While I would like to ask this question of many other games the answer to each system of rules would undoubtedly be vary different and it'd be too heavy-handed a question to address.   So In regards to D&amp;D 3.5 alone, as you gain levels and the BAB raises with your characters thus the number of attacks they are capable of in a turn, and the monsters you face similarly begin gaining more attacks or perhaps in unfortunate situations more tentacles, heads and other nasty bits, are there any simple tricks for keeping combat moving, driving and exciting?   I do enjoy the rp sections of the game and tend to allow the narrative to absorb most of our campaigns (which my group is quite fond of as well), but of course every now and again you need a really good epic battle with a dragon, and it's a shame if the fight with said dragon ends up disappointing because one round took up our entire play session.  Such examples might include: better ways to track the calculations the fighters need to make every time they attack? Quick tricks to save time like rolling attack and damage at the same time. That sort of thing.   I'm sure there are many of you out there who love D&amp;D and have struggled with this and found your own brilliant solutions to such problems without having to put D&amp;D on a bookshelf and play something else.
I'm looking for ways to speed up the combat in my D&amp;D 3.5 game. It seems that turns - both player and non-player - simply take too long. I have thought about giving everyone a 30 second timer, but I'm afraid that might inhibit their ability to come up with strategy and tactics.
Is it safe to eat potatoes that have sprouted? I'm talking about potatoes that have gone somewhat soft and put out shoots about 10cm long.   Other online discussions suggest it's reasonably safe and the majority of us have been peeling and eating soft sprouty spuds for years. Is this correct? Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots. Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.  My attempt:  Consider the polynomial $g=(x-1)(1+x+x^2+x^3+\cdots+x^n)$  As $f\mid g, g$ all the roots of $f$ are roots of $g$.  This means I have to prove the statement:  If $c$ is a multiple root of $f \Longrightarrow c$ is multiple a root of $g$.  Equivalently:  If $c$ is not a multiple root of $g \Longrightarrow c$ is not a multiple root of $f$.  I'll try to prove this last statement:  Suppose $\exists c$ such that $c$ is a multiple root of $g$, then   $$g(c)=0 \implies g'(c)=0$$  $$ g=(x-1)(1+x+\cdots+x^n) \implies  g'=nx^n $$ $$ g'(c)=nc^n=0 \implies c=0 $$ But  $$ g(0)\neq 0 $$ And we've found the contradiction: $\not \exists c: g(c)=g'(c)=0$. Then $f$ has no multiple roots.  Is this correct?   Also: Assuming this is correct, is it possible to prove $\sum_{k=0}^n \frac {x^n}{n!} $ has no multiple roots in a similar manner? There is a remote Windows server on a private network which I can connect to via Remote Desktop Connection.  I would like to be able to make TCP/IP connections from my computer to other computers on that server's network.  Remote Desktop Connection makes it possible to share printers, drives, and other local resources through the connection.  Is there any way to "tunnel" a TCP/IP connection via RDC?  I'd like something similar to the port-forwarding provided by SSH.  I don't see any way to do this via RDC, but I'm hoping the capability is there and I just don't know about it. Applying PCA to data that fail a normality test "Arnold raced out of the door": grammatical or not? I use a Full HD 24" external display besides my 15" MacBook Pro Retina, on which I use the option for most screen estate regarding the Retina display. In the display arrangement screen, the MacBook Pro is therefore a bigger rectangle than the external display. Is there a way to size these rectangles regarding their "real world size" instead of resolution? I'm annoyed by my cursor teleporting vertically whenever I move from one display to another. Getting polygon boundaries of City in JSON from Google Maps API? Prevent VLC from starting a new instance every time a media file is opened Error "Power::infy: Infinite expression 1/0 encountered". Why do I get this? Finding an angle within an 80-80-20 isosceles triangle Heroes of Might and Magic VI seems to have installed a program called "Uplay", which it uses to launch (in addition to Steam, mind!)  What exactly is Uplay supposed to be? I notice that when I stop playing, it captchalogs the highlights of my past session (achievements earned, etc.) - which is kinda cool, but also kinda creepy and big-brotherish. Given that I value my privacy, is there anyway to play Uplay games without using Uplay?  Is this just Ubisoft's infamous "always-on" DRM? Weird display on screen before swipe lock Samsung Galaxy s6 edge plus Sum of two squares modulo p A Question On Euler's Proof Of the Basel Problem I've studied the proof that Euler gave for the famous , and it would seem that while it is technically correct, he does not justify all of his steps properly. Namely, he assumes that   $$\frac{\sin(x)}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots$$  simply because they have the same roots, which is really not a strong enough condition. How do you really show that the equality holds?  He then notices that if you use the above equality, and consider it against the Taylor expansion for $\frac{\sin(x)}{x}$, then you can equate the coefficients of the two infinite expansions at each order, and the result of the Basel problem follows. But how do you know that if you have two different expansions for a function, then their coefficients at each order must be equal?  I would really appreciate if someone could show me how to make these two intuitive, yet informal, steps rigorous. Showing that $f:(0,2\pi]\to S^1$ is not a homeomorphism I want to show that $f:(0,2\pi]\to S^1$ defined by $t\to (\sin t, \cos t)$ is not a homeomorphism. I will do this by showing that its inverse is not continuous.    The inverse is defined by $(\sin t, \cos t)\to t$. So at $(0,1)$ we want to have $\lvert 2(1-\cos t)\rvert &lt;\delta \Rightarrow \lvert t-2\pi\rvert&lt;\epsilon$ for any $\epsilon$. If we take $\epsilon=\pi/2$, no matter how close $\cos t$ and $1$ are there will always be $t$'s close to $0$. So $f^{-1}$ is not continuous.  What do you think? Are there any mistakes? access parent object in javascript How to prevent automatic redirection of 404 errors and "incorrect" URLs? Wordpress has a feature whereby it will automatically redirect your URLs if it percieves them to be written wrongly. Here is an example: I have a page called my-page  If I go to:  www.mysite.com/something/my-page/   it will immediately redirect me to   www.mysite.com/my-page/   as nothing exists at the first URL.   How can I turn this feature off, and instead just get a 404 if incorrect URL's are typed in? For which $f,g \in k[t]$, $k[f,g]$ is integrally closed? Let $f,g \in k[t]$, $k$ is a field of characteristic zero, $\deg(f)=n \geq 2$, $\deg(g)=m \geq 2$.     Is it possible to characterize all such $f$ and $g$ for which $k[f,g]$ is integrally closed in its field of fractions $k(f,g)$?   I do not mind to further assume that $k(f,g)=k(t)$ (but I do not want to assume that $k[f,g]=k[t]$, which makes my question trivial).  Thank you very much!  Edit: (i)  is relevant.  (ii) If also $k(f,g)=k(t)$, then being integrally closed is equivalent to $k[f,g]=k[t]$. Indeed: (a) If $k[f,g]=k[t]$ then $k[f,g]$ is integrally closed (since $k[t]$ is). (b) If $k[f,g]$ is integrally closed then, in particular, since $t \in k(t)=k(f,g)$ (= the field of fractions of $k[f,g]$) and obviously $t$ is integral over $k[f,g]$ (for example, $t$ is a root of $f(T)-f \in k[f,g][T]$) hence $t \in k[f,g]$, so $k[f,g]=k[t]$.  For example: $f(t)=t^2+1$, $g(t)=t^3-4t$. We have, $t= \frac{t(t^2-4)}{t^2+1-5}=\frac{g}{f-5}$, so $k(f,g)=k(t)$. By Abhyankar-Moh-Suzuki theorem, $k[f,g] \neq k[t]$, so $k[f,g]$ is not integrally closed. Similarly, for every $f,g$ such that $k(f,g)=k(t)$ and one of the degrees does not divides the other, then $k[f,g]$ is not integrally closed.  (iii) If also $k(f,g)=k(t)$ and $k$ is algebraically closed, then being integrally closed is equivalent to $H(t)=(f(t),g(t))$ is injective and $H'(t)=(f'(t),g'(t)) \neq (0,0)$ for every $t \in k$. Indeed, $k[f,g]=k[t]$ is equivalent to the new conditions (it is important that $k$ is algebraically closed!!).  Edit: What if we replace $k$ by an integral domain $D$ (which is a $k$-algebra)? See also  question. I am considering buying my first DSLR.  How do I know whether I need a 18-200 lens, 17-85, or 18-55? Or something else? Preserving order with LINQ I use LINQ to Objects instructions on an ordered array.  Which operations shouldn't I do to be sure the order of the array is not changed?