What's the relationship between those two “heaps”?
Why are two different concepts both called "heap"?
Can we show new/anonymous users more Q&A? With left nav, responsive design, and various new-user notices, this is what an anonymous user currently sees (browser window here is 1090px wide):    I realize that a lot of this content -- code of conduct, new-user intro, cookies notice -- is important.  And I get that the left nav is an important design element, and responsive design dictates the width of the right column.  But when you add it all up, Q&amp;A -- what the site is about -- gets a distressingly small portion of the screen.  Can we do better?   Maybe we can make that intro/mini-tour shorter? Maybe we can hide the right column for anonymous users? Maybe some things can wait until the user's viewed a few pages? Maybe other things I haven't thought of?   My concern is that anonymous users -- people coming from Google, people who were referred to a site by users hoping to recruit them, or others who find their way to us but aren't invested in us -- will turn away before seeing how rich and helpful our sites are.  Which would be a real shame, because our sites are awesome.  First impressions matter.  When I'm asking a rabbi to check out Mi Yodeya or a professional chef to check out Seasoned Advice or my vet to check out Pets (all of these have happened), I want to make a good impression.  I want the person to see more than 2.5 questions; I want to increase the chances the person will see something interesting or say "oh I can answer that!" and dig deeper.  Yes, the other stuff is important, but maybe not now?  Is there a better way to "stage" all this information for people we haven't yet hooked? Edit Mail rules manually I know why questions, answers and some other aspects of the site are color-coded and what they mean but I can't find why badges are color-coded. I'll include a snapshot to show what I mean. How can I make my microSD card writable? How common is "thrice"? Calculate business days I need a method for adding "business days" in PHP. For example, Friday 12/5 + 3 business days = Wednesday 12/10.  At a minimum I need the code to understand weekends, but ideally it should account for US federal holidays as well. I'm sure I could come up with a solution by brute force if necessary, but I'm hoping there's a more elegant approach out there. Anyone?  Thanks. Gimp won't allow me to copy and paste a transparent image On the latest version of GIMP on my PC, whenever I try to copy a transparent image from the internet, GIMP will automatically fill in the transparent areas with the background color I have selected.  This is a non-issue on my Mac, but I need to do this work on my PC, so how can I make it recognize the alpha of the image and not fill it with the background color?  The image canvas that I'm working with does have an alpha channel, but that does not seem to make a difference. Is there a way to make rigid body interact with softbody physics? I'm trying to have a surface with soft body hold obects that are falling on it with rigid body physics, but for some reason, no what matter what mass I give it, the objects just fall trough the softbody enabled surface.  Is there a toggle or a value I need to change to make them stay there? tikz relative coordinates Suppose one defines coordinates a and b. One may draw a line connecting these two coordinates (\draw (a) -- (b)). Suppose one desires to draw a line that is shifted upward 1 unit relative to a and b. A first guess is the command \draw (a){}+(0,1) -- (b){}+(0,1). This command, however, connects the point a+(0,1) to the point b, presumably because addition is subsequent to the draw command. How does one obtain the desired result (drawing from a+(0,1) to b+(0,1)?  MWE:  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc} \begin{document} \begin{tikzpicture}  \draw[fill] (0,0) circle (2pt) coordinate (a);  \draw[fill] (5,0) circle (2pt) coordinate (b);  \draw (a) -- (b);  \draw (a){}+(0,1) -- (b){}+(0,1); \end{tikzpicture} \end{document}     Edit: How does one draw coordinates at a point shifted relative to the arithmetic average of the coordinates of two points (something akin to \node at ($(((s1)+(s2))/2)+(0,1)$)). Forward vs Forwards I need something explained or corrected:  In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:  Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 &lt; A \le M$:  Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$  So the $gcd(A,M) &gt; 1.$ Thus $\phi(M) = 1$ ...?   Which is not even and contradicts the theorem that $\phi(N)$ is even for $N&gt;2.$ Therefore there exists an infinite number of primes.  I get confused by the statement "Thus $\phi(M) = 1$"....  Did i possibly copy this proof down wrong? or am I missing something? Thank you in advance.   Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear.   Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it. What's the weight of the earth? What's the weight of the earth?  That is, what's the weight of the earth in the gravitational field of the sun? I imagine it's either   $0$, because there's no net fall toward the sun, so the downward component of acceleration is $0$ and $F=ma=0$, or $Gm_1m_2/r^2\approx7\cdot10^{-11}\cdot6\cdot10^{24}\cdot2\cdot10^{30}/(1.5\cdot10^{11})^2\approx4\cdot10^{22}$ Newtons,   but which, and why? No non-trivial clopen sets in $\mathbb{R}$? How to give a direct proof? How to give a direct proof of the following result?     Let $A$ be a subset of $\mathbb{R}$ such that $A$ is both open and closed. Then $A$ is either empty or all of $\mathbb{R}$?   My work:   If $A$ is not empty, let $u \in A$. Then there is some open interval $(a,b)$ such that $$u \in (a,b) \subset A.$$  Now if $\mathbb{R} - A$ is not empty either, then let $v \in \mathbb{R} - A$. Then there is some open interval $(c,d)$ such that  $$v \in (c,d) \subset \mathbb{R} - A.$$  Suppose that $u &lt; v$.   Since $$\emptyset \subset (a,b) \cap (c,d) \subset A \cap (\mathbb{R} - A) = \emptyset,$$ we must have $$(a,b) \cap (c,d) = \emptyset.$$ So we can conlclude that $$b \leq c.$$ But $b \in A$ and $c \in \mathbb{R} - A$. So we must have $$b &lt; c.$$  What next? Can anybody here please help complete the proof from here on?  An edit based on a comment by Marc Paul:   Let us assume that the set $A$ is a non-trivial clopen set in $\mathbb{R}$. Let us define a function $f \colon \mathbb{R} \to \mathbb{R}$ as follows:  $$f(x) \colon= \begin{cases} 1 \ \mbox{ for } \ x \in A; \\ 0 \ \mbox{ for } \ x \in \mathbb{R} - A. \end{cases}$$   Let $V$ be an open set in the range space $\mathbb{R}$. We show that the inverse image set $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   The following cases arise:   If $0, 1 \in V$, then $f^{-1}(V) = \mathbb{R}$.   If $0 \in V$ but $1 \not\in V$, then $f^{-1}(V) = \mathbb{R} - A$.   If $1 \in V$ but $0 \not\in V$, then $f^{-1}(V) = A$.   And, if $0 \not\in V$ and $1 \not\in V$, then $f^{-1}(V) = \emptyset$.   Thus, $f^{-1}(V)$ is open in the domain space $\mathbb{R}$.   Hence the function $f$ is continuous.   What next? How does this lead to our desired conclusion?   [Yet another edit, again based on valuable comments from Marc Paul: ]  So if both $A$ and $\mathbb{R} - A$ were non-empty, then let's suppose $a \in A$ and $b \in \mathbb{R} - A$, and we can assume without any loss of generality that $a &lt; b$.   Then $f(a) = 1$ and $f(b) = 0$. So by the intermediate value theorem there is a real number $c \in (a,b)$ such that $f(c) = 1/2$, which is a contradiction because the image set of $f$ does not contain $1/2$. How to get text from an element , when getText fails I have the following element element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)  I want to assure the text on this filed is TESTUSER, so I tried:  expect(element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)).getAttribute('value')).toEqual('TESTUSER');  then I got the error: Expected null to equal 'TESTUSER' So I tried: element(by.css(&quot;span[aria-label='Picker ID, TESTUSER']&quot;)).getText().then(function (value) {         console.log(value);       })  And I got this error: Expected '' to be 'TESTUSER'.  How to get the value of this field and expect it is TESTUSER? I have a problem with verb structure In this sentence:     Four items are responsible for three-fourths of the overall spending around the world, the most prominent share being that of food.   Why after the comma, the verb "being" is used? Why it does not use "is"? How does Cycles work and how is it different from Blender Internal?  Which is better? How can my module add xml sitemap entries? I’ve always understood that the phrase two weeks usually turns into two weeks’ when used as a modifier as in   I’m giving my two weeks’ notice.  I get two weeks’ vacation. (“two weeks’ holiday” for Brits)   with an apostrophe on the word weeks’, indicating that the vacation “belongs” to the weeks.   One way to explain this is the phrase “two weeks of vacation” being contracted to “two weeks’ vacation” – the vacation is “of” the weeks; that is, it is possessed by them.  But I’ve seen a lot of people omit the apostrophe in casual writing, and thinking about it, it seems plausible that the noun vacation would be plainly modified by the adjectival phrase two weeks.  But on the other hand, shouldn’t adjectival phrases be hypenated, namely two-weeks vacation?  Yet that seems wrong, too: the hyphenate should be singular, two-week vacation (like two-tone shoes) because .   touches on this (and references a good ) but doesn’t explain why the plural must be possessive in this case.   suggests that two weeks might be adverbial rather than adjectival, and thus resists being used attributively. But that still doesn’t quite explain this particular muddle of pluralization, possession, and attribution.  Given all this, why should “two weeks vacation” be written “two weeks’ vacation” with a possessive apostrophe? Start Ubuntu Unity With Dash Open Is my mobile AMD card being used?