Suppose I have what I think is a great idea for a new Settlers expansion. (Like 5,000 other people, no doubt!) What would be the most effective way to realize this possibility? Write to the makers and outline the idea? Pitch it like a novel manuscript? Or do such things not get publicly sourced? If I want to play it, is my best bet just to make my own set for use by friends and family? This question doesn't have to be about Settlers; it can be read as being about any game in the market. There are so many editions &amp; flavours of all the most popular games, and I guess the core question is: Are they all generated internally or could a community member propose one?
I made a Yu-Gi-Oh card but it is not broken like current broken Yu-Gi-Oh cards and here it is to see but please before you make your decision about it take another look on ''dragun of red eyes''.
When do you use the "this" keyword? I was curious about how other people use the this keyword. I tend to use it in constructors, but I may also use it throughout the class in other methods. Some examples:  In a constructor:  public Light(Vector v) {     this.dir = new Vector(v); }   Elsewhere  public void SomeMethod() {     Vector vec = new Vector();     double d = (vec * vec) - (this.radius * this.radius); } Is there a website that lets you shop for hotels with taxes and fees included in search results? Every time I book a hotel online, the price increases by 25% in "taxes and fees" at checkout, compared to the price featured in search results.  This does not occur when booking flights (I actually think there's a regulation against it for flights).  Is there a website that lets you compare the ACTUAL price of hotels with taxes and fees included, and also lets you book? Where are the review queue counters? The review queue counters have disappeared:        Where are they?  When I did screenshot the Close Votes queue was not empty, if someone think so. Difference between == and === in JavaScript I am looking for a clean and fast method to force GNOME in Ubuntu 14.04 to switch between active apps only on the current workspace when pressing the shortcut alt + tab. Best way to hide a window from the Alt-Tab program switcher? I've been a .NET developer for several years now and this is still one of those things I don't know how to do properly. It's easy to hide a window from the taskbar via a property in both Windows Forms and WPF, but as far as I can tell, this doesn't guarantee (or necessarily even affect) it being hidden from the Alt+↹Tab dialog. I've seen invisible windows show up in Alt+↹Tab, and I'm just wondering what is the best way to guarantee a window will never appear (visible or not) in the Alt+↹Tab dialog.  Update: Please see my posted solution below. I'm not allowed to mark my own answers as the solution, but so far it's the only one that works.  Update 2: There's now a proper solution by Franci Penov that looks pretty good, but haven't tried it out myself. Involves some Win32, but avoids the lame creation of off-screen windows. Determine if the convergence of $f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ is uniform. $f_n(x) = \sqrt{x^2 + \frac{1}{n^2}}$ converges pointwise in a set $E = [0, \infty)$ to $f(x) = x$.   This problem reminds me a lot of how $\frac{x}{n}$ fails to converge uniformly to $f=0$ on $[0,\infty]$, but does converge uniformly on $[0,a]$. However, when we proved that we used the definition of uniform convergence. I do not wish to do that here.  I wish to use the Weierstrass test here. And that is where I'm stuck. I cannot find $M_n = \sup\limits_{x \in E} |f_n(x)-f(x)|$. Can someone help me out here? Prove that there is a bijection between the set of all subsets of $X$, $P(X)$, and the set of functions from $X$ to $\{0,1\}$. Given any set $X$, let $P(X)$ be the set of all subsets of $X$, and let $\{0,1\}^X$ be the set of all functions $X \rightarrow \{0,1\}$. Construct a bijection (and its inverse) between P(X) and $\{0,1\}^X$.    Let us define $f: P(X) \rightarrow \{0,1\}^X$ by $f(\alpha) = g_{\alpha}$ where $\alpha \in P(X)$ and $g_{\alpha} : X \rightarrow \{0,1\}$ is defined by  $$g_{\alpha}(x) =\begin{cases}        1 &amp; x \in \alpha \\       0 &amp; x \in X-\alpha     \end{cases}$$  We need to check if this function is well-defined. Let $\alpha, \beta \in P(X)$. If $\alpha=\beta$, then $\alpha$ and $\beta$ correspond to the same set, and so correspond to the same function since both $g_{\alpha}$ and $g_{\beta}$ send that set to 1 and the ($X-$that set) to 0.   If we have a function $g_{\alpha} \in \{0,1\}$, then $\alpha$ is a set in $X$ that is sent to 1. But since $P(X)$ is the set of all sets of $X$, we must have $\alpha \in P(X)$. So the function is surjective.  Finally, if $g_{\alpha}=g_{\beta}$ then both $g_{\alpha}$ and $g_{\beta}$ send the set $\alpha$ or $\beta$ to 1. So $\alpha$ and $\beta$ must be same set. In other words, the function is injective; and hence bijective.   The inverse function is $f^{-1}: \{0,1\}^X \rightarrow P(X)$ where $f^{-1}(g_{\alpha}) = \alpha$.     Another way to prove this is to first construct the inverse function, show that it is well defined, and then show that $ff^{-1}$ and $f^{-1}f$ are the identity functions, right?  So if we let $g_{\alpha}=g_{\beta}$, then $\alpha=\beta$, because otherwise $g_{\alpha} (\beta)=0$ and $g_{\beta} (\beta)=1$; a contradiction.   Now we have $f \circ f^{-1} (g_{\alpha}) = f(\alpha) = g_{\alpha}$ and $f^{-1} \circ f(\alpha) = f^{-1}(g_{\alpha}) = \alpha$ , and so $f$ and $f^{-1}$ must be bijective. Deleting self-answered question gets "Delete this answered question" warning What are the basics of commands in Minecraft Java Edition? If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. How to choose a predictive model after k-fold cross-validation? What happened to this user's rep, badges, and name? This user is only displaying 56 rep.   But surely (s)he has much more than that.        The user has also never given out any bounties.  Also, badges are missing: How do I get permissions to delete files on Windows 7? I updated my laptop's OS from Windows XP to Windows 7. There are some leftover files from Windows XP on the computer now. If I try deleting them I get the following error:     You need permission to perform this action.      You require permission from S-1-.... to make changes to this folder.   What's weird is that I am logged in with the only user account on this machine and I have administrator privileges. I tried turning UAC off, but I still can't delete the files.  How can I force removal of these files? What happened to Rhaegal? In Game of Thrones S08E03 ("The Long Night"), the Night King's dragon tore at the one Jon Snow rides (Rhaegal). He falls off of it and we don't see the dragon anymore, is it dead? I'm still learning, but I'm trying to follow a tutorial to apply texture to a model. I only want to apply the texture to one object in the scene, so I selected it, unwrapped it, and painted over it. Now I selected it in the Edit Mode (I can't select it in the Object mode because all is locked together) and went into Node Editor, but the nodes are greyed out.  Why? $(\Bbb Z\oplus \Bbb Z)[X] \not\cong \Bbb Z[X]\oplus \Bbb Z[X]\,$? Well i tried to prove it but failed. They are both commutative and have the same set of units. So i can't see a property of one ring which the other ring does not possess. Thank you for any help. In the King James version of the Bible there is a verse like this:     The Lord is my strength, and my fortress, and my song. And He is become my salvation.   Is it still feasible to use "is become" instead of "has become" and what is the semantic difference? Selenium/PhantomJS raises error How to stop the find command after second match? How to make find command, that will show only first two matches and then end, something similiar to  just version for two records.