No shadow with Eevee
In eevee, shadows are clipped by camera distance
When you discard cards at the beginning of the game, where do they go? The appearance is that they get shuffled back into the deck. However, after discarding key cards and not receiving them the whole game, I am now wondering if they get put in the bottom of the deck. Where do they go after being discarded? What is the difference between sector and cluster? I have gone through wikipedia and other sites and here's the information I have obtained.  Sector: Smallest Addressable Unit (mostly 512 bytes) Cluster: Smallest Allocation Unit (equals to 1 or more sectors and the number of clusters depends on the file system).  Let's say in the below example:  sector =512 bytes and a cluster =8 sectors=4096 bytes. Now, I am having a hard time figuring out, when I am trying to create a file what does the file system do? Does it consider 4096 bytes as the smallest write/read unit or 512 bytes?  If the answer is 4096 bytes: When the manufacturer of the disk drives has provided 512 bytes as the smallest addressable unit, why does the file system consider cluster as the smallest unit? I mean, what are the advantages in doing so? Why does an conditionally convergent infinite series, when deranged, have a different limit? $\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit? First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g. \begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*} one gets the sum of first n terms as different functions, but the series's limit remains. One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason. Are there other explanations? Is the use here of 'countably many' is proper? Why is Newton's method not widely used in machine learning? This is something that has been bugging me for a while, and I couldn't find any satisfactory answers online, so here goes:  After reviewing a set of lectures on convex optimization, Newton's method seems to be a far superior algorithm than gradient descent to find globally optimal solutions, because Newton's method can provide a guarantee for its solution, it's affine invariant, and most of all it converges in far fewer steps. Why is second-order optimization algorithms, such as Newton's method not as widely used as stochastic gradient descent in machine learning problems? An instructor has an exam of 5 questions and a question bank of 50 questions. Are there $50\choose 5$ or $(50)_5$ possible exams? Where $(50)_5 = 50! / 45! $ and $50\choose 5$ $= 50!/(5!45!)$   I've narrowed it down to these two possibilities (though even these may be wrong), but I think I'm confusing myself as to which one I use. Can anyone suggest a good way of thinking of this problem that will make it apparent whether I need use use the falling factorial or the binary coefficient?   Edit: thank you for all the answers! Indeed, I meant $50!/45$ as the falling factorial. Also, the question doesn't specify whether it was meant to be ordered or unordered, but I think I understand my confusion now! Animated textures don't work in Blender viewport playback I am trying to animate in 2d. I do it with planes that have textures on them. I recently came up with the idea of having a plane with an image sequence of mouths because animating actual lip syncing by hand takes too long and isn't really necessary for what I'm trying to acomplish. However, when playing back the animation in textured viewport shading the animated texture stays at the last frame until the animation stops. I can use the arrow keys and it works fine as well as scrubbing through the timeline, just not when playing back the animation with alt+a. Can anyone help me with this problem? How to to calculate the hash of an unordered set Suppose I have a set of elements, with known hash (e.g. SHA-2). How can I calculate the hash of the set? With it I mean an unordered set, so the order of elements is undefined and shall not play any role in determining the hash of the set.  In theory any commutative operation applied to the hashes should be fine, but I have the strong feeling that XORing all of them together is quite unsafe. An attacker could add an element to the set to produce a hash collision, although finding an element with that wanted hash is as hard as breaking the hash of the whole set.  Anything better solution than the trivial one of sorting the elements into a list, calculating their hashes and then calculating the hash of the hash list?  I think Merkle trees can't help here if they are based on some sort of ordering of the elements. The new profile shouldn't mistakenly imply that you can earn the Informed badge on a meta site I just joined Sports, and I decided to take a look on Sports Meta. I clicked on my profile there. This is part of what I saw:     Yet the profile shouldn't indicate that it is, in theory. . .  This seems to be a bug. Self-deleted posts and the 3 vote / 60 day rule With regards to  post about  and the part about retaining reputation earned from deleted posts that have a score of 3 or better and were visible on the site for 60 days, does this also apply to self-deleted posts and will it continue to do so? I culled a few of my own answers over the weekend that were less than stellar, and my reputation dropped 1.8k, but the rep is back today after the changes were announced. I thought it would be useful to clarify that this also applies to self-deleted stuff. A man enters a comatose author's dreams to wake him up How do I install Firefox 52 ESR on 16.04? This  asks how to install FF32 bit while keeping the 64 bit.  I want the much more simple thing: is there a canonical way to install firefox 52 ESR as "default" firefox in my 16.04?  I found this  ... but there are no packages; probably no wonder given the fact that the linked blog article dates back to 2012.  So, 2017, Ubuntu 16.04; how to install FF 52 ESR (as I simply need the java plugin to work for me).  ( my current solution: I downloaded the TAR file from mozilla, extracted that to /opt; and then changed /usr/bin/firefox to be a link to that ESR version of firefox. works, but ugly ) Direction arrows on a circle I would like to plot circle (or another curve) with positive direction (and show it wit arrows).  This plots the circle  \documentclass{article} \usepackage{tikz} \RequirePackage{pgfplots} \usetikzlibrary{shadows} \usetikzlibrary{shapes} \usetikzlibrary{decorations} \usepackage{verbatim} \usepackage[active,tightpage]{preview} \PreviewEnvironment{tikzpicture} \setlength\PreviewBorder{5pt} \begin{document} \begin{tikzpicture}     \begin{axis}[             xmin=-1.5,xmax=2.5,ymin=-1.5,ymax=1.5,             axis lines=center,             ticks=none,             unit vector ratio*=1 1 1,            ]                   \addplot [very thick, smooth, domain=(0:2*pi)] ({1.+cos(deg(x))},{sin(deg(x))});        \addplot[only marks,mark=*] coordinates{(1,0)};         \end{axis} \end{tikzpicture} \end{document}   And I want to add arrows like that    Would you advise me how to add such arrows? Incubated but not hatched egg I got the message "oh" and it showed my egg. Before I could do anything the server went down. When I got back on my egg was gone. Is there anything I can do to retrieve my incubated but not hatched egg? How do I change a user name in Windows 7? How do I change extension of multiple files recursively from the command line? If I have the following code:  MyClass pClass = new MyClass(); pClass.MyEvent += MyFunction; pClass = null;   Will pClass be garbage collected? Or will it hang around still firing its events whenever they occur? Will I need to do the following in order to allow garbage collection?  MyClass pClass = new MyClass(); pClass.MyEvent += MyFunction; pClass.MyEvent -= MyFunction; pClass = null; Show $\,p\mid 9^q -1 \implies q\mid p-1$ for odd primes $p$ and $q$ If $p$ and $q$ are odd primes show that $$p|(9^q -1) \implies q|(p-1).$$ ....  I am having difficulty to prove this.  Yes that was the question I have misunderstood between 9 and q!!! Adding boolean yes/no field in ArcGIS Desktop? How does one add a field to an ArcGIS feature class with a boolean data type? That is an attribute where the allowed value is only one of a pair such as 1/0, Yes/No, On/Off, Present/Not-present, etc.   A boolean data type is not available in the Data Type pick-list. Morphing only a part of a full path in tikz This might be a tikz newbie question but I have some trouble solving this.  I want to draw and fill a rectangular path where path morphing should be applied only to the upper side of the shape, more specifically I want to have an upper part with random steps. The rest of the path (left, lower and right sides) should just be straight lines. The shape should be closed and filled with one color.   My problem is that I cannot figure out how to apply the morphing only to the upper side and I suppose I cannot break up the path into different subpaths since I want to fill the final shape. Is there a way to use the morphing only on a part of the full path? Another problem is that the cycle command does not really close the shape at the upper left corner.  I tried to find a solution in the tikz manual but there was nothing that really helped me.  Here is some example code:  \documentclass[tikz, border=10pt]{standalone} \usepackage{tikz}  \begin{document}  \usetikzlibrary{decorations.pathmorphing} \begin{tikzpicture}  \draw[decorate, decoration={random steps,segment length=3.0mm, amplitude=2.0mm}, thick, rounded corners=1ex, fill=brown!70] (0,0) -- (4,0) -- ++(0, -1) -- ++(-4,0) -- cycle;   \end{tikzpicture} \end{document} Nautilus doesn't launch on Ubuntu