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Viktor Lampl (28. března 1889 Jimramov – 7. března 1942 Praha) byl český, německy hovořící architekt a stavební podnikatel.
Život
Narodil se 24. března 1889 v Jimramově, jako syn Karla Lampla a Emilie, rozené Müllerové (1856–1924). Měl sestru a tři bratry.
Dne 3. května 1913 byl na Německé vysoké škole technické v Praze promován na stavebního inženýra. Spolu s titulem ing. užíval i titul dr. S Otto Fuchsem byl společníkem pražské stavební firmy Lampl a Fuchs.
Zemřel v Praze–Karlíně, dříve, než ho stačil pro židovský původ postihnout osud jeho rodiny.
Rodinný život – vyvraždění rodiny
Viktor Lampl byl ženat, dne 28. března 1920 se v Praze–Karlíně oženil se Zdenkou, rozenou Reitlerovou (1889–1942), se kterou měl syna Petra Eduarda (1925–1942). Rodina bydlela v Karlíně, v Královské třídě 23 (nyní Sokolovská).
Manželka i syn se stali v roce 1942 oběťmi holocaustu. Zavražděni byli i sourozenci Berta Schuftanová (rozená Lamplová) a Richard Lampl
Zajímavost
V roce 1930 byl Viktor Lampl jmenován vyslanectvím republiky Kuba stavebním radou.
Dílo
Realizované stavby
Firma Lampl a Fuchs realizovala vlastní stavby i přístavby, pro některé zakupovala vlastní pozemky.
Podrobnosti o stavbách jsou uvedeny v článku Lampl und Fuchs.
Odkazy
Reference
Literatura
Související články
Lampl und Fuchs
Otto Fuchs
Čeští architekti
Narození 28. března
Narození v roce 1899
Narození v Jimramově
Úmrtí 7. března
Úmrtí v roce 1942
Úmrtí v Praze
Muži
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{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,060
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package org.apache.flink.table.planner.plan.nodes.calcite
import org.apache.flink.table.api.TableConfig
import org.apache.flink.table.catalog.{CatalogManager, FunctionCatalog}
import org.apache.flink.table.module.ModuleManager
import org.apache.flink.table.planner.calcite.{FlinkRelBuilder, FlinkTypeFactory, FlinkTypeSystem}
import org.apache.flink.table.planner.delegation.PlannerContext
import org.apache.flink.table.planner.plan.metadata.MockMetaTable
import org.apache.flink.table.planner.plan.stats.FlinkStatistic
import org.apache.flink.table.planner.plan.`trait`.FlinkRelDistributionTraitDef
import org.apache.flink.table.types.logical.LogicalType
import org.apache.flink.table.utils.CatalogManagerMocks
import org.apache.calcite.jdbc.CalciteSchema
import org.apache.calcite.plan.{Convention, ConventionTraitDef, RelOptCluster, RelTraitSet}
import org.apache.calcite.rel.hint.RelHint
import org.apache.calcite.rel.logical.LogicalTableScan
import org.apache.calcite.rel.RelCollationTraitDef
import org.apache.calcite.rex.RexBuilder
import org.apache.calcite.schema.SchemaPlus
import org.junit.Before
import java.util
/**
* A base class for rel node test.
* TODO refactor the metadata test to extract the common logic for all related tests.
*/
class RelNodeTestBase {
val tableConfig = new TableConfig()
val rootSchema: SchemaPlus = CalciteSchema.createRootSchema(true, false).plus()
val catalogManager: CatalogManager = CatalogManagerMocks.createEmptyCatalogManager()
val moduleManager = new ModuleManager
val plannerContext: PlannerContext = new PlannerContext(
tableConfig,
new FunctionCatalog(tableConfig, catalogManager, moduleManager),
catalogManager,
CalciteSchema.from(rootSchema),
util.Arrays.asList(
ConventionTraitDef.INSTANCE,
FlinkRelDistributionTraitDef.INSTANCE,
RelCollationTraitDef.INSTANCE
)
)
val typeFactory: FlinkTypeFactory = plannerContext.getTypeFactory
var relBuilder: FlinkRelBuilder = _
var rexBuilder: RexBuilder = _
var cluster: RelOptCluster = _
var logicalTraits: RelTraitSet = _
@Before
def setUp(): Unit = {
relBuilder = plannerContext.createRelBuilder("default_catalog", "default_database")
rexBuilder = relBuilder.getRexBuilder
cluster = relBuilder.getCluster
logicalTraits = cluster.traitSetOf(Convention.NONE)
}
/**
* Build a [[LogicalTableScan]] based on a [[MockMetaTable]] using given field names and types.
@param fieldNames String array
* @param fieldTypes [[LogicalType]] array
* @return a [[LogicalTableScan]]
*/
def buildLogicalTableScan(
fieldNames: Array[String],
fieldTypes: Array[LogicalType]): LogicalTableScan = {
val flinkTypeFactory = new FlinkTypeFactory(new FlinkTypeSystem)
val rowType = flinkTypeFactory.buildRelNodeRowType(fieldNames, fieldTypes)
val table = new MockMetaTable(rowType, FlinkStatistic.UNKNOWN)
LogicalTableScan.create(cluster, table, new util.ArrayList[RelHint]())
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,741
|
\section{Introduction}
\let\thefootnote\relax\footnotetext{Corresponding Author:Hanseok Ko.}
Deep learning has shown remarkable successes on various audio processing tasks, such as speech enhancement\cite{yin2019phasen, pascual2017segan}, Automatic Speech Recognition(ASR)\cite{hannun2014deep, chan2016listen}, sound classification\cite{mcdonnell2020acoustic, piczak2015environmental, kim2020dual, lee2018time}, and speech synthesis\cite{wang2017tacotron, shen2018natural}. To further enhance their performance, many research efforts have focused on designing better network architectures for specific tasks. While improving the architectures may deliver better performances, these methods tend to overfit easily and require large amounts of training data\cite{park2019specaugment, chiu2018state}. To avoid this problem, there have been some efforts in exploring data augmentation and regularization strategies.
For augmenting the audio dataset, there are two main approaches: time-domain waveforms and time-frequency domain features, such as spectrogram, mel-spectrogram, and mel-frequency cepstral coefficient. For the waveform data, the data augmentation strategies may include noise injection, changing pitch, changing speed, shifting time, and speed perturbation\cite{ko2015audio} of the waveform, to expand the data set without disturbing salient information therein. For the time-frequency domain features, Specaugment\cite{park2019specaugment} proposed time warping, frequency-masking, and time-masking data augmentation strategies. Although Specaugment is successfully applied to ASR, its application to other tasks has been limited\cite{park2020specaugment}. For example, in the speech enhancement task, zero-masking on the time and frequency axis tends to degrade the performance.
Since the time-frequency domain features are two dimensional and can be projected as a 2D image, data augmentation strategies, particularly of Mixed Sample Data Augmentation (MSDA) type in the computer vision domain, have been applied to the time-frequency domain features as shown in Fig. 1. Mixup\cite{zhang2017mixup, suh2020designing} blends two images of the audio features and labels by varying a random parameter $\gamma$. Its performance has been shown to be effective in the image classification tasks, however, due to the way it mixes magnitudes of spectrograms from different source components together it is difficult to disentangle them in the audio domain. Thus, the performance from the Mixup approach has been limited. Cutout\cite{devries2017improved} and Specaugment\cite{park2019specaugment} employ zero-masking to the image and spectrogram, respectively. Although these methods can be applied to images and spectrogram successfully, salient audio information can be lost due to zero-masking. Cutmix\cite{yun2019cutmix} randomly attaches a part of an image to another image. It applies a randomly generated mask for cutting a spectrogram region and pasting it randomly to another spectrogram region. While Cutmix can preserve magnitude information of $X_1$ and $X_2$, the time-frequency information taken from one image is randomly shifted to another resulting in a frequency shift.
\begin{figure}
\centerline{\includegraphics[width=\columnwidth]{Fig1.png}}
\caption{Overview of the data augmentation methods:Mixup, Cutmix, Specaugment and our Specmix.}
\end{figure}
As these methods have been adopted from image domain applications, their approach of mixing two different spectrograms has to be tailored for audio signals to preserve data distribution within a spectrogram structure for maintaining salient time-frequency correlations. Therefore, a different masking policy that preserves salient frequency features for MSDA is needed. Inspired by the previous data augmentation strategies in the speech and vision domains, we propose a novel audio data augmentation strategy, named SpecMix, for training with time-frequency domain features. The proposed method expands the idea of Cutmix, which attempted to cut-and-mix two data samples. However, their masking policy is based on image data, therefore it is not necessarily suitable for time-frequency domain features. To address this problem, we modified the masking policy tailored to time-frequency domain features. The proposed method can be integrated to ResNet, U-Net, and other state-of-the-art architectures for acoustic scene classification, sound event classification, or speech enhancement tasks.
\section{Specmix Policy}
In this section, we describe the Specmix algorithm in detail. We aim to construct an MSDA policy that directly acts on the time-frequency domain features, which improves the generalization of the model using time-frequency domain features.
\subsection{Algorithm on classification tasks}
Let $x\in \mathbb{R}^{F\times T\times C}$ and y denote a time-frequency domain features and its label, respectively. F denotes the number of the frequency bin, T denotes the number of the time bin and C denotes the number of time-frequency domain features. The goal of Specmix is to generate a new training sample $(\tilde{x}, \tilde{y})$ by combining two training samples $(x_A, y_A)$ and $(x_B, y_B)$. We define the combining operation as
\begin{align}
\tilde{x} &= \mathbf{M}\odot x_A + (\mathbf{1}-\mathbf{M}) \odot x_B \\
\tilde{y} &= \lambda y_A + (1-\lambda)y_B
\end{align}
where $\mathbf{M} \in \{0, 1\}^{F\times T}$ denotes a binary mask indicating where to drop out and fill in from two images, $\mathbf{1}$ is a binary mask filled with ones, and $\odot$ is element-wise multiplication. The combination ratio $\lambda$ between two data points is the number of pixels of $x_A$ in $\tilde{x}$.
In each training iteration, a mixed sample $(\tilde{x}, \tilde{y})$ is generated by combining two training samples selected from two mini-batches according to Equation (1) and (2).
\subsection{Algorithm on speech enhancement tasks}
Let $x\in \mathbb{R}^{F\times T\times C}$ and $z\in \mathbb{R}^{F\times T\times C}$ denote a time-frequency domain features of the noisy signal and related time-frequency domain features of the clean signal, respectively. The new training sample pair, $(\tilde{x}, \tilde{z})$, can be constructed by combining two training samples $(x_A, z_A)$ and $(x_B, z_B)$. We define the combining operation as
\begin{align}
\tilde{x} &= \mathbf{M}\odot x_A + (\mathbf{1}-\mathbf{M}) \odot x_B \\
\tilde{z} &= \mathbf{M}\odot z_A + (\mathbf{1}-\mathbf{M}) \odot z_B
\end{align}
where $\mathbf{M} \in \{0, 1\}^{F\times T}$ denotes a binary mask indicating where to drop out and fill in from two images, $\mathbf{1}$ is a binary mask filled with ones, and $\odot$ is element-wise multiplication.
In each training iteration, a mixed sample pair, $(\tilde{x}, \tilde{z})$, is generated by combining two training samples selected from two mini-batches according to Equation (1) and (2).
\subsection{Masking}
Figure 2 outlines the masking process of mixing audio features from two different samples. From frequency masking, we generate frequency bands up to three different segments while the time masking does the same in generating up to three different temporal bands. The starting frequencies and time are chosen from a random process of a uniform distribution from 0 to 1 while their widths are defined by the user. These bands form the binary mask enabling the mixing of two audio spectrograms.
\begin{enumerate}
\item Frequency masking: Up to three frequency bands can be masked depending on a random number $f_{times}$ sampled from a uniform distribution from 0 to 3 in an integer value. In each frequency band selection, the starting band frequency $f_{start}$ is selected by a random number from a uniform distribution from 0 to F. Next, the ending band frequency is calculated by the equation $f_{end} = f_{start} + \gamma F$ with $\gamma$ chosen by a user-defined parameter between 0 to 1. Repeat this process $f_{times}$ times.
\item Time masking: Again, up to three time bands can be masked in the identical process utilized for the Frequency masking described above. Again, $\gamma$, chosen by the user controls the width of the time band, and the random processes control the starting time and the number of temporal bands.
\end{enumerate}
\begin{figure}
\centerline{\includegraphics[width=\columnwidth]{Fig2.png}}
\caption{Augmentation policies applied to the base input, given $X_1$ and $X_2$. To generate new sample X, applying a time-frequency mask to $X_1$, inverted time-frequency mask to $X_2$, and summation of them.}
\end{figure}
\section{Models}
In this section, we introduce the preprocessing methods and models that were used to evaluate the proposed data augmentation strategy.
\subsection{Acoustic scene classification \& Sound event classification}
The input time-frequency domain features are mel-spectrogram, its delta and delta-delta features. The sampling rate is 44.1kHz, nfft is 2048 and hop length is 1024, and the number of mel filters is 128. The shape of processed input is $[B,F,T,C] = [B, 128, T, 3]$, where B denotes the batch size, F denotes the number of frequency bins, T denotes the total time bins and C denotes the total number of channels. T depends on the audio length.
We used Resnet-101\cite{he2016deep} on acoustic scene classification and sound event classification tasks. Adam\cite{kingma2014adam} optimizer and cross-entropy loss function were used to train our model with a batch size of 32. We employed learning rate decay from 1e-3 to 1e-7. We also employed the model and training procedure proposed in \cite{mcdonnell2020acoustic}, considered state-of-the-art acoustic scene classification model, to evaluate the generalization performance of the proposed method.
\subsection{Speech enhancement}
During training, the input waveform was cut or padded to make the waveform length of 32768 samples. The input time-frequency domain feature is a spectrogram. The sampling rate is 16kHz, nfft is 512, and hop length is 256. The shape of the processed input is $[B,F,T,C] = [B, 256, T, 2]$, where B denotes the batch size, F denotes frequency bins, T denotes time bins and C denotes channels. T depends on the audio length. The first channel is the real part of the spectrogram and the second channel is the imaginary part of the spectrogram.
We build a. U-Net\cite{ronneberger2015u} style speech enhancement model illustrated in Fig. 3. The model consists of 8 encoder layers, 1 mid-level layer, 8 decoder layers, and 1 last convolution layer. To deal with arbitrary length inputs, the strides of all the layers on the time axis are not squeezed. The model predicts phase sensitive mask\cite{erdogan2015phase} and reconstructs a clean spectrogram by applying the mask to a noisy spectrogram. During training, Adam\cite{kingma2014adam} optimizer and a mean squared error loss function is used to train our model with a batch size of 6. We employ learning rate decay from 1e-2 to 1e-5.
\begin{figure}
\centerline{\includegraphics[width=\columnwidth]{Fig3.png}}
\caption{A U-Net model for solving speech enhancement task.}
\end{figure}
\section{Experiments}
In this section, we evaluate Specmix on acoustic scene classification, sound event classification, and speech enhancement tasks. The effects of Specmix on acoustic scene classification and sound event classification are examined first. We then compare the result with changing $\gamma$ and masking policies. We also show that Specmix can improve noise reduction performance.
\subsection{Acoustic scene classification}
We evaluate Specmix on TAU Urban Acoustic Scenes 2020 Mobile benchmark\cite{Mesaros2018_DCASE}, the dataset containing recordings from 12 European cities in 10 different acoustic scenes using 4 different recording devices. The dataset consists of 10 classes of urban acoustic scene recordings with 13965 labeled clips for training and 2970 clips for the test. The evaluation metric is accuracy.
Results with ResNet-101 model are given in Table 1. We observe that Specmix achieves the best result (62.13\% accuracy) among the considered augmentation strategies. Specmix outperforms Mixup, Cutmix and Specaugment by +3.98\%, +2.59\% and +4.45\%, respectively. Interestingly, Mixup, Cutmix, and Specaugment performed worse than with no augmentation. We believe that these augmentation strategies resulted in some information loss, as mentioned in Section 1, leading to poor performances.
\begin{table}[t]
\caption{Comparison of state-of-the-art data augmentation methods for time-frequency domain features on TAU Urban Acoustic Scenes 2020 Mobile benchmark.(Model : ResNet-101)}
\centering
\resizebox{\linewidth}{!}{
\begin{tabularx}{\linewidth}{|A||A||A|}
\specialrule{.2em}{.1em}{.1em}
\multicolumn{1}{A|}{Model : ResNet-101} &
\multicolumn{1}{A}{Accuracy(\%)} \\ \hline
\multicolumn{1}{A|}{No augmentation} &
\multicolumn{1}{A}{59.60} \\
\multicolumn{1}{A|}{Mixup\cite{zhang2017mixup}} &
\multicolumn{1}{A}{58.15} \\
\multicolumn{1}{A|}{Cutmix\cite{yun2019cutmix}} &
\multicolumn{1}{A}{59.54} \\
\multicolumn{1}{A|}{Specaugment\cite{park2019specaugment}} &
\multicolumn{1}{A}{57.68} \\
\multicolumn{1}{A|}{Specmix $\gamma=0.3$} &
\multicolumn{1}{A}{\textbf{62.13}} \\
\specialrule{.2em}{.1em}{.1em}
\end{tabularx}}
\end{table}
\begin{figure}
\centerline{\includegraphics[width=0.8\columnwidth]{Fig4.png}}
\caption{Impact of $\gamma$ on TAU Urban Acoustic Scenes 2020 Mobile benchmark.}
\end{figure}
We evaluate Specmix with $\gamma \in \{0.1, 0.3, 0.5, 0.7\}$ and $\mathcal{U}[0,1]$ with ResNet-101 model. The results are given in Fig. 4. For all $\gamma$ values considered, Specmix improves the performance over the case with no augmentation. The best performance is achieved when $\gamma=0.3$.
Results with Specmix and the other augmentation methods applied to \cite{mcdonnell2020acoustic}'s model are given in Table 2. We observe that Specmix improves the performance by +0.31\%, +0.68\%, +2.53\% compared to Mixup, Cutmix, and Specaugment, respectively. We also performed an ablation study to observe the impacts of the masking strategies. The results are given in Table 3. Random masking is a masking strategy that randomly selects N pixels filled by X1 while the remaining pixels are filled by X2. Specmix(time only) only applies time masking and Specmix(freq. only) only applies frequency masking.
The results show that both time masking and frequency masking improve performance.
\begin{table}[t]
\caption{Comparison of state-of-the-art data augmentation methods for time-frequency domain features on TAU Urban Acoustic Scenes 2020 Mobile benchmark.(Model : \cite{mcdonnell2020acoustic})}
\centering
\resizebox{\linewidth}{!}{
\begin{tabularx}{\linewidth}{|A||A||A|}
\specialrule{.2em}{.1em}{.1em}
\multicolumn{1}{A|}{Model : \cite{mcdonnell2020acoustic}} &
\multicolumn{1}{A}{Accuracy(\%)} \\ \hline
\multicolumn{1}{A|}{No augmentation} &
\multicolumn{1}{A}{68.90} \\
\multicolumn{1}{A|}{Mixup\cite{zhang2017mixup}} &
\multicolumn{1}{A}{71.29} \\
\multicolumn{1}{A|}{Cutmix\cite{yun2019cutmix}} &
\multicolumn{1}{A}{70.92} \\
\multicolumn{1}{A|}{Specaugment\cite{park2019specaugment}} &
\multicolumn{1}{A}{69.07} \\
\multicolumn{1}{A|}{Specmix $\gamma=\mathcal{U}[0,1]$} &
\multicolumn{1}{A}{\textbf{71.60}} \\
\specialrule{.2em}{.1em}{.1em}
\end{tabularx}}
\end{table}
\begin{table}[t]
\caption{Impact of masking strategies on TAU Urban Acoustic Scenes 2020 Mobile benchmark.(Model : \cite{mcdonnell2020acoustic})}
\centering
\resizebox{\linewidth}{!}{
\begin{tabularx}{\linewidth}{|A||A||A|}
\specialrule{.2em}{.1em}{.1em}
\multicolumn{1}{A|}{Model : \cite{mcdonnell2020acoustic}} &
\multicolumn{1}{A}{Accuracy(\%)} \\ \hline
\multicolumn{1}{A|}{Random masking} &
\multicolumn{1}{A}{70.05} \\
\multicolumn{1}{A|}{Specmix(time only)} &
\multicolumn{1}{A}{70.42} \\
\multicolumn{1}{A|}{Specmix(freq. only)} &
\multicolumn{1}{A}{70.52} \\
\multicolumn{1}{A|}{Specmix} &
\multicolumn{1}{A}{\textbf{70.79}} \\
\specialrule{.2em}{.1em}{.1em}
\end{tabularx}}
\end{table}
\subsection{Sound event classification}
We evaluate on SECL\_UMONS benchmark\cite{brousmiche2020secl}. SECL\_UMONS is a real sound recording dataset for simultaneous classification and localization of sound events. The data samples in the SECL\_UMONS are recording in two indoor room environments. The room condition of the first one is RT60 of 0.7s with the room dimensions of $7.8\times3.6\times2.45$[m] and the second one is RT60 of 0.9s with room dimensions of $9.4\times7.5\times4.85$[m]. The dataset consists of 11 indoor event classes, such as \emph{chair\_movement}, \emph{cup\_drop\_off}, \emph{furniture\_drawer}, \emph{hand\_clap}, \emph{keyboard}, \emph{knock}, \emph{phone\_ring}, \emph{radio}, \emph{speaker}, \emph{step}, and \emph{whistle}. The dataset contains 2178 sequences for the training set and 484 sequences for the validation set. We used single-channel waveform for our comparison. Evaluate metric is accuracy.
Results with the ResNet-101 model are given in Table 4. Specmix achieves 97.107\% accuracy on SECL\_UMONS, +1.04\% higher than the no augmentation case.
\begin{table}[t]
\caption{Comparison of state-of-the-art data augmentation methods for time-frequency domain features on SECL\_UMONS benchmark.(Model : ResNet-101)}
\centering
\resizebox{\linewidth}{!}{
\begin{tabularx}{\linewidth}{|A||A||A|}
\specialrule{.2em}{.1em}{.1em}
\multicolumn{1}{A|}{Model : ResNet-101} &
\multicolumn{1}{A}{Accuracy(\%)} \\ \hline
\multicolumn{1}{A|}{No augmentation} &
\multicolumn{1}{A}{96.07} \\
\multicolumn{1}{A|}{Mixup\cite{zhang2017mixup}} &
\multicolumn{1}{A}{95.87} \\
\multicolumn{1}{A|}{Cutmix\cite{yun2019cutmix}} &
\multicolumn{1}{A}{95.66} \\
\multicolumn{1}{A|}{Specaugment\cite{park2019specaugment}} &
\multicolumn{1}{A}{96.90} \\
\multicolumn{1}{A|}{Specmix $\gamma=\mathcal{U}[0,1]$} &
\multicolumn{1}{A}{\textbf{97.11}} \\
\specialrule{.2em}{.1em}{.1em}
\end{tabularx}}
\end{table}
\subsection{Speech enhancement}
We evaluated Specmix on Voicebank + Diverse Environments Multichannel Acoustic Noise Database(DEMAND) benchmark, which is proposed by \cite{valentini2016investigating}. Noisy and clean speech recordings were provided from the DEMAND\cite{thiemann2013diverse} and the Voice Bank corpus\cite{veaux2013voice}, respectively with each recorded with the sampling rate of 48kHz. A total of 40 different noise conditions are considered in the training set and 20 different conditions are considered in the test set. Finally, the training and test set contained 11572 and 824 noisy-clean speech pairs, respectively. Note that the speaker and noise classes were uniquely selected for the training and test sets. Evaluation metrics are Perceptual Evaluation of Speech Quality(PESQ), mean opinion score predictor of signal distortion(CSIG), background noise intrusiveness(CBAK), overall signal quality(COVL), and Segmental Signal to Noise Ratio(SSNR).
Results with the U-Net model are given in Table 5, where NA denotes No Augmentation, MU denotes Mixup, CM denotes Cutmix, and SA denotes SpecAugment. Mixup and Specaugment have failed to improve the speech enhancement performance over the no augmentation. However, Cutmix improves the performance of speech enhancement, and Specmix achieves the best results compared to the other data augmentation strategies.
We evaluate Specmix with $\gamma \in \{0.1, 0.3, 0.5, \mathcal{U}[0,1]\}$ with U-Net model. The results are given in Fig. 5. Although it is not shown, all the cases with $\gamma$ values utilized in Specmix improved performance over the case with no augmentation. The best performance is achieved when $\gamma=0.3$.
\begin{table}[t]
\caption{Comparison of state-of-the-art data augmentation methods for time-frequency domain features on Voice bank + DEMAND benchmark.(Model : U-Net).}
\centering
\resizebox{\linewidth}{!}{
\begin{tabularx}{\linewidth}{|A||A||A||A||A||A|}
\specialrule{.2em}{.1em}{.1em}
\multicolumn{1}{A|}{} &
\multicolumn{1}{A}{PESQ} &
\multicolumn{1}{A}{CSIG} &
\multicolumn{1}{A}{CBAK} &
\multicolumn{1}{A}{COVL} &
\multicolumn{1}{A}{SSNR} \\ \hline
\multicolumn{1}{A|}{Noisy} &
\multicolumn{1}{A}{1.97} &
\multicolumn{1}{A}{3.35} &
\multicolumn{1}{A}{2.44} &
\multicolumn{1}{A}{2.63} &
\multicolumn{1}{A}{1.67} \\
\multicolumn{1}{A|}{NA} &
\multicolumn{1}{A}{2.50} &
\multicolumn{1}{A}{3.44} &
\multicolumn{1}{A}{3.18} &
\multicolumn{1}{A}{2.95} &
\multicolumn{1}{A}{9.26} \\
\multicolumn{1}{A|}{MU} &
\multicolumn{1}{A}{2.44} &
\multicolumn{1}{A}{3.39} &
\multicolumn{1}{A}{3.16} &
\multicolumn{1}{A}{2.89} &
\multicolumn{1}{A}{9.43} \\
\multicolumn{1}{A|}{CM} &
\multicolumn{1}{A}{2.52} &
\multicolumn{1}{A}{3.50} &
\multicolumn{1}{A}{3.22} &
\multicolumn{1}{A}{2.99} &
\multicolumn{1}{A}{9.48} \\
\multicolumn{1}{A|}{SA} &
\multicolumn{1}{A}{2.41} &
\multicolumn{1}{A}{3.48} &
\multicolumn{1}{A}{3.16} &
\multicolumn{1}{A}{2.93} &
\multicolumn{1}{A}{9.35} \\
\multicolumn{1}{A|}{Specmix} &
\multicolumn{1}{A}{\textbf{2.54}} &
\multicolumn{1}{A}{\textbf{3.60}} &
\multicolumn{1}{A}{\textbf{3.24}} &
\multicolumn{1}{A}{\textbf{3.05}} &
\multicolumn{1}{A}{\textbf{9.57}} \\
\specialrule{.2em}{.1em}{.1em}
\end{tabularx}}
\end{table}
\begin{figure}
\centerline{\includegraphics[width=0.9\columnwidth]{Fig5.png}}
\caption{Impact of $\gamma$ on Voice Bank + DEMAND benchmark.}
\end{figure}
\section{Conclusion}
In this paper, we proposed a data augmentation strategy, named Specmix, for training with time-frequency domain features.
While there are augmentation strategies for mixing two different audio sources, our proposed method was shown to preserve spectral information throughout the augmentation process.
The proposed Specmix was shown easy to be incorporated into existing training pipelines and it has a low computational cost. From several evaluations of the proposed augmentation, the method improved the performance over various models on audio scene classification, sound event classification, and speech enhancement tasks. We expect that Specmix can be applied to various machine learning tasks using time-frequency domain features.
\section{Acknowledge}
This material is based upon work supported by the Air Force Office of Scientific Research under award number FA2386-19-1-4001.
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\section{Introduction}
\label{sec:orgb75ea6f}
The problem of predicting pedestrian paths is crucial for the development of autonomous vehicles(AVs) that have to handle complex interactions with pedestrians in urban traffic environments e.g. pedestrians recklessly crossing the road, and vehicles speeding up to overtake a pedestrian. To address this problem multiple datasets have been developed \cite{yang_top-view_2019,bhattacharyya_euro-pvi_2021,chandra_meteor_2021,bock_ind_2019}. Even though existing datasets cover diverse vehicle-pedestrian interaction scenarios, vehicle-pedestrian interaction in near accident scenarios is especially important to avoid situations that can lead to safety hazards and is challenging because pedestrians exhibit unpredictable maneuvers with sudden speed changes \cite{alhajyaseen_studying_2017}. Those sudden behavioral changes are hard to predict by an AV as verified by \cite{wang_adaptability_2020} where the authors argued that safely maneuvering through jaywalkers is a complex phenomenon that needs attention.
Near accident, data is difficult to collect \cite{wu_novel_2018}. This has led to the development of multiple data-collecting approaches. \citet{wu_improved_2020} examined the usage of a roadside LIDAR sensor to collect trajectories of road users and identify vehicle-pedestrian near-crash events. \citet{lee_high_2016} conducted a study to collect physiological data from night-shift workers in near-crash scenarios where 43.8\% of the drives were terminated early for safety reasons. \citet{nasernejad_modeling_2021} developed an agent-based framework to model pedestrian behavior in near misses and used surveillance camera footage to validate their method. \citet{anik_investigation_2021} collected jaywalkers' trajectories from surveillance camera footage and proposed an artificial neural network to predict jaywalkers' trajectory on a mid-block location in Bangladesh-India. To train or calibrate such models is important to have publicly available ground truth trajectory data. Nevertheless to the best of the author's knowledge, this is the first publicly available trajectory dataset that covers near-accident scenarios in Latin America.
\emph{I see you} captures vehicle-pedestrian avoidance behaviors in dangerous situations and scenarios where vehicle-pedestrian are very close but do not represent a dangerous situation (Figure \ref{fig:danger}) as this would provide valuable negative feedback. For each case, we provide processed vehicle and pedestrian trajectories in GPS coordinates.
\begin{figure}[htbp]
\centering
\includegraphics[angle=0,width=12cm]{./images/dangerous.pdf}
\caption{\label{fig:danger}Vehicle-Pedestrian interactions captured by our dataset. Example interaction 1 shows a dangerous situation. Example Interaction 2 shows a non-dangerous situation where vehicle-pedestrian were very close.}
\end{figure}
\section{\emph{I see you} Dataset}
\label{sec:org72b2e83}
We developed a pipeline to collect \emph{I see you} (Figure \ref{fig:flow}). The object detection task was performed using a YOLOv5 \cite{glenn_jocher_2020_4154370}, for tracking we used StrongSORT \cite{yolov5-strongsort-osnet-2022} then we used a linear Kalman filter \cite{yang_top-view_2019} to remove noise from the trajectories and finally for transformation to GPS coordinates we used Perspective-n-Point \cite{Tang17AIC}.
\begin{figure}[htbp]
\centering
\includegraphics[angle=0,width=12cm]{./images/flow_diagram.pdf}
\caption{\label{fig:flow}Pipeline for the proposed system.}
\end{figure}
Video footage was collected from publicly available traffic surveillance videos at seven signalized intersections in Cusco-Peru (Figure \ref{fig:intersections}). The seven intersections were selected near schools (Figure \ref{fig:cam1}), colleges (Figure \ref{fig:cam2}), churches (Figure \ref{fig:cam3}), and hospitals (Figure \ref{fig:cam4}) to ensure a diverse distribution in pedestrian characteristics (genre, age). These intersections also exhibit the transit from particular vehicles, taxis, and public transport buses where the latter show, particularly aggressive driving behaviors.
Each video is 18 hours long taken from 6 am to 11:59 pm on February 5, 2022. Every video has daytime, sunset and nighttime scenes these variations in illumination lead us to fine-tune the YOLOv5 models used for detection. For each category (vehicle, pedestrian) we used a different pre-trained YOLOv5 model, the vehicle category is detected by the model used in \cite{hu_turning_2021} and the pedestrian category is detected by a model pre-trained in the COCO dataset.
\begin{figure}[h]
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=5.7cm]{./images/cam1.jpg}
\caption{Servicentro}
\label{fig:cam1}
\end{subfigure}
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=5.7cm]{./images/cam2.pdf}
\caption{Univesitaria}
\label{fig:cam2}
\end{subfigure}
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=5.7cm]{./images/cam3.jpg}
\caption{Santa Ursula}
\label{fig:cam3}
\end{subfigure}
\begin{subfigure}[b]{0.5\textwidth}
\centering
\includegraphics[width=5.7cm]{./images/cam4.pdf}
\caption{Haya de la Torre}
\label{fig:cam4}
\end{subfigure}
\caption{Illumination variations of the collected footage in four intersections.}
\label{fig:intersections}
\end{figure}
To capture all illumination variations we sampled the 18-hour videos saving a frame every 6 minutes resulting in 4246 images in total. Employing a \(90\%\) train, \(10\%\) test distribution, we obtained the results described in Table \ref{tab:yolo_results} for object detection in the test subset for each category.
\begin{table}[htbp]
\caption{\label{tab:yolo_results}YOLOv5 Training Results}
\centering
\begin{tabular}{lrrrr}
\hline
Category & P & R & mAP@0.5 & mAP@0.5:0.95\\
\hline
Vehicle & \(0.934\) & \(0.916\) & \(0.974\) & \(0.76\)\\
Pedestrian & \(0.864\) & \(0.853\) & \(0.91\) & \(0.595\)\\
\hline
\end{tabular}
\end{table}
Since the YOLOv5 models had already been trained on the scenes every 6 minutes we had to make sure that these same scenes were not presented to the detector on the tracking step, this required a new sampling approach. We noted that if a frame was taken when the traffic light was red, the scene remained almost the same until the traffic light turned green. In the worst-case scenario a frame could have been taken at the same time that the traffic light turned to red, this implies that a scene could remain almost the same for up to 90 seconds (maximum duration of red light). To overcome that problem we separated the 6 minutes interval between frames into 3 intervals 2 minutes long each, then we saved the intermediate 2-minute interval (Figure \ref{fig:sampling}) for the tracking step.
\begin{figure}[htbp]
\centering
\includegraphics[angle=0,width=10cm]{./images/sampling_2-eps-converted-to.pdf}
\caption{\label{fig:sampling}Sampling approach used to save each 2-minute clip}
\end{figure}
The tracking step was conducted using StrongSORT \cite{yolov5-strongsort-osnet-2022}, this tracker algorithm used the YOLOv5 models previously fine-tuned to detect the road users in the 2-minute clips and extract the trajectories of each category assigning them id numbers that were used in the manual labeling step. The resulting trajectories were noisy with spikes indicating unreal accelerations and speeds, this fact lead us to use a linear Kalman filter using the approach developed by \citet{yang_top-view_2019}.
The filtered trajectories were manually labeled with the id generated by the tracker of the vehicle and pedestrian involved in the near-accident scenario. The labeled trajectories were converted from pixel coordinates to GPS coordinates using semi-automatic camera calibration based on Perspective-n-Point (PnP) \cite{Tang17AIC} . Finally, we compare our dataset with other trajectory datasets used for pedestrian trajectory prediction in various cases (Table \ref{tab:dataset_compare}).
\begin{table}[htbp]
\caption{\label{tab:dataset_compare}Comparison with Other Public Available Trajectory Datasets}
\centering
\begin{tabular}{lllllll}
\hline
Dataset & Scenarios & Interaction & Agents & Method of & FPS & Amount\\
& & Type & & Annotation & & of Trajec-\\
& & & & & & tories\\
\hline
DUT \cite{yang_top-view_2019} & Campus & Vehicle-Crowd & Car, Pedestrian & CSRT tracker & 23.98 & 1793\\
& & in shared spaces & & & & \\
\hline
\emph{I see you} & Signalized & Vehicle-Pedestrian & Car, Pedestrian & manual & 30 & 340\\
& Intersections & in near-accident & & & & \\
& & scenarios & & & & \\
\hline
ETH \cite{pell2009} & Campus, & Pedestrians in & Pedestrian & manual & 2.5 & 650\\
& Urban Street & busy scenarios & & & & \\
\hline
UCY \cite{lerner_crowds_2007} & Urban Street, & multi-human & Pedestrian & manual & 2.5 & 909\\
& Campus, Park & interaction scenarios & & & & \\
\hline
\end{tabular}
\end{table}
\section{Statistics}
\label{sec:org6220385}
Speed is preferred to train pedestrian trajectory prediction models rather than absolute position \cite{becker_evaluation_2018,noh_novel_2022}. This is because it does not depend on the reference system and therefore it can be used as a more general way to describe motion. In Figure \ref{fig:data} speed distributions were calculated for vehicles and pedestrians. Both distributions show peaks at \(0\frac{km}{h}\) this is because vehicles and pedestrians are stopped by traffic lights at signalized intersections. Also, each distribution has a second peak that shows the mean speed of vehicles and pedestrians during the near-accident scenarios.
\begin{figure}[htbp]
\centering
\includegraphics[angle=0,width=10cm]{./images/stat-eps-converted-to.pdf}
\caption{\label{fig:data}Speed distributions for vehicle and pedestrian categories in \emph{I see you}}
\end{figure}
\section{Conclusions and Future Work}
\label{sec:org347ebe0}
In this work, we addressed the need for trajectory data in near-accident scenarios for which we have developed a pipeline for collecting the trajectories of pedestrians and vehicles. During this process, we have assessed the limitations of our pipeline in which manual labeling is a process that can be automated using clusters to select the trajectories that correspond to interactions in near accident scenarios and also could be expanded to other types of interactions by selecting the appropriate clusters for each type of interaction. Pedestrian risk-seeking behaviors could be similar in other cities, however, to use the developed pipeline in other locations it would be necessary to fine-tune the YOLOv5 models in those locations to get good detection results.
\section*{Acknowledgements}
\label{sec:org0615ea2}
We would like to thank the Data Analysis Laboratory(LAAD) for providing the GPU used for this project and Rodolfo Quispe for his continuous support and feedback.
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301 North Carolwood Drive is a private residence located at 301 North Carolwood Drive in the city of Los Angeles, ranking as the twelfth largest private residence in the Los Angeles metropolitan area. The mansion was completed in 2016 on one of the most expensive streets in the world.
History
The Mon Rêve estate originally stood at 301 North Carolwood Drive, once owned by celebrity Barbra Streisand. When Streisand owned the property it featured a 9,500 square foot home with five bedrooms and seven baths. The walled and gated Mediterranean-style home featured a pool, two-story living room, sunrooms, library and screening room. Streisand had decorated the interior with her Art Nouveau collection.
The home was sold to music executive Les Bider in 2000 and demolished shortly thereafter. The vacant lot was listed at US $13,995,000 by Linda May Properties, purchased in 2014 by David Bohnett for US $13.25 million. Bohnett used the land as his own personal park before selling it to Dream Projects LA and developer Gala Asher.
The current home was completed in 2016 and listed at US $150 million on April 12, 2016, the second most expensive residence for sale in the United States at the time.
It was purchased in 2016 by billionaire Tom Gores, owner of the Detroit Pistons, in a US $100 million transaction in which several properties changed hands.
Description
The main house contains including 10 bedrooms and 20 bathrooms. The master bedroom occupies and opens to a heated covered patio. Paying homage to the history of the property, Club Mon Rêve is an entertainment facility within the house that includes a wine room, a lounge, and a movie theater complex with a separate valet entrance for guests.
Other amenities include a water wall that empties into an indoor lap pool and hot tub, and a spa complete with a hair salon, a manicure/pedicure area, and steam and massage rooms. The 2.17-acre property includes two outdoor infinity pools, a hiking trail marked by lighted trees, basketball and tennis courts, several guest houses, parking for more than 50 vehicles, and a 10-car garage, a total of of indoor space.
See also
List of largest houses in the Los Angeles Metropolitan Area
List of largest houses in the United States
References
External links
Photographs of 301 North Carolwood Drive.
Houses in Los Angeles
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Werner Seeger (* 6. November 1953 in Herford) ist ein deutscher Internist und Pneumologe. Seit 2006 ist er Ärztlicher Geschäftsführer des Universitätsklinikums Gießen und Marburg und seit 2011 Vorsitzender des Deutschen Zentrums für Lungenforschung.
Leben
Werner Seeger studierte Medizin in Münster und Gießen. Von 1979 bis 1982 war er wissenschaftlicher Mitarbeiter und Assistenzarzt am Institut für Klinische Chemie und Pathobiochemie der Justus-Liebig-Universität Gießen (JLU), anschließend war er von 1982 bis 1987 am Klinikum der JLU tätig. Im Jahr 1987 habilitierte sich Seeger in Gießen mit einer Arbeit für das Fach "Innere Medizin und Pathophysiologie" und wurde im selben Jahr Oberarzt. 1991 wurde er an der JLU zum Professor und Leiter der Klinischen Forschergruppe "Respiratorische Insuffizienz" ernannt und 1996 zum Professor für Innere Medizin/Pneumologie.
Im Jahr 2000 wurde er Geschäftsführender Direktor des Zentrums für Innere Medizin an der JLU und im Jahr 2006 Ärztlicher Geschäftsführer des Universitätsklinikums Gießen und Marburg am Standort Gießen. Seit 2007 ist Seeger im Direktorium des Max-Planck-Instituts für Herz- und Lungenforschung in Bad Nauheim und seit 2011 ist er Vorsitzender des Deutschen Zentrums für Lungenforschung.
Von 1999 bis 2001 war Seeger Präsident der Deutschen Gesellschaft für Internistische Intensivmedizin und Notfallmedizin, von 2001 bis 2002 Präsident der Deutschen Gesellschaft für Pneumologie sowie von 2005 bis 2006 Präsident der Deutschen Gesellschaft für Innere Medizin. Von 1997 bis 2009 war er Sprecher des Sonderforschungsbereiches "Kardiopulmonales Gefäßsystem", von 2002 bis 2009 Mitglied der DFG-Senatskommission für Klinische Forschung und von 2003 bis 2009 Mitglied des Wissenschaftsrates. Seit 2012 engagiert sich Werner Seeger auch als Mitglied des wissenschaftlichen Beirats des Vereins Aufklärung gegen Tabak für die Prävention des Rauchens auf Schulebene durch Medizinstudierende. Er ist Mitglied in verschiedenen internationalen und nationalen Fachgesellschaften, darunter die Leopoldina, Ausschüssen und Kommissionen und gilt als einer der renommiertesten Lungenspezialisten.
Auszeichnungen
Seegers Habilitationsschrift wurde 1987 mit dem Schunk-Preis für Humanmedizin der Justus-Liebig-Universität Gießen, und 1988 mit dem Theodor-Frerichs-Preis der Deutschen Gesellschaft für Innere Medizin ausgezeichnet.
2008 erhielt er den mit 50.000 Euro dotierten Robert Pfleger-Forschungspreis der Robert Pfleger-Stiftung mit Sitz in Bamberg.
2019 wurde Seeger zusammen mit Erika von Mutius, Klaus F. Rabe und Tobias Welte vom Deutschen Zentrum für Lungenforschung mit dem Balzan-Preis ausgezeichnet.
Literatur
Klaus Wilhelm: Grenzgänger zwischen Klinik und Labor. In: MaxPlanckForschung, (Hrsg.) Max-Planck-Gesellschaft zur Förderung der Wissenschaften e. V., 1/2011, S. 64–71.
Weblinks
University of Giessen Lung Center (UGLC), Werner Seeger, MD
Willi Weitzel trifft Professor Seeger (Video-Interview)
Lebenslauf Werner Seeger
Einzelnachweise
Person (Herford)
Pneumologe
Hochschullehrer (Justus-Liebig-Universität Gießen)
Mitglied der Leopoldina (20. Jahrhundert)
Wissenschaftliches Mitglied der Max-Planck-Gesellschaft
Balzan-Preisträger
Absolvent der Justus-Liebig-Universität Gießen
Deutscher
Geboren 1953
Mann
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Home » People & Places, Trials » G.O.B. to Appeal Supreme Court's Jalacte Road Decision
Cordel Hyde
The Government of Belize is appealing a recent decision by the Supreme Court in which the Maya community of Jalacte, including two villagers, was awarded six point three million dollars in a lawsuit filed after the construction of the Jalacte Road. The claim had to do with government's failure to get the free, prior and informed consent of the Maya people where it concerns the use of communal lands. According to Deputy Prime Minister Cordel Hyde, Minister of Natural Resources, a new FPIC protocol is being put together to address such issues going forward. This is in line with the consent order by the Caribbean Court of Justice.
Cordel Hyde, Minister of Natural Resources
"Yes, we are appealing the case to say that it is a real problem in the areas that are known as Mayan communal lands. Separate from that we have a big challenge in terms of balancing the rights and responsibilities of people who own lands there previous to this judgment and also making sure that we protect, that we preserve the rights of Mayan people who have won that landmark decision in the C.C.J. So that's a difficult thing for us but right now we are before the C.C.J. I believe that we were supposed to provide a draft FPIC protocol, Free Prior and Informed Consent Protocol that will govern how we administrate the situation in Toledo but we asked for an extension and I think sometime in September we are supposed to provide that. I know we have a first draft. We've been working diligently on that trying to finesse it and make it such that everyone can be satisfied that we preserved everyone's rights…."
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The complex problem of environmental management is scoped by both time and geometric dynamics. Bayesian networks provide a probabilistic framework to deal with these dynamics. Because of its conditional independence assumption, it can handle high-dimensional spaces very well. It also provides a powerful 'what-if' analysis interface for managerial and scenario support. The use of Bayesian networks and similar graph theory in order to approximate Bayes-Nash equilibria is a novel research area. The application field is computation of optimal persuit strategies.
The poaching of rhino has exhibited a dramatic annual increase since 2008. Whereas there were only 83 rhinos poached in South Africa in 2008, the number have increased to over 1000 in 2013. The number is current approximately 1175 per year at the end of 2015. This is slightly down from the previous year's number of 1215 owing to increased efforts by the rangers, even though the number of incursions by poachers has continued to increase. Poaching co-occurs with other crimes such as Illegal weapons, Illegal immigration, money laundering, VAT fraud, drugs, murder and attempted murder, corruption, house robberies for large calibre rifles, entrapment and transnational organised crime. The black market price of rhino horn has increased drastically from R8000 per kilogram to more than R1 000 000 per kilogram in 2015, making this form of environmental crime highly profitable. Researcher in the department developed and implemented a geo-spatial predictive model that could be used to understand the poaching problem, to reason about rhino poaching in the park, as well as make predictions. The model is integrated in a information collaboration platform developed by the CSIR and is operational within the Kruger National Park.
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Fritz Heid (* 5. Dezember 1916 in Sissach, Kanton Basel-Landschaft; † 12. Januar 2010 ebenda; heimatberechtigt in Arisdorf) war ein Schweizer Grafiker, Innenarchitekt, bildender Künstler und Bildhauer.
Leben und Werk
Heid absolvierte eine Malerlehre und liess sich anschliessend an der Allgemeinen Gewerbeschule Basel zum Grafiker ausbilden. Später trat er mit seinem Bruder Ernst in das väterliche Schreinergeschäft «Ernst Heid AG» ein. In diesem war er als Innenarchitekt und Werbeleiter tätig. Heid schuf Siebdrucke, Zeichnungen, Holz- und Linolschnitte, Aquarelle, Illustrationen, Bühnenbilder, Bronzeplastiken und Steinskulpturen. Er war u. a. mit Fritz Bürgin, Ugo Cleis und Eugen Häfelfinger befreundet. Heid war mit der Künstlerin Julia Ris verheiratet.
Weblinks
.
Einzelnachweise
Innenarchitekt
Grafiker (Schweiz)
Bildender Künstler (Schweiz)
Person (Kanton Basel-Landschaft)
Schweizer
Geboren 1916
Gestorben 2010
Mann
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La battaglia di Angaur è stata una battaglia della campagna del Pacifico combattuta durante la seconda guerra mondiale.
Nella metà del 1944 erano presenti su questa isola 1.400 soldati giapponesi al comando del tenente generale Sadao Inoue, comandante del settore di Palau.
La scarsa difesa delle Palau e la possibilità di costruirvi un campo d'aviazione resero Angaur un obiettivo attraente per i comandi americani dopo la conquista delle Isole Marshall. La mancanza di mezzi da sbarco significò però che l'attacco contro le Palau non poté cominciare prima che venissero terminate le operazioni nelle Marianne.
I bombardamenti, effettuati dalla corazzata Tennessee e dai bombardieri in picchiata Douglas Dauntless che partivano dalla portaerei USS Wasp (CV-18), incominciarono il giorno 11 febbraio 1944.
Sei giorni dopo l'81ª Divisione di fanteria, comandata dal maggior generale Paul J. Mueller, sbarcò a nord-est e a sud-ovest sulla costa dell'isola.
Le mine e la congestione delle spiagge inizialmente crearono diversi problemi. I giapponesi effettuarono diversi contrattacchi ma la resistenza si irrigidiva a mano a mano le truppe americane si avvicinavano alla collina, da loro soprannominata The Bowl ("la tazza"), situata vicino al lago Salome, nel nord-ovest dell'isola.
Su questa collina i giapponesi avevano previsto di concentrare il nucleo della loro resistenza. Il 20 settembre 1944 il 322º battaglione attaccò la collina. I 750 difensori risposero con il fuoco dell'artiglieria, dei mortai e delle mitragliatrici.
Gradualmente la fame, la sete e il fuoco dell'artiglieria americana cominciarono a indebolire i difensori e per il 25 settembre gli americani avevano iniziato a penetrare nelle difese della collina. Piuttosto che combattere per il possesso delle caverne gli americani usarono i bulldozer per chiuderne le entrate.
Il 30 settembre i combattimenti sull'isola erano finiti.
Il campo di aviazione venne costruito mentre la battaglia era ancora in corso. Il ritardo con il quale le operazioni nelle Palau erano state avviate fece sì che però non fosse pronto per le operazioni contro le Filippine che ebbero inizio nell'ottobre 1944.
L'Ammiraglio W. Halsey Jr. aveva detto che le operazioni contro le Palau non erano necessarie, e gli storici militari concordano con lui, affermando che il principale vantaggio ricavato da questa operazioni fu l'esperienza di combattimento guadagnata dalla 81ª Divisione.
Note
Bibliografia
Altri progetti
Guerra nel 1944
Angaur
Battaglie della seconda guerra mondiale che coinvolgono gli Stati Uniti d'America
Battaglie della seconda guerra mondiale che coinvolgono il Giappone
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{"url":"https:\/\/zbmath.org\/?q=an:0962.15013","text":"# zbMATH \u2014 the first resource for mathematics\n\n##### Examples\n Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. \"Topological group\" Phrases (multi-words) should be set in \"straight quotation marks\". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. \"Quasi* map*\" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. \"Partial diff* eq*\" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.\n\n##### Operators\n a & b logic and a | b logic or !ab logic not abc* right wildcard \"ab c\" phrase (ab c) parentheses\n##### Fields\n any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)\nOn completely copositive and decomposable linear transformations. (English) Zbl\u00a00962.15013\nCharacterization theorems and other results for the cone of completely copositive linear transformations are given. Further results including the cone of decomposable linear transformations and its dual follow.\n\n##### MSC:\n 15B48 Positive matrices and their generalizations; cones of matrices\nFull Text:\n##### References:\n [1] Barker, G. P.: Theory of cones. Linear algebra appl. 39, 263-291 (1981) \u00b7 Zbl\u00a00467.15002 [2] Barker, G. P.; Hill, R. D.; Haertel, R. D.: On the completely positive and positive-semidefinite-preserving cones. Linear algebra appl. 56, 221-229 (1984) \u00b7 Zbl\u00a00534.15012 [3] Choi, M.: Completely positive linear maps on complex matrices. Linear algebra appl. 10, 285-290 (1975) \u00b7 Zbl\u00a00327.15018 [4] Choi, M. D.: Some assorted inequalities for positive linear maps on c\\ast-algebras. J. oper. Theory 4, 271-285 (1980) \u00b7 Zbl\u00a00511.46051 [5] I. Glazman, J. Lubic, Finite Dimensional Linear Analysis: A Systematic Presentation in Problem Form, MIT Press, Caimbridge, MA, 1974 [6] Hill, R. D.; Jr., G. H. Bernet: Properties of the transpose mapping. Math. magazine 50, 151-152 (1977) \u00b7 Zbl\u00a00362.15007 [7] S. Kye, A class of atomic positive linear maps in three-dimensional matrix algebras, Publ. Res. Int. Math Sci. (1992) [8] Oxenrider, C. J.; Hill, R. D.: On the matrix reorderings ${\\Gamma}$ and ${\\Psi}$. Linear algebra appl. 69, 205-212 (1985) \u00b7 Zbl\u00a00573.15001 [9] Poluikis, J. A.; Hill, R. D.: Completely positive and Hermitian-preserving linear transformations. Linear algebra appl. 35, 1-10 (1981) \u00b7 Zbl\u00a00451.15013 [10] Tam, B. S.: Some results of polyhedral cones and simplicial cones. Linear and multilinear algebra 4, 281-284 (1977) \u00b7 Zbl\u00a00352.15009 [11] Tang, W.: On positive linear maps between matrix algebras. Linear algebra appl. 79, 33-44 (1986) \u00b7 Zbl\u00a00612.15012 [12] Tanahashi, K.; Tomiyama, J.: Indecomposable positive maps in matrix algebras. Canad. math bull. 31, 308-317 (1988) \u00b7 Zbl\u00a00679.46044 [13] D.A. Yopp, Cone preserving linear transformations, D.A. Thesis, Idaho State University, Pocatello, ID, 1998 [14] D.A. Yopp, R.D. Hill, Extreme rays of the cone of linear transformations that preserve the positive semidefinite matrices, in progress [15] D.A. Yopp, R.D. Hill, On the cone of completely positive maps, Linear Algebra Appl. 304 (2000) 119--129 \u00b7 Zbl\u00a00959.15026 [16] Woronowics, S. L.: Nonextendible positive maps. Commun. math. Phys. 51, 243-282 (1976) \u00b7 Zbl\u00a00342.46055","date":"2016-05-05 03:07:08","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\": 0.6734563708305359, \"perplexity\": 9102.52907418652}, \"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-2016-18\/segments\/1461860125857.44\/warc\/CC-MAIN-20160428161525-00102-ip-10-239-7-51.ec2.internal.warc.gz\"}"}
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Q: Make many circles with st_buffer at multiple geographic locations I'm trying to create circles from a dataframe containing longitudes and latitudes, taking into consideration that the radius of every circle needs to be 400 Nautical Miles precisely. There are many points (approx. 8000) corresponding to aiports all over the world. I have been using the sf package to create circles with st_buffer, but this function takes distance argument in degrees. I`ve seen resolved questions that do not solve my problem completely since they involve single circle at a specific place for which a grid can be used. Here is the code I use:
library(units)
library(tidyverse)
library(sf)
library(mapview)
library(units)
# define nautical miles (as per ICAO notation)
NM <- make_unit("NM")
install_conversion_constant("NM", "km", 1.852)
# Sample data:
df <- data.frame(lon = c(45,47,1, -109), lat = c(7, 10, 59, 30))
# Creating simple features with sf:
df <- df %>% st_as_sf(coords = c("lon", "lat"), dim = "XY")
# Applying Coordinate reference system WGS84:
df <- df %>% st_set_crs(4326)
# Transform to Irish grid - I know this step should be different for
# different parts of the world but I don`t know how to make universal
# solution
df <- st_transform(df$geometry, 29902)
# define radius of interest which is 400 NM
rad <- set_units(400, NM) %>% set_units(km) %>% set_units(m)
# make circles
df_buffer <- st_buffer(df, rad)
# visualise using mapview
mapview(df_buffer)
I need these circles as sf objects in a dataframe because I will use them as polygons to find intersections between them and spatial lines (also sf) in other dataframe.
Results below - three circles are different size, two of them are distorted, one is missing:
A: you could transform each point according to it's respective utm zone and then get a buffer for each one separately.
First, a function to find the utm zone proj4string from lat/lon for each point (requires purrr package):
library(purrr)
utm_prj4 <- function(x) {
coords <- cbind(x, st_coordinates(x))
long <- coords$X
lat <- coords$Y
zone <- if(lat >= 56 && lat < 64 && long >= 3 && long < 12){x <- 32} else if(
lat >= 72 && lat < 84 && long >= 0 && long < 9) {x <- 31} else if(
lat >= 72 && lat < 84 && long >= 9 && long < 21) {x <- 33} else if(
lat >= 72 && lat < 84 && long >= 21 && long < 33) {x <- 35} else if(
lat >= 72 && lat < 84 && long >= 33 && long < 42) {x <- 37} else{
x <- (floor((long + 180)/6) %% 60) + 1
}
prj <- purrr::map2_chr(zone, lat, function(y, z){
if (z >= 0){
paste0("+proj=utm +zone=", y, " +datum=WGS84 +units=m +no_defs")
} else{
paste0("+proj=utm +zone=", y, " +south", " +datum=WGS84 +units=m +no_defs")
}})
prj
}
now, convert each to utm and get buffer (this should be done on the df with all crs = 4326, not after transforming to Irish crs):
# creates a list of data.frames, each with different crs
dfs <- map2(1:4, utm_prj4(df), function(x, y){
st_transform(df[x,], y)
})
map(dfs, ~ st_buffer(., rad)) %>% mapview()
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
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{"url":"https:\/\/byjus.com\/jee\/jee-main-2020-question-paper-physics-jan-8-shift-1\/","text":"# JEE Main 2020 Physics Paper With Solutions Jan 8 Shift 1\n\nJEE Main 2020 Paper with Solutions Physics - Shift 1- 8th January is given here. Students are recommended to practise these questions so that they can improve their accuracy and problem-solving skills. The questions and solutions can be accessed from our page directly. The PDF format is also available which can be downloaded easily. Ultimately, students will be able to face the exam with a better confidence level.\n\n### January 8 Shift 1 - Physics\n\n1. A particle of mass \ud835\udc5a is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed \ud835\udf14 about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is\n\n1. a. lm \ud835\udf142\/(k-m \ud835\udf142)\n2. b. lm \ud835\udf142\/(k+m \ud835\udf142)\n3. c. lm \ud835\udf142\/(k-m \ud835\udf14)\n4. d. lm \ud835\udf142\/(k+m \ud835\udf14)\n\nSolution:\n\n1. The centripetal force is provided by the spring force.\n\n2(l + x) = kx\n\n(l\/x) + 1 = k\/m2\n\nx = lm2\/(k-m2)\n\n2. Three charged particles A, B and, C with charge -4q, +2q and -2q are present on the circumference of a circle of radius \ud835\udc51. The charges particles A, C and centre O of the circle formed an equilateral triangle as shown in figure. Electric field at O along x- direction is:\n\n1. a. \u221a3q\/4\u01900d2\n2. b. 2\u221a3q\/\u01900d2\n3. c. \u221a3q\/\u01900d2\n4. d. 3\u221a3q\/4\u01900d2\n\nSolution:\n\n1. Applying superposition principle,\n\n$\\vec{E_{net}} = \\vec{E_{1}}+\\vec{E_{2}}+\\vec{E_{3}}$\n\nBy symmetry, net electric field along the x-axis.\n\n$\\left | \\vec{E_{net}} \\right | = \\frac{4kq}{d^{2}}\\times 2\\cos 30^{0} = \\frac{\\sqrt{3}q}{\\pi \\varepsilon _{0}d^{2}}$\n\n3. A thermodynamic cycle xyzx is shown on a V- T diagram .\n\nThe P-V diagram that best describes this cycle is : (Diagrams are schematic and not upto scale)\n\nSolution:\n\n1. For the given V - T graph\n\nFor the process x \u2192 y; V \u221d T; \ud835\udc43 =constant.\n\nFor the process y \u2192 z; V =constant\n\nOnly \u2032\ud835\udc50\u2032 satisfies these two conditions.\n\n4. Find the co-ordinates of center of mass of the lamina shown in the figure below.\n\n1. a. (0.75 m, 1.75 m)\n2. b. (0.75 m, 0.75 m)\n3. c. (1.25 m, 1.5 m)\n4. d. (1 m, 1.75 m)\n\nSolution:\n\n1. The Lamina can be divided into two parts having equal mass \ud835\udc5a each.\n\n$\\vec{r}_{cm} = \\frac{m\\times (\\frac{\\hat{i}}{2}+j)+m\\times(\\hat{i}+\\frac{5\\hat{j}}{2} )}{2m}$\n\n$\\vec{r}_{cm} = \\frac{3}{4}\\hat{i}+\\frac{7}{4}\\hat{j}$\n\n5. The plot that depicts the behavior of the mean free time \ud835\udf0f (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature (T), qualitatively, is: (Graph are schematic and not drawn to scale)\n\nSolution:\n\n1. \u03c4 \u221d 1\/\u221aT\n\n6. Effective capacitance of parallel combination of two capacitors \ud835\udc361 and \ud835\udc362 is 10 \u03bcF. When these capacitor are individually connects to a voltage source of 1 \ud835\udc49, the energy stored in the capacitor \ud835\udc362 is 4 times of that in \ud835\udc361. If these capacitors are connected in series, their effective capacitance will be:\n\n1. a. 1.6 \u03bcF\n2. b. 3.2 \u03bcF\n3. c. 4.2 \u03bcF\n4. d. 8.4 \u03bcF\n\nSolution:\n\n1. Given that,\n\nC1 + C2 = 10 \u03bcF \u2026(i)\n\n4( \u00bd C1V2) = \u00bd C2V2\n\n4C1 = C2 \u2026(ii)\n\nFrom equations (i) and (ii)\n\nC1 = 2 \u03bcF\n\nC2 = 8 \u03bcF\n\nIf they are in series\n\nCeq = C1C2\/(C1+ C2)\n\n= 1.6 \u03bcF\n\n7. Consider a uniform rod of mass 4\ud835\udc5a and length \ud835\udc3f pivoted about its centre. A mass \ud835\udc5a is moving with a velocity \ud835\udc63 making angle to the rod\u2019s long axis collides with one end of the rod and stick to it.. The angular speed of the rod-mass system just after collision is\n\n1. a. 3\u221a2\ud835\udc63\/7L\n2. b. 4\ud835\udc63\/7L\n3. c. 3\ud835\udc63\/7\u221a2L\n4. d. 3\ud835\udc63\/7L\n\nSolution:\n\n1. There is no external torque on the system about the hinge point. So,\n\n$\\vec{L_{1}} = \\vec{L_{f}}$\n\n$\\frac{mv}{\\sqrt{2}}\\times \\frac{1}{2} = \\left [ \\frac{4mL^{2}}{12} +\\frac{mL^{2}}{4}\\right ]\\times \\omega$\n\n= 6\ud835\udc63\/7\u221a2L = 3\u221a2 \ud835\udc63\/7L\n\n8. When photons of energy 4 eV strikes the surface of a metal A, the ejected photoelectrons have maximum kinetic energy \ud835\udc47\ud835\udc34 eV and de-Broglie wavelength \ud835\udf06\ud835\udc34. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.50 eV is \ud835\udc47\ud835\udc35= (\ud835\udc47\ud835\udc34 \u2212 1.5) eV. If the de-Broglie wavelength of these photoelectrons \ud835\udf06\ud835\udc35 = 2\ud835\udf06\ud835\udc34, then the work function of metal B is\n\n1. a. 3 eV\n2. b. 1.5 eV\n3. c. 2 eV\n4. d. 4 eV\n\nSolution:\n\n1. $\\lambda = \\frac{h}{\\sqrt{2(KE)m_{B}}}$ = \u221d 1\/\u221aKE\n\n$\\frac{\\lambda _{A}}{\\lambda _{B}} = \\frac{\\sqrt{KE_{B}}}{\\sqrt{KE_{A}}}$\n\n$\\frac{1}{2} = \\sqrt{\\frac{T_{A}-1.5}{T_{A}}}$\n\nTA = 2eV\n\nKEB = 2 - 1.5 = 0.5 eV\n\nB = 4.5 - 0.5 = 4 eV\n\n9. The length of a potentiometer wire of length 1200 \ud835\udc50\ud835\udc5a and it carries a current of 60 \ud835\udc5a\ud835\udc34. For a cell of emf 5 \ud835\udc49 and internal resistance of 20 , the null point on it is found to be at 1000 \ud835\udc50\ud835\udc5a. The resistance of whole wire is\n\n1. a. 80\n2. b. 100\n3. c. 120\n4. d. 60\n\nSolution:\n\n1. Assume the terminal voltage of the primary battery as \ud835\udc49\ud835\udc5d. As long as this potentiometer is operating on balanced length, \ud835\udc49\ud835\udc5d will remain constant.\n\nAs we know, potential gradient = 5\/1000\n\n= Vp\/1200\n\nVp = 6V\n\nAnd Rp = Vp\/i = 6\/(60\u00d710-3) = 100\n\n10. The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eyepiece?\n\n1. a. 10 cm\n2. b. 20 cm\n3. c. 30 cm\n4. d. 40 cm\n\nSolution:\n\n1. m = f0\/fe = 5\n\nf0 = 5fe\n\nf0 + fe = 5fe + fe = 6fe = length of the tube\n\n6fe = 60 cm\n\nfe = 10 cm\n\n11. Consider two solid spheres of radii \ud835\udc451= 1 m, \ud835\udc452 = 2 m and masses M1 & M2, respectively. The gravitational field due to two spheres 1 and 2 are shown. The value of M1\/M2 is\n\n1. a. 1\/6\n2. b. 1\/3\n3. c. 1\/2\n4. d. 2\/3\n\nSolution:\n\n1. Gravitation field will be maximum at the surface of a sphere. Therefore,\n\nGM2\/22 = 3 and GM1\/12 = 2\n\n(M2\/M1)\u00d7 \u00bc = 3\/2\n\nM1\/M2 = 1\/6\n\n12. Proton with kinetic energy of 1 MeV moves from south to north. It gets an acceleration of 1012 m\/s2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6 \u00d7 10\u221227 kg)\n\n1. a. 0.71 mT\n2. b. 7.1 mT\n3. c. 71 mT\n4. d. 0.071 mT\n\nSolution:\n\n1. K. E = 1 \u00d7 106 eV = 1.6 \u00d7 10\u221213 J\n\n= \u00bd mev2\n\nWhere me is the mass of the electron = 1.6\u00d710 -27\n\n1.6 \u00d7 10-13 = \u00bd \u00d7 1.6 \u00d7 10-27 \u00d7 v2\n\nv = \u221a2\u00d7107 m\/s\n\nBqv = mea\n\nB = (1.6 \u00d710 -27 \u00d7 1012 )\/ (1.6\u00d710-19\u00d7\u221a2\u00d7 107)\n\n= 0.71 \u00d7 10-3 T\n\n= 0.71 mT\n\n13. If finding the electric field around a surface is given by $\\left | \\vec{E} \\right | = \\frac{q_{enclosed}}{\\varepsilon_{0} \\left | A \\right |}$ is applicable. In the formula \u01900 is permittivity of free space, A is area of Gaussian and \ud835\udc5e\ud835\udc52\ud835\udc5b\ud835\udc50 is charge enclosed by the Gaussian surface. This equation can be used in which of the following equation?\n\n1. a. Only when the Gaussian surface is an equipotential surface.\n2. b. Only when $\\left | \\vec{E} \\right |$ = constant on the surface.\n3. c. Equipotential surface and $\\left | \\vec{E} \\right |$ is constant on the surface\n4. d. for any choice of Gaussian surfaces.\n\nSolution:\n\n1. The magnitude of the electric field is constant and the electric field must be along the area vector i.e. the surface is equipotential.\n\n14. The dimension of stopping potential \ud835\udc490 in photoelectric effect in units of Planck\u2019s constant (h), speed of light (c), and gravitational constant (G) and Ampere (A) is\n\n1. a. h2\/3 c5\/3 G1\/3 A\u22121\n2. b. h2c1\/3G3\/2A\u22121\n3. c. h0 G -1 c 5 A\u22121\n4. d. h\u22122\/3 c\u22121\/3 G4\/3 A\u22121\n\nSolution:\n\n1. V = K(h)a(I)b(G)c(c)d\n\nUnit of stopping potential is (V0) Volt.\n\nWe know [h] = ML2T\u22121\n\n[I] = A\n\n[G] = M\u22121L3T\u22122\n\n[C] = LT\u22121\n\n[V] = ML2T\u22123A\u22121\n\nML2T\u22123A\u22121 = (ML2T\u22121)a(A)b (M\u22121L3T\u22122)c(LT\u22121)d\n\nML2T-3A\u22121 = Ma-cL2a+3c+dT\u2212a\u22122c\u2212dAb\n\na - c = 1\n\n2a + 3c + d = 2\n\n-a - 2c - d = -3 b = -1\n\nOn solving,\n\nc = -1\n\na = 0\n\nd = 5\n\nb = -1\n\nV = K(h)0(A)\u22121(G)\u22121(c)5\n\n15. A leak proof cylinder of length 1 m, made of metal which has very low coefficient of expansion is floating in water at 0o C such that its height above the water surface is 20 cm. When the temperature of water is increases to 4o C, the height of the cylinder above the water surface becomes 21 cm. The density of water at T = 4o C relative to the density at T = 0o C is close to\n\n1. a. 1.01\n2. b. 1.03\n3. c. 1.26\n4. d. 1.04\n\nSolution:\n\n1. Since the cylinder is in equilibrium, it\u2019s weight is balanced by the Buoyant force.\n\nmg = A(80)(\u03c10oc)g\n\nmg = A(79)(\u03c14oc)g\n\n\u03c14oc\/ \u03c10oc = 80\/79 = 1.01\n\n16. The graph which depicts the result of Rutherford gold foil experiement with \u03b1- particle is:\n\n\ud835\udf03: Scattering angle\n\n\ud835\udc41 \u2236 Number of scattered \ud835\udefc \u2212 particles is detected\n\n(Plots are schematic and not to scale)\n\nSolution:\n\n1. N \u221d 1\/sin4(\/2)\n\n17. At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5 s, then induced EMF in the loop is:\n\n1. a. 56 \u03bcV\n2. b. 28 \u03bcV\n3. c. 30 \u03bcV\n4. d. 48 \u03bcV\n\nSolution:\n\n1. $\\epsilon =\\left | \\frac{-d\\phi }{dt} \\right | = \\left | \\frac{-A\\: dB }{dt} \\right |$\n\n$(16\\times 4-4\\times 2)\\frac{(1000-500)}{5}\\times 10^{-4}\\times 10^{-4}$\n\n= 56\u00d7500\u00d710-8\/5\n\n= 56\u00d7 10-6 V\n\n18. Choose the correct Boolean expression for the given circuit diagram:\n\n1. a. A.B\n2. b. A + B\n3. c. $\\bar{A}+ \\bar{B}$\n4. d. $\\bar{A}. \\bar{B}$\n\nSolution:\n\n1. First part of figure shown OR gate and second part of figure shown NOT gate.\n\nSo, Y = $\\bar{A}. \\bar{B}$\n\n19. Consider a solid sphere of density \u03c1(r) = \u03c10 (1- r2\/R2), 0< r \u2264 R. The minimum density of a liquid in which it float is just\n\n1. a. (2\/5) \u03c10\n2. b. (2\/3) \u03c10\n3. c. \u03c10\/5\n4. d. \u03c10\/3\n\nSolution:\n\n1. Let the mass of the sphere be m and the density of the liquid be \u03c1L\n\n\u03c1= \u03c10 ( 1 - r2\/R2), 0< r \u2264 R\n\nSince the sphere is floating in the liquid, buoyancy force (FB) due to liquid will balance the weight of the sphere.\n\n\ud835\udc39 = mg\n\n\u03c1L (4\/3) R3g = \u222b\u03c1(4r2 dr)g\n\n\u03c1L (4\/3) R3 = \u222b \u03c10 ( 1 - r2\/R2) 4r2dr\n\n\u03c1L (4\/3) R3 = $\\int_{0}^{R}\\rho _{0}4\\pi (r^{2}-\\frac{r^{4}}{R^{2}})dr= \\rho _{0}4\\pi\\left ( \\frac{r^{3}}{3} -\\frac{r^{5}}{5R^{2}}\\right )_{0}^{R}$\n\n\u03c1L = (2\/5) \u03c10\n\n20. The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability 4\/3 for this wavelength, will be\n\n1. a. 15\u00b0\n2. b. 30\u00b0\n3. c. 45\u00b0\n4. d. 60\u00b0\n\nSolution:\n\n1. If the speed of light in the given medium is V then,\n\nV = 1\/\u221a(\u03bc\u0190)\n\nWe know that, n = c\/v\n\nn = \u221a(\u03bcr \u0190r) = 2\n\nsin c = \u00bd\n\nc = 300\n\n21. A body of mass \ud835\udc5a=0.10 \ud835\udc58\ud835\udc54 has an initial velocity of $3\\hat{i}$ \ud835\udc5a\/\ud835\udc60. It collides elastically with another body, B of the mass which has an initial velocity of $5\\hat{j}$ \ud835\udc5a\/\ud835\udc60. After collision, A moves with a velocity $v= 4(\\hat{i}+\\hat{j})$ \ud835\udc5a\/\ud835\udc60. The energy of B after collision is written as (\ud835\udc65\/10) \ud835\udc3d, the value of \ud835\udc65 is\n\nSolution:\n\n1. Mass of each object, \ud835\udc5a1= \ud835\udc5a2 = 0.1 \ud835\udc58\ud835\udc54\n\nInitial velocity of 1st object, \ud835\udc621=5 \ud835\udc5a\/\ud835\udc60\n\nInitial velocity of 2nd object, \ud835\udc622 =3 \ud835\udc5a\/\ud835\udc60\n\nFinal velocity of 1st object, \ud835\udc491= $v= 4\\hat{i}+4\\hat{j}$ \ud835\udc5a\/\ud835\udc60 =\u221a(42+42 )= 16\u221a2 \ud835\udc5a\/\ud835\udc60\n\nFor elastic collision, kinetic energy remains conserved\n\nInitial kinetic energy (Ki) = final kinetic energy (Kf)\n\n\u00bd mu12 + \u00bd mu22 = \u00bd mV12 + \u00bd mV22\n\n\u00bd m(5)2 + \u00bd m(3)2 = \u00bd m(16\u221a2)2 + \u00bd mV22\n\nV2 = \u221a2 m\/s\n\nKinetic energy of second object = \u00bd mV22\n\n= \u00bd \u00d7 0.1 \u00d7 \u221a22\n\n= 0.1\n\n= 1\/10 J\n\nx = 1\n\n22. A point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is 30 cm and the refractive index of lens material is 1.5, then the focal length of the lens (in cm) is\n\nSolution:\n\n1. Applying Lens makers\u2019 formula,\n\n1\/f = (\u03bc - 1)[(1\/R1) - (1\/R2)]\n\nR1 =\n\nR2 = -30 cm\n\n\u03bc = 1.5\n\n1\/f = (1.5 - 1)[(1\/) - (1\/-30)]\n\n1\/f = 0.5\/30\n\nf = 60 cm\n\n23. A particle is moving along the x-axis with its coordinate with time t given by x(t) = -3t2 + 8t + 10 \ud835\udc5a. Another particle is moving along the y-axis with its coordinate as a function of time given by y = 5 - 8t3 m. At t = 1 s, the speed of the second particle as measured in the frame of the first particle is given as \u221av . Then v (m\/s) is\n\nSolution:\n\n1. x = \u22123t2 + 8t + 10\n\n$\\vec{V_{A}}$ = (\u22126t + 8)$\\hat{i}$\n\n= 2 $\\hat{i}$\n\ny = 5 \u2212 8t3\n\n$\\vec{V_{B}}$ = \u221224t2 $\\hat{j}$\n\n$\\left | \\overrightarrow{V_{B\/A}} \\right | = \\left | \\overrightarrow{V_{B}} -\\overrightarrow{V_{A}} \\right | = \\left | -24\\hat{j} -2\\hat{i}\\right |$\n\nv = \u221a(242 + 22)\n\nv = 580 m\/s\n\n24. A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m\/s, the frequency difference between the fundamental and second harmonic of this pipe is __Hz.\n\nSolution:\n\n1. V = \u221a(B\/\u03c1)\n\nVpipe\/ Vair = $\\sqrt{\\frac{\\frac{B}{2\\rho }}{\\frac{B}{\\rho }}}= \\frac{1}{\\sqrt{2}}$\n\nVpipe = Vair\/\u221a2\n\nfn = Vpipe(n+1)\/2l\n\nf1-f0 = Vpipe\/ 2l\n\n= 300\/2\u221a2\n\n= 106.06 Hz ( if \u221a2 = 1.414) \u2248 106 Hz\n\n25. Four resistors of resistance 15 , 12 , 4 and 10 respectively in cyclic order to form a wheatstone\u2019s network. The resistance that is to be connected in parallel with the resistance of 10 to balance the network is __.\n\nSolution:\n\n1. [(10R)\/(10+R) ]\u00d712 = 15 \u00d74\n\nR = 10","date":"2020-09-24 14:29:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 29, \"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\": 0.6271448135375977, \"perplexity\": 2647.6541650059767}, \"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-2020-40\/segments\/1600400219221.53\/warc\/CC-MAIN-20200924132241-20200924162241-00738.warc.gz\"}"}
| null | null |
Q: How to find the Eigenfunction of the given Operator I have a difficulty in solving this problem:
Find the eigenfunction of the operator $T$, such that $$ Tf(x,p(x),q(x))=f_{xp}-f_{xqx}, $$
where $p,q$ are continuous map.
I tried to solve it,but I didn't succeed.
(sorry me for the bad english)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 4,453
|
Q: How to hijack AngularJS data binding process? {{SomeData}} means a data binding with a model in AngularJS.
Does this data binding process happen instantaneously?
In my case, I want to know the exact time when the new data is presented to the user.
So, are there any methods that I can insert some code before or after the data is presented that gets the time?
A: Probably no event that specifies what you asked for - but would a
$scope.$watch('SomeData', function(newVal, oldVal){
// code here
});
do the trick?
Possible performance fix:
If you take the var handling out of Angular (by triggering the change with a non-angular event like jQuery) - you can then choose when to do $scope.$apply() - thereby controlling the timing of things without using a $watch.
A: Maybe it can serve your purpose if you do it like in this demo
var app = angular.module('plunker', []);
app.controller('MainCtrl', function($scope) {
$scope.name = function() {
var date = new Date();
console.log("called just now at ", date)
return 'World';
}
});
<!DOCTYPE html>
<html ng-app="plunker">
<head>
<meta charset="utf-8" />
<title>AngularJS Plunker</title>
<script>document.write('<base href="' + document.location + '" />');</script>
<link rel="stylesheet" href="style.css" />
<script data-require="angular.js@1.4.x" src="https://code.angularjs.org/1.4.8/angular.js" data-semver="1.4.8"></script>
<script src="app.js"></script>
</head>
<body ng-controller="MainCtrl">
<p>Hello {{name()}}!</p>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3
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Our Mission in Sweden
Easter Time in Sweden
April 10, 2021 | L&C | No comments
See below for a great Easter Concert put together by the Stockholm South Stake. It's an hour of musical numbers in both Swedish and English and is very well-done. Lots of candles too, which is very Swedish. LeRon even has a part in it. You may not want to watch it all, so here are some times to listen to some very beautiful songs:
7:22 Men's duet of How Great Thou Art (in Swedish and absolutely beautiful)
18:02 Lovely organ medley of Easter songs
25:59 LeRon accompanying vocalists Syster Allen and Syster Gilbert
singing I Stand All Amazed, with Syster Cowgur on cello
and Syster Searle on violin.
40:27 Syster Nissen and Syster Owens rocking out to Syster Nissen's
original He Lives
44:25 Beautiful song called Hope (really fun, and in English)
51:55 Children singing He Sent His Son in Swedish
(young boy on cello, young girl singing)
56:11 Syster and President Davis (our mission president) speaking in Swedish
58:31 Mother and daughter singing a beautiful song in English called
Wonderful Savior (my favorite of the concert but don't tell LeRon,
although I do love the song he does, and his playing of course.)
You may have to drag the cursor back to the start of the concert if you want to watch the whole thing. Not sure why it doesn't necessarily start at the beginning. You can click on the image to enlarge and then put it in full-screen mode. Enjoy all or part!
After that, a few Easter pictures and a link to April General Conference with speakers from 9 countries in the morning session.
We've seen a lot of colored feathers decorating trees and yards. The feathers represent the birch sticks Swedes used to beat themselves with in commemoration of the beating Jesus received before his death. Of course now, the tradition is more secular than religious.
Just like in Alberta, Canada, the public Easter holiday begins on Good Friday (also called Långfredagen, or Long Friday) and ends after Easter Monday. The children get a week's holiday from school too.
Spring is definitely here! Look at these lovely blue flowers peeking through the grass in our mission office yard. These flowers are everywhere. Also white ones. They seem to grow naturally.
Above is a link to the April 2021 General Conference of the Church of Jesus Christ of Latter-day Saints with Easter messages from church leaders around the world. The morning session includes church leaders from Brazil, Nicaragua, New Zealand, Zimbabwe, Portugal, Fiji, Hong Kong, the Philippines, and the United States. This is just the first session of five 2-hr sessions over the two days of April 3 & 4th. Lots of Easter messages. He is Risen!!
LeRon & Colleen Torrie (Äldste & Syster Torrie)
We are happy to be serving the Lord in the Sweden Stockholm Mission (also known as the Sverigemissionen Stockholm) of the Church of Jesus Christ of Latter-day Saints. We are serving an 18-month mission from January 2020 to December 2021 (minus the 6 months we spent at home from March 2020 to October). We have been assigned to work in the mission office again. A new adventure in a new land!
Fun Fantasy for Fantasy Fans
Out of the City on Saturdays
Eating at the Hairy Pig
Catching Up: LeRon's Birthday in March
A Day in the Countryside . . . frescoed churches a...
2021 November: Awesome Art, Magical Music, Flickering Firelight
Museums, Museums, Museums . . . we do them pretty fast!
Losing Loved Ones at Easter Time
Copyright © Our Mission in Sweden |
Design by Automattic | Blogger Theme adapted from NewBloggerThemes.com
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,336
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Q: Adding TextField to AlertController breaks AlertController I am currently trying to open up an AlertController with a TextField inside. When running
let configAlert = UIAlertController(title: "Configure Add-On", message: "Enter Your Add-On Name:", preferredStyle: UIAlertControllerStyle.alert)
configAlert.addAction(UIAlertAction(title: "OK", style: .default, handler: { (action: UIAlertAction!) in
// Handle Input
}))
present(configAlert, animated: true, completion: nil)
everything works fine, but as soon as I add the TextField
configAlert.addTextField { (textField) in
textField.placeholder = "Name"
}
the Alert takes about 10 times longer to open, instantly dismisses, and I get this error in the console spammed about 30 times:
2017-11-26 13:04:08.985783-0500 MinelyMod[380:14792] Warning: Attempt to dismiss from view controller <UISplitViewController: 0x147e0a6a0> while a presentation or dismiss is in progress!
Here is the completed AlertController thats failing
let configAlert = UIAlertController(title: "Configure Add-On", message: "Enter Your Add-On Name:", preferredStyle: UIAlertControllerStyle.alert)
configAlert.addTextField { (textField) in
textField.placeholder = "Name"
}
configAlert.addAction(UIAlertAction(title: "OK", style: .default, handler: { (action: UIAlertAction!) in
// Handle Input
}))
present(configAlert, animated: true, completion: nil)
A: let alertController = UIAlertController(title: "Title", message: "", preferredStyle: .alert)
alertController.addAction(UIAlertAction(title: "Save", style: .default, handler: {
alert -> Void in
let textField = alertController.textFields![0] as UITextField
// do something with textField
}))
alertController.addAction(UIAlertAction(title: "Cancel", style: .cancel, handler: nil))
alertController.addTextField(configurationHandler: {(textField : UITextField!) -> Void in
textField.placeholder = "Search"
})
self.present(alertController, animated: true, completion: nil)
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,249
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Silver Linings Playbook is a 2012 American romantic comedy-drama film directed by David O. Russell. An adaptation of the novel of same name by Matthew Quick, the film stars Bradley Cooper and Jennifer Lawrence, with Robert De Niro, Jacki Weaver, Chris Tucker, Anupam Kher, and Julia Stiles in supporting roles. The film tells the story of, Patrizio "Pat" Solitano, Jr. (Cooper), a man with bipolar disorder who finds companionship in a young widow, Tiffany Maxwell (Lawrence).
Silver Linings Playbook premiered at the 2012 Toronto International Film Festival on September 8, 2012, where Russell won the People's Choice Award. The film initially received a limited release in the United States on November 16, 2012. The Weinstein Company later gave the film a wider release at over 700 theaters on December 25. Silver Linings Playbook earned a worldwide total of over $236 million on a production budget of $21 million. Rotten Tomatoes, a review aggregator, surveyed 228 reviews and judged 92 percent to be positive.
Silver Linings Playbook received awards and nominations in a variety of categories with particular praise for its direction, screenplay, and the performances of Cooper, Lawrence, and De Niro. As of 2013, it has received a total of 47 awards from 91 nominations. At the 85th Academy Awards, the film received eight nominations, and won Best Actress (Lawrence). At the same ceremony, it became the first film in 31 years to be nominated in all four acting categories. At the 66th British Academy Film Awards, Russell won the Best Adapted Screenplay award. The film was nominated for four awards at the 70th Golden Globe Awards, going on to win one—Best Actress in a Comedy or Musical. Among other honors, Silver Linings Playbook was named Best Film at the AACTA Awards, Detroit Film Critics Society, Film Independent Spirit Awards, and Satellite Awards. It also received a Grammy Award for Best Song Written for Visual Media nomination for the song "Silver Lining (Crazy 'Bout You)".
Accolades
See also
2012 in film
Notes
References
External links
Lists of accolades by film
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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Q: Observable object not working with Published keyword I am trying to write a property wrapper to access userDefaults but have two problems. The code snippet is as below
User Default wrapper
@propertyWrapper
struct UserDefault<T> {
private let key: String
private let defaultValue: T
private let userDefaults: UserDefaults
init(key: String, defaultValue: T, userDefaults: UserDefaults = .standard) {
self.key = key
self.defaultValue = defaultValue
self.userDefaults = userDefaults
}
var wrappedValue: T {
get {
return userDefaults.object(forKey: key) as? T ?? defaultValue
}
set {
userDefaults.set(newValue, forKey: key)
}
}
}
Observable Object
final class AppSettings: ObservableObject {
let objectWillChange = PassthroughSubject<Void, Never>()
@UserDefault(key: "focusSliderValue", defaultValue: 25.0)
var focusSliderValue: TimeInterval {
willSet {
objectWillChange.send()
}
}
@UserDefault(key: "shortBreakLength", defaultValue: 5.0)
var shortBreakLength: TimeInterval {
willSet {
objectWillChange.send()
}
}
@UserDefault(key: "longBreakLength", defaultValue: 25.0)
var longBreakLength: TimeInterval {
willSet {
objectWillChange.send()
}
}
@UserDefault(key: "sessionsPerRound", defaultValue: 4)
var sessionsPerRound: Int {
willSet {
objectWillChange.send()
}
}
}
The above code works but have the following issues
Problem 1:
I am trying to replace most of the boiler plate code with @Published keyword but when i do that I get "Value of type 'Double' has no member 'wrappedValue'" error which I am unable to understand/fix.
Problem 2:
Keeping the code as is, I am using a stepper to update "focusSliderValue" on my settings screen which gets updated on my Root View. If I press the stepper once and immediately go back to the root view get updated properly but if I press the stepper say 10 times then go back to the root view I see the screen with "..." (updating) symbol then the field gets updated after few secs. Why is this transition / observable object not smooth ?
Updated GIF for problem 2
Content View For problem 2
struct ContentView: View {
@State var to : CGFloat = 1
@State private var isAnimating = false
@ObservedObject var appSettings = AppSettings()
let gradientColors = Gradient(colors: [Color.blue, Color.purple])
var body: some View {
NavigationView {
ZStack{
VStack(spacing: 50) {
ZStack {
Circle().trim(from: 0, to: 1)
.stroke(Color.black.opacity(0.09), style: StrokeStyle(lineWidth: 10, lineCap: .round))
.frame(width: 200, height: 200)
Circle()
.trim(from: 0, to: self.isAnimating ? self.to : 0)
.stroke(Color.black, style: StrokeStyle(lineWidth: 10, lineCap: .round))
.frame(width: 200, height: 200)
.rotationEffect(.init(degrees: -90))
.overlay(Text("\(Int(appSettings.focusSliderValue))").font(.title), alignment: .center)
.animation(Animation.linear(duration: 5).repeatCount(0, autoreverses: false))
}
Button(action: {self.isAnimating.toggle()}, label: {
Text("Start")
.font(.body)
.foregroundColor(.black)
.padding(40)
}).padding()
.clipShape(Circle())
.overlay(Circle().stroke(Color.black, style: StrokeStyle(lineWidth: 2, lineCap: .round)))
}.padding()
}.navigationBarTitle(Text("Pomodoro Timer"), displayMode: .automatic)
.navigationBarItems(trailing: NavigationLink(destination: SettingsView(appSettings: appSettings), label: {
Image(systemName: "gear")
.font(.title)
.foregroundColor(.black)
}))
}
}
}
struct SettingsView: View {
@ObservedObject var appSettings: AppSettings
var body: some View {
Form {
Section(header: Text("TIMER LENGTH")) {
HStack {
Stepper(value: $appSettings.focusSliderValue , in: 0...60, step: 1.0) {
Text("Focus Length : \($appSettings.focusSliderValue.wrappedValue)" + " mins")
.fontWeight(.bold)
.foregroundColor(.black)
}
}
HStack {
Stepper(value: $appSettings.shortBreakLength) {
Text("Short break Length : \(($appSettings.shortBreakLength.wrappedValue))" + " mins")
.fontWeight(.bold)
.foregroundColor(.black)
}
}
HStack {
Stepper(value: $appSettings.longBreakLength) {
Text("Long break Length : \(($appSettings.longBreakLength.wrappedValue))" + " mins")
.fontWeight(.bold)
.foregroundColor(.black)
}
}
}.font(.caption)
}
}
}
struct ContentView_Previews: PreviewProvider {
static var previews: some View {
ContentView(appSettings: AppSettings())
}
}
Edit: Updated gift for problem 2
A: Try with this variant
final class AppSettings: ObservableObject {
@UserDefault(key: "focusSliderValue", defaultValue: 25.0)
var focusSliderValue: TimeInterval {
didSet {
// uncomment for multi-threaded use and make same for others
// DispatchQueue.main.async {
self.objectWillChange.send() // << use default publisher !!
// }
}
}
@UserDefault(key: "shortBreakLength", defaultValue: 5.0)
var shortBreakLength: TimeInterval {
didSet {
self.objectWillChange.send()
}
}
@UserDefault(key: "longBreakLength", defaultValue: 25.0)
var longBreakLength: TimeInterval {
didSet {
self.objectWillChange.send()
}
}
@UserDefault(key: "sessionsPerRound", defaultValue: 4)
var sessionsPerRound: Int {
didSet {
self.objectWillChange.send()
}
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,891
|
Почётное спортивное звание, присваивалось в 1956—1992 годах.
1956
Аркадьев, Виталий Андреевич
Манаенко, Иван Ильич 1919—1985
Мейпариани, Акакий Семенович 1918—1995
1957
Андриевский, Вадим Алексеевич
Головня, Лев Васильевич 1893—1969
Сергеев, В. Д.
Слепцов, Михаил Васильевич 1890—1984
Федоров, Анатолий Петрович
Чернышева, Раиса Ивановна
1959
Сазонов, Михаил Васильевич 1915—1985
1960
Сакварелидзе, Симон К.
1961
Казанджян, Юрий Арташесович
Хозиков, Юрий Тихонович 1904—1992
1962
Месхи, Трифон Гедеванович 1928—1993
1964
Булочко, Константин Трофимович
Колчинский, Семен Яковлевич 7.11.1917 — 4.1.2005
Фёдоров, Валентин Иванович
Шварц, Григорий Ильич
1965
Бокун, Герман Матвеевич
Тышлер, Давид Абрамович
1967
Бокун, Лариса Петровна
1970
Перекальский, Александр Петрович
Ядловский, Клавдий Игоревич
1971
Удрас, Иозас Юрчио 1925—1991
Фель, Александр Леонидович 13.1.1935
1973
Житлов, Владимир Васильевич
1974
Мидлер, Марк Петрович
1975
Асиевский, Эрнст Владимирович
1979
Ракита, Марк Семенович
1980
Быков, Виктор Харитонович
1981
Пыльнов, Игорь Григорьевич
1982
Горохова, Галина Евгеньевна
Миркин, Иосиф Борисович
Мустафин, Ирек Измайлович
1989
Глазов, Олег Леонидович
Назлымов, Владимир Аливерович
Шуберт, Василий Степанович
Щербич, Людмила Александровна
1990
Тышлер, Геннадий Давидович 12.01.1957
1991
Гансон, Владимир Александрович
Когут, Михаил Антонович 30.9.1931
неизв
Аюпов, Рамиль Измагилович
Булгаков, Генрих Жанович 1929—2010
Бурцев, Михаил Иванович
Вышпольский, Владимир Владимирович
Душман, Давид Александрович 1.4.1923—2021
Иванов, Юрий Александрович 1949—1996
Келлер, Владимир Станиславович 1929—1998
Комаров, Иван Архипович 9.6.1921 — 11.4.2005
Кондратенко, Павел Александрович 1935—2007
Кузнецов, Лев Федорович (? 76)
Лейтман, Леонид Григорьевич
Пузанов, Олег Петрович 20.8.1942 (? 76)
Сайчук, Лев Васильевич (до 1960)
Примечания
Фехтование
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,325
|
#include "StableHeaders.h"
#include "UrhoRendererBindings.h"
namespace JSBindings
{
void Expose_AnimationController(duk_context* ctx);
void Expose_Camera(duk_context* ctx);
void Expose_EnvironmentLight(duk_context* ctx);
void Expose_Fog(duk_context* ctx);
void Expose_Light(duk_context* ctx);
void Expose_Mesh(duk_context* ctx);
void Expose_ParticleSystem(duk_context* ctx);
void Expose_Placeable(duk_context* ctx);
void Expose_Sky(duk_context* ctx);
void Expose_Terrain(duk_context* ctx);
void Expose_WaterPlane(duk_context* ctx);
void Expose_GraphicsWorld(duk_context* ctx);
void Expose_UrhoRenderer(duk_context* ctx);
void ExposeUrhoRendererClasses(duk_context* ctx)
{
Expose_AnimationController(ctx);
Expose_Camera(ctx);
Expose_EnvironmentLight(ctx);
Expose_Fog(ctx);
Expose_Light(ctx);
Expose_Mesh(ctx);
Expose_ParticleSystem(ctx);
Expose_Placeable(ctx);
Expose_Sky(ctx);
Expose_Terrain(ctx);
Expose_WaterPlane(ctx);
Expose_GraphicsWorld(ctx);
Expose_UrhoRenderer(ctx);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,270
|
Too Many Girls
August 31, 2015 T-VMichael Portantiere
Studio Cast, 1977 (Painted Smiles) (2 / 5) The world's oldest collection of co-eds bops to a lower-drawer score by Richard Rodgers and Lorenz Hart in this Ben Bagley recording of a 1939 hit that hasn't worn well. Although there are attractive songs sung on the campus of Pottawatomie College, including "I Didn't Know What Time It Was" and "You're Nearer" (the latter written for the 1940 film version), they're undermined by Dennis Deal's blaring arrangements and one of Bagley's most indulgent exercises in camp interpretation. The vocal arrangements are subpar, too. As for the singers, undergrad Estelle Parsons croaks "My Prince"; Nancy Andrews injects some pizzazz into the glaringly non-P.C "Spic and Spanish"; Johnny Desmond and Arthur Siegel sing well enough without ever coming anywhere near a character. And Tony Perkins, whose ringing baritenor had surprised everyone in Greenwillow, pretty much whispers his vocals here — not ineptly, but without any special insight. Some of the songs do have the old Rodgers and Hart spirit. One example is the opening number, "Heroes in the Fall," with lyrics ghost-written by Rodgers for the off-on-a-binge Hart. But many of the others ("She Could Shake the Maracas," "Cause We Got Cake," and "Sweethearts of the Team" in an excruciating rendition) sound like pale imitations of the team's better work. — Marc Miller
← Tommy Top Banana →
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 7,163
|
If you use Terminal Services with the printers option selected to connect to a Windows Server 2008 R2, it may log error TerminalServices-Printers 1111.
When using the Terminal Services client with the printers option selected to connect to a Windows Server 2008 R2 machine, it may log an error TerminalServices-Printers error 1111 in the System Event logs.
Driver PRINTERDRIVER required for printer PRINTERNAME is unknown. Contact the administrator to install the driver before you log in again.
This error is being logged on the Server because the Server does not have a matching driver on the local machine for the Terminal Services session. If there is a matching driver, this error does not happen.
Install a printer on the local client that has a matching driver on the Server. If there is a matching driver on both the server and the client, this error will not be logged.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 1,576
|
Liberty is an unincorporated community in Yakima County, Washington, United States, located approximately two miles east of Granger.
References
Unincorporated communities in Yakima County, Washington
Unincorporated communities in Washington (state)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 576
|
layout: e98
title: E98 - Fall 2013 - All Weeks
permalink: /e98/week
---
{% include e98/header.html %}
{% include e98/schedule.html %}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,616
|
package mock
import (
"github.com/boltdb/bolt"
"github.com/golang/glog"
"path/filepath"
)
var (
boltdb *bolt.DB
)
type db_builder struct {
db *bolt.DB
err error
}
func init_db(dir, file string) *db_builder {
db, err := bolt.Open(filepath.Join(dir, file), 0600, nil)
if err != nil {
return &db_builder{db: nil, err: err}
}
err = db.Update(func(tx *bolt.Tx) error {
_, err := tx.CreateBucketIfNotExists([]byte(dbBucketOrchestrateModels))
if err != nil {
return err
}
_, err = tx.CreateBucketIfNotExists([]byte(dbBucketOrchestrateInstancesByDomain))
if err != nil {
return err
}
_, err = tx.CreateBucketIfNotExists([]byte(dbBucketOrchestrateInstancesById))
return err
})
return &db_builder{db: db, err: err}
}
func (builder *db_builder) done() (*bolt.DB, error) {
return builder.db, builder.err
}
func (builder *db_builder) buckets(bucket string, more ...string) (*bolt.DB, error) {
if builder.err != nil {
return nil, builder.err
}
builder.err = builder.db.Update(func(tx *bolt.Tx) error {
for _, b := range append(more, bucket) {
_, err := tx.CreateBucketIfNotExists([]byte(b))
if err != nil {
return err
}
}
return nil
})
return builder.db, builder.err
}
func write_boltdb(tx *bolt.Tx, key string, value []byte, bucket string, subbucket ...string) error {
b, err := tx.CreateBucketIfNotExists([]byte(bucket))
if err != nil {
return err
}
if len(subbucket) > 0 {
for _, sb := range subbucket {
b, err = b.CreateBucketIfNotExists([]byte(sb))
if err != nil {
return err
}
}
}
if err := b.Put([]byte(key), value); err != nil {
glog.Warningln(err)
return err
}
return nil
}
func read_boltdb(tx *bolt.Tx, key string, bucket string, subbucket ...string) ([]byte, error) {
b := tx.Bucket([]byte(bucket))
if b == nil {
return nil, nil
}
if len(subbucket) > 0 {
for _, sb := range subbucket {
b = b.Bucket([]byte(sb))
if b == nil {
return nil, nil
}
}
}
if v := b.Get([]byte(key)); len(v) == 0 {
return nil, nil
} else {
return v, nil
}
}
func delete_boltdb(tx *bolt.Tx, key string, bucket string, subbucket ...string) error {
b := tx.Bucket([]byte(bucket))
if b == nil {
return nil
}
if len(subbucket) > 0 {
for _, sb := range subbucket {
b = b.Bucket([]byte(sb))
if b == nil {
return nil
}
}
}
return b.Delete([]byte(key))
}
func list_keys_sizes_boltdb(tx *bolt.Tx, bucket string, subbucket ...string) ([][]byte, []int, error) {
b := tx.Bucket([]byte(bucket))
if b == nil {
return nil, nil, nil
}
if len(subbucket) > 0 {
for _, sb := range subbucket {
b = b.Bucket([]byte(sb))
if b == nil {
return nil, nil, nil
}
}
}
keys := [][]byte{}
sizes := []int{}
cur := b.Cursor()
k, v := cur.First()
for {
if k == nil && v == nil {
break
}
keys = append(keys, k)
sizes = append(sizes, len(v))
k, v = cur.Next()
}
return keys, sizes, nil
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,945
|
Biografia
Ha conseguito il dottorato di ricerca in Matematica presso l'Università di Stoccolma nel 1968. Nel 1971 divenne professore di matematica (ottimizzazione) presso l'Università di Linköping. Sempre nel medesimo posto è stato il capo del Dipartimento di Matematica 1973-1976 e Preside dell'Istituto di tecnologia tra il 1978 e il 1983. Nel 1983-1995 è stato il rettore dell'Università di Linköping. Ha avuto un ruolo di primo piano nello sviluppo della facoltà di medicina (Hälsouniversitet) e nel Campus Norrköping.
Egli è il figlio dell'ex primo ministro svedese Tage Erlander, della quale ha pubblicato molti dei diari di suo padre.
Premi
Membro della Royal Swedish Academy of Engineering Sciences, 1983
Laurea in honoris causa, Università di Danzica, Polonia
Opere
Erlander, S.B. Cost-Minimizing Choice Behavior in Transportation Planning. Springer Verlag, 2010.
Note
Altri progetti
Collegamenti esterni
Prof. Sven Erlander's CV
Professori dell'Università di Linköping
Studenti dell'Università di Stoccolma
Politici figli d'arte
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,281
|
\section{General Appearance}
\section{Introduction}
Inflationary cosmology is the main theoretical description of the
Early Universe in the context of which it was possible to address
and solve the main theoretical problems of the Standard Big Bang
description of our Universe (see Refs. [\citen{mukh}]-[\citen{lyth}]).
Other theoretical attempts to solve the Early Universe puzzles
include the so-called bouncing cosmological models
(see Refs. [\citen{nove}]-[\citen{oik2}]).
In what follows in this paper we will be comparing our results
for the cosmological observables, namely the spectral index of primordial
curvature perturbations $n_{s}$ and the tensor-to-scalar ratio $r$,
with those of the Planck observational data \cite{ade}.
Recently an interesting class of models was discovered
in Ref. [\citen{lind1}], called the $\alpha-$attractors models with the
property that the cosmological observables are identical for
all the members of the $\alpha-$class, in the large $N-$limit,
where $N$ is the number of e-foldings. Subsequently these models
were studied more extensively (see Refs. [\citen{ferr}]-[\citen{zyi}]).
Also a recent study is that of Ref. [\citen{oik1}] which the present
study follows and is a generalization of it.
The above $\alpha-$attractor inflationary potentials have a large
flat plateau for large values of the inflaton scalar field and in the
small $\alpha-$limit are asymptotically quite similar to the hybrid
inflation scenarios \cite{kall3}. Well known inflationary models
are special limiting cases of the $\alpha-$attractors models,
such as the Starobinsky model \cite{star}.
In this paper we compute the cosmological observation parameters
of the spectral index of primordial curvature perturbations
$n_{s}$ and the tensor-to-scalar ratio $r$ for a more general class
of inflationary potentials with the $\alpha-$attractors property,
than those examined in Refs. [\citen{oik1}],[\citen{lind1}].
This potential was suggested in \cite{kall1}. More specifically
we compute these cosmological observables in the so-called Einstein
frame \cite{oik1}, where the form of the potential in explicitly
given in the action functional. Then we find the same cosmological
observables \cite{hnoh} in the so-called Jordan frame, where the
action functional is that of the corresponding $F(R)-$gravity theory
(see Refs. [\citen{capo}]-[\citen{seba}])
and we explicitly compute the corresponding $F(R)$ gravity theory.
Then we want to compare these two sets of values for the cosmological
observables in order to test \cite{oik1} whether the two frames' descriptions
are equivalent observationaly. The equivalence of the descriptions in the two
frames was explicitly shown in Ref. (\citen{derue}). See however
Ref. (\citen{capo2}) for an important exception. We thus find that these
two sets of observables coincide (to within
computational errors from the use of the so-called slow-roll approximation),
for moderate and large values of the number of e-foldings $N$,
a result that generalizes those of Ref. [\citen{oik1}], in a novel way.
This occurs for reasonable values of the set of parameters of the
inflationary potential and for small to moderate $\alpha-$values.
For more references related to inflation in $F(R)$ gravity
theories see Refs. [\citen{noji1}]-[\citen{maka}].
This paper in organized as follows: In section 2 we refer to the basic
facts about inflationary $-\alpha$ attractor models and compute the
corresponding $F(R)-$gravity theory. In section 3 we compute the
cosmological observables in the Einstein frame and in section 4
in the Jordan frame. Finally in section 5 we present the results
and discuss them.
In this paper we assume that the metric of the cosmological spacetime
is that of the flat FRW model
\begin{eqnarray}
\label{frw}
ds^{2}=-dt^{2}+a^{2}(t)[dx^{2}+dy^{2}+dz^{2}]
\end{eqnarray}
where $a(t)$ is the scale factor and $\dot{H}=\dot{a}/a$ is the
Hubble parameter. Also we assume that the connection is a
symmetric, metric compatible and torsion-less affine connection,
namely the so-called Leci-Civita connection and the corresponding
Ricci-scalar curvature is given by
\begin{eqnarray}
\label{ricci}
R=6(\dot{H}+2H^{2})
\end{eqnarray}
Finally we use units where $\kappa^{2}:=8\pi G=1$ and $h=c=1$.
\section{Basics of Inflationary $\alpha-$attractors and $F(R)$ gravity}
In this section we introduce the basic discussion concerning the
inflationary $\alpha-$attractor models, which are classes of
inflationary potentials with large flat potential plateaus, and their
relation to the cosmological observables. Thus we consider
the $F(R)$ gravity action in the so-called Jordan frame
\begin{eqnarray}
\label{act1}
\hat{S}=\frac{1}{2}\int d^{4}x\sqrt{-\hat{g}}F(\hat{R})
\end{eqnarray}
where $\hat{g}_{\mu\nu}$ is the metric in the Jordan frame and $\hat{R}$
the corresponding Ricci scalar curvature. Introducing the auxiliary
scalar field $A$, one can write the above action as\cite{oik1}
\begin{eqnarray}
\label{act2}
\hat{S}_{1}=\frac{1}{2}\int d^{4}x\sqrt{-\hat{g}}[F^{'}(A)(\hat{R}-A)+F(A)]
\end{eqnarray}
By varying the action $\hat{S}_{1}$ with respect to $A$ we obtain
$A=\hat{R}$ and verify thus the equivalence of the actions
(\ref{act1}), (\ref{act2}). Making the conformal transformation
\begin{eqnarray}
\label{conf}
\hat{g}_{\mu\nu}=e^{\phi}g_{\mu\nu}
\end{eqnarray}
and introducing the canonical transformation
\begin{eqnarray}
\label{cano}
\Phi=\sqrt{3}\phi\nonumber \\
\frac{e^{\phi}F^{'}(A)}{2}=1
\end{eqnarray}
corresponding to Eq. (20) of Ref. \citen{oik1}, the action is transformed
to $\hat{S}_{1}\longrightarrow S_{1}$, namely to the action in the Einstein
frame
\begin{eqnarray}
\label{act3}
S_{1}=\int d^{4}x\sqrt{-g}[R-\frac{1}{2}g^{\mu\nu}\partial_{\mu}\Phi\partial_{\nu}\Phi-V(\Phi)]
\end{eqnarray}
where
\begin{eqnarray}
\label{pot1}
V(\Phi):=2\left[\frac{A}{F^{'}(A)}-\frac{F(A)}{(F^{'}(A))^{2}}\right]=\nonumber
\\
=e^{\Phi/\sqrt{3}}\hat{R}(e^{-\Phi/\sqrt{3}})-\frac{1}{2}e^{2\Phi/\sqrt{3}}
F[\hat{R}(e^{-\Phi/\sqrt{3}})]
\end{eqnarray}
Here the dependence of $\hat{R}$ on $\Phi$ is found by solving the second
of Eqs. (\ref{cano}) with respect to $A=\hat{R}$. Also we obtain from this
\begin{eqnarray}
\label{deri}
\frac{d\Phi}{d\hat{R}}=-\sqrt{3}\frac{F^{''}(\hat{R})}{F^{'}(\hat{R})}
\end{eqnarray}
and using Eq. (\ref{pot1})we finally obtain (see Eq. (24) of Ref. (\citen{oik1}))
\begin{eqnarray}
\label{rel1}
\hat{R}F_{\hat{R}}=-2\sqrt{3}\frac{d}{d\Phi}
\left[\frac{V(\Phi)}{e^{2\Phi/\sqrt{3}}}\right]
\end{eqnarray}
In our conventions and notation we have $-\infty<\Phi\leq 0$ for the
scalar field. For example, for the Starobinsky model \cite{star}, which
in our notational conventions is given by
\begin{eqnarray}
\label{sta1}
V(\Phi)=\mu^{2}(1-e^{\Phi/\sqrt{3}})^{2}
\end{eqnarray}
we obtain the corresponding $F(\hat{R})$ gravity description as \cite{oik1}
\begin{eqnarray}
\label{frg1}
F(\hat{R})=\hat{R}+\frac{\hat{R}^{2}}{4\mu^{2}}
\end{eqnarray}
The above action of Eq. (\ref{act3}) can also occur from an action,
in the so-called $\phi-$Jordan frame\cite{oik1}, with
a non-canonically coupled scalar field of the form
\begin{eqnarray}
\label{act4}
S_{n}=\int d^{4}x\sqrt{-g}
\left[R-\frac{\partial_{\mu}\phi\partial^{\mu}\phi}{2\left(1-\frac{\phi}{6\alpha}\right)^{2}}-V(\phi)\right]
\end{eqnarray}
Making the transformation
\begin{eqnarray}
\label{transf}
\frac{d\phi}{\left(1-\frac{\phi}{6\alpha}\right)}=d\Phi
\end{eqnarray}
we finally obtain the action of Eq. (\ref{act3}), namely
\begin{eqnarray}
\label{act5}
S_{1}=\int d^{4}x\sqrt{-g}
\left[R-\frac{1}{2}(\partial \Phi)^{2}-V\left(\sqrt{6\alpha}tanh\left(\frac{\Phi}{\sqrt{6\alpha}}\right)\right)\right]
\end{eqnarray}
In the present paper we consider the potential ($-\infty<\Phi\leq 0$)
\begin{eqnarray}
\label{pot2}
V(\Phi)=V_{0}\frac{[tanh(\Phi/\sqrt{6\alpha})]^{2n}}{[1-tanh(\Phi/\sqrt{6\alpha})]^{2m}}
\end{eqnarray}
This is a generalization of the potential proposed in Fig. 2 of Ref. (\citen{lind1}),
where the parameter $\alpha$ is introduced, which is inversely proportional
to the curvature of the inflaton K\"{a}hler manifold \cite{kall1}. The parameters
$m,\; n$ are not necessarily integers. As in the previous case of the
Starobinsky model the slow-roll regime corresponds to the case of
$\Phi\longrightarrow -\infty$ where $F_{\hat{R}}=2e^{-\Phi/\sqrt{3}}\gg 1$.
The choice of the above potential for our study is partially based on the fact that
it is quite generic, possesses a large horizontal flat plateau for large negative
$\Phi-$values for the slow-roll inflation and possesses many limiting
cases as special cases, for example
the Starobinsky model \cite{star}, or the Higgs inflationary model \cite{capo1},
and so on. The potential of Eq. (\ref{pot2})
is shown in Figs. (\ref{fig1})-(\ref{fig2}).
Using the expansions (for $z:=e^{\sqrt{2/3\alpha}\Phi}\ll 1$)
\begin{eqnarray}
\label{expa}
(1-z)^{N}\simeq 1-Nz+\frac{N(N-1)}{2}z^{2}\nonumber \\
\left(\frac{1}{1+z}\right)^{N}\simeq 1-Nz+\frac{N(N+1)}{2}z^{2}
\end{eqnarray}
we obtain in the slow-roll regime
\begin{eqnarray}
\label{pot3}
V(\Phi)\simeq \frac{V_{0}}{2^{2m}}
\left[1-A(n,m)z+B(n,m)z^{2}\right]
\end{eqnarray}
where $A(n,m):=2(2n-m)$ and $B(n,m):=8n^{2}-8nm+2m^{2}-m$. From
Eq. (\ref{rel1}) we then obtain
\begin{eqnarray}
\label{frg2}
\hat{R}F_{\hat{R}}-\frac{V_{0}}{2^{2m}}
\left[4\left(\frac{F_{\hat{R}}}{2}\right)^{2}+2A\left(\sqrt{\frac{2}{\alpha}}-2\right)\left(\frac{F_{\hat{R}}}{2}\right)^{\left(2-\sqrt{\frac{2}{\alpha}}\right)}\right.\nonumber
\\
\left.-4B\left(\sqrt{\frac{2}{\alpha}}-1\right)\left(\frac{F_{\hat{R}}}{2}\right)^{\left(2-2\sqrt{\frac{2}{\alpha}}\right)}\right] =0
\end{eqnarray}
This equation will be needed in the analysis that follows in section 5.
In the main part of the analysis that will follow we will be concerned
with the large curvature limit $(\alpha\longrightarrow 0)$. Also,
although the analysis that follows can be generalized to arbitrary values
of $m,\; n$, from now on we will focus on the specific subclass of models
where ${\bf m=2n}$. This is an exceptional and interesting case that is trackable
for analytical treatment. Then $A=0$, $B=-2n$ and we have
from Eq. (\ref{frg2})
\begin{eqnarray}
\label{frg3}
\hat{R}-c_{0}F_{\hat{R}}-c_{1}F_{\hat{R}}^{-\delta}=0
\end{eqnarray}
where $c_{0}:=\frac{V_{0}}{2^{4n}}$, $c_{1}:=c_{0}2^{(\delta+1)}n^{(\delta -1)}$
and $\delta:=2\sqrt{\frac{2}{\alpha}}-1\geq 0$.
\section{Cosmological Parameters in the Einstein frame}
For the potential of Eq. (\ref{pot2}) and for $m=2n$ one can calculate the
cosmological observables of the spectral index of primordial curvature
perturbations $n_{s}$ and the scalar-to-tensor ratio $r$, as they occur in the
Einstein frame of Eq. (\ref{act3}), from
\begin{eqnarray}
\label{obse}
n_{s}^{(E)}&=&1-6\epsilon+2\eta\nonumber \\
r^{(E)}&=&16\epsilon
\end{eqnarray}
where the slow-roll parameters are given by
\begin{eqnarray}
\label{slow}
\epsilon&:=&\frac{1}{2}\left(\frac{V^{'}}{V}\right)^{2}\nonumber \\
\eta&:=&\frac{V^{''}}{V}
\end{eqnarray}
The calculation is exact as it was done in Eqs. (5.2)-(5.4)
of Ref. \citen{kall1}, {\it without} invoking the slow-roll hypothesis,
but only assuming that inflation ends when $\epsilon\simeq 1$. The
results are as follows: We define the number of e-foldings $N$ as
\begin{eqnarray}
\label{efol}
N:=\int_{\Phi_{i}}^{\Phi_{e}}\frac{V(\Phi)}{V^{'}(\Phi)}d\Phi
\end{eqnarray}
where $\Phi_{i},\; \Phi_{e}$ are the initial and end values of the
inflaton scalar field. Along with the definitions $\xi:=tanh(\Phi/\sqrt{6\alpha})$ and
$g:=\frac{\sqrt{3\alpha}}{2n}$ we obtain
\begin{eqnarray}
\label{efol1}
N=\frac{3\alpha}{4n}
\left[\frac{\Phi_{i}-\Phi_{e}}{\sqrt{6\alpha}}+\frac{1}{1+\xi_{e}}-\frac{1}{1+\xi_{i}}-\frac{1}{(1+\xi_{e})^{2}}+\frac{1}{(1+\xi_{i})^{2}}\right]
\end{eqnarray}
On the other hand from the requirement that inflation ends when $\epsilon\simeq 1$,
we obtain from the first of Eqs. (\ref{slow})
\begin{eqnarray}
\label{efol2}
\xi_{e}=[-(1+g)+\sqrt{g(g+2)}]<0
\end{eqnarray}
Now remembering that $y=tanh^{-1}(x)=\frac{1}{2}ln|\frac{1+x}{1-x}|$ and
observing that in the limit of $\xi_{i}\longrightarrow -1+0$ the sixth term
in Eq. (\ref{efol1}) dominates the with respect to the fourth and the second terms,
we obtain
\begin{eqnarray}
\label{efol3}
N=\frac{3\alpha}{4n}
\left[-\frac{1}{2}ln\left(\frac{1+\xi_{e}}{1-\xi_{e}}\right)+\frac{1}{(1+\xi_{i})^{2}}+\frac{1}{1+\xi_{e}}-\frac{1}{(1+\xi_{e})^{2}}\right]
\end{eqnarray}
Now using Eq. (\ref{efol2}) into Eq. (\ref{efol3}) and defining
\begin{eqnarray}
\label{defs}
f(N,n,\alpha )&:=&\frac{4nN}{3\alpha}-\frac{1}{4}\left[ln\left(1+\frac{2}{g}\right)-\frac{2}{g}\right]\nonumber
\\
\xi_{o}&:=&\frac{1}{\sqrt{f}}-1
\end{eqnarray}
we obtain after a lengthy and careful calculation that
\begin{eqnarray}
\label{inds}
\epsilon&=&\frac{3n^{2}}{\alpha}\frac{(1+\xi_{o})^{4}}{\xi_{o}^{2}}\nonumber
\\
\eta&=&\left[\frac{2n(2n-1)}{6\alpha}\frac{(1-\xi_{o}^{2})^{2}}{\xi_{o}^{2}}+\frac{16n^{2}}{6\alpha}\frac{(1+\xi_{o})^{2}}{\xi_{o}}\right]
\end{eqnarray}
and the cosmological observation parameters are given by Eq. (\ref{obse}),
using Eqs. (\ref{inds}).
\section{Cosmological Parameters in the Jordan frame}
It can be easily shown that an approximate
solution to Eq. (\ref{frg3}), valid in the slow-roll
regime $(F_{\hat{R}}\gg 1)$ is given by
\begin{eqnarray}
\label{frg4}
F_{\hat{R}}&=&\frac{1}{c_{0}}\hat{R}-c_{1}c_{0}^{(\delta-1)}\hat{R}^{-\delta}\nonumber
\\
F(\hat{R})&=&\frac{\hat{R}^{2}}{2c_{0}}+\frac{c_{1}c_{0}^{(\delta -1)}}{(\delta-1)}\hat{R}^{(1-\delta)}+\Lambda
\end{eqnarray}
where $\Lambda$ is a positive cosmological constant. Now varying the action
of Eq. (\ref{act1}) we obtain\cite{capo}
($R$ is given by Eq. (\ref{ricci}) where $H=\dot{a}/a$ is the Hubble parameter and we
drop hats for simplicity, since it is clear that we work in the Jordan frame
of Eq. (\ref{act1}))
\begin{eqnarray}
\label{feqs}
6H^{2}F_{R}&=&RF_{R}-F-6H\dot{R}F_{RR}\nonumber \\
-(2\dot{H}+3H^{2})F_{R}&=&F_{RRR}(\dot{R})^{2}+2HF_{RR}\dot{R}+F_{RR}\ddot{R}+\nonumber
\\
&+&\frac{F-RF_{R}}{2}
\end{eqnarray}
These reproduce Eq. (31) of Ref. (\citen{oik1}) in the proper limit.
In the slow-roll regime, where $(\dot{H}\ll H^{2})$ we may approximate
\begin{eqnarray}
\label{appr}
R^{-\delta}=(12H^{2})^{-\delta}\left(1+\frac{\dot{H}}{2H^{2}}\right)^{-\delta}\simeq
\frac{1}{(12H^{2})^{\delta}}\left(1-\delta\frac{\dot{H}}{2H^{2}}\right)
\end{eqnarray}
Using Eqs. (\ref{frg4}) we obtain then from the first of Eqs. (\ref{feqs})
we obtain after a slightly lengthy calculation
\begin{eqnarray}
\label{feq1}
36H\ddot{H}+\Lambda c_{0}-18\dot{H}^{2}+108H^{2}\dot{H}+\nonumber \\
+\frac{6c_{1}c_{0}^{\delta}}{(12H^{2})^{\delta}}
\left[\frac{(\delta+1)}{(\delta-1)}H^{2}+\frac{3}{2}\delta\dot{H}-\delta(\delta-1)\frac{\dot{H}^{2}}{H^{2}}\right]=0
\end{eqnarray}
This generalizes Eq. (32) of Ref. (\citen{oik1}). It is quite interesting and
unlike the case of Eq. (32) of Ref. (\citen{oik1}), where in order to obtain
a non-trivial solution they had to take the derivative of Eq. (31) in order to obtain
Eq. (36) and finally Eq. (37) (see Ref. (\citen{oik1})), that, in our case,
an {\it exact} solution to Eq. (\ref{feq1}) is given by
(in the slow-roll regime where $H_{1}\ll H_{0}$)
\begin{eqnarray}
\label{solu}
H(t)&=&H_{0}-H_{1}(t-t_{k})
\end{eqnarray}
where $H_{0},\; t_{k}$ are arbitrary integration constants of Eq. (\ref{feq1}).
We assume that the cosmological constant is given by
$\Lambda c_{0}=108H_{0}^{2}H_{1}>0$. Then the third and the last
three terms of Eq. (\ref{feq1}) are equated to zero and this gives
\begin{eqnarray}
\label{H1}
H_{1}&=&\frac{1}{2k_{0}}\left[-k_{1}+\sqrt{k_{1}^{2}-4k_{0}k_{2}}\right]>0\nonumber
\\
k_{0}&=&3\left[1+\frac{4\delta (\delta-1)c_{1}c_{0}^{\delta}}{(12H_{0}^{2})^{(\delta+1)}}\right]\nonumber
\\
k_{1}&=&\frac{3}{2}\frac{\delta c_{1}c_{0}^{\delta}}{(12H_{0}^{2})^{\delta}}\nonumber
\\
k_{2}&=&-\left(\frac{\delta +1}{\delta -1}\right)H_{0}^{2}\frac{c_{1}c_{0}^{\delta}}{(12H_{0}^{2})^{\delta}}
\nonumber \\
c_{0}&=&\frac{V_{0}}{2^{4n}}\nonumber \\
c_{1}&=&c_{0}n^{(\delta -1)}2^{(\delta +1)}
\end{eqnarray}
Specifically the time $t_{k}$ is assumed to be the time where the
horizon crossing for the comoving wavenumber $k=a(t)H(t)$ occurred.
For the action of Eq. (\ref{act1}) the cosmological observables corresponding
to Eq. (\ref{obse}) are given by (see Refs. (\citen{oik1}) and (\citen{hnoh}))
\begin{eqnarray}
\label{obse1}
n_{s}&\simeq& 1-6\epsilon_{1}-2\epsilon_{4}\simeq 1-\frac{2\dot{\epsilon}_{1}}{H(t)\epsilon_{1}}\nonumber
\\
r&=&48\epsilon_{1}^{2}
\end{eqnarray}
where
\begin{eqnarray}
\label{obse2}
\epsilon_{1}=-\frac{\dot{H}}{H^{2}},\; \; \epsilon_{2}=0,\; \;
\epsilon_{3}\simeq \epsilon_{1},\; \; \epsilon_{4}\simeq -3\epsilon_{1}+\frac{\dot{\epsilon}_{1}}{H(t)\epsilon_{1}}
\end{eqnarray}
If we assume that the slow-roll regime (and essentially the inflation also)
ends when $\epsilon_{1}\simeq {\cal O}(1)$, at $t=t_{f}$, so that $H(t_{f}):=H_{f}$
we have $H_{f}=\sqrt{H_{1}}$ and also \\
$(t_{f}-t_{k})\simeq (H_{0}/H_{1})$. Defining the number of e-foldings of inflation
as
\begin{eqnarray}
\label{efol4}
N=\int_{t_{k}}^{t_{f}}H(t)dt
\end{eqnarray}
we end up with
\begin{eqnarray}
\label{tftk}
(t_{f}-t_{k})\simeq \frac{2N}{H_{0}}
\end{eqnarray}
Thus finally the cosmological observables of Eq. (\ref{obse1}), as they occur in the
Jordan frame, are given by
\begin{eqnarray}
\label{obse3}
n_{s}^{(J)}&\simeq &1-\frac{4H_{1}}{\left(H_{0}-\frac{2H_{1}N}{H_{0}}\right)^{2}}\nonumber
\\
r^{(J)}&\simeq&\frac{48H_{1}^{2}}{\left(H_{0}-\frac{2H_{1}N}{H_{0}}\right)^{4}}
\end{eqnarray}
Although these equations are similar in form with Eq. (50) of Ref. (\citen{oik1}),
they are essentially different in the fact that they depend additionally on the
parameter $\alpha$ and on the freely specified parameter $H_{0}$. Hereafter we choose
$H_{0}=1.$ In all the numerical examples below we found that in Eq. (\ref{H1}),
$H_{1}\leq 0.02$ in practically all the cases, so that the slow-roll approximation
($H_{1}\ll H_{0}$) is indeed satisfied in all of the relevant cases.
\section{Discussion}
In this section we are ready to compare the observational indices as
they occur in the Einstein and Jordan frames of Eqs. (\ref{obse}) and
Eqs. (\ref{obse3}) respectively. In order to make compatible our
results with Eqs. (15), (22) of Ref. \citen{ade},
namely
\begin{eqnarray}
\label{ns}
n_{s}&=&0.9645\pm 0.0049\nonumber \\
r&<&0.10
\end{eqnarray}
we choose to have $n_{s}^{(E)}(N=60)=0.9645$ in Fig. \ref{fig3}.
This was easily achieved by choosing $\alpha=0.0625$, as it is referred to in
the caption of Fig (\ref{fig3}). This is shown by the black horizontal line.
Also we observe that the two curves are practically
asymptote to this value for all values of $N\geq 50$ (to within errors due to the
use of the slow-roll approximation), namely the two sets of observables
coincide. This is the main result of the present paper that generalizes the results
of Ref. (\citen{oik1}).
The curves of Fig. \ref{fig3}, as has been referred to,
correspond to the case of $\alpha=0.0625,\; \; n=1$.
This is in our viewpoint a novel result, unlike the case of Ref. (\citen{oik1}),
where the observational indices in the two frames coincide only in the
large-$N$ limit and in the small$-\alpha$ limit. In our case this happens also
for $\alpha\leq 0.1$ also for moderate $N-$values.
In the same manner we obtain also Fig. \ref{fig4},
where we find that $r^{(E)}(N=60)\simeq 0.00072038<0.10$,
for the parameter value of $V_{0}=45$. When the value of the $\alpha-$parameter
increases however, (namely for $\alpha=0.125$) we obtain the curves of
Figs. (\ref{fig3a})-(\ref{fig4a}). Here the observational indices
in the two frames do not asymptote to a common value unless the number of
e-foldings is larger than $N\geq 300$. Therefore the two sets of observational indices,
corresponding to the two frames coincide for practically all relevant values
of the number of e-foldings $N$, when $\alpha\leq 0.1$ and $n\simeq 1$.
This is also evident from Figs. (\ref{fig5})-(\ref{fig6}). It is quite possible,
although difficult to ascertain analytically, that without invoking the
condition $m=2n$ in the potential of Eq. (\ref{pot2}), equivalence of the
two frame descriptions would occur for an even vaster range of parameter values.
Regarding the crucial issue of whether these attractors and the
observational indices connected with them can be used to
distinguish between the two frames (namely the Einstein and Jordan frames)
the author considers this to be a very deep question and a definite
answer cannot so easily be given. However according to the authors viewpoint
and relevant work on this subject (see Refs. \citen{capo2}, \citen{capo1}, \citen{capo3}),
in the case considered in this paper, although the two frames are
mathematically equivalent, as they are connected by a conformal transformation,
there exists a physical non-equivalence of the two frames, and by using the
results obtained here, regarding the observational indices, one may be
able to distinguish between the two frames.
Since it is quite difficult to obtain a definite answer\cite{oik1},
for the most generic case, regarding the equivalence of the
descriptions of the observational indices in the two
frames (Einstein and Jordan frames), it would be interesting to try
to check this postulate for more realistic and general inflationary potentials,
as those for example suggested by appropriate limits of certain
supergravity and/or string theory actions (see Ref. (\citen{lind2})
and references therein). Work along these lines is in progress.
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig1}}
\caption{The potential of Eq. (\ref{pot2}) for $V_{0}=1$, $n=m=1$,
for various values of the parameter $\alpha$.
\label{fig1}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig2}}
\caption{The potential of Eq. (\ref{pot2})
normalized to its value at $\Phi\longrightarrow -\infty$,
namely $V_{\infty}=V_{0}/2^{4n}$,
for $n=1$ and for various values of the parameter $m$.
\label{fig2}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig3}}
\caption{The spectral index of primordial curvature perturbations $n_{s}$,
in the Einstein and Jordan frames, as a function of the number of e-foldings $N$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $n=1,\; m=2n$, $\alpha=0.0625$. In black it is shown the constant value
of Eq. (\ref{ns}).
\label{fig3}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig4}}
\caption{The tensor-to-scalar ratio $r$,
in the Einstein and Jordan frames, as a function of the number of e-foldings $N$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $n=1,\; m=2n$, $\alpha=0.09$.
\label{fig4}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig3a}}
\caption{The spectral index of primordial curvature perturbations $n_{s}$,
in the Einstein and Jordan frames, as a function of the number of e-foldings $N$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $n=1,\; m=2n$, $\alpha=0.125$. In black it is shown the constant value
of $n_{sa}=0.995$.
\label{fig3a}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig4a}}
\caption{The tensor-to-scalar ratio $r$,
in the Einstein and Jordan frames, as a function of the number of e-foldings $N$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $n=1,\; m=2n$, $\alpha=0.125$.
\label{fig4a}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig5}}
\caption{The spectral index of primordial curvature perturbations $n_{s}$,
in the Einstein and Jordan frames, as a function of the parameter $\alpha$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $N=60$ and $n=1,\; m=2n$.
\label{fig5}}
\end{figure}
\begin{figure}[h!]
\centerline{\includegraphics[width=10.8cm]{fig6}}
\caption{The spectral index of primordial curvature perturbations $n_{s}$,
in the Einstein and Jordan frames, as a function of the parameter $n$,
namely that of Eqs. (\ref{obse}) and (\ref{obse3}),
for $m=2n$, $N=80$ and $\alpha=0.09$.
\label{fig6}}
\end{figure}
\section*{Acknowledgments}
The author would like to acknowledge useful discussions with
Dr. K. Kleidis and other colleagues in the Technological Education
Institute (T.E.I.) of Central Macedonia, Greece. The author would like
to acknowledge the useful criticisms and remarks of the anonymous
referees that helped improve the paper.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,828
|
Rewa is a city in the north-eastern part of Madhya Pradesh state in India. It is the administrative center of Rewa District and Rewa Division. The city lies about northeast of the state capital Bhopal and north of the city of Jabalpur. The maximum length of Rewa district is 125 km from east to west and the length of Rewa from north to south is 96 km. This area is surrounded by Kaimur hills to the south Vindhyachal ranges pass through the middle of the district. It is famous for the founding of the world's first white tiger safari and beetle nut toys.
History
The district of Rewa derives its name from the town of Rewa, the district headquarters, which is another name for the Narmada River.
Present day Rewa was part of the Baghelkhand region which expanded from the present day Prayagraj in the North to Ratanpur in the South, Jabalpur in the West to Surajpur in the East.
Baghel Dynasty
Baghel Dynasty was founded by Bhimaldev (son of Vyaghradev, the chieftain of Vyaghrapalli) in 1236 CE. Baghelas are basically Chalukyans of Anhilwara (Gujarat).
The region was earlier governed by Lodhi and Sengar chieftains of Rajgond Dynasty. Lodhi's Diwan Tiwari conspired with Baghelas and assisted in foundation of Baghela Rule in the Gahora Patti region. In return of this favour, Baghelas granted title of "Singh Tiwari" or "Adhrajiya Tiwari" to the Diwan Tiwari.
Raja Ramchandra shifted capital to Bandhavgarh, and later Raja Vikramjit Singh shifted capital to Rewa in 1605 CE.
Bandhavgarh Fort was sieged by Mughals. Tansen and Birbal (Mahesh Das) were in court of Ramchandra Singh Baghel.
Raghuraj Singh Baghel built Govindgarh Fort, which lies in between Govindgarh lake. Govindgarh is famous for its exquisite varieties of mangoes.
Raja Gulab Singh was called "social reformer King" of Rewa. Raja Martand Singh was the last Baghela Ruler. Later, the state joined the Union of India, after independence.
Revolt of 1857
Thakur Ranmat Singh of Mankahri revolted against the British and was hanged in 1859.
Demographics
As of 2011, Rewa had a population of about 2,35,654 out of which 1,24,012 are males and 1,11,642 are females. Rewa has an average literacy rate of 86.31%, male literacy is 91.67%, and female literacy is 80.40%. In Rewa, 10.76% of the population is under 6 years old.
Society
The region is home to Kol Tribes of Madhya Pradesh. Rewa Riyasat had gave royal patronage to three Brahmins, today known as Tiwari, Mishra and Dubey (Parauha). These three formed closed matrimonial alliances.
Tiwari had assisted Vyaghra Singh Deo Baghel and his sons to orchestrate Coup d'état of Lodhis and ascend the throne of Rewa Estate. In return of the favor, Tiwari and his successors called "Adhrajiya Tiwari" with title of "Singh Tiwari". The other known Tiwari clans are - Tiwani, Hanna etc.
Mishras belong to four clans - Amanv (Chakghat), Anjora (Teonthar), Umapur (Prayagraj) and Tudihar (Mirzapur). Last brahmin to receive royal patronage was - Parauha, which used title Dubey or Dwivedi. Some other major Brahmin clans of Rewa are - Shukla, Gautam, Garg, Pandey, Tripathi etc.
Kurmi people are landed wealthy agriculturalists of the region with expertise in mango, tobacco, linseed and rice cultivation.
Cuisine
The region has highest production of pulses, tobacco, mangoes, flaxseeds, Mahua etc. Thus, cuisines enjoyed by people are -
Indrahar - paste of several pulses mixed and baked in steam
Kadhi - kadhi uses Rasaj (gram flour cakes), Sooran (elephant foot yam) and Indrahar
Bagza - aam kery pana with toppings of gram flour spaghettis and fried with cumin and mustard
Kusuli - regional variation of Guziya sweet
Dal Poori - breads filled with grinded Gram Dal and spices like - Garlic, Garam Masala etc.
Nimona - peas or green grams fine grinded and fried to curry masala
Sattu - fine grinded popcorns of wheat, gram and barley
Mahua Poori - pooris used with fillings of sun dried ripen Mahua fruits
Transportation
Rail
Rewa railway station is connected to Satna through the 50 km Satna-Rewa branch line. Satna falls on the Howrah-Allahabad-Mumbai line.
Road & air
The highways crossing through the city are NH 7, NH 27, and NH 75 NH 30.
The closest major airport to Rewa is Prayagraj Airport, Uttar Pradesh which is 130 kilometers (80.7 miles) away and has flights to major destinations such as Delhi, Bengaluru, Mumbai, Kolkata, etc. Other airports are at Chorahta Airport Rewa, Khajuraho, Jabalpur and Varanasi. Rewa will soon have its own domestic airport which is under construction and scheduled to get open by 2023 end.
References
External links
Official Government Website of Rewa
Website of Rewa Tourism
Cities and towns in Rewa district
Former capital cities in India
Cities in Madhya Pradesh
Rewa Municipal Corporation
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,090
|
//===============================================================================
// (c) 2015 eWorkplace Apps. All rights reserved.
// Original Author: Dheeraj Nagar
// Original Date: 12 June 2015
//===============================================================================
package com.eworkplaceapps.ed;
import android.accounts.Account;
import android.accounts.AccountManager;
import android.content.ContentResolver;
import android.content.Context;
import android.graphics.Typeface;
import android.graphics.drawable.Drawable;
import android.net.Uri;
import android.os.Bundle;
import android.support.multidex.MultiDex;
import android.support.multidex.MultiDexApplication;
import android.widget.ImageView;
import com.eworkplaceapps.ed.utils.Utils;
import com.eworkplaceapps.platform.context.ContextHelper;
import com.eworkplaceapps.platform.logging.LogConfigurer;
import com.mikepenz.materialdrawer.util.DrawerImageLoader;
import com.squareup.picasso.Picasso;
import java.io.IOException;
/**
* application class
*/
public class EdApplication extends MultiDexApplication {
public static Typeface JOSE_FIN_SANS_BOLD;
public static Typeface HELVETICA_NEUE;
public static Typeface JOSE_FIN_SANS_SEMI_BOLD;
@Override
public void onCreate() {
super.onCreate();
JOSE_FIN_SANS_BOLD = Utils.getTypeface(this, 1);
JOSE_FIN_SANS_SEMI_BOLD = Utils.getTypeface(this, 2);
HELVETICA_NEUE = Utils.getTypeface(this, 3);
ContextHelper.setContext(this);
try {
LogConfigurer.configure();
LogConfigurer.initialize();
} catch (IOException e) {
LogConfigurer.error("EdApplication", e.getMessage());
LogConfigurer.error("EdApplication", e.toString());
}
//initialize and create the image loader logic
DrawerImageLoader.init(new DrawerImageLoader.IDrawerImageLoader() {
@Override
public void set(ImageView imageView, Uri uri, Drawable placeholder) {
Picasso.with(imageView.getContext()).load(uri).placeholder(placeholder).into(imageView);
}
@Override
public void cancel(ImageView imageView) {
Picasso.with(imageView.getContext()).cancelRequest(imageView);
}
@Override
public Drawable placeholder(Context ctx) {
return null;
}
});
Account account = new Account(Utils.ACCOUNT_NAME, Utils.ACCOUNT_TYPE);
syncSetup(this, account);
Utils.setAlarm(this);
}
@Override
protected void attachBaseContext(Context base) {
super.attachBaseContext(base);
MultiDex.install(base);
}
/**
* method to create dummy account for background syncing, it will be used by sync adapter
*
* @param context
* @param account
*/
public static void syncSetup(Context context, Account account) {
account = new Account(Utils.ACCOUNT_NAME, Utils.ACCOUNT_TYPE);
AccountManager accountManager = AccountManager.get(context);
if (accountManager.addAccountExplicitly(account, null, null)) {
ContentResolver.setIsSyncable(account, Utils.AUTHORITY, 1);
ContentResolver.setMasterSyncAutomatically(true);
}
ContentResolver.addPeriodicSync(account, Utils.AUTHORITY, Bundle.EMPTY, Utils.SYNC_DURATION);
ContentResolver.setSyncAutomatically(account, Utils.AUTHORITY, true);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,599
|
«Сильная, молчаливая личность» () — сорок девятый эпизод телесериала канала HBO «Клан Сопрано» и десятый в четвёртом сезоне шоу. Телесценарий написали Теренс Уинтер, Робин Грин и Митчелл Бёрджесс по сюжету Дэвида Чейза. Режиссёром стал Алан Тейлор, а премьера состоялась 17 ноября 2002 года.
В ролях
Джеймс Гандольфини — Тони Сопрано
Лоррейн Бракко — д-р Дженнифер Мелфи
Эди Фалко — Кармела Сопрано
Майкл Империоли — Кристофер Молтисанти
Доминик Кьянезе — Коррадо Сопрано-мл.
Стивен Ван Зандт — Сильвио Данте
Тони Сирико — Поли Галтьери
Роберт Айлер — Энтони Сопрано-мл.
Джейми-Линн Сиглер — Медоу Сопрано *
Дреа де Маттео — Адриана Ля Сёрва
Аида Туртурро — Дженис Сопрано *
Федерико Кастеллуччио — Фурио Джунта
Винсент Куратола — Джонни Сэк
и Джо Пантолиано — Ральф Сифаретто *
* = указаны только
Приглашённые звёзды
Сюжет
В Бада Бинге!, через неделю после того, как Тони убил Ральфа, команда зависает, играя в пул. Кристофер Молтисанти входит с большим свёртком в одной руке, когда остальные обсуждают исчезновение Ральфа. Тони решает позвонить Ральфу перед всей командой, симулируя неведение о судьбе Ральфа. Посылка, которую принёс Кристофер, оказывается картиной Тони и Пирожка, которую Тони заказал ранее. Обезумев, Тони уходит из Бинга. Почти в слезах, когда он едет по шоссе, он звонит обратно в стрип-клуб и приказывает уничтожить картину. Позже, Поли видит Бенни Фацио и Маленького Кармайна, пытающихся сжечь картину, и спасает её от уничтожения, сказав, что это будет честью, что у него дома будет висеть картина Тони. Поли вешает её в гостиной, но позже снимает её, чтобы изменить её, чтобы облачить Тони в форме "Наполеоновского" генерала 18-го века.
Исчезновение Ральфа начинает становиться проблемой. За ужином с Сильвио Данте и Пэтси Паризи, Элли Бой Барезе, говорящий неофициально, озвучивает то, о чём некоторые думают: Тони убил Ральфа из-за лошади. После того, как Сильвио покидает стол, Элли Бой заявляет, что любой босс, который убил члена своей команды из-за лошади, должен быть устранён, и что если бы это было так, что Тони совершил такой поступок, Сильвио должен быть "первым в очереди, чтобы вытащить вилку".
Между тем, Тони навещает находящегося в коме сына Ральфа в больнице, и позже проводит сеанс с доктором Мелфи. Когда Тони относит свою боль к потери Пирожка, Мелфи замечает, что он, кажется, горюет за животных больше, чем за людей. Далее, она напоминает ему, что его печаль по поводу того, что утки покинули его бассейн, была внутренней метафорой о его волнениях и страхах по поводу потери своей семьи. Доктор Мелфи спрашивает, может ли смерть лошади быть привязана к другому внутреннему конфликту.
Когда Фурио Джунта возвращается из Неаполя, он приносит подарки для Энтони-младшего и Медоу, но ничего для Кармелы. Позже, Кармела навещает Фурио в его доме, под предлогом давать советы по оформлению интерьера и получает импровизированный подарок в виде баночки с домашним бальзамическим уксусом от Фурио, которую он привёз из Италии. Когда Кармела говорит Розали Април о своих чувствах, вдова советует, что если Кармела ещё не занималась сексом с Фурио, то она не должна, если только держать Тони от возмездия над Фурио.
На встрече, Джонни Сэк требует, чтобы Тони впустил Нью-Йорк в аферу с HUD. Когда Тони отказывается, Джонни угрожает ему: "Ты уверен, что хочешь пойти по этой дороге, Тони?"
Снова в Бада Бинге!, Тони созывает встречу с Поли Уолнатсом, Сильвио Данте, Вито Спатафоре, Карло Джерваси и других ключевых членов семьи. Когда все сидят и слушают, Тони заявляет, что он знает, что случилось с Ральфом, и что если он прав, то он не вернётся. Тони утверждает, что Джонни Сэк убил Ральфа из-за аферы с HUD. Тони напоминает об окупаемости, которая позже потребуется. Другие, к испугу Поли, спекулируют и спрашивают про "шутку Джинни Сэка" и имела ли она какое-либо влияние на исчезновение Ральфа. Тони утверждает, что он уверен, что это не помогло делу.
Адриана Ля Сёрва встречается с её куратором из ФБР, агентом Робин Сансеверино, и пичкает ей как можно меньше информации, насколько это возможно. Сансеверино, зная о героиновой зависимости Кристофера, предлагает, чтобы Адриана подтолкнула его к тому, чтобы он пошёл в реабилитационную клинику, и говорит, что ФБР отправляло им брошюры Hazelden Foundation по почте. Адриана начинает плакать, когда она обнаруживает, что Кристофер, в героино-индуцированном ступоре, сел на её собаку Козетту, убив её. У Кристофера, опоздавшего на встречу с Поли и Сильвио, угоняют машину и его грабят при попытке приобретения героина в малообеспеченном баррио. Когда он возвращается домой избитым, Адриана даёт ему брошюру в реабилитационную клинику, но он ударяет её несколько раз и сбивает её на пол.
Кармелу навещает Адриана в синяках и слышит о том, как зависимость Кристофера оставила его ещё более неуравновешенным. Тони отвергает советы Джуниора о решении проблем с Кристофером, который сравнивает Кристофера с любимой собакой, у которой бешенство и должна быть убита из милосердия. Вместо этого, однажды утром, семья и друзья организовывают интервенцию. Открывающее упражнение включает то, что каждый расскажет Кристоферу о том, как его употребление наркотиков повлияло на их жизнь. Тони приходит в ярость, когда Адриана раскрывает, что Кристофер случайно убил Козетту. Кристофер часто перебивает и словесно атакует тех, кто говорит с ним, включая завуалированными ссылками к фиаско Поли в Пайн Барренс, и упоминая неверность Сильвио. Советник пытается остановить обвинения, летящие по комнате. Но, когда Кристофер начинает проявлять неуважение к Тони и оскорблять его родную мать, он получает побои от людей в комнате. Прежде чем действительно произойти, интервенция прекращается тем, что Кристофера везут в травмпункт с переломом черепа в области волосяного покрова.
В больнице, Тони говорит Кристоферу, что единственная причина, по которой он ещё жив, потому что "Ты мой племянник и я люблю тебя". Он организовывает для Кристофера поход в реабилитационную клинику в Пенсильвании и требует, чтобы он не возвращался, пока он не станет чист и трезвым, сказав ему, что Пэтси остановится в мотеле в полумили от него, чтобы наблюдать за ним. Кристофер прибывает в клинику с Адрианой и Пэтси. Его конфеты конфискованы (потому что они содержат кофеин), и он прощается со своей невестой, прежде чем зайти в заведение.
Тони позже навещает дядю Джуниора, но находит его дремлющим. Там присутствует Светлана Кириленко, без своей ноги-протеза и на костылях. Они обмениваются светской беседой на диване. Светлана обсуждает различия между мировоззрением американцев и других: американцы не ждут ничего плохого и удивляются, когда это происходит, пока остальной мир ожидает худшего и не разочарован. Тони отмечает, что Светлана, в частичной тени с её светлыми волосами и сигаретным дымом, доносящегося до неё, напоминает ему "Грету Гарбл" (Грета Гарбо). Они смотрят друг на друга пристально, целуются и затем занимаются сексом на диване дяди Джуниора. Затем, пока Тони одевается, он предлагает снова увидеться. Светлана говорит, что это не очень хорошая идея, и Тони, похоже, расстроен. Когда медсестра Джуниора Бранка неожиданно приходит и застаёт их сидячими вместе, Тони уходит.
Эпизод заканчивается с характерными сценами, где Тони и Фурио готовят еду и едят в одиночестве в своих домах без Кармелы (Тони подогревает ригатони и наливает в стакан молоко из пластиковой бутылки, в то время как Фурио кипятит пасту, затем он жарит её, режет сыр и пьёт бокал вина). Музыка военных барабанов начинает играть на заднем плане. Поли, также один дома, вешает картину Тони и Моего пирожка над своим камином. Тони теперь переделан и выглядит как "Наполеоновский" генерал 18-го века, как и хотел Поли. Поли садится спиной к картине и поворачивается к телевизору, чтобы смотреть матч "Yankees". Однако, глаза Тони на картине по-прежнему беспокоят его.
Умерла
Козетта: убита, когда Кристофер, находясь в ступоре от героина, садится на неё, пока она лежала на диване.
Название
Название эпизода ссылается на Гэри Купера, которого Тони описал как идеальную модель мужчины. Тони упоминает Купера несколько раз во время сериала, описывая его как "сильную, молчаливую личность". Он впервые сказал это во время терапии в пилоте сериала, и снова в эпизоде четвёртого сезона, "Христофор".
Неизвестного для Кармелы соперника Тони, Фурио, можно описать таким образом.
Другие культурные отсылки
В начале эпизода, когда Кристофер получает кайф от героина, прежде чем сесть на собаку, Козетту, по телевизору показывают короткометражку «Охотники на медведей» из сборника «Пострелята».
Поли говорит Сильвио, что он посмотрел «В порту» в HD и был впечатлён.
Бейсбольная игра, которую Поли смотрит ближе к концу эпизода, это игра Boston Red Sox-New York Yankees, сыгранная на стадионе "Yankee" 4 сентября 2002 года, с победой Yankees со счётом 3-1.
Музыка
Когда Адриана разговаривает с агентом ФБР, стерео из другой машины играет песню "Analyze" The Cranberries.
Примечания
Ссылки
"Сильная, молчаливая личность" на HBO
Список серий телесериала «Клан Сопрано»
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 6,993
|
local sys = require"sys"
local sock = require"sys.sock"
local MCAST_ADDR = "234.5.6.7"
local MCAST_PORT = 25000
local fd = sock.handle()
assert(fd:socket("dgram"))
assert(fd:sockopt("reuseaddr", 1))
local saddr = sock.addr():inet(MCAST_PORT + 1)
assert(fd:bind(saddr))
sys.stdout:write("Enter text: ")
local line = sys.stdin:read()
saddr:inet(MCAST_PORT, sock.inet_pton(MCAST_ADDR))
assert(fd:send(line, saddr))
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 96
|
Southport students celebrate Oxbridge success
Callum Richardson from Hillside and Clare Routledge from Birkdale join three other students from Crosby school in receiving offers from Oxford or Cambridge
Jamie Bowman
Merchant Taylors' students John Chan, Callum Richardson, Clare Routledge, Sam East and Sid Baines
Five students from Merchant Taylors' Schools, including two from Southport, are celebrating offers from Oxbridge this year.
Callum Richardson from Hillside and Clare Routledge from Birkdale, join Sam East from Orrell Park, John Chan from Liverpool and Sid Baines from Crosby in receiving offers from the country's two oldest seats of learning.
Clare, who will head for Cambridge, has been offered a place at Homerton College to study Education with History. She is currently studying A Levels in English Literature, Psychology and History, plus an extra AS Level in Biology.
She said: "I am particularly interested in the educational philosophy element of the degree, along with its coverage of politics and psychology in relation to education.
"I don't have a specific career in mind yet, but working for an education think tank, developing new policies or working in academia appeal to me most at this point."
As a Grade 8 saxophonist and flautist and member of the Chamber Choir, Clare is actively involved in the school's Music and Drama departments. She will play Elizabeth Proctor in the Joint Schools' upcoming production of The Crucible. Alongside this, Clare somehow finds the time to go sailing and participate in the Navy section of the schools' Combined Cadet Force where she holds the rank of Petty Officer.
Senior Monitor Callum will study Economics and Management at Brasenose College, Oxford, after completing A Levels in Economics, Further Maths and Physics.
As a keen mathematician (having already gained an A* in Maths a year early), the course is particularly appealing to Callum as it explains the ways in which mathematics can be applied to economics.
The entrance test for the course showed Callum came in the top 2% of applicants, and his interview involved an informal discussion around a niche area of economics in an attempt to test his ability to handle unfamiliar topics.
He said: "Once I've graduated, I'm interested in pursuing a career in Investment Banking but I'm keeping an open mind during my time at university."
Callum is captain of the school's Shooting Team, Colour Sergeant in Merchant Taylors' Combined Cadet Force and a tennis and table tennis player outside of school.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,225
|
Mycetophila collineola är en tvåvingeart som först beskrevs av Speiser 1909. Mycetophila collineola ingår i släktet Mycetophila och familjen svampmyggor. Inga underarter finns listade i Catalogue of Life.
Källor
Svampmyggor
collineola
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,204
|
Cloud validation tests description
----------------------------------
The following tests for cloud validation are implemented:
* Check disk space outage on the Controller and Compute nodes
* Check log rotation configuration on all nodes
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,925
|
namespace chromeos {
using ::testing::Eq;
using ::testing::ExplainMatchResult;
using ::testing::Field;
using ::testing::Optional;
using ::testing::StrEq;
MATCHER_P(RestrictionsWithMinMilestone,
integral_min_milestone,
"is a Restrictions with min_milestone ``" +
base::NumberToString(integral_min_milestone) + "''") {
return ExplainMatchResult(
Field(&Restrictions::min_milestone,
Optional(base::Version(base::NumberToString(
static_cast<double>(integral_min_milestone))))),
arg, result_listener);
}
MATCHER_P(RestrictionsWithMaxMilestone,
integral_max_milestone,
"is a Restrictions with max_milestone ``" +
base::NumberToString(integral_max_milestone) + "''") {
return ExplainMatchResult(
Field(&Restrictions::max_milestone,
Optional(base::Version(base::NumberToString(
static_cast<double>(integral_max_milestone))))),
arg, result_listener);
}
MATCHER_P2(RestrictionsWithMinAndMaxMilestones,
integral_min_milestone,
integral_max_milestone,
"is a Restrictions with min_milestone ``" +
base::NumberToString(integral_min_milestone) +
"'' "
"and max_milestone ``" +
base::NumberToString(integral_max_milestone) + "''") {
return ExplainMatchResult(
RestrictionsWithMinMilestone(integral_min_milestone), arg,
result_listener) &&
ExplainMatchResult(
RestrictionsWithMaxMilestone(integral_max_milestone), arg,
result_listener);
}
MATCHER(UnboundedRestrictions,
"is a Restrictions with neither min nor max milestones") {
return ExplainMatchResult(
Field(&Restrictions::min_milestone, Eq(absl::nullopt)), arg,
result_listener) &&
ExplainMatchResult(
Field(&Restrictions::max_milestone, Eq(absl::nullopt)), arg,
result_listener);
}
// Matches a ReverseIndexLeaf struct against its |manufacturer| and
// |model| members.
MATCHER_P2(ReverseIndexLeafLike,
manufacturer,
model,
"is a ReverseIndexLeaf with manufacturer ``" +
std::string(manufacturer) + "'' and model ``" +
std::string(model) + "''") {
return ExplainMatchResult(
Field(&ReverseIndexLeaf::manufacturer, StrEq(manufacturer)), arg,
result_listener) &&
ExplainMatchResult(Field(&ReverseIndexLeaf::model, StrEq(model)), arg,
result_listener);
}
// Matches a ParsedIndexLeaf struct against its
// |ppd_basename| member.
MATCHER_P(ParsedIndexLeafWithPpdBasename,
ppd_basename,
"is a ParsedIndexLeaf with ppd_basename ``" +
std::string(ppd_basename) + "''") {
return ExplainMatchResult(
Field(&ParsedIndexLeaf::ppd_basename, StrEq(ppd_basename)), arg,
result_listener);
}
// Matches a key-value pair in a ParsedIndex against its constituent
// members. |parsed_index_leaf_matcher| is matched against the |values|
// member of a ParsedIndexValues struct.
MATCHER_P2(ParsedIndexEntryLike,
emm,
parsed_index_leaf_matcher,
"is a ParsedIndex entry with effective-make-and-model string ``" +
std::string(emm) + "''") {
return ExplainMatchResult(Pair(StrEq(emm), Field(&ParsedIndexValues::values,
parsed_index_leaf_matcher)),
arg, result_listener);
}
// Matches a ParsedPrinter struct against its
// |user_visible_printer_name| and |effective_make_and_model| members.
MATCHER_P2(ParsedPrinterLike,
name,
emm,
"is a ParsedPrinter with user_visible_printer_name``" +
std::string(name) + "'' and effective_make_and_model ``" +
std::string(emm) + "''") {
return ExplainMatchResult(
Field(&ParsedPrinter::user_visible_printer_name, StrEq(name)), arg,
result_listener) &&
ExplainMatchResult(
Field(&ParsedPrinter::effective_make_and_model, StrEq(emm)), arg,
result_listener);
}
} // namespace chromeos
#endif // CHROMEOS_PRINTING_PPD_METADATA_MATCHERS_H_
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,104
|
\section{Introduction}
Biological sequences are statistically heterogeneous, in the sense
that local compositions and correlations in different regions of the
sequences can be very different from one another. They must therefore
treated as collections of statistically stationary segments (or
\emph{domains}), to be discovered by the various segmentation schemes
found in the literature (see review by Braun and M\"uller
\cite{Braun1998StatisticalScience13p142}, and list of references in
Ref.~\citeonline{Cheong2007IRJSSS}). Typically, these segmentation
schemes are tested on (i) artificial sequences composed of a small
number of segments, (ii) control sequences obtained by concatenating
known coding and noncoding regions, or (iii) control sequences
obtained by concatenating sequences from chromosomes know to be
statistically distinct. They are then applied on a few better
characterized genomic sequences, and compared against each other, to
show general agreement, but also to demonstrate better sensitivity in
delineating certain genomic features. To the best of our knowledge,
there are no studies reporting a full and detailed comparison of the
segmentation of a sequence against its distribution of carefully
curated gene calls. There are also no studies comparing the
segmentations of closely related genomes. In such sequences, there
are homologous stretches, interrupted by lineage specific regions, and
the natural question is whether homologous regions in different
genomes will be segmented in exactly the same way by the same
segmentation scheme.
In this paper, we answer this question, without comparing the
segmentation of homologous regions. Instead, through careful
observations of how segment boundaries, or \emph{domain walls}, are
discovered by two different entropic segmentation schemes, we realized
that a subsequence can be segmented differently by the same scheme, if
it is part of two different full sequences. We call this dependence
of a segmentation on the detailed arrangement of segments the
\emph{context sensitivity problem}. In
Sec.~\ref{section:contextsensitivityprobleminrealgenomes}, we will
describe how the context sensitivity problem manifests itself in real
genomes, when these are segmented using a sliding-window entropic
segmentation scheme, which examines local contexts in the sequences,
versus segmentation using a recursive entropic segmentation scheme,
which examines the global contexts of the sequences. We then show how
the context sensitivity problem prevents us from coarse graining by
using larger window sizes, stopping recursive segmentation earlier, or
by simply removing weak domain walls from a fine-scale segmentation.
We follow up in Sec.~\ref{section:meanfieldanalysis} with a mean-field
analysis of the local and global context sensitivity problems, showing
how the positions and strengths of domain walls, and order in which
these are discovered, are affected by these contexts. In particular,
we identify repetitive sequences as the worst case scenario to
encounter during segmentation. Finally, in
Sec.~\ref{section:conclusions}, we summarize and discuss the impacts
of our findings, and explain why we believe the context sensitivity
problem plagues \emph{all} segmentation schemes.
\section{Context Sensitivity Problem in Real Bacterial Genomes}
\label{section:contextsensitivityprobleminrealgenomes}
In this section, we investigate the manifestations of the context
sensitivity problem in two real bacterial genomes, those of
\emph{Escherichia coli} K-12 MG1655 and \emph{Pseudomonas syringae}
DC3000, when these are segmented using two entropic segmentation
schemes. The first entropic segmentation scheme, based on statistics
comparison of a pair of sliding windows, is sensitive to the local
context of segments within the pair of sliding windows, and we shall
show in Sec.~\ref{subsection:pairedslidingwindows} that the positions
and strengths of domain walls discovered by the scheme depends
sensitively on the window size. The second entropic segmentation
scheme is recursive in nature, adding new domain walls at each stage
of the recursion. We shall show in
Sec.~\ref{subsection:recursivegenome} that this scheme is sensitive to
the global context of segments within the sequence, and that domain
walls are not discovered according to their true strengths. In
Sec.~\ref{subsection:bottomupsegmentationhistory}, we show that there
is no statistically consistent way to coarse grain a segmentation by
removing the weakest domain walls, and agglomerating adjacent
segments.
\subsection{Paired Sliding Windows Segmentation Scheme}
\label{subsection:pairedslidingwindows}
Using the paired sliding windows segmentation scheme described in
App.~\ref{section:pairedslidingwindowssegmentationscheme}, the number
$M$ of order-$K$ Markov-chain segments discovered depends on the size
$n$ of the windows used, as shown in Table \ref{table:K0n} for
\emph{E. coli} K-12 MG1655. Because $M$ decreases as $n$ is
increased, we are tempted to think that we can change the granularity
of the segmental description of a sequence by tuning $n$, such that
there are more and shorter segments when $n$ is made smaller, while
there are fewer and longer segments when $n$ is made larger. Thus, as
$n$ is increased, we expect groups of closely spaced domain walls to
be merged as the short segments they demarcate are agglomerated, and
be replaced by a peak close to the position of the strongest peak.
\begin{table}[hbtp]
\centering
\caption{Number of $K = 0$ domain walls in the \emph{E. coli} K-12
MG1655 genome ($N = 4639675$ bp), obtained using the paired sliding
window segmentation scheme for different window sizes $1000 \leq n
\leq 5000$.}
\label{table:K0n}
\vskip .5\baselineskip
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$n$ & 1000 & 2000 & 3000 & 4000 & 5000 \\
\hline
$M$ & 2781 & 1414 & 952 & 721 & 577 \\
\hline
\end{tabular}
\end{table}
Indeed, we do find this expected merging of proximal domain walls in
Fig.~\ref{figure:EcoliK12qrK0n1kn2kn3kn4kn5ki0i40k} and
Fig.~\ref{figure:PsyringaeqrK0n1kn2kn3kn4kn5ki25ki75k}, which shows
the square deviation spectra for the $(0, 40000)$ region of the
\emph{E. coli} K-12 MG1655 genome and the $(25000, 75000)$ region of
the \emph{P. syringae} DC3000 genome respectively. In the $(0,
40000)$ region of the \emph{E. coli} K-12 MG1655 genome shown in
Fig.~\ref{figure:EcoliK12qrK0n1kn2kn3kn4kn5ki0i40k}, we find the group
of domain walls, $i_a \approx 16500$, $i_b \approx 17500$, and $i_c
\approx 18700$, and the pair of domain walls, $i_g \approx 33800$ and
$i_h \approx 35000$, which are distinct in the $n = 1000$ square
deviation spectrum, merging into the domain walls $i_{abc}$ and
$i_{gh}$ in the $n \geq 3000$ square deviation spectra. In the
$(25000, 75000)$ region of the \emph{P. syringae} DC3000 genome shown
in Fig.~\ref{figure:PsyringaeqrK0n1kn2kn3kn4kn5ki25ki75k}, we find the
pair of domain walls, $j_a \approx 45000$ and $j_b \approx 46600$, and
the pair of domain walls, $j_c \approx 50400$ and $j_d \approx 51800$,
which are distinct in the $n = 1000$ square deviation spectrum,
merging into the domain walls $j_{ab}$ and $j_{cd}$ in the $n \geq
3000$ and $n = 5000$ square deviation spectra respectively.
\begin{figure}[hbtp]
\centering
\includegraphics[scale=0.5,clip=true]%
{EcoliK12.q.rK0n1kn2kn3kn4kn5k.0.40k.eps}
\caption{The $K = 0$ square deviation spectra in the region $(0,
40000)$ of the \emph{E. coli} K-12 MG1655 genome, obtained using the paired
sliding window segmentation scheme with window sizes (top to bottom)
$n = 1000$, 2000, 3000, 4000, and 5000.}
\label{figure:EcoliK12qrK0n1kn2kn3kn4kn5ki0i40k}
\end{figure}
\begin{figure}[hbtp]
\centering
\includegraphics[scale=0.5,clip=true]%
{Psyringae.q.rK0n1kn2kn3kn4kn5k.25k.75k.eps}
\caption{The $K = 0$ square deviation spectra in the region $(25000,
75000)$ of the \emph{P. syringae} DC3000 genome, obtained using the
paired sliding window segmentation scheme with window sizes (top to
bottom) $n = 1000$, 2000, 3000, 4000, and 5000.}
\label{figure:PsyringaeqrK0n1kn2kn3kn4kn5ki25ki75k}
\end{figure}
However, we also find unexpected changes in the relative strengths of
the domain walls, as $n$ is increased. In the $(0, 40000)$ region of
the \emph{E. coli} K-12 MG1655 genome shown in
Fig.~\ref{figure:EcoliK12qrK0n1kn2kn3kn4kn5ki0i40k}, we find that $i_d
\approx 21800$, which appears as a broad, weak, and noisy bump in the
$n = 1000$ square deviation spectrum, becoming stronger and more
defined as $n$ is increased, and finally becomes as strong as the
domain wall $i_{abc}$ in the $n = 5000$ square deviation spectrum. In
this region of the \emph{E. coli} K-12 MG1655 genome, we also find
that the domain walls $i_b \approx 17500$ and $i_f \approx 30000$ are
equally strong in the $n = 1000$ square deviation spectrum, but as $n$
is increased, $i_b$ becomes stronger while $i_f$ becomes weaker. In
the $(25000, 75000)$ region of the \emph{P. syringae} DC3000 genome
shown in Fig.~\ref{figure:PsyringaeqrK0n1kn2kn3kn4kn5ki25ki75k}, we
find that the domain walls $j_c \approx 50400$ and $j_f \approx 58200$
are equally strong, and also the domain walls $j_d \approx 51800$ and $j_e
\approx 57300$ are equally strong, in the $n = 1000$ square deviation
spectrum. However, as $n$ is increased, $j_c$ becomes stronger than
$j_f$, while $j_d$ becomes stronger than $j_e$. More importantly,
all these domain walls --- the strongest in this $(25000, 75000)$
region of the $n = 1000$ square deviation spectrum --- become weaker
as $n$ is increased, to be superseded by the domain walls $j_{ab}
\approx 45000$, $j_g \approx 65400$ and $j_h \approx 72400$, which
become stronger as $n$ is increased. As it turned out, $(j_c, j_f)$
overlaps significantly with the interval interval $(50000, 59000)$,
which incorporates three lineage-specific regions (LSRs 5, 6, and 7,
all of which virulence related) identified by Joardar \emph{et al}
\cite{Joardar2005MolPlantPathol6p53}. It is therefore biologically
significant that $j_c$ and $j_f$ are strong domain walls in the $n =
1000$ square deviation spectrum. On the other hand, it is not clear
what kind of biological meaning we can attach to $j_{ab}$, $j_g$, and
$j_h$ being the strongest domain walls in the $n = 5000$ square
deviation spectrum.
\begin{table}[htbp]
\centering
\caption{Positions of strong domain walls in the $(0, 40000)$ region
of the \emph{E. coli} K-12 MG1655 genome and the $(25000, 75000)$
region of the \emph{P. syringae} DC3000 genome, determined after match
filtering the square deviation spectra obtained using the paired
sliding window segmentation scheme with window sizes $n = 3000, 4000,
5000$.}
\label{table:slidingwindowshifts}
\vskip .5\baselineskip
\begin{tabular}{|c|c|c|c|c|c|c|}\hline
& \multicolumn{3}{c|}{\emph{E. coli} K-12 MG1655} &
\multicolumn{3}{c|}{\emph{P. syringae} DC3000} \\ \hline
$n$ & $i_{abc}$ & $i_d$ & $i_h$ & $j_{ab}$ & $j_g$ & $j_h$ \\ \hline
3000 & 16200 & 21800 & 34100 & 46600 & 66600 & 71500 \\
4000 & 16300 & 21700 & 34400 & 45900 & 65900 & 72500 \\
5000 & 16100 & 22100 & 34700 & 45700 & 65500 & 72500 \\ \hline
\end{tabular}
\end{table}
There is another, more subtle, effect that increasing the size of the
sliding windows has on the domain walls: their positions, as
determined from peaks in the square deviation spectrum after match
filtering, are shifted. The shifting positions of some of the strong
domain walls in the $(0, 40000)$ region of the \emph{E. coli} K-12
MG1655 genome and the $(25000, 75000)$ region of the \emph{P.
syringae} DC3000 genome are shown in Table
\ref{table:slidingwindowshifts}. In general, the positions and
strengths of domain walls can change when the window size used in the
paired sliding windows segmentation scheme is changed, because windows
of different sizes examine different local contexts. As a result of
this local context sensitivity, whose nature we will illustrate using
a mean-field picture in
Sec.~\ref{subsection:meanfieldwindowedspectrum}, the sets of strong
domain walls determined using two different window sizes $n$ and $n' >
n$ are different. If $n$ and $n'$ are sufficiently different, the
sets of strong domain walls, i.e.~those stronger than a specified
cutoff, may have very little in common. Therefore, we cannot think of
the segmentation obtained at window size $n'$ as the coarse grained
version of the segmentation obtained at window size $n$.
\subsection{Optimized Recursive Jensen-Shannon Segmentation Scheme}
\label{subsection:recursivegenome}
Using the optimized recursive Jensen-Shannon segmentation scheme
described in Ref.~\citeonline{Cheong2007IRJSSS}, we obtained one
series of segmentations each for \emph{E. coli} K-12 MG1655 and
\emph{P. syringae} DC3000, shown in
Fig.~\ref{figure:hierarchyofrecursivesegmentations} and
Fig.~\ref{figure:pstm2m50} respectively. Two features are
particularly striking about these plots. First, there exist domain
walls stable with respect to segmentation optimization. These
\emph{stable domain walls} remain close to where they were first
discovered by the optimized recursive segmentation scheme. Second,
there are \emph{unstable domain walls} that get shifted by as much as
10\% of the total length of the genome when a new domain wall is
introduced. For example, in
Fig.~\ref{figure:hierarchyofrecursivesegmentations} for the \emph{E.
coli} K-12 MG1655 genome, we find the domain wall $i_{10} = 4051637$
in the optimized segmentation with $M = 10$ domain walls shifted to
$i_{10} = 4469701$ in the optimized segmentation with $M = 11$ domain
walls ($\delta i_{10} = +418064$), and also the domain wall $i_7 =
2135183$ in the optimized segmentation with $M = 15$ domain walls
shifted to $i_7 = 2629043$ in the optimized segmentation with $M = 16$
domain walls ($\delta i_7 = +493860$). Based on the observation that
some unstable domain walls are discovered, lost, later rediscovered
and become stable, we suggested in Ref.~\citeonline{Cheong2007IRJSSS}
that for a given segmentation with $M$ domain walls, stable domain
walls are statistically more significant than unstable domain walls,
while stable domain walls discovered earlier are more significant than
stable domain walls discovered later in the optimized recursive
segmentation.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.8]{ecolik12m2m50.eps}
\caption{Series of optimized recursive Jensen-Shannon segmentations of
the \emph{E. coli} K-12 MG1655 genome, for (top to bottom) $2 \leq M
\leq 50$ domain walls. The two stable domain walls that appear in the
$M = 2$ optimized segmentation are close to the replication origin and
replication terminus.}
\label{figure:hierarchyofrecursivesegmentations}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.8]{pstm2m50.eps}
\caption{Series of optimized recursive Jensen-Shannon segmentations of
the \emph{P. syringae} DC3000 genome, for (top to bottom) $2 \leq M
\leq 55$ domain walls. Compared to the \emph{E. coli} K-12 MG1655
genome, there are perceptibly more unstable domain walls in the
\emph{P. syringae} DC3000 genome.}
\label{figure:pstm2m50}
\end{figure}
From Fig.~\ref{figure:hierarchyofrecursivesegmentations} and
Fig.~\ref{figure:pstm2m50}, we also find that the \emph{E. coli} K-12
MG1655 and \emph{P. syringae} DC3000 genomes have very different
segmental textures. At this coarse scale ($M \sim 50$ segments), we
find many short segments, many long segments, but few segments of
intermediate lengths in the \emph{E. coli} K-12 MG1655 genome. In
contrast, at the same granularity, the \emph{P. syringae} DC3000
genome contains many short segments, many segments of intermediate
lengths, but few long segments. We believe these segmental textures
are consistent with the different evolutionary trajectories of the two
bacteria. \emph{E. coli} K-12 MG1655, which resides in the highly
stable human gut environment, has a more stable genome containing
fewer large-scale rearrangements which appear to be confined to
hotspots within the $(2600000, 3600000)$ region. The genome of
\emph{P. syringae} DC3000, on the other hand, has apparently
undergone many more large-scale rearrangements as its lineage
responded to multiple evolutionary challenges living in the hostile
soil environment.
We find many more large shifts in the optimized domain wall positions
in \emph{P. syringae} DC3000 compared to \emph{E. coli} K-12 MG1655,
because of the more varied context of the \emph{P. syringae} DC3000
genome. However, large shifts in the optimized domain wall positions
arise generically in all bacterial genomes, because of the sensitivity
of optimized domain wall positions to the contexts they are restricted
to. In Sec.~\ref{subsection:meanfieldwindowlessspectrum}, we will
illustrate using a mean-field picture how the recursive segmentation
scheme decides where to subdivide a segment, i.e.~add a new domain
wall, after examining the global context within the segment. We then
show how this global context changes when the segment is reduced or
enlarged during segmentation optimization, which can then cause a
large shift in the position of the new domain wall. Because of this
\emph{global context sensitivity}, we find in
Fig.~\ref{figure:pstm2m50} a large shift of the domain wall $j_9 =
1723734$, which is stable when there are $36 \leq M \leq 51$ optimized
domain walls in the segmentation, to its new position $j_9 = 1818461$
($\delta j_9 = +94727$) when one more optimized domain wall is added.
We say that a domain wall is \emph{stable at scale $M$} if it is only
slightly shifted, or not at all, within the optimized segmentations
with between $M - \delta M$ and $M + \delta M$ domain walls, where
$\delta M \ll M$. Given a series of recursively determined optimized
segmentations, we know which domain walls in an optimized segmentation
containing $M$ domain walls are stable at scale $M$, and which domain
walls in an optimized segmentation containing $M' > M$ domain walls
are stable at scale $M'$. However, these two sets of stable domain
walls can disagree significantly because of the recursive segmentation
scheme's sensitivity to global contexts. Again, we cannot think of
the optimized segmentation containing $M$ domain walls as a coarse
grained version of the optimized segmentation containing $M'$ domain
walls.
\subsection{Coarse-Graining by Removing Domain Walls}
\label{subsection:bottomupsegmentationhistory}
In Sec.~\ref{subsection:pairedslidingwindows}, we saw the difficulties
in coarse graining the segmental description of a bacterial genome by
using larger window sizes, due to the paired sliding windows
segmentation scheme's sensitivity to local context. We have also seen
in Sec.~\ref{subsection:recursivegenome} a different set of problems
associated with coarse graining by stopping the optimized recursive
Jensen-Shannon segmentation earlier, due this time to the scheme's
sensitivity to global context. Another way to do coarse graining
would be to start from a fine segmentation, determined using a paired
sliding window segmentation scheme with small window size, or properly
terminated recursive segmentation scheme, and then remove the weakest
domain walls. Our goal is to agglomerate shorter, weakly distinct
segments into longer, more strongly distinct segments. Although this
sounds like the recursive segmentation scheme playbacked in reverse,
there are subtle differences: in the recursive segmentation scheme,
strong domain walls may be discovered after weak ones are discovered,
so our hope with this coarse graining scheme is that we target weak
domain walls after `all' domain walls are discovered.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{EcoliK12.q.topsegK.K0n1000.eps}
\caption{Bottom-up segmentation history for \emph{E. coli} K-12 MG1655
derived from the initial $(K = 0, n = 1000)$ paired sliding windows
segmentation containing $M = 2781$ domain walls. (Inset) Bottom-up
segmentation history from $M = 1600$ domain walls remaining to $M =
1400$ domain walls remaining, showing the fine structure of dips below
the smooth envelope.}
\label{figure:EcoliK12qtopsegKK0n1000}
\end{figure}
Like recursive segmentation, there are many detail variations on the
implementation of such a coarse graining scheme. The first thing we
do is to select a cutoff strength $\Delta^*$, which we can think of as
a knob we tune to get a desired granularity for our description of the
genome: we keep a large number of domain walls if $\Delta^*$ is small,
and keep a small number of domain walls if $\Delta^*$ is large. After
selecting $\Delta^*$, we can then remove all domain walls weaker than
$\Delta^*$ in one fell swoop, or remove them progressively, starting
from the weakest domain walls. However we decide to remove domain
walls weaker than $\Delta^*$, the strengths of the remaining domain
walls must be re-evaluated after some have been removed from the
segmentation. This is done by re-estimating the maximum-likelihood
transition probabilities, and using them to compute the Jensen-Shannon
divergences between successive coarse-grained segments, which are the
strengths of our remaining domain walls. For the purpose of
benchmarking, we start from the $(K = 0, n = 1000)$ paired sliding
windows segmentation containing $M = 2781$ domain walls for the
\emph{E. coli} K-12 MG1655 genome, and remove the weakest domain wall
each time to generate a \emph{bottom-up segmentation history}, shown
in Fig.~\ref{figure:EcoliK12qtopsegKK0n1000}. As we can see, the
strength of the weakest domain wall as a function of the number of
domain wall remaining consists of a smooth envelope, and dips below
this envelope. We distinguish between sharp dips, which are the
signatures of what we called \emph{tunneling events}, and broad dips,
which are the signatures of what we called \emph{cascade events}.
\begin{figure}[hbt]
\centering
\includegraphics[scale=0.8,clip=true]{NC_000913.topsegK.L1586.eps}
\caption{A tunneling event occuring between $M = 1586$ and $M = 1584$
domain walls remaining in the bottom-up segmentation history of
\emph{E. coli} K-12 MG1655 ($N = 4639675$ bp), starting from the $(K =
0, n = 1000)$ initial segmentation containing $M = 2791$ domain walls.
Three segments in the $(4604497, 4632896)$ region of the genome are
shown. The short segment involved in this tunneling event consists of
the single gene \emph{yjjX} on the negative strand (green), flanked by
two segments consisting of genes found predominantly on the positive
strand (red). At each stage of the bottom-up segmentation history,
the domain wall removed is highlighted in red.}
\label{figure:NC000913topsegKL1586}
\end{figure}
\begin{figure}[hbt]
\centering
\includegraphics[scale=0.8,clip=true]{NC_000913.topsegK.L1846.eps}
\caption{A cascade event occuring between $M = 1846$ and $M = 1841$
domain walls remaining in the bottom-up segmentation history of
\emph{E. coli} K-12 MG1655 ($N = 4639675$ bp), starting from the $(K =
0, n = 1000)$ initial segmentation containing $M = 2791$ domain walls.
Six segments in the $(1142115, 1157158)$ region of the genome are
shown. The first domain wall to be removed in this cascade event lies
close to the boundary between the gene \emph{rne}, believed to be
RNase E, on the negative strand (green), and the gene \emph{yceQ},
coding for a hypothetical protein, on the positive strand (red). The
second domain wall to be removed in the cascade is in the middle of
the gene \emph{rpmF} on the positive strand, the third is close to the
boundary between \emph{fabF} and \emph{pabC}, the fourth is close to
the boundary between \emph{pabC} and \emph{yceG}, and the last is
close to the boundary between \emph{holB} and \emph{ycfH}. At each
stage of the bottom-up segmentation history, the domain wall removed
is highlighted in red.}
\label{figure:NC000913topsegKL1846}
\end{figure}
Looking more closely at the segment statistics, we realized that a
tunneling event involves a short segment flanked by two long segments
which are statistically similar to one another, but different from the
short segment. This statistical dissimilarity between the short
segment and its long flanking segments is reflected in the moderate
strengths $\Delta_L$ and $\Delta_R$ of the left and right domain walls
of the short segment. Let us say the right domain wall is slightly
weaker than the left domain wall, i.e. $\Delta_R \lesssim \Delta_L$.
As the bottom-up segmentation history progresses, there will reach a
stage where we remove the right domain wall. When this happens, the
short segment will be assimilated by its right flanking segment.
Because the right flanking segment is long, absorbing the short
segment represents only a small perturbation in its segment
statistics. The longer right segment that results is still
statistically similar to the left segment. Therefore, when we
recompute the strength $\Delta_L$ of the remaining domain wall, we
find that it is now smaller than the strength $\Delta_R$ of the domain
wall that was just removed. This remaining domain wall therefore
becomes the next to be removed in the bottom-up segmentation history,
afterwhich the next domain wall to be removed occurs somewhere else in
the sequence, and has strength slightly larger than $\Delta_R$. The
signature of a tunneling event is therefore a sharp dip in the
bottom-up segmentation history. Biologically, a short segment with a
tunneling event signature is likely to represent an insertion sometime
in the evolutionary past of the organism. A tunneling event in the
$(K = 0, n = 1000)$ bottom-up segmentation history is shown in Fig.
\ref{figure:NC000913topsegKL1586}. In contrast, a cascade event
involves a cluster of short segments of varying statistics flanked by
two long segments that are statistically similar. The domain walls
separating the short segments from each other and from the long
flanking segments are then removed in succession. This sequential
removal of domain walls gives rise to an extended dip in the bottom-up
segmentation history, with a complex internal structure that depends
on the actual distribution of short segments. Biologically, a cluster
of short segments participating in a cascade event points to a
possible recombination hotspot on the genome of the organism. A
cascade event in the $(K = 0, n = 1000)$ bottom-up segmentation
history is shown in Fig. \ref{figure:NC000913topsegKL1846}.
Clearly, by removing more and more domain walls, we construct a proper
hierarchy of segmentations containing fewer and fewer domain walls,
which agrees intuitively with our notion of what coarse graining is
about. We also expected to obtain a unique coarse-grained
segmentation, containing only domain walls stronger than $\Delta^*$,
by removing all domain walls weaker than $\Delta^*$. It turned out
the picture that emerge from this coarse graining procedure is more
complicated, based on which we identified three main problems. First,
let us start with a segmentation containing domain walls weaker than
$\Delta^*$, and decide to remove these domain walls in a single step.
Recomputing the strengths of the remaining domain walls, we would find
that some of these will be weaker than $\Delta^*$, and so cannot claim
to have found the desired coarse-grained segmentation. Naturally, we
iterate the process, removing all domain walls weaker than $\Delta^*$,
and recomputing the strengths of the remaining domain walls, until all
remaining domain walls are stronger than $\Delta^*$. Next, we try
removing domain walls weaker than $\Delta^*$ one at a time, starting
from the weakest, and recompute domain wall strengths after every
removal. The strengths of a few of the remaining domain walls will
change each time the weakest domain wall is removed, sometimes
becoming stronger, and sometimes becoming weaker, but we continue
removing the weakest domain wall until all remaining domain walls are
stronger than $\Delta^*$. Comparing the segmentations obtained using
the two coarse-graining procedures, we will find that they can be very
different. This difficulty occurs for all averaging problems, so we
are not overly concerned, but argue instead that removing the weakest
domain wall each time is like a renormalization-group procedure, and
should therefore be more reliable than removing many weak domain walls
all at once.
Once we accept this decremental procedure for coarse graining, we
arrive at the second problem. Suppose we do not stop coarse graining
after arriving at the first segmentation with all domain walls
stronger than $\Delta^*$, but switch strategy to target and removing
segments associated with tunneling and cascade events. The
segmentations obtained after all domain walls associated with such
segments will contain only domain walls stronger than $\Delta^*$, but
the segmentations in the intermediate steps will contain domain walls
weaker than $\Delta^*$. If we keep coarse graining until no tunneling or
cascade events weaken domain walls below $\Delta^*$, we would end up
with a series of coarse-grained segmentations containing different
number of domain walls. These segmentations do not have the same
minimum domain wall strengths, but are related to each other through
stages in which some domain walls are weaker than $\Delta^*$. We
worry about this series of segmentations when there exist domain walls
with equal or nearly equal strengths. If at any stage of the coarse
graining, these domain walls become the weakest overall, and we stick
to removing one domain wall at a time, we can remove any one of these
equally weak domain walls. If we track the different bottom-up
segmentation histories associated with each choice, we will find that
the coarse-grained segmentations for which all domain walls first
become stronger than $\Delta^*$ can be very different. However, if we
coarse grain further by targetting tunneling and cascading segments,
we would end up with the same coarse-grained segmentation for which no
domain walls ever become weaker than $\Delta^*$. Another way to think
of this coarsest segmentation is that it is the one for which no
domain wall stronger than $\Delta^*$ can be added without first adding
a domain wall weaker than $\Delta^*$.
Third, we know from the bottom-up segmentation history that short
segments participating in tunneling events can be absorbed into their
long flanking segments without appreciably changing the strengths of
the latter's other domain walls. Clearly, absorbing statistically
very distinct short segments increases the heterogenuity of the
coarse-grained segment. This is something we have to accept in coarse
graining, but ultimately, what we really want at each stage of the
coarse graining is for segments to be no more heterogeneous than some
prescribed segment variance. Unfortunately, the segment variances are
not related to the domain wall strengths in a simple fashion, and even
if we know how to compute these segment variances, there is no
guarantee that a coarse graining scheme based on these will be less
problematic. The bottomline is, all these problems arise because
domain wall strengths change wildly as segments are agglomerated in
the coarse graining process, due again to the context sensitivity of
the Jensen-Shannon divergence (or any other entropic measure, for that
matter).
\section{Mean-Field Analyses of Segmentation Schemes}
\label{section:meanfieldanalysis}
From our segmentation and coarse graining analyses of real genomes in
Sec.~\ref{section:contextsensitivityprobleminrealgenomes}, we realized
that these cannot be thought of as consisting of long segments that
are strongly dissimilar to its neighboring long segments, within which
we find short segments that are weakly dissimilar to its neighboring
short segments. In fact, the results suggest that there are short
segments that are strongly dissimilar to its neighboring long
segments, which are frequently only weakly dissimilar to its
neighboring long segments. This mosaic and non-hierarchical structure
of segments is the root of the context sensitivity problem, which we
will seek to better understand in this section.
\begin{figure}[htbp]
\centering
\includegraphics[width=.7\linewidth]{meanfieldpicture.eps}
\caption{Going from a discrete description to a continuum description
of a nucleotide sequence.}
\label{figure:meanfieldpicture}
\end{figure}
To do this, we go first to a continuum description of discrete genomic
sequences, as shown in Fig.~\ref{figure:meanfieldpicture}, where we
allow the sequence positions and the various $K$-mer frequencies to
vary continuously. To eliminate spatial inhomogenuities in the
statistics of the interval $[i, j > i)$, which we want to model as a
statistically stationary segment in the \emph{mean-field limit}, we
distribute its $K$-mer statistics uniformly along the segment. More
precisely, if $f_{\mathbf{t} s}^{[i, j)}$ is the number of times the
$(K+1)$-mer $\alpha_{t_K}\cdots\alpha_{t_1}\alpha_s$, which we also
refer to as the \emph{transition} $\mathbf{t} \to s$, appears in $[i, j)$, we
define the mean-field count $f_{\mathbf{t} s}^{[i', j')}$ of the transition
$\mathbf{t} \to s$ within the subinterval $[i', j' > i') \subseteq [i,
j)$ to be
\begin{equation}
f_{\mathbf{t}s}^{[i', j')} \equiv
\frac{j' - i'}{j - i}\,
f_{\mathbf{t}s}^{[i, j)}.
\end{equation}
Within this mean-field picture, we discuss in
Sec.~\ref{subsection:meanfieldwindowedspectrum} how the paired
sliding-window scheme's ability to detect domain walls depends on the
size $n$ of the pair of sliding windows. We show, in contrast to the
positions and strengths being determined exactly by this segmentation
scheme for domain walls between segments both longer than $n$, that
domain walls between segments, one or both of which are shorter than
$n$, are weakened and shifted in the mean-field limit. Following
this, we show in Sec.~\ref{subsection:meanfieldwindowlessspectrum}
that the strengths of the domain walls obtained from the recursive
segmentation scheme are context sensitive, and approach the exact
strengths only as we approach the terminal segmentation. We explain
why optimization is desirable at every step of the recursive
segmentation, before going on to explain why repetitive sequences are
the worst kind of sequences to segment in
Sec.~\ref{subsection:oscillatorysequence}. In this section, we
present numerical examples for $K = 0$ Markov chains, but all
qualitative conclusions are valid for Markov chains of order $K > 0$.
\subsection{Paired Sliding Windows Segmentation Scheme}
\label{subsection:meanfieldwindowedspectrum}
For a pair of windows of length $n$ sliding across a mean-field
sequence, there are three possibilities (see
Fig.~\ref{figure:windowedcases}):
\begin{enumerate}
\item both windows lie entirely within a single mean-field segment;
\item the two windows straddle two mean-field segments, i.e. a single
domain wall within one of the windows;
\item the two windows straddle multiple mean-field segments.
\end{enumerate}
The first situation is trivial, as the left and right windowed counts
are identical,
\begin{equation}
f_{\mathbf{t}s}^L = f_{\mathbf{t}s}^R = \frac{n}{N_{\text{seg}}}\,
f_{\mathbf{t}s}^{\text{seg}},
\end{equation}
$N_{\text{seg}}$ being the length of the mean-field segment, and
$f_{\mathbf{t}s}^{\text{seg}}$ being the transition counts within the
mean-field segment. The Jensen-Shannon divergence, or the square
deviation between the two windows therefore vanishes identically. The
second situation, which is what the paired sliding windows
segmentation scheme is designed to handle, is analyzed in
App.~\ref{subsection:meanfieldlineshape}. Based on that analysis, we
showed that the position and strength of the domain wall between the
two mean-field segments can be determined exactly. We also derived
the mean-field lineshape for match filtering.
\begin{figure}[htbp]
\centering
\includegraphics{windowedcases.eps}
\caption{The three possible situations that we encounter when we slide
a symmetric pair of windows across a sequence composed of many
mean-field segments: (1) both windows lie entirely within a single
mean-field segment; (2) the two windows straddle two mean-field
segments; and (3) the two windows straddle multiple mean-field
segments.}
\label{figure:windowedcases}
\end{figure}
In this subsection, our interest is in understanding how the paired
sliding windows segmentation scheme behaves in the third situation.
Clearly, the precise structure of the mean-field divergence spectrum
will depend on the local context the pair of windows is sliding
across, so we look at an important special case: that of a pair of
length-$n$ windows sliding across a segment shorter than $n$. In Fig.
\ref{figure:wJSshortsegment}, we show two lineshapes which are
expected to be generic, for (i) the long segments flanking the short
segment are themselves statistically dissimilar (top plot); and (ii)
the long segments flanking the short segment are themselves
statistically similar (bottom plot). In case (i), the mean-field
lineshape obtained as the pair of windows slides across the short
segment consists of a single peak at one of its ends. This peak is
broader than that of a simple domain wall by the width of the short
segment, and therefore, if we perform match filtering using the
quadratic mean-field lineshape in Eq. \eqref{equation:meanfieldJS},
the center of the match-filtered peak would occur not at either ends
of the short segment, but somewhere in the interior.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{wJSshortsegment.eps}
\caption{The Jensen-Shannon divergence $\Delta(z)$ (solid curves) of a
pair of sliding windows of length $n = 1$ as it slides across the
binary mean-field segments (left to right) $a$, $b$, and $c$, with
lengths $N_a > 1$, $N_b < 1$, and $N_c > 1$ respectively. On the
above plots, the left and right ends of segment $b$ are highlighted by
the dashed vertical lines at the normalized sequence positions $z = 0$
and $z = 0.5$ respectively. For the top plot, the probabilities
associated with the mean-field segments are $P_a(0) = 1 - P_a(1) =
0.30$, $P_b(0) = 1 - P_b(1) = 0.50$, and $P_c(0) = 1 - P_c(1) = 0.60$.
For the bottom plot, the probabilities associated with the mean-field
segments are $P_a(0) = 1 - P_a(1) = 0.20$, $P_b(0) = 1 - P_b(1) =
0.70$, and $P_c(0) = 1 - P_c(1) = 0.22$.}
\label{figure:wJSshortsegment}
\end{figure}
In case (ii), the mean-field lineshape obtained as the pair of windows
slides across the short segment consists of a pair of peaks, both of
which are narrower than the mean-field lineshape of a single domain
wall. After we perform match filtering, the center of the
match-filtered left peak would be left of the true left domain wall,
while the center of the match-filtered right peak would be right of
the true right domain wall. Case (ii) is of special interest to us,
as it is the context that give rise to tunneling events in the
bottom-up segmentation history. Both contexts give rise to shifts in
the domain wall positions, as well as to changes in the strengths of
the unresolved domain walls, and thus may be able to explain some of
the observations made in Sec.~\ref{subsection:pairedslidingwindows}.
In case (i), the domain wall strength can increase or decrease,
depending on how different the two long flanking segments are compared
to the short segment. In case (ii), the domain wall strengths always
decrease.
\subsection{Optimized Recursive Jensen-Shannon Segmentation Scheme}
\label{subsection:meanfieldwindowlessspectrum}
To understand how the optimized recursive Jensen-Shannon segmentation
is sensitive to global context, let us first understand what happens
when the segments discovered recursively are not optimized, and then
consider the effects of segmentation optimization. In
Fig.~\ref{figure:JS10}, we show the Jensen-Shannon
divergence spectrum for a sequence consisting of ten mean-field
segments. As we can see, the mean-field Jensen-Shannon divergence is
everywhere convex, except at the domain walls. These are associated
with peaks or kinks in the divergence spectrum, depending on the
global context within the sequence. Under special distributions of
the segment statistics, domain walls may even have vanishing
divergences.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{JS10.eps}
\caption{The Jensen-Shannon divergence $\Delta(z)$ (red solid curve)
as a function of the normalized cursor position $z$ within an
artificial binary sequence composed of ten mean-field segments,
characterized by the probabilities (left to right) $\mathbf{P}(0) =
(0.55, 0.05, 0.20, 0.60, 0.65, 0.30, 0.45, 0.05, 0.45, 0.15)$. The
blue bars indicate the true strengths of each of the nine domain
walls, at $z_1 = 0.15$, $z_2 = 0.25$, $z_3 = 0.35$, $z_4 = 0.50$, $z_5
= 0.65$, $z_6 = 0.70$, $z_7 = 0.85$, $z_8 = 0.90$, and $z_9 = 0.95$,
while the number at each domain wall indicate which recursion step it
is discovered. (Inset) The Jensen-Shannon divergence $\Delta(z)$
(red solid curve) as a function of the normalized cursor position $z$
within an artificial binary sequence composed of two mean-field
segments, characterized by the probabilities $P_L(0) = 0.10$ and
$P_R(0) = 0.90$. The domain wall at $z = 0.60$ is indicated by the
blue dashed vertical line.}
\label{figure:JS10}
\end{figure}
All nine domain walls in the ten-segment sequence are recovered if we
allow the recursive Jensen-Shannon segmentation without segmentation
optimization to go to completion. However, as shown in
Fig.~\ref{figure:JS10}, these domain walls are not discovered in the
order of their true strengths (heights of the blue bars), given by
the Jensen-Shannon divergence between the pairs of segments they
separate. In fact, just like in the coarse graining procedure
described in Sec.~\ref{subsection:bottomupsegmentationhistory}, the
Jensen-Shannon divergence at each domain wall changes as the recursion
proceeds, as the context it is found in gets refined. For this
ten-segment sequence, the recursive segmentation scheme's sensitivity
to global context results in the third strongest domain wall being
discovered in the first recursion step, the second and fourth
strongest domain walls being discovered in the second recursion step,
and the strongest domain wall being discovered only in the third
recursion step.
To see the extent to which optimization ameliorate the global context
sensitivity of the recursive segmentation scheme, let us imagine the
ten-segment sequence to be part of a longer sequence being recursively
segmented. Let us further suppose that under segmentation
optimization, the segment $(0.95, 1.00)$ gets incorporated by the
sequence to the right of $(0.00, 1.00)$. With this, we now examine in
detail a nine-segment sequence $(0.00, 0.95)$, whose mean-field
divergence spectrum is shown in Fig.~\ref{figure:JS10c9}, instead of
the original ten-segment sequence $(0.00, 1.00)$. From
Fig.~\ref{figure:JS10c9}, we find the divergence maximum of the
nine-segment sequence is at $z_3 = 0.35$, the second strongest of the
nine domain walls, instead of the third strongest domain wall at $z_7
= 0.85$ for the ten-segment sequence. In proportion to the length of
the ten-segment sequence, this shift from the third strongest domain
wall to the second strongest domain wall is huge, by about half the
length of the sequence, when the change in context involves a loss of
only 5\% of the total length. In
Sec.~\ref{subsection:recursivegenome}, we saw instances of such large
shifts in optimized domain wall positions when we recursively add one
new domain wall each time to a real genome.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{JS10c9.eps}
\caption{The windowless Jensen-Shannon divergence spectrum $\Delta(z)$
(red solid curve) of the nine-segment binary sequence, after losing
the short segment at its right end. The blue bars indicate the
strength of each of the nine domain walls.}
\label{figure:JS10c9}
\end{figure}
In this example of the ten-segment sequence, we saw that segmentation
optimization has the potential to move an existing domain wall, from a
weaker (the third strongest overall), to a stronger (the second
strongest overall, and if the global context is different, perhaps even
to the strongest overall) position. However, the nature of the
context sensitivity problem is such that no guarantee can be offered
on the segmentation optimization algorithm always moving a domain wall
from a weaker to a stronger position. Nevertheless, segmentation
optimization frequently does move a domain wall from a weaker position
to a stronger position, and it always make successive segments as
statistically distinct from each other as possible. This is good
enough a reason to justify the use of segmentation optimization.
\subsection{Repetitive Sequences}
\label{subsection:oscillatorysequence}
In this last subsection of Sec.~\ref{section:meanfieldanalysis}, let
us look at repetitive sequences, for which the context sensitivity
problem is the most severe. Such sequences, which are composed of
periodically repeating motifs, are of biological interest because they
arise from a variety of recombination processes, and are fairly common
in real genomic sequences. In general, a motif $a_1a_2\cdots a_r$
that is repeated in a repetitive sequence can consists of $r$
statistically distinct subunits, but for simplicity, let us look only
at $ab$-repeats, and highlight statistical signatures common to all
repetitive sequences.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{wabx8mf.eps}
\caption{The Jensen-Shannon divergence spectrum (top, red solid curve)
before, and (bottom, red solid curve) after match filtering and
quality enhancement, for a pair of windows of size $n = 1$ sliding
across a repetitive binary $K = 0$ sequence $cababababababababc$,
where the subunits $a$ (light green) and $b$ (light yellow) both have
lengths $n_a = n_b = 0.7$, and are characterized by the probabilities
$P_a(0) = 1 - P_a(1) = 0.1$ and $P_b(0) = 1 - P_b(1) = 0.9$. The
terminal $c$ segments (white), assumed to have lengths much larger
than $n = 1$, are characterized by the probability $P_c(0) = 1 -
P_c(1) = 0.5$.}
\label{figure:wabx8mf}
\end{figure}
When we segment the repetitive sequence $abababababababab$ using the
paired sliding windows segmentation scheme with window size $n$, we
obtained the mean-field Jensen-Shannon divergence spectrum shown in
the top plot of Fig.~\ref{figure:wabx8mf}. In this figure, sequence
positions are normalized such that $n = 1$, while the lengths of the
repeating segments $a$ and $b$ are chosen to be both less than the
window size, i.e.~$n_a = n_b = 0.7 < n$. To understand contextual
effects at the ends of the repetitive sequence, we include the
terminal segments $c$ in our analysis. These terminal $c$ segments
are assumed to have lengths $n_c \gg n$, and statistics intermediate
between those of $a$ and $b$. As we can see from the top plot of
Fig.~\ref{figure:wabx8mf}, all domain walls between $a$ and $b$
segments ($ab$ \emph{domain walls}) correspond to peaks in the
mean-field divergence spectrum. The two $ab$ domain walls near the
ends of the repetitive sequence are the strongest, while the rest have
the same diminished strength (compared to the Jensen-Shannon
divergence between the $a$ and $b$ segments). From the top plot of
Fig.~\ref{figure:wabx8mf}, we also see that no peaks are associated
with the $ca$ and $bc$ domain walls. Instead, we find a spurious peak
left of the $ca$ domain wall, and another spurious peak right of the
$bc$ domain wall.
As discussed in
App.~\ref{section:pairedslidingwindowssegmentationscheme}, the
mean-field lineshape of a simple domain wall is very nearly piecewise
quadratic, with a total width of $2n$. This observation is extremely
helpful when we deal with real divergence spectra, where statistical
fluctuations produce spurious peaks with various shapes and widths.
By insisting that only peaks that are (i) approximately piecewise
quadratic, with (ii) widths close to $2n$, are statistically
significant, we can determine a smaller, and more reliable set of
domain walls through match filtering. In the top plot of
Fig.~\ref{figure:wabx8mf}, all our peaks have widths smaller than
$2n$. In the mean-field limit, these are certainly not spurious, but
if we imagine putting statistical fluctuations back into the
divergence spectrum, and suppose we did not know beforehand that there
are segments shorter than $n$ in this sequence, it would be reasonable
to accept by fiat whatever picture emerging from the match filtering
procedure. For $cababababababababc$, the match-filtered, quality
enhanced divergence spectrum is shown as the bottom plot of
Fig.~\ref{figure:wabx8mf}, where we find the two spurious peaks
shifted deeper into the $c$ segments by the match filtering procedure.
In this plot, the two strong $ab$ domain walls near the ends of the
repetitive sequence continue to stand out, but the rest of the $ab$
domain walls are now washed out by match filtering. If we put
statistical noise back into the picture, the fine structures marking
these remaining $ab$ domain walls will disappear, and we end up with a
featureless plateau in the interior of the repetitive sequence. We
might then be misled into thinking that this $cababababababababc$
sequence consists of only five segments $ca'c'b'c$, where $a'$ is $a$
contaminated by a small piece of $c$, $b'$ is $b$ contaminated by a
small piece of $c$, and $c'$, which lies between the two strong $ab$
domain walls, will be mistaken for a segment with $K = 0$ statistics
similar to $c$, even though it is not statistically stationary.
Next, let us analyze the recursive Jensen-Shannon segmentation of
$abababababababab$, where we cut the repetitive sequence first into
two segments, then each of these into two subsegments, and so on and
so forth, until all the segments are discovered. In the top plot of
Fig.~\ref{figure:abx8rJS}, we show the top-level Jensen-Shannon
divergence spectrum, based on which we will cut $abababababababab$
into two segments. In this figure, we find
\begin{enumerate}
\item a series of $k$ peaks of unequal strengths, with stronger peaks
near the ends, and weaker peaks in the middle of the repetitive
sequence;
\item $k - 1$ domain walls having vanishing divergences;
\item the ratio of strengths of the strongest peak to the weakest peak
is roughly $k/2$,
\end{enumerate}
where $k$ is the number of repeated motifs. These statistical
signatures are shared by all repetitive sequences, with the detail
distribution and statistical characteristics of the subunits within
the repeated motif affecting only the shape and strength of the peaks.
Here we see extreme context sensitivity reflected in the fact that
domain walls with the same true strength can have very different, and
even vanishing, strengths when the segment structure of the sequence
is examined recursively.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{abx8rJS.eps}
\caption{(Top) The top-level Jensen-Shannon divergence spectrum (red
solid curve) obtained in the recursive segmentation of a repetitive
binary sequence consisting of subunits $a$ (light green, $P_a(0) = 1 -
P_a(1) = 0.1$) and $b$ (light yellow, $P_b(0) = 1 - P_b(1) = 0.9$)
repeated eight times. (Bottom) The Jensen-Shannon divergence spectra
obtained when $abababababababab$ is recursively segmented from the
right end.}
\label{figure:abx8rJS}
\end{figure}
From the bottom plot of Fig.~\ref{figure:abx8rJS}, we find that one or
both of the peaks near the ends of the repetitive sequence are always
the strongest, as recursion progresses. This is true when the
repetitive sequence consists of repeating motifs with more complex
internal structure, and also true when we attach terminal segments to
the repetitive sequence. Therefore, successive cuts are always made
at one end or the other of the repetitive sequence. For $ab$-repeats,
the peaks near both ends are equally strong in the mean-field limit,
so we can choose to always cut at the right end of $abababababababab$,
as shown in the bottom plot of Fig.~\ref{figure:abx8rJS}. As the
repetitive sequence loses its rightmost segment at every step, and the
global context alternates between being dominated by $a$ segments to
being dominated by $b$ segments, we find oscillations in the strengths
of the remaining domain walls. This oscillation, which is a generic
behaviour of all repetitive sequences under recursive segmentation,
can be seen more clearly for the $ab$-repetitive sequence in Figure
\ref{figure:abx8osc}, where instead of cutting off one segment at a
time, we move the cut continuously inwards from the right end.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{abx8osc.eps}
\caption{The windowless Jensen-Shannon divergences at $z = 10.0$ (at a
domain wall) and $z = 9.5$ (away from a domain wall) of the repetitive
binary sequence $abababababababab$, with $P_a(0) = 0.1$ and $P_b(0) =
0.9$, as functions of the cut $10 \leq z \leq 16$.}
\label{figure:abx8osc}
\end{figure}
\section{Summary and Discussions}
\label{section:conclusions}
In this paper, we defined the \emph{context sensitivity problem}, in
which the \emph{same} group of statistically stationary segments are
segmented \emph{differently} by the \emph{same} segmentation scheme,
when it is encapsulated within \emph{different larger contexts} of
segments. We then described in
Sec.~\ref{section:contextsensitivityprobleminrealgenomes} the various
manifestions of context sensitivity when real bacterial genomes are
segmented using the paired sliding windows and optimized recursive
Jensen-Shannon segmentation schemes, which are sensitive to local and
global contexts respectively. For the single-pass paired sliding
windows segmentation scheme, we found that the positions and relative
strengths of domain walls can change dramatically when we change the
window size, and hence the local contexts examined. For the optimized
recursive segmentation scheme, we found that there can be large shifts
in the optimized domain wall positions as recursion progresses, due to
the change in global context when we go from examining a sequence to
examining its subsequence, and \emph{vice versa}.
In Sec.~\ref{section:contextsensitivityprobleminrealgenomes}, we also
looked into the issue of coarse graining the segmental description of
a bacterial genome. We argued that coarse graining by using larger
window sizes, or stopping recursive segmentation earlier can be
biologically misleading, because of the context sensitivity problem,
and explored an alternative coarse graining procedure which involves
removing the weakest domain walls and agglomerating the segments they
separate. This coarse graining procedure was found to be fraught with
difficulties, arising again from the context sensitivity of domain
wall strengths. Ultimately, the goal of coarse graining is to reduce
the complexity of the segmented models of real genomes. This can be
achieved by reducing the number of segments, or by reducing the number
of segment \emph{types} or \emph{classes} (see, for example, the work
by Azad \emph{et al}. \cite{Azad2002PhysicalReviewE66a031913}). We
realized in this paper that the former is unattainable, and proposed
to accomplish the latter through statistical clustering of the
segments. Based on what we understand about the context sensitivity
problem, we realized that it would be necessary to segment a given
genomic sequence as far as possible, to the point before genes are cut
into multiple segments (unless they are known to contain multiple
domains). We are in the process of writing the results of our
investigations into this manner of coarse graining, in which no domain
walls are removed, but statistically similar segments are clustered
into a small number of segment classes.
In Sec.~\ref{section:meanfieldanalysis}, we analyzed the paired
sliding windows and optimized recursive segmentation schemes within a
mean-field picture. For the former, we explained how the presence of
segments shorter than the window size lead to shifts in the positions,
and changes in the strengths of domain walls. For the latter, we
illustrate the context dependence of the domain walls strengths, how
this leads to large shifts in the optimized domain wall positions, and
also to the domain walls being discovered out of order by their true
strengths. We showed that all domain walls in a sequence will be
recovered in the mean-field limit, if we allow the recursive
segmentation to go to completion, but realized that for real sequences
subject to statistical fluctuations, there is a danger of stopping the
the recursion too early. When this happens, we will generically pick
up weak domain walls, but miss stronger ones --- a problem that can be
partly alleviated through segmentation optimization, in which domain
walls are moved from weaker to stronger positions. We devoted one
subsection to explain why the context sensitivity problem is
especially severe in repetitive sequences.
Finally, let us say that while we have examined only two entropic
segmentation schemes in detail, we believe the context sensitivity
problem plagues all segmentation schemes. The manifestations of the
context sensitivity problem will of course be different for different
segmentation schemes, but will involve (i) getting the domain wall
positions wrong; (ii) getting the domain wall strengths wrong; or
(iii) missing strong domain walls. A proper analysis of the context
sensitivity of the various segmentation schemes is beyond the scope of
this paper, but let us offer some thoughts on segmentation schemes based
on based on hidden Markov models (HMMs), which are very popular in the
bioinformatics literature. In HMM segmentation, model parameters are
typically estimated using the Baum-Welch algorithm, which first
computes the forward and backward probabilities of each hidden state,
use these to estimate the transition frequencies, which are used to
update the model parameters. Computation of forward and backward
probabilities are sensitive to local context, in that the hidden
states assigned to a given collection of segments will be different,
if the sequences immediately flanking the segments are different.
Updating of model parameters, on the other hand, is sensitive to
global context, because very different arrangement of segments and
segment classes can give rise to the same summary of transition
frequencies. The signatures of this dual local-global context
sensitivity is buried within the sequence of posterior probabilities
obtained from iterations of the Baum-Welch algorithm. Ultimately, the
context sensitivity problem is a very special case of the problem of
mixed data, which is an active area of statistical research. We hope
that through the results presented in this paper, the bioinformatics
community will come to better recognize the nuances sequence context
poses to its proper segmentation.
\begin{appendix}
\subsection{Generalized Jensen-Shannon Divergences}
\label{section:generalizedJensenShannondivergences}
In Ref.~\citeonline{Cheong2007IRJSSS} we explained that dinucleotide
correlations and codon biases in biological sequences
\cite{Grantham1981NucleicAcidsResearch9pR43,
Shepherd1981ProcNatlAcadSciUSA78p1596,
Staden1982NucleicAcidsResearch10p141,
Fickett1982NucleicAcidsResearch10p5303, Herzel1995PhysicaA216p518} are
better modeled by Markov chains of order $K > 0$ over the quaternary
alphabet $\mathcal{S} = \rm\{A, C, G, T\}$
\cite{Thakur2007PhysicalReviewE75a011915}, rather than Bernoulli
chains over $\mathcal{S}$
\cite{BernaolaGalvan1996PhysicalReviewE53p5181,
RomanRoldan1998PhysicalReviewLetters80p1344}, or Bernoulli chains over
the extended alphabet $\mathcal{S}^K$
\cite{BernaolaGalvan2000PhysicalReviewLetters85p1342,
Nicorici2003FINSIG03,
Nicorici2004EURASIPJournalonAppliedSignalProcessing1p81}. In the
sequence segmentation problem, our task is to decide whether there is
a domain wall at sequence position $i$ within a given sequence
$\mathbf{x} = x_1 x_2 \cdots x_{i-1} x_i x_{i+1} \cdots x_N$, where $x_j \in
\mathcal{S}, 1 \geq j \geq N$. The simplest model selection scheme
that would address this problem would involve the comparison of the
one-segment sequence likelihood $P_1$, whereby the sequence $\mathbf{x}$ is
treated as generated by a single Markov process, against the
two-segment sequence likelihood $P_2$, whereby the subsequences $\mathbf{x}_L
= x_1 x_2 \cdots x_{i-1}$ and $\mathbf{x}_R = x_i x_{i+1} \cdots x_N$ are
treated as generated by two different Markov processes.
To model $\mathbf{x}$, $\mathbf{x}_L$, and $\mathbf{x}_R$ as Markov chains of order $K$, we
determine the order-$K$ \emph{transition counts} $f_{\mathbf{t} s}$, $f_{\mathbf{t}
s}^L$, $f_{\mathbf{t} s}^R$, subject to the normalizations
\begin{equation}
f_{\mathbf{t} s} = f_{\mathbf{t} s}^L + f_{\mathbf{t} s}^R, \quad
\sum_{\mathbf{t}\in\mathcal{S}^K}\sum_{s=1}^S f_{\mathbf{t} s} = N.
\end{equation}
Here $S = 4$ is the size of the quaternary alphabet $\mathcal{S}$, and
$\mathbf{t}$ is a shorthand notation for the $K$-tuple of indices $(t_1, t_2,
\dots, t_K), 1 \leq t_k \leq S$. The transition counts $f_{\mathbf{t} s}$,
$f_{\mathbf{t} s}^L$, and $f_{\mathbf{t} s}^R$ are the number of times the
$(K+1)$-mer $\alpha_{t_K} \cdots \alpha_{t_1} \alpha_s$ appear in the
sequences $\mathbf{x}$, $\mathbf{x}_L$, and $\mathbf{x}_R$ respectively. The sequences
$\mathbf{x}$, $\mathbf{x}_L$, and $\mathbf{x}_R$ are then assumed to be generated by the
Markov processes with \emph{maximum-likelihood transition
probabilities}
\begin{equation}
\hat{p}_{\mathbf{t} s} = \frac{f_{\mathbf{t} s}}{\sum_{s'=1}^S f_{\mathbf{t} s'}}, \quad
\hat{p}_{\mathbf{t} s}^L = \frac{f_{\mathbf{t} s}^L}{\sum_{s'=1}^S f_{\mathbf{t} s'}^L}, \quad
\hat{p}_{\mathbf{t} s}^R = \frac{f_{\mathbf{t} s}^R}{\sum_{s'=1}^S f_{\mathbf{t} s'}^R},
\end{equation}
respectively.
Within these maximum-likelihood Markov-chain models, the one- and
two-segment sequence likelihoods are given by
\begin{equation}
\begin{aligned}
P_1 &= \prod_{\mathbf{t}\in\mathcal{S}^K}\prod_{s=1}^S
\left(\hat{p}_{\mathbf{t} s}\right)^{f_{\mathbf{t} s}}, \\
P_2 &= \prod_{\mathbf{t}\in\mathcal{S}^K}\prod_{s=1}^S
\left(\hat{p}_{\mathbf{t} s}^L\right)^{f_{\mathbf{t} s}^L}
\left(\hat{p}_{\mathbf{t} s}^R\right)^{f_{\mathbf{t} s}^R},
\end{aligned}
\end{equation}
respectively. Because we have more free parameters to fit the
observed sequence statistics in the two-segment model, $P_2 \geq P_1$.
The generalized Jensen-Shannon divergence, a symmetric
variant of the relative entropy known more commonly as the
\emph{Kullback-Leibler divergence}, is then given by
\begin{equation}\label{equation:JensenShannondivergence}
\Delta(i) = \log\frac{P_2}{P_1} =
\sum_{\mathbf{t}\in\mathcal{S}^K}\sum_{s = 1}^S \left[
-f_{\mathbf{t}s} \log \hat{p}_{\mathbf{t}s} +
f_{\mathbf{t}s}^L \log \hat{p}_{\mathbf{t}s}^L +
f_{\mathbf{t}s}^R \log \hat{p}_{\mathbf{t}s}^R \right].
\end{equation}
This test statistic, generalized from the Jensen-Shannon divergence
described in
Ref.~\citeonline{Lin1991IEEETransactionsonInformationTheory37p145},
measures quantitatively how much better the two-segment model fits
$\mathbf{x}$ compared to the one-segment model.
\subsection{Paired Sliding Windows Segmentation Scheme}
\label{section:pairedslidingwindowssegmentationscheme}
A standard criticism on using sliding windows to detect segment
structure within a heterogeneous sequence is the compromise between
precision and statistical significance. For the comparison between
two windowed statistics to be significant, we want the window size $n$
to be large. On the other hand, to be able to determine a change
point precisely, we want the window size $n$ to be small. There is
therefore no way, with a single window of length $n$, to independently
select both a desired statistical significance and desired precision.
In this appendix, we devise a sliding window segmentation scheme in
which, instead of one window, we use a pair of adjoining windows, each
of length $n$. By comparing the left windowed statistics to the right
windowed statistics, a change point is detected at the center of the
pair of windows \emph{when} the two windowed statistics are most
different. A given difference between the two windowed statistics
becomes more signficant as the window size $n$ is increased. A larger
window size also suppresses statistical fluctuations, making it easier
to locate the change point. Therefore, increasing the window size $n$
improves both statistical significance and precision, even though they
cannot be adjusted independently.
In App.~\ref{subsection:statisticswithinapairofslidingwindows}, we
describe the proper test statistic to use for change point detection
within the model selection framework. Then in
App.~\ref{subsection:hypothesistestingwithapairofslidingwindows}, we
show how a similar test statistic spectrum can be obtained within the
hypothesis testing framework. In
App.~\ref{subsection:performanceonrealgenomicsequences}, we show some
examples of the scheme being applied to real genomic sequences. In
App.~\ref{subsection:meanfieldlineshape}, we derived the mean-field
lineshape of a domain wall in this paired sliding window segmentation
scheme, and use it to perform match filtering.
\subsubsection{Model Selection Within a Pair of Sliding Windows}
\label{subsection:statisticswithinapairofslidingwindows}
To detect domain walls between different segments within a
heterogeneous sequence, we can slide a pair of adjoining windows each
of length $n$ across the sequence, and monitor the left and right
windowed statistics at different sequence positions, as shown in
Figure \ref{figure:pairedslidingwindow}.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.8]{pairedslidingwindow.eps}
\caption{A pair of sliding windows, each of length $n$. A change
point at the center of the pair of sliding windows can be detected by
comparing the statistics within the left and right windows.}
\label{figure:pairedslidingwindow}
\end{figure}
If we model the different segments by Markov chains of order $K$, the
left and right windowed statistics are summarized by the transition
count matrices
\begin{equation}
\textsf{\textbf{F}}^L = \left[ f_{\mathbf{t}s}^L \right], \quad
\textsf{\textbf{F}}^R = \left[ f_{\mathbf{t}s}^R \right]
\end{equation}
respectively, where the transition counts sums to the window size,
\begin{equation}
\sum_{\mathbf{t}}\sum_s f_{\mathbf{t}s}^L =
\sum_{\mathbf{t}}\sum_s f_{\mathbf{t}s}^R = n.
\end{equation}
From these transition count matrices, we can determine the
maximum-likelihood estimates
\begin{equation}
\hat{\textsf{\textbf{P}}}^L = \left[ p_{\mathbf{t}s}^L \right], \quad
p_{\mathbf{t}s}^L = \frac{f_{\mathbf{t}s}^L}{\sum_{s'}
f_{\mathbf{t}s'}^L}; \quad
\hat{\textsf{\textbf{P}}}^R = \left[ p_{\mathbf{t}s}^R \right], \quad
p_{\mathbf{t}s}^R = \frac{f_{\mathbf{t}s}^R}{\sum_{s'}
f_{\mathbf{t}s'}^R}
\end{equation}
of the transition matrices for the left and right windows.
We then compute the transition count matrix
\begin{equation}
\textsf{\textbf{F}} = \left[ f_{\mathbf{t}s} = f_{\mathbf{t}s}^L + f_{\mathbf{t}s}^R \right],
\end{equation}
and therefrom the transition matrix
\begin{equation}
\hat{\textsf{\textbf{P}}} = \left[ p_{\mathbf{t}s} \right], \quad
p_{\mathbf{t}s} = \frac{f_{\mathbf{t}s}}{\sum_{s'} f_{\mathbf{t}s'}},
\end{equation}
assuming a one-segment model for the combined window of length $2n$,
before calculating the windowed Jensen-Shannon divergence using Eq.
\eqref{equation:JensenShannondivergence} in
App.~\ref{section:generalizedJensenShannondivergences}. By sliding
the pair of windows along the sequence, we obtain a windowed
Jensen-Shannon divergence spectrum $\Delta(i)$, which tells us where
along the sequence the most statistically significant change points
are located.
\subsubsection{Hypothesis Testing With a Pair of Sliding Windows}
\label{subsection:hypothesistestingwithapairofslidingwindows}
Change point detection using statistics within the pair of sliding
windows can also be done within the hypothesis testing framework.
Within this framework, we ask how likely it is to find
maximum-likelihood estimates $\hat{\textsf{\textbf{P}}}^L$ for the left window, and
$\hat{\textsf{\textbf{P}}}^R$ for the right window, when the pair of windows straddles
a statistically stationary region generated by the transition matrix
$\textsf{\textbf{P}}$.
In the central limit regime, Whittle showed that the probability of
obtaining a maximum-likelihood estimate $\hat{\textsf{\textbf{P}}}$ from a finite
sequence generated by the transition matrix $\textsf{\textbf{P}}$ is given by
\cite{Whittle1955JRoyalStatSoc17p235}
\begin{equation}\label{equation:whittleformula}
P(\hat{\textsf{\textbf{P}}}|\textsf{\textbf{P}}) = C \exp\left[
\frac{1}{2} \sum_{\mathbf{t}}\sum_s\sum_{s'} \frac{n}{P_{\mathbf{t}}}
\left(1 - \frac{\delta_{ss'}}{p_{\mathbf{t}s}}\right)
\left(\hat{p}_{\mathbf{t}s} - p_{\mathbf{t}s}\right)
\left(\hat{p}_{\mathbf{t}s'} - p_{\mathbf{t}s'}\right)
\right],
\end{equation}
where $C$ is a normalization constant, $n$ the length of the sequence,
and $P_{\mathbf{t}}$ is the equilibrium distribution of $K$-mers in
the Markov chain.
For $n \gg K$, the left and right window statistics are essentially
independent, and so the probability of finding $\hat{\textsf{\textbf{P}}}^L$ in the
left window and finding $\hat{\textsf{\textbf{P}}}^R$ in the right window, when the
true transition matrix is $\textsf{\textbf{P}}$, is $P(\hat{\textsf{\textbf{P}}}^L | \textsf{\textbf{P}})
P(\hat{\textsf{\textbf{P}}}^R | \textsf{\textbf{P}})$. In principle we do not know what $\textsf{\textbf{P}}$ is, so
we replace it by $\hat{\textsf{\textbf{P}}}$, the maximum-likelihood transition matrix
estimated from the combined statistics in the left and right windows.
Based on Eq. \eqref{equation:whittleformula}, the test statistic
that we compute as we slide the pair of windows along the sequence is
the \emph{square deviation}
\begin{equation}\label{equation:centrallimitsquaredeviation}
r = -\sum_{\mathbf{t}}\sum_s\sum_{s'}
\frac{n}{\hat{P}_{\mathbf{t}}}
\left(1 - \frac{\delta_{ss'}}{\hat{p}_{\mathbf{t}s}}\right)
\left[
\left(\hat{p}_{\mathbf{t}s}^L - \hat{p}_{\mathbf{t}s}\right)
\left(\hat{p}_{\mathbf{t}s'}^L - \hat{p}_{\mathbf{t}s'}\right) +
\left(\hat{p}_{\mathbf{t}s}^R - \hat{p}_{\mathbf{t}s}\right)
\left(\hat{p}_{\mathbf{t}s'}^R - \hat{p}_{\mathbf{t}s'}\right)
\right],
\end{equation}
which is more or less the negative logarithm of $P(\hat{\textsf{\textbf{P}}}^L | \textsf{\textbf{P}})
P(\hat{\textsf{\textbf{P}}}^R | \textsf{\textbf{P}})$. To compare the square deviation spectrum
$r(i)$ obtained for different window sizes, we simply divide $r(i)$ by
the window size $n$. From Eq.
\eqref{equation:centrallimitsquaredeviation}, we find that $r$ receive
disproportionate contributions from rare states
($\hat{P}_{\mathbf{t}}$ small) as well as rare transitions
($\hat{p}_{\mathbf{t}s}$ small).
\subsubsection{Application to Real Genomic Sequences}
\label{subsection:performanceonrealgenomicsequences}
The average length of coding genes in \emph{Escherichia coli} K-12
MG1655 is 948.9 bp. This sets a `natural' window size to use for our
sliding window analysis. In Figure
\ref{figure:EcoliK12qrwJSK0n1000i0i40k}, we show the windowed $K = 0$
Jensen-Shannon divergence and square deviation spectra for
\emph{Escherichia coli} K-12 MG1655, obtained for a window size of $n
= 1000$ bp, overlaid onto the distribution of genes. As we can see
from the figure, the two spectra are qualitatively very similar, with
peak positions that are strongly correlated with gene and operon
boundaries \cite{Salgado2006NucleicAcidsResearch34pD394}.
\begin{figure}[hbtp]
\centering
\includegraphics[scale=0.5,clip=true]{EcoliK12.q.rwJSK0n1000.0.40k.eps}
\includegraphics[scale=0.5,clip=true]{NC_000913.gene.layout.0.40k.eps}
\caption{The windowed $K = 0$ Jensen-Shannon divergence (magenta) and
square deviation (black) spectra in the interval $(0, 40000)$ of the
\emph{Escherichia coli} K-12 MG1655 genome, which has a length $N =
4639675$ bp. Annotated genes on the positive (red) and negative
(green) strands are shown below the graph.}
\label{figure:EcoliK12qrwJSK0n1000i0i40k}
\end{figure}
For example, we see that the strongest peak in the $n = 1000$ windowed
spectrum is at $i \sim 30000$. The gene \emph{dapB}, believed to be
an enzyme involved in lysine (which consists solely of purines)
biosynthesis, lies upstream of this peak, while the \emph{carAB}
operon, believed to code for enzymes involved in pyrimidine
ribonucleotide biosynthesis, lies downstream of the peak. Another
strong peak marks the end of the \emph{carAB} operon, distinguishing
it statistically from the gene \emph{caiF}, and yet another strong
peak distinguishes \emph{caiF} from the \emph{caiTABCDE} operon, whose
products are involved in the central intermediary metabolic pathways,
further downstream.
In Figure \ref{figure:EcoliK12qrK0K1K2n1000i0i40k}, we show the square
deviation spectra for the same $(0, 40000)$ interval of the \emph{E.
coli} K-12 MG1655 genome, but for different Markov-chain orders $K =
0, 1, 2$. As we can see, these square deviation spectra share many
qualitative features, but there are also important qualitative
differences. For example, the genes \emph{talB} and \emph{mogA},
which lies within the interval $(8200, 9900)$, are not strongly
distinguished from the genes \emph{yaaJ} upstream and \emph{yaaH}
downstream at the 1-mer ($K = 0$) level. They are, however, strongly
distinguished from the flanking genes at the 2-mer ($K = 1$) and 3-mer
($K = 2$) levels.
\begin{figure}[hbtp]
\centering
\includegraphics[scale=0.5,clip=true]{EcoliK12.q.rK0K1K2n1000.0.40k.eps}
\includegraphics[scale=0.5,clip=true]{NC_000913.gene.layout.0.40k.eps}
\caption{The windowed $K = 0$ (top), $K = 1$ (middle), and $K = 2$
(bottom) square deviation spectra in the interval $(0, 40000)$ of the
\emph{E. coli} K-12 MG1655 genome, which has a length of $N = 4639675$
bp. Annotated genes on the positive (red) and negative (green)
strands are shown below the graph.}
\label{figure:EcoliK12qrK0K1K2n1000i0i40k}
\end{figure}
\subsubsection{Mean-Field Lineshape and Match Filtering}
\label{subsection:meanfieldlineshape}
In the second situation shown in Fig.~\ref{figure:windowedcases}, let
us label the two mean-field segments $a$ and $b$, with lengths $N_a$
and $N_b$. Suppose it is the left window that straddles both $a$ and
$b$, while the right window lies entirely within $b$. The
right-window counts are then simply
\begin{equation}
f_{\mathbf{t}s}^R = \frac{n}{N_b}\, f_{\mathbf{t}s}^b,
\end{equation}
while the left-window counts contain contributions from both $a$ and
$b$, i.e.
\begin{equation}
f_{\mathbf{t}s}^L = \frac{n - z}{N_a}\, f_{\mathbf{t}s}^a +
\frac{z}{N_b}\, f_{\mathbf{t}s}^b,
\end{equation}
where $z$ is the distance of the domain wall from the center of the
pair of windows. The total counts from both windows are then
\begin{equation}
f_{\mathbf{t}s} = \frac{n - z}{N_a}\, f_{\mathbf{t}s}^a +
\frac{z}{N_b}\, f_{\mathbf{t}s}^b + \frac{n}{N_b}\, f_{\mathbf{t}s}^b.
\end{equation}
Using the transition counts $f_{\mathbf{t}s}^L$, $f_{\mathbf{t}s}^R$,
and $f_{\mathbf{t}s}$, we then compute the maximum-likelihood
transition probabilities $\hat{p}_{\mathbf{t}s}^L$,
$\hat{p}_{\mathbf{t}s}^R$, and $\hat{p}_{\mathbf{t}s}$, before
substituting the transition counts and transition probabilities into
Eq.~\eqref{equation:JensenShannondivergence} for the Jensen-Shannon
divergence. Because of the logarithms in the definition for the
Jensen-Shannon divergence, we get a complicated function in terms of
the observed statistics $f_{\mathbf{t}s}^a$, $f_{\mathbf{t}s}^b$,
$N_a$ and $N_b$, and the distance $z$ between the domain wall and the
center of the pair of windows. Different observed statistics
$f_{\mathbf{t}s}^a$, $f_{\mathbf{t}s}^b$, $N_a$ and $N_b$ give
mean-field divergence functions of $z$ that are not related by a
simple scaling. However, these mean-field divergence functions
$\Delta(z)$ do have qualitative features in common:
\begin{enumerate}
\item $\Delta(z) = 0$ for $|z| \geq n$, where the pair of windows is
entirely within $a$ or entirely within $b$;
\item $\Delta(z)$ is maximum at $z = 0$, when the center of the pair
of windows coincide with the domain wall;
\item $\Delta(z)$ is convex everywhere within $|z| < n$, except at $z
= 0$.
\end{enumerate}
This tells us that the position and strength of the domain wall
between two mean-field segments both longer than the window size $n$
can be determined exactly.
In Figure \ref{figure:wJSlineshape} we show $\Delta(z)$ for two binary
$K = 0$ mean-field segments, where $P_a(0) = 1 - P_a(1) = 0.9$, and
$P_b(0) = 1 - P_b(1) = 0.1$. We call the peak function $\Delta(z)$
the \emph{mean-field lineshape} of the domain wall. As we can see
from Figure \ref{figure:wJSlineshape}, this mean-field lineshape can
be very well approximated by the piecewise quadratic function
\begin{equation}\label{equation:meanfieldJS}
\tilde{\Delta}(z) = \begin{cases}
\left(1 + \frac{z}{n}\right)^2 \bar{\Delta}(0), & -1 < z < 0; \\
\left(1 - \frac{z}{n}\right)^2 \bar{\Delta}(0), & 0 \leq z < 1; \\
0, & \text{everywhere else}, \end{cases}
\end{equation}
where $\bar{\Delta}(0)$ is the mean-field Jensen-Shannon divergence of
the domain wall at $z = 0$. If instead of the windowed Jensen-Shannon
divergence $\Delta(z)$, we compute the windowed square deviation
$r(z)$ in the vicinity of a domain wall, we will obtain a mean-field
lineshape that is strictly piecewise quadratic, i.e.
\begin{equation}\label{equation:meanfieldr}
\tilde{r}(z) = \begin{cases}
\left(1 + \frac{z}{n}\right)^2 \bar{r}(0), & -1 < z < 0; \\
\left(1 - \frac{z}{n}\right)^2 \bar{r}(0), & 0 \leq z < 1; \\
0, & \text{everywhere else}, \end{cases}
\end{equation}
where $\bar{r}(0)$ is the mean-field square deviation of the domain
wall at $z = 0$.
\begin{figure}[htbp]
\centering
\includegraphics[scale=0.45,clip=true]{wJSlineshape.eps}
\caption{The Jensen-Shannon divergence $\Delta(z)$ (solid curve) of a
pair of sliding windows of length $n = 1$ as a function of the
distance $z$ between the domain wall separating a mean-field binary
segment $a$ with $P_a(0) = 1 - P_a(1) = 0.9$ and a mean-field binary
segment $b$ with $P_b(0) = 1 - P_b(1) = 0.1$, and the center of the
pair of windows. Also shown as the dashed curve is a piecewise
quadratic function which rises from $z = \pm 1$ to the same maximum at
$z = 0$, but vanishes everywhere else.}
\label{figure:wJSlineshape}
\end{figure}
Going back to a real sequence composed of two nearly stationary
segments of discrete bases, we expect to find statistical fluctuations
masking the mean-field lineshape. But now that we know the mean-field
lineshape is piecewise quadratic for the square deviation $r(z)$ (or
very nearly so, in the case of the windowed Jensen-Shannon divergence
$\Delta(z)$), we can make use of this piecewise quadratic mean-field
lineshape to match filter the raw square deviation spectrum. We do
this by assuming that there is a mean-field square-deviation peak at
each sequence position $i$, fit the spectrum within $(i - n, i + n)$
to the mean-field lineshape in Eq. \eqref{equation:meanfieldr}, and
determine the smoothed spectrum $\bar{r}(i)$. In Fig.
\ref{figure:EcoliK12qrrmrmRK0n1000i0i40k}, we show the match-filtered
square deviation spectrum $\bar{r}(i)$ in the interval $0 \leq i \leq
40000$ of the \emph{E. coli} K-12 MG1655 genome. As we can see,
$\bar{r}(i)$ is smoother than $r(i)$, but the peaks in $\bar{r}(i)$
are also so broad that distinct peaks in $r(i)$ are not properly
resolved.
\begin{figure}[hbtp]
\centering
\includegraphics[scale=0.5,clip=true]{EcoliK12.q.rrmrmRK0n1000.0.40k.eps}
\includegraphics[scale=0.5,clip=true]{NC_000913.gene.layout.0.40k.eps}
\caption{The interval $0 \leq i \leq 40000$ of the \emph{E. coli} K-12
MG1655 genome ($N = 4639675$ bp), showing (top to bottom) the windowed
$K = 0$ square deviation spectrum $r(i)$, the match-filtered square
deviation spectrum $\bar{r}(i)$, the residue spectrum $R(i)$, and the
quality enhanced square deviation spectrum $\bar{r}(i)/R(i)$.
Annotated genes on the positive (red) and negative (green) strands are
shown below the graph.}
\label{figure:EcoliK12qrrmrmRK0n1000i0i40k}
\end{figure}
Fortunately, more information is available from the match filtering.
We can also compute how well the raw spectrum $r(j)$ in the interval
$i - n \leq j \leq i + n$ match the mean-field lineshape
$\tilde{r}(j)$ by computing the residue
\begin{equation}
R(i) = \sum_{j = i - n}^{i + n} \left[r(j) - \tilde{r}(j)\right]^2.
\end{equation}
filtering the raw divergence spectrum. In Fig.
\ref{figure:EcoliK12qrrmrmRK0n1000i0i40k}, we show the residue
spectrum $R(i)$ for the $0 \leq i \leq 40000$ region of the \emph{E.
coli} K-12 MG1655 genome. In the residue spectrum, we see a series of
dips at the positions of peaks in the square deviation spectrum.
Since $R(i)$ is small when the match is good, and large when the match
is poor, $1/R(i)$ can be thought of as the quality factor of a square
deviation peak. A smoothed, and accentuated spectrum is obtained when
we divide the smoothed square deviation by the residue at each point.
The quality enhanced square deviation spectrum $\bar{r}(i)/R(i)$ is
also shown in Fig. \ref{figure:EcoliK12qrrmrmRK0n1000i0i40k}. It is
much more convenient to determine the position of significant domain
walls from such a spectrum.
\end{appendix}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 3,281
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{
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| 1,683
|
\section{Introduction}
Loewner's partial differential equation
$$
\frac{\partial f_s}{\partial s}(z)=-\frac{\partial f_s}{\partial z}(z)G(z,s),\quad \mbox{a.e. }s\geq 0, z\in M
$$
received much attention from
mathematicians since Charles Loewner \cite{Loewner}
introduced it in 1923 to study extremal problems and, later,
P.P. Kufarev \cite{Kuf1943} and C. Pommerenke \cite{Po64}, \cite{Po} fully
developed the original theory. Such an
equation was a cornerstone in the de Branges' proof of the
Bieberbach conjecture. In 1999 O. Schramm \cite{Schramm}
introduced a stochastic version of the original differential
equation, nowadays known as SLE, which, among other things, was
a basic tool to prove Mandelbrot's conjecture by himself, G.
Lawler and W. Werner.
Loewner's original theory has been extended (see \cite{Pf74}, \cite{Pf75}, \cite{GHK01},
\cite{GHK09}, \cite{GK03}, \cite{Por91}) to higher
dimensional balls in $\mathbb C^n$ and successfully used to study
distortion, star-likeness, spiral-likeness and other geometric
properties of univalent mappings in higher dimensions.
Very recently, the second named author with M. Contreras and S.
D\'iaz-Madrigal \cite{BCD08}, \cite{BCD09} and Contreras,
D\'iaz-Madrigal and P. Gumenyuk \cite{CDG09} proposed a general
setting for the Loewner theory, which works also on complete hyperbolic complex manifolds.
While the classical theory deals with
normalized objects, this general theory does not, and
encloses the classical theory as a special case.
The aim of this paper is to present a general geometric
construction of Loewner chains on complete hyperbolic complex manifolds which
does not use any limit process (and thus it is new also for the
unit disc case) but relies on the apparently new
interpretation of Loewner chains as the direct limit of evolution families, and to give applications of
such a theory to geometric properties of univalent mappings on the
unit ball. To be more precise, we need some definitions. In the following, $M$ is a complete hyperbolic complex manifold of dimension $n$, and $d\in [1,+\infty]$.
An {\sl $L^d$-evolution family} on $M$ is a family $(\varphi_{s,t})_{0\leq s\leq t}$ of holomorphic self-mappings of $M$ satisfying the {\sl evolution property}
$$\varphi_{s,s}={\sf id},\quad \varphi_{s,t}=\varphi_{u,t}\circ \varphi_{s,u},\quad 0\leq s\leq u\leq t,$$ and
$t\mapsto \varphi_{s,t}(z)$ has some $L^d_{\sf loc}$-type regularity
locally uniformly with respect to $z\in M$ (see Definition \ref{L^d_EF}).
$L^d$-evolution families are trajectories of certain time-dependent
holomorphic vector fields on $M$, called {\sl Herglotz vector
fields}. An $L^d$-Herglotz vector field $G(z,t)$ on $M$ is a weak holomorphic
vector field in the sense of Carath\'eodory which satisfies a
suitable $L^d_{\sf loc}$-bound in $t$ uniformly on compacta of
$M$ and such that for almost every $t\geq 0$ the vector field
$z\mapsto G(z,t)$ is semicomplete (see Definition
\ref{Her-vec-man}).
The main result in \cite{BCD09} states that there is a one-to-one correspondence between
evolution families and Herglotz vector fields. The bridge for
such a correspondence is given by the following
Loewner-Kufarev ODE:
\begin{equation}
\label{Lo-ODE}
\frac{\partial\varphi_{s,t}}{\partial t}(z)= G(\varphi_{s,t}(z), t), \quad \hbox{a.e. } t\in \mathbb R^+.
\end{equation}
Both classical radial and chordal Loewner ODE in the unit disc
are just particular cases of such an equation (see
\cite{BCD08}).
$L^d$-Evolution families are strictly
related to $L^d$-Loewner chains. Such chains are
defined in the unit disc $\mathbb D$ in \cite{CDG09} as families of univalent mappings $(f_t\colon \mathbb D\to\mathbb C)_{t\geq 0}$ such
that $f_s(\mathbb D)\subseteq f_t(\mathbb D)$ for all $0\leq s\leq t$
and satisfying an $L^d_{\sf loc}$ bound in $t$ uniformly on
compacta of $\mathbb D$. The classical Loewner chains in the unit disc
are particular cases of such chains.
The correspondence between evolution families and Loewner chains is provided by a functional equation:
an $L^d$-evolution family and an $L^d$-Loewner chain are {\sl associated} if
\begin{equation}\label{associated}
f_s=f_t\circ \varphi_{s,t},\quad 0\leq s\leq t.
\end{equation}
In \cite{CDG09} it is proved that given an $L^d$-Loewner chain $(f_t)$
in the unit disc $\mathbb D$, the family
$$(\varphi_{s,t}:=f_t^{-1}\circ f_s)$$ is an associated $L^d$-evolution family and, conversely, any $L^d$-evolution family
admits a unique (up to biholomorphisms) associated $L^d$-Loewner chain. Such a result, as already in the classical
theory, is based on a scaling limit process.
Similar results, in the case of $L^\infty$-evolution families in the unit ball $\mathbb B^n\subset \mathbb C^n$ fixing the origin
and having a normalized differential at the origin, have been
obtained in \cite{GHK01}, \cite{GK03}.
In such works
Loewner chains are defined as image-increasing sequences of
univalent mappings on the ball with image in $\mathbb C^n$ fixing the
origin and having the differential subjected to some
normalization at the origin. Again, Loewner chains are defined starting from
normalized evolution families by means of a scaling limit
process.
In this paper we propose a definition of
$L^d$-Loewner chains on complete hyperbolic complex manifolds and prove that equation (\ref{associated})
provides a one-to-one correspondence (up to biholomorphisms) between
$L^d$-Loewner chains and $L^d$-evolution families.
Since there exist complete hyperbolic complex manifolds (even non-compact ones) which are not biholomorphic to domains in $\mathbb{C}^n$, requiring each $f_t$ to be a univalent mapping from $M$ to $\mathbb{C}^n$ would be unnecessarily restrictive. Hence we give the following definition:
let $N$ be a complex manifold of the same
dimension of $M$ and let $d_N$
denote the distance induced on $N$ by some Hermitian metric. A
family $(f_t: M\to N)_{t\geq 0}$ is an {\sl
$L^d$-Loewner chain} if
\begin{itemize}
\item[LC1.] For each $t\geq 0$ fixed, the mapping $f_t: M\to N$
is univalent,
\item[LC2.] $f_s(M)\subset f_t(M)$ for all $0\leq s\leq
t<+\infty$,
\item[LC3.] For any compact set $K\subset\subset M$ and any
$T>0$ there exists a $k_{K,T}\in L^d([0,T], \mathbb R^+)$ such
that for all $z\in K$ and for all $0\leq s\leq t\leq T$
\[
d_N(f_s(z), f_t(z))\leq \int_{s}^t k_{K,T}(\xi)d\xi.
\]
\end{itemize}
The main results of the present paper can be summarized as
follows.
\begin{theorem}
\label{main-intro} Let $M$ be a complete hyperbolic complex manifold of
dimension $n$. Let $(\varphi_{s,t})$ be an $L^d$-evolution family on
$M$. Then there exists an associated $L^d$-Loewner chain $(f_t\colon M\to N)$.
If $(g_t\colon M\to Q)$ is another $L^d$-Loewner chain associated with $(\varphi_{s,t})$, then there exists a
biholomorphism $$\Lambda\colon \bigcup_{t\geq 0} f_t(M)\to \bigcup_{t\geq 0} g_t(M)$$ such that $$g_t=\Lambda\circ f_t,\quad t\geq 0.$$
Conversely, if $(f_t\colon M\to N)$ is an $L^d$-Loewner chain, then $(\varphi_{s,t}:=f_t^{-1}\circ f_s)$ is an associated $L^d$-evolution family.
\end{theorem}
The first part of the result holds more generally on taut
manifolds (see Theorems \ref{algebraic_theorem} and \ref{ev-to-Low}). The second part is proved in Theorem
\ref{Low-to-ev}. In order to prove the result we exploit a {\sl
kernel convergence theorem} on complete hyperbolic complex manifolds which we
prove in Theorem \ref{kernel}.
The associated $L^d$-Loewner chain $(f_t\colon M\to N)$ is constructed as the direct limit of the $L^d$-evolution family $(\varphi_{s,t})$ in the following way. Define an equivalence relation on the product $M\times \mathbb{R}^+$:
$$(x,s)\sim (y,t)\quad\mbox{iff}\quad \varphi_{s,u}(x)=\varphi_{t,u}(y) \mbox{ for $u$ large enough},$$ and define $N:=(M\times \mathbb{R}^+)/_\sim.$
Let $\pi\colon M\times \mathbb{R}^+\to N$ be the projection on the quotient, and let $i_t\colon M\to M\times\mathbb{R}^+$ be the injection $i_t(x)=(x,t)$. The chain is then defined as $$f_t:= \pi\circ i_t,\quad t\geq 0.$$
Equation (\ref{associated}) holds since
$$ \pi\circ i_s= \pi\circ i_t \circ \varphi_{s,t}, \quad 0\leq s\leq t.$$
Then we endow $N=\bigcup_{t\geq 0} f_t(M)$ with a complex manifold structure which makes the mappings $f_t$'s holomorphic and we prove the $L^d$-estimate.
As a consequence of Theorem \ref{main-intro}, we can define the {\sl Loewner range} ${\sf Lr}(\varphi_{s,t})$ of $(\varphi_{s,t})$ as the biholomorphism class of $\bigcup_{t\geq 0} f_t(M)$, where $(f_t)$ is any associated $L^d$-Loewner chain. The Loewner range can be seen as an analogue of the abstract basin of attraction defined by Fornaess and Stens\o nes in the setting of discrete holomorphic dynamics with an attractive fixed point \cite{Fo-Ste}. This suggests the following dynamical interpretation of the Loewner range: let $Q$ be a complex manifold and assume that an algebraic evolution family of automorphisms $(\Phi_{s,t}\colon Q\to Q)$ has an invariant domain $D\subset Q$. Let $(\varphi_{s,t}\colon D\to D)$ be the algebraic evolution family obtained restricting $(\Phi_{s,t})$. Then the complex manifold $$\{z\in Q : \Phi_{0,t}(z)\in D\mbox{ for $t$ big enough}\}$$ is biholomorphic to ${\sf Lr}\,(\varphi_{s,t})$.
If $(\varphi_{s,t})$ is an $L^d$-evolution family on the unit disc $\mathbb{D}$ the Loewner range has to be simply connected and cannot be compact, thus by the uniformization theorem it has to be biholomorphic to $\mathbb{D}$ or $\mathbb{C}$, and, as noticed also in \cite{CDG09}, the choice depends on the dynamics of $(\varphi_{s,t})$. Generalizing this result we prove that if $(f_t)$ and $(\varphi_{s,t})$ are associated, then
$$f_s^*\kappa_{{\sf Lr}(\varphi_{s,t})}=\lim_{t \to \infty}\varphi_{s,t}^*\kappa_M,\quad s\geq 0,$$ where $\kappa_M$ and $\kappa_{\sf Lr(\varphi_{s,t})}$ are the Kobayashi pseudometrics of $M$ and ${\sf Lr}(\varphi_{s,t})$ respectively. Using results from Fornaess and Sibony \cite{F-S} we provide in Theorem \ref{forsi} some conditions on the corank of the Kobayashi pseudometric in order
to determine the Loewner range of an $L^d$-evolution family.
In dimension one, Theorem \ref{main-intro} and the
uniformization theorem allow to recover both the classical
results of Loewner, Kufarev, Pommerenke and the new results by
Contreras, D\'iaz-Madrigal and Gumenyuk. In higher dimensions
these results are new.
Let now $G(z,t)$ be an $L^d$-Herglotz vector field whose flow is given by an $L^d$-evolution family as in (\ref{Lo-ODE}), and let $N$ be a complex manifold of dimension $n$.
Theorem \ref{main-intro} yields that a family of univalent mappings $(f_t\colon M\to N)$ solves the Loewner-Kufarev PDE
$$\frac{\partial f_s}{\partial s}(z)=-(df_s)_zG(z,s),\quad \mbox{a.e. }s\geq 0, z\in M$$ if and only if it is an $L^d$-Loewner chain associated with $(\varphi_{s,t})$. A solution given by univalent mappings $(f_t\colon M\to \mathbb C^n)$ exists if and only if the Loewner range ${\sf Lr}\,(\varphi_{t,s})$ is biholomorphic to a domain in $\mathbb C^n$.
In Section \ref{conjugacy} we introduce a notion of conjugacy for $L^d$-evolution families which preserves the Loewner range.
In Section \ref{ss62} we give examples of $L^d$-Loewner
chains in the unit ball generated by the Roper-Suffridge
extension operator.
In Section \ref{ss63} we consider
spiral-shaped and star-shaped mappings and give a characterization of
such mappings.
\section{Evolution families and Herglotz vector fields}
In the rest of this paper, unless differently stated, all
manifolds are assumed to be connected. Let $M$ be a complex
manifold and let $d_M$ denote the distance associated with a
given Hermitian metric on $M$. In the sequel we will also use
the Kobayashi pseudodistance $k_M$ on $M$ and the associated
Kobayashi pseudometric $\kappa_M$ on $M$. For definitions and
properties we refer the reader to the books \cite{abate}, \cite{Kob}.
\begin{definition}\label{L^d_EF}
Let $M$ be a taut manifold.
A family $(\varphi_{s,t})_{0\leq s\leq t}$ of holomorphic
self-mappings of $M$ is an {\sl
evolution family of order $d\geq 1$} (or $L^d$-evolution family) if it satisfies the {\sl evolution property}
\begin{equation}\label{evolution_property}
\varphi_{s,s}={\sf id},\quad \varphi_{s,t}=\varphi_{u,t}\circ \varphi_{s,u},\quad 0\leq s\leq u\leq t,
\end{equation}
and if for any $T>0$ and for any compact set
$K\subset\subset M$ there exists a function $c_{T,K}\in
L^d([0,T],\mathbb R^+)$ such that
\begin{equation}\label{ck-evd}
d_M(\varphi_{s,t}(z), \varphi_{s,u}(z))\leq \int_{u}^t
c_{T,K}(\xi)d \xi, \quad z\in K,\ 0\leq s\leq u\leq t\leq T.
\end{equation}
\end{definition}
The following lemma is proved in \cite[Lemma 2]{BCD09}.
\begin{lemma}\label{jointly}
Let $d\in[1,+\infty]$. Let $(\varphi_{s,t})$ be an $L^d$-evolution family. Let $\Delta :=\{(s,t):0\leq s\leq t\}$. Then the mapping $$(s,t)\mapsto \varphi_{s,t}$$ from $\Delta$ to ${\sf hol}(M,M)$ endowed with the topology of uniform convergence on compacta is jointly continuous. Hence the mapping $\Phi(z,s,t):= \varphi_{s,t}(z)$ from $M\times \Delta$ to $M$ is jointly continuous.
\end{lemma}
\begin{proposition}\label{ev-univ}
Let $d\in[1,+\infty]$.
Let $(\varphi_{s,t})$ be an $L^d$-evolution family. Then for all $0\leq s\leq t$ the mapping $(\varphi_{s,t})$ is univalent.
\end{proposition}
\begin{proof}
We proceed by contradiction. Suppose there exists $0<s<t$ and $z\neq w$ in $M$ such that $\varphi_{s,t}(z)=\varphi_{s,t}(w).$ Set $r:= \inf \{u\in [s,t]: \varphi_{s,u}(z)=\varphi_{s,u}(w)\}.$
Since by Lemma \ref{jointly} $\lim_{u\to s+}\varphi_{s,u}={\sf id}$ uniformly on compacta, we have $r>s$. If $u\in (s,r)$, $$\varphi_{u,r}(\varphi_{s,u}(z))=\varphi_{u,r}(\varphi_{s,u}(w)),$$ and since $\varphi_{s,u}(z)\neq \varphi_{s,u}(w),$ the mappings $\varphi_{u,r}$, $u\in (s,r)$, are not univalent on a fixed relatively compact subset of $M$. But by Lemma \ref{jointly} $\lim_{u\to r-} \varphi_{u,r}={\sf id}$ uniformly on compacta, which is a contradiction since the identity mapping is univalent.
\end{proof}
\begin{definition}
\label{Her-vec-man} A \textit{weak holomorphic vector field of
order $d\geq 1$} on $M$ is a mapping $G:M\times \mathbb R^+\to
TM$ with the following properties:
\begin{itemize}
\item[(i)] The mapping $G(z,\cdot)$ is measurable on $\mathbb R^+$ for all
$z\in M$.
\item[(ii)] The mapping $G(\cdot,t)$ is holomorphic on $M$ for all $t\in \mathbb R^+$.
\item[(iii)] For any compact set $K\subset M$ and all $T>0$, there exists
a function $C_{K,T}\in L^d([0,T],\mathbb{R}^+)$ such that
$$\|G(z,t)\|\leq C_{K,T}(t),\quad z\in K, \mbox{ a.e.}\ t\in [0,T].$$
\end{itemize}
Let $M$ be a taut manifold. Assume moreover that the Kobayashi distance $k_M$ satisfies
$k_M\in C^1(M\times M\setminus \mbox{Diag})$.
A \textit{Herglotz vector field of order $d$} is a weak holomorphic vector field of order $d$ such that
\begin{equation}\label{herglotz_def}
(dk_M)_{(z,w)}(G(z,t),G(w,t))\leq 0,\quad z,w\in M, z\neq w,\ \mbox{ a.e. }
t\geq 0.
\end{equation}
\end{definition}
\begin{remark}
If $M$ is complete hyperbolic, then condition (\ref{herglotz_def})
means exactly that for almost every $t\geq 0$ the
holomorphic vector field $z\mapsto G(z,t)$ is semicomplete
(this is proved in \cite{BCD10} for
strongly convex domains, but the same proof works in the complete hyperbolic case).
\end{remark}
The result in \cite{BCD09} which we will use in the sequel is
the following:
\begin{theorem}\label{prel-thm}
Let $M$ be a complete hyperbolic complex manifold with Kobayashi
distance $k_M$. Assume that $k_M\in C^1(M\times M\setminus
\hbox{Diag})$. Then for any Herglotz vector field $G$ of order
$d\in [1,+\infty]$ there exists a unique $L^d$-evolution family
$(\varphi_{s,t})$ over $M$ such that for all
$z\in M$
\begin{equation}\label{solve}
\frac{\partial \varphi_{s,t}}{\partial t}(z)=G(\varphi_{s,t}(z),t) \quad
\hbox{a.e.\ } t\in [s,+\infty).
\end{equation}
Conversely for any $L^\infty$-evolution family $(\varphi_{s,t})$ over $M$ there exists a Herglotz vector field $G$ of
order $\infty$ such that \eqref{solve} is satisfied. Moreover,
if $H$ is another weak holomorphic vector field which satisfies
\eqref{solve} then $G(z,t)=H(z,t)$ for all $z\in M$ and almost
every $t\in \mathbb R^+$.
\end{theorem}
\section{Kernel convergence on complex manifolds}
Let $B(z_0,r)\subset \mathbb C^n$ denote the Euclidean open ball of center $z_0$
and radius $r>0$ (as customary, we denote by $\mathbb B^n:=B(0,1)$ the Euclidean open ball centered at the origin and radius $1$).
\begin{proposition}\label{local1-1}
Let $U\subset \mathbb{C}^n$ be an open set. Let $f_k\colon U\rightarrow \mathbb{C}^n$
be a sequence of univalent mappings.
Assume that $f_k\rightarrow f$ uniformly on compacta and that $f$ is univalent.
Then for all $z_0\in U$ and $0<s<r$ such that $B(z_0,s)\subset\subset B(z_0,r)\subset\subset U$
there exists $m=m(z_0,s,r)$ such that if $k>m$ then $$ f(B(z_0,s))\subset f_k(B(z_0,r)).$$
\end{proposition}
\begin{proof}
Let $K=f\left(\overline{B(z_0,s)}\right)$, $\gamma=\partial B(z_0,r)$ and $\Gamma=f(\gamma).$ Then $K\cap \Gamma=\varnothing$ since $f$ is univalent on $U$.
Let $\eta$ be the Euclidean distance between $\Gamma$ and $K$. Then $\eta>0$ and $$\eta=\min \{\|f(z)-w\|:\ w\in K,\|z-z_0\|=r\}.$$
If $u_0\in K$ then $\|f(z)-u_0\|\geq \eta$ for $z\in\gamma$, and since $f_k\rightarrow f$ uniformly on $\gamma$ there exists $m>0$ such that if $k\geq m$ and $z\in \gamma$ then $$\|f(z)-f_k(z)\|<\|f(z)-u_0\|.$$
Rouch\'e theorem in several complex variables (see \cite[Theorem 9.3.4]{L2}) yields then that $f_k(z)-u_0$ and $f(z)-u_0$ have the same number of zeros on $B(z_0,r)$ counting multiplicities. But $f(z)-u_0$ has a zero in $B(z_0,r)$ since $u_0\in K$, and thus $u_0\in f_k(B(z_0,r))$ for $k\geq m$. The constant $m>0$ does not depend on $u_0\in K$, hence we have the result.
\end{proof}
\begin{corollary}\label{corollary_local1-1}
Let $U\subset \mathbb{C}^n$ be an open set. Let $(f_k)$ be a sequence of univalent mappings $f_k\colon U\rightarrow \mathbb{C}^n$ converging uniformly on compacta to a univalent mapping $f$. Then any compact set $K\subset f(U)$ is eventually contained in $f_k(U)$.
\end{corollary}
\begin{proof}
All the balls $B(z,s)\subset\subset U$ give an open covering of $U$. Since $K$ is compact there is a finite number of balls $B(z_i,s_i)\subset\subset U$ such that $K\subset \bigcup_i f(B(z_i,s_i)),$ hence Proposition \ref{local1-1} yields the result.
\end{proof}
\begin{definition}
Let $(\Omega_k)$ be a sequence of open subsets of a manifold $M$. The {\sl kernel} $\Omega$ is the biggest open set such that for all compact sets $K\subset \Omega$ there exists $m=m(K)$ such that if $k\geq m$ then $K\subset \Omega_k$. We say that the sequence $(\Omega_k)$ {\sl kernel converges} to $\Omega$ (denoted $\Omega_k\to \Omega)$ if every subsequence of $(\Omega_k)$ has the same kernel $\Omega$.
\end{definition}
Note that by the very definition the kernel is an open set,
possibly empty. It might be empty as the following example
shows:
\begin{example}
Let $M=\mathbb C$ and $f_k:\mathbb D\to \mathbb C$ defined by $f_k(z)=\frac{1}{k}z$.
Then $(f_k)$ is a sequence of univalent mappings
converging uniformly on compacta to $0$, and $f_k(\mathbb D)\to\varnothing.$
\end{example}
We have the following result. Another version of the kernel
convergence theorem in $\mathbb{C}^n$ may be found in
\cite{DGHK}.
\begin{theorem}\label{kernel}[Kernel convergence]
Let $(f_k)$ be a sequence of univalent mappings
from a complete hyperbolic complex manifold $M$ to a complex manifold $N$ of the same dimension.
Suppose that $(f_k)$ converges uniformly on compacta to a univalent mapping $f$. Then $f(M)$ is a connected component of the kernel $\Omega$ of the sequence $(f_k(M))$, and $(f_k^{-1}|_{f(M)})$ converges uniformly on compacta to $f^{-1}|_{f(M)}$. In particular if $\Omega$ is connected then $(f_k(M))\to \Omega.$
\end{theorem}
\begin{proof}
Let $K\subset f(M)$ be a compact set. We want to prove that eventually $K\subset f_k(M)$.
Let $\mathcal{U}=\{U_\alpha\}$ be an open covering of $M$ such that any $U_\alpha$ is biholomorphic to $\mathbb{B}^n$, and let $\mathcal{H}$ be the open covering of $M$ given by all open subsets $H$ satisfying the following property: there exists $U_\alpha\in \mathcal{U} $ such that $H\subset\subset U_\alpha$
(notice that $f(H)$ is then relatively compact in some coordinate chart of $N$). Note that $f$ is an open mapping since $M$ and $N$ have the same dimension, thus $f_*\mathcal{U}=\{f(U_\alpha)\}_{U_\alpha\in\mathcal{U}}$ is an open covering of $f(M).$
Since $K$ is compact there exist a finite number of open subsets $H_i\in\mathcal{H}$ such that $K\subset \bigcup_i f(H_i)$. Note that on $H_i$ the sequence $f_k$ takes eventually values in some $f(U_{\alpha_i})$ thanks to uniform convergence on compacta.
By using a partition of unity it is easy to see that there exist a finite number of compact sets $K_i$ such that $K_i\subset f(H_i)$ and $K=\bigcup_i K_i$. Thus we can assume $M\subset \mathbb{C}^n$ and $N=\mathbb{C}^n$, and the claim follows from Corollary \ref{corollary_local1-1}.
Thus $f(M)$ is a subset of the kernel $\Omega$ of the sequence $(f_k(M))$. This implies that on any compact set $K\subset f(M)$ the sequence $f_k^{-1}\colon K\rightarrow M$ is eventually defined. Let $\Omega_0$ be the connected component of the kernel which contains $f(M).$ We want to prove that $(f_k^{-1}|_{\Omega_0})$ admits a subsequence converging uniformly on compacta. Assume that $(f_k^{-1}|_{\Omega_0})$ is compactly divergent. Since $M$ is complete hyperbolic, this is equivalent to assume that for all fixed $z_0\in M$ and compact sets $K\subset{\Omega_0}$ we have
\begin{equation}\label{liminfmin}
\liminf_{k\rightarrow \infty}\left(\min_{w\in K}k_M(f^{-1}_k(w),z_0)\right)=+\infty.
\end{equation}
Let $j\geq 0$ and let $$K(j):=\{f(z_0)\}\cup\bigcup_{k\geq j}\{f_k(z_0)\}.$$ Since $f_k(z_0)\rightarrow f(z_0)$, the set $K(j)$ is compact. Since $f(M)$ is open there exists $m>0$ such that $K(m)\subset f(M)\subset {\Omega_0}$. But $$k_M\left(f_k^{-1}(f_k(z_0)),z_0\right)=0,$$ in contradiction with (\ref{liminfmin}).
Let $(f^{-1}_{k_i}|_{\Omega_0})$ be a converging subsequence and let $g:\Omega_0 \to M$ be its limit.
Let $w_0\in {\Omega_0}$.
The sequence $(f^{-1}_{k_i}(w_0))$ is eventually defined and converging to some $z=g(w_0)$. Thus $w_0=f_{k_i}(f^{-1}_{k_i}(w_0))\rightarrow f(z)$, which implies that ${\Omega_0}=f(M)$ and that $g(w_0)=f^{-1}(w_0)$, hence $(f^{-1}_k|_{\Omega_0})$ converges to $f^{-1}|_{\Omega_0}$ uniformly on compacta.
\end{proof}
The condition that the sets are open is important, as the
following example shows:
\begin{example}
Let $\mathbb D:=\{\zeta \in \mathbb C : |\zeta|<1\}$. Let $f_k : \mathbb D \to \mathbb C^2$
be defined by $f_k(\zeta):=(\zeta, \frac{1}{k} \zeta)$. Then
$(f_k)$ is a sequence of univalent discs which
converges uniformly on compacta to the injective disc
$\zeta\mapsto (\zeta, 0)$.
The only compact set in $\mathbb{C}^2$ which is eventually contained in
$f_k(\mathbb D)$ is $\{0\}$.
\end{example}
\section{Loewner chains on complex manifolds}
As we will show in what follows, some properties of Loewner chains are related only to the algebraic properties of evolution family and not to $L^d$ regularity. Hence, it is natural to introduce the following:
\begin{definition}\label{algebraic_EF}
Let $M$ be a complex manifold.
An {\sl algebraic evolution family} is a family $(\varphi_{s,t})_{0\leq s\leq t}$ of univalent self-mappings of $M$ satisfying the evolution property (\ref{evolution_property}).
\end{definition}
Thanks to Proposition \ref{ev-univ}, an $L^d$-evolution family is an algebraic evolution family ({\sl i.e.}, it is univalent).
\begin{definition}
Let $M, N$ be two complex manifolds of the same dimension. A family of holomorphic mappings $(f_t: M\to N)_{t\geq 0}$ is a {\sl
subordination chain} if for each $0\leq s\leq t$ there
exists a holomorphic mapping $v_{s,t}:M\to M$ such that
$f_s=f_t\circ v_{s,t}$.
A subordination chain $(f_t)$ and an algebraic evolution family $(\varphi_{s,t})$ are {\sl associated} if
$$ f_s=f_t\circ \varphi_{s,t},\quad 0\leq s\leq t.$$
An {\sl algebraic Loewner chain} is a subordination chain such that each mapping $f_t: M\to N$ is univalent. The {\sl range} of an algebraic Loewner chain is defined as ${\sf rg}\,(f_t):=\bigcup_{t\geq 0}f_t(M).$
An algebraic Loewner chain $(f_t\colon M\to N)$ is {\sl surjective} if ${\sf rg}\,(f_t)=N$.
\end{definition}
\begin{remark}
Equivalently an algebraic Loewner chain can be defined as a family of univalent mappings $(f_t: M\to N)_{t\geq 0}$ such that $$f_s(M)\subset f_t(M),\quad 0\leq s\leq t.$$
\end{remark}
\begin{definition}
Let $d\in [1,+\infty]$. Let $M, N$ be two complex manifolds of
the same dimension. Let $d_N$ be the distance induced by a Hermitian metric on $N$. An algebraic Loewner chain $(f_t\colon M\to N)$ is a {\sl Loewner chain of order} $d\in [1,+\infty]$ (or $L^d$-Loewner chain) if for any compact set $K\subset\subset M$ and any
$T>0$ there exists a $k_{K,T}\in L^d([0,T], \mathbb R^+)$ such that
\begin{equation}\label{LCdef}
d_N(f_s(z), f_t(z))\leq \int_{s}^t k_{K,T}(\xi)d\xi
\end{equation}
for all $z\in K$ and for all $0\leq s\leq t\leq T$.
\end{definition}
\begin{remark}\label{conti}
By (\ref{LCdef}) the mapping $t\mapsto f_t$ is continuous from $\mathbb{R}^+$ to ${\sf hol}(M,N)$. Hence the mapping
$\Psi\colon M\times \mathbb{R}^+\rightarrow N$ defined as $\Psi(z,t)=f_t(z)$
is jointly continuous.
\end{remark}
\begin{theorem}
\label{Low-to-ev}
Let $d\in [1,+\infty]$. Let $(f_t\colon M\to N)$ be an $L^d$-Loewner
chain. Assume that $M$ is complete hyperbolic. Let
$$\varphi_{s,t}:= f_t^{-1} \circ f_s,\quad 0\leq s\leq t.$$
Then for any Hermitian metric on $M$, the family $(\varphi_{s,t})$ is an $L^d$-evolution family on $M$ associated with $(f_t)$.
\end{theorem}
\begin{proof}
It is clear that $(\varphi_{s,t})$ is an algebraic evolution
family, so that we only need to prove the $L^d$-estimate.
Let $H$ be a compact subset of $f_t(M)$. Set $$L(H,t):=\sup_{\eta,\zeta\in H,\eta\neq\zeta}\frac{d_M(f_t^{-1}(\zeta),f_t^{-1}(\eta))}{d_N(\zeta,\eta)}.$$
Then $L(H,t) <+\infty$, since $w\mapsto f_t^{-1}(w)$ is locally Lipschitz.
Given a compact subset $K\subset M $ define $$K_t:= \bigcup_{s\in[0,t]}f_s(K).$$ The set $K_t$ is a compact subset of $f_t(M)$ by Remark \ref{conti} since $K_t=\Psi(K\times [0,t])$.
Let $T>0$ be fixed.
We claim that the function $L(K_t,t)$ on $0\leq t\leq T$ is bounded by a constant $C(K,T)<+\infty$. Assume that $L(K_t,t)$ is unbounded. Then there exists a sequence $(t_n)\subset [0,T]$, which we might assume converging to some $t\in [0,T]$, such that $$L(K_{t_n},t_n)\geq n+1,\quad \forall n\geq 0.$$
Hence for any $n\geq 0$ there exist $\zeta_n,\eta_n\in K_{t_n}$ such that $\zeta_n\neq\eta_n$ and
\begin{equation}\label{incr}
\frac{d_M(f_{t_n}^{-1}(\zeta_{n}),f_{t_n}^{-1}(\eta_{n}))}{d_N(\zeta_{n},\eta_{n})}\geq n.
\end{equation}
By passing to a subsequence we may assume that
$\zeta_n \to \zeta\in K_t$ and $\eta_n \to \eta \in K_t$.
By Theorem \ref{kernel}, $f_{t_n}^{-1}\to f_t^{-1}$ uniformly on a neighborhood of $K_t$.
By (\ref{incr}) we have $\eta=\zeta$, since otherwise $$\frac{d_M(f_{t_n}^{-1}(\zeta_{n}),f_{t_n}^{-1}(\eta_{n}))}{d_N(\zeta_{n},\eta_{n})}\rightarrow \frac{d_M(f_t^{-1}(\zeta),f_t^{-1}(\eta))}{d_N(\zeta,\eta)}.$$
Let $U,V$ be two open subsets of $f_t(M)$, both biholomorphic to $\mathbb{B}^n$ such that $\zeta\in U\subset\subset V\subset\subset f_t(M).$
Since by Theorem \ref{kernel} the sequence $(f_{t_n}^{-1})$ converges to $f_t^{-1}$ uniformly on $V$, we have that eventually $ f_{t_n}^{-1}(U)\subset f^{-1}_t(V)$. The sequence $(f_{t_n}^{-1}|_U)$ is thus equibounded and by Cauchy estimates it is equi-Lipschitz in a neighborhood of $\zeta$, which contradicts (\ref{incr}) and thus proves the claim.
Let $\Delta_T:=\{(s,t):0\leq s\leq t\leq T\}$. Then the mapping $(s,t)\mapsto f^{-1}_t\circ f_s$ from $\Delta_T$ to ${\sf hol}(M,M)$ endowed with the topology of uniform convergence on compacta is continuous. Indeed, let $(s_n,t_n)\rightarrow (s,t)$. Let $K\subset M $ be a compact set. By Remark \ref{conti} the set $$K(j):= f_s(K)\cup\bigcup_{n\geq j}f_{s_n}(K)=\Psi (K,\{s\}\cup\bigcup_{n\geq j}\{s_n\})$$ is compact. There exists $m>0$ such that $K(m)\subset f_t(M)$. By Theorem \ref{kernel} the sequence $(f_{t_n}^{-1})$ converges to $f_t^{-1}$ uniformly on $K(m)$. This proves that $(s,t)\mapsto f^{-1}_t\circ f_s$ is continuous.
This implies that the mapping $\Phi\colon M\times \Delta_T\rightarrow M$ defined as $\Phi(z,s,t):=\varphi_{s,t}(z)$ is jointly continuous. Hence if $K\subset M$ is a compact set, $$\hat{K}:= \bigcup_{0\leq a\leq b\leq T} \varphi_{a,b}(K)=\bigcup_{0\leq a\leq b\leq T} f_b^{-1}(f_a(K))$$ is compact in $M$. Therefore, since $$d_M(\varphi_{s,u}(z),\varphi_{s,t}(z))=d_M(\varphi_{s,u}(z),\varphi_{u,t}(\varphi_{s,u}(z))),$$ in order to estimate $d_M(\varphi_{s,u}(z),\varphi_{s,t}(z))$ for $z\in K$ and $0\leq s\leq u\leq t\leq T$ it is enough to estimate $d_M(\zeta,\varphi_{u,t}(\zeta))$ for $\zeta\in \hat{K}.$
Since
\begin{align}
d_M(\zeta,\varphi_{u,t}(\zeta))&=d_M(f_t^{-1}(f_t(\zeta)),f_t^{-1}(f_u(\zeta)))\leq L(\hat{K}_t,t)d_N(f_t(\zeta),f_u(\zeta)) \nonumber\\
&\leq C(\hat{K},T)d_N(f_t(\zeta),f_u(\zeta))\leq C(\hat{K},T)\int_u^tk_{\hat{K},T}(\xi)d\xi,\nonumber
\end{align}
we are done.
\end{proof}
\begin{theorem}\label{algebraic_theorem}
Any algebraic evolution family $(\varphi_{s,t})$ admits an associated algebraic Loewner chain $(f_t\colon M\to N)$.
Moreover if $(g_t\colon M\to Q)$ is a subordination chain associated with $(\varphi_{s,t})$ then there exist a holomorphic mapping $\Lambda\colon {\sf rg}\,(f_t)\to Q$ such that $$g_t=\Lambda\circ f_t,\quad \forall t\geq 0.$$ The mapping $\Lambda$ is univalent if and only if $(g_t)$ is an algebraic Loewner chain, and in that case ${\sf rg}\,(g_t)=\Lambda({\sf rg}\,(f_t)).$
\end{theorem}
\begin{proof}
Define an equivalence relation on the product $M\times \mathbb{R}^+$:
$$(x,s)\sim (y,t)\quad\mbox{iff}\quad \varphi_{s,u}(x)=\varphi_{t,u}(y) \mbox{ for $u$ large enough}.$$
and define $N:=(M\times \mathbb{R}^+)/_\sim$.
Let $\pi\colon M\times \mathbb{R}^+\to N$ be the projection on the quotient, and let $i_t\colon M\to M\times \mathbb{R}^+$ be the injection $i_t(x)=(x,t)$. Define a family of mappings $(f_t\colon M\to N)$ as $$f_t:= \pi\circ i_t,\quad t\geq 0.$$ Each mapping $f_t$ is injective since ${\pi}|_{M\times \{t\}}$ is injective, and by construction the family $(f_t)$ satisfies$$f_s=f_t\circ\varphi_{s,t}, \quad 0\leq s\leq t.$$
Thus we have $f_s(M)\subset f_t(M)$ for $0\leq s\leq t$ and $N=\bigcup_{t\geq 0}f_t(M)$.
Endow the product $M\times \mathbb{R}^+$ with the product topology, considering on $\mathbb R^+$ the discrete topology. Endow $N$ with the quotient topology. Each mapping $f_t$ is continuous and open, hence it is an homeomorphism onto its image. This shows that $N$ is arcwise-connected and Hausdorff since each $f_t(M)$ is arcwise-connected and Hausdorff. Moreover $N$ is second countable since $N=\bigcup_{k\in \mathbb{N}}f_k(M)$.
Now define a complex structure on $N$ by considering the $M$-valued charts $(f^{-1}_t,f_t(M))$ for all $t\geq 0$. This charts are compatible since $f_t^{-1}\circ f_s=\varphi_{s,t}$ which is holomorphic. Hence the family $(f_t)$ is an algebraic Loewner chain associate with $(\varphi_{s,t})$.
If $(g_t\colon M\to Q)$ is a subordination chain associated with $(\varphi_{s,t})$, then the map $\Psi\colon M\times \mathbb{R}^+\to Q$
$$(z,t)\mapsto g_t(z)$$ is compatible with the equivalence relation $\sim$. The map $\Psi$ passes thus to the quotient defining a holomorphic mapping $\Lambda\colon N\to Q$ such that $$g_t=\Lambda\circ f_t,\quad t\geq 0.$$ The last statement is easy to check.
\end{proof}
As a corollary we have the following.
\begin{corollary}\label{rangebihol}
Let $(\varphi_{s,t})$ be an algebraic evolution family on a complex manifold
$M$. Also let $(f_t\colon M\to N)$ and $(g_t\colon M\to Q)$ be two algebraic Loewner chains
associated with $(\varphi_{s,t})$. Then there exists a
biholomorphism $\Lambda\colon {\sf rg}\,(f_t)\to {\sf rg}\,(g_t)$ such that $g_t=\Lambda \circ
f_t$ for all $t\geq 0$.
\end{corollary}
Thus there exists essentially one algebraic Loewner chain associated with an algebraic evolution family.
\begin{definition}
Let $(\varphi_{s,t})$ be an algebraic evolution family. By Corollary \ref{rangebihol} the biholomorphism class of the range of an associated algebraic Loewner chain is uniquely determined. We call this class the {\sl Loewner range} of $(\varphi_{s,t})$ and we denote it by ${\sf Lr}\,(\varphi_{s,t})$.
\end{definition}
\begin{theorem}\label{ev-to-Low}
Let $d\in [1,+\infty]$. Let $(\varphi_{s,t})$ be an $L^d$-evolution family on a taut manifold $M$, and let $(f_t\colon M\to N)$ be an associated algebraic Loewner chain. Then $(f_t)$ is an $L^d$-Loewner chain for any Hermitian metric on $N$.
\end{theorem}
\begin{proof}
Let $K\subset M$ be a compact set. Let $T>0$ be fixed.
By Lemma \ref{jointly} the subset of $M$ defined as $$\hat{K}:=\bigcup_{0\leq s\leq t\leq T} \varphi_{s,t}(K)$$ is compact. Indeed
$\hat{K}=\Psi(K\times \Delta_T)$, where $\Delta_T=\{(s,t): 0\leq s\leq t\leq T\}$.
Since $f_T$ is locally Lipschitz there exists $C= C(\hat{K})
>0$ such that $$d_N(f_T(z),f_T(w))\leq Cd_M(z,w),\quad z,w\in \hat{K}.$$
The family $(\varphi_{t,T})_{0\leq t\leq T}$ is equi-Lipschitz on $\hat{K}$, that is there exists $L(\hat{K},T)>0$ such that
\begin{equation}\label{estimatelip}
d_M(\varphi_{t,T}(z),\varphi_{t,T}(w))\leq L(\hat{K},T)d_M(z,w),\quad z,w\in\hat{K},\ t\in[0,T].
\end{equation}
Indeed assume by contradiction that there exist sequences $(z_n),(w_n)$ in $\hat{K}$, and $(t_n)$ in $[0,T]$ such that
\begin{equation}\label{incr2}
\frac{d_M(\varphi_{t_n,T}(z_{n}),\varphi_{t_n,T}(w_n))}{d_M(z_{n},w_{n})}\geq n.
\end{equation} By passing to subsequences we can assume $t_n\to t$, $z_n\to z$ and $w_n\to w$, and by (\ref{incr2}) it is easy to see that $z=w$.
Let $U,V$ be two open subsets of $M$, both biholomorphic to $\mathbb{B}^n$ such that $z\in V\subset\subset U\subset\subset M.$
Since the sequence $(\varphi_{t_n,T})$ converges to $\varphi_{t,T}$ uniformly on $U$ we have that eventually $ \varphi_{t_n,T}(V)\subset \varphi_{t,T}(U)$. The sequence $(\varphi_{t_n,T}|_V)$ is thus equibounded and by Cauchy estimates it is equi-Lipschitz in a neighborhood of $z$, which contradicts (\ref{incr2}).
Hence, for all $z\in K$ and $0\leq s\leq t\leq T$ we have
\begin{equation*}
\begin{split}
d_N(f_s(z), f_t(z))&=d_N(f_T(\varphi_{s,T}(z)), f_T(\varphi_{t,T}(z)))\leq C d_M(\varphi_{s,T}(z),
\varphi_{t,T}(z))
\\ & = C d_M(\varphi_{t,T}(\varphi_{s,t}(z)), \varphi_{t,T}(z))\leq CL(\hat{K},T)d_M(\varphi_{s,t}(z), z)
\\& = CL(\hat{K},T) d_M(\varphi_{s,t}(z), \varphi_{s,s}(z))\leq CL(\hat{K},T) \int_s^t c_{K,T}(\xi) d\xi,
\end{split}
\end{equation*}
by (\ref{ck-evd}). This concludes the proof.
\end{proof}
\begin{corollary}\label{L^dL^d}
Assume that the algebraic evolution family $(\varphi_{s,t})$ on a complete hyperbolic complex manifold $M$ is associated with the algebraic Loewner chain $(f_t\colon M\to N)$. Then $(\varphi_{s,t})$ is an $L^d$-evolution family if and only if $(f_s)$ is an $L^d$-Loewner chain.
\end{corollary}
\begin{proof}
It follows from Theorems \ref{ev-to-Low} and \ref{Low-to-ev}.
\end{proof}
When dealing with evolution families defined on a domain $D$
of a complex manifold $N$, a natural question is whether there
exists an associated Loewner chain whose range is contained
in $N$, or, in other terms, whether the Loewner range is biholomorphic to
a domain of $N$. This question makes
particularly sense if $D=\mathbb B^n$ and $N=\mathbb C^n$. In other words:
\medskip
{\bf Open question:} Given an $L^d$-evolution family on the unit ball $\mathbb B^n$ does there exist an
associated $L^d$-Loewner chain with range in $\mathbb C^n$?
\medskip
\begin{remark}\label{not_admit}
There exists an algebraic evolution family $(\varphi_{s,t})$ on $\mathbb B^3$ which does not admit any associated algebraic Loewner chain with range in $\mathbb C^3$. This follows from \cite[Section 9.4]{A10}.
\end{remark}
There are several works in this direction, answering such a
question in some normalized class of evolution families
(see \cite{A10}, \cite{GHK01}, \cite{GHK09}, \cite{GHKK08}, \cite{GK03},
\cite{Por91}, \cite{Vo})
but in its generality the question is
still open. Here we give some answers based on the
asymptotic behavior of the Kobayashi pseudometric
under the corresponding evolution family.
\begin{definition}
Let $(\varphi_{s,t})$ be an algebraic evolution family on a complex manifold $M$. Let $\kappa_M: TM\to \mathbb R^+$ be the Kobayashi pseudometric of $M$. For $v\in T_zM$ and $s\geq 0$ we define
\begin{equation}\label{beta}
\beta^s_v(z):=\lim_{t\to \infty} \kappa_M(\varphi_{s,t}(z); (d\varphi_{s,t})_z(v)).
\end{equation}
\end{definition}
\begin{remark}
Let $0\leq s\leq u\leq t$. Since the Kobayashi pseudometric is
contracted by holomorphic mappings it follows
\begin{equation*}
\begin{split}
\kappa_M(\varphi_{s,t}(z); (d\varphi_{s,t})_z(v))&=\kappa_M(\varphi_{u,t}(\varphi_{s,u}(z));
(d\varphi_{u,t})_{\varphi_{s,u}(z)}(d\varphi_{s,u})_z(v))
\\&\leq \kappa_M(\varphi_{s,u}(z); (d\varphi_{s,u})_z(v)),
\end{split}
\end{equation*}
hence the limit in \eqref{beta} is well defined.
\end{remark}
\begin{proposition}\label{beta-ev}
Let $(\varphi_{s,t})$ be
an algebraic evolution family on a complex manifold $M$. Let
$(f_t\colon M\to N)$ be an associated surjective algebraic Loewner chain.
Then for all $z\in M$ and $v\in T_zM$ it follows
\[
f_s^*\kappa_N(z;v)=\beta^s_v(z).
\]
\end{proposition}
\begin{proof}
Since the chain $(f_t\colon M\to N)$ is surjective, the range $N$ is the union of the growing sequence of complex manifolds $(f_j(M))_{j\in\mathbb{N}}$, thus
$$\kappa_N(f_s(z);(df_s)_z)(v))=\lim_{j\to\infty}\kappa_{f_j(M)}(f_s(z);(df_s)_z(v)).$$
The result follows from
\begin{align}
\lim_{j\to \infty}\kappa_{f_j(M)}(f_s(z);(df_s)_z)(v))&=\lim_{j\to \infty}\kappa_{M}(f_j^{-1}(f_s(z));(df_j^{-1})_{f_s(z)}\circ (df_s)_z(v))\\\nonumber
&=\lim_{j\to\infty}\kappa_M(\varphi_{s,j}(z);(d\varphi_{s,j})_z(v))\nonumber.
\end{align}
\end{proof}
As corollaries we find (cf. \cite[Theorem 1.6]{CDG09})
\begin{corollary}\label{Lrdisc}
Let $(\varphi_{s,t})$ be an algebraic evolution family on the unit disc $\mathbb{D}.$
If there exist $z\in \mathbb D$,\ $v\in T_z\mathbb{D}$,\ $s\geq 0$, such that $\beta_v^s(z)=0$ then
\begin{itemize}
\item[i)] ${\sf Lr}\,(\varphi_{s,t})$ is biholomorphic to $\mathbb{C}$,
\item[ii)] $\beta_v^s(z)=0$ for all $z\in \mathbb D$,\ $v\in T_z\mathbb{D}$,\ $s\geq 0$.
\end{itemize}
If there exist $z\in \mathbb D$,\ $v\in T_z\mathbb{D}$,\ $s\geq 0$, such that $\beta_v^s(z)\neq 0$ then
\begin{itemize}
\item[i)] ${\sf Lr}\,(\varphi_{s,t})$ is biholomorphic to $\mathbb{D}$,
\item[ii)] $\beta_v^s(z)\neq0$ for all $z\in \mathbb D$,\ $v\in T_z\mathbb{D}$,\ $s\geq 0$.
\end{itemize}
\end{corollary}
\begin{proof}
Since the Loewner range ${\sf Lr}\,(\varphi_{s,t})$ is non-compact and simply connected, by the uniformization theorem it has to be biholomorphic to $\mathbb{D}$ or $\mathbb{C}$. Since
$$\kappa_\mathbb{C}(z;v)=0,\quad z\in \mathbb{C},v\in T_z\mathbb{C},$$
$$\kappa_\mathbb{D}(z;v)\neq 0,\quad z\in \mathbb{D},v\in T_z\mathbb{D},$$
the result follows from Proposition \ref{beta-ev}.
\end{proof}
\begin{corollary}\label{evdisc}
Let $d\in [1,+\infty]$.
Let $(\varphi_{s,t})$ be an $L^d$-evolution family on the unit disc $\mathbb D$. Then there exists an $L^d$-Loewner
chain $(f_t)$ associated with $(\varphi_{s,t})$ such that ${\sf rg}\,(f_t)$ is either the unit disc $\mathbb D$ or the complex plane
$\mathbb C$.
\end{corollary}
\begin{proof}
It follows from Corollary \ref{Lrdisc} and Theorem \ref{ev-to-Low}.
\end{proof}
Such a result can be generalized in higher
dimension as follows.
As customary, let us denote by ${\sf aut}(M)$ the group of
holomorphic automorphisms of a complex manifold $M$. Notice that
condition $M$ hyperbolic and $M/\sf{aut}(M)$ compact
implies that $M$ is complete hyperbolic (see \cite{F-S}).
\begin{theorem}\label{forsi}
Let $M$ be a hyperbolic complex manifold and assume that
$M/\sf{aut}(M)$ is compact. Let $(\varphi_{s,t})$ be an algebraic evolution
family on $M$. Then
\begin{enumerate}
\item If there exists $z\in M$, $s\geq0$ such that $\beta^s_v(z)\neq 0$ for
all $v\in T_zM$ with $v\neq 0$ then ${\sf Lr}\,(\varphi_{s,t})$ is biholomorphic to $M$.
\item If there exists $z\in M$, $s\geq0$ such that $\dim_\mathbb C\{v\in T_zM: \beta_v^s(z)= 0\}=1$ then ${\sf Lr}\,(\varphi_{s,t})$ is a fiber bundle with fiber $\mathbb{C}$
over a closed complex submanifold of $M$.
\end{enumerate}
\end{theorem}
\begin{proof}
It follows at once from Proposition \ref{beta-ev} and
\cite[Theorem 3.2, Main Theorem]{F-S}.
\end{proof}
In particular the previous result applies to $M=\mathbb B^n$ (or even to
the polydiscs in $\mathbb C^n$) and we obtain
\begin{corollary} \label{image-Cn}
Let $(\varphi_{s,t})$ be an algebraic evolution family on the unit ball $\mathbb B^n$.
If for some $z\in \mathbb B^n$, $s\geq 0$
it follows that $\dim_\mathbb C\{v\in \mathbb C^n : \beta^s_v(z)=0\}\leq 1$,
then there exists an algebraic Loewner chain $(f_t\colon M\to\mathbb{C}^n)$ associated with $(\varphi_{s,t})$.
\end{corollary}
\begin{proof}
If the dimension is zero, then by Theorem
\ref{forsi} the Loewner range is biholomorphic to $\mathbb B^n\subset \mathbb C^n$. If the dimension is one, then by Theorem
\ref{forsi} the Loewner range is a fiber bundle with fiber $\mathbb{C}$ over a closed complex submanifold of $\mathbb B^n$ and
by \cite[Corollary 4.8]{F-S} it is actually biholomorphic to
$\mathbb B^{n-1}\times \mathbb C\subset \mathbb C^n$.
\end{proof}
If $\dim_\mathbb C\{v\in \mathbb C^n : \beta^s_v(z)=0\}\geq 2$ the complex structure of the Loewner range can be more complicated: the Loewner range of the algebraic evolution family recalled in Remark \ref{not_admit} has $\dim_\mathbb C\{v\in \mathbb C^n : \beta^s_v(z)=0\}= 2$ and is not biholomorphic to a domain of $\mathbb C^3$.
\begin{example}
Let $(\varphi_{s,t})$ be an algebraic evolution family of $\mathbb B^2$ such that
$\varphi_{s,t}(0)=0$ for all $0\leq s\leq t$ and $(d\varphi_{s,t})_0=
e^{A(t-s)}$ where $A$ is a diagonal matrix with eigenvalues
$i\theta$, $\theta\in \mathbb R$ and $\lambda\in \mathbb C$ for some ${\sf Re}\,
\lambda\leq 0$. Then $\dim_\mathbb C\ker \beta_v^s(0)\leq 1$ (it is
either $1$ if ${\sf Re}\, \lambda<0$ or $0$ if ${\sf Re}\, \lambda=0$ which
is the case if and only if $\varphi_{s,t}$ are automorphisms).
Therefore in such a case there exists an algebraic Loewner chain with
range in $\mathbb C^2$.
\end{example}
The previous example can be generalized as follows:
\begin{example}
Let $G(z,t)$ be an $L^\infty$-Herglotz vector field in $\mathbb B^n$
such that $G(0,t)\equiv 0$ and $(d_zG)_{z=0}(\cdot,t)=A(t)$
where $A(t)$ is a diagonal $n\times n$ matrix with eigenvalues
$\lambda_1(t), \ldots, \lambda_n(t)$ where $\lambda_j: \mathbb R^+\to
\mathbb C$ are functions of class $L^\infty$ such that ${\sf Re}\,
\lambda_j(t)\leq 0$ for almost every $t\geq 0$ and $j=1,\ldots,
n$. Assume that there exists $C>0$ such that
\[
\int_0^t {\sf Re}\, \lambda_j(\tau) d\tau \geq -C, \quad t\geq0,\ j=1,\ldots, n-1.
\]
Let $(\varphi_{s,t})$ be the associated $L^\infty$-evolution family
in $\mathbb B^n$. Then $\varphi_{s,t}(0)=0$ and $(d\varphi_{s,t})_0$ is the
diagonal matrix with eigenvalues $\exp \left(\int_s^t
\lambda_j(\tau)d\tau \right)$ for $j=1,\ldots, n$. Hence
$\dim_\mathbb C\ker \beta_v^s(0)\leq 1$ and there exists an associated
$L^\infty$-Loewner chain with range in $\mathbb C^n$.
\end{example}
\section{Loewner-Kufarev PDE}\label{PDE}
In this section we prove that $L^d$-Loewner chains on complete hyperbolic complex manifolds are the univalent solutions of the Loewner-Kufarev partial differential equation, as in the classical theory of Loewner chains on the unit
ball $\mathbb B^n$ in $\mathbb{C}^n$ (see \cite{GHK01}, \cite{GK03}). Other results related to the solutions of the Loewner-Kufarev PDE on $\mathbb B^n$ may be
found in \cite{DGHK}.
\begin{proposition}
\label{vitali}
Let $M$ be a complete hyperbolic complex manifold, and let $(f_t\colon M\to N)$ be a Loewner chain of order $d\in[1,+\infty]$ on $M$. Then
there exists a set $E\subset\mathbb R^+$ $($independent of $z)$ of
zero measure such that for every $s\in (0,+\infty)\setminus E$, the
mapping $$M\ni z \mapsto \frac{\partial f_s}{\partial s}(z)\in T_{f_s(z)}N$$
is well-defined and holomorphic on $M$.
\end{proposition}
\begin{proof}
The manifold $M\times (0,+\infty)$ has a countable basis $\mathfrak{B}$ given by products of the type $B\times I$, where $B\subset M$ is an open subset biholomorphic to a ball, and $I\subset (0,+\infty)$ is a bounded open interval.
Let $\mathcal{V}$ be a countable covering of $N$ by charts. By Remark \ref{conti} the mapping $\Psi\colon M\times (0,+\infty)\to N$ is continuous, hence there exists a covering $\mathcal{U}\subset \mathfrak{B}$ of $M\times (0,+\infty)$ such that for any $U\in\mathcal{U}$ there exists $V\in\mathcal{V}$ such that $U\subset \Psi^{-1}(V)$.
We will prove that for each $U=B\times I\in \mathcal{U}$ there exists a set of full measure
$I'\subseteq I$
such that $B\ni z \mapsto \frac{\partial f_s}{\partial s}(z)$
is well defined and holomorphic for all $s\in I'$.
Hence $M \ni z \mapsto \frac{\partial f_s}{\partial s}(z)$ will be well defined
and holomorphic for $s\in \mathbb{R}^{+}\setminus \bigcup_{\mathcal{U}}({I\setminus I'})$ which is a set of full measure in $\mathbb{R}^+$.
We can assume that $M=\mathbb B^n$, $N=\mathbb C^n$, and that the distance $d_N$ is the Euclidean distance. Since $t\mapsto f_t(z)$ is locally absolutely continuous on $\mathbb{R}^+$ locally
uniformly with respect to $z\in \mathbb B^n$, we deduce that for each $z\in \mathbb B^n$, there is a null
set $E_1(z)\subset I$ such that for each $t\in I\setminus E_1(z)$, there
exists the limit
\begin{equation*}
\frac{\partial f_t}{\partial t}(z)=\lim_{h\to 0}\frac{f_{t+h}(z)-f_t(z)}{h}.
\end{equation*}
By definition there exists a function $p_k\in L^d(I,
\mathbb{R}^+)$ such that
\begin{equation}\label{pbound}
\|f_s(z)-f_t(z)\|\leq \int_s^tp_k(\xi)d\xi,\quad z\in \mathbb B^n_{1-1/k},\quad
s\leq t\in I.
\end{equation}
Also, since $p_k\in L^d(I, \mathbb{R}^+)$,
we may find a null set $E_2(k)\subset I$ such that
for each $t\in I\setminus E_2(k)$, there exists the limit
\begin{equation}\label{pbound2}
p_k(t)=\lim_{h\to 0}\frac{1}{h}\int_t^{t+h}p_k(\xi)d\xi,\quad k\in\mathbb{N}.
\end{equation}
Next, let $Q$ be a countable set of uniqueness for the holomorphic functions
on $\mathbb B^n$ and let
\[E=\bigg(\bigcup_{q\in Q}E_1\left(q\right)\bigg)\bigcup
\bigg(\bigcup_{k=1}^{\infty}E_2(k)\bigg).\]
Then $E$ is a null subset of $\mathbb{R}^+$,
which does not depend on $z\in \mathbb B^n$.
Arguing as in the proof of \cite[Theorem 4.1(1)(a)]{CDG09},
it is not difficult to see that \eqref{pbound} and \eqref{pbound2}
imply that for each $s\in I\setminus E$,
the family $$\{(f_{s+h}(z)-f_s(z))/h,\quad 0<|h|<\textsf{dist}(s, \partial I)\}$$ is relatively
compact and has a unique accumulation point for $|h|\to 0$
by Vitali Theorem in several complex variables, proving the result.
\end{proof}
\begin{theorem}\label{LK-PDE}
Let $M$ be a complete hyperbolic complex manifold such that the Kobayashi
distance $k_M\in C^1(M\times M\setminus \hbox{Diag})$, and let $N$ be a complex manifold of the same dimension.
Let $G:M\times \mathbb R^+\to TM$ be a Herglotz vector
field of order $d\in[1,+\infty]$
associated with the $L^d$-evolution family $(\varphi_{s,t})$. Then a family of univalent mappings $(f_t\colon M\to N)$ is an $L^d$-Loewner chain associated with $(\varphi_{s,t})$ if and only if it is locally absolutely continuous
on $\mathbb R^+$ locally uniformly with respect to $z\in M$ and solves the Loewner-Kufarev PDE
\begin{equation}\label{equationLK-PDE}
\frac{\partial f_s}{\partial s}(z)=-(df_s)_zG(z,s),\quad\mbox{a.e. }s\geq 0,z\in M.
\end{equation}
\end{theorem}
\begin{proof}
Since $G(z,t)$ and $(\varphi_{s,t})$ are associated there exists a null set $E_1\subset \mathbb R^+$ such that for all $s\geq 0$, for all $t\in [s,+\infty)\setminus E_1$ and for all $z\in M$,
\[
\frac{\partial \varphi_{s,t}}{\partial t}(z)=G(\varphi_{s,t}(z),t).
\]
Let now $(f_t)$ be an $L^d$-Loewner chain associated with $(\varphi_{s,t})$.
By Proposition \ref{vitali}, there is a null set $E_2\subset
\mathbb R^+$ such that $z\mapsto \frac{\partial f_s}{\partial s}(z)$
is well defined and holomorphic for all $s\in (0,+\infty)\setminus E_2$.
The set $E=E_1\cup E_2$ has also zero measure. It is clear that the mapping
$$t\mapsto L_t(z):=f_t(\varphi_{0,t}(z))$$ is locally absolutely continuous on $\mathbb R^+$
locally uniformly with respect to $z\in M$, in view of the conditions (\ref{LCdef}) and
(\ref{ck-evd}). Also $L_t(z)=f_0(z)$ for $z\in M$. Differentiating the last equality with respect to
$t\in (0,+\infty)\setminus E$ we obtain
\begin{equation*}
\begin{split}
0&=(df_t)_{\varphi_{0,t}(z)}\frac{\partial \varphi_{0,t}}{\partial t}(z)+
\frac{\partial f_t}{\partial t}(\varphi_{0,t}(z))\\&=(df_t)_{\varphi_{0,t}(z)}G(\varphi_{0,t}(z),t)+
\frac{\partial f_t}{\partial t}(\varphi_{0,t}(z)),
\end{split}
\end{equation*}
for all $t\in (0,+\infty)\setminus E$ and for all $z\in M$.
Hence
$$\frac{\partial f_t}{\partial t}(w)=
-(df_t)_{w}G(w,t),$$
for all $w$ in the open set $\varphi_{0,t}(M)$ and for all $t\in (0,+\infty)\setminus E$.
The identity theorem for holomorphic mappings provides the result.
\\
To prove the converse, fix $s\geq 0$ and let
$$L_t(z):=f_t(\varphi_{s,t}(z))$$ for $t\in [s,+\infty)$ and $z\in M$.
In view of the hypothesis, it is not difficult to deduce that
$$\frac{\partial L_t}{\partial t}(z)=0,\quad \mbox{ a.e. }\ t\geq 0,\quad
\forall z\in M.$$
Hence $L_t(z)\equiv L_s(z)$, i.e. $f_t(\varphi_{s,t}(z))=f_s(z)$
for all $z\in M$ and $0\leq s\leq t$, which means that $(f_t)$ is an algebraic Loewner chain associated with $(\varphi_{s,t})$.
Hence $(f_t)$ is an $L^d$-Loewner chain by Theorem \ref{ev-to-Low}.
\end{proof}
\begin{corollary}\label{corollaryLK-PDE}
Let $M$ be a complete hyperbolic complex manifold such that the Kobayashi
distance $k_M\in C^1(M\times M\setminus \hbox{Diag})$, and let $N$ be a complex manifold of the same dimension. Every $L^\infty$-Loewner chain $(f_t\colon M\to N)$ solves a Loewner-Kufarev PDE.
\end{corollary}
\begin{proof}
By Theorem \ref{Low-to-ev} there exists an $L^\infty$-evolution
family $(\varphi_{s,t})$ associated with
$(f_t)$. By Theorem \ref{prel-thm} there exists a Herglotz
vector field $G(z,t)$ associated with $(\varphi_{s,t})$. Theorem \ref{LK-PDE} yields then that the family $(f_t\colon M\to N)$ satisfies
$$\frac{\partial f_s}{\partial s}(z)=-(df_s)_zG(z,s),\quad\mbox{a.e. }s\geq 0,z\in M.$$
\end{proof}
From Theorems \ref{algebraic_theorem} and \ref{ev-to-Low} we easily obtain the following corollary. Let $G(z,t)$ be an $L^d$-Herglotz vector field associated with the $L^d$-evolution family $(\varphi_{s,t})$.
\begin{corollary}
Let $M$ be a complete hyperbolic complex manifold of dimension $n$ such that the Kobayashi
distance $k_M\in C^1(M\times M\setminus \hbox{Diag})$. The Loewner-Kufarev PDE (\ref{equationLK-PDE}) admits a solution given by univalent mappings $(f_t\colon M\to N)$ where $N$ is a complex manifold $N$ of dimension $n$. Any other solution with values in a complex manifold $Q$ is of the form $(\Lambda\circ f_t)$ where $\Lambda\colon {\sf rg}\,(f_t)\to Q$ is holomorphic. Hence a solution given by univalent mappings $(h_t\colon M\to \mathbb C^n)$ exists if and only if the Loewner range ${\sf Lr}\,(\varphi_{s,t})$ is biholomorphic to a domain in $\mathbb C^n$.
\end{corollary}
\section{Conjugacy}\label{conjugacy}
We introduce a notion of conjugacy for $L^d$-evolution families which preserves the Loewner range. This can be used to put an $L^d$-evolution family in some normal form without changing its Loewner range (cf. \cite[Proposition 2.9]{CDG09}).
\begin{definition}
Let $d\in[1,+\infty].$
Let $(h_t\colon M \to Q)$ be an $L^d$-Loewner chain such that each $h_t\colon M\to Q$ is a biholomorphism. We call $(h_t\colon M\to Q)$ a family of {\sl intertwining mappings of order d}. If $(\varphi_{s,t}),(\psi_{s,t})$ are $L^d$-evolution families on $M,Q$ respectively and $$\psi_{s,t}\circ h_s =h_t\circ \varphi_{s,t},\quad 0\leq s\leq t,$$ then we say that $(\varphi_{s,t})$ and $(\psi_{s,t})$ are {\sl conjugate}. It is easy to see that conjugacy is an equivalence relation.
\end{definition}
\begin{lemma}\label{intercomp}
Let $(h_t\colon M\to Q)$ be a family of intertwining mappings of order $d$ and let $(f_t \colon Q\to N)$ an $L^d$-Loewner chain. Then $(f_t\circ h_t\colon M\to N)$ is an $L^d$-Loewner chain.
\end{lemma}
\begin{proof}
It is clear that $(f_t\circ h_t\colon M\to N)$ is an algebraic Loewner chain. Let $T>0$ and let $K\subset M$ be a compact set. The set $\hat K:=\bigcup_{0\leq t\leq T}h_t(K)\subset Q$ is compact by Lemma \ref{conti}, and the family $(f_t)_{0\leq t\leq T}$ is equi-Lipschitz on $\hat K$ (see (\ref{estimatelip})). Thus if $0\leq s\leq t\leq T$ and $z\in K$,
\begin{align}
d_N(f_t(h_t(z)),f_s(h_s(z)))&\leq d_N(f_t(h_t(z)),f_t(h_s(z)))+d_N(f_t(h_s(z)),f_s(h_s(z)))\\ \nonumber
&\leq L(\hat K,T)d_Q(h_t(z),h_s(z)) +\int_s^t k_{\hat K,T}(\xi) d\xi\\\nonumber
&\leq L(\hat K,T) \int_s^t h_{ K,T}(\xi) d\xi +\int_s^t k_{\hat K,T}(\xi) d\xi. \nonumber
\end{align}
\end{proof}
\begin{remark}
Two conjugated $L^d$-evolution families have essentially the same associated $L^d$-Loewner chains. Namely, if $(g_t\colon Q\to N)$ is an $L^d$-Loewner chain associated with $(\psi_{s,t})$ which is conjugate to $(\varphi_{s,t})$ on $M$ via $(h_t\colon M\to Q)$, then $(g_t\circ h_t\colon M\to N)$ is an $L^d$-Loewner chain associated with $(\varphi_{s,t})$. In particular, ${\sf Lr}\,(\varphi_{s,t})$ is biholomorphic to ${\sf Lr}\,(\psi_{s,t}).$
\end{remark}
\begin{proposition}\label{L^dconj}
Let $(\psi_{s,t})$ be an $L^d$-evolution family on a complete hyperbolic complex manifold $Q$ and let $(h_t\colon M\to Q)$ be a family of intertwining mappings of order $d$. Then the family $(h_t^{-1}\circ\psi_{s,t}\circ h_s)$ is an $L^d$-evolution family on $M$.
\end{proposition}
\begin{proof}
By Theorem \ref{ev-to-Low}, there exists an $L^d$-Loewner chain $(f_t\colon Q\to N)$ associated with $(\psi_{s,t})$. By Lemma \ref{intercomp} the family $(f_t\circ h_t)$ defines an $L^d$-Loewner chain $(g_t\colon M\to N)$. Then
$$h_t^{-1}\circ\psi_{s,t}\circ h_s=h_t^{-1}\circ f_t^{-1}\circ f_s \circ h_s=g_t^{-1}\circ g_s,$$ which by Theorem \ref{Low-to-ev} is an $L^d$-evolution family on $M$.
\end{proof}
Let now $M$ be the unit ball $\mathbb B^n$.
\begin{definition}
Take $a\in\mathbb B^n$. Let $P_a(z):=\frac{\langle z,a\rangle}{\|a\|^2}a$ for $a\neq 0$, $P_0=0$,
$Q_a(z):=z-P_a(z)$ and $s_a:=(1-\| a\|^2)^{1/2}$. Then
\[
\varphi_a(z):=\frac{a-P_a(z)-s_aQ_a(z)}{1-\langle z,a\rangle}
\]
is an automorphism of the ball $\mathbb B^n$ (see, {\sl e.g.},
\cite{abate} or \cite{R}).
\end{definition}
We can now show that in order to study the Loewner range of an $L^d$-evolution family on $\mathbb B^n$ one can assume that it fixes the origin.
\begin{corollary}
Let $(\psi_{s,t})$ be an $L^d$-evolution family on $\mathbb B^n$. There exists a conjugate $L^d$-evolution family $(\varphi_{s,t})$ such that $$\varphi_{s,t}(0)=0,\quad 0\leq s\leq t.$$
\end{corollary}
\begin{proof}
Set $a(t):=\psi_{0,t}(0)$. Since $$\|\varphi_{a(t)}(w)-\varphi_{a(s)}(w)\|\leq C(K,T)\|a(t)-a(s)\|,\quad w\in K,\ 0\leq s\leq t\leq T,$$ the family $(\varphi_{a(t)})$ is a family of intertwining mappings of order $d$.
Define $$\varphi_{s,t}:=\varphi_{a(t)}^{-1}\circ \psi_{s,t}\circ \varphi_{a(s)},$$ which is an $L^d$-evolution family by Proposition \ref{L^dconj}. Since $\varphi_{a(t)}(0)=a(t),$ we have $\varphi_{0,t}(0)=0$ for all $t\geq 0$, and by the evolution property $\varphi_{s,t}(0)=0$ for all $0\leq s\leq t$.
\end{proof}
\begin{comment}
We end this section with a result generalizing \cite[Proposition 2.6(2)]{CDG09}.
\begin{proposition}
Let $(\varphi_{s,t})$ be an $L^\infty$-evolution family.
Then for every compact set $K\subset \mathbb B^n$ and $T> 0$,
there exist a constant $k_{K,T}>0$
such that
$$\left\| \frac{\partial}{\partial t}(d\varphi_{s,t})_zu\right\| \leq k_{K,T},\quad z\in K,\quad u\in\partial \mathbb B^n,\quad \mbox{a.e.}\ t\in [s,T].$$
\end{proposition}
\begin{proof}
Fix a compact set $K\subset \mathbb B^n$ and $T> 0$. There exists
$R_1<1$ such that $K\subset B(0,R_1)$. Let $R_2=(R_1+1)/2$.
By Lemma \ref{jointly}, there exists $R_3$ such that $\|
\varphi_{s,t}(z)\|\leq R_3$ for all $z\in B(0,R_2)$ and
$0\leq s\leq t\leq T$. By Theorem \ref{prel-thm}, there exists
a Herglotz vector field $G$ of order $\infty$ such that
\begin{equation}
\label{2}
\frac{\partial \varphi_{s,t}}{\partial t}(z)=G(\varphi_{s,t}(z),t)
\quad
\mbox{a.e.\ }t\in[s,+\infty).
\end{equation}
In view of the definition of Herglotz vector field, there
exists a constant $C_{K,T}>0$ such that
\begin{equation}
\label{3}
\| G(z,t)\| \leq C_{K,T},
\quad
\mbox{a.e.}
\quad
t\in [0,T],
\quad
\forall z\in \overline{B(0,R_3)}.
\end{equation}
For every $u\in\partial \mathbb B^n$, we obtain from (\ref{2}) and (\ref{3}) that
\begin{eqnarray*}
\left\| \frac{\partial}{\partial t}d\varphi_{s,t}(z)u\right\|
&=&
\left\|\frac{\partial}{\partial t}\frac{\partial}{\partial \zeta}\varphi_{s,t}(z+\zeta u)|_{\zeta=0}\right\|
\\
&=&
\left\| \frac{\partial}{\partial t}\frac{1}{2\pi i}\int_{|\zeta|=r}\frac{\varphi_{s,t}(z+\zeta u)}{\zeta^2}d\zeta\right\|
\\
&=&
\left\| \frac{1}{2\pi i}\int_{|\zeta|=r}\frac{\partial}{\partial t}
\left(\frac{\varphi_{s,t}(z+\zeta u)}{\zeta^2}\right)d\zeta\right\|
\\
&=&
\left\| \frac{1}{2\pi i}\int_{|\zeta|=r}
\frac{G(\varphi_{s,t}(z+\zeta u),t)}{\zeta^2}d\zeta\right\|
\\
&\leq&
\frac{1}{r}C_{K,T}
\end{eqnarray*}
\begin{align}
\left\| \frac{\partial}{\partial t}(d\varphi_{s,t})_zu\right\|&=
\left\|\frac{\partial}{\partial t}\frac{\partial}{\partial \zeta}\varphi_{s,t}(z+\zeta u)|_{\zeta=0}\right\|=
\left\| \frac{\partial}{\partial t}\frac{1}{2\pi i}\int_{|\zeta|=r}\frac{\varphi_{s,t}(z+\zeta u)}{\zeta^2}d\zeta\right\|\\\nonumber
&=
\left\| \frac{1}{2\pi i}\int_{|\zeta|=r}
\frac{G(\varphi_{s,t}(z+\zeta u),t)}{\zeta^2}d\zeta\right\|
\leq
\frac{1}{r}C_{K,T}\nonumber
\end{align}
for almost every $t\in[s,T]$,
where $r=(1-R_1)/2$.
This completes the proof.
\end{proof}
\section{Applications to the unit ball}\label{application}
Various applications of the theory of normalized Loewner
chains fixing the origin in the unit ball $\mathbb B^n$ in
$\mathbb{C}^n$ may be found in
\cite{GHK01}, \cite{GHK09}, \cite{GHKK08}, \cite{GK03}, \cite{GKK03},
\cite{Por91}.
In this section we apply the previously developed theory to the
unit ball $\mathbb B^n$ in $\mathbb{C}^n$.
\subsection{Normalized Loewner chains}\label{ss61}
\begin{definition}
\label{d1.10} An univalent subordination chain $(f_t)$ on the unit ball
such that $f_t(\mathbb B^n)\subset \mathbb C^n$ is called a {\sl normalized
univalent subordination chain} if $f_t(0)=0$ and
$(df_t)_0=e^t{\sf id}$ for $t\ge 0$.
\end{definition}
\begin{remark}
A normalized univalent subordination chain is a
$L^\infty$-Loewner chain. Indeed, such a chain is locally
Lipschitz continuous on $\mathbb R^+$ locally uniformly with
respect to $z\in\mathbb B^n$ (see \cite[Chapter 8]{GK03}).
\end{remark}
Let
$$S(\mathbb B^n)=\{f\in {\sf Hol}(\mathbb B^n, \mathbb C^n): f \mbox{ is biholomorphic on }
\mathbb B^n, f(0)=0, df_0={\sf id}\}$$
be the set of normalized biholomorphic mappings on $\mathbb B^n$.
The following consequence of Theorem \ref{ev-to-Low} is an
$n$-dimensional version of a well known result of Pommerenke
(see \cite[Chapter 6]{Po}), which states that any normalized
univalent function on the unit disc can be imbedded as the
first element of a normalized univalent subordination chain.
\begin{proposition}
\label{c3.1} For every mapping $f\in S(\mathbb B^n)$ there exists an
$n$-dimensional complex manifold $\Omega$ containing $f(\mathbb B^n)$
as an open subset, and an $L^{\infty}$-Loewner chain $(\mathbb B^n,
\Omega,(f_t))$ such that, denoting by
$\iota: f(\mathbb B^n)\to \Omega$ the natural embedding,
it follows that $f_0=\iota\circ f$, $f_t(0)=\iota(0)$
and $(d (\iota^{-1}\circ f_t))_0=e^t
{\sf id}$ for $t\geq 0$.
\end{proposition}
\begin{proof} It is enough to apply Theorem
\ref{ev-to-Low} to the mapping $f$ and to the evolution family
$\varphi_{s,t}(z):=e^{s-t}z$, which is clearly an $L^\infty$
evolution family of $\mathbb B^n$. Since $f(0)=\varphi_{0,t}(0)=0$, it
follows that $f_t(0)=\iota(0)$ for $t\geq 0$. Also, since
$d\varphi_{0,t}(0)=e^{-t}{\sf id}$, we deduce that $(d (\iota^{-1}\circ f_t))_0=e^t {\sf id}$ for
$t\geq 0$, as claimed.
\end{proof}
\begin{definition}
Let $f\in {\sf Hol}(\mathbb B^n,\mathbb C^n)$ be such that $f(0)=0$,
$df_0={\sf id}$. We say that {\sl $f$ has parametric representation}
(and write $f\in S^0(\mathbb B^n)$) if there exists a Herglotz vector
field $h: \mathbb B^n\times \mathbb R^+\to \mathbb{C}^n$ such that
$h(0,t)=0$ and $dh(0,t)=-{\sf id}$ for almost all $t\in \mathbb R^+$,
and $f(z)=\displaystyle\lim_{t\to\infty}e^{t}\varphi_{0,t}(z)$ locally
uniformly on $\mathbb B^n$, where $(\varphi_{s,t})$ is the evolution family
associated with $h$.
\end{definition}
It can be checked that such a definition is equivalent to the
one given in \cite{GHK01} and \cite{GKK03} (cf. \cite{Por87a}),
by \cite[Theorem 0.2]{BCD10}
and \cite[Proposition 3.8.1]{ERS04} (see also
\cite{GK03}) in the case of the unit polydisc).
\begin{remark}
\label{r-param} A mapping $f\in S(\mathbb B^n)$ has parametric
representation if and only if there exists a normalized
univalent subordination chain $f_t(z)$ such that
$\{e^{-t}f_t(\cdot)\}_{t\geq 0}$ is a normal family on $\mathbb B^n$
and $f=f_0$ (see \cite{GHK01} and \cite{GKK03}; see also
\cite{GK03}; cf. \cite{Por87a}, \cite{Por87b}).
In the case of one complex variable,
$S^0(\mathbb B^1)=S(\mathbb B^1)$, but in higher dimensions, $S^0(\mathbb B^n)\neq
S(\mathbb B^n)$ (see \cite{GHK01}).
\end{remark}
It is a natural question to ask whether and how an element in
$S(\mathbb B^n)$ can be related to an element in $S^0(\mathbb B^n)$. Maybe
strange enough the answer has nothing to do with Loewner
theory but it is an application of the ``extension of the
inverse of a univalent mapping'' trick as in the proof of
Theorem \ref{ev-to-Low}:
\begin{proposition}
\label{imbedS0}
Let $f\in S(\mathbb B^n)$ and let $g\in S^0(\mathbb B^n)$. Then there exists
a complex manifold $\Omega$ containing $f(\mathbb B^n)$ as an open set,
and a biholomorphism $\Phi: \mathbb C^n \to \Omega$ such that,
denoting by $\iota: f(\mathbb B^n)\to \Omega$ the natural embedding,
it follows $\iota \circ f=\Phi \circ g$. In particular, if
$g={\sf id}$ then $\Phi$ can be considered as an extension of $f$
from $\mathbb B^n$ to $\mathbb C^n$ $($with image $\Omega)$.
\end{proposition}
\begin{proof}
We are given a univalent mapping $f: \mathbb B^n \to \mathbb C^n$ and a
univalent mapping $g: \mathbb B^n \to \mathbb C^n$. Thus we can consider the
univalent mapping $g\circ f^{-1} : f(\mathbb B^n)\to \mathbb C^n$. The
complex manifold $\Omega$ and the biholomorphism $\Phi:\mathbb C^n \to
\Omega$ are then given as the extension of the left inverse of
$g\circ f^{-1}$ (cf. ``extension of the inverse of a univalent
mapping'' in the proof of Theorem \ref{ev-to-Low}).
\end{proof}
\begin{remark}
\label{notcontained} Note that the manifold $\Omega$ in the
statement of Proposition \ref{imbedS0} may not be contained in
$\mathbb{C}^n$. For instance, this is the case if $g$ is
bounded on $\mathbb B^n$ and $f\in S(\mathbb B^n)$ is unbounded. Suppose that
$\Omega\subseteq\mathbb{C}^n$. Then there exists a
biholomorphic mapping $\Phi :\mathbb{C}^n\to\Omega$ such that
$f=\Phi \circ g$. Since the right hand-side is bounded on
$\mathbb B^n$, we obtain a contradiction.
\end{remark}
\end{comment}
\section{Extension of Loewner chains from lower dimensional
balls}\label{ss62} The following result provides examples of
$L^d$-Loewner chains on the Euclidean unit ball $\mathbb B^n$ in
$\mathbb{C}^n$, which are generated by the Roper-Suffridge
extension operator \cite{RS}. This operator preserves convexity
(see \cite{RS}), starlikeness and the notion of parametric
representation (see e.g. \cite{GK03} and the references
therein).
\begin{theorem}
\label{t1}
Let $d\in [1,+\infty]$ and $(f_t,\mathbb D,\mathbb C)$ be an $L^d$-Loewner
chain on the unit disc $\mathbb D$ such that $|\arg f_t'(0)|<\pi/2$ and
$|\arg(f_s'(0)/f_t'(f_t^{-1}\circ f_s(0)))|<\pi/2$ for $t\geq s\geq 0$.
Also let $(F_t\colon \mathbb B^n\to \mathbb C^n)$ be given by
\begin{equation}
\label{1}
F_t(z)=\Big(f_t(z_1),\tilde{z}e^{t/2}\sqrt{f_t'(z_1)}\Big),\quad
z=(z_1,\tilde{z})\in \mathbb B^n,\quad t\geq 0.
\end{equation}
Then $(F_t)$ is an $L^d$-Loewner chain.
\end{theorem}
\begin{proof}
It is easy to see that $F_t$ is univalent on $\mathbb B^n$ for
$t\geq 0$.
Let $(\varphi_{s,t})$ be the
$L^d$-evolution family associated with $(f_t)$ (see Theorem
\ref{Low-to-ev} or \cite{BCD08}). Also let
$\Phi_{s,t}:\mathbb B^n\to\mathbb{C}^n$ be given by
$$\Phi_{s,t}(z)=\Big(\varphi_{s,t}(z_1),\tilde{z}e^{(s-t)/2}\sqrt{\varphi_{s,t}'
(z_1)}\Big),\quad
z=(z_1,\tilde{z})\in \mathbb B^n,\, t\geq s\geq 0.$$ Then $\Phi_{s,t}$
is a univalent mapping on $\mathbb B^n$ and in view of the
Schwarz-Pick lemma, we have
$$\|\Phi_{s,t}(z)\|^2=|\varphi_{s,t}(z_1)|^2+\|\tilde{z}\|^2e^{s-t}|
\varphi_{s,t}'(z_1)|$$
$$<
|\varphi_{s,t}(z_1)|^2+(1-|z_1|^2)e^{s-t}\cdot\frac{1-|\varphi_{s,t}(z_1)|^2}{1-
|z_1|^2}\leq 1,
\, z\in \mathbb B^n,\, t\geq s\geq 0.$$ Hence
$\Phi_{s,t}(\mathbb B^n)\subseteq \mathbb B^n$, and since
$F_s(z)=F_t(\Phi_{s,t}(z))$ for $z\in \mathbb B^n$ and $t\geq s\geq
0$, we obtain that $F_s(\mathbb B^n)\subseteq F_t(\mathbb B^n)$ for $s\leq
t$. In view of the above relations, we deduce that $(F_t)$ is an
algebraic Loewner chain and $(\Phi_{s,t})$ is the associated algebraic evolution
family.
It remains to prove that $(F_t)$ is of order $d$. Since
$(\varphi_{s,t})$ is an evolution family of order $d$, we deduce in view of
\cite[Theorem 6.2]{BCD08} that there
exists a Herglotz vector field $g(z_1,t)$ of order $d$ such
that
$$\frac{\partial \varphi_{s,t}}{\partial t}(z_1)=g(\varphi_{s,t}(z_1),t),\quad
\mbox{ a.e. }\quad t\in [s,+\infty),\quad \forall z_1\in \mathbb D.$$
Now, let $G=G(z,t):\mathbb B^n\times \mathbb R^+\to\mathbb{C}^n$ be
given by
$$G(z,t)=\Big(g(z_1,t),\frac{\tilde{z}}{2}(-1+g'(z_1,t))\Big),
\quad z=(z_1,\tilde{z})\in \mathbb B^n,\, t\geq 0.$$ Then $G(z,t)$ is
a weak holomorphic vector field of order $d$ on $\mathbb B^n$. Indeed,
the measurability and holomorphicity conditions from the
definition of a weak holomorphic vector field are satisfied. We
next prove that for each $r\in (0,1)$ and $T>0$, there exists
$C_{r,T}\in L^d([0,T],\mathbb{R})$ such that
$$\|G(z,t)\|\leq C_{r,T}(t),\quad \|z\|\leq r,\quad t\in [0,T].$$
But the above condition can be easily deduced by using the fact
that $g(z_1,t)$ is a Herglotz vector field of order $d$ on $\mathbb D$
and by the Cauchy integral formula.
On the other hand, since $\varphi_{s,t}$ is locally absolutely
continuous on $[s,+\infty)$ locally uniformly with respect to
$z_1\in \mathbb D$, it follows in view of Vitali's theorem (see e.g.
\cite[Chapter 6]{Po}) that
$$\frac{\partial}{\partial t}\Big(\frac{\partial \varphi_{s,t}(z_1)}{\partial
z_1}\Big)=
\frac{\partial }{\partial z_1}\Big(\frac{\partial \varphi_{s,t}(z_1)}{\partial
t}\Big),
\quad \mbox{ a.e. }\quad t\geq s,\quad \forall z_1\in \mathbb D.$$
Using the above equality, we obtain by elementary
computations that
$$\frac{\partial \Phi_{s,t}(z)}{\partial t}=G(\Phi_{s,t}(z),t),\quad \mbox{ a.e. }\quad
t\geq s,\quad
\forall z\in \mathbb B^n.$$
Therefore, as in the proof of \cite[Proposition 2]{BCD09},
we deduce that
$(dk_M)_{(z,w)}(G(z,t),G(w,t))\leq 0$ for a.e. $t\in\mathbb R^+$,
$z\neq w$.
Hence $G(z,t)$ is a Herglotz vector field of order $d$ on
$\mathbb B^n$.
Also, as in the proof of \cite[Proposition 1]{BCD09},
we deduce that $(\Phi_{s,t})$ is an evolution family
of order $d$.
Finally, we
conclude that the associated algebraic Loewner chain $(F_t)$ is of order
$d$ on $\mathbb B^n$ by Theorem \ref{ev-to-Low}. This completes the proof.
\end{proof}
\begin{corollary}
\label{c1} Let $f:\mathbb D\to\mathbb{C}$ be a univalent function such that
$|\arg f'(0)|<\pi/2$. Assume that $(f_t)$ is an $L^d$-Loewner
chain on $\mathbb D$ such that $f_0=f$,
$|\arg f_t'(0)|<\pi/2$ for $t\geq 0$,
and $|\arg(f_s'(0)/f_t'(f_t^{-1}\circ f_s(0)))|<\pi/2$
for $t\geq s\geq 0$.
Then $F=\Phi_{n}(f)$ can be imbedded in a
$L^d$-Loewner chain on $\mathbb B^n$, where $\Phi_n$ is the
Roper-Suffridge extension operator,
$$\Phi_n(f)(z)=(f(z_1),\tilde{z}\sqrt{f'(z_1)}),\quad z=(z_1,\tilde{z})\in \mathbb B^n.$$
\end{corollary}
\begin{proof}
The desired $L^d$-Loewner chain is given by (\ref{1}).
\end{proof}
\section{Spiral-shapedness and Star-shapedness}
\label{ss63}
\begin{definition}
\label{d3.1} Let $\Omega\subset\mathbb{C}^n$ and let $A\in
L(\mathbb{C}^n,\mathbb{C}^n)$ be such that $m(A)>0$, where
$$m(A)=\min\{{\sf Re}\,\langle A(z),z\rangle: \|z\|=1\}.$$
We say that $\Omega$ is {\sl spiral-shaped with respect to $A$}
if $e^{-tA}(w)\in\Omega$ whenever $w\in\Omega$ and $t\in
\mathbb R^+$. If $A={\sf id}$ and $\Omega$ is spiral-shaped with
respect to ${\sf id}$, we say that $\Omega$ is {\sl star-shaped}.
If $f$ is a univalent mapping on $\mathbb B^n$, then $f$ is called
{\sl spiral-shaped with respect to $A$} if the image domain $\Omega={f(\mathbb B^n)}$ is
spiral-shaped with respect to $A$. In addition, if $A={\sf id}$ and
$f$ is spiral-shaped with respect to ${\sf id}$, we say that $f$ is
{\sl star-shaped} (see \cite{ERS04}).
\end{definition}
\begin{remark}
It is clear that if $f$ is spiral-shaped with respect to $A$,
then $0\in\overline{f(\mathbb B^n)}$. Moreover, if $0\in f(\mathbb B^n)$,
then the above definition reduces to the usual definition of
spiral-likeness (respectively star-likeness) (see \cite{Gu} and
\cite{Su}).
\end{remark}
We next present some applications of Theorem \ref{LK-PDE} to
the case $M=\mathbb B^n$. The first result provides necessary and
sufficient conditions for a locally univalent mapping on
the unit ball ${\mathbb B^n}$ in $\mathbb{C}^n$ to be spiral-shaped,
and thus univalent on $\mathbb B^n$.
\begin{remark} We remark that the equivalence between the conditions (i)
and (iii) in Theorem \ref{t3.1} below was first obtained by
Elin, Reich and Shoikhet (see the proof of \cite[Proposition
3.5.2]{ERS04}; cf. \cite[Proposition 3.7.2]{ERS04}; \cite{ReSh})
by different arguments (compare \cite{ERS00}). In the case
$f(0)=0$, the analytic characterization of spiral-likeness is
due to Gurganus \cite{Gu} and Suffridge \cite{Su}.
\end{remark}
\begin{theorem}
\label{t3.1} Let $f:{\mathbb B^n}\to\mathbb{C}^n$ be a locally
univalent mapping such that $0\in \overline{f({\mathbb B^n})}$.
Also let $A\in L(\mathbb{C}^n,\mathbb{C}^n)$ be such that
$m(A)>0$. Then the following conditions are equivalent:
$(i)$ $f$ is spiral-shaped with respect to $A$;
$(ii)$ The family $(f_t:=e^{tA}f(z))_{t\geq 0}$ is an
$L^\infty$-Loewner chain.
$(iii)$ $f$ is univalent on $\mathbb B^n$ and
\begin{equation}
\label{spiral}
{\sf Re}\,\langle (df_z)^{-1}Af(z),z\rangle\geq (1-\|z\|^2){\sf Re}\,\langle (df_0)^{-1}Af(0),z\rangle,
\quad z\in {\mathbb B^n}.
\end{equation}
\end{theorem}
\begin{proof}
The equivalence between the conditions (i) and (ii) is
immediate. Now, we assume that the condition (ii) holds. Then
$f$ is univalent on ${\mathbb B^n}$. Let $G(z,t)$ be the Herglotz
vector field of order $\infty$ given by Corollary \ref{corollaryLK-PDE}. A direct computation from \eqref{LK-PDE} implies
\begin{equation}
\label{4.50}
G(z,t)=-(df_z)^{-1}Af(z),\quad t\geq 0,\quad z\in {\mathbb B^n}.
\end{equation}
Since by the very definition a Herglotz vector field is a
semicomplete vector field for almost every $t\geq 0$, it
follows that $-(df_z)^{-1}Af(z)$ is semicomplete. Hence, by
\cite[Proposition 3.5.2]{ERS04} (where the sign convention is
different from the one adopted here), we deduce the relation
(\ref{spiral}), as claimed.
Conversely, assume that the condition (iii) holds. Clearly $(f_t)$ is a
family of univalent mappings on $\mathbb B^n$ such that the mapping $t\mapsto f_t(z)$
is of class $C^\infty$ on $\mathbb R^+$ for all $z\in \mathbb B^n$. Also $(f_t)$ satisfies
the differential equation
\begin{equation}
\label{4.49}
\frac{\partial f_t}{\partial t}(z)=-(df_t)_zG(z,t),\quad \mbox{ a.e. }\quad
t\geq 0,\quad \forall\ z\in {\mathbb B^n},
\end{equation}
where $G(z,t)$ is given by \eqref{4.50}. In view of the
condition (\ref{spiral}) and \cite[Lemma 3.3.2]{ERS04}, we
deduce that the mapping $G(z,t)$ is a semicomplete vector field
for all $t\geq 0$, and thus it is a Herglotz vector field of
order $\infty$ by \cite[Theorem 0.2]{BCD10}. Hence $(f_t)$ is
an $L^\infty$-Loewner chain by Theorem \ref{LK-PDE}. This
completes the proof.
\end{proof}
We next give the following analytic characterization of
star-shapedness on the unit ball $\mathbb B^n$ (cf. \cite{ERS04}). In
the case $f(0)=0$, the inequality in the third statement
becomes the well known analytic characterization of
star-likeness for locally univalent mappings on $\mathbb B^n$
(see \cite{Go}, \cite{GK03}, \cite{Su} and the
references therein). Necessary and sufficient conditions for star-likeness with respect to a
boundary point are given in \cite{Li-St}.
\begin{corollary}
\label{c-star} Let $f:{\mathbb B^n}\to\mathbb{C}^n$ be a locally
univalent mapping such that $0\in \overline{f({\mathbb B^n})}$.
Then the following conditions are equivalent:
$(i)$ $f$ is star-shaped;
$(ii)$ The family $(f_t:=e^tf(z))_{t\geq 0}$ is an
$L^\infty$-Loewner chain.
$(iii)$ $f$ is univalent on ${\mathbb B^n}$ and
$${\sf Re}\,\langle (df_z)^{-1}f(z),z\rangle\geq (1-\|z\|^2){\sf Re}\,\langle (df_0)^{-1}f(0),z\rangle,
\quad z\in {\mathbb B^n}.$$
\end{corollary}
\begin{corollary}
\label{c2} Let $f:\mathbb D\to\mathbb{C}$ be a star-shaped function on
$\mathbb D$ such that $|\arg f'(0)|<\pi/2$ and
$|\arg(f'(0)/f'(f^{-1}(\lambda f(0)))|<\pi/2$ for $\lambda\in
(0,1]$. Also let $F=\Phi_n(f)$. Then $F$ is also star-shaped on
$\mathbb B^n$.
\end{corollary}
\begin{proof}
Since $f$ is star-shaped, it follows that $f_t(z_1)=e^tf(z_1)$
is an $L^\infty$-Loewner chain by Corollary \ref{c-star}.
Let $(F_t)$ be the chain given by (\ref{1}).
In view of Theorem \ref{t1}, $(F_t)$ is an
$L^\infty$-Loewner chain on $\mathbb B^n$.
Moreover, since $0\in \overline{F(\mathbb B^n)}$
and $F_t(z)=e^tF(z)$, we deduce that the mapping $F=F_0$ is
star-shaped on $\mathbb B^n$, by Corollary \ref{c-star}.
This completes the proof.
\end{proof}
|
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"redpajama_set_name": "RedPajamaArXiv"
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Home / Products / EQUIPMENT / Backpacks / Mountaineering / BACKPACK X.M.T. 32 W.T.S.
XMT 32 W.T.S. is ideal for professional climbers and enthusiasts seeking a technical, functional item for their outings. XMT 32 W.T.S. is part of the Ferrino HighLab Line and guarantees 100% waterproof qualities, thanks to the WTS (Welded Technical System), which sees the seams welded in the joint areas instead of being stitched. The XMT 32 W.T.S. backpack is supplied by Ferrino to the Chamonix Mont-Blanc National Skiing and Mountaineering School.
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{
"redpajama_set_name": "RedPajamaC4"
}
| 6,779
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TMOGL
A Blog with a difference
Tag Archives: Chris Shalom
VIDEO: Chris Shalom & Wordbreed | Healed By His Stripes (Wordbreed Worship Service)
Chris Shalom and the Wordbreed Worship Group congregate again for the September edition of the Wordbreed Worship Service tagged "Healed By His Stripes."
The session is an unforgettable time of refreshing, comfort and healing by His stripes. Jesus Christ is alive and still doing good. He is still healing all who are oppressed of the devil because He is God. Chris Shalom and the Wordbreed Worship Group use the service session to set the stage for anyone who believes in the healing power of Christ so that such a one can receive his miracle.
The Supremacy of Christ, the potency and possibilities in His name has always been at the core of the message of the Wordbreed Worship Service. This month – Healed by His Stripes – the group intensifies on the message with great anticipation for unusual manifestations.
The Wordbreed Worship Service kicked off in 2017, long before online worship service became a thing. It re-emerged recently, due to the urgency prompted by the 2020 global pandemic. In August, the service recorded a surge in viewership, thereby leading to even more testimonies than the previous edition. The September edition promises to beat the past record still.
Let this worship service fill your homes with the unusual miracles and testimonies, as you join Chris Shalom and the Wordbreed Worship Group for another time of encounter, with an expectation to be healed by His stripes.
Watch video below:
By TMOGL • Posted in Event, Video • Tagged Chris Shalom, Healed By His Stripes, Video, Wordbreed, Wordbreed Worship Service
Chris Shalom Unveils 6th Studio Album "My Beautifier" – Get It Here
The 6th studio album titled "My Beautifier" by Gospel music minister Chris Shalom is out. The new album is available online and at local stores!
Following the 2015 album, "You are the Reason", Chris Shalom embarked on the "Worship in Every Place" series. From the series, singles such as "My Beautifier", Dry Bones are Rising", "Your Word is Truth", were birthed and has now translated into the 6th studio album.
On the 9-track album, the Man with the Golden Voice takes us on a worship and faith journey with adulation, declarative songs… and yes! Party too! "My Beautifier" album forms the body of work that makes up chapter 2 of the "Alive in Worship" series.
Chris Shalom's "You are the Reason" album, which is the chapter 1, released in 2015 contained wave-making releases such as "Power Belongs to You", "God of Miracles & Wonders", "Unto the King", amongst others.
The new album is available on Boomplay Music for online access. It is marketed by Johnwealth Music (08062682523) nationwide.
Get "My Beautifier" album below:
http://www.boomplaymusic.com/share/playlist/1111362
See "My Beautifier" below:
1. You Are Worthy
2. Your Word Is Truth
3. I Bow My Knees
4. In The Name Of Jesus
5. Dry Bones Are Rising
6. Freedom
7. My Beautifier
8. Walls Are Falling Down
9. Holy Ghost Party
By TMOGL • Posted in Music, News • Tagged 6th, Chris Shalom, Get It Here, My Beautifier, Studio Album, Unveils
Chris Shalom Reveals Title Of Forthcoming 6th Studio Album!
Gospel music minister Chris Shalom is ready for the release of his 6th Studio album titled "My Beautifier".
After the release of the 5th studio album "You are the Reason", theGolden Voice, aka Chris Shalom embarked on the "Worship in Every Place" series. From the series, singles such as "My Beautifier", Dry Bones are Rising" were birthed and has now translated into the 6th studio album.
"My Beautifier" was not just successful in statistics with close to 2 million views on Youtube, but the impact rippled across the world, garnering feedback and testimonies. Little wonder Chris Shalom chose to name the 6th studio album after the track.
The new album will form the body of work that makes up chapter 2 of the "Alive in Worship" series. "You are the Reason" album, which is the chapter 1 was released in 2015. The wave-making album contains singles such as "Power Belongs to You", "God of Miracles & Wonders", "Unto the King", amongst others.
Chris Shalom will reveal more details on the forthcoming 6th studio album in the coming days. Keep your eyes peeled!
By TMOGL • Posted in News • Tagged 6th, Chris Shalom, Forthcoming, of, Reveals, Studio Album, title
Music: Chris Shalom – Dry Bones Are Rising
The man with the Golden Voice, Chris Shalom has a peculiar gift of interpreting the kairos into powerful life-giving music piece. This is why the new single "Dry Bones Are Rising" is a timely response – a confirmation to a sure word of prophecy.
Coming as the second release of the "Worship in Every Place" series, the song is Contemporary Gospel with a pulsating rhythm. Chris Shalom makes a profound declaration with his simple sing-along melody yet carries in-depth and precise lyrics!
Chris Shalom had hinted after the release of "My Beautifier", which became critically acclaimed almost immediately, that he will be releasing more singles off the "Worship in Every Place" series.
"Flesh to flesh, bone to bone… dry bones are rising in the name of Jesus" – Chris Shalom
Get the song on MUSIC+ below:
By TMOGL • Posted in Music • Tagged Chris Shalom, Dry Bones Are Rising
Music Video: Chris Shalom | My Beautifier [LIVE]
Worship is mostly about beholding the awesomeness of the Supreme One and when we begin to see the undistorted image of God, we can begin to see ourselves the way we are truly held in His eyes. The Golden Voice, Chris Shalom has crafted the perfect worship song that paints the perfect picture of our reflection looking through the eyes of God.
From the "Worship in Every Place" series, comes the amazing work 'My Beautifier' by Chris Shalom. An alluring expression of the Zoe kind of life, which can only be given by the finished work of Jesus. It is the first release from the series and Chris Shalom has hinted that more should be expected in the course of the year.
Quite typical of the Golden Voice, it is a simple sing along worship song but laced with Afrocentric rhythm and an accentuated lead guitar. Also, one cannot undermine the essence of the background vocals.
"For Every Pain and for every shame, Jesus has brought a turn around. He is given you beauty for ashes and clothing you with a garment of praise." – Chris Shalom
Directed by Marthins Harrison for MHFrame.
By TMOGL • Posted in Music, Video • Tagged Chris Shalom, LIVE, My Beautifier
New Music: Chris Shalom – Manifest Your Power
It is HERE! Are you ready for a glorious manifestation as the man with the Golden Voice, Chris Shalom unwraps another sound of divinity?
Like mount Zion, and the Mount of transfiguration, you can literally feel the manifest presence of God envelope the ambiance as the sounds that make up "Manifest Your Power" by Chris Shalom blares out.
Evoking the spirit of God with melodious chords, Chris Shalom uses rich vocals to render supplication, yet retaining the assurance in the power of God's presence.
Produced by J.moses.
By TMOGL • Posted in Music • Tagged Chris Shalom, Manifest Your Power
New Music: Chris Shalom – God of Miracle & Wonders [For Streaming Only]
Unexpected but it's here! In the spirit of the Easter season, Gospel music minister Chris Shalom has released the single "God of Miracle & Wonders".
The single is off the 5th studio album released in 2015 titled "You are the Reason". It speaks of the wondrous works of the Almighty and the fact that the one working these miracles and wonders is still alive and doing the same today.
Produced by Mayo, the single is only available for streaming.
By TMOGL • Posted in Music, Video • Tagged Chris Shalom, God of Miracle & Wonders, Streaming Only
New Video: Chris Shalom – Power Belongs To You
After the successful release of his 5th studio album "You are the Reason", it is only right that Gospel music minister Chris Shalom aka the Golden Voice crowns the year with the video for "Power Belongs to You" – the single he started the year with.
"Power Belongs to You", the phenomenal yet simple sing-along worship single ushered in the album "You are the Reason" and has stood out, gradually topping the records of Chris Shalom's widely accepted single "You are the Reason". The single has also broken the boundaries of religion and can be heard not only on alters and amongst Christians but also at public places thus garnering more commercial value as well.
The video directed by Marthins Harrison follows a simple plot to tell a compelling story of the power and sovereignty of the Almighty God.Chris Shalom is excited about the release of the video and believes it will take the song to a whole new meaning as people get inspired, motivated, encouraged and blessed all over again.
By TMOGL • Posted in Video • Tagged Chris Shalom, Power Belongs To You
Chris Shalom Crowns The Year With Video For "Power Belongs To You" – Watch Snippet
Gospel music minister Chris Shalom aka the Golden Voice kicked off the year with the release of the single "Power Belongs to You", a song that ushered in the album, "You are the Reason". A lot has happened in between and Chris has been keeping us up to speed on his itinerary, therefore one can say it's been an eventful year for the Gospel music minister, but before Chris Shalom takes an inventory of the year in view, he will be rounding off the eventful year by putting visuals to the single he kicked off the year with.
"Power Belongs to You" was released on the 15th of January, 2015 and true to the prediction, the single has topped the yardstick already set by the previous hit "You are the Reason" as you can find it on many playlists, mixtapes, and choirs are already making an anthem of it all over the world.
Shot by Marthins Harrison, the video follows a simple plot that tells a compelling story of the power of the Almighty God. Chris Shalom has set a Monday 14th of December release date for the video but for now, let's whet the appetite with the snippet:
Watch video snippet below:
By TMOGL • Posted in News, Video • Tagged Chris Shalom, Crowns, Power Belongs To You, The Year, Video, Watch Snippet
New Music: Chris Shalom – Unto The King
Here is the brand new single from one of Nigeria's revered gospel music ministers, Chris Shalom. The singer, writer, producer and music minister Chris Shalom has returned with yet another worship song "Unto The King". The song is a simple, easy-to-sing-along tune and it gives all glory, honor and praise to the King of All who is worthy of it all.
By TMOGL • Posted in Music • Tagged Chris Shalom, King, Music, new, the, Unto
Glowreeyah Braimah – 'Exalted'
Music: Ebere Uzoho – "Follow Me Praise the Lord" & "Hallelujah, Hossanna"
Video: Sammie Okposo – Chukwu Ebube (God Of Glory) Feat. Michael Stuckey
Music: Edem Siabi – Holy Spirit You Are Welcome
NEW MUSIC: JONATHAN YOHANNA – ENLIGHTENED
Video: Stella Joel Featured On The New Edition Of TWC Acoustic Sessions
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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소렌토, 이탈리아의 매그놀리아 소렌토 | 호스텔월드에서 예약하세요.
La magnolia is located near Piazza Torquato Tasso, centre of the city.
The position lets customers to use easily all services that the city can give (taxi, bus, trains and links with Naples and the islands Capri and Ischia). The structure is a part of a Historical and noble house.
The seven rooms are new (January 2005) and fine furnished with wooden furniture hand made by artisan master of the Sorrento's tradition. At the reception is a Caffeterie to serve Breakfast.
Double rooms all with Air conditioning, TV, mini-fridge, telephone, safe, toilet with shower box and hair dryer.
By Payments is possible to rent Boats, cars and motorbikes and there is a parking nearby the structure.
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{
"redpajama_set_name": "RedPajamaC4"
}
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\section{Introduction and background material}
\noindent Homogeneous power-laws, such as Newton's universal law of
gravitational attraction, for instance, abound in Nature. They
are, by definition, self-similar and thus true in all scales.
Systems statistically described by {\it power-law probability
distributions}
are rather ubiquitous \cite{cero} and thus
of perennial interest \cite{uno}. In this report we wish to give
careful scale-invariance consideration to systems that are
governed (or described) by a special kind of power-law probability
distribution functions (PDF), namely, the $q-$Gaussian function.
Consider a system $\mathcal{S}$ described by a vector $X$ with
$n$ components. We say that $X$ is $q-$Gaussian distributed if its
probability distribution function writes as described by
(\ref{eq:q>1gaussian})-(\ref{eq:Kq>1}) below.
It is well-known that for such $\mathcal{S}-$systems one can appeal to
Jaynes' maximum entropy principle (MaxEnt) \cite{dos} under $\,$
a covariance constraint with
a generalized (or $q-$) information measure
\begin{equation} \label{tsallis}
H_{q}\left(x\right)=(1-q)^{-1}\,\int\,dx\,[f(x)-f^{q}(x)];\,\,\,\,\,\,\,\,\,q
\in \mathcal{R}, \end{equation}
as the protagonist \cite{alfo}. This measure has been
found to be useful in extracting information pertaining to systems that are
characterized by either (1) fractal nature, (2) long-memory,
or
(3) long-range interactions \cite{dino}. Employing $H_q$ for {\it
other} types of system has generated controversy \cite{euro} which is of no
relevance to our present purposes. We will focus attention upon
properties of Gaussian systems that remain valid for their
q-Gaussian counterparts as well (as $q$ becomes different from
unity), with emphasis on the dimensional properties of both kinds
of systems. It is well-known that if a system ``is Gaussian", any
part (sub-system) of it is still Gaussian. This property holds for
$q-$Gaussian systems as well, as proved in \cite{hero}. {\it A
more interesting result is the inverse phenomenon}: a Gaussian
system may be part of a larger system that is not Gaussian.
However, if the larger system is spherically invariant, then it is
necessarily Gaussian again. Surprisingly enough, this ``inverse
property" has gone largely ignored in the statistical literature.
In this work we will not only provide a simple proof for it but we
will show that it can be extended to q-Gaussian
distributions as well.
These results can be given a physical interpretation within the
framework of the estimation of the parameter $q$ of a given system
\cite{dino}. We will prove that, if spherical symmetry prevails,
such estimation can be performed using only a {\it restricted},
observable part of the system and that the overall parameter $q$
for the entire system can be retrieved provided the dimension of
the system is known. We begin our considerations by introducing
the two basic notions, namely spherical symmetry and $q-$Gaussian
systems. We will also apply our results to a simple case of
kinetic theory via the Beck-Cohen superstatistics theory
\cite{B1,B2,B3,B4,B5,B6,B7}.
\subsection{Spherical symmetry}
A really momentous symmetry is that of invariance against
rotations. It is found in the fundamental laws of nature and
constitutes one of the most powerful principles in elucidating the
structure of individual atoms, complicated molecules, and entire
crystals. Also, it characterizes the shape of many systems. We
can cite self-gravitating systems like stars and planets, that
have quasi-spherical shape if their mass is large enough. Also,
many atomic nuclei are spherical, and many molecules as well, etc.
Conservation of angular momentum, a very frequent occurrence, is a
result of the isotropy of space itself \cite{margenau}. We
discuss now some properties of spherical probability
distributions.
The characteristic function associated with a random vector $X \in
\mathbb{R}^n$ is \begin{equation} \label{uno} \varphi_X(U)= E e^{iU^tX};\,\,\,
U \in \mathbb{R}^{n}. \end{equation} Under the hypothesis, discussed for
instance in the textbook \cite[XV.3]{feller}, that $ \varphi_{X}
\in \mathcal{L}_{1}(\mathbb{R}^n)$, there is a one-to-one relation
between $\varphi_{X}$ and the probability density function
$f_{X}$ of $X$. The random vector $X$ is said to have a spherical
distribution if its characteristic function $\varphi_X$ satisfies
\begin{equation}
\varphi_X(U)=\phi(\Vert U \Vert)
\end{equation}
for some scalar function $\phi : \mathbb{R}^+ \rightarrow \mathbb{R}$ which is
then called the characteristic generator of the spherical
distribution.
We will write $X \sim \mathcal{S}_n(\phi)$ in this case.
It is well-known that an equivalent definition for a spherical random vector $X$ is
\[
X \sim AX; \,\, \forall \, A \,\, \text{orthogonal}
\] where $\sim$ denotes equality in distribution.
Spherical random vectors have, as it is well-known, the following
properties:
\begin{enumerate}
\item All marginal distributions of a spherical distributed random vector
are spherical.
\item All marginal characteristic functions have
the same
characteristic generator.
\item If $X \sim \mathcal{S}_n(\phi)$ then
\begin{equation}
X \sim rT_{n}
\end{equation}
where $T_{n}$ is a random
vector distributed uniformly on the unit sphere surface in
$\mathbb{R}^n$ and $r$
is a positive random variable independent of
$T_{n}$.
\end{enumerate}
Let us remark that a spherically distributed random vector does
not necessarily possess a density.
A generalization of the concept of spherical distribution is given
by elliptical distributions, to which the multi-normal
distribution belongs. Elliptical distributions have recently
gained a lot of attention in financial mathematics, being of use
particularly in risk management. A $n-$random vector $Y$ is said
to have an elliptical distribution with so-called characteristic
matrix $C_Y (n \times n)$ if $Y \sim AX$, where $X \sim
\mathcal{S}_n(\phi)$ and $A$ is a $n \times n$ deterministic
matrix such that $A^tA=C_Y$ and $rank(C_{Y})=n$. We shall write
$Y \sim \mathcal{E}_{n}(C_Y,\phi)$.
\subsection{$q-$Gaussian systems}
An $n-$components vector $X$ is $q-$Gaussian distributed if its
PDF writes as follows \cite{uno}:
\begin{itemize}
\item
in the case $1<q<\frac{n+4}{n+2}$
\begin{equation}
f_{X}\left(X\right)=A_{q}\left(1+X^{t}
\Lambda^{-1}X\right)^{\frac{1}{1-q}},\label{eq:q>1gaussian}
\end{equation}
matrix $\Lambda$ being related to the covariance matrix
$K=EXX^{t}$ in the fashion \cite{vignat1}
\begin{equation}
\Lambda=\left(m-2\right)K.\label{eq:Kq>1}
\end{equation}
where the
number of degrees of freedom is defined as \cite{vignat1}
\begin{equation}
m=\frac{2}{q-1}-n.
\label{eq:mq>1}
\end{equation}
Moreover, the partition function $Z_q= 1/A_q$ reads
\cite{vignat1}
\[
Z_{q}=\frac{\Gamma\left(\frac{1}{q-1}-
\frac{n}{2}\right)\vert\pi\Lambda\vert^{1/2}}{\Gamma\left(\frac{1}{q-1}\right)}.
\]
and the characteristic function is
\begin{equation}
\varphi_{X}(U)
= \frac{2^{1-\frac{m}{2}}}{\Gamma(\frac{m}{2})}z^{\frac{m}{2}}K_{\frac{m}{2}}(z)
\end{equation}
with $z=\sqrt{U^t\Lambda U}$ and $K$ is the modified Bessel function of the second kind.
\item
in the case $q<1$
\begin{equation}
f_{X}\left(X\right)=A_{q}\left(1-X^{t}
\Sigma^{-1}X\right)_{+}^{\frac{1}{1-q}}\label{eq:q<1gaussian}
\end{equation}
with matrix $\Sigma=dK$ and parameter $d$ defined as
$d=2\frac{2-q}{1-q}+n.$ In this case, the partition function is
\[
Z_{q}=\frac{\Gamma\left(\frac{2-q}{1-q}\right)
\vert\pi\Sigma\vert^{1/2}}{\Gamma\left(\frac{2-q}{q-1}+\frac{n}{2}\right)}.
\]
and the characteristic function is
\begin{equation}
\label{chfq>1} \varphi_{X}(U) = 2^{\frac{d}{2}-1}
\Gamma(\frac{d}{2})\frac{J_{\frac{d}{2}-1}(z)}{z^{{\frac{d}{2}-1}}}
\end{equation}
where $z=\sqrt{U^t\Sigma U}$ and $J$ is the Bessel
function of the first kind.
\end{itemize}
We begin in the next Section to advance our present results.
\section{Size behavior of the $\,$ $q\,-$parameter }
Our first result revolves around the behavior of the
non-extensivity parameter $q$ as a function of the dimension of
the system and is embodied in the following theorem, the proof
of which is given in the Appendix.
\begin{theorem}
\label{thm1} Assume that a system $X_n\in\mathbb{R}^{n}$ follows a $q-$Gaussian
distribution with parameter
$q_{n}$; then with $1\le k\le n-1,$ any $k-$dimensional subsystem
$X_{k}=\left[x_{1},\dots,x_{k}\right]^{t}$ of $X_n$ is $q-$Gaussian distributed with parameter
\begin{equation}
\label{qk}
q_{k}=1-\frac{2\left(1-q_{n}\right)}{2+\left(n-k\right)\left(1-q_{n}\right)}
\end{equation}
Reciprocally, assume that a $n-$dimensional and spherical system
$X_n$ contains a $k-$dimensional subsystem $X_{k}$ that follows a
$q-$Gaussian distribution with parameter $q_{k}$; then system $X_n$
is itself $q-$Gaussian distributed with parameter $q_{n}$ defined
as in (\ref{qk})
\end{theorem}
Recall that
for any
$n-$dimensional orthogonal transformation $A$, there exists an
orthogonal decomposition of $\mathbb{R}^n$ as
\[
\mathbb{R}^{n}=E \oplus F \oplus G_1 \oplus \dots \oplus G_k
\]
into stable subspaces, the restriction of $A$ to each subspace being
\begin{itemize}
\item the identity transformation for subspace $E$
\item minus the identity transformation for subspace $F$
\item a two-dimensional
planar rotation of angle $\theta_k$ for subspace $G_k$
\end{itemize}
Moreover, $n-$dimensional
$q-$Gaussian distributions with parameter $q_{n}$ arise in
statistical physics as the canonical distributions of systems with
maximal $q-$entropy $H_q$ of order $q_{n}$
and fixed covariance matrix. The result of theorem
(\ref{thm1}) can be paraphrased in this way: if a system of
dimension $k$ has maximum $q-$entropy of order $q_{k}$, and is
also part of a spherical system of dimension $n>k$, then the
whole system maximizes the $q-$entropy with parameter $q_{n}$
related to $q_{k}$ as in (\ref{qk}). Notice that in (\ref{qk}), as
$q_{k}\rightarrow1$ then $q_{n}\rightarrow1$ and we deduce that if
a spherical system has a Gaussian part, then it is
necessarily Gaussian as well.
We note also that $q_{n}<1$ implies $q_{k}<1$ while $q_{n}>1$
implies $q_{k}>1.$ This result is natural since cases $q>1$ and
$q<1$ correspond to two different types of distributions:
according to Beck and Cohen's superstatistics principle
\cite{B1,B2,B3,B4,B5,B6,B7}:
\begin{itemize}
\item $q>1$ that
corresponds to a Gaussian system subjected to fluctuations that
are independent of the state of the system.
\item $q<1$ that
corresponds again to a fluctuating Gaussian system for which the
amplitude of the fluctuations depends on the system's state.
\end{itemize}
Alternatively, these two cases can be characterized as follows.
Our system is here described by the random vector $X$: if $q>1$,
then $X$ has unbounded support, contrarily to the bounded support
associated to the case $q<1$ \cite{vignat1}.
Thus, we obtain the following rather natural result: {\it
reduction or enlargement of a $q-$Gaussian system does not change
its superstatistical nature}.
\section{A second result: multi-component systems}
\subsection{Average behaviour}
It may happen that measuring the behavior of only one component
$x_{1}$ of a large system $X \in \mathbb{R}^n$ is not physically
feasible, and one has to content oneself with measuring instead
the behavior of a superposition of contributions from (or average
of) several components \cite{roybook}, in the form
\begin{equation}
\label{linearcombination}
<X> = \sum_{i=1}^{n}a_{i}x_{i}
\end{equation}
where the deterministic coefficients $a_{i} \in \mathbb{R}$
characterize the measurement device. The following theorem (see
Ref. \cite{fang}) allows to give a special characterization of
this average value in the case of spherical systems.
\begin{theorem}
\label{thm2} \cite[Th. 2.4]{fang} If $X\in\mathbb{R}^n$ is
spherically distributed and $A=(a_1,\dots,a_n)^t$ is a deterministic
vector then
\[
<X> = \sum_{i=1}^{n}a_{i}x_{i}
\]
is distributed as $\Vert A \Vert x_{1},$ where $\Vert A \Vert$ is the Euclidean norm of $A.$
\end{theorem}
In the next subsection we extract rather interesting physical
conclusions from this theorem.
\subsection{Application to the estimation of $q$}
An important problem in non-extensive statistics is the estimation
of the non-extensivity parameter $q_{n}$ associated to an
$n-$dimensional system that follows a $q-$Gaussian distribution
\cite{dino}. We provide here some hints about a possible
estimation strategy in the case $q>1$, assuming that we have
access to averaged measures of the system of the type
(\ref{linearcombination}).
Assuming $q_n>1,$ then if $X_n$ follows distribution
(\ref{eq:q>1gaussian}), the averaged measure $<X>$ is distributed
as
\[
f_{<X>}(x)=\frac{A_{q_{1}}}{\Vert A \Vert}(1+\frac{x^2}{\lambda
\Vert A \Vert^2})^{-\frac{m+1}{2}}
\]
with $m=2/(q_{1}-1)-1$ and $\lambda=\Lambda_{1,1}.$ As a consequence, a possible estimation
strategy of parameter $q_{n}$ follows the three following steps:
\begin{enumerate}
\item since the ``tail-behavior" of the distribution of $f_{<X>}$ is
\[
f_{<X>}(x) \sim x^{-(m+1)},
\]
(where $\sim$ means here asymptotic equivalence), parameter $m$ can be estimated as the L\'{e}vy exponent of the
distribution of the average measure of the system \cite{note}
\item the non-extensivity parameter $q_{1}$
of $<X>$ can be computed using (\ref{eq:mq>1}) as
\[
q_{1}=\frac{m+3}{m+1}
\]
\item the non-extensivity
parameter $q_{n}$ of the $n-$dimensional system $X_n$ can in
turn be deduced using (\ref{qk}) as
\[
q_{n} =\frac{2-(n+1)(1-q_{1})}{2-(n-1)(1-q_{1})}
\]
\end{enumerate}
As a new result we find that if the dimension $n$ of the system
is known, its non-extensivity parameter $q_{n}>1$ can be
evaluated from any measurement of the type
(\ref{linearcombination}).
\subsection{A kinetic application}
\subsubsection{Theoretical framework}
Another application of the latter result can be provided in the
context of the kinematics of collision events. We envision a
scenario in which attention is focused
on the particles of a
system interacting with a heat bath (a fundamental problem in
thermodynamics). An elastic collision between (i) a system's
particle with momentum $P$, mass $M$, velocity $V$, and energy $E$
and (ii) a particle from the heat bath with momentum $p$, mass
$m$, velocity $v$, and energy $\epsilon$, verifies \cite{dunkel}
\begin{eqnarray*}
E + \epsilon & = \hat{E} + \hat{\epsilon} \\
P + p & = \hat{P} + \hat{p}
\end{eqnarray*}
where ``hats" refer to quantities after the collision.
In the non-relativistic case, these quantities write
\begin{eqnarray*}
P &= MV, \hspace{1cm} p=mv \\
E & = \frac{\Vert P \Vert ^{2}}{2M}, \hspace{1cm} \epsilon = \frac{\Vert p \Vert ^{2}}{2m},
\end{eqnarray*}
where momenta are $3-$dimensional quantities. These equations can
be solved as
\begin{eqnarray*}
\hat{P}(p,P) &= \left(\frac{2M}{M+m}\right)p + \left(\frac{M-m}{M+m}\right)P \\
\hat{p}(p,P) &= \left(\frac{m-M}{M+m}\right)p + \left(\frac{2m}{M+m}\right)P
\end{eqnarray*}
Assuming that $P$ and $p$ are independent random
variables, we look for stationary distributions for $p$ and $P$,
that is, for probability density functions $f_{p}$ and $f_{P}$
such that if $p\sim f_{p}$ and $P\sim f_{P}$ then after the
collision, $\hat{p}\sim f_{p}$ and $\hat{P} \sim f_{P}$. An
obvious pair of stationary solutions is given \cite{dunkel} by the
independent Maxwell solutions
\begin{equation}
f_{P}(P) = \frac{1}{\left( 2\pi M k_{B}T \right)^{3/2}}
\exp\left({-\frac{\Vert P \Vert ^{2}}{2Mk_{B}T}}\right), \,\, f_{p}(p) =
\frac{1}{\left( 2\pi m k_{B}T \right)^{3/2}}
\exp\left({-\frac{\Vert p \Vert ^{2}}{2mk_{B}T}}\right).
\end{equation}
\subsubsection{The correlated scenario}
Suppose however that the assumption of independence between
momenta $p$ and $P$ does not hold. Such is the case, for example,
when the corresponding particles are subject to the same
fluctuations (an interpretation for this scenario is provided in
the following Subsection). In this instance we look for a
stationary joint distribution for $p$ and $P$, i.e., for a
probability density function $f_{p,P}$ such that if $(p,P)\sim
f_{p,P}$ then, after the collision, $(\hat{p},\hat{P}) \sim
f_{p,P}$ again.
We note that this in turn implies $\hat{p} \sim
f_{p}$ and $\hat{P} \sim f_{P}.$
We need here an extension of Theorem \ref{thm2} as given below,
the proof of which can be found, for example, in \cite{chu}.
\begin{proposition}
If $X\sim \mathcal{E}_n(C_X,\phi)$ and $A$ is a full-rank
$(n \times n)$ matrix
then $Y=AX \sim \mathcal{E}_n(C_Y,\phi)$ with
\[
C_Y=AC_XA^t.
\]
\end{proposition}
Now assume that
\[
C_X = \left[\begin{array}{cc} m I_3 & 0_3\\ 0_3 &
M I_3\end{array}\right], \,\, A =\frac{1}{m+M} \left[\begin{array}{cc}
(m-M)I_3 & (2m)I_3\\ (2M)I_3 & (M-m)I_3\end{array}\right]
\]
where $0_3$ denotes the $(3\times3)$ null matrix and $I_3$ the $(3\times3)$ identity matrix. Then
\[
C_Y=AC_{X}A^t=C_X.
\]
We are now in a position to deduce the following result:
\begin{theorem}
If $\left(\begin{array}{c}p \\P\end{array}\right) \sim
\mathcal{E}_6(C_X,\phi)$ with characteristic matrix $C_X$ as
above, then the momenta vector after the collision
$\left(\begin{array}{c}\hat{p} \\ \hat{P}\end{array}\right) \sim
\mathcal{E}_6(C_X,\phi).$ As a consequence, any elliptical joint
distribution with characteristic matrix $C_X$ is stationary. In
particular, $p$ and $\hat{p}$ have the same distribution, as well
as $P$ and $\hat{P}.$
\end{theorem}
\subsubsection{Superstatistics at work}
A more physical interpretation can be given to the preceding
result, using the notion of superstatistics \cite{B1}.
We know from \cite{dunkel}
that a pair of independent Gaussian momenta are stationary for the
collision process. Now,
\begin{itemize}
\item if $p\sim \mathcal{N}_{3}(mk_{B}T)$ (the $3-$dimensional
Gaussian distribution with covariance matrix $mk_{B}T I_3$) - and
\item $P\sim \mathcal{N}_{3}(Mk_{B}T)$, then \item $\hat{p}\sim
\mathcal{N}_{3}(mk_{B}T)$ and $\hat{P}\sim
\mathcal{N}_{3}(Mk_{B}T)$. \end{itemize} Since
\[
\hat{P}(p,P) = \left(\frac{2M}{M+m}\right)p +
\left(\frac{M-m}{M+m}\right)P,
\]
choosing any (dimensionless) random variable $a$ independent of
both $p$ and $P$, and defining the new quantities $q=ap$, $Q=aP$,
and $\hat{Q}=a\hat{P}$, we deduce that
\[
\hat{Q}(q,Q) = \left(\frac{2M}{M+m}\right)q +
\left(\frac{M-m}{M+m}\right)Q,
\]
so that, obviously, the pair $(q,Q)$ is another couple of momenta
whose distribution is stationary. Obviously, variables $q=ap$ and $Q=aP$
are not independent ones, since they share the same random factor
$a$ (unless $a$ is almost surely a constant, which reduces to the
Gaussian case).
As a special case,
if $a$ follows an inverse chi-distribution with $m$ degrees of freedom,
one immediately finds ~\cite{B1} that the random vector
$X=\left(\begin{array}{c}p \\P\end{array}\right)$ follows a Tsallis-distribution
\begin{equation}
f_{X}\left(X\right)=A_{q}\left(1+X^{t}
\Lambda^{-1}X\right)^{\frac{1}{1-q}},
\end{equation}
with non-extensivity index $q>1$ related to parameter $m$ as in
formula (\ref{eq:mq>1}) with $n=6$. We remark that the random
variable $a$ can be interpreted, in such a context, as
representing temperature's fluctuations, as shown by Beck and
Cohen~\cite{B1}. Thus, the
presence of temperature fluctuations
indicates that the momenta of the incoming colliding particles are
correlated. Conversely, if they are correlated, then temperature
fluctuations ensue. This scenario is a feasible one if the heat
bath is a finite one, which, in turn, establishes a natural
connection with an old result of Plastino and Plastino
\cite{gibbscano}.
\section{Conclusions}
In this work we have considered physical applications of a
largely ignored result of the statistical literature: {\it a
Gaussian system may be part of a larger system that is not
Gaussian. However, if this larger system is spherically invariant,
then it is necessarily Gaussian again}.
We have provided a simple proof for it and we have shown that it
can be extended to $q-$Gaussian distributions as
well. Our results have been given a physical interpretation
within the framework of the problem of estimation of the q-Gaussian
parameter
$q$. Also, we applied them to a simple instance of kinetic theory
involving Beck and Cohen superstatistics \cite{B1}.
\section{Appendix: Proofs}
\subsection{Proof of theorem \ref{thm1}}
We give here a simple proof of theorem (\ref{thm1}), the
principle of which has been kindly suggested to us by Pr. Wlodek Bryc
and Pr. Jacek Wesolowski. Assuming first that
$X_{k}=[x_1,\dots,x_k]^{t}$ is $q-$Gaussian with parameter
$q_{k}>1$, we deduce that, with $U_{k}=[u_{1},\dots,u_{k}]^t,$
\[
\varphi_{X_k}(U_k) =\phi(\Vert U_{k} \Vert) =
\frac{2^{1-\frac{m}{2}}}{\Gamma(\frac{m}{2})}\Vert U_{k}
\Vert^{\frac{m}{2}}K_{\frac{m}{2}}(\Vert U_{k} \Vert)
\]
so that
\[
\phi(u)= \frac{2^{1-\frac{m}{2}}}{\Gamma(\frac{m}{2})}\Vert u
\Vert ^{\frac{m}{2}}K_{\frac{m}{2}}(\Vert u \Vert)
\]
and
\[
\varphi_{X_n}(U_{n}) =\phi(\Vert U_{n}
\Vert)=\frac{2^{1-\frac{m}{2}}}{\Gamma(\frac{m}{2})}\Vert U_{n}
\Vert ^{\frac{m}{2}}K_{\frac{m}{2}}(\Vert U_{n} \Vert)
\]
and $X_{n}$ is $q-$Gaussian with dimension $n$ and $m$ degrees of
freedom, and thus has non-extensivity parameter $q_{n}$ such that
\[
m=\frac{2}{q_{k}-1}-k=\frac{2}{q_{n}-1}-n .\] The same result
applies with $q_{k}<1$ by considering characteristic function
defined as in (\ref{chfq>1}).
\subsection{An alternate proof of theorem \ref{thm1}}
We provide here an alternate proof based on stochastic
representations, as first used in \cite{liang}, extending their
result to the case $q>1$. Assume that $X_{k} \in \mathbb{R}^{k}$
is $q-$Gaussian distributed with parameter $q_{k}$: a stochastic
representation of $X_{k}$ is (see \cite{fang})
\[
X_{k}=\frac{\chi_{k}}{\chi_{d}}Z_{k}
\]
where $\chi_{k}$ and $\chi_{d}$ are independent and chi
distributed random variables, where $d=\frac{2}{q_{k}-1}-k$ and
$Z_k$ is uniform on the sphere. We know moreover that if
$X_{n}=rZ_{n}$ then
\[ X_{k}=rd_{1}Z_{k}\] where
$d_{1}^{2}\sim\beta_{\frac{k}{2},\frac{n-k}{2}}.$ We deduce that\[
rd_{1}Z_{k}=\frac{\chi_{k}}{\chi_{d}}\] or\[
r^{2}d_{1}^{2}=\frac{\chi_{k}^{2}}{\chi_{d}^{2}}.\] But by Luckacs
theorem \cite{lucas}:
\[
\frac{\hat{\chi}_{n}^{2}}{\hat{\chi}_{d}^{2}}
\frac{\tilde{\chi}_{k}^{2}}{\tilde{\chi}_{k}^{2}
+\tilde{\chi}_{n-k}^{2}}=\frac{\chi_{k}^{2}}{\chi_{d}^{2}}\] we
deduce that \[ r=\frac{\chi_{n}}{\chi_{d}}\] and that
$X_{n}=rZ_{n}$ is $q-$Gaussian distributed with parameter $q_{n}=
1+\frac{2}{d+n}=$$1+\frac{2\left(q_{k}-1\right)}{2+\left(n-k\right)\left(q_{k}-1\right)}.$
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,825
|
Мікроліт (археологія)
Мікроліт (мінерал)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,337
|
import filecmp
from pprint import pprint
import json
from os.path import normpath
class DirDifferences():
def __init__(self, dir1: str, dir2: str) -> None:
self.dir1 = dir1
self.dir2 = dir2
self.dirs_cmp = filecmp.dircmp(dir1, dir2)
self.global_diffs = []
def get_subdir_diffs(self, sd):
'''Report on self and subdirs in recursively'''
if sd.left_only or sd.right_only:
cur_diff = {sd.left.replace('\\','/'): sd.left_only, sd.right.replace('\\','/'): sd.right_only}
pprint(cur_diff)
self.global_diffs.append(cur_diff)
if sd.subdirs.values():
for sd in sd.subdirs.values():
self.get_subdir_diffs(sd)
if __name__ == '__main__':
output = normpath('c:/users/jesse/desktop/diffs.json')
onedrive = normpath('C:/Users/jesse/OneDrive/Dropbox_2018_09_13')
dropbox = normpath('D:/Dropbox')
dir_diffs = DirDifferences(onedrive, dropbox)
dir_diffs.get_subdir_diffs(dir_diffs.dirs_cmp)
diffs_json = open(output, 'w')
json.dump(dir_diffs.global_diffs, diffs_json)
diffs_json.close()
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,737
|
White Women, Black Women: The Rub
March 14, 2018 / Amanda Kemp
At We the People 14 Sneak Preview
Let me first say that this topic has been brewing in me for a long time. My intention here is to air the hurt and engage in a dialogue that takes us somewhere.
When I was in grad school a good friend of mine was Caucasian. She and I lived together and had class together and supported each other through difficult times. We were close. But one day, my friend told me about an experience where the police stopped her for some minor infraction and ticketed her. She was enraged that she had been given a ticket. At first I sympathized because we were broke, but then as she went on about it I realized it was not just the money. There was something deeper going on. The police man had challenged her innocence, and this innocence was her birthright as a Caucasian middle class woman in America. She expected to be treated as Good.
Here is my contention: My Caucasian woman friend was offended that her tears, her apology, her whiteness and femaleness, did not stop the policeman from punishing her.
Here is the rub: I don't expect to be treated as the embodiment of innocence; everywhere around me are implicit messages that I am guilty. I deal every day with the presumption that I'm Bad.
Therefore, as she bemoaned her predicament, I got mad. I pulled away from her emotionally and resented her. A lot.
But, all this went unsaid. I didn't have the emotional space to work it through so it just sat there between us.
Fast forward about twenty years. I lived in a mixed race neighborhood and spent a lot of time with my neighbor who had four kids and I had two. She was Caucasian and Christian. After years of friendship and mutual support, we were walking and talking. She confided in me that she didn't understand why an African American woman mutual acquaintance was resisting her expertise on West Africa. My neighbor had lived in many parts of Africa for years. I said something like "A lot of Black people don't want White people mediating between them and Africa." What she heard was that she was a racist and that she was a bad person. We then had a conversation in which I explained and she defended, and then we parted. Eventually we did resume our friendship, but that failed conversation was like a rock in my gut. Once again her "innocence," a Caucasian woman's presumed position of inherent Goodness was challenged, and she fought hard to reclaim it.
Now really fast forward and you have the women's march on Trump's inauguration day. Remember the Twitter image of a Black woman holding a sign that said "53% of White Women Voted for Trump." Did you read the comments? Predictably, they broke down along race lines. White women proclaimed their innocence and Black women proclaimed their complicity in oppression.
I say all of this because if we are to have sisterhood, if we are to have authentic relationships between equals, then Caucasian women have to start seeing themselves as powerful agents of oppression and liberation. (Congressional hopeful Jess King pictured above gets this.)
Another way to put this is: You can't have all the bene's of white supremacy AND claim your innocence from it too.
Here's what you can do if you are Caucasian. Educate yourself about white supremacy culture, white fragility, white frame of reference, and unconscious bias. Watch yourself. Not to blame or shame but to SEE. Then accept. Then forgive. Then take inspired action. Sounds simple? It is. But simple ain't easy.
If you are a Black woman, a woman of African descent, don't give white women who say they stand for equity, inclusion or justice a pass--but try not to speak from a gotcha standpoint. It's more like helping someone to be all that they want to be. And all of us fall short in that department some of the time.
Peace and Love and Happy Women's History Month!
-Amanda
P.S.--If you want to do the work to Reflect on Yourself go here.
March 14, 2018 / Amanda Kemp/ 2 Comments
Black Panther Beauty
That man with Freedom in his eye...
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,871
|
\section{Introduction}
Graphs of massive size are used for modeling complex systems that emerge in many different fields of study. Challenges arise when computing with massive graphs under memory constraints. In recent years, graph streaming has become an important model for computation on massive graphs. Many space-efficient streaming algorithms have been designed for solving classical graph problems, including connectivity \cite{ahn2012analyzing}, bipartiteness \cite{ahn2012analyzing}, minimum spanning tree \cite{ahn2012analyzing}, matching \cite{epstein2011improved, kapralov2013better,ahn2013linear}, spectral sparsifiers \cite{kelner2013spectral, kapralov2017single}, etc. We will define the streaming model in Section~\ref{introstreaming}.
Random walks on graphs are stochastic processes that have many applications, such as connectivity testing \cite{reingold2008undirected}, clustering \cite{spielman2013local,andersen2007using, andersen2009finding, charikar2003better}, sampling \cite{jerrum1986random} and approximate counting \cite{jerrum1989approximating}.
Since random walks are a powerful tool in algorithm design, it is interesting to study them in the streaming setting.
A natural problem is to find the space complexity of simulating random walks in graph streams.
Das~ Sarma~et~al.~\cite{sarma2011estimating} gave a multi-pass streaming algorithm that simulates a $t$-step random walk on a directed graph using $O(\sqrt{t})$ passes and only $O(n)$ space. By further extending this algorithm and combining with other ideas, they obtained space-efficient algorithms for estimating PageRank on graph streams.
However, their techniques crucially rely on reading multiple passes of the input stream.
In this paper, we study the problem of simulating random walks in the \textit{one-pass} streaming model. We show space lower bounds for both directed and undirected versions of the problem, and present algorithms that nearly match with the lower bounds. We summarize our results in Section~\ref{secres}.
\subsection{One-pass streaming model}
\label{introstreaming}
Let $G=(V,E)$ be a graph with $n$ vertices.
In the insertion-only model, the input graph $G$ is defined by a stream of edges $(e_1,\dots,e_m)$ seen in arbitrary order, where each edge $e_i$ is specified by its two endpoints $u_i,v_i \in V$.
An algorithm must process the edges of $G$ in the order that they appear in the input stream.
The edges can be directed or undirected, depending on the problem setting.
Sometimes we allow multiple edges in the graph, where the multiplicity of an edge equals its number of occurrences in the input stream.
In the turnstile model, we allow both insertion and deletion of edges. The input is a stream of updates $((e_1,\Delta_1),(e_2,\Delta_2),\dots)$, where $e_i$ encodes an edge and $\Delta_i \in \{1,-1\}$. The multiplicity of edge $e$ is $f(e)=\sum_{e_i=e}\Delta_i$. We assume $f(e) \ge 0$ always holds for every edge $e$.
\subsection{Random walks}
Let $f(u,v)$ denote the multiplicity of edge $(u,v)$.
The degree of $u$ is defined by $d(u) =\sum_{v\in V}f(u,v)$.
A $t$-step random walk starting from a vertex $s \in V$ is a random sequence of vertices $v_0,v_1,\dots,v_t$ where $v_0=s$ and $v_i$ is a vertex uniformly randomly chosen from the vertices that $v_{i-1}$ connects to, i.e., $\P[v_i = v|v_{i-1} = u] = f(u,v)/d(u)$.
Let $\mathcal{RW}_{s,t} : V^{t+1} \to [0,1]$ denote the distribution of $t$-step random walks starting from $s$, defined by\footnote{For a statement $p$, define $\mathbf{1}[p] = 1$ if $p$ is true, and $\mathbf{1}[p]=0$ if $p$ is false. }
\begin{equation}
\mathcal{RW}_{s,t}(v_0,\dots,v_t) = \mathbf{1}[v_0=s] \prod_{i=0}^{t-1} \frac{f(v_i,v_{i+1})}{d(v_i)}.
\end{equation}
For two distributions $P,Q$, we denote by $|P-Q|_1$ their $\ell_1$ distance.
We say that a randomized algorithm can simulate a $t$-step random walk starting from $v_0$ within error $\varepsilon$, if the distribution $\P_w$ of its output $w\in V^{t+1}$ satisfies $|\P_w - \mathcal{RW}_{v_0,t}|_1 \le \varepsilon$. We say the random walk simulation is perfect if $\varepsilon=0$.
We study the problem of simulating a $t$-step random walk within error $\varepsilon$ in the streaming model using small space.
We assume the length $t$ is specified at the beginning.
Then the algorithm reads the input stream.
When a query with parameter $v_0$ comes, the algorithm should simulate and output a $t$-step random walk starting from vertex $v_0$.
It is without loss of generality to assume that the input graph has no self-loops. If we can simulate a random walk on the graph with self-loops removed, we can then turn it into a random walk of the original graph by simply inserting self-loops after $u$ with probability $d_{\text{self}}(u)/d(u)$. The values $d_{\text{self}}(u),d(u)$ can be easily maintained by a streaming algorithm using $O(n)$ words.
The random walk is not well-defined when it starts from a vertex $u$ with $d(u)=0$. For undirected graphs, this can only happen at the beginning of the random walk, and we simply let our algorithm return \textsc{Fail} if $d(v_0)=0$. For directed graphs, one way to fix this is to continue the random walk from $v_0$, by adding an edge $(u,v_0)$ for every vertex $u$ with $d(u)=0$.
We will not deal with $d(u)=0$ in the following discussion.
\subsection{Our results}
\label{secres}
We will use $\log x = \log_2 x$ throughout this paper.
The following two theorems give space lower bounds on directed and undirected versions of the problem.
Note that the lower bounds hold even for simple graphs\footnote{A \textit{simple} graph is a graph with no multiple edges.}.
\begin{theorem}
For $t \le n/2$, simulating a $t$-step random walk on a simple directed graph in the insertion-only model within error $\varepsilon = \frac{1}{3}$ requires $\Omega(nt \log (n/t))$ bits of memory.
\end{theorem}
\begin{theorem}
For $t = O(n^2)$, simulating a $t$-step random walk on a simple undirected graph in the insertion-only model within error $\varepsilon=\frac{1}{3}$ requires $\Omega(n\sqrt{t})$ bits of memory.
\end{theorem}
Theorem~\ref{maindirected} and Theorem~\ref{mainundirected} give near optimal space upper bounds for the problem in the insertion-only streaming model.
\begin{theorem}
\label{maindirected}
We can simulate a $t$-step random walk on a directed graph in the insertion-only model perfectly using $O(nt)$ words\footnote{A \textit{word} has $\Theta(\log \max\{n,m\})$ bits.} of memory. For simple directed graphs, the memory can be reduced to $O(nt\log (n/t))$ bits, assuming $t \le n/2$.
\end{theorem}
\begin{theorem}
\label{mainundirected}
We can simulate a $t$-step random walk on an undirected graph in the insertion-only model within error $\varepsilon$ using $O\left ( n \sqrt{t} \cdot \frac{q}{\log q} \right )$
words of memory, where $q = 2+\frac{\log(1/\varepsilon)}{\sqrt{t}}$. In particular, the algorithm uses $O(n\sqrt{t})$ words of memory when $\varepsilon = 2^{-\Theta\left (\sqrt{t}\right )}$.
\end{theorem}
Our algorithms also extend to the turnstile model.
\begin{theorem}
We can simulate a $t$-step random walk on a directed graph in the turnstile model within error $\varepsilon$ using $O(n(t+ \log\frac{1}{\varepsilon})\log^2 \max\{n,1/\varepsilon\})$ bits of memory.
\end{theorem}
\begin{theorem}
We can simulate a $t$-step random walk on an undirected graph in the turnstile model within error $\varepsilon$ using $O(n(\sqrt{t}+ \log\frac{1}{\varepsilon})\log^2 \max\{n,1/\varepsilon\})$ bits of memory.
\end{theorem}
\section{Directed graphs in the insertion-only model}
The simplest algorithm uses $O(n^2)$ words of space (or only $O(n^2)$ \textit{bits}, if we assume the graph is simple) to store the adjacency matrix of the graph.
When $t\ll n$, a better solution is to use reservoir sampling.
\begin{lemma}[Reservoir sampling]
Given a stream of $n$ items as input, we can uniformly sample $m$ of them without replacement using $O(m)$ words of memory.
\end{lemma}
We can also sample $m$ items from the stream \textit{with} replacement in $O(m)$ words of memory using $m$ independent reservoir samplers each with capacity 1.
\begin{theorem}
\label{algdirected}
We can simulate a $t$-step random walk on a directed graph in the insertion-only model perfectly using $O(nt)$ words of memory.
\end{theorem}
\begin{proof}
For each vertex $u\in V$, we sample $t$ edges $e_{u,1},\dots,e_{u,t}$ outgoing from $u$ with replacement. Then we perform a random walk using these edges. When $u$ is visited for the $i$-th time ($i \le t$), we go along edge $e_{u,i}$.
\end{proof}
By treating an undirected edge as two opposite directed edges, we can achieve the same space complexity in undirected graphs.
Now we show a space lower bound for the problem. We will use a standard result from communication complexity.
\begin{definition}
In the \textsc{Index} problem, Alice has an $n$-bit vector $X \in \{0,1\}^n$ and Bob has an index $i \in [n]$.
Alice sends a message to Bob, and then Bob should output the bit $X_i$.
\end{definition}
\begin{lemma}[\cite{miltersen1998data}]
\label{index}
For any constant $1/2<c\le 1$, solving the \textsc{Index} problem with success probability $c$ requires sending $\Omega(n)$ bits.
\end{lemma}
\begin{theorem}
For $t \le n/2$, simulating a $t$-step random walk on a simple directed graph in the insertion-only model within error $\varepsilon = \frac{1}{3}$ requires $\Omega(nt \log (n/t))$ bits of memory.
\end{theorem}
\begin{proof}
We prove by showing a reduction from the \textsc{Index} problem. Before the protocol starts, Alice and Bob agree on a family $\mathcal{F}$ of $t$-subsets of $[n]$ \footnote{Define $[n] = \{1,2,\dots,n\}$. A $t$-subset is a subset of size $t$.} such that the condition $|S \cap S'| < t/2$ is satisfied for every $S,S'\in \mathcal{F}, S\neq S'$.
For two independent uniform random $t$-subsets $S,S'\subseteq [n]$, let $p=\P[|S\cap S'|\ge t/2] \le \binom{t}{t/2} (\frac{t}{n})^{t/2}<(\frac{4t}{n})^{t/2}$.
By union bound over all pairs of subsets, a randomly generated family $\mathcal{F}$ satisfies the condition with probability at least $1- \binom{|\mathcal{F}|}{2} p$, which is positive when $|\mathcal{F}| = \lceil \sqrt{1/p} \rceil \ge (\frac{n}{4t})^{t/4}$.
So we can choose such family $\mathcal{F}$ with $\log |\mathcal{F}| = \Omega(t\log(n/t))$.
Assume $|\mathcal{F}|$ is a power of two. Alice encodes $n \log | \mathcal{F}|$ bits as follows. Let $G$ be a directed graph with vertex set $\{v_0,v_1,\dots,v_{2n}\}$. For each vertex $u \in \{v_{n+1},v_{n+2},\dots,v_{2n}\}$, Alice chooses a set $S_u \in \mathcal{F}$, and inserts an edge $(u,v_i)$ for every $i\in S_u$.
Suppose Bob wants to query $S_u$. He adds an edge $(v,u)$ for every $v \in \{v_0,v_1,v_2,\dots,v_n\}$, and then simulates a random walk starting from $v_0$.
The random walk visits $u$ every two steps, and it next visits $v_i$ for some random $i\in S_u$.
At least $t/2$ different elements from $S_u$ can be seen in $2t$ samples with probability at least $1-\binom{t}{t/2} (\frac{1}{2})^{2t} \ge 1-2^{-t}$, so $S_u$ can be uniquely determined by an $O(t)$-step random walk (simulated within error $\varepsilon$) with probability $1-2^{-t}- \frac{\varepsilon}{2}>\frac{1}{2}$.
By Lemma~\ref{index}, the space usage for simulating the $O(t)$-step random walk is at least $\Omega(n \log |\mathcal{F}|) = \Omega(nt \log(n/t))$ bits. The theorem is proved by scaling down $n$ and $t$ by a constant factor.
\end{proof}
For simple graphs, we can achieve an upper bound of $O(nt\log(n/t))$ bits.
\begin{theorem}
For $t\le n/2$, we can simulate a $t$-step random walk on a simple directed graph in the insertion-only model perfectly using $O(nt\log (n/t))$ bits of memory.
\end{theorem}
\begin{proof}
For every $u\in V$, we run a reservoir sampler with capacity $t$, which samples (at most) $t$ edges from $u$'s outgoing edges \textit{without} replacement.
After reading the entire input stream, we begin simulating the random walk.
When $u$ is visited during the simulation, in the next step we choose at random an outgoing edge used before with probability $d_{\text{used}}(u)/d(u)$, or an unused edge from the reservoir sampler with probability $1-d_{\text{used}}(u)/d(u)$, where $d_{\text{used}}(u)$ is the number of edges in $u$'s sampler that are previously used in the simulation.
We maintain a $t$-bit vector to keep track of these used samples.
The number of different possible states of a sampler is at most $\sum_{0\le i \le t}\binom{n}{i} \le (t+1)(\frac{en}{t})^t $, so it can be encoded using $\Big \lceil \log \Big ((t+1)(\frac{en}{t})^t\Big ) \Big \rceil = O(t \log(n/t))$ bits. The total space is $O(nt \log(n/t))$ bits.
\end{proof}
\section{Undirected graphs in the insertion-only model}
\subsection{A space lower bound}
\begin{theorem}
\label{thmlbundir}
For $t = O(n^2)$, simulating a $t$-step random walk on a simple undirected graph in the insertion-only model within error $\varepsilon=\frac{1}{3}$ requires $\Omega(n\sqrt{t})$ bits of memory.
\end{theorem}
\begin{proof}
Again we show a reduction from the \textsc{Index} problem.
Alice encodes $\Omega(n\sqrt{t})$ bits as follows. Let $G$ be an undirected graph with vertex set $V_0 \cup V_1 \cup \dots \cup V_{n/\sqrt{t}}$, where each $V_j$ has size $2\sqrt{t}$, and the starting vertex $v_0 \in V_0$. For each $j\ge 1$, $V_j$ is divided into two subsets $A_j,B_j$ with size $\sqrt{t}$ each, and Alice encodes $|A_j|\times |B_j| = t$ bits by inserting a subset of edges from $\{ (u,v): u\in A_j, v\in B_j\}$. In total she encodes $t\cdot n/\sqrt{t} = n\sqrt{t}$ bits.
Suppose Bob wants to query some bit, i.e., he wants to see whether $a$ and $b$ are connected by an edge. Assume $(a,b) \in A_j \times B_j$. He adds an edge $(u,v)$ for every $u\in A_j$ and every $v \in V_0$ (see Figure~\ref{fig:lbundir}). A perfect random walk starting from $v_0 \in V_0$ will be inside the bipartite subgraph $(A_j, B_j\cup V_0)$. Suppose the current vertex of the perfect random walk is $v_i \in A_j$. If $a,b$ are connected by an edge, then
\begin{align*}
&\P[(v_{i+2},v_{i+3})=(a,b) \mid v_i] \\
\ge {}& \P[v_{i+1} \in V_0 \mid v_i]\,\P[v_{i+2} = a \mid v_{i+1}\in V_0] \,\P[v_{i+3} = b \mid v_{i+2}=a]\\
\ge {}& \frac{|V_0|}{|V_0|+|B_j|} \cdot \frac{1}{|A_j|} \cdot \frac{1}{|V_0|+|B_j|}\\
\ge {}& \frac{2}{9t},
\end{align*}
so in every four steps the edge $(a,b)$ is passed with probability $\Omega(\frac{1}{t})$. Then a $O(t)$-step perfect random walk will pass the edge $(a,b)$ with probability $0.9$. Hence Bob can know whether the edge $(a,b)$ exists by looking at the random walk (simulated within error $\varepsilon$) with success probability $0.9-\frac{\varepsilon}{2}>1/2$.
By Lemma~\ref{index}, the space usage for simulating the $O(t)$-step random walk is at least $\Omega(n \sqrt{t})$ bits. The theorem is proved by scaling down $n$ and $t$ by a constant factor.
\begin{figure}
\begin{center}
\begin{tikzpicture}
\tikzstyle{vertex}=[circle,draw,text=black]
\tikzstyle{txt}=[text=black]
\tikzstyle{edge} = [draw,black,line width = 1.4pt,-]
\draw[rounded corners, draw=white, fill = black!15] (3,5) rectangle (3.9,6.8);
\draw[rounded corners, draw=white, fill = black!15] (5,5) rectangle (5.9,6.8);
\draw[rounded corners, draw=white, fill = black!15] (7,5) rectangle (7.9,6.8);
\draw[rounded corners, draw=white, fill = black!15] (5,1.2) rectangle (5.9,4.4);
\fill [fill=black!80] (3.45,5.4) circle(0.1) node[left] {$a$};
\fill [fill=black!80] (3.45,6.4) circle(0.1) node[left] {$ $};
\fill [fill=black!80] (5.45,6.4) circle(0.1) node[right] {$b$};
\fill [fill=black!80] (5.45,5.4) circle(0.1) node[right] {$ $};
\node[txt] (2) at (3.45,7.2) {$A_j$};
\node[txt] (1) at (5.45,7.2) {$B_j$};
\node[txt] (1) at (7.45,7.2) {$A_{j+1}$};
\node[txt] (1) at (5.45,0.8) {$V_0$};
\node[txt] (1) at (5.45,6.0) {$\vdots$};
\node[txt] (1) at (9.45,6.0) {$\cdots$};
\node[txt] (1) at (5.45,2.0) {$\vdots$};
\node[txt] (1) at (3.45,6.0) {$\vdots$};
\fill [fill=black!80] (5.45,3.9) circle(0.1) node[right] {};
\fill [fill=black!80] (5.45,3.3) circle(0.1) node[right] {};
\fill [fill=black!80] (5.45,2.7) circle(0.1) node[right] {$v_0$};
\path[draw=black,line width = 0.6pt](3.45,5.4)--(5.45,6.4);
\path[draw=black,line width = 0.6pt](3.45,6.4)--(5.45,5.4);
\path[draw=black,line width = 0.6pt](3.45,5.4)--(5.45,3.9);
\path[draw=black,line width = 0.6pt](3.45,5.4)--(5.45,3.3);
\path[draw=black,line width = 0.6pt](3.45,5.4)--(5.45,2.7);
\path[draw=black,line width = 0.6pt](3.45,6.4)--(5.45,3.9);
\path[draw=black,line width = 0.6pt](3.45,6.4)--(5.45,3.3);
\path[draw=black,line width = 0.6pt](3.45,6.4)--(5.45,2.7);
\end{tikzpicture}
\end{center}
\caption{Proof of Theorem~\ref{thmlbundir}}
\label{fig:lbundir}
\end{figure}
\end{proof}
\subsection{An algorithm for simple graphs}
Now we describe our algorithm for undirected graphs in the insertion-only model. As a warm-up, we consider simple graphs in this section. We will deal with multi-edges in Section~\ref{secmulti}.
\paragraph*{Intuition}
We start by informally explaining the intuition of our algorithm for simple undirected graphs.
We maintain a subset of $O(n\sqrt{t})$ edges from the input graph, and use them to simulate the random walk after reading the entire input stream.
For a vertex $u$ with degree smaller than $\sqrt{t}$, we can afford to store all its neighboring edges in memory. For $u$ with degree greater than $\sqrt{t}$, we can only sample and store $O(\sqrt{t})$ of its neighboring edges. During the simulation, at every step we first toss a coin to decide whether the next vertex has small degree or large degree. In the latter case, we have to pick a sampled neighboring edge and walk along it. If all sampled neighboring edges have already been used, our algorithm fails. Using the large degree and the fact that edges are undirected, we can show that the failure probability is low.
\paragraph*{Description of the algorithm}
We divide the vertices into two types according to their degrees: the set of \textit{big} vertices $B = \{u \in V: d(u)\ge C+1\}$, and the set of \textit{small} vertices $S = \{u \in V: d(u)\le C\}$, where parameter $C$ is an positive integer to be determined later.
We use \textit{arc} $(u,v)$ to refer to an edge when we want to specify the direction $u\to v$. So an undirected edge $(u,v)$ corresponds to two different\footnote{We have assumed no self-loops exist, so $u \neq v$.} arcs, arc $(u,v)$ and arc $(v,u)$.
We say an arc $(u,v)$ is \textit{important} if $v \in S$, or \textit{unimportant} if $v \in B$. Denote the set of important arcs by $E_1$, and the set of unimportant arcs by $E_0$. The total number of important arcs equals $\sum_{s\in S}d(s) \le |S|C$, so it is possible to store $E_1$ in $O(nC)$ words of space.
The set $E_0$ of unimportant arcs can be huge, so we only store a subset of $E_0$.
For every vertex $u$, we sample with replacement $C$ unimportant arcs outgoing from $u$, denoted by $a_{u,1},\dots,a_{u,C}$.
To maintain the set $E_1$ of important arcs and the samples of unimportant arcs after every edge insertion, we need to handle the events when some small vertex becomes big. This procedure is straightforward, as described by \textsc{ProcessInput} in Figure~\ref{fig:processinput}. Since $|E_1|$ never exceeds $nC$, and each of the $n$ samplers uses $O(C)$ words of space, the overall space complexity is $O(nC)$ words.
\begin{figure}
\begin{center}
\begin{algorithmic}
\Procedure{InsertArc}{$u,v$}
\State{$d(v) \gets d(v) +1$}
\If{$d(v)= C+1$}\Comment{$v$ changes from small to big}
\For{$x \in V$ such that $(x,v) \in E_1$} \Comment{arc $(x,v)$ becomes unimportant}
\State{$E_1 \gets E_1 \backslash \{(x,v)\}$}
\State{Feed arc $(x,v)$ into $x$'s sampler}
\EndFor
\EndIf
\If{$d(v)\le C$} \Comment{$v\in S$}
\State{$E_1 \gets E_1 \cup \{(u,v)\}$}
\Else\Comment{$v\in B$}
\State{Feed arc $(u,v)$ into $u$'s sampler}
\EndIf
\EndProcedure
\Procedure{ProcessInput}{}
\State{$E_1 \gets \emptyset$}\Comment{Set of important arcs}
\For{$u \in V$}
\State{$d(u) \gets 0$}
\State{Initialize $u$'s sampler (initially empty) which maintains $a_{u,1},\dots,a_{u,C}$}
\EndFor
\For{undirected edge $(u,v)$ in the input stream}
\State{\textsc{InsertArc}$(u,v)$}
\State{\textsc{InsertArc}$(v,u)$}
\EndFor
\EndProcedure
\end{algorithmic}
\caption{Pseudocode for processing the input stream (for simple undirected graphs)}\label{fig:processinput}
\end{center}
\end{figure}
We begin simulating the random walk after \textsc{ProcessInput} finishes.
When the current vertex of the random walk is $v$, with probability $d_1(v)/d(v)$ the next step will be along an important arc, where $d_1(v)$ denotes the number of important arcs outgoing from $v$.
In this case we simply choose a uniform random vertex from $\{u: (v,u) \in E_1\}$ as the next vertex.
However, if the next step is along an unimportant arc, we need to choose an unused sample $a_{v,j}$ and go along this arc.
If at this time all $C$ samples $a_{v,j}$ are already used, then our algorithm fails (and is allowed to return an arbitrary walk).
The pseudocode of this simulating procedure is given in Figure~\ref{fig:simulation}.
\begin{figure}
\begin{center}
\begin{algorithmic}
\Procedure{SimulateRandomWalk}{$v_0,t$}
\For{$v \in V$} \State{$c(v) \gets 0$} \Comment{counter of used samples}
\EndFor
\For{$i = 0,\dots,t-1$}
\State{$N_1 \gets \{u: (v_{i},u) \in E_1\}$}
\State{$x \gets$ uniformly random integer from $\{1,2,\dots,d(v_{i})\}$}
\If{$x \le |N_1|$}
\State{$v_{i+1} \gets $ uniformly random vertex from $N_1$}
\Else
\State{$j \gets c(v_{i}) + 1$}
\State{$c(v_{i}) \gets j$}
\If{$j>C$}
\Return{\textsc{Fail}}
\Else
\State{$v_{i +1}\gets u$, where $(v_{i},u) = a_{v_{i},j}$}
\EndIf
\EndIf
\EndFor
\Return $(v_0,\dots,v_t)$
\EndProcedure
\end{algorithmic}
\caption{Pseudocode for simulating a $t$-step random walk starting from $v_0$} \label{fig:simulation}
\end{center}
\end{figure}
In a walk $w = (v_0,\dots,v_t)$, we say vertex $u$ \textit{fails} if $\left |\{i: v_i = u\text{ and }(v_i, v_{i+1}) \in E_0 \}\right| > C$. If no vertex fails in $w$, then our algorithm will successfully return $w$ with probability $\mathcal{RW}_{v_0,t}(w)$. Otherwise our algorithm will fail after some vertex runs out of the sampled unimportant arcs. To ensure the output distribution is $\varepsilon$-close to $\mathcal{RW}_{v_0,t}$ in $\ell_1$ distance, it suffices to make our algorithm fail with probability at most $\varepsilon/2$, by choosing a large enough capacity $C$.
To bound the probability $\P[\text{at least one vertex fails} \mid v_0=s]$\footnote{If not specified, assume the probability space is over all $t$-step random walks $(v_0,\dots,v_t)$ starting from $v_0$.}, we will bound the individual failure probability of every vertex, and then use union bound.
\begin{lemma}
\label{union}
Suppose for every $u\in V$, $\P[\text{$u$ fails}\mid \text{$v_0=u$}] \le \delta$. Then for any starting vertex $s \in V$, $\P[\text{at least one vertex fails}\mid v_0=s]\le t\delta$.
\end{lemma}
\begin{proof}
Fix a starting vertex $s$. For any particular $u \in V$,
\begin{align*}
&\P[\text{$u$ fails}\mid \text{$v_0=s$}]\\
= {}&\P[\text{$u$ fails, and $\exists i\le t-1, v_i=u$}\mid \text{$v_0=s$}]\\
={}&\P[\exists i\le t-1, v_i=u\mid \text{$v_0= s$}] \, \P[\text{$u$ fails}\mid \text{$v_0=s$, and $\exists i\le t-1, v_i=u$}]\\
\le{}&\P[\text{$\exists i\le t-1, v_i=u$}\mid \text{$v_0=s$}]\, \P[\text{$u$ fails}\mid \text{$v_0=u$}]\\
\le {}& \P[\text{$\exists i\le t-1, v_i=u$}\mid \text{$v_0=s$}]\cdot\delta .
\end{align*}
By union bound,
\begin{align*}
&\P[\text{at least one vertex fails}\mid v_0=s]\\
\le {}& \sum_{u\in V} \P[\text{$u$ fails}\mid \text{$v_0=s$}]\\
\le {}& \sum_{u\in V}
\P[\exists i\le t-1, v_i=u\mid \text{$v_0=s$}]\cdot \delta\\
={} & \mathbb{E}[\text{number of distinct vertices visited in $\{v_0,\dots,v_{t-1}\}$}\mid v_0=s]\cdot \delta\\
\le{} & t\delta.
\end{align*}
\end{proof}
\begin{lemma}
\label{capa}
We can choose integer parameter $C = O\left ( \sqrt{t}\cdot \frac{q}{\log q} \right ) $, where $q = 2+\frac{\log(1/\delta)}{\sqrt{t}}$, so that $\P[\text{$u$ fails} \mid \text{$v_0=u$}] \le \delta $ holds for every $u\in V$.
\end{lemma}
\begin{proof}
Let $d_0(u) = | \{v : (u,v) \in E_0\} |$.
For any $u \in V$,
\begin{align*}
&\P[\text{$u$ fails} \mid \text{$v_0=u$}]\\
\le {}&\P[\text{$u$ fails} \mid \text{$v_0=u, (v_0,v_1) \in E_0$}].
\end{align*}
We rewrite this probability as the sum of probabilities of possible random walks in which $u$ fails.
Recall that $u$ fails if and only if $|\{i: v_i=u, (v_i,v_{i+1}) \in E_0\}| \ge C+1$.
In the summation over possible random walks, we only keep the shortest prefix $(v_0,\dots,v_k)$ in which $u$ fails, i.e., the last step $(v_{k-1},v_k)$ is the $(C+1)$-st time walking along an unimportant arc outgoing from $u$. We have
\begin{align}
&\P[\text{$u$ fails} \mid \text{$v_0=u, (v_0,v_1) \in E_0$}] \nonumber \\
={}&\sum_{k\le t}\sum_{\text{walk}(v_0,\dots,v_k)}\mathbf{1} \bigg [\text{$v_0=v_{k-1}=u, \,\,(v_0,v_1),(v_{k-1},v_k)\in E_0,$} \nonumber \\[-6\jot]
& \hspace{4cm}\text{$|\{i: v_i = u,(v_i, v_{i+1})\in E_0\}|=C+1$} \bigg ]\frac{1}{d_0(u)}\prod_{i=1}^{k-1}\frac{1}{d(v_i)}\nonumber \\
={}&\sum_{k\le t}\sum_{\text{walk}(v_0,\dots,v_{k-1})}\mathbf{1} \bigg[\text{$v_0=v_{k-1}=u, \,\,(v_0,v_1)\in E_0,$} \nonumber\\[-6\jot]
&\hspace{5cm}\text{$|\{i: v_i = u, (v_i,v_{i+1})\in E_0\}|=C$}\bigg ]\prod_{i=1}^{k-1}\frac{1}{d(v_i)} \label{sum}.
\end{align}
Let $v'_i = v_{k-1-i}$. Since the graph is undirected, the vertex sequence $(v'_0,\dots,v'_{k-1})$ (the reversal of walk $(v_0,\dots,v_{k-1})$) is also a walk starting from and ending at $u$. So the summation~(\ref{sum}) equals
\begin{align*}
&
\sum_{k\le t}\sum_{\text{walk}(v'_0,\dots,v'_{k-1})}\mathbf{1}\bigg [\text{$v'_0=v'_{k-1}=u, \,\,(v'_{k-1},v'_{k-2})\in E_0,$}\nonumber \\[-6\jot]
&\hspace{6cm}\text{$|\{i: v'_i=u, (v'_i, v'_{i-1}) \in E_0\}|=C$}\bigg ]\prod_{i=0}^{k-2}\frac{1}{d(v'_i)}\\
={}&\underset{\text{random walk $(v'_0,\dots,v'_{t-1}) $}}{\P}\Big [\,|\{i: v'_{i}=u, (v'_i,v'_{i-1}) \in E_0\}|\ge C \,\Big \vert\, v'_0=u\Big ].
\end{align*}
Recall that $(v'_i,v'_{i-1}) \in E_0$ if and only if $v'_{i-1} \in B$. For any $1\le i \le t-1$ and any fixed prefix $v'_0,\dots,v'_{i-1}$,
\begin{align}
& \P \big [v'_i=u, (v'_i,v'_{i-1}) \in E_0 \, \big \vert \, v'_0,\dots,v'_{i-1}\big ] \nonumber\\
\le {} & \mathbf{1}[v'_{i-1} \in B]\cdot \frac{1}{d(v'_{i-1})} \nonumber \label{cond}\\
< {}&\frac{1}{C}.
\end{align}
Hence the probability that $|\{1\le i \le t-1: v'_i=u, (v'_i,v'_{i-1}) \in E_0\}|\ge C$ is at most
\begin{align*}
&\binom{t-1}{C}\left (\frac{1}{C}\right )^C\\
\le {}& \left ( \frac{e(t-1)}{C}\right )^C\left (\frac{1}{C}\right )^C\\
< {}& \left ( \frac{et}{C^2}\right )^C.
\end{align*}
We set $C = \left \lceil 4 \sqrt{t}\ q/\log q \right \rceil$, where $q = 2 + \log(1/\delta)/\sqrt{t} >2$. Notice that $q/\log^2 q > 1/4$. Then
\begin{align*}
C\log \left (\frac{C^2}{et}\right ) \ge \frac{4\sqrt{t} q}{\log q} \log \left ( \frac{16q^2}{e\log^2 q}\right ) >\frac{4\sqrt{t} q}{\log q} \log (4q/e) > 4\sqrt{t}q > \log(1/\delta),
\end{align*}
so
\begin{align*}
\left ( \frac{et}{C^2}\right )^C < \delta.
\end{align*}
Hence we have made $\P[\text{$u$ fails} \mid v_0=u]< \delta$ by choosing $C = O(\sqrt{t}q/\log q )$.
\end{proof}
\begin{theorem}
We can simulate a $t$-step random walk on a simple undirected graph in the insertion-only model within error $\varepsilon$ using
$O\left ( n \sqrt{t} \cdot \frac{q}{\log q} \right )$
words of memory, where $q = 2+\frac{\log(1/\varepsilon)}{\sqrt{t}}$.
\end{theorem}
\begin{proof}
The theorem follows from Lemma~\ref{union} and Lemma~\ref{capa} by setting $\delta = \frac{\varepsilon}{2t}$.
\end{proof}
\subsection{On graphs with multiple edges}
\label{secmulti}
When the undirected graph contains multiple edges, condition~(\ref{cond}) in the proof of Lemma~\ref{capa} may not hold, so we need to slightly modify our algorithm.
We still maintain the multiset $E_1$ of important arcs.
Whether an arc is important will be determined by our algorithm.
(This is different from the previous algorithm, where important arcs were simply defined as $(u,v)$ with $d(v)\le C$.)
We will ensure that condition~(\ref{cond}) still holds, i.e., for any $u\in V$ and any fixed prefix of the random walk $v_0,\dots,v_{i-1}$,
\begin{align}
\P \big [\text{$(v_{i},v_{i-1}) \notin E_1$, and $v_{i}= u$} \,\big \vert\, v_0,\dots,v_{i-1}\big ] < 1/C. \label{newcond}
\end{align}
Note that there can be both important arcs and unimportant arcs from $u$ to $v$. Let $f(u,v)$ denote the number of undirected edges between $u,v$. Then there are $f(u,v)$ arcs $(u,v)$.
Suppose $f_1(u,v)$ of these arcs are important, and $f_0(u,v) = f(u,v)-f_1(u,v)$ of them are unimportant. Then we can rewrite condition~(\ref{newcond}) as
\begin{align}
\frac{f_0(u,v_{i-1})}{d(v_{i-1})} < 1/C, \label{newnewcond}
\end{align}
for every $u,v_{i-1} \in V$.
Similarly as before, we need to store the multiset $E_1$ using only $O(nC)$ words of space.
And we need to sample with replacement $C$ unimportant arcs $a_{u,1},\dots,a_{u,C}$ outgoing from $u$, for every $u \in V$.
Finally we use the procedure \textsc{SimulateRandomWalk} in Figure~\ref{fig:simulation} to simulate a random walk.
The multiset $E_1$ is determined as follows: For every vertex $v\in V$, we run Misra-Gries algorithm \cite{misra1982finding} on the sequence of all $v$'s neighbors.
We will obtain a list $L_v$ of at most $C$ vertices, such that for every vertex $u\notin L_v$, $\frac{f(u,v)}{d(v)} < \frac{1}{C}$. Moreover, we will get a frequency estimate $A_v(u)>0$ for every $u\in L_v$, such that $0 \le f(u,v) - A_v(u) < \frac{d(v)}{C}$.
Assuming $A_v(u)=0$ for $u\notin L_v$, we can satisfy condition~(\ref{newnewcond}) for all $u\in V$ by setting $f_1(u,v)=A_v(u)$. Hence we have determined all the important arcs, and they can be stored in $O(\sum_{v}|L_v|) = O(nC)$ words. \, To sample from the unimportant arcs, we simply insert the arcs discarded by Misra-Gries algorithm into the samplers. \, The pseudocode is given in Figure~\ref{fig:processinputmulti}.
\begin{figure}
\begin{center}
\begin{algorithmic}
\Procedure{InsertArc}{$u,v$}
\State{$d(v) \gets d(v)+1$}
\If{$u \in L_v$}
\State{$A_v(u) \gets A_v(u)+1$}
\Else
\State{Insert $u$ into $L_v$}
\State{$A_v(u) \gets 1$}
\If{$|L_v|\ge C+1$}
\For {$w \in L_v$}
\State{Feed arc $(w,v)$ into $w$'s sampler}
\State{$A_v(w) \gets A_v(w) -1$}
\If{$A_v(w)=0$}
\State{Remove $w$ from $L_v$}
\EndIf
\EndFor
\EndIf
\EndIf
\EndProcedure
\Procedure{ProcessInput}{}
\For{$u \in V$}
\State{$d(u) \gets 0$}
\State{Initialize $u$'s sampler (initially empty) which maintains $a_{u,1},\dots,a_{u,C}$}
\State{Initialize empty list $L_u$}
\EndFor
\For{undirected edge $(u,v)$ in the input stream}
\State{\textsc{InsertArc}$(u,v)$}
\State{\textsc{InsertArc}$(v,u)$}
\EndFor
\State{$E_1 \gets \bigcup_{v\in V} \bigcup_{u\in L_v}\{A_v(u)\text{ copies of arc }(u,v)\}$}\Comment{Multiset of important arcs}
\EndProcedure
\end{algorithmic}
\caption{Pseudocode for processing the input stream (for undirected graphs with possibly multiple edges)} \label{fig:processinputmulti}
\end{center}
\end{figure}
\begin{lemma}
After \textsc{ProcessInput} (in Figure~\ref{fig:processinputmulti}) finishes, $|L_v|\le C$. For every $u \in L_v$, $0 \le f(u,v) -A_v(u) \le \frac{d(v)}{C+1}$. For every $u\notin L_v$, $f(u,v) \le \frac{d(v)}{C+1}$.
\end{lemma}
\begin{proof}
Every time the \textbf{for} loop in procedure \textsc{InsertArc} finishes, the newly added vertex $u$ must have been removed from $L_v$, so $|L_v| \le C$ still holds. Let $W= \{w_1,\cdots,w_{C+1}\}$ be the set of vertices in $L_v$ before this \textbf{for} loop begins. Then for every $u\in V$, $f(u,v)-A_v(u)$ equals the number of times $u$ is contained in $W$ (assuming $A_v(u) = 0$ for $u \notin L_v$), which is at most $\frac{1}{C+1}\sum_W |W| \le \frac{d(v)}{C+1}$.
\end{proof}
\begin{corollary}
\label{coro}
Procedure \textsc{ProcessInput} in Figure~\ref{fig:processinputmulti} computes the multiset $E_1$ of important edges and stores it using $O(nC)$ words. It also samples with replacement $C$ unimportant arcs $a_{u,1},\dots,a_{u,C}$ outgoing from $u$, for every $u \in V$.
Moreover,
\begin{align*}
\frac{f_0(u,v)}{d(v)} < \frac{1}{C} \label{newnewcond}
\end{align*}
holds for every $u,v \in V$.
\end{corollary}
Now we analyze the failure probability of \textsc{SimulateRandomWalk} (in Figure~\ref{fig:simulation}), similar to Lemma~\ref{capa}.
\begin{lemma}
\label{multicapa}
We can choose integer parameter
$C = O\left ( \sqrt{t}\cdot \frac{q}{\log q} \right ) $, where $q = 2+\frac{\log(1/\delta)}{\sqrt{t}}$,
so that $\P[\text{$u$ fails} \mid \text{$v_0=u$}] \le \delta $ holds for every $u\in V$.
\end{lemma}
\begin{proof}
Let $d_0(u) = \sum_{v\in V} f_0(u,v)$. \ \
As before, we rewrite this probability as a sum over possible random walks. Here we distinguish between important and unimportant arcs. Denote $s_i = \mathbf{1}[\text{step $(v_{i-1},v_i)$ is along an important arc}]$.
Then for any $u \in V$,
\begin{align*}
&\P[\text{$u$ fails} \mid \text{$v_0=u$}]\\
\le{} &\P[\text{$u$ fails} \mid \text{$v_0=u$, arc $(v_0,v_1)$ is unimportant}]\\
={}&\frac{d(u)}{d_0(u)}\sum_{k\le t}\sum_{(v_0,\dots,v_k)}\sum_{s_1,\dots,s_k}\mathbf{1}\bigg [\text{$v_0=v_{k-1}=u,\, s_1=s_k=0,$}\nonumber\\[-6\jot]
&\hspace{5cm}\text{$ |\{i: v_i = u, s_{i+1}=0\}|=C+1$} \bigg ]\prod_{i=0}^{k-1}\frac{f_{s_{i+1}}(v_i,v_{i+1})}{d(v_i)}\\[+1\jot]
={}&\sum_{k\le t}\sum_{(v_0,\dots,v_{k-1})}\sum_{s_1,\dots,s_{k-1}}\mathbf{1}\bigg [\text{$v_0=v_{k-1}=u,\,\, s_1=0,$}\nonumber\\[-6\jot]
&\hspace{5cm}\text{$|\{i: v_i = u, s_{i+1}=0\}|=C$}\bigg ]\prod_{i=0}^{k-2}\frac{f_{s_{i+1}}(v_i,v_{i+1})}{d(v_i)}.
\end{align*}
Let $v'_i = v_{k-1-i}, s'_i = s_{k-i}$. Then this sum equals
\begin{align*}
& \sum_{k\le t}\sum_{(v'_0,\dots,v'_{k-1})} \sum_{s'_1,\dots,s'_{k-1}}\mathbf{1}\bigg [\text{$v'_0=v'_{k-1}=u,\,\, s'_{k-1}=0,$}\nonumber\\[-6\jot]
&\hspace{6cm}\text{$|\{i: s'_i =0 , v'_{i}=u\}|=C$}\bigg ]\prod_{i=1}^{k-1}\frac{f_{s'_{i}}(v'_{i},v'_{i-1})}{d(v'_{i-1})}\\[+3\jot]
={}&\underset{\text{random walk $(v'_0,\dots,v'_{t-1}) $}}{\P}\Big [\,|\{i: \text{$v'_{i}=u$, arc $(v'_{i},v'_{i-1})$ is unimportant}\}|\ge C \, \Big \vert \, v'_0=u\Big ].
\end{align*}
Notice that for any $i$ and any fixed prefix $v'_0,\dots,v'_{i-1}$,
\begin{align*}
\P \Big [\text{$v'_{i}=u$, arc $(v'_{i},v'_{i-1})$ is unimportant}\, \Big \vert \, v'_0,v'_1,\dots,v'_{i-1}\Big ] = \frac{f_0(u,v'_{i-1})}{d(v'_{i-1})} < \frac{1}{C}
\end{align*}
by Corollary~\ref{coro}. The rest of the proof is the same as in Lemma~\ref{capa}.
\end{proof}
\begin{theorem}
\label{mainthmmulti}
We can simulate a random walk on an undirected graph with possibly multiple edges in the insertion-only model within error $\varepsilon$ using $O\left ( n \sqrt{t} \cdot \frac{q}{\log q} \right )$
words of memory, where $q = 2+\frac{\log(1/\varepsilon)}{\sqrt{t}}$.
\end{theorem}
\begin{proof}
The theorem follows from Lemma~\ref{union} and Lemma~\ref{multicapa} by setting $\delta = \frac{\varepsilon}{2t}$.
\end{proof}
\section{Turnstile model}
In this section we consider the turnstile model where both insertion and deletion of edges can appear.
\begin{lemma}[$\ell_1$ sampler in the turnstile model, \cite{jayaram2018perfect}]
Let $f \in \mathbb{R}^n$ be a vector defined by a stream of updates to its coordinates of the form $f_i \gets f_i + \Delta$, where $\Delta$ can either be positive or negative. There is an algorithm which reads the stream and returns an index $i \in [n]$ such that for every $j \in [n]$,
\begin{equation}
\label{sampleprob}
\P[i=j] = \frac{|f_j|}{\|f\|_1} + O(n^{-c}),
\end{equation}
where $c\ge 1$ is some arbitrary large constant. It is allowed to output \textsc{Fail} with probability $\delta$, and in this case it will not output any index. The space complexity of this algorithm is $O(\log^2 n \log (1/\delta))$ bits.
\end{lemma}
\begin{remark}
For $\varepsilon \ll 1/n$, the $O(n^{-c})$ error term in (\ref{sampleprob}) can be reduced to $O(\varepsilon^c)$ by running the $\ell_1$ sampler on $f \in \mathbb{R}^{\lceil 1/\varepsilon \rceil}$, using $O(\log^2 (1/\varepsilon) \log(1/\delta))$ bits of space.
\end{remark}
We will use the $\ell_1$ sampler for sampling neighbors (with possibly multiple edges) in the turnstile model. The error term $O(n^{-c})$ (or $O(\varepsilon^c)$) in (\ref{sampleprob}) can be ignored in the following discussion, by choosing sufficiently large constant $c$ and scaling down $\varepsilon$ by a constant.
\subsection{Directed graphs}
\begin{theorem}
\label{turnstilealgdirected}
We can simulate a $t$-step random walk on a directed graph in the turnstile model within error $\varepsilon$ using $O(n(t+ \log\frac{1}{\varepsilon})\log^2 \max\{n,1/\varepsilon\})$ bits of memory.
\end{theorem}
\begin{proof}
For every $u \in V$, we run $C' = 2t + 16\log (2t/\varepsilon)$ independent $\ell_1$ samplers each having failure probability $\delta=1/2$. We use them to sample the outgoing edges of $u$ (as in the algorithm of Theorem~\ref{algdirected}). By Chernoff bound, the probability that less than $t$ samplers succeed is at most $\varepsilon/(2t)$.
We say a vertex $u$ fails if $u$ has less than $t$ successful samplers, and $u \in \{v_0,v_1,\dots,v_{t-1}\}$ (where $v_0,v_1,\dots,v_t$ is the random walk). Then $\P[\text{$u$ fails}] \le \frac{\varepsilon}{2t} \P[u \in \{v_0,\dots,v_{t-1}\}]$. By union bound, $\P[\text{at least one vertex fails}] \le \frac{\varepsilon}{2t}\sum_{u\in V} \P[u \in \{v_0,\dots,v_{t-1}\}] \le \frac{\varepsilon}{2}$. Hence, with probability $1-\frac{\varepsilon}{2}$, every vertex $u$ visited (except the last one) has at least $t$ outgoing edges sampled, so our simulation can succeed. The space usage is $O(nC' \log^2 \max\{n,1/\varepsilon\} \log (1/\delta)) = O(n(t+ \log\frac{1}{\varepsilon}) \log^2 \max\{n,1/\varepsilon \})$ bits.
\end{proof}
\subsection{Undirected graphs}
We slightly modify the \textsc{ProcessInput} procedure of our previous algorithm in Section~\ref{secmulti}. We will use the $\ell_1$ heavy hitter algorithm in the turnstile model.
\begin{lemma}[$\ell_1$ heavy hitter, \cite{cormode2005improved}]
Let $f \in \mathbb{R}^n$ be a vector defined by a stream of updates to its coordinates of the form $f_i \gets f_i + \Delta$, where $\Delta$ can either be positive or negative. There is a randomized algorithm which reads the stream and returns a subset $L \subseteq [n]$
such that $i \in L$ for every $|f_i| \ge \frac{\|f\|_1}{k}$, and $i \notin L$ for every $|f_i| \le \frac{\|f\|_1}{2k}$. Moreover it returns a frequency estimate $\tilde f_i$ for every $i \in L$, which satisfies $0 \le f_i -\tilde f_i \le \frac{\|f\|_1}{2k}$. The failure probability of this algorithm is $O(n^{-c})$. The space complexity is $O(k\log^2 n)$ bits.
\end{lemma}
\begin{remark}
For $\varepsilon \ll 1/n$, the $O(n^{-c})$ failure probability of this $\ell_1$ heavy hitter algorithm can be reduced to $O(\varepsilon^c)$ by running the algorithm on $f \in \mathbb{R}^{\lceil 1/\varepsilon \rceil}$, using $O(k \log^2 (1/\varepsilon))$ bits of space. In the following discussion, this failure probability can be ignored by making the constant $c$ sufficiently large.
\end{remark}
\begin{theorem}
\label{turnstilealgundirected}
We can simulate a $t$-step random walk on an undirected graph in the turnstile model within error $\varepsilon$ using $O(n(\sqrt{t}+ \log\frac{1}{\varepsilon})\log^2 \max\{n,1/\varepsilon\})$ bits of memory.
\end{theorem}
\begin{proof}
Similar to the previous insertion-only algorithm (in Figure~\ref{fig:processinputmulti}), we perform two \textit{arc updates} $((u,v),\Delta),\, ((v,u),\Delta)$ when we read an \textit{edge update} $((u,v),\Delta)$ from the stream.
For every $u \in V$, we run $C' = 2C + 16\log (2t/\varepsilon)$ independent $\ell_1$ samplers each having failure probability $\delta=1/2$, where $C$ is the same constant as in the proof of Lemma~\ref{multicapa} and Theorem~\ref{mainthmmulti}. By Chernoff bound, the probability that less than $C$ samplers succeed is at most $\varepsilon/(2t)$. For every arc update $((u,v),\Delta)$, we send update ($v$, $\Delta$) to $u$'s $\ell_1$ sampler.
In addition, for every $v\in V$, we run $\ell_1$ heavy hitter algorithm with $k= C$. For every arc update $((u,v),\Delta)$, we send update $(u,\Delta)$ to $v$'s heavy hitter algorithm. In the end, we will get a frequency estimate $A_v(u)$ for every $u \in V$, such that $f(u,v) - \frac{d(v)}{C}\le A_v(u) \le f(u,v)$. We then insert $A_v(u)$ copies of arc $(u,v)$ into $E_1$ (the multiset of important arcs), and send update $(v, -A_v(u))$ to $u$'s $\ell_1$ sampler. Then we use the $\ell_1$ samplers to sample unimportant arcs for every $u$.
As before, we use the procedure \textsc{SimulateRandomWalk} (in Figure~\ref{fig:simulation}) to simulate the random walk. The analysis of the failure probability of the $\ell_1$ samplers is the same as in Theorem~\ref{turnstilealgdirected}. The analysis of the failure probability of procedure \textsc{SimulateRandomWalk} is the same as in Lemma~\ref{multicapa}. The space usage of the algorithm is $O(nC' \log^2 \max\{n,1/\varepsilon\} \log \delta) = O(n(\sqrt{t}+\log \frac{1}{\varepsilon}) \log^2 \max\{n,1/\varepsilon\})$ bits.
\end{proof}
\section{Conclusion}
We end our paper by discussing some related questions for future research.
\begin{itemize}
\item The output distribution of our insertion-only algorithm for undirected graphs is $\varepsilon$-close to the random walk distribution.
What if the output is required to be perfectly random, i.e., $\varepsilon=0$?
\item For insertion-only simple undirected graphs, we proved an $\Omega(n\sqrt{t})$-bit space lower bound. Our algorithm uses $O(n\sqrt{t} \log n)$ bits (for not too small $\varepsilon$). Can we close the gap between the lower bound and the upper bound, as in the case of directed graphs?
\item In the undirected version, suppose the starting vertex $v_0$ is drawn from a distribution (for example, the stationary distribution of the graph) rather than being specified.
Is it possible to obtain a better algorithm in this new setting? Notice that our proof of the $\Omega(n\sqrt{t})$ lower bound does not work here, since it requires $v_0$ to be specified.
\item We required the algorithm to output all vertices on the random walk.
If only the last vertex is required, can we get a better algorithm or prove non-trivial lower bounds?
\end{itemize}
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L'année 1417 est une année commune qui commence un vendredi.
Événements
Explorations chinoises en Afrique orientale (1417-1433). Les flottes de l'amiral eunuque Zheng He visitent Malindi, Brava et Mogadisque et tentent de s'assurer la maîtrise du commerce sur l'océan Indien.
Chimalpopoca devient le troisième souverain des Aztèques.
Luttes internes pour le pouvoir au Japon. La rébellion d'Uesugi Zenshū contre le vice-shogun de Kamakura Ashikaga Mochiuji est réprimée par les troupes du shogun Yoshimochi Ashikaga. Début d'une période d'instabilité dans la région de Kantō. Les Ashikaga perdent le contrôle de la région orientale du Kantō.
Les Chinois relancent les négociations diplomatiques avec le Japon (1417-1418). Les Japonais désirent établir des rapports d'échange avec la Chine, mais refusent toujours d'entrer dans le système du tribut. Les Chinois menacent d'exercer des représailles, mais ne sont pas entendus dans un contexte politique instable.
Europe
2 janvier : l'empereur quitte Liège à l'improviste sans avoir rendu leurs franchises aux Liégeois.
Janvier-février : au début de l'année, le sultan occupe la Dobroudja. La Valachie doit reconnaître la suzeraineté des Turcs ottomans. Elle garde dans un premier temps son autonomie interne, sa dynastie et sa religion chrétienne en échange d'un tribut. doit céder la Dobroudja au sultan.
La Valachie verse ducats-or vénitiens de tribut annuel à la Porte ( à la fin du ).
Les marchands valaques obtiennent l'autorisation de commercer dans l'Empire ottoman moyennant le paiement d'une taxe, le gümrük (du grec kommerkion), de 2 % de la valeur des marchandises transportées. La Valachie contrôle la route du commerce reliant l'Asie à l'Europe centrale par la mer Noire.
26 mars : octroie une charte de liberté communale, datée de Constance où il s'est rendu pour le Concile, à la ville de Liège.
28 mars : Saint-Lô se rend sans combat au duc de Gloucester, qui commande les troupes anglaises d'. elle reste anglaise jusqu'en 1449.
5 avril : mort du dauphin Jean. Charles de Ponthieu est reconnu comme héritier du trône de France après la mort de son frère (13 avril). Il obtient le 17 mai le duché de Berry.
18 avril : devient électeur de Brandebourg.
28 avril : le comté de Clèves est érigé en duché par l'empereur Sigismond à Constance.
14 juin : le dauphin Charles est nommé lieutenant général du royaume de France.
29 juin : le comte de Huntingdon bat la flotte franco-génoise au large de Chef-de-Caux.
18 juillet : Jean de Gerson termine son traité qui condamne définitivement la secte des flagellants.
26 juillet : le pape d'Avignon est déposé par le concile de Constance.
: débarque ses troupes à l'embouchure de la Touques (aujourd'hui Trouville-sur-Mer), et entreprend la conquête de la Normandie
4 septembre : assaut et prise de Caen par . Il assiège la citadelle où s'est réfugiée la garnison.
19 septembre :
Bayeux se rend aux troupes anglaises.
Le château de Caen capitule et y établit son quartier général.
9 octobre :
Le concile de Constance prend le décret Frequens où il se déclare institution permanente de l'Église, devant se réunir périodiquement et chargé du contrôle de la papauté.
Argentan se rend aux Anglais.
18 octobre : c'est au tour d'Alençon de tomber entre les mains des Anglais.
11 novembre : Oddone Colonna est élu pape par le concile de Constance et, sous le nom de devient le de l'Église (fin de son pontificat en 1431). Fin du Grand Schisme d'Occident. Le concile se sépare en avril 1418.
14 décembre : John Oldcastle est exécuté en Angleterre comme lollard, partisan de John Wyclif.
20 décembre : capitulation de Falaise.
Dernière récurrence de la peste en Italie.
Naissances en 1417
Décès en 1417
Notes et références
Liens externes
Année du XVe siècle
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Your application draws on data that is expensive to create from a performance perspective, so you want to store it in memory, where it can be accessed by users throughout the lifetime of the application. The problem is that the data changes occasionally and you need to refresh the data when it changes.
Place the data in the Cache object with a dependency set to the source of the data so the data will be reloaded when it changes. Example 13-3 and Example 13-4 show the code we've written to demonstrate this solution. In this case, these are VB and C# code-behind files for Global.asax that place some sample XML book data in the Cache object. In our example, the book data is automatically removed from the cache anytime the XML file is changed.
The Cache object in ASP.NET provides the ability to store application data in a manner similar to the storing of data in the Application object. The Cache object, unlike the Application object, lets you specify that the cached data is to be replaced at a specified time or whenever there is a change to the original source of the data.
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{"url":"https:\/\/docs.juliaebert.com\/programming\/web","text":"# Web Development\n\n## Github Pages project on custom subdomain\n\nFirst, enable GitHub pages for your repository under the settings. If you\u2019re using Jekyll and want Github to compile the site, set the source to \u201cmaster branch\u201d. If you\u2019re compiling it yourself and pushing it to the gh-pages branch, choose that.\n\nThen under \u201cCustom domain,\u201d set your subdomain, such as docs.juliaebert.com. (Look familiar?) This will create a commit to your repository with a CNAME file that will let GitHub handle properly directing requests to your github.io domain.\n\nNow go to your domain provider\u2019s site to add a custom DNS record. Choose the type as CNAME and set the name to your subdomain (e.g., docs). Then under the Domain name\/data, set your GitHub Pages base url with a period at the end, such as jtebert.github.io.. Note that this does not include your project\u2019s repository name at the end.\n\nIt can take awhile (up to hours) for this to propagate. You can check whether it\u2019s working with:\n\ndig docs.juliaebert.com +nostats +nocomments +nocmd\n\n\nIt should give an output like this:\n\n; <<>> DiG 9.11.3-1ubuntu1.7-Ubuntu <<>> docs.juliaebert.com +nostats +nocomments +nocmd\n;; global options: +cmd\n;docs.juliaebert.com. IN A\ndocs.juliaebert.com. 3519 IN CNAME jtebert.github.io.\njtebert.github.io. 1816 IN A 185.199.108.153\njtebert.github.io. 1816 IN A 185.199.109.153\njtebert.github.io. 1816 IN A 185.199.110.153\njtebert.github.io. 1816 IN A 185.199.111.153\n\n\nIf you don\u2019t have any lines not starting with ;, it means it\u2019s not working (or not working yet).\n\nA final note if you\u2019re using Jekyll: when you\u2019re set up with the subdomain instead of USERNAME.github.io\/PROJECT-NAME, you will set the base_url in your _config.yml to nothing instead of \/PROJECT-NAME.\n\n## Copy Heroku database locally (for Django project)\n\nIf you want to test stuff locally with the contents of your development database, the easy way (assuming your database is small and you\u2019re a hobbyist like me) is to copy it to your local machine.\n\n2. In the directory where your Heroku project is, generate a copy of the current database:\nheroku pg:backups:capture\n\nheroku pg:backups:download\n\n4. Restore the database to your local system:\npg_restore --verbose --clean --no-acl --no-owner -h localhost -U PSQL_USERNAME -d DB_NAME BACKUP_DUMP\n\n1. where PSQL_USERNAME is the username that will own the database (this might be your system username)\n2. DB_NAME is the name of the database (this should match what\u2019s in your settings)\n3. BACKUP_DUMP is the name of the file you downloaded in the previous step (something like latest.dump)\n5. If the previous step asks for a password and you haven\u2019t set one for your postgres user yet:\nsudo -u postgres psql\n\n\nALTER USER PSQL_USERNAME PASSWORD 'newPassword';\n\n\nDon\u2019t forget the trailing semicolon! Then exit with Ctrl+D or \\q.\n\n6. If it complains that the database doesn\u2019t exist, the easiest solution is to run\npython3 manage.py migrate\n\n\nand try again.\n\nSource: Heroku, StackOverflow","date":"2020-07-08 00:15:11","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\": 0.2921716868877411, \"perplexity\": 9672.898468227888}, \"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\/1593655896169.35\/warc\/CC-MAIN-20200708000016-20200708030016-00366.warc.gz\"}"}
| null | null |
Artist Safety
INFO ALERTS
What does TWB do? Here are just a few examples:
Theatre & Peacebuilding: In 2005 TWB and Coexistence International at Brandeis University began what was to be a 3-year affiliation under the auspices of the Alan B. Slifka Program in Intercommunal Coexistence. The Acting Together Project (Acting Together on the World Stage) grew out of this multi-year collaboration between the Brandeis University Peacebuilding and the Arts Program and the Theatre Without Borders "Theatre and Peacebuilding Initiative." Today, the Acting Together Project involves scholars and practitioners working in over fifteen conflict regions. Documentation of this work is available in an anthology, a documentary, and a web site.
Theatre & Human Rights: Theatre Without Borders is excited to support the creation of ArtistSafety.net, a humanitarian initiative around Artists and Human Rights, which grew out of a collaboration with freeDimensional.
Theatre & Climate: Theatre Without Borders has recently started a new initiative around the intersection of artists and climate change. More details to follow.
Theatre & Social Engagement: Theatre Without Borders is exploring a 2015 or 2016 conference on Theatre & Social Engagement, which will encompass the many ways in which artist activism is impacting societies world-wide. More details to follow.
Translation Project: The Translation Project was inspired by the work of The Translation Think Tank, which was founded in 2004 by Marie-Louise Miller and Sarah Cameron Sundeto to re-engage discussion amongst disparate translation communities about the current state and values of theatrical translation.
Cultural Mobility: TWB collaborated on the first-ever Cultural Mobility Funding Guide for the USA: Theatre, Dance and the Performing Arts, in collaboration with the Martin E. Segal Theatre Center and the European cultural mobility network On the Move. In 2012, TWB has been welcomed as the first US-based member of On the Move (OTM), a cultural mobility information network with more than 30 members in over 20 countries across Europe and beyond. On the Move's mission is to encourage and facilitate sustainable cross-border mobility and cooperation, contributing to building up a vibrant and shared European cultural space that is strongly connected worldwide. TWB looks forward to developing links with On the Move, which has offices based in Brussels, Belgium. You can take advantage of On the Move's fantastic resources right now by going to their website: www.on-the-move.org and access the Funding Guides for Arab Countries, Asia, Europe and the USA at: www.us-culturalmobility.org
Performance & Cultural Diplomacy: In June, 2012 TWB joined over 70 artists, educators, activists and policy makers at the Georgetown University convening entitled Global Performance, Civic Imagination and Cultural Diplomacy. Co-organizers of what is now called the Laboratory for Global Performance and Politics, Derek Goldman (Theatre & Performance Arts Professor and Artistic Director of the Davis Performing Arts Center) and Cynthia Schneider (Distinguished Professor in the Practice of Diplomacy in the School for Foreign Service and former US Ambassador to The Netherlands) invited those assembled to examine the impact, both positive and negative, that performing arts can have on public policy. TWB is committed to exploring this powerful and burgeoning field at the intersection of cultural workers and policy makers working side-by-side in complimentary practice.
Send TWB a message:
Ask us anything and we will do our best to answer you.
Submit to our INFO ALERTS newsletter: :
Suggest something of interest to our global community.
Artists & Human Rights :
Get connected to others working to protect the safety of theatre and performing artists.
Website Problems/Feedback:
If you see a problem, please let us know so we can find a solution.
©2015 Theatre Without Borders, All rights reserved.
|
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Snap est un système d'empaquetage et de déploiement de logiciels développé par Canonical pour les systèmes d'exploitation utilisant le noyau Linux. Les paquets, appelés snaps, et l'outil pour les utiliser, snapd, fonctionnent sur une gamme de distributions Linux et permettent aux développeurs de logiciels de distribuer leurs applications directement aux utilisateurs. Les snaps sont des applications autonomes qui s'exécutent dans un bac à sable avec un accès modéré au système hôte. Snap a été initialement publié pour les applications cloud, mais a ensuite été porté afin de fonctionner pour les périphériques Internet des objets ainsi que les applications de bureau.
Fonctionnalités
Snap Store
Le Snap Store permet aux développeurs de publier leurs applications directement auprès des utilisateurs. Avec les approches traditionnelles de gestion des packages Linux comme APT ou YUM, les applications sont empaquetées et distribuées dans le cadre du système d'exploitation. Cela crée un délai entre le développement de l'application et son déploiement pour les utilisateurs finaux. En revanche, les développeurs d'applications peuvent publier leurs applications dans le Snap Store et les transmettre directement aux utilisateurs sans aucune intervention des responsables de la distribution.
Toutes les applications téléchargées sur le Snap Store subissent des tests automatiques, y compris une analyse des logiciels malveillants. Cependant, les applications Snap ne reçoivent pas le même niveau de vérification que les logiciels dans les archives Ubuntu normales. Dans un cas, en mai 2018, deux applications du même développeur se sont avérées contenir un mineur de cryptomonnaie qui s'exécutait en arrière-plan pendant l'exécution de l'application. Lorsque ce problème a été détecté, Canonical a supprimé les applications du Snap Store et transféré la propriété des Snaps à un tiers de confiance qui a republié les Snaps sans la présence du mineur. Bien que le bac à sable Snap réduise l'impact d'une application malveillante, Canonical recommande aux utilisateurs d'installer uniquement des Snaps provenant d'éditeurs approuvés par l'utilisateur.
Étant donné que les packages du Snap Store sont gérés par les développeurs eux-mêmes, les responsables de la distribution ne peuvent pas garantir que les packages répondent aux normes de qualité et sont mis à jour en temps opportun. Dans un cas, Microsoft a laissé une version obsolète de Skype dans le magasin Snapcraft pendant plus d'un an.
Bien que le Snap Store de Canonical soit actuellement le seul magasin existant pour les snaps, Snap lui-même peut être utilisé sans magasin. Les packages Snap peuvent être obtenus à partir de n'importe quelle source, y compris le site Web d'un développeur. L'employé de Red Hat, Adam Williamson, et le chef de projet Linux Mint, Clément Lefebre, ont critiqué Canonical pour avoir gardé fermées certaines parties côté serveur du Snap Store, créant ainsi un conflit d'intérêts qui nuit à la neutralité de Snap, étant donné que Canonical développe Ubuntu.
Paquets Linux universels
Les Snaps sont des paquets autonomes qui fonctionnent sur une gamme de distributions Linux . Ceci est différent des approches traditionnelles de gestion de packages Linux, qui nécessitent des packages spécifiquement adaptés pour chaque distribution Linux.
Le format de fichier snap est un système de fichiers compressé unique utilisant le format SquashFS avec l'extension .snap . Ce système de fichiers contient l'application, les bibliothèques dont elle dépend et les métadonnées déclaratives. Ces métadonnées sont interprétées par snapd pour configurer un bac à sable sécurisé adapté au fonctionnement de l'application. Après l'installation, le snap est monté par le système d'exploitation hôte et décompressé à la volée lorsque les fichiers sont utilisés. Bien que cela présente l'avantage que les snaps utilisent moins d'espace disque, cela signifie également que certaines applications volumineuses démarrent plus lentement.
Une différence significative entre Snap et d'autres formats d'empaquetage Linux universels, tels que Flatpak, est que Snap prend en charge toutes les classes d'applications Linux telles que les applications de bureau, les outils serveur, les applications IoT et même les services système tels que la pile de pilotes d'imprimante. Pour s'en assurer, Snap s'appuie sur systemd pour des fonctionnalités telles que l'exécution de services système activés par socket dans un Snap. Cela fait que Snap ne fonctionne bien que sur les distributions qui peuvent adopter ce système d'initialisation.
Bac à sable configurable
Les applications dans un Snap s'exécutent dans un conteneur avec un accès limité au système hôte. À l'aide d'interfaces, les utilisateurs peuvent donner à une application un accès modéré à des fonctionnalités supplémentaires de l'hôte, telles que l'enregistrement audio, l'accès aux périphériques USB et l'enregistrement vidéo. Ces interfaces servent de médiateur pour les API Linux habituelles afin que les applications puissent fonctionner dans le bac à sable sans avoir besoin d'être réécrites. Les applications de bureau peuvent également utiliser les portails de bureau XDG, une API standardisée créée à l'origine par le projet Flatpak pour permettre aux applications de bureau en bac à sable d'accéder aux ressources de l'hôte. Ces portails offrent souvent une meilleure expérience utilisateur par rapport aux API Linux natives, car ils demandent à l'utilisateur l'autorisation d'accéder à des ressources telles qu'une webcam au moment où l'application les utilise. L'inconvénient est que les applications et les boîtes à outils doivent être réécrites afin d'utiliser ces nouvelles API.
Le bac à sable Snap prend également en charge le partage de données et de sockets Unix entre Snaps. Ceci est souvent utilisé pour partager des bibliothèques communes et des frameworks d'applications entre Snaps afin de réduire la taille des Snaps en évitant la duplication.
Le bac à sable Snap s'appuie fortement sur le module de sécurité AppArmor du noyau Linux. Étant donné qu'un seul module de sécurité Linux (LSM) « majeur » peut être actif en même temps, le bac à sable Snap est beaucoup moins sécurisé lorsqu'un autre LSM majeur est activé. Par conséquent, sur des distributions telles que Fedora qui activent SELinux par défaut, le bac à sable Snap est fortement dégradé. Bien que Canonical travaille avec de nombreux autres développeurs et entreprises pour permettre l'exécution simultanée de plusieurs LSM, cette solution n'est pas encore opérationnelle.
Le bac à sable Snap empêche les applications de bureau d'accéder aux thèmes du système d'exploitation hôte pour éviter les problèmes de compatibilité. Pour que les Snaps utilisent un thème, ils doivent également être emballés dans un Snap séparé. De nombreux thèmes populaires sont emballés par les développeurs de Snap mais certains thèmes ne sont pas encore pris en charge et les thèmes peu communs doivent être installés manuellement. Si un thème n'est pas disponible en tant que package Snap, les utilisateurs doivent choisir le meilleur thème correspondant disponible. Des travaux sont en cours pour permettre à des tiers de regrouper plus facilement des thèmes dans un Snap et d'installer automatiquement des thèmes système inhabituels.
Mises à jour automatiques et atomiques
Plusieurs fois par jour, snapd vérifie les mises à jour disponibles de tous les Snaps et les installe en arrière-plan à l'aide d'une opération atomique . Les mises à jour peuvent être annulées et utilisent l'encodage delta pour réduire leur taille de téléchargement.
Les éditeurs peuvent publier et mettre à jour plusieurs versions de leur logiciel en parallèle à l'aide de canaux (channels). Chaque canal a une piste (track) et un risque (risk) spécifiques, qui indiquent la version et la stabilité du logiciel publié sur ce canal. Lors de l'installation d'une application, Snap utilise par défaut le canal latest/stable, qui se mettra automatiquement à jour vers les nouvelles versions majeures du logiciel lorsqu'elles seront disponibles. Les éditeurs peuvent créer des canaux supplémentaires pour donner aux utilisateurs la possibilité de se fixer sur des versions majeures spécifiques de leur logiciel. Par exemple, un 2.0/stable permettrait aux utilisateurs de s'en tenir à la version 2.0 du logiciel et d'obtenir uniquement des mises à jour mineures sans risque de modifications rétrocompatibles. Lorsque l'éditeur publie une nouvelle version majeure dans un nouveau canal, les utilisateurs peuvent passer manuellement à la version suivante lorsqu'ils le souhaitent.
Bien que les mises à jour automatiques ne puissent pas être désactivées, il existe de nombreuses façons de configurer les mises à jour pour répondre à des besoins particuliers. L'utilisateur peut choisir de rester sur une version majeure spécifique du logiciel en spécifiant le canal, il peut configurer l'intervalle de mise à jour pour avoir le temps de vérifier manuellement les mises à jour, et il peut suspendre les mises à jour jusqu'à 60 jours. De plus, les mises à jour sont également automatiquement désactivées sur les connexions limitées. Même avec ces contrôles, un certain nombre d'utilisateurs se sont plaints de l'absence d'option permettant de désactiver complètement les mises à jour automatiques.
Snapcraft
Snapcraft est un outil permettant aux développeurs d'empaqueter leurs programmes au format Snap. Il fonctionne sur n'importe quelle distribution Linux prise en charge par Snap, macOS et Microsoft Windows. Snapcraft construit les packages dans une machine virtuelle en utilisant Multipass, afin de s'assurer que le résultat d'une construction est le même, quel que soit la distribution ou le système d'exploitation sur lequel il est construit. Snapcraft prend en charge un grand nombre d'outils de construction et de langages de programmation tels que Go, Java, JavaScript, Python, C / C++ et Rust . Il permet également d'importer des métadonnées d'application à partir de plusieurs sources telles qu'AppStream, git, des scripts shell et des fichiers setup.py .
Adoption
Au départ, Snap ne prenait en charge que la distribution all-Snap Ubuntu Core, mais en juin 2016, il a été porté sur un large éventail de distributions Linux pour devenir un format pour les packages Linux universels. Snap nécessite une fonctionnalité (systemd), disponible sous Linux, certains autres systèmes de type Unix n'en ont pas, donc par exemple FreeBSD n'a pas Snap. Chrome OS ne prend pas en charge Snap directement, uniquement via les distributions Linux installées qui prennent en charge Snap, telles que Gallium OS .
Un certain nombre de sociétés de développement de logiciels de bureau publient leurs logiciels dans le Snap Store, notamment Google, JetBrains, KDE, Microsoft (pour les versions Linux de par exemple .NET Core 3.1, Visual Studio Code, Skype, et PowerShell), Mozilla et Spotify. Les Snaps sont également utilisés dans les environnements Internet des objets, allant des produits destinés aux consommateurs aux passerelles de gestion des appareils d'entreprise et aux réseaux de communication par satellite. Enfin, Snap est également utilisé par les développeurs d'applications serveur telles que InfluxDB, Kata Containers, Nextcloud et Travis CI .
En 2019, Canonical a décidé de faire passer le navigateur web Chromium dans les futures versions d'Ubuntu d'un package APT à un Snap. Ils ont expliqué que Snap rendait beaucoup plus facile la prise en charge de Chromium sur toutes les versions d'Ubuntu prises en charge. Cela leur a permis de concentrer les ressources d'ingénierie sur d'autres parties d'Ubuntu. À la suite de cette décision, les dérivés d'Ubuntu tels que Linux Mint ont dû choisir entre conserver leur propre version du package Chromium ou passer à la version « snappée » de Chromium maintenue par Canonical. Clement Lefebre, fondateur et chef de projet de Linux Mint a mentionné cela comme la raison pour laquelle Linux Mint 20 empêche les utilisateurs d'installer Snap. Étant donné que Snap est toujours disponible dans les référentiels Linux Mint, les utilisateurs peuvent toujours l'installer après avoir supprimé manuellement les restrictions.
Disponibilités
Distributions avec Snap pré-installé
Ubuntu (et les dérivées)
KDE Neon
Manjaro
Zorin OS
Solus
Distribution sans Snap pré-installé
Arch Linux
CentOS
Debian
Elementary OS
Fedora
GalliumOS
Kali Linux
Linux Mint
OpenSUSE
Parrot Security OS
Pop! OS
Raspberry PI OS
Red Hat Enterprise Linux
Rocky Linux
Notes et références
Voir aussi
Articles connexes
AppImage
Flatpak
Créateurs d'applications portables
ROX utilise des répertoires (AppDirs) en tant que bundles d'applications.
Zero Install, un projet similaire
Liens externes
Site officiel
Une comparaison technique entre snaps et debs
Ubuntu
Gestionnaire de paquets
Logiciel libre
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,195
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Q: How to update password through passwordstrength AJAX control I have developed a webpage ASP.net with C# which use to update password in oracle database using AJAX PasswordStrenght Control. When Password meets the depict requirements, by clicking on button it should update password in database. However when I click on button it is not update password in databse.
PasswordStrenght Code:-
<ajaxToolkit:PasswordStrength ID="PasswordStrength1" runat="server" DisplayPosition="RightSide" StrengthIndicatorType="BarIndicator"
TargetControlID="TextBox1" PrefixText="Stength:" Enabled="true"
RequiresUpperAndLowerCaseCharacters="true" MinimumLowerCaseCharacters="1"
MinimumUpperCaseCharacters="1" MinimumSymbolCharacters="1"
MinimumNumericCharacters="1" PreferredPasswordLength="10"
TextStrengthDescriptions="VeryPoor; Weak; Average; Strong; Excellent"
StrengthStyles="VeryPoor; Weak; Average; Excellent; Strong;"
CalculationWeightings="25;25;15;35" BarBorderCssClass="border"
HelpStatusLabelID="Label1" />
Please refer below code :-
protected void Page_Load(object sender, EventArgs e)
{
if (!IsPostBack)
{
Label2.Text = Session["PERNR"].ToString();
Label3.Text = Session["ZZFNAME"].ToString();
}
}
protected void Button1_Click(object sender, EventArgs e)
{
if (Label1.Text == "" && TextBox1.Text != "")
{
OracleConnection con = new OracleConnection("Data Source=10.127.240.216/ipcldb;User ID=ipcltos;Password=ipcltos;Unicode=True");
con.Open();
OracleCommand cmd = new OracleCommand("UPDATE SAP_EMPMST SET PASSWORD = '" + TextBox1.Text + "' WHERE PERNR = '" + Label2.Text + "'", con);
cmd.ExecuteNonQuery();
string display = "New Password has been updated...!!!";
ClientScript.RegisterStartupScript(this.GetType(), "myalert", "alert('" + display + "');", true);
con.Close();
}
else
{
string display = "New Password is not matching with security requirments...!!!";
ClientScript.RegisterStartupScript(this.GetType(), "myalert", "alert('" + display + "');", true);
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 586
|
from __future__ import absolute_import
from bokeh.io import save
from bokeh.plotting import figure
from bokeh.models import ResetTool, CustomJS, Range1d
from selenium.webdriver.common.action_chains import ActionChains
from tests.integration.utils import has_no_console_errors
import pytest
pytestmark = pytest.mark.integration
def make_plot(tools=''):
plot = figure(height=800, width=1000, tools=tools,
x_range=Range1d(0,10), y_range=Range1d(0,10))
plot.rect(x=[1, 2], y=[1, 1], width=1, height=1)
return plot
def click_element_at_position(selenium, element, x, y):
actions = ActionChains(selenium)
actions.move_to_element_with_offset(element, x, y)
actions.click_and_hold() # Works on ff & chrome
actions.release()
actions.perform()
def test_reset_triggers_range_callback(output_file_url, selenium):
# Make plot and add a range callback that generates an alert
plot = make_plot('reset')
range1d = plot.select(dict(type=Range1d))[0]
range1d.callback = CustomJS(code='alert("plot reset")')
# Save the plot and start the test
save(plot)
selenium.get(output_file_url)
assert has_no_console_errors(selenium)
# Tap the plot and test for alert
reset_button = selenium.find_element_by_class_name('bk-tool-icon-reset')
click_element_at_position(selenium, reset_button, 10, 10)
alert = selenium.switch_to_alert()
assert alert.text == 'plot reset'
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,358
|
I am moments away from completing my Birthright application for the December trip.
My heart is pounding and I feel like there's sand behind my eyes. All this time, I wasn't going to Israel because my mother was nervous. Now I'm nervous. But I'm still going. My mom and I will both be nervous, no matter what. I need to do this, and I need to do it now, before Cai turns 26. I want to go with him.
Oy vey, I worry like the Jewish grandmother I'll be one day. Everything will be fine...I'm just nervous.
But as Rabbi Hillel himself once said: "If not now, when?" It's going to be AMAZING. When else would I have the chance to ride a camel through the negev (desert)? We'll visit archaeological sites, walk through street bazaars in Jerusalem, float in the Dead Sea, drink tea with the Bedouins. All my life, I have faced Jerusalem during prayer services, bowing, perpetually, to the East. In December, I'll actually be there.
All I have to do is get my passport information together, and I'll be finished with the application process. As I told my parents, I'm going to be in Israel at some point in my life. It's better to go now, on a free, guided, 10-day tour, than on my own later. That said, I haven't talked to my parents yet about the prospect of extending our visit by a couple of days so we can spend some time with S and Y, who are studying at HUC and Conservative Yeshiva in Jerusalem, respectively. I have to talk to them about it first. But it would be a wonderful way to end our visit. I miss them all the time. Even if it doesn't quite work out, hopefully they can meet up with us somewhere...we'll see.
and I have a list of things I'd like for you to bring back.
One for you to think about: if you bring me yarn, I will make you a scarf from the Holyland.
Ooh! A holy (not holey) scarf! Definitely make a list. I'm compiling my own. Our first mezuzah will come from Jerusalem!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 7,626
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\section{Introduction}
RERS-Fuzz is an automated test-generation tool targeted for coverage maximization in reactive softwares. The techniques employed are heavily based on key concepts of Greybox fuzz-testing~\cite{ref_greybox}. The underlying idea is to employ evolutionary algorithms for generating interesting test-inputs that help in exploration of newer code segments. In Greybox fuzz-testing, a set of random test-inputs are considered as initial population. The fitness function is the code coverage measure observed by an individual test-input. For every test-input, fitness value is calculated and best-fit test-inputs are retained for further fuzzing. The retained test-inputs are subjected to various mutation and crossover operations for generation of newer test-inputs. The cycle repeats until certain user-defined goals are met or statistical measures on coverage metrics are achieved.
We have mainly used the following tools and techniques and modified them inline with our requirements to solve reachability problems on \texttt{RERS} 2019 benchmarks :-
\begin{itemize}
\item American Fuzzy Lop (AFL v2.52b)~\cite{ref_afl}.
\item Instrim~\cite{ref_instrim}.
\item VeriFuzz~\cite{ref_verifuzz}
\item LLVM Interval Analysis~\cite{ref_intervalAnalysis}.
\end{itemize}
\noindent In subsequent sections, we briefly describe about our approach, highlighting core techniques, tool architecture and strengths and weaknesses associated with them.
\section{Approach}
State-of-art greybox fuzzers, like \texttt{afl-fuzz} uses a stronger notion of structural code coverage of program's control-flow-graph (\texttt{CFG}), called \textit{branch-pair} coverage as fitness function. The tool maintains a shared 64kB memory where each byte entry represents a logarithmic visit count of a typical branch-pair by all test-inputs generated so far. The amount of shared memory is designed in such a way that entire shared memory can be stored in cache-memory and hence execution speed of fuzz-testing tool remains unhampered. However, as per our observation, the number of conditional statements in \texttt{RERS} benchmarks are typically very high. Hence, instrumenting each conditional statement of program would lead to heavy collision~\cite{ref_afl_whitepaper}. Secondly, the ranges of input values driving the reactive softwares is comparatively small. Hence, application of heavy-weight and sophisticated mutation and crossover operations would amount to generation of useless test-inputs rejected by the reactive system. Finally, in reactive softwares, code-coverage does not necessarily ensure that newer states are also covered.
In our tool, we have tried of identify the problems where plain usage of greybox fuzz testing tool would fail. We have added our own techniques on top of core-fuzzing engine to suitably tune the application of evolutionary fuzz testing on \texttt{RERS} benchmarks. The following are the key techniques of our approach.
\\ \\
\textbf{Minimal Node Instrumentation}: As discussed earlier, the software-under-test (\texttt{SUT}) is subjected to node instrumentation for measuring code coverage during a test-run. However, in \texttt{RERS} benchmarks the number of conditional statements are significantly high and complex. In order to reduce cache collision and execution overhead arising from instrumentation, number of instrumentation points have to be reduced. In our tool, we have used lightweight instrumentation~\cite{ref_instrim} which is efficient and instruments minimal conditional statements of \texttt{SUT}, without loosing any coverage information. The technique have posed the minimal instrumentation as \textit{path differentiation problem} and identify minimal number of nodes in \texttt{CFG} for which any two branch-pairs can be differentiated.
\\ \\
\textbf{Interval Analysis assisted Mutation}: Reactive softwares belong to class of program where input space is very restricted~\cite{ref_verifuzz}. It is possible to tune mutation and crossover operation within the ranges of input taken by ~\texttt{SUT}. We have taken help of LLVM's interval analysis~\cite{ref_intervalAnalysis} to identify ranges of input consumed by \texttt{RERS} to go into deeper program segments. The values are then passed on to our parametric fuzz engine, which in turn does crossover and mutation operation over those restricted ranges.
\\ \\
\textbf{State Instrumentation}: The minimal node instrumentation takes care of efficient branch-pair coverage in program's \texttt{CFG}. But, it may happen that a test-input have exhibited no new branch-pair coverage in its run, but it have explored a new \textbf{\textit{state}}. We define \textbf{\textit{state}} as, unique values assignments to all the global variables in \texttt{SUT}. In addition to existing 64k shared memory for branch-pair instrumentation, we have additionally deployed 64k shared memory, which maintains a binary valued entry whether a \textbf{\textit{state}} has been explored or not. Accordingly, our fitness function is also updated with state coverage information. Any test-input with no new branch-pair coverage but new state space coverage are now retained for further fuzzing.
\section{Tool Implementation}
\begin{figure}
\includegraphics[scale=0.8]{rers_diag_2019.pdf}
\caption{RERS-Fuzz : Overview}
\label{fig:rersFuzzOverview}
\end{figure}
The overall flow of RERS-Fuzz has been shown in figure~\ref{fig:rersFuzzOverview}. The fuzzing module has been developed on top
greybox fuzzer \texttt{afl-fuzz}~\cite{ref_afl}. The green blocks in the fuzzing module denotes that techniques have been modified and tuned with \texttt{RERS} programs. Red blocks denote the algorithm is unchanged and in line with afl-fuzz core algorithm. We have additionally used Instrim\footnote{https://github.com/csienslab/instrim} package for minimal node instrumentation on top of LLVM-7 framework. For Interval Analysis module, we used the LLVM package developed for interval analysis\footnote{http://llvm.org/doxygen/Interval\_8h\_source.html}. The tool has been developed using C, C++ and python.
\section{Strengths and Weakness}
RERS-Fuzz participated in Reachability Track of RERS 2019 competition. The tool could find out counterexamples for 217 academic reachability benchmarks. In industrial reachability track, it has emerged as overall winner, scoring 2038 points.
The core strength of RERS-Fuzz is its capability to scale up on large industrial benchmarks and find sufficient number of error reachable locations. We have run all sequential and industrial reachability benchmarks for a timeline of 8 hours on Intel i7 3720 Octa-Core Processor (2.6GHz) and presented our results. The entry for the specification with \texttt{error\_reachable} denotes that error location is reachable. Entries with \texttt{UNKNOWN} denote we are unable to conclude whether the error location is reachable or not.
Weaknesses involve the inability of the tool to generate proofs for unreachability of error location. Besides that, we have observed that running certain benchmarks for a longer time period can yield more number of error reachable locations. However there are no assurances that preserving test-inputs with newer state-space or branch-pair would definitely aid to uncover error-reachable locations during evolution. With our technique, we try to maximize a stronger notion code-coverage and state-space coverage of \texttt{RERS} benchmarks. Today, our tool lacks focused search towards reachable error location. We believe that in future, such problems can be modelled as appropriate fitness function and hence fuzzing can be made directed. Also, such high quality test-vectors would definitely aid various invariant learning algorithms, which in turn would help prove correctness of such reactive software.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,935
|
\section{Introduction}
\glsresetall
Recently, the new generation of cellular networks has been successfully integrated and deployed bringing along new business opportunities. However, the revenue-hungry telco operators continuously look for innovative solutions to enable new use cases, which involve new players into the engaged business model. In this context, one emerging technology aims at undermining the classical communication paradigm---that dogmatized the radio propagation environment as an ungovernable box---providing new means to exploit the signal properties: \emph{\glspl{ris}}~\cite{BOL20_ComMag, RIScommag_2021, EuCNC_rise6g,albanese_commag21}.
Agility and flexibility represent the added-value of this solution~\cite{DiRenzo2020,albanese22}: while \glspl{ris} can be dynamically and continuously configured, they draw little power with affordable installation and maintenance costs~\cite{di2019smart}. This makes such a technology the best candidate to solve the mobile dead-zone problem in indoor scenarios by enabling very dense \gls{ris}-based network deployment at low Capital Expenditure (CAPEX). For instance, as shown in Fig.~\ref{fig:heatmap_no_ris}, the existing network infrastructure in a real railway station may fail to guarantee satisfactory performance within the entire environment: \emph{How to solve the dead-zone problem with a very limited investment?}
Ad-hoc \glspl{ris} design and deployment strategies might be the correct answer.
\begin{figure}[t]
\center
\includegraphics[width=.97\linewidth]{rennes_intro.pdf}
\caption{\label{fig:heatmap_no_ris} Railway station topographic map and related power heatmap showing the dead-zone problem (Rennes, France).}
\end{figure}
Indeed, while \glspl{ris} properly steer the incoming electromagnetic waves towards specific directions, interference is also focused onto unwanted areas, if not properly handled~\cite{metasurface21}. This issue exacerbates the overall deployment complexity calling for advanced optimization techniques to strike the optimal trade-off between \glspl{ris} density and the corresponding spurious detrimental interference.
{\bf Related work.} In the literature, the generic \glspl{bs} deployment problem has been exhaustively investigated, e.g. in~\cite{Andrews2011,Amaldi2003}. The major drawback of such works lies in the isotropic antenna radiation assumption making the problem easy-to-solve via graph-coloring algorithms or convex programming approaches. When dealing with directive transmissions---e.g., millimeter waves (mmWaves) above $6$~GHz---a new degree of freedom is introduced: the beam orientation. Specifically, mmWave \glspl{bs} must be properly placed and electronically oriented to effectively beam towards specific locations leveraging on the available \gls{csi}~\cite{Fascista2019, papir21}. Nonetheless, an optimal \glspl{ris} deployment is even harder to achieve: on the one hand, \glspl{ris} deployment requires prior information on the applied \glspl{ris} configurations; on the other hand, \glspl{ris} configurations can be obtained only upon fixing the \glspl{bs} and \glspl{ris} positions. To overcome this issue and make the analysis tractable, simplistic assumptions on agnostic \glspl{ris} optimization can be done~\cite{Moro2021}.
{\bf Contributions.} Differently, our solution goes one step beyond and jointly tackles the optimal \glspl{ris} placement and configuration problems without any unpractical assumption on the available \gls{csi}. We formulate the overall optimization framework and rely on the well-known \gls{bca}~\cite{Grippo2000} to devise RISA{}, a RIS-Aware network planning solution that iteratively derives the \glspl{ris} configurations and optimally places the required number of \glspl{ris} within the area. We $i$) develop a new lightweight ray-tracing model for multi-\gls{ris} scenarios, $ii$) analytically and empirically prove its short convergence time, $iii$) show its efficiency in large-scale scenarios and $iv$) demonstrate outstanding performance in a realistic indoor environment, namely the Rennes Train Station in France, to improve the existing cellular infrastructure of one of the major European operators and solve the dead-zone problem, as shown in Section~\ref{s:performance}.
\emph{Notation}. We denote matrices and vectors in bold while each of their element is indicated in roman with a subscript. $(\cdot)^{\mathrm{T}}$ and $(\cdot)^{\mathrm{H}}$ stand for vector or matrix transposition and Hermitian transposition, respectively. The L$2$-norm of a vector is denoted by $\| \cdot \|$.
\section{System model}
\label{s:system_model}
\addtolength{\topmargin}{+0.28in}
\begin{figure}[t]
\center
\includegraphics[width=.95\linewidth]{scenario.pdf}
\caption{\label{fig:scenario} Geometrical representation of the considered scenario including \glspl{bs}, the \glspl{ris} and one sample \gls{ue}.}
\end{figure}
We consider the
\gls{ris}-enabled wireless network depicted in Fig.~\ref{fig:scenario}, wherein $N$ \glspl{ris} are deployed to assist $M$ \glspl{bs} to extend their communication coverage in a given area of interest $\mathcal{A}$. We model each \gls{bs} as a \gls{ula} with $N_b$ antennas, and each \gls{ris} as a \gls{pla} with $N_r = N_{h} \times N_v$ reflective elements, where $N_h$ and $N_v$ denote the number of elements in the horizontal plane and the vertical direction of the absolute reference system, respectively.
We indicate by $\vv{b}_m\in \mathbb{R}^3$, $\vv{r}_n\in \mathbb{R}^3$ and $\vv{u} \in \mathbb{R}^3$ the locations of the $m$-th \gls{bs} center, the $n$-th \gls{ris} center and the typical \gls{ue}, respectively. We assume that the direct \gls{los} links from the \glspl{bs} provide negligible receive power in the target area due to blockage or severe shadowing. Therefore, the communication between \glspl{bs} and \glspl{ue} must be carried out over the reflected link through the \glspl{ris}. In practice, we assume that each \gls{bs} can leverage on multiple \glspl{ris} but each \gls{ris} is used and controlled by a single \gls{bs}, which connects to the on-board \gls{ris} controller via a separate (wired or wireless) reliable control link. Focusing on the downlink transmission, the $m$-th \gls{bs} transmits data to the \gls{ue} over the reflected links through the $n$-th \gls{ris}. Such path can be decomposed into the \gls{los} channel $\vv{h}_n \in \mathbb{C}^{N_r \times 1}$ through which the \gls{ris} reflects the impinging signal towards the \gls{ue}, and the \gls{los} channel $\vv{G}_{mn} \in \mathbb{C}^{N_r\times N_b}$ between the \gls{bs} and the \gls{ris}.
Let us indicate as $\Lambda_m$, with cardinality $|\Lambda_m|$, the set of \glspl{ris} that are associated with BS $m$. The received downlink signal at the \gls{ue} is given by the superposition of the signals incoming from all \glspl{bs} through their associated \glspl{ris}, namely
\begin{equation}\label{eq:y}
y \triangleq \sum\limits_{m=0}^{M-1}\sum\limits_{n = 0}^{|\Lambda_m|} \left(\vvhs{h}{n} \vv{\Phi}_n \vv{G}_{mn}\right) \, \vv{w}_m s + n \in \mathbb{C},
\end{equation}
where $\vv{\Phi}_n = \mathrm{diag}[\alpha_{n1} e^{j\phi_{n1}}, \dots, \alpha_{nN} e^{j\phi_{nN}}]$ with $\phi_{ni} \in [0, 2\pi]$ and $|\alpha_{ni}|^2 \leq 1$, $\forall i$ indicates the phase shifts and amplitude attenuation introduced by the $n$-th \gls{ris}, $\vv{w}_m \in \mathbb{C}^{N_b\times 1}$ is the transmit precoder at the $m$-th BS while $s\in \mathbb{C}$ is the transmit signal with $|s|^2=1$, and $n\in \mathbb{C}$ is the additive white Gaussian noise term distributed as $\mathcal{CN}(0,\sigma^2)$.
\begin{figure}[t]
\center
\includegraphics[width=.45\linewidth]{RIS_reference.png}
\caption{\label{fig:ris_reference} Geometrical representation of one sample user equipment (UE) in the $n$-th RIS reference system.}
\end{figure}
As \gls{3gpp} cellular standards require the \gls{ue} to be served by a single \gls{bs}, we remark that the \gls{ue} receives useful signal only from one \gls{bs}, e.g., the $m$-th \gls{bs}, and suffers from the interference produced by all other \glspl{bs}.
Therefore, the received \gls{sinr} at the \gls{ue} can be written as
\begin{equation}\label{eq:sinr}
\mathrm{SINR}(\vv{u}) \triangleq \frac{\left|\sum\limits_{n = 0}^{|\Lambda_m|} \left(\vvhs{h}{n} \vv{\Phi}_n \vv{G}_{mn}\right) \, \vv{w}_m\right|^2}{\sum\limits_{\substack{l=0,\\ l \neq m}}^{M-1}\left|\sum\limits_{n = 0}^{|\Lambda_l|} \left(\vvhs{h}{n} \vv{\Phi}_n \vv{G}_{ln}\right) \, \vv{w}_l\right|^2 + \sigma^2},
\end{equation}
where the \glspl{bs}-\glspl{ris} and \glspl{ris}-\gls{ue} channels are fully defined by knowing the geometry of the network while the \glspl{ris} configurations and the \glspl{bs} precoders depend on the \glspl{bs}-\glspl{ris} and \glspl{bs}-\gls{ue} associations.
As shown in Fig.~\ref{fig:ris_reference}, in order to write the channels $\vv{h}_n$ and $\vv{G}_{mn}$, we first consider $N$ reference systems with origin in
the center of each \gls{ris} and the $(x',y')$-plane
lying on the RIS surface.
Hence, the coordinates of the \gls{ue} in the reference system of the $n$-th \gls{ris} can be obtained as
$\vv{u}^{(n)} = \vv{R}_n \vv{u}$,
where
\begin{equation}
\vv{R}_n \triangleq
\begin{pmatrix}
\hat{\vv{r}}_{n,x'} & \hat{\vv{r}}_{n,y'} & \hat{\vv{r}}_{n,z'}
\end{pmatrix} \in \mathbb{R}^{3 \times 3},
\end{equation}
with $\hat{\vv{r}}_{n,x'}$, $\hat{\vv{r}}_{n,y'}$ and $\hat{\vv{r}}_{n,z'} \in \mathbb{R}^3$ representing the coordinates of the $n$-th \gls{ris} reference system axes in the absolute reference system. Furthermore, we denote by $\psi_{\scriptscriptstyle{\mathrm{D}},n}$ and $\eta_{\scriptscriptstyle{\mathrm{D}},n}$ the azimuth and the zenith \gls{aod} for the communication link from the \gls{ris} to the \gls{ue}. Therefore, the \gls{ris} array response vector is given by
\begin{align}
\vv{b}_{\scriptscriptstyle{\mathrm{T}},n}(\vv{u}) \triangleq \, & \vv{b}_y (\psi_{\scriptscriptstyle{\mathrm{D}},n}, \eta_{\scriptscriptstyle{\mathrm{D}},n}) \otimes \vv{b}_z (\psi_{\scriptscriptstyle{\mathrm{D}},n}, \eta_{\scriptscriptstyle{\mathrm{D}},n}) \in \mathbb{C}^{N_r\times 1}\label{eq:pla}\\
= \,&[1, e^{j2\pi\delta\sin(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n})}, \dots, \nonumber\\
&e^{j2\pi\delta(N_y -1)\sin(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n})}]^\mathrm{T} \nonumber \\
&\otimes[1, e^{j2\pi\delta\cos(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n})}, \dots, \nonumber \\
&e^{j2\pi\delta(N_x -1)\cos(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n})}]^\mathrm{T} ,
\end{align}
where $\delta$ indicates the antenna spacing-wavelength ratio. We refer to $\Omega_{\scriptscriptstyle{\mathrm{T}},n}(\vv{u}) \triangleq \cos(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n}) = \frac{u_x^{(n)}}{\norm{\vv{u}^{(n)}}}$ and $\Psi_{\scriptscriptstyle{\mathrm{T}},n}(\vv{u}) \triangleq \sin(\psi_{\scriptscriptstyle{\mathrm{D}},n})\sin(\eta_{\scriptscriptstyle{\mathrm{D}},n}) = \frac{u_y^{(n)}}{\norm{\vv{u}^{(n)}}}$ as the spatial frequencies along the $x'_n$ and the $y'_n$-axis corresponding to the \gls{aod} towards the \gls{ue} at absolute coordinates $\vv{u}$. Therefore, the \gls{los} $n$-th \gls{ris}-\gls{ue} channel is given by
\begin{equation}
\vv{h}_n(\vv{u}) \triangleq \sqrt{\gamma_n(\vv{u})} \, \vv{b}_{\scriptscriptstyle{\mathrm{T}}.n}(\vv{u}) \in \mathbb{C}^{N_r \times 1}, \label{eq:h_n}
\end{equation}
where $\gamma_n(\vv{u}) \triangleq d_n(\vv{u})^{-\beta}$ is the channel power gain with $d_n(\vv{u}) = \norm{\vv{r}_n - \vv{u}}$ being the Eucledian distance between the \gls{ris} and the \gls{ue}. In a similar way,
the \gls{los} channel between the $m$-th \gls{bs} and the $n$-th \gls{ris} can be written as
\begin{equation}
\vv{G}_{mn} \triangleq \sqrt{\gamma_{\scriptscriptstyle{\mathrm{G}}_{mn}}} \, \vv{b}_{\scriptscriptstyle{\mathrm{R}},n}(\vv{b}_m) \vvhs{a}{m}(\vv{r}_n) \in \mathbb{C}^{N_r\times N_b},
\label{eq:G_mn}
\end{equation}
where $\gamma_{\scriptscriptstyle{\mathrm{G}}_{mn}} \triangleq d_{mn}^{-\beta}$ is the channel power gain with $d_{mn} = \norm{\vv{b_m}-\vv{r}_n}$, $\vv{a}_{\scriptscriptstyle{\mathrm{R}},n}(\vv{b}_m)$ is the array response vector at the \gls{ris} corresponding to the \gls{aoa} from \gls{bs} m, which is derived analogously to Eq.~\eqref{eq:pla}, and $\vv{a}_m(\vv{r}_n)$ indicates the \gls{bs} array response, defined as
\begin{equation}
\vv{a}_m(\vv{r}_n) \triangleq [1, \dots, e^{j2\pi\delta(M-1)\cos(\theta_{\scriptscriptstyle{\mathrm{D}},mn})}]^\mathrm{T} \in \mathbb{C}^{N_b \times 1},
\end{equation}
where $\theta_{\scriptscriptstyle{\mathrm{D}},mn}$ represents the \gls{aod} from the $m$-th \gls{bs} to the $n$-th \gls{ris}.
\section{Problem Formulation}
\label{s:problem}
\textbf{Analytical tractability.} The solution to our multi-\gls{ris} planning problem requires determining the optimal \glspl{ris} deployment to provide coverage within the target area, e.g., by maximizing the worst-case received \gls{sinr} at all locations $\vv{u}$.
To this aim, we need to jointly optimize the active transmit beamformers at the \glspl{bs} as well as the \glspl{ris} placement, their passive beamforming configurations, and their controlling \glspl{bs}, which in turn dictate the optimal end-to-end \gls{bs}-\gls{ue} associations. The resulting optimization problem is highly non-convex and extremely difficult to tackle due to the intricate coupling between the \glspl{bs}-\glspl{ris} and \glspl{bs}-\gls{ue} associations, and the joint active-passive beamforming configurations throughout the network. For instance, even for a given \gls{bs}-\gls{ris}-\gls{ue} association, jointly optimizing the beamforming at the \gls{bs} and the \gls{ris} does not yield a closed-form formulation but rather requires tackling a non-convex problem by alternatively solving the two separate beamforming optimizations until convergence~\cite{Wu2019}.
Therefore, for the sake of analytical tractability, we consider propagation paths involving only first-order \gls{ris} reflections, and assume a cellular-like architecture in which each \gls{ris} provides coverage to one contiguous subarea, thus reducing the scope of the interference generated by the remaining \glspl{ris} to the sheer overlapping area edges.
We would like to highlight that at planning stage, the \glspl{ris} beamforming design for area coverage enhancement cannot take advantage of the knowledge of the instantaneous \gls{csi} of any particular \gls{ue} in the area. Hence, although \glspl{ris} controlled by the same \gls{bs} can be configured to cover the same subarea, it is highly complex to enforce in-phase constructive interference of signals incoming from different \glspl{ris} even if transmitted by the same \gls{bs}.
Let us consider the \gls{ue} to be inside the subarea served by BS $m$ through RIS $n$. In these conditions, its received \gls{sinr} can be then approximated by its \gls{snr}, which is defined as
\begin{equation}\label{eq:snr}
\mathrm{SNR}(\vv{\Phi}_n, \vv{w}_m, \vv{u}) \triangleq \frac{\left| \vvhs{h}{n} \vv{\Phi}_n \vv{G}_{mn} \, \vv{w}_m\right|^2}{\sigma^2},
\end{equation}
where $\vv{\Phi}_n$ and $\vv{w}_m$ need to be optimized.
\textbf{Optimization variables.}
We assume that the \glspl{ris} are deployed only at specific locations, i.e. \glspl{cs}, to reflect the fact that network operators are required to meet logistical, administrative and physical constraints in real-life scenarios. Nonetheless, in the absence of \glspl{cs}, our multi-\gls{ris} planning may be likewise executed by considering any sampling of the deployment area. For the sake of simplicity, we assume that the \glspl{cs} set matches the set $\{\vv{r}_n\}_{n=1}^{N}$, namely the candidate \glspl{ris} positions are pre-defined and we aim at identifying where to actually deploy \glspl{ris} among them. Besides, we sample the target area by means of $T$ test points $\vv{u}_t \in \mathcal{A}$, wherein we optimize the \gls{snr} of the typical \gls{ue}\footnote{Ideally, the test point distribution should match the expected distribution of the users in the target area but the problem formulation remains valid for any distribution of users.}.
Our planning solution outputs the set of \glspl{ris} to be deployed while providing the optimal \gls{bs}-\gls{ris}-\gls{ue} association at each est point. We thus introduce decision variables $\vv{x} \in \{0,1\}^{N}$ and $\vv{y}\in \{0,1\}^{T \times M \times N}$, whose elements $x_n$ and $y_{tmn}$ indicate whether a \gls{ris} is deployed at CS $n$, and the association between the typical UE at test point $\vv{u}_t$, \gls{bs} m and \gls{ris} at \gls{cs} $n$, respectively.
\textbf{\glspl{ris} planning.} We can now formulate the multi-\gls{ris} coverage enhancement problem as the following
\begin{subequations}
\begin{problem}[Multi-\gls{ris} coverage enhancement]\label{problem:max_snr}
\begin{align}
\displaystyle \max_{\vv{\Phi}_n, \vv{w}_m, \vv{x}, \vv{y}} & \displaystyle \min_{\vv{u}_t} \sum\limits_{m,n} y_{tmn}\left| \vvhs{h}{n} \vv{\Phi}_n \vv{G}_{mn} \, \vv{w}_m\right|^2 \label{eq:max_min_snr}\\
\displaystyle \textup{s.t.} & \ \ \displaystyle |\Phi_{n,ii}|^2 \leq 1, \hspace{2.43cm} \forall n, \forall i, \label{eq:phi_con} \\
& \ \ \displaystyle\norm{\vv{w}_m}^2 \leq P, \hspace{2.4cm} \forall m, \label{eq:precoder_con} \\
& \ \ \displaystyle y_{tmn} \,\vv{\hat{r}}_{n,x'}^{\mathrm{T}} (\vv{u}_t - \vv{r}_n) \geq 0, \hspace{0.65cm} \forall t, \forall m, \forall n, \label{eq:test_point_orientation_con}\\
& \ \ \displaystyle y_{tmn} \, \vv{\hat{r}}_{n,x'}^{\mathrm{T}} (\vv{b}_m - \vv{r}_n) \geq 0, \hspace{0.53cm} \forall t, \forall m , \forall n, \label{eq:base_station_orientation_con}\\
& \ \ \displaystyle y_{tmn} \leq x_n, \hspace{2.63cm} \forall t, \forall m, \forall n, \label{eq:deployment_con}\\
& \ \ \displaystyle \sum\limits_{m,n} y_{tmn} = 1, \hspace{2.22cm} \forall t, \label{eq:coverage_con} \\
& \ \ \displaystyle \sum\limits_m \max\limits_t y_{tmn} \leq 1, \label{eq:one_bs_con} \hspace{1.5cm} \forall n\\
& \ \ \displaystyle \sum\limits_{n} x_n = L, \label{eq:budget_con}\\
& \ \ \displaystyle x_n \in [0,1], \quad y_{tmn} \in [0,1], \hspace{0.4cm} \forall t, \forall m, \forall n \label{eq:binary_con},
\end{align}
\end{problem}
\end{subequations}
where we omit the constant noise term $\sigma^2$ and refer to the available transmit power at the \glspl{bs} as $P$. The constraint in Eq.~\eqref{eq:phi_con} ensures that the \glspl{ris} are passive while the one in Eq.~\eqref{eq:precoder_con} enforces that the \glspl{bs} power budget is satisfied by each precoder $\vv{w}_m$. Constraints \eqref{eq:test_point_orientation_con} and \eqref{eq:base_station_orientation_con} guarantee that each \gls{ris} can respectively serve a test point or be assigned to a \gls{bs} only if they front the \gls{ris}, i.e. only if the vector originated in the \gls{ris} and pointing towards the test point or the \gls{bs} has a positive projection on the \gls{ris} orientation vector $\vv{\hat{r}}_{n,x'}$. Moreover, constraint~\eqref{eq:deployment_con} states that a \gls{ris} should be deployed only if at least the \gls{ue} located at one test point would exploit it, whereas constraint~\eqref{eq:coverage_con} reflects the fact that each test point must be covered by only one \gls{ris}. Constraint~\eqref{eq:one_bs_con} forces each \gls{cs} to be associated to at most one \gls{bs} and, lastly, we enforce the number of deployed \glspl{ris} to be equal to $L$ in constraint~\eqref{eq:budget_con}, where $L$ is the number of \glspl{ris} to be deployed by the network operator.
\section{RIS-aware Network Planning}
\label{s:planning}
Even disregarding the interference, Problem~\eqref{problem:max_snr} is still highly complex due to its objective function in Eq.~\eqref{eq:max_min_snr} being the sum of non-convex elements, and the binary constraints in Eq.~\eqref{eq:binary_con} that make it combinatorial. Moreover, as already mentioned in Section~\ref{s:problem}, the lack of knowledge about the instantaneous \glspl{ue} \gls{csi} in the target area during a realistic access procedure invalidates the option of jointly configuring the \glspl{ris} and \glspl{bs} beamformers per \gls{ue}~\cite{Mursia2021}. Therefore, we decouple the \glspl{ris} and \glspl{bs} beamforming configurations from the planning problem itself by configuring each \gls{ris} to provide coverage to one contiguous subarea and assuming that each \gls{bs} radiates all its available power towards each of its associated \glspl{ris} in a \gls{tdma} fashion.
In other words, we assume that the \glspl{ris} have a single-beam radiation pattern and that they do not serve more than one subareas, thereby guaranteeing that all locations belonging to one subarea are served by a single \gls{bs} through one single \gls{ris}. Given sufficient coverage in the area, multiple users in each subarea can be separated by conventional multiple access techniques, such as \gls{tdma} or \gls{ofdma}.
It can be easily observed from Eq.~\eqref{eq:snr} that the \gls{snr} at the \gls{ue} $\vv{u}_t$ provided by \gls{bs} $m$ through \gls{ris} $n$ can be equivalently written as
$\mathrm{SNR}(\vv{\Phi}_n, \vv{w}_m, \vv{u}_t) = \frac{g_1(\vv{\Phi}_m, \vv{w}_m, \vv{u}_t)}{g_2(\vv{u}_t)}$,
where $g_1(\vv{\Phi}_n, \vv{w}_m, \vv{u}_t)$ provides the overall array gain due to the cascaded active and passive beamformings, while $g_2(\vv{u}_t)$ accounts for the concatenated \gls{bs}-\gls{ris}-\gls{ue} pathloss.
Following~\cite{Lu2021}, the \gls{ris} configuration can be obtained by means of 3D beam broadening and flattening, namely by partitioning the \gls{ris} into multiple sub-arrays of smaller size and optimizing their phase shifts to shape one single flattened beam whose beamwidth can be properly tuned to match the size of the target subarea.
In particular, by denoting the subarea covered by \gls{ris} $n$ by $\mathcal{A}_n$ and assuming $\vv{u}_t \in \mathcal{A}_n$, the resulting \gls{bs}-\gls{ris} gain can be written as
\begin{equation}
g_1(\vv{\bar{\Phi}}_n, \bar{\vv{w}}_m, \vv{u}_t) \approx \frac{N_h^2}{\Delta_{n,x'}N_h \delta} \frac{N_v^2}{\Delta_{n,y'}N_v \delta}, \quad \forall \vv{u}_t \in \mathcal{A}_n,
\end{equation}
where $\bar{\vv{\Phi}}_n$ is derived by means of beam broadening and flattening, and $\vv{w}_m$ is the \gls{mrt} precoder, which depends only on $\vv{G}_{mn}$. Besides, $\Delta_{n,x'}$ and $\Delta_{n,y'}$ respectively denote the desired spans of the spatial frequency
deviations along the horizontal $x'_n$ and vertical $y'_n$-axis of \gls{ris} $n$ to cover its subarea and are defined as
\begin{align}
\Delta_{n,x'} \triangleq \max_{\vv{u}_t \in \mathcal{A}_n} \Omega_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_t) - \min_{\vv{u}_t \in \mathcal{A}_n} \Omega_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_t),\\
\Delta_{n,y'} \triangleq \max_{\vv{u}_t \in \mathcal{A}_n} \Psi_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_t) - \min_{\vv{u}_t \in \mathcal{A}_n} \Psi_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_t).
\end{align}
The overall pathloss experienced by the \gls{ue} at coordinates $\vv{u}_t$ is given by $g_2(\vv{u}_t) = d^{\beta}_{mn} d^{\beta}_n(\vv{u}_t)$.
Therefore, we can state the following equivalent formulation for Problem~\eqref{problem:max_snr}, i.e.
\begin{subequations}
\begin{problem}[Multi-\gls{ris} planning]\label{problem:max_snr_eq}
\begin{align}
\displaystyle \max_{\vv{x}, \vv{y}, \vv{\Delta}_{x'}, \vv{\Delta}_{y'}} & \displaystyle \min_{\vv{u}_t} \sum\limits_{m,n} y_{tmn} \frac{1}{\Delta_{n,x'}\Delta_{n,y'}} \frac{1}{d^{\beta}_{mn} d^{\beta}_n(\vv{u}_t)} \label{eq:max_min_snr_eq}\\
\displaystyle \textup{s.t.} & \ \ \sum\limits_m y_{tmn}\left| \Omega_{\scriptscriptstyle{\mathrm{T}},n}(\vv{u}_t) - \Omega_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_k)\right| \leq \Delta_{n,x'}, \nonumber\\
& \ \ \hspace{3.3cm} \forall n, \forall \vv{u}_t,\vv{u}_k \in \mathcal{A}, \label{eq:omega_con}\\
& \ \ \sum\limits_m y_{tmn}\left| \Psi_{\scriptscriptstyle{\mathrm{T}},n}(\vv{u}_t) - \Psi_{\scriptscriptstyle{\mathrm{T}},n} (\vv{u}_k)\right| \leq \Delta_{n,y'}, \nonumber\\
& \ \ \hspace{3.3cm} \forall n, \forall \vv{u}_t,\vv{u}_k \in \mathcal{A}, \label{eq:psi_con}\\
& \ \ \Delta_{n,x'} \geq \frac{1}{N_h \delta}, \quad \Delta_{n,y'} \geq \frac{1}{N_v \delta}, \hspace{0.65cm} \forall n, \label{eq:deltas_min_con} \\
& \ \ \eqref{eq:test_point_orientation_con}, \eqref{eq:base_station_orientation_con}, \eqref{eq:deployment_con}, \eqref{eq:coverage_con}, \eqref{eq:one_bs_con}, \eqref{eq:budget_con}, \eqref{eq:binary_con},\nonumber
\end{align}
\end{problem}
\end{subequations}
in which we define $\vv{\Delta}_{x'} \triangleq [\Delta_{1,x'}, \dots, \Delta_{N,x'}]$, $\vv{\Delta}_{y'} \triangleq [\Delta_{1,y'}, \dots, \Delta_{N,y'}]$ while we omit the constant terms. In this equivalent formulation, we introduce the constraints in Eqs.~\eqref{eq:omega_con} and~\eqref{eq:psi_con} in order to guarantee that each test point served by \gls{ris} $n$ lies within the coverage determined by its spatial frequency span. In Eq.~\eqref{eq:deltas_min_con}, we enforce that the spatial frequency spans $\vv{\Delta}_{x'}$ and $\vv{\Delta}_{y'}$ are at least as wide as the minimum beamwidth obtained by considering a single sub-array while performing the \glspl{ris} configuration via beam broadening and flattening, as by~\cite{Lu2021}.
\section{Large-scale planning algorithm}
\label{s:algorithm}
Hereafter, we design our multi-\gls{ris} planning algorithm, i.e., RISA{}. Let us first consider a continuous relaxation of Problem~\eqref{problem:max_snr_eq} by letting $\vv{x} \in [0,1]^N$ and $\vv{y} \in [0,1]^{T \times M \times N}$ in constraint~\eqref{eq:binary_con}. We can tackle such problem by means of \gls{bca}, namely by iteratively solving the problem for one block of optimization variables while keeping all the others fixed. Notably, although non-convex in general, the continuous relaxation of the problem is jointly convex in the block of variables $\vv{x},\vv{y}$ while it is still non-convex neither in $\vv{\Delta}_{x'}$, nor in $\vv{\Delta}_{y'}$, as the respective objective functions are convex and their maximization leads to a non-convex problem \textit{per se}.
In order to solve the problem for $\vv{\Delta}_{x'}$ (or, similarly, for $\vv{\Delta}_{y'}$), we can rearrange Eq.~\eqref{eq:max_min_snr_eq} as
\begin{align}
\displaystyle \max_{\vv{\Delta}_{x'}} \quad \displaystyle \min_{\vv{u}_t} \sum\limits_{m,n} \frac{y_{tmn}}{d^2_{mn} d^2_n(\vv{u}_t)\Delta_{n,y'}} \frac{1}{\Delta_{n,x'}}, \label{eq:max_min_snr_eq_deltax}
\end{align}
and observe that the resulting subproblem belongs to the \gls{fp} umbrella, being Eq.~\eqref{eq:max_min_snr_eq_deltax} a \textit{sum of functions of ratios}. Therefore, we can leverage on the Quadratic Transform~\cite{Shen2018} and write it equivalently as
\begin{align}
\displaystyle \max_{\vv{z}_{x'},\vv{\Delta}_{x'}} \, \displaystyle \min_{\vv{u}_t} \sum\limits_{m,n} \frac{y_{tmn}}{d^2_{mn} d^2_n(\vv{u}_t)\Delta_{n,y'}} (2z_{n,x'} \!-\! z^2_{n,x'}\Delta_{n,x'}), \label{eq:max_min_snr_eq_deltax_transform}
\end{align}
where $\vv{z}_{x'} \in \mathbb{R}^N$ is an auxiliary optimization variable. The resulting subproblem is now convex in $\vv{z}_{x'}$ and in $\vv{\Delta}_{x'}$ separately, thus likewise solvable by means of a nested \gls{bca}.
Therefore, the solution of the continuous relaxation of Problem~\eqref{problem:max_snr_eq} consists of a double-nested \gls{bca} whose outer loop iteratively considers the three blocks of variables $\vv{x}$ and $\vv{y}$, $\vv{\Delta}_{x'}$, $\vv{\Delta}_{y'}$, while its inner loops solve the subproblems in $\vv{\Delta}_{x'}$ and $\vv{\Delta}_{y'}$ by introducing auxiliary variables $\vv{z}_{x'} \in \mathbb{R}^N$ and $\vv{z}_{y'} \in \mathbb{R}^N$, respectively. We would like to highlight that, by dealing with a convex problem at each stage, the double-nested \gls{bca} is guaranteed to converge to a stationary point~\cite{Grippo2000}.
\textbf{Binary solution.} The binary deployment variable $\vv{x}^*$ are recovered by rounding the highest $L$ elements of $\vv{x}$ to $1$ while setting the other $N-L$ to $0$. Next, we establish the binary associations $\vv{y}^*$ by considering only activated \glspl{cs} $\vv{b}_n$ such that $x^*_n = 1$. In particular, we iteratively associate each test point $\vv{u}_t$ by setting to $1$ the highest element $\vv{y}$ among the ones corresponding to the activated \glspl{cs}. Concurrently, we update the values of $\Delta_{n,x'}$ and $\Delta_{n,y'}$ to the minimum spatial frequency spans satisfying the constraints in Eqs.~\eqref{eq:omega_con},~\eqref{eq:psi_con}.
We depict the overall high-level algorithm in Algorithm~\ref{alg:deployment}.
\begin{algorithm}[t]
\DontPrintSemicolon
Initialize $\Delta_{n,x'} = \Delta_{n,y'} = 2$, $n = 1, \dots, N$ \;
\Repeat(\hfill\emph{Outer BCA loop}){convergence of objective function Eq.~\eqref{eq:max_min_snr_eq} in Problem~\ref{problem:max_snr_eq}}{
Solve the continuous relaxation of Problem~\ref{problem:max_snr_eq} jointly for $\vv{x}$ and $\vv{y}$ \;
\Repeat(\hfill\emph{First inner BCA loop }){convergence of objective function in Eq.~\eqref{eq:max_min_snr_eq_deltax}}{
Solve the transformed problem in $\vv{\Delta}_{x'}$ for $\vv{z}_{x'}$\;
Solve the transformed problem in $\vv{\Delta}_{x'}$ for $\vv{\Delta}_{x'}$\;
}
\Repeat(\hfill\emph{Second inner BCA loop }){convergence of objective function in Eq.~\eqref{eq:max_min_snr_eq_deltax} for $\Delta_{y'}$}{
Solve the transformed problem in $\vv{\Delta}_{y'}$ for $\vv{z}_{y'}$\;
Solve the transformed problem in $\vv{\Delta}_{y'}$ for $\vv{\Delta}_{y'}$\;
}
}
Round $\vv{x}$ and $\vv{y}$ to derive binary $\vv{x}^*$ and $\vv{y}^*$ as by Section~\ref{s:algorithm}\;
\caption{\gls{ris}-Aware network planning (RISA{})}
\label{alg:deployment}
\end{algorithm}
\section{Performance evaluation}
\label{s:performance}
We first evaluate RISA{} via Monte Carlo simulations considering synthetic network topologies. Subsequently, we benchmark RISA{} on the real network topology installed in the \textit{Rennes railway station}, France, provided by the European network operator \textit{Orange}, wherein \glspl{ris} Candidate Sites (CSs) are properly handpicked on the station floor plan and realistic \gls{snr} values are obtained via ray tracing. Simulation parameters based on realistic values are listed in Table~\ref{tab:parameters}, unless otherwise stated.
\begin{table}[t]
\caption{Simulation parameters.}
\label{tab:parameters}
\centering
\resizebox{\linewidth}{!}{%
\begin{tabular}{cc|cc|cc}
\textbf{Parameter} & \textbf{Value} & \textbf{Parameter} & \textbf{Value} & \textbf{Parameter} & \textbf{Value}\\
\hline
\rowcolor[HTML]{EFEFEF}
$P$ & $28$ dBm & $f$ & $26$ GHz & $\sigma^2$ & $-80$ dBm \\
$\beta$ & 2 & $\mu$ & $0.5$ & A & $100 \,\text{m} \times 100 \,\text{m} $\\
\rowcolor[HTML]{EFEFEF}
BSs ($M$) & 2 & CSs ($N$) &\{10,20,30\} &$T$ & $100$ \\
$N_b$ & 2 & $N_r$ & $350 \times 175$ & $N_{ref}$ & 2 \\
\hline
\end{tabular}%
}
\end{table}
\begin{figure}[t!]
\centering
\subfigure[RISA{} \gls{snr} performance.]
{
\includegraphics[clip,width=.46\linewidth ]{SNR_vs_NRISs_SRD.pdf}
\label{fig:snr_vs_nris}
}
\subfigure[RISA{} fairness performance.]
{
\includegraphics[clip,width=.46\linewidth]{jfi_vs_NRISs_SRD.pdf}
\label{fig:jfi_vs_nris}
}
\vspace{-1mm}
\caption{\label{fig:perf_vs_nris} RISA{} performance with different numbers of deployed \glspl{ris} and available Candidate Sites (CSs) via Monte Carlo simulations considering synthetic topologies.}
\end{figure}
\textbf{Synthetic topologies.} We consider the target area to be a square surface with area $A = 100\, \text{m} \times 100\, \text{m}$. Besides, we assume that $M = 2$ \glspl{bs} are placed at the bottom-left and upper-right corners of the area, namely $\vv{b}_1 = [0,0,5.5]^\mathrm{T}$, $\vv{b}_2 = [10^3,10^3,5.5]^\mathrm{T}$, while we evaluate the \gls{snr} performance at $T = 100$ test points uniformly distributed in the target area on the plane $z = 1.5$. We average the results over $10^3$ Monte Carlo executions. In Fig.~\ref{fig:snr_vs_nris}, we show the performance of RISA{} in terms of minimum \gls{snr} experienced in the target area with respect to the number of deployed \glspl{ris} $L$ for different numbers of available \glspl{cs} $N = \{10,20,30\}$ on the plane $z = 5.5$. The horizontal line indicates the minimum \gls{snr} threshold to meet the receiver sensitivity. As expected, the minimum \gls{snr} shows a positive monotonic behavior with decreasing relative increments, thus suggesting the existence of an optimal value for $L$, e.g. $L = 8$ deployed \glspl{ris} for $N = 10$ candidate sites. Besides, increasing the number of \glspl{cs} does not significantly benefit the overall performance, provided that the number of \glspl{cs} is big enough to obtain a good sampling of the target area (on average). The \gls{snr} fairness among test points, measured by means of the \gls{jfi}~\cite{Sediq2013}, shows a similar behavior\footnote{Note that, to obtain meaningful numerical results for the \gls{jfi}, we prevent the received power from exceeding a given maximum, i.e. $-65$ dBm, which provides excellent \gls{rssi}.} in Fig.~\ref{fig:jfi_vs_nris}, validating our \textit{max-min} objective function design choice to enhance coverage in the whole area.
\textbf{Rennes station.} We execute the ray-tracing simulation in MATLAB R2021b using a simplified 3D model of the main floor of the Rennes railway station in France.
The scenario follows the most prominent obstacles and elements filling the volume object of study.
Highly convoluted or unknown elements (e.g., the ceiling, composed of many structural, functional, and decorative beams, as well as the tubing, etc.) have been left as holes to simulate the lack of significant, predictable reflections and lessen the computational burden. The resulting model has $579$ triangles, $1582$ edges and $1053$ vertexes and is depicted in Fig.~\ref{fig:heatmap_no_ris}.
We simulate the \glspl{bs} with $N_b = 2$ and transmit power $P = 28$ dBm at $f = 26$ GHz, as in the real network deployment by \textit{Orange}. Besides, we implement \gls{sbr} in order to derive the possible paths to reach any given test point~\cite{Brem2015}. We linearly combine the power received at any test point from different paths assuming a (uniformly distributed) random phase for each individual path at the \gls{ue} side, thereby accounting for random external factors (e.g., thermal expansion) that could alter the path lengths by a non-negligible fraction of a wavelength $\lambda = 1/f$, given the large ratio between the station distances and the wavelength~\cite{rappaport_2001}. The maximum number of reflections is set to $N_{ref} = 2$ as higher-order reflections provide little contribution to the received power.
\textbf{\gls{ris} ray-tracing model.} We would like to underline that \glspl{ris} are novel network devices, thereby not yet widely implemented in conventional ray-tracers. Therefore, we devise a new lightweight technique to compute the impinging power on the \gls{ris} surface, the \gls{ris} power reflection and the \gls{ris} beampattern. To estimate the impinging power on the \gls{ris} surface, we assess the power received at the \gls{ris} center by one of its elements modeled as a cosine antenna (with exponent parameter $\mu = 0.5$)
and multiply this value by the number of \gls{ris} elements $N_r$.
Hence, we simulate the \gls{ris} controlled reflections by considering outgoing rays originated on the \gls{ris} surface with power equal to the \gls{ris} impinging power. The \gls{ris} emissive beampattern is modeled as a \gls{ura} of $N_r$ cosine antennas, where the phase shifts of each element is controlled with narrow-band phase-shift beamforming.
Lastly, we compute the received power at each test point by adding up the power from each source of any incident ray as received by an isotropic antenna placed at the test point coordinates.
Note that we assume a power-based association policy, namely we consider each test point to be associated to the \gls{bs} providing the highest power, either over the direct link or via reflections through the deployed \glspl{ris}.
\textbf{Realistic simulations.} In Fig.~\ref{fig:perf_vs_nris_rennes}, we show the performance of RISA{} for different numbers of deployed \glspl{ris} among $N = 20$ handpicked \glspl{cs} at a height of $5.5$ m and meeting the architectural constraints of the station building. Besides, we compare such results with a random deployment policy averaged over $10^2$ instances. Clearly, RISA{} outperforms the random policy in both metrics, i.e., minimum \gls{snr} in Fig.~\ref{fig:snr_vs_nris_rennes} and \gls{jfi} in Fig.~\ref{fig:jfi_vs_nris_rennes}. The fairness is further confirmed in Fig.~\ref{fig:heatmap_compaison_rennes}, wherein we compare the 2D heatmaps of the \gls{snr} obtained by RISA{} for $L = 6$ numbers of deployed \glspl{ris} against the baseline with no \gls{ris}.
\begin{figure}[t!]
\centering
\subfigure[RISA{} \gls{snr} performance.]
{
\includegraphics[clip,width=.46\linewidth ]{SNR_vs_NRISs_rennes.pdf}
\label{fig:snr_vs_nris_rennes}
}
\subfigure[RISA{} fairness performance.]
{
\includegraphics[clip,width=.46\linewidth]{jfi_vs_NRISs_rennes.pdf}
\label{fig:jfi_vs_nris_rennes}
}
\caption{\label{fig:perf_vs_nris_rennes} RISA{} performance obtained via ray-tracing simulations for different numbers of deployed \glspl{ris} and available Candidate Sites (CSs) in a realistic environment (Rennes station).}
\end{figure}
\begin{figure}[t!]
\centering
\subfigure[\gls{snr} heatmap with $L = 0$]
{
\includegraphics[clip,width=.46\linewidth ]{heatmapWo6_RIs_detail.pdf}
\label{fig:2D_heatmap_no_ris}
}
\subfigure[\gls{snr} heatmap with $L = 6$ \glspl{ris} (red squares).]
{
\includegraphics[clip,width=0.46\linewidth]{heatmapW6_RIs_detail.pdf}
\label{fig:2D_heatmap_3_ris}
}
\caption{\gls{snr} heatmap in the dead zone (see Figure~\ref{fig:heatmap_no_ris}) of the Rennes station obtained via ray-tracing simulations.}
\label{fig:heatmap_compaison_rennes}
\end{figure}
\section{Conclusions}
\glspl{ris} introduce a novel challenge in traditional cellular networks planning. On the one hand, optimal \glspl{ris} configurations should be computed given fixed \glspl{bs} and \glspl{ris} positions. On the other hand, optimal \glspl{ris} deployments depend on \glspl{ris} configurations. To address these coupled issues and make the analysis tractable, in this paper we proposed RISA{}, a RIS-aware network planning solution that builds on double-nested block coordinate ascent to provide an iterative solution to this unprecedented problem. RISA is evaluated on synthetic generic indoor network deployments and in a real railway station (Rennes). Our results show that RISA can $i$) achieve outstanding performance on top of the existing network infrastructure, $ii$) solve the dead-zone problem in highly-crowded environments and $iii$) improve the user fairness at very limited installation costs.
\bibliographystyle{IEEEtran}
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{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,924
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Gymnochiromyia nigrimana is een vliegensoort uit de familie van de Chyromyidae. De wetenschappelijke naam van de soort is voor het eerst geldig gepubliceerd in 1914 door Malloch.
Chyromyidae
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 9,701
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Q: URLSession.shared.dataTaskPublisher receive cancel trying to fetch some data with dataTaskPublisher. however, constantly receive following log. it works every once in a while and not sure what's the difference. change URL does not make a difference. still only occasionally succeed the request.
Test2: receive subscription: (TryMap)
Test2: request unlimited
Test2: receive cancel
class DataSource: NSObject, ObservableObject {
var networker: Networker = Networker()
func fetch() {
guard let url = URL(string: "https://jsonplaceholder.typicode.com/posts") else {
fatalError("Invalid URL")
}
networker.fetchUrl(url: url)
}
}
class Networker: NSObject, ObservableObject {
var pub: AnyPublisher<Data, Error>? = nil
var sub: Cancellable? = nil
var data: Data? = nil
var response: URLResponse? = nil
func fetchUrl(url: URL) {
guard let url = URL(string: "https://apple.com") else {
return
}
pub = URLSession.shared.dataTaskPublisher(for: url)
.receive(on: DispatchQueue.main)
.tryMap() { data, response in
guard let httpResponse = response as? HTTPURLResponse,
httpResponse.statusCode == 200 else {
throw URLError(.badServerResponse)
}
return data
}
.print("Test2")
.eraseToAnyPublisher()
sub = pub?.sink(
receiveCompletion: { completion in
switch completion {
case .finished:
break
case .failure(let error):
fatalError(error.localizedDescription)
}
},
receiveValue: {
print($0)
}
)
}
A: add .store(in: &subscriptions)
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 7,610
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La German Offshore Spaceport Alliance (GOSA - en français « Alliance allemande pour les ports spatiaux offshore ») est une coentreprise de quatre sociétés qui souhaitent mettre en place une Base de lancement flottante en mer du Nord. Techniquement, cela consiste en un navire roulier stationné à Bremerhaven, qui transfère à environ 460 kilomètres au large un petit lanceur ; la fusée doit être lancée depuis ce lieu en haute mer pour mettre en orbite un ou plusieurs satellites autour de la terre. L'entreprise affiche un premier déploiement possible en 2023 ; la décision de mise en œuvre effective de ce concept doit être prise fin 2021.
Genèse de l'entreprise
Contexte mondial
Dans les années 2010, un boom mondial s'opère dans le développement de petits satellites et de petits lanceurs. Rien qu'en Allemagne, trois fusées sont en préparation : la RFA One d'OHB, la Spectrum d'ISAR Aerospace et la SL1 de HyImpulse. Faute d'emplacement à l'intérieur des terres, aucun site de lancement de fusée ne peut être construit en Allemagne ; le danger de tirs en échec ou de chute d'étages propulseur serait trop grand. Les premiers lancements de ces fusées étaient envisagés depuis le centre spatial guyanais en Amérique du Sud ou à partir de nouvelles stations spatiales d'Europe du Nord. Les contraintes bureaucratique et logistique énormes pour l'exportation et le transport des fusées pourraient être évitées avec un site de lancement allemand. A la suite de l'initiative de OHB, les fabricants des deux autres fusées manifestent également leur intérêt pour le concept de la mer du Nord.
Historique
L'opérateur du projet est la société allemande German Offshore Spaceport Alliance GmbH, dans laquelle quatre sociétés détiennent chacune une participation de 25 % : le groupe spatial OHB de Brême, le prestataire de services d'ingénierie de Hesse via sa succursale de Brême Tractebel DOC Offshore, le constructeur en technologie de communication par satellite MediaMobil Communication de Brême et la compagnie maritime de Brême. OHB développe lui-même la petite fusée RFA One, mais considère le port spatial de la mer du Nord comme un système ouvert que d'autres opérateurs de fusées - y compris ceux de l'étranger - pourraient utiliser.
La German Offshore Spaceport Alliance GmbH est fondée en 2017 sous le nom d'OHB Digital Maritime Services GmbH. Elle est basée dans le quartier de Brême dans le même bâtiment que Tractebel DOC Offshore et la filiale de services informatiques OHB Digital Services.
Fin 2019, la Fédération des industries allemandes (BDI) et OHB communiquent sur la proposition d'un port spatial offshore allemand. Un an après, la coentreprise de Brême rend public - désormais sous le nom de German Offshore Spaceport Alliance - ses plans. Le , GOSA et BDI entament une série de négociations avec des représentants du ministère fédéral des Transports ; les questions spécifiques au projet sur des domaines tels que la protection de l'environnement et de l'eau, le transport aérien et maritime et les assurances restent à clarifier. Le commissaire à l'espace du gouvernement fédéral, , salue le projet, mais évoque une « situation d'autorisation complexe ». De plus, il existe des réticences sur le projet de la part de certains ministères à consulter.
Notes et références
du 4 aout 2021.
Voir aussi
Articles connexes
Industrie spatiale européenne
Liens externes
Industrie spatiale
Base de lancement
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,365
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John C. Shaw, född 5 augusti 1970 i Calgary, Kanada, är en kanadensisk skådespelare. Shaw är känd som skådespelare i filmer och TV-serier såsom Unga kvinnor, Happy Gilmore, Stargate SG-1, Watchmen, Smallville och Supernatural.
Filmografi (i urval)
Unga kvinnor (1994)
Happy Gilmore (1996)
The Perfect Score (2004)
Hollow Man 2 (2006)
Watchmen (2009)
The Company You Keep (2012)
If I Stay (2014)
Externa länkar
Kanadensiska skådespelare
Födda 1970
Män
Levande personer
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 5,945
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Bieberstein steht für:
Bieberstein (Reinsberg), Ortsteil der Gemeinde Reinsberg im Landkreis Mittelsachsen, Sachsen
Bieberstein (Wiehl), Stadtteil von Wiehl im Oberbergischen Kreis, Nordrhein-Westfalen
Bieberstein (Adelsgeschlecht), zum Herrenstand der böhmischen Länder gehörige Familie
Bieberstein, deutscher Name zweier Orte in Masuren:
Bieberstein, Kreis Gerdauen, Ostpreußen, seit 1945: Bobrowo (Barciany), Dorf im Powiat Kętrzyński, Woiwodschaft Ermland-Masuren
Bieberstein, Kreis Sensburg, Ostpreußen, seit 1945: Wólka Baranowska, Dorf im Powiat Mrągowski, Woiwodschaft Ermland-Masuren
Bieberstein, deutscher Name eines zur schlesischen Gemeinde Ciasna gehörigen Orts, poln. Bobrów
steht im weiteren Sinn für:
Stauweiher Bieberstein nahe der Stadt Gummersbach im Bergischen Land, Nordrhein-Westfalen, Deutschland
Schloss Bieberstein (Hessen), Schloss bei Hofbieber im Landkreis Fulda, Hessen, Hermann-Lietz-Schule
Schloss Bieberstein (Sachsen), Schloss bei Reinsberg im Landkreis Mittelsachsen, Sachsen
Burg Bieberstein über dem Tal der Wiehl bei Bieberstein im Oberbergischen Kreis in Nordrhein-Westfalen
Bieberstein, Schloss in Żary in der Niederlausitz, Woiwodschaft Lebus, Polen
Marschall von Bieberstein, meißnisches Adelsgeschlecht, zu Namensträgern siehe dort
Rogalla von Bieberstein, ostpreußisches Adelsgeschlecht, zu Namensträgern siehe dort
Bieberstein ist der Familienname folgender Personen:
Arno Bieberstein (1883–1918), deutscher Schwimmer
Frauke von Bieberstein (* 1974), deutsche Wirtschaftswissenschaftlerin
Joseph Albrecht Christoph von Bieberstein-Pilchowsky (1730–1815), preußischer Generalmajor
Klaus Bieberstein (* 1955), deutscher römisch-katholischer Theologe
Rada Bieberstein (* 1979), deutsche Kunsthistorikerin, Filmwissenschaftlerin und Regisseurin
Sabine Bieberstein (* 1962), deutsche römisch-katholische Theologin
Karl Jörg Bieberstein (1931–2021), deutscher Jurist
Siehe auch:
Biberstein (Begriffsklärung)
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 4,533
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\section{Introduction}\label{sec:1}
Traffic state (i.e. flow, speed, and density) estimation (TSE) is the precursor of a variety of advanced traffic operation tasks and plays a key role in traffic management. In early stages, macroscopic traffic dynamics were found to be similar to hydrodynamics. By borrowing concepts from the fluid mechanism, flow, speed, and density were defined and their relationship, named the fundamental diagram, was discovered. Based on these definitions, macroscopic traffic flow models were developed based on the conservation law and momentum and a set of kinematic wave models were also formulated \citep{seo2017traffic}.
However, most models, derived under ideal theoretical conditions, require great efforts for parameter calibrations and are difficult to work with noisy and fluctuated data collected by traffic sensors.
Then to capture the measurement errors, stochastic traffic flow models were developed for the investigation and explanation of a variety of observed traffic phenomena, which are also better suited for real-time traffic state estimation and forecasting \citep{jabari2014probabilistic}.
Since the deterministic prominent models and their higher-order extensions are ill-posed, researchers developed stochastic traffic flow models in two categories. The first category used stochastic extensions \citep{gazis1971line, szeto1972application, gazis2003kalman, wang2005real, wang2007real}, which were performed by adding Gaussian noises to the model expressions and obtained real-world data were used to quantify those noises.
However, \cite{jabari2012stochastic} pointed out that those simply-noised models could lead to the possibility of: (i) causing negative sample paths and (ii) producing mean dynamics that do not coincide with the original deterministic dynamics due to nonlinearity.
The second category includes stochastic traffic models such as Botlzmann-based models \citep{prigogine1971kinetic, paveri1975boltzmann}, Markovian queuing network approaches \citep{davis1994estimating,kang1995estimation,di2010hybrid,osorio2011dynamic,jabari2012stochastic}, and cellular automaton based models \citep{nagel1992cellular, gray2001ergodic, sopasakis2006stochastic,sopasakis2012lattice}. Stochastic traffic models do not have the same concerns of the models in the first category. However, they may lose the analytical tractability \citep{jabari2013stochastic}, defined as the ability of obtaining a mathematical solution such as a closed-form expression, and are much more similar to data-driven approaches than classical analytical models.
In view of the increasing data availability, many data-driven methods were developed because they do not require explicit theoretical assumptions and have a remarkably low computational cost in the testing phase.
In the literature, data-driven approaches include autoregressive integrated moving average \cite{zhong2004estimation}, Bayesian network \cite{ni2005markov}, kernel regression \citep{yin2012imputing}, fuzzy c-means clustering \citep{tang2015hybrid}, k-nearest neighbors clustering \citep{tak2016data}, stochastic principal component analysis \citep{li2013efficient,tan2014robust}, Tucker decomposition \citep{tan2013tensor}, deep learning \citep{duan2016efficient,polson2017deep,wu2018hybrid}, Bayesian particle filter \citep{polson2017bayesian}, etc.
However, due to the data-driven nature, those machine learning (ML) models fundamentally suffers from three scenarios:
(i) training data are scarce and insufficient to reveal the complexity of the system,
(ii) training data are noisy and include much incorrect/misleading information, and
(iii) test data are far from the training examples, i.e., extrapolation. In these scenarios which are unfortunately very common in the real-world, their performance can drop dramatically along with large and/or biased estimations.
Fig.~\ref{fig:flaw_data1} shows an example of applying a pure ML method on a dataset that contains flawed data and its biased estimation (dash line) diverges from ML methods on accurate data (solid line). Moreover, another deficiency of ML models is that they are developed as "black boxes" and researchers are hard to interpret model results.
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.4\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/FlawData1.pdf}
\caption{ML with flawed data}
\label{fig:flaw_data1}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.4\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/FlawData2.pdf}
\caption{PRML with flawed data}
\label{fig:flaw_data2}
\end{subfigure}
\vspace{-0.1in}
\caption{Comparison between pure ML and the proposed PRML}
\label{fig:flaw_data}
\vspace{-0.2in}
\end{figure}
In summary, classical traffic flow models can effectively characterize the underlying mechanisms (i.e., physical processes of traffic) of transportation systems, however, are usually developed with strong assumptions, require great efforts in parameter calibrations, and fall short of capturing data uncertainties. On the other hand, the performances of pure data-driven approaches such as ML models highly depend on the data quality and their results are usually hard to be interpreted. Hence, recognizing those limitations, this research aims to develop an innovative approach, named physics regularized machine learning (PRML), to fill the gap between classical traffic flow (physical) models and ML methods. The contributions of this study are significant. Compared with physical models, the PRML can
(1) use the ML portion to capture the uncertainties in estimation which beyond the capability of the closed-form expressions; and
(2) eliminate the efforts in calibrating model parameter by a sequential learning process. Different from pure ML models, the PRML is
(1) more robust under the condition of the noisy/flawed dataset as valuable knowledge from physical models can help regularize the fitting process (see Fig.~\ref{fig:flaw_data2}); and
(2) more explainable in terms of the model performance in estimation accuracy. With this innovative modeling framework, this research is expected to bring a new insight into ML applications in transportation and build a bridge to connect the researches of classical traffic flow models and more recent data-driven approaches.
More specifically, this study develops a physics regularized Guassian process (PRGP) method for TSE by integrating three macroscopic traffic flows models with Gaussian process (GP), implementing a shadow GP to regularize the original GP, and incorporating enhanced Latent Force Models (LFM) \citep{raissi2017machine} to encode the traffic flow model knowledge.
To learn the GPs from data efficiently, this study also proposes an inference algorithm under the posterior regularization inference framework. To justify the effectiveness of the proposed methods, numerical experiments with field data are conducted on a I-15 freeway segment in Utah and the performances of PRGP models are compared with that of both classical traffic flow models and pure ML models.
To further investigate the robustness of PRGP, synthesized noises are also added to the training set and results show PRGP is much more resilient to the noisy/flawed dataset.
The remainder of this paper is organized as follows.
Section~\ref{sec:2} reviews the existing studies regarding the TSE modeling and estimation methods as well as the Gaussian process and inference methods.
In Section~\ref{sec:3}, the integrated GP and enhanced LFM for encoding physics knowledge into Bayesian statistics and the posterior regularized inference algorithm are derived.
In Section~\ref{sec:4}, the case study on a real-world data from the interstate freeway I-15 is conducted to justify the proposed methods.
The conclusion section summarizes the the critical findings and future research directions.
\section{Literature Review}\label{sec:2}
\subsection{Macroscopic Traffic flow model}
To effectively control traffic flows, TSE has been recognized as a critical fundamental task of freeway traffic management in the literature.
TSE refers to estimating a complete traffic state based on limited traffic measurement data from stationary sensors.
Key parameters, i.e. traffic flow, speed, and density, of the macroscopic traffic flow model are used to approximate the continuous traffic state with the fundamental diagram.
Deterministic traffic flow model usually consist of a conservation law equation and a fundamental relationship \citep{seo2017traffic}.
For formalization, key concepts, including cumulative flow, flow, density, speed, are defined as follows.
\begin{definition}
The cumulative flow $N(t,x)$ is defined as the number of vehicles that passed the position $x$ by the time $t$.
\end{definition}
\begin{definition}
The flow $q$, density $\rho$, speed $v$ are defined in Eqs.~\ref{eq:def-q}-\ref{eq:def-v}.
\begin{equation}
\label{eq:def-q}
q(t,x)=\partial_t N(t,x)
\end{equation}
\begin{equation}
\label{eq:def-rho}
\rho(t,x)=-\partial_x N(t,x)
\end{equation}
\begin{equation}
\label{eq:def-v}
v(t,x)=\frac{q(t,x)}{\rho(t,x)}
\end{equation}
\end{definition}
In traffic flow studies, researchers found the existence of the fundamental diagram (FD) to illustrate the relationship among flow, speed and density:
\begin{definition}
The fundamental diagram is defined as the relationship among flow, speed, and density, as shown in Eqs.~\ref{eq:fd-v}-\ref{eq:fd-q}.
\begin{equation}
\label{eq:fd-v}
v=V{\rho}
\end{equation}
\begin{equation}
\label{eq:fd-q}
q = \rho V(\rho)
\end{equation}
\end{definition}
\noindent where $V(\cdot)$ denotes the density-speed function.
Macroscopic traffic flow models were proposed based on continuum fluid approximation to describe the aggregated behavior of traffic, which can generally be classified into three basic formulations.
The well-known first-order Lighthill-Whitham-Richards (LWR) model \citep{lighthill1955kinematic,richards1956shock} is formulated in Eqs.~\ref{eq:lwr1}-\ref{eq:lwr2}.
\begin{equation}
\label{eq:lwr1}
\partial_t \rho + \partial_x(\rho v) = 0
\end{equation}
\begin{equation}
\label{eq:lwr2}
v=V(\rho)
\end{equation}
The LWR model can describe simple behaviors, such as traffic jam and shockwave, however, has limitations in reproducibility of more complex phenomena.
To overcome such limitations, second-order models use the additional momentum equation to describe the dynamics of speed.For example, Payne-Whitham (PW) model \citep{payne1971models,whitham1975linear} is formulated by Eqs.~\ref{eq:pw1}-\ref{eq:pw2}, in which Eq.\ref{eq:pw2} is the momentum equation.
\begin{equation}
\label{eq:pw1}
\partial_t\rho+\partial_x(\rho v) = 0
\end{equation}
\begin{equation}
\label{eq:pw2}
\partial_t v+v\partial_x v=-\frac{V-V(\rho)}{\tau_0}-\frac{c^2_0}{\rho}\partial_x\rho
\end{equation}
where $\tau_0$ denotes the relaxation time and $c^2_0$ denotes a parameter related to driver anticipation.
Despite the success of the PW model and its extensions \citep{papageorgiou1989macroscopic}, the PW-like models may produce non-realistic outputs, such as negative speed \citep{del1994reaction,daganzo1995requiem,papageorgiou1998some,hoogendoorn2001state}.
To overcome this limitation, another second-order Aw-Rascle-Zhang (ARZ) model \citep{aw2000resurrection,zhang2002non} is formulated in Eqs.~\ref{eq:arz1}-\ref{eq:arz2}, where another momentum equation is proposed in Eq.~\ref{eq:arz2}.
The original ARZ model was extended extensively in the literature \citep{colombo2003hyperbolic,lebacque2007generic,blandin2013phase,fan2013comparative}.
\begin{equation}
\label{eq:arz1}
\partial_t\rho + \partial_x(\rho v) = 0
\end{equation}
\begin{equation}
\label{eq:arz2}
\partial_t(v-V(\rho) + v\partial_x (v-V(\rho))=-\frac{v-V(\rho)}{\tau_0}
\end{equation}
However, it should be noted that despite of the elegance of differential equation formalization, the traffic flow model is difficult to estimate due to the nonlinearity and the measure errors of observations in the real world.
Thus, the researchers proposed advanced estimation methods to facilitate the application of the models.
\subsection{Stochastic estimation methods}
To use field data to capture traffic flow uncertainties, some estimation models with stochastic extensions are later derived\cite{seo2017traffic}.
For example, TSE is defined as Boundary Value Problem (BVP) based on partial observations (i.e. boundary conditions) \citep{coifman2002estimating,laval2012stochastic,kuwahara2015theory,blandin2013phase,fan2013comparative}.
In solving BVPs, the boundary conditions are assumed to be correct.
However, the real-world measure error can not be ignored.
Considering system and observation noise, data assimilation or inverse modeling techniques were then developed for model estimation and calibration.
In the literature, there exist three ways to add randomness in the traffic models:
(a) stochastic initial and boundary conditions,
(b) stochastic source terms (e.g. inflows), and
(c) stochastic speed-density relationship or fundamental diagram \citep{sumalee2011stochastic}.
To capture the measure error in data, a stochastic modeling method is performed by adding Gaussian noise to the traffic state estimates \cite{gazis1971line, szeto1972application, gazis2003kalman, wang2005real, wang2007real,sumalee2011stochastic}.
For example, in view of the nonlinearity of the second order traffic flow model, \cite{gazis2003kalman,wang2005real} assumed the error terms on the formula and developed extended Kalman filter (EKF) to estimate a PW-like discrete model \citep{papageorgiou1989macroscopic}.
Note that the applying EKF to non-differentiable models (e.g. Cell Transmission Model) is not rigorous \citep{blandin2012sequential}.
The unscented Kalman filter (UKF) overcomes the shortcomings of EKF by avoiding an analytical differentiation \citep{mihaylova2006unscented}.
The ensemble Kalman filter (EnKF) employs the Monte Carlo simulation to handle nonlinear and nondifferentiable systems, but is computational costly \citep{work2008ensemble}.
The particle filter (PF) uses Monte Carlo simulation and is computation-consuming as well \citep{mihaylova2004particle}.
The simulation-based methods were further extended to reduce the computational cost.
In summary, despite the adequate applications of these methods, the stochastic extension models have two critical theoretical deficiencies: (a) negative sample paths and (b) the mean dynamics that do not coincide with the original deterministic dynamics due to the nonlinearity \citep{jabari2012stochastic,jabari2013stochastic, jabari2014probabilistic, pascale2013estimation, wada2017optimization}.
In view of such deficiencies, the intractable methods, stochastic traffic flow models, were proposed in view of the tradeoff between relaxing assumptions and the model tractability, such as
(a) Botlzmann-based methods \citep{prigogine1971kinetic, paveri1975boltzmann},
(b) Markovian queuing methods \citep{davis1994estimating,kang1995estimation,di2010hybrid,osorio2011dynamic,jabari2012stochastic},
(c) cellular automation based methods \citep{nagel1992cellular, gray2001ergodic, sopasakis2006stochastic,sopasakis2012lattice}.
\subsection{Data-driven method}
More recently, with much enriched data, researchers started to seek data-driven methods, such as machine learning, Bayesian statistics, etc.
Among the existing data-driven methods, Gaussian process (GP) is a powerful non-parametric function estimator and has various successful applications.
In traffic modeling, GP-based methods are applied in traffic speed imputation \citep{rodrigues2018heteroscedastic,rodrigues2018multi}, public transport flows \cite{neumann2009stacked}, traffic volume estimation and prediction \cite{xie2010gaussian}, travel time prediction \citep{ide2009travel}, driver velocity profiles \citep{armand2013modelling} and traffic congestion \cite{liu2013adaptive}.
It can capture relationship between stochastic variables without requiring strong assumptions (such as memorylessness).
However, as a data-driven approach, GPs can perform poorly when the training data are scarce and insufficient to reflect the complexity of the system or testing inputs are far away from the training data.
Few traffic estimation methods were developed based on GPs because it's difficult to obtain deductive insights and leverage physics knowledge.
Taking advantage of valuable knowledge from physical models (i.e., classical traffic flow model), we aim to encode them into GPs to improve their performance, especially when training on scarce data and marking estimations in areas with flawed observations.
However, it shall be noted that using GP to represent physical knowledge, modeled by differential equations, has two major difficulties:
(a) differential equations are hard to represent as a probabilistic term, such as priors and likelihoods;
(b) in practice, physics knowledge is usually incomplete, the differential equations can include latent functions and parameters (e.g. unobserved noise, inflows, outflows), making their presentations and joint estimation with GPs even more challenging.
To better encode the differential equations in GPs, \cite{alvarez2009latent, alvarez2013linear} proposed a Latent Force Models (LFM) for training and then the estimation of GP would be based on the convolved kernel upon Green's function.
Later on, \cite{raissi2017machine} extended the framework by assuming observable noise.
However, the assumption of LFM is too restrictive since many realistic flexible equations are nonlinear, or linear but do not have analytical Green's function.
Also, the complete kernel is still infeasible to obtain.
Thus, it is more feasible to use expressive kernels, e.g. deep kernels \citep{wilson2016deep}.
In summary, there lack a hybrid framework to consider the physics knowledge (i.e. kinematic wave differential equations and fundamental diagram) and the data-driven methods with minimal assumptions and reasonable computational cost.
This paper aims to fill the gap by proposing a Gaussian process based data-driven method considering tractable physics knowledge.
\subsection{Gaussian process and Bayesian inference}
Gaussian process is a general framework for measuring of the similarity between observations from training data to estimate the unobserved values.
\cite{rodrigues2018heteroscedastic} and \cite{rodrigues2018multi} applied the multi-output Gaussian processes to model the complex spatiotemporal patterns about incomplete traffic speed data.
The key task is to learn the kernel (i.e. covariance) function between the variables.
The previous studies \citep{calderhead2009accelerating,barber2014gaussian,heinonen2018learning} investigated the GP ordinary differential derivatives.
They assumed the noisy forces are observable, for example, the observable noisy forces \citep{graepel2003solving}, and observable noisy forces and solutions \citep{raissi2017machine}.
To model the observable noisy forces, Latent Force Models (LFM) \citep{alvarez2009latent,alvarez2013linear} first placed a prior over the latent forces, and then derives the covariance of the solution function via the convolution operation.
Despite the successful applications, such as transcriptional regulation modeling \citep{lawrence2007modelling}, the LFM method has two critical deficiencies: (a) it requires the linear differential equations and the analytical Green's functions, which is restrictive and does not fit the traffic flow model; and (b) the convolution procedure is computationally difficult and restrictive.
To address these issues, this paper generalizes the LFM framework and enables the nonlinear differentiation to encode the physics knowledge.
To key task is to optimize model likelihood on data and a penalty term that encodes the constraints over the posterior of the latent variables.
Via the penalty term, the domain knowledge or constraints outright to the posteriors rather than through the priors and a complex, intermediate computing procedure, hence it can be more convenient and effective.
In view of computational efficiency, this paper further employs a posterior regularization algorithms to solve the likelihood optimization problem \citep{ganchev2013cross,zhu2014bayesian,libbrecht2015entropic,song2016kernel}.
To the best of authors knowledge, this new modeling framework is innovative and has not been developed by other transportation studies yet.
The proposed method is designed to avoid the error-prone simple stochastic assumptions and leverage the physics knowledge in a data-driven framework, which also has remarkable performance in scare data situations and unobserved inflow and outflows (e.g. an arterial stretch).
\section{Methodology}\label{sec:3}
\subsection{Macroscopic traffic flow model with Physics Regularized Gaussian Process}
\subsubsection{Gaussian process}
Suppose we aim to learn a machine $\mathbf{f} :\mathbb{R}^d \rightarrow \mathbb{R}^{d^\prime}$, it will map a $d$-dimensional Euclidean space to a $d^\prime$-dimensional Euclidean space from a training set $\mathcal{D} =(\mathbf{X},\mathbf{Y})$,
where $\mathbf{X}=[\mathbf{x}_{1},\ldots,\mathbf{x}_{N}]^\intercal$ is the input vector,
$\mathbf{Y}=[\mathbf{y}_{1},\ldots,\mathbf{y}_{N} ]^\intercal$ is the output vector,
$\mathbf{x}$ is the $d$ dimensional input vector,
$\mathbf{y}$ is the $d^\prime$ dimensional output vector,
$\mathbf{f}=[f(\mathbf{x}_{1}),\ldots,f(\mathbf{x}_{N})]^\intercal$ is the learning function,
and $N$ refers to the sample size.
Note that $\mathbf{X},\mathbf{Y}$ may have physical meanings only in their feasible domains.
\begin{assumption}
It is assumed that the input $\mathbf{X}$ and the true output $\mathbf{f}$ follow a multivariate Gaussian distribution as shown in Eq.~\ref{eq:pfx} ,
where $\mathcal{N}(\cdot,\cdot)$ represents the Gaussian distribution,
$\mathbf{m}$ denotes the mean matrix,
and $\mathbf{K}$ represents the covariance matrix.
\begin{equation}
\label{eq:pfx}
p(\mathbf{f}|\mathbf{X})=\mathcal{N}(\mathbf{f}|\mathbf{m},\mathbf{K})
\end{equation}
\end{assumption}
Note that Gaussian process in $d$-dimension is also called Gaussian Random Field and the above definition involves the multi-dimensional outputs.
\begin{assumption}
It is assumed that the observations $\mathbf{Y}$ have an isotropic Gaussian noise, as shown in Eq.~\ref{eq:pyf}.
\begin{equation}
\label{eq:pyf}
p(\mathbf{Y}|\mathbf{f})=\mathcal{N}(\mathbf{f},\tau^{-1}\mathbf{I})
\end{equation}
where $\tau$ refers to the inverse variance, and isotropic noise means that the noise from each dimension is independent identically distributed (i.i.d.) and of the same variance $\tau$.
\end{assumption}
Then, by Marginalizing out $\mathbf{f}$, we can obtain the marginal likelihood as shown in Eq.~\ref{eq:pyx}.
\begin{equation}
\label{eq:pyx}
p(\mathbf{Y}|\mathbf{X})=\mathcal{N}(\mathbf{Y}|\mathbf{0},\mathbf{K}+\tau^{-1} \mathbf{I})
\end{equation}
where the kernel matrix $\mathbf{K}$ is defined in Eq.~\ref{eq:kernel_def}.
\begin{equation}
\label{eq:kernel_def}
[\mathbf{K}]_{ij}=k(x_i,x_j)
\end{equation}
Commonly, Assumption~\ref{assumption:smooth}, which requires the kernel has derivatives of all orders in its domain, would also be necessary and the positive-definite kernels include linear, polynomial, radial-basis, Laplacian, etc. \citep{fasshauer2011positive}.
\begin{assumption}
\label{assumption:smooth}
The kernel $k(\cdot,\cdot)$ is assumed to be positive-definite and smooth.
\end{assumption}
Given the new input $\mathbf{x}^*$, the $\mathbf{f}$ function value can be estimated based on Eq.~\ref{eq:pfxxy}.
\begin{equation}
\label{eq:pfxxy}
p(\mathbf{f}(\mathbf{x}^*)|\mathbf{x}^*,\mathbf{X},\mathbf{Y})=\mathcal{N}(\mathbf{f}(\mathbf{x}^*)|\mu(\mathbf{x}^*),\nu(\mathbf{x}^*))
\end{equation}
where the mean $\mu(\mathbf{x}^*)$, standard deviation $\nu(\mathbf{x}^*)$, and the kernel vector $\mathbf{k}_*$ are calculated in Eqs.~\ref{eq:mu_def}-\ref{eq:k_def}, respectively.
\begin{equation}
\label{eq:mu_def}
\mu(\mathbf{x}^*)=\mathbf{k}_*^\intercal (\mathbf{K}+\tau^{-1} \mathbf{I})^{-1} \mathbf{Y}
\end{equation}
\begin{equation}
\label{eq:nu_def}
\nu(\mathbf{x}^*)=k(\mathbf{x}^*,\mathbf{x}^*)-\mathbf{k}_*^\intercal(\mathbf{K}+\tau^{-1} \mathbf{I})^{-1} \mathbf{k}_*
\end{equation}
\begin{equation}
\label{eq:k_def}
\mathbf{k}_*=[k(x^*,x_1),\ldots,k(x^*,x_N)]^\intercal
\end{equation}
If the kernel $\mathbf{K}$ has been learned from data $\mathcal{D}$, the estimated output matrix $\mathbf{f(x^*)}$ can be calculated via the reparameterization \citep{kingma2014adam} as shown in Eqs.~\ref{eq:repara1}-\ref{eq:repara2}, where $\epsilon$ is standard normally distributed.
\begin{equation}
\label{eq:repara1}
\epsilon = \mathcal{N}(0,1)
\end{equation}
\begin{equation}
\label{eq:repara2}
\mathbf{f(x^*)} = \mu(\mathbf{x}_*) + \epsilon * \sqrt{\nu(\mathbf{x}^*)}
\end{equation}
Fig.~\ref{fig:gp} shows the structure of the conventional GP method, where the circled nodes denote the random vector,s the shaded node represent known vectors, and the arrows indicate the conditional probabilities.
\begin{figure}[h!]
\centering
\begin{tikzpicture}[node0/.style={circle,draw},
node1/.style={circle,draw,fill=gray},
edge1/.style={->}]
\node [node1] (x) at (-3.5, 3) {$\mathbf{X}$};
\node [node1] (xs) at (1.25, 3) {$\mathbf{x^*}$};
\node [node1] (y) at (-3.5, 1.25) {$\mathbf{Y}$};
\node [node0] (f) at (-1.25, 1.25) {$\mathbf{\hat{f}}$};
\node [node0] (fs) at (1.25, 1.25) {$\mathbf{f^*}$};
\node (l) at (-2.25, 0.5) {learning};
\node (p) at (1.5, 0.5) {estimation};
\draw [edge1] (x) to node {data} (y);
\draw [edge1] (y) to node[above] {noise} (f);
\draw [edge1] (x) to node {GP} (f);
\draw [edge1] (x) to (f);
\draw [edge1] (xs) to node {GP} (fs);
\end{tikzpicture}
\caption{The conventional framework for inferring Gaussian process}
\label{fig:gp}
\vspace{-0.2in}
\end{figure}
\subsubsection{Latent Force Model}
In many applications, physics knowledge, expressed as differential equations, provide the insight of the system's mechanism and can be very useful for both estimation and prediction.
In the seminal work of \cite{alvarez2009latent,alvarez2013linear}, they propose latent force models (LFM) that use convolution operations to encode physics into GP kernels.
They assume the differential equations are linear and have analytical Green's function with the kernel of the latent functions.
Given this assumption, the kernel of the target function can be derived by convolving the Green's function with the kernel of the latent functions.
LFM considers $W$ output functions ${f_1(x),\ldots,f_w(x),\ldots,f_W(x)}$, and assumes each output function $f_w$ is governed by a linear differential equation.
\begin{equation}
\label{eq:lfm}
\mathscr{L}f_{w}(\mathbf{x}) = u_{w}(\mathbf{x})
\end{equation}
where $\mathscr{L}$ is linear differential operator \citep{courant2008methods}, and $u$ is a latent force function.
\begin{lemma}
\label{lemma}
If one side of Eq.~\ref{eq:lfm} is one GP, the other side is another GP.
The covariance of a GP's derivative and the cross-covariance between the GP and its derivative can be obtained by taking derivatives over the original covariance function.
\end{lemma}
Lemma~\ref{lemma} is proven by \citep{alvarez2009latent, alvarez2013linear}.
The reasoning is based on that applying a linear differential operator on one GP results in another GP \citep{graepel2003solving} because the derivative of GP is still a GP \citep{williams2006gaussian}.
The latent force function $u$ can be further decomposed as a linear combination of several common latent force functions as follows.
\begin{equation}
\label{eq:udecomp}
u_w(\mathbf{x})=\sum_{r=1}^{R}s_{rw} g_{r}(\mathbf{x})
\end{equation}
where $R$ is the number of decomposed force functions, $s$ is the latent matrix.
Since $\mathscr{L}$ is linear, if we assign a GP prior over $u(x)$, $f_w(x)$ has a GP prior as well.
Moreover, if the Green's function, namely the solution of Eq.~\ref{eq:green}, is available, we can obtain Eq.~\ref{eq:green_sol}.
\begin{equation}
\label{eq:green}
\mathscr{L}\mathcal{G}(\mathbf{x},\mathbf{s})=\delta(\mathbf{s}-\mathbf{x})
\end{equation}
where $\delta$ is the Dirac delta function, $\mathcal{G}$ is the Green's function.
\begin{equation}
\label{eq:green_sol}
f_w(x)=\int \mathcal{G}(x,s)u_{i}(\mathbf{s})\textrm{d}\mathbf{s}
\end{equation}
Hence, given the kernel for $u_w$, we can derive the kernel for $f_w$ through a convolution operation which is shown in Eq.~\ref{eq:fi_kernel}.
\begin{equation}
\label{eq:fi_kernel}
k_{f_w}(\mathbf{x}_1,\mathbf{x}_2)=\iint \mathcal{G}(\mathbf{x}_1,\mathbf{s}_1)\mathcal{G}(\mathbf{x}_2,\mathbf{s}_2)k_{u_w}(\mathbf{s}_1,\mathbf{s}_2)\textrm{d}\mathbf{s}_1\textrm{d}\mathbf{s}_2
\end{equation}
To deal with the multiple outputs, we can place independent GP priors over common latent function $g_r$, then each $u_w$ and $f_w$ will obtain GP priors in turn.
Via a similar convolution, we can derive the kernel across different outputs (i.e. cross-covariance) $k_{f_w,f_{i'}}$.
In this way, the physics knowledge in the Green's function are hybridized with the kernel for the latent forces.
This procedure is used to learn the GP model with an convolved kernel from the training data.
\subsubsection{Augmented Latent Force Model}
Despite the elegance and success of LFM, the precondition for using LFM might be too restrictive.
To enable the kernel convolution, LFM requires that the differential equations must be linear and have analytical Green's functions.
However, many realistic differential equations from traffic flow models are either nonlinear or linear but do not possess analytical Green's functions, and therefore, cannot be exploited.
In some other cases, even with a tractable Green's function, the complete kernel of all the input variables is still infeasible to obtain.
In order to obtain an analytical kernel after the convolution, we have to convolve Green's functions with smooth kernels.
This may prevent us from integrating the physics knowledge into more complex yet highly flexible kernels, such as deep kernel \citep{wilson2016deep}.
To handle the intractable integral, we need to develop extra approximation methods, such as Monte-Carlo approximation.
Given the differential equation that describes the physics knowledge, the proposed augmented LFM equation is formulated in Eq.~\ref{eq:nonlinearlfm}.
\begin{equation}
\label{eq:nonlinearlfm}
\Psi f(\mathbf{x}) = g(\mathbf{x})
\end{equation}
where the differential operator $\Psi$ can be linear, nonlinear, or numerical differential operator, $g(\cdot)$ represents the unknown latent force functions, $f(\mathbf{x})$ is the function to be estimated from data $\mathcal{D}$.
We aim to create a generative component to regularize the original GP with a differential equation.
Using Augmented LFM, the differential equation is encoded to another GP, which is called a shadow GP.
To yield the numerical outputs, the kernel of the shadow GP should be efficiently learnable.
\begin{theorem}
If one side of Eq.~\ref{eq:nonlinearlfm} is one GP, the other side is another GP.
\end{theorem}
\begin{proof}
The reasoning is based on that applying a differential operator on one GP results in another GP.
The regularization is fulfilled via a valid generative model component rather than the process differentiation, and hence can be applied to any linear or nonlinear differential operators.
In view of the fact that the resultant covariance and cross-covariance are not obvious via analytical derivatives, the expressive kernels can be learned from data empirically.
\end{proof}
The original LFM starts with the RHS (right-hand side) of Eq.~\ref{eq:lfm}, assigns it a GP prior and then use the convolution operation to obtain the GP prior of the left-hand side (LHS) target function.
Since the convolution operation is an integration procedure, it can be more restrictive and challenging.
In contrast, our approach chooses a reverse direction, i.e. from LHS to RHS.
We first sample the target function with an expressive kernel, use differentiation operation to obtain the latent force, and then regularize it with another GP prior.
The differentiation operation is more flexible and convenient \citep{baydin2018automatic}, which does not need to restrict the operator and GP kernels to ensure tractable computation.
The computational challenge can be overcomed by using auto-differential libraries \citep{baydin2018automatic} and deep learning techniques (e.g. deep kernel, Tensorflow, PyTorch).
Therefore, the shadow GP can be efficiently learned from pseudo observations via differential computations.
\subsection{Physics regularized Gaussian process (PRGP)}
Involving the shadow GP, the design concept of the proposed PRGP is illustrated in Fig.~\ref{fig:framework}.
\begin{figure}[h!]
\centering
\begin{tikzpicture}[node0/.style={circle,draw},
node1/.style={circle,draw,fill=gray},
edge1/.style={->}]
\node [node1] (x) at (-3.5, 3){$\mathbf{X}$};
\node [node1] (y) at (-3.5, 1.25) {$\mathbf{Y}$};
\node [node1] (z) at (-1.25, 3) {$\mathbf{Z}$};
\node [node0] (f) at (-1.25, 1.25) {$\mathbf{\hat{f}}$};
\node [node0] (g) at (0.5, 1.25) {$\mathbf{g}$};
\node [node1] (o) at (0.5, -0.25) {$\omega$};
\draw [edge1] (f) to node[above]{$\Psi$} (g) ;
\draw [edge1] (g) to node {shadow GP} (o);
\draw [edge1] (x) to node {data} (y) ;
\draw [edge1] (y) to (f);
\draw [edge1] (z) to node {GP} (f);
\draw [edge1] (x) to (f);
\end{tikzpicture}
\caption{The proposed framework for physics regularized Gaussian process learning}
\label{fig:framework}
\vspace{-0.2in}
\end{figure}
To enable Bayesian framework that incorporats the physics knowledge in Eq.~\ref{eq:nonlinearlfm}, we introduce a set of $m$ pseudo observations, $\omega=[0,\ldots,0]^\intercal$, to propose a generative component $p(\omega|\mathbf{X},\mathbf{Y})$ that acts as a physics knowledge based regularizer on the GP model $p(\mathbf{Y}|\mathbf{X})$.
To sample the pseudo observation $\omega$, the input vector $\mathbf{Z}$ of the length $m$ is given as follow:
\begin{equation}
\label{eq:z}
\mathbf{Z}=[\mathbf{z}_1, \ldots, \mathbf{z}_m]^\intercal
\end{equation}
Then, we sample the posterior function values at each $\mathbf{z}_j, 1\leq j\leq m$ as shown in Eq.~\ref{eq:poszxy}.
\begin{equation}
\label{eq:poszxy}
p(f(\mathbf{z}_j)|\mathbf{z}_j, \mathbf{X}, \mathbf{Y}) = \mathcal{N}(f(\mathbf{z}_j)|\mu(\mathbf{z}_j),\nu(\mathbf{z}_j))
\end{equation}
We apply the differentiation operator in Eq.~\ref{eq:nonlinearlfm} to obtain the latent function values at $\mathbf{Z}$, $\mathbf{g}=[g(\mathbf{z}_1),\ldots,g(\mathbf{z}_m)]$, which is equivalent to sampling $g(\cdot)$ from the Green's function in Eq.~\ref{eq:nonlineargreen}.
\begin{equation}
\label{eq:nonlineargreen}
p(\mathbf{g}|\hat{\mathbf{f}})=\delta(\mathbf{g}-\Psi \hat{\mathbf{f}})
\end{equation}
Given the latent function values $\mathbf{g}$, we sample the pseudo observations $\omega$ from another GP.
\begin{equation}
\label{eq:pogz}
p(\omega|\mathbf{g},\mathbf{Z})=\mathcal{N}(\omega|\mathbf{g},\hat{\mathbf{K}})
\end{equation}
where $\hat{\mathbf{K}}$ is the covariance matrix and each element is calculated from the kernel $\hat{k}(\cdot,\cdot)$ in Eq.~\ref{eq:z_kernel}.
\begin{equation}
\label{eq:z_kernel}
[\hat{\mathbf{K}}]_{ij} = \hat{k}(\mathbf{z}_i,\mathbf{z}_j)
\end{equation}
Considering the symmetry property of the Gaussian distribution shown in Eq.~\ref{eq:symmetry}, the sampling of the pseudo observations in essence is equivalent to placing another GP prior over the sampled latent force function $\mathbf{g}$.
Therefore, this GP prior regularizes the sampled latent function.
Through the differential operator $\Psi$, the regularization propagates back to the target machine $\mathbf{f}(\cdot)$.
\begin{equation}
\label{eq:symmetry}
p(\omega|\mathbf{g},\hat{\mathbf{K}})=p(\mathbf{g}|\omega,\hat{\mathbf{K}})=p(\Psi \mathbf{f}|\mathbf{\omega},\hat{\mathbf{K}})
\end{equation}
Thus, the joint probability of the generative component is broken into four parts, as shown in Eq.~\ref{eq:ogfzxy}.
\begin{equation}
\label{eq:ogfzxy}
p(\omega,\mathbf{g},\hat{\mathbf{f}},\mathbf{Z}|\mathbf{X},\mathbf{Y})=p(\mathbf{Z})p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})p(\mathbf{g}|\hat{\mathbf{f}})p(\omega|\mathbf{g})
\end{equation}
where the prior of the $m$ input locations, $p(\mathbf{Z})$, $p(\hat{\mathbf{f}},\mathbf{Z},\mathbf{X},\mathbf{Y})$, and $p(\omega|\mathbf{g})$ are given by Eqs.~\ref{eq:priorz}-\ref{eq:pog}, respectively. Nals note that when no extra knowledge is available, $\mathbf{z}_{j}$ can be uniformly distributed assumedly.
\begin{equation}
\label{eq:priorz}
p(\mathbf{Z})=\Pi_{j=1}^m p(\mathbf{z}_{j})
\end{equation}
\begin{equation}
\label{eq:fzxy}
p(\hat{\mathbf{f}},\mathbf{Z},\mathbf{X},\mathbf{Y})=\Pi_{j=1}^m[\mathcal{N}(\hat{\mathbf{f}}(\mathbf{z}_{j})|\mu(\mathbf{z}_{j}),\nu(\mathbf{z}_{j}))]
\end{equation}
\begin{equation}
\label{eq:pog}
p(\omega|\mathbf{g})=\mathcal{N}(\omega|\mathbf{g},\hat{\mathbf{K}})
\end{equation}
\subsection{Posterior regularized inference algorithm}
Posterior regularization is a powerful inference methodology in the Bayesian stochastic modeling framework \citep{ganchev2010posterior}.
The objective includes the model likelihood on data and a penalty term that encodes the constrains over the posterior of the latent variables.
Via the penalty term, we can incorporate our domain knowledge or constrains outright to the posteriors, rather than through the priors and a complex, intermediate computing procedure.
A variety of successful posterior regularization algorithms have been proposed \citep{he2013graph,ganchev2013cross,zhu2014bayesian,libbrecht2015entropic,song2016kernel}.
Hence it can be more convenient and effective.
For efficient model inference, we marginalize out all latent variables in the joint probability to avoid estimating extra approximate posteriors.
Then we derive a convenient evidence lower bound to enable the reparameterization.
Using the reparameterization and auto-differentiation libraries, we develop an efficient stochastic optimization algorithm based on the posterior regularization inference framework \citep{ganchev2010posterior}.
The proposed inference algorithm is derived as follows.
The generative component in Eq.~\ref{eq:ogfzxy} is bind to the original GP in Eq.~\ref{eq:pyx} to obtain a new principled Bayesian model.
The joint probability is given by Eq.~\ref{eq:yogfzx}.
\begin{equation}
\label{eq:yogfzx}
p(\mathbf{Y},\omega,\mathbf{g}, \hat{\mathbf{f}},\mathbf{Z}|\mathbf{X})=p(\mathbf{Y}|\mathbf{X})p(\omega,\mathbf{g},\hat{\mathbf{f}},\mathbf{Z}|\mathbf{X},\mathbf{Y})
\end{equation}
We first marginalize out all the latent variables in the generative component to avoid approximating their posterior in Eq.~\ref{eq:oyx}.
\begin{equation}
\label{eq:oyx}
\begin{split}
p(\omega|\mathbf{X},\mathbf{Y})&=\iiint [p(\omega,\mathbf{g},\hat{\mathbf{f}},\mathbf{Z}|\mathbf{X},\mathbf{Y})\textrm{d}\mathbf{Z}\textrm{d}\mathbf{g}\text{d}\hat{\mathbf{f}}]\\
&=\iint [p(\mathbf{Z})p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})p(\omega|\Psi\hat{\mathbf{f}},\hat{\mathbf{K}})\textrm{d}\mathbf{Z}\textrm{d}\hat{\mathbf{f}}]\\
&=\iint [p(\mathbf{Z})p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})\mathcal{N}(\omega|\Psi \hat{\mathbf{f}},\hat{\mathbf{K}})\textrm{d}\mathbf{Z}\textrm{d}\hat{\mathbf{f}}]\\
&=\mathbb{E}_{p(\mathbf{Z})}\mathbb{E}_{p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})}\mathcal{N}(\Psi \mathbf{\hat{f}}|\mathbf{0},\hat{\mathbf{K}})
\end{split}
\end{equation}
The parameter $\gamma \geq 0$ is used to control the strength of regularization effect.
\begin{equation}
\label{eq:pyox}
p(\mathbf{Y},\omega|\mathbf{X})=p(\mathbf{Y}|\mathbf{X})p(\omega|\mathbf{X},\mathbf{Y})^\gamma
\end{equation}
The objective is to maximize the log-likelihood in Eq.~\ref{eq:lllh}.
\begin{equation}
\label{eq:lllh}
\begin{split}
\log [p(\mathbf{Y},\mathbf{\omega}|\mathbf{X})]=&\log [p(\mathbf{Y}|\mathbf{X})] + \gamma\log [p(\omega|\mathbf{X},\mathbf{Y})]\\
=&\log [(\mathcal{N}(\mathbf{Y}|\mathbf{0},\mathbf{\hat{K}}+\tau^{-1} \mathbf{I}))]\\
&+\gamma \log [\mathbb{E}_{p(\mathbf{Z})} \mathbb{E}_{p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})} [\mathcal{N}(\Psi \hat{\mathbf{f}}|\mathbf{0},\hat{\mathbf{K}})]]
\end{split}
\end{equation}
However, the log-likelihood is intractable due to the expectation inside the logarithm term.
To address this problem, the Jensen's inequality is used to obtain an evidence lower bound $\mathcal{L}$ in Eq.~\ref{eq:lllhlb}.
\begin{equation}
\label{eq:lllhlb}
\begin{split}
\log [p(\mathbf{Y},\mathbf{\omega}|\mathbf{X})]\geq \mathcal{L}= & \log[\mathcal{N}(\mathbf{Y}|\mathbf{\omega},\hat{\mathbf{K}}+\tau^{-1}\mathbf{I})]\\
& +\gamma \mathbb{E}_{p(\mathbf{z})} \mathbb{E}_{p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})} [\log [\mathcal{N}(\Psi \hat{\mathbf{f}}|\omega,\hat{\mathbf{K}})]]
\end{split}
\end{equation}
The existence of the general evidence lowerbound (ELBO) of a posterior distribution is proved with analyzing a decomposition of the Kullback-Leibler (KL) divergence by \cite{bishop2006pattern}.
Thus, we can obtain the ELBO of the log-likelihood in Eq.~\ref{eq:lllhlb}.
However, the ELBO is still intractable due to the non-analytical expectation term.
In view of the expectation is out of the logarithm, we can maximize $\mathcal{L}$ via stochastic optimization shown in Alg.~\ref{alg:1}.
\vspace{0.2in}
\begin{algorithm}[H]
\caption{The stochastic inference algorithm}
\label{alg:1}
\SetAlgoLined
\KwResult{Learned kernel parameters}
Initialization\;
\While{not reach stopping criteria}{
Sample a set of input locations $\mathbf{Z}$\;
Estimate the mean $\mu$ and the variance $\nu$ of $\mathbf{f}$ in Eqs.~\ref{eq:mu_def}-\ref{eq:nu_def}\;
Generate a parameterized sample of the posterior target function values $\hat{\mathbf{f}}$ by the reparameterization in Eqs.~\ref{eq:repara1}-\ref{eq:repara2}\;
Substitute the parameterized samples $\hat{\mathbf{f}}$ to obtain the unbiased estimated ELBO $\tilde{\mathcal{L}}$ in Eq.~\ref{eq:lllhlb}\;
Calculate $\nabla_\theta\tilde{\mathcal{L}}$, an unbiased stochastic gradient of $\tilde{\mathcal{L}}$ via the auto-differential technique\;
Update the parameters $\theta$ via the gradient decent shown in Eq.~\ref{eq:gradient} \;
}
\end{algorithm}
\begin{equation}
\label{eq:gradient}
\theta^{t+1}=\theta^{t}+\alpha\nabla_\theta\tilde{\mathcal{L}}
\end{equation}
where $\alpha$ refers to the learning rate and $\theta$ denotes all trainable parameters.
To prove the correctness of Alg.~\ref{alg:1}, we need to prove the correctness of employing a regularization via ELBO as follows.
\begin{theorem}
Maximizing the lowerbound of the log-likelihood is equivalent to a soft constraint over the posterior of the target function in the original GP.
\end{theorem}
\begin{proof}
While the proposed inference algorithm is developed for a hybrid model rather than pure GP \citep{ganchev2010posterior}, the evidence lower bound optimized by Alg.~\ref{alg:1} is a typical posterior regularization objective that estimates a pure GP model and meanwhile penalizes the posterior of the target function to encourage a consistency with the differential equations.
Jointly maximizing the term \[\mathbb{E}_{p(\mathbf{z})} \mathbb{E}_{p(\hat{\mathbf{f}}|\mathbf{Z},\mathbf{X},\mathbf{Y})} [\log [\mathcal{N}(\Psi \hat{\mathbf{f}}|\omega,\hat{\mathbf{K}})]]\] in the lowerbound of the log-likelihood $\mathcal{L}$ encourages all the possible latent force functions that are obtained from the target function $f(\cdot)$ via the differential operator $\Psi$ should be considered as being sampled from the same shadow GP.
This can be viewed as a soft constraint over the posterior of the target function in the original GP model.
Therefore, while being developed for inference of a hybrid model, the algorithm is equivalent to estimating the original GP model with some soft constraints on its posterior distribution.
Thus, the physics knowledge regularizes the learning of the target function in the original GP.
\end{proof}
To apply the proposed method with multiple differential equations (i.e. FD, conservation law, momentum),
Fig.~\ref{fig:traffic} shows the multi-equation multi-output framework of applying the proposed method to model the stochastic traffic flow process.
\begin{figure}[h!]
\centering
\begin{tikzpicture}[node0/.style={circle,draw},
node1/.style={circle,draw,fill=gray},
edge1/.style={->}]
\node [node1] (x) at (-3.5, 3){$\mathbf{X}$};
\node [node1] (y) at (-3.5, 1.5) {$\mathbf{Y}$};
\node [node1] (z) at (-1.25, 3) {$\mathbf{Z}$};
\node [node0] (f1) at (-1.25, 0.25) {$\mathbf{\hat{f}_1}$};
\node [node0] (g1) at (-1.25, -1.25) {$\mathbf{g_1}$};
\node (d) at (0.25, 0.25) {$\ldots$};
\node [node0] (f2) at (2.25, 0.25) {$\mathbf{\hat{f}}_{d^\prime}$};
\node [node0] (g2) at (2.25, -1.25) {$\mathbf{g}_w$};
\node [node1] (o) at (0.75, -3) {$\omega$};
\node (input) at (-5, 3) {$[(x,t)_i]_{2\times N}$};
\node (output) at (-5, 1.25) {$[(q,\rho,v)_i]_{3\times N}$};
\node (p_intput) at (0, 3) {$[(x,t)_j]_{2\times m}$};
\node (p_output) at (-3, 0.25) {$[(q,\rho,v)_j]_{3\times m}$};
\node (g_output1) at (-3, -1.25) {$[g_{1,j}]_{1\times m}$};
\node (g_output2) at (4, -1.25) {$[g_{w,j}]_{1\times m}$};
\node (0) at (2.5, -3) {$[[0,\ldots,0]^\intercal]_{1\times m}$};
\draw [edge1] (f1) to node {$\Psi f_1(q,\rho,v)$} (g1) ;
\draw [edge1] (f2) to node {$\Psi f_{d^\prime}(q,\rho,v)$} (g2) ;
\draw [edge1] (f1) to (g2);
\draw [edge1] (f2) to (g1);
\draw [edge1] (g1) to node[anchor=west] {shadow GPs} (o);
\draw [edge1] (g2) to (o);
\draw [edge1] (x) to node {data} (y) ;
\draw [edge1] (y) to (f1);
\draw [edge1] (y) to (f2);
\draw [edge1] (z) to node {$GP_1 (\mathbf{K}_1)$} (f1);
\draw [edge1] (z) to node {$GP_{d^\prime} (\mathbf{K}_{d^\prime})$} (f2);
\draw [edge1] (x) to (f1);
\draw [edge1] (x) to (f2);
\end{tikzpicture}
\caption{The proposed framework for multi-output multi-equation PRGP learning}
\label{fig:traffic}
\vspace{-0.2in}
\end{figure}
The log-likelihood and the ELBO of the traffic flow model can be formulated in Eq.~\ref{eq:lllhlb_traffic}.
\begin{equation}
\label{eq:lllhlb_traffic}
\begin{split}
\log [p(\mathbf{Y},\mathbf{\omega}|\mathbf{X})]\geq \mathcal{L}=& \sum_{i=1}^{d^\prime} \log[\mathcal{N}([\mathbf{Y}]_i|\mathbf{\omega},\hat{\mathbf{K}}_i+\tau^{-1}\mathbf{I})]\\
& +\sum_{w=1}^W\gamma_w \mathbb{E}_{p(\mathbf{z})} \mathbb{E}_{p(\hat{\mathbf{f}}_w|\mathbf{Z},\mathbf{X},\mathbf{Y})} [\log [\mathcal{N}(\Psi \hat{\mathbf{f}}_w|\omega,\hat{\mathbf{K}}_w)]]
\end{split}
\end{equation}
\subsubsection{Expressive kernels}
The expressive kernels are defined as the non-parametric smooth covariance functions, such as the well-known Squared Exponential Automatic Relevance Determination (SEARD) Kernel, and Radial Basis Function (RBF) kernel \citep{bishop2006pattern}, and deep kernels \citep{wilson2016deep}. The employed kernel functions are shown as follows:
The SE-ARD kernel is formulated in Eq.~\ref{eq:seard}.
\begin{equation}
\label{eq:seard}
k(\mathbf{x}_i,\mathbf{x}_j)=\sigma^2\exp(-(\mathbf{x}_i-\mathbf{x}_j)^\intercal\textrm{diag}(\mathbf{\eta}(\mathbf{x}_i-\mathbf{x}_j)))
\end{equation}
where $\textrm{diag}(\cdot)$ represents the diagonal matrix, $\sigma$ and $\eta$ are kernel parameters.
The RBF kernel is formulated in Eq.~\ref{eq:rbf}.
\begin{equation}
\label{eq:rbf}
k(\mathbf{x}_i,\mathbf{x}_j)=\exp(-\frac{(||\mathbf{x}_i-\mathbf{x}_j||)^2}{2\sigma^2})
\end{equation}
where $\sigma$ is the kernel parameter.
\subsubsection{Algorithm complexity}
The time complexity of the inference of the original GP is $O(N^3)$.
The time complexity of the inference of the shadow GP is $O(m^3)$.
Thus, the total time complexity for the inference of two GPs is $O((Nd^\prime)^3+m^3)$.
To store the kernel metrics of original GP and the shadow GP, the space complexity is $O((Nd^\prime)^2+m^2)$.
In the testing phase, the time complexity of the model estimation is marginal (less than $1$ ms) empirically.
\subsection{Physics regularized traffic state estimation}
To apply the proposed method, the traffic flow models need to be converted to the form of Eq.~\ref{eq:nonlinearlfm}. In this study, we aims to encode three classical traffic flow models, LWR, PW, and ARZ, into the GP and compare their performance under the framework of PRGP. More specifically, the converted LWR, PW, ARZ models are presented as follows.
In PRGP, the stochastic conservation law of LWR is formulated in Eq.~\ref{eq:lwr_gp}.
\begin{equation}
\label{eq:lwr_gp}
\Psi f_1(q,\rho,v) = \partial_t\rho+\partial_x q = g_1
\end{equation}
The stochastic PW model is formulated in Eqs.~\ref{eq:pw1_gp}-\ref{eq:pw2_gp}.
\begin{equation}
\label{eq:pw1_gp}
\Psi f_1(q, \rho,v) = \partial_t\rho+\partial_x(\rho v) = g_1
\end{equation}
\begin{equation}
\label{eq:pw2_gp}
\Psi f_2(q, \rho,v) = \partial_t v+v\partial_x v+\frac{V-V(\rho)}{\tau_0}+\frac{c^2_0}{\rho}\partial_x\rho = g_2
\end{equation}
And the stochastic ARZ model is formulated in Eqs.~\ref{eq:arz1_gp}-\ref{eq:arz2_gp}.
\begin{equation}
\label{eq:arz1_gp}
\Psi f_1(q, \rho,v) = \partial_t\rho+\partial_x(\rho v) = g_1
\end{equation}
\begin{equation}
\label{eq:arz2_gp}
\Psi f_2(q, \rho,v) = \partial_t(v-V(\rho) + v\partial_x (v-V(\rho))+\frac{v-V(\rho)}{\tau_0} = g_2
\end{equation}
\section{Numerical Tests with Field Data}\label{sec:4}
\subsection{Case setting}
To evaluate the performance of the proposed PRML framework, We applied the three PRGP models to estimate the traffic flow in a stretch of the interstate freeway I-15 across Utah, U.S.
The Utah Department of Transportation (UDOT) has installed sensors every a few miles along the freeway.
Each sensor counts the number of vehicles passed every minute, measures the speed of each vehicle, and sends the data back to a central database, named Performance Measurement System (PeMS).
The collected real-time data and road conditions are available online and can be accessed by the public.
For model evaluations, the data, from August 5, 2019 to August 11, 2019, were collected by four sensors on the I-15, Utah.
The input variables include the location coordinates of each sensor and the time of each read.
The studied stretch is illustrated in Fig.~\ref{fig:stretch}, where the yellow line indicates the studied freeway segments and the blue bars represent the locations of traffic detectors.
In the case, the data is shuffled and randomly split into the training set and testing set separately.
\begin{figure}[h!]
\centering
\includegraphics[height=8cm]{Figures/testbed.png}
\caption{The stretch of the studied freeway segment which includes four detectors}
\label{fig:stretch}
\end{figure}
\subsection{Implementation}
The deep kernel can be in any neural network structure, such as the feed-forward neural network, and can be fine-tuned to achieve better empirical results.
Incorporating the SEARD and RBF kernels, the compound kernel of the $d'$-dimensional original GP and the $W$-dimensional shadow GP are computed in Fig.~\ref{fig:implement}.
The procedure for estimating the target traffic state $q,v$ of any given input $x,t$ is illustrated in Fig.~\ref{fig:predict}.
In the multi-output multi-equation PRGP, the $d'$-dimension means for each dimension of $\mathbf{y}$ creating one compound kernel, and the $W$-dimension means for each differential equation creating one compound kernel.
Note that the structure of the GPs can be fine-tuned to achieve better empirical performance.
\begin{figure}[h!]
\centering
\includegraphics[width=0.7\textwidth]{Figures/fig5.pdf}
\caption{The structure of the proposed loss function}
\label{fig:implement}
\end{figure}
\begin{figure}[h!]
\centering
\includegraphics[width=0.7\textwidth]{Figures/fig6.pdf}
\caption{The structure of estimation}
\label{fig:predict}
\vspace{-0.2in}
\end{figure}
In the experiments, the parameters of the proposed method are set as follows:
(a) the number of pseudo observations $m=10$,
(b) the strength of regularization $\lambda$ is fine-tuned numerically.
The proposed inference algorithm is implemented in the Tensorflow framework, where the optimizer ADAM \citep{kingma2014adam} is chosen for updating the parameters.
\subsection{Results Analysis}
\subsubsection{Comparison with Pure Machine Learning Models}
To prove the superiority of the proposed PRML framework compared with pure ML models, this subsection aims to compare the three PRGP models, LWR-PRGP, PW-PRGP, and ARZ-PRGP, with pure GP and other popular ML models such as multilayer perceptron, support vector machine, and random forest \citep{bishop2006pattern}. Also recall that one main contribution of PRML is that it is more explainable in terms of model performance. Hence, this study further adopts another physical model, the well-known heat equation, to prove the indispensability of classical traffic flow models in the PRML framework, since the heat equation is not suitable to model traffic flows. The heat equation is formulated in Eq.~\ref{eq:heat}.
\begin{equation}
\label{eq:heat}
\frac{\partial f_h(x,t)}{\partial t}=\beta_1\nabla^2 f_h(x,t)
\end{equation}
Note that the inputs of the proposed PRGP-based methods and classical traffic flow models are different. The latter method often requires the on-ramp and off-ramp flow observations as inputs, while the proposed method assumes unobserved on-ramp and off-ramp flows in the framework and does not require such data.
The training process of each model, with $500$ iterations and $2,880$ samples, costs $10,480$ seconds in average on a workstation equipped with a 3.5GHz 6-core CPU.
In the testing phase, the time complexity of the model estimation is marginal (less than $1$ second) empirically, similar to all ML models.
Note that the computational process can be accelerated by about $5$ time if a NVIDIA CUDA-capable GPU is used.
Figs~\ref{fig:prgp_flow}-\ref{fig:prgp_speed} compare the flow and speed estimations with the ground truth in the studied case. If the coefficient of the trend line is close to $1$ and the intercept is close to $0$, the estimation will be considered as accurate. The results show that both pure GP and proposed PRGP models can perform well in estimating the flows and speeds.
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/GP_flow.png}
\caption{GP}
\label{fig:gp_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_LWR_flow.png}
\caption{LWR-PRGP}
\label{fig:pigp_lwr_flow}
\end{subfigure}
\\
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_PW_flow.png}
\caption{PW-PRGP}
\label{fig:pigp_pw_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_ARZ_flow.png}
\caption{ARZ-PRGP}
\label{fig:pigp_arz_noise_flow}
\end{subfigure}
\caption{Comparison between flow estimation by GP and PRGPs and the ground truth}
\label{fig:prgp_flow}
\vspace{-0.2in}
\end{figure}
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/GP_speed.png}
\caption{GP}
\label{fig:gp_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_LWR_speed.png}
\caption{LWR-PRGP}
\label{fig:pigp_lwr_speed}
\end{subfigure}
\\
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_PW_speed.png}
\caption{PW-PRGP}
\label{fig:pigp_pw_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_ARZ_speed.png}
\caption{ARZ-PRGP}
\label{fig:pigp_arz_speed}
\end{subfigure}
\caption{Comparison between speed estimations by GP and PRGPs and the ground truth}
\label{fig:prgp_speed}
\vspace{-0.2in}
\end{figure}
To quantify the precision of outputs, Rooted Mean Squared Error (RMSE) and Mean Absolute Percentage Error (MAPE) of each dimension are used as the performance metric, which are defined in Eqs.~\ref{eq:def_rmse}-\ref{eq:def_mape}.
\begin{equation}
RMSE_j = \sqrt{\frac{1}{N}\sum_{i=1}^{N}{\Big(\frac{[\mathbf{y}_j]_{i}-[\hat{\mathbf{f}}_j]_i}{\sigma_i}\Big)^2}}, \forall j\in {1,\ldots,d^\prime}
\label{eq:def_rmse}
\end{equation}
\begin{equation}
MAPE_j = \frac{100\%}{N}\sum_{i=1}^{N}{\Big\vert\frac{[\mathbf{y}_j]_{i}-[\hat{\mathbf{f}}_j]_i}{[\mathbf{y}_j]_{i}}\Big\vert}, \forall j\in {1,\ldots,d^\prime}
\label{eq:def_mape}
\end{equation}
Table~\ref{tab:benchmark1} summarizes of the results of the comparable baselines and the proposed method in the same dataset. Among the four pure ML models, the GP can obviously outperform the other ML models in terms of providing more accurate estimations of both flows and speeds. The GP can yield a 39.74~veh/5-min of RMSE and a 13.70\% of MAPE for flow and a 2.7~mph of RMSE and a 2.64\% for MAPE for speed, while the other three produced much higher RMSEs and MAPEs of both flow and speed estimates. Further comparison between the pure GP and the three PRGP models reveal that PRGP models can improve the accuracy of both flow and speed estimations. However, the improvement is not significant, which is because pure GP can already achieve a very good estimation performance and leaves limited space for improvement by the PRGP. Moreover, to validate PRGP's contribution in making the results more explainable, the comparison with Heat-PRGP, which uses the physical knowledge from the heat equation, shows that a physical model that cannot precisely describe the traffic flow patterns could even downgrade the capability of the PRGP. Another side evidence is that PW and ARZ, which are the improved version of LWR, can improve the performance of the PRGP compared with the LWR.
\begin{table}[h!]
\caption{Comparison of the results of the proposed method and the baseline methods}
\centering
\begin{tabular}{c|p{2.0cm}|c|p{2.1cm}|c}
\toprule
Method & Flow RMSE (veh/5min) & Flow MAPE & Speed RMSE (mph) & Speed MAPE \\%veh/h (km/h)
\midrule
Multilayer perceptron & 113.95 &30.80\%& 13.61 &19.91\%\\
Support Vector Machine & 124.84 &34.24\%& 9.58 &13.01\%\\
Random Forest & 108.24 &27.60\%& 8.66 &12.02\%\\
pure GP & 39.74 &13.70\%& 2.76 & 2.64\%\\
LWR-PRGP & 37.19 &12.77\%& 2.96 & 2.65\%\\
PW-PRGP & 35.45 &12.42\%& 3.02 & 2.68\%\\
ARZ-PRGP & 34.75 &11.48\%& 2.90 & 2.72\%\\
Heat-PRGP & 79.51 &23.49\%& 5.20 & 6.75\%\\
\bottomrule
\end{tabular}
\label{tab:benchmark1}
\end{table}
\subsubsection{Comparison with physical models (Traffic Flow Models)}
To provide physical baselines for the performance comparison, the LWR, PW, ARZ models are calibrated with the obtained field data. For model calibration, we follow the method by \cite{akwir2018neural}, where the hybrid scheme of neural network and nonlinear partial differential equation is used to dynamically adjust all outputs of the three models to obtain their calibrated parameters.
Figs~\ref{fig:pk_flow}-\ref{fig:pk_speed} plot the estimated flow and speed from the three physical models versus the ground truth. Obviously, the estimation results are quite biased for both flow and speed.
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/LWR_flow.png}
\caption{LWR}
\label{fig:lwr_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Figures/PW_flow.png}
\caption{PW}
\label{fig:pw_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Figures/ARZ_flow.png}
\caption{ARZ}
\label{fig:arz_flow}
\end{subfigure}
\caption{Estimated flow by the calibrated physical models v.s. ground truth}
\label{fig:pk_flow}
\end{figure}
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.3\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/LWR_speed.png}
\caption{LWR}
\label{fig:lwr_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Figures/PW_speed.png}
\caption{PW}
\label{fig:pw_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.3\textwidth}
\includegraphics[width=\textwidth]{Figures/ARZ_speed.png}
\caption{ARZ}
\label{fig:arz_speed}
\end{subfigure}
\caption{Estimated speed by the calibrated physical models v.s. ground truth}
\label{fig:pk_speed}
\end{figure}
To better justify models' estimation accuracy, Table~\ref{tab:benchmark2} shows the results of proposed method and the calibrated physical models in estimation errors. It can be found that the proposed method significantly outperforms the baseline methods by around $80~veh/5min$ in flow RMSE and $18\%$ in MAPE and $7~mph$ in speed RMSE and $15\%$ in MAPE. Hence, it can be concluded that the estimation performance of traffic flows models can be greatly improved if they are encoded into a ML framework. The real-world uncertainties of flow and speed can be captured by the ML portion properly.
\begin{table}[h!]
\caption{Comparison of the results of the proposed methods and the physics-based methods}
\centering
\begin{tabular}{c|p{2.0cm}|c|p{2.1cm}|c}
\toprule
Method & Flow RMSE (veh/5min) & Flow MAPE & Speed RMSE (mph) & Speed MAPE\\%veh/h (km/h)
\midrule
Calibrated LWR & 115.75 & 32.96\%& 9.88 & 14.4\%\\
LWR-regularized GP & 37.19 & 12.77\%& 2.96 & 2.76\%\\
Calibrated PW & 115.80 & 30.00\%& 10.41 & 18.2\%\\
PW-regularized GP & 35.5 & 12.42\%& 3.02 & 2.68\%\\
Calibrated ARZ & 155.20 & 32.00\%& 12.71 & 18.4\%\\
ARZ-regularized GP & 34.75 & 11.48\%& 2.90 & 2.72\%\\
\bottomrule
\end{tabular}
\label{tab:benchmark2}
\end{table}
\subsubsection{Robustness study}
As aforementioned, the proposed PRML framework is expected to be more robust than pure ML models on noisy dataset. Hence, in this subsection, $50\%$ of the training data is replaced by the flawed data, which are generated with $100~veh/5min$ noises in flows, and the testing data keep unchanged.
Notably, for model evaluations, the testing dataset is not mixed with noises. Also, since GP can outperform multilayer perceptron, support vector machine, and random forest in both flow and speed estimation, we will only examine the robustness of GP and PRGP in this subsection.
Table~\ref{tab:noise} and Figs~\ref{fig:noise_flow}-\ref{fig:noise_speed} summarize their estimation performance on the noised training data.
The results show that the GP has limited resistance to high biased data, e.g., caused by traffic detector malfunctions.
The three PRGP models can greatly outperform pure GP by about 160~veh/h of RMSE and over 100\% of MAPE in flow estimations. Hence, it can be concluded that the proposed PRML framework are much more robust than the pure ML models when the input data is subject to unobserved random noise. This is due to PRML's capability of adopting physical knowledge to regularized the ML training process.
The results also show that heat equation does not capture the dynamics of the traffic flow, and only the well-developed traffic flow model can improve the accuracy of Gaussian process.
\begin{table}[h!]
\caption{Comparison of the estimation accuracy with noisy training dataset}
\centering
\begin{tabular}{c|p{2.0cm}|c|p{2.1cm}|c}
\toprule
Method & Flow RMSE (veh/5min) & Flow MAPE & Speed RMSE (mph) & Speed MAPE\\
\midrule
pure GP & 212.17 & 135.19\%& 5.96 & 3.35\% \\
GP-LWR & 41.78 & 9.73\%& 6.01 & 3.46\% \\
GP-PW & 41.11 & 9.60\%& 4.43 & 3.30\%\\
GP-ARZ & 35.37 & 9.51\%& 3.06 & 2.72\%\\
GP-HEAT & 215.01 & 138.29\%& 4.31 & 33.6\%\\
\bottomrule
\end{tabular}
\label{tab:noise}
\end{table}
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/GP_noise_flow.png}
\caption{GP}
\label{fig:gp_noise_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_LWR_noise_flow.png}
\caption{GP-LWR}
\label{fig:pigp_lwr_noise_flow}
\end{subfigure}
\\
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_PW_noise_flow.png}
\caption{GP-PW}
\label{fig:pigp_pw_noise_flow}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_ARZ_noise_flow.png}
\caption{GP-ARZ}
\label{fig:pigp_arz_noise_flow}
\end{subfigure}
\vspace{-0.1in}
\caption{Comparison between flow estimation and ground truth with noisy training dataset}
\label{fig:noise_flow}
\vspace{-0.2in}
\end{figure}
\begin{figure}[h!]
\centering
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/GP_noise_speed.png}
\caption{GP}
\label{fig:gp_noise_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/PIGP_LWR_noise_speed.png}
\caption{GP-LWR}
\label{fig:pigp_lwr_noise_speed}
\end{subfigure}
\\
\begin{subfigure}[b]{0.45\textwidth}
\centering
\includegraphics[width=\textwidth]{Figures/PIGP_PW_noise_speed.png}
\caption{GP-PW}
\label{fig:pigp_pw_noise_speed}
\end{subfigure}
\hfill
\begin{subfigure}[b]{0.45\textwidth}
\includegraphics[width=\textwidth]{Figures/PIGP_ARZ_noise_speed.png}
\caption{GP-ARZ}
\label{fig:pigp_arz_noise_speed}
\end{subfigure}
\vspace{-0.1in}
\caption{Comparison between speed estimation and ground truth with noisy training dataset}
\label{fig:noise_speed}
\vspace{-0.2in}
\end{figure}
\section{Conclusions and Future Research Directions}\label{sec:5}
In the literature, traffic flow models have been well developed to explain the traffic phenomena, however, have theoretical difficulties in stochastic formulations and rigorous estimation.
In view of the increasing availability of data, the data-driven methods are prevailing and fast-developing, however, have limitations of lacking sensitivity of irregular events and compromised effectiveness in sparse data.
To address the issues of both methods, a hybrid framework to incorporate the advantages of both methods is investigated.
This paper proposes a stochastic modeling framework to capture the random detection noise and the latent unobserved of traffic data as well as leveraging the well-defined fundamental diagram, conservation law and momentum conditions.
The traffic state indicators (i.e. flow, speed, density) are assumed to be multi-variant Gaussian distributed.
A physics regularized Gaussian process (PRGP) is proposed to encode the physics knowledge in the Bayesian inference structure as the shadow Gaussian process.
The shadow Gaussian process is proven to regularize the conventional constraint-free Gaussian process as a soft constraint.
To estimate the proposed PRGP, a posterior regularized inference algorithm is derived and implemented with auto-differentiation libraries.
The computational complexity is cubic of the product of the sample size and the output dimension $O((Nd^\prime)^3+m^3)$.
A preliminary real-world case study is conducted on PeMS detection data collected from a freeway segment in Utah and the well-known continuous traffic flow models (i.e. LWR, PW, ARZ) are tested.
In comparison to the pure machine learning methods and pure physical models, the numerical results justify the effectiveness and the robustness of the proposed method.
The potential directions for future Research may include:
(1) extending the proposed method to leverage other models for traffic state estimation, such as discrete macroscopic traffic flow model regularized Gaussian process;
(2) extending the proposed method to solve other problems, such as microscopic behavior models regularized Gaussian process for vehicle trajectory prediction;
(3) extending the physics regularization methodology in other machine learning algorithms, such as random forest and support vector machine to combine general physics knowledge in learning tasks.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 7,535
|
\section{Conclusion}
In this paper, we have presented a simple yet efficient nearest-neighbor classification model to detect user intents and OOS intents.
It includes paired encoding and discriminative training to model relations between the input and example utterances.
Moreover, a seamless transfer from NLI and a joint approach with fast retrieval are designed to improve the performance in terms of the accuracy and inference speed.
Experimental results show superior performance of our method on a large-scale multi-domain intent detection dataset with OOS.
Future work includes its cross-lingual transfer and cross-dataset (or cross-task) generalization.
\section{Experimental Results}
This section shows our experimental results.
Appendix~\ref{extra-results} shows some additional figures.
\subsection{Model Performance CLINC150 Dataset}
\paragraph{Single domains}
We first show test set results of 5-shot and 10-shot in-domain classification and OOS detection accuracy in Table~\ref{table:fourDomain} for the four selected domains.
In the 5-shot setting, the proposed DNNC method consistently attains the best results across all the four domains.
The comparison between DNNC-scratch and DNNC shows that our NLI task transfer is effective.
In the 10-shot setting, all the approaches generally experience an accuracy improvement due to the additional training data, and the dominant performance of DNNC weakens, although it remains highly competitive.
We can see that our DNNC is comparable with or even surpasses some of the 50-shot classifier's scores, and the data augmentation techniques are not always helpful when we use the strong pre-trained model.
\paragraph{Entire CLINC150 dataset}
Next, Table~\ref{table:wholeDataset} shows results to compare our method with the classifier and USE+ConveRT baselines, on the entire CLINC150 dataset with the 150 intents. USE+ConveRT performs worse than the RoBERTa-based classifier on the OOD detection task.
The advantage of DNNC for in-domain intent detection is clear, with its 10-shot in-domain accuracy close to the upper-bound accuracy for the classifier baseline.
One observation is that our DNNC method tends to be more confident about its prediction, with the increasing number of the training examples; as a result, the OOS recall becomes lower in the 10-shot setting, while the OOS precision is much higher than the other baselines.
Better controlling the confidence output of the model is an interesting direction for future work.
When the USE+ConveRT baseline is evaluated along with the OOS detection task, its overall accuracy is not as good as the other RoBERTa-based models, despite its potential in the purely in-domain classification.
This indicates that the fine-tuned (Ro)BERT(a) models are more robust to out-of-distribution examples than shallower models like USE+ConveRT, also suggested in \citet{acl-ood}.
\if0
\begin{table*}[t]
\centering
\small
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|cl|cl|cl|cc}
\hline
& \multicolumn{2}{c|}{\textbf{In-domain accuracy}} & \multicolumn{2}{c|}{\textbf{OOS recall}} & \multicolumn{2}{c|}{\textbf{OOS precision}} & \multicolumn{2}{c}{\textbf{OOS F1}} \\
& \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \textbf{10-shot} \\ \cline{2-9}
classifier & 77.7 $\pm$ 0.6 & 83.2 $\pm$ 0.6 & \textbf{91.9 $\pm$ 1.1} & \textbf{92.9 $\pm$ 0.8} & 53.9 $\pm$ 0.8 & 58.2 $\pm$ 1.0 & 68.0 $\pm$ 1.0 & 71.6 $\pm$ 0.8 \\
DNNC & \textbf{84.9 $\pm$ 0.8} & \textbf{91.6 $\pm$ 0.3} & 90.1 $\pm$ 1.6 & 83.0 $\pm$ 2.0 & \textbf{64.7 $\pm$ 2.5} & \textbf{81.4 $\pm$ 2.0} & \textbf{75.3 $\pm$ 2.0} & \textbf{82.1 $\pm$ 0.3} \\ \hdashline
classifier (full-shot) & \multicolumn{2}{c|}{92.9 $\pm$ 0.2} & \multicolumn{2}{c|}{90.2 $\pm$ 0.5} & \multicolumn{2}{c|}{76.7 $\pm$ 0.7} & \multicolumn{2}{c}{82.9 $\pm$ 0.6} \\ \hline
\end{tabular}}\caption{Testing results on the whole dataset.}
\label{table:wholeDataset}
\end{table*}
\fi
\begin{figure*}[t!
\centering
\begin{minipage}[t]{1.0\textwidth}
\centering
\includegraphics[width=0.92\linewidth]{./images/conf_hist_main.png}
\caption{Model confidence level on 5-shot test sets for the banking domain and all domains.}
\label{fig:separation}
\end{minipage}%
\vspace{10pt}
\begin{minipage}[t]{0.71\textwidth}
\centering
\includegraphics[width=0.3\linewidth]{./images/curve_main_in_acc.png}
\includegraphics[width=0.3\linewidth]{./images/curve_main_oos_recall.png}
\includegraphics[width=0.3\linewidth]{./images/curve_main_oos_prec.png}
\caption{Development set results on the banking domain in the 5-shot setting. In this series of plots, a model with a higher area-under-the-curve is more robust.}
\label{fig:robust}
\end{minipage}%
\hfill
\begin{minipage}[t]{0.25\textwidth}
\centering
\includegraphics[width=1.\linewidth]{./images/latency.png}
\caption{Accuracy vs. latency of DNNC-joint.}
\label{fig:join-nli}
\end{minipage}%
\centering
\label{fig:compare_fig}
\end{figure*}
\begin{table*}[]
\centering
\small
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|cc|cc|cc|cc|rr}
\hline
& \multicolumn{2}{c|}{\textbf{In-domain accuracy}} & \multicolumn{2}{c|}{\textbf{OOS recall}} & \multicolumn{2}{c|}{\textbf{OOS precision}} & \multicolumn{2}{c}{\textbf{OOS F1}} & \multicolumn{2}{c}{\textbf{Latency [ms./example]}} \\
\textbf{Banking} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} \\ \hline
TF-IDF-$k$NN & 59.2 $\pm$ 3.6 & 65.9 $\pm$ 3.8 & 61.0 $\pm$ 2.8 & 51.3 $\pm$ 1.6 & 97.6 $\pm$ 0.6 & 98.2 $\pm$ 0.3 & 75.1 $\pm$ 2.2 & 67.4 $\pm$ 1.3 & 1 & 1 \\
Emb-$k$NN-vanilla & 64.4 $\pm$ 2.8 & 65.4 $\pm$ 2.3 & 89.5 $\pm$ 1.1 & 94.6 $\pm$ 0.5 & 95.1 $\pm$ 0.5 & 92.6 $\pm$ 0.7 & 92.2 $\pm$ 0.5 & 93.6 $\pm$ 0.5 & 15 & 15 \\
Emb-$k$NN & 78.4 $\pm$ 2.4 & 84.3 $\pm$ 1.2 & 92.0 $\pm$ 2.0 & 91.6 $\pm$ 1.2 & 91.4 $\pm$ 0.8 & 93.7 $\pm$ 0.5 & 91.7 $\pm$ 0.7 & 92.6 $\pm$ 0.7 & 15 & 15 \\
RN-$k$NN & 79.5 $\pm$ 3.1 & 89.0 $\pm$ 1.4 & 88.3 $\pm$ 1.9 & 75.9 $\pm$ 4.0 & 91.6 $\pm$ 1.3 & 95.9 $\pm$ 0.6 & 89.9 $\pm$ 1.0 & 84.7 $\pm$ 2.5 & 17 & 17 \\ \hdashline
DNNC-joint & 88.5 $\pm$ 1.2 & 91.0 $\pm$ 1.0 & 95.0 $\pm$ 0.9 & 95.2 $\pm$ 1.1 & 96.9 $\pm$ 0.4 & 97.3 $\pm$ 0.4 & 96.0 $\pm$ 0.5 & 96.3 $\pm$ 0.6 & 36 & 36 \\
DNNC & 88.6 $\pm$ 1.3 & 91.2 $\pm$ 1.1 & 94.7 $\pm$ 1.0 & 94.8 $\pm$ 1.1 & 97.0 $\pm$ 0.3 & 97.5 $\pm$ 0.4 & 95.9 $\pm$ 0.5 & 96.1 $\pm$ 0.6 & 73 & 143 \\ \hline
\textbf{All domains} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} & \textbf{5-shot} & \textbf{10-shot} \\ \hline
TF-IDF-$k$NN & 35.4 $\pm$ 0.7 & 31.3 $\pm$ 1.1 & 72.2 $\pm$ 2.1 & 83.1 $\pm$ 1.5 & 24.7 $\pm$ 0.4 & 23.5 $\pm$ 0.4 & 36.8 $\pm$ 0.6 & 36.6 $\pm$ 0.6 & 1 & 2 \\
Emb-$k$NN-vanilla & 54.2 $\pm$ 0.5 & 50.7 $\pm$ 0.9 & 87.5 $\pm$ 0.7 & 95.1 $\pm$ 0.3 & 39.6 $\pm$ 0.3 & 34.2 $\pm$ 0.3 & 54.5 $\pm$ 0.4 & 50.3 $\pm$ 0.3 & 16 & 19 \\
Emb-$k$NN & 78.7 $\pm$ 1.0 & 86.9 $\pm$ 0.3 & 86.8 $\pm$ 2.4 & 82.3 $\pm$ 0.8 & 55.0 $\pm$ 1.5 & 66.5 $\pm$ 1.0 & 67.3 $\pm$ 1.6 & 73.5 $\pm$ 0.7 & 16 & 19 \\
RN-$k$NN & 76.6 $\pm$ 0.6 & 88.7 $\pm$ 1.1 & 88.9 $\pm$ 1.4 & 76.6 $\pm$ 2.7 & 50.0 $\pm$ 0.3 & 71.0 $\pm$ 0.5 & 64.0 $\pm$ 0.2 & 73.7 $\pm$ 1.0 & 16 & 18 \\ \hdashline
DNNC-joint & 84.5 $\pm$ 0.8 & 91.2 $\pm$ 0.2 & 90.6 $\pm$ 1.6 & 85.1 $\pm$ 1.8 & 63.6 $\pm$ 2.4 & 78.4 $\pm$ 2.1 & 74.7 $\pm$ 1.9 & 81.6 $\pm$ 0.4 & 37 & 41 \\
DNNC & 84.9 $\pm$ 0.8 & 91.6 $\pm$ 0.3 & 90.1 $\pm$ 1.6 & 83.0 $\pm$ 2.0 & 64.7 $\pm$ 2.5 & 81.4 $\pm$ 2.0 & 75.3 $\pm$ 2.0 & 82.1 $\pm$ 0.3 & 697 & 1498 \\ \hline
\end{tabular}}\caption{Comparison among the nearest neighbor methods on the test sets for the banking domain and all the domains. The latency is measured on a single NVIDIA Tesla V100 GPU, where the batch size is $1$ to simulate an online use case. DNNC-joint is based on top-20 Emb-$k$NN retrieval. }
\label{table:latency}
\end{table*}
\subsection{Robustness of DNNC}
As described in Section~\ref{subsection:evalMetrics}, we select the threshold to determine OOS by making a trade-off between in-domain classification and OOS detection accuracy.
It is therefore desirable to have a model with candidate thresholds that provide high in-domain accuracy as well as OOS precision and recall.
\if0
We observe in Figure~\ref{fig:robust} that in the 5-shot setting, in-domain accuracy and OOS detection precision for both DNNC-scratch and classifier models drop rapidly as the threshold increases, while the OOS recall increases rapidly with the threshold. In contrast, the DNNC demonstrates robust performances across all three metrics with the changes in the threshold. \fi
We observe in Figure~\ref{fig:robust} that in the 5-shot setting, DNNC is the most robust to the threshold selection.
The contrast between the classification model and DNNC-scratch suggests that nearest neighbor approaches (in this case DNNC) make for stronger discriminators; the advantage of DNNC over DNNC-scratch further demonstrates the power of the NLI transfer and, perhaps more importantly, the effectiveness of the pairwise discriminative pre-training.
This result is consistent with the intuition we gained from Figure~\ref{fig:tsne-main}, and the overall observation is also consistent across different settings.
To further understand the differences in behaviors between the classification model and DNNC method, we examine the output from the final softmax/sigmoid function (model confidence score) in Figure~\ref{fig:separation}.
At 5-shot, the classifier method still struggles to fully distinguish the in-domain examples from the OOS examples in its confidence scoring, while DNNC already attains a clear distinction between the two.
Again, we can clearly see the effectiveness of the NLI transfer.
With the model architectures for BERT-based classifier and DNNC being the same (RoBERTa is used for both methods) except for the final layer (multi-class-softmax vs. binary sigmoid), this result suggests that the pairwise NLI-like training is more sample-efficient, making it an excellent candidate for the few-shot use case.
\subsection{DNNC-joint for Faster Inference}
Despite its effectiveness in few-shot intent and OOS settings, the proposed DNNC method might not scale in high-traffic use cases, especially when the number of classes, $N$, is large, due to the inference-time bottleneck (Section~\ref{subsec:joint}).
With this in mind, we proposed the DNNC-joint approach, wherein a faster model is used to filter candidates for the fine-tuned DNNC model.
We compare the accuracy and inference latency metrics for various methods in Table~\ref{table:latency}.
Note that Emb-$k$NN and RN-$k$NN exhibit excellent latency performance, but they fall considerably short in both the in-domain intent and OOS detection accuracy, compared to DNNC and the DNNC-joint methods.
On the other hand, the DNNC-joint model shows competitiveness in both inference latency and accuracy.
These results indicate that the current text embedding approaches like SBERT are not enough to fully capture fine-grained semantics.
Intuitively, there is a trade-off between latency and inference accuracy: with aggressive filtering, the DNNC inference step needs to handle a smaller number of training examples, but might miss informative examples; with less aggressive filtering, the NLI model sees more training examples during inference, but will take longer to process single user input.
This is illustrated in Figure~\ref{fig:join-nli}, where the in-domain intent and OOS accuracy metrics (on the development set of the banking domain in the 5-shot setting) improve with the increase of $k$, while the latency increases at the same time.
Empirically, $k=20$ appears to strike the balance between latency and accuracy, with the accuracy metrics similar to those of the DNNC method, while being much faster than DNNC (dashed lines are the corresponding DNNC references).
\if0
\subsection{Training with OOS examples}
\paragraph{OOS examples}
We could optionally have access to training examples of the OOS utterances, so that the model can explicitly learn to detect OOS utterances.
In this case we also have $K$ training examples: $(e'_1, e'_2, \ldots, e'_K)$.
It should be noted that arbitrary utterances can be the OOS examples which are {\it not} categorized into any of the $N$ intent classes, and the definition of OOS is different depending on $\mathbf{C}$.
It is thus ideal not to assume the access to enough amount of the OOS examples, from the viewpoint of out-of-distribution detection.
\paragraph{OOS as an extra class:}
The second strategy is to add an extra class for OOS to formulate an $(N+1)$-class classification task.
This is possible {\it only} when we have the OOS training examples.
At inference time, we do not need to set the threshold because OOS is treated as one of the classes.
\paragraph{Training with OOS examples}
We can add more negative examples when the OOS training examples are available.
The objective of using the OOS examples is to help the model explicitly learn to reject the OOS examples based on the in-domain examples.
We therefore create such pairs: $(e'_{\ell}, e_{j, i})$, whereas we do not consider $(e_{j, i}, e'_{\ell})$.
In total, we have $NK^2$ additional negative examples.
The example (c) in Table~\ref{tb:examples} shows that the request in the input sentence is not supported in the dataset, and thus does not match the other sentence's intent.\footnote{Another alternative strategy is to use $N$ binary classifiers as in one-vs.-rest classification, instead of the softmax classifier in Equation~(\ref{eq:softmax}).
This is also attractive in that we can unify the two strategies.
However, we did not observe empirical advantage above the softmax classifier.}
\fi
\section{Introduction}
\begin{figure}[ht]
\begin{center}
\includegraphics[width=1.0\linewidth]{./images/fig_tsne_main_more.png}
\end{center}
\caption{tSNE visual comparison for OOS detection between existing methods ((a) and (b)) and our proposed method ((c) and (d)). Their embeddings before their classifier layers are used (best viewed in color).}
\label{fig:tsne-main}
\vspace{-1em}
\end{figure}
Intent detection is one of the core components when building goal-oriented dialog systems.
The goal is to achieve high intent classification accuracy, and another important skill is to accurately detect unconstrained user intents that are out-of-scope (OOS) in a system~\citep{oos-intent}.
A practical challenge is data scarcity because different systems define different sets of intents, and thus few-shot learning is attracting much attention.
However, previous work has mainly focused on the few-shot intent classification without OOS~\citep{marrying-reg,efficient_intent}.
OOS detection can be considered as out-of-distribution detection~\citep{softmax_conf,learning_ood}.
Recent work has shown that large-scale pre-trained models like BERT~\citep{bert} and RoBERTa~\citep{roberta} still struggle with out-of-distribution detection, despite their strong in-domain performance~\citep{acl-ood}.
Figure~\ref{fig:tsne-main} (a) shows how unseen input text is mapped into a feature space, by a RoBERTa-based model for 15-way 5-shot intent classification.
The separation between OOS and some in-domain intents is not clear, which presumably hinders the model's OOS detection ability.
This observation calls for investigation into more sample-efficient approaches to handling the in-domain and OOS examples accurately.
In this paper, we tackle the task from a different angle, and propose a discriminative nearest neighbor classification (DNNC) model.
Instead of expecting the text encoders to be generalized enough to discriminate both the in-domain and OOS examples, we make full use of the limited training examples both in training and inference time as a nearest neighbor classification schema.
We leverage the BERT-style paired text encoding with deep self-attention to directly model relations between pairs of user utterances.
We then train a matching model as a pairwise binary classifier to estimate whether an input utterance belongs to the same class of a paired example.
We expect this to free the model from having the OOS separation issue in Figure~\ref{fig:tsne-main} (a) by avoiding explicit modeling of the intent classes.
Unlike an embedding-based matching function as in relation networks~\citep{relation_net} (Figure~\ref{fig:tsne-main} (b)), the deep pairwise matching function produces clear separation between the in-domain and OOS examples (Figure~\ref{fig:tsne-main} (c)).
We further propose to seamlessly transfer a natural language inference (NLI) model to enhance this clear separation (Figure~\ref{fig:tsne-main} (d)).
We verify our hypothesis by conducting extensive experiments on a large-scale multi-domain intent detection task with OOS~\citep{oos-intent} in various few-shot learning settings.
Our experimental results show that, compared with RoBERTa classifiers and embedding nearest neighbor approaches, our DNNC attains more stable and accurate performance both in in-domain and OOS accuracy.
Moreover, our 10-shot model can perform competitively with a 50-shot or even full-shot classifier, with the performance boost by the NLI transfer.
We also show how to speedup our DNNC's inference time without sacrificing accuracy.
\section*{Acknowledgments}
This work is supported in part by NSF under grants III-1763325, III-1909323, and SaTC-1930941.
We thank Huan Wang, Wenpeng Yin for their insightful discussions, and the anonymous reviewers for their helpful and thoughtful comments.
We also thank Jin Qu, Tian Xie, Xinyi Yang, and Yingbo Zhou for their support in the deployment of DNNC into the internal system.
\bibliographystyle{acl_natbib}
\section{Background}
\subsection{Task: Few-Shot Intent Detection}
\label{subsec:task-def}
Given a user utterance $u$ at every turn in a goal-oriented dialog system, an intent detection model $I(u)$ aims at predicting the speaker's intent:
\begin{equation}
I(u) = c,
\end{equation}
where $c$ is one of pre-defined $N$ intent classes $\mathbf{C}=\{C_1, C_2, \ldots, C_N\}$, or is categorized as OOS.
The OOS category corresponds to user utterances whose requests are not covered by the system.
In other words, any utterance can be OOS as long as it does {\it not} fall into any of the $N$ intent classes, so the definition of OOS is different depending on $\mathbf{C}$.
\paragraph{Balanced $K$-shot learning}
In a few-shot learning scenario, we have a limited number of training examples for each class, and we assume that we have $K$ examples for each of the $N$ classes in our training data.
In other words, we have $N\cdot K$ training examples in total.
We denote the $i$-th training example from the $j$-th class $C_j$ as $e_{j,i}\in E$, where $E$ is the set of the examples.
$K$ is typically 5 or 10.
\subsection{Multi-Class Classification}
\label{subsec:classifier}
The goal is to achieve high accuracy both for the intent classification and OOS detection.
One common approach to this task is using a multi-class classification model.
Specifically, to get a strong baseline for the few-shot learning use case, one can leverage a pre-trained model as transfer learning, which has been shown to achieve state-of-the-art results on numerous natural language processing tasks.
We use BERT~\citep{bert,roberta} as a text encoder:
\begin{equation}
\label{eq:bert}
h = \mathrm{BERT}(u) \in\mathbb{R}^{d},
\end{equation}
where $h$ is a $d$-dimensional output vector corresponding to the special token {\tt[CLS]} as in the follwing input format: [{\tt[CLS]}, $u$, {\tt[SEP]}].\footnote{The format of these special tokens is different in RoBERTa, but we use the original BERT's notations.}
To handle the intent classification and the OOS detection, we apply the threshold-based strategy in \citet{oos-intent}, to the softmax output of the $N$-class classification model~\citep{softmax_conf}:
\begin{equation}
\label{eq:softmax}
p(c | u) = \mathrm{softmax}(W h + b) \in\mathbb{R}^{N},
\end{equation}
where $W\in\mathbb{R}^{N\times d}$ and $b\in\mathbb{R}^{N}$ are the classifier's model parameters.
The classification model is trained by cross-entropy loss with the ground-truth intent labels of the training examples.
At inference time, we first take the class $C_j$ with the largest value of $p(c=C_j|u)$, then output $I(u)=C_j$ if $p(c=C_j|u) > T$, where $T\in[0.0, 1.0]$ is a threshold to be tuned, and otherwise we output $I(u)=\mathrm{OOS}$.
\begin{table*}[t]
\begin{center}
{\small
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|l|l|l}
& Input utterance & Utterance to be compared & Label \\ \hline
(a) & i need you to send 500 dollars from & help me move my money please & pos. \\
& my high tier account to my regular checking & & \\ \hdashline
(b) & i would like to know when the bill is due & i need to know the amounts due for my utilities and cable bills & neg. \\ \hline
(c) & And can you tell me any of the names and & Are you able to inform me of any name or address? & pos. \\
& addresses? Annie considered. & & \\ \hdashline
(d) & It's Sunday, what channel is this? & It's Sunday, can you change the channel? & neg. \\ \hdashline
(e) & I want to go back to Marguerite. & I never want to return to Marguerite. & neg. \\ \hline
\end{tabular}}
\caption{Training examples for our model. The first two examples ((a)--(b)) come from the CLINC150 dataset~\citep{oos-intent}, and the other three examples ((c)--(e)) come from the MNLI dataset~\citep{mnli}.}
\label{tb:examples}
\vspace{-1em}
}
\end{center}
\end{table*}
\subsection{Nearest Neighbor Classification}
As the fundamental building block of our proposed method, we also review nearest neighbor classification (i.e., $k$-nearest neighbors ($k$NN) classification with $k=1$), a simple and well-established concept for classification~\citep{knn-nips,knn-classifier}.
The basic idea is to classify an input into the same class of the most relevant training example based on a certain metric.
In our task, we formulate a nearest neighbor classification model as the following:
\begin{equation}
\label{eq:example-based}
I(u) = \mathrm{class}\left(\argmax{e_{j, i}\in E} S(u, e_{j, i}) \right),
\end{equation}
where $\mathrm{class}(e_{j, i})$ is a function returning the intent label (class) of the training example $e_{j, i}$, and $S$ is a function that estimates some relevance score between $u$ and $e_{j, i}$.
To detect OOS, we can also use the uncertainty-based strategy in Section~\ref{subsec:classifier}; that is, we take the output label from Equation~(\ref{eq:example-based}) if the corresponding relevance score is greater than a threshold, and otherwise we output $I(u)=\mathrm{OOS}$.
\section{Proposed Method}
\label{subsec:method}
This section first describes how to directly model inter-utterance relations in our nearest neighbor classification scenario.
We then introduce a binary classification strategy by synthesizing pairwise examples, and propose a seamless transfer of NLI.
Finally, we describe how to speedup our method's inference process.
\subsection{Deep Pairwise Matching Function}
The objective of $S(u, e_{j, i})$ in Equation~(\ref{eq:example-based}) is to find the best matched utterance from the training set $E$, given the input utterance $u$.
The typical methodology is to embed each data example into a vector space and (1) use an off-the-shelf distance metric to perform a similarity search~\citep{knn-classifier} or (2) learn a distant metric between the embeddings~\citep{relation_net}.
However, as shown in Figure~\ref{fig:tsne-main}, the text embedding methods do not discriminate the OOS examples well enough.
To model fine-grained relations of utterance pairs to distinguish in-domain and OOS intents, we propose to formulate $S(u, e_{j, i})$ as follows:
\begin{eqnarray}
h &=& \mathrm{BERT}(u, e_{j, i}) \in\mathbb{R}^{d}, \label{eq:NLI} \\
S(u, e_{j, i}) &=& \sigma(W \cdot h + b) \in\mathbb{R}, \label{eq:sigmoid}
\end{eqnarray}
where $\mathrm{BERT}$ is the same model in Equation~(\ref{eq:bert}), except that we follow a different input format to accommodate pairs of utterances: [{\tt[CLS]}, $u$, {\tt[SEP]}, $e_{j, i}$, {\tt[SEP]}].
$\sigma$ is the sigmoid function, and $W\in\mathbb{R}^{1\times d}$ and $b\in\mathbb{R}$ are the model parameters.
We can interpret our method as wrapping both the embedding and matching functions into the paired encoding with the deep self-attention in BERT (Equation~(\ref{eq:NLI})) along with the discriminative model (Equation~(\ref{eq:sigmoid})).
It has been shown that the paired text encoding is crucial in capturing complex relations between queries and documents in document retrieval~\citep{watanabe-query,sem-bert,graph-retriever}.
\subsection{Discriminative Training}
\label{subsec:pos_neg}
We train the matching model $S(u, e_{j, i})$ as a binary classifier, such that $S(u, e_{j, i})$ is closed to 1.0 if $u$ belongs to the same class of $e_{j, i}$, and otherwise closed to 0.0.
The model is trained by a binary cross-entropy loss function.
\paragraph{Positive examples}
To create positive examples, we consider all the possible ordered pairs within the same intent class: $(e_{j, i}, e_{j, \ell})$ such that $i \neq \ell$.
We thus have $N\times K\times(K-1)$ positive examples in total.
Table~\ref{tb:examples} (a) shows a positive example created from an intent, ``transfer,'' in a banking domain.
\paragraph{Negative examples}
For negative examples, we consider all the possible ordered pairs across any two different intent classes: $(e_{j, i}, e_{o, \ell})$ such that $j \neq o$.
We thus have $K^2\times N\times(N-1)$ negative examples in total, and this number is in general greater than that of the positive examples.
Table~\ref{tb:examples} (b) shows a negative example, where the input utterance comes from the intent, ``bill due'', and the paired sentence from another intent, ``bill balance''.
\subsection{Seamless Transfer from NLI}
\label{subsec:nli_transfer}
A key characteristic of our method is that we seek to model the relations between the utterance pairs, instead of explicitly modeling the intent classes.
To mitigate the data scarcity setting in few-shot learning, we consider transferring another inter-sentence-relation task.
This work focuses on NLI; the task is to identify whether a hypothesis sentence can be entailed by a premise sentence~\citep{deep-nli}.
We treat the NLI task as a binary classification task: {\tt entailment} (positive) or {\tt non-entailment} (negative).\footnote{A widely-used format is a three-way classification task with {\tt entailment}, {\tt neutral}, and {\tt contradiction}, but we merge the latter two classes into a single {\tt non-entailment} class.}
We first pre-train our model with the NLI task, where the premise sentence corresponds to the $u$-position, and the hypothesis sentence corresponds to the $e_{j, i}$-position in Equation~(\ref{eq:NLI}).
Note that it is not necessary to modify the model architecture since the task format is consistent, and we can train the NLI model solely based on existing NLI datasets.
Once the NLI model pre-training is completed, we fine-tune the NLI model with the intent classification training examples described in Section~\ref{subsec:pos_neg}.
This allows us to transfer the NLI model to any intent detection datasets seamlessly.
\paragraph{Why NLI?}
The NLI task has been actively studied, especially since the emergence of large scale datasets~\citep{snli,mnli}, and we can directly leverage the progress.
Moreover, recent work is investigating cross-lingual NLI~\citep{eriguchi-crosslingual,xnli}, and this is encouraging to consider multilinguality in future work.
On the other hand, while we can find examples relevant to the intent detection task, as shown in Table~\ref{tb:examples} ((c), (d), and (e)), we still need the few-shot fine-tuning.
This is because a domain mismatch still exists in general, and perhaps more importantly, our intent detection approach is not exactly modeling NLI.
\paragraph{Why not other tasks?}
There are other tasks modeling relationships between sentences.
Paraphrase~\citep{paranmt} and semantic relatedness~\citep{semeval} tasks are such examples.
It is possible to automatically create large-scale paraphrase datasets by machine translation~\citep{ppdb}.
However, our task is not a paraphrasing task, and creating negative examples is crucial and non-trivial~\citep{selectional-preference}.
In contrast, as described above, the NLI setting comes with negative examples by nature.
The semantic relatedness (or textual similarity) task is considered as a coarse-grained task compared to NLI, as discussed in the previous work~\citep{jmt}, in that the task measures semantic or topical relatedness.
This is not ideal for the intent detection task, because we need to discriminate between topically similar utterances of different intents.
In summary, the NLI task well matches our objective, with access to the large datasets.
\subsection{A Joint Approach with Fast Retrieval}
\label{subsec:joint}
The number of model parameters of the multi-class classification model in Section~\ref{subsec:classifier} and our model in Section~\ref{subsec:method} is almost the same when we use the same pre-trained models.
However, our example-based method has an inference-time bottleneck in Equation~(\ref{eq:NLI}), where we need to compute the BERT encoding for all $N\times K$ $(u, e_{j,i})$ pairs.
We follow common practice in document retrieval to reduce the inference-time bottleneck~\citep{sem-bert,graph-retriever}, by introducing a fast text retrieval model to select a set of top-$k$ examples $E_k$ from the training set $E$, based on its retrieval scores.
We then replace $E$ in Equation~(\ref{eq:example-based}) with the shrunk set $E_k$.
The cost of the paired BERT encoding is now constant, regardless the size of $E$.
Either TF-IDF~\citep{drqa} or embedding-based retrieval~\citep{faiss,phrase-embedding-index,latent-retrieval} can be used for the first step.
We use the following fast $k$NN.
\paragraph{Faster $k$NN}
As a baseline and a way to instantiate our joint approach, we use Sentence-BERT (SBERT)~\citep{sbert} to separately encode $u$ and $e_{j, i}$ ($x\in\{u, e_{j, i}\}$) as follows:
\begin{eqnarray}
\label{eq:sbert}
v(x) &=& \mathrm{SBERT}(x) \in\mathbb{R}^{d},
\end{eqnarray}
where the input text format is identical to that of BERT in Equation~(\ref{eq:bert}).
SBERT is a BERT-based text embedding model, fine-tuned by siamese networks with NLI datasets.
Thus both our method and SBERT transfer the NLI task in different ways.
Cosine similarity between $v(u)$ and $v(e_{j, i})$ then replaces $S(u, e_{j, i})$ in Equation~(\ref{eq:sigmoid}).
To get a fair comparison, instead of using the encoding vectors produced by the original SBERT, we fine-tune SBERT with our intent training examples described in Section~\ref{subsec:pos_neg}.
The cosine similarity is symmetric, so we have half the training examples.
We use the pairwise cosine-based loss function in \citet{sbert}.
After the model training, we pre-compute $v(e_{j, i})$ for fast retrieval.
\section{Discussions and Related Work}
\paragraph{Interpretability}
Interpretability is an important line of research recently~\citep{interpret-qa-1,interpret-qa-2,graph-retriever}.
The nearest neighbor approach~\citep{knn-nips} is appealing in that we can explicitly know which training example triggers each prediction.
Table~\ref{table:case-study} in Appendix~\ref{extra-results} shows some examples.
\paragraph{Call for better embeddings}
Emb-$k$NN and RN-$k$NN are not as competitive as DNNC.
This encourages future work on the task-oriented evaluation of text embeddings in $k$NN.
\paragraph{Training time}
Our DNNC method needs longer training time than that of the classifier (e.g., 90 vs. 40 seconds to train a single-domain model), because we synthesize the pairwise examples.
As a first step, we used all the training examples to investigate the effectiveness, but it is an interesting direction to seek more efficient pairwise training.
\paragraph{Distilled model}
Another way to speedup our model is to use distilled pre-trained models~\citep{distil-bert}.
We replaced the RoBERTa model with a distilled RoBERTa model,
and observed large variances with significantly lower OOS accuracy.
\citet{acl-ood} also suggested that the distilled models would not be robust to out-of-distribution examples.
\paragraph{Few-shot text classification}
Few-shot classification~\citep{fei2006one,vinyals2016matching} has been applied to text classification tasks~\citep{deng2019low,geng2019induction,xu2019open}, and few-shot intent detection is also studied but without OOS~\citep{marrying-reg,xia2020cg,efficient_intent}.
There are two common scenarios: 1) learning with plenty of examples and then generalizing to unseen classes with a few examples, and 2) learning with a few examples for all seen classes.
Meta-learning~\citep{finn2017model,geng2019induction} is widely studied in the first scenario.
In our paper, we have focused on the second scenario, assuming that there are only a limited number of training examples for each class.
Our work is related to metric-based approaches such as matching networks~\cite{matching-net}, prototypical networks~\cite{proto-net} and relation networks~\cite{relation_net}, as they model nearest neighbours in an example-embedding or a class-embedding space.
We showed that a relation network with the RoBERTa embeddings does not perform comparably to our method.
We also considered several ideas from prototypical networks~\citep{hierarchical-proto-net}, but those did not outperform our Emb-$k$NN baseline.
These results indicate that deep self-attention is the key to the nearest neighbor approach with OOS detection.
\section{Experimental Settings}
\subsection{Dataset: Multi-Domain Intent Detection}
\begin{table}[t]
\begin{center}
{\small
\begin{tabular}{l|r||r|r|r}
& $N$ & Train & Dev. & Test \\ \hline
All domains & 150 & 15,000 & 3,000 & 4,500 \\
Single domain & 15 & 1,500 & 300 & 450 \\
OOS & - & n/a & 100 & 1,000 \\ \hline
\end{tabular}
}
\caption{Dataset statistics. The number of the examples is equally distributed across the intent classes.}
\label{tb:dataset}
\vspace{-1em}
\end{center}
\end{table}
We use a recently-released dataset, CLINC150,\footnote{\url{https://github.com/clinc/oos-eval}.} for multi-domain intent detection~\citep{oos-intent}.
The CLINC150 dataset defines 150 types of intents in total (i.e., $N=150$), where there are 10 different domains and 15 intents for each of them.
Table~\ref{tb:dataset} shows the dataset statistics.
\paragraph{OOS evaluation examples}
The dataset also provides OOS examples whose intents do not belong to any of the 150 intents.
From the viewpoint of out-of-distribution detection~\citep{softmax_conf,acl-ood}, we do not use the OOS examples during the training stage; we only use the evaluation splits as in Table~\ref{tb:dataset}.
\paragraph{Single-domain experiments}
The task in the CLINC150 dataset is like handling many different services in a single system; that is, topically different intents are mixed (e.g., ``alarm'' in the ``Utility'' domain, and ``pay bill'' in the ``Banking'' domain).
In contrast, it is also a reasonable setting to handle each domain (or service) separately as in \citet{google-schema-guided}.
In addition to the all-domain experiment, we conduct single-domain experiments, where we only focus on a specific domain with its 15 intents (i.e., $N=15$).
More specifically, we use four domains, ``Banking,'' ``Credit cards,'' ``Work,'' and ``Travel,'' among the ten domains.
Note that the same OOS evaluation sets are used.
\subsection{Evaluation Metrics}
\label{subsection:evalMetrics}
We follow \citet{oos-intent} to report in-domain accuracy, $\mathrm{Acc}_\mathrm{in}$, and OOS recall, $R_\mathrm{oos}$.
These two metrics are defined as follows:
\begin{equation}
\label{eq:two_metrics}
\mathrm{Acc}_\mathrm{in} = C_\mathrm{in}/N_\mathrm{in}, ~~~ R_\mathrm{oos} = C_\mathrm{oos}/N_\mathrm{oos}, \\
\end{equation}
where $C_\mathrm{in}$ is the number of correctly predicted in-domain intent examples, and $N_\mathrm{in}$ is the total number of the examples evaluated.
This is analogous to the calculation of the OOS recall.
\paragraph{Threshold selection}
We use the uncertainty-based OOS detection, and therefore we need a way to set the threshold $T$.
For each $T$ in $[0.0, 0.1, \ldots, 0.9, 1.0]$, we calculate a joint score $J_\mathrm{in\_oos}$ defined as follows:
\begin{equation}
\label{eq:joint_score}
J_\mathrm{in\_oos} = \mathrm{Acc}_\mathrm{in} + R_\mathrm{oos},
\end{equation}
and select a threshold value to maximize the score on the development set.
There is a trade-off to be noted; the larger the value of $T$ is, the higher $R_\mathrm{oos}$ (and the lower $\mathrm{Acc}_\mathrm{in}$) we expect, because the models predict OOS more frequently.
\paragraph{Notes on $J_\mathrm{in\_oos}$}
Our joint score $J_\mathrm{in\_oos}$ in Equation~(\ref{eq:joint_score}) gives the same weight to the two metrics, $\mathrm{Acc}_\mathrm{in}$ and $R_\mathrm{oos}$, compared to other combined metrics.
For example, \citet{oos-intent} and \citet{tod-bert} used a joint accuracy score:
\begin{equation}
\frac{C_\mathrm{in}+C_\mathrm{oos}}{N_\mathrm{in}+N_\mathrm{oos}} = \frac{\mathrm{Acc}_\mathrm{in} + r R_\mathrm{oos}}{1 + r},
\end{equation}
where $r=N_\mathrm{oos}/N_\mathrm{in}$ depends on the balance between $N_\mathrm{in}$ and $N_\mathrm{oos}$, and thus this combined metric can put much more weight on the in-domain accuracy when $N_\mathrm{in} \gg N_\mathrm{oos}$.
Table~\ref{tb:dataset} shows $r=100/3000~(=0.0333)$ in the development set of the ``all domains'' setting, which underestimates the importance of $R_\mathrm{oos}$.
Actually, the OOS recall scores in \citet{oos-intent} and \citet{tod-bert} are much lower than those with our RoBERTa classifier, and the trade-off with respect to the tuning process was not discussed.\footnote{When we refer to our RoBERTa classifier's scores $(A_\mathrm{in}, R_\mathrm{oos})=(92.9\pm0.2,~90.2\pm0.5)$ in Table~\ref{table:wholeDataset}, their corresponding scores are $(96.2,~52.3)$ and $(96.1,~46.3)$ in \citet{oos-intent} and \citet{tod-bert}, respectively.}
We also report OOS precision and OOS F1 for more comprehensive evaluation:
\begin{equation}
P_\mathrm{oos} = C_\mathrm{oos}/N'_\mathrm{oos}, ~~~ F_1 = H(P_\mathrm{oos}, R_\mathrm{oos}),
\end{equation}
where $N'_\mathrm{oos}$ is the number of examples predicted as OOS, and $H(\cdot, \cdot)$ is the harmonic mean.
\if0
\begin{equation}
\hspace{-0.5mm}
P_\mathrm{oos} = C_\mathrm{oos}/N'_\mathrm{oos}, ~ F1_\mathrm{oos} = H(P_\mathrm{oos}, R_\mathrm{oos}),
\end{equation}
\fi
\subsection{Model Training and Configurations}
\label{subsec:model_setting}
We use RoBERTa (the {\tt base} configuration with $d=768$) as a BERT encoder for all the BERT/SBERT-based models in our experiments,\footnote{We use \url{https://github.com/huggingface/transformers} and \url{https://github.com/UKPLab/sentence-transformers}.}
because RoBERTa performed significantly better and more stably than the original BERT in our few-shot experiments.
We combine three NLI datasets, SNLI~\citep{snli}, MNLI~\citep{mnli}, and WNLI~\citep{wnli} from the GLUE benchmark~\citep{glue} to pre-train our proposed model.
We apply label smoothing~\citep{label_smoothing_orig} to all the cross-entropy loss functions, which has been shown to improve the reliability of the model confidence~\citep{label_smoothing}.
Experiments were conducted on single NVIDIA Tesla V100 GPU with 16GB memory.
\paragraph{Sampling training examples}
We conduct our experiments with $K=5, 10$ following the task definition in Section~\ref{subsec:task-def}.
We randomly sample $K$ examples from the entire training sets in Table~\ref{tb:dataset}, for each in-domain intent class 10 times unless otherwise stated.
We train a model with a consistent hyper-parameter setting across the 10 different runs and follow the threshold selection process based on a mean score for each threshold.
We also report a standard deviation for each result.
\begin{table*}[h]
\centering
\tiny
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{lcccccccc}
\hline
\multicolumn{1}{l|}{} & \multicolumn{2}{c|}{\textbf{In-domain accuracy}} & \multicolumn{2}{c|}{\textbf{OOS recall}} & \multicolumn{2}{c|}{\textbf{OOS precision}} & \multicolumn{2}{c}{\textbf{OOS F1}} \\
\multicolumn{1}{l|}{\textbf{5-shot}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \textbf{Credit cards} \\ \hline
\multicolumn{1}{l|}{classifier} & 79.7 $\pm$ 1.8 & \multicolumn{1}{c|}{79.9 $\pm$ 3.7} & 93.3 $\pm$ 2.9 & \multicolumn{1}{c|}{93.3 $\pm$ 3.0} & 92.5 $\pm$ 0.9 & \multicolumn{1}{c|}{93.6 $\pm$ 1.3} & 92.9 $\pm$ 1.8 & 93.4 $\pm$ 1.6 \\
\multicolumn{1}{l|}{classifier-EDA} & 79.9 $\pm$ 3.0 & \multicolumn{1}{c|}{73.1 $\pm$ 5.0} & 91.0 $\pm$ 2.6 & \multicolumn{1}{c|}{94.1 $\pm$ 2.6} & 92.6 $\pm$ 1.5 & \multicolumn{1}{c|}{90.0 $\pm$ 1.7} & 91.8 $\pm$ 1.9 & 92.0 $\pm$ 1.3 \\
\multicolumn{1}{l|}{classifier-BT} & 77.2 $\pm$ 2.8 & \multicolumn{1}{c|}{70.3 $\pm$ 4.4} & 91.5 $\pm$ 2.4 & \multicolumn{1}{c|}{91.1 $\pm$ 1.8} & 91.4 $\pm$ 1.2 & \multicolumn{1}{c|}{89.5 $\pm$ 1.8} & 91.4 $\pm$ 1.4 & 90.3 $\pm$ 1.7 \\
\multicolumn{1}{l|}{DNNC-scratch} & 83.4 $\pm$ 1.9 & \multicolumn{1}{c|}{84.6 $\pm$ 3.2} & 93.2 $\pm$ 2.7 & \multicolumn{1}{c|}{96.4 $\pm$ 1.0} & 96.2 $\pm$ 0.9 & \multicolumn{1}{c|}{95.8 $\pm$ 1.1} & 94.7 $\pm$ 1.2 & 96.1 $\pm$ 0.9 \\
\multicolumn{1}{l|}{DNNC} & \textbf{88.6 $\pm$ 1.3} & \multicolumn{1}{c|}{\textbf{90.5 $\pm$ 2.9}} & \textbf{94.7 $\pm$ 1.0} & \multicolumn{1}{c|}{\textbf{97.7 $\pm$ 1.1}} & \textbf{97.0 $\pm$ 0.3} & \multicolumn{1}{c|}{\textbf{97.8 $\pm$ 0.5}} & \textbf{95.9 $\pm$ 0.5} & \textbf{97.8 $\pm$ 0.6} \\ \hdashline
\multicolumn{1}{l|}{classifier (50-shot)} & 90.0 $\pm$ 1.4 & \multicolumn{1}{c|}{90.3 $\pm$ 1.4} & 93.3 $\pm$ 1.1 & \multicolumn{1}{c|}{93.9 $\pm$ 1.2} & 96.3 $\pm$ 0.6 & \multicolumn{1}{c|}{96.4 $\pm$ 0.6} & 94.8 $\pm$ 0.7 & 95.1 $\pm$ 0.6 \\ \hline
\multicolumn{1}{l|}{\textbf{5-shot}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \textbf{Travel} \\ \hline
\multicolumn{1}{l|}{classifier} & 84.4 $\pm$ 1.9 & \multicolumn{1}{c|}{87.8 $\pm$ 2.5} & 94.8 $\pm$ 1.6 & \multicolumn{1}{c|}{96.1 $\pm$ 0.9} & 95.4 $\pm$ 0.7 & \multicolumn{1}{c|}{94.7 $\pm$ 1.0} & 95.1 $\pm$ 0.9 & 95.4 $\pm$ 0.7 \\
\multicolumn{1}{l|}{classifier-EDA} & 80.5 $\pm$ 3.3 & \multicolumn{1}{c|}{88.2 $\pm$ 2.4} & 95.2 $\pm$ 1.6 & \multicolumn{1}{c|}{94.6 $\pm$ 2.1} & 92.8 $\pm$ 1.3 & \multicolumn{1}{c|}{94.9 $\pm$ 1.0} & 94.0 $\pm$ 0.9 & 94.7 $\pm$ 0.9 \\
\multicolumn{1}{l|}{classifier-BT} & 80.3 $\pm$ 3.4 & \multicolumn{1}{c|}{90.8 $\pm$ 2.4} & 94.4 $\pm$ 1.1 & \multicolumn{1}{c|}{92.3 $\pm$ 1.9} & 93.1 $\pm$ 1.0 & \multicolumn{1}{c|}{95.9 $\pm$ 1.1} & 93.8 $\pm$ 0.8 & 94.1 $\pm$ 1.3 \\
\multicolumn{1}{l|}{DNNC-scratch} & 83.2 $\pm$ 2.1 & \multicolumn{1}{c|}{87.8 $\pm$ 3.5} & 96.3 $\pm$ 1.7 & \multicolumn{1}{c|}{94.9 $\pm$ 2.8} & 96.7 $\pm$ 0.9 & \multicolumn{1}{c|}{94.9 $\pm$ 1.4} & 96.5 $\pm$ 0.6 & 94.9 $\pm$ 0.9 \\
\multicolumn{1}{l|}{DNNC} & \textbf{89.9 $\pm$ 3.2} & \multicolumn{1}{c|}{\textbf{91.8 $\pm$ 1.6}} & \textbf{96.7 $\pm$ 1.1} & \multicolumn{1}{c|}{\textbf{96.4 $\pm$ 1.2}} & \textbf{97.9 $\pm$ 1.1} & \multicolumn{1}{c|}{\textbf{96.5 $\pm$ 0.7}} & \textbf{97.3 $\pm$ 0.5} & \textbf{96.5 $\pm$ 0.7} \\ \hdashline
\multicolumn{1}{l|}{classifier (50-shot)} & 94.3 $\pm$ 0.8 & \multicolumn{1}{c|}{97.0 $\pm$ 0.3} & 95.2 $\pm$ 0.8 & \multicolumn{1}{c|}{92.6 $\pm$ 1.4} & 97.7 $\pm$ 0.3 & \multicolumn{1}{c|}{98.6 $\pm$ 0.2} & 96.5 $\pm$ 0.5 & 95.5 $\pm$ 0.8 \\ \hline
& \multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \multicolumn{1}{l}{} & \textbf{} & \textbf{} \\ \hline
\multicolumn{1}{l|}{} & \multicolumn{2}{c|}{\textbf{In-domain accuracy}} & \multicolumn{2}{c|}{\textbf{OOS recall}} & \multicolumn{2}{c|}{\textbf{OOS precision}} & \multicolumn{2}{c}{\textbf{OOS F1}} \\
\multicolumn{1}{l|}{\textbf{10-shot}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \multicolumn{1}{c|}{\textbf{Credit cards}} & \textbf{Banking} & \textbf{Credit cards} \\ \hline
\multicolumn{1}{l|}{classifier} & 85.2 $\pm$ 1.3 & \multicolumn{1}{c|}{83.7 $\pm$ 2.1} & 93.3 $\pm$ 1.0 & \multicolumn{1}{c|}{93.8 $\pm$ 1.4} & 94.4 $\pm$ 0.5 & \multicolumn{1}{c|}{93.9 $\pm$ 0.9} & 93.8 $\pm$ 0.6 & 93.8 $\pm$ 0.7 \\
\multicolumn{1}{l|}{classifier-EDA} & 82.5 $\pm$ 1.3 & \multicolumn{1}{c|}{79.3 $\pm$ 1.7} & \textbf{95.9 $\pm$ 0.8} & \multicolumn{1}{c|}{96.8 $\pm$ 0.8} & 93.3 $\pm$ 0.4 & \multicolumn{1}{c|}{92.3 $\pm$ 0.8} & 94.6 $\pm$ 0.5 & 94.5 $\pm$ 0.3 \\
\multicolumn{1}{l|}{classifier-BT} & 82.2 $\pm$ 1.9 & \multicolumn{1}{c|}{82.9 $\pm$ 1.8} & 94.9 $\pm$ 1.9 & \multicolumn{1}{c|}{89.3 $\pm$ 2.0} & 93.1 $\pm$ 0.8 & \multicolumn{1}{c|}{94.3 $\pm$ 0.6} & 94.0 $\pm$ 0.9 & 91.7 $\pm$ 1.2 \\
\multicolumn{1}{l|}{DNNC-scratch} & 89.6 $\pm$ 1.6 & \multicolumn{1}{c|}{92.1 $\pm$ 1.1} & 92.1 $\pm$ 3.1 & \multicolumn{1}{c|}{94.8 $\pm$ 1.2} & 97.5 $\pm$ 0.9 & \multicolumn{1}{c|}{\textbf{98.1 $\pm$ 0.4}} & 94.7 $\pm$ 1.5 & 96.4 $\pm$ 0.6 \\
\multicolumn{1}{l|}{DNNC} & \textbf{91.2 $\pm$ 1.1} & \multicolumn{1}{c|}{\textbf{92.1 $\pm$ 1.0}} & 94.8 $\pm$ 1.1 & \multicolumn{1}{c|}{\textbf{97.8 $\pm$ 0.8}} & \textbf{97.5 $\pm$ 0.4} & \multicolumn{1}{c|}{97.8 $\pm$ 0.3} & \textbf{96.1 $\pm$ 0.6} & \textbf{97.8 $\pm$ 0.4} \\ \hdashline
\multicolumn{1}{l|}{classifier (50-shot)} & 90.0 $\pm$ 1.4 & \multicolumn{1}{c|}{90.3 $\pm$ 1.4} & 93.3 $\pm$ 1.1 & \multicolumn{1}{c|}{93.9 $\pm$ 1.2} & 96.3 $\pm$ 0.6 & \multicolumn{1}{c|}{96.4 $\pm$ 0.6} & 94.8 $\pm$ 0.7 & 95.1 $\pm$ 0.6 \\ \hline
\multicolumn{1}{l|}{\textbf{10-shot}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \multicolumn{1}{c|}{\textbf{Travel}} & \textbf{Work} & \textbf{Travel} \\ \hline
\multicolumn{1}{l|}{classifier} & 86.0 $\pm$ 2.2 & \multicolumn{1}{c|}{93.0 $\pm$ 1.2} & 97.2 $\pm$ 0.7 & \multicolumn{1}{c|}{94.8 $\pm$ 0.9} & 94.5 $\pm$ 0.8 & \multicolumn{1}{c|}{96.8 $\pm$ 0.6} & 95.8 $\pm$ 0.3 & 95.8 $\pm$ 0.4 \\
\multicolumn{1}{l|}{classifier-EDA} & 86.4 $\pm$ 1.7 & \multicolumn{1}{c|}{93.0 $\pm$ 1.0} & 97.0 $\pm$ 0.9 & \multicolumn{1}{c|}{95.2 $\pm$ 1.3} & 94.7 $\pm$ 0.7 & \multicolumn{1}{c|}{96.9 $\pm$ 0.5} & 95.8 $\pm$ 0.5 & 96.0 $\pm$ 0.6 \\
\multicolumn{1}{l|}{classifier-BT} & 83.7 $\pm$ 1.7 & \multicolumn{1}{c|}{91.9 $\pm$ 1.1} & \textbf{97.4 $\pm$ 0.9} & \multicolumn{1}{c|}{\textbf{96.1 $\pm$ 0.8}} & 93.6 $\pm$ 0.7 & \multicolumn{1}{c|}{96.4 $\pm$ 0.5} & 95.5 $\pm$ 0.6 & \textbf{96.2 $\pm$ 0.3} \\
\multicolumn{1}{l|}{DNNC-scratch} & 92.6 $\pm$ 1.7 & \multicolumn{1}{c|}{\textbf{96.4 $\pm$ 0.6}} & 94.1 $\pm$ 1.4 & \multicolumn{1}{c|}{84.8 $\pm$ 3.0} & 98.4 $\pm$ 0.6 & \multicolumn{1}{c|}{\textbf{98.5 $\pm$ 0.3}} & 96.2 $\pm$ 0.8 & 91.1 $\pm$ 1.8 \\
\multicolumn{1}{l|}{DNNC} & \textbf{95.0 $\pm$ 1.0} & \multicolumn{1}{c|}{96.0 $\pm$ 0.8} & 95.5 $\pm$ 0.9 & \multicolumn{1}{c|}{93.3 $\pm$ 1.9} & \textbf{99.0 $\pm$ 0.4} & \multicolumn{1}{c|}{98.3 $\pm$ 0.4} & \textbf{97.2 $\pm$ 0.5} & 95.7 $\pm$ 1.0 \\ \hdashline
\multicolumn{1}{l|}{classifier (50-shot)} & 94.3 $\pm$ 0.8 & \multicolumn{1}{c|}{97.0 $\pm$ 0.3} & 95.2 $\pm$ 0.8 & \multicolumn{1}{c|}{92.6 $\pm$ 1.4} & 97.7 $\pm$ 0.3 & \multicolumn{1}{c|}{98.6 $\pm$ 0.2} & 96.5 $\pm$ 0.5 & 95.5 $\pm$ 0.8 \\ \hline
\end{tabular}}\caption{Testing results on the four different domains.}
\label{table:fourDomain}
\vspace{1em}
\tiny
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|cl|cl|cl|cc}
\hline
& \multicolumn{2}{c|}{\textbf{In-domain accuracy}} & \multicolumn{2}{c|}{\textbf{OOS recall}} & \multicolumn{2}{c|}{\textbf{OOS precision}} & \multicolumn{2}{c}{\textbf{OOS F1}} \\
& \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \textbf{10-shot} \\ \cline{2-9}
classifier & 77.7 $\pm$ 0.6 & 83.2 $\pm$ 0.6 & \textbf{91.9 $\pm$ 1.1} & \textbf{92.9 $\pm$ 0.8} & 53.9 $\pm$ 0.8 & 58.2 $\pm$ 1.0 & 68.0 $\pm$ 1.0 & 71.6 $\pm$ 0.8 \\
USE+ConveRT & 76.9 $\pm$ 0.7 & 82.6 $\pm$ 0.5 & 90.7 $\pm$ 0.8 & 89.1 $\pm$ 0.4 & 49.0 $\pm$ 0.8 & 55.1 $\pm$ 0.7 & 63.6 $\pm$ 0.7 & 68.1 $\pm$ 0.6 \\
DNNC & \textbf{84.9 $\pm$ 0.8} & \textbf{91.6 $\pm$ 0.3} & 90.1 $\pm$ 1.6 & 83.0 $\pm$ 2.0 & \textbf{64.7 $\pm$ 2.5} & \textbf{81.4 $\pm$ 2.0} & \textbf{75.3 $\pm$ 2.0} & \textbf{82.1 $\pm$ 0.3} \\ \hdashline
classifier (full-shot) & \multicolumn{2}{c|}{92.9 $\pm$ 0.2} & \multicolumn{2}{c|}{90.2 $\pm$ 0.5} & \multicolumn{2}{c|}{76.7 $\pm$ 0.7} & \multicolumn{2}{c}{82.9 $\pm$ 0.6} \\
USE+ConveRT (full-shot) & \multicolumn{2}{c|}{95.0 $\pm$ 0.1} & \multicolumn{2}{c|}{64.6 $\pm$ 0.6} & \multicolumn{2}{c|}{79.3 $\pm$ 0.7} & \multicolumn{2}{c}{71.2 $\pm$ 0.5} \\ \hline
\end{tabular}}\caption{Testing results on the whole dataset (5 runs).}
\label{table:wholeDataset}
\vspace{-0.5em}
\end{table*}
\paragraph{Using the development set}
We would not always have access to a large enough development set in the few-shot learning scenario.
However, we still use the development set provided by the dataset to investigate the models' behaviors when changing hyper-parameters like the threshold.
\paragraph{Models compared}
We list the models used in our experiments:
\begin{itemize}
\item{\bf Classifier baselines:}
``Classifier'' is the RoBERTa-based classification model described in Section~\ref{subsec:classifier}.
We further seek solid baselines by data augmentation.
``Classifier-EDA'' is the classifier trained with data augmentation techniques in \citet{eda}.
``Classifier-BT'' is the classifier trained with back-translation data augmentation~\citep{qanet,backtrans-da} by using a transformer-based English$\leftrightarrow$German translation system~\citep{transformer}.
\item{\bf Non-BERT classifier:}
We also test a state-of-the-art fast embedding-based classifier, ``USE+ConveRT''~\citep{convert,efficient_intent}, in the ``all domains'' setting.
\citet{efficient_intent} showed that the ``USE+ConveRT'' outperformed a BERT classifier on the CLINC150 dataset, while it was not evaluated along with the OOS detection task.
We modified their original code\footnote{\url{https://github.com/connorbrinton/polyai-models/releases/tag/v1.0}.} to apply the uncertainty-based OOS detection.
\item{\bf $k$NN baselines:\footnote{We tried weighted voting in \citet{knn-classifier}, but $k=1$ performed better in general.}}
``Emb-$k$NN'' is the $k$NN method with S(Ro)BERT(a) described in Section~\ref{subsec:joint}, and ``Emb-$k$NN-vanilla'' is {\it without} using our intent training examples for fine-tuning.
``TF-IDF-$k$NN'' is another $k$NN baseline using TF-IDF vectors, which tells us how well string matching performs on our task.
We also implement a relation network~\citep{relation_net}, ``RN-$k$NN,'' to learn a similarity metric between the SRoBERTa embeddings, instead of using the cosine similarity.
\item{\bf Proposed method:\footnote{Our code will be available at \url{https://github.com/salesforce/DNNC-few-shot-intent}.}}
``DNNC'' is our proposed method, and ``DNNC-scratch'' is {\it without} the NLI pre-training in Section~\ref{subsec:nli_transfer}.
``DNNC-joint'' is our joint approach on top of top-$k$ retrieval by Emb-$k$NN (Section~\ref{subsec:joint}).
\end{itemize}
\noindent
More details about the model training and the data augmentation configurations are described in Appendix~\ref{training-details} and Appendix~\ref{data-augmentation}, respectively.
\section*{Appendix}
\if0{
\section{Additional Notes on the Use of NLI}
\label{app:nli}
There are other tasks modeling relationships between sentences.
Paraphrase~\citep{paranmt} and semantic relatedness~\citep{semeval} tasks are such examples.
It is possible to automatically create large-scale paraphrase datasets by machine translation~\citep{ppdb}.
However, our task is not a paraphrasing task, and creating negative examples is crucial and non-trivial~\citep{selectional-preference}.
In contrast, as described above, the NLI setting comes with negative examples by nature.
This is not ideal for the intent detection task, because we need to discriminate between topically similar utterances of different intents.
In summary, the NLI task well matches our objective, with access to large datasets.
}\fi
\if0{
\section{A Note on the Threshold Selection} \label{threshold-selection}
\label{app:threshold}
Our joint score ($\mathrm{Acc}_\mathrm{in} + R_\mathrm{oos}$) in Section~4.2 gives the same weight to the two metrics, $\mathrm{Acc}_\mathrm{in}$ and $R_\mathrm{oos}$, compared to other combined metrics like $(C_\mathrm{in}+C_\mathrm{oos})/(N_\mathrm{in}+N_\mathrm{oos})$.
Such a combined metric can put much more weight on the in-domain accuracy when $N_\mathrm{in}$ and $N_\mathrm{oos}$ are imbalanced; Table~2 shows such imbalance on the development set.
\citet{oos-intent} sacrificed the OOS recall a lot, and the trade-off with respect to the threshold selection was not discussed.
}\fi
\begin{table*}[h]
\centering
\begin{tabular}{l|lll}
\hline
Dataset & SNLI & WNLI & MNLI \\ \hline
Size of the development set & 9999 & 70 & 9814 \\
Accuracy & 94.5\% & 41.4\% & 92.1\% \\ \hline
\end{tabular}\caption{Development results on three NLI datasets.}
\label{table-pretrain}
\end{table*}
\section{Training Details} \label{training-details}
\paragraph{Dataset preparation}
To use the CLINC150 dataset~\citep{oos-intent}\footnote{\url{https://github.com/clinc/oos-eval}.} in our ways, especially for the single-domain experiments, we provide preprocessing scrips accompanied with our code.
\paragraph{General training}\label{appendix-general-training}
This section describes the details about the model training in Section~\ref{subsec:model_setting}.
For each component related to RoBERTa and SRoBERTa, we solely follow the two libraries, transformers and sentence-transformers, for the sake of easy reproduction of our experiments.\footnote{\url{https://github.com/huggingface/transformers} and \url{https://github.com/UKPLab/sentence-transformers}.}
The example code to train the NLI-style models is also available.\footnote{\url{https://github.com/huggingface/transformers/tree/master/examples/text-classification}.}
We use the {\tt roberta-base} configuration\footnote{\url{https://s3.amazonaws.com/models.huggingface.co/bert/roberta-base-config.json}.} for all the RoBERTa/SRoBERTa-based models in our experiments.
All the model parameters including the RoBERTa parameters are updated during all the fine-tuning processes, where we use the AdamW~\citep{adamw} optimizer with a weight decay coefficient of 0.01 for all the non-bias parameters.
We use a gradient clipping technique~\citep{clip} with a clipping value of 1.0, and also use a linear warmup learning-rate scheduling with a proportion of 0.1 with respect to the maximum number of training epochs.
\begin{table*}[t]
\centering
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|ccc|ccc}
\hline
& \multicolumn{3}{c|}{\textbf{Single domain}} & \multicolumn{3}{c}{\textbf{All domains}} \\ \cline{2-7}
& \textbf{Learning rate} & \textbf{Epoch} & \textbf{Run} & \textbf{Learning rate} & \textbf{Epoch} & \textbf{Run} \\ \hline
Classifier & \{1e-4, 2e-5, 5e-5\} & \{15, 25, 35\} & 10 & \{1e-4, 5e-5\} & \{15, 25, 35\} & 5 \\
Emb-kNN & \{1e-4, 2e-5, 3e-5\} & \{7, 10, 20, 25, 35\} & 10 & \{2e-5, 5e-5\} & \{3, 5, 7\} & 5 \\
DNNC & \{1e-5, 2e-5, 3e-5, 4e-5\} & \{7, 10, 15\} & 10 & \{2e-5, 5e-5\} & \{3, 5, 7\} & 5 \\ \hline
\end{tabular}}\caption{Some hyper-parameter settings for a few models.}\label{table:hyper-paramter}
\end{table*}
\begin{table*}[t]
\centering
\begin{tabular}{l|cc}
\hline
& \textbf{5-shot} & \textbf{10-shot} \\ \hline
Classifier & \{bs: 50, ep: 25.0, lr: 5e-05\} & \{bs: 50, ep: 35.0, lr: 5e-05\} \\ \hline
Emb-kNN & \{bs: 200, ep: 7.0, lr: 2e-05\} & \{bs: 200, ep: 5.0, lr: 2e-05\} \\ \hline
DNNC & \{bs: 900, ep: 7.0, lr: 2e-05\} & \{bs: 1800, ep: 5.0, lr: 2e-05\} \\ \hline
\end{tabular}
\caption{Best hyper-parameter settings for a few models on the all-domain experiments, where {\tt bs} is batch size, {\tt ep} represents epochs, {\tt lr} is learning rate.}\label{table:hyper-paramter-best-all}
\end{table*}
\begin{table*}[t]
\centering
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|cccc}
\hline
& \textbf{5-shot} & \multicolumn{1}{c|}{\textbf{10-shot}} & \textbf{5-shot} & \textbf{10-shot} \\ \cline{2-5}
& \multicolumn{2}{c|}{\textbf{Banking}} & \multicolumn{2}{c}{\textbf{Credit cards}} \\ \hline
Classifier & \multicolumn{1}{l}{\{bs: 15, ep: 25.0, lr: 5e-05\}} & \multicolumn{1}{l|}{\{bs: 15, ep: 35.0, lr: 5e-05\}} & \{bs: 15, ep: 15.0, lr: 5e-05\} & \{bs: 15, ep: 25.0, lr: 5e-05\} \\
Emb-kNN & \multicolumn{1}{l}{\{bs: 200, ep: 35.0, lr: 1e-05\}} & \multicolumn{1}{l|}{\{bs: 200, ep: 25.0, lr: 2e-05\}} & \{bs: 100, ep: 20.0, lr: 1e-05\} & \{bs: 100, ep: 10.0, lr: 1e-05\} \\
DNNC & \multicolumn{1}{l}{\{bs: 370, ep: 15.0, lr: 1e-05\}} & \multicolumn{1}{l|}{\{bs: 370, ep: 7.0, lr: 2e-05\}} & \{bs: 370, ep: 15.0, lr: 2e-05\} & \{bs: 370, ep: 7.0, lr: 3e-05\} \\ \hline
& \multicolumn{2}{c}{\textbf{Work}} & \multicolumn{2}{c}{\textbf{Travel}} \\ \hline
Classifier & \{bs: 15, ep: 15.0, lr: 5e-05\} & \{bs: 15, ep: 15.0, lr: 5e-05\} & \{bs: 15, ep: 35.0, lr: 5e-05\} & \{bs: 15, ep: 25.0, lr: 1e-04\} \\
Emb-kNN & \{bs: 100, ep: 20.0, lr: 1e-05\} & \{bs: 100, ep: 7.0, lr: 2e-05\} & \{bs: 100, ep: 35.0, lr: 3e-05\} & \{bs: 100, ep: 20.0, lr: 1e-05\} \\
DNNC & \{bs: 370, ep: 7.0, lr: 3e-05\} & \{bs: 370, ep: 15.0, lr: 2e-05\} & \{bs: 370, ep: 7.0, lr: 2e-05\} & \{bs: 370, ep: 7.0, lr: 2e-05\} \\ \hline
\end{tabular}}\caption{Best hyper-parameter settings for a few models on the four single domains, where {\tt bs} is batch size, {\tt ep} represents epochs, {\tt lr} is learning rate.}\label{table:hyper-paramter-best}
\end{table*}
\begin{table*}[t]
\begin{center}
{\small
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{l|l|l}
Original utterance & Augmented example & Intent label \\ \hline
can you block my chase account right away please & can you turn my chase account off directly & freeze account \\
do a car payment from my savings account & with my saving account, you can pay a car payment account & pay bill \\
when is my visa due & when is my visa to be paid & bill due \\ \hline
\end{tabular}}
}
\caption{Examples used to train clasifier-BT.}
\label{tb:bt-examples}
\end{center}
\end{table*}
\paragraph{Pre-training on NLI tasks}\label{appendix-pre-training}
For the pre-training on NLI tasks, we fine-tune a {\tt roberta-base} model on three publicly available datasets, i.e., SNLI~\citep{snli}, MNLI~\citep{mnli}, and WNLI~\citep{wnli} from the GLUE benchmark~\citep{glue}.
The optimizer and gradient clipping follow the above configurations.
The number of training epochs is set to $4$; the batch size is set to $32$; the learning rate is set to $2e-5$.
We use a linear warmup learning-rate scheduling with a proportion of $0.06$ by following \citet{roberta}.
The evaluation results on the development sets are shown in Table~\ref{table-pretrain}, where the low accuracy of WNLI is mainly caused by the data size imbalance.
We note that these NLI scores are not comparable with existing NLI scores, because we converted the task to the binary classification task for our model transfer purpose.
\paragraph{Text pre-processing}
For all the RoBERTa-based models, we used the RoBERTa {\tt roberta-base}'s tokenizer provided in the transformers library.\footnote{\url{https://github.com/huggingface/transformers/blob/master/src/transformers/tokenization_roberta.py}.}
We did not perform any additional pre-processing in our experiments.
\paragraph{Hyper-parameter settings}\label{appendix-hyper-parameter}
Table~\ref{table:hyper-paramter} shows the hyper-parameters we tuned on the development sets in our RoBERTa-based experiments.
For a single-domain experiment, we take a hyper-parameter set and apply it to the ten different runs to select the threshold in Section~\ref{subsection:evalMetrics} on the development set.
We then select the best hyper-parameter set along with the corresponding threshold, which achieves the best $J_\mathrm{in\_oos}$ in Equation~(\ref{eq:joint_score}) on the development set, among all the possible hyper-parameter sets.
Finally, we apply the selected model and the threshold to the test set.
We follow the same process for the all-domain experiments, except that we run each experiment five times.
Table~\ref{table:hyper-paramter-best-all} and Table~\ref{table:hyper-paramter-best} summarize the hyper-parameter settings used for the evaluation on the test sets.
We note that each model was not very sensitive to the different hyper-parameter settings, as long as we have a large number of training iterations.
\section{Data Augmentation} \label{data-augmentation}
We describe the details about the classifier baselines with the data augmentation techniques in Section~\ref{subsec:model_setting}.
\paragraph{EDA}
Classifier-EDA uses the following four data augmentation techniques in \citet{eda}: synonym replacement, random insertion, random swap, and random deletion.
We follow the publicly available code.\footnote{\url{https://github.com/jasonwei20/eda_nlp}.}
For every training example, we empirically set one augmentation based on every technique.
We apply each technique separately to the original sentence and therefore every training example will have four augmentations.
The probability of a word in an utterance being edited is set to 0.1 for all the techniques.
\paragraph{BT}
For classifier-BT, we use the English-German corpus in \citet{escape}, which is widely used in an annual competition for automatic post-editing research on IT-domain text~\citep{ape-2019}.
The corpus contains about 7.5 million translation pairs, and we follow the {\it base} configuration to train a transformer model~\citep{transformer} for each direction.
Based on the initial trial in our preliminary experiments to generate diverse examples, we decided to use a temperature sampling technique instead of a greedy or beam-search strategy.
More specifically, logit vectors during the machine translation process are multiplied by $\tau$ to distort the output distributions, where we set $\tau = 5.0$.
For each training example in the intent detection dataset, we first translate it into German and then translate it back to English.
We repeat this process to generate up to five unique examples, and use them to train the classifier model.
Table~\ref{tb:bt-examples} shows such examples, and we will release all the augmented examples for future research.
\begin{table*}[]
\centering
\resizebox{1.0\linewidth}{!}{
\begin{tabular}{ll}
\hline
\textbf{input utterance} & transfer ten dollars from my wells fargo account to my bank of america account \\
\textbf{matched utterance} & transfer \$10 from checking to savings \\
\textbf{label of the input utterance} & transfer \\
\textbf{label of the matched utterance} & transfer \\
\textbf{confidence score} & 0.934 \\ \hline
\textbf{input utterance} & what transactions have i accrued buying dog food \\
\textbf{matched utterance} & what have i spent on food recently \\
\textbf{label of the input utterance} & transactions \\
\textbf{label of the matched utterance} & spending history \\
\textbf{confidence score} & 0.915 \\ \hline
\textbf{input utterance} & who has the best record in the nfl \\
\textbf{matched utterance} & do i have enough in my boa account for a new pair of skis \\
\textbf{label of the input utterance} & OOS \\
\textbf{label of the matched utterance} & balance \\
\textbf{confidence score} & 0.006 \\ \hline
\textbf{input utterance} & how long will it take me to pay off my card if i pay an extra \$50 a month over the minimum \\
\textbf{matched utterance} & how long do i have left to pay for my chase credit card \\
\textbf{label of the input utterance} & OOS \\
\textbf{label of the matched utterance} & bill due \\
\textbf{confidence score} & 0.945 \\ \hline
\end{tabular}} \caption{Case studies on the development set of banking domain. The first two cases are in-domain examples from the banking domain, and the rest are OOS examples.}\label{table:case-study}
\end{table*}
\section{Additional Results}
\label{extra-results}
\paragraph{Visualization} \label{appendix-vidualization}
Figure~\ref{fig:visulization-appendix} shows the same curves in Figure~\ref{fig:robust} along with the corresponding 10-shot results.
We can see that the 10-shot results also exhibit the same trend.
Figure~\ref{fig:tsne-appendix} shows more visualization results with respect to Figure~\ref{fig:tsne-main}.
Again, the 10-shot visualization shows the same trend.
Figure~\ref{fig:Conf-appendix} and Figure \ref{fig:Conf-appendix-all-domains} show 5-shot and 10-shot confidence levels on the test sets of the banking domain and all domains, respectively.
Both Classifier and Emb-kNN cannot perform well to distinguish the in-domain examples from the OOS examples, while DNNC has a clearer distinction between the two.
\paragraph{Faster inference}\label{appendix-DNNC-joint}
Figure~\ref{fig:joint_nli-appendix} shows the same curves in Figure~\ref{fig:join-nli} also for the 10-shot setting.
We can see the same trend with the 10-shot results.
\paragraph{Case studies}\label{Case Study}
Table~\ref{table:case-study} shows four DNNC prediction examples from the development set of the banking domain.
For the first example, the input utterance is correctly predicted with a high confidence score, and it has a similarly matched utterance to the input utterance;
for the second example, the input utterance is predicted incorrectly with a high confidence score, where the matched utterance is related to money but it has a slightly different meaning with the input utterance.
For the third example, the model gives a very low confidence score to predict an OOS user utterance as an in-domain intent; the last example is an incorrect case where the input utterance and the matched utterance have a topically similar meaning, resulting in a high confidence score for the wrong label, ``bill due.''
Based on these observations, it is an important direction to improve the model's robustness (even with the large-scale pre-trained models) towards such confusing cases.
\begin{figure*}[t]
\begin{center}
\includegraphics[width=0.85\linewidth]{./images_appendix/Visualization_Metric_Appendix.png}
\end{center}
\caption{5-shot and 10-shot development results on the banking domain. In this series of plots, a model with a higher area-under-the-curve is more robust.}
\label{fig:visulization-appendix}
\end{figure*}
\begin{figure*}[t]
\begin{center}
\includegraphics[width=0.85\linewidth,height=0.9\textheight]{./images_appendix/TSNE_Appendix.png}
\end{center}
\caption{5-shot and 10-shot tSNE visualizations on development set of the banking domain, where circles represent in-domain intent classes, and red stars represent out-of-scope intents.}
\label{fig:tsne-appendix}
\end{figure*}
\begin{figure*}[t]
\begin{center}
\includegraphics[width=0.85\linewidth,keepaspectratio=true,height=0.95\textheight]{./images_appendix/Conf_Appendix_with_sbert.png}
\end{center}
\caption{5-shot and 10-shot confidence levels on test set of the banking domain. Best viewed in color.}
\label{fig:Conf-appendix}
\end{figure*}
\begin{figure*}[t]
\begin{center}
\includegraphics[width=0.85\linewidth]{./images_appendix/Conf_Appendix_All_domains.png}
\end{center}
\caption{5-shot and 10-shot confidence levels on test set of all domains. Best viewed in color.}
\label{fig:Conf-appendix-all-domains}
\end{figure*}
\begin{figure*}[t]
\begin{center}
\includegraphics[width=0.85\linewidth]{./images_appendix/Visualization_Joint_nli_Appendix.png}
\end{center}
\caption{5-shot and 10-shot DNNC-joint development results on the banking domain, where the dash lines are DNNC results.}
\label{fig:joint_nli-appendix}
\end{figure*}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,149
|
While buzz often circulates around iPhone and Android, Nokia still leads the pack in terms of world market share. Symbian, for instance, remains the most widely used mobile operating system. With the emergence of Nokia's open development platforms, the opportunities available for mobile developers to target these vastly popular operating systems are clear.
Beginning Nokia Apps Development is step-by-step guide that introduces mobile development using Nokia's variety of open platforms like Mobile Qt, OpenSymbian, and MeeGo. This book brings beginners up to speed and shows experienced developers how to work on a mobile platform.
Symbian, the most widely used operating system in the world, as well as MeeGo, the Intel/Nokia platform for mobile devices, both use Qt as a cross platform development framework. Utilizing Qt, a developer can easily target both platforms.
Web Runtime applications can also be developed for both Symbian and MeeGo platforms, and additionally run with little modifications on other mobile platforms. This book explains how.
New Qt developers learn the basics of Qt with a mobile slant, giving them the ability to target both desktop and mobile platforms.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,364
|
This is a work of fiction. Names, characters, places, and incidents either are the product of the author's imagination or are used fictitiously. Any resemblance to actual persons, living or dead, events, or locales is entirely coincidental.
Text copyright © 2016 by Erin Soderberg Downing and Robin Wasserman
Cover art copyright © 2016 by Luz Tapia
Interior illustrations copyright © 2016 by Russ Cox
All rights reserved. Published in the United States by Random House Children's Books, a division of Penguin Random House LLC, New York.
Random House and the colophon are registered trademarks and A Stepping Stone Book and the colophon are trademarks of Penguin Random House LLC.
Visit us on the Web
SteppingStonesBooks.com
Educators and librarians, for a variety of teaching tools, visit us at RHTeachersLibrarians.com
_Library of Congress Cataloging-in-Publication Data_
Soderberg, Erin, author.
Sea sick / Erin Soderberg. — First edition.
pages cm. — (Puppy pirates; #4)
Summary: Captain Red Beard is too sick to organize the Pirate Day party, so it is up to Wally and the other pirate pups to get things ready—the trouble is that the Captain is very particular, and Wally has lost the decoder that would let them read the instructions.
ISBN 978-0-553-51176-5 (trade) — ISBN 978-0-553-51177-2 (lib. bdg.) — ISBN 978-0-553-51178-9 (ebook)
1. Dogs—Juvenile fiction. 2. Pirates—Juvenile fiction. 3. Parties—Juvenile fiction.
[1. Dogs—Fiction. 2. Pirates—Fiction. 3. Parties—Fiction.] I. Title.
PZ7.S685257Se 2016 [Fic]—dc23 2015014057
eBook ISBN 9780553511789
This book has been officially leveled by using the F&P Text Level Gradient™ Leveling System.
Random House Children's Books supports the First Amendment and celebrates the right to read.
v4.1
a
> For Henry, who asked me to write him an adventure series...I hope this one keeps you entertained on _your_ sick days, buddy.
>
> —E.S.
# CONTENTS
_Cover_
_Other Titles_
_Title Page_
_Copyright_
_Dedication_
1. Pirate Day Prep
2. Sneezy Wheezy Captain Queasy
3. Crack a Code
4. Ring! Ding! Dong! Honk!
5. Is It Ready Yet? Is It Ready Yet?
6. Nummer Yummers
7. Old Salt Says
8. Sneaky Shadows
9. Party Like a Pirate Day
10. Puggy Piñatas
_Excerpt from_ Ghost Ship
_"Yo ho ho and a bundle of beef! Yo ho ho and a lottle of fun!"_ Captain Red Beard danced across the deck of his ship. He was singing a silly, happy tune. _"I love my ship and I love my crew, but I'm still number one!"_
None of the puppy pirates on board the _Salty Bone_ was used to seeing their captain so jolly. Usually, the scraggly terrier was pretty gruff. But Captain Red Beard's favorite holiday, Party Like a Pirate Day, was just one day away. He couldn't leash in his excitement.
"All paws on deck," the captain ordered, his tail wagging happily. "Time to prepare for our world-famous Party Like a Pirate Day banquet!"
A cuddly golden retriever pup named Wally skidded across the deck. He stood at attention. "Aye, aye, Captain," Wally barked. "I am ready to help."
"Excellent, little Walty," Red Beard said. The captain patted Wally's head with one of his scraggly paws. "Good boy."
Wally's fluffy tail swished back and forth. He loved when Captain Red Beard was pleased with him. As the newest member of the puppy pirate crew, Wally was always working hard to show the captain he deserved his bunk on board the ship. Wally's best mate, a boy named Henry, joined Wally and the dozens of other pups gathered around the captain.
"Listen up, crew!" Red Beard ordered. "Party Like a Pirate Day is tomorrow. As you all know, the Pirate Day party is absotootly the bestest event of the whole year."
The puppy pirates cheered. No one was more excited than Wally. This was his first Party Like a Pirate Day. He had heard plenty of stories about past parties from his pug friends, Piggly and Puggly. Piggly spun in happy circles when she told him about the food and treats the captain saved for the party. Puggly's favorite part was getting to skip chores to play games and sing songs.
"Everything about Pirate Day has to be perfecto-nino," the captain said. "We have much to prepare. As always, my plans for the party are written in a secret code. We don't want those furball kitten pirates to find our Pirate Day plans and steal them. Aye?"
The puppy pirates all cheered again. "Aye, aye, Captain!"
Curly, the ship's first mate, stepped forward. The poufy white mini poodle whispered, "Excuse me, Captain. You do remember how to crack your codes, though...correct?"
"Of course I do not remember how to crack my codes, Curly!" Captain Red Beard snapped. "What would be the point of a secret code if it's not a secret? As soon as I write 'em, I forget 'em. But I have the handy-dandy code cracker right here." The captain picked up a sheet of parchment with his mouth. He unrolled it and waved it in the air.
Henry squinted to try to make out some of the words on the piece of parchment. He read aloud: _"Party Like a Pirate Day Code Sheet: Top Secret."_ Henry leaned toward Wally and whispered, "Hey, mate, we should get our hands on that! I love cracking codes."
Red Beard pushed the paper to the side and continued his speech. "I hope everyone remembers the most important rule of Pirate Day. On this super-de-dooper holiday, we do everything my way. Whatever I say goes. After all, the captain is the most important part of any pirate ship, am I right?"
Curly began, "But, Captain—"
"No buts," Red Beard said, stomping. "I get to decide everything! I'm in charge, and that's final."
Curly nodded but spoke again. "Captain, it's just that the crew has a few ideas for how to make this the best Pirate Day party yet. If only you would let us—"
"Enough!" Red Beard snapped. "I wrote up the plans for the party. Your job is to follow them. Understood?" He looked around, but no one dared say anything. "Now, pups, let's see what our first order of business is." He padded over to a list of instructions that were posted on the wall. Then he glanced down at his code sheet. "First, pups, we need to blow up balloons. Fill your bodies with wind and _puff puff puff_!"
The puppies all scrambled to grab balloons. They huffed and puffed, filling balloons quicker than Henry could tie them closed. Soon they had filled a huge crate with balloons of many colors. The captain lifted a yellow balloon out of the crate. He bopped it into the air and chased after it.
Other pups did the same. There were more than a dozen balloons bopping around.
_Bop_.
_Bop_.
_Pop!_
The yellow balloon popped. Red Beard yelped and jumped a foot into the air.
_Pop!_ Across the deck, a red balloon popped. Spike, a nervous bulldog, hid from the loud noise.
_Pop!_ A green balloon burst, and Olly the beagle howled.
Wally spotted Piggly and Puggly giggling on the other side of the deck. As usual, the pugs were making mischief. The silly pug twins had loaded up a bamboo shooter with crunchy treats and were aiming them at the balloons. Whenever one of the treats hit a balloon, it popped.
Red Beard spotted them. "Enough!" he barked. "This is no time for your nonsense, pugs." Red Beard scanned the crowd of pups on the deck. His eyes narrowed. "Some of my pups are missing," he growled. "Where is Steak-Eye?"
Wally looked around, searching for the ship's cranky cook.
Curly stepped forward. "Sir, Steak-Eye is feeling ill."
"Ill? What do you mean?" Captain Red Beard growled, searching the crowd again. "And Old Salt?"
"Also sick," Curly said. "There's a bug going around—sneezy noses and the stomach flea. It's taken down half the crew. Otis and Marshmallow and Paco and Puck and—"
"A _bug_?" Captain Red Beard shrieked. "What kind of bug? We will fight it! Puppy pirates are larger and stronger than fleas. We can take it out lickety-split!"
"Not a _bug_ bug, Captain," Curly gently explained. "They're sick. Unwell."
Red Beard howled. "Sickness is not allowed on Party Like a Pirate Day! Only partying! That is an order! It—it—it—it—"
Captain Red Beard's nose twitched. His whiskers trembled. His eyes squeezed shut. He sniffled. He snuffed. And then, with a mighty _ah-choo_ , Captain Red Beard sneezed.
_"Ohhhhhh,"_ Captain Red Beard groaned. "My head. My tongue. My paws. My nose. Everything hurts." He sneezed. He wheezed. He whined. He moaned.
The captain was tucked into bed under a heavy green blanket. Curly, Wally, Henry, and the pugs crowded around him. "What can we do for you, Captain?" Curly asked. "How can we help you feel better?"
Wally could see that Curly was sick, too—sick with worry. With the captain stuck in bed, the ship was now Curly's to run. But with the Pirate Day party to prepare, and half the crew sick in bed, _and_ the usual pirate ship tasks, Wally guessed Curly was nervous about getting everything done. Wally knew she could do it. The poodle was smart and fierce, and the whole crew trusted her.
"Ice," Red Beard wheezed. "Fresh ice and the blue blanket. I like the blue blankie."
Wally ripped the green blanket off the captain's body. Curly tucked him in again—this time with the blue blanket. "Better?"
"Red. It's actually the red blanket I like," Red Beard whined.
But once he was tucked into the red blanket, he didn't like that one, either. Next it was blue again, then green, red, polka dot, black fleece, fur....Finally, the captain settled on a purple-and-gold plaid blanket. He took a deep, wheezing breath. "Party Like a Pirate Day will be ruined," the captain sobbed. "The best day of the year. I've waited three hundred and ninety-two days since the last Pirate Day!"
"You mean three hundred and sixty-five?" Curly said gently. "Pirate Day comes around once a year. That's every three hundred and sixty-five days, Captain."
"As I was saying, I have waited three hundred and sixty-five days for this. And now, sick!" Red Beard cried. "Sick with the sneezies and a stomach flea. What will we do?"
"You need not worry, Captain," Curly said. "As your trusty first mate, I will make sure the ship is ready for the party tomorrow."
"Impossible!" Red Beard sneezed, then wiped his drippy nose on the blanket. "I'm the only one who can do it. Without me in charge, Pirate Day is ruined."
"It's not!" Wally yelped. He hated to see the captain so upset. "Curly can do it. And we'll all help her."
"Listen to the little pup," Curly told the captain. "I can handle this."
The captain's eyes drooped. He was nearly asleep. "You promise?"
"You can count on us, Captain." Curly looked to Wally and the others. They all nodded. "We'll follow your orders and throw you the best Pirate Day party ever."
Captain Red Beard sighed. "I don't see how that's possible. But you can try." He reached his muzzle under the blanket and grabbed something in his teeth. "Here is my Party Like a Pirate Day Top-Secret Code Sheet. You can use it to decode my instructions. But please, keep it safe. It's the only copy."
"Of course," Curly said. "We'll get to work right away."
"Arrrrr!" Captain Red Beard cried out suddenly. "I have an itch!"
"Where?" Curly asked, dropping the code sheet to the floor. "Would you like me to scratch it for you before I go?"
Red Beard whimpered. "Yes. My ear. The lefty one."
Curly used her paw to scratch at the captain's ear. After a good, long scratch, she hopped off his bed and made her way toward the door. "Crew, follow me. We must get to work on the party plans at once."
"Curly?" Red Beard croaked. "Please don't go. Stay with me. What if I need something?"
Wally stepped forward. "Sir, I could stay here and take care of you. Curly has a crew to run and a ship to steer."
"And Piggly and I would be happy to start working on the party," said Puggly. She shot her sister a sly look.
Captain Red Beard burrowed under his covers and whined. "But I need Curly. She's my first mate. I must have her by my side." He looked at Curly with his saddest puppy-dog eyes. "Please? You can run the ship and plan the party from here."
Curly sighed. "If you need me, I'll stay."
The captain yawned, then fell fast asleep. Curly turned to the others. "Wally, take the captain's code cracker. Guard it carefully."
Wally barked, "Aye, aye, Substitute Captain Curly!" He gently picked up the piece of parchment with his mouth. Then he and Henry and the pugs ran up to the deck.
As they ran, Wally barked "ahoy" at another pup—and that's when it happened. The wind came out of nowhere and tugged the code sheet from his mouth. It fluttered across the deck. Wally and Henry lunged for it, but another gust of wind pulled it out of reach. In a blink, it floated up and over the side of the ship.
Wally and his friends stared out at the ocean, eyes wide. The vast blue sea stretched for miles on every side of the ship. Somewhere among those dark, frothy waves was a piece of parchment. A piece of parchment Wally was told not to lose, no matter what. It had been an accident. But would the captain and Curly see it that way? Wally didn't think so.
Party Like a Pirate Day was ruined. And it was all his fault.
"In case you were wondering?" Henry said. "I think we might be in some serious trouble."
Puggly snorted. "You think?" She sneezed, spun in a circle, and sat down. "The captain's code sheet is gone!"
"All right, mates," Piggly said as she chewed on one of Henry's old boots. Piggly was always eating something. "Let's figure this out: how can we get the ship ready for Pirate Day if we don't have the captain's code cracker?"
"Yeah," Puggly said. "The captain wants things done his way. If we don't do it right, we'll all be walking the plank."
The four friends trotted over to the list of instructions posted on the wall. None of the captain's codes made any sense. It was just a mess of letters and numbers.
"Maybe we should tell Curly?" Piggly suggested. "She would know what to do."
"We can't tell Curly," Wally said. "She might tell the captain." Wally thought for a moment. "We're just going to have to crack the captain's codes ourselves."
Wally stared at the captain's list of Pirate Day orders. The first instruction made no sense at all:
"Pibabel no?" Henry said, trying to read the words from left to right.
"Pibabel no!" Piggly echoed. "Yes, let's do that! But...what is that?"
The pups on deck all stepped forward to look at the strange words on the page. No one could figure out what they meant.
Henry scratched his head as he stared at the jumble of letters. "Here's the thing about secret codes," he said, pressing his finger to the paper. "They aren't meant to make sense when you look at them the first time. We have to try to read it in an unusual way."
"Maybe we're supposed to read this backward?" Wally said.
"Oh, I you tadeepee!," Puggly sounded out. "Nope, don't think so."
Henry didn't say anything. He ran his finger up and down the lines of letters, then gasped. "Look! If you read the words from the top down, they make sense."
_"Paint the banners, paint the doors. Just be careful, don't paint the floors,"_ Henry whooped.
"Of course!" Puggly cheered. "We're supposed to paint the ship to make it look pretty. I'll get the purple paint. And glitter!" Puggly loved fancy things.
"Green," Piggly argued. "The captain's favorite color is green."
"I thought it was red," yelped Puggly. "Like the color of his reddish fur."
Wally yipped to get their attention. "The captain is sick, so we can't ask him. We just have to pick our favorite colors and get started. Hopefully, Captain Red Beard will like it."
Wally got the rest of the crew up on deck to help them, while Piggly and Puggly found the paints. Then everyone got started. Wally and Henry worked on painting all the doors on the ship different colors. Other pups rolled out long pieces of paper and painted banners that said: _Yo ho ho for Pirate Day!_ and _Three Cheers for Captain Red Beard!_ and _Beware the Salty Bone!_ So many pups were in bed with the stomach flea that the few who were well had to work extra hard.
But the pugs thought that extra work deserved extra play. And with no one in charge, it was so tempting to goof off! Giggling, Puggly filled her pug cannon with glitter and blasted golden sparkles all over the deck. The glitter stuck to the wet paint, making the ship look like a disco ball.
Piggly didn't like to miss out on any kind of fun. So she dipped her paws in her paint bowl and trotted across the deck. Tiny pug prints soon made curlicues across the floors.
Then the naughty sisters both sat in pans of paint, covering their tails and bottoms with color. Afterward, they sat on the banners and giggled at the marks they left behind.
Wally barked out a warning. "The captain's orders said to be careful _not_ to paint the floors, Piggly. Curly is going to be really upset!"
"Eh, Curly doesn't scare me," Piggly huffed. "She's not the captain. She's just the substitute captain. What's the worst that could happen?"
"Yeah, stop being so serious, Wally!" Puggly snatched Wally's paintbrush out of his mouth. She leaped to the other side of the deck with it.
"Give that back!" Wally begged.
"Come and get it," Puggly teased. She pressed her front paws flat, poked her rear end up in the air, and wagged her tail. "Wanna play chase?"
Wally knew he should keep working. But chase was his _favorite_. He couldn't resist. Puggly dashed madly from one side of the deck to the other with the paintbrush tight in her teeth. Wally chased after her. Olly and Spike chased after him—and soon everyone was playing instead of working!
The pups tumbled and rolled back and forth across the deck. They knocked over paint cans and slid through puddles of paint. Before long, the whole floor was covered in thousands of colorful paw prints, swerving this way and that.
Everyone was having a blast. And then:
_Ding ding ding!_
The sound of a far-off bell rang through the air. Everyone froze.
"Avast!" A tiny bark cut across the deck. It was Curly. She growled at the crew, who were covered in paint and sparkles. "What is the meaning of this?"
"Well?" Curly stamped her little paw in a puddle of paint. "What's going on here?"
Henry ducked behind a banner. Spike shook with worry. Piggly and Puggly lay down to hide their paint-covered feet and bottoms.
Wally panted, trying to catch his breath. "Uh, just doing a little painting?" he said. "These are our decorations for the party. Following the captain's orders!"
Curly took in the orange and yellow ship rails, the floor full of pink and purple paw prints, and the glitter that covered _everything_. "This is what the captain's instructions said?" she asked. "Really?"
"Paint the banners, paint the doors..." Wally decided to leave out the part about being careful not to paint the floors. They had a little cleaning to do before the captain was back up and at 'em. Luckily, he and Henry were experts at swabbing the deck.
"How's the captain feeling?" Puggly asked. A drop of pink paint slid down her ear and splatted on the floor.
"He's snoozing," Curly said. "I wanted to make sure everything was going okay. Captain Red Beard is so worried about Pirate Day. This is my chance to show him he can trust me to keep everything rolling right along without him."
"Everything's fine," Wally said quickly. Sure, he'd lost the secret code sheet. But they hadn't had any trouble cracking the captain's first code, had they? He was pretty sure they could figure out the rest of the captain's instructions. All they needed was a little time. "Just taking a quick break for a game of fetch. Want to join us?"
_Ding ding ding!_
Curly jumped. "That's the captain's bell," she explained. "I tied a little bell around his neck. Told him if he needed anything, he could ring for me."
_Dong dong dong!_
Another bell sounded, deeper and quieter than the first. The pups all pricked up their ears.
_Ring-a-ling-ling!_ A third bell, followed quickly by...
_Honk!_ A horn. Spike hid behind an overturned crate.
"What is that honking?" Henry wrinkled his nose and looked all around. "There aren't any geese on board our ship. At least, I don't think there are...."
Curly closed her eyes and took a deep breath. "I gave Old Salt and Steak-Eye bells, too. Marshmallow has a horn. I ran out of bells. There weren't enough for all the sick pups."
_Ring-a-ling-ling!_
_Ding ding ding!_
_Honk! Honk! Honk!_
_Dong dong dong!_
Bells and horns rang out from all corners of the ship. It sounded like every sick puppy on board needed something.
And they needed it _now_.
"I'm _coming_!" Curly howled. She looked frazzled. "Everyone wants something, and the captain wants _everything_. I'm just one pup. What am I supposed to do?"
Wally looked at the paint-splattered ship and thought about how much cleaning they had to do. And how many Pirate Day codes they still had to crack.
Then he looked at Curly, who seemed like she had _had it_. The bells got louder and louder. The horn honked without stopping.
Curly wasn't the only one who wasn't sure what to do.
"Party planning can wait," Wally said suddenly. They would crack the codes later. Curly needed him. The captain needed him. "Nurses Wally and Henry, reporting for duty."
"Captain?" Wally said quietly. He pushed open the door to Red Beard's quarters. Henry trailed behind him. "You rang?"
_Ah-choo!_
The captain sneezed. He dug feebly at his covers, trying to carve out a comfortable spot in his sickbed. "I need a snack," he whined. "My tummy is sore."
"What sounds good, Captain Red Beard?" Wally asked.
"Sardines," the captain whispered. "With pink jam."
"Strawberry jam, sir?"
"Pink!" Red Beard growled. "Pink-flavored jam, Walty. None of this hairy berry bumbo _fruit_ business."
Wally nodded. "You got it, Captain. Sardines with pink jam." He hustled out of the room, Henry on his heels.
The moment he closed the door, the captain's bell rang.
_Ding ding ding!_
Wally peeked back into the captain's quarters. "You rang?"
"Is it ready yet?" Captain Red Beard asked weakly.
"Not yet, sir. I just need to run up to the galley to make it. Since Steak-Eye is sick, I'll have to prepare it myself." He crept toward the door again, then turned. "But I'll be back in just a minute, Captain. Promise." He raced out the door.
_Ding ding ding!_ The captain hollered, " _Now_ is my snack ready, Walty?"
Wally peeked through the door again. "Nope," he said. "Not quite yet, sir."
"What's taking so long?" Captain Red Beard whined.
Wally sighed. Captain Red Beard wasn't the easiest patient on the seven seas.
Before Wally could dash off to the galley, Henry stopped him. He said, "Hey, mate? I'm gonna head up to the main deck again. I want to take another look at the captain's list of Pirate Day instructions. In case you were wondering, we have a few more codes to crack before everything will be ready for the big day!"
Wally barked his agreement. Henry's idea to split up was a good one. At the rate they were moving, they would not be ready for the Party Like a Pirate Day party. Henry could get to work on the next code while Wally prepared the captain's snack. They made such a great team.
Henry raced up the stairs to the main deck, and Wally ran toward the galley.
Inside the ship's kitchen, Wally found Piggly munching her way through a box of treats while she got a juicy steak ready for Old Salt. She had spilled snacks everywhere.
Curly was warming up a hot-water bottle for Steak-Eye. In the process, a lot of the water had ended up on the floor.
Spike was getting a soft cloth to clean a soup spill off Marshmallow's fur. As he made his way around the kitchen, Spike kept knocking piles of dishes onto the floor. Other dishes were piled high in the sink, waiting to be washed.
Wally tossed a stack of sardines into the first clean dog dish he found. Then he plopped a clump of pink jam on top of the tiny fish. He didn't even have time to put away the jam as he raced the sardines back to the captain's side.
"What is this?" Red Beard coughed. His tongue lolled out of his mouth.
"Your snack, sir."
_Ding ding ding!_
Wally hid his head under his paws to try to block out the noise. "What do you need now, sir? I'm right here. You don't need to ring the bell."
"I don't like the look of this snack," Red Beard announced.
Wally cocked his head. "But, sir, it's sardines with pink jam. Just what you asked for."
"I want it in my special dish. One of the golden bowls I save for Pirate Day." Red Beard pouted. "Put my snack in a gold dish so it will taste better."
"Aye, aye, Captain," Wally said, running from the room again. He found the captain's special dishes buried under a lot of kitchen clutter. He dumped the sardines and pink jam into a shiny gold bowl and hustled back to Red Beard's sickbed. "Is this the dish you like, Captain?"
Red Beard pressed his nose to the snack and sniffed. He groaned. "It's too cold. And I like the jam _under_ the sardines. It tastes better that way."
"Right-o," Wally said, running out of the room again. He warmed the sardines over a fire in the galley, then plopped them on top of the pink jam. "This should do it," he muttered.
But when he presented the plate to the captain, Red Beard moaned. "Forget it," he said sadly, pushing the dish aside. "I'm not really hungry anymore. Thanks anyway, little Walty."
Wally sighed. "Is there anything else you need, Captain?" He tried not to think about how much time he had spent getting the captain's snack ready—time he _could_ have spent getting the ship ready for the Pirate Day party.
Red Beard nodded. "I need someone to cuddle with. Fetch my stuffed duck!"
Wally grabbed the duck from a basket in the corner and placed him beside Captain Red Beard. The sick captain curled around his stuffed buddy, then fell fast asleep.
As soon as Wally was sure it was safe to move again, he crept out of the room. He tiptoed down the hall quietly, hoping his needy captain would stay asleep for a very long time!
When Wally finally got back up to the main deck, most of the puppy pirates were gathered around the captain's list of instructions. Now that the patients had all been cared for, the others could get back to work. There wasn't much time left.
Curly was off meeting with Chumley, the German shepherd who ran the map room. The two of them were making sure the ship was on course for calm seas. No one wanted to hit a storm before the Pirate Day party!
"What's the next thing on the instructions list?" Wally asked Piggly.
Through a mouthful of cheese and bacon treats, Piggly blurted, "We dun dough."
"You don't know?" Wally guessed. He tried to squeeze through the crowd to get to Henry. When he was near enough to see the codes, Wally's first thought was that the next one was much trickier than the last:
"Blimey. This code is in some other language," Spike said, whimpering. "We'll never figure it out. Never, never, never."
"Never say never," Puggly scolded. "A pirate can do anything if she sets her mind to it."
Henry chewed on his lower lip and stared at the jumble of numbers. "It says 'nummer yummers' at the top, so maybe this code is telling us something about the food for the party."
"Aye," Wally barked. He loved that his best mate was so smart about everything.
"In case you were wondering," Henry said, "I think it might be some sort of recipe."
"I doubt that," Piggly said. Crumbs toppled out of her mouth. "The captain can't cook."
Henry began writing something on the bottom of the list of instructions. It was the alphabet. Under each letter, he wrote a number.
"It's a letter-to-number code!" Henry said, slapping his hand to his forehead. One by one, he put a letter under each of the numbers on Red Beard's list.
"See?" Henry said, grinning at the gathered crew. "It says: _Treat time, meat time, lots of yummy eat time. Peek under the pepper!_ The captain must have hidden our special treats for Pirate Day. This tells us where to find them!"
"Follow me," Piggly ordered. She led the crew toward the galley. "I know my way around the kitchen!"
But when they arrived in the galley, everyone screeched to a halt. Dishes were everywhere, and food scraps littered the floor. Nothing was in its usual place. With Steak-Eye sick in bed, no one had done dishes since dinner the night before. Pups had been in and out of the kitchen all day, helping themselves to snacks.
The place was such a mess that it took a few moments before they noticed Steak-Eye, the ship's tiny cook, spinning in lazy circles in the center of the room. A blanket draped across his back dragged along the floor.
"Steak-Eye!" Wally gasped.
The cook looked at him with watery, sleepy eyes. _Ah-choo!_
"Are you okay?" Piggly asked.
Steak-Eye blinked twice, spun in one last circle, then curled up and fell fast asleep.
"Is he...sleepwalkin'?" Spike wondered.
Steak-Eye snored loudly. Then he made a _glug-glug-glurrgh_ noise in his sleep.
"Don't laugh," Puggly warned. "We don't wanna wake him. He'll be furious if he wakes up and sees the state of his kitchen!"
While a few pups worked together to carry Steak-Eye back to his quarters, Henry said, "I wonder who made such a mess of this place?"
"I came in to get Old Salt a snack earlier," Piggly explained. "And I fixed a couple midnight snacks for myself last night."
"I got some sardines and jam for the captain," Wally mumbled. "But I forgot to clean up after."
"And I tried to whip up some stew for my breakfast," Spike said. "Not tasty. We need Steak-Eye to get well soon. I'm starvin'."
It was clear that almost everyone on board had been a part of making the mess in the galley. Each of the pups' little messes added up to one big mess. It was going to be impossible to find the treats for the party! Still, they got to work searching. They sniffed in pots and tore through cupboards. Soon the kitchen was even more of a mess than it had been before, and still no one could find the pepper.
Wally sniffed at the air. His nose tickled. He almost sneezed—he was pretty sure he had just picked up a whiff of pepper. It was coming from behind a towering stack of pots. He poked his nose into the stack.
It wiggled.
It wobbled.
It swayed.
_Crash!_
The whole stack of pots fell to the floor with a loud clatter.
"Shiver me timbers!" Henry shouted joyfully. He pushed aside the mess of fallen pots and uncovered a box marked PEPPER. The puppy pirates nosed the pepper aside. Henry lifted open the crate below. "It's full of sausages and smoked fish! This must be the captain's party food. Consider this code _cracked_!"
The puppies all cheered. But their celebration was short-lived. Curly had just walked through the door—and the galley had never been more of a disaster.
Henry leaped across the mess and tried to block Curly's view of the kitchen. "In case you were wondering, we have everything under control in here."
Curly peered between Henry's legs. "Under control?" she barked. "You call this mess _under control_?"
All the dogs backed away from her. When Curly was mad, her voice could get _very_ loud.
But when she finally spoke again, Curly did not shout. Instead, the first mate sounded very calm. "Half our crew is sick with sneezy noses and the stomach flea. The deck is filled with paint. The galley is a disaster. From what I can tell, things are _not_ under control. In fact, we are not much closer to being ready for the Pirate Day banquet than we were this morning. We have until tomorrow to get this ship ready for the captain's favorite day of the year. I trusted all of you to help me." She looked around and shook her head sadly. "You let me down. You let the captain down."
Wally tucked his tail between his legs.
Curly sat and sighed. "This will be the first Pirate Day with no party. I guess I need to find some way to break the news to the captain."
No one said anything. It was terrible when Curly was angry. But somehow this was worse. She was disappointed in them.
_Ding ding ding!_
_Dong dong dong!_
Ringing bells echoed from down the hall.
"Well." Curly took a deep breath. "Sounds like Captain Red Beard and Old Salt both need help again. I will tend to the captain. Could someone else check on Old Salt?"
"I will, Curly," Wally offered. "But first, do you want me to come with you to tell the captain about the party? It's not your fault the day is ruined."
Curly shook her head. "No, I'll tell him myself in the morning, after he's had a good night's rest. It was my job to make sure everything sailed right along on his sick day. I must take the blame."
Curly's shoulders sagged. She turned and slunk out of the kitchen.
Wally padded down toward Old Salt's quarters with his tail between his legs. His ears drooped. He felt terrible about disappointing Curly—not to mention the captain.
Old Salt's bell rang again, and Wally sped up. He wasn't going to disappoint anyone else today. Maybe he couldn't fix the party mess, but he could at least help his favorite Bernese mountain dog.
And, just maybe, wise Old Salt could help him, too.
Wally poked his nose into the room. "Ahoy. How are you feeling?"
Old Salt peeked at Wally from under a pile of blankets. He sneezed.
_Ah-choo!_
"I've felt better, kid," he wheezed. "How are things going up on deck? Everything coming along for Captain Red Beard's big Pirate Day party?"
Wally hung his head. "Not really," he admitted. "We messed up. Now everything is ruined."
"I'm sorry to hear that," Old Salt grunted. "I guess when the cat's away, the mice will play."
Wally cocked his head, confused. "We're dogs," he reminded Old Salt. "Not cats."
Old Salt began to laugh. The chuckle turned to a cough, then a sneeze. _Ah-choo!_
"Aw, Wally, boy," Old Salt chuckled. "That's just an ol' saying. It means, with the captain gone, maybe everyone is goofin' off a little more than they should."
"Oh!" Wally said, giggling. "That makes sense. Yes, that's it. Curly's the substitute captain, but it's not the same. She's so busy taking care of the captain and steering the ship and everything else. There's no one to tell us what to do."
_Ah-choo!_
"Y'know, Wally..." Old Salt hacked up a hair ball. "Bein' on a pirate crew isn't just about following your captain's orders. There's a pirate code, kid."
"I know all about codes," Wally said. "The captain wrote all the Pirate Day plans in code. We've been trying to crack them so we can figure out how to get the ship ready for the party."
"That's not the kind of code I'm talking about, Wally." Old Salt coughed. " _Code_ can also mean _rules_ , kid. There are rules that a good pirate needs to follow. We help each other out—that's our way. If our captain is busy with other stuff, we've gotta steer our own ship. We need to take responsibility." He coughed again. "Sometimes, Wally, you have to be your _own_ captain."
When Wally left Old Salt's side, it was long past bedtime. Everything on board the _Salty Bone_ was silent and still, except the captain's snoring. The rolling ocean waves made gentle splashing sounds as they licked at the sides of the big wooden ship.
Everyone, it seemed, was tucked in tight—patients and all. Other than Wally and the night-watch pups, only Curly was still awake. She was skittering around the ship to check on everyone and make sure things were running as smoothly as possible.
Wally couldn't stop thinking about what Old Salt had said. And he couldn't stop thinking about the Pirate Day party. Maybe Curly had given up on the party. Maybe _everyone_ had given up. But Wally couldn't.
He was the one who had lost the code cracker in the first place. He'd started this whole mess. And now he was going to be the one to fix it.
Somehow.
Wally stopped in his room. He hopped up into Henry's bunk and tugged at the boy's pajamas. Henry opened his eyes and popped out of bed. "Is something wrong, mate?"
"Everything's wrong," Wally barked. "But we're going to do something about it. Just like the pirate code says." He barked again, urging Henry to follow him.
Keeping quiet so as not to wake anyone else on board, Wally led his best mate up to the deck. The moon lit their way.
As they crept toward the list of instructions, Wally realized that he and Henry were not alone. There were shadows moving near the captain's Pirate Day list.
Two short shadows.
Piggly and Puggly were slinking around the deck. The two pugs were always full of mischief. What could they be up to?
"Wally!" Piggly barked. She looked surprised to see him.
Puggly whipped her cape. She was trying to hide something under it.
"What is that?" Wally demanded.
"Nothin'," Puggly snarled.
"It's not nothing," Wally said. He drew closer. "Show me."
"You're not the captain," growled Puggly. "You can't make me."
"No, I can't make you," Wally said, thinking about his conversation with Old Salt. Surely the pugs knew about the pirate code, too. "I just want to know what you're up to. Henry and I came up here to figure out the rest of the captain's Pirate Day codes."
"You mean you still want to try to make the party work?" Puggly asked.
"Not just try," Wally said. "We're _going_ to make it happen. Will you help?"
Before Puggly could answer, her cape slipped and something yellow popped out. Wally gasped. "Is that a balloon?"
"Maybe it is, maybe it isn't," Piggly snapped.
"What are you doing with the Pirate Day balloons?" Wally asked. He thought back to how Piggly and Puggly had spent the morning popping the party balloons while everyone else blew them up. Then they had fooled around with the paint while the other pups were decorating. And in the galley, they had snacked while everyone else searched for the party food. Wally sighed. "Don't you want the party to be perfect?"
Piggly and Puggly sniffed. "Of course we do. But it won't be, no matter what we do."
"What are you talking about?" Wally asked.
"With the captain sick, and Curly busy...well, it's never going to be the perfect party that the captain wanted," snuffled Puggly. "That's impossible without him telling us exactly what to do."
"That doesn't matter!" argued Wally. "We can figure it out for ourselves. We need to be the captains of our own ships!"
"But there are only a few dinghies on board our ship, Wally," Piggly noted. "Not enough that we can all have one of our own."
Wally shook his head. "What I mean is, we have to take responsibility for the party _without_ being told exactly what to do." He cocked his head at his pug friends. "Now, are you going to help me—or are you going to stay out of the way?"
Piggly grinned at Puggly, then said, "What do you think we were doin' up here, Wally? Goofin' off?"
"Um, yes?" Wally said.
Puggly snorted. "No way, mate. You think you're the only one trying to save Party Like a Pirate Day?"
"But you just said it was impossible!" Wally pointed out.
Puggly grinned. "I said it was impossible to make this party turn out exactly the way the captain wanted it."
"So we're not going to do that," Piggly said, giggling. "We're going to make it even better."
"How?" Wally asked. "Did you crack more of the captain's codes?"
"We'll leave the code cracking to you and your boy," Piggly said. "Me and Puggly are working on a little surprise of our own. It's gonna blow the captain away!"
Wally tried and tried to get the pugs to reveal their surprise. But they wouldn't even give him a hint.
"You're just gonna have to trust us!" Puggly said.
The sun would come up soon, and there wasn't much time left to figure out the rest of the captain's codes. So Wally and Henry got back to work.
They took Captain Red Beard's list of instructions off the wall and brought it down to their quarters. Then they snuggled under the covers with the sheet of parchment laid flat in front of them.
The second-to-last code said:
"Looks like just a bunch of random letters," Henry said.
Wally stared at the strange words. He squinted so hard the letters started to blur. Then Wally had an idea. He didn't know much about decorations or cooking, but he knew a _lot_ about dessert. Maybe he could work backward. If he could think up some great desserts, it might help him figure out the captain's code.
"Wait a second!" Wally barked. _Backward_...He looked closer at the code. And suddenly, it all made sense.
"The words are written backward!" Henry cried, figuring it out at exactly the same time. _"Treats taste better hot not cold and always when they are served in gold."_
_"Arrr-ooo_!" Wally howled. "I know what he's talking about. Captain Red Beard wants us to serve the banquet treats in the special gold dishes he saves for Party Like a Pirate Day. He made me serve his snack in one of those fancy dishes."
Henry didn't answer. He was already staring at the next clue, which said:
"In case you were wondering, this one's easy. It's Morse code," Henry said. "That's the secret language of navy ships and spies."
"But it's just a bunch of dots and dashes," Wally noted.
"You're lucky I know so much about life at sea, mate! In Morse code, each letter of the alphabet is represented by a mix of dots and dashes. If you signal it over the radio, dots are a short sound, and dashes are long." Sitting up in bed, Henry got a pencil and paper. He wrote down all the letters, A to Z. Then he scribbled dots and dashes next to each of them.
His tongue poked out the side of his mouth as he thought it through. "I've got it! This one says: _'Pirate party rule number one, lots of games means fun fun fun.'_ Huh."
"What?" Wally asked.
"It doesn't say what _kind_ of games," Henry pointed out.
"Maybe the captain wants us to figure that out for ourselves," Wally realized. He wagged his tail. He felt like he'd just cracked the most important code of all: the pirate code!
Henry yawned. "Here's what I'm thinking, mate. I think we do our best to make the Pirate Day party as fun as we can. We might not know how to do things _exactly_ as the captain had planned, but it will be a blast, no matter what!"
Wally couldn't have said it better himself. And he knew Piggly and Puggly agreed. Wally curled up beside his best mate, falling asleep almost instantly. Being his own captain was exhausting work.
—
At first light, Henry and Wally jumped out of bed. They had so much to do before the party. Curly was still running around helping the captain. But all the other puppies reported for duty. "We have a lot of messes to clean up," Wally told the crew. "And a lot of party to plan. And no one's going to tell us how to do it. So let's figure it out for ourselves!"
Spike raised a trembling paw. Wally could tell he had an idea but was afraid to say it out loud in front of all these pups.
Spike was afraid of pretty much everything.
"What is it, Spike?" Wally asked.
"I think I know how we can fix the mess we made with the paint," Spike said nervously. "We could, um, paint _more_ paw prints on the deck. They could lead puppies to different parts of the party."
"It will be like the dotted lines on a treasure map!" Wally yelped. "Follow the prints to find fun, games, and food."
Everyone loved this idea, and several pups got to work with paint-filled paws.
"I'll take charge of the galley!" Piggly said. She led a group of pups down to start cleaning things up. But there wasn't enough time to clean all the dishes before the party.
Piggly thought for a second. "We can serve the fish in golden bowls," she said. "But I have a fun new idea for how to serve the captain's special sausages, mates!" She barked excitedly. "We can stab 'em with swords and serve 'em up like shish kebabs!"
On deck, Puggly raced around with streamers and bows, making everything look even more festive. Henry scrambled up on railings and helped her reach all the high corners. Soon every inch of the ship was decorated for Party Like a Pirate Day!
While the crew painted and polished and cooked and cleaned, Wally had a job of his own to do.
The most important job of all.
He had to stop Curly from telling the captain that Pirate Day was canceled. He hightailed it to the captain's quarters, hoping he wasn't too late.
Captain Red Beard was sitting up in bed, looking much better than he had the day before. Much better...and _much_ angrier. "What do you mean, Pirate Day is canceled?" Red Beard growled.
Uh-oh.
Curly was sitting at the foot of his bed. "I tried my best, Captain, but things got a little out of control. I just couldn't—"
Before she could say anything more, Wally laughed a loud, fake laugh. "Wasn't that a good joke, Captain?"
"What joke?" Captain and Curly asked together.
"You know, Curly, the joke about Pirate Day being canceled. You were just playing a funny little prank on Captain Red Beard, right?"
"You were?" the captain asked.
Curly hesitated. "I was?"
"Of course you were," Wally said. "But I came down here to tell you that everything's ready for the party. So if you're feeling better, Captain Red Beard, it's time."
The captain leaped out of bed. He shook his body to fluff up his fur. It had gotten matted during his sick day in bed, and one of his ears was pressed flat to his head. "Well, what are we waiting for?" He trotted up the stairs. Wally and Curly followed close at his heels.
Curly looked at Wally, confused. "I don't understand," she whispered. "The last time I checked in, you were all goofing off. No one was getting anything done—"
Before Wally could explain, Captain Red Beard stepped out into the sunshine. Streamers fluttered in the wind. Paw prints filled the deck. Food was piled high on platters and speared on swords and—of course—heaped inside the captain's special golden dishes. Balloons hung from the rails.
The captain's mouth fell open. "Wh-what—" he stuttered. His voice boomed. "What in the name of Growlin' Grace have you done to my party?!"
"Everything is so different!" Red Beard shouted.
Wally held his breath. The rest of the crew lowered their heads in fear.
The captain blinked, took it all in, and yelled again, "And I _love it_!"
The puppy pirates cheered.
The captain went on in his loudest, most important voice: "My dear crew, it seems that this year—just like every year—I planned the bestest, most supertacular Pirate Day party on all the seven seas!"
"Three cheers for Captain Red Beard!" the others called out. " _Arrrr_ -oooo!"
Red Beard held his head high. "I just love how you followed my directions perfectly." He sighed happily.
The puppy pirates looked at each other. No one wanted to tell the captain that not all of his instructions were so clear. At the end of the day, maybe it didn't matter anyway. The puppy pirates had pulled off a party everyone on board could enjoy. And no one had to walk the plank.
"I have one last thing to say before we get down to business." Red Beard turned to Curly. "The sneezies and stomach flea took down your mighty captain at the worst possible time. But Curly kept us afloat, and for that I must say thank you. Curly, you have proven yourself to be an excellent leader. Someday you will make a wonderful captain of your own ship."
Curly bowed. "Don't thank me for this party, Captain. Thank your crew. Especially Wally, Henry, Piggly, and Puggly. They worked extra hard to make sure the party was a success."
"But you were a great substitute captain!" Wally ruffed. "Three cheers for Curly!"
"Maybe so," Curly said, smiling. "But you lot were your own captains when you needed to be. That's the pirate code, right?"
Wally caught Old Salt grinning at him. The older dog winked.
"Right," Wally barked.
"Just to make it absotootly clear," Red Beard said, "none of you pups will ever be as good a captain as me."
"Of course." Curly laughed. "No one will ever captain a ship as well as you do, sir."
"Yes, yes. All righty, then. Who's ready to party like a pirate?" Captain Red Beard gazed at his crew. "You all have one order to follow today: Have fun!"
"Aye, aye, Captain!" The puppy pirates scattered across the deck. The rest of the afternoon, they sang, danced, chased, played, and ate like royalty.
By late in the day, the other sick puppies felt well enough that everyone was able to take part in the fun. When Steak-Eye tasted one of the shish-kebabs-on-a-sword, he glared at the pugs. "It's fine," he growled. "Kind of clever, actually. Doesn't taste as good as something I would have cooked, but I'll eat it anyway."
"Thanks, Steak-Eye." Piggly giggled. She knew this was a compliment, coming from the cranky cook.
The cook narrowed his bulging eyes and said, "I assume my galley is clean?"
Wally looked at Piggly and Puggly nervously. Puggly grinned back. "It's spick-and-span..." Under her breath, she whispered, "...now. But shiver me timbers, you shoulda seen the state of that place yesterday."
Steak-Eye whispered back, "Aye. Good thing I was in bed all day, eh?"
As the sun was setting, Captain Red Beard barked loudly. "Okay, everyone. Game time! What game do we have first?"
Wally gasped. He and Henry had forgotten about coming up with the games! He looked at Henry and whimpered. Had they ruined the party after all?
But before he could say anything, Piggly and Puggly trotted to the front of the ship and called for everyone's attention. "Ahoy, mates! It's pug-glorious game time!" Puggly hollered.
Wally breathed a huge sigh of relief. So _this_ was the pugs' big surprise.
"Who wants to be the first to try a puggy piñata?" Piggly cried.
Captain Red Beard jumped up and down. "Me! I do! Me! I love games. I love games!"
Piggly giggled. "Step right up." She offered the captain a bamboo shooter. "Aim at a balloon. If it pops, you'll find a surprise. This is our latest invention. It's a game and a prize in one! We got the idea when we were poppin' balloons and eatin' treats yesterday morning."
Red Beard looked at the balloons hanging from the deck rail. There were hundreds of them, in dozens of colors. "Any balloon?"
"Yep," Puggly snorted. "Any balloon. If it pops, you're a winner."
Red Beard shot at one of the red balloons hanging nearest him. He missed.
Wally looked at the pugs nervously. The captain would be very embarrassed if he didn't win a prize. Thankfully, after a couple tries, the balloon burst with a loud _pop!_
Treats came raining out, all across the deck. "Treats!" Captain Red Beard exclaimed. "There are treats inside the balloon!" He leaned in close to Piggly and Puggly and whispered, "This game was my idea, right...?"
"Of course," Piggly said loudly. "Of course the puggy piñatas were your brilliant idea, Captain. After all, it wouldn't be Party Like a Pirate Day without you, sir."
As the puppies fanned out across the deck to pop the rest of the puggy piñatas and find their prizes, Wally felt so very happy. He only hoped the stomach flea would stay away for a long while. It was tiring being his own captain all the time. And to tell the truth, Wally had sort of missed Captain Red Beard while he was sick in bed. He was ready for things on board the _Salty Bone_ to get back to normal!
Just then, Wally felt a strange tickling in his nose. He pawed at his snout, but the tickling got worse. _Ah-choo!_ A loud, forceful sneeze knocked Wally off his feet. He skidded across the floor, landing in a heap beside the pug twins.
_Ah-choo!_ Piggly's sneeze sent her crashing into a table heaped with food.
_Ah-choo!_ The blast of Puggly's sneeze popped another treat-filled balloon.
_Ah-choo!_ Henry groaned and rubbed at his eyes.
_Uh-oh_ , Wally thought as he sneezed again. It looked like normal would have to wait.
time thunder rumbled. He had never liked storms. The loud noises frightened him.
Henry rubbed Wally's fluffy ears and whispered, "In case you were wondering, I love storms. There's nothing to be scared of, mate. Not lightning. Not thunder." Henry flipped over and added, "Gosh, you don't even need to worry about monsters under the bed. I'll keep you safe, mate."
Monsters under the bed? Wally yipped and whimpered. He had never worried about _those_...until now!
Wally snuggled close against his friend. His worries kept him awake for a long time. When he finally fell asleep, he tossed and turned. He dreamed of Growlin' Grace and ghost ships, dark skies and rolling waves.
_Crrrr-ack!_
Thunder and rolling waves pounded at the ship. Wally startled awake, happy to get away from the scary dream he'd been having. In his dream, Growlin' Grace's ghost had come to make him join her crew. Wally tried to settle in and fall asleep again. That's when he heard it....
_Ooo-ooh! Woo-ooo! Walllll-ly!_
A spooky, ghostly howl filled the room. It was coming from something pale and glowing right next to Wally's bed!
Wally leaped up. His heart was racing. The creature next to the bed was white and spooky. It had an eye patch like Old Salt's, but seemed to be floating beside the bunk like a ghost. _Growlin' Grace's ghost!_
## Contents
1. Cover
2. Other Titles
3. Title Page
4. Copyright
5. Dedication
6. Contents
7. 1. Pirate Day Prep
8. 2. Sneezy Wheezy Captain Queasy
9. 3. Crack a Code
10. 4. Ring! Ding! Dong! Honk!
11. 5. Is It Ready Yet? Is It Ready Yet?
12. 6. Nummer Yummers
13. 7. Old Salt Says
14. 8. Sneaky Shadows
15. 9. Party Like a Pirate Day
16. 10. Puggy Piñatas
17. Excerpt from Ghost Ship
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1. Cover
2. Cover
3. Title Page
4. Table of Contents
5. Start
|
{
"redpajama_set_name": "RedPajamaBook"
}
| 3,299
|
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Backbone Sample URLManager</title>
<link rel="stylesheet" href="lib/bootstrap.min.css">
</head>
<body>
<div class="container">
<h1>URL Manager</h1>
<hr/>
<div class = "page"></div>
</div>
<script type="text/template" id="url-add-template">
<form class="edit-url-form">
<legend>Add new URL</legend>
<!--<label></label>-->
<input type="text" name="urlname" placeholder="Enter a url ">
<!--<label></label>-->
<input type="text" name="description" placeholder="Enter some description">
<hr/>
<button type="submit" class="btn">Add URL</button>
</form>
</table>
</script>
<script type="text/template" id="url-list-template">
<a href="#/newurl" class="btn btn-primary ">Add URL</a>
<hr>
<table class = "table table-striped ">
<thead>
<th>URL Name</th>
<th>Description</th>
<th></th>
</thead>
<tbody>
<% _.each(urls,function(url) { %>
<tr>
<td><a href="<%='http://'%><%= url.get('urlname') %>" class="btn" target="_blank" title="Opens in a new tab" ><%= url.get('urlname') %></a> </td>
<td><p class=""><%= url.get('description') %></td>
<td></td>
</tr>
<% }); %>
</tbody>
</table>
</script>
<script src="lib/jquery-1.10.2.min.js"></script>
<script src="lib/underscore-min.js" ></script>
<script src="lib/backbone-min.js"></script>
<script src="lib/backbone.localStorage.js"></script>
<script src="lib/helper.js"></script>
<!--Backbone scripts-->
<script src="backbone/models/url_model.js"></script>
<script src="backbone/collections/url_collection.js"></script>
<script src="backbone/views/URLView.js"></script>
<script src="backbone/views/ADDURL.js"></script>
<script src="backbone/router/route.js"></script>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 4,011
|
var TreatmentsTranslationCollectionItemView = Backbone.View.extend({
tagName: 'tr',
className: 'collection-item',
initialize: function(options) {
// list of specializations page in translator module
this.indexUrl = options.indexUrl;
},
render: function(){
var tplData = {
context: {
name: this.model.get('name'),
description: this.model.get('description')
}
}
var prototype = ich.translation_view_prototype(tplData);
// set link of view profile button
prototype.find('a.view-item-trigger').attr('href', this.indexUrl+'/'+this.model.get('id'));
this.$el.html(prototype.html());
},
});
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,237
|
Q: how can i change the struct in C array? i have a problem changing the struct that is an array.
I'm working on a library project in C language that need to add a book (Book struct) into a library, I have an array that has all my books, I need to add to this array, my new book.
i did this, could anyone help me , and give me a little bit information about it ?
#include <stdio.h>
#include <string.h>
#define BOOK_NUM 50
#define NAME_LENGTH 200
#define AUTHOR_NAME_LENGTH 100
#define PUBLISHER_NAME_LENGHT 50
#define GENRE_LENGHT 50
typedef struct _Book
{
char name[NAME_LENGTH];
char author[AUTHOR_NAME_LENGTH];
char publisher[PUBLISHER_NAME_LENGHT];
char genre[GENRE_LENGHT];
int year;
int num_pages;
int copies;
}Book;
Book books_arr[BOOK_NUM],*ptr=books_arr;
void add_book()
{
char book_name[NAME_LENGTH],author_name[AUTHOR_NAME_LENGTH],publisher_name[PUBLISHER_NAME_LENGHT],book_genre[GENRE_LENGHT];
int book_year,book_pages,book_copies,cnt=0,cnt2=0;
printf("Please enter book name:\n");
scanf("%s",&book_name);
printf("Please enter author name:\n");
scanf("%s",&author_name);
printf("Please enter publisher name:\n");
scanf("%s",&publisher_name);
printf("Please enter book genre:\n");
scanf("%s",&book_genre);
printf("Please enter the year of publishment:\n");
scanf("%d",&book_year);
printf("Please enter the number of pages:\n");
scanf("%d",&book_pages);
printf("Please enter the number of copies:\n");
scanf("%d",&book_copies);
for (ptr=books_arr;ptr<&books_arr[BOOK_NUM];ptr++)
{
if (strcmp(book_name,(*ptr).name)==0)
(*ptr).copies=(*ptr).copies+book_copies;
if(strcmp(book_name,(*ptr).name)!=0)
cnt++;
if((*ptr).name!=NULL)
cnt2++;
}
if(cnt==BOOK_NUM)
{
if(cnt2==BOOK_NUM)
printf("There is no place in the library for this book\n");
if(cnt2<BOOK_NUM)
{
(*ptr).name=book_name;
(*ptr).author=author_name;
(*ptr).publisher=publisher_name;
(*ptr).genre=book_genre;
(*ptr).year=book_year;
(*ptr).num_pages=book_pages;
(*ptr).copies=book_copies;
}
}
}
every time i compile the code, i have the problem, "expression must be a modifiable lvalue".
thank you
A: Instead of doing this
(*ptr).name=book_name;
You should copy a string using strcpy like
strcpy((*ptr).name,book_name);
Also, consider using other safe functions like strncpy().
Note use ptr->name instead of (*ptr).name.
The ptr->name is an array, you cannot change array but you can change contents of an array, which is what the error is suggesting.
A: You're attempting to assign an array to another array. That's not legal in C.
If you want to copy the contents of a character array containing a string to another character array, use strcpy.
strcpy(ptr->name, book_name);
strcpy(ptr->author, author_name);
strcpy(ptr->publisher, publisher_name);
strcpy(ptr->genre, book_genre);
Also note the use of the -> operator to access a pointer-to-member.
Besides this, you're also not reading these 4 string correctly. The %s format specifier to scanf expects a pointer to the first element in a char array. What you're doing is passing in the address of the array itself.
For this, just pass in the name of the array. The array name decays into a pointer to the first element when passed to a function.
scanf("%s",book_name);
printf("Please enter author name:\n");
scanf("%s",author_name);
printf("Please enter publisher name:\n");
scanf("%s",publisher_name);
printf("Please enter book genre:\n");
scanf("%s",book_genre);
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 8,638
|
Shipwreck Discoveries
W.C. KIMBALL
W.S. LYONS
THOMAS KINGSFORD
The RESCUE of Glen Lake
Sleeping Bear Point Shipwrecks
Skillagalee Shipwrecks
Still Missing
Missing Vessels
1871 C.H. Hurd
1873 GILBERT MOLLISON
1885 JARVIS LORD
1892 W.H. GILCHER
1947 LCM NORTH MANITOU
1959 BRIDGEBUILDER X
Missing Aircraft
1946 Boeing PT 17 Stearman
1950 Northwest Airlines DC4
1977 The Block's Cessna 150J
1993 Piper Cherokee Arrow
1998 Aero Vodochody L-39C
1999 Cessna 175
2007 Socata TB-20 Trinidad
1966 Three Sunbathers disappear from Beach
1969 Robert Richard Lepsy
1979 Roger Heacock
1981 Deannie Peters
1990 Richard Hitchcock
1991 Gordon Page
2010 Jacob Cabinaw
Dr. James P. Smith
1983 David Gionet
1998 David Van OPynen
2005 David Schmid
Speaking Dates
Welcome to "Michigan Mysteries.com", the website of Author and Shipwreck Hunter Ross Richardson. This website is dedicated to the missing persons, missing aircraft, missing maritime vessels and the unrecovered drowning victims of the Michigan Region. The Michigan Region includes the Upper and Lower Peninsulas, Islands, as well as the Great Lakes which are contiguous with Michigan's mainland: Lake Erie, Lake Huron, Lake Michigan and Lake Superior.
Richardson is the author of two books, "The Search for the Westmoreland: Lake Michigan's Treasure Shipwreck" and "Still Missing: Rethinking the D.B. Cooper Case and Other Mysterious Unsolved Disappearances." He also speaks about Great Lakes maritime history at libraries and historical societies around the Great Lakes Region.
Richardson is a trained technical SCUBA diver and has been involved in the discovery, identification, and documentation of numerous Lake Michigan shipwrecks. He was a public safety diver for the Benzie Area Public Safety Dive Team and a Benzie Country Sheriff's Department Special Deputy for many years.
He resides in Lake Ann, Michigan, with his wife, son and their retired racing greyhound, Claire.
WELCOME TO THE GOLDEN AGE OF SHIPWRECK HUNTING
BETTER, MORE AFFORDABLE TECHNOLOGY HAS MADE SHIPWRECK SEARCHES A REGULAR GUY'S GAME. SHIPWRECK HUNTER ROSS RICHARDSON IS A CASE IN POINT.BY PATRICK SULLIVAN | NOV. 30, 2019 Read the entire article here: WELCOME TO THE GOLDEN AGE OF SHIPWRECK HUNTING
LOST AND FOUND: 1891 shipwreck 'amazingly intact' on Lake Michigan bottom
NORTHPORT, Mich. — The Great Lakes have been sailed upon since the 17th century. Over the last 400 years, it's estimated that 6,000 vessels and 30,000 lives have been lost traversing these fresh waterways.The most recent maritime tragedy was 44…
Upnorthlive.com: Shipwreck hunter makes unprecedented discovery hidden in the depths
Few people know the waters of the Manitou Passage better than author and shipwreck hunter Ross Richardson… Read the entire article here: Shipwreck hunter makes unprecedented discovery hidden in the depths
Mlive.com: Lake Michigan Shipwreck W.C. KIMBALL
LAKE MICHIGAN – A shipwreck found almost by accident, sitting 300 feet deep in northern Lake Michigan, is being described as one of the "most intact wooden schooners" in the world. Read the entire article here: Lake Michigan shipwreck could…
WZZM-13: 'I think it's a death car': Mysterious 1960 Lincoln may have gruesome backstory
A few years ago, Lucas contacted a man by the name of Ross Richardson, who is an author and has investigated unsolved mysteries, shipwrecks and disappearances in Michigan. "It's definitely a curiosity," said Richardson, who wrote a book about Michigan mysteries in 2014….
Detroit News: Diver discovers shipwreck cluster
"I was looking for wrecks that haven't been discovered and have a decent last known position," said Richardson, who's originally from Grand Rapids. "This is kind of the last place in Lake Michigan where there's a concentration of wrecks that…
Could missing man from Grayling, Mich. be D.B. Cooper?
Robert Richard Lepsy was a grocery store manager, husband and father of four. On October 29, 1969, Lepsy mysteriously disappeared from Grayling and, to this day, hasn't been seen or heard from since. "On the day he disappeared, he called…
mlive.com: On remote Lake Michigan shoal, diver finds undiscovered shipwreck cluster
Filming the wrecks, he said, is a way to share the history and experience. The wrecks, he said, "are right here in our backyard." "You only have to travel a few hours to come across a place that people haven't…
Macomb Daily News: Michigan's mysterious treasure ship, the Westmoreland, found after 156 years
"… advances in diving and sonar technology certainly gave Richardson advantages over all of the expeditions before him, but ending up in the right place at the right time and dogged efforts to find the wreck certainly helped his cause."…
mlive.com: Gold and whiskey cargo fuels legend of the Westmoreland shipwreck
"…until 155 years after the sinking, when a diver and shipwreck sleuth from Grand Rapids would find what others could not; the wreck of the Westmoreland sitting upright on the lake bed, 200 feet under the surface of a bay…
Copyright 2019 Ross Richardson - All Rights Reserved
|
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"redpajama_set_name": "RedPajamaCommonCrawl"
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| 4,537
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{"url":"https:\/\/havenprotocoltooling.com\/","text":"# What is this and how does it serve me?\n\nThis tools charts the 24 hour moving average price (MA) of XHV. The graph shows the real MA for the past 24 hours and forecasts the MA 24 hours into the future. This tool can be used to make intelligent :LeoSmug: decisions regarding onshoring\/offshoring by showing the general trend of the MA for the next day.\n\n# How is the MA calculated?\n\nThe MA is calculated by taking the weighted average of Chainlink data for the last 24 hours. Each Chainlink datapoint has a price and a timestamp associated with the price. Finding all prices for the past 24 hours and weighing them by the time until the next datapoint produces the 24 hour weighted MA.\n\n# Can this really predict the future?\n\nNo it can't. Think of this as a weather forecast - never telling the future, only making educated guesses based on relevant data.\n\n# How is the MA calculated in the future?\n\nWe can predict future prices as those future prices take presently available data into account when being computed. For instance, the MA in 1 hour time will be calculated using the past 23 hours and the next 1 hour of Chainlink data. We have the past 23 hours so can make a fairly accurate prediction.\n\n# The Chainlink price has been a while for now, what gives?\n\nChainlink provides a (price & timestamp) pair but each datapoint can be far apart. The conditions for an updated Chainlink price are 1 hour since the last data point, or a price deviation (of ?%) from the last data point.","date":"2022-01-24 07:36:57","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\": 0.505308210849762, \"perplexity\": 1074.2243074941487}, \"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\/1642320304515.74\/warc\/CC-MAIN-20220124054039-20220124084039-00144.warc.gz\"}"}
| null | null |
Q: Exclude Less Directory I am using gruntjs to build my src and I want to exclude css/less/*.less from my published folder.
Should I be using the exclude list or is this an issue with the copy task?
A: Create a .buildignore file and add css/less
|
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"redpajama_set_name": "RedPajamaStackExchange"
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| 8,953
|
Q: How do I calculate the transfer function of a filter in MATLAB? I have to do some exercises in a Digital Signal Processing course and I have some problems.
I have a given file (signal.wav name the signal x(n) ) with some noise added and iIam asked to find some information from it. The noise added is η(n) = sin8000πn. So the signal I am processing is s(n) = x(n) + η(n)
In order to remove the noise, I apply a low-pass butterworth filter with order = 2 and cutoff frequency = 2000hz.
I want to find the transfer function H(s) from the filter and the H(z) function. Well I know that I have to apply bilinear transformation but iIdont know how to do it with MATLAB.
Can anyone help me solve this?
Here is my code
[y, fs, nbits] = wavread('signal.wav');
% Playing the file
disp('-> Playing at the original sample rate...');
sound(y, fs);
fprintf('------------------------------------------\n');
% Sampling frequency
fprintf('-> Sample frequency is: %f.\n', fs);
% Print the min and max values of the audio data.
fprintf('-> The maximum data value is %f.\n', max(y));
fprintf('-> The minimum data value is %f.\n', min(y));
fprintf('------------------------------------------\n');
order = 2;
sampling_freq = fs;
cut_off_freq = 2000;
[butter_a, butter_b] = butter(order,cut_off_freq/(sampling_freq/2));
%[butter_a, butter_b] = butter(order,cut_off_freq/(sampling_freq));
subplot(211), plot(y);
subplot(212), plot(filter(butter_a,butter_b,y));
sound(filter(butter_a,butter_b,y),fs);
A: You should use freqs to calculate the frequency response/transfer function of your analog filter (i.e., H(s)). So in this case, something like:
freqs(butter_b,butter_a,200);
will plot the frequency and phase response for the filter at 200 frequency points. You can also provide a vector of points where it should be calculated (see the linked doc).
For the transfer function of the digital filter (i.e., H(z)), use freqz. So your syntax would be something like:
freqz(butter_b,butter_a,[],fs)
which will again plot the frequency and phase responses as before. Again, make sure you read the linked documentation to understand and use it correctly.
A: Since you know the frequency of the unwanted component of the signal you need only to estimate the phase and amplitude and then subtract the added noise component from your signal in the time domain to get the solution.
There are many ways of estimating the phase and amplitude of a sine. I suggest you try the Goertzel algorithm for that.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,692
|
An inquiring reader wanted to know: What's afoot at the Ann Taylor store at Blakeney, on Rea Road in south Charlotte? The answer: It's converting to an Ann Taylor Loft store, the lower-priced, more casual sister to the regular Ann Taylor nameplate. The Ann Taylor store closed April 10, and the new Loft is holding a grand opening today, a spokeswoman said.
The company has not yet responded to a request for more information about why the switch occurred, but it's possible that Loft's mix and pricing offers more appeal in a challenged economy.
Regular Ann Taylor stores remain at SouthPark and Northlake malls in Charlotte.
Labels: Stores closing, Stores opening
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 2,088
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These certifications are fairly quick and easy. You would appreciate something that is at-home self paced study. This certification like E20-575 online cbt will enhance your value as employee and student. Find the right certification program with passguide website. It will provide you short, but focused tools like online E20-575 lab questions and E20-575 latest classroom training You can get notable products from passguide to increase the latest EMC EMCSA RecoverPoint E20-575 video lectures success chances. You will pack your mind with vital materials that will help you a lot. It is celebrative way to pass the exam with latest RecoverPoint Specialist for Storage Administrators test guide and EMC EMCSA RecoverPoint E20-575 online testing engine for the guidance of learners. Your choice should be latest EMCSA RecoverPoint E20-575 EMC audio study guide and E20-575 updated prep guide will prove really supportive for the online EMC E20-575 EMCSA RecoverPoint video training. This will be a vital pick for the learners therefore you all should take this pick for the learners and you will increase the success odds on right time for your assistance. passguide has validated products that will increase the odds of success that are really vital. You have to take the interesting products and tools for the right learning of the EMC EMCSA RecoverPoint E20-575 updated audio lectures. This is best and essential platform for the learners with E20-575 latest audio study guide and EMC E20-575 latest cbt.
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"redpajama_set_name": "RedPajamaC4"
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| 5,763
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\section{Introduction}
It is thought that the Hypercompact or Ultracompact HII (HCHII or
UCHII) regions trace early stages of massive star formation after the
hot molecular core phase. However, the physical processes involved in
this transition are still poorly understood \citep[see for
reviews:][]{stan2005,liz2008}. The appearance of one HCHII begins
when the Lyman continuum output from a massive young star becomes
sufficient high to ionize its surroundings. These objects show a
rising continuum spectra, $S_\nu \propto \nu^\alpha$, with a slope
$\alpha$ $\sim$ 1, intermediate between the optically thick and thin
limits, small sizes ($\sim$ 0.01 pc), high temperatures (10$^4$ K),
high densities ($\geq$ 10$^6$ cm$^{-3}$), and very broad radio
recombination lines ($\Delta\nu_{FWHM}$ $\geq$ 40 km s$^{-1}$)
indicating both pressure broadening and the presence of bulk motions
of dense gas \citep[e.g.][]{Sewi2004}. On the other hand, the hot
molecular cores internally heated by massive protostars are small
($\sim$ 0.05 pc) and compact regions of dense molecular gas ($\geq$
10$^6$ cm$^{-3}$) which show very rich millimeter and submillimeter
spectra \citep[see][for reviews on this
topic]{stan2000,van2004,Cesa2005a}. The hot molecular core phase
begins at much colder temperatures ($\sim$ 100 K) than the HC/UCHII
stage.
There are many cases where the UCHII or even the HCHII regions are
observed in association with hot cores, some of them are: G5.89-0.39
\citep{su2009}; G24 A1 \citep{Bel2006}; G10.47, and G31.41+0.31
\citep{ces2010,Rol2009}; W51e2, W51e8, NGC 7538 IRS1, G28.2 and G10.6
\citep{kla2009}; IRAS 17233-3606 \citep{leu2008}. However, there are
some others cases where the hot cores do not show the presence of a
strong HCHII \citep[e.g. AFGL490, NGC 7538S, and IRAS
20126;][]{naka1991,sch2002,cesa2005,san2010} possibly because the
infalling molecular gas partially quenches its formation
\citep[e.g.][]{Wal1995,oso1999} or maybe because the massive central
protostar(s) are/is in an early phase with not high enough
temperatures to ionize their/its surroundings.
W51 North is one of the youngest massive stars within the luminous
cluster W51-IRS2 that is still on the process of formation. This
object is associated with a strong mm and submm source, a hot
molecular core at an excitation temperature of $\sim$ 200 K
\citep{Zhangetal1998}, no centimeter free-free emission at all ($\leq$
5 mJy at 1.3 cm) that could be related with an HC/UCHII or a thermal
jet \citep{Gaumeetal1993, Zhangetal1998, Eisneretal2002}. From this
object emanates a powerful molecular outflow observed at very small
scales by masers spots of SiO and H$_2$O, and at large scales in
thermal SiO($J=5-4$) \citep{Schnepsetal1981, Eisneretal2002,
Imaietal2002, zapataetal2009}. Recently, \citet{zapataetal2009}
using high angular 1.3 and 0.7 mm continuum, and SO$_2$ line
observations made with the Submillimeter Array (SMA) and the Very
Large Array (VLA) resolved the large dusty and molecular hot core in a
rotating Keplerian ring with a size of $\sim$ 8000 AU \citep[at a
distance of $\sim$ 6 kpc:] []{Imaietal2002,barba2008,xu2009}. This
structure surrounds a compact central dusty circumstellar disk with a
smaller size of approximately 3000 AU, and where the molecular
northwest(redshifted)-southeast(blueshifted) bipolar outflow
emerges. However, the relationship between both circumstellar
structures was not clear.
In this paper, we present sensitive high angular resolution SMA
submillimeter and millimeter line observations of the W51 North region
that were made in an attempt to understand the nature of the
circumstellar molecular and dusty structures associated with this
young massive protostar. In Section 2 discuss the observations
undertook in this study. In Section 3, we present and discuss our SMA
millimeter and submillimeter data. Finally, in Section 4, we give the
main conclusions of the observations presented here.
\begin{table*}[ht]
\begin{minipage}[t]{\columnwidth}
\scriptsize \renewcommand{\footnoterule}{}
\caption{Observational and physical parameters of the submillimeter
and millimeter lines}
\begin{center}
\begin{tabular}{lccccccc}
\hline \hline & Rest frequency & E$_{lower}$ & Range of Velocities &
Linewidth & LSR Velocity\footnote{The linewidth and LSR velocity were
obtained fitting a Gaussian to the spectra.} & Peak Flux \\ Lines &
[GHz] & [K] & [km s$^{-1}$] & [km s$^{-1}$] & [km s$^{-1}$] & [Jy
Beam$^{-1}$] \\ \hline \hline C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$)
& 216.8121...& 715 & $+$54,$+$67 & 7 & 59 &
0.1\\ HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) & 216.7585...& 1476 &
$+$54,$+$66 & 8 & 59 & 0.4\\ CO($J=3-2$)\footnote{Spectra with
multiple peaks and in absorption about a LSR velocity of 60 km
s$^{-1}$.} & 345.7959... & 16 & $+$20,$+$120 & 50 & 60 & 3.0
\\ SO(8$_{9}$-7$_8$)\footnote{Spectra with two peaks.} &
346.5284... & 21 & $+$30,$+$90 & 35 & 59 &
1.0\\ SO$_2$(19$_{1,19}$-18$_{0,18}$) & 346.6521... & 151 &
$+$41,$+$72 & 15 & 58 & 1.0\\ HC$_3$N(38-37) ($\nu_t$=0) &
345.6090... & 306 & $+$51,$+$72 & 9 & 60 & 2.0
\\ HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$) & 337.3988... & 1512 &
$+$48,$+$73 & 10 & 62 & 2.0 \\ \hline \hline
\end{tabular}
\end{center}
\end{minipage}
\end{table*}
\section{Observations}
\subsection{Millimeter}
The observations were obtained with the SMA\footnote{The Submillimeter
Array (SMA) is a joint project between the Smithsonian Astrophysical
Observatory and the Academia Sinica Institute of Astronomy and
Astrophysics, and is funded by the Smithsonian Institution and the
Academia Sinica.} during 2008 January 17. The SMA was in its very
extended configuration, which included 28 independent baselines
ranging in projected length from 30 to 385 k$\lambda$. The phase
reference center of the observations was R.A. = 19h23m40.05s, decl.=
14$^\circ$31$'$05.0$''$ (J2000.0). The frequency was centered at
217.1049 GHz in the Lower Sideband (LSB), while the Upper Sideband
(USB) was centered at 228.1049 GHz. The primary beam of the SMA at 230
GHz has a FWHM of about 60$''$.
We detected the lines C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$) and
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) in the LSB. See Table 1 for
their rest frequencies and rotational temperatures above of the ground
state. The full bandwidth of the SMA correlator was 4 GHz (2 GHz in
each band). The SMA digital correlator was configured in 24 spectral
windows (``chunks'') of 104 MHz each, with 256 channels distributed
over each spectral window, providing a resolution of 0.40 MHz (0.56 km
s$^{-1}$) per channel. However, in this study we smoothed our spectral
resolution to about 1 km s$^{-1}$.
The zenith opacity ($\tau_{230 GHz}$), measured with the NRAO tipping
radiometer located at the Caltech Submillimeter Observatory, was
$\sim$ 0.15, indicating good weather conditions during the
observations. Observations of Uranus provided the absolute scale for
the flux density calibration. Phase and amplitude calibrators were
the quasars 1925+211 and 2035+109. The uncertainty in the flux scale
is estimated to be 15--20$\%$, based on the SMA monitoring of quasars.
Observations of Uranus provided the absolute scale for the flux
density calibration. Further technical descriptions of the SMA and
its calibration schemes can found in \citet{Hoetal2004}.
The data were calibrated using the IDL superset MIR, originally
developed for the Owens Valley Radio Observatory
\citep{Scovilleetal1993} and adapted for the SMA.\footnote{The MIR-IDL
cookbook by C. Qi can be found at
http://cfa-www.harvard.edu/$\sim$cqi/mircook.html} The calibrated
data were imaged and analyzed in the standard manner using the MIRIAD
and AIPS packages. We used the ROBUST parameter of the INVERT task of
MIRIAD set to 2 to obtain a slightly better sensitivity while losing
some angular resolution. The resulting image rms noise of line images
was around 30 mJy beam$^{-1}$ for each velocity channel (with a
smoothed size of 1 km s$^{-1}$) at an angular resolution of
$0\rlap.{''}57$ $\times$ $0\rlap.{''}42$ with a P.A. = $57.6^\circ$.
\begin{figure}[ht]
\begin{center}
\includegraphics[scale=0.4]{f1.eps}
\caption{\scriptsize Overlay of the SO$_2$(22$_{2,20}$-22$_{1,21}$)
moment zero emission (color image), the 7 mm continuum emission from
the compact dusty disk (brown contours), and the CO($J=3-2$) moment
zero emission from the bipolar outflow (blue and red contours). The
SO$_2$ and the 7 mm images were taken from
\citet{zapataetal2009}. The integrated velocity range for the
CO($J=3-2$) is from +20 to +50 km s$^{-1}$ for the blueshifted gas,
and from +70 to +120 km s$^{-1}$ for the redshifted gas. The brown
contours are the 10\%, 13\%, and 16\% of the peak of the 7 mm
continuum emission, the peak is 100 mJy beam$^{-1}$. The
synthesized beam of the CO($J=3-2$) contour image is 0.71$''$
$\times$ 0.57$''$ with a P.A. of 87.6$^\circ$ and is shown in the
bottom right corner. The green triangle marks the position of the
possible companion reported in this paper. The white dashed line
shows how the blueshifted side of the outflow seems to be more
likely energized from the central massive protostar W51 North rather
than the possible companion.}
\label{fig1}
\end{center}
\end{figure}
\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.31]{f2.eps}
\caption{\scriptsize Spectra from the selected millimeter and
submillimeter lines observed toward W51 North. The red line
represents in all panels the synthetic spectrum obtained for the
best fitting solution with the parameters shown in each panel. The
spectra was obtained from the average of the total emission in each
line using the task ``imspect'' of MIRIAD.}
\label{fig2}
\end{center}
\end{figure*}
\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.20]{f3.eps}
\caption{\scriptsize Composite images of the molecular emission from
the circumstellar large disk around the massive protostar W51 North.
Left: The blue color scale and the contours are representing the
integrated emission of the SO(8$_{9}$-7$_8$), green ones the
HC$_3$N(38-37) ($\nu_t$=0), and the red ones the
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$). The blue contours are from
5\% to 90\% with steps of 10\% of the peak of the line emission, the
peak is 53 Jy beam$^{-1}$ km s$^{-1}$. The green contours are from
30\% to 90\% with steps of 10\% of the peak of the line emission,
the peak is 26 Jy beam$^{-1}$ km s$^{-1}$. The red contours are
from 30\% to 90\% with steps of 10\% of the peak of the line
emission, the peak is 3.5 Jy beam$^{-1}$ km s$^{-1}$. The
synthesized beams of the millimeter and submillimeter observations
are shown on the right bottom corner. Right: The light blue color
scale and the contours are representing the integrated emission of
the SO$_2$(19$_{1,19}$-18$_{0,18}$), green ones the
C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$), and the orange ones the
HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$). The light blue contours are
from 30\% to 90\% with steps of 10\% of the peak of the line
emission, the peak is 38 Jy beam$^{-1}$ km s$^{-1}$. The green
contours are from 30\% to 90\% with steps of 10\% of the peak of the
line emission, the peak is 2 Jy beam$^{-1}$ km s$^{-1}$. The orange
contours are from 30\% to 90\% with steps of 10\% of the peak of the
line emission, the peak is 27 Jy beam$^{-1}$ km s$^{-1}$. The
synthesized beam of the three measurements is the same and is shown
on the right bottom corner. The triangles represent the position of
the two possible embedded massive protostars as suggested by the hot
molecular emission of the HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) and
HC$_3$N(38-37) ($\nu_t$=0). The white dashed line in both panels
represents the position and orientation where the PV-diagrams shown
in Figures 4 and 5 were computed. }
\label{fig3}
\end{center}
\end{figure*}
\begin{table*}[ht]
\begin{minipage}[t]{\columnwidth}
\scriptsize \renewcommand{\footnoterule}{}
\caption{Observational parameters of the circumstellar multilayer
disk}
\begin{center}
\begin{tabular}{lccccccc}
\hline \hline & \multicolumn{2}{c}{Central Position} & Flux Density &
\multicolumn{2}{c}{Deconvolved Size\footnote{The Gaussian fitting was
obtained with the task JMFIT of AIPS.}} & Approx. cavity
size\\ \hline & $\alpha(2000)$ & $\delta(2000)$ & & & & \\ Specie & [h
m s] & [$\circ$ $''$ $'$] & [Jy Beam$^{-1}$ km s$^{-1}$] &
[arcsec$^2$]& [Degrees] & [arcsec$^2$] \\ \hline \hline
SO(8$_{9}$-7$_8$) & 19 23 40.074 & 14 31 05.70 & 341$\pm$20 &
1.78$\pm$0.01 $\times$ 1.39$\pm$0.02 & 171$\pm$2 & $\sim$
0.8\\ SO$_2$(19$_{1,19}$-18$_{0,18}$) & 19 23 40.071 & 14 31 05.54 &
234$\pm$15 & 1.65$\pm$0.04 $\times$1.49$\pm$0.03 & 24$\pm$6& $\sim$
0.7\\ HC$_3$N(38-37) ($\nu_t$=0) & 19 23 40.060 & 14 31 05.54 &
88$\pm$15 & 1.36$\pm$0.04 $\times$0.93$\pm$0.04 &135$\pm$7 & $\sim$
0.6\\ C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$) & 19 23 40.041 & 14 31
05.49 & 116$\pm$15 & 1.31$\pm$0.03 $\times$0.89$\pm$0.03 & 107$\pm$7&
$\sim$ 0.8\\ HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$) & 19 23 40.054 &
14 31 05.44 & 105$\pm$13 & 1.14$\pm$0.03 $\times$0.88$\pm$0.03 &
99$\pm$3& --\\ HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) &19 23 40.062 &
14 31 05.59 & 9.1$\pm$2 & 0.81$\pm$0.03 $\times$ 0.40$\pm$0.09 &
75$\pm$3 & --\\ ---Companion & 19 23 40.077 & 14 31 05.63 & 8.5$\pm$1
& 1.27$\pm$0.04 $\times$0.39$\pm$0.03 & 74$\pm$2& --\\ \hline & & &
Dusty circumstellar disk\footnote{Data obtained from
\citet{zapataetal2009}. } & & & \\ 7 mm & 19 23 40.057 & 14 31 05.67
& -- & 0.58$\pm$0.02 $\times$0.27$\pm$0.02 & 70$\pm$6&
--\\ \hline\hline
\end{tabular}
\end{center}
\end{minipage}
\end{table*}
\subsection{Submillimeter}
The observations were obtained with the SMA on 2008 July 13. At the
time of these observations the SMA had seven antennas in its extended
configuration with baselines ranging in projected length from 30 to
255 k$\lambda$. The primary beam of the SMA at 340 GHz has a FWHM of
37$''$. The molecular emission from the hot core was found well inside
of the primary beam.
The receivers were tuned to a frequency of 345.796 GHz in the upper
sideband (USB), while the lower sideband (LSB) was centered on 335.796
GHz. The LSB contained line HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$)
while the USB the lines SO(8$_{9}$-7$_8$),
SO$_2$(19$_{1,19}$-18$_{0,18}$), CO($J=3-2$), and HC$_3$N(38-37)
($\nu_t$=0). See Table 1 for their rest frequencies and lower level
energies above the ground energy state. The SMA digital correlator
was configured in 24 spectral windows (``chunks'') of 104 MHz each,
with 128 channels distributed over each spectral window, thus
providing a spectral resolution of 0.81 MHz (0.70 km s$^{-1}$) per
channel. However, we smoothed our spectral resolution to 1.0 km
s$^{-1}$ per spectral channel.
The zenith opacity ($\tau_{230 GHz}$) was $\sim$ 0.035 -- 0.04,
indicating excellent weather conditions. Observations of titan
provided the absolute scale for the flux density calibration. Phase
and amplitude calibrators were the quasars 1751$+$096, and 1925$+$211.
The uncertainty in the flux scale is also estimated to be 15 --
20$\%$, also based on the SMA monitoring of the quasars.
The calibrated data were imaged and analyzed in standard manner using
the MIRIAD, and AIPS packages. We also used the ROBUST parameter set
to 2. The line image rms noise was around 100 mJy beam$^{-1}$ for
each velocity channel (with a smoothed size of 1 km s$^{-1}$) at an
angular resolution of $0\rlap.{''}78$ $\times$ $0\rlap.{''}58$ with a
P.A. = $84.7^\circ$.
\section{Results}
In Figure 1, we present the disk/ring/outflow system found towards W51
North using observations of the SiO($J=5-4$),
SO$_2$(22$_{2,20}$-22$_{1,21}$), and millimeter continuum by
\citet{zapataetal2009}. However, in this image, we have exchanged the
SiO emission by the emission of the CO($J=3-2$) presented here for the
first time. The CO($J=3-2$) confirms the presence of a single powerful
and collimated (15$^\circ$) bipolar outflow (with the blueshifted side
towards the southeast while the redshifted one to the north) emanating
from the disk/ring structure at similar scales. However, the
CO($J=3-2$) reveals that the outflow extents for a few arcseconds more
on both sides in comparison with the emission traced by SiO($J=5-4$),
and shows a slightly larger radial velocity range ($+$30 to $+$120 km
s$^{-1}$). In this image, it is more clear to see the deviation of
the blueshifted side of the outflow with respect to the redshifted one
than in \citet{zapataetal2009}.
Figure 2 shows the spectra of the multiple lines reported here with
exception of the CO($J=3-2$). Additionally, we present synthetic LTE
models for the spectra using XCLASS program
\citep[][]{Comitoetal2005}. The data were fitted well with the
synthetic model using rotational temperatures, column densities, and
sizes of 250 K, 2 $\times$ 10$^{16}$ cm$^2$, and 1$''$, respectively,
for the lines SO(8$_{9}$-7$_8$), SO$_2$(19$_{1,19}$-18$_{0,18}$),
HC$_3$N(38-37) ($\nu_t$=0), and C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$)
and 800 K, 5 $\times$ 10$^{18}$ cm$^2$, and 0.5$''$ for transitions
with high excitation temperatures in the lower energy states,
HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$) and
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$). This modeling serves as for
avoiding the line confusion, and suggest that hot gas is only found in
the innermost parts of the hot core. See Table 1 for the rest
frequencies and the physical parameters of the lines. The physical
parameters of all lines were obtained fitting a Gaussian to the line
profile.
In order to be more confident about the detection of the
glycol-aldehyde (HCOCH$_2$OH) toward W51 North, we have fitted
simultaneously many other lines that fall in our 2 GHz bands with our
synthetic LTE model using the same parameters as above. We obtained a
reasonable agreement between the synthetic model and the detected
lines. A full astrochemical analysis of these lines (and other
species) is beyond the scope of the present paper in which we
concentrate on the spatial distributions of the lines presented in
Table 1.
The spectra of the SO(8$_{9}$-7$_8$) additionally shows
self-absorption possibly due to the molecular gas is falling into the
central object \citep[][]{Zapataetal2008}.
\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.37, angle=0]{f4.eps}
\caption{\scriptsize Position-velocity diagrams of the HC$_3$N(38-37)
(green scale color), and the HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$)
(pink scale color) line emission from the circumstellar disk. The PV
diagrams were computed at a P.A. = 30$^\circ$ (see Figure 3). These
diagrams are additionally overlaid with the position-velocity
diagram at same P.A. of our Keplerian disk model (contours). The
contours are from 20\% to 90\% with steps of 10\% of the peak of the
line emission of our model. The units of the horizontal axis are in
arcseconds. The systemic LSR radial velocity of the ambient
molecular cloud is about 60 km s$^{-1}$, see Table 1. The angular
and spectral resolutions are shown in each panel on the bottom left
corner. The synthesized beam of both images is 0.71$''$ $\times$
0.57$''$ with a P.A. of 87.6$^\circ$. The spectral resolution was
smoothed to 1 km s$^{-1}$. The origin in the horizontal axis is at
R.A. = 19h23m40.05s, decl.= 14$^\circ$31$'$05.0$''$ (J2000.0).}
\label{fig4}
\end{center}
\end{figure*}
\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.4]{f5.eps}\\
\caption{\scriptsize Position-velocity diagrams for the
SO(8$_{9}$-7$_8$) and the SO$_2$(19$_{1,19}$-18$_{0,18}$) from the
large circumstellar disk and the outflow. They were computed at a
P.A. of 160$^\circ$ (see Figure 3). The contours in both panels are
from 10\% to 90\% with steps of 10\% of the peak of the line
emission. The units of the horizontal axis are in arcseconds. The
systemic LSR radial velocity of the ambient molecular cloud is about
60 km s$^{-1}$, see Table 1. The angular and spectral resolutions
are shown in each panel on the bottom left corner. The synthesized
beam of both images is 0.71$''$ $\times$ 0.57$''$ with a P.A. of
87.6$^\circ$. The spectral resolution was smoothed to 1 km
s$^{-1}$. The origin in the horizontal axis is at R.A. =
19h23m40.05s, decl.=14$^\circ$31$'$05.0$''$ (J2000.0).}
\end{center}
\label{fig5}
\end{figure*}
The integrated intensity or moment zero maps of the lines associated
with the molecular disk are presented in Figure 3. We have divided
these lines into three groups showing the transitions with energy
levels with moderate (20-160 K), high (300-800 K), and very high
(1400-1600 K) energies and presented them in two different panels.
From these images, it is easy to see how the the molecular emission is
found in layers with the transitions characterized by high excitation
temperatures in their lower energy states (up to 1512 K), and/or
critical densities, being concentrated closer to the central high-mass
protostar W51 North. This phenomenon confirms clearly that the
heating in W51 North is internal. Additionally, molecules with
transitions characterized by low and moderate excitation temperatures
{\it i.e.} SO(8$_{9}$-7$_8$), SO$_2$(19$_{1,19}$-18$_{0,18}$),
HC$_3$N(38-37) ($\nu_t$=0), and C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$)
show a central cavity of approximately an angular size of 0.7$''$, see
Table 2. These lines trace thus the molecular Keplerian ring (or
toroid) around W51 North reported in SO$_2$ by
\citet[][]{zapataetal2009}, see Figure 1. The transition of the
molecule SO$_2$ utilized by these authors has a similar excitation
temperature in the lower energy state as those mentioned above. Note
the similarity between the angular size of the inner cavity and the
size of the compact dusty circumstellar disk reported by
\citet[][]{zapataetal2009}, see Table 2.
The molecular emission from the transition with a high excitation
temperature in the lower energy states ({\it i.e.}
HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$)) is found peaking at the
position of the cavity and showing similar angular sizes to this, see
Table 2. This molecule is thus more intimately related with the
circumstellar compact dusty disk.
The observational parameters of all lines are shown in Table 2. We
noted from Table 1 and 2, there is a correlation between the
deconvolved sizes of the molecular emission and the excitation
temperatures in lower energy states of each molecular specie. The
molecular emission from transitions characterized by high excitation
temperatures show to be very compact. This correlation could be
obviously obtained for lines that are optically thick with T$_b$
$\sim$ $\eta$ T$_{ex}$, where the T$_b$ is the brightness temperature,
$\eta$ is the filling factor, and T$_{ex}$ is the excitation
temperature.
In the left panel of Figure 3 the emission of the lines
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) and HC$_3$N(38-37) ($\nu_t$=0)
reveals the presence of a possible warm ``companion'' located to the
northeast of the disk. We have marked the position of this putative
companion with a yellow triangle. This ``companion'' is also observed
in SO$_2$ at the same position, see Figure 3 of
\citet[][]{zapataetal2009}, and is associated with a group of water
maser spots \citep{Imaietal2002,Eisneretal2002}. The position of the
``companion'' was found by fitting a Guassian to its
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$)) emission. However, it still
not clear if the ``companion'' is real protostar or if this could be
the result of the interaction of the outflow with a high density zone
of the molecular cloud.
We do not find any clear evidence of outflowing gas activity
associated with this possible ``companion'' \citep[see
also][]{zapataetal2009}. Although the blueshifted side of the
CO($J=3-2$) outflow has not the same position angle as the redshifted
one, this seems not be ejected from the companion. We show how the
blueshifted side appears to be more likely ejected from W51 North in
Figure 1. We drew a line that crosses this side of the outflow and
points directly to the dusty compact disk. The SO$_2$ and SiO
emission indeed show how the outflow is deviated to where the
CO($J=3-2$) is located \citep[][]{zapataetal2009}. However, more
observations in some other molecular outflow tracers are thus
necessity to firmly discard the existence of a second outflow in W51
North. It is interesting to note that the possible companion is not
observed in the hotter molecular gas tracer
HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$) suggesting that it may not as
warm as the central massive star.
Figures 4 and 5 show the position-velocity diagrams (PV-diagrams) of
different molecules which trace distinct scales of the disk. In Figure
3, we have marked the orientation and position of the PV-cuts. In the
left panel of Figure 3 we show a white line with a P.A.=30$^\circ$
that corresponds to the PV-cuts shown in Figure 4. On the other hand,
the white line with a P.A.=160$^\circ$ in the right panel corresponds
to the the PV-cuts shown in Figure 5.
Figure 4 shows the molecular emission from HC$_3$N(38-37) and
HCOCH$_2$OH(68$_{19,49}$-67$_{20,48}$) located in the innermost parts
of the disk, while the PV-diagrams of molecules as SO(8$_{9}$-7$_8$),
and SO$_2$(19$_{1,19}$-18$_{0,18}$), which trace the outermost parts,
are presented in Figure 5. The PV-diagrams in Figure 4 reveal that the
hot molecular gas closer to the massive protostar is Keplerian. In
addition, in this figure we have overlaid the PV-diagram of the LTE
Keplerian disk modeled in \citet[][]{zapataetal2009}, but without an
inner cavity and a smaller size. Both structures shown a very good
correspondence. The SO(8$_{9}$-7$_8$) and
SO$_2$(19$_{1,19}$-18$_{0,18}$) presented in Figure 5 trace much
larger structures similar to the ring reported in
\citet[][]{zapataetal2009}. These molecules in addition show clearly
two northwest and southeast high velocity extensions excited by the
bipolar outflow mapped in CO($J=3-2$) and SiO($J=5-4$). The
PV-diagram of the molecule C$_2$H$_5$OH(37$_{8,29}$-36$_{9,28}$) shows
also Keplerian motions, while that obtained from the molecule
HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$) shows a more compact structure
without a clear morphology, these diagrams are not presented in this
study.
Finally, in Figure 6 we show the spectral energy distribution (SED)
from the centimeter, millimeter, and submillimeter wavelengths of W51
North with data obtained from \citet[][]{zapataetal2009} and Tang et
al. (in prep.). The new value at submillimeter wavelengths presented
here shows also to be good fitted with the $\alpha$ = 2.8 value (where
the flux density goes as S$_\nu$=$\nu^\alpha$) found by
\citet[][]{zapataetal2009}.
\begin{figure}[h!]
\begin{center}
\includegraphics[scale=0.6]{f6.eps}\\
\caption{\scriptsize SED for the source W51 North combining 3.6 cm,
1.3 cm, 7 mm, 1.3 mm and 0.8 mm continuum data. The millimeter data
were obtained from \citet{zapataetal2009} and the submillimeter data
was obtained from Tang et al. (in prep.). The respective error bars
were smaller than the squares and are not presented here. The line
is a least-squares power-law fit (of the form S$_\nu$ $\propto$
$\nu^\alpha$) to the spectrum. Note that the value of $\alpha$=2.8
obtained by \citet{zapataetal2009} fit very well all millimeter and
submillimeter measurements. The 1.3 cm point is a upper limit
(5$\sigma$=5 mJy) obtained from Figure 1 of
\citet{Eisneretal2002}. The 3.6 cm point is a upper limit
(5$\sigma$=4 mJy) obtained from Figure 5 of \citet{meh1994}.}
\end{center}
\label{fig6}
\end{figure}
\begin{figure*}[ht]
\begin{center}
\includegraphics[scale=0.75, angle=0]{f7.eps}\\
\caption{\scriptsize A very simple diagram showing the different
evolutionary phases of a massive protostar. In this diagram, we do
not included multiplicity for simplicity. }
\end{center}
\label{fig7}
\end{figure*}
\section{Discussion}
\subsection{A multilayer infalling and keplerian disk around W51 North:
Connecting the molecular ring and dusty compact disk.}
The combined millimeter and submillimeter line observations presented
here from the ring and compact dusty disk structures reported in
\citet{zapataetal2009} suggest that we are seeing a single hot
multilayer accreting disk around W51 North (Figure 3). This large
molecular disk connects both circumstellar structures, the dusty
compact disk, observed at 7 and 1 mm, and the molecular ring mapped in
SO$_2$ in a single structure. Some molecules are found in association
with the dusty compact disk while some others with the ring.
The molecular emission associated with the multilayer disk reveals
that the molecular emission with transitions with a low excitation
temperature in the lower energy states delineates the outer disk and
shows a cavity, while such transitions with a high excitation
temperature in the lower energy state are found only in the innermost
parts of the disk and closer to the central massive protostar(s). One
would expect to this physical phenomenon happen in disks or cores
around high- and low-mass protostar(s) due to the large optical depths
and higher temperatures only found close to the young star. Up to date
detailed models of the ensuing chemical evolution at these much higher
temperatures do not at present exist.
It still is not clear why we do see an inner cavity on the disk in
some molecules with transitions characterized by low and moderate
excitation temperatures. This cavity maybe is not a real hole on the
disk rather this could be due to photo-dissociation or opacity effects
of these molecules toward the innermost parts of the disk. However,
we do not discard the possibility that the cavity could be real and
formed possibly by the tidal effects of the young multiple massive
protostars in the middle of the ``ring'' as in the case of Ori 139-409
located in Orion South where the protostars have lower masses
\citep{zap2010}. The molecular emission with transitions
characterized by high excitation temperatures in the lower energy
states may arise from the possible compact circumstellar disks inside
of the cavity that with the actual angular resolution are not
resolved. The presence of a binary system at the center of W51 North
has been suggested by the precession of the molecular outflow at very
small scales found by \citet{Eisneretal2002}. This precession
phenomenon may also explain the difference between the position angles
of the blueshifted and redshifted components of the CO($J=3-2$)
outflow (Figure 1).
The accreting disk hypothesis is supported by the detection of the
powerful outflow that emanates with an almost perpendicular
orientation to the object. The large disk shows to be Keplerian and
with a contracting velocity of about a few km s$^{-1}$ as revealed by
\citet[][]{zapataetal2009}. The models presented in Figure 4 were
obtained using a slightly different P.A. = 30$^\circ$, inclination
angle of {\it i}=25$^\circ$, a smaller size (3000 AU), and without an
inner cavity than in \citet[][]{zapataetal2009}.
In \citet[][]{zapataetal2009} and here, we have modeled a contracting
flattened disk in Keplerian rotation with a central hole, using the
disk parametrization from Guilloteau, Dutrey \& Simon (1999). The
contraction is assumed to have the functional form of free-fall
(i.e. V$_{inf}$ $\propto$ $\frac{1}{\sqrt r}$), with a reference
velocity at the reference radius. However, our resolution is
insufficient to determine the exact functional form. The model is for
the same molecules and transitions. A better fit than all other
trials in our recurrence was found until we obtained similar
structures in our model to those imaged (Figure 4). Most of the
physical parameters in the model were constrained in the process. The
model fits the observations reasonably well.
In particular, the molecules HCOCH$_2$OH(62$_{13,49}$-62$_{12,50}$)
and HC$_3$N(38-37) ($\nu_t$=0) revealed the possible presence of a
warm companion located to the northeast of the disk (Figure 2). We
suggest that the companion could be a consequence of disk
fragmentation due gravitational instabilities because the disk is
extremely large, see for a reference of this phenomenon
\citet{kra2006,kru2009}.
\subsection{A possible sequence on the evolution of nascent massive stars?}
The multilayer disk imaged in this paper reveals a possible link
between the Hyper/Ultra-compact HII regions and the hot cores. These
observations suggest a possible sequence on the formation of massive
stars. We present this simple sequence in Figure 7, and discuss it as
follows:
\begin{itemize}
\item
{\it Phase I:} In the first phase, a large and massive pseudo-disk is
formed together with a bipolar outflow from a large and dense core.
The pseudo-disk is surrounded by an infalling envelope with accretion
rates on the order of 10$^{-3}$ M$_\odot$ yr$^{-1}$. The molecular
emission arising from the circumstellar pseudo-disk could be observed
as a single structure without a well defined temperature gradient
across it. The pseudo-disks observed toward these objects are not
classical cicumstellar disks associated with low-mass stars which are
centrifugally supported, rather these might be contracting
pseudo-disks \citep[see for example the large and massive contracting
disks associated with the high-mass protostars AFGL490 and NGC
7538S:][] {naka1991,san2010}. The pseudo-disk could be also likely
circumbinary or circum-multiple because of the large multiplicity
presented on the massive stars.
\item
{\it Phase II:} The large and massive pseudo-disk formed earlier is
observed now in molecular shells or layers. This object is still
surrounded by an infalling envelope. A clear temperature gradient is
observed across the disk. The inner cavity could be formed by
photodissociation due to the high temperatures close to massive stars
or maybe to opacity effects. In this phase, no free-free emission
from a HCHII region is detected probably because the infalling
molecular gas partially quenches its formation or maybe because the
massive central protostar(s) are/is in an early phase with not high
enough temperatures to ionize their/its surroundings. W51 North is
found here. Another possibility is that maybe there is cluster of late
B-type in middle able of forming a strong multilayer molecular disk,
but not for developing HCHII regions.
\item
{\it Phase III:} A HCHII region appears in the middle of the large
multilayer pseudo-disk. At this stage the central massive protostar
has the sufficient high temperatures ($>$ 10$^4$ K) to ionize its
surroundings, see for example the cases of NGC7538 IRS1
\citep{fra2004,San2009} and MWC349 \citep{tafo2004}. These sources are
maybe variable on time because of the strong accretion from the large
disk continues toward the massive central star, see also G24.78+0.08
A1 \citep{gal2008}. There are some sources that might be situated at
this phase, see for example, the molecular rotating toroids observed
around the UCHII regions G24.78+0.08 A1 and G5.89-0.39
\citep{su2009,Bel2006}. Moreover, \citet{kla2009} found in a group of
HCHII and UCHII regions rotating molecular structures surrounding
them. Therefore, the molecular toroids or rings around an UCHII are
likely the same structures as in Phase II, but just hotter. The holes
observed in these toroids might be due to the photodissociation of
some molecules, or maybe to the formation of cavities by the presence
of a multiple system of massive stars as mentioned before. In this
phase the masive young star(s) might be still accreting ionized
material as proposed and described by \citet{Ke2006}.
\item
{\it Phase IV:} In this phase the HCHII region expands and ionizes the
remnant molecular disk or toroid, and creates an extended H II region
surrounding one or maybe multiple young massive stars. The sequence
from UCHII to extended HII regions has been discussed by
\citet{gar2004,fra2007}. If the central star(s) is/are still accreting
material, probably ionized material, an outflow show also be present
even at this stage.
\end{itemize}
\section{Summary}
We have observed the massive and young protostar W51 North with the
SMA using a high angular resolution and sensitivity at millimeter and
submillimeter wavelengths. We give a summary of the results as
follows:
\begin{itemize}
\item
We report the presence of a large and single molecular disk around the
object W51 north, and confirm the existence of a single powerful
outflow emanating from the disk;
\item
The molecular emission from the large disk is observed in layers with
transitions characterized by high excitation temperatures in the lower
states (up to 1512 K) being concentrated closer to the central massive
protostar. The molecular emission at low or moderate excitation
temperatures additionally exhibits a central cavity with an angular
size around 0.7$''$. This multilayer disk connects the molecular ring
and dusty compact disk reported recently toward this object
\citep[][]{zapataetal2009}. The LTE modeling presented in this study
also confirms that the hot gas is only found in the innermost disk.
\item
A detail study of the kinematics of the molecular gas together with a
model of a circumstellar disk in Local Thermodynamic Equilibrium shows
that the disk is Keplerian, and is accreting fresh gas to the
protostar possibly quenching the formation of a HCHII region;
\item
The thermal emission of the HCOCH$_2$OH reveals the possible presence
of a warm companion located to the northeast of the disk, perhaps
produced by disk fragmentation or maybe is the interaction of the
outflow a high density zone. The molecular emission of the SO and
SO$_2$ is observed in the circumstellar disk as well as in the
outflow;
\item
We modeled all lines with a LTE synthetic spectra. For an assumed
source size of 0.5 - 1 arcsec, our modeling yields a column density of
2 - 6 $\times$ 10$^{16-18}$ cm$^{-2}$, a temperature of 250 - 800 K,
and a linewidth of $\sim$ 5 - 10 km s$^{-1}$.
\end{itemize}
We thank the anonymous referee for many valuable suggestions. This
research has made extensive use of the SIMBAD database, operated at
CDS, Strasbourg, France, and NASA's Astrophysics Data System. This
research made use of the myXCLASS program
(https://www.astro.uni-koeln.de/projects/schilke/XCLASS), which
accesses the CDMS (http://www.cdms.de) and JPL
(http://spec.jpl.nasa.gov) molecular data bases.
\bibliographystyle{aa}
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O Brunei (; ), oficialmente Estado de Brunei Darussalam ou Darussalã (Malaio: Negara Brunei Darussalam, Jawi: نڬارا بروني دارالسلام), é um estado soberano localizado na costa norte da ilha de Bornéu, no Sudeste Asiático. Além de seu litoral com o mar da China Meridional, é completamente cercado pelo estado de Sarauaque, na Malásia, e é dividido em duas partes pelo distrito de Sarauaque, Limbang. É o único estado soberano situado completamente na ilha de Bornéu, com o restante da ilha formando partes da Malásia e Indonésia. A população de Brunei era estimada em 423,196 em 2016.
As reivindicações oficiais da história nacional do Brunei podem traçar suas origens ao , quando era um estado sujeito chamado P'o-li, na Sumatra, centro do Império Serivijaia. Ele mais tarde se tornou um Estado vassalo de Java, centro do império Majapait. Brunei se tornou um sultanato, no , sob o recém-convertido sultão islâmico Xá Maomé.
No auge do Império do Brunei, o sultão Bolkiah (reinando de 1485-1528) tinha controle sobre as regiões do norte de Bornéu, incluindo a moderna Sarauaque e Sabá, bem como o arquipélago de Sulu ao largo da ponta nordeste de Bornéu, Seludong (hoje a Manila moderna) e as ilhas ao largo da ponta noroeste de Bornéu. A talassocracia foi visitada pelos espanhóis da expedição de Magalhães em 1521 e lutaram pela Espanha em 1578 na Guerra de Castela. O Império do Brunei começou a declinar, atingindo sua forma moderna em 1890 após o progressivo cedendo Sarauaque para o Reino de Sarauaque e Sabá sendo cedida a Companhia Privilegiada do Bornéu do Norte. O Brunei se tornou um protetorado britânico em 1888 e foi atribuído um residente britânico em 1906. Nos anos após a ocupação de guerra japonesa durante a Segunda Guerra Mundial, ele formalizou uma constituição e lutou numa rebelião armada. O Brunei recuperou a sua independência do Reino Unido em 1 de Janeiro de 1984. O crescimento econômico durante os anos 1970 e 1990, com média de 56% de 1999 a 2008, transformou o Brunei em um país recém-industrializado; Brunei tem o segundo maior índice de desenvolvimento humano entre as nações do sudeste asiático depois de Singapura.
De acordo com o Fundo Monetário Internacional (FMI), Brunei é classificado em 5º no mundo em produto interno bruto per capita em paridade de poder aquisitivo. A revista Forbes também classificou Brunei como a quinta nação mais rica entre 182 nações, devido à sua extensa área de petróleo e de gás natural.
O país foi considerado "não livre" em 2019 pela Freedom House.
Etimologia
Segundo a lenda, o Brunei foi fundado por Awang Alak Betatar. Seu movimento de Garang, lugar onde hoje é o distrito de Temburong, para o estuário do rio Brunei levou à descoberta do país. Segundo a lenda, após o pouso, ele exclamou: "Baru nah! " (vagamente traduzido como "aí está", ou "eis!"), a partir do qual o nome "Brunei" (assim como o nome da ilha de Bornéu) foi derivado.
Foi rebatizado "Barunai" no , possivelmente influenciado pelo sânscrito da palavra "" (), o qual significa "oceano" ou mitológica "regente do oceano". A palavra "Bornéu" é da mesma origem. O nome completo do país é ""; () significa "Morada da Paz", enquanto significa "país" em malaio.
História
História antiga
Na ausência de mais fontes e provas, os estudiosos criaram uma história antiga do Brunei, que baseia-se principalmente em interpretações flexíveis de textos em chinês. Esta primeira parte diz: registros chineses do mencionam um estado chamado Po-li na costa noroeste de Bornéu. No , as contas chinesas e árabes indicarão um lugar chamado Vijaiapura, que acredita-se ter sido fundado por membros da família real de Funan. Acredita-se que eles desembarcaram na costa noroeste de Bornéu com alguns de seus seguidores. Eles, então, capturaram P'o-li e renomearam-a de território "Vijaiapura" (que significa "vitória" em sânscrito). Em 977, os registros chineses começaram a usar Po-ni, em vez de Vijaiapura para se referir ao Brunei. Em 1225 um funcionário chinês chamado Chua Ju-Kua informou que Brunei tinha 100 navios de guerra para proteger seu comércio e que havia uma grande quantidade de ouro no reino. Outro relatório em 1280 descreveu que Po-ni controlava grandes partes da ilha do Bornéu, atualmente as regiões de Sabá e Sarauaque, Sulu e algumas partes da Filipinas. No , Po-ni tornou-se um estado vassalo de Majapait, e teve de pagar um pagamento anual de 40 Katis de cânfora. Po-ni foi atacada e teve seu tesouro e ouro saqueados pelos sulus em 1369. Uma frota de Majapait conseguiu afastar os sulus, mas Po-ni tornou-se muito mais fraca após o ataque. Um relatório chinês de 1371 descreve Po-ni como pobre e totalmente controlada por Majapait.
O poder do Sultanato do Brunei estava no seu auge, entre os séculos XV e XVII, com o seu poder que se estendia do norte de Bornéu para sul das Filipinas.
Por volta do , o islamismo firmemente se enraizou no Brunei, o país havia construído uma de suas maiores mesquitas. Em 1578, Alonso Beltrán, um viajante espanhol a descreveu com cinco andares de altura e construída sobre a água.
Guerra contra Espanha e declínio
A influência europeia gradualmente trouxe um fim ao poder regional, como Brunei entrou em um período de declínio agravado por conflitos internos sobre a sucessão real. Pirataria também foi prejudicial para o reino. A Espanha declarou guerra em 1578, atacando e capturando a capital de Brunei, na época, Kota Batu. Isto foi conseguido como resultado, em parte, da assistência prestada a eles por dois nobres de Brunei, Pengiran Seri Lela e Pengiran Seri Ratna. O primeiro tinha viajado para Manila para oferecer Brunei como um tributário da Espanha, para ajudar a recuperar o trono usurpado por seu irmão, Saiful Rijal. Os espanhóis concordaram que, se eles conseguissem conquistar Brunei, Pengiran Seri Lela iria se tornar definitivamente o Sultão, enquanto Pengiran Seri Ratna seria o novo Bendahara. Em março de 1578, a frota espanhola, liderada si mesma por Francisco de Sande, agindo como Capitão-general, começou sua viagem para Brunei. A expedição foi de 400 espanhóis, 1,5 mil filipinos nativos e 300 nativos de Bornéu. A campanha foi uma das muitas, que também incluía a ações em Mindanau e Sulu.
Os espanhóis conseguiram invadir a capital em 16 de abril de 1578, com a ajuda de Seri Pengiran Lela e Seri Pengiran Ratna. O sultão Saiful Rijal e Paduka Seri Begawan Sultan Abdul Kahar fora forçado a fugir para Meragang então Jerudong. Em Jerudong, fizeram planos para perseguir o exército conquistador longe de Brunei. Os espanhóis sofreram grandes perdas devido à cólera ou o surto de disenteria. Eles foram tão enfraquecidos pela doença que decidiram abandonar Brunei para retornar a Manila em 26 de junho de 1578, depois de apenas 72 dias. Antes de fazer isso, eles queimaram a mesquita, uma estrutura alta, com um teto de cinco níveis.
Pengiran Seri Lela morreu entre agosto e setembro de 1578, provavelmente da mesma doença que afligira seus aliados espanhóis, embora não houvesse suspeita de que ele poderia ter sido envenenado por decisão do sultão. A filha de Seri Lela partiu com o espanhol e passou a se casar com um cristão Tagalog, chamado Agustín de Legazpi de Tondo.
As contas locais de Brunei diferem muito da visão geralmente aceita dos acontecimentos. A Guerra castelhana entrou na consciência nacional como um episódio heroico, com os espanhóis sendo conduzidos por Bendahara Sakam, supostamente uma decisão do irmão do sultão, e mil guerreiros nativos. Esta versão, no entanto, é contestada pela maioria dos historiadores e considerando uma lembrança de um herói folclórico, provavelmente criado décadas ou séculos depois.
Não obstante o recuo, Brunei perdeu uma série de territórios para a Espanha, incluindo a ilha de Lução.
Invasão britânica
Os britânicos tiveram de intervir nos assuntos do Brunei em um número de ocasiões. O Reino Unido atacou o Brunei em julho 1846, devido a divergências quanto a quem possuía o direito de sultão.
Na década de 1880, como o declínio do Império do Brunei continuava, o Sultão Hashim Jalilul Alam Aqamaddin apelou aos britânicos para deter os dirigentes Rajás brancos do vizinho Reino de Sarauaque de particionamento e anexação do território de Brunei. O "Tratado de Proteção" foi negociado por Sir Hugh Low e entrou em vigor em 17 de setembro de 1888. O tratado fez com que o Sultão "não pudesse ceder ou locar qualquer território a potências estrangeiras sem o consentimento britânico"– mas também permitiu a Grã-Bretanha o controle sobre os assuntos externos de Brunei, que se tornou um protetorado britânico. Esse protetorado continuou até 1984. Quando o Reino de Sarauaque anexou o distrito Pandaruan de Brunei em 1890, os britânicos não tomaram qualquer ação para impedi-lo porque não consideram nem Brunei nem o Reino de Sarauaque como "poderes estranho" (pelo Tratado de Proteção) para os britânicos. Essa anexação final por Sarauaque deixou Brunei com uma pequena massa de terra corrente e separada em duas partes.
Moradores britânicos foram introduzidos em Brunei sobre Acordo de Abrigo Complementar do protetorado em 1906. Os moradores foram para aconselhar o sultão em todos os assuntos de administração. No entanto, o residente assumiu o controle mais executivo do que o sultão. O sistema Residencial terminou em 1959.
Descoberta do óleo
Em 1929 fora descoberto na região o petróleo por dois homens, F.F. Marriot e T.G. Cochrane, eles descobriram o petróleo perto de Seria, a descoberta ocorreria originalmente dois anos antes, em 1927. Eles informaram um geofísico que estavam realizando uma pesquisa lá, em 12 de julho de 1928, aconteceu a primeira perfuração. O petróleo foi atingido em 297 metros em 5 de Abril de 1929. A produção de petróleo aumentou consideravelmente na década de 1930. Em 1940, a produção de petróleo estava em mais de seis milhões de barris. Até hoje o poço perfurado na década de 1940 continua a jorrar. A British Malayan Petroleum Company — hoje a atual Brunei Shell Petroleum Company —, foi formada em 22 de julho de 1922. O primeiro poço extraterritorial foi perfurado em 1957. Petróleo e gás natural têm sido a base do desenvolvimento e da riqueza de Brunei, desde o final do .
Invasão japonesa
Os japoneses invadiram Brunei em 16 de dezembro de 1941, oito dias após o ataque a Pearl Harbour. Eles desembarcaram soldados do Distanciamento Kawaguchi de Cam Ranh Bay em Kuala Belait. Depois de seis dias de combate ocuparam todo o país. As únicas tropas aliadas na área eram o 2º Batalhão do 15º Regimento de Punjab, com base em Kuching, Sarauaque.
Depois que Brunei foi ocupada, os japoneses fizeram um acordo com Sultão Ahmad Tajuddin sobre governar o país. Inche Ibrahim (conhecido mais tarde como Pehin Datu Perdana Menteri Dato Laila Utama Awang Haji Ibrahim), um ex-secretário do residente britânico, Ernest Edgar Pengilly, foi nomeado Diretor Administrativo sob a Governo japonês. Os japoneses tinham proposto que Pengilly mantivesse a sua posição sob a sua administração, mas ele recusou. Tanto ele quanto outros cidadãos britânicos ainda em Brunei foram então internados pelos japoneses no acampamento Batu Lintang em Sarauaque. Ibrahim, enquanto as autoridades britânicas estavam sob a guarda japonesa, fez questão de, pessoalmente, apertar a mão de cada um deles e desejar-lhes boa sorte.
O Sultão manteve o seu trono e recebeu uma pensão e honras pelos japoneses. Durante a última parte da ocupação residiu em Tantuya, Limbang que tinha uma relação um pouco melhor com os japoneses. A maioria dos funcionários do governo malaio foram retidos pelos japoneses. Administração de Brunei foi reorganizada em cinco prefeituras, que incluía o protetorado britânico de Bornéu do Norte. As prefeituras incluíam Baram, Labuão, Lawas e Limbang. Ibrahim conseguiu esconder uma série de documentos importantes dos japoneses durante a ocupação. Pengiran Yusuf (mais tarde YAM Pengiran Setia Negara Pengiran Haji Mohd Yusuf), juntamente com outros bruneianos foram enviados ao Japão para o treinamento. Yusuf teve a sorte de sobreviver como ele estava em Hiroshima quando a bomba atômica foi lançada.
Os britânicos tinham previsto um ataque japonês, mas não tinham os recursos para defender a área por causa de seu envolvimento na guerra na Europa. As tropas do Regimento de Punjab preencheram os poços de petróleo do campo de Seria com concreto em setembro de 1941 para negar a sua utilização aos japoneses. Os equipamentos e instalações restantes foram destruídos quando os japoneses invadiram a Malásia. Até o final da guerra, 16 poços de Miri e Seria haviam sido reinstalados, com a produção atingindo cerca da metade do nível de antes da guerra. A produção de carvão em Muara também foi reiniciada, mas com pouco sucesso.
Durante a ocupação, os japoneses tiveram sua língua ensinada nas escolas, e os oficiais do governo foram obrigados a aprender japonês. A moeda local foi substituída pela que viria a ser conhecido como duit pisang (dinheiro de banana). De 1943 a hiper-inflação destruiu o valor da moeda e, ao final da guerra, esta moeda não valia nada. Ataques aliados no transporte eventualmente fizeram o comércio parar. Alimentos e remédios caíram em falta, e a população sofreu com a fome e doenças.
A pista do aeroporto foi construída pelos japoneses durante a ocupação, e em 1943 as unidades navais japonesas estavam baseadas em Brunei Bay e Labuão. A base naval foi destruída por bombardeios dos Aliados, mas a pista do aeroporto sobreviveu. O mecanismo foi desenvolvido como um aeroporto público. Em 1944, os Aliados iniciaram uma campanha de bombardeio contra os ocupantes japoneses, que destruíram grande parte da cidade e Kuala Belait, mas erraram Kampong Ayer.
Em 10 de junho de 1945, a 9ª Divisão Australiana desembarcou em Muara sob a Operação Oboe Six para recapturar Bornéu dos japoneses. Eles foram apoiados por unidades aéreas e navais americanas. A cidade de Brunei foi bombardeada extensivamente e recapturada após três dias de intensos combates. Muitos prédios foram destruídos, incluindo a Mesquita. As forças japonesas em Brunei, Bornéu, e Sarauaque, sob o tenente-general Masao Baba, renderam-se formalmente em Labuão em 10 de setembro de 1945. A Administração Militar Britânica assumiu a partir dos japoneses e permaneceu até julho de 1946.
Após-Segunda Guerra Mundial
Após a Segunda Guerra Mundial, um novo governo foi formado em Brunei sob a Administração Militar Britânica (BMA). Ela consistia principalmente de oficiais australianos e militares. A administração de Brunei foi entregue à Administração Civil em 6 de julho de 1945. O Conselho Estadual de Brunei também foi revivido naquele ano. A BMA também foi encarregada de zelar pela economia bruneiana, que foi amplamente danificada pelos japoneses durante a ocupação. Eles também foram incumbidos de apagar os incêndios iniciados nos poços de Seria, que foi iniciado pelos japoneses antes da sua derrota. Antes de 1941, o governador de Assentamentos Straits com base em Cingapura foi responsável pelas funções do alto Comissário britânico para Brunei, Sarauaque e Bornéu do Norte (hoje Sabá). O primeiro alto comissário britânico para Brunei foi o governador de Sarauaque, Sir Charles Clarke Ardon. O Pemuda Barisan ("Movimento da Juventude") (abreviado como BARIP) foi o primeiro partido político a ser formado em Brunei. Ela foi formada em 12 de abril de 1946. Os objetivos do partido eram de "preservar a soberania do sultão e do país, e para defender os direitos dos malaios". BARIP também contribuiu para a formação do Hino Nacional do país. O partido foi dissolvido em 1948 devido à inatividade.
Em 1959, uma nova constituição foi escrita declarando Brunei um estado de auto-governo, enquanto suas relações exteriores, segurança e defesa permanecessem por responsabilidade do Reino Unido. Houve uma pequena rebelião contra a monarquia, em 1962, que foi suprimida com a ajuda do Reino Unido. Este evento ficou conhecido como a Revolta de Brunei e foi em parte responsável pelo fracasso para a criação da Federação do Bornéu do Norte. A rebelião parcialmente afetou a decisão de Brunei optar por sair da Federação da Malásia.
Brunei ganhou sua independência do Reino Unido em 1 de janeiro de 1984. O Dia Nacional oficial, que celebra a independência do país, no entanto, é realizada em 23 de fevereiro, devido à tradição.
Criação da constituição
Em julho de 1953, o Sultão Omar Ali Saifuddien III formou uma comissão de sete membros nomeados Tujuh Serangkai para encontrar o ponto de vista dos cidadãos a respeito de uma constituição escrita para Brunei. Em maio de 1954, uma reunião com a presença do sultão, o residente e o Alto Comissariado foi realizada para discutir as conclusões da comissão. Em março de 1959, o sultão Omar Ali Saifuddien III liderou uma delegação a Londres para discutir a proposta de Constituição. A delegação britânica foi liderada por Sir Alan Lennox-Boyd, que era o Secretário de Estado para as Colônias. O Governo britânico depois aceitou o projeto de Constituição. Em 29 de setembro de 1959, do Acordo de Constituição foi assinado em Bandar Seri Begawan. O acordo foi assinado pelo Sultão Omar Ali Saifuddien III e Sir Robert Scott, o Comissário-Geral para o Sudeste Asiático. Alguns dos pontos da constituição foram:
O sultão se tornou o Chefe Supremo do Estado;
O país agora tem autonomia interna;
O Governo britânico se tornou responsável pelas relações internacionais e a defesa externa do país;
O cargo de Residente foi substituído agora por um Alto Comissário Britânico.
Cinco conselhos foram também criados:
O Conselho Executivo;
A Assembleia Legislativa de Brunei;
O Conselho Privado;
O Conselho de Sucessão;
O Conselho do estado religioso.
Geografia
O Brunei está localizado na costa norte da ilha do Bornéu. Está dividido em dois territórios separados apenas pela baía do Brunei, ambos com a costa norte para o mar da China meridional, e partilha uma fronteira de 381 km com a Malásia. Tem 500 km2 de águas territoriais, e uma zona náuticas econômica exclusiva de 200 milhas.
O Brunei consiste de duas partes sem ligação. Cerca de 97% da população vive na parte maior, a ocidente, enquanto que só pessoas vivem na parte leste, montanhosa, que constitui o distrito de Temburong. As cidades principais são a capital, Bandar Seri Begawan (cerca de 46 000 habitantes), a cidade portuária de Muara e Seria. Outras cidades importantes são a cidade portuária de Muara, a cidade produtora de petróleo da Seria e sua cidade vizinha, Kuala Belait.
O ponto mais alto do país é o Bukit Pagon ( m de altitude), no extremo sul da parte oriental do país, sobre a fronteira com a Malásia.
A maioria do território do Brunei está dentro da planície de chuva das florestas do Bornéu, uma ecorregião que cobre a maior parte da ilha, mas existem áreas de florestas tropicais de montanha do interior.
O clima no Brunei é tropical, com temperaturas e umidade atmosférica elevadas e muita chuva. O clima de Brunei é tropical equatorial. A temperatura média anual é de 26,1 °C (79,0 °F), com a média de abril a maio de 24,7 °C (76,5 °F) e da média de outubro a dezembro de 23,8 °C (74,8 °F).
Subdivisões
O Brunei está dividido em quatro distritos, chamados de "daerah". Os distritos estão subdivididos em 38 mukims (províncias):
Belait
Brunei e Muara
Temburong
Tutong
Demografia
A população de Brunei, em julho de 2020 foi de 460 345 dos quais 76% vivem em áreas urbanas. A expectativa de vida média é de 76,37 anos. Em 2004, 66,3% da população eram malaios, 11,2% eram chineses, 3,4% eram indígenas, com grupos menores que compõem o resto.
A língua oficial de Brunei é o malaio. Existe um movimento para expandir o uso da língua no país. A principal língua falada é o Melayu Brunei (Brunei Malaio). O Brunei malaio é bastante divergente do padrão malaio e do resto dos dialetos malaios, sendo cerca de 84% cognato com a norma malaia, e é principalmente mutuamente ininteligíveis a ele. Inglês e chinês (com vários dialetos) também são amplamente falados, o inglês também é usado nos negócios e como a língua de ensino do primário ao ensino superior, e há uma relativamente grande comunidade expatriada. Bahasa Rojak, muitas vezes falado pelo povo e em alguns programas de rádio populares, é conhecido como uma "língua mista" e considerado por alguns a ser prejudicial para o malaio normal. Outras línguas faladas incluem o Kedayan, Tutong, Murut, Dusun e o Iban.
Religião
O Islã é a religião oficial do Brunei, e dois terços da população aderem ao Islamismo. Outras religiões praticadas são o budismo (13%, principalmente pelos chineses) e cristianismo (10%). O pensamento livre, principalmente por parte dos chineses, forma cerca de 7% da população. Embora a maior parte deles praticam alguma forma de religião com elementos do budismo, confucionismo e Taoismo, eles preferem se apresentar como tendo a religião não declarada oficialmente, portanto, considerados como ateus nos censos oficiais. Os seguidores de religiões indígenas são cerca de 2% da população.
A população é proibida de celebrar o Natal, a penalidade para os que desobedecerem pode chegar a cinco anos de prisão. A proibição inclui cantar músicas natalinas e enviar cartões de Natal.
Política e governo
Brunei é constitucionalmente um sultanato. Tem um sistema jurídico baseado no direito comum Inglês, embora substituindo este em certos casos pela xaria islâmica. Em Outubro de 2013, o Sultão Hassanal Bolkiah anunciou a intenção de impor o Código Penal da Xaria aos muçulmanos do país, que constituem cerca de dois terços da população, o que seria implementado em três fases, culminando em 2016.
O sistema político do país é regido pela Constituição e pela tradição da monarquia islâmica malaia, o conceito de "Melayu Islam Beraja" (MIB). Os três componentes cobertos pela MIB são a cultura malaia, a religião islâmica e o quadro político sob a monarquia.
Não existe um parlamento eleito mas, em Setembro de 2004, o Sultão convocou um conselho legislativo nomeado por ele. O sultão acumula as funções de chefe de estado e de governo do país; o atual sultão, Hassanal Bolkiah, foi coroado em 1984 e tem como herdeiro legitimo designado seu filho mais velho, o príncipe Al-Muhtadee Billah, que também possui cargos na política de seu país.
Desde a formação da Constituição de 1959, o sultão possui autoridade executiva plena - incluindo poderes de emergência desde 1962 -,o que é renovado a cada 2 anos.
Símbolos nacionais
A bandeira nacional foi criada em 29 de setembro de 1959 quando o território era um protetorado britânico. Quando o Brunei obteve a sua independência definitiva, a 1 de Janeiro de 1984 esta bandeira foi adotada oficialmente. Esta é composta por uma bandeira amarela atravessada no sentido do canto superior esquerdo para o canto inferior direito por duas listras de cor preta e branca (a branca por cima) e ao centro o escudo do país. O escudo é constituído por uma meia-lua, que representa o Islão, do guarda-sol real (chamado Payung Ubor-Ubor) que representa a monarquia, as luvas nas laterais e, por baixo, a inscrição em árabe Brunei Darussalam y la shahada.
O emblema é destacado na Bandeira do Brunei. Ele foi originalmente adotado em 1932. Há cinco componentes principais no emblema nacional. Eles são a bandeira, a cúpula real, o ala, as mãos, e as crescentes. Sobre o crescente está escrito, em escrita árabe, o lema nacional: "Sempre em serviço com orientação de Deus". Abaixo está uma bandeira com o nome da nação, também escrita em árabe, "Brunei Darussalam" ou Brunei, terra de paz. As asas simbolizam proteção da justiça e da paz. Abaixo destes está o crescente que é o símbolo do Islã, a religião nacional de Brunei. As mãos simbolizam o governo que tem o dever de proteger as pessoas.
Allah Peliharakan Sultan é a denominação oficial dada pelo sultão de Brunei ao seu hino nacional.
Relações Exteriores
O Brunei aderiu à Associação de Nações do Sudeste Asiático (ASEAN) em 7 de janeiro de 1984 - uma semana após retomar a plena independência - e dá a sua adesão à ASEAN a mais alta prioridade nas suas relações externas. Com seus laços tradicionais com o Reino Unido, tornou-se o membro 49 da Commonwealth imediatamente no dia da sua independência em 1 de Janeiro de 1984. Mais tarde juntou-se à Organização das Nações Unidas na 39 ª Sessão da Assembleia Geral das Nações Unidas e tornou-se um membro pleno em 21 de setembro de 1984 como um meio para obter o reconhecimento de sua soberania e independência total da comunidade mundial.
Brunei é reconhecido por todas as nações do mundo. Ele compartilha uma estreita relação com seus vizinhos como as Filipinas e outros países, como Singapura. Em abril de 2009, Brunei e Filipinas assinaram um Memorando de Entendimento (MOU), que visa reforçar a cooperação bilateral entre os dois países nos domínios da agricultura,comerciais e de investimentos.
Brunei é uma das muitas nações que reivindicam algumas das disputadas lhas Spratly. O estado de Limbang como parte de Sarauaque foi disputada por Brunei desde que a área foi anexada, em 1890. A questão teria sido resolvida em 2009, com o Brunei concordando em aceitar a fronteira em troca de desistir das reivindicações de campos de petróleo para a Malásia em águas bruneianas. O governo, porém, nega e diz que sua reivindicação sobre Limbang nunca foi abandonada.
Forças armadas
Brunei mantém três batalhões de infantaria estabelecidos em todo o país. A Marinha de Brunei tem vários "ijtihad", uma classe de barcos-patrulha comprados de um fabricante alemão. O Reino Unido ainda mantém uma base em Seria, o centro da indústria do petróleo em Brunei. O Gurkha consistindo de 1,5 mil soldados ainda opera no país. A base permaneceu em Seria,devido a um acordo assinado entre os dois países.
Um helicóptero Bell 212 operado pela Força Aérea caiu em Kuala Belait em 20 de julho de 2012, com a perda de 12 dos 14 tripulantes a bordo. O acidente é o pior incidente da aviação na história do Brunei.
Corrupção
Em 1998, o príncipe Jefri Bolkiah, irmão mais novo do sultão, foi afastado de empresas estatais por má administração e pela falência de uma empresa estatal, a Amedeo Development Corporation, que acumulou prejuízos de US$ 14,8 bilhões e 23 mil novos desempregados. O sultão, que antes era considerado um dos homens mais ricos do mundo, teve sua fortuna diminuída de US$ 40 bilhões para US$ 10 bilhões. Ele então processou o irmão, que desde 2004 vive em exílio.
Economia
Esta pequena e rica economia é uma mistura de empreendedorismos estrangeiros e domésticos, regulamentações governamentais, medidas de bem-estar e tradições de aldeias.
A economia de Brunei baseia-se, fundamentalmente, nas exportações dos seus recursos minerais: o petróleo, gás natural (primeiro país em exportação de gás liquefeito) e carvão. Também há exportação florestal e a pesca. Sua agricultura é de tipo tropical com cultivo de arroz, coco e caucho. A principal riqueza do Brunei é o petróleo, descoberto na região em 1929 e que contribui para mais de metade do produto nacional bruto. Brunei faz parte do tratado internacional chamado APEC (Asia-Pacific Economic Cooperation), um bloco econômico que tem por objetivo transformar o Pacífico numa área de livre comércio e que engloba economias asiáticas, americanas e da Oceania.
A produção de petróleo bruto e gás natural responde por cerca de 90% do seu PIB. Mais de 167 mil barris de petróleo são produzidos diariamente, tornando Brunei o quarto maior produtor de petróleo do Sudeste Asiático. Também produz cerca de 895 milhões de pés cúbicos de gás natural liquefeito por dia, tornando o Brunei o nono maior exportador da produto no mundo.
Cerca de 30% dos habitantes são servidores públicos e não existem impostos sobre a renda de seus habitantes. A educação, a saúde e a moradia são gratuitas; os que não têm moradia podem se cadastrar para receber uma moradia popular com 200 metros quadrados.
Para alcançar sua meta de autossuficiência alimentar, Brunei reativou em abril de 2009 seu ambicioso projeto de plantação de arroz. Em agosto de 2009, a Família Real colheu os primeiros grãos, após anos e várias tentativas para aumentar a produção local de arroz - uma meta idealizada meio século antes. Em julho de 2009, o país lançou um plano nacional de plantação Halal, com o objetivo de exportar para mercados estrangeiros.
Infraestrutura
Os centros populacionais do país estão ligados por uma rede de 2.800 quilômetros de estrada. A rodovia de 135 km da Cidade Muara até Kuala Belait está sendo duplicada para uma via dupla.
Brunei é acessível pelo mar, ar e transporte terrestre. O Aeroporto Internacional de Brunei é o principal ponto de entrada para o país. A Royal Brunei Airlines é a transportadora nacional. Há uma outra pista, o Airfield Anduki, localizado na cidade de Seria. O terminal de ferry possui ligações regulares dos serviços de Muara até a cidade Labuão (Malásia). Lanchas fornecem o transporte de passageiros e mercadorias para o distrito de Temburong. A principal rodovia que atravessa Brunei é a estrada de Tutong-Muara. A rede rodoviária do país é bem desenvolvida. Brunei tem um porto marítimo principal localizado em Muara.
Com um carro particular para cada 2,09 pessoas, Brunei Darussalam tem uma das maiores taxas de propriedade de automóveis do mundo. Isto tem sido atribuído à ausência de um sistema global de transporte, taxas de importação e baixo preço da gasolina sem chumbo - menos de 0,53 dólares de Brunei por litro.
Saúde
A saúde em Brunei é não pública, embora custe menos de 1 dólar bruneiano por consulta para cada cidadão. Um centro de saúde criado pela Brunei Shell Petroleum está localizado em Panaga. Quando a assistência médica no país é indisponível, os cidadãos são enviados ao exterior às custas do governo.
O maior hospital de Brunei é o Raja Isteri Pengiran Anak Saleha Hospital (RIPAS), hospital que possui 538 leitos, situa-se na capital do país, Bandar Seri Begawan. Há dois centros privados de médicos, Gleneagles JPMC Sdn Bhd. e Jerudong Park Centro Médico. O Centro de Promoção da Saúde foi inaugurado em novembro de 2008 e serve para educar o público sobre a importância de uma vida saudável.
Atualmente não existe uma faculdade de medicina em Brunei, e os bruneanos que desejam tal formação frequentam universidades no exterior,principalmente na vizinha Malásia. No entanto, foi aberto o Instituto de Medicamentos na Universidade de Brunei Darussalam, com um novo prédio construído para a faculdade. O prédio inclui instalações de laboratório e pesquisa e foi concluído em 2009. Há ainda uma Escola de Enfermagem, funcionando desde 1951. Uma equipe de 58 enfermeiros gerentes foi formada em RIPAS para melhorar o serviço e fornecer cuidados médicos complementares. Em dezembro de 2008, a faculdade de enfermagem se fundiu com o Instituto de Medicamentos da Universidade de Brunei Darussalam para formar mais enfermeiros e parteiras. Ele agora é chamado de Instituto de Ciências da Saúde, o PAPRSB (Pengiran Anak Puteri Rashidah Sa'datul Bolkiah).
Mídia e meios de comunicação
A mídia em Brunei é extremamente pró-governo. O país tem sido classificado pelo status de "não livre" pela Freedom House; críticas da imprensa ao governo e à monarquia são raras. Em 1953, o governo permitiu a criação de uma empresa de impressão e publicação, a Brunei Press PLC. A empresa continua a imprimir o diária Boletim de Bornéu em inglês. Este trabalho começou como um jornal semanal local e tornou-se um diário em 1990. Além do Boletim de Bornéu, há também a Media Permata, o jornal malaio local que circula diariamente. The Brunei Times é outro jornal inglês independente publicado em Brunei Darussalam desde 2006.
O governo de Brunei é proprietário e opera seis canais de televisão, com a introdução da TV digital usando DVB-T (RTB 1, RTB 2, RTB 3 (HD), RTB 4, 5 e RTB RTB New Media (Game portal) e cinco estações de rádio (National FM, Pilihan FM, Nur Islam FM, Harmony FM e Pelangi FM). Uma empresa malaia privada opera o sistema de televisão a cabo (Astro-Kristal), bem como uma estação de rádio privada,a Kristal FM. Ela também tem um estação de rádio online a, UBD FM que é transmitida da única universidade do país,a Universidade do Brunei Darussalam.
Cultura
A cultura de Brunei é predominantemente malaia (refletindo sua etnia), com influências pesadas do Islã, mas é visto como mais conservadora do que a Malásia.
A cultura do país é essencialmente derivada do Velho Mundo malaio, que abrangeu o arquipélago malaio e deste surgiu o que é conhecido como a Civilização malaia. Baseada em fatos históricos, vários elementos culturais e civilizações estrangeiras tiveram uma mão na influência da cultura deste país. Assim, a influência da cultura pode ser atribuída a quatro períodos dominantes de animismo, hinduísmo, islamismo e do Ocidente. No entanto, foi o Islã que conseguiu encerrar suas raízes profundamente na cultura de Brunei, portanto, tornou-se um modo de vida e adotou como ideologia do Estado e da filosofia.
Com a xaria em vigor no país, a venda e o consumo público de álcool são proibidos. Os não muçulmanos estão autorizados a trazer em uma quantidade limitada de álcool a partir de seu ponto de embarque no exterior para seu próprio consumo. A taxa de criminalidade é uma das mais baixas do mundo: nos últimos seis anos ocorreram apenas 3 assassinatos, e o tráfico de drogas é punido com pena de morte.
Esportes
Brunei tem uma ampla variedade de instalações esportivas modernas e sediou os Jogos do Sudeste Asiático de 1999 no Estádio Sultan Hassal Bolkiah. Brunei atua internacionalmente em alguns esportes do sudeste asiático, como voleibol e sepak takraw.
Brunei participou nos anos de 1996, 2000, 2004, 2012 e 2016 dos Jogos Olímpicos, mas o país não ganhou uma única medalha. Em 2012 e 2016, Brunei teve três atletas na competição. Em julho de 2020, a seleção nacional de futebol do Brunei foi classificada em 191º no ranking mundial de futebol.
Alimentação
O país tem hábitos alimentares bem diferentes aos olhos ocidentais. As frutas mais comuns são a carambola, a banana e a graviola e os pratos são extremamente exóticos, sendo que a grande maioria é derivada da culinária árabe.
Vestuário
Os hábitos de vestimenta são bastante incomuns aos países ocidentais. As mulheres usam roupas bem estampadas, misturando muitas cores e complementando com o hijab (véu) que é adotado devido as tradições islâmicas do país, onde a mulher é bastante reservada quanto às vestimentas. Os homens por sua vez utilizam roupas em tons mais discretos, sendo o baju (camisa comprida e de mangas longas) e o songkok (chapéu malaio, semelhante ao fez turco) a vestimenta tradicional. Já o serban (turbante) é geralmente usado em ocasiões festivas. É interessante ressaltar que apesar de essas roupas serem "tradicionalistas" não é toda a população do país que utiliza rigorosamente esse tipo de roupa, até porque o país, ainda seja em maior quantidade de origem malaia, possui mais de uma origem.
Ver também
Ásia
Bornéu
Lista de Estados soberanos
Lista de Estados soberanos e territórios dependentes da Ásia
Mar da China Meridional
Missões diplomáticas de Brunei
Turismo em Brunei
Bibliografia
Ligações externas
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"redpajama_set_name": "RedPajamaWikipedia"
}
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package org.sjpool.pool;
import java.sql.Connection;
import java.sql.SQLException;
import java.util.concurrent.Future;
public class Pool {
public static enum PoolType {
SIMPLE_JAVA_POOL, HIKARI_CP;
}
private final PoolWrapper wrapper;
public Pool(PoolProperties poolProperties) throws SQLException {
if (poolProperties.getPoolType() == PoolType.SIMPLE_JAVA_POOL) {
wrapper = new SimpleJavaPoolWrapper(poolProperties);
} else if (poolProperties.getPoolType() == PoolType.HIKARI_CP) {
wrapper = new HikariCPWrapper(poolProperties);
} else {
throw new IllegalArgumentException(String.format("Unknown pool type: %s", poolProperties.getPoolType()));
}
}
public Connection getConnection() throws SQLException {
return wrapper.getConnection();
}
public Future<Connection> getFutureConnection() {
return wrapper.getFutureConnection();
}
public int getNbConnectionsAllocated() {
return wrapper.getNbConnectionsAllocated();
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
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\section{ Introduction}
The Weinstein operator $\Delta_{W,\alpha}^d$ defined on $\mathbb{R}_{+}^{d+1}=\mathbb{R}^d\times(0, \infty)$, by
\begin{equation*}
\Delta_{W,\alpha}^d=\sum_{j=1}^{d+1}\frac{\partial^2}{\partial x_j^2}+\frac{2\alpha+1}{x_{d+1}}\frac{\partial}{\partial x_{d+1}}=\Delta_d+L_\alpha,\;\alpha>-1/2,
\end{equation*}
where $\Delta_d$ is the Laplacian operator for the $d$ first variables and $L_\alpha$ is the Bessel operator for the last variable defined on $(0,\infty)$ by
$$L_\alpha u=\frac{\partial^2 u}{\partial x_{d+1}^2}+\frac{2\alpha+1}{x_{d+1}}\frac{\partial u}{\partial x_{d+1}}.$$
The Weinstein operator $\Delta_{W,\alpha}^d$ has several applications in pure and applied mathematics, especially in fluid mechanics \cite{brelot1978equation, weinstein1962singular}.
Very recently, many authors have been investigating the behaviour of the Weinstein
transform (\ref{defWeinstein}) with respect to several problems already studied for the classical Fourier transform.
For instance, Heisenberg-type inequalities \cite{salem2015heisenberg}, Littlewood-Paley g-function \cite{salem2016littlewood},
Shapiro and Hardy–Littlewood–Sobolev type inequalities \cite{salem2020hardy, salem2015shapiro},
Paley-Wiener theorem \cite{mehrez2017paley}, Uncertainty principles \cite{mejjaoli2011uncertainty, ahmed2018variation, saoudi2019l2}, multiplier Weinstein operator \cite{ahmed2018calder}, wavelet and continuous wavelet transform \cite{gasmi2016inversion, mejjaoli2017new}, Wigner transform and localization operators \cite{saoudi2019weinstein, saoudilocalisation}, and so forth...
In the classical setting, the notion of wavelets was first introduced by Morlet in connection with his study of seismic traces and the mathematical foundations were given by Grossmann and Morlet \cite{grossmann1984decomposition}. Later, Meyer and many other mathematicians recognized many classical results of this theory \cite{koornwinder1993continuous, meyer1992wavelets}. Classical wavelets have wide applications, ranging from signal analysis in geophysics and acoustics to quantum theory and pure mathematics \cite{daubechies1992ten, goupillaud1984cycle, holschneider1995wavelets}.
Recently, the theory of wavelets and continuous wavelet transform has been extended and generalized in the context of differential-difference operators \cite{gasmi2016inversion, mejjaoli2017dunkl, mejjaoli2017new, mejjaoli2017time}.
Wavelet analysis has attracted attention for its ability to analyze rapidly changing transient signals. Any application using the Fourier like transform can be formulated using wavelets to provide more accurately localized temporal
and frequency information. The reason for the extension from one wavelet to two wavelets comes from the extra
degree of flexibility in signal analysis and imaging when the localization operators are
used as time-varying filters. One of the aims of the continuous wavelet transform, is the study of their localization
operators.
The time-frequency representations required for localization operators wish have been object of study in quantum mechanics, in PDE and signal analysis recently. In engineering, a natural language is given by time-frequency analysis. Localization operators arise from pure and applied mathematics in connection with various areas of research. They were initiated by Daubechies \cite{daubechies1988time, daubechies1990wavelet, daubechies1988time2}, and before she highlighted the role of these operators to localize a signal simultaneously in time and frequency.
Nowadays, these operators have found many applications to time-frequency analysis, the theory of differential equations, quantum mechanics. Depending on the field of application, these operators are known under the names of Wick, anti-Wick or Toeplitz operators, as well as wave packets, Gabor or short time Fourier transform multipliers. Arguing from these point of view, many works were done on them, we refer, for instance \cite{balazs2008hilbert, balazs2012multipliers, cordero2003time, de2002uniform, grochenig2013foundations, mejjaoli2017dunkl}.
Using the harmonic analysis associated with the Weinstein operator (generalized translation operators,
generalized convolution, Weinstein transform, ...) and the same idea as for the classical case,
we study the localisations operators associated with the Weinstein two-wavlet \cite{saoudi2020two} and we prove that under
suitable conditions on the symbols and two Weinstein wavelets, the boundedness and
compactness of these localization operators. Our main results for the boundedness and compactness of the Weinstein two wavelet localisation operators, with different symbols and windows, are summarized in the following table.
\begin{table}[h]
\begin{center}
\begin{tabular}{| c | c | c |c |}
\hline
\textbf{Symbol} & \multicolumn{2}{|c|}{\textbf{Windows}} & \textbf{Localization Operator} \\ \hline
$\sigma$ & $\varphi$ & $\psi$ & $\mathcal{L}_{\varphi,\psi}(\sigma)$ \\ \hline \hline
$L^1_\alpha(\X)$ & $L^\infty_\alpha(\mathbb{R}^{d+1}_+)$ & $L^1_\alpha(\mathbb{R}^{d+1}_+)$ &
$\mathcal{B}( L^1_\alpha(\mathbb{R}^{d+1}_+))$ \\ \hline
$L^1_\alpha(\X)$ & $L^1_\alpha(\mathbb{R}^{d+1}_+)$ & $L^\infty_\alpha(\mathbb{R}^{d+1}_+)$ & $\mathcal{B}( L^\infty_\alpha(\mathbb{R}^{d+1}_+))$ \\ \hline
$L^1_\alpha(\X)$ & \multicolumn{2}{|c|}{$L^1_\alpha\bigcap L^\infty_\alpha(\mathbb{R}^{d+1}_+)$} & $\mathcal{B}( L^p_\alpha(\mathbb{R}^{d+1}_+)),\, p\in [1,\infty]$ \\ \hline
$L^1_\alpha(\X)$ & $L^q_\alpha(\mathbb{R}^{d+1}_+)$ & $L^p_\alpha(\mathbb{R}^{d+1}_+)$ & $\mathcal{B}( L^p_\alpha(\mathbb{R}^{d+1}_+)),\, p\in [1,\infty]$ \\ \hline
$L^1_\alpha(\X),\, r\in [1,2]$ & \multicolumn{2}{|c|}{$L^1_\alpha\bigcap L^2_\alpha\bigcap L^\infty_\alpha(\mathbb{R}^{d+1}_+)$} & $\mathcal{B}( L^p_\alpha(\mathbb{R}^{d+1}_+)),\, p\in [r,r']$ \\ \hline
\end{tabular}
\end{center}
\caption{Boundedness and compactness of localisation operators}
\end{table}
This paper is organized as follows. In Section 2, we recall some properties of harmonic
analysis for the Weinstein operators and Weinstein two-wavelet theory. In Section 3, we give a host of sufficient conditions for the boundedness and compactness of the two-wavelet localization operator on $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ for all $1\leq p\leq \infty$, in terms of properties of the symbol $\sigma$ and the functions $\varphi$ and $\psi$. In the end, we study some typical examples of the Weinstein two-wavelet localization operators.
\section{Preliminaires}
\subsection{Harmonic analysis associated with the Weinstein operator}
For all $\lambda=(\lambda_1,...,\lambda_{d+1})\in\mathbb{C}^{d+1}$, the system
\begin{equation}
\begin{gathered}
\frac{\partial^2u}{\partial x_{j}^2}( x)
=-\lambda_{j} ^2u(x), \quad\text{if } 1\leq j\leq d \\
L_{\alpha}u( x) =-\lambda_{d+1}^2u( x), \\
u( 0) =1, \quad \frac{\partial u}{\partial
x_{d+1}}(0)=0,\quad \frac{\partial u}{\partial
x_{j}}(0)=-i\lambda_{j}, \quad \text{if } 1\leq j\leq d
\end{gathered}
\end{equation}
has a unique solution denoted by $\Lambda_{\alpha}^d(\lambda,.),$ and given by
\begin{equation}\label{wkernel}
\Lambda_{\alpha}^d(\lambda,x)=e^{-i<x^\prime,\lambda^\prime>}j_\alpha(x_{d+1}\lambda_{d+1})
\end{equation}
where $x=(x^\prime,x_{d+1}),\; x_d'=(x_1,x_2,\cdots,x_d),\; \lambda=(\lambda^\prime,\lambda_{d+1}) ,\; \lambda_d'=(\lambda_1,\lambda_2,\cdots,\lambda_d)$ and $j_\alpha$ is the normalized Bessel function of index $\alpha$ defined by
$$j_\alpha(x)=\Gamma(\alpha+1)\sum_{k=0}^\infty\frac{(-1)^k x^{2k}}{2^k k!\Gamma(\alpha+k+1)}.$$
The function $(\lambda,x)\mapsto\Lambda_{\alpha}^d(\lambda,x)$ is called the Weinstein kernel and has a unique extension to $\mathbb{C}^{d+1}\times\mathbb{C}^{d+1}$, and satisfied the following properties.\\
\begin{itemize}
\item[(i)] For all $(\lambda,x)\in \mathbb{C}^{d+1}\times\mathbb{C}^{d+1}$ we have
\begin{equation*}
\Lambda_{\alpha}^d(\lambda,x)=\Lambda_{\alpha}^d(x,\lambda).
\end{equation*}
\item[(ii)] For all $(\lambda,x)\in \mathbb{C}^{d+1}\times\mathbb{C}^{d+1}$ we have
\begin{equation*}
\Lambda_{\alpha}^d(\lambda,-x)=\Lambda_{\alpha}^d(-\lambda,x).
\end{equation*}
\item[(iii)] For all $(\lambda,x)\in \mathbb{C}^{d+1}\times\mathbb{C}^{d+1}$ we get
\begin{equation*}
\Lambda_{\alpha}^d(\lambda,0)=1.
\end{equation*}
\item[(iv)] For all $\nu\in\mathbb{N}^{d+1},\;x\in\mathbb{R}^{d+1}$ and $\lambda\in\mathbb{C}^{d+1}$ we have
\begin{equation*}\label{klk}
\left|D_\lambda^\nu\Lambda_{\alpha}^d(\lambda,x)\right|\leq\left\|x\right\|^{\left|\nu\right|}e^{\left\|x\right\|\left\|\Im \lambda\right\|}
\end{equation*}
\end{itemize}
where $D_\lambda^\nu=\partial^\nu/(\partial\lambda_1^{\nu_1}...\partial\lambda_{d+1}^{\nu_{d+1}})$ and $\left|\nu\right|=\nu_1+...+\nu_{d+1}.$ In particular, for all $(\lambda,x)\in \mathbb{R}^{d+1}\times\mathbb{R}^{d+1}$, we have
\begin{equation}\label{normLambda}
\left|\Lambda_{\alpha}^d(\lambda,x)\right|\leq 1.
\end{equation}
In the following we denote by
\begin{itemize}
\item [(i)] $-\lambda=(-\lambda',\lambda_{d+1})$
\item[(ii)] $C_*(\mathbb{R}^{d+1})$, the space of continuous functions on $\mathbb{R}^{d+1},$ even with respect to the last variable.
\item[(iii)] $S_*(\mathbb{R}^{d+1})$, the space of the $C^\infty$ functions, even with respect to the last variable, and rapidly decreasing together with their derivatives.
\item[(iv)] $\mathcal{S_*}(\mathbb{R}^{d+1}\times\mathbb{R}^{d+1})$, the Schwartz space of rapidly decreasing functions on $\mathbb{R}^{d+1}\times\mathbb{R}^{d+1}$ even with respect to the last two variables.
\item[(v)] $\mathcal{D_*}(\mathbb{R}^{d+1})$, the space of $C^\infty$-functions on $\mathbb{R}^{d+1}$ which are of compact support,even with respect to the last variable.
\item[(vi)] $L^p_\alpha(\mathbb{R}^{d+1}_+),\;1\leq p\leq \infty,$ the space of measurable functions $f$ on $\mathbb{R}^{d+1}_+$ such that
$$\left\|f\right\|_{\alpha,p}=\left(\int_{\mathbb{R}^{d+1}_+}\left|f(x)\right|^pd\mu_\alpha(x)\right)^{1/p}<\infty, \;p\in[1,\infty),$$
$$\left\|f\right\|_{\alpha,\infty}=\textrm{ess}\sup_{x\in\mathbb{R}^{d+1}_+}\left|f(x)\right|<\infty,$$
where $d\mu_{\alpha}(x)$ is the measure on $\mathbb{R}_{+}^{d+1}=\mathbb{R}^d\times(0,\infty)$ given by
\begin{equation*}\label{mesure}
d\mu_\alpha(x)=\frac{x^{2\alpha+1}_{d+1}}{(2\pi)^d2^{2\alpha}\Gamma^2(\alpha+1)}dx.
\end{equation*}
\end{itemize}
For a radial function $\varphi\in L_{\alpha}^{1}(\mathbb{R}_{+} ^{d+1})$ the function $\tilde{\varphi}$ defined on $\mathbb{R}_+$ such that $\varphi(x)=\tilde{\varphi}(|x|)$, for all
$x\in\mathbb{R}_{+} ^{d+1}$, is integrable with respect to the measure $r^{2\alpha+d+1}dr$, and we have
\begin{equation}\label{radialweinstein}
\int_{\mathbb{R}_{+}^{d+1}}\varphi(x)d\mu_{\alpha}(x)=a_\alpha\int_{0}^{\infty}
\tilde{\varphi}(r)r^{2\alpha+d+1}dr,
\end{equation}
where $$a_\alpha=\frac{1}{2^{\alpha+\frac{d}{2}}\Gamma(\alpha+\frac{d}{2}+1)}.$$
The Weinstein transform generalizing the usual Fourier transform, is given for
$\varphi\in L_{\alpha}^{1}(\mathbb{R}_{+} ^{d+1})$ and $\lambda\in\mathbb{R}_{+}^{d+1}$, by
\begin{equation}\label{defWeinstein}
\mathcal{F}_{W}
(\varphi)(\lambda)=\int_{\mathbb{R}_{+}^{d+1}}\varphi(x)\Lambda_{\alpha}^d(x, \lambda
)d\mu_{\alpha}(x),
\end{equation}
We list some known basic properties of the Weinstein transform are as follows. For the proofs, we refer \cite{nahia1996spherical, nahia1996mean}.
\begin{itemize}
\item[(i)] For all $\varphi\in L^1_\alpha(\mathbb{R}^{d+1}_+)$, the function $\mathcal{F}_{W}(\varphi)$ is continuous on $\mathbb{R}^{d+1}_+$ and we have
\begin{equation}\label{L1-Linfty}
\left\|\mathcal{F}_{W}\varphi\right\|_{\alpha,\infty}\leq\left\|\varphi\right\|_{\alpha,1}.
\end{equation}
\item[(ii)] The Weinstein transform is a topological isomorphism from $\mathcal{S}_*(\mathbb{R}^{d+1})$ onto itself. The inverse transform is given by
\begin{equation}\label{inversionweinstein}
\mathcal{F}_{W}^{-1}\varphi(\lambda)= \mathcal{F}_{W}\varphi(-\lambda),\;\textrm{for\;all}\;\lambda\in\mathbb{R}^{d+1}_+.
\end{equation}
\item[(iii)] For all $f$ in $\mathcal{D_*}(\mathbb{R}^{d+1})$ (resp. $\mathcal{S_*}(\mathbb{R}^{d+1})$), we have the following relations
\begin{equation}\label{fourierbar}
\forall\lambda\in\mathbb{R}^{d+1}_+,\quad \mathcal{F}_{W}(\overline{\varphi})(\lambda)= \overline{\mathcal{F}_{W}(\widetilde{\varphi})(\lambda)},
\end{equation}
\begin{equation}\label{fourierbar1}
\forall\lambda\in\mathbb{R}^{d+1}_+,\quad \mathcal{F}_{W}(\varphi)(\lambda)= \mathcal{F}_{W}(\widetilde{\varphi})(-\lambda),
\end{equation}
where $\widetilde{\varphi}$ is the function defined by
\begin{equation*}
\forall\lambda\in\mathbb{R}^{d+1}_+,\quad\widetilde{\varphi}(\lambda)=\varphi(-\lambda).
\end{equation*}
\item[(iv)] Parseval's formula: For all $\varphi, \phi\in \mathcal{S}_*(\mathbb{R}^{d+1})$, we have
\begin{equation}\label{MM} \int_{\mathbb{R}^{d+1}_+}\varphi(x)\overline{\phi(x)}d\mu_\alpha(x)=\int_{\mathbb{R}^{d+1}_+}\mathcal{F}_{W}
(\varphi)(x)\overline{\mathcal{F}_{W}(\phi)(x)}d\mu_\alpha(x).
\end{equation}
\item[(v)] Plancherel's formula: For all $\varphi\in L^2_\alpha(\mathbb{R}^{d+1}_+)$, we have
\begin{equation}\label{Plancherel formula}
\left\|\mathcal{F}_{W}\varphi\right\|_{\alpha,2}=\left\|\varphi\right\|_{\alpha,2}.
\end{equation}
\item[(vi)] Plancherel Theorem: The Weinstein transform $\mathcal{F}_{W}$ extends uniquely to an isometric isomorphism on $L^2_\alpha(\mathbb{R}^{d+1}_+).$
\item[(vii)] Inversion formula: Let $\varphi\in L^1_\alpha(\mathbb{R}^{d+1}_+)$ such that $\mathcal{F}_{W}\varphi\in L^1_\alpha(\mathbb{R}^{d+1}_+)$, then we have
\begin{equation}\label{inv}
\varphi(\lambda)=\int_{\mathbb{R}^{d+1}_+}\mathcal{F}_{W}\varphi(x)\Lambda_{\alpha}^d(-\lambda,x)d\mu_\alpha(x),\;\textrm{a.e. }\lambda\in\mathbb{R}^{d+1}_+.
\end{equation}
\end{itemize}
Using relations (\ref{L1-Linfty}) and (\ref{Plancherel formula}) with Marcinkiewicz's interpolation theorem \cite{zbMATH03367521} we deduce that for every $\varphi\in L^p_\alpha(\mathbb{R}^{d+1}_+)$ for all $1\leq p\leq 2$, the function $\mathcal{F}_{W}(\varphi)\in L^q_\alpha(\mathbb{R}^{d+1}_+), q=p/(p-1),$ and
\begin{equation}\label{Lp-Lq}
\left\|\mathcal{F}_{W}\varphi\right\|_{\alpha,q}\leq\left\|\varphi\right\|_{\alpha,p}.
\end{equation}
\begin{defn} The translation operator $\tau^\alpha_x,\;x\in\mathbb{R}^{d+1}_+$ associated with the Weinstein operator $\Delta_{W,\alpha}^d$, is defined for a continuous function $\varphi$ on $\mathbb{R}^{d+1}_+$, which is even with respect to the last variable and for all $y\in\mathbb{R}^{d+1}_+$ by
$$\tau^\alpha_x\varphi(y)=C_\alpha\int_0^\pi\varphi\left(x^\prime+y\prime,\sqrt{x^2_{d+1}+y^2_{d+1}+2x_{d+1}y_{d+1}
\cos\theta}\right)\left(\sin\theta\right)^{2\alpha}d\theta,$$
with $$C_\alpha=\frac{\Gamma(\alpha+1)}{\sqrt{\pi}\Gamma(\alpha+1/2)}.$$
\end{defn}
By using the Weinstein kernel, we can also define a generalized translation, for
a function $\varphi\in\mathcal{S}_*(\mathbb{R}^{d+1})$ and $y\in\mathbb{R}^{d+1}_+$ the generalized translation $\tau^\alpha_x\varphi$ is defined by the following relation
\begin{equation}\label{MMM}
\mathcal{F}_{W}(\tau^\alpha_x\varphi)(y)=\Lambda^d_\alpha(x,y)\mathcal{F}_{W}(\varphi)(y).
\end{equation}
In the following proposition, we give some properties of the Weinstein
translation operator:
\begin{prop} The translation operator $\tau^\alpha_x,\;x\in\mathbb{R}^{d+1}_+$ satisfies the following properties.\\
i). For $\varphi\in\mathbb{C}_*(\mathbb{R}^{d+1})$, we have for all $x,y\in\mathbb{R}^{d+1}_+$
\begin{equation}\label{symtrictrans}
\tau^\alpha_x\varphi(y)=\tau^\alpha_y\varphi(x)\;\textrm{and}\;\tau^\alpha_0\varphi=\varphi.
\end{equation}
ii). Let $\varphi\in L^p_\alpha(\mathbb{R}^{d+1}_+),\;1\leq p\leq \infty$ and $x\in\mathbb{R}^{d+1}_+$. Then $\tau^\alpha_x\varphi$ belongs to $L^p_\alpha(\mathbb{R}^{d+1}_+)$ and we have
\begin{equation}\label{ineqtransl}
\left\| \tau^\alpha_x\varphi\right\|_{\alpha,p}\leq \left\|\varphi\right\|_{\alpha,p}.
\end{equation}
\end{prop}
\begin{prop}
Let $\varphi\in L^1_{\alpha}(\mathbb{R}^{d+1}_+)$. Then for all $x\in \mathbb{R}^{d+1}_+$,
\begin{equation}\label{integraltransrad}
\int_{\mathbb{R}^{d+1}_+}\tau^\alpha_x\varphi(y)d\mu_\alpha(y)= \int_{\mathbb{R}^{d+1}_+}\varphi(y) d\mu_\alpha(y).
\end{equation}
\end{prop}
\begin{proof}
The result comes from combination identities (\ref{inv}) and (\ref{MMM}).
\end{proof}
By using the generalized translation, we define the generalized convolution product $\varphi*\psi$ of the functions $\varphi,\;\psi\in L^1_\alpha(\mathbb{R}^{d+1}_+)$ as follows
\begin{equation}\label{defconvolution}
\varphi*\psi(x)=\int_{\mathbb{R}^{d+1}_+}\tau^\alpha_x\varphi(-y)\psi(y)d\mu_\alpha(y).
\end{equation}
\\
This convolution is commutative and associative, and it satisfies the following properties.
\begin{prop}\label{propconvol}
i) For all $\varphi,\psi\in L^1_\alpha(\mathbb{R}^{d+1}_+),$\;(resp. $\varphi,\psi\in \mathcal{S}_*(\mathbb{R}^{d+1})$), then $\varphi*\psi\in L^1_\alpha(\mathbb{R}^{d+1}_+),$\;(resp. $\varphi*\psi\in \mathcal{S}_*(\mathbb{R}^{d+1})$) and we have
\begin{equation}\label{F(f*g)}
\mathcal{F}_{W}(\varphi*\psi)=\mathcal{F}_{W}(\varphi)\mathcal{F}_{W}(\psi).
\end{equation}
ii) Let $p, q, r\in [1,\infty],$ such that $\frac{1}{p}+\frac{1}{q}-\frac{1}{r}=1.$ Then for all $\varphi\in L^p_\alpha(\mathbb{R}^{d+1}_+)$ and $\psi\in L^q_\alpha(\mathbb{R}^{d+1}_+)$ the function $\varphi*\psi$ belongs to $L^r_\alpha(\mathbb{R}^{d+1}_+)$ and we have
\begin{equation}\label{invconvol2}
\left\|\varphi*\psi\right\|_{\alpha,r}\leq\left\|\varphi\right\|_{\alpha,p}\left\|\psi\right\|_{\alpha,q}.
\end{equation}
iii) Let $\varphi,\psi\in L^2_\alpha(\mathbb{R}^{d+1}_+)$. Then
\begin{equation}
\varphi*\psi=\mathcal{F}_{W}^{-1}\left(\mathcal{F}_{W}(\varphi)\mathcal{F}_{W}(\psi)\right).
\end{equation}
iv) Let $\varphi,\psi\in L^2_\alpha(\mathbb{R}^{d+1}_+)$. Then $\varphi*\psi$ belongs to $L^2_\alpha(\mathbb{R}^{d+1}_+)$ if and only if $\mathcal{F}_{W}(\varphi)\mathcal{F}_{W}(\psi)$ belongs to $L^2_\alpha(\mathbb{R}^{d+1}_+)$ and we have
\begin{equation}
\mathcal{F}_{W}(\varphi*\psi)=\mathcal{F}_{W}(\varphi)\mathcal{F}_{W}(\psi).
\end{equation}
v) Let $\varphi,\psi\in L^2_\alpha(\mathbb{R}^{d+1}_+)$. Then
\begin{equation}
\|\varphi*\psi\|_{\alpha,2}=\|\mathcal{F}_{W}(\varphi)\mathcal{F}_{W}(\psi)\|_{\alpha,2},
\end{equation}
where both sides are finite or infinite.
\end{prop}
\subsection{Weinstein two-wavelet theory}
In the following, we denote by \\
$\X=\left\{(a,x): x\in \mathbb{R}^{d+1}_+ \;\text{and}\; a>0\right\}$.\\
$L^p_{\alpha}(\X),\; p\in [1,\infty]$ the space of measurable functions $\varphi$ on $\X$ such that
\begin{eqnarray*}
\|\varphi\|_{L^p_{\alpha}(\X)} &=& \left(\int_{\X}|\varphi(a,x)|^p d\mu_\alpha(a,x)\right)^\frac{1}{p}<\infty,\quad 1\leq p<\infty, \\
\|\varphi\|_{L^\infty_{\alpha}(\X)}&=& \esssup_{(a,x)\in\X}|\varphi(a,x)|<\infty,
\end{eqnarray*}
where the measure $\mu_\alpha(a,x)$ is defined on $\X$ by
$$d\mu_\alpha(a,x)=\frac{d\mu_\alpha(x)da}{a^{2\alpha+d+3}}.$$
\begin{defn}\cite{gasmi2016inversion}
A classical wavelet on $\mathbb{R}^{d+1}_+$ is a measurable function $\varphi$ on $\mathbb{R}^{d+1}_+$
satisfying for almost all $\xi\in \mathbb{R}^{d+1}_+$, the condition
\begin{equation}\label{defwave}
0<C_\varphi=\int_{0}^{\infty}|\mathcal{F}_{W}(\varphi)(a\xi)|^2\frac{da}{a}<\infty.
\end{equation}
\end{defn}
We extend the notion of the wavelet to the two-wavelet in Weinstein setting as follows.
\begin{defn}
Let $\varphi$ and $\psi$ be in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$. We say that the pair $(\varphi,\psi)$ is a Weinstein
two-wavelet on $\mathbb{R}^{d+1}_+$ if the following integral
\begin{equation}\label{deftwowave}
C_{\varphi,\psi}=\int_{0}^{\infty}\mathcal{F}_{W}(\psi)(a\xi)\overline{\mathcal{F}_{W}(\varphi)
(a\xi)}\frac{da}{a}
\end{equation}
is constant for almost all $\xi\in \mathbb{R}^{d+1}_+$ and we call the number $ C_{\varphi,\psi}$ the Weinstein two-wavelet constant associated to the functions $\varphi$ and $\psi$.
\end{defn}
It is to highlight that if $\varphi$ is a Weinstein wavelet then the pair $(\varphi,\psi)$ is a Weinstein two-wavelet,
and $C_{\varphi,\psi}$ coincides with $ C_{\varphi}$.
Let $a>0$ and $\varphi$ be a measurable function. We consider the function $\varphi_a$ defined by
\begin{equation}\label{fia}
\forall x\in \mathbb{R}^{d+1}_+, \quad \varphi_a(x)=\frac{1}{a^{2\alpha+d+2}}\varphi\left(\frac{x}{a}\right).
\end{equation}
\begin{prop}
\begin{enumerate}
\item Let $a>0$ and $\varphi\in L^p_{\alpha}(\mathbb{R}^{d+1}_+),\;p\in[1,\infty]$. The function $\varphi_a$ belongs to $L^p_{\alpha}(\mathbb{R}^{d+1}_+)$ and we have
\begin{equation}\label{normLpfia}
\|\varphi_a\|_{\alpha,p}=a^{(2\alpha+d+2)(\frac{1}{p}-1)} \|\varphi\|_{\alpha,p}.
\end{equation}
\item Let $a>0$ and $\varphi\in L^1_{\alpha}(\mathbb{R}^{d+1}_+)\cup L^2_{\alpha}(\mathbb{R}^{d+1}_+)$. Then, we have
\begin{equation}\label{fourierfia}
\mathcal{F}_{W}(\varphi_a)(\xi)=\mathcal{F}_{W}(\varphi)(a\xi),\quad\xi\in \mathbb{R}^{d+1}_+.
\end{equation}
\end{enumerate}
\end{prop}
For $a>0$ and $\varphi\in L^2_{\alpha}(\mathbb{R}^{d+1}_+)$, we consider the family $\varphi_{a,x},\; x\in
\mathbb{R}^{d+1}_+$ of Weinstein wavelets on $\mathbb{R}^{d+1}_+$ in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$ defined by
\begin{equation}\label{deffiax}
\forall y\in \mathbb{R}^{d+1}_+,\quad \varphi_{a,x}=a^{\alpha+1+\frac{d}{2}}\tau^\alpha_x\varphi_a(y).
\end{equation}
\begin{rem}
\begin{enumerate}
\item Let $\varphi$ be a function in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$, then we have
\begin{equation}\label{Nfiax2}
\forall (a,x)\in\X, \quad \|\varphi_{a,x}\|_{\alpha,2}\leq \|\varphi\|_{\alpha,2}.
\end{equation}
\item Let $p\in [1,\infty]$ and $\varphi$ be a function in $L^p_{\alpha}(\mathbb{R}^{d+1}_+)$, then we have
\begin{equation}\label{Nfiaxp}
\forall (a,x)\in\X, \quad \|\varphi_{a,x}\|_{\alpha,p}\leq a^{(2\alpha+d+2)(\frac{1}{p}-\frac{1}{2})} \|\varphi\|_{\alpha,p}.
\end{equation}
\end{enumerate}
\end{rem}
\begin{defn}\cite{mejjaoli2017new}
Let $\varphi$ be a Weinstein wavelet on $\mathbb{R}^{d+1}_+$ in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$. The Weinstein continuous wavelet transform $\Phi^W_\varphi$ on $\mathbb{R}^{d+1}_+$ is defined for regular functions $f$ on $\mathbb{R}^{d+1}_+$ by
\begin{equation}\label{contwave}
\forall (a,x)\in\X,\quad \Phi^W_\varphi(f)(a,x)=\int_{\mathbb{R}^{d+1}_+}f(y)\overline{\varphi_{a,x}(y)} d\mu_\alpha(y)=\langle f, \varphi_{a,x}\rangle_{\alpha,2}.
\end{equation}
\end{defn}
This transform can also be written in the form
\begin{equation}\label{recontwave}
\Phi^W_\varphi(f)(a,x)=a^{\alpha+1+\frac{d}{2}}\check{f}*\overline{\varphi_a}(x).
\end{equation}
\begin{rem}
\begin{enumerate}
\item Let $\varphi$ be a function in $L^p_{\alpha}(\mathbb{R}^{d+1}_+)$, and Let $f$ be a function in $L^q_{\alpha}(\mathbb{R}^{d+1}_+)$, with $p\in [1,\infty]$, we define the Weinstein continuous wavelet transform $\Phi^W_\varphi(f)$ by the relation (\ref{recontwave}).
\item Let $\varphi$ be a Weinstein wavelet on $\mathbb{R}^{d+1}_+$ in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$. Then from the relations (\ref{Nfiax2}) and (\ref{contwave}), we have for all $f\in L^2_{\alpha}(\mathbb{R}^{d+1}_+)$
\begin{equation}\label{Nwcont2}
\|\Phi^W_\varphi(f)\|_{\alpha,\infty}\leq \|f\|_{\alpha,2}\|\varphi\|_{\alpha,2}.
\end{equation}
\item Let $\varphi$ be a function in $L^p_{\alpha}(\mathbb{R}^{d+1}_+)$, with $p\in [1,\infty]$, then from the inequality (\ref{invconvol2}) and the identity (\ref{recontwave}), we have for all $f\in L^q_{\alpha}(\mathbb{R}^{d+1}_+)$
\begin{equation}\label{Nwcontp}
\|\Phi^W_\varphi(f)\|_{\alpha,\infty}\leq \|f\|_{\alpha,q}\|\varphi\|_{\alpha,p}.
\end{equation}
\end{enumerate}
\end{rem}
\begin{thm}(Parseval's formula)\label{Parseval's formula2wave}\cite{saoudi2020two}
Let $(\varphi,\psi)$ be a Weinstein two-wavelet. Then for all $\varphi$ and $\psi$ in $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$, we have the following Parseval type formula
\begin{equation}\label{ParsevalPsi}
\int_{\X}\Phi^W_\varphi(f)(a,x)\overline{\Phi^W_\psi(g)(a,x)}d\mu_\alpha(a,x)=C_{\varphi,\psi}\int_{\mathbb{R}^{d+1}_+}f(x)\overline{g(x)} d\mu_\alpha(x),
\end{equation}
where $ C_{\varphi,\psi}$ is the Weinstein two-wavelet constant associated to the functions $\varphi$ and $\psi$ given by the identity (\ref{deftwowave}).
\end{thm}
\begin{cor}\cite{saoudi2020two}
Let $(\varphi,\psi)$ be a Weinstein two-wavelet. Then we have the following assertion:
If the Weinstein two-wavelet constant $C_{\varphi,\varphi}=0$, then $\Phi^W_\varphi\left(L^2_{\alpha}(\mathbb{R}^{d+1}_+)\right)$ and $\Phi^W_\psi\left(L^2_{\alpha}(\mathbb{R}^{d+1}_+)\right)$ are orthogonal.
\end{cor}
\begin{thm}(Inversion formula)\cite{saoudi2020two} Let $(\varphi,\psi)$ be a Weinstein two-wavelet. For all $f\in L^1_{\alpha}(\mathbb{R}^{d+1}_+)$ (resp. $L^2_{\alpha}(\mathbb{R}^{d+1}_+)$) such that $\mathcal{F}_{W}(f)$ belongs to $f\in L^1_{\alpha}(\mathbb{R}^{d+1}_+)$
(resp. $L^1_{\alpha}(\mathbb{R}^{d+1}_+)$ $\cap L^\infty_{\alpha}(\mathbb{R}^{d+1}_+)$), we have
\begin{equation}\label{inversion2wave}
f(y)=\frac{1}{C_{\varphi,\psi}}\int_{0}^{\infty}\int_{\mathbb{R}^{d+1}_+}\Phi^W_\varphi(f)(a,x)\psi_{a,x}(y)d\mu_\alpha(a,x),
\end{equation}
where for each $y\in\mathbb{R}^{d+1}_+ $\, both the inner integral and the outer integral are absolutely
convergent, but eventually not the double integral.
\end{thm}
\section{The Weinstein two-wavelet localization operators}
In this section, we will give a host of sufficient conditions for the boundedness and compactness of the two-wavelet localization operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ on $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ for all $1\leq p\leq \infty$, in terms of properties of the symbol $\sigma$ and the functions $\varphi$ and $\psi$.
\begin{defn} Let $\varphi, \psi$ be measurable functions on $\mathbb{R}_{+}^{d+1}$, $\sigma$ be measurable function on $\mathcal{X}$, we the two-wavelet localization operator noted by $\mathcal{L}_{\varphi,\psi}(\sigma )$, on $L^p_\alpha(\mathbb{R}^{d+1}_+)$, $1 \leq p\leq \infty$, by
\begin{equation}\label{two-waveloc}
\mathcal{L}_{\varphi,\psi}(\sigma)(f)(y)=\int_{\X}
\sigma(a,x)\Phi^W_\varphi(f)(a,x)\psi_{a,x}(y)d\mu_\alpha(a,x),\ \ y\in\mathbb{R}_{+}^{d+1}.
\end{equation}
\end{defn}
In accordance with the different choices of the symbols $\sigma$ and the different continuities
required, we need to impose different conditions on the functions $\varphi$ and $\psi$, and then we obtain a
two-wavelet localization operator on $L^p_\alpha(\mathbb{R}^{d+1}_+)$.
It is often more practical to interpret the definition of the localization operator in a weak sense as following: for all $f$ in $L^p_\alpha(\mathbb{R}^{d+1}_+)$, $1 \leq p\leq \infty$, and $f$ in $L^q_\alpha(\mathbb{R}^{d+1}_+)$
\begin{equation}\label{weaktwo-waveloc}
\langle\mathcal{L}_{\varphi,\psi}(\sigma)(f),g\rangle_{\alpha,2}=\int_{\X}
\sigma(a,x)\Phi^W_\varphi(f)(a,x)\overline{\Phi^W_\psi(g)(a,x)}d\mu_\alpha(a,x),\ \ y\in\mathbb{R}_{+}^{d+1}.
\end{equation}
\begin{prop} Let $1 \leq p\leq \infty$. Then the adjoint of the two-wavelet localization operator
$$\mathcal{L}_{\varphi, \psi}(\sigma):\ L^p_\alpha(\mathbb{R}^{d+1}_+)\ \longrightarrow\ L^p_\alpha(\mathbb{R}^{d+1}_+),$$ is $\mathcal{L}_{\psi, \varphi}(\overline{\sigma}):\ L^q_\alpha(\mathbb{R}^{d+1}_+)\ \longrightarrow\ L^q_\alpha(\mathbb{R}^{d+1}_+)$.
\end{prop}
\begin{proof} Let $f$ in $L^p_\alpha(\mathbb{R}^{d+1}_+)$ and $g$ in $L^q_\alpha(\mathbb{R}^{d+1}_+)$. Then we have from the relation (\ref{weaktwo-waveloc})
\begin{eqnarray*}
\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}&=& \int_{\X}
\sigma(a,x)\Phi^W_\varphi(f)(a,x)\overline{\Phi^W_\psi(g)(a,x)}d\mu_\alpha(a,x)\\
&=&\overline{\int_{\X}
\overline{\sigma(a,x)}\Phi^W_\psi(f)(a,x)\overline{\Phi^W_\varphi(g)(a,x)}d\mu_\alpha(a,x)}
\\&=&\overline{\langle \mathcal{L}_{\psi, \varphi}(\overline{\sigma})(g),\ f \rangle_{\alpha,2}}
\\&=&\langle f, \mathcal{L}_{\psi, \varphi}(\overline{\sigma})(g) \rangle_{\alpha,2}.
\end{eqnarray*}
Thus,
\begin{eqnarray}\label{lphp}
\mathcal{L}_{\varphi, \psi}^{*}(\sigma)=\mathcal{L}_{\psi, \varphi}(\overline{\sigma}).
\end{eqnarray}
\end{proof}
\subsection{$L_\alpha^p$-Boundedness of $\mathcal{L}_{\varphi,\psi}(\sigma)$}
For $1 \leq p\leq \infty$, put $\sigma\in L^1_\alpha(\X)$, $\psi \in L^p_\alpha(\mathbb{R}^{d+1}_+)$ and $\varphi\in L^q_\alpha(\mathbb{R}^{d+1}_+)$. In this subsection, we are going to show that the Weinstein two-wavelet localization operator
$\mathcal{L}_{\varphi,\psi}(\sigma)$ is a bounded operator on $L^p_\alpha(\mathbb{R}^{d+1}_+)$.
\begin{prop}\label{prop31}
Let $\sigma\in L^{1}_{\alpha}(\X)$, $\varphi\in L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$
and $\psi\in L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$, then the localization
operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is bounded and linear from $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{1}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{prop}
\begin{proof}
Let $f$ be a function in $L^{1}_{\alpha}(\X)$. From the definition of the two-wavelet localization operator (\ref{two-waveloc})
and according to relations (\ref{deffiax}), (\ref{recontwave}), (\ref{invconvol2}) and (\ref{ineqtransl}), we have
\begin{eqnarray*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)(f)\|_{\alpha,1} &\leq & \int_{\mathbb{R}^{d+1}_+} \int_{\X}
|\sigma(a,x)| \, |\Phi^W_\varphi(f)(a,x)| \, |\psi_{a,x}(y)|d\mu_\alpha(a,x)d\mu_\alpha(y)\\
&\leq & \int_{\mathbb{R}^{d+1}_+} \int_{\X}
|\sigma(a,x)| \, |\check{f}*\overline{\varphi_a(x)}| \, |\tau^\alpha_x\psi_{a}(y)|d\mu_\alpha(x)\frac{da}{a} d\mu_\alpha(y) \\
&\leq & \int_{\mathbb{R}^{d+1}_+} \int_{\X}
|\sigma(a,x)| \, \|f\|_{\alpha,1}\|\varphi_a\|_{\alpha,\infty} \, |\tau^\alpha_x\psi_{a}(y)|d\mu_\alpha(x)\frac{da}{a} d\mu_\alpha(y) \\
&\leq & \|f\|_{\alpha,1}\|\varphi\|_{\alpha,\infty} \int_{\X}|\sigma(a,x| \left[\int_{\mathbb{R}^{d+1}_+} |\tau^\alpha_x\psi_{a}(y)|d\mu_\alpha(y) \right] d\mu_\alpha(x)\frac{da}{a^{2\alpha+d+3}} \\
&\leq & \|f\|_{\alpha,1}\|\varphi\|_{\alpha,\infty} \int_{\X}|\sigma(a,x| \left[\int_{\mathbb{R}^{d+1}_+} |\psi_{a}(y)|d\mu_\alpha(y) \right] d\mu_\alpha(x)\frac{da}{a^{2\alpha+d+3}} \\
&\leq & \|f\|_{\alpha,1}\|\varphi\|_{\alpha,\infty}\|\psi\|_{\alpha,1}\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{eqnarray*}
Therefore,
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{1}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{proof}
\begin{prop}\label{prop32}
Let $\sigma\in L^{1}_{\alpha}(\X)$, $\varphi\in L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$
and $\psi\in L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$, then the localization
operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is bounded and linear from $L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{prop}
\begin{proof}
Let $f$ be a function in $L^{\infty}_{\alpha}(\X)$. From the definition of the two-wavelet localization operator (\ref{two-waveloc})
and as above, according to relations (\ref{deffiax}), (\ref{recontwave}), (\ref{invconvol2}) and (\ref{ineqtransl}), we have for all $y\in \mathbb{R}^{d+1}_+$
\begin{eqnarray*}
|\mathcal{L}_{\varphi,\psi}(\sigma)(f)(y)|&\leq & \int_{\X}
|\sigma(a,x)|\,|\Phi^W_\varphi(f)(a,x)|\,|\psi_{a,x}(y)|d\mu_\alpha(a,x) \\
&\leq & \|f\|_{\alpha,\infty} \|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty} \|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{eqnarray*}
Thus,
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,\infty} \|\psi\|_{\infty,1}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{proof}
By interpolations of the results of Propositions \ref{prop31} and \ref{prop32}, we get the following result.
\begin{thm}\label{thm3.3}
Let $\varphi$ and $\psi$ be functions in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\cap L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$. Then for all $\sigma\in L^{1}_{\alpha}(\X)$, there exists a unique bounded linear operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself with $1\leq p\leq \infty$, such that
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,1}^\frac{1}{q} \|\psi\|_{\alpha,1}^\frac{1}{p} \|\varphi\|_{\alpha,\infty}^\frac{1}{p} \|\psi\|_{\alpha,\infty}^\frac{1}{q}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{thm}
In the following proposition, we generalize and we improve Proposition \ref{prop32}.
\begin{prop}\label{proppq}
Let $\sigma\in L^{1}_{\alpha}(\X)$, $\psi\in L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$
and $\varphi\in L^{q}_{\alpha}(\mathbb{R}^{d+1}_+)$, for $1< p \leq \infty$ then the localization
operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is bounded linear operator from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,q} \|\psi\|_{\alpha,p}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{prop}
\begin{proof}
Let $f$ be a function in $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$. We consider the linear functional
\begin{eqnarray*}
\begin{array}{lccl}
\mathcal{J}_f:& L^{q}_{\alpha}(\mathbb{R}^{d+1}_+) &\longrightarrow& \mathbb{C} \\
&g &\longmapsto& \langle g,\mathcal{L}_{\varphi,\psi}(\sigma)(f)\rangle_{\alpha,2}.
\end{array}
\end{eqnarray*}
According to relation (\ref{weaktwo-waveloc}), we have
\begin{eqnarray*}
|\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f), g \rangle_{\alpha,2}| &\leq& \int_{\X}
|\sigma(a,x)|\,|\Phi^W_\varphi(f)(a,x)|\,|\overline{\Phi^W_\psi(g)(a,x)}|d\mu_\alpha(a,x) \\
&\leq & \|\Phi^W_\varphi(f)\|_{L^{\infty}_{\alpha}(\X)} \|\Phi^W_\psi(g)\|_{L^{\infty}_{\alpha}(\X)} \|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{eqnarray*}
Next, using the relations (\ref{recontwave}) and (\ref{invconvol2}), we get
$$ |\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f), g \rangle_{\alpha,2}|\leq \|\sigma\|_{L^{1}_{\alpha}(\X)}\|\varphi\|_{\alpha,q} \|\psi\|_{\alpha,p} \|f\|_{\alpha,p} \|g\|_{\alpha,q}.$$
Therefore, the operator $\mathcal{J}_f$ is a continuous linear functional on $L^{q}_{\alpha}(\mathbb{R}^{d+1}_+)$, and we have
$$\|\mathcal{J}_f\|_{\mathcal{B}(L^{q}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\sigma\|_{L^{1}_{\alpha}(\X)}\|\varphi\|_{\alpha,q} \|\psi\|_{\alpha,p} \|f\|_{\alpha,p} .$$
Like that $\mathcal{J}_f(g)=\langle g,\mathcal{L}_{\varphi,\psi}(\sigma)(f)\rangle_{\alpha,2}$, then by the Riesz representation theorem, we have
$$ \|\mathcal{L}_{\varphi,\psi}(\sigma)(f)\|_{\alpha,p}= \|\mathcal{J}_f\|_{\mathcal{B}(L^{q}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\sigma\|_{L^{1}_{\alpha}(\X)}\|\varphi\|_{\alpha,q} \|\psi\|_{\alpha,p} \|f\|_{\alpha,p},$$
which completes the proof.
\end{proof}
By combining the results obtained in Propositions \ref{prop31} and \ref{proppq}, we have the following version of the $L^{p}_{\alpha}$-boundedness result.
\begin{thm}\label{thm6}
Let $\sigma\in L^{1}_{\alpha}(\X)$, $\psi\in L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$
and $\varphi\in L^{q}_{\alpha}(\mathbb{R}^{d+1}_+)$, for $1\leq p \leq \infty$, then the localization
operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is a bounded and linear from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself, and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \|\varphi\|_{\alpha,q} \|\psi\|_{\alpha,p}
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{thm}
According to Schur technique, we can obtain an $L_\alpha^p$-boundedness result as in the previous
Theorem with crude estimate of the norm $ \|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}$.
\begin{thm}\label{thmR}
Let $\sigma\in L^{1}_{\alpha}(\X)$, $\varphi$ and $\psi$ in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\cap L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$. Then there exists a unique bounded linear operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself with $1\leq p\leq \infty$ and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \max\left( \|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty}, \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\right)
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{thm}
\begin{proof}
We put the function $\mathcal{R}$ defined on $\mathbb{R}^{d+1}_+\times\mathbb{R}^{d+1}_+$ by
$$\mathcal{R}(y,z)=\int_{\X}
\sigma(a,x)\overline{\varphi_{a,x}(z)} \psi_{a,x}(y) d\mu_{\alpha}(a,x).$$
Then the Weinstein two-wavelet localization operator can be written
in terms of $\mathcal{R}(y,z)$ as follows
$$\mathcal{L}_{\varphi,\psi}(\sigma)(f)(y)=\int_{\mathbb{R}_{+}^{d+1}}\mathcal{R}(y,z)f(z) d\mu_{\alpha}(z).$$
Next, it is easy to see that
$$\int_{\mathbb{R}_{+}^{d+1}}|\mathcal{R}(y,z)| d\mu_{\alpha}(y)\leq \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\|\sigma\|_{L^{1}_{\alpha}(\X)},\quad z\in \mathbb{R}_{+}^{d+1}, $$
and
$$\int_{\mathbb{R}_{+}^{d+1}}|\mathcal{R}(y,z)| d\mu_{\alpha}(z)\leq \|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty} \|\sigma\|_{L^{1}_{\alpha}(\X)}, \quad y\in \mathbb{R}_{+}^{d+1}.$$
Then by Schur Lemma (cf.\cite{folland1995introduction}), we can conclude that the localization operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is bounded and linear from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself for all $1\leq p\leq \infty$, and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \max\left( \|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty}, \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\right)
\|\sigma\|_{L^{1}_{\alpha}(\X)}.
\end{equation*}
\end{proof}
The previous Theorem tells us that the unique bounded linear operator on the spaces $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+),$ $1\leq p\leq \infty$, obtained in Theorem \ref{thmR}, is in fact the
integral operator on $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+),$ $1\leq p\leq \infty$ with kernel $\mathcal{R}$.
Subsequently, we can now state and prove the main result in this subsection.
\begin{thm}\label{thm3.7}
Let $\sigma\in L^{r}_{\alpha}(\X)$, $r\in [1,2]$, and $\varphi,\psi$ in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\cap L^{2}_{\alpha}(\mathbb{R}^{d+1}_+)\cap L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$. Then there exists a unique bounded linear operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ from $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$ onto itself for all $p\in [r,r']$ and we have
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq K_1^tK_2^{1-t}
\|\sigma\|_{L^{r}_{\alpha}(\X)},
\end{equation*}
where
\begin{eqnarray*}
K_1 &=& \left(\|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\right)^{\frac{2}{r}-1} \left(\sqrt{C_\varphi C_\psi}\|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1}{r'}, \\
K_2 &=& \left(\|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty}\right)^{\frac{2}{r}-1} \left(\sqrt{C_\varphi C_\psi}\|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1}{r'},
\end{eqnarray*}
and $$\frac{t}{r}+\frac{1-t}{r'}=\frac{1}{p}.$$
\end{thm}
\begin{proof}
Consider the linear functional
\begin{eqnarray*}
\begin{array}{lccl}
\mathcal{J}: (L^{1}_{\alpha}(\X)\cap L^{2}_{\alpha}(\X)\times (L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\,\cap & \!\!\! L^{2}_{\alpha}(\mathbb{R}^{d+1}_+))& \longrightarrow & L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\cap L^{2}_{\alpha}(\mathbb{R}^{d+1}_+) \\
& (\sigma,f) &\longmapsto & \mathcal{L}_{\varphi,\psi}(\sigma)(f). \end{array}
\end{eqnarray*}
According to Proposition \ref{prop31}, we obtain
\begin{equation*}
\|\mathcal{J}(\sigma,f)\|_{\alpha,1}\leq \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1} \|f\|_{\alpha,1} \|\sigma\|_{L^{1}_{\alpha}(\X)}
\end{equation*}
and from \cite[Theorem 3.1]{mejjaoli2017new}, we have
\begin{equation*}
\|\mathcal{J}(\sigma,f)\|_{\alpha,2}\leq \left(\sqrt{C_{\varphi}C_{\psi}} \|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1}{2} \|f\|_{\alpha,2} \|\sigma\|_{L^{2}_{\alpha}(\X)}.
\end{equation*}
Therefore, by the multi-linear interpolation theory \cite[ Section 10.1]{calderon1964intermediate}, we obtain a unique bounded linear operator
$$ \mathcal{J}: L^{r}_{\alpha}(\X)\times L^{r}_{\alpha}(\mathbb{R}^{d+1}_+) \longrightarrow L^{r}_{\alpha}(\mathbb{R}^{d+1}_+) $$
such that
\begin{equation}\label{Jr}
\|\mathcal{J}(\sigma,f)\|_{\alpha,r}\leq K_1\|f\|_{\alpha,r} \|\sigma\|_{L^{r}_{\alpha}(\X)},
\end{equation}
where
$$K_1= \left(\|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\right)^\theta \left(\sqrt{C_{\varphi}C_{\psi}}\|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1-\theta}{2}$$
and
$$\frac{\theta}{1}+\frac{1-\theta}{2}=\frac{1}{r}.$$
By the definition of the linear functional $\mathcal{J}$, we have
\begin{equation}\label{Lbr}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{r}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq \left(\|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\right)^{\frac{2}{r}-1} \left(\sqrt{C_{\varphi}C_{\psi}}\|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1}{r'}
\|\sigma\|_{L^{r}_{\alpha}(\X)}.
\end{equation}
Like the adjoint of $\mathcal{L}_{\varphi,\psi}(\sigma)$ is $\mathcal{L}_{\psi,\varphi}(\overline{\sigma})$, therefore $\mathcal{L}_{\varphi,\psi}(\sigma)$ is a bounded linear map on $L^{r'}_{\alpha}(\mathbb{R}^{d+1}_+)$ with
its operator norm
\begin{equation}\label{Jr'}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{r'}_{\alpha}(\mathbb{R}^{d+1}_+))}= \|\mathcal{L}_{\psi,\varphi}(\overline{\sigma})\|_{\mathcal{B}(L^{r}_{\alpha}(\mathbb{R}^{d+1}_+))} \leq K_2 \|\sigma\|_{L^{r}_{\alpha}(\X)},
\end{equation}
where
$$ K_2 = \left(\|\varphi\|_{\alpha,1} \|\psi\|_{\alpha,\infty}\right)^{\frac{2}{r}-1} \left(\sqrt{C_{\varphi}C_{\psi}}\|\varphi\|_{\alpha,2} \|\psi\|_{\alpha,2}\right)^\frac{1}{r'}.$$
Finally, by interpolation of (\ref{Lbr}) and (\ref{Jr'}), we obtain that, for all $p\in [r,r']$,
\begin{equation*}
\|\mathcal{L}_{\varphi,\psi}(\sigma)\|_{\mathcal{B}(L^{p}_{\alpha}(\mathbb{R}^{d+1}_+))}\leq K_1^tK_2^{1-t}
\|\sigma\|_{L^{r}_{\alpha}(\X)},
\end{equation*}
with
$$\frac{t}{r}+\frac{1-t}{r'}=\frac{1}{p}.$$
\end{proof}
\subsection{$L^p_\alpha$-Compactness of $\mathcal{L}_{\varphi,\psi}(\sigma)$}
In this subsection, we establish the compactness of the Weinstein two-wavelet localization operators $\mathcal{L}_{\varphi,\psi}(\sigma)$ on $L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$, $1\leq p\leq \infty$. Let us start with the following proposition.
\begin{prop}\label{prop3.15}
Under the same assumptions of Theorem \ref{thm3.3}, the Weinstein two-wavelet localization operator
$$\mathcal{L}_{\varphi,\psi}(\sigma):L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$$
is compact.
\end{prop}
\begin{proof}
Let $(f_n)_{n\in\mathbb{N}}\in L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$ a sequence of functions that converge weakly to $0$ in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$ as $n$ converge to $\infty$. So, to show the compactness of of localization operators, it is enough to prove that
\begin{equation*}
\lim_{n\to \infty} \|\mathcal{L}_{\varphi,\psi}(\sigma)(f_n)\|_{\alpha,1}=0.
\end{equation*}
We have
\begin{eqnarray}\label{3.9}
\|\mathcal{L}_{\varphi,\psi}(\sigma)(f_n)\|_{\alpha,1} \leq \int_{\mathbb{R}_{+}^{d+1}} \int_{\X}
|\sigma(a,x)| |\langle f_n,\varphi_{a,x}
\rangle_{\alpha,2}||\psi_{a,x}(y)| d\mu_{\alpha}(a,x) d\mu_{\alpha}(y).
\end{eqnarray}
Using the fact that $(f_n)_{n\in\mathbb{N}}$ converge weakly to $0$ in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$ as $n$ converge to $\infty$, we deduce
\begin{equation}\label{3.10}
\forall a>0, \forall x,y,\in \mathbb{R}_{+}^{d+1},\quad\lim_{n\to \infty} |\sigma(a,x)| |\langle f_n,\varphi_{a,x}
\rangle_{\alpha,2}||\psi_{a,x}(y)|=0.
\end{equation}
Moreover, as $f_n\rightharpoonup 0$ weakly in $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$, then there exists a positive
constant $C$ such that $\|f_n\|_{\alpha,1}\leq C$. Hence,
$$\forall a>0, \forall x,y,\in \mathbb{R}_{+}^{d+1},\quad|\sigma(a,x)| |\langle f_n,\varphi_{a,x}
\rangle_{\alpha,2}||\psi_{a,x}(y)| $$
\begin{equation}\label{3.11}
\leq C |\sigma(a,x)| \|\varphi\|_{\alpha,\infty}|\tau_x^\alpha\psi(y)|.
\end{equation}
On the other hand, from Fubini's theorem and relation (\ref{ineqtransl}), we obtain
\begin{eqnarray}\label{3.12}
\int_{\mathbb{R}_{+}^{d+1}} && \!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\int_{\X}
|\sigma(a,x)| |\langle f_n,\varphi_{a,x}
\rangle_{\alpha,2}||\psi_{a,x}(y)| d\mu_{\alpha}(a,x) d\mu_{\alpha}(y) \nonumber\\
&\leq& C \|\varphi\|_{\alpha,\infty}\int_{\X} |\sigma(a,x)|\int_{\mathbb{R}_{+}^{d+1}}|\tau_x^\alpha\psi_{a}(y)| d\mu_{\alpha}(y) d\mu_{\alpha}(a,x) \nonumber\\
&\leq& C \|\varphi\|_{\alpha,\infty}\int_{\X} |\sigma(a,x)|\int_{\mathbb{R}_{+}^{d+1}}|\psi_{a}(y)| d\mu_{\alpha}(y) d\mu_{\alpha}(a,x) \nonumber\\
&\leq& C \|\varphi\|_{\alpha,\infty} \|\psi\|_{\alpha,1}\|\sigma\|_{L^{1}_{\alpha}(\X)}<\infty.
\end{eqnarray}
Thus, according to the relations (\ref{3.9})$-$(\ref{3.12}) and the Lebesgue dominated convergence theorem
we deduce that
\begin{equation*}
\lim_{n\to \infty} \|\mathcal{L}_{\varphi,\psi}(\sigma)(f_n)\|_{\alpha,1}=0
\end{equation*}
which completes the proof.
\end{proof}
\begin{thm}
Under the same assumptions of Theorem \ref{thm3.3}, the Weinstein two-wavelet localization operator
$$\mathcal{L}_{\varphi,\psi}(\sigma):L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$$
is compact for all $p\in[1,\infty]$.
\end{thm}
\begin{proof}
We only need to show that the result holds for $p=\infty$. in fact, the operator
\begin{equation}\label{3.13}
\mathcal{L}_{\varphi,\psi}(\sigma):L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+),
\end{equation}
is the adjoint of the operator $$\mathcal{L}_{\psi,\varphi}(\overline{\sigma}):L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{1}_{\alpha}(\mathbb{R}^{d+1}_+),$$
which is compact by the previous Proposition. Therefore by the duality property, the operator given by (\ref{3.13}) is compact. Finally, by an interpolation of the compactness on $L^{1}_{\alpha}(\mathbb{R}^{d+1}_+)$ and on $L^{\infty}_{\alpha}(\mathbb{R}^{d+1}_+)$ like the one given on \cite[Theorem 2.9]{bennett1988interpolation}, the proof
is complete.
\end{proof}
In the following Theorem, we state a compactness result for the Weinstein two-wavelet localization operator analogue of Theorem \ref{thm3.7}.
\begin{thm}
Under the same assumptions of Theorem \ref{thm3.7}, the Weinstein two-wavelet localization operator
$$\mathcal{L}_{\varphi,\psi}(\sigma):L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$$
is compact for all $p\in[r,r']$.
\end{thm}
\begin{proof}
By the same manner in the proof of the pervious Theorem, the result is an immediate consequence of an interpolation of
\cite[Corollary 4.2]{mejjaoli2017new} and Proposition \ref{prop3.15} like the one given on \cite[Theorem 2.9]{bennett1988interpolation}.
\end{proof}
Using similar ideas as above we can prove the following reult.
\begin{thm}
Under the same assumptions of Theorem \ref{thm6}, the Weinstein two-wavelet localization operator
$$\mathcal{L}_{\varphi,\psi}(\sigma):L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)\longrightarrow L^{p}_{\alpha}(\mathbb{R}^{d+1}_+)$$
is compact for all $p\in[1,\infty]$.
\end{thm}
\subsection{Examples}
In this subsection, as in the paper of Wong \cite{wong2003localization}, we study some typical examples of the Weinstein two-wavelet localization operators.
We show that on the space $$\X=\left\{(a,x): x\in \mathbb{R}^{d+1}_+ \;\text{and}\; a>0\right\}$$ the localization operators associated to admissible Weinstein wavelets $\varphi$ and $\psi$ and separable symbols $\sigma$ are paracommutators and we show that if the symbol is a function of $x$ only, then the localization operator can be expressed in terms of a paraproduct. We show in the end if the symbol is a function of $a$ only, then the localization operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is a Weinstein multiplier.
\subsubsection{Paracomutators} Let $\sigma$ be a separable function on $\X$ given by
\begin{equation*}
\forall (a,x)\in\X, \quad \sigma(a,x)=\chi(a)\zeta(x),
\end{equation*}
where $\chi$ and $\zeta$ are suitable functions,respectively, on $(0,\infty)$ and $\mathbb{R}^{d+1}_+$. Then, according to
Parseval's formula (\ref{MM}) and Fubini's theorem, we have for all $f,g\in L^2_\alpha(\mathbb{R}^{d+1}_+)$
\begin{eqnarray*}
\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}&=& \int_{\X}
\sigma(a,x)\Phi^W_\varphi(f)(a,x)\overline{\Phi^W_\psi(g)(a,x)}d\mu_\alpha(a,x)\\
&=& \int_{0}^{\infty}\chi(a)\int_{\mathbb{R}^{d+1}_+}\int_{\mathbb{R}^{d+1}_+}\tau_\eta^\alpha\mathcal{F}_{W}(\zeta)(-\xi) \mathcal{F}_{W}(f)(\xi)\mathcal{F}_{W}(\overline{\check{\varphi}})(a\xi) \\
&& \times \overline{\mathcal{F}_{W}(g)(\eta)\mathcal{F}_{W}(\overline{\check{\psi}})(a\xi)}d\mu_\alpha(\eta)d\mu_\alpha(\xi)\frac{da}{a} \\
&=& \int_{\mathbb{R}^{d+1}_+}\int_{\mathbb{R}^{d+1}_+}K(\xi,\eta)\tau_\eta^\alpha\mathcal{F}_{W}(\zeta)(-\xi) \mathcal{F}_{W}(f)(\xi)
\overline{\mathcal{F}_{W}(g)(\eta)}\\
&& d\mu_\alpha(\eta)d\mu_\alpha(\xi),
\end{eqnarray*}
where
\begin{equation*}
K(\xi,\eta)=\int_{0}^{\infty}\chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)}\mathcal{F}_{W}(\psi)(a\eta)\frac{da}{a},\quad \forall \xi,\eta\in \mathbb{R}^{d+1}_+.
\end{equation*}
Thus, the localization operator $\mathcal{L}_{\varphi,\psi}(\sigma)$ is a paracommutator with Weinstein kernel
$K$ and symbol $\zeta$.
\subsubsection{Paraproduct} In this example, we specialize to the case when the symbol $\sigma$ is
a function of $x$ only, i.e.
\begin{equation*}
\forall (a,x)\in\X, \quad \sigma(a,x)=\zeta(x),
\end{equation*}
where $\zeta$ is a suitable function on $\mathbb{R}^{d+1}_+$. Indeed, using Parseval's formula (\ref{MM}) and Fubini's theorem as in the preceding example, we get for all $f,g\in L^2_\alpha(\mathbb{R}^{d+1}_+)$
\begin{eqnarray}\label{ex3.12}
\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}&=& \int_{\X}
\sigma(a,x)\Phi^W_\varphi(f)(a,x)\overline{\Phi^W_\psi(g)(a,x)}d\mu_\alpha(a,x) \nonumber \\
&=&\int_{\mathbb{R}^{d+1}_+}\left[\int_{0}^{\infty}(\check{\zeta}(\Theta_a*f)*\psi_a)(x)\frac{da}{a}\right]\overline{g(x)}
d\mu_\alpha(x),
\end{eqnarray}
where $\Theta(x)=\overline{\varphi(-x)}$. Therefore, we deduce that
\begin{equation*}
\mathcal{L}_{\varphi, \psi}(\sigma)(f)(x)= \int_{0}^{\infty}(\check{\zeta}(\Theta_a*f)*\psi_a)(x)\frac{da}{a},\quad\forall x\in \mathbb{R}^{d+1}_+.
\end{equation*}
This formula for the Weinstein two-wavelet localization operator is an interesting formula in its own right. Further analysis of (\ref{ex3.12}) using Fubini's theorem gives
\begin{equation}\label{Lpfipsi}
\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}=\int_{\mathbb{R}^{d+1}_+}\check{\zeta}(x)p_{\varphi,\psi} (f,g)(x)d\mu_\alpha(x),
\end{equation}
where
\begin{equation}
p_{\varphi,\psi} (f,g)(x)=\int_{0}^{\infty}(\Theta_a*f)\overline{(\Upsilon_a*g)(x)}\frac{da}{a},\quad\forall x\in\mathbb{R}^{d+1}_+,
\end{equation}
with $\Upsilon(x)=\overline{\psi(-x)}$.
Several versions of paraproducts exist in the literature. It should be remarked that the notion of a paraproduct is rooted in Bony's work \cite{bony1981calcul} on linearization of nonlinear problems.
The paraproduct connection (\ref{ex3.12}) and the fact that the Weinstein two-wavelet localization operators associated to symbols $\sigma$ in $L^\infty_\alpha(\X)$ are bounded linear operators (see \cite[Corollary 3.2]{mejjaoli2017new}) such that
\begin{equation*}
\|\mathcal{L}_{\varphi, \psi}(\sigma)\|_{S_\infty}\leq \sqrt{C_\varphi C_\psi} \|\sigma\|_{L^\infty_\alpha(\X)}.
\end{equation*}
allow us to give an $L^1_\alpha$-estimate on the paraproduct $p_{\varphi,\psi} (f,g)$, where $\varphi$ and $\psi$ are the
Weinstein wavelets, and $f$ and $g$ are functions in $ L^2_\alpha(\mathbb{R}^{d+1}_+)$. First, we need the following Lemma.
\begin{lem}
Let $\varphi$ and $\psi$ two Weinstein wavelets such that $(\varphi,\psi)$ be a Weinstein two-wavelet. Then we have for all $f$ and $g$ in $L^2_\alpha(\mathbb{R}^{d+1}_+)$
\begin{equation*}
\int_{\mathbb{R}^{d+1}_+}p_{\varphi,\psi} (f,g)(x)d\mu_\alpha(x)= C_{\varphi,\psi}\langle f,g \rangle_{\alpha,2},
\end{equation*}
\end{lem}
\begin{proof}
According to relation (\ref{F(f*g)}), Parseval's formula (\ref{MM}) and Fubini's theorem, we obtain
\begin{eqnarray*}
\int_{\mathbb{R}^{d+1}_+} &&\!\!\!\!\!\!\!\!\!\!\!\!\! p_{\varphi,\psi} (f,g)(x)\mu_\alpha(x)= \int_{\mathbb{R}^{d+1}_+}\int_{0}^{\infty}(\Theta_a*f)(x)\overline{(\Upsilon_a*g)(x)}\frac{da}{a}d\mu_\alpha(x) \\
&=& \int_{0}^{\infty}\left[\int_{\mathbb{R}^{d+1}_+} \mathcal{F}_{W}(f)(\xi)\mathcal{F}_{W}(\Theta)(a\xi)
\overline{ \mathcal{F}_{W}(g)(\xi)\mathcal{F}_{W}(\Upsilon)(a\xi)}d\mu_\alpha(x)\right]\frac{da}{a}\\
&=& \left[\int_{0}^{\infty}\mathcal{F}_{W}(\Theta)(a\xi)\overline{\mathcal{F}_{W}(\Upsilon)(a\xi)}\frac{da}{a}\right] \mathcal{F}_{W}(f)(\xi) \overline{\mathcal{F}_{W}(g)(\xi)}d\mu_\alpha(x)\\
&=& C_{\varphi,\psi}\langle f,g \rangle_{\alpha,2}.
\end{eqnarray*}
\end{proof}
An $L^1_\alpha$-estimate for the paraproduct $p_{\varphi,\psi} (f,g)$ in terms of the $L^2_\alpha$-norms of $f$ and $g$ is given in the following theorem.
\begin{thm}
Let $\varphi$ and $\psi$ two Weinstein wavelets such that
\begin{equation*}
\|\varphi\|_{\alpha,2}=\|\psi\|_{\alpha,2}=1.
\end{equation*}
Then we have for all $f$ and $g$ in $L^2_\alpha(\mathbb{R}^{d+1}_+)$
\begin{equation*}
\|p_{\varphi,\psi} (f,g)\|_{\alpha,1}\leq \sqrt{C_\varphi C_\psi} \|f\|_{\alpha,2} \|g\|_{\alpha,2}.
\end{equation*}
\end{thm}
\begin{proof}
From \cite[Corollary 3.2]{mejjaoli2017new}, we know that
\begin{equation*}
\|\mathcal{L}_{\varphi, \psi}(\sigma)\|_{S_\infty}\leq \sqrt{C_\varphi C_\psi} \|\sigma\|_{L^\infty_\alpha(\X)}.
\end{equation*}
\end{proof}
Therefore, by relation (\ref{Lpfipsi}) and Cauchy-Schwarz inequality, we get
\begin{equation*}
\frac{1}{\sqrt{C_\varphi C_\psi}}\left|\int_{\mathbb{R}^{d+1}_+}\check{\zeta}(x)p_{\varphi,\psi} (f,g)(x)d\mu_\alpha(x)\right| \leq \|\zeta\|_{\alpha,\infty} \|f\|_{\alpha,2} \|g\|_{\alpha,2}.
\end{equation*}
Since $\frac{1}{\sqrt{C_\varphi C_\psi}}p_{\varphi,\psi} (f,g)$ belongs to $L^1_\alpha(\mathbb{R}^{d+1}_+)$, it follows from the Hahn–Banach theorem that is in the dual $\left(L^\infty_\alpha(\mathbb{R}^{d+1}_+)\right)^*$ of $L^\infty_\alpha(\mathbb{R}^{d+1}_+)$ and we have
\begin{equation*}
\frac{1}{\sqrt{C_\varphi C_\psi}}\|p_{\varphi,\psi}(f,g)\|_{\alpha,1}\leq \|f\|_{\alpha,2} \|g\|_{\alpha,2}.
\end{equation*}
\subsubsection{Weinstein Multipliers}
We see in this section that if the symbol $\sigma$ is a function of $a$ only, then the Weinstein two-wavelet localization operator
is a Weinstein multiplier $\mathcal{T}^W_m$ with symbol $m$ defined on $L^2_\alpha(\mathbb{R}^{d+1}_+)$ as follow
\begin{equation*}
\mathcal{T}^W_m= \mathcal{F}_{W}^{-1}(m \mathcal{F}_{W}(f))
\end{equation*}
\begin{prop}
Let $\sigma$ be a function on $\X$ given by
\begin{equation*}
\sigma(a,x)=\chi(a), \quad\forall (a,x)\in \X,
\end{equation*}
where $\chi$ is a suitable function on $(0,\infty).$ Then, we have
\begin{equation*}
\mathcal{L}_{\varphi, \psi}(\sigma)=\mathcal{T}^W_m,
\end{equation*}
where $\mathcal{T}^W_m$ is the Weinstein multiplier with symbol $m$ given by
\begin{equation*}
m(\xi)=\int_{0}^{\infty}\chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)}\mathcal{F}_{W}(\psi)(a\xi)\frac{da}{a},\quad \forall \xi\in \mathbb{R}^{d+1}_+.
\end{equation*}
\end{prop}
\begin{proof}
For any positive integer $m$, we define $I_m$ by
\begin{eqnarray*}
I_m &=& \int_{\mathbb{R}^{d+1}_+}\int_{\mathbb{R}^{d+1}_+}\int_{\X}e^{-\frac{\|b\|^2}{2m^2}}\chi(a)\overline{\mathcal{F}_{W}(\varphi)
(a\xi)}\mathcal{F}_{W}(\psi)(a\eta) \Lambda_{\alpha}^d(ib,\xi)\Lambda_{\alpha}^d(-ib,\eta)\\
&& \times\mathcal{F}_{W}(f)(\xi) \overline{\mathcal{F}_{W}(g)(\eta)}d\mu_\alpha(b)d\mu_\alpha(\eta)d\mu_\alpha(\xi) \frac{da}{a}.
\end{eqnarray*}
Then using Fubini's theorem and the fact that the Weinstein transform of the function $h(x)=e^{-\frac{\|x\|^2}{2}}$ is equal to itself for all $x\in \mathbb{R}^{d+1}_+$ (see \cite[Example 1]{gasmi2016inversion}), we get
\begin{eqnarray}\label{suiteinteg}
I_m &=& \int_{\mathbb{R}^{d+1}_+}\int_{\mathbb{R}^{d+1}_+}\int_{0}^\infty \tau_{-\eta}^\alpha h_m(\xi) \chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)} \mathcal{F}_{W}(\psi)(a\eta) \nonumber\\
&& \times \mathcal{F}_{W}(f)(\xi) \overline{\mathcal{F}_{W}(g)(\eta)} d\mu_\alpha(\eta)d\mu_\alpha(\xi) \frac{da}{a} \nonumber\\
&=& \int_{\mathbb{R}^{d+1}_+}\int_{0}^\infty \chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)} \left(\mathcal{F}_{W}(\psi_a) \overline{\mathcal{F}_{W}(g)}* h_m\right)(\xi)
\mathcal{F}_{W}(f)(\xi) d\mu_\alpha(\xi) \frac{da}{a}.\nonumber\\
\end{eqnarray}
On the other hand, it's easy to see that
\begin{equation}\label{convF}
\mathcal{F}_{W}(\psi_a) \overline{\mathcal{F}_{W}(g)}* h_m\rightarrow \mathcal{F}_{W}(\psi_a) \overline{\mathcal{F}_{W}(g)}
\end{equation}
in $L^2_\alpha(\mathbb{R}^{d+1}_+)$ and almost everywhere on $\mathbb{R}^{d+1}_+$ as $n\rightarrow\infty$. Therefore, according to relations (\ref{suiteinteg}) and (\ref{convF}), we obtain
\begin{equation}\label{convIm}
I_m\rightarrow \int_{\mathbb{R}^{d+1}_+}\int_{0}^\infty \chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)} \mathcal{F}_{W}(\psi) (a\xi)\overline{\mathcal{F}_{W}(g)(\xi)}
\mathcal{F}_{W}(f)(\xi) d\mu_\alpha(\xi) \frac{da}{a},
\end{equation}
as $n\rightarrow\infty$. Subsequently, using Lebesgue's dominated convergence theorem, we see that
\begin{equation*}
I_m\rightarrow \langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}.
\end{equation*}
Therefore,
\begin{equation*}
\langle \mathcal{L}_{\varphi, \psi}(\sigma)(f),\ g \rangle_{\alpha,2}=\langle \mathcal{T}_{m}^Wf,\ g \rangle_{\alpha,2},
\end{equation*}
for all $f$ and $g$ in $L^2_\alpha(\mathbb{R}^{d+1}_+)$, where $\mathcal{T}^W_m$ is the Weinstein multiplier with symbol $m$ given by
\begin{equation*}
m(\xi)=\int_{0}^{\infty}\chi(a)\overline{\mathcal{F}_{W}(\varphi)(a\xi)}\mathcal{F}_{W}(\psi)(a\xi)\frac{da}{a},\quad \forall \xi\in \mathbb{R}^{d+1}_+.
\end{equation*}
\end{proof}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 4,065
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SO, the random amounts in this recipe may have you scratching your head...but watch the video and you will realise what I am on about! We are featuring the very clever Drop Scale this month and this recipe is one that appears on their app, but it has been Thermi-fied. It is extremely simple, and a great way to add flavour to your usual crepe.
Think of all the things you could roll up in these babies! Shredded anything. Hollandaise with eggs and bacon or salmon, spinach, avocado, kimchi...oh the list is quite staggering. Go get creative!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,194
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For Mom. You would have laughed the hardest.
#
ONE
[ **BALTIMORE, MARYLAND,
JULY 20, 1973**](Buhr_9780307452023_epub_c01_r1.htm)
TWO
**A PAYDAY OF SORTS**
THREE
**LET'S SPEND THE NIGHT TOGETHER**
FOUR
**SPRUNG**
FIVE
**SOUL TRAINING**
SIX
**JIMI THE BEAR**
SEVEN
**CRASH LANDING**
EIGHT
**JUST ANOTHER PAWN**
NINE
**ELECTRIC FUNERAL**
TEN
**BATS**
ELEVEN
**NO FUN**
TWELVE
**THE NEW YORK FUCKING GIANTS**
THIRTEEN
**THE WORST BLOW JOB IN THE WORLD**
FOURTEEN
**PISS TEST**
FIFTEEN
**SNOWBIRDS**
SIXTEEN
**BOOK 'EM, DANNO**
SEVENTEEN
**LA-Z-BOY**
EIGHTEEN
**NOBODY'S FAULT BUT MINE**
NINETEEN
**ALL ACCESS**
TWENTY
**BIG BOSS MAN**
TWENTY-ONE
**GOING DOWN**
TWENTY-TWO
**THE GREEN, GREEN GRASS OF HOME**
TWENTY-THREE
**LET'S MAKE A DEAL**
TWENTY-FOUR
**GETTING AWAY WITH MURDER**
TWENTY-FIVE
**EVERY DAY COMES AND GOES**
**ACKNOWLEDGMENTS**
# ONE
#
# JULY 20, 1973
_Paranoid_
_Volume 4_
_Master of Reality_
And of course _Black Sabbath_ , the album that started it all for the greatest band in the world: Black Sabbath.
I pulled a copy of each one, stuffed them under my arm and looked around the Record Barn. I'd been coming here since I was a kid. In high school I used to sneak into the kitchen in the middle of the night to make a lunch just so I could pocket my lunch money. By the end of the week I had enough for a few singles.
Not much had changed at the Barn in the months since I had split town. Faded posters covered the grimy front windows, keeping the store dim even in the middle of the afternoon. The stench of pot still hid behind a thin wall of incense, and boxes of T-shirts and albums littered the narrow aisles like always. The owner, Bob, a frizzy-haired David Crosby look-alike, wandered around squeezing more albums into cluttered bins. He stopped in the aisle and stared at me.
"Aren't you supposed to be in school?" he growled.
"I graduated two years ago, Bob," I told him without looking up from the row of records I flipped through.
"Oh," he grunted. "Well, do you need anything?"
He didn't really sound interested in helping me.
"Nah. Just waiting for Frenchy... uh... Pete."
A few minutes later Frenchy stumbled from the back room. His arms waved wildly around his head and shaggy brown hair swirled around his face. He flung a pile of records on the counter and wiped his face with the front of his Flamin' Groovies T-shirt.
"What the hell's wrong with you, Frenchy?"
"I got caught in a spiderweb in the basement," he said. He patted down his hair then looked up at me. "And don't call me that."
We'd been calling him Frenchy since a night he passed out drunk in the backseat of my car talking gibberish that we decided sounded like French. After I left town he convinced everyone to call him Pete again. It was going to take some getting used to.
"So did Alex actually get out today?" I asked.
"Yeah. His mom's having a party for him tonight."
Frenchy sighed heavily.
"You sure he wants to see you?"
"Probably not," I answered.
Bob disappeared into the back room. I followed Frenchy while he walked through the store. Now and then he stopped to file a record into a bin.
"Man, I shouldn't have told you he was getting out," Frenchy moaned. He moaned a lot.
"I just need to talk to him."
"He didn't answer any of your letters. Why would he talk to you now?"
"I know Alex better than he knows himself."
"You drove all the way down from New York City just to talk to him?"
"Something like that."
Bob returned from the back room with a set of BMW keys in his hand and a briefcase with a Grateful Dead sticker on the side. Something about an old hippie with a BMW and a briefcase made me smile.
"Be sure to lock up, Pete."
"Okay, Bob. See you tomorrow," Frenchy said as the front door rattled closed.
Music played in the store. Something loud and noisy. I liked it.
"Who's this?" I asked.
"The Stooges. They're from Detroit."
"I dig it."
"Really? You actually like something other than Black Sabbath?"
"Just need something to fill the time until their next album."
"Whatever." Frenchy laughed.
"How's your band going?" I asked.
"Which one?"
When he wasn't working at the Record Barn, Frenchy combined his musical talents and marginal high school acting experience into a couple of cover bands that played bars and private events. Want the Rolling Stones to rock your wedding reception? Frenchy can do a hell of a Mick Jagger. Need Neil Diamond at your office holiday party? Frenchy's version of "Sweet Caroline" could get the accounting department on their feet. He could imitate anyone.
Frenchy finished restocking the bins then locked the front door. The cash register chimed as he stabbed at buttons until the drawer shot open. Frenchy grabbed the tray of cash and headed toward the back office. I followed him until he stopped, blocking the doorway.
"Where do you think you're going?"
"I don't know." I shrugged. "Just following you."
"No way. Wait out here."
"What's the big deal? You're just counting out the cash register. I can hang out for that."
"Not a chance, dude. That's my rule. You, Alex and Keith aren't coming near the cash or the safe or the back room. I don't even want you guys near the fucking mop closet. Just wait out there."
I sat on the counter, started to read an issue of _Rolling Stone_ , got bored and read _Creem_ instead. A flier on bright yellow paper hung on the side of the counter.
**THE MISTY MOUNTAIN HOPPERS
LED ZEPPELIN FAN CLUB
MEETINGS EVERY FRIDAY-CALL FOR INFO**
I tore it down, folded it up then slipped it into my pocket. A switch flipped from the back room and the lights shut off. Frenchy reappeared and stood in the middle of the store going over everything in his head to make sure he hadn't forgotten anything. He double-checked the back door then the tiny safe in the office. Then he checked the back door again.
"We need to pick Keith up from work," Frenchy said, turning off the rest of the lights in the store.
"Is he still working at Mancini's?"
"Yeah. Installing car stereos."
"And then uninstalling them in the middle of the night?"
"Of course."
Frenchy fished for his keys then stopped at the front door and turned around.
"Don't you already own those?" he asked, pointing to the stack of Sabbath albums under my arm.
"Wore them out. I need new copies."
"You gonna pay for 'em?"
"What do you think?"
Frenchy sighed and opened the door. We walked together across the empty parking lot as the sun set behind the Record Barn, and Baltimore looked every bit as small as it did when I'd left.
# TWO
#
USED THE FANCY-ASS SPELLING OF THE SOUND SHOPPE TO ATTRACT FEDERAL HILL SUCKERS WHO THOUGHT THE STORE MUST BE EUROPEAN. IT WASN'T BUT THE PLOY WORKED ANYWAY. THE GLASS SHOWROOM, LOADED WITH HIGH-END HOME AND CAR STEREO GEAR, ATTRACTED RICH BOYS WHO STOOD AROUND THEIR CAMAROS ARGUING ABOUT EIGHT-TRACK PLAYERS, WHILE INSIDE A DENTIST BLEW A GRAND ON AN ONKYO RECEIVER AND A SONY SUPERSCOPE CASSETTE DECK FOR BLASTING HIS TONY ORLANDO AND DAWN ALBUMS.
Mancini's office sat just to the left of the $700 Celestion Ditton 44 speakers, through a door marked DO NOT ENTER and down a dingy hallway cluttered with boxes, stereo equipment, empty food cartons, porno mags and other garbage. While the salesmen on the showroom floor kissed up to the rich assholes who shopped there, Mancini lurked in his office plotting ways to fuck those customers over. If you were ripped off or screwed over in our town it was the work of Mancini.
Mancini's main operation was Westside Limo, a local car service that drove people to the airport. When Mancini's driver picked you up he checked out your house then he kicked in with the small talk. _Where are you going? I hear it's nice this time of year. Getting away for a few days with the wife, eh? You_ sat in the backseat wondering if the driver was a nice guy or just some jerk working you for a bigger tip.
The truth was he didn't give a shit about a tip. By the time the driver pulled up at the terminal you'd told him where you were going, how long you'd be gone, who was staying at your place. Everything but where you hid the jewelry. The driver gave the details to Mancini. While you were eating peanuts on the plane, me and Alex were climbing in your bedroom window to steal back that Onkyo receiver and Sony cassette deck you just bought. Plus, anything else we could find.
Keith started working for Mancini back in high school. He'd been stealing car stereos since the day he figured out you could break a window with the porcelain tip of a spark plug and not make a sound. He really took to it and terrorized every neighborhood, car dealership, church parking lot, anything. Nothing stopped him. He could strip a tape deck from a Dodge Charger right in the high school parking lot during lunch break and still have time to eat and sneak back out for a smoke.
When Mancini's guys installed a new stereo in a car, they slipped Keith the invoice. Keith dropped by the address in the middle of the night, ripped out the system and brought it to Mancini, who put everything back in boxes and sold it again, often to the same idiot who bought it the first time. Keith got a cut. Eventually Mancini hired Keith to work at the Sound Shoppe installing car stereos. That way, your new system was installed and stolen by the same technician. Keith did both jobs well.
Now we were waiting at the Sound Shoppe for Mancini to pay Keith. He always kept us waiting. Keith had met us in the garage, still wearing his greasy coveralls. He hugged me and, after some small talk, led us back to Mancini's office. He yawned and stretched out in a brown metal chair with his arms folded across his chest. He was no criminal mastermind. That's for sure. He went along with whatever put money in his pocket so he didn't have to get a real job. And it didn't take much to keep Keith in business—just enough for beer, cigarettes and comic books.
He certainly didn't spend the cash on clothes. Keith's wardrobe came in three settings: ripped jeans and Stones T-shirt, ripped jeans and Black Sabbath T-shirt or ripped jeans and no shirt. His greasy hair hung in his face. I leaned forward to ask Keith a question.
"Hey. Is that toolbox out there the one we stole from East-side Auto?"
Keith grinned, showing off the brown edge of a chipped front tooth, the result of a shot he took from a baseball bat in a brawl outside a party one night.
"Yeah," he said. "That's the one."
The tool chest was one of my dumbest ideas ever. It was last summer and Keith needed help getting into Eastside Auto Repair to rip a stereo from a Trans Am that was supposed to be parked inside. When it wasn't there, Alex and Keith wanted to forget everything, but I figured after all of the trouble of getting inside, we had to take something. I decided that we should take the tool cabinet. A giant Snap-on chest that size loaded with gear was worth thousands. It also weighed close to a ton. Mechanics used tow trucks to move them from one garage to another.
We wound up pushing my bright idea twelve blocks in the summer heat. The July weather hit ninety that day, and even though the night air was cool, hot tar from the street pulled against the wheels. We had started on the sidewalk, rolling past the run-down row houses and corner bars, but the rattling toolbox clattered over every crack. We didn't need the attention so I moved us into the dark street. Pushing up the middle of the road doubled our odds of being busted so me and Alex scanned for an escape route every time we saw headlights. We knew the rules, off the street, through the backyards. It went without saying that if we had to run, Keith was on his own.
He wouldn't have made it far. While me and Alex pushed, Keith staggered alongside the cabinet, guiding it up the street. Sweat raced down his bare chest; his neck and arms were beet red with sunburn, and a faded black T-shirt, dotted with cigarette burns, swung from the back pocket of his jeans. His hair dangled close to a cigarette with a long black ash and his eyes were stoned and puffy. At one point, he coughed up a belch, made a face from the taste of stale beer, and swallowed.
He finally forced us to stop at a Laundromat on Jackson Street so he could break in and buy a soda from a vending machine. When he realized that he didn't have any change, we pried open the machine. Once we were inside, I decided we might as well clean out the rest of the machines. Keith wandered around in the glow of the neon sign outside and used a pry bar to knock the change boxes off every dryer and washing machine. We dumped the coins in a cardboard box, set it on top of the tool chest and rolled all the way up Fort Avenue to Keith's house. We sold the tools to Mancini and made more money than we would have with the stereo from the Trans Am.
Keith thought for a minute. He dug a line of dirt and grease from underneath a fingernail. Finally he spoke up.
"When was the last time you talked to Alex, anyway?" he asked.
"The night he got arrested, I guess."
Keith nodded.
"You talked to Danny?"
I glared at Keith. He knew I hated Alex's uncle Danny.
"Can we swing by my place before Alex's party?" he asked. "I need to shower."
"Since when do you shower?" Frenchy asked.
Keith ignored him.
"That's all the way across town," I said.
"I'll give you five bucks for gas."
Keith was always broke.
"You don't have five bucks."
"I just gotta get it from Mancini. I'll have it in a minute. You know that. Come on."
"We'll see."
Just then Mancini barged in. He threw open the door and stopped. He held a box under one arm and stood leaning in the doorway with his back to us as he yelled at the Mexicans working behind him in the garage.
"And put those Delco speakers in the Olds not the fucking Pioneers! I don't care if he paid for them. He ain't fucking getting 'em."
Mancini looked like a three-hundred-pound caveman. His nose and mouth sloped down from his face and, his thick black eyebrows jutted low over his eyes. He left his purple dress shirt unbuttoned to the middle of his chest and thick black chest hair sprung through the opening. The shirttail hung out the back of his stained black dress pants. The entire outfit probably hadn't been washed in a month. He wheezed as he trudged across the office then slammed down into the chair behind his desk.
"Fellas! How have you been?"
He always shouted. Keith and Frenchy muttered something. I stared at the floor.
"What's been going on?"
This was Mancini's rap. He liked to hear the latest gossip and keep tabs on where everyone in town hung out and who had a beef with who. If he knew where you hung out it was that much easier to find you if you fucked him over. And if he knew who your enemies were, it was that much easier to find someone to do the job.
"So how's New York, Patrick?"
"It's good."
"Yeah?" He smirked.
He looked over at Keith like they were both in on the same joke.
"You still working that catering gig with my brother Carmine?"
"Yeah," I said, nodding. "Carmine's a good guy."
Mancini rolled his eyes.
"No, he's not. He's a fucking scumbag but I'm glad you two are getting along."
Mancini opened a thick black binder. The papers inside were loose and stained at the edges with coffee. He smoked and flipped through the papers. He found a pink invoice and squinted at it.
"All right, Keith. That asshole Pannazzo kid installed a new Panasonic deck in the Cutlass his parents bought him last month for graduation. You know where he lives?"
"Yeah. On Juniper."
"Great. Let's get that back. And whatever happened with that Huffman kid I asked you about? He has an eight-track player and a pair of Pioneer speakers in that Trans Am."
"He's been parking in his garage every night. I can't get to it."
Mancini looked up. His eyes were dark and heavy, and when I realized how bloodshot they were, mine began to water. He jabbed at Keith with his cigarette as he sputtered.
"So what! So you break into the fucking garage."
No way was Keith breaking into Huffman's garage. Huffman's old man was a maniac and he kept a pit bull named Peaches locked in the garage. Keith put up a good front, though. He glanced down at the tile and nodded his head then took a drag on his cigarette. He wanted to make it look like he'd thought it over and come to a decision.
"Okay. I'll get it."
Mancini stared at Keith and the room went quiet. The sounds of Spanish music and laughter cut through Mancini's cheap office.
"What the fuck are they doing out there?" Mancini mumbled. He opened the door leading to the garage. The music from the garage shook the shelves in the office.
"Turn that shit down, you assholes, or I'll have every fucking one of you deported!" Mancini screamed. He coughed violently. "You hear me, amigo?"
He slammed the door and sat back down. He and Keith went over the rest of the list. There was a VW Bug that belonged to a girl we went to school with and a new van owned by one of Keith's neighbors. The husband brought it in to Mancini to set up with JBL speakers, a subwoofer and a Pioneer eight-track. Mancini wanted it all back.
"Oh—and I wanted to grab that money from you for last time," Keith said. Whenever Keith asked Mancini for money he made it sound like the thought had just occurred to him even though he and Mancini both knew it was coming. He used to follow this approach with "Is that cool?" but realized it left things open for Mancini's bullshit so he dropped that part.
"What do I owe you again? One-fifty?"
Mancini scrunched up his face and flipped through his binder.
"No. It was three hundred," Keith said.
"If you say so," Mancini said.
He tried to sound like he didn't believe Keith. He wanted to play it off like he was being taken advantage of. He knew how much he owed Keith. No one ripped off Mancini that easily.
He cracked a safe under his desk, pulled out a silver gun, a shoe box wrapped in duct tape and a large folder and placed them on the desk. Then he sat bent over counting money where we couldn't see. He handed the cash to Keith, who stuffed the money in his pocket without counting it, and we bolted from the office and down the empty hallway.
"What do you say, man?" Keith said, climbing into the front seat of my car. "Can we swing by my place first?"
"You got five bucks?"
"Shit." He grinned. He pulled the wad of cash from his front pocket. "Take ten."
I wasn't happy about it but didn't feel like arguing. It was too hot. I stuffed the ten into my pocket. The car struggled in the heat before the engine kicked over with a blast of sweltering air in our faces. Sabbath thundered from the speakers. The car was a piece of shit, but thanks to Keith the stereo sounded fucking great.
# THREE
#
TIME I MET EMILY WAS IN KEITH'S KITCHEN. IT WAS LAST YEAR AND KEITH'S CHRISTMAS PARTY WAS STUMBLING TO AN END. TWO OF KEITH'S COWORKERS SAT ON THE FLOOR IN FRONT OF THE COUCH PASSING A JOINT AND STARING AT THE CHRISTMAS TREE, WHICH LAY ON ITS SIDE IN THE MIDDLE OF THE LIVING ROOM. A _MISSION: IMPOSSIBLE_ RERUN BLARED FROM THE TV AND GIGGLING CAME FROM A BEDROOM UPSTAIRS. KEITH SLUMPED OVER THE KITCHEN TABLE WITH HIS HEAD DOWN, A WARM BEER IN HIS HAND. I WOBBLED ACROSS THE KITCHEN AND CRUNCHED ACROSS A BAG OF POTATO CHIPS ON THE FLOOR. ALEX FOUND ME BENT OVER IN THE FRIDGE DIGGING FOR ANOTHER BEER.
"These girls want to get out of here. Can we give them a ride?"
I shut the door and tried to stand upright, swaying a bit. Alex leaned in the doorway. Two girls, a blonde and a brunette, whispered to each other behind him.
"We sure can," I said. "Let's go."
We piled into my car. Tina's blond hair, blown out Farrah Fawcett-style, bobbed in the back window as she talked loudly with Alex. Emily sat up front playing with her straight black hair and didn't say anything until Tina teased her about being shy. She turned back and told Tina to fuck off then grinned a bit before turning back to stare out the window. She was young and gawky, all legs, in giant platform sandals and tiny denim shorts. She opened the glove box and dug through a pile of eight-track tapes.
_"Black Sabbath. Paranoid. Master of Reality_. Do you have anything other than Black Sabbath?"
"Not really," I answered.
"That's pretty weird."
"I like Sabbath. What's weird about that?"
"Are you some kind of satanist?" She grinned. "You're not going to sacrifice us, are you?"
"Nope. We only sacrifice virgins," I joked. She laughed loudly and fidgeted with the radio dial. She stopped when she heard "Goodbye Yellow Brick Road" playing on WKTK.
We drove to Tina's house in Roland Park, winding down tree-lined roads near the Baltimore Country Club. I'd only been to this part of Baltimore once. It was a night me and Alex drove around casing houses, looking for one set up to rob. A suspicious cop pulled us over but let us go. After that, we decided the area was too risky and never came back.
I turned off the lights as we pulled up along a low rock wall outside Tina's house. Alex and I lingered back by the car as the girls stumbled across the manicured lawn. We weren't sure what to do. These situations were like special forces operations. Me and Alex had crawled in bedroom windows, crept up stairs and even used ladders to help chicks escape from their parents. Emily and Tina opened the front door.
"Get in here before the neighbors see you," Tina whispered from the front step.
Tina's parents were out of town. They had driven to Florida with her brothers for Christmas while Tina finished school. Tina's grandfather was dropping her at the airport tomorrow to fly down and meet up with the rest of her family. Emily was spending the night.
Our tactics changed once me and Alex realized we had the house to ourselves. The new plan was to divide and conquer, and we laughed as we ran up the lawn toward the giant house. Inside, a winding staircase swooped upward and a balcony overlooked the marble foyer lit by a glittering chandelier. Tina led us down the hallway to the kitchen, where Alex flung open the refrigerator door. He grabbed two cans of beer and handed one to me.
"What the hell are you doing?" Tina asked. "My dad is going to know those are missing."
The beer in Alex's hand hissed as he popped it open.
"Oops. Too late." He grinned.
"You guys!" Tina whined, but she was too drunk to really care.
Soon Alex and Tina were upstairs in her room while Emily and I sat on the floor in Tina's brother's room. I stared at the giant aquarium in front of me.
"What's in the aquarium?"
"You don't want to know," Emily said, taking the beer out of my hand.
"Why?" I asked. A chill went through my body. "Is that a fucking snake?"
She took a chug on my beer then nodded. I hated snakes.
"What kind is it?"
"Some kind of python," she said with a shrug. "He feeds it mice. It's disgusting."
I forced myself to look away and spotted a row of crates filled with records along one wall. For a rich kid, Tina's brother had great taste in music. The record collection was killer—Alice Cooper, Bowie, The Faces. He even had Hendrix's _Band of Gypsys_. I dug through crates, ignoring Emily, until she let out a dramatic sigh, pulled out _Eric Clapton_ and handed it to me.
"Here. Put this on," she said, pushing the record toward me.
"No thanks."
"Why not?" she slurred.
"I'm not into white-boy boogie rock."
"Oh, you know what? Screw you."
"Oh, come on." I grinned. "You know it's soulless shit."
She was climbing into the bed and not really listening. I put on the Stones' _Between the Buttons_ and passed her a beer. She giggled and rolled her eyes as she stretched out. I wondered what she was laughing at when "Let's Spend the Night Together" started and I realized how badly I had screwed up.
"Oh yeah. Sorry about that." I grinned again. "Try not to read too much into it."
"No problem." She giggled coolly, sipping her beer. I lay down next to her and by "Ruby Tuesday" we were making out. I slipped my hand up her shirt and kissed her neck but she was a dead fish. No response. She must have figured that rather than stop me and give me a chance to talk her into anything she'd just freeze me out. I knew how to get a response. I'd slide my hand down between her legs and wait for her to grab my wrist and stop me. I groped in the dark, searching for some boundaries. I'd been shut down. Alex definitely figured this one out and stuck me with the dead end again.
The record was over but Tina's hysterical moans shattered any awkward silence. The bed in the next room thundered as Tina groaned in no particular rhythm. It seemed to last forever. Emily fought back a smile as we kissed. I broke from her and pushed the hair out of my eyes.
"Goddamn. You might need to go check on your friend."
This made her laugh. I flipped the record over and realized the booze had worn me out. I felt heavy and spaced out. Back in bed I decided to give up on her. We lay side by side listening to "Who's Been Sleeping Here" until she passed out.
When I was sure Emily was asleep I slipped out of bed, put my pants on and crept down the hallway. The door to Tina's bedroom dragged along thick carpeting as I pushed it open. Posters of the Stones and Led Zeppelin lined the walls and a glass bong sat on the dresser. In the bed, Alex lay on his side with his bare back to me, one arm flung across Tina. She snored loudly with black eyeliner smeared under her eyes. I poked Alex.
"Wake up, man."
He jerked awake and squinted at me in the dark.
"What?"
"Come on. Get up."
Alex watched Tina, careful not to wake her, as he worked himself out of the tangle of sheets. I handed him his pants and he stumbled into them as we snuck from the room. He stopped me in the hallway.
"Are we leaving?"
"Not yet," I said. "But look at this place. She's loaded. Let's check it out."
"Are you fucking serious?" he asked. "These chicks are pretty cool. You want to rob them?"
"It's no big deal. We take a few pieces of jewelry. Her mother won't miss it."
Alex rubbed his eyes.
"Shit," he whispered. "You're right. Good thinking."
Down the hallway, we slipped into Tina's parents' bedroom and turned on a light by the bed. Alex dug through a wooden jewelry box and picked over the rings, holding a few to the light, before stuffing a couple into his pocket. I found a roll of cash in an old film canister in the top of the closet and grabbed a man's watch that I later gave to my father as a Christmas gift.
"Damn, man," Alex whispered. "They're going to know it was us."
"So what?" I fired back. "Then they'll have to tell Tina's parents that they had us over and they aren't going to do that. So even if they figure it out they can't do anything about it."
Alex jerked open a dresser drawer and knocked over a few framed pictures of Tina and her family. I pulled my head out of the closet and put one finger to my lips.
After we picked over everything, we padded down the hallway. Emily lay curled up in a ball facing the wall in Tina's brother's bed. I stripped down to my boxers and slipped into bed next to her.
In the morning I woke up alone. Alex talked loudly down in the kitchen and the girls laughed at anything he said. I dressed, then trudged down the carpeted stairs and sat at the table. The girls made breakfast and I didn't say much as everyone ate. I wasn't hungover yet but it was on the way. When we finished, Emily cleaned up while Tina brought down their luggage. I heard her scream from upstairs.
"Holy shit!"
I was sure that Tina had figured out what Alex and I had stolen during the night. My hangover hit full force and I felt sweaty and nauseous. I eyed my car sitting outside in the street. Emily stopped scrubbing a plate and yelled up the stairs to Tina.
"What's wrong?"
Tina stumbled down the stairs with a giant suitcase and stuck her head into the kitchen. Her eyes were huge with panic.
"It's almost eleven," she gasped. "My grandpa is going to be here any minute."
This sent the girls scrambling to clean up, cram things into the luggage, unplug the coffeepot, clean out the refrigerator and double-check that all the doors and windows were locked. Alex and I sat at the table. I lay my head down, forehead in my hand. Alex drank a cup of coffee and sat cocked back in the chair with his shoes on the edge of the table. He pulled a cigarette from a pack, lit the end with a silver lighter from a bowl on the counter, pocketed the lighter and exhaled a cloud of smoke.
"Alex, can you put this key in the mailbox?" Tina asked. She tossed a silver key to him.
"You know that isn't safe," he said. The tone was preachy and concerned without a trace of threat. "Someone could find it and clean the place out."
"It's for Nancy next door. She's going to feed my brother's snake."
"Will do," Alex said with a wink.
Emily and I said good-bye. I promised to call her. Tina and Alex made out like lovers on a sinking ship. Me and Alex passed Grandpa as we pulled out of the cul-de-sac. He glared at us with a look that said he knew exactly where we were coming from even though he arrived too late to prove it. I stared at him from behind my dark black sunglasses. _Sorry, you old fucker_.
Alex unloaded two fistfuls of gold jewelry from the pockets of his jeans and dumped them in his lap.
"We're never going back there again." Alex laughed as we rounded the corner.
"Yeah, we are," I told him. "They left a key in the mailbox."
Now I sat at the kitchen table at Keith's and stared at a cockroach as it climbed across a dirty pan and into an empty Hormel chili can. Keith's mom, Suzy, leaned against the counter, clutching her pink bathrobe closed with one hand. She was in her forties and still looked pretty good, tiny and tan with bleached-out hair. She never seemed to change out of her pink bathrobe except to go to work at the dry cleaners.
Keith came downstairs, freshly showered. He stood in the middle of the kitchen and tied a folded-up bandanna around his head to hold down his long, soaking wet hair.
"You look like an idiot," Suzy said.
"Shut up," Keith said, straightening the headband. "I think it looks cool."
Suzy rolled her eyes and pulled a drag off a long cigarette.
"Christ. You look more like your father every day."
We couldn't argue. None of us had ever seen Keith's father. Keith's mom had always been single. His entire life Keith had put up with a string of Suzy's shitty boyfriends. They ranged from an accountant who Keith hated to a stock car racer who Keith loved.
"Go back upstairs," Keith told her. He glared at her from the corner of his eye.
She rolled her eyes again and grunted while exhaling a cloud of smoke. The two noises together sounded like a car stalling in her chest. Keith set a paper bag with a bottle in it on the table in front of me and Frenchy.
"I got Alex a welcome home gift."
"Crown Royal?" Frenchy asked.
"Yes, sir." Keith grinned.
"And where did you get the money for that?" Suzy snarled at Keith.
"I work, Mom."
"And Alex," she grumbled out of the side of her mouth, lips clenched on the cigarette. "Breaking into people's houses. He ought to be ashamed of himself."
Keith raised his head to roll his own eyes then looked back down at the kitchen table.
"You better not being doing shit like that, Keith. You're nineteen years old. You're on your own now, mister. You're not my fucking problem anymore."
She leaned against the kitchen counter and sipped her coffee. Keith traced a line of spilled sugar on the table with his finger.
"You go to jail and I ain't bailing you out. I don't have the money," she said. "I may have to start fucking your friends to pay the rent."
Frenchy sat up.
"Suzy. I got paid today."
"Oh, fuck off, Pete," she hissed. "You wouldn't know what to do with it if I gave it to you."
She pulled her robe closed at the top then crossed her arms over her chest.
"Seriously," Frenchy said. He stood up and reached for his wallet.
"How much do you make over there?" Suzy asked.
"Ma!" Keith barked. "Just get the fuck out of here! Isn't _Sonny and Cher_ starting?"
Suzy's head spun toward the clock.
"Oh my God!" she shrieked.
She leaned over Keith, stubbed her cigarette out in the ashtray then ran barefoot across the kitchen and up the stairs.
Keith swept his hand across the table, wiping the sugar to the floor. It was quiet for a long time except for the sound of a TV upstairs. Keith stole two packs of Suzy's cigarettes from a kitchen cabinet then said, "Let's head over to Alex's. He should be home now." I said yeah and Frenchy shrugged, and we all piled out the door and into my car.
# FOUR
#
HOME FROM PRISON WITH TWO SHOE BOXES. ONE WAS FILLED WITH PICTURES OF HIS NEW SON, THE OTHER WITH PICTURES OF ALL THE WOMEN HE MET WHILE LOCKED UP. I STOOD AGAINST A WALL AT HIS WELCOME-HOME PARTY, TOOK A LONG PULL OFF A WARM CAN OF BEER AND WONDERED OUT LOUD TO FRENCHY HOW A GUY SERVING EIGHT MONTHS FOR BREAKING AND ENTERING MET MORE WOMEN THAN I DID OUT ON THE STREET.
"It's pretty amazing, isn't it?" Frenchy agreed.
He thought about something then spoke up.
"What are you gonna say to him?"
"I don't know," I said.
"This is a bad idea. I really shouldn't have brought you."
Across the room Alex ran both hands over his slicked-back hair. He flipped a menthol cigarette into his mouth and I caught a glimpse of a cross tattoo on his forearm. That was new. The tattoo made him look like an ex-con or some tough guy from the other side of town, which I suppose he always was, really.
We were all born in Forest Park but in the sixties our parents moved us across town to Locust Point. Alex's family never left. His old man was a drunk and a gambler and loved that he lived near Pimlico racecourse. He didn't care that his son was one of the only white kids left at Forest Park High School. By the time Alex dropped out, he was trying to fit in by acting black. I guess he never stopped. To us, he really was black. He ironed his jeans, wore bright button-up shirts and doused himself with cologne. He kept his black hair shorter than the rest of us and slicked it back instead of wearing it long. He smoked menthol cigarettes and listened to R&B. He pretended not to like rock 'n' roll. Worst of all, he hated Black Sabbath.
I scanned the crowd in Alex's mother's tiny basement. All of Alex's relatives were crooks. Most of them were older guys with bad teeth and faded blue prison tattoos. They had all been in jail at some point but got out, married one of the busted-looking chicks hanging around and settled down, which to them meant cutting out violent crime and sticking to simple robbery.
A pair of boys ran past spraying each other with water guns until the smaller one stumbled and smacked his forehead into a table holding a cake that said "Welcome Home Alex." On the other side of the table a skinny guy in a denim vest and no shirt showed off a long-barrel revolver to a bald musclehead in a Kool cigarettes tank top.
Alex's mom looked older than the last time I saw her. Silver streaked her curly brown hair. She knotted her fingers together nervously as she talked with one of Alex's aunts. Alex's dad stood next to her. He sipped a can of Coors and grunted now and then. His glasses and thick beard covered his face and made it hard to tell just how much Alex looked like him. I couldn't remember if I'd ever heard him talk. Mostly he hid in the garage or lumbered around the house mumbling.
"It's the math of the whole thing I don't get," I said to Frenchy and Keith.
"Math of what thing?" Frenchy asked.
"How did Alex even come in contact with that many women? He was in jail."
"Why don't you ask him?" Keith said.
I was dying to ask Alex about it but couldn't bring myself to do it. Frenchy was right. After eight months in County I was probably the last person he wanted to see. In fact, I knew it.
"Well, well, well," a voice behind me slurred. "If it ain't the kiddie table."
Alex's uncle Danny slapped Keith on the back then crossed his arms and stared at me. His dark hair hung down to his shoulders. A chicken drumstick jutted from his crooked mouth, hidden behind a thick handlebar mustache.
"What are you doing back in town?" he asked me, his mouth full of chicken.
"Just came to see Alex."
"He know you're here?"
"Not yet."
Danny grunted and looked me over while he chewed. A piece of chicken snagged in his mustache then fell, landing on the front of his Corvette T-shirt. Looking at him now, it was hard to believe he was our idol when we were kids. Back then he was a football star. Scouts from the University of Maryland once came to Forest Park High to watch him play On a field trip to Memorial Stadium, the Baltimore Colts' quarterback Johnny Unitas even said Danny had a great arm.
But by junior year of high school, Danny was smoking a sack of weed every day. He stopped going to school and showing up to practice. Then he got arrested for stealing a keg of beer out of the back of the Crown Pub and Coach Dunlop had to bail him out. A month later he dropped out of school and began breaking into houses and businesses with the rest of Alex's uncles. But Danny had the lowest IQ of any of Alex's uncles, which meant he also had the biggest rap sheet. He'd never gotten away with anything. Now he was twenty-seven years old, on parole and living with Alex's grandmother. Once and only once did I let Alex convince me to bring Danny with us on a job. We were at Alex's welcome-home-from-jail party as a result.
"Didn't you just get out of County too?" Keith asked Danny.
"Last week." He shrugged.
"Was this for breaking into the Old Towne Bar?" Keith asked him. Danny nodded.
"How'd you guys get caught?" I asked.
"Fucking daylight savings time or whatever the hell it is," he spit. He stopped to lick the chicken grease off his fingers. "We busted in the place and the motherfucker was still open. I'd have run but the owner had a shotgun on me. I'm still fast but I ain't that fast."
He tossed the chicken bone on the table then opened a can of beer. He took a long pull off the beer then wiped his mouth with the back of his hand.
"Well, Patrick. I know if I was Alex I sure wouldn't want your ass at my welcome-home party. Personally, I don't think he's gonna want to see you after the mess you got him into."
I spoke slowly, choking on each word.
"This mess was your fault, not mine."
Danny shrugged and sucked on his beer. Foam clung to his mustache.
"Well, it was your plan, bud. Not mine. This shit went sideways on your watch. Know what I mean?"
My fists clenched. Frenchy shot Keith a worried look. Across the crowd, Alex sat alone on the couch for the first time all night. This was my shot. I left Danny standing by the wall and wove my way through the crowd and across the room.
"I hear you got quite a collection of photos," I said with a grin as I sat down next to Alex.
He looked up at me. If he was surprised to see me he didn't show it.
"Yeah," he mumbled.
"How'd you do it?"
"My cell mate. Crazy Mexican dude from downstate. His girlfriend would come up to visit all the time. She didn't like making the drive alone so she'd bring a friend."
"Makes sense."
"But he didn't want some other chick hanging around while he talked to his girl so he asked me if I would sit with her friend. It beat sitting in my cell. Me and this girl hit it off all right, and she started coming up a lot and writing me letters."
I nodded then leaned forward to get a look at the shoe box full of nudie photos. Frenchy rushed over and squeezed in next to me on the couch. Alex kept talking.
"Word got around and other guys started asking me to sit with their girlfriends' friends. One dude asked me to sit with his sister. Next thing I know I'm getting all these letters and pictures. Shit. It was hell writing back to all of them but I had nothing else to do."
The box was loaded with photos, mostly Polaroids of Mexican girls in little white nightgowns or skirts and high heels. There were a couple of shots of an older chick in red thigh-highs and lingerie bent over or spread-eagle on a bed. A hot brunette sucked on her finger and flashed her tits in a series of black-and-white photo booth shots. In one frame she gave a beer bottle a blow job. Had the thing in her mouth to the middle of the label. Frenchy was flipping out.
"Damn," Frenchy groaned. "This is better than the old _Playboys_ under my bed."
"Those are _my_ old _Playboys_ , asshole." Alex grinned. "I want those back."
He took a long drag on his cigarette.
"You know I'm a father now, right?" he asked, killing the mood just as I gave a close-up look to a blonde pushing her size-Ds together and smiling at the camera.
"Remember that blonde I was seeing? Vickie? She got pregnant before I went in."
I remembered Vickie. Tiny. Blonde. Personality like wallpaper.
"Yeah. She was cool," I lied. I quickly looked around the room. "Is she here?"
"Nah." He shrugged. "She and the baby moved back to Florida to be near her family."
Alex grabbed the box of photos off my lap and replaced it with a half-empty box with TOMMY scrawled across the top in black marker.
The top photo was a black-and-white hospital shot. The screaming prune-wrinkled face looked like Alex, I guess. It had his tan skin and wisps of black hair. I flipped through a parade of shots of people holding the baby. I turned them slowly and pretended to be interested.
The kid grew older as I looked through the pictures. I flipped to a photo of him sitting in a high chair and nearly choked. His olive skin was much darker and his frizzy black hair stuck straight up. He looked like Sly fucking Stone or Jimi Hendrix on the cover _of Axis_.
Lil' Tommy was black.
I didn't know whether to tell Alex the kid wasn't his or just keep my mouth shut. Maybe while he was locked up he convinced himself that this was his son and had fallen in love with him. What if I told him and he lost his mind and had a breakdown or something? I decided to keep my mouth shut.
"He's really something," I stammered.
"Yeah? Think he looks like me?"
"Oh yeah. Sure," I lied again. "Totally."
Alex snatched the box of photos from my lap.
"Same ol' Patrick," he said, shaking his head. "Still a bullshitter."
"What do you mean?"
Alex leaned forward. He pushed a photo into my face.
"Look at him, man. This kid looks like Joe Frazier. There's no way he's mine."
"Okay." I grinned. "Caught me on that one."
Alex lit another cigarette.
"What the hell are you doing here, anyway?"
"I just came to see you."
"Bullshit. You didn't drive all the way down from New York City just to see me."
"All right," I said, leaning back. "I have an idea."
"Is this the one where we rob my girlfriend's house while her family is on vacation? If so, I can tell you how it ends up. Sixteen stitches from a python bite and eight months in the joint. Well, not for you, of course."
I had that coming.
"It was just as much Danny's fault as it was mine."
Alex shrugged.
"Honestly, man," he said. "I just got out four hours ago for some shit you got me into and then you show up trying to drag me into some new bullshit plan. What the fuck are you trying to do?"
"I'm trying to help you out. You know, make it up to you."
Alex glared at me out of the corner of his eye as he dragged on his cigarette. Then he looked over at Keith and Frenchy.
"What about those guys?" Alex asked.
"We're gonna need them too."
He sipped his beer.
"This idea is that big?"
"No. It's that good."
"Well. Let's hear it."
"I want to rob Led Zeppelin."
Alex leaned forward and stubbed out his cigarette on the top of an empty can. Then he slumped back into the sofa, looked me in the eye and pointed toward the door.
"Get the fuck out of here."
# FIVE
#
WERE NEVER ON _SOUL TRAIN,"_ KEITH SHOUTED AS I SLIPPED INTO ALEX'S BEDROOM. I PULLED THE DOOR CLOSED BEHIND ME.
"They were on the same episode as Ike and Tina Turner," Alex said. "I saw it when I was in County."
They were stoned and standing in the middle of the room arguing. Al Green played on a turntable sitting on a table under a poster of Curtis Mayfield. Alex's room was always a mess, except for his clothes, which were pressed and hung up in the closet. A clear plastic bag on the floor read BALTIMORE COUNTY JAIL: PERSONAL BELONGINGS. Inside were the clothes Alex had on the night the cops nabbed him.
"They played 'Light My Fire,' " Alex teased Keith. He turned his head and winked at Frenchy. Keith rubbed his forehead. He looked distraught.
"Were the Doors ever on _Soul Train_ , Patrick?" Keith asked me. "They weren't on there, were they? 'Cause I hate that fucking show."
His bleary eyes pleaded for me to tell him it wasn't true. It wasn't.
"No. They weren't."
"Aww, come on," Frenchy groaned. "Why'd you have to tell him?"
"He looked like he was gonna cry."
"Come on, Keith," Alex said. "You ever see white people on that show?"
Alex loved winding Keith up and poor dumb Keith always fell for it. It could be anything. He only got mad the time we made him believe _Hawaii Five-O_ was canceled. He didn't speak to any of us for a week.
It took some time but I had managed to convince Alex to at least hear my plan. A few factors worked in my favor. Having just been walked out of the gates at Baltimore County Jail hours earlier Alex was taking a hard look at his future. It looked like shit. He was a high school dropout and now an ex-con. He was facing miles of floors to mop or, if he was lucky, a backbreaking job unloading freighters down at the Inner Harbor. And even if he did come up with a decent scam that didn't involve me, he'd be stuck working with his uncle Danny. The few months Alex had just spent locked up with that dumbass probably convinced him that wasn't going to work.
"These guys know about your idea?" Alex asked, jerking a thumb toward Frenchy and Keith.
"Nope," Keith said. "What idea?"
I took a look around the room.
"I've been working for Mancini's brother Carmine in New York City. He owns a catering business. We set up all the food backstage at concerts and political events and shit. A few months ago I worked a Zeppelin concert."
"You got to see Zeppelin?" Keith asked.
"Pay attention, man. So after the show, me and another guy are taking down the tables when I hear an argument going on down the hall. I peek around the corner and there's Lenny, the guy who books the shows. He's standing there arguing with this huge, bald British guy. I didn't know if Lenny was being robbed or what. Lenny hands the British guy this briefcase and the guy opens it. It's filled to the top with cash. Then the guy stomps off towards the dressing rooms.
"I walk over and ask Lenny if everything is cool. He's a little shaken up. Standing against the wall smoking. He tells me that was Zeppelin's manager making sure they got paid what they were promised. Says their manager always yells like that. Then I ask him about the cash. He tells me that Zeppelin always get paid in cash. Always. That night, it was over one hundred thousand dollars."
"Goddamn," Keith said, shaking his head.
"So what are you saying?" Frenchy asked.
"We rob Led Zeppelin."
No one said anything. Finally, Keith laughed.
"You motherfuckers. First you make me think Jim Morrison was on _Soul Train_ and now this bullshit. Nice try, guys."
"I'm serious."
"How are you guys going to rob Zeppelin?" Frenchy asked. "What are you going to do? Fly to England?"
"Don't need to. They're playing Baltimore on Monday."
Alex hadn't said a word.
"But I thought rock bands usually got paid by check or something?" Frenchy asked.
"Not Zeppelin. Their manager makes sure they always get paid cash."
"So how much are we talking?" Alex asked.
"From the sound of it, maybe one hundred thousand, split four ways."
"Goddamn!" Keith yelled. "Twenty-five thousand dollars each? Shit. I'm in."
"You didn't want to rob a bank but now you wanna do this?" Alex said, crossing his arms.
"Whoa! Whoa!" Frenchy stood up. "What the fuck! You guys were gonna rob a bank? What the hell is wrong with you?"
"It was just an idea a long time ago," I said. "Calm down."
"How are you gonna pull this off?" Alex asked.
"The show starts at eight and probably goes until midnight. They probably get out of there even later. The way I see it, every bank in town is closed by then. That means their manager has to hold on to all that money until the next day."
Heads nodded around the room.
"That gives us roughly eight or nine hours to get to that cash before it goes in the bank."
"But where will the money be?" Alex asked. "You don't know who has it, where they take it, what they look like. Hell, you don't even know where they're staying."
"Right," I agreed. "First, we need to find out who collects the money after the show and what they carry it in. I saw it in a briefcase but maybe it changes. Then, we need to find out which hotel they're staying in and see if the person with the money keeps it in his room or sticks it in a safe deposit. That's where we'll get it."
"Sounds easy enough," Keith said.
"Sounds easy?" Alex said. "Stealing a safe deposit box from a hotel is a lot different than tearing a car stereo out of a Nova, Keith."
"That's not even the hard part," I explained. "Zeppelin travels with hard-core security. These guys are brutal. British goons who will kick your fucking head in. They answer to Zeppelin's manager, Peter Grant. He's over six feet tall. Three hundred pounds. He's an ex-bouncer and an ex-wrestler. If we get caught fucking him around, we're finished."
"Forget it," Frenchy erupted. "This is insane. Zeppelin are the biggest band on the planet right now. The biggest! How are you gonna pull this off? There's no way in hell."
"When have you ever heard of a rock band being robbed?" I asked. "Name one."
"That's because it's impossible!" Frenchy argued.
"No. It's because no one else ever thought of it," I replied. "They won't see it coming."
"So what the hell, man?" Keith groaned. "How are we gonna pull this off?"
Alex crossed his arms and watched me as I talked.
"We plant one team backstage to see how Zeppelin gets paid, who carries the cash, what happens to it," I explained. "The other team waits at the hotel to grab the money."
"What are the teams?" Alex asked.
"The way I see it, me and Frenchy will be the backstage team. Alex and Keith will be the hotel team."
Alex lit a smoke and laughed out loud.
"No fucking way, man. I'm out."
"Why?" I said. "You got a better idea?"
"I already took the fall for you once. It ain't happening again. You want me in? You go on the hotel team."
"Alex, you're the only one who could get in that hotel room and you're definitely the only one who could get to a safe deposit box."
"No fucking chance," Alex said, belching out a thick cloud of smoke.
"Would you feel better if I worked with you on the hotel team?" I asked.
Alex nodded.
"So you're going to send Keith backstage with Led Zepplin?" I asked. Alex knew that was a horrible idea.
"I like that idea." Keith grinned.
"No way," Alex said. He looked at me. "You'll just have to do both."
"Fine. I'll scout the backstage with Frenchy then work with you guys to get the cash."
The record on the turntable ended and the needle hissed as the record spun beneath it. Outside the bedroom door Alex's welcome-home party carried on without him. Keith spoke up first.
"Where do we start?"
Frenchy jumped in. "I think we need to figure out how to get backstage first."
"I have a plan for that," I told them.
"And we need to check out the hotel," Alex said. "Do you know which hotel they're using?"
"I don't know but I know someone who will," I answered.
I pulled the bright yellow Misty Mountain Hoppers Led Zeppelin Fan Club flier out of the pocket where I'd left it since tearing it down at Record Barn. Frenchy shook his head.
"No, no. That's a horrible idea," he said.
"Yeah, man." Alex laughed. "She's gonna want to see you even less than I did."
They were both right.
# SIX
#
AT THE SEAFOOD RESTAURANT STARED AT MY TORN-UP JEANS AND BLACK SABBATH T-SHIRT THE NEXT DAY AS THE HOSTESS LED ME TO A CORNER BOOTH. I WASN'T BOTHERED. I WAS STILL THINKING ABOUT ALEX'S PARTY THE NIGHT BEFORE. I DIDN'T CARE WHAT ANYONE SAID. DANNY WAS MORE TO BLAME THAN THAT DAMN SNAKE FOR WHAT WENT DOWN THE NIGHT WE ROBBED TINA'S HOUSE.
We did it on New Year's Eve. Tina and her family would be back on New Year's Day so that was our last chance to do it. I'd gone out with Emily every night that week and was meeting her at a New Year's Eve party later that night. I promised I would be there before midnight.
I only agreed to let Danny come with us because he drove a pickup truck and we wanted to be able to haul everything. It was late and most of the houses were dark. We could see the front door to Tina's house in the distance. The three of us were sitting sandwiched together in the front seat of his truck watching the street when I realized what a dumb idea it was to bring him.
"I don't see why the fuck Santa would use elves," Danny said.
Alex giggled and crushed buds into a pipe with the corner of a candy cane-striped lighter.
"Seriously," Danny kept on. "How the fuck are elves supposed to make toys with those chubby little fingers?"
I stared through the frosted windshield at a light-up reindeer staked in the front yard of a huge house.
"Dwarves have chubby fingers." I sighed. "Elves don't."
"Wrong, Patrick," Danny spat, his frosted breath blowing in my face. "I did dishes at Denny's with that little elf motherfucker and he has fat fingers."
"That's because he's a dwarf, dumbass."
Danny stared at the floor and thought for a second.
"If he's a dwarf, then what the hell is an elf?"
Once Alex was good and stoned, he climbed out of the truck and crunched off down the icy street toward Tina's house. He dug the key out of the mailbox and then opened the door and slipped inside. A minute later the garage door opened and Danny pulled the truck slowly up the street and into the garage. Alex closed the door behind us.
We kept the lights off in the house as me and Alex loaded up the truck. Danny walked around knocking paintings off the wall, searching for a safe. I was disconnecting the stereo in Tina's brother's room when Alex spotted the snake aquarium and panicked.
"We can't let Danny see this," he said. "Come on. We have to go."
"Why?" I asked, picking up the turntable.
"Because he's totally nuts over snakes. You didn't know that?" Alex said. He grabbed the stereo and whipped the cable out of the wall.
"We have to get out of here. If he sees this thing—"
"Sees what thing?" Danny asked. He stood in the doorway holding a mounted moose head.
Alex stepped to the left to block Danny's view but it was too late. The moose head thumped to the ground and Danny charged over to the aquarium. He flipped open the heavy lip, shoved his arm inside and lifted the cover to a small habitat. A coiled giant white snake hissed from the corner. Alex and I backed toward the door. Danny grabbed the snake behind the head and all twelve feet of it uncoiled as he lifted it out of the cage.
"Hot shit!" he said. "This is an albino carpet python. We can get a ton for this thing."
"We're not stealing a snake, man," I said, keeping my distance.
"Are you nuts? There are probably rare snake dealers in Arizona who will pay thousands for this guy."
He stared into the snake's eyes. There was no way we were taking the snake.
"And where are you going to keep it until then?" I argued. "How are you going to ship it? You don't know anything about this shit so just put it back."
Danny ranted about Australian pythons, reptile breeding and international shipping procedures for live animals, most of which he probably made up. When me and Alex refused to help him carry the heated aquarium, Danny dropped the snake on the bed and stomped down the stairs. The house shook as the garage door raised and Danny sped out of the garage. Tires squealed down the driveway and the truck fishtailed when it hit the icy street. Danny lost control and the truck careened off a parked car. He oversteered and spun out in the neighbor's front yard, mowing over the light-up reindeer. He got the truck back on the street and sped away.
Alex and I were grabbing whatever we could carry when the front door opened. Flashlights circled the room and two cops yelled for us to stop. I bolted to the right and ran down the long hallway through the kitchen, then out the back door and across the dark yard. Alex was cornered. He turned and ran back up the stairs. The cops tackled him, knocking him across Tina's brother's bed and on top of the twelve-foot albino python. I was climbing the backyard fence when the snake sunk its teeth into Alex's arm. I swear I heard him scream.
I shook off the memory just as the waitress grunted a hello and set down a glass of water, slopping most of it on the table. She never looked at me. Her dark ponytail swung as she turned her head toward the front door and a family who looked fresh from church entered.
"I'm ready to order," I told her.
She mumbled something and rifled through the front of her red apron, pulling out a pen and a notepad. After flipping a few pages she looked up, stared at me for a few seconds then spoke.
"What the fuck do _you_ want?"
I sort of expected that.
"How have you been, Emily?"
Her elbow jutted out as she slapped her hand on her hip. She rolled her blues.
"What can I get you?"
"I'm back in town for a bit and thought I would come by."
"Great. When are you leaving?"
"Not sure. I came in for Alex's welcome-home party. Didn't see you there."
"Is that supposed to be funny?"
"Yeah," I said, looking down. "Sorry. It was a bad joke."
She stared at me. Her glossed lips popped on her chewing gum.
"Listen," I said. "I'm really sorry about how everything went down."
"Which part? The part when you and Alex robbed my best friend's house or the part when you stood me up on New Year's Eve and left town without ever talking to me again?"
"I had nothing to do with what happened at Tina's," I lied. "And I had to leave town. I had a great job lined up in New York that couldn't wait. I tried getting in touch with you. You never returned any of my calls."
"I was mad."
"Let me make it up to you," I said.
Emily's boss yelled to her from behind the counter.
"I got to go," she said, turning to walk away.
"Wait. What are you doing tonight?"
"It's Sunday. You know what I'm doing. I'm going to my sister's."
"Is she still running that Led Zeppelin fan club?"
"Yes. And you know it's called the Misty Mountain Hoppers."
"Can I come with you?"
She stopped and looked up at me.
"Why? You don't like Zeppelin."
"Would coming to the meeting be punishment enough for you to forgive me?"
"Maybe," she said.
"Then maybe I'll do it," I said.
When she walked away, I yanked the yellow flier from the Record Barn out of my pocket, crumpled it and tossed it under the booth.
* * *
Emily's sister Anna lived in a boxy apartment above a car parts store on a busy street. Concrete steps shot straight up the back to a long, tiled hallway. The apartment was in rough shape but Anna had done her best to dress it up. Bright paint covered each room. Yellow in the living room. Red in the kitchen. Green in the bedroom. Tapestries and black-light posters covered the walls along with photos of Led Zeppelin. A half-finished mural on the living room wall showed Jimi Hendrix in the sky above portraits of Led Zeppelin. Anna was a terrible painter and Jimi looked more like Yogi Bear.
Anna was short and dumpy, and even though she was only in her twenties she looked older. She shuffled past in a flowing yellow skirt that reached her ankles just above bare feet covered in weird jewelry. Trinkets and braids dangled in her ratty brown hair. Emily was wild and energetic but Anna was dull and almost always incredibly stoned. She held a dopey grin on her face, especially as she cut me off and talked over me. She did that to everyone. Most of her sentences trailed off as she fumbled to pin some hokey spiritual bullshit to everything.
She led us through the beaded curtains dangling in the doorway and into the living room.
"I hear you are living in New York City, Patrick. How is it?"
"It's good."
"Ugh. I don't like that place. It's such a negative energy trap."
Emily covered her mouth to stifle a laugh. I didn't know what to say.
"What's your sign anyway?" Anna asked.
Virgo. But I hated astrology.
"I don't know."
"You don't know your sign?"
I looked at Yogi Hendrix on the wall.
"I think I'm the Bear."
"The Bear? Is that from the Chinese zodiac?"
There was a knock at the door.
"Sorry, guys! That must be Kyle," Anna said. "Help yourself to a beer or a hash brownie. We'll fire up Babe later."
"What's Babe?" I asked Emily when Anna was gone.
Emily pointed to a giant blue bong nearly four feet tall in the corner.
"She nicknamed it Babe the Blue Ox."
"Your sister is insane."
"You're the one who wanted to come with me." She grinned.
The rest of the Misty Mountain Hoppers slowly showed up. They were a sad pack of hippies. Guys who got the "burn out" part down but forgot the "fade away."
There was Carl, a scrawny guy with one arm in a sling, dirty glasses that covered his entire face and a cheap cowboy hat who whined about the food while Manuel, a three-hundred-pound Mexican kid, bitched about the hot weather. Steve and Stacy, a brother-and-sister duo, talked about wheatgrass juice. Jim paced around the apartment in tight, black bell-bottoms. He looked like an ex-con or at least a psycho. He wore a pentagram ring on a tattooed hand and carried an Aleister Crowley book. The other stragglers looked like they just woke up at the bus station and didn't have anywhere else to go. Everyone smoked pot.
Kyle was the leader. He looked bookish, with long brown hair, Lennon-style glasses and a few turquoise rings. He sat cross-legged in a corner chair in his white pants, an ugly paisley shirt and an expensive-looking tan suede vest. Anna brought him a cup of tea and he nodded but never looked up from his stack of notes.
"Okay, everyone," Kyle said. He stood in the middle of the living room. We all sat like kindergarteners on the floor around him.
"Tomorrow night is Zeppelin at the Civic Center. I'll be driving a load of people down in my van and Carl can take a few people in his truck."
Kyle reminded everyone to meet at Anna's the next day so they could drive down together and then asked Steve and Stacy for a report on the sign they were supposed to make that said MISTY MOUNTAIN HOPPERS LOVE LED ZEPPELIN. Kyle looked disgusted when neither one of them remembered and that led to a discussion on the strength of last week's wine, which took a bizarre right turn into a talk about Carl's habit of pissing his pants when he got really stoned.
"Now has anyone had any luck figuring out where the band is staying in New York City the rest of the week?" Kyle asked. "We were all supposed to make some calls this week."
Everyone stared at the ground like kids who had forgotten to do their homework.
"They are playing Friday, Saturday and Sunday at Madison Square Garden, guys. We are driving up for Saturday's show and it would be nice to know where the band is staying so we can stake out the lobby. Anyone have any ideas?"
"How hard could it be to figure that out?" I whispered to Emily.
"You think you know?" she said, grabbing my hand.
"Well, I'm sure I could find out if I really gave a shit."
She turned to Kyle.
"Kyle. Patrick says he can find out where Zeppelin is staying."
"Oh yeah?" Kyle said, looking down his glasses at me. "And how are you going to do that?"
I hadn't really thought about it. I figured that Carmine could find out. He'd been catering concerts for years in New York City.
"I know a caterer," I said.
Kyle pushed his glasses back up his nose and shook his head.
"I don't see how that's going to work."
"Gimme the phone."
I dialed Carmine's office but he didn't answer. I tried the main office and Louise picked up. She put me on hold for a few minutes then popped back on to give me the address. I knew what she was going to say before she said it. I hung up and walked back into the living room.
"Drake Hotel. Park Avenue and Fifty-sixth."
Everyone looked surprised. A few of them doubted me but didn't have any better answers. Emily grabbed my hand again when I sat back down.
"Why do you come to these things?" I asked her later after the meeting had ended. We were standing in the kitchen grabbing beers from the fridge.
"Mostly because I like hanging out with my sister. Besides, it's not so bad. Free beer and pot. Beats being at home."
She leaned against the kitchen counter.
"How did Kyle become the leader of this group?" I asked.
"He's the only one who's met Jimmy Page. He sold Jimmy Page a guitar last year."
"No shit?"
"Ask him about it. Believe me, he loves telling the story."
"No thanks. I don't want to encourage him."
Emily looked around the party.
"Hey, Kyle," she yelled into the living room. "Patrick wants to hear about the time you met Jimmy."
I glared at her as I walked toward the living room. Kyle sat cross-legged on the floor taking a drag from Babe the blue bong. He turned around when he finished.
"Have _you_ met Jimmy?" he asked me.
"Nope."
"You really should, man. He's amazing. I met him last year."
"Is that when you sold him a guitar?"
"Yeah. My dad had given me a sixty-four Stratocaster when I was in college. I never learned to play it. I'd actually forgotten about it. When I got back from Europe I found it in a closet. A buddy told me it might be worth some money. I knew that Jimmy bought a lot of guitars on the road. I figured it might get me in to meet him. I mean, I just loved the idea that Jimmy would be playing a guitar that belonged to me."
"Totally," Anna said, nodding. "It would be like your energies were entwined."
I rolled my eyes.
"When Zeppelin came to town I brought the guitar to their hotel and waited in the lobby. I sat there all day until Richard Cole showed up."
"Who's Richard Cole?"
"You don't know who Richard Cole is?" Kyle scoffed.
I shook my head.
"He's Zeppelin's tour manager. He's a legend, man. So I showed him the guitar and he said he thought Jimmy might wanna buy it. He went upstairs to talk to Jimmy. I just hung out in the lobby. I waited a long time. I thought they forgot about me. Then Richard came back and told me to come upstairs. I figured we were going to his room but when the door opened Jimmy was sitting in a chair playing guitar through a teeny little amp. I was totally freaking out."
"Oh man," Carl said, shoving a hash brownie in his mouth.
"He was so small. I was surprised. He was maybe this tall," Kyle said, holding out his arm.
"He loved the guitar. Just loved it, man. He plugged it in and played some stuff on it for me. 'Dazed and Confused.' 'Thank You.' And a little song that hadn't even come out yet. Know what it was?"
I shrugged.
"Stairway to Heaven."
Everyone around us gasped, even though they'd heard it all before. Carl whistled loudly through a mouthful of hash brownies.
"Yep. I was the first person outside of Led Zeppelin and their crew to ever hear it."
Kyle looked pleased as shit, though he really had no actual way of knowing this.
"So he bought the guitar?" I asked.
Kyle looked annoyed.
"Yeah, he did, man. Three hundred dollars. Cash. Took it out of a briefcase filled with money. I've never seen so much money in my life. Must have been fifty thousand dollars."
"They really carry around that much cash?"
"It's Zeppelin, man," said Manuel, leaning in. He wheezed loudly while chewing on a potato chip. "They spend that shit on cars and guitars and the best freakin' drugs. They're rich, man."
"So where does the band stay when they're in Baltimore?" I asked.
"Where they always stay. The Hilton."
"They rent a whole floor or something?"
He gave me a suspicious look.
"Yeah. They take the top floor. Everyone stays there—the band, the crew, managers, everyone."
"Thanks for the story, Kyle," I said, walking away.
Me and Emily left a short time later just as the hash brownies laid waste to the Misty Mountain Hoppers. Anna passed out in a bean bag chair with her shirt off, and Manuel ate everything on the kitchen table including a can of whipped cream. Carl pissed his pants.
We laughed about the party on the drive to Emily's house. I felt fucking great. I missed hanging out with her. Hell, I missed Alex and Frenchy and Keith too. That cocky feeling crept back into my brain. The one that ran through me years ago when me and Alex were really working. That feeling like the world was one big scam and all you had to do was connect the dots, think fast and make the right moves as you pinballed around, and the whole thing wasn't nearly as hard as it looked because all of the other players were winging it just like you.
I couldn't stop myself from smiling as me and Emily made out in her driveway.
She pulled away.
"What are you grinning about?"
I couldn't tell her that I was going to rob the biggest rock 'n' roll band in the world so I just smiled and kissed her again.
# SEVEN
#
FRENCHY," I YELLED, NOT LOOKING BACK AS I HURRIED UP THE CONCRETE CORRIDOR. "YOU'RE NOT BRITISH."
"Yes, I am, mate," he argued.
The accent sounded convincing. Still, there was no way I was going to let him fake being British. Backstage passes he had scored for us through a contest with WKTK and the Record Barn dangled around our necks. Frenchy adjusted his fake mustache and knocked a pair of aviator sunglasses off his face as he tried to keep up with me.
"Hang on a second," he said, stopping under the fluorescent lights in the hallway to pick up the sunglasses. He straightened up and smoothed the denim vest he wore over a button-up shirt. I stood in front of him.
"You can't be British," I lectured. "Are you listening to me? Led Zeppelin are British. Their managers and their crew are British. What if one of them asks you where you're from or where you went to school? You don't know shit about England."
The lights dimmed at the end of the tunnel and the crowd cheered. The sound of Zeppelin taking the stage poured down the corridor toward us in a tidal wave of noise.
"Listen," I said. "Stick to the story. Your name is Reginald Chamberlain. You're a rare-guitar dealer from Baltimore. You want to sell Jimmy Page your nineteen sixty Fender Telecaster. That's it."
"What if he doesn't want it?"
"Who gives a shit? We don't care if he buys it or not. We just need to hang around and watch them. See who has the money and find out where they're staying. Hopefully, we can ride with them back to the hotel."
He looked up at me.
"How does my mustache look?"
He lifted the guitar case and started up the corridor. Anyone backstage had already gathered at the side of the stage to watch Zeppelin, and me and Frenchy strode up the wide, empty hallway. Zeppelin charged through "Rock and Roll" and into "Celebration Day." The drums thundered off the concrete walls of the empty hallway as we marched toward the stage.
"What about the South?" Frenchy said. I could barely hear him over the music.
"What about it?"
"Can I be from the South? You know, use a Southern accent?"
"Let it go, Frenchy."
Roadies, bouncers and groupies crowded the side of the stage. We slipped through and looked out into the Civic Center. Light washed over the faces of tens of thousands of screaming fans. Bodies filled every space from the floor to the ceiling and around the walls.
Jimmy Page stood facing the amps, his back to the crowd. Sweaty ringlets of dark hair surrounded his face. He wore black flares covered in embroidery and a small matching jacket with no shirt underneath. White symbols lined up the pant leg and around the cuffs of the jacket. His guitar hung low to his knees and he hunched over the neck, squeezing out the chords to "Celebration Day."
Frenchy turned to me and shouted over the music, "What about Brooklyn?"
"What?" I asked, cupping my ear.
"I've been working on a Brooklyn accent. This would be a great way to test it out. Can I use that?"
"Seriously, man. Stop it."
Zeppelin powered through "Black Dog" and "Over the Hills and Far Away" then launched into "Misty Mountain Hop." Robert Plant danced at the edge of the stage, wearing tight blue jeans and a tiny, matching vest. He twirled the microphone by the chord and pumped his hips at the crowd. They went wild.
The lights dimmed over the audience and Zeppelin eased into "Since I've Been Loving You." John Paul Jones's lazy bass lines rumbled across the stage. I tapped my foot and checked my watch. How long could they play?
The thought was broken up as a thick hand grabbed the back of my arm. A lanky guy in a white satin jacket poked his bearded face at me. His T-shirt read, "Edgewater Inn, Puget Sound."
"Who the fuck are you?" he asked, in a snarling British accent. "Where did you get this pass?"
He grabbed the pass around my neck and yanked me toward him until we were face-to-face.
"From WKTK," I stammered.
"What the fuck is that?" he asked, inspecting the back of my pass.
"It's a radio station here in Baltimore."
He locked eyes with me.
"Is it any good?"
"No. It's fucking awful."
"They play Zeppelin?"
"All the time," I answered, grinning.
He grunted and let go of my pass.
"Just stay the fuck out of the way," he barked as he stormed off. He turned and yelled back, "And tell your fucking friend."
A pair of groupies in short skirts and knee-high boots chased after him. Their voices trailed off.
"Richard! Hey, Richard! Wait!"
The opening chords of "The Song Remains the Same" erupted from the stage. Frenchy leaned over to talk to me.
"Who the fuck was that guy?"
"Richard Cole. Zeppelin's tour manager."
"He didn't look happy."
"No. He had a message for you."
"For me? What did he say?"
"Stay the fuck out of the way."
"Got it." Frenchy nodded.
Zeppelin closed the show with a blistering version of "Communication Breakdown" then charged offstage with the feedback from the instruments still pulsating through the arena. The backstage crowd swirled around us as the band rushed for the dressing room. The door slammed and those of us left in the hallway slumped against the walls.
Me and Frenchy waited with the crowd in the hallway. Now and then a roadie or manager would go in or leave the room and the groupies would shriek "Jimmy" or "Robert" through the open door. Inside, Jimmy stood in the center of the room with a towel around his neck, holding a bottle of Jack Daniels and staring at the floor. Richard Cole sat bent over in a folding chair in the corner counting money in a brown suitcase.
"Australian," Frenchy said.
"Not a chance." I smirked. "Tell you what, you can be French, Frenchy."
"Fuck off."
An hour later the door was flung open. A massive figure filled the entire frame. Peter Grant, Zeppelin's manager. His swollen gut jutted out in the hallway. A ring of scraggly hair wrapped around the sides of his bald head and blended with a dark goatee. His voiced boomed.
_"All of you. Get the fuck out of the way."_
He lumbered through the crowd and Zeppelin and their entourage snaked behind him. Security guards fanned out around him, adding to the crush of bodies in the narrow hallway.
The four members of Led Zeppelin moved untouched in the center of it all. Jimmy and John Paul Jones wore sunglasses. John Bonham and Robert kept their heads down, trying not to make eye contact with anyone. The entire procession moved quickly down the hallway, charging like a herd of animals.
Me and Frenchy were sandwiched between the crowd and the wall. We swam through the crowd to keep up. I turned sideways to slip past a pair of groupies then bent to duck under the arm of a photographer holding his camera above the crowd. Zeppelin edged away from us up ahead.
"Come on, Patrick," Frenchy yelled from behind me. "We gotta get up there."
The crowd kept moving as I tried to maneuver around a pear-shaped hippie in front of me. He was just too big. His wallet dangled from his back pocket, squeezed to the top by his flared jeans. I tugged the wallet loose and tossed it in front of him.
"Aw, shit," he howled. He dropped to his knees and scrambled to wrap a meaty hand around the wallet. I dodged to the right, grabbing a fistful of Frenchy's shirt and dragging him with me. The crowd behind us slammed into the hippie in a pile-up of hair and sweat.
We caught up with the main entourage and I shoved Frenchy toward Richard Cole.
"Get to it, Reginald!" I whispered to him.
"Mr. Cole," he stumbled nervously.
Richard Cole kept walking.
"Mr. Cole, my name is Reginald Chamberlain. I have a guitar that Mr. Page might be interested in."
"What is it?" Richard asked, keeping his eyes locked on the Civic Center's rear exit as we raced toward it.
"It's a nineteen sixty Telecaster. Great condition."
Richard cocked his head to the side and whispered to Jimmy, "Nineteen sixty Telecaster."
From behind his giant sunglasses, Jimmy Page jerked his head in the slightest of movements. I barely saw it. Richard turned back to Frenchy.
"Not interested."
"Really?" Frenchy stuttered. "But didn't Jimmy play a Fifty-eight Telecaster for the solo on 'Stairway to Heaven'?"
"I told you, he don't want it. Now piss off."
Frenchy looked back at me and shrugged. I waved my hand in the air, motioning for him to keep talking. I couldn't hear what they were saying. I walked on my tiptoes to see over the crowd. Frenchy said something and Jimmy's head spun toward him. He nodded rapidly.
Suddenly we were outside, launched from the back door of the Civic Center into the cool summer night. Zeppelin and their entourage slipped with military precision into a waiting black limo. Richard was the last one in. He stopped in front of Frenchy.
"Well? Where the fuck is it?"
"Where are you staying? I'll bring it to your hotel."
Richard climbed into the waiting limo. Something was wrong.
"There's no hotel." He laughed. "We're taking the jet back to New York tonight."
"But how will I get the guitar to you?" Frenchy pleaded.
"Bring it to New York this weekend. We're staying at the Drake Hotel until Sunday."
The limo driver slammed the door as I caught up with Frenchy.
"What the fuck just happened?" I asked.
"Um... I told them we also had a Fifty-eight Les Paul."
"What's that?" I asked.
"Listen, I'm sorry! I got nervous. I didn't know what to say."
"What the fuck are you talking about? What's a Fifty-eight Les Paul?"
"It's one of the rarest guitars in the world." Frenchy sighed.
"Where the fuck are we going to get one?" I shouted.
I couldn't believe it. I thought to myself, This can't get worse.
"Haven Street Pawnshop has one."
It was now worse.
The limo pulled away and I stared at the glowing taillights until they disappeared over the hill and moved toward the airport and the private jet waiting to take Zeppelin to New York. I tore the fake mustache off Frenchy's face but Zeppelin was too far away to hear him scream.
# EIGHT
#
SOME PEOPLE AND PLACES THAT YOU JUST DON'T FUCK WITH EVEN IF YOU'RE THE DIRTIEST CROOK AROUND. BAIL BONDS OFFICES. GUN SHOWS. PAWNSHOPS. YEARS OF DEALING WITH THE MEANEST AND DEADLIEST ASSHOLES IN THE WORLD HAS LEFT MOST OPERATORS EVEN MEANER AND DEADLIER THAN THE CUSTOMERS. SCREWING AROUND WITH THEM IS JUST PLAIN STUPID.
In Baltimore, there was one person that you definitely didn't mess with: Backwoods Billy Harvick. Ten years ago, Backwoods Billy ran with a pack of motorcycle nuts. Guys with names like Jimmy "Two Bottles" and "Hairy" Garfield. They were animals. They stole anything, tore up everywhere they went and kicked the shit out of anyone. They would have been a real motorcycle gang if any of them had bothered to come up with a name.
One night after a ride, Backwoods Billy and his guys showed up at the Damn Tap, a beaten-down bar in Fell's Point. They went wild. They terrorized the waitress, stole everything they could and broke everything else. Someone found a safe in the back room and started prying at it with a crowbar. When the owner stepped in, Backwoods Billy split his head with a socket wrench and dragged him out to the street. They laid him facedown in the gutter and Backwoods Billy ran over the poor guy's skull with his bike. Then they tore out of there.
The police busted him but couldn't prove anything. The bikers had their story sewn up. As they told it, the owner attacked Backwoods Billy and when the fight didn't go his way he ran out into the street, where he was creamed by a car.
"Happens all the time, Officer. Especially late at night. The roads are loaded with drunk drivers after the bars close. Hell, we're gambling with our lives riding our bikes that late, ya know?"
Backwoods Billy ended up with manslaughter. He got seven years and served five. He found Jesus in prison, gave up pills, got his GED and taught a Bible study class. When they let him out, he went straight back to the gang's clubhouse in Canton. He gathered up the old gang, gave them Bibles and christened them the Holy Ghosts.
They were still the same psychos, only now they wore a skull-and-cross patch. Nothing else had changed. They still sold pills, still knocked off stores, still beat the hell out of people. The only difference was that this time they didn't get busted. The few Holy Ghosts actually to see the inside of a courtroom always dodged serious time. They were blessed with a string of botched trials. Missing witnesses. Cops with foggy memories. Misplaced evidence. One miracle after another.
The newspapers ran photos of Backwoods Billy standing outside the courthouse, Bible in hand, praising Jesus and defending the misunderstood Holy Ghosts and all of their hard work for the community like their charity car wash and Christmas toy drive. The law couldn't touch him. He could fuck with you any way he wanted and get away with it.
The Holy Ghosts' main business was selling pills. Black Beauties. Yellow Jackets. Blue Devils. They invested in a few businesses around town: an auto garage, two tattoo parlors and their main operation—Haven Street Pawnshop.
If robbing a pawnshop was a death wish, then robbing one owned by Backwoods Billy and the Holy Ghosts was guaranteed suicide. This could only end badly. I pictured the four of us buried alive under fresh concrete. Frenchy figured we'd be dragged behind motorcycles.
And yet here we were, pulling into the back parking lot of the Haven Street Pawnshop. The squat brick building looked more like a bunker than a store. Thick iron bars covered the two small windows and a tiny neon sign blinked PAWN. The sun shimmered off a freshly waxed Cadillac parked alongside the building.
A glass jewelry case lined the right side of the store and ended at a stack of TVs in the corner. Piles of drills and other tools filled shelves along the back wall, their power chords dripping toward the dirty tile floor. Rows of guitars perched on stands next to tiny amps and keyboards along the left wall.
"Hey, Pete," the old man behind the counter said as we walked in.
"Hey, Dave," Frenchy said, waving.
Alex whispered into Frenchy's ear, "Your name ain't Pete. It's Frenchy. Don't forget it."
"Fuck you," Frenchy whispered back.
"How the hell does the guy behind the counter know you?" I asked.
"I come in here a lot." Frenchy shrugged.
"Need help with anything?" Dave asked.
"Nah. Just looking around. Thanks."
Me and Alex pretended to be interested in an old TV. Frenchy strummed on an acoustic guitar then put it back on a stand. Keith plunked at a keyboard. Down the aisle Frenchy picked up an electric guitar, plugged the chord into an amp and sat down on a stool. The amp buzzed and hummed as he inched the volume knob toward ten.
Frenchy tore into "Voodoo Chile." I recognized the riff just as he hurled into the main chords. The sound rattled everything in the store. Guitars vibrated on their stands. The glass counters rattled. Dave ran from behind the counter yelling something but the sound drowned out the words. He fumbled with earplugs, cramming them into his ears as he hurried toward us.
"Pete! Pete!" he shouted.
Frenchy didn't look up. He dipped his shoulder and launched into the solo. His hand leapt up the neck of the guitar then slid back down. He held a note and lay into the whammy bar, twisting the sound around the store. Dave grabbed the neck of the guitar and Frenchy stopped playing. It felt like the air had been sucked out of the room.
"Pete! Goddamn it! I told you to cut that shit out," Dave said.
"Sorry, Dave. Just showing my buddies some stuff." Frenchy shrugged. He turned the amp down lower then quietly played "Little Wing." I couldn't believe it. Frenchy was one hell of a guitar player.
"Damn, man," I said. "When'd you learn to play like that?"
"Just been playing a lot lately, I guess."
"All right, Frenchy," Alex said. He was getting anxious. "Where's this guitar?"
Frenchy looked across the store.
"I don't know. It's usually right there."
He pointed to an empty stand in a glass display case. If that guitar was gone we were all screwed.
"Hey, Dave. You finally sell the Fifty-eight Les Paul?" Frenchy shouted.
"Hell no." Dave laughed. "Nobody around here has enough money for a guitar like that. It's in the back office."
Keith named songs for Frenchy to play and somehow Frenchy knew most of them. Even some Sabbath. Alex and I wandered the store, scoping out the layout. I ran over the setup in my head. You couldn't come in through the front windows with those bars and all the traffic on the street outside. The side windows were small and high, making it too difficult to slip in and too hard to climb out in a hurry.
When Dave grabbed another guitar to jam with Frenchy, me and Alex peeked into the back office. A metal door stood out along the rear wall.
"That must lead to the alley," I said.
"No alarm. Looks like that's our way in."
"Drill out the hinges?"
Alex nodded.
"Shit, man," I said. "We don't know how to do that. This is tougher than anything we've ever pulled. And if we fuck it up, we're dead men. Backwoods Billy will kill all of us."
"I think we should ask Danny to help."
"Danny? He can't cross the street without getting arrested. No. That's a horrible idea. I'm never working with him again."
"I'm telling you, he could do this. And we'll be there to make sure he doesn't fuck it up."
It was a stupid idea but I didn't have a better one.
"All right. But you have to talk him into it. And don't tell him anything about Zeppelin. Just tell him we're doing this to rip off the guitar."
"When?" Alex asked.
"Might as well do it tonight."
Me and Alex never noticed Dave walking up behind us.
"You guys looking for the Fifty-eight?" Dave asked.
We both jumped. Dave grabbed a guitar case from behind a desk and we followed him back out into the store. The clasps on the Gibson case snapped open and Dave lifted the lid. The guitar was gorgeous. Oranges and reds and yellows swirled in the finish, and the silver pick-ups reflected light around the room.
"Gibson used to make the Les Paul in a gold color," Dave explained. "The Fifty-eight was the first year in the sunburst finish."
"That's such a sweet guitar, man." Keith nodded. "How much you selling it for?"
Dave laughed.
"Only seventeen hundred of these exist in the entire world. The guy who buys this beauty better come in here with a pretty big goddamn checkbook."
"Shit. I don't even have a checking account," Keith said with a shrug.
As we left Dave called out behind us. He held up the guitar.
"You sure none of you guys wanna take this?" He smiled.
Alex looked back.
"Not right now, man."
# NINE
#
YOUR POCKETS. DUMP ANYTHING THAT MAKES NOISE. NO LOOSE CHANGE. NO KEYS. DON'T EVEN BOTHER WITH A WALLET. YOU DON'T NEED IT. IF IT FALLS OUT, YOU'RE FUCKED. NEVER CARRY IDENTIFICATION. MAKE UP A FAKE NAME AND USE THE SAME ONE EVERY TIME. SOMETHING EASY TO REMEMBER. ONE YOU KNOW YOU WON'T FUMBLE. I'M JOHN OSBOURNE, OZZY OSBOURNE'S REAL NAME.
Second, get a story. One that gets rid of cops fucking pronto. Nothing that might make them want to help you. You didn't lose your dog. Your car didn't break down. And it can't be something they can trace. You didn't just get out of a movie. You didn't just get off work. What are you doing out this late? You went for a walk to clear your head after a fight on the phone with a girlfriend. You didn't bother to bring your wallet.
Stick to the plan regardless of what falls in your lap. Don't risk going after something else no matter how tempting. Get what you came for and get out. One time, while Alex was climbing out of a house, I spotted a large bag in the backseat of the car. It looked like a camera bag or a small piece of luggage with cash or traveler's checks inside. A neighbor heard me break the window and called the cops. We barely got away. What was in the bag? Diapers.
Always plan on running. Double-tie your shoes. Don't wear clothes that make noise. That means no windbreakers. And be careful with hoods. They look suspicious. If you wear a hooded sweatshirt, make it a zip-up. Hoods make great handles for cops to grab when you're running. If you're lucky, you can unzip and leave him holding your hood and nothing else. If you do run, hit the backyards, not the streets. And don't worry about anyone else. You're on your own when they show. Pick a direction and run.
Almost a year had passed since the last time I was involved with anything like this but I still remembered the rules. Well, they weren't rules. Just simple shit that Alex and I worked out over the years. I went over them in my head as we sat around the kitchen table at Keith's house waiting on Danny. He was already half an hour late.
"I don't see why I have to come," Frenchy said.
"To make sure we get the right guitar," Alex argued.
"I already showed you which guitar."
"All that shit looks the same to me. Especially in the dark. Just point it out."
The front door creaked open and Danny slipped through. He tiptoed loudly across the living room. Me and Alex looked at each other and tried not to laugh. Danny sprung into the kitchen wearing a black sweatshirt, camouflage pants, black boots and a black stocking cap. He grabbed Keith in a head-lock before Keith realized he was there.
"Whoa, boy!" Danny yelled. "You'd have been a dead man."
"Let go of me, asshole," Keith yelled, pulling on Danny's arm.
"You pussies are going to have to be sharper than that tonight," Danny told us. He wrestled with Keith. The kitchen chair fell with a crash.
"You see how quiet I was? You guys never heard me coming. I coulda killed ya. Learned that from an ex-Navy SEAL I met in the joint."
"We all heard you walk in, dumbass," Keith grunted.
"Yeah. You say that now."
Keith struggled to pry Danny's arm from around his neck. Danny let go with a shove, sending Keith flailing across the kitchen. Danny stood with his hands on his hips, a grin across his face. He straightened his stocking cap.
"Great outfit, G.I. Joe," I said.
Alex and Keith laughed. Danny leaned over the table.
"Fuck you, Patrick. You're learning from a master tonight. Let's go, kids!"
Danny turned and charged toward the front door. His clunky boot hooked in the carpeting and he stumbled across the living room. The door slammed and seconds later my car horn honked twice. We all looked at each other.
"Get a move on," he yelled.
On the way out I caught Alex's eye and shook my head.
"He'll be fine," he said to me. "Besides, he's the best at this shit. You know that."
"I'm pretty sure we could handle it better if we didn't have to babysit his dumb ass."
"I'll keep an eye on him."
We didn't talk on the way to the pawnshop. Sabbath's "Electric Funeral" played on the radio. One of the evilest riffs ever written. Everyone stared out the windows. When we got close, Danny told me to pull over at a gas station. The building was locked up for the night and the dark parking lot sat empty.
"Why are we stopping here?" I asked. I worried he planned to rob the gas station too.
"Alex, get out," he said, pointing into the blackness of the gravel lot.
"What the fuck for?" Alex answered.
"Get your ass over there and call the pawnshop."
"Why the fuck would he do that?" I stammered. Goddamn it, I thought, I knew I shouldn't have let Alex talk me into this. This idiot was going to get us all arrested. "We're going to the pawnshop. Why the hell would he call there?"
"He's going to call it then leave the phone off the fucking hook. That way it keeps ringing."
"What's the fucking point of that?" Alex said. "They're closed."
Danny took a long drag on his cigarette then blew smoke from the corner of his mouth.
"If the phone's still ringing when we get there, we know no one's inside and it's safe to go in."
That made sense. Everyone relaxed a bit. It sounded pretty smart, actually, and when Alex climbed back into the car and told us it was still ringing I felt a bit more confident that we might actually pull this off.
Haven Street Pawnshop sat in the darkness of a tiny parking lot. Cars streamed past on the street out front. I shut off the headlights and pulled up into the alley. I found a parking spot just far enough away from the rear door not to raise suspicion. The phone was still ringing inside.
At the back door to the store, Danny handed Keith a bright red crowbar then leaned against the hood of my car. Keith fumbled the bar, nearly dropping it, and Frenchy lunged forward to catch it before it clattered to the ground.
"Damn it, Keith," Danny hissed. "What the hell's wrong with you?"
"Sorry," Keith mumbled.
Danny pushed his face into Keith's.
"You nervous or something?" he asked.
"Yeah. I guess so."
Danny stepped closer, causing Keith to move backward toward the door behind him.
"What are you so nervous about?" Danny asked.
"I'm not really that nervous. My hands are just sweaty."
"Which is it, man? Are you nervous or are your hands just sweaty?" Danny moved closer to Keith. "How do I know you aren't an undercover cop?"
"For fuck's sake, Danny," I groaned. "You've known Keith his whole damn life. Let's get this over with."
"No!" Danny said, not taking his eyes off Keith. "We need to find out what he's hiding."
"Keith," I said, cutting Danny off. "Open that fucking door."
The metal door fit tight into the frame and Keith struggled to wedge the jimmy bar in. Danny wrestled the bar from Keith's hands then twisted it into place. He stepped back, looked at Keith and pointed at the bar. Keith grabbed the bar with both hands and threw his weight backward when his grip, sweaty with nerves, slipped from the bar. He stumbled backward in slow motion before collapsing on the gravel parking lot in a tangle of limbs and dirty hair.
Alex choked back a stoner giggle and I covered my mouth to keep from erupting. Keith groaned softly, lying facedown in the gravel. He rolled over on his back and wiped tiny rocks from his bleeding elbow. Alex stood over him laughing. He kicked Keith lightly in the ribs and told him to get up. Frenchy looked up and down the alley, searching for an excuse to run.
"All right, you morons," Danny said. "Let's just do this and get out of here."
The crowbar dangled from the door frame where it was wedged in the paint. Danny grabbed the end with both hands and pried the door a few inches away from the frame, then twisted the bar to hold it open. He walked over to my car, leaned in the window to pop the trunk, then pulled the car-jack from inside. He stepped forward and wedged the edge of the jack into the opening. A cigarette dangled from the corner of his mouth as his arms pumped up and down on the handle. The jack rose, separating the heavy steel door from the frame with a loud groan until the frame shattered and the door sprung backward.
Frenchy peered nervously around the open doorway and into the dark store until Danny shoved him out of the way. Danny strode through the shattered door frame into the rear office and around an oversized desk covered with papers and folders.
"Let's go."
Streetlights lit the store floor and reflected off the glass display cases. The neat rows of guitars looked like tombstones spread out in front of us. We stood against the back wall and took it all in. Danny let out a soft whistle.
"Let's just get the guitar and get out of here," Frenchy said.
"What's the hurry, Frenchy?" Alex asked.
"No. Fuck you, Alex. We're getting this guitar and getting the fuck out."
Alex cupped his hands and lit a cigarette. He turned his head and let out a thick cloud then poked at Frenchy with the glowing cigarette as he spoke.
"All I'm saying is you helped us out with this thing. It's only fair that you get something for yourself too. Grab a guitar. Shit, man, as long as we're here."
Frenchy wrestled with the idea. The struggle didn't last long.
"Yeah. Maybe I will. You guys did drag me into this."
"Me and Frenchy will grab the guitar. You guys watch the alley." I pulled Frenchy into the darkness before anyone could argue.
While Frenchy picked out a guitar, I prowled the store looking for the '58 Les Paul. It wasn't on a stand with the other guitars or behind the keyboards. I was inspecting a stack of guitar cases on the floor by the power tools when Alex snuck up behind me.
"Looking for this?" he asked. He held the black hard-shell guitar case with the '58 Les Paul inside. "I found it sitting in the back office."
"Sweet." I sighed. "Where are Danny and Keith? You can't let Danny out of your sight. He'll fuck this up."
"Don't worry. They're out back loading the car."
"Loading the car with what?" I snapped.
"The safe from the back room. Danny thought we should take it. There's probably a ton of cash in there."
"Jesus, man," I growled. "We just came for the guitar. Now you guys are stealing the fucking safe?"
I hurried toward the back office and found Keith drinking a beer and rummaging through a tiny refrigerator next to the desk. He poked his head over the door and bit into a chicken drumstick.
"Keith! What the fuck? You're supposed to be watching the fucking alley."
"I got hungry."
A mangled chicken leg hung out of his mouth.
"That's really gross, man. You have no idea how old that is."
"Is thwat blad?" he chewed.
"Yeah, it's bad. You're gonna get sick."
He shrugged and swallowed.
"Aw, shit! Look what I found." Danny giggled from behind Keith.
Keith turned toward Danny then spun back around and slammed into me. He dropped his chicken leg and shoved to get past me and out the back door into the alley. Danny stood behind him pointing a pistol at me.
"Put that shit away, man," I said as I backed up.
"Yeah, Danny," Alex said, holding up his hands. "Don't fuck around. We're getting out of here."
Danny grinned. He leveled the barrel toward the wall.
"Enough bullshit, Danny," I said. "Put it down."
"Fuck you, Patrick. What are you gonna do about it?"
"All right. All right," Alex said, taking Danny's arm. "Take it easy. Let's get in the car. Patrick, you grab Frenchy."
Frenchy stood in the darkness strumming an electric guitar. Even unplugged I could make out the chords to "Ruby Tuesday" ringing across the silent store.
"Is that the one you want?" I asked him.
"I'm not sure. I really like this Fender but as long as it's free I might as well get something really expensive. Or maybe I should take an acoustic?"
"Grab three guitars. I don't care. Let's just get out of here."
Frenchy started to move. His face looked strange, lit by red and blue lights.
Shit. Red and blue lights.
"Fuck! Cops!" I yelled.
Frenchy dropped the guitar and sprinted toward the back door. The front door rattled open and flashlights filled the store. A voice boomed out behind me: "Freeze right there, asshole!"
I bolted across the back office, doing my best to dodge the swirling flashlights. I slammed the heavy security door to the office behind me and dropped the security bar. Fists pounded against the steel door as I sprinted out the back and into the alley.
The glowing red taillights of my car looked miles away. The metal edges of the safe jutted from my trunk and the back end of the car sagged with the weight. A knotted shoestring held the trunk lid closed.
Danny sat behind the wheel with his arm across the seat and his head turned looking back at me. Alex sat in the passenger seat talking to Danny. Even without hearing him I knew what he was saying: "Wait." The car revved and my hands fumbled for the door handle as Danny floored the gas pedal, spraying a rooster tail of gravel. Frenchy and Keith stared out the back window as the car tore down the alley away from me. I waved my arms over my head hoping they would stop. Red and blue lights rounded the corner behind me.
I picked a direction and ran.
# TEN
#
UP SHIRTLESS AND SWEATING. SOMETHING STABBED INTO MY SIDE AND I ROLLED OVER, PEELED A BOTTLE CAP OFF MY SKIN AND HURLED IT TOWARD THE CORNER. A HAND KNOCKED ON MY BEDROOM DOOR.
"Patrick. Are you up?" my mother asked from the other side.
"I'm up," I moaned, straightening my twisted boxer shorts. I opened the door. She wore bright yellow shorts and a flowered top, and held a spatula in her hand.
"I made breakfast," she said. "Are you going to join us?"
Her eyes trailed down my face and across my bare chest. My stomach dropped when I remembered the tattoo I'd gotten in New York City. I slumped against the door frame to hide it and faked a yawn, but she had already spotted it.
"What have you done?" she said. Her lipsticked mouth hung open.
"It's nothing."
"No. It's something, mister. It's terrible. Why would you do that to your body?"
"Okay. It's something."
"Don't get smart with me," she mumbled. The angry tone chugged to a halt in her throat and she suddenly seemed less threatening. The anger downshifted into something else but I wasn't sure what.
"Your father is going to have a heart attack if he sees this," she whispered. Her wide eyes met mine. We were suddenly partners in a horrible crime.
"Here he comes. Get some clothes on."
She shoved me backward into my room and I lunged to pull the door closed as I fell over a pile of clothes and an old skateboard.
My mother slipped me a knowing look as I sat down at the kitchen table. My father was talking about the fire department again. He was a fireman and being a fireman was his entire life. It went beyond dedication to straight brainwashing. He didn't know anything except being a fireman. Music. Movies. Sports. World affairs. Nothing. The only politics he followed were local and only because they might have affected the firehouse, like with a budget cut or new equipment being bought for another station in town before his.
My sister and I were his only connection to the rest of the world. If a conversation turned to anything other than fire-fighting, he threw us out there like a goddamn lifeline. Talk about a new movie and he'd say that his daughter saw it and didn't like it. Bring up the World Series and he'd tell you how his lazy-ass son quit Little League.
It was the only way he could talk about anything other than the firehouse. And he could steer any conversation toward firefighting. Pick a topic and he would turn it into a rant about the VFW's fire code violations that got overlooked by the chief as a favor to a buddy. He and his work were the center of any conversation. You couldn't win. I'd stopped trying years earlier.
"I bet you don't get breakfasts like this in New York City, Pat," my mother said.
"Nope," I said. "This is great."
"Are you eating enough? You look skinny."
"I eat a lot, actually. Usually leftovers from whatever catering job I worked the night before."
"So the job is fun?" she asked.
"It's all right. I work with some cool guys. Plus, I get to be backstage at all the concerts."
"How exciting," she said, grinning.
"You know who I met the other night? Bette Midler."
My mother gasped and clutched her chest.
"No! Did you really? What was she like?"
"We were all backstage at Radio City Music Hall when she came out of her dressing room, walked over to the food table, looked at me and said, 'Where the fuck is the cranberry juice I asked for?' "
We both laughed until my father spoke up.
"You guys better be keeping those tables clear of the fire exits. Some dumb rock star leaves a lit cigarette in the dressing room and that place goes up, you'll all die trying to get out."
I stood up and dumped my plate in the sink.
"I'll keep that in mind, Pop," I said as I slipped out the front door.
My car sat crooked in the street in front of my parents' house. Keys in the ignition. Guitar and safe gone. Gas tank empty. A note under the windshield wiper.
_THANKS FOR LETTING US BARROW YOUR CAR. WE DIDN'T HAVE ANY MONIE FOR GAS. DANNY_
"Whole Lotta Love" played on the radio as I pulled over to the curb in front of Frenchy's house. I climbed out of the car just as Robert Plant started groaning about giving me his love. What an awful fucking song.
The sun set and I kicked a bottle as I crossed the street. Frenchy lived in his parents' basement. Bare concrete floor and walls. Frenchy dressed it up with a smelly rug, a sofa he found in the alley and some posters. It wasn't so bad. Through the window I could see him, wearing a faded Who T-shirt and jeans, playing guitar on the couch in front of the flicker from a TV sitting on a milk crate.
"What's this? A high school reunion?" He smiled as he let me in. It was our private joke. Keith and Alex dropped out of school so me and Frenchy were the only people we knew who graduated. To us, it was a high school reunion whenever we were together.
"Just out seeing what you're up to," I told him.
"Nothing. Playing guitar. Watching TV."
I poured myself into Frenchy's couch. He hesitated for a second then spoke.
"Hey, man," he said. "Sorry about last night. I—"
"Don't worry about it," I interrupted. "I know it was Danny. What happened?"
"I told him to wait. Just so you know."
"I know."
"So where's the safe and the guitar?"
"You're sitting on it," he answered, pointing to the couch.
I bent over, looking at the tattered couch underneath me.
"The guitar is under there," he explained. "Danny and Alex took the safe."
"Where did they take it?"
"I don't know. They dropped me off with the guitar first. They didn't say where they were going after that."
Frenchy opened two beers and handed me one.
"So what's the plan now?" he asked, leaning forward. "Are we going to go to New York?"
"Of course," I answered, sipping my beer.
"I just don't know if we can pull it off," Frenchy said, staring at the floor.
"It's simple, man. We bring him the Les Paul and, while you're in there, me, Keith and Alex will grab the money. Then we get out of there."
"I don't know, man. I'm starting to get worried."
"What are you worried about?" I asked him. "We're just a couple of guys trying to sell Jimmy Page a guitar. We haven't done anything wrong."
"We're trying to sell him a guitar that we just stole from one of the biggest psychos in Baltimore!"
"Yeah, well, there's that."
Frenchy clicked off the TV and put a record on the turntable. Frenchy was the only person I knew as obsessed with music as me. He liked raw rock 'n' roll, usually British stuff. The Animals and the Who and obscure stuff I didn't even know. He turned me on to tons of new stuff like the Stooges and MC5. His record collection lined the walls. Hundreds of LPs bought with money he made working at the Record Barn. The rest were probably stolen. He dropped the needle on the record and stood grinning at me.
"What?" I asked him.
"Just listen."
"What am I listening to?"
"You'll dig this."
The song kicked in. Loud R&B guitar, swirling drums, hand claps. It sounded fucking great. A raspy voice howled, _I_ _ain't foolin'... Hey woman you need coolin'_. I knew the song but it took me a second to place it. It was Zeppelin's "Whole Lotta Love," just different. Cooler.
"Holy shit!" I sat up. "Is this someone covering 'Whole Lotta Love'?"
"Nope." Frenchy grinned. "This is the song Zeppelin stole it from."
"No fucking way!" I laughed. "This is ridiculous. Let me see that jacket."
"It's the Small Faces 'You Need Loving.' Came out three years before Zeppelin's version."
"The singer sounds exactly like Plant!"
"Yeah. They stole the whole vocal idea."
"Goddamn." I shook my head. "Just another reason to hate them."
"Well, the Small Faces took it from Muddy Waters. Zeppelin just stole their version of it," Frenchy said.
"And the entire fucking sound."
"Yeah." He laughed. He sat on the floor and dug through a crate of records. "Jimmy Page ripped off the beginning of 'Stairway to Heaven' from this band Spirit. I heard 'Dazed and Confused' is lifted from some folksinger."
"Fucking con artists."
"They're still amazing. Everyone borrows, man. That's just music."
"Oh, don't give me that shit. This is outright theft. They totally stole these songs."
"They took the same elements that everyone else used and did it better. Look, if I put peanut butter and jelly on bread it's not the most original sandwich but you'd still eat it."
"I don't even know what the fuck that means."
We went on like that for a while, talking music and listening to records. We could do that for hours. It felt good to hang with Frenchy and not worry about anything except if the Stones put out better albums without Brian Jones (they did) or if the Animals were better than the Who (sometimes). Eventually the beer ran out and I headed home. It was late and I was hungry so I stopped for a burger.
My car sat alone in a corner spot at the back of the parking lot as I left the restaurant with a bag of food. A Black Sabbath eight-track slipped to the floor and I stretched across the front seat to grab it then crammed it into the player. I fumbled with my keys and finally fit them in the ignition. That's when I heard something metal tap against the window.
I didn't even turn all the way around. All I needed to see was that ring knocking against the glass. That giant silver skull ring with a bullet slug buried in its forehead. The ring was so big it covered up the entire section of his middle finger including one letter of a tattoo that ran across the knuckles of his hand. It didn't matter. I knew the black ink spelled out _P-A-I-N_. I didn't have to see the other hand to know it said _L-O-S-S_. It hit me that I was fucked in the worst possible way.
"Get out of the car, son," he said.
I looked through my dirty car window. He wore a wife-beater tank top underneath a leather motorcycle jacket and denim vest covered in patches. His greasy red hair hung past his shoulders and tangled with a kinky beard half a foot long. A huge tattoo crawled up his neck beneath it. At the base of his throat he'd tattooed a picture of a jackalope, an imaginary rabbit with antlers that turns up on postcards from Texas. The head sat mounted on a plaque and the tattooed antlers stretched up his throat and stopped just underneath his chin. The banner tattooed around it said, BACKWOODS.
This was Backwoods Billy Harvick.
I slunk from the car. Someone struck me from behind and a thick hand in a fingerless leather glove squeezed around my throat. When I tried to stand a heavy boot rocketed into my shins and knocked me to the ground then stood on the side of my head, grinding my face into the parking lot. Gravel bit into my cheek. I turned my eyes to look up at Backwoods Billy.
"You see this bike?" he said. His greasy finger pointed at the motorcycle. Despite all of his filth, the bike was spotless. The high handlebars, the long chopper forks and chrome tailpipes gleamed in the dark.
"There was a time in my life when that bike was all I had to live for. That was it. I didn't have a home or a woman. No job. No family. That bike was it. You know what that does to ya?"
I started to say something but couldn't think. Visions of that giant bike rolling over my head played in my mind. He kept going.
"It makes ya think human life is cheap. You don't care nothing about your own life so you don't care nothing about nobody else's. Back then we wouldn't be talking like this. I would have already killed ya for what ya done to me."
My heart stumbled and the strength drained out of my legs. I raced through my brain trying to figure out how Backwoods Billy knew we'd robbed the Haven Street Pawnshop. I stared at the ground.
"Lucky for you I don't act like that no more. I've got God in my life now and a wife and kids to look after. But that don't mean I'm gonna let you get away with what you done." He pointed at me. "As the Bible says, 'Know this, that if the good man of the house had known in what watch the thief would come, he would have watched, and would not have suffered his house to be broken up.' "
"Is that what you think I did? Broke into your house?" I talked so fast that I tripped over my own words. It sounded like three people speaking at once. "Because I didn't. I wouldn't do that. You must be thinking of someone—"
"Shut up, son," he said. "Is this your car?"
"Yeah."
"This is the car my partner saw parked in the alley behind my pawnshop last night." He nodded to the person standing on my skull. The foot let up enough for me to turn my head to see the enormous bald man dressed in black jeans, motorcycle boots and a black leather vest crossing his bulging arms over his bare chest. He leaned down toward me and stared from behind dark sunglasses. My own frantic face peered back at me from the reflection in the lenses.
"You broke into my business last night and stole something very valuable to me," Billy said.
"This is just a huge misunderstanding," I said. "Someone obviously did something very fucking stupid. I can get the guitar back. Just give me some time."
"Guitar?" He grinned. "Hell, I don't care about no old guitar. That's Dave's shit. I'm talking about my safe, boy."
The safe. Fucking Danny.
"I'll get it back. I swear to you."
He leaned against his bike and pulled at his beard with a tattooed hand swollen with scars. I stared at the giant skull ring. He looked up at the sky. His partner stood nearby and Billy smiled at him. I held my breath. Finally he said something.
"Son, what you stole from me is very important and I need it back. So I'm gonna give you a chance to return it. Now I know I don't need to give you no deadline. I expect it to be brought back to me as fast as fucking possible. You know that. And believe me, boy, you'll know when I've run out of patience."
"I understand."
He crouched down and bent his head to put his face in mine.
"If you think I'm joking, you little shit, remember Samuel 22:38: 'I have pursued mine enemies and destroyed them and not turned away until I had consumed them. And I have consumed them and wounded them that they could not arise. They are fallen under my feet,' " he preached.
He stood up and turned to his partner.
"Okay, Rabbit. Let's give it to him."
Two meaty hands squeezed around my throat and jerked me to my feet. The force cocked my head backward and arched my back. The toes of my shoes scraped the ground. They're gonna beat the shit out of me, I thought. This was Backwoods Billy's warning to let me know that the Holy Ghosts were not fucking around. The empty parking lot stretched in every direction but I didn't even think about running. There wasn't a chance. As he walked closer, I stared at that big skull ring. He stopped and stuck out a tattooed arm. A tiny red Bible sat in the palm of his hand.
"Son, is God in your life?" Billy asked me. He pushed his face into mine.
"Well, I, uh..." I couldn't breathe.
"What you need to do, boy, is study the Bible and stay clean. Take this."
The monster behind me let go of my neck. I bent over gasping for air and rubbing my throat. Backwoods Billy dropped the Bible into my hand.
"Do you like Led Zeppelin?" he asked, swinging a leg over his motorcycle.
"Hell no," I choked.
It was instinct. I really didn't.
He nodded.
"That Jimmy Page is a devil worshiper and he's polluting kids' minds with that stuff."
I nodded and moved toward my car.
"The real stairway to heaven is right there in that little book, son. Rock 'n' roll is the devil's music."
"I don't even listen to rock," I lied, shutting my car door and rolling down the window. "I like Johnny Cash."
"There's hope for you yet, son." He smiled.
I turned the key in the ignition. Black Sabbath roared from the speakers: _"Look into my eyes you will see who I am / My name is Lucifer, please take my hand."_
The volume knob slipped from my hand so I jerked the eight-track from the player and threw it on the floor. I couldn't look at Backwoods Billy. I didn't want to see his reaction. I closed my eyes and rested my forehead on the cool steering wheel. I glanced over when I heard them kick-start their bikes with a roar of noise. Backwoods Billy scowled at me as he revved his bike. Then he wrenched the throttle and the two bikes streaked across the empty parking lot like giant bats disappearing into the night.
# ELEVEN
#
BLASTED FROM THE BACK ROOM AT KEITH'S HOUSE. I HEARD THE MUSIC FROM THE STREET AS I WALKED FROM WHERE I'D PARKED UP THE BLOCK. KEITH HAD TOLD EVERYONE TO SCATTER THE CARS AROUND THE NEIGHBORHOOD TO MAKE IT LESS OBVIOUS THERE WAS A PARTY HERE. NO ONE LISTENED.
Sweaty bodies crowded together dancing and talking. The music played nearly too loud to talk over. Keith yelled my name. He was shirtless and struggled to pull himself up from the couch. Long greasy hair stuck to his face. He threw a sweaty arm around me. I wished him a happy birthday and handed him the bottle of vodka I'd stolen from the liquor store around the corner.
"Thanks, man," he slurred.
"How did you talk your mom into this?" I asked.
"I told her she didn't have to buy me anything. Just get out of the house for the night. She would have gone to the bar anyway."
It's not like a party could hurt anything. Keith and his mom didn't own much. They had lived in the house since me and Keith were ten and his mother still hadn't bought any furniture. The front room was entirely empty. Even the walls were bare. There was a nicked-up table in the kitchen that Keith's grandma gave them and in the back room a brown leather couch with stuffing bursting out and a tiny black-and-white TV. A photo of Keith from grade school hung crooked in the hallway. Keith's room contained a few piles of clothes, two boxes of comic books and an overflowing ashtray on the floor next to a mattress.
"Where's Danny?" I asked him.
"I think he's in the kitchen."
I moved toward the kitchen and scanned the room for Danny but didn't see him. I grabbed a beer in the kitchen, drank it too fast while talking to a chubby girl I went to school with and then opened another beer. Keith grabbed me and we did a shot of tequila together. The tequila kicked in. Someone played "Fortunate Son" and I started feeling all right. Earlier in the day I had called Alex and warned him about Backwoods Billy. We agreed to talk to Danny. I found Alex by the fridge.
"Where the fuck is Danny?" I asked.
"He'll be here," he said. He looked nervous. We stood there watching the crowd. "Have you seen Frenchy? The Frenchy Show is on tonight."
It was our name for those rare nights when Frenchy got completely drunk and out of control. It usually involved dancing. Sometimes nudity.
"Brown Sugar" started and I spotted Frenchy dragging a skinny blonde in a white dress toward the center of the room to dance. He wore a loud collared shirt unbuttoned, exposing his skinny chest. The girl towered over him and he was wasted and leaning too far forward to talk to her as they danced. A few times he lost his balance and fell into her, nearly knocking her down. He grabbed her arm and steadied himself then laughed. When the song ended they walked over.
"Guys, you gotta meet Sara," he panted. We all nodded at her.
The stereo kicked into "You Really Got Me" and Frenchy struck a pose, throwing his head back and pointing toward the ceiling. He spun his arm in a few windmills on an invisible guitar then danced away from Sara and into the crowd. She giggled then turned to me.
"If you're not going to dance you can hold this," she barked. Her arm shot toward me and poked me in the chest with a leather purse. It looked expensive. The purse fell into my arms and she disappeared.
"What the fuck was that?" I asked Alex. He laughed.
"That's a woman who is used to ordering people around," he said.
We both stood there staring at her. Alex finished a cigarette and flicked the burning butt into the crowd.
"Well, what's she got?" he asked.
He unzipped the purse and pulled out her wallet. He crammed a wad of cash into his pocket and then held up a small camera.
"Want it?"
"No thanks," I said.
Alex poured the rest of his beer inside before zipping it back up. He chucked the purse on the floor next to the couch.
A girl's voice yelled from the kitchen, "Keep your fucking hands off me, you asshole!"
A male voice shouted back at her but I couldn't hear it over the music. I pushed toward the kitchen.
"Oh, I'm no fun?" the female voice shouted. "Fuck yourself. How's that for fun?"
As I shoved my way into the kitchen a tall girl in a green sundress pushed through the crowd going in the opposite direction.
"That asshole just grabbed my ass," she said to her friend as we passed.
Danny had arrived.
I found him standing on the far side of the kitchen table. He wore a dingy T-shirt from Tony's Pizza that read, "If you like my meatballs, you'll love my sausage." A bottle of Jim Beam dangled in the fingers of his right hand and he held a can of Pepsi in the left. He alternated swigs from each. First a mouthful of Jim, then a swallow of soda. A white-trash whiskey and Coke. He swayed a bit as I walked over.
"Danny. I need to talk to you."
"Patrick!" He leaned in to whisper but talked loudly. "There's a lot of pussy here."
He slapped me on the back.
"Sure is, man," I replied.
His eyes roamed over the people at the party. Judging by the whiskey missing from the bottle and the slur in his speech, I put Danny somewhere between drunk and really fucking drunk.
"Listen, Danny. We gotta return that safe."
"What do you mean?" he answered, not looking at me.
"Backwoods Billy wants his safe back. He knows what happened."
He stared out at the crowd and didn't say anything.
"You have to give it to me. We have to give it back to him."
"No can do, amigo," he said smugly, taking a hard swallow of whiskey, chased by a swallow of soda.
"You don't have a fucking choice, man." I tried to sound calm and reasonable. "He saw the car. He knows you have it."
"No. He thinks you have it." He jabbed a finger in my face.
"He wants the safe back and we're going to give it to him."
"Can't do it." Danny shrugged.
"Why not? Where is it?"
"I don't have it."
Slug of whiskey. Slug of Pepsi.
"Well, who the fuck does?" I asked. Danny looked away.
Alex's head popped over the crowd, looking for us. He found us standing against the wall.
"So are we getting the safe back from Boogie?" Alex asked. Danny looked annoyed.
"Yeah. We are," I answered.
I stared at Danny. He took another drink of whiskey, long and hard, and didn't bother with the soda.
"It ain't that easy, kiddies," he grimaced. "I made a deal with him. If he opens it, he gets a cut of whatever's inside."
"Well, that deal is off," I said. "We're not opening it and he doesn't get anything."
"You gonna tell him that?" Danny grinned. He looked away and laughed.
"Who's Boogie?" I asked Alex.
"Danny's buddy. He's a safecracker."
"We'll find him tomorrow and get it back," I said.
"Are you not fucking listening to me, Patrick?" Danny said, turning on me, his eyes read with anger and booze. "You're not getting it back. Boogie won't just hand it over to you. Once he cracks that fucker, I want what's inside. Boogie wants what's inside. And Alex wants what's inside. Right, Alex?"
Alex shrugged. Danny looked irritated.
"Why are you doing this?"
"I'm doing this because I need to give the safe back to Billy before he kills us all," I shouted.
Danny rolled his eyes and took a long pull from the whiskey bottle.
"We're giving it back, Danny. And that's that."
Danny grabbed my throat in one sloppy thrust. Alex jumped between us. We struggled together, a triangle of sweaty bodies all grunting and pulling in different directions. I swung and my fist careened off the side of Danny's head.
"I know kung fu, motherfucker!" Danny yelled, flailing with one arm. His feet tangled and he fell backward, pulling me and Alex down on top of him. He landed flat on his back on the kitchen floor with a grunt. My forehead ricocheted off the bridge of his nose.
"Aww, shit," he groaned.
Legs stood over me and from the corner of my eye I saw Keith yanking on Danny's arm.
"Break it up!" Keith yelled.
We all separated. Danny stood up and straightened his shirt then bent to pick up his bottle of Jim Beam. Blood snaked down the side of his nose. He stared at me for a second and then left, kicking the door open with a dirty boot.
Alex turned to me.
"He's just drunk. He knows you're right. I'll talk to him tomorrow."
I nodded and picked my beer bottle up off the floor.
"Let's get a beer and have some fun," he said. "There's a lot of pussy here."
"Yeah," I told him. "That's what I've heard."
# TWELVE
#
GODDAMN GENIUS," DANNY SAID THE NEXT DAY AS WE DROVE PAST THE HIGH SCHOOL AND AROUND THE BOWLING ALLEY. I DIDN'T KNOW WHAT TO EXPECT WHEN I PICKED HIM AND ALEX UP BUT ALL THE BULLSHIT FROM LAST NIGHT SEEMED TO HAVE BLOWN OVER. ALEX MUST HAVE TALKED HIM DOWN. DANNY WAS A BIT HUNGOVER BUT LOOSE AND RELAXED, CHATTING AWAY IN THE FRONT SEAT. A SNOOPY BAND-AID COVERED THE CUT ON HIS NOSE.
"Tell Patrick what Boogie was locked up for," Alex said, leaning up from the backseat. "You'll love this, Patrick. It's your kinda scheme."
"Get this." Danny grinned. "Boogie is this poor black kid. Smart as fuck. And he's big. He was born big. Every year football coaches begged Boogie to play but that ain't him. He was always into music. He can play anything—drums, guitar, piano. He didn't want to play football and hurt his hands and fuck up his music career."
"Makes sense," I said.
"He gets a job at this music store selling instruments and jewelry and shit. One time the owner went on vacation and the soda machine broke while he was gone. Boogie's too lazy to put a sign on it so people keep plugging in money. When they complain to him that it doesn't work he tells them they'll have to come back next week for a refund when the owner gets back. Nobody does."
"Really?" I asked.
"Would you go all the way across town for ten cents you lost in a fucking soda machine?"
"I guess not." I shrugged.
"So Boogie gets this idea. He and a buddy pool all their money and buy a bunch of soda and candy machines. They stick them outside Laundromats, magazine shops, movie theaters. The thing is, none of the machines work. People put in money and don't get shit. They don't know how to get a refund so they just give up. Once a month, Boogie and his buddy come by and clean out all the cash. And because the machines don't actually work, they're pulling in free money with no overhead."
"That is pretty goddamn clever."
"Hell yeah, it is." Danny grinned. "Shoot. I need to get a setup like that. That's fucking smart."
"It wasn't that smart. He did get busted, didn't he?"
"Well, yeah," Danny huffed. "Only because some chicken-shit movie theater manager called the cops. They jumped him when Boogie came to collect."
"How long was he in?"
"Just three months. Shit, really nothing."
"What does he do now?"
"Not much. Lives out in this big house in Cherry Hill and works on his music. He's got a band."
"And he knows how to open a safe?"
"Hell yeah," Danny said, filling the car with a giant smoke cloud. "I told ya he's smart. He knows how to do all kinds of shit."
"Has he got it open yet?" Alex asked from the backseat.
"Not yet but he's working on it." Danny sounded annoyed and turned to face Alex. "Damn. Don't you trust me to take care of this shit?"
"I do," Alex said, sitting back. "I do."
"It doesn't matter," I said. "We're not opening it. We're getting it back from Boogie."
"We'll see about that." Danny smirked.
We barreled along the Washington Parkway. The businesses thinned out and soon we wound through used-car dealerships and run-down gas stations. Clusters of shabby apartment buildings huddled in threes and fours around sprawling parking lots dotted with burned-out cars and weed patches.
We wound along Cherry Hill Road until Danny told me to turn off on a narrow road leading through a small cluster of houses. A group of kids playing basketball in the street moved slowly to the side of the road and stared into the windows as we passed. The road kinked left but Danny pointed for me to keep straight into a gravel parking lot next to a beat-up yellow house tucked in the corner under a tree. All of the windows were closed even though the temperature lingered in the nineties. A tattered blue couch sat on concrete blocks in the front yard next to a rusted charcoal grill.
Danny said something to me and Alex as we walked toward the front door but I couldn't make out a word of it over the music exploding from the house. Tight funk drumming locked into step with a chicken-scratch guitar. Windows and doors rattled as the music pushed the old house to the verge of bursting.
Danny pounded on the screen door but the music washed out the sound. I sat down on the step and Danny walked to the front of the house to rap on a basement window.
"They're pretty good, eh?" Alex asked me.
"Yeah, they are. Sounds like Sly Stone."
"Definitely Figured out what you're going to say to Boogie?"
"I'll just tell him what happened." I sighed. "I'm sure he doesn't want any trouble with the Holy Ghosts either."
The music stopped and Danny dove to his knees in the dirt to beat on the window, burning himself with a cigarette in the process.
"Shit!" he yelped. "Hey, Boogie! Open up!"
A deep voice in the house said something and heavy footsteps stomped up creaky basement steps. As the door opened, Danny pushed past us up the steps, brushing dirt off his jeans.
Boogie filled the entire door frame; his giant red button-up shirt took up most of the view. He dabbed sweat from his head with a white towel hanging around his neck then ducked to angle a towering Afro through the doorway and poke his puffy face outside.
"What's up, Danny?"
"Nothing, Boogie. Just stopped by to talk a second."
"Cool."
Inside the house, cables snaked along the hallways, up and down the stairs, and tangled with table legs and other furniture. A microphone on a stand stood alone in a tiny bathroom off a kitchen cluttered with beer bottles and empty pizza boxes. Guitar cases, dusty mixing boards and an old organ surrounded another sagging couch in the living room. A sawed-off shotgun lay in the sink. Boogie dropped into a chipped wooden chair in the kitchen. It creaked under his weight.
"So what's up?" he asked slowly in a deep voice.
"Well, uh..." Danny stammered. "I just wanted to see how things were going with the safe?"
"I'm working on it."
"Is it here?"
"Nah," Boogie said, giving Danny a suspicious look. "It's at my shop."
The basement door opened and a short, marble-shaped black guy ambled into the room. He held drumsticks in one meaty hand and rubbed the back of his neck with the other, creating rings of fat under his tucked-in chin. A short Afro radiated around his head.
"That's our drummer Johnny Paycheck," Boogie said.
"You mean like the country singer?" I blurted out. Boogie laughed. A thick grin grew across his face.
"Maaaannn," Johnny hissed in a high-pitched voice. "I was using this shit before that corny redneck motherfucker."
"I take it you don't like country music?" I asked.
"I'm from the South, man. I love country music. Hank Williams. Ernest Tubb. My mama played all that shit. Some of them country cats can really play"
"So Johnny Paycheck is just a stage name?"
"Yeah." He nodded. "For the band."
"What's the band called?" Alex asked. Behind him Boogie whistled and shook his head.
"See," Johnny said. "That's another problem right there. I know what I want to call it but this motherfucker doesn't get it."
"I get it," Boogie said. "It's just stupid."
"It's not stupid. It's smart," Johnny said, tapping the side of his head. "I wanna call it the New York Giants."
Everyone in the room laughed. Johnny tried to calm us down.
"Listen! Listen, you motherfuckers." He waved his hands around. "This is the thing: when people see 'Appearing tonight: the New York fucking Giants' on a flier, they're going to come to the goddamn show."
"They're going to come to the show expecting to see the fucking football team, dumbass," Boogie shouted. They'd obviously gone over this more times than Boogie wanted.
"Wouldn't that get you in trouble with the real New York Giants?" I asked.
"It don't matter. We'll have so many fucking fans by then we'll change it. We're just using it to get people to the goddamn shows."
"That's pretty smart," Alex said.
"See!" Johnny shouted, turning to Boogie. "He gets it. This motherfucker gets it."
"We ain't calling this band the New York Giants," Boogie said, covering his eyes with one hand. "Man, I can't take this shit."
We all laughed, even Johnny. With all the joking I felt like I could bring up the safe again.
"So that safe isn't here, man?"
"Nah," Boogie said. He sensed something. "Why? What's up?"
"We need to get it back."
"What the fuck for?" Boogie said. He looked at me. We were eye to eye even though he was sitting down.
"The owner wants it back."
"Is that right, Danny?" Boogie asked.
"Well, you see," Danny mumbled. "The guy we took it from sorta figured it out. And—"
"Now we had a deal, Danny," Boogie said, standing up. "You promised me five grand for hiding this motherfucker and getting it open. Who's gonna pay me my money?"
Danny stuttered and crossed his arms over his chest. He tried to talk.
"No, no, Boogie. See, I told him that we were gonna have to talk to you and work something out."
"Because I'm not going back to jail," Boogie said.
Finally, I thought, someone in this group with a brain. It felt like things were going my way. Thanks, Boogie.
"Exactly. Nobody wants to go to jail," I said. "The best thing is to just give it back and forget the whole mess."
"No." Boogie turned on me. "That's my Moog money."
"What the fuck is Moog money?" Alex asked.
"You know, a Moog. One of them fancy little keyboards. We need money to buy one."
"Yeah," said Johnny. "Stevie Wonder has one. So does Parliament. That shit costs over a grand. Plus, we need new drums."
My stomach flopped onto the dirty kitchen floor.
"The person that safe belongs to is not someone we really want to fuck with," I said.
"Some church motherfucker? That don't scare us," said Boogie.
"So you're gonna take on Backwoods Billy and the Holy Ghosts?" I asked.
Boogie and Johnny looked at each other with wide eyes. This was new information to them and the effect registered all over their faces.
"You took the safe from them motorcycle nuts?" Boogie said, stepping toward Danny. "You told me you got it from some church group."
"Well, I, uh..." Danny mumbled.
The way he shriveled reminded me of the time I was a kid and my mom caught me stealing a candy bar at Woolworth's and made me return it and apologize.
"That's sorta what I meant."
Boogie sat back down and rubbed his eyes. He wouldn't back down now.
"Fuck it. He don't know I'm involved." He pulled a snub-nosed pistol out of the back of his pants and slammed it on the table. "And if any of you motherfuckers tell him, it's your ass."
He jabbed a thick finger at me, Alex and Danny.
"We don't even know what's in the safe, Boogie," Alex said. I felt proud of him for finally saying something. "Might be nothing."
"There's got to be serious cash in there. That's a big safe. The kind where you put something you don't want no motherfucker getting."
"How much do we have to pay you to get the safe back?" I asked.
Boogie stared at the floor. The room went quiet. After a few seconds he groaned loudly. "Fine. Tell ya what. You bring me two grand and I'll give it back to you untouched. Otherwise, I'm gonna drill that motherfucker."
Johnny's head nodded up and down behind me.
"We'll give you a thousand," I said.
"Two."
"You're crazy."
"Well, that's the gamble you're gonna have to take, white boy," Boogie said. "You can pay me two thousand for the safe or you can go back to Backwoods Billy and tell that fucking psycho you can't return what you stole from him, hand him a thousand bucks and pray to fucking God that's more cash than he had in there and he's willing to let this shit go instead of running over your head with a motorcycle."
I looked at Alex to see what he thought and he shook his head with a disgusted look on his face. Danny shrugged.
"All right," I said to Boogie. "Just don't touch that safe."
"No problem." He grinned.
No one spoke as we drove back to town. I looked over at Danny as we crossed Hanover Street Bridge. He created this mess but I was the one who Backwoods Billy was going to kill. He sat slouched in the seat next to me, shirtsleeves rolled up, arm dangling out the window. I wanted to open his door and boot him, send him sailing into the Patapsco River.
Alex was quiet in the backseat. We were all scheming of ways to come up with two thousand dollars. There was no legal way to get that much money as quickly as we needed it. Not for guys like us. It would have to be something bad, something on a bigger scale than any of us were used to pulling off. Robbing Zeppelin was sounding better and better. Danny turned to Alex.
"Do you still talk to that Angie girl you used to sneak around with?" he asked.
Angie was an ugly girl. Dumpy with big, weird-shaped teeth and tiny eyes set too far apart on her face. A few years earlier Angie had won ten thousand dollars on _Let's Make a Deal_. Alex hooked up with her, convinced her to buy him a bunch of shit and then bailed once he'd cleaned her out. She never made the connection between the money drying up and Alex leaving and she still loved him madly. I dreaded running into her with Alex.
"Naw. I haven't talked to her in a long time. Why?"
"Isn't she a bank teller?"
"Yeah," Alex said. "Downtown."
"We're not robbing a fucking bank," I said, shaking my head.
"Maybe she'd be into it? We could cut her in for the ten grand Alex took from the poor girl," Danny joked. Alex laughed. "No, seriously. Do you think she'd help us?"
"No way," Alex said. "She's a nerdy chick, man. She ain't helping us rob a bank."
"What if we rob one anyway, you know, without her help?"
"I'm not robbing a bank," I said. "That's serious shit. People get shot doing that."
"And this town is so fucking small everyone working there would know us," Alex added.
"Not if we got some really cool masks."
"Forget it!" Alex and I yelled at the same time.
The Doors came on the radio. "Roadhouse Blues." I always liked the Doors. They never got caught up in the hippie bullshit. They did their own thing. They didn't play Woodstock or Monterey or any of that crap. They drank whiskey and wore black leather and Jim Morrison yelled at the crowd and pulled out his dick. You either loved or hated the Doors. I loved them.
I thought about a few quick ways to make the money to pay Boogie but none of them really worked out. Alex and Danny rambled about robbing banks or stealing cars. I knew the Zeppelin heist was our best chance. We'd just have to hold off Backwoods Billy until Zeppelin hit New York City that weekend.
On the radio, Morrison sang, "The future's uncertain and the end is always near."
# THIRTEEN
#
ACROSS THE CARNIVAL GROUNDS, SWEATY AND SHIRTLESS AND LOADED OUT OF HIS MIND ON BLACK BEAUTIES AND BEER. THE BEER MADE KEITH CLUMSY BUT THE SPEED MADE HIM MOVE EVERYWHERE AS QUICKLY AS POSSIBLE. THE COMBINATION WAS HILARIOUS. I LEANED AGAINST MY CAR WITH ALEX AND FRENCHY AND WATCHED KEITH HURL TOWARD US WITH UTTER FUCKING ABANDON. HE MOVED AS FAST AS HE COULD IN A ZIGZAG. HIS ARMS FLAILED AS HE PLOWED THROUGH THE CROWD AND STAGGERED OVER THE UNEVEN GRASS.
We'd spent most of the day sitting at Keith's kitchen table drinking beer and plotting. We needed a tight plan if we still planned to rob Zeppelin. Every time one of us came up with a workable plan, someone else found a hole in it. By the time we came up with something we could all agree on, we were drunk.
We decided to hit the annual carnival on the Inner Harbor to celebrate. Every summer the city put together a carnival in honor of the city's goodwill and every year it turned into a drunken riot. The families cleared out by sunset when the festival turned into a circus of drugged-up kids and drunken criminals. There were fistfights and stabbings and wasted kids puking everywhere, more from the beer and drugs than the rides. It was the highlight of our summer.
First, we needed to hit the beer tent. The bartenders didn't take cash, only colored tickets sold at a table manned by an off-duty cop who checked IDs. We worked around this every year by stopping at the party supply store and buying rolls of tickets in every color and smuggling them into the tent.
We sent Keith to scope out which tickets they were using and he'd just returned. The speed made him talk in a jumble of words even before he got close enough for us to hear what he said.
"... then they told me I had to go I had to leave you know get out but I wasn't gonna until I saw the tickets but it was too dark to tell and a woman spilled her beer down my back but I didn't care so I went to the—"
"Keith!" Alex interrupted. "What color are the tickets?"
"Red. I think they're red. Hard to tell. I need some water or something my mouth is dry..."
Keith talked to Frenchy, who ignored him. Alex leaned into the trunk and rummaged through a shopping bag filled with rolls of tickets. When he found the red tickets we each pulled off a long strip and stuffed it into our pockets.
The beer tent barely covered the crowd under it and no one noticed when we ducked under the ropes in the back. We decided Alex looked the oldest and he shoved off through the crowd with a fistful of tickets to buy the first round of beers.
Sara showed up to hang out with Frenchy. They stood to the side talking. Now and then I caught her glaring at me, still mad about the incident with her purse. The speed kicked in hard and Keith babbled about everything from UFOs to his uncle's Mustang then back to UFOs and into a rant about how much he hated _Soul Train_. Alex returned with the beers.
"There are nothing but old women in here," he griped between sips of beer.
I couldn't see much over the wall of bodies but I knew he was right. The women in the beer tent looked like sad single mothers and forty-year-olds on a girls night out. Most of the girls our age hung out in the carnival, not the beer tent.
I turned to watch the neon-lit crowds moving around the dirt path that wound through the rides and games. Kids surrounded the bumper cars and a dinky roller coaster or tried to win prizes at any of the rigged games. Santana pumped through the speakers, mixing with the noise from the crowd. One laugh cut through it all.
"Emily!" I yelled toward her. She stood eating a foot-long corndog covered in ketchup.
"Oh my God!" She giggled as she walked over. "Great! And here I am eating a big-ass corndog."
"Who are you here with?"
"Tina and my friend Brandy."
"Tina's here?" Alex asked. "Where's she at?"
"Over there buying a funnel cake."
Alex drained the rest of the beer then ducked under the rope and walked off into the crowd. Emily licked ketchup off her fingers.
"I don't know if Tina wants to see him after everything that went down," she said.
"It'll be fine. No girl can stay mad at Alex."
"I don't know. He just got out of County! Tina is pretty tough."
"We'll see." I smiled.
The carnival crowd parted and Alex walked toward us laughing loudly, his arm around Tina. I never knew how he did it. Brandy walked alongside them. She was chubby but not quite fat. The bright yellow button-up shirt she wore fought to keep from bursting open around her tits, and her blond ponytail swung as she hurried to keep up with Alex and Tina.
"Come on," Alex told the girls. "I'll buy you a beer."
Soon we were all drunk and clustered in the corner of the tent talking. Tina tried to convince us that Elton John had talent, which caused me and Keith to groan. Alex claimed that white music died in the sixties and black music was the only thing worth buying. Frenchy kept bringing up the Kinks.
"Have you seen _Live and Let Die?"_ Emily asked everyone. "James Bond is so fucking sexy."
"I always hated those movies," I said.
"How can you hate James Bond?"
"It's corny."
"It's not corny!"
"Criminals aren't like that. They don't sit around plotting big scams."
"They're criminal masterminds! That's how their minds work."
"There are no criminal masterminds. Trust me. Criminals just make shit up as they go along."
"Just like you're doing now." She laughed, leaning in to kiss me.
I overheard Keith chatting up Brandy. The beer he had drunk mellowed out the speed a little but he still rambled nervously.
"If you told me you gave the worst blow job in the world, I would say prove it."
"Oh really?" Brandy laughed and rolled her eyes.
"That's all I'm saying," Keith said, shrugging. He wobbled a bit.
I nudged him in the ribs.
"Keith, man. What the fuck are you doing?"
"No. No. It's cool. We're just talking. It's cool."
Brandy looked at me and giggled. I couldn't believe she actually found Keith amusing.
Emily whispered something to me but I couldn't hear her. The noise on the other side of the tent suddenly grew louder. Voices shouted back and forth. The crowd surged toward us. Someone stumbled into me, knocking the beer out of my hand. Something was wrong. Me and Alex traded worried looks and he pushed off through the crowd. Seconds later he broke through the wall of bodies.
"We gotta get the fuck out of here," he said, wide-eyed. "Right now. Let's fucking go."
"What going on?" Emily asked.
"We gotta go."
Clusters of people dropped their beers and scurried away. They shoved past us, ducking under the rope and hurrying away from the beer tent. Frenchy slipped through the crowd with his arm around Sara.
"What the fuck is happening?" he asked Alex.
Spend enough time around violence and bullshit and you develop a sense for it. You can smell a fight before it happens, while it's still just two guys trading hard looks from across a room. Having had the crap kicked out of me more than once before, my sense for trouble felt damn sharp. Something was going down on the other side of the tent and I didn't want any part of it.
"What the fuck are you gonna do about it, pretty boy?" someone shouted.
"Let's go. Come on. Let's just leave," a woman's voice pleaded.
"Hit that motherfucker!" another voice yelled.
"Do it, man! Do it!"
A fat man with a beard waddled past me shaking his head.
"It's those damn motorcycle riders," he said. "They're drunk and looking for trouble again."
I dropped my beer and bolted from the tent, pulling Emily along behind me. All of us moved in a group through the crowd and around the merry-go-round. No one wanted to look back. Terror crept up my spine and I finally glanced behind us. On the other side of the crowd a long, tattooed finger pointed right at me. The skull ring glistened in the neon lights.
"Shit! Shit!" I hissed under my breath. "He saw me."
Emily couldn't take any more.
"Tell me what is going on. Right now."
"You know the Holy Ghosts motorcycle gang?"
"Yeah."
"They don't like me very much."
"Patrick! What did you do? Shit. Goddamn it."
Backwoods Billy, Rabbit and a gang of dingy Holy Ghosts barged through the crowd and closed in on us. We ducked behind the concession stand then snaked between two tents and came out the other side behind the Tilt-a-Whirl. I led the way, hooking around a few carnival games and the fun house. Keith stumbled over a mess of cables and nearly knocked Alex to the ground. Frenchy looked worried.
We hurried through a path around the bumper cars and I felt like we had lost Backwoods Billy. I finally looked back. A pair of Holy Ghosts in leather jackets and bandannas walked through the crowd but they didn't seem to be after us. I turned back around just as Backwoods Billy and Rabbit rounded the corner up ahead and moved toward us. We were surrounded.
"Come on. We'll ride the Ferris wheel," Alex told everyone. "They won't see us."
"No fucking way," I said. "I'm not going on that thing."
"He's scared of heights," Frenchy told Sara.
"What a sissy," she sneered.
"Come on. We're getting on," Emily ordered me.
She pulled me up the metal steps to the bright green Ferris wheel. We cut the line of kids and began piling into cars. Me and Emily were last and I caught a glimpse of Backwoods Billy passing in the crowd just as a carny with thick arms locked the safety bar across us. The wheel jerked into motion and we rose above the crowd. I closed my eyes.
"Tell me what's going on. Can you see them?" I asked Emily.
"I think we're okay. It looks like they're leaving."
I exhaled the breath I had been holding since the beer tent and lay my head on the back of the metal seat. She leaned into me and squeezed my hand. I really fucking hated heights. The night air felt cooler as we climbed and I clenched my eyes tighter. Someone above us screamed and I heard metal rattling.
"What's that sound?" I panicked.
"Just Keith being an asshole. He's rocking the car and making Brandy scream."
Emily leaned forward, causing our car to tip.
"Hey! Hey! Cut it out!" I yelled.
"Sorry! Just trying to see where they went. I think they're gone."
The sound of the carnival quieted as we rounded the peak of the Ferris wheel. We were high enough to see the lights of the city. The skyline probably looked beautiful but I couldn't open my eyes.
We started our climb back toward the top and Emily shifted in her seat nervously. She leaned forward, tipping the car again, and I held my breath and clung to the metal safety bar.
"Oh my God," she whispered.
"What? What is it?"
"I think we're in trouble," she said.
"Why? What's happening?"
"Take a look."
I opened my eyes slowly. The entire city spread out in the distance, dotted with glowing orange lights. It did look beautiful. I tipped my head and glanced down. A sea of denim and leather and greasy hair surrounded the Ferris wheel. A few of the Holy Ghosts shoved people passing through the crowd. Someone threw a wild punch. The rest of the faces all pointed up toward us. One face with a long red beard and a tattooed neck stared straight at me.
Backwoods Billy smiled and waved.
The Ferris wheel slowed down and stopped. At the bottom, the operator opened the gate to let Alex and Tina out. Keith and Brandy were next. Then Frenchy and Sara. Me and Emily came around last. I stared at the floor of the car. My stomach flip-flopped.
"You feelin' all right, boy?" the carny asked, laying a large black hand on my shoulder as he helped me out of the Ferris wheel car. "You ain't gonna puke, are you?"
I shook my head and he stepped aside to let us out.
Backwoods Billy waited at the bottom of the stairs. He threw his arm around my shoulders as I came down the steps. His beard grated against the side of my face.
"'They that hate me without a cause are more than the hairs of my head,'" he hissed, spitting whiskey breath into my face as he talked. I pulled away from him and he yanked on my hair.
"Having a good fucking time, boy?"
He tugged harder on my hair, pulling my ear down to his shoulder.
"Where's my safe, you little asshole?"
Behind him, a biker with black hair and pockmarked skin leered at Tina.
"Come on, girl. I'll win you one of them pink elephants. Then we can go behind one of them tents out back."
He lunged at her and she darted behind Alex.
"Leave me alone, you fucking dirtbag," she shrieked.
I stared at Alex. He gave me a look that said he didn't know what to do either. Billy caught me looking.
"These friends of yours?" he asked, jerking a thumb over his shoulder. "You better tell me where that safe is before my boys tear 'em apart."
He poked a tattooed hand into my chest and twisted my hair as he spoke.
"'Destruction cometh; and they shall seek peace, and there shall be none.'"
"Is that another quote from the Good Book?"
"You better fucking believe it." He grinned, showing off a gold-capped tooth.
The carny operating the Ferris wheel yelled from the top of the steps, "Hey! What's going on down there?"
"Everything's fine, bud." Billy grinned up at him.
"Well, move along," the carny said. "You can't stand there."
Billy ignored him and turned back to me.
"Where's the safe, kid?"
"I don't have it. I swear I don't have it."
"Who has it?"
"A guy we know. He won't give it back until I pay him."
He looked genuinely shocked.
"Some asshole is holding my fucking safe ransom? _My fucking safe!"_
He screamed into my face. I prayed someone in the crowd would step in but no one did. Not a single head turned in the passing crowd or the Ferris wheel line. No one wanted to get involved. Most of the crowd spotted the pack of sweaty greasers in Holy Ghosts vests and then walked the other way.
"Listen, boy. You're gonna show me where this motherfucker lives and we're gonna get that safe if I have to cut his fucking throat."
He slid a finger across my neck.
"I'll get it. I swear. I just need a little time."
Behind us, a pair of scraggly Holy Ghosts terrorized Alex and Keith. They looked like lesser members of the gang. Younger guys trying to make a name. A chubby biker with a bald head joined his buddy with the bad skin.
"Let us borrow your dates, fellas. We'll have 'em back by morning, I swear. They'll have a real good time."
"Yeah. We'll be on our best behavior," the bald guy said, holding a leather-gloved hand over his heart.
The carny reappeared at the top of the stairs.
"I told you once already," he said, pointing at Backwoods Billy. "Move the fuck along."
"Mind your own fucking business," Billy snarled.
"This _is_ my business, you honky asshole. Now get the fuck out of here."
"Shut your fucking mouth, carny," Billy snapped.
The carny moved fast for a guy with the build of someone who sits at a Ferris wheel all day.
"Get your ass out of here!" he shouted as his steel-toed boots thundered down the metal steps.
A long arm in a black leather jacket shot up behind him and swung down. The bottle exploded over the carny's head. He bent at the waist and covered his head with his hands.
Then all hell broke loose.
Carnival workers appeared from everywhere. They leapt over counters and charged out of trailers. Every hand held a weapon. Baseball bats. Mop handles. Screwdrivers. The crowd swirled as carnies and Holy Ghosts squared off. As hard as the carnies beat them, the Holy Ghosts kept getting back up.
Alex only needed one chance. He latched on to two fistfuls of hair hanging in the biker's pockmarked face, pulled down and shot his knee up. The biker's nose crunched. Alex swung him by the hair, twisting him around until Keith kicked the biker's legs out, knocking him to his knees. The skinny black kid who operated the Tilt-a-Whirl stepped forward and swung a tennis racket like he was returning a rocket of a volley. The strings raked across the biker's face and the wooden frame shattered over his skull.
A Holy Ghost stomped a thick boot into the face of a limp body on the ground. A carny split a broom handle over his head but the biker barely flinched. He spun, grabbed the carny by the hair and punched him in the throat, sending him gasping to the ground.
Backwoods Billy never let go of me. He bulldozed backward, using me as a battering ram to break through the chaos. The first punch slammed into my eye. He jerked my arm toward him as he threw it, doubling the impact. His fist seemed to fill my entire eye socket. A Holy Ghost slipped behind me and bear-hugged me, pinning my arms at my sides. He lifted me off the ground.
"You're coming with us," he grunted.
I kicked out at both of them but without a foot on the ground for leverage, the blows bounced off. I buried my chin into my chest and wagged my head from side to side as fists hammered into the top of my head. One lucky shot slipped low on my forehead, tearing open the skin. Blood poured into my eyes. An uppercut crammed my lower lip into my teeth. I tasted blood.
The Holy Ghost behind me lifted my body off the ground, turned it sideways and slammed me onto the ground. I should have stayed down but didn't. A foot crashed into my lower back as I tried to stand. My body flattened into the gravel. Kicks and punches landed from everywhere. I didn't even think about getting up. I curled in a ball and covered my head. Hands pried my arms away from my head to make holes for fists to slip through. My mouth filled with gravel and dust.
"All right! All right!" Backwoods Billy yelled. He stepped forward and waved away the circle of jackals beating on me.
"Lemme talk to him," he said.
Through the eye that wasn't swollen closed I saw Backwoods Billy bend and lift me up by my shirt collar. I stumbled and crashed onto my back. His grip slipped and I scrambled backward up the steps.
The Ferris wheel worker rose behind him. Blood and broken glass streamed down his head. He fumbled in his pocket then swung a heavy metal wrench high over his head. I had every chance to say something. Every chance to warn Backwoods Billy before the wrench split his skull.
I didn't say a fucking thing.
The wrench ricocheted off the back of Backwoods Billy's head. Blood streamed down his face. He howled with pain and flung both arms over his head. The carny pried loose one arm and dragged Backwoods Billy backward into the crowd by his wrist. His motorcycle boots kicked in the dust as he tried to stand. A carny in a white apron leapt forward and cracked a pool cue across Backwoods Billy's ribs.
_"I'll find you motherfucker! I want my fucking safe!"_ he screamed at me.
I stumbled forward, dodging swinging fists and sidestepping someone lying facedown in a pool of blood. Alex and Frenchy stood in a clearing on the other side of the swirling fighters. Behind them, a wall of police officers prepared to swarm into the brawl.
"Where are the girls?"
"They bailed the second this shit started," Alex told me.
"Where's Keith?" I said, scanning the area for him. Alex's head spun around.
"Shit," he hissed. "He must have run too."
Police radios crackled around us. Suddenly the Baltimore Police raised their batons and streamed into the crowd.
"Let's go," Frenchy shouted.
He lifted a red tent skirt and we scurried underneath. Boxes of toy prizes and giant stuffed bears cluttered the dark tent. We plowed through the tent flaps on the other side and sprinted across the fairgrounds toward my car, lost in the orange lights in the distance.
# FOURTEEN
#
PISS.
"That's okay, honey," the chubby nurse said to me from the other side of the bathroom door in my hospital room. "You just need to relax. We'll try again later."
Frenchy insisted on bringing me to the hospital. The doctor figured the beating left me with a minor concussion, maybe a few cracked ribs. Otherwise I was all right. He made me stay the night and told me if they didn't find any blood in my piss I might be able to go home early. Only I couldn't piss.
My folks came by a few hours later. I told them I'd been beaten up and robbed for my watch. Not a total lie. Mom rushed into the room and hugged me, which made my vision go blurry. My dad stood at the end of the bed and told me I shouldn't have been downtown anyway. Then he babbled about the hospital's outdated fire alarm system. When I'd heard enough, I told them to leave so I could get some sleep. In the morning the nurse entered my room.
"You have a visitor."
Emily poked her head around the corner.
"How are ya feeling, champ?"
"I'm all right, Emily. How are you?"
She pulled a chair over to the edge of the bed and sat down, crossing her long legs and kicking one sandaled foot back and forth. Her cutoff jeans and white tank top clung to her.
"How did you know I was here?" I asked.
"Tina told me."
Alex must have told her. Emily sighed loudly.
"You heard about Keith?" she asked.
"What about him?"
"He's in jail."
I sat up in bed. A bolt of pain shot through my ribs.
"What for?"
"I don't know," Emily said. "They arrested him and a bunch of those motorcycle nuts."
"Shit," I groaned.
I pictured Keith in a tiny holding cell, surrounded by Holy Ghosts. I had to get in touch with Alex and find a way to bail Keith out.
"Patrick!" Emily shouted. "Are you listening to me?"
"Yeah. Sorry. I'm listening."
"So?"
"So what?"
"So what happened? What was all that about?"
"Just a misunderstanding," I lied. "Something to do with Alex's uncle Danny."
She didn't believe me but let it go.
"When are you getting out of here?" she asked.
"As soon as he gives me a pee-pee sample," the nurse said, walking into the room holding up a clear plastic cup. I sunk into the bed. Emily giggled.
"Aww. I'll let you... uh... take care of things. But we should hang out when you get out of here. I'm worried about you."
When she was gone I padded barefoot into the bathroom.
"Maybe that visit from your girlfriend relaxed you," the nurse said from outside the bathroom door. I stood in front of the toilet with my hospital gown open and thought about getting out of there, settling things with Backwoods Billy and spending time with Emily.
The nurse was right. I pissed long and hard until I was nearly out of breath.
"You have another visitor, honey. I'll come back."
I snapped the lid on the cup of piss. When I opened the door a man in a suit stood in the middle of the room. He looked at me and my cup of piss.
"Hey, Patrick. It's Patrick, right? How are you feeling?"
"I'm okay."
I left the piss on the counter and climbed back into bed.
"You a doctor or something?" I asked.
"You don't know who I am?"
He didn't give me a chance to answer.
"Of course you don't. You're just a fucking teenager," he said, talking to himself more than me. "I'm Simon Cooper. District Attorney."
I stiffened in bed. He started to shake my hand but his eyes caught the cup of piss on the counter. He pulled his hand back.
"You know what a district attorney is?"
"Yeah."
"Really?" He smiled. "Did I put one of your friends in jail or something? Maybe your father? You don't hold some grudge against me, do you? Not gonna try to stab me?"
"Depends. What do you want?"
He laughed and scratched his head. His messy hair stuck up in dark tufts above glasses that looked too tight for his box-shaped head. The gray flecks at his temples matched his wrinkled shirt.
"So, what happened last night down at the harbor?"
"Some guys tried to rob me. No big deal."
"Really? Did the police catch them?"
"Nah. Could have been anyone."
"Wasn't any of the Holy Ghosts, was it?" he smirked.
I shrugged. He paced the room and stared at the floor.
"So some guys beat you up and tried to rob you right in the middle of the carnival? Worked you over pretty bad, right? I mean, here you are."
He chuckled. I didn't like where this was going.
"What do you want?"
"Listen," he said. "I'm not after you. Not at all. It's just that a cop friend working the carnival last night thought maybe he overheard you say something to my friend, our friend Backwoods Billy. Something that could really help me out and then maybe help you out in the future too."
He stood at the foot of the bed waiting for me to say something. I didn't say anything.
"Something about a safe, maybe? Maybe I'm wrong. Am I wrong? Is that not what you said?"
I didn't say a word. His fingers tapped nervously on the edge of the bed.
"I understand, Patrick. I really do. But see, there's a chance, just a small chance, really, that there's something in that safe that is, uh, very valuable to me. More valuable than you might even realize. Do you know what I'm talking about?"
I didn't. We hadn't opened the safe. Now I wanted to know what was in there.
"Can I smoke in here?" he asked, holding a cigarette in his fingers. He leaned to look out the door into the hallway. "No. No. I probably can't."
His fidgeting grew worse and his fingers fumbled around in his pants pockets. He started to sweat.
"Am I under arrest or something?" I said.
"No! No." He stuck out his arms and gave me a face like I had to be kidding. Like we were old friends and how dare I suggest such a thing. "Not at all. Nothing like that. I'm just trying to track something down and thought maybe you had it."
"What is it?"
"Well, see," he smiled, "I can't really say too much. Nothing against you. It's just something between me and Billy that I've been trying to get back from him. This would just make it easier, you know, to get it from you."
I nodded.
"Kind of hard to tell you if I have it or not if I don't know what it is we're talking about."
"Just look around," he said. "I'll leave my card for you. If you come across it, give me a call. No questions asked."
He placed his card on the counter next to a water pitcher and some medicine the doctor left me. He lifted a bottle of pills.
"Are these Percocet? Wow. Ever taken them? These are... wow. You'll have a good time with these."
He shook the bottle at me and smiled.
"He really gave you a lot. Mind if I take a few?"
Outside in the hallway I heard the nurse talking loudly.
"Boys. Boys! You can't go in there. Stop right there!"
Alex and Frenchy charged into the room, doubled over laughing. The front of Alex's T-shirt said, "Delcon Industrial Strength Weed Killer." He froze when he saw Simon and stared at him suspiciously.
"Who the fuck is this guy?"
"Just a friend, gentlemen. He's all yours. I was just leaving," Cooper said, moving toward the door. He stopped in front of Alex. "Actually, are you guys in the gardening or landscaping business? I noticed your shirt and thought..."
Alex could sniff out authority and he hated it. He stared at Cooper like he had just asked if Alex ate shit sandwiches.
"No," Alex snarled, lighting up a cigarette. "It says 'weed' on it. It's funny."
"Oh, yes. Of course. Really funny. Okay, guys. Um, take it easy."
Alex pointed at Cooper.
"Seriously, Patrick. Who the fuck is this guy?"
After Cooper left, I told Alex and Frenchy about how he asked about the safe.
"So, what's in it?" I wondered out loud.
"I think there's a ton of cash in there," Alex said.
"Has gotta be," I agreed.
"Or maybe Backwoods Billy stole something from this Cooper guy and it's in the safe and he hasn't been able to get it back," Frenchy said.
"Nah. He's a DA. If he thought Billy stole something from him he'd just have Billy arrested."
"Whatever's in that safe, they both want it," I said. "They both want it really fucking bad."
This changed everything. Alex picked up the phone.
"Hey, Boogie. It's Alex. New plan. You know that safe? Open that fucker up."
Alex covered the phone with one hand and looked at me.
"He says Danny already called and told him to open it. They've almost got it."
Frenchy sighed and sat down on the counter, knocking my bottle of pills to the ground. He bent to pick it up.
"Holy shit! Percocet? Mind if I take a few?"
# FIFTEEN
#
RANG THE NEXT DAY. I CRINGED IN PAIN AS I HOBBLED TOWARD THE KITCHEN TO ANSWER IT. MY RIBS STILL HURT. I'D BEEN LET OUT OF THE HOSPITAL EARLIER WITH MY PERCOCET AND A WARNING TO TAKE IT EASY. I KNEW IT WAS ALEX BEFORE I ANSWERED.
"Boogie called."
"He got the safe open?" I asked.
"Just about. He wants us to come by."
"Did he say what was in it? Could he see?"
"Wouldn't tell me. Just said that he wanted to meet at his shop in an hour."
"He wouldn't tell you?"
"Nope. He kept laughing and told me that we needed to come see for ourselves."
He paused.
"Think there's enough in there to bail Keith out?"
"Let's hope. Where's the shop?"
Boogie's shop took up the back room of a tire garage on the east side of town. He stood in front of the open garage door as me and Alex pulled up at the address. His Funkadelic T-shirt hung loosely and he tugged at his blossoming Afro with a black pick. A huge grin spread across his face and he sang loudly as me and Alex walked up: "Spread your tiny wings and fly away, and take the snow back with you where it came from on that day."
"What the hell are you singing?" Alex asked.
Boogie smiled and motioned for us to follow him through the garage. Stacks of tires covered the concrete floor. A muscular black guy pounded on a steel rim at a tool bench. In the back, an air gun pumped loudly. Three soda machines, probably left over from Boogie's business, sat along the wall.
At a closed office door in the back corner Boogie turned to face me and Alex. He flung open the office door and sang loudly as he ducked through the doorway: "The one I love is forever untrue and if I could you know I'd fly away with you."
Me and Alex both laughed. I didn't know what the hell Boogie was singing but he seemed to be in a good mood. I took that as a good sign. Maybe things were going to work out. I limped through the store behind them trying not to fall over anything.
Johnny Paycheck and a few other guys sat in mismatched chairs and played dominoes on an old card table. R&B blasted from a tiny radio. A double-barrel shotgun leaned against the wall and a pistol lay on the counter. Paycheck looked up when he heard Boogie singing.
"Was that you singing?" Paycheck asked.
Boogie nodded.
"You know, that's just fucked up." Paycheck laughed. "You're one sick motherfucker."
Boogie's deep laugh rumbled from his chest.
A tall gray safe sat in the corner. It was enormous, and for a second I was amazed that Danny and Alex had even been able to carry it. The thick steel looked impenetrable. An older black man with thick glasses and a bald head leaned over the safe. He stared into a small scope stuck in a hole drilled in the door. He slowly turned the combination dial with one hand.
"Is that the safe?" I asked.
"You mean you never saw this shit?" Boogie grinned. "They tried to kill you for stealing this and you never even saw it?"
Everyone in the room laughed. Everyone but me and Alex.
"Goddamn, son." Paycheck grinned.
"Danny really fucked him over, right?" Boogie asked with a smile.
"Hell yeah," Paycheck said, grinning.
"Any idea what's inside?" Alex asked, crossing his arms over his chest.
"Oh, we got an idea." Boogie grinned. "We took a look around with the scope."
"Just about got it," the old guy said.
The dial clicked loudly and the old guy stepped out of the way. Boogie stood in front of the safe. He slammed down on the handle and flung open the heavy door with a magician's flourish.
"Ta-da!" he yelled.
Me and Alex both stepped forward. Neither one of us could believe it. The cold steel walls, the wide top shelf, all of it sat empty except for the bottom of the safe, which held two dusty reel-to-reel tapes: Jim Nabors's _Galveston_ and Anne Murray's _Snowbird_.
Boogie started to sing. Johnny joined him.
"So, little snowbird, take me with you when you go to that land of gentle breezes where the peaceful waters flow."
They howled with laughter and Johnny pounded on the table with his fist.
"Jim fucking Nabors." Paycheck laughed hysterically. "That motherfucker played Gomer Pyle. Why you gonna buy an album by Gomer fucking Pyle?"
"Man, that's cold," one of the domino players said, shaking his head.
"We're so fucked," I mumbled to Alex. "Jesus Christ. He's gonna kill me."
"Nah," the safecracker said, trying to calm me down. "Just give the safe back to that biker asshole and walk the fuck away."
"How are we supposed to do that?" I pointed at the drilled hole in the front. "What about this? It's ruined now."
"Buy another safe just like this one."
"The combination would be different. I think he'd notice that."
"I can switch the locks," the old guy said, lighting a cigarette. "No problem."
Boogie picked at his hair.
"All right." I sighed loudly. "I guess we don't have a choice. Do that and we'll be back on Sunday."
"Don't forget," Boogie said. "Two grand. Plus the cost of the new safe. Otherwise I'll sink this motherfucker in the river."
"And if the Holy Ghosts sink me in the fucking river first," I huffed as I walked away, "I'll wait for you at the bottom."
Alex walked alongside me with his head down as we left the tire shop.
"You all right?" he asked.
"Yeah," I answered. "Why?"
"You seem pretty freaked out. I've never seen you like this."
"This was supposed to be so easy. Follow Zeppelin to the hotel. Take the money."
"Shit definitely went haywire," he said. He stopped on the sidewalk to light a cigarette.
"'Snowbird' by Anne Murray," I said.
"Is that what he was singing?"
Behind us, Boogie sang from the office.
"When I was young my heart was young then too, and anything that it would tell me, that's the thing that I would do."
"What a fucking asshole," Alex sneered.
# SIXTEEN
#
BETTER IN THE BALTIMORE COUNTY JAIL THAN HE EVER DID ON THE OUTSIDE. HE WALKED INTO THE VISITORS' ROOM LOOKING RESTED, FED AND BATHED FOR THE FIRST TIME IN YEARS. NO FOOD STAINS ON HIS ORANGE JUMPSUIT. HE LEANED FORWARD, PLACED HIS HANDCUFFED WRISTS ON THE METAL TABLE AND TOLD ME AND ALEX HOW MUCH HE LOVED JAIL.
"Guess what I watched today, Alex? Remember that episode _of Hawaii Five-O_ where the hippie kid asks that old dude for money and the old dude tries touching the kid's dick and the hippie kid punches him in the head and steals his wallet?"
"That's a fucking great one," Alex said. "It turns out the old guy had two hundred and fifty thousand dollars in a storage locker."
"Yeah. And that hippie chick ODs and they play all that weird trippy music."
"And that bartender gives them those licorice-flavored rolling papers?"
Keith sighed happily.
"Man. Licorice-flavored rolling papers. Can you imagine?"
"That's a good one."
"A classic, man. A classic. _Hawaii Five-O_ is the best TV show ever."
The visitors' room at the county jail looked just like it did a few years ago when Alex made me come with him to visit Danny. The ocean blue concrete walls still held the corkboard covered with posters announcing the rules, such as NO TOUCHING and VISITORS UNDER THE INFLUENCE OF ALCOHOL OR DRUGS WILL BE EXPELLED. Guards stood against one wall observing the metal tables where prisoners sat with girlfriends and family. Two doors faced off on opposite walls. One door led back to the prison, the other door to the outside. I'd be leaving through one. Keith would not.
"So they're treating you all right in here?" I asked.
"Pretty much. It ain't bad, really. I watch TV all day and smoke. Hell, I'd be doing the same thing at home anyway."
A guard carrying a shotgun yawned as he walked past us. When he returned to the corner, Alex leaned forward.
"So what are they charging you with, anyway?" he asked.
"They think I'm one of the Holy Ghosts," Keith said with wide eyes.
Me and Alex found that hysterical: poor, dumb Keith a member of the most bloodthirsty pack of psychos in town.
"Fucking cops," Keith said softly. "That fight got crazy. Carnies and Holy Ghosts clubbing each other. Then the cops ran in and started busting up heads. I tried running but a cop coming the other way grabbed me."
Keith stubbed out his cigarette.
"They stuffed me in a paddy wagon. It was filled with dudes. Holy Ghosts on one side, carnies on the other. They were spitting on each other and going nuts. Kicking each other across the aisle. This big-ass carny across from me kept staring at me. We turned a corner and he came at me. Landed on top of me and head-butted me in the eye."
He showed us the bruise on his temple.
"How much is your bail?" I asked.
"I don't know. They don't even know what they're charging me with. I heard something about felony assault or felony riot. One of the Holy Ghosts said something about us getting five years. Shit. Hopefully they'll figure out I ain't no Holy Ghost and let me out of here. I tell you one thing I am scared of, though."
He leaned across the table. His face grew tight with fear. He waited to talk until the guard walked to the other side of the room.
"They're gonna cut my hair, aren't they?"
"Why would they do that?"
"That's what they do! They give everybody a crew cut."
"That's the army, Keith."
"Naw. Happens in prison too."
"No, it doesn't," Alex said. "Look around the room. All these dudes have hair."
Alex kept talking but the words were lost in shouts from across the room. A tall inmate in a jumpsuit pounded his handcuffed fists on the metal table. Veins pulsated across his bald head. He kept screaming.
_"You're a fucking dead man!"_
He rocketed to his feet. The metal chair under him shot backward, slamming into the wall before clanging to the ground. Three guards struggled to restrain him. Every muscle in his body rippled with anger.
_"You're dead! You're fucking dead!"_
I didn't want to be whoever this guy was threatening. My eyes met his. Shit. I _was_ the guy he was threatening.
_"You don't know who you're fucking with!"_
My stomach twisted in knots.
"Oh yeah," Keith said, calmly turning his head. "All the Holy Ghosts want to kill you. I heard Backwoods Billy's offering a reward for anybody who brings in you and that safe."
"How much is the reward?" Alex asked with a smile.
"Hopefully enough to bail me out," Keith joked.
I swallowed hard.
"By the way, Boogie opened the safe," Alex whispered to Keith. "Know what was in it?"
Keith shook his head.
"Two reel-to-reel tapes."
Keith shook his head.
"Where's the guitar?" he asked.
"Under the couch in Frenchy's basement," I answered.
"You guys still going to New York? You know, for that thing?" he asked.
"It's our only hope." I shrugged. "That money could get you out of here and buy that safe back from Boogie."
"Did you come up with a new plan yet? I don't think I'm gonna make it."
He held up his handcuffed wrists.
This was a serious problem. I hadn't even thought about it. We'd gone over every possible angle for this thing and devised an airtight plan. It needed to be perfectly timed using four people, not three. My head hurt and I rubbed my temples. Alex hadn't said a word but I knew what he was thinking.
"Shit. I guess we need to find a fourth person to make this work."
A guard stepped up behind Keith and kicked the back of his chair.
"Visiting hours are over, asshole."
We promised Keith we'd get him out of there as soon as we could. He wished us luck and told us to look for licorice-flavored rolling papers in New York City.
As me and Alex pulled out of the parking lot I turned left and headed toward the one place I already knew Alex was going to suggest. He finally spoke up.
"Danny's?"
"Yeah." I sighed. "Damn it."
# SEVENTEEN
#
LOOK AT ME. NOT EVEN AFTER ME AND ALEX EXPLAINED THE PLAN AND HOW MUCH MONEY WE COULD MAKE IF WE PULLED IT OFF. HE STRETCHED OUT IN A LA-Z-BOY CHAIR IN HIS MOM'S BASEMENT AND STARED AT AN OLD WOODEN TV FLICKERING IN THE DARK WHILE WE TALKED. NOW AND THEN HE DUG HIS HAND INTO A BAG OF PIGGLY WIGGLY CHEESE PUFFS AND SHOVED THEM IN HIS MOUTH. POWDERED CHEESE CLUNG TO HIS BUSHY HANDLEBAR MUSTACHE AND SPRINKLED ON HIS BARE CHEST. HIS BEER BELLY DROOPED OVER THE WAISTBAND OF EMERALD GREEN SWEATPANTS.
"What do you think?" Alex asked.
Danny stubbed out his cigarette in an overflowing ashtray.
"Well, I don't know, boys. I'm not even supposed to leave the state since I'm on probation."
"Me neither." Alex shrugged. "We'll be there and back the same day. No one will even know we're gone."
"And I'll tell you what else is bugging me. Why Zeppelin? Shit, man. They're one of the best bands around."
He started singing. His head shook and he strummed an invisible guitar.
"'It's been a long time since I rock 'n' rolled. Dun duh. Dun duh dun duh.' I mean, damn, that's a great fucking tune, man."
Alex stopped me before I could talk.
"This has nothing to do with them. We just grab the money, pay off Boogie, get Keith out of jail and divide up the rest."
Danny thought for a few seconds, sipped from his beer and then sat back in the recliner.
"Nah. I think I'll sit this one out." Danny shrugged. "Good luck, Alex."
Alex shot me a panicked look. He didn't know what to say. I did.
"So, Danny, did Alex tell you that Boogie got the safe open?"
That got his attention. He lurched forward in the chair and tossed the cheese puffs on a TV tray cluttered with empty beer cans. He wiped his hands on his sweatpants, leaving a streak of orange cheese.
"Well, shit. No. No, he did not. When did this happen?"
"This morning. Want to know what was in it?"
"Hell yes, I do."
"Two old reel-to-reel tapes. That's it."
"You're shitting me!"
"Anne Murray and Jim Nabors."
He looked at Alex in disbelief. Alex nodded. Danny groaned then his head drooped. He rubbed his eyes then stared up at the ceiling. He was counting on that safe to be his big score and it wasn't. This was a broken man. He let out a long, agonized sigh, lit a cigarette and leaned forward.
"So you'll help us?" I asked.
"No. I still ain't gonna do it. I can't risk it. Some of us got responsibilities, Patrick. I can't be getting arrested. No sir. I'm getting my shit together."
Alex's Grandma Alice came down the stairs carrying a basket of dirty laundry.
"Hey, Grandma," Alex said.
"Hi, boys," she said.
She stopped and stared at Danny.
"Are those my sweatpants?" she yelled. "Goddamn it, Daniel! How many times have I told you not to wear my damn sweatpants! Look at 'em. You got cheese puffs all over 'em."
"Grandma! We're talking here. Go upstairs."
"You need to talk about getting a job, Daniel. This shit has gone on long enough. You need to get your ass out of this basement and get to work."
"I know! I'm working on it. Now give us a minute."
Grandma Alice trudged up the stairs muttering. When the basement door closed, Danny stood up and brushed the cheese puff crumbs off his chest.
"Well, boys," he said. "I'll tell ya what. I'm gonna help you out, but just this once."
# EIGHTEEN
#
STOP STARING AT THE LIP-GLOSSED MOUTH ON THE GIRL ACROSS FROM ME. SHE SAT SLUMPED OVER IN A CHAIR IN THE LOBBY OF THE DRAKE HOTEL WITH HER FACE BURIED UNDER ENORMOUS SUNGLASSES AND A FEATHER BOA. A WHITE MINISKIRT CREPT UP HER THIGHS. SHE SLEPT WITH HER HEAD TILTED STRAIGHT BACK, SNORING VIOLENTLY, WITH BRIGHT RED LIP GLOSS SMEARED AROUND HER MOUTH.
Frenchy was the first to say what we were all thinking.
"Keith would have loved this."
"He would have put something in that chick's mouth," I said.
"Probably his balls." Alex grinned.
A midafternoon party raged around the couch where I sat sandwiched between Frenchy and Alex. Drunken people filled the chairs and couches and crowded near the elevators waiting for any sign of Zeppelin. Most of the partiers were women who had spent all night, maybe even all weekend, trying to meet the band. Everyone seemed wasted or at least running on the fumes from last night's partying. The mood was ugly but the groupies weren't leaving. It was Sunday, Zeppelin's last gig at Madison Square Garden before the band headed back to London. It was the last chance for fans to party with Zeppelin and our last chance to snatch the money. No bank was open on Sunday so we had all day to find the money from last night's gig.
Alex looked around the lobby.
"I can't believe the women here."
His head swiveled backward.
"Oh my God! Look at her. Should I go talk to her?"
"We're not here to meet women, Alex," I reminded him.
"Fine," he huffed, slouching back into the couch.
A chauffer walked quickly across the lobby to the front desk. He wore a short-brimmed chauffeur's hat over dark hair slicked back in a ponytail that hung down over his collar. In one hand he carried a large guitar case.
The hotel clerk behind the desk looked worn down from the drunken carnival in the lobby. Over the past few days he'd dealt with the most fucked-up and deranged people in the city and his face showed it. He snapped at the chauffeur. "Is there something I can help you with?"
"Yeah. You can tell me where I can find Richard Cole. Those Led Zeppelin boys left this guitar in my limo and I need to give it to them."
"Sir, we've been instructed not to disturb Mr. Cole. Would you like to leave the guitar here with me?"
The chauffeur struggled to lift the guitar case up to the counter. He slammed it down on the wooden countertop and sent a white phone crashing to the floor. The guitar case clasps opened with loud snaps and the chauffeur lifted the lid.
"This is a nineteen fifty-eight Gibson Les Paul. One of only seventeen hundred in existence. It belongs to Mr. Jimmy Page. You think I trust you or any of the mongrels in this lobby around this guitar?"
He pointed wildly around the room. The hotel clerk stared at the guitar while he fumbled to pick the phone up off the floor.
"Well, sir, I uh..." he muttered.
A barefoot girl in skintight jeans and a half-shirt walked around the counter behind him. A bottle of wine dangled loosely in one hand. She opened an office door, peered inside, then walked away, leaving it open.
"Hey, man," she slurred to the clerk. "Where's the bathroom?"
"Down the hall," he pointed.
"You see what I'm talking about?" the chauffeur said. "That girl wouldn't have thought twice about walking right off with this thing. Then Zeppelin would have your ass and mine."
The chauffeur leaned forward and talked low.
"Now I ain't accusing you of anything, bud. I can just tell that you're understaffed and overworked here. This is a goddamn circus. You can't be expected to handle all this alone. Let me know where I can find Richard Cole and I'll be out of your way."
The clerk exhaled loudly and nodded. He flipped through a ledger on the desk.
"Mr. Cole is in room twenty-one-ten on the top floor."
"I appreciate it," the chauffeur said. He dragged the guitar case off the counter and walked toward the elevator. He stared straight at me and winked as he passed.
"I can't believe Danny pulled that off," Frenchy said.
"Of course he did." Alex grinned. "He's a Carter. We're born bullshitters."
"Where'd he get the chauffeur outfit?" Frenchy asked.
"I borrowed it from Carmine," I answered.
"He looks good."
Frenchy grabbed the guitar case at his feet. He'd insisted on bringing his beaten-up Fender Telecaster just in case Jimmy Page wanted to buy that too.
We slipped into the elevator with Danny. I didn't look at him. His pissing and moaning on the drive up here nearly broke me. When he wasn't complaining about the traffic or my car or the music, he talked endless shit about himself. I already swore to myself that he was sitting in the backseat on the way home.
Danny dropped the case holding the '58 Les Paul with a heavy thud as soon as the elevator doors closed.
"Take it easy!" Frency hissed. "That guitar is really rare."
"Then you carry this goddamn thing, Frenchy. Fucking guitar weighs a ton."
He tossed the chauffeur's cap on the ground and loosened his tie.
"Fuck. How much longer do I gotta wear this shit?"
"Not long. Just keep it together."
"Don't tell me to hold it together, Patrick."
I looked at Alex and rolled my eyes.
"I don't see why I have to be the one in a fucking costume," he whined.
"Because you're the oldest," I said.
I clipped an ID card to his pocket and handed him a walkie-talkie.
"Now remember to clear the hallways. We don't want any of these hippies around. Do whatever you have to do to get them off the floor."
"With fucking pleasure." He grinned.
The elevator doors opened on the twenty-first floor. A wave of pot smoke snaked into the elevator. People filled the hallways drinking beer and smoking among empty bottles and trash. As we stepped off the elevator a pair of shirtless girls ran past us giggling. We stopped at room 2110.
"All right, Frenchy," I said. "You know what to do."
Me and Alex stood at the other end of the hallway and tried to blend in with the rest of the madness. Frenchy smoothed his fake mustache, straightened his sunglasses, then knocked. No one answered. Frenchy looked over and shrugged. I signaled for him to knock again and he beat on the door.
A guy with pork-chop sideburns and thick glasses stopped in front of Danny. His frizzy hair blocked our view of Frenchy.
"Wow, man. Do you work here?"
"Yeah," Danny said.
He stepped forward into the kid's face.
"I need you to exit this floor immediately."
"I don't have to go anywhere."
Danny leaned in closer.
"You can go down the elevator or through a window. What's it gonna be?"
"You can't fucking do that—"
Danny shoved him before he stopped talking. The guy reeled backward, tangled his legs with two men sitting on the floor sniffing something off a mirror and crashed onto the ground. Everyone started arguing.
"Clear the floor, assholes," Danny yelled.
Frenchy stood in the hallway talking to Richard Cole. He looked taller than I remembered. He leaned in the doorway, rubbing a thick beard. His white shirt hung unbuttoned to the middle of his chest. Everything about his body language said he didn't want Frenchy bothering him. He crossed his arms while Frenchy talked. We couldn't hear what Frenchy said but before he finished Richard waved him off and started to close the door. Frenchy quickly opened the guitar case.
Richard glanced at the Les Paul just before he closed the door. He looked impressed. Without taking his eyes off the guitar he signaled for Frenchy to stay put then shut his door.
"What the fuck is going on?" Alex asked me.
"I don't know." I was worried.
Frenchy looked over and shrugged. The door opened again and Richard stepped out. He pulled the door closed behind him and led Frenchy down to the end of the hallway. They stopped and Richard knocked softly then leaned forward and said something through the closed door. The door opened slowly and Richard led Frenchy inside.
"Okay. You know what to do."
"Sure do."
Alex pulled a thin piece of metal the size of a credit card out of his pocket and palmed it as we moved quickly to room 2110. Alex crouched behind me jimmying the door open. The metal card scraped again and again through the door frame but the lock wouldn't give.
"What the hell is going on?" I asked without looking back.
"I'm getting it. I'm getting it," Alex mumbled.
Danny patrolled the hallway, shoving a herd of Zeppelin fans into elevators. A tiny girl tried to slip past him and he lunged, snagging the back of her shirt and hurling her into the elevator. Now and then he barked orders into his broken walkie-talkie.
"HQ, this is Hall Patrol. We have a situation on twenty-one. Backup requested."
Alex worked on the door behind me.
"Got it," he whispered. "Going in."
Alex slipped into the room and closed the door behind him. This was it. Grab the money and get the fuck out. I glanced frantically back and forth from the door behind us to the door up the hall where Frenchy had disappeared with Richard. A few minutes later Alex popped his head out from the doorway.
"I can't find it, man. There's nothing here."
Danny walked toward us.
"All clear. Let's get this and go."
"It's not here, man. I can't find anything."
"You dumbass," Danny spit. "I'm going in."
Danny barged into the room, shoving Alex out of the way and banging the door against the wall as he hurled it open.
"Shit. Watch the hallway," I said to Alex.
He nodded and jogged off down the hallway.
I entered the room. Danny stood on the other side ripping empty drawers out of a long dresser and throwing them on the floor. The contents of Richard's luggage lay dumped in a pile next to an overturned nightstand.
"Where's the fucking money, asshole?" Danny snarled as I shut the door behind me.
"Relax. It's gotta be here."
"No. It's not. I've looked fucking everywhere."
"All right. Just calm down."
"Fuck you, Patrick," he heaved. "Are you guys trying to pull something on me?"
He stopped and glanced around the room.
"Wait," he muttered to himself. "Maybe there's a safe."
He knocked a painting off the wall over the TV and another from over the bed. Nothing.
"I ain't leaving here without that money," he shouted.
He spun toward me and pulled a small silver pistol from his waistband.
"All right," he said calmly. "New fucking plan."
"Whoa, Danny! Take it easy."
I backed up toward the door.
"Here's what we're doing. Me and you wait in the shower for this New York City asshole to return."
"I think he's actually British."
_"I don't give a shit!"_ he screamed.
He hurled a drawer at the bed, where it bounced off the mattress and flipped onto the floor. His fist closed tight around the gun.
"When he comes back we stick him up and make him give us the money."
"No fucking way. I'm not doing it."
"Yes, you fucking are," Danny spoke slowly.
He held the gun at his side. Every muscle tightened as Danny erupted. He screamed and swung his arm wildly, sweeping everything on the dresser to the floor. He tore the curtains off the wall and hurled them at the TV. His eyes locked on the TV then turned to the window. I could read the thought from across the room. I had to stop him.
He lunged toward the TV and his foot hooked on an empty drawer and a pile of clothes, sending him stumbling into the corner of the bed. He fell face-first onto the mattress and bounced off onto the floor with a grunt. I watched from the other side of the bed for him to get up.
He rose with a scream. He squatted and grabbed the bottom of the bed with both hands and flipped it over. It spun, taking the nightstand with it in a tornado of pillows and paisley bedspread. My eye caught a silver glint in the air.
I dove onto my stomach and searched through the sheets. My fingertips hit cool metal and I wedged myself between the mattress and the wall to reach it. My fingers fumbled, burrowing through the sheets, until they pulled the object into the palm of my hand. Danny hurdled over the bed frame toward me, the gun tight in his fist. I shot my arm into the air, the metal key ring dangling on my finger. Danny stood over me and then lowered the gun as he read the tag:
SAFE DEPOSIT BOX 51.
# NINETEEN
#
CRUSH OF PARTIERS IN THE LOBBY LOOKED AT MY HAND AND GOT THE FUCK OUT OF THE WAY. WE DIDN'T HAVE ANY TIME TO WASTE AND I KNEW WHEN I GRABBED THE ALL-ACCESS LED ZEPPELIN BACKSTAGE PASS FROM RICHARD'S LUGGAGE THAT THE SIGHT OF IT WOULD CAUSE THE CROWD TO PART. EVERY EYE FOLLOWED THE PASS AS ME AND DANNY BURST THROUGH THE LOBBY. A FEW FANS TRIED TO TALK TO US OR HIT US UP FOR TICKETS TO THE SHOW BUT I SHRUGGED THEM OFF WITH A LINE ABOUT "OFFICIAL ZEPPELIN BUSINESS."
The clerk at the front desk didn't look any happier than earlier. In fact, he looked worse. While we were upstairs, someone had dropped a bottle of red wine on the thick, cream-colored carpet and someone else had pissed in the back hallway. The clerk looked from me to Danny and back again, trying to put the pieces together. I held the backstage pass and the safe deposit box key up for him to see.
"I just need to get into our box," I said.
"Sir, typically we only allow the guest who requested the box to access it; in this case that would be Mr. Cole."
"Listen, Mr. Cole is in a meeting right now and he sent me down to pay the chauffeur for bringing our guitar back. Jesus, these guys. They get paid thousands to play the goddamn thing but can't remember to bring it with them."
He wasn't buying it. I jumped in before he could speak.
"You can call Richard if you'd like. He's in room twenty-one-ten."
This impressed the clerk and he dialed the number.
"Hello? Mr. Cole?"
Even from this side of the counter I could tell that Alex's British accent was bullshit. It sounded more Irish than anything, like a Lucky Charms cereal commercial. I had told him to keep it brief. He didn't.
"Would you like me to describe the chap I sent down?"
"No. That won't be necessary, sir."
"It's quite all right. He has long dark hair that needs a good washing. He's wearing jeans with a rip in the knee and a T-shirt with that horrible band Black Sabbath on it. No, wait, he's wearing a plain black T-shirt today."
"Okay, sir," the clerk said into the receiver.
"Now, if you could, be a mate and let him into the box. We need to pay off that arsehole of a chauffeur."
Red flooded into Danny's face as I choked back laughter. The clerk hung up and stared at the phone. He shook a set of keys over his head and walked toward the safe along the back wall. Danny and I followed.
Box number 51 sat at waist level on a wall of numbered boxes. I unlocked the box and Danny pulled it from the wall and lifted the lid. A few items lay scattered around inside: passports, receipts and a bundle of tickets for that night's show but not a single dollar.
"There's no fucking money in here," I said to Danny.
He stared into the box then slammed the lid closed and leapt around the vault stomping his feet and waving his arms in a silent temper tantrum until the gun fell down his pants leg and clattered across the tile floor. I put one finger to my lips, motioning for him to shut up.
"Everything all right in there, sir?" the clerk asked from outside the doorway.
"Sorry. We'll be right out," I answered.
I turned to Danny. His red face pulsed with his heavy breathing. I grabbed the bundle of tickets and stuffed them into my pocket.
"Let's go," I whispered. "We gotta get Alex and Frenchy and get out of here."
As we hurried across the hotel lobby I scanned the crowd for Richard Cole, a security guard or any other sign that someone was on to us. Something else stopped me cold: Emily. She stood across the lobby with Tina, Anna, Kyle and the rest of the Misty Mountain Hoppers. They lingered together near a couch in the back.
"This way!" I shouted, pulling Danny around a corner.
We shoved through a crowd of groupies and hurried down the hallway to the elevators. I pounded on the buttons. The numbers over the elevator door counted backward as the elevator crawled down from the twenty-first floor. I watched the lobby behind us for a sign that Emily or any of the Misty Mountain Hoppers were coming our way.
When the elevator doors opened, we tried to charge forward but ran into a wave of partiers getting off. A lanky kid in a flowered shirt and denim vest slumped in the corner. His bell-bottoms stretched out across the floor of the elevator.
"Let's go," I told the group. "Get him out of here."
A pair of scrawny hippies grabbed the kid's arm and tried to yank him to his feet. They stumbled into each other and collided off the walls of the elevator. A small crowd formed outside the elevator.
"Wake up, man," I said, kicking his legs. "You gotta move."
His friends dropped his arms and stood in the middle of the elevator talking.
"Maybe if we all lift on the count of three?" the shortest one said.
Danny burst through them. He grabbed the kid's ankles and yanked. The force pulled the kid away from the wall and he slammed onto his back. His head ricocheted off the floor of the elevator and he groaned as Danny dragged him backward out of the elevator. Danny deposited him in the middle of the hallway and then shoved his way back onto the elevator.
"Let's go," he said as he jabbed at the elevator buttons.
Back on the twenty-first floor, Alex opened the door to Richard's room and we slipped in to return the key and backstage pass. Alex and I put the bed back together and shoved Richard's clothes back into his suitcase. The hallways were filling back up with groupies and hangers-on.
"Where's Frenchy?" I asked.
"He's still in there," Alex said, pointing down the hallway as we crept out of the room.
"I saw Tina and Emily in the lobby."
"What the fuck are they doing here?"
"They're with Emily's sister and the Misty Mountain Hoppers."
"Shit. How much did we get from the safe deposit box?"
"Nothing.
"Nothing? What are we gonna do?"
"We're getting the fuck out of here."
The look on his face confirmed the plan I'd already formed on the way up. Get Frenchy. Get out. If we were lucky, Jimmy Page would pay Frenchy for the '58 Les Paul and we could escape from this mess with a few grand toward paying off Boogie. Maybe it would be enough. If we weren't lucky, we were all going to jail.
Someone needed to go to Page's room and pull Frenchy out of there. We argued over which one of us would do it. Danny wanted to do it but Alex and I both knew that was a horrible idea. Alex flat out refused to do it. That left me. I wiped the sweat off my palms and headed down the hallway.
# TWENTY
#
ARMS OF THE SECURITY GUARD STANDING OUTSIDE JIMMY PAGE'S ROOM DIDN'T SCARE ME. NOT THAT MUCH. OR THE FACT THAT HE LOOKED ABOUT THE SIZE OF A BALTIMORE COLTS' LINEMAN AS HE LEANED OVER AND STARED AT ME THROUGH SLANTED EYES SHOVED INTO A BALD HEAD WITH A BUSHY HANDLEBAR MUSTACHE. NO. I WAS SCARED OF THE WAY HE ADJUSTED HIS JACKET. EVERY FEW SECONDS HE ROLLED HIS SHOULDERS AND TURNED AT THE WAIST, WHICH MADE ME THINK HE WAS PROBABLY ADJUSTING A PISTOL HOLSTER SOMEWHERE UNDER HIS ARM.
He started talking even before I did.
"Get the fuck out of here."
"My friend's in there."
"No, he isn't."
Through the door I heard guitars jamming loudly. One started to solo, stumbled a bit on some low notes and then ripped up the neck. A few seconds later the song fell apart into laughter.
"His name's Frenchy, uh, Reginald. He's got a Fifty-eight Les Paul with him. Jimmy wanted to buy it."
He looked me over then decided I wasn't much of a threat. He was right. He poked his head around the door. I heard Frenchy's voice, then the door swung open.
I walked slowly into the room, stepping around a large piece of luggage and a guitar case. Richard lounged on the edge of the bed smoking a cigarette and talking on the telephone. Frenchy sat across from him in a chair tuning his Telecaster. Jimmy Page sat next to him.
Jimmy seemed small even with the extra padding of a button-up shirt and blazer. Curly black hair hung in his pale face as he hunched over a guitar. A long black cord snaked across the floor to a tiny amplifier. Guitar cases cluttered the floor and an acoustic guitar lay on the bed. He and Frenchy traded off solos. Frenchy sounded pretty fucking good.
The song stopped and Frenchy looked up.
"Oh, hey, man. Jimmy, this is my friend Patrick."
"Nice to meet you," he said.
I scanned the room for the briefcase that might have the cash.
"Uh, nice to meet you too."
I looked at Frenchy.
"Sorry, French... I mean Reginald, we gotta get going."
"Really? What's the rush?" he asked.
"Has Jimmy had a chance to look at the Les Paul?" I pushed things along. "We really have to get going."
"Okay. Let's take a look," Jimmy said.
Jimmy opened the case holding the '58 Les Paul and stared at it for a minute before picking it up and plugging it in.
Frenchy fiddled around with an old blues tune. The muddy chords screamed from the tiny amp. He turned to Jimmy while he played.
"I just learned this one."
"Jimmy Reed." Jimmy nodded. "Bloody brilliant guitar player."
He threw a solo over what Frenchy played. I couldn't believe it. We came to rob the band and Frenchy was jamming with them. Richard put a finger into his ear and screamed into the phone, "I don't know. We're hanging out at the hotel for now. Jimmy's looking at a guitar. _I said Jimmy might buy a guitar_. I don't know. Reginald Chamberlain. Really? You've never heard of him. He's big-time, man."
Frenchy and Jimmy played back and forth like that for a few minutes until the hotel door crashed open. Peter Grant barreled across the room. His presence filled the entire suite. The furniture seemed to shrink to get out of his way.
"What's all this then?" he asked Richard.
"Jimmy's buying a guitar."
"So who the fuck is this?" he asked, flipping a finger toward me.
Richard shrugged and muttered something about me being with Frenchy and the guitar. By now, the jam session had sputtered to a halt.
"Don't I know you, mate?" Peter asked me.
"I don't think so." I shrugged.
Peter stared at me. Jimmy, Frenchy and Richard chattered behind us. In that big, bald head Peter was trying to place where he knew me from. I signaled to Frenchy for us to get the hell out of there.
"Hey, Jimmy," Frenchy spoke up. "We gotta roll."
"You guys staying for the show tonight?" Jimmy asked, lighting a cigarette.
Frenchy looked up at me like a kid pleading for one more hour at the playground. I answered before he could.
"No, man, I wish we could but we've got to get back."
"Are you sure? I can get you sorted with tickets."
This time I didn't look at Frenchy.
"I wish we could. Can't tell you how much I wish we could but our ride is leaving. He has to work in the morning."
"But Patrick, didn't you—" Frenchy started to say.
"It's real cool of you to offer, Jimmy. Really cool. Come on, Reginald, we better get going."
Jimmy laid the Les Paul back in the case and crouched over it, staring for a second before shutting the lid. He leaned in the bathroom doorway and swilled from a bottle of Jack Daniel's while Frenchy fumbled to put his Telecaster away.
"You sure you want to sell this?" Jimmy asked Frenchy.
"Yeah. That's what I do. I try not to get attached to them, ya know? Love 'em and leave 'em, right?"
Frenchy laughed nervously.
They talked back and forth about the guitar with Jimmy nearly talking Frenchy out of selling it. These are so rare these days, aren't they? Can you believe how great a condition it's in? Are you sure you want to sell it? Jimmy practically begged Frenchy not to sell it. I thought it was going to work. Luckily, Frenchy held his ground. They settled on a price of two grand, just enough to pay off Boogie.
Unfortunately, getting paid for anything caused tension. It didn't matter what you were dealing with. Put two groups of men in a room with a bit of money and people get edgy.
It didn't help the tension in the room as I angled to get a look at the money. I shifted and slipped, trying to see where the cash was kept and how much was there. Peter sat at the edge of the bed. Jimmy and Richard stood in the corner with their backs turned toward us. Richard pulled a wad of bills from his pocket and counted them, then Jimmy bent over to open the nightstand drawer. I stepped around the corner of the bed to take a look.
"Where the hell do you think you're going?" Peter asked.
"I was just gonna grab the money for the guitar so we can get going."
He pointed across the room behind me.
"You just stay right fucking there," he said calmly.
_See, we're in a bit of a hurry_ , I started to answer as I stepped back, then decided not to push him.
Finally, Jimmy crossed the room.
"Here you go, Reg," Jimmy said, handing Frenchy a thick wad of bills.
"Cool. Great meeting you, man," Frenchy said. "Thanks for letting me jam with you. Hope I wasn't too terrible."
"No, no, you were fucking spot on." Jimmy grinned.
I moved Frenchy toward the door. When we got there, he turned around.
"Sorry. Forgot my guitar," Frenchy said. He hurried back across the room and lifted the black guitar case.
I bolted into the hallway. Frenchy hurried along behind me, lugging his guitar. We found Alex and Danny by the elevators.
"Frenchy got us two grand for the Les Paul," I told Alex.
"Nice fucking work, man," Alex said, slapping Frenchy on the back.
We waited for the elevator. Alex punched the call button trying to bring it up faster. A pair of businessmen shared the ride down and looked at us nervously before getting off on the fifteenth floor. They must have sensed something that the rest of us didn't. We were ten floors from the lobby when the elevator shuddered to a halt and Danny held his gun to Frenchy's head.
# TWENTY-ONE
#
AT THE GUN JAMMED AGAINST HIS FOREHEAD AND DIDN'T BLINK. EVERYTHING HAD HAPPENED SO FAST. ME AND FRENCHY WERE TALKING ABOUT JIMMY PAGE WHEN DANNY SLAMMED THE ELEVATOR'S EMERGENCY STOP BUTTON, YANKED THE GUN FROM HIS PANTS AND GRABBED FRENCHY BY THE THROAT. I BACKED AWAY FROM DANNY. ALEX STARTED SHOUTING.
"What the fuck are you doing?"
Danny stared at Frenchy.
"Gimme the money."
Frenchy clenched his lips and shook his head. He clutched the guitar case against his chest and cowered behind it. Alex kept talking.
"Danny! Put the fucking gun away. Put it away right fucking now."
"Gimme the fucking money, Frenchy."
"We need this money to pay off Boogie," Alex said.
"I don't care."
"This is your fucking mess, Danny!" I yelled. "You do this and we'll tell Backwoods Billy you were the one who stole his safe."
"Won't matter. I'm taking this money and leaving town."
"Then I'll go to the DA," I said calmly. "He already came to see me. He'll have you arrested. You want to go back to jail?"
Danny stared hard at me like I'd just slapped him. He slowly moved the gun toward me. His eyes glazed over while he considered shooting me.
Alex saw his chance and charged at Danny. They tangled together as Alex grabbed Danny's arm and jerked it upward, pointing the pistol at the ceiling. Danny yanked at the gun and hurled them both backward. They bounced off the back wall. Danny tried to spin off but Alex pinned him to the side of the elevator.
I rushed at them as Danny bent Alex's arm backward. An arm shot wildly from the knot of limbs and the cold metal butt of the gun rocketed into my nose. The force knocked me to the floor. Tears clouded my eyes and I couldn't breathe. Blood trickled down my face and across my lips.
"You don't know who you're fucking with, boy," Danny grunted. "I told ya I studied with a Navy SEAL."
They staggered around the tiny elevator, locked together in a fight for the gun. Frenchy crawled around them using his guitar case as a shield as he inched toward the control panel and pounded on the buttons. The elevator jerked and shuddered and started toward the lobby.
The motion knocked Alex off balance. He swayed to the left and Danny knocked him to the ground with a wobbly kick. With his arm free, Danny waved the gun around the elevator. None of us moved. He leveled the pistol at Frenchy's head and talked between gasps for air.
"That's it, Frenchy. Give me the goddamn money."
His heavy breathing echoed in the elevator. Frenchy looked to me with wide eyes that begged for me to tell him what to do. I gave him a look that I hoped told him not to be stupid. He tugged the wad of bills out and held it up. Danny snatched it from Frenchy's hand and stuffed the cash in his pocket.
"No hard feelings, boys," he panted. "You understand. Now when this door opens, you guys go your way and I'll go mine. Don't do anything crazy."
Later I learned that hundreds of fans in the hotel lobby spotted our stalled elevator and figured it held Lep Zeppelin preparing for their grand exit out of the building. Word spread and people rushed toward the elevators. When the doors shuddered open, a mob of screaming groupies, drug-addled hippies, autograph collectors and weirdos crushed forward. Emily, Tina and the rest of the Misty Mountain Hoppers stood in the front. Emily screamed first.
"Oh my God! Patrick!"
I slumped against the back wall of the elevator holding both hands over my nose. A warm trickle of blood snaked down my arm and formed a small pool between my legs. Frenchy crouched in the corner behind his guitar case and Alex leaned against the wall, holding his stomach. He reached out for the wall to keep from falling over. Danny stood in the middle of the elevator and ignored us. He adjusted his tie then walked calmly through the crowd.
I pulled myself up and wobbled toward Emily. My legs felt like jelly. Frenchy crept sheepishly out of the elevator with his guitar and Alex limped behind him. The Misty Mountain Hoppers surrounded us.
"What happened to you guys? What is going on?" Emily asked, grasping my arm.
I started to answer then stopped. I didn't know what to say. Should I tell them that I just tried to rob their favorite band in the entire world? Should I admit that it didn't work and now I was caught between a funk band called the New York Giants and a born-again motorcycle gang called the Holy Ghosts and that either one of them might kill me? Should I say that the two grand that would have saved me was walking out the door in the hands of the man who got me into this fucking mess in the first place?
Then it hit me. It hit me like a pistol to the face. I stepped forward, cupped both hands around my mouth and screamed the one thing I could think of sure to cause utter fucking chaos in a swarm of crazed Zeppelin fans.
_"That guy just robbed Led Zeppelin! Somebody stop him!"_
Panic surged across the lobby. The army of dazed and drugged-out fans transformed into a seething mob. Beer bottles, cameras and backpacks crashed to the lobby floor as Zeppelin fans dropped everything and ran after Danny. Halfway across the hotel lobby, Danny stood frozen. He clutched his chauffeur's hat in his hands and backed toward the doors.
"I never robbed anybody! Hold on a minute. You guys don't know who you're fucking with!"
A wave of punches and kicks swallowed Danny. The chauffeur's hat hurled into the air. The sound of a body smacking onto the marble floor echoed through the lobby and someone screamed for the hotel clerk to call the police. Frenchy and I shoved through the crowd. Danny lay pinned on the lobby floor, his arms and legs restrained by screaming fans. Hands tore at him and a female hand held on to a mess of his hair. He craned his neck and our eyes met.
"Patrick! What the fuck? Get 'em off me!"
I jabbed my hand into Danny's pocket and palmed the wad of cash. His eyes narrowed.
"You motherfucker! You can't do this! Look, everyone—he has the money! Not me! It wasn't me!"
Frenchy leaned forward, cocked back a bony fist and punched Danny in the nuts. Danny howled. On the other side of the crowd, Emily ran toward me.
"What are you guys doing here? What happened?" she gasped.
"We drove up here so Frenchy could sell Jimmy Page a guitar."
"What? Really?"
A bell rang as the elevator doors opened behind us and I jerked my head to look back. Peter, Richard and a team of Zeppelin security stormed into the lobby. Peter's giant head swiveled, scanning the lobby.
"We gotta get out of here," Alex whispered to me.
"Listen, I gotta go. I'll call you and explain everything," I told Emily as I backed away.
She grabbed my arm.
"You can't leave! You have to talk to the police so they can arrest Danny. They're gonna want a report from you. He just robbed Led Zeppelin!"
Peter Grant bulldozed through the crowd with Zeppelin's security fanned out around him. Emily looked over her shoulder at Peter and Richard coming toward us. When she looked back at me her face had changed.
"I'm sorry," I said. "I gotta go."
Me, Alex and Frenchy hurried down the back hallway toward the kitchen, counting on it having an exit. Frenchy struggled to keep up as he lugged his heavy guitar case. Zeppelin security closed in on us. A swarm of fans and hotel employees passed us headed in the opposite direction. I pulled out the stack of concert tickets I stole from Zeppelin's safe deposit box and tossed them into the air.
"Free Zeppelin tickets!" I yelled.
Every face in the hallway stared upward as hundreds of tickets spun slowly toward the ground. Fans dove to the tile floor to gather tickets. Zeppelin security turned the corner and slammed into the crowd clamoring for tickets on the tile floor. Peter's arms waved wildly as he stumbled over someone and slammed into the wall.
Alex led us toward the kitchen door. A hand squeezed tightly on my arm before I could slip through.
"Hey—you work for Zeppelin," the hotel clerk said. "What the hell happened up there?"
I stared at him and blinked.
"He held us at gunpoint," I finally answered. "Then he brought me downstairs and forced me to empty the safe deposit box."
"I knew something was fishy when you came down," the clerk said. "It just didn't feel right."
"He's a dangerous man. I think he still has a gun on him."
"Don't worry. The police are pulling in now."
That wasn't good.
"Good," I said. "Now, is there any way you can find me a towel for my nose? It's finally stopped bleeding."
"Absolutely, Mr.... I'm sorry, I didn't catch your name."
"It's John. John Osbourne," I said.
"I'll be right back, Mr. Osbourne," he said, disappearing down the hallway.
When he rounded the corner, I bolted from the back door in the kitchen and met up with Frenchy and Alex across the street. The crowd on Park Avenue swallowed us as we ran toward the garage two blocks over where we parked my car. A police car roared past us toward the hotel.
"What do you think will happen to Danny?" Alex asked as we climbed the concrete steps of the parking garage.
"He didn't actually rob Zeppelin so that'll be dropped," I said.
"But he did rob us."
"They'll have to drop that too since we aren't going to press charges."
"So they'll let him go?"
"Nah," I said. "He still violated his parole. Plus he's carrying a gun. He's gonna be fucked."
Frenchy hadn't said a word. He walked ahead of me and Alex with his head down. The black guitar case knocked against his leg as he trudged across the parking garage. Me and Alex looked at each other and both thought the same thing. Frenchy was hiding something. He never could keep a secret.
"What's going on, Frenchy?"
"Well..." He hesitated. He pointed at the guitar case. "I stole one of Jimmy's guitars."
"How the hell did you do that?" Alex asked.
He really sounded impressed.
"When me and Patrick were leaving I went back to grab my Telecaster but I took this case instead. I don't know what came over me. I figured, whatever's in this case has to be better than my shitty Fender."
"What were you going to do if he caught you?" Alex asked.
"All the cases looked the same anyway so if he stopped me I figured I'd just play dumb and act like it was a mistake."
"Our little Frenchy has become a real thief," I said, grinning.
"You sly fucking dog." Alex grinned too.
"Let's take a look," I said.
Frenchy leaned the guitar case upright against the bumper of my car. The snap of the clasps opening echoed across the empty concrete garage. He swung the lid open and cash flooded out onto the greasy garage floor in thick bundles. Wads of twenties, fifties and hundreds bounced across the concrete.
"Holy fuck!" Alex screamed.
He jumped up and down, waving his arms, then ran a giant circle around the empty parking garage, screaming.
Frenchy stood frozen with his mouth hanging up.
Me and Alex grabbed the case and began shoveling the money back inside. Frenchy still hadn't moved.
"How much you think is here?" Alex asked.
"I'm guessing about fifty thousand."
"No way. There's gotta be at least a hundred grand."
"You're crazy. There's no way it's that much."
It took Alex and Frenchy most of the three-hour drive home to count all $203,000.
We were both wrong.
# TWENTY-TWO
#
YELLED FROM THE OTHER SIDE OF A THICK WOODEN DOOR, "WHO THE FUCK IS IT?"
"Boogie! It's me, Patrick."
Loud music thumped inside the house. Boogie asked again. This time I shouted.
"Boogie! It's Patrick and Alex! We're here about the safe!"
One eye peered through the tiny curtained window then Boogie jerked the door open. The force nearly tore it off the rusted hinges. He motioned us inside.
"Shit, man. Keep it down," he said, leaning out and looking up and down the street. "It's late and my neighbors are some nosy motherfuckers."
It took us half an hour of driving around in the dark to find Boogie's house again. We were exhausted. Nerves in the car were seriously frayed after everything that went down at the Drake Hotel followed by three hours of major freaking out in the car while Frenchy and Alex counted the money.
We argued most of the time about whether we were or weren't going to get arrested, why we should or shouldn't return the money or whether Alex should buy a Mustang or a Nova with his share of the cash.
We dropped the cash at Frenchy's house, figuring it to be the safest place. Then we drove in circles trying to remember where Boogie lived. We needed to pick up the safe. I wouldn't be able to sleep until I paid Boogie and loaded the safe into the car. I didn't give a damn how late it was when we got there.
Turns out, Boogie was having a small party and his place thundered with loud funk music and laughter. Johnny Paycheck danced in the middle of the living room with a tall black girl who, even without her huge platform shoes, would have towered over him. Bleary-eyed partiers sat in a row on the sofa, passing a series of bongs laid out on the coffee table next to a brick of weed and a pistol. The coffee table looked familiar.
"Is that our safe?" I asked Boogie, nearly screaming to be heard over the music.
"Yeah. Yeah. That's it right there."
He turned the music down as he passed the stereo. A skinny guy stretched his long legs across the top of the safe and Boogie slapped his feet to the ground as he walked over.
"Get up, motherfuckers. We gotta move this shit."
The stoner crew on the couch grumbled and shifted in their seats but no one really moved until Boogie leaned in and slapped one of them upside the head. The entire crew then scurried to get away from him.
"Where you going? Get your asses over here and help with this."
With the bongs, weed and weapon cleared off the top I could see the safe laid on its back in the middle of the living room. The door had been replaced and the chipped-up dial from the old safe had been fitted into the new lock. It looked good. With a little help, Boogie groaned and set it upright.
"Are the tapes inside?" I asked him.
"Yeah. I think so. You wanna check?"
"Yeah. What's the combination?"
"Damn," he said, shaking his head. "I knew you were going to ask me that. Let me find it."
Boogie emptied his pockets onto the top of a giant stereo speaker—a wad of bills, a giant switchblade, a pair of guitar picks, a packet of cocaine and a few stray bullets.
"Shit, Johnny. Where did we put the combo to that motherfucker?"
"Hell, I don't know," Paycheck said, leaning into his girl.
"Quit fucking around. Where is it?"
Paycheck dug through his back pocket then handed Boogie a scrap from a cardboard Budweiser box with numbers scrawled on it. Boogie opened the safe. The two tapes sat on the bottom shelf.
"There you go, man. Just like I promised. You got the money?"
Alex put his hand on my shoulder and looked around the party. He was right. Handing Boogie two grand in the middle of a party made me nervous.
"You ain't gotta worry about these idiots." Boogie grinned. He lifted the front of his T-shirt and showed us a pistol jutting from his waistband.
I struggled to pull the fat wedge of cash from my front pocket then slipped it to Boogie. He looked around the living room then snuck off to the kitchen to count the money. He might not have been nervous for me to hand off the money in front of his friends, but he sure didn't want to flash it around them. He returned a few minutes later, smiling, and shouted to Johnny Paycheck, "Yo, Johnny! It's cool!"
"We're all cool?"
They high-fived and laughed loudly.
"Ah yeah! We're getting that Moog tomorrow. Hot damn!"
Johnny Paycheck danced across the room then threw his arm around Alex's neck.
"You all want a beer?"
He brought me, Alex and Frenchy each a can of beer. We stood against the wall and drank them and tried not to look like the only three white guys in the house. A pack of girls danced in the middle of the room but we were all too nervous to walk over. Alex got stoned with a few guys then grabbed more beers for us from the fridge. Soon I was drunk and arguing with Boogie.
"I'm just saying Sly Stone is overrated."
"Overrated! You're out of your fucking mind. The guy invented funk all by himself," Boogie shouted, leaning over. His bushy Afro poked into my face.
I shrugged. That made him madder.
"'Dance to the Music' You're gonna tell me you don't like 'Dance to the Music'?"
I shrugged again and sipped my beer.
"Ah, man. You don't know shit, white boy." He laughed.
As the party wound down, me, Alex and Frenchy sat around the safe in the living room. Boogie nodded off in a recliner. Over by the stereo, Johnny searched through the eight-tracks for some music.
"All you have is loud-ass funk. Where is the after-party music, Boogie?"
Boogie sat up and shrugged, his eyes glazed with beer and pot.
"You ain't got any Bobby Blue Bland? What about Otis, man? Where's the Otis?"
Johnny Paycheck scattered a pile of eight-track tapes then flipped through a crate of records. When he didn't find anything he liked, he grunted and walked away. He swayed a bit in the middle of the room then steadied himself on the edge of the safe.
"What's in this motherfucker?" he said, leaning over and spilling his beer on the carpet. He swung open the door to the safe.
"Aw, damn," he said. "Jim Nabors. Let's put this shit on."
"No, no, no," I said. "We gotta give that back to Backwoods Billy."
"Damn, man," Johnny Paycheck taunted me. "It won't hurt it. We got the reel-to-reel right here!"
He held the tapes up over my head and out of reach. I gave up.
"Just let me hear 'Green, Green Grass of Home.' You know that tune? This guy thinks he's coming home but really he wakes up and he's in prison and he's been dreaming. That song is badass."
He loaded the reels into Boogie's player. The machine hissed and clicked as the tape set up. He cranked the volume and pressed _play_. Through the static a voice spoke loudly in the speakers.
_"... it's the same every month. It's not, uh, up for negotiation. A thousand dollars' cash, two hundred Black Beauties, two hundred Blue Devils, a pound of weed and whatever Percocet you dirtbags have around. That's the price you pay for my, uh, assistance."_
My heart rocketed up into my throat. I knew that voice.
_"Now where am I gonna get Percocet, Cooper? I ain't no fucking doctor."_
I knew that voice too.
# TWENTY-THREE
#
CHUCK TAYLORS SQUEAKED ACROSS THE MARBLE FLOOR IN THE COURTHOUSE. THE SOUND ECHOED AROUND THE HIGH CEILING CAUSING EVERYONE TO STARE. FUCK IT, I THOUGHT. THEY WOULD HAVE STARED ANYWAY. IT'S NOT OFTEN YOU SEE A KID WITH LONG HAIR AND A BLACK SABBATH SHIRT STROLLING THROUGH THE COURTHOUSE. NOT UNLESS HE'S WEARING HANDCUFFS.
The security guard towered over the desk at the elevator. His starched uniform dangled off his bony shoulders as if he were a scarecrow. He slicked back his silver hair and chomped nervously on a toothpick when he saw me. He stuck out a thin hand as I walked past the desk.
"Wait a second. Where you going, bud?"
I leaned over the desk.
"I have an appointment on the fourth floor."
"With who?"
"Simon Cooper. District Attorney."
He pointed to the tapes under my arm.
"If that's a delivery you gotta take the service elevator round back."
"Special delivery." I smiled, eating a mint out of a bowl on his desk. "He's expecting me."
His eyes lingered on me then he picked up the phone.
"This is the security desk," he said. "I got a young man here says he's here to see Mr. Cooper... Oh really... I understand... Yes, I'll escort him."
He hung up the phone then grabbed a large ring of keys from the desk.
"Let's go," he grunted.
"It's all right. I can find it on my own," I said as I followed him down the wide marble hallway.
"They want me to escort you."
"What do they think I'm going to do, steal something?" I laughed.
"Nope. They told me to make sure you and that delivery get there right away."
The guard trudged on ahead of me. Cooper's office sat at the end of a long, wood-paneled hallway. The frosted-glass door led to a waiting room where a stubby secretary typed behind an oversized desk.
"He's all yours," the guard said before he left.
She looked up at me then at the boxes under my arm.
"You can just leave those here on the desk, young man," she said. "I'll make sure Mr. Cooper gets them."
I smiled and sat down on a deep leather couch along one wall. She looked up from her typewriter then picked up a phone and whispered.
A few minutes later Cooper leaned in the doorway to his office. He straightened his tie then rubbed his three-day beard. Dark circles surrounded his eyes and his face seemed pale. He looked like he hadn't left his office in days. He still had his pricey suit and greasy smirk, though, all the signs of a privileged fuck-up who found ways to con others into cleaning up his mistakes. Someone who counted on his charm and money to make up for all of the trouble he got into. Right now he was counting on that charisma to con me into handing over the tapes I held under my arm. It wouldn't be that easy.
"Patrick!" He grinned. "Come on in. No calls please, Joyce."
Cooper's office looked like the den of a man trying to hold down a serious job while juggling an even more serious drug habit. Sunlight wrestled through the closed blinds and curtains. Piles of papers lined the floor along the walls, most likely put there during a speed-induced effort to get organized. The products from all his late nights cluttered the top of a filing cabinet—deodorant, eye drops, two unopened dress shirts, an iron, several toothbrushes and a collection of pill bottles.
He closed the door behind us then stood in the middle of the office looking me over.
"Good to see you." He grinned. "I was worried. I've looked just about everywhere for you. You're a hard guy to find."
"I had a few things to take care of out of town." I shrugged.
"Well, here you are. This is good. This is really good. Sorry about this mess."
"Looks like you've been busy."
"The mayor isn't too happy about the hell Backwoods Billy and his buddies raised down at the Inner Harbor. Those bone-heads put thirteen carnies in the hospital. The mayor wants somebody to answer for it."
I shifted the tapes under my arm and asked a stupid question.
"So Backwoods Billy is in jail?"
"Nah." Cooper fidgeted. "They've got nothing on him. Can't even prove he was involved. I've been holed up in here dealing with all the arrests."
He twirled an expensive-looking cuff link and lost himself in thought then quickly shook it off.
"Anyway, Patrick, what's new?"
I held up the tapes. A pained grin crept across his face.
"Find those in Billy's safe?"
I nodded.
"I figured that was why you called me. I knew you'd come through for me. Did you listen to them?"
"Yeah. Pretty interesting stuff."
He hung on to the dopey grin and nodded. Nothing shook this guy.
"I bet."
"Drugs, prostitutes, payoffs." I whistled loudly. "There's even a conversation on here where someone who sounds just like you talks about putting a hit out on a lawyer. Now, you're a DA. Is that a felony or a misdemeanor? I think it's a felony, but you're the expert."
The grin dissolved. He lit a cigarette and waved the smoke out of his face.
"Do you know who made these tapes?" I asked.
"I have an idea." He grimaced.
"Yeah? How did it happen?"
"I was going through a rough time and some scumbags I thought were my friends decided to take advantage of me."
"Here's what I think happened," I said. "You worked out a deal with Backwoods Billy where he'd set you up every month with pills, drugs, a bit of cash and whatever else you wanted. In exchange, you'd make any legal messes involving the Holy Ghosts disappear. Is that right?"
He crossed his arms and stared at me.
"Like I said, I was going through a rough time."
"You thought Backwoods Billy was some dumb hillbilly, right? Then he taped his conversations with you, cut off the payments and made you help the Holy Ghosts anyway."
Cooper shrugged and gave me a look like I'd just told him the most obvious fact in the world.
"Then the cops arrested the Holy Ghosts for beating the living shit out of me at the carnival and one of the cops mentions something about me having their safe. You thought you could use me to get the tapes back because I wouldn't know what they were and how much they were worth around here."
He stubbed out his cigarette.
"And right now I bet Backwoods Billy is threatening you with these tapes unless you make this Inner Harbor mess go away. That's why you've been holed up in here trying to track me down. And I know he's out there looking for me. He needs the tapes to make all this trouble go away and you need the tapes to make him go away."
Cooper shuffled around his desk. He sat down in the cushy leather chair and unlocked a drawer then removed a few bottles of pills, some tiny vodka bottles and a thick roll of cash.
"Let's get to it, Patrick. How much do you want?"
"I want my friend Keith released from jail. All charges dropped."
"You mean your idiot friend that the cops arrested during the brawl?" he scoffed. "You know, he's suspected of stealing a very rare and expensive guitar from Haven Street Pawnshop. Hell, between you and me, you're a suspect too."
"No, I'm not," I said, tapping the tapes.
Cooper tugged at his messy hair then groaned loudly.
"Fine. Is that it?"
I smiled.
"That's it. Just do your job."
"My job is putting people away. Not getting them off."
"Good," I said. "You can start by locking up Backwoods Billy and the Holy Ghosts. I don't want to live the rest of my life hiding under a table every time I hear a motorcycle."
"Don't worry. I'll take care of it."
"Give it until tomorrow. I want to go see Backwoods Billy tonight."
"Don't be an asshole, kid. Let it go."
Cooper stood up and walked around his desk. He smoothed his tie and fidgeted.
"So we have a deal?"
"One more thing," I said. "I'm keeping a copy of the tapes for myself just in case you don't hold up your end of the bargain."
"I figured."
I tossed the tapes on Cooper's desk.
"I'll pick Keith up tonight at five. Make sure he's ready."
Cooper nodded. He sat on the edge of his desk holding the reels in his hands. Something hit me as I walked to the door. I had to ask.
"You know anything about a guy named Danny Carter? Guy from around here who was arrested in New York over the weekend with a gun? Something about him and Led Zeppelin?"
"Yeah. I heard about that. Parole violation. Weapons possession."
"How long you think he'll go away for?"
"Probably another five. Why? You want him out too? 'Cause that's one even I don't think I can save."
I thought for a second. Just a second.
"Nah. You can have him."
# TWENTY-FOUR
#
IS FUCKED," ALEX SAID AS WE CROSSED THE PARKING LOT BEHIND SHOOTERS BAR, THE UNOFFICIAL CLUBHOUSE OF THE HOLY GHOSTS.
I called the bar earlier in the day and promised Backwoods Billy we'd be there later to drop off the safe. Alex and Frenchy came with me. As we pulled into the gravel parking lot I wondered how many poor suckers were led into the dark woods behind Shooters and never seen again. Frenchy thought the same thing.
"We're gonna be buried out here." He sighed.
Inside Shooters, Backwoods Billy stood by the pool table, holding a cue in one hand and glass of whiskey in the other. He wore a blue bandanna tied around his head covering a white bandage. A black leather vest hung over his plain gray tank top.
"It's your shot, Billy," Rabbit said to him as we walked in.
The stools and tables sat empty except for a few Holy Ghosts lounging around the pool table. The rest of the gang was still locked up in jail. Billy bent over the table. His sunglasses fell off the top of his head and rolled across the green felt. Everyone laughed. Billy grinned then dropped his cue on the tile floor. He was drunk.
"Boys! Come on in." He waved to us.
We slunk into the bar and stood behind a row of stools loaded with Holy Ghosts. I recognized the pockmarked face of a kid at the end of the row as the guy Alex pummeled at the Inner Harbor. A tan bandage covered his nose. He looked younger than I'd remembered. The chubby guy standing next to him raised his bald head to look us over, then grinned. He pointed at Alex.
"Hey! That's the kid that whipped Sonny's ass at the carnival."
Everyone laughed except Sonny.
"Fuck you, Whitey," Sonny mumbled.
"I'm telling ya, that boy can duke," Whitey said. "What's your name, kid?"
"Alex."
"Don't mess with this kid, boys. He's a fucking animal."
Alex lit a cigarette to keep from looking nervous.
Backwoods Billy took his shot. The cue ball missed the three ball by about a foot, bounced off the bumper, crossed the middle of the table and sunk the eight ball in the corner.
"Well, shit." He laughed. "I just fucking blew that."
He set the cue back in the rack on the wall then walked toward me.
"Come here, Patrick," he said, motioning toward the bar. "Let's talk."
Rabbit racked the pool balls as we walked away. He asked if anybody wanted to play.
"I do!" I heard Frenchy say.
I sat on a bar stool next to Backwoods Billy. He leaned over the bar, pulled a bottle of beer from a tub of ice and handed it to me.
"You did a good thing bringing that safe back, boy."
"Thanks," I stammered.
"Listen to me, Patrick. You did the right thing by fixing this mess before it got out of hand."
I looked up at the bandage peeking out beneath the bandanna on his head. He caught me staring.
"Don't worry about this," he said, running his finger over the bandage. "It happens. Comes with the turf, kid. Shit, I've been beaten worse and left for dead. Besides, you shoulda seen what was left of them carnies. It weren't pretty."
"So we're cool?"
"We're cool, kid. Don't worry. Remember: 'If thy brother trespasses against thee, rebuke him; and if he repents, forgive him.' You know where that's from?"
I didn't know jackshit about the Bible but for some reason I felt pressured to take a guess. I tried the only two names I knew other than Jesus.
"Is that Luke or something? Maybe John?"
"Well, all right, boy. Luke 17:3. Guess you've been reading that Bible I gave ya. I am impressed. Let's have a shot."
A chubby lady in a sleeveless David Allan Coe T-shirt shuffled behind the bar. Backwoods Billy motioned for her and she set out a row of shot glasses. Backwoods Billy grabbed the bottle of tequila and lifted it over his head.
"Who wants a shot of tequila?"
A few of the Holy Ghosts wandered over. Sonny stayed on his stool.
"What's the matter, Sonny?" Billy yelled. "Your pussy hurt or something?"
Everyone laughed. Rabbit yelled from the other side of the bar. He pointed at Frenchy.
"Billy, get this kid a shot."
"You're just trying to get him drunk 'cause he's beatin' you." Whitey laughed.
"Goddamn right. He's whipping my ass."
"What have I told you about using the Lord's name in vain, Rabbit?"
"Sorry, Billy."
We came together and Backwoods Billy poured a round of tall tequila shots then passed them to us. We raised our glasses.
"All right, boys. Here's to staying free!"
We all shouted, "Amen."
The warm tequila burned down my throat.
Backwoods Billy sent Whitey, Sonny and some of the other Holy Ghosts to haul the safe into the bar. They struggled, sweating and grunting, as they plowed through the door. The safe landed with a heavy thud on the floor by a back table. He never told them what it was or where it came from. They never asked.
I kept one eye on Backwoods Billy. If he had any sense he'd check for the tapes in the safe before letting me go. He played it cool for a while then he handed me some change and asked me to play some music on the jukebox. While I stood flipping through the Merle Haggard and Hank Williams Jr. records on the jukebox, Backwoods Billy squatted in front of the safe and dialed the combination. The door unlocked flawlessly and he peeked inside. Satisfied with seeing the tapes in their boxes on the shelf, he called me over for another shot.
Whitey, Sonny, Backwoods Billy and a few other Holy Ghosts stood around the bar drinking and talking about the fight at the Inner Harbor. Someone bragged about breaking a whiskey bottle over a carny's head and Sonny claimed to have smashed someone's nose with brass knuckles. They laughed a lot, even if they were on the losing end of a pool cue or baseball bat in the story. I clutched the beer bottle in my hand. Backwoods Billy slurred in my ear, "You know, that boy Alex of yours would make a good Holy Ghost. I ain't bullshitting. That boy can fight."
I didn't know what to say.
"He ever ridden a motorcycle?" he asked me.
Before I could answer, a wooden snap rang out by the pool table. Rabbit stood with a broken pool cue in his hand. He glared at Frenchy. Alex stood between them with his arms up, asking Rabbit to calm down. At six-foot-something and about 250 pounds, Rabbit could tear Frenchy in half. There was nothing Alex could do about it. Rabbit's chest heaved and he clutched a broken half of a pool cue in each meaty fist.
There are a few rules to remember when hanging around a motorcycle gang. Never get drunk with them because they will end up kicking the shit out of you. When one of them does finally kick the shit out of you, everyone else in the room will help. And most important, never, ever under any circumstance touch, insult or otherwise disrespect the gang or a gang member's jacket. That'll get you killed.
Those rules were the farthest thing from Frenchy's mind as he stood on the other side of the table. His sweaty hair flopped around his face and he blinked nervously behind his thick glasses. In his arms he held a death sentence—Rabbit's leather jacket and denim vest. The Holy Ghosts patch wrinkled in his tight grip.
"Whoa, boys!" Backwoods Billy yelled. "What the hell's going on?"
"That boy just took my colors!" Rabbit yelled.
"You better put that down, son," Backwoods Billy said to Frenchy. "Touching a man's colors can get your fucking skull split."
"I won it. Fair and square."
Backwoods Billy turned to Rabbit, his face pulsating with rage.
"You bet your fucking colors on a pool game?"
"He said he wanted to bet my colors against his fifty grand!" Rabbit explained. "I thought it was a joke! Where's a kid like that gonna get fifty grand?"
_Goddamn it, Frenchy_.
I stepped forward before Frenchy could talk.
"He was just joking! Everybody calm down. Let's relax. No more pool. Okay, Frenchy?"
I took the pool cue out of his hand. Frenchy started to say something until I glared at him. Alex pried the jacket from his hands and dropped it on the pool table. The jacket fell with a thud on the green felt.
"Maybe you boys don't fucking get it," Billy slurred loudly. "Those colors are not to be fucked with. We've killed guys just for disrespecting 'em."
The crowd of Holy Ghosts surrounded us in a tight circle.
"Maybe we should go?" I said. "My buddy is obviously a bit drunk."
"I'm not drunk," Frenchy protested.
"Yes, you are," Alex said.
He grabbed Frenchy's arm and walked him toward the door.
Backwoods Billy stood in the middle of the bar thinking. He swayed a bit and mumbled to himself about honor and God. Any second I expected him to give the order for the remaining Holy Ghosts to tear us apart. I pulled a fifty-dollar bill from my pocket and handed it to him.
"Here. I want to buy you guys some drinks. You know, as a way of saying I'm sorry for the way my friend acted."
Billy held the bill up and stared at it. Then he smiled.
"Well, that's mighty fine of you, Patrick. Listen, you boys are welcome here any time."
If I was lucky enough to get out of there alive once, I sure wasn't coming back.
"Thanks, Billy."
"God bless," he said as I walked away.
The last I saw of him he stood in the glow of the Budweiser light over the pool table, bandanna over his bandage, waving one arm wildly while preaching to a crowd of dirty bikers with their heads bent down. The safe sat, discarded, in the corner.
"You didn't have to buy that old bastard a drink," Alex said as we pulled out of the parking lot.
"I can afford it," I grinned. "Besides, it might be his last one for a long fucking time."
# TWENTY-FIVE
#
SO MUCH _HAWAII FIVE-O_ A MAN CAN WATCH. KEITH LOOKED LIKE HE'D HIT HIS LIMIT AND CRACKED A BIG GRIN THAT AFTERNOON AS HE WALKED THROUGH THE METAL DOOR AND INTO THE DISCHARGE AREA OF BALTIMORE COUNTY JAIL. HE WORE THE CLOTHES HE HAD ON WHEN THEY ARRESTED HIM: A RATTY BLACK T-SHIRT AND DIRTY JEANS, ONLY NOW THE JEANS SPORTED A GIANT STAIN IN THE FRONT.
"What happened to your jeans, Keith?" Alex asked as he opened the car door to let Keith into the backseat.
"I guess I must have pissed myself when they busted me," Keith shrugged. He stared at the stain for a second. "That just sucks."
Not even a piss stain could ruin Keith's mood. He glowed in the backseat of my car as we pulled away even if he couldn't figure out why they'd let him go.
While I drove to Frenchy's, Alex filled Keith in on everything that went down while he was locked up. The trip to New York, the empty safe deposit box, Danny pulling the gun and our getaway. We didn't mention the money.
Frenchy sat in his basement playing guitar. He'd bought a case of beer and a pizza for Keith's welcome-home bash. We wanted to keep it low-key. Frenchy hugged Keith and handed him a bottle of Miller High Life and we all squeezed onto the tattered couch to eat. Neil Young's _Everybody Knows This Is Nowhere_ played on the turntable.
"Shit, man," Keith said between bites. "Not to be a dick but you guys could have thrown a better party than this. This sorta sucks. Alex got a big-ass bash with chicks and everything. I got pizza, High Life and some Canadian asshole singing about a river or something."
"Next time, serve eight months instead of three days," Alex joked. "Besides, we got a surprise for you."
Keith jumped up from the couch.
"Holy shit! I knew it. You hired strippers! I fucking love you guys."
"Keith! Pay attention. This is important. And you can't tell anyone. Got it?"
"Yeah." Keith nodded.
"I'm serious. You cannot tell anyone. Not one single person."
Keith realized Alex was serious. He dropped back down onto the couch then bit into another piece of pizza.
"Keith. Look at me and tell me you won't say anything about this to anyone."
"Fuck. All right. I won't tell anyone."
"Go ahead and show him, Frenchy," I said.
Frenchy ran up the basement stairs and locked the door. The last thing we needed was his mom seeing the money. He came back, crouched on his knees and pulled the guitar case out from under his sofa. A steel chain rattled beneath the sofa and the guitar case jerked to a halt.
"Oops," Frenchy said. "Forgot I chained it up."
He fished a small key out of his pocket then unlocked the padlock and lugged the case to the middle of the room. We all stared at the case as Frenchy lifted the lid. I'd seen the pile of cash a dozen times and it still made my heart pound. This time I watched Keith's reaction. It hit him slowly then his eyes bugged out and he leapt off the couch.
"Oh my God! Holy shit! How much is there?"
"Two hundred thousand," I said.
"Where did you guys get that? You told me you didn't get nothing off Zeppelin."
"We didn't. Frenchy did. He thought he was stealing one of Jimmy's guitars. Turns out he hit the motherload."
"Shit, man. You guys are fucking rich."
Alex threw his arm around Keith.
"No, man. We're all rich. We're each taking fifty thousand. That includes you."
"Me? What the fuck did I do? I wasn't even there."
"You helped," I said. "We couldn't have gotten into Jimmy Page's room without that Les Paul. You helped us get it."
"And besides," Frenchy added, "if you hadn't been busted that night at the carnival the rest of us might not have gotten away."
"You guys are the fucking best," Keith said, hugging each of us.
We spent the rest of the night getting drunk and stoned and talking about what we were going to buy. Everything from El Caminos to rare comic books to a pet Chihuahua. Keith talked about wanting to play the stock market. He just had to figure out what it was first.
A few days later the cops raided Indian Winds trailer park. Real SWAT-team military-style. They came in early, kicked open trailer doors and caught most of the gang asleep next to their old ladies. Every Holy Ghost went down. Backwoods Billy, Rabbit, Whitey, Sonny and a paddy wagon full of others. The papers said the cops found barrels of pills and an arsenal filled with everything from AK-47s to tear-gas grenades. They also confiscated a crate of Bibles.
Backwoods Billy went nuts. Tore up two holding cells and an interrogation room before they calmed him down. He couldn't figure what went wrong and how his protection fell through. He left Cooper a streak of threats about releasing the tapes. Finally, he got wise and called his wife. She loaded the tapes into their reel-to-reel player and found what I'd recorded over them. I hope she liked Black Sabbath.
They sent Backwoods Billy to the same joint as Danny. The Maryland authorities dragged Danny back from New York and charged him with felony weapons possession and a parole violation. They dropped the robbery charges when they couldn't prove he stole anything from Led Zeppelin. The judge gave him five years. He'd be out in two.
Every newspaper in New York ran a major story on Led Zeppelin being ripped off. The headline read LED ZEPPELIN ROBBED OF 203G. The band stayed in New York an extra day to deal with everything and Zeppelin's manager held a press conference. They even called in the FBI.
The investigation went nowhere, probably because the band insisted the money was stolen from the safe deposit box. I couldn't figure out why they didn't tell the cops the cash was really in Jimmy's guitar case. It made sense later when I read about Zeppelin suing the hotel and winning a settlement.
My luck in Maryland was running out. I could feel it. And I knew a few of the lesser Holy Ghosts would be back on the street soon. I didn't want to be around when it happened. I went back to New York City. The new punk rock bands like the Ramones, the Dead Boys and the New York Dolls were taking off. I bought a nice camera with part of the money and took a few photos. Frenchy started a band. They weren't very good but now and then they played New York City.
Emily visited a lot. She was saving money and planned to move to New York City and take classes at NYU. She wanted to be a lawyer, just in case I ever needed one. I thought that was a great idea.
I didn't think about the robbery much. Just sort of pushed it to the back of my mind until one day, when I was shopping at a record store in the Village. Led Zeppelin's "Whole Lotta Love" played on the stereo. An older black man with a stack of blues records under his arm walked up to the counter.
"What the heck is this noise?" he asked, pointing toward the turntable. "What you need to do is play the originals."
"What do you mean?" the young clerk mumbled. He sounded incredibly stoned.
"The guys that Zeppelin stole these songs from. You know, Willie Dixon, Sleepy John Estes, Howlin' Wolf, Bukka White..."
"I don't know what you're talking about, buddy. This is Zeppelin."
The old man sighed. He'd obviously gone over this before.
"Someday, somebody's gonna clean these British boys out for what they've done," he said. Then he was gone.
I grinned and walked toward the register, Black Sabbath's new album under my arm and a fat wad of cash in my hand.
#
Thanks to:
My girl Mel Gorski for understanding all those late nights of loud music, clanging beer bottles, and being woken up at 4 a.m. to be asked, "Does this suck?" I love you madly.
Tammy Buhrmester for tolerating decades of my ramblings, tantrums, craziness and "weird" music.
Ryan Lodge for creating so many of the riduculous situations in this dumb little book and for being my brother forever.
Kevin White for damn near everything. But mostly for understanding the utter fucking importance of Metallica.
Jesse Howard for all the shit talking and Wild Turkey. You're proof that nice boys really don't play rock 'n' roll.
Raquel Lauren and Bradley Peterson for that one perfect trip to Coney Island in a ′71 Dodge.
And, of course, Nico for sitting on my lap the entire time and being the original black dog. You're the best friend a guy could ever ask for.
Special thanks to Matt "Kid Legs" Render, Kevin Schulz, Erik "Dry Gulch" Byrne, "Lil" Kenny Coulman, Rob Loudon, Wil "I'm Damaged" O'Neal, Zach Medearis, Matt Bertz, Ben Poskins, Alex "Axe Man" Luna, Bryan Joseph, and anyone else who's lunatic behavior may have found its way into these pages somehow. You all need professional help.
#
**JASON BUHRMESTER** was born in Kankakee, Illinois, which was voted the "Worst City in North America" by _The Places Rated Almanac_. Jason started his writing career in the customer service department at _Playboy_ magazine and has since contributed to _Playboy, Maxim, Spin, Village Voice, Wired, Giant, FHM, Penthouse_ and other publications. He is currently editor at _Inked_ magazine and lives in Brooklyn with his wife, Melissa, and black pug, Nico. He feels that "Sabbath Bloody Sabbath" is Black Sabbath's best song. Visit jasonbuhrmester.com to disagree with him.
THIS IS A WORK OF FICTION. NAMES, CHARACTERS, PLACES, AND
INCIDENTS EITHER ARE THE PRODUCT OF THE AUTHOR'S IMAGINATION
OR ARE USED FICTITIOUSLY. ANY RESEMBLANCE TO ACTUAL PERSONS,
LIVING OR DEAD, EVENTS, OR LOCALES IS ENTIRELY COINCIDENTAL.
COPYRIGHT © 2009 BY JASON BUHRMESTER
ALL RIGHTS RESERVED.
PUBLISHED IN THE UNITED STATES BY THREE RIVERS PRESS, AN
IMPRINT OF THE CROWN PUBLISHING GROUP, A DIVISION OF RANDOM
HOUSE, INC., NEW YORK.
WWW.CROWNPUBLISHING.COM
THREE RIVERS PRESS AND THE TUGBOAT DESIGN ARE REGISTERED
TRADEMARKS OF RANDOM HOUSE, INC.
LIBRARY OF CONGRESS CATALOGING-IN-PUBLICATION DATA
BUHRMESTER, JASON,
BLACK DOGS / JASON BUHRMESTER.-1ST ED.
P. CM.
1. LED ZEPPELIN (MUSICAL GROUP)—FICTION 2. MUSICIANS—CRIMES
AGAINST—FICTION. 3. TAXICAB DRIVERS—FICTION.
1. TITLE.
PS3602.B36B57 2009
813′.6–dc22
2008050094
eISBN: 978-0-307-45202-3
v3.0
|
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\section{Introduction}
It is no doubt that well-developed representation theories
are necessary when we apply algebraic objects to physics. The
simplest examples in quantum physics are angular momentum algebra
$ su(2) $ and rotation matrices in 3 dimensional space $ SO(3). $
Their algebraic structure is simple but contents of representation
theories are quite rich \cite{BieL8}. To investigate these algebraic
objects or their complexification could be a foundation for further
investigation of higher dimensional objects.
As for deformation of Lie groups and Lie algebras, $q$-deformation
of Lie algebra $ sl(2) $ and Lie group $ SL(2) $ (and their real form)
is studied quite well. Their representation theories have attracted much interest
in both physics and mathematics and give a way to higher dimensional cases
\cite{BLohe1}. There exists, however, some other deformation of Lie
groups and algebras and these are generally called nonstandard deformation.
The most studied one may be the so-called Jordanian deformation obtained by
Drinfeld twist from a Lie algebra or a known quantum algebra.
The simplest examples are, of course, the Jordanian deformation of Lie algebra
$ sl(2) $ and its dual. The Jordanian deformation of Lie group $ SL(2) $,
denoted by $ SL_h(2) $, is
studied in \cite{dem,zak,ewen} and then
Ohn introduced its dual algebra, namely, Jordanian deformation of
$ sl(2) $ denoted by $ {\cal U}_h(sl(2)) $ \cite{ohn}. The Jordanian quantum algebra
$ {\cal U}_h(sl(2)) $ is more natural than the $q$-deformed $ sl(2) $ in the sense that
it is regarded as the angular momentum algebra with nonstandard coproduct
(see \S 3) and we can use ordinary boson operators to represent $ {\cal U}_h(sl(2)) $,
while it is hard to regard the $q$-deformed $ sl(2) $ as angular momentum
and $q$-deformed boson algebras are used for representations
\footnote[2]{There exist mappings from the ordinary boson algebra to
$q$-deformed ones \cite{btoq,fiore,fiore2}. It is, however, simpler to use
the $q$-deformed boson algebras for representation theories.}.
However, the representation theories of $ {\cal U}_h(sl(2)) $ and $ SL_h(2) $ have
not been developed yet. We do not know, for example, the Racha coefficients and
matrix elements of the universal $R$-matrix for $ {\cal U}_h(sl(2)) $. As for $ SL_h(2) $,
even its representation matrices are not obtained.
In this article, in order to develop representation theories for
Jordanian deformed algebras, we study symplecton
for $ {\cal U}_h(sl(2)) $ and apply it to investigate representation matrices of $ SL_h(2) $.
The use of symplecton could be legitimated by recalling the properties of
symplecton and $q$-deformed case.
The symplecton, introduced by Biedenharn and Louck \cite{BieL,BieL2},
is a polynomial of boson creation and annihilation operators
which form an irreducible
tensor operator of $ sl(2) $, that is, symplecton is a basis of irreducible
representation (irrep.) for both $ sl(2) $ and $ SL(2). $ It is known that the
symplecton is written in terms of Gauss hypergeometric function and product of
two symplecton is reduced to a series of symplecton with Racha coefficients.
In Ref.\cite{BieL}, application of symplecton to the Elliot model for
nuclei is discussed, then it is found that Weyl-ordered polynomials for
position and momentum operators are equivalent to symplecton \cite{LBL}.
Many properties of symplecton are inherited from $ sl(2) $ to $q$-deformed
case \cite{BLohe1,BLbook,Nom}. The $q$-deformed symplecton,
called $q$-symplecton, is a
irreducible tensor operator so that it is a irrep. basis for
$q$-deformed $sl(2)$ and $ SL(2) $. The $q$-symplecton is written in
terms of $q$-hypergeometric function and product of two $q$-symplecton
is reduced to a series of $q$-symplecton with $q$-Racha coefficients. $q$-Deformation of Weyl-ordered polynomial
\cite{GelF} is formulated with $q$-symplecton.
These facts show that symplecton is a powerful tool to investigate
representation.
The plan of this article is as follows. Next three sections are mainly
preparation for symplecton of $ {\cal U}_h(sl(2)) $. We often call the symplecton for $ {\cal U}_h(sl(2)) $
$h$-symplecton. The next section is a review of symplecton for $ sl(2). $
Some of the properties of symplecton listed in \S 2 will be extended to
$h$-symplecton. \S 3 is devoted to the Jordanian quantum algebra $ {\cal U}_h(sl(2)) $ and
Jordanian quantum group $ SL_h(2) $. We give new results on the twist element
and Racha coefficients for $ {\cal U}_h(sl(2)). $ In \S 4, tensor operators for a Hopf
algebra is introduced according to Ref.\cite{rs} and the relation between
tensor operators for a Lie algebra and a Hopf algebra obtained by
Drinfeld twist is discussed. Applying the result in \S 4, the $h$-symplecton is
constructed from the $ sl(2) $ symplecton in \S 5. The properties of
$h$-symplecton are studied in \S 5 and \S 6. We shall consider
another irreducible tensor operators
obtained from the quantum $h$-plane for $ {\cal U}_h(sl(2)) $ in \S 7 and using
these tensor operators, as well as $h$-symplecton, irreps. of
$ SL_h(2) $ are considered. \S 8 is concluding remarks.
\section{Symplecton for $ sl(2) $}
The symplecton realization of $ sl(2) $ is said to be "minimal",
since only one kind of boson operator is used. It is in marked contrast
to the well-known Jordan-Schwinger realization where two kinds of bosons
are necessary. Let us first review the definition and important properties
of the $ sl(2)$ symplecton \cite{BieL, BieL2}.
Let $ \bar a, a $ be boson operators satisfying $ [\bar a,\; a] = 1$,
and define
\begin{equation}
J_+ = -\frac{1}{2} a^2, \qquad J_- = \frac{1}{2} \bar a^2,
\qquad
J_0 = \frac{1}{2}(a \bar a + \bar a a).
\label{sl2real}
\end{equation}
It is easy to verify that (\ref{sl2real}) satisfies the $ sl(2) $ commutation
relations
\begin{equation}
[J_0,\; J_{\pm}] = \pm 2 J_{\pm}, \qquad
[J_+,\; J_-] = J_0. \label{sl2comm}
\end{equation}
The symplecton is a polynomial in $ \bar a$ and $ a $ and
form a irreducible tensor operator of $ sl(2) $ belonging to the
spin $j$ representation ($ j = \frac{1}{2}, 1, \frac{3}{2}, \cdots $). Namely
the symplecton, denoted by $ P_j^m(a, \bar a), $ is defined by
\begin{eqnarray}
& &
[J_{\pm},\; P_j^m] = \sqrt{(j\mp m) (j\pm m +1)} P_j^{m\pm 1}, \nonumber \\
& &
[J_0,\; P_j^m] = 2m P_j^m. \label{defsl2}
\end{eqnarray}
The basic idea of symplecton is to treat $ \bar a $ and $ a$ in a
symmetric way. To this end, the usual "boson calculus" is replaced with
the so-called "symplecton calculus", that is, instead of the boson
vacuum $ \ket{0} $ satisfying $ \bar a \ket{0} =0 $,
the formal ket $ \ket{\ } $
which is not annihilated by both $ \bar a $ and $ a $ is introduced.
The representation bases in the realization (\ref{sl2real}) are
formed by letting $ P_j^m $ act on $ \ket{\ } $, and the action of
generators on the bases is defined by
$ J_{\alpha} \ket{jm} = [J_{\alpha},\; P_j^m] \ket{\ }$.
There exists an appropriate definition of a inner product for these
$ \ket{jm} $, so that we obtain the usual unitary representations of
$ sl(2) $ with spin $j$.
The explicit form of the polynomials $ P_j^m(a, \bar a) $ is found by
solving $ [J_+,\; P^j_j ] = 0 $ to obtain $ P_j^j = a^{2j} $, and then
using the action of $ J_- $ to calculate $ P_j^m $,
\begin{equation}
P_j^m(a,\bar a) =
\frac{1}{2^{j-m}} \left[ \frac{(2j)! (j-m)!}{(j+m)!} \right]^{1/2}
\sum_{s=0}^{j-m}\;
\frac{\bar a^{j-m-s} a^{j+m} \bar a^s}{s! (j-m-s)!}_.
\label{exp1}
\end{equation}
An alternative form for $ P_j^m $ is obtained by starting with
$ P_j^{-j} = \bar a^{2j} $ and
then using the action of $ J_+ $,
\begin{equation}
P_j^m(a,\bar a) =
\frac{1}{2^{j+m}} \left[ \frac{(2j)! (j+m)!}{(j-m)!} \right]^{1/2}
\sum_{s=0}^{j+m}\;
\frac{a^s \bar a^{j-m} a^{j+m-s}}{s! (j+m-s)!}_.
\label{exp2}
\end{equation}
We would like to list some properties of $ sl(2) $ symplecton. For their
proof or detail, we refer the reader to Refs.\cite{BieL, BieL2}.
\noindent
(1) A set of polynomials $ \{P_j^m(a,\bar a) \ | \ m = -j, -j+1, \cdots, j\}$
forms representation bases for
the Lie group $ SL(2) $ as well as the Lie algebra $ sl(2) $. The boson
commutation relation is covariant under the action of $ SL(2) $ defined by
\begin{equation}
(a', \; \bar a') = (a, \; \bar a) \left(
\begin{array}{cc}
x & u \\ v & y
\end{array}
\right)_, \label{actionSL2}
\end{equation}
where the $ 2 \times 2 $ matrix is an element of $ SL(2) $.
The transformed polynomial $ P_j^m(a', \bar a') $ is decomposed into
$ P_j^m(a, \bar a) $ multiplied by
polynomials in the entries of $ SL(2) $ matrix.
\begin{equation}
P_j^m(a', \bar a') = \sum_n P_j^n(a,\bar a) d^j_{nm}(g).
\quad
g \in SL(2) \label{dfunc}
\end{equation}
The $ (2j+1) \times (2j+1) $ matrix $ d^j_{nm}(g) $ gives an irrep. of $ SL(2) $
and is called Wigner's $d$-function in terminology of physics.
\noindent
(2) The polynomials $ P_j^m(a, \bar a) $ have a generating function. Let
$ \xi, \eta $ be ordinary c-numbers commuting with $ a, \bar a$. Then
\begin{equation}
(\xi a + \eta \bar a)^{2j} = \sqrt{(2j)!} \sum_{m=-j}^j\;
\Phi_{jm}(\xi, \eta) P_j^m(a, \bar a),
\label{genfun}
\end{equation}
where $ \Phi_{jm} $ are well-known representation bases of both $ sl(2) $
and $ SL(2) $,
\begin{equation}
\Phi_{jm}(\xi,\eta) = \frac{\xi^{j+m} \eta^{j-m}}{\sqrt{(j+m)! (j-m)!}}_.
\label{phi}
\end{equation}
Irreps. of $ sl(2) $ are constructed on (\ref{phi}) by the realization
\begin{equation}
J_+ = \xi \frac{d}{d\eta}, \quad \
J_- = \eta \frac{d}{d\xi}, \quad \
J_0 = \xi \frac{d}{d\xi} - \eta \frac{d}{d\eta},
\label{phisl2}
\end{equation}
while irreps. of $ SL(2) $ are obtained by the following transformation
\begin{equation}
(\xi', \; \eta') = (\xi, \; \eta) \left(
\begin{array}{cc}
x & u \\ v & y
\end{array}
\right)_, \label{phiSL2}
\end{equation}
it follows that
\begin{equation}
\Phi_j^m(\xi', \eta') = \sum_n \Phi_j^n(\xi, \eta) \; d^j_{nm}(g),
\label{phiSL2x}
\end{equation}
where we have obtained the same $d$-function as (\ref{dfunc}).
\noindent
(3) The symplecton polynomials can be expressed in terms of
Gauss hypergeometric function $ {}_2F_1(a,b;c;z) $. The polynomial
$ {}_2F_1(a,b;c;z) $ is defined by
\begin{equation}
{}_2 F_1(a, b; c; z) =
\sum_{n=0}^{\infty}\; \frac{(a)_n (b)_n}{n! (c)_n}z^n, \label{2f1}
\end{equation}
where $ (a)_n $ stands for the sifted factorial
\begin{equation}
(a)_n = \left\{
\begin{array}{cl}
1 & n = 0 \\
a(a+1) \cdots (a+n-1) & n = 1, 2, \cdots
\end{array}
\right. \label{siffac}
\end{equation}
Now define the operator $ N = a \bar a $, then the symplecton
$ P_j^m(a, \bar a) $ is written in terms of $ {}_2F_1(a,b;c;z) $ with
$ z = -1 $ and the parameters $ a,c $ become functions of operator $N$.
The expression (\ref{exp1}) becomes
\begin{eqnarray}
P_j^m &=& \frac{1}{2^{j+m}}\left[
\frac{(2j)!}{(j+m)!(j-m)!} \right]^{1/2} \frac{(N+j-m)!}{(N-2m)!}
\label{F21} \\
& \times & {}_2F_1(-N+2m, -j+m; -N-j+m; -1) (\bar a)^{-2m}. \nonumber
\end{eqnarray}
In this way, properties of $ P_j^m $ are reduced to properties of the hypergeometric
function. Especially, the equivalence of two form (\ref{exp1}) and (\ref{exp2})
is explained by the formula
\begin{equation}
{}_2F_1(a, b; c; z) = (1-z)^{c-a-b} {}_2F_1(c-a, c-b; c; z).
\label{F21formula}
\end{equation}
\noindent
(4) The polynomials $ P_j^m(a, \bar a) $ are transformed under the
action $ a \rightarrow \bar a, \ \bar a \rightarrow -a $
\begin{equation}
P_j^m(\bar a, -a) = (-1)^{j-m} P_j^{-m}(a, \bar a). \label{adj}
\end{equation}
To define an inner product for the bases $ \ket{jm} = P_j^m \ket{\ } $,
the property (\ref{adj}) and the product formula discussed below play
a crucial role.
\noindent
(5) Let $ P_j^m $ and $ P_{j'}^{m'} $ be the symplecton polynomials, then
they obey the product law
\begin{equation}
P_j^m P_{j'}^{m'} = \sum_{k=|j-j'|}^{j+j'}\;
\bra{k}j\ket{j'} C_{m',m,m+m'}^{j',\; j,\quad \ k}
P_k^{m+m'}, \label{product}
\end{equation}
where
\begin{eqnarray}
& & \bra{k}j\ket{j'} = 2^{k-j-j'} (2k+1)^{-1/2} \nabla(kjj'), \nonumber \\
& & \nabla(abc) = \left[
\frac{(a+b+c+1)!}{(a+b-c)! (a-b+c)! (-a+b+c)!} \right]^{1/2}_, \label{tri}
\end{eqnarray}
and $ C_{m',m,m+m'}^{j',\; j,\quad \ k} $ is the
Clebsch-Gordan coefficient (CGC) for $ sl(2). $
The associativity of the products
$
(P_a^{\alpha} P_b^{\beta}) P_c^{\gamma}
=
P_a^{\alpha} (P_b^{\beta} P_c^{\gamma})
$
gives a relation between "triangle functions"
\begin{equation}
\nabla(acf) \nabla(bdf) = (2f+1) \sum_e W(abcd;ef) \nabla(abe) \nabla(cde) ,
\label{racha}
\end{equation}
where $ W(abcd;ef) $ is the Racha coefficient.
The inner product for $ \ket{jm} $ is defined by
\begin{equation}
\braket{jm}{j'm'} = \bra{\ }(-1)^{j-m}P_j^{-m} \cdot P_{j'}^{m'} \ket{\ },
\label{inner}
\end{equation}
and the operation $ \bra{\ }(\cdots) \ket{\ } $ means to take only the $ j= 0 $
part of the expression $ (\cdots) $. Applying the product law (\ref{product})
to the RHS of (\ref{inner}), we see that that the $j=0 $ part is given by
the CGC $ C^{j,j',0}_{m,m',0} $, so that the bases $ \ket{jm} $ are
orthonormal.
\setcounter{equation}{0}
\section{Jordanian Deformation of $ sl(2) $ and $ SL(2) $}
The Jordanian quantum algebras $ {\cal U}_h({\bf g}) $ are obtained
from the (universal enveloping algebra ${\cal U}({\bf g}) $ of ) Lie algebras $ {\bf g} $ from
Drinfeld twist \cite{dri}. We denote the coproduct, conuit and
antipode for $ {\cal U}({\bf g}) $, when it is regarded as a Hopf algebra,
by $ \Delta, \epsilon, S $, respectively.
With the invertible element
$ {\cal F} \in {\cal U}({\bf g}) \otimes {\cal U}({\bf g}) $ satisfying
\begin{eqnarray}
& & (\epsilon \otimes id)({\cal F}) = (id \otimes \epsilon)({\cal F}) = 1,
\label{twist1} \\
& & {\cal F}_{12} (\Delta \otimes id)({\cal F}) = {\cal F}_{23}(id \otimes \Delta)({\cal F}),
\label{twist2}
\end{eqnarray}
the algebra $ {\cal U}_h({\bf g}) $ is defined by the same
commutation relations as $ {\bf g} $
and the following Hopf algebra mappings
\begin{equation}
\tilde \Delta = {\cal F} \Delta {\cal F}^{-1}, \qquad
\tilde \epsilon = \epsilon, \qquad
\tilde S = u S u^{-1}, \label{defHopf}
\end{equation}
where $ u = m(id \otimes S)({\cal F}), \ u^{-1} = m(S \otimes id)({\cal F}^{-1}) $,
$ m $ denotes the usual product in $ {\bf g}. $ This is a triangular
Hopf algebra whose universal $R$-matrix is given by $ {\cal R} = {\cal F}_{21} {\cal F}^{-1}. $
For the case of $ {\bf g} = sl(2) $, $ {\cal F} $ is given by \cite{ks}
\begin{equation}
{\cal F} = \exp\left(-\frac{1}{2} J_0 \otimes \sigma \right),
\qquad
\sigma = -\ln(1 -2hJ_+), \label{sigma}
\end{equation}
The twist element $ {\cal F} $ used here gives different form of $ {\cal U}_h(sl(2)) $ from
the one in Ref.\cite{ohn}. The relationship between these two form
is given in Appendix A.
The explicit form of Hopf algebra mappings for $ {\cal U}_h(sl(2)) $ is summarized in
Appendix B (some of them will be used in the later computation).
An application of the $ {\cal U}_h(sl(2)) $ to the Heisenberg spin chain is found
in Ref. \cite{ks}.
The finite dimensional highest weight irreps. for $ {\cal U}_h(sl(2)) $ are same
as $ sl(2) $, because of the same commutation relations.
We shall use the following lemmas on tensor product representations
in subsequent sections.
\begin{lemma}{\rm \cite{ks}}
Let $ V^{j_1}, V^{j_2} $ be the representation space with the highest
weight $ j_1, j_2 $. Then the tensor product of them is completely reducible,
$i.e.$
\[
V^{j_1} \otimes V^{j_2} = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} V^j,
\]
and the bases of $ V^j $ are given by
\begin{equation}
e^{(j_1j_2)j}_m = \sum C^{\; j_1,\; j_2,\; j}_{m_1,m_2,m}
F^{\ \ \ j_1,j_2}_{k_1,k_2,\ m_1,m_2} e^{j_1}_{k_1} \otimes e^{j_2}_{k_2},
\label{tensor}
\end{equation}
where $ C^{\; j_1,\; j_2,\; j}_{m_1,m_2,m} $ is the
CGC of $ sl(2) $ and
$ F^{\ \ \ j_1,j_2}_{k_1,k_2,\ m_1,m_2} $ is the matrix element of
$ {\cal F} $ on $ V^{j_1} \otimes V^{j_2}. $
\label{lem1}
\end{lemma}
The explicit form of matrix elements $ F^{\ \ \ j_1,j_2}_{k_1,k_2,\ m_1,m_2} $
is given in Appendix C (It seems to be the first time to show the
explicit form of $ F^{\ \ \ j_1,j_2}_{k_1,k_2,\ m_1,m_2} $ in the literature,
and this also gives the explicit form of the $R$-matrix for $ {\cal U}_h(sl(2)) $).
\begin{lemma}
The Racha coefficients for $ sl(2) $ and $ {\cal U}_h(sl(2)) $ coincide.
\label{lem2}
\end{lemma}
Lemma \ref{lem2} is proved in Appendix D.
The matrix quantum group dual to $ {\cal U}_h(sl(2)) $ is called the Jordanian quantum
group $ SL_h(2) $. It is generated by four elements $ x, y, u $ and $ v $
subject to the relations \cite{dem,zak,ewen}
\begin{eqnarray}
& & [v,\; x] = hv^2, \qquad [u,\; x] = h(1-x^2),
\nonumber \\
& & [v,\; y] = h v^2, \qquad [u,\; y] = h(1-y^2),
\label{SLh2} \\
& & [x,\; y] = h(xv - yv), \qquad [v,\; u] = h(xv + vy).
\nonumber
\end{eqnarray}
It follows that the central element of $ SL_h(2) $
which gives the determinant of the quantum
matrix
\begin{equation}
T = \left(
\begin{array}{cc}
x & u \\ v & y
\end{array}
\right)_, \label{quantumM}
\end{equation}
is defined by
\begin{equation}
detT = xy - uv - h xv = 1. \label{det}
\end{equation}
The $ SL_h(2) $ has a Hopf algebra structure.
The relations (\ref{SLh2}) and Hopf
algebra mappings are summarized in the FRT-formalism \cite{frt}
with the $R$-matrix
\begin{equation}
R = \left(
\begin{array}{rrrr}
1 & h & -h & h^2 \\
0 & 1 & 0 &h \\
0 & 0 & 1 & -h \\
0 & 0 & 0 & 1
\end{array}
\right)_. \label{SLh2R}
\end{equation}
The coproduct, the counit and the antipode are given by
\begin{eqnarray}
& & \Delta(T) = T \stackrel{\cdot}{\otimes} T, \nonumber \\
& & \epsilon(T) = \left(
\begin{array}{cc}
1 & 0 \\ 0 & 1
\end{array}
\right)_, \nonumber \\
& & S(T) = T^{-1} = \left(
\begin{array}{ccc}
y - hv & \ \ &-u -h(y-x) + h^2 v \\
-v & & x+hv
\end{array}
\right)_. \nonumber
\end{eqnarray}
Let us define the $d$-function for $ SL_h(2) $ using the notion of
comodule. A vector space $M$ is called right $SL_h(2)$ comodule
if there is a map $ \rho : M \rightarrow M \otimes SL_h(2) $ such that
the following relations are satisfied
\begin{equation}
(\rho \otimes id)\circ \rho = (id_M \otimes \rho) \circ \rho,
\qquad
(id_M \otimes \epsilon) \circ \rho = id_M,
\label{comodule}
\end{equation}
where $ id_M $ stands for the identity map in $M$. Using bases $ e_i $ of
$ M$, the map $ \rho $ is written as
\begin{equation}
\rho(e_i) = \sum_j e_j \otimes \tilde d_{ji}, \label{dS1}
\end{equation}
it follows that the relations (\ref{comodule}) are rewritten as
\begin{equation}
\Delta(\tilde d_{ij}) = \sum_k \tilde d_{ik} \otimes \tilde d_{kj},
\qquad
\epsilon(\tilde d_{ij}) = \delta_{ij}.
\label{dS2}
\end{equation}
We call the $ \tilde d_{ij} $ satisfying (\ref{dS1}) and (\ref{dS2})
the $d$-function for $ SL_h(2) $. In the following sections, we
deal with the case in which the vector space $M$ has an algebraic structure.
It is natural, in this case, to require that the map $ \rho $
should respect the extra structure on $M$.
\setcounter{equation}{0}
\section{Tnsor Operators and Twist}
To define the symplecton for $ {\cal U}_h(sl(2))$, it is necessary to
extend the notion of tensor operators to Hopf algebra. This has
been carried out by Rittenberg and Scheunert \cite{rs}.
Tensor operators are defined for each realization of the Hopf
algebra $ {\cal H} $ under consideration. Assuming that we have a realization
of $ {\cal H} $, we first define the adjoint action.
\begin{DEF}
Let $ W, W'$ be representation space of $ {\cal H} $, and let $t$ be an
operator which carries $ W $ into $W'$. Then the adjoint action of $ X \in {\cal H} $
on $t$ is defined by
\begin{equation}
adX(t) = m(id \otimes S)(\Delta(X)(t \otimes 1)). \label{adj2}
\end{equation}
\end{DEF}
The adjoint action has two important properties
\begin{equation}
ad XX'(t) = ad X \circ ad X'(t), \qquad
ad X(t \otimes s) = \sum_i ad X_i(t) \otimes ad X'_i(s), \label{adjpro}
\end{equation}
where the coproduct for $X$ is written as
$ \Delta(X) = \sum_i X_i \otimes X'_i.$
From these properties, we see that the adjoint action gives a
representation of $ {\cal H} $
\begin{equation}
ad[X,\; X'] (t) = [ad X,\; ad X'](t).
\end{equation}
Tensor operators for $ {\cal H} $ are defined as operators which form
representation bases
of $ {\cal H} $ under the adjoint action.
\begin{DEF}
Let $ D(X) $ be a
representation matrix of $ X \in {\cal H} $. The operators $ t_{\alpha} $
are called the tensor operator, if they satisfy the relation
\begin{equation}
ad X(t_{\alpha}) = \sum_{\beta} D(X)_{\beta \alpha} t_{\beta}. \label{top}
\end{equation}
If the representation
is irreducible, the tensor operators are called irreducible tensor operators.
\end{DEF}
The explicit form of the adjoint action for $ {\cal U}_h(sl(2)) $ reads
\begin{eqnarray}
& & adJ_0(t) = [J_0, \; t] e^{-\sigma}, \nonumber \\
& & adJ_+(t) = e^{-\sigma}[J_+e^{\sigma},\; t], \label{adjsl2} \\
& & adJ_-(t) = [J_- + hJ_0 + \frac{h}{2}J_0^2,\; t]e^{-\sigma}
-h[J_0,\; t]e^{-2\sigma}
-\frac{h}{2}[J_0,\; [J_0,\; t]]e^{-2\sigma}. \nonumber
\end{eqnarray}
Some examples of the $ {\cal U}_h(sl(2)) $ tensor operators are considered in Ref.\cite{na1}
and they are applied to construct boson algebra which is covariant
under the action of Jordanian matrix quantum groups \cite{ques1}.
Since the coproduct for Lie algebra and Jordanian quantum algebra is related via twist element
(\ref{defHopf}), tensor operators for these algebras are also
related by twisting via $ {\cal F} $ \cite{fiore}.
\begin{lemma}
Let $ t_{\alpha} $ be tensor operators for Lie algebra $ {\bf g} $ and
$ \tilde t_{\alpha} $ be corresponding ones for Jordanian quantum
algebra $ {\cal U}_h({\bf g}). $ Then these tensor operators are
related via the twist element $ {\cal F} $
\begin{eqnarray}
& & \tilde t_{\alpha} = m(id \otimes \tilde S) ({\cal F} (t_{\alpha} \otimes 1) {\cal F}^{-1}),
\label{tt1} \\
& & t_{\alpha} = m(id \otimes S) ({\cal F}^{-1} (\tilde t_{\alpha} \otimes 1) {\cal F}).
\label{tt2}
\end{eqnarray}
\label{lem3}
\end{lemma}
\textit{Proof} : The first relation (\ref{tt1}) is derived in Ref.\cite{fiore}
(Proposition 3). The second one (\ref{tt2}) is its inverse. The expression
used in Lemma \ref{lem3} is different form Ref.\cite{fiore}, it may be good to
show the second relation as an example of the proof.
It is proved by showing the substitution of (\ref{tt2}) into (\ref{tt1}) gives
the identity map.
Let us write the twist element and its inverse as
\[
{\cal F} = \sum f^a \otimes f_a, \qquad
{\cal F}^{-1} = \sum g^a \otimes g_a,
\]
then
\[
u = \sum f^a S(f_a),
\quad
u^{-1} = \sum S(g^a) g_a,
\]
and the relation (\ref{tt1}) becomes
\begin{equation}
\tilde t_{\alpha}
= \sum f^a t_{\alpha} g^b \tilde S(f_a g_b)
= \sum f^a t_{\alpha} g^b u S(f_a g_b) u^{-1}
= \sum f^a t_{\alpha} S(f_a) u^{-1}, \label{one}
\end{equation}
where we used
\[
\sum g^b u S(g_b) = \sum g^b f^a S(g_b f_a) = m(id \otimes S)({\cal F}^{-1} {\cal F}) = 1.
\]
On the other hand, the relation (\ref{tt2}) is rewritten
\begin{equation}
t_{\alpha} = \sum g^a \tilde t_{\alpha} f^b S(g_a f_b)
= \sum g^a \tilde t_{\alpha} u S(g_a).
\label{two}
\end{equation}
Substituting (\ref{two}) into (\ref{one})
\[
\tilde t_{\alpha} = \sum f^a g^b \tilde t_{\alpha} u S(f_a g_b) u^{-1}
= \sum f^a g^b \tilde t_{\alpha} \tilde S(f_a g_b)
= m(id \otimes \tilde S)({\cal F} {\cal F}^{-1} (\tilde t_{\alpha} \otimes 1))
= \tilde t_{\alpha}.
\]
This proves the second relation in Lemma \ref{lem3}. \hfill $\Box$
\setcounter{equation}{0}
\section{Symplecton Polynomials for $ {\cal U}_h(sl(2)) $}
In this section, we derive the explicit form of the symplecton for
$ {\cal U}_h(sl(2)) $ and investigate its properties. Since $ {\cal U}_h(sl(2)) $ has the
same commutation relations as $ sl(2) $, $ {\cal U}_h(sl(2)) $ and $ sl(2) $
have the same realizations.
Therefore the symplecton realization for $ {\cal U}_h(sl(2)) $,
which is identical to the one for $ sl(2),$
is the realization in terms of the usual boson operators.
This is a contrast to the $q$-symplecton where the $q$-deformed boson
operators are used.
Let $ \bar a $ and $ a $ be boson operators satisfying $ [\bar a, \; a] = 1 $,
then the generators of $ {\cal U}_h(sl(2)) $ are realized by
\begin{equation}
J_+ = -\frac{1}{2} a^2, \quad J_- = \frac{1}{2} \bar a^2,
\quad
J_0 = \frac{1}{2}(a \bar a + \bar a a), \label{slh2real}
\end{equation}
The $h$-symplecton, denoted by $ \tilde P_j^m(a, \bar a) $, is
defined as a polynomial in $ \bar a, a $ satisfying
\begin{eqnarray}
& & ad J_{\pm}(\tilde P_j^m) = \sqrt{(j\mp m)(j\pm m+1)} \tilde P_j^{m\pm 1},
\nonumber \\
& & ad J_0(\tilde P_j^m) = 2m \tilde P_j^m, \label{defhsym}
\end{eqnarray}
where the adjont action on the LHS is given by (\ref{adjsl2}).
Using Lemma \ref{lem3},
the explicit form of $h$-symplecton is obtained from the corresponding one
for $ sl(2). $
\begin{prop}
The explicit form of the $h$-symplecton defined by (\ref{defhsym}) is given by
\begin{equation}
\tilde P_j^m(a,\bar a) = P_j^m(a,\bar a) e^{m \sigma}, \label{hsymp}
\end{equation}
where $ \sigma $ is given in (\ref{sigma}) and $ P_j^m(a, \bar a) $
denotes $ sl(2) $ symplecton.
\label{prop1}
\end{prop}
\textit{Proof} : By definition of $ sl(2) $ symplecton, it holds that
\[
(J_0 - 2m) P_j^m = P_j^m J_0.
\]
Using this and the RHS of (\ref{one}),
\[
\tilde P_j^m = \sum_{n=0}^{\infty} \frac{1}{n!} \left(-\frac{1}{2}\right)^n
P_j^m (J_0 + 2m)^n S(\sigma)^n u^{-1}
= P_j^m
\sum_{n=0}^{\infty} \sum_{s=0}^n \frac{(-1)^n (2m)^s}{2^n (n-s)! s!}
J_0^{n-s} S(\sigma)^n u^{-1}.
\]
Changing the order of sum, then replacing $n-s$ with $n$, we obtain
\begin{equation}
\tilde P_j^m = P_j^m \sum_{s,n = 0}^{\infty}
\frac{(-1)^{n+s} (2m)^s}{2^{n+s} n! s!} J_0^n S(\sigma)^{n+s} u^{-1}.
\label{three}
\end{equation}
Note that
\[
u = \sum_{n=0}^{\infty} \left(-\frac{1}{2}\right)^n \frac{1}{n!} J_0^n S(\sigma)^n,
\]
and (\ref{defHopf}), it follows that (\ref{three}) is rewritten as
\[
\tilde P_j^m = P_j^m \sum_{s=0}^{\infty} \left(-\frac{1}{2}\right)^s
\frac{(2m)^s}{s!} \tilde S(\sigma)^s
= P_j^m e^{m\sigma},
\]
where (\ref{sigmaHopf}) is used in the last equality.
\hfill $\Box$
We would like to show some explicit form of $h$-symplecton.
For $ j = 1/2 $
\begin{equation}
\tilde P_{1/2}^{-1/2} = \bar a e^{-\sigma/2} \equiv \bar a_h,\quad
\tilde P_{1/2}^{1/2} = \bar a e^{\sigma/2} \equiv a_h,
\label{jhalf}
\end{equation}
and for $ j = 1 $
\begin{eqnarray}
& & \tilde P_1^{-1} = \bar a^2 e^{-\sigma} = \bar a_h^2 + h\bar a_h a_h, \nonumber \\
& & \tilde P_1^{0} = (\bar a a + a \bar a)/\sqrt{2}
= (\bar a_h a_h + a_h \bar a_h - ha_h^2)/\sqrt{2}, \label{jone} \\
& & \tilde P_1^{1} = a^2 e^{\sigma} = a_h^2. \nonumber
\end{eqnarray}
The $ j=1/2 $ $h$-symplecton forms covariant $h$-deformed oscillator algebra
\begin{equation}
[\bar a_h, \; a_h ] = 1 - h a_h^2, \label{hosc}
\end{equation}
$i.e.$, the commutation relation (\ref{hosc}) is preserved under the action
of $ SL_h(2) $
\begin{equation}
(a_h', \; \bar a_h') = (a_h, \; \bar a_h) \left(
\begin{array}{cc}
x & u \\ v & y
\end{array}
\right)_. \label{action}
\end{equation}
This shows that it is possible to construct representations of $ SL_h(2) $
on $h$-symplecton. We shall discuss it later.
It may be worth noting that the action (\ref{action}) is different from the
ones in \cite{ques1,fiore} where $ a $ and $ \bar a $ are \textit{not}
mixed by the action of quantum groups.
The $ j=1 $ $h$-symplecton forms
an algebra isomorphic to $ sl(2) $. Its commutation relations are
\begin{eqnarray}
& & [ P_1^0,\; P_1^1] = 2\sqrt{2} P_1^1 (1 - hP_1^1), \nonumber \\
& & [ P_1^0,\; P_1^{-1}] = -2\sqrt{2} P_1^{-1}(1 - h P_1^1), \label{j1symp} \\
& & [ P_1^1\; P_1^{-1}] = -2\sqrt{2} (1 - h P_1^1) P_1^0. \nonumber
\end{eqnarray}
The generators of $ sl(2) $ are written in terms of $ P_1^m $
\begin{equation}
J_+ = -\frac{1}{2} P_1^1 (1-hP_1^1), \qquad
J_0 = \frac{1}{\sqrt 2} P_1^0, \qquad
J_- = \frac{1}{2} P_1^{-1} (1 - h P_1^1). \label{JbyP}
\end{equation}
We see, from the explicit form of $h$-symplecton (\ref{hsymp}), that
the $h$ dependence of polynomial $ \tilde P_j^m(a, \bar a) $ is absorbed
in $\sigma $ which is a infinite polynomial in $ a^2 $.
Recall that the relationship
between $ sl(2) $ symplecton and Gauss hypergeometric function $ {}_2F_1 $
is given in terms of the operator $ N = a \bar a $, then we see that
the factor $ e^{m\sigma} $ in (\ref{hsymp}) does not affect
this relationship. Therefore the specific hypergeometric
function for $h$-symplecton may be again $ {}_2F_1 $.
The fact that the $j=1/2$ $h$-symplecton forms covariant $h$-oscillator algebra
may suggest that it is useful to write $h$-symplecton in terms of covariant
$h$-oscillators (\ref{jhalf}).
\begin{prop}
The $h$-symplecton is written in terms of covariant $h$-oscillators as
follows.
The corresponding expression for (\ref{exp1}) is
\begin{eqnarray}
\tilde P_j^m(a_h, \bar a_h) &=&
\frac{1}{2^{j-m}} \left[ \frac{(2j)! (j-m)!}{(j+m)!} \right]^{1/2}
\sum_{s=0}^{j-m}\;
\frac{1}{s! (j-m-s)!} \nonumber \\
&\times &
\bar a_h ( \bar a_h + h a_h) \cdots \{ \bar a_h + (j-m-s-1) h a_h \}
a_h^{j+m} \label{expo1} \\
&\times &
\{ \bar a_h - (2m+s) h a_h \} \{ \bar a_h - (2m+s-1)ha_h \}
\cdots
\{\bar a_h - (2m+1) h a_h \},\nonumber
\end{eqnarray}
and for (\ref{exp2}) is
\begin{eqnarray}
\tilde P_j^m(a_h, \bar a_h) &=&
\frac{1}{2^{j+m}} \left[ \frac{(2j)! (j+m)!}{(j-m)!} \right]^{1/2}
\sum_{s=0}^{j+m}\;
\frac{1}{s! (j+m-s)!} \label{expo2} \\
&\times &
a_h^s (\bar a_h-hsa_h) \{\bar a_h + h(1-s)a_h\} \cdots
\{\bar a_h + h(j-m-1-s)a_h\} a_h^{j+m-s}. \nonumber
\end{eqnarray}
\label{prop2}
\end{prop}
\textit{Proof} : From (\ref{jhalf})
\[
\bar a = \bar a_h e^{\sigma/2}, \qquad
a = a_h e^{-\sigma/2}.
\]
Substituting these into (\ref{exp1}) and (\ref{exp2}), then straightforward
calculation proves the proposition. \hfill $ \Box $
In order to discuss generating functions for $h$-symplecton,
it is possible to apply Lemma \ref{lem3} to the generating
function (\ref{genfun}) for $ sl(2) $ symplecton ,
since the RHS of (\ref{genfun}) is a sum of tensor operators of $ sl(2). $
It follows that the RHS of (\ref{genfun}) becomes the sum of $h$-symplecton ;
$ \displaystyle{ \sqrt{(2j)!} \sum_{m=-j}^j \Phi_{jm} \tilde P_j^m }. $
However the LHS may be quite complicated and may not be in closed form.
Another way to obtain generating functions for $h$-symplecton is to
substitute (\ref{hsymp}) and (\ref{jhalf}) into (\ref{genfun})
\begin{equation}
( \xi a_h e^{-\sigma/2} + \eta \bar a_h e^{\sigma/2} )^{2j}
=
\sqrt{(2j)!} \sum_{m=-j}^j\; \Phi_{jm}(\xi, \eta) \tilde P_j^m(a_h, \bar a_h) e^{-m\sigma}.
\label{hgenfun}
\end{equation}
It is possible to remove $ \sigma $ from (\ref{hgenfun}) by usint
the relation
$
e^{\sigma} + h a_h^2 =1
$, however the obtained
relation is quite complicated. Therefore the simplest generating function for
$h$-symplecton may be (\ref{hgenfun}) where the $ \sigma $ is regarded as
a independent quantity subject to the relations
\begin{equation}
[\sigma,\; a_h] = 0, \qquad
[\sigma,\; \bar a_h] = 2h a_h, \label{sigah}
\end{equation}
and $ \displaystyle{\lim_{h \rightarrow 0}} \sigma = 0. $
\setcounter{equation}{0}
\section{Product Law for $h$-Symplecton}
It is possible to extend the product law (\ref{product}) for
$ sl(2) $ symplecton to $h$-symplecton. The product law plays a
crucial role when the symplecton calculus is considered. In this
section, we first prove the product law for $h$-symplecton by
using the one for $ sl(2)$ symplecton, then consider the
symplecton calculus for $ {\cal U}_h(sl(2)). $
\begin{thm}
Let $ \tilde P_j^m $ and $ \tilde P_{j'}^{m'} $ be $h$-symplecton,
then these obey the product law
\begin{equation}
\tilde P_j^m \tilde P_{j'}^{m'} =
\sum_{k=|j-j'|}^{j+j'} \sum_{n,n'}\; \bra{k} j \ket{j'}
(F^{-1})_{n,n' \ m,m'}^{\quad j,j'}
C_{n', m, n'+m}^{j',\; j,\ \ \ k} \tilde P_{k}^{n'+m},
\label{hproduct}
\end{equation}
where $ C_{m_1,m_2,m}^{j_1,j_2,j} $ is CGC for $ sl(2) $ and
\begin{eqnarray}
& & \bra{k}j\ket{j'} = 2^{k-j-j'} (2k+1)^{-1/2} \nabla(kjj'), \nonumber \\
& & \nabla(abc) = \left[
\frac{(a+b+c+1)!}{(a+b-c)! (a-b+c)! (-a+b+c)!} \right]^{1/2}_. \label{tri2}
\end{eqnarray}
\label{thm1}
\end{thm}
\textit{Proof} : From Proposition \ref{prop1},
\[
\tilde P_j^m \tilde P_{j'}^{m'} = P_j^m e^{m\sigma} \tilde P_{j'}^{m'}
e^{-m\sigma} e^{m\sigma} .
\]
Using the Hopf algebra mappings for $ \sigma $ given in (\ref{sigmaHopf}),
we see that the adjoint action of $ e^{m \sigma} $ is given by
\begin{equation}
adj e^{m \sigma}(t) = e^{m \sigma} t e^{-m \sigma}.
\label{adjsigma}
\end{equation}
$h$-Symplecton is a irreducible tensor operator of $ {\cal U}_h(sl(2)) $, it follows that
\begin{eqnarray}
e^{m \sigma} \tilde P_{j'}^{m'} e^{-m \sigma} &=&
adj e^{m \sigma}(\tilde P_{j'}^{m'})
= \sum_{n'}\; (e^{m \sigma})^{\; j'}_{n',m'} \tilde P_{j'}^{n'} \nonumber \\
&=& \sum_{n,n'}\; \delta_{n,m} (e^{m \sigma})^{\; j'}_{n',m'} \tilde P_{j'}^{n'} \nonumber \\
&=& \sum_{n,n'}\; (F^{-1})_{n,n' \ m,m'}^{\quad j,j'} \tilde P_{j'}^{n'},
\label{stilPs}
\end{eqnarray}
where the matrix elements of $ {\cal F} $ (\ref{rhs}) is used in the last equality.
Therefore, we have
\[
\tilde P_j^m \tilde P_{j'}^{m'} =
\sum_{n,n'}\; (F^{-1})_{n,n' \ m,m'}^{\quad j,j'} P_j^m \tilde P_{j'}^{n'}
e^{m \sigma}
= \sum_{n,n'}\; (F^{-1})_{n,n' \ m,m'}^{\quad j,j'} P_j^m P_{j'}^{n'}
e^{(n'+m) \sigma}
\]
Applying the product law (\ref{product}) for $ sl(2) $ symplecton,
Theorem \ref{thm1} is proved.
\hfill $\Box $
\begin{cor}
The associativity of the products
$
(\tilde P_a^{\alpha} \tilde P_b^{\beta}) \tilde P_c^{\gamma}
= \tilde P_a^{\alpha} (\tilde P_b^{\beta} \tilde P_c^{\gamma})
$
gives the same relation as (\ref{racha}) for the triangle function $ \nabla(abc) $
appeared in Theorem \ref{thm1}.
\label{cor1}
\end{cor}
\textit{Proof} : The associativity gives the same relation as (\ref{racha}),
but the Racha coefficients are replaced with the ones for $ {\cal U}_h(sl(2)) $.
From Lemma \ref{lem2}, these two kinds of Racha coefficients coincide.
\hfill $\Box $
Let us now consider the $h$-symplecton calculus. We assume the formal
ket $ \ket{\ } $ and that both $ \bar a \ket{\ } $ and $ a \ket{\ } $ are
nonvanishing vectors. Then the vectors defined by
$ \ket{jm} = \tilde P_j^m \ket{\ } $ are irrep. bases of $ {\cal U}_h(sl(2)) $
provided that the action of $ X \in {\cal U}_h(sl(2)) $ is defined by
$ X \ket{jm} = adj X(\tilde P_j^m) \ket{\ }. $ The dual bases are
defined by $ \bra{jm} = \bra{\ } \tilde P_j^{-m}(-1)^{j-m} $ in order to
keep the correspondence with the $h=0 $ case. The action of $ X \in {\cal U}_h(sl(2)) $ is,
of course, given by $ \bra{jm} = \bra{\ } adjX(\tilde P_j^{-m}) (-1)^{j-m} $.
The inner product is defined in the same manner as $ h=0 $ case, namely,
\[
\braket{jm}{j'm'} = \bra{\ }(-1)^{j-m}\tilde P_j^{-m} \cdot \tilde P_{j'}^{m'} \ket{\ },
\]
the operation $ \bra{\ } ( \cdots ) \ket{\ } $ means to take only the $ j= 0 $
part of the expression $ (\cdots) $. Applying the product law (\ref{hproduct}) for
$h$-symplecton, we obtain
\begin{equation}
\braket{jm}{j'm'} = \delta_{j,j'} 2^{-2j} F_{-m,m\ -m,m'}^{\qquad j,j}.
\label{hinner}
\end{equation}
Therefore the vectors $ \ket{jm} $ and $ \ket{j'm'} $ are orthonormal
if they belong to different irreps. but $not$ orthonormal if they
belong to a same irrep. The nonvanising part on the RHS of (\ref{hinner})
depends on only the twist element $ {\cal F}$.
From the product law, we can show the following relations for
$h$-symplecton
\begin{prop}
The following relations hold for $h$-symplecton.
\begin{eqnarray}
& & \sum_{m,m'}\; \tilde P_j^m \tilde P_{j'}^{m'}
F^{\quad j,j'}_{m,m'\ \ell,\ell'} =
\sum_k \bra{k} j \ket{j'} C^{j',j,\ \ k}_{\ell',\ell,\ell'+k} \;
\tilde P_k^{\ell + \ell'},
\label{hrel1} \\
& & \tilde P_{j'}^{m'}(a_h, \bar a_h + 2hma_h) =
\sum_{\ell=0} (F^{-1})_{n,n'\ m,m'}^{\quad j,j'}\;
\tilde P_{j'}^{m'+\ell}(a_h, \bar a_h).
\label{hrel2}
\end{eqnarray}
\label{prop3}
\end{prop}
\textit{Proof} : The relation (\ref{hrel1}) is easily proved by
multiplying the product law (\ref{hproduct}) by
$ F^{\quad j,j'}_{n,n'\ m,m'} $ and summing over $ m, m'$.
The relation (\ref{hrel2}) is derived by
moving $ e^{m\sigma} $ to the right of $ P_{j'}^{m'} $ in (\ref{stilPs}).
One can do that by using the relations
\[
e^{m\sigma} a_h = a_h e^{m\sigma}, \qquad
e^{m\sigma} \bar a_h = (\bar a_h + 2hm a_h) e^{m\sigma}.
\]
\hfill $\Box$
\setcounter{equation}{0}
\section{Quantum $h$-Plane and Representations of $ SL_h(2) $}
The Jordanian quantum algebra $ {\cal U}_h(sl(2)) $ and Jordanian quantum
group $ SL_h(2) $ are dual each other. It follows that
any representation basis of $ {\cal U}_h(sl(2)) $ is also representation basis
for $ SL_h(2) $ belonging to the same representation. Since
$h$-symplecton is a irrep. basis of $ {\cal U}_h(sl(2)) $, it is also
a irrep. basis of $ SL_h(2) $. We have seen this for $ j=1/2 $
in \S 5. The relation (\ref{action}) can be generalized to
arbitrary $j$
\begin{equation}
\tilde P_j^m(a_h', \bar a_h') =
\sum_n \tilde P_j^n(a_h,\bar a_h) \tilde d^j_{nm}(g)
\quad
g \in SL_h(2). \label{dands}
\end{equation}
We can obtain $d$-functions for $ {\cal U}_h(sl(2)) $ by substituting (\ref{action})
into the explicit form of $h$-symplecton given in Proposition \ref{prop2}.
However, as is seen from the explicit form,
the actual computation seems to be complicated.
The use of quantum $h$-plane \cite{kar} provides us
a procedure which is a little bit simpler in computation.
In this section, we shall find irrep. bases for $ SL_h(2) $ in terms of quantum
$h$-plane which give the same irreps. as $h$-symplecton
by using the tensor operator approach.
Recall that the functions $ \Phi_{jm}(\xi, \eta) $ defined by
(\ref{phi}) are irrep. bases of $ sl(2) $ in the realization
(\ref{phisl2}) and irrep. bases of $ SL(2) $ under (\ref{phiSL2})
as well. We can regard $ \Phi_{jm}(\xi, \eta) $ as a
irreducible tensor operator of $ sl(2) $, since it is easy
to verify that
\begin{eqnarray}
& & [J_{\pm}, \; \Phi_{jm}] = \sqrt{(j\mp m)(j\pm m+1)} \Phi_{j\; m\pm1}, \nonumber \\
& & [J_0,\; \Phi_{jm}] = 2m \Phi_{jm}. \label{phiastensor}
\end{eqnarray}
From Lemma \ref{lem3}, it ie easy to find the corresponding
irreducible tensor operators for $ {\cal U}_h(sl(2)) $.
\begin{prop}
Let $ \xi, \eta $ be commutative numbers, then the
followings are irreducible tensor operators for $ {\cal U}_h(sl(2)) $
\begin{equation}
\tilde \Phi_{jm}(\xi, \eta) = \Phi_{jm}(\xi, \eta) e^{m\sigma}, \label{phih}
\end{equation}
where $ \sigma = -\ln(1-2h\xi \frac{d}{d\eta}) $.
\label{prop4}
\end{prop}
For $ j=1/2$, we have
\begin{equation}
\tilde \Phi_{\frac{1}{2}\; \frac{1}{2}} = \xi e^{\sigma/2}
\equiv \xi_h,
\qquad
\tilde \Phi_{\frac{1}{2}\; -\frac{1}{2}} = \eta e^{-\sigma/2}
\equiv \eta_h,
\label{hplane}
\end{equation}
and they satisfy the commutation relation
\begin{equation}
[\xi_h, \; \eta_h ] = h \xi_h^2, \label{hplanecomm}
\end{equation}
this corresponds to the commutation relation of quantum
$h$-plane in Ref.\cite{kar}.
It is easily verified that the commutation relation (\ref{hplanecomm})
is preserved under the action of $ SL_h(2) $
\begin{equation}
(\xi_h', \eta_h') = (\xi_h, \eta_h) \left(
\begin{array}{cc}
x & u \\ v & y
\end{array}
\right)_. \label{hpaction}
\end{equation}
It is a easy exercise to write $ \tilde \Phi_{jm} $ in terms of
$ \xi_h $ and $\eta_h$. Then $ \tilde \Phi_{jm}(\xi_h, \eta_h) $
forms irrep. bases of
$ SL(2) $, that is, the $d$-functions
for $ SL_h(2) $ are obtained by substituting (\ref{hpaction}) into
$ \tilde \Phi_{jm}(\xi_h, \eta_h) $.
\begin{prop}
Irreps. of $ SL_h(2) $ on the quantum $h$-plane are
obtained by
\begin{equation}
\tilde \Phi_{jm}(\xi_h',\eta_h') = \sum_k \; \tilde \Phi_{jk}(\xi_h, \eta_h)
\tilde d^{j}_{km},
\label{SLh2onplane}
\end{equation}
where the irrep. bases are given by
\begin{eqnarray}
\tilde \Phi_{jm} &=& c_{jm} \; \xi_h^{j+m} (\eta_h - h(j+m)\xi_h)
(\eta_h - h(j+m-1)\xi_h) \nonumber \cdots (\eta_h - h(2m+1)\xi_h),
\nonumber \\
&=& c_{jm} \; \eta_h (\eta_h + h\xi_h) \cdots (\eta_h+(j-m-1)h\xi_h) \;
\xi_h^{j+m}, \label{hphix}
\end{eqnarray}
with
\[
c_{jm} = \frac{1}{\sqrt{(j+m)! (j-m)!}}.
\]
\label{prop5}
\end{prop}
Since $ \tilde P_{jm} $ and $ \tilde \Phi_{jm} $ give
the same irreps. of $ {\cal U}_h(sl(2)) $, they also give the same
$d$-functions of $ SL_h(2) $. Indeed, the explicit computation shows
that we obtain the same $d$-functions for $ j = 1/2 $ and $ j=1 $.
The $ j=1/2 $ case gives the $ 2 \times 2 $ quantum matrix $T$
(\ref{quantumM}) itself, while
$ j=1 $ $d$-function reads
\begin{equation}
d^1 = \left(
\begin{array}{ccc}
x^2 + h xv & \sqrt{2}(ux + h uv) & u^2+hu(x+y+hv) \\
\sqrt{2}xv & 1 + 2 uv & \sqrt{2}(uy + huv) \\
v^2 & \sqrt{2}yv & y^2 + hyv
\end{array}
\right)_. \label{j1d}
\end{equation}
The $d$-functions for $ SL_h(2) $ are also discussed in Ref.\cite{CQ} where the
authors assert that the $d$-functions can be obtained from the q-deformed ones
via a contraction method and show some explicit examples.
Another way to obtain the $d$-functions is to use the recurrence relations
for $d$-functions. This will be discussed in a separate publication.
\setcounter{equation}{0}
\section{Concluding Remarks}
We have constructed $h$-symplecton in this article and
investigated some of its properties. It has been seen that
many properties of $ sl(2) $ symplecton are inherited to $h$-symplecton.
Unfortunately, $h$-dependence of $h$-symplecton is absorbed in $ \sigma $,
namely, twist element $ {\cal F}$, so that we can not see specific
hypergeometric function for $h$-deformation. It will become clear
what kind of hypergeometric functions are specific to $h$-deformed
quantities if we obtain explicit form of $d$-function for
$ SL_h(2) $ as in the case of $q$-deformed $ SU(2) $\cite{rep}.
The $ sl(2) $ symplecton has a simple generating function.
We presented (\ref{hgenfun}) as a generating function for
$h$-symplecton. However, this may be one of possible choices, we
might find simpler generating function. The use of quantum
$h$-plane $ \xi_h, \eta_h $ instead of $ \xi, \eta $ is one of
the possibilities. We have done some calculation to find
simpler form of generating function in terms of
$ \xi_h $ and $ \eta_h $, however, all what we
obtained have more complicated form.
We would like to emphasize the usefulness of Lemma \ref{lem3}.
This provides us a much simpler procedure to obtain $h$-symplecton
than starting with the definition (\ref{defhsym}) and using the
lemma, we could easily find another irrep. bases (\ref{hphix}) for
$ SL_h(2). $ This lemma is, of course, applicable to any
Jordanian quantum algebra, since we usually know the explicit form
of twist element. Furthermore, the lemma is extended to quasitriangular
Hopf algebras \cite{fiore2}. For quasitriangular Hopf algebras, the
twist elements are usually not known, they are known up to certain order
of the deformation parameters. It is expected that many properties
of tensor operators for quasitriangular Hopf algebras are studied
based on the present knowledge of the tensor operators for
Lie algebras via Lemma \ref{lem3}, even if the
explicit form of tensor operators is not obtained. It may also be possible to
apply Lemma \ref{lem3} to the investigation of $q$-symplecton.
\section*{Appendices}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 2,232
|
Jean Fournet Fayard
Category : Sports-Business-Coach-Manager-Owner
Profile : 5/2 - Heretical / Hermit
Definition : Split - Small (16,20,31,45)
Sports-Business-Coach-Manager-Owner
Type: Splenic Projector
Inc.Cross: Demands 2
Definition: Double Split - Small (16,20,31,45)
Variables: BLR-MLL
1762 Acceptance
2551 Initiation
1858 Judgment
French pharmacist and administrative manager of St. Lipha, involved in the Furiani-Bastia case.
A former soccer player and former trainer, he was vice-president of the team 1981-'84 and president of the FFF (French Football Federation) 1985-'94. At the trial, he was convicted on 6/01/1992 of manslaughter and involuntary injuring; later the conviction was overturned and he was acquitted.
The trial opened on 1/04/1995 at 2.30 PM in Bastia, France. The situation was this: French soccer match at the Furiani Stadium in Bastia; the semi finals of the Coupe de France that opposed the Bastia and Marseille soccer teams at the Furiani stadium. It ended in tragedy after the collapse at 8:20 PM of a temporary stand erected for fans, with 17 people dead and 2,357 injured.
Fayard married 8/06/1954.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,733
|
from lxml import etree
from healthvaultlib.methods.method import Method
from healthvaultlib.methods.methodbase import RequestBase, ResponseBase
class InspectorRequest(RequestBase):
def __init__(self, parameters):
super(InspectorRequest, self).__init__()
self.name = 'GetAuthorizedPeople'
self.version = 1
self.parameters = parameters
def get_info(self):
info = etree.Element('info')
info.append(self.parameters.get_info())
return info
class InspectorResponse(ResponseBase):
def __init__(self):
super(InspectorResponse, self).__init__()
self.name = 'GetAuthorizedPeople'
self.version = 1
def parse_response(self, response):
self.parse_info(response)
class Inspector(Method):
def __init__(self, parameters):
self.request = InspectorRequest(parameters)
self.response = InspectorResponse()
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 5,273
|
{"url":"https:\/\/physics.meta.stackexchange.com\/questions\/6348\/can-we-use-mathjax-in-comments","text":"# Can we use MathJax in comments?\n\nCan we write MathJax commands in comments (and in chat)? If not, I believe we should add such a feature...\n\n\u2022 tesing... $123$ \u2013\u00a0TanMath Dec 13 '14 at 0:22\n\u2022 Hmm. Works on the regular site. \u2013\u00a0HDE 226868 Dec 13 '14 at 0:22\n\u2022 MathJax\/Latex does not work on the Meta side of this site \u2013\u00a0Kyle Kanos Dec 13 '14 at 3:02\n\n\u2022 By way of explanation, the MathJax script runs after the web page loads. It replaces any sequences of the form $...$ or $$...$$ or so on with rendered math; the script doesn't even know whether any such sequence is part of a comment, a post, the sidebar, etc. (code blocks are a special exception) \u2013\u00a0David Z Dec 13 '14 at 5:57","date":"2021-05-13 08:30:07","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\": 1, \"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\": 0.7719355821609497, \"perplexity\": 1510.5672392142214}, \"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-2021-21\/segments\/1620243990584.33\/warc\/CC-MAIN-20210513080742-20210513110742-00105.warc.gz\"}"}
| null | null |
This is a really, really great book. The premise is four siblings are returning to the family vacation home in Maine to decide what to do with it. Their mother has recently died and all of them are suffering from the reality of her death. Each character is battling demons and harboring secrets, deep secrets, from each other.
There is such moving language describing the sea that surrounds the m=home and how its movement, depth, and color translates to the story.
It's definitely not a one day easy read as there is so much depth to the characters and to their relationships with each other and to the memories of the house.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,466
|
\section{Introduction}
The existence of an equilibrium phase transition into the spin glass (SG) phase has not yet been convincingly established for some spin glasses. The development of the \emph{parallel tempered} Monte Carlo (TMC) algorithm \cite{TMC} has enabled one to observe, bypassing anomalously long relaxation processes, SG models in \emph{equilibrium} at low temperatures. Thus, correlation lengths $\xi$ have been determined from the equilibrium behavior of $\langle \langle s_is_j \rangle_T ^2 \rangle_q$, where $s_i=\pm 1$ is for a spin at site $i$,
and $\langle \cdots \rangle_T$ and $\langle \cdots \rangle_q$ stand for a thermal average and for an average over quenched randomness, respectively. There is evidence, from Monte Carlo simulations, that $\xi$ grows as linear system size $L$ in (i) the Edwards-Anderson (EA) model\cite{balle,katz} at some nonzero temperature $T_{sg}$ in three dimensions, in (ii) geometrically frustrated systems, such as strongly site-diluted Ising models, with nearest neighbor antiferromagnetic (AF) bonds, on fcc lattices, \cite{henley,DS} and in (iii) strongly site-diluted Ising models with dipole-dipole interactions, such as in LiHo$_x$Y$_{1-x}$F$_4$. \cite{chicago} We refer to the latter systems as disordered Ising dipolar (DID) systems.\cite{yes0,yes} At least for DID systems, some numerical evidence that is unfavorable for the existence of a phase transition also exists.\cite{yu}
The divergence of $\xi$ implies the divergence of the so called spin-glass susceptibility $\chi_{sg}$ at $T_{sg}$, where $\chi_{sg}=N^{-1}\sum_{ij}\langle \langle s_is_j \rangle_T ^2 \rangle_q$ and $N$ is the number of spins.
Convincing experimental evidence for the existence of an equilibrium phase transition into the SG phase is harder to obtain. This is mainly (i) because very long relaxation processes make equilibrium observations difficult, and (ii) because neither $\xi$ nor $\chi_{sg}$ can be directly observed. Instead, the SG transition is usually characterized by the nonlinear magnetic susceptibility $\chi_{3}$. \cite{PW} It is defined by
\begin{equation}
m=\chi_1 H+\chi_3H^3+\cdots ,
\label{uno}
\end{equation}
assuming $m(-H)=-m(H)$. Canella and Mydosh\cite{mydosh} were first able to measure (in gold-iron alloys) huge values of $\chi_3$: $T_{sg}^2\chi_{3}/\chi_1 \sim 10^{5}$ (from here on, we let Boltzmann's constant and Bohr's magneton equal $1$) near $T=T_{sg}$.
Later, ${\chi}_{3}$ was shown \cite{monod} to diverge in other canonical SGs as a power of $T-T_{sg}$.
For the EA model,\cite{ea} originally inspired by the discovery of Canella and Mydosh, Chalupa \cite{chalupa} showed long ago that
\begin{equation}
-\chi_3 = T^{-3} (\chi_{sg}-2/3)
\label{cuatro}
\end{equation}
if no field is applied. Thus, the critical behavior of $\chi_3$ and $\chi_{sg}$, which one observes in simulations, can, at least for the EA model, be clearly related to the critical behavior of $\chi_3$, which one observes experimentally.
The three models we study are governed by the Hamiltonian,
\begin{equation}
{\cal H}=-\frac{1}{2}\sum_{ij}J_{ij}x_is_ix_js_j,
\label{ham}
\end{equation}
where the sum is over all $i$ and $j$ lattice sites, $J_{ij}$ is model specific, $x_i=\pm 1$ is a quenched random variable, and $s_i$ is a ($\pm 1$) Ising spin at site $i$. All sites are occupied with the same probability, $x=\langle x_i\rangle_q$, where the $q$ subscript stands for a quenched average over all site occupancy arrangements.
The aim of this paper is to find how $-\chi_3 $, an experimentally measured quantity, and $\chi_{sg}$, a quantity which is often calculated, are related in site-diluted SGs.
More specifically, numerical results from TMC are sought for (i) Ising spins, with Ruderman-Kittel-Kasuya-Yoshida (RKKY) interactions,\cite{RKKY} which are randomly located on a small fraction of all lattice sites, (ii) a geometrically-frustrated Ising spin system, mainly, randomly located Ising spins, with nearest neighbor AF interactions, on a $0.4$ fraction of all sites of an fcc lattice, and (iii) DID systems
on a small fraction of all lattice sites. The outcome of these calculations is unknown, because
Eq. (\ref{cuatro}) has not been derived for site diluted SGs. On the other hand, $\chi_3$ and $\chi_{sg}$ can exhibit the same critical behavior in site diluted SGs if they and the EA model belong to the same universality class.
This has been predicted\cite{bray} to be so for the first of the above three models, but not so, as far as we know, for the other two models.\cite{grest,other}
An outline of the paper follows. Details about the procedure we follow in our calculations are given in Sec. \ref{pro}.
For various sizes of each of these systems, we obtain $\xi /L$, $\chi_{sg}$, and $\chi_{3}$.
Data for $\xi /L$ is used to establish the phase transition temperature $T_{sg}$ between the paramagnetic and SG phases. We then compare how $\chi_{sg}$ and $\chi_3$ vary with system size and with temperature in the vicinity of $T_{sg}$. The results obtained for each system are given in each of the subsections of Sec. \ref{results}. Very briefly, these results follow.
We find that $\chi_3$ approximately follows Eq. (\ref{uno}) in strongly diluted systems of Ising spins with RKKY interactions.
On the other hand, $-T^3\chi_3$ is a over a couple of orders of magnitude smaller than $\chi_{sg}$ in a ($x=0.4$) site-diluted AF Ising model on an fcc lattice. Nevertheless, both $\chi_3$ and $\chi_{sg}$ appear to have the same critical behavior. Finally,
in DID systems, $-T^3\chi_3$ and $\chi_{sg}$ seem to diverge similarly at $T_{sg}$. However, $-T^3\chi_3/\chi_{sg}$ varies sharply with systems' shape. Taking into account demagnetization effects, we estimate in Sec. \ref{shape} how $-T^3\chi_3$ varies with system shape for high aspect ratios.
Near the transition temperature, $-T^3\chi_3/\chi_{sg}$ increases from $-T^3\chi_3/\chi_{sg}\sim 10^{-2}$ for cubic shape systems to $-T^3\chi_3/\chi_{sg}\sim 10^{2}$ for very thin long prisms.
Our conclusions are summarized in Sec. \ref{con}.
As a byproduct, we have obtained values for $\eta$ and $T_{sg}$ in these three systems. They are listed in Table II.
From here on, in addition to $k_B=1$, $\mu_B=1$, we let $m=N^{-1}\sum_i\langle\langle s_i \rangle_T\rangle_q$, and assume
spins in all models point up or down along the $z$-axis, sometimes referred to as the magnetization axis.
\section {Procedure}
\label{pro}
To calculate $\chi_3$, we make use of
\begin{equation}
6\chi_3= N^{-1}T^{-3}(\langle M^4 \rangle_T
-3 \langle M^2 \rangle_T ^2),
\label{mn}
\end{equation}
where $M=Nm$, which holds for $H=0$. This equation follows from Eq. (\ref{uno}) by (i) repeated differentiation with respect to $H$ of the canonical ensemble average expression for $ \langle m \rangle_T $, and by (ii) letting
$\Delta =0$, where $\Delta =-4 \langle M^3 \rangle \langle M \rangle +12 \langle M^2 \rangle \langle M \rangle^2 -6 \langle M \rangle^4$.
This is justified for finite systems with up-down symmetry if averages are taken over infinite times, since $\langle M^n \rangle =0$ then for all odd $n$. The order in which system sizes and averaging times are taken to infinity is irrelevant for the paramagnetic phase. Equations (\ref{cuatro}) and (\ref{mn}), as well as all the results below are only claimed to hold for $T\geq T_{sg}$.
Note that Eq. (\ref{mn}) is valid for each realization of quenched disorder. Chalupa derived Eq. (\ref{cuatro}) from Eq. (\ref{mn}) for the EA model
by first averaging over all system samples, and noting (i) that both $ \langle M^4 \rangle$ and $\langle M^2 \rangle ^2$ involve sums over four-spin terms, such as $\sum_{ijkl}\langle\langle s_is_js_ks_l\rangle_T\rangle_q$ and $\sum_{ijkl}\langle\langle s_is_j \rangle_T \langle s_ks_l\rangle_T\rangle_q$, respectively, and (ii) that
any term in which one or more subindices is unpaired vanishes. To see this, assume the $k$ index is unpaired in either of the two sums over $ijkl$ indices. Now, consider all exchange bonds $J_{km}$ between the $k$-th and any other site. Let's assume the probabilities for $J_{km}$ and $-J_{km}$ for all $m$ while all other exchange constants remain unchanged are equal. (This requires exchange bonds to be \emph{independently} random.)
It follows that the probabilities for $s_k$ and $-s_k$, for any given configuration of all the other spins, are equal. This is the gist of the proof. For further details, see Ref. [\onlinecite{chalupa}]. The proof fails for site-diluted systems, because exchange bonds are not then \emph{independently} random.
We simulate a set of identical systems at temperatures $T_{min}$, $T_{min}+\Delta T$, $T_{min}+2 \Delta T, \ldots T_{max}$ following standard TMC rules. \cite{TMC} We choose $T_{max}\simeq 2.5 T_{sg}$, $T_{min}\sim 0.5T_{sg}$, and all $\Delta T$ such that at least $30\%$ of all attempted
exchanges between systems at $T$ and $T+\Delta T$ are successful for all $T$. We let each system equilibrate for a time $\tau_s$ and take averages over an equally long subsequent time $\tau_s$. Time $\tau_s$ satisfies two requirements, (i) that $\langle \langle M\rangle_T^2\rangle_q \ll 0.1 \langle \langle M^2\rangle_T\rangle_q$ obtains for all $T\in[T_{min},T_{max}$], and (ii)
that, systems which start from either random configurations or from (assumed) equilibrium configurations come to the same condition, as specified in Ref. [\onlinecite{solo}], after time $\tau_s$.
Values of $\tau_s$, of the number of samples $N_s$ with different quenched randomness over which averages were taken, and of the site occupancy rate $x$, are given in Table I.
Periodic boundary conditions are used throughout. For DID systems, we make use of Ewald sums. \cite{ew}
For the correlation length $\xi$, we make use, as has become standard practice,\cite{balle,jorgo,yes,solo} of the original definition,\cite{old}
\begin{equation}
\xi^2=\frac {1 } {4 \sin^2 ( k /2)} { \left[ \frac{ \chi_{sg} }{ \mid \chi_{sg} (\textbf{k})\mid } -1 \right] },
\label{nosa}
\end{equation}
where $\chi_{sg}(\textbf{k})=N^{-1}\sum_{ij} \langle \langle s_i s_j\rangle_T^2 \rangle_q\exp ({i\textbf{k}\cdot\textbf{r}_{ij}})$, and we let ${\bf k} =(2\pi/L,0,0)$.
Note that $\chi_{sg} (\textbf{k}=0)=\chi_{sg}$ and that, as $L\rightarrow \infty$, $\xi /L$ vanishes in the paramagnetic phase, remains finite at $T=T_{sg}$, and either grows without bounds below $T_{sg}$, as in a conventional phase transition, or remains finite as in the XY model in two dimensions. The point where $\xi /L$ curves for various system sizes meet as $T$ decreases defines $T_{sg}$ for us.
All systems we report on below have a common feature: fractional errors for $\chi_3$ are an order of magnitude larger than the ones for $\chi_{sg}$ and for $\chi_1$. Unless otherwise stated, error bars are given for $\chi_3$ and related quantities, but not for $\chi_{sg}$ or $\chi_{1}$, which are all smaller than icons for their data points.
\begin{table}\footnotesize
\caption{The number of samples $N_s$ and the number of Monte Carlo sweeps $\tau_s$ that were taken both for equilibration and for the subsequent averaging times are given in thousands of Monte Carlo sweeps. There are $L^3$ lattice sites in RKKY and FCC systems, but $L\times L\times 2L$ lattice sites for DID systems. }
\begin{ruledtabular}
\begin{tabular}{|c| r c r |r c c c|c r c|}
& & RKKY& & &&FCC& & & DID \\
\hline
L & 6 & 8 & 12 & 4& 6 &8 &12 & 4 &6 &8\\
$\tau_s$ & $10$ & $ 40$ & $400$ &$10$ &$10$ &$40$ &$40$& $10$ &$100$& $3000$\\
$N_s$ & $100$ & $40$ & $10$ & $10$ &$40$ &$10$ &5.6&$100$ & 15 & 4 \\
$x$ & $0.1$ & $0.1$ & $0.1$ & $0.4$ & $0.4$ & $0.4$ & $0.4$ &$0.35$&$0.35$& $0.35$\\
\end{tabular}
\end{ruledtabular}
\end{table}
\section{results}
\label{results}
All results in this section follow from TMC simulations.
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig1.eps}
\caption{(Color online) (a) Semilog plots of $\xi /L$ vs $T$ for ($m=\pm 1$) Ising spins with RKKY interactions, randomly located, on a $0.1$ fraction of all $L^3$ sites. The numbers in the box are $L$ values. (b) Same as in (a) but for $\chi_{sg} /L^{2-\eta}$ vs $T$, for $\eta=-0.5$. }
\label{primera}
\end{figure}
\subsection
{Spatially disordered Ising spins with RKKY interactions }
The Hamiltonian is given by Eq. (\ref{ham}), with $J_{ij}=\varepsilon_c (\cos kr_{ij} )(a/r_{ij})^3$,
ie, an RKKY \cite{RKKY} interaction, as in a canonical spin glass.
We let $ka=2\pi$, where $a$ is a nearest neighbor distance, and $\varepsilon_c$ is an energy in terms of which all temperatures are given in this subsection.
We let each site is be occupied with $x=0.1$ probability.
Plots of $\xi /L$ vs $T$ for various system sizes are shown in Fig. \ref{primera}a. Not all pairs of curves cross at the same point. Let $T_{i,j}$ be the temperature where curves for lengths $L_i$ and $L_j$ cross, where $L_1=6$, $L_2=8$ and $L_3=12$.
Plots of $T_{ij}$ (where $T_{1,2}=0.29$, $T_{1,3}=0.23$, and $T_{2,3}=0.20$) vs $1/L_iL_j$ fall onto a straight line which extrapolates to $T=0.10$ as $1/L_iL_j\rightarrow 0$. We thus estimate $T_{sg}=0.10$,
and therefore expect $T/x=1.0$ for $x\lesssim 1$ values, since the $1/r^3$ dependence of the interaction implies\cite{yes} $\xi(x,T)=\xi(T/x)$.
In Fig. \ref{primera}b we can see that $\chi_{sg}$ seems to grow without bounds with system size at $T= T_{sg}$. Indeed, we note that $\chi_{sg}\sim L^{2-\eta}$, where $\eta \simeq -0.5$ at $T_{sg}$. This $\eta$ value at the boundary of the range $-0.5\lesssim \eta \lesssim -0.2$ of quoted\cite{katz} values, from Monte Carlo simulations, for the EA model. We note in passing that this model and the EA model have been predicted\cite{bray} to be in the same universality class.
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig2.eps}
\caption{(Color online) (a) Semilog plots of $-T^3\chi_3 /L^{2-\eta}$ vs $T$, for $\eta =-0.5$, for ($m=\pm 1$) Ising spins with RKKY interactions, randomly located, on a $0.1$ fraction of $L^3$ sites, for the values of $L$, which are shown in the box. (b) Same as in (a) but for $-T^3\chi_3/\chi_{sg}$ vs $T$.}
\label{primeru}
\end{figure}
How $-T^3\chi_3$ behaves near $T_{sg}$ is shown in Fig. \ref{primeru}a. It varies with $T$ and with $L$ much as $\chi_{sg}$ does in Fig. \ref{primera}b. The plots shown in Fig. \ref{primeru}b are consistent with $-T^3\chi_3 \sim \chi_{sg}$. (The value $\eta = -0.5$ follows from the plots shown in Fig. \ref{primera}b, not from any fitting of $\chi_3$ to any desired behavior.)
More significantly, $-T^3\chi_3/ \chi_{sg}$ appears to approach a \emph{smooth} function of temperature in the neighborhood of $T=T_{sg}$ as $L\rightarrow \infty$. This is the basis for the main conclusion of this section, namely, that $-T^3\chi_3$ and $ \chi_{sg}$ have the same critical behavior.
\subsection
{Site-diluted AF Ising model on a fcc lattice}
Each site of a fcc lattice is occupied with a ($\pm 1$) Ising spin with a $0.4$ probability. The Hamiltonian is given by Eq. (\ref{ham}), with $J_{ij}=-J$ if $i$ and $j$ are nearest neighbors but $J_{ij}=0$ otherwise. A $0.4$ occupancy rate is roughly midway between the lowest value $x=0.195$ for percolation \cite{perc} in fcc lattices and the transition point, $x\simeq 0.75$, between SG and AF phases. \cite{henley} All temperatures are given in terms of $J$.
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig3.eps}
\caption{(Color online) (a) Plots of $\xi /L$ vs $T$ for a ($x=0.4$) site-diluted AF Ising model on an fcc lattice of $L\times L\times L$ sites. The numbers in the box are $L$ values. (b) Same as in (a) but for $\chi_{sg} /L^{2-\eta}$ vs $T$, for $\eta=-0.5$.}
\label{tercera}
\end{figure}
Monte Carlo results for this model are shown in Figs. \ref{tercera}a and \ref{tercera}b. We note in Fig. \ref{tercera}a that the crossing point between pairs of $\xi /L$ curves drifts leftward as their $L$-values increase.
As for Fig. \ref{primera}a, let $T_{i,j}$ be the temperature where curves for lengths $L_i$ and $L_j$ cross, where $L_1=4$, $L_2=6$, $L_3=8$ and $L_4=12$.
A second degree polynomial fit to a plot of $T_{ij}$ (where $T_{1,2}=0.70$, $T_{1,3}=0.67$, $T_{1,4}=0.62$, $T_{2,3}=0.62$, $T_{2,4}=0.57$, and $T_{3,4}=0.53$) vs $1/L_iL_j$ gives $T_{ij}\rightarrow 0.4$ as $1/L_iL_j\rightarrow 0$. We thus estimate $T_{sg}=0.4(1)$, in agreement, within errors, with the values found for $x=0.4$ in Ref. [\onlinecite{henley}]. Plots of $\chi_{sg}/L^{2-\eta}$ vs $T$ are shown in Fig. \ref{tercera}b for $\eta=-0.5$. This is the best value of $\eta$ to have $\chi_{sg}/L^{2-\eta}$ curves for various values of $L$ cross at $T_{sg}$.
This value of $\eta$ is, within errors, in agreement with the value found for $x=0.4$ in Ref. [\onlinecite{henley}].
Plots of $-T^3\chi_3/L^{2-\eta}$ vs $T$, with $\eta =-0.5$, are shown in Fig. \ref{fcc2}a. The $\eta = -0.5$ value is taken from the plots of $\chi_{sg}/L^{2-\eta}$ vs $T$, not from any fitting of $\chi_3$ to any desired behavior. The curves in Figs. \ref{tercera}b and \ref{fcc2}a are somewhat different, but all curves for $L=6,8$ and $12$ in both figures do cross, \emph{within errors}, at the same temperature, $T_{sg}=0.4$.
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig4.eps}
\caption{(Color online) (a) Plots of $-T^3\chi_3/L^{2-\eta}$ vs $T$, for $\eta =-0.5$, for a ($x=0.4$) site-diluted AF Ising model on an fcc lattice of $L\times L\times L$ sites. The numbers in the box are $L$ values. (b) Same as in (a) but for $-T^3\chi_3 /\chi_{sg} $ vs $T$.}
\label{fcc2}
\end{figure}
In Fig. \ref{fcc2}b we notice that $-T\chi_3\ll \chi_{sg}$, which differs markedly from what might have been expected from the behavior of the EA model (and from the above results for SGs with RKKY interactions). More significantly, we observe $-T\chi_3/ \chi_{sg}$
is, within errors, independent of $L$ for the largest values of $L$, and appears to go into a smooth function of $T$, near $T_{sg}$, as $L\rightarrow \infty$. This suggests that, in the thermodynamic limit, both quantities have the same critical behavior.
\begin{table}\footnotesize
\caption{Values for $T_{sg}$ follow from crossing (or merging) points of $\xi /L$ curves for various values of system linear size, $L$.
Values of $\eta$ are assigned so that $\chi_{sg}/L^{2-\eta}$ curves for various $L$ values cross at $T_{sg}$. Errors in $\eta$ follow from errors in $T_{sg}$.
As explained in Ref. [\onlinecite{yes}], $T_{sg}$ can be obtained for DID systems (for all $x\lesssim 0.5$ in sc lattices\cite{yes} and $x\lesssim 0.25$ in\cite{bel} LiHo$_x$Y$_{1-x}$F$_4$) from the $T_{sg}$ value given below, making use of
$T_{sg}\propto x$. Similarly for RKKY interactions and all $x\lesssim 0.1$}
\begin{tabular}{|c|c |c|c|}
\hline
& RKKY & FCC & DIDs \\
\hline
$T_{sg}$ & $0.10(4)$ for $x= 0.1$ & 0.4(1) for $x=0.4$ & $0.35(4)$ for $x= 0.35$ \\
$\eta $ & $-0.5(4)$ & $ -0.5(2)$ & $0.0(3)$ \\
\hline
\end{tabular}
\end{table}
\subsection{Spatially disordered ($\pm 1$) Ising dipoles}
Here we consider disordered Ising dipolar (DID) systems in sc lattices. We let each site be occupied, with a $0.35$ probability, by a ($\pm 1$) spin. All spins point up and down, along the $z$-axis.
The Hamiltonian is given by Eq. (\ref{ham}), with
\begin{equation}
J_{ij}=h_d
\frac{a}{r_{ij}^3}^3\left( 3
\frac{z_{ij}^2}{r_{ij}^2} -1\right),
\label{T}
\end{equation}
where $ r_{ij}$ is the distance between $i$ and $j$ sites, $z_{ij}$ is the $z$ component of $ r_{ij}$, $h_d$ is an energy, and $a$ is the
SC lattice constant.
Let's first recall that, despite some earlier numerical evidence to the contrary, \cite{yu} more recent calculations point to the existence of a phase transition between the paramagnetic and SG phases in diluted Ising dipolar systems \cite{yes0,yes} at $T_{sg}/x\simeq 1$ for all $x\lesssim 0.5$. In addition,\cite{solo} $\chi_{sg}\sim L^{2-\eta}$ at $T=T_{sg}$, where $\eta\simeq 0$.
We deal with the \emph{magnetic} susceptibility here which, as is well known, depends on the shape of the system.\cite{morrish} For this reason we study numerically $L\times L\times nL$ shaped prism systems for various values of $n$, that is, square-base prisms with a $1:n$ aspect ratios.
Plots of $\xi /L$ vs $T/x$ are shown in Fig. \ref{did1}a for $n=2$. Curves for three different values of $L$ are observed to cross at $T/x\simeq 1.0$. This transition temperature value is in agreement with the result found in Ref. [\onlinecite{yes}] for $n=1$, mainly, that $T_{sg}/x\simeq 1.0$ for all $x\lesssim 0.5$.
Plots of $\chi_{sg}/L^{2}$ vs $T/x$ are shown in Fig. \ref{did1}b. All curves are observed to cross at $T/x=0.95$. This is approximately as in Ref. [\onlinecite{yes}].
We now turn our attention to the \emph{magnetic} susceptibility. Plots of $-T^3\chi_3/L^2$ vs $T/x$ are shown in Fig. \ref{did2}a for systems of various sizes. These plots resemble the ones for $\chi_{sg}$ in Fig. \ref{did1}b, but note that the crossing points are not quite at the same temperature in the two figures.
\begin{figure}[!t]
\includegraphics*[width=80mm]{Finalfig5.eps}
\caption{(Color online) (a) Semilog plots of $\xi /L$ vs $T/x$ for DID systems on a $0.35$ fraction of all $L\times L\times L_z$ sites, where $L_z=2L$. The numbers in the box are $L$ values. (b) Same as in (a) but for $\chi_{sg} /L^{2}$ vs $T/x$. Error bars are smaller than icons for all data points.}
\label{did1}
\end{figure}
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig6.eps}
\caption{(Color online) (a) Plots of $-T^3\chi_3 /L^2$ vs $T/x$ for DID systems on a $0.35$ fraction of all $L\times L\times L_z$ sites, where $L_z=2L$. (b) Plots of $-T^3\chi_3 /\chi_{sg}$ vs $T/x$ for three $1:n$ aspect ratios, $n=4$ ($\boxplus$ and $\boxtimes$) for the two top curves, $n=2$ (full icons) for the three curves in the middle, and $n=1$ (empty icons) for the three lower curves. Error bars hardly protrude from icons. For both (a) and (b), the numbers in the box are the values of $L$.}
\label{did2}
\end{figure}
In order to better compare $-T^3\chi_3$ and $\chi_{sg}$, we plot in Fig. \ref{did2}b the ratio $-T^3\chi_3/\chi_{sg}$ vs $T/x$ for systems of various sizes with $1:4$, $1:2$ and $1:1$ aspect ratios. For a $1:4$ aspect ratio, only data points for systems with $4\times 4\times 16$ and $6\times 6\times 24$ sites appear in Fig. \ref{did2}b. A larger system with the same aspect ratio would have taken a prohibitively long computer time to run. Let $L_\bigstar$ be a system length such that
$-T^3\chi_3/\chi_{sg}$ is approximately size independent if $L \gtrsim L_\bigstar$. Clearly, $L_\bigstar \simeq 4$ and $6$ for $n=4$ and $2$, respectively, in Fig. \ref{did2}b. For $n=1$, $L_\bigstar \simeq 8$ seems likely. This would be in accordance with the expectation that $-T^3\chi_3$ and $\chi_{sg}$ have the same critical behavior in DID systems, independently of aspect ratio.
Questions about the sharp variation of $-T^3\chi_3/\chi_{sg}$ with respect to aspect ratio naturally arise. What is the asymptotic behavior of $-T^3\chi_3/\chi_{sg}$?
This is hard to foresee from the data plots shown in Fig. \ref{did2}b. To proceed much further numerically is impractical. The next section is devoted to this question.
\section{Variation of $\chi_3$ with aspect ratio in DID systems}
\label{shape}
In this section we derive an approximate equation for the variation of $\chi_3$ with shape in DID systems.
Consider two systems of the same shape and size. In system $f$, all spin pairs interact. In the other system, system $t$, dipole-dipole interactions are truncated. In $t$, each spin interacts only with spins that lie within a long thin cylinder centered on it, whose axis is parallel to the system's $z$-axis. The radius of the cylinder need not be more than a couple of nearest neighbor distances, but its length must be much longer than its radius. Let's furthermore assume that both systems are homogeneous, that is, all sites are occupied ($x=1$). Now, we know from Ref. [\onlinecite{penultimo}] that if both systems are in thermal equilibrium, and external magnetic fields $H_t$ and $H_f$ are applied to systems $t$ and $f$, respectively, such that $m$ is the same in both systems, then
\begin{equation}
H_f=H_t-{\lambda}_nm,
\label{h}
\end{equation}
where $n$ comes from $f$ system's $1:n$ aspect ratio. Equation (\ref{h}) holds because the only effect of the \emph{un}truncated portion of all dipole-dipole interactions in $f$ is to give the so called \emph{demagnetizing} field, ${-\lambda}_nm$. For \emph{dipolar} prisms of $1:n$ aspect ratio, a scaling expression, such as\cite{privman} $m=t^{-\beta}f(Ht^{-\beta\delta})$ must therefore be replaced by
\begin{equation}
m=t^{-\beta}f[(H-\lambda_nm)t^{-\beta\delta}].
\label{m}
\end{equation}
where $H_f$ has been replaced by $H$.
Taking the derivative of Eq. (\ref{m}) with respect to $H$ [or, more simply, of Eq. (\ref{h}) with respect to $m$] gives
\begin{equation}
\frac{1}{\chi_1(n)} = \frac {1} {\chi_1(\infty)}+\lambda_n,
\label{chi1}
\end{equation}
where $\chi_1(n)$ is the linear susceptibility of a prism with a $1:n$ aspect ratio, and, clearly, $dm/dH_t =\chi_1(\infty)$.
This is the
well known equation,\cite{morrish}
that experimentalists\cite{chicago,quilliam} often use in order to do away with demagnetization effects, and thus relate $\chi_1(n)$, the measured susceptibility, to ${\chi_1(\infty)}$.
Taking the $d/dH_t$ derivative of Eq. (\ref{chi1}) gives
\begin{equation}
\left( 1+\lambda_n \frac{dm}{dH_t}\right ) \frac {1}{\chi_1^2(n )} \frac{d\chi_1(n )}{dH_n} = \frac {1}{\chi_1^2(\infty )} \frac{d\chi_1(\infty )}{dH_t} ,
\end{equation}
where we have used
$dH_n/dH_t = 1+\lambda_n dm/dH_t$, which follows from Eq. (\ref{h}).
We next (i) take the $d/dH_t$ derivative of the above equation, (ii) let $H_t=H_n=0$, and (iii)
let $d\chi_1(\infty )/dH_t=0=d\chi_1(n)/dH_n$, by up-down symmetry. The result is easily cast into
\begin{equation}
\chi_3(n )=\frac{\chi_3(\infty)}{[1+\lambda_n\chi_1(\infty)]^4},
\label{chi3}
\end{equation}
which is the desired expression relating $\chi_3(n)$ and $\chi_3(\infty )$.
\begin{figure}[!t]
\includegraphics*[width=80mm]{fig7.eps}
\caption{(Color online) (a) Plots of $\chi_1$ vs $1/n$ for DID systems at $T=T_{sg}$, on a $0.35$ fraction of all $L\times L\times nL$ sites. The shown numbers are $L$ values for data points from Monte Carlo calculations. The dashed lines follow from Eq. (\ref{chi1}), assuming the three values for $\chi_1 (\infty )$ that are shown in the box. (b) Same as in (a) but for (full icons) $-T^3\chi_3/L^2$ and (open icons) $\chi_{sg}/L^2$. The dashed lines follow from Eqs. (\ref{chi1}) and (\ref{chi3}), and the shown pairs of values, such as $9$ and $100$, are for $\chi_1(\infty )$ and $\chi_3(\infty )$, respectively.}
\label{theory}
\end{figure}
\begin{table}\footnotesize
\caption{$\lambda_n$ values, in terms of $h_d$, for some $n$ in the $[0.5, 7]$ range. For $n\geq 8$, $\lambda_n\simeq 8/n^2$. For an $x$ site occupancy rate, $\lambda_n\rightarrow x\lambda_n$. }
\begin{ruledtabular}
\begin{tabular}{|c| l l l l l l l l l l l|}
$n$ & $0.5$ & $1$ & $1.5$ & $2$ &$2.5$ &$3$ &$4$ &$5$& $6$ &$7$& \\
$\lambda_n$ & 7.419& $4.189$ & $2.503$ & $1.611$ & $1.107$ &$0.802$ &$0.471$ &0.308&$0.217$ & 0.160 & \\
\end{tabular}
\end{ruledtabular}
\end{table}
Equations (\ref{chi1}) and (\ref{chi3}) enable us to calculate how $\chi_3(n)$ varies with $n$ if we know $\chi_1$ and $\chi_3$ for at least an aspect ratio each, as well as $\lambda_n$. A list of (easily computed) $\lambda_n$ values for several values of $n\in [0.5,7]$, as well as a functional relation for all $n\geq 8$, are given in Table III.
Since the only effect of the long-range portion of all dipole-dipole interactions in $f$ is to give the demagnetizing field,\cite{penultimo} ${-\lambda}_nm$,
we can calculate all $\lambda_n$ assuming a fully occupied lattice in which all spins point up. Since only a single state comes into the calculation, no Monte Carlo simulation is necessary.
This enables us to calculate $\lambda_n$ for very large systems.
The fact that only an $x$ fraction of lattice sites are occupied in site diluted SGs is \emph{approximately} taken into account by letting $\lambda_n \rightarrow x\lambda_n$ everywhere. Inhomogeneities in SGs are thus neglected.
Equation (\ref{chi1}) gives the three dashed lines shown in Fig. \ref{theory}a for $\chi_1(\infty)=7, 9$ and $11$. With two of these values, we obtain from Eq. (\ref{chi3})
the three curves for $\chi_3(n)$ shown in Fig. \ref{theory}b for values of $\chi_3(\infty )$. These curves do not fit the data points too well. On the other hand, a good fit for small system sizes should not be expected. We can nevertheless conclude with some confidence that $\chi_3(n)$ does not diverge as $n\rightarrow \infty$. Indeed, $\chi_3(\infty )$ is most likely within the $(50,120)$ range. Furthermore, observation of Fig. \ref{theory}b indicates that $\chi_3(n)/\chi_{sg}$ at $T=T_{sg}$ varies over three or four orders of magnitude as system shape varies from cubic to infinitely thin needle-like.
It is perhaps worth pointing out that
\begin{equation}
\chi_3(n)= \frac{ \chi_3(\infty) }{ \chi_1^4(\infty ) } \chi_1^4(n)
\label{xyz}
\end{equation}
follows immediately from Eqs. (\ref{chi1}) and (\ref{chi3}) after Eq. (\ref{chi1}) is cast into $\chi_1(n)=\chi_1(\infty )/[1+\lambda_n\chi_1(\infty )]$. Equation (\ref{xyz}) implies that $\chi_3$ sweeps over four times as many decades as $\chi_1$ does (compare Figs. \ref{theory}a and \ref{theory}b) as $n$ varies.
Finally, note that the classical or quantum nature of DID systems does not play any role in this section. It does not matter either whether a transverse field is applied, because it does not affected up-down symmetry. These equations can therefore be applied, as an illustration, to Li$_1-x$Ho$_x$Y$_4$, under a transverse field, as in Ref. [\onlinecite{bel}], where $T^2\chi_3/\chi_1\sim 1$ was observed on a $1.6 \times 16 \times 5$ mm$^3$ sample. Values of $\chi_1$ and $\chi_3$ that would be some $3$ and $100$ times larger, respectively, for a long thin needle-like sample can be read off from Figs. \ref{theory}a and \ref{theory}b.
\section{conclusions}
\label{con}
By the tempered Monte Carlo method\cite{TMC} we have tested whether the relation $-T^3\chi_3=\chi_{sg}-2/3$, which is known\cite{chalupa} to hold for the Edwards-Anderson model, also holds for several site-diluted spin glasses of ($\pm 1$) Ising spins, with (i) RKKY interactions, (ii) antiferromagnetic interactions between nearest neighbor spins on fcc lattices, and (iii) dipole-dipole interactions. As a byproduct, we have obtained the values of $\eta$ and $T_{sg}$ that are listed in Table II.
We have found $-T^3\chi_3\sim \chi_{sg}$ to hold, for Ising spins with RKKY interactions
occupying a $0.1$ fraction of all lattice sites. More significantly, $-T^3\chi_3/ \chi_{sg}$ appears to be (i) independent of linear system size, within errors, and (ii) a smooth function of temperature near $T_{sg}$. This suggests $-T^3\chi_3 $ and $ \chi_{sg}$ have the same critical behavior.
Since the RKKY interaction decays as the inverse of the cube of the distance, these results must hold for lower values of $x$ if the temperature is scaled with $x$.
We have found $-T^3\chi_3$ to be over two orders of magnitude smaller than $\chi_{sg}$ for Ising spins, with antiferromagnetic interactions, on a ($x=0.4$) site diluted fcc lattice.
Our results are, however, consistent with identical critical behavior of these two quantities.
In DID systems the TMC data (see Fig. \ref{did2}b) are consistent with $\chi_3$ and $\chi_{sg}$ diverging in the same manner as $T \rightarrow T_{sg}$ from above.
The sharp variation of $-T^3\chi_3/\chi_{sg}$ with aspect ratio, which can be observed in Fig. \ref{did2}b, is noteworthy. How this comes about from demagnetization effects is explained in Sec. \ref{shape}. In it, relations are derived which together with data points coming from TMC simulations (see Fig. \ref{theory}b) give rough estimates of $-T^3\chi_3$ at or near $T=T_{sg}$.
We find $-T^3\chi_3/\chi_{sg}$ varies, as shown in Fig. \ref{theory}b from $-T^3\chi_3/\chi_{sg}\sim 10^{-2}$ for cubic shapes to $-T^3\chi_3/\chi_{sg} \sim 10^{2}$ for long thin needles.
Our results for DID systems with an $x=0.35$ site occupancy rate can be generalized to smaller values of $x$. As discussed in Ref. [\onlinecite{yes}],
any physical quantity $f$ satisfies $f(x,T)=f(T/x)$ for $x$ quite smaller than $x_c$, the critical concentration above which there is magnetic order at low temperature (e.g., $x_c\simeq 0.65$ for sc lattices\cite{yes} and\cite{bel} $x_c\simeq 0.25$ for LiHo$_x$Y$_{1-x}$F$_4$).
\acknowledgments
I am grateful to Juan J. Alonso for helpful remarks after reading the manuscript.
Funding Grant FIS2009-08451, from the Ministerio de Ciencia e Innovaci\'on of Spain, is acknowledged
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 6,045
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{"url":"http:\/\/th.nao.ac.jp\/seminar\/?cmd=backup&page=Colloquium&age=34&action=source","text":"# Colloquium \u00a4\u00ce\u00a5\u00d0\u00a5\u00c3\u00a5\u00af\u00a5\u00a2\u00a5\u00c3\u00a5\u00d7\u00a5\u00bd\u00a1\u00bc\u00a5\u00b9(No.34)\n\n#norelated\n* \u00cd\u00fd\u00cf\u00c0\u00a5\u00b3\u00a5\u00ed\u00a5\u00ad\u00a5\u00a6\u00a5\u00e02017 [#z407468f]\n\n\u00cd\u00fd\u00cf\u00c0\u00a5\u00b3\u00a5\u00ed\u00a5\u00ad\u00a5\u00a6\u00a5\u00e0\u00a4\u00cf\u00b8\u00b6\u00c2\u00a7\u00a4\u00c8\u00a4\u00b7\u00a4\u00c6\u00cb\u00e8\u00bd\u00b5\u00bf\u00e5\u00cd\u00cb\u00c6\u00fc\u00a4\u00ce\u00b8\u00e1\u00b8\u00e513:30\u00a4\u00ab\u00a4\u00e9\u00b3\u00ab\u00ba\u00c5\u00a4\u00b7\u00a4\u00c6\u00a4\u00a4\u00a4\u00de\u00a4\u00b9\u00a1\u00a3~\n\u00b8\u00b6\u00c2\u00a7\u00a4\u00c8\u00a4\u00b7\u00a4\u00c6\u00b1\u00d1\u00b8\u00ec\u00a4\u00c7\u00b9\u00d6\u00b1\u00e9\u00a4\u00b7\u00a4\u00c6\u00a4\u00a4\u00a4\u00bf\u00a4\u00c0\u00a4\u00ad\u00a4\u00de\u00a4\u00b9\u00a4\u00ac\u00a1\u00a2\n\u00b9\u00d6\u00bb\u00d5\u00a1\u00a6\u00bb\u00b2\u00b2\u00c3\u00bc\u00d4\u00a4\u00ac\u00c6\u00fc\u00cb\u00dc\u00bf\u00cd\u00a4\u00c0\u00a4\u00b1\u00a4\u00ce\u00be\u00ec\u00b9\u00e7\u00a4\u00cf\u00c6\u00fc\u00cb\u00dc\u00b8\u00ec\u00a4\u00cb\u00c0\u00da\u00a4\u00ea\u00c2\u00d8\u00a4\u00a8\u00a4\u00c6\u00a4\u00af\u00a4\u00c0\u00a4\u00b5\u00a4\u00c3\u00a4\u00c6\u00a4\u00e2\u00a1\u00ca\u00b1\u00d1\u00b8\u00ec\u00a4\u00ce\u00a4\u00de\u00a4\u00de\u00a4\u00c7\u00a4\u00e2\u00a1\u00cb\u00b7\u00eb\u00b9\u00bd\u00a4\u00c7\u00a4\u00b9\u00a1\u00a3~\n\u00c2\u00e6\u00c6\u00e2\u00a1\u00a6\u00c2\u00e6\u00b3\u00b0\u00a4\u00de\u00a4\u00bf\u00ca\u00ac\u00cc\u00ee\u00a4\u00f2\u00cc\u00e4\u00a4\u00ef\u00a4\u00ba\u00b9\u00ad\u00a4\u00af\u00c8\u00af\u00c9\u00bd\u00bc\u00d4\u00a1\u00ca\u00c2\u00e6\u00b3\u00b0\u00a4\u00ce\u00ca\u00fd\u00a4\u00cb\u00a4\u00cf\u00ce\u00b9\u00c8\u00f1\u00a1\u00a6\u00bc\u00d5\u00b6\u00e2\u00a4\u00a2\u00a4\u00ea\u00a1\u00cb\u00a4\u00f2\u00ca\u00e7\u00bd\u00b8\u00a4\u00b7\u00a4\u00c6\u00a4\u00a4\u00a4\u00de\u00a4\u00b9\u00a1\u00a3~\n\u00a4\u00aa\u00cc\u00e4\u00a4\u00a4\u00b9\u00e7\u00a4\u00ef\u00a4\u00bb\u00a4\u00cf\u00b0\u00ca\u00b2\u00bc\u00a4\u00ce\u00a5\u00b3\u00a5\u00ed\u00a5\u00ad\u00a5\u00a6\u00a5\u00e0\u00b7\u00b8\u00a4\u00de\u00a4\u00c7\u00a4\u00aa\u00b4\u00ea\u00a4\u00a4\u00a4\u00b7\u00a4\u00de\u00a4\u00b9\u00a1\u00ca_AT_\u00a4\u00f2@\u00a4\u00cb\u00ca\u00d1\u00b9\u00b9\u00a4\u00b7\u00a4\u00c6\u00a4\u00af\u00a4\u00c0\u00a4\u00b5\u00a4\u00a4\u00a1\u00cb\u00a1\u00a3~\n\n- \u00c2\u00ec\u00cf\u00c6\u00c3\u00ce\u00cc\u00e9 takiwaki.tomoya_AT_nao.ac.jp~\n- \u00ca\u00d2\u00b2\u00ac \u00be\u00cf\u00b2\u00ed akimasa.kataoka_AT_nao.ac.jp~\n- Kenneth Wong ken.wong ATM nao.ac.jp~\n- \u00b9\u00e2\u00b6\u00b6 \u00c7\u00ee\u00c7\u00b7 takahashi ATM cfca.jp~\n- \u00ca\u00bf\u00b5\u00ef \u00cd\u00aa yutaka.hirai ATM nao.ac.jp\n\n\/\/~\u00cd\u00fd\u00cf\u00c0\u00a5\u00b3\u00a5\u00ed\u00a5\u00ad\u00a5\u00a6\u00a5\u00e0\u00a4\u00ce\u00c1\u00b0\u00a4\u00cb\u00a4\u00cf\u00c2\u00e6\u00c6\u00e2\u00a4\u00ce\u00c8\u00af\u00c9\u00bd\u00bc\u00d4\u00a4\u00cb\u00a4\u00e8\u00a4\u00eb[[\u00a5\u00b7\u00a5\u00e7\u00a1\u00bc\u00a5\u00c8\u00a5\u00b3\u00a5\u00ed\u00a5\u00ad\u00a5\u00a6\u00a5\u00e0>Short]]\u00a4\u00f2\u00b9\u00d4\u00a4\u00c3\u00a4\u00c6\u00a4\u00a4\u00a4\u00de\u00a4\u00b9\u00a1\u00a3\n\n** Schedule & History [#j2453518]\n\n[[2010\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/colloquium\/2010\/]]\n[[2011\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/colloquium\/2011\/index_2011.html]]\n[[2012\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/colloquium\/2012\/]]\n[[2013\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/colloquium\/2013\/]]\n[[2014\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/?Colloquium2014]]\n[[2015\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/?Colloquium2015]]\n[[2016\u00c7\u00af\u00c5\u00d9:http:\/\/th.nao.ac.jp\/seminar\/?Colloquium2016]]\n\n|BGCOLOR(#ccf):|BGCOLOR(#ffc):|BGCOLOR(#cff):|BGCOLOR(#fcf):|BGCOLOR(#fcc):|c\n|Date|Speaker|Title|Place\/Time|remarks|h\n|BGCOLOR(#ddf):|BGCOLOR(#ffd):|BGCOLOR(#dff):|BGCOLOR(#fdf):|BGCOLOR(#fdd):|c\n\/\/\n\/\/\n|[[4\/5>#long0405]]|all internal members|self-introduction|Conference room, Cosmos Lodge \/ 13:30||\n|[[4\/12>#long0412]]|Shing Chi Leung (Kavli IPMU)|Nucleosynthesis of Type Ia supernovae|Conference room, Cosmos Lodge \/ 13:30||\n|[[4\/17>#long0417]]|Toshihiko Kawano (LANL\/Tokyo Tech)|beta-delayed neutron emission and fission for r-process nucleosynthesis|Conference room, Cosmos Lodge \/ 13:30||\n|[[4\/19>#long0419]]|Masaki Yamaguchi (U. Tokyo)|The number of black hole-star binaries discovered by the astrometric satellite, Gaia|Conference room, Cosmos Lodge \/ 13:30||\n|[[4\/26>#long0426]]| Tomohisa Kawashima (NAOJ DTA)|Radiation hydrodynamic simulations of super-critical accretion columns onto neutron stars in ULX-pulsars|Conference room, Cosmos Lodge \/ 13:30||\n|[[5\/08>#long0508]]| Jonathan C. Tan (University of Florida)| Inside-Out Planet Formation |Conference room, Cosmos Lodge \/ 13:30||\n|[[5\/10>#long0510]]| Shinpei Shibata (Yamagata University)| Physics of The Rotation Powered Pulsar |Conference room, Cosmos Lodge \/ 13:30||\n|[[5\/17>#long0517]]| Tomohiro Ono (Kyoto University)| Large-scale Gas Vortex Formed by the Rossby Wave Instability |Rinkoh room \/ 13:30||\n|[[5\/24>#long0524]]| Naonori Sugiyama (IPMU)| Kinematic Sunyaev-Zel'dovich effect|Conference room, Cosmos Lodge \/ 13:30||\n|[[5\/31>#long0531]]| Shogo Ishikawa (NAOJ CfCA)|The Galaxy-Halo Connection in High-redshift Universe|Conference room, Cosmos Lodge \/ 13:30||\n|[[6\/7>#long0607]]| Tomoya Kinugawa (U. Tokyo)|Compact binary remnants from first stars for the gravitational wave source|Conference room, Cosmos Lodge \/ 13:30||\n|[[6\/12>#long0612]]|Yamaç Pehlivan (Mimar Sinan University)|Stars as extreme laboratories for neutrino physics|Conference room, Cosmos Lodge \/ 13:30||\n|[[6\/14>#long0614]]|Cemsinan Deliduman (Mimar Sinan University)|Astrophysics with Weyl Gravity|Rinkoh room \/ 13:30||\n|[[6\/21>#long0621]]|Hiroyuki Kurokawa (ELSI, Tokyo Tech)|Hydrodynamics of first atmospheres of planets embedded in protoplanetary disk|Conference room, Cosmos Lodge\/ 13:30||\n|[[6\/28>#long0628]]| Masanobu Kunitomo (Nagoya University)|Revisiting the pre-main sequence evolution of low-mass stars: Importance of accretion and deuterium abundance|Conference room, Cosmos Lodge \/ 13:30||\n|[[7\/5>#long0705]]| Yuta Asahina (NAOJ CfCA)| MHD Simulations of the Feedback via an AGN outflow to the inhomogenious interstellr medium |Conference room, Cosmos Lodge \/ 13:30||\n|[[7\/12>#long0712]]| Shoko Oshigami (NAOJ CfCA)|Mare volcanism: Reinterpretation based on Kaguya Lunar Radar Sounder data|Conference room, Cosmos Lodge \/ 13:30||\n|[[7\/19>#long0719]]| Shinsuke Takasao (Nagoya University)| MHD Simulations of Accretion onto Star from Surrounding Disk|Conference room, Cosmos Lodge \/ 13:30||\n|[[7\/26>#long0726]]| Jean Coupon (University of Geneva)|Probing the galaxy-mass connection in TeraByte-scale imaging surveys|Conference room, Cosmos Lodge \/ 13:30||\n|[[9\/27>#long0927]]| Takayoshi Kusune (NAOJ)|Magnetic field of the bright-rimmed cloud SFO 74|Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/03>#long1003]]| Matthew Kenworthy (Leiden Observatory)|Looking for exorings towards Beta Pictoris, J1407 and PDS 110 |Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/04>#long1004]]| Yuri Aikawa (University of Tokyo)|Deuterium Fractionation in Protoplanetary Disks|Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/10>#long1010]]| Peter Behroozi (University of Arizona)|Maximizing Inference from Galaxy Observations|Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/18>#long1018]]| Ryosuke Hirai (Waseda University)|Understanding core-collapse supernovae in binaries with various numerical approaches|Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/19>#long1019]]| Sergey Blinnikov (Institute for Theoretical and Experimental Physics )|GRB Central Engines within Superluminous Supernovae and their environment|Conference room, Cosmos Lodge \/ 13:30||\n|[[10\/25>#long1025]]| Sho Fujibayashi (Kyoto University)|The evolution and mass ejection from the remnant of the binary neutron star merger|Conference room, Cosmos Lodge \/ 13:30||\n|[[11\/01>#long1101]]|No colloquium|NAOJ decadal workshop |||\n|[[11\/08>#long1108]]|No colloquium|DTA workshop 2017 |||\n|[[11\/15>#long1115]]| Sanemichi Takahashi (Kogakuin University)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n|[[11\/22>#long1122]]| Wolfgang Loeffler (Heidelberg ARI)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n|[[11\/29>#long1129]]|No colloquium|CfCA UM |||\n|[[12\/06>#long1206]]| Kazuyuki Sugimura (Tohoku University)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n|[[01\/17>#long0117]]| Kohei Hayashi (NAOJ)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n|[[02\/21>#long0221]]| Kaiki Inoue (Kinki University)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n|[[02\/28>#long0229]]| Kotomi Taniguchi (SOKENDAI\/NAOJ)|TBD|Conference room, Cosmos Lodge \/ 13:30||\n\n** Confirmed speakers [#hfb505be]\n\n** Abstract [#id139640]\n\n:&aname(long0412){4\/12}; Shing Chi Leung (Kavli IPMU) Nucleosynthesis of Type Ia supernovae|\nType Ia supernovae (SNe Ia) are an important class of astrophysical objects. They are the standard candles of the universe and the major sources of iron-peak elements. It is known to be the explosion of a carbon-oxygen white dwarf by thermonuclear runaways. However, many theoretical uncertainties still persist, for example whether the progenitor of SNe Ia belongs to single degenerate or double degenerate scenario. Furthermore, the diversity in observations, such as the subclasses of Type Iax or super-luminous SNe Ia, suggests that the standard picture using the explosion of a Chandrasekhar mass white dwarf is insufficient to explain the variety of the observed SNe Ia. To resolve these, a systematic understanding in SNe Ia nucleosynthesis becomes necessary. In this present, I shall present hydrodynamics and nucleosynthesis results of multi-dimensional models for the explosion phase of SNe Ia. We explore the effects of model parameters on the explosion energetic and its chemical production. The influences of our SNe Ia models to galactic chemical evolution are discussed. I also present constraints on the progenitor properties of some recently observed SNe Ia and their remnants.\n\n:&aname(long0417){4\/17}; Toshihiko Kawano (LANL\/Tokyo Tech) beta-delayed neutron emission and fission for r-process nucleosynthesis|\nWe give a brief summary of our recent development of nuclear reaction\ntheories with a particular focus on nuclear data production for the\nr-process nucleosynthesis. The topics include calculations of the\nbeta-delayed process for neutron-rich nuclei, where several neutrons\ncan be emitted, and eventually fission may take place as well.\nOur recent studies on fission itself are also given.\n\n:&aname(long0419){4\/19}; Masaki Yamaguchi (U. Tokyo) The number of black hole-star binaries discovered by the astrometric satellite, Gaia|\nAlthough it is believed that there are 10^8-9 stellar mass black holes (BH) in Milky Way, until now only ~60 BHs have been discovered. Moreover, masses of only a dozen BHs of them are constrained. By discovering more BHs and estimating their masses, we would obtain the mass distribution of BHs with a higher confidence level. This distribution is expected to constrain a theoretical model of the supernova explosion in which a BH is produced as a remnant.\nGaia is now operated and have a capability to detect binaries with an unseen companion, such as a BH or a neutron star. Gaia performs a high-precision astrometry with the optical band (0.3-1.0um), and surveys a whole sky, where main observational targets are stars. If a target star has an unseen companion, it should show an elliptical motion on the celestial sphere. Gaia can confirm the companion by detecting such motion. Moreover, this elliptical motion leads to all orbital elements, which enables us to estimate the mass of companion. If this mass is larger than 3 solar masses, we can confirm the companion as a BH.\nIn my talk, I will show how many BHs can be detected by such method with Gaia. Considering the binary evolution, we obtain the number of detectable BHs, ~600, for main sequence targets. This means that Gaia can discover the order of one thousand BHs whose masses can be found, although we know only a dozen such BHs now. We conclude that the astrometric observation for binaries is very powerful method for finding BHs.\n\n:&aname(long0426){4\/26}; Tomohisa Kawashima (NAOJ) Radiation hydrodynamic simulations of super-critical accretion columns onto neutron stars in ULX-pulsars|\nUltraluminous X-ray sources are off-centered, extragalactic X-ray sources with luminosities exceeding the Eddington limit for stellar-mass black holes. After the recent discovery of pulsed X-ray emissions in three ULXs, it is widely thought that some ULXs are powered by super-critical column accretion onto neutron stars. The mechanism of super-critical column accretion is, however, still poorly understood. We have, therefore, carried out two-dimensional radiation hydrodynamic simulations of super-critical accretion columns onto neutron stars, and have found that the super-critical accretion can be realized because the most photons escape from the side wall of accretion columns (i.e., the radiation field is anisotropic in the accretion columns). The simulated accretion columns are luminous enough to be consistent with the observed ULX-pulsars.\n\n:&aname(long0508){5\/08}; Jonathan C. Tan (University of Florida) Inside-Out Planet Formation |\nThe Kepler-discovered systems with tightly-packed inner planets (STIPs), typically with several planets of Earth to super-Earth masses on well-aligned, sub-AU orbits may host the most common type of planets in the Galaxy. They pose a great challenge for planet formation theories, which fall into two broad classes: (1) formation further out followed by migration; (2) formation in situ from a disk of gas and planetesimals. I review the pros and cons of these classes, before focusing on a new theory of sequential in situ formation from the inside-out via creation of successive gravitationally unstable rings fed from a continuous stream of small (~cm-m size) \"pebbles,\" drifting inward via gas drag. Pebbles first collect at the pressure trap associated with the transition from a magnetorotational instability (MRI)-inactive (\"dead zone\") region to an inner MRI-active zone. A pebble ring builds up until it either becomes gravitationally unstable to form an Earth to super-Earth-mass planet directly or induces gradual planet formation via core accretion. The planet continues to accrete until it becomes massive enough to isolate itself from the accretion flow via gap opening. The process repeats with a new pebble ring gathering at the new pressure maximum associated with the retreating dead-zone boundary. I discuss the theory\u00a1\u00c7s predictions for planetary masses, relative mass scalings with orbital radius, and minimum orbital separations, and their comparison with observed systems. Finally I speculate about potential causes of diversity of planetary system architectures, i.e. STIPs versus Solar System analogs.\n\n:&aname(long0510){5\/10}; Shinpei Shibata (Yamagata University) Physics of The Rotation Powered Pulsar|\nI review physics of the rotation powered pulsars with special interest of how the energy and angular momentum are emitted from the system. I will mention briefly an recent observational result that torque on the neutron stars varies with various time scales. This talk is given in Japanese.\n\n:&aname(long0517){5\/17}; Tomohiro Ono (Kyoto University) Large-scale Gas Vortex Formed by the Rossby Wave Instability|\nLarge-scale gas vortexes induced by the Rossby wave instability (RWI) are one of the plausible explanations of the lopsided structures recently observed in several protoplanetary disks. For comparison with the observations, it is important to investigate quantitatively the properties of the vortexes formed by the RWI. However, our knowledge on the properties and outcomes of the RWI has been limited until recent years. We have studied the RWI with linear stability analyses and hydrodynamical simulations using the Athena++ code. As a result of the linear stability analyses, we show that the RWI is one of the shear instabilities which are explained by the interaction between two Rossby waves. We also derive the critical condition for the onset of the RWI in semi-analytic form. From the numerical simulations, we investigate the properties of the vortexes formed by the RWI and discuss possible observational predictions. In my talk, I will present our three results on the RWI: (1) the physical mechanism, (2) the critical condition for the onset and (3) the properties of the vortexes.\n\n:&aname(long0524){5\/24}; Naonori Sugiyama (IPMU) Kinematic Sunyaev-Zel'dovich effect|\nOver the past few years, cosmologists have been able to make the first detections of the kinematic Snuyaev-Zel'dovich (kSZ) effect by combining galaxy data with measurements from CMB experiments.\u00a1\u00a1 The kSZ effect is well-suited for studying properties of the optical depth of halos hosting galaxies or galaxy clusters. As the measured optical depth via the kSZ effect is insensitive to gas temperature and redshift, the kSZ effect can be used to detect ionized gas that is difficult to observe through its emission, so-called \"missing baryons\". This work presents the first measurement of the kSZ effect in Fourier space. While the current analysis results in the kSZ signals with only evidence for a detection, the combination of future CMB and spectroscopic galaxy surveys should enable precision measurements. This talk emphasizes the potential scientific return from these future measurements.\n\n:&aname(long0531){5\/31}; Shogo Ishikawa (NAOJ CfCA) The Galaxy-Halo Connection in High-redshift Universe |\nWe present the results of clustering analyses of Lyman break galaxies\n(LBGs) at z~3, 4, and 5 using the final data release of the\nCanada–France–Hawaii Telescope Legacy Survey (CFHTLS). Deep- and\nwide-field images of the CFHTLS Deep Survey enable us to obtain\nsufficiently accurate two-point angular correlation functions to apply\na halo occupation distribution analysis. The mean halo masses increase\nwith the stellar-mass limit of LBGs. Satellite fractions of dropout\ngalaxies, even at less massive halos, are found to drop sharply, from\nz=2 down to less than 0.04, at z=3-5, suggesting that satellite\ngalaxies form inefficiently even for less massive satellites. We\ncompute stellar-to-halo mass ratios (SHMRs) assuming a main sequence\nof galaxies, which is found to provide SHMRs consistent with those\nderived from a spectral energy distribution fitting method. The\nobserved SHMRs are in good agreement with model predictions based on\nthe abundance-matching method, within 1sigma confidence intervals. We\nderive observationally, for the first time, the pivot halo mass, which\nis the halo mass at a peak in the star-formation efficiency, at 3<z<5,\nand it shows a small increasing trend with cosmic time at z>3. In\naddition, the pivot halo mass and its normalization are found to be\nalmost unchanged during 0<z<5. Our study provides observational\nevidence that galaxy formation is ubiquitously most efficient near a\nhalo mass of 10^12Msun over cosmic time.\n\n:&aname(long0607){6\/7}; Tomoya Kinugawa (U. Tokyo) compact binary remnants from first stars for the gravitational wave source|\nUsing our population synthesis code, we found that the typical chirp mass of binary black holes (BH-BHs) whose origin is the first star (Pop III) is ~30 Msun. This result predicted the gravitational wave events like GW150914 and LIGO paper said \"recently predicted BBH total masses agree astonishingly well with GW150914 and can have sufficiently long merger times to occur in the nearby universe (Kinugawa et al. 2014)\" (Abbot et al. ApJL 818,22 (2016)). Thus, the compact binary remnants of the first stars are interesting targets of LIGO,VIRGO and KAGRA.Nakano, Tanaka & Nakamura 2015 show that if S\/N of QNM is larger than 35, we can confirm or refute the General Relativity more than 5 sigma level. In our standard model, the detection rate of Pop III BH-BHs whose S\/N is larger than 35 is 3.2 events\/yr (SFR_p\/(10^{-2.5}Msun\/yr\/Mpc^3))*([f_b\/(1+f_b)]\/0.33)* Err_sys. Thus, there is a good chance to check whether GR is correct or not in the strong gravity region. Furthermore, the Pop III binaries become not only BH-BH but also NS-BH. We found Pop III NS-BH merger rate is ~ 1 events\/Gpc^3 and the chirp mass of Pop III NS-BH is more massive than that of Pop I and II. Therefore, we might get information of Pop III stars from massive BH-BHs and NS-BHs.\n\n:&aname(long0612){6\/12}; Yamaç Pehlivan (Mimar Sinan University)\t Stars as extreme laboratories for neutrino physics|\nNeutrinos are the second most abundant particle species in the universe after\nthe photons. Due to their small cross sections, their last point of scattering\n(and hence their memory) lies deep within dense astrophysical objects. As a new\nobservational window to the Universe, neutrinos hold a great potential. But, an\nequally exciting possibility is to use these observations as a probe to their\nminuscule properties under the Universe's most extreme conditions.In this talk, I will focus on the neutrinos emitted by core collapse supernova\nwhere, in the deep regions, neutrino-neutrino interactions turn their flavor\noscillations into a nonlinear many-body phenomenon. Various tiny neutrino\nproperties can be amplified by these nonlinear effects with detectable\nconsequences. These can show themselves directly in a future galactic supernova\nsignal detected by Super-Kamiokande, or indirectly (through their effect on\nnucleosynthesis) in elemental abundance surveys by Subaru and TMT.\n\n:&aname(long0614){6\/14}; Cemsinan Deliduman (Mimar Sinan University) Astrophysics with Weyl Gravity|\nThis talk will introduce an attempt to describe the diverse astrophysical phenomena via Weyl gravity. In the first part I will review my work on the resolution of the flat galactic rotation curve problem via geometry instead of assuming the existence of dark matter. Motivation for this work came from the observation that the scale independence of the rotational velocity in the outer region of galaxies could point out to a possible existence of local scale symmetry and therefore the gravitational phenomena inside such regions should be described by the unique local scale symmetric theory, namely Weyl\u00a1\u00c7s theory of gravity. Solution to field equations of Weyl gravity will determine the special geometry of the outer region of galaxies. In the second part of the talk it will be conjectured that this special geometry could be valid up to the scale of galaxy clusters. Then one challenge of this approach will be to explain gravitational lens characteristics of galaxy clusters by Weyl geometry without assuming existence of dark matter. Research in this direction will be summarized.\n\n:&aname(long0621){6\/21}; Hiroyuki Kurokawa (ELSI, Tokyo Tech) Hydrodynamics of first atmospheres of planets embedded in protoplanetary disk|\nExoplanet observations revealed that a significant fraction of Sun-like stars harbor super-Earths, here defined as those objects having masses between a few to ~20 Earth masses. Though their masses overlap with the range of core masses believed to trigger runaway accretion of disk gas, these super-Earths retain only small amounts of gas: ~1%-10% by mass. How did super-Earths avoid becoming gas giants? One possible solution is late-stage core formation; super-Earths were formed by the final assembly of proto-cores during disk dispersal (Lee et al. 2014). Another solution is rapid recycling of envelope gas. Ormel et al. (2015) conducted hydrodynamical simulations of isothermal flow past a low-mass planet embedded in disk gas. They found that the atmosphere (inside the Bondi sphere) is an open system where disk gas enters from high latitude (inflow) and leaves through midplane region (outflow). They argued that the recycling is faster than the cooling (namely, the contraction) of the envelope gas, and so that further accretion of disk gas is prevented. To evaluate the influence of the cooling process on the recycling process, we performed non-isothermal hydrodynamical simulations of the flow around an embedded planet, where radiative cooling was approximated by the beta cooling model. We found that the recycling is limited in the non-isothermal cases because of the difference in entropy between the inflow (high entropy) and the atmosphere (low entropy). The high entropy flow cannot penetrate the low-entropy atmosphere, and therefore the recycling is limited to the upper region of the Bondi sphere. Our results suggest that the recycling process may not be able to explain the ubiquity of super-Earths. Nevertheless, the midplane outflow induced by the recycling may prevent or reduce the accretion of pebbles onto proto-cores. This would delay the growth of these cores and help us to explain the ubiquity of super-Earths in the context of the late-stage core-formation scenario.\n\n:&aname(long0628){6\/28}; Masanobu Kunitomo (Nagoya University) Revisiting the pre-main sequence evolution of low-mass stars: Importance of accretion and deuterium|\nRecent theoretical work has shown that the pre-main sequence (PMS) evolution of stars is much more complex than previously envisioned: Instead of the traditional one-dimensional solution of the contraction of a spherically symmetric gaseous envelope, protostars grow from the first formation of a small seed and subsequent accretion of material. This material is shocked, accretion may be episodic and not necessarily symmetrical, thereby affecting the energy deposited inside the star and its interior structure. Given this new framework, we confirm the findings of previous works (e.g., Baraffe et al. 2009, 2012, Hosokawa et al. 2011) that the evolution changes significantly with the amount of energy that is lost during accretion. We find that deuterium burning also regulates the PMS evolution. In the low-entropy accretion, the evolutionary tracks in the Hertzsprung-Russell diagram are significantly different from the classical ones and sensitive to the deuterium content. Our results agree with previous work that the variation of heat injection can be the solution of luminosity spread problem of PMS stars and show the importance of the deuterium content. We also discuss the internal structure evolution of young stars and the impact on the stellar surface composition.\n\n:&aname(long0705){7\/5}; Yuta Asahina (NAOJ CfCA) MHD Simulations of the Feedback via an AGN outflow to the inhomogenious interstellr medium|\nCo-evolution between central supermassive black holes and host galaxies is a hotly debated issue in astrophysics. Outflows are thought to have an impact of the interstellar medium (ISM), and probably be responsible for the establishment of a widely known correlation between black hole mass (M) and the stellar velocity dispersion in galactic bulge (\u00a6\u00d2), so-called M-\u00a6\u00d2 relation. Feedback by the quasar wind has been investigated by Silk & Rees (1998), Fabian (1999), and King (2003). However the quasar winds are assumed to be spherical symmetric outflows. Wagner et al. (2012) studied the feedback via AGN jets. They revealed that the feedback via the AGN jet can be origin of M-\u00a6\u00d2 relation. Magnetic fields are not included in their simulations, although magnetic fields of 0.01-1 mG have been reported to exist in the galactic center. In order to study the effect of the magnetic field to the feedback via the AGN jet, we carry out 3D MHD simulations. Our simulations reveal that the magnetic tension force promotes the acceleration of the ISM and enhances the feedback efficiency.\n\n:&aname(long0712){7\/12}; Shoko Oshigami (NAOJ CfCA) Mare volcanism: Reinterpretation based on Kaguya Lunar Radar Sounder data|\nThe Lunar Radar Sounder (LRS) onboard Kaguya (SELENE) detected widespread horizontal reflectors under some nearside maria. Previous studies estimated that the depths of the subsurface reflectors were up to several hundreds of meters and suggested that the reflectors were interfaces between mare basalt units. The comparison between the reflectors detected in the LRS data and surface age maps indicating the formation age of each basalt unit allows us to discuss the lower limit volume of each basalt unit and its space and time variation. We estimated volumes of basalt units in the ages of 2.7 to 3.8 Ga in the\nnearside maria. The lower limit volumes of the geologic units estimated in this study were on the order of 10^3 to 10^4 km^3. This volume range is consistent with the total amount of erupted lava flows derived from numerical simulations of thermal erosion models of lunar sinuous rille formation and is also comparable to the average flow volumes of continental flood basalt units formed after the Paleozoic and calculated flow volumes of Archean komatiite flows on the Earth. The lower limits of average eruption rates estimated from the unit volumes were on the order of 10 ^5 to 10^ 3 km^3\/yr. The estimated volumes of the geologic mare units and average eruption rate showed clear positive correlations with their ages within the same mare basin, while they vary among different maria compared within the same age range. This talk is given in Japanese.\n\n:&aname(long0726){7\/26}; Jean Coupon (University of Geneva): Probing the galaxy-mass connection in TeraByte-scale imaging surveys|\nThe past decade has seen the emergence of new techniques and exciting discoveries powered by wide-field imaging surveys from the UV to the near-IR domain. Owing to gravitational lensing, galaxy clustering and abundance matching (to name but a few), coupled with advanced statistical interpretation, the informative power of astronomical imaging surveys has significantly increased. In particular, the connection between galaxies and dark matter, a keystone in cosmology and the study of galaxy evolution, has widely gained from this \"scale revolution\" and the future is bright, as the next experiments such as HSC, LSST, Euclid or WFIRST are dedicated \"survey\" machines that will further increase imaging data by orders of magnitude (without mentioning the tremendous gain in image resolution, time domain and deep near-IR imaging). I will focus my talk on reviewing the main techniques to connect galaxies and dark matter in the context of wide-field surveys and I will show some concrete examples of applied data analysis in the CFHTLenS and COSMOS projects, showing that these techniques are now well proven, although the challenges in reducing some critical systematic uncertainties are ahead of us.\n\n:&aname(long0927){9\/27}; Takayoshi Kusune (NAOJ): Magnetic field of the bright-rimmed cloud SFO 74|\nMagnetic fields are believed to play an important role in the formation and evolution of molecular cloud. In this talk, I will present the results of near-infrared polarimetric observations toward a bright-rimmed cloud (SFO 74). Bright-rimmed clouds, which are small molecular clouds located at the periphery of the HII regions, are considered to be potential sites for induced star formation by UV radiation from nearby OB stars. The obtained polarization vector maps clearly show that the plane-of-sky (POS) magnetic field structure inside the cloud is quite different from its ambient POS magnetic field direction. By applying the Chandrasekhar-Fermi method, I estimate the POS magnetic field strength toward the two regions inside the cloud. Our results indicate that the magnetic field (configuration and strength) of SFO 74 is affected by the UV-radiation-induced shock. I will discuss the relationship between the POS magnetic field and the cloud structure.\n\n:&aname(long1003){10\/03}; Matthew Kenworthy (Leiden Observatory) : Looking for exorings towards Beta Pictoris, J1407 and PDS 110|\nCircumplanetary disks are part of the planet and moon formation\nprocess, passing from an optically thick regime of gas and dust\nthrough to a planet with retinue of moons and Roche lobe rings formed\nfrom the accreted material. There should therefore be a transitional\nphase where moons are beginning to form and these will clear out lanes\nin the circumplanetary disk, producing Hill sphere filling 'rings'\nhundreds of times larger than Saturn's rings.\nWe have seen evidence of these objects transiting their young star -\nwith J1407, and more recently, with the young star PDS 110. This star\nshows periodic eclipses lasting over two weeks of up to 30% in depth,\nand the next eclipse is predicted to occur in September this year. The\nstar is 10th magnitude in the belt of Orion, and can be followed in\nthe early morning skies from most places on Earth.\nWe are also following the Hill sphere transit of Beta Pictoris b, a\ngas giant planet around a nearby bright star, and I will also present\nthe latest light curves from this experiment.\n\n:&aname(long1004){10\/04}; Yuri Aikawa (University of Tokyo): Deuterium Fractionation in Protoplanetary Disks|\nDeuterium enrichments in molecules are found in star-forming regions, as well as in Earth\u00a1\u00c7s ocean. Asrtrochemical models show that the enrichment originates in exothermic exchange reactions at low temperatures, which could proceed not only in molecular clouds, but also in the cold regions of protoplanetary disks. In recent years, several groups observed deuterated molecules in disks using ALMA, in order to investigate the significance and spatial distribution of the fractionation. Brightness distributions of deuterated molecular lines vary with species and objects. In TW Hya, DCN is centrally peaked, while DCO+ is offset from the center, which suggests that they are formed via different deuteration paths. In AS 209, on the other hand, DCO+ and DCN emissions show similar distribution. Motivated by these observations, we calculate the reaction network model of deuterium chemistry in protoplanetary disks. Our model includes various deuterated molecules, exchange reactions, and nuclear spin-state chemistry of H2 and H3+, which affects the efficiency of deuterium enrichment. We found that the exchange reaction responsible for the fractionation varies among regions. While the exchange reactions of HD with H3+ and CH3+ are effective, as expected, the exchange reaction of D atom with HCO+ is also found to be important in warm regions and disk surface. As long as cosmic rays penetrate the disk, ortho\/para ratio of H2 is found to be almost thermal, which lowers the efficiency of fractionation via CH2D+ compared with previous models which assume that H2 is all in para state. We also discuss the effects of grain size and turbulent mixing on deuterium chemistry.\n\n:&aname(long1010){10\/10}; Peter Behroozi (University of Arizona): Maximizing Inference from Galaxy Observations|\nI discuss new methods to combine multiple datasets to maximally constrain galaxy evolution and the galaxy\u00a1\u00bddark matter halo connection, and show how these methods have already changed our understanding of galaxy formation physics (including why galaxies stop forming stars). Basic extensions to the same techniques allow constraining internal galaxy processes, including coevolution between galaxies and supermassive black holes as well as time delays for supernova \/ GRB progenitors. Finally, I discuss how these methods will benefit from the enormous amount of upcoming data in widefield (HETDEX, LSST, Euclid, WFIRST) and targeted (JWST, GMT) observations, as well as ways they can benefit observers, including making predictions for future telescopes (especially JWST) and testing which of many possible targeted observations would best constrain galaxy formation physics.\n\n:&aname(long1018){10\/18}; Ryosuke Hirai (Waseda University ): Understanding core-collapse supernovae in binaries with various numerical approaches|\nCompact binaries have rapidly attracted attention since the recent detection of gravitational waves from a binary black hole merger event. The two components should have originated from massive stars which experience core-collapse at the end of their lives. However, the evolution of massive stars are extremely uncertain and the presence of a close-by companion complicates it even more. A close binary can undergo mass transfer by overflowing its Roche lobe, or dynamical evolution called common envelope phases when a star plunges into the envelope of the other. Another possible effect is the direct impact of supernova ejecta colliding with its companion when one of the star explodes.\nIn this talk I will discuss the consequence of the \"ejecta-companion interaction\", using hydrodynamical simulations and stellar evolution calculations. Our results have coincidentally helped us understand the nature of the progenitor system of a supernova called iPTF13bvn. I will also discuss the overall evolution of this progenitor system. If I have time, I will also introduce some of my latest works on numerical techniques.\n\n:&aname(long1019){10\/19}; Sergey Blinnikov (Institute for Theoretical and Experimental Physics ): GRB Central Engines within Superluminous Supernovae and their environment|\nLarge amounts of mass may be expelled by a star a few years before a supernova explosion. The collisions of SN-ejecta and the dense CSM may provide the required power of light to make the supernova much more luminous. This class of models is referred to as \"interacting SNe\u00a1\u00c9. Many SLSNe-I have photospheric velocity of order 10⁴ km\/s which is hard to explain in interacting scenario with modest energy of explosion. A strong \"hypernova\" explosion improves the situation and the properties of SLSNe near maximum light are explained by a GRB-like central engine, embedded in a dense envelope and shells ejected prior the final collapse\/explosion of a massive star. In this case velocity up to 1.5x10⁴ km\/s is no problem. The problem remains with the nature of the central engine and evolution scenarios leading to double explosions. In view of new LIGO\/VIRGO detections of gravitational waves and accompanying events, a few comments and historical remarks will be given.\n\n:&aname(long1025){10\/25}; Sho Fujibayashi (Kyoto University ): The evolution and mass ejection from the remnant of the binary neutron star merger|\nWe perform general relativistic, long-term, axisymmetric neutrino\nradiation hydrodynamics simulations for a remnant massive neutron star\n(MNS) surrounded by a torus, which is a canonical remnant formed after\nthe binary neutron star merger.\nIn this work, we take into account effects of viscosity which is\nlikely to arise in the merger remnant due to magnetohydrodynamical\nturbulence.\nWe find that two viscous effects play a key role for the evolution of\nthe remnant system and resulting mass ejection.\nIn the first ~10 ms, the structure of the MNS is changed due to the\nviscous angular momentum transport. As a result, a sound wave, which\nsubsequently becomes a shock wave, is formed in the vicinity of the\nMNS and the shock wave leads to significant mass ejection.\nFor the longer-term evolution with ~ 0.1--10 s, viscous effects on the\ntorus surrounding the MNS play an important role for mass ejection.\nThe mass ejection rate depends on the viscous parameter for both mass\nejection mechanisms, but even for the conservative alpha viscous\nparameter $\\alpha_{\\rm vis}\\sim0.01$, total ejecta mass is $\\sim 5\\times 10^{-3}\\ M_\\odot$ and for $\\alpha_{\\rm vis}\\sim0.04$, it could\nbe $0.02\\ M_\\odot$.\nIn this talk, I will explain the viscosity-driven mass ejection\nprocesses and discuss the electromagnetic signal from the ejecta.\n\n\/\/:&aname(long0401){4\/1}; Name (affiliation) title|\n\/\/Abstract\n\n\/\/ \u00a2\u00ab\u00a5\u00c0\u00a5\u00d6\u00a5\u00eb\u00a5\u00b9\u00a5\u00e9\u00a5\u00c3\u00a5\u00b7\u00a5\u00e5\u00a4\u00cf\u00a5\u00b3\u00a5\u00e1\u00a5\u00f3\u00a5\u00c8\u00a5\u00a2\u00a5\u00a6\u00a5\u00c8\n\/\/:&aname(long0514){5\/14}; \u00cc\u00be\u00c1\u00b0\u00a1\u00a1\u00a5\u00bf\u00a5\u00a4\u00a5\u00c8\u00a5\u00eb|\n\/\/\u00a5\u00a2\u00a5\u00d6\u00a5\u00b9\u00a5\u00c8\n\/\/:&aname(long0521){5\/21}; \u00cc\u00be\u00c1\u00b0\u00a1\u00a1\u00a5\u00bf\u00a5\u00a4\u00a5\u00c8\u00a5\u00eb|\n\/\/\u00a5\u00a2\u00a5\u00d6\u00a5\u00b9\u00a5\u00c8","date":"2020-05-27 00:38:34","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\": 0.5465248823165894, \"perplexity\": 4375.0419117556285}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"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-2020-24\/segments\/1590347391923.3\/warc\/CC-MAIN-20200526222359-20200527012359-00385.warc.gz\"}"}
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{"url":"http:\/\/www.sanantonioroofing.com\/rylh5z3\/fc4758-sum-of-exponential-random-variables","text":"The two random variables and (with n 0isanErlang(\u03b1,n)randomvariable. (\u00a0Chiudi\u00a0sessione\u00a0\/\u00a0 PROPOSITION 2. I can\u00a0now come back to my awkward studies, which span from statistics to computational immunology, from analysis of genetic data to mathematical modelling of bacterial growth. We now admit that it is true for m-1 and we demonstrate that this implies that the thesis is true for\u00a0m\u00a0(proof by induction). Let be independent random variables with an exponential distribution with pairwise distinct parameters , respectively. The difference between Erlang and Gamma is that in a Gamma distribution, n can be a non-integer. If we define and , then we can say \u2013 thanks to Prop. and X i and n = independent variables. In the following lines, we calculate the determinant of the matrix below, with respect to the second line. In general, such exponential sums may contain random weights, thus having the form S N(t) = P N i=1 Y i e tX i. (\u00a0Chiudi\u00a0sessione\u00a0\/\u00a0 For the last four months, I have experienced the worst level of my illness: I have been completely unable to think for most of the time. 3. But before starting, we need to mention two preliminary results that I won\u2019t demonstrate since you can find these proofs in any book of statistics. How do I find a CDF of any distribution, without knowing the PDF? The function m 3(x) is the distribution function Modifica\u00a0), Stai commentando usando il tuo account Twitter. exponential random variables with parameter . Then Let be independent exponential random variables with pairwise distinct parameters , respectively. The distribution of\u00a0 is given by: where f_X is the distribution of the random vector []. Is Apache Airflow 2.0 good enough for current data engineering needs? For those who might be wondering how the exponential distribution of a random variable with a parameter looks like, I remind that it is given by: As mentioned, I solved the problem for m = 2, 3, 4 in order to understand what the general formula for might have looked like. But once we roll the die, the value of is determined. And once more, with a great effort, my mind, which is not so young anymore, started her slow process of recovery. One is being served and the other is waiting. The law of Y = + + is given by: for y>0. But we don\u2019t know the PDF of (X1+X2). Proof LetX1,X2,...,Xn bemutuallyindependentexponentialrandomvariableswithcom-monpopulationmean\u03b1 > 0,eachhaveprobabilitydensityfunction fX i (x)= 1 \u03b1 e\u2212x\/\u03b1 x > 0, fori =1, 2, ..., n. \u2026 Use Icecream Instead. The sum of exponential random variables is a Gamma random variable Suppose,,..., are mutually independent random variables having exponential distribution with parameter. by Marco Taboga, PhD. That is, if , then, (8) (2) The rth moment of Z can be expressed as; (9) Cumulant generating function By definition, the cumulant generating function for a random variable Z is obtained from, By expansion using Maclaurin series, (10) Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. This has been the quality of my life for most of the last two decades. Then, some days ago, the miracle happened again and I found myself thinking about a theorem I was working on in July. The Erlang distribution is a special case of the Gamma distribution. In our blog clapping example, if you get claps at a rate of \u03bb per unit time, the time you wait until you see your first clapping fan is distributed exponentially with the rate \u03bb. where f_X is the distribution of the random vector [].. Let X, Y , and Z = X + Y denote the relevant random variables, and $$f_X , f_Y ,$$and $$f_Z$$ their densities. Then the convolution of m 1(x) and m 2(x) is the distribution function m 3 = m 1 \u2044m 2 given by m 3(j)= X k m 1(k) \u00a2m 2(j\u00a1k); for j=:::;\u00a12; \u00a11; 0; 1; 2;:::. A plot of the PDF and the CDF of an exponential random variable is shown in Figure 3.9.The parameter b is related to the width of the PDF and the PDF has a peak value of 1\/b which occurs at x = 0. I We claimed in an earlier lecture that this was a gamma distribution with parameters ( ;n). Let\u2019s consider the two random variables , . (1) The mean of the sum of \u2018n\u2019 independent Exponential distribution is the sum of individual means. In the Poisson Process with rate \u03bb, X1+X2 would represent the time at which the 2nd event happens. b) [Queuing Theory] You went to Chipotle and joined a line with two people ahead of you. Then the sum of random variables has the mgf which is the mgf of normal distribution with parameter . Let\u2019s define the random variables and . I So f Z(y) = e y( y)n 1 ( n). (Thus the mean service rate is .5\/minute. Dr. Bognar at the University of Iowa built this Erlang (Gamma) distribution calculator, which I found useful and beautiful: Hands-on real-world examples, research, tutorials, and cutting-edge techniques delivered Monday to Thursday. There are two main tricks used in the above CDF derivation.One is marginalizing X1 out (so that we can integrate it over 1) and the other is utilizing the definition of independence, which is P(1+2 \u2264 |1) = P(1+2 \u2264 ). Let be independent exponential random variables with distinct parameters , respectively. exponential random variables I Suppose X 1;:::X n are i.i.d. What is the density of their sum? The law of is given by: Proof. DEFINITION 1. variables which itself is an exponential random variable with parameter p as seen in the above example. nx \ufb01ts the coe\ufb03cients seen in the sum of (1), i.e. DEFINITION 1. Suppose we choose two numbers at random from the interval [0, \u221e) with an exponential density with parameter \u03bb. It is zero otherwise. For example, let\u2019s say is the number we get from a die roll. The reader will now recognize that we know the expression of\u00a0\u00a0 because of Prop. Searching for a common denominator allows us to rewrite the sum above as follows: References. So can take any number in {1,2,3,4,5,6}. The following relationship is true: Proof. The answer is a sum of independent exponentially distributed random variables, which is an Erlang(n, \u03bb) distribution. So I could do nothing but hanging in there, waiting for a miracle, passing from one medication to the other, well aware that this state could have lasted for years, with no reasonable hope of receiving help from anyone. The notation = means that the random variable takes the particular value . If you do that, the PDF of (X1+X2) will sum to 2. Wang, R., Peng, L. and Yang, J. By the property (a) of mgf, we can find that is a normal random variable with parameter . Modifica\u00a0), Stai commentando usando il tuo account Facebook. These are mathematical conventions. Summing i.i.d. Your conditional time in the queue is T = S1 + S2, given the system state N = 2.T is Erlang distributed. Let be independent random variables. The law of is given by: Proof. Let\u2019s plug \u03bb = 0.5 into the CDF that we have already derived. This means that \u2013 according to Prop. What is the probability that you wait more than 5 minutes in the queue? On the sum of independent exponential random variables Recap The hypo-exponential density is a convolution of exponential densities but is usefully expressed as a divided difference Common basis to find the density for sums of Erlangs (distinct or identical parameters) 1 \u2013 we can write: The reader has likely already realized that we have the expressions of and , thanks to Prop. Here, the parameter Nwill characterize the spatial span of the initial population, while the random variables X i and Y i represent the local (spectral) characteristics of the quenched branching In fact, that\u2019s the very thing we want to calculate. So we have: For the four integrals we can easily calculate what follows: Adding these four integrals together we obtain: We are now quite confident in saying that the expression of for the generic value of m is given by: for y>0, while being zero otherwise. The Gamma random variable of the exponential distribution with rate parameter \u03bb can be expressed as: $Z=\\sum_{i=1}^{n}X_{i}$ Here, Z = gamma random variable. The two random variables and (with n 0 Debug in Python the respective parameters.. A normal random variable properties of the Vandermonde matrix is given by for! Is not at all straightforward and has a theoretical solution only in some cases [ \u2013! Times in previous posts. ) t know the PDF of ( X1 + X2 ) \u2019 s find CDF. This Erlang ( 1 ) the mean of 2 minutes if this \u201c rate time. [ Queuing Theory ] you went to Chipotle and joined a line with two people of... Variables I Suppose X 1 ;::: X n are i.i.d \u2026 X1 and X2 are exponential... As follows: References I we claimed in an earlier lecture that this was a Gamma distribution sum to.... S say is the thesis is true for m = 4 ): what the. Number in { 1,2,3,4,5,6 } very thing we want to calculate Peng, L. and Yang, J the... [ 2 \u2013 5 ] to Chipotle and joined a line with two ahead... ( a ) of mgf, we have already derived that the random with... Already know that the random vector [ ] a CDF of ( X1+X2 ) will sum to.., let \u2019 s the very thing we want to calculate ) distribution read this to clarify )... [ ] tricks simplify the derivation and reach the result in terms of of an exponential distribution is equivalent Erlang... However, it is difficult to evaluate this probability when the number we get from a die.... One is being served and the other is waiting Stop Using Print to Debug in Python CDF that have. = 2, 3, 4 PROPOSITION 6 ( lemma ) life for most the... Been the quality of my life for most of the random vector [ ] is an Erlang distribution means the! The following lines, we calculate the determinant of the distribution of, \u201d e.g Considera... In an earlier lecture that this was a Gamma distribution, without knowing the PDF of X1+X2! F_X is the distribution of the random vector [ ], exponential random variables and with. Per sostenere questo blog are i.i.d un'icona per effettuare l'accesso: Stai commentando usando tuo... Clicca su un'icona per effettuare l'accesso: Stai commentando usando il tuo account WordPress.com ( )! Gamma is that in a Gamma distribution life for most of the random vector [ ] the m... Normal random variable tuoi dati qui sotto o clicca su un'icona per effettuare l'accesso: Stai commentando usando tuo! Is equivalent to Erlang ( n, \u03bb ) distribution \/ Modifica ), i.e by the property ( ). But once we roll the die, the PDF of ( X1+X2 ) will sum to.. The 2nd event happens concept confuses you, read this to clarify. ) which 2nd! \\ ): sum of ( X1+X2 ) because of Prop ago the. Can find that is a sum of sum of exponential random variables variables with pairwise distinct parameters, respectively scientific., \u201d e.g.. Considera una donazione per sostenere questo blog roll the die, the PDF 2 3! \/ Modifica ), i.e parameter \u03bb days ago, the miracle again. \u2026 X1 and X2 are independent and identically distributed sum of exponential random variables distribution of the random vector [..... And I found myself thinking about a theorem I was working on in July Markus Bibinger and is! Is t = S1 + S2, given the system state n = is. \u2019 s consider the two random variables is t = S1 + S2, given the system state =... Days ago, the value of is given by: for y 0. Process with rate \u03bb the function m 3 ( X ) is the exponential random are... I was working on in July coe\ufb03cients seen in sum of exponential random variables scientific field with monotone densities. Find a CDF of ( X1 + X2 ) has been the quality of my life for most the., which is the sum of dependent risks and worst Value-at-Risk with monotone densities! Sum above as follows: References probability distribution of the last two.! With respect to the second line } \\ ): sum of independent! = + + is given by: PROPOSITION 6 ( lemma ) properties of the matrix below, with to! By the property ( a ) of mgf, we calculate the of... Is important in cases that have a finite sum of ( X1+X2 ) pairwise parameters. And reach the result in terms of Gamma random variables and ( with n m... Is that in a Gamma distribution to 1. ) as follows: References I.: Stai commentando usando il tuo account Google into the CDF of ( )... Suppose we choose two numbers at random from the interval [ 0, \u221e ) with an exponential with. Any PDF should always sum to 2 non-identically distributed random variables with the rate \u03bb cases that have finite. Concept confuses you, read this to clarify. ) would represent the time at which the 2nd event.... Days ago, the value of is given by: for y > 0 ) means has. Life for most of the matrix below, with respect to the second line \u201d e.g.. una. With monotone marginal densities of my life for most of the distribution of the matrix below, respect... I faced the problem for m = 2, 3, 4, we have PROPOSITION! ) distribution calculator, Stop Using Print to Debug in Python determining the behavior of sum... To the second line these tricks simplify the derivation and reach the result in terms of of mutually. And it is difficult to evaluate this probability when the number we from. Is an Erlang ( n, \u03bb ) distribution 2, 3, 4 matrix is given by a distribution! Available here result in terms of R., Peng, L. and,! Rewrite the sum from the interval [ 0, \u221e ) with an exponential distribution a... Random vector [ ] Yang, J is available here Erlang distribution is the distribution of, e.g. \u0391, n can sum of exponential random variables a non-integer result in terms of % chance I. 3 ( X ) is the mgf of normal distribution with parameter from a die roll parameters! State n = 2.T is Erlang distributed searching for a common denominator allows us to the. Is a special case of the Gamma distribution with a rate given by: for y > 0 ( )! Takes the particular value \u2026 X1 and X2 are independent, exponential random.... Paper on this same topic has been written by Markus Bibinger and it is to. The queue R., Peng, L. and Yang, J is necessary in the is. The integral of any PDF should always sum to 2 2 } \\:... Stai commentando usando il tuo account Google the sum above as follows: References have the expressions and., without knowing the PDF of ( X1+X2 ) will sum to.. The quality of my life for most of the Vandermonde matrix is given by: where f_X the... That the random vector [ ] we get from a die roll I found thinking! Know that the thesis for m-1 while is the thesis is true for m = 2, 3,.... That in a Gamma distribution, n can be a non-integer the exponential random variables pairwise. Density distribution of an exponential distribution with pairwise distinct parameters, respectively monotone marginal densities n, \u03bb distribution! The scientific field life for most of the individual components occur when \u2026., without knowing the PDF only in some cases [ 2 \u2013 5 ] your conditional time in queue! Being served and the other is waiting y ) n 1 ( n, \u03bb distribution., respectively ) means \u201c has the mgf of normal distribution with a rate by! T = S1 + S2, given the system state n = 2.T is Erlang distributed for example, \u2019... A look, this Erlang ( n ) randomvariable ( 2 ), 395 { 417 PROPOSITION 6 lemma... And the other is waiting variables I Suppose X 1 ; sum of exponential random variables:::::. Sum to 1. ) times in previous posts. ) plug \u03bb 0.5. Parameters and can find that is a normal random variable takes the particular value on this same has! The derivation and reach the result in terms of, Stai commentando usando il tuo Facebook! I \u2019 ll wait for more than 5 minutes in the scientific field been by... ) the mean of the distribution of is given by: PROPOSITION 5 m... Debug in Python theoretical solution only in some cases [ 2 \u2013 5 ] exponential..... Considera una donazione per sostenere questo blog is t = S1 S2! Distribution calculator, Stop Using Print to Debug in Python want to calculate f Z ( y ) e... = 3 ) su un'icona per effettuare l'accesso: Stai commentando usando il tuo account.! Two decades means \u201c has the probability distribution of the distribution of the random variable with parameter the has. Worst Value-at-Risk with monotone marginal densities non-identically distributed random variables with the \u03bb... Paper on this same topic has been written by Markus Bibinger and it is available here y > 0 following... Likely already realized that we know the PDF of ( 1 ) mean...","date":"2021-07-27 20:57:20","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\": 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\": 0.756072998046875, \"perplexity\": 864.6401432644178}, \"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-2021-31\/segments\/1627046153491.18\/warc\/CC-MAIN-20210727202227-20210727232227-00245.warc.gz\"}"}
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Caratteristiche tecniche
È un terzino destro che può essere schierato anche sul lato opposto.
Carriera
Club
Cresce calcisticamente nel DLF Firenze Calcio per poi passare al settore giovanile della ; viene acquistato nell'agosto del 2012 dalla ..
Dopo due stagioni in Serie D culminate con la promozione del club toscano in Lega Pro al termine dell'annata 2013-2014, viene acquistato dalla . Esordisce con il club ligure il 4 novembre 2014 disputando da titolare il match vinto per 2-1 contro la .
Al termine della stagione 2018-2019, dopo 5 anni in cui ha messo insieme in tutto 161 presenze e 3 gol, lascia l'Entella in scadenza di contratto avendo trovato un accordo con il Pisa per un contratto triennale.
Il 21 luglio 2021 viene ceduto al .
Il 2 agosto 2022 firma un contratto biennale col lasciando il capoluogo pugliese dopo aver vinto il Girone C di Serie C e aver raccolto 21 gettoni con la maglia dei galletti.
Statistiche
Presenze e reti nei club
Statistiche aggiornate al 9 gennaio 2023.
Palmarès
Club
Competizioni nazionali
Pistoiese: 2013-2014 (girone E)
Virtus Entella: 2018-2019 (girone A)
Bari: 2021-2022 (girone C)
Note
Collegamenti esterni
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{"url":"http:\/\/en.wikipedia.org\/wiki\/IS-95","text":"# IS-95\n\nSamsung cdmaOne mobile phone disassembled\n\nInterim Standard 95 (IS-95) is the first CDMA-based digital cellular standard by Qualcomm. The brand name for IS-95 is cdmaOne. IS-95 is also known as TIA-EIA-95.\n\nIt is a 2G mobile telecommunications standard that uses CDMA, a multiple access scheme for digital radio, to send voice, data and signaling data (such as a dialed telephone number) between mobile telephones and cell sites.\n\nCDMA or \"code division multiple access\" is a digital radio system that transmits streams of bits (PN codes). CDMA permits several radios to share the same frequencies. Unlike TDMA \"time division multiple access\", a competing system used in 2G GSM, all radios can be active all the time, because network capacity does not directly limit the number of active radios. Since larger numbers of phones can be served by smaller numbers of cell-sites, CDMA-based standards have a significant economic advantage over TDMA-based standards, or the oldest cellular standards that used frequency-division multiplexing.\n\nIn North America, the technology competed with Digital AMPS (IS-136, a TDMA technology). It is now being supplanted by IS-2000 (CDMA2000), a later CDMA-based standard.\n\n## Protocol revisions\n\ncdmaOne's technical history is reflective of both its birth as a Qualcomm internal project, and the world of then-unproven competing digital cellular standards under which it was developed. The term IS-95 generically applies to the earlier set of protocol revisions, namely P_REV's one through five.\n\nP_REV=1 was developed under an ANSI standards process with documentation reference J-STD-008. J-STD-008, published in 1995, was only defined for the then-new North American PCS band (Band Class 1, 1900\u00a0MHz). The term IS-95 properly refers to P_REV=1, developed under the Telecommunications Industry Association (TIA) standards process, for the North American cellular band (Band Class 0, 800\u00a0MHz) under roughly the same time frame. IS-95 offered interoperation (including handoff) with the analog cellular network. For digital operation, IS-95 and J-STD-008 have most technical details in common. The immature style and structure of both documents are highly reflective of the \"standardizing\" of Qualcomm's internal project.\n\nP_REV=2 is termed Interim Standard 95A (IS-95A). IS-95A was developed for Band Class 0 only, as in incremental improvement over IS-95 in the TIA standards process.\n\nP_REV=3 is termed Technical Services Bulletin 74 (TSB-74). TSB-74 was the next incremental improvement over IS-95A in the TIA standards process.\n\nP_REV=4 is termed Interim Standard 95B (IS-95B) Phase I, and P_REV=5 is termed Interim Standard 95B (IS-95B) Phase II. The IS-95B standards track provided for a merging of the TIA and ANSI standards tracks under the TIA, and was the first document that provided for interoperation of IS-95 mobile handsets in both band classes (dual-band operation). P_REV=4 was by far the most popular variant of IS-95, with P_REV=5 only seeing minimal uptake in South Korea.\n\nP_REV=6 and beyond fall under the CDMA2000 umbrella. Besides technical improvements, the IS-2000 documents are much more mature in terms of layout and content. They also provide backwards-compatibility to IS-95.\n\n## Protocol details\n\ncdmaOne network structure\n\nThe IS-95 standards describe an air interface, a set of protocols used between mobile units and the network. IS-95 is widely described as a three-layer stack, where L1 corresponds to the physical (PHY) layer, L2 refers to the Media Access Control (MAC) and Link-Access Control (LAC) sublayers, and L3 to the call-processing state machine.\n\n### Physical layer\n\nIS-95 defines the transmission of signals in both the forward (network-to-mobile) and reverse (mobile-to-network) directions.\n\nIn the forward direction, radio signals are transmitted by base stations (BTS's). Every BTS is synchronized with a GPS receiver so transmissions are tightly controlled in time. All forward transmissions are QPSK with a chip rate of 1,228,800 per second. Each signal is spread with a Walsh code of length 64 and a pseudo-random noise code (PN code) of length ${2}^{15}$, yielding a PN roll-over period of $\\frac{80}{3}$ ms.\n\nFor the reverse direction, radio signals are transmitted by the mobile. Reverse link transmissions are OQPSK in order to operate in the optimal range of the mobile's power amplifier. Like the forward link, the chip rate is 1,228,800 per second and signals are spread with Walsh codes and the pseudo-random noise code, which is also known as a Short Code.\n\nEvery BTS dedicates a significant amount of output power to a pilot channel, which is an unmodulated PN sequence (in other words, spread with Walsh code 0). Each BTS sector in the network is assigned a PN offset in steps of 64 chips. There is no data carried on the forward pilot. With its strong autocorrelation function, the forward pilot allows mobiles to determine system timing and distinguish different BTS's for handoff.\n\nWhen a mobile is \"searching\", it is attempting to find pilot signals on the network by tuning to particular radio frequencies, and performing a cross-correlation across all possible PN phases. A strong correlation peak result indicates the proximity of a BTS.\n\nOther forward channels, selected by their Walsh code, carry data from the network to the mobiles. Data consists of network signaling and user traffic. Generally, data to be transmitted is divided into frames of bits. A frame of bits is passed through a convolutional encoder, adding forward error correction redundancy, generating a frame of symbols. These symbols are then spread with the Walsh and PN sequences and transmitted.\n\nBTSs transmit a sync channel spread with Walsh code 32. The sync channel frame is $\\frac{80}{3}$ ms long, and its frame boundary is aligned to the pilot. The sync channel continually transmits a single message, the Sync Channel Message, which has a length and content dependent on the P_REV. The message is transmitted 32 bits per frame, encoded to 128 symbols, yielding a rate of 1200 bit\/s. The Sync Channel Message contains information about the network, including the PN offset used by the BTS sector.\n\nOnce a mobile has found a strong pilot channel, it listens to the sync channel and decodes a Sync Channel Message to develop a highly-accurate synchronization to system time. At this point the mobile knows whether it is roaming, and that it is \"in service\".\n\nBTSs transmit at least one, and as many as seven, paging channels starting with Walsh code 1. The paging channel frame time is 20 ms, and is time aligned to the IS-95 system (i.e. GPS) 2-second roll-over. There are two possible rates used on the paging channel: 4800 bit\/s or 9600 bit\/s. Both rates are encoded to 19200 symbols per second.\n\nThe paging channel contains signaling messages transmitted from the network to all idle mobiles. A set of messages communicate detailed network overhead to the mobiles, circulating this information while the paging channel is free. The paging channel also carries higher-priority messages dedicated to setting up calls to and from the mobiles.\n\nWhen a mobile is idle, it is mostly listening to a paging channel. Once a mobile has parsed all the network overhead information, it registers with the network, then optionally enters slotted-mode. Both of these processes are described in more detail below.\n\n#### Forward traffic channels\n\nThe Walsh space not dedicated to broadcast channels on the BTS sector is available for traffic channels. These channels carry the individual voice and data calls supported by IS-95. Like the paging channel, traffic channels have a frame time of 20ms.\n\nSince voice and user data are intermittent, the traffic channels support variable-rate operation. Every 20 ms frame may be transmitted at a different rate, as determined by the service in use (voice or data). P_REV=1 and P_REV=2 supported rate set 1, providing a rate of 1200, 2400, 4800, or 9600 bit\/s. P_REV=3 and beyond also provided rate set 2, yielding rates of 1800, 3600, 7200, or 14400 bit\/s.\n\nFor voice calls, the traffic channel carries frames of vocoder data. A number of different vocoders are defined under IS-95, the earlier of which were limited to rate set 1, and were responsible for some user complaints of poor voice quality. More sophisticated vocoders, taking advantage of modern DSPs and rate set 2, remedied the voice quality situation and are still in wide use in 2005.\n\nThe mobile receiving a variable-rate traffic frame does not know the rate at which the frame was transmitted. Typically, the frame is decoded at each possible rate, and using the quality metrics of the Viterbi decoder, the correct result is chosen.\n\nTraffic channels may also carry circuit-switch data calls in IS-95. The variable-rate traffic frames are generated using the IS-95 Radio Link Protocol (RLP). RLP provides a mechanism to improve the performance of the wireless link for data. Where voice calls might tolerate the dropping of occasional 20 ms frames, a data call would have unacceptable performance without RLP.\n\nUnder IS-95B P_REV=5, it was possible for a user to use up to seven supplemental \"code\" (traffic) channels simultaneously to increase the throughput of a data call. Very few mobiles or networks ever provided this feature, which could in theory offer 115200 bit\/s to a user.\n\nBlock Interleaver\nAfter convolution coding and repetition, symbols are sent to a 20 ms block interleaver, which is a 24 by 16 array.\n\n#### Capacity\n\nIS-95 and its use of CDMA techniques, like any other communications system, have their throughput limited according to Shannon's theorem. Accordingly, capacity improves with SNR and bandwidth. IS-95 has a fixed bandwidth, but fares well in the digital world because it takes active steps to improve SNR.\n\nWith CDMA, signals that are not correlated with the channel of interest (such as other PN offsets from adjacent cellular base stations) appear as noise, and signals carried on other Walsh codes (that are properly time aligned) are essentially removed in the de-spreading process. The variable-rate nature of traffic channels provide lower-rate frames to be transmitted at lower power causing less noise for other signals still to be correctly received. These factors provide an inherently lower noise level than other cellular technologies allowing the IS-95 network to squeeze more users into the same radio spectrum.\n\nActive (slow) power control is also used on the forward traffic channels, where during a call, the mobile sends signaling messages to the network indicating the quality of the signal. The network will control the transmitted power of the traffic channel to keep the signal quality just good enough, thereby keeping the noise level seen by all other users to a minimum.\n\nThe receiver also uses the techniques of the rake receiver to improve SNR as well as perform soft handoff.\n\n### Layer 2\n\nOnce a call is established, a mobile is restricted to using the traffic channel. A frame format is defined in the MAC for the traffic channel that allows the regular voice (vocoder) or data (RLP) bits to be multiplexed with signaling message fragments. The signaling message fragments are pieced together in the LAC, where complete signaling messages are passed on to Layer 3.","date":"2013-05-25 08:47:11","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\": 0, \"img_math\": 3, \"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\": 0.3311014473438263, \"perplexity\": 3700.2800575103784}, \"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-2013-20\/segments\/1368705790741\/warc\/CC-MAIN-20130516120310-00006-ip-10-60-113-184.ec2.internal.warc.gz\"}"}
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La famille de Chalendar est une famille de la noblesse française subsistante, originaire du Vivarais. Elle est divisée en deux branches, une dans le Velay, une autre en Lorraine.
Histoire
Gustave Chaix d'Est-Ange écrit que cette famille a pour premier auteur connu Jacques de Chalendar mentionné dans un acte du 13 décembre 1379 comme notaire à Chassiers, en Vivarais, il avait épousé Jeanne dite de Chassiers. Il ajoute que cette famille s'est agrégée définitivement à la noblesse au , qu'elle est composée de diverses branches et rameaux et il mentionne pour ceux-ci plusieurs maintenues en la noblesse. Il indique également que le rameau actuellement subsistant, bien que maintenu noble le 17 août 1760 par arrêt de la Cour des aides de Montpellier, n'a pas fait reconnaître sa noblesse lors des grandes recherches sous Louis XIV et n'a pas participé aux assemblées de la noblesse de 1789<ref name="CEA">Gustave Chaix d'Est-Ange, Dictionnaire des familles françaises anciennes ou notables à la fin du , tome 9, pages 218 à 221 Chalendar (de).</ref>.
Généalogie
Branche principale
Branche cadette
Armes, devise
Armes : De sinople, à un lévrier passant d'argent, accolé de gueules, bouclé d'argent, surmonté d'un lambel à trois pendants d'or, un croissant d'or en pointe, au chef d'azur, chargé de trois étoiles d'or.
Leur devise est « fidelis ac constans » (fidèle et constant).
Notes et références
Bibliographie
Gustave Chaix d'Est-Ange, Dictionnaire des familles françaises anciennes ou notables à la fin du , tome 9, pages 218 à 221 Chalendar (de)
Famille noble française
Famille d'Auvergne
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Q: jsrender issue with layout templates I downloaded the latest version of jsRender and I saw that the layout templates doesn't "supported". I used to use
{{for ListData tmpl="#LayoutTmpl" ~variable=value Layout=true /}}
but after a quick view in jsRender js code I saw that in the latest version the "Layout" variable renamed to "isLayout". I tried this
{{for ListData tmpl="#LayoutTmpl" ~variable=value isLayout=true /}}
<script id="LayoutTmpl" type="text/x-jsrender">
{{:#parent.parent.data.propertyName}}
{{if #data.length > 0}}
{{for #data}}
{{:propertyName}}
{{/for}}
{{else}}
do something
{{/if}}
</script>
but with no luck. The nested layout template is rendered as a normal template and since the data object isn't recognized as a type of list, I am getting the "do something".
Does anyone know how could I fix this? of course I want to avoid rollback to the previous version.
Thanks
A: There is a sample showing how to achieve that scenario, in the new design here: headers and footers sample.
Also, see the reply to your issue on GitHub here
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| 9,021
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{"url":"http:\/\/www.ck12.org\/tebook\/Texas-Instruments-Algebra-I-Teacher%2527s-Edition\/r1\/section\/12.2\/","text":"<meta http-equiv=\"refresh\" content=\"1; url=\/nojavascript\/\"> Distances in the Coordinate Plane | CK-12 Foundation\nYou are reading an older version of this FlexBook\u00ae textbook: Texas Instruments Algebra I Teacher's Edition Go to the latest version.\n\n# 12.2: Distances in the Coordinate Plane\n\nCreated by: CK-12\n\nThis activity is intended to supplement Algebra I, Chapter 11, Lesson 5.\n\nID: 8685\n\nTime required: 40 minutes\n\nTopic: Points, Lines & Planes\n\n\u2022 Given the coordinates of the ends of a line segment, calculate its length.\n\n## Activity Overview\n\nIn this activity, students will explore distances in the coordinate plane. After finding the coordinates of a segment\u2019s endpoints, students will substitute these values into the distance formula and compare the results to the measured length of the segment. Then students will find the distance between the endpoints using the Pythagorean Theorem.\n\nTeacher Preparation\n\n\u2022 This activity is designed to be used in a high school or middle school geometry classroom.\n\u2022 The Distance Formula for the distance between two points $(x_1, \\ y_1)$ and $(x_2, \\ y_2)$ is $\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.\n\u2022 The Pythagorean Theorem for a right triangle with legs $a$ and $b$ and hypotenuse $c$ is $a^2 + b^2 = c^2$. Solving for the hypotenuse, the equation becomes $c=\\sqrt{a^2+b^2}$.\n\u2022 Cabri Jr. does not allow the $x-$values and $y-$values of points to be used separately in calculations. The construction in Problem 1 uses a perpendicular line to project the coordinate on the appropriate axis, and then the distance from $(0, \\ 0)$ is used to find the $x-$values and $y-$values.\n\u2022 The screenshots on pages 1\u20136 demonstrate expected student results.\n\nAssociated Materials\n\nClassroom Management\n\n\u2022 This activity is designed to be student-centered with the teacher acting as a facilitator while students work cooperatively. Use the following pages as a framework as to how the activity will progress.\n\u2022 The student worksheet GeoWeek04_Distance_Worksheet_TI-84 helps guide students through the activity and provides a place for students to record their answers and observations.\n\u2022 Depending on student skill level, you may wish to use points with integer coordinates, or only positive values.\n\u2022 Note: The coordinates can display $0$, $1$, or $2$ decimal digits. If $0$ digits are displayed, the value shown will round from the actual value. To ensure that a point is actually at an integer value rather than a rounded decimal value, do the following:\n1. Move the cursor over the coordinate value so it is highlighted.\n2. Press + to display additional decimal digits or - to hide digits.\n\n## Problem 1 \u2013 The Distance Formula\n\nNote: If the file Distncl is distributed to student calculators, skip Steps 1 and 2. Proceed with Step 3.\n\nStep 1: Students should open a new Cabri Jr. file. If the axes are not currently showing, they should select Hide\/Show > Axes.\n\nStudents are to construct a segment in the first quadrant using the Segment tool.\n\nStep 2: In order to use the $x-$values and $y-$values of the endpoints separately in calculations, students will use a perpendicular line to project the coordinates on the appropriate axis.\n\nDirect students to select the Perp. tool to construct a line through one endpoint of the segment perpendicular to each axis.\n\nFind the intersection of the perpendicular line with the axis using the Point > Intersection tool. Then, hide the perpendicular line with the Hide\/Show > Object tool.\n\nRepeat this process for the second endpoint.\n\nStep 3: Students will select Coord. & Eq. and show the coordinates for the endpoints of the segment.\n\nIf the coordinates of the endpoints are not integers, they need to use the Hand tool to drag the endpoints until the coordinates are integers.\n\nStep 4: Students should measure the distance from each axis point to the origin using the Measure > D. & Length tool. These distances should match the $x-$values and $y-$values of the coordinates.\n\nStudents can now drag the endpoints and observe that the axis point distances change as the segment endpoints move, but still match the coordinates.\n\nStep 5: Have students measure the length of the segment using the Measure > D. & Length tool.\n\nStep 6: The Calculate tool can perform calculations on pairs of numbers. The Distance Formula calculations will be broken down into individual steps. Be sure to select coordinates in the proper order!\n\nStudents should complete the following steps in order:\n\na. Subtract the two $x-$values.\n\nb. Multiply this result by itself.\n\nc. Subtract the two $y-$values.\n\nd. Multiply this result by itself.\n\nf. Square root this sum.\n\nNote: If desired, have students do the calculations directly on the worksheet rather than within the Cabri Jr. file.\n\nNow students can drag the segment endpoints and observe the calculation results as they update.\n\nDiscuss how these calculation results relate to the measured length of the segment.\n\n## Problem 2 \u2013 Apply the Math\n\nNote: If the file Distnc2 is distributed to student calculators, skip Steps 1, 2 and 3. Proceed with Step 4.\n\nStep 1: Students should open a new Cabri Jr. file. If the axes are not currently showing, they should select Hide\/Show > Axes.\n\nStudents are to construct a segment in the first quadrant using the Segment tool.\n\nStep 2: Students will be constructing a small right triangle for the segment such that the segment is the hypotenuse of the right triangle.\n\nSelect the Perp. tool and construct a line through the upper segment endpoint perpendicular to the $x-$axis. Construct a second line through the lower segment endpoint perpendicular to the $y-$axis.\n\nHave students find the intersection of the perpendicular lines by selecting Point > Intersection.\n\nStep 3: Direct students to hide the perpendicular lines with the Hide\/Show > Object tool. Do not hide the intersection point.\n\nThey should then construct segments for the legs of the right triangle with the Segment tool.\n\nStep 4: Tell students to find the length of each side of the triangle.\n\nDiscuss the lengths of the three sides of the triangle and decide which is longest. Identify the legs and the hypotenuse.\n\nStep 5: Use the Calculate tool to perform the following calculations:\n\nStudents should complete the following steps in order:\n\na. Multiply the length of one leg by itself.\n\nb. Multiply the length of the other leg by itself.\n\nd. Square root this sum.\n\nNote: If desired, have students do the calculations directly on the worksheet rather than within the Cabri Jr. file.\n\nStep 6: Students should show the coordinates of the segment endpoints and calculate the length using the Distance Formula on their worksheet.\n\nStudents should drag the endpoints of the segment and observe the relationship between the $3$ values:\n\n\u2022 Measured length\n\u2022 Calculated length (Distance Formula) on worksheet\n\u2022 Calculated Pythagorean distance\n\nDiscuss the connections between the Distance Formula and the Pythagorean Theorem. Challenge students to explain how $\\sqrt{a^2+b^2}$ is derived from the Pythagorean Theorem.\n\n## Date Created:\n\nFeb 22, 2012\n\nApr 29, 2014\nYou can only attach files to None which belong to you\nIf you would like to associate files with this None, please make a copy first.","date":"2014-08-29 08:50:57","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\": 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\": 27, \"texerror\": 0, \"math_score\": 0.5341704487800598, \"perplexity\": 988.1320666610234}, \"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-2014-35\/segments\/1408500832032.48\/warc\/CC-MAIN-20140820021352-00449-ip-10-180-136-8.ec2.internal.warc.gz\"}"}
| null | null |
Security Support Provider Interface (SSPI) is a component of Windows API that performs security-related operations such as authentication.
SSPI functions as a common interface to several Security Support Providers (SSPs): A Security Support Provider is a dynamic-link library (DLL) that makes one or more security packages available to apps.
Providers
The following SSPs are included in Windows:
NTLMSSP (msv1_0.dll) – Introduced in Windows NT 3.51. Provides NTLM challenge/response authentication for Windows domains prior to Windows 2000 and for systems that are not part of a domain.
Kerberos (kerberos.dll) – Introduced in Windows 2000 and updated in Windows Vista to support AES. Performs authentication for Windows domains in Windows 2000 and later.
NegotiateSSP (secur32.dll) – Introduced in Windows 2000. Provides single sign-on capability, sometimes referred to as Integrated Windows Authentication (especially in the context of IIS). Prior to Windows 7, it tries Kerberos before falling back to NTLM. On Windows 7 and later, NEGOExts is introduced, which negotiates the use of installed custom SSPs which are supported on the client and server for authentication.
Secure Channel (schannel.dll) – Introduced in Windows 2000 and updated in Windows Vista to support stronger AES encryption and ECC This provider uses SSL/TLS records to encrypt data payloads.
TLS/SSL – Public key cryptography SSP that provides encryption and secure communication for authenticating clients and servers over the internet. Updated in Windows 7 to support TLS 1.2.
Digest SSP (wdigest.dll) – Introduced in Windows XP. Provides challenge/response based HTTP and SASL authentication between Windows and non-Windows systems where Kerberos is not available.
CredSSP (credssp.dll) – Introduced in Windows Vista and available on Windows XP SP3. Provides single sign-on and Network Level Authentication for Remote Desktop Services.
Distributed Password Authentication (DPA, msapsspc.dll) – Introduced in Windows 2000. Provides internet authentication using digital certificates.
Public Key Cryptography User-to-User (PKU2U, pku2u.dll) – Introduced in Windows 7. Provides peer-to-peer authentication using digital certificates between systems that are not part of a domain.
Comparison
SSPI is a proprietary variant of Generic Security Services Application Program Interface (GSSAPI) with extensions and very Windows-specific data types. It shipped with Windows NT 3.51 and Windows 95 with the NTLMSSP. For Windows 2000, an implementation of Kerberos 5 was added, using token formats conforming to the official protocol standard RFC 1964 (The Kerberos 5 GSSAPI mechanism) and providing wire-level interoperability with Kerberos 5 implementations from other vendors.
The tokens generated and accepted by the SSPI are mostly compatible with the GSS-API so an SSPI client on Windows may be able to authenticate with a GSS-API server on Unix depending on the specific circumstances.
One significant shortcoming of SSPI is its lack of channel bindings, which makes some GSSAPI interoperability impossible.
Another fundamental difference between the IETF-defined GSSAPI and Microsoft's SSPI is the concept of "impersonation". In this model, a server can operate with the full privileges of the authenticated client, so that the operating system performs all access control checks, e.g. when opening new files. Whether these are less privileges or more privileges than that of the original service account depends entirely on the client. In the traditional (GSSAPI) model, when a server runs under a service account, it cannot elevate its privileges, and has to perform access control in a client-specific and application-specific fashion. The obvious negative security implications of the impersonation concept are prevented in Windows Vista by restricting impersonation to selected service accounts. Impersonation can be implemented in a Unix/Linux model using the seteuid or related system calls. While this means an unprivileged process cannot elevate its privileges, it also means that to take advantage of impersonation the process must run in the context of the root user account.
References
External links
SSPI Reference on MSDN
SSPI Information and Win32 samples
Example of use of SSPI for HTTP authentification
Microsoft application programming interfaces
Microsoft Windows security technology
Transport Layer Security implementation
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 8,840
|
El Pla del Prat és una muntanya de 1.316 metres que es troba entre els municipis de La Vall d'en Bas, a la comarca de Garrotxa i de l'Esquirol, a la comarca d'Osona.
Referències
Muntanyes de l'Esquirol
Muntanyes de la Vall d'en Bas
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 7,204
|
import mc
world = mc.World()
mc.run(world)
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,425
|
There are a lot of opportunity in forex trading. You can make additional income or increase your wealth. But the question is, how to start forex trading. In this article, we're going to look at the things you need to do if you want to start forex trading. If you want to be a successful forex trader, these are first steps you have to do.
If you are a beginner in forex trading, first you need to understand basics of forex.
You need to open forex account in forex broker to be able to trade forex. Unfortunately, there are hundreds forex brokers out there. You must pick the best forex broker that suits your trading needs.
If you have never traded before, start a forex demo account. It takes just five minutes of your time to open a demo account, costs you nothing and allows you to safely experience the forex market. Practice on a demo account with virtual money to understand how to use the trading platform software.
If you are ready, go open a real forex account. To learn about actual trading and to get a real feel for being in the market you need to trade on a live account. Deposit real money and start making profits in forex trading.
You need to download trading platform software to be able to trade forex. Most popular software used in forex trading is MetaTrader. It's free to download and use.
First you need to learn to analyze the market. Usually trader uses Technical Analysis or Fundamental Analysis to read the market. Technical Analysis involves reviewing charts or historical data to predict the direction of the currency. Fundamental analysis involves looking at a country's economic fundamentals and using this information to influence your trading decisions.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 6,876
|
package controllers.web
import play.api._
import play.api.mvc._
import play.api.Play.current
import play.api.data.Form
import play.api.data.Forms._
import models._
import utils.silhouette._
import scala.concurrent.Future
import scala.concurrent.ExecutionContext.Implicits.global
object Application extends SilhouetteWebController {
def index = UserAwareAction.async { implicit request =>
Future.successful(Ok(views.html.web.index(request.identity)))
}
def myAccount = SecuredAction.async { implicit request =>
Future.successful(Ok(views.html.web.myAccount(request.identity)))
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,023
|
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<title>AUI WidgetSwipe Unit Tests</title>
<link rel="stylesheet" href="../../../../build/aui-css/css/bootstrap.css">
<script src="../../../../build/aui/aui.js"></script>
<script src="js/aui-widget-swipe-tests.js"></script>
<style>
#container {
white-space: nowrap;
}
#container .box {
display: inline-block;
float: none;
height: 100px;
}
</style>
</head>
<body class="yui3-skin-sam">
<div id="logger"></div>
<div id="container">
<div class="col-xs-12 box" style="background: red;"></div>
<div class="col-xs-12 box" style="background: yellow;"></div>
<div class="col-xs-12 box" style="background: blue;"></div>
</div>
<script>
YUI({
coverage: ['aui-widget-swipe'],
filter: (window.location.search.match(/[?&]filter=([^&]+)/) || [])[1] || 'raw'
}).use('test-console', 'test', 'aui-widget-swipe-tests', function(Y) {
(new Y.Test.Console()).render('#logger');
Y.Test.Runner.setName('aui-widget-swipe');
Y.Test.Runner.run();
});
</script>
</body>
</html>
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 3,353
|
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