On a recent review audit on SU, I clicked "Edit" and was greeted with the plain text "This post was deleted and cannot be edited". It wasn't in a popup box or anything; the actual page consisted of just that text, with no formatting (as though it were the only content of the HTML page). This makes it really obvious I'm dealing with an audit.
I just got a question in the review queue that, though poor, I thought was potentially salvageable via editing.  However when I clicked the Edit button I got the deleted post error.  Since the post was gone, I clicked I'm Done (what is there do do with a deleted post?) only to get the review audit failed message instead.  The current behavior is also problematic in that, now that I know about it, it's a risk-free way to check if a question is an audit and not an actual review. That allows me or anyone else who knows about it to detect audits without having to recognize that the question is bad.
How to assign symbols to functions Point reflection over a line I'm having trouble understanding the solution presented below (It's from a textbook). I tried to get something similar, but to no avail. Help me find the way he derived those a,b,x2 and y2 expressions.   Point p is a point which we want to reflect (It's a simple structure with two integers x and y representing coordinates on a 2d plane); (x0,y0) and (x1,y1) are coordinates of endpoints of a line.  Expected result is a new point with reflected coordinates.   Point mirror(Point p, int x0, int y0, int x1, int y1) {     double dx,dy,a,b;    long x2,y2;    Point p1; //reflected point to be returned      dx  = (double) (x1 - x0);    dy  = (double (y1 - y0);     a   = (dx * dx - dy * dy) / (dx * dx + dy*dy);    b   = 2 * dx * dy / (dx*dx + dy*dy);     x2  = Math.round(a * (p.x - x0) + b*(p.y - y0) + x0);     y2  = Math.round(b * (p.x - x0) - a*(p.y - y0) + y0);     p1 = Point((int)x2,(int)y2);      return p1;  } Next badge select dialog broken The "Select your next badge" dialog box is not displaying correctly. It looks like this:    Freehand circles, as requested:    It seems to be an issue with box-sizing, because when it's content-box, it looks fine. I think the width and height rules for .popup-badges .all-badge-progress .badge-progress should be increased a little, for example:  .popup-badges .all-badge-progress .badge-progress {   width: 214px;   height: 96px; }   Tested on the latest Chrome and Firefox on Linux Mint. In Android, is it possible to block apps from accessing the Internet? Many apps will often connect to remote servers in the background, including Google's own apps or system services. Over time this can eat away at a user's data limit set by a mobile carrier. Does Android allow an app to act as a firewall whereby the user can select which apps have access to the Internet? Or does the device have to be rooted to have that ability? Booking a flight without passport info I have been looking to book a bundle (hotel and flight) with Expedia to Puerto Vallarta MX. When I get to the booking section it does not ask for my passport number. Do they ask for my passport after I book? Or later when the date gets near? My passport is currently being renewed and I want to book my hotel and flight before they sell out but I am afraid it'll ask me for my passport number. Can anyone inform me about this situation? Differentiable but not continuously differentiable. Given $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as $$f(x)=\left\{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&amp;,x\neq 0\\ 0&amp;,x=0\end{array}\right\}.$$ I am trying to prove $f$ is differentiable at $x=0$ but not continuously differentiable there. Maximum and minimum of $f(x)=\cos(\sin(x))-\sin(\cos(x))$ Given the function: $$f(x)=\cos(\sin(x))-\sin(\cos(x))$$ it has absolute maxima at $x=(2k+1)\pi$ with $k=0,1,..N$ and relative maxima at $x=2k\pi$. It is not clear where are the minima. Putting the derivative to zero doesn't help. Any suggestion on how to find the minimum value and where is it? Thanks. Book about actress forced to disguise alien murderer I read this book once when I was in middle school. A girl solves a murder or a mystery on a spaceship. It was a thin paperback books.   All I can remember is that there was an alien, and there was a movie star or actress or famous makeup artist on the ship traveling with another woman. By the end of the book the murderer was the woman the actress was traveling with who forced her to use her makeup skills to disguise her, and she was really some kind of alien. 'longtable' caption without incrementing 'table'? For reasons both complex and confusing, I would like to caption a longtable without incrementing the table counter in the process.  I want to have two consecutive longtable environments to be captioned with identical table numbers; i.e. one longtable captioned "Table x.y: Data Set (Part A)" and one captioned "Table x.y: Data Set (Part B)", with x.y being identical for both tables.  Apparently however, even \caption*{} command bumps table. I am a bit at a loss here.  Note: I know that the xtab package would enable me to have one long table with switches of headers and captions mid-table. However I got a (different) problem with that package, akin to  but sadly not solved by egreg's answer there; so this question should be considered longtable-only as xtab, supertabular etc. don't float my boat right now. :-( An exciting new feature is described in the :        There is a new NPC housing interface that can be used to assign living areas to NPC’s. This interface will also tell you if a room is missing any requirements.      How can this interface be accessed? How to convert PDF to (La)TeX? I know that hardly any information is passed to the PDF when a .tex file is compiled.  But is there a tool that can convert a PDF document back to (La)TeX? What's the best way to write Mathematical Equations on the Web? This question led to a new package:      I'm giving my students an old exam and barely remembered to change the date at the top. A colleague and I joked about how long we could reuse the same exam, and it made me wonder if I could use LaTeX to create an exam that looked like it was written in 1963 on a typewriter.  Here's what I'd like to see:   Typewriter font even in math mode. (Several ways to accomplish this; is there a "best" way?) Slight inconsistencies in the printing of each letter. I attempt this below adapting code taken from . Each word has a random vertical shift, though I'd like to see it reduced to each letter having its own vertical shift. Other typewriter-like irregularities would be great. It would be great if math-mode delimeters \left(, \right}, etc., can be created in a piece-meal fashion as they did in days of old. Fraction bars, tops of square-root symbols -- any horizontal line, really -- is made up of a series of en-dashes that don't quite match up.   Here is what I've come up with so far:  \documentclass[10pt]{article}% This is a document class providing more font size options \usepackage{graphicx} \usepackage{ifthen} \usepackage[textwidth=7.5in,textheight=9.5in]{geometry} \pagestyle{empty}  % thanks to Bruno Le Floch: https://tex.stackexchange.com/q/9331/4012 % and in his comments to https://tex.stackexchange.com/a/29458/4012 \usepackage[first=-1,last=1]{lcg}% you can play around with these values \makeatletter \newcommand{\globalrand}{\rand\global\cr@nd\cr@nd} \makeatother  \newcommand{\randomvshift}[1]{\globalrand\raisebox{\value{rand}pt}{#1}}  %%% thanks to Martin Scharrer: https://tex.stackexchange.com/q/11598/4012 \makeatletter \def\typewriter#1{%     \@typewriter#1 \@empty } \def\@typewriter#1 #2{%    \randomvshift{#1}\space    \ifx #2\@empty\else     \expandafter\@typewriter    \fi    #2% } \makeatother \begin{document}\tt  \typewriter{Math 101 Fall 1963}  \begin{enumerate} \item   \typewriter{$\frac{\texttt{d}}{\texttt{dx}}\texttt{(x)}^2$}  \item \typewriter{State the Fundamental Theorem of Calculus.}  \end{enumerate}  \end{document} Lagrange Multipler KKT condition I am looking at SVM, and reviewing Lagrange multiplier. Let's say with the constraint function $$g(x,y) = x^2 + y^2 -1 &gt; 0, $$ I am maximizing $$f(x,y) = x + y -1 .$$ Intuitively, since the constraint function does not provide a finite constrained region, there is no solution, and $f(x,y)$ can infinitely increase.  However, if I just plainly proceed with Lagrange multiplier calculation,  $$L(x,\lambda) = (x + y -1) +\lambda(x^2 + y^2 -1)$$ I get the solution: $$y,x = -\sqrt{2}/2, \lambda = 1/\sqrt{2}$$ And this still satisfies KKT conditions although the solution is wrong.  $$g(x) =&gt; 0 , \lambda =&gt;      0, \lambda g(x) = 0.$$  I expected that I would have some result that conflicted KKT conditions, but nothing was violated.   Does this mean that Lagrange multiplier under KKT conditions can produce a solution that cannot be solved, without giving any indication of intractability?  If functions are complicated, what's the systematic way to confirm that the optimization with inequality constraint functions is solvable using Lagrange multiplier? How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$? I am curious about the practical implementation of a binary split in a decision tree - as it relates to levels of a categorical predictor $X{j}$.  Specifically, I often will utilize some sort of sampling scheme (e.g. bagging, oversampling etc) when building a predictive model using a decision tree - in order to improve its predictive accuracy and stability. During these sampling routines, it is possible for a categorical variable to be presented to a tree fitting algorithm with less than the complete level set.  Say a variable X takes on levels {A,B,C,D,E}. In a sample, maybe only levels {A,B,C,D} are present. Then, when the resulting tree is used for prediction, the full set may be present.  Continuing from this example, say a tree splits on X and sends {A,B} to the left and {C,D} to the right. I would expect the logic of the binary split to then say, when faced with new data: "If X has value A or B, send to the left, otherwise, send this case to the right". What seems to happen in some implementations is "if X has value A or B, send to the left, if X has value C or D send to the right". When this case takes on value E, the algorithm breaks down.   What is the "right" way for a binary split to be handled? It seems the much more robust way is implemented often, but not always (see Rpart below).  Here are a couple examples:  Rpart fails, the others are ok.  #test trees and missing values  summary(solder) table(solder$PadType)  # create train and validation set.seed(12345) t_rows&lt;-sample(1:nrow(solder),size=360, replace=FALSE) train_solder&lt;-solder[t_rows,] val_solder&lt;-solder[-t_rows,]  #look at PadType table(train_solder$PadType) table(val_solder$PadType) #set a bunch to missing levels(train_solder$PadType)[train_solder$PadType %in% c('L8','L9','W4','W9')] &lt;- 'MISSING'   #Fit several trees, may have to play with the parameters to get them to split on the variable  ####RPART mod_rpart&lt;-rpart(Solder~PadType,data=train_solder) predict(mod_rpart,val_solder) #Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = attr(object,  :  #factor 'PadType' has new level(s) D6, L6, L7, L8, L9, W4  ####TREE mod_tree&lt;-tree(Solder~PadType,data=train_solder,split="gini") predict(mod_tree,val_solder) #works fine  ####ctree mod_ctree&lt;-ctree(Solder~PadType,data=train_solder,control = ctree_control(mincriterion = 0.05)) predict(mod_ctree,val_solder) #works fine How to make grass appear on top of ground not in middle? Find the limit of $\frac{n^{x}}{(1 + x)^{n}}$ Let $x&gt;0$. Find $$\lim_{n\to\infty}\frac{n^{x}}{(1 + x)^{n}}.$$   What I was thinking involved an inequality, by reason of considering the fact that  $$(1 + x)^{n} &gt; C(n, k) x^{k}$$ for some $k$ such that $n&gt;k$. Expanding the binomial coefficient and taking the $n$ common from each bracket, we have something like $$(1 + x)^{n} &gt; C(n, k) x^{n} &gt; \frac {x^{k}n^{k}}{k! 2^{k}}.$$ But now, I am totally clueless. By intuition, I know that the limit has to be zero, so I just want to prove using the concept of inequalities that what I thought was the right approach (being the sandwich theorem). Any help would be appreciated! How to cancel a jquery.load()? In GLM regression I have always been told to set the reference level of categorical/ordinal/dummy variables to the level with the most exposure (level with most data), because this somehow makes the model more stable. Is there any statistical reason behind this, or just historical practice without statistical reason?  I can understand that one wants a "baselevel" (reference level) in the model that are most likely to be correct, but if you are not sure about the other levels, then they surely should not be included in the regression in the first place?