In Episode IV we see the remnants of Luke's aunt and uncle who are really his father (Vader's) step-family.     Does Vader know his stormtroopers shot them up? Did he authorize it? If he found out after the fact how did he react? Looking for current Disney cannon answer if possible, EU expanded universe as an after thought.
The  EU book The Rise and Fall of Darth Vader reveals that Vader personally gave the stormtroopers the order to kill Owen and Beru Lars, and that he knew who he was condemning to death before he gave the order:  “Lord Vader,” said one of the sandtroopers, adjusting a control on his helmet so that only Vader could hear his voice.  “The Jawas sold a protocol droid and an astromech to these moisture farmers, but both droids are gone.” Moisture farmers? Intrigued, Vader examined the holograms of the kneeling couple and said, “The farmers’ names?” “Owen and Beru Lars, sir,” the sandtrooper responded. “They say they don’t know where the droids are, but it looks like a landspeeder is missing from their garage.” Owen and Beru, Vader recalled... “Tell Mr. and Mrs. Lars that they seem to have trouble keeping protocol droids on their property.”1 Not certain if he had heard correctly, the sandtrooper said, “Sir?” “Then you may extend to them every courtesy that you showed the Jawas before you continue your search. Establish checkpoints to detain any droids entering Mos Espa or Mos Eisley spaceports. And one more thing.” “Yes, Sir?” “Do not stop transmitting until I break the connection.” “Understood,” said the sandtrooper. Vader watched the sandtroopers carry out his orders on their helpless victims. He found the sight of rising flames — even holograms of flames burning millions of light years away — to be most satisfying.  I had always assumed that the stormtroopers killed the Larses on their own initiative, or in accordance with a previously given order.  The version of events in this book is interesting, but it raises the question of why it took Vader so long to figure out who Luke was. In canon, who gave the order to kill Owen and Beru Lars?  If Vader was involved, did he recognize the couple before he gave the order?  1 A reference to the fact that Anakin took C-3PO from the Lars homestead the last time he was on Tatooine, without informing the Larses that he would do so.
I am curious if it is possible to have a VPS that has data on it that is not readable by the hosting provider, but is still usable on the VPS.  Obviously there are some things that you could do to prevent them from reading anything...   You could change all the passwords, including root. But then, they could still use some alternate boot to reset the password, or they could just mount the disk another way. So, you could encrypt the disk or at least some of the contents on the disk. But then it seems that if you decrypted the content, they could still "peer in" to see what you were doing at the console, because after all, the virtualization platform should allow this. And even if you could stop that, it seems they could just read the RAM of the VPS directly.   Of course, the VPS can store data on it and as long as the key is not on the VPS and the data is never decrypted there, then the host cannot get the data.  But it seems to me that if any point the data on the VPS is decrypted...for use on the VPS...then the hosting provider can get the data.  So, my two questions are:   Is this correct? Is it true that there is no way to 100% secure data on a VPS from a host from seeing it, while keeping it accessable by the VPS? If it is possible to make it 100% secure, then how? If it is not possible, then what is the closest you can get to hiding data from the web host? How do I use the Cherokee keyboard layout to type English characters? Wow, I've really done it this time. I've locked myself out of my machine.  In preparing for a Cherokee language course, I added the Cherokee keyboard layout to my options. However, I was unable to generate any Cherokee characters by selecting that input device.  I found an article on the web which instructed me to use the dpkg-reconfigure keyboard-configuration command in a terminal.  Once I did that, I rebooted as instructed. Once I was presented with the login screen, I entered my password as usual only for it to fail.  After failing a few times, I gave my attention to what is usually a mundane exercise (i.e., logging in).  While I carefully entered my password, each time, I get the message, "Sorry that didn't work..."  So, I brought up the onscreen keyboard... But the password box displays an asterisk for only the first character of my password.  I attempted to use the direct Unicode input method (ctrl-shift-Uxxxx) but the Cherokee keyboard has no "U" key.  Can anyone please help me troubleshoot this? I have the Blender 2.8 version from the 21st Sep 2018 I was looking forward to render a test sequence but cannot find a render button anywhere? Does anyone know where to find it? 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 ? My grandparents bought a flight on United Airlines that departs on Feb 10 2019, as the one and same ticket, from Toronto, to ORD with 1 55-minute layover, then Maui. Please ping me if I ought post their return flight on Feb 24's details.   advises:     There should be no customs and immigrations formalities at ORD, as YYZ is a , and CBP matters will be handled in Toronto.    I'm assuming that  persists during their vacation to Maui.   reported some TSA agents' refusal to work.  Will  refuse to work, or any other federal workers pertinent to travelling to and from the US? This is a problem from Introduction to Representation Theory by Etingof et al. I have no idea how to approach it. My friend wrote a program that uses Gaussian elimination to find a basis, which starts with x, y, [x, y], [y, [x, y]], ... but freezes once it gets to elements with 9 or more occurrences of x, y.  I expect that there is some way to characterize the nonzero elements of this Lie algebra (i.e., elements that are "immune" to being shown to be zero through repeated application of the Jacobi identity) which could be turned into a way to enumerate them. I don't see what it is, though.  It's also totally unclear to me how to show that some set we know spans the Lie algebra (proven using some criterion as described in the previous paragraph) is actually linearly independent, unless we find a faithful representation... but I have no idea how to find a faithful representation for an infinite-dimensional Lie algebra.  Hints would be greatly appreciated.  Edit: After looking at the partial output, we can form the "straightforward" conjecture that:   If $n \geq 4$ is even, then there is one basis element of grade $n$, namely $\mathrm{ad}([x, y])^{(n-4)/2}[y, [y, [y, x]]]$. If $n \geq 5$ is odd, then there are two basis elements of grade $n$, namely $[x, z]$ and $[y, z]$ where $z$ is the single basis element of grade $n-1$.   I also have a different conjecture, which is mutually exclusive with the above: that if we flatten out all nonzero elements into the form ${\rm ad}(e_1) {\rm ad}(e_2) \ldots {\rm ad}(e_{n-1}) e_n$ using the Jacobi identity, then in the sequences $e_1, \ldots, e_{n-1}, e_n$ and $e_1, \ldots, e_n, e_{n-1}$, any span containing two x's must contain at least two y's and any span containing five y's must contain at least two x's, otherwise we can use the Jacobi identity to rearrange the element somehow to show that it vanishes. If this is true, then the long-range behaviour must be:   For $n \geq 8$ there is exactly one basis element of grade $n$, namely $[... [y, [x, [y, [y, [x, y]]]]] ...]$ (the pattern $y x y$ repeats).   I still don't see an approach to proving that none of the elements in this sequence actually vanishes, though.  Edit 2: Here is the hint given by Etingof in an email:     No, there are no published solutions, since these problems are used as   a homework for classes. This is indeed a nontrivial problem.      This is the positive part of the affine Kac-Moody algebra of type   A_2^2, which is the twisted loop algebra for sl(2). So what you need   to do is to relate this Lie algebra to loops into sl(2). I see from   your post that you are on the right track. Why Oracle discontinued Oracle Java (JDK) Installer PPA? How can we differentiate between change and progress in the area of history and natural sciences? Remove rogue app from menu bar At the moment I'm taking an algebraic topology course. In our current exercise we have to prove the following: Consider any topological space $X$ and any two maps $f, g : X \to S^n$, such that $f(x) \neq -g(x)$. It follows that $f$ is homotopic to $g$.  My idea is to use the mapping cylinders $M_f, M_g$ to construct a homotopy via $$M := M_f \amalg M_g/ (x, 0)_f \sim (x,0)_g$$ But this doesn't seem to work. In addition I don't know how $f(x) \neq -g(x)$ is going to help me (I can see how this could help if $n=1$, but I have no clue how this would work for $n&gt;1$).  I would appreciate any hints towards a solution.  Thanks in advance. Mathematica won't give eigenvectors but Wolfram Alpha will? What am I doing wrong? cmd has wildcard bug? Files in folder:  a.j ab.jp abc.jpg abcd.jpeg   dir command results:  Command       Result (Files shown) ----------    -------------------- dir ?.*       a.j dir ??.*      a.j, ab.jp dir ???.*     a.j, ab.jp, abc.jpg dir ????.*    a.j, ab.jp, abc.jpg, abcd.jpeg   So we can see single ? it means "0 or 1 letter".  Now more dir command results:  Command       Result (Files shown) ----------    -------------------- dir *.?       a.j dir *.??      a.j, ab.jp dir *.???     a.j, ab.jp, abc.jpg, abcd.jpeg --&gt; What is this?! dir *.????    a.j, ab.jp, abc.jpg, abcd.jpeg   In 3rd command why ??? is showing jpeg? Can you explain? Is this bug in cmd? I got a bash script to synchronize data between two computers. It works fine but I have to type my password every time the rsync command is called.   #!/bin/bash  sourceIP="192.168.178.128" sourceUser="user1" destinationUser="user2"  function sync() {      rsync --archive --progress -v -e "ssh -l $sourceUser " $sourceIP:/home/$sourceUser/$1/ /home/$destinationUser/$1 }  sync Pictures sync Music sync Videos sync Documents   How to store the password (via prompt) in a variable and pass it to rsync/ssh? More fine grained syntax highlighting in Emacs/AUCTeX "They interviewed several candidates who/whom he thought had the experience he required." They interviewed several candidates who he thought had the experience and qualifications he required.   My test prep book says this should be "who" because of the subordinate clause's predicate:     They interviewed several candidates who he thought had the experience and qualifications he required.   I feel like it should be "whom" as it's the object of the main clause.     The interviewed several candidates whom he thought had the experience and qualifications he required.   Who's right? Meta Meta Privilege I just took a look at meta today and noticed something weird in the meta privileges. If you look at the meta privilege list, you can see that at . I know that there is no meta for meta, and the link they give you in the privilege page leads you to meta.stackexchange.com, right where you started. Why do they have this privilege? I have a feeling that this is a bad recursive joke. You must be  member of meta to get the privilege of being a member of meta! :) Why can't the Polynomial Ring be a Field? I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.   The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so? Upgraded to 13.10 now clock settings are all disabled and clock does not display I've upgraded to 13.10 and now I don't have the standard menu clock, which I need for work.  I checked and 'indicator-datetime' is installed.  I even uninstalled/reinstalled that applet with no luck.  Also my "clock" settings under System Preferences are all disabled.  I can't change anything.  Does anyone know how I can get the old menu clock back?  Alternatively is there another menu clock app I can download?  Edit 1  Thanks for the responses    I've restarted several times and that didn't fix the issue. I just ran the 13.10 update today.  But I ran it again a few minutes ago.  I got   about 200KB in random updates.  The issue is still present after a reboot. # apt-cache policy indicator-datetime indicator-datetime: Installed: 13.10.0+13.10.20131016.2-0ubuntu1 Candidate: 13.10.0+13.10.20131016.2-0ubuntu1 Version table: *** 13.10.0+13.10.20131016.2-0ubuntu1 0     500 http://us.archive.ubuntu.com/ubuntu/ saucy/main amd64 Packages     100 /var/lib/dpkg/status   Edit 2  I believe the issue is related to 'rarings' changing the systray somehow...   Or , although I haven't got that solution to work yet.  Edit 3  Because of a suggestion from RolandiXor I've check the binaries.  My system is missing 'libical.so.0' but it has 'libical.so.1'.    ldd /usr/lib/x86_64-linux-gnu/indicator-datetime-service libical.so.0 =&gt; not found   Does anyone know how to get the libical.so.0 binary that the datetime service needs but doesn't install? Hi is it possible to exclude traffic from a certain country in google analytics?  If yes, how do I do it? 'this is' or 'these are'? I have the following sentences.        This is our first arrival in the UK.   These are our first arrival in the UK.      I am not sure, which sentence is correct?