I've been having some trouble aligning multiple figures. The problem is present with png figures and tikz figures, so I guess I am missing something.  tikz figures not properly aligned:   the particular bit of code:  \begin{figure}     \centering     \input{figs/grf_left_fb.tikz}     \input{figs/grf_fb.tikz}        \input{figs/phase_fb.tikz}     \input{figs/zap_fb.tikz}     \caption{......}     \label{fig:feedback} \end{figure}   and with png figures (both vertical and horizontal alignment needed):     \begin{figure}     \centering     \includegraphics[scale=0.25]{figs/s1.eps}     \includegraphics[scale=0.25]{figs/s2.eps}     \includegraphics[scale=0.25]{figs/s3.eps}     \includegraphics[scale=0.25]{figs/s11.eps}     \includegraphics[scale=0.25]{figs/s28.eps}     \caption{S..........}     \label{fig:data_good_bad} \end{figure}
I have several very similar images except some have some extra graphics that enlarge them slightly. This causes some pictures to be offset and not aligned visually when it's clear they should be.  How can I align several tikzpictures along the the same origin they were created and center vertically and horizontally properly?   First, I want to align the 3 large boxes regardless of the what the small boxes are doing Second, I want to be able to center them all without adding additional spaces the \begin{center} seems to do Third, I want to be able to reproduce the exact same results but everything in landscape mode without using the landscape package if possible. (as you can see, the problem is that TeX is putting one rectangle at a time which is not what I want) The methods should be easy and quick. I have many rectangles stacked and some need to be grouped arbitrarily. What these means is I want the methods to sort of be "smart" in that they can figure out themselves so I don't have to go change a lot of stuff each time I have to change the grouping. Basically I would like the first rectangle on the page to define the alignment location and the others on the same page to use it unless I tell them not to.   I've tried creating a custom bounding box as shown in  but it didn't do anything unless I make it too large which then causes other problems. (I can end up doing the alignment but it seems more complex than necessary)  \documentclass[10pt]{book} \usepackage{tikz}  \begin{document} \begin{tikzpicture}     \draw[line width=30pt] (0,0) -- (7,0); \end{tikzpicture}  \begin{tikzpicture}     \draw[line width=30pt] (0,0) -- (7,0);     \draw[line width=20pt] (-0.4, 0) -- (-0.1, 0); \end{tikzpicture}  \begin{tikzpicture}     \draw[line width=30pt] (0,0) -- (7,0);     \draw[line width=20pt] (-0.4, 0) -- (-0.1, 0);     \draw[line width=20pt] (-0.8, 0) -- (-0.5, 0); \end{tikzpicture}    \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (7,0); \end{tikzpicture}% \end{center}%  \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (7,0);     \draw[line width=20pt] (-0.4, 0) -- (-0.1, 0); \end{tikzpicture}% \end{center}%  \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (7,0);     \draw[line width=20pt] (-0.4, 0) -- (-0.1, 0);     \draw[line width=20pt] (-0.8, 0) -- (-0.5, 0); \end{tikzpicture}% \end{center}%     \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7); \end{tikzpicture}%   \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7);     \draw[line width=20pt] (0,-0.4) -- (0,-0.1); \end{tikzpicture}%  \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7);     \draw[line width=20pt] (0,-0.4) -- (0,-0.1);     \draw[line width=20pt] (0,-0.8) -- (0,-0.5); \end{tikzpicture}%    \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7); \end{tikzpicture}% \end{center}   \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7);     \draw[line width=20pt] (0,-0.4) -- (0,-0.1); \end{tikzpicture}% \end{center}%   \begin{center}% \begin{tikzpicture}%     \draw[line width=30pt] (0,0) -- (0,7);     \draw[line width=20pt] (0,-0.4) -- (0,-0.1);     \draw[line width=20pt] (0,-0.8) -- (0,-0.5); \end{tikzpicture}% \end{center}%   \end{document}
I am looking for either an online tool or one that I can download to convert DD to DMS.  For example, I want to convert from:  41.590833, -93.620833 to 41° 35′ 27″ N, 93° 37′ 15″ W ...  Those are the coordinates, according to , for Des Moines. I find that at the end of a long day, I sometimes have a large number of programs running. All which I will need to use tomorrow. Normally, this isn't an issue, I can simply lock the machine and come back tomorrow. My problem arrises when windows update launches in the middle of the night and force restarts my computer. That in turns closes all my open software. I of course save everything regularly so I don't loose anything, but I waste time reopening all of those resources whenever there is a restart.   [EDIT]  I should clarify that I still want to be able to restart my computer when an update comes down. Preventing the restart only delays the problem until later. I should have been more specific in that I want to be able to recover my working environment after a restart for any reason. Things like scheduled maintence, power loss, updates, and software installs.  [EDIT]  I can't simply have them setup to launch at startup becasuse those files change from week to week. So I need something that monitors what I have open, and gives me the option to "recover" those software sessions when I log back in.  Anyone have any suggestions on what I can do? I'd even be willing to purchase software to do this for me if that is the only option.  Thanks Can we show new/anonymous users more Q&A? With left nav, responsive design, and various new-user notices, this is what an anonymous user currently sees (browser window here is 1090px wide):    I realize that a lot of this content -- code of conduct, new-user intro, cookies notice -- is important.  And I get that the left nav is an important design element, and responsive design dictates the width of the right column.  But when you add it all up, Q&amp;A -- what the site is about -- gets a distressingly small portion of the screen.  Can we do better?   Maybe we can make that intro/mini-tour shorter? Maybe we can hide the right column for anonymous users? Maybe some things can wait until the user's viewed a few pages? Maybe other things I haven't thought of?   My concern is that anonymous users -- people coming from Google, people who were referred to a site by users hoping to recruit them, or others who find their way to us but aren't invested in us -- will turn away before seeing how rich and helpful our sites are.  Which would be a real shame, because our sites are awesome.  First impressions matter.  When I'm asking a rabbi to check out Mi Yodeya or a professional chef to check out Seasoned Advice or my vet to check out Pets (all of these have happened), I want to make a good impression.  I want the person to see more than 2.5 questions; I want to increase the chances the person will see something interesting or say "oh I can answer that!" and dig deeper.  Yes, the other stuff is important, but maybe not now?  Is there a better way to "stage" all this information for people we haven't yet hooked? Is there a convergent, alternating series that fails the AST? The alternating series test (AST) says, briefly, that if   $a_k&gt;0$ $a_k \geq a_{k+1}$, and $a_k \to 0$ as $k \to \infty$   then $\sum_k (-1)^k a_k$ converges.  This seems to be a one-way test (that is, if an alternating series fails the test, we don't know that it diverges).   document says so explicitly. It then gives an example of an alternating series which fails the AST. That doesn't prove the series is divergent (but it turns out to be divergent, anyway, by $n$th term test).  Is there a convergent, alternating series that fails the AST?  Of course, if the series fails condition 3, then it also fails the $n$th term test, and must diverge. And if it fails the first condition, then it's not strictly alternating, anyway.  So it must be a series that is not getting smaller (condition 2), but still converges. Exclusive lighting Is (or will be) there any sort of light linking in Cycles, or a method to get a light to affect only certain objects that doesn't involve compositing two or more different render layers? Sort of what 3Dstudio does with the Include/Exclude methods. Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$ I want to prove this equation, $$ \binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a} $$ I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.   (Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross. Recover Data from Card that cannnot be read I am trying to help a friend who is very concerned that she has just lost all of the data on her 32GB SanDisk microSD card.  I told her I would post this and see if anyone could offer help.  Here are the details:   The SanDisk microSD card is less than two years old.  Probably only about a year old.  The store will take it back, but that doesn't really help.  She wants her data. She used the card in her Samsung Galaxy S4 smartphone for over a month.  The phone was never exposed to water or temperature extremes. Today, her phone suddenly said something like "Your SD card is blank or an unsupported format". Her phone was then unable to read the SanDisk microSD card at all.   Here's what she has tried:   Rebooting her phone.  Had no effect. Taking out the SanDisk microSD card and putting it inside a full-size SD card adapter, and inserting it into a Windows 7 SP1 computer.  The computer did not recognize its presence at all. Putting the SanDisk microSD card into a USB 2.0 external card reader and inserting the card reader into a USB port on a Windows 7 SP1 computer.  The computer recognized the card reader and installed drivers for it.  The card reader shows up as 'Removable Disk' in Windows Explorer.  Clicking on 'Removable Disk' in Windows Explorer results in the error "Insert disk: Please insert a disk into the Removable Disk". After trying #3, tried opening the 'Removable Disk' in FreeCommander.  Doing so results in the error "The device is not ready".   She does not have access to a linux box, but does have access to her Android phone.  What can she do to recover the data on her 32GB SanDisk microSD card? Is it a weakness to use a single shared secret for protecting messages using a Encrypt-then-MAC scheme?    Assuming a system is using AES-256-CBC and a SHA1-HMAC and the same secret key for both operations.  Upon intercepting one of these messages and attacker could brute force this secret key by trying to solve the HMAC or by successfully decrypting the first block of the ciphertext.  Performing  a decryption on a single block of AES-256 should be a lighter operation than performing a SHA1-HMAC on a potentially large message.  Therefore AES-256 is ideal for a massively parallel brute-force attack and the use of the SHA1-HMAC doesn't hurt the system.  Is this approach sound, or is there an attack that has not been taken into consideration? Story featuring smart people sent on a space mission to nowhere The story in question featured, as the vague title suggests, a "space mission" that is really a cover for forcing a bunch of really smart people to be in close proximity to each other for long enough that they can truly evolve their collective intelligence. I recall them developing new maths and, e..g, "farseeing" and eventually figuring out that while there was no planet at the end of their journey as they were told, they were able to construct one for themselves. Ubuntu on 32-bit UEFI (only) based tablet pc I've created a bootable USB of Ubuntu 16.04 64-bit with Rufus, setting the drive to be a GPT for UEFI. Then I booted my tablet PC and I ran the UEFI BIOS, from which I turned secure boot off. I then restarted and went again on the BIOS to boot from my key, but all I got was a black screen that lasted few seconds and then it brought me back to the BIOS. I read about a bootia32.efi file and other procedures that can be followed, but I didn't find a good explanation of what I need to get done. My tablet PC runs an Intel Core Atom Z3735F and has 2 GB of RAM. If you have the answer please let me know. Why with two classes, LDA gives only one dimension? I am working with dimensionality reduction algorithms. Linear Discriminant Analysis (LDA) is a supervised algorithm that takes into account the class label (which is not the case of PCA for example). I am using Python to do a comparative study between some algorithms.  Why with two classes (k = 2), regardless of the data dimensionality, LDA gives one dimension, i.e. the new subspace is composed of only one dimension? For example if I try to project my dataset (of 100 attributes and 2 classes) to 10-dimensional space by using LDA(n_components=10) I only obtain 1 dimension?  For k&gt;2 I can obtain the number of discriminant vectors that I specified as n_components. How can I get n_components dimensions for k=2? When are "if" and "whether" equivalent? Are if and whether equivalent in sentences like the ones below?     How to determine if my saddle is too high?   How to determine whether my saddle is too high?         We should check if everything is okay now.   We should check whether everything is okay now. In summer, should I close curtains during the day? I'm trying to connect 32 I²C "root of trust"  devices.    I'm thinking about a good way that does that, I mean what kind of communication should I use? A digital multiplexer that selects a device? but based on what? switching on the VDD source? but that needs reinitialization of the device and that takes time. Suppose that $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$. How can I show that if $y_1$ and $y_2$ vanish at the same point then they are not linearly independent?   Here is my attempt to prove the problem. Since $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$, it is enough for me to show that the Wronskian of $y_1$ and $y_2$ denoted by $W(y_1,y_2)$ is zero. Now since $y_1$ and $y_2$ vanish at the same point say $p$ we have $y_1(p)=0$ and also $y_2(p)=0$.   Solving for $W(y_1,y_2)(p)$ we have: $y_1(p)y_2'(p)-y_2(p)y_1'(p)=0$ since $y_1(p)=0$ and also $y_2(p)=0$.   Am I correct? Thanks My Tier 5 YMS visa is set to start on September 30th, 2016 however, I want to get there earlier.  Can I enter the UK as a tourist, if I explain to border control/immigration that I am entering as a tourist and do not plan to work or do anything really until my Tier 5 visa starts?  I can find no official contact or information source on this with UK Visas Information, the British High Commission, etc. How do you make a torus 'telescope' open? Remove application from Google Launcher In "The Matrix Revolutions" How Does Neo Stop the Machine Weapons in the Real World? Can two different planes share the same flight number?