val('');
}
},
error : function(data){
cf.showMessage(data, container, 'error');
}
});
},
createFeedBackTemplate : function(message, type){
return $('<div></div>', {
text : message,
class : 'alert_box_' + type
}).hide();
},
showMessage : function(message, container, type){
var form = container;
if(form.hasClass('showed_message')) return false;
var alert = this.createFeedBackTemplate(message, type);
form.append(alert).addClass('showed_message');
alert.slideDown(300).delay(3000).slideUp(300, function(){
$(this).remove();
form.removeClass('showed_message');
});
}
},
/**
** Blog's layout type
**/
tableLayoutType : function(){
$('.layout_type').on('click', '[data-table-layout]', function(event){
event.preventDefault();
var $this = $(this),
container = $($this.parent().data('table-container'));
$this.addClass('active').siblings().removeClass('active');
container.removeClass('grid_view list_view list_view_products').addClass($this.data('table-layout'));
});
},
/**
** Reset filter
**/
resetFilter : function(){
if(!window.startRangeValues) return;
var startValues = window.startRangeValues,
min = startValues[0].toFixed(2),
max = startValues[1].toFixed(2);
$('.filter_reset').on('click', function(){
var form = $(this).closest('form'),
range = form.find('.range');
console.log(startValues);
// form.find('#slider').slider('option','values', startValues);
form.find('#slider').slider('values', 0, min);
form.find('#slider').slider('values', 1, max);
form.find('.options_list').children().eq(0).children().trigger('click');
range.children('.min_value').val(min).next().val(max);
range.children('.min_val').text('$' + min).next().text('$' + max);
});
},
sellerInfoDropdown: function(){
$('.seller_info_wrap').on('click', '.seller_name', function(){
$(this).next('.seller_info_dropdown').fadeToggle();
});
},
acceptCookies: function(){
$('.accept_cookie').on('click', function(event){
event.preventDefault();
$(this).closest('.cookie_message').fadeOut();
});
}
},
helpers : {
/**
** find the maximum height
**/
sameheight : function(){
var $this = $(this), max = 0;
$this.find('.owl-item').children().css('height','auto').each(function(){
max = Math.max( max, $(this).outerHeight() );
}).promise().done(function(){
$this.find('.owl-item').children().css('height', max);
});
},
/**
** Get first and last visible elements in carousel
**/
owlGetVisibleElements : function(){
var $this = $(this);
$this.find('.owl-item').removeClass('first last');
$this.find('.owl-item.active').first().addClass('first');
$this.find('.owl-item.active').last().addClass('last');
}
},
responsiveTopBar : {
/**
** initialize events for the responsive version of mega menu in the side menu
**/
init : function(){
this.container = $('#header').find('.topbar');
this.menu = this.container.find('.submenu');
this.createResponsiveButton(this.container);
this.checkViewPort();
$(window).afterResize(this.checkViewPort.bind(this),300);
},
/**
** activated the desired handler relative to window size
**/
checkViewPort : function(){
var wW = $(window).width();
// landscape
if(wW > 767 && wW < 1192){
this.activateTablet();
}
// mobile && portrait
else if(wW <= 767){
this.activateMobile();
}
// desktop
else{
this.deactivateResponsive();
}
},
/**
** Reset tablet handler and add mobile handler
**/
activateMobile : function(){
this.deactivateResponsive();
this.menu.add(this.container).hide();
this.menu.parent().off('click.tablet').on('click.mobile', this.mobileHandler);
},
/**
** Reset mobile handler and add tablet handler
**/
activateTablet : function(){
this.deactivateResponsive();
this.menu.parent().off('click.mobile').on('click.tablet', this.tabletHandler);
},
/**
** Reset all handlers and return default state
**/
deactivateResponsive : function(){
$('.tb_toggle_menu').removeClass('active');
this.container.add(this.menu).show();
this.menu.parent().removeClass('tablet_active mobile_active');
this.container.find('.prevented').removeClass('prevented');
this.menu.parent().off('click.mobile click.tablet');
},
/**
** Mobile handler
** @param event - event handler
**/
mobileHandler : function(event){
var link = $(this).children('a'),
menu = $(this).children('.mega_menu, .submenu');
if(!link.hasClass('prevented') && menu.length){
link.addClass('prevented');
menu.stop().slideDown();
$(this).addClass('mobile_active').siblings().removeClass('mobile_active').children('a').removeClass('prevented').next().stop().slideUp();
event.preventDefault();
}
},
/**
** Tablet handler
** @param event - event handler
**/
tabletHandler : function(event){
var link = $(this).children('a'),
menu = $(this).children('.mega_menu, .submenu');
if(!link.hasClass('prevented') && menu.length){
link.addClass('prevented');
$(this).addClass('tablet_active').siblings().removeClass('tablet_active').children('a').removeClass('prevented');
event.preventDefault();
}
},
/**
** Create button for main navigation
**/
createResponsiveButton : function(container){
var el = $('<button></button>',{
class: 'tb_toggle_menu'
}).insertBefore(container);
el.on('click.responsiveButton', function(){
$(this).toggleClass('active').next().slideToggle();
});
}
},
responsiveSideMegaMenu : {
/**
** initialize events for the responsive version of mega menu in the side menu
**/
init : function(){
this.menu = $('.mega_menu, .submenu');
this.menu.data('in_main_nav', this.menu.closest('.main_navigation').length ? true : false);
this.createResponsiveButton();
this.checkViewPort();
$(window).afterResize(this.checkViewPort.bind(this),300);
},
/**
** activated the desired handler relative to window size
**/
checkViewPort : function(){
var wW = $(window).width();
// landscape
if(wW > 991 && wW < 1400 && Core.ISTOUCH){
this.activateTablet();
}
// mobile && portrait
else if(wW <= 991){
this.activateMobile();
}
// desktop
else{
this.deactivateResponsive();
}
},
/**
** Reset tablet handler and add mobile handler
**/
activateMobile : function(){
this.deactivateResponsive();
this.menu.add($('.main_navigation')).hide();
this.menu.parent().children('a').off('click.tablet').on('click.mobile', this.mobileHandler);
},
/**
** Reset mobile handler and add tablet handler
**/
activateTablet : function(){
this.deactivateResponsive();
this.menu.parent().children('a').off('click.mobile').on('click.tablet', this.tabletHandler);
},
/**
** Reset all handlers and return default state
**/
deactivateResponsive : function(){
$('.toggle_menu').removeClass('active');
this.menu.add($('.main_navigation')).show();
this.menu.parent().removeClass('tablet_active mobile_active');
this.menu.siblings('.prevented').removeClass('prevented');
this.menu.parent().children('a').off('click.mobile click.tablet');
},
/**
** Mobile handler
** @param event - event handler
**/
mobileHandler : function(event){
var link = $(this),
menu = link.next('.mega_menu, .submenu');
if(!link.hasClass('prevented') && menu.length){
link.addClass('prevented');
menu.stop().slideDown();
$(this).parent().addClass('mobile_active').siblings().removeClass('mobile_active').children('a').removeClass('prevented').next().stop().slideUp();
event.preventDefault();
}
},
/**
** Tablet handler
** @param event - event handler
**/
tabletHandler : function(event){
var link = $(this),
menu = link.next('.mega_menu, .submenu');
if(!link.hasClass('prevented') && menu.length){
link.addClass('prevented');
$(this).parent().addClass('tablet_active').siblings().removeClass('tablet_active').children('a').removeClass('prevented');
event.preventDefault();
}
},
/**
** Create button for main navigation
**/
createResponsiveButton : function(){
var el = $('<button></button>',{
class: 'toggle_menu'
}).insertBefore($('.main_navigation'));
el.on('click.responsiveButton', function(){
$(this).toggleClass('active').next().slideToggle();
});
}
},
stickyMenu : {
/**
** Initialize variables
**/
initVars: function(){
this.NAVIGATION = $('#main_navigation_wrap, .sticky_part');
if(!this.NAVIGATION.length) return false; // temp
this.HEADER = $('#header');
this.updatesInfo();
this.needToBeSticky();
$(window).afterResize(this.needToBeSticky.bind(this), 300);
},
/**
** Initialize sticky menu
**/
initializeSticky: function(){
this.NAVIGATION.addClass('sticky_initialized');
this.activateSticky();
$(window).on('scroll.sticky', this.activateSticky.bind(this));
},
/**
** Checks if sticky menu need to initialize
**/
needToBeSticky: function(){
if($(window).width() > 991 && this.NAVIGATION.hasClass('sticky_initialized')){
this.updatesInfo();
this.activateSticky();
}
else if($(window).width() > 991 && !this.NAVIGATION.hasClass('sticky_initialized')){
this.initializeSticky();
}
else if($(window).width() <= 991){
this.destroy();
}
},
/**
** Method that checks scrollbar position and adds/removes
** fixed class on main navigation wrapper element
**/
activateSticky: function(){
if($(window).scrollTop() >= this.navTopOffset && !this.NAVIGATION.hasClass('sticky')){
this.NAVIGATION.addClass('sticky');
this.headerSizeCompensation(true);
}
else if($(window).scrollTop() < this.navTopOffset && this.NAVIGATION.hasClass('sticky')){
this.NAVIGATION.removeClass('sticky');
this.headerSizeCompensation(false);
}
},
/**
** Returns main navigation wrapper element to default position
**/
resetStickyPosition: function(){
this.NAVIGATION.removeClass('sticky');
this.headerSizeCompensation(false);
},
/**
** Updates information about main navigation wrapper element (height, offset)
**/
updatesInfo: function(){
this.resetStickyPosition();
this.navTopOffset = this.NAVIGATION.offset().top;
this.navHeight = this.NAVIGATION.outerHeight();
},
/**
** Add padding-bottom to header for compensation sticky menu size
**/
headerSizeCompensation: function(on){
if(on){
this.HEADER.css('padding-bottom', this.navHeight);
}
else{
this.HEADER.css('padding-bottom', 0);
}
},
/**
** Destroy sticky menu
**/
destroy : function(){
this.NAVIGATION.removeClass('sticky_initialized');
this.resetStickyPosition();
$(window).off('scroll.sticky');
this.headerSizeCompensation(false);
}
}
};
Core.extend();
$(document).ready(function(){
Core.afterDOMReady();
});
$(window).load(function(){
Core.afterWindowLoaded();
});
return Core;
}(Core || {})); | {
"redpajama_set_name": "RedPajamaGithub"
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\section{Introduction}
One of the most important cornerstones for our understanding of critical phenomena is the concept of self-similarity and scale invariance~\cite{Cardy1996}. These terms describe a situation in which the physical state of a system is in some sense invariant under a change of scale of the supporting geometry. Scale-invariant properties are often found to be universal, i.e. they depend only on the symmetries of the system but not on the specific microscopic realization.
On a lattice, a scale transformation can be carried out by coarse-graining the elementary degrees of freedom. One of the best known examples is Wilson's block-spin renormalization of the Ising model, where several spins are grouped into blocks~\cite{Fisher,Wilson,Kadanoff}. If the Hamiltonian in terms of these block variables has the same form as the original one (possibly with different parameters), this procedure maps the system onto itself, forming a group of renormalization transformations. The analysis of the renormalization group (RG) flow and the corresponding fixed points reveals essential information that characterizes the critical behavior.
In most studies, scale invariance is used as a mathematical tool for the analysis of critical phenomena. The aim of the present study is to view scale transformations from a different perspective, namely, as part of the physical process itself. More specifically, we will consider time-dependent processes in which infinitesimal scale transformations are continuously carried out as part of the dynamics. This would correspond to a self-inflating (or self-deflating) supporting geometry on which the process takes place. If the process itself is scale-invariant, it would be interesting to study how it responds to the ongoing change of scale of the underlying support. As a possible motivation, such self-inflating scale-free processes may be viewed as toy models for critical phenomena taking place in an expanding universe.
\begin{figure}
\centering\includegraphics[width=130mm]{figures/demo/demo}
\label{fig:rw}
\caption{\textbf{Random walk on a self-inflating support.} Left: Ordinary random walk in a static two-dimensional plane. The changing color scale indicates the advance of time. Right: The corresponding random walk with exactly the same sequence of microscopic displacements taking place on a self-inflating two-dimensional plane (see text). Shown is the situation at the end of the simulation. Consequently, earlier steps have undergone a stronger expansion compared to later ones.}
\end{figure}
As a simple example let us consider a random walk in two dimensions, as shown in Fig.~\ref{fig:rw}. The left side of the figure shows an ordinary random walk on a static background consisting of $1000$ statistically independent displacements in random directions. The right side shows the same random walk with an identical sequence of steps on a self-inflating background, dilating the underlying space during each step by the factor $1.002$. Illustratively stated, this could be the random path of an ant on an inflating balloon.
It is important to note that the random walk on the right side consists of exactly the same sequence of microscopic displacements as the one shown on the left side. However, early moves are significantly stretched by the ongoing dilatation, leading in total to a completely different trajectory of the random walker. Obviously, a random walker on a self-inflating support moves faster away from the origin than in the static case. Are the resulting distances still normally distributed? If so, how do they grow with time?
The purpose of this paper is to investigate these questions in detail. To this end we first discuss the phenomenological scaling properties of a self-similar stochastic process on a homogeneously inflating background. In Sect.~3, we recall how random walks can be analyzed in terms of cumulants. We then analyze a random walk on an expanding background both numerically and by analytical arguments. Here, we consider two particular types of inflation, namely, exponential and power-law driven expansion. The paper ends with concluding remarks.
\section{Phenomenological scaling on an expanding background}
Consider a physical Euclidean space which is expanding homogeneously and isotropically by itself. Choosing one of its points as the origin, any other point at location $\vec x(t)$ will move away from the origin. More specifically, the corresponding position vector $ \vec{x}(t) $ will grow according to the differential equation
\begin{equation}
\label{eq:selfinflation}
\frac{{\rm d}}{{\rm d} t}{\vec x}(t) = H(t)\,\vec x (t)\,,
\end{equation}
where the scalar quantity $H(t)$ is the (possibly time-dependent) expansion rate. Alluding to cosmology, this function will be referred to as the Hubble parameter. For a given Hubble parameter, we can define the dimensionless scale factor $a(t)$. This scale factor evolves by the differential equation
\begin{equation}
H(t)=\frac{\dot a(t)}{a(t)}
\end{equation}
with the solution
\begin{equation}
a(t) \;=\; a(t_0) \,\exp \Bigl[ \int_{t_0}^t {\rm d} t'\, H(t') \Bigr].
\label{eq:scalefactor}
\end{equation}
The scale factor $a(t)$ tells us by which factor the length scales have been inflated with respect to a certain reference time $t_0$. Since we want to start from a non-stretched system, we set $ a(t_0)=1 $.
Let us now consider a scale-free dynamical critical phenomenon on this self-inflating space. The scaling hypothesis states that the physical state of such a system is invariant under rescaling of space and time by $\vec x \to a \vec x$ and $t \to a^z t$, where $a>0$ is a scale factor and $z=\nu_\parallel/\nu_\perp$ is the so-called dynamical exponent. In addition, for a random walk the probability density $p(\vec x,t)$ has to be rescaled by $p \to a^{-1}p$ in order to preserve normalization.~\footnote{More generally, for an arbitrary nonequilibrium critical phenomenon, the order parameter $\rho$ has to be rescaled by $\rho\to a^{-\beta/\nu_\perp}\rho$, where $\beta$ is the critical exponent associated with the order parameter.}
Assuming that the random walk is still self-similar even on a self-inflating support, we expect it to be invariant under the scale transformation
\begin{equation}
\vec x \to a(t)\, \vec x\,, \qquad t \to a(t)^z t \,, \qquad p \to a(t)^{-1} p\,.
\label{eq:scaling-transformation}
\end{equation}
As a result, the probability distribution of the random walker should still be given by a Gaussian stretched by the ongoing expansion. This could be used to 'accelerate' the buildup of the correlation length in such a process. However, it is intuitively clear that we cannot accelerate this buildup as much as we like because the expansion does not only stretch the large-scale structure but also the small-scale deviations caused by lattice effects. It could even be that the underlying geometry expands faster than the correlation length of the random walk itself, freezing its current structure on large scales.\footnote{This may have happened, for example, with quantum fluctuations during the inflationary period of the early Universe, which now manifest themselves as frozen irregularities on large scales in the microwave background~\cite{Liddle2009}.} In fact, as we see below, this will depend on the expansion characteristics, i.e., on the choice of the function $a(t)$.
\section{Characterizing a random walk by its cumulants}
\subsection*{Cumulants as a measure of non-Gaussianity}
Does a random walk on a self-inflating background still obey the central limit theorem? To answer this question we study the \textit{cumulants} of the corresponding probability densities. These quantities were first introduced by Thiele (1889) and have been given the name \textit{cumulants} by Fisher and Wishart (1932). A comprehensive account of their derivation and properties can be found in many standard textbooks on statistics, such as e.g., Kendall (1969) and can be summarized as follows:
Recall that for a random variable $X$ with real values $x \in \mathbb R$ distributed according to the probability density $P(x)$ the cumulants are defined by
\begin{equation}
\label{eq:cumulants}
\kappa_n \;:=\; (-i\partial_k)^n \ln \varphi(k) \Bigr|_{k \to 0}\,,
\end{equation}
where
\begin{equation}
\varphi(k) \;=\; \int_{-\infty}^{\infty} {\rm d} x\, e^{i k x} P(x)
\end{equation}
is the so-called \emph{characteristic function}. The cumulants are combinations of moments which are chosen in such a way that for a normal distribution $P(x)=\frac{1}{\sigma\sqrt{2 \pi}} e^{-(x-\mu)^2/2\sigma^2}$ all cumulants except for the mean $\kappa_1=\mu$ and the variance $\kappa_2=\sigma^2$ are equal to zero. This property is unique to the normal distribution. Therefore, in order to check if a certain distribution is Gaussian it is sufficient to show that all cumulants $\kappa_n$ for $n>2$ vanish identically. To this end, it is convenient to study the dimensionless cumulant ratios
\begin{equation}
R_n :=\frac{\kappa_n}{\kappa_2^{n/2}} \quad\quad\mbox{for}\quad n\geq2
\label{eq:cumulant_ratio_definition}
\end{equation}
as a measure for the non-Gaussianity of a distribution. The latter are especially useful in cases where the bare cumulants of a sum of random variables diverge, since they are standardized by the width of the underlying distribution.
The cumulants feature some useful properties. At first, it can easily be shown from Eq.~\eqref{eq:cumulants} that in the case of a sum of two statistically independent random variables $ X_1+X_2 $, where the corresponding characteristic functions multiply according to the convolution theorem, the cumulants simply add up, i.e.
\begin{equation}
\kappa_n[X_1+X_2] = \kappa_n[X_1] +\kappa_n[X_2] \qquad n \in \mathbb{N}.
\end{equation}
Secondly, the $ n $-th cumulant is a homogeneous function of degree $ n $, meaning that
\begin{equation}
\kappa_n[cX]=c^n \kappa_n[X]
\end{equation}
for all $ c>0 $. Both relations will be used in what follows.
\subsection*{Cumulants of an ordinary one-dimensional random walk on a static background}
As mentioned before, a random walk will be asymptotically Gaussian if all higher cumulants $ \kappa_3,\kappa_4,\ldots $ (or cumulant ratios $R_3,R_4,\ldots$) tend to zero as the number of steps increases. Let us first verify this criterion in the case of an ordinary (non-inflating) symmetric one-dimensional random walk
\begin{equation}
\label{eq:OrdinaryRW}
x_{j+1} := x_j + r_j \qquad \Rightarrow \qquad x_N = \sum_{j=0}^{N-1} r_j
\end{equation}
with the initial condition $x_0=0$ and statistically independent random displacements $r_j \in \mathbb R$. For simplicity let us assume that all displacement are drawn from a flat distribution between $-1$ and $1$, i.e. their probability density is given by
\begin{equation}
P(x) \;=\; \mathcal{U}_{(-1,1)}(x)\,=\,
\left\{\begin{array}{cl}
\frac{1}{2} & \mbox{ for }\; -1< x < 1, \\
0 & \mbox{ otherwise. }
\end{array}\right.\,
\label{eq:uniform_distribution}
\end{equation}
With the corresponding characteristic function $\varphi(k) = \frac{\sin k}{k}$ it is straight-forward to calculate the cumulants~(\ref{eq:cumulants}) of a single random walker step, the first few non-vanishing being
\begin{equation*}
\kappa_2=\frac13, \quad \kappa_4=-\frac2{15}, \quad \kappa_6=\frac{16}{63}, \quad \kappa_8=-\frac{16}{15},\,\quad \ldots
\end{equation*}
or, as a general expression
\begin{equation}
\kappa_n= \frac{2^n B_n}{n} \quad\quad \mbox{for} \quad n\geq 2,
\end{equation}
where $ B_n $ denotes the $ n $-th \emph{Bernoulli number}.
In order to prove the central limit theorem, we consider the sum over $ N $ subsequent steps of the random walker
\begin{equation*}
Y = X_0+X_1+ ... + X_{N-1},
\end{equation*}
where the random variables $ X_0, X_1, \ldots $ denote the single moves. Since all $ X_j $ are drawn from the same distribution, the index $ j $ can be omitted and the cumulant ratio after $ N $ steps, $ R_n^{(N)}:=R_n[Y] $, reads
\begin{eqnarray}
R_n^{(N)}& = \frac{\kappa_n[Y]}{\kappa_2[Y]^{n/2}}
= \frac{\sum_{j=0}^{N-1}\kappa_n[X_j]}{\left(\sum_{j=0}^{N-1}\kappa_2[X_j]\right)^{n/2}}
= \frac{\sum_{j=0}^{N-1}\kappa_n}{\left(\sum_{j=0}^{N-1}\kappa_2\right)^{n/2}}\nonumber\\
&= \frac{N\kappa_n}{(N\kappa_2)^{n/2}}
= \frac{R_n}{N^{n/2-1}}
= \left\{
\begin{array}{cl}
1 &\quad\mbox{for}\quad n=2, \\
0 &\quad\mbox{for}\quad n>2\quad\mbox{and}\quad N\rightarrow\infty.
\end{array}\right.
\end{eqnarray}
Hence, in the limit $ N\rightarrow\infty $, the sum $Y$ exhibits a symmetric Gaussian distribution with variance $ \sigma_Y^2=N \kappa_2 $, reproducing the central limit theorem for an ordinary random walk with bounded displacements.
\section{Self-inflating random walk}
Let us now turn to a random walk on a self-inflating background. To this end we first rewrite the dynamical equation of an ordinary random walk (\ref{eq:OrdinaryRW}) in the form
\begin{equation}
\frac{{\rm d}}{{\rm d} t} x(t) \;=\; \sum_{j=0}^{N-1} \, r_j \, \delta(t-t_j)\,.
\end{equation}
where the $ t_j $ denote the time instances when the jumps take place. In this equation the effect of self-expansion can be incorporated by simply including the term on the r.h.s. of Eq.~(\ref{eq:selfinflation}), i.e.
\begin{equation}
\frac{{\rm d}}{{\rm d} t} x(t) \;=\; H(t)x(t) + \sum_{j=0}^{N-1} \, r_j \, \delta(t-t_j)\,.
\label{eq:equation_of_motion}
\end{equation}
Introducing the rescaled position $s(t)={x(t)}/{a(t)}$, one can easily show that the Hubble parameter drops out, leading to
\begin{equation}
\frac{{\rm d}}{{\rm d} t} s(t) \;=\; \sum_{j=0}^{N-1} \, \frac{r_j}{a(t_j)} \, \delta(t-t_j)\,.
\end{equation}
Hence, the rescaled position of the random walker at time $t_N$ will be given by the sum
\begin{equation}
s_N \;:=\; s(t_N) \;=\; \sum_{j=0}^{N-1} r_j\, w_j
\label{eq:final_rescaled_position}
\end{equation}
of statistically independent random numbers $r_j$ weighted by a factor
\begin{equation}
w_j:=w(t_j)=\frac{1}{a(t_j)}.
\end{equation}
On an inflating space, where $a(t)$ increases with time, this means that the weights decrease from move to move. In fact, as we saw in Fig.~\ref{fig:rw}, the stretch is maximal for early moves.
\begin{figure}
\centering\includegraphics[width=0.65\textwidth]{figures/weighted_distributions/distributions}
\caption{\textbf{Weighted probability density.} Starting from the symmetric uniform distribution of Eq.~\eqref{eq:uniform_distribution}, the succeeding distributions become narrower with time (lighter colors) in the rescaled frame of reference. Drawn are the first ten steps, with constant time steps $ \Delta t=1 $. }
\label{fig:distributions}
\end{figure}
\subsection{Cumulants of a self-inflating random walk}
As pointed out in the preceding section, the random walk on an inflating background is equivalent to a random walk on a static background with weighted steps. Following Eqs.~\eqref{eq:uniform_distribution} and \eqref{eq:final_rescaled_position}, the probability density of a single step $ X_j $ in the rescaled frame of reference now reads
\begin{equation}
P_{X_j}(s)=\mathcal{U}_{(-w_j,w_j)}(s) = \left\{
\begin{array}{cl}
\frac{1}{2w_j}&\quad\mbox{for}\quad -w_j<s<w_j,\\
0 &\quad\mbox{otherwise}.
\end{array}\right.
\end{equation}
Fig.~\ref{fig:distributions} depicts the time evolution of $ P_{X_j} $ schematically.
It turns out that the random variable $ X_j $, which determines the $ j $-th step of the random walker, is obviously given by $ X_j=w_j X_0 $, whereat $ X_0 $ denotes the random variable of the initial step ($ w_0=1 $). However, in order to avoid any misunderstandings, it is important to notice that for the particular realizations $ X_0=r_0 $ and $ X_j =r_j $ in general $ r_j\neq w_j r_0 $ holds since the $ r_j $ are independently generated random numbers. We can now easily calculate the $ n $-th cumulant of $ X_j $ using the homogeneity property
\begin{equation}
\kappa_n[X_j] = \kappa_n[w_j X_0] = w_j^n \kappa_n[X_0] = w_j^n\, \frac{2^n\,B_n}{n}.
\end{equation}
Furthermore, the cumulants after $ N $ subsequent steps,
\begin{equation}
Y=X_0+X_1+...+X_{N-1},
\label{eq:nonstandardized_sum}
\end{equation}
can be obtained by simply summing up all single cumulants
\begin{equation}
\kappa_n^{(N)}:=\kappa_n[Y]= \sum_{j=0}^{N-1}\kappa_n[X_j].
\label{eq:cumulants(N)}
\end{equation}
If we calculate the cumulant ratio \eqref{eq:cumulant_ratio_definition} of the $ j $-th step which we label $ R_n[X_j] $, the weights cancel, thus giving
\begin{equation}
R_n[X_j] = R_n[X_0] = \frac{\kappa_n}{\kappa_2^{n/2}} = \frac{(2\sqrt{3})^n\,B_n}{n}.
\end{equation}
Finally, the cumulant ratio after $ N $ steps is given by the following expression
\begin{equation}
R_n^{(N)} = \frac{\kappa_n^{(N)}}{\left(\kappa_2^{(N)}\right)^{n/2}} = \frac{\kappa_n}{\kappa_2^{n/2}}\; \frac{\sum_{j=0}^{N-1}w_j^n}{\left(\sum_{j=0}^{N-1}w_j^2\right)^{n/2}}.
\label{eq:cumulant_ratio(N)}
\end{equation}
\subsection{Exponentially expanding support}
We now consider the special case of an exponential expansion $ a(t)=a(t_0)\,\mathrm{e}^{\,\mu (t-t_0)} $ where $ \mu $ controls the expansion time scale. This corresponds to a choice of $ H(t)=\mu=\mathrm{const} $ for the Hubble parameter. We calculate the sum \eqref{eq:nonstandardized_sum} of $ N $ subsequent random walker steps. For convenience, we set $ t_0=0 $, $ a(t_0)=1 $ and use a time step of $ \Delta t=1 $, thus giving weight factors of $ w_j =1/\mathrm{e}^{\,\mu j} $. The cumulants \eqref{eq:cumulants(N)} then read
\begin{equation}
\kappa_n^{(N)}=\frac{2^n\,B_n}{n}\sum_{j=0}^{N-1}\frac{1}{\mathrm{e}^{n\mu j}} =\kappa_n\; \frac{1-\mathrm{e}^{-n\mu N}}{1-\mathrm{e}^{-n\mu}},
\end{equation}
which, as $ N\rightarrow \infty$, results in
\begin{equation}
\kappa_n^\infty = \frac{\kappa_n}{1-\mathrm{e}^{-n\mu}} = \mathrm{const.} \neq 0 \quad \forall \mu> 0 \quad\mbox{and}\quad \forall n \quad\mbox{even}.
\end{equation}
Consequently, also the cumulant ratio $R_n^\infty\neq0$ and the central limit theorem fails in this case, indicating that on an exponentially inflating support, the random walk loses its characteristic behavior. Moreover, for large $ \mu $, the above expression becomes $ \kappa_n^\infty =\kappa_n $ which is nothing else than the uniform distribution of the first step. This is a reasonable result, since for very large $ \mu $ the footprint of the initial step dominates all remaining moves.
\subsection{Algebraically expanding support}
\label{sec:Algebraically_expanding_support}
As exponential background growth is obviously too fast to conserve the normal distribution of a random walker, we study now the behavior in the case of an algebraical expansion, i.e. the expansion is directed by a power law. According to Equation~\eqref{eq:scalefactor}, the expansion rate $H(t)=\lambda/t$ leads to a scale factor $a(t)=(t/t_0)^\lambda$. Hence the cumulants~\eqref{eq:cumulants(N)} are given by
\begin{equation}
\kappa_n^{(N)}=\frac{2^n\,B_n}{n}\sum_{j=1}^{N}\frac{1}{j^{n \lambda}} =\kappa_n \sum_{j=1}^{N}\frac{1}{j^{n \lambda}},
\label{eq:cumulant_algebraic(N)}
\end{equation}
using time steps of $\Delta t=1$ and setting $ t_0=1 $. In order to determine for which choices of $ \lambda $ the central limit theorem is still valid, we consider the cumulant ratio \eqref{eq:cumulant_ratio(N)}. It reads
\begin{equation}
R_n^{\infty}=\frac{\kappa_n}{\kappa_2^{n/2}}\frac{\sum_{j=1}^{\infty}\frac{1}{j^{n \lambda }}}{\left(\sum_{j=1}^{\infty}\frac{1}{j^{2\lambda}}\right)^{n/2}}.
\end{equation}
By definition, $R_2^{\infty}=1$ holds for the second cumulant ratio, whereas for $n>2$ we have to consider three different cases:
\begin{itemize}
\item For $0\leq\lambda\leq1/n$ the denominator diverges faster than the numerator so that the cumulant ratios $R_n^\infty$ vanish. This can be seen if one compares both series, since for each two corresponding summands the inequality $\frac{1}{j^{2\lambda}}>\frac{1}{j^{n\lambda}}$ is fulfilled.
\item In the range $1/n<\lambda\leq1/2$ the numerator converges whereas the denominator diverges, thus $R_n^{\infty}\rightarrow0$ still holds.
\item For $\lambda>1/2$ one can write
\begin{equation}
R_n^{\infty}=\frac{\kappa_n}{\kappa_2^{n/2}}\frac{\zeta(n \lambda)}{\zeta(2\lambda)^{n/2}},
\end{equation}
where $\zeta(x)=\sum_{m=1}^{\infty}m^{-x}$ denotes a representation of the Riemann zeta function. It is monotonously decreasing for $x\in(1,\infty)$ with
\begin{equation}
\lim\limits_{x\rightarrow 1}\zeta(x)=\infty \quad\mbox{and}\quad\lim\limits_{x\rightarrow \infty}\zeta(x)=1.
\label{eq:limitzeta}
\end{equation}
Therefore, it is easy to see that $0< R_n^{\infty}< \frac{\kappa_n}{\kappa_2^{n/2}}$ for $n>2$ and $\lambda>1/2$.
\end{itemize}
Summing up, we have
\begin{equation}
R_n^{\infty}=\left\{\begin{array}{cl}
0 & \quad\mbox{ if}\quad 0\leq\lambda\leq\frac{1}{2} \\
\frac{\kappa_n}{\kappa_2^{n/2}}\frac{\zeta(n \lambda)}{\zeta(2\lambda)^{n/2}}\neq 0 & \quad\mbox{ if}\quad \lambda>\frac{1}{2}
\end{array}\right.\,\quad\quad\mathrm{for\ even}\ n>2.
\end{equation}
For this reason, we conclude that the central limit theorem still holds if $\lambda\leq\frac{1}{2}$, but fails for stronger background growth. Apparently, the expansion process prevails over the random walk in the latter case, since the underlying support expands faster than the length scale $\xi\sim t^{1/2}$ of the random walk itself.
Furthermore, we see from Eqs.~\eqref{eq:cumulant_algebraic(N)} and \eqref{eq:limitzeta}, that in the limit of large $ \lambda $
\begin{equation}
\lim\limits_{\lambda\rightarrow\infty}\kappa_n^\infty =\lim\limits_{\lambda\rightarrow\infty}\kappa_n \sum_{j=1}^\infty\frac{1}{j^{n \lambda}} = \kappa_n\lim\limits_{\lambda\rightarrow\infty}\zeta(n\lambda)=\kappa_n
\end{equation}
holds. Consequently, for very strong algebraic expansion the distribution becomes again uniform, analogously to the exponential case. This behavior was confirmed by numerical simulations (see left panel of Fig.~\ref{fig:numerical_results}), where the shape of the distribution differs more and more from a Gaussian as $ \lambda $ is increased. Furthermore, the right panel of Fig.~\ref{fig:numerical_results} shows the mean quadratic distance from the origin (i.e. the correlation length squared) for different choices of the expansion control parameter in both the exponential (dashed lines) and the algebraic case (solid lines). These numerical calculations support our analytical results, since it can be seen that only the solid lines for $ \lambda=0, 0.1 $ and $ 0.5 $ stay parallel, whereas all other curves (especially the exponential ones) diverge faster.\\
\begin{figure}[t]
\centering\includegraphics[width=0.49\textwidth]{figures/resulting_distributions/resulting_distributions}
\centering\includegraphics[width=0.49\textwidth]{figures/correlation_length/correlation_length}
\caption{\textbf{Numerical results.} Left panel: Resulting distributions on algebraically expanding backgrounds for different values of the expansion control parameter, binned from $ 10^7 $ Monte Carlo runs \`{a} $ 10^3 $ steps. Right panel: Mean quadratic distance from the origin over time for random walks on differently strong expanding backgrounds. Each data point was averaged over $ 10^6 $ Monte Carlo simulations. The solid straight lines denote algebraic expansion, with $ \lambda=0,\,0.1\,,0.5\,,1.0\,,2.0\,,5.0 $ in ascending order. The dashed lines denote exponential expansion with $ \mu=0.001,\, 0.01,\, 0.1,\, 1$, again in ascending order.\vspace{3mm} }
\label{fig:numerical_results}
\end{figure}
\subsection{Dynamical exponent}
The previous results indicate that the overall dynamics of the system can be split into a \emph{diffusion} and an \emph{inflation} component. When interested in the dynamical exponent of the system, we may therefore evaluate it for both sub-processes separately. It is commonly known that for ordinary diffusion the reference length of the system $ \xi $ scales as $ \xi \sim t^{1/z} $ with a dynamical exponent of $ z_{\mathrm{Dif}}=2 $. For the inflation however, the dynamical exponent is simply given by $ z_\mathrm{Inf} = 1/\lambda $ for a power-law driven expansion, since the length scale of the system scales as $ \xi \sim a(t)= t^{\lambda} $ in this case.
In Fig.~\ref{fig:dynamical-exponent} the inverse dynamical exponent is plotted against $ \lambda $ for both competing processes. There, the solid curve indicates the effective dynamical exponent of the full system, whereas the dashed lines denote the inferior processes. Altogether, we found that for the algebraically expanding support the dynamical exponent is given by
\begin{equation}
z = \min\left(z_\mathrm{Dif},z_\mathrm{Inf}\right)= \min\left(2,\lambda^{-1}\right)
= \left\{\begin{array}{cl}
2\quad &\mbox{for}\quad \lambda\leq \frac{1}{2}, \\
\frac{1}{\lambda}\quad &\mbox{for} \quad\lambda> \frac{1}{2}.
\end{array}\right.\,
\end{equation}
Finally, note that for exponential growth the dynamical exponent is meaningless since the inflation as the dominating sub-processes behaves non-algebraic in this case.
\begin{figure}[h]
\centering\includegraphics[width=0.49\textwidth]{figures/dynamical-exponent/z-lambda}
\caption{\textbf{Dynamical exponent.} Shown is the inverse dynamical exponent as a function of the expansion parameter $\lambda$. For $\lambda>1/2$, the inflation starts to dominate the dynamics and the system shows super-diffusive behavior.}
\label{fig:dynamical-exponent}
\end{figure}
\section{Conclusion}
In the present work we have studied the behavior of a random walk on a self-inflating background which expands isotropically and homogeneously with time. The expansion is described by the Hubble parameter $H(t)=\dot{a}(t)/a(t)$, where $a(t)$ is the associated scale factor.
Two cases have been considered, namely, an exponential inflation $H(t)=\mu=\mathrm{const}$ and an expansion with power-law characteristics where $H(t) = \lambda/t$. For exponential inflation, it turns out that the key properties of the random walk, most notably its scale invariance, is destroyed after some time since the random walker cannot propagate as fast as the background expands. For a power-law driven expansion, however, the situation depends on the value of the exponent $\lambda$. For $0<\lambda \leq \frac12$ the central limit theorem still holds so that a Gaussian distribution is obtained. For $\lambda> \frac12$, however, the expansion is so fast that higher cumulant ratios do not vanish. Here, the process is initially governed by diffusion with the dynamical exponent $z=2$, crossing over to an accelerated expansion with $z=1/\lambda < 2$ where the rescaled probability density is essentially frozen. We have only considered the case of a random walk in one dimension, since a generalization to a multi-dimensional setting (i.e. isotropic random walk as well as isotropic expansion) is straightforward as our results can be applied to each dimension individually.
This study has been focused on the random walk as the simplest example of a self-similar stochastic process. It would be interesting to see how other self-similar stochastic processes such as systems at the critical point of a phase transition behave on a self-inflating background.
\section*{References}
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,484 |
\section{Introduction}
We consider a mixture of $I$ monatomic gases, labeled with $\mathcal{A}_1, \dots, \mathcal{A}_I$. In the kinetic theory framework, each species of the mixture $\mathcal{A}_i$ is statistically described with its own distribution function $f_i:=f_i(t,x,v)$, that in general depends on time $t\geq0$, space position $x\in \mathbb{R}^3$ and velocity of molecules $v \in \mathbb{R}^3$ (in this manuscript we restrict ourselves to the spatially homogeneous case, that is, we drop dependence on space position $x$). The distribution function $f_i$ changes due to binary interactions (or collisions) with other particles. {\color{black} In the mixture setting, these particles can belong to other species $\mathcal{A}_j$, $j\neq i$. Therefore, the evolution of each $f_i$ involves not only the particle-particle interaction of specie $\mathcal{A}_i$, but also interactions between $\mathcal{A}_i$ and $\mathcal{A}_j$, $j\neq i$.}
In the mixture framework, the evolution of each distribution function $f_i$ describing the mixture component $\mathcal{A}_i$, is governed by the Boltzmann-like equation, that traditionally introduces collision operator as a measure of its change. Now, one has multi-species collision operators and their transition probabilities, or cross sections, between the different distribution functions describing each component of the mixture \cite{Sir62}. Since all species are considered simultaneously in a system of species with binary interactions, one is led to introduce a vector valued set of distribution functions $\mathbb{F}=\left[f_i\right]_{1\leq i \leq I}$, whose evolution is governed by a vector of collision operators, whose $i-$th component (that describes precisely evolution of $f_i$) is $\left[\mathbb{Q}(\F)\right]_{i}=\sum_{j=1}^I Q_{ij}(f_i,f_j)$. In this formula, operator $Q_{ij}(f_i,f_j)$ describes influence of species $\mathcal{A}_j$ for the distribution function $f_j$ on species $\mathcal{A}_i$ with the distribution function $f_i$. Note that summation over all $j=1,\dots,I$ is in the spirit of taking into account influence of all species $\mathcal{A}_j$, $j=1,\dots,I$, on the considered species $\mathcal{A}_i$.
From a mathematical viewpoint, the challenging situation occurs when masses of species molecules are not equal (i.e. $m_i\neq \text{m}_j$). In such a situation, underlying binary collisions between molecules lose some symmetry properties which can dramatically change mathematical treatment, for instance in order to study diffusion asymptotics when one needs to show the compactness of a part of linearized Boltzmann operator \cite{BGPS}. In the mixture framework, a linear system of linearized Boltzmann equations has been recently studied in \cite{BriantDaus16}, corresponding to the perturbative setting of our model when the non-linear system is linearized near Maxwellian states corresponding to each species. In this case authors showed existence, uniqueness, positivity and exponential trend to equilibrium. \\
In this work, we give the first existence and uniqueness result for the non-linear system of spatially homogeneous Boltzmann equations for multi-species mixtures with binary interactions in a suitable Banach space.
We also emphasize that our approach for solving the Cauchy problem for the Boltzmann equation with variable hard potentials relies on some specific conditions on the initial moments, without requesting entropy boundedness. The hard potentials assumption correspond to collision cross sections {\color{black} related to the species $\mathcal{A}_i$ and $\mathcal{A}_j$} proportional to the local relative speed with a power exponent $ { \gamma_{ij}} \in(0,1]$, and {\color{black} $L^1-$integrable} angular part $ {\color{black}b_{ij}} $, as function of the scattering direction.
In addition, the existence and uniqueness of a vector value solution $\mathbb{F}(t,v)$ need to assume that initially its scalar zero and second moment (i.e. the scalar number density and energy of the mixture) are strictly positive and finite, and additionally that this function has at least an upper bounded ${k_*}$-polynomial moments, {\color{black} where ${k_*}:= \max\{{\overline{k}}, 2+2 { {\color{black}\overline{\gamma}}} \}$, for {\color{black} ${\overline{k}}=\max_{1\leq i, j \leq I}\{ k^{ij}_*\}$ and $ { {\color{black}\overline{\gamma}}} = { {\color{black}\max_{1\leq i, j \leq I}\gamma_{ij}}} $,} is sufficiently large to ensure the prevail of the polynomial moments of loss term with respect to those same moments of the gain term. Each
$k^{ij}_* $ depends on the angular transition rate $b_{ij}$ as well as on the two-body mass fraction $r_{ij}:= m_i/(m_i+m_j)$ associated to each component on the vector solution.
All these parameters are defined in the next Section~\ref{Section notation} dedicated to notation, preliminaries and main results.}
\\
The result is obtained following general ODE theory that studies differential equations in suitable Banach spaces \cite{Martin}. In the context of (single) Boltzmann equation, this theory was proposed as a main tool in \cite{Bressan} for solving the Cauchy problem with hard spheres in three dimensions and constant angular transition probability kernel. However, the notes \cite{Bressan} do not completely verify all conditions of general ODE theory for the Boltzmann equation. This was motivation for \cite{GambaAlonso18} to revise the application of ODE theory from \cite{Martin} in the case of Boltzmann equation with more general hard potentials and integrable angular cross section, and in particular, to provide a complete proof of sub-tangent condition.
One very interesting new aspect from this approach is that the ODE flow in a suitable Banach space without imposing initial bounded entropy condition yields an alternative approach that allows for a rather general theory for gathering estimates where one can apply a rather general result in order to find solutions to the Cauchy problem for Boltzmann type flows where there is no classical entropy that is dissipated, or even some conservation laws may not be satisfied. Such problems have already been solved in for polymers kinetic problems \cite{AlonsoLods18}, quantum Boltzmann equation for bosons in very low temperature \cite{GambaAlonsoTran17} and more recently to study the weak wave turbulence models for stratified flows \cite{GambaSmithTran18}.\\
After proving the existence and uniqueness of the vector value solution $\mathbb{F}$ to the Boltzmann system, we turn to the study of generation and propagation of scalar polynomial and exponential moments of its solution $\mathbb{F}$.
The techniques we use in this manuscript are adaptations or extensions of results that have been developed for scalar Boltzmann type equations models.
In the case of the classical Boltzmann equation for the single elastic monatomic gas model, polynomial moments have been exclusively considered, for instance, in \cite{Des93} and \cite{wennberg97} for hard potentials where propagation and generation of such moments was proved. About the same time, Bobylev introduced in \cite{Bob97} the concept of exponential moment as a measure of the distribution solution tail, referred as to {\sl tail temperature},
by showing that solutions to the Boltzmann equation for monatomic gases, modeled by elastic hard spheres (i.e. {\color{black} power exponent } $\gamma=1$)
in three dimensions with a constant angular dependent cross-section as a function of the scattering direction, have inverse Maxwellian weighted moments, globally in time, whose tail decay rate depend on moments of the initial data. His proof consists in showing that infinite sums of renormalized polynomial moments are summable whose limit is proportional to a $L^1$- Gaussian weighted norm for the unique probability density function solving the initial value problem associated to the Boltzmann equation, whose rate depends on the initial data that must also be integrable with a Gaussian weight. These techniques of understanding moments summability in order to obtain high energy tail behavior for the solution of the Boltzmann equation was extended to inelastic interactions with stochastic heating sources, shear flows or self-similarity scalings to obtain non-equilibrium statistical stationary (NESS) states \cite{GambaBobPanf04} where the exponential rates did not necessarily correspond to Gaussian weighted moments.
This concept in the elastic case was further extended by \cite{GambaPanfVil09} to collision kernels for hard potentials (i.e. $\gamma\in (0,1]$) for any angular section
with $L^{1+}$-integrability. Further, generation of exponential moments of order $\gamma/2$ with bounded angular section were shown in \cite{Mouhot06}.
By then it became clear that the study of general forms of exponential moments resulted as a by-product of the analysis of polynomial moments (or tails), and so a spur of work arose for the improvement of conditions and results that will allow to estimate, globally in time.
These results were extended to collision kernels for hard potentials with $\gamma\in(0,2]$ for any angular section
with just $L^{1}$-integrability by a new approach using partial sums summability techniques, rather than using summability studies by power series associated to
renormalized moments as proposed in \cite{Bob97, GambaBobPanf04, GambaPanfVil09,Mouhot06}. The generation results were improved to obtain exponential moments of order $\gamma$, while Gaussian moments were propagated for any initial data that would have that property,
independent of $\gamma$. All these results were extended to the angular non-cutoff regime (lack of angular integrability) in \cite{GambaTask18, LuMouhot12} still for hard potentials with $\gamma\in(0,2]$, and in \cite{BobGamba17, PT} for pseudo-Maxwellian and Maxwellian case ($\gamma=0$).
In the later referenced work, these non-Gaussian tailed moments are called Mittag-Leffler moments as in fact the summability of partial sums is shown to converge to an $L^1$-Mittag-Leffler function weighted norm for the unique probability density function solving the initial value problem associated to the Boltzmann equation, whose order and rate depend on the initial data as much as on the order of singularity in the angular section.
A very important tool for the success of summability properties for polynomial moments relies on the fact that such moments are both created and propagated depending on how moments of the collision operator can be estimated: the positive part of the (gain) collision operator must have a decay rate with respect to the moment order while the negative part of such moments prevails in the dynamics, when sufficiently many moments are taken into account.
This is indeed a key step, arising as a consequence of an {\em angular averaged Povzner lemma}. In the case of single gas components,
these estimates are based on integration of the collision operator against polynomial test functions on the pre-collisional velocities in the sphere.
While these objects were originally introduced by Povzner \cite{Pov} in 1960s, a sharper form that uses the conservation of energy and angular averaging was introduced in \cite{Bob97} for the case of hard spheres in three dimensions with a constant angular cross section, where the polynomial test functions are proportional to even powers of the velocity magnitude. Later this technique was extended in \cite{GambaBobPanf04} for the inelastic collision with heating sources,
in \cite{GambaPanfVil09} to the elastic case with hard potentials with $L^{1+}$ integrable angular cross section, as well as in \cite{Gamba13} for the case with just $L^{1}$ integrable angular cross section. Further, the approach was enlarged to hard spheres with non-integrable angular cross section in \cite{LuMouhot12} and \cite{GambaTask18} for hard potentials. { {\sl All of these estimates were developed for the mono-component model. } }
Hence, the angular averaged Povzner lemma is our starting point in the case of mixtures as well. However, it requires a subtle modification of the polynomial weight that define {\emph{the scalar moment for the mixture}}, to be defined in \eqref{SPM} next section, that renormalizes the polynomial test function from just even powers of the magnitude of the velocity vector to a dimensionless bracket form independent of mass density units, as the mono-component treatment to obtain moment estimates from \cite{Bob97} for the elastic case, or from \cite{GambaPanfVil09} for inelastic hard sphere interactions, can not be directly extended to the mixture case, when masses are possibly different.
This facts enticed us to introduce a new approach that relies on the way to
rewrite collisional rules and scalar polynomial moments in such dimensionless, independent of mass density units form that is very convenient to obtain a convex combination form between the conserved local quantities for a binary interaction, namely, local center of mass and energy.
As a consequence, we conclude that averaging over the $S^2$-sphere yield decay properties as a function of the moment order for as long as angular kernel is {\color{black}$L^1-$integrable} on $S^2$. {\color{black} In particular, these decay properties will be significantly influenced by the fact how much species masses are disparate. It will be shown that as much as renormalized species masses deviates one from each other, the decay rate will be more slowly.\\
}
The paper is organized as follows. In Section \ref{Section notation} we introduce notation and preliminaries, and state the main results, namely the Existence and Uniqueness Theorem for the vector value solution of the homogeneous Boltzmann system, and then generation and propagation of both scalar polynomial and exponential moments. Then in Section \ref{Section Kinetic Model} we describe in details kinetic model that we use. Section \ref{Section prelim Lemmas} contains two preliminary Lemmas that we need for further work, including Povzner lemma. Sections \ref{Section Ex Uni proof}, \ref{section proof generation of poly} and \ref{Section gen prop exp mom} are devoted to proofs of our main results. A final Appendix contains some auxiliary calculations relevant to our estimates.
\section{Notation, Preliminaries and Main Results}\label{Section notation}
\subsection{Notation and Preliminaries}
In this paper, we consider mixture of $I$ gases, and we label its components with $\mathcal{A}_1$, $\dots$, $\mathcal{A}_I$. Each component of the mixture $\mathcal{A}_i$, $i=1,\dots,I$, is described with its own distribution function, denoted with $f_i:=f_i(t,v)\geq0$, that, in this manuscript, depends on time $t>0$ and velocity $v\in \mathbb{R}^3$. Fixing some $i\in \left\{1,\dots,I\right\}$, distribution function $f_i$ satisfy Boltzmann like equation, which now, in the mixture context, has to take into account influence of all other components of the mixture on species $\mathcal{A}_i$. In the kinetic theory style, this is achieved by defining collision operator $Q_{ij}$ for each $j=1,\dots,I$ that measures interaction between species $\mathcal{A}_i$ that we fixed and all the others $\mathcal{A}_j$, $j=1,\dots,I$, including itself $\mathcal{A}_i$. If the species $\mathcal{A}_j$ are described with distribution functions $f_j$, then the evolution of $f_i$ is described via
\begin{equation}\label{boltzmann i}
\partial_t f_i(t,v) = \sum_{j=1}^I Q_{ij}(f_i,f_j)(t,v), \qquad i=1,\dots,I.
\end{equation}
The form of $Q_{ij}$, for {\color{black} distribution functions $f$ and $g$ and } any $i,j=1,\dots I$, is given by the non-local bilinear form
\begin{equation}\label{QijJ}
Q_{ij}(g,h)(v)= \int_{\mathbb{R}^3} \int_{S^2} \left( \frac1{\mathcal{J}} \, g(v'_{ij}) h(v'_{*ij}) - g(v) h(v_*) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \, \mathrm{d} \sigma \, \mathrm{d}v_*,
\end{equation}
where pre-collisional quantities $v'_{ij}$ and $v'_{*ij}$ depend on post-collisional ones $v$, $v_*$ and parameter $\sigma$, {\color{black}as much as on the masses $m_i$ and $m_j$ mass of colliding particles of species $\mathcal{A}_i$ and $\mathcal{A}_j$ respectively}, in the following manner
\begin{equation}\label{collisional rules intro}
v'_{ij} =\frac{m_i v + m_j v_*}{m_i + m_j} + \frac{m_j}{m_i + m_j} \left| v-v_* \right| \sigma, \quad v'_{*ij} =\frac{m_i v + m_j v_*}{m_i + m_j} - \frac{m_i}{m_i + m_j} \left| v-v_* \right| \sigma.
\end{equation}
{\color{black}The collisional rules \eqref{collisional rules intro} can be written in scattering direction coordinates (or in a center of mass reference framework) by introducing the velocity of center of mass $V_{ij}$ and relative velocity $u$ of the two colliding particles,
\begin{equation}\label{center-of-mass}
V_{ij}:=\frac{m_i v + m_j v_*}{m_i + m_j}, \qquad u:=v-v_*,
\end{equation}
as follows
\begin{equation}\label{cof-rv-coor}
v'_{ij} =V_{ij}+ \frac{m_j}{m_i + m_j} \left| u \right| \sigma, \quad v'_{*ij} =V_{ij} - \frac{m_i}{m_i + m_j} \left| u \right| \sigma,
\end{equation}
or equivalently, introducing the two-body mass fraction parameter $r_{ij}= \frac{m_i}{m_i + m_j}\in (0,1)$, associated to one of the particles, say $m_i$ without loss of generality,
\begin{equation}\label{cof-rv-coor-2}
v'_{ij} =V_{ij}+ (1-r_{ij})\left| u \right| \sigma, \quad v'_{*ij} =V_{ij} - r_{ij} \left| u \right| \sigma.
\end{equation}
\begin{remark}\label{eliminate subindex ij}
For simplicity of notation, from now on, we will eliminate subindices $i,j$ from $v'_{ij}$, $v'_{*ij}$, $V_{ij}$ and $r_{ij}$.
\end{remark}
}
The transition probability rates or collision cross section terms $\mathcal{B}_{ij}$ are positive functions supposed to satisfy the following micro-reversibility assumptions
\begin{equation}\label{Bij micro}
\mathcal{B}_{ij}(v,v_*,\sigma) = \mathcal{B}_{ij}(v',v'_*,\sigma')=\mathcal{B}_{ji}(v_*,v,-\sigma),
\end{equation}
where {\color{black} $\sigma=u'/\left|u'\right|$ and $u'=v'-v'_*$ (note that then $\sigma'=u/\left|u\right|$). }
The factor in the positive non-local binary term {\color{black} ${\mathcal{J}}= \left| \det J_{(v',v'_*,\sigma')/(v,v_*,\sigma)} \right|$ is the absolute value of } determinant of the Jacobian associated to the exchange of velocity variables transformation \eqref{collisional rules intro} from pre to post for the given binary interaction. {\color{black} The Jacobian of this transformation can be easily computed by passing to the scattering direction coordinates i.e by considering the following mappings $(v',v'_*,\sigma') \mapsto (u',V',\sigma')\mapsto (\left|u'\right|, \frac{u'}{\left|u'\right|} , V', \sigma') \mapsto (\left|u\right|, \frac{u}{\left|u\right|} , V, \sigma) \mapsto (u, V, \sigma) \mapsto (v, v_*, \sigma)$, with the notation \eqref{center-of-mass} and using Remark \ref{eliminate subindex ij}. The first mapping is of unit Jacobian from definition of $u$ and $V$, the second one is passage from Cartesian to spherical coordinates for $u'$. Since from the collisional rules \eqref{collisional rules intro} it follows $\left|u'\right| = \left|u\right|$ and $V'=V$ the passage from primes to non-primes described in the third mapping is of unit Jacobian. Then we pass from spherical to Cartesian coordinates for $u$ and finally go back to the original variables $(v,v_*,\sigma)$. Thus, the Jacobian is computed as the decomposition of the mentioned mappings,
\begin{equation*}
\mathcal{J} = 1 \cdot \frac{1}{\left|u'\right|^2} \cdot 1 \cdot \left| u \right|^2 \cdot 1 = 1,
\end{equation*}
since $\left|u'\right| = \left|u\right|$.
Therefore, } each $Q_{ij}$ from \eqref{QijJ} simple becomes,
\begin{equation}\label{Qij}
Q_{ij}(g,h)(v)= \int_{\mathbb{R}^3} \int_{S^2} \left(g(v') h(v'_*) - g(v) h(v_*) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \, \mathrm{d} \sigma \, \mathrm{d}v_*\, .
\end{equation}
\bigskip
Since we consider a mixture as a whole, it will be convenient to introduce the following vector notation. We put all distribution functions $f_i$, $i=1,\dots, I$ into vector of distribution functions
\begin{equation}\label{distribution function vector form}
\mathbb{F}=[f_i]_{1\leq i \leq I}.
\end{equation}
Moreover, a vector value collision operator is defined
\begin{equation}\label{collision operator vector form}
\mathbb{Q}(\F)=\left[\sum_{j=1}^I Q_{ij}(f_i,f_j)\right]_{1\leq i \leq I}.
\end{equation}
Then the system of Boltzmann equations \eqref{boltzmann i} can be written in a vector form
\begin{equation}\label{BE intro}
\partial_t \mathbb{F}(t,v) = \mathbb{Q}(\F)(t,v).
\end{equation}
\begin{definition}[\sl Bracket forms for the mixture's dimensionless polynomial moments independent of mass density units]\label{SPM}
Let $\mathbb{F}=[f_i]_{1\leq i \leq I}$ be a suitable vector value distribution function. Let mixture's bracket forms be denoted by
\begin{equation}\label{vi}
\left\langle v \right\rangle_i :=\sqrt{ 1+ \frac{m_i}{\sum_{j=1}^I m_j} \left| v \right|^2}, \qquad v\in \mathbb{R}^3.
\end{equation}
\emph{Scalar polynomial moments independent of mass density units} of order $q\geq 0$ for $\mathbb{F}$ is defined with
\begin{equation}\label{poly moment}
\mathfrak{m}_q[\F](t) = \sum_{i=1}^I \int_{\mathbb{R}^3} f_i(t,v) \left\langle v \right\rangle_i^q \mathrm{d}v.
\end{equation}
In particular, we define scalar polynomial moment of zero order for each species $\mathcal{A}_i$
\begin{equation*}
\mathfrak{m}_{0,i}[\mathbb{F}](t) = \int_{ \mathbb{R}^3} f_i(t,v) \, \mathrm{d}v, \quad i=1,\dots,I,
\end{equation*}
having in mind that $\sum_{i=1}^I \mathfrak{m}_{0,i}[\mathbb{F}] = \mathfrak{m}_0[\F]$.
\emph{Scalar exponential moment}, or exponential weighted $L^1-$forms, for $\mathbb{F}$ of a rate $\boldsymbol{\alpha}:=(\alpha_1,\dots,\alpha_I)$, $\alpha_i>0$, and an order $\mathbf{s}:=(s_1,\dots,s_I)>0$, $0<s_i\leq 2$, is defined by
\begin{equation}\label{exp moment}
{\color{black}\mathbb{\mathcal{E}}_{ \mathbf{s}}[\F](\boldsymbol{\alpha},t) } = \sum_{i=1}^I \int_{\mathbb{R}^3} f_i(t,v) e^{\alpha_i \left\langle v \right\rangle_i^{s_i}} \mathrm{d}v.
\end{equation}
The case $s_i =2$, $\forall i$, is referred to as inverse Maxwellian (or Gaussian) moment, otherwise they are super exponential moments (some authors referred as stretched exponentials though this concept usually refers to exponential times).
\end{definition}
\begin{remark}
It can be noticed that such both dimensionless polynomial and exponential moments for the mixture are defined as a sum of the resulting moments corresponding to each species independent of mass density units. In particular, when $\mathbb{F}$ solves the Boltzmann system of equations \eqref{BE intro}, then $\mathfrak{m}_{0,i}[\mathbb{F}]$ is interpreted as number density of the species $\mathcal{A}_i$, for any $i=1,\dots,I$, while the zeroth scalar moment $\mathfrak{m}_0[\F]$ is the total number density of the mixture. Moreover, the second scalar moment $\mathfrak{m}_2[\F]$ represents total energy of the mixture.
\end{remark}
\begin{remark}
If, for given exponential moments individually for each species $\mathcal{A}_i$, we seek for the maximum value of both their rate and order, i.e.
\begin{equation}\label{max rate max order}
\hat{\alpha}= \max_{1\leq i \leq I} \alpha_i, \quad \hat{s}= \max_{1\leq i \leq I} s_i,
\end{equation}
then $$ {\color{black}\mathbb{\mathcal{E}}_{ \mathbf{s}}[\F](\boldsymbol{\alpha},t) } \leq \sum_{i=1}^I \int_{\mathbb{R}^3} f_i(t,v) \, e^{\hat{\alpha} \left\langle v \right\rangle_i^{\hat{s}}} \mathrm{d}v =: \mathcal{E}_{\hat{s}}[\F](\hat{\alpha},t)$$
Therefore, finiteness of the exponential moment $ {\color{black}\mathbb{\mathcal{E}}_{ \mathbf{s}}[\F](\boldsymbol{\alpha},t) }$ is
{\color{black} a consequence of} the finiteness of $\mathcal{E}_{\hat{s}}[\F](\hat{\alpha},t)$, with $\hat{\alpha}$ and $\hat{s}$ as in \eqref{max rate max order}, for any time $t\geq 0$.
\end{remark}
\subsubsection{Functional space}
We work in $L^1$ space weighted polynomially in velocity $v$ and summed over all species, that is
\begin{equation}
\begin{split}\label{space L_k^1}
L_k^1 &= \left\{ \mathbb{F}= [f_i]_{1\leq i \leq I} \ \text{measurable}: \sum_{i=1}^I \int_{\mathbb{R}^3} \left| f_i(v) \right|\left\langle v \right\rangle_i^k \mathrm{d}v < \infty, \ k\geq 0 \right\}\\
\end{split}\end{equation}
{where} the polynomial weight was defined in \eqref{vi} by $\left\langle v \right\rangle_i ={\left( 1+ \frac{m_i}{\sum_{j=1}^I m_j} \left| v \right|^2 \right)^{1/2}}.$
Its associated norm is
\begin{equation}\label{norm I}
\left\| \mathbb{F} \right\|_{L_k^1} = \sum_{i=1}^I \int_{\mathbb{R}^3} \left| f_i(v) \right|\left\langle v \right\rangle_i^k \mathrm{d}v.
\end{equation}
Note that if $\mathbb{F}\geq 0$, then its norm in $L_k^1$ is precisely its polynomial moment of order $k$, i.e.
$
\left\| \mathbb{F} \right\|_{L_k^1} := \mathfrak{m}_k[\F].
$
Sometimes we will consider species separately, i.e. fix some component of the mixture $\mathcal{A}_i$. We define a space together with its norm
\begin{equation*}
L_{k,i}^1 = \left\{ g \ \text{measurable}: \int_{\mathbb{R}^3} \left|g(v)\right| \left\langle v \right\rangle_i^{k} \mathrm{d}v < \infty, k\geq 0 \right\}, \left\| g \right\|_{L_{k,i}^1} = \int_{\mathbb{R}^3} \left|g(v)\right| \left\langle v \right\rangle_i^{k} \mathrm{d}v.
\end{equation*}
Note that the norm of $\mathbb{F}$ in $L_k^1$ is related to the norm of its components $f_i$ in the space $L_{k,i}^1$ via
$
\left\| \mathbb{F} \right\|_{L_k^1} = \sum_{i=1}^{I} \left\| f_i \right\|_{L_{k,i}^1}.
$
Finally, since we use bracket forms $\left\langle \cdot \right\rangle$ defined in \eqref{vi}, the monotonicity property holds, i.e.
\begin{equation}\label{monotonicity of norm}
\left\| f_i \right\|_{L_{k_1,i}^1} \leq \left\| f_i \right\|_{L_{k_2,i}^1} \ \text{and} \ \left\| \mathbb{F} \right\|_{L_{k_1}^1} \leq \left\| \mathbb{F} \right\|_{L_{k_2}^1}, \text{whenever} \ 0\leq \ k_1 \leq k_2.
\end{equation}
\subsection{Main Results}
We study the Cauchy problem for system of spatially homogeneous Boltzmann equations for the mixture of gases $\mathcal{A}_1$, $\dots$, $\mathcal{A}_I$:
\begin{equation}\label{Cauchy problem}
\left\{ \begin{split} & \partial_t \mathbb{F}(t,v) = \mathbb{Q}(\F)(t,v), \quad t>0, \ v \in \mathbb{R}^3, \\& \mathbb{F}(0,v) = \mathbb{F}_0(v),\end{split} \right.
\end{equation}
where $\mathbb{F}$ is a vector of distribution functions $\mathbb{F}=[f_i]_{1\leq i \leq I}$, $f_i$ being distribution function of the component $\mathcal{A}_i$, $i=1,\dots,I$, as defined in \eqref{distribution function vector form}, and $\mathbb{Q}(\F)=\left[\sum_{j=1}^I Q_{ij}(f_i,f_j)\right]_{1\leq i \leq I}$ is a collision operator introduced in (\ref{Qij}, \ref{collision operator vector form}).
We consider the particular case when the transition probability terms $\mathcal{B}_{ij}$, $i,j=1,\dots,I$ are assumed to take the form
{\color{black}
\begin{equation}\label{cross section}
\mathcal{B}_{ij}(v,v_*, \sigma) =\left|u\right|^{ { \gamma_{ij}} } \, {\color{black}b_{ij}} (\sigma \cdot \hat{u}), \ { \gamma_{ij}} \in (0,1], \ \text{and} \ {\color{black} {\color{black}b_{ij}} (\sigma \cdot \hat{u})\in \Ljed},
\end{equation}
}
where $u:=v-v_*$, $\hat{u}:=u/\left|u\right|$. This form of cross section corresponds to variable hard potentials with {\color{black} an integrable angular part.} {\color{black} In the mixture setting, both potential $ { \gamma_{ij}} $ and angular kernel $ {\color{black}b_{ij}} $ may depend on species $\mathcal{A}_i$ and $\mathcal{A}_j$. In order to satisfy micro-reversibility assumptions \eqref{Bij micro}, it is supposed that
\begin{equation*}
{ \gamma_{ij}} = \gamma_{ji}, \quad \text{and} \quad {\color{black}b_{ij}} (\sigma \cdot \hat{u}) = b_{ji}(\sigma \cdot \hat{u}),
\end{equation*}
for any choice $i,j=1,\dots,I$. Moreover, let $ { {\color{black}\overline{\gamma}}} $ denote the maximum value of potentials $ { \gamma_{ij}} $, i.e.
\begin{equation}\label{gamma max}
{ {\color{black}\overline{\gamma}}} = { {\color{black}\max_{1\leq i, j \leq I}\gamma_{ij}}} .
\end{equation}
}
\subsubsection{Povzner lemma by angular averaging} The essential ingredient of this manuscript is the Povzner lemma obtained by averaging in the scattering angle representation of the collision kernel, originally introduced in \cite{Bob97}, \cite{GambaBobPanf04}, for the case of elastic and inelastic collisions.
It estimates the positive contribution of the collision operator after integration against $\sigma \in S^2$, that is crucial for all further proofs.
\begin{lemma}[Povzner lemma by angular averaging for the mixing model]\label{Povzner intro} Let the angular part $ {\color{black}b_{ij}} (\sigma \cdot \hat{u})$ of the cross-section be {\color{black}integrable in $\sigma$ variable (that is $ {\color{black}b_{ij}} \in \Ljed$)}, $\hat{u}$ being the normalized relative velocity $u=v-v_*$.
Let $v'$ and $v'_*$ be functions of $v,v_*, \sigma$ as in \eqref{collisional rules intro}, with $m_i, m_j >0$. Then the following estimate holds for any {\color{black} fixed $i, j$},
\begin{equation}\label{Povzner estimate gain}
\int_{S^2} \left( \left\langle v' \right\rangle_i^k + \left\langle v'_* \right\rangle_j^k \right) \, {\color{black}b_{ij}} (\sigma \cdot \hat{u}) \, \mathrm{d}\sigma \leq \boldsymbol{\mathcal{C}}^{ij}_\frac{k}{2} \left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 \right)^\frac{k}{2},
\end{equation}
where constant {\color{black} $\boldsymbol{\mathcal{C}}^{ij}_\frac{k}{2}$ tends to zero as $k$ grows
and moreover
\begin{equation}\label{povzner constant prop}
\boldsymbol{\mathcal{C}}^{ij}_\frac{k}{2} - \bijnjed <0, \qquad \text{for any} \ k\ge k^{ij}_*, \ \ \ 1\leq i, j \leq I,
\end{equation}
{ where each $k^{ij}_*$ depends on $ {\color{black}b_{ij}} $ and $r_{ij}$.} }
\end{lemma}
The proof of Lemma~\ref{Povzner intro} genuinely reflects difference between single and multicomponent gas, with an accent on writing collisional rules in a convex combination form for mixtures, in contrast to symmetric or ``half-half" writing for the single component gas. It turned out that single component case due to symmetry had a lot of room for estimates and further simplification, presented in \cite{GambaBobPanf04} for example. For mixtures, this is not the case any longer, and writing should be exact as much as possible: we use Taylor expansion of second order with a reminder in the integral form, and estimates are done only in the reminder.
{\color{black} A very important consequence of the Povzner lemma is the ability to estimate moments of the collision operator. In particular, averaging over the sphere yields decay properties of the gain term polynomial moment with respect to its order. This decay allows polynomial moments of loss term to prevail in dynamics, when sufficiently many moments are taken into account. In a single component gas, it suffices to take $2+$ order of polynomial moment, that is slightly more than energy, to obtain this property \cite{GambaBobPanf04}. Mixtures bring great novelty in this aspect, too: decay properties of the constant issuing from the Povzner lemma strongly depend on the two-body mass fraction parameter $r_{ij}$. We study this issue in detail in the case $ {\color{black}b_{ij}} \in \Linf$ when it is possible to explicitly calculate the constant $\boldsymbol{\mathcal{C}}^{ij}_{k/2}$ from \eqref{povzner constant prop}. It will be shown that when $r_{ij}=1/2$ (which corresponds to $m_i=m_j$), we recover the same decay properties of the constant $\boldsymbol{\mathcal{C}}^{ij}_{k/2}$ as in the case of single gas component. However, when mixtures are considered, we observe that as much as $r_{ij}$ deviates from $1/2$, the larger $k^{ij}_*$ that ensures \eqref{povzner constant prop} is, or larger and larger order of moment that guarantees prevail of loss term moment is.
}
\subsubsection{Existence and uniqueness theory}
In this manuscript, we discuss existence and uniqueness for the vector value solution $\mathbb{F}$ to the initial value problem \eqref{Cauchy problem} of space homogeneous Boltzmann equations for monatomic gas mixtures, with transition probabilities (or collision kernels) {\color{black} associated to species $\mathcal{A}_i$ and $\mathcal{A}_j$, $i,j \in \left\{1,\dots,I\right\}$} having hard potential growth of order $\left|u\right|^ { \gamma_{ij}} $ for $ { \gamma_{ij}} \in (0,1]$ and an {\color{black} integrable angular part} $ {\color{black}b_{ij}} $, with an initial total mixture number density and energy bounded below (i.e. the initial data can not be singular measure), and {\color{black} have at least a ${k_*}$ (scalar) polynomial moments,
\begin{equation}\label{kstar}
{k_*} \geq \max\{ {\overline{k}} , 2+2 { {\color{black}\overline{\gamma}}} \} \ \qquad\ \text{for} \ \
{\overline{k}} = \max_{1\leq i, j \leq I}\{ k^{ij}_*\} \ \ \text{and}\ \ { {\color{black}\overline{\gamma}}} = { {\color{black}\max_{1\leq i, j \leq I}\gamma_{ij}}} ,
\end{equation} }
{\color{black} chosen to ensure the inequality \eqref{povzner constant prop} holds for any $i,j=1,\dots,I$.}
Such a study fits into an abstract framework of ODE theory in Banach spaces, which can be found in \cite{Martin}. For the Boltzmann equation, the application of this theory was clarified in \cite{GambaAlonso18}, after being recognized in \cite{Bressan}. The formulation of theorem that we apply in this manuscript is given in Appendix \ref{Appendix exi and uni}. As for the choice of Banach space, it is known that natural Banach space to solve the Boltzmann equation is $L^1$ polynomially weighted, or in mixture setting space $L_k^1$ defined in \eqref{space L_k^1}. More precisely, here we take $k=2$, because the norm in that space is related to energy whose conservation is exploited. \\
In order to apply Theorem \ref{Theorem general}, we need to find an invariant region $\Omega \subset L_2^1$ in which collision operator $\mathbb{Q}: \Omega \rightarrow L_2^1$ will satisfy \emph{(i) H\"{o}lder continuity}, \emph{(ii) Sub-tangent} and \emph{(iii) one-sided Lipschitz} conditions.
To that end, we first study the map $\mathcal{L}_{ \gmax, k_*}:[0,\infty) \rightarrow \mathbb{R}$, defined with $$\mathcal{L}_{ \gmax, k_*}(x)=- Ax^{1+ \frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}}+ B x,$$
where $A$ and $B$ are positive constants, $ { {\color{black}\overline{\gamma}}} \in (0,1]$ and {\color{black} $k_*$ defined in \eqref{kstar}}. This map has only one root, denoted with $x^*_{ \gmax, k_*}$, at which $\mathcal{L}_{ \gmax, k_*}$ changes from positive to negative. Thus, for any $x\geq0$, we may write
\begin{equation*}
\mathcal{L}_{ \gmax, k_*}(x) \leq \max_{0\leq x \leq x^*_{ \gmax, k_*}} \mathcal{L}_{ \gmax, k_*}(x) =: \mathcal{L}^*_{ \gmax, k_*}.
\end{equation*}
Define
\begin{equation}\label{c dva plus dva gama def}
C_{\ks} := x^*_{ \gmax, k_*} + \mathcal{L}^*_{ \gmax, k_*}.
\end{equation}
Now, we are in position to define the bounded, convex and closed subset $\Omega \subset L_2^1$,
\begin{multline*}
\Omega = \Big\{ \mathbb{F}(t, \cdot) \in L_2^1: \mathbb{F}\geq 0 \ \text{in} \ v, \sum_{i=1}^I \int_{\mathbb{R}^3} m_i v \, f_i(t,v) \mathrm{d}v=0,
\\
\exists \, c_{0}, C_0, c_{2}, , C_{2}, C_{2+\varepsilon} >0, \text{and} \ C_0 < c_{2}, \text{ such that} \ \forall t\geq0,
\\
c_{0} \leq \mathfrak{m}_0[\F](t) \leq C_0, \ c_{2} \leq \mathfrak{m}_2[\F](t) \leq C_{2},
\\
\mathfrak{m}_{2+\varepsilon}[\F](t) \leq C_{2+\varepsilon}, \ \text{for } \ \varepsilon>0,
\\
\mathfrak{m}_{\ks}[\F] (t) \leq C_{\ks}, \ \text{with} \ C_{\ks} \ \text{from } \eqref{c dva plus dva gama def} \ \Big\},
\end{multline*}
where
\begin{equation*}
\mathfrak{m}_{2+\varepsilon}[\F](t) = \left\| \mathbb{F} \right\|_{L_{2+\varepsilon}^1}= \sum_{i=1}^I \int_{\mathbb{R}^3} \left| f_i(t,v) \right|\left\langle v \right\rangle_i^{2+\epsilon} \mathrm{d}v,
\end{equation*}
for any $\epsilon>0$, which can be arbitrary small.
Then, existence and uniqueness theory of a vector value $\mathbb{F}$ solution to the Cauchy problem \eqref{Cauchy problem} fits into the study of ODE in a Banach space $(L_2^1, \left\| \cdot \right\|_{L_2^1})$ and its bounded, convex and closed subset $\Omega$. The collision operator $\mathbb{Q}$ is viewed as a map $\mathbb{Q}: \Omega \rightarrow L_2^1$. We will show that it satisfies H\"{o}lder continuity, sub-tangent and one-sided Lipschitz conditions, which will enable us to prove the following Theorem.
\begin{theorem}[Existence and Uniqueness]\label{theorem existence uniqueness} Assume that $\mathbb{F}(0,v)=\mathbb{F}_0(v) \in \Omega$. Then the Boltzmann system \eqref{Cauchy problem} for the cross section \eqref{cross section} has the unique solution in $\mathcal{C}\left(\left[0, \infty \right), \Omega \right) \cap \mathcal{C}^1\left(\left(0,\infty \right), L_2^1 \right)$.
\end{theorem}
\begin{remark} Let us point out that for the existence and uniqueness result no conditions on initial entropy are necessary. However, if the initial data has finite entropy, then the entropy inequality implies that it will remain bounded for all times. Let us give a sketch of the proof. Definition of the entropy and entropy inequality is taken from \cite{DesMonSalv}, Proposition 1.
\begin{definition}[Mixture entropy and entropy production]
Let $\mathbb{F}$ be a vector value distribution function as in \eqref{distribution function vector form}. The (mixture) entropy is defined as
\begin{equation}\label{entropy def}
\eta(t) = \sum_{i=1}^I \int_{ \mathbb{R}^3} f_i \log f_i \mathrm{d}v,
\end{equation}
while the (mixture) entropy production is given with
\begin{equation}\label{entropy production def}
D(\mathbb{F})= \sum_{i=1}^I \int_{ \mathbb{R}^3} \left[\mathbb{Q}(\F)\right]_i \log f_i \mathrm{d}v.
\end{equation}
\end{definition}
Then the following Proposition holds.
\begin{proposition}[Entropy inequality or the first part of the H-theorem, \cite{DesMonSalv}]
Let us assume that the cross section terms $\mathcal{B}_{ij}$, $1\leq i, j \leq I$, are positive almost everywhere and that $\mathbb{F}\geq0$ is such that both collision operator $\mathbb{Q}(\F)$ and entropy production are well defined. Then the entropy production is non-positive, i.e.
$
D(\mathbb{F}) \leq 0.
$
\end{proposition}
As an immediate consequence, we get from the Boltzmann equation that $\partial_t \eta \leq 0$, or in other words, $\eta(t)\leq \eta(0)$, for any $t\geq0$. Therefore, we conclude that the entropy inequality implies that mixture entropy remains bounded at any time if initially so.
\end{remark}
\subsubsection{Generation and propagation of polynomial moments} The second part of the manuscript is devoted to the study of generation and propagation of scalar polynomial moments associated to the solution of the Boltzmann system \eqref{Cauchy problem} for the cross section \eqref{cross section}, that initially belongs to $\Omega$.
First, in the following Lemma, we derive from the Boltzmann system \eqref{Cauchy problem} an ordinary differential inequality for polynomial moment of order $k$, $\mathfrak{m}_k[\F](t)$, for {\color{black} large enough $k$}, that relies on the Povzner estimate from Lemma~\ref{Povzner intro}, {\color{black} uniformly in each pair $i,j$. }
\begin{lemma}[Ordinary differential inequality for polynomial moments]\label{poly moments}
Let $\mathbb{F}=\left[f_i\right]_{i=1,\dots,I}$ be a solution of the Boltzmann system \eqref{Cauchy problem} {\color{black} with the cross section \eqref{cross section}-\eqref{gamma max} }. Then the polynomial moment \eqref{poly moment} satisfies the following Ordinary Differential Inequality
\begin{equation}\label{pomocna 100}
\frac{\mathrm{d}}{\mathrm{d} t} \mathfrak{m}_k[\F](t) = \sum_{i=1}^I \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^k \mathrm{d} v
\leq - A_k \, \mathfrak{m}_k[\F](t)^{1+\frac{ { {\color{black}\overline{\gamma}}} }{k}}+ B_k \, \mathfrak{m}_k[\F](t),
\end{equation}
for {\color{black} large enough $k$ to ensure \eqref{kstar}}, and some positive constants $A_k$ and $B_k$.
\end{lemma}
The proof of this Lemma follows from comparison principles for ODE's, which yields the generation and propagation estimates
stated in the following Theorem that is proved in Section \ref{section proof generation of poly}.
\begin{theorem}[Generation and propagation of polynomial moments]\label{theorem bound on norm}
Let $\mathbb{F}$ be a solution of the Boltzmann system \eqref{Cauchy problem} with
a cross section \eqref{cross section}-\eqref{gamma max} and an initial data $\mathbb{F}(0,v)=\mathbb{F}_0(v) \in \Omega$. \\
\begin{itemize}
\item[1.](Generation) There is a constant $\mathfrak{C}^{\mathfrak{m}}$ such that {\color{black} for any $k>k_*$ defined in \eqref{kstar}} ,
\begin{equation}\label{bound on norm}
\mathfrak{m}_k[\F](t) \leq \mathfrak{C}^{\mathfrak{m}} \left(1-e^{-\frac{ { {\color{black}\overline{\gamma}}} B_k t}{k} } \right)^{-\frac{k}{ { {\color{black}\overline{\gamma}}} }}, \qquad \forall t>0,
\end{equation}
where constants $\mathfrak{C}^{\mathfrak{m}}$ depend on $A_k$, $B_k$ {\color{black} from \eqref{pomocna 100} and $ { {\color{black}\overline{\gamma}}} $ }. \\
\item[2.](Propagation) Moreover, if $\mathfrak{m}_k[\F](0)<\infty$, then
\begin{equation}\label{poly propagation}
\mathfrak{m}_k[\F](t) \leq \max\{\mathfrak{C}^{\mathfrak{m}}, \mathfrak{m}_k[\F](0)\},
\end{equation}
for all $t\geq 0$.
\end{itemize}
\end{theorem}
Finally, we show that, under the assumed conditions on collision kernel form \eqref{cross section},
the renormalized series of moments is summable depending on the moments of the initial data {\color{black}yielding} the following result on generation and propagation of exponential, or Mittag-Leffler moments.
\subsubsection{Generation and propagation of exponential moments} With bounds on polynomial moment at hand, one can deal with exponential moments. We prove the following Theorem.
\begin{theorem}[Generation and propagation of exponential moments]\label{theorem gen prop ML}
Let $\mathbb{F}$ be a solution of the Boltzmann system \eqref{Cauchy problem} with
a cross section \eqref{cross section}-\eqref{gamma max} and an initial data $\mathbb{F}(0,v)=\mathbb{F}_0(v) \in \Omega$. \\
\begin{itemize}
\item[(a)](Generation) There exist constants $\alpha>0$ and $\mathfrak{B}^{\mathcal{E}}>0$ such that
\begin{equation*}
\mathcal{E}_{ { {\color{black}\overline{\gamma}}} }[\mathbb{F}](\alpha \min\left\{t,1\right\}, t) \leq \mathfrak{B}^{\mathcal{E}}, \quad \forall t\geq 0.
\end{equation*} \\
\item[(b)](Propagation) Let $0<s\leq 2$. Suppose that there exists a constant $\alpha_0 >0$, such that
\begin{equation}\label{initial data exp prop}
\mathcal{E}_{s}[\F]({\alpha_0},0) \leq M_0 < \infty.
\end{equation}
Then there exist constants $0<\alpha \leq \alpha_0$ and $\mathfrak{C}^{\mathcal{E}}>0$ such that
\begin{equation}\label{exp propagation p}
\mathcal{E}_{s}[\F]({\alpha},t) \leq \mathfrak{C}^{\mathcal{E}}, \qquad \forall t\geq 0.
\end{equation}
\end{itemize}
\end{theorem}
\section{Kinetic Model}\label{Section Kinetic Model}
\subsection{Study of collision process}
In our setting molecules are assumed to interact via elastic collisions. Let us fix two colliding molecules; one of the species $\mathcal{A}_i$ having mass $m_i$ and pre-collisional velocity $v'$ and the another one belonging to the species $\mathcal{A}_j$ with mass $m_j$ and pre-collisional velocity $v'_*$ {\color{black} (note that we here immediately adopted the simplicity of notation pointed out in Remark \ref{eliminate subindex ij})}. If the post-collisional velocities are denoted with $v$ and $v_*$, respectively, than the momentum and kinetic energy during the collision are conserved
\begin{align}
m_i v' + m_j v'_* &= m_i v + m_j v_*, \nonumber\\
m_i \left| v' \right|^2 + m_j \left| v'_* \right|^2 &= m_i \left| v \right|^2 + m_j \left| v_* \right|^2. \label{CL micro energy}
\end{align}
As usual, we parametrize these equations with a parameter $\sigma \in S^2$, in order to express pre-collisional velocities in terms of post-collisional ones,
\begin{equation}\label{collisional rules bi}
v' =\frac{m_i v + m_j v_*}{m_i + m_j} + \frac{m_j}{m_i + m_j} \left| v-v_* \right| \sigma, \quad v'_* =\frac{m_i v + m_j v_*}{m_i + m_j} - \frac{m_i}{m_i + m_j} \left| v-v_* \right| \sigma.
\end{equation}
Note that if $m_i=m_j$, then the collisional rules simplify and take the usual single component gas form
\begin{equation}\label{collisional rules equal masses}
v' =\frac{v + v_*}{2} + \frac{1}{2} \left| v-v_* \right| \sigma, \quad v'_* =\frac{ v + v_*}{2} - \frac{1}{2} \left| v-v_* \right| \sigma.
\end{equation}
\begin{figure}
\begin{center}
\begin{tikzpicture}[scale=2]
\node (v) at (1.93,0.52) [preaction={shade, ball color=red},pattern=vertical lines, pattern color=black,circle,scale=2.5] {};
\node (vs) at (-1.93,-0.52) [preaction={shade, ball color=white!70!gray},circle,scale=1, ,pattern=crosshatch dots, pattern color=black] {};
\node (V) at (1.158,0.312) [circle,scale=1.7, shade,ball color=blue] {};
\node (vp) at (1.72,-0.25) [preaction={shade,ball color=red},circle,scale=2.5, ,pattern=vertical lines, pattern color=black] {};
\node (vps) at (-1.1,2.57) [preaction={shade,ball color=white!70!gray}, circle,scale=1,pattern=crosshatch dots, pattern color=black ] {};
\node (V old) at (0,0) [circle,scale=1, shade,ball color=blue] {};
\node (start) at (-1,-3) [circle,scale=0.1, gray] {};
\node (vp old) at (1.414,-1.414) [circle,scale=1, shade,ball color=green!30!black!70!white] {};
\node (vps old) at (-1.414,1.414) [circle,scale=1, shade,ball color=green!30!black!70!white] {};
\begin{scope}[>=latex]
\draw[->,line width=0.6ex,gray!80!black] (V old) -- (0.7071,-0.7071) node[midway, above,black] {$\sigma$};
\draw[->,line width=0.2ex] (start) -- (vs) node[left,pos=0.98] {$v_*$};
\end{scope}
\begin{scope}[>=latex, on background layer]
\draw[densely dotted,color=green!30!black!70!white,line width=0.2ex] (0,0) circle (2cm);
\draw[densely dotted,->,color=green!30!black!70!white,line width=0.2ex] (vps old) -- (vp old) node[densely dotted,color=green!30!black,pos=0.3,above right=-0.1cm,line width=0.2ex] {$ \left|u \right| \sigma$};
\draw[->,line width=0.2ex] (start) -- (v) node[pos=0.96,right=0.1cm] {\color{red}$v$};
\draw[->,line width=0.2ex] (vs) -- (v) node[pos=0.3,sloped, black,above left] {$u$};
\draw[densely dotted,->,color=green!30!black!70!white,line width=0.2ex] (start) -- (V old) node[color=green!30!black,pos=0.9,below left=-0.1cm] {$\frac{v+v_*}{2}$};
\draw[->,dashdotted,color=blue,line width=0.2ex] (vps)--(vp) node[color=green!30!black,above right=-0.1cm,midway] {\color{blue}$u'$};
\draw[densely dotted,->,color=green!30!black!70!white,line width=0.2ex] (start) -- (vp old) node[color=green!30!black,below right,line width=0.2ex] {$ \frac{v + v_*}{2} + \frac{1}{2} \left|u \right| \sigma$};
\draw[densely dotted,->,color=green!30!black!70!white,line width=0.2ex] (start) -- (vps old) node[color=green!30!black, left=0.1cm] {$ \frac{v + v_*}{2} - \frac{1}{2} \left|u \right| \sigma$};
\draw[->,densely dashed,blue,line width=0.6ex] (V old) -- (V) node[midway, sloped,above, blue] {$\frac{m_i-m_j}{2(m_i+m_j)}u$};
\draw[dashdotted,->,color=blue,line width=0.2ex] (start) -- (V) node[pos=0.98,below right=-0.05cm] {\color{blue}$V_{ij}$};
\draw[dashdotted,->,color=blue,line width=0.2ex] (start) -- (vps) node[pos=0.98,left] {\color{black}$v'_*$};
\draw[dashdotted,->,color=blue,line width=0.2ex] (start) -- (vp) node[pos=0.98,below right=-0.1cm] {\color{red}$v'$};
\end{scope}
\end{tikzpicture}
\end{center}
\caption{Illustration of the collision transformation, with notation $V_{ij}:=\frac{m_i v + m_j v_*}{m_i + m_j}$, $u:=v-v_*$, $u':=v'-v'_*$. The displacement of the center of mass with respect to a single component elastic binary interaction is given by $(r_{ij}-\frac 12)u = \frac{m_i-m_j}{2(m_i+m_j)}u$, if $m_i>m_j$.
Solid lines denote vectors after collision, or given data. Dash-dotted vectors represent primed (pre-collisional) quantities that can be calculated from the given data, and compared to the case $m_i=m_j$, represented by dotted vectors. Dashed vector direction is the displacement along the direction of the relative velocity $u$ proportional to the half difference of relative masses, (which clearly vanishes for $m_i = m_j$, reducing the model to a classical collision). Note that the scattering direction $\sigma$ is preserved as the pre-collisional relative velocity $u'$ keeps the same magnitude as the post-collisional $u$, $u'$ is parallel the reference elastic pre-collisional relative velocity $|u|\sigma$. }
\label{picture}
\end{figure}
Figure \ref{picture} illustrates the collision transformation \eqref{collisional rules bi} and aims at explaining its difference with respect to the collision transformation \eqref{collisional rules equal masses} when masses are equal. Namely, for given $v$, $v_*$, $\sigma$ and $m_i$, $m_j$, we calculate center of mass $V=\frac{m_i v + m_j v_*}{m_i + m_j}$, and velocities $v'$ and $v'_*$ according to \eqref{collisional rules bi}. One can notice that the magnitude of the relative velocity does not change during the collision, i.e. $\left|v-v_*\right|= \left|v'-v'_*\right|$, as it is when masses are the same. Difference comes with the vector of center of mass: the vector of center of mass for equal masses $\frac{v+v_*}{2}$ displaces by adding a quantity that is proportional to the difference of masses $m_i-m_j$ and thus is peculiar to the mixture case. More precisely,
$$
V= \frac{v+v_*}{2} + \frac{m_i-m_j}{2(m_i+m_j)}u,
$$
with $u:=v-v_*$.
\subsection{Collision operators} Collision operators $Q_{ij}$, as defined in \eqref{Qij}, describe binary interactions between molecules of species $\mathcal{A}_i$ and $\mathcal{A}_j$, $i,j=1,\dots,I$.
Fix the species $\mathcal{A}_i$ for any $i=1,\dots,I$, and let its distribution function be $g$. On the other hand, let distribution function $h$ describe species $\mathcal{A}_j$.
Note that each $Q_{ij}$ for a fix $(i,j)$-pair has its corresponding counterpart, $Q_{ji}$, that describes interaction of molecules of species $\mathcal{A}_j$ with molecules of species $\mathcal{A}_i$
\begin{equation}\label{collision operator ji}
Q_{ji}(h,g)(v) = \int_{\mathbb{R}^3} \int_{S^2} \left( h(w') g(w'_*) - h(v) g(v_*) \right) \mathcal{B}_{ji}(v,v_*,\sigma) \, \mathrm{d} \sigma \, \mathrm{d}v_*,
\end{equation}
where pre-collisional velocities $w'$ and $w'_*$ now differ from the previous ones given in \eqref{collisional rules bi} by an exchange of mass $m_i \leftrightarrow m_j$, i.e.
\begin{equation}\label{collisional rules symmetry}
w' =\frac{m_j v + m_i v_*}{m_i + m_j} + \frac{m_i}{m_i + m_j} \left| v-v_* \right| \sigma, \quad w'_* =\frac{m_j v + m_i v_*}{m_i + m_j} - \frac{m_j}{m_i + m_j} \left| v-v_* \right| \sigma.
\end{equation}
When $m_i=m_j$, although primed velocities are the same, $Q_{ij}$ and $Q_{ji}$ still defer, because of the cross section.
\subsection{Weak form of collision operator}
Testing the collision operator against some suitable test functions $\psi(v)$ and $\phi(v)$ yields
\begin{multline*}
\int_{\mathbb{R}^3} Q_{ij} (g,h)(v) \psi(v) \mathrm{d} v \\ = \iiint_{\mathbb{R}^3 \times \mathbb{R}^3\times S^2} g(v) h(v_*) \left( \psi(v') - \psi(v) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v,
\end{multline*}
and
\begin{multline*}
\int_{\mathbb{R}^3} Q_{ji} (h,g)(v) \phi(v) \mathrm{d} v \\ = \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} h(v_*)g(v) \left( \phi(v'_*) - \phi(v_*) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v,
\end{multline*}
where now $v'$ and $v'_*$ are denoting the post-collisional velocities as defined by \eqref{collisional rules bi}. Therefore, looking at these two integrals pairwise, meaning that each time when $Q_{ij}$ is considered we add his pair $Q_{ji}$, we have
\begin{multline}\label{weak form ij+ji}
\int_{\mathbb{R}^3} \left( Q_{ij} (g,h)(v) \psi(v) + Q_{ji} (h,g)(v) \phi(v) \right) \mathrm{d} v \\ = \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} g(v) h(v_*) \left( \psi(v') + \phi(v'_*) - \psi(v) - \phi(v_*) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v,
\end{multline}
with $v'$ and $v'_*$ are now given by the post-collisional velocities as defined by \eqref{collisional rules bi}. \\
Some choice of test function leads to annihilation of the weak form. Namely, from the conservation laws during collision process, we see
\begin{equation}\label{collision invariants bi mass}
\int_{\mathbb{R}^3} Q_{ij} (g,h)(v) \mathrm{d} v = 0,
\end{equation}
as well as
\begin{equation}\label{collision invariants bi}
\int_{\mathbb{R}^3} \left( Q_{ij} (g,h)(v) \left( \begin{matrix} m_i v \\ m_i \left| v \right|^2 \end{matrix} \right) + Q_{ji} (h,g)(v) \left( \begin{matrix} m_j v \\ m_j \left| v \right|^2 \end{matrix} \right) \right) \mathrm{d} v = 0.
\end{equation}
Therefore, if we consider distribution function $\mathbb{F}=\left[f_i\right]_{1\leq i \leq I}$, then the weak form \eqref{weak form ij+ji} yields
\begin{multline}\label{weak form any test}
2 \sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} Q_{ij} (f_i,f_j)(v) \psi_i(v) \mathrm{d} v = \sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} f_i(v) f_j(v_*) \\ \times \left( \psi_i(v') + \psi_j(v'_*) - \psi_i(v) - \psi_j(v_*) \right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v.
\end{multline}
\subsection{Conservation laws} Weak forms of collision operator imply some its conservative properties. More precisely, for any suitable $\mathbb{F}$, \eqref{collision invariants bi mass} implies
\begin{equation}\label{cons operator mass}
\int_{\mathbb{R}^3} \left[\mathbb{Q}(\F) \right]_i\mathrm{d}v
= 0, \qquad \text{for any} \ i=1,\dots,I,
\end{equation}
and moreover, from \eqref{weak form any test} choosing $\psi_\ell(x)=m_\ell \left|x\right|^2$, and $\psi_\ell(x)=m_\ell x$, $x\in \mathbb{R}^3$, one has
\begin{equation}\label{cons operator energy}
\sum_{i=1}^I \left[ \mathbb{Q}(\F) \right]_i m_i \left|v\right|^2 \mathrm{d} v=0,
\end{equation}
and
\begin{equation*}
\sum_{i=1}^I \left[ \mathbb{Q}(\F) \right]_i m_i v\, \mathrm{d} v=0,
\end{equation*}
for any time $t\geq 0$.
If $\mathbb{F}$ is a solution to the Boltzmann system \eqref{Cauchy problem}, then these properties imply conservation laws for number density of each species $\mathcal{A}_i$, $i=1,\dots,I$, and total energy of the mixture. Indeed,
\begin{equation}\label{conservation mass energy}
\partial_t \mathfrak{m}_{0,i}[\mathbb{F}](t)=0, \ \forall i=1,\dots,I, \qquad \partial_t \mathfrak{m}_2[\mathbb{F}](t)=0.
\end{equation}
\section{Proof of Povzner lemma~\ref{Povzner intro} }\label{Section prelim Lemmas}
The proof of Povzner lemma~\ref{Povzner intro} by angular averaging for the mixing model entices to obtain estimates for the quantity $ \left\langle v' \right\rangle_i^k + \left\langle v'_* \right\rangle_j^k $ integrated over sphere $S^2$, that represents the gain part of \eqref{weak form any test} for $\psi_i(x) = \left\langle x \right\rangle_i^k$. The usual techniques used in \cite{Gamba13} for example, can not be directly adapted when $m_i\neq m_j$. This becomes clear when one writes local kinetic energies of each colliding molecule pair. When $m_i\neq m_j$, these energies can be written as a certain convex combination, while single component case (or in the same fashion when $m_i=m_j$) correspond to the ``middle" of this convex combination, or to the ``halfs" (see Remark \ref{Remark single vs mixture} below). Single component situation (or when $m_i=m_j$) is therefore ``symmetric", in a sense, and the techniques for proof of a sharper Povzner lemma by angular averaging, as done by \cite{Bob97} or \cite{GambaPanfVil09},
can not be extended to the mixture case in a straight forward form.
Indeed, in the mixture setting when $m_i\neq m_j$, the proof of the Povzner
lemma~\ref{Povzner intro} in the cases of non-linear gas mixture system uses a non-trivial modification of a powerful energy identity in scattering angle coordinates. This identity is needed in order to compute moment estimates that clearly show positive moments from the gain collision operator part are dominated by the moments of the corresponding loss part, which yields a very sharp estimate sufficient to obtain not only moments propagation and generation, but also their scaled summability that prove propagation and generation of exponential moment estimates as well. An energy identity in scattering angle coordinates was first developed in \cite{Bob97,GambaBobPanf04} for the elastic and inelastic case for scalar Boltzmann binary models. While such identity is rather easy in the elastic single species setting, where local energies are just the sum of the collision invariant $|v|^2$ and just its interacting counterpart $|v_*|^2$, in the mixing case under consideration the problem becomes highly non-trivial and the local energies to be estimated now depend on binary sums
of $\left\langle v \right\rangle_i^2 $ and $\left\langle v_* \right\rangle_j^2$ and their corresponding post collisional sum of $\left\langle v' \right\rangle_i^2 $ and $\left\langle v'_* \right\rangle_j^2$.
\begin{lemma}[{Energy identity in scattering direction coordinates for the {\color{black} $(i,j)-$}pair of colliding particles}]\label{energy-identity-mixture}
Consider any {\color{black} $(i,j)-$}pair of interacting velocities $v$ and $v_*$ corresponding to particles masses $m_i$ and $m_j$, respectively, {\color{black}with $i, j$ fixed}.
Let their local micro energy be $E_{\color{black}{ij}} = \left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2$, with $\left\langle v \right\rangle_i^2$ and $\left\langle v_* \right\rangle_j^2$ defined according to \eqref{vi}, and recall the two-body mass fraction parameter $r_{\color{black}{ij}}=\frac{m_i}{m_i+m_j}$ introduced in \eqref{cof-rv-coor}.
Then, there exists a couple of functions $p_{\color{black}{ij}}=p_{\color{black}{ij}}(v,v_*,m_i,m_j) $ and $q_{\color{black}{ij}}=q_{\color{black}{ij}}(v,v_*,m_i,m_j)$ such that, $p_{\color{black}{ij}}+q_{\color{black}{ij}} = E_{\color{black}{ij}}$ and
the following representation holds
\begin{equation}\label{EI1}
\langle {\color{black} v'_{ij} }\rangle_i^2= p_{\color{black}ij} + \lambda_{\color{black}ij}\, \sigma \cdot \hat{V}_{\color{black}ij}, \ \ \ \qquad
\langle {\color{black} v'_{*ij} }\rangle_j^2= q_{\color{black}ij} - \lambda_{\color{black}ij} \, \sigma \cdot \hat{V}_{\color{black}ij}.
\end{equation}
where $\lambda_{\color{black}ij}:= 2 \sqrt{r_{\color{black}ij} (1-r_{\color{black}ij}) (sE_{\color{black}ij} -1)((1-s_{\color{black}ij})E_{\color{black}ij}-1)}$ with $s_{\color{black}ij}=s_{\color{black}ij}(v,v_*,m_i,m_j)\in [0,1]$. In particular, this representation preserves the local energy identity
\begin{equation}\label{EI2}
\langle {\color{black} v'_{ij} }\rangle_i^2 + \langle {\color{black} v'_{*ij} }\rangle_j^2\ =\ p_{\color{black}ij}+q_{\color{black}ij}\ =\ E_{\color{black}ij} \ = \ \left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2.
\end{equation}
Moreover, the following inequalities hold
\begin{equation}\label{key point}
p_{\color{black}ij} +\lambda_{\color{black}ij} \leq E_{\color{black}ij} , \quad q_{\color{black}ij} + \lambda_{\color{black}ij} \leq E_{\color{black}ij} ,
\end{equation}
for any velocities $v, v_* \in \mathbb{R}^3$ and any masses $m_i, m_j >0$.
\end{lemma}
{\color{black} As we mentioned earlier in Remark \ref{eliminate subindex ij}, we eliminate subindex $ij$ from $E_{ij}$, $p_{ij}$, $q_{ij}$, $\lambda_{ij}$, $s_{ij}$ as we did in Remark \ref{eliminate subindex ij} for $v'_{ij}$, $v'_{*ij}$, $V_{ij}$ and $r_{ij}$. }
\begin{proof}[Proof of Lemma~\ref{energy-identity-mixture}] As anticipated, we represent the exchange of coordinates at the interaction using the center of mass and relative velocity reference frame \eqref{collisional rules intro} (with its symmetric form \eqref{collisional rules symmetry})
where the angular integration if performed in the scattering direction corresponding to the post-collisional relative velocity $\sigma=\hat{u'}$. Thus, let's denote with $V$ the vector of center-of-mass and with $u$ the relative velocity as in \eqref{center-of-mass},
\begin{equation*}
V=\frac{m_i v + m_j v_*}{m_i + m_j}, \qquad u=v-v_*.
\end{equation*}
Then, taking the squares of the magnitudes of the post-collisional velocities given in \eqref{collisional rules bi}, one obtains
\begin{equation*}
\begin{split}
\left|v'\right|^2 &= \left|V\right|^2 + \frac{ m_j^2}{(m_i+m_j)^2} \left|u\right|^2 + \frac{2 m_j}{m_i + m_j} \left|u\right| \left|V\right| \sigma \cdot \hat{V},\\
\left|v'_*\right|^2 &= \left|V\right|^2 + \frac{m_i^2}{(m_i+m_j)^2} \left|u\right|^2 - \frac{2 m_i }{m_i + m_j} \left|u\right| \left|V\right| \sigma \cdot \hat{V},
\end{split}
\end{equation*}
where $\hat{V}$ denotes the unit vector of $V$. Passing to $\langle \cdot \rangle$ bracket forms from \eqref{vi}, implies
\begin{equation}\label{kinetic energies squares}
\begin{split}
\left\langle v' \right\rangle_i^2 &= 1+ \frac{m_i}{\sum_{\ell=1}^I m_\ell} \left|V\right|^2 + \frac{m_i m_j^2}{(m_i+m_j)^2 \sum_{\ell=1}^I m_\ell} \left|u\right|^2 \\ &\hspace*{5cm}+ \frac{2 m_i m_j}{(m_i + m_j)\sum_{\ell=1}^I m_\ell} \left|u\right| \left|V\right| \sigma \cdot \hat{V},\\
\left\langle v'_* \right\rangle_j^2 &= 1+ \frac{m_j}{\sum_{\ell=1}^I m_\ell} \left|V\right|^2 + \frac{m_j m_i^2}{(m_i+m_j)^2\sum_{\ell=1}^I m_\ell} \left|u\right|^2 \\ &\hspace*{5cm}- \frac{2 m_i m_j}{(m_i + m_j)\sum_{\ell=1}^I m_\ell} \left|u\right| \left|V\right| \sigma \cdot \hat{V}.
\end{split}
\end{equation}
Let us introduce the total energy $E$ of two colliding particles in $\langle \cdot \rangle$ bracket forms, which is conserved during collision process by \eqref{CL micro energy},
\begin{equation*}
\begin{split}
E := \left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 = \left\langle v' \right\rangle_i^2 + \left\langle v'_* \right\rangle_j^2.
\end{split}
\end{equation*}
Using the above equations \eqref{kinetic energies squares}, the energy $E$ can be written in $u-V$ notation, as well,
\begin{equation}\label{convex-E}
\begin{split}
E &= 2 + \frac{m_i+m_j}{\sum_{\ell=1}^I m_\ell} \left| V \right|^2 + \frac{m_i m_j}{(m_i+m_j)\sum_{\ell=1}^I m_\ell} \left| u \right|^2.
\end{split}
\end{equation}
\\
The aim is to represent the squares of the post-collisional velocities $\left\langle v' \right\rangle_i^2$ and $\left\langle v'_* \right\rangle_j^2$ as a scalar convex combination of different ``parts" of the energy $E$. This is achieved by introducing two quantities,
\begin{itemize}
\item[i)] the parameter $r\in(0,1)$, that distributes masses in the following convex pair
\begin{equation}\label{parameter-r}
r=\frac{m_i}{m_i+m_j} \ \ \text{and}\ \ 1-r = \frac{m_j}{m_i+m_j},
\end{equation}
\item[ii)] the function $s\in [0,1]$ that convexly partitions the energy $E $ into two components, one related to $\left| u \right|^2$ and another to $\left| V \right|^2$,
using the above identity \eqref{convex-E} as follows
\begin{equation}\label{variable-s}
sE= 1+ \frac{m_i m_j}{(m_i+m_j)\sum_{\ell=1}^I m_\ell} \left| u \right|^2 \ \ \text{and}\ \ (1-s)E= 1 + \frac{m_i+m_j}{\sum_{\ell=1}^I m_\ell} \left| V \right|^2.
\end{equation}
\end{itemize}
Finally, each of the post-collisional quantities $\left\langle v' \right\rangle_i^2$ and $\left\langle v'_* \right\rangle_j^2$ as written the representation as in \eqref{kinetic energies squares}, can be {\color{black}recast}
through the energy $E$ and the dot product between center of mass vector $V$ and the scattering direction $\sigma$ as follows
\begin{equation}\label{kin energy mixture}
\begin{split}
\left\langle v' \right\rangle_i^2&= r (1-s) E + (1-r) s E + 2 \sqrt{r (1-r) (sE -1)((1-s)E-1)} \, \sigma \cdot \hat{V},\\
\left\langle v'_* \right\rangle_j^2&= r s E + (1-r) (1-s) E - 2 \sqrt{r (1-r) (sE -1)((1-s)E-1)} \, \sigma \cdot \hat{V},
\end{split}
\end{equation}
which yields the important relation that expresses the post - collisional local micro energy $E$ as a rotation of factors of $E$ and $V\cdot\sigma$, while preserving the local energy itself.
Indeed, denoting
\begin{equation*}
\begin{split}
& p= r (1-s) E + (1-r) s E, \\ & q=E-p= r s E + (1-r) (1-s) E, \\ &\lambda= 2 \sqrt{r (1-r) (sE -1)((1-s)E-1)} \\& \phantom{\lambda}= 2 \sqrt{r (1-r)}\frac{\sqrt{m_i m_j}}{\sum_{\ell=1}^I m_\ell} \left| u \right| |V|,
\end{split}
\end{equation*}
clearly $p+q=E$ and the representation \eqref{EI1} follows while preserving the binary micro energy relation \eqref{EI2}
\begin{equation*}
\left\langle v' \right\rangle_i^2= p + \lambda\, \sigma \cdot \hat{V}, \quad
\left\langle v'_* \right\rangle_j^2= q - \lambda \, \sigma \cdot \hat{V},
\end{equation*}
which completes the proof of the energy identities \eqref{EI1} and \eqref{EI2}.
Moreover, it follows
\begin{equation*}
\frac{1}{E} \left( p+ \lambda\right) \leq \left( \sqrt{r(1-s)} + \sqrt{(1-r)s} \right)^2 \leq 1,
\end{equation*}
since
\begin{equation*}
\max_{\substack{ 0<r<1 \\ 0\leq s \leq 1}} \left( \sqrt{r(1-s)} + \sqrt{(1-r)s} \right) =1.
\end{equation*}
Similarly,
\begin{equation*}
\frac{1}{E} \left( q+ \lambda\right) \leq \left( \sqrt{rs} + \sqrt{(1-r)(1-s)} \right)^2 \leq 1,
\end{equation*}
{\color{black} uniformly in any $(i,j)-$pair,}
which concludes the proof of Lemma.
\end{proof}
\begin{remark}\label{Remark single vs mixture} Let us elaborate more on the difference between writing kinetic energies \eqref{kinetic energies squares} when $m_i\neq m_j$ versus $m_i=m_j$. In order to be more precise, we will put a bar on a quantity when assuming the same masses. For instance, total energy of the two colliding particles of the same masses $m_i$ is
\begin{equation*}
\bar{E}= \left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_i^2 = 2 + \frac{2 m_i}{\sum_{j=1}^I m_j} \left|V\right|^2 + \frac{ m_i}{2 \sum_{j=1}^I m_j} \left|u\right|^2.
\end{equation*}
When $m_i=m_j$ then the parameter $r=1/2$, and consequently for $\bar{p}:=p(v, v_*, m_i, m_i)$, $\bar{q}:=q(v, v_*, m_i, m_i)$ and $\bar{\lambda}$ we have
\begin{equation*}
\bar{p} = \bar{q} = \frac{1}{2} \, \bar{E}, \quad \bar{\lambda} = \frac{ m_i}{\sum_{j=1}^I m_j} {\left|u\right| \left|V\right| },
\end{equation*}
which gives the squares of the magnitudes of the post-collisional velocities when $m_i=m_j$,
\begin{equation}\begin{split}\label{kin energy single}
&\left\langle v' \right\rangle_i^2 = \bar{E} \left( \frac{1}{2} + \frac{ m_i}{\sum_{j=1}^I m_j} \frac{ \left|u\right| \left|V\right| }{\bar{E}} \sigma \cdot \hat{V} \right), \\
&\left\langle v'_* \right\rangle_i^2 = \bar{E} \left( \frac{1}{2} - \frac{ m_i}{\sum_{j=1}^I m_j} \frac{ \left|u\right| \left|V\right| }{\bar{E}} \sigma \cdot \hat{V}\right).
\end{split} \end{equation}
Now, the difference between \eqref{kin energy mixture} as a convex combination writing in the mixture setting and \eqref{kin energy single} as its special ``middle point", or ``half", case in single component case (or mixture for $m_i=m_j$) is clear.\\
Another important aspect to be pointed out is the comparison of inequalities \eqref{key point} in the case $m_i\neq m_j$ versus $m_i=m_j$. When $m_i=m_j$, simply performing Young's inequality we get
\begin{equation}\label{s=1/2}
\frac{\bar{\lambda}}{\bar{E}} \leq \frac{1}{2},
\end{equation}
which yields
\begin{equation*}
\frac{1}{\bar{E}} \left( \bar{p}+ \bar{\lambda}\right) = \frac{1}{\bar{E}} \left( \bar{q}+ \bar{\lambda}\right) \leq 1.
\end{equation*}
This inequality is an analogue of \eqref{key point} for $m_i = m_j$. Note than when masses are the same we can make use of the Young inequality, while in the case of different masses, we had to be more precise since both $\frac{1}{{E}} \left( {p}+ {\lambda}\right)$ and $ \frac{1}{{E}} \left( {q}+ {\lambda}\right)$ attain 1 as a maximal value for some values of their arguments, and therefore there is no room for any inequality. In particular, this inequality will be of decisive importance for the success of the Povzner lemma that will guarantee decay of the gain term with respect to the number of moments.
\end{remark}
\medskip
\begin{proof}[Proof of Povzner lemma~\ref{Povzner intro}]
In order to compute the angular average estimate \eqref{Povzner estimate gain} we use the representation \eqref{EI1} and \eqref{EI2} from the energy identity Lemma~\ref{energy-identity-mixture} raised to power $k/2$.
Then, the left hand side integral of \eqref{Povzner estimate gain} becomes
\begin{multline}\label{pomocna 2}
\int_{S^2} \left( \left\langle v' \right\rangle_i^k + \left\langle v'_* \right\rangle_j^k \right) {\color{black} {\color{black}b_{ij}} (\sigma\cdot \hat{u})} \mathrm{d}\sigma \\ = \int_{S^2} \left( \left( p + \lambda\, \sigma \cdot \hat{V} \right)^\frac{k}{2} + \left( q - \lambda \, \sigma \cdot \hat{V} \right)^\frac{k}{2} \right) {\color{black} {\color{black}b_{ij}} (\sigma\cdot \hat{u})} \, \mathrm{d}\sigma.
\end{multline}
{\color{black}
Now we use polar coordinates for $\sigma$ and $\hat{V}$ with zenith $\hat{u}$. Namely, denoting with $\theta$ the angle between $\sigma$ and $\hat{u}$, we decompose $\sigma$ as
\begin{equation}\label{sigma decomposed}
\sigma = \cos\theta\, \hat{u} + \sin\theta\, \omega, \ \text{with} \ \hat u\cdot\omega=0 \ \text{and} \ \omega=(\cos\varphi, \sin\varphi), \ \theta\in[0,\pi), \varphi\in[0,2\pi).
\end{equation}
In the same fashion we decompose $\hat{V}$, by denoting with $\alpha \in [0,\pi)$ the angle between $\hat{V}$ and $\hat{u}$,
\begin{equation*}
\hat{V} = \cos\alpha\, \hat{u} + \sin\alpha\, \Phi, \ \text{where} \ \Phi\in S^1 \ \text{with} \ \hat u\cdot\Phi=0.
\end{equation*}
Then the scalar product $\sigma \cdot \hat{V}$ becomes
\begin{align*}
\sigma \cdot \hat{V}
&=\cos\theta\cos\alpha + \Phi\cdot\omega \sin\theta\sin\alpha.
\end{align*}
Defining $\tau:= \cos\theta$ and expressing $\sin\theta=\sqrt{1-\tau^2}$, since $\sin\theta \geq 0$ on the range of $\theta$, this scalar product reads
\begin{equation}\label{define_mu}
\sigma \cdot \hat{V} = \tau\cos\alpha + \Phi\cdot\omega \sqrt{1-\tau^2} \sin\alpha=:\mu =\mu(\tau,\alpha, \Phi\cdot\omega).
\end{equation}
In the integral \eqref{pomocna 2}, we first express $\sigma$ in its polar coordinates \eqref{sigma decomposed} and then change variables $\theta\mapsto\tau=\cos\theta$, which yields
\begin{multline*}
\int_{S^2} \left( \left( p + \lambda\, \sigma \cdot \hat{V} \right)^\frac{k}{2} + \left( q - \lambda \, \sigma \cdot \hat{V} \right)^\frac{k}{2} \right) b_{ij}(\sigma\cdot \hat{u})\, \mathrm{d}\sigma
\\
= \int_{0}^{2\pi} \int_{0}^{\pi} \left( \left( p + \lambda\, \sigma \cdot \hat{V} \right)^\frac{k}{2} + \left( q - \lambda \, \sigma \cdot \hat{V} \right)^\frac{k}{2} \right) b_{ij}(\cos\theta) \, \sin\theta \mathrm{d}\theta\mathrm{d}\varphi
\\
= \int_{0}^{2\pi} \int_{-1}^{1} \left( \left( p + \lambda\,\mu \right)^\frac{k}{2} + \left( q - \lambda \, \mu\right)^\frac{k}{2} \right) b_{ij}(\tau) \mathrm{d}\tau\mathrm{d}\varphi.
\end{multline*}
}
For $k\geq 4$, Taylor expansion of $ \left( p + \lambda\, \mu \right)^{k/2}$ and $\left( q - \lambda \, \mu \right)^{k/2} $ around $\mu=0$ up to second order yields:
\begin{equation*}
\begin{split}
\left( p + \lambda\, \mu \right)^\frac{k}{2} &= p^\frac{k}{2} + \tfrac{k}{2} p^{\tfrac{k}{2}-1} \lambda \mu + \tfrac{k}{2} \left(\tfrac{k}{2}-1\right) \lambda^2 \mu^2 \int_{0}^{1} (1-z) (p+\lambda \mu z)^{\frac{k}{2}-2} \mathrm{d}z,\\
\left( q - \lambda \, \mu \right)^\frac{k}{2} &= q^\frac{k}{2} - \tfrac{k}{2} q^{\frac{k}{2}-1} \lambda \mu + \tfrac{k}{2} \left(\tfrac{k}{2}-1\right) \lambda^2 \mu^2 \int_{0}^{1} (1-z) (q - \lambda \mu z)^{\frac{k}{2}-2} \mathrm{d}z.
\end{split}
\end{equation*}
For $2<k<4$, we stop at the first order and proceed similarly.
Now, let us analyze the integrands. By the Young inequality, for $\lambda$ the following estimates hold
\begin{equation}\label{lambda estimate}
\pm \lambda\leq q -1 \leq q, \quad \text{and} \quad \pm \lambda \leq p-1 \leq p.
\end{equation}
We recall definition of $p$ and $q$,
\begin{equation*}
p= \left(r(1-s)+(1-r)s\right) E, \quad q= \left(rs+(1-r)(1-s)\right)E,
\end{equation*}
for $r\in (0,1)$ and $s\in [0,1]$. Considering $r$ as parameter, for both coefficients maximum with respect to variable $s$ is achieved on the boundary, i.e. for either $s=0$ or $s=1$, and moreover the following estimate holds for both coefficients
\begin{equation*}
r(1-s)+(1-r)s \leq \max\{r,(1-r)\}, \quad rs+(1-r)(1-s) \leq \max\{r,(1-r)\}.
\end{equation*}
Denoting
\begin{equation}\label{rmax}
{\overline{r}}=\max\{r, 1-r\},
\end{equation}
we conclude on upper bound for both $p$ and $q$,
\begin{equation*}
p \leq {\overline{r}} E, \qquad q\leq {\overline{r}} E.
\end{equation*}
Moreover, for $p$ and $q$ it holds
\begin{equation*}
p = r (1-s)E + (1-r) sE \geq {\underline{r}} E
\quad \text{and} \quad q \geq {\underline{r}} E
\end{equation*}
where we have denoted
\begin{equation}\label{rmin}
{\underline{r}}=\min\{r, 1-r\}.
\end{equation}
Taking into account inequalities above, one has
\begin{equation*}
p + \lambda \mu z \leq p+ q \mu z= E - q(1-\mu z) \leq E (1- {\underline{r}}(1- |\mu| z)),
\end{equation*}
and similarly
\begin{equation*}
q - \lambda \mu z \leq E (1- {\underline{r}} (1-|\mu| z)).
\end{equation*}
Therefore,
\begin{multline*}
\left( p + \lambda\, \mu \right)^\frac{k}{2} + \left( q - \lambda \, \mu \right)^\frac{k}{2} \leq p^\frac{k}{2} + q^\frac{k}{2} + \tfrac{k}{2} \mu \left( p^{\frac{k}{2}}+q^{\frac{k}{2}} \right) \\ + k \left(\tfrac{k}{2}-1\right) {\overline{r}}^2 \mu^2 E^{\frac{k}{2}} \int_{0}^{1} (1-z) (1- {\underline{r}}(1-|\mu| z))^{\frac{k}{2}-2} \mathrm{d}z.
\end{multline*}
Then
{\color{black}
\begin{equation*}
\int_{0}^{2\pi} \int_{-1}^{1} \left( \left( p + \lambda\,\mu \right)^\frac{k}{2} + \left( q - \lambda \, \mu\right)^\frac{k}{2} \right) b_{ij}(\tau) \, \mathrm{d}\tau\,\mathrm{d}\varphi \leq P_1 + P_2 + P_3,
\end{equation*}
with
\begin{equation*}
\begin{split}
P_1 &:= \left(p^\frac{k}{2} + q^\frac{k}{2}\right) \int_{0}^{2\pi} \int_{-1}^{1} b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi ,\\
P_2 &:= \tfrac{k}{2} \left( p^{\frac{k}{2}}+q^{\frac{k}{2}} \right) \int_{0}^{2\pi} \int_{-1}^{1} \mu \, b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi, \\
P_3 &:= k\left(\tfrac{k}{2}-1\right) {\overline{r}}^2 E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 \mu^2 \\ &\qquad \qquad \qquad \qquad \times \left(\int_{0}^{1} (1-z) (1- {\underline{r}}(1-|\mu| z))^{\frac{k}{2}-2} \mathrm{d}z \right) b_{ij}(\tau)\, \mathrm{d}\tau\, \mathrm{d}\varphi.
\end{split}
\end{equation*}
}
\subsubsection*{Term $P_1$}
Introducing constant $\tilde{C}_n$
\begin{equation}
\tilde{C}_n = {\overline{r}}^n, \quad 0<{\overline{r}}<1,
\end{equation}
which clearly decays in $n$, we have
{\color{black}
\begin{equation*}
P_1 = \bijnjed \left(p^\frac{k}{2} + q^\frac{k}{2} \right) \leq \bijnjed 2 \tilde{C}_\frac{k}{2} E^\frac{k}{2}.
\end{equation*}
\subsubsection*{Term $P_2$}
Taking into account definition of $\mu$ from \eqref{define_mu}, the parity arguments yield
\begin{equation*}
P_2 = \tfrac{k}{2} \left(p^\frac{k}{2} + q^\frac{k}{2} \right) \int_0^{2\pi} \int_{-1}^1 \tau \cos \alpha \, b_{ij}(\tau) \, \mathrm{d} \tau \, \mathrm{d}\varphi,
\end{equation*}
after bounding $\cos\alpha \leq 1$.
Using the estimate above for $P_1$ and the fact that $\tau \cos \alpha\leq1$, we finally obtain
\begin{equation*}
P_2 \leq \bijnjed k \, \tilde{C}_\frac{k}{2} E^\frac{k}{2}.
\end{equation*}
Since the constant $ \tilde{C}_\frac{k}{2}$ has power decay in $k$, the constant $ k \, \tilde{C}_\frac{k}{2} $ also decreases in $k$.
\subsubsection*{Term $P_3$}
We can compute explicitly the integral with respect to $z$
\begin{multline*}
\int_{0}^{1} (1-z) (1- a (1-A z))^{n-2} \mathrm{d}z \\= \frac{1}{a^2 A^2} \frac{1}{n(n-1)} \left( \left(1+ a(A-1)\right)^n - (1-a)^n - a A(1-a)^{n-1} n\right),
\end{multline*}
for any $0<a<1$ and $A>0$. If $A=0$, then we easily obtain
\begin{equation*}
\int_{0}^{1} (1-z) (1- a )^{n-2} \mathrm{d}z = \frac{1}{2} (1-a)^{n-2}.
\end{equation*}
In our case $a={\underline{r}}$ and $A=|\mu|$, $\mu$ being a function of variables of integration $\tau$ and $\varphi$ defined in \eqref{define_mu} that satisfies $|\mu|\leq 1$, and thus $P_3$ becomes
\begin{multline*}
P_3 =2\frac{{\overline{r}}^2}{ {\underline{r}}^2} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 \left( \left(1+ {\underline{r}}(|\mu|-1)\right)^{\frac{k}{2}} \right. \\ \left. - (1-{\underline{r}})^{\frac{k}{2}} - {\underline{r}} |\mu|(1-{\underline{r}})^{\frac{k}{2}-1} \tfrac{k}{2}\right) b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi\\
=: P_{3_1} + P_{3_2} + P_{3_3},
\\ \leq E^{\frac{k}{2}} \left( \check{C}^{b_{ij}}_{\frac{k}{2}} + \bijnjed \left( \bar{C}_{\frac{k}{2}} + \hat{C}_{\frac{k}{2}} \right) \right),
\end{multline*}
where we have denoted
\begin{equation*}
\begin{split}
P_{3_1} &:= 2\frac{{\overline{r}}^2}{ {\underline{r}}^2} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 \left(1+ {\underline{r}}(|\mu|-1)\right)^{\frac{k}{2}} b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi,\\
P_{3_2} &:= - 2\frac{{\overline{r}}^2}{ {\underline{r}}^2} (1-{\underline{r}})^{\frac{k}{2}} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi,\\
P_{3_3} &:= - \frac{{\overline{r}}^2}{ {\underline{r}}} \, k\, (1-{\underline{r}})^{\frac{k}{2}-1} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 |\mu| \, b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi.\\
\end{split}
\end{equation*}
\subsubsection*{Term $P_{3_1}$} We rewrite term $P_{3_1}$,
\begin{equation*}
P_{3_1} = \check{C}^{b_{ij}}_{\frac{k}{2}} E^{\frac{k}{2}},
\end{equation*}
by introducing the constant $\check{C}^{b_{ij}}_n$
\begin{equation}\label{Povzner C check}
\check{C}^{b_{ij}}_n = 2\frac{{\overline{r}}^2}{ {\underline{r}}^2} \int_{0}^{2\pi} \int_{-1}^1 \left(1+ {\underline{r}}(|\mu|-1)\right)^{n} b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi.
\end{equation}
In order to study its properties, we first note that $1+{\underline{r}}(|\mu|-1) \leq 1$, since $|\mu|\leq 1$, and the equality holds only when $|\mu|=1$ (or $\sigma = \left\{ \pm \hat{V} \right\}$). Therefore, the sequence of functions
\begin{equation*}
A_n(x):= \left(1+ {\underline{r}}(x-1)\right)^{n}
\end{equation*}
decreases monotonically in $n$ and tends to 0 as $n\rightarrow \infty$ for every $x\in (0,1)$ up to a set of measure zero. Finally, we conclude by monotone convergence Theorem that ${\color{black} \check{C}^{b_{ij}}_{\frac k2} }\searrow 0$ as $k\rightarrow \infty$.
When $ {\color{black}b_{ij}} \in \Linf$, we can obtain the explicit decay rate of the constant ${\color{black} \check{C}^{b_{ij}}_{\frac k2} }$, since in this case the integral \eqref{Povzner C check} significantly simplifies. The rate will be calculated in the Remark \ref{Remark Povzner constant Linf} below.
\subsubsection*{Term $P_{3_2}$} For the term $P_{3_2}$ we immediately obtain
\begin{equation*}
P_{3_2} = \bijnjed \bar{C}_{\frac{k}{2}} E^{\frac{k}{2}},
\end{equation*}
with the constant
\begin{equation*}
\bar{C}_{n} = - 2 \frac{{\overline{r}}^2}{ {\underline{r}}^2} (1-{\underline{r}})^n.
\end{equation*}
\subsubsection*{Term $P_{3_3}$} We first estimate the term $P_{3_3}$ using $|\mu|\leq 1$,
\begin{equation*}
P_{3_3} \leq \frac{{\overline{r}}^2}{ {\underline{r}}} \, k\, (1-{\underline{r}})^{\frac{k}{2}-1} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi.
\\ \leq \bijnjed \hat{C}_{\frac{k}{2}} E^{\frac{k}{2}},
\end{equation*}
and the constant is defined with
\begin{equation}\label{Povzner C hat}
\hat{C}_n = 2 \frac{{\overline{r}}^2}{ {\underline{r}}} \, n\, (1-{\underline{r}})^{n-1}.
\end{equation}
Gathering estimates for $P_1$, $P_2$ and $P_3$ completes the proof of \eqref{Povzner estimate gain} with
{\color{black}
$$\boldsymbol{\mathcal{C}}^{ij}_{n} = \bijnjed \left( (2n+2) \tilde{C}_n + \bar{C}_n + \hat{C}_n \right) + \check{C}^{b_{ij}}_n, n>2,$$
and $\boldsymbol{\mathcal{C}}^{ij}_{n} = \bijnjed 2 \tilde{C}_n$, if $1<n\leq2$. Thus, the constant $\boldsymbol{\mathcal{C}}^{ij}_{\frac k2} $ issuing from Povzner lemma satisfies {\color{black} $\boldsymbol{\mathcal{C}}^{ij}_{\frac k2} \rightarrow 0$, as $k\rightarrow \infty$, and so there exists $k^{ij}_*=k^{ij}_*(r_{ij}, b_{ij})$ for which $\boldsymbol{\mathcal{C}}^{ij}_{\frac k2} <\bijnjed$, for $k>k^{ij}_*$}.
\end{proof}
{\color{black}
\begin{remark}[The case $b_{ij} \in \Linf$]\label{Remark Povzner constant Linf} When the angular kernel is assumed bounded, some calculations are simpler. Pulling out the $L^\infty$ norm of $b_{ij}$, we have
\begin{equation*}
\int_{0}^{2\pi} \int_{-1}^{1} b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi \leq 4\pi {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } ,
\end{equation*}
and so terms $P_1$ and $P_{3_2}$ become
\begin{equation*}
P_1 \leq 8\pi {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \tilde{C}_\frac{k}{2} E^\frac{k}{2}, \qquad P_{3_2} = 4\pi {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \bar{C}_{\frac{k}{2}} E^{\frac{k}{2}}.
\end{equation*}
Moreover, when $b_{ij}$ is assumed bounded, the starting integral \eqref{pomocna 2} do not depend $\sigma \cdot \hat{u}$ anymore, so we may instead of $\hat{u}$ take $\hat{V}$ as a zenith of polar coordinates in \eqref{sigma decomposed}, which amounts to take $\alpha=0$ in \eqref{define_mu} that implies $\mu=\tau$. In this case, thanks to the parity arguments, term $P_2$ vanishes, and term $P_{3_3}$ can be explicitly calculated, without using any estimate,
\begin{equation*}
P_{3_3} = - \frac{{\overline{r}}^2}{ {\underline{r}}} \, k\, (1-{\underline{r}})^{\frac{k}{2}-1} E^{\frac{k}{2}} \int_{0}^{2\pi} \int_{-1}^1 |\tau| \, b_{ij}(\tau) \, \mathrm{d}\tau \, \mathrm{d}\varphi\\
= - 2 \pi\, {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \hat{C}_\frac{k}{2} E^{\frac{k}{2}},
\end{equation*}
with the constant $ \hat{C}_\frac{k}{2}$ from \eqref{Povzner C hat}.
Finally, let us compute explicitly the constant $ \check{C}^{b_{ij}}_{n}$ from \eqref{Povzner C check} when $b_{ij}(\sigma \cdot \hat{u}) \in \Linf$. Namely, pulling out the $L^\infty$ norm of $b_{ij}$ from the integral and using $\mu=\tau$, we get
\begin{multline*}
\check{C}^{b_{ij}}_n = 2\frac{{\overline{r}}^2}{ {\underline{r}}^2} {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \int_{0}^{2\pi} \int_{-1}^1 \left(1+ {\underline{r}}(|\tau|-1)\right)^{n} \mathrm{d}\tau \, \mathrm{d}\varphi
\\
=8 \pi \, \frac{{\overline{r}}^2}{ {\underline{r}}^3} {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \left( \frac{1}{n+1} - \frac{(1-{\underline{r}})^{n+1}}{n+1} \right),
\end{multline*}
that shows its decay rate.
To summarize, the constant from the Povzner lemma in the case of bounded angular part reads
\begin{equation}\label{c povzner}
\boldsymbol{\mathcal{C}}^{ij}_n = 4 \pi {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } \boldsymbol{\mathcal{C}}^{\infty}_n(r),
\end{equation}
where we have denoted
\begin{equation}\label{c povnzer renorm}
\boldsymbol{\mathcal{C}}^{\infty}_n(r) = 2 \tilde{C}_n + \bar{C}_n - \frac{1}{2} \hat{C}_n + 2 \frac{{\overline{r}}^2}{ {\underline{r}}^3} \left( \frac{1}{n+1} - \frac{(1-{\underline{r}})^{n+1}}{n+1} \right), \quad n>2,
\end{equation}
and $\boldsymbol{\mathcal{C}}^{\infty}_n(r) = 2 \tilde{C}_n $ if $1< n \leq 2$,
recalling \eqref{rmin} and \eqref{rmax}. Moreover, it satisfies {\color{black} $\boldsymbol{\mathcal{C}}^{ij}_{\frac k2} < 4 \pi {\color{black} \left\| b_{ij} \right\|_{L^\infty(\mathrm{d}\sigma)} } $, or equivalently $\boldsymbol{\mathcal{C}}^{\infty}_n(r) <1$, for sufficiently large $k^{ij}_*$ depending on $r_{ij}$ and $b_{ij}$.}
\end{remark}
{\color{black}
\subsection{Study of the Povzner constant for $b_{ij}(\sigma \cdot \hat{u}) \in \Linf$} In this paragraph we study in detail the constant \eqref{c povzner} from the Povzner lemma \ref{Povzner intro} in the case of bounded angular part. More precisely, we study its normalized part \eqref{c povnzer renorm}
\begin{equation}\label{c povnzer renorm 2}
\boldsymbol{\mathcal{C}}^{\infty}_n(r) = 2 {\overline{r}}^n - 2 \frac{{\overline{r}}^2}{ {\underline{r}}^2} (1-{\underline{r}})^n - \frac{{\overline{r}}^2}{ {\underline{r}}} \, n\, (1-{\underline{r}})^{n-1} + 2 \frac{{\overline{r}}^2}{ {\underline{r}}^3} \left( \frac{1}{n+1} - \frac{(1-{\underline{r}})^{n+1}}{n+1} \right),
\end{equation}
for $n>2$ and $\boldsymbol{\mathcal{C}}^{\infty}_n(r) = 2 {\overline{r}}^n$ if $1< n \leq2$, with ${\overline{r}} = \max\left\{r, 1-r \right\}$ and ${\underline{r}} = \min\left\{r,1-r\right\}$,
and elaborate more on its decay rate in $n$ depending on $r$.
First, taking $r=\frac{1}{2}$ we expect to recover the same properties as for the single gas when decay rate of the Povzner constant \cite{GambaAlonso18} was $\frac{2}{n+1}$, that monotonically decreases and tends to zero in $n>1$. In our case,
\begin{equation*}
\boldsymbol{\mathcal{C}}^{\infty}_n\left(\frac{1}{2}\right) =
\begin{cases}
\frac{4}{n+1} - \left(\frac{1}{2}\right)^n \left( n+ \frac{2}{n+1} \right), \text{if} \ n>2,\\
2 \left(\frac{1}{2}\right)^n, \text{if} \ 1<n\leq2.
\end{cases}
\end{equation*}
that keeps the same properties as for the single gas, which can be illustrated as in Figure \ref{figure-povzner-single}.
For general $r\in(0,1)$ decay properties of the constant issuing from the Povzner lemma \eqref{c povnzer renorm} strongly depend on $r$ or on the fact how much species masses $m_i$, $i=1,\dots,I$ are disparate. It is clear that, since $0<r<1$, the constant $\boldsymbol{\mathcal{C}}^{\infty}_n(r)$ will tend to zero as $n$ goes to infinity. Here we are interested in a more subtle question: determine $n_*$ such that it holds $\boldsymbol{\mathcal{C}}^{\infty}_n(r)<1$ for $n\geq n_*$ and any fixed $0<r<1$. Converge of $\boldsymbol{\mathcal{C}}^{\infty}_n(r)$ in $n$ towards zero for any $0<r<1$ ensures existence of such $n_*$. It can be observed that $n_*$ grows as much as $r$ is deviated from $\frac{1}{2}$, since the constants in $\boldsymbol{\mathcal{C}}^{\infty}_n(r)$ with power decay rate will decay more slowly as $r$ deviates from $\frac{1}{2}$. This behavior is illustrated in Figure \ref{figure-povzner-mixture}. We can reformulate the question: for some fixed value of $n$ determine the interval of $r$ for which it holds $\boldsymbol{\mathcal{C}}^{\infty}_n(r)<1$, that is illustrated in Figure \ref{figure-povzner-mixture-2}.
}
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\caption{Comparison of the Povzner constant for $r=\frac{1}{2}$ in the mixture setting and the single component gas for $n>1$.}
\label{figure-povzner-single}
\end{figure}
\begin{figure}
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\caption{Constant $\boldsymbol{\mathcal{C}}^{\infty}_n(r)$ from Povzner lemma \ref{Povzner intro} for some fixed value of $r=:r_*$. This figure illustrates the non-monotonic behavior in $n$ variable, and the growth of $n$ needed to ensure that $\boldsymbol{\mathcal{C}}^{\infty}_{n}(r_*)<1$ caused by a deviation of $r$ with respect to $\frac{1}{2}$.}
\label{figure-povzner-mixture}
\end{figure}
\begin{figure}
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\end{center}
\caption{Constant $\boldsymbol{\mathcal{C}}^{\infty}_n(r)$ from Povzner lemma \ref{Povzner intro} for some fixed value of $n=:n_*$. This figure illustrates the interval of $r$ for which it holds $\boldsymbol{\mathcal{C}}^{\infty}_{n_*}(r)<1$.}
\label{figure-povzner-mixture-2}
\end{figure}
\section{Proof of Existence and Uniqueness Theorem \ref{theorem existence uniqueness}}\label{Section Ex Uni proof}
{\color{black}
Before proving Theorem \ref{theorem existence uniqueness}, we first study a property of the collision operator that is a consequence of the Povzner lemma \ref{povzner constant prop} and lemma \ref{lemma lower bound}.
}
\begin{lemma}\label{lemma Q za tangent cond}
Let $\mathbb{F}=\left[f_i\right]_{i=1,\dots,I} \in \Omega$ {\color{black} and $k_*$ as defined in \eqref{kstar}. } Then, the following estimate holds
\begin{equation}\label{Q za tangent cond}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v
\leq - A_{k_*} \, \mathfrak{m}_{\ks}[\F] (t)^{1+ \frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}}+ B_{k_*} \, \mathfrak{m}_{\ks}[\F] (t),
\end{equation}
with positive constants
\begin{equation}\label{const Ak Bk}
\begin{split}
A_{k_*} &= \min_{1\leq i, j \leq I} \left(\bijnjed - \cpov_{\frac{\ks}{2}} \right) \frac{c_{lb}}{\max_{1\leq i \leq I} m_i} \left(I C_0 \right)^{-\frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}}, \\
B_{k_*}&= 2 \, C_{2} \max_{1\leq i, j \leq I} \left( \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^ { \gamma_{ij}} \cpov_{\frac{\ks}{2}} \right) {\color{black} \sum_{\ell=1}^{\lfloor\frac{{k_*} + 1}{2}\rfloor} \left( \begin{matrix}
{k_*} \\ \ell
\end{matrix} \right)},
\end{split}
\end{equation}
{\color{black}
where $C_0$ and $C_{2}$ are from the characterization of the set $\Omega$, $c_{lb}$ is from the lower bound \eqref{lower bound}, and $\cpov_{\frac{\ks}{2}}$ is a constant from the Povzner lemma \ref{Povzner intro} with {\color{black} ${k_*} > {\overline{k}}$, as defined in \eqref{kstar}, ensuring the property \eqref{povzner constant prop} for any pair $(i,j)$ that yields positivity of the constant $A_{k_*}$}.
}
\end{lemma}
\begin{remark}\label{coercive_constant}
It is important to notice that the strict positivity of the constant $A_{k_*}$ can be view as a {\bf coercive condition} that secures global in time solutions, without the need to require boundedness of entropy.
\end{remark}
\begin{proof}
We start with the weak form \eqref{weak form any test}. Taking test function $\psi_i(x)=\left\langle v \right\rangle_i^{{k_*}}$, and cross section \eqref{cross section}, we have
\begin{multline} \label{pomocna 28}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v =\sum_{i=1}^I \sum_{{j=1 }}^I \int_{\mathbb{R}^3} \left\langle v \right\rangle_i^{{k_*}} Q_{ij}(f_i,f_j) \, \mathrm{d}v\\ = \frac{1}{2}\sum_{i=1}^I \sum_{{j=1}}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \left|v-v_*\right|^ { \gamma_{ij}} f_i(v) f_j(v_*) \\ \times \left( \left\langle v' \right\rangle_i^{{k_*}} + \left\langle v'_* \right\rangle_j^{{k_*}} - \left\langle v \right\rangle_i^{{k_*}} - \left\langle v_* \right\rangle_j^{{k_*}} \right) {\color{black}b_{ij}} (\sigma \cdot \hat{u}) \,\mathrm{d}\sigma \mathrm{d}v_* \mathrm{d}v,
\end{multline}
where collisional rules are \eqref{collisional rules bi}. The primed quantities integrated over sphere $S^2$ are estimated via Povzner lemma. Indeed, by Lemma \ref{Povzner intro} , \eqref{pomocna 28} becomes
\begin{multline}\label{poly moment start}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v \leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \iint_{\mathbb{R}^3 \times \mathbb{R}^3} f_i (v) f_j(v_*) \left|v-v_*\right|^ { \gamma_{ij}} \\ \times \left( \cpov_{\frac{\ks}{2}} \left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 \right)^{\frac{{k_*}}{2}} - \bijnjed \left( \left\langle v \right\rangle_i^{{k_*}} + \left\langle v_* \right\rangle_j^{{k_*}} \right) \right) \mathrm{d} v_* \mathrm{d}v,
\end{multline}
where $\cpov_{\frac{\ks}{2}}$ is a constant from Povzner lemma \ref{Povzner intro} {\color{black} with $k_* \geq {\overline{k}}=\max_{1\le i,j\le I}\{ k^{ij}_*\} $ chosen large enough to ensure \eqref{povzner constant prop}} uniformly in $i,j$-pairs. On one hand,
we use polynomial inequalities from Lemmas \ref{binomial} and \ref{moment products}
\begin{equation*}
\begin{split}
\left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 \right)^{\frac{{k_*}}{2}} & \leq \left(\left\langle v \right\rangle_i + \left\langle v_* \right\rangle_j\right)^{{k_*}} \\& \leq \left\langle v \right\rangle_i^{k_*} + \left\langle v_* \right\rangle_j^{k_*} + \sum_{\ell=1}^{\ell_{{k_*}}} \left( \begin{matrix}
{k_*} \\ \ell
\end{matrix} \right) \left( \left\langle v \right\rangle_i^{\ell} \left\langle v_* \right\rangle_j^{{k_*}- \ell}+ \left\langle v \right\rangle_i^{{k_*}- \ell} \left\langle v_* \right\rangle_j^{\ell} \right),\\
& \leq \left\langle v \right\rangle_i^{k_*} + \left\langle v_* \right\rangle_j^{k_*} + \left( \left\langle v \right\rangle_i \left\langle v_* \right\rangle_j^{{k_*}- 1}+ \left\langle v \right\rangle_i^{{k_*}- 1} \left\langle v_* \right\rangle_j \right) \left( \sum_{\ell=1}^{\ell_{{k_*}}} \left( \begin{matrix}
{k_*} \\ \ell
\end{matrix} \right) \right)
\end{split}
\end{equation*}
with $\ell_{k_*} = \lfloor\frac{{k_*} + 1}{2}\rfloor$,
and therefore
\begin{multline}\label{pomocna 30}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v
\leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \iint_{\mathbb{R}^3 \times \mathbb{R}^3} f_i (v) f_j(v_*) \left|v-v_*\right|^ { \gamma_{ij}} \\ \times \left\{ - \left(\bijnjed - \cpov_{\frac{\ks}{2}} \right) \left(\left\langle v \right\rangle_i^{{k_*}} + \left\langle v_* \right\rangle_j^{{k_*}}\right) \right. \\ \left. + \cpov_{\frac{\ks}{2}} \left( {\color{black} \sum_{\ell=1}^{\ell_{{k_*}}} \left( \begin{matrix}
{k_*} \\ \ell
\end{matrix} \right) } \right) \left( {\color{black} \left\langle v \right\rangle_i \left\langle v_* \right\rangle_j^{{k_*}- 1}+ \left\langle v \right\rangle_i^{{k_*}- 1} \left\langle v_* \right\rangle_j }\right) \right\} \mathrm{d}v_* \mathrm{d}v.
\end{multline}
On the other hand we use upper and lower bound of the non-angular cross section $\left|v-v_*\right|^ { \gamma_{ij}} $. For the upper bound, from \eqref{estimate on u^g} it follows
\begin{equation*}
\left|v-v_*\right|^ { \gamma_{ij}} \leq \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^ { \gamma_{ij}} \left(\left\langle v \right\rangle_i^ { \gamma_{ij}} + \left\langle v_* \right\rangle_j^ { \gamma_{ij}} \right) \leq \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^ { \gamma_{ij}} \left(\left\langle v \right\rangle_i^ { {\color{black}\overline{\gamma}}} + \left\langle v_* \right\rangle_j^ { {\color{black}\overline{\gamma}}} \right),
\end{equation*}
for $ { {\color{black}\overline{\gamma}}} = { {\color{black}\max_{1\leq i, j \leq I}\gamma_{ij}}} \in(0,1]$. For the lower bound, we use Lemma \ref{lemma lower bound}, but
we first check that all assumptions are satisfied from the fact that $\mathbb{F} \in \Omega$. Indeed, bounds on $\mathfrak{m}_{0}$ implies
\begin{equation*}
c_{0} \min_{1\leq i \leq I} m_i \leq \sum_{i=1}^I \int_{ \mathbb{R}^3} m_i \, f_i \, \mathrm{d}v \leq C_0 \max_{1\leq i \leq I} m_i.
\end{equation*}
From the other side, bounds on $\mathfrak{m}_2$ yield
\begin{equation*}
\left(c_{2} - C_0 \right) \sum_{j=1}^I m_j \leq \sum_{i=1}^I \int_{ \mathbb{R}^3} m_i \left|v\right|^2 f_i \, \mathrm{d}v \leq \left(C_{2} - c_{0} \right) \sum_{j=1}^I m_j.
\end{equation*}
Therefore, for constants $c$ and $C$ from assumptions of Lemma \ref{lemma lower bound} we can choose
\begin{equation*}
\begin{split}
c&:=\min\left\{ c_{0} \min_{1\leq i \leq I} m_i, \left(c_{2} - C_0 \right) \sum_{j=1}^I m_j \right\}, \\
C&:= \max\left\{ C_0 \max_{1\leq i \leq I} m_i, \left(C_{2} - c_{0} \right) \sum_{j=1}^I m_j \right\}.
\end{split}
\end{equation*}
Note that positivity of $c$ is guaranteed by the definition of the set $\Omega$. Finally, since it can be estimated
\begin{equation*}
\sum_{i=1}^I \int_{ \mathbb{R}^3} m_i \left|v\right|^{2+\epsilon} f_i \, \mathrm{d}v \leq \mathfrak{m}_{2+\varepsilon} \left( \sum_{j=1}^I m_j \right)^{1+\frac{\varepsilon}{2}} \max_{1\leq i \leq I} m_i^{-\frac{\varepsilon}{2}},
\end{equation*}
we can choose
\begin{equation*}
B:= C_{2+\varepsilon} \left( \sum_{j=1}^I m_j \right)^{1+\frac{\varepsilon}{2}} \max_{1\leq i \leq I} m_i^{-\frac{\varepsilon}{2}}.
\end{equation*}
Then \eqref{lower bound} implies
\begin{equation*}
\sum_{i=1}^I \int_{ \mathbb{R}^3} f_i(v) \left|v-v_*\right|^ { \gamma_{ij}} \mathrm{d}v \geq \frac{1}{\max_{1\leq i \leq I} m_i} c_{lb} \left\langle v_* \right\rangle_j^ { {\color{black}\overline{\gamma}}} ,
\end{equation*}
and
\begin{equation*}
\sum_{j=1}^I \int_{ \mathbb{R}^3} f_j(v_*) \left|v-v_*\right|^ { \gamma_{ij}} \mathrm{d}v \geq \frac{1}{\max_{1\leq j \leq I} m_j} c_{lb} \left\langle v \right\rangle_i^ { {\color{black}\overline{\gamma}}} .
\end{equation*}
With these estimates, \eqref{pomocna 30} becomes
\begin{equation*}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v \leq - D_{k_*} \mathfrak{m}_{\ks+\gmax} + E_{k_*} \left( \mathfrak{m}_{1+\gmax} \, \mathfrak{m}_{\ks-1} + \mathfrak{m}_{\ks-1+\gmax} \, \mathfrak{m}_{1} \right),
\end{equation*}
where $D_{k_*}$ and $E_{k_*}$ are positive constants
\begin{equation*}
\begin{split}
D_{k_*}&= \min_{1\leq i, j \leq I} \left( \bijnjed - \cpov_{\frac{\ks}{2}} \right) \frac{c_{lb}}{\max_{1\leq i \leq I} m_i} , \\
E_{k_*}&= \max_{1\leq i, j \leq I} \left( \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^ { \gamma_{ij}} \cpov_{\frac{\ks}{2}} \right) {\color{black} \sum_{\ell=1}^{\ell_{{k_*}}} \left( \begin{matrix}
{k_*} \\ \ell
\end{matrix} \right) }.
\end{split}
\end{equation*}
{\color{black}
In particular, $D_{k_*}$ is positive since, by assumption, {\color{black} ${k_*}\ge {\overline{k}}$ defined in \eqref{kstar} large enough ensuring \eqref{povzner constant prop} for the constant $\cpov_{\frac{\ks}{2}}$ from Povzner lemma \eqref{Povzner estimate gain}.}
}
Arriving in moment notation, we can use monotonicity of moments \eqref{monotonicity of norm}, together with an estimate on $\mathfrak{m}_2$ from characterization of set $\Omega$, to get the following estimate
\begin{equation*}
\sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v \leq - D_{k_*} \, \mathfrak{m}_{\ks+\gmax} + 2 E_{k_*} C_{2} \, \mathfrak{m}_{\ks} .
\end{equation*}
It remains to use a control from below derived in \eqref{jensen} for the highest order moment $\mathfrak{m}_{\ks+\gmax}$, {\color{black} taking $k={k_*}$, $\lambda= { {\color{black}\overline{\gamma}}} $ and $C_{\mathfrak{m}_{0}}=C_0$ there},
$$
\mathfrak{m}_{\ks+\gmax} \geq \left(I C_0 \right)^{-\frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}} \mathfrak{m}_{\ks} ^{1+\frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}},
$$
which yields final estimate \eqref{Q za tangent cond}.
\end{proof}
{\color{black}
We turn to the proof of Existence and Uniqueness Theorem \ref{theorem existence uniqueness}.
}Our proof follows the one given in \cite{GambaAlonso18} for the single Boltzmann equation. In particular, our aim is to apply Theorem \ref{Theorem general} from a general ODE theory in Banach spaces. In order to do so, we first show that the collision operator is a mapping $\mathbb{Q}: \Omega \rightarrow L_2^1$. Indeed, take any $\mathbb{F} \in \Omega$. Then,
\begin{equation}\label{pomocna 32}
\left\| \mathbb{Q}(\F) \right\|_{L_2^1} = \sum_{i=1}^I \int_{ \mathbb{R}^3} \left| \left[\mathbb{Q}(\F) \right]_{i}(v) \right| \left\langle v \right\rangle_i^2 \mathrm{d}v \\ \leq \sum_{i=1}^I \sum_{j=1}^I \int_{ \mathbb{R}^3} \left| Q_{ij}(f_i,f_j)(v) \right| \left\langle v \right\rangle_i^2 \mathrm{d}v.
\end{equation}
The absolute value $\left| Q_{ij}(f_i,f_j)(v) \right|$ is written with the help of sign function and shorter notation
$$
\left| Q_{ij}(f_i,f_j)(v) \right| = Q_{ij}(f_i,f_j)(v) \, s_{ij}(v), \quad s_{ij}(v):= \text{sign}\left( Q_{ij}(f_i,f_j)(v) \right).
$$
Then $s_{ij}(v) \left\langle v \right\rangle_i^{\color{black} 2}$ in \eqref{pomocna 32} are viewed as test functions, so the weak form \eqref{weak form any test} implies
\begin{multline*}
\left\| \mathbb{Q}(\F) \right\|_{L_2^1} \leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} f_i(v) f_j(v_*) \, \mathcal{B}_{ij}(v,v_*,\sigma) \\ \times \left( s_{ij}(v') \left\langle v' \right\rangle_i^{\color{black} 2} + s_{ji}(v'_*) \left\langle v'_* \right\rangle_j^{\color{black} 2} - s_{ij}(v) \left\langle v \right\rangle_i^{\color{black} 2} - s_{ji}(v_*) \left\langle v_* \right\rangle_j^{\color{black} 2} \right) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v.
\end{multline*}
{\color{black} Since the sign function is upper bounded by 1, we obtain}
\begin{multline*}
\left\| \mathbb{Q}(\F) \right\|_{L_2^1} \leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} f_i(v) f_j(v_*) \, \mathcal{B}_{ij}(v,v_*,\sigma) \\ \times \left( \left\langle v' \right\rangle_i^{\color{black} 2} + \left\langle v'_* \right\rangle_j^{\color{black} 2} + \left\langle v \right\rangle_i^{\color{black} 2} + \left\langle v_* \right\rangle_j^{\color{black} 2} \right) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v.
\end{multline*}
Using conservation of energy \eqref{CL micro energy}, together with the form of cross section \eqref{cross section}, implies
\begin{multline*}
\left\| \mathbb{Q}(\F) \right\|_{L_2^1} \leq \sum_{i=1}^I \sum_{j=1}^I \bijnjed \iint_{\mathbb{R}^3 \times \mathbb{R}^3 } f_i(v) f_j(v_*) \, \left|v-v_*\right|^ { \gamma_{ij}} \\ \times \left( \left\langle v \right\rangle_i^{\color{black} 2} + \left\langle v_* \right\rangle_j^{\color{black} 2} \right) \mathrm{d}v_* \mathrm{d} v.
\end{multline*}
Finally, using upper bound \eqref{estimate on u^g product}, we obtain the estimate in terms of norms,
\begin{multline*}
\left\| \mathbb{Q}(\F) \right\|_{L_2^1} \leq \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \\ \times \sum_{i=1}^I \sum_{j=1}^I \iint_{\mathbb{R}^3 \times \mathbb{R}^3 } f_i(v) f_j(v_*) \, \left\langle v \right\rangle_i^ { {\color{black}\overline{\gamma}}} \left\langle v_* \right\rangle_j^ { {\color{black}\overline{\gamma}}} \left( \left\langle v \right\rangle_i^{\color{black} 2} + \left\langle v_* \right\rangle_j^{\color{black} 2} \right) \mathrm{d}v_* \mathrm{d} v\\
= 2 \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \left( \left\| \mathbb{F} \right\|_{L_{2+\gmax}^1} \left\| \mathbb{F} \right\|_{L_{\gmax}^1} \right).
\end{multline*}
Since $\mathbb{F} \in \Omega$ the right hand side is bounded, and therefore $\mathbb{Q}(\F) \in L_2^1$. \\
The next task is to show that the mapping $\mathbb{F} \mapsto \mathbb{Q}(\F)$, when restricted to $\Omega$ satisfies
(i) H\"{o}lder continuity, (ii) sub-tangent and (iii) one-sided Lipschitz conditions. Indeed, the proof is divided into proofs of these three properties.
Assume that $\mathbb{F},\mathbb{G} \in \Omega$ and cross section $\mathcal{B}_{ij}$ is given in \eqref{cross section}. Then, the following three properties hold
\begin{itemize}
\item[(i)] H\"{o}lder continuity condition
\begin{equation}\label{holder estimate}
\left\| \mathbb{Q}(\F) - \mathbb{Q}(\G) \right\|_{L_2^1} \leq C_H \left\| \mathbb{F}-\mathbb{G} \right\|_{L_2^1}^{\frac{1}{2}},
\end{equation}
\item[(ii)] Sub-tangent condition:
\begin{equation*}
\lim\limits_{h\rightarrow 0+} \frac{\text{dist}\left(\mathbb{F} + h \mathbb{Q}(\F), \Omega \right)}{h} =0,
\end{equation*}
where
\begin{equation*}
\text{dist}\left(\mathbb{H},\Omega\right)=\inf_{\omega \in \Omega} \left\| \mathbb{H}-\omega \right\|_{L_2^1}.
\end{equation*}
\item[(iii)] One-sided Lipschitz condition:
\begin{equation*}
\left[\mathbb{Q}(\F)-\mathbb{Q}(\G), \mathbb{F}-\mathbb{G}\right] \leq C_L \left\| \mathbb{F}-\mathbb{G} \right\|_{L_2^1},
\end{equation*}
where, by Remark \ref{Lip remark},
\begin{multline*}
\left[\mathbb{Q}(\F)-\mathbb{Q}(\G), \mathbb{F}-\mathbb{G}\right] =\lim_{h\rightarrow 0^-}\frac{\left(\left\| \left(\mathbb{F}-\mathbb{G}\right) + h \left(\mathbb{Q}(\F)-\mathbb{Q}(\G)\right) \right\|_{L_2^1} - \left\| \left(\mathbb{F}-\mathbb{G}\right) \right\|_{L_2^1} \right)}{h} \\\leq \sum_{i=1}^I \int_{ \mathbb{R}^3} \left(\left[\mathbb{Q}(\F)\right]_i(v) - \left[\mathbb{Q}(\G)\right]_i(v) \right) \, \text{sign}\left(f_i(v)-g_i(v)\right) \left\langle v \right\rangle_i^2 \mathrm{d}v.
\end{multline*}
\end{itemize}
Constants $C_H$ and $C_L$ depend on $\bijnjed$, number of species $I$ and their masses $m_{i}$, $i=1,\dots,I$, and constants from characterization of the set $\Omega$.
\begin{proof}[Proof of (i) H\"{o}lder continuity condition]
Let $\mathbb{F}=\left[f_i\right]_{1\leq i \leq I}$ and $\mathbb{G}=\left[g_i\right]_{1\leq i \leq I}$ belong to $\Omega$. We need to estimate the following expression
\begin{equation}\label{pomocna 4}
I_{H}:=\left\| \mathbb{Q}(\F) - \mathbb{Q}(\G) \right\|_{L_2^1}
= \sum_{i=1}^I \int_{\mathbb{R}^3}\left| \sum_{j=1}^I \left( Q_{ij}(f_i,f_j) - Q_{ij}(g_i,g_j) \right) \right| \left\langle v \right\rangle_i^2 \mathrm{d} v.
\end{equation}
Using the binary structure of collision operator \eqref{boltzmann i}, it follows
\begin{equation}\label{collision operator diff and sum}
Q_{ij}(f_i,f_j) - Q_{ij}(g_i,g_j) = \frac{1}{2} \left(Q_{ij}(f_i-g_i, f_j + g_j) + Q_{ij}(f_i+g_i, f_j-g_j) \right).
\end{equation}
Therefore, using properties of absolute value, \eqref{pomocna 4} becomes
\begin{equation}\label{pomocna 7}
I_{H} \leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} \left( \left| Q_{ij}(f_i-g_i, f_j + g_j) \right| + \left| Q_{ij}(f_i+g_i, f_j-g_j) \right| \right) \left\langle v \right\rangle_i^2 \mathrm{d} v.
\end{equation}
The absolute value of collision operator will be written with the help of sign function, using $\left|\cdot\right|= \cdot \,\text{sign}(\cdot)$. Since, at the end, all sign functions will be bounded by 1, we will not go deeply into details of its structure. So, let us for the moment denote
\begin{equation*}
\text{sign}(Q_{ij}(f_i-g_i, f_j + g_j)) = s_{ij}^{-+}, \qquad \text{sign}(Q_{ij}(f_i+g_i, f_j - g_j)) = s_{ij}^{+-}.
\end{equation*}
Then, \eqref{pomocna 7} becomes
\begin{multline}\label{pomocna 8}
I_{H} \leq \frac{1}{2} \sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} \left( Q_{ij}(f_i-g_i, f_j + g_j) s_{ij}^{-+}\left\langle v \right\rangle_i^2 \right. \\ \left. + Q_{ij}(f_i+g_i, f_j-g_j) s_{ij}^{+-}\left\langle v \right\rangle_i^2 \right) \mathrm{d} v.
\end{multline}
Now we use the weak form \eqref{weak form ij+ji}, and in order to do so, we have to match pairs. Indeed, we notice that the pair for $ij$-th element of the first sum is the $ji$-th element of the second sum. That is, \eqref{weak form ij+ji} implies, after dropping the sign function,
\begin{multline*}
\int_{ v \in \mathbb{R}^3} \left( Q_{ij}(f_i-g_i, f_j + g_j) s_{ij}^{-+}\left\langle v \right\rangle_i^2 + Q_{ji}(f_j+g_j, f_i-g_i) s_{ji}^{+-}\left\langle v \right\rangle_j^2 \right) \mathrm{d}v
\\
\leq \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \left|f_i(v)-g_i(v)\right| (f_j(v_*) + g_j(v_*)) \\ \times \left( \left\langle v' \right\rangle_i^2 + \left\langle v'_* \right\rangle_j^2 + \left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2\right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v
\\
= 2 \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \left|f_i(v)-g_i(v)\right| (f_j(v_*) + g_j(v_*)) \\ \times \left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2\right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v,
\end{multline*}
the last equality is due to the conservation law at the microscopic level \eqref{CL micro energy}. Therefore, \eqref{pomocna 8} becomes
\begin{multline*}
I_{H} \leq \sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \left|f_i(v)-g_i(v)\right| (f_j(v_*) + g_j(v_*)) \\ \times \left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2\right) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d} v.
\end{multline*}
Now we use the form of cross section \eqref{cross section}.
Inequality \eqref{estimate on u^g} yields the following upper bound of the previous expression
\begin{multline*}
I_{H} \leq \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \\ \sum_{i=1}^I \sum_{j=1}^I \iint_{\mathbb{R}^3 \times \mathbb{R}^3 } \left|f_i(v)-g_i(v)\right| (f_j(v_*) + g_j(v_*)) \\ \times \left(\left\langle v \right\rangle_i^{2+ { {\color{black}\overline{\gamma}}} } + \left\langle v \right\rangle_i^{2} \left\langle v_* \right\rangle_j^ { {\color{black}\overline{\gamma}}} + \left\langle v_* \right\rangle_j^2 \left\langle v \right\rangle_i^ { {\color{black}\overline{\gamma}}} + \left\langle v_* \right\rangle_j^{2+ { {\color{black}\overline{\gamma}}} }\right) \mathrm{d}v_* \mathrm{d} v
\\ \leq \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \left( \left\| \mathbb{F} - \mathbb{G} \right\|_{L_{2+\gmax}^1} \left\| \mathbb{F}+\mathbb{G} \right\|_{L_0^1} \right. \\ \left. + \left\| \mathbb{F} - \mathbb{G} \right\|_{L_2^1} \left\| \mathbb{F}+\mathbb{G} \right\|_{L_{\gmax}^1} + \left\| \mathbb{F}- \mathbb{G} \right\|_{L_{\gmax}^1}\left\| \mathbb{F}+\mathbb{G} \right\|_{L_2^1}+ \left\|\mathbb{F}-\mathbb{G} \right\|_{L_0^1}\left\|\mathbb{F}+\mathbb{G} \right\|_{L_{2+\gmax}^1} \right).
\end{multline*}
Monotonicity of the norm \eqref{monotonicity of norm} yields
\begin{multline*}
I_{H} \leq 2 \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \\ \times \left\| \mathbb{F} - \mathbb{G} \right\|_{L_{2+\gmax}^1} \left( \left\| \mathbb{F}+\mathbb{G} \right\|_{L_2^1}+\left\|\mathbb{F}+\mathbb{G} \right\|_{L_{2+\gmax}^1} \right).
\end{multline*}
By the interpolation inequality \eqref{interpolation inequality vector}, it follows
\begin{multline}\label{pomocna 15}
I_{H} \leq 2 I \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \\ \times \left\| \mathbb{F} - \mathbb{G} \right\|_{L_2^1}^{1/2} \left\| \mathbb{F} - \mathbb{G} \right\|_{L_{2+2\gmax}^1}^{1/2} \left( \left\| \mathbb{F}+\mathbb{G} \right\|_{L_2^1}+\left\|\mathbb{F}+\mathbb{G} \right\|_{L_{2+\gmax}^1} \right).
\end{multline}
Then we can bound term by term:
\begin{equation*}
\left\| \mathbb{F} - \mathbb{G} \right\|_{L_{2+2\gmax}^1}^{1/2} \leq \left\| \mathbb{F} \right\|_{L_{2+2\gmax}^1}^{1/2} +\left\| \mathbb{G} \right\|_{L_{2+2\gmax}^1}^{1/2} \leq 2 \, C_{2+2\gmax}^{1/2},
\end{equation*}
and in the same fashion
\begin{equation*}
\left\| \mathbb{F}+\mathbb{G} \right\|_{L_2^1} \leq 2 C_{2}, \qquad \left\| \mathbb{F}+\mathbb{G} \right\|_{L_{2+\gmax}^1} \leq 2 C_{2+\gmax},
\end{equation*}
since both $\mathbb{F}$ and $\mathbb{G}$ belong to $\Omega$. Therefore, \eqref{pomocna 15} becomes
\begin{equation*}
I_{H} \leq 8 \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) C_{2+2\gmax}^{1/2} \left( C_{2} + C_{2+\gmax} \right) \left\| \mathbb{F} - \mathbb{G} \right\|_{L_2^1}^{1/2},
\end{equation*}
which concludes the proof of H\"{o}lder continuity.
\end{proof}
\begin{proof}[Proof of (ii) sub-tangent condition]
In order to prove sub-tangent condition, we first observe that, since we are in cut-off case, it is possible to split collision operator $\mathbb{Q}(\F)$ into gain and loss term. Namely,
\begin{equation*}
\left[ \mathbb{Q}(\F) \right]_i = \left[ \mathbb{Q}^+(\mathbb{F}) \right]_i - f_i(v) \left[ \nu(\mathbb{F}) \right]_i,
\end{equation*}
where $\mathbb{Q}^+$ is a positive operator, and collision frequency $\nu(\mathbb{F})$, for any component $1\leq i\leq I$ reads
\begin{equation*}
\left[ \nu(\mathbb{F}) \right]_i=\sum_{j=1}^{I} \iint_{\mathbb{R}^3 \times S^2 } f_j(v_*) \mathcal{B}_{ij}(v,v_*,\sigma) \mathrm{d}\sigma \mathrm{d}v_* \geq 0.
\end{equation*}
In our case, $\nu(\mathbb{F})$ is finite whenever $\mathbb{F} \in \Omega$. Indeed, for the cross section \eqref{cross section}-\eqref{gamma max}, and since {\color{black} $\left|v-v_*\right|^ { \gamma_{ij}} \leq \left|v-v_*\right|^ { {\color{black}\overline{\gamma}}} $, for $\left|v-v_*\right| \geq 1$ and }$\left|v-v_*\right|^ { {\color{black}\overline{\gamma}}} \leq \left|v\right|^ { {\color{black}\overline{\gamma}}} + \left|v_*\right|^ { {\color{black}\overline{\gamma}}} $,
\begin{multline*}
0 \leq \left[ \nu(\mathbb{F}) \right]_i(v) \leq {\color{black} \left( \max_{1\leq i, j \leq I} \left\| b_{ij} \right\|_{L^1(\mathrm{d}\sigma)} \right) } \sum_{j=1}^{I} \int_{\mathbb{R}^3 } f_j(v_*) \left|v-v_*\right|^ { \gamma_{ij}} \mathrm{d}v_*\\
{\color{black} \leq {\color{black} \left( \max_{1\leq i, j \leq I} \left\| b_{ij} \right\|_{L^1(\mathrm{d}\sigma)} \right) } \left( \sum_{j=1}^{I} \int_{\left|v-v_*\right|<1} f_j(v_*) \, \mathrm{d}v_* \right.}
\\ {\color{black} \left.
+ \sum_{j=1}^{I} \int_{\left|v-v_*\right|\geq 1} f_j(v_*) \left|v-v_*\right|^ { {\color{black}\overline{\gamma}}} \mathrm{d}v_* \right) }\\
\leq {\color{black} \left( \max_{1\leq i, j \leq I} \left\| b_{ij} \right\|_{L^1(\mathrm{d}\sigma)} \right) } \left( C_0 + \left|v\right|^ { {\color{black}\overline{\gamma}}} C_0 + \left( \frac{\sum_{i=1}^I m_i}{\min_{1\leq j \leq I} m_j} \right)^{ { {\color{black}\overline{\gamma}}} /2}\left\| \mathbb{F} \right\|_{L_{\gmax}^1} \right) \\
\leq K \left( 1+\left|v\right|^ { {\color{black}\overline{\gamma}}} \right),
\end{multline*}
where
\begin{equation}\label{constant c tangency}
K= {\color{black} \left( \max_{1\leq i, j \leq I} \left\| b_{ij} \right\|_{L^1(\mathrm{d}\sigma)} \right) } \left( 2 C_0 + \left( \frac{\sum_{i=1}^I m_i}{\min_{1\leq j \leq I} m_j} \right) C_{2} \right).
\end{equation}
\begin{proposition}\label{prop tangency}
Fix $\mathbb{F} \in \Omega$. Then, for any $\varepsilon>0$ there exists $h_1>0$, such that $B(\mathbb{F}+h\mathbb{Q}(\F), h \varepsilon) \cap \Omega \neq \emptyset$, for any $0<h<h_1$.
\end{proposition}
\begin{proof}
Set $\chi_R(v)$ the characteristic function of the ball of radius $R>0$ and introduce the truncated function $\mathbb{F}_R(t,v)=\chi_R(v)\mathbb{F}(t,v)$. Let
\begin{equation}\label{wR}
\mathbb{W}_R = \mathbb{F} + h \mathbb{Q}(\mathbb{F}_R).
\end{equation}
The idea of the proof is to find $R$ such that from on one hand $\mathbb{W}_R \in \Omega$, and on the another hand $\mathbb{W}_R \in B(\mathbb{F}+h\mathbb{Q}(\F), h \varepsilon)$, with $h$ explicitly calculated.
\subsubsection*{Step 1.} We first show that it is possible to find an $h_1$ such that $\mathbb{W}_R$ remains non-negative for as long $0<h<h_1$. Indeed, for any $\mathbb{F} \in \Omega$ its truncation $\mathbb{F}_R \in \Omega$ as well. Since $\mathbb{Q}^+$ is a positive operator, we have
\begin{equation*}
\left[\mathbb{W}_R\right]_i = f_i + h \left[\mathbb{Q}^+(\mathbb{F}_R)\right]_i - h \left[\mathbb{F}_R\right]_i \left[\nu(\mathbb{F}_R)\right]_i \geq f_i\left( 1 - h \, K\, \left( 1 + R^ { {\color{black}\overline{\gamma}}} \right) \right) \geq 0,
\end{equation*}
for any $0<h<\frac{1}{K(1+R^ { {\color{black}\overline{\gamma}}} )}$, and $1\leq i \leq I$,
with $K $ from \eqref{constant c tangency}.
\subsubsection*{Step 2.} Since $\mathbb{F}_R \in \Omega$, we use conservative properties of the collision operator detailed in \eqref{cons operator mass} and \eqref{cons operator energy}, to obtain
\begin{equation*}
\sum_{i=1}^I \int_{\mathbb{R}^3} \left[ \mathbb{Q}(\mathbb{F}_R) \right]_i\mathrm{d}v = 0, \qquad \sum_{i=1}^I \left[ \mathbb{Q}(\mathbb{F}_R) \right]_i \left\langle v \right\rangle_i^2 \mathrm{d} v=0.
\end{equation*}
From \eqref{wR}, we get
\begin{equation*}
\mathfrak{m}_{0}[\mathbb{W}_R] = \mathfrak{m}_{0}[\mathbb{F}], \qquad \mathfrak{m}_2[\mathbb{W}_R] = \mathfrak{m}_2[\F],
\end{equation*}
independently of $R$, which yields all needed lower and upper bounds on this quantities.
\subsubsection*{Step 3} Finally, we need to show that $L^1_{k_*}$ norm of $\mathbb{W}_R$ is bounded
Let the map $\mathcal{L}_{ \gmax, k_*}:[0,\infty) \rightarrow \mathbb{R}$, be defined with $\mathcal{L}_{ \gmax, k_*}(x)=- A_{k_*}x^{1+ \frac{ { {\color{black}\overline{\gamma}}} }{{k_*}}}+ B_{k_*} x$, where $ { {\color{black}\overline{\gamma}}} \in (0,1]$ and ${k_*}$ as defined in \eqref{kstar}
that yields positivity of constants $A_{k_*}$ and $B_{k_*}$. It has only one root, denoted with $x^*_{ \gmax, k_*}$, at which $\mathcal{L}_{ \gmax, k_*}$ changes from positive to negative. Thus, for any $x\geq0$, we may write
\begin{equation*}
\mathcal{L}_{ \gmax, k_*}(x) \leq \max_{0\leq x \leq x^*_{ \gmax, k_*}} \mathcal{L}_{ \gmax, k_*}(x) =: \mathcal{L}^*_{ \gmax, k_*}.
\end{equation*}
Now, Lemma \ref{lemma Q za tangent cond} implies
\begin{equation*}
\sum_{i=1}^I \int_{ \mathbb{R}^3}\left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v \leq \mathcal{L}_{{ \gmax, k_*}}\left( \mathfrak{m}_{\ks}[\F] \right) \leq \mathcal{L}^*_{ \gmax, k_*}.
\end{equation*}
Define
\begin{equation*}
\xi_{{ \gmax, k_*}} := x^*_{ \gmax, k_*} + \mathcal{L}^*_{ \gmax, k_*}.
\end{equation*}
For any $\mathbb{F} \in \Omega$ we have two possibilities: either $\mathfrak{m}_{\ks}[\F] \leq x^*_{ \gmax, k_*}$ or $\mathfrak{m}_{\ks}[\F] > x^*_{{ \gmax, k_*}}$.
For the former, it follows that
\begin{equation*}
\mathfrak{m}_{\ks} [ \mathbb{W}_R] \leq x^*_{{ \gmax, k_*}} + h \left( \sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\mathbb{F}_R) \right]_i \left\langle v \right\rangle_i^{{k_*}} \mathrm{d} v \right) \leq x^*_{{ \gmax, k_*}} + h \, \mathcal{L}^*_{{ \gmax, k_*}} \leq \xi_{{ \gmax, k_*}},
\end{equation*}
where we have assumed, without loss of generality, that $h\leq 1$. For the latter, we choose $R=R(\mathbb{F})$ sufficiently large such that $\mathfrak{m}_{\ks} [\mathbb{F}_R]> x^*_{{ \gmax, k_*}}$, and therefore,
\begin{equation*}
\mathcal{L}_{{ \gmax, k_*}}\left(\mathfrak{m}_{\ks} [\mathbb{F}_R]\right) \leq 0.
\end{equation*}
As a consequence,
\begin{equation*}
\mathfrak{m}_{\ks} [\mathbb{W}_R] \leq x^*_{{ \gmax, k_*}} \leq \xi_{{ \gmax, k_*}}.
\end{equation*}
Therefore, we constructed a constant $C_{\ks}$ from characterization of the set $\Omega$, that is $ \xi_{{ \gmax, k_*}}$.\\
The conclusion is that $\mathbb{W}_R \in \Omega$ for any $0<h<h_*$, where
\begin{equation*}
h_*=\min\left\{1, \frac{1}{K \left(1 + R(\mathbb{F})^ { {\color{black}\overline{\gamma}}} \right)}\right\},
\end{equation*}
and $K$ is from \eqref{constant c tangency}.\\
Now, H\"{o}lder estimate \eqref{holder estimate} implies
\begin{equation*}
h^{-1} \left\| \mathbb{F}+ h \mathbb{Q}(\F) - \mathbb{W}_R \right\|_{L_2^1} = \left\| \mathbb{Q}(\F) - \mathbb{Q}(\mathbb{F}_R) \right\|_{L_2^1} \leq C_H \left\| \mathbb{F} - \mathbb{F}_R \right\|_{L_2^1}^\frac{1}{2} \leq \varepsilon,
\end{equation*}
for $R:=R(\varepsilon)$ sufficiently large. Then, for this choice of $R$, $\mathbb{W}_R \in B(\mathbb{F}+ h \mathbb{Q}(\F),h \epsilon)$. \\
Finally, choosing $R=\max\{R(\mathbb{F}),R(\epsilon)\}$ and $h_1$ as
\begin{equation}\label{h1}
h_1=\min\left\{1, \frac{1}{K \left(1 + R^ { {\color{black}\overline{\gamma}}} \right)}\right\},
\end{equation}
with $c$ given in \eqref{constant c tangency},
one concludes that $\mathbb{W}_R \in B(\mathbb{F}+h \mathbb{Q}(\F), h \epsilon) \cap \Omega$.
\end{proof}
Once the Proposition \ref{prop tangency} is proved, it immediately follows
\begin{equation*}
h^{-1} \text{dist}\left(\mathbb{F}+ h \mathbb{Q}(\F), \Omega \right) \leq \varepsilon, \qquad \forall \, 0<h<h_1,
\end{equation*}
with $h_1$ from \eqref{h1}, which concludes the proof of tangency condition.
\end{proof}
\begin{proof}[Proof of (iii) one-sided Lipschitz condition]
From definition and representation \eqref{collision operator diff and sum}, we have
\begin{multline*}
I_L:=\left[ \mathbb{Q}(\F) - \mathbb{Q}(\G), \mathbb{F}-\mathbb{G} \right]
\\ \leq \sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} \left( Q_{ij}(f_i,f_j) - Q_{ij}(g_i,g_j) \right) \text{sign}(f_i(v)-g_i(v)) \left\langle v \right\rangle_i^2 \mathrm{d}v
\\= \frac{1}{2}\sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} \left(Q_{ij}(f_i-g_i, f_j + g_j) + Q_{ij}(f_i+g_i, f_j-g_j) \right) \text{sign}(f_i(v)-g_i(v))\left\langle v \right\rangle_i^2 \mathrm{d}v.
\end{multline*}
Changing $i\leftrightarrow j$ in the second integral, we precisely obtain binary structure of the weak form \eqref{weak form ij+ji} that yields
\begin{multline*}
I_L\leq \frac{1}{2}\sum_{i=1}^I \sum_{j=1}^I \int_{\mathbb{R}^3} \left(Q_{ij}(f_i-g_i, f_j + g_j)\: \text{sign}(f_i(v)-g_i(v)) \:\left\langle v \right\rangle_i^2 \right. \\ \left.+ Q_{ji}(f_j+g_j, f_i-g_i) \: \text{sign}(f_j(v)-g_j(v) ) \: \left\langle v \right\rangle_j^2 \right) \mathrm{d}v
\\
=\frac{1}{2}\sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \mathcal{B}_{ij}(v,v_*,\sigma) \left( f_i(v) - g_i(v)\right) \left( f_j(v_*) + g_j(v_*)\right) \\ \times \left( \text{sign}(f_i(v')-g_i(v')) \:\left\langle v' \right\rangle_i^2 + \text{sign}(f_j(v'_*)-g_j(v'_*)) \:\left\langle v'_* \right\rangle_j^2 \right. \\ \left. - \text{sign}(f_i(v)-g_i(v)) \:\left\langle v \right\rangle_i^2 - \text{sign}(f_j(v_*)-g_j(v_*) ) \: \left\langle v_* \right\rangle_j^2 \right) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d}v.
\end{multline*}
Using the upper bound of the sign function, one has
\begin{multline*}
I_L \leq \frac{1}{2}\sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \mathcal{B}_{ij}(v,v_*,\sigma) \\ \times \left( \left| f_i(v) - g_i(v)\right| \left( f_j(v_*) + g_j(v_*)\right) \left( \left\langle v' \right\rangle_i^2 + \left\langle v'_* \right\rangle_j^2 \right) \right. \\ \left. - \left|f_i(v)-g_i(v)\right| \left( f_j(v_*) + g_j(v_*)\right) \:\left\langle v \right\rangle_i^2 \right. \\ \left.+ \left|f_i(v)-g_i(v) \right| \left( f_j(v_*) + g_j(v_*)\right) \left\langle v_* \right\rangle_j^2 \right) \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d}v.
\end{multline*}
Then, conservation of energy implies
\begin{multline*}
I_L \leq\sum_{i=1}^I \sum_{j=1}^I \iiint_{\mathbb{R}^3 \times \mathbb{R}^3 \times S^2} \mathcal{B}_{ij}(v,v_*,\sigma) \\ \times \left| f_i(v) - g_i(v)\right| \left( f_j(v_*) + g_j(v_*)\right) \left\langle v_* \right\rangle_j^2 \mathrm{d}\sigma \mathrm{d}v_* \mathrm{d}v.
\end{multline*}
Now, specifying the collision cross section \eqref{cross section} and using \eqref{estimate on u^g product}
\begin{equation*}
\left|v-v_*\right|^ { \gamma_{ij}} \leq \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^ { \gamma_{ij}} \left\langle v \right\rangle_i^ { \gamma_{ij}} \left\langle v_* \right\rangle_j^ { \gamma_{ij}} \leq {\color{black} \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^\gij } \left\langle v \right\rangle_i^ { {\color{black}\overline{\gamma}}} \left\langle v_* \right\rangle_j^ { {\color{black}\overline{\gamma}}} ,
\end{equation*}
we obtain
\begin{equation*}
I_L \leq \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) \left\| \mathbb{F} - \mathbb{G} \right\|_{L_{\gmax}^1} \left\| \mathbb{F} + \mathbb{G} \right\|_{L_{2+\gmax}^1}.
\end{equation*}
Thanks to the monotonicity of norms \eqref{monotonicity of norm}
\begin{equation*}
\left\| \mathbb{F} - \mathbb{G} \right\|_{L_{\gmax}^1} \leq \left\| \mathbb{F} - \mathbb{G} \right\|_{L_2^1},
\end{equation*}
we finally obtain
\begin{equation*}
I_L \leq 2 \max_{1\leq i, j \leq I}\left( {\color{black}\bijnjed} {\color{black}{\cub}} \right) C_{2+\gmax} \left\| \mathbb{F} - \mathbb{G} \right\|_{L_2^1},
\end{equation*}
which completes the proof of one-sided Lipschitz condition.
\end{proof}
\section{Proof of Theorem \ref{theorem bound on norm} (Generation and propagation of polynomial moments)}\label{section proof generation of poly}
The proof consists of several steps. First, once the existence and uniqueness of vector value solution $\mathbb{F}$ to the Boltzmann system \eqref{Cauchy problem} is proven, we can derive from the Boltzmann system an ordinary differential inequality for the scalar polynomial moment $\mathfrak{m}_k[\F](t)$. Then, the comparison principle for ODEs will yield estimates that guarantee both generation and propagation of these polynomial moments.
\subsubsection*{Step 1. (Ordinary Differential Inequality for the polynomial moment)}
\begin{lemma}\label{ODI poly}
Let $\mathbb{F}=\left[f_i\right]_{i=1,\dots,I}$ be a solution of the Boltzmann system \eqref{Cauchy problem}. Then the polynomial moment \eqref{poly moment} satisfies the following Ordinary Differential Inequality
\begin{equation}\label{ODI F}
\frac{\mathrm{d}}{\mathrm{d} t} \mathfrak{m}_k[\F](t)
\leq - A_k \, \mathfrak{m}_k[\F](t)^{1+\frac{ { {\color{black}\overline{\gamma}}} }{k}}+ B_k \, \mathfrak{m}_k[\F](t),
\end{equation}
for {\color{black} $k\ge {k_*}$ as defined in \eqref{kstar},
} with positive constants $A_k$ and $B_k$ as defined in Lemma~\ref{lemma Q za tangent cond}, equation \eqref{const Ak Bk}, after replacing ${k_*}$ by $k\ge{k_*}$.
\end{lemma}
\begin{proof}
Consider $i-$th equation of the Boltzmann system \eqref{Cauchy problem},
\begin{equation*}
\partial_t f_i(t,v) = \sum_{{j=1}}^I Q_{ij}(f_i,f_j)(t,v), \qquad i=1,\dots,I.
\end{equation*}
Integration with respect to velocity $v$ with weight $\left\langle v \right\rangle_i^k$, $k\geq0$, and summation over all species $i=1,\dots,I$ yields
\begin{equation} \label{ode}
\frac{\mathrm{d}}{\mathrm{d}t} \mathfrak{m}_k[\F](t) = \sum_{i=1}^I \sum_{{j=1 }}^I \int_{\mathbb{R}^3} \left\langle v \right\rangle_i^k Q_{ij}(f_i,f_j)(t,v)\mathrm{d}v,
\end{equation}
after recalling definition \eqref{poly moment} of polynomial moment.
{\color{black}
Using results from Lemma \ref{lemma Q za tangent cond} for $k\ge {k_*}$ as defined in \eqref{kstar}, we conclude the estimate \eqref{ODI F}.
}
\end{proof}
\subsubsection*{Step 2. (Comparison principle)}
The starting point is the inequality \eqref{ODI F}. We associate to it an ODE of Bernoulli type
\begin{equation}\label{associated ode}
y'(t)
= - a \,y(t)^{1+c}+ b \, y(t),
\end{equation}
whose solution will be an upper bound for $\mathfrak{m}_k[\F](t)$. Indeed, solution to \eqref{associated ode} reads
\begin{equation}\label{ode comparison}
y(t) = \left(\frac{a}{b}\left(1-e^{-c \, b\, t}\right) + y(0)^{-c} e^{-c\,b\,t} \right)^{-\frac{1}{c}}.
\end{equation}
\subsubsection*{Step 3. (Generation of polynomial moments)}
Dropping initial data in \eqref{ode comparison} yields
\begin{equation*}
y(t) \leq \left(\frac{a}{b}\left(1-e^{-c \,b \,t}\right) \right)^{-\frac{1}{c}}, \qquad \forall t > 0.
\end{equation*}
Setting $y(t):= \mathfrak{m}_k[\F](t)$, $a:= A_k$, $b:=B_k$ and $c:= { {\color{black}\overline{\gamma}}} /k$ this implies generation estimate \eqref{bound on norm} with
$$\mathfrak{C}^{\mathfrak{m}} = \left(\frac{A_k}{B_k}\right)^{-\frac{k}{ { {\color{black}\overline{\gamma}}} }}, \qquad {\color{black} \text {for any} \ k\ge k_*}.$$
\begin{remark}
For later purposes, we derive also the following inequality by approximating the last result. Namely, for $t<1$, we may write
$$
\left(1-e^{-c \, b \, t}\right)^{-\frac{1}{c}} = \left(c \, b \, t\right)^{-\frac{1}{c}} \left(1 + \frac{b}{2} t + o(t)\right) \leq \left(c \, b \right)^{-\frac{1}{c}} e^{\frac{b}{2} t} \ t^{-\frac{1}{c}} \leq \left(c \, b \right)^{-\frac{1}{c}} e^{\frac{b}{2} } \ t^{-\frac{1}{c}}.
$$
On the other hand, for $t\geq 1$, it follows
\begin{equation*}
\left(1-e^{-c \,b \,t} \right)^{-\frac{1}{c}} \leq \left(1-e^{-c \,b } \right)^{-\frac{1}{c}}.
\end{equation*}
Therefore,
\begin{equation}\label{generation poly decay}
y(t) \leq \left(\frac{a}{b} \right)^{-\frac{1}{c}} \begin{cases}
\left(c \, b \right)^{-\frac{1}{c}} e^{\frac{b}{2} } \ t^{-\frac{1}{c}}, & t<1 \\
\left(1-e^{-c \,b } \right)^{-\frac{1}{c}}, & t\geq 1.
\end{cases}
\end{equation}
In other words, plugging $y(t):= \mathfrak{m}_k[\F](t)$, $a:= A_k$, $b:=B_k$ and $c:= { {\color{black}\overline{\gamma}}} /k$, it yields
\begin{equation}\label{poly gen max t}
\mathfrak{m}_k[\F](t) \leq \mathfrak{B}^{\mathfrak{m}} \ \max\{ 1, t^{-\frac{k}{ { {\color{black}\overline{\gamma}}} }}\}, \quad \forall t>0,
\end{equation}
where the constant is $$\mathfrak{B}^{\mathfrak{m}} = \mathfrak{C}^{\mathfrak{m}} \max\left\{ \left(\frac{ { {\color{black}\overline{\gamma}}} }{B_k k }\right)^{-\frac{k}{ { {\color{black}\overline{\gamma}}} }} e^{\frac{B_k}{2}}, \left( 1- e^{-\frac{ { {\color{black}\overline{\gamma}}} }{B_k k }}\right)^{-\frac{k}{ { {\color{black}\overline{\gamma}}} }} \right\}, \qquad {\color{black} \text {for any} \ k\ge k_*}.$$
\end{remark}
\subsubsection*{Step 4. (Propagation of polynomial moments)}
For propagation result, when $y(0)$ is assumed to be finite, we first notice that $y(t)$ is a monotone function of $t$, which approaches to $y(0)$ as $t\rightarrow 0$ on one hand, and converges to $ (a/b)^{-1/c}$ when $t\rightarrow\infty$ on the other hand. Therefore, $$y(t) \leq \max\{y(0), (a/b)^{-1/c}\},$$ for all $t\geq 0$.
Again, taking $y(t):= \mathfrak{m}_k[\F](t)$, $a:= A_k$, $b:=B_k$ and $c:= { {\color{black}\overline{\gamma}}} /k$, {\color{black} for any $k\ge k_*$ }, implies the propagation estimate \eqref{poly ODI prop}.
\section{Generation and propagation of exponential moments}\label{Section gen prop exp mom}
Let $\mathbb{F}$ be a solution of the Boltzmann system \eqref{Cauchy problem}. In this section we prove both generation and propagation of exponential moment \eqref{exp moment} related to $\mathbb{F}$.
The proof strongly relies on generation and propagation of polynomial moments stated in Theorem \ref{bound on norm}. Moreover, it uses polynomial moment ODI, but written in a slightly different manner than in Section \ref{ODI poly}, which we make precise in the following Lemma.
\begin{lemma}
Let $\mathbb{F}$ be a solution of the Boltzmann system \eqref{Cauchy problem}. Then there exists positive constants $K_1$ and $K_2$ such that the following two polynomial moments ODI hold
\begin{itemize}
\item ODI needed for propagation of exponential moments
\begin{multline}\label{poly ODI prop}
\frac{\mathrm{d}}{\mathrm{d} t} \mathfrak{m}_{s k}[\mathbb{F}](t) \leq - K_1 \mathfrak{m}_{s k+ \gmax }[\mathbb{F}](t) \\+ K_2 {\color{black}\left( \max_{1\leq i, j \leq I}\boldsymbol{\mathcal{C}}^{ij}_{\frac{s k}{2}} \right)} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k\\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{s \ell+\gmax}[\mathbb{F}](t) \ \mathfrak{m}_{s k- s \ell}[\mathbb{F}](t) + \mathfrak{m}_{s k- s \ell+\gmax}[\mathbb{F}](t) \ \mathfrak{m}_{s \ell}[\mathbb{F}](t)\right).
\end{multline}
\item ODI needed for generation of exponential moments
\begin{multline}\label{poly ODI gen}
\frac{\mathrm{d}}{\mathrm{d} t} \mathfrak{m}_{\gmax k}[\F](t) \leq - K_1 \mathfrak{m}_{\gmax k+ \gmax }[\mathbb{F}](t) \\+ K_2 {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}}\right)} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k\\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{\gmax \ell+\gmax}[\mathbb{F}](t) \ \mathfrak{m}_{\gmax k- \gmax \ell}[\mathbb{F}](t) + \mathfrak{m}_{\gmax k- \gmax \ell+\gmax}[\mathbb{F}](t) \ \mathfrak{m}_{\gmax \ell}[\mathbb{F}](t)\right).
\end{multline}
\end{itemize}
\end{lemma}
\begin{proof}
We briefly point out that the main steps in the proofs are adaption of the proof given in \cite{GambaTask18}. Let us consider polynomial moment
\begin{equation*}
\mathfrak{m}_{\delta q}[\mathbb{F}](t)=: \mathfrak{m}_{\delta q}, \quad 0<\delta\leq 2, \ q\geq 0, {\color{black} \ \ \text{ with}\ \delta q>k_* ,}
\end{equation*}
{\color{black} with $k_*$ as defined in \eqref{kstar}, and derive an ODI for it starting from \eqref{poly moment start}
so that $\boldsymbol{\mathcal{C}}^{ij}_{\frac{\delta q }{2}} < \bijnjed$ holds uniformly for any pair $i,j=1,\dots,I$, with $\boldsymbol{\mathcal{C}}^{ij}_n$ being the constant from Povzner lemma \ref{Povzner intro}}. Once we derive it, \eqref{poly ODI prop} will follow setting $\delta := s$, and \eqref{poly ODI gen} will follow with $\delta:= { {\color{black}\overline{\gamma}}} $. Indeed, from \eqref{poly moment start} we get
\begin{multline*}
\mathfrak{m}_{\delta q}' = \sum_{i=1}^I \int_{ \mathbb{R}^3} \left[ \mathbb{Q}(\F) \right]_i \left\langle v \right\rangle_i^{\delta q} \mathrm{d} v \leq \frac{1}{2}\sum_{i=1}^I \sum_{j=1}^I \iint_{\mathbb{R}^3 \times \mathbb{R}^3} f_i (v) f_j(v_*) \left|v-v_*\right|^ { \gamma_{ij}} \\ \times \left( \boldsymbol{\mathcal{C}}^{ij}_{\frac{\delta q}{2}} \left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 \right)^{\frac{\delta q}{2}} - \bijnjed \left( \left\langle v \right\rangle_i^{\delta q} + \left\langle v_* \right\rangle_j^{\delta q} \right) \right) \mathrm{d} v_* \mathrm{d}v.
\end{multline*}
Before applying Lemma \ref{binomial}, we first estimate, since $(\delta/2) \leq 1$,
\begin{equation*}
\left(\left\langle v \right\rangle_i^2 + \left\langle v_* \right\rangle_j^2 \right)^{\frac{\delta q}{2}} \leq \left(\left\langle v \right\rangle_i^{\delta} + \left\langle v_* \right\rangle_j^{ \delta} \right)^{{ q}},
\end{equation*}
and then apply it, which gives the following
\begin{multline*}
\left(\left\langle v \right\rangle_i^{\delta} + \left\langle v_* \right\rangle_j^{ \delta} \right)^{{ q}} \leq \left\langle v \right\rangle_i^{\delta q} + \left\langle v_* \right\rangle_j^{\delta q} \\ + \sum_{\ell=1}^{\ell_{q}} \left( \begin{matrix}
q\\ \ell
\end{matrix} \right) \left( \left\langle v \right\rangle_i^{ \delta \ell} \left\langle v_* \right\rangle_j^{ \delta q - \delta \ell}+ \left\langle v \right\rangle_i^{ \delta q- \delta \ell} \left\langle v_* \right\rangle_j^{\delta \ell} \right),
\end{multline*}
with $\ell_q = \lfloor\frac{q + 1}{2}\rfloor$. The bound from above and below of the non-angular part of the cross-section, $ \left|v-v_*\right|^ { \gamma_{ij}} $, is used as in Section \ref{ODI poly}. This implies a polynomial moment ODI
\begin{equation*}
\mathfrak{m}_{\delta q}'(t) \leq - K_1 \mathfrak{m}_{\delta q + \gmax} + \boldsymbol{\mathcal{C}}^{ij}_{\frac{\delta q}{2}} \, K_2 \sum_{\ell=1}^{\ell_{q}} \left( \begin{matrix}
q\\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{ \delta \ell + \gmax} \mathfrak{m}_{\delta q -\delta \ell} + \mathfrak{m}_{\delta q -\delta \ell +\gmax} \mathfrak{m}_{ \delta \ell } \right),
\end{equation*}
where $K_1$ and $K_2$ are positive constants {\color{black} since $\delta q \ge k_*$, with $k_*$ as defined in \eqref{kstar}, }
\begin{equation*}
\begin{split}
K_1 &= \min_{1\leq i, j \leq I} \left(\bijnjed - \boldsymbol{\mathcal{C}}^{ij}_{\frac{\delta q }{2}}\right) \frac{c_{lb} }{\max_{1\leq i \leq I} m_i}, \\
K_2 &=\frac{1}{2} \left( \max_{1\leq i, j \leq I} {\color{black} \left( \frac{\sum_{i=1}^I m_i}{\sqrt{m_i m_j}} \right)^\gij } \right),
\end{split}
\end{equation*}
which completes the proof.
\end{proof}
\section{Proof of Theorem \ref{theorem gen prop ML} (b) (Propagation of exponential moments)}
Using Taylor series of an exponential function, one can represent exponential moment as
\begin{equation*}
\mathcal{E}_{s}[\F]({\alpha},t) = \sum_{k=0}^\infty \frac{\alpha^k}{k!} \, \mathfrak{m}_{s k}[\mathbb{F}](t).
\end{equation*}
{\color{black} We will show that the exponential rate $\alpha=\alpha(k_*)$, that is, depending on the $k_*$ parameter defined in \eqref{kstar}.}
We consider its partial sum an a shifted by $ { {\color{black}\overline{\gamma}}} $ one, namely,
\begin{equation}\label{pomocna 18}
\mathcal{E}_{s}^n[\F]({\alpha},t) = \sum_{k=0}^n \frac{\alpha^k}{k!} \, \mathfrak{m}_{s k}[\mathbb{F}](t), \quad \mathcal{E}_{s;\gmax}^n[\F]({\alpha},t)= \sum_{k=0}^n \frac{\alpha^k}{k!} \, \mathfrak{m}_{s k+ \gmax }[\mathbb{F}](t).
\end{equation}
In order to have lighter writing, we will drop from moment notation dependence on $t$ and $\alpha$, and relation to $\mathbb{F}$, and we will instead write
\begin{equation*}
\mathcal{E}_{s}^n[\F]({\alpha},t)=:\mathcal{E}_{s}^n, \qquad \mathcal{E}_{s;\gmax}^n[\F]({\alpha},t):=\mathcal{E}_{s;\gmax}^n, \qquad \mathfrak{m}_{s k+ \gmax }[\mathbb{F}](t)=:\mathfrak{m}_{s k+ \gmax }.
\end{equation*}
When it will be important to highlight dependence on $t$ and $\alpha$, we will also, for example, write $\mathcal{E}_{s}^n(\alpha,t)$ instead of $\mathcal{E}_{s}^n$.\\
The idea of proof is to show that the partial sum $\mathcal{E}_{s}^n$ is bounded uniformly in time $t$ and $n$. To this end, we first derive ordinary differential inequality (ODI) for it.
\changelocaltocdepth{1}
\subsection*{ODI for $\mathcal{E}_{s}^n$}
Taking derivative with respect to time $t$ of \eqref{pomocna 18}, we get
\begin{align*}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n
= \sum_{k=0}^{k_0-1} \frac{\alpha^{ k}}{k!} \mathfrak{m}_{s k}' + \sum_{k=k_0}^n \frac{\alpha^{ k}}{k!} \mathfrak{m}_{s k}',
\end{align*}
where $k_0$ is an index that will be determined later on. We use a polynomial moment ODE \eqref{poly ODI prop} for the second term that yields
\begin{multline}\label{pomocna 10}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n \leq \sum_{k=0}^{k_0-1} \frac{\alpha^{k}}{k!} \mathfrak{m}_{s k}' - K_1 \sum_{k=k_0}^n \frac{\alpha^{ k}}{k!} \, \mathfrak{m}_{s k+ \gmax } \\ + K_2 \sum_{k=k_0}^n {\color{black}\left( \max_{1\leq i, j \leq I}\boldsymbol{\mathcal{C}}^{ij}_{\frac{sk}{2}}\right)} \frac{\alpha^{ k}}{k!} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k \\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{s \ell+\gmax} \mathfrak{m}_{s k- s \ell} + \mathfrak{m}_{s k- s \ell+\gmax} \mathfrak{m}_{s \ell}\right)\\
=: S_0 - K_1 S_1 + K_2 S_2.
\end{multline}
We estimate each sum $S_0$, $S_1$ and $S_2$ separately.
\subsubsection*{Term $S_0$} Propagation of polynomial moments \eqref{poly propagation} ensures bound on $\mathfrak{m}_{s k}$ uniformly in time, which implies from \eqref{ODI F} bound on its derivative, i.e. there exist a constant $c_{k_0}$ such that
\begin{equation}\label{pomocna 9}
\mathfrak{m}_{s k}, \mathfrak{m}_{s k}' \leq c_{k_0} \quad \text{for all} \ k\in\{0,1,\dots,k_0\}.
\end{equation}
For $S_0$ this yields
\begin{equation}\label{S0 estimate}
S_0 \leq c_{k_0} \sum_{k=0}^{k_0-1} \frac{\alpha^{k}}{k!} \leq c_{k_0} e^{\alpha} \leq 2 \, c_{k_0},
\end{equation}
for $\alpha$ small enough to satisfy
\begin{equation}\label{alpha first cond}
e^{\alpha} \leq 2.
\end{equation}
\subsubsection*{Term $S_1$} We complete first the term $S_1$ to make appear shifted partial sum $\mathcal{E}_{s;\gmax}^n$ by means of
\begin{equation*}
S_1 = \sum_{k=k_0}^n \frac{\alpha^{k}}{k!} \, D_k \, \mathfrak{m}_{s k+ \gmax } = \mathcal{E}_{s;\gmax}^n - \sum_{k=0}^{k_0-1} \frac{\alpha^{ k}}{k!} \, D_k \, \mathfrak{m}_{s k+ \gmax }.
\end{equation*}
From the bound \eqref{pomocna 9} we can estimate $\mathfrak{m}_{s k+ \gmax }$ as well,
\begin{equation*}
\mathfrak{m}_{s k+ \gmax } \leq c_{k_0}, \quad k=0,\dots,k_0-1,
\end{equation*}
which together with considerations for the term $S_0$ yields
\begin{equation}\label{S1 estimate}
S_1 \geq \mathcal{E}_{s;\gmax}^n - 2 c_{k_0}.
\end{equation}
\subsubsection*{Term $S_2$} Term $S_2$ can be separated into two terms, namely
\begin{equation*}
S_2 = \sum_{k=k_0}^n {\color{black}\left( \max_{1\leq i, j \leq I}\boldsymbol{\mathcal{C}}^{ij}_{\frac{sk}{2}} \right)}\frac{\alpha^{ k}}{k!} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k \\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{s \ell+\gmax} \mathfrak{m}_{s k- s \ell} + \mathfrak{m}_{s k- s \ell+\gmax} \mathfrak{m}_{s \ell}\right)=: S_{2_1} + S_{2_2}.
\end{equation*}
Their treatment is the same, so let perform an estimate on $S_{2_1}$. Rearranging we can write
\begin{equation*}
S_{2_1} = \sum_{k=k_0}^n {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{sk}{2}} \right)} \sum_{\ell=1}^{\ell_{k}}\frac{\alpha^\ell \, \mathfrak{m}_{s \ell+\gmax}}{\ell!} \frac{\alpha^{k-\ell} \, \mathfrak{m}_{s k- s \ell}}{(k-\ell)!} \leq {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{sk_0}{2}}\right)} \mathcal{E}_{s;\gmax}^n \, \mathcal{E}_{s}^n,
\end{equation*}
{ \color{black} the last inequality is due to the decreasing property of $\boldsymbol{\mathcal{C}}^{ij}_k$ in $k\ge k_*$, uniformly for any $i,j$, with $k_*$ defined in \eqref{kstar}. } Therefore, we can estimate
\begin{equation}\label{S2 estimate}
S_{2} \leq 2 {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{s{k_0}}{2}} \right)} \mathcal{E}_{s;\gmax}^n \, \mathcal{E}_{s}^n.\\
\end{equation}
Finally, desired ODI for $\mathcal{E}_{s}^n$ is obtained from \eqref{pomocna 10} gathering all estimates \eqref{S0 estimate}, \eqref{S1 estimate} and \eqref{S2 estimate}. Namely,
\begin{equation}\label{pomocna 12}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n \leq - K_1 \mathcal{E}_{s;\gmax}^n
+ 2 c_{k_0} (1+K_1) + 2\, K_2 {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{s{k_0}}{2}} \right)} \mathcal{E}_{s;\gmax}^n \, \mathcal{E}_{s}^n.
\end{equation}
\subsection*{Bound on $\mathcal{E}_{s}^n$} For each $n \in \mathbb{N}$ we define
\begin{equation*}
T_n := \sup\{ t\geq 0: \mathcal{E}_{s}^n(\alpha,\tau) \leq 4 M_0, \ \forall \tau \in [0,t] \},
\end{equation*}
where $M_0$ is a bound on initial data in \eqref{initial data exp prop}. We will show that $\mathcal{E}_{s}^n(t)$ is uniformly bounded in $t$ and $n$ by proving that $T_n=\infty$ for all $n\in \mathbb{N}$.\\
The sequence $T_n$ is well-defined and positive. Indeed, since $\alpha \leq \alpha_0$, at time $t=0$ we have
\begin{equation*}
\mathcal{E}_{s}^n(\alpha,0) = \sum_{k=0}^n \frac{\alpha^{ k}}{k!} \mathfrak{m}_{s k}(0) \leq \sum_{k=0}^n \frac{\alpha_0^{ k}}{k!} \mathfrak{m}_{s k}(0) \leq \mathcal{E}_{s}({\alpha_0},0) < 4 M_0,
\end{equation*}
uniformly in $n$, by assumption \eqref{initial data exp prop}. Since each term $\mathfrak{m}_{s k}(t)$ is continuous function of $t$, so is $\mathcal{E}_{s}^n(\alpha, t)$. Therefore, $\mathcal{E}_{s}^n(\alpha, t)< 4M_0$ on some time interval $[0,t_n)$, $t_n>0$. Thus $T_n$ is well-defined and positive for every $n\in \mathbb{N}$.\\
For $t\in [0,T_n]$ it follows $\mathcal{E}_{s}^n(\alpha, t)\leq4 M_0$, which from \eqref{pomocna 12} implies
\begin{equation}\label{pomocna 20}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n \leq - \mathcal{E}_{s;\gmax}^n \left( K_1 - 8 \,K_2 {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{s k_0}{2}} \right)} M_0 \right) \\ + 2 c_{k_0} \left( 1 + K_1 \right).
\end{equation}
Since $\boldsymbol{\mathcal{C}}^{ij}_{\frac{s k_0}{2}}$, {\color{black} for any $i,j$, }converges to zero as {\color{black} $\frac{s k_0}{2}> k_*$} goes to infinity we can choose {\color{black} $k_0> 2\frac{k_*}{s}$ } such that
\begin{equation*}
K_1 - 8 \,K_2 {\color{black}\left( \max_{1\leq i, j \leq I}{\color{black} \boldsymbol{\mathcal{C}}^{ij}_{k_*} }\right)} M_0 > \frac{K_1}{2},
\end{equation*}
{\color{black} or, equivalently,
\begin{equation}\label{K_1kstar}
K_1 < 16 \,K_2 {\color{black}\left( \max_{1\leq i, j \leq I}{\color{black} \boldsymbol{\mathcal{C}}^{ij}_{k_*} }\right)} M_0,
\end{equation}
with $K_1$ depending on $k_*$ as defined in \eqref{kstar}. }
Hence, \eqref{pomocna 20} becomes
\begin{equation}\label{pomocna 21}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n \leq -\frac{K_1}{2}\, \mathcal{E}_{s;\gmax}^n+ 2 {\color{black} c_{k_*}} \left( 1 + K_1 \right).
\end{equation}
Next step consists in finding lower bound for $\mathcal{E}_{s;\gmax}^n$ in terms of $\mathcal{E}_{s}^n$. Indeed,
we can estimate
\begin{multline*}
\mathcal{E}_{s;\gmax}^n = \sum_{k=0}^n \frac{\alpha^{ k}}{k!} \mathfrak{m}_{s k+ \gmax } \geq \sum_{k=0}^n \frac{\alpha^{ k}}{k!} \sum_{i=1}^I \int_{\left\{ \left\langle v \right\rangle_i \geq \alpha^{-1/2} \right\}} f_i(t,v) \left\langle v \right\rangle_i^{sk+ { {\color{black}\overline{\gamma}}} } \mathrm{d}v
\\ \geq \alpha^{- { {\color{black}\overline{\gamma}}} /2} \left( \mathcal{E}_{s}^n - \sum_{k=0}^n \frac{\alpha^{ k}}{k!} \sum_{i=1}^I \int_{\left\{ \left\langle v \right\rangle_i < \alpha^{-1/2} \right\}} f_i(t,v) \left\langle v \right\rangle_i^{sk} \mathrm{d}v \right)
\\ \geq \alpha^{- { {\color{black}\overline{\gamma}}} /2} \left( \mathcal{E}_{s}^n - \sum_{k=0}^n \frac{\alpha^{k (1-\frac{s}{2})}}{k!} \mathfrak{m}_{0}(0) \right) \geq \alpha^{- { {\color{black}\overline{\gamma}}} /2} \left( \mathcal{E}_{s}^n - \mathfrak{m}_{0}(0) e^{\alpha^{1-\frac{s}{2}}} \right).
\end{multline*}
Plugging this result into \eqref{pomocna 21} yields
\begin{equation*}
\frac{\mathrm{d} \, }{\mathrm{d} t}\mathcal{E}_{s}^n \leq -\frac{K_1}{2}\, \alpha^{- { {\color{black}\overline{\gamma}}} /2} \mathcal{E}_{s}^n + \frac{K_1}{2} \alpha^{- { {\color{black}\overline{\gamma}}} /2} \mathfrak{m}_{0}(0) e^{\alpha^{1-\frac{s}{2}}} + 2 c_{k_0} \left( 1 + K_1 \right).
\end{equation*}
By the maximum principle for ODEs, it follows
\begin{multline}\label{pomocna 22}
\mathcal{E}_{s}^n(\alpha, t) \leq \max\left\{ \mathcal{E}_{s}^n(\alpha, 0), \mathfrak{m}_{0}(0) \, e^{\alpha^{1-\frac{s}{2}}} + \frac{4 c_{k_0} \left( 1 + K_1 \right)}{ K_1 \, \alpha^{- { {\color{black}\overline{\gamma}}} /2}} \right\} \\ \\
\leq M_0 + \mathfrak{m}_{0}(0) \, e^{\alpha^{1-\frac{s}{2}}} + \alpha^{ { {\color{black}\overline{\gamma}}} /2} \ \frac{4 c_{k_0} \left( 1 + K_1 \right)}{ K_1 \, },
\end{multline}
for any $t\in [0,T_n].$
On the other hand, since $s \leq 2$, the following limit property holds
\begin{equation*}
\mathfrak{m}_{0}(0) \, e^{\alpha^{1-\frac{s}{2}}} + \alpha^{ { {\color{black}\overline{\gamma}}} /2} \ \frac{4 c_{k_0} \left( 1 + K_1 \right)}{ K_1 \, } \rightarrow \mathfrak{m}_{0}(0), \quad \text{as} \ \alpha\rightarrow 0,
\end{equation*}
and $\mathfrak{m}_{0}(0) < \mathcal{E}_{s}^n(\alpha_0,0)$ for any $\alpha_0$, and therefore, by \eqref{initial data exp prop}, $\mathfrak{m}_{0}(0)< M_0$. Thus, we can choose sufficiently small $\alpha = \alpha_1$ such that
\begin{equation}\label{alpha second cond}
\mathfrak{m}_{0}(0) \, e^{\alpha^{1-\frac{s}{2}}} + \alpha^{ { {\color{black}\overline{\gamma}}} /2} \ \frac{4 c_{k_*} \left( 1 + K_1 \right)}{ K_1 \, }< 3 M_0,
\end{equation}
for any $s\leq 2$ and {\color{black} $K_1=K_1(k_*)$ from \eqref{K_1kstar}}. In that case, inequality \eqref{pomocna 22} implies the following strict inequality
\begin{equation}\label{pomocna 23}
\mathcal{E}_{s}^n(\alpha, t) < 4 M_0,
\end{equation}
for any $t\in [0,T_n]$ and
{\color{black}$0<\alpha({k_*}) \leq \alpha_1$, with $\alpha$ depending on $k_*$ defined in \eqref{kstar}}.\\
\subsection*{Conclusion I} If $k_0$ is chosen such that \eqref{pomocna 21} holds, and the choice of $\alpha$ is such that $0< \alpha \leq \alpha_0$ and \eqref{alpha first cond}, \eqref{alpha second cond} are satisfied, which amounts to take $\alpha =\min \left\{ \alpha_0, \ln 2, \alpha_1 \right\}$, then we have strict inequality \eqref{pomocna 23}, $\mathcal{E}_{s}^n(\alpha, t) < 4 M_0$, that holds on the closed interval $ [0,T_n]$ uniformly in $n$. Because of the continuity of $\mathcal{E}_{s}^n(\alpha,t)$ with respect to time $t$, this strict inequality actually holds on a slightly larger time interval $[0, T_n + \varepsilon)$, $\varepsilon >0$. This contradicts the maximality of $T_n$ unless $T_n = + \infty$. Therefore,
$\mathcal{E}_{s}^n(\alpha, t) \leq 4 M_0$ for all $t\geq 0$ and $n\in \mathbb{N}$. Thus, letting $n\rightarrow \infty$ we conclude
\begin{equation*}
\mathcal{E}_{s}[\F]({\alpha},t) = \lim\limits_{n\rightarrow \infty} \mathcal{E}_{s}^n[\F]({\alpha},t) \leq 4 M_0, \quad \forall t\geq 0,
\end{equation*}
i.e. the solution $\mathbb{F}$ to system of Boltzmann equations with finite initial exponential moment of order $s$ and rate $\alpha_0$ will propagate exponential moments of the same order $s$ and a rate $\alpha$ that satisfies $\alpha =\min \left\{ \alpha_0, \ln 2, \alpha_1 \right\}$. {\color{black} It is also very interesting to note that the rate $\alpha$ depends on the $k_*$ parameter from \eqref{kstar}, which depends on uniform in the $i,j$ pairs upper bounds for the intermolecular potentials $\gamma_{ij}$ and for controls of the $k_*^{ij}$ as defined in \eqref{povzner constant prop} in the Povzner lemma~\ref{Povzner intro}.}
\section{Proof of Theorem \ref{theorem gen prop ML} (a) (Generation of exponential moments)}
We consider an exponential moment of order $ { {\color{black}\overline{\gamma}}} $ and rate $\alpha t$, {\color{black} where $\alpha$ depends on $k_*$ from \eqref{kstar},} for the solution $\mathbb{F}$ of the Boltzmann system, namely
\begin{equation*}
\mathcal{E}_{\gmax}[\F](\alpha t,t)= \sum_{i=1}^I \int_{\mathbb{R}^3} f_i(t,v) \, e^{\alpha t \left\langle v \right\rangle_i^{ { {\color{black}\overline{\gamma}}} }} \mathrm{d}v =\sum_{k=0}^\infty \frac{(\alpha t)^k}{k!} \mathfrak{m}_{\gmax k}[\F](t).
\end{equation*}
Consider its partial sum, and a shifted one
\begin{equation*}
\mathcal{E}^n_{\gmax}[\F](\alpha t,t)= \sum_{k=0}^n \frac{(\alpha t)^k}{k!} \mathfrak{m}_{\gmax k}[\F](t), \qquad \mathcal{E}^n_{\gmax; \gmax}[\F](\alpha t,t)= \sum_{k=0}^n \frac{(\alpha t)^k}{k!} \mathfrak{m}_{\gmax k+ \gmax }[\mathbb{F}](t).
\end{equation*}
As usual, we will most of the time relieve notation by omitting explicit dependence on time $t$ and relation to $\mathbb{F}$, and write
\begin{equation*}
\mathcal{E}^n_{\gmax}[\F](\alpha t,t)=: \mathcal{E}^n_{\gmax}, \quad \mathcal{E}^n_{\gmax; \gmax}[\F](\alpha t,t):= \mathcal{E}^n_{\gmax; \gmax}.
\end{equation*}
Fix $\alpha$ and $ { {\color{black}\overline{\gamma}}} $ and define
\begin{equation*}
\bar{T}_n := \sup\left\{ t\in [0,1]: \mathcal{E}^n_{\gmax}[\F](\alpha t,t)\leq 4 \bar{M}_0 \right\}.
\end{equation*}
$ \bar{T}_n $ is well defined. Indeed, taking $\bar{M}_0:= \sum_{i=1}^I f_i(t,v) \left\langle v \right\rangle_i^2 \mathrm{d}v = \sum_{i=1}^I f_i(0,v) \left\langle v \right\rangle_i^2 \mathrm{d}v$, for $t=0$, we get $\mathcal{E}^n_{\gmax}(0,0) \leq \mathcal{E}_ { {\color{black}\overline{\gamma}}} (0,0)=\mathfrak{m}_{0}(0) < 4 \bar{M}_0$. By continuity of partial sum $\mathcal{E}^n_{\gmax}$ with respect to $t$, $\mathcal{E}^n_{\gmax}(\alpha t,t) \leq 4 \bar{M}_0$ on a slightly larger time interval $t\in[0,t_n)$, $t_n>0$, and thus $\bar{T}_n>0$.
\subsection*{ODI for $ \mathcal{E}^n_{\gmax}$}
Taking time derivative of $ \mathcal{E}^n_{\gmax}$ yields
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} = \alpha \sum_{k=1}^n \frac{(\alpha t)^{k-1}}{(k-1)!} \mathfrak{m}_{\gmax k} + \sum_{k=0}^{k_0-1} \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k}' + \sum_{k=k_0}^n \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k}'.
\end{equation*}
For the first term we simply re-index the sum and use definition of shifted partial sum, and for the last one we use polynomial moment ODI \eqref{poly ODI gen}, which together implies
\begin{multline}\label{pomocna 11}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} \leq \alpha \, \mathcal{E}^n_{\gmax; \gmax} + \sum_{k=0}^{k_0-1} \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k}' - K_1 \sum_{k=k_0}^n \frac{(\alpha t)^{k}}{k!}
\mathfrak{m}_{\gmax k+ \gmax }\\+ K_2 \sum_{k=k_0}^n \frac{(\alpha t)^{k}}{k!} {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}} \right)} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k\\ \ell
\end{matrix} \right) \left( \mathfrak{m}_{\gmax \ell+\gmax} \mathfrak{m}_{\gmax k- \gmax \ell} + \mathfrak{m}_{\gmax k- \gmax \ell+\gmax} \mathfrak{m}_{\gmax \ell}\right)
\\ =: \alpha \, \mathcal{E}^n_{\gmax; \gmax} + S_0 - K_1 S_1 + K_2 \left(S_{2_1} + S_{2_2} \right).
\end{multline}
\subsubsection*{Term $S_0$} From polynomial moment generation estimate \eqref{poly gen max t} we can bound polynomial moment of any order, as well as its derivative by means of \eqref{ODI F}. In particular,
\begin{equation*}
\mathfrak{m}_{\gmax k} \leq \mathfrak{B}^{\mathfrak{m}} \ \max_{t>0}\{ 1, t^{-{k}}\}, \quad \mathfrak{m}_{\gmax k}' \leq B_{ { {\color{black}\overline{\gamma}}} k} \mathfrak{B}^{\mathfrak{m}} \ \max_{t>0}\{ 1, t^{-{k}}\}.
\end{equation*}
Denote
\begin{equation*}
\bar{c}_{k_0} := \max_{k\in \{0,\dots k_0-1\}} \left\{ \mathfrak{B}^{\mathfrak{m}}, B_{ { {\color{black}\overline{\gamma}}} k} \mathfrak{B}^{\mathfrak{m}} \right\}.
\end{equation*}
For $S_0$ taking $t\leq 1$, we have $ \mathfrak{m}_{\gmax k}' \leq \bar{c}_{k_0} t^{-k}$ and therefore
\begin{equation*}
S_0 := \sum_{k=0}^{k_0-1} \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k}' \leq \bar{c}_{k_0} \sum_{k=0}^{k_0-1} \frac{\alpha^{k}}{k!} \leq \bar{c}_{k_0} e^{\alpha} \leq 2 \bar{c}_{k_0},
\end{equation*}
for $\alpha$ such that
\begin{equation}\label{gen alpha 1}
e^\alpha \leq 2.
\end{equation}
\subsubsection*{Term $S_1$} Using boundedness of $\mathfrak{m}_{\gmax k+ \gmax }$, we can write
\begin{equation*}
S_1:= \sum_{k=k_0}^n \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k+ \gmax } = \mathcal{E}^n_{\gmax; \gmax} - \sum_{k=0}^{k_0-1} \frac{(\alpha t)^{k}}{k!} \mathfrak{m}_{\gmax k+ \gmax }
\geq \mathcal{E}^n_{\gmax; \gmax} - 2 \bar{c}_{k_0} \frac{1}{t},
\end{equation*}
for $\alpha$ chosen as in \eqref{gen alpha 1}.
\subsubsection*{Term $S_2$} Terms $S_{2_1}$ and $S_{2_2}$ are treated in the same fashion. We will detail calculation for $S_{2_1}$. We first reorganize the terms in sum and get
\begin{multline*}
S_{2_1}:=\sum_{k=k_0}^n \frac{(\alpha t)^{k}}{k!} {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}} \right)} \sum_{\ell=1}^{\ell_{k}} \left( \begin{matrix}
k\\ \ell
\end{matrix} \right) \mathfrak{m}_{\gmax \ell+\gmax} \mathfrak{m}_{\gmax k- \gmax \ell}
\\
= \sum_{k=k_0}^n {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}} \right)} \sum_{\ell=1}^{\ell_{k}} \frac{(\alpha t)^{\ell} \mathfrak{m}_{\gmax \ell+\gmax}}{\ell!} \frac{(\alpha t)^{k-\ell}\mathfrak{m}_{\gmax k- \gmax \ell} }{(k-\ell)!} \leq {\color{black}\left( \max_{1\leq i, j \leq I} \boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k_0}{2}} \right)} \mathcal{E}^n_{\gmax; \gmax} \, \mathcal{E}^n_{\gmax}..
\end{multline*}
since constant $\boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}} $ decays with respect to $k$, {\color{black} for any $i,j$ {\color{black} and large enough $k_0\ge 2\frac{k_*}{ { {\color{black}\overline{\gamma}}} }$, with $k_*$ from \eqref{kstar} to ensure \eqref{povzner constant prop}, } } and therefore {\color{black} $\boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k}{2}} \leq \boldsymbol{\mathcal{C}}^{ij}_{ k_*} $.}
Gathering all estimates together, \eqref{pomocna 11} becomes
\begin{equation}\label{pomocna 25}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} \leq \alpha \, \mathcal{E}^n_{\gmax; \gmax} + 2 \bar{c}_{k_0} - K_1 \left( \mathcal{E}^n_{\gmax; \gmax} - 2 \bar{c}_{k_0} \frac{1}{t} \right) + K_2 {\color{black}\left( \max_{1\leq i, j \leq I} {\color{black} \boldsymbol{\mathcal{C}}^{ij}_{k_*}} \right)} \mathcal{E}^n_{\gmax; \gmax} \, \mathcal{E}^n_{\gmax},
\end{equation}
for $\alpha$ satisfying \eqref{gen alpha 1}.
\subsection*{Bound on $ \mathcal{E}^n_{\gmax}$} Consider $t\in[0,\bar{T}_n]$. On this interval, $\mathcal{E}^n_{\gmax}(\alpha t,t)\leq 4 \bar{M}_0$, as well as since $\bar{T}_n\leq 1$ yields $t^{-1}\geq 1$, which implies for \eqref{pomocna 25} the following estimate
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} \leq - \mathcal{E}^n_{\gmax; \gmax} \left( - \alpha + K_1 - K_2 {\color{black}\left( \max_{1\leq i, j \leq I} {\color{black} \boldsymbol{\mathcal{C}}^{ij}_{k_*} } \right)} 4 \bar{M}_0\right)+ {\color{black} \frac{ 2 \bar{c}_{k_*} }{\left(1+ K_1 \right){t}}. }
\end{equation*}
Since $\boldsymbol{\mathcal{C}}^{ij}_{\frac{ { {\color{black}\overline{\gamma}}} k_0}{2}} $ converges to zero as {\color{black} $k_0\ge 2\frac{k_*}{ { {\color{black}\overline{\gamma}}} }$, uniformly $i, j$, so choosing such large $k_0$ and small enough $\alpha$ such that }
\begin{equation*}
- \alpha + K_1 - K_2 \, {\color{black}\left( \max_{1\leq i, j \leq I}{\color{black} \boldsymbol{\mathcal{C}}^{ij}_{k_*}} \right)} 4 \bar{M}_0 > \frac{K_1}{2}.
\end{equation*}
{\color{black} with $K_1=K_1(k_*)$ },
yields
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} \leq - \frac{K_1}{2} \mathcal{E}^n_{\gmax; \gmax} + \frac{ K_3}{t},
\end{equation*}
for {\color{black} $K_3(k_*) :=2 \bar{c}_{k_*} \left(1+ K_1(k_*) \right)$}. Finally, shifted moment can be bounded as follows
\begin{equation*}
\mathcal{E}^n_{\gmax; \gmax}(\alpha t,t) = \sum_{k=1}^{n+1} \frac{ (\alpha t)^k \mathfrak{m}_{\gmax k}(t) }{k!} \frac{k}{\alpha t} \geq \frac{1}{\alpha t} \sum_{k=2}^{n} \frac{ (\alpha t)^k \mathfrak{m}_{\gmax k}(t) }{k!} \geq \frac{\mathcal{E}^n_{\gmax}(\alpha t,t) - \bar{M}_0}{\alpha t},
\end{equation*}
that yields
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax} \leq - \frac{K_1}{2\alpha t} \left( \mathcal{E}^n_{\gmax} - \bar{M}_0 - \frac{2\alpha}{K_1} { K_3} \right).
\end{equation*}
Now we choose $\alpha$ small enough so that
$$
\bar{M}_0 + \frac{2\alpha}{K_1} { K_3} < 2 \bar{M}_0, \quad \text{or, equivalently} \quad {\color{black} \alpha=\alpha({k_*})}<\frac{{\color{black} K_1(k_*)} \bar{M}_0}{2 {\color{black} K_3(k_*)}},
$$
which implies
\begin{equation*}
\frac{\mathrm{d}}{\mathrm{d} t} \mathcal{E}^n_{\gmax}(\alpha t,t) \leq - \frac{K_1}{2\alpha t} \left( \mathcal{E}^n_{\gmax}(\alpha t,t) - 2\bar{M}_0 \right).
\end{equation*}
As in \cite{GambaTask18}, integrating this differential inequality with an integrating factor $t^{\frac{K_1}{2\alpha}}$, yields
\begin{equation}\label{pomocna 26}
{ {\mathcal{E}^n_{ { {\color{black}\overline{\gamma}}} }(\alpha t,t)}} \leq \max\left\{ \mathcal{E}^n_{\gmax}(0,0), 2\bar{M}_0 \right\} \leq 2 \bar{M}_0, \qquad \forall t\in [0,\bar{T}_n],
\end{equation}
since $\mathcal{E}_ { {\color{black}\overline{\gamma}}} (0,0)=\mathfrak{m}_{0}(0)<2\bar{M}_0$.
\subsection*{Conclusion II} From \eqref{pomocna 26} the following bound on
${ {\mathcal{E}^n_{ { {\color{black}\overline{\gamma}}} }(\alpha t,t)}}$ holds
\begin{equation*}
\mathcal{E}^n_{\gmax}(\alpha t,t) \leq 2 \bar{M}_0 < 4\bar{M}_0, \qquad \forall t \in [0,\bar{T}_n].
\end{equation*}
Exploring the continuity of the partial sum
${ {\mathcal{E}^n_{ { {\color{black}\overline{\gamma}}} }(\alpha t,t)}}$
this inequality holds on a slightly larger interval, which contradicts maximality of $\bar{T}_n$, unless $\bar{T}_n=1$. Therefore, we can conclude $\bar{T}_n=1$ for all $n \in \mathbb{N}$, or in other words
\begin{equation*}
{ {\mathcal{E}^n_{ { {\color{black}\overline{\gamma}}} }(\alpha t,t)}} \leq 4\bar{M}_0, \qquad \forall t \in [0,1], \quad \forall n \in \mathbb{N}.
\end{equation*}
Letting $n\rightarrow \infty$, we conclude
\begin{equation}\label{pomocna 27}
{ {\mathcal{E}^n_{ { {\color{black}\overline{\gamma}}} }(\alpha t,t)}} \leq 4\bar{M}_0, \qquad \forall t \in [0,1].
\end{equation}
In particular, for time $t=1$, \eqref{pomocna 27} can be seen as an initial condition for propagation \eqref{initial data exp prop}, and thus the exponential moment of the order $ { {\color{black}\overline{\gamma}}} $ and a rate {\color{black} $0<\bar{\alpha}\leq \alpha(k_*)$} stays uniformly bounded for all $t>1$, {\color{black} for $k_*$ as in \eqref{kstar}}.
\bigskip
\section{Acknowledgements.}
The authors would like to thank Professor Ricardo J. Alonso for fruitful discussions on the topic. This work has been partially supported by NSF grants DMS 1715515 and RNMS (Ki-Net) DMS-1107444. Milana Pavi\'c-\v Coli\'c acknowledges the support of Ministry of Education, Science and Technological Development, Republic of Serbia within the Project No. ON174016. This work was completed while Milana Pavi\'c-\v Coli\'c was a Fulbright Scholar from the Univerisity of Novi Sad, Serbia, visiting the Institute of Computational Engineering and Sciences (ICES) at the University of Texas Austin co funded by a JTO Fellowship. ICES support is also gratefully acknowledged.
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{"url":"https:\/\/infoscience.epfl.ch\/record\/170458","text":"Infoscience\n\nJournal article\n\n# Sampling High-Dimensional Bandlimited Fields on Low-Dimensional Manifolds\n\nConsider the task of sampling and reconstructing a bandlimited spatial field in $\\Re^2$ using moving sensors that take measurements along their path. It is inexpensive to increase the sampling rate along the paths of the sensors but more expensive to increase the total distance traveled by the sensors per unit area, which we call the path density. In this paper we introduce the problem of designing sensor trajectories that are minimal in path density subject to the condition that the measurements of the field on these trajectories admit perfect reconstruction of bandlimited fields. We study various possible designs of sampling trajectories. Generalizing some ideas from the classical theory of sampling on lattices, we obtain necessary and sufficient conditions on the trajectories for perfect reconstruction. We show that a single set of equispaced parallel lines has the lowest path density from certain restricted classes of trajectories that admit perfect reconstruction. We then generalize some of our results to higher dimensions. We first obtain results on designing sampling trajectories in higher dimensional fields. Further, interpreting trajectories as $1$-dimensional manifolds, we extend some of our ideas to higher dimensional sampling manifolds. We formulate the problem of designing $\\kappa$-dimensional sampling manifolds for $d$-dimensional spatial fields that are minimal in manifold density, a natural generalization of the path density. We show that our results on sampling trajectories for fields in $\\Re^2$ can be generalized to analogous results on $d-1$-dimensional sampling manifolds for $d$-dimensional spatial fields.","date":"2016-12-04 04:48:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5528371334075928, \"perplexity\": 405.97941971417026}, \"config\": {\"markdown_headings\": true, \"markdown_code\": false, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2016-50\/segments\/1480698541187.54\/warc\/CC-MAIN-20161202170901-00502-ip-10-31-129-80.ec2.internal.warc.gz\"}"} | null | null |
{"url":"https:\/\/doc.dataiku.com\/dss\/latest\/preparation\/processors\/split.html","text":"Split column\u00b6\n\nSplit a column into several columns on each occurrence of the delimiter. The output columns are numbered: The first chunk will be in prefix_0, the second in prefix_1, and so on.\n\nExamples\u00b6\n\n\u2022 Split col=a\/b\/c using \/ as the delimiter and chunk as the output column prefix\n\n\u2022 Output: chunk_0=a, chunk_1=b, chunk_3=c\n\n\u2022 Split col=a\/b\/c using \/ as the delimiter, chunk as the output column prefix, and keep 2 columns from the beginning\n\n\u2022 Output: chunk_0=a, chunk_1=b\n\nOptions\u00b6\n\nDelimiter\n\nSeparates values from each input column within the output.\n\nOutput columns prefix\n\nAdd a prefix to identify the output columns.\n\nOutput as\n\nOutput the result(s) of the split as separate columns or as an array (A-B[\"A\",\u201dB\u201d]).\n\nTruncate\n\nLimit the number of output columns and keep only the first N columns or the N last columns.\n\nKeep empty chunks\n\nPreserve empty chunks between consecutive delimiters. (App, delimiter p[\"A\", \u201c\u201d, \u201c\u201d])","date":"2023-02-05 01:32:41","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.23469102382659912, \"perplexity\": 9519.734707791005}, \"config\": {\"markdown_headings\": false, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2023-06\/segments\/1674764500158.5\/warc\/CC-MAIN-20230205000727-20230205030727-00577.warc.gz\"}"} | null | null |
{"url":"https:\/\/math.stackexchange.com\/questions\/2988140\/finding-the-100-th-term-of-1-3-4-9-10-12-13-dots-powers-of-3-or-sums-of","text":"# Finding the $100$-th term of $1,3,4,9,10,12,13\\dots$ (powers of $3$, or sums of distinct powers of $3$)\n\nThe increasing sequence $$1,3,4,9,10,12,13\\dots$$ consists of all positive integers which are power of $$3$$ or sums of distinct powers of $$3$$. Find the $$100$$-th term of this sequence.\n\n\u2022 What's a hundred in base $2$? \u2013\u00a0Lord Shark the Unknown Nov 7 '18 at 4:46\n\u2022 @Lord Shark the Unknown: Bet you think that you are clever. Well, you are. So there! \u2013\u00a0marty cohen Nov 7 '18 at 5:40\n\nIt's clear the numbers are:\n\n$$1, 3, 3^2, 3^3, .....$$\n\n$$1+3, 1+ 3^2, 1 + 3^3, ...$$\n\n$$1 + 3 +3^2, 1 + 3 + 3^2,....$$\n\nthe only issue is what order do we put them in?\n\nThe thing to note is that we can't have anything $$+ 3^k$$ until we have $$3^k$$ first and $$1 + 3 + 3^2 + .... + 3^k < 3^{k+1}$$ so we must got through everything $$+ 3^k$$ before we can get to $$3^{k+1}$$.\n\nSo we have a rather recursive pattern.\n\n$$k = 0$$; First term: $$1$$\n\n$$k = 1$$;Term 2-3; $$3; 3+1$$\n\n$$k = 2$$; Term 4-7: $$3^2; 3^2 + 1; 3^2 + 3; 3^2 + 3 + 1$$\n\n$$k = 3$$; Term 8-15: $$3^3; 3^3 + 1; 3^3 + 3; 3^3 + 3 + 1; 3^3 + 3^2; 3^3 + 3^2 + 1; 3^3 + 3^2 + 3; 3^3 + 3^2 + 3 + 1$$.\n\n....etc.\n\nSo of the group $$k=k$$ we have stuff $$+3^k$$ and the \"stuff\" can be for each $$0\\le i < k$$ that we either add $$3^i$$ or .. we don't. So there are $$2^{k}$$ in the $$k$$ group. And $$2^{k-1}$$ in the $$k-1$$ group.\n\nAnd before the $$k$$th group we have $$1 + 2 + 2^2 + ... + 2^{k-1} =2^k - 1$$ terms. So the $$k$$th group starts and $$2^k$$ and goes to for $$2^k$$ terms to end at $$2^k + (2^k -1) = 2^{k+1} -1$$.\n\nAnd we stuff the $$k$$ group in order of fitting the subgroups of lower values than $$k$$ in them first.\n\n$$3^k, 1+3^k, 3+3^k, 1+3 + 3^k , 3^2 + 3^k, etc;$$\n\nAnd each sum in the $$k$$ group is of the form $$b_0*1 + b_1*3 + b_2*3^2 + ..... + b_{k-1}*3^{k-1} + 3^k; b_i = \\{0,1\\}$$ with the $$b_0*1 + b_1*3 + b_2*3^2 + ..... + b_{k-1}*3^{k-1}$$ terms ordered as they were in the $$k-1$$ group.\n\nSo as $$64 < 100< 128$$ so the 100th term is $$3^6 + something$$.\n\nSo the $$1$$st term is $$1$$. The second is $$3$$. The $$4$$th is $$3^2$$ and the $$64$$ is $$3^6$$. Then $$64 + 32=96$$th term is $$3^6+3^5$$ and $$64+32 + 4=100$$ is $$3^6 + 3^5 + 3^2$$.\n\nThen we give ourselves a giant dope slap to the head because this is clearly just binary representation.\n\nThat is: If $$n = \\sum_{i = 0}^m b_i*2^i; b_i=\\{0,1\\}$$ be the binary representation of $$n$$ then $$a_n$$ is $$\\sum_{i=0} b_i 3^i$$ And that preserves order.\n\nNote that, if you represent the numbers in base $$3$$, that you get the numbers $$1, 10, 11, 100 ...$$ which are numbers in base $$3$$ such that each of the digits is either $$1$$ or $$0$$.\n\nTherefore, we just need to find the $$100$$th number in the series and convert it from base $$3$$ to base $$10$$. This is equivalent to finding the $$100$$th binary number, which is $$1100100$$. Therefore, because we express this number in base 3 the $$100$$th term in the series is $$1100100$$ in base 3, which is $$981$$.\n\n\u2022 you are talking about base 3 then using binary ? \u2013\u00a0maveric Nov 7 '18 at 7:12\n\u2022 Andin last line it should not be 11001003 but 1100100 \u2013\u00a0maveric Nov 7 '18 at 7:18\n\nTry this question first: how many numbers can be derived with $$3^i$$ for $$i$$ from $$0$$ to $$n$$ using the way described in OP? Or a different but very similar question how many such numbers are there that are less then $$3^n$$\n\nYou have\n\n$$3^0, 3^1, 3^2, \\cdots, 3^n$$\n\nThe nature of the numbers present in the sequence determines that however you pick numbers from it, a different combination (of the numbers you picked) will result in a different sum. For each number in the sequence, you either pick it or leave it. So numbers derived from the sequence, including numbers already in it, are $$2^n-1$$, minus $$1$$ because we are not allow to picking no numbers at all.\n\nThis is what Lord Shark the Unknown meant by the binary representation of 100.\n\n\\begin{align} \\text{Seq}\\hspace{1cm} &\\text{Binary} & n\\text{th Number}&\\\\ 1\\hspace{1cm}& 1& 1\\hspace{0.5cm}&\\\\ 2\\hspace{1cm}& 10 &3\\hspace{0.5cm}& \\text{bin.rep 1 with 1 trailing zero}=3^1\\\\ 3\\hspace{1cm}& 11 &4\\hspace{0.5cm}& \\\\ 4\\hspace{1cm}& 100 &9\\hspace{0.5cm}& \\text{bin.rep. 1 with 2 trailing zero} =3^2\\\\ 5\\hspace{1cm}& 101 &10\\hspace{0.5cm}& 10=9+1\\\\ 6\\hspace{1cm}& 110 &12\\hspace{0.5cm}& \\text{binary 100+binary 10, sum of corresponding numbers } 9+3=12\\\\ 7\\hspace{1cm}& 111 &13\\hspace{0.5cm}& \\text{binary 100+binary 11, sum of corresponding numbers }9+4=13\\\\ 8\\hspace{1cm}& 1000 &27\\hspace{0.5cm}& \\text{binary rep 1 with 3 trailing zero}=3^3\\\\ \\end{align}\n\n$$100=64+32+4=2^6+2^5+2^2$$ so the number you are looking at is $$3^6+3^5+3^2$$\n\nThis technique also applies if a different base is used.","date":"2019-07-21 03:26:39","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 82, \"wp-katex-eq\": 0, \"align\": 1, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.9314337372779846, \"perplexity\": 595.6807354169508}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-30\/segments\/1563195526818.17\/warc\/CC-MAIN-20190721020230-20190721042230-00172.warc.gz\"}"} | null | null |
Boasting a garden private pool and sea views Tower Villa is a villa in Capilungo. Guests staying at this villa have access to a fully equipped kitchen and a patio. The villa is equipped with a TV. Free private parking is available at the villa. Tower Villa offers an outdoor pool. Lecce is 47 km from the accommodation while Gallipoli is 14 km away. Brindisi - Salento Airport is 80 km from the property. | {
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\section{APPENDIX}
\input{convergence}\label{App:convergence}
\iffalse
\section{Coordinate Descent for solving lasso problem}\label{App:CD}
Given a data point $\mathbf{X} = (\mathbf{x}_1, \mathbf{x}_2, \cdots, \mathbf{x}_n) \in \mathbb{R}^{m\times{N}}$ and a dictionary $\mathbf{D} \in \mathbb{R}^{m\times t}$, the lasso problem is given as follows:
\begin{equation}
\min_\mathbf{Z} f(\mathbf{Z}) = \frac{1}{2}||\mathbf{DZ} - \mathbf{X}||_2^2 + \lambda||\mathbf{Z}||_1,
\label{eq:A1}
\end{equation}
where $\mathbf{Z}=(\mathbf{z}_{1}, \cdots, \mathbf{z}_{t}) \in \mathbb{R}^{N\times{t}}$.
Suppose we freeze all columns of $\mathbf{Z}$ except the $j$-th column $\mathbf{z}_j$ in Eq.~\ref{eq:A1}. Let $\mathbf{d}_j$ denote the $j$-th column of $\mathbf{D}$ and $d_{i, j}$ is the element of $i$-th row and $j$-th column in $\mathbf{D}$. Therefore, we have
\begin{align*}
& \argmin_{\mathbf{z}_j}\frac{1}{2} (d_{i, j}\mathbf{z}_j - \mathbf{x}_i)^2 + \lambda|\mathbf{z}_j|\\
= &\argmin_{\mathbf{z}_j}\frac{1}{2} (\mathbf{z}_j^2 - 2b_j\mathbf{z}_j + {b_j}^2) + \lambda|\mathbf{z}_j| \\
= &\argmin_{\mathbf{z}_j}\frac{1}{2} (\mathbf{z}_j - b_j)^2 + \lambda|\mathbf{z}_j|,
\end{align*}
where $b_j = \sum_{i=1}^m(\mathbf{x}_i - \sum_{k\neq j} d_{ik}\mathbf{z}_k)d_{ij}$ and we use the condition that each column of $D$ is unit norm. Then $z_j$ has an optimal solution: $\mathbf{z}_j = h_{\lambda}(b_j)$, where $h_\lambda$ is a soft thresholding shrinkage function so called the proximal operator of the $\ell_1$ norm~\citep{Combettes2005-MMS}. The definition of $h_\lambda$ is as follows:
\small
\[
h_{\lambda}(v)=\left\{
\begin{array}{ll}
v + \lambda, &v < -\lambda \\
0, &-\lambda\leq v \leq \lambda\\
v - \lambda, &v < \lambda
\end{array}
\right.
\]
Note that $b_j = \mathbf{d}_j^T\mathbf{x}_i - \mathbf{d}_j^T\mathbf{Dz}_j + (\mathbf{d}_j^T\mathbf{d}_j)\mathbf{z}_j = \mathbf{d}_j^T(\mathbf{x}_i - \mathbf{D}\mathbf{z}_j) + \mathbf{z}_j$. Therefore, the computational cost of updating the $j$-th coordinate $\mathbf{z}_j$ depends on computing the vector $\mathbf{x}_i - \mathbf{Dz}$ and the inner product $\mathbf{d}_j^T(\mathbf{x}_i - \mathbf{Dz})$.
\section{Updating the dictionary}\label{App:dictionary}
It is known that updating the sparse code (step 2) is the most time consuming part~\citep{balasubramanian2013smooth}. CD~\citep{TTW08a} is known as one of the state-of-the-art method for solving this lasso problem. Given an input vector $\mathbf{x}_i$, CD initializes $\mathbf{z}_i^0 = 0$ and then updates the sparse code many times via matrix-vector multiplication and thresholding. Empirically, the iteration may take tens of hundreds of steps to converge. However, we observe that after a few steps, the supports of the coordinates are very accurate and usually take less then ten steps. Moreover, since the original sparse coding involves an alternating updating, we do not need to run the CD to final convergence during this updating procedure. Therefore, we propose to update the sparse code $\mathbf{z}_i^{k-1}$ by using only a few steps of CD and $\mathbf{z}_i^{k-1}$ is an initial sparse code for updating $\mathbf{z}_i^{k}$.
After updating the sparse code, we get its supports to update the dictionary. One of our key insights is that we only need to focus on the supports of the dictionary instead of all columns of the dictionary. Let $z_{i, j}^k$ denote $j$-th entry of $\mathbf{z}_i^k$ and $d_{i,j}^k$ denote the $j$-th column of the dictionary $\mathbf{D}_{i}^k$. If $z_{i, j}^k = 0$, then $\nabla_{d_{i,j}^k} f_i(\mathbf{D}_{i}^k, \mathbf{z}_i^k) = (\mathbf{D}_{i}^k \mathbf{z}^k_i - \mathbf{x}_i) z^k_{i,j} = 0$. Therefore, $d_{i,j}^k$ does not need to be updated. If $z_{i, j}^k \neq 0$, we can update $d_{i+1,j}^k$ (the $j$-th column of the dictionary $\mathbf{D}_{i+1}^k$) as follows:
\small
\begin{equation}\label{eq:update-dict}
d_{i+1, j}^k \leftarrow d_{i, j}^k - \eta_{i,j}^k \nabla_{d_{i, j}^k} f_i(\mathbf{D}^k_i, \mathbf{z}_i^k) = d^k_{i,j} - \eta_{i,j}^k z_{i,j}^k(\mathbf{D}^k_i\mathbf{z}_i^k - \mathbf{x}_i).
\end{equation}
\normalsize
Note that $\mathbf{z}_i^k$ is a sparse vector, therefore computing $\mathbf{D}^k_i\mathbf{z}_i^k$ is very efficient. $\eta_{i,j}^k$ is the learning rate of the $j$-column for $i$-th input in $k$-th epoch. The computational cost will be significantly reduced when there are limited supports. In contrast, ODL usually has to update all columns of the dictionary. It is because that ODL uses the averaged gradient, which means the supports of the dictionary is itself. Therefore, one has to update all columns of the dictionary and it is time consuming especially when the dictionary size is very large.
When the dataset is very large, the learning rate $\eta^k_{i,j}$ will be very small after going through large number of input vectors. In this case, the dictionary will not change very much and the efficiency of the training will decrease. Therefore, we use an adaptive learning rate in this paper. We aim to design a learning rate with the following two principals. The first one is that for different columns of the dictionary, we may use different learning rates. The second is that for the same column, the learning rate should decrease, otherwise the algorithm might not converge. To obtain the learning rate, we use the Hessian matrix of the objective function. It can be shown that the following matrix provides an approximation of the Hessian:
$
\mathbf{H} = \sum_{k,i} \mathbf{z}^k_i(\mathbf{z}^k_i)^T,
$
when $k$ and $i$ go to infinity. According to the second order SGD, we should use the inverse matrix of the Hessian as the learning rate.
However, computing a matrix inversion problem is computationally expensive. In order to obtain the learning rate, we simply use the diagonal element of the matrix $\mathbf{H}$. Note that if the columns of the dictionary have low correlation, $\mathbf{H}$ is close to a diagonal matrix. Specifically, we first initialize $\mathbf{H} = 0$. Then update the matrix $\mathbf{H}$ as follows:
\small
\begin{eqnarray}
\mathbf{H} &\leftarrow& \mathbf{H} + \mathbf{z}^k_i(\mathbf{z}_i^k)^T,
\end{eqnarray}
\normalsize
when updating the $j$-th column for the $i$-th input vector $\mathbf{x}_i$, we replace $\eta_{i, j}^k$ in Eq.~\ref{eq:update-dict} by $1/h_{jj}$, where $h_{jj}$ is the $j$-th diagonal element of $\mathbf{H}$. In this way, we do not have to tune the learning rate parameter. It might be worth noting that we do not have to store the whole matrix of $\mathbf{H}$ but only its diagonal elements.
\fi
\section{Conclusions and Future Works}
Here we present a hyperbolic sparse coding with ring-shaped patch selection algorithm, which may improve the
accuracy for AD diagnosis and prognosis with sMRI biomarkers. In the future, we will explore whether the proposed framework will work with other shape statistics, such as spherical harmonics and radial distance. In our previous preclinical AD study~\citep{dong2019applying}, we found APOE-e4 does effects on hippocampal morphometry of cognitively unimpaired subjects. In our future work, we will further explore whether the proposed FPSBS and HSC methods are useful for such preclinical AD study.
\section{Convergence Analysis of Stochastic Coordinate Coding}
\label{App:converge}
Here we show that our algorithm is convergent. The objective function ${\cal F}$ in Eq.~\ref{eq:opt} can be re-written as follows:
\small
\begin{equation}\label{eq:obj-scc-ca}
\mathcal{F}(\mathbf{D}, \mathbf{z}_1, \cdots, \mathbf{z}_n) = \frac{1}{n}\sum_{i=1}^n (\frac{1}{2}\| \mathbf{Dz}_i - \mathbf{x}_i \|^2_2 + \lambda \| \mathbf{z}_i \|_1).
\end{equation}
\normalsize
It is clear that ${\cal F}(\mathbf{D}, \mathbf{z}_1, \cdots, \mathbf{z}_n)$ is a nonnegative continuous function over a bounded set $\mathbf{D}\in \mathbb{C}$ and
$\|\mathbf{z}_i\|\le M$ for a real number $M<\infty$, ${\cal F}(\mathbf{D}, \mathbf{z}_1, \cdots, \mathbf{z}_n)\to \infty$ if $\|\mathbf{z}_i\|_1
\to \infty$. Thus, the minimization problem (Eq.~\ref{eq:obj-scc-ca}) has a solution. As the minimization functional ${\cal F}$ is not a convex function, the problem (Eq.~\ref{eq:obj-scc-ca}) may have multiple solutions and we show our algorithm convergence under certain conditions.
The proof is divided into three parts. We first show the convergence analysis for updating sparse codes (CD step) and updating dictionary (SGD step), respectively. We then combine these two parts to show the convergence of the stochastic coordinate coding (SCC).
\subsection{Convergence Analysis of the CD Step}
First, we analyze the convergence of CD step. Suppose we update the $j$-th coordinate $z^k_{i, j}$ of $\mathbf{z}^k_i$ in our cyclic selection approach. It is clear that
\small
\[
z^k_{i, j} = \arg\min_{\mathbf{z}} f_i(\mathbf{D}^k_i, z^{k}_{i, 1}, \cdots, z^{k}_{i, j-1}, z_{i, j}^k , z^{k-1}_{i, j+1}, \cdots, z^{k-1}_{i, t}).
\]
\normalsize
Therefore, after going through the whole cycle, we obtain the following result.
\normalsize
\begin{Prop}\label{prop:cd--z-dec}
\textbf{For any $k$ and $i$, we have}
\small
\begin{equation}\label{eq:cd-z-dec}
f_i (\mathbf{D}^k_{i}, \mathbf{z}_i^{k}) \leq f_i (\mathbf{D}^k_{i}, \mathbf{z}^{k-1}_i ).
\end{equation}
\normalsize
\textbf{Further, we have $f_i (\mathbf{D}^k_{i}, \mathbf{z}_i^{k}) \leq f_i (\mathbf{D}^k_{i}, \mathbf{z}_i^{k-1}) - \gamma\|\mathbf{z}_i^k- \mathbf{z}_i^{k-1}\|^2$ for a constant $\gamma>0$.}
\end{Prop}
\emph{Proof.}
\textbf{For simplicity, we let $f_i(\mathbf{D}^k_{i},\mathbf{z}^k_i) = \ell(\mathbf{D}^k_{i},\mathbf{z}^k_i) + \lambda \|\mathbf{z}_i^k \|_1$. Then we can use Taylor expansion of $\ell(\mathbf{D}^k_{i},\mathbf{z}^k_i)$ at $\mathbf{z}_i^{k-1}$ to rewrite $\ell(\mathbf{D}^k_{i}, \mathbf{z}_i^k) $ into}
\small
\begin{equation}
\begin{aligned}
\label{eq:CD1}
&\ell(\mathbf{D}^k_{i}, \mathbf{z}_i^k) = \ell(\mathbf{D}^k_i,\mathbf{z}_i^{k-1}) + \langle \nabla \ell(\mathbf{D}^k_{i},\mathbf{z}_i^{k-1}), \mathbf{z}_i^k-\mathbf{z}_i^{k-1}\rangle + \frac{1}{2}\|\mathbf{D}_i^k (\mathbf{z}_i^k-\mathbf{z}_i^{k-1})\|^2
\\ \leq & \ell(\mathbf{D}^k_i,\mathbf{z}_i^{k-1}) + \langle \nabla \ell(\mathbf{D}^k_{i},\mathbf{z}_i^{k-1}), \mathbf{z}_i^k-\mathbf{z}_i^{k-1}\rangle
+ \frac{\| \mathbf{D}_i^k\|^2 }{2}\|\mathbf{z}_i^k-\mathbf{z}_i^{k-1}\|^2
\end{aligned}
\end{equation}
\normalsize
\textbf{As shown in~\citep{lin2014stochastic}, $\mathbf{z}_i^k$ is a minimizer of Eq.~\ref{eq:cd-obj-a}. Therefore, we have the following equation:}
\small
\begin{equation}
\begin{aligned}
\label{eq:CD2}
&\mathrm{CD}( \mathbf{z}_i^{k}, \mathbf{D}^k_i, \mathbf{x}_i, \mathbf{z}_i^{k-1} )
= \langle \nabla \ell(\mathbf{D}^k_{i},\mathbf{z}_i^{k-1}), \mathbf{z}_i^k-\mathbf{z}_i^{k-1}\rangle + \frac{L}{2}\|\mathbf{z}_i^k- \mathbf{z}_i^{k-1}\|^2 + \lambda \| \mathbf{z}_i^k \|_1 \\
\leq &\mathrm{CD}( \mathbf{z}_i^{k-1}, \mathbf{D}^k_i, \mathbf{x}_i, \mathbf{z}_i^{k-1} )
= \lambda \| \mathbf{z}_i^{k-1}\|_1.
\end{aligned}
\end{equation}
\normalsize
\textbf{We then add Eq.~\ref{eq:CD1} and Eq.~\ref{eq:CD2} together, and simplify the inequality, we have}
\small
\begin{align*}
\ell(\mathbf{D}^k_{i},\mathbf{z}_i^k) + \lambda \| \mathbf{z}_i^k \|_1& \le \ell(\mathbf{D}^k_{i},\mathbf{z}_i^{k-1}) + \lambda \| \mathbf{z}_i^{k-1} \|_1 - \frac{L-\| \mathbf{D}_i^k \|^2}{2} \|\mathbf{z}_i^k- \mathbf{z}_i^{k-1}\|^2 \\
f(\mathbf{D}^k_{i},\mathbf{z}_i^{k}) &= f(\mathbf{D}^k_{i},\mathbf{z}_i^{k-1}) - \gamma \|\mathbf{z}_i^k- \mathbf{z}_i^{k-1}\|^2,
\end{align*}
\normalsize
\textbf{where $L > ||\mathbf{D}_i^k||^2$ is the Lipschitz constant, so $\gamma = (L-\| \mathbf{D}_i^k \|^2)/2 > 0$. }
If there are $q$ steps using CD, we should have
\small
\begin{equation}
\label{keystep1}
f_i (\mathbf{D}^k_{i}, \mathbf{z}_i^{k+q}) \leq f_i (\mathbf{D}^k_{i}, \mathbf{z}^{k-1}_i ).
\end{equation
\hfill{$\square$}
\normalsize
When $q$ is sufficiently large, we know that $f_i(\mathbf{D}_i^k,\mathbf{z}_i^{k+q})$ decreases to the minimum value $f_i^* = \min_{z} f_i(\mathbf{D}_i^k,\mathbf{z})$.
Since $f_i(\mathbf{D}_i^k, \mathbf{z})$ is a convex function, $f^*_i \le f_i(\mathbf{D}_i^k,0) = \dfrac{1}{2}\|\mathbf{x}_i\|^2 =1/2$. It follows that $\|\mathbf{z}_i^{k+q}\|_1 \le \frac{1}{2\lambda}$. This analysis works for all $i=1, \cdots, n$ and for any $k, q$. We take enough steps so that $f_i(\mathbf{D}^k_i, \mathbf{z}_i^{k+q}) \le 1/2$. Therefore, we have the following results for the output $\mathbf{z}_i^k$:
\begin{Prop}
\label{bound}
If the number of iterative steps for the CD is sufficiently large, all $\mathbf{z}_i^k$ are uniformly bounded for $i=1, \cdots, n$ and $k\ge 1$.
\end{Prop}
It might be worth noting that in practice performing only a small number steps of CD is sufficient to guarantee that all $\mathbf{z}_i^k$ are uniformly bounded.
\subsection{Convergence Analysis of the SGD Step}
Second, we study the SGD step. Note that we can always re-index $\mathbf{D}^k_i$ as $\mathbf{D}_{(k-1)n+i+1}$ for convenience. To simplify the notation, we omit the superscript $k$ on the dictionary $\mathbf{D}$ and the learning rate $\eta$ in this section.
Our SGD step in Eq.~\ref{eq:sgd} is equivalent to the following two sub-steps by using proximal gradient method~\citep{parikh2014proximal}.
\small
\begin{equation}\label{eq:sgd-step1}
\hat{\mathbf{D}}_{i+1} = \arg \min_\mathbf{D} g_i(\mathbf{D}_i) + \langle \nabla g_i(\mathbf{D}_i), \mathbf{D} - \mathbf{D}_i \rangle + \frac{1}{2\eta_i} \| \mathbf{D} - \mathbf{D}_i \|^2_2,
\end{equation}
\begin{equation}\label{eq:sgd-step2}
\mathbf{D}_{i+1} = P_{\mathbb{C}} (\hat{\mathbf{D}}_{i+1}) = \arg\min_{\mathbf{D} \in \mathbb{C}} \| \hat{\mathbf{D}}_{i+1} - {\mathbf{D}} \|.
\end{equation}
\normalsize
For simplicity, we let $g_{i}(\mathbf{D}) \equiv \frac{1}{2} \| \mathbf{Dz}^k_i - \mathbf{x}_i \|^2_2 $. Next we show that $g_i$ decreases after performing SGD.
\begin{Prop}
$g_i(\hat{\mathbf{D}}_{i+1}) \leq g_i(\mathbf{D}_i)$ if $\eta_i \le \frac{1}{L}$.
\end{Prop}
\emph{Proof.}
We use Taylor expansion of $g_i$ at $\mathbf{D}_i$ and the Lipschitz differentiation of $g_i(\mathbf{D})$, we have
\small
\[
\begin{aligned}
g_i(\hat{\mathbf{D}}_{i+1}) &= g_i (\mathbf{D}_i) + \langle \nabla g_i (\mathbf{D}_i), \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \rangle + \frac{L}{2} \| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \|^2 \\
&\leq g_i (\mathbf{D}_i) + \langle \nabla g_i (\mathbf{D}_i), \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \rangle + \frac{1}{2\eta_i} \| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \|^2 \\
&\leq g_i (\mathbf{D}_i) + \langle \nabla g_i (\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}_i \rangle + \frac{1}{2\eta_i} \| \mathbf{D}_i - \mathbf{D}_i \|^2 \\
&=g_i(\mathbf{D}_i),
\end{aligned}
\]
\normalsize
where the third inequality is due to the optimality condition of Eq.~\ref{eq:sgd-step1} and $L$ is Lipschitz constant.\hfill $\square$
\begin{Prop}\label{claim:sgd-dec}
$g_i({\mathbf{D}}_{i+1}) \leq g_i(\mathbf{D}_i)$ if $\eta_i \| \mathbf{z}^k_i \|^2\le 1$.
\end{Prop}
\emph{Proof.}
Note that $ \| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_{i+1} \| \leq \| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \|$ by Eq.\ref{eq:sgd-step2}. By a direct computation of CD in Eq.\ref{eq:sgd-step1}, we have
$
\hat{\mathbf{D}}_{i+1} = \mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i)
$
because $\hat{\mathbf{D}}_{i+1}$ is the proximation of $\mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i) $.
Thus
\small
\[
\begin{aligned}
\label{eq:above}
\| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_{i+1} \|^2 = \| \mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i) - \mathbf{D}_{i+1} \|^2 \leq \| \hat{\mathbf{D}}_{i+1} - \mathbf{D}_i \|^2 = \| \eta_i \nabla g_i(\mathbf{D}_i) \|^2.
\end{aligned}
\]
\normalsize
Then, we expand the left side of the above inequality and have
\small
\begin{equation}\label{eq:inequ-norm-inner}
\| \mathbf{D}_{i+1} - \mathbf{D}_i \|^2 \leq -2 \eta_i \langle \mathbf{D}_{i+1} - \mathbf{D}_i, \nabla g_i(\mathbf{D}_i)\rangle .
\end{equation}
\normalsize
According to the definition of $g_i$, we have
\small
\begin{eqnarray*}
g_i(\mathbf{D}_{i+1}) &=& \frac{1}{2} \| \mathbf{D}_{i+1} \mathbf{z}_i^k - \mathbf{x}_i \|^2 \\
&=& \frac{1}{2} \| (\mathbf{D}_{i+1} - \mathbf{D}_i) \mathbf{z}_i^k + \mathbf{D}_i\mathbf{z}_i^k - \mathbf{x}_i \|^2 \\
&=& \frac{1}{2} \| \mathbf{D}_{i+1} - \mathbf{D}_i\|^2 \| \mathbf{z}_i^k\|^2 + \langle (\mathbf{D}_{i+1} - \mathbf{D}_i) \mathbf{z}_i^k, \mathbf{D}_i\mathbf{z}_i^k - \mathbf{x}_i \rangle + g_i(\mathbf{D}_i) \\
&\leq& -\eta_i \| \mathbf{z}_i^k\|^2 \langle \mathbf{D}_{i+1} - \mathbf{D}_i, \nabla g_i(\mathbf{D}_i) \rangle + \langle (\mathbf{D}_{i+1} - \mathbf{D}_i) \mathbf{z}_i^k, \mathbf{D}_i\mathbf{z}_i^k - \mathbf{x}_i \rangle + g_i(\mathbf{D}_i) \\
&=& -\eta_i \| \mathbf{z}_i^k\|^2 \langle \mathbf{D}_{i+1} - \mathbf{D}_i, \nabla g_i(\mathbf{D}_i) \rangle + \langle \mathbf{D}_{i+1} - \mathbf{D}_i , \nabla g_i(\mathbf{D}_i)\rangle + g_i(\mathbf{D}_i) \\
&=& (1 -\eta_i \| \mathbf{z}_i^k\|^2) \langle \mathbf{D}_{i+1} - \mathbf{D}_i, \nabla g_i(\mathbf{D}_i) \rangle + g_i(\mathbf{D}_i)
\end{eqnarray*}
\normalsize
By Eq.~\ref{eq:inequ-norm-inner}, we get $ \langle \mathbf{D}_{i+1} - \mathbf{D}_i, \nabla g_i(\mathbf{D}_i) \rangle \leq -\frac{1}{2\eta_i}
\| \mathbf{D}_{i+1} - \mathbf{D}_i \|^2 \leq 0$. If $\eta_i \leq \frac{1}{\| \mathbf{z}_i^k \|^2}$, we have $g_i(\mathbf{D}_{i+1}) \leq g_i(\mathbf{D}_i)$.
\hfill $\square$
Since $\|\mathbf{z}_i^k\|_1, i\ge 1, k\ge 1$ are uniformly bounded as in Proposition~\ref{bound}, $g_i(\mathbf{D}_{i+1}) \le G$ for a positive constant $G$ independent of $i$. Furthermore, we have $\nabla g_i(\mathbf{D}_i^k) = (\mathbf{D}_i^k \mathbf{z}_i^k -\mathbf{x}_i)(\mathbf{z}_i^k)^T$ and it is easy to see that $\|\nabla g_i(\mathbf{D}_i^k)\|\le C^2$ for a positive constant $C$. Thus, we have
\begin{Prop}
Suppose $\|\nabla g_i(\mathbf{D}_i^k) \|^2 \leq C^2$ for all $i$ and $k$, then
$
\| \mathbf{D}_{i+1} - \mathbf{D}_i \| \leq 2\eta_i C.
$
\end{Prop}
\emph{Proof.}
With Eq.~\ref{eq:inequ-norm-inner}, we have
\small
\begin{align*}
\| \mathbf{D}_{i+1} - \mathbf{D}_i \|^2 &\leq -2 \eta_i \langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_{i+1} - \mathbf{D}_i \rangle \\
&\leq 2\eta_i \| \nabla g_i(\mathbf{D}_i) \| \, \|\mathbf{D}_{i+1} - \mathbf{D}_i \| \\
&\leq 2\eta_i C \|\mathbf{D}_{i+1} - \mathbf{D}_i \|.
\end{align*}
\normalsize
The conclusion follows dividing both sides by $\|\mathbf{D}_{i+1} - \mathbf{D}_i\|$ if it is not zero. If it is zero, then the inequality in the proposition is obviously valid. \hfill $\square$
\iffalse
One can further prove that $\mathbf{D}_i$ is square summable, we then have the following
\begin{Coro}
\label{coro:di-d}
$$
\sum_{i=1}^\infty \| \mathbf{D}_i- \mathbf{D}_{i+1}\|^2 \le 4 C^2 \sum_{i=1}^\infty \frac{1}{(i+1)^2}<4C^2.
$$
\end{Coro}
\emph{Proof.}
Indeed, we have
\small
\begin{align*}
\| \mathbf{D}_{i+1} - \mathbf{D}_i \|^2 &\leq 2\eta_i \| \nabla g_i(\mathbf{D}_i) \| \, \|\mathbf{D}_{i+1} - \mathbf{D}_i \| \\
&\le 2C^2 \eta_i^2 + \frac{1}{2} \|\mathbf{D}_{i+1} - \mathbf{D}_i \|^2
\end{align*}
\normalsize
by using Cauchy-Schwarz inequality.
$
\|\mathbf{D}_{i+1} - \mathbf{D}_i \|^2 \le 4 C^2 \eta_i^2.
$
Summing over $i\ge 1$ concludes the desired inequality.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{$\square$}
\fi
\subsection{Convergence Analysis of the HSC}
Now we can give the critical decreasing proposition.
\begin{Prop}\label{prop:exp-dic-dec}
For any epoch $k\ge 1$, we have
\small
\[
\frac{1}{n} \sum_{i=1}^n f_i(\mathbf{D}^k_{i+1}, \mathbf{z}_i^k) \leq \frac{1}{n} \sum_{i=1}^n f_i(\mathbf{D}^k_i, \mathbf{z}_i^{k-1}).
\]
\normalsize
\end{Prop}
\emph{Proof.}
With Proposition \ref{prop:cd--z-dec} we have
\small
\begin{equation}\label{eqn:fidizifidizik-1}
f_i(\mathbf{D}_{i}^k, \mathbf{z}_i^k) \leq f_i (\mathbf{D}_i^k, \mathbf{z}_i^{k-1}),
\end{equation}
\normalsize
With Proposition~\ref{claim:sgd-dec}, we have
\small
\begin{equation}
g_i(\mathbf{D}_{i+1}^k, \mathbf{z}_i^k) + \lambda \| \mathbf{z}_i^k \|_1\leq g_i (\mathbf{D}_i^k, \mathbf{z}_i^k) +\lambda \| \mathbf{z}_i^{k} \|_1,
\end{equation}
\normalsize
It is equivalent to
\small
\begin{equation}\label{eqn:gibyprop5}
f_i(\mathbf{D}_{i+1}^k, \mathbf{z}_i^k) \leq f_i (\mathbf{D}_i^k, \mathbf{z}_i^k).
\end{equation}
\normalsize
Combining Eq.~\ref{eqn:fidizifidizik-1} and Eq.~\ref{eqn:gibyprop5} and summing from $i=1$ to $n$, we get the desired inequality. \\\hfill $\square$
\iffalse
\begin{Prop}
\label{prop:di-d*}
Suppose that $\mathbf{D}^*\in \mathbb{C}$ is a local minimizer such that the mean value ${E}(\nabla g_i(\mathbf{D}^*))=0$ over random variables $(\mathbf{z}_i, \mathbf{x}_i)$. Suppose $ g_i(\mathbf{D})$ is strongly convex, i.e. we use parameter $\beta>0 $. Let $\eta_i = \frac{1}{(i+1)\beta }$ for any $i$ and $k$, then the mean of the difference $\|\mathbf{D}_i - \mathbf{D}^*\|$ is ${E}(\| \mathbf{D}_{i} - \mathbf{D}^* \|) \leq \frac{1}{i}(1+ \frac{C^2}{\beta^2}).$
\end{Prop}
\emph{Proof.}
Due to the non-expansive property of the projection $P_{\mathbb{C}}$, we have
\small
\begin{align*}
&\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2 = \| P_{\mathbb{C}}(\mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i)) - P_{\mathbb{C}}(\mathbf{D}^*) \|^2 \\
\leq &\|\mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i) - \mathbf{D}^* \|^2 \\
= &\| \mathbf{D}_i - \mathbf{D}^* \|^2 + \eta_i^2 \| \nabla g_i(\mathbf{D}_i) \|^2 - 2\eta_i \langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle \\
\leq & \| \mathbf{D}_i - \mathbf{D}^* \|^2 + \eta_i^2 C^2 - 2\eta_i \langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle.
\end{align*}
\normalsize
Due to the strong convexity~\citep{boyd2004convex} of $g_i$, we compute the mean of $\langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle$ over $(\mathbf{z}_i, \mathbf{x}_i)$ so that
\small
\begin{align*}
{ E}(\langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle) & = { E}( \langle \nabla g_i(\mathbf{D}_i) - \nabla g_i(\mathbf{D}^*), \mathbf{D}_{i}-\mathbf{D}^*\rangle) \\
&\ge \frac{\beta}{2} { E}( \| \mathbf{D}_i - \mathbf{D}^* \|^2).
\end{align*}
\normalsize
Combining the above inequalities, we have
\small
\begin{align*}
{ E}(\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2) &\leq { E}(\| \mathbf{D}_i - \mathbf{D}^* \|^2) + \eta_i^2 C^2 - \eta_i\beta{ E}( \|\mathbf{D}_i - \mathbf{D}^*\|^2)\\
&= (1-\beta \eta_i) { E}(\| \mathbf{D}_i - \mathbf{D}^* \|^2) + \eta_i^2 C^2.
\end{align*}
\normalsize
By induction, it follows that
\small
\begin{align*}
{ E}(\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2) &\le \prod_{j=1}^i (1-\beta\eta_j) {E}(\|\mathbf{D}_1- \mathbf{D}^*\|^2)\cr
&\quad + \sum_{j=1}^i \prod_{k=j+1}^i (1-\beta\eta_k) C^2 \eta_j^2.
\end{align*}
\normalsize
Thus, ${ E}(\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2 ) \leq \frac{1}{i+1} {E}(\|\mathbf{D}_1- \mathbf{D}^*\|^2) + \frac{C^2}{\beta^2}\frac{1}{i+1},$
where
$\prod_{j=1}^i(1-\frac{1}{j+1})=\frac{1}{i+1}$.{$\square$}
\fi
\begin{Prop}
\label{prop:di-d*}
Suppose that $\mathbf{D}^*\in \mathbb{C}$ (recall $\mathbb{C}$ is the feasible set of $\mathbf{D}$ in Eq.~\ref{eq:sgd}) is a local minimizer such that the mean value ${\cal E}(\nabla g_i(\mathbf{D}^*))=0$ over random variables $(\mathbf{z}, \mathbf{x})$.
Suppose $ g_i(\mathbf{D})$ is strongly convex~\citep{boyd2004convex}, so there exist $\eta_i = \frac{1}{(i+1)\beta } (\beta > 0)$ for any $i$ and $k$. Then the mean of the difference $\|\mathbf{D}_i - \mathbf{D}^*\|$ is
\[
{\cal E}(\| \mathbf{D}_{i} - \mathbf{D}^* \|) \leq \frac{1}{i+1}(1+ \ln(i+1) \frac{G^2}{\beta^2}),
\]
where in Prop.~\ref{bound}, $g_i(\mathbf{D}_{i+1}) \le G$ for a positive constant $G$ independent of $i$.
\end{Prop}
\begin{proof}
Due to the non-expansive property of the projection $P_{\mathbb{C}}$, we have
\begin{align*}
\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2 &= \| P_{\mathbb{C}}(\mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i)) - P_{\mathbb{C}}(\mathbf{D}^*) \|^2 \\
&\leq \|\mathbf{D}_i - \eta_i \nabla g_i(\mathbf{D}_i) - \mathbf{D}^* \|^2 \\
&= \| \mathbf{D}_i - \mathbf{D}^* \|^2 + \eta_i^2 \| \nabla g_i(\mathbf{D}_i) \|^2 - 2\eta_i \langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle \\
&\leq \| \mathbf{D}_i - \mathbf{D}^* \|^2 + \eta_i^2 C^2 - 2\eta_i \langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle.
\end{align*}
Due to the strong convexity of $g_i$, we compute the mean of $\langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle$ over $(\mathbf{z}_i, \mathbf{x}_i)$ so that
\begin{align*}
{\cal E}(\langle \nabla g_i(\mathbf{D}_i), \mathbf{D}_i - \mathbf{D}^* \rangle) = {\cal E}( \langle \nabla g_i(\mathbf{D}_i) - \nabla g_i(\mathbf{D}^*), \mathbf{D}_{i}-\mathbf{D}^*\rangle) \ge
\frac{\beta}{2} {\cal E}( \| \mathbf{D}_i - \mathbf{D}^* \|^2).
\end{align*}
Combining the above inequalities, we have
\begin{align*}
{\cal E}(\| \mathbf{D}_{i+1} - \mathbf{D}\|^2) &\leq {\cal E}(\| \mathbf{D}_i - \mathbf{D} ^*\|^2) + \eta_i^2 C^2 - \eta_i\beta{\cal E}( \|\mathbf{D}_i - \mathbf{D}^*\|^2)\\
&\leq (1-\beta \eta_i) {\cal E}(\| \mathbf{D}_i - \mathbf{D}^*\|^2) + \eta_i^2 C^2.
\end{align*}
By induction, it follows that
\begin{equation*}
{\cal E}(\| \mathbf{D}_{i+1} -\mathbf{D}\|^2)\le \prod_{j=1}^i (1-\beta\eta_j) {\cal E}(\|\mathbf{D}_1- \mathbf{D}^*\|^2) + \sum_{j=1}^i \prod_{k=j+1}^i (1-\beta\eta_k) C^2 \eta_j^2.
\end{equation*}
Thus, we have
\[
{\cal E}(\| \mathbf{D}_{i+1} - \mathbf{D}^* \|^2 ) \leq \frac{1}{i+1} {\cal E}(\|\mathbf{D}_1- \mathbf{D}^*\|^2) + \frac{C^2}{\beta^2}\frac{\ln(i+1)}{i+1},
\]
where we have used $\displaystyle \sum_{j=1}^{i+1}\frac{1}{j} > \ln (i+1) + O(1/i+1)$~(\cite{boyd2004convex}).
\end{proof}
\begin{Prop}\label{prop:lai}
Let $a_k, k\ge 1$ be a positive sequence. If $a_k, k\ge 1$ satisfy the following
\small
\begin{equation}
\label{eq:ak}
a_{k+1}\le a_k + \frac{1}{k^{1+\epsilon}}, \quad \forall k\ge 1,
\end{equation}
\normalsize
then the sequence $a_k, k\ge 1$ converges.
\end{Prop}
\emph{Proof.}
For convenience, let $\epsilon=1/2$. Because $k \ge 1$ in the assumption, so that we can have $\sqrt{k(k+1)}(\sqrt{k}+\sqrt{k+1}) \ge \sqrt{kk} (\sqrt{k}+\sqrt{k})=2k\sqrt{k}$. Then when $\epsilon=1/2$, we can rewrite Eq.~\ref{eq:ak} to $a_{k+1}\le a_k + \frac{1}{k^{1+\frac{1}{2}}} = a_k + \frac{1}{k\sqrt{k}}, \quad \forall k\ge 1.$
It follows that
\begin{align*}
a_{k+1}\le a_k + \frac{1}{k\sqrt{k}} &\le a_k + \frac{2}{\sqrt{k(k+1)}(\sqrt{k}+\sqrt{k+1})} = a_k+ \frac{2}{\sqrt{k}}- \frac{2}{\sqrt{k+1}}.
\end{align*}
\normalsize
Let $b_n= a_n+ \dfrac{2}{\sqrt{k}}$. Thus, $0\le b_{k+1}\le b_k$ and hence, $b_k, k\ge 1$ are convergent. Let us say $b_k\to a^*$, that is, $a_k+ \dfrac{2}{\sqrt{k}}\to a^*$. Similarly, we can verify that it is true for any $\epsilon>0$. Therefore, it follows that $a_k\to a^*$. \hfill $\square$
Now we are ready to give the main result of the SCC convergence analysis.
\begin{Them}
Suppose $\mathbf{D}^*\in \mathbb{C}$ is a local minimizer such that the mean value ${E}(\nabla g_i(\mathbf{D}^*))=0$ over random variables $(\mathbf{z}_i, \mathbf{x}_i)$ according to Proposition~\ref{claim:sgd-dec} and Proposition~\ref{prop:di-d*} such that
\begin{align*}
M_k= &\frac{1}{n}\sum_{i=1}^n ( \langle( \mathbf{D}^*\mathbf{z}_i^{k-1} -x_i)\mathbf{z}_i^{k-1}, (\mathbf{D}_{i}^k- \mathbf{D}^*)\rangle +
\frac{1}{2}\|\mathbf{z}_i^{k-1}\|^2\|(\mathbf{D}^k_{i}-\mathbf{D}^*)\|^2 \\
-&(\langle (\mathbf{D}^*\mathbf{z}_i^k - x_i)\mathbf{z}_i^k, (\mathbf{D}_{i+1}^k- \mathbf{D}^*)\rangle +
\frac{1}{2} \|\mathbf{z}_i^k\|^2\|\mathbf{D}^k_{i+1}-\mathbf{D}^*\|^2 )) = O(1/k^{1+\epsilon}),
\end{align*}
as $k\to \infty$ for a positive $\epsilon>0$. Due to Proposition~\ref{prop:di-d*}, we know that $\mathbf{D}^k_i \to \mathbf{D}^*$ in the following fashion as $\mathbf{D}_i - \mathbf{D}^* =O(\frac{1}{i}), \hbox{ as } i\to \infty.$ Then our algorithm converges.
\end{Them}
\emph{Proof.}
It is worth noting the $M_k$ can be rewrite into
\begin{align*}
M_k= &\frac{1}{n}\sum_{i=1}^n ( \lang \nabla g_{i}(\mathbf{D}^*), \mathbf{D}_{i}^k- \mathbf{D}^*\rangle + \frac{1}{2}\|(\mathbf{D}^k_{i}-\mathbf{D}^*)\mathbf{z}_i^{k-1}\|^2 \\
-&(\langle\nabla g_{i+1}(\mathbf{D}^*), \mathbf{D}_{i+1}^k- \mathbf{D}^*\rangle +
\frac{1}{2}\|(\mathbf{D}^k_{i+1}-\mathbf{D}^*)\mathbf{z}_i^k\|^2 )).
\end{align*}
We have the Taylor expansion of $\sum f_i(\mathbf{D}^k_{i+1}, \mathbf{z}_i^k)$ at $D^*$ as follows:
\small
\begin{align*}
& \frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^k) + \frac{1}{n}\sum_{i=1}^n (\langle \nabla g_{i+1}(\mathbf{D}^*), \mathbf{D}_{i+1}^k- \mathbf{D}^*\rangle + \frac{1}{2} \|(\mathbf{D}^k_{i+1}-\mathbf{D}^*)\mathbf{z}_i^k\|^2 ).
\end{align*}
\normalsize
Similar for $\sum f_i(\mathbf{D}^k_i, \mathbf{z}_i^{k-1})$, we also have Taylor expansion
\small
\begin{align*}
&\frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^{k-1}) +\frac{1}{n}\sum_{i=1}^n (\langle \nabla g_{i}(\mathbf{D}^*), \mathbf{D}_{i}^k- \mathbf{D}^*\rangle + \frac{1}{2} \|(\mathbf{D}^k_{i}-\mathbf{D}^*)\mathbf{z}_i^{k-1}\|^2).
\end{align*}
\normalsize
Therefore, we mainly use Proposition~\ref{prop:exp-dic-dec} for the following proof. For each epoch $k$,
\small
\begin{align*}
\frac{1}{n} \sum_{i=1}^n f_i(\mathbf{D}^k_{i+1}, \mathbf{z}_i^k) \leq \frac{1}{n} \sum_{i=1}^n f_i(\mathbf{D}^k_i, \mathbf{z}_i^{k-1}).
\end{align*}
\normalsize
When $k$ goes infinity, we can rewrite $\mathbf{D}_i^k = \mathbf{D}_{(k-1)*n+i+1}=\mathbf{D}^*+ O(\frac{1}{kn})$ as $k\to \infty$.
Thus, we have
\small
$$
\frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^k) \le \frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^{k-1}) + M_k,
$$
\normalsize
where $M_k$ is the one as in the assumption of this theorem.
Given $M_k = O(1/{k^{1+\epsilon}})$ as $k\to \infty$ for $\epsilon>0$ and Proposition~\ref{prop:lai}, the new sequence $\dfrac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^k) $ converges to limit ${\cal F}(\mathbf{D}^*, \mathbf{z}_1^*, \cdots, \mathbf{z}_n^*)$, where $\mathbf{z}_i^*$ is the limit of a sub-sequence of $\mathbf{z}_i^k$ for $i=1, \cdots, n$ as $\mathbf{z}_i^k, k\ge 1$ are bounded by Proposition~\ref{bound}.
Therefore, from the above discussion,
$\frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}_{i+1}^k, \mathbf{z}_i^k) - \frac{1}{n}\sum_{i=1}^n f_i(\mathbf{D}^*, \mathbf{z}_i^{k}) \to 0$ and when $k\to \infty$, we conclude that $\lim_{k\to \infty} \frac{1}{n}\sum_{i=1}^n f_i (\mathbf{D}^k_{i+1}, \mathbf{z}_i^{k}) = {\cal F}(\mathbf{D}^*, \mathbf{z}_i^*, \cdots, \mathbf{z}_n^*)$ so that our algorithm converges. \hfill $\square$
\section{Discussion}
The current work presents our initial efforts to develop efficient machine learning methods to work with brain sMRI features computed from general topological surfaces. We validate our proposed FPSBS and HSC methods on two datasets and the preliminary experimental results demonstrate that the proposed algorithms outperform some other works on
classification accuracy. By reducing the dimension of hyperbolic TBM features with the novel HSC algorithm, the present study is capable of applying the low-dimensional HSC measures to diagnose AD and its prodromal stages. In Dataset I, the proposed system successfully distinguishes the ventricular HSC measures of MCIc subjects from MCIs subjects with a higher accuracy ($>85\%$) than the classification systems using whole ventricular volume, area and surface-based biomarkers.
In Dataset II, the proposed system has an outstanding performance to discriminate the cortical HSC measures of AD, MCI and CU groups (accuracy$>80\%$).
These experimental results are consistent with our hypothesis that the lower-dimensional TBM statistics (or HSC measures) may outperform volume, area and shape-based structural measures on discriminating kinds of symptomatic groups related with AD.
Our study has two main findings. First, in order to better initialize the dictionary for sparse coding on the hyperbolic parameter domain, we propose a ring-shaped patch selection algorithm -- FPSBS -- to capture the surface features. The extracted patch structures help reduce feature noises and enhance statistical power of the computed surface TBM features. Second, we propose an efficient sparse coding and dictionary framework defined in the hyperbolic parameter space -- HSC. To the best of our knowledge, HSC is the first sparse coding framework which is designed for general topological surfaces admitting the hyperbolic geometry. Our work may shed new lights on how to extract important surface features from hyperbolic geometry analyses.
In this paper, similar to our prior work on Euclidean parameter domain~\citep{Zhang:ISBI2016} and spherical parameter domain~\citep{zhang2017empowering} , we propose patch-based sparse coding feature selection method to leverage the surface analysis for early disease diagnosis. Differently, our current ring-shaped patch selection algorithm, farthest point sampling with breadth-first search (FPSBS) method, captures the surface features within the hyperbolic space. The proposed FPSBS method can better capture surface features to initialize the dictionary for sparse coding. It is worth noting that the FPSBS is a general algorithm and can also be applied to any Riemann surfaces. The current application to surfaces with general topology is remarkable because, different from Euclidean and spherical parameter domains, it is difficult to generate regular grids on the hyperbolic parameter domain. Our method provides a convenient way to take advantage of the intrinsic geometry structure of hyperbolic parameter domain and generate efficient patch coverage for surfaces with general topology. Therefore, the ring shaped patch selection is an important component of the current brain image analysis system.
We present an efficient sparse coding and dictionary learning method which works on features extracted from surfaces with general topology admitting the hyperbolic geometry. The work is a generalization and enrichment of existing SCC research~\citep{lin2014stochastic,LvSparse,lv2015holistic,Lvtask,zhang2016hyperbolic,zhang2017multi,zhang2018multi,Zhang:ISBI17}. Overall, our work show that the proposed HSC
may extract critical shape features for classifying different stages of AD. Our work should be very useful and meaningful in brain image analysis, shape analysis and machine learning research. HSC has the potential to be a useful tool to study 3D surfaces with general topology structures.
To identify patients as early as possible in the course of MCI, ADNI subdivided MCI into ``early'' and ``late'' stages based on the Wechsler Memory Scale-Revised (WMS-R) Logical Memory II Story A score~\citep{petersen2010alzheimer}. Studies under this MCI staging framework reveal distinguished biomarkers of "early" MCIs (EMCL) as compared to other clinical groups~\citep{wee2019cortical,tosun2014multimodal}. We also applied our proposed method on EMCI and "late" MCI (LMCI) in our recent work, and achieved 84\% accuracy, on ADNI2-dataset with 37 LMCI and 73 EMCI~\citep{zhang2017empowering}. However, studies of~\citep{brooks2007substantial,brooks2008potential,de2005accidental} show that it is overinclusive to use this ADNI staging method based on a single test score, the study of~\citep{edmonds2019early} revealed that ADNI's early MCI group included a large proportion (56\%) of false-positive diagnostic errors.
Another important problem in AD diagnosis research is to identify the probability of MCI patients converting to dementia. Identifying MCI converters from the MCI stable is critical for clinical management. Based on MRI data,~\cite{shi2015studying} successfully revealed significant ventricular morphometry differences between MCI converter group and MCI stable group using a novel ventricular morphometry analysis system. Using the same MRI cohorts, this work further proposes HSC to extract effective structural features and classify 71 MCI converters versus 62 nonconverters. The AUC is 0.85.~\cite{moradi2015machine} proposed a low density separation scheme to learn aggregate biomarkers to discriminate 164 MCI converters from 100 stable MCI patients. They achieved a 0.7661 AUC. The study of~\citep{sorensen2016early} applied hippocampal texture features and support vector machine (SVM) classifier to predict 8 MCI converters versus 17 nonconverters. The best prognostic AUC was 0.83.~\cite{chincarini2011local} applied a medial temporal lobe intensity and textural features and SVM classifier to separate 136 MCI converters from 166 MCI non converters with AUC=0.74.~\cite{collij2016application} applied SVM classifier to predict arterial spin labeling perfusion maps of 12 patients with MCI diagnosis converted to AD versus 12 subjects with stable MCI. The AUC was 0.71. Table~\ref{tab:5} presents the AUC values of this work and the above studies. Compared to other single modality neuroimaging-based classifiers, our proposed system has a larger or comparable AUC on predicting MCI converters versus nonconverters. There are also studies demonstrating that multimodality machine learning models have superior performances than single modality classifiers~\citep{varatharajah2019predicting,rathore2017review,moradi2015machine}. We have developed a series of surface-based biomarkers of various brain structures for AD research~\citep{dong2020applying,Wang:NIMG10,wang2011surface,dong2019applying,fan2018tetrahedron}. Our latest work~\citep{dongintegrating} has indicated that combining these biomarkers could empower the prediction of AD progression. In future work, we expect to improve the MCI conversion prediction performance by introducing these effective multiple statistics. We also note that the study~\citep{edmonds2019early} proposed a neuropsychological approach to improve the reliability of staging early and late MCI. We expect that our biomarkers for predicting MCI converters versus nonconverters also work well with reliable neuropsychological tests for staging early and late MCI. We will study it in our future work.
\begin{table}[t]
\centering
\begin{tabular}{cccc}
\hline
Method & Subjects (MCIc/MCInc) & Feature & AUC \\ \hline
HSC (this paper) & 71/62 & \begin{tabular}{@{}c@{}}Ventricular surface\\ TBM features\end{tabular} & 0.85 \\
\begin{tabular}{@{}c@{}}Low density separation\\ ~\citep{moradi2015machine} \end{tabular} & 164/100 & Gray matter density & 0.77 \\
\begin{tabular}{@{}c@{}}Support vector machine (SVM)\\~\citep{sorensen2016early} \end{tabular}& 8/17 & Hippocampal texture & 0.83 \\
SVM~\citep{chincarini2011local} & 136/166 &\begin{tabular}{@{}c@{}} Medial temporal lobe\\ intensity and textural\\ MRI-based features\end{tabular} & 0.74 \\
SVM~\citep{collij2016application} & 12/12 &\begin{tabular}{@{}c@{}} Arterial spin labeling\\ perfusion maps\end{tabular} & 0.71 \\ \hline
\end{tabular}
\caption{Studies to distinguish MCI converters (MCIc) from nonconverters (MCInc).}\label{tab:5}
\end{table}
There are three important caveats when applying the proposed framework to AD diagnosis and prognosis. First, because of the overlapping patch selection and max-pooling scheme, we generally cannot visualize the selected features and it decreases the comprehensibility although we may always visualize statistically significant regions in our prior group difference studies~\citep{shi2013surface,wang2013applying}. However, our recent work~\citep{zhang2018multi} made some progress which may potentially better address this problem. In our recent work~\citep{zhang2018multi}, instead of randomly selecting patches to build the initial dictionary, we use group lasso screening to select the most significant features. Therefore, the features used in sparse coding may be visualized on the surface map. In future, we will incorporate this idea into our current framework to improve its interpretation ability. Second, our current work, similar to several other work~\citep[e.g.][]{fan2007compare,colliot2008discrimination,kloppel2008automatic,gerardin2009multidimensional,magnin2009support,cuingnet2011automatic,liu2011combination,shen2012detecting,ahmed2015classification}, uses clinical diagnoses as the ``ground truth'' diagnoses for training and cross-validation. However, some recent work~\citep[e.g.][]{Beach:JNEN12} has reported that neuropathological diagnoses only had limited accuracy values (e.g. only 80 - 90\% of the labels are correct) when confirmed with AD histopathology. Under this limitation, we should be cautious when making inferences and conclusions on our work for the AD diagnosis since our discovered features are not necessarily real AD biomarkers. Even so, our recent work~\citep{wu2018hippocampus} has studied hippocampal morphometry on a cohort consisting of $A\beta$ positive AD ($N=151$) and matched $A\beta$ negative cognitively unimpaired subjects ($N=271$) where $A\beta$ positivity was determined using mean-cortical standard uptake value ratio (SUVR) with cerebellum as the reference region over the amyloid PET images. With our Euclidean SCC work~\citep{Zhang:ISBI2016} integrating the proposed HSC and MP methods, we achieved an accuracy rate of 90.48\% in that task~\citep{wu2018hippocampus}. The results demonstrate that our proposed framework may potentially help discover pathology-confirmed AD biomarkers. Third, as our initial attempt to integrate geometry analysis and machine learning method for AD diagnosis, the current work reports very limited experimental results since we mainly reuse the data in our prior published work~\citep{shi2015studying,shi2019hyperbolic}. The geometry analysis part involves multiple steps, including image segmentation, surface registration, surface parameterization, etc. Our ongoing work, e.g.~\citep{Mi:AAAI20}, is developing novel approaches which will make the whole process more automatic and more accurate. Once they are available, we will apply the proposed method to analyze more longitudinal ADNI data.
\section{Results}
\subsection{Dataset I: MCI Converter vs. MCI stable Subjects}
In Dataset I, we try to use ventricular morphometry features to discriminate between MCIc and MCIs subjects. To extract hyperbolic surface features, we automatically segment lateral ventricular volumes with the multi-atlas fluid image alignment (MAFIA) method~\citep{chou2010ventricular} from each MRI scan. We then use a topology-preserving level set method~\citep{han2009a} to build surface models and the marching cube algorithm~\citep{lorensen1987marching} is applied to construct triangular surface meshes (Fig.~\ref{fig:pipeline} (b)). After the topology optimization, we apply hyperbolic Ricci flow method and conformally map the ventricular surface to the Poincar\'{e} disk and register them via harmonic map~\citep{shi2015studying}. We finally compute the surface TBM features~\citep{shi2015studying} and smooth them with surface heat kernel method~\citep{chung2005cortical}.
\iffalse
\begin{table}[!t]
\centering
\begin{tabular}{m{2cm}<{\centering}|c|cccc }
\hline
Methods & Steps & 1000 & 1500 & 3000\\
\hline
\multirow{4}{*}{ODL} & {Updating $Z$} & 5.40 & 15.28 & 38.75\\
& {Updating $D$} & 11.07 & 40.34 & 73.59\\
& {Total} & 16.47 & 55.62 & 112.34\\
& {OF Value} & 0.298 & 0.270 & 0.244\\
\hline
\multirow{4}{*}{HSC} & {Updating $Z$} & 0.158 & 0.235 & 0.550\\
& {Updating $D$} & 0.030 & 0.033 & 0.043\\
& {Total} & 0.188 & 0.278 & 0.593\\
& {OF Value} & 0.299 & 0.270 & 0.245\\
\hline
\end{tabular}
\caption{Computational time (hours) and objective function (OF) values of the ODL~\citep{mairal2009online} and HSC for different dictionary sizes.}\label{tab:time}
\end{table}
\fi
Here we select $2,000$ ring-shaped patches (300-vertex, Fig.~\ref{fig:3}) by the FPSBS method (Alg.~\ref{alg:FBS}) on each side of ventricle for each subject and finally have $n = 532, 000$ ring-shaped patches. The generated patches are consistent across all subjects since surfaces are registered already~\citep{shi2015studying}. Surface TBM is a scalar feature defined on each vertex on the hyperbolic domain so the feature number of each subject is $1,200,000$ ($m=300$, $p=4,000$ in notations of Sec.~\ref{sec:sparsecoding},~\ref{sec:maxpooling}). We initialize the dictionary via selecting random patches~\citep{coates2011importance}, which has been shown to be a very efficient method in practice. Then we learn the dictionary and sparse codes by HSC using the initial dictionary. All our experiments involve training for ten epochs with a batch size of one. After the optimization, all subjects use the same dictionary ($m=300$, $t=2,000$ in notation of Sec.~\ref{sec:sparsecoding}). With the sparse coding, we obtain $4,000$ samples each of which has $2,000$ sparse codes per subject. After that, max-pooling is adopted to choose the maximum value for each sparse code as a feature on these $4,000$ samples. Our final feature dimension for classification is $2,000$ per subject.
\iffalse
\subsubsection{Computational Efficiency}
Comparisons of computational time as well as objective function values are given in Table~\ref{tab:time}. We show the time to update the dictionary $D$ and the sparse code $Z$, respectively, together with the total running time. Table~\ref{tab:time} reports the computational time on three different dictionary sizes, i.e., $1000 \times 300$, $1500\times 300$ and $3000\times 300$.
Note that when the size of the dictionary increases, the computational time of ODL~\citep{mairal2009online} increases rapidly. However, for HSC the computational time increases much slower compared to ODL, especially on updating the dictionary. The speedup of updating $Z$ is from 34 times up to 70 times when we increase the dictionary size. Therefore, HSC has a better scalability when dealing with large size dictionaries. In addition, it is worth noting that HSC archives comparable objective function values with ODL. When the dictionary sizes increase, the objective function values decrease, indicating that the dictionary representation ability improves. The convergence analysis and the computational efficiency analysis demonstrate HSC is one potential strategy to apply hyperbolic TBM statistics in the classification analysis of AD diagnosis and prognosis.
\fi
We take a nested cross-validation approach by pre-separating training, validation and testing sets. Specifically, we use the ratio of 7:1:2 for training, validation and testing. We select the hyper parameters based on the validation set and test all methods on the same testing set.
Besides, we also compare our work with some other measures and methods. We compute bilateral ventricular volumes and surface areas, which are used as MRI biomarkers in AD research. We also compare HSC with a ventricular surface shape method in~\citep{pmid18228600} (\emph{Shape}), which automatically generates comparable meshes of all ventricles. The deformations based on the morphometry model are employed with repeated permutation tests and then used as geometry features. With our ventricle surface registration results, we follow the \emph{Shape} work~\citep{pmid18228600} for selecting biomarkers and use support vector machine (SVM) for classification on the same dataset. We implement the low-rank shared dictionary learning (LRSDL) method, based on the paper~\citep{vu2017fast} and the github source code~\footnote{ https://github.com/tiepvupsu/DICTOL\_python}. We select the hyper-parameter for LRSDL by using the same strategy as HSC on the training set. We run LRSDL 50 iterations with $\lambda_1=0.13$, $\lambda_2=0.1$, $\eta=0.05$, $k=20$, $k_0=10$. Same as HSC, we apply the LRSDL on ring-shape patches and apply max-pooling as post processing on the learnt sparse codes. The same classifier is applied on the learnt features on the same test set as HSC. We test \emph{HSC, Shape, volume, area and LRSDL} measures on the left, right and whole ventricle, respectively. Accuracy (ACC), Sensitivity (SEN), Specificity (SPE) and compute Area Under The Curve (AUC) are computed to evaluate classification results. Table~\ref{tab:4} shows classification performances of four methods.
\begin{table}[t]
\centering
\begin{tabular}{ccccccc}
\hline
Dataset I & & HSC & Shape & Volume & Area & LRSDL \\ \hline
\multirow{4}{*}{whole} & ACC & 85.19\% & 70.37\% & 66.67\% & 59.26\% & 77.78\% \\
& SEN & 76.92\% & 100.00\% & 85.71\% & 78.57\% & 75.00\% \\
& SPE & 81.25\% & 57.89\% & 63.16\% & 57.89\% & 80.00\% \\
& AUC & 0.8516 & 0.7857 & 0.6731 & 0.5934 & 0.7775 \\
\hline
\multirow{4}{*}{left} & ACC & 74.07\% & 81.48\% & 62.96\% & 59.26\% & 70.37\% \\
& SEN & 76.92\% & 76.92\% & 61.54\% & 83.33\% & 76.92\% \\
& SPE & 71.43\% & 75.00\% & 56.52\% & 52.63\% & 64.29\% \\
& AUC & 0.7418 & 0.8159 & 0.6264 & 0.5907 & 0.706 \\
\hline
\multirow{4}{*}{right} & ACC & 70.37\% & 66.67\% & 62.96\% & 59.26\% & 70.37\% \\
& SEN & 53.85\% & 84.62\% & 84.62\% & 78.57\% & 69.23\% \\
& SPE & 78.57\% & 57.14\% & 57.89\% & 57.89\% & 71.43\% \\
& AUC & 0.7088 & 0.6703 & 0.6319 & 0.5879 & 0.7033 \\
\hline
\end{tabular}
\caption{The comparison results by nested cross-validation on Dataset I.}\label{tab:4}
\end{table}
From the experimental results, we can find that the best accuracy (85.19\%) and the best specificity (81.25\%) are achieved when we use TBM features on ventricle hyperbolic space of both sides (whole) for training and testing. The comparison shows that our new framework selects better features, and achieves better and more meaningful classification results.
The HSC algorithm with whole ventricle TBM features achieves the best AUC (0.8516). The comparison demonstrates that our proposed algorithm may be useful for AD diagnosis and prognosis research.
\subsection{Dataset II: ADNI Baseline Cortical Surfaces}
\label{data1}
\begin{figure}[t]
\centering
\includegraphics[width=5in]{Figure/fig7_v3.eps}
\caption{Modeling cortical surface with hyperbolic geometry. (a) shows six identified open boundaries, $\gamma_1, \cdots, \gamma_6$. (b) shows the hyperbolic parameter space, which is the Poincar\'{e} disk model}
\label{fig:7}
\end{figure}
Many researches have analyzed that the cortial surface morphometry is a valid imaging biomarker for AD~\citep{shi2019hyperbolic,thompson2004mapping,chung2008tensor}. In Dataset II, we apply HSC to analyze cortical morphometry for AD related clinical group classification. We use the left hemispheric cerebral cortices and follow~\citep{shi2019hyperbolic} to preprocess cortical surface data. We first use FreeSurfer software~\citep{fischl2012freesurfer} to preprocess the MRIs of 115 subjects and reconstruct their left cortical surfaces. The Caret software~\citep{van2012cortical} is then used to automatically label six major brain landmarks, which include the Central Sulcus, Anterior Half of the Superior Temporal Gyrus, Sylvian Fissure, Calcarine Sulcus, Medial Wall Ventral Segment and Medial Wall Dorsal Segment. Fig.~\ref{fig:7} (a) shows an example of the landmark curves on the left cortical surface, where the six landmark curves are modeled as open boundaries and denoted as $\gamma_1, \cdots, \gamma_6$. The fundamental group of paths are computed by connecting boundary $\gamma_5$ to every other boundary and the path is denoted as $\tau_1, \tau_2, \tau_3, \tau_4, \tau_6$. Fig.~\ref{fig:7} (b) shows that they are embedded into the Poincar\'{e} disk. After we cut the cortical surfaces along the delineated landmark curves, the cortical surfaces become genus-0 surfaces with six open boundaries. We finally randomly select the left cortical surface of a CU subject, who is not in the studied subject dataset, as the template surface, and perform the processing steps described in Sec.~\ref{sec:HRF} and Sec.~\ref{sec:sTBM} to get the hyperbolic surface TBM features.
For Dataset II, since there are only 115 subject for three classes, we use the same hyperparameter as what we used in Dataset I for training. We apply five-fold cross-validation to evaluate our algorithm, which guarantees the model is tested on all subjects.
All experiments are trained for $k= 10$ epochs with a batch size of 1. The regularization parameter $\lambda$ is set to $0.10 \approx 1.2/\sqrt{m}$, $1/\sqrt{m}$ is a classical normalization factor~\citep{bickel2009simultaneous} and the constant 1.2 has been shown to produce about 10 non-zero coefficients. We select $p = 2,000$ ring-shaped patches as shown in Fig.~\ref{fig:8} by FPSBS on the cortical surface and we have $n = 230,000$ ring-shaped patches for Dataset II. Fig.~\ref{fig:8} (right) is the visualization of cortical morphometry on the hyperbolic parameter domain and Fig.~\ref{fig:8} (left) projects the selected patches on the hyperbolic parameter domain back to the original cortical surface. Our FPSBS patch selection algorithm can maintain the same topological structure as the parameter domain.
\begin{figure}[t]
\centering
\includegraphics[height=4cm]{Figure/cortical_surface_patch_v4.eps}
\caption{Visualization of computed image patches on the cortical surface (left) and hyperbolic space (right). Each patch has a unique color. The zoom-in pictures show some overlapping areas between image patches.}
\label{fig:8}
\end{figure}
After learning the sparse codes via HSC, we apply max-pooling~\citep{boureau2010theoretical} for further dimension reduction. Finally, we employ the Adaboost~\citep{rojas2009adaboost} to do the binary classification and distinguish individuals from different groups. We report the classification results of (1) AD vs. CU, (2) AD vs. MCI and (3) MCI vs. CU in Table~\ref{tab:3}.
\begin{table}[t]
\centering
\begin{tabular}{cccc}
\hline
Dataset II & AD vs. CU & AD vs. MCI & MCI vs. CU \\ \hline
ACC & 88.57\% & 82.67\% & 80\% \\
SEN & 89.29\% & 84.00\% & 79.17\% \\
SPE & 83.33\% & 70.00\% & 84.44\% \\
AUC & 0.879 & 0.8111 & 0.7944 \\ \hline
\end{tabular}
\caption{The classification results by five-fold cross-validation on Dataset II.}\label{tab:3}
\end{table}
In our prior work~\citep{shi2019hyperbolic}, we have shown that the hyperbolic surface features are significantly associated with the diagnostic disease severity. However, it is difficult to directly use hyperbolic surface features for different stages of disease diagnosis classification due to the large amount of features and limited subject numbers. Table~\ref{tab:3} shows that HSC overcomes the above issue and FPSBS has a good generalization capability to capture the meaningful features from ring-shaped patches. HSC works well on even more subtle difference classification problem (CU vs. MCI) compared with AD vs. CU. Our new framework makes meaningful and high performances on different groups and may be useful for AD diagnosis and prognosis researches
\section{Introduction}
Alzheimer's Disease (AD), an irreversible neurodegenerative disaese, is the most common cause of dementia among older adults. It is generally agreed that accurate presymptomatic diagnosis and preventive treatment of AD could have enormous public health benefits. Brain structural magnetic resonance imaging (sMRI) analysis has the potential to provide valid diagnostic biomarkers of the preclinical stage as well as symptomatic AD~\citep{pmid20139996}. For example, a single-valued sMRI-based atrophy is used as a neurodegeneration marker in the recently proposed AD descriptive "A/T/N" (amyloid, tau, neurodegeneration) system~\citep{jack2016t} to clinically define AD. Tosun et al. proposed MRI-based approaches to impute Abeta status~\citep{tosun2014multimodal,tosun2016amyloid}. Their results demonstrated that sMRI can be used to predict the amyloid status of MCI individuals and mild AD patients. Recently, brain morphology measures have been integrated with machine learning algorithms to classify individual subjects into different diagnostic groups~\citep[e.g.][]{sun2009elucidating,pmid18228600,wang2013applying,li2014abnormal}. It offers a promising approach to computer-aided cognitive decline prediction by leveraging both sensitive brain image features and powerful machine learning techniques.
\if 0
Deformation-based morphometry (DBM) and tensor-based morphometry (TBM) are well studied in the analysis of brain imaging in structure volumes and shapes. Prior work has demonstrated that surface-based analyses~\citep{thompson2000growth,fischl2012freesurfer} can offer advantages over volume measures, due to their sub-voxel accuracy and the capability of detecting subtle subregional changes. DBM \citep{ashburner1998identifying,Chung2001a,wang2003changes,chung2003deformation} uses deformations obtained from the nonlinear registration of brain images to a common anatomical template, to infer 3D patterns of statistical differences in brain volume or shape. TBM \citep{thompson2000growth,chung2008tensor} is a related method, which examines spatial derivatives of the deformation maps registering brains to a common template. Morphological tensor maps are used to derive local measures of shape characteristics such as the Jacobian determinant, torsion or vorticity. DBM, by contrast, analyzes 3D displacement vector fields encoding relative positional differences in anatomical structures across subjects, after mapping all brain images to a common stereotaxic space \citep{thompson1997detection,cao1997new}. The advantage of TBM for studying brain structure is that it derives local derivatives and tensors from the deformation for further analysis.
\fi
Although most brain sMRI analysis approaches use cortical and subcortical volumes \citep[e.g.][]{pmid12552040,pmid18054253,pmid20375138,pmid20382238}, recent research has demonstrated that surface-based analyses, \citep[e.g.][]{pmid15772166,thompson2004mapping,pmid18228600,pmid20129863,pmid21272654} can offer advantages over volume measures, due to their sub-voxel accuracy and the capability of detecting subtle subregional changes. In surface-based brain imaging research, a practical approach to model brain landmark curves is to model them as surface boundaries by cutting open cortical surfaces along these landmarks. Thus they are modeled as open boundaries to be matched across subjects \citep{Tsui:IPMI13,shi2019hyperbolic} or be used as shape indices~\citep{Zeng2013,shi2017conformal}. Similarly, adding open boundaries have been proved to be useful in modeling ventricular surfaces which have a concave shape and complex branching topology~\citep{Wang:NIMG10,shi2015studying}. We call these genus-zero surfaces with more than two open boundaries as \emph{general topological surfaces} and hyperbolic geometry has been demonstrated to be useful to model general topological surfaces. However, most of current hyperbolic space-based brain imaging methods have been focused on studying group difference between diagnostic groups. To develop brain imaging methods for personal medicine research, it would be advantageous to design powerful machine learning methods that work on general topological surface features for the identification of AD symptoms on an individual basis.
One of the major challenges to directly apply vertex-wise surface features, such as surface tensor-based morphometry (TBM)~\citep{thompson2000growth,chung2008tensor}, to cognitive decline prediction research is that the surface feature dimension is usually much larger than the number of subjects, the so-called \emph{high dimension-small sample problem}.
Existing feature dimension reduction approaches include feature selection~\citep{fan:miccai05,jain1997feature}, feature extraction~\citep{saadi2007optimally,guyon2008feature,scholkopft1999fisher,jolliffe2011principal} and sparse coding-based methods~\citep{vounou2010discovering,donoho2006,wang2013applying}. In most cases, information is lost when mapping high-dimenstional features into a lower-dimensional space. However, by defining a better lower-dimensional subspace, sparse coding~\citep{lee2006efficient,mairal2009online} may limit such information loss. Sparse coding has been previously proposed to learn an over-complete set of basis vectors (dictionary) to represent input vectors efficiently and concisely~\citep{pmid16576749}. It has shown to be efficient for many tasks such as image deblurring~\citep{yin2008}, super-resolution~\citep{yang2010image}, classification~\citep{mairal2009online}, functional connectivity~\citep{pmid29993466,lv2015holistic,Lvtask,pmid26466353,LvSparse} and structural morphometry analysis~\citep{zhang2017multi,li2017transcriptome}. However, solving sparse coding remains a
computationally challenging problem, especially when dealing with large-scale datasets and learning large size dictionaries~\citep{lin2014stochastic}.
To generalize sparse coding to process general topological surface features~\citep{mairal2009online},
we propose a novel pipeline to extract sparse hyperbolic features for classification termed hyperbolic stochastic coding (HSC), consisting of our unique farthest point sampling with breadth-first search (FPSBS) method for ring-shaped surface patches extraction, stochastic coordinate coding (SCC) and max-pooling methods for feature dimension reduction, to extract critical low-dimensional shape features from the hyperbolic TBM maps. We call such features as HSC measures. Then the AdaBoost classifier~\citep{freund1997decision} is further adopted on these HSC measures for AD clinical group classification and cognitive decline prediction. We hypothesize that our HSC measures may outperform volume, area and shape-based cortical structural measures on discriminating clinical groups related with AD~\citep{li2014abnormal,pmid18228600,jack1999prediction,leung2010automated}
We validate our system in a publicly available brain image dataset, Alzheimer's Disease Neuroimaging Initiative (ADNI) cohort~\citep{ADNI}. With the sMRI baseline data of 133 mild cognitive impairment (MCI) subjects, consisting of 71 MCI converter (MCIc) and 62 MCI stable (MCIs) subjects~\citep{shi2015studying} and 115 subjects (30 AD, 44 MCI and 40 cognitively unimpaired (CU) subjects)~\citep{shi2019hyperbolic}, we set out to test our hypothesis by performing
classification accuracy comparison with three other popular structural measures (volume, area, and shape-based biomarkers).
\section{Subjects and Methods}
\subsection{Subjects}
Data for testing the performances of our proposed HSC are obtained from the ADNI database (adni.loni.usc.edu). The ADNI was launched in 2003 as a public-private partnership, led by Principal Investigator Michael W. Weiner, MD. The primary goal of ADNI is to test whether biological markers such as serial MRI and positron emission tomography (PET), combined with clinical and neuropsychological assessments can measure the progression of MCI and early AD. Determination of sensitive biomarkers aids researchers and clinicians to develop new treatments and monitor their clinical effectiveness, as well as lessen the time and cost of clinical trials. The initial ADNI (ADNI-1) database recruited 800 subjects from over 50 sites across the U.S. and Canada and it has been followed by ADNI-GO and ADNI-2. To date, these three databases have recruited over 1500 adults, ages 55 to 90, consisting of elderly cognitive unimpaired individuals, people with early or late MCI, and people with early AD. The follow up duration of each subject is specified in their corresponding protocols for ADNI-1, ADNI-2 and ADNI-GO. Subjects of ADNI-1 and ADNI-GO had the option to be followed in ADNI-2. For up-to-date information, see www.adni-info.org.
We use two ADNI datasets to validate our system. They are the same datasets used in our prior work~\citep{shi2015studying,shi2019hyperbolic} which mainly studied group differences between different clinical groups. As a generalization of our prior work, the current work studies personalized diagnosis with the same datasets. Studies indicate that ventricular enlargement is an important measure related with AD progression~\citep{shi2015studying,pmid15275931}. In Dataset I, we select 133 subjects from the MCI group in the ADNI-1~\citep{ADNI} baseline dataset as~\citep{shi2015studying,zhang2016hyperbolic}. All subjects have both sMRI and fluorodeoxyglucose positron emission tomography (FDG-PET) data. They include 71 subjects (age: $74.77 \pm 6.81$) who develop incident AD during the subsequent 36 months, which we call the MCI converter (MCIc) group, and 62 subjects (age: $75.42 \pm 7.83$ years) who do not during the same period, which we call the MCI stable (MCIs) group. These subjects are chosen on the basis of having at least 36 months of longitudinal data. If a subject developed incident AD more than 36 months after baseline, it is assigned to the MCIs group. All subjects undergo thorough clinical and cognitive assessments at the time of acquisition, including the Mini-Mental State Examination (MMSE) score, Alzheimer's disease assessment scale – Cognitive (ADAS-COG)~\citep{rosen1984new} and Auditory Verbal Learning Test (AVLT)~\citep{Rey:1964}. The demographic statistical information of this dataset is shown in Table~\ref{tab:dataset1}.
\begin{table}[b]
\centering
\begin{tabular}{ccccc} \\
\hline
Group &Gender (F/M) & Education & Age & MMSE\\\hline
MCIc&26/45&15.99$\pm$2.73&74.77$\pm$6.81&26.83$\pm$1.60\\
MCIs&18/44&15.87$\pm$2.76&75.42$\pm$7.83&27.66$\pm$1.57\\
\hline
\end{tabular}
\caption{Demographic statistic information of Dataset I.}
\label{tab:dataset1}
\end{table}
In Dataset II, we study cortical morphometry for tracking AD progression. Dataset II has 115 T1-weighted MRIs from the ADNI-1~\citep{ADNI} baseline dataset, including 30 AD patients, 45 MCI subjects and 40 CU subjects~\citep{shi2019hyperbolic}. All subjects underwent through MMSE~\citep{folstein1975mini}. The demographic statistics with matched gender, education, age and MMSE are shown in Table~\ref{tab:dataset2}.
\begin{table}[t]
\centering
\begin{tabular}{ccccc} \\
\hline
Group&Gender (F/M) & Education & Age & MMSE\\\hline
AD&15/15&15.22$\pm$2.61&76.22$\pm$7.34&23.07$\pm$2.02\\
MCI&19/26&16.11$\pm$2.56&73.86$\pm$8.20&26.95$\pm$1.34\\
CU &18/22&17.25$\pm$1.90&76.53$\pm$6.02&29.11$\pm$1.03\\
\hline
\end{tabular}
\caption{Demographic statistical information of Dataset II.}
\label{tab:dataset2}
\end{table}
\begin{figure}[!ht]
\centering
\includegraphics[height=0.7\textheight]{Figure/MEDIA_graphicabstract01.eps}
\caption{The major processing steps in the proposed framework.}
\label{fig:pipeline}
\end{figure}
\subsection{Overview of the Proposed Framework}
The major computational steps of our proposed work are illustrated in Fig.~\ref{fig:pipeline} where we take a left ventricular surface as an example. There are two major stages in the process. In the first stage, we perform ventricular surface reconstruction from MRI data, surface registration and surface TBM feature computation. The second stage is for HSC measure computation. Specifically, we build ring-shaped patches on the hyperbolic parameter space by FPSBS to initialize the original dictionary. Dictionary learning and max-pooling are performed for feature dimension reduction. Following that, Adaboost is adopted to diagnose different clinical groups and predict future AD conversions.
\subsection{Brain Surface Registration with Hyperbolic Ricci Flow and Harmonic Map}
\label{sec:HRF}
Taking a left ventricular surface $S$ as an example, the corresponding framework is summarized in Algorithm\ref{alg:HWDC}~\citep{shi2015studying} and Fig.~\ref{fig:pipeline} (c). Its critical steps are shown in Fig.\ref{fig:2}. Following our prior work~\citep{shi2015studying}, three horns of a ventricular surface are identified and three cuts $\{\gamma_1,\gamma_2,\gamma_3\}$ are made on these horns (Fig.\ref{fig:2} (a)). We term this step as \emph{topology optimization}. As a result, each ventricular surface becomes a topologically multiply connected surface and admits the hyperbolic geometry. We apply the hyperbolic Ricci flow method to compute its discrete hyperbolic uniformization metric. With the hyperbolic uniformization metric, we can embed {$S$} onto the Poincar\'e disk. In the obtained Poincar\'e disk, we apply the geodesic curve lifting algorithm~\citep{shi2015studying} to obtain a canonical parameter space (Fig.\ref{fig:2} (b)). Furthermore, we convert the Poincar\'e disk to the Klein model. It converts the canonical fundamental domains of the ventricular surfaces to a Euclidean octagon, as shown in Fig.\ref{fig:2} (c). Then we compute surface harmonic map with the Klein disk as the canonical parameter space for the following surface morphometry analysis~\citep{shi2015studying}.
\begin{algorithm}[t]
\caption{Brain surface registration with hyperbolic Ricci flow and harmonic map}
\label{alg:HWDC}
\KwIn{Brain surface $S$ with more than 2 open boundaries.}
\KwOut{Klein model of $S$}
Compute the hyperbolic uniformization metric of $S$ with hyperbolic Ricci Flow.
Compute the fundamental group of paths on $S$ and, together with original boundaries, obtain the simply connected domain $\bar{S}$.
Embed $S$ onto the Poincar\'e disk with its hyperbolic metric and its simply connected domain $\bar{S}$, we obtain the fundamental domain of $S$.
Tile the fundamental domain of $S$ with its Fuchsian group of transformations to get a finite portion of the universal covering space of $S$.
Compute the positions of the paths in the fundamental group as geodesics in the universal covering space. By slicing the universal covering space along the geodesics, we obtain the canonical fundamental domain of $S$.
Convert the canonical Poincar\'e disk to the Klein model and construct the harmonic map between $S$ and a selected template surface.
\end{algorithm}
\begin{figure}[t]
\centering
\includegraphics[width=5in]{Figure/Figure2.eps}
\caption{Modeling ventricular surface with hyperbolic geometry. (a) shows three identified open boundaries, $\gamma_1$, $\gamma_2$, $\gamma_3$, on the ends of three horns. After that, ventricular surfaces can be conformally mapped to the hyperbolic space. (b) and (c) show the hyperbolic parameter space, where (b) is the Poincar\'{e} disk model and (c) is the Klein model.}
\label{fig:2}
\end{figure}
\subsection{Surface Tensor-Based Morphometry}
\label{sec:sTBM}
Suppose $\phi = S_1 \rightarrow S_2$ is a map from surface $S_1$ to surface $S_2$. The derivative map of $\phi$ is the linear map between the tangent spaces $d\phi:TM(p)\rightarrow TM(\phi(p))$, induced by the map $\phi$, which also defines the Jacobian matrix of $\phi$. The derivative map $d\phi$ is approximated by the linear map from one face $[v_1, v_2, v_3]$ to another one $[w_1, w_2, w_3]$. First, we isometrically embed the triangles $[v_1, v_2, v_3]$ and $[w_1, w_2, w_3]$ onto the Klein disk, the planar coordinates of the vertices are denoted by $v_i, w_i, i=1, 2, 3$, which represent the 3D position of points
Then, the Jacobian matrix for the derivative map $d\phi$ can be computed as $J=d\phi = [w_3 - w_1, w_2 - w_1][v_3 - v_1, v_2 - v_1]^{-1}.$
Based on the derivative map $J$, the surface TBM is defined as $\sqrt{det(J)}$, which measures the amount of local area changes in a surface with the map $\phi$ (Fig.~\ref{fig:pipeline} (d)). As pointed out in \citep{chung2005cortical}, each step in the processing pipeline including MRI acquisition, surface deformation, etc., are expected to introduce noise in the deformation measurement. The deformation is applied to map each subject's surface to a template surface. The Jacobian matrices of the transformation were used per subject. To account for the noise effects, we apply surface heat kernel smoothing algorithm proposed in~\citep{chung2005cortical} to improve SNR in the TBM features and boost the sensitivity of statistical analysis. The vertical-wise surface TBM features are used as the inputs for dictionary learning. We use the coordinates of these vertices to localize the ring-shaped patches and we use 3-dimentional TBM features as the feature map of ring-shape patches.
\subsection{Ring-Shaped Patch Selection}
The hyperbolic space is different from the original Euclidean space. The common rectangle patch construction developed in Euclidean space~\citep{zhang2017multi} cannot be directly applied to the hyperbolic space. Therefore, we propose FPSBS on hyperbolic space to initialize dictionaries for sparse coding (Fig.~\ref{fig:pipeline} (e)). The intuition of the algorithm is that we want to select patches without losing the geometry information and all vertices on the hyperbolic space selected at least once. This will guarantee we learn complete information from the hyperbolic space. Fig.~\ref{fig:3} (right) is the visualization of patch selection on the hyperbolic parameter domain. And Fig.~\ref{fig:3} (left) projects the selected patches on the hyperbolic parameter domain back to the original ventricular surface, which still maintains the same topological structure as the parameter domain. In Fig.~\ref{fig:3}, each patch has a unique color and patches may overlap with each other. Together all patches cover the entire surfaces. In the following paragraph, we explain how these patches are selected.
We first randomly select a patch center point $c_1$ on the hyperbolic space $V$, where $c_1 \in V$ and $V$ is the set of all discrete vertices on the hyperbolic space. We then find all $u$ vertices connected with the center point $c_{1, i} (i = 1, 2, ..., u)$ and $c_{1, i}$ is the $i$-th vertex connected with $c_1$. The procedure is called breadth-first search (BFS)~\citep{patelcomparison}, which is an algorithm for searching graph data structures. It starts at the tree root and explores the neighbor nodes first, before moving to the next level neighbors. We use the same BFS procedure to find all connected vertices with $c_{1, i}$, which are $ c_{1,i_j} (j = 1, 2, \cdots, w_i)$. $w_i$ is the number of connected vertices with center point $c_{1, i}$. Finally, we get a vertex set (no duplicate vertices) $\mathbf{x}_1$ as follows, we call it a selected ring-shape patch on hyperbolic space and the patch center is $c_{1}$.
\begin{equation}
\mathbf{x_1} =\{c_{1}, c_{1,1}, \cdots, c_{1,1_{w_1}}, \cdots, c_{1,u}, \cdots, c_{1,u_{w_u}} \}
\label{eqn:onepatchset}
\end{equation}
The dimension of $\mathbf{x}_1$ is $u+w_1+\cdots +w_u = m$ and $\mathbf{x}_1 \in \mathbb{R}^m$. We construct the topological patches based on hyperbolic geometry and the edge connections among different points from $\mathbf{x}_1$. $\mathbf{x}_1$ is the first selected patch. To select the second patch center, we sample the farthest point with $c_1$, s.t. radius $r = \max_{c_{v} \in V} d_V(c_{v}, c_1)$. We now find the second patch center $c_2 \in V$ with the farthest distance $r$ of $c_1$. We follow the farthest point sampling scheme~\citep{moenning2003fast}, the sampling principle is repeatedly placing the next sample point in the middle of the least known area of the sampling domain, which can guarantee the randomness of the patch selection.
\begin{figure}[t]
\centering
\includegraphics[height=4cm]{Figure/vent_combine.eps}
\caption{Visualization of computed image patches on the ventricle surface (left) and hyperbolic space (right). The zoom-in pictures show some overlapping areas between image patches.}
\label{fig:3}
\end{figure}
Here, $d$ is the hyperbolic distance in the Klein model. Given two points $v'$ and $v''$, draw a straight line between them; the straight line intersects the unit circle at points $a$ and $b$, so $d$ is defined as follows:
\begin{equation}
d(v', v'')=\frac{1}{2}(\log{\frac{|av''||bv'|}{|av'||bv''|}})
\end{equation}
where $|av''|>|av'|$ and $|bv'|>|bv''|$.
$V_r$ denotes the set of selected patch centers ($V_r=\{c_1\}$ when we compute $c_2$). After selecting the second patch $x_2$, we add $c_{2}$ into $V_r$ ($V_r=\{c_1, c_2\}$). We iterate the patch selection procedure $p$ times to get $p/2$ patches which cover every surface vertex at least once ($p/2$ patches on each side result in $p$ patches per subject). The details of FPSBS are summarized in Algorithm~\ref{alg:FBS}.
\begin{algorithm}[t]
\caption{Farthest Point Sampling with Breadth-first Search (FPSBS)}
\label{alg:FBS}
\KwIn{Hyperbolic parameter space.}
\KwOut{A collection of different amount overlapped patches on topological structure.}
Start with $V_r = \{c_{1}\}$, $V$ denotes all discrete vertices on the hyperbolic space and $V_r$ denotes the set of selected patch centers.\\
\For{T$=$1 to $n$}
{\For {$r$ determine sampling radius}
{Find set $\mathbf{x}_{T}$ by following Eq.~\ref{eqn:onepatchset} and two times BFS.\\
$r = \max_{c_{v} \in V} d_V(c_{v}, c_{T})$\\
\If{ $r \leq 10e^{-2}$ } {STOP}
\textbf{Find the farthest point $c_{T+1}$}\\
Add $c_{T+1} = \arg\max_{c_{v} \in V}dr(c_{v}, V_r)$ to $V_r$
}
}
\end{algorithm}
\subsection{Sparse Coding and Dictionary Learning}\label{sec:sparsecoding}
We model surface TBM features as a sparse linear combination of atoms selected from a dictionary which is initialized by FPSBS on the hyperbolic parameter space. This modeling procedure is known as sparse coding~\citep{mairal2009online}. Our aim is to reduce the original surfaces dimension with the over-complete dictionary and find a linear combination of the dictionary bases to reconstruct the original surface statistics. The problem statement of sparse coding is described as below.
Given a finite training set of ring-shaped patches (as the description in Sec II. C) $\mathbf{X}=(\mathbf{x}_1, \mathbf{x}_2, \cdots, \mathbf{x}_n) \in \mathbb{R}^{m\times{n}}$, and $\mathbf{x}_{i}\in\mathbb{R}^m$, $i = 1, 2,\cdots, n$, where $m$ is the dimension of each ring-shaped patch and $n$ is the total number of patches. In this paper, we use superscript to represent $k$-th epoch and use subscript to represent $i$-th coordinate. We use boldface lower case letters $\mathbf{x}$ to denote vectors and use boldface upper case letters $\mathbf{X}$ to denote matrices. We then learn dictionary and sparse codes for these input patch features $\mathbf{x}_i$ using sparse coding.
We use $f_i(\cdot)$ to represent the optimization problem of sparse coding for each patch $\mathbf{x}_i$:
\small
\begin{equation}~\label{eqn:optimizationequation}
\min_{\mathbf{D}\in \mathbb{R}^{m \times t}, \mathbf{z}_i\in \mathbb{R}^t}{f_i}(\mathbf{D}, \mathbf{z}_i) = \frac{1}{2}||\mathbf{Dz}_i - \mathbf{x}_i||^2_2 + \lambda||\mathbf{z}_i||_1
\end{equation}
\normalsize
where $\lambda$ is the regularization parameter, $||\cdot||_2^2$ is the standard Euclidean norm and $||\mathbf{z}_i||_1 = \sum_{j=1}^{t}|z_{i, j}|$. In Eq.~\ref{eqn:optimizationequation}, each input vector will be represented by a linear combination of a few basis vectors of a dictionary. The first term of Eq.~\ref{eqn:optimizationequation} is the reconstruction error, which measures how well the new feature represents the input vector. The second term of Eq.~\ref{eqn:optimizationequation} ensures the sparsity of the learned feature $\mathbf{z}_i$. Each $\mathbf{z}_i$ is often called the \textit{sparse code}. Since $\mathbf{z}_i$ is sparse, there are only a few entries in $\mathbf{z}_i$ which are non-zero. We call its non-zero entries as its \textit{support}, i.e., supp($\mathbf{z}_i$) $= z_{i, j}: z_{i, j}\neq 0, j=1, \cdots, t$. $\mathbf{D} = (\mathbf{d}_1, \mathbf{d}_2, \cdots, \mathbf{d}_t) \in \mathbb{R}^{m\times{t}}$ is so called \textit{dictionary}, each column represents a basis vector.
Specifically, suppose there are $t$ atoms $\mathbf{d}_j \in \mathbb{R}^m, j = 1, 2, \cdots, t$, where the number of atoms is much smaller than $n$ (the total number of training image patches) but larger than $m$ (the dimension of the image patches). $\mathbf{x}_i$ can be represented by $\mathbf{x}_i = \sum_{j=1}^{t}z_{i, j}\mathbf{d}_j$. In this way, the $m$-dimensional vector $\mathbf{x}_i$ is represented by a $t$-dimensional vector $\mathbf{z}_i = (z_{i,1}, \cdots, z_{i, t})^T$ ($\mathbf{Z} = (\mathbf{z}_1, \cdots, \mathbf{z}_n) \in \mathbb{R}^{t\times n}$). To prevent an arbitrary scaling of the sparse codes, the columns $\mathbf{d}_i$ are constrained by $\mathbb{C}\overset{\triangle}{=}\{\mathbf{D}\in \mathbb{R}^{m\times{t}}$, s.t. $\forall j = 1, \cdots, t, \mathbf{d}_j^T\mathbf{d}_j \leq 1\}$.
Thus, we use ${\cal F}(\cdot)$ to represent the sparse coding problem for $\mathbf{X}$, we then rewrite ${\cal F}(\cdot)$ as a matrix factorization problem:
\small
\begin{equation}
\min_{\mathbf{D}\in \mathbb{C}, \mathbf{Z}} {\cal F}(\mathbf{D}, \mathbf{Z})\equiv\frac{1}{n}\sum_{i=1}^{n} f_i(\mathbf{D}, \mathbf{z}_i) =\frac{1}{2}||\mathbf{X} - \mathbf{D}\mathbf{Z}||^2_F+\lambda||\mathbf{Z}||_1
\label{eq:opt}
\end{equation}
\normalsize
where $||\cdot||_F$ is the Frobenius norm. Eq. \ref{eq:opt} is a non-convex problem. However, it is a convex problem when either $\mathbf{D}$ or $\mathbf{Z}$ is fixed. When the dictionary $\mathbf{D}$ is fixed, solving each sparse code $\mathbf{z}_i$ is a Lasso problem~\citep{Tibshirani94regressionshrinkage}. Otherwise, when the $\mathbf{Z}$ are fixed, it becomes a simple quadratic problem. Here we adopt the SCC method~\citep{lin2014stochastic} to optimize Eq.~\ref{eq:opt}, which has been studied in a number of prior work~\citep[e.g.,][]{lin2014stochastic,LvSparse,lv2015holistic,Lvtask,Zhang:ISBI17,zhang2016hyperbolic,zhang2017multi,zhang2018multi}. Following~\cite{lin2014stochastic}, we update $z_i^k$ via one or a few steps of coordinate descent (CD)~\citep{TTW08a}:
\begin{equation}\label{eq:cd-obj-a}
\mathbf{z}_i^k = \mathrm{CD}( \mathbf{D}^k_i, \mathbf{z}_i^{k-1},\mathbf{x}_i )
\end{equation}
The updated sparse code is then denoted by $\mathbf{z}^{k}_i$. A detailed derivation of CD utilizing software thresholding shrinkage function~\citep{Combettes2005-MMS} can be found in \cite{lin2014stochastic}. We then update the dictionary $\mathbf{D}$ by using stochastic gradient descent (SGD)~\citep{bottou1998online}:
\begin{equation}\label{eq:sgd}
\mathbf{D}^k_{i+1} = P_{\mathbb{C}}(\mathbf{D}_{i}^k - \eta_{i}^k \nabla_{\mathbf{D}_{i}^k} f_i(\mathbf{D}_{i}^k, \mathbf{z}_i^k))
\end{equation}
where $P$ is the shrinkage function, $\mathbb{C}$ is the feasible set of $\mathbf{D}$ and $\eta_i^k$ is the learning rate of $i$-th step in $k$-th epoch. We set the learning rate as an approximation of the inverse of the Hessian matrix $\mathbf{H}$. We illustrate the algorithmic framework in Fig.~\ref{fig:alg-frame}. At each iteration, with a ring-shaped patch $\mathbf{x}_i$, we perform one step of CD to find the supports of the sparse code $\mathbf{z}_i^{k-1}$. Next, we perform a few steps of CD on the supports to obtain a new sparse code $\mathbf{z}_i^{k}$. Then we update the supports of the dictionary by the second order SGD to obtain a new dictionary $\mathbf{D}^k_{i+1}$.
\begin{figure}[t]
\centering
\includegraphics[width=4in]{Figure/fig_HSC_v2.eps}
\caption{Illustration of hyperbolic stochastic coding (HSC) framework.}
\label{fig:alg-frame}
\vspace{-4mm}
\end{figure}
\iffalse
\begin{algorithm}[t]
\caption{Hyperbolic Stochastic Coding (HSC)}\label{alg:scc}
\KwIn{A collection of overlapped patches}
\KwOut{$\mathbf{D} \in \mathbb{R}^{m\times t}$ and $\mathbf{Z} = (\mathbf{z}_1 \cdots \mathbf{z}_n) \in \mathbb{R}^{t \times N}$}
Initialize $\mathbf{D}_1^1$ by selecting random ring-shape patches~\citep{coates2011importance} from the output of Algorithm~\ref{alg:FBS}, $H = 0$, $\mathbf{z}_i^0 =0$ and $i=1,\ldots, N$\\
\For{$k=1$ to $\kappa$}{\For{$i=1$ to $N$}{Get an input vector $\mathbf{x}_i$\\
Update $\mathbf{z}_i^k$ via one or a few steps of CD:\\\quad\quad$\mathbf{z}_i^k \leftarrow CD(\mathbf{D}_i^k, \mathbf{z}_i^{k-1}, \mathbf{x}_i)$\\Update the Hessian matrix and the learning rate:\\\quad\quad$\mathbf{H} \leftarrow \mathbf{H} + \mathbf{z}^k_i(\mathbf{z}_i^k)^T, \quad \eta_{i, j}^k = 1/h_{jj}$\\Update the supports of the dictionary via SGD:\\\quad\quad$d_{i+1, j}^k \leftarrow d^k_{i,j} - \eta_{i,j}^k z_{i,j}(\mathbf{D}^k_i\mathbf{z}_i^k - \mathbf{x}_i)$\\\If{$i=n$}{$\mathbf{D}^{k+1}_1 = \mathbf{D}^k_{n+1}$}}}
\end{algorithm}
\fi
\subsection{Max-Pooling and Adaboost Classifier}\label{sec:maxpooling}
With a trained dictionary $\mathbf{D}$, for a set of ring-shaped patches from a new subject, $\mathbf{x}_i$, $i=1, 2, ..., p$, $p$ is the patch number of an individual subject, we can learn its sparse features $\mathbf{z}_i$, $\mathbf{z_i}\in \mathbb{R}^t, i=1, 2, ..., p$. In theory, one could use all learned features as input data of a classifier but it poses intractable computational challenges. Thus, to describe our surface features efficiently, one natural approach is to aggregate statistics of these features at various locations. A key component of deep learning models, max-pooling~\citep{boureau2010theoretical} takes the most responsive node of a given region of interest. In our system, we borrow the idea of max-pooling and apply it to the extracted sparse coding surface features (sparse codes) from HSC (Fig.~\ref{fig:pipeline} (g)). Specifically, one could compute the max value for each feature ($t$ features obtained from HSC) over all patches ($p$ patches per subject), which is equivalent to applying a high-pass filter to the learned sparse codes. These summary statistics are much lower in dimension $t$ compared to using all of the learned surface patch features and reduce over fitting. Finally, Adaboost~\citep{rojas2009adaboost} classifier is used for binary classification, as shown in Fig.~\ref{fig:pipeline} (h).
| {
"redpajama_set_name": "RedPajamaArXiv"
} | 2,954 |
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