Strange ConcurrentModificationException behavior
Why am I not getting a java.util.ConcurrentModificationException in this example?
Wrong numeration in toc longer then one page I tried anything I could find with \cleardoublepage and \newpage and so on; mostly they use \addcontentsline. I don't need to use it as listof=totoc already puts it into the toc so even if I use \addcontentsline instead I get the same result.  \documentclass[     fontsize=11pt,     DIV10,     ngerman,     paper=a4,      twoside=false,     titlepage,     parskip=half,     listof=totoc,     bibliography=totoc,     index=totoc,     captions=tableheading,     foottotes=multiple,     final ]{scrartcl}  \usepackage[     automark,     headsepline,     ilines ]{scrpage2}  \usepackage[ngerman]{babel} \usepackage[babel,german=quotes]{csquotes} \usepackage[dvips]{graphicx}  \begin{document}  \ofoot{\pagemark} \pagenumbering{Roman} \clearpage \tableofcontents \clearpage \listoffigures \clearpage \pagenumbering{arabic} \section{Test} \label{sec:Test}  Test  \subsection{Test} \label{sec:Test}  \begin{figure}[H]  \centering \caption{Testfig} \label{fig:testfig} \end{figure}  \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test} \subsection{Test} \label{sec:Test}  \end{document} Is it acceptable to edit (correct) grammar or spelling? Some individuals may be sensitive to edits of their posts on the basis of spelling or grammar. Some editors may be concerned about what to put in as a reason/explanation for the edit. New editors, in general, are likely uncertain of what the accepted "etiquette" is, on these sites, for such corrections. Embed .mov file via 'Add Media' not working A logic puzzle involving a balance. Possible Duplicate:          You have 12 balls and you know that they all weigh the same except for 1 which is heavier or lighter than all the others (you don't know which though). How can you make sure you know which ball is the heaviest/lightest in only 3 weighings?  The way I approached it was to split up the 12 balls into three sets of 4 and weigh two of the sets. If the sets balanced the scale, then I know the ball I am looking for must be in the set of 4 balls not weighed, else, I disregard said set and arbitrarily choose the heaviest set of 4 (as opposed to choosing the lightest set). I split the heaviest set of 4 balls into 2 and weigh that... etc.  Repeating this process until all 3 tries have been "used up", even if everything just so happened to be in your favor (the arbitrary choice you have in choosing the heaviest or lightest set happens to be the correct choice) in the end you still end up having to choose between 2 balls. A 50% chance is good, but I am wondering, is there a way to make sure 100%? How was Jon Snow able to kill the White Walker in S05E08? Reading list in topological QFT Conjugate Variables, Noether's Theorem and QM How do I get an XBOX 360 controller working? I have a Microsoft XBOX 360 gamepad with the wireless adapter and I was wondering what I would have to install to use them with Ubuntu I do know that when using this on Windows I have to manually install the drivers for the wireless receiver (if that is relevant.) How to execute sp_send_dbmail while limiting permissions Is there a way to provide access to users in my database to execute  msdb.dbo.sp_send_dbmail without needing to add them to the MSDB database and the DatabaseMailUserRole?  I've tried this:  ALTER PROCEDURE [dbo].[_TestSendMail] (   @To NVARCHAR(1000),   @Subject NVARCHAR(100),   @Body NVARCHAR(MAX) ) WITH EXECUTE AS OWNER AS      BEGIN         EXEC msdb.dbo.sp_send_dbmail @profile_name = N'myProfile',             @recipients = @To, @subject = @Subject, @body = @Body     END   But I get this error:  The EXECUTE permission was denied on the object 'sp_send_dbmail', database 'msdb', schema 'dbo'.   Thanks! Are there any studies on the use of justified text in web apps? How does justification affect screen legibility?  Example of justified text: Should I always use a comma after "e.g." or "i.e."? How to build a difference distribution table? What is the chance that at least two people were born on the same day   of the week if there are 3 people in the room?   I'm wondering if my solution is accurate, as my answer was different than the solution I found:  Probability that there are at least 2 people in the room born on the same day = 1 - (No one was born on the same day) - (Exactly one person was born on the same day)  There are (3 choose 2) different pairs of couples. Each couple has the same birthday as another couple with the chances of 1/7 and different with chances 6/7. Thus:  $$1 – (6/7)^3 – 3(1/7)(6/7)^2 = 0.0553$$  Thanks for any help! How can I make Firefox open magnet-links in Transmission? How to make an embedded question from "What time is it?" How to render as only wireframe in Cycles? How to use a decimal range() step value? Is there a way to step between 0 and 1 by 0.1?   I thought I could do it like the following, but it failed:  for i in range(0, 1, 0.1):     print i   Instead, it says that the step argument cannot be zero, which I did not expect. Beamer keeping line position constant on itemize I want to have a slide in beamer with itemize exposing new line on each click while at the bottom part of the slide there are some equations. This is my attempt  \begin{frame}   \begin{overprint}{\textwidth}     \leavevmode     \begin{itemize}[&lt;+-&gt;]     \item &lt;1-&gt; Poisson Boltzman equation (PB)     \item &lt;2-&gt; Debye-H\"uckle equation  (DH)     \item &lt;3-&gt; The naive approach - applying (DH) equation to PE     \end{itemize}   \end{overprint}   \begin{tikzpicture}     \node at (0,0) {};         \node&lt;1,2&gt; [text width = 20cm]at (10.5,-1) {$ \nabla^2 \phi =\sum_i \frac{n_i q_i\text{e}^{-q_i \phi/kT}}{4 \pi \epsilon}$};         \node &lt;2&gt; [text width = 20cm]at (10.5,-2) {$q_i \phi/kT \ll 1 \Rightarrow \nabla^2 \phi \simeq \kappa^2 \phi$};         \node &lt;2&gt; [text width = 20cm, font=\footnotesize]at (10.5,-2.5) {$\kappa^2 = \lambda^{-1/2} =  \sum_i \frac{n_i q_i}{4 \pi \epsilon}$};         \node &lt;3&gt; [text width = 20cm] at(10.5,-1) {$\frac{F_{el}}{VkT} = - \left ( \frac{ 4 \pi e^2}{\epsilon kT}     \sum_{i=1} n_i z_i^2 \right )^{3/2} \left ( 12 \pi \right )^{-1}$};   \end{tikzpicture} \end{frame}   However, the itemized line jump up and down between slides. I've tried to use itemize without any environment surrounding it. I've also tried to use overlayarea instead of overprint. How can I keep the line position constant?  I've uploaded a gif animation to display this. On my machine it run really fast and I can't fix this. Hope the problem is understood.   When I run this, I get I can prove the converse of it, but I cannot do this one. Here is the problem:     Prove that the implicit function theorem implies the inverse function theorem. Subdividing a beveled mesh creates new vertices in wrong places