I've noticed some time ago that some of my shortcuts in Ubuntu 14.10. stopped working. I hoped that update to 15.04. will fix it but unfortunately, that was not the case. I have a problem only with few shortcuts, not all of them. For instance:   default shortcuts to move windows to left or right (CCtrl + Super + left/right). I've tried to do this with mouse by dragging window but it stoped working as well... default shortcuts for managing workspaces (Ctrl + Alt + Arrow, Ctrl + shift + alt + arrow).   I've been googling this already but I didn't find any solution. It seems to be unity problem but I don't have any other ideas to fix it.
So, my update to 13.04 didn't go very well. Due to the lack of a decent driver (Yes, i'm unlucky to have a legacy driver which doesn't play nice with the new Xorg.), I had to mess around with stuff to first troubleshoot and afterwards "fix" my system.  I've finally got unity back, but now CtrlAltarrow and CtrlAltShiftarrow don't work anymore. They are bound in "keyboard" > "shortcuts", but nothing happens when I try it. A strange thing is that other keyboard shortcuts, like ctrlaltt to show terminal are still working.  I noticed that CtrlAltarrow doesn't show in the list with "unity shortcuts" (the list that shows when you hold the super key).  Does anyone know how to re-enable this way of workspace switching?  I enabled workspaces in the appearance menu. It doesn't have any effect though. Switching via the workspace-switcher in the unity dash works fine though.  CCSM is installed, but I can't seem to find a way to add keyboard bindings there.     System: ubuntu 13.04 64bit
How to plot and visualize a single linear vector in 3D? How do I get started playing Dungeons and Dragons 4th edition by myself? I've been wanting to start playing D&amp;D for some time, and I was wondering how I could get started by myself with relative ease. I own the first two player handbooks, and a set of dice. Why (...) doesn't spawn a new child process when run in background? Check out the screenshot below, I want to extend my C partition (where Windows is installed) into the unallocated free space shown. This space was cleared up by shrinking the D partition. I can extend the D partition if I want to, but not the C.  Why not? How can I specify JUnit test dependencies? Print two files in two columns I want to print two files in two columns -- first file on the left side and second on the right side. paste doesn't do the job, because it can only insert a character as delimiter, so if first file lines have different length output will be twisted: $ cat file1 looooooooong line line $ cat file2 hello world $ paste file1 file2 looooooooong line   hello line    world  If it was a command to add trailing spaces like fmt --add-spaces --width 50 the problem would be solved: $ paste &lt;(fmt --add-spaces --width 50 file1) file2 looooooooong line                                 hello line                                              world  But I don't know a simple way to do this. So how to merge and print several files horizontally without twisting? Actually, I just want to look at them simultaneously.  UPD: command to add trailing spaces does exist (for example, xargs -d '\n' printf '%-50s\n') But solution like $ paste &lt;(add-trailing-spaces file1) file2  does not work as expected when file1 has fewer lines than file2. Javascript && operator as if statement? Two Clash of Clans accounts on one device I have been playing on two clash of clans accounts on one device for a long time but just recently i cannot switch from my lvl17 to my lvl75. The switch villages option is not popping up for me and I don't know why. Can anyone help me get my main account back? A Regex that will never be matched by anything default permissions for /var/www I was trying to open a file and write to it with PHP at /var/wwwfolder but it wasn't working so I did  sudo chmod 777 /var/www   now I want to set the permissions for /var/www to the default. what are the default permissions for /var/www? Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with probability function:  $$p_X(x)=\begin{cases} 1/2 &amp;x=-1,1 \\ 0 &amp; \text{otherwise} \end{cases}$$  Now if we define the average   $\bar{X}= \frac{1}{n} \sum X_i$, I need to show that that for an odd $n$ the probability function for $\bar{X}$ is:  $$p_{\bar{X}} (x)= \frac{\binom{n}{\frac{n}{2} (x+1)}}{2^n}$$ for $x=\pm 1/n,\pm 3/n,\ldots,\pm 1$, $0$ otherwise.    Since $P[\sum X_i=k] =P [\bar{X}=k/n ]$, I thought it would be easier to first derive the probability function for the sum. Since they are still $2^n$ n-tuples and each one is equilikely, I need to count the ones whose sum is $k$ for:  $k=\pm 1,\pm 3,\ldots \pm n-2,\pm n$ (since even numbers are not a feasible combination)  The problem here though is that I do not immediately recognize how I could count all these tuples. Could you please help me with this? If it is easier to proceed another way, I am of course all ears. Why is the Star Trek 2009 reboot called a reboot? Calculating the optimal number of bins for severely skewed data Removing intercept from GLM for multiple factorial predictors only works for first factor in model I am running a binomial logistic regression with a logit link function in R. My response is factorial [0/1] and I have two multilevel factorial predictors - let's call them $a$ and $b$ where $a$ has 4 factor levels  $(a_1,a_2,a_3,a_4)$ and $b$ has 9 factor levels $(b_1,b_2,\dotsc,b_9)$. Therefore:  mod &lt;- glm(y ~ a + b, family = binomial(logit), data=pretend) summary(mod)   The model output would then show all the information about the model as well as the coefficients.  There is a factor level for both a and b missing (a1 and b1) from the summary output. I understand that it is constrained in the "intercept" of the model. I have read that if I want to remove the intercept term and see the estimates for those factor levels I can just add -1 to the model formula, i.e.:  mod2 &lt;- glm(y ~ a + b - 1, family=binomial(logit), data=pretend) summary(mod2)   In the new model (mod2) the intercept term is then gone and variable a's factor-level a1 is given amongst the list of coefficients. But, variable b's factor-level b1 is still missing and given that there is no intercept term anymore, how can I interpret the odds-ratio for that factor level then?  Could someone please explain to me how to get the coefficient for b1 too and why this is happening? Can I have my iPhone play sound for MacBook? Say I am playing a video on my MacBook Pro, and I don't want to distract others with the sound. The problem is, I don't have a 3.5 mm Headphone or AirPods at my hand, but I have an iPhone 8 Plus and its EarPods with Lightning Connector. Can I play the sound from the video playing on macOS with the iPhone, preferably without connecting the iPhone to the Mac so that I can plug in the EarPods to the iPhone's socket?  This is basically the reverse of K-means is a widely used method in cluster analysis. In my understanding, this method does NOT require ANY assumptions, i.e., give me a dataset and a pre-specified number of clusters, k, and I just apply this algorithm which minimizes the sum of squared errors (SSE), the within cluster squared error.  So k-means is essentially an optimization problem.  I read some material about the drawbacks of k-means.  Most of them say that:   k-means assumes the variance of the distribution of each attribute (variable) is spherical;  all variables have the same variance; the prior probability for all k clusters is the same, i.e., each cluster has roughly equal number of observations;   If any one of these 3 assumptions are violated, then k-means will fail.  I could not understand the logic behind this statement. I think the k-means method makes essentially no assumptions, it just minimizes the SSE, so I cannot see the link between minimizing the SSE and those 3 "assumptions". Positive operator is bounded For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.   I don't know how to do that, maybe it's an application of the closed graph theorem? Identification of particles and anti-particles The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become the normal particle and the rare and unusual positrons become the .  My question is, when you have made the choice of the electron and positron as particle and anti-particle does this automatically identify every other particle (every other fermion?) as normal or anti?  For example the proton is a particle, or rather the quarks inside are. By considering the interactions of an electron with a quark inside a proton can we find something, e.g. a conserved quantity, that naturally identifies that quark as a particle rather than an antiparticle? Or do we also just have to extend our convention so say that a proton is a particle rather than an antiparticle? To complete the family I guess the same question would apply to the neutrinos. Equipartition of energy Let $u$ solve the initial-value problem or the wave equation in one dimension: $$\begin{cases}u_{tt}-u_{xx}=0 &amp; \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h &amp; \text{on } \mathbb{R} \times \{t=0\}. \end{cases}$$ Suppose $g,h$ have compact support. The kinetic energy is $k(t) := \frac 12 \int_{-\infty}^\infty u_t^2(x,t) \, dx$ and the potential energy is $p(t) := \frac 12 \int_{-\infty}^\infty u_x^2(x,t) \, dx$. Prove (a) $k(t)+p(t)$ is constant in $t$, (b) $k(t)=p(t)$ for all large enough times $t$.  This is Chapter 2, Exercise 24 of PDE Evans, 2nd edition. I am only doing part (a) right now; my work is shown below: Define $$e(t):=k(t)+p(t)=\int_{-\infty}^\infty u_t^2+u_x^2 \, dx.$$ Then $$e'(t)=\frac 12 \int_{-\infty}^\infty 2u_tu_{tt}+2u_{x}u_{xt} \, dx= \int_{-\infty}^\infty u_tu_{tt}+u_{x}u_{xt} \, dx.$$ Now, I want to get $e'(t)=0$ so that $e(t)$ is constant. How can I go about doing this? I do know that I can use $u_{tt}-u_{xx}=0$, if $t &gt; 0$. With $t&gt;0$ into mind, do I have to consider the $t=0$ case separately? Or can I treat both cases together as $t \ge 0$, since $g$ and $h$ have compact support? Mathematical Basis behind inflation of Standard errors of Regression estimates due to multicollinearity We know that due to multi-collinearity, the standard errors of beta estimates get inflated. But what is the mathematical basis to it?   I am looking for some mathematical relationship or something to explain this.   Like I understand if standard error of betas goes up, then t-statistics goes down and we might not be able to reject the null or the variables would appear non-significant.   But what is the mathematical relationship between multicolllinearity and inflation in variance of coefficients?