I have a flow chart that has all of the nodes that I want, there is just one path that I can't get.       \begin{figure}[H] \begin{center}  \tikzstyle{decision} = [diamond, draw, fill=blue!20,      text width=4.5em, text badly centered, node distance=3cm, inner sep=0pt] \tikzstyle{block} = [rectangle, draw, fill=blue!20,      text width=5em, text centered, rounded corners, minimum height=4em] \tikzstyle{line} = [draw, -latex'] \tikzstyle{cloud} = [draw, ellipse,fill=red!20, node distance=3cm,     minimum height=2em]  \begin{tikzpicture}[node distance = 3cm, auto]     % Place nodes     \node [block] (init) {Set Quotient to 0};     \node [block, below of=init] (identify) {Subtract Divisor from Dividend};     \node [block, below of=identify] (evaluate) {Add 1 to Quotient};     \node [decision, below of=evaluate] (decide) {is Result Register Greater than 0?};     \node [block, below of=decide, node distance=3cm] (stop) {Stop - Quotient is Correct};     % Draw edges     \path [line] (init) -- (identify);     \path [line] (identify) -- (evaluate);     \path [line] (evaluate) -- (decide);     \path [line] (decide) -- node {no}(stop);     \end{tikzpicture}  % use \figcap to build the caption. DO NOT use the default \caption command \figcap{Control scheme for Repeated Subtraction Division} % put the caption of your figure here \end{center} \end{figure}   I would like to add a path that loops from the last block (stop) to the block (identify). Whenever I try to do this I get a path that does not go around the blocks, but rather through them.   Any ideas?
Say I have a code to draw a flow chart, I want now a arrow from node 3 to node 1, how can I do that?  1&lt;--| |   | 2   | |   | 3----   MWE  \documentclass[class=article,border=0pt]{standalone}  \usepackage{tikz} \usetikzlibrary{shapes,arrows}  % Define block styles \tikzset{     block/.style={rectangle, draw, line width=0.5mm, black, text width=5em, text centered,                   rounded corners, minimum height=2em},     line/.style={draw, -latex} }% &lt;- if you insist in using this in the document add this % here. \begin{document} \begin{tikzpicture}[node distance = 1cm, auto]     % Place nodes     \node [block] (BLOCK1) {a};     \node [block, below of=BLOCK1] (BLOCK2) {b};     \node [block, below of=BLOCK2, node distance=1cm] (BLOCK3) {c};     % Draw edges     \path [line] (BLOCK1) -- (BLOCK2);     \path [line] (BLOCK2) -- (BLOCK3); \end{tikzpicture} \end{document}
I want to better understand why the construction something + [adjective] makes grammatical sense. Background: I was recently reading a Washington Post article and came across the following sentence.  So if you fall in that category, [the movie] feels more like magic than repetitive.  (Note: I've copied the relevant spoiler-free sentence above; following the link to the article will open you up to movie spoilers! ) The phrase more like magic than repetitive struck me as odd, presumably because it was non-parallel (more like [noun] than [adjective]). I would have expected &quot;...more like magic than something repetitive.&quot; (Actually, that needs to be rewritten to flow better, but the point is that at least it appears to be grammatical.) Thus, one would have the noun phrase &quot;something repetitive,&quot; of the form &quot;[noun] [adjective].&quot; This got me to thinking about the use of adjectives after nouns. There is a great Stack Exchange article  that involves the characteristics of the adjective that indicate it should come after the noun. But I haven't been able to find examples of how the word &quot;something&quot; requires a postponed adjective. Consider the following sentences:  I want to buy something red. I am looking for something sweet to eat.  You cannot say:  *I want to buy red something. *I am looking for sweet something to eat.  But you can say:  I want to buy something (that is) red. I am looking for something (that is) sweet to eat.  So, my question:  Do the above examples mean that the &quot;something [adjective]&quot; construction is an example of the ellipsis of &quot;that is&quot;?  Is there something special about the word &quot;something&quot; that requires the postponed adjective? Note that I am NOT looking for details about adjectives that are postponed; I'm interested in the word that precedes the adjective (either &quot;something&quot; or similar words).   By the way, I am not interested in the following constructions, which are different:  I want to paint the house red. I tried to build the brick house taller.  Thanks in advance! How does one go about proving that the function $x \sin x$ is not uniformly continuous on the set of real numbers ?  Any method that uses the sequential criterion for discontinuity would be preferred.  Thanks in advance ! An awful lot of people seem to use the phrase "substitute X for Y" to mean "replace X with Y", while I've always used and understood it as "replace Y with X". This makes sense to me, given that a substitute is the replacement, not the thing which has been replaced.  I thought I might be able to figure things out from a dictionary, but now after seeing two sorts of definitions, I'm even more confused:   to put or use in the place of another to take the place of; replace   The first one suggests that the object will be the substitute (replacement), the thing put into place, while the second suggests that the object will be the thing which has been replaced.  Is one of these usages more correct in any sense? (I know this is probably a pointless question, since people will continue to speak the way they speak.) Is there perhaps a difference between American and British English?   And most importantly, is there any reliable way to tell what someone means when they ask how to substitute honey for sugar, or is the verb simply guaranteed to cause confusion? How do I avoid the "S to Skip" message on boot? Showing $\sum\limits_{p \in P} \frac{1}{p}$ where $P$ is the set of all primes is divergent TikZ: remove margin, padding, border around nodes containing images I want to create a simple figure where images are connected with arrows. Here is what I did:  \begin{tikzpicture} \node[](A) {\includegraphics[width=0.2\textwidth]{a}}; \node[right=1mm of A](B) {\includegraphics[width=0.2\textwidth]{b}}; \draw[-&gt;] (A) to (B); \end{tikzpicture}     But I want the arrow to start and end right at the images with no space between. How can I do that? When would you use the Builder Pattern? What are some common, real world examples of using the Builder Pattern?  What does it buy you?  Why not just use a Factory Pattern? I absolutely hate it when I watch a video on YouTube and it automatically loads another video page after it's done playing.  This is especially annoying when I'm in the middle of typing a comment as it will cause me to lose what I have typed.  I'd like the video to simply stop and for no further action to take place once it is finished playing.  Is there any way to accomplish this? Does anybody know of a method to trigger an event in Prototype, as you can with jQuery's trigger function?  I have bound an event listener using the observe method, but I would also like to be able to fire the event programatically.  Thanks in advance When using unicode-math, how do I get the original calligraphic alphabet back? I am happy with the great majority of what unicode-math does. But at times I find that I prefer a symbol the way it was before. With \varnothing and \complement, for example. In those cases I was able to save them under a different name before loading unicode-math, but this doesn't always work.  Right now I'm trying to access the \mathcal alphabet from plain LaTeX. How might I accomplish this?    Additional Question  I've been looking through unicode-math, trying to figure out where exactly the original \mathcal alphabet gets discarded, but I eventually gave up.  It would be interesting if someone could explain:   how/where the original \mathcal font is defined anyway, and what kind of 'switch' is thrown by unicode-math that makes it unavailable. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? Let $G(r, n)$ be the Grassmannian of the set of all $r$-planes in a $n$-dim vector space. How to show that the cohomology of a Grassmannian has a basis consisting of the equivalent classes represented by Schubert cycles? I am confused since I don't know how to compute the cohomology of a variety (how to construct the co-chain complex). I am really appreciate if you can compute the cohomology of, for example, $G(2,4)$ and show that it has a basis consisting of the equivalent classes represented by Schubert cycles. Graphics horribly corrupted on Ubuntu 16.04 in VirtualBox, showing random strips of pixelated colors Fine-tuning "Open safe files after downloading" Integral $\int_0^\infty \frac{\arctan(x) dx}{x(1+x^2)}$? How would one evaluate the integral $$\int_0^\infty \frac{\arctan(x) \,dx}{x(1+x^2)}$$?  I was told it had a nice closed form and could have been solved with differentiation under the integral sign; however, I tried to set $$I(\alpha) = \int_0^\infty \frac{\arctan(\alpha x) \,dx}{x(1+x^2)}$$ and got nowhere (the resulting integral was very messy). Is there a much more clever substitution that can be used to tackle the integral? Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.   $$\int (\cos^3x)(\sin^2x)dx  = \int(\cos x)(\cos^2x)(\sin^2x)dx  = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$  let $u = \sin x$, $du = \cos xdx$  $$\int(1-u^2)u^2du = \int(u^2-u^4)du  = \frac{u^3}{3} - \frac{u^5}{5} +C$$  Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Cannot add PPA: 'ppa:~darklin20/ubuntu/bomi'.     ERROR: '~darklin20' user or team does not exist.   Showing this message when suing this command:    sudo add-apt-repository ppa:darklin20/bomi   please help me. Show that $x_{n+1}-\dfrac{x_n}{2}$ converges to zero implies that $(x_n)_n$ also converges to zero. Let $(x_n)_n$ be a sequence of real numbers such that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0.$$ Show that $(x_n)_n$ must also converge to zero.     $\textbf{My Attempt:}$   My original idea was to use the triangle inequality and the fact that $$\lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0$$   to show that the sequence $(x_n)_n$ is Cauchy and hence converges to some real number $a$, which would give me $$\lim\limits_{n\rightarrow \infty}x_n=a=\lim\limits_{n\rightarrow \infty}x_{n+1}\implies a-\dfrac a 2= \lim\limits_{n\rightarrow \infty}\left( x_{n+1} - \dfrac{x_n}{2}\right)=0\implies a=0.$$ However, I'm having trouble showing that $(x_n)_n$ is Cauchy.   Any suggestions/hints would be greatly appreciated. Thanks! evaluate $\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})$ I can't get grub menu to show up during boot