Hatched a Grimer which caused the game to crash during the hatch-animation. Clicking on Grimer afterwards, results in a freeze at the stats-screen of Grimer. It also results in freezes for every pokemon I caught after Grimer.  Crashed happened initially on the non-updated version. Tried a re-install (and now am updated to the latest version, i.e. 0.31.0) and still not working.  The phone I was playing on is a Sony Xperia Compact Z1.   Any ideas how to get rid of that bug ?  Bug summary: clicking on Grimer (either in the Pokédex or in the list with all your Pokémons) results in a freeze, so does clicking on the Pokémon left or right (before and behind), likely because the app is already trying to load the image of Grimer.  Update 1: I can click on any pokemon except on Grimer now without a crash but none of the images of any pokemon is loading. The images are only there when the pokemon stats are automatically showing up directly after you caught them. (see Update 3 for an explanation).  Update 2: Android Version 4.4.4  Update 3: After a few crashes and restarts, the pokemon images are back.  Update 4: After a little bit of research, I found that for  (did not work for me and for most of the other users.). Also, I , just to let them know (they might read it, maybe...).  Update 5: Bug is still there in current version (0.33.0)  Update 6: Version 0.35.0, bug is still there  Update 7 (as requested from @Vemonus): in my opinion, it is not a duplicate of  since that question was specifically addressing the arenas (when somebody else put a Grimer there) whereas here we are talking about the situation when you own a Grimer (although the origin of the bug is, of course, the same: Grimer's heavy animation).
Today i was at a arena. A player from the red team put a Grimer there, which causes my game to crash every time I select the arena (Samsung Galaxy S4 mini). All other pokemons i saw so far did not make any problems  I searched a bit, and found out that this is a known bug, e.G.:   Now my question, is there a way to battle in the arena or even select it without crashing my app? Some option like "hide model in arena",changing some smartphone-settings or another trick? Or do i have to wait till a bugifx from niantec comes or somebody else which has a handy which dont crashes with grimer clears the arena, before selecting the arena?  Because somebody could place him there i guess this is only a bug for specific smartphones, but this could be used to exclude a lot of players from arenafighting by only placing grimers in arenas to let the game crash for them, are there some official notes about this from niantec or how to avoid the crash?
If $f^2$ and $f^3$ are smooth, does it follow that $f$ is smooth? Let $f: \mathbb{R} \to \mathbb{R}$ be given. Assume that the square and cube of $f$ are smooth. Is $f$ smooth? That is if $f \cdot f \in C^{\infty}$ and $f \cdot f \cdot f \in C^{\infty}$, does it follow that: $f \in C^{\infty}$   I got this from another question on SE.   So, $f(x)^2$ is infinitely differentiable, and so is: $f(x)^3$   Also, realize:  $$f(x) = \frac{f(x)^3}{f(x)^2}$$  But what can I do? Is there a utility to replace one block type with another within a region? I'd like to be able to select a region (either with a graphical utility or command line) and be able to switch one block type with another, e.g. change all cobblestone in the region to stone. Does a program/mod exist that can do this? A gerund or use a regular noun? Which of these is correct or more natural?        One of my greatest strengths is critically analyzing.      One of my greatest strengths is critical analyzing.     One of my greatest strengths is critical analysis. I am looking for an option to require a passcode to power off my iPhone. That way, if the phone is lost, the person who finds it cannot simply turn it off, rendering Find My iPhone useless. Is there a way to accomplish this? Is it possible to make armour that can survive /kill? /kill is said to be infinity damage but I have a question can you really survive it in Java edition. However, it is possible that you can pretty much survive this if you are on Minecraft Java Edition in 1.8 or lower and you can just have resistance 5 and you will survive this but can you make armour that helps you survive in 1.16 and in java-edition. I have the following equation, where there is a lot of stuff under the sum symbol:  \begin{equation} d(\vec{x},\vec{y})=   \sum_{Z_{xy}\in\vec{Z}_{xy},\forall x\in\vec{x},\forall y\in\vec{y}}      f(Z_{xy}) \end{equation}   In the resulting document, I find it kind of hard to read. Is there a way to write the equation to make the result more readable, e.g. putting the stuff under the sum symbol on different lines? Age replacement policy for hard disks and SSDs at servers I'm planing an age replacement policy for our storage and servers. Most of them are for DBs and some for images (static content) so yes, they have an huge I/O everytime.  Also, we use Samsung 840 Pro SSDs for the RAID Controllers (PERC H700i) as CacheCade.  Are you guys managing the replacement of old hard drives and solid state drives? Manipulating Partial Derivatives of Inverse Function In lectures we're told:  $$\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}$$  as long as the same variables are being held constant in each partial derivative.  The course is 'applied maths', i.e non-rigorous, so don't confuse me.  But anyway, if we have:  $$\xi = x - y \qquad \eta = x$$  Then $\dfrac {\partial x} {\partial \xi} = 0$ and $\dfrac {\partial \xi} {\partial x} = 1$.  The rule presumably fails because one of the partial derivatives is $0$.  But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact $0$, but then I've worked out the partial derivative manually anyway. Do virtual particles actually physically exist? I have heard  pop in and out of existence all the time, most notable being the pairs that pop out beside black holes and while one gets pulled away. But wouldn't this actually violate the conservation of energy? What is the recommended way to empty a SSD? I've just received my new SSD since the old one died. This Intel 320 SSD supports TRIM. For testing purposes, my dealer put Windows on it, but I want to get rid of this and install Kubuntu on it.  It does not have to be a "secure wipe", I just need the empty the disk in the most healthy way. I believe that dd if=/dev/zero of=/dev/sda just fills the blocks with zeroes and thereby taking another write (correct me if I'm wrong).  I've seen the answer , but it looks like it's suited for clearing empty blocks, not wiping the disk.  hdparm seems to be the program to do it, but I'm not sure if it clears the disk OR cleans empty blocks. From its manual page:     --trim-sector-ranges           For  Solid State Drives (SSDs).  EXCEPTIONALLY DANGEROUS. DO NOT           USE THIS OPTION!!  Tells the drive firmware to discard  unneeded           data  sectors,  destroying  any  data that may have been present           within them.  This makes those sectors available  for  immediate           use  by  the firmware's garbage collection mechanism, to improve           scheduling for wear-leveling of the flash  media.   This  option           expects  one  or  more  sector range pairs immediately after the           option: an LBA starting address, a colon, and  a  sector  count,           with no intervening spaces.  EXCEPTIONALLY DANGEROUS. DO NOT USE           THIS OPTION!!              E.g.  hdparm --trim-sector-ranges 1000:4 7894:16 /dev/sdz   How can I make all blocks appear as empty using TRIM? Which is correct: "has died" or "died"? Migrated question's answers can be upvoted twice by the same user How can I get a views exposed filter to persist across multiple views? I have 2 separate views that return a set of results for each of my content types. The fields are different, but the filters are exactly the same - Sort by Location / Distance. This Distance / Proximity filter is exposed, and it is set to 'remember' the last entry. I followed the tutorial here  to be able to filter results by distance. It works great.  In views, to remember the last entry, Views uses sessions. But, each view has its own individual $_SESSION variable. That means what I set for one view does not carry over to the next, and even each view can have different values, which is not what I want. Here is a dump of the session after they have been set:  [views] =&gt; Array     (         [search_pages] =&gt; Array             (                 [default] =&gt; Array                     (                     )              )          [baby_clothes_pages] =&gt; Array             (                 [default] =&gt; Array                     (                         [distance] =&gt; Array                             (                                 [postal_code] =&gt; 11211                                 [search_distance] =&gt; 100                                 [search_units] =&gt; mile                             )                      )              )          [baby_gear_pages] =&gt; Array             (                 [default] =&gt; Array                     (                         [distance] =&gt; Array                             (                                 [postal_code] =&gt; 22343                                 [search_distance] =&gt; 100                                 [search_units] =&gt; mile                             )                      )              )      )   What I would like to do is create a small module or snippet of code that will simply allow someone to type in their 5 digit zip code and then it update all the values across all my pages with this filter. If it is changed on any page, then it changes it for the rest of the pages as well. It also needs to work so that it affects the results on the next page load so that it correctly shows them filtered.  Here is a snippet of code I was experimenting with to try to set this value, but it's the wrong direction.   $all_views = views_get_all_views();            // get an array of all views in use on the site  foreach ($all_views['category_pages']-&gt;display as $key =&gt; $value) {  //check to see if the structure exists I am looking to modify if(array_key_exists('postal_code', $all_views['category_pages']-&gt;display[$key]-&gt;display_options['filters']['distance']['value'])) {  //if so, set its value $all_views['category_pages']-&gt;display[$key]-&gt;display_options['filters']['distance']['value']['postal_code'] = '28152';  //try setting the session instead $_SESSION['views']['category_pages'][$key]['distance']['postal_code'] = '28152';     }   What's the correct direction? How do I get this value changed? Could it be done in the template.php file with preprocess function? Should I do it in a module, and if so, what do I need to alter?  Very lost at this point. Thanks for the direction! Creating all possible line segments between all points using QGIS After cruising through a lot of material online, and answers over here, my understanding of contravariant and covariant vectors are, in a finite-dimensional vector space, suppose we have a vector, which is contravariant to its basis. However, using the metric tensor, we can map this vector to a one form in the dual space, which acts as a vector space for this example. This one form, or covactor, varies similarly to the basis in our original vector space. However, the covectors have a different basis, which is the dual basis, and are expanded in terms of those. Here's where my confusion arises : Many diagrams over the internet give a geometric picture of the scenario, by claiming that contravariant components of an 'arrow' are found by drawing parallel lines along the axes, and checking where they intersect. Co-variant components could be found by sketching the perpendicular lines on these axes. So, these materials are treating covariant and contravariant as different representations of the same arrow, while I'm inclined to believe they are completely different. However, if we let go of rigour, in exchange for a 'more' geometric understanding, I think we are allowed to do this. The diagrams are usually of the form :  However, even if we assume that contravariant and covariant are different representations of the same object, this particular diagram still seems wrong to me. Particularly the locations of $x_1,x_2$. In this diagram, they are located on the span of the original basis. Shouldn't covector components be located on the span of the dual basis ? This diagram suggests to me, that co-variant components have the same basis, as the contravariant ones. Shouldn't the diagram look more like this ?  This second diagram seems to fit the concept better according to me. This is because the dual vector must be 'contravariant' with respect to the 'dual basis'. This means, their components must be found out by sketching parallel lines along the span of the dual basis. These lines intersect the original axis at a right angle, which is to be expected, as dual bases are orthogonal to the original bases. Moreover, this second picture can also show in a much better way, how scaling up of the original basis, scales down the 'contravariant components' and the 'dual basis', which in turn scales up the 'covariant' components.This shows that the components of the dual vector are covariant to the original basis. This is something that is not readily visible in the first diagram. So, am I correct in assuming this second 'geometrical' representation is correct ? I know this doesn't make much sense, because as the mathematicians said, that vectors are completely different from one forms, and one forms should be represented using the number of hyperplanes intercepted by our arrow. However, I've seen most course material refer to it in this way, and frankly it is easier to visualize. However, most of this material used the first picture. Can anyone point out my mistakes, if any, and tell me if the first picture is correct, or the second one ? Or in this case 'less-flawed'. What is the benefit of Krivine's notation? I saw some people uses Krivine's notation for function application when presenting the syntax for the $\lambda$-calculus.  For example, the $\lambda$-term $\lambda f . \lambda x . \lambda y . f\ x\ y$ (with the normal convention that function application associates to the left, so it actually means $\lambda f . \lambda x . \lambda y . ((f\ x)\ y)$) is written $\lambda f . \lambda x . \lambda y . (f)\ x\ y$ (with a similar convention that it actually means $\lambda f . \lambda x . \lambda y . ((f)\ x)\ y$).  I do not see the point of having another pair of parentheses around the innermost $f$.  Why do people use Krivine's notation instead of the usual one? Is a citizen of Canada required to apply for a US visitor visa after visiting Iran? Under new rules, non-US citizens from Visa Waiver Program countries who have visited Iraq, Syria, Iran, Sudan, Libya, Somalia, or Yemen are required to apply for a proper visa because they are no longer eligible for the VWP.   Does the above apply to citizens of Canada also? Canada is not a Visa Waiver Program country, but instead Canadian citizens can enter the US visa-free. If a Canadian citizen has visited Iran for example, is the entry procedure to the US any different?  I have looked through  but that page does not mention Canada. I am no native speaker and always confused about the comma in introductory phrases, in particular in prepositional phrases. Is there any hard rule when a comma must be set? If I make a google search for certain phrases, I often find both variants. Typical examples where I am not sure whether a comma must be set are:        First[ly][,] I must phone mother   In this case[,] we must...   For simple problems[,] the algorithm...   From Lemma 1.2[,] we obtain...   For a typical user[,] the algorithm...   In our theory[,] we...   A dark line on the horizon[,] the mountain range we were headed for seemed somehow not as threatening as when we had started out on the trip.   Avidly reading her novel[,] Jane did not notice that her stop was approaching.   To be sure[,] it is not easy to see a solution.   When you get to Rome[,] give me a ring. If $\alpha$ is an algebraic element and $L \subset K$ are both field, does the polynomial ring $L[\alpha]$ is also a field?  I am trying to prove that the ring of fraction $L(\alpha)$ is equal to $L[\alpha]$. To do this, I use an exercise done in class : Let $K$ a field, $\alpha \in K$ and $L$ a subfield of $K$. Then $L(\alpha)$ is the smallest subfield of $K$ containing $L$ and $\alpha$.  Is anyone could help me? How do I find peaks in a dataset? If I have a data set that produces a graph such as the following, how would I algorithmically determine the x-values of the peaks shown (in this case three of them):