This is a follow-up question to        If you want to redo the same command with sudo !! after doing a command like this:  echo "something"&gt;/path/file   You can use the global replace syntax to recall the command:  !!:gs/&gt;/|sudo tee -a /   Use a space after the -a parameter.  This is the equivalent to sudo !! but helps you bypass sudo restrictions for the &lt; and >. Because sudo does not allow you to use [&lt;, >].  How can I put this into an alias on my .bashrc so I can use it simple?  I tried  alias redo='!!:gs/&gt;/|sudo tee -a /'   But this doesn't work, I get the error:  $ echo "sdfdsf"&gt;/path/file bash: /path/file: no permissions  $ redo bash: /: Is a directory tee: /: Is a directory
I'm trying to set an alias for sudo !! in Bash. I tried alias sbb='sudo !!', but it interprets that as a literal  !! and prints   sudo: !!: command not found   If I use double quotes, it substitutes the double bang in the string itself, so that doesn't work.  Is there any way to make this work? Or an alternative alias? `
Case insensitive regular expression without re.compile? In Python, I can compile a regular expression to be case-insensitive using re.compile:  &gt;&gt;&gt; s = 'TeSt' &gt;&gt;&gt; casesensitive = re.compile('test') &gt;&gt;&gt; ignorecase = re.compile('test', re.IGNORECASE) &gt;&gt;&gt;  &gt;&gt;&gt; print casesensitive.match(s) None &gt;&gt;&gt; print ignorecase.match(s) &lt;_sre.SRE_Match object at 0x02F0B608&gt;   Is there a way to do the same, but without using re.compile. I can't find anything like Perl's i suffix (e.g. m/test/i) in the documentation. How to find range $a_{75}$ of the term of the series $a_n=a_{n-1}+ {1 \over {a_{n-1}}} $ If $a_1=1$ and for n&gt;1$$a_n=a_{n-1}+ {1 \over {a_{n-1}}} $$ $a_{75}$ lies between (a) (12,15) (b) (11,12) (c) (15,18)  Now , in this question, I rewrote, $a_n-a_{n-1} = {1 \over {a_{n-1}}}$, to make a telescopic type and summed the terms to get $$a_n-a_{1}= {1 \over {a_{n-1}}}+ {1 \over {a_{n-2}}}...{1 \over {a_{1}}}$$ And as ${1 \over {a_{r}}} &lt;1$ for all r , I got $a_{75} &lt;76$ , but it's of no help as , the upper bound is much higher. I got no further ideas , on how to solve it, please help. Dark Souls 2 coop/invading problem I was playing Dark Souls II on my pc a lot. After about 200 hours of ingame time I found that I cant summon or invade anybody. I tried new characters and reinstalling but it didn't help. So any tips, and reasons. P.S. I didn't used any cheats. How to control Windows 7 snap feature with two monitors? There is an  feature in Windows 7. However, it seems to not work when dragging a window to the edge that is adjoined to the second monitor. How to make it work properly? Maybe some regedit modification or something? How do I uninstall Pandoc in Ubuntu 19.10 (and install most recent version) I have an old version of Pandoc installed, but I'd like to install the latest version of Pandoc.   To begin I've run the following to get rid of the old version:  sudo apt-get autoremove --purge pandoc  But when I check that everything is uninstalled by running: pandoc --version Then pandoc 2.2.3.2 pops up.  I can't seem to get rid of this old version of pandoc so that I can install a new one.  Could you advise me on how I can get rid of the old version and install the new one please. Why does the new review page sometimes shows questions to vote to close without anybody having voted to close it? I was reviewing  for , and noticed that nobody voted to close it, even though the page reports a duplicate.      Why does this happen? I a chat discussion it was suggested that \hspace*{\fill} (with a space before it to allow for a line break) was sufficient to always get the text following aligned to the right. And indeed, it is aligned to the right.    However, in one case it is inserting an additional line  break where one should not be required as illustrated by the last example here.  I want the text bumped to the next line if and only if (iff ) there is no room:    If I remove the space before the \hspace*{\fill}, there are cases (second and third) where a new line should have been inserted but is not:    Related Question:      I have tried the suggestions at the related question and:   Martin Scharrer's suggestion of using \hbox{}\hfill results in the text being on the left for the last case. Herbert's \When macro exhibits a similar problem for the last case. Thorsten Donig's \signed macro does yield the correct number of lines, but portions of the file name for the middle two cases show up on the second line when they easily fit in the first line.     Notes:  I believe that this is unrelated to the problem, as I was able to reproduce the problem without it.  But, am mentioning it here in case it causes an issue with the proposed solution.  I am using \texttt for the file name. This resulted in an additional complication which I initially thought was the cause of the problem.  So, my actual use case is commented out below and it incorporates a solution from        Code:  \documentclass{article} \usepackage[paperwidth=7.0cm,showframe]{geometry} \usepackage{xcolor} \usepackage{xparse}   \newcommand*{\FillLine}[2]{% \noindent\parbox{\linewidth}{%     \raggedright     File:~#2 % &lt;-- Need space here     %Space before \hspace* allows for a break before it     \hspace*{\fill}\texttt{\small CONFIDENTIAL}%  }% }  %\newcommand*{\EnableHyphenationInTexttt}{\hyphenchar\font=45\relax} % %\newcommand*{\FillLine}[2]{% %    \noindent\parbox{\linewidth}{% %       \raggedright %       File:~\texttt{\EnableHyphenationInTexttt#2} % &lt;-- Need space here %       %Space before \hspace* allows for a break before it %       \hspace*{\fill}\texttt{\small CONFIDENTIAL}%  %    }% %}  \begin{document} \FillLine{Name}{Value}  \bigskip \FillLine{Name}{A-Long-Value}  \bigskip \FillLine{Name}{iiiiiii-iiiii-iiiiiii}  \bigskip \FillLine{Name}{Some-Really-Really-Long-Value} \end{document} How many syllables are in the word 'hour'? Does the word 'hour' have 1 or 2 syllables? Some lands have an ability to become a creature, but they usually say they are still a land. If a land is a 3/3 creature and a land, can you target and kill it with a  to send it to the graveyard, or does it remain in play as a land? Does Stack Exchange contain any edit suggestion or bug submission system, or such type of discussion forum to inform SO or its administrators about suggestions and bug reports that relate to SO itself? What is the "Event Horizon" of a black hole So I took an introductory abstract algebra course a few semesters ago, and we were shown that groups and rings can both be made into products, i.e. if I have some group $G$ (resp. ring $R$) and some indexing set $I$, then I can make a group $G^{I}$ (resp. ring $R^{I}$) with the appropriate cardinality. However, it was also demonstrated that for fields, you cannot keep the multiplicative inverse under a product structure, i.e. if I have fields $F_{1}, F_{2}$, then $F_{1} \times F_{2}$ will still be a ring, but you will not (not just "not in general", but won't) have multiplicative inverses for all non-zero (i.e. not $(0, 0)$) elements, as you could pick $(0, x)$, where $x \neq 0$. According to Wikipedia, an ultraproduct will preserve field structure, but I'm not sure what the cardinality of $\prod_{i \in I} F_{i} / \mathscr{U}$ would be in general.  I saw  that said you could make a field of arbitrarily large cardinality by extending a given field through throwing in a whole bunch of "transcendental" elements. The example given was that if I had $\mathbb{Q}$, I could start by dumping in the complexes, but then I suppose after that I'd just start throwing in dogs and cats or something; I'm really not sure what that responder meant, and ceteris paribus, my dogs tend to stay inside, so I'd rather keep them out of my fields, particularly the large ones the thread was interested in.  Moreover, the thread mentioned earlier seemed more concerned with just making the fields big. My question is a bit more nuanced: Given any cardinality $\kappa &gt; 1$, can I generally construct a field $F$ of a cardinality $\kappa$? Moreover, how vague do I have to be about it (i.e. do I have to use some really choicy methods, or can I make it a bit more straightforward)?  Thanks.  EDIT: Sorry. I am aware that for finite fields, you are limited to powers of primes. I meant for infinite cardinals. Integral of periodic function over the length of the period is the same everywhere $\sqrt{1-x^2}$ is not a rational function If $k$ is a field in which $1 + 1 \neq 0$, prove that $\sqrt{1-x^2}$ is not a rational function. Hint. Mimic the classical proof that $\sqrt{2}$ is irrational   It seems rather odd to discuss the square root function in the context of an arbitrary field, but here goes nothing!  Proof:   Suppose that $\sqrt{1-x^2}$ is a rational function in $k(x)$. Then there exist polynomials $p(x)$ and $q(x) \neq 0$ that are relatively prime with $\sqrt{1-x^2} = \frac{p(x)}{q(x)}$; clearly we may take $p(x) \neq 0$. Then $1-x^2 = \frac{p(x)^2}{q(x)^2}$ or $q(x)^2(1-x^2) = p(x)^2$, which says $q^2 | p^2$, and since $q | q^2$, we can infer $q | p^2$. Moreover, since $p|p^2$ and $(p,q)=1$, then $pq|p^2$ or $q|p$, which contradicts the fact that the polynomials are relatively prime. Hence $\sqrt{1-x^2}$ cannot be a rational function.    How does this sound? Aside from the problem of discussing $\sqrt{~~}$ in the context of an arbitrary field, I am worried about not using the fact that $1+1 \neq 0$, at least not explicitly. Where exactly is this assumed used, if at all?  EDIT:  Suppose that $q(x) =1$, and let $f(x) = a_n x^n + ... a_1 x + a_0$. Then $1-x^2 = f(x)^2$. If $x=0$, $1 = a^2_0$ and therefore $a_0$ must be a unit. Letting $x=1$ we get $0 = (a_n + ... + a_1 + a_0)^2$; and letting $x=-1$ we get $0 = (-a_n - ... - a_1  + a_0)^2$. Since we are working in a field, there can be no nonzero nilpotent which means that $a_n + ... + a_1 + a_0$ and $-a_n - ... - a_1  + a_0$ are both zero. Adding the two equations together yields $2a_0 = 0$, and since $1+1 \neq 0$, $a_0 = 0$ which contradicts the fact that $a_0$ is a unit.   How does this sound?  Another attemtpt:  If $f(x)^2 = 1-x^2$, then $2 = \deg(f^2) = 2 \deg (f)$ implies $\deg(f) = 1$. Yet if $x=-1$, we get $f(-1)^2 = 1 - 1 = 0$ and therefore $f(-1)=0$ since there are no nonzero nilpotent elements in a field; similarly, $f(1)=0$. Since $1+1 \neq 0$, $f$ has two distinct roots in $k$ yet is only a $1$-st degree polynomial, a contradiction. How do you run Ubuntu Server with a GUI? Is it possible to cross the Wagan border by car? I was planning a trip by car when I found something a bit weird in :if you try to cross from Lahore to India, it tells that you have to 6457 km detour.  Is it possible to cross this border driving? If not, is there another way to cross without having to make more than 6000 km? Is it safe to fully cook a stuffed turkey the night before thanksgiving? Why is quantum entanglement considered to be an active link between particles? From everything I've read about quantum mechanics and quantum entanglement phenomena, it's not obvious to me why quantum entanglement is considered to be an active link. That is, it's stated every time that measurement of one particle affects the other.  In my head, there is a less magic explanation: the entangling measurement affects both particles in a way which makes their states identical, though unknown. In this case measuring one particle will reveal information about state of the other, but without a magical instant modification of remote entangled particle.  Obviously, I'm not the only one who had this idea. What are the problems associated with this view, and why is the magic view preferred? How do I fix 'Invalid JSON' errors? I've been at this for almost two hours and no one seems to have the same problem as I and I doubt this will be solved but I'll try anyway.  this is the code  /tellraw @p ["Would you like to teleport?",{"text":"Yes","color":"green","bold":"true","clickEvent":{"action":"run_command","value":"/tp @p ~ ~4 ~ "},"hoverEvent":{"action":"show_text","value":{"text":"Confirm"}},{"text":"No","color":"red","bold":"true","clickEvent":{"action":"run_command","value":"/tellraw @p{"text":"Come back when you want to teleport"}"}}}]   this part is fine  /tellraw @p ["Would you like to teleport?",{"text":"Yes","color":"green","bold":"true","clickEvent":{"action":"run_command","value":"/tp @p ~ ~4 ~ "},"hoverEvent":{"action":"show_text","value":{"text":"Confirm"}}   it's just when I add this  ,{"text":"No","color":"red","bold":"true","clickEvent":{"action":"run_command","value":"/tellraw @p{"text":"Come back when you want to teleport"}"}}}]   Minecraft gives me the error:     Invalid JSON at line 1 column 203   I've tried to space it out and figure out what column 203 is but I can't, I don't know what I did. Anyone got any ideas? How do I loop through or enumerate a JavaScript object?