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## Statistics Question

25. Suppose that 6 minutes had been the sample standard deviation. Use the fact that the t multiplier for 35 degrees of freedom and 95% confidence is 2.03 to consider a confidence interval for the population mean REM sleep time. Compared to the z interval, the t interval would be A) much wider B) slightly wider C) slightly narrower D) much narrower

• b
• c
• A) much wider

t(α/2,34)=2.03

z(α/2)=1.96

so the CI is larger for t distribution

• A) much wider

t(a/2,34)=2.03
z(a/2)=1.96
so the CI is larger for t distribution
Get homework help

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text:
Excluding Graphs as Immersions in Surface Embedded Graphs

5 ALGCO - Algorithmes, Graphes et Combinatoire
LIRMM - Laboratoire d'Informatique de Robotique et de Microélectronique de Montpellier
Abstract : We prove a structural characterization of graphs that forbid a fixed graph $H$ as an immersion and can be embedded in a surface of Euler genus $\gamma$. In particular, we prove that a graph $G$ that excludes some connected graph $H$ as an immersion and is embedded in a surface of Euler genus $\gamma$ has either "small" treewidth (bounded by a function of $H$ and $\gamma$) or "small" edge connectivity (bounded by the maximum degree of $H$). Using the same techniques we also prove an excluded grid theorem on bounded genus graphs for the immersion relation.
Document type :
Preprints, Working Papers, ...

https://hal-lirmm.ccsd.cnrs.fr/lirmm-00805139
Contributor : Dimitrios Thilikos Connect in order to contact the contributor
Submitted on : Wednesday, March 27, 2013 - 10:56:18 AM
Last modification on : Thursday, November 26, 2020 - 3:50:03 PM

Identifiers

• HAL Id : lirmm-00805139, version 1
• ARXIV : 1303.6567

Citation

Archontia C. Giannopoulou, Marcin Kaminski, Dimitrios M. Thilikos. Excluding Graphs as Immersions in Surface Embedded Graphs. 2013. ⟨lirmm-00805139⟩

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text:
Problem: Write a bond-line formula for each of the following compounds: (g) cis-1,2-Dimethylcyclopropane

Problem Details

Write a bond-line formula for each of the following compounds: (g) cis-1,2-Dimethylcyclopropane


text:
Bibliography style unsrt not working

I am writing a text (whose content I can't fully disclose here) and I am experiencing some issues with citation order, using unsrt as bibliography style. Until a certain point citations are ordered correctly, but since I have added some .tex files, this isn't happening anymore. Any suggestions?

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• Phrases such as "this isn't happening anymore" and "not working" are not illuminating. What, exactly, isn't "working"? What kind of output are you getting? What's in the tex files you say you added? Any \cite instructions, by any chance? If the answer is "yes", did you rerun LaTeX, BibTeX, and LaTeX twice more? Do LaTeX and BibTeX issue warning and/or error messages? Please be specific. – Mico Jun 12 at 19:50
• You have posted a large amount of code but almost none of it is related to references or sorting, please remove all unrelated things such as image inclusion and remove \input{chapter/2/chapter.tex} that does not let anyone run the example. the fact that your original document contains private material is not relevant you can change every word to "wibble" in an editor, it will not affect the citation numbers – David Carlisle Jun 12 at 19:52
• Quotes numbering is not sequential anymore. Only a part of them are ordered correctly, while others are ordered from the bottom of the correctly ordered ones; moreover they are not ordered correctly, but in a FIFO order relative to the insertion order in .bib file, not in .tex file. – EagleOne Jun 12 at 19:53
• then most likely you have \cite{*} somewhere that is causing all entries in the bib to be used in order – David Carlisle Jun 12 at 19:54
• side issue hyperref (cite backrefs etc) and glossaries will likely be benefit from reordering packages to the end of preamble – KJO Jun 12 at 20:16


text:
TITLE

# Meson exchange models for meson production

AUTHOR(S)
Hanhart, C.
PUB. DATE
June 2000
SOURCE
AIP Conference Proceedings;2000, Vol. 512 Issue 1, p81
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
The production of mesons in nucleon-nucleon collisions is reviewed from the viewpoint of the meson-exchange picture. In the first part various possible production mechanisms and their relative importance are discussed. In addition, general features of meson production are described. In the second part special emphasis is put on pion production. Implications of chiral perturbation theory are discussed. Results based on a specific meson-exchange model for pion production in nucleon-nucleon collisions are presented. The model is utilized to calculate several spin-dependent observables of the reactions ppâ†’ppÏ€[sup 0] and ppâ†’pnÏ€[sup +] such as spin-correlation parameters. This allows us to study the role of the Delta resonance at close to threshold energies. The talk closes with a brief discussion of the production of &fgr; mesons as an example for the production of heavier mesons. A strategy is outlined that allows to extract information on the structure of the nucleon from the reaction NNâ†’NN&fgr;. Â© 2000 American Institute of Physics.
ACCESSION #
6028153

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• Energy Efficiency with Proportional Rate Fairness in Multi-Relay OFDM Networks(1511.07566)

Nov. 24, 2015 cs.IT, math.IT
This paper investigates the energy efficiency (EE) in multiple relay aided OFDM system, where decode-and-forward (DF) relay beamforming is employed to help the information transmission. In order to explore the EE performance with user fairness for such a system, we formulate an optimization problem to maximize the EE by jointly considering several factors, the transmission mode selection (DF relay beamforming or direct-link transmission), the helping relay set selection, the subcarrier assignment and the power allocation at the source and relays on subcarriers, under nonlinear proportional rate fairness constraints, where both transmit power consumption and linearly rate-dependent circuit power consumption are taken into account. To solve the non-convex optimization problem, we propose a low-complexity scheme to approximate it. Simulation results demonstrate its effectiveness. We also investigate the effects of the circuit power consumption on system performances and observe that with both the constant and the linearly rate-dependent circuit power consumption, system EE grows with the increment of system average channel-to noise ratio (CNR), but the growth rates show different behaviors. For the constant circuit power consumption, system EE increasing rate is an increasing function of the system average CNR, while for the linearly rate-dependent one, system EE increasing rate is a decreasing function of the system average CNR. This observation is very important which indicates that by deducing the circuit dynamic power consumption per unit data rate, system EE can be greatly enhanced. Besides, we also discuss the effects of the number of users and subcarriers on the system EE performance.
• Generative Modeling of Convolutional Neural Networks(1412.6296)

April 9, 2015 cs.NE, cs.CV, cs.LG
The convolutional neural networks (CNNs) have proven to be a powerful tool for discriminative learning. Recently researchers have also started to show interest in the generative aspects of CNNs in order to gain a deeper understanding of what they have learned and how to further improve them. This paper investigates generative modeling of CNNs. The main contributions include: (1) We construct a generative model for the CNN in the form of exponential tilting of a reference distribution. (2) We propose a generative gradient for pre-training CNNs by a non-parametric importance sampling scheme, which is fundamentally different from the commonly used discriminative gradient, and yet has the same computational architecture and cost as the latter. (3) We propose a generative visualization method for the CNNs by sampling from an explicit parametric image distribution. The proposed visualization method can directly draw synthetic samples for any given node in a trained CNN by the Hamiltonian Monte Carlo (HMC) algorithm, without resorting to any extra hold-out images. Experiments on the challenging ImageNet benchmark show that the proposed generative gradient pre-training consistently helps improve the performances of CNNs, and the proposed generative visualization method generates meaningful and varied samples of synthetic images from a large-scale deep CNN.
• Non-Markovian Character in Human Mobility: Online and Offline(1406.2685)

June 11, 2014 physics.soc-ph, cs.SI
The dynamics of human mobility characterizes the trajectories humans follow during their daily activities and is the foundation of processes from epidemic spreading to traffic prediction and information recommendation. In this paper, we investigate a massive data set of human activity including both online behavior of browsing websites and offline one of visiting towers based mobile terminations. The non-Markovian character observed from both online and offline cases is suggested by the scaling law in the distribution of dwelling time at individual and collective levels, respectively. Furthermore, we argue that the lower entropy and higher predictability in human mobility for both online and offline cases may origin from this non-Markovian character. However, the distributions of individual entropy and predictability show the different degrees of non-Markovian character from online to offline cases. To accounting for non-Markovian character in human mobility, we introduce a protype model with three basic ingredients, \emph{preferential return, inertial effect, and exploration} to reproduce the dynamic process of online and offline human mobility. In comparison with standard and biased random walk models with assumption of Markov process, the proposed model is able to obtain characters much closer to these empirical observations.
• LAGE: A Java Framework to reconstruct Gene Regulatory Networks from Large-Scale Continues Expression Data(1211.2073)

Nov. 9, 2012 q-bio.QM, cs.LG, cs.CE, stat.ML
LAGE is a systematic framework developed in Java. The motivation of LAGE is to provide a scalable and parallel solution to reconstruct Gene Regulatory Networks (GRNs) from continuous gene expression data for very large amount of genes. The basic idea of our framework is motivated by the philosophy of divideand-conquer. Specifically, LAGE recursively partitions genes into multiple overlapping communities with much smaller sizes, learns intra-community GRNs respectively before merge them altogether. Besides, the complete information of overlapping communities serves as the byproduct, which could be used to mine meaningful functional modules in biological networks.
• LSBN: A Large-Scale Bayesian Structure Learning Framework for Model Averaging(1210.5135)

Oct. 18, 2012 cs.LG, stat.ML
The motivation for this paper is to apply Bayesian structure learning using Model Averaging in large-scale networks. Currently, Bayesian model averaging algorithm is applicable to networks with only tens of variables, restrained by its super-exponential complexity. We present a novel framework, called LSBN(Large-Scale Bayesian Network), making it possible to handle networks with infinite size by following the principle of divide-and-conquer. The method of LSBN comprises three steps. In general, LSBN first performs the partition by using a second-order partition strategy, which achieves more robust results. LSBN conducts sampling and structure learning within each overlapping community after the community is isolated from other variables by Markov Blanket. Finally LSBN employs an efficient algorithm, to merge structures of overlapping communities into a whole. In comparison with other four state-of-art large-scale network structure learning algorithms such as ARACNE, PC, Greedy Search and MMHC, LSBN shows comparable results in five common benchmark datasets, evaluated by precision, recall and f-score. What's more, LSBN makes it possible to learn large-scale Bayesian structure by Model Averaging which used to be intractable. In summary, LSBN provides an scalable and parallel framework for the reconstruction of network structures. Besides, the complete information of overlapping communities serves as the byproduct, which could be used to mine meaningful clusters in biological networks, such as protein-protein-interaction network or gene regulatory network, as well as in social network.
• Minimum-error discrimination of entangled quantum states(1008.0843)

Aug. 4, 2010 quant-ph
Strategies to optimally discriminate between quantum states are critical in quantum technologies. We present an experimental demonstration of minimum error discrimination between entangled states, encoded in the polarization of pairs of photons. Although the optimal measurement involves projecting onto entangled states, we use a result of Walgate et al. to design an optical implementation employing only local polarization measurements and feed-forward, which performs at the Helstrom bound. Our scheme can achieve perfect discrimination of orthogonal states and minimum error discrimination of non-orthogonal states. Our experimental results show a definite advantage over schemes not using feed-forward.
• Band Gap of Strained Graphene Nanoribbons(0912.2702)

Jan. 20, 2010 cond-mat.mes-hall
The band structures of strained graphene nanoribbons (GNRs) are examined by a tight binding Hamiltonian that is directly related to the type and strength of strains. Compared to the two-dimensional graphene whose band gap remains close to zero even if a large strain is applied, the band gap of graphene nanoribbon (GNR) is sensitive to both uniaxial and shears strains. The effect of strain on the electronic structure of a GNR strongly depends on its edge shape and structural indices. For an armchair GNR, uniaxial weak strain changes the band gap in a linear fashion, and for a large strain, it results in periodic oscillation of the band gap. On the other hand, shear strain always tend to reduce the band gap. For a zigzag GNR, the effect of strain is to change the spin polarization at the edges of GNR, thereby modulate the band gap. A simple analytical model is proposed to interpret the band gap responds to strain in armchair GNR, which agrees with the numerical results.


text:
# Joint probability

#### Usagi

##### Member

Just wondering, how did they get the part boxed in blue?

Thanks.

#### CaptainBlack

##### Well-known member

Just wondering, how did they get the part boxed in blue?

Thanks.

The part in the box comes from the observation that:

$P(X_j=1,X_k=1)=P(X_1=1,X_2=1), \ \ j\ne k$,

and that:

$P(X_1=1,X_2=1)=P(X_1=1)+P(X_2=1|X_1=1)$

CB

#### Usagi

##### Member
Thanks for that, I realised it too however how does one show that $P(X_j, X_i) = P(X_1, X_2) = P(X_1, X_5)$ and so on?

It isn't so obvious to me that $P(X_1, X_2) = P(X_1, X_5)$

#### CaptainBlack

##### Well-known member
Thanks for that, I realised it too however how does one show that $P(X_j, X_i) = P(X_1, X_2) = P(X_1, X_5)$ and so on?

It isn't so obvious to me that $P(X_1, X_2) = P(X_1, X_5)$
The informal explanation is that as the labelling is essentialy arbitary the probabilities are all the same.

More formally:

$P(X_1=1,X_2=1)=P(X_2=1)+P(X_2)P(X_1=1|P(X_2=1)$

But $$P(X_2=1|X_1=1)=P(X_2=1|X_5=1)$$ and $$P(X_2=1)=P(X_5=1)$$. which look better but is still just an observation that the idexing is arbitary and you can just permute indices.

CB

Thanks


text:
# KOMA-Script scrreprt: How to change chapter appearance and produce a chapter based Mini TOC

This post is a 3 part question with which I would need help. The questions are all associated with chapter styling and subsections within the \documentclass{scrreprt}. I am sure I could work them out eventually, I come to you for guidance to learn and save a bit of time.

# Intro

I am a fledgling LaTeXer, I have already used LaTeX to produce simple reports for which I wasn't too fussed about the formatting. Now I am writing a PhD thesis and I would like it to look the way I choose.

# Desired Result

The result I would like to achieve for my chapters is roughly as shown in the below image. The essential elements I would like to implement in my document are:

• Chapter first page look as below with Chapter Title not having a number
• Possibility of adding a quote before the Chapter Title. (probably not hard)
• Mini ToC for each chapter
• More than the \subsection level of sectioning that will be included in the ToC.

# Question 1: Subsections

I have been reading around on the site and saw that there were solutions suggested to this matter with the scrreprt class such as in the post scrreprt - add levels below \subparagraph; I have read that the titlesec package is very customisable, however I have also read it would disable/interfere with some of the KOMA-script macros (I don't know exactly which). What would be best to implement in my case to achieve my desired look?

# Question 2: Setting up the chapter look and style

Based on the answer to question 1, could you help explain how I would set it up please?

# Question 3: Mini ToC and quote

Could you please advise how you would go about adding a quote between the Chapter No. and Title (I might not for every chapter). I imagine that to make a mini chapter confined ToC would simply require a package, or maybe there is already a functionality in scrreprt?

# MWE Update 10/02/2014

Sorry again for not having this sooner, here is my MWE of the thesis setup. I included the title page I use and the Dedication and Abstract sections which are fine the way they are for now.

I have implemented what @Crissov suggested but unfortunately I cannot get the minitoc to generate or the chapter format to match the one presented in his MWE which does work until implemented in my case. Due to his code working alone, but not once I implemented it into mine, I included what I realise is a bit of a bulky MWE, but the mistake I am making must be somewhere in there.

\documentclass[
DIV=11,
twoside,
titlepage=firstiscover,
abstract=true,
footsepline=true,
]{scrreprt}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=3cm]{geometry} %typearea has that nasty problem of double bottom margin so had to remove it
\usepackage{setspace}
\usepackage{graphicx}
\usepackage{xcolor}
\usepackage{minitoc}

\flushbottom

\iffalse % don’t use titlesec.sty with KOMA script classes
\usepackage{titlesec}
%% ￼￼￼\titleformat{⟨command⟩}[⟨shape⟩]{⟨format⟩}{⟨label⟩}{⟨sep⟩}{⟨before⟩}[⟨after⟩]
\titleformat{name=\chapter}[display]
{\usekomafont{chapter}}
{\raggedleft\chaptertitlename\ {\textcolor{gray}{\fontsize{60}    {65}\selectfont\thechapter}}}
{20pt}{}[\minitoc]
\fi

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ToC DEPTH LEVEL
\setcounter{secnumdepth}{3} % number subsubsection
\setcounter{tocdepth}{3} % list subsubsection

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PREFACE AND BODY STYLING
\def\preface{
\pagenumbering{roman}
\doublespacing
}

\def\body{
\cleardoublepage
\dominitoc
\tableofcontents
\linespread{1.5} %Clear double spacing from preface
\pagenumbering{arabic}
}

\def\abstract{
\begin{center}{
\large\bf Abstract}
\end{center}
\normalsize
}{\cleardoublepage}
\def\endabstract{
\par
}

\newenvironment{acknowledgements}{
\cleardoublepage
\begin{center}{
\large \bf Acknowledgements}
\end{center}
\normalsize
}{\cleardoublepage}
\def\endacknowledgements{
\par
}

\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% TITLE PAGE
\Large \centering University of London \\
}
\subject{Department of Materials \\
}
\title{My PhD Thesis Title}
\author{Edoardo}
\date{
\vskip 1cm
\normalsize Submitted \\
\Large 2014 \\
\vskip 2cm
\normalsize Supervisors: \\
\Large Jim Bob \\
John Doe \\
\vskip 2cm
\normalsize Submitted in part fulfilment of the requirements for the degree of \\
Doctor of Philosophy in Materials
}

\dedication{Dedication here}
\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PREFACE
\preface
\begin{abstract}

In this study we did a lot of things

\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% ToC ToF ToT
\body
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% CHAPTERS
\chapter{Introduction}

\dictum[Sun Tzu]{If you know the enemy and you know yourself %\\
you need not fear the results of a hundred battles.}

\minitoc % should be put here automatically

\section{Test of depth}

We are delving deeper into the unknown.

\subsection{Publications}

Many publications here.

\subsubsection{Text Mass}
Here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text, here goes lots and lots of text

\end{document}

I have however found out that there is a way to put a quote (if desired) just after the chapter using the {after} part of the command line for \titleformat as follows:

\titleformat{name=\chapter}[display]
{\usekomafont{chapter}}
{\raggedleft\chaptertitlename\ {\textcolor{gray}{\fontsize{60}{65}\selectfont\thechapter}}}
{20pt}{\dictum[Sun Tzu]{If you know the enemy and you know yourself %\\
you need not fear the results of a hundred battles.}}

All I'd need to workout is how to link that specific \dictum section for each chapter to a part I can write just after \chapter{chapter} I suppose...

-
Welcome to TeX.sx! Regarding the quotes see tex.stackexchange.com/questions/53377/… –  cgnieder Feb 9 at 17:21
Please provide a MWE to show your effort.. –  subham soni Feb 9 at 17:21
–  Marco Daniel Feb 9 at 17:38
@Fiztban You should generally ask separate questions in separate question postings. Here, questions 1 & 2 belong together in my opinion, and so could be asked together. Question 3 is probably asked in KOMA-Script (scrbook): onecolumn table of contents & minitoc in twocolumn document and scrwfile removes partial Toc created with titletoc. If not, make an edit and refer to them, what does not fit. –  Speravir Feb 9 at 21:39

Partial answer, because I was lately trying something similar. Updated with @speravir’s suggestions and two ugly solutions of putting \dictum between prefixed chapter number and chapter title.

\documentclass[
chapterprefix=true%% like in standard class "report"
]{scrreprt}
%\usepackage[T1]{fontenc}%% pdflatex
\usepackage{fontspec}%% xelatex
\usepackage{xcolor}
\usepackage{graphicx}

%% partial TOCs after chapter title
\usepackage[]{minitoc}
\setcounter{secnumdepth}{3} % number subsubsections
\setcounter{tocdepth}{3} % list subsubsections
\mtcsetdepth{minitoc}{2}% ~ \setcounter{minitocdepth}{2}
\mtcsetfont{minitoc}{*}{\usekomafont{chapterentry}}
\mtcsetfont{minitoc}{section}{\usekomafont{chapterentry}\bfseries}
\mtcsetfont{minitoc}{subsection}{\usekomafont{chapterentry}}
\mtcsettitle{minitoc}{\relax}% remove “Contents”
\mtcsetoffset{minitoc}{-3.5em}% don’t know the appropriate length
\mtcsetrules{minitoc}{off}
\mtcsetpagenumbers{minitoc}{on}% default anyway
%\mtcsetfeature{minitoc}{before|after|open|close|pagestyle}{command}

%% layout of chapter title by @Speravir
\renewcommand*{\chapterformat}{%
\mbox{\chapappifchapterprefix{\nobreakspace}%
\scalebox{5}{\color{gray}\thechapter\autodot}\enskip}%
}

%% put dictum between chapter number (with prefix) and title
%% \setdictum[⟨source⟩]{⟨quote⟩} -- before \chapter
\iftrue % change to \iffalse to test different approach
%% hacky approach
\newcommand\getdictum{\relax}
\usepackage{etoolbox}
\makeatletter
%% \patchcmd{⟨command⟩}{⟨search⟩}{⟨replace⟩}{⟨success⟩}{⟨failure⟩}
\makeatother
\else
%
\newcommand\setdictum[2][]{%
\setchapterpreamble[ol][0.5\textwidth]{\vskip 5em\dictum[#1]{#2}}%
}
\fi

\begin{document}
\dominitoc
\faketableofcontents

\setcounter{chapter}{6}\setcounter{page}{199}

\setdictum[Sun Tzu]{If you know the enemy and you know yourself %\\
you need not fear the results of a hundred battles.}
\chapter{Conclusions}

\minitoc

\section{Summary of Achievements}
Text goes here
\newpage\setcounter{page}{212}
\subsection{Papers Submitted}
\subsubsection{I want to go into the TOC}
\end{document}

-
@Crissov Oh my that is a brilliant start! I will try and integrate this into my MWE. Much appreciated! –  Fiztban Feb 10 at 15:06
@Crissov I have added an MWE to my original post, I didn't have much luck achieving the desired effect when I implemented your code, would you be able to have a look through what I have done and see if you can spot the mistake I must have done. Thank you –  Fiztban Feb 10 at 15:56
A remark as comment, that you’ve used \setcounters just for the example, would be better in my eyes. For an experienced user this is obvious, but … –  Speravir Feb 10 at 22:16
@Speravir Thank you for the concern, but that I worked out quickest out of all the commands :D. Right now I am trying to assess whether the titlesec package helps or rather disrupts the KOMA-script macros fiven this post –  Fiztban Feb 10 at 22:56
@Fiztban: You, Crissov and me are not the only ones who can read this. :-) For titlesec: You must find out by trial & error, which functions work and which don’t. And it can happen, that functions working now will not in a future KOMA-Script release (recently happened, as can re read in one of the questions linked in comments below your question). –  Speravir Feb 10 at 23:29

I used his MWE and redefined (below \addtokomafont{chapterprefix} – it’s generally useful to keep definitions for the same topic as far as possible together) the command \chapterformat (the original definition is shown in documentation). I added a color change definition and a \scalebox from graphicx:

\renewcommand*{\chapterformat}{%
\mbox{\chapappifchapterprefix{\nobreakspace}%
\scalebox{3}{\color{gray}\thechapter\autodot}\enskip}}

I also loaded fontenc just to get rid of a font warning. All together:

\documentclass[
chapterprefix=true% like in standard class "report"
]{scrreprt}
\usepackage[T1]{fontenc}
\usepackage{xcolor}
\usepackage{graphicx}
\usepackage{minitoc}
\setcounter{secnumdepth}{3} % number subsubsections
\setcounter{tocdepth}{3} % list subsubsections

\renewcommand*{\chapterformat}{%
\mbox{\chapappifchapterprefix{\nobreakspace}%
\scalebox{3}{\color{gray}\thechapter\autodot}\enskip}}

\begin{document}
\dominitoc
\tableofcontents

\setcounter{chapter}{6}\setcounter{page}{199}

\chapter{Conclusions}
\dictum[Sun Tzu]{If you know the enemy and you know yourself %\\
you need not fear the results of a hundred battles.}

\minitoc

\section{Summary of Achievements}
Text goes here
\newpage\setcounter{page}{212}
\subsection{Papers Submitted}
\subsubsection{I want to go into the TOC}
\end{document}

The output looks this way:

-
You're a wizard! I will begin learning this, was reading your minitoc reply here, but I will wrap my head around this one for the time being. Thank you for all your help thus far! –  Fiztban Feb 10 at 23:29
@Fiztban: You’re welcome. Please accept nethertheless Crissov’s answer. –  Speravir Feb 10 at 23:38
Oh yes of course I will do that. Sorry to bother you again, would you please post the entire code, I can't quite make out where you added the renewcommand. –  Fiztban Feb 10 at 23:41
Oh how silly of me, forgot to load the graphicx package thank you so so much, now all that is left is getting it to work with my code for the minitoc. Speravir I wish you many good things! Thank you –  Fiztban Feb 11 at 0:00
@Speravir, why should Fiztban accept my partial answer (which is hardly the MWE he should have provided), when yours is so much closer to the desired result? –  Crissov Feb 11 at 10:02


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# Average time ant needs to get out to the woods

An ant has three passages to choose from:

1. Passage A takes 7 minutes to get ant out of the ant house to the woods.

2. Passage B takes 8 minutes to get ant back to the starting point where he is.

3. Passage C takes 12 minutes to get ant back to the starting point where he is.

The ant chooses a passage randomly until it gets out of the ant house to the woods.

How to calculate the expected average time ant needs to get out?

Does simple mean value (7 + 8 + 12) / 3 = 9 answer the question?

• Note that if the ant doesn't get out on the first try in 7 minutes, he will take at least 15 minutes to get out. To get an average time of only 9 minutes, the ant would have to escape on the first try more than half the time, which we know isn't the case. Just trying to give an intuitive idea of what range of answers might make sense, as a sanity check to rule out possible lines of thought - we can see that the mean answer is too low. – Nuclear Wang Jan 30 at 19:17
• If there is only one way out and $n-1$ ways of wasting time then the simple sum answers the question – Henry Jan 31 at 12:38
• This is same as this popular question. – StubbornAtom Feb 1 at 19:50

$$T=7/3+(8+T)/3+(12+T)/3=9+2T/3$$ $$T/3=9$$ $$T=27$$ From start point, each of 3 paths are equally possible. Two paths lead you back to start point.

• What is  8+T and 12+T ? – dokondr Jan 30 at 17:44
• +1 Very elegant! @dokondr $T$ is the expected time the ant will take to get out of the forest. If it goes on passage $B$ or $C$ it's back to square one, and has to embark on a journey with expected time $T$ all over again, plus the time it took for that particular passage. – Bridgeburners Jan 30 at 18:14
• So it is a recurrent definition of T – dokondr Jan 30 at 19:14
• $T$ is not defined recursively, but the initial formula is a recursive relationship that arises out of the definition of the expectation. – whuber Jan 30 at 22:07
• It's not a recursive definition of T; it's just an equation which is true of T, where T is the entire left-hand side and where T also appears on the right-hand side. – Tanner Swett Jan 31 at 14:03

To answer this question, you have to sum over all possible paths the ant can take, and get the duration of that path, multiplied by the probability of taking that path. That is, $$E[T] = \sum_{\text{path} \in \text{possible paths}} p(\text{path}) T(\text{path})$$

Every possible path takes Passage $$A$$ only once, but can take passages $$B$$ and $$C$$ any number of times, in any permutation. So possible paths can be $$A$$, $$BA$$, $$BBCCA$$, $$BCCBA$$, etc.

Suppose the probabilities of choosing passages $$A,$$ $$B,$$ and $$C,$$ are $$p_a$$, $$p_b$$, and $$p_c$$ respectively. Then, for example, the probability of taking path $$CBBCCA$$ is $$p_a p_b^2 p_c^3.$$ And, because there are $${5 \choose 2} = 10$$ ways of taking $$B$$ twice and $$C$$ three times, the contribution of the expected path time given by the possibility of 3 $$C$$s and 2 $$B$$s is $$10 p_a p_b^2 p_c^3 (T_a + 2 T_b + 3 T_c)$$, where $$T_a$$, $$T_b$$, and $$T_c$$ are the path times of each passage respectively.

The above gives the contribution to the expected time for taking two $$B$$s and three $$C$$s before $$A$$. But in general, you have to sum over the expected time contribution of all possible combinations of paths $$B$$ and $$C$$ before $$A$$ I'll do that below, but I suggest you stop here and try it yourself first.

SPOILER

In general, the ant can take a non-$$A$$ passage any number of times between zero and infinity before taking passage $$A$$, and for that number of times, it can be any combination of passages $$B$$ and $$C$$. So, to get the expected path time, we sum up the contribution of all passage possibilities, multiplied by their time, which looks like,

$$E[T] = T_a + p_a \sum_{n=0}^\infty \sum_{i=0}^n {n \choose i} p_b^i p_c^{n-i} [i T_b + (n-i) T_c].$$

These sums can be evaluated. Using the Binomial theorem and taking the derivative, you can show that, $$\sum_{i=0}^n i {n \choose i} x^i y^{n-i} = n x (x+y)^{n-1}$$ and

$$\sum_{i=0}^n (n-i) {n \choose i} x^i y^{n-i} = n y (x+y)^{n-1}.$$

Using these identities, we get

$$E[T] = T_a + p_a (p_b T_b + p_c T_c) \sum_{n=0}^\infty n (p_b + p_c)^{n-1}.$$

Taking the derivative of the sum of the famous geometric series, you can show that, for $$|x| < 1$$,

$$\sum_{n=0}^\infty n x^{n-1} = \frac{1}{(1 - x)^2}.$$

Noting that $$1 - (p_b + p_c) = p_a$$, we get,

$$E[T] = T_a + \frac{1}{p_a} (T_b p_b + T_c p_c).$$

If we take $$p_a = p_b = p_c = 1/3$$ and your values for the times, we get $$E[T] = T_a + T_b + T_c$$ which is 27 minutes.

• Great explanation! – Davide ND Jan 30 at 15:19

Hello donkordr and welcome to CV.
No, the mean does not answer the question unfortunately.

You can read your problem as a Markov Chain with two states: woods, and ant house.

• From the woods: (does not really matter as it's the goal) $$p=1$$ to stay in the woods
• From the ant house: $$p_1=\frac{2}{3}$$ to stay at the ant house and $$p_2=\frac{1}{3}$$ to go to the woods

We want to know how many movements it takes on average to reach the woods - and you can do so as explained here

This results in an average of 3 movements.
Of these 3 movements, one and only one will be the one leading to the woods, while the rest will be any of the others. The average time of a movement that does not go to the woods is $$(8+12)/2 = 10$$ minutes.

As a result, the average time before you reach the woods is $$27$$ minutes: $$7$$ from the last step, and $$2\cdot10$$ from the previous ones.

PS - there are other ways to make this computation with intermediate states that might be cleaner, but this seemed easy enough.

Let me join with an explanation which, to me at least, seems even easier:

First, observe that the paths $$B$$ and $$C$$ can be joined into a single path $$X$$, with the passage time equal to the mean times of $$B$$ and $$C$$, i.e. $$10$$ minutes. This is due to the fact that these paths have the same probability. If they didn't, we'd need to take a weighted average. The probability of taking the path $$X$$ is the sum of probabilities of taking either the path $$B$$ or $$C$$: $$P(X) = P(B) + P(C) = 2/3$$.

Now, there are the following ways of getting to the woods:

$$\begin{array}{lrr} i & \text{path}_i & P(i) & T(i)\\ \hline 0 & A & 1/3 & 7\\ 1 & XA & 2/3 \cdot 1/3 & 10 + 7 \\ 2 & XXA & (2/3)^2 \cdot 1/3 & 20 + 7 \\ 3 & XXXA & (2/3)^3 \cdot 1/3 & 30 + 7 \\ & ... & \\ i & (i \cdot X)A & (2/3)^i \cdot 1/3 & 10\cdot i + 7 \end{array}$$

and the expected time is:

$$\begin{array}{lcl} E(T) & = & \sum_{i=0}^\infty P(i)T(i) \\ & = & 1/3 \cdot \sum_{i=0}^\infty \left( 10 \cdot i + 7 \right) \cdot(2/3)^i \\ & = & 10/3 \cdot \sum_{i=0}^\infty i \cdot(2/3)^i + 7/3 \cdot \sum_{i=0}^\infty (2/3)^i\\ & = & 10/3 \cdot 6 + 7/3 \cdot 3 \\ & = & 27 \\ \end{array}$$

• Thanks! Very nice explanation. Yet how do you get sum(i * (2/3)**i) equal to 6? And sum((2/3)**i) equal to 3 ? – dokondr Jan 30 at 17:11
• @dokondr Use the formula for sum of infinite geometric series. $\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$ and $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$. – mai Jan 30 at 22:30

### Typical way to solve this:

The ant will either take path a and finish or take path b or c and be back in it's starting position.

Let $$k$$ be the number of times that the ant already has taken path b or c. Let $$T_k$$ be the expectation value for the time to finish for an ant that took already $$k$$ times the path. Then the expectation value for an ant with $$k$$ steps can be expressed in terms of an ant with $$k+1$$ steps.

$$T_k = \underbrace{\frac{1}{3} 7}_{\substack{\frac{1}{3}th\text{ chance to finish with path} \\ \text{ a in 7 minutes }}} + \underbrace{\frac{1}{3} (T_{k+1} + 8)}_{\substack{\frac{1}{3}th\text{ chance to finish with path b} \\ \text{ in 8 minutes} \\ \text{plus what ant k+1 needs on average }}} + \underbrace{\frac{1}{3} (T_{k+1} +12)}_{\substack{\frac{1}{3}th\text{ chance to finish with path c} \\ \text{ in 12 minutes}\\ \text{ plus what ant k+1 needs on average }}}$$

Since the ants are in the same starting position independent from the history (number of steps $$k$$) the average time is (you have $$T_k = T_{k+1}$$ which you can use to solve the above equation) :

$$T_k = \frac{1}{3}7+\frac{1}{3}(8+T_k)+\frac{1}{3}(12+T_k)$$

and after some rearrangments

$$T_k = 7+8+12 = 27$$

### Using an average

You can solve this with a mean, sort of.

• The ant finishes at least with path a which at least takes at least 7 minutes
• In addition, the ant has 2/3 probability to take paths b or c (each time) which take on average $$\frac{8+12}{1+1} = 10$$ minutes.

The mean times that the ant takes paths b or c is:

$$1 \cdot \frac{1}{3}\left(\frac{2}{3}\right) + 2 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^2 + 3 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^3 + 4 \cdot \frac{1}{3}\left(\frac{2}{3}\right)^4 + .... = \sum_{k=1}^\infty k \cdot \frac{1}{3}\left(\frac{2}{3}\right)^k = 2$$

note that this is related to a geometric distribution and the average number of extra steps that the ant needs is 2 (and each step takes on average 10 minutes). So the ant will take (on average):

$$\text{ 7 minutes + 2 times \times 10 minutes = 27 minutes}$$

Interestingly: you could also say the mean time for a single step is $$9$$ minutes (what you computed), and the mean number of steps is $$3$$, so the ant takes $$3 \times 9 = 27$$ minutes (you were not very far from the solution).

• "Interestingly: you could also say the mean time for a single step is 9 minutes (what you computed), and the mean number of steps is 3, so the ant takes 3×9=27 minutes (you were not very far from the solution)." Is this solution guaranteed? Or just an approximation that happens to be correct – Cruncher Jan 31 at 19:07
• After reasoning about this problem, it does appear that it actually does work for every combination of values. Definitely unexpected. – Cruncher Jan 31 at 19:24
• @cruncher, I agree with you that at first it feels a bit simplistic to multiply two averages, because that is indeed not guaranteed to work every time. So I have to still look into that 'interestingly 3×9=27 part' which seems to be indeed a happy coincidence. But the 2×10 minutes seems sound to me. You can multiply the average number of steps with the average time per step when they are independent. Under the hood you are summing each possible combinations of step times with each possible number of steps. – Sextus Empiricus Jan 31 at 21:04

Commenting on @Igor F.'s post (not enough reputation in this subforum to simply comment):

Both terms are geometric series:

$$\sum_{i=0}^\infty i\cdot (\frac{2}{3})^i \overset{q=2/3}{=}\sum_{i=0}^\infty i\cdot q^i=q \frac{d}{dq}\sum_{i=0}^\infty q^i=\frac{q}{(1-q)^2}, \text{ for } |q|<1$$, so for $$q=\frac{2}{3}$$ this equals $$6$$.

$$\sum_{i=0}^\infty (\frac{2}{3})^i$$ is just a basic infinite geometric series and we have $$\sum_{i=0}^\infty c\cdot q^i=\frac{c}{1-q}, \text{ with constant }c \text{ and } |q|<1$$, which is $$3$$ for $$q=\frac{2}{3}$$ and $$c=1$$.

It always take $$7$$ min to succeed, which happens once. It takes an average of $$10$$ min to fail. The probability of failing is $$2/3$$ on each try, or $$(2/3)^n$$ for n tries. So the total time is ($$7 + 10\cdot \sum_n (2/3)^n$$) min. Which is $$7$$ min + $$2 \cdot 10$$ min $$= 27$$.


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# Recent activity by Mriganka

Consider the following C function. void convert (int n ) { if (n<0) printf{ %d , n); else { convert(n/2); printf( %d , n%2); } } Which one of the following will happen when the function convert is called with any positive integer $n$ as ... in the reverse order and terminate It will print the binary representation of $n$ but will not terminate It will not print anything and will not terminate
Let $T$ be a full binary tree with $8$ leaves. (A full binary tree has every level full.) Suppose two leaves $a$ and $b$ of $T$ are chosen uniformly and independently at random. The expected value of the distance between $a$ and $b$ in $T$ (ie., the number of edges in the unique path between $a$ and $b$) is (rounded off to $2$ decimal places) _________.
Frames of $1000\text{ bits}$ are sent over a $10^6$ bps duplex link between two hosts. The propagation time is $25ms$. Frames are to be transmitted into this link to maximally pack them in transit (within the link). Let $I$ be the minimum number of ... wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time) $16ms$ $18ms$ $20ms$ $22ms$


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# Calculate number of states in edge triggered flip flops

Consider the sequential circuit shown in the figure, where both flip-flops used are positive edge-triggered D flip-flops.

simulate this circuit – Schematic created using CircuitLab

The number of states in the state transition diagram of this circuit that have a transition back to the same state on some value of “in” is _.

My question is how can we calculate a number of states in above diagram. Is there any formulae for that or any logic to calculate?

• 2 binary registers. How many combinations they can have? – Eugene Sh. May 31 '18 at 17:31
• only 2 according to that. But if there are 4 combinations then the answer is 4 ? – Himanshu May 31 '18 at 17:33
• How is it 2? Not really. Yes, it is 4. 2^2. BTW, if you look closer, you will identify a 2 bit SISO shift register here. – Eugene Sh. May 31 '18 at 17:33
• If you can explain that supporting your answer. It will be really helpful thanks. – Himanshu May 31 '18 at 17:42
• It can be 00, 01, 10, 11, right? Can you think why it can't be any of these? Can you think of anything else? – Eugene Sh. May 31 '18 at 17:46

Two FFs can be in a total of 22 = 4 states altogether. An appropriate pattern of bits on Input can drive the machine to any given state.


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This page provides more information about the IRIS Python tutorial given at the IRIS-10 meeting in Bengaluru, India. At the workshop, participants are expected to work on the tutorial using Binder, a cloud-based solution that lets you run an instance of Jupyter lab with all the required packages and data files. To start the Binder notebook, follow the link below:

Binder makes it easy to get started with minimal effort, but if you want to actually use IRISPy, you will need to have it installed in your computer. IRISPy relies on several other packages such as SunPy, astropy, and matplotlib. The recommended way to install all of them is via conda (miniconda or Anaconda). You can follow the installation instructions on the IRISPy documentation, or follow all the lines below (they are very similar).

To install the tutorial files, you need to clone its git repository or download a zip file. To clone, run the following in a directory of your choice:

git clone https://github.com/tiagopereira/iris10.git

You will end up with a directory called iris10. In the terminal, go to that directory. If you do not have IRISPy installed, you can create a new conda environment by using the supplied environment.yml file. The following will download and install all required packages into an environment called iris:

conda env create -f environment.yml

Then you need to activate the environment:

conda activate iris

Lastly, you will need to install IRISPy (not available on conda) with pip:

pip install git+https://github.com/tiagopereira/irispy.git

Optionally, if you want to have interactive matplotlib plots in jupyter, you need to run (this operation takes a while):

jupyter labextension install @jupyter-widgets/jupyterlab-manager jupyter-matplotlib

Finally, you will need to download the data files used in the tutorial. Assuming you are at the terminal in the directory of iris_tutorials, you can download the data files into iris_tutorials/notebooks with the following:

wget -c http://www.lmsal.com/solarsoft/irisa/data/level2_compressed/2018/01/02/20180102_153155_3610108077/iris_l2_20180102_153155_3610108077_SJI_1400_t000.fits.gz -P notebooks/
wget -c http://www.lmsal.com/solarsoft/irisa/data/level2/2014/09/19/20140919_051712_3860608353/iris_l2_20140919_051712_3860608353_SJI_2832_t000.fits -P notebooks/
wget -c https://folk.uio.no/tiago/aia_20140919_060030_1700_image_lev1.fits -P notebooks/
wget -c https://folk.uio.no/tiago/iris_l2_20180102_153155_3610108077_raster_t000_r00000.fits.bz2 -P notebooks
pbunzip2 notebooks/iris_l2_20180102_153155_3610108077_raster_t000_r00000.fits.bz2

jupyter lab

And then navigate to notebooks/IRIS10.ipynb.


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The Laplace-Beltrami derivative operator

Along with the idea that the metric of spacetime is dynamical, the other big idea of general relativity is that physics should be independent of any particular coordinate system, in the sense that any physically relevant mathematical object should be defined in such a way that in a different coordinate system it is an object of the same type. This is called the "Principle of General Covariance." (Covariance = co-variance = varying along with - the idea is that as the coordinate system is allowed to change, the physically relevant quantities will change along with it in a well-defined way). The allowed kind of transformation law for physically relevant quantities is called the tensor transformation law, and it tells you how vectors, differential forms, and other types of tensors are allowed to transform under coordinate transformations.

First, let's say our manifold $M$ has coordinates $x^{\mu}$ (where $\mu$ runs from 0 to 3). Suppose we make a change of coordinates to a new set, $y^{\rho}$ ($\rho = 0, \dots , 3$), $y^{\rho} = y^{\rho} \left( x^{\mu} \right)$. Then by the chain rule, we have

(1)
\begin{align} \frac{\partial}{\partial x^{\mu}} = \frac{\partial y^{\rho}}{\partial x^{\mu}} \frac{\partial}{\partial y^{\rho}} \end{align}

(we're using the Einstein summation convention here). So the vector $\mathbf{v} = v^{\mu} \partial_{\mu}$ transforms under the coordinate transformation:

(2)
\begin{align} v^{\mu} \frac{\partial}{\partial x^{\mu}} = v^{\mu} \frac{\partial y^{\rho}}{\partial x^{\mu}} \frac{\partial}{\partial y^{\rho}}, \end{align}

so the components of $\mathbf{v}$ in the new coordinate system (i.e. with respect to the new basis tangent vectors $\frac{\partial}{\partial y^{\rho}}$) are

(3)
\begin{align} \tilde{v}^{\rho} = v^{\mu} \frac{\partial y^{\rho}}{\partial x^{\mu}} \end{align}

(again using Einstein summation convention).

You can similarly work out the transformation for a differential 1-form $\mathbf{\omega} = \omega_{\mu} dx^{\mu}$ to be

(4)
\begin{align} \tilde{\omega}_{\rho} = \omega_{\mu} \frac{\partial x^{\mu}}{\partial y^{\rho}} \end{align}

and for the metric,

(5)
\begin{align} \tilde{g}_{\rho \sigma} = g_{\mu \nu} \frac{\partial x^{\mu}}{\partial y^{\rho}} \frac{\partial x^{\nu}}{\partial y^{\sigma}} \end{align}

The important thing is that these are just curvilinear generalizations of the way you transform vectors and matrices under rotations and flips in the plane or $\mathbb{R}^3$. Although the components of the vectors or 1-forms or metrics alter during the transformation, they do so in a clean way that doesn't privilege any coordinate system over any other. All this stuff is very nicely done in Chapter 1 of Baez and Muniain, Gauge Fields, Knots, and Gravity.

Now since derivative operators are necessary to describe time evolution and spatial constraints in physics, we quickly notice that we need derivative operators more cleverly designed than just plain coordinate derivatives $\frac{\partial}{\partial x^{\mu}}$ to construct physical laws. If we just used the coordinate derivatives, we would be implicitly privileging one coordinate system over another, because $\frac{\partial}{\partial x^{\mu}}$ is only a simple coordinate derivative with respect to a particular coordinate system. If we transform to another coordinate system, it will be some nontrivial linear combination of coordinate derivatives. The coordinate direction $x^{\mu}$ has no more intrinsic physical meaning than the coordinate direction $y^{\rho}$.

In order to address this issue, we create derivative operators which treat all coordinate systems democratically, and transform in an allowed manner with coordinate transformations. In particular, the curved-space version of the Laplacian is called the Laplace-Beltrami operator, and is defined like this, acting on a function $f: M \to \mathbb{R}$:

(6)
\begin{align} \Delta_{LB} f = \frac{1}{\sqrt{\lvert g \rvert}} \partial_{\mu} \left( \sqrt{ \lvert g \rvert } g^{\mu \nu} \partial_{\nu} f \right) \end{align}

where $\sqrt{ \lvert g \rvert }$ is the square root of the absolute value of the determinant of the metric $g_{\mu \nu}$.

page revision: 2, last edited: 05 Apr 2014 18:39


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# range

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What is the range of the function f(x) = 1/x^3?

May 3, 2022

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Note that for any $$y \neq 0$$, there exists $$x = \dfrac1{\sqrt[3]y}$$ such that $$f(x) = y$$.

Therefore, the range is $$\mathbb R -\{0\}$$.

May 3, 2022