Trying to add a NOT NULL constraint to a table with 1 billion rows. I cannot afford a table lock for more than a couple of seconds. Is there a way to prevent a full table scan during the alter table statement? I created an index on the column hoping it would be used but that doesn't seem to work. May be a check constraint? Other options? Thank you!
When adding NOT NULL columns with a DEFAULT value - does PostgreSQL optimize this operation?  In case the table has n rows, a not-optimized alter-table-add-column would yield n writes of the default value - which could be very painful, obviously. With the optimization the DB would instantaneously create the new column, store just one copy of the default value which would be returned when no non-default value is found for that column in a suitable index data structure.  For example  such an .
What is "my guy syndrome" and how do I handle it? Is there any text editor which supports Markdown (Ask Ubuntu's formatting style)? Normed vector spaces over finite fields Normed vector spaces are typically defined over the reals or complex numbers.  Is there any "standard," well-behaved construction that generalizes this to a vector space over a finite field, such as $\Bbb F_2$?  I'm looking for something kind of like the class of $\ell_p$ norms, except designed with finite fields in mind.  Ideally, something that has deep fundamental properties making it well-behaved in the same way that the Euclidean norm is. Improving Review Queues - Design overview II: Changing review bans and other new features In MK (Morse-Kelley) set theory life is easy: $\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)$. If $X=\left\{\right\}$ then $\bigcap X=U$, where $U$ is the universal class. So the (unary) intersection of the empty set is the class that contains all sets as elements. In ZF (Zermelo-Fraenkel) set theory, instead, proper classes are not allowed. So, how can I define $\bigcap X$ in ZF? I tried with the following definitions:   $\forall X\left(X\not=\left\{\right\}\rightarrow\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)\right)$. This means that $\bigcap\left\{\right\}$ is undefined, which is not that good. $\forall X\forall y\left(y\in\bigcap X\leftrightarrow X\not=\left\{\right\}\land\forall x\left(x\in X\rightarrow y\in x\right)\right)$. This means that $\bigcap\left\{\right\}=\left\{\right\}$, which is the opposite of MK.   I couldn't find any other valuable definition. Any ideas? Thank you. Let's say I have Table A: BookingsPerPerson  Person_Id    ArrivalDate    DepartureDate 123456       2012-01-01     2012-01-04 213415       2012-01-02     2012-01-07   What I need to achieve with a view is the following:  Person_Id    ArrivalDate    DepartureDate    Jan-01    Jan-02    Jan-03    Jan-04    Jan-05    Jan-06    Jan-07 123456       2012-01-01     2012-01-04       1         1         1         1 213415       2012-01-02     2012-01-07                 1         1         1         1         1         1   The system is for events, so each hotel booking could take anything between 1 to 15 days but no more than that. Any ideas would be very much appreciated. Suppose that f is continuous on $(a,b)$ and $\lim_{x\to a^+}  f(x) = \lim_{x\to b^-} f(x) = \infty$. Prove that $f$ has a minimum on all of $(a,b)$.  I was thinking of applying the extreme value theorem, but I realized that it applies only to closed intervals. So I was thinking of defining a closed interval that is in $(a,b)$, and using the definition of one sided limits to go about the problem, but I don't know how to proceed. Help please? Benchmark (hash rate) for SHA-3 (SHAKE128) To estimate the time for brute-force attack we need to compute keyspace size divided by hash rate, where the hash rate (hash/second) varies depending on the computer's capabilities.  The website  shows the hash rate for different computers.  Can you explain the following two confusing phenomena?   The hash rate decreases as the message size increases. Core i3 has a higher hash rate than Core i7 for SHAKE128. Check if macro is fully expandable Question  Is there a way to check whether a macro is fully expandable (or rather "safe in an expansion-only context" )?    Consider this code:  \def\a{Just a string} \def\b{\a} \def\c{\def\unsafe} \def\d{\c}   How could I check which of the macros (a-d) are safe in an expansion-only context? By looking at them I know that a and b are whereas c and d are not but if I wanted to know the same for a macro I haven't written myself this could get quite useful.      Background  I am working on a way to detect whether some input is a valid number in PGF. For this I developed  which makes use of passing the input into \pgfmathfloatparsenumber. The problem I have run into is that said macro appears to somehow manages to expand the input until there is an error (if the input is in fact not safe in an expansion-only-context). I tried using protected, noexpand and similar but somehow PGF manages to circumvent those.    So the idea is to check whether the input is safe before actually passing it to PGF. The problem is: I don't know how I'd go about that... Do I need to play previous Witchers before Witcher3 I'm thinking about buying The Witcher 3 and I know it's heavily based on a story/narrative.   For the story/narrative to make sense, do I need to play the previous 2 Witcher games - or are the stories for each title independent of one another? d'Alembert-like functional equation: $f(x+y)+g(x-y)=\lambda f(x)g(y)$ I'm trying to create a command, which calls on an environment, which defines a custom column definition with one argument using \newcolumntype, and defines a second command, containing parameters for a table column.  Then, still within the first command, I begin the tabular environment and call the custom column, with the second command as its argument. I always get an undefined control sequence error, though. Any thoughts on why that might be?  I'm sure that was clear as mud, but I've cut it down to a minimal example, which follows:  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl}  \newenvironment{tablecolor}{     \newcommand{\tlight}{\columncolor{SkyBlue!30}\color{black}}     \newcolumntype{C}[1]{&gt;{##1}c} }{}  \newcommand{\tablegoals}{ \begin{tablecolor} \begin{tabular}{l|C{\tlight}} Goal 7  &amp; \\ \end{tabular} \end{tablecolor} }  \begin{document}     \tablegoals \end{document}     Here's my second attempt. Different approach, same undefined control sequence error with \columncolor.  \documentclass{article} \usepackage{array} \usepackage[table,dvipsnames]{xcolor} \usepackage{colortbl} \usepackage{ifthen}  \newenvironment{tablecolor}{     \newcolumntype{C}[1]{&gt;{         \ifthenelse{\equal{dark}{##1}}{             \columncolor{Cyan}\color{white}             }{             \columncolor{SkyBlue!30}\color{black}         }     }c} }{ }  \newcommand{\tablegoals}[1]{     \begin{tablecolor}         \begin{tabular}{l|C{dark}}             Goal 7  &amp;   #1         \end{tabular}     \end{tablecolor} }  \begin{document}     \tablegoals{Check mark} \end{document} Objects, shadows and reflections not rendering in EEVEE I started to work with EEVEE short time ago and I was amazed by the capabilities of this engine. However, I noticed something strange when I was rendering some tests. Some objects/reflections/shadows dissapear when they're off canvas. I imagine that maybe this is for save memory data, but it can be not suitable to some projects. To make my point graphical, I made a simple scene with simple objects. The main object is the sphere with a mirror-like shader. Out of focus there are different objects with different shaders. A yellow cone, a blue cylinder and a white cube light. But surprisingly, those objects dissapear when the composition is rendered. See the rendered animation below:  Notice that the blue cylinder is not even showing at all in the final render. And the blooming effect from the light cube also dies. I would like to improve this bahaviour and render the objects that are still out of canvas using EEVEE. Do I missing something like memory configuration or reflection/shadow tweaks? Is this caused by the EEVEE engine or by my GPU config (GTX1060)? Thank you in advance! tikz: Distribute evenly and randomly circles I want to fill a part of a picture with more or less evenly and randomly distributed circles which should not overlap. It would be okay if circles are only partly inside the shape.  The density should be like what you see on this page  with values 80-100. (But I do not need so many circles, 30 should be enough, I could repeat the pattern).   I cannot use a grid and "wobble" the circles a bit: The circles are too large and it simply does not look randomly.   Currently I am thinking about a random attempt: Get some random point, measure its distance to existing circles, either paint or throw the point away. Try 100 times and hope that you get the right number of circles (and that it does not take too long). But perhaps someone has a better idea.   The example draws the circles simply manually:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{backgrounds} \begin{document}   \begin{tikzpicture}[framed,gridded,radius=0.5cm]  \draw[red](0,0) rectangle (5,5);   \draw (1,1) circle;   \draw (2.3,1.1) circle ;   \draw (4.5,0.8) circle ;   \draw (5.1,1.8) circle ;   \draw (0.4,3.3) circle ;   \draw (2.1,2.8) circle ;   \draw (3.8,3.5) circle ;   \draw (4.8,4.2) circle ;   \draw (0.8,4.9) circle ;   \draw (2.1,4.1) circle ;   \draw (3.8,2.0) circle ;   \draw (3.5,0.6) circle ;   \draw (3.0,5.0) circle ;      \draw (4.1,5.1) circle ;             \draw (0.9,2.1) circle ;  \end{tikzpicture}  \end{document} Showing that $\int_0^\infty x^{-x} \mathrm{d}x \leq 2$. If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$  I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$  I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions? Can't login with Stack Exchange OpenID when HTTPS-Everywhere plugin is active - apparent protocol mismatch I can't seem to log in with my Stack Exchange OpenID. When I try, I get the error message:     Unable to log in with your OpenID provider:      The openid.return_to parameter (.) does not match the actual URL (.) the request was made with.   It works if I disable the HTTPS-Everywhere plugin in Firefox, but if I then re-enable the plugin and go to a site on which I am not logged in (in which case login is normally nearly seamless), it fails again with the same error message, or at least something very similar.  I'm not aware of having done anything that might have caused this, although I will admit I don't recall seeing https:// in the address bar on Stack Exchange before (I do now, and that's a good thing IMO!). I was using the site  and it worked fine then, but not now... Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists). Invariance of a maximally entangled state under unitary operation $U \otimes U^\dagger$ Apparently the (d-dimensional) maximally entangled state, $|E \rangle = \sum_{i} |ii\rangle /\sqrt{d}$ is invariant under operations of the form $U \otimes U^\dagger$. I want to prove this result, which amounts to showing that  $$\left(\sum_i U | i \rangle \otimes V |i \rangle = \sum_i | ii \rangle \right)\Rightarrow (V = U^\dagger )$$  I have no idea how to even start. I suppose it's some simple linear algebra result, but I don't see it. A hint would be appreciated. My system is not dual-boot, I run a standard Ubuntu desktop system "on the metal" (I think running it in a VM is the same).  Pressing c while booting does not cause the GRUB menu to appear (which I believe it does for a dual-boot system).  How can I get the GRUB-menu to present itself on a single-boot system?