Referenced Objects in List Change on Contents Change, but Not Setting to NULL
Is Java "pass-by-reference" or "pass-by-value"?
Ubuntu 18: Top bar for all monitors I just installed Ubuntu 18. Since the top bar is only on the primary monitor, I can't use the mouse to grab and move my Chrome window (the top of the screen looks like below). Trying to do so results in grabbing a single tab and opening it in a new window.     Is there a way around this? Ideally, I'd like to have the top bar  across all screens. Another obvious disadvantage of only having the top bar on one screen is that the window menu of any window, not just Chrome, is displayed on the left upper corner of my left screen even when the window is in the lower right corner of the right screen. What techniques are available for collapsing (or pooling) many categories to a few, for the purpose of using them as an input (predictor) in a statistical model?    Consider a variable like  (discipline chosen by an undergraduate student). It is unordered and categorical, but it can potentially have dozens of distinct levels. Let's say I want to use major as a predictor in a regression model.  Using these levels as-is for modeling leads to all sorts of issues because there are just so many. A lot of statistical precision would be thrown away to use them, and the results are hard to interpret. We're rarely interested in specific majors -- we're much more likely to be interested in broad categories (subgroups) of majors. But it isn't always clear how to divide up the levels into such higher-level categories, or even how many higher-level categories to use.  For typical data I would be happy to use factor analysis, matrix factorization, or a discrete latent modeling technique. But majors are mutually exclusive categories, so I'm hesitant to exploit their covariance for anything.  Furthermore I don't care about the major categories on their own. I care about producing higher-level categories that are coherent with respect to my regression outcome. In the binary outcome case, that suggests to me something like linear discriminant analysis (LDA) to generate higher-level categories that maximize discriminative performance. But LDA is a limited technique and that feels like dirty data dredging to me. Moreover any continuous solution will be hard to interpret.  Meanwhile something based on covariances, like multiple correspondence analysis (MCA), seems suspect to me in this case because of the inherent dependence among mutually exclusive dummy variables -- they're better suited for studying multiple categorical variables, rather than multiple categories of the same variable.  edit: to be clear, this is about collapsing categories (not selecting them), and the categories are predictors or independent variables. In hindsight, this problem seems like an appropriate time to "regularize 'em all and let God sort 'em out". Glad to see this question is interesting to so many people! byte[] to hex string How do I convert a byte[] to a string? Every time I attempt it, I get      System.Byte[]    instead of the value.  Also, how do I get the value in Hex instead of a decimal? The Fibonacci sequence is defined recursively by $F_1 = 1, F_2 = 1, \; \&amp; \; F_n = F_{n−1} + F_{n−2} \; \text{ for } n ≥ 3.$ Prove that $2 \mid F_n \iff 3 \mid n.$   Proof by Strong Induction : $\bbox[5px,border:1px solid green]{\color{green}{n = 1 }}$ $2 \mid F_1$ is false. Also, $3 \mid 1$ is false. The implication [False $\iff$ False] is vacuously true.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Induction Hypothesis}}}$ Assume that $2 \mid F_i \iff 3 \mid n$ for every integer $i$ with $1 ≤ i ≤ k$.  $\bbox[5px,border:1px solid green]{\color{green}{k + 1 \text{th Case}}} \;$  To prove:  $\quad 2 \mid F_{k + 1} \iff 3 \mid k + 1.$   $\bbox[5px,border:1px solid green]{\color{green}{n = k + 1 = 2}} \;$ $2 \mid F_2$ is false. Also, $3 \mid 2$ is false. So [False $\iff$ False] is vacuously true.  Hence assume that $k + 1 ≥ 3.$ We now consider three cases:  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 1: } k + 1 = 3q}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 \require{cancel}\cancel{\mid} (k − 1)$. By the ind hyp, $3 \require{cancel}\cancel{\mid} k    \iff F_k$ odd &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Since $F_{k+1} = F_k + F_{k−1}$, thus $F_{k+1}$ = odd + odd = even.  $\bbox[5px,border:1px solid green]{\color{green}{\text{Case 2: } k + 1 = 3q + 1}}$  Thus $3 | k$ and $3 \require{cancel}\cancel{\mid} (k − 1).$ By the ind hyp, $3 | k \iff F_k$ even &amp; $3 \require{cancel}\cancel{\mid} (k − 1) \iff F_{k - 1}$ odd. Thus $F_{k+1}$ odd.  $\bbox[5px,border:1px solid green]{\color{green}{{\text{Case 3: }} k + 1 = 3q + 2}}$ Thus $3 \require{cancel}\cancel{\mid} k$ and $3 | (k −1).$ By the ind hyp, $3 \require{cancel}\cancel{\mid} k \iff F_k$ odd and $3 \mid (k − 1) \iff F_{k - 1}$ even. Thus $F_{k+1}$ odd. $\blacksquare$  $\Large{1.}$ Does the proof clinch the $(\Leftarrow)$ of the $(k + 1)$th case?   $\Large{2.}$ Since the recursion contains $n, n - 1, n - 2$, thus the recursion "time lag" is $3$ here. So shouldn't $3$ base cases be checked?   $\Large{3.}$ Further to #2, shouldn't "assume $k + 1 \geq \cancel{3} 4$" instead?   $\Large{4.}$ Shouldn't the $n = k + 1 = 2$ case precede the induction hypothesis?  I referenced . Source: Exercise 6.35, P152 of Mathematical Proofs, 2nd ed. by Chartrand et al    Supplement to peterwhy's Answer:  $\Large{1.1.}$ I wrongly believed that all 3 Cases proved the $\Leftarrow$. I now see that Case 1 is $\Leftarrow$ via a Direct Proof. Cases 2 and 3 are $\Rightarrow$ via a Proof by Contraposition. Nonetheless, how would one foreknow/prevision to start from $3 \mid n$  for both directions of the proof? Delaunay triangulation to generate abstract networks and nodes Is it possible to use Delaunay triangulation to generate an image like the below from a raster or vector of an object (in this case, a human heart)? What software and techniques would be required to make something like this? My intention is to make and use something like this in Inkscape without having to draw all the nodes/lines/effects manually. I'm trying to create globally-unique identifiers in JavaScript.  I'm not sure what routines are available on all browsers, how &quot;random&quot; and seeded the built-in random number generator is, etc. The GUID / UUID should be at least 32 characters and should stay in the ASCII range to avoid trouble when passing them around. Why does the standard error of the intercept increase the further $\bar x$ is from 0? The standard error of the intercept term ($\hat{\beta}_0$) in $y=\beta_1x+\beta_0+\varepsilon$ is given by $$SE(\hat{\beta}_0)^2 = \sigma^2\left[\frac{1}{n}+\frac{\bar{x}^2}{\sum_{i=1}^n(x_i-\bar{x})^2}\right]$$ where $\bar{x}$ is the mean of the $x_i$'s.  From what I understand, the SE quantifies your uncertainty- for instance, in 95% of the samples, the interval $[\hat{\beta}_0-2SE,\hat{\beta}_0+2SE]$ will contain the true $\beta_0$. I fail to understand how the SE, a measure of uncertainty, increases with $\bar{x}$. If I simply shift my data, so that $\bar{x}=0$, my uncertainty goes down? That seems unreasonable.  An analogous interpretation is - in the uncentered version of my data, $\hat{\beta}_0$ corresponds to my prediction at $x=0$, while in the centered data, $\hat{\beta}_0$ corresponds to my prediction at $x=\bar{x}$. So does this then mean that my uncertainty about my prediction at $x=0$ is greater than my uncertainty about my prediction at $x=\bar{x}$? That seems unreasonable too, the error $\epsilon$ has the same variance for all values of $x$, so my uncertainty in my predicted values should be the same for all $x$.  There are gaps in my understanding I'm sure. Could somebody help me understand what's going on? Are more megapixels good?  Are more megapixels bad?  Do more megapixels increase detail? Do they make my images sharper? On the other side, is there a point that is too much? Do megapixels cause increased noise and other problems? How does print and viewing size come into it?  Widespread Internet-forum wisdom used to be that 6 megapixels was the sweet spot — below that just wasn't enough, but above that, there wasn't much benefit. File sizes got larger, but detail was lost to noise and other problems. The arguments are that cramming too many pixels into a small sensor makes each pixel too small to provide any real benefit, and that higher-megapixel sensors outresolve cheap lenses anyway. (Substitute something like 12 instead of 6 if we're discussing APS-C DSLRs, or 24 for full-frame.)  By 2011,  was in the range of 12-18 megapixels. In 2016 most were are in the 16-24 megapixel range. Certainly none are at 6 anymore; even smartphones come with 8 megapixel cameras. Does this offer any real improvement over that old "sweet spot"? Has technology improved to the point where the "wisdom" needs to be updated, or are we all suffering for marketing? Or have we gone past the sweet spot in some ways, but it's okay because of previously-un-argued points. (For example: more noise, sure, but more detail as well.)  Within the 16-24 megapixel common-today range, for the same sensor size, is there any actual benefit to the higher end? How do megapixels directly affect image quality with today's technology? What are the benefits and when do they apply? What are the drawbacks, and when do they apply? How should I adjust my technique (and expectations) based on my camera's megapixel count? How OPTIONS are settled can be an eye opener and I hope I can get some answers The only reason why I put it here is to find out whether I am a victim or just ignorant of the way brokers deal with option positions that were exercised. Whether I am a victim or just ignorant depends on the answer that I posted here. The facts are: I had a 2 SHORT QS Option at 75 which was exercised against me on the 23 Dec 00.45 AM and I later exercised my 2 LONG QS Option at 70 on 23 Dec 9.31 AM. The broker charged me Borrow Fee for 24-27 Dec for $1102, which means instead of getting $1000 profit I turn this into a loss of $102 plus comm. My question:  Since both long and short options were exercised on the same day one at 45 mins on 23 Dec and the later at 9.31 am on the same day, how can there be a SHORT POSITION in the stocks since both are net-off in my account ?  The answer I got from my broker and this is the point that I want to clarify.  The response I got from IB was that the initial exercise at 45 mins on 23 Dec was actually on 22 Dec (not 23 as I alleged). They refused to show me how they got 22 Dec.  This is my explanation to my Broker in reply.  Even if the short call and long call was exercised on different dates 22 and 23 Dec respectively, IB can only charge me for the SHORT position for one day and not from 24-27 Dec by ignoring my long call position was not there.  Further Facts:  My Broker ignored my answer and insisted the dates are correct and therefore I was charged to borrow shares that was need to satisfy the short position on exercised. So I am lost as I do not know whether the broker is correct and they can do this by ignoring my long call position after exercised or there is something they had not tell me. That is the reason I came to this site soliciting answers from those who know more than I do or had experienced the same.  I hope I have explained this clearly now and why would I want to give an opinion when I am looking for an answer ? Delete commits from a branch in Git Computing: $\int_0^1 \sqrt[3]{x\log{\frac{1}x}} \, \mathrm dx$ Cross-referencing problem if counter of theorem-like environments is shared The problem is when i share numbering the \cref can't show name, even by \crefname and \autoref show all name Equation.  Here is an MWE:  %&amp;latex \documentclass{amsart}  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % If you want to insert other packages. Insert them here \usepackage{amsthm} \usepackage{hyperref} \usepackage[nameinlink]{cleveref} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  \begin{document}  \numberwithin{equation}{section}  \newtheorem{The}[equation]{Theorem} \crefname{The}{Theorem}{Theorems}  \newtheorem{Dfn}[equation]{Definition} \newtheorem{Lem}[equation]{Lemma} \newtheorem{Prs}[equation]{Proposition} \newtheorem{Crl}[equation]{Corollary} \newtheorem{Rmk}[equation]{Remark}  \begin{The}\label{T1}    Assume ... \end{The}  ... \begin{Dfn}\label{D1}    Assume ... \end{Dfn}  ... \begin{Lem}\label{L1}    Assume ... \end{Lem}  ... \begin{Prs}\label{P1}    Assume ... \end{Prs}  ... \begin{Crl}\label{C1}    Assume ... \end{Crl}  ... \begin{Rmk}\label{R1}    Assume ... \end{Rmk}  \begin{equation}\label{E1}    Assume ... \bigskip \end{equation}  \noindent \cref{T1} \\ \cref{D1} \\ \cref{L1} \\ \cref{P1} \\ \cref{C1} \\ \cref{R1} \\ \cref{E1} \\  \noindent \autoref{T1} \\ \autoref{D1} \\ \autoref{L1} \\ \autoref{P1} \\ \autoref{C1} \\ \autoref{R1} \\ \autoref{E1} \\  \end{document}     How can I solve it? Returning anonymous functions from lapply - what is going wrong? How do I make the Super key open the activities overview in Ubuntu 20.04 LTS? Classifying whether a collection of objects is a set I'm wondering whether or not there is a definite criterion for determining whether a collection of objects is a set. In the literature, I've recently encountered some obscure examples (in Riemann Surface Theory, to be precise) of naturally-arising collections of objects which are not sets.   Here is some pseudo-rigor to make the question more precise:   Let $S$ be a collection of objects. Is there some property $P$ so that $S$ is a set iff $S$ has property $P$? Proving convergence of sums: If $\sum_{n=1}^{\infty}a^2_n$ and $\sum_{n=1}^{\infty}b^2_n$ converge then $\sum_{n=1}^{\infty}a_nb_n$ converges In grade school we learn to rationalize denominators of fractions when possible. We are taught that $\frac{\sqrt{2}}{2}$ is simpler than $\frac{1}{\sqrt{2}}$.  says that "there is a bias against roots in the denominator of a fraction". But such fractions are well-defined and I'm failing to see anything wrong with $\frac{1}{\sqrt{2}}$ - in fact, IMO it is simpler than $\frac{\sqrt{2}}{2}$ because 1 is simpler than 2 (or similarly, because the former can trivially be rewritten without a fraction).  So why does this bias against roots in the denominator exist and what is its justification? The only reason I can think of is that the bias is a relic of a time before the reals were understood well enough for mathematicians to be comfortable dividing by irrationals, but I have been unable to find a source to corroborate or contradict this guess. A continuous function that attains neither its minimum nor its maximum at any open interval is monotone Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f$ attains neither its minimum nor its maximum at any open interval $I \subseteq  \mathbb R$ , then how to prove that $f$ is monotone ? For any triangle with sides $a$, $b$, $c$, show that $a^2b(a-b)+b^2c(b-c)+c^2a(c-a)\ge 0$ How can I work with file uploads during a Webdriver test? This is a canonical question, intended to provide a comprehensive answer to many related questions  During the course of a Webdriver test, I want to upload files. However, this opens a native file picker, which Webdriver doesn't seem to support interacting with. How do I interact with it?