On the last move, after only 2 train pieces remains, is it possible to claim a 3-carriage route, if you have the 3 required color cards but only 2 plastic trains left to place on the board?
A recent quandary came up with some friends in a game of Ticket to Ride.  If, in the last turn, you have the cards to play a route longer than the number of trains that you have left, can you do so?  Or are you limited by your number of trains?
How to test @future methods with setup objects and avoid MIXED_DML_OPERATION My dad has an old Sears KS-2 + 50mm f/1.7, which from a bit of research seems to use a Pentax K-mount.  I've looked around a bit and found a , which says it supports af-confirmation, and costs ~$30 US.  Has anyone tried using an adapter ring like this, and if so, would you recommend doing so?  I am perfectly happy with shooting manual, but I'm curious about the quality of older lenses.  I'm assuming that if I use older primes the optical quality will be fine, and it looks like the build quality of the older lenses is much better that similarly priced new lenses. I have written code like below and all works fine, but while submitting the form I get the error below. I am trying to filter the second dropdown based on the first drop down selection.     An illegal choice has been detected. Please contact the site administrator.   How do I overcome that error?       function dynamic_location_dropdown_form_alter(&amp;$form, $form_state, $form_id) {   if($form_id == 'product_node_form') {    $location_options = array();    if(isset($form['field_destination']['und']['#default_value'][0])) {     $destination = $form['field_destination']['und']['#default_value'][0];   }   else {     $destination = 0;   }    $location_options = dynamic_location_dropdown_locations($destination);    $form['field_destination']['und']['#ajax'] = array(     'event' =&gt; 'change',     'wrapper' =&gt; 'squadron-wrapper',     'callback' =&gt; 'dynamic_location_dropdown_ajax_callback',     'method' =&gt; 'replace',   );   $form['field_product_location']['#validated'] = true;   $form['field_product_location']['und']['#prefix'] = '&lt;div id="squadron-wrapper"&gt;';   $form['field_product_location']['und']['#suffix'] = '&lt;/div&gt;';   $form['field_product_location']['und']['#options'] = $location_options;   } }  function dynamic_location_dropdown_ajax_callback($form, $form_state) {   $country_id = $form['field_destination']['und']['#value'];   $form['field_product_location']['#validated'] = true;   $form['field_product_location']['und']['#options'] =  dynamic_location_dropdown_locations($country_id);    return $form['field_product_location']; }  function dynamic_location_dropdown_locations($destination_id) {   $nodes = array();   $nodes[''] = '- None -';   if($destination_id != '') {     $select = db_query("       SELECT node.title AS node_title, node.nid AS nid, node.created AS node_created       FROM {node} node       LEFT JOIN {field_data_field_location_country} field_data_field_location_country        ON node.nid = field_data_field_location_country.entity_id        AND (field_data_field_location_country.entity_type = 'node'        AND field_data_field_location_country.deleted = '0')       WHERE (( (node.status = '1')        AND (node.type IN  ('location'))        AND (field_data_field_location_country.field_location_country_nid = $destination_id)))       ORDER BY node_title ASC     ");     $nodes[''] = '- None -';     foreach ($select as $node) {       $nodes[$node-&gt;nid] = $node-&gt;node_title;     }   }    return $nodes; } What is serial voting and how does it affect me? I just noticed that I lost a bunch of points from my reputation score, and I used the "reputation" tab on my user profile page to try and track down the cause.  During my investigation, I noticed there was an unusual event of type "reversal". In the normal place of a question title, it says "voting corrected".     What does this mean, and what caused it?  Did I do something wrong? Why did I lose all of that reputation? Is the system punishing me for leaving too many good answers?  Is there any way to earn it back? Where did newtonian size setting go in 2.79a Area 51 login no longer redirects nor logs in Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks. Does removing an application from Software Center do a remove or a purge When I remove an application already installed through the Software Center. Does it remove the app the same way as doing a purge to it (aptitude purge) or does it remove it like a "remove" (aptitude remove). In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on.  The first issue I have is that my book began with the statement that given  $$L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  the problem $L[y] \ = \ f$ can be re-cast in the form $L[y] \ = \ \lambda y$.  Now it could be a typo on their part but I see no justification for the way you can just do that!  More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows:  Given a linear second order ode   $$F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)$$  it is an exact equation if it is derivable from a differential equation of one order lower, i.e.  $$F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').$$  The equation is exact iff  $$a''(x) - b'(x) + c(x) = 0. $$  If $F$ is not exact it can be made exact on multiplication by a suitable integrating factor $\alpha(x)$.  This equation is exact iff  $$(\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 $$  If you expand this out you get the Adjoint operator  $$L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 $$  If you expand $L^*$ you see that we can satisfy $L \ = \ L^*$ if $a'(x) \ = \ b(x)$ &amp; $a''(x) \ = \ b'(x)$ which then turns $L[y]$ into something of the form  $$L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).$$  Thus we seek an integratiing factor $\alpha(x)$ so that we can satisfy this &amp; the condition this will hold is that $\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}$  Then we're dealing with:  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)$$  But again, by what my book said they magically re-cast this problem as  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)$$  Then calling  $$\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0$$  a Sturm-Liouville problem.  My question is, how can I make sense of everything I wrote above? How can I clean it up &amp; interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues &amp; finding full solutions - what's going on? I want to be able to look at $a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)$ &amp; know how &amp; why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness &amp; integrating factors, thanks for reading! What is the white smoke-like fog called? It's not really fog, mist or haze. I used to think that it's mist, but from what I've read here, it turns out that it isn't.  It's the white smoke-like fog that is close to the ground. I've seen it recently above a couple of small ponds (and ground) in a damp forest. It was almost as white as smoke, you can barely see through it, but it is close to the ground, from 20cm to a meter or so.  Since it isn't fog (which is a cloud that you can see, say from a mountaintop, and if it's not mist (which is more like 'rain'), then what is it? 'took several years to be built' or 'took several years to build' – which one is correct? Which one of these sentences is correct?  The palace took several years to build.  or  The palace took several years to be built.  I heard the first one while I was watching Discovery Channel and I think it is not correct because the palace did not build on its own – someone had to build it. So the passive form seems the correct sentence. What home printer would allow for exact alignment of front and back? Note: I'm not a graphic designer, but this seemed like the most relevant place to post. Feel free to send me off somewhere else if I'm not in the right place.  I would like to make some notebooks with double-sided grid paper that I've already designed, but my printer seems like it doesn't allow for absolutely exact alignment. Even if one side prints nice and straight, the other side will be slightly skewed left to right by one to two millimeters, which I read from an earlier post is a pretty reasonable variance for a home printer. So I suspect it's not possible to get both sides completely aligned on this printer – but if I'm wrong, please let me know how.  What kind of printer would allow for absolutely exact alignment? Is there any difference between inkjet and laser in this regard, or a more expensive printer versus a cheaper one? What is the Total Reputation field on Area51? What does the total reputation field on Area51 represent?  It's obviously not the total of the user's reputation on the various sites, as that is the number listed as user reputation. options for [for /f "usebackq" %%f in (`find "" path`) do] sentence in batch I symplified my problem: @echo off set res=  for /f &quot;usebackq&quot; %%f in (`find &quot;a&quot; list.txt`) do set res=%%f echo %res%  pause&gt;nul and that things.txt file in the code is like:  list.txt file in the code is: apple pitch pear melon mango  As you expect, the result of this code is mango  because the line 'mango' is located below the line 'apple' and 'pear' But the result I want is apple  or pear  Specifically, I want to put each of the other values above mango such as apple and pear into the variable res using option keys provided. is there any way to do that? How can I port forward with iptables? I'm writing my final report in LaTeX for my bachelor's degree. I'm aiming at having a document similar to the .doc template used normally.   I want to have my customs headers and footers applied on pages when I'm starting a new chapter.  I'm using documentclass report with fancyhdr and lastpage packages.  My headers are defined like this:  \lhead{\myTitle} \rhead{\today} \cfoot{\thepage\ of \pageref{LastPage}}   My chapters are created using standard \chapter{title} macro.  On pages with new chapters, the only headers/footers I see is the standard page number at the bottom.  What should I do?  EDIT: I've finally gone with Peter Breitfeld's solution. I'Ve gone a step further and created my own chapter command:  \newcommand{\myChapter}[1]{     \chapter{#1}     \thispagestyle{fancy} }   Thanks for the help! Strange floating-point behaviour in a Java program In my program I have one array with 25 double values 0.04 When I try to sum these values in a loop I get following results:  0.0 + 0.04 = 0.04 0.04 + 0.04 = 0.08 0.08 + 0.04 = 0.12 0.12 + 0.04 = 0.16 0.16 + 0.04 = 0.2 0.2 + 0.04 = 0.24000000000000002 0.24000000000000002 + 0.04 = 0.28 0.28 + 0.04 = 0.32 0.32 + 0.04 = 0.36 0.36 + 0.04 = 0.39999999999999997 0.39999999999999997 + 0.04 = 0.43999999999999995 0.43999999999999995 + 0.04 = 0.4799999999999999 0.4799999999999999 + 0.04 = 0.5199999999999999 0.5199999999999999 + 0.04 = 0.5599999999999999 0.5599999999999999 + 0.04 = 0.6 0.6 + 0.04 = 0.64 0.64 + 0.04 = 0.68 0.68 + 0.04 = 0.7200000000000001 0.7200000000000001 + 0.04 = 0.7600000000000001 0.7600000000000001 + 0.04 = 0.8000000000000002 0.8000000000000002 + 0.04 = 0.8400000000000002 0.8400000000000002 + 0.04 = 0.8800000000000002 0.8800000000000002 + 0.04 = 0.9200000000000003 0.9200000000000003 + 0.04 = 0.9600000000000003   Why on earth could that happen?! Is it posible to add iframe directly in lightning components? Is it possible to add iframe directly in a lightning component as below to visible in community sites ?  &lt;aura:component implements="forceCommunity:availableForAllPageTypes" access="global" &gt;     &lt;iframe width="800" height="600" style="border: 1px solid" src="---" /&gt; &lt;/aura:component&gt; I have functionality on a cloud page that depends on the cloud page url. I know how to Requestparameter(), but how do I get the entire url using ampscript or SSJS? I've just installed MariaDb on a fresh Ubuntu Gnome and ran  afterwards where I set a decent admin password, removed the anonymous user etc.  Afterwards I realized some strange behavior regarding the admin password:   If I try to login from my normal user account using the command mysql -u root -p I always get an error: ERROR 1698 (28000): Access denied for user 'root'@'localhost' I'm pretty sure I entered the correct password which I set with  earlier... When I run the command from root using sudo mysql -u root -p, I always get access to the database, no matter which password I actually enter...   Is this normal behavior, am I doing anything wrong or have I somehow screwed up the installation?