As the title says, every time when I open Nautilus from unity launcher, it opens with all files being shown. Normally it opens with hidden files being hidden, how can I change it back to that behavior?   My system is Ubuntu 14.04 LTS with Unity, no other WM/DM's are installed.
Whenever I open my files, my hidden files show up. I can hide them but when I open up my files again, they're showing again. How do I make the hidden files invisible permanently?
New phone: How to transfer game progress? I'm planning to buy a new Android phone and I'd like to be able to transfer the progress I have made at several games to the new device, so I won't have to play all the levels again from scratch.  I currently have an HTC Desire and will buy an HTC Evo 3D.  There's probably no simple/general solution to this.  But I was wondering if anyone did something similar and could provide some advice (e.g. where to look for files that might hold game progress information). $X_1, X_2, \ldots, X_n$ are $n$ i.i.d. uniform random variables. Let $Y = \min(X_1, X_2,\ldots, X_n)$. Then, what's the expectation of $Y$(i.e., $E(Y)$)?  I have conducted some simulations by Matlab, and the results show that $E(Y)$ may equal to $\frac{1}{n+1}$. Can anyone give a rigorous proof or some hints? Thanks! I have a shapefile with 3800 polygons representing the smallest administrative areas in Ireland. Many of these, in rural areas, have tiny populations, under 100 people. In urban areas, all would have adequate populations.  For each area I have a population figure, and various other attributes, all of which can be intelligently summed to give valid results for areas produced by combining the smaller areas.  I need to merge the areas to get say, 500 or 600 areas of reasonable population. I need to leave the larger (in population) areas alone, where possible. I would greatly prefer to do this automatically. This is necessary for model fitting purposes, primarily.  I've looked at a number of the dissolve questions here:   'Batch Dissolve Polygons based on Attributes in QGIS or FWTools' I think would require me to know which areas to merge, when this is what I need to calculate. 'Practical way of managing polygons and unions of them' and 'How do dissolve polygons from shapefile using open source tools?' have the same issue.   Any suggestions? What exactly is a bad backlink that is Disavow worthy? I have been reading Webmasters SE and many other blogs on this topic but I could not understand how do we truly evaluate a bad link that we must disavow in GWT. Everybody seem to have a different opinion on this.  A little background  I am a WordPress developer and 3 years ago I gave away a WordPress theme to my blog visitors which has a dofollow link back to my business website. Like this.  WordPress theme designed by &lt;a href="http://domain.com"&gt;MyBusinessName&lt;/a&gt;   From last 2 years I was not paying any attention on organic traffic to my business website because of other projects.  That free WordPress theme helped us getting ~10K backlinks from variety of domains and websites. All most 99% backlinks are from non-related websites and in many different languages.  Now  Last week it came to my attention that in October 2013, organic traffic from Google started decreasing and in just 2 months it became as low as 25% of October. Before October it was gradually increasing ~5-10% per month.  So I researched and draw to the conclusion that it could be because of Penguin Update. Since  was introduced at the same time.  Now I need information on what a bad/toxic link is.  I was about to disavow almost all links but after reading many topics on Webmasters SE I think I need more information on how to identify a bad link.   Does credit link in footer consider as spam because it build thousands of backlinks? Do foreign language website backlinks consider as spam? Should I remove all backlinks from theme users because almost all are from non-related content domains? Although some links from non-related content domains have nice DA/PA. Should I remove those too? I understand foreign language anchor text backlink could be a spam but what if it is from related foreign language content.   The question is how exactly one can find a link that is bad and must be removed. How can I run an external program from C and parse its output? I've got a utility that outputs a list of files required by a game.  How can I run that utility within a C program and grab its output so I can act on it within the same program? UPDATE: Good call on the lack of information.  The utility spits out a series of strings, and this is supposed to be portable across Mac/Windows/Linux.  Please note, I'm looking for a programmatic way to execute the utility and retain its output (which goes to stdout). How can I test whether a command's output is an empty string while not   interfering with buffering?  Example:  command | ??? if [ $? = 0 ] &amp;&amp; [ -n "$output_from_command"]; then    ...   I want the output from the code above to be the same as when running command. Normal subgroup of prime index Generalizing the case $p=2$ we would like to know if the statement below is true.  Let $p$ the smallest prime dividing the order of $G$. If $H$ is a subgroup of $G$ with index $p$ then $H$ is normal. I'm an Egyptian citizen, who holds Schengen visa and stay permit in Italy. Can I go to UK for some days with my stay permit and Schengen visa? Below, the  of second-order logic is used.    My question is about a second-order analogue of $ZFC$ other than the usual "second-order $ZFC$." Rather than define the latter, I'll just state that its (set-sized) models are exactly the $V_\kappa$s with $\kappa$ strongly inaccessible (or if you prefer, the uncountable Grothendieck universes).  Specifically, I'm interested in the second-order theory consisting of the usual (first-order) axioms of Pairing, Extensionality, Union, Choice, Infinity, and Powerset, and the Separation and Replacement schemes ranging over second-order formulas. I'll call this "$ZFC_2^{scheme}$."    Specifically, for each formula $\varphi(x, y_1,...,y_k)$ of second-order logic with only the displayed variables, each of which is first-order, we have a corresponding Separation and Replacement instance. We do not allow second-order parameters here, so formulas with second-order free variables don't yield Separation/Replacement instances.   The passage from $ZFC$ to $ZFC_2^{scheme}$ has a nice uniformity: it's an example of a more general operation $\mathcal{ZFC}$ which spits out a $ZFC$-analogue given a logic at least as strong as first-order logic.    This is the reason for the parameter restriction above: I want $ZFC_2^{scheme}$ to look just like $\mathcal{ZFC}(\mathcal{L})$ for any other logic $\mathcal{L}$, and not take into account the particular feature that we now have a new kind of variable at our disposal - and in particular, take as "syntax-free" an approach to logics as we can get away with. (This is a choice of arguable value, but it's the one I'm making for now.)   It's easy to show that every model of $ZFC_2^{scheme}$ is well-founded, so to understand its models we only need to look at transitive sets. A natural hope at this point is that $ZFC_2^{scheme}$ is just second-order $ZFC$ in disguise, that is, that for a transitive set $M$ we have $M\models ZFC_2^{scheme}$ iff $M=V_\kappa$ for some strongly inaccessible $\kappa$. However, this isn't at all obvious to me (although it is easy to show the right-to-left direction).     Question. What are the (set-sized) models of $ZFC_2^{scheme}$?   (Note by contrast that the "arithmetic analogues" $PA_2^{scheme}$ and second-order $PA$ are equivalent since each pins down $\mathbb{N}$ up to isomorphism - although that leads to .)  (I'm happy to drop Choice if that would help.)  EDIT: As Hanul Jeon pointed out below,  shows that consistently $ZFC_2^{scheme}$ (there called "$ZFC_2^{def}$") has countable models, so we have a partial answer. However, I don't see at the moment a way to get an outright $ZFC$ proof that $ZFC_2^{scheme}$ is strictly weaker than second-order $ZFC$. induction with 2 recursive sequnces I'm having trouble solving this problem. I have relation for two sequences of natural numbers.   $$a(n)+\sqrt3\cdot b(n)=\left(1+\sqrt3\right)^n$$  and I have to prove that recursions:  $$\begin{align*} &amp;a(n+1)=a(n)+3b(n)\\ &amp;b(n+1)=a(n)+b(n) \end{align*}$$  will satisfy relation mentioned above. I tried solve it using induction but I failed. Can anybody help with some other ideas? How to create a good glass reflection with Principled BSDF Not getting Rep Notification for new up votes So just a little while ago i was flipping out because i was 2 rep away from my 10k rep goal    but then all of a sudden it was then 10,008 rep because someone upvoted me. however i didn't get any notification about this +10  [    now as you can see in the screenshots above, over the past 2 days i got 2 upvotes to a question i posted, however the notifications only shows 1 +10 notification. and they system knows i got the upvote because i knew that i now have access to mod tools.  I am on a Windows 10 Laptop using Chrome 78.0.3904.97 (Official Build) (64-bit) Why are relays so frequently driven by optocouplers? cvc-complex-type.2.4.a: Invalid content was found starting with element 'init-param' Evaluating the limit of a sequence given by recurrence relation $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$. Is my solution correct? Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct? I've got Ubuntu 10.10 installed to use my whole hard-disk, but would like to use the 130GB of free space I've got to create a new partition and do some development on Unity and Natty.  Can I use gparted to do this?  If so how? In order to protect a portal in the nether I have an obsidian room with an iron door, which works off a switch in the room. The problem is that if I leave the room and someone else on the server closes the door from the inside I can't get back to that portal.  Powering red stone on the outside of the door doesn't seem to work, as the switch inside is already powering it, so I'm guessing something like an XOR gate will be the solution, however I am finding it difficult to see how to make this work in practice. Why does the reputation need to be publicly displayed? Looking for a curly script capital A A math book I'm studying uses a symbol and I'd like to be able to reproduce it.  I used detexify to search for it, but was unsuccessful. Most likely it is the letter 'a' in some font I'm not familiar with.   Here's a zoomed in image of the symbol:    Here it is in context: Completely uninstall an app on Lollipop I am having a problem with uninstalling apps on Android 5.0 (Nexus 5). I am developing an Android app, so I debug/run the app directly on my device. For production, I exported the same app with my keystore, uninstalled the debug APK that I ran earlier, tried to install the signed APK, and I got this error:      An existing package by the same name with a conflicting signature is already installed.   Is it because the bytecodes are still in the ART runtime? Anyone has any idea how to solve this?