I've been battling the castle guards for ages and I have been getting no where! just losing a bunch of candies from all the health potion making. What is the quickest way to beat all the guards?
I already have the pink and red leather gloves, the armor, the crown, and the monkey's staff (and they're all enchanted). Regardless of how many health potions I use, I always die. Help!
Windows 10 - disable reopening programs after restart/startup Which files shouldn't be checked in into version control in Android Studio The age of the universe Counting the Lotto number draws which contain consecutive numbers What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime?  Consider the following sample code - inside the Example() method, what's the most concise way to invoke GenericMethod&lt;T&gt;() using the Type stored in the myType variable?  public class Sample {     public void Example(string typeName)     {         Type myType = FindType(typeName);          // What goes here to call GenericMethod&lt;T&gt;()?         GenericMethod&lt;myType&gt;(); // This doesn't work          // What changes to call StaticMethod&lt;T&gt;()?         Sample.StaticMethod&lt;myType&gt;(); // This also doesn't work     }      public void GenericMethod&lt;T&gt;()     {         // ...     }      public static void StaticMethod&lt;T&gt;()     {         //...     } } How to completely disable auto-save and versions in Mac OS X Lion? I hate the automatic saving and versioning in Mac OS X Lion.  Is there a way to completely turn it off?  (This has been asked before , but the answer and the linked article are completely wrong, so I thought I'd try again.) I have only seen infinite vector spaces and the one finite vector space i.e the trivial vector space $\{0\}$. Is there any other finite vector space? Which one is the formula of angular velocity? Is it $\omega= vr$? If not then why does the source below use $mvr=nh/2\pi$, where $mvr$ is angular momentum and $vr$ is angular velocity? Again $\omega=v/r$ shows radian per second. I saw $\omega= vr$ in one book and $\omega= v/r$ in another! In my custom Drupal 8 plug-in I perform such query:  $select = db_select('my_table', 'e'); $select-&gt;addField('e', 'field1');  $select-&gt;condition('field3', 'NULL', '!='); $select-&gt;condition('field4', array(11, 12, 31, 32, 33), 'IN');  $select-&gt;groupBy("e.field1"); $select-&gt;orderBy("e.field1"); $select-&gt;orderBy("e.field5"); $entries = $select-&gt;execute()-&gt;fetchAll(\PDO::FETCH_ASSOC);    And it works fine. It returns me all records grouped by field1 that meets the condition, orderered. But it returns only field1 column.  I wanted to get more columns, so I have added another $select-&gt;addField('e', 'field2'); just after first one:  $select = db_select('my_table', 'e'); $select-&gt;addField('e', 'field1'); $select-&gt;addField('e', 'field2');  $select-&gt;condition('field3', 'NULL', '!='); $select-&gt;condition('field4', array(11, 12, 31, 32, 33), 'IN');  $select-&gt;groupBy("e.field1"); $select-&gt;orderBy("e.field1"); $select-&gt;orderBy("e.field5"); $entries = $select-&gt;execute()-&gt;fetchAll(\PDO::FETCH_ASSOC);    This time I get error:     Drupal\Core\Database\DatabaseExceptionWrapper: SQLSTATE[42000]: Syntax error or access violation: 1055 Użyto 'my_database.e.field2' bez umieszczenia w group by:   Which means "my_database.e.field2 has been used without putting in into group by". But I don't want to group by every column I select.  Unfortunatelly Drupal 8 added ONLY_FULL_GROUP_BY :( Then, how can I use groupBy and select more then one column at the same time? Autosegmental Representation in tikz-qtree This is what I want to achieve    and this:    But all I've been able to make with tikz-qtree is the following:  \documentclass[12pt,a4]{article} \usepackage[utf8]{inputenc} \usepackage{tipa,tikz,tikz-qtree} \begin{document}  \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ] [.$\mu$ : ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ]  ] ] \end{tikzpicture}   And this:  \begin{tikzpicture}[baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ l ] ] [.R [.$\mu$ a ] [.$\mu$ l ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ : ] ] [.R [.$\mu$ a ] [ l ] ] ] \end{tikzpicture} \end{document}   Which renders:    and    The important part here is the linking from the leaves to more than one parent (it would be nice if I had the option of making the line dashed too). Now, pst-asr doesn't quite achieve what I want it to. It's important that I have the moras ('mu's). (Also it's a pstricks package, which means I must typeset it in DVIPSPDF.) I have tried a bit of the forest-package (especially forest-GP1), but I can't make that work as I want either. Lastly, some have recommended xyling, but I find that package very difficult to use.  Any recommendations?  Thanks. How does evolving relate to CP? proving that elements of a sequence are pairwise coprime For the sequence $a_n = 6^{2^n} + 1$:   How can I prove that the elements of this sequence are pairwise coprime, i.e. prove that if $m$ is not equal to $n$ then $\gcd(a_m, a_n) = 1$. (begin by proving that $a_n∣ (a_{n+1} - 2)$) Show that f(x) = inf{g(x1)+h(x2)} is convex subject to x1+x2=x and where g(.) and h(.) are convex functions.  Can I just go about this by using the regular definition of a convex function or which properties should I use? I know that the sum of convex functions is convex, but the infimum part confuses me. Thank you I have the formula $x! + y! = z!$ and I'm looking for positive integers that make it true. Upon  inspection it seems that x = y = 1 and z = 2 is the only solution. The problem is how to show it.   From the definition of the factorial function we know $x! = x(x-1)(x-2)...(2)(1)$  So we can do something like this:  $$ [x(x-1)(x-2)...(2)(1)] + [y(y-1)(y-2)...(2)(1)] = [z(z-1)(z-2)...(2)(1)]$$   we can then factor all of the common terms out on the LHS.  $$ [...(2)(1)][x(x-1)(x-2)... + y(y-1)(y-2)...] = [z(z-1)(z-2)...(2)(1)]$$ and divide the common terms out of the right hand side  $$[x(x-1)(x-2)...] + [y(y-1)(y-2)...] = [z(z-1)(z-2)...]$$  I'm stuck on how to proceed and how to make a clearer argument that there is only the one solution (if indeed there is only the one solution).   If anybody can provide a hint as to how to proceed I would appreciate it. Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".  Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.  Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by  $$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.  A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if $\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.  The exercise as stated:  Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.  Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.  I'm having problems with the $(\Rightarrow)$ direction.  My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ &amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ &amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ &amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\ &amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*}  Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.  Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.  Thanks for any help you can give. Are the Yongnuo clones of Canon flash equipment and lenses legal to sell/buy? Can't install MySQL on Mountain Lion finite and infinite product of discrete space Non constant analytic function from $\{z\in\mathbb{C}:z\neq 0\}$ to $\{z\in\mathbb{C}:|z|>1\}.$ Adding DC voltages of PSU I need to add two voltages together that share a common ground (power supply from computer). I tried using bridge rectifiers and Op-Amps.