Similar to the question (and MWE) asked in the link below, I would like to organise my navigation bullets horizontally per section. However, in addition, I would like to group the bullet points by subsection.   hence, instead of:  Section 1  ooooooooooo  I would like to have  Section 1  ooo ooo ooo oo  Where the dots are grouped by subsection. I am using the Singapore theme.
I'm using the beamer miniframes theme (with subsection=false) and the compress option. Currently the navigation shows the dots for each subsection side-by-side. I would prefer if there would be a horizontal space (e.g. with the width of a dot) between the dots of each subsection. Is this possible? Right now it looks like:  Section 1 oooooo  And I would prefer it to look like:  Section 1 ooo ooo  MWE: \documentclass[compress]{beamer}   \useoutertheme[subsection=false]{miniframes}   \begin{document}   \section{Sec 1}   \subsection{}   \begin{frame}[t]     hi   \end{frame} \subsection{}   \begin{frame}[t]     hi   \end{frame} \end{document}`
Every time since a few days ago after I caught Mespirit, when I go to the portal in the Nameless Cavern and I press A to interact. All that happens is a short dialog sequence and nothing else. I know the requirements are along the lines of having 3 max happiness Pokemon with you in your party which I do. I have also downloaded the recent update/patch from the eShop if that helps at all. How can I make Uxie and Azelf appear in the cavern? Debug JavaScript within IE6 Limit of $n/\ln(n)$ without L'Hôpital's rule I am trying to calculate the following limit without L'Hôpital's rule:     $$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$   I tried every trick I know but nothing works. You don't have to prove it by definition. I just saw a user named: .yahoo.co.jpaqwsykcj3aulh3h1k0cy6nzs3isj  Could new users be named according to their user-number, or something other than "Unknown"? Perhaps User1234567 instead? It would be less awkward to address them in answers and comments.   Additionally, I often times have to check profiles to make sure users aren't abusing the system, since I find many of the same questions from "unknown (google)" from time to time. That, or I'll mistakenly think I've helped the user on a similar issue in the past, and assume the user has knowledge that he or she does not have. I'm trying to show that  $$\delta\big(f(x)\big) = \sum_{i}\frac{\delta(x-a_{i})}{\left|{\frac{df}{dx}(a_{i})}\right|}$$  Where $a_{i}$ are the roots of the function $f(x)$. I've tried to proceed by using a dummy function $g(x)$ and carrying out:  $$\int_{-\infty}^{\infty}dx\,\delta\big(f(x)\big)g(x)$$  Then making the coordinate substitution $u$ = $f(x)$ and integrating over $u$. This seems to be on the right track, but I'm unsure where the absolute value comes in in the denominator, and also why it becomes a sum.  $$\int_{-\infty}^{\infty}\frac{du}{\frac{df}{dx}}\delta(u)g\big(f^{-1}(u)\big) = \frac{g\big(f^{-1}(0)\big)}{\frac{df}{dx}\big(f^{-1}(0)\big)}$$  Can any one shed some light? Wikipedia just states the formula and doesn't actually show where it comes from. How to get a combined ticket in IRCTC (Indian Railway) Suppose I want to go from Station A to Station C but Station C is not a stoppage of some train say train 01, so I can get a combined ticket from station A to Station B  through train 01 plus Station B to station C through a different train say 02, so my total fare will be of same ticket as from Station A to Station C.  We can get Combined ticket from the Booking Window but how's that possible on the IRCTC website? What torque for caliper brakes onto carbon-alloy forks? I've recently got a new bike with a . This comes with deep-drop calliper brakes (Shimano R451 Dual Pivot). The first time out I bumped one of the brakes and it twisted around the single attachment bolt and rubbed on the rim.  I managed to easily twist it back by hand and carry on, however I thought the bolt probably needs tightening. The question is: How tight can it be?  As the fork is part carbon I've so far only dared to take it up to 4Nm, but it still seems possible to twist the whole calliper around with a bump of the hand. Maybe that's normal? Perhaps it's easier to bump around because of the longer deep-drop lever. Should I be aiming to get it really secure?    I'm not sure whether the brake mount in the fork is all carbon, or some of it is alloy underneath, but either way I'd like to know what a sensible torque is. As far as I can see from the  there is no such specification. How can I find this out? This question led to a new package:      I'm giving my students an old exam and barely remembered to change the date at the top. A colleague and I joked about how long we could reuse the same exam, and it made me wonder if I could use LaTeX to create an exam that looked like it was written in 1963 on a typewriter.  Here's what I'd like to see:   Typewriter font even in math mode. (Several ways to accomplish this; is there a "best" way?) Slight inconsistencies in the printing of each letter. I attempt this below adapting code taken from . Each word has a random vertical shift, though I'd like to see it reduced to each letter having its own vertical shift. Other typewriter-like irregularities would be great. It would be great if math-mode delimeters \left(, \right}, etc., can be created in a piece-meal fashion as they did in days of old. Fraction bars, tops of square-root symbols -- any horizontal line, really -- is made up of a series of en-dashes that don't quite match up.   Here is what I've come up with so far:  \documentclass[10pt]{article}% This is a document class providing more font size options \usepackage{graphicx} \usepackage{ifthen} \usepackage[textwidth=7.5in,textheight=9.5in]{geometry} \pagestyle{empty}  % thanks to Bruno Le Floch: https://tex.stackexchange.com/q/9331/4012 % and in his comments to https://tex.stackexchange.com/a/29458/4012 \usepackage[first=-1,last=1]{lcg}% you can play around with these values \makeatletter \newcommand{\globalrand}{\rand\global\cr@nd\cr@nd} \makeatother  \newcommand{\randomvshift}[1]{\globalrand\raisebox{\value{rand}pt}{#1}}  %%% thanks to Martin Scharrer: https://tex.stackexchange.com/q/11598/4012 \makeatletter \def\typewriter#1{%     \@typewriter#1 \@empty } \def\@typewriter#1 #2{%    \randomvshift{#1}\space    \ifx #2\@empty\else     \expandafter\@typewriter    \fi    #2% } \makeatother \begin{document}\tt  \typewriter{Math 101 Fall 1963}  \begin{enumerate} \item   \typewriter{$\frac{\texttt{d}}{\texttt{dx}}\texttt{(x)}^2$}  \item \typewriter{State the Fundamental Theorem of Calculus.}  \end{enumerate}  \end{document} Removing all \n\r characters from a node XSLT? wondered if you could help me please? I have node in xml that is as followed  $LOG: 08880xbpnd $ fhdsafidsfsd df sd fsd f sd fsd  I was wondering is there anyway to make all the text go on to one line so that it then can be passsed through to a javascript function? so it would turn out like this   $LOG: 08880xbpnd $fhdsafidsfsddfsdfsdfsdfsd Record Animation In BGE Should questions include "tags" in their titles? I've seen questions edited to remove the language name from the title. Won't search engines - and folks using search engines - have an easier time finding them if these "tags" are kept as part of the title? Calculate the limit including binomial coefficients Would a female Jedi be referred to as Jedi Mistress instead of Jedi Master? Likewise would a Sith female be referred to as a Sith Lady instead of a Sith Lord? Set margins in maketitle How do I work recursively through a directory tree and execute a specific command on each file, and output the path, filename, extension, filesize and some other specific text to a single file in bash. How to check if Node connected to another node input in Python I am doing a process to some switch links in materials. I need to know if something is ever linked or not before to process. I can't find out how to :/ The script works well if something is linked only.  import bpy for m in bpy.context.object.material_slots:     mat = m.material     nodes = mat.node_tree.nodes      # get node input (socket)     alphasocket = mat.node_tree.nodes.get("Principled BSDF").inputs[18]     # get node link that connects into the socket     lnk = (link.from_socket for link in mat.node_tree.links if link.to_socket == alphasocket)      #add emission node     emi= nodes.new('ShaderNodeEmission')     emi.name= "emibak"     outepute= mat.node_tree.nodes.get("Material Output").inputs[0]      #make new link destination     #here 's where I cant find out how to check     if lnk != None:         print(mat.name + " alphanod found")        # mat.node_tree.links.new(lnk, emi.inputs[0])     else :         print(mat.name + " alphanod missing")      mat.node_tree.links.new(emi.outputs[0], outepute)     here is another way I am searching...  import bpy  for m in bpy.context.object.material_slots:      mat = m.material     nodes = mat.node_tree.nodes          print("**** Material " + mat.name+ " ****")      for mat_node in mat.node_tree.nodes:         if mat_node.type == 'BSDF_PRINCIPLED':             print("From " + mat_node.name )                         alphasocket = mat_node.inputs[18]             #print(alphasocket)             node_in = mat_node.inputs             #print(node_in)                    for n in alphasocket.links:                                print("Alpha connected to " + n.from_node.name)   This show me what is connected to the alpha channel but if nothing is connected I want to connect a new node. Algorithm for computing Hadamard product of two rational generating functions If I have two generating functions $A(x) = \sum_na_nx^n$ and $B(x) = \sum_n b_nx^n$   then the Hadamard product is $(A \star B)(x) = \sum_{n} a_nb_nx^n$.  Now when $A$ and $B$ are both rational functions ($P(x)/Q(x)$ with polynomials $P, Q$) I've seen non-constructive proofs that $A \star B$ is also rational.  Given two rational generating functions $A$, $B$, is there an algorithm that efficiently computes $A \star B$? Or at least a constructive proof you could use to write an algorithm? Why is menstruation in wild animals not a disadvantage to organismal survival? When I watch wilderness specials for more than a few minutes, I notice a familiar pattern: predators are depicted as being alerted by the scent of blood. Wounded animals seem to make the best prey and predators seem to have a heightened awareness for the smell of blood.   It would seem that it would then be a disadvantage to evolutionary survival for female organisms to menstruate. How do females of a species not attract predators to the signal of menstrual blood?   Do female organisms in the wild actually menstruate? If so, is there a difference in the composition of menses that does not alert predators? How would this not at the very least give predators an idea of the general vicinity of the female?  If wild animals excrete their menses, do they have ways in which they are able to mask the evidence of its release or are they somehow internalizing the tissue so that they are not released?  If female organisms in the wild do menstruate, then why is this not a survival disadvantage? Why aren't variable-length arrays part of the C++ standard? How can I solve this quantum mechanical "paradox"? Let a (free) particle move in $[0,a]$ with cyclic boundary condition $\psi(0)=\psi(a)$. The solution of the Schrödinger-equation can be put in the form of a plane wave. In this state the standard deviation of momentum is $0$, but $\sigma_x$ must be finite. So we find that $\sigma_x\sigma_p=0$. Is something wrong with the uncertainty principle?