If I'm launching a  (that requires the sacrifice a Goblin) and my opponent uses an  on the Goblin (or similar effects like the ability of a ) does this effectively counter my spell?
I have a question that bothers me after reading a part of the big book of rules :).  It's about the rule 717, illegal actions.     717.1. If a player realizes that he or she can’t legally take an action after starting to do so, the entire action is reversed and any payments already made are canceled. No abilities trigger and no effects apply as a result of an undone action. If the action was casting a spell, the spell returns to the zone it came from. The player may also reverse any legal mana abilities activated while making the illegal play, unless mana from them or from any triggered mana abilities they triggered was spent on another mana ability that wasn’t reversed. Players may not reverse actions that moved cards to a library, moved cards from a library to any zone other than the stack, or caused a library to be shuffled.   For example, player A plays a .     As an additional cost to cast Goblin Grenade, sacrifice a Goblin.   Goblin Grenade deals 5 damage to target creature or player.   It goes on the stack. Player B gains priority and casts a  on the only goblin of Player A.     Destroy target nonartifact, nonblack creature. It can't be regenerated.   The terror goes on the stack. Then both players agree to resolve the stack. The terror kills the last goblin standing and then the Goblin Grenade can't be cast because of the additional cost which can't be paid. So in my opinion that leaves two possibilities:   The Goblin Grenade returns to the zone it was previously from (the hand of Player A) The Goblin Grenade becomes illegal and gets countered, so it goes to the graveyard   Based on rule 717 I think it goes back to the hand but I'm not quite sure. :)
How do distributed transactions work (eg. MSDTC)? -Bash command not found appearing every time I start terminal every time I start my terminal, it always says "-bash:  .: command not found" twice.  Does anyone know how I can get rid of that? Why does centering not work with calligra? How to remove the whitespace BEFORE itemize/enumerate? I'm somewhat of a LaTeX noob, and this is driving me completely nuts.  When I type  \documentclass{article} \usepackage{enumitem}  \begin{document}     Some text     \begin{itemize}[noitemsep]         \item Item 1         \item Item 2         \item Item 3     \end{itemize} \end{document}   I don't want the space which appears in between the text "Some text" and the itemization to be there:    When I search around on Google all I see is tons of long, complicated advice on this site, StackOverflow, forums, etc. on how to not fix the problem ("parskip is evil", etc.), or tons of vague/confusing answers on what package to look at, with no actual example for me to learn from.  So while this question is already asked many times, I have never found a single concrete answer to this question.    The question couldn't be simpler: How do I "properly" get rid of this extra space? A commercial airport being any airport that has regular scheduled passenger flights.  The mathematics of  got me thinking: is the network of scheduled flights a connected graph?  That is, is there a pair (or triplet or even larger set) of airports such that they only have scheduled flights with each other?  Or can you get from any airport to any other?  Ignore vagaries like cargo and general aviation. I'm looking for two airports that I can shuttle between, but would have to drive (or sail) to get to one or the other of them. Why does subsurf modifier cause weird deformation at the end of a cylinder? My problem is simple. I think that the strange deformations on the top are there because there are so many vertices in one face. How do I fix this while maintaining nice topology?  Although I triangulated the top polygon, the problem still persists. I was working on a problem in group theory, which asks about the automorphism group of a direct product of groups.  Okay, so I know that if $G,H$ are two groups whose orders are relatively prime, then $\operatorname{Aut}(G\times H)\cong\operatorname{Aut}(G)\times \operatorname{Aut}(H)$.  I also know that if $G,H$ are abelian and simple and isomorphic, then $\operatorname{Aut}(G\times H)\cong \operatorname{GL}_2(\mathbf{Z}_p)$ where $p$ is the order of $G$.  What I don't know is the following: what if $G,H$ are simple, yet not abelian? Let us for now focus on a more restricted case where $G\cong H$, in particular, $G=H$. What can we say about the group $\operatorname{Aut}(G\times H)$? (I heard something about this group being related to what is called "wreath product", which I do not know...)  Thanks bunch in advance! How do I connect Android devices to a WiFi AP and mobile data network simultaneously? When I turn WiFi on, my 3G connection shuts down, which is fine in most cases. However, I need to connect to a LAN at work that is not connected to the internet and I'd prefer not to lose my data connection when I connect to the work intranet. How can I keep both connections active? Ideally, I could limit this desired behavior to this one SSID because I want it to continue to operate with the default behavior for all other WiFi connections.  If that can't be done, prefer a solution that is easy to toggle on and off. I'm using a Samsung Galaxy Nexus (Verizon Wireless) on CyanogenMod 9. A while back, I read a story in a collection, but I don't remember very much. I know that it was about some missionaries on Mars. Eventually they found a bunch of spheres that apparently have escaped sin by leaving the material world. I don't remember anything else about it. Someone told me that it was in "The Martian Chronicles" by Ray Bradbury, but I checked my copy and couldn't find. I'm pretty sure that it was by Ray Bradbury though. Help would be greatly appreciated. How do we prove that something is unprovable? Should you use a plural verb if you refer to a fraction?     Nine tenths of the pillar have rotted away.   I think has should be used here, but I'm not sure. If $f:D\to D’$ is analytic and $u: D'\to R$ is harmonic then the composition of $u$ and $f$ is harmonic in $D$ If $f:D\to D’$ is analytic and $u: D'\to R$ is harmonic then the composition of $u$ and $f$ is harmonic in $D$.  How can I show that the above statement is true/false? Can anyone help me? How do you use a variable in a regular expression? I would like to create a String.replaceAll() method in JavaScript and I'm thinking that using a regex would be most terse way to do it.  However, I can't figure out how to pass a variable in to a regex.  I can do this already which will replace all the instances of "B" with "A".  "ABABAB".replace(/B/g, "A");   But I want to do something like this:  String.prototype.replaceAll = function(replaceThis, withThis) {     this.replace(/replaceThis/g, withThis); };   But obviously this will only replace the text "replaceThis"...so how do I pass this variable in to my regex string? Neg Binomial and the Jeffreys' Prior I'm trying to obtain the Jeffreys' prior for a negative binomial distribution. I can't see where I go wrong, so if someone could help point that out that would be appreciated.  Okay, so the situation is this: I am to compare the prior distributions obtained using a binomial and a negative binomial, where (in both cases) there are $n$ trials and $m$ successes. I get the right answer for the binomial case, but not for the negative binomial.   Let's call the Jeffreys' prior $\pi_J(\theta)$. Then,  $$ \pi_J(\theta)\propto [I(\theta)]^{1/2}. $$  Under the regularity conditions (fulfilled as we are dealing with the exponential family),  $$ I(\theta)=-E\left(\frac{\partial^2 \log L(\theta|x)}{\partial \theta^2}\right) $$ where for the negative binomial $n$ is $x$ in the above expression (the total number of successes $m$ is fixed, $n$ is not). The distribution --I think-- is  $$ p(m|\theta)\propto\theta^m(1-\theta)^{n-m} $$ since $\theta$ is defined as the probability of success and $m$ is the number of successes. This is also the likelihood, since $m$ is a scalar and not a vector. Hence,  $$ L(\theta|n)\propto\theta^m(1-\theta)^{n-m}\\ \log L(\theta|n)=m\log\theta +(n-m)\log (1-\theta)\\ \frac{\partial\log L(\theta|n)}{\partial \theta}=\frac{m}{\theta}-\frac{n-m}{1-\theta}\\ \frac{\partial^2\log L(\theta|n)}{\partial \theta^2}=-\frac{m}{\theta^2}-\frac{n-m}{(1-\theta)^2} $$ so the Fisher information is  $$ I(\theta)=-E\left(\frac{\partial^2\log L(\theta|n)}{\partial \theta^2}\right)=\frac{m}{\theta^2}+\frac{E(n)-m}{(1-\theta)^2}=\frac{m}{\theta^2}+\frac{\frac{m\theta}{1-\theta}-m}{(1-\theta)^2}\\ =\frac{m(1-\theta)^2+\frac{m\theta^3}{(1-\theta)}-m\theta^2}{\theta^2(1-\theta)^2}=\frac{m(1-2\theta)+\frac{m\theta^3}{(1-\theta)}}{\theta^2(1-\theta)^2}\\ =\frac{m(1-2\theta)(1-\theta)+m\theta^3}{\theta^2(1-\theta)^3}=\frac{m(1-3\theta+2\theta^2+\theta^3)}{\theta^2(1-\theta)^3}\\ \propto\frac{1-3\theta+2\theta^2+\theta^3}{\theta^2(1-\theta)^3} $$  This, however, does not give me the correct answer. The correct answer is  $$ \pi_J(\theta)\propto \frac{1}{\theta(1-\theta)^{1/2}} $$ which means that the Information I get should be   $$ I(\theta)=\frac{1}{\theta^2(1-\theta)} $$ since the prior should be proportional to the square root of the information.   Can anyone find any mistakes? I wouldn't be surprised if I screwed something up with the set up of the distribution (successes vs failures with their respective probabilities, etc).   I used the expected value from  and I know the correct answer from . Dispatcher.BeginInvoke: Cannot convert lambda to System.Delegate I cannot respond to comments on my own question (no "add a comment" link anywhere) I joined a new community and asked a question ()  People are commenting &amp; asking questions, but I cannot answer them back. Even "Comment Everywhere" privilege requires 50 reputations and I have 101 reputations.  Am I missing something? How can I synchronize my two Mac Computers? This problem is from Romanian G.M. and although it is very short, it is also (I think) very hard. Let $x, y, z$ real non-negative numbers such that $x+y+z=3$. Prove that   $$27 \leq (x^2+2)(y^2+2)(z^2+2) \leq 44 $$  I didn't succeed in any of the inequalities, I only managed to find the equality cases ($x=y=z=1$ for the first one and one of them $3$ and the others $0$). Any hint/idea/solution is welcome. What does the term saturating nonlinearities mean? I was reading the paper  and in section 3 were they explain the architecture of their Convolutional Neural Network they explain how they preferred using:     non-saturating nonlinearity $f(x) = max(0, x). $   because it was faster to train. In that paper they seem to refer to saturating nonlinearities as the more traditional functions used in CNNs, the sigmoid and the hyperbolic tangent functions (i.e. $f(x) = tanh(x)$ and $f(x) = \frac{1}{1 + e^{-x}} = (1 + e^{-x})^{-1}$ as saturating).   Why do they refer to these functions as "saturating" or "non-saturating"? In what sense are these function "saturating" or "non-saturating"? What do those terms mean in the context of convolutional neural networks? Are they used in other areas of machine learning (and statistics)? I'm seeing an odd character like a very flat U, where there should be a space, with the listings package. See example below. Does anyone have any idea why this is happening, or how to get rid or work around it?  \documentclass{standalone} \usepackage{underscore} \usepackage{listings} \usepackage[usenames,dvipsnames, rgb, svgnames, x11names]{xcolor} % Text layout \topmargin 0.0cm \oddsidemargin 0.5cm \evensidemargin 0.5cm \textwidth 16cm \textheight 21cm  \lstnewenvironment{cmd}[1][]{%   \lstset{language=sh, backgroundcolor=\color{LightSteelBlue1}, #1}} {}  \begin{document} \begin{cmd}[caption=Configuration file section including filters] [affy6] [...]     [[hapmap]]     phenofiles = "data/Hapmap6.0/Pheno/hapmappheno_909.csv",     genofile = "/data/snppy/Hapmap6.0/Geno/birdseed-v2.calls.txt"     anno_filter = "chromosome = 5"     idlink_filter = "studyid = 'CEU'"     some_other_stuff = "foo = bar" \end{cmd} \end{document}