Strange behavior of symbol::messageName in debugger
Strange bug with "MessageName[a,b] is not of the form symbol::name ", Debugger & cursor location
Adobe Reader and xcolor within a table breaks the line render I'm using the xcolor package to colour the top row in a table. This is displayed properly using PDF viewers such as Preview or Skim (Mac). But using Adobe Reader 10.1.3 (either Windows or Mac) doesn't display the table line properly.   But removing the colouring for the table fixes it. Here is the example that works as expected:  \begin{table} \caption{Table showing Feature 5's Functional Tests} \hspace*{-0.5in} \begin{tabular}[h!]{|p{0.6cm}|p{4cm}|p{5cm}|p{4cm}|p{1.5cm}|} \hline {\bf Test ID } &amp; {\bf Task } &amp; {\bf Expected} &amp; {\bf Actual} &amp; {\bf Outcome} \\ \hline    5.1 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.2 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.3 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.4 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline \end{tabular}  \end{table}   And here is the example with a coloured row:  \begin{table} \caption{Table showing Feature 5's Functional Tests} \hspace*{-0.5in} \begin{tabular}[h!]{|p{0.6cm}|p{4cm}|p{5cm}|p{4cm}|p{1.5cm}|} \hline \rowcolor{gray!25} {\bf Test ID } &amp; {\bf Task } &amp; {\bf Expected} &amp; {\bf Actual} &amp; {\bf Outcome} \\ \hline    5.1 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.2 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.3 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline    5.4 &amp; Something &amp;  Something &amp; Something &amp;  Success \\ \hline \end{tabular}  \end{table}   Removing the \rowcolor{} command replacing it with \cellcolor{} creates the same view. Any idea how to fix this?  I've tried printing it out of Adobe and the lines print fine, but as this will be read on as a soft copy it's pretty important. .  Any idea how to fix this? Why not to extend the set of natural numbers to make it closed under division by zero? We add negative numbers and zero to natural sequence to make it closed under subtraction, the same thing happens with division (rational numbers) and root of -1 (complex numbers).  Why this trick isn't performed with division by zero? Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$ I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.  The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).  For example one easy case, it's when $a&gt;0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3}  =  - \root 3 \of 3  $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a&lt;0$ , for example  $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?  $\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple. What are the primary differences between the rules of D&D 3.5 and Pathfinder? What are some of the rule changes that Pathfinder brings in comparison to D&amp;D 3.5?  I am curious what changes were made so I can explain to my group what it has to offer. If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:     If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is   also an integer, does it implies that $k=1$?   This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$. Why do we need both using namespace and include directives in C++ programs?  For example,   #include &lt;iostream&gt;  using namespace std;  int main() {  cout &lt;&lt; "Hello world"; }   Why is it not enough to just have #include &lt;iostream&gt; or just have using namespace std and get rid of the other?  (I am thinking of an analogy with Java, where import java.net.* will import everything from java.net, you don't need to do anything else.) How to compute the  formula $\sum  \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$ how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$   Thank you Why doesn't xpath work when processing an XHTML document with lxml (in python)? How do you switch from Wayland back to Xorg in Ubuntu 17.10? Some of my applications don't work on Ubuntu 17.10 Wayland. How can I switch back to Xorg? I am working on a web application where I want the content to fill the height of the entire screen.  The page has a header, which contains a logo, and account information. This could be an arbitrary height. I want the content div to fill the rest of the page to the bottom.  I have a header div and a content div. At the moment I am using a table for the layout like so:  CSS and HTML    #page {     height: 100%; width: 100% }  #tdcontent {     height: 100%; }  #content {     overflow: auto; /* or overflow: hidden; */ } &lt;table id="page"&gt;     &lt;tr&gt;         &lt;td id="tdheader"&gt;             &lt;div id="header"&gt;...&lt;/div&gt;         &lt;/td&gt;     &lt;/tr&gt;     &lt;tr&gt;         &lt;td id="tdcontent"&gt;             &lt;div id="content"&gt;...&lt;/div&gt;         &lt;/td&gt;     &lt;/tr&gt; &lt;/table&gt;     The entire height of the page is filled, and no scrolling is required.  For anything inside the content div, setting top: 0; will put it right underneath the header. Sometimes the content will be a real table, with its height set to 100%. Putting header inside content will not allow this to work.  Is there a way to achieve the same effect without using the table?  Update:  Elements inside the content div will have heights set to percentages as well. So something at 100% inside the div will fill it to the bottom. As will two elements at 50%.  Update 2:  For instance, if the header takes up 20% of the screen's height, a table specified at 50% inside #content would take up 40% of the screen space. So far, wrapping the entire thing in a table is the only thing that works. Fixing a code to highlight formulas and text on several lines In  I've found the following amazing code to highlight text on several lines.  The problem is that code dosen't work well with formulas : for example, fraction using the mode display are partially out of the highlighting frame.   Here is the code with one content showing the problem.  \documentclass{article}     \usepackage{soul}     \usepackage{tikz}     \usetikzlibrary{calc}     \usepackage{amsmath}     \usepackage{xcolor}      \makeatletter         \newcommand{\defhighlighter}[3][]{%             \tikzset{every highlighter/.style={color=#2, fill opacity=#3, #1}}%         }          \defhighlighter{yellow}{.5}          \newcommand{\highlight@DoHighlight}{             \fill[                 outer sep = -15pt, inner sep = 0pt,                  every highlighter, this highlighter              ](                 $(begin highlight)+(0,8pt)$) rectangle ($(end highlight)+(0,-3pt)$             );         }          \newcommand{\highlight@BeginHighlight}{             \coordinate (begin highlight) at (0,0) ;         }          \newcommand{\highlight@EndHighlight}{             \coordinate (end highlight) at (0,0) ;         }          \newdimen\highlight@previous         \newdimen\highlight@current          \DeclareRobustCommand*\highlight[1][]{%             \tikzset{this highlighter/.style={#1}}%             \SOUL@setup             %             \def\SOUL@preamble{%                 \begin{tikzpicture}[overlay, remember picture]                     \highlight@BeginHighlight                     \highlight@EndHighlight                 \end{tikzpicture}%             }%             %             \def\SOUL@postamble{%                 \begin{tikzpicture}[overlay, remember picture]                     \highlight@EndHighlight                     \highlight@DoHighlight                 \end{tikzpicture}%             }%             %             \def\SOUL@everyhyphen{%                 \discretionary{%                     \SOUL@setkern\SOUL@hyphkern                     \SOUL@sethyphenchar                     \tikz[overlay, remember picture] \highlight@EndHighlight ;%                 }{}{%                     \SOUL@setkern\SOUL@charkern                 }%             }%             %             \def\SOUL@everyexhyphen##1{%                 \SOUL@setkern\SOUL@hyphkern                 \hbox{##1}%                 \discretionary{%                     \tikz[overlay, remember picture] \highlight@EndHighlight ;%                 }{}{%                     \SOUL@setkern\SOUL@charkern                 }%             }%             %             \def\SOUL@everysyllable{%                 \begin{tikzpicture}[overlay, remember picture]                     \path let \p0 = (begin highlight), \p1 = (0,0) in \pgfextra                     \global\highlight@previous=\y0                     \global\highlight@current =\y1                     \endpgfextra (0,0) ;                     \ifdim\highlight@current &lt; \highlight@previous                         \highlight@DoHighlight                         \highlight@BeginHighlight                     \fi                 \end{tikzpicture}%                 \the\SOUL@syllable                 \tikz[overlay, remember picture] \highlight@EndHighlight ;%             }%             \SOUL@         }     \makeatother   \begin{document}  Lorem ipsum \highlight{dolor sit amet, consectetur adipis-icing elit, sed do eiusmod tempor} incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation \highlight[red]{ullamco $laboris$ nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit} in voluptate velit esse cillum dolore eu fugiat nulla pariatur.  Excepteur sint occaecat \highlight[green, draw=blue]{cupidatat non proident  $\dfrac{789}{7889}$ % The frame is not good for that formula ! suntinculpaquiofficiadeseruntmollitanimidestlaborum. Loremipsumdolorsitametconsecteturadipisicingelitseddoeiusmodtemporincididuntutlabore-etdoloremagnaaliqua.} I suppose I could write some more text here.  \end{document}     Is someone could help me to fix it? What's the nice 'trick' to showing that the following expression is irrational?  $2^{1/3} + 2^{2/3}$ What causes the 'Connection Refused' message? This is a  about Connection Refused    We see a lot of questions to the effect      When I try to connect to a system I get a message      Connection refused      Why is this ? After upgrading from 10.04 to 12.04 I am trying to install different packages.  For instance ia32-libs and skype (4.0).  When trying to install these, I am getting the 'Unable to correct problems, you have held broken packages' error message.   Output of commands:  sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded.   After running this:  sudo dpkg --configure -a foo@foo:~$ sudo apt-get install -f Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. (Sorry to re-post this, I have lost access to my original account.)  I've recently installed Kubuntu 18.04.3 on a new AMD Ryzen 3000-series-based system. The system is installed on a SATA SSD, specifically on an Ext4 partition, with a SATA HDD formatted in NTFS mounted for storage. The system has been in use for a few days, with a maximum uptime of two days (so far).  My issue is this: the HDD indicator light on the case flashes every 1-3 seconds, even when the system is idle, and there is no obvious reason to access the disk. The flashes are not regular: sometimes the light is dark for as long as 3 seconds (obviously it is difficult to time such short intervals -- I am approximating), sometimes there are a few quick pulses less than a second apart. The flashes seem to vary in duration as well -- sometimes very brief, sometimes nearly a second. Unmounting the secondary HDD does not appear to have any effect.  Some searching lead me to try iotop. I ran sudo iotop -Pao, and waited until 10 flashes of the HDD light had occurred. This is the result:   6887 be/4 root          0.00 B      0.00 B  0.00 %  0.02 % [kworker/u64:1-events_power_efficient]   433 be/3 root          0.00 B      0.00 B  0.00 %  0.02 % [jbd2/sda2-8]  6302 be/4 root          0.00 B      0.00 B  0.00 %  0.01 % [kworker/u64:0-events_unbound]   The "kworker" processes seem to switch between various states, such as "events_power_efficient", "events_unbound", and "events_freezable_power_". The jdb2 process does sometimes write to disk, though that was not observed here (even though, as I said, 10 flashes of the HDD light occurred). The computer was otherwise idle during this test.  My searches also lead me to check /var/log. The result of ls -lth | head is as follows:  -rw-r-----  1 syslog adm             951K Aug 27 19:49 kern.log -rw-r-----  1 syslog adm             594K Aug 27 19:49 syslog -rw-r-----  1 syslog adm             528K Aug 27 19:49 ufw.log -rw-r-----  1 syslog adm              53K Aug 27 19:48 auth.log -rw-r--r--  1 root   root             40K Aug 27 19:15 Xorg.0.log -rw-rw-r--  1 root   utmp             79K Aug 27 18:29 wtmp -rw-r--r--  1 root   root              55 Aug 27 16:28 prime-offload.log -rw-r--r--  1 root   root              30 Aug 27 16:28 prime-supported.log -rw-r--r--  1 root   root            1.2K Aug 27 16:28 gpu-manager.log   I have no experience interpreting log files, so I don't know if anything here is out of the ordinary.  If the kernel version is relevant, the output of uname -r is 5.0.0-25-generic.  My questions are, 1. Why is my HDD light flashing, even though there is no obvious disk activity, 2. Does this indicate some kind of problem with my system that needs to be fixed, and 3. If so, how do I fix it? The light itself does not bother me -- if this is just a sign of a normal, healthy Ubuntu install, then no problem. My concern is that something is wrong, and if so, I would like to fix it.  Thanks in advance for your time.  In response to heynnema (in the original posting of this question), the issue occurs even at the log-in screen, before I log in. So, it is not tied to my user account. However, it does not occur in the BIOS, suggesting that it is specific to Kubuntu (or at least, to the SSD Kubuntu is installed on). So I installed vlc daily built snap package. It's working great, no problems there. However I noticed that after upgrade the system has 2 versions of vlc (7 and 9). That can be seen when browsing /snap/vlc/ directory and using system monitor. This is also the case with other snaps. Is there a way to automatically delete older versions of snaps? Or do I just delete them using plain &quot;rm&quot; command? Is it safe to delete them that way? Help us test the alpha version of our Android app We are in the early stages of development of the official Stack Exchange Android app and we need your help with testing. To set a few expectations: there will be bugs, crashes, and missing features, all of which will be fixed in .    We are initially targeting Android devices running  (Android 4.1 or higher) starting with phones (up to ~5 inch) and eventually moving our way to tablets.    To register for alpha testing .  After you register we will send you an invite to a private Google community where you can find a link to download our app directly from the  store. The invitations will be sent out in waves, so if you don’t get an invite immediately, be patient.  Crash reports will be automatically reported back to us using . Bugs or feature requests should be reported here on meta and tagged . Please do not use the Google community to report bugs.  As always, we welcome your feedback and are committed to building the best mobile app possible for browsing the Stack Exchange sites. Question of remainder on dividing by 7 Question :  What is the remainder when   $$ 10^{10} + 10^{10^2} +10^{10^3} + \ldots + 10^{10^{100}} $$  is divided by $7$? I have a query pattern that must be very common, but I don't know how to write an efficient query for it.  I want to look up the rows of a table that correspond to "the most recent date not after" the rows of another table.  I have a table, inventory say, which represents the inventory I hold on a certain day.  date       | good | quantity ------------------------------ 2013-08-09 | egg  | 5 2013-08-09 | pear | 7 2013-08-02 | egg  | 1 2013-08-02 | pear | 2   and a table, "price" say, which holds the price of a good on a given day  date       | good | price -------------------------- 2013-08-07 | egg  | 120 2013-08-06 | pear | 200 2013-08-01 | egg  | 110 2013-07-30 | pear | 220   How can I efficiently get the "most recent" price for each row of the inventory table, i.e.  date       | pricing date | good | quantity | price ---------------------------------------------------- 2013-08-09 | 2013-08-07   | egg  | 5        | 120 2013-08-09 | 2013-08-06   | pear | 7        | 200 2013-08-02 | 2013-08-01   | egg  | 1        | 110 2013-08-02 | 2013-07-30   | pear | 2        | 220   I know one way of doing this:  select inventory.date, max(price.date) as pricing_date, good from inventory, price where inventory.date &gt;= price.date and inventory.good = price.good group by inventory.date, good   and then join this query again to inventory. For large tables even doing the first query (without joining again to inventory) is very slow.  However, the same problem is quickly solved if I simply use my programming language to issue one max(price.date) ... where price.date &lt;= date_of_interest ... order by price.date desc limit 1 query for each date_of_interest from the inventory table, so I know there is no computational impediment.  I would, however, prefer to solve the whole problem with a single SQL query, because it would allow me to do further SQL processing on the result of the query.  Is there a standard way to do this efficiently? It feels like it must come up often and that there should be a way to write a fast query for it.  I'm using Postgres, but an SQL-generic answer would be appreciated. Find the remainder of $\dfrac{2^{2014}}{7}$   I am new to the modular arithmetic, Any suggestions to solve this question?