I'm using a Dell Inspiron 5530 laptop with 8 GB RAM, integrated 4 GB Intel graphics and a 1 TB HDD. I upgraded Ubuntu from 18.04 to 20.04 yesterday. On starting my laptop, the desktop wallpaper has many blackish-yellow squares all over the screen, as shown here:  Also, some of the icons on desktop disappears and sometimes the laptop freezes. What should I do? Is Ubuntu 20.04 still unstable or is it some issue of my laptop. Can someone help?
I have recently upgraded Ubuntu 19.10 to latest LTS Ubuntu 20.04. After updating, I get serious artifacts on my screen. Every time I reboot (or log-in), my screen is fuzzy or scrambled (picture attached):     This is temporary. The fuzzy screen goes away when I change the background. But it reappears the moment I reboot or log back in.   I have Intel HD Graphics 520 (Skylake GT2).  The output of lspci -nn |grep -E 'VGA|Display' is:  00:02.0 VGA compatible controller [0300]: Intel Corporation Skylake GT2 [HD Graphics 520] [8086:1916] (rev 07)   Update:   On enabling Wayland in Ubuntu 20.04 and logging in to, these artifacts are however cured.   This is okay for now, but I am not sure about Wayland as I have not yet used it before.   Is there a way to correct the graphic issue with normal Ubuntu startup?
Use HERE file and redirect output of command I have the following code in a batch script:  mpirun -np 6 ./laplace &lt;&lt;END 100 100 100 0.01 100 3 2 1 END | tail -n 1 &gt; output   But it isn't working. What I want it to do is to use the HERE file as input to the mpirun command, and then pipe the output to the tail command. However, I think the HERE file and tail output things are getting confused.  How should I write this so that it does what I want? Let $a,b,c$ be $3$ distinct integers, and let $P$ be a polynomial with integer coefficients.Show that in this case the conditions $$P(a)=b,P(b)=c,P(c)=a$$ cannot be satisfied simultaneously.  Any hint would be appreciated. Vertically align text next to an image? I'm not sure if there's already an existing term for this, so I'm inventing my own. (tl;dr: I call them &quot;chameleon questions&quot; because they change every time you submit or edit an answer.  If you're already intimately familiar with the phenomenon, please skip past the first set of bullet points to where I ask for recommendations.)  I think many of us have experienced this on occasion:  You open up a narrow-looking question, usually dealing with some specific error message.  The question probably contains a snippet of sloppy but not quite awful code; code indicating that the author, typically a new/low-rep user, possibly a , doesn't really know what they're doing.  I'm not trying to be condescending here, they're probably very intelligent and hardworking, but in this specific case, they're clearly in way over their head.  However, as a question, it's not really too bad.  The goal and problem are both clearly stated.  The code is formatted (sort of).  Mort was polite and checked their spelling.  The question looks answerable, although from a rep perspective* it's an obvious dead-end, it's had maybe 3 views in the past 30 minutes.  So you decide, you know what, this person took the time to write up the question properly, they deserve an answer, put on your good citizen hat and throw them a bone.  You slog through the ugly code, find the offending line, direct their attention to it and explain why it's wrong and how it needs to be fixed.  Later, they leave a comment saying that yes, that was indeed the problem and your answer helped them fix it (great!).  Oh, but, now there's another problem (uh-oh...). Something else is broken.  After letting out a long sigh, you read the comment and/or edited question, roll up your sleeves, and write up Take 2, hoping that's the end of it.  And of course it's not the end.  Now maybe the error is gone, but it's not doing what it's supposed to.  If you're anything like me, your patience is now starting to wear thin. It's lousy code and you didn't volunteer to help them rewrite their whole app.  And of course, the question still only has 12 views, and nobody has upvoted either the question or your answer, including the person you're helping, so this is turning out to be a genuinely thankless task in all respects.  Welcome to &quot;Family Tech Support&quot;, Web 2.0 Edition.   * Which is of course just an arbitrary number that none of us care about. Not really, anyway. Not very much, at least.  So, here's where I get to my question.  At this point, do you:  Leave a comment, stating as politely as possible that you've done your best to help them answer the original question they asked, and if they have a new problem then they should start a new question?  (I did this once, and actually ended up with an accept, but I felt kind of guilty afterwards, I don't think anybody answered their 2nd question).  Bite the bullet and edit your answer a 3rd, 4th, 5th time, however many times it takes to resolve their issue, knowing full well that they may not even remember to accept the answer once you're done? (I may work for free on SO, but that doesn't mean my time is worthless.)  Just abandon the question, hoping that it will force Mort to start thinking for himself?  Something else?   I've tried all three of the above and none of them ever quite feel right. I do want to help, but at the same time there are a lot more &quot;lucrative&quot; questions I could be answering. And I don't mind spending a lot of time on an answer either, if the question is interesting or important to me, but debugging is dull work. I'd prefer not to reference specific questions here; I'm pretty sure that this is a shared experience, and my objective isn't to point fingers at &quot;misbehaving&quot; members.  I'm more looking for some practical advice on how to dig myself out of these pits of despair. Anybody have any other ideas? How to navigate to C drive in bash on WSL-Ubuntu? Mutineers force captain to record instructions to spaceship's computer but he leaves out "please" The crew mutinies on a spaceship whose captain controls it with verbal orders. They test whether they can control the ship with recordings of the captain's voice with an order to increase orbital speed (currently orbiting a planet). It works, so they force the captain to record all the possible orders they may need, and execute him. Only then do they realize that the ship obeys only orders couched in polite language, and only the test recording was like this, but the other recordings are not. They will die once the ship's faster orbit somehow causes it to run out of fuel, or something like that. I seem to remember the ship's &quot;computer&quot; following the captain around on robotic legs. I read the story circa 1980, but it might have been published much earlier. I know that I can create/burn bootable CD/DVD or live USB and can boot/install from it. But suppose I am currently running GNU/Linux and I've ISO file of another GNU/Linux that I want to install on my hard disk, then Can I directly boot from ISO from hard disk and try/install that GNU/Linux operating system? How do I use subprocess.Popen to connect multiple processes by pipes? How do I execute the following shell command using the Python  module?  echo "input data" | awk -f script.awk | sort &gt; outfile.txt   The input data will come from a string, so I don't actually need echo.  I've got this far, can anyone explain how I get it to pipe through sort too?  p_awk = subprocess.Popen(["awk","-f","script.awk"],                           stdin=subprocess.PIPE,                           stdout=file("outfile.txt", "w")) p_awk.communicate( "input data" )   UPDATE:  Note that while the accepted answer below doesn't actually answer the question as asked, I believe S.Lott is right and it's better to avoid having to solve that problem in the first place! I just installed the lastest version of . I works fine but in order to use it, I have to enter the following in the terminal: cd ugene-1.11.5/ then execute ./ugene -ui to launch UGENE GUI.   This takes too long. I would like to build a .desktop file for UGENE to be used in Ubuntu 12.04 so that I can simply open it from the unity launcher.  Not sure if this is helpful but the website says: By default ugene script launches the command-line version of UGENE.  Here's what I have (not sure if it's right):  [Desktop Entry] Version=1.0 Name=my Unipro UGENE GenericName=Integrated Bioinformatics Suite Comment=Unipro UGENE is a cross-platform visual environment for DNA and protein sequence analysis. MimeType=text/english;text/plain;text/x-makefile;text/x-c++hdr;text/x-c++src;text/x-chdr;text/x-csrc;text/x-java;text/x-moc;text/x-pascal;text/x-tcl;text/x-tex;application/x-shellscript;text/x-c;text/x-c++; Exec=/home/samuel/ugene-1.11.5 ./ugene -ui -c %F Icon=/usr/share/icons/hicolor/scalable/apps/ugene.svg Type=Application Terminal=true Categories=Utility;Development; StartupWMClass=UGENE   I added the launcher named my Unipro UGENE to my unity bar and tried opening it but it doesn't appear. I must be missing something. I don't know what the StartWMCClass field is suppose to be nor do I know whether I entered the correct MimeType. In which way does quantum mechanics disprove determinism? In  I prove, using generating functions, a statement equivalent to  $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even.  (Clearly the sum is $0$ when $n$ is odd.)  The nice expression on the right-hand side indicates that there should be a pretty combinatorial proof of this statement.  The proof should start by associating objects with even parity and objects with odd parity counted by the left-hand side.  The number of leftover (unassociated) objects should have even parity and should "obviously" be $2^n \binom{n}{n/2}$.  I'm having trouble finding such a proof, though.  So, my question is     Can someone produce a combinatorial proof that, for even $n$, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}?$$   Some thoughts so far:   Combinatorial proofs for $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$ are given by  and by .  The proofs are basically equivalent.  In Phira's argument, both sides count the number of paths of length $2n$ starting from $(0,0)$ using steps of $(1,1)$ and $(1,-1)$.  By conditioning on the largest value of $2k$ for which a particular path returns to the horizontal axis at $(2k,0)$ and using the facts that there are $\binom{2k}{k}$ paths from $(0,0)$ to $(2k,0)$ and $\binom{2n-2k}{n-k}$ paths of length $2n-2k$ that start at the horizontal axis but never return to the axis we obtain the left-hand side.  With these interpretations of the central binomial coefficients $2^n \binom{n}{n/2}$ could count (1) paths that do not return to the horizontal axis by the path's halfway point of $(n,0)$, or (2) paths that touch the point $(n,0)$.  But I haven't been able to construct the association that makes these the leftover paths (nor do all of these paths have even parity anyway).  So perhaps there's some other interpretation of $2^n \binom{n}{n/2}$ as the number of leftover paths.    Update. Some more thoughts:  There's another way to view the identity $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$.  Both sides count the number of lattice paths of length $n$ when north, south, east, and west steps are allowed.  The right side is obvious. The left side has a similar interpretation as before: $\binom{2k}{k}$ counts the number of NSEW lattice paths of length $k$ that end on the line $y=0$, and $\binom{2n-2k}{n-k}$ counts the number of NSEW lattice paths of length $n-k$ that never return to the line $y =0$.  So far, this isn't much different as before.  However, $2^n \binom{n}{n/2}$ has an intriguing interpretation: It counts the number of NSEW lattice paths that end on the diagonal $y = x$ (or, equivalently, $y = -x$).  So maybe there's an involution that leaves these as the leftover paths.  (Proofs of all of these claims can be found on , for those who are interested.) I need some advice as I am struggling with the following combinatorics exercise.  Let $k$ be a given positive integer. Show that any non-negative integer $N$ can be written uniquely in the form $$N=\binom{x_k}{k}+\binom{x_{k-1}}{k-1}+\dots+\binom{x_1}{1},$$ where $0 \leq x_1 &lt; \dots &lt; x_{k-1} &lt; x_k$.  The exercise is accompanied by this advice:  Let $x$ be such that $\binom{x}{k} \leq N &lt; \binom{x+1}{k}$. Then any possible representation has $x_k = x$. Now use induction and the fact that $N-\binom{x}{k} &lt; \binom{x}{k-1}$ to show the existence and uniqueness of the representation.  I am really stuck and thus grateful for the slightest hint on how to solve it. Why is the change in Hamiltonian for an active infinitesimal canonical transformation defined the way it is? I'm trying to understand infinitesimal canonical transformations and conservation theorems (section 9.6 Goldstein ed3). My specific problem is with understanding eq 9.104,         $\partial H = H(B) - K(A^{'}) $, where $\partial H$ represents the change in the hamiltonian under an infinitesimal canonical transformation (from the active view point ofc). The issue only arises for generating functions that explicitly depend on time.  Here's what I've thought of so far. I understand what $K$ is from the passive viewpoint (or so I think), its the hamiltonian that would furnish Hamilton's equations for the transformed variables $Q=q+ \delta q$ and $P = p + \delta p$. In the specific case of the generating function being a time dependent hamiltonian, $K = H + \delta t \frac{\partial H}{\partial t} = H + \delta t\dot H$, which makes sense because once you transform the variables to what they'll be in time $\delta t$, you'll also need to add on a term that comes from the Hamiltonian being at a later time $t+ \delta t$.   In the book, this is explained as a transformation depending on time changes the hamiltonian, so $H(A) \rightarrow K(A^{'}) $. Which I sorta get. But why not for $H(B)$? And what about the case when the Hamiltonian is time dependent and it is the generating function? Is $H(B)$ at a later time then?   Note: $A$ represents the original variables, $B$ is the transformed variables in the active sense, so in the same phase space as $A$, and $A^{'}$ is the same point as A but in the transformed phase space (passive). equal size numerator and denominator I am trying to size the numerator and denominator of the below equation so that they are equal in size.  I have not had much luck suggestions are appreciated.  \begin{equation} \sum_{i=1}^{n} \frac{\frac{cf_{n}}{(1+i)^n}} {\displaystyle\sum_{i=1}^{n}\frac{cf_n}{(1+i)^n}} , n_{i} \end{equation} How are java.lang.Object's protected methods protected from subclasses? The keyword protected grants access to classes in the same package and subclasses ().  Now, every class has java.lang.Object as superclass ().  Hence I conclude that every class may access java.lang.Object's methods even if they are protected.  Take a look at the following example:   public class Testclass {   public Object getOne() throws CloneNotSupportedException {     return this.clone();   }   public Object getTwo() throws CloneNotSupportedException {     return ((Object) this).clone();   } }   While getOne() compiles fine, getTwo() gives   Testclass.java:6: clone() has protected access in java.lang.Object         return ((Object) this).clone();   I neither understand why getTwo() doesn't compile nor what's the difference (regarding the access of java.lang.Objects members) with getOne(). How can I compile, run and decompile C# code in Ubuntu terminal? Steward badge can now be awarded multiple times -- backfill details As we announced in our post on  to the Review Queues, the Steward badge can now be earned more than once per queue. The purpose of this post is not to debate the merits of this decision ( already exists), but to explain the nuts and bolts of how it will be backfilled. This is how it will appear on your profile if you have earned it multiple times:  Once we push the visual design changes live across the network, this is how badge attainment will look on /review/close/stats:  For those of you who have already completed enough reviews to earn multiple Steward badges, we will be retroactively awarding them to you, starting today. The backfill script will run about twice an hour, and your achievements trophy will light up as each one is awarded. For those of you who have done thousands and thousands of reviews, things may be a bit noisy, so we thank you in advance for your patience. It may take up to two days for all of your badges to show up. How do I force a column-break in a two-column document? One divided by Infinity? Find Max Value for each month for the last 3 months, properly