I need to create layered PDF output from QGIS Map Composer so that one can turn the layers in PDF on and off. And I have no clue how to do it.   Is it possible at all and if so, how could I achieve it?
In ArcMap, when you export data to PDF it preserves the layering within the PDF so that you can turn on/off individual map layers (within the PDF itself).  When I export from QGIS to PDF, I don't appear to have the option for this.  Does QGIS support preserving layers when exporting to PDF?  Is there a special setting needed to enable this capability?  Or has anyone wrote a plugin for this feature?
Short story with aliens who hear their own thoughts as god-like voices I would have read this about 20-25 years ago.  I believe the subject was an alien individual, but it could have been a non-standard human instead.  The hook of the story was that the subject lacked an organ like the  that allows the halves of the brain to communicate efficiently, allowing the illusion of a single consciousness.  The subject is conscious, and has thoughts, but is subject to hearing commands like "Wash your hands!" "Fetch water!" within its head.  I recall the subject (and others like the subject) believing that the voices are their gods talking to them.  To the reader it becomes clear that this is a separate part of the subject's own mind that is responsible for long-term planning and simply can't communicate with the rest of the mind in the way a "normal" human is accustomed to. The average of inverse square radius $r^{-2}$ for any state quantum $|n\ell\rangle$ using the radial Schrödinger equation "method driver /usr/lib/apt/methods/https could not be found" update error sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found. michael@Ubuntu:~$ sudo apt-get clean michael@Ubuntu:~$ cd /var/lib/apt michael@Ubuntu:/var/lib/apt$ sudo mv lists lists.old michael@Ubuntu:/var/lib/apt$ sudo mkdir -p lists/partial michael@Ubuntu:/var/lib/apt$ sudo apt-get clean michael@Ubuntu:/var/lib/apt$ sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found. E: Could not get lock /var/lib/dpkg/lock - open (11: Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/), is another process using it? michael@Ubuntu:/var/lib/apt$ sudo apt-get update E: The method driver /usr/lib/apt/methods/https could not be found.   Why am I getting these errors when doing an apt-get update? Example of non-isomorphic structures which are elementarily equivalent I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).  I am using the notaion of David Marker's book "Model theory: an introduction". What is Elrond's position/status, as compared to Galadriel's? In the Lord of the Rings books, Galadriel of Lorien is even more exalted than in the movies.  She is essentially a queen, and the deference shown to her far exceeds that shown to Elrond of Rivendell.  Yet in the films, we see her trying to persuade Elrond to intervene in the battle of Helm's Deep by sending Elf troops to reinforce Theoden's forces.  Eventually, Elrond relents, and orders the Elves into battle.  But the troops in question are Galadriel's, not Elrond's - they are led by Haldir, one of Galadriel's guards; the trip from Rivendell to Helm's Deep (500 miles as the crow flies) is about 2 times longer than the trip from Lorien to Helm's Deep (250 miles as the crow flies), far too long for the reinforcements to arrive in time.  But then Haldir greets Theoden of Rohan by saying he has come bearing a message from Elrond - not Galadriel.     I bring word from Lord Elrond of Rivendell. An Alliance once existed between Elves and Men. Long ago we fought and died together. We come to honor that allegiance.    This seems odd for a few reasons. The troops are Galadriel's, not Elrond's, and they come from Lorien, not Rivendell. She clearly wanted to send them to Helm's Deep - she was cajoling, prodding, and guilt-tripping him into doing just that.  She is also a queen, and could presumably order them into battle regardless of what Elrond says.    So why does she have to talk Elrond into letting her send her own troops into battle, especially when he is too far away to contribute manpower (elfpower) or materiel to the battle of Helm's Deep?  Why not say, "I'm not going to sit here doing nothing while Saruman runs roughshod over Rohan;  you can support me or not, Elrond, but if you aren't willing to help, you can at least get the hell out of the way"?  Why does her aide Haldir "bring word from Elrond" when the troops - and the impetus for sending them - came from Galadriel, and Elrond resisted sending them out for as long as he could?  And again, what is Elrond's official position/status/rank, and why would a queen need his permission to deploy her own troops? Is it possible to prove $g^{|G|}=e$ in all finite groups without talking about cosets? Find the number of zeroes (counting multiplicity) of $p(z)=z^6+z^3+10z^2+4z+3$ inside the annulus $1&lt;|z|&lt;2$. I think this can be solved using Rouché’s theorem. First consider $|z|\leq 1$ on the boundary $|z|=1$ and $|z^6+z^3|&lt;|10z^2+4z+3|$ which can be shown using triangle inequalities, the left side is at most $2$ while right side is at least $3$. We could also consider $|z|\leq 1+ \epsilon$ and our proof still works. Thus $p(z)$ has 2 roots inside $|z|\leq 1$. Also for $|z|= 2$ we have $|z^6+z^3|&gt;|10z^2+4z+3|$. Thus we have 6 zeroes of $p(z)$ inside $|z|&lt;2$ and therefore 4 of them are in $1&lt;|z|&lt;2$. Is this correct? The only &quot;tricky&quot; part is ruling out that there are zeros on $|z|=1$. How come the year doesn't show here? Are there any disadvantages to using your companions for transporting Power Armor Find all natural numbers $n &gt; 1$ and $m &gt; 1$ such that $1!3!5!\cdots(2n - 1)! = m!$   I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(n, m)$, however the best I have been able to find so far are $m \ge 2n - 1$ and $m \lt n^2$, which is clearly not enough.  Update. The question has tag combinatorics as it is from a book about combinatorics, so there must be at least partly combinatorial solution. Display data from geoserver on click map with openlayers I tried to display data with popup from Geoserver or Postgres to my map, but it dosen't work at all. Can anyone please help me? I used this code:      function Info(e)          {                 if (e.features &amp;&amp; e.features.length)                  {                      //console.log(e.features[0].attributes);                     var popupId = e.xy.x + "," + e.xy.y;                     var popup = popups[popupId];                     if (!popup || !popup.map) {                     popup = new OpenLayers.Popup.FramedCloud(                     popupId,                      map.getLonLatFromPixel(e.xy),                     null,                     " ",                     null,                     true,                     function(e) {                         delete popups[this.id];                         this.hide();                         OpenLayers.Event.stop(e);                     }                 );                 popups[popupId] = popup;                 map.addPopup(popup, true);                 }                popup.setContentHTML(popup.contentHTML + "nom_user:" +e.features[0].attributes.nomservice +" region_user :"+ e.features[0].attributes.region);             popup.show();                                    }       selectcltr = {         click: new OpenLayers.Control.WMSGetFeatureInfo({             url: 'http://localhost/geoserver/wms',              title: 'Identify features by clicking',             layers: [ser],             queryVisible: true         }),         hover: new OpenLayers.Control.WMSGetFeatureInfo({             url: 'http://localhost/geoserver/wms',              title: 'Identify features by clicking',             layers: [ser],             hover: true,             // defining a custom format options here             formatOptions: {                 typeName: 'projet',                  featureNS: 'http://www.openplans.org/topp'             },             queryVisible: true         })     } In the following snippet:  public class a {     public void otherMethod(){}     public void doStuff(String str, InnerClass b){}     public void method(a){         doStuff("asd",             new InnerClass(){                 public void innerMethod(){                     otherMethod();                 }             }         );     } }   Is there a keyword to refer to the outer class from the inner class?  Basically what I want to do is outer.otherMethod(), or something of the like, but can't seem to find anything. How do I wrap text in a pre tag? Where can I find a complete reference for the $PS1 variable? The default PS1 variable on my machine (Kubuntu 13.10) is this:  \[\e]0;\u@\h: \w\a\]${debian_chroot:+($debian_chroot)}\u@\h:\w\$   I'm looking for a reference for how the $PS1 variable works that, at a minimum, allows me to understand the above PS1 variable. Why on average does each bootstrap sample contain roughly two thirds of observations? How to perform Stream functions on an Iterable? Rails lists have .first and .second – is there a .hundredth or .sixty_nineth ? How can I resize a partition with Disk Utility? I have 2 partitions, I reduced the size of one and want to expand the other to fill the now available space.   How do I expand the Lion Partition into the empty grey space beneath Macintosh HD partition? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers? How can I find $\gcd(a^m+1,a^n+1)$ with $a,m,n$ positive integers?    I have this idea:  Let $d=\gcd(m,n)$. Then there exist positive integers $x,y$ such that $mx-ny=d$ (WLOG). We shall find $G=\gcd(a^m+1,a^n+1)$.  If $m,n$ are odd, then $d$ is odd, therefore one and only one of $x,y$ is even. We have:  $$a^{ny}(a^d+1)=a^{mx}+a^{ny}=(a^{mx}-1)+(a^{ny}+1)=(a^{mx}+1)+(a^{ny}-1).$$  If $x$ is even and $y$ is odd, then $a^{m}+1\mid a^{mx}-1$ and $a^{n}+1\mid a^{ny}+1$, therefore $G\mid a^{ny}(a^d+1)$. If $x$ is odd and $y$ is even, then $a^{m}+1\mid a^{mx}+1$ and $a^{n}+1\mid a^{ny}-1$, thus $G\mid a^{ny}(a^d+1)$. However, since $\gcd(a^m+1,a^{ny})=\gcd(a^n+1,a^{ny})=1$, so $\gcd(G,a^{ny})=1$, hence $G\mid a^d+1$. We also have $a^d+1\mid a^{m}+1$ and $a^d+1\mid a^{n}+1$, so $a^d+1\mid G$. Thus $G=a^d+1$.  If $v_2(m)=v_2(n)=v_2(d)=k&gt;1$, then there exist some odd numbers $m_1,n_1,d_1$ such that $m=2^km_1,n=2^kn_1,d=2^kd_1$. We shall have $m_1x-n_1y=d_1$, so one and only one of $x,y$ is even, and we can use the same argument when $m,n$ are odd, so $G=a^d+1$.  However, if $v_2(m) \neq v_2(n)$, I cannot find any solutions for this. I think that $G \in \{1,2\}$, but I cannot prove or disprove it. How can I find $G=\gcd(a^m+1,a^n+1)$ if $v_2(m) \neq v_2(n)$ ? Moreover, is there anything from my arguments that can be improved ? I'm searching how to remove the chapter+section number on the theorem numbering with a section counter.  With no section counter it's like:  \newtheorem{thm}{Title} Chapter1 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 3 --------thm 4 Chapter2 ----Section 1 --------thm 5 --------thm 6 ----Section 2 --------thm 7 --------thm 8   And with the section counter:  \newtheorem{thm}{Title}[section] Chapter 1 ----Section 1 --------thm 1.1.1 --------thm 1.1.2 ----Section 2 --------thm 1.2.1 --------thm 1.2.2 Chapter 2 ----Section 1 --------thm 2.1.1 --------thm 2.1.2 ----Section 2 --------thm 2.2.1 --------thm 2.2.2   And I want numbering my theorems like this:  \newtheorem???????? Chapter 1 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 1 --------thm 2 Chapter 2 ----Section 1 --------thm 1 --------thm 2 ----Section 2 --------thm 1 --------thm 2   I know it's always a pain to remove such things, thanks for whoever will help me ^^