I'm going to Bulgaria in a week. I'm an Iranian passport holder, with an Italian residence permit. I read that I don't need to apply for a visa because I have a Schengen residence permit. Is that correct?
I have looked it up and the resources all talk about schengen visa, travelling with eu national family member etc. so I would like to get a more assuring answer about my situation:   I am a non-EU national with an Italian residence permit. My father is an Italian citizen. I will be travelling to Sofia and Bucharest next month with my girlfriend who is a British citizen. Do I need to get a visa to enter these two cities?
Idiomatic phrase for small type words at end of a document Is there a phrase in English that that can be used in the following situation:  You read an article and from the title you understand that this is something positive, but if you read it all, very carefully, you will see that "small words" at the end reveal the reality.  The same situation is when signing contracts, when something negative for the customer is indicated in the end with small letters How do I get an animated gif to work in WPF? What control type should I use - Image, MediaElement, etc.? Transactions in .net What are the best practices to do transactions in C# .Net 2.0. What are the classes that should be used? What are the pitfalls to look out for etc. All that commit and rollback stuff. I'm just starting a project where I might need to do some transactions while inserting data into the DB. Any responses or links for even basic stuff about transactions are welcome. Does there exist a holomorphic function with the following property? Possible Duplicate:          In a question, I see that each time I refresh the page, the answers were placed in a different order. The question has 3 answers each with zero votes. The users' reputation was something like 6, 1200, and 7300 but these numbers didn't affect the order.  What is the criteria of the sort order of the answers? What is a "good fit" Brier score and Harrell's C Index Are more megapixels good?  Are more megapixels bad?  Do more megapixels increase detail? Do they make my images sharper? On the other side, is there a point that is too much? Do megapixels cause increased noise and other problems? How does print and viewing size come into it?  Widespread Internet-forum wisdom used to be that 6 megapixels was the sweet spot — below that just wasn't enough, but above that, there wasn't much benefit. File sizes got larger, but detail was lost to noise and other problems. The arguments are that cramming too many pixels into a small sensor makes each pixel too small to provide any real benefit, and that higher-megapixel sensors outresolve cheap lenses anyway. (Substitute something like 12 instead of 6 if we're discussing APS-C DSLRs, or 24 for full-frame.)  By 2011,  was in the range of 12-18 megapixels. In 2016 most were are in the 16-24 megapixel range. Certainly none are at 6 anymore; even smartphones come with 8 megapixel cameras. Does this offer any real improvement over that old "sweet spot"? Has technology improved to the point where the "wisdom" needs to be updated, or are we all suffering for marketing? Or have we gone past the sweet spot in some ways, but it's okay because of previously-un-argued points. (For example: more noise, sure, but more detail as well.)  Within the 16-24 megapixel common-today range, for the same sensor size, is there any actual benefit to the higher end? How do megapixels directly affect image quality with today's technology? What are the benefits and when do they apply? What are the drawbacks, and when do they apply? How should I adjust my technique (and expectations) based on my camera's megapixel count? In how many different ways can we place $8$ identical rooks on a chess board so that no two of them attack each other? In how many different ways can we place $8$ identical rooks on a chess board so that no two of them attack each other?  I tried to draw diagrams onto a $8\times8$ square but I'm only getting $16$ ways.  Does that sound right?  Thanks for the help! A ring without the Invariant Basis Number property I was reviewing my homework and it seems I overlooked something crucial while proving some ring has no Invariant Basis Number property. This is exercise VI.1.12 in Aluffi's Algebra: Chapter 0  The setup: $V$ is a $k$-vector space and let $R = \mathrm{End}_{k}(V)$.   Prove that $\mathrm{End}_{k}(V\oplus V) \cong R^4$ as an $R$-module Prove that $R$ doesn't satisfy the IBN property if $V = k^{\oplus \mathbb N}$.   For the first, I used to the fact that $V \oplus V$ is both the product and coproduct (in $k$-Vect) of $V$ with itself to get the isomorphism. What I just realized is I only showed that the two are isomorphic as groups not $R$-modules. So what would be the $R$-module structure on $\mathrm{End}_{k}(V \oplus V)$?   For the second, I used the fact that $V = k^{\oplus \mathbb N}$ implies $V \cong V \oplus V$ which in turn implies $R = \mathrm{End}_{k}(V) \cong \mathrm{End}_{k}(V \oplus V)$. Again, I just realized that I only showed the latter two are isomorphic as groups.   It may be obvious (and maybe why my professor let it pass?) but I can't come up with a good $R$-module structure that makes the two group isomorphisms $R$-linear.  Edit:  Explicitly, these are the isomorphisms I'm dealing with. Let $\pi_j, i_j$ be the natural projection/inclusion maps of the $j$-th factor resp. and $\psi: k^{\oplus \mathbb N} \oplus k^{\oplus \mathbb N} \to k^{\oplus \mathbb N}$ the isomorphism given by $\psi(e_i, 0)=e_{2i-1}$ and $\psi(0, e_i)=e_{2i}$.  Then the first isomorphism $\mathrm{End}_k(V \oplus V)\to R^4$ is given by $\varphi \mapsto (\pi_1\varphi i_1,\pi_2\varphi i_1,\pi_1\varphi i_2,\pi_2\varphi i_2)$  The second isomorphism $R \to \mathrm{End}_k(V \oplus V)$ is given by $\alpha \mapsto \psi^{-1} \alpha \psi$  The composition doesn't seem to be $R$-linear if I use the obvious $R$-module structure on $R$ and $R^4$. No non-trivial clopen sets in $\mathbb{R}$? How to give a direct proof? How to give a direct proof of the following result?     Let $A$ be a subset of $\mathbb{R}$ such that $A$ is both open and closed. Then $A$ is either empty or all of $\mathbb{R}$?   My work:   If $A$ is not empty, let $u \in A$. Then there is some open interval $(a,b)$ such that $$u \in (a,b) \subset A.$$  Now if $\mathbb{R} - A$ is not empty either, then let $v \in \mathbb{R} - A$. Then there is some open interval $(c,d)$ such that  $$v \in (c,d) \subset \mathbb{R} - A.$$  Suppose that $u &lt; v$.   Since $$\emptyset \subset (a,b) \cap (c,d) \subset A \cap (\mathbb{R} - A) = \emptyset,$$ we must have $$(a,b) \cap (c,d) = \emptyset.$$ So we can conlclude that $$b \leq c.$$ But $b \in A$ and $c \in \mathbb{R} - A$. So we must have $$b &lt; c.$$  What next? Can anybody here please help complete the proof from here on?  An edit based on a comment by Marc Paul:   Let us assume that the set $A$ is a non-trivial clopen set in $\mathbb{R}$. Let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ as follows:  $$f(x) \colon= \begin{cases} 1 \ \mbox{ for } \ x \in A; \\ 0 \ \mbox{ for } \ x \in \mathbb{R} - A. \end{cases}$$   Let $V$ be an open set in the range space $\mathbb{R}$. We show that the inverse image set $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   The following cases arise:   If $0, 1 \in V$, then $f^{-1}(V) = \mathbb{R}$.   If $0 \in V$ but $1 \not\in V$, then $f^{-1}(V) = \mathbb{R} - A$.   If $1 \in V$ but $0 \not\in V$, then $f^{-1}(V) = A$.   And, if $0 \not\in V$ and $1 \not\in V$, then $f^{-1}(V) = \emptyset$.   Thus, $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   Hence the function $f$ is continuous.   What next? How does this lead to our desired conclusion?   [Yet another edit, again based on valuable comments from Marc Paul: ]  So if both $A$ and $\mathbb{R} - A$ were non-empty, then let's suppose $a \in A$ and $b \in \mathbb{R} - A$, and we can assume without any loss of generality that $a &lt; b$.   Then $f(a) = 1$ and $f(b) = 0$. So by the intermediate value theorem there is a real number $c \in (a,b)$ such that $f(c) = 1/2$, which is a contradiction because the image set of $f$ does not contain $1/2$. What's the best way to use matching methods with multiple treatment groups? I'm assessing the impact of an intervention on an outcome. For my first analysis, I used the MatchIt package (see code below) to match treatment and control groups and to compare outcomes for the treated vs control. The Treat variable is 0 or 1.  nn_match &lt;- matchit(treat ~ cov1 + cov2+ cov3 ,                      method = "nearest", data = df)   In my second analysis, I'd like to assess the effect of different treatment levels (for example, I have Treatment A, Treatment B, and Treatment C) on the outcome, but I'm stuck on the best way to proceed. Does it make sense to run separate matching analysis for each treatment group while excluding the observations from the other treatments groups?   nn_treatA_match &lt;- matchit(treat_A ~ cov1 + cov2+ cov3 ,                      method = "nearest", data = df_minus_other_treatBC) nn_treatB_match &lt;- matchit(treat_B ~ cov1 + cov2+ cov3 ,                      method = "nearest", data = df_minus_other_treatAC)   Or is there a way to run the analysis with the entire dataset and match controls to each treatment at the same time? Other posts have mentioned the twang package, but I'm not sure if that's what I should go with as I'd like to compare outcomes for each treated group to the control and not just across treated groups. I tried to change network interfaces name on this new Ubuntu 16.04 LTS version but doesn't have the /etc/udev/rules.d/70-persistent-net.rules.  So, I tried to use /lib/udev/write_net_rules but it not exist.  Why do I need this modification? Because I'm using a tool to Simulate virtual platforms that use flex licensing and the authentication need to be in eth0 interface name.  Any suggestion?  The command ip link returns:  user@laptop:~$ ip link  1: lo: &lt;LOOPBACK,UP,LOWER_UP&gt; mtu 65536 qdisc noqueue state UNKNOWN mode DEFAULT group default qlen 1 link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00 2: enp6s0: &lt;NO-CARRIER,BROADCAST,MULTICAST,UP&gt; mtu 1500 qdisc pfifo_fast state DOWN mode DEFAULT group default qlen 1000 link/ether b8:2a:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff 3: wlp7s0: &lt;BROADCAST,MULTICAST,UP,LOWER_UP&gt; mtu 1500 qdisc mq state UP mode DORMANT group default qlen 1000 link/ether 5c:e0:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff  user@laptop:~$ ifconfig  enp6s0    Link encap:Ethernet  HWaddr b8:2a:xx:yy:xx:yy   lo        Link encap:Local Loopback   wlp7s0    Link encap:Ethernet  HWaddr 5c:e0:xx:yy:xx:yy     (Some information were ignored and suppressed)  I already tried some links: Why does my wildcard SSL certificate cause a domain mismatch error on a second level subdomain? 2D random walk within a bounded area EXR Transparency Border with bloom and glare was solved by saving to EXR, however when loading that EXR into the VSE over a background a dark border appears around the transparent text that cuts through the glare and bloom. I'm using Blender 2.92.0. EEVEE with bloom Render layers -&gt; Glare -&gt; Composite (with &quot;UseAlpha&quot; checked) Tikz, Tree without overlaps Storing datetime values in Shapefiles using QGIS? I have a shapefile with date in a string field. I create a new field with date type and use the calculator to make the conversion. But the result only provide the date part, the time part is removed. Even when the preview show the correct conversion.  Here you can see the original string field time and the result date time2. And the use of the conversion function to_datetime() and the preview result at the end    If you go to the layer properties you can change the edit mode to date/time. Then if go to the data and try to edit time2 you get the current date/time. So is possible save date and time in the same field. But the update tool still doesn't work as expected      As J.R said. Even when you can change QGIS edit to allow Date/Time the save to shapefile will discard the time. Winter 18  does not get values from a parent record I am testing the new  &lt;lightning:datatable&gt;, it is really good, and is a good replacement for most of my use cases which today I use a custom list component that I have developed.  I am getting data from my server through apex method, to populate it into the &lt;lightning:datatable&gt; component. I can bind the returned list directly to the data attribute and if the column names are same as the field names, all data is bound correctly.  The problem occurs when I get data from a parent record. For example - I want the Account name of my contact.  This example can show the problem:  Component Markup    &lt;aura:handler name="init" value="{!this}" action="{!c.doInit}"/&gt;  &lt;aura:attribute name="mydata" type="Object"/&gt; &lt;aura:attribute name="mycolumns" type="List"/&gt; &lt;aura:attribute name="sortedBy" type="String"/&gt; &lt;aura:attribute name="sortedDirection" type="String"/&gt;  &lt;lightning:datatable data="{! v.mydata }"                      columns="{! v.mycolumns }"                      keyField="Id"                      onrowselection="{! c.getSelectedName }"                      onsort="{!c.updateColumnSorting}"                      sortedBy="{!v.sortedBy}"                      sortedDirection="{!v.sortedDirection}"                      hideCheckboxColumn="{!false}" /&gt;        Component Controller  ({     doInit: function (component, event, helper) {         component.set('v.mycolumns', [                     {label: 'Id', fieldName: 'Id', type: 'id', sortable:"true"},                     {label: 'First Name', fieldName: 'FirstName', type: 'text', sortable:"true"},                     {label: 'Last Name', fieldName: 'LastName', type: 'text', sortable:"true"},                     {label: 'First Name Hebrew', fieldName: 'First_name_Hebrew__c', type: 'text', sortable:"true"},                     {label: 'Last Name Hebrew', fieldName: 'Last_name_Hebrew__c', type: 'text', sortable:"true"},                     {label: 'Account', fieldName: 'Account.Name', type: 'text', sortable:"true"},                     {label: 'Email', fieldName: 'Email', type: 'email', sortable:"true"},                     {label: 'Phone', fieldName: 'Phone', type: 'phone', sortable:"true"}                 ]);          var action = component.get("c.getRecordsFromDb");         action.setParams({             sObjectName: "Contact",             commaDelimitedFieldNames: "Id,FirstName,LastName,First_name_Hebrew__c,Last_name_Hebrew__c,Account.Name,Email,Phone",             filter: ""         });         action.setCallback(this, function(response){             var state = response.getState();             if (component.isValid() &amp;&amp; state === "SUCCESS") {                 var data = response.getReturnValue();                  component.set("v.mydata", data);             }         });         $A.enqueueAction(action);     } })   APEX Aura Enabled Method  @AuraEnabled public static List&lt;sObject&gt; getRecordsFromDb(String sObjectName,         String commaDelimitedFieldNames,         String filter) {      system.debug('\n\n\n\ngetRecordsFromDb sObjectName: ' + sObjectName + ', commaDelimitedFieldNames: ' + commaDelimitedFieldNames + ', filter: ' + filter + '\n\n\n\n');      // create a dynamic soql to get the value of the field     String sql = 'SELECT ' + commaDelimitedFieldNames + ' FROM ' + sObjectName;      if (!String.isBlank(filter))         sql += ' WHERE ' + filter;      system.debug('\n\n\n\ngetRecordsFromDb sql: ' + sql + '\n\n\n\n');     List&lt;sObject&gt; records = Database.query(sql);     system.debug('\n\n\n\ngetRecordsFromDb: ' + records + '\n\n\n\n');      return records; }   The problem is with the Account.Name field - the &lt;lightning:datatable&gt; component does not display the column data.   I know that the data is returned ok, and I know that in the object that is returned, the Account is contained like an object (probably this is the problem):    Is there a good way to show parent record data with this component? Too many blanks between summation symbol and its associated expression I'm having trouble restarting a 2D game when the user presses the "New Game" button.  Structure of my game:  The game has several 'screens': Opening screen, instructions screen, main game screen, and a game-over screen.  Each screen is an instance of a class: OpeningScreen class, GameScreen class, etc.  There is a class named Main. This class has three roles.   It's the JFrame (the window) of the program. It has the main method - first class to run. It has a method that change's screens.   When the program starts, this class runs. When it first runs:   Creates a new instance of itself (it's a JFrame-type class). Creates an instance of every 'screen-class'. The constructor of each 'screen-class' takes the instance of Main as an argument: (aka: GameOverScreen gameOverScreen = new GameOver(this)). Creates a new JPanel named mainPanel, and adds it to the JFrame. Any screen to be displayed will be contained in this JPanel (The screens are also JPanel-type classes).   How the changeScreen() method works:  The signature of changeScreen() is this: changeScreen(JPanel from, JPanel to).  Whenever a screen needs to be changed, the current screen that is displayed calls this method, like so: main.changeScreen(this, main.InstructionsScreen). It removes the previous screen from mainPanel and adds the new screen to mainPanel. As said, these screens are created when the program launches.  This is the method that changes screens:   public void changeScreen(JPanel from, JPanel to){      mainPanel.remove(from);     mainPanel.add(to);      mainPanel.revalidate();     mainPanel.repaint();  }   My problem:  After a game is finished, the Game class calls the changeScreen method: main.changeScreen(this, main.GameOverScreen) .  This replaces the game screen with a game over screen. Pressing the "New Game" button in the game-over screen, calls main.changeScreen(this, main.GameScreen) .  The problem: Pressing the "New Game" button in the game-over screen, indeed displays once again the instance of the Game class. But this is the old instance, so the game is not restarted. It's already finished.  I tried to do the following changes in changeScreen():  public void changeScreen(JPanel from, JPanel to){      mainPanel.remove(from);      if(to.equals(gameScreen)){         gameScreen = null;         gameScreen = new GameScreen(this);     }     mainPanel.add(gameScreen);      mainPanel.revalidate();     mainPanel.repaint();  }   This should reset the gameScreen object. But this creates a delay of about 10 seconds when pressing the "New Game" button.  What would be a good way of restarting the game?  Thank you